Churchill, R. v. - Operational Mathematics

Operational Mathematics Third Edition

Ruel V. Churchill Professor Emeritusof Mathematics Universityof Michigan

McGraw-HiII Book Company New York St. Louis San Francisco Düsseldorf Johannesburg Kuala Lumpur London Mexico Montreal New Delhi Panama Rio de Janeiro Singapore Sydney Toronto

I

Operational

Mathematics

Copyright O 1958, 1972 by MeGraw-Hill, lne. Al! rights reserved. Copyright 1944 by MeGraw-Hill, Ine. Al! rights reserved. Printed in the United States of Ameriea. No part of this publieation may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronie, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Library ofCongress

Catalog Card Number

70-174611

07-010870-6

78910

KPKP

7832109

This book was set in Times Roman, and was printed and bound by the Kingsport Press. The designer was Jo Jones; the drawings were done by John Cordes, J. & R. Technical Services, Ine. The editors were Howard S. Aksen and Madelaine Eiehberg. Matt Martino supervised produetion.

Contents

Preface

xi

Chapter 1. The Laplace TransformatioR 1. Introduction

9. The Solution of Simple Differential Equations 10. Generation of the Transformation

1 I 3 5 7 lO 14 16 17 20 24

Chapter 2. Further Properties of the Transformation

27

2. Definition of the Laplace Transformation 3. Sectionally Continuous Functions. Exponential Order 4. Transforms of Derivatives 5. Examples. The Gamma Function 6. The Inverse Transform 7. A Theorem on Substitution 8. The Use of Partial Fractions (Table 1)

11. Translation

27 29

of F(t)

12. Step Functions 13. The Impulse 14. Integrals

Symbol c5(t

Containing

-

to)

a Parameter

15. Improper Integrals 16. Convolution 17. Properties

of Convolution

18. Differential

and Integral

19. Derivatives

of Transforms

Equations

20. Series of Transforms 21. Differential

Equations

22. Integration

of Transforms

with Variable

Coefficients

33 39 41 43 46 50 54 57 61 65 v

vi

CONTENTS

23. Periodic Functions

66

24. Partial ,Fractions 25. Repeated Linear Factors 26. Quadratic Factors

70 73 75 78

27. Tables of Operations and Transforms Chapter 3.

EIemeotary

28. Free Vibrations 29. Forced Vibrations

ApplicatioDS

of a Mass on a Spring without

Damping

30. Resonance 31. Forced Vibrations with Damping 32. A Vibration Absorber 33. Electric Circuits 34. Evaluation

of lntegrals

35. Exponential-

and Cosine-integral

36. Static Deftection

Functions

of Beams

37. The Tautochrone 38. Servomechanisms 39. Mortality

Chapter 4.

of Equipment

Problems in Partial Differeotial EquatioDS

40. The Wave Equation 41. Displacements in a Long String 42. A Long String under lts Weight 43. The Long String lnitially Displaced 44. A Bar with a Prescribed Force on One End 45. 46. 47. 48. 49.

Equations of Diffusion Temperatures in a Semi-infinite Solid Prescribed Surface Temperature Temperatures in a Slab A Bar with Variable End Temperature

50. A Cooling Fin or Evaporation Plate 51. Temperatures in a Composite Solid 52. Observations on the Method Chapter 5. FunCtiODS01 a Complex Variable 53. Complex Numbers 54. Analytic Functions 55. 56. 57. 58.

Exponential and Trigonometric Functions Contour lntegrals Integral Theorems Power Series

59. Singular Points and Residues

85 85 88 91 95 96 102 106 110 113 115 117 119 123 123 126 130 133 135 143 145 146 151 153 153 155 159 162 162 163 166 168 170 171 173

vii

CONTENTS

60. Branches of Multiple-valued Functions 61. Analytic Continuation 62. Improper Cauchy Integrals

Chapter 6.

The Inversion Integral

63. Analytic Transforms 64. Permanence

of Forms

65. Order Properties 66. The Inversion

of Transforms

Integral

67. Conditions

onf(s)

68. Conditions

on F(t)

69. Uniqueness

of Inverse Transforms

70. Derivatives

of the Inversion

71. Representation

Integral

by Series of Residues

72. Residues at Poles 73. Validity ofthe 74. Alterations

Chapter 7.

Representation

of the Inversion

by Series Integral

Problems in Heat Conduction

75. Temperatures in a Bar with Ends at Fixed Temperatures 76. The Solution Established 77. The Series Form Established 78. Properties of the Temperature Function 79. Uniqueness ofthe Solution 80. Arbitrary End Temperatures 81. Special End Temperatures 82. 83. 84. 85.

Arbitrary Initial Temperatures Temperatures in a Cylinder Evaporation from a Thick Slab Duhamel's Formula

Chapter 8. Problems in Mechanical Vibrations 86. A Bar with a Constant Force on One End 87. Another Form of the Solution 88. Resonance in the Bar with a Fixed End 89. Verification of Solutions 90. 91. 92. 93.

Free Vibrations of a String Resonance in a Bar with a Mass Attached Transverse Vibrations of Beams Duhamel's Formula for Vibration Problems

Chapter 9.

GeneraUzed Fourier Series

94. Self-adjoint

Differential

Equations

177 179 181

186 186 188 189 193 195 198 201 202 206 208 210 213 219 220 222 224 226 228 230 232 237 240 245 247 253 253 256 258 259 264 266 268 270 276 277

viii

CONTENTS

95. Green's Functions 96. Construction of Green's Function

105. Steady Temperatures in a WalI 106. Verification ofthe Solution

278 281 283 288 291 292 293 294 296 299 305 308

107. Singular Eigenvalue Problems

309

97. 98. 99. 100.

Orthogonal Sets of Functions Eigenvalue Problems A Representation Theorem The Reduced Sturm-Liouville System

101. A Related Boundary Value Problem 102. The Transform y(x,s) 103. Existence of Eigenvalues 104. The Generalized Fourier Series

Chapter 10. General Integral Transforms 108. Linear Integral Transformations 109. Kernel-product

Convolution

Properties

110. Example 111. Sturm-Liouville

Transforms

112. Inverse Transforms 113. Further

Properties

114. Transforms

of Certain

Functions

115. Example

of Sturm- LiouvilIe Transformations

116. Singular

Cases

117. A Problem 118. Other

in Steady Temperatures

Boundary

Value Problems

Chapter 11. Finite Fourier Transforms 119. Finite Fourier

Sine Transforms

120. Other Properties

of S.

121. Finite Cosine Transforms 122. Tables of Finite Fourier

Transforms

123. Joint Properties of C. and S. 124. Potential in a Slot 125. Successive Transformations 126. A Modified

Sine 'Fransformation

127. Generalized

Cosine Transforms

128. A Generalized

Sine Transform

129. Finite Exponential 130. Other Properties

Transforms

E.{F}

of E.

Chapter 12. Exponential Fourier Transforms 131. The TransformationE.{F}

317 317 319 320 325 327 329 332 333 336 341 343

348 348 350 354 356 357 360 362 368 370 372 376 378 383 383

ix

CONTENTS

385 388 391

132. The Inverse Transformation 133. Other Properties

of E.

134. The Convolution Integral 135. Convolution Theorem

for E.

393 396 397

136. Tables of Transforms 137. Boundary

Value Problems

Chapter 13. Fourier Transfonns 08 the HaJf Line 138. Fourier

Sine Transformsf.(a)

139. Fourier

Cosine Transformsfc(a)

140. Further

Properties

141. Convolution

of S. and C.

Properties

142. Tables of Sine and Cosine Transforms 143. Steady Temperatures in a Quadrant 144. Deflections in an Elastic Plate 145. A Modified 146. Convolution

Fourier

Transformation

1;.

for 1;.

147. Surface Heat Transfer

Chapter 14. Hankel Transfonns 148. lntroduction 149. Finite Hankel Transformations 150. 151. 152. 153. 154.

lnversion of H.j Modified Finite Transformations H.h A Boundary Value Problem Nonsingular Hankel Transformations Hankel Transformations H." on the Half Line (x> O)

155. Further Properties of H." 156. Tables ofTransforms H..{F} 157. Axially Symmetric Heat Source Chapter 15.

Legendre and Other Integral Transfonns

158. The Legendre Transformation 1;,on the Interval (-1,1) 159. Further Properties of 1;, 160. Legendre Transforms on the lnterval (0,1) 161. Dirichlet Problems for the Sphere 162. Laguerre Transforms 163. Mellin Transforms

401 401 404 405 406 407 408 410 414 416 418 420 420 421 423 424 426 431 432 434 436 437 441 442 444 446 448 452 453

Bibliography

456

Appendixes AppendixA Tables of LaplaceTransforms

458

x

CONTENTS

Table A.l Operations Table A.2 Laplace Transforms Appendix B Tables of Finite Fourier Transforms Table B.l Finite Sine Transforms Table 8.2 Finite Cosine Transforms Appendix C Table of Exponential F ourier Transforms Appendix D Tables of Fourier Sine and Cosine Transforms Table D.l Sine Transforms on the Half Line Table D.2 Cosine Transforms on the Half Line

Index

458 459 467 467 469 471 473 473 475

477

Preface

This is an extensive revision of the second edition of "Operational Mathematics" published in 1958. Chapters have been added on general integral transforms, finite Fourier transforms, exponential Fourier transforms, Fourier transforms on the half line, Hankel transforms, and on Legendre and other integral transforms. The presentation of theory and applications of the Laplace transformation has been revised. Tables of several of the most useful transforms now appear in the Appendix or in the text. Additional problems illustrate applications of the various integral transformations. The book is designed as a text and a reference on integral transforms and their applications to problems in linear differential equations, to boundary value problems in partial differential equations in particular. It presents the operational properties of the linear integral transformations that are useful in those applications. The selection of a transformation that is adapted to a given problem, by observing the differential forms and boundary conditions that appear in the problem, is emphasized in this edition. The Laplace transformation receives special attention because of its many useful operational properties and the large class of problems to which it applies, including applications outside the field of differential equations. The applications to problems in physics and engineering are kept on a fairly elementary level. They include problems in vibrations or displacements in elastic bodies, in diffusion or heat conduction, and in static potentials. No previous preparation in the subject of partial differential equations is required of the reader. This book is a companion volume to "Fourier Series and Boundary Value Problems" and "Complex Variables and Applications." The three books cover, respectively, these principal methods ofsolving linear boundary xi

xii

PREFACE

value problems in partial differential equations: the operational methods of integral transforms, separation of variables and Fourier series, and conformal mapping. Generalized Fourier series and their applications are presented in Chapter 9 here, with the aid of the theory of the Laplace transformation. A summary of useful theory of functions of a complex variable is given in Chapter 5. All three books are intended to present sound mathematical analysis as well as applications. Conditions of validity of analytical results are kept on a simple and practicallevel. More elegant conditions may call for training in analysis beyond the level of advanced calculus. The first four chapters are designed to serve as a text for a short course in real Laplace transforms and their applications. The impulse symbol or "delta function" is introduced in Seco13 in an elementary and careful manner. No theory of distributions, or generalized functions, is included in the book. A satisfactory presentation of the theory would require considerable space and the introduction of concepts not needed elsewhere in the texto Neither can the author justify a presentation of the abstract theory of linear spaces in this book. An intuitive approach from vectors to functions (Secs. 10 and 97) serves as a guide for writing inner products of functions as integrals. In preparing the three editions of this book the author has taken advantage of improvements suggested by many students and teachers. He is grateful to them for that assistance; also to authors referred to in the Bibliography and footnotes, whose publications have influenced the selection of material. Ruel V. Churchill

1 The Laplace Transformation

1

INTRODUCTION

The operation of differentiating functions is a transformation from functions F(t) to functions F'(t). If the operator is represented by the letter D, the transformation can be written D{F(t)} = F'(t). The function F'(t) is the image, or the transform, of F(t) under the transformation; the function 3t2, for example, is the image of the function t3. Another transformation of functions that is prominent in calculus is that of integration, I{F(t)} = I: F(t) dt. The result of this operation is a functional f(x), the image of F(t) under the transformation. A simpler transformation of functions is the operation of multiplying all functions by the same constant, or by a specified function.

2

OPERATIONAL MATHEMATICS

SECo 1]

In each of the above examples inverse images exist; that is, when the image is given, a function F(t) exists which has that image. A transformation T{F(t)} is linear if for every pair of functions Fl(t) and F2(t) and for each pair of constants C 1 and C2, it satisfies the relation (1) Thus the transform of a linear combination of two functions is the same linear combination of the transforms of those functions, if the transformation is linear. Note the special casesof Eq. (1) when C2 = Oand when C 1 = C2 = 1. The examples cited above represent linear transformations. The class of functions to which a given transformation applies must generally be limited to some extent. The transformation D{F(t)} applies to all differentiable functions, and the transformation 1{F(t)} to all integrable functions. Linear integral transformations of functions F(t) defined on a finite or infinite interval a < t < b are particularly useful in solving problems in linear differential equations. Let K(t,s) denote some prescribed function of the variable t and a parameter s. A general linear integral transformation of functions F(t) with respect to the kemel K(t,s) is represented by the equation

r

(2) T{F(t)} = K(t,s)F(t) dt. It represents a function f(s), the image,or transform,of the function F(t). The class offunctions to which F(t) may belong and the range ofthe parameter s are to be prescribed in each case. In particular, they must be so prescribed that the integral (2) exists. We shall seethat with certain kemels K(t,s) the transformation (2), when applied to prescribed linear differentialforms in F(t), change&those forms into algebraicexpressionsinf(s) that involvecertain boundary values ofthe function F(t). Consequently, classes of problems in ordinary differential equations transform into algebraic problems in the image of the unknown function. If an inversetransformation is possible,the solution of the original problem may be determined. Boundary value problems in partial differenÍial equations can be simplifiedin a similar way. The operational mathematics presented in this book i:¡,the theory as well as the application of such linear integral transformations that bears on the treatment ofproblems in ordinary or partial differential equations. Later on, we shall return to the general transformation (2)and to the question of deciding upon the special cases that may apply to a given problem in differential equations. First we present the special case that is of greatest general importance, the operational mathematics of the Laplace transformation. Other prominent cases include the various Fourier transformations, to be presented later.

THE LAPLACE TRANSFORMATION

[SECo 2

3

When a = O and b = 00 and K(t,s) = e-sr, the transformation (2) becomes the Laplace transformation. The direct application of this transformation replaces the earlier symbolic procedure known as Heaviside's operational calculus.1 The development of the transformation and the accompanying operational calculus was begun before Heaviside's time; Laplace (1749-1827) and Cauchy (1789-1857) were two of the earlier contributors to the subject,2 In this chapter we present the basic operational property ofthe Laplace transformation, the property that gives the image of differentiation of functions as an algebraic operation on the transforms of those functions. In the following chapters further properties of the transformation will be derived and applied to problems in engineering, physics, and other subjects. Applications to boundary value problems in partial differential equations will be emphasized. Our study of the Laplace transformation leads to the theory of expanding functions in series of the characteristic functions of Sturm-Liouville systems. Such expansions form the basis of the method of solving boundary value problems by separation of variables, a classical method of great importance in partial differential equations.3 Furthermore, we can use that theory to adapt the integral transformation (2) to certain types of linear boundary value problems. 2

DEFINITION

Ir a function

OF THE LAPLACE TRANSFORMATION

F(t), defined for all positive values of the variable t, is multiplied

by e-sI and integrated with respect to t from zero to infinity, a new function f(s) of the parameter s is obtained; that is, {O e-SIF(t) dt = f(s). As indicated in the preceding section, this operation on a function F(t) is called the Laplaee transformation of F(t). It will be abbreviated here by the symbol L{F}, or by L{F(t)}; thus L{F} = foaJe-SIF(t) dt. The new function f(s) is called the Laplaee transform, or the image,ofthe objeet funetion F(t). Wherever it is convenient to do so, we shall denote the I Oliver Heaviside,

English electrical

engineer,

1850-1925.

2 For historical accounts see J. L. B. Cooper, Heaviside and the Operational Gazette, vol. 36, pp. 5-19, 1952, and the references given there. 3 That method 2d ed., 1963.

is presented

in the author's

"Fourier

Series and Boundary

Calculus,

Math.

Value Problems,"

4

OPERATlONAL MATHEMATICS

SECo 2]

object function by a capital letter and its transform by the same letter in lowercase. But other notations that distinguish between functions and their transforms are sometimes preferable; for example, tfJ(s) =

or

L{ f(t)}

.9(s) = L{y(t)}.

For the present, the variable s is assumed to be real. Later on, we shall let it assume complex values. Limitations on the character of the function F(t) and on the range of the variable s will be discussed soon. Let us note the transforms of a few functions. First, if F(t) = 1 when t > O, then

L{F}

=

hence, when s > O,

1 e-sI dt = --e-st o s 1 L{I} = -. s

f

]; o

Ir F(t) = elctwhen t > O,where k is a constant, then

L{F} hence, when s > k,

=

1

fo ~e-st

dt

]

= -e-(s-k)1 k-s

o

;

L{e"'}=~.s-k

With the aid of elementary methods of integration, the transforms of many other functions can be written. For instance,

k L{ sin kt} = S2 +

s

.

'),

L{cos kt}

= S2 +

., ')

when s > O; but soon we shall have still simpler ways to obtain those transforms. It follows from elementary properties of integrals that the Laplace transformation is linear in the sense defined by Eq. (1), Seco1. We can illustrate the use of this property by writing L

!e"I- !e-lct =!~ 2 } 2s-k

{2

- !~.

2s+k'

when s> k and s> -k; that is,

~

L{sinhkt} = s -

(s > Ikl).


[SEC.3

5

PROBLEMS

1. Use the linearity property and known transfonns to obtain these transfonnations, where a, b, and e are constants: (s > O);

(a) L{a + bt} = asS2 +b (b) L{a + bt + et2} = L{(a + bt) + et2} as2+bs+2e

(s > O);

S3

a + bs (e) L{asint + bcost}

= S2

(s > O);

+ 1

s (d) L{cosh et}

(e) L{e'"

-

ebt}

= S2 -

(s> lel);

e2

(s> a and s> b).

= (s _aa~sb- b)

2. Use trigonometric identities, such as 2 COS2 t = 1 + cos2t, 2 sinat sinbt = cos (a - b)t - cos (a + b)t, and known transfonns to find f(s) when s > O, in case F(t)is (a) cos2 t; (b) sin2 t; (e) sin t sin 2t; (d) sin t cos t; (e) sin3 t

s2+2

Ans.(aL,

~2

.

2

,,;

(b)s(S2+ 4);

=W-

cos 2t) sin t.

4s

(e)(r + l)(r + 9); 1 ; (d)-r-s +4

(e)

6 .

"

..

-.'

3. Show that the linearity property (1),Seco1, can be extended to linear combinations of three or more functions when all the transfonns existo 4. If for all functions of some class and for every constant C a transfonnation T satisfies the two conditions T{F(t)

+ G(t)} =

T{F(t)} +T{G(t)},

T{CF(t)} = CT{F(t)}, prove that the transformation is linear. 3

SECTIONALL y CONTINUOUS ORDER

FUNCTIONS. EXPONENTIAL

A function F(t) is seetionally continuous on a bounded interval a < t < b if it is such that the interval can be divided into a finite number of subintervals interior to each of which F is continuous and has finite limits as t approaches

either end point of the subinterval

from the interior.

6


SECo 3]

Thus the values of such a function may take at most a finite number of finite jumps on the interval (a,b). The class of sectionally continuous functions includes functions that are continuous on the closed interval a ~ t ~ b. The integral of every function of this class, over the interval (a,b), exists; it is the sum of the integrals of the continuous functions over the subintervals. The unit step function Sit)

=O =1

when O < t < k, when t > k,

is an example of a function that is sectionally continuous on the interval O < t < T for every positive number T (Fig. 1). The Laplace transform of this function is oo

f

.

Sk(t)e-SI dt

o

=

f

1

oo

e-sI dt

k

= --e-sI;J S

oo k

thus whenever s > O, e

-ks

L{Sit)} = s' A function F(t) is of exponential order as t tends to infinity, provided some constant IXexists suchthat the product e-"'/IF(t)1 is bounded for all t greater than some finite number T. Thus ¡F(t)1does not grow more rapidly than M e"'las t -+ 00, where M is some constant. This is also expressed by saying that F(t) is of the order of e"'/,or that F(t) is ((J(e"'/). The function Sk(t) above, as well as the function tn, is of the order of e"'las t -+ 00 for any positive IX;in fact, for the first function and, when n = O,for the second, we may write IX= O. The function él is of exponential order (IX~ 2); but the function el1is not of exponential order. Ir a function F(t) is sectionally continuous on each bounded interval O < t < T and if, for some constant a, F is ((J(e"'f),.then the Laplace transform of F exists whenever s > IX.This follows from a well-known comparison test

o

(k,O)

(T,O)

Fig. 1


.

[SECo 4

7

for the convergenceofimproper integrals (seeProbo14,Seco5). For in view of the sectional continuity of F, and consequently of the product e-S'F(t), that product is integrable over every bounded intervalO < t < T. AIso, since F is C9(ea'), a constant M exists such that for all positive t le-S'F(t)1 < Me-(s-a),. But the integral from O to 00 of M e- (s-a)' exists when 8 > OC;consequently,

not only the convergence but also the absolute convergence of the Laplace integral .

is ensured when 8 > OC.

The above conditions for the existence of the transform of a function are adequate for most of our needs; but they are 8ufficient rather than necessary conditions.

The function F may have an infinite discontinuity

at

t = Ofor instance, that is, IF(t)1 --+00 as t --+O,provided that positive numbers m, N, and T exist, where m < 1, such that IF(t~ < N/tm when O < t < T. Then if F otherwise

satisfies the above conditions,

its transform

still exists

because of the existence of the integral

{T e-S'F(t) dt. For example, when F(t) = Ct, its transform can be written, after the substitution of x for fit, in the form

(1)

f

'"

t-te-s,

dt = -

'"

l

JS o

o

The last integral has the value fi/2

2

e-x2 dx

(8 > O).

(Prob. 10, Seco5); hence (8 > O).

4

TRANSFORMS OF DERIVATIVES

By a formal integration by parts wehave L{F'(t)} = 1'" e-S'F'(t)dt = e-S'F(t)J~

+ 8 {'" e-S'F(t)dt.

8


SECo 4]

Let F(t) be of order of elZlas t approaches infinity. Then whenever s > oc,the bracketed term becomes F(O),and it follows that

-

(1)

L{ F'(t)} = sf(s) - F(O),

where f(s)

=

L{ F(t)}.

Therefore in our correspondence between functions differentiation of the object function corresponds to the multiplication ofthe result function by its variable s and the addition of the constant - F(O).Formula (1)thus gives the fundamental operational property of the Laplace transformation, the property that makes it possible to replace the operation of differentiation by a simple algebraic operation on the transformo As noted above, formula (1) was obtained only in a formal, ormanipulative, manIler. It is not even correct when F(t) has discontinuities. The following theorem will show to what extent we can rely on our formula. Theorem 1 Let the function F(t) be continuous with a sectionally continuous derivative F'(t), over every finite intervalO;;;¡;t ;;;¡;T. Also let F(t) be of order of elZlas t -. oo. Then when s > oc,the transform of F'(t) exists, and (2)

L{F'(t)} = sL{F(t)}

- F(O).

Since F(t) is continuous at t = O,the number F(O)here is the same as F( +0), the limit of F(t) as t approaches zero through positive values. To prove this theorem, we note first that

f

L{F'(t)} = lim

T

e-SIF'(t) dt,

T oo o

ifthis limit exists. We write the integral here as the sum ofintegrals in each of which the integrand is continuous. For any given T, let ti' t2," ., tn denote those values of t between t

=

O and t

=

T for which F'(t) is discontinuous

(Fig. 2). Then

f

T

o

e-SIF'(t)dt =

.F1t)

f

l2

ll

o

e-stF'(t)dt +

I I I I I I I

01

(tl,O)

(t2,OI Fig.2

I

(T,O)

f

I1

e-SIF'(t)dt +oo. +

f

T

In

e-SIF'(t)dt.


[SECo 4

9

After integrating each of these integrals by parts, we can write their sum as

]o

+ e-stF(t)

] tI

Now F(t) is continuous

(3)

T

t2

tl

e-stF(t)

+

]

... + e-stF(t)

tn

T

+s

f

o

e-StF(t)dt.

so that F(t 1 - O) = F(t 1 + O),etc., and hence

{T e-stF'(t)dt= -F(O) + e-STF(T)+ s{T e-stF(t)dt. Since IF(t)1< Meat for large t for some constants O(and M, it follows

that le-sTF(T)1 < Me-(s-a)T,

and since s > 0(,this product vanishes as T ~ oo. AIso the last integral in Eq. (3) approachesL{F} as T ~ '00 because F is continuous and lP(eat). Hence the limit as T ~ 00 of the right-hand member of Eq. (3) exists and equals

-

F(O) + sf(s); therefore,

the same is true of the left-hand

member.

Thus Theorem 1 is pro ved. If F is continuous when t ~ O except for a finite jump at to, where to > O,the other conditions remaining as stated in the theorem, the above proof is easily modified to show that our formula (2) must be replaced by the formula (4)

L{ F'(t)} = sf(s)

-

F(O) - [F(to+ O)- F(to - O)]e-sto.

The quantity in brackets is the jump of F at to. We use the symbol F' here and in the sequel to denote the function whose value is the derivative of F wherever the derivative exists. In the case of our step function Sk(t), for instance, S~(t) = O when O < t < k and when t > k, but S~(k)has no value. To obtain the transform of the derivative F" of the second order, we apply Theorem 1 to the function F'. Let both F and F' be continuous when t ~ O and lP(ea.t);also let F" be sectionally continuous on each bounded interval. Then L{F"(t)} = sL{F'(t)} - F'(O) = s[sL{ F(t)} - F(O)]- F'(O). Hence we have the transformation (5)

L{ F"(t)} = s2f(s)

-

sF(O)

- F'(O).

When Theorem 1 is applied to F1n-1)(t)to write L{Fln)(t)} = sL{Fln-l)(t)} - F1n-l)(0) and again to write L{Fln-l)(t)} in terms of L{Fln-2)(t)}, and so on, the following result is indicated.

10


SECo 5]

2 Let F and each of its derivatives of order up to n - 1 be continuous functions when t ~ O and &(eC%t); also let F(n)(t)be sectionally continuous on each bounded intervalO < t < T. Then the transform of the derivative F(n)(t)exists when s > ex,and it has the following algebraic expressio~ in terms of the transform f(s) of F(t):

Theorem

(6)

L{F(n)(t)} = snf(s) - sn-1F(0) - sn-2F'(0) - s"-3F"(0) - ... - F(n-l)(O)

(s > ex).

Theorem 2 can be pro ved by using Theorem 1 and induction. Under the conditions stated, suppose that formula (6) is valid when n is replaced by some integer k where O < k < n. But Theorem 1 expresses L{ F(k+1)} in terms of L{ F(k)}to which (6)now applies to show that formula (6)is true when n is replaced by k + 1. But the formula is true when k = 1, according to Theorem 1; it is therefore true when k = 2, hence when k = 3,. . . , n. The Laplace transformation resol ves the differential form F'(t) in terms of f(s), s, and the initial value F(O) when F satisfies the conditions stated in Theorem 1. As a consequence ofTheorem 1, Theorem 2 shows how that transformation resolves the iterates F"(t), F"'(t),... of that form in terms of f(s), s, and initial values of F and its deriva tives. 5

EXAMPlES.

THE GAMMA FUNCTION

In order to gain familiarity with the above fundamental operational property of the transformation, let us first use it to obtain a few transforms. Example 1 Find L{t}. The functions F(t)

= t and

F'(t) = 1 are continuous, and F is

&(eC%t) for any positive ex..Hence,

L{F'(t)} = sL{F(t)} or

L{l}

-

(s > O),

F(O)

= sL{t}.

Since L{ 1} = l/s, it follows that

1 L{t} = s2

(s > O).

Example 2 Find L{sin kt}. The function F(t) = sin kt and its deriva tives are all continuous and bounded, and therefore of exponential order, where ex= O. Hence L{ F"(t)} = S2L{ F(t)} or

-

sF(O)

-

F'(O)

-k2 L{sin kt} = s2L{sin kt} - k.

(s > O),


[SECo 5

11

Solving for L{sin kt}, we see that

~

(s > O).

L{sinkt} = s + Example 3 Find L{ tm} where m is any positive integer. The function F(t)

=

tm satisfies all the conditions

of Theorem

2

for any positive ex.Here F(O)

= F'(O)= .. . = F(m-l)(O)= O,

F(m)(t)= m!,

F(m+l)(t) = O.

Applying formula (6) when n

= m + 1, wefind that

L{F(m+l)(t)} = O = sm+lL{tm} - m!, and therefore (s > O).

(1)

This formula can be generalized to the case in which the exponent is not necessarily an integer. To obtain L{tk} where k> -1, we make the substitution x = st in the Laplace integral, giving (s > O). The integral on the right represents the gamma function, or factorial function, with the argument k + 1. Hence (2)

L{tk}

= r(ks

+ 1)

(k > -1, s > O).

Formula (1) is a special case of (2) when k is a positive integer (see Probo 13). Example 4 Find L{fh F(r) dT} when F is sectionally continuous and of exponential order. The function (3)

= {F(T) dT G(O)= O.AIso G'(t) =

G(t)

is continuous (Sec. 14),and F(t), except for those values of t for which F(t) is discontinuous; thus G'(t) is sectionally continuous on each finite interval. Ir the function G(t) is also l!J(e"t), then according to Theorem 1, L{ G'(t)} = sL{ G(t)} = L{ F(t)}

(s> ex);

12


SECo 5]

thus, if a > Oso that s > O,

(4)

L{{

F(T)dT} = ~f(S)

(s > a > O).

To show that the integral (3) represents a function of exponential order when F is sectionally continuous and of exponential order, we first note that constants a and M exist such that jF(t)1< M eatwhenever t ~ O, and if the number a is not positive, it can be replaced by a positive number. Then t

IG(t)1~

IF(T)I dT

SO

t M < M S ea'dT = -(eat - 1) o a

(a > O),

and therefore (a > O). This establishes the exponential order of the function (3). PROBLEMS 1.

State why each of the functions (a) F(t)

= te2t

and (b) G(t)

- 1 when = t+ 1 {1

O< t < 2

when t > 2,

is sectionally continuous on every interval.O < t < T and (!)(e"t)as t -+ 00, where a > 2 for F and a ~ Ofor G. 2. State why neither of the functions (a)(t - 1)-1 or (b)tan t is sectionally continuous on the intervalO < t < 3. 3. Ir Sk(t)is the unit step function (Sec. 3),draw graphs of the following step functions and find their transforms when s> O: (a) F(t) = 1 Sl(t); (b) G(t) = Sl(t) - S2(t). AIso, note that both F and G vanish except on bounded intervals; thus their Laplace integrals become definite integrals that exist for all s ( - 00 < s < 00). Show that /(0) = 1 and g(O) = 1,and that

-

(a)les) = (1 - e-")/s (s:;' O); (b) g(s)= (e-S e-2s)/s (s:;' O).

-

4.

Obtain these transforms with the aid of Theorem 2:

5.

(s> O); (a) L{coskt} = S/(S2 + p) (b) L{sinh kt} = k/(S2 - k2) (s> IkJ). Given the transform of e"', use Theorem 1 to show that

1 L{te'"} = (s - k)2

(s > k).


[SECo 5

13

-

6. If G(t)= OwhenO;;¡; t ;;¡;k and G(t) = t k whent ;¡;;k, drawgraphs of Gand G'. Giventhe transformof Sit), (a)apply Theorem 1to prove that g(s)= s-2e-ks (s> O); (b) show that G(t) = f~Slr) d! and find g(s) from formula (4), Seco5. 7. Prove that the function F(t) = sin (eI2)is of exponential order (IX ;¡;;O)and that its derivative F' is not of exponential order. Show that Theorem 1 ensures the existence of the Laplacetransformof F' when s > O,in this case whereF' is not of exponential order. 8. (a) Derive Eq. (4),Seco4. (b)Illustrate that equation by using it to find the transform of the unit step function Sk(t). 9. (a) If a function F and its derivative F' are both (J)(e"')and continuous when t ;¡;;O except possibly for finite jumps at a point to, and if F" is sectionally continuous on each intervalO < t < T, apply Eq. (4), Seco4, to derive the formula L{F"(t)} = s2f(s) - sF(O) - F'(O) - se-SIO[F(to+ O)- F(to - O)] - e-srO[F'(to + O) - F'(to - O)]

(s > IX,to > O).

(b) Show that the formula applies to the function F(t) = sin t when O ;;¡; t to give f(s) 10.

;;¡; 1t,

F(t)

= O when

t ;¡;;1t

= (1 + e-~S)/(s2+ 1)when - 00 < s < oo.

Let J denote the second integral in Eq. (1), Seco3; then J2 = fo'" e-x2 dx t'" e-y2 dy = fo'" fo'" e-(X2+y2)dx dy.

Evaluate the iterated integral here by using polar coordinates and show that J = .;:;r/2. 11. Prove that each linear combination AF(t) + BG(t) of two functions F and G of exponential order is also of exponential order. 12. Use properties of continuous functions to show that iftwo functions are sectionally continuous on an interval (a,b) then (a) each linear combination of the two is also sectionally continuous on that interval; (b)the product ofthe two functions is sectionally continuous on the interval.

13. As noted in Seco5, the gammafunction is definedfor positivevaluesof r by the formula r(r) = t'" x,-Ie-x dx

(r >0).

(a) Integrate by parts to show that the function has the factorial property f(r + 1) = rf(r). (b) Show that f(1) = 1,and hencethat f(n + 1)= n! whenn = 1,2,.... (e) From the value of the integral J found in Probo 10, show that r(t) = .;:;r. and formula (2)to show that L{Jt} = t.Jir/si. Then use the factorial property to find 14. Improper integrals Consideronly functions that are sectionally continuous on each intervalO < t < T.

nt)

14

OPERATIONAL

SEC.6]

MATHEMATICS

(a) If O ;;;¡¡g(t) ;;;¡¡h(t) whenever t > O and if S~ h(t) dt exists, use the definition of the improper integral

['" g(t) dt = lim [T g(t) dt Jo T-",Jo to prove the existence of that integral. [Note that the value of the last integral is nondecreasing as T increases. It never exceeds 1 where 1 = S;> h(t) dt because S~ g(t) dt ;;;¡¡ h(t) dt ;;;¡¡1, so it has a limit.] . (b) Iff;> Ip(t)1dtexists,provethatf;> p(t)dtexistsby writingp(t) = (P(t)+ Ip(t)l]-

f{;

Ip(t)l. Note that O ;;;¡¡p(t) + Ip(t)1;;;¡¡2Ip(t)1 and apply the test in part (a) to those two components of p(t). r dt exists, (e) Prove this comparison test: If Iq(t)1;;;¡¡r(t) whenever t > O and if then the improper integral S'" q(t) dt is absolutely convergent, and the integral itself

S;

exists.

6

THE

o

INVERSE

Let the symbol

Thus if

L

TRANSFORM

- 1 {f(s)} denote a function whose Laplace transform is f(s). L{ F(t)} = f(s),

then

F(t) = L -l{f(S)}.

Using two of the transforms obtained in the foregoing sections, we can write, for instance, L -1

~

{s - k }

= ekl,

L -1

k = sinkt: S2 + k2 } {

This correspondence between functions f(s) and F(t) is called the inverse Laplace transformation, F(t) being the inverse transform of f(s). In the strict sense of the concept of uniqueness of functions, the inverse Laplace

transform

is not unique.

The function

F1(t)

= ekt

is an inverse

transform of 1/(s - k); but another, for instance, is the function (Fig. 3) F2(t) =

=

ekt

when O < t < 2, or t > 2,

1

when t = 2.

For the transform of Fz{t) is foa)e-srF2(t)dt = f: e-s1ekldt + ta) e-srél dt, and this is the same as L{ ekl}. The function F2(t) could have been chosen equally well as one that differs from Fl(t) at any finite set ofvalues oft, oreven at such an infinite set ast = 1,2,3,....


[SECo6

FZ(tI!

15

../ I

I

f(2,1) I I I

O Fig.3

A theorem on uniquenessof the inversetransform will be proved later (Sec. 69). To state it, we first define a c1ass of functions of exponential order. Let the class tff denote the set of all functions F(t) defined on the half line t > O,sectionally continuous on each bounded interval, and defined at eachpoint to whereF is discontinuousas the mean value ofits limitsfrom the right and left, F(to) = t[F(to + O)+ F(to - O)]

(to > O);

also, for each individual function F of the set, let constants M and IXexist such that IF(t)1< M ea.twhen t > O. As we have seen, F then has a transform f(s) defined on some half line s > IX. The theorem states that no two functions of class tff can have the same transforms. Thus if f(s) is the transform of some function F(t) of c1ass tff, then F(t) is the unique inverse transform L -1 {J(s)} in tff. For'example, the only function L -1{e-S/s} ofc1ass rf is the unit step function S 1 (t) defined in Seco3 if S 1(1) is defined to be t. As another example, the only function L - 1 {1/(s k)} of c1ass tffis ekt.

-

It is well to note here that not everyfunction of s is a transformo The

.

kind of functions f(s) that are transforms of functions F(t) of broad c1asses are limited, as we shall see (Chap. 6), by conditions of regularity that include requirements that f be continuous on a half line s > IXand that f(s) -+ Oas s -+ oo. We have noted that if L{ F} and L{ G} exist, then (1)

L{ AF(t) + BG(t)}

= Af(s)

+ Bg(s)

whenever A and B are constants. Let us restrict our functions of t to those of c1ass tff, and functions of s to transforms of such functions. Then unique inverse transforms exist, and the linearity property (1) can be written (2)

L -1{Af(s)

+ Bg(s)} = AF(t) + BG(t) = AL

that is, L - 1 is also

-1{J(S)} + BL -1{g(S)};

a linear transformation

of functions.

16

SECo 7]

OPERATIONAL

MATHEMATICS

The most obvious way of finding the inverse transform of a given function of s consists of reading the result from a table of transforms. A fairIy extensive table is given in Appendix A. But we shall take up methods of obtaining inverse transforms of certain combinations and modifications of functions of s, as well as methods of resolving such functions into those listed in the tables. With the aid of such procedures, we shall be able to make much use of the transformation. In addition, there are explicit formulas for L -l{f(S)}. The most useful of these formulas involves an integral in the complex planeo To use this integral, we must let s be a complex variable and we must be prepared to employ some theorems in the theory of functions of a complex variable. 7

A THEOREM ON SUBSTITUTION

Let a function F(t) be such that its Laplace integral converges when s > IX.Then, replacing the argument of the transform f(s) by s - a,wherea is a constant, we have

when s

-

a > IX.Therefore

(1)

(s > IX+ a).

f(s - a) = L{ e'"F(t)}

Let us state this simple but important property as a theorem. Theorem 3 The substitution of s a for the variable s in the transform corresponds to the multiplication of the object function F(t) by the function e"', as shown in formula (1). ~

To iIlustratethis property, let us recall that mI 5"'; 1

Hence

= L{ tm}

m! 's - a)m+ 1

= L{ tme"t}

(m = 1,2,... ; s > O). (s > a).

As another iIlustration,

s L{coskt} = S2+ and therefore

s+a L{e-a, cos kt} = (s + a)2 + k2

(s > O),

(s> -a).


8

[SECo 8

THE USE OF PARTIAL FRACTIONS

17

(TABLE 1)

A few examples will show how the theory of partial fractions can be used in finding inverse transforms of quotients of polynomials in s. In the next chapter, a more systematic use of this procedure will be introduced.

Example 1 Find L -l{(S

+ 1)/(S2 + 2s)}.

The denominator of the function of s here is of higher degree than the numerator and has factors that are linear and distinct.. Therefore constants A and B can be found such that

s+l -=-+s(s + 2)

A

B

s

s +2

for all valuesof s exceptOand - 2. Clearing fractions, we have s + 1 = (A + B)s + 2A, and this is an identity if A + B = 1and 2A = 1. Thus A = B = !, and hence s+l 11 1 1 +-- 2 s 2 s + 2' .

Since we know the inverse transforms of the two functions on the right, we have the result L-

{

1

S

S2

+ 1 =! ! - 2/ + 2s 2 + 2 e .

}

The procedure can be shortened for such a simple fraction by writing

s + 1 = !(s + 2) + !s, s+l

1 1 +-S(S+ 2) - 2 s 2 s + 2'

and hence

Example 2 Find L -lL(s

11

:

a)2}'

In view of"the repeatedlinear factor, we write a2 s(s + a)2

A B =-+-+ s s+ a

e (s + a)2'

Clearing fractions and identifying coefficients of like powers of s as before, or else by noting that

a2 = (s + a)2 - s(s + a) - as,

Table 1

A short table of transforms

F(t)

f(s)

1

1

-1

2

e'"

-

3

t" (n = 1,2,...)

n! 5"+1

4

t"e'" (n = 1,2,...)

5

sin kt

6

cos kt

7

sinh kt

a (s > a) O

s

1

a

s-a

O

n!

a

(5- a)"+ 1 k S2

+

O

k2

s S2 + k2

O

k 1'"

S2

-

S2

- k2

(5

+ a)2 +

P

5

8

coshkt

9

e-al sinkt

10

e-al cos kt

Ikl

k

.

k2

5+a

-a -a

(5 + a)2 + k2

11

fi

Ji

O

2.j?

12

1 Ji

13

t1(k > -1)

r(k + 1) 7+1

O

14

t1e"'(k>-1)

r(k + 1) (5 - a)1+1

a

15

SJt) (Sec. 3)

-e-u s

O

16

e'" - ebl(a > b)

a-b (s - a)(s - b)

a

17

-sinat--sm 1 1 . bt a b

18

cos at

O

-

cos bt

b2 - a2 (s2 + a2)(52 + b2)

-

(b2 a2)s (S2 + a2)(52 + b2)

O O


[SEC.8

19

we find that a2

1

1

s(s + af

= -;-

a

s + a - (s + a)2.

Referring to Table 1, we can now write the result a2

L-l

{ s(s + a)2}

=

1

-

- ate-al.

e-al

Example 3

Since s

S

(S2 + a2)(s2 + b2)

= a2 -

(S2 + a2) b2

(S2

-

(S2

+ b2)

+ a2)(s2 + b2)

Example 4 Find F(t) if

=.

f(s)

Ss + 3

~. I '"

In view of the quadratic factor, we write Ss + 3 A -~+ (s - 1)(s2+ 2s + 5) - s - 1

Bs + e S2+ 2s + 5.

After clearing fractions and identifying coefficients of like powers of s,

wefindthat A = 1,B = -1, and e = 2; thus 1 s- 2 f(s) = -s - 1 1

s+l

s- 1

(s + 1)2+ 4

=--

+-

3

(s + 1)2+ 4.

Referring to Table 1, or to Theorem 3, we see that F(t) = el

- e-'(cos 2t - ! sin 2t).

20

9


SECo 9]

THE SOLUTION OF SIMPLE DIFFERENTIALEQUATIONS

The application of the Laplace transformation to the solution of linear ordinary differential equations with constant coefficients, or systems of such equations, can now be made clear by means of examples. Such problems can of course be solved also by methods studied in a first ~ourse in differential equations. Later on, when we have developed further properties ofthe transformation, we shall solve problems of this sort with greater efficiency. We shall also be able to solve more difficult problems, especially in partial differential equations. Example 1 Find the general solution of the differential equation (1)

Y"(t) + k2 Y(t) = O. Let the value of the unknown function at t

the constant

= O be

A and the value of its first derivative

at t

denoted by

=

O by the

constant B; that is,

Y'(O)= B.

Y(O)= A,

In view of the differential equation, we can write L{ Y"(t)} + P L{ Y(t)} = O,

assuming that Yand Y" have transforms. If the unknown function Y satisfiesthe conditions in Theorem 2, then L{ Y"(t)} = s2y(S) - As

-

B,

where y(s) = L{ Y(t)}. Hence y(s) must satisfythe equation s2y(S)

-

As

- B + k2y(s)= O,

which is a simple algebraic equation. Its solution is clearly s y(s)

= A s2 +

B

k2

_

k

+ k s2 + - . ~

Now Y(t) = L -l{y(s)},and the inversetransformsofthefunctions on the right in the last equation are known. Hence (2)

Y(t)

= A cos kt

+

~ sin

kt,

= A cos kt + B' sin kt, where A and B' are arbitrary constants since the initial values Y(O)and Y'(O)are not prescribed. To verify our formal result given by formula (2), we need only to find Y"(t) from that formula and substitute into Eq. (1) to see that the


[SECo 9

21

differential equation is satisfied regardless of the values of A and B'. Thus it is not necessary to justify the use of Theorem 2. However, our function A cos kt + B' sin kt does satisfy all conditions in that theorem, and the order of the steps taken above can be reversed to show in another

way t)lat our function

equation. These

satisfies the differential

remarks on verifying the solution apply also to the other examples and problems that follow in this section.

Example 2 Find the solution of the differential equation (3)

Y"(t)

-

Y'(t)

-

6Y(t)

=2

satisfying the initial conditions Y(O) =

(4)

1,

Y'(O) = O.

Applying the transformation to both members of the differential equation, and letting y(s) denote the transform of Y(t), we obtain formally the algebraic equation 2 s2y(S) - S - sy(s) + 1 - 6y(s) = -, s where we have used the initial conditions in writing the transforms of Y"(t) and Y'(t). Hence S2 - s + 2 (S2 - S - 6)y(s) = , s or

y(s) =

S2

-

s+2

s(s - 3)(s + 2)

A

B

e

s

s- 3

s+ 2

=-+-+-.

Evaluating the coefficients A, B, and e as in the preceding section, we find that 11 8 1 4 1 y(s) = ---+--+--. 3s 15 s - 3 5 s + 2 Hence

Y(t)

= -j + /se3t + !e-2t.

It is easy to verify that this function Y satisfies the differential equation (3) and both conditions (4). Example 3 Find the functions Y(t) and Z(t) that satisfy the following system of differential equations:

-

Y"(t)

2 Y"(t) Y(O)

-

Z"(t)

Z"(t)

+ Z'(t) - Y(t) =

-

et

-

2Y'(t) + Z(t) = - t,

= Y'(O)= Z(O) = Z'(O) = o.

2,

22

OPERATIONAL

SEC.9]

MATHEMATICS

Let y(s)and z(s)denotethetransformsof Y(t)and Z(t),respectively. Then in view of the differential equations and the initial conditions, those transforms formally satisfy the following simultaneous algebraic equations: 1 2 s2y(S) - S2Z(S) + sz(s) y(s) = s- 1 - -, s

_

-

2s2y(s)- S2Z(S)- 2sy(s)+ z(s) =

1 s

-2"'

These equations can be written .

s-2 s(s - 1)2' 1 S2(S- 1)"

(s + l)y(s) - sz(s) = 2sy(s) - (s + l)z(s)

=

Eliminating z(s), we find that s2-2s-1

(S2- 2s - l)y(s) = s(s -

W.

With the aid of partial fractions, we then find that

1

B e y(s) = s(S - 1)2 = S + s - 1 + (s 1 =---+

s

Therefore

A

1

1

s - .1

(s - 1)2'

Y(t)

W

= 1 - e' + te'.

Likewise we find that 2s-1 z(s)

1

= -'-'- -

1)2 =

j

and therefore

Z(t)

-2"S

1

+ ;--

H?>

= -t + te'.

Example4 Solvethe problem Y"'(t)

Y(O)= o, Let e

-

2Y"(t)

+ 5Y'(t) = O,

Y'(O)= 1,

denote the unknown initial value Y"(O). Then

S3y(S)

-

s - e - 2s2y(s)+ 2 + 5sy(s)= O,


so

[SECo 9

23

that

C-2+s y(s)

= S(S2 -

2s + 5)

C- 2 1

[-:;-

= SC Thus

Y(t)

-

2

= --s-

s

-

1

(s - 1)2 +

]

C+ 3

4 + 10

2 (s - 1)2 +

4'

c + 3 C- 2 + e'( 10 sin2t - Scos 2t) .

Since Y(n/8) = 1,it followsthat C- 2 e"/8 1 = -+-(C + 3 - 2C + 4), 5 10j2 or that C = 7. Hence the solution is Y(t)

= 1+

e'(sin2t - cos 2t).

PROBLEMS 1. Use Theorem 3 to (a) obtain entry 9 from entry 5 in Table 1 and (b) show that L -l{(S - a)-t} = e"'(1tt)-t, given L -l{S-t}. 2. Use partial fractions to find the inverse transforms of a

a2

(a) s(s + a);

(b) s(S2+ a2);

a3 (e) s(s + a)3'

Ans. (a) 1 - e-a,; (b) 1 - cos at; (e) 1 - (1 + at + ta2t2)e-a,. 3. Use partial fractions to obtain the inverse transforms shown in (a) entry 16 of Table 1; (b) entry 17 of Table 1.

Solvethe followingproblemsand verifythat your solutionssatisfythe differential equatioñs and any accompanyingboundary conditions. 4. Y"(t)- k2Y(t)= O(k # O). Ans. Y(t) = C¡e'"+ C2e-kt. 5. Y"(t)- 2kY'(t) + k2Y(t) = O. Ans. Y(t) = e"'(C¡+ C2t). 6. Y"(t)+ k2Y(t) = a. Ans. Y(t) = C¡ sinkt + C2coskt + a/k2. 7. Y"(t)+ 2Y'(t) + 2Y(t) = O,Y(O)= O,Y'(O)= 1. Ans. Y = e-' sin t. 8. Y"(t)+ 4Y(t)= sint, Y(O)= Y'(O)= O. Ans. Y(t)= sint - i sin2t. 9. Y"(t)+ Y'(t) = t2 + 2t, Y(O)= 4, Y'(O)= -2. Ans. 3Y(t) = t3 + 6e-' + 6. 10. Y"'(t)+ Y'(t) = lOe2',Y(O)= Y'(O)= Y"(O)= O. Ans. Y(t) = e2'- 5 - 2 sint + 4 cos t.

t

11.

y(4/(t) = Y(t), Y(O) = Y'(O) = Y"(O) = O, Y"'(O) = 2.

12.

X'(t) + Y'(t) + X(t) + Y(t) = 1, Y'(t)

Ans. Y(t) = sinh t - sin t.

.

-

2X(t)

-

Y(t) = O,X(O)= O,Y(O)= 1.

Ans. X(t)

= e-r - 1, Y(t)= 2 - e-'.

24


SECo 10]

13. Y"(t)+ 2Z'(t)+ Y(t) = O,Y'(t) - Z'(t) - 2Y(t) + 2Z(t) = 1 - 2t, Y(O)= Y'(O)= Z(O)= O. Ans. Y(t) = 2(1 - e-' - te-'), Z(t) = Y(t) - t. . 14. Y"(t)+ Y(t) = t, Y'(O)= 1, Y(n) = O. Ans. Y(t) = t + n cos t. 15. y"(x) + y(x) = 1,y(O)= 1,y(tn) = O. Ans. y(x) = 1 - sinx. 16. y"(x)+ 2y'(x)+ y(x) = O,y(O)= O,y(l) = 1. 10

GENERATION OF THE TRANSFORMATION

We can show in a manipulative way why a linear integral transformation of functions F(t) defined on the half line t ~ O,

(1)

T {F(t)} =

L CX)K(t,s)F(t)

dt = f(s),

must be essentially the Laplace transformation if it is to have the basic operational property of that transformation, namely, that it replaces the simple differentialform F'(t) by an algebraicform inf(s), s and the initial value F(O)of the function F. Let primes denote differentiation with respect to t. Then by a formal integration by parts we can write (2)

T {F'(t)} = SoCX) K(t,s)F'(t) dt

= K(t'S)F(t)J~ -

SoCX)K'(t,s)F(t)dt

- K(O,s)F(O)-

SoCX) K'(t,s)F(t)

=

dt,

provided that K(t,s)F(t) -> O as t -> oo. The kernel K must involve some parameter s if the transform f is to correspond to a unique function F of some large class; otherwise, f is merely the numerical value of the integral (1) of the product K(t)F(t), a number that is the same for many functions F. In order that the final integral in formula (2) will representf(s) except for a factor A.(s),where A.is some function of s, we require that K'(t,s) = - A.(s)K(t,s).

(3)

Thus K(t,s) = ce-J.'. But it is convenient to write c = 1 and choose A.(s)as the parameter, so we write A.(s)= s here. Then

(4) .

K(t,s) = e-s"

T{F(t)}=

LCX)

e-stF(t)dt = L{F(t)},

THE

LAPLACE

TRANSFORMATlON

[SECo 10

26

and formula (2) is the basic operational property of the Laplace transformation: (5)

T{F'(t)} = -F(O) + sT{F(t)}.

A corresponding procedure will be used later on for the generation of integral transformations that will reduce other differential forms in functions defined on some prescribed interval, in terms of the transform of the function itself and prescribed boundary values of the function or its derivatives at the ends of the interval. Readers acquainted with the concept of a function F(t) (a ~ t ~ b) as a generalized vector1 can see that a linear integral transformation (6)

T{F(t)} =

f

K(t,s)F(t)dt = /(s)

is a generalization o/ linear vector trans/ormations. Our transformation (1) is a special case of (6) in which a = Oand b is infinite. A generalized vector, or function, F has an infinity of components consisting of the values F(t) for each t; that is, the components are all the ordinates of points on the graph of F(t) over the interval a ~ t ~ b. The generalized scalar product, called the inner product, of two such vectors F and G is the generalized sum of products of corresponding components, namely

r

F(t)G(t) dt.

Thus for each fixed s, the integral (6) is the inner product of the generalized vectors K(t,s) and F(t). We begin with ordinary vectors F in three-dimensional space having rectangular cartesian components F1,F2,F3, which may be written F(l), F(2), F(3). A general linear transformation of F into a new vector f can be written in terms of components and constants Kij (i,j = 1,2,3). For the first component we write /1

= KllF1

+

K12F2

+

K13F3

= K1 . F

where K1 is a fixed vector with components Kll, K12, K13, and the dot denotes the scalar product ofvectors. Similarly, K2 and K3 are fixed vectors and /3 = K3 F. /2 = K2 F,

.

.

In space of n dimensions the general linear vector transformation is determined by a familyof n fixed vectors K¡: (7) 1

¡; = K¡ . F

(i = 1,2,..., n).

Churchill, R. V.:"Fourier Series and Boundary Value Problems," 2d ed., chap. 3, 1963.

26


SECo 10]

The extension to generalized vectors, or functions, is now fairly natural.

A fixed function of the variable t and a parameter s, K(t,s), is a family of generalized veetors, one veetor for eaeh fixed value of s of some set, say e ~ s ~ d. The inner produet of K and F representsthe s component of the transformed veetor f(s); that is,

(8)

f(s)

=

f

K(t,s)F(t) dt.

This is the linear integral transformation (6).

Functions of certain elasses,as well as vectors, are elements of linear veetor spaees. The abstract theory of linear spaees is independent of the kind of elements involved. In that respeet the treatment of funetions as veetors is not merely an intuitive approach or an analogy. But the abstraet theory specifies properties of inner products without suggesting formulas for them. The theory will not be needed here.

2 Further Properties of the Transfor.mation

11

TRANSLATION OF F(t)

There are several further operational properties of the Laplace transformation that are important in the applications. Those properties whose derivations and applications do not necessarily involve the use of complex variables will be taken up in this chapter. We begin with an analog of Theorem 3 of the first chapter. According to that theorem, the multiplication of the object function by an exponential function corresponds to a linear substitution for s in the transformo Now let us note the correspondence arising from the multiplication of the transform by an exponential function. Let F(t) ha ve a transform,

f(s) = {X) e-S'F(t) dt. Then 27

28

OPERATIONAL MATHEMATlCS

SECo 11]

where b is a constant, assumed to be positive. Substituting t + b = T, we can write the last integral in the form

{X>e-StF(T- b)dT=

s: O+ {X>e-StF(T -

b) dT.

Thus if we define a function Fb(t)as follows, Fb(t) = O

(1)

when O < t < b,

= F(t we see that

b)

when t > b,

f(s)e-bS = Lco e-S'Fb(T) d'C.

The following property is therefore established. Theorem 1 If f(s) = L{F(t)}, thenfor any positive constant b, (2)

e-bSf(s) = L{Fb(t)},

where Fb(t)is the function defined by Eq. (1).

The function Fb(t)is illustratedin Fig. 4. Its graph is obtainedby translating the graph of F(t) to the right through a distance of b units and making Fb(t)identically zero between t = Oand t = b. We can refer to Fb(t) as the translated function. Our unit step function Sb(t) is the translation of the function So(t) = 1 (t > O). It serves as a familiar illustration of the above theorem, since its transform is s-le-bs. This step function can be used to describe the translation of any function F(t) by writing. . (t > O), provided that F(t - b) is defined where t > O; that is, provided that F(t) has numerical values for those values of t in the range t > - b. For example, the function F(t) = sin kt is defined for all t, so in this case we can write k -bs (t > O,b ~ O). Fb(t) = Sb(t)sin k(t - b) = L- 1 { S2 e+ k2 } Rt)

o

~

o Fig.4

I L (b,O¡

FURTHER PROPERTIES OF THE TRANSFORMATION

[SECo 12

29

On some occasions it is convenient to define F(t) as zero for all negative values of t. When that is done, the graph of Fb(t) is simply a translation of the graph of F(t), and

for all t. Now consider the linear substitution esfor s in f(s), where e is a positive constant. If the transform f(s) of a function F(t) exists whenever s > IX,then whenever

es > IX.

f(es) = {O e-CStP('r)d't The substitution t = e't enables us to write that formula as

f(es) = ~foooe-~p(~) dt in terms of the transform of F(t/e), to establish

Theorem 2 [f L{ P(t)} then

= f(s)

this theorem:

whenever s > IX,and if e is a positive eonstant,

(3) We can write a = l/e to obtain an alternate form of (3): (4)

L{F(at)}

(a > O,s > alX).

= ~f(~)

Given, for example, that L{ cos t} = S/(S2+ 1) when s > O,it follows that L {cos at} = -1

s/a

a (s/a)2+ 1

s

= 2 2 s +a

(s > O).

The effectof a generallinear substitution for s can beseenfrom formula (3) and Theorem 3, Chap. 1, since (5)

12

(a > O).

STEP FUNCTIONS

When t ~ O,the bracket symbol [t] is used in mathematics to denote the . . . , that does not exceed the number t. Thus [n] = 3

greatest integer, O, 1,2,

= [3]. The function [t]is therefore a step function of the type that is sometimes called a staircase function with unit rise and run :

30

SECo 12]

OPERATIONAL

[t]

=O =1 =2

The

staircase

beginning [1 +

t/h] or To

function

at the 1 +

obtain

with

[t/h]. The the

an

t =

origin

(O

;;;¡; t <

1),

(1

;;;¡; t <

2),

(2

;;;¡; t <

3),..

arbitrary

positive

..

ron

h, and

with

O (Fig. 5), is then represented function

transform

c[l +

t/h] has

of the function

MATHEMATlCS

the rise e and

shown

a unit rise

by the symbol the

(1)

may

order

describe together

the

with

function an

by

+

[1

means

>

O),

=

i]

of a difference

initial condition, Y(t)

equation

of the first

as follows.

-

Y(t

+ 1

h)

(t

=0 The

h.

(h

Y(t) = we

ron

in Fig. 5,

function

in view

Y(t -

of Theorem

h) is then 1, the

~

O),

(t < O).

the same

as the translated

transform

y(s) of the

function

function

Yh(t) and,

Y(t) satisfies the

equation

y(s)

=

e

1

h'y(S)

s

+

(s > O).

Therefore the transform ofthe staircase function (1) is

(2)

y(s)

=L

l+

{[

Another

!.

h] }

= !--.!

s1 - j

(s

useful step function is the unitfinite impulse function

1 (3) ¡(h,

t

-

to)

= h when

to <

= O when

t <

t <

to and

10,1)

O

Ih,OI

Fig.5

to

+

when

h,

t >

to

+

h,

>

O).


[SECo 12

31

l(h,t-fo) 11

I '1 II I¡¡ I 'hl 1-:J (fo,O)

o

Fig.6

illustrated in Fig. 6. Let us introduce a variation of our unit step function Sk(t), one that is defined to be zero when t is negative; namely, (4)

So(t

-

to)

=O

when t < to,

= 1

when t > to.

Then our function 1 can be written (5) It follows at once from the transform of So(t

-

to) that

1

(6)

L{I(h, t - to)} = e-sto -, e

-hs

(to ~ O,h > O,s> O).

It is clear from the definition (3) that (7)

-

lim I(h, t

h

O

to) = O

when t :F to.

AIso, the area under the graph of the function 1 is unity for every positive h; hence (8)

lim h

O

f

et) I(h,

t

-et)

-

to) dt

= 1.

The order of the positions of the limit and the integral here is important, for in view of Eq. (7)

f

et)

-et)

lim I(h, t h

-

to) dt

= O.

O

It is interesting to note the behavior of the transform of the finite unit impulse function as h tends to zero. By evaluating the limit ofthe right-hand member of Eq. (6), we find that (9)

limO L{I(h, t

h

-

to)} = e-sto

But note that L{fim h OI(h, t

-

to)}

= L{O} = O.

(to

~ O,h > O,S > O).

32

-

(9) presents the exponential function e -sto, or exp ( sto), not as the

Equation

transform

I(h, t


SECo 12]

-

of a function,

but as a limit of transforms

of functions of the set

to) consisting of one function for each positive value ofthe parameter

h. Thus exp(- sto)approximatesthe transformof the impulsefunction I(h, t

-

to) for small positive values of h, when to ~ O.

Example A partic1e of mass m, initially at rest at the origin X = O,is subjected to a force in the form of a unit finite impulse I(h,t) beginning at the instant to = O,in the direction of the X axis. 1ts displacements X(h,t), which satisfy the conditions (10)

mX"(h,t) = I(h,t),

(h > O),

X(h,O) = X'(h,O) = O

where X' = dX/dt, could be found here by integration. But if we use transforms to solve for X(h,t), we write

(11)

ms2x(h,s)= L{l(h,t)}=

-hs

1 -e

.

,

assuming that X and X' are continuous functions of t of exponential order. Thus mx(h,s)

=

1 2 2h ( S3

-

2

-hs

S3e

)

and therefore (12) that is (13)

1 2 X(h,t) = 2mh t t2

=

- (t 2mh

when O ~

h)2

t ~ h,

1 ;;; ( t - 2.h) when t ~ h. 1

=

That function satisfies all conditions (10). Now if h -+ O,the second of formulas (13) applies when t > Oto give the limiting displacements X(t):

(14)

X(t)

= h-+O lim X(h,t) = ~ m

(t > O).

The displacements X(t) correspond to the idealized unit impulse given to m at die instant t = O. As a verificationof the solution (14)of that idealized case, we note that X"(t) = Owhenever t > O,and X( +0) = O.


[SECo 13

33

Furthermore, the particle has the momentum mX'(t) = 1 whenever t > O; thus if it is initially at rest, then its momentum jumps suddenly from O to 1 at the instant t = O. The instantaneous impulse has the same effect as an initial velocity l/m.

13 THE IMPULSESYMBOL 15(t- to) The above example is a simple illustration of problems in differential equations involving instantaneous impulses. In order to reduce the number of steps in formal solutions of such problems, we shall occasionally use the unit impulse symbol c5(t- to) to a limited extent indicated below. Let a function F(t), defined for all real t, be continuous on some interval to ~ t ~ to + k. Then according to the basic law of the mean for integrals to each h(O < h < k), there corresponds a number 0(0 < O < 1) such that

f

1

a>

- a>I(h, t

-

to)F(t) dt

'O+h

f

=h

lo

F(t) dt

= F(to +

Oh).

Consequently the impulse function 1 has the property lim

(1)

h"'O

f

a>

-a>

I(h, t

-

to)F(t) dt

= F(to),

sometimes called the sifting property of selecting that value F(to) of F(t). We abbreviate that property by writing (2)

f:a> c5(t- to)F(t) dt = F(to),

where the entire left-hand member is a symbol that denotes the operation represented by the left-hand member of Eq. (1). That symbol is selected so as to suggest an integral of a function, because if c5(t- to) is replaced by I(h, t to), the symboI does represent an integral whose value is approximately F(to) when h is a small positive number. If to ~ O,then I(h, t - to) = Owhen t < Oand

-

lim

(3)

f

a>

h"'O o

I(h, t

-

to)F(t) dt

= F(to)

(to

~ O,h > O)

according to Eq. (1); that is, in symbolic form (4) foa> c5(t

When

(5)

F(t) = 1, formula

.

-

to)F(t) dt

=

F(to)

(2) becomes

f~a> c5(t- to)dt = 1,

(to ~ O).

34


SECo 13]

the symbolic form of Eq. (8), Seco 12. When F(t) = e-sI and to ~ O, the symbolic integration formula (4) can be written (6) which is the symbolic fomi of Eq. (9), Seco12. The letter f>used in our symbolic integrals has further significance in differential equations. We usethe example treated in Seco12 as an illustration. Let us write

(7)

mX"(t) = f>(t);

X(O) = X'(O) = O,

to signify that X is the limit as h -+ Oof the solution X(h,t) of problem (10) there, the problem with f>(t)replaced by [(h,t). In the transformed problem (11),Seco12,we may let h tend to zero and assumethat x(h,s)-+ x(s) where x(s) = L{ X(t)} ; theresultis (8) But that result is obtained at onceby formally applying the operator L to terms in problem (7)if L{ t>(t)}= 1,in accordance with formula(6),while thesymbolL{ X"(t)} meansthe limit ash -+Oof the transformof X(h,t) so that we may be justified in replacing it by S2X(S).Thus mx(s) = 1/s2,andour result mX(t) = t followso That symbolic method gives limits of solutions, as h -+ O, of other problems in differentialequations with [(h, t - to) asa forcingfunction.We shall use the method occasionally in the sequel. Results obtained by such manipulations cleariy need to be verified if they are to be relied upon. More often we sacrifice the brevity gained by using the f>symbol and useinstead the function [(h, t - to), then find the limit of the solution as h -+ 00 The symbol t>(t - to), often called the Dirac delta function, is not a function. In particular, it is not the limit of [(h, t - to) as h -+ O; that limit is a function with value zero everywhere except at the one point t = to, a function whoseintegral is zero (Sec.12). The symbolt>(t- to)may, however, be replaced by the function [(h, t - to), when h is small, to give results that approximate the effect of t>. Generalizations of functions, called generalized functions or distributions, have been developed since 1950 which include the symbol f>(t to) and its generalized derivatives. There are now a variety of theories that give sound developments of distributions. But even the introductory presentations of the theory of distributions seem too lengthy to present here.1

-

1 See, for instance; the artic1e entitled From Delta Functions to Distributions, by A. Erdélyi, in "Modero Mathematics for the Engineer," Second Series, 1961, or A.-E. Danese, "Advanced

Calculus,"

vol. 2, pt. 6, 1965, or references

listed in those books.


[SECo 13

35

PROBLEMS

1 . With the aid of our theorems and known transforms find the inverse transforms F(t) tabulated below and draw a graph of F(t). F(t)

I(s) e-bs (a) ---¡-(b > O) s

0(0 < t < b);t - b(t > b)

(b) fis-te-2'

e -,.. (e) S2 +

- S.(t) sin t

1

sint(O < t < n);O(t > n) se -s/2 (e) S2 + n2 4 (f) (2s + 1 4 (g)--

2. From the Maclaurin series that represents (1 - X)-l when Ixl < 1, or from the sum of an infinite geometric series, show that 1

1

1

e-los

e-2h.

00 e-.Ios

-s 1 - e-Ios=-+-+-+"'= s s s

L- s .=0

(h > O, s > O).

Apply the inverse transformation term by term to this infinite series, formally, to obtain the result shown in Eq. (2), Seco12; namely, -1 1

L 3.

1

t

{-;1 - e-los } = 1 + h]

[

(h > O).

Show that Eq. (2), Seco12,can be written in the form

L{[1 + ~]} = ~( 1 + coth ~).

4.

(a) Use the formal method indicated in Probo 2 to find F(t) when I(s)

1

2

= -sl+e' =-

(s > O).

Draw the graph of F to see that F can be written in terms of the bracket symbol [t] in the form F(t)

= 1 + (-1)1'1.

36


SECo 13]

(b) Show that the function F in part (a) is described by the following difference equation of the first order, when t > O: F(t) + F(t

-

1) = 2

where F(t)

=O

when t < O.

Transform that difference equation to verify the formula given in part (a) for j(s). 5. Apply Theorem 2 to the transformation found in Problem 4 to show that L 1 + (- di] } {

= !si +2..e M

(h > O, s > O).

6. Find the function Y(t) that satisfies the following difference equation of the first order and the accompanying initial condition.

Y(t) - eY(t - h) = F(t) when t < O,

Y(t) = O

where e and h are constants and h > O,and where F(t) = Owhen t < O. The expansion of (1 - ce-hS)-l in powers of ce-hs is helpful here (compare Probo 2). Show that the solution can be written 00

Y(t)

= n=O I c"F(t -

nh);

but for each fixed value of t this series is a finite series because F(t t - nh < O. Thus an altemate form of the solution is

-

nh) = O when

Y(t) = F(t) + cF(t - h) + c2F(t - 2h) + ... + cmF(t- mh), wherem = O,1,2,... , whenmh < t < (m+ l)h. Verifythe solution. 7. Find the function Y(t)that satisfiesthe followingdifferenceequation of the second order and the accompanyinginitial condition. Y(t) - (a + b)Y(t - h) + abY(t - 2h) = F(t) Y(t) = O

when t < O,

where F(t) = Owhen t < O. The constants a and b are such that a # b, and the constant h is positive. Note that the solution of the transformed problem can be simplified, with the aid of partial fractions in the variable e-M, to j(s) y(s) =

a

(

b

)

a - b 1 - ae-M- 1 - be-M' 1

00

Ans. Y(t) = -a - b n=O I (an+l - bn+l)F(t- nh). 00

= a.

8.

Solve Probo 7 when b

9.

Solve the difference-differential equation Y'(t)

Ans. Y(t) =

-

a Y(t

-

1) =

F(t),

I (n + n=O

l)a"F(t -

nh).


where F(t) when

= b when t >

Oand F(t)

= Owhen t

[SECo 13

< O,under the ~ondition that Y(t)

=O

t ;;;; O.

Ans Y(t) .

=b

[

a t + -(t - 1)2 +... + -(t a"

2!

- n)"+l

(n+1)!

where n ;;;; t ;;;;n 10.

37

J'

+ 1 and n = O,1,2,.. . .

Initially a particle of mass m moving along the X axis is at the origin X

= O with

a velocity Vo. The only external force that acts on that particle is an instantaneous impulse Poat time to, in the X direction. Thus the displacement X(t) satisfies the following symbolic equation and initial conditions: mX"(t) = Poc5(t- to),

X(O) = O,X'(O) = Vo.

Find the formal solution (writing L{X"} = S2X - vo) X(t)

= vot + PO(t m -

to)So(t

-

to)

(t ~ O,to > O)

and show X(t) graphically. To verify the solution, show that X"(t) = O when t # to, that X(O) = Oand X'(O) = Vo,and that the momentum mX'(t) undergoes a jump Po at the instant to. 11. Replace the instantaneous impulse Poc5(t- to) in Probo 10 by the finite impulse PoI(h,t - to) and then find the displacements X(h,t). Show that the limit of X(h,t) as h Ois the function X(t) found before. 12. Problem 10 can be stated in terms of functions as folIows:

mX"(t)= Owhen t # to,

X(O)= O,

mX'(to + O)- mX'(to

-

X'(O)= Vo

O)= Po

(to > O),

where X and X' are continuous except for the specified jump in X'. Use our formula (Prob. 9, Seco5) for the transform of the second derivative in this case to obtain the formula found before for X(t). 13.

Show formalIy that the solution of the symbolic problem X"(t)

+ k2 X(t) = Po c5(t -

to),

X(O) = X'(O) = O,

(to ~ O)

where k and Po are constants, is (if we write L{X"} = S2Xhere)

14. Show that the formal solution of the symbolic problem (if we write L{ X'} = sx here) X'(t) = c5(t- to),

X(O)= O

-

(to ~ O)

is X(t) = So(t - to). In this sense, then, c5(t to) corresponds to the derivative of the unit step function So(t

-

to) as we could anticipate from the symbolic formula

38


SECo 13]

15. Let Al denote the change in /(h, t - to) when !J.t = function of t represented by the difference quotient Al

T-

-

- to) - /(h, t - h -

/(h, t

-h.

Draw the graph of the

to)

h

If F and F' are continuous on an interval that includes the points to and to + 2h, show . that Al -F(t) -a) h

f

a)

dt = --

1

f

'OH

F(-r:

h 'o 1

+ h) - F(-r:) d-r: h

'O+h

f

= -¡¡ 'o

F'(-r:+ 8h)d-r:

where O < 8 < 1, and hence that a) lim -Al F(t) dt = - F'(to)' , 0 -00 h

f

A symbolic form of that formula is

-

f:", <5'(t

to)F(t)dt

= -F'(to);

which is a symbolic integration by parts: f:a) <5'(t- to)F(t)dt

=

- f:a) <5(t

to)F'(t)dt= -F'(to)'

In particular, if to ~ O,note that L{<5'(t

-

to)} =

- ddt e-sr] '='0 = se-sro.

16. LetFbecontinuous(t ~ O)exceptforajumpjoatto(to> O).ThenF(t)- joSo(t- to) is continuouswhenevert ~ Oif F(to)is properlydefined. If F is exponentialorder and F' is sectionallycontinuous,then L{fc[F(t)

- joSo(t -

to)]}

= {f(S) -

jo

e~sro]

- F(O).

Show that this formula agrees with formula (4), Seco4, and note that its symbolic form may be written (Prob. 14)

17. Since /(h,t) = l/h when 0< t < h and vanishes when t < O and when t > h, describe the function /(h, - t) and show that

!~L: F(t)/(h, - t) dt = F(O)


39

[SECo 14

when F is defined for all t and continuous over some intervalltl < a. Symbolically,

f'", F(t)b(-t) dt = F(O)= f:", F(t)b(t)dt; thus b has the symbolic property b( - t) = b(t).

14

INTEGRALS CONTAINING

A PARAMETER

We now review some properties of integrals that will be useful. Let x denote a para meter and f(x,t) a continuous function of the two variables x and t together in a region a(x) ~ t ~ fJ(x),e ~ x ~ d, where a and fJ are continuous functions. Then the integral (l(X)

f

(1)

f(x,t)

a(x)

dt = g(x)

(e

~ x ~ d)

represents a continuous function of x, which we have called g. Ifthe derivatives a'(x) and fJ'(x) exist, and if of /ox as well as f is continuous, then the derivative of the integral (1) is given by the generalized Leibnitz formula,

f

(l(X)

(2)

g'(x) =

o

a(x) a X f(x,t)dt

+ f[x,fJ(x)]fJ'(x) - f[x,a(x)]a'(x).

The foregoing two properties of the integral (1) are established in advanced calculus from fairly elementary considerations. Let us indicate here how the continuity requirements on the integrand can be relaxed in establishing certain properties. Let a, b, and e denote constants, and let the integrand of the integral aX+b

(3)

f

(O ~ x ~ e)

f(x,t) dt = h(x)

o

be such that the region O~ t ~ ax + b, O~ x ~ e can be divided into polygonal subregions by a finite number of straight lines (Fig. 7) t

and

= a¡x + b¡

(i

= 1,2,...,m)

(j = 1,2,...,

n),

where f(x,t) is continuous over each subregion up to its boundary; that is, for each subregionf can be definedon the boundary in such a way that it is continuous in x and t together over the closed subregion. We then say that f is continuous by subregions.

40


SECo 14]

e

01

When x 1

~

a sum ofintegrals

X

x

Fig.7

~ X2 in Fig. 7, for instance, the integral can be written as

ofthe type (1), such as aX+b

f

(4)

f(x, t) dt

a2X+ b2

in which the integrand is continuous so that the integral represents a continuous function of x(x1 ~ x ~ X2). Similarly when x is between any two adjacent x/s, h(x) is continuous. Consequently, the integral (3) represents a seetionally eontinuous funetion h(x) on the interval (O,e)when f is continuous by subregions. As a special case, when the integrand is a sectionally continuous function f(t), independent of x, the integral (3) represents a continuous function of x on the closed interval [O,e]. The integral of the sectionally continuous function (3) e

(5)

f

O h(x) dx

e

=

ff

ax+b

O O

f(x,t)

dt dx

exists. It can be written as the sum of iterated integrals over each of the subregions in which f(x,t) is continuous. When the order of integration is interchanged in each subregion, the sum of the integrals represents the iterated integral, with order interchanged, over the entire region. Thus our condition of continuity by subregions permits the interchange of order of integration with respect to t and x. In particular, when a = O so that the limits are constants, (6) 5: s: f(x,t) dt dx = s: 5: f(x,t) dx dt. These conclusions can be reached by the same procedure for regions of more general shapes. The argument used above applied, for instance, when the lower limits ofthe integrals (5) and (6) are constants other than zero.

II FURTHER PROPERTIES OF THE TRANSFORMATION

15

[SECo 15

41

IMPROPER INTEGRALS

Corresponding

properties can be established for convergent improper

integrals with the aid of a condition of uniform convergence with respect to the parameter x. The concept of uniform convergence of integrals, which applies only to improper integrals that involve a parameter, is illustrated in the following example. For each value of x(x ;¡; O)the particular Laplace integral (1) exists; it has the values p(O) = O

(2)

p(x)

=

when x > O.

1

It can be written as a definite integral plus a remainder,

p(x) =

f: xe-xt dt + {X)xe-xt dt = 1 -

e-xT + R(x,T),

whereR(x,T) = f~ xe-xt dt; then R(O,T) = O

(3)

when x > O.

R(x,T) = e-xT

Uniform convergence of the integral for all x in a prescribed interval means not only that the integral exists for each x in the interval, but also that the remainder can be made arbitrarily small in absolute value by taking T sufficiently large, uniformly for all x in the interval. Thus when x ;¡; 1 and € is any small positive number, we see that

IR(x,t)1= e-xT ~ e-T <

€

when T> log (1M, a value that is independent of x, so that the integral converges uniformly with respect to x when x ;¡; 1. In the same way we can see that it converges uniformly when x ;¡; Xl whenever Xl > O. But the convergence is not uniform in the range x ;¡; Oor x > O,because the remainder (3) when x > O is small only if xT is large. Since x can be arbitrarily small, the value of T that makes R(x, T) small must depend on x. A consequence of this lack of uniform convergence in the range x ;¡; Ois the discontinuity of the function p(x) at x = Othat is exhibited by Eq. (2). The Weierstrass test for uniform convergence of the integral (4)

{X' f(x,t) dt = q(x)

(O~ x ~ e)

j

42


SECo 15]

can be stated as follows. Letf(x,t) be continuous by subregions, in the sense described in Seco14, over the region O ;;:;x ;;:;c, O ;;:;t ;;:;T, for each positive constant T. Ir a sectionally continuous function M(t), independent of x, exists such that (O ;;:;x ;;:;c, t > O)

If(x,t)1 ;;:; M(t)

and such that the integral f;;>M(t) dt exists, then the integral (4) is uniformly convergent with respect to x(O ;;:;x ;;:;c). This test is established by first noting that the integral (4) exists according to the comparison test (Probo 14, Seco5),then that the absolute value of the remainder for the integral, for all x, in the interval, do es not exceed the remainder f,;?M(t) dt in the integral of M. The latter remainder is independent of x, and it tends to zero as T -. 000 The Weierstrass test sta tes useful sufficient, but not necessary, conditions for uniform convergenceo Under those conditions the integral (4) is also absolutely convergent. Ir the integral (4) is uniforrnly convergent with respect to x and if its integrand f(x,t) is continuous by subregions over the rectangle O ;;:;x ;;:;c, O ;;:;t ;;:;T for each positive T, then the integral represents a sectionally continuous function q(x); moreover (5)

I: q(x) dx

= I: {" f(x,t) dt dx = foooI: f(x,t) dx dt;

that is, the order of integration of the definite and improper integrals here can be interchanged. In particular, let f(x,t) be continuous in each rectangle O ;;:;x ;;:;c, O ;;:; t ;;:; T, except possibly for finite jumps across a set of lines t

= t¡(i =

1,2,. . o, n). Then q(x) is continuous ifthe integral (4) is uniforrnly convergent. Suppose that éJf/éJxas well as f satisfies those continuity requirements and that the integral f;;>(of/éJx)dt converges uniformly and the integral (4) exists. Then the derivative of the latter integral exists and the order of differentiation with respect to x and integration with respect to t can be interchanged: d

(6)

oo

dxf

f

oo

o

o f(x,t) dt = o oxf(x,t) dt

(O < x < c).

The continuity of q(x) can be seen by writing q(x)

= foTf(x,t)

dt + R(x,T)

(O ;;:; x ;;:;e).

For each fixed T the integral on the right is a sum of integrals of the type

"

f"-1 f(x,t) dt


[SECo 16

43

with continuous integrands; hence it is a continuous function of x so that the change in its value is small when the change L1xin x is small. But IR(x,T)1and IR(x + L1x,T)I are small for all x and x + L1xin the interval (O,e)as long as T is taken sufficiently large, beca use of the uniform convergence of the integral (4). Thus by taking Tlarge first, then lL1xlsmall, the continuity of q(x) follows; that is, Iq(x

+ L1x) -

q(x)1 is arbitrarily

small when lL1xlis sufficiently small.

The proof of property (5) follows the plan commonly used in advanced calculus, where f(x,t) is generally assumed to be continuous. The details are left to Probo 11 at the end of Seco17. Property (6)can be established in the following manner. Because ofthe uniform convergence ofthe integral on the right in Eq. (6) and the continuity properties of 8f/8x and f, for each r(0 < r < e) we can write (7)

I: fooo

:

dt dx

= fooo

I:: dx dt

= fooof(r,t) dt -

fooof(O,t)dt,

where we have used property (5) to invert the order of integration. Since f:;' (8f/8x) dt is a continuous function of x, differentiationofthe first and last membersof Eqs. (7)with respect to r leads to the equation d oo oo 8 (O < r < e), f(x,t) dt = dr o f(r,t) dt o 8x x='

f_ ]

f

which is the same as Eq. (6) with x replaced by r. 16

CONVOLUTION

We now determine the operation on two functions of t that corresponds to multiplying their transforms together. This convolution operation, which gives the inverse transform ofthe product oftwo transforms directly in terms of the original functions, is one of primary importance in operational mathematics. Let F(t) and G(t) denote any two functions that are sectionally continuous on each finite intervalO ;;¡;t ;;¡;T and (9(elXt), and write

f(s) = L{F(t)},

g(s) = L{ G(t)}

(s > a).

It will be convenient to define G(t) to be zero when t < O; but our final result depends only on the values of F(t)and G(t)when t > O. According to Theorem 1, for each fixed -r('r;¡¡:;O),

e-Sfg(s)= L{G(t- -r)}= fooo e-stG(t- -r)dt,

44


SECo 16]

where s > .CX.Hence

that is, T

(1)

f(s)g(s) = lim

ff

T-+oo o

OO F(T)e-srG(t - T) dt dT. o

Since f(s) and g(s) exist when s > cx,the limit here exists. The integrand ofthe inner integral in Eq. (1)is continuous by subregions

(Sec.14),over the rectangular region O ;;;¡;T ;;;¡;T, O ;;;¡;t ;;;¡;R, for each pair of positive constants T and R. For if the jumps of F(T)and G(t) occur at T = T~i = 1,2,..., m) and t = tJ{j = 1,2,..., n), then the jumps of the integrand occur at the lines T = TIand t

=T+

tj' In view of Eq. (5), Seco15,

the order of integration in Eq. (1) can be interchanged, provided that the improper integral there is uniformlyconvergentwith respectto its parameter T(O ;;;¡;T ;;;¡; T). The uniform convergence is seen by noting that, owing to the exponential order of F and G, a constant N exists such that (2) where M(t) = N exp [-(s - cx)t]. The function M(t) satisfies the conditions ofthe Weierstrass test (Sec.15); that is, M(t) is independent of Tand integrable from zero to infinity. Equation (1) can now be written in the form (3)

f(s)g(s)

=

f

ooe-sr lim T-+oo o

f

T

o

F(T)G(t - T)dTdt

= T-+oo lim [11(T)+ lz(T)], where

11(T) = foT e-sr IoTF(T)G(t - T)dTdt,

lz(T) = Irooe-sr I:

F(T)G(t - T)dTdt.

In view of condition (2) 11z(T)1< N therefore

fT

ooe-(S-",)r

NT fo dTdt = s-cx T

lim001z(T) = O. T-+

e-(s-",)T;


[SEC. 16

for the integral 11(T) is the square O ~ ~ T. But G(t - 1:)= Owhen 1:> t. Therefore

The region of integration

O~

t

f: {

11(T) =

e-sI

F(1:)G(t

f(s)g(s) = lim 11(T) =

f f

-

1: ~

45

T,

1:)d1: dt

and Eq. (3) reducesto (4)

T-ro

eo e-SI

l

o

o

F(1:)G(t

-

1:)d1:dt.

The convolutionF * G of the functions F(t) and G(t) is defined as the function (5)

F(t) * G(t) = f~ F(1:)G(t - 1:)d1:,

so that Eq. (4) can be written (6)

f(s)g(s) = L{ F(t)

* G(t)}.

We summarize our result as follows. 1ff(s) and g(s) are the transforms oftwofunctions F(t) and G(t) Theorem 3 that are sectionally continuous on each interval O ~ t ~ T and of the order of eal as t tends to infinity, then the transform of the convolution F(t) * G(t) exists when s > IX; it is f(s)g(s). Thus the inverse transform of the product f(s)g(s) is given by the formula (7)

L -l{J(S)g(s)}

= F(t)

* G(t).

The functions F(t) = t and G(t) = eal,for example, satisfy the conditions of Theorem 3. Consequently

Partial fractions can also be used to obtain that resulto When G(t) = F(t), we have the formula (8)

(f(s)JZ = L{F * F}.

As an example that will be useful in finding inverse transforms with the aid of

46

SECo 17]

OPERATlONAL

MATHEMATICS

partial fractions, we note that (9)

L-l{(S2:

P)2}

=

:2sinkt*sinkt

=

:2

{ sin k. sin k(t -

.) d.

= 2~3(sin kt - kt cos kt). The conditions stated in Theorem 3 are narrower than necessary for the

=

validity of formula (7). If F(t)

t-t for example, F is not sectionally

con-

tinuous on an intervalO < t < T, but the Laplace integral of F(t) is absolutely convergent and formula (7) is still valid if G(t) satisfies the conditions stated in the theorem.

Thus

1

L -1 { ~(s

-

If we make the substitution r

= .J"i

*e' - 2t! r' e-<~

- jñ Jo

2.fi.

here, we find that

2e' ./l

1 (10)

~

- 1)} - Fe

L-1 { .¡;s(s-l)

}

= y1tr=

f

o e-r2dr=e'erf(.jh

where the error function erf (x), also callect the probability integral, is a tabulated function defined by the equation

(11)

2 (X erf(x) = jñJoe-r2dr.

It was shown in Probo 10 at the end of Seco5 that erf(oo) = 1. Since the substitution of s + 1 for s in a transform corresponds to multiplication of the object function by e-', it follows from Eq. (10) that

(12)

17

L -1

~

{ s,s+l }

= erf(Ji).

PROPERTIES OF CONVOLUTION

By substituting the new variable of integration A = t - 1"in the convolution integral (5), Seco 16, we find that the convolution operation is commutative; that is, (1)

F(t) * G(t)= G(t)* F(t) = {F(t - A)G(A)dA.

FURTHER PROPERTlES OF THE TRANSFORMATION

[SECo 17

47

The operation is clearIy distributive with respeet to addition: (2)

F(t) * [G(t) + H(t)]

= F(t)

+ F(t) * H(t).

* G(t)

AIso, F * (kG) = k(F * G) if k is a eonstant. Properties (1) and (2) are valid whenever the funetions are sectionalIy eontinuous over an interval (O,T) that eontains the point t. For sueh funetions the operation is also associative: (3)

F(t) * [G(t) * H(t)]

=

[F(t) * G(t)] * H(t).

To prove this we first write, in view of Eq. (1), F*(G*H)

= f~ F(-r:) f~-t G(t - T - A.)H(A.)dA.dT

and observe that the iterated integral here represents an integration over tbe triangular region shown in Fig. 8. When the order of integration is reversed, the integral becomes

{

{-

H(A.)

- A.-

A F(T)G(t

T) dT dA.,

whieh represents H * (F * G) or (F * G) * H. Note that for eaeh fixed value of t the integrand of the iterated integral is a funetion of T and A.that is

continuous by subregíonsover tbe regíon O ;;;¡;A. ;;;¡;t - T, O ;;;¡;T ;;;¡;t, because of the sectional eontinuity of the funetions F, G, and H. Tbe interehange of order of integration is thereforejustified (Sec. 14). Tbe eonvolution of a seetionalIy eontinuous funetion F(t) and our unit step funetion Sk(t)can be written (4)

F(t) * Sk(t)

=

f~ Sir:)F(t

-

T) dT

'

=

i k

=

O

when

O ;;;¡;t ;;;¡;k,

'-k

F(t

-

T)dT

Fig.8

=

i

o

o

F(A.)dA.

when t ;¡;;k.

48


SECo 17]

Since the integral of a sectionally continuous function F(A.)is a continuous function of its upper limit when the lower limit is a constant (Sec. 14) and since the function (4) is continuous at t = k, it followsthat F * Sk is continuous on the intervalO ;;;¡;t ;;;¡;T ir F(t) is sectionally continuous there. A proof, based on the continuity of F * Sk>that the convolution of any pair of sectionally continuous functions is a continuous function, will be left to the problems. If two functions F(t) and G(t) are sectionally continuous on each interval (O,T) and of the order of e'" as t tends to infinity, then for some constant M 1 (t > O). But ir ( is any positive number and ir M 2 represents the maximum value of the function te-d when t ;¡:;O, then

where M = M 1 M 2' Consequently F * G is of exponential order: IF(t) * G(t)1< Me(a+<)1

(5)

«( > O,t > O).

Now, if three functions F(t), G(t), and H(t) satisfy the conditions of Theorem 3, then the function G * H satisfies those conditions and its transform is g(s)h(s). Therefore f(s)g(s)h(s)

= L{F(t)

* [G(t) * H(t)]};

but since the convolution operation is associative, this equation can be written in the form (6)

f(s)g(s)h(s)

=

L{F(t) * G(t) * H(t)}.

A similar formula can be written for the product of n transforms. When G(t) = 1 in Theorem 3, we have a result noted earlier (Sec. 5): Theorem 4 Division of the transform of a function by s corresponds to integration of thefunction between the limits Oand t: (7) L -1Hf(S)}

= I~ F(T)dT,

(8) L -1 {s12f(S)}

=

{I:

F(A.) dA. dT,

etc., for division by sn, provided F(t) is sectionally continuous and of the order of eal(ex> O),wheres > ex.


[SEC. 17

49

As examples, we note that L - 1 L(s2

:

k2)}

= {sin

L -1 {S2(S2 k+ k2)} =

k1:d7:= t(1

-

cos kt),

{s: sin kA.dA d7: = :2 (kt -

sin kt).

PROBLEMS 1.

Use transfonns to show that tS (a) 1 * 1 * 1 = tt2; (e)

2.

L 't2COS(t -

(b) t * t * t

't)d't

= 2t -

= 5!;

2sint.

Show that an alternate fonn of the transfonnation (8) Seco17,is

L -1 {~f(S)} = t * F(t) = t

{F('t)d't - L

'tF('t) d't.

3. Showthat 1

1

)S-I=s-l+

1

1

( JSl+s-l

)

and in viewof fonnula (10),Seco16,that L -1

= el+ ~ + e'erf(Jt). { --dv' s - 1} v' 1[t

4. Generalize fonnula (10), Seco 16, by replacing s by sla2 to obtain transfonn 40, Appendix A, Table A.2. 5. Generalize fonnula (12), Sec. 16, to obtain transfonn 44, Appendix A, Table A.2. 6. Find L -I{(S + 1)-1s-t} and generalize to obtain transfonn 41, Appendix A, Table A.2. 7. Obtain the inverse transfonn

L-I{SS~\}

=e'erf(Jt)-2~.

8. With the aid of transfonnation (9), Seco16, show that L -1 {(s211 )3}

= 3 (sin t -

t cos t) - t2 sin t.

9. Let t = h and t = k be the points of discontinuity of a sectionallycontinuous functionG(t)on an intervalO < t < T, and let jh andA denotethejumpsin thevalue of G(t)at those points. With the aid of a graph of G(t),note that the function (O~ t ~ T)

60


SECo 18]

is continuous if its values at t = h and t = k are properly defined. Write F * G in tenns of F * G, and of convolutions like the one in Eq. (4), Sec. 17, to show that if F(t) is sectionally continuous, then F * G is continuous. AIso note how the result can be generalized to pennit G(t) to have n jumps in the interval. 10. Show that the integral

x

00

= 1o

u(x) has the values

u(x)

=

-n/2

I + x 2 t 2 dt

when x < O, u(x) = n/2 when x > Oand u(O) = O. Examme

the remainder R(x, T) for this improper integral and show that the integral is not unifonnly convergent with respect to x over any interval that contains the point x = O in its interior or as an end point. 11. Prove property (5), Sec. 15, under the conditions stated there. This can be done by first showing that for each positive number T

[

q dx

= [s: f

dt dx + [R(X, T) dx

and hencethat, correspondingto each positivenumber E,a number T.existssuch that when T > T,

12. Extend property (5),Sec. 15,for the interchange of order of integration of definite and improper integrals, to the following case: I: Loo
f

+

Loo I:
where b 6; Oand e, and f:>
18

whenb < x < e and t > x,

= O

wheneitherx < b or t < X.

DIFFERENTIALAND INTEGRAL EQUATIONS

General nonhomogeneous linear differential equations with constant coefficients can be solved with the aid of the convolution property. Example 1 (1)

Find the general solution of the differential equation Y"(t)

+

k2 Y(t)

= F(t)

in terms of the constant k and the function F(t). We assume for the present that F(t) is sectionally continuous and of exponential order, that the derivative Y"(t) of the unknown


[SECo 18

51

function satisfiesthose conditions, and that Y'(t) is continuous and of exponential order. Then the equation in the transforms of Y(t) and F(t) can be written S2y(S)

-

sY(O)

-

Y'(O) + k2y(s) =

I(s),

where Y(O) and Y'(O) are arbitrary constants. Hence 1

y(s) =

k S2

k s Y'(O) + kd(s) + Y(O)S2 + k2 + k

k ?

.

.?,

and therefore Y(t) = ~ (sin kt) * F(t) + Y(O)cos kt + Y~O)sin kt. This general solution of Eq. (1) can be written (2)

Y(t)

=~

{

sin k(t ...:. T)F(T) dT

+ C 1 cos kt + C2 sin kt,

where CI and C2 are arbitrary constants. The function CI cos kt + C2 sin kt is the general solution of the homogeneous equation that arises when F(t) = O in Eq. (1). To verify the solution (2), it is therefore sufficient to show that the first term on the right is a solution ofEq. (1). If F(T)hasjumps at T = ti' t2"'" t" when O < T < t, that term can be written 1 Z(t)

=k

"

fo sin k(t -

'l

T)F(T) dT

+ . .. +

k

'

f~ sin k(t - T)F(T)dT.

The Leibnitz formula (2), Seco 14, applies to each of the integrals in this equation; thus ' '.

Z'(t) =

i

o

cos k(t - T)F(T)dT+ ... +

f~ cos k(t -

T)F(T)dT + O.

By applying that formula again to the integrals here and collecting the resulting integrals, we find that Z"(t) =

-k

{

sin k(t

-

T)F('r) dT

+

F(t)

or Z"(t) = -k2Z(t) + F(t). Therefore Z(t) satisfies Eq. (1), and the function (2) is verified as the general solution of that equation for all values of t. The function Y(t) given by Eq. (2) is continuous, together with Y'(t), when F(t) is sectionally continuous, in view of the above expressions for Z(t) and Z'(t). The condition of exponential order on

52

SECo 18]


F(t) was not used in verifying the solution (2); the condition that F(t) be sectionally continuous, on an interval containing all values of t to be considered, was sufficient. An equation in which the unknown function occurs inside an integral is called an integral equation. In certain applied problems, to be illustrated in the following chapter, the integral in the equation is a convolution integral. Such íntegral equations of the convolution type transform into algebraic equations. Example 2 Solve the integral equation

=

Y(t)

+ {Y(-r) sin (t - -r)d-r.

at

We can write this equation in the form

= at + Y(t) * sin t

Y(t)

and transform both members to get the algebraic equation a 1 y(s) = S2 + y(s)S2 + l' whose solution is y(s) = a(s~ + s~}.

Therefore

Y(t) = a(t + it3),

which can be verified directly as the solution of the above integral equation. The general integral equation of convolution type has the form (3)

Y(t)

= F(t) +

{

G(t - -r)Y(-r)d-r,

where the functions F(t) and G(t) are given and Y(t) is to be found. Since

the transformed equation is y(s)

= f(s) + g(s)y(s),

the transform of the unknown function is (4)

f(s) y(s)

=

1

-

g(s)"

Even if Eq. (3) is modified by replacing Y(t) by linear combinations of Y(t) and its derivatives, the transform of the modified equation is an

[SECo 18

FURTHER PROPERTIES OF THE TRANSFORMATlON

53

algebraic equation in y(s). For instance, the integrodifferential equation (5)

aY(t)

+ bY'(t) = F(t) +

{

G(t -

T)Y(-r)dT,

where a and b are constants, givesrise to the transformed equation (a + bs)y(s) - b Y(O)

= f(s) + g(s)y(s),

which is easily solved for y(s). The equation (6)

F(t)

(O < b < 1),

= {(t - T)-bY'(T)dT

is known as Abel's integral equation. The unknown function could of course be considered here as Y'(t) instead of Y(t); but no advantage is gained by doing so. The solution of that equation is Y(t) = Y(O) + r(b)r(11 - b)tb-l * F(t),

(7)

valid when F(t) satisfies certain conditions of continuity. The formal derivatiODof this solutioD is left to the problems. PROBLEMS Solve the following differential equations. 1. 2.

Y"(t) - k2 Y(t) = F(t), if Y(O)= Y'(O)= O(k # O). Ans. 2kY(t) = e'''f~e-k'F(r)dt Y"(t) - 2k Y'(t) + k2 Y(t) = F(t).

- e-k'f'o~rF(t)dt.

Ans. e-kIY(t) = CI + C2t + f~ (t - t)e-krF(t)dt. 3. 2Y"(t) - 3Y'(t) - 2Y(t) = F(t), if Y(O)= Y'(O)= O. Verify that your function satisfies the differential equation and the initiaI conditions. 4. Y"(t) + 4Y/(t) + 5Y(t) = F(t), if Y(O)= Y(7t/2)= O. Ans. e2IY(t) = sin t f~/2 F(t)e2rcos t dt - COS t f~F(t)e2' sin 1:dt. 5. Y///(t)- Y'(t) = F(t). Verifyyour result. 6. Y///(t)- Y"(t)+ Y'(t) - Y(t) = F(t),if Y(O)= Y'(O)= Y"(O)= O.

Ans. 2Y(t)= f~ F(t- t)(e' - cost 7.

Show that the solution of the system of differentiaI equations X/(t)

under the conditions

-

X"(t) - Y"(t) + Y(t) = O,

2Y'(t) = F(t),

X(O) = X'(O) = Y(O)= Y'(O)= O,so that F(O)= O,is

X(t) = f~ F(t) dt

Y(t)=

-

2

{F(t) cos (t -

- L F(t)cos(t -

t) dt.

t) dt,

~

sint) dt.

54

8.

SEC, 19]

OPERATIONAL

MATHEMATICS

Solve the following system and verify your result:

Y'(t) + X(t) = 1,

X'(t) + Y(t) = F(t),

X(O)= 1,

Y(O)= o.

9. Solvefor Y(t) and verifyyour solution:

{

Y(t) dt

-

Y'(t) = t,

Y(O)= 2. Ans. Y(t) = 1 + cosht.

10. Derive the solution (7) of Abel's integral equation (6). 11 . Solve the integral equation Y(t)

= a sin t -

2

{

Y(t)cos (t

- t)dt.

Ans. Y(t)

= ate-'.

12. Find the solution of the integralequation Y(t) = a sinbt + e

{ Y(t) sin b(t -

t) dt

(a) when b2 > be; (b) when b = e. Ans. (a) Y(t) = ab(b2 - be)-t sin (tJb2 - be); 13. Solve the nonlinear integral equation 2Y(t) + {Y(t)Y(t

14.

-

t)dt

=t+

2.

(b) Y(t) = abt.

Ans. Y(t) = 1.

Write Y"(t) = V(t). (a) Ir Y(O)= a and Y'(O)= b show that Y'(t) = 1* V(t) + b,

Y(t) = t* V(t) + bt + a;

(b) then convert the initial-value problem Y"(t)

+

a(t)Y'(t)

+

P(t)Y(t)

= F(t),

Y(O) = Y'(O) = O,

in Y(t) into the following integral equation in V(t): V(t) + {[cx(t) + P(t)(t -

19

t)]V(t)

dt

=

F(t).

DERIVATIVESOF TRANSFORMS

When the Laplace integral (1)

f(s) = 1'" e-S'F(t)dt

is formally differentiated with 'respect to the parameter s by carrying out the differentiation inside the integral sigo, the formula

f'(s) = 1'" e-st(-t)F(t)dt = L{ -tF(t)}


[SECo 19

55

is obtained. Another formal manipulation with the integral (1) indicates that f(s) ~ O as s ~ oo. We shall establish conditions under which those

formulas are valido First, we note that if F(t) is ofthe order of (!'"as t ~ 00,then the function tnF(t), where n = 0,1,2,. . ., is of exponential order. Let € be a positive number. Then constants N 1 and N 2 exist such that (t > O), where N 2 represents the maximum value of the function tne-" when t > O. tnF(t) is of the order of ellOI,where CXo= cx + €.

Thus the function

Also let F(t) be sectionally continuous on each intervalO < t < T. Then tnF(t) has that property. The absolute value of the integrand of the Laplace integral of tnF(t)satisfiesthe condition (2)

(n = O, 1,2, . . . ),

where N denotes a constant. Consequently when s > cxo, (3) therefore L{~F(t)} according

~

to condition

Oas s ~ (2),

oo. Moreover, if s ~ CXlwhere CXl> cxo,then

ItnF(t)e-SII<

= M(t),

Ne-(Il,-IlO)1

.

where this eXPQnential function M(t) is independent of s and integrable from zero to infinity. It follows from the Weierstrass test (Sec. 15) that the Laplace integral (n = o, 1,2,. . . ) IX) ~F(t)e-SI

dt

converges uniformly with respect to s when s ~ CXl> CXo. Theorem 5 1f F(t) is sectionally continuous and of the order of (!"I,then each ofthe Laplace integrals L{F(t)}, L{tF(t)}, L{t2 F(t)},.. ., is uniformly convergent when s ~ CXIwhere CXI> cx;moreover (4)

lim f(s) = O

and

lim L{~F(t)}

S-'co

=O

(n

= 1,2,...).

The function F(t)e-SI and its partial derivative of each order, with

respect to s, satisfy our conditions for the validity of formula (6), Seco 15. Hence differentiation with respect to s can be performed in Eq. (1) inside the integral sign, and the following theorem is established.

56


SECo 19]

6 Differentiation of the transforrn of a function corresponds to the multiplication of the function by t :

Theorem

-

(n = 1,2,...);

(5)

moreover, pnl(s) -. O as s -. oo. Those properties hold true whenever F(t) is sectionally continuous and ofthe order of e"', if s > IXinformula (5). Since a function is continuous wherever its derivative exists, it is true that f(s) and each of its derivatives is continuous when s > IX. To illustrate the last theorem, we can note that since k S2 + ,-2 = L{sin kt}

(s > O),

it follows that -2ks (S2 + '.2\2 =

L{- t sinkt}.

Thus we have a formula that is useful in finding inverse transforms with the aid of partial fractions: 2ks . (s > O). (6) k} { L t S10 t = (S2 + k2)2 Transformation (6) can be obtained by another method that is sometimes useful, that of differentiating a Laplace integral with respect to a parameter k, independent of s. When s > O,the Laplace integral

f

OO

(7)

L{cos kt}

=

e-s'cosktdt = ~s

o

s +

converges, and the integral (8) L{ :k cos kt} = fooo:k(e-st cos kt) dt = - 100e-S't sin kt dt converges uniformly with respect to k for all real k, according to the Weierstrass test. AlI integrands here are continuous functions of k and t. . Hence

formula (6), Sec. 15,applies to show that the integral (8) represents

the derivative, with respect to k, of the integral (7); that is, o

-L{tS1Okt}

a = ak(

and this is the transformation (6).

S ?

.

) .? =

2ks

-,?

. .?,?,


20

SERIES

OF

[SEC. 20

57

TRANSFORMS

A useful method of finding inverse transforms L -1 {J(s)} is that of representing the functionfby an infinite series ofknown transforms, then applying the operator L -1 to the terms of the series. For certain types of series in powers of l/s, conditions for the validity of the procedure will now be established. Theorem 7 Let f(s) denote the sum of an infinite series of positive integral powers of l/s which is absolutely convergent when s > IX,where IX~ O: (1)

1 an¿ n=O ~+l

ao

ex>

f(s) =

= -

S

al

+ - + ...

(s > oc~ O).

S2

Then the power series in t obtained by applying the operator L - 1 to that series term by term converges to a function F(t) whose transform is f(s), when s > IX: ex>

tn

¿

(2)

F(t) = n=Oan,n. = L -l{J(S)}

(t

~ O),

where O! = 1. A Iso, F is continuous when t ~ O, and of exponential order (!.J(e"lt) whenever OCl> oc. Since series (1) is absolutely convergent when s ~ OClwhenever OCl> a ~ O,it follows that lanl/sn+1 ~ lanl/a1+1 -+ O as n -+ oo. Therefore a

constant M, independent of s, exists such that lanl/~+1 < M; that is, (3)

lanl < M~+

1

(n

= O,1,2,.. . ; s ~ al > a ~ O).

Ir we write s = al here it follows that, when t ~ O, tn

(4) I

ann! -

tn

I

(IXlt)n

= la1n n! < = Mal- n! .

The series of terms (altnn! converges to e"lt, so by the comparison test series (2) is absolutely convergent, and its sum is of exponential order, (t ~ O,al > a). AIso, F is continuous because it is the sum of a convergent power series in t. Let s be fixed (s > IX)and let T be a positive number. The series of the ternis ane-Sttn/n! is uniformly convergent with respect to t over the intervalO < t < T according to the Weierstrass test for infinite series of functions, for in view of condition (3), tn

(sT)n

lanle-st, n. < Ms-.-n.

(s > a,O < t < T);

58


SECo 20]

the numbers on the right are independent of t, and they are terms of a convergent series. The series of the continuous functions ane-slr'fn! can therefore be integrated term by term over the bounded interval (O,T) (cf. Probo 15, Seco21); thus

I

T

a

00

o e-SIF(t) dt

= n=ono L~

T

I

o e-sIr' dt

f - n. fT

= n=O I: anrL{r'} noL = I:

n=O( S'a: 1

a~

ooe-str' dt

]

T

.

ooe-sIr' dt )

By subtracting the last series from series (1) which converges to f(s), we find that

I

T

(5)

-

f(s)

o

00

e-SIF(t)dt =

a

oo

L ~ fT

n=on.

e-sIr' dt.

We are to prove that the difference (5) vanishes as T -+ 00; that is,

to each positive number such that (6)

€

we are to exhibita correspondingnumber T.

/f(S)

-

f: e-SIF(t)dtl<

€

wheneverT> T.o

It will follow that f(s) = L{F(t)}; then Theorem 7 will be provedo The series in Eq. (5) is absolutely convergent because

f.

lanl 00 -st.,. d t n., T e,

(~

::::;lanl L {r'}

- n.I

"

= lanl ..n + 1

~

and series (1)is absolutely convergent. Hence if N = 1,2,. oo, (8)

f(s) I

-

I

T

o

e-stF(t)dt

~

Ni:.l1a~1

I

n=O n.

i

ooe-sIr' dt T

+ RN(T)

where, in view of condition (7),

i

ooe-str' dt ~ I: la:'l' RN(T) = n=N I: la~1 n. T n=NS'

Since series (1) is absolutely convergent, it now follows that to each

€ (€

> O)

there corresponds a number N.. independent of T, such that RN(T) < !€

wheneverN > N..

Now consider the sum from n = O to N - 1 in condition (8). Let N be some fixed integer greater than N. By successive integrations by parts o


[SECo 20

59

we find that 1

f

co

Tn

e-ST

Tn-l

Tn-2

1

e-stl"dt = - + + + ... + - . n! T s [ n! (n - l)!s (n - 2)!S2 snJ Since Iani < Ms"+l, it follows that, if sT > 1 and n ~ N - 1,

f

lanl coe-Sltn dt < Me-ST n! T ~ Me-ST(ST)n

[

(ST)n

+ (sT)n-l +... + 11

n!

(n - 1)!

~ + (n - 1 1)! + ...

J

+ 1 < Me-sT(sT)NN.

[n!

J

Thus the sum in condition (8) is less than MN2e-sT(sT)N, which vanishes as T -+ co for that fixed N. Therefore, there is a number T. such that MN2e-sT(sTt < f/2 when T> T.. and condition (6) follows to complete the proof of the theorem. Theorem 7 can be generalized to the following theorem: Theorem 8 Let g(s) be represented by an absolutely convergent series when s > a (a ~ O) of this type:

1

co

(9)

g(s)

= n=O L

an-l

(O< k ~ 1,s > a ~ O).

sn+k

Then L{ G(t)} = g(s) when s > a, where the function G is continuous when t > O, is l!1(ea,,)if al > a, and is represented by the convergent series of inverse transforms: co

(10)

a

G(t) = n~o L -1 {

co

tn+k-l

;~: } = n~o an-l

(t > O).

A proof can be based on Theorem 7 by writing

a_l gs( ) =-+~

1 ~ an-l 1.. -=-+~n=l sn

a_l ~

1 ~ an 1..~n=Os"+l'

The last series here represents the transform of the function CO

(11)

tn

F(t) = L an" n=O n.

a function that is continuous (t ~ O), and l!1(ea,,)if al > a. AIso we know that (Sec. 5) (k > O,s > O).

60


SECo 20]

Since k ;¡; 1, the function rc-1 is bounded when t ~ T> o. Then ..k-l -l a-l..k-l (O< k ;¡; 1). { } tU

L

g(s)

= r(k) t

+ r(k) * F(t)

= O to t = € plus an to t, we can modify the proof in Seco 17 to show that

By writing the convolution here as an integral from t

integral from

€

rc-l * F(t), and hence L -1{g(S)}, is 19(e"")if (Xl> (X. In view of Eq. (11) we can write (12)

rc-1 * F(t)

= E (t

- t)k-l F(t) dt

a

«)

'

= n=On. L~ =

I

o (t - tt-1tndt

«) a

L ...!!...tk-l* tn, n=On!

where the term-by-term integration is valid because the power series for F(t) converges uniformly with respect to t over the interval O < t < t (Prob. 15, Seco21). By using known transforms, we find easily that

..k-l * tn =

tU

r(k)n! ("+k r(n + 1 + k)

and therefore rc-l -

«)

r(k) * F(t)

= n=O L an-

tn+k

.-'

It now follows that L -l{g} is the function G represented by series (lO). That convergent series is the product of rc-l, or t-(1-k), by a power series in t. Hence G is continuous whenever t > O,and Theorem 8 is proved. " Example To establish transformation 78, Appendix A, Table A.2, when k = 1 there, we write the absolutely convergent representation g(s)

= s-te-l/s

= n=O ~

(-l)n n!

~s"+t

(s > O).

According to Theorem 8 and the form 6, Appendix A, Table A.2, of the elementary transformation L -l{s-n-t}, then, L{G} = g where

G(t) = -

1

~

L«) -(-l)n

n=O n!

2n+ l("+t

(1)(3)(5)... (2n + Ir


[SECo 21

The terms of that series reduce to (- 1f(2Jt)2"

therefore

+ 1j(2n

61

+ 1)! and

1

e-l/S

G(t)= L- 1 { sJS} = .fi sin(2Jt). 21

DlFFERENTIALEQUATIONS WITH VARIABLE COEFFICIENTS

We have seen that d" L{t"Y(t)} = (-1)" ds"L{Y(t)} = (-l)"y(")(s), and therefore we can write the transform of the product of t" and any derivative of Y(t) in terms of y(s); for instance, d2 L{t2Y'(t)} = ds2[sy(s) - Y(O)] = sy"(s)+ 2y'(s),

d L{tY"(t)} = - dS[S2y(S)- sY(O)- Y'(O)] = -s2y'(S) - 2sy(s)+ Y(O). A linear differential equation in Y(t) whose coefficients are polynomials in t transforms into a linear differential equation in y(s) whose coefficients are polynomials in s. In case the transformed equation is simpler than the original, the transformation may enable us to find the solution of the original equation. If the coefficients are polynomials of the first degree, the transformed equation is a linear equation of the first order, whose solution can be written in terms ofan integral. To find the solution ofthe original equation, however, the inverse transform of the solution of the new equation must be obtained. Example 1

Find the solution of the problem Y"(t) + tY'(t)

-

Y(t)

= O,

Y(O) =

O,

The transformed equation is 2 s y(s) -

or

d 1 - ds[SY(s)] - y(s) = O,

y'(s)+ (~ - s)y(S)= -~,

Y'(O)= 1.

62

SECo 21]


which is a linear equation of the first order. An integrating factor is exp[f (~- s) dSJ = exp (2logs - !S2) = s2e-ts2, so the equation can be written

Integrating, we have

l e y(s)= -S2 + -ets2 S2 ' where e is a constant of integration. But e must vanish if y(s)°isa transform since y(s)must vanish as s tends to infinity. It followsthat Y(t)

= t,

and this is readily verified as the solution. Example 2

Solve Bessel's equation with index zero, t Y"(t)

+ Y'(t) + t Y(t) = O

under the conditions that Y(O) = 1 and Y(t) and its derivatives have transforms. The point t = Ois a singular point of this differentialequation such that one of the solutions is a function that behaves like log t near that singular point, and the Laplace transform of the derivative of the function does not exist. The transformed equation is

d

,

2

d

--[s ds y(s)- s - Y (O)]+ sy(s) - l - -y(s) ds = O, or

(S2

+ I)y'(s) + sy(s) = O.

Separating variables, we have dy

y

s ds

=-

+ l'

S2

and upon integrating and simplifying, we find that

e

(2) y(s)

= p-:¡::l'

where e is a constant of integration.

PROPERTlES OF THE TRANSFORMATION

FURTHER

[SEC. 21

63

Expanding the functi<;mfor y(s) by the binomial series,we have, when s > 1, y(s)

.

I+ =~ S (

-

e

[

~

-t =

S2 )

ex>

~ I - ~~ - ~ -~ ~ S 2 S2 2 2 2 !S4

[

(2n

-

2nn.s, 1 2n

.

n(I)(3)...

- -S 1 + n=1 ¿ (- 1)

... J

)

(

-

I)

ex> (-I)n(2n)!

L (2nn.1)2?H'" " s J - e n=O

where O! = 1. The ratio test shows that series in positive powers of l/s are absolutely convergent when s > 1. Theorem 7 therefore applies to show that the operator L - 1 can be applied term by term

to that series to represent the continuous function whose transform is y(s) as a convergent series in powers of t: L -1 { ()}

(3)

y S

=

e

~

(-I)n

n~O (2nn!)2t

2n

.

It is not difficultto verify that for all t the power series (3) is a solution of Bessel's equation (1). Ir that function is to satisfy the condition Y(O)= 1, then e = 1,and the solution(3)can be written (4)

Y(t)

= J o(t)

where J o is Bessel' s function of the first kind with index zero: (5)

Jt= o()

t 2n

ex>(-1)n

n~o (n!)2 ( 2 )

t2

- 1

22 +-~

We have shown above that, when s > 1, 1 (6)

L{Jo(t)}=

t4

J S2 +

..~.

-... .

1'

a result that is actually valid whenever s > O, because J o can be represented by an integraP that shows that IJo(t)1~ 1. The differential equation (7)

t2 Y"(t)

+ t Y'(t) + (t2 - n2) Y(t) = O

is Bessel's equation with index n. It can be verified that the function t n+2k ex> (-It (n = 0,1,2,...), (8) Jn(t) = k=O kl(. n + k)I.( _2)

L

known as Bessel's function of the first kind with index n, is a solution of that equation. In the problems to follow we shall establish the transforms of J1(t) and tnJn(t). 1

ChurchilI, R. V.. "Fourier Series and Boundary Value Problems," 2d ed.. p. 175, 1963.

64


SECo 21]

PROBlEMS 1. Use the transformation (6) and Theorem 2 to find L{J o(at)} listed in Appendix A, Table A.2 (transform 55). 2. If Y, Y', and Y" are of exponential order and Y, Y' are continuous (t ~ O)while Y" is sectionaJly continuous, and if L{ Y(t)} = y(s), show that L{t2Y"(t)} = s2y"(S)+ 4sy'(s) + 2y(s). Solve the following differentiaJ equations for Y if Y and its derivatives are to have transforms.

3.

Y"(t)+ atY'(t) - 2aY(t) = 1, Y(O)= Y'(O)= O,a > O.

4. 5. 6. 7.

tY"(t) + (t - l)Y'(t) + Y(t) = O,Y(O)= O. tY"(t) + (2t + 3)Y'(t) + (t + 3)Y(t) = 3e-'. t2Y"(t) - 2Y(t) = 2t, Y(2) = 2. When k is a constant, show that the equation

Ans. Y(t) = t2/2. Ans. Y(t) = Ct2e-'. Ans. Y(t) = (C + t)e-'.

Ans. Y(t) = t2 - t.

t2Z"(t) + 2tZ'(t) + kZ(t) = O leads to the same differentiaJ equation in the transform z(s). 8. With the aid of transformation (6) show that s L{Jó(t)} = ~1. S2

+ l

Use formula (8) to show that Jó(t) = -J l(t), and hence that

L{J¡(t)} = p+t-

p+t

9.

s.

Use the transforms of Jo and J¡ (Prob. 8) to show that (a)

(b)

i

-

' Jo(-r)Jo(t- -r)d-r= sint;

o

í

JO(-r)Jl(t

-

"C)d"C= Jo(t)

- costo

10. Expand the function S-l exp(-s-l) to verify that

in powers of S-1 and apply Theorem 7

L -1 He-l/.} = Jo(~, and thus obtain the transformation 75, Appendix A, Table A.2. Apply Theorem 8 to verify the following transformations: 1.

11.

L-l{)se-l/.}

= ~CoS(~. (Cf. transformation 76, Appendix A, Table A.2.)

12. % ¡¡ i

L -1

J

{ Js

e1/.

}

=~

.fii

cosh (~. (Cf. transformation 77, Appendix A, Table A.2.)


13

.

1

-1

L

{ (S2

f'J .(t)

65

)

+ 1).+t} = (1)(3)(5)...(2n - 1)

(n = 1,2,... . (Cf.transformation57,AppendixA, TableA.2.)

1 14.

[SECo 22

L -1 { 1-

1""t"k 1} = t);:l r(nk)

(k> O,s > 1).

15. Given that a power series L':=oA.-r. converges uniformly over an interval O ~ -r ~ t to a sum F(-r),that is, the series converges to F(-r)there and N-l

F(-r)

= "=0 L A.-r. +

(O ~ -r~ t)

R~-r)

where RN(-r) Ouniformly with respect to -r as N oo. Ir O~ e < 1, prove that "" t-C* F(t) = A.ec * t",

L

11=0

a result that was used to obtain Eq. (12), Seco20. Suggestion: Write N-l t-C * F(t) - L A.t-C * t. = t-C * RN(t) n=O

and prove that the right-hand member vanishes as N fixed.

22

00, when t and e are kept

INTEGRATIONOF TRANSFORMS

When a function F(t) is sectionally continuous and of the order of ¿'t, then its Laplace integral f(x) = La>e-xtF(t) dt is uniformly convergent with respect to x in every interval x;¡;; OCI,where OCl> oc,according to Theorem 5. It follows from formula (5), Seco 15, that when r > s > oc,

f

f(x)

dx

= f La>e-xtF(t) dt dx = La>F(t) f e-xt dx dt.

If the function F is such that F(t)jt has a limit as t tends to zero, then the latter function is also sectionally continuous and of exponential order. Under those conditions the last equation can be written .

dt f' f(x) dx = Jo F(t)e-st t fa>

s

fa>~-,t

Jo

t

dt

= g(s) -

g(r),

where g(s) = L{F(t)jt}. But g(r) ~ Oas r ~ 00 (Theorem 5); hence (1)

dt fs f(x) dx = Jofa>F(t)e-st t a>

(s > oc)

66

SECo 23]


and we have established the following theorem: Theorem 9 Division of the function F(t) by t corresponds to integration of the tran~formf(s), in this manner: (2) Sufficient conditions for the validity of formula (2) are that F(t) be sectionally continuous and of the order of eaJ,that s > exin formula (2), and further

that the limil of F(t)/t exists as t

-. + O.

The function F(t) = sin kt, for example, satisfies the above conditions when ex

= O;in particular,t-l sin kt

-

(3)

L{

t }

=

f

sinkt -. k as t -. O. Hencewhens > O, k dx

""

x2 +

s

1t k2

S

= -2 - arctank

k

= arctan-.

s

Recalling how integration with respect to t corresponds to division by s, we can now write the transform of the sine-integral function 'Sint . (4) SI t = -dt. o t

i

This function is of some importance in applied mathematics. Its values are tabulated

in the more extensive

mathematical

tables.

If k

=

l in Eq. (3),

it follows from Eq. (4) that (5)

= ~arccots = ~arctan~ s s s

L{Sit}

(s > O).

As another illustration of Theorem 9 we note that e-al_e-bl

L

{

t

""

}

=

f

s

1

1

---

( x+a

x+b )

dx=log-

x+a

x+bs ]

""

,

when s > -a and s> -b. Hence (6)

e-al - e-bt L

{

t

S+ b = log-.s + a }

When a = O and b = 1, we have the special case (7)

23

(s > O).

PERIODIC FUNCTIONS

Let a function F be periodic with period a over the half line t > O and sectionally continuous over a period O < t < a. That periodicity can be


[SECo 23

67

described by either of the two statements F(t + a)

= F(t)

or

when t > O,

F(t) = F(t - a)

when t > a.

The function is bounded over the half line and sectionally continuous over each bounded subinterval, so it has a transformf(s) when s > O. For convenience in examining the transform, we write

F(t) = O

when t < O

and introduce a function Fo that is the same as F when O < t < a and zero elsewhere; thus Fo(t)

= [1 -

80(t

-

a)]F(t),

Then F is described for all t by the difference equation (1)

F(t)

-

F(t

- a) = Fo(t),

as we can see either directIy from graphs of F(t) and F(t noting that when t > a, the equation becomes F(t) = F(t t < a, it becomes F(t) = Fo(t). Therefore

f(s) - e-asf(s) = fo(s)

and

- a) or else by - a), and when

f(s) = I fo(s) - e-aso

Theorem 10 1f F(t) is periodic with period a when t > O, and sectionally continuous, then

(2)

(s > O).

Let us applythat formulato the function M(c,t)= I = -1

when O < t < e, when e < t < 2c,

M(c, t + 2c) = M(c,t). This is sometimes called the square-wave function (Fig. 9). Sin~

jf

i ~

68


SECo 23]

Mle,t)

(0,1)

I I I .I

o

1 I I 1 I I I

(2e,0) I I I

L-

Fig.9

the transform of M(e,t) is (1 - e-CS)2 1 - e-CS s(1 - e-2C")= s(1 + e-c"}' Hence

1 (3)

L{M(e,t)}

es

(s > O).

= stanh"2

The integral ofthe function M from Oto t is the function H(e,t) defined as follows:

when O< t < e,

H(e,t) = t =2e-t H(e, t

+ 2e) =

when e < t < 2e,

H(e,t).

This function, whose graph is the triangular wave shown in Fig. lO, has the transform 1 (4)

es

L{ H(e,t)} = S2tanh"2

(s > O).

Let G denote an antiperiodic function when t > O:

G(t + e)= -G(t)

when t > O,

or

G(t) = -G(t - e)

when t > e,

which is sectionally continuous over the interval O< t < c. Then Hle,t) (O,e)

Fig.10


G(t +

- G(t + e) = G(t); that

2e)=

[SECo 23

69

is, G is periodic with period 2e. M(e,t)

is an example of sueh a funetion, as is sin t when e = n, 3n, Sn, . .. , Ir Go(t) = G(t) when O < t < e and Go vanishes for aIl t outside that interval, and if we write G(t) = Owhen t < O,we can see that G is deseribed by the difference equation (5)

G(t) + G(t - e)

= Go(t).

The transform g(s) therefore satisfies the equation g(s) + e-esg(s) = go(s) =

I: e-s'G(t)

dt,

and we have established the foIlowing theorem: Theorem 11 G(t

If G is seetionally eontinuous over an interval (O,e) and

+ e) = - G(t) when t > O, then

(6)

g(s) =-

f~ e-s'G(t)dt

(s > O).

For the periodie function G1 with period 2e sueh that G1(t) = G(t)

when O < t < e, G1(t) = O

when e < t < 2e,

where G is the above antiperiodic funetion, formula (2) beeomes (7)

g e-s'G(t)dt-

g(s) l - e-2es - I - e-es

L{ G1(t)}

(s > O).

For example, when G(t) = sin t and e = n, g(s) = (S2 + 1)-1 and the funetion G1 shown in Fig. II has the transform (8)

L {G (t)} = L 1

{

sin t + ¡sinti = 2

}

I (S2

+ 1)(1 - e-'IS)

(s > O).

SimilarIy, the periodic funetion G2 sueh that Git)

= G(t)

when O < t < e, G2(t + e) = G2(t) where G(t + e) =

o

(271',0) Fig.11

- G(t),

-

70


SECo 24]

has the transform

therefore (s > O).

(9) When G(t)

(10)

I !

= sin t, for example,

then G2(t) = Isin ti if c = 1t and

L{lsin ti} = S2 ~ 1 coth

~

(s > O).

In case the antiperiodic function G has nonnegative values over the interval (O,c),then G(t) ;;¡;Owhen c < t < 2c and the function G1 described above is the half-wave rectificatíon of G. It replaces the negative values of G by zero. Also, in that case G2 is the full-wave rectification of G, and we can write G1(t) = HG(t) + IG(t)lJ,

24

Git) = lG(t)1.

PARTIAL FRACTIONS

We shall now systematize the procedure of finding inverse transforms of quotíents of polynomíals in s. Let p(s) and q(s) denote polynomials in the variable s with no factor in common, and let the degree of p(s) be lower than that of q(s). We shall see that the inverse transform of the function f(s) = p(s)/q(s)exists and that it can be found when the elementary factors of q(s) can be determined. Consider first the case in which q(s) has a linear factor s - a, not repeated. Let tP(s)denote the function that is left after removing that factor from the denominator of f(s); that is,

(1)

f(s) = p(s) = q(s)

cP(s)

s

- a

.

Note that cP(s)may be a quotient of polynomials. According to the theory .

of partial fractions, a constant e exists such that (2)

cP(s) = ~ s-a s-a

+ h(s),

where h(s) represents the sum of the partial fractions that correspond to the other linear and quadratic factors of q(s), any of which may be repeated.


[SECo 24

In order to determine the value of e, we multiply

71

both members of

Eq. (2) by s - a, when s ::j: a, to obtain the equation 4>(s)= e + (s - a)h(s),

which is satisfied identicaIly for aIl values of s in a neighborhood of the point s = a,exceptpossiblyat that point. But both membersofthe equation are continuous functions of s at that point; thus their limits as s ~ a are the same as their valueswhen s = a. Thereforee = 4>(a).The inverse transform of the partial fraction corresponding to the factor s - a, or the term in F(t) corresponding to that factor, is 4>(a)eat. In view of Eq. (1), we can also write S

lim 4>(s)= lim

s-a

where we have evaluated

of the derivatives of s

s-a

-

[

p(s)-

the limit of (s

-

1

a

q(S) ]

-

' = p(a)-,q (a)

a)/q(s) as the limit of the quotient

a and q(s), since q(a) = O and q'(a) ::j:O because s = a is a simplezero of q(s). ConsequentIy 4>(a)= p(a)/q'(a). When aIl the factors of q(s) are linear and not repeated and when q(s) is written in the form (3) where all the constants an are distinct, we can write the inverse transform of f(s) in fuIl. Let qn(s) denote the product of aIl the factors on the right of Eq. (3) except the factor s - an, so that the function 4>(s)corresponding to that factor is p(s)/qn(s). Then

I

L- 1 P(S) = p(an) ~nt { q(s) } n= 1 qn(an)

(4)

=

I

p(an) eant

n= 1 q'(an)

.

The second sum here is sometimes caIled Heaviside's expansion. The principal results are stated in the foIlowing theorem. Theorem 12 lf f is the quotient p(s)/q(s) of two polynomials suchthat q(s) has the higher degree and contains thefactor s - a which is not repeated, then the term in F(t) corresponding to that factor can be written in either of these two forms : p(a) at ,1,. ""t (5) ( ) 'f' a t:: or q'(a{ , '

where 4>(s)is the quotient of p(s) divided by the product of all factors of q(s) except s-a. Theorem 12 is valid when the constant a is any complex number.

72


SECo 24]

For complex arguments the exponential function is defined by the equation (Sec. 55) ¿+iJl = ¿(cos y + i siny) (x and y real).

Example

Find F(t) when 152

-

4s

= (2s + l)(s2 + 1)"

f(s)

We display the factors of the type s - a by writing f(s)

S2

-

15

= (s + t)(s

-

i)(s + i)'

Using the first of the two forms (5), or the first of the expansions (4), we find that -1 + 2i -it ( ) -.i -t/2 - l + 2i it F t - 4~e ("1"2 + 1)2'e + ( -1"2. + 1)( - 2")e 1 1

= e-t/2 -

2

eit - e-it

2i

= e-t/2 -

2 sin t.

PROBLEMS 1. Show that Theorem 9 applies and find the transfonn of: (a) (1 - cos at)/t; (b) (1 - cosh at)/t; (e) (e' - cos t)/t. Ans. (a) (b) See transfonns 105 and 106,Appendix A. Table A.2; (e) log [(s2 + 1)/(s - 1)2](s > 1). 2. The condition that F(t)/t has a limit as t -+ +0 was used in proving Theorem 9. Ir it is replaced by the condition that for some positive constants k and M, IF(t)1< Mr< over an interval O < t < T, then g(s), the transfonn of F(t)/t, exists and vanishes as s -+ oo. Thus fonnula (2) in the theorem is still valido Use this fact to establish transfonn 36, Appendix A, Table A.2: el" ti" (s > a and s > b), L ---=--- =~-~

t

{

zF

}

by showingthat 11ft- i!"1/0 < M00n

j I

i

I ,

an interval (O,T). 3. (a) Use Theorem 11 to derive the transfonn (3), Seco23, of M(c,t). (b) Apply fonnula (7), Sec. 23, to obtain the transfonn 68, Appendix A, Table A.2, of the half-wave rectification of M(k,t). (e) Use fonnula (9), Seco23, to verify that the transfonn of the full-wave rectification of M(c,t) is l/s. 4. Sketch the graph of the periodic function F for which F(t) = t when -1 < t < 1 and F(t + 2) = F(t) for all t, and show that the transfonn of F is l l f(s)=--~ (s > O). S2 s smh s


73

[SECo 25

5. Let F be the periodic function such that F(t) = t when 0< t < 1 and F(t + 1)= F(t), and let G be this antiperiodic function: G(t) = t whenO< t O),

(a) f(s) = SI - s(1 - eI

2

I

~

(b) g(s) = SI

- (SI+ s) I + e-s

(e) gl(S) = I

f(s) + e-s

(s > O), (s > O).

6. Use Theorem 12 to find the inverse transforms tabulated below, where a, b, and e are distinct constants.

-e"'-e'" a-b

s (b) (s - a)(s

ae'" - bé" a-b

-1

(e) 7.

F(t)

f(s) I (a) (s - a)(s - b)

(b - c)e'"

+ (e - a)e'"+ (a - b)e<'

(a - b)(b - c)(c - a)

Use Theorem 12 to find the inverse transforms of the following functions. 4s + I

(a) f(s) = .

.

. ...

.. .

Ans. F(t) = e'/z- e-'/z

- l. s

s (b) f(s) = SZ- kZ'

(el)f(s) =..

+ e-'

(e) f(s) = SZ +

3sz

kZ'

.. .

Use Theorem 12in solving the following differential equations. 8.

Y"(t) - Y(t) = I + e3'.

Ans. Y(t) = C1e'+ Cze-' - I + te3'.

9.

Y"'(t) + Y"(t) - 4Y'(t) - 4Y(t) = F(t), if Y(O)= Y"(O)= Oand Y'(O)= 2. Ans. Y(t) = sinh 2t + lzF(t) * (eZ'+ 3e-z,- 4e-'). 10. y<4'(t)- 2Y"'(t) - Y"(t) + 2Y'(t) = 6F(t), if Y{t) and its first three derivatives are zero when t = O. 25

REPEATED LINEAR FACTORS

We now consider partial fractions for the case in which the polynomial q(s)

(1)

contains

a repeated

linear

factor (s -

f(s) = p(s) -

a)n+ 1. We write

cp(s) q(s)- (s - at+1'

¡;,

74


SECo 25]

where cp(S)is the quotient of polynomials obtained by removing the factor (s af+ 1 from the denominator of the fraction p(s)/q(s). As before, the degree of the polynomial p(s) is assumed to be lower than the degree of q(s). Note that cp(s)and its derivatives are continuous functions at the point s = a. The representation of f(s) in partial fractions now has the form

-

(2)

cp(s) Ao n+l =-+ s-a (s-a )

AlAr (s-a )2 +...

An + (s-a )'+1 + ... + (s-a )n+l +h(s),

where the numbers Ar are independent of s, and h(s) is the sum of the partial fractions corresponding to the remaining factors of q(s). It follows from Eq. (2) that (3) cp(s)= Ao(s

-

a)n + ... + Ar(s - af-r + ... + An + (s - a)n+lh(s)

in a deleted neighborhood of the point s = a. When s -. a, we see that An = cp(a). To find the remaining coefficients A" we differentiate both members of Eq. (3) with respect to s, n - r times, in order to isolate the number Ar. When s -. a in the resulting equation, we find that cp(n-r)(a) = (n

-

r)!Ar.

Equation (2) can now be written in the form n

(4)

f(s)

l r.)r (s - ay+ 1 + h(s),

cp(n-r)(a)

= r=O L (n

-

where O! = l and cp(O)(a) = cp(a). Ir H(t) = L -1{h(s)}, it follows that the inverse transform off(s) is n

(5)

F(t)

= r=O L

cp(n-r)(a) 'u uH,t'¿>t+H(t).

This equation can be simplified by recalling the formula for the derivative of order n of the product of two functions u(s) and v(s), namely,

dn n -(uv) = L ds"

When u

form

(6)

= cp(s)and

r=o(n

,

-

n.

r)!r!

u(n-r)(s)v(r)(s).

v = tT',then arv/as' = t'tT' and Eq. (5) reduces to the

F(t) = ~!{:n[CP(S)e"]} s=a + H(t).

This result can be stated as follows.


Theorem 13

[SECo 26

75

1f f(s) is the quotient p(s)/q(s) of two polynomials such that

q(s) has the higher degree and contains the factor (s

-

a)n+ 1, then the

term in F(t) corresponding to that factor is c1Iia,t),where (7)

1 an c1In(s,t)= n! a~[4>(s)e"]

and 4>(s)is the function

indicated by Eq. (1).

The term in F(t) corresponding to a factor (s (8)

c1I1(a,t) = [4>'(a)

+

-

a)2in q(s),for instance, is

4>(a)t]e"t,

and the term corresponding to a factor (s - a)3 is (9)

c1I2(a,t) = H4>"(a)

+

24>'(a)t

+

4>(a)t2]eat.

As an example, if

1 f(s) = (s - l)(s - 2)2' to the factor s - 1 is é. To correspond with the factor (s - 2)2, we have 4>(s)= (s - 1)- 1 and 4>'(s)= - (s - 1)- 2 so that 4>(2)= 1 and 4>'(2)= - 1. In view of formula (8) the term in F(t) is (-1 + t)e2t. Hence

then the term in F(t) corresponding

F(t)

= é + (t - l)e2t.

Since the number a may be imaginary and since a factorization of every polynomial into linear factors, real or imaginary, exists, Theorems 12and 13providea systematicway offinding inversetransformsof quotients of polynomials in all cases where the factors of the denominator can be determined. Ir, however,imaginaryfactors are present,the results are given in terms of imaginary functions. The reduction of the results to real forms is sometimes tedious. To obtain the real form of the inverse transform directly, and to observe the character of that function in general, we may proceed as follows. 26

aUADRATIC FACTORS

In this section we assume that the polynomials p(s) and q(s) have real coefficients.The imaginaryzeros of q(s) then occur in pairs, each pair consisting of some complex number a + ib and its conjugate. The corresponding linear factors of q are (1)

ql

=

s

-

a

-

ib,

q2 = s - a + ib

(b ::/=O),

where a and b are real numbers. The product of those factors is the real quadratic factor (s - a)2 + b2.

,~

76


SECo 26]

Let q have the factors (1), not repeated,and let
f(s) = p(s) =

(2)

q(s)

-

= s -
According to Theorem 12 the sum of terms in F(t) corresponding to those two linear factors is

(3)

This is the component of F corresponding to the quadratic factor (s

-

a)l + b1. We shall represent it in terms of real-valued functions. The rational function
In a neighborhood of that point excluding the point itself
(4)

-

a - ib)f(s).

Similar1y,

-

ib and

= (s - a + ib)f(s)

throughout a neighborhood of that point with that point deleted. From elementary properties of complex conjugates we can see that the conjugate of f(s), a rational function with real coefficients, is f(S). Consequently
= (s - a + ib)f(s) = (s -

a

-

ib)f(s)

=

In view of the continuity of
cOS

bt

-

i sin bt, it is the conjugate of eibt. Thus formula

G(t) = eQt[

ib)=

rleiOI

= rl(cos81 + isin(1),

we can write the component G in the real form (8)

G(t) = 2eQtr1 cos (bt + (1)

(b =FO,rl > O).

Ir a = O, the component G(t) is the simple periodic function 2rl cos (bt + 8¡), and if a < O, it is a damped periodic function. Those components, corresponding to cases a ~ O,represent stable oscillationsin


[SECo 26

77

theory of control of mechanical and electrical systems where t denotes time. Stable oscillations are bounded as t -+ oo. When a> O, the component G(t) represents an unstable oscillation. To illustrate the use of formula (8), we find F(t) when

the

f(s)

2s - 2 2s - 2 l)(s2 + 2s + 5) = (s + I)[(s

= (s +

=

2s - 2
where

' -4 + 4i 1 - i ( 1 2) '1'1 - + I = 2i(4i) = ~

Th

A..

en

.J2 -i,,/4

= le

and G(t)= .J2e-t cos (2t - nJ4)= e-t(cos 2t + sin 2t). After adding the term corresponding to the linear factor s + 1,we find that F(t) = e-t(cos 2t + sin 2t - 1). In case q contains the square of the quadratic factor (s - a)2 + b2, we write (9)

f(s) = p(s) = q(s)

(s - a

-

ib)2

=

(s

- a + ib)2

and apply formula (8), Seco25, to get the corresponding component of F(t): G(t) = e"t[
+ eat[
(lO)

G(t) = 2e"tRe [

(11) where rl #=O because
ib) #=O. Then formula (10) becomes

G(t) = 2e"t[Plcos(bt + ifJl)+ rltcos(bt + 81)]

(rl > O).

Qur results can be summarized as follows. Theorem 14 Whenf(s) = p(s)Jq(s)where p(s) and q(s) are polynomials with real coefficients and q has the higher degree and contains a nonrepeated real quadratic factor (s a)2 + b2, where b #=O, the component of F(t) corresponding to thatfactor is thefunction G(t) given by formula (8).

-

78


SECo 27]

If q contains the square of that quadratic factor, the component of F(t) is given by formula (12). We note that the component (12) represents a stable oscillation only if the real part of the imaginary zeros a :t ib of q is negative: a < O. If a = O, the component contains the term 2r1 t cos (bt + ()1) representing an unstable oscilIation. When the quadratic factor appears to a degree n + 1(n = 2,3, . . . ),we can see from the procedure we used when n = 1 and from formula (5), Seco25, that the component in F will contain a term of the type Cea'tncos (bt + ()1), where C is a constant. When p and q are not polynomials in s, partial fractions in s cannot be used. Such cases will be treated in Chap. 6 with the aid of residues and contour integrals. 27

TABLES OF OPERATIONS AND TRANSFORMS

Appendix A, Table A.1, contains a list of operations on F(t) with corresponding operations on f(s). That table of operations summarizes several of our resu1ts on the theory of the Laplace transformation. The table of Laplace transforms (Appendix A, Table A.2) gives a fairly extensive list of transforms of particular functions. Derivations of a number of them have been presented above. References to some more extensive tables can be found in the Bibliography. Transforms 82 to 84, Appendix A, Table A.2, of functions that are prominent in problems of diffusion and conduction of heat, will now be derived. A more direct derivation will be made later on in Chap. 6 with the aid of contour integral s and functions of a complex variable; but we wish to use those transforms earlier, in Chap. 4. First we perform formal manipulations that will suggest the inverse transforms of the two functions (1)

y(s)

=

~

e-k./S

(k ;?; O,s > O),

Js

z(s) = e-k,(S

(2)

(k > O,s > O).

We see that (.Jiy)' = -kz/(2.Ji) and 2z' = -ky, so that y and z satisfythis system of differentialequations with coefficientsthat are linear in s: (3)

15y'(s)

+ y(s) + kz(s) = 2z'(s)+ ky(s) = O.

The corresponding system Jor the inverse transforms Y(t) and Z(t) is therefore 2(- t Y)' + Y + kZ = O,

- 2tZ + k Y = O.


79

[SECo 27

Thus -2tY' - Y + kZ = Oand k2

(4)

.

k

=

2t Y(t).

1

2 Y'(t) = ( 2t2

-

t ) Y(t),

Z(t)

The solution of system (4) is C

Ck

k2

= -fi exp ( - 4t ) '

Y(t)

k2

Z(t) = 2.jt3 exp ( - 4t ) .

But when k = O, then y(s) = 1/j""i, and we know that Y(t) is then 11Ft, so if C is independent of k, it follows that C = 1/.j;., and our formal results can be written

- k2

(5)

L ~exp { Ft

(6)

L ~exp

4t)}

(

-

{ 2.jit3

(

= ~e-kJS .Js

k2

4t )}

(k ~ O),

= e-kJS

(k > O).

Let us now prove that transformations (5) and (6) are correct when

k

= 1 and s > O. The function

-~

Y1(t)= ~exp

(7)

Ft

(

4t)

when t > O,Y(O) = O,

is continuous when t ~ O,and bounded. Hence when s > O, eo

.j;.Yl(S)

1

f

dt

eo

1

f [

2

dt

( 4t) -fi = e-../S o exp - (JSi - 20 ) J -fi' = o e-S1exp-

We substitute1: for 2-fi to write (8)

fie../SY1(S) = Leo exp

and make a further substitution tJ;r 2 eo

[- (f1:

- ~rJ d1:,

= 1/).to see that

.JS

[- (2). = .fi f

1

2

d)'

o exp - ¡ ) J ).2' By first adding corresponding members of equations (8) and (9), then substitutingx fort0). 1/)" we can write

(9)

.j;.e../SY1(s)

-

.

2.j;.e-Isy

.fi [ ( (s) = .fi o exp - 2). 2

1

f r: f

2

=v

eo

eo

1

2

.fi

- ¡ J (2 + )

~

exp (_X2) dx = 2 -. s -eo S

1 ).2 ) d)'

.--

80


SECo 27]

Therefore

Yl(S) =

L{J;r exp( - ~t)} = j-se-,¡s

(S> O),

and when k> O,transformation (5) follows by writing Yl(k2s) and applying Theorem 2. Formula (5) was established earlier when k = O. Ir k > O, the funetion Z, defined as follows,

k

Z(t)

= -exp

(

'2~

--

k2 when t > O,Z(O) = O,

4t)

is continuous and bounded when t ~ O; also Z

Y(t) = ~exp

.jid

- k2 (

= tkt - 1 Y where

and

(S > O).

4t )

Theorem 9 therefore applies to give the transformation k

f

z(s) ="2

dx

a:>

s

e-k./X

~

(s > O),

= e-k./S

which is transformation (6). Finally, in view of transformation (6), we note that L -l

I

' e -k2/( 4f) . -i..- d. = - 2 l -k-.Ís = - k -e {s } 2.;:;c o fi

f

a:>

1 e -J.2 dJI.

k/(2../i)

Therefore (10)

(k ~ O,s > O),

where erf(x) is the error function defined in Seco 16. Equation (10) can be written

(11)

(k ~ O,s > O),

where the complementary error function erfe (x) is defined as 2

(12) erfc (x)

o,

= 1-

erf(x) = fi

f

a:>

x e-J.2dA.


[SECo 27

81

PROBLEMS

1. Find the inversetransformstabulated below: f(s) s+a (a) (s + b)(s + e)2

F(t)

-b

a - b -c' (b- e)2e + [ b - et- (b- e)2e J a

-b'

a- e

s+3 (b) (2s + l)(s2 + 2s + 2) 2b4 (e) (S2+ b2) s2-b2

t cos bt

(d) (S2 + b2)2 2Ss2 (e) (s + 1)2(S2+ 4)

(St - 8)e-' + 6 sin 2t + 8cos 2t

2. Without finding F(t), determine whether the oscillations of that function are stable or unstable when its transform is Ans. Stable.

(a) f(s) = - .152_+.1 (b) f(s)

3.

s

= .- .

-~~.

Ans. Unstable.

Solve the following differential equations: (a) Y"(t) - 2Y'(t) + Y(t) = 1.

Ans. Y(t) = (CI + C2t)e'+ 1. (b) Y"(t)+ Y(t) = 2 sint, if Y(O)= Oand Y'(O)= -1. Ans. Y(t) = - t cos t. (e) 4Y"'(t) + 4Y"(t) + Y'(t) = F(t). (d)y(4)(t) + 2Y"(t) + Y(t) = O, if Y(O)= O, Y'(O) = 1, Y"(O)= 2, and .

Y"'(O)",; -3. Ans. Y(t) = t(sin t + cos t). 4. In Seco26, ifwe write
-

PI sin bt).

5. With the aid of transformation (S), Sec. 27, find L -1{s-texp(-kJS)}listed as transformation 8S in Appendix A, Table A.2. 6. Solve for Y(t): (a) (t t2)Y"(t)+ 2Y'(t) + 2Y(t) = 6t, Y(O) = Y(2) = O. Ans. Y(t) = t2 - 8t-l(t - 1)380(t - 1). . (b) Y(t) + 2 J~ Y(x) cos (t - x) dx = ge2'.

.

-

Ans. Y(t) = Se2'+ 4e-' - 6te-'. 7. Draw the graph of the ramp function Y such.that Y(t)

and show that Y(t) = t

-

=t-

Y(t

-

1)

and

Y(t)

=O

when

t ;;;! O,

n when 2n;;;!t ;;;!2n + 1, Y(t) = n + 1 when 2n + I ;;;!

82


SECo 27]

2n + 2, n = O, 1,2, .. . , and that

l L{Y(t)} = s2(1+ e-') 8.

When F, G, and H are sectionallycontinuous, show that F(t) * G(t) * H(t) =

9.

(s > O).

E F(t -

x) J: G(x

- y)H(y) dy dx.

When F is sectionally continuous and G and its derivative G' are continuous over

an interval O;;:¡;t ;;:¡;T, show that, at each point t(0 < t < T) where F is continuous, d dt[F(t) 10.

* G(t)]

= F(t)

* G'(t) + G(O)F(t).

Under the following set of conditions the limit of the integral

(x > e), ~(x) = focof(x,t) dt

-

as x 00, is the same as the integral ofthe limito (a) Letf(x,t) be continuous over each rectangle e ;;:¡; x ;;:¡; e, O ;;:¡; t ;;:¡; T, except possibly for finite jumps across a finite number oflinest ti in each rectangle,and (b)let the above integralconvergeuniformlywith

=

respect to x when x> c. Also, (e) letfhave a limit

= x~co lim f(x,t)

F(t)

uniformly with respect to t(t ,p ti) on each interval O < t < T; that is, for each positive E there is a number N. independent of t such that IF(t) - f(x,t)1 <

whenever

E

x> N.

(O < t < T),

and (d) let f~ F(t) dt exist. Then prove that

lim q(x) = X-CIO

l

co F(t) dt.

o

Note: The difference of the last integral and q(x) can be written

I: [F(t) - f(x,t)]

dt +

f;

F(t)dt

-

f;

f(x,t)dt

= tsq.

Take T large first, independent of x, then x large, to make Itsc¡lsmall. 11. Use a remainder to prove that the integral q(x) .

=

l

CO

x

~dt o t + x

is not uniformly convergent with respect to x on the half line x > l (cf. Probo 10). 00, Show that lim q(x) = rr/2 while the integral of the limit of the integrand, as x is zero.

x~co

-

12. If F(t) ~ G(t) when t > O,then f(s) ~ g(s)for every real value of s for which the transformsf and g existo Prove that property, and iIIustrate it with some particular functions. Note that the case F(t) = 2 sin t, G(t) = sin 2t shows that the converse of the property is not always valido


[SEC. 27

83

13: Inilial-value theorem Ir F is continuous over the half line t ~ O except possibly for a finite number of finitejumps, while F' is sectionally continuous over each bounded interval there, and if F and F' are of exponential order, then the limit of the product

sf(s), as s

- 00, exists and equals the initial value of F: lim sf(s) = F(O).

s~oo

Use a formula for L{ F'} and Theorem 5 to prove that theorem.1 14. Illustrate the initial-value theorem (Prob. 13) with these particular functions: (a) F(t) = A; (b) F(t) = So(t - to); (e) F(t) = t; (d) F(t) = cos kt; (e) F(t) = 11"[1 - So(t - to)]. 15. Irf(s) = p(s)/q(s)where p and q are polynomials whose terms of highest degree are a.s' and b.s"+1, respectively, use the initial-value theorem (Prob. 13) to prove that F(O)= a.lb.. 16. Let F and F' be continuouswhileF" is sectionallycontinuousovereach interval O~ t ~ T, and let all three functions be of exponential order. Then if s2f(s) has a limit as s

-

00, prove that

lim s2f(s) =

F'(O)

s~oo

and

F(O)= o.

17. Illustrate the modified initial-value theorem presented in Probo 16 in case F(t) is (a) sin kt; (b) at + bt2; (e) 1 - cosh t. 18. Ajinal-valueIheorem Let both F and F' be bounded over the half line t ~ Oand continuous there except possibly for a finite number of finite jumps; also let 1F'(t)1 be integrable from t = O to t = oo. Then lim sf(s)

3-+0

Prove that theorem.

In the formula

= t-Q'J lim F(t).

for L{F'} note that our conditions

ensure the

continuity of that Laplace integral with respect to s when s ~ O (Sec. 15). 19. Illustrate the final value theorem (Prob. 18) in case (a) F(t) = A + Be-'; (b) F(t) = So(t - to); (e) F(t) = t - (t - I)So(t - 1). 20. Generalizedconvolulion For a function F(t;r) of two variables, defined over the quadrant t > O,-r > O,letf(s,-r) denote its transform with respect to t. Using the same parameter

s, let](s,s)

denote the iterated transform:

J(s,s) = Loo e-s,!(s,-r) d-r = Loo e-S
.

Indicate formally why J(s,s) is the transform of the generalized convolution F * (t) of F(t.-r) defined below: /

J(s,s) = L{F

* (t)}

where F * (t)

= E F(t

- -r,-r)d-r.

It is convenient to write F(t,-r)= Owhen either t < Oor -r < o. (See Probo 21 for conditions ofvalidity ofthe generalized-convolution property.) When F(t,-r)= F(t)G(t),note that the property reduces to the convolution property (6), Seco16. 21. Prove that the generalized-convolution property written in Probo 20 is valid when s> IXunder these conditions on F: IF(t,-r)1< M exp[IX(t+ -r)]for all positive I

Similar theorems with conditions on I(s) rather than F(I) will be found in Chap. 6.

84

OPERATlONAL MATHEMATlCS

SECo 27]

t and " and F is continuous by subregions (Seco14) over each rectangle, O ~ t ~ T, O ~ . ~ R (cf. Seco 16). 22. When F(t,.) = 80(t - .) in Probo 20, show that J(s,s) = ts-2 when s > O, also that F * (t) = tt, and thus verify the generalized-convolution p~rty in this caseo 23. When F(t,.) = J o(2Jh), we found (Probo 10,Seco21) th¡itf(s,~ = S-l exp (- ./s)o Apply the generalized convolution property (Probo 20) to derive the integration formula (t ~ 0)0 24. When. is fixed and positive, let Z(t,.) denote the solution of the differential equation with constant coefficients (independent of t and -r) m

Z(t,-r)

dn

+ J1 andt"Z(t,.)=

F(-r)

such that Z and dnZ/dtn(n = 1,2, ,m - 1)all vanishwhen t = O. Write the corresponding equation for the iterated transform z(s,s) of Z(t,.) (Probo 20) and compare it with the equation for the transform y(s) of Y(t), where o o o

m Y(t) +

L any,n)(t) = F(t),

Y(O)

= Y'(O)=

o

o

o

= y'm- 1)(0)= O,

n= 1

to show that y(S) = SZ(S,S).Thus deduce the following formula for the solution of the problem in Y in terms of the solution of the simpler problem in Z: Y(t) 25.

=

io

' -Z(t - " .)d.. oot

Use the formula in Probo 24 to find the solution of the problem Y"(t) - Y(t) = F(t),

in the form Y(t) = J~F(r)sinh(t the differential equation in Z(t,-r)o

Y(O)= Y'(O)= O,

Note that - F(-r)is a particular solution of -r)d-ro

3 Elementary Applications

The properties of the Laplaee transformation that we have derived up to this point enable us to solve many problems in engineering and physies involving ordinary linear and partial differential equations. In this ehapter we shall solve a number of problems in elastie vibrations involving ordinary differential equations. They are problems in whieh our method is very eonvenient, although not essential. We shall also treat some simple applieations of integral equations. The next ehapter eontains applieations that involve partial differential equations. The solution of problems of this type is a primary objeetive of this book. In later ehapters we shall extend our treatment of sueh problems. 28.

FREE VIBRATIONS OF A MASS ON A SPRING

Let a body of mass m attaehed to the end of a eoil spring (Fig. 12)be given an initial displaeement and an initial veloeity and allowed to vibrate. The other end of the spring is assumed to be kept fixed, and the spring is assumed to 85

.

86

OPERATiONAL MATHEMATICS

SECo 28]

..x

Fig.12

obey Hooke's law, SOthat the force exerted by the free end is proportional to the displacement of that end. The factor k of proportionality is called the spring constant. We also assume that the mass of the spring can be neglected in comparison with the mass m and that no frictional forces or external forces act on m. Let X denote the displacement of m from the position of equilibrium; that is, let the origin O denote the position of m when the spring is not deformed. Then according to Newton's second law of motion, d2X (1) (k > O). m dt2 = kX

-

Let Xo denote the initial displacement and Vothe initial velocity, so that the function X(t) satisfies the conditions X(O) = Xo,

(2)

X'(O) = Vo.

We can determine the function X(t) by applying the Laplace transformation to both members of Eq. (1) and using the conditions (2). Thus if x(s) denotes the transform o~X(t), it follows that m[s2x(s)

-

sXo

-

vo] = -kx(s),

5

and therefore X(5) = Xo

2

.

n.

1 L'

+

Vo 52

+ (k/m)"

Hence (3)

X(t)

= Xocos wot + ~

Wo

= jxo2 + where .

Wo =

A m -,

sin wot

(~:r sin (wot +

ex),

xowo.

tan ex=

-, Vo

WoXo

smex =

y,Wo 2Xo 2

+

Vo2'

The motion described by formula (3) is a simple vibration with angular frequency wo, called the natural frequency of this system, and phase angle ex, and with the amplitude [xo2 + (VO/wo)2]t. If a viscous damping force proportional to the velocity also acts upon the mass m, as indicated by the presence of a dashpot c in Fig. 13,the equation of motion becomes (4)

mX"(t) = -kX(t) - cX'(t),

ELEMENTARY APPLICATIONS

[SEC. 28

87

Fig.13

where the eoefficient of damping e is a po sitive constant. Let the mas s start from the origin with initial velocity Vo: X(O) = O,

X'(O)

= Vo.

The equation in the transform x(s) becomes ms2x(s)

-

mvo

= -

-

kx(s)

esx(s)

or, if we write 2b = e/m, and again write illo2 = k/m, (5)

x(s) =?

-.Vo

? =

Vo .

n1

Ir b2 < illo2, that is, if the coefficient of damping

1

-

l'

is smalI enough that

e2 < 4km,

then the formula for the displacement is (6) In the case of critical damping, that is, when illo = b, or e2 = 4km, it folIows from Eq. (5) that (7) We can see from this formula that the mass m moves in the direction of Vo until the time t = l/b, then reversesits direction and approaches O as t tends to infinity. When e2 > 4km, a similar motion of the mass takes place. The discussion of this case and the case of other initial conditions is left to the problems. The mathematical problem treated in this section can be interpreted also as a problem in electric circuits. This welI-known analogy between problems in vibrations of mechanical systems and electric-circuit theory will be observed for other problems in this chapter. NaturalIy, the notation and terminology differ in the two types of problems. In the electric circuit shown in Fig. 14,let Q be the charge accumulated in the capacitor e at time t, and 1 the current in the circuit, so that (8)

./

I(t) = Q'(t).

88


SECo 29]

Fig.14

This equation requires that I(t) be positive when Q(t) is increasing. Ir the positive sense of ftow I(t) of positive charges is taken in the clockwise direction in Fig. 14, then Q(t) measures the charge on the upper plate of the capacitor. The circuit has a resistance R and a coil of inductance L. Since the sum of the three voltage drops in the circuit is zero in this case of no impressed voltage, then 1 (9)

LI'(t)

+ CQ(t)

+ RI(t) = O.

The system of first-order differential equations (8) and (9) can be used directly to determine I(t) and Q(t) in terms of initial values of those functions. By eliminating I(t) from the two equations, however, we see that (10)

1 LQ"(t) + RQ'(t) + CQ(t) = O.

Except for the notation used, this equation is the same as Eq. (4). When the resistance is negligible, R = O,the equation reduces to our Eq. (1). The initial conditions in the electrical problem can be made the same as those in the mechanical problem. For example, if the capacitor has an initial charge Qo and if the initial current is 10, then Q'(O) = 10,

Q(O) = Qo,

which are the same as the initial conditions (2) in our first mechanical problem. 29.

FORCED VIBRATIONS WITHOUT DAMPING

Let an external force F(t) act upon the mass in the mechanical system of the last section, assuming there is no damping (Fig. 15). The displacement X(t) of the mass m then satisfies the differential equation

mX"(t) = Ir the initial conditions are

(1)

(2)

- kX(t)

X(O) = Xo,

+ F(t).

X'(O)= Vo,


[SEC. 29

~

Fig.15

89

m Ftt) O~x

the equation in the transform x(s) becomes

m[s2x(s)- sXo - vo] = -kx(s) + f(s), where f(s) is the transform of the force function F(t). Let Wo again denote the natural frequency of the system,

wo=A. Then we can write (3)

x(s)

xos + Vo

1

1

= s2 + Wo2 + -m f(s)---z.. s + Wo

Hence the displacement for any F(t) can be written, with the aid of the convolution, as (4)

X(t) = Xocos wot +

~ sin wot + ~ ' sin wo(t Wo mwo o

f

- -r)F(r) d-r,

a result that satisfies conditions (1) and (2) above. But the motion of the mass under particular external forces F(t) is more interesting than the general formula (4). In these special cases it is often easier to refer to the transform (3) than to formula (4). When F(t) is a constant F o, as in the case when the X axis is vertical and the force of gravity acts on m, Eq. (1) can be written

mX"(t)= -k[ X(t) - ~J. If Y = X - Fo/k,this becomes mY" = -kY; so the motion is the same as freevibrations if displacementsare measuredfrom a new origin at a distance of Fo/kunitsfromO. Note that Fo/k is the staticdisplacementXo of the free end of the spring due to the applied forceFo, sinceFo = kX o. In case (5)

F(t) = Fo =0

then

f(s) = F{~ - exp(;tos1

when O < t < to, when t > to,

1 ..

90


SECo 29]

Ir Xo =

Vo

= O,it follows from Eq. (3) that

(6)

x(s)

=

FO m S(S2

[

- exp(-toS)

1

+

(02)

S(S2

+

(02)

].

Now

and ifwe write

tf¡(t) = sin2 !wot

when t > O,

=0

when t < O,

it follows from Eq. (6) that

(7)

2Fo X(t)

=k

[tf¡(t) - tf¡(t

- to)].

The graph of this function can be drawn easily by composition of ordinates. When to is approximately !n/wo, the graph is the full-drawn curve in Fig. 16. When to = 2n/wo, it follows from Fig. 16 that the mass m performs one oscillation and then remains at the origin (Fig. 17). The step function (5) is proportional to the unit finite impulse function (3), Seco12, with impulse starting at t = O.The area under the graph of our function is Foto. Ir this is kept fixed, Foto = Mo, while to -+ O and Fo -+ 00, the force F(t) becomes the impulse represented formally by the equation (8)

F(t)

=

M ol5(t),

where l5(t)is the unit impulse symbol (Sec. 13). The constant M o represents the increase in momentum of the mass m at the instant t = O.

X(t)

,

Fig.16

ELEMENTA.RY APPLICATIONS

[SECo 30

91

X¡t¡

lO, 2FO)

"

o Fig.17 The formal transform of M o <5(t)was found to be M o. This suggests that, in view of Eq. (3), when Xo = Vo= O, Mo 1 x(s) = 2 " m s + Wo and hence that (9)

X (t )

Mo.

= -SInWoto mwo

It will be left as an exercise to show that formula (9) follows rigorously as a limiting case of formula (7). But the formal solution can be verified as well by noting that, according to formula (9),the momentum of the mass m, M(t) = mX'(t) = Mo cos wot, satisfies the condition M( +0) momentum 30.

= Mo. Ir the mass started from rest, then its

jumped to the value M o at the instant t

=

00

RESONANCE

Let the external force in the problem of the last section be F(t)

= Fo sin wt,

where Fo and w are positive constants. Then according to Eq. (3), Seco29, w (1) and, if w ::f:wo, (2)

l

X(t)

= Xocos wot + -Wo

[

Vo

+

.

Fow m (w

2

-

Wo

-

2

)J

SIn wot

Fo 20 m(w - Wo2)SInwto

92


SECo 30]

That is, the motion is the superposition of two simple harmonic motions, one with frequency Wo and known as the natural component of vibration, and the other with frequency w which is called the forced component of the vibration. Note that the natural vibrations are not present in case Xo

= O,

Vo=

Fow

2 m(Wo

-

2

'

w )

However, if the frequency of the periodic force F is the same as the natural frequency of the oscillator, w = wo, then (3) The presence of the repeated quadratic factor in the denominator here shows that X(t) will contain an unstable component (Sec. 26) having the form ofthe product of t by a cosine function. In fact, (4)

X(t)

l Fo . Fo t cos wot. SIRwot - = Xo cos wot + ---"2 vowo + _ 2 2 Wo ( m) mwo

In view of the last term here, the amplitude of the oscillations of m increases indefinitely. In this case the force F(t) is said to be in resonance with the system. We note in particular that if Xo = O and Vo = - Fo/(2mwo) the resonance type of motion reduces to X(t)

Fo

= .-,.2mwo -

t cos wot,

shown in Fig. 18.

X(t) "'"

.....---

o

(.:. ,o/ ---u

---

-- --

--

"'".....-

'

-- --Fig.18

t

ELEMENTARY APPLlCATIONS

[SEC. 30

93

PROBLEMS

1. A good mechanical oscillator for demonstrating vibrations can be constructed by selecting a length of coil spring such that when a weight of! lb H = mg,g = 32ft/sec2) hangs at rest from the free end, the spring is extended by 0.2 fl. Show that the spring constant then has the value k = Ib/ft, and that the natural frequency of the oscillator is Wo = 4JiO = 12.65rad/sec, or about 2 cycles/sec,and hence the period T of oscillation, T = 2n/wo, is approximately ! seco 2. Let the supported upper end of the spring in Probo 1 be moved vertically with displacement Z(t), where Z(O)= O. Let X(t) denote the vertical displacement of the mass m from its position on the lower end of the spring at time t = Owhen the spring has its natural length. AIso let Z and X be positive for displacements downward. If damping forces are neglected, note why X(t) then satisfies the equation

t

mX"(t)=

- k[X(t) -

Z(t)]+ mg.

Let Y(t) be the displacement of the mass m from its static position when supported by the spring while Z(t) = o. Show that

mY"(t) = -k[Y(t)

- Z(t)]

= -kY(t)

+ kZ(t),

and hence that Y also represents the displacements in a horizontal oscillator with an external force kZ(t) applied to the mass. 3. The upper end of the spring in the oscillator described in Probs. l and 2 is suddenly lowered l in. at the instant t = O and kept in that position during twice the natural period 2T = 4n/wo sec (about l sec), then suddenly returned to its initial position, so that

Z(t) = -h[1 - So(t - 2T) Ifthe mass m (ofweight

(t > O)

t lb) initially hangs at rest, show that it then makes two oscilla-

tions 2 in. downward from and back to its initial position, then remains there after the instant t = 2T,since Y(t) 4.

In Probo 2 if Z(t)

1 . 2 nt = "6sm T

=

whenO;;¡¡ t;;¡¡ 2T,

Y(t) = O

when t ~ 2T.

vot and Y(O) = Y'(O)= O,show that

Y(t) = vot -

~

Wo

sin wot

and desocibe the motion of the upper end of the spring and the mass. 5. Let the upper end of the spring in the oscillator described in Probs. l and 2 be moved vertically by means of a rotating eccentric or cam with the periodic displacement Z

= l2 sin wot

ft, with the natural

frequency

Wo

= 4JiO rad/sec of that oscillator.

If Y(O)= Y'(O)= O,show that l. Wo Y(t ) = 24 smwot - 24 tcoswot. Note that the vibration of the mass is unstable under that periodic 2-in. oscillation of the support, and that within only four cycles of the oscillations (t = 8n/wo < 2 sec), displacements of the mass mount to approximately l ft.

94

6.


SECo 30]

If the coefficient of damping for the oscillator shown in Fig. 13 is so large that = Oand X'(O) = vo,(a) show that

e2 > 4km and if X(O)

:

V. X(t)

=

et smh (at) exp ( - 2m')

p-4km

wherea =

2m

.

(b) Find X'(t) and show that the mass m moves in the direction of vo until the instant t

=

11

e+2am - 2am'

2a og e

when it turns and approaches the origino 7. Let the mass m in the damped oscillator shown in Fig. I3 be released from rest with an initial displacement X(O) = xo, and write 2b = e/m. (a) If e2 < 4km and WI = (W02 - b2)t, show that

. b wh eresm(X=-, Wo

1t O;;¡¡(X< "2;

(b) if e2 = 4km show that X(t)

= xoe-bt(l + -bt),

and hence that the mass never moves across the origino 8. When the force on the undamped oscillator shown in Fig. 15is the periodic function F(t) = Fo sin (wt + 8),where 8 is a constant, show that resonance occurs when w = Wo. 9. When t = O. let the current in the circuit shown in Fig. 14 be zero and let the capacitor have a positive charge Qo. If R > 2jL!C, derive the formula t Rt . (Xt 2Qo 2 4L l(t)= --exp -smh(X= R -(XC 2L) 2L C) ( (

[

for the current.

Show that l(t) ;;¡¡O and 1(00)

]

= O.

10. The current 1, and the charge Q on the capacitor in the circuit shown in Fig. 19, are functions of t that satisfy the conditions L-dI + -Q + RI = E dt C o'

Q = f~ I(T)dT,

1(0)= O,

when Q and 1 are initial1y zero, where t is the time after cIosing the switch K, and the electromotive force Eo is constant. If b = R(2L)-1 and Wl2 = (LC)-l - b2 > O, derive the formula

1 = EOe-bl sin wlt. WIL

Fig.19


31.

[SEC. 31

95

FORCED VIBRATIONS WITH DAMPING

When an external force F(t) acts parallel to the X axis on the mass in the damped oscillator shown in Fig. 13,the equation of motion becomes

mX"(t) =

(1)

- kX(t) -

eX'(t)

+ F(t).

Ir X(O)= X'(O) = O,the solution of the transformed equation is 1

(2)

x(s)

1

= -m f(s)

Again, consider the periodic force F(t)

= Fo sin wt

and the case e2 < 4km. Then b < wo, and Eq. (2) becomes Fo

(3)

x(s)

= -;;(S2 +

w w2)[(S + b)2 +

(W12 Wl2]

= W02 - b2).

It follows from Theorem 14, Seco26, that X(t) consists of terms of the types A cos (wt + e) and Be-bt cos (wI + el), where A, B, e, and el are constants. ConsequentIy the component of oscillation with frequency WI is nearIy damped out after a sufficiently long time, and the periodic forced component of vibration

(4) remains. 4>1(S)

(A > O)

Xp(t) = A cos (wt + e)

=

Its (s

-

amplitude A is, according iw)x(s). Thus

A

(5)

= IFol

to

Seco 26, 214>1(iw)1 where

.

1

m I(iw+ b)2 + wl21

The square of the absolute value in the denominator can be written as IW02 - W2

+ 2ibwl2,which is (W2-

W02)2

+ 4b2w2,and by expanding and

completing the square, we can write that polynomial in w as (W2 + 2b2 Wo 2)2 + 4b2(wo 2 - b2). Thus (6)

A=

-

~

u~

IFol

~-~

-

~_..

The amplitude A of the periodic vibration depends on the frequency w of the force Fo sin wt, and it varies directIy with the amplitude IFol of that force. When Fo is fixed, the value of w for which A is greatest is the resonance frequency wr of our damped oscillator under that periodic force. Let the coefficient of damping be small enough that e2 < 2km. Then 2b2 - W02 < O, and we can see from formula (6) that A is greatest when

96

002


SECo 32]

=

0002

- 2b2. Thus the resonance frequency in that case is

(7)

00, = Jooo~ - 2b2

(2b = ;,e2

< 2km),

a frequencyonly slightly less than 000when e is small. When 00= 00" the amplitude A of the undamped component of vibration has the value

~

A = = IFol , 2bmool eOOl'

(8)

We can seethat the expression(3) for X(5)cannot contain a repeated

factor of the type (52 + (2)2 in its denominator unless e = Oso that b = O. Hence no value of the frequency w of our periodic force will induce a com-

ponent of X(t) of type t cos (wt + 9) as it did for the undamped oscillator. The amplitude A of the forced component of vibration is still given by formula (5) when the force Fosin wt is replaced by a force Fosin (wt + IX) (Prob. 3, Seco32). In case e2 ~ 2km, it tums out that the amplitude A of the forced vibration tends toward an upper bound IFol/kas 00tends to zero. When w is small and t is large enough that wt is near an odd multiple of n/2, the magnitude

of the force Fo sin wt is then near IFol. Under static conditions

the force IFol would displace the end of the spring a distance IFol/k. The formula for X(t) when e2 = 2km, found in Probo 4, Sec. 32, shows that when

w is small the mass may oscillate slowlywith an amplitude near IFol/k. 32.

A VIBRATION ABSORBER

We have seen that for the simple damped oscillator with an excitingforce Fo sin wt the forced component of vibration A cos(wt + O) remains undamped. Let another spring and mass be connected in series with the original mass (Fig. 20), where that second oscillator is undamped. We shall see that if the spring constant and mass of the auxiliary oscillator are chosen so that the natural frequency of that oscillator coincides with the fixed frequency w of the exciting force, then the forced vibrations of the first mass will be eliminated.

F!tl mI A

m

k

kl I OL

Fig.20

..x

ELEMENTARY APPLlCATlONS

[SECo 32

97

Let X and X 1 denote the displacements of the masses m and mi' respectively, from the positions they have when both springs have their naturallengths. If the exciting force is F(t)

= Fo sin rot,

and if m and mi are initially at rest at their respective origins, then the displacements X(t) and X i(t) satisfy the following system of differential equations:

. m dt2 = -kX + k1(X1- X) - c-;¡¡ + Foslnrot d2X

d2X1 mld(2

dX

= -k1(X1 - X),

X(O) = X'(O) = X 1(0) = X'I(O) = o. The transforms x(s) and XI(s) of X(t) and X l(t) therefore satisfy the simultaneous algebraic equations Foro (ms2 + cs + k + k1)x(s) - k1xi(S) = -y'l' S +ro

kix(s) - (m1s2 + k1)Xi(S) =

o.

Eliminating x1(s), we find that (1)

mls2 + ki ' x(s) = Foro(s+rops 2 2) ( )

where (2) In view ofthe presence ofthe quadratic factor S2 + ro2in the denominator of the right-hand member of Eq. (1), it follows that X(t) will contain a term of the type (3)

ecos (rot + O)

unless kdml = ro2,in which case the numerator ofthe above fractioh cancels with the.factor in the denominator, leaving (4)

1 x(s) = FOromlp(s)'

The inverse transform X(t) of the function (4) represents a damped oscillation of the mass m if each of the four roots, real or imaginary, of the

98


SECo 32]

equation p(S)= Ohas

a negative real part (Secs.24 to 26). Now the poly-

nomial p(s) is the determinant of this system of homogeneous equations in x and Xl:

(ms2+ es + k + kl)x - klxl = O,

(5)

-klx + (mls2 + kl)XI = O.

When the determinant vanishes, that system is satisfied by numbers x and Xl' both differentfrom zero. Let s be any one of the four roots of the equation p(s) = O and let

z,z1 denote a pair of nonvanishing roots of the system(5)that corresponds to that value of s, where z and z1may be either real or imaginary numbers. We now write X = z and Xl

= Zl in Eqs. (5), then

multiply the members of

those equations by the complex conjugates Zand Zl' respectively,and add to obtain the equation (mzz + mlzlzl)s2 + ezzs + B = O,

(6) where

B = kzz + kl(zz - ZZI- ZIZ + ZIZ¡)

= klzI2 Ifwe write A

+ kllz - zl12 > O.

= mlzI2 + mllzl12,then Eq. (6)becomes As2 + elzl2s + B = O,

(7)

where the coefficients A, clzl2, and B are positive numbers that depend on the value of s. However, the number s is given in terms of those positive coefficients by the quadratic formula 1 (8)

s

= 2A (-

clzl2 :f:

.J

e21z14

- 4AB).

Whether the radical here is real or imaginary, it follows from formula (8) that the real part of s is negative. This completes the proof that the oscillation X(t) is damped. Thus the forced component of the vibration of the main mass m is eliminated by the system mI' kl, if the natural frequency of that system coincides with the frequency of the exciting force:

[k; = OJ.

(9)

V~

Since all components of the vibration of m are then damped, that mass approaches a fixed position as t increases. This is the principIe of the Frahm vibration absorber, which has been used in such practical appliances as electric hairclippers.l Note that, in 1 Den Hartog,

J. P., "Mechanical

Vibrations,"

4th ed., pp. 87 fr., 1956.


[SECo 32

99

view of Eq. (9), the absorber is designed for a fixed frequency ro of the exciting force Fo sin rot.

By solving the above equations for XI (s),we can see that the mass mi has an undamped component of vibration of the type (3). PROBlEMS 1. If the initialconditionsX(O)= X'(O) = Oused in Seco31 for the damped oscillator with exciting force F(t) are replaced by the conditions X(O)= XO,X'(O)= vo, show that the additional terms in the formula for X(t) represent damped oscillations. 2. When F(t) = FoD(t)and X(O)= X'(O) = Ofor the damped oscillator considered in Seco31, where e2 < 4km, show formally that X( t)

Fo

= -e-

mrol

b..

s!nrolt

(b

2 e 2 k = -,rol =- - b ,

2m

m

)

and note that the jump in the momentum of the mass m at the instant t = Ois Fo. 3. Show that formula (6), Seco31, for the amplitude A of the forced component of vibration for the damped~scillator is valid (a) when the exciting force F(t) is Focos (J)f; (b) when F(t)

=

Fo sin (rot + IX).

4. If e2 = 2km for the damped oscillator considered in Seco31, and if F(t) = Fosin wt and X(O)= X'(O) = O,show that X(t) = ~[COS 92cos (rot+ 9¡) - e-b. cos 91 cos (bt + (2)] where 2b

=

e/m and 91 and 92 are these arguments 91

= arg [ - 2bro +

of complex numbers:

¡(ro2 - 2b2)],

Hence when ro is small and rot is large, show that X(t) is approximately (Fo/k)sin rol. 5. In Seco31, if e2 = 4km, F(t) = Fosinrot, and X(O)= X'(O)= O,show that X(t) has the form A cos (rot + IX)+ (B + Ct)e-bt, and when ro is small that IAI is near its upper bound IFol/k.

t

6. Let the oscillator describedin Probs. 1 and 2, Seco30,for whichk = Ib/ft and mg = lb, be subject to a damping force - eY'(t) where e = f4Ib/ft/sec. (a) If the upper end of the spring is fixed so that Z(t) = O,and if Y(O)= Yoand Y'(O)= O,show that after about nine cydes of oscillation of the mass the amplitude of the displacements Y(t) is reduced to approximately o.llyo!. (b) When the upper end is oscillated so that Z(t) = Zosin rot, show that resonance occurs if ro = ror where ror = )159.5 rad/sec, a frequency slightly less than roo(cf. Seco31). (e) When ro = rorin part (b),show that the amplitude A ofthe undamped forced compone~t ofvibration exceeds 121zol. 7. Let the two masses in the system treated in Seco32 have arbitrary initial displacements and velocities. When condition (9) is satisfied by the eIements of the absorber, show that the vibration of the main mass again contains no undamped component. 8. (a) If the exciting force Fosin rot in the system of Seco32 is replaced by the force Fosin (rot + IX),where IXis any constant, show that when condition (9) is satisfied the vibration of m is again entirely damped.

t

I

I

I !

i

I 1 [¡

100


SECo 32]

(b) If the exciting force is replaced by Al sin rolt + A2 sin ro2t,where the A's and ro's are constants and rol f= ro2' show that the values of mi and kl cannot be adjusted so that alI undamped vibrations of m are absorbed. 9. Let the force F(t) be removed from the mass m in Fig. 20 and let the end A of the spring k be moved horizontalIy so that its distance from the walI is F(t)/k. Show that the differential equations of motion of m and mi are then the same as in the original problem. 10. A system of two masses and two springs in series, with no damping, is shown in Fig. 21. The weights of the two masses are mlg = 8lb and m~ = 2lb (g = 32 ft/sec2). The spring constants have the values kl = 24lb/ft and k2 = 8lb/ft. If a periodic force F = Focos 2nct acts on m2' find the frequencies c (cycles per second) of the force for which resonance occurs in the vibration of mi' Assume zero initial displacements and velocities. Also show tltat resonance occurs in the vibration of m2for those same values of e. Ans. e = 4n-I, 4 J3 n-l cycles/sec.

Fig.21 11. For the system shown in Fig. 21, if mi = m2 = ls, kl = ?s, k2 = ?s, and if both masses start from rest in their equilibrium positions, find the formula for the displacement X l(t) of mi when the force F(t) on m2 is arbitrary. Ans. X l(t)

12.

=

(6 sin t

- J6 sin t J6) * F(t).

InitialIy the two masses shown in Fig. 22 are at rest and the spring has its natural

length. ThenaconstantforceF

=

Fo(t > O)actsonml'

Ifthesystemisfreefromdamp-

ing, find the formula for the displacement of m2 and note that the displacement consists of a simple harmonic motion superimposed upon a uniformly accelerated motion. Also show that ifFo > O,the spring is always compressed by an amount proportional to 1 - cosrot,wherero2= k/ml + k/m2' Ans. 2w2(ml + m2)X 2(t) = Fo(ro2t2 - 2 + 2 cos rot).

Fig.22

13. If the force F = O in the system shown in Fig. 22 and if the masses are initialIy released at rest from positions Xl = al and X 2 = a2' show that the frequency of vibration of the masses has the value ro = (k/ml + k/m2)t. When a2 = -alml/m2' show that XI = al cos rot and X 2 = a2 cos rot and describe the vibration. 14. Let the force in the system shown in Fig. 22 be the periodic force F = Fosin rot, where ro2 = k/m2' and let a viscous dampingforce -eXí(t) act on mi' Find the steadystate vibration of m2 and note that it does not depend on c. Ans. - Fok- 1sin rot. 15. In Fig. 23 the end E of the first spring has a periodic displacement Y = A sin rot. The two springs are identical. Find the value of ro for which this damped system is in resonance, when e2 < 4km. Ans. 2m2ro2= 4km - e2.

ELEMENTARY APPLlCATlONS

[SEC. 32

E

~ I

m

k

I

~y

Fig.23

101

I

k

~X

16: In Fig. 24 the two masses are unit masses (m=:=1)and the two springs are identical. Initially, the springs have their naturallength, the first mass has velocity vo, and the second mass is at rest. Find the undamped component of vibration of those masses. Ans. !(vo/.jk) sin t .jk.

Fig.24 17. For the undamped system of two equal masses and three identical springs shown in Fig. 25, the initial conditions are XI(O) = al' X2(O)= a2, X'I(O)= Xí(O)= O. (a) If lall * la21,show that the components of vibration of each mass have frequencies

-fl0ñ

and J3k/m.

(b) If a2 = al'

show that both masses vibrate

in unison with

frequency -fl0ñ. (e) If a2 = - al' showthat the massesvibrate in oppositedirections with frequency J3k/m.

f

Fig.25

I

L-..

Xl

L-. X2

18. A dampedabsorber Figure 26 shows an absorber with elements mi, kl, and e in which the damping is located in the absorber itself. When a force Fosin wt acts on the main mass m, the undamped vibrations of m cannot be completely absorbed here, but the coefficient of damping e can be adjusted to give an optimum range of the amplitudes ofthose vibrations. Let mand mI be initially at rest in their positions of equilibrium. Derive the formula x(s)

=~

mls2 + es + k¡ S2 + W2(mls2 + es + k¡)(ms2 + es + k + k¡) - (es + k¡)2

for the transform of the displacement X(t) of m, then show that the amplitude A of the forced component A cos (wt + O)of X(t) is given by the formula (m¡w2 - k¡)2

A2

F? = [(mw2 -

k)(m¡w2

-

k¡)

-

m¡k¡w2]2

+ e2w2 + e2w2[(m+ m¡)w2

- kp.

(A graphical examination of A2 as a function of W2 would indicate a value of e for

whichthe rangeof valuesof A2willbe as smallas possiblefor all w.¡) ¡ See Den Hartog, op. cit., pp. 93 ff.,ror a detailed discussion.

.1

102


SECo 33]

Fig.26

33. ELECTRIC CIRCUITS In Seco28 we used Kirchhoff's voltage law: for each closed circuit in an electrical network the impressed voltage equals the sum of the voltage drops across the elements in that circuit. His law 01 currents states that at each junction point of branches of a network the total current into the point equals the total current away from the point. Let us now present additional applications of these laws for instantaneous behavior of voltage and current, in setting up differential equations for networks. An electrical analog of the Frahm vibration absorber discussed in Sec. 32 is indicated in Fig. 27, where the impressed voltage is V

=

Va sin rot.

According to the law of currents, applied at the junction P, (1) We apply the voltage law to the circuit on the left to see that l (2)

l

V = LI'(t) + RI(t) + CQ(t) + CI QI(t),

where Q'(t) = I(t) and Q'l(t) = II(t); then to the circuit on the right to see that

(3)

~l QI(t) = LII;(t). p Ll

Fig.27


[SECo 33

103

Let Q2(t) denote the difference petween quantities of charge on the two capacitors, so that (4)

Q~(t) = 1(t) - 11(t) = 12(t).

In terms of Q and Q2 Eqs. (2) and (3) take the form

(5)

LQ"(t) = - ~Q(t) + ~ [Q2(t) - Q(t)] - RQ'(t) + Vosin wt, 1 1 LIQ2(t)

=-

CI [Qz(t) - Q(t)].

Except for differences in notation the system (5) of differential equations in Q(t)and Q2(t)is precisely the system written in Seco32 for the displacements X(t) and Xl (t) of the masses m and mi in the absorber (Fig. 20). Since all the electrical coefficients here, R, L, etc., are positive constants, the signs of the

coefficientsin Eqs. (5) match those in the equations for X and X l' The initial conditions for the network match those chosen for the absorber, and the analogy is complete if (6)

Q(O) = Ql(O)

= 1(0) = 12(0) = O.

Under conditions (6) the transforms of 1(t)and 12(t)are sq(s) and SQ2(S), which correspond to sx(s) and SXl(S) for the absorber. According to the results found in Seco 32 therefore, the current 1(t) has an undamped component of type B cos (wt + IX)unless 1 = W2. LICI

(7)

But when the elements ofthe LI C 1circuit satisfy condition (7)for a prescribed frequency w of the impressed voltage V,the current 1(t)contains only damped components; thus 1(t) ~ O as t ~ oo. The currents 11(t) and 12(t) in the LI C1circuit, however, do have undamped components oftype B cos (wt + IX). Note that Eqs. (1) to (3) with conditions (6) can be solved directly for the currents either by replacing

Q(t) by

f~ 1(.) d.

or q(s) by i(s)/s, etc. Thus

the transformations of Eqs. (2) and (3) gives the equations

(8)

(LS + R + ~s)i(S) + cljl(S) = v(s), LISi(s)

-

(LIS + Clls)il(S)

= O.

Upon eliminating il(s), we find that (9)

i(s) = v(s)y(s)

if

y(s)

= s(LIs2

+ CI-I)

p(s)

,

104


SECo 33]

where P(s)is the polynomial introduced in Seco32: p(s) = (L1S2 + Cl-l)(Ls2 + Rs + C-l + Cl -1) - Cl -2. The function y(s) depends only upon the characteristics of the network, not upon the impressed voltage. Since the formal transform of the unit impulse symbol b(t) is unity, it follows from Eq. (9) that y(s) = i(s) when V(t) = b(t) formally; that is, Y(t) represents the current 1 produced by the voltage b(t). When the network is considered as a system with input V(t) and output I(t), then, in the language of systems analysis, y(s) is the transfer function for the system; in view of Eq. (9) the output corresponding to any input V(t) is given by the formula (lO)

I(t)

= V(t)

* Y(t).

PROBLEMS 1. If 11(O)= 12(0)= Oin the network shown in Fig. 28, while the capacitors have the same initial charge Qo, show that

wherero = (CL)-t.

e 11

R

12 Fig.28

2. Ifl1(0) = 12(0) = Oin thenetwork shown in Fig. 28 and ifQI(O) = alandQ2(0) = a2, find the undamped

component

Í 1 (t) of the current 11(t), the steady-state

current through

the first inductance coil. Ans. Í 1 = -!(al + a2)wsin rot,ro = (CL)-t. 3. Show that the network indicated in Fig. 28 represents an electrical analog of the mechanical system shown in Fig. 24 where QI and Q2 correspond to the displacements of the two masses. 4.

In Fig. 29 the currents 11(t) and 12(t)and the charge Q(t) are initially zero. (a) Show that the current 11(t) is unstable under the impressed voltage V = Vosin (rot + ex)if ro2 = (L1q-l + (L2Q-I. (b) Show that the system indicated in Fig. 22 is a mechanical analog to the network here with X I(t) and X 2(t)corresponding to Ql(t) and Q2(t),where Q'l(t) = 11(t), Qí(t) = 12(t), and Q(t) = Ql(t)

- Q2(t).

Fig.29


[SECo 33

105

5. A prescribed current l(t) is supplied to the network shown in Fig. 30. The initial values of 1I(t)and Q(t)are zero. Use the following notation: V(t) is the resulting potential drop from the junction A to the junction B, m02 = (LC)-I, 2b = (RC)-I, and ml2 = m02 - b2, given that b2 < m/. (a) Show that the transfer function from input l(t) to output 1¡(t) is y( s)

-

-

S2

m o2 + 2bs + mo2'

thus thatI ¡(t) = l(t) * Y(t), where Y(t) = m/ml -Ie-br sin mi t. (b) Find V(t) when l(t) = lo, a constant. Ans. V(t) = 10(Cm¡)-¡e-br sin mlt.

Fig.30 6. In a simple LC circuit with inductance coil and capacitor in series, let the impressed voltage be a periodic function, with period T, of the type N

V(t) = ao +

L (a. cos nmt + b. sin nmt), n=1

where m = 2n/T. Show that the current l(t) in the circuit is unstable ifthe frequency m of V(t) has any one ofthe values

m-

l

-

1

(m = 1,2,..., N),

-myU

providedthat amand bmare not both zero. 7. In the network shown in Fig. 31 all currents and chargesare zero at the instant t

= Owhen the switch K

V

= Vosin mt,

is closed. If the impressed voltage is given by the equation

where m2 = 2(LC)-I,

find the steady-state

value 12(t) of the current

12(t)and note that it is independent of the value of the resistance R. Ans. 12(t)

=-

VoCmcos mt.

Fig.31 8. In the system shown in Fig. 25 apply an exciting force to the first mass and viscous damping to the second mass. Show that the resulting system is a mechanical analog ofthe network shown in Fig. 31.

106


SECo 34]

9. Initially the currents and the charge on the capacitor are zero in the network shown in Fig. 32. The voltage drop from Al to B1 caused by the mutual inductance of the two coils is MI;(t), and the drop from A2 to B2 caused by that mutual inductance is Mlí(t),whereM2 < L1L2. Theimpressedvoltageisgivenbytheequation V = Vocoscot. When the values of C and L2 are adjusted so that L2C = co-2, show that the current Il(t) in this idealized resistance-free circuit has a simple periodic variation at all times with a frequency greater than co.

Fig.32 10.

Show that the network in Fig. 33 represents an analog of the damped absorber

shownin Fig. 26 if initial currents and charges are zero, where charges Q and Q correspond Fo sin cot.

to displacements

X and Xl and the voltage

V corresponds

-

Q¡

to the force

Q

~c

1

Ql R

11 Fig.33

34.

EVALUATION

OF INTEGRALS

Laplace transforms can be used to evaluate some types of integrals containing a parameter. The manipulative procedure is often direct and simple, but, as the examples and exercises here will indicate, proofs of the reliability of those formal results can be challenging. In our proofs we use this special case of the theorem on uniqueness stated in Seco6: A function f(s) cannot have more than one inverse transform F(t) that is continuous over the half line t ~ Oand of exponential order. Example 1 Evaluate the integral (1)


[SECo 34

107

Here F(t) is the Fourier cosine transform (Chap. 10) of (X2 + a2)-1. Note that F is bounded: ¡F(t)1;;;¡; f~ (X2 + a2)-1 dx = n/12al; also, F( t) = F(t). Let us first proceed formalIy to transform with respect to t and interchange the order of integration with respect to x and t: eo dx eo s dx

-

(2)

I(s)

= fo

L{costx} eo

s

=

-

.

f

o

2

x +a

2

=

fo

1 ( x2

+

1 a2

-

x2

+

d S2 )

x

n

1

= 2a s +

if a > Oand s > O. Thus the integral has the value n (3) F(t) = - e-al 2a

a

(a > O,t E:;;O).

TOsee that the formal step (2)is sound, we first note that, whenever T and s are positive constants, T T eo cos tx eo 1 (4) e-sI 2 2 dx dt = 2 2 e-sI cos tx dt dx, o ox+a ox+a o

f

f

f

f

because the absolute value of the entire integrand does not exceed (X2 + a2) - 1, a function

independent

of t whose integral from x

=

O

to x = 00 exists. The integral F(t)e-st therefore converges uniformly with respect to t when t E:;;O by the Weierstrass test and (Sec. 15) the interchange of order of integration with respect to x and t made in step (4)is valido Also, F is continuous (t E:;;O)and bounded so, according to the theorem on uniqueness, if L{ F} is given by formula (2), then F must be the exponential function (3). The final integral in Eq. (4) can be evaluated by integration by parts, and the equation can be written (5)

fOTe-SIF(t)dt = Leog(x,T)dx,

where g(x,

T) - e-SI(xsin Tx - s cos Tx) + s (X2 + a2)(x2 + S2) .

This continuous function of x and T has a limit g(x,oo), g(x,oo) = T-+eo lim g(x, T) = S(X2 + a2)-I(x2 + S2)-I, uniformly with respect to x when x E:;;Osince

Ig(x, T) -

g(x,oo)1 < e -sT

X

+s

2 2 2 2 (x +a )( x +s )

< = M e -sT ,

where M is the maximum value of the quotient of polynomials in X.

108


SECo 34]

To eachpositive number f there corresponds a number T.,independent of x, such that Me-sT < f when T> T.. Moreover, the second integral in Eq. (5) converges uniformly with respect to T, according to the Weierstrass test. The limit of the integral (5) as T ~ 00 is therefore fO' g(x,oo)dx (Prob. 10, Seco 27), since the latter integral exists; thus Eq. (2) follows from Eq. (4). Example 2

Evaluate the integral

(6)

=

F(t)

sin tx dx.

(a)

Jo

x

The formal procedure is simple. When t > Oand s > O

=

f(s)

(7) hence F(t)

= n/2.

.

i

a)

L{smtx}-

o

dx x

=

i

dx

a)

n

2

2

oX+s

; =_ 2 s

In view of Eq. (6), F(t) is an odd function:

F( - t)

=

- F(t), and F(O)= O. Therefore n

(8)

F(t)

=

n F(t)

2" (t > O);

F(O) =

= -2" (t < O);

O.

The Weierstrass test cannot be used here to justify step (7). In fact, if our formal result (8) is correct, then F is discontinuous at the origin t = O,and the integral (6) cannot converge uniforrnlyover an interval that includes the origino Now the function S(r)

=

sin rlr, r =1= O, S(O) = 1, that is, r2

r4

S(r)= 1 - -3! + -5! - . . .

(9)

(-oo
is continuous, and IS(r)1~ 1, for all real r. Note that the sum of the alternating seóes (9) is positive and not greater than unity when Irl

~ 1, and if Irl > 1, then Isinrlrl < 1. To see that the integral (6)

exists, we note that its integrand tS(tx) is continuous in t and x and, if to, xo, and XI are positive constants, an integration by parts shows that

f

XI

Xo

sin tx dx x

= cos txo txo

- cos tx 1 tXI

f

cos:x dx tx

XI

Xo

and this has a limit as XI ~ oo. Thus F(t) exists because XO

F( t)

=

fo

sin tx

-

x

dx+--cos txo txo

and because F(O)= Oand F( - t) = - F(t).

f

a)

XI)

cos tx

y- dx

tx

(t > O),


109

[SEC. 34

When t > O,the substitution r = tx showsthat our integral has a constant value, F(t) = SO'r-l sin r dr. Hence F is bounded, continuous when t > O, and F( + O) exists. Ir we establish formula (7) for /(s), then F must be the step function (8). The remainder for integral (6) can be written sin tx -dx x x

cos tX = tX

f

co

R(t,X) =

cos tx r dx x tx

f

co

and we can seethat, when t ~ to > O,for eachpositive number € there is a number X independent of t such that (

l

1

2

when X> X(. ::;;< € , - tX tX - toX Thus integral (6) converges uniformly with respect to t (t ~ to > O),and wecanwrite, with the aid of anelementaryintegration, T T co 1

IR(t X )I ::;;-

f

e-SIF(t) dt =

lO

+-

f f -

o X

e-sI sin tx dt dx

lo

= Lco h(T,x) dx

-

Lco

h(to,x) dx

where h is this continuous function: e-sI

h(t,x) = - x 2 +s

2

sin tx sttx (

+ cos tx

)

(5

> O).

The integral

SO' h(to,x) dx converges uniformly with respect to < 1) according to the Weierstrass test, so it is continuous in to at the point to = O; thus

to(O

;;;;;

to

T

(lO)

f

o

co

e-SIF(t) dt =

f

o

h(T,x) dx

+

f

dx

co

-y-

ox+s

(5 > O),

Let N(s) denote the maximum value of(sT + l)e-sT. Then (11)

Ih(T,x);;;;;I

N(s) x

2

+

s

2

and

Ih(T,x) I =< sT z-+ 1 e -sT . s

The uniform convergence of the integral SO' h(T,x) dx and the fact that h(T,x) -. Oas T -. 00, uniform1y with respect to x, foIlow from con-

ditions (11). Thus as T -. 00, that integral vanishes, and Eq. (10) leads to step (7), so the bounded continuous function F has the transform tn/s, and therefore F(t) = tn when t > O. This completes the proof. . IncidentaIly, we have shown that Si(00) = tn, for when t > O,we found that F(t) = !1t, and Eq. (6) can be written co sin r . 1t (12) F(t)= o -;:-dr=Sl(oo)="2'

f

110

35.


SECo 35]

EXPONENTIAL- AND COSINE-INTEGRAL

FUNCTIONS

Convenient forms of the exponential-integral function are oo

oo -y

f

E1(t) =

(1)

=-dy = y

I

f

-Ix

~dx

1

(t > O).

X

Formally, the transform of the second integral is oo

(2)

dx

oo 1 1 x 1 - - dx = -log X + s) s x+s (X

1

J,1 x(x +

=s) s f 1

00

]

(s > O). 1

The proof that expression (2) represents L{El(t)} is left to the problems. The basic form of transform 100, Appendix A, Table A.2, is, therefore, 1 (s > O). (3) L{E1(t)} = -log(s + 1) s In Seco 22 we found that the transform of the sine-integral function, Si t, is S-1 arccot s. Now let us wríte the transform of the cosine-integral function Ci t

(4)

=-

f.I

oocos r dr = -

(00cos tx dx

r

J1

(t > O).

x

An integration by parts, first over a bounded interval (l,X), shows that . sin t Clt=--t

(5)

1 t

f

oo sin tx

-

1

dx

(t > O);

X2

thus Ci t is continuous when t > O,and Ci (00) = O. The procedure used in Example 2, Sec. 34, can be applied to the final integral in formula (4) to prove that oo

(6)

dx

L{Ci t} = - s J,1 xx(2

1

+s 2)

=-

_2s log(s2 + 1)

(s > O).

Next let us derivethe transform (No. 98,AppendixA, Table A.2)of the function cos t Si t cost

,-dr sinr

1o

r

-

sin t Ci t which can be written

.

+ smt

f.I

oocos r -dr

r

= cost

oo

sin r

fo -drr,

-

f

oo

dr

(sinrcost - cosrsint)-

r

when t > O. By replacing Si ((0) here by tn (Sec. 34) and substituting tx for r - t, we can write the last expression in the form (7)

1t -cos t -

2

oo sin

f,r

(r

-

t)

dr

1t

= -cos t 2

f

oo sin

tx

-dx. ox+l


[SEC. 35

111

When we transfonn that final fonn (7), we find that

.

L{ cos t S1t

(8)

.

-

Sin

'

t C1 t } =

log s

zs +

1

(s > O).

In like manner we can derive the transformation (No. 99, Appendix A, Table A.2) L{ cos t Ci t + sin t Si t} = - S2 s log + S1

(9)

(s > O).

Finally, let us evaluate the Laplace integral oo

(10)

g(x)

=

-tx

fo t:+ 1dt

Its transfonn with respect to x is found to be n s logs g(s) = 2 S2 + 1 - S2 + 1 A

(x > O).

(s > O).

In view of formula (8) we have the integration fonnula (11)

g(x)

= (tn - Si x) cos x + Ci x sin x,

which is transformation 120, Appendix A, table A.2: (12)

LL2

~ I} = (~n - Sis) cos s + Ci s sins

(s > O).

.

PROBLEMS Establish the integration formulas in Probs. l to 3, where t ~ O. 1.

l

'" sin tx o x,(x2:1- I)dx

1t

= 2(1 -

-/ e ),

the Fourier sine transform of X-1(X2 + 1)-1 (Chap. 13). '" 1 - eos2tx '" sin2 tX d = 1tt. dx = 2. -r,x -'" x o x

l

f

3.

l

'" l

o

-

exp ( - tx2) d 2 X x

-- y'1tt. c:

Obtain formally the integration

4. 5.

'"

f-'"

formulas in Probs. 4 to 7.

. x sm tx X2 + a2 dx = 1te-aI

f.o.'"(-r +e-X< 1)fi, d,= 1te' erfe fi

(ef. transformation 111, Appendix A, Table A.2.)

(a > O, t > O).

(x ~ O).

112

6.


SECo 35]

f

fi

o

n 1-

eo

sin tx

eo

dx

[Note that y4 +

f

= 2 o sin ty2 dy = ( 2t) S2 = (y2 + S)2 - 2si = (y2 - y.j2s + s)(y2+ y.j2s + s).] 1

eo

7.

fo exp(-tx2)cos2txdx

/1r

(t > O).

V-¡e-'

=2

(t > O).

8. WhenT > to> Oands > O,provethat

-s,

T

f'o e f

eoe-'X

-

¡

dx d t =

X

f

eo exp

[-(x + s)to] - exp [-(x + s)T] d . x,

¡

x(x

+ s)

then prove that the limit of that integral as to -> Oand T -> 00 is the integral (2),Seco35, which represents L{E¡(t)}. 9. Use the transformation (3), Seco35, to show that, when a > O, . logs - loga L{e"'E¡(at)} = (s> a; cf. transformatlOn 97, Appendix A, Table A.2). s-a Obtain formally the transforms given in Probs. 10 to 12, if a > O and b> O (see Probo 9). eo

10.

-sr

I { t+a }

=L -

fo t+a ~dt

= easE¡(as)

(cf.transformation118, AppendixA,Table A.2.) 11. L{(t /a~t b+b)} = ebSEI(bs)- easE¡(as) 12. Ltt:

(a * b).

a)2} = 1 - aseasE¡(as)(cf.transformation119,AppendixA,Table A.2.)

13. Prove that the remainderfor the integral Ci t = - SI x-¡ COS tx dx satisfiesthe condition IR(t,X)1< 2J(toX)when t ~ to-> O and hence that the integral converges uniformlywhent ~ to. Then completethe proof of transformation(6),Seco35. When a > Oand t > O,obtain formallythe Fourier transformationsof(x + a)-¡ stated in Probs. 14and 15. eo

14.

n

sin tx

..

.

fo x + adx = (2 - SI at) cos at + Ci at Sin ato. eo cos

tx

n

d

.

S

.

C

.

1.

fo x + a x =

16. 17. 18.

Complete the derivation of formula (11), Seco35. Obtain formally the transformation (9), Seco35. The beta function is defined as follows:

5

(

2-

1at ) Sin at

-

¡at cos ato

B(x,y) = f1 r"-¡(1 - r)Y-¡ dr , Jo

(x > O,Y > O).


Since L{r-I}

113

[SECo 36

= r(x)s-X, note that L{tX-I * tY-I}

=

r(x)r(y)s-X-y

= r(x r(x)r(Y)L{tX+Y-l}. + y)

Write the convolution here in tenns of the integral representing B(x,y) to show that the beta function can be written in tenns of gamma functions (Sec. 5) as r(x)r(y) B(x,y) = r(x + y) 36.

(x > O, Y > O).

STATIC DEFLECTION OF BEAMS

Let Y(x) denote the static transverse displacement of a point at distance x from one end of a uniform beam, caused by a load distributed in any manner along the beam (Fig. 34). Under certain idealizing assumptions, primarily that displacements Y(x) and slopes Y'(x) are small, it is shown in mechanics that the internal bending moment M(x) exerted by any span of the beam upon an adjacent span, through their common cross section Ax, is proportional to the curvature of the beam at position x. In fact (1)

M(x) = ElY"(x),

where E is Young's modulus, the modulus of elasticity in tension and compression for the material, and 1 is the moment of inertia of the area Ax with respect to the neutral axis of the cross section, the line in Ax about which Ax turns when the beam bends. Note that Y"(x) is approximately the curvature since Y'(x) is small. Ir F(x) denotes the internal shearing force at A.x>it can be seen that F(x) dx = dM(x); that is (Prob. 14, Seco39), F(x) = M'(x),or (2)

F(x) = ElY"'(x).

Let W(x) represent the transverse load per unit length along the beam. Then W(x) dx = dF(x), and it follows from Eq. (2) that (3)

y(4)(X)

= aW(x)

(a-1

r I I

0.---------------------I ., x-e y Fig.34

Wo

= El).

l I I.

x-2e

.."

114


SECo 36]

The shear F(x) is a continuous function except at points where concentrated loads or supports acto WhenW(x) is sectionally continuous then, in view of Eqs. (2) and (3), Y"'(x) is continuous and y(4)(X) is sectionally continuous. These are just the continuity conditions that are implied by our formula for the transform, with respect to x, of y(4)(X). Although the simple differential Eq. (3) can be solved by successive integrations, the requirement that the solution Y(x) and its derivatives of the first three orders be continuous at all points often involves tedious labor when W(x) is sectionally continuous. The simplicity of Eq. (3) enables us to use the Laplace transformation even though that transformation is not adapted to the two-point boundary conditions which will accompany that equation. Certain Fourier transformations specified by the boundary conditions and the differential form d4Y/dx4 are properly adapted to such problems (Chap. ll). Let us determine the displacements Y(x) in a beam whose end x = Ois built into a rigid support while the end x = 2e is free, or unsupported (Fig. 34). The load W(x) per unit length is zero over the span O < x < e and a constant Woover the span e < x < 2e. The total load Wo is woe. Equation (3) can now be written (4)

y(4)(X)

= aWoSe(x)

(O< x < 2e),

where Se(x) is our unit step function. The end conditions are (5)

Y(O)

Y"(2e) = Y"'(2e) = O,

= Y'(O) = O,

since no bending or shear acts on the end x = 2e.. Let y(s) denote the Laplace transform of Y(x) when Y(x) satisfies Eq. (4) on the semi-infinite range x > O, together with the first pair of boundary conditions (5). In view of the continuity conditions to be satisfied by Y(x) and its derivatives, we may expect that S4y(S)

-

S3(0)

-

S2(0) - sY"(O)

-

Y"'(O)= aw)e-es. s

We have used a convenient extension, awoS.(x) when x > 2e, ofthe function aW(x) in the right-hand member of Eq. (4) where O < x < 2e. The particular extension used is immaterial because we first seek a function Y(x)

that satisfiesEq. (4)when O < x < 2e and conditions Y(O)= Y'(O)= O,and contains two arbitrary constants A = Y"(O)and B = Y"'(O).Since y (S)

A

= 3" S

B

1

+ "4 + awose S S

-es

,

a function that satisfies those requirements is (6)

Y(x)

= tAx2 + iBx3 + l4awO(x- e)4Se(x).


When X>

[SECo 37

115

c, it follows from Eq. (6) that

Y"(x) = A + Bx + tawo(x - C)2,

Y"'(x) = B + awo(x - c).

The last two of conditions (5) are then satisfied if (7)

B

= -awoc,

and the displacements in the beam are given by the equation (8)

Y(x)

= aWo(ic2x2 -

!CX3

+ f.rtx -

C)4S.(X)]

(O;;;; X ;;;; 2c).

This function satisfies Eqs. (4) and (5) and the continuity conditions. Since Y"'(O)= B = -awoc, it follows from Eq. (2) that the shear at x = O has the value - woc,or F(O)= Wo as we should expect, since the magnitude of the vertical force exerted by the support must be the same as the totalload on the beam. Actually, the displacements Y(x) for this cantilever beam can be found by first noting that the shear - Y"'(x)/a at each section is the totalload on the span to the right of that section. The bending moment at the end x = Ois Y"(O)/aor Ala:

-

M(O)= ~cWo

(Wo= cWo)'

If the end x = 2c were pin-supported (hinged or simply supported) so that it can rotate freely about a fixedaxis, then the conditions at that end become Y(2c) = Y"(2c)= O. 37.

THE TAUTOCHRONE

We shall now discuss a problem in mechanics that leads to a simple integral equation of the convolution type. The problem is that of determining a curve through the origin in a vertical xy plane such that the time required for a particle to slide down the curve to the origin is independent of the starting position. The particle slidesfreelyfrom rest under the action of its weight and the reaction of the curve on which it is constrained to move. The required curve is called the tautochrone. Let a denote the length of arc of the curve, measured from the origin O, and let (x,y) be the starting point and (~,tI)any intermediate point (Fig. 35). Equating the gain in kinetic energy to the loss of potential energy, we have I da 2 2m( dt ) = mg(y ti),

-

where m is the mass of the particle and t is time. Thus

da = - J2gJY='"iidt,

116


SECo 37]

y

o

%

Fig.35 and upon separating variables and integrating from r¡ = y to r¡ = 0,we have "=y da

TJ2i=

S,,=0.j y

- r¡,

where T is the fixed time of descent. Now a = H(r¡), where the function H(r¡)depends upon the curve, and therefore (1)

T J2i =

I

(y - r¡)-tH'(r¡)dr¡.

This is an integral equation of convolution type in the unknown function H'(y). We may write it in the form T

fii

= y-t

* H'(y).

Let h(s)be the Laplace transform of H(y) with respect to the variable y. Since H(O) = O,it followsformallyfrom Eq. (1)that

T

Thatis,

fii~ = Sh(S)L{h}

= Sh(S)~.

sh(s)= Tf! ~;

hence (2)

H'(y) =

~fii ~. 1t VY

We can see that this function does satisfy our integral equation (1) by substituting it into that equation and performing the integration. Since

H'(y) = da = dy

)

1+

dX 2, ( dy )

[SECo 38


117

the differential equation of the curve in terms of the variables x and y is, according to Eq. (2), dx

1+

2

( dy )

where a

= 2gT2jn2.

= 2gT2 =~, n2y

y

Separating variables here, we have

~ dx = V---ydY, and the necessary integration can be performed easily by substituting y = a sin2 !O, for we then find that 1 a =

dx

a COS2 20 dO = 2(1

+ cos O) dO.

Noting that x = Owhen y = O,we see that the parametric equations of the tautochrone are therefore

a (3)

x

= ~(O + sinO),

y

= 2(1

-

cos O).

These equations represent the cycloid generated by a point P on a circle of radius !a as the circle rolls along the lower side of the line y

= a.

The parameter Ois the angle through which the radius drawn to the point P has tumed, where the initial position of P is at the origino Qur tautochrone is of course just one arch of this cycloid. Since a = 2gT2jn2, the diameter of the generating circle is determined by the time T of descent. The above problem can be generalized in various ways so as to lead to other interesting questions; in fact, it was a generalization of the problem of the tautochrone that led the great Norwegian mathematician Niels Abel (1802-1829) to introduce the subject of integral equations.1 Ir the time T of descent is a function F(y), for example, our integral equation (1) becomes (4)

38.

j2gF(Y) = I: (y - 17)-tH'(17)d17. SERVOMECHANISMS

Simple integral equations as well as differential equations arise in the theory of automatic control. As a special case we consider servomechanisms that force the angle of tum 0o(t) of a rotating shaft to follow closely the angle of tum 0¡(t) of a pointer or indicator, where t denotes time. The shaft and 1 See Bócher, M., "Integral

Equations,"

p. 6, 1909.

118

OPERATIONAL

SECo 38]

MATHEMATICS

material rigidity attached to it have a total moment of inertia 1 that is much greater than that of the pointer. Mechanisms of that general type are used, for instance, in directing antiaircraft guns and in aircraft control. Let (t) be the angle of deviation between shaft and pointer or the difference between output angle and input angle:

=

(1)

(t)

00(t)

-

0¡(t).

An auxiliary system or servomechanism can be designed to measure (t) and feed back to the shaft a component of torque that is proportional to the deviation (t).The servomechanism may contain its own source of power, motors, generators, and other electrical equipment. In order to provide damping in the system, let the servo also supply a component of torque proportional to the rate of deviation '(t).Then, since the product of 1 by the angular acceleration of the shaft equals the torque applied to the shaft, (2) 10~(t) = k(t) - e'(t),

-

.

where k and e are positive constants. Ir the shaft is initially at rest and its angle of tum is measured from the initial position, then 00(0) = 0~(0) = O. AIso, in view of Eq. (1), <1>(0) = - 0¡(0) and in terms of transforms Eq. (2) becomes 1s280(s)

= -(k + es)ljJ(s) - e0¡(0),

where we have assumed that 00(t), 0~(t), and (t), and therefore 0¡(t), are continuous when t ;E; O. Since80(s)= ljJ(s) + 8¡{s),it followsthat (3)

ljJ(s)

= -

+ e0¡(0). 1S2 + es + k

1s2(J¡(s)

Under an input 0¡(t) = At, it followsfrom Eq. (3)that e A.

= - ~ e-bt smwt

( b = 21'W2

(t)

k

=1-

e2

412) .

Ir k > c2/(4I), the angle of deviation (t) has a damped oscillation with initial value zero. Since Isin wt/(wt)1 < 1 and (be)-l is the maximum value of te-bt, that oscillation is small at all times when b is large, because I(t)1< IAlte-bt

~ I~.

Ir in addition to the two components of torque shown in Eq. (2), a component proportional to the accumulated angle of deviation is produced by the servo, then

(4)

10~(t) = - k(t) - c'(t) - b

{

(-r)d-r,


119

[SECo 39

where b is a positive constant. When E>0(0)=

1S2[
-

E>~(O)

(k+ es + ~)

= Otherefore, -

eE>;(O).

In terms of positive numbers B, C, and K, where lC = e,

lK2 = k,

lB3 = b,

the last equation can be written (5)

-- -

s3

s38;(s) + CsE>;(O)

+ Cs2 + K2S + B3'

In the special case E>;(t)= At, K2 = BC, Eq. (5) becomes (6)

As

=-

and since the )?olynomials in the numerator and denominator here are of degrees one and three in s, respectively, it folIows from our earlier observations (see the problems, Seco27) that c1I(O) = O; that is, the initial value ofthe angle of deviation is zero. When C > B, all values of s that make the poIynomial in the denominator vanish ha ve negative real parts, and consequently (Sec. 26) c1I(t)contains only damped components. But when C < B, c1I(t)has an unstable oscillation. 39. MORTALlTYOF EQUIPMENT Let the function F(t) denote the number of pieces of equipment on hand at time t, where the number is large enough that we can consider it as a continuous variable instead of a variable that takes on only integral values. The equipment wears out in time, or is lost from service for other reasons, so that, out of no pieces of new equipment introduced at time t = O,the number N(t) in service at time t is given by the fonnula (1)

N(t) = noH(t),

where H(t) is a function that determines the surviving equipment after t units of time. Note that H(O)

= 1,necessarily.

If R(r) is the total number of replacements up to time 't',then R'('t')d't' is the number of replacements during the time interval from t = to 't'

t = 't' + d't'and the number of survivals at any future time t, out of these replacements, is R'('t')H(t - 't')d't'.

The total amount

of equipment

in service at time t is the sum of these

survivals from the replacements during every time interval d't' between

120


SECo 39]

= O and T = t, increased of course by the survivals from the new equipment on hand at time t = O. Therefore

T

(2)

F(t) = F(O)H(t) +

{ R'(T)H(t

-

T) dT.

We have assumed here that the equipment F(O)on hand at time t = Ois all new; then R(O) = O. Ir the amount F(t) that must be in service at each instant is known and if the survival factor H(t) is known, then Eq. (2) is an integral equation of convolution type in R'(t). Its solution gives the formula by which replacements must be made. The equation is an integral equation in the survival factor H(t) when F(t) and R(t) are known. In either case, the transformed equation is (3)

. f(s)

=

F(O)h(s)

+ s r(s)h(s).

Then

- f(s) - F(O)h(s) () , rs s h(s)

(4)

and R(t) is the inversetransform of that function. Suppose the mortality is exponential in character so that H(t) = e-kt and that the amount of equipment on hand is to be a constant,

F(t) = b. Then h(s)= l/(s + k) and f(s) = bIs,and it followsfrom Eq. (4)that 1 r(s)

= bkz. s

Therefore replacements must be made at such arate that the total equipment replaced up to time t is R(t) = bkt, a result that is easily verified as the solution of Eq. (2). Thus replacements must be made at the rate of bk pieces per unit time. PROBLEMS 1.

In the example solved in Seco 36 let the end x

=

2c of the beam

be pin-supported,

rather than free, with no other changes in conditions. Find the vertical force exerted by the beam on the pin, and the vertical force and bending moment exerted on the Ans. *wo;HWo;ncWo' supportatx = O.


[SECo 39

121

2. Both ends x = Oand x = 2e of a beam are pin-supported. Find the vertical force on each support when a transverse load Wo is distributed uniformly over the span e < x < 2e. Ans. Wo/4; 3Wo/4. 3.

Solve Probo 2 if both ends are built in rather than pin-supported. Ans. f¡,Wo;tiWo' 4. The end x = O of a beam is built in and the end x = 2e is pin-supported. The load per unit length is bx on the span O < x < e, and b(2e - x) on the span e < x < 2e. Find the vertical force on each support in terms of the total load Wo on the beain. Ans. ~~Wo;gwo.

5. In addition to a distributed load W(x) a single concentrated load W¡ acts between the ends of a beam, at position x = b. Then the shear F(x) is continuous except for a jump W¡ at x = b. Apply formula (4), Seco4, to the function Y"'(x) to show that the transform of Eq. (3),Seco36, is S4y(S)

.Compare

-

-

s3y(0)

s2y'(0)

-

sY"(O)

-

Y"'(O)

- aW¡e-bs= aw(s).

this equation in y(s) with the one obtained formally by replacing W(x) in Eq. (3), Seco36, by W(x) + W¡t5(x - b) and transforming as if Y"'(x) were continuous,

where t5(x) is the unit impulse symbol.

6. In Seco37, let the time T of descent be proportional to.JY, T.j2g = 2BJY, whereB > 1. Showthat the curveof descentis the linex = yJB2=l.

7. Ire = k = 21 in Eq. (2), Seco38, find the output angle 80(t) corresponding to the constant input angle 8¡{t) = l(t > O), and compare them graphically. Assume that 80(0) = 80(0) = O. AIso, note that the value of the output lags behind that of the input until t = 3n/4. Ans. 80(t) = 1 - J2e-t sin(t + n/4). 8. For the servomechanism corresponding to Eqs. (4) and (5), Seco38, consider this special case: 8¡{t) = At, B3 = CK2. Ir a represents the argument of the complex number K + Ci, derive the formula CI>(t)=

2

~ rf2[Ce-Ct - JC2

+ K2sin(Kt

+ a)]

for the angle of deviation, and note the undamped component of Cl>(t). 9. Let the servomechanism discussed in Seco 38 supply only a corrective torque proportional to the deviation angle CI>(t) while the shaft itself is subject to a damping torque proportional to its angular velocity. Ir 80(0) = 80(0) = O,show that k 8Js), 80(s) = rs2 + es + 'o where e, k, and 1 are positive constants. When 8;(t) = A, showthat Bo(t)approaches A as t increases. 10. When H(t) = e-Ict,where H(t) is the survival factor in Seco39 and k is a positive constant, (a) find the replacement function R(t) corresponding to an arbitrary amount F(t) of equipment on hand; (b) find R(t) when F(t) = At + B. Ans. (b) R(t) = (A + Bk)t + !Akt2.

-

11. When H(t) in Seco39 is the step function l Sk(t) so that every piece of new equipment survives for k units of time, (a) show that the number of replacements R(t)

122


SECo 39]

up to time t required to maintain F(t) pieces in serviee at that time is R(t)

= F(t) -

F( +0) + F(t

-

k) + F(t

-

2k) + F(t

-

3k) +

...

if we define F(t) to be zero when t < O. (b) When F(t) = A(t > O),show that R(t) = A[t/k], where [t] is the bracket symbol, and draw the graph of R(t). 12. A particle mass m moves on a vertical X axis under two forees: the force of gravity and a resistance proportional to the velocity. If the axis is taken positive downward, the equation of motion is

mX"(t) = mg - kX'(t). Show that its solution, under the conditions X(O) = O,X'(O)= vo, is

where b = k/m, and discuss the motion. 13. Each one of a set of radioactive elements El, E2' E3' and E4 disintegrates into the succeeding one at arate proportional to the number of atoms present, exeept for the end product E4 which is a stable elemento If N¡(t), where i = 1, 2, 3, 4, denotes the number of atoms of element E¡ present at time t and if the distinct positive constants CI, C2' and C3are the respective coefficients of decay of the first three elements, then

N~(t)= -cINI(t),

N3(t)=

Ní(t) = -c2N2(t) + cINI(t),

- c3N3(t)+ c2N2(t),

N~(t)= c3N3(t).

When onlyM atoms of El are presentinitially,derivethe formula N4(t)

M

14.

=1-

clc2e-c3t c2c3e-Qt clc3e-c2t (C2- CI)(C3- CI) (CI- C2)(C3 - C2) (CI- C3)(C2 - C3r

In Seco36 the shear F(x) is the internal force in the direction of the Yaxis exerted

at a cross section Ax of the beam upon the span to the right of that section, and M(x) is the bending moment exerted on that span there. If F(O)= Fo, M(O)= Mo, and W(x) is the load per unit length, apply equilibrium conditions to a span extending x units to the right ofthe section Ao, at x = O,to show that Fo +

Mo + Fox +

J: (x -

I: W(~) d~ ~)W(~)d~

-

F(x)

= O,

M(x)

= O.

Thus derive these relations between shear, load, and bending moment: F'(x) = W(x),

M'(x) = F(x).

4 Problems in Partial Differential Equations

40

THE WAVE EQUATION

Several functions in physics and engineering satisfy the partial differential equation (1) known as the wave equation in two independent variables x and t. We shall use literal subscripts to indicate partial derivatives; then Eq. (1) can be written

Brief derivations of this equation for some elementary physical functions will now be given. The derivations are helpful in writing modifications of the equation and in setting up boundary conditions for specific problems. 123

124

SECo 40]


y

-H

"

...-..::::-.I I I I Y I I I I

o Fig.36

First, let Y(x,t) denote the displacement at time t away from the x axis of a point (x, Y) of a string in the x Y plane under tension P (Fig. 36). The string is assumed to be flexible enough that all bending moments transmitted between its elements can be neglected; thus each element pulls tangentially on an adjacent element with a force of magnitude P. Suppose further than conditions are such that the magnitude H of the x component of the tension remains essentially constant at all times for all points (x, Y); in particular all displacements Y(x,t) are assumed small compared with the length of the string. Finally let the slope angle IXremain small in order that each small element of length of the string may be approximated by the length of its projection Llx on the x axis. All these idealizing assumptions are satisfied, for instance, by strings of musical instruments under ordinary conditions of operation. The vertical component of tension is the vertical force Vexerted by the part ofthe string to the left ofpoint (x, Y)on the part to the right ofthe point. It is proportional

(2)

to the slope of the string, V(x,t)

-

VI H

=

tan IX(Fig. 36); that

is,

= - HY,Jx,t).

This is the basic formula for deriving the equation of motion. Now consider an element oflength ofthe string whose projection on the x axis is Llx. If p denotes the mass per unit length, the mass of the element is approximately p Llx. If no external forces act on the string, the application of Newton's second law to the element gives, in view of formula (2), (3)

pLlxY,,(x,t) = -Hy"(x,t) + HY,,(x + Llx,t)

approximately, when Llx is small; that is, t) - y"(x,t) y.II(x,t) = H y,,(x + Llx, A . P l.lX When we let Llx approach zero, this becomes Eq. (1), where H a2 =-. (4) p

PROBLEMS IN PARTlAL DIFFERENTIAL EQUATIONS

[SECo 40

125

If in addition to the internal force a force F(x,t) per unit of massacts in the Y direction along the string, then the additional term p L\x F(x,t) appears on the right in Eq. (3). The resulting modification ofEq. (1) is (5)

Y,r(x,t) = a2Yxx(x,t) + F(x,t).

When the Yaxis points vertically upward and the weight of the string is to be taken into account, for instance, F(x,t) = - g, where g is the acceleration of gravity. As another physical example consider the longitudinal displacements in a cylindrical or prismatic elastic bar. The values of the variable x are marked on the bar so as to designate the cross section that is x units from one end when the bar is neither stretched nor compressed (Fig. 37). For each value of x the longitudinal displacement Y(x,t) is measured from a fixed origin outside the bar, an origin in the plane occupied by the cross section at x when the bar is unstrained and in some position of reference. Thus, if the bar is moved lengthwise as a rigid body, Y(x,t) is a constant at each time t. Since Y(x + L\x, t) is the displacement of the cross section at x + L\x, an element of the bar whose naturallength is L\x is stretched by the amount Y(x + L\x, t) - Y(x,t) at time t. According to Hooke's law the force exerted by the bar upon the left-hand end ofthe element to produc~ that extension is

-

AE Y(x + L\x,t) - Y(x,t).

whereA is the area ofthe crosssectionand E is Young's modulus of elasticity of the material, the modulus in tension and compression. When L\x tends to zero, it follows that the internal force from left to right at the cross section is (6)

F(x,t) = -AE~(x,t).

This basic formula corresponds to Eq. (2) for the string. Let p denote the mass of the material per unit volume. When we apply Newton's second law to an element ofthe bar, (7)

pA L\xY,ix,t) = -AEYx(x,t)

I

O

+ AEYx(x + L\x, t),

j

" Fig.37

"

126


SECo 41]

o G)---%---t~

~

'-y Fig.38

we find as before that Y(x,t) satisfiesEq. (1),where E a2 =-. (8) p When the elastic bar is replaced by a column of air, Eq. (1) has further applications in acoustics.1 Again, the function Y(x,t) may represent the angle of turn of a cross section x units from one end of an elastic cylindrical shaft under torsion (Fig. 38). Let 1 denote the moment of inertia of the cross section with respect to its axis, Es the modulus of elasticity of the material in shear, and p the density of the material. Then, by steps analogous to those used above, we find that the internal torque -r acting through a cross section at position x is (9)

-r(x,t) = - lEs Yix,t)

and that Eq. (1) is satisfied by the angle Y(x,t), where a2 =-.Es p

(10)

Finally, it should be remarked that Eq. (1) is a special case of the telegraph equation (11)

Yxx(x,t) = KLY,,(x,t) + (RK + SL)Y,(x,t) + RSY(x,t),

where Y(x,t) represents either the electric potential or the current at time t at a point x units from one end of a transmission line or cable.2 Here the elements of resistance, inductance, etc., are distributed along the cable or through the circuit, in contrast to the lumped elements in elementary circuits (Sec. 33) that lead to ordinary differential equations in the currents. The cable has resistance R, electrostatic capacity K, leakage conductance S, and self-inductance L, all per unit length. When R and S are so small that their effect can be neglected, Eq. (11) reduces to Eq. (1), where a2 = (KL)-l. 41

DISPLACEMENTS IN A LONG STRING

Let Y(x,t) represent the transverse displacements of the points of a semiinfinite stretched string, a string having one end fixed so far out on the x axis 1

Lord Rayleigh.

"Theory

of Sound,"

vols. I and 2, Dover,

2A derivation of Eq. (1I) is outlined in Probo 6, Seco93.

1945.


127

[SECo 41

that the end may be considered infinitely far from the origin, and having its other end looped around the Y axis. The loop, initially at the origin, is moved in some prescribed manner along the Yaxis (Fig. 39) so that Y = F(t) when x = O and t ~ O, where F(t) is a prescribed continuous function and

F(O) = O. Ir the string is initiallyat rest on the x axis, let us find the formula for Y(x,t). The above conditions on Y(x,t) can be written (x > O,t > O),

(1)

Yrr(x,t)= a2 Yxx(x,t)

(2)

Y(x,O) = Yr(x,O)= O

(x > O),

(3)

Y(O,t)= F(t),

(t

lim Y(x,t) = O

x->oo

~ O),

where a2 = H/p (Sec.40). The equation ofmotion (1) implies that no external forces act along the string. A problem composed of such conditions is called a boundary value problem in partial differential equations. We shall use a formal procedure to solve the problem and then show how our result can be verified as a solution. Ir y(x,s) is the Laplace transform of Y(x,t), then, in view of the initial conditions (2), L{ Yrr} = s2y. AIso,

if the function e-srY(x,t) satisfies conditions under which the indicated interchange of order of integration with respect to t and differentiation with respect to x is valid (Sec. 15). When both members of the partial differential equation (1) are transformed and conditions (2) are used, we therefore obtain the equation s2y = a2yxx in the transform of our unknown function. From conditions (3) we find that y(O,s)= I(s), where I(s) is the transform of F(t), and limx->00y(x,s) = O, provided that the order of integrating with respect to t and taking the limit as x --+oc>can be interchanged. The y

Ji1t)

o Fig.39

128

.

SECo 41]


transformed boundary value problem in y(x,s) is therefore (4)

(5)

d2y

S2

-dX2 - -y =O a2

.

(x > O),

lim y(x,s) = O.

y(O,s) = f(s),

x-oc

Here we have used the symbol for ordinary rather than partial differentiation because s is only a parameter in the problem ; no differentiation with respect to s is involved. A convenient form of the general solution of Eq. (4) is y(x,s) = Cle-sx/a + C2e'x/a, where CI and C2 may be functions of s. This solution could of course be obtained by transforming the members of Eq. (4) with respect to x. We consider s as positive, since Laplace transforms generally exist for all s greater than some fixed number. Then C2 = O,if y(x,s) is to approach zero as x tends to infinity. The first of conditions (5) is also satisfied if C 1 = f(s), and the solution of the transformed problem is (6)

y(x,s)

= e-(x/a)s¡(s).

The translation property (Sec. 11) enables us to write the inverse transform of y(x,s) at once: (7)

Y(x,t) = F( t - ~)

=0

when t >

~

= a'

when t ~~.a

Since F(t) is continuous and F(O) = O,we can see that the function Y(x,t) described by formula (7) is continuous when x ~ O and t ~ O, including points on the line x = at in the xt planeo The function clearly satisfies the boundary conditions (2) and (3). Any function of t - x/a is easily seen to be a solution of the wave equation (1) when its derivative of the second order exists. Let F(t) satisfy these additional conditions: F'(t) and F"(t) are continuous when t ~ O except possibly for finite jumps at t = ti (i = 1, 2, . . .). The function Y(x,t) described by formula (7) then satisfies Eq. (1) except possibly at points (x,t) on the lines t - x/a = ti and t - x/a = O in the quadrant x > O,t > O. In this sense, then, the function (7) is verified as a solution of our boundary value problem regardless of the validity of formal steps taken to obtain that function. In case F"(t) is continuous whenever t ~ O, and F(O)= F'(O)= F"(O)= Oso that Y(x,t) and its partial derivatives of first and second order are also continuous on the line t = x/a, then the function (7)satisfiesEq. (1)


[SEC. 41

129

with no exceptions when x > O and t > O and represents a solution of the boundary value problem in the ordinary sense. Note that conditions of exponential order are not involved in the verification of our solution. According to formula (7), a point of the string x units from the origin remains at rest until the time t = x/a. Starting at that time, it executesthe same motion as the loop at the left-hand end. The retarding time x/a is the time taken by a disturbance to travel the distance x with velocity a. Since a = (H/p)t, note that it has the physical dimensions ofvelocity. The vertical force from left to right at a point (Sec. 40) is -HYix,t); thus the vertical force exerted on the loop to make it move in the prescribed manner, as found from formula (7), is

H V(O,t)= -F'(t) = JPijF'(t). a Finally, let us observe instantaneous positions ofthe string corresponding to the loop movement (8)

F(t)

= sin nt

when O ~ t ~ 1,

=0

when t ~ 1;

thus the loop is lifted to the position Y = 1 and returned to Y = O, where it remains after time t = 1. In this case F(t - x/a) = Owhen t - x/a ~ 1,that

is, when x ~ a(t - 1).AIsoF(t - x/a) = sin n(t - x/a) when O ~ t - x/a ~ 1,that is, when a(t - 1) ~ x ~ atoThenfromformula(7)wefindthat (9)

when x ~ a(t - 1)or when x ~ at,

Y(x,t) = O = sin n(t - x/a)

when a(t - 1) ~ x ~ ato

Thus the string coincides with the x axis except on an interval of length a, if t > 1, where it forms one arch of a sine curve ending at x = ato As t increases, the arch moves to the right with velocity a (Fig. 40). It is interesting to note that for each fixed x in formula (9) the function Yr(x,t)has a jump n at t = x/a and at t = 1 + x/a and hence s2y(X,S)is not the transform of Yrt(x,t). On the other hand it can be shown that d2y/dx2 is not the transform of y"'ix,t), because of the discontinuities of Yx(x,t); but the transform

of Yrt - a2 Yxx is zero.

(t>l) O alt-l)-O Th"'b Fig.4O

"

130

SECo 42]


The function Y(X,t) above can also represent the longitudinal displacements in a semi-infinite elastic bar (Sec. 40), initially at rest and unstrained.

The distant

end of the bar is held fixed and the end x

= O is

displaced in a prescribed manner, Y(O,t) = F(t). When F(t) is given by formula (8), the solution (9) or Fig. 40 shows that at time t the section of the bar fram x

=

a(t

-

=

1) to x

at is strained while the remainder

is unstrained

and at resí.

42

A LONG STRING UNDER ITS WEIGHT

Let the semi-infinite string be stretched along the positive half of a horizontal x axis with its end x = O fastened at the origin and with its distant end looped around a vertical support that exerts no vertical force on the loop (Fig. 41). In view of formula (2), Seco40, for the vertical force at points of the stretched string, Yx(x,t) vanishes at the distant end. The string is initially supported at rest along the x axis. At the instant t = O the support is removed and the string moves downward under the action of gravity. Let us find the displacements Y(x,t). As noted in Seco40, Eq. (5), the equation of motion is (1)

y,/(x,t) = a2 Yxx(x,t)

-

g

(x > O,t > O).

The boundary conditions are (2)

Y(x,O) = Y,(x,O) = O

(3)

Y(O,t)= O,

(x

lim Yx(x,t) = O

x'" 00

The transformed problem is found formally to be (4) (5)

y(O,s) = O,

lim y'(x,s) = O, x'"

~ O

I I

00

%

lat.-'lgt2)

Fig.41

, I I I I I { I

~ O),

(t ~ O).

PROBLEMS IN PARTIAL DIFFERENTIAL EQUATIONS

131

[SEC. 42

where the primes denote derivatives with respect to x. Since the constant g/S3is a particular solution of Eq. (4), the general solution ofthat equation is

-

When conditions (5) are applied, we find that (6)

y(x,s) = _gL13 - :3e-
(7)

Y(X,t)

= - 2a2(2axt - x

2 )

when x ~ at, when x ~ at.

The details of this step and the verification of the solution (7) are left to the problems. An instantaneous position of the string is shown in Fig. 41. We note that up to time t the segment of the string to the right of the point x

=

at

has moved like a freely falling body. PROBLEMS Solve these boundary value problems and verify your results: 1.

YAx,t) + x Y;(x,t)= O; Y(x,O) = O, Y(O,t)= t. Ans. Y = Oif t ;;;;!X2, y = t

2.

Yx(x,t)+ 2x Y;(x,t)= 2x; Y(x,O) = Y(O,t)= 1. Ans. Y = 1 + t if t ;;;;x2, y = 1 + x2 if t ~ x2. xYx(x,t) + Y;(x,t)+ Y(x,t) = xF(t); Y(x,O) = Y(O,t)= O.

3.

4.

5.

-

!X2

if t ~ !X2.

Wxx(x,t) + W,Ax,t) - 2 W,,(x,t) = O(x > O,t > O); W(x,O) = W,(x,O)= limx_ooW(x,t) = O, W(O,t)= F(t). Ans. W(x,t) = F(t - 2x), where F(r) = O if r < o.

~Ax,t) - 2W,x(x,t) + W,,(x,t) = O(O < x < 1, t > O); W(x,O) = Jv,(x,O)= W(O,t)= O; W(l,t) = F(t) (t > O). Ans. W(x,t) = xF(x + t - 1),where F('r)= O if T < o. 6. In the problem on the semi-infinite string falling under its own weight (Sec. 42), (a) give the details in deriving formula (7) for displacements Y(x,t) and verify the result fully; (b) use formula (7) to verify that the vertical force at time t exerted on the string by the support at the origin equals the weight gpat of the curved segment of the string. 7. In Seco42let the weight ofthe string be replaced by a general vertical force of F(t) units per unit mass of string, so that another special case of Eq. (5), Seco40, is involved.

132.


SECo 42]

Show that the displacements are

Y(x,t) = G(t)

-

G( t

- ~)

where G(t) = J~J~ F(-r) d-rdr if t ~ O, G(t) = O if t ;;;¡¡O. 8. The end x = Oof a semi-infinite stretched string is looped around the Yaxis which exerts no vertical force on the loop, and the distant end is fixed on the x axis. The string is displaced

into the position

Y = e - X(x ~ O) and released from

rest in that position

at the instant t = O. If no external forces act on the string, set up the boundary value problem for the displacements Y(x,t) and find its solution in the form Y = e-a, cosh x if x ;;;¡¡ato Y = e-X cosh at ifx ~ ato Suggestion: Find A(s) so that A(s)e-X is a particular solution of the nonhomogeneous equation a2y"

-

s2y = -se-X in y(x,s).

9. The force per unit area on the end x = O of a uniform semi-infinite elastic bar (Fig. 42) is F(t): - EYx(O,t)= F(t). Ifthe infinite end is fixed and the initial displacement and velocity of each cross section are zero, set up the boundary value problem for longitudinal displacements Y(x,t) and derive the solution

Y(x,t) = iG( t -

~)

where G(r) = J~ F(-r) d-r if r ~ O, G(r) = O if r ;;;¡¡O; thus Y(x,t) = O when t ;;;¡¡x/a. Note that the displacement of the end x = Ois Y(O,t)= (a/E)G(t).

iIfi

~~:::~

w

-r+-r--t

-

f3 ~,

~~

~

1

O

Fig.42

10. When the pressure F(t) in Probo 9 is a finite impulse: F(t) = EFo when t < to. F(t) = Owhen t > to, show that Y = Owhen t ;;;¡¡x/a, y = (at - x)Fo when x/a ;;;¡¡t ;;;¡¡ to + x/a, and Y = Foato when t ~ to + x/a. Study this function Y(x,t) graphically. 11. Under the instantaneous impulse ofpressure F(t) = 1<5(t) in Probo 9 show formally that al Y(x,t) =

E

if x < at,

Y(x,t) = O

if x > ato

Note that in this hypothetical case a part of the bar is displaced into a region already occupied by another parto 12. Let the pressure F in Probo 9 be constant, F(t) = Fo. (a) Show formally that Ey(x,s) = aFos-2 exp( -sx/a) and hence Y(x,t) = FoE-l(at - x)So(at- x). Verify that result as a solution of the boundary value problem.


[SECo 43

133

(b) Draw graphs of Y(X,t) here as a function of x with t fixed, and as a function of t with x fixed. Note the jumps in Yxand Y;where x = at and the removable discontinuities in the functions Yxxand Y;,there. 13. For the function Y(x,t) and its transform y(x,s) found in Probo 12, show that L{ Y;,}is zero, not S2y as assumed in the formal solution; also that L{ YxJ is zero, not d2y/dx2. Note, however,that L{Y;, - a2yxx}= O= s2y- a2 d2y/dx2. 14. If a rigid mass m is attached to the end x = O of the semi-infinite bar in Probo 9 (Fig. 42), when the bar has unit cross-sectional area, and the force F(t) is a constant Fo, show why the boundary condition at that end becomes

mY;,(O,t)= EYx(O,t)+ Fo. Write b = E/am and derive the formula

Y(x,t) =

x

xlI b(' Xla¡ -aFo o t- - . E [ t - -a - -b + -eb a) Js (

15. An unstrained semi-infinite elastic bar is moving lengthwise with velocity - Vo when, at the instant t = O, the end x = O is suddenly brought to rest, the other end remaining free (Fig. 43). Set up and solve the boundary value problem for the longitudinal displacements Y(x,t). AIso show that the force per unit area exerted by the support

at x

= O upon

the end of the bar is Evo/a. Ans. Y = -vot when t ~ x/a, Y = -vox/a when t ~ x/a.

-

Vo

~

.

,I ~

x

I

Ag.~

16.

Let the end x

= O of the bar

in Probo 15 meet an elastic support at the instant

t = O,such that the pressureexertedby that support upon the end is proportional to the displacement of that end:

- EYAO,t) = - BY(O,t),or :fx(O,t)=

(b = B/E).

b Y(O,t)

Show that the velocity ofthat end is -voe-ab' since abY(O,t) = -vo(1 - e-ab'). 43

THE LONG STRING INITIALLY DISPLACED

Let the ends of a semi-infinite

string stretched

along the positive x axis be

kept fixed, and let the string be given some prescribed displacement Y = «1I(x) initially a~d released from that position with initial velocity zero. Here «11(0) = «11(00) = O. Then th~ boundary value problem in the transverse displacements Y(x,t) is (x > O,t > O),.

134

(1)


SECo 43]

Y(x,O) = cI>(X),

Y,(x,O) = O,

Y(O,t)= O,

lim Y(X,t) = O.

x-+00

The problem in the transform y(x,s) is therefore (2) (3)

s2y(X,S)

-

scI>(x)

y(O,s)= O,

= a2yxx(x,s)

(x > O),

lim y(x,s) = O,

x-+ 00

where yxx(x,s) = d2y/dx2. We shall solve the ordinary differential equation (2) by using the Laplace transformation with respect to x. Let u(z,s) denote that transform of y(x,s); that is,

u(z,s) = foooe-ZXy(x,s)dx. Since y(O,s)= O,when we transform both members ofEq. (2), we obtain the equation

where q>(z)is the transform of cI>(x).Let the unknown function of s,yx(O,s)be denoted by C. Then the solution of the last equation can be written C u(z,s) =

Z2

-

s

(s2/a2)

-

a2q>(Z) Z2

-

1 (s2/a2)'

and performing the inverse transformation with respect to z, with the aid of the convolution, we ha ve (4)

y(x,s) = aC sinh sx

s

a

I

x cI>(~)sinh ~ (x ~ a o a

-

- ~)d~.

In view ofthe condition requiring y(x,s) to vanish as x tends to infinity, it is necessary that the coefficient of ¿x/a on the right of Eq. (4) should vanish as x becomes infinite. Writing the hyperbolic sines in terms of exponential functions, we find that coefficient to be aC 2s

-

~

I

2ao

x cI>(~)e-(s~/a) d~.

Since the limit of this function is to be zero as x -+ 00, we have (5)


[SECo 44

135

Substituting this into Eq. (4), we can write the result in the form 2ay(x,s) = LX' cI>(~)e-[s(~-xllla d~ - {XJW(~)e-[s(x+~lIIa d~ +

J: W(~)e-[S(x-~)]/a

d~.

The integralshere can be reduced to Laplace integrals. In the first one

we substitute. = (~- x)/a, in the second. = (~+ x)/a, and in the third . = (x - ~)/ato get (6)

2y(x,s) = IJ:>W(x + a.)e-St d. ""

-

f

W(-x + a.)e-St d. +

x/a

x'a W(x - a.)e-St d..

i

O

In order to combine the last two integrals, let W¡(x) represent the odd extension of the function W(x): (7) ifx ~ O, tPl(X) = W(x)

= thus Wl( -x)

-W( -x)

ifx ;;;¡;O;

= -Wl(X) for all x. Then Eq. (6) can be written 2y(x,s) = L{W(x + at)} + L{W¡(x - at)}

and therefore Y(x,t) = t[W(x + at) + Wl(X - at)].

(8)

It is easily seen that a twice-differentiable function of the variable x + at, or of the variable x - at, as well as each linear combination of such functions, satisfies the linear homogeneous wave equation Y" = a2yxx. Qur continuous function (8)therefore satisfiesthe wave equation for all x and t where W"(x + at) and Wl(x - at) existo It satisfies the condition Y,(x,O)= O when x > Owherever W'(x)exists, and it satisfies the remaining conditions of problem

(1) because

W¡(x)

= W(x)

when x > O, W1(-at)

= -W(at)

and

W(oo,t) = O. . Instantaneous positions of the string can be sketched by adding ordinates as suggested by formula (8). When t is fixed, the graph of !¡(x. at), for instance, is obtained by translating the graph of the function !w¡(x), defined for all real x, to the right through the distance al. 44

A BAR WITH A PRESCRIBED FORCE ON ONE END

Let the end x = O of an elastic bar of length e be kept fixed, and let F(t) denote a prescribed force per unit area acting parallel to the bar at the end x = e (Fig. 44). Ir the bar is initially unstrained and at rest, the boundary

136


SECo 44]

--+ o

%

%

Fig.44

value problem in the longitudinal displacements Y(x,t) is the following, Y,,(x,t) = a2Yxx(x,t)

(O< x < e, t > O),

Y(x,O) = Y,(x,O)= O,

Y(O,t)= O,

EYx(e,t) = F(t),

The transform of Y(x,t) therefore satisfies .the conditions s2y(X,S) = a2yxx(x,s),

.y(O,s)= O,

EYx(e,s)= f(s),

and the solution of this transformed problem is readily found to be a

(1) CONSTANT

sinh (sx/a)

y(x,s) = Ef(s) s cosh (se/a)" FORCE

Consider first the case F(t)

(2)

= Fo.

Then (3) and, when x = e,

aFo sinh (sx/a) y(x,s) =

E

S2 cosh (se/a)

aFo 1 se y( e ,S) = - E s"2tanh -.a

In Seco23 we found that S-2 tanh (bs/2) is the transform of the triangular wave function H(b,t) of period 2b (Fig. 10). Therefore the displacement of the end x = e is aFo 2e

(4)

Y(e,t) = EH ( ~,t ) ;

that is, the end moves byjerks as indicated in Fig. 45.

PROBLEMS IN PARTIAL DIFFERENTIAL EQUATlONS

~~

[SECo 44

~

(0.2C;0) Q

137

(4cti ,01 Fig.45

To find the displacement of an arbitrary point using formula (3) we may write, when s > O, sinh (sx/a) cosh (se/a)

e-[(c-x)sJ/a- e-[(c+x)sJ/a

=

1+

e-(2cs/a) 00

= [e-[(C-X)sJ/a-

L (_1)ne-(2ncs/a),

e-[(c+X)sJ/a]

n=O

00

since (1 + z)-1

=L (o

1tzn when O < z < 1. Therefore

1

aFo 00

[

y(x,s)=E~(-lt

(2n+

{ s2exp -s

l)e

-

a

X

]

- :2 exp

[-

s(2n + ~)e + x]}.

Formally applying the inverse transformation to the terms of the infinite

series,and using braces to denote this translation of the linear function -r: (5) we obtain the formula (6)

{-r - k} = (-r- k)So(-r- k),

F.

Y(x,t)= ;({at - e + x} - {at - e - x} - {at - 3e + x} + {at - 3c - x} + {at - Se+ x} - {at - Se- x} - .. .).

For any fixed t the series (6) is finite since each ofthe braces is to be replaced by zero when the quantity inside is negative. The number of nonvanishing terms in the series increases as t increases. The function (6) can be verified directly as the solution of our problem. Its graph, for a fixed value of x, is shown in Fig. 46. .Y(x.t)

3c-x ti

k ti Fig.46

138


SECo 44]

A GENERAL FORCING

FUNCTION

When the pressure on the end x G(t) = {F(r)

= e is F(t), let us write

dr if t ~ O,

G(t) = O if t ;;¡;O.

Then L{G} = J(s)/s, and formula (1) can be written y(x,s) =!!:. (s)sinh (sx/a) o E g cosh (se/a) Thus if we write qn(x) = [(2n + l)e - xJ/a then, as in the above case, a co

y(x,s) =

E n~o (-lt{exp

[-sqn(X)] - exp [-sqn( -x)]}g(s)

and formally, (7)

a co

Y(x,t) = -

I

(-lt{G[t - qn(x)J- G[t- qn(-x)]}o

E n=O

AN IMPULSE OF PRESSURE

In the hypothetical case where (8)

F(t)

= 1 ()(t),

J(s) = 1 in formula (1) and

(9)

(x,s) = al sinh(sx/a) y E s cosh (se/a) o

The displacements Y(x,t) can be represented by a seriesof the step functions So{t - [(2n + l)e:!: xJ/a}. Those displacementsare not continuous. To see the behavior of the end x = e, we note that al 1 se y(e,s)= -E -s tanh-.a The displacement of that end is therefore represented by the square-wave function (Seco23, Figo9), al (10)

2e

Y(e,t) = E M ( a,t

)

o

Thus the end jumps suddenly back and forth between two fixed positions. It is possible to demonstrate a close approximation to that behavior by substituting for the bar a loosely wound coil spring. Ir, when the spring is hanging from one end, the free lower end is given a sharp tap, the lower end tends to move as indicated.


[SECo 44

139

PROBLEMS

1. A string is stretched between two fixed points (0,0)and (c,O)oIf it is displaced into the curve Y = b sin (nxjc) and released from rest in that position at time t = O,set up and solve the boundary value problem for the displacements Y(x,t). Verify the result fully and describe the motion of the string. Ans. Y(x,t) = b cos (natjc) sin (nxjc). 2. If the initial displacement of the string in Probo 1 is changed to Y(x,O) = b sin nnx e

(O~ x ~ e),

where n is any integer. derive the formula

nnat

Y(x,t) = bcos-sm-. e

nnx

o

e

Note that the sum of two or more of these functions with different values of n and b is a solution of the equation of motion that satisfies aIl the boundary conditions if the initial displacement Y(x,O) corresponding to such a superposition of solutions is a certain linear combination of the functions sin (nnxjc). 3. As a special case of the initiaIly displaced semi-infinite string considered in Seco43, where Y(O,t)= Y(co,t) = O, let the end segment O~ x ~ 1 be plucked so that the initial displacement has this form:

if O~ x ~ 1,

Y(x,O) = sin nx

if x ~ 1,

Y(x,O)= O

where again Y,(x,O)= O when x > O. Show the displacement Y(x,t) graphically as a function of x when t is fixed and at > 1, and note that the displacement is a wave formed by letting one cycle of the curve Y = t sin nx, -1 < x < 1, move to the right with velocity a. (Boys have used such plucking of a kite string to dislodge parachutes hooked onto the string and blown toward the kite.) 4. A long string stretched along the entire x axis is released at rest from a prescribed initial displacement. Thus ( Y(x,O) = (x),

x-lim - 00Y(x,t) = O,

Y,(x,O)= O

- co <

x < 00, t > O),

(-00 < x < 00),

lim Y(x,t) = O x-oo

(t ~ O).

Assuming "(x) exists for all x, derive the formula

-

Y(x,t) = t[(x+ at) + (x at)]

and verify that solution. 5. A semi-infinite string stretched along the positive x axis with ends fixed, Y(O,t)= limx_ooY(x,t) = O,is given an initial velocity Y,(x,O)= g(x), where g is sectionally continuous on each bounded interval, limx_oog(x) = O,and g is integrable from zerotoinfinity. (a)Ifg¡ istheoddextensionofg,sothatg¡(r) = g(r)andg¡(-r) = -g(r) if r > O,show that the transverse displacements can be written

_f 1

Y(x,t) = 2 a

x+ar

x-al

_1

g¡(r)dr = 2 [G(x+ at) - G(x - at)], a

140


SECo 44]

where G(r) = f~gl(P)dp [cf.Eqs. (2),(3),and (6),Seco43]. Verifythat resulto(b)Show Y(x,t) graphically as a function ofx when at > 1in case g is the step function 1 - So(x - 1). 6. In Seco44, when the force per unit area at the end x = e of the bar (Fig. 44) is constant, F(t) = Fo, and Y(x,O) = Y;(x,O)= Y(O,t)= O, show that the force per unit area exerted by the bar on the support at the end x = Ois the function shown in Fig. 47. Note that the force becomes twice the applied. force during regularly spaced intervals of time. EYz(O,t)

n 2Fo

o

r-I I I I

II 3c a

Fig.47

7.

Let the constant force Fo in Probo 6 be replaced by a finite impulse of duration 4e/a: F(t)

4e . IfO < t<-, a

= Fo

4e . If t>-. a

F(t) = O

(a) Show that the force per unit area exerted by the bar on the support at the end x

= Ois that shown in Fig. 47 up to the time t = 3e/a after which time the force is zero.

(b) Show that the end x = e moves with constant velocity Vo = aFo/E during the time intervalO < t < 2e/a, then with velocity - Voduring an equal time interval and that Y(e,t) = Owhen t ~ 4e/a.

8. A constant longitudinal force Fo per unit area is applied at the end x = e of an elastic bar (Fig. 48) whose end x = Ois free and which is initially at rest and unstrained. Set up the boundary value problem for the longitudinal displacements Y(x,t) and show that Ey(x,s) = aFos- 2 cosh (sx/a)/sinh (se/a). (a) Derive the formula Y;(O,t)= Vo

f So[t - (2n + 1)~J

.;0

a

2aFo. where

Vo

= E'

hence note that the free end moves with velocity Vo from time t = e/a until t = 3e/a, then with velocity 2vo until t = Se/a,etc. (b) In terms ofthe function {T k} = (r k)So(T k), show that

-

F.

-

-

00

Y(x,t) = ; .~o [{at + x - e(2n+ 1)}+ {at - x - e(2n+ 1)}].

o

e Fig.48


9.

[SECo 44

141

In Probo 8, let the force Fo be replaced by the force F(t) = Fo cos rot,

where ro = naJ(2e). Show that [formula (10), Seco23] nEY(e,t) = 2eFoIsinrotl. 10. The end x = O of a bar or heavy coil spring (Fig. 49) is free. The end x = e is displaced in a prescribed manner, Y(c,t) = G(t). If the bar is initially unstrained and at rest, find the transform y(x,s) of longitudinal displacements.

J 01 Fig.49 (a) When G(t) = Fot, show that the displacement Y(O,t) of the free end is repre-

sented by the function obtained by integrating the function EYx(O;r)in Fig. 47 from Oto t and show the displacement graphically. (b) If G(t) = bt when t ;;¡¡4eJa and G(t) = 4beJa when t ;¡¡;4eJa, show that the free end moves with a uniform velocity 2b to a new position and remains there, as shown

in Fig. 50. Y(O,t)

o Fig.50 11. An unstrained elastic bar is moving lengthwise with velocity Vo when at the instant t = O its end x = e meets and adheres to a rigid support (Fig. 51). (a) If A is the area of the cross section of the bar and M denotes the square-wave function (Fig. 9), show that the force on the support can be written AEvo

2e

-AEy"(e,t) = ---;¡-M ( -;i',t)

(a2= ~).

- Vo [ O Fig.51

1-.

142


SECo 44]

(b) In terms ofthe function {t 1

- X t - {t a- } - { t e

VoY(x,t) =

-

k}

= (t -

k)So(t - k), show that

3e -

e+ X

+ {t -

a- }

3e + X

X

a- }

+ {t -

a- } -. .. ,

and hence that Y(x,2e/a)= Oand Y,(x,2e/a)= -Vo for all X (O< X < e. Thus if the = e is free to leave the support, the bar wiII move after time t = 2e/a as a rigid unstrained body with velocity - Vo . end X

12. A steel bar 10 in. long is moving lengthwise with velocity Vowhen one end strikes a rigid support squarely (Fig. 51). Find the length of time of contact of the end of the bar with the support and note that the time is independent of vo' For steel take E to be 30 x 106Ib/in.2,

and mass per unit volume

p such that gp

= 0.28Ib/in.3, where

g = 384 in./sec2. (See Probo 11.) Ans. 0.0001 seco 13. An unstrained cylindrical shaft is rotating with angular velocity úJwhen it ends x = :t e are suddenlycIamped(Fig.52). Derivethe formula

[-

8(x s) = ~ 1 , S2

cosh (sx/a) cosh (se/a)]

for the transform of the angular displacements 0(x,t) of the cross sections, where a2 = EJ p (Sec. 40). (a) Ir H denotes the triangular-wave function (Fig. 10), show that the displacement of the middle cross section is

ift ~ ~ - a'

0(0,t) = úJt 2e

úJe

0(0,t) = -a - úJH-,t ( a

e

- -a )

. e ¡ft f;-.a

(b) Describea stretched-stringanalogfor 0(x,t) here,as seenfromthe boundary valueproblem.

(

-e .

r

I, O

+

-

"

~-

e

Fig.52

14. Let M denote the square-wave function (Fig. 9). Show that the torque acting on the support at the end x = e of the shaft in Probo 13 is a-1 E.IúJM(2e/a, t). 15. The end x = Oof a cylindrical shaft is kept fixed. The end x = e is rotated through an angle 80 and, when all parts have come to rest, that end is released; thus 0(x,0) = 8ox/e (Fig. 53). Show that the angular displacement of the free end at each instant can be written a 2e 0(e,t) = 80 - 80cH ( -;' t) , where H is the triangular-wave function (Fig. 10).


[SECo 45

143

~ e

ofFig.53

45

EQUATIONS

OF DIFFUSION

Let U(x,y,z,t) denote the temperature at a point (x,y,z) within a solid at time t. Let S be aplane or curved surface passed through that point and let the coordinate n represent directed distance along the line normal to S at the point. Then the flux of heat, the quantity of heat per unit area per unit time transferred across S at the point by conduction, is given by the formula (1)

d
-

K dn U(x,y,z,t),

where the coefficient K is the thermal conductivity of the material. The negative sign that appears with this directional derivative of U gives the sense of flow, where positive flux signifies flow in the positive direction of the normal. Formula (1) is an empiricallaw or basic postulate for the conduction of heat in solids. It is likewise a basic postulate for the flux of a substance undergoing simple diffusion into a porous solid if U represents the concentration of the diffusing substance and K the coefficient 01 diffusion. Ir the temperature or concentration is a function U(x,t) of x and t only, then the flux in the x direction is (2)

When the temperature function has the simple form U(x,t), consider a portion of the solid in the shape of a prism parallel to the x axis (Fig. 54). Since there is no temperature variation with respect to distance normal to the lateral surface of the prism, no heat is conducted across that surface. According to formula (2), an element of the prism extending from position

Fig.54

144


SECo 45]

x to x + ~X receives heat by conduction through its two bases at the rate -AKUx(x,t)

+ AKUx(x + ~x,t),

where A is the area of the cross section of the prism. This is the quantity of heat gained by the element per unit time if no heat is generated or lost inside the element. Another expression for that rate can be written in terms of the coefficient e of heat eapacity per unit volume, the quantity of heat required to raise the temperature of a unit volume of the material by one degree. Then when ~x is small, eA ~x U(x,t) is a measure ofthe instantaneous heat content of the element, and (3)

eA ~x U,(x,t) = A[KU x(x + ~x, t) - KU Ax,t)].

Ir K depends on x, then the first factor K in Eq. (3)is actually K(x + ~x, t) and the second, K(x,t). When we divide by ~x and let ~x tend to zero, we obtain the partial differential equation (4)

eU,(x,t) = [K(x,t)U Ax,t)]x'

When K is either constant or independent of x, the equation becomes (5)

U,(x,t) = kU xx(x,t),

where k = K/e and k is called the thermal diffusivity of the material. Equation (5) is the simple form of the heat equation or the equation of diffusion. When U(x,t) represents the concentration of a diffusing substance in weight per unit volume, Eq. (3) follows as before if e = 1 and K is the coefficient of diffusion. Thus, when K is constant, U(x,t) satisfies Eq. (5) where k= K. The heat equation or equation of diffusion for U(x,y,z,t) (6)

U, = k(Uxx + Uyy + Uzz),

when coefficients are constant, can be derived in a similar manner by considering a three-dimensional element of volume ~x ~x ~Z.1 Modifications of the heat equation are easily written. For example, let U(x,t) represent temperatures in a slender wire along the x axis. Let heat be generated at a time rate R(x,t) per unit volume of the wire and let heat loss from the lateral surface take place at arate proportional to the difference between the temperature of the wire and the temperature Uo of the surroundings. The modification of Eq. (3) then shows that (7)

1 U,(x,t) = kUxix,t) + -R(x,t) - h[U(x,t) - Uo], e

where h is a positive eoefficient of surfaee heat transfer. 1

See chap. 1 of the author's "Fourier Series and Boundary Value Problems," 2d ed., listed in

the Bibliography,

for a full derivation

of the heat equation.


[SEC. 46

145

In the problems to follow we assume that the coefficients K, k, e, and h are constants. AIso, we often consider semi-infinite solids, where results are relatively simple. Such bodies are not necessarily large in any absolute sense. In the diffusion of hardening materials into steel, for instance, test bars of length 1 in. or even less may be represented quite accurately as semi-infinite solids.

46 TEMPERATURES IN A SEMI-INFINITE SOLIO Let us now derive a formula for the temperatures U(x,t) in a semi-infinite solid x ~ O,initially at temperature zero, when a constant flux of heat is maintained at the boundary x = O (Fig. 55). In this idealized case of a thick slab of material, we shall substitute for the thermal condition at the right-hand boundary the condition that U tends to zero as x tends to infinity. The boundary value problem is then (1)

(x > O,t > O),

Ulx,t) = kUxx(x,t)

(x > O),

(2)

U(x,+O) = O

(3)

-KU,,(O,t) =
lim U(x,t) = O

(t -> O).

Let u(x,s) be the transform, with respect to t, ofthe temperature function U(x,t). Transforming the members of Eqs. (1) and (3), we have the following problem in ordinary differential equations which u(x,s) must satisfy: (x > O),

su(x,s) = ku",,(x,s)

- Ku,,(O,s)=

lim u(x,s) = O.

x-oo

The solution of that problem is u(x,s) =
Ksys

Accordingto formula (5),Seco27, we can write

L-1 {fie - M/S/k} =~

exp( - ::J

-'

4>0-!

-10

-1

Fig.55

x

-

146


SECo 47]

and in view of the factor l/s in our formula for u(x,s) it follows that 4>

U(x,t) = ~ K

,

~

4>ox

d.

X2

exp --( 1t1o 4h

f

)

.Ji

1

oo

2

= Kv ¡:1t x/(2./kt¡A2exp (- A )dA, where the second integral is obtained from the first by the substitution A = x/(2.)kT). Upon integrating the last integral by parts, we find that U(x,t) =

4>~ r2Jkt Kv 1tL

-

exp

(

2 X

4k t)

-

2x

f

oo exp (- A(!)dA . ] x/(2./kt¡

We can therefore write our formula in terms of the complementary error function (Sec. 27) in the form

-~

x erfc ~ . U(x,t) = 4>0 2 {kiexp ( ) ( 2Jkt)J K V -; 4kt We can show that the function (4) satisfies all our conditions (1), (2), and (3). We observe that (4)

[

-

1

K 1t U(O,t)= 24>1f

fi.

Thus the temperature of the face of the solid must vary as

the fluxof heat through the faceshall be constant. 47

Jt in order that

PRESCRIBED SURFACE TEMPERATURE

Let the temperature of the face of a semi-infinite solid x ;;;;Obe a prescribed function F(t) of time. If the initial temperature is zero, the temperature function U(x,t) is the solution of the boundary value problem U,(x,t) = kUxx(x,t) U(x,O) = O U(O,t)= F(t),

lim U(x,t) = O

x""

00

(x> O,t > O), (x > O), (t > O).

The transform u(x,s) of U(x,t), therefore, satisfies the conditions su(x,s) = kuxx(x,s) u(O,s) = f(s),

lim u(x,s) = O,

x"" 00

where f(s) is the transform of F(t). It follows that

(1)

u(x,s) = f(s)e-xJ;ik.

(x > O),


[SECo 47

147

Let us first study the flux of heat through the face of the solid, ~(t) = -KU",(O,t).

- Ku",(O,s),according

The transform of this function,

K (2)

K

cp(s)= y'k.fif(S)

to formula (1), is

1

= y'ksf(s) .fi'

Since s-t = L{(nt)-t} and sf(s) = L{F'(t)} + F(+0), assuming that F is a continuous function,then cp(s)=

~ [F(+0) + L{F'(t)}~ ] y'k .fi .fi

and with the aid of the convolution it follows that (3)

~(t)

=~

pjt [

fo~

r F'(t

F(+0) +

- 't) d't

]

,

when F is continuous. Ir F( + O)# O,the flux is infinite initially, of the order of t-t as t ~ O. In particular if F(t) = Fo,a constant, then ~(t) = KFo(nkt)-t. The total amount of heat that has been absorbed by the solid through a unit area of the face at time t is Q(t)

= {~('t)

1

Hence

d't.

K

q(s) = -cp(s) =

s

1

r;: /:f(s),

ykys

and therefore (4) For example, when (5)

F(t) = Fo if t < to

and

F(t)

= O if t >

to,

it follows from formula (4) that

Q(t) = 2KFo r: Pyt

when t ;;¡;to,

= 2KFo

when t ~ to.

p(jt-~)

148

OPERATIONAL MATHEMATIC!:

SECo 47]

Q

o

f

Fig.56

This function is shown in Fig. 56. Its greatest value is

(t;

2KFoVnk'

which is assumed at the instant t = to. Returning to formula (1)for u(x,s) and noting that, according to formuJ¿ (6), Seco27,

, exp -~ ( 4kt)} we can write,with the aid of the convolution, e-x./Sik = L

X

U(x,t) = 2#

x

{ 2v1nkt3

i t

o

F(t

-

X2

T)

Ti

exp ( - 4h

)

dT.

Substitutinga new variable of integration, we have for the general temperature formula (7)

2

oo

2

fi f

U(x,t) = -

F t x txl../ki ( - 4k,F ) e-;.2 dA..

When the temperature of the surface is constant, (8)

F(t) = Fo,

the temperature within the solid is therefore (9) U(x,t) =.Fo erfc I (2~).

Since this is a function of Fo and x/Jki only, it follows that the rapidity o heating is proportional to k; for if k is increased and t decreased so that k is unchanged, the temperature at any given distance x from the face is thc same. It is also interesting to note that for a fixed k two points Xl and x2 \Vi have equal temperatures at times ti and t2 provided x¡/-.¡¡; = X2/.jt;, thaJ is, if (10)


[SECo 47

149

This is sometimes called the law of times in the conduction of heat in semiinfinite solids. Ir Fo is positive, formula (9) assigns a positive value to U(x,t) for each x and t beca use the complementary error function has only positive values. At each interior point of the solid U(x,t) is small when t is sufficiently small, but the temperature does not retain its initial value U(x,O)= O during an interval of time after t = O. That instantaneous though highly attenuated transfer of the thermal disturbance at the surface can be noted in other cases. In Seco46, for example, the flux

-KUjx,t)

= 4>0erfc

(~fit)

is clearly different from zero immediately after the thermal disturbance took place at the surface x = O. The instantaneous transfer is a consequence of the basic postulate on the flux: <1>= - K dU/dn (Sec. 45). PROBLEMS

.

1 The initial temperature of a semi-infinite solid x ~ Ois zero. The surface temperature is this step function: U(O,t)= Fo ifO < t < to,

U(O,t)= O

if t > to.

Obtain the temperature distribution formula x U(x,t) = Foerfc" r,:; ~kt XX

= Fo erf - erf[ 2Jk(t - to) 2.jki J

= 0.15 cgs (centimeter-gram-second) unit is initially at O°C throughout. Its surface is suddenly heated to a temperature of 500°C and maintained at that temperature for 5 min, after which the surface is kept chilled to O°C. (See Probo 1.) Find the temperature to the nearest degree at a depth of 10 cm below the surface (a) at the end of 5 min; (b) at the end of 10 mino Ans. 146°C; (b) 82°C. 3. Solve Probo 2 if the slab is made of firebrick with diffusivity k = 0.007 cgs unit. Ans. (a)O°C; (b) O°c. 2. A thickslab ofiron withthermal diffusivityk

4.

The surface of a thick slab of concrete for which k

= 0.005

cgs unit, initially at O°C,

undergoes temperature changes described in Probo 2. Show that at each instant the temperature at any depth x ¡ in the concrete slab is the same as the temperature at the depth X2

= J30x¡ in the iron slab. Generalizethis result for materialswith diffusivitiesk¡

and k2 and any common time interval to of heating the surfaces of the slabs. 5. At time t = O,the brakes of an automobile are applied, bringing the automobile to a stop at time to. Assuming that the rate of generating heat at the surface of the brake

150

SECo 47]


bands varies linearly with the time, then Ux(O,t)= A(t

-

to)

where A is a positive constant and x is the distance from the face of the bando Ir to is not large, the band can be assumed to be a semi-infiníte solid x ~ O. Ir the initial temperature of the band is taken as zero show that the temperature at the face is

fk r;,

2A

U(O,t)= 3 V nV t(3to Hence show that this teniperature

2t)

is greatest at the instant t

= tto

and that thís maximum

temperatureis j2U(O,to). . 6. Theinitialtemperatureofa semi-infinitesolidx ~ Ois zero. The inwardfluxofheat through the facex = Ois a prescribed function 4I(t)oftime and the distant face is kept at temperaturezero. Derivethe temperatureformula (x > O). Also show that the temperature of the face can be written 1 (k U(O,t)= K.,Jn

r

i

O 4I(1:")(t -

1:")-t d1:".

7. An impulse of heat of quantity Qo per unít area is introduced through the face x = O of a semi-infinite solid x ~ Oat the instant t = Oand the face is ínsulated when t > O,so that -KUx(O,t) = Qo c5(t).(That instantaneous source ofheat may be realized approximately by burning a layer of highly combustible fuel at the face.) Ir U(x,O)= O when x > O,show formally that the subsequent temperature is X2

U(x,t)

= ~QK /ff¡ -1ftexp --4kt (

(t > O).

)

(See Probo 8 for a verification.) Observe that U(x,t) > Owhenever t > O. 8. The function I/I(x,t)= t-t exp [-x2/(4kt)] and its derivative I/Ix are known as fundamental svlutivns vf the heat equativn I/Ir= kl/lxx' (a) Verify that 1/1satisfies that equation when t > O,that 1/1 iO,t) = Oand I/I(oo,t) = O when t > O,and that I/I(x,+O)= O when x > O; then complete a verification of the formal solution of Probo 7 by showing that the total heat content of the solid per unit area offace, f~ cU(x,t) dx, relative to the initial heat content, is Qo. (b) Show that the functions 1/1and I/Ixare unbounded in a neíghborhood of the point x = t = O when x > O and t > O. Note that functions AI/I(x,t), where A is an . arbitrary constant, could be added to

the solution (4),Seco46, to producea familyof

solutions of the problem in that section if unbounded solutions were permitted. 9. For a solid whose temperature is a function U(x,t) show that the total quantity of heat transmitted

across a unit area of a plane x

= Xo can

be written, formally,

PROBLEMS IN PARTlAL DIFFERENTIAL EOUATIONS

151

[SECo 48

For the solid in Seco47 show that the formal result is

= -/CKs-o limJif(s)

Q(xo)

(K

= ck).

For the solid in Probo 7, show that Q(xo) = Qo. 10.

The initial temperature distribution of a solid x ~ Ois prescribed: (x > O).

U(x,O) = g(x) If U(O,t) = O when t > O and UAoo,t) = O, show formally that, when t > O,

1

OJ

(r

U(x,t)= 2¡;;ki fo g(r){exp [ -~ 11.

-

X)2

[

(r + X)2

J - exp -"4kt

J } dr.

If the initial temperatureis uniformin Probo10,g(x) = Uo' showthat U(x,t)= Uoerf

(~).

12. Show that the sum of the temperature function found in Probo 10 and the function (7), Seco47, represents the temperature in the solid x ~ Owith initial temperature g(x) and face temperature F(t).

48

TEMPERATURES IN A SLAB

The initial temperature of a slab of homogeneous material bounded by the planes. x = Oand x = 1is Uo. Let us find a formula for the temperatures-in this solid after the face x = O is insulated and the temperature of the face x = 1is reduced to zero (Fig. 57). The temperature function U(x,t) satisfies the following conditions Ut(x,t) = kUxAx,t)

(O< x < 1,t > O), (O < x < 1),

U(x,O) = Uo U AO,t) = O,

U(1,t)= O

(t > O).

The transform therefore satisfies the conditions (1) (2)

su(x,s)

-

Uo

= kuxx(x,s),

u(1,s)= O.

U(x,O)-UQ

o Fig.57

u-o x

152


SECo 48]

The solution of the ordinary differential equation (1) that satisfies the first of the conditions (2) is

u(x,s) = USo + e cosh xA, where e is determined by the second of conditions (2). Thus (3)

u(x,s) = uo

[! - ! coshx# ]. s

Let us write

s cosh1#

q=f¡,

and note that 1 coshxq coshIq = e-lq(eXq+ e-Xq)1 + e 21q
+ e-(I+x)q]

=

[e-(l-X)q

=

L (-I)"{exp "=0

Lo (-I)"e-2",q

[-q(ml - x)]

+ exp [ -q(ml + x)]},

where m = 2n + 1. We have seen(Sec.27) that (ex ~ O).

Therefore it follows formally from Eq. (3) that
(4) U(x,t)

= Uo

-

L

Uo "=0 (-1)"

(2n

{ [ erfc

+

1)12 kt

X

Jki ~

+ erlC

] [

(2n

+

1)1+ x r,::

2ykt

]} .

We shall not take up the verification of this formula sincea complete discussionwould be lengthy. However, it is not difficult to show with the aid of the ratio test that the series converges uniformly with respect to x and t and that the series can be differentiated term by termo Since the value of the complementary error function here decreases rapidly as n increases, the convergence of the seriesis rapid, especiallywhen t is small. Moreover the error function is one that is tabulated so that the series is well adapted to computation.


[SECo 50

153

Fig.58

49

A BAR WITH VARIABLE END TEMPERATURE

Let us determine a formula for the temperature V(x,t) in a bar with its lateral surface insulated against the flow of heat when the initial temperature is zero and one end is kept at temperature zero while the temperature of the other end is a prescribed function of t. If we take the unit of length as the length of the bar (Fig. 58) and select the unit of time such that (l/k) av/at' becomes av/at, that is, so that t = kt' where t' is the original and t the new variable, our boundary value problem can be written as follows: (O< x < 1, t > O), Vt(x,t) = V xJx,t) (O< x < 1),

V(x,O) = O

V(l,t) = F(t)

V(O,t) = O,

(t > O).

The solution of the transformed problem is found to be

u(x,s)

= f( s) sinhx-fi1: . sinhv s

Proceeding as in the last section, and using the convolution, we find that 2 (1)

'"

'"

V(x,t) = V~L 7tn=O

{f

'" t(m +x)/./l

-

F[t

When the temperature of the facex (2)

F(t)

our formula can be written '" (3)

4A2 J e-A2 dA}

(m

= 2n + 1).

= 1 is constant,

= Fo, 2n + 1 - x

2n + 1 + X

t [erf(

V(x,t) = Fo

X)2

4A2 J e-A2dA

(m + X)2

f

-

-

(m

F tt(m-x)/./l [

2jt

)

- erf(

2.Ii

)J

.

Details are left to the problems. 50

A COOLlNG

FIN OR EVAPORATION

PLATE

A thin semi-infinite plate occupies the space x ~ O,O~ Z ~ Zo, - 00 < y < oo. Heat transfer takes place at the faces Z = O and Z = Zo,into a medium at

temperature zero,accordingto the linear lawofsurfaceheat transfer(Sec.45);

154


SECo 50]

t

u-o

t

S-::'::==::Y :c

+ + U{x.O)-O Fig.59

but the thickness Zoofthe plate is small enough that variations oftemperature with z can be neglected. Ir the initial temperature is zero and the end x = O is kept at temperature Fo (Fig. 59), the boundary value problem for the temperature function becomes Ut(x,t) = kU xAx,t) - hU(x,t) -t

(1)

U(x,O)= O,

U(O,t)= Fo,

~i. (x >

O, t > O),

lim U(x,t) = O.

Here U(x,t) may also represent the concentration of moisture in the plate when the plate is initially dry and evaporation takes place into a dry medium, if the concentration at the end x = Ois a constant Fo. In that case k = K, the coefficient of diffusion (Sec. 45), and the positive constant h is a coefficient of evaporation. In terms of transforms, our problem (1) becomes kUxAx,s) - (s + h)u(x,s)= O Fo

u(O,s)= -,s

(x > O),

lim u(x,s) = O. x-""

The solution of this problem is

r;+h

Fo (2)

u(x,s) =

-¡ exp ( - Xv --¡;- ) .

Knowing the inverse transform of e - xJiik, we can write, with the aid of our property on substitution of s + h for s (Sec. 7), xe-ht

L -1

{exp (-xPi)}

X2

= 2~ nkt3 exp ( - 4kt) .

It follows from formula (2) that F. x

(3)

t

i

X2

dt

U(x,t) = ~ e-ht exp --( 2~ o 4kt ) tt "" 2F. hx2 = '\! n Jtx¡../ki exp ( -A2 - 4k12 A ) dA,

~

where the second integral is obtained by the substitution A = tx/~.


155

[SECo 51

The formula (3)can be ehanged to a more usefulform with the aid of the integration formula (see Probo 14 below) (4)

~

100

exp

(

-AZ - ~~) dA = eZaerfe (r +~) + e-Za erfe (r - ~).

This formula can be verified by noting that its two members have the same derivative with respeet to the parameter r and that both members vanish as r tends to infinity. With the aid offormula (4) the solution (3) of our problem becomes (5) U(x,t) =

~[ebXerfc (2~ + Jht) +

e-bx erfe

(2~- Jht)1

whereb =.Jhik. Note that U(x,t) > Ofor eaeh x whenevert > O,if Fo > O. In obtaining Eq. (5) we have found a desirable form of the inverse transform of the funetion (2). The verifieation of our formal solution of problem (1)is left to the problems. 51

TEMPERATURES IN A COMPOSITE SOLlD

Let us write a formula for temperatures U(x,t) in a solid x ~ O,eomposed of a layer O < x < a of material initially at uniform temperature A in perfeet thermal contact with a semi-infinite solid x > a of another material initially at temperature zero, when the face x = O is kept insulated (Fig. 60). If the thermal eonduetivity and diffusivity are K1 and k1, respeetively, in the first part and Kz and kz in the second part, the boundary value problem is the following one: (1)

Ur(x,t) = k1 Uxx(x,t)

(O< x < a, t > O),

(2)

Ur(x,t) = kz Uxx(x,t)

(x > a, t > O),

(3) (4)

U(x,O)= A(O< x < a), UAO,t)= O,

U(x,O)= O

lim U(x,t) = O

(x > a), (t > O),

x-oo

U(%,O)-A

o

a

Kl.kl Fig.60

%

156


SECo 51]

(5)

U(a

(6)

Kl UAa

- O,t) = U(a + O,t) - O,t) = K2UAa + O,t)

(t > O), (t> O).

Condition (5) states that the temperature at the interface x = a is the = Owhen the point approaches the interface from either direction. Condition (6)states that the flux of heat out of the first part through the interface must be equal to the flux into the second part, at each instant. The problem in the transform of U(x,t) is then

same after t

(7)

-

xu(x,s)

(8)

A

su(x,s)

u(a

(10)

K1ux(a

(O < x < a), (x > a),

= k2ux::.(x,s)

lim u(x,s) = O,

uiO,s) = O,

(9)

= k1uxx(x,s)

x-+00

- O,s) = u(a + O,s), - O,s) = K2ux(a + O,s).

The solution of Eq. (7) that satisfies the first of conditions (9) is

fS

A

u(x,s)= Cl coshx-vI; + -;

(O~ x < a),

and the solution of (8) that satisfies the second of conditions (9) is

(x > a).

u(x,s) = C2exp (-xJf)

By applying the conditions (10) to these functions, the values of Cl and C2 are easily found. The formulas for u(x,s) can then be written

-

1

A

(11)

u(x,s)- -; ( 1 -~.

(12)

u(x,s) =

-

+ e-ala+X» 1 - Ae-2aa )

A e-ala-x)

A(l + A)e-aplx-a)- e-aI2a+px-pa)

(O< X < a),

(x > a),

where

r; u

= vk;'

(k; Jl

= -VI<;'

and therefore UJl= .;;¡k; and IAI< 1. Equation (11) can be written in the form

u(x,s) = A

!s -

12

A

f

A"(e-alma-x)

"=0 s

+

e-alma+x»

(O< X < a),


where m

= 2n +

157

.

1, and (12) can be written

u(x,s) = 1 + A A

[SEC. 51

2

f

n=O

-

An (e-a(2na+/lx-/la)

s

e-a(2na+2a+/lx-/la))

(x > a).

It therefore follows that (13)

1 - A00

(2n + l)a - x ~ An{erfe [ 2.jk;i ]

U(x,t)= A - A-y-

+ erfe (2n + l)a + x

~

[

(14)

2yll.lt

2na + Jl(x- a)

1 + A 00

U(x,t)= A-y-

(O< x < a),

]}

~ An{ erfe [

2.jk;i

- erfe (2n + 2)a + Jl(x - a)

[

~

]}

2yll.lt

] (x > a).

In sueh problems if U(x,t) represents the eoneentration of a substance diffusing in a eomposite porous medium, the eoneentrations in adjaeent layers of different media may maintain some ratio a (a # 1), the ratio of equilibrium concentrations in the two media (see Probo 13 below). InterfaCe eondition (5) would then be replaeed by the eondition U(a + O,t) = aU(a - O,t).

(15) PROBLEMS

1. Completethe derivationofformula (1),Seco49. 2. Whenx = 1,showthat thegeneraltermofthe seriesinformula(3),Seco49,becomes

and hence that U(I,t) = Fo. 3. In Seco50, verify the solution (5) ofproblem (1). 4. In the problem (1) of Seco50, show that the flux into the plate at the end x given by the formula

(O,t) = KFo(~e-ht 5.

= Ois

+ AerfJht).

In the problem (1) of Seco50, make the substitution V(x,t) = ehIU(x,t)

and show that the resulting problem in V(x,t) is a special case of one solved in Seco47.

158

OPERATIONAl

SECo 51]

MATHEMATICS

= Oand x = 1 of a slab of material

for whieh k = 1 are kept at temperaU = 1, respeetively, until the temperature distribution beeomes U = x. After time t = Oboth faces are held at temperature U = O. Derive the temperature formula

6.

The faces x

tures

U

=O

and

oo

U(x,t)= x - J;o ( erfe

2n+l+x

2n+1-x 2.1t

(t > O).

2../i

erfe

)

7. In Prob. 6, let the face x = 1 be insulated when t > Owhile U = Oas before at the face x = O. Find the flux of heat outward through the face x = O.

2n + l

00

[

Ans. K 1 - 2 "~o(-1)" erfe 2../i

J.

8. Heat is generated in a long bar x $;;Oat arate R(t) units per unit volume [Seco45, Eq. (7)],and the lateral surface of the bar is insulated. Ir the initial temperature is zero and the end x = Ois kept at that temperature while the infinite end is insulated (Fig. 61), derive the temperature formula X2 2 oo U(x,t) = Q(t) - r= Q( t 4U2 ) exp(-'¡'2)dl, V 1t

l

!:x;¡./fi

-

l t Q(t) = R(t) dt. e o

i

where

u-0-c~;;;;:::;;;;~~:::r:;:~!;::;:::) -1/4O UIz,O¡-O --.,.7, Fig.61 9.

In Probo 8, let the rate of generation of heat be a eonstant B and show that

B

i

t

~dt.

U(x,t) = e oerf 2../fT 10.

x

The lateral surface of a bar of unit length is insulated while its ends x

= O and

= 1are kept at temperaturezero. Heat is generatedthroughoutthe bar at a eonstant

rate of B units per unit volume. Ir the initial temperature is B(x

-

x2)/(2K), show that

the bar retains that initial temperaturedistributionwhent > O. 11. Units of time and distance are ehosen so that k = 1 and a given bar has unit length. The lateral surface and the end x

= O are

insulated.

The initial temperature

of

the bar is zero, and the end x = 1 is kept at that temperature. Ir heat is generated throughout the bar at arate of R(t) units per unit volume, derive the temperature formula cU(x,t)= 00

where

E R(t)dt - I: R(t -

[

t)E(x,t) dt,

2n + 1 -

2n + 1 + X

E(x,t)= ~ (-1)" erfe(

2..!t

)

+ erfe(

2,!t

X

)J .

PROBlEMS IN PARTIAl DIFFERENTIAl ENQUATIONS

159

[SECo 52

12. Let the semi-infinite evaporation plate (Sec. 50, Fig. 59) have a uniform initial concentration of moisture U(x,O) = uo, and let the end x = O be kept dry, U(O,t) = O, while the infinite end is impervious to moisture. Derive, and verify, this formula for the concentration :

-h, f x er rv;.

U(x,t) = uoe

2yKt 13. The entire surfaceof a long porous cylinder is impervious to moisture. The cylinder consistsof two semi-infiniteparts, the first ( - 00 < x < O)of material with a coeffieientof diffusion of moisture K ¡ and the second (O < x < 00) with a eoefficient K 2 (Fig. 62). Initially, the concentration of moisture is zero in the part x < O and Uo in the part x > O.Let IXbe the ratio of equilibrium concentrations in the two parts so that, as indicated in Seco51, the concentration U(x,t) satisfies the eonditions U( +O,t) = IXU(-O,t), at the interface x = O. Derive the formula

U(x,t) = IX

=

where p = .JK¡/K2'

-x

Uo

+

p erfe

- Uo

whenx < O,

rr;;-;

( 2yK¡t )

X

whenx > O,

+ p erf" rr;;-;) IX + p( ¿.yK2t IX

Show that the ratio ofthe limiting coneentrations, as t

-

00, in

two parts is 0(.

1

~UI#IJ)-O ~UI#IJ)-" L---G:r~~ K~::::::~ O

Fig.62

14.

Derive the integration formula (4), Seco50. Suggestion:Write

z(x) =

~ro exp [- (l- I)]dl;

then the integral (4), Seco 50, is e - 2Dz(a).Show that z'(x) = 4e4xerfe(r + xlr) and

z(0)= 2erfer, thenuseintegrationbypartsto findz(x). 52

.

OBSERVATIONS ON THE METHOD

All the problems treated in this chapter involve partial differential equations and boundary conditions that are linear,that is, offirst degree in the unknown function and its derivatives. The limitation of our operational method to the treatment of such linear boundary value problems is a natural one, since we have presented no formula giving the Laplace transform of the product

/

160

SECo 52]

of two

functions in terms of the transforms of the individual functions. It is


known that the transform of the product of two arbitrary functions can be expressed by a convolution integral of the two transforms, where the integration is one in the complex plane of s. But it is safe to say that no advantage can be anticipated in replacing nonlinear differential forms by complex nonlinear integral forms. We have solvedproblems with constant coefficients. Ir the coefficients are functions of t, the variable with respect to which the transformation is made, the transformed problem is not likely to be simpler than the original one. For even when the coefficients are polynomials in t, the transformed problem involves derivatives with respect to s in place of the derivatives with respect to t present in the original one. Ir the coefficients are not functions of t, the transformed problem will be simpler. We have made the transformation with respect to time t in all our problems in partial differential equations. Ir the physical problem involved the first derivative U" the initial value U(x,O) was prescribed; if it involved Y,,, then Y(x,O) and Y,(x,O)were both prescribed. Consequently, when we applied the formula for the transformation of those derivatives, u(x,s) or y(x,s) was the only unknown function arising. But suppose the Laplace transformation with respect to x had been applied in the temperature problems. Then if L{ U(x,t)} = u(z,t),

L{ Uxx(x,t)} = Z2U(Z,t)- zU(O,t) and not both of the functions

-

U)O,t),

U(O,t) and U)O,t) could be prescribed,

since

both the temperature and the flux of heat at the surface x = O cannot be prescribed. Thus one of those unknown functions must be determined with the aid of other boundary conditions, and the procedure becomesunwieldy; the Laplace transformation is not the integral transformation that is adapted to the boundary conditions with respect to x here. The operational method of solving partial differeQ#ialequations is of coursenot limitedto equations of the second order. We shall soon take up a more powerful method of obtaining inverse transforms, and then we can attack problems whose solutions depend upon more involved inverse transformations than those in this chapter. As we have illustrated here and in the preceding chapter, the operational method is well adapted to the solution of many problems in differential equations in which some of the given functions or their derivativesare dis. continuous. That is one of the remarkable features of the method. In case the boundary value problem involves more than two independent variables, say x, y, and t, a Laplace transformation with respect to t stillleaves us with a partial differential equation with independentvariables x and y. The new problem may be adapted to solution by one ofthe Fourier transformations or another integral transformation with respect to x or y,


[SEC. 52

161

depending on the differential forms and boundary conditions involved there. That procedure of using successive transformations will be illustrated in

Chap. 11. As the problems in the. present chapter have indicated, transformation methods help to show whether the number and type of boundary conditions accompanying a given partial differential equation are adequate to determine a definitesolution of the boundary value problem. A physicalinterpretation of the problem servesas another guide for setting boundary conditions that lead to just one solution. A complete treatment or rigorous solution of a boundary value problem, however, consists first of proving that the formal.solution does satisfy the differential equation and all boundary conditions and continuity requirements, then showing that the function is the only one that satisfies all those conditions. That procedure of establishing a unique solution. is apt to be lengthy. We shall illustrate it in later chapters.

5 Functions of a Complex Variable

We give now a synopsis of theorems and definitions from the basic theory of functions of a complex variable, material needed to develop fu¡ther the theory ofthe Laplace transformation. For complete treatments ofthe topics, including proofs of statements and theorems, the reader may refer to books on the subject.l Derivations of extentions of the Cauchy integral formulas are given in the final section of this chapter. 53

COMPLEX NUMBERS

We shall not review the algebra of complex numbers except to remind the reader of properties of absolute values and conjugates. We write z 1

= x + iy = r(cos 9 + i sin 9),

See, for instance, the author's "Complex Variables and Applications," 2d OO.,1960, and

references

162

listed in the Bibliography

in that book.

FUNCTIONS OF A COMPLEX VARIABLE

x = Re z,

[SECo 54

and

y = 1m z, r = Izl

163

e ;;., arg z.

Ir Zl and Z2 are complex numbers, then if Z2 "" O.

The triangle inequalities are IZ1 :t z21 ;¡¡; Iz11

+

Iz21,

Neither the statement z 1 > z 2 nor z 1 < z2 has meaning unless z 1 and z 2 are both real numbers.

Ir m and n are integersand z "" O,then

. me

me znfn = r"'fn ( ros The

conjugate

--;¡-+ i SID --;¡-) ,

operation

on

z, Z = x

Izmfnl= Izlmfn.

-

iy,

has

these

distributive

properties: Zl + Z2 = Zl + Z2,

ZlZ2 = ZlZ2,

AIso zz = Iz¡Z.. A neighborhood of a point Zoin the complex plane is a two-dimensional circular domain, or disk, consisting of all points z such that Iz - zol < (, where ( is some positive number. 54

ANAL YTIC FUNCTIONS

Let w denote the values of a single-valued function j of the complex variable z, defined on some set of points in the z planeo We write w = j(z) = u(x,y) + iv(x,y) where u and v are real-valued functions of two real variables. Ir w = Z2 for all z, for example, then j(z) = (x + iy)2 = X2 - y2 + 2xyi; thus u(x,y) = x2 - y2 and v(x,y) = 2xy. Ir w = Re z then u(x,y) = x and v(x,y) = O. Ir w is a single-valued function z! such that z! = .j;(cos~

+ iSin~)

(r>

O, -1t < e < 1t),

then it is convenient to write the components u and v as functions of r and e: u(r,e) =

~ cos!O,

v(r,e)= ~ sin!O (r > O,-1t < e < 1t).

164


SECo 54]

Here u and v are single-valued in the domain consisting of all points of the complex plane except those for which y

=

O and x ~ O.

A function f has a limit Woat a point Zo, lim f(z) = WO,

%-+%0

if and only if for each positive number If(z)

-

lO

there is some number ~ such that whenever Iz - zol < ~ and z :1=Zo.

wol < lO

Thus the value of f is arbitrarily close to Wo throughout some deleted neighborhood of Zo, a two-dimensional neighborhood with Zo excluded. The definition requires f to be defined throughout some neighborhood of the point Zo, except possibly at the point itself. Ifwe write Re Wo= uo, 1mWo= vo, and Zo= Xo + iyo, then it tums out that the condition lim f(z)

%-%0

= %-%0 lim (u(x,y) + iv(x,y)]= Uo+ ivo

is satisfied if and only if the two-dimensional limits of the real-valued functions u and v at the point (xo,Yo) are Uoand vo, respectively. Since f is continuous at Zo only if its limit Wois the same as its value f(zo), then it is continuous there if and only if both u and v are continuous at the point (xo,Yo). The derivative of f at a point z is defined as

f'(z) = 6:lim0 f(z + ~z) - f(z) ~z

Jo

provided the limit exists, that is, provided that for each positive number there is a number ~ such that

I

f(z +

~; -

f(z)

-

f'(z) < I

lO

lO

whenever I~zl < ~(~z :1=O). The two-dimensional limit here with respect to the variable ~z (Fig. 63) requires that, iff'(z) exists, the partial derivatives offirst order of the functions u and v exist at the point (x,y) and satisfy the Cauchy-Riemann conditions

au -=-,

av

ax ay

au ay

-=--,

av ax

at the point. AIso, when f'(z) exists, we find that

f'(z) = ux(x,y)+ ivx(x,y)= Vy(x,y)- iuy(x,y).

FUNCTlONS OF A COMPLEX VARIABLE

[SECo 54

165

y

Fig.63

o

Suppose that U and V are real-valued functions of x and y whose partial derivatives of first-order are continuous at a point (x,y).! Then in order for the function f = u + iv to have a derivative f'(z) at the point z = x + iy, it is necessary and sufficient that u and v satisfy the CauchyRiemann conditions at the point. For example, the components u and v of the function Z2satisfy those conditions everywhere; hence

d dz (Z2) =

o oX (X2

o -

y2) + i ox (2xy) = 2(x + iy)

= 2z.

In contrast, the components of the function Izl2 = X2 + y2, whose partial derivatives are continuous everywhere, satisfy the Cauchy-Riemami conditions at the origin only, where 2x = Oand 2y = O. Thus the derivative of Izl2 exists at z = Oonly, where its value is zero. A function f is analytic at a point Zoif its derivative f'(z) exists at every point in some neighborhood of zo. Analyticity implies continuity, but not conversely. The functions Re z, z, and Iz12,for instance, are everywhere continuous, but nowhere analytic because their derivatives do not exist throughout any neighborhood. Formulas for derivatives of sums, products, quotients, and powers of functions and of composite functions g[f(z)J correspond to those in calculus. Consequently the sum, product, or quotient of two analytic functions is analytic except, in the case of the quotient, at those points where the denominator vanishes. An analytic function g(w) of an analytic function w = f(z) is analytic for values of z in a domain (a connected open region) where w is analytic and where the values of w are interior to a domain where g(w) is analytic. The function l/z is an example of a quotient of two functions that are everywhere analytic. The function is analytic at all points except the origino Its derivative is -1/z2 if z =1= O. An entire function is one that is analytic for all finite Z. Every polynomial in z is an entire function. 1The continuity of those partial derivatives ensures the continuity of the functions u and v themselves, at the point.

166

55


SECo 55]

EXPONENTIALAND TRIGONOMETRIC FUNCTIONS

The exponential function, written exp z or eZ,can be defined by the equation (1)

exp z

=

eX(cos y + i sin y)

(z

= x + iy),

where y is the radian measure of the argument of the trigonometric functions. Its components u and v have partial derivatives which are continuous and satisfy the Cauchy-Riemann conditions everywhere. Therefore exp z is an entire function. Its derivative is the function itself: d (2) dz exp z = uX O for all z; thus ¿ is never zero. When z = x, the function expz becomesthe real-valued function ~. The polar representation of z can now be written reie:

I

z

= r(cos lJ +

i sin lJ)~ r exp (ilJ).

The hyperbolic cosine and sine are these linear combinations of the entire functions ¿ and e-Z:

(3) hence they are entire. Their components are shown by the formulas cosh z

= cosh x cos y + i sinh x sin y,

sinh z = sinh x cos y + i cosh x sin y. Each of the remaining hyperbolic functions of z,

tanh z = sinh z cosh z'

l coth z = tanh z'

l sechz = cosh z'

l csch z = sinh z'

is analytic except at those points where the denominator on the right-hand side vanishes. The circular functions of z can be defined as follows:

.

(4)

2

smz

=

eiz

.

- e-1Z 2i

tanz=sinz cosz'

the remaining three being the reciprocals of these. The functions cos z and sin z are entire. In view of these definitions, it follows that cos z = cosh iz = cos x cosh y

sinz =

-

i sin x sinh y,

- i sinh iz = sinx cosh y +

i cosh x sinh y.


[SECo 55

167

AIso, sin z and cos z are periodic with period 2n, while sinh z and cosh z are periodic with period 2ni. The forms of the trigonometric identities are the same as those for functions with real arguments; for instance, sin2 z + COS2Z = 1,

cos 2z = COS2Z

sin(zl + Z2) = sin Zl cos Z2 + COSZl

-

sin2z,

sin Z2'

The same is true for the relations between the six hyperbolic functions. Furthermore, the formulas for the derivatives of all those functions retain the same form when the variable is real. Thus for all z in a region where w(z) is analytic, the composite functions sin w(z) and cos w(z) are analytic functions of z and

d

-sm dz

.

w = cosw-

dw dz'

d -cosw dz

. dw = -sm w-. dz

The absolutevalues of sin z and cos z are not bounded as Iyl -. the problems to show that

+ sinh2 y,

(5)

IsinZl2 = sin2 x

(6)

Isinhzj2 = sinh2x

+ sin2 y,

oo. It

is left to

Icos Zl2 = COS2x + sinh2 y, Icosh Zl2 = sinh2 x + COS2y.

From those formulas we can see that the zeros of sin z and cos z are just the real zeros of those functions, while the zeros of sinh z and cosh z are pure imaginary numbers. PROBLEMS

1.

(a) Interpret the triangle inequalities (Sec.53) in tenns oflengths of sidesoftriangles. (b) Extend the first triangle inequality to apply to n complex numbers as follows: IZ1+ Zz +

... + z.1 ;;; Iz11 + Izzl + ... + Iz.l.

2.

Ir Izzl =FIz31,prove that

3. 4.

Ir n = 2, 3, . . . , show that zll. has n distinct values. Ir a. and b. are real numbers, prove that the function f(z)

(z = x + iy)

= alx + bly + e + i(a2x + bzy)

is an entire function if and only if the coefficients are such that f(z)

5. Provethat the functionf(z) = XZ + iyZis nowhereanaIytic. 6. State whythe functionssin(¿) and exp(cosz)are entire.

= (al

- ibl)z + c.

168

SECo 56]

OPERATIONAl

MATHEMATICS

7. Show that (a) e-Z = lje"; (b) exp (Zl + Z2) = (exp zl)(exp Z2);(e) lexp (-z2)1 is not bounded for all z. 8.

In Seco55, prove (a) identities (5); (b) identities (6).

9.

Prove that (a) sin z

= O if and

only if z

:tmri, wheren = O,1,2,.... I I

10.

=

:tmr; (b) sinh z

Prove that tanh z is analytic except at the points z

=

= O if and

:t(n

-

only if z

=

!)ni, where n = 1,

2,... . 11.

Show that (a) tan (z + n)

=

tan z; (b) tanh (z + ni)

=

tanh Z.

12. 13.

Establish the inequalities sinh Ixl ~ Isinh zl ~ cosh X. Use the definition of f'(z) when dz = dx, then when dz = idy, to show that f' = Ux+ ivx= Vy- iuywhen f'(z) exists. Hence deduce that the components u and v necessarily satisfy the Cauchy-Riemann conditions at points wheref'(z) exists.

I I

14. In polar coordinates, where z = reiO,r > O, verify that the Cauchy-Riemann conditions take the form ou

1 ov,

or = -;: 00'

I

and that f'(z) = e-iO[u,(r,O)+ iv,(r,9»).

56

CONTOUR INTEGRALS

Consider first a complex-valued function G of a real variable t, with components V and V, G(t) = V(t) + iV(t).

We call it sectionallycontinuous over a bounded interval if both the realvalued functions V and V have that property. Its derivativeis V'(t) + iV'(t) wherever V' and V' exist. Its integral (1)

f

G(t) dt

=

f

V(t) dt + i

f

V(t) dt

has the properties of integrals of real-value functions as to integrals of sums, multiplication by a constant, reversal of limits of integration, and integration by parts. Also, when G is sectionally continuous on the interval a < t < b, it is absolutely integrable over the interval and (2)

If G(t)dtl

~ f'G(t)'dt.

The limit b ofthe integral (1)can be replaced by

[SECo 56

169

G is sectionally continuous over each bounded subinterval of the half line t > a and if IG(t)1~ M(t) where the real-valued function M is integrable from ato 00, then G is integrable from a to oo. In particular, G is integrable there if IGIis integrable,and a proof1of inequality (2)can be extendedto the improper integral to write (3)

Ir" G(t) dtl ~

LX)

IG(t)1dt.

Now let e denote either an are of a smooth curve

x = q,(t),

(4)

y = Ijt(t),

where q" Ijt, q,', and Ijt' are continuous and the derivatives do not vanish simultaneously for any of the values of t, or a curve formed by joining a finite number of such ares endwise. Then e is called a contour. If e is closed and do es not intersect itself, it is called a closed contour; boundaries of triangles, rectangles, and circles are examples. Ir the parametric equations (4) represent a contour e over which a ~ t ~ b and if j(z) is sectionally continuous over e, that is, if the function j[q,(t) + iljt(t)], a < t < b, is sectionally continuous, then the contour integral of j over e can be defined in terms of an integral of type (1), namely (5)

fe j(z) dz = lb f[ q,(t) + iljt(t)][q,'(t) + iljt'(t)]dt.

The integrand is a sectionally continuous function of t. Inequality (2) enables us to write (6)

lfe j(z) dzl ~ fe If(z)lldzl~ ML,

if If(z)1~ M for all z on e, L is the lengthof e, and Idzlis the elementof are length of e :Idzl= Iq,'(t) + W(t)1 dt = ds. Using the components u(x,y) and v(x,y) ofj(z), we can write the contour integral (5) in terms of real-line integrals over e as follows: (7)

fe j(z) dz

= fe (udx -

v dy) + i fe

a result that is arrived at formally by writingf

(u dy + v dx),

= u+

iv and dz

then expanding the productj(z) dz. I Churchill,

R. V., "Complex

Variables

and Applications,"

2d ed., p. 96, 1960.

= dx

+ i dy,

170

57


SECo 57]

INTEGRALTHEOREMS

We now state basic theorems on contour integrals of analytic functions. Cauchy integral theorem lf afunctionfis and on a closed contour C, then

(1)

analytic at all points interior to

fe j(z) dz = O.

A simply connected domain D is an open-connected region such that every closed contour within it endoses only points of D. The interior of a closed contour and the entire z plane are examples, but the exterior of a closed contour is not simply connected, nor is the domain between two concentric cirdes. The Cauchy integral theorem permits C to be any closed contour within a simply connected domain throughout which f is analytic. The theorem is easily extended so as to permit C to be replaced by the boundary B of certain multiply connected regions, as follows. Let Co denote a closed contour, and CI, C2, ..., Cn a finite number of closed contours within Co with no interior points in common (Fig. 64). Let R be the closed multiply connected region consisting of all points on and within Co except for points interior to Cj(j = 1,2,..., n). Let B denote the boundary of R, consisting of Co and all Cj described in such a direction that the regio n R lies on the left of the boundary. Then iff is analytic at all points of R, it is true that (2)

fB f(z) dz = O. The value offat each point interior to a closed contour C is determined by the values on C by the following theorem. Cauchy's integral formula Let f be analytic everywhere within and on a closed contour C. Then if Zois a point interior to C, 1 f(z) f(zo) = ---: dz, (3) 2m e z - Zo

i

where the integration is counterclockwise around C. y

o

Fig.64


[SECo 58

171

Here, too, e can be replaced by the boundary B of a multiply connected region under the conditions prescribed for formula (2). Derivatives of the integral (3) with respect to Zo exist and are given by the integral of the derivative of the integrando In particular,

" f (Zo) --

(4) Consequently

the derivative

f'

1

f

f(z)

3 dz. ni e (z - zo)

is analytic

wherever

f

is analytic;

thus the

derivative ¡
analytic, then a contour integral between the points,

f

(5)

z ZO

f(~) d~ = F(z),

has a value F(z) independent of the contour e as long as e

líes entirely

within D. Moreover that single-valuedfunction F is analytic over D. It is an antiderivative,or indefiniteintegral, off: (6)

for all z in D.

F'(z) = f(z)

Any two antiderivatives off differ by a constant. In terms ofan antiderivative F of the analytic functionf, we can write ZI

f

f(z) dz = F(Zl)

-

F(zo)

ZO

over any contour interior 58

to D from point Zo to point z1.

POWER SERIES

Let f be analytic over the interior of a circle Iz - zol = ro. Then at each interior point z it is true that (1)

-

that is, Taylor's series converges whenever Iz zol < ro and its sum is f(z). That theorem is a consequence of Cauchy's integral formula. The region of convergence of any power series 00

(2)

L an(z n=O

zo)n,

where an are complex constants, is always a disk about the point Zo, Iz zol < ro, unless the series converges only when z = Zoor for all z. Let f(z) denote the sum of the series. The series can be differentiated term by

-

172

SEC, 58]


term to give a power series that converges over that disk to j'(z).

Thus series (2) represents an analytic function over that disk; in fact, it is Taylor's series representation, about the point Zo, of its sum f(z). Thus l an = ,f(nJ(zo) n.

(n = 1,2,. ..).

The power series (2) converges absolutely and uniformly over each interior disk Iz - zol ~ rl, where rl < ro.

with respect to z

Since ¿ is an entire function, its Maclaurin series (zo = O), (3)

e% =

l

ex> l _zn n= 1 n!

+ ¿

(lzl < (0),

represents that function for all z. Consequently,the function g(z) = exp (Z2) has the power series representation

(lzl < (0), which must be Maclaurin's series for g, so it follows that g(2nJ(0)= (2n)!jn! and g(2n-1)(0) = O.

With the aid of Taylor's series we find that the zeros of an analytic function are isolated. More precisely, if a function f is analytic at a point Zo and not identically zero over some neighborhood of that point, then there is a neighborhood of Zosuch that f(z) :F Oat any of its points except possibly Zo itself. A generalization of Taylor's series also follows from Cauchy's integral formula. If f is analytic over an an,nular region R, r1 < Iz zol < ro, it is represented there by Laurent' s series:

-

ex>

(4)

f(z) =

L-

n=

co

Aiz - zot

where, if e is any closed contour described counterclockwisearound the annulus R (Fig. 65), (5)

.

A

l

J -f(z)zot+1

n = 2ni c(z

d Z

(n

= O,:t 1, :t2,.. .).

We note in particular that

_1tlJe f(z)dz.

A_1 = 21 .

Taylor's and Laurent's series can be integrated term by term within their domains of convergence. It follows that a representation of f by a


[SECo 59

173

y

Fig.65

o

x

convergent series of form (4) is necessarily the Laurent series, and hence the values of the integrals (5) can be seen from such a representation. For example, from Maclaurin's series (3)for eZwe can write this Laurent series representation: (6)

Here

exp A_1

~=

( Z)

1+

~Z + ~2! ~Z2 + ~3! ~Z3 +

...

(lzl > O).

= 1, A_2 = 1, and Zo= O. According to formula (5), if e is a

closed contour described counterclockwise around the origin, then

fe exp (~)

dz

= 2ni,

fe z exp (~)

dz

= ni.

The region of convergence of a series of type (4) is always an annulus with center at Zo. The series represents f as a sum of two functions: the sum of the series of nonnegative powers of z - Zoand the sum of the series of negative powers. The first function is analytic when Iz - zol < ro and the second one when Iz - zol > r1. Two power series of type (2) in which Zo is the same for both can be added, multiplied, or divided to produce series of that type which converge to the corresponding combinations of the sums of the series, under natural restrictions on the regions. Those operations apply to some types of Laurent series too,ofcourse,as wecansee by replacingz - Zoin series (2)by(z - ZO)-1 or by multiplying the series by a negative power of z - Zo. But in the general case the product of two Laurent series is a double series.1 59

SINGULAR POINTS AND RESIDUES

Ir a function is analytic at some point in every neighborhood of a point Zo, but not at Zo itself, then Zo is called a singular point of the function. Ir the function is analytic at all points except Zo, in some neighborhood of Zo, then 1

For properties

pp. 114 ff., 1959.

of double series see, for example, E. Hille, "Analytic

Function

Theory,"

vol. 1,

174


SECo 59]

Zo is an isolated singular point. The function (Z2 + 1)-1, for example, is analytic everywhere

except for the two isolated singular points Z =

:t i.

About an isolated singular point Zo a function always has a Laurent series representation: (1)

f( Z)

A-I

= -Z -

Zo

+

A_2 2

(z - Zo)

+ . .. + Ao + Al (Z - Zo) +... (O< Iz - zol < ro),

where ro is the radius ofthe neighborhood in whichfis ana1ytic except at Zo. The coefficients An are given by formula (5), Seco58, where the inner radius r 1 of the annu1us, around which C is described, is now zero. In particular,

A_1= _21. f

(2)

1t¡)c

f(z)dz,

where C can be any dosed con tour, described counterdockwise, containing Zoin its interior and such that, except for Zo,f is analytic within and on C. The complex number A - 1 ,the coefficientof(z - zo)- 1 in the expansion

(1),is called the residueoff at the isolated singular point zo; 21tiA- 1 is the value of the integral of f in the positive direction around a contour that endoses no other singular points but Zo. The expansion (6), Seco 58, for instance, shows that the residue of the function exp (I/z) at the point z = O has the value A-I = 1. Ir C is a cIosed contour within and on which f is analytic except for a finite number of singular points Z1, Z2, .. . , Zm,interior to the region bounded by C, the residue theorem states .that (3)

fe f(z) dz = 21ti(P1 + P2 + . . . + pm>,

where Pn denotes the residue off at Znand where the integration is in the positive direction around C. Note that the singular points Znare necessarily isolated because of their finite number. In the representation

(1) the series of negative

powers of (z

-

zo) is

called the principal part of f(z) about the isolated singular point Zo. The point Zois an essential singular point offifthe principal part has an infinite number ofnonvanishingterms. It is a poleofordermirA - m =1= Oand A-n = O when n > m. It is called a simple pole when m = 1; thus, if Zo is a simple pole, a number ro exists such that

(4)

(A_1 =1= O,0< Iz - zol < ro).


[SECo 59

175

The function exp (I/z) has an essential singular point at the origin, in view of its representation (6), Seco58. The function

cos z

-=z2

l

00

L:

(- I)n

l

l

Z2

-Z2n=___+-+... Z2 2!

Z2n=O (2n)!

4!

(lzl

> O)

has a pole of order 2 at z = O,where the residue of the function is zero. If a function is not analytic at Zobut can be made so by merely assigning a suitable value to the function at that point, then Zois a removable singular point of the function. Thus the function sin z

Z2

-=1--+--... z 3!

Z4

(lzl > O)

5!

is analytic when z :Fo. Ir we define its value to be unity when z = O,the function is represented by the above convergent power series for all z and is therefore entire. Consequently z = O is a removable singular point of Z-1 s.inz. Whenfhas a pole of order m at Zo, let us write (5)

(O< Iz - zol < ro).

From the Laurent expansion (1) offabout Zoit follows that Zois a removable singular point of 4>,that 4>is analytic at Zoif we make the definition (6) Then
-

zol < ro, and it can be

(7) This is a useful formula for the residue off at a pole. For a simple pole (m = 1) it becomes (8)

A -1

= 4>(zo) = lim (z -

zo)f(z).

%~%o

-

-

We also note that If(z)1 OCJas z Zo whenever Zo is a pole. Conversely, for a givenj, suppose that a positive integer m exists such that the function 4>,defined by Eq. (5), is analytic at Zowhen 4>(zo)is suitably defined and that this value 4>(zo):F O. Then Taylor's series for 4>shows that f has a pole of order m at Zoand that the residue offthere is given by formula (7), or by (8) if m = l. For example, if

e-Z

e-Z

-

l

f(z) = Z2+ n2 = z + ni z - ni

176


SECo 59]

then corresponding to the singular point z = ni, 4>(z)=

e-Z ~

z+m

andm = I,sothatniisasimplepole. Here4>(ni) ==e-1tij(2ni);thustheresidue of j at this pole is ij(2n). Let functions p and q be analytic at Zo, where p(zo) # O. Then the function

= p(z) =

j(z)

(9)

q(z)

p(z)

q(zo) + q'(zo)(z

has a simple pole at Zo if and only if q(zo)

-

zo) + . . .

= O and

q'(zo) # O. The residue

of j at. the simple pole is given by the formula (10)

A-l

p(zo) . q Zo)

= -,-(

Ir q(zo) = q'(zo) = O and q"(zo) # O, then Zo is a pole of j of order 2, and conversely. SimilarIy for poles of higher order. Since formulas for residues in terms of p and q are awkward when m > 1, we rely on formula (7) or the Laurent expansion instead. PROBLEMS 1 . When e is the boundary of a square with opposite vertices at the points z = Oand z = 1 + i, evaluate the integral f e Z2dz directly in terms of realline integrals to show that its value is zero. 2. When e is the circle Izl = 2 described counterclockwise, use Cauchy's integral theorem or Cauchy's integral formula to prove that (a) (e) 3.

r~

dz = O; Jcz + 9 Z2 + 4

i_ cZ-

1 dz

= IOni;

(b) fe sec~dz = O; (d)

i_

sinhz .dz = n. 2 cZ + 1tI

Establish these expansions in the regions indicated:

1 ao = I (-l).z" 1 + z .=0

(a) -

1

(b) ~= 1 z

-

ao

I .=0

Z2.

ao

(e) sinhz = Z

(d) z

(Izl < 1);

Z2.-¡

I n=1 ao

1

- 1 = .~o?

(lzl < 1);

(lzl<

(0);

(lzl> 1).


[SECo 60

177

4. Showthat the functiontan z is analyticexceptfor simplepoles at z = :t(n - t)1t and that its integral around the square bounded by the lines x = :t 2, Y = :t 2 has the

value -41ti. 5.

Show that the integral of tanh z around the circle Izl

= 21t has

the value 81ti.

6. Showthat the functionssin(I/z) and cos(l/z) are analyticexceptat z = Oand that z

= Ois an essential singular point.

7.

Ans. 1; O.

Find the residues of these functions at their singular points:

(a) zz + - l.l' 8.

Find the residues there. 1 (d) Z3cosh?;

1 . (b) sinz'

Ans. (a) 2; (b) :t 1; (e) -sin Zo; (d) t; (e) O. If e is the circle Izl = 8 described counterclockwise, show that

_21 '

i

eZl

-:--

1tI e Sinh z

dz

= 1-

2 cos 1tt + 2 cos 21tt.

9. If e is the circleIzl = 4 described counterclockwise, show that

10.

Show that the singular point z

= Oofthe

function 1

f(z) = sin(1tlz) .

is not isolated. 11. With the aid of the Cauchy-Riemann conditions prove that the components u and v of an analytic function f are harmonic functions, that is, they are continuous with continuous partial derivatives up to the second order and satisfy Laplace's equation

60

BRANCHES

OF MULTIPLE-VALUED

FUNCTIONS

When z :f=O,the function

(1)

zt = ~(cos~

+ iSin~)

has one value corresponding to a particular choice of Oand a second value when that argument is increased by 2n. Those two values of zt, which differ only in algebraic sign, are the only possible values for a given z; thus the function (1) is double-valued. A branch of a multiple-valued function f is a single-valued function that is analytic in some region and whose value at each point there coincides

178

SECo 60]


with one ofthe values oflat the point. The function (2)

11(Z)

= ~(CoS~ + iSin~)

(-n < O< n, r > O),

for example, is a branch of the double-valued function (1). The definition of the derivative can be used to show that 11 is analytic everywhere exce~t at the origin and points on the negative real axis. Since 11(z) tends to i'¡-;' when O-. n and to - i~ when O-. -n, the function has no limit as z -. -r (r> O). Thusl1 cannot be defined on the negative real axis so as to make the function continuous there, and the ray O = n must be excluded from the region of analyticity (Fig. 66). The negative real axis is called a branch cut of 11' a boundary that is needed to define the branch in the greatest possible region. Note that each point of the branch cut is a singular point of 11 that is not isolated. The origin is a branch point, a point associated with the multiple-valued function in this way: each branch of the function has a branch cut running from that point. Since the function (1) is double-valued unless the range of Ois limited . to the value 2n or less, its branches must have cuts running out from the ongm. A second branch of the function (1) with the same cut as 11 is (3)

liz)

=

~(cos ~ + i sin~)

(n < O< 3n, r > O).

Here 12(Z)= -11 (z). The number of branches is unlimited. A branch with the positive real axis as its branch cut is

(4)

liz) = ~(cos~ + iSin~)

(O< O < 2n, r > O);

one that has the positive y axis as a cut is the function (1) wi;:hthe restriction n/2 < O < 5n/2, r > O; etc. A branch of the n-valued function zl/n (n = 2,3, . . .) with a ray O = 00 as the branch ~ut is described by the conditions (5)

zl/n =

~(cos ~ + i sin~)

y

o

x

Fig.66

(00 < O < 00 + 2n, r > O).

FUNCTIONS

OF A COMPlEX

VARIABLE

[SECo 61

179

For this function it can be shown that d

1 ZI/.

_(ZI/') = -n -.z dz

The logarithmic function log z, the inverse of the exponential function, is infinitely multiple-valued. It can be written (6)

log z

= log

(reiO)

= Log r + iO,

where Log r denotes the real natural logarithm of the positive number r. The branch known as the principal value of log z is the function (7)

Log z = Log r + i(}

(

-n<

()

< n, r > O),

whose branch cut is the negative real axis. Its derivative is l/z. When e is a complex number, we define zCthus: (8)

ZC= exp (clog z)

(z :/=O).

Unless e is an integer, the function ZCis multiple-valued. Ir e = m/n where m and n are integers and n :/=O, :t 1, ZCbecomes the n-valued function zm/.. Branches of ZCcan be written by using branches of log z in definition (8). The inverse trigonometric and hyperbolic functions, which can be written in terms of logarithms, are further examples of multiple-valued functions.

61

ANAL YTIC CONTINUATION

Ir a function is single-valued and analytic throughout a region, it is uniquely determined throughout the regio n by its valúes over an are, or over a subregion, within the given region. That theorem can be proved with the aid of Taylor's series. Letfl denote a given function analytic in a region R l' and let R be a greater region containing R1. A functionf, analytic throughout R and equal to fl whenever z is in R 1, may exist. Ir so, there is only one such function in view ofthe foregoing theorem. The functionfis called the analytic continuation of fl into the larger region R.

As an example, let fl be definedby this power series: 00

fl(Z)

= .=0 I z'

(lzl< 1).

Then f¡ is analytic in the region Izl < 1,the region of convergence ofthe series; but the function is undefined for all other values of z because the series diverges whenever Izl ;;;;;1. The series is the Mac1aurin series representing the function (1 - Z)-1 in the region; thus f¡(z) = (1 - Z)-1 when Izl < 1.

180

OPERATIONAl

SECo 61]

MATHEMATICS

This second representation of f1 discloses the analytic continuation 1 f(z)

(z ~ 1)

= 1- z

off1 into the entire z plane, excluding the point z = 1,becausefis everywhere analytic except at that point and f(z) = f1 (z) when Izl < 1. As another example, consider the Laplace transform f1(Z)

= {O e-zl dt = 1'" e-XIcos yt dt - i {O e-J
The Laplace integral here exists only if x > O,where its value is l/z. Thus f1(Z) = l/z (x > O),and the function is analytic in that right half planeo But since l/z is ana1ytic everywhere except at the origin, the ana1ytic continuation of f1 out of the half plane is the function 1

f(z)

(z~

=z

O).

We also note that ifthe Laplace integral were known to represent an analytic function in the half plane and to have the value l/x when z

=

x there, then

the integral represents l/z there because the latter function is analytic and has the values l/x on the x axis. Finally, consider the branch (2), Seco 60, of the function zt. In the half plane y > O it is given by the conditions

f1(Z)

(O < 8 < n, r > O).

= fi(cos~ + iSin~)

It is not analytic on the negative real axis. The function

[(z)

=

fi(cos ~ + i sin~)

(O < 8 < 2n, r > O)

is ana1ytic except on the positive real axis, and its values coincide with those off1 in the half plane y > O. Thus f is the analytic continuation of f1 downward across the negative real axis. The following principie of rejlection is easily established with the aid of the Cauchy- Riemann conditions. Let a function w = f(z) be analytic in .

some region R that includes a segment of the x axis and is symmetricwith respect to that axis. Hf(x) is real whenever x is a point on that segment, then (1)

f(z) = w

wheneverz is in R. Conversely,if conditionl!lis satisfied,thenf(x) is real. Condition (1) can be written f(z) = f(z), or f(z) = f(z).


[SECo 62

181

As examples, the entire functions Z2, eZ, and sin z are real when z is real, and the complex conjugate of each function is the same as the function of Z. But the entire functions iz and Z2 + i are not real when z = x; their conjugates are different from iz and Z2 + i. 62

IMPROPER CAUCHY INTEGRALS

In the following chapter we shall use extensions of the Cauchy integral formula and Cauchy's integral theorem in which the closed contour C is replaced by a straight line x = y, the boundary of a half plane x ~ y over which the functionf is analytic. The following proof of those extensions will serve to illustrate a prominent type of application of contour integration. First we introduce the notion of order of magnitude of a function of z for large Izl. Order conditions serve our purpose more fully than the concept of analyticity at the infinite point of the complex planeo A function f is of the order of i', written (!J(i'), as Izl -+ ex:>in a specified part ofthe z plane, if positive numbers M and ro exist such that Iz-kf(z)1 < M there whenever Izl > ro; that is, when Izlis sufficiently large,

Theorem 1 Let f be analytic over a right half plane Re z ~ y and (!J(Z-k) there as Izl -+ 00, where k > O. Then ifRe Zo > y, f(zo) is given by the improper Cauchy integralformula (1)

f(zo)

=

l

Y+;oo f(z) -dz

2m fY-;OOz ;

Zo

= --

l

oo

f(z)

-dy, 21tf-00 Z - Zo

where the integration is along the line x = y; thus z = y + iy. To prove the theorem, let CR denote the arc x ~ y of a circle Izl = R where R > Iyl and R > IZol,so that the point Zo is interior to the region bounded by the arc and the line x = y (Fig.67). Ifwe write p = JR2 - y2, y -y+i/3

R

Fig.67

-y-i/3

x

182


SECo 62]

then according to Cauchy's integral formula, (2)

f(zo) =

~ [f 2m

f(z)

CRz

-

dz

-

Zo

j

],

Y+ifJ f(z) dZ y-ifJ Z - Zo

where C R is described in the positive sense.

Whenpoint z is on CR,then Iz - zol~ R - IZol and, since If(z)1< Mlzl- k when Izl = R and R is sufficiently large, it follows that f(z)

I

z

-

< Zo

1

M RkR

1

- IZoI

whenR>ro.

The integrands of the integrals in formula (2) are continuous. CR is less than 2nR; therefore (Sec. 56)

I

f

f(z)

-

CRz

-

2nRM

d

Zo z

The length of

2nM

< Rk(R- IZol)- Rk(l - IZolR -1)

I

and the last member clearly tends to zero as R -+ 00,since k > O. It also

tends to zero as p -+ 00, because R2 = p2 + y2. Sincef(zo) is independent of p, we see from formula (2) that

(3)

f(zo) = ---;

.

1

bm

j

Y+ifJ f(z)

-

dz;

2m fJ~oo y-ifJ Z - Zo

1 = --

.

l 1m

2n fJ~ 00

fJ

[J

f(y + iy) d . y

o Y + IY - Zo

+ Jp

f(y

.-

iy) d y

o Y - IY - Zo

].

The integrands of the last two integrals are of the order of 1/1+ 1 as y -+ 00,and sincek + 1 > 1, the improper integrals, as p -+ 00, exist and (4)

f(zo) =

-~

2n

[i

oo

f(Y. + iy) dy +

o y + IY - Zo

I

o

]

f(Y. + iy) dY

-00 y + IY - Zo

,

This is the extension (1) of Cauchy's integral formula. Formula (1) states that the integral of the function cP(z)=

z

f(z)

-

Zo

along the line x = y is the residuef(zo) of cPat its singular point Zo. The proof was based upon the analyticity of cP(z)(z - zo) throughout the half plane and the fact that cPis (f}(l/zk+ 1) there.

But Taylor's

series.,representation

of the

analytic functionf about the point Zo shows that Zo is a removable singular point of cPif and only iff(zo) = O. Thus if cP is also analytic at zo, its integral along the line x = y is zero, the value f(zo), where f(z) = (z - zo)cP(z), and we have established the following extension of the Cauchy integral theorem.


[SECo 62

183

Theorem 2 Let cjJbe analytic over a half plane x ~ y and (9(Z-k) as Izl -+ 00 there, where k > l. Then Y+iOJ

OJ

f

(5)

f

y-iOJ cjJ(z)dz=i

-OJcjJ(Y+

iy)dy=O.

The proof of formula (5) applies as well when cjJis analytic and (9(Z-k) in the left half plane x ~ y, if k > 1. Also, the residue theorem (Sec. 59) can be extended in the same manner to the case in which cjJhas a finite number of singular points interior to either half planeo For the left half plane that extension of the residue theorem can be stated as follows. Theorem 3 Let cjJ be analytic when x ~ y except for n singular points z 1 , Z 2, . . . , zn interior to that half plane, where cjJhas residues P 1 , P2 , . . . , Pn. Then if for some real numbers M, k, and ro, where k > I and ro exceeds the greatest of the numbers Iz11,Iz21,. .. ,Iznl, cjJsatisfies the order condition IcjJ(z)1< Mlzl-k when Izl > ro in that half plane, it is true that Y+iOJ

f

(6)

.

Y-IOJ

cjJ(z) dz = 2ni(pl + P2 + . . . + Pn).

As examples, consider first the function f(z) = e-z/z whichsatisfies the conditions in Theorem 1 when y = 1, for it is analytic over the halfplane x ~ I and (9(I/z) there because I

e-1

when x ~ 1.

~ -g-

If(z)1 = Gje-x

Thus formula (1) applies to f, giving the integration formula

f

l+iOJ

e-Z

1- iOJz(z - zo)

The function

y

cjJ(z)

= 1; therefore

f

(7)

= z-2

-zo

when Re Zo > l.

satisfies the conditions in Theorem 2 when

1+iOJdZ

l-iOJ

dz = --e2ni Zo

"

2 = Z

I

f

dy

OJ

-OJ (1

. 2 = O. + IY)

When t is real and t ~ Oand y = 1, the conditions in Theorem 3 are satisfiedby the function eZl/(Z2+ 1)with simple poles :!:i; therefore (8)

1

1 + OJ

eZt

eit

-dz=---=smt 2ni f 1- OJZ2 + 1 2i

-

e

- it 2i

.

(t ~ O).

As we shall see in the following chapter, the left-hand member of formula (8) is an integral representation of the inverse Laplace transform of 1/(s2+ 1).

184


SECo 62]

PROBLEMS 1.

Use Maclaurin's series for cos Z to prove that the function cos.j;

2.

Point out why the function

and entirewhendefinedas l at z = O.

gl(Z)

= exp(-~cos~)[

is single-valued

cos(~sin~) - isin (~sin~)J

where r > O and O< () < 2n, is a branch (Sec. 60) of the double-valuedfunction exp(-.j;) that is analyticeverywhereexceptat pointson the realaxiswherex ~ O. 3. The branchg1ofexp(-.j;) describedin Probo2 is not analyticalongthe positive real axis. Showwhy its analyticcontinuation from the first quadrant into the fourth quadrant is the branch g2(Z)

= exp(-~cos~)[

cos(~sin~) - isin (~sin~)J

wherer > Oand -!n ~ ()~!n. 4. (a)If)/ > O,provethat the branch g2ofexp( -.j;) describedin Probo3 is (!)(z-~as IzI--. ex)in the half plane x ~ )/ for each fixedpositivenumber k. [Note that Ig2(z)1 < exp(-..fifi) there.] (b)ApplyTheorem2, Seco62,to showthat

f

Y+lOO z"g2(Z) y-loo

5.

dz

=

(n = 0,1,2,...;)/ > O).

O

Show that the complex-valued Laplace transform of sin t, (Re z > O),

fl(z) = L"" e-Z'sintdt

is analytic over the halfplane x > Oand that its analytic continuation into the rest ofthe

z plane,excludingthe points :t i, is the functionfez) = 1/(Z2+ 1). 6.

.

Use Theorem 1, Seco62, to prove that, if k is real and positive,

f-100~-_?:!. Z2- k2-

(k > O).

lOO

k

Verifythat resultby integratingwith respectto y. 7. In the half planex ~ l showthat coshx -< Itanhzl -< -:-h

-

s10 x-tan

-

l

h l'

and that the function Z-1 tanh z satisfies the conditions in Theorem 1, so that

--

l

f

l +ioo

tanh z

dz=-tanh

2ni 1-100z(z- zo)

Zo

Zo

ifRe Zo > 1.

8. ApplyTheorem2 to prove that I +100e"

S1-100

2dz Z

= O

ift ~ O.

FUNCTlONS

OF A COMPlEX

VARIABLE

185

[SECo 62

9. Show that the integral (8), Seco62, vanishes if t ~,O. 10. Use Theorems 2 and 3 to prove that, if'Y> O, 1 ;

f

Y+iOO

Z

2m y-ioo(z

e"

z dz

+ 1)

-

-

ift ~ O, ift ~ O.

!sin t - tt cos t {O

Thus when t ;¡;;O,the left-hand member represents L -1{(SZ + l)-Z}. 11.

Prove Theorem 1, Seco62, by using open rectangular contours CfJwith sides

along

the lines x

= p, y =

::!:P (Fig. 68) in place

of the circular

arcs C R .

y

.

+iP lP+iP

o

'Y

+-. c¡¡

{3-i{3

Fig.68 12. Use the residue theorem (Sec. 59) and give details of the proof of Theorem 3, Seco62.

6 The Inversion Integral

We shall now extend our theory ofthe Laplace transformation by letting the letter s in the transform f(s) represent a complex variable. As before F(t) represents a real-valued function of the positive real variable t; but the transformf(s) can assume complex values. We shall see that properties ofthe transformation airead y obtained by assuming that s is real carry over to the case in which s is complex. 63

ANALYTIC TRANSFORMS

When s is a complex variable,

s = x + iy, the Laplace transform of a real-valuedfunction F(t), (1) 186

THE INVERSION INTEGRAL

[SECo 63

187

can be written in the form

(2)

f(s)

= u(x,y) + iv(x,y),

where the components u and v are real integrals:

(3)

u(x,y) = {O e-XI cos ytF(t) dt, v(x,y) =

-

{O e-xI sin ytF(t) dt.

Since the integrands of the integrals (3) are the real and imaginary coefficientsof the integrand e-SIF(t)of the Laplace integral (1),we note that (4)

le-XI cos ytF(t)1 ;;¡:;; le-SIF(t)l, 1- e-xI sin ytF(t)1 ;;¡:;; le-SIF(t)1.

These inequalities will be useful in discussing the convergence of the Laplace integral. Under broad conditions on F its transform f(s) is analytic over a right half plane and bounded there. Let F be sectionally continuous over each

intervalO < t < T and of exponential order as t and a exist such that IF(t)1< Mea.1when t x ~ xo, where Xo > a, (5)

~

- oo. Then constants M

O. Consequently

whenever

le-SIF(t)1= e-X/IF(t)1< M exp [-(xo - a)t].

The integrands of the real-valued integrals (3)are continuous functions of x, y, and t except for possible finitejumps of F(t). Their absolute values are less than M exp [ -(xo - a)t] when x ~ Xo, in view of inequalities (4) and (5). According to the Weierstrass test, then, the integrals (3) converge uniformly with respect to x and y on the half plane x ~ Xo. Thus they represent con-

tinuous functions u(x,y)and v(x,y) there (Sec.15). When F is (!)(~/), we noted earlier (Sec. 19) that tF(t) is (!)(expXlt) whenever Xl > a. It follows that (6)

uix,y) =

{D e - XI(

(x > a), t) cos ytF(t) dt

because the integral here converges uniformly with respect to x and y when x ~ xo(xo > a); also Uxis continuous there. From the second of Eqs. (3) we see that the integral (6) also represents v.(x,y); hence u and v satisfy the first Cauchy-Riemann condition over the half plane x > a. In the same manner we find that the second condition Uy = - Vxis satisfied there. Those conditions along with the continuity of the first-order partial deriva tives of u and v establish the analyticity of f over the half plane x > a (Sec. 54).

188


SECo 64]

Now l' = Ux+ ivx wherever f is analytic; thus when X> rx

I'(S) =

-

foO:> te-xt(COS yt - i sin yt)F(t) dt

= L{ -

tF(t)}.

Since tF(t) is also sectionalIy continuous and of exponential order, our result applies as welIto that function; thus (x>

rx),

and so on for ¡ o:> M l/(s)1

~

f.o

e-xtIF(t)1 dt < M

f.o

e-(xo-cr)t dt

= -.

Xo

-

rx

SimilarIy, we find that II'(s)1 < M/(xo - rx)2,etc., so the derivatives of f are bounded over the half planeo FinalIy we note that f(s) is real when s = x because F(t) is real, and it folIows from the principie of reflection (Sec. 61), or directIy from Eqs. (2) and (3), that f(s) = f(s). Our results can be stated as folIows. Theorem 1 Let a real-valued function F be sectionally continuous over each intervalO < t < T and (9(ec1t) as t -+ oo. Then its transform f(s)

= {O:>e-stF(t) dt

= L{F}

(s = x + iy)

is an analytic function of s over the half plane x > rx. Its Laplace integral is absolutely and unlformly convergent over each half plane x ~ xo(xo > rx),and f and its derivatives are bounded there. Also, (7) (8)

¡
(n = 1,2,...; x > rx), (x>

rx).

Formula (7) was derived in Seco19 when s is real. The conditions in Theorem l can be relaxed. For instance F may become infinite at t

= to in such

a way that (t

-

to)kF(t) remains bounded

as

t -+ to, where k < 1. Then the conclusions in the theorem are still valid. As .an example, the transform of t-t is analytic in the half plane x > O, and formula (7) applies to it.

64 PERMANENCEOF FORMS We have seen that the Laplace integral of F(t) represents a functionf(s) that is analytic in a half plane x > ex,where s = x + iy. Ir the integration is


[SECo 65

189

when s = X, a real function cp(x)is obtained that is identical with J(s) to the right of point s = exalong the real axis; that is, cp(x)= J(x) when x > ex.Ir cp(s)is an analytic function in the half plane, then it must be identical with J(s), since two different analytic functions cannot be identical along a line (Sec. 61). It follows that transforms can be found by carrying out the integration as if s were a real variable. That the function J(s) so found is analytic when Re s > excan be seen in the particular cases; but it is true in general because the integration formulas are the same whether the parameter in the integral is complex or real. The transform of t2, for instance, was found to be 2s- 3 when s is real. Now t2 is (9(e O.Therefore, L{ t2} = 2s - 3 for all complex s in the half plan e x > O. All our transforms of particular functions, tabulated in Appendix A, Table A.2, are valid when s is complex. We seldom need the value of ex which determines the half plane in which s lies; the existence of the number ex usually suffices. The operational properties of the transformation developed in the first two chapters and tabulated in Appendix A, Table A.t, are likewise valid when s is a complex variable in some half plane x > ex. For the sake of simplicity, we may make an exception of operation 10, Appendix A, Table A.l:

performed

LHF(t)}

= 1(%)J(J.)dJ.,

and agree that s is real here. The permanence of the forms of those properties is again a consequence ofthe fact that the steps used in their derivations are independent ofthe real or complex character of the parameter s. However, the derivations could be rewritten when s = x + iy by taking the corresponding steps with the real integrals (3), Seco63, that represent u(x,y) and v(x,y). 65

ORDER PROPERTIES OF TRANSFORMS

When Isl -+-00 in a half plane of convergence of the Laplace integral, the behavior of the transformJ(s) is governed by regularity properties of the object function F. Theorem Z Let a jÜnctíon F as we/l as íts derívatíve F' be sectíonally continuous over each bounded intervalO < t < T and let F itself be (9(elXt). Then in any half plane Re s ~ Xo where Xo > ex,its transJormJ(s) tends to zero as Isl -+- 00: (1)

lim J(s) = O

Isl-+(%)

(Res ~ Xo > ex).

190

SECo 65]


For points s

=x

each positive number

+ iy in the half plane x ;¡;;Xo we are to prove that for E

there is a number '. such that

(2)

/f(s)1 <

wheneverIsl> '. and x ;¡;;Xo.

E

The Laplace integral of F converges uniformly with respect to s in the half plane x ;¡;;Xo (Theorem 1), so the remainder for that improper integral can be made uniformly small in absolute value for all s there. That is, to the given number E there corresponds a number T. independent of s such that

(3)

ILooe-SIF(t)dt!

;;; I

f:' e-SIF(t)dtl

+ IR(s,T.)1

and IR(s,T.)I < tE for all s in the half planeo Now we shall determine '. such that the first term on the right of condition (3) is less than tE when Isl > '.. The interval (O,T.) consists of a finite number of subintervals interior to each of which F and F' are continuous and have limits from the interior at the end points. Consider a subinterval (O,TI) as a sample. Over it we can apply integration by parts to the complex-valued integral (Sec. 56); thus TI

-s

fo

TI

e-SIF(t)dt

(TI

= e-S'F(t)J o - Jo e-S'F'(t)dt.

The absolute value ofthe right-hand member does not exceed the function (TI

IF(O)I

+ exp (-xT.)IF(TI)1 + Jo e-xtlF'(t)1dt

and, since x ;¡;;Xowhere Xo may be negative, and F' is bounded, IF'(t)1< NI, this function is less than MI, where MI = IF(O)I+ exp (lxolT.) [IF(T¡)I+ N ¡T¡]. Similarly for the remaining subintervals of (O,T.). Thus a number M exists such that (4)

IsllLT e-S'F(t)dt/ < M

when x ;¡;;Xo.

Then from condition (3)weconc1udethat iflsl > 2M E

'

that is, condition (2) is established by selecting r. as 2M/E. Weaker-order properties onf(s), such as the boundedness of xf(x + iy), can be derived by assuming only that F itself is sectionally continuous and of exponential order (Probs. 7 and 8). Stronger ones will now be noted in case F or its derivatives are continuous and of exponential order.


191

[SECo 65

Theorem 3 If F is continuous and F' and F" are sectionally continuous over each intervalO ;;¡;t ;;¡;T while both F and F' are (9(e"'t),then lim sf(s) = F(O)

(5)

1.1

when x ~ Xo > oc;

00

in particular,f is &(1/s) in that half plane: (6)

(x

Isllf(s)1 < M

~

Xo

>

oc).

Under the conditions stated we know that (7)

L{F'(t)} = sf(s)

-

F(O).

Since F' satisfies the conditions on F in Theorems 1 and 2, then L{ F'} is bounded over the ha1f p1ane x ~ Xl and tends to zero as Isl -> 00 there. Therefore sf(s) is bounded and, since it equa1s L{F'} + F(O), its limit as Isl -> 00 exists and is given by formula (5). . As illustrations the functions 1,.cos t and t satisfy the conditions in Theorem 3 whenever oc> O. Their transforms l/s, S/(S2+ 1), and 1/s2 are (9(l/s) and satisfy condition (5). In fact the 1ast of those transforms is (9(1/s2), a conc1usion that follows from properties of the function t with the aid of the following extension of Theorein 3. Theorem 4 If F and F' are continuous and F" and F'" are sectionally continuous over each intervalO ;;¡;t ;;¡;T while F, F', and F"are (9(e"'t),then s2f(s) - sF(O)is bounded over any half plane x ~ Xo where Xo > oc;in fact (8)

lim [s2f(s) 1.1

-

sF(O)] = F'(O)

(x ~ Xo > oc),

00

and the additional condition F(O) = O is then adequate and necessary for f(s) to be (9(1/S2)in the half planeo Since our formula for L{ F"} is valid here, then L{F"(t)} + F'(O)= s2f(s) - sF(O)

In viewofTheorems 1 and 2 the function on the left is bounded and has the limit F'(O)as Isl-> 00, so the same is true for the function s2f(s) - sF(O). The second component -sF(O) of that function is bounded if and only if F(O) = O, in which case s2f(s) is bounded

over the half planeo

Extensions ofTheorem 4 are evident. For instance,f(s) is (9(1/s3)over the half plane if F, F', and F" are continuous and the next two derivatives are sectionally continuous, while F and its first three derivatives are (9(e"'t) and F(O)= F'(O)= O;alsos3f(s)-> F"(O)as Isl -> 00 in the half planeo

192


SECo 65]

PROBlEMS 1 . State why none of the following functions can be Laplace transforms of real-valued sectionally continuous functions of exponential order (Theorem 1). (a) s;

1 (e)~. S+I

1 (ti) 7; S

(b) ssins + l'.

2. Use Theorem 2 to show that none of the following functions can be Laplace transforms of a function F of exponential order such that F and F' are sectionally continuous. (a) 1;

s (b) s - 2;

3. Determine an order property of the transform f(s) from the character of the function F in each of the following cases, and verify by writing f(s). (a) F(t) = sin t; (b) F(t)

= cosh

t; (e) F(t)

=

t sin t; (ti) F(t)

= (t

- 1)80(t - 1).

l l 1 Ans. (a) (!J2' ~ Xo > O. s x ~ Xo > O; (b) (!J-,x s ~ Xo > 1; (e) (!J3'X s 4.

Ir a function F is sectionally continuous over a bounded interval O < t < T and

F(t)

= O whenever

t > T, state why F(t) is (!J(e''')for every real IX,and why its transform

f(s) is an entire function of the complex variable s. 5. The step function l - 80(t - 1) satisfies the conditions on F in Probo 4. Show directly that its transform can be written f(s)

= l - s e-s

whenever s # O,and feO) = 1;

then represent f as a power series in s convergent for all s to verify that f is entire. 6. Ir a function F is sectionally continuous over some interval O < t < T but continuous when t ~ T, and ifF' is sectionally continuous over every such interval and both F and F' are (!J(e"'),use the formulafor L{F'} to prove that the transformf(s) is (!J(I/s) over the half plane Re s ~ Xo when Xo > IX. 7. Ir F is sectionally continuous over each interval O < t < T and (!J(e"'), prove that its transform f(s) satisfies the condition that xf(x + iy) is bounded over the half plane x ~ Xowhen Xo > IX;consequently lim f(x + iy) x-a>

= o.

8. According to the Riemann-Lebesgue theorem in the theory of convergence ofFourier series,l if a function G is sectionally continuous over an interval a < t < b, then lim y

co

f G(t)cos yt dt = lim f b a

y

b oo a

G(t)sin yt dt = O,

a result that is plausible if graphs of the integrands for large values of y are considered.

, See the

author's

"Fourier

Series and Boundary

Value Problems,"

2d OO.,p. 87, 1963.


[SECo 66

193

(a) If G is sectionallycontinuous over each intervalO < t < T, and absolutely integrable from Oto 00, extend the above result to improper integrals by showing that

i

lim

«> e-i)'tG(t)dt

1,1-«>o

= O.

(b) Write G(t)= e-X'F(t)where F is sectionallycontinuous over each interval (O,T) and l!I(eat)to prove that the transform f(s) of F satisfies this condition when x is fixed : lim

f(x + iy)

1,1-«>

66

=O

(x > a).


According to our extension ofCauchy's integral formula (Sec. 62), a function f(s) that is analytic and (!J(S-k) over a half plane Re s ~ y, where k > O,can be represented in terms of its values on the boundary of the half plane by a line integral: 1 Y+iOOf(z) (1) f(s) = ---; dz (Res> y).

I

2m y-ioo s - z

Supposef(s) = L{F(t)}. Ifwe formally apply the inverse Laplace transformation to the functions of s in formula (1) and interchange the order of the operation L - 1 and the integration, we find that oo . 1 Y+iOO eyl (2) F(t) = ---; e'Zf(z)dz = e'Y'f(y + iy) dy. 2m

I

f

y-ioo

21t -

.

This is an integral transformation of f(s) that may represent in a direct manner the inverse Laplace transform of that function. The improper integral(2)existsifand onlyifthe integrals of eiy'f(y + iy) from - 00to Oand from Oto 00 both existoFor some transformsf(s) those improper integral s fail to exist, at least for certain values of t, while the Cauchyprincipalvalueof the integral (2),

f

(l.

eyl

_ 2

(3)

lim

e'Y'f(y+ iy) dy,

1t (1 00 -(1

does existo This is illustrated in Probo 7, Seco70. Ir the improper integral (2) does exist, its principal value exists and is equal to the improper integral. The principal value (3) is called the complexinversionintegral for the Laplace transformation. We use the symbol Li - 1 for that linear integral transformation of f(s), a symbol that is intended to suggest an integration as well as an inverse Laplace transformation; thus (4)

1

Li

Y+i(l

- 1 {J(s)} = ---: lim 1 i(l e'Zf(z)dz. 2m (1 Y-

194

SECo 66]


y -y+i(3

z=-y+iy

o

x

-

-y i(3

Fig.69

We assume that f(5) = f(s), a condition that is satisfiedby transforms of real-valued functions F(t) (Sec. 63). Then although the inversion integral is an integral in the complex plane along the line x = y (Fig. 69), we can write it as a real improper integral.

Since z

=

y + iy, where y is fixed, the integral

in Eq. (4) can be written as

.

ieyt[f~f/ éY'f(y + iy)dy + I: eiy'f(y + iY)dY]. When the variable of integration

in the first integral here is replaced by

the expressioninside the brackets becomes

I: [e-iy'f(y

- y, .

- iy) + eiY'f(y + iy)Jdy.

The integrand of this last integral is 2 Re eiy'f(y + iy), because the complex conjugate of eiy'f(y + iy) is e-iy'f(y - iy). We write f(x + iy) = u(x,y) + iv(x,y);

then Re [eiY'f(y + iy)J = u(y,y) cos yt - v(y,y) sin yt and therefore eyt

(5)

Li -1 {J(s)}

=-

oo

f

¡¡; o

[u(y,y)cos yt - v(y,y)sinytJdy.

When f(s) satisfies certain conditions, we shall see that the value of the real integral (5) is independent of y for all values of y greater than a prescribed number. Even for simple functions f, however, the integration involved in the real form (5) is generally difficult. To evaluate the inversion integral, we shalI use the complex form (4) in conjunction with the theory of residues or processes of changing the path of integration.


[SECo 67

195

In the following sections conditions on either J or F will be established under which the inversion integral represents the inverse transform F(t) of J(s). Other formulas for the inverse transformation are known; 1 but the inversion integral (4) has greater general utility than the others. 67

CONDITIONS

ON I(s)

The following theorem gives conditions on a function J that are sufficient to ensure that L¡- 1 {J(s)} exists and represents a function F whose transform is J(s). Useful properties of F(t) also follow from the conditions on J(s), properties that give the initial value F(O), the continuity and exponential order of F. Theorem 5 Let J be any Junction oJ the complex variable s that is analytic and (9(S-k) for all s(s = x + iy) over a half plane x ~ ex,wherek > 1;

alsolet J(x) bereal whenx ~ ex.ThenJorall real t theinversionintegral oJJ(s) along any Unex = y, where y ~ ex,convergesto a real-valued Junction F that is independentoJy, (1)

(Id < ex),

whose Laplace transJorm is the givenJunction J(s): (2)

L{F(t)}

= J(s)

(Re s > ex).

Furthermore F(t) is (9(e"t),it is continuous ( - ex)< t < ex), and (3)

F(t)

=O

when t

~ O.

The integrand etzJ(z) of the inversion integral, where z = y + iy and y ~ ex,is everywhere continuous in y and t sinceJis analytic when Re z ~ ex. Thus the integral along each bounded segment Iyl ~ Yo of the line x = y exists as a continuous function of t. According to the order condition on J, positive constants M and Yo exist such that (4)

when Iyl > Yo.

Since k > 1, the improper integral s of Mlyl-k from Yo to ex)and from - ex) to - Yo exist, so the inversion integral exists as an improper integral, and it is equal to its principal value. 1

See books by Doetsch and by Widder listed in the Bibliography.

196


SECo 67] AIso,

since f(x) is real, then, according to the reflection principIe for

analytic functions,f(z) = f(z), and the inversion integral takes its real form 00

eyt

(5)

L¡-l{J(S)} = [u(y,y)cos ty - v(y,y) sin ty] dy. 1t 1o

Now the components u(y,y) and v(y,y) of f(y + iy), as well as f itself, satisfy condition (4). Thus the absolute value of the integrand of integral (5) is less than 2My-k when y > Yo, a function that is integrable from Yo to 00 and independent of t. Consequently, integral (5) converges uniforrnly with respect to t, and it follows that L¡-1{J} is a continuous real-valued function F(t) for all rt.al t. To show that F(t) is independent of y when y ~ el, we use a second path ofintegration x = y',where y' > y. Sinceetzf(z)is analytic when x ~ y, the integral of that function around the boundary of the rectangle ABCD in Fig. 70 vanishes, according to Cauchy's integral theorem. On sides BC and AD, Izl ~ fJand, since If(z)1< Mlzl-\ letz' (z)1 <

M

etx-

J

N(t) fJk

<-

fJk =

where N(t) is the greater ofthe two numbers Mety and Mety'. The length of those sides is y' y; hence the absolute value of the integral of etzf(z) along

-

either side BC or AD does not exceed N(t)(y'

- y)fJ-k,a quantity

that vanishes

as fJ -+ oo. The vanishing of the contour integral around the boundary can be expressed in the form

f

etzf(z) dz

AB

+

i

etzf(z)dz =

CD

f

AD

y

o

x

'Y

A

I

DI'Y'-iP I

Fig.70

etzf(z) dz

- JBC r etzf(z) dz.


197

[SECo 67

The left-hand member tends to zero as P-+ 00 because the member on the right does so. But we have seen that the limit as P-+ 00 of each of the integrals on the left exists when k > 1. It follows that Y+iP

lim p

Y'+IP

f

f

e'Zf(z)dz = lim

oo y-IP

p

e'Zf(z)dz,

oo y'-IP

so the function F defined by Eq. (1) is independent of y.

= IXin

Let us write y

l

f

oo

e-CZ/IF(t)1= -

2n

Eq. (1). Then

I

.

- 00

eUYf(1X+

iy) dy ~ -

l

oo

f

2n - 00

I

If(1X+ iy)1dy.

The last integral exists because f is lP(y-k) as Iy/ -+ 00 where k > 1, and f is continuous. Its value is independent of t. Therefore F(t) is lP(eCZ/) as t -+ oo. To evaluate F when t ~ O,it is convenient to select a positive number for y; this can be done since y can be any number in the range y ~ IX.Then over the half plane x ~ y, le'zl = e'" ~ l when t ~ Oand where k > 1.

le'Zf(z)1< Mlzl-k

It follows from our extension of Cauchy's integral theorem (Theorem 2, Sec.62)that Y+loo

f

.

(t

e'Zf(z) dz = O

~ O);

Y-JC()

that is, F(t) = Owhen t ~ O. In particular, F(O)= O. Finally, the transform of F(t) must exist when Re s > IXbecause F is continuous and lP(eCZ/), so l

_ f

L{F(t)} = 2 lim nT oo

oo

T

o

e-sI

f

-00

e'Zf(z) dy dt

(Re s > IX),

when z = IX+ iy. Since the convergence of the inversion integral is uniform with respect to t, the order of integration can be interchanged: oo T l (6) L{F}=-lim f(z) e-(S-z)ldtdy 2nT

f f f- [

00

o

-00

oo

= .2- lim 2n T 00

00

f(z) - f(z) e-(S-Z)T s - z s- z

]

dy.

The first term inside the brackets here is independent of T and, in view of the extended Cauchy integral formula (Theorem 1, Seco 62), its integral exists and has the value 2n.f(s): (7)

~ 2n

f

oo

-

00

f(rx

s

+ iy!

- rx- ,y

dy = f(s)

(Res>

rx).

198


SECo 68]

The limit as T

-. 00 of the

integral of the second term inside the brackets = a + ib, where a >
in Eq. (6) is zero. For if we write s

f-",s-z '"

I

f(z) e-(s-Z)T dy l

~ e-(~a)T a (Xf'"-'" If(z)1 dy.

The member on the right vanishes as T along the line x

-.

00 becausefis

absolutely integrable

= (x. Thus (Re s > (X).

L{ F(t)} = L{L¡-1[f]} = f(s)

(8)

The proof of Theorem 5 is now complete. 68

CONDITIONS ON F{t)

The foregoing conditions under which the inversion integral formula is valid are severe. They are not satisfied, for example, by the function l/s, which is @(l/sk) where k = 1, nor by transforms of functions F that are discontinuous or for which F(O):f: O. By using a Fourier integral theorem and qualifying the function F instead of f, we can relax the conditions so that the inversion integral formula can be seen to apply in nearly all cases of interest to uso In fact we shall see that our formula is only a modified form of the Fourier integral formula. Let G be a function defined for all real t, sectionally continuous over each bounded interval ofthe taxis, and absolutely integrable from - 00 to oo. We agree to define G(t) at each point to where G is discontinuous as the mean value of its limits from the right and left at to: (1)

G(to)

= ![G(to

+ O)+ G(to - O)].

Also let the derivative G' be sectionally continuous over each bounded interval; this condition actually implies that G itself is sectionally continuous (Prob. 10, Seco70). Then G(t) is represented in terms of trigonometric functions of t by the F ourier integral formula

(2)

1

1 "' '" G(t) = G('r)cos y(t - .) d. dy

f f

1t o 1

= - lim

-00

(-oo
p oo.

f. f -'" 21tP-+00o

. [G(.)e,y(t-f)+ G(.)e-,y(t-f)]d. dy.

Since the integral of IG(.)I from - 00 to 00 exists, the inner integralof each term within the brackets converges uniformly with respect to y to a con1

See, for instance, the author's

115 Ir., 1963.

"Fourier

Series and Boundary

Value Problems,"

2d ed., pp. 83,


[SECo 68

199

tinuous function of y whose integral from y = O to y = p exists. Thus the iterated integral in the second line of formula (2) can be written as the sum p eiyt

f.o

f

co -co

G(r)e-iy
f

o

eiYI

-P

f

co

G(1:)e-iYt d1: dy.

-co

Hence the Fourier integral formula has the exponential form

(3)

G(t) =

~

lim

21tp-co

f f p

eiYI

-P

( - 00 < t < 00).

co G(1:)e-iyt d1:dy -co

Note that the existence of the principal value of the improper integral with respect to y from - 00 to 00 is ensured here, but not the convergence of the improper integral itself. Now let F denote a function defined when t ;¡;;O and (9(eat)as t -+ 00, such that F', and therefore F itself, is sectionally continuous over each interval O < t < T. If we write

(4)

when t > O,

G(t)= O

when t < O,

where y > IX,then G satisfies the sufficient conditions stated above for the representation (3), so

(5)

G(t)=

~

lim

f

p

eiYI

21tP-co -P

f.ocoe-(Y+iy)tF(1:)

d1:dy

(Itl < 00).

The inner integral represents f(y + iy), where f(s) is the Laplace transform of F(t). Ir z = y + iy, it follows that, for all real t, (6)

eY1G(t) =

Y+iP

I

: lim

2m P-co

Jy-ip

e'Zf(z) dz

= Li -1{J(S)}.

When t < O, G(t) = O,and therefore Li -1{J(S)} = O. When t > O,eY1G(t)= F(t), and Eq. (6) represents F(t) as the inversion integral Li-1 {J(s)}. But sincethe function G representedby formula (5)has the mean value (1) at eachpoint to of discontinuity, the inversion integral converges to F(to) at a point to where F has a jump, provided we agree that (7)

F(to) = t[F(to + O)+ F(to - O)]

(to > O).

When t = O,the value of the inversion integral is ![F(+ O)+ O]. The existence of the inversion integral (5)in the form of the principal value of the improper integral along the line x = y was established here, but the improper integral itself may not exist under the conditions stated. The following theorem is now established. Theorem 6 1f f(s) is the Laplace transform of a function F of exponential order (9(eat)whose derivative F' is sectionally continuous over each

200

OPERATIONAL

SECo 68]

MATHEMATlCS

intervalO < t < T, then the inversion integral of f along any line Re s = y, where y > oc,exists and represents F(t):

(t > O).

(8)

At a point to(to > O) where F is discontinuous,the inversion integral has the mean value (7) of F(t); when t = O, it has the value tF( + O), and whent < O,it has the valuezero. The conditions here on F can be relaxed because the conditions on G for the validity of the Fourier integral formula (2) can be modified in various ways.l For instance, on each bounded interval let the function G have at most a finite number of infinite discontinuities but still be absolutely integrable over the entire taxis, where each interval can be subdivided into a finite number of open intervals such that G is monotonic (nondecreasing or nonincreasing) over each one. Then the Fourier integral expression (2) has the value t[G(t + O)+ G(t - O)] at each point t where G(t + O)and G(t - O) exist. As a consequence of those m9difiedconditions, for instance, (y > O,t > O), because the function G(t) = e-ytct when t> O, G(t)= O when t < O, is represented by its Fourier integral formula when t > O and y > O, and because L{t-i} = (n/s)i when Re s> O. Even though its conditions are on F, the function usually sought, rather thanf, Theorem 6 serves as a useful supplement to Theorem 5. The following example and corollary illustrate this. Example Let,j'; denote the branch J;eiOI2, r > O, 101< n/2, where s = reiO.Show that the inversionintegral formula applies,when y > O, to the function

f(s) = On a half plane Re s

f

.

sys+l

~ oc> O, the branch ,j';

s,j'; + 1 never vanishes sin<::eRe (s + there but only (9(1/s), which fails to Theorem 5. By carrying out one step write f as a sum of two components, 1 (9) f(s) = - - S

S2,j';

is analytic

and

l/,j';) > O. Thus f is analytic satisfy the order condition in of a division, however, we can 1

+s

1 See, for instance, H. S. Carslaw, "Fourier's Series and Integrals," Titchmarsh, "Theory of Fourier Integrals," p. 13, 1937.

chap. 10, 1930, and E. C.

"


201

[SECo 69

such that Theorem 5 applies to the second, which is (9(S-5/2)and analytic over the half plane, and real when s is real, while Theorem 6 applies to the first to give Li-l{l/s} = 1 when t> O. Thus Li-l{f} = F(t) where F(t)

= Li-l

~

{s }

+

Li-l

;/

{ S2 s + s}

= 1 + Fl(t)

(t > O),

and F1 is continuous (t ~ O); also F1(0) = O and F1 is (9(e O. ConsequentIy F is continuous when t > Oand (9(e
lf f(s) can be written in the form

(10)

(Res

~

IX)

wherefl satisjies the conditions in Theorem 5 andf2 is the transform of a function F2 satisfying the conditions in Theorem 6, then Li -1{f} is the inverse transform F(t) of f(s). Moreover F is (9(e
UNIQUENESS OF INVERSE TRANSFORMS

In order to make a simple analysis of uniqueness of the inverse Laplace transformation, we consider the class S of real-valued functions on the half Une t > O, defined in Seco6. The class consists of all functions F that are sectionally continuous over each bounded interval, where F(t) is dejined as the mean value of its Umitsfrom right and left at each point of discontinuity, and such that each function F is (9(e«t)for some IX.Then F has a transform f(s) when Re s > IX.Our theorem on uniqueness of inverse transforms can be stated as follows.

Theorem 7 Amongallfunctions of class S no two distinctfunctions canhave the same Laplace transform over a right half plane Re s > ex,or for all real s on a half Une s > IX. Since transforms of functions of class S are analytic functions of s over some right half plane, any two transforms are the same over a half plane if they are identical for all real s on a half line s > IX(Sec. 64). In particular, the theorem shows that no two functions of exponential order that are continuous when t ~ Ocan have identical transforms for all real s on a half line s > a. To prove the theorem, first let S' denote the subclass of all functions F in class S for which F' is also sectionally continuous over each bounded

202


SECo 70]

interval on the half line t > O. According to Theorem 6 if F belongs to e', it is represented by the inversion integral of its transform f That is, when

F(t) is (9(e"'),then F(t) = Li -1 {J(s)} when t > Oregardless of the value ofy in the inversion integral as long as y > IX.In case a function F in e' has the vanishing transform f(s) = O for all s in some right half plane, it follows that F(t) = Ois the only function of class e' whose transform vanishes over a right half planeo Next let G and H be two functions of c1ass $, so that no conditions are imposed on their derivatives. Let both be (9(e"/)and suppose that they have identical transforms g(s): (1)

L{ G(t)} = L{ H(t)} = g(s)

(Re s > IX).

Now the function (2)

F(t) =

f~[G(r)

-

H(!)] d!

(t ~ O)

is continuous and (9(eP/)whenever {3> IX(Sec. 17). Moreover, on each interval where G and H are continuous, (3)

F'(t) = G(t)

-

H(t);

thus F' is sectionally continuous over each bounded interval, and F is therefore in c1ass e'. We can choose {3as a positive number. Then from condition (1) and the representation (2) of F as a special convolution, it follows that 1 L{F(t)} = -[g(s) - g(s)] = O s

(4)

(Re s > {3),

and since F is in e', we conc1ude that F(t) = O when t> O. Therefore F'(t) = O(t> O);that is, wherever G and H are continuous, (5)

G(t)

-

H(t)

= O.

At each point where G or H has a jump, it followsthat the one-sidedlimits of G(t) - H(t) vanish; thus

!(G(t + O)+ G(t - O)] = t[H(t + O)+ H(t - O)]

(t > O).

Consequently G(t) = H(t) whenever t > O,and Theorem 7 is proved.1 70

DERIVATIVES OF THE INVERSION INTEGRAL

When the solution of a boundary value problem is found in the form of an inversion integral Li - 1{J}, it is often possible to verify the solution by noting 1

For theorems on uniqueness that permit L -1{f} to have certain int'inite discontinuities, see

G. Doetsch, "Handuch

der Laplace-Transformation,"

vol. 1, pp. 72 ff., 1950.


[SECo 70

203

properties of the function f The two theorems in this section are useful for that purpose. Their proofs foIlow from Theorem 5 and properties of uniformly convergent integrals (Sec.15). When the inversion integral in Theorem 5 is differentiated with respect to t under the integral sign, we obtain L¡ -1 {sf(s)} :

_1

(1)

f

21/:

oo

-

a

I

oo

f

-;-[etzf(z)]dy = 2 _ 00ut 1/:

etzzf(z) dy, - 00

where z = y + iy. If sf(s) asweIl asf(s) satisfiesthe conditions imposed on fin that theorem, the integral (1) converges uniformly with respect to t and represents F'(t), the derivative with respect to t of L¡ -1 {f(s)}. Then F' satisfies the conditions stated for the function F in the theorem. The additional condition needed here is that sf(s) is @(S-k)where k > 1; that is, that f(s) is @(S-k-l). If we replace f(s) above by sf(s), we see that F"(t) is represented by L¡ -1{S2f(s)}, and so on, giving the foIlowing theorem.

Theorem 8 Let f be a function of s that is analytie and @(s-k-m) over a half plane x 6 ex,where s = x + iy, k > 1, and m is a positive integer. Also let f(x) be real (x 6 ex). Then the inversion integral off along any Une x = y, where y 6 ex,converges to the inverse transform off, a real-valued function F, (2)

(t 6 O),

L¡-1{f(S)} = F(t)

and the derivatives of F are given by the formula (n = 1,2,... , m);

(3)

furthermore, F and each of its derivatives (3) are continuous when t 6 O and @(e"'t), and their initial values are zero: (4)

F(O) =

F'(O)= .. . =

F(m)(o)

= O.

Consider again, as an example, the inverse transform F1 of the second function on the right in Eq. (9), Seco68; F1(t) = L¡-I

{ ;/+ }. S2

s

s

Its transform is @(s-5/2),so m = 1, and it foIlows from Theorem 8 that both F1 and F~ are continuous when t 6 Oand both are @(e"'t)when ex> O,and that F1(O) = F~(O)=

O. .

It wilI be recaIled that formula (3) cannot be valid unless either F or f is subjected to rather severe restrictions. For according to our basic property

204


SECo 70]

of transfonning derivatives,

p 1 and the constants M, k, and IXare independent of p. Also let f(p,x) be real when x ~ IX. Even if p has an unbounded range of values, let f(p, IX+ iy) be bounded for all p and y. Then the inverse transform off with respect to s, which can be written

Theorem 9

(y is a continuous real-valued function

~

IX, t

~ O),

of its two variables p and t(t ~ O)

for whicha constantN independentof p exists suehthat IF(p,t)1< N¿zJ; in particular, F is a bounded funetion of p for eaehfixed t. 1f the partial derivative J;,(p,t) also satisfies the conditions imposed here on f, then

-

(5)

:p F(p,t) = Li - 1{:pf(P,S)},

and IFp(p,t)1< N(/'t. The theorem can be applied to J;,(p,s)to obtain corresponding results for Fpp(P,t), and so on for derivatives of higher order. Applications of Theorems 8 and 9 are made in the folIowing chapters.

PROBLEMS 1. Apply Theorem5 to the functions(a)f(s)

= (S2

- 1)-1 and (b)g(s)= s-2e-s to

show that their inverse transforms F(t) and G(t) are represented by inversion integrals, also to find order and continuity properties of F and G and to show that F(O)= G(O)= O. Verify those properties by writing F and G as known inverse transforms.

2. ApplyTheorem8 to the functionf(s) = S-4 to deduce that the inverse transform F(t) and its derivatives offirst and second order at least are continuous when t ;¡; O,and 19(e"")if IX> O, and that F(O) = F'(O) = F"(O) = O. Verify those properties

that F(t)

= t3/6.

by recaIling


205

[SECo 70

3. Apply Theorem 6 to the transform of the function e-' and simplify the inversion integral, when }' = O,to establish the integration formula

l

eo cos

o

ty + y sin ty d -, y=ne 2 Y + 1

(t > O).

4. When f(s) in Theorem 5 is the transform (s + 1)-2 of te-', simplify the inversion integral, when )' = O,to pro ve the integration formula

l

eo

O - y2)cos ty + 2y sin ty d = nte-t

o

(y2 + 1)2

(t ;;;;O).

Y

~

5. Let denote the branch j;eiBI2, -n < IJ< n, r > O,where s = reiO,an analytic function over each half plane Re s ;;;;IX> O. If f(s) = exp (-~), prove that If(s)1 < exp(-.¡;¡i) overthe halfplane and thatf(s) is l!I(s-m)there for every positive integer m. Show that Theorem 8 applies to establish f(s) as the transform of a function F(t) with continuous derivatives of all order (t ;;;;O)such that F(O) = O

and

(n = 1,2,.. .).

F(n)(o)= O

[Compare the transformation (6), Seco27.] 6.

Use power series to prove that the function g, where

g(z)= 1 - ze-S

if z i' O, g(O) = 1,

is entire and hence bounded over a neighborhood of the origin in the z plane. Then apply our corollary, Seco68, to the function

f(s) = !e-lls = !s - sO - S2e-l/') s to prove that L -1{J} is a continuous function F(t) when t > O,l!I(e.')if IX> O,and that F(+O) = 1. State why the inversion integral formula applies when}' = 1to the known transform for all t; but show that when t = 1 the Cauchy principal value Li-l{J} must be used because the improper integral itself fails to exist for that value of t. AIso verify that, when t = 1, Li-l{J(S)} = t. 8. The entire functiong(s) = exp (S2)is not l!I(S-k)over a right half plane for any positive k. But prove that Li-l{g}, integrated along the imaginary axis, is the function 7.

f(s) = s-le-s

F(t) A known integration

l

= -n

eo

f

o

exp (- y2) cos ty dy.

formula [Eq. (7), Seco74] shows that F(t)

= tn-t

exp (- t2/4), and

thus (transformation 109,Appendix A, Table A.2)

L{F}= L{2fiexp

(-~)}= texp(s2)erfcs

i'

g(s).

.

Thus the inversion integral may exist without representing the inverse transformo

206

9.


SECo 71]

Show why the function G, where when t > O,

G(O)= t,

G(t)= O

when t < O,

satisfies all conditions for its representation by the exponential form of the Fourier integral formula. When t = O,however, show that the Cauchy principal value (3),Seco68, must be used in that representation, for the improper integral itself,with respect to y, does not existo 10. Prove that F(t) itself is sectionally continuous over an interval if its derivative F'(t) is sectionally continuous there. Note that F' is continuous over each of a finite number of subintervals, such as a ~ t ~ b, when properly defined at the end points, so F is continuous when a < t < b and, when a < to < b, then

r' F'('r:) dt

= F(t) - F(to)

(a < t < b).

J,o

Deduce that F(a + O)and F(b - O)exist.

71

REPRESENTATIONBY SERIES OF RESIDUES

Throughout this and the following two sections, f(s) denotes a function that is analytic for all finite values of the complex variable s except for a set of isolated singular points

confined to some left half plane Re s < y. We assume also thatf(s) satisfies conditions under which its inversion integral along the line x = y converges to the inverse transform F(t), say the conditions in either Theorem 5, Seco67, or Theorem 6, Seco68. Then F(t) can be represented formally by a series, finite or infinite, depending on the number of singular points, and we shall establish practical conditions under which the representation is valido Since etz is an entire function of z, the singular points z = Snoff(z) are the singular points of the integrand etzf(z) of the inversion integral. Let Pn(t) denote the residue of the integrand, for any fixed t, at the isolated singular point Sn: (1)

Pn(t) = residue of etzf(z) at z = Sn'

According to the residue theorem, the integral of etzf(z) around a path encIosing the points SI' S2, . . . , SN has the value 21ti[Pl(t)

+

P2(t)

+ . . . + PN(t)].

Let the path be made up of the line segment joining the points y - iPN, Y + iPN,and some contour CNbeginning at the second and ending at the


[SECo 71

207

y

52

O

Fig.71

first of these two points and Iying in the half plane x ~ y (Fig. 71). Then (2)

_2l .f nI

Y+¡PN

y-¡PN

l

etzf(z)dz + _ 2nI

'

N

f

etzf(z) dz

CN

=

L Pn(t). n=1

As PN- 00, the value of the first integral here tends to L¡-1{J(S)}, sincethe inversionintegralis the limitofthe correspondingintegralinvolving P,as P 00 in any manner. Let the numbers PN(N = 1,2,...) be selected so that PN 00, and let the curves CNtogether with the line x = y endose

--

the points S1, S2, ... , SN, if the number of singular points is infinite. If the number is finite, let all of them be endosed when N is greater than some fixed number. Then if j(z) and CNsatisfy additional conditions under which (3) it follows, by letting N

_1 . lim

f

2nI N-oo CNetzf(z) dz = O,

- 00 in Eq. (2), that 00

L¡-1{J(S)} =

L Pn(t), n= 1

or by a finite series in case the number of singular points is finite. The series on the right is necessarilyconvergent because the limit, as N 00, of the left-hand member of Eq. (2) exists. Since the inversion integral represents F(t) by hypothesis, the inverse transform off (s)is represented as the series of residues of etzf(z):

-

00

(4)

F(t)

= n= L1 Pn(t).

It is not essential that the line x = y and the curve CNendose exactIy N of the poles, of course. F or example, if two poles are induded in the ring

208

SECo 72]


between CN and CN+1, the residues at these two poles are simply grouped as a single term in the series. It is sometimes convenient to use the series representation (4) directly and formally, without regard to conditions under which the inversion integral represents F(t) and the integral over CN tends to zero. The function F so obtained may be such that its transform can be shown to be f(s), or such that it satisfies all conditions in a problem whose solution was sought by the transformation method. As we shall see in the problems at the end of this chapter, however, there are simple cases in which such a formal procedure leads to incorrect results. 72

RESIDUES AT POLES

At singular points Sn' which are poles of the functionf, useful formulas for the residues Pn(t) of e'Zf(z) can be found. Let Sn be a pole of f of order m. Over a neighborhood of that pole

with the point Snitself excIuded (O< Iz Laurent series (1)

f( ) Z

A_1

= -Z -

Sn

A_2

+ (Z -

< cJ, f is represented by a

snl

~

A-m

+ ... +

Sn)2

(Z - Snr

where A-m =FO and the A's depend on Sn' Thus (z

.

{

+ j=f.J O AJ,z -

-

,;y S

,

Sn)'''f(z)is represented

there by the power series IX)

(2)

A_1(z - snr-1

+ A_2(z - snr-2 + ... +

A-m

+

¿

j=O

AJ{z - sny+m.

Taylor's series expansion ofthe entire function e'z about the point Snis (3)

~-1

[

e'z. = exp(snt) 1 + t(z

-

Sn)

]

+ ... + (m - 1)!(Z - Snr-1 + ... .

The coefficientof (z - sn)-l in the Laurent expansion of e'zf(z) over the domain O< Iz - snl < cn, the residue Pn(t) of that function at Sn' is the coefficientof (z - Snr- 1 in the product of the power series(2)and (3). Thus we find that ~

Pn(t)

~-1

[

= exp(snt) A-1 + tA_2 +...

]

+ (m - 1)!A-m .

Formula (4) has a useful feature. It presents the residue of e''i(z) at a pole Snof order m as the inverse Laplace transform, with respect to z, of the principal part of the Laurent expansion (1) of the function f When sn is a simple pole (m = 1) of f (5)

Pn(t)

=

exp

(snt)A"::

1

= exp (snt) lim [(z - sn)f(z)]. z s,.


[SECo 72

Note that A_1 is the residue offitself (6)

209

at sn. In particular, if

fez) = p(z) q(z)

where p and q are analytic at Snand q(sn) = Owhile q'(sn) :i=O and p(sn) :i=O, then P(Sn) exp (snt). Pn(t) = ~ q (Sn)

(7)

Ir all singular points off are simple poles andf has the fractional form (6), then formula (4),Seco71,can be written (8)

F(t) = L -

P« S)

1

{

)}

qS

=

f

Sn) P,«

n= 1q Sn)

exp (snt).

In caseP and q are polynomials, the number of poles offis finite, and formula (8) becomes Heaviside's expansion (4), Seco24. Assume now that f(z) = f(z) except at the singular points off, all of which are isolated, a condition that is satisfied if f(z) is real over the real axis except at singular points (Sec. 61). Let SI = a + ib, where b :i=O, be a pole off of order m. Then over a domain O < Iz - sIl < c1, (9)

f (z) = - A_l Z - SI

~ + ... + (Z A-m m + ~ - SI) j=O

{

AJ,z

-

\i

(A-m

Sl1

:i= O).

The operation of taking complex conjugates is distributive to the terms of a convergent series; that is, if an infinite series ~zn converges to a sum S, then the series ~zn converges to S, as we can see by considering remainders. It follows from representation (9) and the condition f(z) = f(z) that (10) when 0< Iz - sIl < Cl, that is, when 0< Iz - sIl < Cl. By writing S for z here, we see that f(s) has a Laurent series expansion in the domain O < Is - sIl < Cl with the A's as coefficients, and, since A-m :i=O, SI is also a pole off of order m. According to formula (4) then, the sum of the residues of etzf(z) at the poles S1 = a + ib and S1is P1(t) + P1(t), where

-

(a+ib)t

Pl(t) - e

[

A_l

+ tA_2 +

...

m-l

t

+ (m -

I)!A-m

]

.

Therefore the component of F(t) corresponding to the pair of poles a :t ib (b :i=O) of f(s) is the real-valued function

(11) P1(t)+ P1(t)= 2~tRe {eibt[ A_l

+ tA_2 +... + (~:-11)!A_mJ}.

210


SECo 73]

Let us write the residue A - 1 off at a pole SI A_1

(12)

= r1 exp

=a+

ib in its polar form as

(iOl).

Ir s1is a simple pole, formula (11) then gives the component of F corresponding to the poles a :t ib as (13)

Pl(t) + Pl(t) = 2e"trl cos (bt + (1)

(b :F O,r > O),

a result that includes the component G(t) obtained in Seco26, wherefwas a quotient of polynomials. In case the singular points off consist only of the simple poles O, :t inro (n = 1,2,.. .), wherefhas residue ro at z = Oand residues r" exp (i0,,)at inro, formula (4), Seco71, becomes 00

(14)

F(t)

= ro + 2 ,,=1 L r"cos(nrot

+ O,,);

then F is a periodic function of t with period 21C/ro. Ir SI = a + ib is a second-orderpole off, we write

(Pl > O).

(15) Then if b :F O,according to formulas (11) and (12), (16)

Pl(t)

+ Pl(t) = 2e"t[rlcos(bt + (1) + p1tcos(bt + "'1)]'

Whena = O,the term 2p 1t cos (bt + '" ¡) is one of resonance type representing an unstablecomponent of F(t), since Pl > O. . Similarly if ib is a pole of order m off(z) then F(t) contains an unstable

component of the type Ctm- 1 cos (bt + O)where C :FO. In fact we can see from the above formulas that F(t) contains an unstable component iff(z) has a real or imaginary pole a + ib of any order such that a > O, or a pole z = O or z = ib of order m = 2, 3, . . .. This is a generalization of results found in Seco26 where partial fractions could be used. 73

VALlDITYOF THE REPRESENTATIONBY SERIES

We now establish sufficient conditions on f under which the integral s of etzf(z) over certain useful contours CN tend to zero as N -+ 00 as assumed in Sec.71. Since the integral s must exist, the contours should not pass through singular points off, nor should they become arbitrarily close to such points as N increases if the limit of the integrals is to exist. The ease of finding an order of magnitude of If(z)1 when z is on CN depends on the particular function f and the type of paths chosen for CN' We treat three types of paths CN that are especialIy useful. In each case we assume thatf(z) is (!)(Z-k)on CN where k > O; that is, constants M


and

211

[SECo 73

k, independent of N, exist such that M

(1)

(k > O),

for points z on CN

If(z)1 < Izlk

and we shall prove that, since PN-+ 00 as N -+ 00, lim

(2)

N-
f

et%f(z)dz

If k> 1, statement (2) holds also when t RECTANGULAR

=O

(t > O).

CN

= O.

PATHS

Order properties of many functions on horizontal and vertical lines are easily determined. Let CN be an open rectangie with sides along lines y = :tPN and x = - PN as indicated in Fig. 72, where PN -+ 00 as N -+ oo. Then iffsatisfies the order condition (1) on CN, the integrand ofthe inversion integral satisfies the condition M let%f(z)1< -éx Izlk

The integrals over the sides y Y

(3)

II-/IN'

= :t PN are then such that

M et%f(z)dx < 1

when z is on CN.

pNk

M

I y

-/IN

éXdx

= -t

ety

- e-t/lN

PNk

when t > O. Thus they tend to zero as N -+ oo. The length ofthe third side of CN is 2PN,and the absolute value ofthe integrand et%f(z)is less than M exp ( PNt)PN-k there, so when t > O, the integrals over those sides vanish as N -+ oo. Thus condition (2), where t > O, is established. Moreover, if k > 1,

-

then

ScNf(z)

dz -+ O as N -+ 00, so that condition

(2) holds when t

=

O.

y "(+i(3N

'PN+iPN CN -(3N

"(

o

Fig.72

(3N-i(3N

%

"(-i(3N

212

OPERATIONAl

SECo 73]

CIRCULAR

MATHEMATICS

ARCS

Let CN be ares Izl = RN, x ~ y. Then RN2= {3N2+ y2 and RN-+ 00 as N -+ oo. Ir y > O, the length of eaeh of the two ares of CN in the strip O ~ x ~ y tends to y as N -+ oo. Sineefsatisfies eondition (1),then le'Zf(z)1< Me'YRN-k on those ares, and the integrals over them tend to zero as N -+ oo. Regardless of the sigo of y, the integral over the are CN in the region x ~ O satisfies the eondition 3"/2

M (4)

e'Zf(z)dz < ---¡

IfCN

1

f

exp (tRNeos O)RNdO

RN "/2 2M ,,/2 = ---¡ exp (- tRN sin
The graph of sin 2
2M "/2 A,. nM 1 ---¡¡ exp - 2tR N- Nd'1' = RN o n)R t (

f

-

exp (-tRN) k RN

when t > O,and this quantity vanishes as N -+ oo. Therefore eondition (2) is satisfied when CN are those circular ares. Moreover, if k> 1, eondition (2) follows when t = Obeeause

[ f(z) dz < Mk2nRN -+ O

(5) I

JCN

I

as

RN

N -+ oo.

PARABOllC ARCS

Transforms involving

fi

arise in solutions of heat equations. It is often

~

desirable to seleetexpanding paths CNon whieh is bounded away from zero. But Re = eos 0/2 which vanishes on the negative real axis

~ .fi

~

(O = n). We may ehoose paths on whieh 1m never vanishes by taking CN as ares of eurves sin 0/2 = ~, where aN -+ 00 as N -+ oo. Then CNis a parabolie are

.fi

2aN

(6)

r

= 1-

eos O

(x ~ y)

with vertex at x = -aN,Y = Oand foeus at the origin, resemblingthe are shownin Fig.71. Its equation can be written in the form (7)

y2

= 4aN(x+

aN)

from whieh we find that

1+

dy 2 ( dx)

= x + 2aN. x + aN

[x ~ y, {3N2= 4aN(y+ aN)]


Since Izl

[SECo 74

213

~ aN and f satisfy condition (1) on CN,

r eIZf(z)dz

(8)

JCN

I

 21yl. Then when t > O,

(9)

f

-taN

+ 2aN

jWf; +

Ix

e -aN

dx

< exp --

aN

X

ta N

2

(

2 aN

)~f

-taN -aN

-

dx

Jx + aN

and when we evaluate the last integral, we find that the right-hand member of condition (9) vanishes as N -+ oo. AIso, 1

e aN f-taN

(10)

---¡¡

y

IX!

J

Ix y + 2aN y d d xx+aN + 2aN x< aNkJaN/2 -taN e x

f

which tends to zero as N -+ 00 if t > O, so condition (2) is satisfied on the parabolic ares. Again we find that the right-hand member of condition (8) vanishes as N -+ 00 when t = Oin case k > l. Our results can be summarized as follows. Theorem 10 Let f be analytic everywhere except for isolated singular points sn(n = 1,2,...) in a half planex < y and such that its inversionintegral along the line x = y represents the inverse Laplace transform F(t) of f(s). On contours CN(N = 1,2,...) in the left hand plane x ;;¡;y, let f satisfy the order condition If(z)1 < Mlzl-k where M and k are positive constants independent of N, and CN is either (a) the rectangular path shown in Fig. 72, (b) circular arcs Izl = (PN2+ y2)t, x ;;¡;y, or (c) the parabolic arcs (7), where PN -+ 00 as N -+ oo. Then whenever t > O,the series of residues Pn(t)of eIZf(z) at sn converges to F(t), 00

(11)

F(t) =

L Pn(t)

n= 1

(t > O),

if its terms corresponding to points sn within the ring between successive paths CNand CN+1are grouped as a single terms of the series. Also, if k > 1 and Li - 1{J} represents iF( + O)when t = O,then 00

(12)

74

iF( + O)=

L Pn(O). n= 1

ALTERATIONSOF THE INVERSION INTEGRAL

When the singular points of f(s) are not all isolated, it is often possible to reduce the inversion integral to a desirable form by introducing a new path of

214


SECo 74]

integration.

Weillustrate the procedure by findingan inversetransformthat

was obtained by another method in Seco27. Let us find F(t) when 1 (1) f(s) = - exp (-st), s where s = reioand st is the single-valued function (2) st

(r > 0, -n < (}< n).

= "¡;efi° = ~(cos~ + iSin~)

Since the function (2) is analytic everywhere except on the branch cut (} = n, r ~ 0, f is analytic in the same region. In the half plane Re s ~ y, wherey is a positive constant, -!n < (}< !n and cos!lJ > l/./i; therefore

l,skf(s)1= ,-1<-1exp(-"¡;cos~)

< ,-I<-le-.¡;]2.

This last function of r is continuous when r ~ y and vanishes as r -+ 00; hence it is bounded, and therefore f is 19(s-~ in the half plane for every value of the constant k, and in particular if k > 1. According to Theorem 5, the inversion integral along the line x = y therefore represents the inverse transform F(t), (3)

1

F(t)

= --:

lim

f

Y+i{l

2m {I-+oo y-I{I

dz

(P exp ( - zt)-

(t

~ O).

Z

The sum of the integral in formula (3) and the integral along the path ACDD'C' A', consisting of the circular ares and line segments shown in Fig. 73, is zero, since the integrand °isanalytic except on the nonpositive real axis. Thus if IAc denotes the integral of e'=j(z) over the arc AC, etc., we can write (4) Let R and ro denote the radii of the large and small cicular arcs; thus p2 so that P -+ 00 when R -+ oo. Along the arc AC, z = Rel°, dz = iReiOdO, and zt = JRei0/2. Hence the integrand of the integral is a R2

= y2 +

continuous function of (}whenever f. ~ 0, wheref. is the anglebetweenDC or D'C' and the negative real axis. For any fixed R, the limit ofthe integrals IAC and IA'c', as f.-+ 0, therefore exists. Likewise for any fixed positive ro the limits of the integrals over the other parts of the path exist. Since formula

(4)is true for everypositive f. and the integral on the left is independent of f., it follows that we can let each of the integrals on the right have their limiting

valuesas f. -+ 0, and consider hereafterthe path in Fig. 74.


[SECo 74

215

y 'Y + ifJ

c x

c'

'Y-ifJ

Fig.73

Fig.74

Write J AC = lim o JAC, and so on, for the integrals over the other arcs and lines. We shall now let ro approach zero and R tend to infinity, so that the left-hand member ofEq. (4),which is incidentally independent of ro, becomes the inversion integral or F(t), except for a factor

- 2ni.

When z is on the circle r = ro, Z = roei9,and the integral over that circle can be written

.

JDD, = i L-" exp(troei9 and the integrand is a continuous

-

Áefi9)

dO,

function of Oand ro when r o ;¡¡;O. Therefore

-" lim JDD.= i

ro-+O

fn

dO = -2ni.

On the line CD, z = rei("-f) and zt = '¡;ei("-fl/2; Z ~ -r

ifi

and zt ~

thus as € ~ O,

On the limiting position of D'C', however,z = -r

and zt = - i'¡;. Therefore + J D'C' =

JCD

f

ro

. dr e-tre-''¡; - +

R '

= 2¡

r R

fro

.

and

hm (JCD + JD,c')

ro O

e

R

. dr

e-tre¡,f;-

ro

r

,¡;d r,

-tr sin

-

I

r

.

= 21 f. R e _trsin'¡; -dr o

= 4i f.

r

./R sin JL exp ( - tJL2)dJL. o JL

Consequently we can write, in view offormula (4), (5)

F(t)

=

1

.

: hm (JAC 2m R-+oo

+

JC'A')

2 oo 2 sin JL + 1 - - f. exp(-tJL )-dJL. n o

JL

216

OPERATlONAl

SECo 74]

MATHEMATICS

Now on ares AB and A' B', Re (tz - zt) ~ ty; thus the integrand of the integrals J AB and J B'A' satisfies the condition

Iz-l exp (tz - zt)1~ R-le'Y. Since the length of each are tends to y as R -+ 00, those integrals tend to zero as R -+ oo. Finally, on ares BC and B'C', Re (tz zt) ~ tR cos so that

-

"

IJBe!~

e

"/2

I

exp (tR cos e) de = ,,/2

f.o exp (- tR sin cp)dcp,

and JC'B' satisfies the same condition. The value of the last integrand does not exceed exp ( - 2tRCP/n),as pointed out under Circular Ares, Seco73, and an elementary integration then shows that J BCand JC'B'tend to zero as R -+ 00, if t > O. The limit in formula (5) therefore vanishes, and 2 '" sinp. F(t) = l - exp (-tp.2)-dp. f. n o P.

(6)

(t > O).

When we integrate both members of the known 1 formula '" (7)

f.o

l

exp (- tp.2)COSocp.dp.

= - A- exp --

(

2 t

O(2

4t )

(t > O),

with respect to oc,from zero to 1, we find that 2 '"

- f. n o

exp ( -

sin p.

tp.2)-

P.

l

1

OC2

dp.= e f. exp ( -4 ) doc V nt o t l/(2 jij

2

= fi f.o

exp (_).2) d)..

Thus we can write our result in the form

(8)

1

F(t) = 1 -

erf ( 2.fi }

-

-

erfc (

l

2-fi

. }

I See, for instance, R. V. Churchill, "Complex Variables and Applications," 2d ed., p. 171, exercise 7, 1960. Formula (7) also follows from the result in Probo 7, Seco35.


[SECo 74

217

PROBLEMS 1 . Instead of simply using partial fractions, show that Theorems 5 and 10 apply to 1(s)when 1 (b) f(s) = S2(S+ 1)'

2s + 1 (a) f(s)

= S(S2 + 1)'

and find F(t) as a sum of residues. Ans. (a) F(t) = 1 - cos t + 2 sin t. 2. Use identities (6),Seco55, to prove that the function tanh z is bounded over each of the half planes x ~ y and x ~ -y when y> O, and on the lines y = INn(N = 1, 2, . . .) uniformly

with respect

to N.

Thus

show that

f(s)

= tanh s S2

the function

satisfies all conditions in Theorems 5 and 10 such that F(t) is represented by a series of residues as follows: F(t)

f

= 1 - .! n2 n =1 (2n

1

.. ,

-

co (2n

s

(t ~ O).

I)nt

Note that, according to Seco23, F is the triangular-wave function H(2,t) shown graphically in Fig. lO. 3. In Probo 4, Seco24, we found that the function 1 f(s)

= S2 -

1 s sinh s

is the transform of the periodic function F for which F(t)

=t

when -1 < t < 1,F(t + 2) = F(t).

Show that s = Ois a removablesingular point of f, that f satisfiesall conditionsin Theorems6 and 10,and thus that F has the sine seriesrepresentation 2 00 (-1)"+1 F(t) = 1t - n= 1 n

¿

sin nnt.

4. Use' Maclaurin's series for coshz to show that the function g(s)= cosh ¡;, s #- O,g(O)= 1, is an entire function of s for any branch of function f(s)

=-

¡;.

Then show that the

1

s cosh

.jS

is analytic except for simple poles s = O,s = -(n - t)2n2and, with the aid of one of identities (6), Seco5}. that f satisfies all conditions in Theorems 5 and 10 when CN are ares of parabolas r sin 8/2 = Nn. Thus find L -1{J} in the form

v

F(t)

~ (-I)n = 1 + -4 '-exp nn=12n-1

[

(2n

4

1)2n2

]

t

(t > O).

218

5.


SECo 74]

Write F(t) fonnally as a series of residues when f(s)

= 4s2 2 tanh + n2s .

2 . nt nt '" 1 (2n - l)nt Ans. n F(t) = ntstn2' + oos2' - n~2n(n- l)ooS 2 .

6. Whenf(s) = s-2e-s, show that F(t) is representedby the inversionintegral of f, where y > O,but that the residueof e/% f(z) at the lone singularpoint off represents F(t)only whent ~ 1. This exampleand those in Probs. 7 and 8 belowiUustratehow - the fonnal procedureof writing the sum of residuesPn(t)may fail to givethe inverse transfonn.

Note that in this case lel%f(z)1-+ OCJas x -+ - OCJif t < 1.

7. StatewhyL,-l{s-le-'} = 80(t- 1)ifY> O,but showthat the sumof residues

Pn(t),oonsisting ofjust one tenn, fails to represent 80(t - 1)when t < I. 8. Apply the real fonn of the inversion integral, fonnula (5), Seco66, to the entire function

= l - s e-S

f(s)

to find F(t) directly. Here we can write y

= Oand

[s "p 0,f(0) = 1], use the integration

fonnula (8), Seco34.

Note that the residue procedure can represent F(t) only where F(t) = Obecause e'"f(z) is entire. 9. From properties ofthe transfonn ofthe function erfc (tt-i) noted in Seco74, prove that the function and each ofits derivatives tends to zero as t -+ +0. 10. When,JS = ~i912( -n < (J< n, r > O),use the procedure in Seco74 to prove that L;l

-

1

{ ,JS }

=-

2

n0

i

'"

o

exp(-12) d1 =-

1

(t > O)

jitt

11 . Generalize Theorem 10 for rectangular paths by letting the sides of CNlie on lines x = -CPNand y = IPN and usingonlya oonditionof boundednessIf(z)1 O,where e and B are positive oonstants independent of N. 12. Prove Theorem 10 when CN are semicircles Iz - yl = PN'x ~ y, and If(z)1< Mlzl-k on CN. -

7 Problems in Heat Conduction

We shall now illustrate the use of the theory just developed in solving further boundary value problems in the conduction of heat in solids. We present examples of problems that cannot be fully treated with the more elementary theory used in Chap. 4.1 The formal solution of the problem in the next section is followed by a complete mathematical treatment of that problem. The purpose is to illustrate a means of rigorously establishing the solutions of such problems. Since the procedure is lengthy, the reader is advised to use it sparingly in his workon the sets of problems that follow. A clear understanding of the formal method of solution is of primary importance. 1AItemative methods for solving a few of the problems in this and the following chapter are presented in chap. 7 of the author's "Fourier Series and Boundary Value Problems," 2d ed., 1963. 219

220


SECo 75]

75 TEMPERATURES IN A BAR WITH ENDS AT FIXED TEMPERATURES Let U(x,t) denote the temperature at any point in a bar (Fig. 75) with insulated lateral surface and with its ends x

=

1 kept at temperatures

Ur(x,t) = Uxx(x,t)

zero

U(+O,t)= O,

U(l

(O< x < 1, t > O), (O < x < 1),

U(x,+O) = O

-

O,t) = Fa

(t > O),

where Fa is a constant. The problem in the transform of U(x,t) is su(x,s) = uxx(x,s) u( +O,s) = O,

(O< x < 1),

u(1 - O,s) = Fa. s

Since this problem in ordinary differential equations has a solution that is at x = O and x = 1, u( +O,s) = u(O,s) and u(l - O,s) = u(l,s). The solution is continuous

(1)

u(x,s) = Fasinh xjS

s sinhjS , where the symbol fi denotes some branch ofthe double-valued function sto As long as represents the same branch, regardless of which one, in both

fi

,I L i i

=

O and x

and Fa, respectively, when the initial temperature is zero throughout. In Seco49 we obtained a formula for U(x,t) in the form of a series of error functions, a series that converges rapidly when t is small. We shall now obtain another series representation of this temperature function. This series will converge rapidly for large t. Let us proceed formally to the solution here, leaving the fuIl justification of our result to the foIlowing sections. We have taken the unit of length as the length of the bar, and we observed earlier that, by a proper choice of the unit of time, we can write k = 1 in the heat equation, where k is the diffusivity. The boundary value problem in U(x,t) is then

{:::::::::~:~::::}o . O x-l

Fig.75

PROBLEMS IN HEAT CONDUCTION

[SECo 75

221

numerator and denominator, the quotient of hyperbolic sines can be written

(2)

x-fi -

sinh

sinh

x-fi + (x-fi)3/3! + . .. - x + (x3s/3!)+ ... - 1+ s/3!+ ... '

-fi - -fi+ (-fi)3/3! + ...

-fi = O,namely,

except at those points where sinh

s = O,

(3)

(n = 1,2,.. o).

The final member of Eqs. (2) is the quotient of two power series in s that converge

for every

s.

It follows that u(x,s) is an analytic function of s

except for isolated singular points (3)0 It also follows from Eqso (1) and (2) that lim su(x,s) = Fox; s-o

(4)

x =1-' O, the function u(x,s) has a simple pole at s = O with residue Fox. We note that u is real-valuedwhen s is real.

therefore when

Now let the branch cut of .jS be taken as the positive real axis, so

that -fi is analytic on the negative real axis. Then u(x,s)has the fractional form p(x,s)/q(s),where the functions p(x,s) = Fosinh x-fi, s

q(s)= sinh-fi

are analytic at s = - n2n2, and

Hence s = - n2n2 are simple poles, where the residues of e"u(x,s) are (5)

P (x,-n

22

n

)

_"2,,2,

e q ( -n n ) '

2 2

=

.' F.01 SIn nnx 2 2

-n n

" 2 2 ( - 1) 2nm exp ( - n n t) '

o

At least formally then (Sec. 71) the inverse transform of u(x,s) is the sum of the residues (4) and (5) of e"u(x,s) at the poles (3); that is, (6)

[

2 00 (-1)" n "= 1 n

U(x,t) = Fo x + -

L

]

exp ( - n2n2t) sin nnx .

This formal solution can be verified by showing that the function defined by formula (6) satisfies all the conditions of our boundary value problem (see Probs. 5 and 6, Seco81). But we shall now see that the theory in the precedingchapter enables us to make the verificationin another way that

has some advantages.

222

76


SECo 76]

THE SOLUTION ESTABLlSHED

We have seen that the function

sinh x.jS u(x,s) = F.o 1: s sinhy s is analytic with respect to s in any half plane Re s ~ y where y > O. To examine its order in that half plane, let us write -

thenRe.jS = .fi cos (()/2)>

~

~ .JYi2. Thus

1: 1 - exp ( - 2X.jS) smh.jS 1 - exp(- 2y's) ] 1 + exp(- 2x.jYji) r.:¡; exp[-(1 - x)y r/2]~ M exp[-(1 - x) 02], ) exp y/2 y'l'"< 1 (2Ji72

sinhx.jS .

I

I

I

exp [(x

-

l)y s]

[

I

-

= 2/[1 exp (- 2.JYi2)]. Thus when XI < 1, constants Mm' independent of x over the interval O~ x ~ Xl> exist such that for each integer m = 2, 3,. . .

where M -

=

~lu(x,s)1< FoM~-lexp[-(1 that is, u satisfiesthe order condition (1)

- XI).fi1i] < Mm

(Res ~ y);

(m = 2,3, .. ., Re s ~ y, O~ x ~ XI)' According to Theorem 8, Seco70, the inversion integral of u along the

line Re s = y, therefore, represents

-

the inverse transform

U(x,t) = L¡- l {u(x,s)}

(2)

U(x,t) of u(x,s), (t

~ O,O ~ x < 1),

and U is a real-valued continuous function of t that satisfies the required initial condition (3)

U(x, + O) = U(x,O) = O

(O~ x < 1)

Ut(x,t) = L¡ - l {su(x,s)}

(O ~

and also the condition (4)

x < 1).

Now in view of Theorem 9, Seco70, U is a continuous function of x and t together when t ~ Oand O ~ x ~ XI; thus (5)

U( + O,t) = U(O,t) = L¡ - l {u(O,s)} = O


[SECo 76

223

since U(O,S)= O. Likewise, the deriva tives cosh

u..,(x,s)

= Fo !:.

x.jS

!:'

uxx(x,s) = su(x,s),

V s smh V s

are l!'i(s-m)in the half plane, for each integer m, uniformly with respect to x

when O ;;;¡;x

;;;¡;XI

< l. This is evident by comparing those functions with

u itself. Therefore

ux..,(x,t) =

L¡ - l {ux..,(x,s)} = L¡ - l {su(x,s)}

(O< X < 1),

and it follows from Eq. (4) that U satisfies the heat equation (6)

(O< x < 1).

U,(x,t) = Uxx(x,t)

We have now shown that our function (2) satisfiesall conditions of our boundary value problem exceptthe end condition (7)

U(I

-

(t > O).

O,t) = Fo

It is evident that U satisfies the condition

U(I,t) = L¡-I{U(I,s)} = L¡-I{~O} = Fo

(t> O);

but this does not assure us that U(x,t) -. Fo as x -. 1, which is the condition the temperature function should satisfy. By writing hyperbolic sines in terms of exponential functions and carrying out one step of a division, we find that

sinh x.Js

(8)

sinh.Js

!: = exp[ - (1 - x)V s] +

exp[-(2 - x).Js] - exp(-x.Js) . - exp (.Js) - exp(-.Js)

It is the first term on the right that has a weak order with respect to s when x = 1. Now we can write

(9)

u(x,s) = Foexp [ -(1 - x).jS] + g(x,s) s

where (10)

g(x,s) = ~oexp [-(1 + x).jS]exp[1 -2(1 x).jS] - 1. - exp-(-2.jS) On the half plane Re s ~ y we can see that IF.I Ig(x,s)1<

r ~exp(-J12)1

2

- exp(- .fiY 2y)

(O ;;;¡;x ;;;¡; 1)

so that g is l!'i(s-m)there for each integer m, uniformly with respect to

I ~

I

I

I ~

224


SECo 77]

x (O;;¡;X ;;¡;1). Consequently, Li l{g} represents a function G(x,t) that is continuous in x and t and, since g(l,s) = O, G(I

!

- O,t) = G(l,t) = L¡-l{g(1,S)} = O

(t ;¡;;O).

With the aid of the inversion integral evaluated in Seco 74, we find that

= Foerfc(12~)

(11) L¡-l{~O exp [-(1 - X)j;]}

(O ;;¡;x ;;¡; 1, t > O).

It now follows from Eq. (9) that

1.

U(x,t) = Foerfc ( 2.Ji

(12)

X

(O ;;¡;x ;;¡; 1, t > O), )

+ G(x,t)

- O,t) = Fo + G(I,t) = Fo when t > O. Our function (2) thus satisfies the end condition (7), and it is therefore completely established as a solution of our boundary value problem. Moreover, we have shown that the transform of our temperature function is the function u(x,s) from which we obtained U(x,t). Some interesting properties of U will follow from the order properties of U. We still have to prove that the series obtained in the last section represents our solution (2).

and U(l -

-

77

THE SERIES FORM ESTABLlSHED

We have seen that the function .

u(x,s) =

F.o

sinh xj; s sinh

-

j;

is analytic except for the poles s = Oand s = - n2n2 and that its inversion integral converges to a function U(x,t) that is a solution of our boundary value problem. The series representation of U(x,t), given in Seco75, is valid provided the integral

1Cn e'%u(x,z)dz, taken along a curve Cn of a family of curves (n

= 1,2,...) between the poles,

tends to zero as n tends to infinity (Sec. 71). To select curves Cn so that lu(x,z)1has a suitable order property whenever z is on Cn, we first note that Isinh ;;12

=

sinh2

(-fi cos~) + sin2(-fi sin~).


Hence if

[SECo 77

225

fi sin!8 = an,where

(1)

(n = 1,2, .. .),

then (2)

Isinhj;12 = sinh2(fi cos~) + 1 ~ 1.

Let us therefore take Cn as the arc of the parabola an2 r-- 2an2 - sin2(}/2- 1 - cos()

(3)

that líesto the leftof the line~ = y, wherey is any positiveconstantand z = ~+ ir¡(Fig. 76). Then when z is on Cn (4) I

-zu(x,z) 2 = sinh2(xfi cos (}/2)+ sin2(xfi sin (}/2)<1 Fa sinh2(fi cos (}/2)+ 1 = , 1

because O~ x ~ 1 and sinh2y increases when Iyl increases. Thus u(x,z) is O(z- 1)on Cnand, according to Theorem 10,Seco73, the seriesof residues of eZ1u(x,z)converges to the inversion integral for all positive values of t. The series (6),Seco75, therefore does represent a solution of our problem; that is, our solution can be written (5)

[

2

00 (-I)n L ¡¡;n= 1 n

U(x,t)= Fo x + -

]

exp (- n2'nh) sin n¡¡;x

-a~

Fíg.76

(t

> O).

226


SECo 78]

We have established the sum of series (5) as a solution of our boundary value problem by identifying it with our solution L¡- 1{u}. Owing to the uniforrn convergence of such series when t ~ to > O (Prob. 5, Seco 81) however, we find easily that series (5) represents a function that does satisfy the end conditions U(I - O,t) = Fo, U(+0, t) = O and the heat equation. Hence a simpler verification consists of those observations coupled with our brief proof that L¡I{U} satisfies the initial condition (3), Seco76, after establishing the series representation of the inversion integral. We found in Seco 76 that L¡-I{U} = O when t = O and O~ x < 1. Certain order conditions satisfied by u on Cn are actualIy adequate to show that the series representation

(5) of L¡ -1 {u} is valid when t

=

O and con-

sequently that 2 ex> (- 1)" . L1 -smn1rx x= -- 1tn= n

(6)

(O ~ x < 1).

For if € is a smalI positive angle and z is on either part of any of the parabolic paths

Cn on which

181 < 1t

-

€, it can be seen from the Eq. (4) that for each

fixed x (x < 1)a number M. independent ofn exists such that lz1'u(x,z)1< M.. for any constant k. This order property, together with the inequality (4) satisfied on the remaining part of Cn, is sufficient to show that the representation (5) is valid when t = Oand x < 1. The proof, made by first selecting € smalI and then n large, is left to the problems. It establishes the Fourier sine series representation (6) of the function x on the interval O ~ x < 1. We shalI use a similar procedure in Chap. 9 to establish a generalization of Fourier series representations of arbitrary functions. 78

PROPERTIES OF THE TEMPERATURE FUNCTION

It was shown in Seco 76 that for any integer m the transforrn u(x,s) of our temperature function U(x,t) is (i)(s-m)in a right half plane of the variable s, uniforrnly for all x in any interval O ~ x

~ Xl where Xl < 1. The derivatives

of u with respect to x also satisfy that order property. As a consequence, our temperature function possesses the folIowing properties, according to Theorems 5, 8, and 9 of the preceding chapter. The function U(x,t) is a continuous function of both x and t when t ~ Oand O ~ x < 1, and each of its derivatives with respect to x or t has this continuity property. At any interior point of the bar, the temperature begins to change very slowly at the time t = O,since U(x,O)= Oand (1)

U,(x,O) = Utt(x,O)= Uttt(x,O)= ... = O

(O~ x < 1).

Of course, U(x,t) is not identicalIy zero because u(x,s) is no1. In fact, from


the representation

[SECo 78

227

(3), Seco 49, of U(x,t) as a series of error functions, we can

see that l F:o U(x,t) > O

(2)

whenever t > Oand O < x < l.

The flux of heat through any section x = xo, cI>(xo,t)= -KUixo,t), where K is thermal conductivity, has the transform

- - K coshxo';; Fo r:. r:

,/.,

'1'(xo,s) -

if O ~ Xo < 1.

v s smhv s

From the order of cjJwe can see that the flux also changes very slowly from its initial value cI>(xo,O) = O,since (3)

cI>,(xo,O)= cI>rr(xo,O) =

... =

O

when O ~ Xo < 1.

In Seco 76 we found that

l

-

U(x,t) = Foerfc ( 2ft

X

)

+ G(x,t)

(O~ x ~ 1,t > O),

where G and its derivativesvanish as t -+ O. Therefore,the flux through the right-hand faceof the bar can be written (4) and (5)

lim

,-o

[

cI>(I,t)

+

F~

J v 1I:t

= o.

That is, the flux at that face is (!)(I/.Ji) as t -+ O; it becomes infinite as t -+ Olike - FoKIFt. The total quantity of heat that has passed through a unit area of the face x = l up to time t can be written

in view offormula (4). Thus

228


SECo 79]

Since Gx(l,t) is continuous when t ~ O, (6)

lim Q(1,t) = O, t-O

a condition that would not be satisfied if there were an instantaneous source of heat over the surface x = 1 at t = O. Such a source is an idealization of an actual situation in which a large quantity of heat is generated over a surface in a very short time interval, by combustion, for instance. 79

UNIQUENESS OF THE SOLUTION

Qur treatment of the boundary value problem is not strictly complete until we have shown that our solution is the only one possible. The physical problem of the temperatures in a bar with prescribed initial temperature and prescribed thermal conditions at the boundary must have just one solution. If we have completely stated the problem as one in mathematics, that problem must also have a unique solution. The conditions we have imposed on U(x,t), namely, (1)

U,(x,t)

(2)

U(x,+O)

(3)

U(+O,t)= O,

= U xX
(O< x < 1, t > O),

= O

(O< x < 1),

U(1 - O,t) = Fo

(t > O),

are not sufficient to ensure just one solution. They do not exclude the possibility of instantaneous sources of heat at the ends of the bar at t = O. The equation of conduction (1) is the statement that heat distributes itself interior to the bar after"the time t = O,by conduction. In the derivation of that equation, it is assumed that the functions U, U" Ux, and Uxx are continuous with respect to the two variables x and t, interior to the solid after conduction begins. We shall therefore require our solution to have those properties of continuity. Physically, the presence of heat sources interior to the bar when t > Ois then prohibited. Let the required temperature function satisfy the conditions (1), (2), and (3) and the following continuity and order conditions.

l. U is a continuous function of x and t when t > Oand O ;;;¡; x

;;;¡;1,

and U, Ut, and U x are continuous when t ~ Oand O ;;;¡;x < l. 2. Constants exand M exist such that IU(x,t)1 < M eat over the strip t> O, O < x < 1; also for each Xl (O < Xl < 1), a constant Ml exists such

that the functions Ut and Ux are less in absolute value than M 1eat strip t > O,O < x < Xl'

over the

We could write ex= Ohere, simply requiring the functions to be bounded over the regions specified; but it is easier to show that our functions are

PROBLEMS IN HEAT CONDUCTlON

[SECo 79

229

&(e'tt)than to pro ve they are bounded for all t. Note that V xx satisfies the eonditions

=

imposed on VI because V,

We have seen that the funetion (4)

V(x,t)

V xx'

= L¡ - 1 Fosi~hxfi { ssmhfi }

satisfies conditions (1), (2). and (3) and that it is eontinuous, together with V, and Vx, when t ~ Oand O ~ x < 1. We saw also that l

-

X

V(x,t) = Foerfe ( 2Jt

)

+ G(x,t)

(O~ x ~ 1,t > O),

\vhere G is continuous and &(e«')when t ~ O and O~ x ~ 1, if IX> O, aecording to the eharaeter of g(x,s). The error funetion here is eontinuous in x and t when t > Oand O~ x ~ l. Hence the funetion (4) satisfiesour eontinuity requirements 1. The first of eonditions 2 is satisfied beeause that error funetion is bounded over the strip t.> O,O < x < 1, and G is l!J(e«')there. The order condition there on V, and V x when O < x < Xl is satisfied according to the representation (4). Qur funetion (4) therefore satisfies eonditio~s (1), (2), (3), and the eontinuity and order eonditions. If a seeond funetion V does so, then the funetion (5)

W(x,t) = V(x,t) - V(x,t)

also satisfies those eontinuity and order eonditions, and also these homogeneous eonditions: (6)

W;(x,t) = JtYxx(x,t)

(7)

W(x,O) = O

(8)

W(O,t) = O,

(O< x < 1, t > O), (O< x < 1),

W(I,t) = O

(t > O).

In writing eonditions (8), we have used the faet that W is eontinuous when t > Oand O ~ x ~ 1. We shall prove that this problem in W has only the

trivial solution W(x,t) == O in the class of funetions that satisfy the continuity and order eonditions above. Sinee W satisfies those eonditions, its transform w(x,s) exists when Re s> IXas well as the transforms w'(x,s) and w"(x,s) of JtYxand Wxx, where the primes denote derivatives with respeet to x. Moreover, those transforms are eontinuous funetions of x and s when O ~ x < l. In view of condition (7) the transform of W; is sw. It follows from conditions (6) and (8) that for

everys in the right half plane Re s > (9)

IX

sw(x,s) = w"(x,s)

(O< x < 1),

230


SECo 80]

w(O,s) =

(10) X

=

w(l,s)= O.

O,

The continuity of w with respect to X actually extends to the point 1. To prove it, we write 'O

W(X,S)=

GO

i

O e-S'W(x,t) dt

+

i .

'0 e-S'W(x,t) dt

(to > O),

where the second integral is a continuous function of X when O ;;¡; x ;;¡; 1 that vanishes when x = 1. Since IW(x,t)1 < M eal and Re s > 0(, then le-S'W(x,t)1 < M throughout the strip t > O, O < x < 1. Thus we can take to small enough to make the absolute value of the first integral arbitrarily small independent of x when O < x < 1. Then for that to the absolute value of the second integral is arbitrarily small when x is sufficiently close 1; that is, w(l - O,s) = O= w(1,s). to 1. Hence w(x,s) Owhen x We have now shown that for each s in a right half p1ane,wis a continuous function of x (O;;¡;x ;;¡;1)that satisfies the linear ordinary differential equation (9) with constant coefficients, also that w' is continuous when O ;;¡; x < 1, and w(O,s)= O. In the theory of linear differential equations it is shown that, when the value w'(O,s)of the derivative at x = Ois also prescribed, such an

-

-

equation has one and only one solution satisfying those continuity conditions. In our case that solution is w'(O,s)

w(x,s) =

-fi

.

r:.

smh Xv s

) (O ;;¡;x ;;¡; 1 .

But w(l,s) = O,and sincesinh -fi = O,only whens = -n2n2 (n = 0,1,2...) and not for all s in a right halfplane, then w'(O,s)= O; hence (11)

w(x,s)= O.

Since w(x,s) is the transform of a continuous function W of exponential order and Jv,is continuous, ourTheorem 7, Seco69, on uniqueness ofinverse transforms applies to show that W(x,O) = O; that is,

(12)

V(x,t) = U(x,t)

(O;;¡;x < 1).

The proof that the problem consisting of conditions (1), (2), (3), and the continuity and order conditions has just one solution is now complete.

80

ARBITRARY END TEMPERATURES

. Let the temperature of the end x

= 1 of the

bar be a prescribed function

F(t) (Fig. 58). The temperature function U then satisfiesthe heat equation U, = Un, the initial condition U(x,+O)= O,and the end conditions (t > O). U(l - O,t)= F(t) U( +O,t) = O,


[SECo 80

231

As noted in Seco49. the solution of the transformed problem is (1)

( ) uX,s

- f( ) sinhx.fi - s r:. sinh v s

Let V(x,t) denote the temperature sections when F(t) = 1. Then

function found in the preceding

(2)

u(x,s) = sf(s)v(x,s)

since

v(x,s) = sinh x.fi.

ssinh.fi

Now sv(x,s)is the transform of V,(x,t)when O~ x < 1. In view of the convolution property, it follows from Eq. (2) that (3)

U(x,t) = {F(t

- T)V,(x;r)dT.

It was shown that V(x,t) is represented by a series: V(x,t) = x

2

(-1)"

+ -11:,,=L1-

n

The series obtained by differentiating

exp ( - n211h)sin mr:x.

this series term by term with respect

to t does not converge when t = O; but it was shown that the function V,(x,t)is continuous when t ~ Oand O~ x < 1and that

(O~ x < 1).

V,(x,O) = O

The differentiated series simply fails to represent V,(x,t)at t = O. To arrive at another form of the temperature function U, we assume

F continuous, F' sectionallycontinuous, and F of exponential order. Then L{ F'(t)} = sf(s) - F( + O),

and

u(x,s) = F( +O)v(x,s) + L{F'(t)}v(x,s).

Consequently we have the formula (4)

U(x,t) = F( +O)V(x,t) +

{

F'(t

-

T)V(X,T) dT.

The two formulas (3) and (4) give the temperature U(x,t) in terms of the temperature V(x,t) corresponding to a fixed surface temperature. They are two forms of Duhamel's formula (Sec. 85). The above series for V(x,t) can be substituted into formula (4), and it can be shown that the temperature

232

SECo 81]


function can be written

+ 2F(+ O) ~ t..,-(-

(5) U(x,t) = xF(t)

n

n= 1

1)" n

mrx

'

200(-1)"

+ -nn=l L -sinn 81

e-n2"2I Sin.

mrx

i

o

F'(t - "t)e-n2,,2fd"t.

SPECIAL END TEMPERATURES

When the end temperature F(t) is a specific function, a convenient formula for U(x,t) may be found directIy from the transform u(x,s). For example, let

F(t) = At

(1)

in the problem of the last section, where A is a constant. Then

sinh xfi u(X,s) = A 1:' S2sinh v s a functionwith a poie ofthe second order at s = O.Wenotedin Seco75that sinhxfi sinh

- x + (x3s13!)+ (xSs215!)+ ... 1 + (sI3!) + (s215!)+ ...

fi

By carrying out the indicated division here, the first two terms are found to be x + X(X2 - 1)s3!; hence u(x,s) has the following representation in a neighborhood

of s

=

O: X

X(X2

u(x,s)= A ["2 . s +

1)

3'-.S

00

+ n=O L an(x)~J .

The residue of eZlu(x,z)at z = O is therefore (Sec. 72) x(x2

[

A xt +

-

1)

J.

3!

The residue of ¿IU(X,Z)at the simplepole z = -n2n2 is 2A( _1)n-l . ¿' sinh x'¡-; z z cosh z J z=-n2,,2 2A.¡-;.¡-; = n3n3 Sin (mrx) exp (-n

2 2

n t).

ConsequentIythe formula for the temperatures can be written (2)

X3

U(x,t) = A -

[

6

x

2 + xt + "3 n

00

L n=l

(_1)n-l 3

n

exp ( - n2n2t) sinmrx .

J


233

[SECo 81

That function can be verified completely as a solution of the boundary value problem by just the same procedure that was used in Secs.76 and 77. But the procedure can be simplified in this case in view of the fact that u(x,s) is (9(S-2) in a right halfplane and on the parabolas Cn, uniformly with respectto x when O ;;¡;x ;;¡; 1. ConsequentIyU is a continuousfunctionof its two variables for all x and t (O ;;¡;x ;;¡; 1, t ~ O); also the series representation (2) is valid at t = O. Since U(x,O) = O,it follows from Eq. (2) that

12 ex>(_I)n-l x which

is the Fourier ;;¡; x ;;¡; 1.

X3 = "3

¿

1t n=1

n

sine series expansion

3

(O ;;¡; x ;;¡; 1).

sin n1tx of the function

x

-

X3 on

the

intervalO

PROBLEMS

1. A bar with its lateral surface insulated is initialIy at unifonn temperature A

= O and x = 1are then kept at constant temperatures B and C, respectivelyo Show fonnalIy that its temperature distribution can be written (Figo 77)0 Its ends x

A - B + (-l)n(c U(x,t ) = B +-xC - B + -2 ~ L.., 1 7tn=1 n

-

A)

o

n27t2kt

n7tX

sm-exp 1

(

--,:-

1

)

o

Fig.77

2. A slab of iron with diffusivity k = 0.15 cgs unit, 20 cm thick, is initially at O°C throughout,and one faceis kept at that temperatureoIts other faceis suddenlyheated to 500°C and kept at that temperatureo Use the fonnula in Probo 1 to compute the temperature of the midsection of the slab after 5 min, to the nearest degree, and compare the result with a corresponding temperature found in Probo2, Seco47, for a semi-infinite solid of the same material. Ans. 145°C. 3. For fixed values of A, B, C, and k, note that the temperature U(x,t) in Probo 1 is detennined by the two numbers xii and tl120For two slabs with those same fixed values and widths 11and 12,show why the temperature ofa section x = XI in the first at time ti is the same as the temperature

of the corresponding

section x

= x2(x2112 = xdl¡)

of

the second at time t2 if

Thus the slowness of heating varies as the square of the widtho

= 1= 1 for the bar in Probo 1, let Uo(x,t) denote the temperatures in case W(x,t) the temperatures when C is replaced by F(t), so that W(l,t) = F(t),

4. When k C

= Oand

~ ~

,.

234


SEC. 81]

W(O,t) = B, and W(X,O)= A. Ir V(X,t) is the function (5), Seco80, so that V, = V xx' V(x,O) = V(O,t) = Oand V(I,t) = F(t), show by superposition that W(x,t) = V o(x,t) + V(x,t).

5.

When t ~ to > O and O ~ x ~ 1, prove that the series (6), Seco75, and the series obtained by differentiating that series once or twice with respect to x or once with respect to t are unifonnly convergent with respect to x and t. The series then represent continuous functions of x and t, and tennwise differentiation is valid, since the terms themselves are continuous functions. As a consequence, show that the function V(x,t) defined by the series satisfies the heat equation V, = Vxx when O < x < l and t > O, and end conditions V( +O,t) = O, V(I - O,t) = Fo, when t > O.

6. According to Abel's test for uniform convergence,

¡ a series 1::",. ¡ An T,,(t) converges

unifonnly with respect to t if the series 1::'=1 An converges, and if functions T,,(t) are bounded unifonnly with respect to t and n and such that T,,+¡(t) ~ T,,(t). Given that the Fourier series representation (6), Seco77, is valid, use Abel's test when O ~ t ~ to to show that the function defined by Eq. (6), Seco75, is continuous with respect to t when t ~ Oand that it satisfiesthe initial condition V(x,+O) = Ofor eachfixedx such that O < x < l. This, together with the result in Probo 5, again establishes the solution (6), Seco 75.

7.

Give details of the proof outlined in Seco 77 of the Fourier

(6) ofthe

function

x(O ~ x < 1). Suggestion:

arc Cn and 181< 1t

-

(, then

ZU(X,Z) I

Fo

sine series representation

Show first that when z is on the parabolic

~ cosh(x-fi cos8/2) < 2 exp[-(1 - x>-fi sin (/2]. I

- cosh(-fi cos8/2)

l

+ exp(- 2-fi sin(/2)

8. As in Seco80 let V(x,t) denote temperatures in the slab with faces x = O and x = l when V(x,O) = V(O,t) = O, V(l,t) = 1, and k = 1, and write V(x,t) = O when t < O. Let V(x,t) be the temperatures under the same conditions except that the face x = l is kept at temperature A from t = O ~o t = to and thereafter at temperature zero; thus

-

V(I,t) = A

ASo(t

-

to)

(t > O).

(a) Show that

V(x,t) = A[V(x,t)

-

V(x, t - to)].

(b) Note that the total quantity of heat conductedacrossa unit areaof a section x = xo, Q = - K So Vx(xo,t)dt, is the value of L{ Vx(xo,t)} when s = O if that transfonn exists when s = O. Thus for the above slab with temperatures V show fonnalIy that Q

= -AKto.

9. Obtainthesolutionof the temperatureproblem V,(x,t) = V xx(x,t) V(x,O) = 1, I

For a proofsee the author's "Fourier

1963, which

also treats

uniqueness

(O< x < 1,t > O),

V(O,t) = V(I,t) = O,

Series and Boundary Value Problems," 2d ed., chap. 10,

of solutions.

PROBLEMS IN HEAT CONDUCT/ON

235

[SECo 81

in the fonn

~

U(x,t ) = -4 L..- sin(2n 1t n= 1

10.

-

2n -

1)1tx 2 2 exp [ - (2n - 1) 1t t] .

1

Derive the fonnula U(

x,t

)

-

I -

-

~ ~ (_l)n-I 1t /;;'1 2n

(2n - 1)1tx - I cos 21 exp

[-

(2n - 1)21t2t

412

J

for the temperatures in a wall with its face x = O insulated and its face x = 1 kept at temperature U = 1, if the initial temperature is zero (Fig. 78) and if k = l.

UI%,OI-O

U-l

o

%

Fig.78

11.

Establish the fonnula in Probo 10 as a solution of the boundary value problem.

12.

Find U in Probo 10 as a series of error functions.

13.

Ans. U = I - U o(x,t), where U o is the function (4), Seco48, when Uo = k = l. Obtain the solution of the problem in Seco48 in the fonn U(

(2n- l)1tx ) - 4uo ~ (-Irl x,t - 1t nf¡ 2n - I cos 21 exp

[-

(2n

-

1)21t2kt

412

J.

14. Let the temperatureof the facex = 1of the wall in Probo10 be F(t),where F is continuous,F' is sectionallycontinuous,and F(O)= O. Derive the temperaturefonnula 4

00 (-l)n-I

U(x,t) = F(t) - -1t n= L1 2n where

' Gix,t) =

15.

U

Jo F'(t -

r) exp

The face x = O of a slab is insulated.

cos

.

[

(2n - l)1tx ..., Gn(x,t),

(2n - 1)21t2r

-

412

J dr.

The temperature

= t(t ~

of the face x

= 1t

is

O). Ir the slab is initially at temperature zero throughout and k = 1. (a) derive the fonnula

U(x,t) = t

1t2 - X2 - ---z-

f

- .1tn=1 16 (-I)"cOS (n - tlx exp -(2n - 1)2~ (2n-l) 4J [ (b) Verify this solution of the problem.

(O~ x ~ 1t,t ~ O).

. ..

236

16.


SECo 81]

Solve the boundary value problem cU,(x,t) = KU xx(x,t) + R(t),

Ux(O,t)= U(I,t) = U(x,O)= O,

for the temperatures in an internally heated bar (cf. Probo 11, Seco51) in the form

U(x,t)=

L R(t -

-r)G(x;t)d-r,

4 00 (_1)n+ 1 (2n - I)nx I cos 2 exp

G(x,t)= -nc n=1 L 2n-

[

-

(2n - 1)2n2kt 4 J

where k = K/c. Note that G(x,t) is formally the temperature distribution when R(t) = b(t). 17. Ir heat is introducedthrough the face x = 1 of a slab at a uniform rate A per unit area while the face x = O is kept at the initial temperature zero of the slab, the temperature function U satisfies the conditions U,(x,t) = kU xx(x,t)

(O< x < 1,t > O),

U(x,O)= U(O,t)= O,

Writem = n -

t and derive the solution A 2 00 (-I)n.. U(x,t)= _ L mr K x + 2n n=

[

1

s10 (mnx) exp (-m2n2kt)

J.

18. At the face x = Oof a wall, heat transfer takes place into a medium at temperature zero according to the linear law of surface heat transfer, so that Ux(O,t)= hU(O,t) (h > O). Units are chosen so that k = 1,and the wall has unit thickness. Ir the other conditions are those indicated in Fig. 79, set up the boundary value problem for temperatures U(x,t) and show that

- I h sinhx-fi + -fi coshx-fi . uX,s ( )s

h sinh

-fi

+

JS cosh -fi

, ,

U(.x.O)-O

0'-

U(1,t)-l %

-10

-

-----

Fig.79

Obtain formally the solution

+ I U(x,t) -- hx h+ I

- 4 L~ 2sinan(1-. 2x) exp ( - an2t), n=1 an - s10 an


where

IXI, 1X2, . . . are

[SECo 82

237

the positive roots of the equation IX tan IX=

- Ii'

Show how those roots can be approximated graphicalIy when the value of h is known, and note that IX.is only slightly greater than (n - t)71:when n is large. To show that s = Oand s = -IX/ are the only singular points of u(x,s),write p(z) = h sinh z + z cosh z and z = A + ip.and prove that p(j';) = O only if s = O or s is real and negative by showing that 2Ip(z)1 ~

Ie\/(h + A)2+ p.2- e-AJ(h - Af + p.21> O

if A > O or if A < O. Hence A = O and z

82

=

ip. when p(z)

= o.

ARBITRARY INITIALTEMPERATURES

Let the initial temperature distribution in a bar or slab be a prescribed function g(x) of the distance from one face. When the lateral surface of the bar is insulated and the ends are kept at temperature zero (Fig. 80), units can be selected so that the boundary value problem for the temperature in the bar

.

becomes

(O< x < 1, t > O),

U,(x,t) = Uxx(x,t)

(O< x < 1),

U(x,+O) = g(x) U(+O,t) = U(l - O,t) = O

(t > O).

The problem in the transform of U(x,t), (1) (2)

u"(x,s)

- su(x,s) = - g(x)

(O < x < 1),

u(O,s) = u(l,s) = O,

can be solved by any one of several methods, including a Laplace transformation with respect to x, or the process ofusing a Green's function to be described in Chap. 9. Regardless of the method used, the solution can be written in terms of Green's function R for the problem, in the form (see Probs.9 to 11, Seco83)

(3)

u(x,s)= -

f

R(x,~,s)g(~)d~,

O' Fíg.80

o

U(x,O)-g(x)

O'

l.

238


SECo 82J

where R is this symmetric funetion of x and ~:

R(x,~,s)=

(4)

-

sinh [(1

-

x)JS] sinh~JS

JS sinhJS

- - sinh[(1- ~)JS] sinhxJS

-

JS sinhJS

(O~ ~ ~ x),

(x ~ ~ ~ 1).

JS

By writing Maclaurin's series in powers of for the hyperbolie sines and properIy defining R(x,~,O), we see that, regardless of whieh branch of st is represented by ./S, R is an analytie funetion of s exeeptfor the singular

points (n = 1,2,.. .).

(5)

Ir g is seetionally eontinuous, the order of integration with respect to ~ in Eq. (3)and differentiationwith respect to s can be interchanged,and hence u(x,s) is analytic except for the singular points (5). By taking the branch

JS

JS

cut of along the positive real axis and selecting q(s) as sinh in formula (7), Seco 72, we find that the points (5) are simple poles of u(x,s), where the residues of e'lu(x,s) are % exp ( - n2n2t)

-

[f.o g(~)sin nn(1 -

1

"leos nn

x) sin nn~d~

+ Since sin nn(1

-

f

When t

-

~) d~ 1

gR) sin nn~ d~.

therefore, the solution of our problem is ~

(6)

g(~) sin nnx sin nn(1

x} = - cos nn sin nnx, the residues can be written 2 exp ( - n2n2t) sin nnx

Formally,

f

l

U(x,t) = 2 n~1 exp (- n2n2t) sin nnx f. o g(~) sin nn~ d~.

=

O, the series here is the Fourier

sine series for the function g(x),

on the intervalO < x < 1. In fact, the boundary value problem in U(x,t) here is especially well adapted to the classical method of solution by using separation of variables and Fourier series, a method we shall discuss in Chap. 9. VERIFICATION

Solution (6) can be established by the procedure used in Secs. 76 and 77. Since some variations are needed, we outline the method here. Over any


[SECo 82

239

half plane Res ~ y > O we find that IR(x,~s)1< ME(x,~,r)f0-' where E(x,~,r)= exp (-Ix - ~1j;12), r = Isl, and M is a constant. Since g is

H

g

bounded and E d~ and E d~ are 19(r-t), it follows from Eq. (3) that u is 19(r-l). This is not sufficient to show that the inversion integral applies to u. For the sake of brevity we assume g" continuous and g'" sectionally continuous (O~ x ~ 1). We integrate both integrals by parts in the expresslOn

u(x,s)= -

I: R(x,~,s)gR) d~ -

f

R(x,~,s)g(~)

d~,

and introduce an integral P of R with respect to ~,where

(7)

-

- x)j;] cosh~j;

(O~ ~ < x),

- cosh[(1- ~)j;] sinhxj; .

(x < ~ ~ 1),

P(x,~,s)=

sinh [(1

-

ssinhj;

ssinhj;

and thus isolate the term In u of weak order by writing

g(x)

(8)

U

=-

s

l

fo Pg'(~)d~.

+ g(O)P(x,O,s) - g(1)P(x,l,s)+

The integral in Eq. (8) is 19(r-3/2)in the half plane Re s ~ 1'. From Seco 76 we find that the inversion integrals of P(x,O,s)and P(x,l,s) exist and satisfy certain boundary conditions. Since Li -1{g(x)fs} = g(x), we can conclude Li - 1{u} represents

that

a function

U(x,t) that

satisfies

the conditions

U(x, +0) = g(x) and U(+0, t) = U(l - O,t) = O. From Eq. (8) we can show that u' = g(OW,,(x,O,s)- g(l)P..,(x,l,s) + fg'(~)Px(X,~,S) d~ and, by an integration by parts here, that u'(x,s) is 19(r-3/2)when O < x < 1.

That u"(x,s)also has this same order followsfrom the fact that u" = su - g and integration by parts twice in Eq. (8). Thus Uxx s(u gjs) is 19(r- 3/2), then

-

= Li - 1{u"} and,

Li -1 {s( U- ~)} = [U(x,t) - g(x)], = U,(x,t) = Uxix,t)

since

(O< x < 1). .

The function (9)

U(x,t) = Li - 1 {u(x,s)}

is therefore established as a solution of our problem.

(y > O)

240


SECo 83]

From Eqs.J.3) and (4) it can be shown that u(x,s) is O(r-t) when s is on r sin (J/2 = (n - t)n. Therefore the series (6) represents the solution (9) when t > O. In addition to this, we can see from Eq. (8) that the inversion integral of the function u(x,s) - g(x)/s converges to zero when t = O and that it is represented by its series of residues when t = O, provided O< x < 1. It follows that

the parabolas

V

l

00

f

(lO)

2 n~l sin nnx o g(~)sin nn~d~

-

g(x)

=O

(O< x < 1),

which is the Fourier series expansion mentioned above. The results hold true if g or its derivatives are sectionally continuous instead of continuous. The proof is longer, since it involves the writing of each integral as the sum of integral s over intervals on which the functions are continuous.

83

TEMPERATURES IN A CYLlNDER

Let us derive formally the temperature function U(r,t) for a solid circular cylinder of infinite length whose initial temperature is zero and whose surface is kept at unit temperature (Fig. 81). Units of time and length can be chosen so that the boundary value problem becomes I U,(r,t) = Urr(r,t)+ -Ur(r,t) r

(1)

(O~ r < 1, t > O),

U(r,O) = O

(O~ r < 1)

U(I,t) = 1

(t > O),

where the cylindrical coordinate r is the distance from the axis ofthe cylinder.

I I I I

U-l

U(r,OI-OI I I I

~~--T---,,"

r I

1 .......

Fig.81

....


[SECo 83

241

The function U is to be continuous interior to the cylinder, at r = O in particular. We can assume that U is bounded over the domain O < r < 1, t > O. Then the formal problem in the transform u(r,s) is su =-

(2)

1 u(l,s) = s'

(3)

d2u

l du +-dr2 r dr

< 1,Re s > O),

(O ;;¡; r

u(r,s) continuous

(O ;;¡;r ;;¡; 1, Re s > O).

Let J1.denote a complex constant. The substitutionst = J1.Zand y(z) = Y(J1.z)in Bessel's equation (7), Seco21, leads to the form (4)

z2y"(Z) + zy'(z) + (J1.2z2- n2)y(z) = O

The solution

(4), which is continuous

of Bessel's equation

(n = O, 1,2,. . .). at z

=

O as well

as elsewhere, is y = CJ n(J1.z),where C is any constant. Equation (2) is a special case of that equation in which n = O, J1.2 = -s, and z = r, so its solution that satisfies our continuity

requirement

is

u(r,s) = CJo(irj;)

(5)

where (Sec. 21) for each branch of j; 00

(6)

=

Jo(irj;)

1

k~o(k!)2

r

2k

( 2)

~.

Since this power series in s converges for all complex s, it represents an entire function of s for each fixed value of r. The condition u(l,s) = l/s determines C, so that

(7)

u(r,s) = Jo(irj;). sJo(ij;)

Now the zeros ofBessel's function Jo(z) are all real and simple; they consist of an infinite sequence of positive numbers (Xn(n= 1,2,...) together with their negatives - (Xn,such that (Xn-. 00 as n -. 00.1 Their values are tabulated ; in particular, to four significant figures. (Xl

Thus

= 2.405, JO(Í(Xn)

(X2

= 5.520,

= O and

(X3

= 8.654,

(X4

=

11.79.

JÓ(Í(Xn) =FO. In view of formula (7), the

singular points of u(r,s)in the plane of the complex variable s consist of the simple poles (8)

s

= O,

s = -(X/

(n = 1,2,.. .),

1For such properties of Bessel functions see, for instance, chap. 8 of the author's "Fourier Series and Boundary

Value Problems,"

2d 00., 1963. The function Jo(iz) is aIso written as 10(z).

242

since


SECo 83]

s

=

-IX/ whenever ifi

= :f:IX", for

= .jGiei812where O <

(J < 21t.

e"1 Jo(irfi)

]

s dJo(ifi)/ds

that is analytic

=-

$=_,,"2

2 Jo(-IX"r)

-IX" J~(-IX,,)

say the branch

- IX"2 it is

The residue of u(r,s)e"1 at s = Ois 1,and at s = -

fi

J - IX"2 is defined,

over the negative real axis so that

fi

a branch of

exp ( -

2

t) " ,

IX

where J~(z)= dJo(z)/dz= -J1(z) = J1(-z). FormalIy, then, J (IXr) IX"~ 1 IX", exp(-IX,,2t).

CO

(9)

U(r,t)= l

- 2 ,,=1L

(

To write this formula in terms of standard units of length and time, centimeters and seconds, for example, let P denote the radial distance and T the time in such units. Ir the radiusof the cylinderis Poand the thermal diffusivity of the material is k, then to transform the heat equation in problem (1) into Uf = k(Upp + Up/p) (O~ P < Po),we put P r = -, Po

kT t = -Z, Po

where r and t are the variables used in formula (9). AIso write V(P,T) = A U(r,t) so that the constant surface temperature is arbitrary: V(Po,T) = A.

Our temperature formula then takes the form (10)

V(P,T)

=

A

[ - ,,=f I

2

JO(IX"p/PO)exp

1 IX"J 1(IX,,)

(

-

IX/~T

.

}J

Po

PROBLEMS 1 . The initial temperature of a slab is U(x,O)= Ax. Ir the faces x = O and x = I are kept at temperature zero, derive the temperature formula

2AI U(x,t)= 2.

n

. n1tx

n21t2kt

00 (-1)"-1

L

1t ,,=1

exp

(

-

y-

I

)

Sln-.

I

Derive the following formula for the temperature function in Probo 1:

~ U(x,t)= Ax - Al ,,=0 L- { en

(2" + 1)1+ X

[

2, /ktkt

] - erf[

(2n + 1)/- x '" ~Ktr¡¡:. ]} .

3. Ir the slab in Probo 1 is 20 cm thick and is made of iron for which k = 0.15 cgs unit, and if the initial temperature varies uniformly through the slab from O to 100°C, find to the nearest degree the temperature at the center after the faces have been kept at O°C(a) for 1min, (b) for 100mino Ans. (a) 48°C; (b) O°C.


[SECo 83

243

4. The facesx = Oand x = Iof a slab are insulated (Fig. 82)and the initial temperature distribution is prescribed: U(x,O)= g(x). Find the transform of the temperature function U(x,t) in the form

'

i

l u(x,s) = - k og@R(x,~,s) d~, where

R(x,~,s)= - coshx# cosh(1- ~># #sinh/#

when x ~ ~ and R(x,~,s) = R(~,x,s). Then obtain the formula

2

where

i'

mr:x

(n = O, 1, 2,

an= 1 o g(x)cos¡dx

. . .).

U(x.O)-g(x)

o

%

1Fig.82 5.

In Probo4, if g(x) = A when O < x < 1/2and g(x) = Owhen 1/2 < x < 1,show that

-

U(x,t ) -- A 2

2A

~

(_l)n-¡ (2n - 1)1tx 1 cos I exp

+-'2 1tn=¡n-

[

-

(2n - 1)21t2kt

12

J

.

6. The facex = Oof a slab is kept at temperature zero while the face x = 1 is insulated. If k = 1 and the initial temperature is U(x,O)= x, derive the temperature formula

-~ U(x,t)

00 (-1)"+

- 1t2n~¡ (2n -

¡ . (2n - l)1tx

1)2SIn

2

exp

[-

(2n

-

1)21t2t

4

J.

7. The initial temperature of a cylinder of infinite length is zero. If the surface r = 1 is kept at temperature A from t = Oto t = to and at temperature zero thereafter, and if k = 1, derive the following formula for the temperatures in the cylinder: W(r,t) = A[U(r,t)

-

U(r, t

-

to»),

where U(r,t) is the function defined by formula (9), Seco83, when t 6; Oand U(r,t) = O when t < O.

244


SECo 83]

8. The flux of heat into a cylinder of infinite length through its surface r = l is a constant, so that U,(I,t) = Ao If k = l and the initial temperature is zero, derive the temperature formula

where p¡, P2'

are the positive roots ofthe equation J ¡(fJ) = O,given that the equation has only real roots and that J'¡(Pn)"# 00 Note that according to Bessel's equation -J;;(x) = Jo(x) - J¡(x)/x, and since -J;;(x) = J~(x) it follows that J~(Pn)= JO(Pn)o 9. Use symbolic transforms, as in Seco13,to indicate formally that the solution of the problem o

o

o

R(O,~,s)= R(I,~,s) = O, where O < ~ < 1, is Green's function (4), Seco820 10. Multiply the members of all equations in Probo 9 by - g@ and integrate from ~ = Oto ~ = l. Formally interchange the operations of differentiation with respect to x and integration with respectto ~, as might be justifiedif ó(x - ~ in Probo 9 were

replacedby l(h, x

-

~),to indicatethat the function u(x,s)=

-

f

R(x,~,s)gR)d~

is the solution of the problem (Seco82) d2u ~-su= dx

-g(x) '

u(O,s) = u(l,s) = 00

Write R in the form (4), Seco82, and verify the solutiono 11. (a) Note that Green's function R defined by Eqso (4), Seco82, is a continuous function of x (O~ x ~ 1) when O < ~ < 1, but show that its derivative R" has a unit jump at x = ~: (O< ~ < 1)0 (b) Integrate the members of the equation d2R

-dX2 with respectto x from ~

sR

= ó(x -

~)

(O <

~<

1)

-

( to ~ + (and let ( tend to zero to indicate formally that R" has a unit jump with respect to x at x = ~o

12. Show that the derivative R,,(x,~,s) of Green's function R in Probo 4 has a unit jump with respect to x at the point x = ~(O< ~ < 1);also showthat R itselfis a con-

tinuous functionof x (O~ x ~ 1)that satisfiestheseconditions: d2R k- - sR = O (x "# 0, dX2


84

[SECo 84

245

EVAPORATION FROM A THICK SLAB

We now illustrate a type of problem that is especiallywell adapted to treatment by the Laplace transformation in contrast with other methods.

Let C(x;r)denote the concentration of moisture at time 't" in a porous semi-infinite solid (x ~ O)with a uniform initial concentration Co (Fig. 83). Evaporation of moisture takes place at the face x = O into a dry adjacent medium according to the linear law tbat the outward flux is E[C(O,'t")- O], where the positive constant E is a coefficient of evaporation. A boundedness condition on C will take the place of a condition of diffusionat a right-hand boundary. The condition that C(x,'t")is 19(eat)uniformly with respect to x, whenever ex> O,will be sufficient. Then if K is the coefficient of diffusion of moisture in the solid, C(x,'t")satisfies the conditions Ct = KCxx,

(x > O,'t"> O);

C(x,O) = Co,

(x > O);

(1) KCx(O,'t") = EC(O,'t"),

lC(x,'t")1< M~t.

In terms of new variables t, V, and a positive constant h, where (2)

t = K't",

E h= K'

V(x,t) = C(x,'t") Co '

the problem in the relative concentration V(x,t) becomes (3)

VI = Vxx,

(x> O,t > O);

Vx(O,t)= hV(O,t),

V(x,O) = 1,

IV(x,t)1<

(x > O);

Ne{J1

for some constant N and each positive constant p. The problem in the transform of V becomes,formally, (4)

su(x,s) u'(O,s) = hu(O,s),

-

1

= u"(x,s)

(x > O,Re s ~ y > P),

N lu(x,s)1< y-p'

O

----10

C(X, O) = Co

--

Fig.83

x

246

SECo 84]

Ir.j';=jr


ei912where -11: < (J < 11:and s

can be written as

I

(5)

= rei9,the solution of problem (4)

h

u(x,s) = -s - s(h + .j';s) exp(- x.j';). SinceRe.j'; < Owhen e =F11:, then h + .j'; =FOand u is an analytic function of s everywhere except on the nonpositive real axis. The ray (J = 11: is the branch cut of.j'; and of u. The inversion integral on the line Re s = y applies to both terms on the right in Eq. (5) since the second term is (Il(S-3/2), uniformly with respect to x. According to our theorems in Chap. 6, it represents a function (6)

V(x,t) = I - Li - 1

{ s(h +h .j';s) exp (- x.j'; )}

(t > O,x ~ O)

that is (Il(eY')uniformly in x where y can be any positive number. Moreover that function is continuous in x and t, and V(x, + O)= l. When x ~ Xo > O, the function inside the braces is (Il(s-m)when Re s ~ y, for each integer m, and it follows readily that V, = V xx when t > O and x > O. Thus the function (6) satisfies all the conditions (3) except possibly the condition V x(O,t) = hV(O,t).

The procedure in Seco74 of altering the path of integration of the inversion integral can be applied to the last term in Eq. (6) with no essential change. In particular we find that the integral around the small circle Isl

=

ro in Figs. 73 and 74,

-

-x.¡s

9="

1

211:i

f

e lf'ds 9= -" s(h + .j';)

tends to 1 as ro -. O. The integráIs over the rays s = re-i" and s = rei" constitute the rest of the value of V(x,t) - 1, and the result is this more usefulrepresentation of the function (6): (7)

V( x,t )

=~

J

a)

hsinx~

e-" dr

+ ~cosx~ 2

r (h +r )

11:o

(t > o, x ~ O).

Since Iy-l sin yl O,we can see that the integrand in formula (7)satisfiesthe condition .

I

when

O ;;¡;¡ x

xh sinx~/(x~) ~(h2

+ cos x~

e-" < xoh + 1 e-rt

+ r)

h2~

I

< xo. Thus the integral (7) converges uniformly with respect

to x(O;;¡;¡ x < xo)when t > Oand V( + O,t)=

h

V(O,t)= -

f

a)

11: o

e-rt

h- 2

dr r.:

+ryr

(t > O).


[SECo 85

247

The integral of the derivative with respect to x is uniformly convergent in x when t > Oand x ~ O,so that Ux( + O,t) =

h2

Ux(O,t)= -

f

co e-rt

1t o

-h2

dr

+ryr

1:

(t > O).

Thus Ux(O,t)= hU(O,t) when t > O,so the function U with representations (6) and (7) satisfies all conditions in boundary value problem (3). That function has the transform (5); thus u(Os)

,

l

h

=- s

s(h

+

JS)

=-

l

s - h2

-

h

JS(s - h2)

and it follows from Probo 4, Seco17,or transform 40, Appendix A, Table A.2, that 2

fco

(8) U(O,t)= exp (h2t) erfc (h.jt) = ~Jo

exp(-y2

- 2h.jty)dy

~ l.

Using formulas (7) and (8) and the substitutions (2), we can write the concentration C in the slab and at the face as 2hCo

-

f

co

h sin xy + y cos xy 2 exp ( - K -ry ) dy, 2 2 y(y + h )

(9)

C(X,-r= )

(10)

C(O,-r)= Coexp (h2K-r)erfc(h~).

1t

o

Another representation of C(x,-r),in terms of error functions, can be written from our formula for u(x,s) (Prob. 1, Seco85). 85

DUHAMEL'S FORMULA

In Seco80 we obtained a formula for the temperatures in a bar with variable end temperature, in terms ofthe temperature function when the end temperature is constant. We now obtain a more general formula that simplifies heat conduction problems in the same way. Let U(x,y,z,t) be the temperatures in any solid, filling a regio n R, that is initialIy at temperature zero throughout. Let the temperature at every point of some part S of the boundary be a prescribed function F(t) of time, and let the remainder S' of the boundary be kept at temperature zero (Fig. 84). Ir A(U) represents the linear differential form

(1)

l

a au

a au

a au

A(U)=- - K+- K+- K, e[ ax( ax ) ay( ay ) az( az )]

248

OPERATIONAL

SECo 85]

MATHEMATICS

S'

Fig.84

the equation of conduction can be written (2)

Vt

= A(V)

[(x,y,z) in R, t > O].

We assume that the thermal coefficients K and e are either constants or independent of t. The boundary value problem in V consists of Eq. (2) and the conditions (3)

V(x,y,z,O) = O

V=O

(4)

= F(t)

interior to R, on S' onS.

The transform u(x,y,z,s) then satisfies the conditions (5) (6)

su

= A(u)

inR, on S'

u=O

= f(s)

onS.

Let V(x,y,z,t) be the temperature function V when F(t) = 1; that is, V satisfies the heat equation (2), the initial condition (3), and the surface conditions onS' V=O (7) on S. = 1 Then the transform v(x,y,z,s) satisfies the differential equation (5) and the conditions (8)

v=O

onS'

-1 s

onS.

Since s is a parameter in the linear homogeneous differential equation (5), the product of sf(s) by the solution v is also a solution. But according to conditions (8),

sf(s)v = O = f(s)

onS' onS;


[SECo 85

249

thus the function sf(s)v also satisfies all the conditions (5) and (6). Ir the problem consisting of conditions (5) and (6) is to have but one solution, then sf(s)v must be the same as the solution u: (9)

u(x,y,z,s) = sf(s)v(x,y,z,s).

Since sv is the transform of

v"

it follows from Eq. (9) with the aid of

the convolution that

I: F(t - T)1í;(X,y,z,T) dT o ' = ot oF(t - T)V(X,y,Z,T) dT.

(lO)

U(x,y,z,t) =

f

These are two cases of Duhamel's formula representing temperatures U corresponding to variable surface temperatures in terms of temperatures V in the simpler problem where surface temperatures are constant. A more general case is given in Probo 14. Given that Vis the solution of the simpler problem, the solution (10) ofthe problem in U can be verified directly under reasonable conditions on F. PROBlEMS 1. From formula(5),Seco84, for u(x,s)and transform 86,AppendixA, Table A.2, obtainthisformulafor the concentrationofmoisturein the slabx ;¡;;° withevaporation at the surface:

C(X,f) = c{ erf(~)

+ exp (hx + h2 Kf) erfc

(~

+

h~) J.

2. Identify the two expressions in formula (8), Seco85, for the concentration U(O,t) at the face of the semi-infinite solid and thus show that ° < U(O,t)~ 1 when t ;¡;;O. 3. A layer of finite thickness extending from the face of the slab in Seco84 is initially dry. The initial concentration of the rest of the semi-infinite slab is uniformo Choose units so that the relative concentration U(x,t) satisfies the conditions

U, = U""',X> O,t > 0,

U(x,O)= So(x

-

1),

u ,.(O,t) = hU(O,t),

and a boundedness condition; also U and Ux are to be continuous when t > 0, at

x = 1 in particular. Showformallythat 2su(x,s)

= exp [-(1 - x)../S] - exp[-(1 + x)j;]h - Jf:.

h + yS and use the tablesof transformsto get theseparticular results: U(O,t) = exp (h + h2t) erfc

(zfi+ hJt),

U(l,t)= ~erf(jr) + exp(2h+ h2t)erfc (jr + hfi).

(O~ x ~ 1),

250

A'-

-


SECo 85]

U(%.O)-O

Fig.85 4. The face x = ° of a semi-infinite solid (Fig. 85) is exposed to a medium at constant temperature A. Heat is transferred from that medium to the face of the solid according U(O,t)],where U(x,t) is the temperature in the to the law that the flux of heat is E[A solido Thus the boundary condition at the face becomes

-

U..(O,t)= h[U(O,t)- A],

where h = E/K. Ir the initial temperature is zero, derive the followingformulawith the aid of the tablesin AppendixA:

U(x,t)

= A[ erfc

(~)

- exp(hx+ h2kt)erfc(hJkt + ~)

J

5. Let V(x,t) be the temperaturefunctionfor the solid in Probo4 when A = 1. Let W(x,t) be the temperature of the solid when the constant A is replaced by a function II>(t), so that the medium to which the face is exposed has a variable temperature. Derive the formula W(x,t) = f~ II>(t- t)V,(x,t)dt. 6. The temperature of the faceof a semi-infinitesolid x ~ ° varies in the following manner: U(O,t)= A sin rot. Taking the initial temperature as zero, for convenience, show that, when t is large, the temperature U(x,t) is approximately V(x,t), where

V(x,t) = A sin (rot

- x~) exp(-x~),

a simple periodic function of time. Note that the closed contour corresponding to the

one shownin Fig.73willendose two simplepoless = :!:iwofu(x,s)in this case. (Also, see Probo 7.) 7. Use the following elementary method to obtain the formula in Probo 6 for the undamped component V of the temperature function U. Assume V periodic in t with frequency ro, for each x, and write V(x,t) = 1m W(x,t) where W(x,t) = F(x)ei"", F is complex-valued, and Jv. = kWn, W(co,t) = 0, W(O,t)= Aei..,. Then V, = kVn, V(co,t) = 0, and V(O,t)= A sin rot. Thus write the problem in ordinary differential equations in the function F and show that F(x) = A exp[-(1 + i)xJW/J2k].


[SECo 85

251

8. The diffusivity k of the earth 's soil in a certain locality is 0.005 cgs unit. The temperature of the surface of the soil has an annual variation from 8 to 22°C. Assumingthe variation is approximately sinusoidal (Prob. 6), show that the freezing temperature will penetrate to a depth of approximately 170cm (considerably less, because of the latent heat of freezing).

-

9. In Probo 8 find the approximate depth at which the variation of temperature with time is 6 months out of phase with the variation of the surface temperature. Show that the amplitude of the variation at that depth is less than 1°C. Ans. x = 705 cm = 23 n, approximately. 10. Let the functions V(x,t) and W(y,t) satisfy the heat equations V. = kJ.~x and ~ = k»jy, respectively. Prove by direct substitution that the product of those temperature functions, U(x,y,t) = V(x,t)W(y,t), satisfies the heat equation

u,

= k(Uxx

+

Uyy).

Ir in addition V(O,t) = V(a,t) = O and W(O,t) = W(b,t) = O, and if V(x,O) = f(x) and W(y,O) = g(y), then show that U(x,y,t) represents the temperatures in a rectangular plate (Fig. 86) with insulated faces, if the edges are at temperature zero and the initial temperature is U(x,y,O) = f(x)g(y).

U-o U-o

U(x,y,O)-

o

Fig.86

fTx)g(y)

U-o

(a,b) U-o

x.

11. Use the product of solutions (Prob. 10) to obtain the following formula for the temperaturesin an infiniteprism with a squarecrosssection,if the initial temperature is A and the surfacetemperatureis zero, if the unit of length is the side of the square, and k = 1: V(x,y,t) = A U(x,t)U(y,t), where U is the temperature function found in Probo 9 of Seco81. 12.

With the aid of Probo 10, derive the formula 4 U(x,y,t)

x

00

= -7t erf ( 2y r.t.) n=L 1

sin(2n -

I)lty

2n - l

exp[ - (2n -

l )2lt2t]

for the temperatures in the semi-infiniteslab x ;¡;;; O,O;;;¡; Y ;;;¡;1 with its boundary at temperature zero, if U(x,y,O)= 1 (Fig. 87) and k = 1.

t A

252

OPERATIONALMATHEMATICS

SECo85]

YI

U-O

"'''1 O

U-O

U(x,y,Oj -1 U-O

\

"

Fig.87

13. Generalize the method of Probo 10 to the case of three dimensions, and give an illustration of its use in finding temperatures in a cube. 14. Use the generaJized convolution property found in Probo 20, Seco27, to make a formal derivation ofthe following more general Duhamel formula. Let A(U) denote the differential form (1), Seco85, P the point (x,y,z), n the distance normal to the boundary surface S or S', and let a(P), b(P),H, F, and G represent prescribed functions. The temperature function U(P,t) in the region R is to satisfy the conditions

U, = A(U) + H(P,t),

U(P,O)= G(P)

dU a(P)U+ b(P)dn = F(P,t)

(P in R); (P on S or S').

When the functions H(P,t) and F(P,t) are replaced by H(P,t') and F(P,t') respectively, where t' is any fixed vaJue of t, V(P,t,t') denotes the solution of the problem. Write the problem in the transform v(P,s,t')and transform its members with respect to t', using the same parameter s, to obtain a problem in the iterated transform D(P,s,s)and thus derive this generaJization of formula (10):

a f' U(P,t) = ai Jo V(P, t

-

T, T) dT.

Note that if a(P) = F(P,t) = Owhen P is on S', then the surface S' is insulated. Thus the generalized formula could be applied to Probo 14, Seco81.

8 Problems in Mechanical Vibrations

This chapter contains further illustrations of the uses of those properties ofthe Laplace transformation that involve complex variables. The problems taken as illustrations deal with vibrations and resonance in continuous mechanical systems-systems in which the mass and elastic characteristics are distributed over the system. Consequently these problems are boundary value problems in partial differential equations, of the type treated in Chap. 4. It is the intention here to present fairly simple physical problems in their mathematical form, although the mathematical problems may have more important physical interpretations. Some electrical analogs of mechanical vibrations, involving transmission lines, are included among the exercises. 86

A BAR WITH A CONSTANT FORCE ON ONE END

In Seco 44 we derived a formula for the longitudinal displacements in an elastic bar in the form of a prism, when one end of the bar is fixed and a 253

254

OPERATIONAL

SECo 86]

MATHEMATICS

Fíg.88

constant force Fo per unit area acts parallel to the bar on the other end (Fig. 88). Let all parts of the bar be initially at rest and unstrained. The displacements Y(x,t) then satisfy the conditions in the boundary value problem Y,,(x,t) = a2 Yxx(x,t)

(1)

(O< x < c, t > O),

Y(x,O) = Y,(x,O) = O,

Y(O,t)= O,

Ey"(c,t)= Fo,

where a2 = E/p, E is Young's modulus of elasticity, and p is the mass per unit volume of the material. Let us obtain another formula for Y(x,t) here. From problem (1)we found this transform of Y(x,t): aFo sinh (sx/a) y (x,s ) = , E S2cosh (sc/a)

(2)

an analytic function of s everywhere except at the origin and at the zeros :tsn of cosh (sc/a), where (2n

(3)

Sn=

- l)na. 2c

1

(n = 1,2,...).

When s ::/=O and s ::/=:tsn, y is continuous in x and s, real-valued when s is real, and y(O,s)= O. Now the quotient of hyperbolic functions in formula (2), Q(x,s) = sinh (sx/a) cosh (sc/a) fails to enhance the order property of y over a right or left half plane of the variable s. For if s = A + ip., then (Sec. 55) (4)

IQ(

x,s

)12 = sinh2 (Ax/a) + sin2 (p.x/a)

sinh2 (Ac/a) + COS2(p.c/a)

sinh2 (Ax/a) 1 ~ sinh2 (Ac/a) + sinh2 (J..c/a)

(J..::/=O).

Thus Q is bounded over either half plane 1..1.1 ;¡;;y if Y > O, since IQI2~ 1 + 1/sinh2 (ye/a). But Q does not always vanish there as Isl-+ OCJbecause

PROBLEMS IN MECHANICAL VIBRATIONS

IQI2~ sinh2(Ax/a)/cosh2 (AC/a), and

[SECo 86

255

this quotient remains fixed when A is

constant and Jl --+ :t oo. Therefore y(x,s) is (9(1/s2) over half planes A ~ Y and A ~ -y, when y > O, uniformly with respect to x, and L¡ -l{y} is a continuous function Y(x,t) with transform y(x,s). AIso, Y(x,O)= Y(O,t) = O. But the order properties of the derivatives Yx and Yxxare not adequate for the representation of derivatives of Y by corresponding inversion integrals. We can see from Eq. (4) that Q is bounded on horizontal lines Jl = :t mea/cbetween the singular points :t Snsince IQI2

::; cosh2(Ax/a)::; 1 - cosh2 (Ac/a) -

for all points s on those lines. Thus y is (9(1/s2)on those lines, and L¡-l{y} is represented by the infinite series of residues of esty(x,s) (Theorem 10, Seco73). We can write

a function that is analytic at the point s = O. In viewof Eq. (2) then the origin is a simple pole of (f'y(x,s) when x =1= O, and the residue there is (aFo/E)(x/a), or Fox/E. The residue of y itself at the simple pole Sncan be written

-Foa2 sinh (snx/a) =-4Foc (-lt Ecs; sinh (snc/a) n2E (2n -

(5)

. (2n- l)nx

1)2

sIn

2c'

It is real-valued. Thus according to formula (13), Seco72, in which (}l = O

and a = O,the sum of the residues of (fty(x,s) at the pair of poles :tsn, or Snand sn, is

-

(6)

Pn

+

8Foc (-l)n . (2n - l)nx (2n - l)nat Pn = n2E (2n - 1)2 SIn 2c COS 2c .

Finally, then, the inversion integral L¡-1{y(X,S)}is a function Y(x,t) represented by this infinite series: (7)

Y

-

~

Fo 8c (_1)n . (2n - l)nx (2n - 1)1tat - E [ x + n2 n~l (2n - 1)2SIn 2c COS 2c ].

E;very term in that convergent series is periodic in t with period 4c (8)

To = -a =

fE 4cyE:

so the function Y itself has that periodicity: Y(x, t + To) = Y(x,t). Note,

256

SECo 87]

however,


that the series is not twice differentiablewith respect to x or t.

Neither the series nor the inversion integral representation can be used directly to verify that our function Y satisfies the wave equation. We shall sum the series (7)in order to represent Y in a form that enables us to complete the verification of our solution of problem (1). 87

ANOTHER FORM OF THE SOLUTION

In establishing the inversion integral representation of our function Y, we noted that Y(x,O) = O. Since the function is also represented by series (7), Seco86, it follows that (1)

x

= -- 8e ~ i.J

( - 1)n

sm (2n o

n2 n=1 (2n - 1)2

l)nx

(O ~

2e

x ~ e).

Each term of series (1) is an odd function of x, as is x itself, so the series represents the function x over the interval - e ~ x ~ eo When x is replaced by x + 2e, each term merely changes sign so the sum of the series is an antiperiodic function for all x, with period 2e; consequendy it is periodic with period 4e. Thus for all real x the series represents the triangular-wave function H shown in Fig. 89, described by the two conditions

(2) H(x) = x

(-e ~ x ~ e),

H(x + 2e) = -H(x)

( - co < x < co)o

'The function H also has these properties for all x: (3)

H(x)

=

-H( -x)

= H(x + 4e) = H(2e -

x).

Now the sum of the series in Eq. (1) can be written (4)

8e

~

(-I)n-1

. o (2n -

2i.J 2sm .n n=1(2n - 1)

l)nx

2e

= H(x )

(- co < x < co).

The series is, in fact, the Fourier series representing the odd periodic function H for all Xo We write m = (2n l)nf(2e) and note that

-

2 sin mx cos mat = sinm(x + at) + sinm(x - at). H[x)

Fig.89


[SECo 87

257

Then our series representation (7), Seco86, of Y(x,t) can be written F.

y=

;

4c a) (- 1)n-l

n~1 (2n - 1)2[sin m(x + at) + sin m(x {x - 71:2

-

at)]}

and, in view of formula (4), Y has this representation: Y(x,t) = Fo[x E - !H(x + at) - !H(x - at)].

(5)

We can use formula (5) to complete the verification of the solution of our boundary value problem (1), Seco86. We first note that H(x) has deriveverywhere except at the points x = :!:(2n - l)c (n = 1,2,.. .). Since any twice-differentiablefunction of x + at or x - at satisfies the linear homogeneous-wave equation Y;,= a2y"x,and x itself satisfiesit, our atives

function (5) satisfies that equation except when x and t are such that x :!: at = :!:(2n - 1)c, for which values the partial derivatives of Y do not exist. To see that the condition Ey"(c,t) = Fo is satisfied by our function (5), we note that (6)

EYx(c,t)

= Fo[1 - fH'(c + at) - !H'(c - at)].

But since H(x) = H(2c - x), then H'(x) = -H'(2c - x). Therefore H'(e + at) = - H'(2e - e - at) = - H'(e - at) and, exceptfor the values of t such that e :!: at

=

:!:(2n - l)e, Eq. (6) reduces to EYx(e,t) = Fo.

The condition Y;(x,O)= O is easily verified from formula (5). This completes the verificationof our solution. Formula (5)is well adapted to graphical descriptions of the variation of y with either x or t. The graph of H(x + at) for a fixed value of t, for example, is obtained by translating the graph of H(x) to the left through a distance ato The displacement at the end x

= e can

be written

F.

(7)

Y(e,t) =

;

[e

- !H(at

+ e) + fH(at - e)]

Fo

= -[e E + H(at - e)] since H(at + e) = - H(at + e - 2e). Y(e,t) is shown in Fig. 45. The reader can show that the force per unit area on the fixedsupport, Ey"(O,t),assumes the values 2Foand zero periodically. Incidentally,we have established this transformation here: (8)

L

-1 2a sinh (sx/a) = 2x - H(x + at) - H(x - at). { s2 cosh (se/ a) }

r:

258

88


SECo 88]

RESONANCEL'N

THE BAR WITH A FIXED END

Let the end x = O of the bar again be fixed while a simple periodic force

(ro > O)

F(t) = A sin rot

per unit area acts lengthwise at the end x = e. Ifthe bar is initially unstrained and at rest, only the end conditionat x = e in problem (1), Seco86, need be changed to read (1)

EYx(e,t) = A sin rot.

The transform of the displacement now becomes (2)

Aaro

Y(x,s) =

2 7>,

.

~

sinh (sx/a) n'

2

S cosh (se/a)

.

Here y is an analytic function of s except at the origin s = O, a removable singular point, and the points s = :t iro and s = :t Snwhere s = n

(2n - I)na. I 2e

(n = 1,2,...).

If iro is not equal to any one of the numbers Sn' that is, if

ro#--

(n

-

!)na

e

(n = 1,2,...),

the singular points :t iro and :t Snare all simple poles unlessthe value of x

case they serve as simple poles

is such that some are removable, in which

with residue zero. According to formula (13), Seco72, the component of Y corresponding to the pair of poles :t iro of y is a-term of the type b cos(rot + O); here the real numbers b and Odepend on x. Similar components correspond to a pair of simple poles :t Sn' Thus Yis represented formally by a series of this type: 00

(3)

Y(x,t)

+ ,,=1 L bn(x)cos[ro"t + On(x)],

= bo(x)cos[rot + Oo(x)]

where ro" = (n - !)na/e. We shall find the b's and O's in Seco89 and establish

formula (3) as a solution of our boundary value problem in Y. The series shows that each section of the bar moves as a superposition of two periodic motions, one with frequency ro and the other with frequency rol = na/(2e), or period TI = 4e/a. But if the frequency ro of the external force coincides with one of the frequencies ro", say ro = ro, for some positive integer r, then the function se q(s) = (S2 + ro,2)cosha


[SEC.89

259

in the denominator of expression (2) for y(x,s) is such that q(iw,) = q'(icD,)= O while q"(iw,) =FO. Consequently the points s = :t iw, are poles'of y óf the second order, and corresponding to that pair of poles, according to formula (16), Seco72, Y(x,t) contains an unstable component of the type (4)

tb,(x)cos[w,t

+ I/I,(x)],

where b,.(x) =Fo.

The remaining component of Y is periodic with frequency Wl>the common frequency of all its periodic terms. That unstable oscillation of sections of the bar is called resonanee. The periodic external force is in resonance with the bar when its frequency w coincides with any one of the resonance frequencies

(5)

w, = (r - e!)1la

(r =

1,2,...).

The frequenciesw, depend on the physical properties of the bar and the mannerin whichthe bar is supported. If the end x = O of the bar is free or elastically supported, for instance, we find that the set of resonance frequenciesis differentfrom the set (5). To produce resonance, the external force need not be restricted to the simple form A sinw,t. For any prescribed force F(t) at the end x = e, the transform of Y(x,t) is (6)

a sinh (sx/a) y(x,s) = "Ef(s) s cosh (se/a)'

Consequently if F(t) contains a term A, sin w,t or B, cos w,t, where w, is a number of the set (5), then y(x,s) will contain a term with the product (S2 + w/) cosh (se/a) in the denominator, and Y(x,t) will have an unstable component of type (4). In fact, if F is any periodic function with a frequency w" then resonance will occur, as can be seen from the form in Seco 23 of the transform of a periodic function. 89

VERIFICATIONOF SOLUTIONS

When the formal solution of a boundary value problem can be written in terms of a finite number of simple functions, that form is usually best for verifying the solution, as we illustrated in Seco87. But to establish a solution represented only by an infinite series or by an improper integral, the procedure based on properties of its transform, used in Chap. 7, may be useful. Let us illustrate a combination of the two methods for the problem in the preceding section. The following transform of displacements Y(x,t) in a bar with its end x = Ofixed and with pressure A sin wt applied at the end x = e was found

260


SECo 89]

formally: B sinh (sx/a) y(X,S)= S2+ ro2s cosh (se/a)

(1)

The quotient of hyperbolic functions here was shown (Sec. 86) to be bounded

uniformly in x over half planes Re s ~ ')1and Re s ;;¡;-

')1

when ')1 > O;

cosh (sx/a)/cosh (se/a) is also bounded there. Hence y is l!J(S-3)and Yx is l!J(s- 2) over those half planes, uniformly in x, and L¡- 1 {y} along a line Re s = ')1represents a continuous function Y(x,t) such that Y(x,O) = Y(O,t) = O; also Y has continuous derivatives of first order,

y':(x,t)= L¡- 1 {yx(x,s)},

Y,(x,t)= L¡- 1{sy(x,s)},

such that Y,(x,O)= Oand Ey':(e,t) = A sin rot. Hence Y satisfies all boundary conditions in the problem, but the order of y is not adequate to show directly that Y/t =

a2yxx.

To prove that Y does satisfy the wave equation, we write 1 1 ro2 S2 + ro2 = S2 -

(2)

B sinh (sx/a)

u(x s) = -

,

S S2 cosh (se/a)'

Then

S2(S2

v(x s) ,

+ ro2)'

- Bro2sinh (sx/a)

= S3(S2 + ro2) cosh (se/a).

y(x,s) = u(x,s) + v(x,s)

where v is l!J(s-S)and Vxxis (9(S-3). It follows readily that L¡-l{V} satisfies the wave equation. The function L¡- 1{u} can be written as a linear combination of known solutions of the wave equation since (Sec. 87)

2au(x,s)= Bs-1 L{2x - H(x + at) - H(x - at)} where H is the triangle-wavefunction shown in Fig. 89. Thus L¡-l{U} is the function B

U(x,t)

'

fo [2x - H(x + a-r)- H(x - a-r)]d-r = _B 2xt - -l H@ d~ . 2a [ af ] = 2a

x+al

x-al

In terms of the periodic function (3) G(r) = we can write U in the form B (4)

I: H(~)d~

(-
U(x,t) = 2a2[2axt - G(x + at) + G(x - at)],


[SECo 89

261

fram which we see that V" = a2V"",oTherefore our function -1

(5)

Y(X,t) = L¡

B sinh (sx/a)

-1 {y(X,S)} = L¡

{ s(s2 + ro2) cos h (se/)a }

satisfies the wave equation and all the boundary conditionso We shall use Theorem 10, Seco73, to prave that the series of residues of e"y(x,s) converges to the inversion integral for each frequency ro. Then we shall have a series representation of our solution (5). Qur function y is analytic in s except for the poles (6)

wherem =- (n

and

:tiro

-

n

!)n

e

(n = 1,2,.. ')0

As noted in Seco86, the quotient sinh (sx/a)/cosh (se/a) is bounded on the horizontal lines 1ms = :t ann/eo Thus y(x,s) is (!)(s- 3) on all such lines for which ann/e > ro. Since y has that order over the left half plane Re s < -y, the infinite series of residues of esty(x,s) converges to L¡-l{y} when t ~ O and O ~ x ~ eo Let us write that series explicitly when ro is not one of the resonance frequencies amn. Then the poles of y are all simple, and formula (13), Seco72, gives the sum of the residues of esty at each of its pairs of conjugate poles. The residue of y itself at the poles s = iro and s = iamn are easily found to be B sin (rox/a) B( -I)n sin mnX -1 an-I d (7) 2ro2 cos (roe/a) emn(amn2 - ro2)' o

o

respectively. The sum of the residues of e"y at the pair of poles :t iro is therefore B sin (rox/a)

cos rot ro2cos (roe/a) (

-

n

- . 2) '

at the pair :t iamn the sum is 2B(

-

1)" sin mnx 2

( emnamn

-

n

cos amnt - _ 2) ro ) ( 2

o

Thus we have this representation of our solution (5): (8)

Y(x,t) = Bsin (rox/a) sin rot ro2 cos (roe/a)

2n-1 ( mn=~n

)

o

When ro coincides with one of the resonance frequencies am" then y has a conjugate pair of poles of order two, and the series for Y contains a resonance term of type (4), Seco88.

262


SECo 89]

PROBLEMS

1. The end x = O of an elastic bar is free while a constant longitudinal force Fo per unit area acts longitudinally at the end x = c (Fig. 48). The bar is initially at rest and unstrained. One formula for the longitudinal displacements Y(x,t) in the bar was found in Probo 8, Seco44. Derive this altemate formula: Y = -Fo

2

22

2

12c2

~ (_I)n

mrx

mrat

x + 3a t - c r '-- ---y-cos-cos6Ec[3 n n=¡ n c 2. Reducethe solutionof Probo1 to the form

c ]

.

.

F.

Y(x,t) = 12~c[6x2+ 6a2t2-

2C2

+ Q(x + at) + Q(x - at)]

where Q is this periodic function: (-00 < x < 00).

Q(x + 2c) = Q(x)

(-c
Use that form to verify the solution of the boundary value problem. 3. An unstrained elastic bar or heavy coiled spring is clamped along its length c so as to prevent longitudinal displacements, then hung from its end x = O (Fig. 90). At the instant t = O,the clamp is releasedand the bar or coil vibrateslongitudinally because of its own weighí. Thus Y,,(x,t) = a2Yxx(x,t)+ g

(O< x < c, t > O),

where g is the acceleration of gravity. Complete the boundary value problem, write m = n - 1. and derive the formula g

[

4C2 1

. mnx

mnat

Y(x,t) = x(2c - x) - - "-sm-cos2a2 n3 nf'¡m3 c

c ]

.

Fig.90

4. Reducethe solution of Probo3 to the form Y(x,t) = 4:2 [2P(x)- P(x + at) - P(x

-

at)],

where the function P is described by the conditions P(x)

= x(2c -

x)

(O~ x ~ 2c),

P(x + 2c) = -P(x)

Use that form to verify the solution of the problem.

(-00 < x < 00).


[SECo 89

263

5. The ends x = :t e of an unstrained shaft spinning about its axis with angular velocity w are brought to rest at the instant t = O (Fig. 52, Probo 13, Seco44). Derive this formula for the angular displacements of its cross sections: O( ) O' x,t

8cw ~ (_1)n-1 (2n - 1)1tx. SIn (2n - 1)1tat . 2 '-2 cos 2e 2e 1t a "=1 (2n - l )

--

6. Reducetheformulafore in Probo5 to the form w

e (x,t) =

2a [H(x + at)

-

H(x

-

at)],

where H is the triangular-wave function shown in Fig. 89 and represented by series (4), Seco87. Use this simple representation of e to verify the solution of the boundary value problem in Probo 5. 7.

The end x = O of an elastic bar is fixed while a force proportional to t2 acts

10ngitudinalIy at the end x = l so that Yx(l,t) = At2, where Y(x,t) denotes longitudinal displacements. The unit of time is such that a l. If the bar is initially at rest and

=

unstrained, derive the formula Y(x,t)

=A [

X(t2

x3

- 1)+ _ 3

4

00 (-1)" ¿

-"4 ~ 1t n=1 m

sin m1txcos mnt

where m = n-t. Verifythe solutionfullywith the aid of the representation of Y. (Also, seeProbs.8 and 9, below.)

8.

]

inversionintegral

Reduce the solution of Probo 7 to the form

l

X3

Y(x,t)

= A [ X(t2 -

1) + 3"

l

+ '2R(x + t) + '2R(x-

]

t)

where X3 R(x) = x - 3"

(-1 ~ x ~ 1),

R(x

+ 2) =

(-00 < x < (0).

-R(x)

Show R graphically. Use this simple representation of Y to verify the solution of Probo7.

9. Showthat the function Y found in Probo 7 and its derivatives of first and second order are everywhere continuous in x and t and (9(e'")uniformly with respect to

x (O~ x ~ 1). Prove that no other function of that type can satisfy alI conditions in the boundary value problem. (Cf. Seco79; but here the proof of uniqueness is simpler

becauseof the favorablecontinuityconditions.) 10.

The end x

= O of

an elastic bar is fixed while the end x

=e

is displaced longi-

tudinalIy in the manner Y(c,t) = A sin wt. Show that the resonance frequencies are w = w" where

w,--- r1ta e

(r = 1, 2, . .. ).

Note that the lowest frequency W1 is twice as great as the lowest in the set (5), Seco88.

.

11 If the end x = O of the bar in Probo 10 is free, rather than fixed, show that the resonance frequencies are the same as the set (5), Seco88.

264

OPERATIONAl

SECo 90]

MATHEMATICS

12. A simple periodic longitudinal force acts uniformly on all elements of an elastic bar with its end x = O fixed and its end x = e free, so that the displacements Y(x,t) satisfy the conditions 1';,(x,t) = a2Y"",(x,t) + B sin (Ot, Y(O,t) = Y,,(c,t) = Y(x,O) = 1';(x,O)= O.

Show that resonance occurs when (O= (o" where (0=

,

(r - t)na e

(r = 1,2,...).

13. A simple periodic transverse force acts uniformly on all elements of a stretched string of length e with fixed ends, so that the transverse displacements Y(x,t) satisfy the equation 1';, = a2y= + B sin (Ot. Let the string be initially at rest in its position of equilibrium. Show that resonance occurs only when (O= (o,where (O,

= (2r -e 1)1ta

(r= 1,2,...).

14. The supports of the ends of a stretched string vibrate in a transverse direction so that Y(O,t)= A sin at,

Y(c,t) = B sin pt.

The string is initially at rest on the x axis. In general, show that resonance takes place p has one of the values n1ta/c. But if p = IX,show that the resonance frequencies are 2n1ta/c when B = -A, and (2n - l)na/c when B = A, where

if either a or

n = 1,2,

Hint: When p = a and B = -A, showthat y( ) x,s -

90

~ sinh [s(c - 2x)/(2a)] S2 + a2 sinh [sc/(2a)] .

FREE VIBRATIONS OF A STRING

A string stretched between the origin and a point (e,O)is given a prescribed initial displacement Y(x,O) = g(x) and released from rest in that position. To find its transverse displacements Y(x,t), we solve the problem Y,,(x,t) = a2Yx.x
(O < x < e, t > O; a2 = ~),

Y,(x,O) = Y(O,t) = Y(e,t) = O.

The function g must naturally be continuous (O~ x ~ e), and vanish when x = Oand when x = e. Let g' be sectionallycontinuous. The solution of the transformed problem in y(x,s), a2y"

-

s2y

= -sg(x),

y(O,s)= y(e,s)= O,


[SECo 90

265

can be written (cf. Seco82) p(x,S) y(x,s) = a sinh (sela when s ,,¡:0, where

.

p (x,s ) = SInh

(e - x)s

.

a

xs

+ SInh a

f.o g."( ) SIn

"

c

f

" . h -~sa.""

x

.

g." ( ) SInh

a

(e - ~)sa" .". a

x

When s = 0, the solution is y(x,O)= o. Now p is an entire function of s, and the singular points of y are the simple poles s = :t Sn' where s =-

iml:a

n

(n = 1,2,...).

e

Ir we write (1)

bn

2 c

. ml:~

= - f. g(~)SIn-a~, e o e

then p(x,sn)=tbne cos mI:sin (ml:xfe),and the residue ofy at Snis tbn sin (ml:xfe). Thus the sum ofthe residues of esty(x,s) at the pair of poles :tsn is (Sec. 72)

.

nnx e

nnat e

bnSIn-COS-.

The formal solution of the boundary value problem is then

" en

.

Y(x ,t) -- f..J bn sln -cos-

(2)

nnx

nnat

e

n= 1

e '

where the coefficients bn are given by formula (1). Since Y(x,O)= g(x), formula (2)implies that en

nnx

L bn sin- e = g(x) n= 1

(O ~ x ~ e).

Each term of this series is an odd periodic function of x with period 2e. Consequently, for all x the series should represent the odd periodic extension G of g defined by the conditions G(x) = g(x)

(3) G(-x)

= -G(x),

G(x

+

when ° ~ x ~ e,

2e) = G(x) for all x.

~ ,.

266


SECo 91]

Then G is everywhere continuous, and CO

L bnsinn= 1

(4)

mtx

( - 00 < x < (0).

= G(X)

e

Since the terms in series (2) can be written as 1

. n7t(x + at)

2[b n Sin

+

e

b . n7t(x- at) n Sin 2 ]'

our formal solution has the simple representation (5)

Y(x,t) = t[G(x + at) + G(x

- at)].

The function (5) is easily verified as a solution of our boundary value problem. 118form is a convenient one to use in studying the motion of the string. 91

RESONANCE IN A BAR WITH A MASS ATTACHED

The end x = ° of an elastic bar is fixed. To the end x = e a rigid mass M is attached, and a periodic 10ngitudinal force B sin rot ac18 on that mass (Fig. 91). The elastic displacements within the mass itself are assumed negligible, and the bar is assumed to be too heavy to be considered as a coiled spring with negligible mass. We shall find the frequencies ro for which resonance takes place. Let A denote the area of the cross section of the bar, m the mass of the bar, and Y(x,t) the longitudinal displacements in the bar. The force exerted by the bar on the attached mass Mis then -AEYic,t), and so the end conditions are (1)

Y(O,t)= 0,

MY,t(c,t) = -AEy"(c,t)

+ B sin wt.

Ir the bar is initially at rest and unstrained, the remaining conditions in the boundary value problem are (2)

Y,t(x,t) = a2 y"ix,t)

(3)

Y(x,O) = Y,(x,O) = O.

(O < x < e, t > 0, a2 = ~) ,

0L!BSintl1tFig.91

PROBlEMS

VIBRATIONS

IN MECHANICAl

[SECo 91

267

This is also the problem of the torsional vibrations in a shaft with one end fixed and with a flywheel, on which a periodic torque acts, attached to the other end. Note that one end condition here involves a derivative of the second order. An interpretation of the problem in terms of torsional vibrations in a propeller shaft is given in the problems at the end of the chapter. The problem in the transform y is a2y"(x,s)

(4)

-

S2y(X,s)= O,

AEy'(c,s) + Ms2y(c,s) = s2Bw + w..

y(O,s) = O,

Since m = Acp = AEc/a2, we can write the solution of problem (4) in terms of a function q, where (5) in this form:

q(z) = Mz sinh z + m cosh z, y(

(6)

x,s

)-

-

Bcw sinh (sx/a)

as(S2 + w2)q(sc/a)'

= :1::iw and at the zeros + iJ1.,we find from Eq. (5) that

Thus y is analytic in s except at the points s of q(sc/a). But if we write z 2Iq(z)1

=

=

..1.

I(Mz + m)eZ - (Mz - m)e-ZI

~ IIMz + mieA - IMz - mle-AI

= 1[(M..1. +

m)2 + M2J1.2]té - [(M..1.- m)2 + M2J1.2]te-AI,

from which we can see that Iq(z)1 > O if ..1.> O or if ..1.< O. Therefore q(z) can vanish only if ..1.= O,and thus z = iJ1.,where J1.is a root of the equation M J1.sin J1. + m cos J1. = O; that is,

-

(7)

m 1 tanJ1. = --o MJ1.

The graphs of the functions tan J1.and m/(MJ1.)show that Eq. (7) has an infinite sequence of roots :l::J1.n where O< J1.1< n/2 and J1.n - (n - l)n ~ O

as n ~ ex)(Fig.92). Ir w = aJ1.n/c, where n is some positive integer,then the

Fig.92

268

SECo 92]


two points s = :t. iafJ.n/care poles of y of the second order and Y contains a term of resonance type bn(x)t cos [wnt + 8n(x)] with the resonance frequency

(8)

Wn = a:n

(tan fJ.n= ::fJ.n' n = 1,2,...)

.

The representation of Y(x,t) as a series of residues can be written in the usual way. Except for a term of resonance type in case W = Wn, each term of the series is periodic in t, but the terms have no common periodo

92

TRANSVERSE

VIBRATIONS

OF BEAMS

Let Y(x,t) denote the transverse displacement of a point at distance x from one end of a bar or beam, at time t. As in the case of static displacements discussed in Seco36, the instantaneous bending moment transmitted through a cross section is EIYxx(x,t), where 1 is the moment of inertia of the cross section with respect to its neutral axis. The shearing force at a cross section is Ela3Y/ax3. When the cross section is uniform and no external force acts along the beam, the displacement function Y(x,t) satisfies the equation a2y a4y at2 + a2 ax4 = O

(1)

under certain idealizing assumptions, where Ap is the mass of the beam per unit length. Let the end x = Obe hinged, so that no bending moment is transmitted across the section at x = O and the displacement is zero there. Let the end x = e be hinged on a support which moves parallel to the Yaxis in a simple harmonic manner (Fig. 93). Ir the beam is initially at rest along the x axis, the boundary conditions that accompany Eq. (1) are then Y(x,O) = Y,(x,O)= O, Y(O,t) = Y",X
Yxx(c,t) = O.

Let us find the frequencies w at which resonance will occur. y

11

lB sin c"t I 1(e,O)

Fig.93

PROBLEMS IN MECHANICAL VIBRATlONS

[SECo 92

269

The problem in the transform y(x,s) is d4y dX4

a2 -

(2)

+ s2 y

= O ' Bw

y(O,s) = YxiO,s) = Yxie,s) = O,

y(e,s) = S2 + w"

It wilI be convenient to write (3)

Then (4)

S

S2

.

2

.

= ¡aq

= - a2q4, and the general solution of Eq. (2) can be written

y(x,s) = CI sin qx + C2cosqx + C3sinh qx + C4coshqx,

where the C's can be functions of the parameter s. When those constants are determined so that the boundary conditions on y(x,s) are satisfied, the solution (4) becomes (5)

y (x,s )

=

Bw 2 sin qx sinh..qe + sinh qx sin qe . 2 s +w 2 sm qe smh qe

The Maclaurin series representations of the sine and hyperbolic sine functions show that the final fraction in Eq. (5) is an analytic function of q2, and therefore of s, except at points where the denominator vanishes. The point s = Ois a removable singular point. The denominator vanishes when qe = :i:n1t and qe = :i:in1t; that is, whenever q2e2 = :i:n21t2. In view of Eqs. (3)and (5)then, the singular points of y(x,s) are s

(6)

=

:i:iw,

s

=

in21t2a e

:i:---y-

(n = 1,2,...).

Ir the value of w is distinct from all the numbers (n = 1,2,...),

(7)

the derivative of the entire denominator in the expression (5) for y(x,s) does not vanish at any of the points (6),nor does the numerator exceptfor particular values of x. Those points are therefore simple poles of y(x,s). In this case the formula for the displacements has the form co

Y(x,t)

=

bo(x) cos [wt + Oo(x)]+

L bn(x) cos [wnt + n=1

On(x)].

Ir the valueof w coincides with one of the numbers Wngiven by formula (7), then y has poles ofthe second order at the points s = :i: iwn, and a term of resonance type appears in Y(x,t). Hence the resonance frequencies are the frequencies (7).

270


SECo 93]

In case the hinge at x = e is kept fixed and a simple harmonic bending moment acts on that end of the beam, the conditions at x = e have the form Y(e,t) = O,

Yxx(e,t)= B sin rot.

The reader can show that the expression for y(x,s) then has the same denominator as it does in Eq. (5). The resonance frequencies are again those given by formula (7). Other cases are included in the problems at the end of the chapter. 93

DUHAMEL'S

FORMULA

FOR VIBRATION

PROBLEMS

As in the case of problems in heat conduction (Sec. 85), the convolution property of the transform displays a relation between the solutions of problems in vibrations with variable boundary conditions and corresponding problems with fixed conditions. Consider for example the transverse displacements Y(x,t) in a string. If both damping and an elastic support are present (Fig. 94), the equation of motion has the form

(1)

Y;lx,t) = a2 ~ix,t)

-

b Y;(x,t)

- h Y(x,t).

To permit the end x = O to be elasticallysupported or kept fixed or to slide freely along the Yaxis, we can write the condition (2)

A,l(Y)

=O

at x

= O,

where A,l(Y) = h1Y - k1~. If a prescribed force F(t) acts on the end x = e, a fairly general boundary condition is (3)

A,2(Y) = F(t)

(4)

Y(x,O) = :Y;(x,O)= O,

the transform y(x,s) satisfies the conditions (5) (6)

(S2 + bs + h)y(x,s) = a2Yxix,s), A,¡[y(O,s)]= O,

A,2[y(e,s)]= f(s).

y

.F(t¡

o

(e,O)

Fig.94

x

at x = e,


[SECo 93

271

Let Z(X,t) represent the displacement Y(x,t) in the special case m which F(t)

= l.

Then z(x,s) satisfies Eq. (5) and the boundary conditions 1 A.¡[z(O,s)] = 0,

A.2[Z(C,S)] = -.

s

It follows that the product sf(s)z(x,s) satisfies the conditions (5) and (6), and therefore (7)

y(x,s) = sf(s)z(x,s).

In view of the convolution property then,

o

(8)

t

I

Y(x,t) = ot o F(t - t)Z(x;r)dt,

or (9)

Y(x,t)=

E F(t -

t)Zt(X,t) dt.

These are two forms ofDuhame1's formula for the resolution ofthe problem in Y(x,t) with a variable end condition into one with a fixed end condition. The derivation of these formulas can be extended easily to other problems. In the case of transverse vibrations of bars, for instance, the derivative of the fourth order with respect to x replaces Yxx(x,t),and additional boundary conditions are involved. In the case of transverse displacements in a membrane, the Laplacian of Y replaces Yxx in our problem. In these cases, the steps in the derivation of Duhamel's formula are the same as in the case treated above.

PROBLEMS 1. An elastic bar with its end x = O kept fixed is initially stretched so that its longitudinal displacements are Y(x,O) = Ax (O~ x ~ e), then released from rest in that position at the instant t = Owith its end x = e kept free. Derive these two formulas for its displacements: Y(x,t)

-- -=r 8Ac ~ (-Ir 'n n= 1 (2n -

1 . (2n - I)nx COS (2n - I)nat 2e 2e 1)2 Sin

= tA[H(x + at) + H(x

-

at»),

where H is the triangular-wave function (2), Seco87 (Fig. 89). Verify the solution. 2. The end x = O of an elastic bar is elastically supported (Fig. 95) S9 that the longitudinal force exerted on that end is proportional to the longitudinal displacement: y"(O,t) = h Y(O,t)

(h > O).

272


SECo 93]

Fig.95

When a longitudinal periodic force F(t) = A sin rot acts at the end x = e, show that resonance takes place if ro has any one of the values aa../e (n = 1,2,...), where a.. is a positive root of the equation tan a. = he/a.. Show those roots graphically

(cf. Fig. 92).

3. A string stretched from the origin to a point on the x axis is initially straight with a uniform velocity in the direction of the Y axis, as if a moving frame supporting the end points is brought to rest at the instant t = O. There is a smalldampingforceper unit length proportional to the velocity. Select units so that the transverse displacements satisfy the conditions

Y,,(x,t)= Yxx(x,t)- 2bY,(x,t)

(O< x < n, t > O,O< b < 1),

Y(O,t)= Y(n,t) = Y(x,O)= O,

Y,(x,O)= Vo.

1[p is a branch of the function (S2 + 2bs)t, show that L{Y} = y(x s) = Vocosh (np/2) - cosh (x

,

p2

cosh (np/2)

-

n/2)p

and obtain formally this formula for displacements:

f

Y(x,t) = 4voe-h' sin (2n - l)x sin mt n .=1 (2n-l)m 4. A string is stretched from the origin to the point (e,O)and given an initial veiocity Y,(x,O)= g(x) but no initial displacement. Derive this formula for its displacements: 2

00

I . nnx . nnat

L -sm-sm-e na.=ln

Y(x,t) = -

e

i

. nn~

e

o

g(~)sm-d~. e

5. In Probo 4 let G(x) be the odd periodic extension, with period 2e, of the velocity function g(x) and show formaIly that 2 Y,(x,t) = G(x + at) + G(x and

I

Y(x,t) = -2

-

at)

x+ar

f

a x-at

G(~)d~.

Verify that integral form of Yas a solution of the problem. 6. Let I(x,t) denote the current and V(x,t) the voltage at distance x from one end of a transmission line, at time t. FoIlowing the notation used in the telegraph equation (11), Seco40, and indicated in Fig. 96, let R, L, S, and K represent resistance, inductance, leakage conductance, and capacitance to ground, respectiveIy, per unit length of line. Then the functions I(x,t) and V(x,t) satisfy the system of partial differential equations (a)

-Ji"=RI+LI,,

-Ix=SV+KYr,

called the transmission Uneequations. By differentiating the members of equations (a) and eliminating one of the functions and its derivatives, show that either 1 or V satisfies


[SECo 93

" Source

R

L

273

---- x-e

load

1

1 Fig.96

the telegraph equation. Note that equations (a) are better adapted to initial and end conditions that involve values of both I and V, however, than is the telegraph equation (see Probs. 7 to 9).

7. Supposethat R = S = O in the transmission line shown in Fig. 96, and that the current and voltage are zero initiaUy. If one end is kept grounded, V(O,t)= O, and the current at the end x = e is proportional to t, find the transform v(x,s) of V(x,t) using the transmission line equations (a), Probo 6. Thus show that V(x,t) is an eIectrical analog of the displacements Y(x,t) in the bar in Seco86, where a2corresponds to (KL)- 1. 8. In the transmission line shown in Fig. 96, R = S = O and I(x,O)= V(x,O)= O. If V(O,t)= A sin rot and the end x = e is grounded, find the values of ro for which the voltage and current become unstable, using the transmission line equations (a), Prob.6. Ans. ro = mrc-l(KL)-t (n = 1,2,...). 9.

Solve Prob. 8 when the condition that the end x

=

e is grounded

is replaced

by the

condition that the circuit is open at that end. Ans. ro = (n - t)1tc-l(KL)-t (n = 1,2,...). 10. One end of a beam is built into a rigid support (Fig. 97); thus if Y(x,t) denotes transverse displacements, Y(O,t)= YJO,t) = o. y;- .

Fig.97 The pin-support

of the other end x

= e vibrates,

Y(c,t) = B sin rot,

o

x

so that

Yxx(c,t)= O.

Initially the beam is at rest along the x axis. If a2 is the coefficient in Eq. (l), Seco92, show that resonance occurs in the vibration of this beam if ro = aa.//C2,where ex.is any positive root of the equation

tan ex= tanh ex. Show how the roots ex. can be

approximated graphically.

.

11 One end of a bar of length e is free. The other end is built into a support that undergoes a transverse vibration Y = B sin rot (Fig. 98). The bar is initially at rest with no displacements. Show that resonance takes place in the transverse vibrations of the bar when ro = aa..2/C2,where ex.is any positive root of the equation cos ex= -sechex and a2 is the coefficient in Eq. (1), Sec. 92. Also show how the roots ex.can be approximated graphically.

274


SECo 93]

y YlO,tl-B sin (lit

Fig.98

12. Both ends of a beam of length e are built into rigid supports. The beam is initially at rest with no displacements. A simple periodic force per unit length acts along the entire span, in a direction perpendicular to the beam, so that the transverse displacements Y(x,t) satisfy the equation

Show that resonance occurs when w = 4aa.//C2,where ex.is any positive root of the equation

tan ex=

- tanh

ex.

13. A membrane, stretched across a fixed circular frame r = e, is initiallyat rest in its position of equilibrium. Ir a simple periodic force per unit area in a direction perpendicular to the membrane acts at all points, the transverse displacements Z(r,t) satisfy an equation of the type

Ztt

= b2( Z" + ~Zr) + A sinwt.

Show that the resonance frequencies are w = bcx./e,where ex.is any positive root of the equation J o(ex)= O. 14. The end x = Oof the propeller shaft for a ship is connected rigidly to a massive flywheel that turns with unifonn angular velocity w. The end x = e supports a propeller whose moment of inertia with respect to the axis of the shaft is l. In addition to the steady-state torque transmitted between shaft and propeller, the action in the water induces a perturbation in the nature of a torque proportional to sin 4wt exerted on the propeller. Let 0(x,t) denote the corresponding perturbation in the angular displacements of the circular cross sections of the shaft. Thus 0 (x,t) = Owhen the shaft

operatesunder steady-stateconditionswithangularvelocityw. Ir 1p is the polar moment of inertia of the cross section of the shaft, of radius ro (1p = tro 4), and if E. is the modulus of elasticity in shear for the material in the shaft, show that 10tt(e,t) =

-

E.J p0 ",(e,t) + B sin 4wt,

and that the boundary value problem in o (x,t) is an analog of the problem in Seco91. where a2 now represents EJ p and p is mass of the material per unit volume. Show that torsional resonance takes place when w = aA..I(4e),where A..is any positiveroot of the equation tan A. = k/A.and where k = epl p/l.


[SECo 93

275

15. The propeller shaft in Probo 14 is 150 ft long with a diameter of 10in., and made of steel. The moment of inertia 1 of the propeller equals that of a steel disk 4 in. thick and 4 ft in diameter. The steel weighs 0.28 lb/in.3 (gp = 0.28),and Es = 12 X 106 lb/in. 2 Show that the lowest ftywheel speed at which torsional resonance takes place is approximately 135 revolutions per minute.

9 Generalized Fourier Series

We shall use theory developed in Chap. 6, on the inversion integral for the Laplace transforrnation, to assist us in establishing some basic theory of eigenvalue problems in differential equations. The theory shows that the solutions of certain types of homogeneous boundary value problems in ordinary differential equations form orthogonal sets of functions, and that an arbitrary function can be represented by a series of functions of such a set, a generalized Fourier series. A classical method of solving some types of linear boundary value problems in partial differential equations begins with a separation ofvariables, then requires a representation of a prescribed function by a generalized Fourier series. The orthogonal functions in the series are the solutions of an eigenvalue problem that arises from the separation of variables. The method will be illustrated at the end of this chapter. In the following chapter we shall see how such eigenvalue problems are encountered in a process of designing an integral transformation to fit 276

...

GENERALIZED FOURIER SERIES

[SECo 94

277

the requirements of a given boundary value problem. Then the inverse transformation is represented by the corresponding generalized Fourier series.

94

SELF-ADJOINT

DIFFERENTIAL

EQUATIONS

We begin with some concepts and an existence theorem from the theory of linear ordinary differential equations. The adjoint D!y of the linear differential form (1) of the second order is this related form : (2)

D!J

= (IX2(X)y(X)]" -

(IXl(X)y(X)]' + IXO(X)y(X).

It is useful in constructing forms that represent derivatives of other forms. With the aid of a second function v(x), for instance, a form that is an exact derivative can be written (3)

vD2y

-

yD!v = d~(rJ.2VY'- (rJ.2V)'Y + IXIVY].

The Lagrange identity (3) is easily verified. We find that the form (4)

~2Y = (r(x)y'(x)]' - Q(x)y(x)

is self-adjoint, ~!y = ~y. For it, Lagrange's identity becomes

(5)

V~2Y - y~2V

= :x(Vy' - v'y)r].

When both members of the differential equation (6)

y"(x) + Pl(X)Y'(x) + Po(x)y(x) = f(x)

are multiplied by an integrating factor (7)

r(x)

= exp [f Pl(X)

dx J.

the equation assumes its self-adjoint form (8)

(r(x)y'(x)]'

-

Q(x)y(x) = F(x),

where Q(x) = - r(x)Po(x) and F(x) = r(x)f(x). Adjoints of differentialforms of order higher than 2 are defined in a manner that corresponds to the definition (2) (Prob. 3, Seco97). But linear differential equations of higher order do not always have self-adjoint forms.

..

278


SECo 95]

An existence theerem for solutions ofthe homogeneous form ofEq. (6), in which the coefficient Po(x) may involve a parameter A in a lin~ar manner, will now be cited. The theorem can be proved by a method of successive approximations.l THEOREM

Let A, B, and e be continuous functions of x over some closed interval 1 and let Xo be a fixed point on 1. Let A denote a complex-valued parameter and Yo and Yl two prescribed constants independent of A. Then the initial-value problem y" + A(x)y' + [B(x) + AC(X)]Y= O, (9) y(xo,A.)= Yo,

y'(xo,A.)= Yl

in the complex-valued function y(X,A) has one and only one solution y = u(X,A.)such that u and u' are continuous functions of x and A.for all A and all x on the interval 1 and, for each x on 1, u and u' are entire functions of the complex parameter A. In Seco96 we shall see that further restrictions are needed to ensure a unique solution of a two-point boundary value problem in linear ordinary differential equations. This fact is illustrated by the simple problem (10)

y" + AY = O,

y(O,A) = O,

y(l,A.) = O,

which clearly has a solution Y(X,A.)==O. But when the parameter A has any one ofthe values n2n2 (n = 1,2,.. .),problem (10)has the additional solutions

(11)

y

= k sin nnx,

where the constant k is arbitrary. AIso, note that the problem (12)

y"(x) + n2y(x) = O,

y(O)= O,

y(l) = 1,

has no solution, since the function k sin nx that satisfies the first two conditions fails to satisfy the condition y(1) = 1 for any value of the constant k. 95

GREEN'S FUNCTIONS

On an interval a ~ x ~ b let r, Q, and F denote prescribed continuous functions of x such that r' is also continuous and r(x) > O. In the boundary value problem (1)

[r(x)y'(x))' - Q(x)y(x) = F(x),

y(a) = y(b) = O,

1 See, for instanee, E. L. Inee, "Ordinary Di!ferential Equations," pp. 72 !f., 1927, also, p. 123 for more on adjoints, or E. C. Titehmarsh, "Eigenfunetion Expansions," p. 6, 1946.

GENERALlZED FOURIER SERIES

[SECo 95

279

. the unknown function y can be interpreted as the static transverse displacements in a stretched string attached to an elastic foundation. The elastic support may have variable properties so that it contributes a transverse force - Q(x)y(x) per unit length as well as a variable tension r(x) in the string. An external transverse force - F(x), per unit length, acts along the string. (Cf. Secs. 40 and 93.) The ends are fixed at the points (a,O)and (b,O). We make the interpretation to assist us in writing the solution of problem (1), at first formally, in terms of solutions of simpler problems. Let F(x) be replaced

formally

by the unit impulse symbol <5(x-

~),

where a < ~ < b. This corresponds to the application of a negative unit transverse force at the point x = ~ (Fig. 99). When the string has that concentrated load, we write G(x,~) for the displacements y(x). Since the vertical force from left to right is - r(x)y'(x), we note that the value of - ry' must jump by the amount -1 at x = ~; that is, 1 (2)

GA~ + o, ~) - Gi ~ - O,~) = r(~r

The displacement function G(x,~) itself should be continuous in the region a ;;¡;x ;;¡;b, a ;;¡; ~ ;;¡;b, and its derivative Gx(x,~) should be continuous there except for the jump prescribed by Eq. (2) at the diagonal x =~. Since <5(x- ~) is zero except when x = ~, square

(x # ~);

(3)

therefore, in view ofthe conditions imposed on r and Q, the function Gxix,~) is continuous except at the diagonal x = ~. AIso, G(x,~) satisfies the boundary conditions (4) A function that satisfies conditions (2) to (4) and the stated continuity conditions is called Green's function for problem (1). Specifically, and for a problem with general linear homogeneous boundary conditions, we describe the function as follows. y-G(x,f¡

o

(a,O)

(b,O)

Fig.99

x

J

280


SECo 95]

I

I I

I

DEFINITION

Green's function G(X,~)for this boundary value problem with self-adjoint differential equation: (5) (r(x)y'(x)]' - Q(x)y(x) = F(x), (6)

Aly(a) + A2y'(a) = O,

Bly(b) + B2Y'(b) = O,

is a solution of the homogeneous differential equation (3) when a < x < ~ and when ~ < x < b that satisfies the following conditions. It is continuous

in x and ~ together over the square a ;;¡¡;x ;;¡¡;b, a ;;¡¡;~ ;;¡¡;b (Fig. 100). Its derivativeGxiscontinuousovertheclosedtriangleT¡{a;;¡¡;

x;;¡¡; b,a;;¡¡; ~;;¡¡; x)

when properly defined on the diagonal x = ~,and continuous over the closed triangle T2when again properly defined on the diagonal, but Gx has the jump l/r(~) from TI to T2 at x = ~ as specifiedby condition (2). AIso, when ~ is fixed (a < ~ < b), G satisfies boundary conditions (6): (7)

AIG(a,~) + A2Gx(a,~) = BIG(b,~) + B2Gx(b,~)= O.

Of course, one of the A's and one of the B's here must not vanish. A convenient way to specify this is to write (8)

Al

= cosa,

A2 = sina,

BI = cosp,

B2 = sinp,

where a and pare constants. Note that Gxx has the continuity properties of Gx, in view of Eq. (3). We now indicate how the solution of problem (1) may be written as a weighted superposition of solutions G of the problem in which the load F(x) is replaced by concentrated

loads 15(x -

~).

The impulse symbol has the formal properties 15(- x) F(x)

=

f

F(~)15(~

-

x) d~

=

f

F(~)15(x

= 15(x)and

- ~) d~;

that is, F(x) is a superposition of impulses. If all members of the symbolic equations (r(x)Gx(x,~)]x

-

Q(x)G(x,~) = 15(x -

G(a,~) = G(b,~) = O,

~ (O,b)

10,a)

o

(b,O)

x

Fig.100

~),


[SECo 96

are multiplied by F(~) and integrated with respect to ~ from ~= a to the result is the three statements in problem (1) when we write (9)

=

y(x)

f

281

~ = b,

F(~)G(x,~)d~

and identify (ry')' with the symbol S: F(~)(r(x)Gx(x,~)]xd~. In order to verify the formal solution (9), we first note that y(a) = y(b) = Obecause G(x,~)satisfies the boundary conditions (4). When we write y(x) =

r

F@G(x,~) d~ +

f

F(~)G(x,~) d~,

the integrand of each integral and its first two partial derivatives with respect to x are continuous functions of (x,~) in the triangular regions 1;. and T2, respectively, in Fig. 100. Differentiation by the Leibnitz formula (Sec. 14) is therefore valido We find that

since G is continuous, and that (10)

(ry')' =

f

F(~)[r(x)Gx(x,mx

d~ + F(x)r(x)J(x),

where J(x) = G..,(x,x - O)- Gx(x,x + O). Thus J(x) is the jump in the value of Gx as the point (x,~) moves across the diagonal from triangle TI into triangle ~ (Fig. 100). According to Eq. (2), J(x) = l/r(x). In view of Eq. (3), Eq. (10) can now be written (ry')' = Q(x)

f

F(~)G(x,~)d~ + F(x)

= Qy + F;

that is, the function (9) also satisfies the differential equation in problem (1). The solution (9) of problem (1) is now established if Green's function G exists. In fact the solution is valid if F is a sectionally continuous function. For modifications of the procedure when the boundary conditions are type (6), see Probs. 8 and 9, Seco97. 96

CONSTRUCTION OF GREEN'S FUNCTION

A function f(x) is of class C' on an interval a ~ x ~ b if both f and f' are continuous over the interval. Let r be of class C' over that interval, where r(x) > O, and let Q be continuous there. Then, according to the existence theorem stated in Seco94,

282


SECo 96]

the initial-value problem (1)

[r(x)u'(x)J' - Q(X)U(X)= 0,

u(a) = 0,

u'(a) = 1,

has a unique solution u(x) of class C' on that interval. Let v(x) be the unique solution, of that class, of the initial-value problem

(2)

[r(x)v'(x)J' - Q(x)v(x) = 0,

v(b) = 0,

v'(b) = ~.

Then on the square region a ~ x ~ b, a ~ ~ ~ b, the function (3)

G(x,~) =

A(~)u(x)

(x ~ ~)

{ B(~)v(x)

(x ~ ~),

where A and B are continuous functions, satisfies the self-adjoint differential equation (3), Seco95, and the boundary conditions G(b,~) = O.

G(a,~)= 0,

It is continuous there, on the diagonal x =

~ in particular,

and its derivative

Gxhas the jump prescribed by Eq. (2), Seco95, at x = ~ if A.and B satisfy the simultaneousequations (4)

B( ~)v(~)

-

1 A( ~)u(~) = 0,

B(~)v'(~)

-

A(~)u'(~) = r(~)'

According to Lagrange's identity (5), Seco 94, the determinant - v(~)u'(~)+ u(~)v'(~)of system (4) satisfies the condition d (5) --[(vu' - uv')r] = -[(ru'Y - Qu]v + [(rv'Y - Qv]u. d~ The right-hand member vanishes since u and v satisfy the differential equationsinproblems(1)and(2).

Hence(vu'

-

uv')risconstantand,sincev(b) =°

and v'(b) = 1, -[v(~)u'R)

- u(~)v'(~)]r(~) = u(b)r(b).

Then in case u(b) =F0, we find that the solution of system (4) is

u(~) B(~) = u(b)r(b)'

v(~) AR) = u(b)r(b)'

and formula (3) for Green 's function becomes (6)

G(x,~) = u(x)v(~) u(b)r(b)

when x ~ ~,

= u(~)v(x) u(b)r(b)

when x ~ ~.


Note

(7)

[SECo 97

283

that G is symmetric in its two arguments: G(x,~) = G(~,x),

a consequence

of the self-adjoint

character

of the differential equation.

In case u(b) = O then, since u(a)= O and u'(a)= 1, the function w = u(x) is not identically zero, and it is a solution of class e of the homogeneous

problem

-

w(a)= w(b)= O. Since v satisfies the same differential equation and v(b) = O,it followsthat

(8)

(rw')'

Qw = O,

v(x) = ku(x) when u(b) = O,where ku'(b) = 1. The following theorem is now established. Theorem 1 1J the homogeneous problem (8) has no solution oJ class e on the interval a ~ x ~ b other than w(x) ==O, then Green's Junction G(x,~) exists Jor the boundary value problem (9)

(r(x)y'(x»)'

-

Q(x)y(x)

= F(x),

y(a) = y(b) = O,

where r is oJ class e, Q is continuous, F is sectionally continuous, and r(x) > O, on the interval a ~ x ~ b. Green's Junction is expressed in terms oJ solutions oJ the initial-value problems (1) and (2) by Jormula (6); it is symmetric in x and~. The solution, oJ class e, oJ problem (9) is

(10)

y(x)

=

f

F(~)G(x,~) d~.

Although the initial-value problems (1) and (2) are simpler than the boundary value problem (9), they can be solved explicitly only for limited types of coefficients r and Q. Under the conditions as()umed in Theorem 1, suppose that there is. another solution z(x) of problem (9), in addition to y(x). Then the function w = y - z satisfiesthe homogeneousproblem (8)whichhas only the solution w(x) ==O. Thus z = y, and we ha ve the following theorem on uniqueness. Theorem 2 Under the conditions stated in Theorem 1, the solution (10) oJ problem (9) is the only one oJ class C. Examples and additional material on Green's functions are given in the problems following Seco97.1 97

ORTHOGONAL SETS OF FUNCTIONS

Consider now real-valued functions of x that are continuous over an interval a ~ x ~ b. A function J may be called a generalized vector, with one component J(x) corresponding to each value of x on the interval (Sec. 10). 1

For further theory on Green's functions for ordinary and partial differential equations see, for instance,

R. Courant

and D. Hilbert,

"Methods

of Mathematical

Physics,"

vol. 1, 1953.

284


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A generalized scalar product, the inner product (J,g), of two functions f and g is defined as the sum of products of corresponding components: (1)

(J,g) =

The norm of a function (fJ)t; that is, (2)

f

f(x)g(x) dx = (g,f).

J, the generalized

IIfII =

concept of the length of a vector, is

{f If(xW

dx

y.

Two functions f and g are orthogonal on the interval if (J,g) = O. This signifies nothing about actual perpendicularity, but instead that the product of the functions changes sign in such a way that

f

(3)

f(x)g(x) dx

= O.

Corresponding to units of length that differ from one coordinate axis to another in vector analysis, we introduce a nonnegative weight function p and define the inner product with respect to p, of two functions, as

(4)

(J,g) =

f

p(x)f(x)g(x) dx

= (gj)

[P(x)~ O].

The norm 11 fII is then (fJ)t; and f and g are orthogonal if (J,g) = O. Let a set of functions Yn(n = 1,2,...) be orthogonal on the interval, with weight function p. It is convenient to normalize the set by dividing each function by its norm. Thus the set

(a ~ x ~ b, n = 1,2,...)

(5)

is orthogonal with unit norms, an orthonormal seto It corresponds to a set of mutually perpendicular unit reference vectors. A prescribed function f may possibly be represented as a generalized linear combination, an infinite series, of those orthonormal functions, 00

(6)

f(x) = :¿ cncPn(x) n=1

(a < x < b).

Ir so, and if it is true that (cPmJ)= :¿:= 1 cn(cPm,cPJ,we can determine the constants cn. F or (cPm,cPn) = Owhen n "" m, and 11 cPm11= 1. Thus (J,cPm)= cm; that is (7)

(n = 1,2,.. .).


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285

Series (6) with coefficients (7) is the generalized Fourier series for f corresponding to the orthonormal set of functions
=

A sin nx

(O< x < n, n = 1,2,.. .).

F or this set series (6) becomes the Fourier sine series: co

f(x) =

A-nn=l¿

2 Cnsin nx = -

"

co

¿

nn=l

5.o

f(~) sin n~d~sin nx,

which converges to f(x) when O < x < n at each point where f is continuous, if f' is sectionally continuous on the interval.l PROBLEMS 1. Verify Lagrange's identities (3) and (5), Seco94. 2. Use Lagrange's identity to show that v(x) is an integrating factor for the differential equation

if v is not identically zero and satisfies the homogeneous equation D~v = 0, where D! is the adjoint of D2. Thus show that D2y = f reduces to this equation of first order: a].vy' + [al v - (a2v)']y= 3.

f

v(x)f(x) dx:

The adjoint of the linear differentialform D"y = ao(x)y(x) + al(x)y'(x) + a2(x)y"(x) + ... + an(x)y(n)(x)

of order n, is the form D:y = aoy - (aly)' + (a2Y)"- ... + (-l)n(any)(n). When n ==3, establish the Lagrange identity

1

See the author's "Fourier Series and Boundary Value Problems," 2d ed., 1963; also, ortho-

gonal

sets are treated

somewhat

more

fully in Chap.

3 of that

book.

286


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where B3 is this bilinear form in v and y: B3

4.

= CXIVY

+

-

[CX2Vy'

-

+ [CX3VY"

(CX2V)'Y]

(CX3V)'Y'+ (CX3V)"Y].

Show that this special case of D.y in Probo 3, D4y = cxo(x)y(x)+ y(4)(X),

is self-adjoint, and that it satisfies the Lagrange identity d vD4y

5.

-

yD4v = dx(vy'"

-

v'y" + v"y'

-

v"'y).

Find Green's function for the elementary problem y(O) =

y"(x) = F(x),

O,

y(1)= O,

and solve for y in the form y(x) = (x - 1)

I: ~F(~)d~ +

x

f

(~ - I)F(~)d~.

Ans. G(x,~) = x(~ - 1) when x ~ ~, G(x,~) = G(~,x). 6.

Obtain Green's function (4), Seco82, for the problem u"(x,s)

- su(x,s) = - g(x),

u(O,s) = u(l,s) = O

(s > O).

7. Let F be a sectionally continuous function, with only one jump at a point Xo, for the sake of simplicity, on the interval a < x < b. Carry out the verification of the solution (Sec. 95) y(x)

=

f

F(~)G(x,~)d~

of the boundary value problem [r(x)y'(x)]'

-

Q(x)y(x) = F(x),

y(a) = y(b) = O

(x #
where G is Green's function for the problem. Note that the solution y(x) is of class C' if y'(xo) is defined

as

f: F(~)G x<.x'o'~)d~.

8. Let the constants cxandP be real and let w = u(x) and w = v(x) be the solutions of the homogeneous self-adjoint equation (rw')' - Qw = Othat satisfy these initial conditions: u(a) =

sin cx,

v(b)= sinp,

u'(a) = -cos

CX,

v'(b)= -cosp.

ConstructGreen's functionfor this problem(Sec.95): [r(x)y'(x)]'- Q(x)y(x)= F(x) y(a) cos cx+ y'(a) sin cx= O,

y(b) cos

P+

y'(b) sin P = O,

in case e #
= -r(b)[u(b)cosP + u'(b)sinP],

(a < x < b),


287

[SECo 97

in the form when

x

= u(~)v(x) e

when

~ ;;¡¡ x.

e

9.

~

G(x,~) = u(x)v(~)

;;¡¡

When F is continuous(a ;;¡¡ x ;;¡¡b), verify the solution y(x)

=

f

F(~)G(x,~) d~

of the boundary value problem in y for which Green's function G was constructed in Probo 8. 10. When the constant k is neither zero nor an integer, use Green's function (Probs. 8 and 9) to solve this problem for y:

Ans. y(x)k sin k7t =

11.

y'(7t)= o.

y'(0)= O,

y"(x) + ey(x) = F(x),

I: F(~) cos k~ d~ cos k(7t -

x) +

f

F@

COS k(7t

-

~)d~ cos kx.

Write the differential equation in the problem

(x + l)y"(x) + y'(x) = F(x),

y(O)= O,

y'(I) = O,

in self-adjoint form and use Green's function (Probs. 8 and 9) to solve the problem in the form y(x)

=-

f: F(~)log(~ + l)d~ -

log(x + 1)

f

F(~)d~.

12. Interpret the symmetry property G(x,xo) = G(xo,x) of Green's function in terms of displacementsofthe stringconsideredin Seco95 and concentratedloads,at pointsx and Xo.

13. Supposea function w of c1assC' is not identicalIyzero and satisfiesthe homogeneous problem [r(x)w'(x)]'

-

Q(x)w(x)

= O,

w(a)

=

w(b)

=

O.

Then if the problem [r(x)y'(x)]' - Q(x)y(x) = F(x),

y(a) = y(b) = O,

has a solution

of cIass C', use Lagrange's identity to show that F must be orthogonal to w on the interval (a,b):

f

F(x)w(x) dx

= O.

IIIustrate that result using the problem y"(x)

+ y(x) = -

3 sin 2x,

y(O) = y(7t) = O,

whichhas the familyof solutionsy = e sin x + sin 2x.

288

14.


SECo 98]

Show that the set of functions

1

mrx

.fi'

e e A-cos-

(n =

1,2,...)

is orthonormal on the interval (O,e)with unit weight function, and that the generalized Fourier series for a functionJ, with respect to the set, is the Fourier eosine series l

-

2 '"

98

mrx

mr.;

[ f(.;)d.; + -e ¿ cos- e [o f(.;)cos-d.;e

e o

n= 1

(O < x < e).

EIGENVALUE PROBLEMS

The homogeneoustwo-point boundary value problem (1)

[r(x)Y'(X,A)]'- [q(x) + Ap(X)]Y(X,A)= O A1y(a,A) + A2Y'(a,A) = O,

(a < x < b)

B1y(b,A) + B2Y'(b,A)= O,

involving a parameter A in the manner indicated, is called a Sturm-Liouville problem or system.l The constants Al, A 2, B l' and B2 are real and independent of A. The functions p, q, r, and r' are real-valued and continuous, and p(x) > Oand r(x) > O,over the entire interval a ~ x ~ b. Problem (1) has the trivial solution y = O,for everyvalue of A. But for certain values Anof A,called eigenvalues or characteristic numbers,the problem has a non trivial solution Y(X,An) = Yn(x),

an eigenfunction or characteristic function. Note that y = Ym(x)and y = y.(x) are solutions of different problems obtained by writing Amand An,respectively, for A in problem (1). Also note that if Yn(x)is an eigenfunction, then CYn(x)is also one, where C is any constant other than zero. The Sturm-Liouville problem is a particular type of eigenvalue problem. Such problems arise in the process of solving boundary value problems in partial differential equations by the classical method of separation of variables, a method we shall illustrate in this chapter.2 Since eigenfunctions satisfy the Sturm-Liouville differential equation, their deriva tives of second order are to exist over the open interval (a,b). We assume the functions are of class C' (Sec. 96) on the interval a ~ x ~ b,and show later on that eigenfunctions of that class do exist. AIso, we write the Sturm-Liouville equation in terms of our self-adjoint operator ~2 in the form ~2Y - APY = O, where ~2Y = (ry')' - qy. 1

The first extensive development

of the theory of such systems was published

by 1. C. F. Sturm

and 1. Liouville in the first three volumes of Journal de mathématique, 1836-1838. 2 See also the author's "Fourier Series and Boundary Value Problems," 2d ed., 1963.

GENERAlIZED FOURIER SERIES

[SECo 98

289

Let y and z denote any two eigenfunctions of problem (1) corresponding to eigenvalues A and J1.,respectively. Then both functions satisfy the first boundary condition, (2)

Aly(a) + AzY'(a)= O,

Since Al and A2 are not both zero, the determinant ofthis pair of simultaneous linear equations in Al and A2 must vanish; that is, W(a) = O,where W is the Wronskian of the functions y and z: (3)

y(x)

z(x)

y'(x)

z'(x)

W(x) =

= y(x)z'(x) - y'(x)z(x), I

I

and similarly for the second boundary condition. must be such that

W(a) = O,

(4)

AIso, they satisfy the Sturm-Liouville

Thus the functions y and z

W(b) = O. equations

~2Y = Ap(X)y(x),

~2Z = J1.p(x)z(x),

and from these and Lagrange's identity (5), Seco94, it follows that (5)

(J1.- A)PYZ= y~2Z - Z~2Y d = d)Yz'

d - y'z)r) = d)W(x)r(x)).

In view of conditions (4), therefore, (6)

(J1.- A)

f

pyz dx = W(b)r(b)

...,.W(a)r(a)

= O.

In case A.=f.J1.,it follows that y and z are orthogonal on the interval (a,b), with weight function p:

(7)

f

(A =f. J1.). p(x)y(x)z(x)

dx = O

If the eigenfunctions y and z correspond to the same eigenvalue, then J1.= A in Eq. (5), and so (Wr)' = O and Wr is constant; but the constant is zero because W(a) = O. Since r(x) > O,it follows that the Wronskian of y and z vanishes : (8)

W(x) = y(x)z'(x) - y'(x)z(x)

==

O

(a

In that case we now prove y and z linearly dependent. Consider the linear combination of y and z, (9)

X(x) = [A2z(a) - Al~'(a))y(x) - [A2y(a) - A¡y'(a))z(x),

~ x ~ b).

290

OPERATIONAL

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MATHEMATICS

which also satisfies the homogeneous equation f!)2X = A.x. The reader can X(a) = Oand X'(a) = O because W(a) = O. Therefore X(x) == O, the unique solution of that equation under those conditions. Moreover, the coefficient of y(x) in formula (9) cannot vanish, in view of the second of boundary conditions (2) and because z(a) and z'(a) cannot both vanish, for the determinant of the pair of equations

verify that it satisfies the initial conditions

A2z(a)

-

Alz'(a) = O,'

A1z(a)+ A2z'(a)= O

in z(a) and z'(a) has the value A22 + Al 2. That value is positive for our real

numbers Al and A2. Likewise, the coefficient of z(x) in formula (9) is a nonzero constant.

SinceX(x)

==

O,it followsthat there is a constant k such that

(10)

z(x)

= ky(x)

(k*O,a~x~b).

Finally, we show that all eigenvalues are real and have corresponding real-valued eigenfunctions. For if y(x) is an eigenfunction corresponding to an eigenvalue ..1.,then by writing complex conjugates of all terms of the Sturm-Liouville problem (1), we can find that f!)2Y= Áy,

Aly(a) + A2Y'(a) = B¡ji(b) + B2y'(b) = O.

Thus y is an eigenfunction corresponding to Á and, according to Eq. (6), it follows that

f

(Á - ..1.) p(x)y(x)y(x) dx

= O.

But y is not identically zero, and yy = IYl2~ O,while p(x) > O,so the integral

here has a positive value; therefore, Á - A.= O. Thus Á = ..1., so A.is real. Now since y and y are eigenfunctionscorresponding to the same real eigenvalue ..1.,then y and y are linearly dependent; that is, for some constant k, y = ky. In case k = -1, then y = - y and iy = iy, so iy is a real-valued

eigenfunction. If k * -1, then (1 + k)y is the eigenfunction y + y, or 2 Re y, which is real-valued. That is, each eigenfunction can be made real-valued by multiplying it by an appropriate complex constant. We have now established the following properties of characteristic numbers and functions. Theorem 3 Each eigenvalue ..1.n of the Sturm-Liouville problem (1) is real. The eigenfunction Yn corresponding to ..1.nis unique up to an arbitrary constant factor k other than zero, and k can be chosen so that the eigenfunction is real-valued. Eigenfunctions Ym and Yn corresponding to different eigenvalues ..1.m and A.nhave the orthogonality property

f

p(x)Ym(x)Yn(x) dx = O

(A.m* ..1.n)'


99

291

[SECo 99

A REPRESENTATIONTHEOREM

In the sections to follow, the existence of an infinite sequerice oí eigenvalues Anofthe Sturm-Liouville system will be established. We then 'show that the generalized Fourier series of orthonormal eigenfunctions, for each function f of a broad class, does converge to f(x) on the interval (a,b). Let Yndenote real-valued eigenfunctions of the system (1)

[r(x)y'(x)]'

-

[q(x)

+ Ap(x)]y(x) = O

A1y(a)+ A2J'(a)= O,

(a < x < b),

B1y(b)+ B2J'(b)= O,

in case eigenfunctionsexist. Then the normalized eigenfunctions (2) are orthonormal on the interval (a,b) with weight function p: b

Oifmi=n

fa p(x)~m(x)~n(x)dx = { 1 if m = n

(3)

(m,n = 1,2,

o

o 0)0

The generalized Fourier series of those functions, for a function f on the interval,is 00

(4)

L en~n(x) n= 1

where en

=

f

p(x)f(x)~n(x) dxo

Our theorems on representing the inversion integral by a series of residues are useful in establishing the following representation theorem for the series (4)0 Theorem 4 Let p, q, r, r', and (pr)" be eontinuous real-valued funetions of x sueh that p(x) > O and r(x) > O, when a ~ x ~ b, and let the realvalued eonstants Al, A2, B l' and B2 be independent of A. Then the Sturm-Liouville problem (1) has an irifinite sequenee of eigenvalues An(n= 1,2,.. .), all real, and at most afinite number of them are positive. The eorresponding real-valued orthonormal eigenfunetions ~n are of class C' on the interval a ~ x ~ b; also ~n(x) and, when Ani= O, An-t~~(x) are bounded with respeet to both x and An° Eaeh seetionally eontinuous funetion f with seetionally eontinuous derivatives l' and f" is represented by itsgeneralized Fourier series(4)at all points x (a < x < b) where f is eontinuouso1 The proof of the theorem is lengthyo We carry it out fully for the case A2 = B2 = O and indicate in the problems some variations needed to 1 The theorem is true without the condition sectionaIly continuous.

onf",

but in our proofit

is convenient

to assumef"

292


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prove the general case. The proof exhibits interesting devices for finding properties of solutions of differential equations when the equations are too general to solve. 100 THE REDUCED STURM-LlOUVILLE SYSTEM Under the conditions stated in Theorem 4 on p, q, and r, the Sturm-Liouville problem can be reduced to a simpler form by substituting certain new variables t and z for x and y. In the Sturm-Liouville differential equation we first write y(x) = v(x)X(x) and x = x(t) or t = t(x), where v and t(x) are yet to be determined, then write X(x) = z(t). Ir dots denote deriva tives with respect to t, then the equation (ry')' = (q + Ap)y takes the form (1)

2rvl + r'v .

.. z+

X , PX2 x - -; z. -+ 1\.- z, x) r ) (Q

rv

(

where Q(t) is independent of A. Equation (1) can now be simplified by determining t(x) or x(t) so that px2jr = e2, where e is a constant, then v(x) so that the coefficientof i vanishes. Details are left to the problems. The results are (2)

t

= .!. e

x

f

a

P(~)

[

r(~)J

d~

t

,

z(t) = (pr)*y(x),

and, to reduce the interval a < x < b to O < t < 1, we write b

e=

(3)

P(~) t d~. r(~)J

[

f

a

Then the Sturm-Liouville

(4)

equation z(t)

-

takes the reduced form

[Q(t)

+ Jl]z(t) = O

(O< t < 1),

where Jl = e2 A and z is a function of t and Jl. Under the substitutions (2) the boundary conditions in the SturmLiouville system retain their linear homogeneous forms. Thus (5)

alz(O) + a2i(0)

= O,

b1z(l) + b2i(1) = O,

where the a's and b's are real-valued constants related to the A's and B's. In particular, if A2 = B2 = O so that y(a) = y(b) = O, then z(O)= z(l) = O, and therefore a2 = b2 = O. The reduced Sturm-Liouville problem consisting of the simpler Sturm-Liouville equation (4) and conditions (5) transforms back into the original problem in y under the substitutions (2). Orthogonality of eigenfunctions zit) on the interval (0,1) with unit weight function corresponds to orthogonality of Yn(x) on (a,b) with weight function p. The reduced system can be used to prove Theorem 4.


101

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293

A RELATED BOUNDARY VALUE PROBLEM

To shorten the analysis, we now consider the case where a2 = b2 = Oin the boundary conditions of our reduced Sturm-Liouville problem. Except for changes in notation, that problem then reads (1)

X"(x)

-

[A + q(x)]X(x) = O,

X(O) = X(l) = O.

We introduce, to serve as a guide to the analysis, a boundary value problem in partial differential equations whose solution is associated with problem (l). Let Y(x,t) denote transverse displacements in a string stretched between points (0,0)and (1,0)when the string is attached along its span to an elastic medium that may be nonuniform. If the string starts from r:est with a prescribed initial displacement Y = f(x), and if the unit of time is properly chosen, then (2)

Y,r(x,t)= y':x(x,t) - q(x) Y(x,t)

(O < x < 1, t > O),

Y(O,t)= Y(l,t) = Y,(x,O)= O,

Y(x,O) = f(x).

The differential equation and both two-point boundary conditions art homogeneous. Hence the method of separation of variables may apply to problem (2). First, to find all functions not identically zero of type X(x)T(t) that satisfy al! homogeneous conditions, we can write

XT" = X"T - qXT, X(O)= X(l) = T(O) = O, T"(t) X"(x ) q(x )X (x ) = =A Consequently T(t)

X(x)

,

where the parameter A is independent of x and t, because T"jTis independent of x and equal to a function that is independent of t. The function X therefore satisfies all conditions in problem (1). Thus A must be an eigenvalue Anof that Sturm-Liouville problem and, except for an arbitrary constant factor, X(x) must be the corresponding orthonormal eigenfunction 4>n(x).The function T(t) then satisfies the system T" = AnT, T(O) = O. The functions X.T can therefore be written (3)

(n

= 1, 2, . . .),

where the constallts Cnare arbitrary. A sum of functions (3) also satisfies all homogeneous conditions in problem (2). (Note that the sum itself is not a product of a function of x alone by a function of t alone.) Formally, the series

(4)

Y(x,t) =

L

n=l

cn4>n(x) cos r';=¡

satisfies all conditions in the problem, including the condition Y(x,O) = f(x),

294


SECo 102]

if the numbers Cnare the Fourier constants of f(x) corresponding to the orthonormal functions 4Jn(x),so that .

(5)

IX)

f(x)

= n=L1 cn4Jn(x)

(O <

x < 1),

where (n = 1,2,.. .).

(6)

Thus the Sturm-Liouville expansion (5) is needed to complete the formal solution of problem (2) by the method of separation of variables. Ir we now apply the Laplace transformation with respect to t to the boundary value problem (2) in Y(x,t) and establish order properties of the transform y(x,s), we may be able to show that L¡-l{y} = f(x) when t = O, or that when t

L¡-l{y(X,s) - f~X)} = O

= O.

Then wecan anticipate that the sum of residuesofy may bejust the series(5), and that carefulanalysiswill show that the seriesconvergesto f(x). 102

THE TRANSFORM y(x,s)

Formally, the Laplace transform of the function Y(x,t) in problem (2), Seco101, satisfies the conditions y"(X,S)

(1)

-

[S2

+ q(x»)y(x,s) = -sf(x), y(O,s) = y(1,s) = O.

To construct Green's function for problem (1) here, we introduce, as in Seco96, the unique solutions u(x,s) and v(x,s) of these initial-value problems: (2)

U" - (S2

+ q)u = O,

u(O,s) = O,

u'(O,s) = 1,

(3)

v" -

+ q)v= O,

v(l,s) = O,

v'(l,s) = 1.

(S2

Note that u and v are actually functions of x and S2,hence even functions of s; they are entire functions of S2 (Sec. 94) and therefore of S. Then for all s such that u(l,s) :F O,Green's function can be written as u(x,s)v(~,s)

u(l,s)

(4) G(x,~,s) = {

u(~,s)v(x,s) u(l,s)

(x ~ ~)

(x ~ ~).


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295

Thus ifjis sectionally continuous, the solution of problem (1) is h(x,s)

l

y(x,s) = -s fo j(~)G(x,~,s)d~ = -s u(l,s)'

(5) where (6)

h(x,s) = v(x,s)

f: f(~)u(~,s)

d~ + u(x,s)

f

j(~)v(~,s)

d~.

For each fixed x, h is an entire function of s. This follows from the CauchyRiemann conditions, because the products uv are entire functions of s. Consequently y is an analytic function of s except at the zeros of the entire function u(l,s), which are necessarily isolated. But when u(1,s) = O, we see from problem (2) that u(x,s) is a solution of our Sturm-Liouville problem (1), Seco101, in which A.= S2. Ir A.nis the corresponding eigenvalue and denotes one of the two square roots of that real number, then s = :t Sn where Sn =..¡¡::,. Thus if ljJnis the normalized eigenfunction, u(x, :t sn) = kljJn(x)for some nonzero constant k. Since u'(O,sn)= 1, it follows

A

that ljJ~(O)=1:O and (7)

But v(l,:tsn) = O so when u(l,:tsn) = O, then v(x,:tsn) = Ku(x,:tsn), where K =1:Oand, since v'(l,:tsn) = 1, ljJn(x)

(8)

v(x,:tsn) = ljJ~(1)' The function ljJn'not identically

zero, satisfies the conditions

(9) From Eq. (6) to (8) we find that (10) ljJ~(1)h(x,:tsn)

= :~~~;

f

j(~)ljJn(~)

d~

ljJn(x) =

Cn ljJ~(O) ,

where Cnis the Fourier constant for our functionj (11)

f

ljJ~(O)

u(x,:tsn)ljJix)dx =

f

Also, note that

[ljJn(x)]2dx

Eliminating q between Eqs. (2) and (9), we find that

= 1.

296


SECo 103]

so, in view of the boundary conditions on u and
(12) (S2

-

s/)

f

u(x,s)
Consequently formula (5) can be written y(x,s) =

1

I

s

s 2 -s

o u(x,s)
~

.

n

It follows that I snare simple poles of y. If Sn=1= O,the residue of y at

I

Snis,

in view of formulas (10)and (11),

and the sum of those equal residues is cn
L cn
n=1

We have yet to prove that an infinite sequence of eigenvalues exists and that the series of residues converges to f(x). 103

EXISTENCE OF EIGENVALUES

Again, let u(x,s) be the unique solution of the initial-value problem (1)

u" - (S2

+ q)u = O,

u(O,s)= O,

u'(O,s)= 1.

Then the solution of this problem in z(x,s):

z" -

(2)

S2Z

= q(x)u(x,s),

z(O,S) = O,

z'(O,S)= 1

can be written in the form

(3)

z(x,S) =

! sinh sx + ! s

I x

s o

q(~)u(~,s)sinh s(x

- ~)d~

when s =1= o, as is easily verified or derived (cf. Probo 17, Seco 104). The solution (3) is unique because the difference w - z of two solutions w and z of problem (2) must satisfy a completely homogeneous initial-value problem, so that w z ==O. But problem (1) shows that z = u(x,s)is the solution of

-


[SECo 103

297

problem (2). Therefore u satisfies the integral equation (4)

su(x,s) = sinh sx +

f: q(~)u(~,s) sinh s(x -

~) d~,

where s is any complex number. Likewise, for all s the solution v ofthe initial value problem (3), Seco102, satisfies the integral equation (5)

sv(x,s) We write p

=

sinh s(x

= Isl and

- 1) +

f

-

q(~)u(~,s) sinh s(~

x) d~.

(J = IRe si. Then

when O~ x ~ 1; thatis,e-uXlsinh sxl ~ 1ande-uxlcosh sxl ~ 1whenO ~ x ~ 1. Forafixeds let M(s) denote the maximum value of the continuous function plu(x,s)le-UX on the interval O ~ x ~ 1. The function assumes its maximum value at some point x = k on the interval. When we write x = k in formula (4) and introduce the factor e-uk there, we find that M(s) ~ 1 +

f: Iq(~)u(~,s)sinh s(k -

~)Ie-ukd~.

When the integrand here is written in the form

we can see that M(s) ~ 1 + qo M(s), p

where

Therefore (6)

whenever p > 2qo. A similar argument applies to v, using Eq. (5); thus

(7)

plu(x,s)le-UX< 2,

plv(x,s)le-U(1-X) < 2

whenever p > 2qo.

From Eq. (4) we now find that (8)

su(x,s) = sinh sx + s-leUXR1(x,s),

(9)

u'(x,s) = cosh sx + s-leUXR2(x,sy,

298


SECo 103]

where R 1 and R2 are continuous functions whose absolute values are less than 2qo when p > 2qo. In particular, (10)

su(l,s)e-a = e-a sinh S +

where

1

-R1(l,s), S

if

We found that u(l,s) cannot vanish unless S has real or pure imaginary values :t.JI,;. If S is real and not zero, S = :tu, then e-a sinh S approaches :ti as o' -+ 00, and if follows from Eq. (10) that a number')l exists such that u(l,s) =FO when o' ~ ')l. All real zeros of the entire function u(1,s),therefore,

líe on a bounded interval of the real axis; hence they are finite in number. When S is apure imaginary number, o' = Oand s = ir¡,so r¡u(l,ir¡)= sin '1 - .!.R1(1,ir¡) '1

('1 real).

Note that r¡u(l,ir¡)is an odd function ofr¡ since u(x,s) is an even function of s. AIso u(x,ir¡)is real-valued because itis the solution ofthe initial-value problem (1) in which the coefficients are all real, so that ü satisfies the same problem and hence ü = u. Thus the function r¡-lR1(1,ir¡)is real-valuedand odd. It is continuous and vanishes as '1 -+ m), or simple poles :tsn = :t.JI,; of y(x,s),therefore exists. When n > m, then sn = ir¡n'where 'Inis real and, according to Eqs. (8) and (9), (12)

(13) With the aid of the last two equations we find that the norm tends to 1/J2 as n -+ 00, and that a constant K, independent of n, exists such that the normalized eigenfunctions l/Jnsatisfy conditions 11r¡nu(x,sn)

(14)

11

14>n
Since u is of class C' in x, then 4>nis also of that class. Qur principal results are stated as follows.


[SECo 104

Theorem 5 The reduced Sturm-Liouville

299

problem (l), Seco 101, has an

than afinite number ofthem are A.n. positive. The value of - A.nand is arbitrarily infinite set (ll) of eigenvalues All eigenvalues arjeal, no more close to an integral multiple of 1t when n is sufficiently large. The corresponding orthonormal eigenfunctions
THE GENERALlZED

FOURIER SERIES

To prove that the series of residues of y converges to f(x}, we first find order

properties of Green's function G and our function y. Againwewriteu = IResi and p = Isl. When x

~ ~,

) - u(x,s}vR,s} - [se-"Xu(x,s}][se-"(l-~)v(~,s)]

G(x,~,s-

u(l,s) -

s2e-"u(1,s)e"(~-X) .

In view of conditions (7) and Eq. (10), Seco103, then 4

(1)

IG(x,~,s)1 < pe1 -" sms .h

(1 + s -l R l'S

)1

This condition is valid also when x ~ ~, owing to the sYlOlOetryof Green's function. The denominator vanishes only at points lying along the axis of ilOaginaries or at points on a segment u < y of the real axis, since the zeros :t Snof u(l,s) are so distributed. In case all the zeros are pure ilOaginary nulObers, we let y denote any positive number. Throughout either of the two half planes u ~ y, the denominator in condition (1) is bounded away frolO zero, since e-"Isinh si ~ !<1 - e-2y) > O and s-lR1(1,s) -t Oas p -t oo. Therefore G is (I)(p-l) over those halfplanes. For points s = :tu + il] on lines 1]= :t(N - t)1t (N = 1,2,.. .), e-"Isinh si = e-" cosh u = t(1 + e-2") >. t.

It follows frolO condition (1) that G is (I)(p- 1) on those lines as p -t oo. Hence for all points s in a right half plan e or on the open rectangular paths CN used in Theorem 10, Seco 73, in which PN = (N - t)1t, a constant M independent of x, ~, and N exists such that M (2) IG(x,~,s)1< when Re s ~ y or when s is on CN. p It follows that our function (3) y(x,s) = -s ff(~)G(x'~'S)d~ is at least bounded over the halfplane and on CN.

300


SECo 104]

Since for a fixed x, Green's function satisfies the homogeneous differential equation of which u(~,s) and vR,s) are solutions, we can make the substitution (~ :F x) into Eq. (3) to write 1 l 1 l y(x,s) + f(~)G~~(x,~,s)d~ = f(~)q(~)G(x,~,s) d~. s o s o

f

f

The right-hand member here is &(s- 2) on the half plane Re s ~ 'Yand on the paths CN, in view of condition (2). The left-hand member of the equation therefore satisfies the same conditions, and so its inversion integral along the line Re s

=

'Y has

the value zero when t

= O; also

(Sec. 73) its integral

over CN vanishes as N -. oo. That is, if CN denotes the closed rectangle with the side along Re s = 'Yadded to CN, then (4)

ICN

]

[

y(x,s) + ~ I: f(~)G~~(x,~,s) d~ ds -. O

as N

-. oo.

Now let us assume 1", and therefore f' and J, sectionally continuous. It will suffice to let f have just one jump of an amount jo at a point XO (O< Xo < 1). Let x be a fixed point of the interval at whichf is continuous, say O < x < Xo. The jump in G~ at the point ~ = x is 1, as we can see from the expression for G in terms of u and v. Then x

l

fo

fG~~ d~

=

fo

l

xo

fG~~

d~+

f

x

fG~~

d~ +

J~

fG~~d~

= fG~lo + fG~I~o + fGd;o - I: f'G~

d~

= - f(x) - f(O)G~(x,O,s)- joG~(x,xo,s)

+ f(l)G~(x,l,s) -

I: f'G~

d~,

if we write f(O) for f( + O) and f(l) for f(1 - O). Note that G~x,O,s)is u'(O,s)v(x,s)/u(l,s)and u'(O,s)= 1, etc. Thus in terms of the function (5)

(x s) - f(O) v(x,s) g , su(l,s)

+ . u(x,s)v'(xo,s) - f(l) u(x,s) lo

su(l,s)

the bracketed integrand in Eq. (4) is the sum (6)

su(l,s)


[SECo 104

301

The final member of the sum (6) is (!J(s-Z)on CNas we can see by integrating the sectionally continuous function l' G~by parts and recalling that G itself and therefore such terms as j¡G(x,x¡,s) and gf"G d~ are (!J(S-l). Thus the integral around CN of that member tends to zero as N -+ oo. We find, corresponding to Eqs. (8) and (9), Seco103, that (7)

sv(x,s)= -sinh s(l - x) + S-lea(1-X)R3(x,s),

(8)

v'(x,s) = coshs(l - x) + s-lea(1-X)R4(x,s),

where R3 and R4 are bounded

when Isl > 2qo and O ~

x ~ 1.

We write the coefficient of jo in the sum (5) in the forms 1 ~ 1 [sinh sx + s-leaXR1(x,s)][cosh s(l s u( ,s)

-

xo) + S-lea(1-xo)R4(xo,s)]

= e-a sinh sx cosh s(l - xo) se-a sinh s + R1(1,s) + gl(X,XO's), where it is readily verified that gl is (!J(s-z) on CN. Since l/(a + b) can be written as l/a (b/a)/(a + b), that coefficient is

-

sinh sx cosh . s(l s smh s

-

xo) + R1(1,s)sinh sx cosh s(l ;--.

- xo) +

s smh s se-a smh S + R1(1,s )

g 1(X,Xo,S)

- sinh s(l - x - xo) + gz(x,xo,s) + gl(X,XO's), = sinh s(l + x - xo) 2ssinhs where gz is (!J(s-Z)on CN. Note that 0< 1 + x -

Xo < 1 and 11 -

x - xol

< 1. Thus the two terms of weak order here are of the type tf¡(e,s)= si~h es s smh s

where

- 1 < e < 1.

Actually, tf¡ is the transform of a square-wave function which vanishes over an interval containing the origin t = O(Prob. 9, Seco104). An examination ofthe integral of 1tf¡1 around CN shows that JCNtf¡ds -+ O as N -+ 00 (Prob. 11 below). Similarly, the integrals around CN of the coefficients of f(O) andf(l) in the sum (5) vanish as N -+ 00; the reduction

in terms of the function tf¡is simplerfor those coefficients.Thus JCNgds -+ O

as N -+ oo.

Consequently, condition (4) reduces to the equation (9)

i[

. f(X) 11m y(x,s) - ds = O; N oo CN S]

302


SECo 104]

that is, the sum of the residues of the integrand at the poles inside CN converges to zero as N -+ 000 But the sum of residues of y were found in Seco102 to be 'I:.cnl/Jn(x);therefore 00

L cnl/Jn{x) n= 1

(lO)

f{x) = O,

where Cn =

f

f{x)l/Jn{x)dx.

There are no essential changes in the proof of formula (lO) when Xo < x < 1. Thus :rheorem 4 is established for the reduced form of the Sturm-Liouville problem with boundary conditions X{O) = X(1) = O. A similar procedure, but more involved, can be followed to prove the theorem when the more general boundary conditions of type (5), Seco 100, are usedo Theorem 4 as stated in Seco 99 then follows by transforming the reduced problem back to the original Sturm-Liouville problem with the substitutions (2), Seco 100.

PROBLEMS 1 . Prove that the eigenvalues A..of the Sturm-Liouville problem y"(x) - (A.+ h)y(x) =

O,

O< x < e;

where h is real and constant, consist of the numbers

y(0) = y(c) = O,

-h-

n2n2/el

(n = 1,2,0 o0)0 Note

that not more than a finite number of them are positive, and that - h is not an eigenvalueo Show that the normalized eigenfunctions 4J. are .j2jC sin (nnx/c) and that according to Theorem 4, iff" is sectionally continuous (O< x < e), then f(x) is represented at each point wherefis continuous by its Fourier sine series on the interval (O,e): 2 QO nnx sin -

=-

f(x)

L

e .= 1

e

l

c

o

f(~) sin

nn~ - e d~

(O< x < e).

2. Writef(x) = 1 (O < x < e) in Probo 1 to show that

in .=f 12n ~sin 3.

(2n

1

-e l)xx = 1

(O< x < c)o

Find all A..and 4J.(x)for the Sturm-Liouville problem y"(x)

- A.y(x)= O,

y'(0) = y'(e) = O,

and obtain the Fourier cosine series expansion of f(x), 1

f(x)

= -e

l

2

c

o

f(~) d~ +

QO

nnx

-e.=L1cos - e

l

c

o

nn~

f(~) cos -

e

d~

(O< x < e),

that when A. = Othe general solution ofthe SturmLiouville equation is y = Ax + B, where A and B are arbitrary constantso 4. Find all characteristic numbers and functions of the systems (a) y"(x) - A.y(x)= O, y(0)= y'(c)= O, (b) y"(x) - A.y(x)= O, y(0) = O, y(l) + hy'(1) = O,

as a specialcaseofTheorem 4. Note


[SECo 104

303

where h is a positive constant. Note the orthogonality of the characteristic functions as given by Theorem 3. Anso A. = -a/, y.(x) = sina.x, where in (a), a. = (n - !)7tc-l, and in (b), a. > ° and tan a. = -ha. o 5.

Use the substitutions (2), Seco100, to reduce the problem

-

(X4y')'

(A

- 2x2)y = 0,

y(1)= 0,

y'(2)= 0,

to the form z"(t)

-

JlZ{t)

= 0,

z(O)= 0,

z(1)

-

z'(1) = O.

Then use them to derive the following expressions for the eigenvalues and eigenfunctions of the problem in y: Al

= 0,

A.= -4a.2

where a. are the positive roots of the equation tan a

(n = 2,3,.. o)

= a; (n = 2, 3, . o.).

6.

For the fourth-order eigenvalue problem d4y dX4

-

Ay =

0,

y(0) = y'(0) = y(c) = y'(e) = 0,

prove that two eigenfunctions y and z and their corresponding eigenvalues A and ¡.L must satisfy the condition (A - ¡.L) J: y(x)z(x) dx

= 0,

and hence that ifA '# ¡.L,y is orthogonal to z on the interval (O,e);also that all eigenvalues must be real. 7. and 8. 9.

In Seco 100, complete the derivation of form (1) of the Sturm-Liouville equation of the substitutions (2) that produce the reduced form (4). Derive formulas (7) and (8), Seco104, for sv(x,s) and v'(x,s). Let'l' be this periodic square-wave function:

'I'(c,t)= °

if° < t < 1 - e or if 1 + c < t < 2,

'I'(c,t)= 1

if 1 - e < t < 1 + c,

'I'(c,t + 2)= 'I'(t)

whent > 0,

-

where ° ;;¡;e < 1; also '1'( c,t)= - 'I'(e,t). Show '1' graphicallyas a function of t. Use Theorem 10, Seco23, to prove that its transform is the function t/J(e,s)used in Seco104: sinh es L{'I'(e,t)} = t/J(c,s)= s sinh s

(-1 < e < 1, Re s > O).

10. Apply Theorem 6, Seco68, to the square-wave-function 'I'(e,t) in Probo 9 to prove that L¡-I{t/J} = Owhent = O.

304

11

.


SECo 104]

If we write s

on the lines v

=

=v+

i", the sides of the cIosed rectangle CN used in Seco104 lie - PN, and" = IPN, where PN = (N - t)7t, N = 1,2,....

y(y > O), v =

For the function sinh es tf¡(e,s)= s sinh s

(-1 < e < 1)

establish this property, used in Seco104: lim

f

N-C() JCN

tf¡(c,s)ds

= O,

with theaid ofthe result found in Prob.lO. 12. If the function f is continuous (O~ x ~ 1) and satisfies the boundary conditions in the reduced Sturm-Liouville problem (1), Seco 101, so that f(O) = f(l) = O,and if!" is sectionally continuous, prove that condition (9), Seco104, is satisfied uniformly in x; hence that

f(x)

= n=l L

(O~ x ~ 1)

c.4>.(x)

and this generalized Fourier series for f converges uniformly. 13.

Write the reduced system (1), Seco101, in the form

X"(x) - q(x)X(x) = AX(X),

X(O)= X(l) = O,

and show that, if zero is not an eigenvalue of the system, each eigenfunction X. is a solution of the homogeneous integral equation

= A. fX.(~)G(x,~,O)d~

X.(x)

where A.is the eigenvalue corresponding to X. and G is Green's function (4), Seco102. Notation used in Probs. 14 to 18: Let N.[X] and N¡¡[X]and their derivatives with respect to the real-valued parameters exand Pdenote these boundary values on the interval O~ x ~ 1: N.[X] = X(O)cosex+ X'(O)sin ex,

N¡¡[X]= X(l)cosP + X'(l)sinp,

N~[X] = -X(O) sin a + X'(O)cosex, Np[X] = -X(l)sinp

+ X'(1)cosP.

Also, u(x,s) and v(x,s) denote the solutions of these initial-value problems, so arranged that N.[u] = N¡¡[v] = O: (a) u" - (S2 +

(b) v" -

(S2

q)u = O,

+ q)v = O,

u(O,s) = sin ex,

u'(O,s)= -cos ex,

v(1,s) = sin p,

v'(l,s) = -cosp,

and y(x,s) is the solution of the problem (e) y"

-

(S2

+ q)y = -sf(x),

The following problems display several steps in proving Theorem 4 for the SturmLiouville problem (d) X"(x) - [A+ q(x)]X(x) = O,


[SECo 105

305

14. Use Green 's function (Probs. 8 and 9, Seco97) to write the solution of the boundary value problem (c) above in the form

+

Y(x,s) = N~U][V(X'S) f: f(~)u(~,s)d~

15.

Let A. and l.

u(x,s)

f

1

f(~)v(~,s)d~

be eigenvalues and normalized eigenfunctions of Probo (d) above

andwrites. = .J A..Notewhy:ts. arezerosofNp[u] and singularpointsofthe function y (Prob. 14). Recall that (Prob. 8, Seco97) Np[u] = -N Av] and that (Sec. 98) eigenfunctions corresponding to the same eigenvalue are linearly dependent; then show that, corresponding to Eqs. (7) and (8), Seco102, 4>.(x)

4>.(X)

- N~[4>.],

u(x,:ts.) = 16.

v(x,:ts.) =

-

Np[4>.r

Verify that (cf. Seco 102 and Probo 15)

N p[u]NíM.]

=-

(S2

-

s/)

f

u(x,s)4>.(x) dx.

Then deduce that :t s. are simple poles of y(x,s) and that the sum of the residues of y at the two poles :ts. is c.4>¡,(x)if s. :¡!:O, and if s. = O, the residue is c.4>.(x). 17.

Verify or derive the solution of the initial-value z"

-

S2Z

= q(x)u(x,s),

problem

z(O,s) = sin a,

z'(O,s)=-cos

a,

where u is the solution of Probo (a) above, in the form

z(x,s)= sinacosh sx

- !cosasinhsx+ !s f.ox q(~)u(~,s)sinhs(x s

~)d~.

Hence show that u satisfies the integral equation su(x,s) = s sin a cosh sx 18.

cos a sinh sx +

f: q(~)u(~,s) sinh

s(x

-

~) d~.

Prove that [ef.condition (6), Seco103,and see Probo 17] Isu(x,s)le-<7X

105

-

< 2(1ssin al + Icosal + Icosal)

STEADY TEMPERATURE IN A WALL

Let U(x,y) be the steady-state temperatures in a semi-infinite wall bounded by the planes x = O,x = 1, and y = O(Fig. 101),for which the face x = O is insulated and surface heat transfer takes place at the face x = 1,whilethe face y = Ois kept at temperature F(x), so that the boundary value problem becomes (1) (2) (3)

u xx(x,y)+ Uyy(x,y)= O (O< x < 1, Y > O), UAO,y)= O, Ux(1,y)= -hU(l,y) (y> O), U(x,O)= F(x), lim U(x,y) = O (O< x < 1). y ro

306


SECo 105]

,,-O' -

o

U-}1x1 (1,0)

Fig.101

The constant h is positive, and the function F is prescribed. Although the variable y has a semi-infinite range, the problem is not well adapted to solution by using the Laplace transformation for reasons indicated in Seco52. The method of separation of variables does apply. To find all functions of type X(x) Y(y), other than zero itself, that satisfy all homogeneous conditions in the problem, we first note that, from condition (1), X"(x) Y(y) + X(x) Y"(y) = O. X"(x)

Therefore

Y"(y)

X(x) = - Y(y) = A,

where the parameter A is independent of x and y (cf. Seco101). Then all the homogeneous conditions are satisfied if X(x) is a solution of the SturmLiouville problem (4)

X"(x)

-

AX(X) =

O,

X'(O)= O,

hX(1) + X'(1) = O,

and if Y satisfies the conditions (5)

Y"(y) + AY(y) = O,

lim Y(y)= o. Y"" 00

When A = O,the general solution ofthe differential equation in problem (4) is X(x) = Ax + B, and this satisfiesthe two boundary conditions in the problem only if the constants A and B are both zero. Consequently zero is not an eigenvalue. When A :F O, the function that satisfies the first two of conditions (4) can be written X(x)

= e coshx..ji.

It satisfiesthe third condition if (6)

h cosh

..ji + ..ji sinh ..ji

= O.

According to Theorem 3, all eigenvalues of problem (4) are real. Ir A > O,


[SECo 105

307

the left-handmember of (6)has a positive value. All eigenvaluesof problem (4) are therefore negative. We write A.= -CX2. Then Eq. (6) becomes h (7) tan cx= -, cx

an equation that has an infinite set of roots :!:CXn (n = 1,2,. . .) that can be approximated graphically. In fact the number CXn is only slightly greater than mI:when n is large. The complete set of eigenvalues of problem (4) is therefore (n = 1,2,. ..), where the numbers cxnare the positive roots of Eq. (7),and the corresponding eigenfunctions are

=

Xn(X)

COS CXnX.

Let Pndenote the norm IIXnll. Then

Pn2 =

f l

2 COS CXnXdX

I

= _2 1 +

sin

CXncos

CXn

CXn

)

(

o

and, since CXn is a root of Eq. (7), we find that (8)

Pn = 2-t( 1 + h2

: cx/r

The orthonormal eigenfunctionsof problem (4)are therefore 1 (9) 4Jn(x)= - COS CXnX (n = 1,2, .. .). Pn

When A.= -cx/, the solution ofproblem (5) is Y(y) = cn exp ( - cxny), where the constant

Cnis arbitrary,

and therefore

cos cxx

= Cn~exp

X(x)Y(y)

(-cxny)

(n

= 1,2,...).

The sum of any number of these functions also satisfies all homogeneous conditions in problem (1) to (3); but unless F(x) is a linear combination of the functions cos CXnX, no finite sum will satisfy the remaining condition U(X,O) = F(x) The function represented
L

by the infinite series C ..!!. COS CXnX exp

n= 1 Pn

(-cxnY)

(O < x < 1).

308


SECo 106]

formally satisfies the homogeneous conditions. According to Theorem 4, this function reduces to F(x) when y = O,provided F" is sectionally continuous, if the numbers Cnare the Fourier constants for F with respect to
(1 1 (1 Cn= Jo F(x)
Note that the series used to represent F(x) here, G() e

(11)

¿ cos n=1Pn

F(x) =

...!!.

(O< x < 1),

(l(nX

is a generalized Fourier series, not the Fourier cosine series for the interval O
(12)

U(x,y) =

¿ exp ( - IXnY) COS IXnX, n= 1 Pn ...!!.

where the constants Pnand en are given by formulas (8) and (10). 106

VERIFICATIONOF THE SOLUTION

In the preceding temperature problem F", and hence the prescribed function F itself, is sectionally continuous (O< x < 1). At each fixed point x where F is continuous, the Sturm-Liouville series (11), Seco105, therefore converges to F(x):

e

G()

(1)

¿ ...!!.COS(l(nX = n= 1 Pn

(O < x < 1).

F(x)

We order the positive numbers (l(nso that (l(n+1 > (l(n'Then the functions exp (-(I(nY) in our formal solution (12), Seco 105, are bounded, and nonincreasing with respect to n: 0< exp(-(I(nY) ~ 1,

exp(-IXn+1Y)~ exp(-(I(nY)

(y

~ O).

According to Abel's test (Prob. 6, Seco81), therefore, the series G()

(2)

e

n~1 P: cos (l(nXexp (-(I(nY)

(y ~ O,O < x < 1)

converges uniformly with respect to y when y ~ O. Since its terms are continuous functions of y, its sum U represents a continuous function of y when y ~ O. Qur formal solution therefore satisfies the condition U(x, + O) = U(x,O) = F(x)

(O< x < 1).


[SECo 107

309

From Eq. (8), Seco105, we seethat Pn > l/fi. Also, a fixed number M exists such that ICnl/Pn< M. Whenevery ~ Yowhere Yo> O,the absolute values of the terms in series (2) are less than the constants M exp ( - anyo); also the absolute values of the deriva ti ves of those terms, of first and second order, with respect to x and y, are less than Man exp (-anYO) and M an2 exp ( - anyo), respectively. The series of those constant terms converges,

-

according to the ratio test, since an - me O as n-oo. Hence series (2) and the series of derivativesof its terms convergeuniformlywith respect to x and Y over each strip O ~ x ~ 1, Y ~ Yo > O. The terms are continuous functions, so the series converge to a continuous function U(x,y) and its continuous partial derivatives. In particular, U and Ux are continuous functions of x on the interval O ~ x ~ 1,when Y > O. They satisfy the homogeneous boundary conditions Ux(1,y) + hU(l,y) = O

because the terms in series (2) satisfy those conditions. The terms also satisfy Laplace's equation (1), Seco 105; hence the sum U satisfies that equation when Y > Oand O < x < 1. FinaIly, we note that the sum U(x,y) satisfies the remaining condition U O and y 00 because each term satisfies that condition and the remainder after N terms is arbitrarily smaIl in absolute value, uniformly with respect to x and y(y ~ Yo),when N is sufficientlylarge. Thus formula (12),Seco105, is established as a solution ofthe boundary value problem in the temperature function U.

-

-

107 SINGULAR EIGENVALUE PROBLEMS' If the differential equation in an eigenvalue 'problem has a singular point on the interval, or if the interval is unbounded, the eigenvalue problem is singular. A requirement of continuity or boundedness of eigenfunctions replacesa boundary condition at a singular point or at an infinitelydistant end of the interval. The representation theory for such problems may paralIel that used for the Sturm-Liouville system, but the singular aspects of the probems require variations in the development of the theory.2 Some singular problems have discrete eigenvalues like those in SturmLiouville problems. Others have continuous eigenvalues, consisting of aIl numbers on a bounded or unbounded segment of the axis of reals. When eigenvalues are continuous, the representation of an arbitrary function in terms of eigenfunctions takes the form of an integral with respect to A,instead 1

Further details, inc1uding proofs or references to proofs of representation theorems, and

applications, are given in the author's 8, and 9, 1963. 2 See E. C. Titchmarsh,

"Eigenfunction

"Fourier

Series and Boundary

Expansions,"

1946.

Value Problems,"

chaps. 6,

310


SECo 107]

of a series, of eigenfunction~. Prominent examples of singular problems follow.: :' BESSEL'S

EQUATlON

An example of an eigenvalue problem with Bessel's differential equation of index zero, on an interval (O,e),is [xX'(x)]'

(1)

X(O)= O,

- ..txX(x) = O

(O < x < 1),

X and X' continous

(O~ x ~ e).

Here the differentialequation is a Sturm..Liouvilleequation, but r(x) = x and reO)= O. When we write it in the form (2)

X"(x) + .!.X'(x) - ..tX(x) = O x

(O< x < e),

we see that one of the coefficients in that form is discontinuous at the end = Oof the interval, so that end point is a singular point of the equation, and the eigenvalue problem is singular. The solution of Eq. (2) that is continuous, to~her with its derivative, over the interval O ~ x ~ e, is (Sec. 83) X = J o(x"'¡ - ..t),except for a constant factor. Thus X(e) = O if Jo(e.¡::i) = O.As in Seco83, let an(n= 1,2,...) denote the positive zeros of J oCa).Then the eigenfunctions and eigenvalues of problem (1) are

point x

(3)

Xn

= Jo( CXn~)'

The orthogonality of the eigenfunctions on the interval (O,e) with weight function p(x) = x can be established by the procedure used in the Sturm-Liouville theory. A representation theorem states that if.f' is sectionally continuous on the interval, the generalized Fourier series of those eigenfunctions converges to f(x) at each interior point wherefis continuous. The representation can be written (4)

(O < x < e).

There are similar results if the differential equation in the problem is Bessel's equation with a positive index v, (xX')' or when the boundary

condition

-

(..tx

(O < x < e),

+ ~)X = O

at the end point x

hX(e) + X'(e) = O

= e reads (h ~ O).


[SECo 107

311

But if the interval does not contain the origin X = O as an end point or as an interior point, these eigenvalue problems in Bessel's equation, with a homogeneous boundary condition at each end of the interval, are nonsingular special cases of the Sturm-Liouville problem. LEGENDRE'S

EQUATION

The eigenvalue problem

[(1 - X2)X'(X)]' - AX(X) = O

(5)

(-1 < x < 1),

- 1 ;;;¡; x ;;;¡; 1,

X and X' continuous on the closed interval

is another example of a singular problem. Legendre's differential equation here has singular points at both end points x = :!: 1 of the interval. Its solu-

tion satisfies the continuity condition only if A = - n(n+ 1),wheren = O,1,2, .... Corresponding to those eigenvalues, the eigenfunctions are the Legendre polynomials

(6)

P( x--L.)- 1

n

~ (-1}i

2nj=O j!

(2n - 2j)!

(n - 2j)!(n - j)!

x

n-2j

(O!= 1)

wherem = tn ifn is even and m = t(n - 1)ifn is odd. They form an orthogonal set with weight function p(x) = 1 on the interval (-1,1). The representation of a functionj on that interval, such thatf' issectionallycontinuous, as a generalized Fourier series of the eigenfunctions, can be established rather easily by using properties

of Pn. It can be written

(7)

FOURIER

(-1 < x < 1).

INTEGRALS

The singular eigenvalue problem (8)

X" - AX = O,

x> O;

X'(O) = O,

IX(x)1< M (O< x < ex»),

where M is some constant, is one in which the interval is unbounded. The solution of the differential equation that satisfies the condition X'(O) = O is, except for a constant factor,X = cos x~, for any value of A.,including zero. That solution is bounded over the half line x > O if and only if A is real and A ;;;¡;O. We write A = - rx2,where rx~ O. Then the eigenfunctions and eigenvalues are

(9)

X(x) = cosrxx,

(rx~ O).

312


SECo 107]

In this example the eigenvalues are continuous, A ~ O, rather than discreteo The eigenfunctions lack an orthogonality property, but a representation theorem follows from the Fourier integral formula (Sec. 68) when the function G there is even. Ir!, is sectionally continuous on each bounded interval of the half line x > Oand iflis absolutely integrable over that half line, then '" 2 '" (10) (x ~ O). I(x) = cos ax IR) cos
f

Thus lis represented as an integral of the eigenfunctions cos
The steady-state temperature problem (cf. Seco105) u ",,(x,y) + Uyy(x,y)

= O,

IU(x,y)1< M,

(O< x < 1,Y > O)

U x(O,y) = U x~,y)

= O,

U(x,O)= F(x)

(O< x < 1),

where M is some constant. Show that A.= Ois an eigenvalue of the associated SturmLiouville problem and obtain the formal solution in the form 1

U(x,y) =

,.

1o F@d~

""

1

+ 2 ft~1e-m'Ycosn7tx 1o F@cosn7t~d~.

Note that as y -+ 00, U tends to the mean value of F. 2. The problem in Seco90 on displacements Y(x,t) in a string when Y(x,O) = g(x), Y,(x,O)= Y(O,t)= Y(c,t) = O. 3. Probo 1, Seco93, in the longitudinal displacements Y(x,t) of a stretched bar, where Y(x,O)= Ax, Y,(x,O)= Y(O,t)= Yx(l,t) = O. 4. Prob.4, Seco93, for displacements Y(x,t) in a string with initial velocity Y,(x,O)= g(x),O < x < C. 5. (a) The problem in Seco82 on temperatures U(x,t) in a bar when U(x,O)= g(x), O < x < 1, U(O,t)= U(l,t) = O,and k = 1. (b) When g' is sectionally continuous, verify the solution of the boundary value problem in part (a) using the series representation "" 1 U(x,t) = 2 ft~1exp(-n27t2t)sinn7tx

1o g(~)sinn7t~d~.


6.

313

[SECo 107

Give a physical interpretation of U(x,y) in the problem u x(O,y)= U(l,y) = O,

-KU,(x,O) = F(x),

where U is bounded over the region O < x < 1,Y > O,and K > O. Write m = (n - t)1t

and derive the solution in the form

-2

l

1

00

U(x,y)= K L -e-my cosmx n= I m

f o

F(~)cosm~d~.

7. Let U(x,y) be the steady-state temperatures in a thin plate in the shape of a semiinfinite strip, from which heat is transferred at the faces into a medium at temperature zero, so that

,

(O< x < 1,Y > O,b > O).

Uxx+U,,-bU=O

If U is bounded and satisfies the conditions

U(x,O)= 1

Ux(1,y) = - hU(l,y),

U(O,y) = O,

(O< x < 1).

and h > O,derive the formula 00

U(x,y) = 2h

L

A --!! exp [

n= 1 an

- y(b +

where An = (1 -

(b > O, h > O),

(Xn2)*] sin (XnX

cos (Xn)/(h+ cos2 (Xn)and (XI' (X2,. .. equation tan (X= -(X/h.

are the positive roots of the

8. Let U(x,y)be the steady'state temperaturesin an infiniteprism bounded by the planesx = O,Y = O,x = 1,andy = 1. If U = 1onthefacey = 1andifUx = - hU at x = 1,whereh > O,and U = Oon the other two faces(Fig.102),derivethe formula

where the numbers An and (Xnare those described in Probo 7. Y/

U=1

0

U=O

Fig.102 9. Writem = (n

0,1)

= -O.

.-

O U=O--X -

- t)1t and obtain the solution of the problem (t + l)U,(x,t) = Uxx(x,t) U(O,t)= Ux(1,t)= O,

(O< x < 1, t > O),

U(x,O)= 1

in the form

U(x,t)= 2

f

..!..(t+ Wm2 sin mx. n=1m

Verify that this function satisfies the condition U(x,+O) = 1(0 < x < 1).

(O< x < 1),

314


SECo 107]

Fig.103

10. The end x = 1 of a stretched string is elastically supported (Fig. 103)so that the transverse displacement Y(x, t) satisfies the condition y"(l,t) = -hY(1,t), where h > O. Also, let Y satisfy the conditions

Y(O,t)= O,

Y(x,O) = bx,

y'(x,O) = O.

Show that the solution of the equation y',(x,t) = y"x(x,t)is then '"

sin IX"sin 1X"x

Y(x,t) = 2bh(h+ 1).=L1 IX.2(h + cos2 IX.

\

COSlX.t,

1X1,1X2,. . . are the positive roots of the equation tan IX= -IX/h. 11. The longitudinal displacements Y(x,t) in a certain nonuniform bar satisfy the conditions

where

(O< x < 1, t > O), Y(O,t)= Y(l,t) = y'(x,O)= O,

Y(x,O) = F(x)

(O< x < 1).

Derive the formula ao

Y(x,t) =

L c.l/J.(x)cos P.t, n=1

where P. = n7te/(e - 1) and l/J.(x)= j!jsin

[P.(1 - e-X)],

12. The electrostatic potential V(r,z) in the space bounded by the cylinders r = a and r = b above the plane z = O satisfies Laplace's equation (rV.)r+ rv.. = O and the boundary conditions V(a,z) = V(b,z) = O V(r,O)= F(r)

(z > O), (a
also V(r,z) is bounded. Derive the formula ao

V(r,z) =

L A. 11=1

exp (-IX.z)t/!(rx.,r),

- Jo(aa)Yo(lXr).Jo and Yo are Bessel functions of the first and second kind, linearly independent solutions of Bessel's equation. The numbers IX.

where t/!(ex,r)= Jo(rxr)Yo(aa)


[SECo 107

315

are the roots ofthe equation tJ¡(a,b) = O,all reaP am > O,and An are given by theformula

f

An

r[tJ¡(an,r)]2

dr =

f

rF(r)tJ¡(a.,r)

dr.

13. The temperature V(r,t) in a solid cylinder r ~ 1 of infinite length satisfies the conditions

(O~ r < 1,t > O),

V(r,O)= 1,

V(l,t) = O;

also V and V. are continuous when O ~ r ~ 1 and t > O. Show that the method of separation of variables involves a singular eigenvalue problem of type (1), Seco 107. With the aid of the integration formula

[XJo(X)dX

= cJ¡(c),

solve for V in the form

~

J o(a.r)

2

exp(- a. t), V(r,t) = 2 L.. - J .= ¡ a. ¡(a.) where the a;s are the positive 14.

The temperature

zeros of J o(a).

U(x,t) in a semi-infinite

function is a step function,

solid x ~ O, whose initial

temperature

satisfies the conditions

u,= kUxx;

U(x,O) = 1 - So(x - 1),x> O;

U x(O,t) = O,

and U is bounded. Show that the solution by separation of variables involves the singular eigenvalue problem (8), Seco107, and derive the formula 2 "" sin a U(x,t) = -cosaxexp 7t1o a 15.

(-a2kt)

da.

Prove that the eigenvalues and eigenfunctions of the singular problem X" - J.X = O,O < x < co;

X(O) = O;

IX(x)1~ l,x > O;

are J. = -a2(a > O) and X = sin ax. If l' is sectionally continuous and absolutely integrable over the half line x > O,note that, according to the Fourier integral formula (Sec. 68) for odd

functions,

f

is represented

in terms of those eigenfunctions

by the

F ourier sine integral formula

2

f(x) =

""

""

- 1 sinax 1 f(~) sin a~d~ 7t o o

(x > O).

16. Use the results found in Probo 15 to solve the problem in Seco43 for the displacements Y(x, t) of an initially displaced semi-infinite stretched string, namely,

x > O,t > O; Y;(x,O) I

= Y(O,t)= O,

Y(x,O) = (x), x > O;

IY(x,t)1 < M,

1(x)1 <

M,

See H. S. Carslaw and J. C. Jaeger, "Conduction of Heat in Solids," 2d ed., p. 206, 1959, and

their

tables,

p. 493.

316


SECo 107]

in the form y = ~ (00[sin ex(x+ at) + sin ex(x- at)]

nJo

(00 cI>(~) sin ex~ d~ dex.

Jo

Show how this reduces to the form (8), Seco 43. 17.

Prove that the eigenfunctions

-

X"(x)

of the singular problem

AX(X) =

O,

IX(x)1~ 1

(-00 < x < (0)

are X = cosex(x- ~), where A = -exz and ex~ Oand the parameter ~ has any real value. Note that accordingto the Fourier integralformula(2),Seco68, f(x) 18.

1

=-

oo

if

n o

a>

f(~) cos ex(x- ~)d~dex

(-00

<

x < (0).

-00

Note that the first few Legendre polynomials are Po(x) = 1,

P1(x) = x,

Pz(x) = !(3xZ

-

1),

P3(x) = !(5X3 - 3x).

Verify that they satisfy Legendre's equation in problem (5),Seco107,when A = -n(n + 1),n = 0,1,2,3. 19.

Legendre's polynomials have the property

p.(-x) = (-l)"P.(x)

(n = O, 1,2,

. . .),

so PZnand PZn+lare even and odd functions, respectively. Deduce that X = PZ.+l(X) - 2(2n + l)(n + 1)are eigenfunctions and eigenvalues ofthis singular problem on the interval (0,1):

and A =

[(1 - XZ)X')'

- AX = 0,0 < x < 1;

X(O)= O,

where X and X' are continuous on the c10sed interval O ~ x ~ 1. 20.

If the condition X(O)

= O is replaced

by the condition X'(O) = O in Probo 19, show

that A = -2n(2n + 1) are eigenvalues and X = pz.(x) are eigenfunctions.

10 General Integral Transforms

108

LINEAR INTEGRAL TRANSFORMATIONS

Let (a,b) denote a bounded or unbounded interval of the x axis and let K(x,A.) be a prescribed function of x on that interval and of a parameter A.. Then, as we pointed out in Seco 1, a general linear integral transformation of functions F on the interval is (1)

T{F} =

f

F(x)K(x)) dx

= f(A.).

The resulting function f(A.) is the transform of F under that transformation with kernel K. Of course the class of functions from which F and K are chosen must be restricted and a set A of values of the parameter A.must be specified so as to ensure not only the existence ofthe integral (1) but also the uniqueness of the inverse transformo That is, not more than one function F should correspond to a given transform f(A.). 317

318

SECo 108]


The class of functions, the function space, to which F belongs is to be a linear space in order that AF(x) + BG(x) will belong to that space whenever F and G belong to it, for each pair of constants A and B. Then T is a linear transformation : (2)

T{AF + BG} = AT{F} + BT{G} = Af().) + Bg().).

Ir inverse transforms T-1 {J} are unique, the linearity condition (2) can be written (3) that is, the inverse transformation is also linear. When a = O,b = 00, and K(x,).) = e-J.x, the transformation T is the Laplace transformation L{ F} = f().). In that case we noted that the function space of F may consist of all real-valued functions of exponential order on the half line x > O which are sectionally continuous on each bounded interval, and the set A of values of)' may be all complex numbers in a right half planeo We restricted the function space somewhat further in order to ensure uniqueness of L - 1 {f}. Also, we found a general formula, the inversion integral, for the inverse Laplace transformation. We shall soon indicate how the transformation T may be specified so that it transforms a given linear differential form in F into an algebraic form in f().), ). and prescribed boundary values of F at the ends of the interval (a,b). That gives the basic operational property of T for the reduction of boundary value problems in differential equations. As in the case of the Laplace transformation, where that operational property is L{ F'} = ).f().) - F(O),some further restrictions on the space of the functions F are needed to ensure the validity of the property. The operational calculus of an integral transformation becomes more effective when its basic operational property and its inversion process are supplemented by other operational properties and tables of transforms. A convolution property of T is one that gives a representation of T-1 {J().)g().)} directly in terms of the functions F and G whose transforms are f and g. The representation is often simpler than one found by applying a formula for inverse transformations. In some cases the property is modified byreplacingfg bya weighted product oftransforms to give T-1{w().)f().)g().)}, where the weight function w is fixed for a given transformation. We ha ve seen that the convolution is an important property of the Laplace transformation. Convolution properties are also prominent operational properties ofvarious Fourier transformations. But for some ofthe less common integral transformations such properties are either unknown or so complicated that they are tedious to use.

GENERAL INTEGRAL TRANSFORMS

109

"KERNEL-PRODUCT

[SECo 109

AND CONVOLUTION

For the Laplace transformation product property

(1)

the kemel K(x)

319

PROPERTIES

= e-.I.x has

the simple

K(x,).)K(xO,A) = K(x + Xo, A).

Then if f(A) = L{F} and F(x) = O when x < O,it followedthat L has the translation property (Sec. 11) (2)

f(A)K(xO,A) = L{ F(x

-

xo)}

(xo > O).

This formula represents the product of the transform of F by the kemel as the transform of a function related to F. We call it a kernel-product property of the Laplace transformation. That property led us to the convolution property

(3)

(Sec. 16)

f(A)g(A)= L{S; F(y)G(x

-

Y)dY}

= L{F* G},

wheref(A) = L{F} and g(A) = L{G}. Now, in the case of another transformation of the type (4)

T{F} =

f

F(x)K(x,A) dx

= f(A),

suppose we have a modified or weighted kemel-product property such that

for some specifiedfunction W(A)and for each function F there is a function PF(x,xo) for which

(5)

(a < xo < b).

Then a modified convolution property follows, formally. For ifJ(A) = T{F} and g(A)= T{G}, we can write (6)

f(A)g(A)W(A) =

=

=

f f ff

g(A)w(A)K(y,A)F(y) dy

T{PG(x,y)}F(y)dy

PG(x,y)K(X,A)dx F(y) dy.

When the order of integration is reversed, the iterated integral becomes

f

K(X,A)

f

F(y)PG(x,y) dy d~

320


SECo 110]

which is the transform of the function (7)

XFG(X)

=

f

F(y)PG(x,y)dy.

We call the function X FG the convolutionof F and G for the transformation T, with respect to the weight function w. It is the inverse transform of the product fgw. That is, (8)

f(A)g(A)W(A) = T{XFG(x)} = T{XGF(x)},

where the last form T {XGF}follows by interchanging the functions F and G, and their transforms, in Eqs. (6). Thus we have indicated how a convolution property (8)follows from a kernel-product property (5). The convolution is described by Eq. (7) in terms ofthe function PFin property (5). An example is given in the following section. Other examples will be found in the problems and in the chapters that follow. A corresponding result for a generalized convolution property for iterated transforms of functions H(x,y) will also be noted in the problems (Prob. 8, Seco 110). Ir we replace F in the transformation (4) by the delta symbol, we ~nd, formally, (9)

if a < Xo < b.

When F is replaced by l5(x

(10)

-

xo), the convolution

X.JG(x)= PG(x,xo)

(7) becomes

[15= l5(x - xo), a < Xo < b],

and the convolution property (8) reverts to our kernel-product property written in terms of g and G: (a < Xo < b).

110

EXAMPLE

We illustrate the results in the preceding section by deriving a convolution formula for the Fourier sine transformation on the half line x > O, (1)

SA{F} = 1'" F(x) sin AXdx = f.(A)

(A > O).

Other operational properties of SA will be developed in Seco116and Chap. 13. We assume that our functions F are sectionally continuous on each bounded interval and absolutely integrable over the half line x > O. It is convenient to introduce the odd extension Fl ofF defined over the fullline - 00 < x < 00: (2)

Fl(X)

= F(x) if x> O,

Fl ( - x) = - Fl (x) for all X.


We

(3)

[SECo 110

321

introduce also the continuous function

H(x)

= foo Ft(~)d~ = foo

+ I: Ft(~)d~

Ft(~)d~

(-00 < x < (0) and note that it is even, H( - x) = H(x) or H(lxl) = H(x),and that it vanishes as x --+ :t 00; also, H'(x) = Ft(x). To find a icernel-product property of SJ..,we write 2

oo

1

lJ.
=l f

-00

Ft(x) sin AXdx

=

-00

] f

= -H(x)

-

A

x-H'(x)dx

oo

00

sin AX

[

f

sin AX

oo

H(x)cos AXdx.

-00-00

The bracketed term here vanishes. Thus when Xo > O, ~.f.(A)sin AXo = - f:oo H(x) cos AXsin A.xodx

=

t f:oo

H(x)[sin A(X - xo) - sin A(X + xo)] dx'

=

t f:

[H(y + xo) - HOy - xol)] sin AYdy.

00

The integrand of the final integral is an even function of y. Thus 2 l.f.(A) sin AXo =

oo

Y+xO

f

f

o sin AY Iy-xolFt(~) d~dy.

Since F1(~) = F(~) when ~ > O, a kernel-product property for the transformation SJ..can be written (4)

(Xo > O).

In terms of the notation used in formula (5), Seco109, then 2

(5)

m(A)=

I

when K(X,A) = sin AX, a = O,b = 00, and T is SJ... The corresponding convolution of two functions F and G is oo

(6)

XFG(X) =

f o

oo

F(y)PG(x,y)dy =

f o

X+Y

fIx-yl G(~)d~dy,

F(y)

322


SECo 110]

according to formula (7), Seco 109. The convolution property reads (oo

fs(A.)g.(A.)~= SA{ Jo

(7)

F(y)

{y+x

G(~)d~dY

J,y-xl

}

(A.> O),

wherefs(A.) = SA{F} and g.(A.) = SA{G}. ILLUSTRATION

OF FORMULA

(7)

In the particular case (8)

F(x)

= G(x) = e-ax

(o

> O),

an elementary integration shows that (9)

fs(A.)

also,

=

g.(A.)

x+y

PG(x,y) =

=-

A.

(A.> O);

1

- e-a(x+y)]. f.Ix-yl e-a~ d~ = -[e-alx-yl o

Then the convolution (6) becomes

1 e-aYe-a(x+y)dy = -xe-ax. o o o 1

--

f

oo

Thus we have this useful Fourier sine transformation (10)

(o > O).

PROBLEMS 1.

Use the convolution property (7), Seco110,when F(x) = e-ax,

G(x) = e-bx

(a > 0, b > 0, a '# b)

to obtain this Fourier sine transformation: (a> O,b > O,a '# b). (This result can be found also by using partial fractions.) Evaluate the integral transformation on the right to verify the result. 2. Interchange the order of integration in formula (6), Seco 110, to show that the convolution X FGfor the Fourier sine transformation SA is symmetric in F and G: XFG(x) = XGP(x).


[SECo 110.

323

3. The exponentialFourier transformationof functions F on the entÍre x a~is is (Chap 12)

(-00 <). < 00).

= f:", F(x)e-UX dx = f.().)

EA{F}

We consider functions F that are sectionalIy continuous on each bounded interval, absolutely

integrable

-

from

00 to 00 and bounded.

Show that EA has the kernel-

product property

= EA{F(x -

f.().)exp(-iho)

( - 00 < Xo < 00)

xo)}

and, formalIy, that the corresponding convolution is XFG(x)

=

f:

F(Y)G(x

-

y)dy = XGF(x),

so that if ge().)= EA{G}, then f.().)ge().)= EA{XFG}= EA{f:", F(y)G(x - y)dY}. 4. Let So(x) be the unit step function: So(x) = Owhen x < O,So(x) = 1 when x > O. Note that the exponential Fourier transforms of the two functions

= e-XSo(x), G(x) = ~So( -x) ge().)= (1 - i).)-I. Use the convolution

(- 00 < x < 00)

F(x) are f.().)

= (1 +

i).)-1 and

property of EA in

Probo3 to derivethis transformation: 1 ).2

-

+ 1-

E {1 -lxl } A '[e .

Verifythe result by evaluatingthe integral EA{e-lxl}. 5. When K(x,).)= cos ).x, a = O, and b = 00, our transformation T becomes the Fourier cosine transformation eA on the half line (Chap. 13), (). > O).

eA{F} = L'" F(x) cos h dx = fc().)

Let the functions F be sectionalIy continuous on each bounded interval and absolutely

integrableover the half line. If F2 is the even extension of F, so that F2(X)= F(lxl) when 00 < x < 00,then

-

fc().)cos ho =

t L"'", F2(x) COSh

cos ho dx.

Show that a kernel-product property of eA is fc().)cos).xo

= eA{t[F(x + xo) + F(lx -

xol)]}

then that the corresponding convolution property is fc().)gc().) = eA{

t 50'"F(y)[G(x+ y) + G(lx-

yl)] dY};

(xo > O);

324


SECo 110]

that is, the convolution for C¡ when W(A.)= 1 is XFG(x) =

t f'

F(y)[G(x + y) + G(lx - yl)]dy.

6. Prove that the convolution for C¡ found in Probo 5 is symmetric in F and G: XFG(x) = XGF(x). 7. Let F be the step function 1 - So(x - k), where k> O,and let G be e-ax (a > O). If C¡ is the cosine transformation in Probo 5, then C¡{F} = sinA.H ' Show that the convolution of F and G for C¡ is the function XFG(x) = a-l(1 - e-akcoshax)

when x ;;¡;k

= a-le-axsinhak

when x ~ k,

so that we have the transformation

= C¡{XFG(x)}.

asinH

8. Generalized convolution For a function H(x,y) of two variables, let h(A.,y)denote the transform T{H} with respect to x,

h(A.,y)=

f

H(x,y)K(x,A.)dx

(a < y < b),

and let h(A.)denote the transform of h(A.,y)with respect to y when the same parameter A. is used,

h(A.)=

.

f

h(A.,y)K(y,A.)dy,

if H and T are such that this iterated transform exists. In case T has a kernel-product property h(A.,y)w(A.)K(y,A.) =

f

QH(x,y)K(x,A.)

dx,

show formally that it has the generalized convolution property

h(A.)w(A.)

=

f [f QH(X,y)dY]K(X,A.)dX.

Note that this reduces to property (8), Seco 109, in case H(x,y) = F(x)G(Y); then QH = PF(x,y)G(y). Examples are given in Probs. 9 and 10. 9. When the transformation in Probo 8 is the exponential Fourier transformatión E¡ describedin Probo3 and w(A.)= 1, show that the generalizedconvolution property becomes


where

Ji(l)

=

[SECo 111

f:

e-iAv I-OOXJH(x,y)e-iAx dx

325

dy.

10. Ir H(x,y) = 1 when x > O and y > O and x + y < 1, while H(x,y) = O for all other real x and y, show that the iterated exponential Fourier transfonn of H is Ji(l) = (1 + iA)e-iA - 1 l and verify that this agrees with the convolution property in Probo 9. 111

STURM-LlOUVILLE

TRANSFORMS

We are now ready to present a method of determining a linear integral transformation T that simplifies a given linear boundary value problem in a function F. The problem is to be such that in the differential equation the differential form in F with respect to some one of the independent variables x can be isolated and, in the boundary conditions, boundary values with respect to x can be isolated. The transformation T with respect to x is to be designed so as to replace that differential form by an algebraic form in the transform T {F} and the prescribed boundary values on F at the ends of the interval over which the variable x ranges. We begin with an important case that can be treated with the aid of the Sturm-Liouville theory in Chap. 9. Let the differential form with respect to x be of order 2, (1)

D2F = A(x)F"(x) + B(x)F'(x) + C(x)F(x)

(a < x < b),

and let the interval (a,b) be bounded. For the sake of simplicity we assume that aIl functions of x here, together with their derivatives of {he first and second order, are continuous over the cIosed interval a ~ x ~ b, although those conditions can be relaxed. Also, we assume that A(x) > O over that cIosed interval. In applications to problems in partial differential equations, F will be a function of x and the remaining independent variables, while A, B, and e will be functions of x alone. The boundary values with respect to x, to be prescribed as functions of the remaining variables, are (2)

NaF = F(a) cos ex+ F'(a) sin ex,

(3)

N (iF = F(b) cos /3 + F'(b) sin /3,

where the constants exand P are real. In order to design a transformation T{F}

=

f

F(x)K(x,A)dx = J(A)

326


SECo 111]

that transforms D2F into an algebraic form in f(A), NaF, and NpF, it is convenient to write D2F in terms of a self-adjoint differential form (4)

!!)2F = [r(x)F'(x)]' - q(x)F(x).

We do that by writing D2F = Ar-1(rF" + A-1BrF') + CF, where

. X

(5)

r(x)

= exp [Ia

B«() A«() d(

]

so that r' = A -1 Br, and introducing the functions r(x) p(x) = A(x)'

(6)

q(x) = - p(X)C(X).

Then 1 !!)2F= ![(rF')' - qF]. px( ) . p

D2F= _

(7)

Note that r(a) = 1 here. Corresponding representations in terms of self-adjoint forms are not always possible for forms of order higher than 2'In such cases the original linear differential form may be used; then the adjoint of that form arises in the process of designing T. We now write the kernel of our transformation T as K(x,A.) =

p(X)eI>(X,A)

where the weight function p is rlA, and the kernel el>and the range of the parameter A are yet to be determined. Then the transform of D2F becomes f~eI>!!)2F dx and, either by integration by parts or by using Lagrange's identity (5), Seco94, according to which we can write (8)

T{D2F}

=

f

F(X)!!)2e1>(X,A)dx

+

[(eI>F'

-

eI>'F)rJ:.

In order to express F(a) and F'(a) in terms of the prescribed boundary value NaF, we introduce the derivative of NaF with respect to a, N~F = - F(a) sin a + F'(a) cos a, and solve this equation simultaneously with Eq. (2) to get F(a) = NaFcos a

- N~Fsin a,

F'(a) = N~Fcos a + NaFsin a.

At the lower limit the bracketed term in Eq. (8) can now be written

[eI>(a,A.)F'(a) - eI>'(a,A.)F(a)Jr(a) = -NaFN~eI>+ N~FNael>.


The

[SECo 112

327

value of this term is determined by NaF and $ if N/1> = O. Likewise

at x

=

b we find that

- $'F)r]x=b= -r(b)NpFN'p$

($F' if N p$ = O, where

N'p$ = - $(b).) sin P + $'(b,A)cosp. The integral in Eq. (8) becomes ,if(A) if $ satisfies the Sturm-Liouville equation ~2$ = Ap$. Hence we have the desired transformation (9) provided that $ is a non trivial solution of the Sturm-Liouville problem (10)

(r$')'

-

(q

+ Ap)$ = O,

Np$=O.

Since problem (10) has nontrivial solutions when and only when A is an eigenvalueAn,the kernel $ must be an eigenfunction$n corresponding to the eigenvalueAn;then (11)

=

T{F}

f

F(x)p(x)$n(x) dx

= f(An)

(n

= 1,2,...).

We call this the Sturm-Liouville transformation. Its parameter A ranges through the discrete real values A.l,A.2,. . ., the set of all the eigenvalues of the Sturm-Liouville problem (10). The transformf(An) is also an infinite sequence of numbers. The basic operational property (9) of T now reads (12)

112

INVERSE TRANSFORMS

We select some convenient complete set of real-valued eigenfunctions $n, not necessarily normalized, as the kernel in our Sturm-Liouville transformation (11) above. The generalized Fourier series for a function F on the interval (a,b) can be written ex>$n(X) n~l

II$nll

l

b

a F(~)p(~)

$n(~) 11$n 11d~

ex> $n(X)

=

n~l II$nlld(An),

where II$nIl2= s: p(~)[$n(~)F d~. When F is of class C", that is, when F, F', and F" are continuous over the closed interval a ~ x ~ b, that series converges to F(x) over the open interval. Thus the series represents F in terms of its transform f(An).

328


SECo 112]

A formula for the inverse Sturm-Liouville transform off(An) is therefore co

(1)

F(x) =

L f(AnWl>nllzcI>n(x)= T-1{f(An}} n= 1 .

(a

< x < b).

Uniqueness of inverse transforms among functions of class C" follows from that inversion formula. For if two such functions F and H have identical

F(x)

-

transforms, H(x) == O.

then

T{F

- H} = O, and formula (1) shows that

Alternate ways of tinding inverse transforms are furnished by tables of transforms, by devices for representingf(An) in terms of known transforms, or by operational properties of T. The transform of any particular eigenfunction cI>m(x), where m is some positive integer, can be written at once from the detinition of T and the orthogonality of the eigenfunctions. Thus (2)

if n = m;

if n :/=m,

that is, the transform mllz, O,O,.... We present now an operational property that gives T-l{f(An)/An} in terms of the function F and a solution of a differential equation associated with the Sturm-Liouville An :/=O; then

transformation

T. We write Y(An) = f(An)/An if

(3) Suppose tirst that T is such that no eigenvalue Anis zero. According to our basic operational property, Eq. (3) is true if Y(An)= T{Y} and Y is a solution of class C" of the boundary value problem (4)

Dz[Y(x)]

= F(x),

N"Y = NpY = O.

The notation is that used in Seco111; thus ~z Y

"" Y -- (r Y'Y Dz Y -- P - 1 JJz

-

= pF since qy

p

.

The homogeneous problem corresponding to problem (4) is the SturmLiouville problem in which A = O; ~z y = O,N"Y = N p Y = O. Since zero is not an eigenvalue, that homogeneous problem has only the solution

Y(x) == O. Therefore Green's function G(x,~)for problem (4) exists,where (rGx)x - qG

=O

(x :/=~),

N"G = NpG = O,

and Gx has the jump l/r(~) at the point x =~. problem (4) is then Y(x)

=

f

(a < ~ < b)

The unique solution of

F(~)p(~)G(x,~)d~

(a ~ x ~ b).


[SECo 113

Thus when zero is not an eigenvalue of the Sturm-Liouville

329

problem

associated with T, then

(5)

T-I

{fi~n)}=

f

F(~)p(~)G(x,~) d~;

that is, T {Y} = f()"n)/Anwhere Y is the solutionof problem(4). 1f one of the eigenvalues is zero, say Al = O,then AnY(An)= f(An) when

n = 2,3,... ; that is, if NaY = NpY = O and Y is of cIass C", then T{Dz[Y] According to transformation conditions

- F} = O

when

n = 2, 3,

... .

(2), then, for some constant k, Y satisfies the

Dz[Y(x)] - F(x) = kI(X), AnY(An) - f(An) = kT{¡} and, when n = 1, it follows that -f(O) = klllllz. Thus Y, the inverse transform of f(An)/An, n i= 1, with f(AI)/AI undefined, is a solution of class C" of the problem

Therefore

(6)

NaY=

NpY=

O (Al = O),

if that problem has such a solution.

Recall, however, that Green's function

does not exist for that problem, nor does the since A <1> 1(x) is a solution of the homogeneous

problem have a unique solution, form of the problem

when A

is an arbitrary constant. Thus if Y = Z(x) is one solution of problem (6), then Y = Z(x) + AI(X)is a family of solutions.

113

FURTHER PROPERTIES

Until special cases of Sturm-Liouville transformations are treated, cases in which the kernels are fairly elementary functions, only a few operational properties and transforms of particular functions can be presented. In the general case, however, the basic operational property can be extended easily to iterates of the differential form DzF. Let the function DzF as well as F itself belong to the class C"(a ~ x ~ b). Then

Let D/ (1)

denote the iterated differential operator here. Then

T{D/F}

= A/f(An) +

AnNiF]N~" - A"r(b)Np[F]Np"

+ Na[DzF]N~" - r(b)Np[DzF]Np".

330

SECo 113]


Thus T replaces the differential form D22F of fourth order by an algebraic form in the transform j()"n) of F and the four boundary values NaF, Na[D2F], NpF, and Np[D2F]. Further iterations of D2 can be treated in like manner. Our transformation was developed in Seco 111 by expressing the differential operator D2 in terms of a self-adjoint operator ~2' a procedure that can be followed for operators of the second order. To proceed directly without introducing the self-adjoint operator, we first write, as before,

T {F} =

f

F(x)K(x,A) dx = f(A)

and D2F = A(x)F"(x) + B(x)F'(x) + C(x)F(x). Then the transform T{D2F} = fK(X,A)D2[F(X)] dx is to be written in terms Ofj(A) and the two boundary values NaF and NpF. In terms of the adjoint D! of D2, Lagrange's identity (3), Seco94, shows that d KD2F - FD!K = d)AKF'

where r(x) = exp are of class C",

f: [B(~)/AR)] d~, and r' =

- (AK)'F + BKF]

rB/A. Therefore when F and K

(2) The remaining steps correspond to those used in Seco 111, where we expressed the values of F and F' at the end points x = a and x = b in terms of NaF, N~F, NpF, and N'pF. Details are left to the problems. We find that Eq. (2) can be written as the basic operational property (3) provided that K = Kn(x) and A = An,where Kn and Anare the characteristic functions and numbers of the Sturm-Liouville problem

(4)

D!K - AK = O,


[SECo 113

331

Note that

(5)

Na[ A:] = {AK cos a + [(AK)' - BK] sin a}x=a,

N{ A:] = [r(b)r1{AKcos P + [(AK)' -

BK] sinP}x=b'

Then the Sturm-Liouville transformation has the form (6)

(n

= 1,2,...).

Formulas (3) to (6) follow also from those in Seco 111 when the latter are written in terms of the kernel Kn and the coefficients A, B, and e in D2, where Kn = pCl>n. The basic operational formula for T{D2F} must be modified in case F or F' has jumps. Suppose, for instance, that both functions are eontinuous over the interval a ;;;¡; x ;;;¡;b exeept at one interior point x = e where they have jumps jc and j~:

jc = F(e + O)- F(e - O),

j~ = F'(e + O)- F'(c - O) (a < e < b),

and that F" is sectionally continuous. Then if we write

T{D2F}

=

f

Cl>Q¿2F dx

+

f

Cl>Q¿2F dx

and use Lagrange's identity, we find that (7)

T{D2F}

= Anf(An) + N~[CI>n]NaF- r(b)Np[CI>n]NpF + r(c)[jcCl>~(e)-

j~CI>n(e)],

where T{F} = f~FpCl>ndx and Cl>nand An are the characteristic functions and numbers of the Sturm-Liouville problem (10), Seco111. We cite one more property of Sturm-Liouville transforms f(An). When F is sectionally continuous, its Fourier constants en

=

IICI>nll-l

f

F(x)p(x)CI>n(x)dx

(n = 1, 2, . .. )

with respect to the orthonormal set of functions Cl>n/llCl>nll on the interval (a,b) tend to zero as n -+ oo. That is a consequence of Bessel's inequality, C12+ e/ + ... + Cm2;;;¡;IIFII2

(m = 1,2,...),

332


SECo 114]

for orthonormal sets.l Thus the transform of each sectionally continuous function F must satisfy the condition

(8)

114

TRANSFORMS OF CERTAIN FUNCTIONS

As noted in Seco 112, for a fixed positive integer m the Sturm-Liouville transform of the kernel $m is zero when n "Í' m, and JI$mJl2when n = m. Thus if f()on) is the transform of a function F and k is a constant, when n "Í' m, when n = m.

(1)

Suppose Tis such that no eigenvalue Anis zero. Then Green's function

G(x,c)for problem (4), Seco112,exists. It is continuous, but its derivative Gx has thejumpj~ = [r(c)r 1at the point x = c (a < c < x); also, (2)

(x"Í' c),

According to the modified operational property (7), Seco113, then 0=

AnT{G(x,c)} - r(c)[r(c)r1$n(c).

Thus the Sturm-Liouville transform of G(x,c) is 1 (3) T{G(x,c)} = :f$n(c) n

(a < c < b).

The continuity of G(x,c)and the sectional continuity of its first derivative are sufficient (Theorem 4, Seco 99) for the validity of our inversion formula (1), Seco 112. Therefore we have the representation (4)

(a < x < b,a < c < b),

known as the bilinearformula for Green's function. From transform (3) and property (5), Seco112, we find another SturmLiouville transform when no eigenvalue is zero : (5)

T{f

P(~)G(X,~)G(~'C)d~}= A:2$n(C)

(a < c < b).

A further consequenceof formula (3)is (6) I

T{:cG(X,C)} = ;n$~(C)

(a < c < b).

Churchill, R. V., "Fourier Series and Boundary Value Problems," 2d ed., p. 60, 1963.


[SECo 115

333

The basic operational property can be used in various ways to obtain transforms of particular solutions of differential equations involving our differential operator D2. For instance, when a function H(x) of cIass C" satisfies the differential equation D2H = O, it follows from property (12), Seco 111, that its transform is

(7)

;

h(An) =

n {r(b)Np[n]NpH

-

N~[n]N",H}

If constant values, not both zero, are assigned to N",Hand NpH here, then

H is the solution of the differential equation D2H = O that satisfies those boundary conditions. We note that in case IX= P = O,formula (7)becomes (8) T{H} = An-l[r(b)~(b)H(b) - ~(a)H(a)]. If F(x)

reduces to (9)

== 1 (a

~ x ~ b), then D2F = C(x) and our formula for T{D2F}

T{C(x)} = AnT{l} + N~[n]COSIX - r(b)Np[n]cosP.

Thus when n,An, and either of the two transforms T{C(x)} or T{I} are known, the other transform can be found. If C(x) = Co, a constant other than one of the eigenvalues,then T{1}can be found,since

(10) 115

EXAMPLE OF STURM-LlOUVILLETRANSFORMATIONS

Let the differential form D2F be the simple self-adjoint form F" on the

interval (O,n),and let the prescribed boundary values be F(O)and F'(n).

Then (1)

~2F = F"(x),

O < x < n;

N",F = F(O),

NpF = F'(n);

thus IX = O,P = n/2, r(x) = p(x) = 1, and q(x) = O. In this case the kernel n of the Sturm-Liouville transformation is the family of all eigenfunctions of the problem (2)

"(x)-

A(X)

= O,

<1>(0)= '(n)= O.

We find that A = O is not an eigenvalue, and that (3)

n(x)= sin (n

-

t)x,

Let Sm denote the transformation

modifiedfinite Fourier transformation (4)

Sm{F}=

f

F(x) sin (n

An =

-

(n

-

t)2

T here, where m

-

t)xdx =fs(m)

(n = 1, 2, .. . ).

= n - t.

It is the

(n = 1,2,...).

334

SECo 115]

Since

N~
OPERATIONAl

MATHEMATlCS

= m and N'p
tional property of Smfor functions of class C" is Sm{F"} = - m2f.(m) + mF(O)- (- l)nF'(n)

(5)

(m = n - !).

-

We find that II
!.

¿

Transforms of a number of functions can be found by evaluating the integral Sm{F} or by applying property (5) or formula (6). For instance, (7)

(m = n - !).

For other particular transforms see Probs. 5 to 8, Seco 116, and Table 2, Seco 126. We can derive kernel-product and convolution properties for Sm, whose kernel is an odd function that is antiperiodic with the period 2n. Let F3 be the extension of F which is odd and antiperiodic with period 2n;

~~

.

(8)

O < x < n;

F3(x) = F(x),

all x.

F3(X)= -F3( -x) = -F3(x + 2n),

Then F3(X)sin mx, where m = n - !, is even and periodic with period 2n. We introduce the integral

~

~w=[~m~=[~m~-[~m~

which is even and antiperiodic with period 2n; also Fó(x) Fo(n) = Fo(-n) = O. Then

= - F3(x) and

2Sm{F}= f" F3(X)sinmxdx = - f" Fó(x)sinmxdx. An integration by parts here shows that 2Sm{F}

= 2f.(m) = m f" Fo(x) cos mx

dx.

If Xo is a constant (O< Xo ~ n), then f.(m)3... sin mxo = ~

m

I

" Fo(x)[sin m(x + xo) 2 -"

-

sin m(x

-

xo)] dx


[SECo 115

336

beca use the integrand is periodic with period 2n. AIso, the integrand is an even function of ~, SOthe last equation is the kernel-product property (m

(10)

= n - t).

When O < Xo ~ n and O ~ x ~ n, we find that

(11)

if x + Xo~ n if x + Xo~ n.

When Xo = n, Eq. (10) becomes

(12) In terms of the notation used in Seco 109, w(A.)= 2/m and PF(x,xo) is the function inside the braces in property (10). Hence the convolution of two functions H(x) and F(x) corresponding to our kernel-product property is XHF(X)

=

f

H(y)[Fo(x

-

y) - Fo(x + y)]dy,

which can be written (13)

X HF(X) =

" H(y)

5.o

f" F(~)d~ [ )Ix-yl

I

" F3(~)d~ dy.

x+y

J

A convolution property for Smis therefore (14)

(m

=n-

t),

where hs(m)= Sm{H}. Similar derivations of convolution properties can be made for other Fourier transforms (cf. Seco110 and the chapters to follow). Properties of Smare summarized in Table 2, Seco126. Our transformation Sm reduces boundary value problems in partial differential equations in which the differential form in the unknown function U with respect to a variable x is Uxx' and the values of U and Ux are prescribed at boundaries x = Oand x = n, respectively.

336


SECo 116]

This problem in U(x,y) is an illustration: a(y)U"" + b(y)U + c(y)Uyy = Q(x,y)

U(O,y)= O,

Ux(n,y)= p(y),

(O< x < n, y > O),

U(x,O)= Uy(x,oo)= O,

assuming that the prescribed functions a, b, e, Q, and pare such that the problem has a solution. Let us(m,y)be Sm{U}with respectto x. Then according to our formula for Sm{D2U},the transformed problem is, formally, a(y)[-m2us - (-l)"p(y)] + b(y)us+ c(y)u;= qs(m,y)

(y > O)

us(m,O) = u~(m,oo) = O,

where qs(m,y) = Sm{Q} and u~ = duJdyo If this problem in ordinary differential equations can be solved for us(m,y), then U(x,y) can be written with the aid of our inversion formula, or by using known transforms and our convolution property. If a term 04U/OX4,an iteration of the form Uxx' is added to the above partial differential equation in U, and if the boundary conditions are augmented by prescribing UxX
04U Sm{ OX4} =

-m2Sm{Uxx}

+ mUxx(O,y)- (-l)"UxxX
and Sm{U",,} = - m2us(m,y) - (-l)"p(y)o Note that neither the Laplace transformation nor separation of variables applies directly to these problems in U. 116

SINGULAR CASES

Our method of setting up an integral transformation that applies to some specified differential form and boundary values may require the kernel to satisfya singular eigenvalue problem (Seco 107). Singular problems arise if the interval of our variable x is unbounded, or if the differential equation in the eigenvalue problem has a singular point on the interval, usually at an end point of the interval. Conditions on the order of magnitude or regularity of functions at singular points or infinitely distant end points may replace prescribed boundary values at those points. We present two examples to illustrate the procedureo Example 1 Let the interval be unbounded, O < x < 00, and let the differential form and the prescribed boundary value and order condition be these: (1) D2F = F"(x),

x> O;

N"F = F(O), F(oo) = F'(oo) = O.


[SECo 116

337

Here D2 is self-adjoint. The integral transformation (2)

fOF(x)eI>(x,..1.)dx

T{F} =

= f(..1.)

requires additional conditions on F and el>to ensure the existence of the improper integral. We assume that F is of class C" and absolutely integrable over the half line x ~ O,and that F((0) = F'((0) = O. Formula (8), Seco111, obtained either by using Lagrange's identity or by integrating by parts, now becomes

{O F(x)eI>"(x,..1.)dx +

T {F"(x)} =

= ..1.IX) F(x)eI>(x,..1.) dx

if el>satisfies the singular eigenvalue (3)

eI>"= ..1.eI>,x

> O;

+

[eI>F'

-

eI>'F]~

eI>'(O,..1.)F(O)

problem

eI>(O,..1.) = O, el>and
(x > O).

Ir k is a constant, the solution of problem (3) is (k :F O),

c1>(x,..1.) = k sinh xfi

provided that ..1.is such that
..1.=

fi

- Jl2,Jl > O,and if we write ik = 1, then

(4)

..1.= _Jl2

(jl > O).

Hence T is the Fourier sine transformation (Sec. 110) (5)

T{F} = SIl{F} = IX) F(x) sin JlXdx = f.(jl)

(jl > O)

with the basic operational property (6) Its inverse transformation is given by the Fourier sine integral formula (Prob. 15, Seco 107): (7)

F(x)

=

2 -

f

""

non

2 f.(jl) sin JlX dJl = -Sx{f.(Jl)}.

For other properties of Sil see Chap. 13 and Appendix D.

338


SECo 116]

Example 2 As an illustration of a differential form with the end point x = Oof the bounded interval 0-< x < b as a singular point, consider the form 1 (8) (O < x < b). D2F = F"(x) + -F'(x) x

The form arises when the Laplacian operator is written in terms of cylindrical coordinates. Let the prescribed boundary value and regularity conditions be (9)

F of class C"(O~ x ~ b).

NpF = F'(b),

In terms of the self-adjoint form ~2F = (xF')' and the notation used in Seco 111, we write D2F = x-l(xF')', T{F} =

p(x) = r(x) = x,

f:

/3= n/2,

F(x)xCl>(x,A)dx = f(A).

Then formula (8), Seco 111, for T{D2F} becomes

I:

(xF')'CI> dx =

I:

(xCI>')' F dx

+

[x(CI>F'

-

CI>'F)]g,

since r(x) = X. Ir CI> satisfies the singular eigenvalue problem (lO)

(xCl>')'= AXCl>, CI>'(b)) = O,

CI> in class C" (O~ x ~ b),

then T{D2F} = Af(A) + bCl>(b))F'(b). The differential equation xCI>"+ CI>'- AXCl> = O in problem (10) is Bessel's equation with index n = O, whose solution of class = kJo(x~), where k is a constant (Sec. 83). C" (O~ x ~ b) is CI> The condition CI>'(b))= O requires A to be zeros of the function J1(b~). We write b~ = J1..The odd function J1(J1.)has an infinite sequence of zeros J1.1= O, ::!:J1.2,::!:J1.3"'" all real; its nonnegative zeros J1.¡ yield all the eigenfunctions. Thus the eigenvalues are ,1= -J1.¡2/b2,and CI>(x))= Jo(J1.¡~)

[J 1(J1.¡)= O, J1.¡~

O,i = 1,2,...].

Our transformation on the bounded interval (O,b) is then the particular finite H ankel transformation "

(11)

T{F}

= I: F(X)XJo(J1.¡~)dx

= fo(J1.¡},


where J 1(f.l¡) = (12)

[SECo 116

O,f.l¡ ~ O,and f.lj-+ 00 as

339

i -+ oo. It has the property

T {F" + ~F'} = -b-2f.l/fo(P¡)+ bJo(Pj)F'(b).

Recall that f.ll = O and J 0(0) = 1. We shall treat Hankel transforms more fully in Chap. 14 where we shall note that Fourier-Bessel series furnish inversion formulas for the finite transformations. PROBLEMS 1. When D2F = F"(x), O< x < b, N.F = F(O), and NpF = F(b), show Stunn-Liouville interval (O,b),

that

the

transformation becomes the .finite Fourier sine transformation on the b

T{F} = Sn{F} =

mrx

f

o F(x) sin Tdx

= fs(n)

(n = 1,2,. .. );

show also that if F of cIass C" (O ~ x ~ b), then

Sn{F"}

= - (n;f.r.(n) + nb1t [F(O)- (-l)nF(b»),

and the inversion formula (1), Seco112,is the Fourier sine seriesrepresentationof F

on the interval (O,b). 2. (a) If F is of cIassC(4)(O~ X ~ b), show that the transformation Snin Probo l has the property Sn{F(4)(x)}= (n;)~(n)

- (n;f[F(O) - (-l)nF(b»)

+ nb1t[F"(O)- (-l)nF"(b»). x(X2

-

(b) If F(4)(X) = O, F(O) = F(b) = F"(O) = O, and F"(b) = 6b, show that F(x) b2) and fs(n) = 6b4( -lf/(n1t)3.

=

3. When D2F= F"(x), O< x < b, N.F = F'(O),and NpF = F'(b), show that the Sturm-Liouville transformation becomes the .finite Fourier cosine transformation on the interval (O,b), Cn{F} =

f

b

n1tX

o F(x) cosT

dx

= fc(n)

(n = O, 1, 2, . . . );

show also that if F is cIass C" (O~ x ~ b), then

Cn{F"}= -(n;ffc(n) - F'(O)+ (-l)nF'(b)

(n = O,1,2, . . . )

and that Cn-l{fc} is given by the Fourier cosine series F(x)

1 2 00 n1tX = bfc(O)+ b n~lfc(n)cosT

(O< x < b).

340


SECo 116]

4. Find the Sturm-LiouvilIe transformation on the interval (0,1) that resolves the differential form F"(x) in terms of the transform of F and the boundary values F(O) and hF(I) + F'(l), where h is a positive constant, and show that it has the operational property

- F(O) + [hF(l)+ F'(l)]cOS/l.,

T{F"} = -/l/T{F}

wherep. (n = 1,2,... ) are the positive roots of the equation tan /l

=

Ans. T{F}

f

= h//l

F(x) cOS/l.X dx

(cf. Seco 105).

(n = 1,2,...).

5. Show that Green's function for the problem Y"(x) = 0, Y(O)= Y'(1t)= O is the function (O~ x ~ e),

G(x,e) = - x

G(x,e) = - e

(e ~ x ~ 1t);

then apply formulas (3) and (6), Seco114, when T is the modified finite Fourier transformation Smdefined in Seco115, where m = n - t. to obtain the inverse transforms 1 sin me Sm- {--;;¡r-}

l cosme

= -G(x,e),

Sm- { ;;;- } = So(x- e),

whereO< e < 1tand Sois the unit step function. 6. Showthat the transformation(Sec.115) Sm{F} =

s: F(x) sin

mx dx

= f.(m)

(m = n -

t)

has the property

Sm

-1{ f~7)} = J: r F(t) dt dr = J: tF(t) dt + x f F(t) dt,

and that S -1 m

~

{ m3 }

= 1tX -

X2

2'

7. - When f.(m) = e-my where m = n - t and y > O and independent of x, inversion formula (6), Seco115, for the modified finite Fourier transformation Smcan be written F(x,y) =

~ I 1t.=1

e-my sin mx

I

=~ 1m 1t .=1

(exp iz).-t,

where z = x + iy and lexp izl < 1. Sum the series here to show that Sm-l{e-my}

= .!.Re 1t

(

ese:' 2)

=

1 (sinx/2)(coshy/2) 1tcosh2 y/2 - cos2 x/2

when y > O. [Since F(x,y) is the imaginary component of an analytic function of z, it is a harmonic function. See Probo 2, Seco118, where e-my arises. Also note that F satisfies conditions (8), Seco115, on F3.]


[SECo 117

341

8. For the transformation S", in Seco115, where m = n - t, show that if y> independent of x and z = x + iy, then

O and

-

e-my

{- }

S", 1

m

2 = - 1m 7t

L (expizr m 00

n= 1

2 2 1 + exp iZ/2 sin x/2 = - I m l og = - arctan 7t 1 - exp iz/2 ) 7t ( ( sinh y/2 )

(y>

O).

9. Complete the derivation of the basic operational property (3), Seco113, for the Sturm-Liouville transformation in the form f:FK dx. 10. Write F(x) = e-kX(k > O) in the basic operational property (6), Seco116, of the Fourier sine transformation SIl to obtain the transform J.I S/l{e-kX}

(k> O).

= J.l2 + k2

Differentiate with respect to k to show that S/l{xe-kX} = 2kJ.l(U2+ k2)-2

(k> O).

11. Derive the integral transformation on the half line x ;¡;;O that reduces F"(x) in terms of the boundary value F'(O) when F is of cIass C", absolutely integrable over the half line, and F(oo) = F'(oo) = O.This is the Fourier cosine transformation

(¡.t~ O)

C/l{F} = {') F(x) cos J.lXdx = fc(J.I) (Prob. 5, Seco110 and Chap 13). Show that C/l{F"(x)} = _J-l2fc(J-l)- F'(O).

12.

Derive the integral transformation that reduces the form F"(x) + X-l F'(x) on

the interval(0,1)in termsof the boundary value F(1)when F is of cIassC"(O~ x ~ 1). This is the finite Hankel transformation (Chap. 14) T{F}

= Jo(1 F(x)xJo(¡.tr)dx

(j = 1,2,... ),

=f(J.I.)

J

where the real numbers J-ljare the positive roots of the equation Jo(¡.t)= O. Since Jó(¡.t) =

-

J 1 (¡.t), show that

T{F" + x-1F'} = -J-l/f(J-lj) + J.I/l(J-l)F(l). 117

A PROBLEM IN STEADY TEMPERATURES

As an example ofboundary value problems that are well adapted to solution by specific Sturm-Liouville transformations, we present the following one on steady-state temperatures in a semi-infinite slab in which heat is generated at a steady rate while surface heat transfer takes place at one face. Let the temperature function U(x,y) satisfy the conditions , (1)

u xx(x,y) + Uyy(x,y) + Q(x,y) = O U(O,y) = P(y),

U(l,y)

+ hUx(1,y)= O,

(O< x < 1, Y > O) U(x,O) = O,

342

SECo 117]


where h is a constant (h ~ O),the functions P and Q are bounded, and we require U to be bounded. . Here the differential form with respect to x is Uxx' and the prescribed boundary values at x = O and x = 1 are Na.U = U(O,y)a~d NpU = U(l,y)cosP + Ux{l,y)sinpwheretanp = h(O ~ P < n/2). Hencethekemel $n(x) and the parameter An of the Sturm-Liouville transformation that applies to problem (1) are the characteristic functions and numbers of the problem (2)

$"

-

A$

= O,

$(1) + h$'(l,A) = O.

$(O,A) = O,

Thus we find that tan J.ln= -hJ.ln

(3)

(J.ln> O),

and we can see from the graphs of the functions tan J.l and - hJ.lthat J.ln- (n - !)n -+ Oas n -+ 00 when h > O; when h = O,J.ln= nn. Our transformation is (4)

T{ U}

=

f

U(x,y) sin J.lnXdx = un(Y)

(n = 1,2,...).

Now Na.U = P(y), IX= O,and N~$n = $~(O)= J.ln;also NpU = O,and An= - J.l/. Therefore T { U xx} = - J.ln2un(y) + J.lnP(y)

and if qn(Y) = T{ g(x,y)}, the transformed problem is u:(y) - J.ln 2un(y)+ J.lnP(y)+ qn(y) = O,

(5)

where Unis bounded on the half line y > O. With the aid of the equation tan J.ln=

- hJ.ln,we find that lI$nll2 = !(l + h COS2J.ln)'

After solving problem (5) for Un,we have the formal result (6)

U(x,y) = 2

f

un(y)sin ~nx .

n=11 + hcos J.ln

The form of the solution can be improved in special cases. When P(y)

= O and

Q(x,y) ==Qo, a constant,

'

then

and the solution of problem (5) becomes (7)

un(y) = Qo

1

- cosJ.ln 3 J.ln

- Qo

1 - cosJ.ln 3 J.ln

exp (- J.lnY).


The first fraction is -T{l}/An. - T{Z} where Z"(x) = 1,

[SECo 118

343

According to formula (4), Seco112, this is . Z(O) = O,

Z(1)

+ hZ'(l)

= O.

Hence we find that Z(x) =

-~ 1 + 2h + 21+h

!X2 = -T-l

2

l - COSJl.n

{

Jl./

}

and the inverse of transform (7) can be written (8)

- Qo 1 + 2h - 2 U( X,Y) x 2 1 + hx

[

+4 ~ L...

n= 1

1 - cos Jl.n exp (- Jl.nY).

1 + h cos 2 Jl.n

3

Jl.n

].

SIn Jl.nx

Form (8) can be verified fully as the solution of problem (1) in this special caseP = O,Q = Qo. Note that if only a finite number of terms of the infinite series is used in formula (8), the resulting function satisfies all conditions in problem (1) except the condition at the base y = O. Also, we see that U has the property

.

11m U(x,y) = y-oo

Qo l + 2hx - x 2 . -2 ( 1+ h )

The differential form Uyy and boundary conditions in problem (1) are such that the Fourier sine transformation with respect to y, on the half line y > O, applies to the problem provided that the transforms of both the functions P and Q exist; but that is not the case when Q(x,y) = Qo. In case the second method does apply, we have an advantage of obtaining a different representation of the function U. Note that problem (1) is not adapted to a Laplace transformation. Since the partial differential equation is not homogeneous, the method of separation of variables does not apply. For further properties of transformation (4) see Seco 128.

"' 118

OTHER BOUNDARYVAlUE PROBlEMS

Other special and singular cases of Sturm-Liouville transformations will be applied to boundary value problems in the chapters to follow. Operational properties of some of them are adequate to represent solutions of problems in more than one form, sometimes in quite simple forms. But in each application of our process an eigenvalueproblem must be solved to determine the kernel of the transformation; then the transformed problem must be solved to find the transform of the unknown function. Failure of the transformedproblem to have a unique solution suggeststhat conditions are missing or incorrect in the boundary value problem.

344


SECo 118]

Tomake some observations on types of problems to which the method applies, we consider here problems in two independent variables and SturmLiouville transformations. But when boundary conditions are properly modified, our observations apply to other cases including singular cases, differential forms of higher order, and problems in more than two independent variables. In contrast with the method of separation of variables the method of integral transformations is not restrieted to problems with homogeneous dijJerential equations or with eertain homogeneous boundary eonditions. Let V(x,y) denote the unknown function in our problems and let the differential form and prescribed boundary values on x be D2 V = A(x)Vxx(x,y)

+ B(x)Vx(x,y) + C(x)V(x,y)

(a < x < b)

N aV = V(a,y) cos rx + VAa,y) sin rx, N (JV = V(b,y) cos /3 + VAb,y) sin /3. Note that the coefficientsA, B, and real and constant. Then as before

T {V} =

f

e are functions of x alone; rxand /3are

(n = 1,2,...). V(x,y)p(x)c1>n(x) dx = vn(y)

where ~n are the eigenfunctions of the Sturm-Liouville problem corresponding to the operators D2, Na, and N(J' To indicate some types of boundary value problems in V to which transformation T applies, let L¡,y, i = 1, 2, 3, denote linear homogeneous differential operators with respect to y whose coefficients are functions of y alone. Those operators may be of any order; if the order is zero, the form L¡,yVis simply a product E(y)V(x,y). Consider a linear boundary value problem consisting of a part1al differential equation and boundary conditions

+ L2,y[D2 V]

(1)

Ll,y[V]

(2)

NaV = G(y),

= F(x,y),

N(JV= H(y),

and some boundary conditions with respect to y of type (3)

av amv el V + e2 - + .. . + e - = P(x e)

ay

m ayrn

,

when y = e,

where the e's are constants. When Tis applied to Eq. (1) and (3) with interchanges of operations with respect to x and y, then, in view of conditions (2), the resulting differential equation is


[SECo 118

345

wheref,,(y) = T{F(x,y)}. If Pn(C)= T{P(x,c)}, the boundary conditions on Vnare of type dVn

(5)

d"'vn

C1Vn+ C2 dy + ... + Cmdym

=

h Pn(C)

w en y

= c.

Then V(x,y) = T-1{vn(Y)}. Unless tables or operational properties of T can be used to simplify the inverse transformation, we may write the formal solution as ex>

(6)

V(x,y) =

L IIcI>nll-2vn(y)cI>n(x). n=l

The transformation T may apply if the problem involves D2 and some of its iterations. For example, the partial differential equation may have the form (7) if it is accompanied by boundary conditions at x = a and x = b that prescribe Na.V,NpV, Na.[D2V],and Np[D2V]. Then even ifEq. (7) and those boundary conditions are homogeneous, the usual method of separation of variables may not apply. A special case of Eq. (7) is the biharmonic equation 04V

(8)

04V

04V

oy4 + 20y2ox2 + OX4 = O

satisfied by the static transverse displacements V(x,y) in an unloaded elastic plate. Here D2V = Vxx, Ll.,y = o4joy4, L2,y = 202joy2, and L3,y = 1. If displacements Vand bending moments are given along the edges x = a and x = b, then V and Vxxare prescribed on those edges. PROBLEMS 1.

Note why the transformation (Sec. 115) Sm{U(x,y)} =

J: U(x,y) sin mx dx = u.(y)

(m= n -

t)

applies to this problem in steady temperatures: U",,(x,y) + Uyy(x,y)= O, U bounded,

U(O,y)= U",(x,y)= O, Show that u.(y)

= m-le-my

U(x,O)= 1

(OO), (O< x < x).

and thus (Prob. 8, Seco116) U(x,y) =

-2x arctan

sin x/2

--=

sm h y /2

(O< x < x, y > O).

346

SECo 118]

OPERATIONAl

MATHEMATlCS

2. Use the transformation Sm(Sec. 115) with respect to x to find the solution of the problem

uxx(x,y) + Uyy(x,y)= O, U bounded, U(O,y)= Ux(1t,y)= O,

(O< x < 1t,Y > O),

Uix,O) = -1,

in the form U(x,y) = Sm-1{m-2e-my}; thus show that x

x

U(x,y)= -log 1t

i

cosh y/2 + cos x/2 cos h y /2 - cos x/2 + o tF(t,y)dt

where (Probs. 6 and 7, Seco116)

F(t, ) = .!. Recsc t + iy =.!. (sint/2)(coshy/2) . y

2

1t

1tcosh2 y/2

-

cos2t/2

3. Determine the Sturm-Liouville transformation T, with respect to x, that is adapted to the problem (O< x < 1, t > O),

(t + l)U,(x,t) = Uxx(x,t) U(x,O)= O,

Ux(O,t)= -1,

U(l,t) = O.

Find T-1{1/ln} by solving and transforming the elementary problem Y"(x) = O, Y'(O)= 1, Y(l) = O; then derive the formula (t + l)-N (2n - l)1tX U(x,t) = 1 - x - 2 ~ L.. cos-

n=1

N

2

4. Determine the integral transformation (a) with respect to y, (b) with respect to x, that applies to the problem U xx(x,y)

U(O,y)= G(y),

(O< x < 1, O < y < 1),

+ Uyix,y) + 2Ux(x,y) = O

U(l,y) = O,

Uy(x,O)= O,

Uy(x,l) = H(x),

and write the transformed problem in each case. Ans. (a) uC
(b) u(ln,y) =

f~ U(x, y) cos n1ty dy

(n = O, 1,2, . . . ); u; + 2u~ - n21t2uc = (_l)n+1 H(x), uC
f~ U(x,y)~ sinn1txdx

(ln = -1 - n21[2, n = 1,2,...);

u" - (1 + n21t2)u= -n1tG(y), u'(ln'O)= O,u'(ln,l) = h(ln)' 5.

The line y = O is a singularity of the equation

Vxx(x,y)+ y2Vyix,y) + y~(x,y) + F(x) = O. If V is to be bounded over the rectangle O < x < 1t, O < Y < b, solve that partial differential equation under the conditions VAO,y)= V(1t,y)= V(x,b) = O. n

Ans. V(x,y) =

f

x

i '

o

2 f(l

) y

L -+ -b 1t n= 1 m

F(r)dr dt - -

m

cos mx,

( }

where f(ln) = So"F(r) cos mr dr and m = n - í.


[SECo 118

347

6. Note why the Fourier sine transform (Sec. 110) on the half line x > O applies to this problem on transverse displacements of a stretched string with prescribed initial velocity (Prob. 5, Seco44): (x > O,t > O)

Y,,(x,t) = a2 Y",,(x,t) Y(x,O) = O,

Y,(x,O) = G(x),

Y(O,t)= O.

Use property (4), Seco110, to show that 1 "+a' Y(x,t) = 2a ',,-a" G(r)dr

i

and verify that solution of the problem. Without solving the following problems for V(x,y), determine procedures that are adapted to their solution.

7.

V",,(x,y)+ Vyy(x,y)- V(x,y)= O,

(x> ,O < Y < 1),

V(O,y)= 1, Vy(x,O)= O, Vy(x,l)= - V(x,l) + e-". Ans. Fourier sine transformation with respect to x (Sec. 110); the SturmLiouvilletransformationin Probo4, Seco116,with respectto y, where h = 1. 8.

V",,(x,y)+ X-l V,,(x,y)- Vyy(x,y)= Q(x,y)

(O< x < 1,y > O),

V(l,y) = O, V(x,O)= O, Vy(x,O) = O. Ans. The finiteHankel transformationin Probo12,Seco116;the Laplacetransformation with respect to y.

11 Finite Fourier Transforms

119

FINITE FOURIER SINE TRANSFORMS

A useful special case of the Sturm-Liouville transformation is that in which the differential form is the simple self-adjoint form F"(x), and the prescribed boundary values are the values of F at the ends of a bounded interval (Prob. 1, Seco116). It is convenient to choose the origin and the unit of length so that the interval is (0,1t). Then in terms of the notation used in Chap. 10, O
N"F = F(O),

NpF = F(1t);

here oc= p = O,p = r = 1,and q = O. The kernel of the transformation T consists of the eigenfunctionsof the problem <1>" -

A.=

O,

<1>(0)= (1t)= O.

Thus A.= -n2(n = 1,2,.. .), <1> = sin nx, and T becomes the finite Fourier 348

FINITE FOURIER TRANSFORMS

[SECo 119

349

sine transformation (1)

Sn{F}

(n = 1, 2, .. .).

= J: F(x) sin nx dx = f.(n)

The basicoperationalproperty(12),Seco111,becomes (2)

+ n[F(O) - (-l)nF(n)]

Sn{F"(x)} = -n2f.(n)

when F is of class C" (O~ x ~ n), or even if F and F' are continuous while F" is sectionally continuous over the interval. lt follows that Sn{F(4)(X)}= n4f.(n) - n3[F(0)-

(3)

(-l)nF(n)]

+n[F"(O) - (-l)nF"(n)] if F is of class C(4). Formulas for transforms ofhigher-ordered iterates ofthe form F" can be written easily. Transforms of many elementary functions can be found by evaluating the integral Sn{F}. Some can be found readily from property (2); as examples, when F(x) ==1, or when F is x, x(n - x), or ¿x, that formula gives these transforms :

(4)

Sn{~(n- X)} = 1 - (;-l)n, Other entries in Appendix B, Table B.l, can be derived by those methods. Some operational properties of Sn are included in that table of finite sine transforms. Following the procedure used in Seco 112, we write f.(n)fn2 = ys(n) where ys(n) = Sn{Y(x)}. Then n2Ys(n) = f.(n), and this is true if Ysatisfiesthe conditions Y"(x) = - F(x),

Y(O)= Y(n) = O.

Thus we find that, when F is sectionally continuous, (5) Sn -1

;

{f~~)} = J:

(n

-

r)F(r) dr

-

I: (x

-

r)F(r) dr.

For Sn the series representation (1), Seco 112, of the inverse SturmLiouville transformation becomes the Fourier sine series representation of F(x): 2

(6)

Sn

DO

L f.(n) sin nx n n= 1

-1{f.(n)}= F(x)= -

(O < x < n),

" 1: l. Ij

~

350


SECo 120]

= n/2.

This formula for the inverse sine transformation is valid over the interval if F and F' are sectionally continuous there and if, at each point Xowhere F is discontinuous, F(x) is defined as the mean value of its one-sided limits, since Ilsin nxll2

(7)

F(xo) = !-[F(xo + O)+ F(xo - O)]

(O < Xo < n).

For all real x the series (6) converges to this odd periodic extension F1 of F, with period 2n: (8)

F1(x) = F(x)

O < x < n,

when

F1(0)= F1(n)= O; ( - 00 < x < 00),

F1(x + 2n) = F1(x)

F1(-x) = -F1(x),

because sin nx is odd and periodic with period 2n. 120

OTHER PROPERTIES OF SR

Let F1 be the odd periodic extension, with period 2n, of a function F that is sectionally continuous over the interval (O,n),and let k be real and constant. Then =

2f.(n) cos nk

f"

F1(x) sin nx cos nk dx

since the integrand is an even function of x. The integral can be written in these forms :

t f" F1(x)sin n(x + =

t f" F1(r -

k) dx +

k)sinnrdr

t f" F1(x) sin n(x +

t f" F1(r +

-

k) dx

k)sinnrdr

because the last two integrands are periodic functions of r with period 2n. The sum of those integrands is an even function, so 2f.(n) cos nk =

f

[F1(r - k)

+

F1 (r

+ k)] sin nr dr;

that is, if F is sectionally continuous and Sn{F} = f.(n), then

(1)

2f.(n) cos nk = Sn{F1(x - k) + F1(x + k)}.

When k = n, then, since F1(x- n) = -F(n - x) when O < x < n and F1(x + n) = F1(x - n),property (1)becomes (2)

(n = 1,2,.. .).


[SECo 120

361

For example, since S"{x} = 1tn-l( -1)"+ 1, it folIows that (3)

Kernel-product and convolution properties can be written for S". Again let Fl be the odd periodic extension of F with period 21t. Then the function Fo(x) =

(4)

J: Fl(r)dr

(-00 < x < (0)

is periodic with period 21t and even, Fo(x) = Fo(lxl), and Fó(x) = Fl(X) wherever Fl is continuous. Therefore (5)

fs(n) =

f

Fó(x) sin nx dx = - n fa" Fo(x) cos nx dx.

If Xo is a constant, then

~f.(n) n

1

sin nxo

=_ 2

I It -It

Fo(x)[sin n(x

-

xo)

- sin n(x + xo)]dx,

and this kernel-product property folIows easily: (6)

x+xO

2

- f.(n) sin nxo = S" n

{f

Fl(r) dr

}

x-~

.

The corresponding convolution of two sectionalIy continuous functions F and H is (Sec. 109) X+Y

lt

(7)

XFH(X)

=

i

o H(y)

f

x-y

F1(r)drdy,

with the convolution property, with weight function 2/n, 2 S"{XFH(X)}= -f.(n)h.(n) n

(8)

where h. = S"{H}. The integral (7) can be written in terms of H and the function F (Prob. 13). In case F is such that Fl is everywhere continuous and F' is sectionalIy continuous, then Fl(:!::n) = Oand It l It (9) 2f.(n) = -It Fl(x)sinnxdx = -;; -It F;(x)cosnxdx.

I

I

Thus a kernel-product property with weightfunction w namely, (10)

= 2n can be obtained,

352


SECo 120]

For Sn' property

(8), Seco 113, becomes lim.fs(n)

= O

whenever F is sectionally continuous. Some further properties of Sn' and transforms of particular functions, are given in the problems. PROBLEMS 1.

Use properties of S. to obtain these transforms: (a) S.{X3}

= n(-1)"(:3 - :2) ;

(b) S.{sinkx}

= (-1)"+I~ n -

n (e) S.{coshex} = ~[1 n +e

if k =F:t 1, :t 2, .. .;

k2sinkn

- (-1). coshen].

2. (a) If F(x) = sinmx wherem is a positiveinteger,showthatf.(n) = Owhenn =Fm, andf.(m)= nj2. (b) Derive: S.-I{:3(-1).+I}

= :n(n2 - X2).

(e) Derive: S.-1 {:3} = :n (X2- 3nx + 2n2). 3. If F(n - x) = F(x),showthatf.(n) = Owhenn is even. 4. Let F and F' be continuouswhenO~ x ~ n exceptthat F' has a jump b at a point x = e interiorto the interval,and let F" be sectionallycontinuous.Prove that

+ n[F(O)- (-1).F(n)] - bsinne. 5. Note the graph of the functionF(x) = Isin 2xl and use the results found in Probs. S.{F"} = -n2f.(n)

3 and 4 to show that

f.(n) = O

if n = 2,4, . .. ,

f.(n) = 4 4sin- nnj2 n2

if n = 1,3,5,.. ..

6. Derivethe inversetransform

.

S -1 nsinne { n2 }

=

(n

-

e)x

.{ (n - x)e

ifO ~ x ~ e < n if e ~ x ~ n,

(a) with the aid of formula (3), Seco114, when T is S.; (b) with the aid ofthe kernel-product property (6),Seco120,whenf.(n) = n-l. 7. If the real numbers al' a2, . . . are such that the series al + a2 + ... is absolutely convergent, prove that

(n = 1,2,...).


[SECo 120

8. When y> O,the numbers am = m-I(-I)m+le-my, conditions in Probo 7. If z = x + iy, showthat '" (-I¡m+1

L

m=1

where m = 1,2,...,

'" (-I¡m+1

= 1m m=1 L

e-my sin mx

m

m

353

satisfy the

. (elZ¡m

= 1m [Iog (1 + eiZ)],

and thus derive the inverse transform ( -1)n+ 1 -ny S -1 -e n { n }

9.

2 n

= - arctan

sin x

if y> O.

( eY+cosx )

From the transform found in Probo 9 note that

2 sinx 1 Sn-1 -e-ny = -arctann eY- cosx } {n when y > O; then show that S -1 n {

2 sin x (-I)n e-ny = -arctan-,.-

1-

n

}

n

if Y > O.

smh y

10. When -1 < b < 1 and -n/2 < arctan t < n/2, use transforms found in Probs. 8 and 9 to show that bn 2 b sin x - = S -arctan n nn{ l-bcosx

(-1 < b < 1); }

then differentiate with respect to b to get the transform bn 11.

=

2

S

n{

bsinx

n1 + b2 - 2b cos x }

(-1 < b < 1).

Given that, when c =PO,the function Y( ) = sin cx sinh c(2n - x) - sin c(2n - x) sinh cx , x 4c2(sin2cn + sinh2 cn)

satisfies the conditions Y(O)

=

Y(n) = Y"(n) = O,

Y"(O)= -1,

showthat n Sn{Y(x)} = n4 + 4c"

12.

If Y1is the odd periodic extension of the function Y in Probo 11, with period 2n,

find the solutionof the problem

.

2(4)(X)

+ 4c42(x) = 2F(x),

2(0) = 2(n) = 2"(0) = 2"(n) = O,

354

SECo 121]


in the form

"

Z(x)

=

x+'

i i o F(t)

Y¡(r) dr dt.

X.-/

13. Express the convolution integral (7), Seco120, in terms of the functions H and F by the formula

i I i f

Jo" H(y)

x+y

x-y

F¡(r)drdy

= J "-x o

H(y)

+ " H(y) 2"-X-Y F(r)drdy "-X

-

o

Jox+y F(r) dr dy

i i x

o

H(Y)

X-Y

o

F(r)drdy

H(y) S:-X F(r)dr dy.

14. Ir F¡, the odd periodic extension of F with period 2n, is everywhere continuous and if F'¡ and the function H are sectionally continuous, show that a convolution property for S" corresponding to the kernel-product property (10), Seco120, is

a "

i

Sn-¡{2nf.(n)h.(n)} = ax o H(y)[F¡(x - y)

-

F¡(x + y)] dy.

15. Establish the following relation between the Fourier sine transformation on the interval (O,b)introduced in Probo 1, Seco116, and OUTtransformation Snon the interval (O,n): b . nnx b b Jo F(x) SIDTdx = Sn{ ( )} .

,/;rx

121

FINITE COSINE TRANSFORMS

The Sturm-Liouville transformation that resolves the differential form F"(x) on the interval (O,n)in terms of the transform of F and the boundary values F'(O)and F'(n) has a kernel which satisfies the eigenvalue problem "(x)-

A(X)

= O,

<1>'(0) =

<1>'(n) = 00

Thus A = - n2 and (x) = cos nx, where n = O,1,2, . o.. In this case

= 1 is an eigenfunction corresponding to the eigenvalue A = transformation is the finite Fourier cosine transformation (x)

(1) Cn{F(x)}

= I: F(x)

cos nx dx

= fc(n)

00 The

(n = O, 1,2,. o.)

with the operational property (2)

Cn{F"(x)} = -n2fc(n) - F'(O)+ (-1)nF'(n)

(n = O, 1,2, o. o)

when F and F' are continuous, and F" is sectionalIy continuous, over the interval O~ x ~ no

I


[SECo 121

355

When F is of cIass C(4), it folIows that (3)

=

Cn{F(4)(X)}

n4fc(n)

+ n2[F'(O) - (-lfF'(x)]

- F"'(O)+"( -lrF"'(x).

The formula for Cn{F(6)}can be written by replacing F by F" in formula (3), and so on for transforms of the higher even-ordered derivatives. The boundary values that arise are those of the odd-ordered derivatives. Since IIcosnxll2 = x/2 when n = 1,2,..., then if m = 1,2,...,

(4)

Cn{cosmx} = O

ifn.p m, Cm{cosmx} = !x;

(5)

Cn{1}

ifn = 1,2,...,

=O

Co{1}=x.

Ir A is a constant andfc(n) = Cn{F},then (6)

when n = 1,2,... when n = O.

C"{F(x) + A} = fc(n) = fc(0) + 7tA

AIso, fc(n) -. O as n -. 00 when F is sectionalIy continuous. Note that fc(O)/x is the mean value of F on the interval (O,x). Thus Co{x}/x = x/2, while property (2) shows that C"{x}

= -[1 - (-1)n]n-2

when n = 1,2,....

When F is sectionalIy continuous, a function Y whose transform yc(n)is J.(n)/n2when n = 1,2,..., can be found by writing n2Yc(n)= fc(n). That equation is satisfied,according to properties (2)and (6),if Y"(x) = -F(x) + B, Y'(O)= Y'(x) = O, where B is to be determined by the conditions on Y'. We find that

cxr Y(x)

1

= Jo J, F(r)drdt + 2xfc(O)(x-

X)2

+ C.

The constant C is determined by the value assigned to y.(O). We can simplify the iterated integral to write fc(n) (7)

7

1

X

= c" {Ix (x

when n = 1,2,...,

-

r)F(r)dr + 2xJ.(O)(x-

where A = C +

I

x

'0

X)2

+ A}

rF(r) dr.

For Cn the series representation of the inverse Sturm-Liouville transform is the Fourier cosine series (8)

1

2

co

C"-1 {fc(n)} = - fc(0) + - L: fc(n)cos nx = F(x) x 7t"=1

when O < x < x, valid if F and F' are sectionalIy continuous and if F is defined as its mean value at each point of discontinuity. For alI real x the

L.

356


SECo 122]

series represents the even periodic extension F2 of F with period 2n: F2(X) =

(9)

F(x)

when O< x < n,

= F2(lxl) = F2(x + 2n) for all x. This kernel-product property for Cn can be written by using the even periodic extension (9) of F: (10)

2fc(n) cos nxo = Cn{F2(x

Ir° ~

Xo

- xo) + F2(x + xo)}

(n = O,1,2,. . .).

~ n, then F2(x - xo) = F(lx - xol) here. It follows that

(11)

=

fc(n)( -lt

Cn{F(n

-

x)}

(n

= 0,1,2,.. .).

The convolution property corresponding to formula (10) is (12)

Cn -1{2fc(n)hc(n)}

= s:

[F2(x

- y) + F2(x + y)]H(y) dy,

where F and H are sectionally continuous. The integral here will be written in terms of H and the function F in Seco 123. 122

TABLES OF FINITE FOURIER TRANSFORMS

Tables of Cn{F} and Sn{F} will be found in Appendix B. Transforms of particular functions are found by evaluating the integrals Cn{F} or Sn{F}, by using partial fractions or other decompositions of transforms, by differentiating or integrating with respect to a parameter, by using operational. . properties of the transformations, or by summing the series that represents the inverse transformo Relations between sine and cosine transforms, given in the following section, also aid in extending the tables. An example of the use of the Fourier cosine series will be given here, corresponding to Probs. 7 and 8, Seco120. Suppose 1 (1) fc(n,y) = n-e-ny when n = 1,2,..., and fc(O,y)= 0, where y > O. Then the series with termsfc(n,y) is absolutely convergent and fc(n,y) = Cn{F(x,y)} when 2001 2001. F(x,y) = n=ln L: -e-ny cos nx = n-Re n=ln L: -(e'z)n

n

where z

= x + iy and y > O.It followsthat 2

F(x,y) = --Re[log(l n

.

1

- e'Z)]= --log(1 + en

1

= --log[2e-Y(coshy n

- cosx)].

2 y

- 2e-ycosx)

1


[SECo 123

357

Thus we find that (2)

nF(x,y) = y - log 2 - log (cosh y - cos x)

(y > O).

The method used to derive formula (2)is not sound if y = O. But in that case the function F is still defined: (3) F(x,O) =

-~log [2(1 - cos x)] = -~ log (2 sin~),

and our results suggest that ifn (4)

=O

if n = 1, 2, . . . .

The improper integral Cn{F(x,O)}can be evaluated with the aid ofthe theory of residues1 to prove that formula (4) is correct. 123 JOINT PROPERTIES OF Cn AND Sn Some operational properties of our two finite Fourier transformations have simple forms when stated in terms of both transformations. Ir F is continuous and F' is sectionally continuous, integration by parts shows that, when n = O,1, 2, . . . , (1)

Sn{F'(x)} = -nCn{F(x)},

(2)

Cn{F'(x)} = nSn{F(x)} - F(O)+ (-l)nF(n).

Alternate forms of those properties are (3)

Sn{H(x)} = -nCn{J: H(r)dr},

(4) Cn{H(X)

-

~hc(O)}

= nSn{f: H(r)dr- ;hc(O)},

when H is sectionally continuous. Let F1 and F2 be the odd and even periodic extensions, with period 2n, of a function F that is sectionally continuous on the interval (O,n), and let k be real and constant. Then by the method used to derive property (1), Seco120, we find that

=

+ k) -

- k)},

(5)

2f.(n) sin nk

(6)

2.fc(n)sin nk = Sn{F2(x - k) - F2(x + k)}.

I Cf. seco 73 of the author's

"Complex

Cn{Fl(X

Variables

Fl(X

and Applications,"

2d ed., 1960.

358

OPERATIONAl

SECo 123]

MATHEMATICS

When P and Q are sectionally continuous on the interval ( -n,n) periodic with period 2n, the function (7)

* Q(x)

P

=

f" P(x -

and

r)Q(r) dr

is periodic with period 2n and continuous. Let us call it the faltung integral of P and Q on the interval (-n,n). We find that

(8)

P

* Q(x) =

Q * P(x),

and that the faltung is an even function if P and Q are both even or both odd; it is odd if either P or Q is even and the other odd. Now let F2 and H 2 be the even periodic extensions, with period 2n, of two functions F and H that are sectionally continuous on the interval (O,n). By writing the faltung of F2 and H 2 on ( -n,n) as a sum of integrals and introducing new variables of integration, we find that

"

(9)

F2 * H 2(X) =

4

f

+ F2(x + r)]H(r) dr =

o [F2(x - r)

¡~l l¡(x),

when O< x < n, where 1¡are theseintegrals involving F and H: 11 =

(lO)

s: F(r)H(x

13= I:-

x

- r) dr,

12 =

+ r) dr,

F(r)H(x

f

F(r)H(r - x) dr,

14 = f,,"-x F(r)H(2n - x - r) dr.

In view of Eq. (9) the convolution property (12), Seco121, takes the form

(11) Similarly, if Fl and H 1are the odd periodic extensions of F and H, with period 2n, it turns out that (12)

Fl

* H1(x) =

I: [Fl(X

- r) - Fl(X

= 11(X) - 12(x) - 13(x) (13)

Fl * H2(x) =

I: [F1(x.-

r)

+ r)]H(r)dr + 14(x),

+ Fl(X + r)]H(r)dr

= Io" F(r)[H2(x - r) - H2(x + r)] dr

i I

I


[SECo 123

359

Then froID the product properties (5) and (6) we find these joint convolution properties: (14)

-2f.(n)hs(n)

(15)

= Cn{F¡ * H¡(x)},

2f.(n)hc(n) = Sn{F¡ * H2(x)}.

PROBLEMS 1.

Find these transforms, with the aid of properties of Cn: n2 (a) Cn{x} = (-1~: - 1 if n = 1,2,..., Co(x)= 2; n3 2n (b) Cn{(n - X)2} = "2n if n = 1,2,..., Co{(n - X)2} = 3; (e) Cn{coskx}

= ksink

(d) Cn{coshe(n - x)} 2.

=

(_1)n+1 n r--- k 2 e sinh en n2

+ e2

if k #: O,:tI, :t2,...; if e#: O.

With the aid of transform (b), Probo 1, show that if 1 yc(n) = 4' n

-1

then

Cn 3.

n3 n 2 1 3 1 4 {yc(n)} = 45 - 6'X + 6X - 24nX .

If a #: Oand b #: O,derive the transforms a

(a2 - b2)(_1)n

( ) (n2+ a2)(n2+ b2) (b)

3'

h2

2a SIn

=C

cosh bx n { b sinhbn

-

cosh ax

a sinhan}

(-1)" C -I an n { (n2 + a2)2 }

= (an cosh 4.

when n = 1, 2, . . . , and Yc(O)= O,

an + sinh an) cosh ax

-

ax sinh an sinh ax.

We noted in Probo 9, Seco120, that when y> O, sin x n

-e-ny = S 2arctan

n

n{

eY-cosx }

(n = 1,2,.. .).

Use formula (2), Seco103, and its special case Co{F'(x)} = F(n:)- F(O)to prove that, when n = O,1,2, . . . , and y > O,

if -1 < b < 1 and b #: O.

,.. 360

5.

OPERATIONAl

SECo 124]

(a) From Sn-1{n-le-n,} 1

MATHEMATICS

given in Probo 4, deduce that

2 x

sin r

if n = 1,2,

le-n, = Cn -- L arctan dr n eY-cosr } { n o

o

o

o

, and y > 00

(b) Differentiate with respect to y in formula (a), Probo 4, to show that ne 6.

-n, .!. cos x cosh y - 1 - Cn{n (cosx - cosh y )2}

when y > 00

Use Cn to find the solution of the problem Y"(x) - e2Y(x) = - F(x),

in the form 2e sinh en Y(x) solution to the form

= F2* H 2(X)where

e sinh enY(x)

= cosh ex

f

(e #- O)

Y'(O)= Y'(n) = O, H(x)

= cosh e(n -

x), and reduce the

F(r) cosh e(n - r) dr

+ cosh e(n - x) I: F(r) cosh er dr. 7.

If F and F' are continuous

when O ;;;;x ;;;;n except at an interior point x

=

e of

that interval, where F(x) and F'(x) have jumps b and b', respectively, and if F" is sectionally continuous, prove that Cn{F"} = -n2fc(n) - F'(O)+ (-I)"F'(n) - bn sin ne - b' cos ne. 8.

When p is not necessarily an integer, let us write

fc(p)=

s: F(x) cos px dx,

f.(p) =

s: F(x) sin px

dxo

When k is a constant, show that (a) 2Sn{F(x)coskx}

= f.(n + k) + f.(n -

(b) 2Cn{F(x) sin kx} = f.(n + k)

-

f.(n

-

k); k);

(e) 2Cn{F(x)coskx}= fc(n - k) + fc(n+ k); (d) 2Sn{F(x) sin kx}

124

= fc(n -

k)

-

fc(n + k).

POTENTIALIN A SLOT

Let V(x,y) denote the electrostatic potential in a space bounded by the planes x = O,x = n, and y = O,in which there is a uniform distribution of space charge of density h/(4n). Then the function V satisfies Poisson's equation (1)

v"x(x,y)+ V,,(x,y)= -h

(O< x < n, y > O).


[SECo 124

361

y

Fig.104

V=o

h 4r

V=A

o

V=O

" (IT,O)

Let the planes x = Oand y = Obe kept at potential zero and the plane x at another fixed potential V = A (Fig. 104). Then

V(O,y)= O,

(2)

V(n,y)= A

(y > O), (O< x < n),

V(x, O) = O

(3)

=n

IV(x,y)1 < M

throughout the region, where M is some constant. The determination of V(x,y) is a type of problem that arises in the subject of electronics. The function V here can also be interpreted as steady-state temperatures in a semi-infinite slab containing a uniform source of heat with strength proportional to h. Our boundary value problem [(1) to (3)] is not adapted to either the method of separation of variables or solution by Laplace transforms. The finite Fourier sine transformation with respect to x does apply, because the differential form in that variable is Vxxand the prescribed boundary values on the intervalO < x < n are V(O,y)and V(n,y). We write v.(n,y) = Sn{V(x,y)} and transform the members of Eq. (1), using conditions (2), to find formally that d2 -n2v.(n,y) - An( -1)" + dy2v.(n,y) = -hSn{1}. We also write conditions (3) in terms of transforms. Thus the problem in v.(n,y) becomes d2v. dy2

-

2 n v.

= An(-1)" - hSn{1}

v.(n,O) = O,

(n = 1,2,.. .),

Iv.(n,y)1 < Mn.

The solution of that problem is (4)

V.(n,y )

= hSn{1} - 2An( -1)" n

(1 - e -ny)

= [h1 - ~;1)" + A(-l:+](1

- e-ny).

362


SECo 125]

The formula for the potential can therefore be written 2 (5)

V(x,y)

G()

L

vs(n,y)sin nx.

= nn=1

In addition to that form of the solution in terms of an infinite series we can derive a closed form of the solution. B.l, we find that S n

-1

l - (-l)n

{

n3

Sn

1

}

- ~( - 2 n

e-ny

{

( - l)n+ 1 -

n

x

}

Referring to Appendix B, Table

),

(-l)n+l n {

Sn-l 2

= - arctan n

}

=

~ n'

sin x

eY

+ cos X ,

and Sn-l{l

-

~_l)n e-ny}

= a(x,y),

where 2 sin x a(x,y) = - arctan -:-n SInh y

(6)

From our formula for Sn- 1{.fs(n)/n2} it follows Sn -1

{

l - (-1)" n3

e

-ny

}

(y

~ O).

that

= U( x,y ),

where

(7)

X

"

x

f

U(x,y) = (n - r)a(r,y) dr n o

f

o

(x - r)a(r,y) dr.

Our potential function is therefore X

(8)

[

V(x,y) = h -(n 2 - x) - U(x,y)

]

A

+ -n ( x

sin x - 2 arctan ey

+ cos x ) ,

where U(x,y) is given by Eq. (7) in terms of the function (6). Both inverse tangent functions in volved here are known solutions of Laplace's equation; consequently it is not difficult to verify that the function (8) satisfies all conditions in our boundary value problem [(1) to (3)]. 125

SUCCESSIVE

TRANSFORMATIONS

A semi-infinite slab of finite thickness is initially at temperature zero, and its base is kept at that temperature. One of its parallel faces is insulated and the other is subjected to a prescribed constant and uniform inward flux of


[SECo 125

363

heat. Units are chosen so that the boundary value problem in the temperature function V(x,y,t) becomes (O< x < n, y > O,t > O), (1)

V(x,y,O) = O,

V x(O,y,t)= -1,

V x(n,y,t) = O,

V(x,O,t) = O;

also IVI < Mt throughout the slab, for some constant M. The presence of the differential form V xx together with prescribed values of V x at x = O and x = n indicates that the finite Fourier cosine transformation, with respect to x, can be used here. The form VI and a prescribed

value of V at t

=

O indicate

the Laplace

transformation

with

respect to t. Let us begin with the cosine transformation. The transform with respect to x, (2)

(n = O, 1,2, . . .),

W(n,y,t) = Cn{V(x,y,t)}

satisfies this problem in partial differential equations: (3)

~ = - n2W + 1 + J-Y,y,

W(n,y,O) = W(n,O,t) = O,

where IW(n,y,t)1 < Mnt. The Laplace transform w(n,y,s) of W, with respect to t, therefore satisfies a corresponding boundedness condition and the conditions d2w 1 w(n,O,s) = O. (4) (s + n2)w - =-, dy2 s The solution of this problem can be written as (5)

w(nys) = , ,

1

s(s + n2)

_! exp(-y~). s

s + n2

With the aid of the inverse transform of S-l exp (- y,J;) and the operational property on replacing s by s + n2, we find that L-l

! exP(-y~) {s

s

+ n2

}

= En(y,t,)

where y ~ Oand

(6)

y

1

En(y,t) =

1o exp (- n2.) erfc 2y.;: d.

(n = O, 1,2, . . .),

The inverse Laplace transform of the function (5) is therefore 1

W(n,y,t)= 2" n

-

exp ( - n2t) - En(y,t) n?

(7) W(O,y,t) = t

-

Eo(y,t)

(n # O,Y ~ O), (y ~ O).

364


SECo 125]

The inverse cosine transform offc(n) = n-2,fc(0) = O,is!(x - n)2/nn/60 Thus Cn- 1{W} can be written (8)

U(x y t) =

"

(x - n)2 2

n

t

Eo(y,t)

- -6 + n- - - n

2 exp (-n2t)

- -n n=1 L

[

n

]

+ En(y,t) cosnx.

2

An integration by parts in formula (6) shows that En(y,t) is of the order of n-4 for large n, when t > Oand y > O. The series in our solution (8) can be differentiated termwise, and the solution satisties all conditions in the boundary value problem (1). PROBLEMS 1 . A steady-state temperature function V satisfies the conditions (O < x < n, y > O),

Vxx(x,y)+ y,y(x,y) = O

V(O,y)= O,

(y > O),

V(n,y)= A

V(x,O)= B

(O< x < n);

also V(x,y)is boundedoDerivethe formula A 2A sin x V(x,y) = -x , n - -n arctan e+cosx

2B

sin x

+ -x arctan-=-smyh

o

2. A boundedharmonicfunctionU satisfiesthe conditions Uxx(x,y) + U,,(x,y) = O U(O,y)= U(n,y) = O,

U(x,+O) = F(x)

(O< x < n, y > O), when O < x < 110

Show formally that u.(n,y) = f.(n)e-ny = 2f.(n)qc(n,y),where e -1 { n } =

n

qc(,y)

~

sinhy

2n coshy - cos x

when y > O,

an even periodic function. of x with period 2110Use our joint convolution property to represent U in this form, when y > O: 1 " sinhy sinhy F

i [

Ux = r ro (,y) 2n o () coshy - cos (r - x) coshy - cos (r + x)Jd

3. When F(x) = A sin x in Probo 2, note the values of u.(n,y) to show that U(x,y)= Ae-Y sinx. Verifythat solutiono 4. Solvethis problemfor V whenO~ x ~ 1land y > O: Vxx(x,y)+ Y,y(x,y) = G(x) V(O,y)= V(n,y) = V(x,+O) = O,

(O< x < n, y > O),


365

[SECo 125

where V is bounded. Obtain these formulas for V: 2

V(x,y)= F(x) + -

1

00

L "2g.(n)e-n, sin nx

11:n= 1 n

= F(x) -

U(x,y),

where U is the solutionof Probo2 and

i

x

F(x) = 5.

o

-

(x

X

r)G(r) dr

"

i

- -11: o

(n

-

r)G(r) dr.

Let V(x,y) satisfy the conditions (Fig. 105)

(O< x < n, O< Y < Yo), V(n,y)= 1,

V(O,y) = O,

V,(x,O) = V(x,Yo) = O.

Derive the formula x

2 00 (-l)n

cosh ny sin nx.

V(x,y)= ñ + ir n~l n coshnyo

YI

v=o

=0

Fig.105

VD O

(1T',):) V=lo -

"

6. The steady-state temperature U(x,y) in a semi-infinite slab with insulated faces satisfies the conditions Uxx + U"

= O and IU(x,y)1< M when O< x < n,y > O, Ux(O,y)= Uin,y) = O,U(x,+ O)= F(x).

(a) Obtain this formula for U(x,y): U

= -1

"

sinhy

sinhy

f F(r)[coshy - cos(x - r) + coshy - cos(x + r)Jd r. "

21t o

(b) When F(x) = A, note the valuesof u«n,y)and deducethat U(x,y)= A. (e) When F(x) = A cos x, note the values of u«n,y); thus show and then verify that U(x,y) = Ae-' cos x.

= Oand x = n of a semi-infinite slab O ;;;¡;x ;;;¡;n, y ;¡;; O, are insulated = O is kept at temperature V = O. A steady source of heat KQ(x) with

7. The facesx and the base y

mean value zero, J~Q(x)dx = O, is distributed throughout the slab; thus Vxx+ Yy, + Q = O,where V(x,y) is the steady-state temperature. If Vis bounded, derive the formulas 2 00 1

V(x,y) = F(x)

- -n n=1n L ,q«n)e-n, cosnx

" = F(x) - U(x,y),

366


SECo 125]

where F(x) = f;(X - r)Q(r)dr and U is the temperature function in Prob. 6 corresponding to this function F. 8. Ir V is a bounded harmonic function in the semicircular region r < a, O < () < n, whose values are prescribed as F«() on the semicircIe and as zero on the bounding diameter (Fig. 106),then r2v,.,(r,() + rf;(r,() + V¡¡8(r,()= O,

(r < a, O < () < n),

lV(r,()1 < M

V(r,O)= V(r,n) = O,

V(a

-

O, () = F(().

Derive this Poisson integral formula 1 for V:

1

n

i

V(r,() = 2n O F(a)[P(r, a

-

()

- P(r, a + ()] da,

when r < a and O ;;¡;() ;;¡;n, where P is Poisson's kernel, a2

-

r2

P(r,() = a2 + r2 - 2ar cos

x

Fig.106

9. In Prob. 8, if the condition on the bounding diameter of the semicircular region is replaced by the condition that the normal derivative of V vanishes there, V¡¡(r,O)

=

V¡¡(r, n) = O,

derive this Poisson integralformula for V: 1 "

i

(r < a, O ;;¡; () ;;¡; n).

V(r,() = 2n o F(a)[P(r,a - O)+ P(r, a + ()] da

10. The transverse displacement Y(x,t) in a stretched string with a steady transverse force distributed along it satisfies the conditions Y,,(x,t) = a2y':x(x,t) + F(x)

whenO< x < n and t > O,

Y(O,t)= Y(n,t) = Y(x,O)= Y,(x,O)= O. Find the solution, and verify it, in the form 1 Y(x,t) = 2a2[2G(x) - G1(x + at)

-

G1(x - at)]

where G1 is the odd periodic extension, with period 2n, of the function x x " G(x) = (n - r)F(r)dr (x - r)F(r)dr. n o o

I

I

Cf. seco 101 of the author's

"Complex

I

Variables

and Applications,"

2d ed., 1960.

,


[SECo 125

367

11. The ends of a stretched string are fixed at the origin and the point (n,O). The string is initially at rest along the horizontal x axis, then it drops under its own weight. Thus the vertical displacements Y(x,t) satisfy the equation Yr,= a2Yxx + g, where g is the acceleration of gravity. If Q(x) denotes the odd periodic extension of the function tx(n - x), where O < x < n, with period 2n, derive the formula

Y(x,t) = 2:2[2Q(X)- Q(x - at) - Q(x + at»). 12. If a steady transverse load is distributed along a beam, the transverse displacements Y(x,t) satisfy an equation 82y

af

84Y

= -a2 8x4 + F(x).

If the ends x = Oand x = n are hinged so that Yand Yxxvanish there, and if the initial displacement and velocity are zero, derive the formula 1 a

2 «> f,(n) . L ~ cos n2at SInnx, na 0=1 n

Y(x,t) = zG(x) - ~

where G(4)(X)= F(x) and G(x) = G"(x) = Oat x = Oand at x = n. 13.

Note that for the elementary problem (O < x < n, t > O),

0(x,0) = 1,

0(0,t) = 0(n,t) = O,

the temperature function can be written

0(x,t) = where

2

«>

-n 0=1 L 8s(n,t)sin nx,

8.(n,t) = So{l} exp(-n2t).

Use Fourier transforms to show that the problem U,(x,t) = f(t)U xx(x,t) + g(t)

(O< x < n, t > O),

U(x,O)= U(O,t)= U(n,t) = O, a problem not adapted to the Laplace transformation, can be solved formally in terms of 0 in the form

where 14.

U(x,t)=

L g(t)0[x,

F(t,t) =

f

F(t,t)] dt,

f(r) dr.

Obtain the solution of the diffusion problem U,(x,t) = Uxx(x,t)

-

h(t)U(x,t)

+

U(O,t)= U(n,t) = U(x,O)= O,

A

(O< x < n, t > ( .,

368


SECo 126]

in the fonn of a series. Also derive the fonn A

'

i

U(x,t) = H(t) o H(t)0(x, t - 't) d't, where 0(x,t) is the function described in Probo 13, and H(t) = exp [E h(r)dr]. 15.

Derive the fonnula for the temperatures U(x,y,t) when (x > O, O < Y < 11:,t > O),

u, = UJCJC + Uyy + A

U(x,y,O)= U(O,y,t)= U(x,O,t)= U(x,1I:,t)= O, and U(x,y,t) is bounded, in the fonn

i

r

X

U(X,y,t) = A O0(y,'t) erf 2J"i d't, where 0 is the function described in Probo 13.

126

A MODIFIED SINE TRANSFORMATION

In Seco115 we presented the modified finite Fourier sine transformation (1)

Sm{F(x)}

= I: F(x) sin mx dx = f.(m)

t

where m = n - and n = 1,2,..., as an example of Sturm-Liouville transformations. It resolves F"(x) on the interval (O,n)in terms of f.(m) and the boundary values F(O)and F'(n), when F is of class C": (2) We introduced the odd antiperiodie exteflsion F3 of F with period 2n; that is, (3)

F3(X)= F(x)

F3(- x) = - F3(x),

when O < x < n,

F3(x+ 2n)= - F3(x) forall x.

In terms of F3 the properties of Smrepresented by Eqs. (10), (13), and (14), Seco 115, can be written x+e 2 (4) - f.(m) sin me = Sm F3(r)dr , m x-e }

{f

(5)

2

-

m

x+y

lt

f.(m)hs(m)

= Sm

{fo

H(y)

fx-y F3(r)dr dy} .

Table 2 summarizes properties of Sm and lists transforms of some functions.

Table 2

Modified

sine transforms'

!,(n

- !)

(m = n - t,n = 1,2,...)

J.(m)

F(x)

s: F(x) sin mx dx = Sm{F}

F(x) 2

(O < x < n)

00

2

I

J.(m) (m = n - t)

-1t 11-1 L f.(m) sin mx

3

I

F"(x)

4

I

-m2f.(m)+ mF(O)- (-I)"F'(n) 1 -mf.(m)(-1)"+ 1

5 6

I I

f-x

f

r

1 2m f.(m)

8

-1

9 I

--m

F3(r) dr

x-e

O

¡'F(r)dr+ x

"

2 -mf.(m)h.{m)

7

F(r) dr

x+e

2 . - J.(m) Sinme m

f

H(y)

O

m cosme

r

f

F3(r) dr dy

O ir O
(O
{1

I

1 _(_1)"+1 m2

x

11

I

sin me (O < e < n) m

X ir O~x~e e { ir e~x~n

12

I

2 m3

13

I

6 -( -1)" m4

14

I

15

I

sinhex coshen

(-lr+1

m - ee""(-I)" m2 + e2

16

k(-lr+1 m2 -p

17

e-my (y > O)

18

1 -e-my m

I

F(r)dr

x-y

10

m2 + e2

x

x+y

sin kx cos kn

(k,p :!:t.:!:f,...)

1 (sin x/2)(cosh y/2)

;

-

cos2 x/2

2 -arctan -sin x/2 n sinhy/2

(y > O)

1 For details see Secs. 115 and 126 and the problerns

cosh2 y/2

rollowing

Seco 116.

369

370


SECo 127]

Note that the modified cosine transformation Cmthat corresponds to Smis not essentially different from Smbecause (6)

Cm{F(x)} =

s: F(x)cosmxdx

= (-lt+lSm{F(n

- x)}.

Thus Cm transforms F"(x) in terms of the boundary values reO) and F(n). That variation of boundary values is taken care of by Sm itself when we replace the coordinate x by n-x. 127

GENERALlZEDCOSINE TRANSFORMS

The transformation (1)

Ch{F(x)}

=

f

F(x) cos qnx dx = fc(qn)'

where qn (n = 1,2,...) are the po sitive roots of the equation h and tan q =h > O, (2) n

qn

is the special case (Prob. 4, Seco 116) of the Sturm-Liouville transformation on the interval (0,1) that has the property (3)

Ch{F"(x)} = -q/fc(qn) - F'(O)+ [hF(l) + F'(l)] cos qn'

The boundary value hF(l) + r(1) arises in applications involving elastic supports or surface heat transfer. Since qnsin qn = h cos qn, we find that (1 h + sin2 q n. IIcosqnxll2 = Jo cos2qnxdx = Thus the inversion formula becomes 00 cos qnx (4) F(x) = 2h n=1 fc(qn)h + sm . 2 qn

I

(O< x < 1).

From Eq. (2) we can see graphically that qn - nn -+ O as n -+ oo. Let F2 denote the even periodic extension of F with period 2: (5)

F2(x) = F(lxl) if -1 < x < 1,

When O ~ Y ~ 1, the formulas 2fc(qn)cos qnY

=

Ch{F2(x + y) + F2(x - y)} -2 sin qn I: F(x + 1 - y) sin qnx dx,

(6) sin qn

i

y

o

F(x) sin qnx dx

=h

fl-y l

e-hx

fl-yx ehrF(l -

r) dr cos qnx dx


[SECo 127

371

can be derived or verified by making several elementary manipulations. From them we can find this kernel-product property: (7) where z

=2 -

x

R(z) =

- y and

f

e-hrF(r) dr

if z < 1,

R(z) = O if z > 1.

The corresponding convolution property for Ch is (8) in terms of this convolution of F and G: (9)

XFG(x) =

f

G(y)[F2(x + y) + F2(x - y)] dy

- 2hG(1 - x) * F(l - x) * e-hx. Here the asterisk denotes the convolution integral used with Laplace transforms (Sec. 16); thus (10) G(l - x) * F(l - x) * e-hx = e-hx

s: ehtG(l -

t) s:-t ehrF(l

-

r) dr dto

Unfortunately, our convolution is not a simple one.1 When y = 1, formula (7) becomes (11) fc(qn) cos qn

=

Ch{ F(l

-

x) - he-hx

s: ehrF(l -

r) dr}

.

From property (3) we find in the usual way that (12)

J;(q )

(1

1

r

~J = Ch{ Jo ( 1 + h - r) F(r)dr + Jo (r - x)F(r)dr} . These transforms can be found in various ways:

(13)

(14) 1Another derivation of the property can be found in the author's paper, Generalized Finite Fourier Cosine Transforms, Micho Matho Jour., vol. 3, pp. 85-94,1955-56.

372


SECo 128]

A RELATED TRANSFORMATION

Since the above kernel cos qnx is an even function of X, the transformation (15)

Ch{F(x)} = fl

F(x)cosqnxdx

=!c(qn)

of functions defined on the interval (-1,1), where qn (n = 1,2,...) are again the positive roots of Eq. (2), is expressed in terms of Ch by the equation (16) It follows from property (3) that, when F is of class C" on the interval -1 ;;;;x ;;;; 1, Ch resolves F"(x) in terms of the boundary values hF( -1) F'( -1) and hF(l) + F'(l): (17)

Ch{F"(x)} = -qn2!c(qn)

+ [hF(-l) - F'(-l) + hF(l) + F'(l)]cosq..

Since h > O,the kernel ofCh is the set ofall eigenfunctions oftheproblem ~"(x) = A~(X),

h~( -1)

- ~'( -1) = h~(l)

+ ~'(1) =

O,

corresponding to the eigenvalues A = - qn2. The inversion formula for Ch is

(-1 < x < 1).

(18)

128

A GENERALlZED SINE TRANSFORM'

The transformation (1)

Sk{F(x)}

where Pn(n = 1,2,...) (2)

=

s:

F(x)

sin PnX dx

= !s(Pn),

are the positive roots of the equation Pn

tan Pn =

- k

and

k> O,

was used in Seco117 to solve a problem involving surface heat transfer. It has the property (3) Sk{F"(x)} = - P/!s(Pn) + F(O)Pn+ [kF(l) + F'(l)] sin Pn, and the inversion formula (4) 1 The

C()

¿

F(x) = 2k n=1 fs(Pn)k

sin PnX

+ cos 2 Pn

(O< x < 1).

transformations in Secs.126to 128are among the generalizedfinite Fourier transforma-

tions treated by Ida Roettinger (Kaplan) in Quart. Appl. Math., voL 5, pp. 298-319, 1947, and in lour. Math. Physics, vol. 27, pp. 232-239, 1948.


s -I k

J.(Pn)

{ p/ }

=

1-

~

l

f

-Sk{F(x)}= Pn

(7) Sk{F(l

-

(1 rF(r) dr

1 + k ) Jo

(

1

(6)

o

+

I (x

Sx

-

r)F(r)dr,

l

cosPnx

f

F(r)drdx, x

- ke-kxI: érF(l - r)dr} = sinPn

x)

373

has the further properties

The transformation

(5)

[SECo 128

f

F(x)cos p,.xdx.

Let FI be the odd periodic extension of F with period 2:

= F(x) ifO < x < 1, for all x. Fl(x) = - Fl ( - x) = FI (x + 2) FI(x)

(8)

When O~ Y ~ 1, we can obtain the integration formula x+y

2

- J.(Pn)sin PnY=

(9)

Sk

Pn

{fx-y

Fl (r) dr

}

+ 2 sin Pn (y cos pnX Jo

f

l F(r) dr dx, x+l-y

from which we find the kernel-product property 2

-

(10)

x+y

J.(Pn)sin PnY= Sk

Pn

{fx- y Fl (r)dr + 2éz R(z)}

where z = 2 - x - y and R(z) =

r

e-krF(r) dr if z < 1,

R(z) = O if z > 1.

The corresponding convolution property is 2

(11)

p/'
when g.(Pn)= Sk{G(x)}and .

(12)

l

XFG(x) =

f

o

x+y

G(y)

f

x-y

Fl(r) dr dy + 2G(1 - x) * F(l

-

x) * e-kx,

where the final term, the iterated convolution used with Laplace transforms, is given by formula (10), Seco 127.

374


SECo 128]

We note some particular transforms. (13) (14)

Pn

=

Sk{e-kx}

.

Sk{sin ¡rx} = ¡r SInPn

+ p/'

k2

¡r2

-

Pn

-

.

PROBLEMS 1.

Use the transformations Sm(Sec. 126) to solve for Y(x) when d4y dX4 - y

= 2x,

Y(O)= Y"(O)= Y'(n) = Y"'(n) = O.

Ans. Y(x) = sechn sinh x - 2x - sinx. 2. Let the steady-statetemperatures V(x,y)in a semi-infinitewall O;;¡;x ;;¡;n, y ;;;O, satisfythe conditions lV(x,y)1 < M

y"x(x,y) + y,y(x,y) = O, V(O,y)= O,

y"(n,y) = 1,

(O< x < n, y > O),

V(x,+O) = O,

where M is some constant. Find the solution of that boundary value problem in the forms 2

ex>(-1)'

V(x,y)= x + -n ,= L1 my- e-my sin mx

(m = n - !)

2 x sin r/2 dr. arctan --;-n n-x ( SInh Y/2)

=x - -

I

3. A harmonic function V(r,O)in the semicircular region r < 1, O < O< n, satisfies these conditions : r2Y.r(r,0) + ry'(r,O) + J.oók,O)= O

V(r,O)= 1,

J.oó(r,n) =

O,

(r < 1, O < O< n),

V(l,O) = O;

also, Vis bounded over the region. Derive the formula

V(r,O)= 1 - ~ sin0/2 n arctan 20 1-r' 4. Ir V(r,O)satisfies the conditions stated in Probo 3 in the part of the upper half plane exterior to the semicircIe, the unbounded region r > 1, O < O< n, show that V(r,O)= 1 -

~ n arctan

20 r-1 sin(0/2)

(r > 1, O < O < n).

I I

J


5.

[SECo 128

375

For the transformation Ch in Seco127 show that, if O ~ e ~ 1, cos qnc (a ) Ch- 1 -{ q/ }

1 + -l - e h

-

1

{

if O~ x ~ e,

1

. f

+¡¡-x

I c~x~l;

if O < x < c, ifc
Y.,(O,t) =

- B,

Yx(I,t)= -hY(I,t),

(O< x < 1, t > O), Y(x,O)

=

Y,(x,O) =

O,

where h > O. Derive the formula

[

Y(x,t) = B 1 +

!h - x -

2h

f

n=1

cos ¡qnat) C~S}qnX) ' qn

(h + sm qn) J

where ql' q2'... are the positive roots of the equation tan qn = h/qn' Also,use transform (a) in Probo 5 to show that, when at ~ 1,

Y(x,t) = B(at - x) if x ~ at,

Y(x,t) = O if x ~ ato

7. Use the transformation Ch in Seco127 to solve the following problem for steadystate temperatures in a semi-infinite wall with heat transfer at one face into a medium at uniform temperature A : (O< x < 1, Y > O), vxx(x,y) + Y,y(x,y) = O

Vx(O,y) = O,

- Vx(l,y)= h[V(I,y)- A]

(y > O), (O< x < 1),

V(x,O)= O, Y,(x,oo)= O whereh > O. Obtain the solution in the form V(x,y)= A 1 - 2h2

[

f cosqn~o~qnX exp(-;qnY>l,J h + sm qn

n= 1

qn

where tan qn = h/qn and qn > O. 8. Let the temperatures U(x,t) in a slab with surface heat transfer at one face, and with heat generated within the slab, satisfy the conditions

(O< x < 1, t > O),

U,(x,t) = Uxx(x,t) + F(x)

-

U(O,t) = O,

Ux(I,t) = bU(I,t),

U(x,O) = O,

where b > O. Derive the formula 1

U(x,t) = (

-~

f

l

1 + b) o

rF(r)dr

+

I

I (x x

-

r)F(r)dr

~ f.(PJ sin p"x2 exp ( - Pn2t), - 2b Lo..~ n=1 Pn b + cos Pn

376


SECo 129]

where p¡, P2'... are the positive roots of the equation tan P.

J.(P.)= 129

f

= - pJb and

F(x) sin P.Xdx.

FINITE EXPONENTIALTRANSFORMS E.{F}

A linear integral transformation of functions F(x) on an interval (-n,n) that resolves F'(x) in terms of the cyclic boundary value F(n) - F( -n) is often useful in solving problems in which x is an angle, such as the polar coordinate e, or in finding periodic solutions of problems in differential equations. For a transformation T{F(x)}

=

f" F(x)«I>(x..l)dx = f(..l)

we find that, when F and el>are of class C' (-n ~ x ~ n), T{F'(x)} = Fc1>]':." - f"

Fc1>' dx = ..lf(..l)+

if «I>and ..l satisfy the homogeneous (1)

«I>'(x) =

-

..lc1>(x),

c1>(n,..l)[F(n)

- F(-n)]

conditions «I>(n,..l)= «I>(-n).

This last equation is a periodic boundary condition on «I>.The non trivial solutions of problem (1) are «I>(..l,x) = e-inx and ..l= in where n = O,Il, I 2, . . .. Thus T is the finite Fourier exponential transformation (2) En{F(x)}

= f" F(x)e-inxdx = fe(n)

(n = O,:t 1, :t2,.. .),

with the basic operationalproperty (3) E.{F'(x)} = inEn{F(x)} + (-l)n[F(n) - F(-n)]. When F is of class en on the interval - n ~ x ~ n, or even if F and F' are continuous there and F" is sectionalIycontinuous, we find, by replacing F and F' in formula (3),that (4)

En{F"(x)}= -n2fe(n) + in(-l)n[F(n) - F(-n)] +( -l)n[F'(n) - F'(-n)].

The formula can be extended easily to transforms of derivatives of higher order in terms offe(n) and cyclicboundary values of F and its derivatives. The set of functions e- inx(n = O, Il, I2,...) is orthogonalin the Hermitian sense on the interval (-n,n); that is, when m = O,:t 1, :t2,..., (5)

f" exp (-inx)exp (-imx) dx = f" ei(m-n)xdx = O ifm i=n.


[SECo 129

377

That set is the basis for the exponential form of the Fourier series on that intervaJ.l In terms offe(n) that series representation of F(x) can be written as 1 (6)

=

F(x)

2n

n=m

l~

( - n < x < n),

n];.m fe(n)einx

valid if F and F' are sectionally continuous over the interval and F is defined as its mean value from the right and left at each point of discontinuity. Note that the sum in Eq. (4) ineludes

the term corresponding

to n

equation is a formula for En-l{fe(n)}. In case the infinite series of terms lfe(n)l, where n runs from converges, then

(7)

= O. The

-

00 to 00,

(-n
that is, the sum of the series here exists, and so it is the same as the principal value used in formula (6). In fact, whenever the series of terms lfe(n)1converges, then series (7) converges uniformly to a continuous sum F(x) whose transform is fe(n). In terms of Fourier sine and cosine transforms

En{F(x)}

=

f

[F(-x)

+ F(x)] cos nxdx

+ ¡s: [F(-x) - F(x)]sin nx dx where n = O,:t 1, :t 2,. . .. In particular, when F is an evensectionallycontinuous function on the interval (-n,n), we have a useful formula for obtaining exponential transforms from cosine tran~orms: (8)

if F( -x)

= F(x)

(n = O,:t 1, :t2,.. .).

When m = :t 1, :t 2, . . . , we find easily that (9)

(n = O, :t 1, :t 2, . . .).

From formula (5) and the linearity of En it follows that (10) (11)

if n

=

:t 1, :t 2, . . . ,

En{F(x)+ A} = fe(n)

but if n .¡. O,

Eo{F(x) + A) = fe(O)+ 2nA. I Churchill, R. V., "Fourier Series and Boundary Value Problems," 2d ed., seco43, 1963.

.

378


SECo 130]

From property (3) we find that, when F is sectionally continuous, if n =FO.

(12)

130

OTHER PROPERTIES OF En

Let F" denote the periodic extension

(1)

F,,(x) = F(x)

of F with period 21t,

F,,(x+ 21t)= Fix)

if -1t < x < 1t,

for all x.

In case F" is of cIass C' on every interval, or if F" is everywhere continuous and F' is sectionally continuous over the interval (-1t,1t), then according to property (3), Seco 129, (2)

(n = O, :t 1, :t 2, .. .).

Ir F" has a continuous derivative of order m everywhere, then

(m = 1,2,.. .).

(3) This kernel-product property for En{F} is easily found: "+Y

(4)

e-inYfe(n)=

f-,,+yF(x - y)e-inXdx = En{F,,(x- y)}

for all real y. When y = 1t, it becomes (5)

(-1)nfe(n)

= En{F,,(x-

1t)}

[2" = Jo F(x - 1t)e-inxdx.

Note that F,,(x - 1t)= F(x + 1t)if -1t < x < O,and F,,(x - 1t)= F(x - 1t) if O < x < 1t. The corresponding convolution property is

(6)

fe(n)ge(n)= En{XFG(x)}

where gin) = En{G(x)} and (7)

XFG(x)

= f" G(y)F,,(x- y)dy.

In terms of G and F directly, that convolution has the form "+X " (8) when

f-"

- y) dy +

I

XFG(x)

=

-

< O; but when O < x < 1t,

1t < X

G(y)F(x

"h

G(y)F(x

XFG(x)= f~,,"h G(y)F(x- y - 21t)dy+ f"h

-

y + 21t)dy

G(y)F(x- y)dy.


In

[SECo 130

379

form (7)the convolution has the properties

X FG(X)= X GF(X) = X FG(X + 2n). Table 3 lists transforms of several functions, transforms that can be derived by using our properties of En including the relation (8), Seco 129, between Enand Clnlwhen F is even. Sóme ofthe properties of Enare included in the table. (9)

Table 3

Finite exponential f.(n)

transforms

E.{F:

(n = O, i: 1, i: 2, . . .)

r

-n F(x)e-inXdx = En{F(x)}

2

f.(n)

3 4 5

J~(n - m) (m = i: 1, i:2,...)

6

f.(n)g.(n)

7 8 9

(n = O, i: 1, i:2,...)

inf.(n)+ (-I)n[F(7f)- F(-n)] e-in'f.(n)

F(x)

(-n

< x < n)

F(x) 1 -1'

m

2n..~

n=L-m f.(n)einx

F'(x) eimxF(x) F.(x y) (Sec. 130)

-

f.

G(y)F.(x- y)dy

1

o ir n,p O,f.(O)= 2n o ir n,p i: l,f.(i: 1)= n o ir n,p i:1,f.(1) = -ni,f.(-I) = ni

cos x sin x

10

(-I)n 21ti ¡¡-

x

11

2ni n ir n,p 0,f.(0) = o

X+ n { x-n

12

(-1). 2 :- (e ,p O, - 1, - 2,.. .) e - In

2 sinh en

13

2 sinhen (e2,p O,-1, - 2, . . .) e - in

{ e-".e'" ir O< x < n

14

(-Ir 4n~n

15

(-I)n n2 + e2

16

(_I)n+1 n2 - e2

17 18

ir n,p 0,f.(0) = o

eCX

e<.ec;" ir - n < x < O

ir n,p 0,f.(0)= ~n3 cosh ex (e2 ,p O, -1, -2,...) (e,p O, i: 1, i:2,...)

(_l)n+1

~

ir -n
ir n,p i: I,f.(i: 1) = O

blnl (-1 < b < 1, b ,p O)

2e sinh en cos ex 2e sin en

1 . -(2x SIOx + cosx) 4n 1 1 - b2 2n 1 + b2 - 2bcos x

.

380


SECo 130]

Applications of En are illustrated in the set of problems that follows. Finally, we note that by writing x = (2n/c)t (10)

En{F(x)}= (-lt

- n,

s: e-inw1G(t)dt

where úJ = 2n/c and G(t) = úJF(úJt- n). The integral in Eq. (10) is the

exponentialtransformation of G(t)on the intervalO < t < c. Intervals other than (-n,n) have been usedas a basisfor finiteexponentialtransformations.l PROBLEMS 1. Ir F(x) is sectionally continuous over the interval (-n,n) andJ.(O) = O,prove that, when n = :t 1, :t 2, . . . , f.(n)

-::r = n

X

En

{f-n

(r

X

-

x)F(r) dr

K

- _21tf-n rF(r) dr} .

2. Use En with respect to t to find the periodic solution Y(t), with period 2n, of the equation Y'(t) - eY(t) + P(t) = O, where P(t + 2n) = P(t) for all t, and e2 =f.O,-1, - 2, . . . . '+2K

i

Ans. Y(t)(l - e-2
P(r)e-
3. Ir the external force acting on the mass m shown in Fig. 15 (Sec. 29) is periodic, the unit of time can be chosen so that the displacement X(t) of m satisfies an equation

where

mX"(t)= -kX(t) + P(t)

P(t + 2n) = P(t) and k > O.

Write úJo= jki;. When Wo =f.1,2,..., and P is sectionally continuous over its period, obtain these formulas for the periodic displacements with period 2n: X( t )

-- ~

~ p.(n) '-2 2mnn=_""WO -n

2e

in'

--

f:K P(t + n - . t)

COSúJot dt

2mwosmwOn

'+2n

f,

= (2mwosinwon)-¡ 4.

P(r)coswo(t+ n - r)dr.

Let P(x) be a prescribed periodic function with period 2n such that P" is everywhere

continuous. Use En to find the periodic function Y(x) of period 2n that satisfiesthe integral equation f~K Y(x

-

r) cosh er dr

= P(x)

Ans. 2eY(x)sinhen = e2p(x - n)

wheree > O.

- P"(x -

n).

¡ Kaplan, W., "Operational Methods for Linear Systems," chap. 4, 1962; Doetsch, G., Integration von Differentialgleichungen vermittels der endlichen Fourier-Transformation, Mathematische Annalen, vol. 112, pp. 52-68, 1935.


5.

[SECo 130

If P(t) and Y(x,t) are both periodic in t with period

211:, derive

381

the solution of the

problem Y(O,t) = P(t),

y"(1I:a,t) = O

in the form

6. In Probo 6, Seco85, let the unit of time be chosen so that the temperature V(x,t) in the semi-infinite solid x ~ Osatisfies the conditions V(O,t)= A sin t,

V(oo,t) = O;

also V(x,t + 211:) = V(x,t). Showthat the exponentialtransformve(x,n)vanishesexcept when n

=

j: 1, and that V(x,t) = A exp

(

-

p) sin(t- p)'

7. Heat is generated at a steady rate uniformlyalong the radius 9 = 11:of a thin, circular plate r < 1, - 11:< 9 ~ 11:(Fig. 107)whose faces are insulated. If the edge r = 1 is kept at temperature zero, the steady temperatures U(r,9) in the plate are such that r2 Urr(r,9) + rU,(r,9) + Uoir,9) = O 1 -[Uir,1I:)Uir, -11:+ O)]= A, U(r,1I:)= U(r, -11: + r

(r < 1, - 11:< 9 < 11:), O),

U(l,9) = O.

Also, U is bounded over the region r < 1, - 11:< 9 < 11:,and U and its partial derivatives of the first order are continuous there. Show that, with respect to 9, ifn *" j:1,

A = -rlogr 2

ifn = j:l.

Thus derive the formula A

r

co (-1)"

[ - r9 sin 9 - (2- r log r) cos 9 U(r,9) = 211:1

-

2 "~2 n2

-

].

1r"cos n9

Fig.107

I I'

382

8.


SECo 130]

If V is a bounded harmonic function interior to the circle r

= a that

assumes

values F(O)on the circle,then r2 Vrr(r,O) + ¡.y'(r,O) + V88(r,O) = O V(a

-

O,O) = F(O)

- 1t ~ O~ 1t), when - 1t < O< 1t;

(r < a,

also, V and its partial derivatives of first and second order are continuous everywhere inside the circle and thus periodic in Owith period 21t. Show that, with respect to O,

when F is sectionally continuous. Thus derive Poisson's integralformula for the harmonic function in the disk (cf. Probs. 8 and 9, Seco125),

x V(r,O)= ~ 21t

I -x .

(a2- r2)F(IX) L. . -. ua

(r < a, -1t ~ O~ 1t).

12 Exponential Fourier Transforms

The exponential Fourier transformation is a prominent singular linear integral transformation of functionsF(x) on the entire x axis. Conditions of validity on its operational properties are severe. For most properties weuse

-

the requirement that F be absolutely integrable from x =
THE TRANSFORMATION E.{F}

Let F denote a real- or complex-valued function of a real variable x such that F(x) is defined over the entire x axis. A linear integral transformation of such 1

See, for inslance, books by Wiener and Titchmarsh on the theory of Fourier integrals, listed

in the Bibliography.

383

I L

384


SECo 131]

functions

is T{F} = f:oo F(x)K(X,A.)dx = f(A).

Let us determine the kernel K and the domain of the parameter A so that T transforms derivatives F' into products .if(A). For the purpose of designing the transformation so that it has that property, we shall assume that, for each fixed value of A.,the kernel K has a continuous derivative K' with respect to x and that K and K' are bounded for all x. We assume that the functions F are everywhere continuous with sectionally continuous derivatives F'(x) on each bounded interval (a,b), that F(:!:(0) = O,and that F is absolutely integrable from 00 to oo. Then

-

f

F'(x)K(x,A) dx = F(x)K(x,A)]~

-

f

F(x)K'(X,A)

dx,

and the right-hand member here has a limit as a -. - 00 and b integral on the left has the same limit, and it followsthat

-. 00, so the

T {F'} = f:oo F'(x)K(x,A.)dx = - f:oo F(x)K'(X,A) dx. Thus T{ F'} = Af(A) if K is a solution other than zero of the horno geneous equation K'(X,A) = - AK(x,A) and if K satisfies our requirements of continuity and boundedness. We may select the solution K(x,A.) = e-'¡'x. This function is bounded for all real x only if A = iexwhere exis any real number; then K' is also bounded with respect to x. Thus T becomes the exponential Fourier transformation (1)

E,,{F(x)} =

Loooo

F(x)e-iOtX dx = fe(ex)

(-

00

< ex< (0),


E,,{ F'(x)}

= iexfe(ex).

If F' is everywhere continuous and F" is sectionally continuous, and if F(:!:(0) = F'(:!:(0) = O and f~ ooIF(x)1dx exists, then (3)

f

f

F"(x)e-i"X dx = [e-iOtX(F' + iexF)]~- ex2

F(x)e-i"x dx

and, when a -. - 00 and b -. 00, we find that (4)

(- 00 < ex< (0).

..

EXPONENTIAL FOURJER TRANSFORMS

[SECo 132

385

Likewise, the operational property (2) can be extended as follows: Theorem 1 If F(x) is absolutely integrable from - 00 to 00, if its derivative F(m-l) is everywhere continuous while F(m) is sectionally continuous on each bounded interval, and if F(:too)

= F'(:too)

=...

= F(m-l)(:too)

= O,

then

(m = 1,2,. ..).

(5)

In case F has a jump discontinuity at a point x = e but otherwise satisfies our conditions for the validity of property (2), we find that (6)

E..{F'(x)} = iafia) - e-iCZC[F(c + O)- F(c - O)].

The literature on exponential Fourier transforms contains minor variations in the choice of the kernel and in notation. The kernels eiax and (21t)-teiax, for instance, are sometimes used.

132

THE INVERSE TRANSFORMATION

In Seco68 we stated conditions under which a function is represented by its Fourier integral formula 1

(1)

f f p

F(x) = - lim 21t(J-+00

oo

o

e"'"

-p

o

F(-r)e-IIItd-rda

(-

00

< x < (0).

-00

Since the inner integral here is.fe(a),the formula represents the inverse ofthe exponential Fourier transformation: (2)

F(x) = -

1

f

(J

lim 21t (J-+oo

o

.fe(a)e"'"da = E..-l{.fe(a)}.

-(J

Sufficient conditions for its validity can be stated as follows, even if F is a complex-valued function (Probs. 9 and 10, Seco133). Theorem 2 Let F(x) be absolutely integrablefrom - 00 to 00 and let F' be sectionally continuous over each bounded interval. AIso let F(x) be defined at each point e of discontinuity as t[F(c + O) + F(c - O)]. Then its traniform fJ..a)exists, and the inversionformula (2) is validfor all real x. In casef J..a)is also absolutely integrable from - 00 to 00, or in case the da exists, formula (2) can be written integral J~oo.fe(a)eiax (3)

1 F(x) = -

21t

f

oo

-00

1 .fe(a)eiu:da = - E-A.fe(a)} 2 1t

(-00

< x < (0).

386


SECo 132]

By replacing a by - a in the integral here, we can write 1 F(x) = 2nExCfe(-a)}.

(4)

Thusunderappropriateconditionson F andfe the inverseof the operator Ea can be written in terms of Ea itself. That the principal value of the improper integral E_x{f.(a)} in the inversion formula (2) may be needed is iIIustrated by the following example. Consider the function F such that F(x)

=O

if x < O,

F(O)= t,

F(x) = e-X if x> O,

which satisfies the conditions in Theorem 2. Here F'(x) = - F(x) when x =FOand, according to formula (6),Seco131,

- fe(a) = iafe(a)- 1; integrable from - 00 to 00 since 1 - ia

Ea{F'} = thus fe is not absolutely

1 fe(a) = 1 + ia = a2 + 1

1 lfe(a)1= ¡:===--;,. 1 + a2

and

When x = O, the integral E-Afe(a)} does not exist because (a2 + 1)-1 is integrable from - 00 to 00 while a(a2 + 1)-1 is not. But - i

-

f

fJ

a

=O

~da 21t - fJa + 1

and

i

lim fJ-+00 21t

f

a

fJ

--y--

+1

- fJ a

da

= O. '

therefore ti

lim

_21 f

fJ-+oo 1t

fJ

-fJ

_1 f

da

oo

fe(a)da = 2 n

-ooa

~

+

1

1 =_ = F(O). 2

No two functions F and G that satisfy the conditions on F in Theorem 2 can have identical transforms. For if F and G are absolutely integrable from - 00 to 00, then F - G is also, since IF GI ;;¡;IFI + IGI. Thus we see that the function F - G satisfies all those conditions. But Ea{F - G} = O if F and G ha ve identical transforms and, according to the inversion formula (2), F(x) - G(x) = Ofor all X. In that cIass of functions, therefore, there is a one-to-one correspondence between functions and their transforms. Theorem 2 gives conditions on F that are sufficient for the validity of inversion formula (2). We now establish sufficient conditions on the function fe(a) instead, conditions that also establish fe as a transform and as an inverse transform; but one of our conditions is apt to be difficult to apply.

-

Theorem 3 Let a function f(a) have a sectionally continuous derivative f' over each bounded interval, and let f be absolutely integrable from - 00

EXPONENTIAL FOURIER TRANSFORMS

[SECo 132

387

to OO.Then its transform ExU} exists. We write (5)

27t4>(x) = ExU(a)}

( - 00 < x < (0).

= f~oo f(-r)e-irxd-r

lfin addition 4>(-X) has a transformfor all real a, then (6) and,Jor all real x, the inversionformula 1

(7)

4>(-x)

=-

27t

I

1

oo

f(a)ei«xda = -E_xU(a)}

-

27t

is valid. Also,Jrom Eq. (5) it follows that

(8)

= 27t(a).

E«U(x)}

Under the conditions on f the integral (5) converges uniformly to a function 27t4>(x)that is continuous for all real x. Since the integral Ea{4>(- x)} exists, by hypothesis, we can write it as E_«{(x)}, or in the forms

f

oo

1

.

(x)el«~ dx

-

= -27t

f

oo

.

el«x

-

f

.

oo

f(-r)e-ItX d-r dx.

- IX)

But the right-hand member here representsf(a) for all real a, according to the Fourier integral formula (1), sincefsatisfies the conditions ofvalidity ofthat formula. Thus f(a)

= f~oo<1>( -x)e-i«x dx = E«{( -x)}.

Equation (5) shows that 4>(-x) is represented by the inversion formula (7). To illustrate Theorem 3, let us find the inverse transform of

= exp (-Ial),

f(a)

a function that is everywhere continuous, absolutely integrable from - 00 to 00,for whichf' is continuous except for a jump at the point a = O.For that function

The integrals here can be written as simple Laplace transforms (s = 1),and we find that (x)=

1

1

--z 7t1+x

= (-x).

I

388


SECo 133]

Since ct>is absolutely integrable over the x axis, Ea{ct>}exists and (9)

{! ~ }.

exp ( -Ial) = Ea

nl+x

AIso, according to formula (8) 2 (10)

Ea{exp (-Ixl)}

133

OTHER PROPERTIES OF E.

= 1

+ é'

We now point out several properties of the exponential Fourier transformatíon that follow readily from properties of integrals noted in Secs.14, 15, and 56. It is convenient to assume throughout this section that F(x) is seetionally eontinuous over eaeh bounded interval on the x axis. Suppose that, for all real a, the integral

Ea{F(x)} = f~", F(x)e-iax dx = fe(a) exists. Then if e is real and constant, we find that (1) (2)

E.{F(x

-

e)} = e-ic.fe(a),

Ea{F(-x)} = f( -a),

(3)

if e ::f=O, Ea{F(ex)} = 1~lfe(~)

(4)

Ea{eicxF(x)} = fe(a - e).

AIso the complex conjugates offe and F are related: (5) Ir F is an even or an odd function, its transformfe is even or odd, respectively, andfe can be written in terms ofFourier cosine or sine transforms (Chap. 13):

(6) (7)

fe(a) = 2 1'" F(x) cos ax dx = fe( -a) fe(a) = -2i 1'" F(x) sin ax dx = -fe( -a)

if F(-x) = F(x), if F(-x) = -F(x).

Property (1) is the kernel-produet property of Ea that we noted in Probo 3,

Seco110. Suppose now that F is absolutely integrable from - 00 to oo. Then the integral Ea{F} converges uniformly with respect to a, and we find that (8)

fia) is eontinuous for all real a.


[SECo 133

389

Also,fe is bounded; in fact, the Riemann-Lebesgue theorem noted in Probo 8, Seco65, shows that

lim firx) = o.

a-;t co

- 00 to 00, then E~{ - ixF(x)}

In case IxF(x)1 is also integrable from f~(rx) =

(10)

since the member on the right represents a uniformly convergent integral (Sec. 15). Example

Ir F(x) = exp (- X2),then

fe(rx) = f:ex> exp

= exp ( -

[-

(X2

+ irxX)]dx

:) f:ex> exp ( - r2) dr = ~

exp ( - :),

since the integral of exp ( - Z2)over each horizontalline z = x + irx/2in the complex plane is equal to the integral over the real axis. Thus (11)

E~{exp(_X2)} =

~

exp (

- :).

Then from properties (3) and (10) we find these transforms: E~{exp ( - C2X2)}

(12)

=

¡; exp

E~{xexp(_X2)} = -if

(

- ::2)

(e real, e =1= O),

rxexp ( - rx:).

PROBLEMS 1.

If F(x)

= 1 when

-b < x O,we found in Seco132that 1

l-ia.

E«{e-XSo(x)}= 1 + ia = 1 + 11.2

390


SECo 133]

Use properties of E. in Seco133 to deduce that ife > O; 1 . (b) E.{xe-XSo(x)} = (1 + io,? , 1 (e) E.{ e"So(- x)}

= 1 - ia'

3. Use transform(10),Seco132,of exp(-Ixl) to prove that 2e (a) E.{exp(-elxl)} =--r-2 e +a

ife > O;

2 e-ic. (b) E.{exp (-Ix - el)} = 1 + a2

if e is real;

-4ia (e) E.{xexp(-Ixl)} = (1 + (2)2' 4. From transform(9),Seco132,of (1 + X2)-1 deducethat (a) E. a 2 +a x 2 { } (b) E. 5.

{(x-e

~2

= n exp(-alal)

+a 2 }

= nexp(-alal-

Apply Theorem 3 to the function f(a)

(a) 2fiexp(-a2) then deduce that

(b)

fi

exp

(

ifa > O; if a > Oand e is real.

ica)

= exp (-(2)

to establish the transform

= E.{exp (- x;)};

- ~) = E.{exp(- X2)}.

6. (a) Show formally that

E.{xF'(x)} =

- ;a[a/.(a)].

(b) Apply E. formally to find a solution Y, for which E.{ Y} exists, ofthe ditTerential equation

2Y"(x) + xY'(x) + Y(x) = O. Verify that solution. 7. Ir a > O and F(x) = COS2ax

Ans. Y(x) = A exp(-x2f4).

whenalxl ~ I'

and

F(x) = O

whenalxl~ I'

[SECo 134


391

show that F"(x) = 2a2[1 - 2F(x)] when alxl < 1t/2and hence that

8. When c > Oand F(x) = Oir Ixl> c while F(x) =

-1

and

ir -c < x < O,

F(x) = 1

ifO
showthat 2 E«{F(x)} = -;-(1 IIX - cos CIX).

9. Write F(x) = U(x) + iV(x) where U and V are real-valued functions of x (- 00 < x < (0). Let F be sectionally continuous on each bounded interval of the x axis (Sec. 56). If F is absolutely

integrable

from

-

00 to 00, point out why U and V

also have that property. 10. Theorem 2, Seco132, gives conditions under which a function F is represented by the Fourier integral formula, conditions that were stated in Seco68 when F is realvalued. With the aid of the observation in Probo 9, show why Theorem 2 applies when F(x) = U(x) + iV(x) where U and V are real-valued functions.

134

THE CONVOLUTION

INTEGRAL FOR E«

The kernel-product property (1), Seco133, for eaeh fixed real t is e-itl'fe(rx)= E«{F(x - t)}. As noted in Probo 3, See. 110, the eorresponding eonvolution of two funetions F and G with transforms .fe(rx)and girx) is (1)

XFG(x)

=

t:

F(t)G(x

-

t) dt

=

(-00

< x < (0).

F * G(x)

Note that the eonvolution or faltung operation here, F * G(x), is not the one used with Laplaee transforms sinee the integral has limits - 00 and 00 rather than O and X. Both eonvolutions are the same in case F(x) = G(x) = O whenever x < O. Before establishing eonditions on F and G under whieh Ea{XFG} = Je(rx)ge(rx), we need some properties of the operation F * G and information on the nature of the function X FG(X). We assume in this seetion that our functions F(x) and G(x) are sectionally continuous on each bounded interval of the x axis, and that they are bounded and absolutelyintegrableJrom - 00 to oo. Sineea eonstant M exists such that lG(x - t)1< M, then IF(t)G(x - t)1<

MIF(t)l. But MIF(t)1is independent of x and integrable from - 00 to oo.

"

392


SECo 134]

Hence the convolution integral (1) converges absolutely and uniforrnlyfor all real x. Moreover,

f

F(t)G(x

-

t) dt

=

r~: G(y)F(x

-

y) dy

and since the first integral tends to XFG(X)as a -+ - 00 and b -+ 00, then the second one has the same limito It follows that the operation F * G is commutative : (2)

XFG(x)

=

f:

G(y)F(x

-

y) dy

=

XGF(x).

If Fl and F2satisfyour conditions on F, it is clear that (3)

[Fl(X) + F2(X)]* G(x)

= Fl

* G(x)

+

F2 * G(x).

To establish the continuity of XFG(X),first assume that G(x) is continuous on an interval Ixl ~ 2e. Then G(x - t) is continuous in x and t together when - e ~ x ~ e and - e ~ t ~ e. But F(t) is continuous over the intervalltl ~ e except possibly for a finite number of jumps at points ti' Thus the integral (4)

le(x)

= fe F(t)G(x -

t) dt

can be written as a sum of integrals from -e to tI' tI to t2, etc., whose integrands are continuous in x and t. ConsequentIy le is continuous when

- e ~ x ~ e. Next let G be the unit step function So(x - k) where - 2e < k < 2e. - t) = So(x - t - k) = 1 ift < x - k, and it vanishesift > x-k. If - e ~ t ~ e, then t ~ x - k when e ~ x - k, and so

Then G(x

le(x) = fe F(t)So(X- t - k)dt = fe F(t)dt Since So(X - t

-

k)

= Owhen

t >

- e> x

-

if x ~ k + e.

k, then if x ~ k-e.

Thus IAx) has constant values on each of the half lines x ~ k + e and x ~ k-e. On the remaining interval of the x axis, k - e < x < k + e, we note that both t and x - k lie on the interval (- e,e),and thus e "-k

f

f

(k - e ~ x ~ k + e). le(x) = -e F(t)So(x- k - t)dt = -e F(t)dt This continuous function of x assumes the above constant valuesat the end

-

points k e and k + e of the interval. Therefore le is continuous for all x when G(x) = So(x - k).


[SECo 135

393

Finally, let G be sectionally continuous over the interval (- 2e,2e), with jumpsjl,j2,...,jn at points k1,k2,...,kn (k1 < k2 <...
- kd - ... - jnSo(x- kn)

is continuous on the interval when properly defined at the points x = k¡. Thus G is a linear combination of the continuous function Gnand unit step functions. It follows that the function le represented by the integral (4) is continuous when - e ~ x ~ e if F and G are sectionally continuous over each bounded interval. The continuity ofthe convolution integral (1)for all x now follows from the uniform convergence of that improper integral. We write (5)

X FG(X) =

fe F(t)G(x -

t) dt

+ Re(x)

and the absolute value of the remainder RAx) can be made arbitrarily small, uniformly for all x, by making e sufficiently large. Let x be fixed and its increments .1x be such that l.1xl < b, where b is some positive constant, and select e such that e > Ixl + b. Then IXFG(x + .1x) - XFG(x)1~ Ile(x + .1x) - le(x)1+ IRe(x + .1x)1+ IRe(x)1 and the right-hand member can be made arbitrarily small by first taking e large, then l.1xl small because le is a continuous function. We ha ve now established these properties of F * G(x): Theorem 4 Let twofunetions F and G be bounded and absolutely integrable from - 00 to 00, and seetionally eontinuous over eaeh bounded intervalo Then their eonvolution integral

F * G(x) = J:CX}F(t)G(x

-

t) dt

(-oo
converges absolutely and uniformly to afunetion XFG(x) that is bounded and eontinuous for all real X. Also, F * G(x) has the eommutative and distributive properties (2) and (3). 135

CONVOLUTION THEOREM

We now prove that the product of two transforms is the transform of the convolution of the original functions. Theorem 5 The eonditions on funetions F and G stated in Theorem 4 are sufficient not only for the existenee of the transforms fe(a.)and ge(a.),but

394


SECo 135]

also for the existence of the transform of the convolution (1) X FG(X)

= f~

-

F(t)G(x

00

t) dt

(- 00 < X < (0),

and the transform of that convolution is the product fe(a)ge(a):

( - 00 < a < (0).

(2)

Also, X FG(X)is absolutely integrable from - 00 to oo. The existence and even the continuity of the transformsfe and ge were noted earlier (Sec. 133). Since the functions IFI and IGI satisfy the conditions on F and G in Theorem 4, the convolution IFI * IG(x)1is a continuous function (3)

Q(x)

=

t:

lF(t)IIG(x

-

(

t)1dt

- 00 <

x < (0),

and the integral here converges uniformly with respect to x. With an inversion of order of integration (Sec. 15),we can write rb Q(x) dx =

Jo

f-00

oo IF(t)1 rb IG(x

Jo

-

t)1dx dt.

That integral is bounded for all positive numbers b, since

s: Q(x) dx

= {0000IF(t)1

~

f:'

IG(Y)I dy dt

f~oo IF(t)1dt f~oo IG(Y)Idy.

It is a monotone nondecreasing function of b because Q(x) ~ O. Therefore it it has a limit as b -> oo. Likewise or a~i~ooLO Q(x) dx, exists, and hence the improper

(4)

f~

00

Q(x) dx

f:

00

Q(x) dx,

integral

= f~ f~ 00

00

IF(t)IIG(x

-

t)1dt dx

converges. Since le-iuXFG(x)1 ~ Q(x) for all a, it follows that the Fourier integral (5)


[SECo 135

395

converges absolutely and uniformly with respect to tl. In particular, X FGis absolutely integrable. To prove that the transform (5) is fJ..tl)gJ.tl), we may use the kernelproduct property e-i"'ge(tl) = E,,{G(x - t)}

(6)

(-00 < t < 00),

which is satisfied for each fixed t under our conditions on F and G. Then

fe(tl)gJ.tl) = f~", F(t)e-i"'ge(tl) dt = f:",

F(t)E,,{G(x - t)} dt.

The integral here is the limit as e -+ 00 of the integral

(7)

fe f:", F(t)G(x - t)e-i"X dx dt.

Here the integral with respect to x converges uniformly in t when Itl < e = b to 00 satisfies the conditions

because F is bounded, IF(t)1 < M, and the remainderfrom x

I

L'" F(t)G(x- t)e-iaxdx = L~, F(t)e-¡"'G(y) dy I I

I

~ M L~e IG(Y)Idy. The last member is independent of t and vanishes as b -+ 00; similarly for the remainder from - 00 to x = a as a -+ - oo.

It follows (Sec. 15) that the order of integration can be interchanged to write the iterated integral (7) as f:", q(c,x)dx where q is the function (8)

q(c,x) = e-iax fe F(t)G(x - t) dt.

The integral here, and hence q, is a continuous function of x (Sec.134). We have now shown that (9)

fe(tl)ge(tl)= !~~ f~", q(c,x) dx.

We note that q(c,x)-+ e-i"xXFG(x) as e -+ 00, and that limit is uniform with respect to x because the improper integral XFGconverges uniformly for all x.

396


SECo 136]

Since

Iq(c,x)1~

fe lF(t)G(x -

t)1dt

~ f~oo IF(t)G(x - t)1dt = Q(x) -

and Q is integrable from - 00 to 00, then the integral in Eq. (9) converges uniformly with respect to e (e > O). Our earlier observation [Eq. (5)] that Ea{XFG} exists completes the facts that will enable us to interchange the limit and the integral in formula (9). To prove that lim c-+oo

f

oo -00

q(c,x)dx=

f

oo e-iaxXFG(x)dx,

-00

we write the difference of the two integrals here in the form

f

(10)

-

fa

[e-iaxXFG(X) - q(c,x)]dx

-00

+ foo e-iaxXFG(x)dx

q(c,x)dx+ oo e-iaxXFG(x)dx - ooq(c,x)dx. Jb Jb

The improper integrals here converge absolutely; those involving q converge uniformly with respect to c. Thus their absolute values can be made small, independendy of e, by taking a large and negative and b large and positive. Then by taking e large and positive, the absolute value of the first integral can be made small beca use the integrand tends to zero uniformly

in x as e ~ oo. This completesthe proof of Theorem 5 since it followsthat fe(a)ge(a) = f:oo e-iaxXFG(x)dx = Ea{XFG(x)}. The convolution property here was illustrated in Probo 4, Seco110. In Probs.9 and 10,Seco110, we noted and illustrated a generalized convolution property for Ea, one that represents the iterated transform h(a) of a function H(x,y) of two variables as the transform of the function Loo:(X

136

- t, t)dt.

TABLES OF TRANSFORMS

Appendix e consists of a table that summarizes properties of Ea and lists exponential Fourier transforms of several functions. Tables listed in the Bibliography under the names Erdélyi (vol. 1) and Oberhettinger give exponential Fourier transforms of some other functions.

[SECo 137


397

AIso, chap. 5 of W. Kaplan's "Operational Methods for Linear Systems," 1962, presents short tables of transforms of functions and generalized func- . tions and a few additional properties of Ea' Tables of Laplace transforms furnish a source of exponential Fourier transforms. In correspondence with the Laplace transform of a particular function, we can list a certain exponential Fourier transformo For if F(t) is defined for all real t and has a Laplace transform L{ F(t)} = f(s) when s = y + irx,then f(y + irx) = f" e-(y+ia)IF(t)dt

= 5:00 e - ¡alSo(t)e - ylF(t) / dt.

Thus, in terms of the Laplace transformf(s)

of F(t),

Ea{So(x)e-YXF(x)} = f(y + irx). 137

BOUNDARY

VALUE PROBLEMS

Problems in differential equations that are adapted to solution by means of the exponen tial Fourier transformation are singular in that the variable, called x here, with respect to which the transformation is made, ranges from - 00 to oo. But they are singular boundary value problems with rather severe boundary conditions since the formula for the transform of a derivative with

respect to x assumes that the function itself vanishes as x ~ :t oo. That condition is more severe than the condition of exponential order as t ~ 00 used in initial-value problems treated by the Laplace transformation. Conditions of integrability with respect to x from - 00 to 00 further restrict the variety of problems to which Ea applies. Of course, formal procedures may yield verifiable solutions of some problems in which the implied conditions are not satisfied. But the variety of problems that can be solved by using Ea is small compared with the varieties that are adapted to solution by finite Fourier or Laplace transformations. Here and in the set of problems to follow we give some examples of problems that can be solved readily by using Ea. Example 1 Let F(x) be such that F' is continuous and F itself is absolutely integrable, from - 00 to 00, and such that F(:t 00) = O. To find a solution of the boundary value problem

(1)

Wx(x,y) + Jv,,(x,y) + 2yW(x,y) = F(x) W(x,O) = O,

(- 00 < x < 00,y> O),

W(:t oo,y) = O,

398


SECo 137]

we let we(oc,y)denote the exponential Fourier transform of W with respecttoxandwritefe(ex) = Ea{F(x)}. Then,formally,Ea{Wx} = iexwe, and the transform of problem (1) becomes

d

.

(2)

(iex+ 2y)we(ex,y) + dy we(ex,y)= fe(ex) We(ex,O) =

(y > O),

O.

The solution of problem (2)in linear ordinary differentialequations of first order is (3)

I: !e(ex)eia(t-y)exp (t2)dt.

we(ex,y)= exp(-y2)

According to property (1), Seco133, the inverse transform offe(ex)ei'x(ty) is F(x + t - y). By formally inverting the order of the operators Ea-1 and integration with respect to t, we can write (4)

W(x y) = exp ( - y2)

I: F(x -

y + t) exp (t2) dt.

It is not difficultto verifythat the function (4)satisfiesall conditions in problem (1). Example 2 To obtain a formula for the bounded harmonic function V(x,y) in the halfplane y > Othat assumes values F(x) on the boundary y = O,we solve this problem: (5)

Vxix,y) + Vyy(x,y)= O

(- 00 < x < 00, y > O),

V(x, +0) = F(x), lV(x, y)1 < M, where M is some constant and, over the entire x axis, the function F is bounded and continuous except possibly for a finite number of finite jumps. To solve the problem using the transformation Ea with respect to x, we shall assume some auxiliar y conditions on F, conditions associated with our method of solution but not necessarily required in order that the resulting formula for V(x,y) will be valido We assume that F(x) is such that its transform f..(ex)exists and that the problem will have a solution V(x,y) whose transform vJex,y)with respect to x exists and is bounded when y > O for each fixed ex. Further conditions on V are needed to write Ea{Vxx} = -rx2ve and to justify steps in transforming problem (5) into the problem 2

(6)

-

d2

+ dy2ve(rx,y)= O ve(rx,O)= fe(rx), Ive(rx,y)1< N(rx),

where N is independent

ex vAex,y)

of y.

(y > O),


[SECo 137

399

The solution of problem (6) is

(7)

V.,(a,y) = f.,(a.) exp ( - yla.l)

We found earlier that exp ( - yla.1)=

E", !

2

Y

when y > O.

2

+y } Thus our convolution property indicates that 1 '"' yF(t) (8) V(x,y) = -1t -'" (X - t)2 + y 2dt

{ 1tX

(y > O).

J

That formal solution of problem (5) is known as the Poisson integral formula for the half plane y > O,or the Schwarz integralformula. It can be fully verified as a solution under the conditions on F first stated with problem (5), without the auxiliary conditions.l If F(x) is a constant Fo, for instance, it satisfies the conditions stated in problem (5), butJ.,(a.) does not exist unless Fo = O. In that case formula (8) reduces to '" Fo 1t 1t F. t - x

(9)

V(x,y) =

:

arctany

]-'" = -; 2 + 2 = Fo, (

)

i I

which is c1earlya solution of problem (5). In case F(x) = So(x), formula (8) becomes 1 '" X y dt 1 1t + arctan 2 (10) V(x,y) = -1t o (t-x )2 +y 2 = -1t( _ Y)

i

.

This function is bounded and harmonic in the half plane y > O; it vanishes as y -+ +0 if x < O,and V(x,y) -+ 1 as y -+ +0 if x> O. PROBLEMS 1. Let F(x) be sectionally continuous over an interval -c < x < c, while F(x) = O when x < -c and when x > c. Use the exponential Fourier transformation to find a solution of the differential equation Y"(x)

-

Y(x)

+ 2F(x) = O

(-00 < x < (0)

such that Y' is everywhere continuous and Y(i: (0) = Y'( i: (0) = O,in the form Y(x) = e-X fe e'F(t)dt + e" [ e-'F(t)dt. Verify that solution. Note that the first integral here vanishes when x < -c and the second one vanishes when x > c, since F(t) = O when Itl > c. 1

For a full verification see the author's "Complex Variables and Applications," 2d ed., seco

l.

108, 1960.

I

q

400

2.


SECo 137]

Use E. and the inversion formula to solve the problem

=O

( - 00 < x < 00, y > O),

lV(x,y)1 < My

when y > O,

Vxx(x,y)+ Yy,(x,y) + x exp ( V(x,O) = O,

X2)

where M is some constant, in the form V(x y)

=-

1

.

f

(2

(1 - e-.y)sm ax 2¡-ic o a exp ( -"4 ) da.

3. Let h(y) be continuous when y ;¡; O,and let F(x) have a continuous derivative F' for all x and vanish as x --+i: oo. Derive the solution of the problem ( - 00 < x < 00, y > O),

Wx(x,y) + J-Y,(x,y)+ h(y)W(x,y) = O

W(x,O)= F(x),

W(i: oo,y)= O,

in the form

W(x,y) = F(x

-

y) exp

[- I:

h(t) dtJ.

Verify that solution. 4. The transverse displacements Y(x,t) in a string stretched along the entire x axis satisfy the conditions ( - 00 < x < 00, t > O), Y(x,O) = F(x),

Y,(x,O) = O;

also F(x) --+O and Y(x,t) --+O as x --+i: oo. Transform with respect to x to derive the formula

Y(x,t) = t[F(x - at) + F(x + at)]. 5. Transformwithrespectto x to solvethisproblemformallyfortemperaturesU(x,t): U,(x,t) = kU xx(x,t)

(- 00 < x < 00, t > O),

U(x,O)= F(x).

Ans. U(x,t)=

~

2.jidZi

f-

F(~) exp

1 =

[-

(x

- ~)2 d~ 4kt J

O>

f

j1r - F(x

+ 2r.jki) exp(- r2)dr.

13

Fourier Transforms on the Half Line

Fourier sine and cosine transforms of functions defined over the half line x > O were introduced in Chap. 10 as examples of general integral transforms. They transform the differential form F" in terms of the transform of F itself and the initial values F(O)and F'(O),respectively. Operational properties of the two transformations are noted here in greater detail and summarized in Appendix D; some applications are given. AIso, we present a modified transformation that resolves F" in terms of a boundary value F'(O) - hF(O). 138

FOURIER SINE TRANSFORMS f.((1,)

In Seco116 we derived a linear integral transformation that transforms F"(x) in terms of the boundary value F(O)and the transform of F when F is defined over the positive x axis. It is the Fourier sine transformation

(1)

Sa{F(x)}=

La:> F(x)

sin exxdx

= f.(ex)

(ex> O), 401

402


SECo 138]

whose basic operational property is (2)

Ser{F"(x)}

= _(1.21.((1.) + (1.F(O).

The inverse transformation Ser-l{J.} can be expressed in terms ofthe transformation itself: 2 (3)

F(x) = -

i

OO

non

2 1.((1.)sin (1.Xd(1.= -S,,{f.(a)}

(x > O).

The transformf.(a) exists as a continuous function of the parameter ex if F is sectionalIy continuous over each bounded intervalO < x < e and absolutely integrable from O to oo. Our conditions are sufficient, but not necessary for the existence ofJ.(ex);for example, Ser{1/x}= n/2 when a > O. Under those conditions it folIows from the Riemann-Lebesgue theorem (Prob. 8, Seco65) that

lim f.(a) = o.

(4)

er-+oo

The operational property (2) is valid if F" is sectionalIy continuous over each bounded interval while F and F' are continuous when x ~ Oand vanish as x -+ 00, and if F itself is absolutely integrable from Oto oo. The condition that F be absolutely integrable can be replaced by the condition that J.(a) exists for all positive a because integration by parts shows that, whenever e> O, (5)

s: F"(x) sin ax dx

= -a2

s: F(x)

sin ax dx

+ [F'(x)sinax - aF(x)cosax]ó. As e -+ 00, the bracketed term tends to (1.F(O);thus the integral on the left has the limit Ser{F"}when the integral on the right has a limit. By replacing F by F" in Eq. (5) and integrating by parts, we find that (6) under these conditions : F and its derivatives up to F'" are continuous when x ~ Oand vanish as x -+ 00, F(4)is sectionalIy continuous on each bounded interval, and J.(ex)exists; similarly for transforms of further iterates of the operator d2/dx2. Sufficient conditions for the inversion formula (3), the Fourier sine integral formula, are that F be absolutely integrable from O to 00 and that F' be sectionalIy continuous on each bounded interval. Then the formula represents F(x) where F is continuous, and the mean value of F(x + O)and F(x - O)where F has a jump.

FOURIER TRANSFORMS ON THE HALF LlNE

[SECo 138

403

The sine transformation is a special case of the exponential Fourier transformation. For if F1 denotes the odd extension of F to the entire x axis, that is,

ifx>O,

F1(x) = F(x)

for all x,

then as noted in Seco133, Eer{Fl(X)}= -2i {"> Fl(X) sin ocxdx.

Therefore (7) Some properties of Serfollow from properties of Eer. In particular, according to Theorem 3, Seco132, the inversion formula (3) is valid iff~(oc)is sectionally continuous on each bounded interval, if!s is absolutely integrable from O to 00, and if Sx{fs(oc)}exists. Property (2) is modified in case F and F' have jumps b and b', respectively, at a point x = c: (8)

Ser {F"(x)}

= - oc2fs(OC) + ocF(O) +

ocb cosocc

-

b' sin occ

(c > O).

From property (2) and the inversion formula we find that

x

oc Ser{e-X}= 1 + OC2'

1t Ser{ 1 + X2} = "le-er.

From property (8) we find that the function G(x)

=c-

x

if O ~ x ~ c,

G(x) = O

ifx;¡;; c,

has this transform : Ser {G(x)}

=

oc- 2(OCC-

sin occ).

Under our conditions for the existence of fs(oc)we find that (9)

if k > O.

Ir in addition x2F(x) is absolutely integrable from Oto 00,we can differentiate integral (1) with respect to (Xto see that (10) It is convenient to introduce the cosine transformation on the half line before taking up further properties of Ser.

404

139


SECo 139]

FOURIER COSINE TRANSFORMS 'c(a.)

In Probo 11, Seco116, we noted that the integral transformation that resolves F" on the half line x > O in terms of the transform of F and the boundary value F'(O)is the Fourier cosine transformation (1)

Ca{F(x)}

=

100

F(x) cos ax dx

= fc(a)

(a > O).

Its basic operational property is (2) The Fourier cosine integral formula furnishes an inverse transformation Ca-1 in terms of CAfc(a)}: 2 00 2 (x > O). (3) F(x) = fc(a)cos ax da = -Cx{fc(a)} 1t 5.o

1t

Sufficient conditions for the existence offc(a), or for the validity of the inversion formula (3) or the operational properties of Ca, are those used for the corresponding formulas for Sa in the preceding section. We find, corresponding to property (6), Seco 138, that (4) Property (2) can be extended easily to formulas for the transform of higher derivatives of F of even order. In case F and F' are continuous except for jumps b and b', respectively, at a point x = c, we find that property (2) is modified as follows: (5)

Ca{F"(x)} = -a2fc(a) - F'(O)

-

ab sin ac

-

b' cos ac

(c > O).

We find that, corresponding to other properties of Sa, (6) (7) (8)

if k > O, fc(ak) = ~Ca{F(I)} f;(a) = Ca{ -x2F(x)}, lim fc(a) = O.

The relation between Ca and the exponential Fourier transformation (Sec. 133) can be written (9) where F2 is the even extension of F over the entire x axis. Conditions of validity of the kernel-product property (10)

2fc(a)cos ak = Ca{F(x + k) + F(lx - kl)}

(k > O)

FOURIER TRANSFORMS ON THE HALF UNE

[SECo 140

405

were noted in Probo 5, Seco 110. The corresponding convolution property for Ca will be established in Seco 141. 140

FURTHER PROPERTIES OF S. AND C.

Some useful operational properties of the sine and cosine transformations involve the two transformations jointly. If F is continuous when x ~ O and F((0) = O,if F' is sectionallycontinuous on each bounded interval, and if either F or F' is absolutely integrable from O to oo~then upon integrating by parts between limits O and e and letting e tend to infinity, we find that (1)

Sa{F'(x)} = -ocCa{F(x)},

(2)

Ca{F'(x)} = ocSa{F(x)} - F(O).

If G is sectionally continuous and absolutely integrable from O to 00, then the function

(x ~ O)

F(x) = {") G(r)dr satisfies our conditions since F'(x) = (1) can be written

-

G(x) and F( (0)

= O. Thus

property

(3) By differentiating the integrals Sa{F} and Ca{F} with respect to oc,we find that (4)

/;(oc) = Ca{xF(x)},

(5)

/;((1.) = Sa{ -xF(x)},

assuming the sectional continuity of F and the absolute integrability of xF(x) from O to oo. Let Fl and F2 be the odd and even extensions of F: X Fl (x)

= IxlF(lxl),

F2(X) = F(lxl),

when -00 < x < oo.

Assuming the sectional continuity and absolute integrability of F, we find these properties corresponding to the kernel-product property (10) of the preceding section: 2fs(oc)sinock = Ca{F1(x + k) - F1(x - k)}, (6) 2fs(oc)cos ock = Sa{F1(x + k) + Fl(X - k)}, 2fc(oc)sin ock = Sa{F2(x

where k is a real-valued constant.

-

k)

- F2(x + k)},

/

406

SECo 141)

OPERATIONAL

MATHEMATICS

The existence of the trbnsforms!s and h for all positive a, and the natural agreement that fs(-;a) = - fs(a) and fc(-a) = fc(a), enables us to write the relations (7)

fs(a

+ k) + fs(a - k) = Scz{2F(x)cos kx},

fs(a + k) - fs(a - k) = Ccz{2F(x)sin kx}, (8)

fc(a + k) + fc(a fc(a

141

-

k)

-

-

k)

= Ccz{2F(x)cos kx},

fc(a + k) = Scz{2F(x) sin kx}.

CONVOLUTION PROPERTIES

Let F and G be two bounded functions that are sectionally continuous on each intervalO < x < e and absolutely integrable from O to oo. Then

where F2(x) = F(lxl) and G2(x) = G(lxl) when Theorem 5, Seco 135, the product

-

00

< x < oo. According to

is the exponential Fourier transform of the convolution

That integral is easily written in the form

which is an even function of x, so its exponential transform is twice its cosine transformo It follows that Cczhas the convolution property (1) 2fc(a)gc(a) = Ccz{L'"F(r)[G(x

+ r) + G(lx - rl)]dr}.

Under the same conditions on F and G, and in the same manner, we can establish these joint properties of sine and cosine transforms: (2)

2fs(a)g.(a) = Ccz{L'" F(r)[G(x

+ r) - Gt(x - r)]dr} ,

FOURIER TRANSFORMS ON THE HALF LINE

[SECo 142

(3) 2.fs(IX)gc(lX) = S..{foa>F(r)[G(lx - rl) - G(X

= S..{foa>G(t)[F(x

407

+ r)] dr}

+ t) + F1(x -

t)]dt}

where F1 and G1 are the odd extensions of F and G. To establish a convolution property for the sine transformation alone, we impose an additional condition on the function G, namely, that its integral (x

H(x) = la> G(t) dt

~ O)

be absolutely integrable from x = O to x = oo. Now H is continuous when x ~ O,and H(oo) = O. According to property (3),Seco140,then gs(lX)= IXC..{H(x)} = IXhc(lX) and it follows from formula (3) that 2f.(IX)gs(lX)= 21Xf.(IX)hc(lX)

= IXs..{foa>F(r)[H(lx Thus we have the convolution Seco110:

property that was obtained formally in

- f.(IX)gs(lX)= S..

{f.o

IX

142

x+r

a>

2

(4)

- rl) - H(x + r)] dr}

F(r)

i

G(t) dt dr Ix-rl

}.

TABLES OF SINE AND COSINE TRANSFORMS

Appendix D consists of tables that summarize properties of Fourier sine and cosine transformations on the half line and list transforms of particular functions. Some transforms can be found by evaluating integrals with the aid of Laplace transforms (Chap. 3). The results shown in Probs. 14 and 15,Seco35, for example, can be written as (1)

(2)

s..{c ~ x} = (~-

C..L ~ A

SiIXC)COSIXC + CilXCsinlXc,

= (~ - Si IXC)sin IXC- Ci IXCcos IXC,

--

408

OPERATIONAL

SECo 143]

MATHEMATICS

where e > O. The evaluation of improper integrals by using contour integrals and residue theory, illustrated in seco72 ofthe author's "Complex Variables and Applications," 2d ed., 1960, is another method of deriving some of the transforms listed in Appendix D. Other transforms can be found by direct integration, by using operational properties of Sa.and Ca.,or by differentiating or integrating a known transform with respect to a parameter. Extensive tables of sine and cosine transforms will be found in vol. 1 of the "Tables of Integral Transforms" edited by A. Erdélyi. Those tables, listed in tbe Bibliography, give transforms of several hundred special functions. The book by Ditkin and Prudnikov, also listed in the Bibliography, includes extensive tables of those transforms. 143

STEADY TEMPERATURES IN A QUADRANT

Let V(x,y) denote steady temperatures in a solid that fills the space x > O, Y > O, - 00 < Z < 00 (Fig. 108). Ir the face x = O is insulated, if V(x,O)= F(x), and if a steady source ofheat varying only with x is distributed throughout the solid, then (1)

Vxx(x,y) + v,y(x,y) + H(x) = O

(2)

J-:(O,y)= O,

(x> O,Y > O),

V(x,O) = F(x);

also, Vis to satisfy boundness conditions as x -+ 00 or y -+ oo. To help keep temperatures steady, we assume that the total rate of generation of heat throughout the solid is zero, that is,

(3)

{' H(x) dx = o.

Let us assume also that whenever x ~ O, the integral

(4)

P(x) = {>OI H(t) dt dr

exists and that P has a cosine transformo Since P"(x) = - H(x) and P'(O) = O, the cosine transforms of P and H satisfy the relation épc(a)

(5) y ...-;

-Hb:)

O

V=F(x)

x

Fig.108

= h.(a).


[SECo 143

409

Our problem in Vis adapted to the Fourier cosine transformation with respect to x on the half line x > O, in view of the differential forro Vxx in Eq. (1) and the prescribed boundary value Vx(O,y). Let vc(ex,y)denote the transform Ca{V} where the transformation is made with respect to x, and writefc(ex)= Ca{F(x)}. Then, formally, (6)

d2 - ex2viex,y) + dy2 vc(ex,y)+ hc(ex)= O,

where y > O. The bounded solution of problem (6) is

vc(ex,y)= :2hiex) + [fc(ex)

-

:2hc(ex~

e-ay

or, in view of Eq. (5), (ex> O,Y > O).

(7) From Appendix D, Table D.2, we see that e-ay = C

~

y

a { 7t X2

+

y2 }

and, according to the convolution property (1), Seco 141, it follows fram Eq. (7) that y

(8) V(x,y) = P(x) + -

ex>

f

7t O

[

[F(r)- P(r)J

(x+r

1 )

2

1 +y

2+

2

(x-r

) +y

]

2 dr.

We note some special cases. First, if F(x) = P(x), then V is a function of x alone: (9)

V(x,y) = P(x) = {ex>L H(t) dt dr,

a result that is easily verified as a solution. In case no source is present, then H(x) = O,and I

(10)

1

fex>

[

Y

Y

]

V(x,y)= ; Jo F(r) (x + rf + y2 + (x - r)2+ y2 dr

when x ~ Oand y > O. That is an integralformula for the harmonic function V in the quadrant such that Vx(O,y)= O and V(x,+O) = F(x). In this case where H(x) = O, the problem can be solved also by using the sine transformation with respect to y. When H(x) = Oand F(x) = 1,wefindfromformula(10)that V(x,y) = 1, and this is cIearIy a solution of the problem even thoughfc(x) does not exist.

11

410

SECo 144] ln

X

cas.::

H(x)

OPERATIONAL MATHEMATICS =

D, and

.F(x)

~

1

W'h.::n

O <' x

<' e

W'hil.:: F(x)

-

O

ovhen

> e, formula (10)gives the verifiablesolution 1 e + x eV(x,y) = - arctan+ arctann( y y

144

X

. )

DEFLECTIONSIN AN ELASTIC PLATE

As noted in Seco 118, the sta tic deftections or transverse displacements Z(x,y) in a thin elastic plate with no transverse load satisfy the biharmonic equation V2(V2Z)

= V4Z =

°

where V2Z = Zxx + ZYY' The bending moment about an edge x = e is proportional to Zxx(c,y), so if that edge is simply supported, then Z(c,y) = ° and Zxx(c,y) = 0, and similarly for the other edges.

Let the plate have the form of a semi-infinitestrip x ;¡;;0, ° ;;;¡; y ;;;¡; n. H there is no load and the edgesy = ° and y = n are simplysupported, whilethe edge x = ° has a deftectionbsiny but no bendingmomentis applied about the axis x = 0, then deftections Z(x,y) satisfythe conditions 04Z 04Z -+2-+-=0 OX4 ox20y2

(1) (2)

Z(O,y) = b sin y,

(3)

04Z oy4

(x > 0, ° < y < n),

Zxx(O,y) = 0,

Z(x,O) = Zyy(x,O)= Z(x,n) = Zyy(x,n) = O;

also Z and Zx should tend to zero as x -+ oo. The differential operators with respect to x in this problem are 02/0X2 and its iterate 04/0X4, on the half line x > 0, and the prescribed values at the boundary

x

=°

are the values of Z and 02Z/0X2.

formation with respect to x may apply. We write zs(et,y) = S",{Z(x,y)};

then, formally,

Thus the sine trans-


411

[SEC. 144

Then our boundary value problem in Z transforms into the following

problem in z.(a,y): d4z.

(4)

2d2z.

4

3'

b

dy4 - 2a dy2 + a z. = (2a + a ) SinY

(O< Y < n, a >

d2z

=-1=0 . dy

when y

z

= Oand

when y

O)

,

= n.

We find that A sin y is a particular solution of the differential equation here if (1 + (2)2A = b(2a + (3), and that particular solution satisfiesthe conditions at the edges y = Oand y = n. Thus (5) Referring to Appendix D, Table D.1, we see that

Z(x,y) = b(l + tx)e-X sin y.

(6)

That function does satisfy conditions (1), (2), and (3) and it vanishes, together with its derivatives, as x ~ oo. Such problems in Z are also adapted to the finite Fourier transformation Snwith respect to y, or to successive transformations S. with respect to x and Sn with respect to y. PROBLEMS 1. 2.

Derive properties (6), Seco140. For the function

= c - x when O~ x ~ c, (1 - cos !XC)a-2.

G(x)

G(x) prove that C.{ G(x)} =

=O

when x ~ c,

3. The static transverse displacements Y(x) in a string stretched upon an elastic support along the positive x axis, with its end x = Ofree to slide along the Y axis and with a uniform force over the span O < x O),

Y(
where F(x) = A when O < x b, A, b, and h being positive constants. Also, Yand Y' are to be continuous when x ~ O. Show formally that A sin!Xb

C.{ Y(x)}

= - h2 ( ---¡;--

!X

h2

.

+ !X2 SIn!Xb

)

412


SECo 144]

and hence that

-h2y(x)

={

when O ;§; x ;§; b,

A(1 - e-bh cosh hx)

when x ~ b.

Ae-h" sinh bh

Verify that solution. 4.

Use a transformation with respect to x in the boundary value problem (t > O,x > O),

Y,,(x,t) = Yxx(x,t)

Y.,{O,t)= -1,

Y( oo,t) = O,

Y(x,O) = y,(x,O) = O,

to derive the solution t

-

x

when O ;§; x ;§; t,

Y(x,t) = { O

whenx ~ t.

5. Use a transformation with respect to x to find the solution of the boundary value problem (t > O,x > O), Y(x,O) = «II(x),

Y,(x,O) = O,

Y(O,t)= O,

Y(oo,t) = O,

in the form (8), Seco43. Also, use superposition of the solutions of this problem and Probo 6, Seco118, to write the solution when the initial conditions are Y(x,O) = «II(x),

6.

(x > O).

Y,(x,O) = G(x)

Find the solution of the problem (t > O,x > O),

Y,,(x,t) = Y",,(x,t)+ 2F(x) Y(x,O) = Y,(x,O)= Y(O,t)= O, in terms of the odd extension Flor F, in the form

7. (a) Transform Laplace's equation Yx" + Y,p= O with respect to x to derive the integral formula y y 1 f'" dr V(x,y) = -;)0 F(r) (r - X)2 + y2 - (r + X)2 + y2

[

]

for the harmonic function in the quadrant x ~ O, Y > O, such that V(O,y)= O and V(x,+O) = F(x). (b) In case F(x) = 1 whenever x> O,then, even thoughf.(cx)does not exist, show that the formula in part (a) gives the valid solution

V(x,y) = ~ 1tarctan ~ y = 1- ~ 1t1m [Log (x + iy»).


[SECo 144

413

(e) Verify the solution given by the integral formula in part (a) when F(x) = 1 if O < x < 1, 8.

and

F(x) = O

if x > 1.

The steady temperatures W(r,x) in a semi-infinite cylinder satisfy the conditions 1 J-Y,,(r,x)+ -r J-Y,(r,x)+ "Yxx(r,x)= O (O< r < 1, x > O),

W(l,x) = O,

W(r,+O) = 1,

and W is to be continuous when O ~ r ~ 1 and x > O. Use a transformation with respect to x to derive the formula 2 "> sin (XX 2 ">Jo(iar) sin ax d a = 1 -da. ( Wrx)=1-- JO(iar) -, 1t o J o(ia)] a 1t o Jo(ia) a

i[

9.

i

Temperatures in a semi-infinite solid satisfy the conditions U,(x,t) = Uxx(x,t) + H(x,t) U(x,O) = F(x),

(t > O,x > O),

U(O,t) = G(t).

Derive formalIy the solution 2 "> U(x,t) = u.(a,t)sin (XXda 1t o where

i

u.(a,t)exp (a2t) = f.(a) + {[aG(-r) + h.(a,t)) exp (a2T)dT, f. and h. being sine transforms of F and H with respect to x. 10. (a) When G(t) = H(x,t) = Oin Probo9, obtain the formulafor U(x,t)in the form given in Probo lO, Seco47. (b) When F(x) = H(x,t) = O in Probo 9, obtain the formula for U(x,t) in the form derived in Seco47.

:

i

I l'

11. The static deflections Z(x,y) in a thin elastic plate in the form of a quadrant satisfy the conditions 04Z 04Z 04Z (x > O,Y > O), OX4 + 2ox2oy2+ oy4 = O Z(O,y) = Zxx(O,y)= O bx Z(x,O) = 1 + x2'

Zyy(x,O) = O

(y > O), (x > O);

I

also Z and its derivatives are to vanish as x -> 00 and y -> oo. Derive the formulas b ("> Z(x,y) = '2Jo (2 + ay) exp [- (1 + y)et)sin ax da

I

} .

bx + bx 1+ y . - X2 + (1 + y)2 y [X2+ (1 + yf)2 The first form is helpful in verifying the solution.

I I I I

414

12.


SECo 145]

In Probo 11 let the conditions on the edge y

= O be replaced

by the conditions

Z(x,O)= O,

(x > O),

then obtain the solution in the forros b

Z(x,y) =

""

-: f.

o exp [-(1 + y)cx]sin cxxdcx

- b xy - -¡ x2+ (1+ y)2' 145

A MODIFIED FOURIER TRANSFORMATION

We follow the procedure

T,.

used in Chap. 10 to design a transformation

T {F(x)} =

1X>

F(x)cf)(x,A)dx

= f(A)

of functions F on the half line x > O that resolves F" in terms of f(A) and the boundary value F'(O) hF(O),where the constant h is po sitive. Suppose that F is of class C" and absolutely integrable over the half line x ~ Oand that F(oo) = F'(oo) = O. If for each fixed ..1.the kernel <1> and its derivative <1>' with respect to x are bounded and if <1> is of class C", we can integrate by parts from Oto e and let e tend to infinity to write

-

T{F"} =

1X> F"dx

i

+

-

[F'

F']O'

F'(O)

oo"

=

o F(x) (x,A) dx

-

/

-

<1>'(0,..1.) - h(O)

Thus if <1>andA satisfy the singular eigenvalue (1)

"(x) = A(x)

where <1>and <1>'are bounded,

hF(O)

when x > O,

F(O) <1>(0,..1.) I .

problem

- h(O,A)= O, <1>'(0,..1.)

then

T{ F"(x)} = .lf(A) - (O)[F'(O) - hF(O»). A solution of that eigenvalue problem positive, and (x) = ..(x) where

is ..1.= _0(2, where O(is real and

(2)

+

,,(x) = O(COS o(x

h sin
(O(> O).

Our transformation T is therefore the modified Fourier transformation T..: (3)

(O(> O),


[SECo 145

415


1;,{F"(X)}= -a2fy{a) - a[F'(O)- hF(O)].

From formulas (2) and (3) we see that 1'..can be expressed in terms of Fourier cosine and sine transformations: (5)

1;,{F(x)} = aC«{F(x)} + hS«{F(x)};

that is,

fT(a)

=

+

afc(a)

hJ.(a).

Thus particular transforms and some properties of 1;,can be obtained from tables and properties of C« and S«; in fact, property (4) can be verified by writing the transformations there in terms of sine and cosine transformations. If we write (6)

a O(a) = arctan

11:

h

where

0< O(a)< 2'

then 41«= Jh2 + a2 sin (ax + O),and 1;,takes the form (7)

1;,{F(x)} = .Jh2 + 11.2100F(x) sin [ax + 0(11.)] dx. In Seco 140 we noted conditions

on F under which

aC«{F(x)} = -S«{F'(x)}. Under those conditions form (5) for 1;,can be written (8)

1;,{F(x)} = fT(a) = S«{hF(x) - F'(x)}.

Formally then we can invert the sine transformation to write (9)

hF(x)

2

-

F'(x) = -

oo

f

11: o

fT(a)sin ax da

(x > O).

An inverse of the linear differential operation hF(x)

(10)

-

F'(x) = P(x)

is found by solving the differential equation here for F. The particular inverse that is bounded when P is bounded can be written

(11)

F(x) = ehx

r

Jx

e-hrP(r) dr = (00e-htp(x + t) dt. Jo

When P(x) is the right-hand member of Eq. (9), then 2

F(x)

=-

oo

f

11:o

oo

e-ht

fo fT(a)sina(x + t)dadt.

416

I f


SECo 146]

We invert the order of integration here and use the known Laplace transform of sin a(x + t) with respect to t, with parameter h, to write this formula for the inversetransformation T,.-l{JT(a)} : (12)

F(x)

=~ nJo

flX)

fT(a) a cos ~~ + a. h~sinax da. (x > O).

The inversion formula (12) is valid when F satisfies our conditions (Sec. 138) under which F(x) is represented by its Fourier sine integral formula. 1 Note that if h = O,formula (12)reducesto the inversionformula for the Fourier cosine transformation. 146

CONVOLUTION FOR r.

The following development of kernel-product and convolution formulas for our modified Fourier transformation is formal. The formulas are more complicated than those for the sine and cosine transformations, so they are apt to be troublesome to apply. We illustrate in the following section how our expression (8), Seco145, for T,.in terms of S" may be used to circumvent our convolution formula for T,.. The expression just referred to can be written if

(1)

P(x)

= hF(x) -

F'(x),

wherefT(a) = T,.{F(x)}. Now P(x) = -Q'(x) if (2)

Q(x) = F(x) + h {IX)F(r) dr

and, since S,,{-Q'} = aC,,{Q}, we have this alternate representation T,.{F} :

of

(3) To obtain a kernel-product formula, we write ~fT(a)~ik) = 2aqia) cos ak + 2hqJa) sin ak a 1 The proof, with the aid of Fourier

sine and cosine integral

formulas,

a somewhat more general integral formula. See R. V. Churchill, Formulas, Mich. Math. Jour., vol. 2, pp. 133-139, 1953-54.

is not difficult even for

Generalized

Fourier

Integral

I j


417

[SECo 146

where k > O,and use the kernel-product properties (AppendixD) 2qc(a.)cosa.k =

Ccx{Q(lx

-

kl) + Q(x

+ k)},

2qc(a.)sin a.k = Scx{QUx- kl) - Q(x + k)}, to write 2 (4) -fT(a.)ik) = J:{Q(lx- kl)} + a.Ccx{Q(x + k)} - hScx{Q(x+ k)}. a. In formulas (1) and the definition offT we can replace h by

-

h and F

by Q.Thus a.qc(a.)- hqs(a.) = -Scx{hQ(x) + Q'(x)}. Ir we replace h by - h in the definition (2) of Q and write

= F(x) -

R(x)

h L~ F(r) dr,

we find that hQ(x) + Q'(x) = F'(x) + h2 L~ F(r) dr = - hR(x) + R'(x).

I

Then, in view of Eqs. (1),

. L

and the right-hand member of Eq. (4) becomes

-

kl) + R(x + k)} =

J:{PF(x,k)}

where PF(x,k)= FUx- kl) + h

F(r)dr + F(x + k) - h

oo JIx-kl

f

oo x+k

F(r) dr.

Qur kernel-product formula is therefore (5)

(k > O),

where X+k

(6)

PF(x,k) = F(lx

-

kl) + F(x + k) + h

JIx-kl F(t)dt.

The corresponding convolution (Sec. 109) for J: is

(7)

X FG(X) = 100 G(r)PF(x,r) dr =

[G(r)

Jo

I

I ¡

-Scx{hQ(x) + Q'(x)} = J:{R(x)},

J:{Q(lx

(

[

F(lx

-

rl) + F(x + r) + h

x+r F(t) dt JIx-rl ]

dr,

418


SECo 147]

and if gT(a) = T.,{G(X)}, the convolution formula is (8)

147

SURFACE HEAT TRANSFER

Let V(x,y) denote

steady temperatures

in a quadrant

x ~ O,Y ~ O,

- 00 < Z < 00,whose two facesare subject to the linear law of surfaceheat

transfer. Ifthe medium outsidethe face x outside

(1)

the face y

= O is at

a temperature

= Ois at temperature

that varies with x, then

Vxx(x,y)+ y"y(x,y)= O V",(O,y)= hV(O,y),

zero and that

(x > O,Y > O),

y,,(x,O)= kV(x,O) - F(x);

also, V and F are to satisfy appropriate condítions of regularity. The constants h and k are positive. Our transformation T.,with respect to x applies here in view of the differential form Vxx,on the halfline x > Oand a prescribed value of V", hV at the boundary x = O. If vT(a,y)= T.,{V(x,y)}and fT(a) = T.,{F(x)},then problem (1) transforms into the problem

-

d2 -a2vT(a,y)

(y > O),

+ dy2VT(a,y) = O

(2) dVT

-

dy

= kv -

when y = O.

fT(a)

y

The bounded solution of this problem can be written as (3)

VT(a,y) = fT(a) e-ay = fT(a.)éY k+a.

f

oo e-(k+a)t y

d7:.

According to our inversion formula (12), Seco145, then (4)

V(x,y) =

2 roo a.cos ax + h sina.x ; Jo fT(a)e-ay (k + a.)(h2+ a2) da.

That form of the solution of problem (1) can be verified. To present another form ofthe solution, let us express the transforms in terms of sine transforms. We write (5)

U(x,y) = hV(x,y) - Vx(x,y).

Then according to formula (1), Seco146, (6)


[SECo 147

419

SimilarIy,

if P(X)

(7)

= hF(x) -

F'(x).

We use formula (11), Seco145, to recover V and F ftom U and P: V(x,y)

(8)

=

{X' e-h~U(x

+ ~,y) d~,

F(x) = f" e-h~p(x + ~)d~ The second representation of Vr in Equation (3) can now be written in the form (9) Since 2 t - 1t a{ t2 +

e-at - -C

X2 } '

it followsfrom the joint convolution property (3),Seco141,that e-atps(ex)= ~Sa {IX' P(r)[t2 + (:

-

r)2

-

t2

+ (: + rfJ dr}.

Formally then, the inverse of the sine transform (9) is eky

(10)

oo

oo

f f

U(x,y) = -1ty

t

e-kt o P(r)[ t+x-r 2 (

t

)2

2 ( t+x+r

)2J dr dt.

Formula (8) then gives a representation of V(x,y). The solution represented by Eqs. (8) and (lO) can be simplified and verified in case F(x) = 1,in whichcase P(x) = h, eventhough the transforms fT(ex)and p.(ex)do not exist.

14 Hankel Transforms

148

INTRODUCTION

Transforms produced by linear integral transformations whose kernels are Bessel functions are called Hankel transforms.1 They are sometimes referred to as Bessel transforms. There is a great variety of those transformations, first because of the variety of Bessel functions, solutions of Bessel's differential equation (1) containing two parameters J1and v. Furthermore, as in the case of Fourier transforms, the transformation may be made over either a bounded interval and involve various boundary conditions, or over a half line. 1 Friedrich W. Bessel (1784-1846), German (1839-1873), German mathematician.

420

aslronomer

and mathemalician;

Hermann

Hankel

HANKEL TRANSFORMS

[SECo 149

421

The origin is a singular point of Bessel'sdifferential equation. Ir v ~ O, the solution of the equation that is continuous everywhere, including the origin, is (2)

y(x) = CJv(jlx)

where C is any constant and Jv is Bessel's function of the first kind with the series representation t V+2j
(3)

+j +

Jv(t) = j~O j!r(v

1)( 2)

.

In the more common applications the parameter v is a nonnegative integer. Then Eq. (3) becomes t n
= ( 2) j~Oj!(n + j)! ( 2)

We restrict our attention to that case and first treat finite transformations, then transformations on the half line x > 0.1 Useful operational properties of Hankel transformations are much fewer than those for Fourier and Laplace transformations. But again, applications to the solution of linear boundary value problems do not depend on the homogeniety of differential equations or boundary conditions. 149

FINITE HANKEL TRANSFORMATIONS

We consider functions on a bounded interval extending to the right of the origin and take our unit of length as the length of the interval, and this differential form in such functions : (1)

D2F = F"(X)

+ .!.F'(x) x

- n:F(X) x

(O< x < 1),

or a positive integer. Thát form arises when the Laplacian operator V2 is written in terms of cylindrical coordinates; in that case our variable x here is the radial coordinate usually called r or p (cf. Seco83).

where n is zero

In terms of a

self-adjointform we write 1f

(2)

D2F

=

x1[xF'(x»)' -

n2

x-F(x)}

(O< x < 1,n = 0,1,2,...),

so in the notation used in Seco 111, p(x) = r(x) = x and q(x) = -n2jx. 1

As noted in Seco 83, basic properties

and representations

of Jn are developed

in the author's

"Fourier Series and Boundary Value Problems." For extensive treatments ofBessel functions see G. N. Watson, "A Treatise on the Theory of Bessel Functions," 2d ed., 1944, or A. Gray, G. B. Mathews, and T. M. MacRobert, "A Treatise on Bessel Functions and Their Applications to Physics," 2d ed., 1952. For a shorter account see F. Bowman, "Introduction to Bessel Functions," 1958.

422


SECo 149]

Let us first find the integral transformation (3)

T{F}

f

=

F(x)X(X,A)dx = f(A),

with weight function p(x) = x, that transforms D2Fin terms of A,j(A), and the boundary value F(l). According to Eq. (8), Seco111, we can write

T {D2F} =

f

F(X{ (x')'-

:

<1>]dx

+

[x(F'

-

<1>' Fm.

Thus if <1> and A are such that n2

(4)

(x')'

- -<1> x = AX(X))

and

<1>(1)) =

O,

then (5)

T{ D2F} = Af(A) - '(l,A)F(l).

We assume, say, that <1> and F are of cIass C" on the interval O ~ x ~ 1. It is convenient to write A = - J1.2.The differential equation (4) is Bessel's equation with index n, (6)

x2<1>" + x'+ (J1.2X2- n2) = O.

Its solution of cIass C" (O~ x ~ 1) is <1> = Jn(J1.x),and the condition (l,A) = Ois satisfied if J1.is such that Jn(J1.)= O. For each fixed n (n = 0,1, 2,. . .), the equation Jn(J1.)= Ohas an infinite sequence of roots, all real, and only the positive roots need be considered beca use they correspond to the full set of linearly independent eigenfunctions of the singular eigenvalue problem (4). Let J1.j,or J1.j(n),denote those roots: (7)

Jn(J1.j)= O

where j = 1,2,..., J1.j> O,and J1.jdepends on n.

Our transformation T then becomes the finite Hankel transformation on the interval (0,1),

(8)

HnAF(x)} =

f

F(x)xJiJ1.jx)dx = fH(J1.j)

(j

= 1,2,.. .)

with the operational property HnA D2F}

= - J1./fH(J1.)- J1./~(J1.)F(l).

Here J~denotes the derivative of Jn with respect to its argument, so dJn(J1.jx)¡ dx = J1./~(J1.jx).Note that for each fixed n the transforrnfH(J1.)is a sequence of real numbers.

HANKEL TRANSFORMS

[SECo 150

423

In view of the known formula

tJ~(t) = nln(t)

(9)

-

tJn+l(t)

and the fact that JnVJ)= O,it foIlowsthat J~(/l)= basic operational property of Hnj can be written (10)

+ ~F'(X)

Hnj{F"(X)

-

::F(X)} =

-Jn+l(J1.j)'Thus the

-/l/fH(J1.j)

+ /l/n

+ 1(J1.j)F(l),

when F is of cIass C" (O~ x ~ 1) and /l1, /l2,'" are the positive zeros of J n(J1.).Numerical values of the first four zeros of J o(J1.) were given in Seco83. When n = O, D2[1]= O and D2[X2 1] = 4. Thus these particular transforms foIlow from formula (10):

-

1

(11)

HOj{l} = -J1(J1.j), /lj

(12)

HoA1 - X2} = ~Hoj{l}

= ~Jl(J1.j),

/lj

/lj

where Jo(/lj) = O and /lj > O. Also from property (10) we find that (13)

Hnj{xn} = .lJn+l(J1.j), /lj

(14)

/l' J

Hnj{Jn(cx)}= Jn(c) 2 /lj

- e

(e > O),

2Jn+l(J1.j)

where /lj are the positive zeros of Jn(/l) when n is some nonnegative integer, and e is not one of those zeros.

150

INVERSION

OF H.j

For a fixed n the set of functions Jn(J1.jx),where j = 1,2,..., is orthogonal on the interval (0,1) with weight function x, and norms of the functions are known.

We can express that orthogonality

in this form:

(1)

(i

= 1,2,...) ifj = i,

where /lb /l2,' .. are the positive zeros of Jn(/l). For a function F on the interval the generalized Fourier series in terms of the functions of that set is the Fourier-Bessel series

424


SECo 151]

That series converges to F(x) on the intervalO < x < 1 if F' is sectionally continuous there and if F is defined as the mean value of its limits from the right and left at each point where F is discontinuous. Then we can write 00

(2)

F(x)

=2

¿

[Jn+t(.uj)r2fH(.uj)Jn(.ujx)

(O< x < 1),

j= t

wherefH is our Hankel transform of F corresponding to the zeros J.ljof Jn(.u). Equation (2) is an inversion formula for the transformation Hnj' Kernel-product and convolution properties are not known for finite Hankel transformations. Some general properties of Sturm-Liouville transforms noted in Chap. 10 can be extended to this singular case. The basic formula for Hnj{D2F} when F or F' has a jump on the interval O < x < 1, for instance, can be written by the method used to obtain formula (7), Seco 113 (see Probo 3, Seco 152, for the case n = O). The following is another such property. ~ IrfH(J.lj)= Hnj{F(x)},thenfH(.uj)/J.l/is the transform YH(J.lj) ofa function Y(x) such that D2Y = -F(x) and Y(l) = O,according to our basic operational property for Hnj' Thus (3)

x2Y"(x)

Y(l) = O,

+ xY'(x) - n2y(x) = -x2F(x),

and Yand Y' are to be continuous and Y" sectionallycontinuous over the interval O ~ x ~ 1. A particular solution of the homogeneousform of the differentialequation here is xn. If we write Y = xnZ, problem (3)is reduced to one of first order in Z'. In that way we find this formula for Y(x), the inverse transform offH(.uj)/J.l/: (4) where J n(.uj)= O,J.lj > O,n = O, 1,2,. . . , a result that is valid if F is sectionally continuous on the interval (0,1). 151

MODIFIED

FINITE TRANSFORMATIONS

Hnh

Ir our differential form D2F is to be reduced by the transformation on the interval (0,1) in terms of a boundary value hF(l) + F'(1), where h ~ O, we denote the transformation by Hnh' The special case n = h = O was treated briefly in Seco 116. Ir (x,A.) is the kernel, with weight function x, then as in Seco 149 we can write

i, f ¡ _J

1

I HANKEL TRANSFORMS

We now require

[SECo 151

426

that 2

(1)

(X~')'

and

- ~~ = A.X~, x

when x = 1,

so tha t (2) If we write A.= - p2, a solution of the singular eigenvalue problem (1) on the interval (0,1) is ~ = Jn(PiX)where Pl' P2"" are the roots ofthe equation (3)

where n is somefixed nonnegative integer. For each pair of the numbers h and n the roots of Eq. (3) are all real. A full set of eigenfunctionsof the singular problem (1) arisesfrom the nonnegative roots, roots that make up an infinite sequence Pl' P2' . .. where O ~ Pl < P2 < The sequence depends on the values of h and n. In case n stant

=

h = O, we find that Pl = O, then the first eigenfunction is the con= 1; otherwise Pl > O. Our transformation is then the modified finite H ankel transformation

Jo(O)

(4)

1

,

I!

(i = 1, 2, .. .),

where Pi are the nonnegative zeros of the function hJiP) + pJ~(P). The

transformfH(P¡)is a sequenceof numbers, a sequencethat depends on the valuesgivento n and h. The basicoperationalproperty(2)now takes the form (5)

- ::F(X)} = -P/fH
Hnh{F"(X) + ~F'(X)

+ JiP¡)[hF(l) + F'(l)], valid if F is of class C" on the interval O ~ x ~ 1. Again, for each fixed pair of constants n and h, where n = O,1,2, ... and h ~ O,the set of functions Jn(PIX),Jn(P2X),,,,,is orthogonal on the interval (0,1),with weight function x. Using known formulas for norms of Jn(PiX),we can write O if i ~ j, where j = 1,2,. . . , (6)

Hnh{Jn(PjX)} = P/ + h2 -

{ except that in case n (7)

=

2P/

h = O,then

Hoo{l}

=

n2

2

P1 = O and

f

if i = j,

[Jn(Pj)]

xJo(Pix)dx

J 0(0)

=!

=

1 so

when i = 1.

~

I

426


SECo 152]

The Fourier-Bessel series, the generalized Fourier series of those orthogonal functions corresponding to a function F, converges to F(x) under the conditions stated for the representation (2~ Seco 150, and serves as the inversion formula

(8) but if n

F(x)

- 2 ;t, Po' +~: - n' [~:gJ],J.(P,X)

r

< x < 1);

= h = O,then P1 = O,and the first term in the summation is the

constant 18(0)

= f~ tF(t)

dt.

Some further properties of finite Hankel transforms and transforms of particular functions are included in the problems folIowing Seco 152. The development of Hankel transforms has only minor modifications when the integer n is replaced by an arbitrary positive number V. 152

A BOUNDARV VAlUE PROBlEM

In a half cylinder of semi-infinite length with internal heat generation, let steady temperatures U(r, O), (2)

U(l,
U(r,O,z) = U(r,1t,z) = U(r,
Equation (1)is Poisson's equation V2U = -Q(r) in cylindricalcoordinates r, O. Let u,(r,n,z)denote the finite Fourier sine transform of U with respect to

(3)

1 ou ur

+ -r

n2

T

02U

- zU, + uZ ~ 2' + Q(r)Sn{l}= O r (O < r < 1, z > O),

z (%,y,z)

z y -----------------%

Fig.109

HANKEL TRANSFORMS

(4)

[SECo 152

us(1,n,z) = Sn{l},

where Sn{l} = [1 - (-l)"]n-l

427

u.(r,n,O)= O,

and n = 1,2,....

The differential form with respect to r in Eq. (3) and the bound~~ = 1 are adapted to the finite Han~ transformation Hnj with respect to r. We write

condition at the cylindrical surface r

usH{Jlj,n,z)= Hnj{us(r,n,z)}

(j = 1,2,.. .),

where Jlj are the positive zeros of Jn{Jl). Then according to property (10), Seco 149, d2 -Jl/UsH

+ Jl!n+l{Jlj)Sn{l}

+ dz2USH + qH{Jlj)Sn{l}

=

O

when z > O, and usH{Jlj,n,O) = O, where qH{Jlj)is the transform Hnj{Q(r)}. The bounded solution of this problem is (5)

qH{Jlj) + JljJn+l{Jlj) 1.. ) Sn{l} [l - exp ( - JljZ)]. usH\fA'j,n,z= 2 Jlj

The inverse transformations are represented by series:

then

~ ~

,1,. 4 usH{Jlj,n,z) ) ) ',1,. U(r,'I"z = -1tn=lj=l L.J L.J [Jn+l (Jlj WJ n{Jljr SIO n'l'

(6)

where UsHis given by formula (5) and Jlj are the positive zeros of J n{Jl),which depend on the value of n. In case the condition U(l,cjJ,z)= 1 is replaced by Ur(1,cjJ,z)= -hU(1,cjJ,z)

+

P(cjJ,z),

a condition of heat transfer at the cylindrical surface, the modified finite Hankel transformation, treated in Seco151, can be made with respect to the variable r. The more common applications of Hankel transforms involve the special case n

=

O and

the Bessel

function

Jo.

PROBLEMS 1. Derive the finite Hankel transfonns (13) and (14), Seco 149, of the functions X' and Jn(cx). 2. Obtain the finite Hankel transfonns (a) HoAx2}

J.ll-4

= -r-J¡(J1.j), J.lj

I

d

428


SECo 152]

(b) HOj{l

-

X4}

= 1~HoAx2}, J-Ij

where JO(JJ.j) = Oand J-Ij> O (j = 1,2,.. .). 3. If F and F' are continuous on the interval O ~ x ~ 1 except possibly for finite jumps at an interior point x = e, and if F" is sectionally continuous on the interval, show that

HOj{F"(X)

+ ~F'(X)}

= -J-I/fH(JJ.j) + J-I/l(JJ.j)F(l)

- eJ-l/l(JJ.}.c)[F(e + O)- F(e - O)] - eJo(JJ.}.c)[F'(e + O)- F'(e - O)] where J-Ijare the positive zeros of Jo(JJ.)and

fH(JJ.j)

=

= s:

HOj{F}

(j = 1, 2, . . .).

F(x)xJo(JJ.r)dx

4. Let G(x,e) be Green's function (Sec. 95) for the differential form D2 when n such that

(O< x < 1, x ,p e),

xGxx(x,e) + Gx(x,e) = O

G(l,e)= O,

= O,

1 Gx(e+ O,e) - GAe- O,e) =- c

and G is continuouswhen O~ x ~ 1, whereO< e < 1. Showthat G(x,e)= logc

whenO~ x ~ e,

= logx

when e ~ x ~ 1;

then use the operational property found in Probo 3 to obtain the transform 1 HOj{G(x,e)} = -~Jo(JJ.}.c) J-Ij

(j = 1, 2, .. .),

where Jo(JJ.j)= O and J-Ij> O. Compare formula (3), Seco114. 5. If n and h are not both zero and show that

Pl'

P2,

. . . are the positive zeros of hJ .
h+ n

l

fo x.+ lJ.
pJ~(P),

(n = O, 1, 2, . . .).

When n = O and h > O and p¡ are the positive zeros of the function hJo
pJ'o
fH
s: F(x)xJo
derive the formula 1

l

p/fH
y

t

f.o yF(t)dtdy

1

l

+ h f.o tF(t)dt } .

(i = 1,2,. ..),

HANKEL TRANSFORMS

[SECo 152

429

Also, from the transform HOh{l} found in Probo 5, show that h

1

1 - X2

P/Jo(/J¡)= HOh { 2h +

¡- } .

7. When n = h = O in the transformation Hnh, note that HOO{F"(X)+ ~F'(X)} = -P/Hoo{F} + F'(l)Jo(/J¡) = F'(l) ifi = 1,

ifi = 2,3,..., since p¡ = O,

where P¡, P2, . . . are the nonnegative zeros of J ¡(/J). Show that Hoo{1}

=O

if i

= 2, 3, . . . ,

if i = 1;

ifi = 2,3..., if i = 1. 8. If e> O and Jo(e) t= O, show that the function Y of class C" (O~ x ~ 1) that satisfies the conditions 1 Y"(x) + -x Y'(x) + e2 Y(x) = x2

-

1

Y(l)

= O,

has the transform

9. Solvethe problem 1 Y"(x) + -x Y'(x)

-

e2Y(x) + F(x) = O,

O
Y(l) = O,

in the form

fHÚJ.j)=

I:

(j = 1,2, . . .). F(t)tJ

oÚJ.¡t) dt

In case F(x) = JoÚJ.¡x),show that the solution becomes

L

. 430

SECo 152]


~ /

10. Use the Hankel transformation HOj with respect to r to derive, in only a few steps, the formula (9), Seco83, for temperatures U(r,t) in a circular cylinder of infinite length, when 1

U,(r,t) = Urr(r,t)

+ -Ur(r,t) r

U(r,O)= O,

U(I,t) = 1.

(O ;;i; r < 1, t > O),

11. Use results found in Probo 7 to show that the appropriate Hankel transform of U(r,t), where 1 U,(r,t)

= Urr(r,t)

U(r,O)= O,

(O;;i;r < 1, t > O),

+ -¡:Ur(r,t) Ur(l,t) = A,

has the value

ifi=2,3,o.., 1[P~zJo(fJ¡)

- ¡zJo(fJ¡)exp(-p/t)] if i = 1,

4[~ +M4t +~)] and thus obtain the formula for U(r,t) found in Prob:S, Seco 830

12. In Probo 8, Seco144, an integral representation was found for the temperature distribution W(r,x) in a semi-infinite circular cylinder whose axis is the positive x axis, when ~

(O< r < 1,x > O), W(I,x) = O

ifx > O,

W(r,O)= 1

ifr < 1;

also, W is to be continuous and bounded when O ;;i;r ;;i; 1 and x > 00 Use a Hankel transformation to derive this series representation of W: W(r,x)

..,

1 J o(p.jr)

= 2 L --exp(-JljX) j=¡ Jlj J¡(p.j)

when O ;;i; r < 1 and x > O,where Jo(Jlj) = O and Jlj > 00 13. Let transverse displacements Z(r,t) in a membrane stretched over a circular frame satisfy the conditions 1 Ztt = Zrr + -Zr - 2kZ, (O;;i;r < 1, t > O), r

Z(I,t) = Z(r,O)= O,

Z,(r,O) = Vo

o

Ifthe coefficient of damping 2k is such that O < k < Jl¡ where Jl¡, Jlz," oare the positive zeros of J o(p.)in order of Íncreasing magnitude, derive the formula Z(r,t)

-

-

-k' ;, Jo(p.I)sin[t(p./ -

2voe

L. j= ¡ Jll¡(p.j)

kZ)t] z zt (p.j - k )

o

HANKEL TRANSFORMS

14.

[SECo 153

431

Let Jlj denote the positive zeros of J ¡ (J1.)and write

=

H¡AF(r)}

f

. u = 1, 2, .. . ).

= fH(J1.j)

F(r)rJ1(J1.jr)dr

Derive the formula

f ro !.P 2F(t)dtdP } '

= Hlj { , ~ P

~fH(J1.j) Jl)

and since H1j{r}

= Jlj-IJ2(J1.j), according 1 -J2(J1. j ) Jlj3

{ }

to transform

r-

= Hl" { )

(13), Seco 149, show that

r3

8

}

.

15. If Y = r sin ,p and V2V + Ay = O iriside a half cylinder 0< r < 1, O < ,p < n, z > O,and if V = Oon the boundaries,then V(r,4>,z) satisfiesthe conditions 1 1 "~ v" + -v, + -V4>4> + V.Z+ ArslD'I' = O r r2

(O< r < 1, O < ,p < n, z > O),

V(I,,p,z) = V(r,O,z)= V(r,n,z) = V(r,,p,O)= O. Reduce the problem to one in W(r,z) by writing V(r,,p,z) = A sin ,p W(r,z)

/ and, withthe aid ofthe transformfoundin Probo14,derivethe solution r-r3

where

153

J ¡(J1.j)

= O and

«)

g- - 2j=IJlj ¿

W(r,z)=

J1(J1.y) 3 J 2Jlj ( )

exp ( -

JljZ)

Jlj > O.

NONSINGULAR

HANKEL TRANSFORMATIONS

Ir our differential form

1

(1)

D2F = F"(x)

+ -F'(x) x

n2

- zF(x), x

where n is a fixed nonnegative integer, applies to functions F on a bounded intervalO < a ~ x ~ b, one that does not have the origin as an end point or an interior point, then D2 has no singular point on the interval. The linear integral transformation that resolves D2F in terms of the transform of F itself and prescribed values of F or F', or of a linear combination of F and F',

at the points x = a and x = b is a nonsingular special case oJ the SturmLiouville transJormation introduced in Chap. 10. The kernel eI>(x, - Jl2)satisfiesBessel'sequation

(2)

n2

(xel>')'

- -el> x = - p.2xel>

(a < x < b)

\i

i ~

432


SECo 154]

and homogeneous boundary conditions at the end points of the interval. The weight function p(x) in the integral transformation is x itself. To satisfy both end conditions, the general solution of Eq. (2), (3)

<1>= AJ n(Px)

+ B y"(JLx)

is needed, where Y"is Bessel's functiOir of the second kind.1 The ratio of the constants A and B, and the discrete values J1.1,J1.2,... of J1.,are determined by the homogeneous boundary conditions on <1>. In Probo 12, Seco107, we used a generalized Fourier series to solve the following problem for the electrostatic potential V(r,z) in the charge-free space between cylindrical surfaces r = a and r = b, above the plane z = O. 1 V,r(r,z) + -r V,(r,z)+ ~z(r,z) = O (O< a < r < b, z > O), (4) V(a,z) = V(b,z) = O, V(r,O) = F(r).

The solution found there can be obtained by using the nonsingular Hankel transformation (5)

H o{F(r)} =

where

f

F(r)rtf¡(ai,r)dr

(i = 1,2,. ..),

(6)

and ai are the positive zeros of the function tf¡(a,b).But the transformation H o applies also to problems in which the differential equation and all boundary conditions in problem (4) are made nonhomogeneous. 154

HANKEL TRANSFORMATIONS

H.. ON THE HALF UNE (x > O)

The best known of the various Hankel transformations formations of the type

are singular trans-

T {F} = 1"" F(x)X(X,A) dx = f(A);

that is, they apply to functions F(x) defined over the positive half of the x axis. The transformation T corresponding to a fixed integer n (n = O,1, 2, . . .) replaces our differential form D2F I

= F"(x)

+ !F'(x) x

-

n: x F(x)

= x~ (XF')' - n2x F [ ]

The function y"(x)is of the order of x-o as x ->Oif n = 1,2,..., while Yo(x)is of the order

of log x as x -> O. A surnrnary of representations and properties of Bessel functions, including y., with references, is given by A. Erdélyi et al., "Higher Transcendental Functions," vol. 2, chap. 7, 1953.

I

J

HANKEL TRANSFORMS

[SECo 154

433

on the half line x > O by A.f(A.). That operational property is the principal one for applications of such transforms to problems in differential equations. To construct T so that it has that property, we write

(1)

T{D2F}

=

LX)

= {X>

[(XF')'

-

[(~')'

-

: :

FJ~dX

~JF dX-:' [x(~F'- ~'F)]o'

We assume, say, that F and ~ are of class C" on the half line x ~ O and that the functions F and ~ are such that the last bracketed term vanishes as x -+ 00 and the second improper integral exists. Then we require thar2

(2)

(x~/)'

- ~~ x = A.~

(x > O)

so that T{D2F} = A.T{F}. Equation (2) is Bessel's equation. We write A.= - rx.2 ; then if rx.is real, the bounded solution of that equation that is of class C" when x ~ Ois ~ = AJn(rx.x).Alllinearly independent solutions are represented if A

=

1 and rx.~

O.

Thus T becomes the Hankel transformation (3)

Hna{F(x)} = LX>F(x)xJn(ax)dx

where n has one of the values O, 1,2,

(rx.~

= fH(rx.)

O),

. . .. Its basic operational property is

(4) Here the Hankel transformfH(rx.)is a function of rx.that depends on the choice of n. The notationfHn(rx.) is more descriptive

but rather cumbersome.

An asymptotic representationl of Jn(t) for large values of t shows that Jit) is 19(ct) as t -+ 00; that is, .JtIJn(t)1 is bounded. From the formula (5)

tJ~(t)

=

nJn(t)

-

tJ n+ l(t)

(n

= O,1,2,. . .)

it follows that J~(t) has that same order as t -+ oo. Now in Eq. (1), ~ = Jn(rx.x) and d~/dx = rx.J~(rx.x)where J~ is the derivative of Jn with respect to the argument of Jn; also, (x~')' - n2x-l~ = -rx.2xJn(ax). Then if F(x) is

19(X-k)as x -+ 00,where k >

t, the integrand

ofthe second improper integral

in that equation is 19(x-k+t)and, sincek - ! > 1, that integral is absolutely convergent. Ifin addition F' is 19(X-k+l),the final term in Eq. (1)vanishes; thus the right-hand member of the equation exists, and so T{D2F} exists and is equal to -rx.2Hna{F}. I

Erdélyi, A., et al., "Higher Transcendental Functions;; vol. 2, p. 85.

434


SECo 155]

Operational property (4) is therefore valid if F is class C" when x ~ O, if F is (9(X-k), and if F' is (9(X-k+l) as x -+ 00, where k > !. The Hankel transform (3) of F exists if F is sectio~lIy continuous on each bounded interval of the positive x axis and if F ís (9(X-k) as x -+ 00, where k>

t.

The Fourier-Bessel integral representation of F(x) corresponding to the Fourier sine or cosine integral formula, furnishes an inverse transformation (6)

(x >,0).

It is valid if F(x) is (9(X-k) as x -+ 00, where k > !, if F' is sectionally continuous over each bounded interval, and if F(x) is defined as the mean value of its limits from the right and left at each of its points of discontinuity. Our conditions under which the above formulas are valid are sufficient. They can be relaxed. The transform (3) of F(x) may exist, for instance, when F is unbounded at the origin but such that xn+1F(x) is (9(XC)as x -+ O,where e > -1. No major changes are involved when the integer n in this section is replaced by v where v > - t. 155

FURTHER PROPERTIES OF Hna

Ir fH(a) = Hn".{F(x)},we find that when k > O.

(1)

A property on division of Hankel transforms fH(a) by a2 follows from our basic operational property (4), Seco 154, by solving the differential equation

Y"(x) + !. Y'(x) - x2 n2 Y(x) = - F(x) X for Y(x), since -a2YH(a) (2)

(x > O)

= - fH(a). The formula is

1 a2 fH(a) = Hna

A

{f x

x (

n

r'

t n+ 1

y) JB ( Y)

F(t) dt dy}

where n is a nonnegative integer and A and B are nonnegative constants or infinity, to be chosen if possible, so that the transform on the right exists. Ir F satisfies our conditions under which the operational property (4), Seco 154, is valid except for finite jumps jc = F(c + O) - F(c - O)and j; = F'(c + O)- F'(c - O) in F and F' at a point x = e, the property takes the

HANKEL TRANSFORMS

[SECo 155

436

modified form

(3)

Hna;

1

{

F"

n2

A kernel-product

(4)

}

+ ~F' - x2F = -a.2fH(a.)+ a.cJ~(a.c)jc - cJ;;cac)j~. property for Hna;arises from the formula1

J (x)J (y) = n

x"y"

n

2n~r(n

+ t)

i

n Jn[R(x,y,O)] sin 2nO dO

o [R(x,y,o)]n

where R(x,y,O) is the length of the side of a triangle opposite the angle O included

(5)

between

two sides of

length

--

x and y; that is

R(x,y,O) = (X2 + y2 - 2xy cos 0)+.

If we use formula (4), the product Jn(ay)Hna;{F(x)} can be written as an iterated integral with respect to x and O. It is an integral over the upper half plane if x and Oare interpreted as polar coordinates. When that integral is written in terms of R and ,polar coordinates about the point (y,0) on the axis O = O, we obtain the kernel-product property

1

(6)

a"yJ..{a.Y)fH(a.)

= Hna;{PF(x,y)}

wherefH(a.) = Hna;{F(x)}and PF(x,y) =

(7)

x"y"+1 r" F[R(x,y, )] sin ~2nr(n + t) lo [R(x,y,
2n d.

The corresponding convolution XFG(x)of two functio~s F and G such that

(8) is then formally, as indicated in Seco 109, (9)

X FG(X)= {ex>F(y)PG(x,y)dy.

Formulas (4) to (9) are presented here to display their complexity. In case n = O,the results are included in Table 4, but even in that case the kernel-product and convolution properties are difficult to use. The integer n can be replaced by v, where v ~ O,in those formulas.2 1 Watson,

G. N., op. cit., p. 367.

2 Derivations and conditi~ns of validity of those kemel-product and convolution properties are inc1uded in J. S. Klein's doctoral dissertation, "Some Results in the Theory of Hankel Transforms," University of Michigan, 1958. Earlier, J. Delsarte, J. Math. Pures Appl. (9), vol. 17, pp. 213-231,1936, and Acta Math., vol. 69, pp. 259-317,1938, gave convolution properties associated with Hankel transforms.

436

156

SECo 156]

OPERATlON\l

MATHEMATICS

TABlES OF TRANSFORMS H",,{F}

Table 4 summarizes properties of the transformation Hna on the half line x > O and lists transforms of some particular functions, for the important special case n = O. Entry 5, giving a possible inverse transform of fH(rx)jrx2, is the special case of formula (2), Seco155, when n = O,A = 00, and B = oo. Derivations of some of the transforms listed in the table are indicated in problems following Seco 157. Table 4

Hankel transforms

(n = O) F(x)

fH(a) (a > O, n = O)

f"" o F(x)xJo(ax)dx = Ho.{F} 2

I F"(x) + 'XF'(x) fH(ka) (k > O)

5 6

F(x)

{"" fH(a)aJO(ax)da = Hox{fH}

3 4

-

1tJo(ak)fH(a)

(k>

O)

7

9

"" ""t -F(t)dtdy x y Y

f f

r o

8

F[(X2 + k2 "-.

-

2kx cos O)t]dO

fo""yG(y) s: F[(X2 + y2 - 2xy cos O)t]dOdy

I

1 a

x I

J a2 + e2

(e> O)

I -ex -e x

10 II

(e> O)

12

(e> O)

.jX2 + e2 13

(x> O)

(e> O)

14

(e> O)

15

(e> O)

1 -Iog 2 I O

{

.j X2 + e2 - e .j X2 + e2 + e ir O< x < e ir x > e

~Jo(~)

HANKEL TRANSFORMS

[SECo 157

437

An extensive table of Hankel transforms on the half line x> O,given by A Erdélyi el al., "Tables of Integral Transforms," voL 2, 1954. In those tables the transformation is written, except for variations in the letters used, as {X>G(x)~Jv(ax)dx

= g(a;v).

/

Thus in terms of the notation we use here, G(x)

F(x) =

.fi

.

and

g(a;v).

fH(a) =

.fi

Those tables list several properties of transforms g(a; v)in which different values of the index v occur in the same property. Such properties follow from relations such as Eq. (5), Seco 154, between J~, Jn, and Jn+l' or the recurrence relation

The tables list transforms

of particular

functions

in cases v

=

O and v

= 1,

and for the general index v. Many of the functions and transforms involve higher transcendental functions. Another table will be found in the book by Oitkin and Prudnikov; other treatments and applications of Hankel transforms are included in books by 1. N. Sneddon ("Fourier Transforms") and C. J. Tranter, alllisted in the Bibliography. 157

AXIALLYSYMMETRIC HEAT SOURCE

The following is an example of a boundary value problem for which a Hankel transformation Ho",with respect to r is the natural method ofsolution. Find a formula for steady temperatures V(r,z) in a semi-infinite solid z ~ O throughout which heat is generated at a steady rate that varies only with the radial coordinate r, when the face z = O is kept at temperature zero. Then Y,.r(r,z) +

1

-r y"(r,z)+ ~%(r,z)+ F(r) = O (r > O,O ~

(1)

< 2n, z > O),

V(r,O) = O,

where r,
-.

00 such that a solution exists.

/'

438

OPERATIONAL

SECo 157]

MATHEMATICS

Let vH(a,z)denote the Hankel ttansform of V with respect to'r, for which n = O:

(2)

v~a,z) =

HOIX {V(r,z)} =

LX)V(r,z)rJ o(ar)dr

(a > O),

and writefH(a) = HoIX{F(r)}.Then the transformed problem is, formally, d2 -a2VH(a,z) + dz2 VH(a,z)+ fH(a) = O

(a > o, z > O),

(3)

I The bounded solution of that problem is

(4)

1 - e-IX:

vH(a,z)= fH(a) a

.

Thus a formal solution of problem (1) is

f

a>

(5)

V(r,z)=

o

1 - e-IX:

fH(a)

a

Jo(ar)da.

Another form of the solution can be written by using Table 4, first to represent the inverse transform of a-2fH(a) as an iterated integral in terms of F, then by using the convolution integral (entry 7) to write an inverse transform of either the product [a-2fH(a)]e-:IXor fH(a)(a-2e-:IX). But those forms are complicated, even when F is specified as the simple step function F(r) = A when O< r < e, F(r) = O when r > e, representing the case in which heat is generated uniformly throughout a cylindrical core of the s9lid. In the special case k (6)

F(r)

= (r2 + k2)3/2

where k > O,

thenfH(a) = e-k«, and formula (5)can be verified as a solution ofproblem (1). But another form of the solution can be written since (7)

VH(a,z)= ~e-kIX - ~exp [-(z + k)a]. a a

From Table 4 it follows that (8)

W (r O) - 1 W(r,z) - 1 , - log 2V(r,z) = log W(r,O) + 1 W(r,z) + 1

where r2

(9)

W(r,z) = [ (z + k)2 + 1J

1/2

..

/'

HANKEL TRANSFORMS

[SECo 157

439

Formula (8) can be verified as a solution of problem (1) when F is the function (6). According to that formula, the temperatures on the z axis tend to infinity as z -. 00, since

V(+O,z) = logZ +k k.

(10)

PROBLEMS

1. UsetheknownLaplace transformation L{ J O(lXt)} =

+

(S2

/ (s > O, IX> O)

1X2) -1/2

to establish the transform H o. {x -1 e -ex} listed in Table 4. Also, differentiate with respect

to the parametere to showthat, if e > Oand m = 1,2,...,

2. Note that our conditions of validity of the inversion formula (6), Seco 154, are satisfied by the function F(x) = e-ex whene> O.Thus when x > O. From the transform Ho.{e-ex} established in Probo 1, deduce transform Ho.{e(x2+ e2)-3/2} listed in Table 4. 3. Use transformations 5 and 14 in Table 4 to indicate that, when e > O,

where

e2

G(x)= O {

-

x2

+ 2c210g~ e

if O< x ~ e, if x ~ e.

4. Let V(r,z) be bounded in the half space z > O, r ~ O, O ~ f/>< 21t, where r,z, and f/>are cylindrical coordinates, and satisfy the conditions

1 V2V(r,z)= v" + -v, r + V.z= O

(r > O, z > O),

1 V(r,+O) = 'r2 + e2,1I2' where e > O.Derive and verify the solution V(r,z) = [r2 + (z + e)2r 1/2. 5.

If V(r,f/>,z) satisfies Poisson's equation v" + !v, r + r12V",,,,+ V.Z+ F(r)sin f/>= O

J

/

440

OPERATIONAl

SECo 157]

MATHEMATICS

in the region r > O,O < cP < n, Z > O,and if V = Oon the boundaries and Z = O,derive formally the formula V(r,cP,z)= sincP

a:>1 -

f

o

e-U ex

cP

= O,cP = n,

a:>

f

J1(exz)

o

F(P)J1(exp)dpda..

6. Let V(r,z) denote steady temperatures in a slab r ;?;O, O ~ cP< 2n, O ~ z ~ 1, where r, cP,and z are cylindrical coordinates. If the face z = O is kept at temperature V

= O and

the face z

= 1 is

insulated

except that heat is supplied through a Ci{cular

region, such that if O c,

y'(r,l) = { ~

then V2V = Owhen r ;?;Oand O < z < 1, and V(r,O)= O. Derive the formula

-

V(r,z) -c

f

a:>

sinh

exz J 1(exC)

-o cosh z

ex

J otXr ( ) dex.

15 Legendre and Other Integral Transforms

We have presented the operational calculus of various integral transformations, with emphasis on properties that are useful in solving problemNn differential equations. We gave spedal attention to the Laplace transformation because a great variety of prominent problems, especialIy in partial differential equations, can be solved or simplified by using Laplace transforms. A different variety, not so extensive, can be solved by using various Fourier transforms, and still more special types of problems are adapted to Hankel transformations. Of course those and other transformations may be useful outside the field of differential equations, including applications to theory of functions, integral equations, or difference equations. Unless a transformation has a substantial variety of applications, a development of its operational properties here is hardly justified. Procedures given in Chap.l0 may be used to design transformations needed for particular problems. Legendre transforms, treated below, have some good applications, especialIy to problems in potential theory that involve spherical polar coordinates. 441

442

158


SECo 158]

THE LEGENDRETRANSFORMATION T. ON THE INTERVAL(-1,1)

Let r, tjJ,and e denote the sphericalcoordinatesof points (X, y,z) such that (Fig. 110) y Z = reos e X = tan tjJ, (o~e~1t ) - "--.-

.

In terms of those coordinates the Laplacian of a function V(r,tjJ,O)takes the form

2 1 1. 2 1 (V¡¡Sin O)0 V V = 2r (r V,)r+ ~Sin OV.. + --=-Sin e ]

[

(1)

In case V is independent of

that is, V = V(r,O)so the values of V have

tjJ,

axial symmetry about the Z axis, the differential that appears in V2 V is

1.

o

sin O(V¡¡Sin 0)0 = sin 000

Ir we write x

= cos O, the

Ov(r,e)

2

[

form with respect to O

(1 - cos O)sin O00]

.

form becomes

~2 V

(2)

.

o = -ox

2 OV x )-

[(1

(-1 < x < 1).

ox]

That self-adjoint form ~2 V is singular at the points x

=

:!:1. We can

construct a transformation of functions F on the interval (-1,1), T{F(x)} =

fl

F(X)~(X,A)dx

= f(A),

such that T {~2F} = Af(A) by first writing

T{~2F}

=

f

= fl F~2~dx+

[(1

=

f

1 ~~2F dx

z ¡cx,y,z) : I I

i I

y

x

Fig.110

1 ~(x,A)[(l

-

x2)(~F'

-

x2)FT dx

-

~'F)]~l'

LEGENDRE AND OTHER INTEGRAL TRANSFORMS

[SEC. 158

443

then requiring that ~2~ = A~ and that ~ and F be of class C" on the interval - 1 ~ x ~ 1. Thus

(3)

[(1

-

X2)'(X,A)]' = A~(X,A)

- 1<

when

x< 1

and ~ is to be of class C" on the closed interval - 1 ~ x ~ 1. That singular eigenvalue problem in Legendre's differential equation (3) has as its set of eigenvalues A = -n(n + 1)

(4)

(n

= O, 1,2, . . . ).

"

The corresponding eigenfunctions are the Legendre polynomials Pn(x) of degree n: (5)

- 1 m (-lY ~(X,A) - Pn(X)- 2 n ]=0 -:-¡J.

L

(n

(2n - 2j)! -2] 2")1 '.] .(n - J.")Ix"

-

I

-

where m = !n if n is even or zero and m = !(n 1) if n is odd.1 Thus T {F} is the Legendre integral transformation (6)

T,,{F(x)} =

f

(n 1

F(x)Pn(x) dx = fL(n)

. 1I

= O, 1, 2, . . . )

on the interval (-1,1), with the operationalproperty (7)

T,,{[(1

-

x2)F'(x)]'}= -n(n + l)fL(n).

According to Eq. (5), the first few Legendre polynomials are Po(x)

= 1,

P1(x)= x,

P3(x) = !(5X3 - 3x),

P2(x)= !(3X2 - 1),

P4(x) = !(35x4 - 30x3 + 3).

We note that Pn is an even or odd function according as n is even or odd: (n = O, 1,2, . . . ).

In case F and F' are continuous on the interval -1 ~ x ~ 1 except at an interior point x = e where they have jumps jc and j~, respectively, while F" is sectionallycontinuous, property (7)is replaced by (8) The set of funccions Pn is orthogonal on the interval (-1,1). That orthogonality and the norms of Pn are expressed by the transform of Pm, where m has one of the values O,1,2,. .. : if n :p m (9)

T,,{Pm(x)}= { ~m + !)-l

if n = m.

1 For basic properties of Legendre polynomials see the author's "Fourier Series and Boundary Value Problems," 2d ed., chap. 9,1963.

444

SECo 159]


The corresponding generalized Fourier series representation of a function F on the interval is Legendre's series 00

F(x) = L: (n + t)fdn)Pn(x)

(10)

(-1 < x < 1).

n=O

That inversionformula for T,.is valid if F' is section\lIy continuous over the interval and if F is defined as the mean value of its limits from the left and right at each point where F is discontinuous. When F is a polynomial of degree m, it is a linear combination of Po, P1,..., and Pm,represented by series (10) in whichfL(n) = Owhen n > m.

= cos e, our transformation

Under the substitution x T,.{F(cos

e)}

=

f

F(cos e)Pn(cose) sin e de = fL(n)

transforms the <;lifferentialform (O < e < n) into -n(n + l)ji:,
FURTHER PROPERTIES OF T.

By replacing n(n + 1) in property (7) above by (n + t)2 (1)

(n + t)2fdn)

= T,,{!F(x)-

- l,

we find that

[(1 - x2)F'(x)]').

Since Po(x) = 1 and Po is orthogonal to Pnwhen n = 1,2,..., then, if C is a constant, when n i= O

fdn)

(2)

{ fL(O)+ 2C

T,.{F(x) + C} =

when n = O.

Suppose that a sectionally continuous function F has been adjusted by the addition of a constant so that fL(O)= O;that is,

f1 F(x)dx

= O.

Then by solving the differential equation [(1

-

X2)Y'(X)]' = -F(x)

for Yand requiring Yand Y' to be continuous on the interval -1 ~ x ~ 1, we obtain the property (3)

A(n) = n(n + 1)

T,.

F(t) dt ds + s - 1 5-1 {fo +X

s

C

}

(n = 1, 2, .. . ).


[SECo 159

445

The value of the transform on the right when n = O depends on F and the choice of the constant C. Legendre polynomials have the product propertyl Pn(cos f3)Picos O) =

(4)

1

-

n

i

1t o

Pn(COSv) da

where (5)

~v=~f3~O+~f3~O~~

The following convolution property of 1'"has been established with the aid of that property.2 Let F(x) and G(x) be sectionally continuous functions with Legendre transforms fL(n) and gL(n). Then (6)

A(n)gL(n)

=

T,,{XFG(x)}

=

50" XFG(cos O)Pn(cos O) sin O dO

where XFGis this convolution of F and G: (7)

X FG(COSO) =

1

-

n

i

1t o

n

i

F(cos 13)sin 13

o

G(cos v) da df3,

cos v being given by formula (5). The convolution here is not a simple one to use. Some useful transforms of particular functions can be found from the representation 00

(8)

(1

-

2tx + t2)-t

= n=O L t"Pn(x)

(-1

< t < 1)

of a generating function for Pn. When t is fixed, the series converges uniformly with respect to x, and it follows that (9)

2tn T.{(1 2tx + t2)-t} = n 2n+1

= :t 1.

We can differentiatethe members of the equation with respect to t to write when -1 < t < 1. Equation (9) is also valid when t

I MacRobert, T. M., "Spherical Hannonics," p. 138, 1927. z Churchill, R. V., and C. L. Dolph, Inverse Transforms of Products of Legendre Transforms, Proc. Amer. Math. Soc., vol. 5, pp. 93-100, 1954. In formula (5), p, 6, and v can be interpreted as sides of a spherical triangle on the hemisphere XZ + yZ + zZ = 1, Z ?; O,and a as the angle between the first two sides.

446

SECo 160]

OPERATIONAL

MATHEMATICS

Then with the aid of transform (9) it follows that (10)

-

T,,{(1

2t"

2tx + t2)-t} = _ 1 -t

2

(-1 < t < 1).

Transforms of a few other functions and some applications of T" are given in the author's paper, The Operational Calculus of Legendre Transforms, Jour. Math. Physics, vol. 33, pp. 165-178, 1954. Brief treatments of Legendre transforms are given in the books by Sneddon and Tranter listed in the Bibliography. 160

LEGENDRETRANSFORMS ON THE INTERVAL(0,1)

Let F(x) be defined over the interval (0,1) and let F1 be the odd extension of F to the interval (-1,1). Then the product F1(X)Pix) is an even function when n is an odd integer; it is odd if n is zero or even. Thus the Legendre transform of F1 on the interval (-1,1) can be written T2n+¡{F1(x)}

=2

f

F(X)P2n+l(X)dx,

T2n{F1(x)} = O,

wheren = 0, 1,2, . . .. Let us use the symbol1;n+ 1 {F} to denote the integral transformation of F that appears here; that is, (1)

T2n+l{F(x)} =

f

F(x)P2n+ l(X) dx = JL(2n + 1) (n = 0, 1, 2, .. . ).

We ca1lJL(2n + 1) the Legendre transform of F on the interval (0,1) with odd indices 2n + 1. In terms of T" on the interval (-1,1)

(2) Since T2n{F.} = 0, the Legendre series representation (10), Seco 158, of the function F1 becomes 00

F1(x)

=

L

n=O

(2n + 1 + !)T2n+1{F.}P2n+1(X)

This becomes an inversionformulafor

(-1 < x < 1).

1;n+1:

00

(3)

F(x) =

L (4n + 3)JL(2n +

n=O

1)P2n+1(x)

(O <

x < 1),

because F1(x) = F(x) when ° < x < 1. The formula is valid if F' is s~ctionally continuous over the interval (0,1).


[SEC. 160

The orthogonality of the set of polynomials

P211+1(X) on

447

the interval

(0,1) follows from Eq. (2) by writing F(x) = P2m+l(X) = Fl(X) and recalling that the set PII(X)is orthogonal on the interval (-1,1).

In caseF and F' are continuous on the intervalO

;:¡;

x ;:¡;1, then at

the point x = O the function F1 has the jump 2F(0), but the jump in Fí there is zero; elsewhere on the interval -1 ;:¡;x ;:¡; 1 the functions F1 and Fí are continuous. Then if F" is sectionally continuous, formula (8), Seco158, shows that . (4)

T,,{~2F1} = -n(n + 1)T,,{F1} + 2P~(0)F(0),

where ~2F1 = [(1 - x2)F'1]' = (1 - x2)F~ - 2xFí. Since ~2F1 is an odd function, then T211+d~2F1}= 2T211+1{~2F},while both members ofEq. (4) vanish when n = O,2, 4, .... It follows that

From the formula (5)

(n

= 1,2,...)

we find that Píll+ 1(0) = (2n + l)P211(0). Therefore

(6)

'1'211+ 1 {[(1

- x2)F'(x)]') = -(2n - 1)(2n + 2)JL(2n + 1)

+ (2n +

1)P2n(0)F(0).

Formula (6), the basic operational property of '1'211+ 1>shows how the form ~2F on the interval (0,1) is transformed in terms of T211+dF}and the initial value F(O). In like manner we find that the Legendre transformation with even indices, on the interval (0,1), (7)

(n = O, 1, 2, . . . ),

is related to the transformation of the even extension F2 of F on the interval (-1,1) by the formula (n = O, 1,2, . . . ),

(8)

and T211+ 1{F2} = O. It follows that '1'211 has the inversion formula 00

(9)

F(x) =

L

n=O

(4n

and the basic operational property (10)

+ 1)JL(2n)P211(x)

(O < x < 1),

448

SECo 161]


that resolves the differential form ~2F on the interval (0,1) in terms of 1;n{F} and the initial value F'(O) of F'. AIso, the set of functions P2n is orthogonal on the interval (O,1). 161

DIRICHLETPROBLEMS FOR THE SPHERE

The problem of determining the harmonic function in a region which assumes prescribed values on the boundary is called a Dirichlet problem. If instead the normal derivative of the harmonic function is prescribed on the boundary, the problem is called a N eumann problem. Consider the Dirichlet problem for the region interior to the unit sphere r = 1 such that the harmonic function V(r,cos ()) is independent of the spherical coordinate
~

V2V = r12 (r2v,)r+ sin () (Sin [

()J.-9)o

=O

(r < 1, O < () < n),

J

and V -> F(cos ()) as r -> 1; also V and its partial derivatives of first and second order are to be continuous interior to the sphere. We write x = cos (). Then (1)

[r2 V,(r,x)]r + [(1

(2)

V(1

- x2)Vx(r,x)]x= O

- O,x) = F(x)

(r < 1), (-1 < x < 1).

Ir vL(r,n) is the Legendre transform of V with respect to x on the interval (-1,1),

(3)

vdr,n) =

f

1 V(r'X)Pn(X)dx = T,,{V}

(n = O,1,2, . . . ),

and fL(n) = T,,{F(x)}, the above problem transforms into the problem (4)

2 d2vL

r dr2

dVL

+ 2r-¡¡: - n(n + l)vL = O,

In problem (4) VLis to be a continuous function of r when O ~ r < 1. The solution with that continuity is (5)

(O~ r < 1, n = O,1,2, . . .),

and so the formal solution of the problem in V is 00

(6)

V(r,x) =

¿ (n + t)fL(n)rnPix) n=O

(r

~ 1, -1 < x < 1).

This form of the solution is also obtained when the problem in Vis solved by the method of separation of variables.


[SEC. 161

449

Another representation of the solution follows from our convolution

property for T". According to transform (lO), Seco159, 1 - r2

r" = ~T,,{(1

+ r2

(O ~ r < 1).

- 2rxt3/2}

The productfL(n)r" in Eq. (5) is the transform of the convolution (7), Seco159, of the functions F and T"-1{r"}; that is, 1 - r2

(7) V(r,cosO)= where

cos v = cos

"

2n

I

P cos

O

o

"

F(cosP)sin P

+

sin

p sin

I

o

(1 + r2 - 2r cos v)- 3/2da dp

O cos a.

Formula (7) is known as the Poisson integral formula for the harmonic function interior to the sphere r = 1 in this special case where V is prescribed on the boundary as a function F (cos O) of O only. It is of some interest to note, especially when V is interpreted as steady temperatures in a solid sphere r ~ 1,that either representation (6) or (7) gives the value of V at the center r = O of the sphere as the mean value of F over the surface. In the corresponding Dirichlet problem for the bounded harmonic function W(r,cos O) in the region r > 1 exterior to the unit sphere, where W(l + O,cos O) = F(cos O),the solution of the transformed problem is l

1

n

(n = O, 1,2, . . . ).

wL(r,n) = -;:f(n) ( -;:)

i

Therefore, in terms of the function V found above,

I

(8)

(r > 1).

W(r,cosO)= ~V( ~,cosO)

I

PROBLEMS

1. Using

I

properties of Pn(x)on the interval (-1,1), display the orthogonality and norms for the set of polynomials P2n+I(X),n = 0,1,2,..., on the interval (0,1) by showing that for each m (m = O,1,2, . . .) orthogonality

-

O

T2n+1{P2m+1(x)}

= { (4m+

3)-1

if n # m ifn = m

where(Sec.160) T2n+I{F} = f~F(x)P2n+l(X)dx. 2. Display the orthogonality and normsoftheset ofpolynomialsP2n(x),n = O,1,2,..., on the interval (0,1) by showing that for each m(m = 0,1,2,.. .)

-

O

T2n{P2m(x)} = { (4m + 1)-1 where (Sec. 160) T2n{F} = gF(X)P2n(x)dx.

ifn#m ifn = m

I

450

3.

Use properties (6) and (10), Seco160, to show that P2n(0) (a) T2n+l{1} = 2n + 2' (b) T2nX { }=

4.


SECo 161]

P2n(0) .-

-.'

The Neumann problem in a function U(r,cos O) independent of the spherical

coordinate C/>, for the interior of the unit sphere,is when r < 1, v2 U(r,x) = O Ur(l

-

O,x) = F(x)

(-l
where x = cos O. Since U is to be continuous interior to the sphere, at the center r = O in particular, its Legendre transform udr,n), with respect to x, must be continuous when O ~ r < 1. Show that, as a consequence, the prescribed values F(x) of the normal derivative of U on the surface r = 1 must be such thatfL(n) = O when n = O; that is, the mean value of F(x) must be zero, or (a)

I: F(cos O)sin OdO =

O.

Also if U is interpreted as steady temperatures and Ur(l,x) as flux of heat, note why condition (a) is to be expected in order that temperatures remain steady. Under condition (a) derive the formulas ifn

1 1""2n+l U(r,cosO)= -2 C + 2 L -

_n=1

n

= 1,2,...,uL(r,0) = C,

fL(n)r"Pn(cosO)

where r < 1, O ~ O~ n, and C is an arbitrary constant. 5. In the Dirichlet problem in V(r,x) represented by Eqs. (1) and (2), Seco 161, let F(cos O) satisfy condition (a) in Probo 4 above; then vL(r,O)= O, and vdr,n) = fdn)r" when n = 1, 2, . . .. From the transform of the solution U(r,x) of the corresponding Neumann problem with the same function F, found in Probo 4, show that

i

r1

uL(r,n)=

-vL(p,n) dp op

if n = 1,2,..., uL(r,O) = C,

and deduce this relation between U and V: r 1 1 U(r,cos O)= op- V(p,cos O)dp + '2C

i =i t 1

O

1

1 V(rt,cos O)dt + '2C

J

.

(r < 1),

where C is an arbitrary constant. The relation also follows from the series representations of U and V. It can be verified even if F, U, and V are not independent of C/>, provided

that

I:

F(c/>,cos O) sin O dO = O.

"I

.EGENDRE AND OTHER INTEGRAL TRANSFORMS

[SEC. 161

451

i. Let U(r,cos O) be harmonic interior to the sphere r = 1 and satisfy this condition It the surface:

U,(1,x) + (k + 1)U(I,x) = F(x)

vhere x = cos Oand the constant k + 1 is positive. Show that the Legendre transform 'L(r,n) on the interval -1 < x < 1 is related to the transform vdr,n) = fL(n)r" in the :orresponding Dirichlet problem (Sec. 161) in V(r,cos O) by the equation r-k :r[~+ luL(r,n)] = vL(r,n).

fhus deduce the relation

U(r,cos O)=

f

V(rt,cosO)tkdt,

Nhich can be verified even when F, U, and V are not independent of the spherical :oordinate cp. 7. Heat is generated at a steady and uniform rate throughout a solid hemisphere zero. Thusthe ) ~ r < c, ° ~ O< n/2. The entireboundaryis keptat temperature ¡teady temperatures V(r,x), where x = cos O,satisfy the conditions

V2V(r,x) + A = °

(r
V(c,x) = V(r,O)= 0, where A is a constant. With the aid of the transform 1'2n+1 {1}found in Probo 3, derive the formula

'" V(r,cos O) = Nhere r

I (4n n=O

+ 3)¡iL(r,2n

+ I)P2n+ I(COSO)

~ e, ° ~ O~ n/2 and r 2

Ae2p (O)

e

¡iL(r,2n+ 1) = 4(n + 1)(2/':r 3n - 1)[( )

r 2n+1

-

( e)

J

.

~.

Let U(r,cos O) be the steady concentration of a substance diffusing through a )orous hollow hemisphere e < r < 1, ° < O < n/2 whose hemispherical boundaries ife kept at zero concentration. Ir the flux of the substance through the base O = n/2, ; < r < 1, is uniform and constant, then OU

= sinO ~ = roOJ 8=.12 r sinOoOJ 8=.12 Nherex = cosOand B is a constant. Thus

_!U (rO)= B r x ,

(e < r < 1,° < x ~ 1),

Ux(r,O)= -rB.

.

'lote that 1'2n{x} is given in Probo 3 and show that

T { U(r,x) + Brx } 2n

=

BT (x ) 2n (1 - e2n+2)r2n , An+ 1

[

.

,

~ , '11

V2U(r,x) = ° U(I,x) = U(e,x) = 0,

J

,

,

f

e

+ (e - e2n)-r

()

2n+1

J

.

452


SECo 162]

In the special case e = O,show that 00

U(r,cos 6) =

162

- Brcos 6 +

B

¿ (4n + n=O

1)T2.{x}r2.P2.(cos6).

LAGUERRE TRANSFORMS

Operational properties have been developed for another integral transformation whose kernel is a family of orthogonal polynomials, the Laguerre polynomials.1 The Laguerre transformation is one over the half line x > O, with weight function e-x. Its properties are presented only briefty here because the differential form associated with this transformation is not one that is frequently encountered in differential equations. Laguerre polynomials of degree n can be written

eXd. . (-lYn!xj L.(x)= -. d .L ('1)2( n. ",,(x.e-X) = }=o J. n - J.')1'

(1) where n

= O,1,2, . . .. In particular Lo(x) = 1,

L1(x) = 1 - x,

They satisfy the differential equation (2)

xL;(x) + (1 - x)L~(x)+ nLix) = O

which has the self-adjoint form d dx[xe-XL~(x)] = -ne-XLn(x). Thus Ln(x) are eigenfunctions and A.= -n are eigenvaluesof the singular eigenvalue problem consisting of the Sturm-Liouville equation [xe-Xy'(x)]'

- A.e-Xy(x)= O

(x > O)

and the requirement that y be of class C" on the half line x ~ O. We find that the Laguerre integral transformation oo

(3)

Gn{F(x)} =

fo F(x)e-xL.(x) dx = fG(n)

~

(n = 0,1,2,...)

has the operational property (4)

Gn{é(xe-XF')'} = Gn{xF"(x) + (1 - x)F'(x)} = -nfG(n),

valid if F is of class C" when x ~ Oand if F is (!)(e
-. 00, where IX<

1.

McCully, Joseph, The Laguerre Transform, SIAM Review, vol. 2, pp. 185-191, 1960. The

author

includes a convolution

property

for the transformation.


The set of

[SECo 163

453

polynomials Lix), n = O,1,2,..., is orthonormal on the

half line x > O with weight function e - x; that is, when m has one of the values O,1, 2, ... , then

The corresponding generalized Fourier series representation of a function F can be written "" (5) (x > O). fG(n)Ln(x) F(x) =

L

n=O

That inversion formula for the transformation Gn is valid where F is continuous if F' is sectionally continuous on each bounded interval and if F is C9(eU) as x --+00, where IX< t. The convolution of two functions, whose transform is the product of the transforms, can be represented by an iterated integral; but the integral is a complicated one. 163

MElLlN TRANSFORMS

Our formal procedure of designing an integral transformation shows that if the differential form xF'(x) on the half line x > O is to transform into - SfM(S),where fM(S) is the transform of F itself, then the kernel of the transformation can be taken as the function XS- 1. The transformation is the M ellin transformation

(1)

Ms{ F(x)} = 1"" F(x)xS-1 dx = fM(S)

with the operational property

(2)

Ms{xF'(x)} = -sMs{F}

= -SfM(S).

Formally, from property (2) we find that (3)

Ms{x(xF')'} = Ms{x2F"(x) + xF'(x)} = s2fM(S)

and so (4) For further iterations of the operator x d/dx, property (2) leads to the formula (5)

(n = 1,2,...).

454


SECo 163]

The differential form p(p Vp)pwith respect to p can be isolated in Poisson's equation in V(p,cf» where p and cf>are polar coordinates in the planeo The forms r2v,r and rV, with respect to r can be isolated in Poisson's equation in V(r,cf>,lJ) where r, cf>,and lJare spherical coordinates. Thus the MeIlin transformation with respect to the radius vector p in the first case, or which respect to r in the second, will remove those differential forms, provided that p and r range through all positive values. That range of values of p and r and conditions under which the MeIlin transforms exist restrict the types of problems to which M. applies. Let us assume as usual that F is at least sectionally continuous over each bounded interval of the positive x axis. Ir s = a + ia where a and a are real, then x' = exp (s log x) = exp (alog x) exp (ia log x)

and thus Ix'l = xa when x > O. The integrand of the Mellin integral (1) is then 19(xa- 1) as x -+ O and we shall assume a > O to ensure the convergence

of the integral at its lower limit. Suppose that F(x) is 19(X-k) as x -+ 00, where k> O. Then the integrand is 19(I/xk-a+l) as x -+ 00, so the Mellin transform J.(a + ia) exists when k - a > O and a > O, that is, when O < a < k, provided that F(x) is 19(x-k) as x -+ 00, where k > O. When F is continuous (x ~ O) and F' is sectionally continuous, an integration by parts shows that M.{xF'} exists and that property (2) is valid if O < (T< k. The Mellin transformation can be described in terms of the exponential Fourier transformation Ea. Writing s = (T+ ia and substituting x = e-t, we find that

that is, (6) Thus, known exponential transforms furnish corresponding Mellin transforms, and some properties of M. follow from properties of Ea. ButJhe relationship also suggests the substitution x = e-r in order to change the differential form xF'(x) and its iterates into forms to which Ea applies. According to relation (6) and our inversion formula for Ea (Sec. 132), when F' is sectionaIly continuous, then 1

F(e-r)e-ar

=-

lim

f

p

2np-oo -p

if IF(e-r)le-ar is inte.grable from t

=-

.

fM(a + ia)e"arda

00 to t = 00, and we find this is true

lIIIII!IMI"'f'I'I!'!'!!'ftlH!I"!!It!H"

""

f


[SEC. 163

455

if F(x) is (9(X-k) as x --+00 and ifO < u < k, where x = e-l. Thus by writing e-I = x, we have an inversionformula for Ms: (7)

p 1 lim fM(U 2n p oo -{I

F(x) = -

=

1

_ . lim

f i

2nI p '"

+

. iex)x-u-,,,

dex

U+iP

fM(S)X-Sds.

u-iP

The kernel-product property

(y > O)

(8)

is easily derived. The corresponding convolution of two functions F(x) and G(x) is (Chap. 10)

(9)

XFG(x) = {'" G(Y)F(~) d;

such that, if Ms{ G} = gM(S),then (lO)

Ms{XFG(x)} = fM(S)gM(S),

Tables of Mellin transforms, including someof the properties of Ms, are given in A. Erdélyi et al., "Tables ofIntegral Transforms," vol. 1. Other treatments of the theory of the transformation can be found in E. C. Titchmarsh, "Theory of Fourier Integrals" and G. Doetsch, "Handbuch der Laplace-Transformation," vol. 1. Other books listed in the Bibliography, under authors Sneddon and Tranter, include some applications of the transformation.

Bibliography

Boehner, S.: "Lectures on Fourier Integrals," Prineeton University Press, Princeton, N.J., 1959. Carslaw, H. S., and J. C. Jaeger: "Operational Methods in AppliOOMathematies," Oxford University Press, London, 1941. -: "Conduetion ofHeat in Solids," 2d oo., Oxford University Press, London, 1959. Churehill, R. V.: "Fourier Series and Boundary Value Problems," 2d oo., MeGrawHill Book Company, New York, 1963. Doetsch, G.: "Theorie und Anwendung der Laplace-Transformation," SpringerVerlag, Berlin, 1937. -: "Handbueh der Laplace-Transformation," vol. 1, Verlag Birkhauser, Basel, 1950; vol. 2,1955; vol. 3, 1956. -: "Einfúhrung in Theorie und Anwendung der Laplace-Transformation." 2d ed., Verlag Birkhauser, Basel, 1970. Kaplan, W.: "Operational Methods for Linear Systems," Addison-Wesley Publishing Company, Ine., Reading, Mass., 1962. Lighthill, M. J.: "Fourier Analysis and Generalized Funetions," Cambridge University Press, London, 1958. Mikusinski, J.: "Operational Caleulus," Pergamon Press, New York, 1959. van der PoI, B., and H. Bremmer: "Operational Caleulus BasOOon the Two-sidOO Laplace Integral," 2d ed., Cambridge University Press, London, 1955. Sneddon, 1. N.: "Fourier Transforms," MeGraw-Hill Book Company, New York, 1951. -: "Funetional Analysis," Handbueh der Physik, vol. 2, pp. 198-348, SpringerVerlag, Berlin, 1955. Titchmarsh, E. c.: "Theory of Fourier Integrals," Oxford University Press, London, 1937. ' -: "Eigenfunetion Expansions," Oxford University Press, London, 1946. Tranter, C. J.: "Integral Transforms in Mathematical Physics," 2d oo., Methuen and Co., Ltd., London, 1956. Voelker, D., and G. Doetsch: "Die Zweidimensionale Laplace-Transformation," . Verlag Birkhauser, Base!, 1950. Widder, D. V.: "The Laplace Transform," Prineeton University Press, Princeton, NJ., 1941. Wiener, N.: "The Fourier Integral," Cambridge University Press, London, 1933. 456

1

BIBLIOGRAPHY

457

Tablas

Campbell, G. A., and R. M. Foster: "Fourier Integrals for Practical Applications," D. Van Nostrand Company, Ine., New York, 1948. Ditkin, V. A., and A. P. Prudnikov: "Integral Transforms and Operational Calculus," Pergamon Press, New York, 1965. Doetsch, G., H. Kniess, and D. Voelker: "Tabellen zur Laplace-Transformation und Anleitung zurn Gebrauch," Springer-Verlag, Berlin, 1947. Erdélyi, A., W. Magnus, F. Oberhettinger, and F. Tricomi: "Tables of Integral Transforms," vols. 1 and 2, MeGraw-HilI Book Company, New York, 1954. Oberhettinger, F.: "Tabellen zur Fourier Transformation," Springer-Verlag, Berlin, 1957.

...

Appendix A Tables of Laplace Transforms .

Table A.1

Operations'

f(s)

(Res>

F(t)

0:)

(t > O)

F(t)

2

f(s)

I -lim 2ni P-.,

3

sf(s) - F(O)

F'(t)

4

5"f(s)-

r

I F(O) - 5"-2F'(O)- .. . - F(n-I)(O)

I 5 I -f(s) s

I

,-1' et%f(z)dz

F(n)(t)

f~ F(T)dT

6 I e-bY(s) (b > O)

F(t

{O

- b) if t > b ift
7 I f(s)g(s)

f

8 I /,(s)

- tF(t)

9 I pn)(s)

(-1)"tnF(t)

10 l.

~

Y+IP

o F(T)G(t - T)dT

r

I -F(t) t

f(x) dx

~~ II 12

I

13 I

f(s f(es)-(ea)> O) f" e-"F(t)dt o I -e -as

>O)

F(t) if F(t + a)

I See Chaps. 1, 2, and 6 for details and other operations. 458

=

F(t)

-¡

459

APPENDIX A

(Res>

I(s)

14 15

f

ex)

F(t) F(t) if F(t + a)

"o e-si F(t) dt (a > O) 1 + e -as

00 a 2: .~o s"

~

Tabla A.2

(t > O)

= - F(t)

(k > O)

Laplaca transforms' F(t)

I(s)

(t > O)

s 2 3

1 s"

(n = 1.2

)

(n

4

5

-

1)!

1

fi

J:;ü

1 s,fi

2jf

6

(n = 1.2

) 1 x 3 x 5...(2n - l)fi

7 8

r(k)

rk-I

(k > O)

T

e"'

s-a 1

9

te"'

(s - a)2 1

10

(s - ar r(k)

11

(n = 1.2

(n

- l)(-Ie"I

\

(k> O)

(s

-

(s

- a)(s - b)

a¡t

1

)

1

1 -(e"'-eI") a-b

s

1 -(ae"' a-b

(s - a)(s - b)

- beb')

I For more extensive tables of Laplace transforms see the books by A. Erdélyi et al.. vol. l. G. Doetsch et al.. or V. A. Ditkin and A. P. Prudnikov. listed under Tables in the Bibliography. 2

Here a. b. and (in entry 14) e represent

distinct

constants.

460


Tabla A.2

(continuad) F(l)

f(s)

14

15

1

(b

1

S

+

cosat 1 . - smhal a

S

18

cosh at

- a2

S2

1

19

1 2"(1 - cos at) a

s(s2+ a2)

1 . "3(at - smat) a

1 S2(S2+ a2) 1

1 . 3(sm al - al cosal) 2a

(S2+ a2)2

1 . -smat 2a

S

22

(S2 + a2)2

1 -(sin at + at cosat) 2a

S2

23

(S2 + a2)2 S2 (S2

- a2 + a2j2

t cos at cos at

S

25 (S2

+ a2)(s2 + b2) 1

26

-

(s

a)2

(s - a)2

+ a3 4a3

S4

+ 4a4

S4

+ 4a4

S

30

b2

-

cos bt a2

1 -e"'sinbt b

e'"cosbt

+

3a2 s3

29

31

+ b2

s-a

27

28

+ (e - a)e" + (a - b)e" e)(e - a)

(a - b)(b -

a2

17

24

O)

1 . - sm al a

S2+ a2

S2

21

e)e'"

(s - a)(s - b)(s - e)

16

20

-

(1)

b2

e-a' - e"'/2

(

atJ3

cos~

r;.

sin al coshal - cosal sinh al I ~2a sin al sinh al 1 32a (sinh al - sin al)

at Ij

- ...;3sm-T

)

1 APPENDIX

461

A

I(s) s 32 1 s-¡---¡ - a 33 1 34

1 22a (eosh al

eos at)

e' dn Ln(/) = - -(tne-') n! dtn

t-T

--

s

s

1

(s-a)

)mea'(1

36 1

2fit3(eh' - e"') 1

1 ,fi+a

)m - ae""erfe(ajt)

38 1

,fi s - a2

39 I

,fi s + a2

40 1

1 JS
2 -e-a" a..fi

JS
+ 2al)

1

-

371-

- a2)(b + ,fi)

e""[b

fajl J,o eA'dA

-

a erf(ajt)]

-

beh" erfe (bJi)

1

43 1 441

O)

8a3s2

35 I

42 1

-

(1)

(S2 + a2)3

'1 s

41 1

F(/)

e"" erfe (aJi)

JS<,fi + a) 1 (s + a).¡;+b b2

45 1 JS
462 1 (1

-

J

- a2

a2)(';;

+ b)

n!

- sr

s"+t

(2n)!)mH 2n(Ji) n!

47 I -(1 - s)

,

sn+l,fi 1 Ln(t) is the Laguerre

1t(2n + 1)!

polynornial

of degree n (See. 162).

2Hn(x) is the Herrnite polynornial, Hn(x) = e"'(dn/dxn)(e-X').

H

~

2n+ 1('" t)

462


Tabla A.2

("continuad)

/(s)

(t > O)

F(t)

Js + 2a - 1

-fi

49

50

1

~fi+b r(k) (s + 4(s + b)k

~.fi+b
55

56

(~ (~

.fi~

(k >0)

(v> -1)

1

J

S2

+

a2

(~-st

(v> -1)

+

aYJ y(at)

a2

1 (S2 + a2t'

(k > O)

58

(k> O)

59

(v> -1)

60

Ik_t(a ~ bt)

te-tI4+bIt[Io(a~ bt) + Il(a ~ bt)]

+ b)

(a- bY' + fi+b)2k + .fi)-2Y

JS2

57

~ br\-tt4+b1.

~-.fi ~+.fi

53

54

fiL

1

51

52

(k >0)

1 (S2- a2t'

(k> O)

aY1y(at)

!... k-t It-t(at) r(k) ( 2a) Sk(t) =

61

-

fi

O

O when O < t < k

{1 when t > k

when O < t < k

62

e-U SZ

{t - k when t > k

63

e-o' s"

O when O< t < k (t - k)"-' when t > k { r(Jl)

(Jl > O)

II.(x) = i-.J.(ix), where J. is Bessel's function of the first kind.

I

APPENDIX A

463

F(t)

I(s)

64

1-

e-b s

I when O < t < k

{O when t > k

65

1 I + coth !ks s(1 - e-Os)= ~

66

1 s(eh - a)

67

68

69

(t > O)

1 + [t/k] = n when (n - I)k < t < nk (n = 1,2,...) (Fig.5) O when O < t < k 1 + a + a2 + ... + a"-I

{ when nk < t < (n + I)k (n -- 1,2,...) M(2k,t) = (-1)"- I

1 - tanh ks s

when 2k(n

-

1) < t < 2kn

(n = 1,2,... )(Fig. 9) 1 1 1 - (-1)" -M (kt ) + - =

2

s(1 + e-k') 1 "2 s tanh ks

'2 2 when(n - I)k < t < nk

H(2k,t) (Fig. 10)

s sinh ks

F(t) = 2(n - 1) when(2n - 3)k < t < (2n - I)k (t > O)

5 cosh k5

M(2k,t + 3k) + 1 = 1 + (-1)" when(2n - 3)k < t < (2n - I)k (t > O)

72

: coth k5

F(t)

73

k -cothS2+p

70 71

5

1tS 2k

(52 + 1)(1

- e-ns)

Isin ktl

Hsin t + Isin ti)

1 -e-Ik/')

I

5

1

76

1"

.Jii cos 2ft 1 .Jii cosh2ft

77

1

78

"j;k sin2ft 1

79 80

1 when 2k(n - 1) < t < 2kn

1

74 75

= 2n -

J;k sinh2ft 1 ~e-Ik/')

(/l > O)

t I.-I)/2 ( k}

J.-I(2ft)

I

464

OPERATlONAL

Tabla A.2

(continuad) I(s)

81

(JI> O)

82

(k> O)

83

erfe (2~) 1

(k ~ O)

jmexp 85

89

\ 90

( -41 )

V;

(

4t)

-kerfe

~ ( 2Jr )

(k ~ O)

-e"ke""erfe(a.jÍ + 2~) + erfe(2~)

s(a + ./S) e -k./S

(k ~ O)

fl.a + ./S) e- k,/$+.i

e"ke""erfe (aJr" + 2~) O

when O < t < k

Js(s + a)

{ e-t"loHaJt2=k2)

when t > k

{ ~o(aJt2=k2)

when O < t < k when t > k

{~ o(aJt2=k2,

when O < t < k when t > k

~ e-k./sl-G1

J S2

91

P

2 f!...exp-~

(k ~ O)

86

88

(t > O)

F(t)

(k ~ O)

84

87

MATHEMATICS

e-k(~-S)

~

a2

(k ~ O)

92

when O < t < k

!

when t > k I

1

93

when O < t < k

J

94

when t > k

(v> -1) ~(~+s)V

95

I -Iogs s

96

1 l< log s

!

1

-

f'(1) (k> O)

k- 1

t

log t r'(k)

{[r{kJF -

[f'(I) = -0.5772] log t f(k)

}

APPENDIX

465

A

f(s) 97'

F(t)

logs (a>O) s-a

e"'[loga + E1(at)]

logs s2 + 1

eos I Si I - sin I Ci I

s logs S2 + I

-sinlSil

100'

1 -Iog(l s

101

logd s-b

¡.

(t > O)

- eoslCil

(k> O)

+ ks)

1 J>t r -(e- - e") I

102 103

(a> O)

104

\

(a > O)

210g a - 2 Ci (ar) 2 -[al log a + sin al - al Ci (al)] a

105

2 -(1 - eos ar) I

106

2 -(1 - eoshae) e 1 .

k

-s

107

aretan

108

1 k -aretan s s

109

e"'" erfe (ks)

-Sin e

kI

Si (ke) 1 (k> O)

(2

-exp

(

k-fi 110

~e"'" erfe (ks) s

111

e'" erfe .Jks

112

1

fierfe

113

(k > O)

- 4k2)

erf (;k)

I

-fi

(k> O)

nJt k

\

{

(.Jks)

1

1 (k> O) Jse'" erfe (.Jks)

Jn(t + k)

I The exponential-integral funetion El(t) is defined in See. 35. For tables of this funetion and other integral funetions, see, for instance, Jahnke and Emde, "Tables of Funetions." 2The eosine-integral funetion is defined in Seco35. Si t is defined in Seco22.

/

I L..

466

Tabla


A.2 (continuad) I(s)

F(t)

(t > O)

1

;t sin (2kfi)

114 I erf (~)

1, /Serfc - k

115 I -é

Js

(

Js

-e1 )

116 I ne-kslo(ks)

Fe

-2e"".'t

[t(2k - t)]-t when O < t < 2k when t > 2k

{O

k

117

118

e-ksll(ks)

e'''EI(as)

, 119 I -1 - se"EI(as) a 120 I 121 I

-

t

{ ~kJt(2k -

t) when when 0< t > 2k t < 2k

1 t + a (a > O) 1 (t + a)2 (a > O)

( - Sis) coss + Ci s sin s

1 t2 + 1

(i - Sis) sin s - Ciscoss

t t2 + 1

J

/

Appendix B Tables of Finite Fourier Transforms

Table 8.1

Finite sine transforms

(n = 1.2....)

f/.n)

r

= S"{F}

o F(x)sinnxdx

2

I

1

F(x) F(x) 2 '"

- I

f.(n)

1t n= 1

3 I -n2f.(n)

+ n[F(O) - (-I)"F(n)]

1 4 I 2f.(n) n 5

I

(-Ir+

(O< x < n)

f.(n) sin nx

F"(x)

"

x

x

f

(n - r)F(r) dr n ~ 11.(11)

f

(x - r)F(r) dr

o

F(n - x)

6 I 2f.(n)cosnc

F1(x + e) + F1(x - e)

2 7 I - f.(n)h.(n) n

f f

8 I

n-x

" H(Y) x+, F1(r)drdy

o

n

1 9 I -(-lr+1 n

x-y

x n

1 10 I -[1 n - (-1)"] 11 I

n

cosnc

12 I ..; n sin nc 1

-x

(O< e < n) (O< c < n)

ifx
{n-xifx>c - c)x ifx ~ c {(n - x)c if x ~ e (n

See Chap. 11 for details and operations not tabulated here. 467

OPERAT/ONAL MATHEMATICS

468

1I Tabla B.1

,I

(continuad)

(n = 1,2,...)

fs(n)

(O< x < n)

F(x)

6n 13 1 n3

x(n - x)(2n - x)

2 14 I -[1 n3

x(n - x)

- (-1)"] .

15 I n2 -(-1)" n

+I

2

--[1-(-1)"] n3

16 I 1t(-I)" - -n ( n3 6 n2) 17 1 181

n2 + el

[I - (-O"eCO]

n

sinh e(n - x)

(e #
n + c

19 I 201

n + c

cosh ex

[1 - (-1)" cosh cn]

n (k #< O, + 1, + 2,.. .)

n -

n 21 I Oirn #
n n-k

I

sinh en

n

[1 - (-l)"coskn]

sin k(n

-

x)

sin kn sinmx

(m = 1, 2,. ..)

cos kx

(k ,¡, :t 1, :t 2, .. .)

1 - (-O"'+'. 23 I n n-m 2 2 Irn #
cos mx

2kn 24 1 (n2\- k2)2

(k #< O,:t 1, :t 2, . . .)

!... sin k(n ak sin kn

25 I

(e #
!... sinh e(x - n) oe sinh en ]

2cn (n2 + C2)2

26 1 4c2(sin2cn + sinh2 en)n n4 + 4e4 27 I 2k2n(-1)" 4 4 n-k 28 1

e2(- O" n(n2 + e2)

29 I

k2( -1)" n(n2 - k2)

30 I b"

(e #
(e #
(-1 < b < 1)

[

X)

]

[

(k #
/ (k

(m = 1, 2, .. .)

sin ex sinh e(2n

-

x)

- sin e(2n - x) sinh tx sinh kx --sinh kn

sin kx sin kn

sinh ex --sinh en

x n

x ---

sin kx

7t

sin k7t

2 bsinx 7t 1 + b2 - 2b cos x

APPENDIX

469

B

fJ.n)

j

e-ny

bn 32 I n

(n = 1,2,...)

n cosh y

2

cos x bsinx

2 sinx - arctan n e'-cosx

(y > O)

35 I 1 - (-I)n e -n, n

-

-arctan n 1 - bcosx

(-1 O)

1 33 I -e-ny n

F(x)

2 2bsinx - arctann 1 - b2

(-I
----

2 sinx - arctann sinhy

(y> O)

Table B.2 Finite cosine transforms fM

(n

= O,1,2.00 .)

I

(O< x < n)

F(x)

I: F(x)cosnxdx = Cn{F} 2

F(x)

.fc(n)

.fc(0) +

1t

~

~

.fc(n) cos

1t..=1

nx

3 I -n2/.,(n)- F'(O)+ (-lrF'(n)

F"(x)

4 I

. /.,(0) f. (x - r)F(r)dr+ -(x~. - n)2+ A

n

/.,(n)ifn = 1,2....

5

(-Ir/.,(n)

F(n - x)

6

2.fc(n)cos ne

Fix + e) + F2(x- e)

7

28n {F} sin ne

F¡(x + e) - F¡(x - e)

8

1 -Sn{F} irn = 1,2,...

9

2/.,(n)ht(n)

10

"

- s: F(r)dr+ A I:. F2(x- r)H2(r)dr

/.,(n) ir n .¡. 0;/.,(0) + An ir n = O

1

11 I O irn = 1,2'00' ;(O) = n I 12 I - ir n = 1,2.... ;.fc(0) = O n 131

2 -n sin ncir n .¡.O;/.,(0)= 2c

F(x) + A

-

n

-~ log (2 sin~) l ifO < x < e { - 1 ire < x < n

I J

470


f,(n) 14

nI (-1)"n2 - 1 ifn 'fi 0;.fc(0) = 2"

15

2( - 1)"if n 'fi O;h(O) = -

16 17

21t

19

n

-

n

241t --:¡-(-I)n n

cosh e(n - x) e sinh en

~'fi~ (-I)"e'n-I n2 + e2

1 -e'X e

~'fi~

1

-

22

(-I)m+n-I n2 - m2

(k #< O, :t 1, :t2,...)

¡fn 'fi m;.fc(m) = O 1

n2 - 1 ¡fn 'fi l,h(l) (-Ir!

sinkx k sinmx m

(m = 1,2,...)

1 .

-x Sin X 1t

= -4

cos mx

Oifn'fim;.fc(m)=z

2e(_I)n+! (n2 + e2)2 ~

27

a

~'fi~

(-1 

~>~

O)

1 --Iog(l 1t

+ b2 - 2bcosx)

y - log(2coshy - 2cos x)

1a

sinh y

1t éiy( cos x

31

= 1,2,...)

sinhy 1tcoshy - cos x

bn

-n

(m

OC( e sinh

;

~>~

28

30

k sin k1t

1t

24

26

cos k(1t - x)

(k 'fi O. :t 1, :t 2, . . .)

k2

(-1)" cos k1t - 1 n2 - p

25

n 6

¡fn 'fi 0;.fc(0) = O

21

23

x

(n --- - X)2 2n

2" If n 'fi O; .fc(0) = O

n2

(O < x < n)

3

1.

20

I

F(x)

1t3

18

f

(n = O, 1,2,...)

n bn -2 -n! if n -# O.' j,(0) e = 1t

-

cosh y )

exp (b cos x) cos (b sin x)

I

j

Appendix e Table of Exponential Fourier Transforms

Table C.1

1

Experimental Fourier transforms

(-oo
j.(rx)

F(x) F(x)

f"" F(x)e-1OXdx= E.{F}

lim -

2 I J.(rx)

I

r

1 -p p"" 27t

(m = 1,2,...) 43

.

xF(x) (e real)

6 I J.(rx - e) 7 I 1.(crx)

J.(rx)e'OXdrx

P"')(x)

(irx)'"I.(rx) if(rx)

5 I eic"fe(rx)

(-oo
F(x + e)

(e real)

e'uF(x)

(e real, e #c-O)

F( ) F(-x)

981

f.(-rx) 1.(-rx)

F(x)

r

10 I J.(rx)ge(rx) 112 12

131

1

-"" F(t)G(x

1 e + irx

-

1

e - irx (e + irx)2

(e> O)

e-e",so(x)

(e > O)

I é"'So(-x)

(e> O)

I

See Chap. 12 for details and Sec. 136 for references

2S0(X)

= O if x

-

t) dt

---

xe-e",So(x)

to other tables.

< O, So(x)= 1 if x> O. 471

472


Tabla C.1

(continuad)

(-oo
f.(a)

14

15

1 (e - ia)2 2e

(e> O)

- xe""So(- x)

(e> O)

exp (-elxl)

16

17

4ica (e2 + a2)2

18 19

n!

(e real)

exp(-Ix - el)

(e> O)

-x exp (-elxl)

e (e> O)

7t exp (-elal)

e2 + X2

(e> O)

i7ta exp (

-

2ex elal)

(e> O)

22

-

(e2 + x2)2

(e> O)

23

2 . -sIDea a

24

21-cosea ia

25

(-oo
(n = 1,2,...)

(Rep > O)

(p + ia)n+1

20 21

F(x)

(e> O)

So(x + e) - So(x - e)

(e> O)

(e> O)

-1 if -e < x < O, ifO < x < e, O¡flxl > e

{1

COS2ex iflxl < ~ iflxl >2e

{O

a1 i

i i

J

Appendix D Table. of Fourier Sine and

Cosine Transforms

Table D.1 Sine transforms

on the half IIne

(a> O)

f.(a)

(' F(x)sinax dx = S.{F(x)} 2

F(x)

(x> O)

F(x) 2 '"

-Itof

f.(a)

3

2

f.(a) sin ax da

= -SAf.(a)} It

F"(x)

4

i -E.{ 2 F.(xJ}

5

f.(ak)

6

!;la) 2

7

1

(E. in Appendix C)

F(x)

(k> O) -x2F(x)

- I.
a

f'"o F(r)r+r I%-rlG(r)dr dr

8

2f.(a) cos ak

F.(x + k) + F.(x - k)

9

f.(a + k) + f.(a

-

k)

2F(x) cos kx

10

f').a)

11

a/.{a)

-F'(x)

12

~.(a) sin :rk

F2(x- k) - F2(x+ k)

13

f.(a

(1. in Table D.2)

-

k)

-

fcta

+ k)

-xF(x)

2F(x)sin kx

. For details and propertics not listed here see Chap. 13. Rererences to other tables are given in Seco 142 473

474


Tabla D.1

(continued) f.(a)

14

(a>

O)

2f.(a)g,(a)

(x> O)

F(x)

{O F(r)[GUx

-

rl)

= {O G(t)[F(x 15 16

17

18 19 20

2

+ t) + F¡(x

-

x

a e2 + é

(e> O)

a 1 + a4 2ea

(e> O)

(e2 + é)2 8c3a (e2 + (2)3

(e> O)

x(l + ex)e-CX

1

fil

.fi

,r~

fi

2 (e> O)

22

25

a

26

4e

x

-;

(e2 + x2j2

4x 3e2 - X2

(e> O)

24

x

;-e2 + X2

(e> O)

23

1t (e2 + X2)3

(e> O)

--2

e2

(e> O)

2 e - aretan -

1tX e2

+ x2

1t

X

(e> O)

27

(e> O)

a 1

- eosea a

=

2Jc

{Oifx > e 4

sin ea a

x

erfe

1 ifO < x < e

(e> O)

29 30

G(x + r)]dr

1t

21

28

-

(e> O)

x

I x +e -Iog1t Ix - el

/

t)] dt

APPENDIX D Table

D.2

475

Cosine transforms

on the half Une

(rx> O)

fc(rx)

F(x)

fO o F(x)cosrxxdx = C.{F(x)}

F(x)

2

fc(rx)

-non s

3

-rx2fc(rx) - F'(O)

F"(x)

4

tE. {F(lxIH

F(x)

5

fc(rxk)

2

O>

(x> O)

2 fc(rx)cosrxxdrx = -Cx{fc(rx)}

(k> O)

- x2 F(x)

6 7

2fc(rx)gc(rx)

8

2fc(rx)cos rxk

9

fc(rx+ k) + fc(rx- k)

2F(x) cos kx

10

f~(rx)

xF(x)

11

1 -f.(rx) rx

{O>

(k > O)

(fs in Table D.I)

2f.(rx)sin rxk

13

f.(rx + k)

14

2f.(rx)g,(rx)

-

f.(rx

-

F(x + k)

-

2F(x) sin kx

F1(x

-

k)

o F(r)[G(x+ r) - G1(x- r)]dr.

so>

15

e (e> O)

e2 + rx2 2

16

(1 + rx2)2 17

18

(l + x)e-X

1 1 + rx4

exp

(

- fi)

20 21

sin

(¡ +

fi)

(¡ +

fi)

rx2 1 + rx4

[21

exp (

19

rl)] dr

F(r) dr

(k > O) k)

-

F(lx - kl) + F(x + k)

fO> x

12

F(r)[G(x + r) + G(lx

1

.fi.

fi)

cos

V~vÍx (e> O)

2

e

~'e2

+ X2

2 n (l

1

-

+

X2 X2)2

/ 476


Table D.2 (continued) fc(a) 22 I

e-a. - e-ea. a

23 I

-1 -e

(a> O)

1

(e> O)

24 I exp (-ea2)

e2

-Iog 1t

(

1 + -X2

1

(e> O)

1 . 25 I -e-' SIna a

)

X2

c: exp -...¡1te ( 4c) 1 2 - arctan 1t

X2

l ifO < x < c

1 . 26 I -SIn ea a

(e> O)

1 27 I - arctan a a 28 I arctan

(x> O)

(e> O)

-c«

a

F(x)

2

{O if x > c

E.(x) (E. in Seco 35)

2 -e-X sin x

x

Index

Abel, Niels, 117 Abel's integral equarion, 53 Abel's test, 234, 308 Adjoint, 277 Analytic continuation, 179 Analytic function, 165 Analytic transforms, 188 Angular displacements (see Vibrating shaft) Antiperiodic function, 68

Bar (see TemperatUre¡ Vibrating bar) Basic operational propeny, 8, 318 Beam: deflection of, 113, 120-122 shearing force in, 113 vibrating (see Vibrating beam) Bessel, F. W., 42On. Bessel's equation, 62, 63, 420 Bessel's functions, 63, 421 product of, 435 zeros of, 241,422,425 Biharmonic equation, 345,410 Bilinear formula, 332 80undary value problem, 127, 397 linear, 159 in ordinary differential equations, 278-290 uniqueness tbeorem for, 283 superposition of solUtions of, 139, 234 uniqueness of solution of, 228-230,23411., 263. verification of solutions of, 128, 161, 222226,238-240,257,259,308 (See also Diffusion; Temperature¡ under Vibrating)

I I

Branch of function, Branch cut, 178

177

Cauchy, A. L., 3 Cauchy-Riemann conditions, 164 Cauchy's integral formula, 170 extensions of, 181 Cauchy's integral tbeorem, 170, 182 Characteristic function (see Eigenfunctions) Characteristic number, 288 ' (See also Eigenvalues) ClassC', C", 281,327 Concentration, 144 (See also Diffusion) Continuity: sectional, 5 by subregions, 39 Contour integral, 169 Convolution, 318-320 for cosine transforms, 406 for exponential transform, 391-396 for finite cosine transforms, 356, 358, 371 for imite exponential transform, 378 for imite sine transforms, 335, 351, 354, 359, 373 generalized, 83, 324 for Hankel transform, 435 - joint, 359, 406 for Laguerre transform, 452n. for Laplace transform, 45-48 477

478

INDEX

Convolution:

Exponential function, 32, 166 Exponential-integral function, 110 Exponential order, 6

for Legendre transform, 445 for Mellin transform, 455 for modified Fourier transform, for sine transform, 407

417

Cosine-integral function, 110 Cylinder, diffusion in, 159 (See also Temperature)

Damping, viscous, 86 Difference equation, 30, 36 Differential equation, 20-24, 50-54 boundary value problem in, 278 existence of solution of, 278 initial-value problem in, 278 partial, 123 (See also Boundary value problem) self-adjoint, 277 with variable coefficients, 61-64 Differential form, 277 adjoint of, 277, 285 transform of, 2, 325 Diffusion, 143, 144 in composite solid, 157, 159 equation of, 144 in hemisphere, 451 in plate, 154, 159 in solid, 245, 249 Diffusivity, 144 Dirac delta function, 34 Dirichlet problem, 448, 450, 451 Distributions, 34 Duhamel's formula, 249, 252, 271

Eigenfunctions, 288 orthogonality of, 290 Eigenvalue problem, 288, 303 singular, 309 Eigenvalues,288,290,296 continuous, 309, 312 Elastic displacements (see Vibrating bar; Vibrating beam; Vibrating membrane; Vibrating sbaft; Vibrating string) Elastic support, 133, 270, 279 Elecnic circuit, 87, 102-106 Kirchhoff's laws for, 102 Entire function, 165 Error function, 46 complementary,80 Evaporation, 154, 159, 245

FaltUng integral, 358 Final-value theorem, 83 Flux of heat, 143 Formal derivation, 8 Fourier-Bessel integral, 434 Fourier-Bessel series, 423, 426 Fourier constaMS, 285 Fourier integral formula, 198, 312, 315,391 Fourier series, 285, 288, 302 generalized, 285 representation theorem for, 291 Fourier transform, 2 cosine, 323, 341,40+408 table, 475, 476 exponential, 323, 3~397 table, 471, 472 finite cosine, 354-360 table, 469, 470 finite exponential, 376 table, 379 finite sine, 339, 348-354 table, 467-469 generalized cosine, 370 generalized sine, 372 on half !ine, 4Oi-418 modified, 414 modified finite, 333 convolution for, 325 modified sine, 368 table, 369 sine, 320, 337,401-407 table, 473, 474 Function: even extension of, 405 of exponential order, 6 generalized, 34 multiple-valued, 177 odd extension of, 135, 405 periodic extensioyof, 378 sectionally continuous, 5

Gamma function, 11 Green's function, 237, 244, 279-283 construction of, 281, 286 transform of, 332

INDEX

479

Hankd, Hermann, 42On. Hankd transform, 420-437 finite, 338, 341,422 on half line, 432-437 modified

imite,

nonsingular, table, 436 Harmonic

425

431

function,

in circular regions, 366, 382 in half plane, 398 Heat conduction, 143,219

Heaviside's expansion, Hooke's law, 86 Hyperbolic functions,

sine, 337,402 Sturm-Liouville, 328 Inversion integral, 193-201 aIteration of, 213-216 derivatives of, 202-204 series representation of, 206-213 Iterated transforms, 83, 252

177

(See also Temperature) Heat equation, 144 fundamental solutions Heaviside, Oliver, 3

Inverse transformation:

Kernd, 2, 317

.

Kernel-product property, Kirchhoff's laws, 102

319-324

of, 150 71 166, 167

Image, 3 Impulse function, 30 Impulse symbol, 33-39 lnitial-value problem, 278 existence theorem for, 278 Initial-value theorem, 83, 191 Inner product, 25, 284 Integral equation, 52-54, 116, 297 of convolution type, 52 Integral transformation, 2 general linear , 317 kernd of, 2, 317 singular, 336 Integrals: evaluation of, 106-112 improper, 13,41,50 limit of, 82 principal value of, 193 uniform convergence of, 41 Inverse transformation, 14 cosine, 404 exponential, 385-387 finttecosine,339,355,370 imtteexponential,377 finite Hankd, 424, 426 finite sine, 334, 349, 372 Hankel, 434 Laguerre, 453 Laplace, 14-16, 193 Legendre, 444, 446, 447 Mdlin, 455 modified Fourier, 416

Lagrangeidentky,277,285 Laguerre polynomials, 452 Laguerre transform, 452 Laplace, P. S., 3 Laplace transformarion, 3 basic operational property convolution for, 45-48 of derivatives, 8-10 generation of, 24 i[¡verse of, 14-17, 193-202

of, 8

operational propenies of, table, 458, 459 translation property of, 28 Laplace transforms, 3 analytic, 186-189

derivativesof, 54-56, 188

.

integrarion of, 66 order properties of, 189-193 series of, 57-60 tables, 18,458-465 Laplace's equation, 177 Laplacian in spherical coordinates, 442 Laurent's series, 172 Legendre polynomials, 311, 316,443 product property of, 445 Legendre transform, 442-448 on interval (-1,1), 443 ./ on interval (0,1),446,447 Legendre's equation, 311,443 Legendre's series, 444 Leibnitz formula, 39 Linear combinarion, 2 Logarithmic function, 179

Mellin transform,

453-455

Mortaltty of equipment, 119-122 Motion of a particle, 11 5, 122

INDEX

480 NatUral frequency, 86 Neumann problem, 448, 450 Norm, 284

/

Object function, 3 Orthogonal functions, 284 Orthonormal set, 284 Oscillation: stable, 76 unstable, 210

Partial differential equation, 123 (See also Boundary value problem) Partial fractions, 17, 70-78 Periodic funcrion, 66,265, 356 rectification of, 70 transform of, 67-70 Plate, deflections in, 410, 413, 414 Poüson~equabon,360,426 Poüson's integral fomrula, 366, 382, 399 Pole, 174 Potential: in cylindrical region, 314 in slot, 360 Producr: of temperatUre functions, 251 of transforms (see Convolution)

Reflection, principIe of, 180 for Laplace transforms, 188 Replacement of equipment, 119 Residuetheorem, 174,183 Residues, 174 series of, 206-213 Resonance, 92, 259 in bar, 258-266, 273 in beam, 269, 273, 274 in membrane, 274 in propeUer shaft, 274, 275 in string, 264 in vibrating spring, 92, 96,100 Resonance type, 210 Riemann-Lebesgue theorem, 192

Schwarz integral formula, 399 Sectionally continuous derivative, 198, 206 Sectionally continuous function, 5 Self-adjoint form, 277

Semi-inf'mite solid: diffusion in, 245 temperarure in, 145-151, 155, 250, 413 Separation of variables, 293, 306 Servomechanisms,117,121 Sbear, modulus Es' 126, 275 Sifting propeny, 33 Sine-integral function, 66 Singular point, 62, 173, 310, 311 isolated, 174 Slab, temperarure in (see Temperarure) Square-wave funcrion, 67 Stable oscillations, 76 Staircase funcrion, 30 Steady temperarure, 312 in cylinder, 413, 426 in hemisphere, 451 in plate, 313, 381 in prism, 313 in quadrant, 408,.418 in sIab, 341, 364, 365, 374, 375, 440 in solid, 437 in sphere, 4:49 in waIl, 305 Srep function, 6, 31 String, transverse displacements in, 124 static,279,411 (See also Vibrating string) StUrm-Liouville expansion, 294 StUrm-Uouville problem, 288, 304 reduced fnrm of: 292 Srurm-Liouville rransfonn, 327 inverse of, 328 Successive transformations, 161, 362, 427 Superposition of solutions, 139, 234, 293 Surface heat rransfer, 144

/

Tautochrone,115 Taylor's series, 171 Telegraph equation, 126, 272 Temperarure, 143 in bar, 153, 158, 220, 233 in composite solids, 155, 159 in cooling f'm, 153 in cylinder, 240-244, -315, 430 periodic, 250 in slab, 149, 151,158, 233-237, 242, 243, 368 in soU, 251 in salid$, 145-149, 220-252, 315,400, 413

j

I 481

INDEX

t r

¡ ! j

1

l.

TemperatUre: steady-state (see Steady temperatUre) TemperatUre functions: product of, 251 propenies of, 226-228 Torque, 126 Transfer function, 104 Transform, 2 Transformation of functions: Iinear,2 linear integral, 2, 317 Translated function, 28 Transmission line equations, 272 Triangular wave function, 68

Uniqueness: of inverse transform, 1S, 201, 328 of solutions, 229, 278, 283 Unit step function, 6, 31 Units, 153, 242 Unstable oscillations, 210

Vibrating bar: with mass attached, 266 resonance in, 259-2~ Vibrating beam, 268, 367 resonance in, 269, 273, 274 Vibrating membrane, 274,430 resonance in, 274 Vibrating shaft, 126, 142,263,272 resonance in, 274, 275 Vibrating spring, 85-101 Vibrating string, 126-135,264, 314, 366, 367, 400 equations for, 124, 125 resonance in, 264 Vibration, forced, 92, 95 Vibration absorber, 96-98,101

Wave equation, Weierstrass

123-126

test, 41

Weight function,

284

I Vibrating bar, 132-142, 253-268, 314, 375 equations for, 125

Young's modulus, 113 for steel, 142

Churchill, R. v. - Operational Mathematics

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