Contents Notation
5
1 Laplace Transform
7
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
1.2 Proper perties of Laplace Transform . . . . . . . . . . . . . . . . . . . . .
8
1.3 1.3 Furt urther her Prop Propeertie ties of the the Lapl Laplac acee tran transf sfor orm m . . . . . . . . . . . . . .
14
1.4
Dirac Delta Function . . . . . . . . . . . . . . . . . . . . . . . . . . .
16
1.5 Differentiation of Transform . . . . . . . . . . . . . . . . . . . . . . .
17
2 Vector Calculus
19
2.1 Rectangular Coor oordinates In 3-Space . . . . . . . . . . . . . . . . . . .
19
2.2 Surfaces In 3 Dimensions . . . . . . . . . . . . . . . . . . . . . . . . .
20
2.2.1
Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20
2.2.2
Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
2.2.3
Cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22
2.2.4
Quadric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . .
23
2.2.5
Graph aphs Of Two-variable Functions . . . . . . . . . . . . . . .
28
2.3 Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29
1
2.3. 2.3.11
Some Some Ap App plica licattions ions Of Do Doub uble le Inte Integr gral alss . . . . . . . . . . . . .
29
2.3. 2.3.22
Doub Do uble le Inte Integr gral alss over Rect Rectan angu gula larr Regio egions ns . . . . . . . . . . .
29
2.3. 2.3.33
Doub Do uble le Inte Integr gral alss over NonNon-re rect ctan angu gula larr Regi Region onss . . . . . . . . .
30
2.3. 2.3.44
Doub Do uble le Inte Integr gral alss in Polar olar Co Coor ordi dina nattes . . . . . . . . . . . . . .
32
2.4 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
33
2.5 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35
2.5.1
Line Integrals of Vector fields . . . . . . . . . . . . . . . . . .
40
2.6 Green’s Theorem in the Plane . . . . . . . . . . . . . . . . . . . . . .
44
2.7 Surface Area and Surface Integrals . . . . . . . . . . . . . . . . . . .
46
2.7.1
Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46
2.7.2
Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . .
47
2.7.3
Surface Integrals als of Vector Fields . . . . . . . . . . . . . . . .
48
2.8 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50
2.9 Gauss’ Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . .
52
3 Z Transform
55
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55
3.2 Prope Propert rtie iess of of Th Thee Z Transform . . . . . . . . . . . . . . . . . . . . .
58
3.3 More More Propert Properties ies of The Z Transform . . . . . . . . . . . . . . . . . .
59
3.4 Th Thee Inv Inverse erse Z Transform . . . . . . . . . . . . . . . . . . . . . . . . .
60
3.5 Solving Difference Equations . . . . . . . . . . . . . . . . . . . . . . .
61
3.6 A Table Of Z Transform . . . . . . . . . . . . . . . . . . . . . . . . .
63
4 Complex Analysis
65 2
2.3. 2.3.11
Some Some Ap App plica licattions ions Of Do Doub uble le Inte Integr gral alss . . . . . . . . . . . . .
29
2.3. 2.3.22
Doub Do uble le Inte Integr gral alss over Rect Rectan angu gula larr Regio egions ns . . . . . . . . . . .
29
2.3. 2.3.33
Doub Do uble le Inte Integr gral alss over NonNon-re rect ctan angu gula larr Regi Region onss . . . . . . . . .
30
2.3. 2.3.44
Doub Do uble le Inte Integr gral alss in Polar olar Co Coor ordi dina nattes . . . . . . . . . . . . . .
32
2.4 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
33
2.5 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35
2.5.1
Line Integrals of Vector fields . . . . . . . . . . . . . . . . . .
40
2.6 Green’s Theorem in the Plane . . . . . . . . . . . . . . . . . . . . . .
44
2.7 Surface Area and Surface Integrals . . . . . . . . . . . . . . . . . . .
46
2.7.1
Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46
2.7.2
Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . .
47
2.7.3
Surface Integrals als of Vector Fields . . . . . . . . . . . . . . . .
48
2.8 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50
2.9 Gauss’ Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . .
52
3 Z Transform
55
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55
3.2 Prope Propert rtie iess of of Th Thee Z Transform . . . . . . . . . . . . . . . . . . . . .
58
3.3 More More Propert Properties ies of The Z Transform . . . . . . . . . . . . . . . . . .
59
3.4 Th Thee Inv Inverse erse Z Transform . . . . . . . . . . . . . . . . . . . . . . . . .
60
3.5 Solving Difference Equations . . . . . . . . . . . . . . . . . . . . . . .
61
3.6 A Table Of Z Transform . . . . . . . . . . . . . . . . . . . . . . . . .
63
4 Complex Analysis
65 2
4.1 Revision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
65
4.2 Loci oci and Regions of the Complex Plane . . . . . . . . . . . . . . . . .
68
4.3 Functions of a complex variable . . . . . . . . . . . . . . . . . . . . .
69
4.4 More Elementary Functions . . . . . . . . . . . . . . . . . . . . . . .
73
4.5 Complex Integration . . . . . . . . . . . . . . . . . . . . . . . . . . .
78
4.6 Two Integration Methods . . . . . . . . . . . . . . . . . . . . . . . .
80
5 The Fourier Integral and Fourier Transforms
89
5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
89
5.2 Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
93
5.3 Some Prope operties of the Fourier transform . . . . . . . . . . . . . . . .
95
5.4 A table of Fourier transforms . . . . . . . . . . . . . . . . . . . . . .
99
6 Partial Differential Equations
101
6.1 Revision : Ha Half-range Expansions . . . . . . . . . . . . . . . . . . . . 101 6.2 6.2 Revi Revisi sion on : 2nd Orde Orderr HLDE HLDE With With Co Cons nsta tan nt Co Coeffi effici cien ents ts . . . . . . . 102 102 6.3 Partial Differential Equations . . . . . . . . . . . . . . . . . . . . . . 104 6.4 6.4 Solv Solvin ingg Part artial ial Diff Differen rential ial Equa Equati tion onss . . . . . . . . . . . . . . . . . . 106 106 6.5 Eigenvalue Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 6.6 Separation Of Variables Method . . . . . . . . . . . . . . . . . . 109
Bibliography
113
3
4
Notation
denotes
“since”, “because of”.
∴
denotes
“therefore”, “thus”, “hence”.
denotes
“there exists”, “there is a/an”, “there are some”.
denotes
“there does not exists”.
denotes
“for all”, “for every”, “for each”.
denotes
“belongs to”.
denotes
“does not belongs to”.
∃ ∃ ∀ ∈ ∈/
∈ A means “x is an element of the set A” A ⊆ B means “A is a subset of the set B” x
R
=the set of real numbers
R2 R3 C
= (x, y) x, y
{ | ∈ R} denotes the set of all ordered pairs of real numbers. = {(x,y,z) | x,y,z ∈ R} denotes the set of all ordered triples of real numbers.
=the set of all complex numbers.
{ ···} =the set of all natural numbers (positive integers). Z = {0, ±1, ±2, ···} =the set of all integers. p ⇒ q means “ p implies q” N
= 1, 2, 3,
5
6
Chapter 1 Laplace Transform 1.1
Introduction
Definition 1.1.1. Let f (t) be defined for t
∞
0
≥ 0. Then
e−st f (t) dt = lim b→∞
b
e−st f (t) dt
0
is called the Laplace transform of f , provided that the improper integral exists.
Note 1. (a) Usually we denote the Laplace transform of f by L f (t) or F (s).
{ }
(b) The domain of the transform F (s) is taken to be all values of s for which the improper integral exists. (c) If L f (t) = F (s), then f (t) = L−1 F (s) is called the inverse Laplace trans-
{ }
form of F.
{
}
7
Example 1. (a) Let f (t) = 1, t
≥ 0. Then L{f (t)} = 1s for s > 0. In general, L{a} = as , s > 0
where a is a constant.
≥ 0. Then L{f (t)} = s1 for s > 0. Using mathematical induction, 1 n! t we can show that L{t } = or L− . = (n)! s s
(b) Let f (t) = t, t
2
n
1
n+1
(c) Let f (t) = eat , t
1.2
n
n+1
≥ 0. Then L{f (t)} = s −1 a for s > a.
Properties of Laplace Transform
Theorem 1.2.1 (The Linear Property Of The Laplace Transform) . Let f and g be functions whose Laplace transforms exist, and let a and b be constants. Then L af (t) + bg(t) = aL f (t) + bL g(t) .
{
}
{ }
{ }
Note 2. The inverse LT is also linear. Example 2.
{ } s +ω ω s (b) Show that L{cos ωt} = s +ω (a) Show that L sin ωt =
2
2
for s > 0.
2
2
for s > 0.
Example 3. Find the inverse LT of F (s) = Answer. f (t) = 2 + 3e5t
5s s2
− 10 . − 5s
Reading Assignment 1. Find the inverse LT of G(s) = Answer. By partial fractions, 5s 4 A Bs + C = + 2 = G(s) = s(s2 + 9) s s +9 Taking inverse Laplace transform, 4 5 4 + sin(3t) + cos(3t) g(t) = 9 3 9
−
·· · = − 49 1s + 53 s
5s 4 . s3 + 9s
−
3 4 s + 2+9 9 s2 + 9
−
8
Definition 1.2.1 (Sufficient Conditions for Existence of L f (t) ).
{ } (a) A function f is said to be piecewise continuous on [0, ∞) if , in any interval 0 ≤ a ≤ t ≤ b, there are at most a finite number of points t , k = 1, 2, . . . , n , k
at which f has finite discontinuities and is continuous on each open interval tk−1 < t < t k .
(b) A function f is said to be of exponential order ect if there exist constant c and ct
positive constants M and T such that f (t)
| | ≤ Me
for all t > T.
Notes : This says that f (t) does not grow faster than the exponential function Mect .
Theorem 1.2.2 (Existence of the Laplace Transform) . If f (t) is piecewise continuous on [0,
ct
∞) and of exponential order e
, then the Laplace transform of f (t) exists for
all s > c.
Theorem 1.2.3 (L.T of the derivatives of f ). (a) If f (t) is continuous for all t
γt
≥ 0 and is of exponential order e
, and if f (t)
exists and piecewise continuous on every finite interval in the range t the Laplace transform of f (t) exists for all s > γ and L(f ) = sL(f )
− f (0).
(b) If f (t) is piecewise continuous , then L(f ) = s2 L(f )
− sf (0) − f (0).
(c) Hence, by induction L(f (n) ) = sn L(f )
n 1
n 2
(n 1)
− s − f (0) − s − f (0) − · · · − f
9
− (0).
≥ 0, then
Example 4. Use the Laplace Transform method to solve the initial value problem (a) y (t) + 2y(t) = e−t , y(0) = 2. (b) y (t) + 4y(t) = 5e−t , y(0) = 2, y (0) = 3.
Answer. (a) y = e−t + e−2t (b) y = 2 sin(2t) + cos(2t) + e−t Reading Assignment 2. Using Laplace transforms, solve the IVP : y
− 5y + 6y = 4, y(0) = −3, y(0) = 7.
Answer. Apply LT to the DE, [s2 Y
− sy(0) − y(0)] − 5[sY − y(0)] + 6Y = 4s
Solve for Y,
− 7] − 5[sY + 3] + 6Y = 4s 4 (s − 5s + 6)Y = 22 − 3s + s 22 − 3s 4 + Y = (s − 2)(s − 3) s(s − 2)(s − 3) [s2 Y + 3s 2
Using partial fractions, 21 18 43 1 = + Y = 3s s 2 3 s 3
·· ·
− −
−
Invert using the Laplace Transform table, 2 43 18e2t + e3t y= 3 3 Theorem 1.2.4 ( LT of Integrals). If f (t) is piecewise continuous and is of exponen-
−
tial order eγt ,then
t
L
0
for s > 0, s > γ.
1 f (τ )dτ = L(f ) s
Theorem 1.2.5 (First Shifting Theorem(s-shifting) . If f has Laplace transform F (s) when s > γ , then eat f (t) has Laplace transform F (s L eat f (t) = F (s
− a)
or L−1 10
− a) when s − a > γ . That is, F (s − a) = e f (t).
at
Example 5. Find L eat cosh bt if L cosh bt =
{
Answer.
}
{
}
s
s2
s a . (s a)2 b2
−
−
−b . 2
−
Reading Assignment 3. Find L eat tn if L tn =
{
}
{ }
n! sn+1
. n!
Answer. Let f (t) = tn so that F (s) = Then L eat tn = L eat f (t) = F (s
{
}
sn+1
.
− a) = (s −n!a)
n+1
.
Example 6. Solve y + 2y + 10y = 0, y(0) = 2, y (0) = 1. Answer. Y =
· ··
3 + 2(s + 1) = = (s + 1)2 + 32
3 s = 2 + 2 s + 32 s2 + 32 s→s−(−1)
···
y = [sin(3t) + 2 cos(3t)] e−t
Exercise 1. Find L−1
Answer. 2e4t cos(2t) +
2s + 7 s2 8s + 20
−
.
15 4t e sin(2t) 2
Definition 1.2.2 (The Convolution Integral). The convolution of f (t) and g(t) is the function
t
(f g)(t) =
∗
f (τ )g(t
0
− τ )dτ.
t
Remark 1. (f g)(t) = (g f )(t) =
∗
∗
g(τ )f (t
0
− τ )dτ
Example 7. Find the convolution of f (t) = sin t and g(t) = t. Answer. t
− sin t.
Theorem 1.2.6 ( Convolution Theorem). L (f g)(t) = L f (t) L g(t) = F (s)G(s) or L−1 F (s)G(s) = (f g)(t).
{ ∗
}
{ }{ } Example 8. Find L{t ∗ sin t}.
{
11
}
∗
Answer.
1 s2 (s2 + 1)
Example 9. Find L−1 Answer. e−t
− e−
2t
1 (s + 1)(s + 2)
by convolution.
Example 10. Solve the IVP y + y = f (t),
y(0) = 0,
y (0) = 1
where f (t) is a continuous function that has a transform.
Note : Unlike the undetermined coefficients method, the Laplace transform method can solve this IVP even if f (t) is unknown.
Answer. y(t) = sin t + (sin t) f (t)
∗
Exercise 2. Find the inverse Laplace transforms of the following functions using convolution. 8 (a) 2 2 s (s + 4)
(b)
9 (s + 1)(s
2
− 2) Answer. (a) 2t − sin2t (b) (3t − 1)e
2t
(a)
10s (s2 + 4)(s2 + 9)
+ e−t (c) 2 cos(2t)
− 2cos(3t)
Definition 1.2.3. An equation of the type t
y(t) = f (t) +
0
y(τ )g(t
− τ )dτ = f (t) + y(t) ∗ g(t)
where f and g are known functions is called Volterra integral equation
12
Example 11. Solve the given Volterra integral equation: t
y(t) = t +
y(τ ) sin(t
0
− τ ) dτ.
t3 Answer. t + 6 t
Reading Assignment 4. Solve the integral equation y = 3 + e−t
0
t
Answer. y = 3 +
e−(t−τ ) y(τ )dτ = 3 + e−t y(t)
∗
0
Apply LT, 3 Y Y = + s s+1 Solve for Y, 3 s Y = s+1 s 3(s + 1) 3 3 = + Y = s2 s s2 Taking the inverse Laplace transform, y = 3 + 3t
Exercise 3. Solve the solution of the given equation: t
t
(a) y(t) = te +
τ y(t
0
− τ ) dτ.
t
(b) y + 3y + 2
y(τ ) dτ = 2t,
y(0) = 3.
0
Answer. (a) y = (b) y = 7e−2t
− 18 e− + 18 e
− 5e−
t
t
t
3 1 + tet + t2 et 4 4
+1 13
eτ y(τ )dτ
1.3
Further Properties of the Laplace transform
Definition 1.3.1. If a H (t
≥ 0, then the unit step function or Heaviside function
− a) has a jump size 1 at t = a and is defined by H (t
− a) =
0
,
t
1
,
t > a.
Note that H (t) = 1.
Theorem 1.3.1 (Second Shifting Theorem-t-shifting). If f has Laplace transform F (s), then the function f (t
− a)H (t − a) =
f (t
0
,
t
− a)
,
t > a.
has Laplace transform e−as F (s), i.e.
L f (t
− a)H (t − a)
= e−as F (s) or L−1 e−as F (s) = f (t
Note 3. It is convenient to think of the effect of H (t
− a) as to ”switch on” f (t − a)
at time t = a, or to shift f (t) by an amount a along the t-axis.
Example 12. Consider a pulse height c from t = a to t = b f (t) = c H (t
− a) − H (t − b)} i.e. switch on at t = a and off at t = b. Find L{f (t)}. c − − Answer. F (s) = e e− s {
as
bs
Example 13. Find L g(t) if g(t) =
{ }
Answer.
1 −s (e s2
− e−
2s
)
t
0
,
0
−1
,
1
,
t
1
14
− a)H (t − a).
≤t≤1
≥2
Example 14. Find the inverse LT of F (s) =
Answer. (t
− 2)e−
e−2s . (s + 1)2
(t 2)
− H (t − 2)
Example 15. For an RC electric circuit, the current i(t) satisfies t
1 Ri(t) + C
i(τ )dτ = v(t)
0
where the resistance is R ohms and the capacitance is C farad and the electromotive force is v volts. Assuming the circuit is initially quiescent and that R = 100, C = 0.1, use the Laplace transform to find the current i(t) given that v(t) =
− e−
Answer. i(t) = (1
0 10(t
− 1)
,
t<1
,
t>1
0.1(t 1)
Exercise 4. Let f (t) =
− )H (t − 1)
2t
3
, 0
−5
,
≤t<4 t≥4
(a) Express f (t) in terms of the unit step functions.
(b) Find L f (t) . (c) Obtain the response of the harmonic oscillator x + x = f (t) to such a forcing function, given that x = 1 and
Answer. (a) f (t) = 3[1
− H (t − 4)] + (2t − 5)H (t − (c) x(t) = 3 − 2cos t + 2[t − 4 − sin(t − 4)]H (t − 4) Exercise 5. Find L−1
Answer. 2e4(t−5)
(2s + 7)e−5s s2 8s + 20
− 15 cos2(t − 5) + e 2
dx = 0 when t = 0. dt 3 1 4) (b) L f (t) = + 2e−4s 2 s s
.
4(t 5)
− sin2(t − 5) H (t − 5) 15
·
1.4
Dirac Delta Function
Mechanical systems are often acted upon by an impulsive force (or emf in an electrical circuit) of large magnitude that acts only for a very short period of time. The function f k (t) =
1/k , a 0
,
≤ t ≤ a+k otherwise
could serve as a mathematical model for such a force. (a) The impulse of such a force is I =
∞
a+k
f k (t)dt =
a
−∞
1 dt = 1. k
1 e−as e−(a+k)s (b) L f k (t) = . k s s (c) The limit δ(t a) = lim f k (t) is called the Dirac delta function.
{
}
−
−
k
→0+
(d) Note that the Dirac delta function is not a proper function. It is a generalized function characterized by the two properties (i) δ(t
− a) =
(ii) I =
∞
∞
, t=a
δ(t)dt = 1
−∞
{ − a)} = e−
(e) L δ(t
0
, t=a
as
Example 16. Solve y + 2y + 5y = 50t 50 10 You are given that 2 2 = 2 s (s + 2s + 5) s Answer. y = 10t
− 4 − 12 e−
− δ(t − 2), y(0) = −4, y(0) = 10. −2 . − 4s + s 4s + 2s + 5 2
(t 2)
− sin(2t − 4)H (t − 2)
Exercise 6. Find the solution of the equation y + 2y + 10y = 6δ(t
− 2),
y(0) = 3,
y (0) = 0.
Answer. y = e−t sin(3t) + 3e−t cos(3t) + 2e−(t−2) sin3(t 16
− 2)H (t − 2)
1.5
Differentiation of Transform
−
Theorem 1.5.1 (Differentiation of Transform) . L tf (t) = tf (t).
1
−F (s) or L− −F (s)
=
dn Note : It can be shown by induction that L t f (t) = ( 1) F (s), n = 1, 2, 3, . . . . dsn n
Example 17. Find L t cos ωt . s2 ω 2 Answer. 2 (s + ω 2 )2
−
Exercise 7. Find L t2 sin3t .
{
}
18s2 54 Answer. (s2 + 9)3
−
Example 18. Find L−1 Answer.
1 bt (e t
at
−e
s ln s
−a −b
.
)
Exercise 8. Find g(t) = L−1 ln e−3t + e2t Answer. g(t) = t
s s2 + s
−6
−1
17
.
n
Table of Laplace Transforms f (t)
F (s) = L f (t)
1
1/s n! (n = 1, 2, 3, . . .) sn+1 1 s a n! (n = 1, 2, 3, . . .) (s a)n+1 ω s2 + ω 2 s 2 s + ω2 sL f f (0)
{ }
tn eat
− −
tn eat sin ωt cos ωt f
{ }− s L{f } − sf (0) − f (0) F (s − a) 2
f eat f (t) f (t
e−as F (s)
− a)H (t − a)
−F (s)
tf (t)
t
(f g)(t) =
∗ δ(t − a)
0
f (τ )g(t
− τ )dτ
F (s)G(s) e−as
18
Chapter 2 Vector Calculus 2.1
Rectangular Coordinates In 3-Space
The three mutually perpendicular coordinate axes (the x-,y- and z-axes) form a 3dimensional rectangular or Cartesian coordinate system. Their point of intersection is called the origin of the coordinate system. Each pair of coordinate axes determines a plane, called a coordinate plane. The three coordinate planes are called the xy-plane (with equation z = 0); the xz-plane (with equation y = 0) and the yz-plane. The coordinate planes divide 3-space into eight octants. The first octant is the one for which the three coordinates are positive. In this rectangular coordinate system, a point P is space can be described by an ordered triple (x,y,z) where x = directed distance from P to the yz-plane y = directed distance from P to the xz-plane z = directed distance from P to the xy-plane 19
2.2 2.2.1
Surfaces In 3 Dimensions Planes
The linear equation ax + by + cz = d ((a,b,c) = (0, 0, 0)) represents a plane in space.
Figure 2.1: Plane:2x
20
− 5y + z = 4
2.2.2
Spheres
A sphere with center (a,b,c) and radius r is the set of all points (x,y,z) such that the distance between (x,y,z) and (a,b,c) is r. It can be represented by the (standard) equation (x
− a)
2
+ (y
− b)
2
+ (z
− c)
2
= r2 .
Figure 2.2: Sphere :x2 + y2 + z 2 = 1
Example 19. Some examples are : Equation
Graph
x2 + y2 + z 2 = 1
Sphere with center (0,0,0) and radius 1
(x
2
− 1)
+ (y + 2)2 + (z
x2 + 2x + y2
− 4y + z
2
2
2
− 3) = 4 − 6z + 5 = 0
Sphere with center (1,-2,3) and radius 4 Sphere with center (-1,2,3) and radius 3
21
2.2.3
Cylinders
An equation that contains only 2 of the variables x,y, and z represents a cylinder in space (parallel to the axis of the missing variable).
Remark 2. The cylinder can be obtained by graphing the equation in the coordinate plane of the 2 variables that appear in the equation and then translating that graph parallel to the axis of the missing variable.
Example 20. Sketch the graph of x2 + y2 = 1 in R3 . Answer. In R2 , this equation represent a circle. In R3 , x2 + y2 = 1 is a circular cylinder: it is made of lines parallel to the z axis that passing through the circle x2 + y2 = 1 on the xy plane.
−
Figure 2.3: Circular Cylinder :x2 + y2 = 1
22
−
2.2.4
Quadric Surfaces
The equation of a quadric surface in space is an equation of the form
Ax2 + By 2 + Cz 2 + Dxy + Exz + F yz + Gx + Hy + Iz + J = 0. The following are the 6 basic types of quadric surfaces: x2 y2 z 2 (a) Ellipsoid 2 + 2 + 2 = 1 a b c x2 y2 z 2 (b) Hyperboloid of One Sheet 2 + 2 =1 a b c2 z 2 x2 y2 (c) Hyperboloid of Two Sheet 2 =1 c a2 b2 x2 y2 z 2 (d) Elliptic Cone 2 + 2 =0 a b c2 x2 y2 (e) Elliptic Paraboloid z = 2 + 2 a b 2 y x2 (f) Hyperbolic Paraboloid z = 2 =1 b a2
− − −
−
−
Remark 3. How to sketch a surface? The shape of a surface can be obtained by considering the curves of intersection between the surface and some well-chosen planes. The curve of intersection between the surface and a plane is called the trace of the surface in the plane.
23
Example 21. Sketch and name the surface z = 4 + x2 + y2 .
Figure 2.4: Circular Paraboloid:z = 4 + x2 + y2
24
Figure 2.5: Paraboloid
Figure 2.6: Ellipsoid
25
x2 y2 Figure 2.7: Elliptic Cone:z = 2 + 2 3 8 2
Figure 2.8: Hyperbolic Paraboloid :z = y 2
26
2
−x
Figure 2.9: Hyperboloid of 1-Sheet:x2 + y 2
Figure 2.10: Hyperboloid of 2-Sheet:z 2
27
2
−z
2
−x −y
2
=1
=5
2.2.5
Graphs Of Two-variable Functions
Definition 2.2.1. The graph of the 2-variable function f (x, y) is the set of points (x,y,z) for which z = f (x, y) and (x, y) is in the domain of f. It is also called the surface z = f (x, y).
Example 22. The graph of the function f (x, y) = x+2y+3 is the plane z = x+2y+3. Example 23. The graph of the function f (x, y) = 4 z=4
2
2
− x − 4y .
2
− x − 4y
2
is the paraboloid
Example 24. The graph of the function f (x, y) = y2 is the parabolic cylinder z = y2 .
28
2.3
Double Integrals
Definition 2.3.1. The double integral of f over a closed region R is defined as n
lim
n
→∞
f (xk , yk ) xk yk
k=1
if the limit exists. If the limit exists, then f is said to be integrable over R and we denote this limit as
f (x, y)dA where dA = dxdy or dydx.
R
2.3.1
Some Applications Of Double Integrals
(a) If f (x, y)
≥ 0 and f is continuous on the rectangle R, then the volume of the
solid that lies above R and under the surface z = f (x, y) is given by V =
f (x, y)dA.
R
(b) In particular, if f (x, y) = 1 in R, then
f (x, y)dA = A(R), the area of R.
R
2.3.2
Double Integrals over Rectangular Regions
Theorem 2.3.1. Let R = (x, y) : a
{
region. If f is continuous on R. then
d
f (x, y)dA =
R
b
c
≤ x ≤ b, c ≤ y ≤ d} be a closed rectangular b
f (x, y)dxdy =
a
a
3
Example 25. Evaluate the iterated integral
2
0
d
f (x, y)dy dx.
c
(4x + 6y)dxdy.
1
Answer. 45 Example 26. Find the volume V of the solid under the plane x + y + z = 4 and over the rectangular region R : 0
≤ x ≤ 2, 0 ≤ y ≤ 1 in the xy-plane.
Answer. 5 29
2.3.3
Double Integrals over Non-rectangular Regions
Theorem 2.3.2. Let f be continuous in the region R. (a) If R = (x, y) : a
{
≤ x ≤ b, g(x) ≤ y ≤ h(x)}, then
b
f (x, y)dA =
R
(b) If R = (x, y) : g(y)
{
a
g(x)
d
f (x, y)dA =
y=
f (x, y)dy dx.
≤ x ≤ h(y), c ≤ y ≤ d}, then R
Example 27. Evaluate
h(x)
h(y)
c
f (x, y)dxdy.
g(y)
6xy dA over the region R enclosed between the curves
√x, 2y = x, x = 2 and x = 4. R
Answer. 11 Example 28. Use double integral to find the volume of the tetrahedron bounded by the coordinate planes and the plane 3x + 6y + 4z = 12.
Answer. 4 Example 29. Find the volume of the solid that lies under the paraboloid z = x2 + y2 and above the region bounded by the x-axis, the y-axis, and the line x + y = 1.
Answer. 1/6 Reading Assignment 5. Find the volume of the solid that lies under the plane 2x + 2y + z = 18 and above the triangular region R bounded by the lines y = x,
Answer. The volume V = 2
18y
0
− 2xy −
2x y2 x
(18
2
dx =
R
(18x
0
y = 2x, and the line x = 2. 2
− 2x − 2y)dA = 2
− 5x )dx = 30
2x
− 2
9x
0
x
5 3 x 3
(18
− 2x − 2y)dydx =
=
68 3
2 0
Example 30. Reversing The Order Of Integration Sketch the region of integration for the integral π
π
sin y dy dx y
0
x
and write an equivalent integral with the order of integration reversed. Then evaluate the integral.
Answer. 2 1
Reading Assignment 6. Evaluate the double integral I =
0
versing the order of integration.
Answer. The region of integration R = (x, y) : y2
{
1
2
yex dxdy by re-
y2
≤ x ≤ 1, 0 ≤ y ≤ 1} .
Sketch R. (Do it yourself!) Describe R in another way : 0
√x
1
Then I =
0
0
2
ye x dy dx =
≤ y ≤ √x, 0 ≤ x ≤ 1. 1
2
0
y 2
√x
2
ex dx =
1
0
0
31
2
x x ex e dx = 2 4 2
1
= 0
e
−1 4
2.3.4
Double Integrals in Polar Coordinates
Theorem 2.3.3. To evaluate a double integral using polar coordinates, we use
f (x, y)dA =
R
f (r cos θ, r sin θ)rdrdθ,
G
where G denotes the region of integration in polar coordinates.
Example 31. Use polar coordinates to evaluate the double integral
√ 1
I =
−1
1 y2
−
√ − −
(x2 + y2 )dxdy.
1 y2
Answer. π/2
√16−x
4
Example 32. Use polar coordinates to evaluate the double integral I =
0
0
2
dy dx . (9 + x2 + y 2 )3/2
Answer. π/15 Reading Assignment 7.
√ √ − −
1 y2
1
Use polar coordinates to evaluate the double integral I =
0
Answer. Let R be the the region of integration (x, y) : In polar coordinates, R = (r, θ) : 0
{
π
1
I =
6(r cos θ) r sin θrdrdθ =
0 π
=
π
2
0
0
6 5 r 5
1 2
π
cos θ sin θdθ =
0
0
1 y2
−
y2
1
6x2 ydxdy.
≤x
≤ r ≤ 1, 0 ≤ θ ≤ π} .
Using the substitution, x = r cos θ, y = r sin θ, r =
−
−
1
0
−
≤ − 1
y2, 0
x2 + y2 , dA = rdrdθ,
6r4 cos2 θ sin θdrdθ
0
6 cos2 θ sin θdθ = 5
6 cos3 θ 15
π
0
=
4 5
Exercise 9. Let R be the annular region lying between the two circles x2 + y 2 = 1 2
2
and x + y = 9. Evaluate the integral
(x + 2y 2 )dA.
R
2π
Answer.
3
0
1
r cos θ + 2 (r sin θ)2 rdrdθ =
32
··· = 40π
≤y
≤
1 .
2.4
Triple Integrals
Definition 2.4.1. The triple integral of f over a bounded solid region D is defined as
n
lim
n
→∞
f (xk , yk , zk ) xk yk zk
k=1
provided the limit exists. If it exists, then f is said to be integrable over D. We will denote the limit as
f (x,y,z)dV where dV = dxdydz or dxdzdy or . . . .
D
Note 4. The volume of D =
dV.
D
Theorem 2.4.1. Fubini’s Theorem Let B = (x,y,z) : a
{
≤ x ≤ b, c ≤ y ≤ d, g ≤ z ≤ h} be a closed rectangular box.
If f is continuous on R. then
h
f (x,y,z)dV =
B
d
b
g
c
f (x,y,z)dx dy dz.
a
Note : The iterated integral on the right can be replaced by any of the five other iterated integrals by changing the order of integration.
Example 33. Evaluate
xy2 z 3 dV over the rectangular box
B
B : 0
≤ x ≤ 3, −2 ≤ y ≤ 2, 0 ≤ z ≤ 1.
Answer. 6 Theorem 2.4.2. Let R be a closed region in the xy-plane and let g(x, y) and h(x, y) be continuous functions such that g(x, y) over the region D = (x,y,z) :
{
D
≤ h(x, y) for all (x, y) ∈ R. If f is integrable (x, y) ∈ R, g(x, y) ≤ z ≤ h(x, y)}, then h(x,y)
f (x,y,z)dV =
R
f (x,y,z)dzdA.
g(x,y)
Note 5. Determining The Limits Of Integration 33
To find the limits for a particular order of integration, it is advisable to first determine the innermost limits, which may be functions of the outer two variables. After integrating f with respect to the innermost variable, we end up with a double integral over the projection of D onto the coordinate plane of the outer two variables. Then, you can determine the remaining limits of integration by methods used for double integrals.
Example 34. Using a triple integral to find volume Find the volume of the solid formed by the intersection of the cylinder y = x2 and the two planes given by z = 0 and y + z = 4.
Answer. V = 256/15 Example 35. Using a triple integral to find volume Find the volume of the solid enclosed by the paraboloids z = x2 +y2 and z = 18 x2 y2 .
− −
Answer. 81π Exercise 10. Find the volume of the region in the first octant bounded above by the cylinder z = 1
−y
2
and lying between the vertical planes x + y = 1 and x + y = 3. 1 y2
Answer. The volume V = 3
− y,
0
≤ y ≤ 1.
R
−
dzdA =
0
34
··· = 43 where R
: 1
−y ≤ x ≤
2.5
Line Integrals
Definition 2.5.1. (a) A plane curve can be described by a pair of parametric equations x = f (t), y = g(t), t
∈ I
where f and g are continuous functions on the interval I. The variable t is a
parameter for the curve. (b) A space curve can be described by a pair of parametric equations x = f (t), y = g(t), z = h(t), t
∈ I
where f ,g and h are continuous functions on the interval I. A curve can also be specified by giving the position vector r(t) of a point P = P (t) = (f (t), g(t), h(t)). That is,
−→
r(t) = OP = (f (t), g(t), h(t)) = f (t)i + g(t) j + h(t)k. The tip of r(t) traces out the curve as t varies over I.
Example 36. The parametric equations for the line passing through the point (x0 , y0, z0 ) and parallel to the nonzero vector v = (a,b,c) are x = x0 + at,y = y0 + bt,z = z0 + ct.
Example 37. (a) Parametrize the line segment joining the point A(1, 4, 3) and B( 2, 5, 6).
−
(b) Write down the corresponding vector equation for the line segment.
Answer. (a) x = 1
− 3t, y = 4 + t, z = 3 + 3t, 0 ≤ t ≤ 1 (b) r(t) = (1 − 3t)i + (4 + t) j + (3 + 3t)k, 0 ≤ t ≤ 1 35
Example 38. A Parametrization Of The Circle x2 + y2 = a2 The following is a parametrization for the circle x2 + y 2 = a2 : x = a cos t, y = a sin t, 0
≤ t ≤ 2π.
Note that the circle is traced out counterclockwise as t increases from t = 0 to t = 2π. x2 y2 Example 39. A Parametrization Of The Ellipse 2 + 2 = 1 a b x = a cos t, y = b sin t, 0
≤ t ≤ 2π.
Note that the ellipse is traced out counterclockwise as t increases from t = 0 to t = 2π.
Example 40. Find parametric equations for the portion of the parabola y = x2 joining A( 1, 2) and B(3, 6), oriented from A to B.
−
−x
Example 41. (a) Eliminate the parameter to find the Cartesian equation of the curve with parametric equations x = 3t2 , y = 5t + 2, 0
≤ t ≤ 2.
(b) Sketch the curve represented by the parametric equations. Indicate with an arrow the direction in which t increases.
Answer. x =
3 (y 25
2
− 2) , 2 ≤ y ≤ 12
Exercise 11. Find parametric equations for the line segment joining the points A(2, 3, 4) and B( 5, 0, 1).
−
−
Write down the corresponding vector equation for the line segment. Draw coordinate axes and sketch the line segment, indicate the direction of increasing t for your parametrization.
Answer. x = 2
− 7t, y = 3 − 3t, z = −4 + 5t, 0 ≤ t ≤ 1; r = (2 − 7t)i + (3 − 3t) j + (−4 + 5t)k, 0 ≤ t ≤ 1; The line segment with this parametrization moves from A to B. 36
Exercise 12. Find parametric equations for the portion of the parabola x = y2 joining A(4, 2) and B(9, 3), oriented from A to B.
−
Answer. x = t2 , y = t,
−2 ≤ t ≤ 3.
Definition 2.5.2. Let r(t) be the position vector of a curve C defined on an interval I. We call C a smooth curve or r(t) a smooth function if r (t) is continuous and
r (t) = 0 for any value of t
∈ I (except possibly at any endpoints of I ).
Note : The tangent vector r (t) for a smooth curve varies ”continuously” without abrupt changes in direction as t increases. We can think of a smooth curve as a curve with no ”sharp corners” (called cusps).
Definition 2.5.3. A curve that is made up of a finite number of smooth curves is called piecewise smooth.
Definition 2.5.4. The length of a smooth curve C : r(t) =< x(t), y(t), z(t) > , a
≤ t ≤ b, that is traced exactly once as t increases from a to b is given by
b
L=
a
dx dt
2
dy + dt
dz + dt
2
2
b
dt or L =
r (t) dt.
a
Definition 2.5.5. The arc length parameter s for the curve C with base point P = P (a) is the function
t
s(t) =
a
dx dt
2
dy dt
+
2
+
dz dt
2
dt.
Note : By the Fundamental Theorem of Calculus, ds = dt
dx dt
2
dy + dt
2
dz + dt
2
.
Definition 2.5.6. If f is defined on a smooth curve C : x = x(t), y = y(t) a t
≤ b, then the line integral of f along C is
C
n
f (x, y)ds = lim n
→∞
k=1
if the limit exists. 37
f (xk ∗ , yk ∗ ) sk
≤
Theorem 2.5.1. If f is a continuous function, then
f (x, y)ds =
C
dx dt
2
0, then
≥
dx dt
f (x(t), y(t))
a
(a) Recall that ds = (b) If f (x, y)
b
dy + dt
2
2
+
dy dt
2
dt.
dt.
f (x, y)ds represents the area of the ”curved curtain”
C
whose base is C and whose height above the point (x, y) is f (x, y).
Example 42. Evaluate the line integral I = t2 , 0
xds, where C is the curve x = t, y =
C
≤ t ≤ 1.
Answer. I =
√125 − 1 12
Reading Assignment 8. Evaluate the line integral
(2 + x2 y)ds, where C is the upper half of the unit circle
C
x2 + y2 = 9.
Answer. A parametric form for C is x = 3 cos t, y = 3 sin t, Hence, ds =
− dx dt
2
+
dy dt
π
2
(2 + x y)ds =
C
0
≤ t ≤ π.
2
dt =
( 3sin t)2 + (3 cos t)2 dt = 3dt and
2
2 + (3 cos t) (3sin t) 3dt = 3 2t
0
3
− 9cos
t
π 0
= 6π + 54
Theorem 2.5.2. If C is a piecewise-smooth curve; i.e., C is the union of a finite number of smooth curves C 1 , C 2 , . . . , Cn , where the end point of C i−1 is the initial point of C i . Then
f (x, y)ds =
C
C 1
f (x, y)ds +
f (x, y)ds +
C 2
··· +
f (x, y)ds.
C n
Note 6. The line integrals of f with respect to x and y are defined by
C
n
f (x, y)dx = lim n
→∞
k=1
f (xk ∗ , yk ∗ )xk and 38
C
n
f (x, y)dy = lim n
→∞
k=1
f (xk ∗ , yk ∗ ) yk
provide the limits exist.
≤ t ≤ b then dx = x(t)dt,dy = y(t)dt and
Note :If x = x(t), y = y(t), a
b
f (x, y)dx =
C
f (x(t), y(t))x (t)dt,
a
b
f (x, y)dy =
C
Example 43. Evaluate the line integral I =
f (x(t), y(t))y (t)dt.
a
xydx + (x
C
− y)dy, where C consists
of the line segments from (0, 0) to (2, 0) and from (2, 0) to (3, 2).
Answer. I = 17/3 Example 44. Integrate f (x,y,z) = xy + y + z over the path C : r(t) = 2ti + t j + (2
− 2t)k, 0 ≤ t ≤ 1.
Answer. 13/2 Reading Assignment 9. Evaluate the line integral I =
(2+x yz)ds, where C is the polygonal path consisting
−
C
of the circular arc x2 + y2 = 9, z = 2 from (3, 0, 2) to (0, 3, 2) and the line segment from (0, 3, 2) to (3, 7, 2).
Answer. (i) C = C 1 + C 2 where
≤ t ≤ π2 C : x = 3t, y = 3 + 4t, z = 2, 0 ≤ t ≤ 1. (ii) I = I + I where I = (2 + x − yz)ds and I = C 1 : x = 3cos t, y = 3 sin t, z = 2, 0 2
1
2
− − 1
2
C 1
dx dt
(iii) On C 1 : ds = 3dt,
2
+
dy dt
2
+
dz dt
C 2
(2 + x
− yz)ds.
2
dt =
( 3sin t)2 + (3 cos t)2 + (0)2 dt =
π/2
hence I 1 =
(2 + 3 cos t
2(3 sin t))3dt = 3 2t + 3 sin t + 6 cos t
0
39
π/2
0
= 3π
−9
(3)2 + (4)2 + (0)2 dt = 5dt,
(iv) On C 2 : ds =
1
hence I 2 =
−
(2 + 3t
− 2(3 + 4t))5dt = 5 83 (v) I = (3π − 9) + (−65/2) = 3π − 2 0
2.5.1
4t
−
5 2 t 2
1
0
=
− 652
Line Integrals of Vector fields
Definition 2.5.7. (a) A vector field is a function F that assigns to each point (x1 , x2 , . . . , xn ) in its domain D a unique vector F(x1 , x2 , . . . , xn ). (b) A vector field on R3 is a function of the form
F(x,y,z) = M (x,y,z)i + N (x,y,z) j + P (x,y,z)k. F is continuous if its component functions M, N and P are continuous ; differentiable if functions of M, N and P are differentiable, and so on. (c) If the partial derivatives of M, N and P all exist, then (i) the curl of F is the vector field defined by ∂P curl F = ∂y
−
∂N i ∂z
−
∂P ∂x
−
∂M ∂N j + ∂z ∂x
Remark 4. A physical interpretation of curl F
−
∂M k. ∂y
Let the vector field F represents the velocity field in fluid flow. Particles near (x,y,z) tend to rotate about the axis that points in the direction of curl F and curl F is a measure of how quickly the particles move around the axis.
If curl F = 0 at a point P , then the fluid is free from rotations at P and F is said to be ir-rotational. (ii) The divergence of F is the function defined by div F =
∂M ∂N ∂P + + . ∂x ∂y ∂z
40
(d) By introducing the differential operator div F =
· F and curl F =
×
= i ∂x∂ + j ∂y∂ + k ∂z∂ , we may write
i ∂ F= ∂x M
A physical interpretation of div F
j ∂ ∂y N
k ∂ . ∂z P
Let F be the velocity of a fluid, then div F represents the net rate of change (with respect to time) of the mass of fluid flowing from the point (x,y,z) x,y,z ) per unit volum volume. e. That That is, div F measures the tendency of the fluid to diverge form the point (x,y,z) x,y,z ) (div F > 0 ) or accumulate toward (x,y,z) x,y,z ) (div F < 0 ). If div F = 0, then F is said to be incompressible.
Example 45. Sketch the vector field F(x, y) =
Answer. F r = 0 and F =
·
has length equal to
1 2
| |
1 2
− 12 yi + 12 x j.
|r| ⇒ F is tangent to a circle centered at (0, (0, 0) and
the radius of that circle.
Suppose a spherical spherical object object of mass M is centered at the origin. Derive Example 46. Suppose the formula for the gravitational field of force F(x,y,z) x,y,z) exerted by the mass on an object of mass M located at a point P ( P (x,y,z) x,y,z) in space. Then sketch this field.
Answer. F =
GM m − GMm |r| r 3
3
Exercise 13. Let F = 2xyi
− 5y z j + 4xz 4xz k. Find the divergence and the curl of F. 15y z + 4x 4x; 5y i − 4z j − 2xk Answer. 2y − 15y Exercise 14. Let F = 2yz i − 5y x j + 4yz 4yz k. Find the divergence and the curl of F. 15y x + 8zy 8zy;; 4z i + 2y 2y j − (5y (5y + 2z 2 z )k Answer. −15y 2
3
3
2
2
2
3
41
Definition 2.5.8. Let C : x = x(t), y = y (t), z = z (t), a
≤ t ≤ b be a smooth
curve with unit tangent vector T. Then the work done by a force F in moving a particle along C from t = a to t = b is W =
F T ds.
C
·
Remarks Different ways to write the work integral: b
(a) W =
F(r(t)) r (t)dt =
·
a
(b) W =
C
F dr
·
M dx + N dy + P dz if F(x,y,z) x,y,z ) = M (x,y,z) x,y,z)i + N (x,y,z) x,y,z ) j + P ( P (x,y,z) x,y,z )k.
C
Example 47. Find the work done by the force field F(x, y) = x2i
− xy j in moving a particle counterclockwise along the quarter-circle r(t) = cos ti + sin t j, 0 ≤ t ≤ π/2 π/ 2.
Answer.
− 23
Example 48. Find the work done by the force field F(x,y,z) x,y,z) = yi + z j xk in moving
−
a particle along the twisted cubic C : x = t, y = t2 , z = t3 from (0, (0, 0, 0) to (1, (1, 1, 1). 1).
Answer.
− 601
42
Reading Assignment 10. Calculate the work done by F(x,y,z) x,y,z ) = i
− y j + xyz k
in moving a particle from (0, (0, 0, 0) to (1, (1, 1, 1) along the curve of intersection of the cylinder y =
2
−x
and the plane z = x.
−
Answer. Let C be the curve of intersection. Method 1 We can describe C using the parametric equations x = t, y =
2
−t , z = t.
At (0, (0, 0, 0) : t = x = 0 while at (1, (1, 1, 1) : t = x = 1.
−
On C :
F = i + t2 j dr dx i+ = dt dt dr =1 F dt
4
−t k
dy dz j + k = i dt dt 3
− 2t j + k
4
− 2t − t
·
The work done =
C
Method 2
1
F dr =
·
0
dr F dt = dt
·
1
0
(1
− 2t − t )dt = · · · = 103 3
4
2
y=
−x , z = x ⇒ dy = −2xdx, xdx, dz = dx The work done = F · dr = dx − ydy + xyzdz 3 = )(x)dx = (1 − 2x − x )dx = · · · = dx − (−x )(−2xdx) xdx) + x(−x )(x 10 Exercise 15. Find the work done by the force F = −3xz i + 2y j + 4k in the displacement along the line x = −2z, y = 3z from (4, (4, −6, −2) to (0, (0, 0, 0). 0). Answer. −60 1
C
2
C
1
2
0
0
43
3
4
2.6
Green’s Theorem in the Plane
Definition 2.6.1. A plane curve C : r = r(t) (a
≤ t ≤ b) has an initial end point
at r(a) and a final end point at r(b). C is said to be simple if it does not intersect itself anywhere between its endpoints.
Example 49. A circle is a simple curve but the figure 8 is not. Theorem 2.6.1. Green’s Theorem Let C be a positively oriented, piecewise-smooth simple closed curve in the plane and let D be the region bounded by C. If P and Q have continuous partial derivatives on an open region that contains D, then
P dx + Qdy =
C
D
∂Q ∂x
−
∂P dA ∂y
Remarks (Other formulas for Green’s Theorem) Let F = P i + Q j. Then (a)
−
P dx + Qdy =
C
(b)
(curl F) k dA
D
Qdx + P dy =
C
D
·
∂P ∂Q + dA = ∂x ∂y
Example 50. Evaluate the line integral
− (3y
div F dA
D
sin x
e
)dx + (7x +
C
C is the circle x2 + y2 = 9.
y4 + 1)dy, where
Answer. 36π Example 51. Verify that Green’s Theorem is true for the line integral
xydx + x2 dy,
C
where C is the triangle with vertices (0, 0), (1, 0), and (1, 2).
44
Answer. 2/3 Reading Assignment 11. Verify that Green’s Theorem is true for the line integral
y2 dx + x2 dy,
C
where C is the boundary curve of the region R lying between y = x and 4x = y2 .
Answer. Let I 1 =
P dx + Qdy =
C
(2x
R
− 2y)dA.
2
2
y dx + x dy and I 2 =
C
R
∂Q ∂x
−
∂P dA = ∂y
We have to verify that I 1 = I 2 . Calculate I 2 =
(2x
R
2
R:y
≤ x ≤ y4 ,
I 2 =
4
0
·· · = − 32 15
0
− 2y)dA :
≤y≤4
y
4
(2x
y 2 /4
Calculate I 1 =
− 2y)dxdy =
2
x
0
− 2yx
y y 2 /4
4
dy =
− 0
y2 dx + x2 dy :
C
C = C 1 + C 2
C 1 : y = x from x = 0 to x = 4 On C 1 : dy = dx
4
2
2
y dx + x dy =
C 1
x2 dx + x2 dx =
0
··· = 128/3
C 2 : 4x = y2 from y = 4 to y = 0 1 On C 2 : 4dx = 2ydy dx = ydy 2 0 1 2 2 = 224/5 y dx + x dy = y2 ydy + (y2 /4)2 dy = 2 C 4 32 + = (128/3) + ( 224/5) = = I 2 as expected. I 1 = 15 C C
⇒
2
1
2
−
· ·· −
−
45
y4 y 3 + 16 2
−y
2
dy =
2.7
Surface Area and Surface Integrals
2.7.1
Surface Area
Let R be the region on the xy-plane. Let S be the surface above R with equation z = f (x, y). If f has continuous first partial derivatives on R, then the area of S is A(S ) =
f x 2 + f y 2 + 1 dA.
R
Example 52. Find the area of the surface S if it is the part of the plane x+y +z = 1 that lies in the first octant.
√3/2
Answer.
Reading Assignment 12. Find the surface area of the portion of the paraboloid z=4
2
−x −y
2
lying above the xy-plane.
⇒
1 + f x2 + f y2 dA.
Answer. The area of the surface S is A(S ) =
R
2
f (x, y) = 4
2
2
2
1 + 4x + 4y dA − x − y ⇒ f = −2x, f = −2y A(S ) = The vertical projection of S onto the xy−plane is the circular region R = {(x, y) : x + y ≤ 4}. In polar coordinates, R = {(r, θ) : 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π}. x
y
R
2
Using the substitution x2 + y2 = r2 , dA = rdrdθ,
√ 2π
A(S ) =
0
2
0
1 + 4r 2
1 (1 + 4r2 )3/2 rdrdθ = 2π 12
46
2
0
= π(173/2
− 1)/6
2
2.7.2
Surface Integrals
Let R be the region on the xy plane. Let S be the region above R with equation z =
−
f (x, y). If f has continuous first partial derivatives on R and g(x,y,z) is continuous on S, then the surface integral of g on S is defined as
g(x,y,z) dS =
S
g(x,y,f (x, y)) 1 + f x 2 + f y 2 dA.
R
Remarks (a)
dS = area of S.
S
(b) dS =
1 + f x 2 + f y 2
dA.
Example 53. Evaluate the surface integral
√ cylinder z = 1 − x
2
z 2 dS where S is the part of the
S
that lies above the square with vertices ( 1, 1), (1, 1), ( 1, 1)
and (1, 1).
Hint : You may assume
√ − 1
x2 dx =
√
x 1 2
− −
2
−x
+
− −
1 −1 sin x + C 2
Answer. π Example 54. A curved lamina is the portion of the paraboloid z = x2 + y2 below the
√
plane z = 1 and has density σ(x,y,z) = 6 1 + 4z. Find the mass of the lamina.
Answer. 18π
47
Reading Assignment 13.
Let S be the portion of the cone z = 4 2 x2 + y2 between z = 0 and z = 4. Evaluate
−
− ≤ ≤ − − − − 5x2 + 5y2 dS.
3z
the surface integral
S
Answer. S : z = 4 2 x2 + y2 , 0 2x 2y zx = , zy = x2 + y2 x2 + y 2 dS =
1+
I =
zx2
+
x2
3z
zy2 dA +
y2
=
4
4x2 4y2 1+ 2 + dA x + y2 x2 + y2
dS =
3 4
S
=
z
2
x2
+
y2
5x2
R
+ 5y2
4x2 4y2 1+ 2 + dA x + y 2 x2 + y2
2 x2 + y2 3 x2 + y2 dA
3 4
R
Using polar coordinates, R:0
≤ r ≤ 2, 0 ≤ θ ≤ 2π 9(4r − 2r )drdθ = ·· · = 48π I = ··· = 2π
2
0
2.7.3
2
3
0
Surface Integrals of Vector Fields
Definition 2.7.1. Oriented Surfaces A surface S is orientable if it is possible to choose a unit normal vector n at each point (x,y,z) on S (except possibly at any boundary points) in such a way that n varies continuously (have no abrupt changes in direction) as we traverse along any curve on S . Once n has been chosen, we say that we have oriented the surface.
Note: A surface can be oriented only if it has 2 sides, the process of orientation consists essentially in choosing which side we will call ”positive” and which ”negative”. (If the surface is closed, then it is natural to speak of the ”inside” and ”outside”.)
Example 55. The 2 possible orientations of a sphere are by inward unit normals or outward unit normals. 48
Example 56. A Mobius strip is a non-orientable surface since it has only 1 side.
Note : We will exclude the non-orientable surfaces from further consideration. Hence forth, whenever we say ”surface”, we mean an ”orientable surface”.
49
Definition 2.7.2. If F is a continuous vector field defined on an oriented surface S with unit normal vector n, then the surface integral of F over S is
S
This integral is also called the flux of F across S.
F n dS.
·
Note 7. (a) Recall that the gradient vector of a scalar function h(x,y,z) at a point P (x0 , y0 , z0 ) is the vector
h(x , y , z ) = h (x , y , z )i + h (x , y , z ) j + h (x , y , z )k. (b) At any point (x , y , z ) in the domain of h(x,y,z), h(x , y , z ) is orthogonal 0
0
0
0
0
x
0
0
0
y
0
0
0
0
z
0
0
0
0
0
0
to the surface h(x,y,z) = h(x0 , y0 , z0 ).
Example 57. Evaluate the flux integral I =
F n dS for the vector field
S
F =
−yi + x j + 3zk, where S is the hemisphere z =
orientation.
·
− 16
x2
−y
2
with upward
Answer. 128π
2.8
Stokes’ Theorem
Definition 2.8.1. A surface S : G(x,y,z) = C is smooth if the normal vector G is continuous and never vanishes on S . (It has no ”corners”.)
It is said to be piecewise smooth if it consists of a finite number of smooth parts ”joined together” For example, the surface of a sphere is smooth whereas the surface of a cube is piecewise smooth.
Theorem 2.8.1. Stokes’ Theorem Let S be an oriented piecewise-smooth surface that is bounded by a simple, closed, piecewise-smooth boundary curve C with positive orientation. If F is a continuously 50
differentiable vector field on an open set that contains S, then
F dr =
C
·
curl F n dS.
·
S
Example 58. Use Stokes’ Theorem to compute the line integral F(x,y,z) = 4y i
C
F dr, where
·
− 3z j + xk and C is the triangle having vertices at (1, 0, 0), (0, 1, 0),
and (0, 0, 1), counterclockwise when viewed from above.
Answer.
−1
Remark 5. If S 1 and S 2 are oriented surfaces with the same oriented boundary curve C and both satisfy the hypotheses of Stokes’ Theorem, then
curl F n dS =
·
S 1
F dr =
·
C
curl F n dS.
·
S 2
This fact is useful when it is difficult to integrate over one surface but easy to integrate over the other.
Example 59. Let S be the part of the paraboloid z = 5
2
−x −y
2
that lies above
the plane z = 1, oriented upward. Verify Stokes’ Theorem for the vector field F = 2z i + 3x j + 5yk.
Answer. Both integrals are equal to 12π. Reading Assignment 14. Let F = (z
− y)i + (x − z) j + (x − y)k.
Use Stokes’ Theorem to evaluate the surface integral I = is the hemisphere z =
− 1
S
x2
2
−y .
Answer. By Stokes’ Theorem, I =
F dr
C
51
·
curl F n dS, where S
·
where C is the circle x2 + y 2 = 1, z = 0. C : x = cos t, y = sin t, z = 0,
0
≤ t ≤ 2π
⇒ r = − sin ti + cos t j F = (z − y)i + (x − z) j + (x − y)k = (0 − sin t)i + (cos t − 0) j + (cos t − sin t)k F · r = sin t + cos t = 1 F · r dt = I = dt = 2π r = xi + y j + xk
2
2π
2π
0
2.9
2
0
Gauss’ Divergence Theorem
Theorem 2.9.1. Gauss’ Divergence Theorem Let S be a sectionally smooth surface. Let V be the solid enclosed by S. Let F be a vector field defined on S . If the component functions of F have continuous first derivatives in V , then
S
F n dS =
·
div F dV
V
where n is the outward unit normal.
Example 60. Let F = 3xyi +y2 j x2 y4 k, and S be the surface of the tetrahedron with
−
vertices (0, 0, 0), (1, 0, 0), (0, 1, 0), and (0, 0, 1). Use the divergence theorem to evaluate the surface integral
S
F n dS, where n is the outward-pointing unit normal.
·
Answer. 5/24 Example 61. Verify the divergence theorem for F = xi + y j + z k over the sphere x2 + y2 + z 2 = a2 .
Answer. Both integrals are equal to 4πa3 .
52
Reading Assignment 15. Verify the divergence theorem for F = x2 i + y2 j + z 2 k over the unit cube bounded by x = 1, y = 1, z = 1, and the coordinate planes.
Answer. Let I 1 =
S
F n dS and I 2 =
·
div F dV.
V
We have to verify that I 1 = I 2 . ∂ 2 ∂ 2 ∂ 2 (i) divF = x + y + z = 2x + 2y + 2z ∂x ∂y ∂z
· ··· · 1
I 2 =
1
(2x + 2y + 2z)dV =
V
(ii) I 1 =
1
0
F ndS +
(2x + 2y + 2z)dzdydx =
0
··· = 3
F ndS,
+
S 1
0
S 6
··· + S , where S , . . . , S are the 6 faces of the cube. On S : x = 0, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1, we have n = outward unit normal to S = −i S = S 1 +
6
1
6
1
1
F = x2 i + y2 j + z 2 k = y 2 j + z 2 k F n=0
·
Hence, J 1 =
F ndS = 0
· On S : x = 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1, we have S 1
2
n = outward unit normal to S 2 = i F = i + y2 j + z 2 k F n=1
·
Hence, J 2 =
dS = area of S 2 = 1.
S 2
The remaining calculations are summarized in the following table: S
S 1 : x = 0 S 2 : x = 1
S 3 : y = 0 S 4 : y = 1
S 5 : z = 0 S 6 : z = 1
n
−i
i
i + y2 j + z 2 k x2 i + z 2 k
−k
k
y2 j + z 2 k
− j
x2 i + j + z 2 k x2 i + y2 j
x2 i + y 2 j + k
0
1
1
0
1
1
0
1
F F n
· ·
0
1 0 F ndS 0 ∴ I 1 = 0 + 1 + 0 + 1 + 0 + 1 = 3 = I 2 as expected. S
53
j
54
Chapter 3 Z Transform 3.1
Introduction
Difference equations are used to model discrete-time systems. They are of growing importance as more and more engineering systems now contain a microprocessor or computer. For example, many factories have a production control computer to schedule production. Such as the Laplace transform was used to solve differential equations, we will develop the z-transform and use it to solve difference equation.
Definition 3.1.1. A sequence is an ordered set of complex numbers
{x }∞−∞ = {. . . , x− , x− , x , x , x , . . .}. k
2
1
0
1
2
xk is called the kth term of the sequence.
{x } is called a causal sequence if x k
k
= 0 for k < 0.
Definition 3.1.2. The z transform of a given causal sequence , xk ∞ 0 , is defined
{ }
as X (z) =
∞
k=0
xk . zk
whenever the sum exists and where z is a complex variable, as yet undefined. 55
Some simple examples of z transforms of common discrete signals are now given.
Example 62. Determine the z-transform of the unit impulse sequence δk ∞ 0 =
{ }
{1, 0, 0, . . .}. Answer. 1
Example 63. Determine the z-transform of the unit step sequence uk = 1 ∞ 0 =
{1, 1, 1, . . .}. Answer.
z
{
}
z
−1
Example 64. Determine the z-transform of the exponential sequence xk = ak ∞ 0
{
where a is a constant.
}
Answer. Example 65. Determine the z-transform of the unit ramp sequence k Answer. Z k =
{}
{}
z (z
− 1)
2
Reading Assignment 16. Determine the z-transform of the sequence xk = kak−1
{
where a is a constant.
}
Answer. From the previous example, we see that k
{ }=
Z a
∞
k=0
ak z = zk z a
−
Differentiating wrt a : ∞ k −1 ∞ k −1 ka ka z = = (z a)2 zk zk k=1 k=0 z Z kak−1 = (z a)2
⇒ {
}
−
−
Example 66 (Sampling a continuous signal). Consider the continuous signal f (t) = e−t defined for t
≥ 0. Suppose we sample or measure the values of f (t)
at intervals of time, T , we obtain a discrete signal defined by the sequence xk = f (kT ) = e−kT , for k = 0, 1, 2, . . . . Find the z transform of the resulting signal. 56
z z e−T
Answer. The transform is Z (e−T )k =
{
}
−
Reading Assignment 17. The continuous-time signal f (t) = 5 sin 5t, π sampled at equal intervals of t = . 10 (a) Write down the general term of the sequence of samples.
t
≥ 0 is
kπ , 2
k =
(b) Find the z transform of the sequence.
Answer. (a) The general term of the sampled sequence xk = f k 0, 1, 2, . . . . 5z (Table) (b) X (z) = 2 z +1
Exercise 16. The continuous-time signal f (t) = 4t, 2k,
π 10
k = 0, 1, 2, . . . .
t
≥
(a) Write down the general term of the sequence of samples. (b) Find the z transform of the sequence.
Answer. xk = 8k, k = 0, 1, 2, . . . ; X (z) =
8z (z 1)2
57
−
= 5sin
0 is sampled at t =
3.2
Properties of The Z Transform
The z-transform has many important and useful properties which are useful for manipulating and solving problems.
Theorem 3.2.1 (Linearity). Let xk and yk be two sequences with z transforms X (z) and Y (z) respectively. The z-transform operator Z is a linear operator. That is , Z axk + byk = aZ xk + bZ yk = aX (z) + bY (z)
{
}
{ }
{ }
for any constants a and b.
{ } = z −z a
Example 67. Use the fact Z ak
and the linear property to find the z
transform of the causal sequence (a) cos(ak) (b) sin(ak)
1 iak 1 iak Hint : cos(ak) = e + e−iak , sin(ak) = e 2 2i z(z cos a) z sin a (b) 2 Answer. (a) 2 2z cos a + 1 2z cos a + 1 z z
−
−
− e−iak
−
Theorem 3.2.2. ( The First Shift Theorem) 1 Z xk−m = m Z xk where m is a positive integer. z
{
}
{ }
Proof. Z xk−m =
{
m) = z −m
∞
k=0
}
∞
k=0
∞
xk−m xk−m −m = z , k −m zk z k=m
(since xk−m = 0 for k m < 0 or k <
−
1 xk = Z xk zk zm
{ }
Example 68. The causal sequence xk is generated by
{ }
xk = ak , k
≥ 0.
where a is a constant. Determine the z transform of the shift sequence xk−m where
{
m is a positive integer. 58
}
Answer. Z xk−m =
{
}
1 z zm z a
−
Theorem 3.2.3 (The Second Shift Theorem) . (a) Z xk+1 = zZ xk
{ } { } − zx = zX (z) − zx (b) Z {x } = z X (z) − z x − zx − (c) Z {x } = z Z {x } − x z − where a is a positive integer. k+2
k+a
0
0
2
2
a
0 a 1
k
1
a n
n
n=0
∞
∞
∞
xk+1 xk+1 xm Proof. (a) Z xk+1 = = = =z z z k k+1 m z z z m=1 k=0 k=0 = z[Z xk x0 ] = zZ xk zx 0 = zX (z) zx 0
∞
xm zm m=0
{ } −x { }− { }− − (b) Z {x } = zZ {x } − zx ⇒ Z {y } = zZ {y } − zy , y = x ⇒ Z {x } = zZ {x } − zx = z(zZ {x } − zx ) − zx = z X (z) − z x − zx k+1
k
k+2
0
k+1
k+1
k
1
0
k
0
k
1
0
k+1
2
2
0
1
(c) Prove by induction.
Example 69. Use the fact Z ak =
{ }
Z ak+2 .
{
}
z z
−a
and the Second Shift Theorem to find
a2 z Answer. . z a
−
3.3
More Properties of The Z Transform
Theorem 3.3.1 (Multiplication by ak ). If Z xk = X (z), then for a constant a, Z ak xk = X (a−1 z) or
{
}
{ } Z {a− x } = X (az). k
k
This is also referred to as the change of scale or damping rule.
{ } = (z −z 1)
Example 70. Use the fact Z k az Z kak = . (z a)2
{ }
−
59
2
and the damping rule to show that
Answer.
az (z a)2
−
Theorem 3.3.2 ( Multiplication by k).
{ } −z dzd [X (z)]. z cos θ − 2z + z cos θ z(z − cos θ) if Z {cos kθ } = Example 71. Show that Z {k cos kθ } = . (z − 2z cos θ + 1) z − 2z cos θ + 1 Z kxk =
3
2
2
Answer.
3.4
2
2
z 3 cos θ 2z 2 + z cos θ (z 2 2z cos θ + 1)2
−
−
The Inverse Z Transform
Definition 3.4.1. If Z xk = X (z), then xk = Z −1 X (z) is called the inverse z
{ }
transform of X (z).
{
}
Theorem 3.4.1. The inverse z transform is also a linear operator. Example 72. Find Z −1 Answer. xk = k7k−1 , k
z (z
− 7)
10 k−1 2 3
.
≥0
Example 73. Find xk = Z −1 Answer. xk =
2
3z + 4 . (z + 1)(z 2)
−
− 13 (−1) − , k ≥ 1; x k 1
0
=0
Example 74. Invert the z transform Y (z) = Answer. yk == ak−1 sin 12 kπ,k
≥0
60
z where a is a real constant. z 2 + a2
Reading Assignment 18. Find yk = Z −1 Answer. Let Y = Z yk
{ }
36z 2 12z . (z + 1)(z 3)2
−
−
36z 2 12z = . (z + 1)(z 3)2
−
−
(a) By partial fractions, 36z 12 Y A B C = = + + (z + 1)(z 3)2 z z + 1 z 3 (z 3)2 3 24 3 = = + z 3 (z 3)2 z + 1 3z 24z 3z + Y = . z 3 (z 3)2 z + 1 (b) By using the Table,
−
···
⇒
−
− − − −
− −
yk = 3(3)k + 8k(3)k
3.5
−
k
− 3(−1) ,
k
−
≥ 0.
Solving Difference Equations
Example 75. Solve the first order difference equation yk+1
Answer. yk = 7(5)k , k
− 5y
k
= 0, y0 = 7.
≥0
Example 76. Solve the second order difference equation yk+2
− 5y
Answer. yk = 5(2)k + 2(3)k , k
k+1
+ 6yk = 0, y0 = 7, y1 = 16.
≥0
61
Reading Assignment 19. Solve yk+2 + 6yk+1 + 9yk = 2k with y0 = y1 = 0. Answer. [z 2 Y
− z y − zy ] + 6[zY − zy ] + 9Y = z −z 2 2
0
1
0
⇒ z Y + 6zY + 9Y = z −z 2 2
z ⇒ Y = (z − 2)(z + 3)
2
1 1 1 1 1 1 1 ⇒ Y z = (z − 2)(z − − = · ·· = + 3) 25 z − 2 25 z + 3 5 (z + 3) 1 z 1 z 1 ⇒ Y = 25 z − 2 − 25 z + 3 − 5 (z +z 3) ⇒ y = 251 (2) − 251 (−3) + 151 k(−3) , k ≥ 0 2
2
k
k
k
k
Reading Assignment 20. Solve the difference equation yk+2
− 2y
k+1
+ yk = 2k
given that y0 = 2 and y1 = 1.
− z y − zy ] − 2[zY − zy ] + Y = z −z 2 z [z Y − 2z − z] − 2[zY − 2z] + Y = z−2 z (z − 2z + 1)Y = 2z − 3z + z−2 2z − 3 1 Y = + (z − 1) (z − 2)(z − 1) z 2 1 1 1 1 = ··· = + − − − z − 1 (z − 1) z − 2 z − 1 (z − 1) 1 1 2 = + − z − 2 z − 1 (z − 1) 2z z z − + Y = z − 2 z − 1 (z − 1) y = 2 + 1 − 2k, k ≥ 0 Answer. [z 2 Y 2
2
0
1
0
2
2
2
2
2
2
2
2
k
k
62
2
2
3.6
A Table Of Z Transform
The z transforms of some common functions are listed in the following table where a and b are constants. xk (k δk =
≥
0)
Z xk = X (z)
1
,
k=0
0
,
k>0
uk = 1 a
{ }
1 z
−z 1 z−a z (z − 1) az (z − a) z(z − cos a) z − 2z cos a + 1 z sin a z − 2z cos a + 1 1 z
k
k
2
ka k
2
cos ak
2
sin ak
2
X (z) za zX (z) zx 0
xk−a xk+1
− z X (z) − z x − zx −z dX dz 2
xk+2 kxk ak xk
2
0
1
X (z/a)
Exercise 17. Use the given table to find the z transforms of (b) ( 1)k ek (c) kek z z ez (b) (c) Answer. (a) (z e)2 z+1 z+e Exercise 18. Find the inverse z transform of 6z 5z z z (a) (b) 2 (c) + 3z + 1 z 1 (ez + 1)2 z z +1 2 kπ Answer. (a) 2( 1/3)k + 5 (b) sin 3 3 (a) cos kπ
−
−
− −
(c)
−
k e
− 1 e
k
−
(d)
√
1 ( 1/5)k where k = 0, 1, 2, . . . (d) + 6 30
−
63
5z 2
z2 4z
− −1
64
Chapter 4 Complex Analysis 4.1
Revision
Definition 4.1.1. (a) A complex number can be written in the Cartesian form as a + bi,
∈ R and i = √−1, i = −1. Note that i = −1, i = −i, i = 1, i = i i = i and so on. 2
where a, b
2
3
4
5
4
(b) If z = a + bi, then the real numbers a and b are respectively called the real part and imaginary part of z. We denote them as a = Re z and b = Im z. (c) The number 0 + bi, b = 0 is called a pure imaginary number.
(d) Graphical Representation of Complex Numbers A complex number a + bi can also be considered as an ordered pair (a, b). So we can represent it by a point in an xy-plane called the complex plane or Argand
diagram. We call the x-axis as the real axis and the y-axis as the imaginary axis. 65
(e) The number z¯ = a
− bi is called the complex conjugate of z = a + bi. It is the
reflection of the point z about the real axis. (f) Let z = a + bi where a, b
∈ R. √ (i) The number |z | = a + b is called the magnitude or modulus of z. Note : z z¯ = (a + bi)(a − bi) = a + b = |z | b (ii) The argument of z, written arg z, is the angle θ = tan− , a = 0, which a 2
2
2
2
2
1
is multi-valued.
(iii) Define the principal argument or principal value of the argument as Arg z where
−π < Arg(z) ≤ π. Example 77. Find Arg(z) if (a) z = i, (b) z = −1, (c) z = 1 + i π π Answer. Arg(i) = ,Arg(−1) = π,Arg(1 + i) = . 2 4
66
Theorem 4.1.1. A complex number z = x + yi can be written in polar form as z = r(cos θ + i sin θ) = rcis θ where r = z =
||
Note 8.
x2 + y2 and θ = arg(z).
(a) By considering the power series of the exponential function, we can show that eiθ = cos θ + i sin θ. This is called Euler’s formula. Thus we can now write in an exponential
form z = reiθ . (b) z¯ = e−iθ
Theorem 4.1.2 (De Moivre Formula) . (cos θ + i sin θ)n = cos nθ + i sin nθ, n = 0, 1, 2, 3, . . . .
Note : This formula also holds for any real number n. For example, θ θ cos + i sin . 2 2
√cos θ + i sin θ =
Definition 4.1.2 (Roots of Complex Numbers) . (a) A number w is called an nth root of a complex number z if wn = z, and we write w = z 1/n . (b) From De Moivre Formula, we can show that if n z
1/n
1/n
= [r(cos θ+i sin θ)]
=r
1/n
∈ N, then
θ + 2kπ θ + 2kπ cos +i sin , k = 0, 1, 2, . . . , n 1. n n
−
Notice that there are n different values for z 1/n , i.e. n different nth roots of z provided z = 0.
(c) The solutions of the equation z n = 1 where n unity and are given by z = cos
2kπ 2kπ + i sin =e n n 67
2kπ
n
i
∈ N are called the nth roots of
, k = 0, 1, 2, . . . , n
− 1.
(d) If we let ω = cos
2π 2π + i sin =e n n
2π
n
i
, then the nth roots are
1, ω , ω2 , . . . , ωn−1 . Geometrically, they represent the n vertices of a regular polygon of n sides inscribed in the unit circle x2 + y2 = 1.
Reading Assignment 21. Find the cube root of 1, i.e. solve ω 3 = 1. 3
Answer. ω = 1 = cis0 ∴
ω = 1,
1 2
− ±
√3 2
1/3
⇒ω=1
0 + 2kπ cis , k = 0, 1, 2 3
⇒ ω = cis(0), cis
i.
2π 4π , cis 3 3
Reading Assignment 22. Find the complex numbers z which satisfy z 2 = 4i. π + 2kπ Answer. z = 4i = 4cis 2 π + 2kπ 1/2 z = 4 cis 2 , k = 0, 1 2 5π 5π π π + i sin z = 2 cos + i sin , 2 cos 4 4 4 4 2(1 + i). z= 2
⇒ ⇒ ⇒
√ ±
4.2
Loci and Regions of the Complex Plane
Definition 4.2.1. (a) A circle with radius r and center z0 can be represented by the equation
|z − z | = r. 0
It can also be represented in the parametric form z = z0 + reit , 0 (b) The set of points z centered at z0 .
{ ∈C
:
≤ t < 2π.
|z − z | < r} is called an open disc of radius r 0
68
(c) The set of points z centered at z0 .
{ ∈C
:
|z − z | ≤ r} is called an closed disc of radius r 0
(d) Any open disc about z0 is called a neighborhood of z0 . (e) z0 is called an interior point of a set S if there exists a neighborhood of z0 which contains only points of S. (f) z0 is called a boundary point of a set S if every neighborhood of z0 contains both points in S and points not in S. (g) A set S is open if every point of S has a neighborhood containing entirely of points in S . (h) A set S is closed if its complement (all points not in S ) is open. (i) An open set S is connected if any 2 points of S can be joined by a polygonal line, consisting of a finite number of line segments joined end to end, that lies entirely in S . (j) An open set that is connected is called a domain. (k) A set is bounded if it can be enclosed by a circle of finite radius. Otherwise, it is unbounded.
Example 78. (a) The interior of a circle is open. (b) The set of all points outside the open disc of radius r about z0 , i.e. z : z z0
{
r is a closed set.
}
| − |≥
(c) The set z : 0 < z
| | ≤ 1} is neither open nor closed. (d) The open set |z | < 1 is connected . The annulus 1 < |z | < 2 is also connected. {
4.3
Functions of a complex variable
Definition 4.3.1. Let S be a set of complex numbers. A function f : S defined on S is a rule that assigns to each z 69
∈ S a complex number f (z).
→C
Note : We can write f (z) = f (x + iy) = u(x, y) + iv(x, y) where u(x, y) = Ref (z), v(x, y) = Im f (z) are real-valued functions of the real variables x and y.
Example 79. Find Re(f ) and Im(f ) if f (z) = z 2 . Answer. Re(f ) = x2
2
− y , Im(f ) = 2xy
Definition 4.3.2 (Polynomial and Rational Functions). (a) Let n be a nonnegative integer. The function P (z) = an z n + an−1 z n−1 +
··· + a z + a , 1
0
∈ C, is called a polynomial function.
where a0 , . . . , an
(b) R(z) = P (z)/Q(z) where P and Q are polynomials is a rational function.
Definition 4.3.3 (Limit). We say that the limit of f (z) as z approaches z0 exists and is equal to a number L, written, lim f (z) = L,
z
→z
0
if for every > 0 there exists δ > 0 such that
|f (z) − L| < whenever 0 < |z − z | < δ. 0
Remark 6. z may approach z0 from any direction. Definition 4.3.4 (Continuity). A function f (z) is said to be continuous at a point z0 if f (z0 ) exists, lim f (z) exists and z
→z
0
lim f (z) = f (z0 ).
z
→z
0
70
Definition 4.3.5 (Differentiability) . f is differentiable at z0 if the limit f (z) z →z z
f (z0 ) = lim
0
− f (z ) or lim f (z → −z 0
0
δz
0
0
+ δz) δz
− f (z ) 0
exits.
Note 9. The differentiation rules of real calculus are also valid in complex differentiation.
Exercise 19. Use the definition to show that f (z) = 2z if f (z) = z 2 . Example 80. Show that f (z) = z¯ = x
− iy is not differentiable everywhere.
Definition 4.3.6. [Analyticity] (a) A function f is analytic at a point z0 if f is defined and has a derivative at every point in a neighborhood of z0 . (b) f is analytic in a domain D if it is analytic at every point of D.
Note 10. A point at which f is not analytic is called a singular point or singularity. Example 81. f (z) =
1
− 4 has a singularity at z = 4. 1 has singularities at z = ±2i. Example 82. f (z) = z +4 z
2
Example 83. The polynomial functions are analytic everywhere. That is, they are what we called the entire functions.
Example 84. f (z) = z
||
2
is differentiable only at z = 0, so it is nowhere analytic.
Theorem 4.3.1. ( Properties of analytic functions) If f and g are analytic in a domain D, then f
± g , f g , and f ◦ g are analytic in D. Similarly, f /g is analytic in D provided g(z) = 0 for all z ∈ D. In particular, the rational function
p(z) ( p and q are polynomials) is analytic in any q(z)
domain throughout which q(z) = 0.
71
3z + 5 is analytic everywhere except at the points z2 + 4 where z 2 + 4 = 0 i.e., at the points z = 2i.
Example 85. The rational function
±
Theorem 4.3.2. [Cauchy-Riemann Equations] Let f (z) = u(x, y) + iv(x, y) be defined and continuous in some neighborhood of a point z = x + iy. If f is differentiable at z, then at this point, the first-order partial derivatives of u and v exist and satisfy the Cauchy-Riemann equations ∂u ∂v ∂u = and = ∂x ∂y ∂y
∂v − ∂x .
(4.1)
Corollary 4.3.3. If f is analytic in a domain D, then the first-order partial derivatives of u and v exist and satisfy the Cauchy-Riemann equations at all points of D.
Corollary 4.3.4. In this case, we can calculate using f (z) = ux + ivx or f (z) =
−iu
y
+ vy .
Theorem 4.3.5. If the real-valued functions u(x, y) and v(x, y) have continuous first partial derivatives that satisfy the Cauchy-Riemann equations in some domain D, then f (z) = u(x, y) + iv(x, y) is analytic in D.
Definition 4.3.7. [Harmonic Functions] A function h(x, y) is said to be a harmonic function if satisfies the Laplace’s equation 2
h=h
xx
+ hyy = 0.
Theorem 4.3.6. If f (z) = u(x, y) + iv(x, y) is analytic in a domain D, then u and v are harmonic functions.
Note 11. The functions u(x, y) and v(x, y) are sometimes called conjugate harmonic functions. Given one we can find the other (within an arbitrary additive constant) so that u + iv = f (z) is analytic . 72
Example 86 (Finding a conjugate harmonic function). Verify that v(x, y) = 2xy is harmonic and obtain the most general conjugate harmonic function u(x, y) of v.
Answer. u(x, y) = x2
−y
2
+ C where C is an arbitrary constant
Example 87 (Reading Assignment). (a) Show that u(x, y) = sin x cosh y is harmonic. (b) Find a harmonic conjugate v(x, y) for u.
Answer. (a) We need to show that uxx + uyy = 0. ux = cos x cosh y , uxx = sin x cosh y u(x, y) = sin x cosh y uy = sin x sinh y , uyy = sin x cosh y uxx + uyy = sin x cosh y + sin x cosh y = 0
⇒
⇒
−
−
(b) To find a harmonic conjugate v(x, y), we solve the C-R equations: vy = ux = cos x cosh y vx =
−u
y
=
(4.2)
− sin x sinh y
(4.3)
⇒ v = cos x sinh y + φ(x), (where φ(x) is a function of x) ⇒ v = − sin x sinh y + φ(x) ⇒ φ(x) = 0 (by comparing with (3)) ⇒ φ(x) = C, a constant ⇒ v = cos x sinh y + C (2)
x
4.4
More Elementary Functions
Definition 4.4.1. (a) An exponential function is defined by ez = ex+iy = ex (cos y + i sin y) = ex eiy .
Note : We want ez to be consistent in the case z = x R and insist that it d z satisfies the properties (P1) (e ) = ez and (P2) ez +z = ez ez . dz
∈
1
2
1
2
Example 88. Use the Cauchy-Riemann equations to show that f (z) = ez is entire, i.e. analytic everywhere. Hence, find f (z). 73
Answer. Show that it satisfies the C-R equations. Then f (z) = ux + ivx =
z
··· = e .
Definition 4.4.2 (Trigonometric Functions). We define the trigonometric functions in terms of exponential functions as follows: (a) sin z =
eiz
iz
− e−
2i e + e−iz (b) cos z = 2 sinz (c) tan z = cos z cos z (d) cot z = sin z iz
We will state the following theorems without proofs.
Theorem 4.4.1 (Properties of Trigonometric Functions) . (a) ez and e−z are entire functions cos z and sin z are entire functions, d d d (b) We also have (sin z) = cos z, (cos z) = sin z, (tan z) = sec2 z and so dz dz dz on.
⇒
−
(c) sin(z1
± z ) = sin z cos z ± cos z sin z (d) cos(z ± z ) = cos z cos z sin z sin z (e) cos2z = cos z − sin z, sin2z = 2 sin z cos z 2
1
2
1
2
1
2
1
2
1
2
2
2
Note 12. (a) From the definition, sin z is odd whereas cos z is even. (b) Complex trigonometric functions have all the standard properties of real trigonometric functions. 1 Definition 4.4.3 (Hyperbolic Functions). (a) cosh z = (ez + e−z ) 2 1 (b) sinh z = (ez e−z ) 2 sinh z (c) tanh z = cosh z 1 (d) sech z = cosh z
−
74
Note 13. All the standard formulas apply, for example d d (cosh z) = sinh z, (sinh z) = cosh z, cosh2 z dz dz
2
− sinh
Definition 4.4.4 (Complex Logarithmic Functions). w = ln z 0 as ew = 0).
z = 1.
⇔
z = ew (z =
Thus the natural logarithmic function is the inverse of the exponential function and can be defined by ln z = ln z + iarg z = ln z + i(Arg z + 2nπ), n = 0, 1, 2, . . . .
||
||
± ±
Note that ln z is a multi-valued. However, the principal value of ln z defined as Ln z = ln z + iArg z,
||
− π < Arg ≤ π
is a well defined (single-valued) function.
Note 14. If z = x > 0 (real and positive), then Arg z = 0 and ln z = ln x, a real logarithmic function.
Example 89. Find all the values for (a) ln 2i (b) ln( 3). What are their principal
−
values?
π π Answer. (a) ln 2i = ln 2 + i 2nπ ; Ln(2i) = ln 2 + i (b) ln( 3) = ln 3 + i π 2 2 2nπ , n = 0, 1, 2, . . . ;Ln( 3) = ln 3 + iπ
−
±
−
Example 90 (Reading Assignment). Find all the values for ln(2 principal value?
Answer. (a) Let w = ln z where z = 2
− 2i. Then
| | ··· = √8 and π (ii) arg(z) = ··· = − ± 2kπ,k = 0, 1, 2, . . . . 4 (i) z =
(b) Hence all the values of w are ln(2
−√ 2i) = ln z = ln |z| + iarg(z) π = ln 8 + i − ± 2kπ , k = 0, 1, 2, . . . . 4
75
±
− 2i). What is its
√
(c) Its principal value is Lnz = ln 8 (recall that π < Arg(z)
−
≤ π.)
− i π4
Definition 4.4.5. Any z for which f (z) = 0 is called a solution or a root of that equation or a zero of f.
Example 91. The roots of the quadratic equation az 2 + bz + c = 0 are given by
√ − b ± b − 4ac z= . 2
2a
Example 92. Solve for z : ln z = Answer.
−i
π 2
−i
Exercise 20. Solve z 2 Answer. (i
− 2(i − 2)z − 4(i + 2) = 0.
− 2) ± √11
Exercise 21. Find the zeros of f (z) = sin z. Answer. z = nπ,n = 0, 1, 2, . . . .
± ±
Definition 4.4.6 (General Powers of Complex Numbers) . Define z c = ec ln x where z = 0, c are any complex numbers.
This is in general multi-valued (since ln z is). We define its principal value as ecLn z .
Example 93. Find all the values for ii . π − Answer. e 2
± 2nπ
, n = 0, 1, 2, . . .
√
Example 94 (Reading Assignment). Find all the values for ( 1 + i 3)1/2 .
−
√
√
1 Answer. (a) By definition, w = ( 1 + i 3)1/2 = exp[ ln( 1 + i 3] 2
√ (b) ln(−1 + i 3) = ·· · = ln 2 + i
−
2π 3
± 2kπ 76
−
, k = 0, 1, 2, . . .
√ i( ±kπ 1 π (c) w = exp[ ln2 + kπ ] = eln 2 e 2 3 π π = 2 cos kπ + i sin kπ 3 3 π π = 2 cos cos kπ+i sin cos kπ using the identities cos(A+B) = 3 3 b) = π π = 2 cos + i sin , cos(kπ) = ( 1)k = 1 3 3 1 3 = 2 +i 2 2 2 6 = [ +i ] 2 2
π 3
± √ ± ± √ √··· ± − √ √ ± ±
√
√
77
±
·· · , sin(A+
4.5
Complex Integration
Definition 4.5.1. (a) A curve C in the complex plane may be described by a parametric representation z(t) = x(t) + iy(t), a
≤ t ≤ b.
dz at each point. dt (c) C is piecewise smooth if it consists of finitely many smooth curves joined end
(b) C is smooth if it has a continuous and nonzero derivative
to end.
Definition 4.5.2. Let f (z) be defined on a piecewise smooth curve C : z(t) = x(t) + iy(t), a
≤ t ≤ b.
Partition C into n parts by the points z0 = a, z1 , . . . , zn−1 , zn = b such that the greatest
|z | = |z − z − | → 0 as n → ∞. Then the line integral of k
f (z) along the curve C, denoted
k
k 1
f (z)dz, is defined as
C n
lim
n
→∞
f (zk ) zk
k=1
provided the limit exists.
Theorem 4.5.1 (Existence Theorem) . The line integral exists if C is piecewise continuous and f (z) is continuous on C.
Note 15. (Connection between real and complex line integrals) If f (z) = u(x, y) + iv(x, y), then
C
f (z)dz =
C
udx
− vdy + i
udy + vdx
C
where both of the integrals on the right are real line integrals. 78
Example 95. Evaluate the line integral I =
z¯ dz along the line segment C from 0 to 2i.
C
Answer. 2 Reading Assignment 23. Evaluate the line integral
z 2 dz along the line segment C from 1 + 2i to 2 + 4i.
C
Answer. C : y = 2x, 1
≤ 2 ⇒ dy = 2dx 2xydx+(x − y )dy] z dz = (x+iy) (dx+idy) = [(x − y )dx − 2xydy]+i [(x (2x) )dx 2x(2x)2dx] + i [2x(2x)dx + (x − (2x) )(2dx)] 77 14 ( 11x )dx i (2x )dx = − − i 3 3
≤ − − − − 2
C
= =
2
C
2
2
1
x
2
2
1
2
C
2
2
2
2
2
C
2
2
2
1
2
1
Theorem 4.5.2 (Some Properties of
f (z)dz).
C
(a) If C = C 1 + C 2 , then
f (z)dz =
C
f (z)dz +
C 1
f (z)dz.
C 2
(b) Reverse the sense of integration, reverses the sign of the integral:
f (z)dz =
−C
(c)
−
f (z)dz
C
k1 f 1 (z) + k2 f 2 (z) dz = k1
C
constants.
f 1 (z)dz + k2
C
C
79
f 2 (z)dz where k1 and k2 are
4.6
Two Integration Methods
Theorem 4.6.1 (First Method : Indefinite Integration of Analytic Functions). Let f be analytic in a simply-connected domain D. Then there is an indefinite integral F of f in D, i.e. an analytic function F such that F (z) = f (z) in D, and for every path C in D joining two points z0 and z1 in D, we have
z1
f (z)dz =
C
z1
f (z)dz = F (z)
z0
z0
= F (z1 )
− F (z ). 0
Note 16. (a) A connected region D in the xy-plane is simply connected if every simple closed curve in D encloses only points in D. Roughly speaking, a connected region D simply connected if it has no holes. The interior of a circle is simply connected whereas the annulus 1 < z < 2 is
||
not. (b) This method is restricted to analytic functions. (c) The value of the integral
f (z)dz depends only on the initial point z0 and the
C
terminal point z1 but not on the choice of the path of integration C in D. We say, the integral is independent of path in D.
Example 96. ( Integral of an analytic function) Evaluate I =
2z dz if C is any path from 0 to 2i.
C
Theorem 4.6.2. ( Second Method : Integration Along A path ) Let C : z(t) = x(t) + iy(t) be a piecewise smooth path parametrized by t with a
≤ t ≤ b. Then
b
f (z)dz =
C
f (z(t))z (t)dt.
a
Note 17. This method is not restricted to analytic functions, but applies to any continuous complex function. 80
Example 97. ( Integral of a non-analytic function) Evaluate I =
z¯dz if C is the counterclockwise unit circle.
C
Comment : As f (z) = z¯ is not analytic, we cannot use the first method to evaluate this integral.
Answer. 2πi Example 98. Let f (z) = (z I =
−z ) 0
m
where m is an integer. Evaluate the line integral
f (z)dz where and C is the (anticlockwise) circle of center z0 and radius ρ.
C
Answer. I = 2πi, if m =
−1; 0 otherwise.
Theorem 4.6.3 (Cauchy’s Integral Theorem ) . If f (z) is analytic in a simply connected domain D, then for every simple closed path C in D,
f (z)dz = 0.
C
Example 99. ( No singularities) Evaluate I =
ez dz for any closed path C .
C
Example 100 (Singularities outside contour). Evaluate
C
unit circle.
dz z2
−4
where C is the
Theorem 4.6.4 (Some Consequences of Cauchy’s Integral Theorem) . (a) Let f (z) be analytic in a simply connected domain D. Let C be a path in D with initial and final points, say A and B respectively. If C is another path in D with the same initial and final points, then
f (z)dz =
C
That is, the integral
f (z)dz.
C
f (z)dz is independent of path.
C
(b) The Principle Of Deformation Of Path If C 1 is obtained from C by continuous deformation of path without passing through any point where f (z) is not analytic, then
C
81
f (z)dz =
C 1
f (z)dz.
(c) The above principle can be extended to the case where f (z) has a finite number of singularities, say, z1 , . . . , zn inside C . By introducing n circles C 1 , . . . , Cn to enclosed each singularity, we can show that
f (z)dz =
C
Example 101. Evaluate I = (a) containing the origin;
f (z)dz +
C 1
··· +
f (z)dz.
C n
1 dz around any simple closed curve z C
(b) not containing the origin.
Answer. (a)2πi (b) 0
82
Theorem 4.6.5. [ Cauchy’s Integral Formula] If f (z) is analytic in a simply connected domain D, then for any point z0 any simple closed path C in D which encloses z0
∈ D and
f (z) dz = 2πif (z0 ) z0 C z
−
where C is taken anticlockwise. By differentiating the above equation repeatedly n times, we can extend Cauchy’s Integral Formula to
2πi (n) f (z) = dz f (z0 ). z0 )n+1 n! C (z
−
Note 18. This formula provides a method for evaluating the integral around a closed path of a function g(z) which has a single simple pole at z = z0 , by writing g(z) = f (z) where f is analytic within and on C. z z0
−
Example 102. Evaluate
sin z π dz if C : z = 3 , oriented counterclockwise. C z 2
||
−
Answer. = 2πi. Example 103. Let a be a nonzero constant. Find
C
2z z2
− a dz where C is any simple closed curve 2
(a) enclosing both the points a and a;
−
(b) enclosing a but excluding a.
−
Answer. (a) 4πi (b) 2πi
83
Theorem 4.6.6 (Taylor’s Theorem). If f (z) is analytic in a domain D and z0 is any point in D, then there exists a unique power series (the Taylor series) representing f (z), i.e. f (z) =
∞
an (z
n=0
n
−z ) , 0
where an = f (n)(z0 )/n! and this representation is valid in (at least) the largest open disc of center z0 lying in D. In general, the theorem is valid up to the nearest singularity of f (z).
Example 104. The Maclaurin expansion of ez is z
e =
∞
n=0
zn z2 z3 = 1+z+ + + 2! 3! n!
Example 105. The Maclaurin expansion of 1 1
−z
=
It is valid for z < 1.
||
∞
··· .
1
1
− z is the geometric series
zn = 1 + z + z2 +
n=0
··· .
Theorem 4.6.7 (Laurent Series Theorem) . If f (z) has an isolated singularity at z = z0 , then it has a Laurent series f (z) =
∞
n=0
valid for 0 < z
an (z
−z ) 0
n
+
b1 (z
−z ) 0
+
b2 + (z z0 )2
−
· ·· ,
| − z | < d where d is the distance to the nearest singularity. 0
Note 19. A singularity z0 of f is called an isolated singularity if there exists a neighborhood of z0 which does not contain any other singularity of f. π 3π Example 106. f (z) = tan z has isolated singularities at z = , ,.... 2 2 ez Example 107. The Laurent expansion of f (z) = 3 about z = 0 is z
± ±
f (z) =
∞
n=0
1 1 11 1 z n /n! z z2 = 3+ 2+ + + + + 2! z 3! 4! 5! z3 z z
84
··· .
Corollary 4.6.8 (Corollary to Laurent Series Theorem) . If the contour C lies in the region of validity of the Laurent series for f (z) about an isolated singularity z0 , namely f (z) =
∞
an (z
n=0
then
−z ) 0
n
+
b1 (z
−z ) 0
+
b2 + (z z0 )2
−
··· ,
f (z)dz = 2πi b1 = 2πi residue of f (z) at z0 .
·
C
·
Thus, calculation of residues is equivalent to evaluation of integrals.
Theorem 4.6.9. ( Calculation of Residues) (a) Simple pole at z0 b1 = lim (z
− z )f (z) (b) A pole of order m ≥ 2 at z = z 1 d − lim (z − z ) b = (m − 1)! → dz − z
→z
0
0
m 1
1
z
z0
Note 20.
m 1
0
0
m
f (z) . (note : OK for m = 1 too.)
If f (z) has a pole of order m at z = z0 , then its LS about z0 is bm bm−1 + + f (z) = (z z0 )m (z z0 )m−1
−
Example 108. f (z) = at z = i.
Answer.
−
·· ·
∞
b1 + + an (z (z z0 )1 n=0
−
n
−z ) 0
(4.4)
9z + i has a simple pole at z = i. Find the residue of f (z) z(z 2 + 1)
−5i
Example 109. Determine the residues of f (z) = Answer. Res(1) = 1/4, Res( 1) =
−
(z
−1/4
−
z at each pole. 1)(z + 1)2
Theorem 4.6.10 (Residue Theorem). Let f (z) be analytic on and inside a simple closed curve C except for a finite number of isolated singularities of f (z) at points z1 , z2 , . . . , zk inside an anticlockwise C . Then
C
k
f (z)dz = 2πi
Residuef (z) at z = z j = 2π i(sum of residues inside C ) .
j=1
85
Example 110. Evaluate the integral such that
4 3z dz counterclockwise around any simple closed path C 2 z z C
− −
(a) 0 and 1 are inside C (b) 0 is inside, 1 is outside
Answer. (a) 2πi( 4 + 1) =
−
−6πi (b) −8πi
Reading Assignment 24. 1 at each pole. z 3 (z 2 + 2z + 2)
Determine the residues of f (z) =
Answer. Find the poles of f (z) :
−
22 4(1)(2) = 1 i. z (z 2 That is, f has a pole of order 3 at z = 0, and two simple poles at z = 1 1 1 d2 d2 Res[f (z), z = 0] = lim 2 z 3 f (z) = lim 2 2 = 2! z →0 dz 2! z →0 dz z + 2z + 2 (z 2 + 2z + 2)2 + 2(z + 1)(z 2 + 2z + 2)(2z + 2) 1 = lim = . z →0 (z 2 + 2z + 2)4 4 3
2
−2 ± + 2z + 2) = 0 ⇒ z = 0, z =
−
Res[f (z), z =
−1 + i]
=
z
lim
→−1+i
1 1 = = ( 1 + i)3 (2i) (2i + 2)(2i)
−
since (i
− ±
3
− 1)
[z
− (−1 + i)]f (z) −1 − i =
1 4 + 4i
−
=
−1 ± i. ···
1 = z →−1+i z 3 (z + 1 + i) lim
8
= i3 + 3i2 ( 1) + 3i( 1)2
− − 1 = −i + 3 + 3i − 1 = 2i + 2. 1 Res[f (z), z = −1 − i] = lim [z − (−1 − i)]f (z) = lim →− − →− − z (z + 1 − i) 1 1 1 −1 + i = = = (−1 − i) (−2i) (2 − 2i)(−2i) −4 − 4i 8 since (−i − 1) = (−i) + 3(−i) (−1) + 3(−i)(−1) − 1 = i + 3 − 3i − 1 = 2 − 2i −
z
1 i
3
3
3
2
2
86
z
1 i
3
=
Reading Assignment 25. Evaluate the integral I =
C
(a) z = 0.5
|| (b) |z | = 1.5 (c) |z | = 2.5
(z
−
1 dz if C is the counterclockwise circle 1)(z + 2)2
Answer. 1 has a simple pole at z = 1 and a double pole at z = (z 1)(z + 2)2 The residue at z = 1 is r1 = = 1/9 f (z) =
−
−2.
· ·· The residue at z = −2 is r = ··· = −1/9 (a) Both of the poles are outside C : |z | = 0.5, so f is analytic inside C. Hence I = 0. (b) Only the pole z = 1 is inside C : |z | = 1.5, so 2
I = 2πi(the residues of the poles inside C ) = 2πir1 =
2πi 9
(c) Both the pole are inside C : z = 2.5, so
||
I = 2πi(the residues of the poles inside C ) = 2πi(r1 + r2 ) = 0
Exercise 22. Show that
sin(3z) dz = 9πi cosh(3) 4 (z + i) Γ if Γ is the anticlockwise triangle having vertices 0, 1 2i and 1
−
− −
Hint : cos(z) = sin(x + iy) = cos(x) cosh(y)
− 2i.
− i sin(x) sinh(y).
Exercise 23. Evaluate the integral I =
Re
C
Answer.
−4π
z dz if C is the counterclockwise circle z = 2. i
||
Exercise 24. Evaluate the integral I =
6z 2 cos(z 3 )dz if C is any smooth curve from 0 to i.
C
Answer.
−2i sinh(1)
87
88
Chapter 5 The Fourier Integral and Fourier Transforms 5.1
Introduction
Recall that we can represent a periodic function in terms of the sines and cosines by means of a Fourier series. We cannot do that to a non-periodic function (eg. the noise and pulse signals). But, we can still represent g in terms of the sines and cosines using an integral (instead of a summation!) called the Fourier integral. The complex form of Fourier integral will then give us the Fourier transform. Both the Fourier integral and Fourier transform are majors tools use to solve boundary and initial value problems in engineering and other fields.
89
Definition 5.1.1. The Fourier integral of a function f defined on ( given by
∞
[A(ω)cos ωx + B(ω)sin ωx]dω,
0
where
1 ∞ A(ω) = f (x)cos ωxdx π −∞
1 ∞ B(ω) = f (x)sin ωxdx. π −∞
Example 111. Find the Fourier integral of
f (x) =
Answer. f ∗ (x) =
∞ 1
ωπ
0
0
,
x<0
1
,
0
0
,
x>2
sin2ω cos ωx + (1
− cos2ω)sin ωx dω
Reading Assignment 26. Find the Fourier integral of
f (x) =
Answer. Using the results,
0
,
x<0
1
,
x=0
2e−x
,
x>0
eax [a cos bx + b sin bx] + C e cos bxdx = 2 a + b2
ax
eax [a sin bx e sin bxdx = 2 a + b2 ax
1 ∞ 1 = A(ω) = f (x)cos ωxdx = π −∞ π x − ∞ 2 e = ( cos + sin ωx ω ωx) 0 π 1 + ω2 2 1 2 1 = 0 ( 1) = 1 + ω2 π π 1 + ω2
−
···
−
−
∞
0
90
− b cos bx] + C
2e−x cos ωxdx
−∞, ∞) is
1 ∞ 1 ∞ −x = 2e sin ωxdx B(ω) = f (x)sin ωxdx = π −∞ π 0 ∞ 2 2 e−x 1 2 ω = ( sin cos = 0 (0 = ωx ω ωx) ω) . 0 1 + ω2 π 1 + ω2 π π 1 + ω2 2 ∞ cos ωx + ω sin ωx ∗ The Fourier integral f (x) = = dω. 1 + ω2 π 0
−
−
··· − ·· ·
−
Definition 5.1.2. Even and Odd Functions (a) A function f is called even if f ( x) = f (x)
−
(i) The functions 1, x2 , x4 , . . . are even.
∀ x ∈ D(f ).
(ii) cos x is even. (b) A function f is called odd if f ( x) =
−
−f (x) ∀ x ∈ D(f ).
(i) The functions x, x3 , x5 , . . . are odd. (ii) sin x is odd.
Remark 7. It follows from the definitions that (a) the product of two even functions is even (b) the product of two odd functions is even (c) the product of an even and an odd function is odd
Theorem 5.1.1. Integral Properties of Even and Odd Functions (a) If f is an even integrable function defined on [ L, L], then
−
L
L
f (x) dx = 2
−L
f (x) dx.
0
(b) If f is an odd integrable function defined on [ L, L], then
−
L
−
f (x) dx = 0.
L
91
Definition 5.1.3. (a) If f is an even function, then its Fourier integral is the Fourier cosine integral
∞
A(ω)cos ωxdω,
0
2 ∞ where A(ω) = f (x)cos ωxdx. π 0 (b) If f is an odd function, then its Fourier integral is the Fourier sine integral
∞
B(ω)sin ωxdω,
0
2 where B(ω) = π
∞
f (x)sin ωxdx.
0
Example 112. Find the Fourier integral of f (x) = 2 Answer. f ∗ (x) = π
∞ cos ωx sin ω
0
ω
dω.
92
1
,
0
,
|x| ≤ 1 |x| > 1
5.2
Fourier Transform
Definition 5.2.1. (a) Suppose
∞
|
f (t) dt converges. Then the Fourier transform of f (t) is defined to
|
−∞
be the function F[f (t)](ω) = F (ω) =
∞
f (t)e−iωt dt.
−∞
ˆ (b) Sometimes we will also write F (ω) as f (ω).
(c) The variable ω is called the frequency of the signal f. (d) The Fourier transform of the signal f (t) is also called its frequency spectrum. (e) The magnitude of the Fourier transform , F (ω) , gives the spectrum ampli-
|
|
tude, while its argument arg(F (ω)), the spectrum phase. ∞ 1 1 − (f) f (t) = F [F (ω)] = F (ω)eiωt dω is called the inverse Fourier trans2π −∞ form of F (ω).
Example 113. Find the Fourier transform of the function f (t) = e−at H (t), where 0 , t<0 is the unit step (or Heaviside a is a positive constant and H (t) = 1 , t 0 )function.
Answer. F (ω) = Note : lim t
→∞
−
≥
1 a + iω
e−(a+iω)t = lim e−at = 0, for a > 0 ⇒ lim −e−(a+iω)t = 0 t→∞ t→∞
Example 114. Find the Fourier transform of the rectangular pulse f (t) =
A
,
0
,
where T is a positive constant. T
Answer. F (ω) =
−T
Ae−iωt dt = 2A
sin ωT ω 93
|t| ≤ T |t| > T
Reading Assignment 27. Let a be a positive constant. Find the Fourier transform of f (t) = e−a|t| . You may assume the results e−(a+iω)t e(a−iω)t lim = 0 and lim =0 t→∞ (a + iω) t→−∞ (a iω)
−
Answer.
(a) f (t) = e−a|t| ⇒ f (t) =
e−at
,
t
eat
,
t<0
≥0
(b) The Fourier transform is F (ω) =
f (t)e−iωt dt,
−∞ eat e−iωt dt +
0
=
∞
∞
− −∞
(a iω)t
(by definition)
e−at e−iωt dt
0
e−(a+iω)t ∞ = + (a iω) −∞ (a + iω) 0 1 1 2a = + = 2 a iω a + iω a + ω2 2a That is, F e−a|t| = 2 . a + ω2 e(a−iω)t e(a−iω)t = lim Remark 8. lim t→−∞ (a t→−∞ (a iω) iω) at e = lim = 0 (since e−iωt = 1, a > 0 and a t→−∞ (a iω) (a−iω)t e lim =0 t→−∞ (a iω) e−(a+iω)t Similarly, lim =0 t→∞ (a + iω) e −
0
−
−
⇒
{
}
− −
−
|
94
−
|
| − iω| = 0)
5.3
Some Properties of the Fourier transform
Theorem 5.3.1 ( Linearity). If f and g are functions of t , then F[af (t) + bg(t)] = aF[f (t)] + bF[g(t)]
for any constants a and b. dn f = (iω)n F (ω) Theorem 5.3.2 (Time-differentiation property) . F n dt 1 Proof. f (t) = F−1 {F (ω)} =
∞
2π −∞
1 F (ω)e dω ⇒ f (t) = iωt
f (t) is the inverse Fourier transform of iωF (ω) follows by induction.
df dt
2π −∞
⇒F
∞
iωF (ω)eiωt dω
⇒
= (iω)F (ω) the result
Example 115. Determine the Fourier transform of the time signal y(t) satisfying the differential equation y + 3y + 4y = e−5t H (t).
Answer. Y (ω) =
(4 + 3iω
−
1 ω 2 )(5 + iω)
Theorem 5.3.3 ( First Shifting Theorem (Frequency-shift property) . F eiω t f (t) = F (ω
{
−ω )
}
0
0
Note This property indicates that multiplication by eiω t shifts the spectrum of f (t) 0
so that it is centered on the point ω0 in the frequency domain. iω0 t
Proof. F e
{
f (t) =
}
∞
iω0 t
e
f (t)e−iωt dt =
−∞
∞
−∞
95
f (t)e−i(ω−ω )t dt = F (ω 0
−ω ) 0
Example 116. Note 21. Amplitude modulation is a technique that allows audio signals to be transmitt transmitteed as electro electromagnet magnetic ic waves. The maximum frequency frequency of audio signals is typically 10 kHz. We would need to use a very large antenna to transmit these signals dire directly. To overcome overcome this problem problem,, we will wil l mix the audio signal (usually called called the whichh has has a highe higherr fre frequen quency cy.. As a modulation modulation signal signal) with a carrier carrier signal signal whic higher frequency signal has a shorter wavelength, we could now use a reasonable sized antenna. Determine the frequency spectrum of the amplitude-modulated signal g (t) = f ( f (t)cos ωc t if f ( f (t) = e−2t H (t). 1 Hint : cos ωc t = (eiω t + e−iω t ) 2 Remark 9. cos ωc t is the carrier signal and f ( f (t) is the modulation signal. 1 1 1 1 Answer. + 2 2 + i(ω ωc ) 2 2 + i(ω + ωc ) Theorem 5.3.4 (Second Shifting Theorem ( Time-shift property)) . F f ( f (t c
c
−
{ − τ ) τ )} =
e−iωτ F ( F (ω )
Remark Remark 10. This property says that delaying a signal by a time τ causes its Fourier transform to be multiplied by e−iωτ . Proof. F f ( f (t τ ) τ ) =
{ − }
e−iωτ F ( F (ω )
∞
−∞
f ( f (t−τ ) τ )e−iωt dt = e−iωτ
∞
−∞
iω (t−τ ) τ ) f ( f (t−τ ) τ )e−iω( dt = e−iωτ
∞
f ( f (v )e−iωv dv =
−∞
Example Example 117. Use the result of Example 114 and the Second Shift Theorem to determine the Fourier transform of the rectangular pulse g(t) =
A
,
0
≤ t ≤ 2T
0
,
elsewhere
Answer. Note that the midpoint of the pulse occurs at t = T . We need to shift the graph T units to the left to center the pulse at 0. Thus we can write g (t) = f ( f (t A , t T where f ( F g (t) = e−iωT F ( f (t) = F (ω ) = e−iωT 2A sinωωT 0 , t > T
||≤ ||
⇒ { } 96
− T ) T )
Example 118. Find the inverse Fourier transform of G(ω ) = 3(t−2) Answer. g (t) = e−3(t H (t
e−2iω . 3 + iω
− 2)
Theorem Theorem 5.3.5 5.3.5 (The symmetry (or duality) property) . If f ( f (t) and F ( F (ω ) form a Fourier transform pair, then F ( F (t) and 2πf ( πf ( ω ) also form a Fourier transform pair
−
i.e., F F ( F (t) (ω ) = 2πf ( πf ( ω ).
{
}
−
Proof. (a) F F ( F (t) (ω) =
{
}
∞
F ( F (t)e−iωt dt, by the definition of the Fourier transform.
−∞
∞
1 F ( F (ω )eiωt dω 2π −∞ ∞ 1 f ( f (ω ) = F ( F (t)eitω dt, t 2π −∞ ∞ 1 f ( f ( ω ) = F ( F (t)e−iωt dt 2π −∞
(b) f ( f (t) =
⇒ ⇒ − ⇒ 2πf ( πf (−ω ) =
∞
−∞
↔ω
F ( F (t)e−iωt dt = F F ( F (t) (ω )
{
97
}
Example 119. Given that the function f ( f (t) =
1
,
0
,
|t| ≤ 1 |t| > 1
has Fourier trans-
sin ω . Use the symmetry symmetry property property to deduc deducee the Fourier transform transform of ω sin t the signal g(t) = . t
form F ( F (ω ) = 2
sin ω = 2g(ω ). ω Thus, by the symmetry property F[F ( )](ω ) = 2πf ( F (t)](ω πf ( ω ). π , That is, 2G(ω ) = 2πf ( πf ( ω ) G(ω ) = πf ( πf ( ω ) = 0 ,
Answer. Given that F ( F (ω ) = 2
− ⇒
−
−
3y = δ (t Example 120 (Solving ODE). Solve y + 3y [Hint:F[δ (t
− a)] = e−
|ω| ≤ 1 |ω| > 1
2). − 2).
iωa
]
3(t−2) Answer. y = e−3(t H (t
− 2)
Remark 11. There is no arbitrary constant in this solution because the Fourier trans form has returned the only solution that is continuous and bounded for all real t. Boundedness is assumed when we use the transform because of the required convergence of
∞
| −∞
y(t) dt.
|
Reading Assignment 28. Use the Fourier transform to solve the differential equation y
− 4y = H (t)e−
where H is the Heaviside function.
Answer. Apply Fourier transform on the DE, 4t
F y
{ } − 4F{y} = F{H (t)e− } 1 (from table) iωY − 4Y = 4 + iω
Solve for Y : 1 1 2(4) = Y = (4 + iω)( 8 42 + ω 2 iω)(iω iω 4) 1 −4|t| (from table). y= e 8
−
−
−
98
4t
Example 121. Use the Fourier transform to solve the differential equation y + 3y + 2y = H (t)e−3t . 1 Answer. y = H (t)e−t 2
5.4
1 + H (t)e−3t 2
A table of Fourier transforms F[f (t)] = F (ω)
f (t) e−at H (t) (a > 0) te−at H (t) (a > 0) e−a|t| (a > 0) e−at
2t
− H (t)e−
2
dn f dtn f (t τ )
−
1 a + iω 1 (a + iω)2 2a 2 a + ω2 π −ω /4a e a
2
(iω)n F (ω) e−iωτ F (ω)
eiω t f (t)
F (ω
δ(t
e−iωa
0
− a)
F (t)
−ω ) 0
2πf ( ω)
−
Exercise 25. Find the Fourier transform of the signal 3 f (t) = 16 + t2 Answer.
3π exp( 4 ω ) 4
−| |
99
100
Chapter 6 Partial Differential Equations 6.1
Revision : Half-range Expansions
Definition 6.1.1. Let f (x) be a function defined only on the finite interval (0, L).
(a) The half-range cosine series expansion of f is the cosine series f ∗ (x) = a0 +
∞
an cos
n=1
nπx L
1 L 2 L nπx where a0 = f (x) and an = f (x)cos dx,n = 1, 2, . . . . L 0 L 0 L (b) The half-range sine series expansion of f is the sine series
f ∗ (x) =
∞
n=1
2 where bn = L
L
0
f (x)sin
bn sin
nπx L
nπx dx,n = 1, 2, . . . . L
101
Exercise 26. For the function f (x) = 1 defined only in the interval 0 < x < π, obtain (a) a half range cosine series expansion (b) a half range sine series expansion
4 1 1 Answer. (a) f 1 ∗ (x) = 1. (b) f 2 ∗ (x) = sin x + sin3x + sin5x + 3 5 π
6.2
···
.
Revision : 2nd Order HLDE With Constant Coefficients
Consider the 2nd order homogeneous linear differential equation (HLDE) with constant coefficients ay + by + cy = 0
(1)
where a, b and c are constants. Equation (1) has solutions eλx where λ satisfies the characteristic equation aλ2 + bλ + c = 0. There are 3 possible forms for the general solution of (1) depending on the roots λ1 and λ2 of the characteristic equation:
Case 1. If λ1 and λ2 are real and distinct, then the general solution of (1) is y = c1 eλ x + c2 eλ x . 1
2
Case 2. If λ1 = λ2 = λ are real and equal, then the general solution of (1) is y = c1 eλx + c2 xeλx .
Case 3. If λ1 and λ2 are complex conjugates, say, λ1 = α + iβ,λ2 = α general solution of (1) is y = eαx c1 cos βx + c2 sin βx .
102
− iβ , then the
Exercise 27. Find a general solution to (a) y
− 4y = 0 (b) y − 4y + 4y = 0 (c) y + 2y + 4y = 0
√
Answer. (a) y = c1e2x + c2 e−2x . (b) y = c1 e2x + c2 xe2x . (c) y = e−x [c1 cos( 3x) +
√
c2 sin( 3x)].
103
6.3
Partial Differential Equations
Definition 6.3.1. A partial differential equation is an equation that contains partial derivatives.
Example 122. Some classical examples of PDEs. (a) 1-D Wave Equation utt = c2 uxx (b) 1-D Heat Equation ut = c2 uxx (c) Laplace Equation uxx + uyy = 0
Definition 6.3.2. A solution of a PDE in some region R is a function that possesses all the derivatives occurring in the equation and satisfies the equation.
Note : PDEs have a large number of solutions and general solutions are not always readily obtained. Typically, further conditions are imposed to obtain a unique solution.
Example 123. [Some PDEs can be solved like an ODE] (a) Find a solution u(x, y) of the PDE ux = 0. (b) Solve for u(x, y) if uy + 2yu = 0.
104
Reading Assignment 29. Solve for u(x, y) if uxy = 0.
∂ ∂u = 0. Answer. uxy = 0 ∂y ∂x By integrating with respect to x, we get ux = f (x).
⇔
By integrating with respect to y, we obtain the solution u(x, y) = f (x) + g(y), where f (x) and g(y) are arbitrary functions of x and y, respectively.
Reading Assignment 30. Solve for u(x, y) if uxx
− u = 0.
Answer. As the PDE contains the derivatives of one of the variables only , we can treat it like an ODE u
x
− u = 0, which has a GS of the form u = Ae−
+ Be x . So the
required solution is u(x, y) = A(y)e−x +B(y)ex where A and B are arbitrary functions of y.
Exercise 28. Verify that u(x, y) = a ln(x2 +y 2 )+b is a solution of Laplace’s equation and determine a and b so that u satisfies the boundary conditions u = 0 on the circle x2 + y2 = 1 and u = 5 on the circle x2 + y2 = 9.
Answer. a = 5/(ln 9), b = 0 Definition 6.3.3. The general second-order linear partial differential equation in two variables x and y is an equation of the form Auxx + Bu xy + Cuyy + Dux + Eu y + F u = G where A , B , C , D , E , F and G are given functions of x and y. A second-order linear partial differential equation in two variables x and y that is not of the above form is a nonlinear PDE. (a) The equation is homogeneous if G = 0 for all x and y in the domain of the equation; otherwise, it is called nonhomogeneous (b) If the coefficients A,B,C,D,E, and F are all constants, then the equation is said to have constant coefficients; otherwise, it has variable coefficients. 105
Example 124. Examples of linear and nonlinear equations are (a) ut = α2 uxx (linear) (b) uxx + uyy + u = sin x (linear) (c) uux + yuy + u = 0 (nonlinear) (d) uxx + uyy + u2 = 0 (nonlinear)
Definition 6.3.4. Three Basic Types of Linear Equations The PDE Auxx + Bu xy + Cuyy + Dux + Eu y + F u = G is called (a) parabolic if B 2
− 4AC = 0. Parabolic equations often describe heat flow and
diffusion phenomena, such as heat flow through the earth’s surface. (b) hyperbolic if B 2
− 4AC > 0. Hyperbolic equations often describe wave motion
and vibrating phenomena, such as violin’s strings and drumheads.
(c) elliptic if B 2 4AC < 0. Elliptic equations are often used to describe steady state
−
phenomena and thus do not depend on time. Elliptic equations are important in the study of electricity and magnetism.
Example 125. (a) ut 4( 1)(0) = 0.
−
−u
xx
= 0 is a parabolic equation since B 2
(b) ux x + uyy = 0 is an elliptic equation since B 2 (c) utt
6.4
−u
xx
2
− 4AC = 0 −
2
− 4AC = 0 − 4(1)(1) = −4. = 0 is a hyperbolic equation since B − 4AC = 0 − 4(−1)(1) = 4. 2
2
Solving Partial Differential Equations
One way to solve a PDE with given initial and boundary conditions is to find ”enough ” solutions of the PDE that can be ”pieced together” to produce a solution of the PDE that also satisfies the initial and boundary conditions. A few methods that are commonly used are (a) Separation of Variables : This method reduces a PDE in n independent variables to n ODEs. 106
(b) Integral Transforms : This method reduces a PDE in n independent variables to a PDE in n
− 1 independent variables. A few commonly used transforms are
Fourier, Mellin, and Hankel.
Note : In particular, a PDE in two variables would be transformed into an ODE. After solving the ODE, we apply the inverse transform to obtain the solution to the PDE. (c) Numerical Methods : These methods (often) change a PDE to a system of difference equations that can be solved by using a computer. For nonlinear equations, this is often the only technique that will provide (approximate) solutions.
Theorem 6.4.1. Superposition Principle If u1 and u2 are any solutions of a linear homogeneous PDE in some regiond R, then the linear combination u = c1 u1 + c2 u2 is also a solution of the PDE.
6.5
Eigenvalue Problems
An important class of problems known as eigenvalue problems arise in solving PDEs using the method of separation of variables. A typical eigenvalue problem would be to find all real values of λ, called eigenvalues , such that the boundary value problem X + λX = 0, 0 < x < L; X (0) = X (L) = 0 has a non-zero solution. The non-zero solution(s), if there are any, are called eigen-
functions.
Example 126. Find the eigenvalues and eigenfunctions of the problem X + λX = 0, 0 < x < L X (0) = X (L) = 0 where λ is a parameter. 107
(6.1) (6.2)
Answer. Consider the 3 cases: case 1. Suppose λ < 0, say, λ =
2
−µ .
Then X = 0 is the only solution of the BVP, i.e. if λ < 0 we do no obtain non-trivial solutions.
case 2. Suppose λ = 0. case 3. Suppose λ > 0, say, λ = p2 . Then the GS of the BVP is X (x) = c1 cos px + c2 sin px. BCs X (0) = X (L) = 0
⇒c
1
= 0, c2 sin pL = 0.
For non-0 solutions, we need to have c2 = 0 and sin pL = 0, i.e. pL = nπ, n = 0, 1, 2, . . .. Thus
± ±
X (x) = cn sin
nπx , (cn arbitrary constants) L
n2 π 2 are non-0 solutions if λ is restricted tothe values λ = p = 2 , n = 1, 2, 3, . . .. This L is because 2
(i) n = 0
⇒ p = 0, which is impossible as p > 0
nπx (ii) negative n give the same values for λ and the same solutions since sin = L nπx sin . L nπx Conclusion : The eigenfunctions of the BVP are X n (x) = sin (put cn = 1) with L n2 π 2 2 the corresponding eigenvalues λ = p = 2 , n = 1, 2, 3, . . .. L
−
−
108
6.6
Separation Of Variables Method
Example 127. The Vibrating String Problem Solve the initial-boundary-value problem utt = c2 uxx , 0 < x < L, t > 0
(6.3)
with boundary conditions u(0, t) = 0, t > 0
(6.4)
u(L, t) = 0, t > 0
(6.5)
and initial conditions u(x, 0) = f (x), 0 < x < L
(6.6)
ut (x, 0) = 0, 0 < x < L
(6.7)
Answer. (i) [ Separate variables] Substituting u(x, t) = X (x)T (t) into (6.3), we obtain two homogeneous ODEs of X and T X
− kX T − c kT 2
= 0, X (0) = 0, X (L) = 0
(6.8)
= 0, T (0) = 0
(6.9)
where k is a constant. (ii) [ Solve the equations (6.8) and (6.9)] (6.9) is a second order ODE with only one IC
⇒ it has non-0 solutions ∀ k.
(6.8) has non-0 solutions only when k < 0, in which case, the non-0 solutions are
nπx corresonding to k = X n = sin L With these values of k, (6.9) becomes
−
n2 π2 T + 2 T = 0, T (0) = 0. L
109
n2 π 2 . L2
cnπt Solving this IVP, we obtain T n = cos . L nπx cnπt Thus un (x, t) = X n (x)T n (t) = sin cos , n = 1, 2, . . . . This satisfies all L L the conditions except (6.6). (iii) [ Solution to the entire problem ]
∞
∞
nπx cnπt cos . This linear combination L L n=1 n=1 also satisfies (6.3), (6.4), (6.5), and (6.7). We find An so that it also satisfies ∞ nπx (6.6) i.e. u(x, 0) = f (x) or u(x, 0) = = f (x) i.e. An are the An sin L n=1 Fourier coefficients of the half range expansion of f . Let u(x, t) =
An un (x, t) =
An sin
Reading Assignment 31. Use the method of separation of variables to solve the boundary value problem ut = uxx , u(0, t) = 0,
0 < x < π, t > 0 t>0
(6.11)
ux (π, t) = 0,
t>0 1 5 u(x, 0) = 3 sin x sin x, 2 2
−
(6.10)
(6.12) 0
≤x≤π
(6.13)
Record all your arguments in detail. You may assume the result that the eigenvalues of the eigenvalue problem X
− kX = 0, X (0) = X (π) = 0
are k=
2
−(2n − 1) /4, n = 1, 2, . . .
and that X n = sin
(2n
− 1)x , 2
are the corresponding eigenfunctions.
Answer. Sub u(x, t) = X (x)T (t) into (6.10): X T = =k XT = X T X T
⇒
110
n = 1, 2, . . .
⇒ X − kX = 0, T − kT = 0 (6.11) and (6.12) ⇒ X (0) = X (π) = 0 ⇒ X
− kX T − kT
= 0, X (0) = X (π) = 0
(6.14)
= 0
(6.15) 2
−(2n − 1) /4, (6.15)⇒ T + (2n −4 1) T = 0 ⇒ T = exp − (2n −4 1) t 2
With k = n
Let u(x, t) = (6.13)
2
∞
⇒
An = 0.
∞
An sin
(2n
n=1
An sin
(2n
n=1
2
− 1)x exp − (2n − 1) t . 2 4
− 1)x = 3sin 1 x − sin 5 x ⇒ A 2
1 The solution is u(x, t) = 3e−t/4 sin x 2
2
2
− e−
25t/4
1
= 3, A3 =
−1, and other
5 sin x. 2
Exercise 29. Solve the boundary-value problem ut = uxx , 0 < x < π, t > 0 ux (0, t) = ux (π, t) = 0, t > 0 u(x, 0) = 3
2
− sin
x, 0
≤x≤π
Note : This BVP can be thought of as a model of heat conduction in a bar of length π. The bar is insulated at both ends. In this case, the solution u(x, t) represents the temperature of the bar when the initial temperature is 3
Answer.
5 1 −4t + e cos(2x). 2 2
111
2
− sin
x.
Reading Assignment 32. Solve the PDE ut = 4uxx by the method of separation of variables.
Answer. Using u(x, t) = X (x)T (t), X T we obtain XT = 4X T or = = k, a constant. 4T X That is, we obtain the equations X kX = 0, T 4kT = 0.
−
−
We consider the following 3 cases: (a) If k = 0, then we obtain X = 0 and T = 0. Integrating, we obtain X = c1 + c2 x and T = c3 . Hence, the solution u(x, t) = c3 (c1 + c2 x) = A + Bx, where A = c1 c3 and B = c2 c3 are arbitrary constants. (b) If k = p2 > 0, we obtain X
2
2
− p X = 0 and T − 4 p T = 0.
The solutions for these two equations are 2
X = c1 e px + c2 e− px and T = c3 e4 p t . 2
2
Hence, u(x, t) = [c1 e px + c2 e− px ]c3 e4 p t = [Ae px + Be − px ]e4 p t . (c) If k = p2 < 0, we obtain X
−
2
2
− p X = 0 and T + 4 p T = 0.
The solutions for these two equations are 2
X = c1 cos px + c2 sin px and T = c3 e−4 p t . 2
2
Hence, u(x, t) = [c1 cos px + c2 sin px]c3 e−4 p t = [A cos px + B sin px]e4 p t .
Exercise 30. Solve the PDE by the method of separation of variables. (a) ut = 4uxx (b) yux + xuy = 0 2
Answer. (a) k = 0 : u = Ax + B; k = p2 > 0 : u = [Ae px + Be − px ]e4 p t ; k = p2 < 2
2
0 : u = [A cos px + B sin px]e4 p t (b) Aek(x −y
112
2
)/2
−
