iθ
ax +i D t#en t#e to t#e nearest singular point. . /f t#e roots are a=b= a , y =( Ax+ B ) e x i D x D x −i D x ")7O87> : @A<@)RM9 ")7O87> . 5et @ be asmople close% solution is y =e ( A e + B e ) . or into forms li0e ax ax curve wit# a continuousl turning tangent except possibl at a y =e ( c1 sin D x +c 2 cos D x )= y =c e sin (D x+ E ) . 97@O6 finite number of points (t#at is, we allow a finite number of O878 5/67A8 7;
: calle% t#e principal part of t#e laurent series. ∞ −& − pt 1 f ( z ) dz f ( z ) dz ! ( f )= f ( t ) e dt = " ( p ) . 1 1 bn = f ( x ) sin nx dx . )ere is an example for it. * is ' from an = , bn= ' 2 & i ( z − z ') n+ 1 2 & i ( z − z ')−n+ 1 /f all t#e bMs are & & ∞ t bn an% put @O6IO5<"/O6: pi +x+ ' an% 1 from '+x+pi.an% '+x+pi.an% %o t#e same for ero, f( is analtic at S', an% we call S' a regular point. /f # ( p ) $ ( P )= e− p t g ( t − J ) % ( J) d J J dt −& ' t = ' J= ' bn ≠ ' but all t#e bMs after after bSn are ero, f( is sai% sai% to #ave a 1 t t#em toget#er in 1:an = '⋅cos nxdx + 1⋅cos nx dx pole of or%er n at S'. S'. /f n1, we sa t#at f( # #as as a simple pole. & ' & ! g ( t − J) % ( J) d J J w#ic# is eual to . /f t#ere are an infinite number bMs %ifferent from ero, f( #as an ' COMPLEX FORM OF FOURIER SERIES: essential singularit at S' . "#e coefficient bS1 of 1U(SS' Vs n= ∞ /8A@ 75"A *<6@"/O6: ix 2i x in x − ix ∞ calle% t#e resi%ue of f( at S'. . or in ot#er f ( x )=c'+ c1 e +c −1 e +c 2 e ... = cn e n ( n) (n) K( x) L ( x− a) dx =(−1) K ( a) . 9O>7 *O8><5A9 ")7 879/<7 ")7O87>: 5et S' be an isolate% singular point n =−∞ ∞ − −& f ( z ) dz aroun% a 1 of f(. 4e are going to fin% t#e value of − inx 1 x> a c n= f ( x ) e dx %o t#e same t#is as %one in wa L . u ( x −a )= /6IO5I/6F . 2& & simple close% curve @ surroun%ing S' but inclosing no ot#er ' x< a 2 singularities. singularities. 5et f( be expan%e% in laurent series about S' INTERVALS t#e coefficients , t#e can example 1: On OTHER INTERVALS a x L ( x )=' ,b x L ' ( x )=−L ( x ) , c x L ' ' ( x )=2 L ( x ) . 2& −& t#at converges near S'. 1 1 @O6IO5<"/O6 ")7O87>: f ( x )cos nx dx bn = f ( x )sin nxdx be written a n= t t f ( z ) dz = 2 & i⋅resid residueof ueof f ( z ) at t%esingular point insideC) insideC) && & ' g ( t − J) % ( J) d J d J= g ( J ) % ( t − J) d J d J 2& 9um of resi%ues if we #ave more t#an one. /n general we write it 1 ' ' *( z ' )=lim z → z −( z − z ') f ( z ) w#en S' is a simple pole. c n= f ( x ) e− inx dx . or u can write t#em in form an% t#e "8A69*O8>9 O* 78/IA"/I79 O* : 2& ' ><5"/?57 ?O579: >ultipl f( b (S'$m, w#ere m is an l ! ( y ' )= p& − y ' 1 n & x integer greater t#at or eual to t#e or%er n of t#e pole, %ifferentiate 2 bn an% c n ./f an = f ( x ) cos dx an% same for ! ( y' ' )= p & − p y '− y' ' l − l l t#e result m1 times, %ivi%e b (m1! An% evaluate t#e resulting ( ! y ' ' ' )= p 3 & − p 2 y '− p y ' ' − y '' ' expression at S'. a function is even li0e x$2 or cos x, w#ose grap# for negativ x is KAP . CAL CULUS OF VARIATIONS: ust a reflection in in axis of its grap# grap# for positive x d ∂ " ∂ " ( x ) an% t#e opposite for o%% func. f ( − x )= f ( − =' 7<578 7;
∫
{
∫
{
l
2 b n= l
o%%,
{
a n=
2 l
∫ '
∫ '
( x ) sin n & x dx f ( l a n ='
s− 1
if f(x is even t#en.
y ' = s a' x
relation between t#e average of t#e suare of f(x an% t#e coefficients coefficients in t#e fourier series for f(x t#at t#e suare is finite. ∞ ∞ 1 f ( x )= a '+ a n cos nx+ bnsin nx .EFINITION OF 2 1 1
∑
∞
∫∞
i x x
f ( x )= g ( ) e
d
−
∞
1 − i x g ( )= f ( x ) e dx . FOURIER SINE 2 & −∞ TRANSFORMS. 4e %efine f s ( x ) an% g s ( ) fourier sine transforms representing o%% functions.
∫
√ √∫
f s ( x )= g s ( )=
2 &
2 &
'
xd ∫ g ( ) sin xd s
,
' '
f s( s ) sin xdx
. FOURIER COSINE
f x ( x ) an% g s ( ) fourier cosine transform representing even function. ' 2 f s ( x )= g ( ) sin xd xd & ' s
TRANSFORM. 4e %efine
g c ( )=
2 &
a pair of
'
f c ( s ) cos x dx
.
'
KAP ". ORINAR# IFFERENTIAL E$UATIONS. 5/67A8 */89"O878 7;?57: dy =− P dx → ln y =− P dx y ' + Py =Q $%&! y − Pdx . 97@O6O878 97@O6O878 5/67A8 y =e− Pdx + C = A e ∫ y ' ' +A ' +B y =' , 7;?57: 4e #ave 2 2 dy d dy d y = y' , D y= = = y ' ' Dy= D =d / dx . dx dx dx d x 2
∫
2
2
D1 y + y y+ B= ' →( D + D+B ) y =' t#is is x$2 expression ( D+1)( D+ B) y= ' →( D+ B ) y =' C ( D+1 ) y =' −B x − x − B x − x . /n general it is: y =c1 e , y = c2 e , → y =c1 e +c 2 e ax
bx
y =c1 e +c 2 e solution of of ( D−a )( D −b ) y ='
∑ (n+ s)( n+ s−1) a x
n + s − 2
n
n= '
KAP 1. PARTIAL IFFERENTIAL E$UATIONS: 5A?5A@7M9 7;
√∫ √∫
n + s −1
n
y ' ' = s ( s −1) a ' x s − 2+( s +1) s a 1 x s −1 +( s+2)( s +1 ) a 2 x s
. PARSEVAL'S PARSEVAL'S THEOREM: A
FOURIER TRANSFORM.
s +1
n ='
b n='
∑
s
+( s+ 1) a a x +( s +2) a 2x
∞
'
n + s
n
n='
l
∫ f ( x ) cos n &l x dx
∞
∑ a x ∞ ...=∑ ( n + s ) a x
y= a ' x s +a1 x s +1+ a2 x s +2 ...=
/f we #ave
KAP 18: FUNCTIONS OF COMPLEX VARIA6LE: A function f( is analtic in a region of t#e complex plane if it #as a %erivative at ever point of region. "#e statement Pf( is analtic at a point aQ means t#at f( #as a %erivative at ever point insi%e some small circle about a. f ( z )=u ( x , y )+iv ( x , y ) is analtic in a ")7O87> 1: /f ∂ u = ∂ v =− ∂ u region, t#en in t#at region . "#e euations are ∂ x ∂ x ∂ y calle% t#e @A<@)R8/7>A66 on con%itions. ")7O87> 2: /f u(x, an% v(x, an% t#eir partial %erivatives wit# respect to x an% are continuous an% satisf t#e @auc# 8ienmann con%itions con%itions in a region, t#en f( is analtic at all points insi%e t#e region. 9O>7 7*/6/"/O69: A regular point of f( is a p oint at w#ic# f( is analtic. A singular A singular point or singularity or singularity of f( is a point at w#ic# f( is not analtic. /t is calle% an isolate% singular point if f( is analtic everw#ere else insi%e some small circle about t#e singular point. ")7O87> 3. /f f( is analtic in a region 8, t#en it #as %erivatives of all or%ers at points insi%e t#e region an% can be expan%e% in a talor series about an point S' insi%e t#e region. "#e power series converges insi%e t#e circle about S' t#at exten%s
CAUCH#'S FORMULA: f ( ( ) d( f ( z ) dz 1 1 ∨ f ' ( a )= f ' ( z )= 2 2 2 &i 2 & i ( ( − z ) ( z −a ) ( ) ( ) f ( d( f z dz ( n) ( n ) n + n + ∨ f ( a )= f ( z )= ∮ ∮ 2 & i ( ( − z )n+ 1 2 & i ( z − a)n+ 1
p
&
nxy . "#e final result can be writter as a 4e 0now t#e value of = + 2 2 2 2i θ 1+ z 1+ x 1+ p e n & x n & y C ' − p sin# un ( x , y )= B n sin . 4e will fin% now a solution a a & no matter #ow large p becomes since t#e contour integral is x o⩽ x< a t#ere are no ot#er singular points besi%es i in t#e upper #alf u ( x ,b )= w#ic# also satisfies t#e con%itions . ' x =a p →∞ plane.5et t#en t#e secon% integral on t#e rig#t of t#e 12 ( ) f z = LAURENT SERIES EXAMPLE: euation above ten%s to ero since t#e numerator contains p an% 8emember t#at suare on t#e xaxis. 4e call t#is new con%ition z ( 2 − z )( 1+ z ) a t#e %enominator p$2. "#us t#e first term on t#e rig#t ten%s to n & b = 2 g ( x ) sin n & x dx "#is function #as singular points at ', 2, an% 1. 4e write it b sin# g(x. g(x x from ' to a. n & (t#e value of t#e contour integral as p →∞ - an% we a a ' a ∞ in a form. dx a −1 =& . .$/ 0.$2D CA3 B 4/D .2 #ave 2 n&b 2 n & x −∞ 1 + x bn = sin# x sin dx integrerer og fZr ∞ a a' a P ( x ) a dx 5A!4A. A3& 3.#*A! 2" .$ "*20) &n x & n x 2 Q( x ) a sin a x cos −∞ −1 a a n & b 2 ) /f ?(x an% ;(x are polnomials wit# t#e %egree of ; at least two − bn = sin# 2 2 a a &n & n greater t#an t#e %egree of ?, an% if ;( #as no real eros. /f t#e ' n+ 1 −1 2 integran% ?(xU;(x is an even function, t#en we can also fin% t#e n & b 2 a (−1) bn = sin# . "#en t#e solution is integral from ' to infinit. COSX; IS AN EVEN FUNCTION a a &n AN SINX; IS AN O FUNCTION. n −1 +1 ∞ 2 a (−1) n&b n & x n & y ⋅sin# ⋅sin sin# u ( x , y )= . a &n a a n= 1 SOME IMPORTANT IFFERENTIAL R ELATIONS:
∮
∫
∮
dz
∫
dx
∫
& ( y )=& n( y )=sin#
i θ
p i e d θ
{
∫
∫
∫
4#en we #ave ( D 2+( 2) y =' 4e get D=±i ( an% t#e i(t − i(t solution can be written in two forms: or y = Ae + B e y =c1 sin ( t +c 2 cos ( t .
A SECON ORER IFFERENTIAL EXAMPLE:
)[
(
]
∫
[
) ( ) ) [ ] ∑( )
( (
( )
]
HERE IS ANOTHER EXERCISE FROM O6LI3 2: ∂2 u( x , y ) ∂ u ( x , y ) = 47 #ave a partial %ifferential: y wit# 2 ∂ y ∂ x u ( x ,o )=L ( x ) . 4e will tr to solve t#is t#e con%ition euation b use of fourier transform, >)-< x a0 a,)a?le a*+ @ a0 pa,ame-e,. 4e use now t#e fourier transform of t#e %erivative. 2 " [ f ' (t )]=i ( " [ f ( t )] , " [ f ' ' ( t )]=−( " [ f ( t )] . 4e ta0e t#e fourier transform of t#e euation. ∂ 4 ( x ,t ) 2 − y ( 4 ( ( ,t )= → − y ( 2 4 ( ( , y ) dy=4 ( ( , y ) ∂ y 1 − y ( 4 ( ( , y ) dy= 4 ( ( , y )⇒ 4 ( ( , y )= A( ( ) e y( 2 ∂ 4 (( , y ) + y ( 24 ( ( , y)= ' ⇒ 4 ( ( , y )= A( ( ) e−1 /2 ( y ∂ y ∞ 1 − i( x u ( x , y )= 4 ( ( , y ) e dx .we 0now t#at 2 & −∞ ∞ 1 ∞ x =' − i( x L ( x )= 4 (( , ' )= L ( x ) e b ' x≠a 2 & −∞
∫
∫
2
2
2
2
∫
{
∫
∫ L (∞ x ) f ( x ) dx= f ( ') #ence for <(w, ' we get 1 4 (( , ')= ∫ u( x ,') e dx = 21& = A( ( ) t#us we #ave 2& ∞
%efinition
− i( x
−
4 (( , y )= u ( x , y )=
1 − 1/ 2( y e 2& ∞ 1 1 2
2
e √ 2 & ∫∞ √ 2 & −
an% t#e final result is i(x
2
e
−1/ 2 y (
2
d(
vi integrerer og bru0er
formelen fra rottman. 8esten er en0elt.
FOURIER TRANSFORM TIL 9 VISE: ∞
∫ '
cos ( x
a 2 +( 2
=
& 2a
e −a∣( x )∣
.
LAPLACE TRAN SFORM OF ERIVATIVES: /t is for t#e first %erivative ! [ f ' ( t ) ]= s ! [ f ( t )]− f ( ') *or 2 t#e secon% %erivative ! [ f ' ' ( t )] = s ! [ f ( t )]− s f ( ')− f ' ( ') SOME OTHER METHOS FOR FININ3 RESIUES: 2&
2&
∫
d t
1 = 1+b cos2 t 2
=
'
dt =
1
i z 1 = 2i
2 ib
dz
∮ ∮(
we set
∫ '
d t i t . → dz =i e d t =i zd t 1 +b cos2 t it
z =e , cos t =
d t b b z +( +1 ) z 2+ B 2 B z d z b
1 2
( z + z − 1)
or
an% get
#ere is not so muc# place so ..
B
z W=(−b−2±2 √ b+1 ) /b . *ire 2 2 2 2 z − z W )( z − z ( ) ± z W er innenfor @ og begge en0le poler, bare polene / 1 b & = resi%ene er . ette gir = 2( z 2W− z 2 ) X √ b+1 √ 1+b =& blir integran%en 1, og 2
∞
RESIUE THEOREM SIMPLER 5A#:
−∞
∞
consi%er
∫∞ 1dz + z −
∫
2
dx 1+ x 2
. 4e
5e
∂ u = ∂2 u ∂ x2 ∂ y2
. on t#e 4#ere @ is close% '⩽ x⩽ a , ' ⩽ y⩽ b me% grensebetingelser intervall boun%ar of t#e semi u (', t )=u (', y )= u( a , y )=' . 4e will use separation of circle s#own #ere. *or variables to fin% t#e general euation. an pY1, t#e semicircle 6 ' ' ( x ) & ' ' ( y ) incloses t#e singular point + =' u ( x , y )= 6 ( x ) & ( y )=' Or 6 ( x ) & ( y ) i an% no ot#ersH t#e − 6 '' ( x ) & ' ' ( y ) resi%ues of t#e integran% at i is : = = (constant ) #ence we #ave 1 1 6 ( x ) & ( y ) ( *( i )= z −i ) = . "#en t#e value of z → i n & x ( z −i )( z +i ) 2 i 6 ' ' + 6 =' =(' =(a' 6 ( x )= 6 n ( x )=sin 2 & i ( 1/ 2i )=& . /ntegral in two parts.(1 t#e contour integral is a 2 2 an integral along t#e xaxis from p to pH f or t#is part xH (2 an n & = n= 2 iθ & ' ' − & = ' , R(' ' . , integral along t#e semicircle, w#ere . "#en we #ave z = p e a
lim
IT >a0 -
2 p3( p +2)
.
USIN3 LAPLACE TRANSFORM TO O6TAIN SOLUTIONS FOR COUPLE E$UATIONS: 2 x¨ ( t )+2 n x˙ ( t )+n x ( t )= ' we will use laplace transfrom 2 y¨ ( t )+2n y ˙ ( t )+ n y ( t )=[ x˙ ( t )
/n @ase]3: t#e solution is given b a a y =c1 x cos( D ln( x ))+c 2 x sin (D ln ( x )) )ere is an example for Freen function.
to obtain t#e solution of x(t an% (t. /nitial con%itions x('(''. x˙ ( t )=\ 4e call 5( R an% 5(x =. 2 x ' ' +2 nx ' +n x= ' "ransforms of %erivatives og . 2 y ' ' + 2 ny ' +n y =[ x' 2 ! ( y' )= p& − y' , ! ( y ' ' )= p − p y '− y '' we insert t#is into e 2
2
2
6 ( p +2 n p + n )− \ =' ⇒ 6 ( p + n) =\ 2
2
2
& ( p + 2 n p+n )−[ p 6 =' ⇒ & ( p +n) =[ p 6 \ p [ \ \ 1 ⇒ !−1 & = , 6 = = \ !−1 B 2 2 2 ( p+ n ) ( p+n ) ( p+n ) ( p+ n) 1 1 −1 !−1 ⋅ = ! [ # ( p )⋅ $ ( p ) ]= g ∗% b 5S3B ( p + n) ( p +n) 1 1 ⇒ ! [ g ( t )] = ⇒ g ( t )=e −n t ⇒ g ( t −J )=e−n( t −J ) # ( p )= p +n p+n 1 1 ⇒ ! [ $ ( p )]= ⇒ e−nt ⇒ %( J)=e −n J $ ( p )= ( p +n) p +n
[
]
t
∫
\ e
[ p \ p+n
−1
!
] [
]
t
−n (t − J )
e
'
& =
[
B
−n J
∫
d J =
[
⇒ [\ !− 1
− nt
\ e
d J ⇒ x ( t )=\ t e
'
p 2
⋅
1 2
( p+n ) ( p+ n)
[ # ( p )⋅ $ ( p )] = g ∗ %⇒ #( p )= 1
−n t
⇒ x˙ ( ' )= \
] p 2
( p+n )
[ ] t
2
3
n t
⇒ % ( t )=t e ⇒ y ( t )=[\ e−n t − 2 2 T ( p +n ) THIS IS VER# IMPORTANT FOR IFFERENTIAL E$. y ' ' + y' −2 y =B sin 2 x . instea% of tac0ling t#is problem $ ( p )=
− nt
⇒( 1 −n t ) e−nt = g ( t )
%irectl, we are first going to solve t#e euation . 9ince exp(2ix 2i x cos 2x W / sin 2x is complex, t#e solution & ' ' +& ' −2 & =B e & =& *+i & is euivalent to R ma be complex also. "#en 2i x & * ' ' +& * ' ' −2 & *= ℜ B e =B cos 2x two euations 9ince t#e 2i x & ' ' +. ' − 2& = ℑ B e =B sin 2x secon% euation above is t#e same as t#e uestion, we see t#at t#e solution of t#e uestion is t#e imaginar part of R. "#us to fin% y p for t#e uestion we fin% & p for t#e euation above an% ta0e its imaginar part. 4e observe t#at 2i is not eual to eit#er of t#e roots of t#e auxiliar euation in R euation. *ollowing t#e met#o% of t#e last paragrap#, we assume a solution 2 ix & p = C e of t#e form an% subsitute it into R euation to get: B (−2i −T) −1 2i x 2i x (−B+2i −2) C e =B e ⇒ C = B = = ( i +3 ) 2 i −T B' A −1 2 i x & p= ( i +3 ) e ta0ing t#e imaginar part of RSp we fin% A −1 3 y p = cos 2x− sin 2x . A A
EN INHOMO3EN LI3NIN3EN. y ' ' + P x y ' +Q x y = * x Ii mer0er oss at #vis y % ( x ) er en l^sning av %en #omogene ligningen 8(x ', og Sp (x en eller annen l^sning av %en in#omogene (parti0ul_r y ( x )= y% ( x )+ y p ( x ) en l^sning av %en l^sning, sZ er ogsZ: in#omogene ligningen. 9p^rsmZlet er %a om %enne l^sningen y ' ( x ') , passer me% gitte grensever%ier for y ( x') og sli0 at vi fZr %en ^ns0e%e ent%ige l^sningen. >e% y %( x )=c1 y 1( x )+ c2 y 2( x ) fZr vi betingelsene
c1 y 1( x ')+ c2 y 2( x')= y ( x ')− y p( x' ) c 1 y 1( x' )+c2 y ' 2( ')= y' ( x ')− y p( x ')
som 0ent at %eterminanten til ligningssettet mZ v_re uli0 null, % vs. y ( x ) y 2 ( x ' ) 7 = 1 ' ≠ ' . Og %ette vet vi fra f^r er o0 y ' 1 ( x') y' 2 ( x ) si%en y 1( x , y 2( x er line_rt uav#engige. ()vis #^re si%en / ligningsettet ovenfor s0ulle v_re li0 null, #ar vi %ire0te at y p ( x ) er en l^sning som tilfresstiller 0ravene, og er %erme% %en s^0te ent%ige l^sningen erme% er problemet formelt l^st. en generelle l^sningen av %en in#omogene %iff ligningen #ar formen y ( x )= c1 y 1( x )+ c2 y 2( x )+ y p ( x ) 6este problem blir y p ( x ) . )er er %et imi%lerti% Z finne en parti0ul_r l^sning flere meto%er som 0an bru0es. 7t vanlig tilfelle er %iff ligninger #vor venstre si%en #ar 0onstante 0oeffis%ienter. y ' ' + ay ' +by = * ( x )er 0an vi 0omme langt me% lit strategis0 getting. rx rx 1. *( x )= A e , pr8v y p ( x )= B e sin cos * ( x )= A rx+ B rx , pr8v y p ( x )=C sin rx + D cos rx 2. 3. * ( x )= polynomav grad 3 , pr8v y p( x )= polynom avgrad 3 B x x * ( x )=e ( A sin rx+ B cos rx ) , pr8v y p ( x )=e (C sin rx + D cos rx ) rx 70semple: . 5^sning av %en #omogen lig y ' ' − 2' + y= A e rx x x vi pr^ver og finner y %( x )=c1 e +c 2 x e y p( x )= B e A A B= y p( x )= og %erme% . ( r −1 )2 ( r −1)2 y ' ' + B y ' + B= cos# x 4e want to fin% Anot#er example is: t#e general solution to t#e %ifferential euation, t#at is we want to fin% t#e solution to t#e #omogeneous %ifferential euation. y ' ' + B y' + B= ' *rom t#e c#aracteristic euation 2 2 \ + B\ +B=( \ +2 ) = ' ⇒ \ =−2, −2 9ince we #ave repeate% root, we #ave to intro%uce a factor of x for one solution to ensure − 2 x − 2 x linear in%epen%ence. 9o we obtain "#e u1 =e ,u 2= x e wrons0ian of t#ese two functions is
∣
THE CAUCH#EULER E$UATION. 2
x2 d y2 +a x d y +b y =' dx dx m
y = x 2
d y
4e assume a trial solution given b
, %ifferentiating, we #ave. m− 2
m− 1 dy =m x dx
An%
=m ( m−1) x . 9ubstituting into t#e original euation 2 dx m− 2 m−1 m 2 x ( m( m−1) x )+ a x ( m x )+b( x )='
0rav til l^sning er
∣
− 2 x
∣
− 2 x
∣
2 − 2x − 2 x − 2 x −2 x e x e 8earranging gives : m +( a−1) m+b=' we t#en can solve for =−e e ( 2 x −1)+2x e e −2 x − 2 x −2 e −e (2 x −1) m. "#ere are t#ree particular cases of interest. m1 an% m2 @ase ]1: "wo %istinct roots, −e −Bx ( 2 x−1)+2 x e− B x =(−2 x+1+2x) e− Bx=e− Bx Necause t#e @ase ]2: One real repeate% root, m wrons0ian is non ero, t#e two functions are linearl in%epen%ent, a ±D i . @ase ]3: @omplex roots, so t#is is in face t#e general solution for t#e #omogeneous %iff m m /n @ase]1, t#e solution is given b: y =c1 x +c2 x euation. 4e see0 functions A(x an% N(x so m m A ( x ) u 1 + B ( x ) u2 is a general solution of t#e non /n @ase]2: t#e solution is given b y =c1 x ln( x )+c 2 x #omogeneous euation. 4e nee% onl calculate t#e integrals 1
2
∫ ∫
∫
∫
∫
∫
FROM CHAPTER : EULER E$UATION EXERCISES. x 2
∫
= x √ 1+ y ' 2 dx
" ( x , y , y ' )= x √ 1+ y' dx 2
4e #ave
x 1
∂ " =', ∂ y
∂ " = ∂ ( x √ 1+ y' 2) ⇒ u = y' 2+1 ⇒ d √ u = 1 ∂ y' ∂ y ' du 2 √ u 2 2 d ( y ' +1) d ( y ' ) d + (1) x x
∂ 1+ y' 2 ⇒ dy ' = dy dy x √ 2 2 ∂ y ' 2√ y ' +1 2 √ y ' +1 ∂ " = x y ' from 7uler euation we #ave partia y' √ y' 2+1 x y' ∂ " =' ∂ ∂ " − ∂ " =' ⇒ ∂ =' ∂ x ∂ y ' ∂ y ∂ x √ y ' 2+1 ∂ y x y ' =c ( constant ) ( x y ' )2=c2 ( y' 2+1) √ y' 2+1 c dy c 2 2 2 2 2 x y ' = y ' c +c ⇒ y ' = 2 2 , y' = dx = d x √ x −c √ x 2−c2
∫ dy=∫
c
√ x
2
−c
∫ arc cos# ( x )=
2
∫
dx ⇒ y =
1
√ x 2−1
y − =c cos# −1( x / c )⇒ −
1
a = !b = c c points x1! x 2
x c
=cos#
c
√
x 2 c
2
dx −1
( ) y − c
∫
1 2 ' + f ( y ) dy=const) y @A97 (% : An euation of t#e form 2 a2 x 2 d y2 +a 1 x d y +a ' y= f ( x ) @alle% an 7uler Or @auc# dx dx euation, can be re%uce% to a linear euation wit# constant coefficients b c#anging t#e in%epen%ent variable from x to d y d y z = x w#ere. x= e *or t#e we #ave an% d x d z 2 2 x2 d y2 = d y2 − d y dx d z d z @A97 (e 8e%uction of or%er. "o fin% a secon% solution of y ' ' + f ( x ) y ' + g ( x )=' given one solution u(x, substitute y =u ( x ) v ( x ) into euation above an% solve for v(x. get
1 1 A ( x )=− u ( x ) B ( x ) dx , B ( x )= u b ( x ) dx i.e. 7 2 7 1 1 2 x − 2x A( x )=− − Bx x e cos# x dx=− xe cos# x dx e 1 x 2 x A( x )= e (( V( x −1)+e (3 x−1 ))+ C ) −1X 1 − 2x 1 x 2 x 2 x B( x )= e cos# xdx = e cos# x dx= e ( 3+ e )+C − B x T e
A IFFERENTIAL EXERCISE 6# USE OF LAPLACE ' 1
f ( t )=
t <' t >'
y '= y '' = '
T
∫
∫
∞ 2 1 −cos ( (a) cos ( ( x ) d( 2 & −∞ (
∫
ANOTHER FOURIER TRANSFORM EXAMPLE: FINN FOURIERREKKA FOR FUNKSONEN (x; 3ITT
⇒ ax=cos# ( ay+b )
constants c an% 0 are %etermine% from given .
HERE IS ANOTHER $UESTION SAME LIKE A6OVE: x 2
∫ (1+ y y ' )2 dx ⇒ " ( x , y , y ' )=(1+ y y ' )2
=
{
1 ( !+ x ) − !⩽ x <' 2 f ( x )= tilZfin% 1 ( ! − x ) ' < x ⩽ ! 2 a' ∞ n & x ∞ n & x + f ( x )= + an cos bn sin 2 n=1 ! ! n= 1 −
∑
∂ " ∂ " =2 ' ( 1 + y y' ) ! =2 y ( 1+ y y' ) ∂ y ∂ y ' d ∂ " ∂ " d − =' ⇒ ( 2 y (1+ y y ' ) )−2 y ' ( 1+ y y' )=' d x ∂ y ' ∂ y dx t#e %erivation ca n #appen wit# pro%uct t#ing. d d v ( u v )= v d u +u An% at last we get. dx dx dx 2 2 2 2⋅( y ⋅ y '' + 2 y⋅ y' + y ' )− 2 y ' ( 1+ y⋅ y ' )=' ⇒ 2 y ( y ' ' − y )=' 1 2 −c' x= y − an% we #ave to fin% t#e x w#ic# is eual to . 2c 2c
.
n =1
∑
Og 0oeffisientene
!
t#en
x 1
∞
∑ sinnn x
an=
finnes av
1 n & x f ( x ) cos dx ! − ! !
∫
og bSn
!
1 n & x f ( x ) sin dx Ii mer0er oss at f(x / %enne ! − ! ! an =' for alle n. oppgaven er en o%%efun0son, sli0 at vi #ar b n si%en f(x er o%%e fun0son. Ii%ere finner vi for
∫
b n=
bn=
2 !
!
∫ ( ! − x ) sin n & x ! dx
*or Z bestemme bSn be#^ver vi
'
!
∫
f^lgen%e integraler: og
sin
− ! n & x dx= ! n&
n
[(−1 ) −1 ]
og
A long rectangular metal plate #as its two long si%es an% t#e far en% at ' @ an% t#e dx n − ! n & x RESIUE EXAMPLE: %en blir til base at 1'' @. "#e wi%t# of t#e plate is ( 1 −B )( 1 +B z ) x sin dx= (−1 ) ( delvisintegras;on ) til C ! n& 1'cm. *in% t#e stea% state temperature i pi x ' e ∇ . =' %istribution insi%e t#e plate. dz er integrasonsveien @ er en lu00et 2 2 ! 2 2 C ( 1 − B )( 1 +B z ) b n= sammen finnes %a og fourierre00a for f(x blir ∂ . + ∂ . =' n& Or "o solve t#is 0urve ra ` 8 til 8 pZ %en reelle a0sen og en #alvsir0el me% ra%ius ∞ ∂ x 2 ∂ y 2 ! 1 n & x 8 / ^vre #alvplan. 7tter or%ans lemma vil integralet over f ( x )= sin setter !=& og #ar %a summen euation, we are going to tr a solution of * →∞ . 9i%en integran%en #ar #alvsir0elen gZ mot null nZr & n ! . ( x , y )= 6 ( x ) & ( y ) . t#e form. n =1 poler pZ %en reelle a0sen for 1U2 og , mZ vi ogsZ legge 2 2 1 fra oppgaven. d 6 smZ #alvsir0ler run%t %isse / ^ vre #alvplan. Ii%ere er %et / ^vre = 1 d & 2 =const =− 2 0Y' . 6 ' ' =− 2 6 An% 2 IFFERENTIAL EXERCISE: 6 d x & d y #alvplan en pol for iU2. ?rinsipal ver%ien av integralet er %a gitt 2 ∞ 2 x y ' ' − xy ' + y = x Sle )-D cos & x & ' ' = & "#e constant 0$2 is calle% t#e separation constant. P dx ve% som er li0 y y ( 1 −Bx )( 1+ Bx ) −∞ sin x sin x e e ⇒ 4e first = = = = 6 , & . 6& Alle − y 2 & i *es ( z = i /2 )+ & i [ *es ( z =− 1 / 2 )+ *es ( z = 1 / 2 ) ] cos x cos x e e− y & y −1 2 %iscar% t#e solutions containing since we are given polene er en0le, og vi #ar %en an%re e *es ( z =i / 2)= ie X . → ' as y → ∞ . 6ext %iscar% solutions containing −i − y cos(0x since "', w#en x'. "#is leaves us ust e sin x , *es ( z =−1/ 2)= *es( z = 1/ 2 )= resultatet blir %a. X but t#e value of 0 is still to be %etermine%. 4#en x1', we are to −& cos pi x & 2 n & dx = ( 1+e ) 2 2 #ave "', t#is will be tru if sin(1'0', t#at is if = for X C ( 1 − B )( 1 +B z ) 1' FOURIER TRANSFROM EXAMPLE: n1,2,... t#us for an integral n, t#e solution n & x − n& y / 1' −a ⩽ x ⩽' 2x +a sin . = e 9atisfies t#e given b oun%ar con%itions f ( x )= −2x + 2a '⩽ x⩽ a 1' en fourier transformerte er %a. ' ellers on t#e t#ree "' si%es. *inall we must #ave "1'' w#en ' ∞ n & x − n& y /1' . = bn e sin for ' we must #ave "1'' 1' n= 1 ∞ n & x =1'' for f(x 1'' wit# l1'. . y= '= bn sin 1' n= 1
∫
cos 2
pi x
2
∫
'
!
∫
2
∑
∫
2
{
{
2
{ "{ "
∫
∑
∑
bn= . =
OTHER SECON ORER E$UATIONS:
ANOTHER FRO6ENIBS METHO EXAMPLE.
y ' = p , y ' ' = p' @A97 (a: epen%ent variable missing: @A97 (b : /n%epen%ent variable x missing: dp dp dy d p = = p y ' = p , y ' ' = dx dy dx dy @A97 d: "o solve y ' ' + f ( y )=' multipl b M. y ' y' ' + f ( y ) y' =' or y ' dy' + f ( y ) dy=' t#en integrate to
2 1'
B'' n
1'
∫ 1'' sin n1'& x dx= '
e−& y /1' sin
& x 1'
+
1 3
{
B'' n& '
e−3 & y / 1'sin
odd n
" becomes t#en
even n 3& n 1'
+...
)
