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Classical Electrodynamics

Chapter 3Boundary-Value Problems in

Electrostatics: II

A First Look at Quantum Physics

2011 Classical Electrodynamics Prof. Y. F. Chen

A First Look at Quantum Physics

2011 Classical ElectrodynamicsProf. Y. F. Chen

Contents§3.1 ▽2 in Different Coordinates§3.2 Laplace Equation in Spherical Coordinates§3.3 Legendre Function§3.4 Ex. Two Hemispheres at Equal and Opposite Potentials, Find the

Potential Outside the Sphere§3.5 Associated Legendre Function §3.6 Spherical Harmonics Function§3.7 Hypergeometric Function§3.8 ▽2Φ in cylindrical coordinates§3.9 Solution of Bessel equation§3.10 General Solution of ▽2Φ=0 in Cylindrical System§3.12. Fourier Analysis of Periodic Function§3.13 Fourier-Bessel series

§3.15 Green Function in Spherical System §3.14 Boundary value problems in cylindrical coordinates

§3. 1 ▽2 in Different Coordinates

Cartesian coordinates

General coordinates

2 2 22

2 2 2x y z

, ,x y z

1 2 3, ,q q q

(1) Gradient

1 1 2 2 3 3

1 1 2 2 3 3

1 1 1dl h dq h dq h dq

h q h q h q

1 2 31 2 3

d dq dq dq dlq q q

d dl

1

1q 2q 3q

1q 2q 3q


2 2 3 1 3 1 2

1 2 3 1 1 1 2 2 2 3 3 3

1 h h h h h hh h h q h q q h q q h q

(2) Gauss’ theorem → flux of field

1 2 3 2 3

2 1 3 1 3

3 1 2 1 2

ds h h dq dqds h h dq dqds h h dq dq

1 1 1 1 1 2 3 2 3 2 3, ,E ds E q dq q q h h dq dq

flux a

1 1 1 1 2 3 2 3 2 3, ,E ds E q q q h h dq dq

flux b

2 3 11 2 3 1 2 3

1 2 3 1

1

d

h h E h h h dq dq dqh h h q

2 3 1 1 3 1 1 2 11

1 2 3 1 2 3

1 h h E h h E h h Etotal flux d E dh h h q q q

2(3) in general coordinates

2

From eq. (1) and (2): 1E

1E

1 1h dq2 2h dq

3 3h dqE1

flux aflux b

1 2 3, ,q q q


2

2 22 2

1 1sin sinsin sin

rr r r

2

2

22

2 2 2 2

1

1 1 1 1 0sinsin sin

rr r

rr r r r

, ,let r R r P Q

2 2

22 2 2 2 2

1 1 1 0sinsin sin

P QrR PQ RQ RPr r r r

2 22

2 2 2 2 2

2 2

2 2 2 2

1 1 1 1 1 1

1 1 1 1 1 1 1

0

sinsin sin

sinsin sin

P QrRR r r P r Q r

P QrRR r r r P Q

divide by R r P Q

§3. 2 Laplace Equation in Spherical Coordinates


22

2

2

2 2

1 0 1 2

1 1 1 1 1 0 1 2

; , , ,

sin ; , , ,sin sin

Q m mQ

letP Q l l l

P Q

(1) 2

22

imd Q m Q Q ed

(2) 2

2

1 1sinsin sin

d d m P l l Pd d

cos , sinlet x dx d

22

21 1 01

:ml

d dP mx l l Pdx dx x

where P x P x associated Legendre functions

0if m

21 1 0

:l

d dPx l l Pdx dxwhere P x P x Legendre functions

3


(1) Generating function for Legendre function: from point charge

˙

˙

˙

q

Px nr

x nr

2 2

1 1 1n n 2

,? cosr r r r rr

x xx - x -

2

2

1

1 2

1

1 2

:cos

:cos

if r rr rr r r

if r rr rr r r

2 2

1 1 11 2

1 2

,

cos

tt txr rr r r

Generating function: 2

0

11 2

, ll

lG x t P x t

t tx

§3. 3 Legendre Function


(2) Recursion relation(a) Differentiate about ,G x t

0

12 2

0

11 2 1 2

,

ll

l

ll

l

P x t

dG x t x tlP x t

dt t tx t tx

2 1

0 0

1 1 1

1 2

1 2 1

l ll l

l l

l l l l l

x t P x t t tx lP x t

xP P l P lxP l P

1 12 1 1l l ll xP l P lP 4

0

2 20

11 2 1 2

,

ll

l

ll

l

P x t

dG x t t P x tdx t tx t tx

2

0 0

1 2 1

0 0 0 0

1 2

2

l ll l

l l

l l l ll l l l

l l l l

t P x t t tx P x t

P x t P x t P x t x P x t

1 2 12l l l lP P P xP 5

t

Differentiate about ,G x t x


(b) Recursion formula 1:

Differentiate eq. (4) by : 1 12 1 2 1 1l l l ll P l xP l P lP

multiply eq. (5) by : l 1 12l l l llP lxP lP lP

6

7

16 7 1 l l ll P xP P

(c) Recursion formula 2:

multiply eq. (5) by : 1l 9

16 9 l l llP xP P

8

1 11 2 1 1 1l l l ll P x l P l P l P

10

(d) Recursion formula 3:

1 18 10 2 1 l l ll P P P

x

11


(3) Prove that satisfy lP x

1l l in eq. (8) : 1 1l l llP xP P

multiply eq. (10) by : x 21l l lxlP x P xP

12

13

2112 13 1 0l l lx P xlP lP

14

Differentiate eq. (14) about : 211 0l l l lx P lP xlP lP

15

Use eq. (10)

21 1 0l lx P l l P

21 1 0d dPx l l Pdx dx

x


(4) Orthogonality condition: 1

1

02

2 1l m

l mP x P x dx

l ml

(a) if l m

multiply eq. (2) by and integrate: mP x

1 12 2

1 11 1 1 1 0m l l l m l mP x P l l P dx x P P l l PP dx

16

l m in eq. (16)

1 2

11 1 0l m l mx P P m m PP dx

17

1

116 17 1 1 0l ml l m m P x P x dx

1

10l mP x P x dx if l m


(b) if l m

1 12 1 1l l ll xP l P lP

1 22 1 1l l ll xP lP l P

eq. (4):

1l l in eq. (4): 18

118 2 1 ll P

21 1 1 12 1 2 1 2 1 1 2 1l l l l ll l xPP l l P P l l P

1 1 1 2

1 1 1 11 1 1

0

2 1 2 1 2 1 1 2 1l l l l ll l xPP dx l l P P dx l l P dx

1 1 2

1 11 12 1 2 1 2 1l l ll l xPP dx l l P dx

19

21 22 1 2 1 2 1 2 1 1l l l l ll l xPP l l P l l PP

1 1 2

11 12 1 2 1 2 1l l ll l xPP dx l l P dx

20

1 12 2

11 1

2 119 202 1l llP dx P dxl

1 1 12 2 2

2 01 1 1

2 1 2 3 2 1 2 3 22 1 2 1 2 1 2 1 2 1l ll l l lP dx P dx P dxl l l l l

118 2 1 ll P


(5) Evaluate the integral 1

0 lP x dx

1 1

1 10 0

12 1l l lP x dx P x P x dxl

1 1 1 11 1 1 0 0

2 1 l l l lP P P Pl

21

1 1

0

0ll

PP x dx

l

1 1

1 114 0 0.

l

l l

Plfrom eq P P

l

Substitute into eq. (21)

(a)

(b) Evaluate 1 0lP

Generating function at :0x

2

0

10 01

, ll

lG t P t

t

2

21

1 1 3 2 1 112 2 21 !

n

n

n tnt

2 222

1 1

1 2 1 1 21 1

2 2

!! !! !

n nn n

n nn n

n nt t

n n

22 1

2!

!!!n

nn

n

23

24

22


Compare eq. (23) and eq. (24)

2

2

0

0 1

2 2

!

!

l

l

l

if l odd

P lif l even

l

(c) From eq. (25), we known that must be odd in eq. (22)

25

l

Let 2 1l n

121 1

20 1

0 1 11

12 2

!

!

l

ll

l

P lP x dx

l l l

22 2

1 2 21 0 1 22 1 2 1

!, , ,

!

n

n

nn

n n

26


(6) Legendre series representation:

0

1

1

2 12

l ll

l l

f x A P x

lwhere A f x P x dx

Ex.

0

1

1

11

1 01 0

for xf x

for x

0

l ll

f x A P x

1 0 1

1 1 0

2 1 2 12 2l l l l

l lA f x P x dx P x dx P x dx

1

10

12

21

22 2

2 12 1 0

2 1 1 1

12 2

4 3 1 2 20 1 2

2 1 2 1

!

!

!, , ,

!

l l

l

l

n

n

ll P x dx P

l

l ll l

n nn

n n


2 222 2

4 3 1 2 22 1 2 1

!

!

nn

l n

n nB a V

n n

V

V

a 0

2

2

,V

aV

10

1, cosl lll

r B Pr

General solution:

At :r a 10

1, cosl lll

a B Pa

Multiply in both side of eq. (27) and integrate

27

coslP

22 2

1

1 0 1

0 1 1

1 0 0

264 3 1 2 22 1 2 1

2 1 2 12 2

2 1 2 12

.!

!

, cos sin ,

n

n

ll ll

l l l

from eqn n

Vn n

B l la P d a P x dxa

l V P x dx P x dx l V P x dx

2 2 2

2 1 121 2 20 0

4 3 1 2 21 32 1 22 1

!, cos cos cos

!

n n

l l nl nl l

n n a ar B P V P V Pr n r rn

(1) Method I:

§3.4 Ex. Two Hemispheres at Equal and Opposite Potentials, Find the Potential Outside the Sphere


1

0

1

0

1 1 2 14 4

1 2 14

, , , , cos , sin

cos , sin

lD

lS Sl

l

lSl

G ar r da l P a d dr r

al P a d dr

(2) Method II : use Green’s function

2 2

2

1 1 1n n

n n,

??

DaG r r

r r r ra r ar r r rr a r

2 22 2

12

0 0

1 1

1 2 1 2

1 1

: ,

cos cos

cos cos

D

ll

l ll l

for r a G r rr r rr a arr r a rr rr

r aP Pr r a rr

1 21 2

10 0

1 1

1 10 0

1 11n

1

cos cos

cos cos

l llD D

l lll lr a

l l

l ll ll l

G G a al P l Pr r a r a

a al P l Pr r

1

10

2 1 cosl

lll

al Pr

(a)


cos cos cos sin sin cos

Set 0cos

cos cos cos

cos cos cos

cosl l

l ll

P P

A P

Set in eq.(28), we get 0 0 1l

l lA

Set in eq.(28), we get0

29

cos cos cosl l lP A P

1cosl l l lP A P A

cos cos cos cos cosl l l lP P P P

31

(b) Evaluate coslP

30

Finally, from eq. (29) and eq. (30)

32


(c) Substitute eq. (32) into eq. (28)

1

0

12

0 00

1

0

1 2 14

1 2 14

1 2 12

, , cos , sin

cos cos , sin

cos cos ,

l

lSl

l

l ll

l

l ll

ar l P a d dr

al P P a d dr

al P P ar

0sin d


22

21 1 01

:ml

dP xd mx l l P xdx dx x

where P x P x associated Legendre functions

(1) Prove that 2 21m m

ml lm

dP x x P xdx

2 21m

mlP x x y x

12 22 21 12

m mm

ld mP x x y x x y xdx

2 2 12 2 2 22 2 22 2 1 2 1 1

m m mm

ld P x m m x x y x mx x y x x y xdx

34

33

21 34 2 33 1 mlx x l l P x

21 2 1 1 1 0x y m x y m m y l l y 35

(a) Let

§3. 5 Associated Legendre Function


(c) Compare eq. (35) and eq. (36)

m

m

ldy x P xdx

2 2

2 2

1

1

mm

l

m m

lm

P x x y x

dx P xdx

(b) Differentiate eq. (35) by m times

0

m k m kmmkm k m k

k

d d duse f g C f gdx dx dx

2 1 1

2 1 121 2 1 2 2 1 0m m m m m m

m m m m m m

l l l l l ld d d d d dx P mx P m m P x P m P l l Pdx dx dx dx dx dx

2 1

2 121 2 1 1 1 0m m m m

m m m m

l l l ld d d dx P m x P m m P l l Pdx dx dx dx

36


(2) Other properties

2 2 22 2

1

1

11 1 12

1

22 1

!!!

!!

m mm l m lml lm l l m

mm ml l

m ml l ll

d dP x x P x x xdx l dx

l mP x P x

l m

l mP x P x dx

l l m


2 1

4!

, cos!

m imlm l

l mlY P el m

(1)

2 1

0 1

1, , ,

, , cos

ml m lm

lm l m ll mm

Y Y

Y Y d d

(2) Addition theorem

˙

˙

˙P

P

x

y

z

42 1

cos , ,l

l lm lmm l

P Y Yl

if z-axis is along : n , ,l

lm m lmm l

Y B Y

Let 0 0: and m

0 0 02 104

, ,lm llY B Y B

0 0, ,lm lB Y Y d 37

38

《Prove》

§3. 6 Spherical Harmonics Function


Similarly, as z-axis is along , can be expressed asn coslP

04

2 1cos , ,

l

l l lm lmm l

P Y b Yl

04

2 1, cos , ,lm lm l lm lb Y P d Y Y d

l 39

From eq. (37), eq. (38), and eq. (39)

04 4

2 1 2 14

2 1

,

cos , , ,

lm lm

l l

l lm lm lm lmm l m l

b B Yl l

P b Y Y Yl


§3. 7 Hypergeometric Function

2

21 1 0d dz z u z z u z u zdz dz

General solution:

0

01 2 1 1

, , ;!

,

, , ; :

kk k

k k

k

u z F z zk

where k

F z Hypergeometric Function

Hypergeometric equation:

(1) Solve hypergeometric equation

Let 0

s kk

ku z z C z

2 1 0z u zu u zu u z

1

0 01 1 1 0s k s k

k kk k

s k s k s k C z s k s k s k C z

1

0 01 0s k s k

k kk k

s k s k C z s k s k s k C z

10 1

0 11 1 1 0s k s s k

k kk k

s k s k C z s s s C z s k s k s k C z

01 0

1s or

s s ss

40


(a) As 0s

11 1k kk k C k k k C

11 1 1k kk k C k k C 1k k

1

1 11k k

k kC C

k k

Choose 0 1C

0 1 2

1 2 1

, , ,!

k kk

k

k

C kk

where k

10

, , ;!

, , ; :

kk k

k k

u z z F zk

F z Hypergeometric Function

41


(b) As 1s

1

1 1 1 11 2k k

k kC C

k k

Choose

1 10 1 2

2

1 2 1

, , ,!

k kk

k

k

C kk

where k

2

0

1 11 1 2

2, , ;

!kk k

k k

u z z F zk

From eq. (41)

11 1 1 1 1 2k kk k C k k C

0 1C


(2) Relation between hypergeometric equation and

Legendre equation

2

221 2 1 0d dx P x x P x v v P x

dx dx Legendre equation:

2

111 2

2 114

xzxz

xz z

2

21 1 2 1 0d dz z P z z P z v v P zdz dz

43

Compare eq. (42) and eq. (43)

11

vv

2

21 1 0d dz z u z z u z u zdz dz

Hypergeometric equation: 42

2

0

11 1 1112 2 2

, , ;!

kk k

vk

v vx x xP z F v vk

Let in Legendre equation


2(1 ) " 2 ' ( 1) 0: (cos ) 0

x P xP Pwith boundary condition P

Legendre's differential equation :

1 20

( ) ( 1)1 1( ) ( , 1,1; )2 ! ! 2

kk k

k

x xP x Fk k

0

( ) ( 1) 1 cos(cos ) cos 0! ! 2

kk k

k

Pk k

P

z

(3) The fields in a Conical hole or near a sharp point:

→ Azimuthal symmetry Problems

General solution:

The power series form has the problem of divergence


1222

0

20

221 0

0 2

2210

0 2

2

2 sin , 0,1,!

( ) ( 1) 1 cos(cos )( !) 2

( ) ( 1)2 1 cos sin! 2

( ) ( 1)2 1 cos sin! 2

2 1 1 cos, 1, ; sin2 2

k k

kk k

k

kkk k

k kk

k k

k k

d kk

Pk

dk

dk

F

20

2 1 2 11 ( 1 ) ( 1 ), 1, ; ( )2 2 1

1( ) (1 ) ( 2 ) ( 1) 02

d

w w w wF w f ww

f w satisfy w w f w f f

(a) Another useful forms of Legendre:


2 2 2

2

2 1 2 1 12 2 2 2 2

2 2

1 12 2

2

sin sin sin ,2 2

1 cos sin sin sin sin2 2 2

1 1 sin cos2 2

(cos sin ) (cos sin ) cos[( ) ]1, 1, ;2 2cos cos

cos[( ) ] co2(cos )cos

let

w i

w

i iF w

P

22

02

12

12 22

02 2

122

2 202 2

122

0

ssin cos

cos[( ) ] sin2 12sin sin

cos[( ) ]22 sin sin

cos[( ) ]22cos 2cos

d

d

d

d

Integral form of Legendre

2011 Classical Electrodynamics

(b) Asymptotic form of Legendre:

12 1(cos ) sin (1 ( )) 0sin 2 4

1 1 3 1, 1,2 . . 1,2 4 4 2

P O

n n e g n

(c) Electric potential and electric field :

1

1

1

( , ) (cos )

0, ( , ) (cos )

(cos )

1 sin (cos )

k

kkk

r

r A r P

at r r Ar P

E Ar Pr

E Ar Pr


(cos ) versusP

Difference between the three forms of Legendre


§3. 8 ▽2Φ in cylindrical coordinates

2 22

2 2 2

1 1 0z

, ,let z R Q Z z

2 22

2 2 2

1 1 0R QZ Q RZ Z RQz

2 22

2 2 2

1 1 1 1 1 0R Q ZR Q Z z

Divide by R Q Z z

22

2

22

2

1 0 1 2

1

; , , ,Q m mQ

letZ k

Z z


(1) 2

22

imd Q m Q Q ed

(3)2

22

1 1 0mR kR

2 22

2 2

1 0d R dR mk Rd d

let x k

2 2 2

2 2 22 2 2

1 1 0 0d R dR m d R dRR x x x m R Bessel equationdx x dx x dx dx

2

22

kzd Z k Z Z z edz

(2)


§3. 9 Solution of Bessel equation

2

2 2 22 0d y dyx x x v y

dx dx

Let 0

s kk

ky x x C x

2 2

0 01 0s k s k

k kk k

s k s k s k v C z C z

2 2 10 1

22

2 2

1 1 1

1 0

s s

s k s kk k

k k

s s s v C z s s s v C z

s k s k s k v C z C z

21 0:sz term s s s v s v

(1) Series solution


(a) As s v

2k K

Choose

01 1 12 1

1 1 1

:

, !v

k

x Gamma function

C where n n n nv

n k n n

2 2

112 1!

KK K vC

K v k

2

10

11 2!

:

K v K

vK

v

xy x J xK v K

J x Bessel Function

221 k kv k v k v k v C C

21

2k kC Ck k v

2 2 2 2 22

02 2 2

02

1 12 2 2 2

1 1 12 2 1 1 2 1

112 1 2 1!

K K K

KK

C C CK K v K K v

CK K v K K v v

CK K v K v v v

Prof. Y. F. Chen

(b) As s v

2

20

11 2!

K v K

vK

xy x J xK v K

Similarly

(2) Properties of solution

if v m integer

1 1

1

2 2

0 0 1 1

1 11

2 2! ! ! !

K Km K m Km

mK K

x xJ xK K m K K m

1 mmJ x

1let K m K

m mJ x and J x are linear dependent

mWe must find another solution which is linear independent to J x

Define Neumann function:

cossin

m mm

m J x J xN x

m

2

10

2

20

11 2

11 2

!

!

K v K

vK

K v K

vK

xy x J xK v K

xy x J xK v K

Solution of Bessel equaiton

(a)


(b) Asymptotic form of Bessel and Neumann function

22 4

22 4

cos

sin

m

m

nJ x xx

As xnN x x

x


Two linear independent solution

(3) Summary in solution of Bessel equation

2

10

2

20

11 2

11 2

!

!

K v K

vK

K v K

vK

xy x J xK v K

xy x J xK v K

(a) if v integer

Two linear independent solution

2

10

2

12! !

cossin

K m K

mK

m mm

xy x J xK m K

m J x J xy x N x

m

(b) if v integer m

Asymptotic form

22 4

22 4

cos

sin

m

m

nJ x xx

nN x xx


(c) Figure of Bessel and Neumann function


§3. 10 General Solution of in Cylindrical System2 0

˙

x

y

z

a

L

(1) Boundary condition:

2 22

2 2 2

1 1 0z

2 22

2 2 2

1 1 1 1 1 0R Q ZR Q Z z

, , z R Q Z z

0

02

, ,

, ,

a z

Lz

22

2

22

2

2 22

2 2

1 0

;

m

d Q m Qd

d Z k Zdz

d R dR mk Rd d

R J k


(2) Boundary condition:

0

02

, ,

, ,

a z

Lz

22

2

22

2

2 22

2 2

1 0

;

m m

m

d Q m Qd

d Z k Zdz

d R dR mk Rd d

R J ik I k

where I x is modified Bessel function

2 2

0 0

12 2! ! ! !

K m K m Km

mK K

ix i xJ ixK m K K m K

Modified Bessel function

2

0

12! !

m Km

m mK

xI x i J ixK m K

mI x


2

0

11 2!

K m K

mK

xJ xK m K

2

0

12! !

K m K

K

xK K m

(1) Recursion relation

44

22

0

22

0

1 12

1 12

! !

! !

K m Km Km

mK

K m Km K

mK

x J x xK K m

x J x xK K m

2 2 12 1

0 0

1

2 2 12 1

0 0

1 2 112 1 2

1 2 112 1 2

! ! ! !

! ! ! !

K Km K m Kmm Km

mK K

mm

K Km K m Kmm K

mK K

m K xd xx J x xdx K K m K K m

x J x

K xd xx J x xdx K K m K K m

1 2 1

10

11 2! !

K m Kmm

mK

x x x J xK K m

45

46

§3. 11 Bessel Function


(a) Recursion formula 1:

From eq. (45):

11

m m m mm m m m

d x J x mx J x x J x x J xdx

1m m mm J x J x J xx

(b) Recursion formula 2: From eq. (46):

11

m m m mm m m m

d x J x mx J x x J x x J xdx

1m m mm J x J x J xx 48

47

(c) Recursion formula 3:

1 1247 48 m m mm J x J x J xx

(d) Recursion formula 4: 1 147 48 2 m m mJ x J x J x

49

50


(2) Orthogonality condition:

220

1

0

2

a

m mn m mnm mn

n nJ k J k d a J k a n n

2

22

1 0m mn mn m mnd d mJ k k J k

d d

Bessel equation:

multiply Bessel equation by and integrate:

n n in eq. (51)

52

m mnJ k

2

220

0a

m mn m mn mn m mn m mnd d mJ k J k k J k J k d

d d

00

22

20

0

22

20

0

0

aa

m mn m mn mn m mn m mn

a

mn m mn m mn

a

m mn mn m mn mn mn m mn m mn

a

mn m mn m mn

d dJ k J k k J k J k dd d

mk J k J k d

J k a ak J k a k k J k J k d

mk J k J k d

0

22

200

a

m mn mn m mn mn mn m mn m mn

a

mn m mn m mn

J k a ak J k a k k J k J k d

mk J k J k d

51

integrate by part

Prof. Y. F. Chen

2 2

00

a

mn m mn m mn mn m mn m mn mn mn m mn m mnk aJ k a J k a k aJ k a J k a k k J k J k d

(a) if n n

2 2

00

0a

mn m mn m mn mn m mn m mn mn mn m mn m mnk aJ k a J k a k aJ k a J k a k k J k J k d

0

0a

m mn m mnJ k J k d (b) if n n

2 20

a mn m mn m mn mn m mn m mnm mn m mn

mn mn

k aJ k a J k a k aJ k a J k aJ k J k d

k k

0

2 2

2 2

2

2

2

lim

lim

mn mn

mn mn

mn mn

a

m mn m mnk k

mn m mn m mn mn m mn m mn

k kmn mn

mn m mn m mn m mn m mn mn m mn m mn k k

mn

m mn m mn

J k J k d

k aJ k a J k a k aJ k a J k ak k

k a J k a J k a aJ k a J k a k a J k a J k a

k

a J k a J k a

0

53

51 52

0


1 12 0m mn m mn m mnmn

m J k a J k a J k ak a

1 1 12 2m mn m mn m mn m mnJ k a J k ax J k a J k a

eq. (49):

eq. (50): 54

Substitute eq. (54) into eq. (53)

2

210 2

a

m mn m mn m mnaJ k J k d J k a


(3) Generating function of Bessel function:

2

0

12

,! !

K m Km m

mm m K

xG x t J x t tK K m

Let 0:m K K K

0 0 0 0

0 0

22

1 12 2 2 2

1 12 2

,! ! ! !

! !

K K K K K

K K K K

K K

K K

xxtt

xt x xt xG x tK K t K K t

xt xK K t

e e

1

2,x t

mtm

mG x t e J x t

55


(4) Integral representation of Bessel function

1

2,x t

mtm

mG x t e J x t

Let i

x kt e

2 sini ik e e ik im

mm

e e J k e

2 2 2

0 0 0

2

sin

mm

ik im im im im imm m

m me e d J k e e d J k e e d

Multiply in both side of eq. (56) and integrateime

56

2

0

12

sinik immJ k e e d


(5) Expansion of a plane wave in terms of Bessel function

˙

,x yk k k

,r x y

x

y

cosx yk r k x k y k

plane wave: cosx yi k x k y ikik re e e

Let and substitute it into i

x k

t ie

2 cos,

i ik ie ie mik imm

mG x t e e J k i e

,G x t

cos mik imik rm

me e J k i e


(6) Completeness relation of cylindrical wave:

21

2x y x yi k x k y i k x k y

x yx x y y e e dk dk

2

0

12

m imm m

m

x x y y

J k J k i e kdk

22 cos cos

0 0

22

0 0

22

0 0

2

12

12

12

mm

ik ik

m mim imm m

m m

m m i m m i m mm m

m m

e e kdkd

J k i e J k i e kdkd

J k J k i e kdk e d

2

0

12

m imm m

mJ k J k i e kdk

0 0

12

im

m

m m m m

e

k kJ k J k kdk J k J k d

k

Prof. Y. F. Chen

§3. 12 Fourier Analysis of Periodic Function(1) periodic function:

2 ni xL

nn

f x f x L f x F e

57

22

2 2

2 2

n nnL Li x i xL L

n nL Ln

f x e dx F e dx F L

2

22

2

1

ni xL

nn

nL i xL

n L

f x F e

F f x e dxL

(2) Helmholtz equation

with periodic B. C.

2

22 0d k x

dx

x x L

General solution: 21 ni x

L

nx e

L

(3) 21 ni x xL

nx x e

L

12

in

ne

2;x L


(4) Fourier integral

2 2 2 2 2

2 2

2 2

1 2 12

n n n n nL Li x i x i x i x i xL L L L L

n L Ln n n

f x F e f x e dx e f x e dx eL L

As 2

2n

dkLL

n kL

12

n n

n

ik x ik x ikx

n

F k

f x dk f x e dx e F k e dk

12

ikx

ikx

f x F k e dk

F k f x e dx


(5) Discrete Fourier Transform (DFT)

2

1 0 1 2 1, , , ,i nN NZ Z e n N

2 11 1 1 0N NZ Z Z Z Z

0

22 1

1

1 0 1 2 1, , ,i sN N

s

Z e

Z Z Z root is Z e s N

58

Let from eq. (58) s n n

21

0

1 0,N i m n n

Nnn

me n n N

N

01

2

3

45

67

8

9

1N

2 2 21 1 1 1 1

0 0 0 0 0

1 1

m

N N N N Ni m n n i mn i mnN N N

n n nn n nn n m m n

F

f f f e f e eN N

21

0

21

0

1

N i mnN

n mm

N i mnN

m nn

f F e

F f eN

59

unit circle


21

0

21

0

1

N i mnN

n mm

N i mnN

m nn

f F e

F f eN

Discrete Fourier Transform (DFT) Fourier Series

2

22

2

1

ni xL

nn

nL i xL

n L

f x F e

F f x e dxL

n

n n

LdxN

x nL Nf x f


22

2

1 0

0:

mn m mn

m mn

d d mk J kd d

with boundary condition J k a

2

21

0 0 0

2 m mn nn

a a a

m mn n m mn m mn n m mn m mnn n

a J k a

f J k d F J k J k d F J k J k d

2 2 01

2

n m mnn

a

n m mnm mn

f F J k

F f J k da J k a

Arbitrary function: n m mnn

f F J k

(1) Expansion of by a series of Bessel function f

§3. 13 Fourier-Bessel series


(2) 2-D Fourier-Bessel series

0 1

, sin cosmn mn m mnm n

V A m B m J k

2

0 0

2

0 00 1

2 2

0 00 0

0

, sin

sin sin sin cos

sin sin sin cos

mn mm

a

m mn

a

mn mn m mn m mnm n

mn mnm m

A

V m J k d d

A m m B m m J k J k d d

A m m d B m m d

01

01

22

12

a

m mn m mnn

a

mn m mn m mnn

mn m mn

J k J k d

A J k J k d

aA J k a

2

2 2 0 01

2 , sina

mn m mnm mn

A V m J k d da J k a

(a) coefficient mnA


0 1

2

2 2 0 01

2

2 2 0 01

2

2

, sin cos

, sin

, cos

mn mn m mnm n

a

mn m mnm mn

a

mn m mnm mn

V A m B m J k

A V m J k d da J k a

B V m J k d da J k a

60

(b) coefficient mnB

2

0 0

2

0 00 1

, cos

cos sin cos cos

a

m mn

a

mn mn m mn m mnm n

V m J k d d

A m m B m m J k J k d d

similarly

2

2 2 0 01

2 , cosa

mn m mnm mn

B V m J k d da J k a


˙

x

y

z

aL

2 22

2 2 2

1 1 0

0

0 0

, ,

: , ,

, , ,

z

a z

with boundary condition z

z L V

(1) General solution: 0 1

, , sin cos sinhmn mn m mn mnm n

z A m B m J k k z

0 1

, , , sin cos sinhmn mn m mn mnm n

z L V A m B m J k k L

mn mnA and B can be obtained from eq. (60)

61

§3. 14 Boundary value problems in cylindrical coordinates


Fourier series → Fourier transform

Fourier-Bessel series → Hankel transform

(2) Hankel transform

In eq. (61), let

:sinh

mn

kzmn

ak k

zk z e

0

0, , sin cos kz

m m mm

z A k m B k m J k e dk

0

0, sin cosm m m

mV A k m B k m J k dk

2

0 0

2

0 0 00

2 2

0 00 0

0

, sin

sin sin sin cos

sin sin sin cos

m mm

m

m m m mm

m mm m

A k

V m J k d d

A k m m B k m m J k dk J k d d

A k m m d B k m m d

0 0 m mJ k J k d dk

(a) coefficient mA k


2

0 0, sinm m

kA k V m J k d d

(b) coefficient mB k

2

0 0

2

0 0 00

, cos

cos sin cos cos

m

m m m mm

V m J k d d

A k m m B k m m J k dk J k d d

Similarly

2

0 0, cosm m

kB k V m J k d d

0 0 0

mm m m m

k kk

k k A kA k J k J k d dk A k dk

k k

62

00

2

0 0

2

0 0

, sin cos

, sin

, cos

m m mm

m m

m m

V A k m B k m J k dk

kA k V m J k d d

kB k V m J k d d


2 4,G x x x - x

2 2

2 2 2 2 2

1 1 1 1 4sin , cos cossin sin

r G r rr r r r

x x

We know that 0

, , cos cosl

lm lml m l

Y Y

63

, , , ,l

l lm lml m l

G g r r Y Y

x x

Substitute into eq. (63) ,G x x

2

2 2 2

11 4,l

l l r rd r g r rr dr r r

§3. 15 Green Function in Spherical System


(1) at : r r

2

2 2

1

11 0,

: ,

l

lll l l

l ld r g r rr dr r

B rgeneral solution g r r A r r

r

→ Consider three cases:

(a) Outside the sphere:

a

r

1

1

: ,

: ,

lll l l

ll l

B ra r r g r r A r r

rB r

r r g r rr

2 1

1

1

: ,

: ,

ll

l l l

ll l

aa r r g r r A r rr

B rr r g r r

r

0,lfrom g r a r

2 1

1 1

1: ,l

ll l l l

ageneral form g r r C r rr r

64


(b) Inside the sphere:

r

b

1

0 : ,

: ,

ll l

lll l l

r r g r r A r r

B rr r b g r r A r r

r

1 2 1

0

1

: ,

: ,

ll l

l

l l l l

r r g r r A r r

rr r b g r r B rr b

0,lfrom g r b r

1 2 1

1: ,l

ll l l l

rgeneral form g r r C r rr b

(c) Concentric sphere (spherical shell):

a

b

65

From eq. (64) and eq. (65)

2 1

1 1 2 1

1: ,ll

ll l l l l

rageneral form g r r C rr r b


(2) at : r r

2

2 2

11 4,l

l ld r g r r r rr dr r

Consider the cases of concentric sphere

2

2

14,

r r

lr r

l l r rd r g r r dr drdr r r

0

2

2

1 4

,

, ,

r

lr

r r

l lr r

d rg r rdr

l ld rg r r dr g r r drdr r r

Upper limit: ,r r

r rr r

2 1 1

1 2 11,

l ll

l l l l lr

d a d rrg r r C rdr r dr r b

66

2 1

1 1 2 1

1 1l l

ll l l l

a rC r l lr r b

67


lower limit: ,r r

r rr r

2 1 1

12 1

1,l l

ll l l l l

r

d d a rrg r r C rdr dr r r b

2 1 1

1 2 1

11l l

ll l l l

a rC l r lr r b

68

From eq. (66), eq. (67), and eq. (68)

2 1

4

2 1 1l l

Calb

2 1

1 1 2 12 10

142 1 1

, , , ,

, ,

l

l lm lml m l

llllm lm l

l l lll m l

G g r r Y Y

Y Y rarr r bal

b

x x

Finally, we obtain


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