Chapter 4.Problems of Boundary Values

in Static Electric Fields1) Separation of variables in system of rectangular coordinate

2) Separation of variables in system of cylindrical coordinate

3) Separation of variables in system of spherical coordinate

4) Image method ( 镜像法不作要求 )

◇ The boundary value problem of electrostatic and steady fields can be solved by Laplace or Poission equation under given boundary conditions

回忆两无源介质间 ( 无电荷无电流 ) 的边界条件

1 Separation of variables in system of rectangular coordinate

Separation of variablesSeparation of variables is a classical differential equation method, which is suitable for a sort of boundary value with ideal boundary conditions. ◇ General steps

Select the coordinate according to the geometric shape and field distribution of the boundary, and list the differential equations and boundary conditions. Separate the variables. Solve the equation (general solution: the linear combination of all special solutions) Determine the integral constant (based on the given boundary conditions) to obtain the final solution

2

0

+ 0k

2

2 2

2

2

电磁学（电动力学）中有三个重要方程：

Lapl ace方程(无源)： ＝

Poi sson方程（有源）： ＝－ ，如对于点电荷， ＝-e r

Hel mhol tz方程(无源+波动)： ＝

它们都带有Lapl ace算符 。这些方程与它们的解具有典型性（普遍适用性），在电动力学、量子力学、流体力学、声学、天体物理中均会出现。

An example:

Determine the potential function in 2D rectangle space.

xa

y

b0U

0 0

0

?

2

0

0 0

0 0 ,0

0

0,

xx a

y y b

x a y b

U

According to the problem, the potential function and its boundary conditions are given as follows:

In rectangular coordinate, can be expressed as:2 0

2 2

2 20

x y

(1)

Separate the variables as follows:

,x y f x g y (2)

Substituting equation (2) into (1), one can arrive at

2 2

2 2

d d0

d d

f x g yg y f x

x y (3)

The two sides of the above equation is divided by

2 2

2 2

d d1 10

d d

f x g y

f x x g y y (4)

Only depends on variable x Only depends on variable y

So the two terms on the left side in Eq. (4) can be equal to two constant numbers, respectively, i.e.,

22

2

22

2

d1

d

d1

d

x

y

f xk

f x x

g yk

g y y

(5)

g y f x

2 2 0x yk k

Where

(6)

If k x is a real number and k y an imaginary number

1 2sin cosx xf x A k x A k x (7)

Thus the solution of the equation (5) is:

1 2

' '1 2

sinh cosh

x x

x x

x x

f x B x B x

f x B e B e

Or (8)

If k x is an imaginary number

1f x C x C

If k x=0

(9)

Analyzing the boundary conditions, we can obtain

(7)

Since =0 when x=a and 0

1 2sin cosx xf x A k x A k x

2

0

0 0

0 0 ,0

0

0,

xx a

y y b

x a y b

U

x

nk

a

where is called the eigenvalue of the present boundary problem:

(8)

Substituting the boundary conditions into Eq. (7), we can get

1 sinn

f x A xa

(9)

(10)

1 2sinh coshy yg y B y B y

Thus, g(y) is

Note:

2 2

22

0

0

x y

x y

y x

k k

k j

k

1 sinhn

g y B ya

Hence,

1

, sin sinhnn

n nx y C x y

a a

(11)

The general solution is given by

00

0

0

42sinh sin d

2 1

4

sinh 2 1

a

n

n

Un nC b U x x

a a a n

UC

nb n

a

Employing the theory of Fourier series expansion

(12)

0

1

4, sin sinh

sinh 2 1n

U n nx y x y

n a ab na

So the final solution is:

(13)

2 Separation of variables in system of cylindrical coordinate

2 2

2 2 2

1 10r

r r r r z

The Laplace equation in cylindrical coordinate system can be written as:

(14)

,r f r g (16)

If the field is two-dimensional field and is independent of z, the Laplace equation can be rewritten (simplified) as

2

2 2

1 10r

r r r r

Assume the solution (r, ) can take the following form

(15)

Substitution of (16) into (15) yields

(17) 2

2 20

g f r f r gr

r r r r

(18)

Function of Function of r

2r

f r g Simplify Eq. (17) by multiplying

2

2

10

f r grr

f r r r g

(19)

So,

2

22

1f r grr

f r r r g

(20)

(21)

2

22

d0

d

gg

Then,

and we obtain

sin cosg A B

(22)

As the potential function should satisfy..., this means ＝ n (integer)

2

sin cosg A n B n

Therefore,

(23)

2

2 22

d d0

d d

f r f rr r n f r

r r

(24)

(25)

The left side of Eq. (19) can be rewritten as

i.e.,

2dd0

d d

f rr r n f r

r r

（ Euler equation ）

(26)

The solution to Eq. (25) is

0 0

0

ln 0

n nf r Cr Dr n

f r C D r n

(27)

The general solution of in the two-dimensional cylindrical coordinate is

1

sin cos sin cosn nn n n n

n

r A n B n r C n D n

Example: an infinitely long dielectric cylinder with radius a and permittivity is placed in an external uniform electrical field E0

vertical to cylinder. If the field is directed along x-axis and the axis of cylinder is z-axis, determine the potential functions inside and outside the cylinder.

a

x0E

, ,r z

0

o

If the potentials inside and outside the cylinder are 1 and 2, respectively, the equations and boundary conditions are given as follows

In cylindrical coordinate, x=r cos, The corresponding potential of E0 is:

0 0 cd osx

x x Ex r

e e (28)

1

1

sin cos

sin cos

nn n

nn n n

r A n B n

r C n D n

So,

(30)

11 1 cos cos

DB r

r

and1 0B E

From condition 1 0 cosr rE

(31)

(32)

(29)有限

21

22

2

1 0

1 2

1 20

0

0

0

cos

r a r a

r a

r a

r

r rE

a a

r r

Finite

2

1

' sin 'cos

'sin 'cos

nn n

nn n n

r A n B n

r C n D n

(33)

Since 1 2a a , we have

1

1

1 0

0

1

' sin 'cos

cos

' cos cos c

o

o

c

s

s

nn n

n

DB a E a

a A n B n

DE a

a

a

(35)

21

' sin 'cosnn n

n

r A n B n

(34)

2 0r Because is finite

11 0 0 2' cos cos cos

DB E

a

1 20 r a r ar r

Because

Thus,

20 01 0 1

0 0

2, ' cosD a E B

(36)

(37)

The potentials are given by

201 0 0

0

1cos cosE r a E

r

02 0

0

2cosE r

(38)

(39)

3 Separation of variables in spherical coordinate system

The Laplace equation in spherical coordinate can be written as

(40)

22

2 2 2 2 2

1 1 1sin 0

sin sinr

r r r r r

The potential function can be expressed as:

(41) ,r f r g Without considering

Separate variables (in the similar fashion), we can get the general solution to equation (14)

(42) 1

0

, P cosmmm m m

m

r f r g A r B r

2

0

12 2

0 1

1 dP 1

2 ! d

P cos P cos sin d 0

2P cos sin d P d

2 1

mm

m m m

m n

m m

x xm x

x xm

Pm x Legendre polynomial:◇

where

Example: a spherical dielectric with radius a and permittivity is placed in a uniform electrical field E0. If the field is along z-axis, determine the potentials inside and outside the sphere.

a

z

0E , ,r

0

o

0 0 cd osz

z zE z E r

e e

In spherical coordinate system, z=r cos . The corresponding potential of E0 can be expressed as

Assume the potentials inside and outside the sphere are 1 and 2, respectively. The equations and boundary conditions are as follows

Because 1 0 cosr rE

21 1 1 cosA r B r

and 1 0A E

The general solution is

11

0

, P cosmmm m m

m

r A r B r

有限

21

22

2

1 0

1 2

1 20

0

0

0

cos

r a r a

r a

r a

r

r rE

a a

r r

finite

12

0

, ' ' P cosmmm m m

m

r A r B r

is finite 2 0r Because

20

, ' P cosmm m

m

r A r

2

20

20 1

2

, ' P c

'c

os

c

o

o

s

s

mm m

m

r A r

E a

A

B a

Since 1 2a a

4 Method of Image ( 镜像法不作要求 )

According to the uniqueness theorem, in some appropriate position outside of the field under consideration, we can utilize some dummy ( 虚拟的 ) charges (image charge) to replace the induced charges of the conductor interface or the polarization charges on the media interface. Transform the original problem into the equivalent problem with the same boundary conditions.

The principles to select image charges:

1. Image charge should be outside of the region to be investigated (The image charge must be external to the volume of interest)

2. The boundary conditions should not be changed after introducing the image charges.3. Image method can only be suitable for those special boundaries (such as plane, circular and spherical boundaries).

● 镜像法基本思路：在所研究的场域外的某些适当位置，用一些虚拟电荷等效替代导体分界面上的感应电荷或媒质分界面上的极化电荷的影响。

● 也即：在研究区域之外，用一些假想的电荷分布代替场问题的边界。——将原问题转为与原问题边界条件相同的等效问题。

● 镜像法求解电位问题的理论依据：唯一性定理。待求区域的电位由其电荷分布与边界条件共同决定。

● 等效电荷一般位于原电荷关于边界面的镜像点处，故称为镜像电荷。大多是一些点电荷或线电荷。

● 镜像电荷位置选择原则：1 、镜像电荷必须位于求解区域以外的空间。

2 、镜像电荷的引入不能改变原问题的边界条件。

x

z

hq

Conductor

1. plane image

Infinite grounding conductor plane (z=0), where a point charge is placed at z=h. Determine the potential distribution of the upper half space.

q

h

z

0

h

q

'R

R

, ,P x y z

x

Because the original charge is placed at z=h, its mirror image should be at z=-h with the quantity being -q. Thus the potential at boundary z=0 keeps zero, and the conductor plane at z=0 can be removed in the equivalent system.

1 1( , , ) ( )

4 '

qx y z

R R

2 2 2 2 2 2

1 1( )( 0)

4 ( ) ( )

qz

x y z h x y z h

So the potential in z>0 is written as

(43)

0s n n

z

D En z

2 2 2 3/ 22 ( )

qh

x y h

The total induced charges in infinite conductor is

(44)

2 2 2 3/ 22 ( )in ss

qhq ds dxdy q

x y h

(45)

The quantity of original and induced charges are equal to each other.

2. Spherical face image

A grounding conductor sphere has the radius a. A point charge is placed at P1 point. The distance between charge and spherical center O is d1. Determine the potential distribution outside the sphere.

2S

2d

1d

1R

1S

2Ra

o2P

P

r

2q 1P

1q

Because the potential outside the sphere is decided by point charge and induced charge on the conductor surface, the latter of which can be replaced by the image charge q2 if the potential keep zero on spherical surface.

1 2

0 1 2

1

4

q q

R R

So, the potential outside the sphere can be written as

(46)

1 22

0 1 2

1 21

0 1 2

1: 0

4

1: 0

4

q qS

a d a d

q qS

d a a d

In order to determine the position and quantity of the image charge, we can consider two special points, say, S1 and S2, inside the sphere, where the potentials are zero, namely,

(47)

The solutions to the above two equations are given by

(48)2 1 2 1

1

2

2 2 11

(

aq q q q

d

ad d d

d

或 舍去）or

given up

(49)

(50)

1 2 1

0 1 2 0 1 1 2

1 1

4 4

q q q a

R R R d R

Thus, the potential outside the sphere can be rewritten as

12 2 2

1 1 1

12 2 2

2 2 2

2 cos

2 cos

R r d rd

R r d rd

where

( 以下内容根据课时情况决定增删 )

Laplace Equations in Spherical and Cylindrical Coordinate Systems

球坐标和柱坐标系中的拉普拉斯方程

2

0

+ 0k

2

2 2

2

2

电磁学（电动力学）中有三个重要方程：

Lapl ace方程(无源)： ＝

Poi sson方程（有源）： ＝－ ，如对于点电荷， ＝-e r

Hel mhol tz方程(无源+波动)： ＝

它们都带有Lapl ace算符 。这些方程与它们的解具有典型性（普遍适用性），在电动力学、量子力学、流体力学、声学、天体物理中均会出现。

Laplace 方程的解（勒让德函数、贝塞尔函数等）称为“特殊函数”。这些解都是确定的、固定的，已经在 100多年前被英美数学家求出，主要用了幂级数展开法（ power series expansion ） . 求解过程是很复杂的，但今天的我们的任务就是利用这些解来表示任何一个边值问题的解，因为任何一个边值问题的解均可以用这些特殊函数来展开。我们的目的就是利用边界条件来定出这些展开系数。

下面的内容来自 Jackson 经典电动力学“ Classical Electrodynamics”(John Wiley & Sons, 1998, pp.95-119) ( 可见复印的资料 ).

建议自学。

33

Boundary-Value Problems

1 Laplace’s Equation in Spherical Coordinates

2 0V

In spherical coordinates ( , , )r 1 1 1

02

2 2 2 2

2

2r rrV

r

V

r

V

( )sin

(sin )sin

We can solve this eq. by separation of variables, let VU r

rP Q

( )( ) ( )

2 2

2 2 2 2sin 0

sin sin2

d U UQ d dP UP d QPQ + + =

dr r d d r d

34

r UPQ2 2sin /Multiplying 2 2

2 22

1 1 1sin sin

sin2 2

d U d dP d Qr θ + ( θ ) =

U dr Pr θ dθ dθ Q d

1 2

22

Q

d Q

dm

We have

0sin

)(sinsin

112

2

2

22

m

d

dP

d

d

Pdr

Ud

Ur

Similarly, we separate andP( ) U r( )

imeQ Clearly the solution is

In order that Q be single valued , m must be an integer 2

35

2

2

2

2

1sin 1 0

sin sin

10

2

d dP mθ + l(l + ) P =

θ dθ dθ θ

d U l(l + )U =

dr r

Introducing another real constant )1( ll

( , , )l 0 1 2

U Ar Brl l 1For r we have,

mlP

For m=0, - ----- Legndre Polynomials Pl (cos )

Let x=cos 01

)1()1(2

22

x

mll

dx

dPx

dx

d

Its solutions are the associated Legendre functions

0)1()1( 2

ll

dx

dPx

dx

d

36

P x P x dxll l ll' '( ) ( )

2

2 11

1

P xl

d

dxxl l

l

ll( )

![ ( ) ]

1

212

)13(2

1)(,)(1)(,1)1( 2

210 xxPxxPxPPl

Orthogonal:

ml

imml eP

r

rUrV

,

)(),,(

So the general solutions of Lapace eq. in Spherical Coordinates

llm

llm rBrArU 1)(

(Rodrigues formula)

37

2 Boundary-value problems with azimuthal Symmetry

V r A r B r Pl

ll

ll

l( , ) [ ] (cos )( )

0

1

A Bl l,

A problem possessing azimuthal symmetry m=0, then

As usual, the coefficients can be determined by the BCs.Suppose Vis the potential on S of sphere of radius a. We want to know the potential V inside the sphere.If no charge at the origin,

V r( )0 -- finite, leading to 0lB for all l.

0

)(cos)(l

ll

l PaAV

From the relation'

1

1

12

2cos)(cos)(cos llll l

dPP

we have Al

aV P dl l l

2 1

2 0( ) (cos ) sin

38

If, for example, in hemisphere,

VV

V( )

/

/

0 2

2

V r Vr

aP

r

aP( , ) [ (cos ) ( ) (cos ) ]

3

2

7

813

3

To find V outside the sphere

( / ) ( / )r a a rl l 1

)0,,( ll Arr

The series is a unique expansion with BCs. So from a knowledge of V in a limited domain, namely on the symmetry axis, we may derive the solution

e.g. z=r V z r A r B rll

ll( ) [ ]( ) 1

Valid for positive z.

If this potential function can be expanded in a power series in z=r , with known coefficient, the solution is derived by multiplying by )(coslP

39

Let us see hemisphere again. We have obtained

V z r Vr a

r r a( ) [ ]

12 2

2 2

This can be expanded in powers of a r2 2/

V z rV j j

ja rj

j

j( ) ( )( / ) ( / )

!( / )

1

2 1 2 1 21

1

2

By comparison withV z r A r B rl

ll

l( ) [ ]( ) 1

only l j 1 2 , i.e., terms with odd l are non-zero.

V rV j j

j

a

rPj

j

jj( , ) ( )

( / ) ( / )

!( ) (cos )

1

2 1 2 1 21

1

22 1

This is the same as the previous result.

40

An important expansion is that the potential at due to a unit point charge at

r

11

0| ' |(cos ) r r

r

rP

l

l ll

)( rrWhere is the smaller (larger) of | '||, | rr

,

-- the angle between

.

| '||, | rr

This can be proved by rotating axis so that r '

lies alone the z axis. Then the potential satisfies Lapace’s eq., possesses azimuthal symmetry, and can be expanded to

except at point r r '

)(cos]['

1 )1(

0

ll

ll

ll

PrBrArr

41

rIf is on the z axis, the RHS reduces to

[ ]( )

ll

ll

lA r B r

0

1

While the LHS becomes

rrrrrrrrrr

1

| '|

1

)cos 2 (

1

| |

12/122

1 1

1

10r r

rr

r

rl

l( )( )

For points off the axis , it is only necessary to multiply each term in the above eq. by

Pl (cos )

1

)1(l

ll

ll

l r

rrBrA

The general result is proved.

11

0| ' |(cos ) r r

r

rP

l

l ll

42

Another example is the potential due to a total charge q uniformlydistributed around a circular ring of radius a, located as shown in theFigure, with its axis the z axis and its center at z=b.

V z rq

r c cr( )

cos

2 2 2

The inverse distance AP, 1/AP can be expanded by using

)()(cos)(0

1crP

r

cqrzV

lll

l

)()(cos)(0

1crP

c

rqrzV

lll

l

The potential at a point P on the axis

11

0| ' |(cos ) r r

r

rP

l

l ll

43

The potential at any point in space is then derived by multiplyingeach member of the series by the Legendre polynomials

V r qr

rP P

l

l l ll

( , ) (cos ) (cos )

1

0

Where is the smaller (larger) of r and c)( rr

44

01

)1()1(2

22

x

mll

dx

dPx

dx

d

3 Associated Legendre Polynomials and the Spherical Harmonics

0)1()1( 2

ll

dx

dPx

dx

d

)(xPlLegendre’s polynomials

So far, we have considered the azimathal symmetry m=0, these involve only ordinary Ledgendze polynomials.

ml

imml eP

r

rUrV

,

)(),,(

45

The general potential problem can, however, have variations so that inm 0 Q e im

In this case, we need the generalization of mll PP )(cos

It can be shown that in order to have finite solution on the interval 1 1x

m l l l l , ( ) ( ),1 1

)()1()1()( 2/2 xPdx

dxxP lm

mmmm

l

lml

mlm

l

mm

l xdx

dx

lxP )1()1(

!2

)1()( 22/2

the parameter l must be zero or a positive integer and the integer mcan take on only

(+|m|,-|m|; Rodrigues’formula)

positive m

46

Since differential eq. for depends only on and m is an integer, Plm

m2

Pl

m and mlP are proportional

P xl m

l mP xl

m mlm

( ) ( )( )!

( )!( )1

For fixed m, Plm

form an orthogonal set in the index l on the interval

1 1x

1

1

'' )!(

)!(

12

2)()( ll

ml

ml ml

ml

ldxxPxP

The sol. of Laplace eq. was decomposed into a product of factors for

r, ,

Q emim( ) form a complete set of orthogonal in the index m

on the interval 0 2

47

Plm

form a similar set in the index l for each m for 1 1cosTherefore,

)()(cos mm

l QP will form a complete orthogonal set on the surface of the unit sphere in the two indices l, m.

From the normalization condition, it is clear that suitable normalized function, denoted by Ylm ( , ) called spherical harmonics

Yl l m

l mP elm l

m im

2 1

4 ( )!

( )!(cos )

It can be seen

*, )1()(

)!(

)!()1(

)!(

)!(

4

12lm

mimml

mml YexP

ml

ml

ml

mllY

g A Ym l

m l

lm lml

( , ) ( , )

0

48

The normalization and orthogonality conditions are

d d Y Yl m lm l l m m sin ( , ) ( , )' '

*' '

00

2

The complete relation is

)'cos(cos)'(),()','(*

0

lmlm

l

lm

lm

YY

l Y 01

400,

lY e

Y

i

1

3

83

4

11

10

sin

cos

49

Note that, for m=0,

)(cos4

12),(0

ll P

lY

Since the spherical harmonics form a complete orthogonal set of function on the surface of the unit sphere, then an arbitrary function can be expanded in spherical harmonics

g( , )

g A Ym l

m l

lm lml

( , ) ( , )

0

ddgYA lmlm sin),(),(*

Where the coefficients

which recovers the ordinary Legendre functions.

50

A point of interest to us is the form for 0

[ ( , )]gl

All

0 0

0

2 1

4with

Al

d P gl l0

2 1

4

(cos ) ( , )

The general sol. for a Boundary-value problem in m 0

spherical coordinates can be written

V r A r B r Ylml

lml

lmm l

m l

l( , , ) [ ] ( , )( )

1

0

If V is specified on a spherical surface, the coefficient can be determined by this BC.

51

6 Laplace’s Equation in Cylindrical Coordinates; Bessel Function

In cylindrical coordinates ( , , ) z

2 0V

2

2 2

2

2

2

2

1 10

V V V V

z

The separation of variables V z R Q Z z( , , ) ( ) ( ) ( )

d Z

dzk Z

2

22 0

d Q

dv Q

2

22 0

0)(1

2

22

2

2

Rv

kd

dR

d

Rd

We get three ordinary differential equations:

52

The solutions of the first two eqs. are elementary

Z z e kz( ) Q e iv( )

For the potential o be single valued, v--integer. But except (barring) some boundary-condition requirement in the z direction, the parameter k is arbitrary. For the present we will assume k is real. The radial eq. can be presented in a standard form by the change of variable x=k

d R

dx x

dR

dx

v

xR

2

2

2

2

11 0 ( )

This is the Bessel’s Eq., and the solutions are called Bessel functions of order

bygiven solution seriespower a isit ,)(xJ v

jj

j

vv x

vjjxxJ 2

0)2/(

)1(!

)1()2/()(

jj

j

vv x

vjjxxJ 2

0

)2/()1(!

)1()2/()(

53

N xJ x v J x

vvv v( )( ) cos ( )

sin

solution.t indenendenlinearly

another find toneed We).()1()( have we(integer), If

function)(Neumann kind second theoffunction Bessel (2)

solutions. independ

linearly ofpair a form )(),( integer,an not is If x.of valuesfinite allfor

converge series The .order first theof functions Bessel called are solutions These

kindfirst theof functions Bessel (1)

)( xJxJm

xJxJ

mm

m

vv

integeran not is ifeven t independenlinearly are )( and )(

kind). second theoffunction (Besselfunction Neumann is This

xJxN vv

eq. Bessel the tosolutions ofset lfundamenta a form functions Hankel The

)()()(

)()()(

:as defined are functions, Hankel called kind, third theoffunction Bessel The )3(

2

1

xiNxJxH

xiNxJxH

vvv

vvv

54

Fourier-Bessel series

Bessel function has an infinite number of roots,

3,2,1,0)( nxJ vnv

. ofroot nth theis (x)Jx vn

:integralion normalizat The

.0 inteval on theset orthogonalan form ,....,2,1,0 fixedfor

),/( Functions kind.first theof functions Besselonly consider We

anv

axJ vnv

nnvnvvnvvnv

a

xJa

axJ

axJd

21

2

0

)]([2

)()(

f A J xavn v vn

n( ) ( )

1

Aa J x

f J xa

dvnv vn

v vn

a

2

21

20( )

( ) ( )

:series Bessel-Fourier ain 0 interval on the )(function

arbitraryan expandcan wecomplete, is functions Bessel ofset that theAssuming

af

Where the coefficients:

55

Modified Bessel functions

)(2

)(),()( )1(1 ixHixKixJixI vv

vvv

v

011

becomes )R(for eq. the,k-by replaced is kconstant sparation theif :Note

2

2

2

2

22

R

x

v

dx

dR

xdx

Rd

The solutions of this equation are called modified Bessel functions; They arejust Bessel functions of pure imaginary argument. The usual choices of linearlyindependent solutions are denoted by

56

7 Boundary Value Problems in Cylindrical Coordinates

Consider the specific Boundary - value problem in Fig.

0)()0(),,()( aVzVVLzV

The cylinder has a radius a and a height L, the top and bottom surfaces being at z=L and z=0. The potential on the side and the bottom of the cylinder is zero, while the top has a potential V=VWe want to find the potential at any point inside the cylinder.

V z R Q Z z( , , ) ( ) ( ) ( )

In order that V be single valued and vanish at z=0

Q A m B m( ) sin( ) cos( )

kzececzZ kzkz sinh}{)( 21

where v=m is an integer and k is a constant to be determined.

57

The radial factor is

)()()( kDNkCJR mm

If V is finite at D should be equal to 0.

0)/2()(

05772.0)2/ln(2

)(,1vx

v

vxxNx

vv

.0)( of roots theare where

,........2,1 ,0)(

: valuesspecial only thoseon can takek

thatmeans as vanish potential t that therequiremen The

mnmmn

mnmnmnm

xJx

na

xkakJ

a

58

Combining all these conditions, we find that the general form of the solution is

0 1

)]cos()sin()[sinh()(),,(m

mnmnmnmnmn

mBmAzkkJzV

At z=L, we are given the potential as V There we have

0 1

)]cos()sin()[sinh()(),(m

mnmnmnmnmn

mBmALkkJV

a),(0 )( and

)2(0 )cos(,)(sin of properties orthogonal theFrom

.in Fourier -Bessel a and in seriesFourier a is This

mnm kJ

mm

nnmnmmnmmnm

a

mm

akJa

dkJkJ

dmm

'2

1

2

'

0

2

0

'

)(2

)()(

)'sin()sin(

59

2

0 02

12

sin)(),()(

)(cos2 a

mnmmnm

mnmn mkJVdd

akJa

LkechA

2

0 02

'2

cos)(),()(

)(cos2 a

mnmmnmn

mnmn mkJVdd

akJa

LkechB

integral.an intoover goes series the,a If a.0 ,in interval finite afor

eappropriat isresult The series. in the 2/ use should we0,mFor 0

nB

be shouldsolution general Then the ).,V( be to0z plane whole theand

0)V(z when 0zfor potential theknow want to weexample,For

V z dke J k A k m B k mkzm m m

m( , , ) ( )[ ( ) sin( ) ( ) cos( )]

00

V dkJ k A m B mm m mm

( , ) ( )[ sin( ) cos( )]

00

The coefficients are determined by

60

2

0 0

')'(

)'()'(

cos

sin),(

1dk

kB

kAkJd

m

mV

m

mm

The variation in is just a Fourier series. Consequently the coefficients are separately specified by the integral relations:

Since)'(

1)'()(

0

kkk

dxxkJkxJx mm

A

Bk

d d V J km

mm

mm

0

2

0( , ) ( )

sin

cos

series. in the 2/)( usemust we0,mfor usual, As 0 kB

:0plane theof area wholeover the integralsby determined are tscoefficien

that thefind we,over gintegratin and )(by sidesboth gMultiplyin

z

kJm