Chapter 4.
Problems of Boundary Values
in Static Electric Fields
1) 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 variables
Separation 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
电磁学（电动力学）中有三个重要方程：
Lapl ace方程( 无源) ： 2＝0
Poi sson方程（有源）： 2＝－，如对于点电荷，2＝- e  r 
Hel mhol t z方程( 无源+波动) ： 2 +k 2＝0
它们都带有Lapl ace算符2。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.
y
b
 0
  U0
 ?
 0
According to the problem,
the potential function and
its boundary conditions are
given as follows:
 0
a
x
2  0


 x 0  0
 x a

 y 0  0, 
 0  x  a,0  y  b 
y b
 U0
In rectangular coordinate, 2  0 can be expressed as:
 2  2

0
2
2
x
y
(1)
Separate the variables as follows:
  x, y   f  x  g  y 
(2)
Substituting equation (2) into (1), one can
arrive at
d2 f  x 
d2 g  y 
g  y
 f  x
0
2
2
dx
dy
(3)
The two sides of the above equation is
divided by g  y  f  x 
2
2
1 d f  x
1 d g  y

0
f  x  dx 2
g  y  dy 2
Only depends on variable
x
(4)
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.,
 1 d2 f  x 
2

k

x
2
 f  x  dx

2
 1 d g  y  k2
y
 g  y  dy 2

(5)
Where
k x2  k y2  0
(6)
Thus the solution of the equation (5) is:
f  x   A1 sin  k x x   A2 cos  k x x 
(7)
If k x is a real number
and k y an imaginary
number
Or
f  x   B1 sinh  x x   B2 cosh  x x 
f  x   B1' e x x  B2' e  x x
(8)
If k x is an imaginary number
f  x   C1 x  C
If k x=0
(9)
Analyzing the boundary conditions, we can obtain
f  x   A1 sin  k x x   A2 cos  k x x 
(7)
Since =0 when x=a and 0
Substituting the boundary conditions into Eq. (7), we can get
 n 
f  x   A1 sin 
x
 a 
where k x 
n
a
(8)
is called the eigenvalue of the present boundary problem:
2  0


 x 0  0
 x a

 y 0  0, 
 0  x  a,0  y  b 
y b
 U0
Thus, g(y) is
g  y   B1 sinh  y y   B2 cosh  y y 
(9)
k x2  k y2  0
Note:
 k x2   j y   0
2
  y  kx
(10)
Hence,
 n 
g  y   B1 sinh 
y
 a 
The general solution is given by

 n 
 n 
x  sinh 
y
 a 
 a 
  x, y    Cn sin 
n 1
(11)
Employing the theory of Fourier series expansion
4U 0
 n 
 n  2
x  dx 
Cn sinh  b    U 0 sin 
 2n  1 
 a 
 a  a0
4U 0
 Cn 
 n 
sinh  b   2n  1 
 a 
a
(12)
So the final solution is:
4U 0
 n 
 n 
sin  x  sinh 
y
n

a 


 a 
n 1
sinh  b   2n  1  
 a 

  x, y   
(13)
2 Separation of variables in system of cylindrical coordinate
The Laplace equation in cylindrical coordinate system can be written as:
1     1  2  2
r
 2 2  2 0
r r  r  r 
z
(14)
If the field is two-dimensional field and  is independent of z,
the Laplace equation can be rewritten (simplified) as
1     1  2
r
 2 2 0
r r  r  r 
Assume the solution (r, ) can take the
following form
  r,   f  r  g  
(15)
(16)
Substitution of (16) into (15) yields
g     f  r   f  r   2 g  
0
r
 2
r r  r 
r
 2
(17)
r2
Simplify Eq. (17) by multiplying
f  r  g  
2
  f  r  
1  g  
0
r

f  r  r  r  g    2
r
Function of r
(18)
Function of 
So,
2
  f  r  
1  g  
2


r

f  r  r  r 
g    2
r
(19)
Then,
d 2 g  
2


g    0
d 2
and we
obtain
g    Asin    B cos  
(20)
(21)
As the potential function should satisfy..., this means ＝n (integer)
    2      
(22)
Therefore,
g    Asin  n   B cos  n 
(23)
The left side of Eq. (19) can be rewritten as
r
i.e.,
d  df  r   2
r
  n f r  0
dr  dr 
d2 f  r 
df  r 
2
r

r

n
f r  0
2
dr
dr
(24)
2
(25)
（Euler equation）
The solution to
Eq. (25) is
f  r   Cr n  Dr  n
f  r   C0  D0 ln r
 n  0
 n  0
(26)
The general solution of  in the two-dimensional cylindrical
coordinate
is

  r n  An sin  n   Bn cos  n    r  n Cn sin  n   Dn cos  n  
n 1
(27)
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.
 r,  , z 
0
E0

o
x
a
In cylindrical coordinate, x=r cos, The corresponding potential of E0 is:
x
   ex  0  ex dx   E0 r cos 
(28)

If the potentials inside and outside the cylinder are 1 and 2, respectively,
the equations and boundary conditions are given as follows
21  0  r  a 

2 2  0  r  a 

Finite
 2  r  0  有限

1  r      rE0 cos 

1  a    2  a 

 2
 1


0
r a
r a

r
 r
So,
r n  An sin  n   Bn cos  n  
1     n

Cn sin  n   Dn cos  n  
n 1   r

(29)

(30)
From condition 1  r    rE0 cos
1  B1r cos  
and
B1   E0
D1
cos 
r
(31)
(32)
n

r
  An 'sin  n   Bn 'cos  n   
2    n

C
'sin
n


D
'cos
n



n 1   r




n
 n




(33)
Because 2  r  0 is finite

 2  r n  An 'sin  n   Bn ' cos  n  
(34)
n 1
Since
1  a   2  a  , we have

a
n 1
n

 An 'sin  n   Bn ' cos  n  
  E0a cos  
D1
cos 
a
 B1 ' a cos    E0a cos  
D1
cos 
a
(35)
Because  0
1
r
r a  
 2
r
r a
D1
cos 
2
a
(36)
2 0
cos 
  0
(37)
  0 2 1
a E0 cos 
  0
r
(38)
 B1 'cos    0 E0 cos  
Thus,
D1 
  0 2
a E0
  0
, B1 ' 
The potentials are
given by
1   E0 r cos  
2  
2 0
E0 r cos 
  0
(39)
3 Separation of variables in spherical coordinate system
The Laplace equation in spherical coordinate can be written as
1   2  
1
 
 
1
 2
0
r
 2
 sin 
 2 2
2
2
r r  r  r sin   
  r sin  
(40)
The potential function can be expressed as:
Without
considering 
  r,   f  r  g  
(41)
Separate variables (in the similar fashion), we can get the general
solution to equation (14)



  r,   f  r  g     Am r m  Bm r  m 1 Pm  cos 
m 0
where
◇ Pm  x 
Legendre polynomial:
m
1 dm 2
Pm  x   m
x

1
 
2 m ! dx m

 P  cos P  cos  sin  d  0
m
n
0

1
2
P
cos

sin

d


P
x
d
x










0  m
1  m 
2m  1
2
2
(42)
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.
z
 r, ,  
E0
0

o
a
In spherical coordinate system, z=r cos . The corresponding potential
of E0 can be expressed as
z
   ez E0  ez dz   E0r cos

Assume the potentials inside and outside the sphere are 1 and 2,
respectively. The equations and boundary conditions are as follows
21  0  r  a 

2 2  0  r  a 

 2  r  0  有限
finite

1  r      rE0 cos 

1  a    2  a 

 2
 1


r a
 0 r r a
r
The general solution is



1  r,    Am r m  Bm r  m 1 Pm  cos 
m 0
Because 1  r    rE0 cos
1   A1r  B1r 2  cos
and
A1   E0



 2  r,    Am ' r m  Bm ' r  m1 Pm  cos  
m 0
Because 2  r  0 is finite

 2  r,    Am ' r m Pm  cos  
m 0
Since
1  a   2  a 

 2  r,    Am ' r m Pm  cos 
m 0
   E0a  B1a 2  cos 
  2  A2 ' cos
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。
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.
P  x, y , z 
z
z
q
h
 q

h
x
R
R'
 0
x
Conductor
h

q
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.
So the potential in z>0 is written as
 ( x, y , z ) 

q
4
(
q
1 1
 )
4 R R '
1
x  y  ( z  h)
2
2
2

(
1
x  y  ( z  h)
2
2
2
)( z  0)
(43)
The total induced charges in infinite conductor is
s  Dn   En  
The quantity of
original and
induced charges are
equal to each other.
z 0
qh
2 ( x 2  y 2  h 2 )3/ 2

 qin    s ds  
s


 
n
z





(44)
qh
dxdy  q
2
2
2 3/ 2
2 ( x  y  h )
(45)
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.
P
r

S2
a
o
P2
q2
R2

R1
S1
P
1
q1
d2
d1
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.
So, the potential outside the sphere can be written as

1  q1 q2 



4 0  R1 R2 
(46)
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,

1  q1
q2 
S
:

 2

0
4 0  a  d1 a  d 2 


 S : 1  q1  q2   0
 1 4  d  a a  d 
0  1
2 

(47)
The solutions to the above two equations are given by
a

q


q1  q2   q1
 2
d1


given
up
或
(舍去）
or


2
d  a
d  d
2
2
1


d1
(48)
Thus, the potential outside the sphere can be rewritten
as
q1  1
1  q1 q2 
a 

  
 

4 0  R1 R2  4 0  R1 d1R2 
(49)
where
R1   r  d  2rd1 cos  
2
2
1
1
2
R2   r  d  2rd 2 cos  
2
2
2
1
2
(50)
(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
电磁学（电动力学）中有三个重要方程：
Lapl ace方程( 无源) ： 2＝0
Poi sson方程（有源）： 2＝－，如对于点电荷，2＝- e  r 
Hel mhol t z方程( 无源+波动) ： 2 +k 2＝0
它们都带有Lapl ace算符2。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。
Boundary-Value Problems
1 Laplace’s Equation in Spherical Coordinates
V0
, )
In spherical coordinates (r,
2
 






 


2
2
1
1
V
1
V
(
r
V
)

(
s
i
n
)


0
2
2
2
22
r
r r
s
i
n
r
s
i
n
We can solve this eq. by separation of variables, let
U
(
r
)
V
 PQ
(

)()

r
d 2U
UQ d 
dP 
UP d 2Q
PQ 2 + 2
=0
 sin
+ 2 2
2
dr
r sin d 
d  r sin  d
33
Multiplying
2 2
r
s
i
n

/U
P
Q
 1 d 2U
1
d
dP 
1 d 2Q
r sin θ 
+ 2
( sinθ
) = 
2
2
U
dr
Pr
sin
θ
dθ
dθ
Q
d



2
We have
2
2
1dQ
2

m
2
Qd

Clearly the solution is
Qeim
In order that Q be single valued
, m must be an integer

2

Similarly, we separate P( ) and U (r )



2
2
1
d
U
1
d
dP
m
2
r

(sin
)


0
2
2
U
dr
P
sin
d dsin
34
Introducing another real constant l (l  1) (
l
0
,,
1
2

)
 1 d 
dP  
m2 
P= 0

 sinθ
 + l(l +1 ) 
dθ  
sin 2θ 
 sinθ dθ 
 2
 d U l(l +1 )

U =0
2
2

r
 dr
For r we have,
Let x=cos
l

1

l
U

A
r
B
r
2


d
m
 2dP

(
1

x
)

l
(
l

1
)


0


2


dx
dx
1

x




Its solutions are the associated Legendre functions
For m=0, -
P
cos)
l(
Pl
m
----- Legndre Polynomials
d
 2dP

(
1

x
)

l
(
l
1
)

0


dx
dx


35
l
1d
2 l
P
(
x
)

[
(
x

1
)
]
l
l
l
2
l
!
d
x
1
Orthogonal:

2
l
l
'
2
l

1
P
x
)
P
(
x
)
d
x

l
'(
l

1
(Rodrigues formula)
1
2
P
(
1
)

1
,
P
(
x
)

1
P
(
x
)

x
,
P
(
x
)

(
3
x

1
)
l
0
1
2
2
So the general solutions of Lapace eq. in Spherical Coordinates
U
(
r
)m
im

V
(
r
,

,

)

P
e

l
r
l
,
m
l
1

l
U
(
r
)
A
r

B
r
lm
lm
36
2 Boundary-value problems with azimuthal Symmetry


A problem possessing azimuthal symmetry m=0, then

l

(
l

1
)

l
l
l
l

0
V
(
r
,
)[

A
r

B
r]
P
(
c
o
s
)
As usual, the coefficients Al , Bl
can be determined by the BCs.
Suppose Vis 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,
Vr
( 0)-- finite, leading to B  0
l
for all l.

l
V
(

)

A
a
P

)

l
l(cos
l

0
1
From the relation
  
2
P
(cos
)
P
(cos
)
d
cos
 ll
l
l
'

2
l

1

1
we have
 


2
l

1
A

V
(
)
P
(
c
o
s
)
s
i
n
d
l
l
l
2
a
0
37
If, for example, in hemisphere,

 
V0


/
2

V
()



V /
2







3
r
7
r
3
V
(
r
,
)

V
[(
P
c
o
s
)

(
)
P
(
c
o
s
)


]
1
3
2
a 8
a
To find V outside the sphere
l

1
(
r
/)
a
(
a
/)
r
l
l
(

r


,r


,A
0
)
l
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
l

()
l

1
V
(
z

r
)

[
A
r

B
r
] Valid for positive z.

l
l
If this potential function can be expanded in a power series in z=r , with
known coefficient, the solution is derived by multiplying by Pl (cos )
38
Let us see hemisphere again. We have obtained
2 2
r

a
V
(
zr

)

V
[
1 2 2
]
rr

a
This can be expanded in powers of a2 / r2

 (/
V
2
j

1
2
)

(
j

1
/
2
)2
j

1
j
V
(
z

r
)

(

1
)
(
a
/
r
)

j

1
j
!
By comparison with
only
l12j
l

()
l

1
V
(
z

r
)

[
A
r

B
r
]

l
l
, i.e., terms with odd l are non-zero.




V
(
2
j

1
/
2
)

(
j

1
/
2
)
a
j

1
2
j

V
(
r
,
)

(

1
)
(
)(
P
c
o
s
)

2
j

1
j

1
j
! r
This is the same as the previous result.
39
An important expansion is that the potential at
due to a unit point charge at

r
l
1 r

P
(
c
o
s

)


l

1l
0
|rr

'| l
r

Where r

(r ) is the smaller (larger) of



| r |, | r ' |

-- the angle between | r |, | r' |

,
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

.
1  l

(
l

1
)

[
A
r

B
r
]
P
(cos
)


l
l
l
r

r
' l

0
 
except at point r  r '
40
If r is on the z axis, the RHS reduces to


(
l

1
)
[
A
r
B
r

l
l
l
l

0
]
While the LHS becomes

1
1
1 1




22
1
/
2
|
r

r
|(
r

r

2
rr
cos
) |
r

r
'
|r

r


1 1
1r

 
()l
r
r
rl0r
(
1
)  
r

For points off the axis , it is only necessary to multiply each term in the above eq. by
P
cos) The general result is proved.
l(
l
1 r


P
(
c
o
s

)

 l
l

1l
|rr

'| 0
r

l


r
l

(
l

1
)




A
r

B
r

l
l
l

1


r


41
Another example is the potential due to a total charge q uniformly
distributed around a circular ring of radius a, located as shown in the
Figure, with its axis the z axis and its center at z=b.
The potential at a point P on the axis
q

V
(
z

r
)

2 2
rc


2
c
r
c
o
s
The inverse distance AP, 1/AP can be expanded by using
l
1 r

P
(
c
o
s

)


l

1l
0
|rr

'| l
r

l


c
V
(
z

r
)

q
P
(cos
)(
r

c
)

l
l

1
r
l

0
l


r
V
(
z

r
)

q
P
(cos
)(
r

c
)

l
l

1
c
l

0
42
The potential at any point in space is then derived by multiplying
each member of the series by the Legendre polynomials

l



l
l

1
l

0


r
V
(
r
,)

q P
(
c
o
s
)
P
(
c
o
s
)
l
r
Where r (r )
is the smaller (larger) of r and c
43
3 Associated Legendre Polynomials and the Spherical Harmonics
U
(
r
)m
im

V
(
r
,

,

)

P
e

l
r
l
,
m
2


d
m
 2dP

(
1

x
)

l
(
l

1
)
2

0



dx
dx
1

x




So far, we have considered the azimathal symmetry
m=0, these involve only ordinary Ledgendze
polynomials.
d
 2dP

(
1

x
)

l
(
l
1
)

0


dx
dx


Pl (x)
Legendre’s polynomials
44
The general potential problem can, however, have variations so that
im

m 0 in Qe
In this case, we need the generalization of
P
)
P
l(cos
m
l
It can be shown that in order to have finite solution on the interval 
1 x1
the parameter l must be zero or a positive integer and the integer m
can take on only m


l
,

(
l

1
)

(
l

1
)
,
l
m
d
P
(
x
)

(

1
)
(
1

x
) m
P
(
x
) positive m
l
dx
m
l
m
2
m
/
2
m
l

m
(

1
)
d
m
2
m
/
2
2 l
P
(
x
)

(
1

x
)
(
x

1
)
(+|m|,-|m|;
l
l
l

m
2
l
!
dx Rodrigues’formula)
45
2
Since differential eq. for P m depends only on m and m is an integer,
l
Pl  m
and Pl m
are proportional
(
lm

)
!m
P
(
x
)

(

1
)
P
x
)
l(
(
lm

)
!

m
l
For fixed m,

1 x1
Pl
m
m
form an orthogonal set in the index l on the interval

2
(
l

m
)!
P
(
x
)
P
(
x
)
dx

ll
'

2
l

1
(
l

m
)!
1
m m
l
'
l

1
The sol. of Laplace eq. was decomposed into a product of factors for
r,,

i
m

form a complete set of orthogonal in the index m
Q
()


e
m
46
2

on the interval 0
Pl m
form a similar set in the index l for each m for

1

c
o
s

1
Therefore,
m
P
)Q
()will form a complete orthogonal set on the surface of
l (cos
m
the unit sphere in the two indices l, m.
From the normalization condition, it is clear that suitable normalized
function, denoted by Y
(
,
)called spherical harmonics
lm

2
l

1
(
lm

)
!
m
i
m

Y

P
(
c
o
s
)
e
l
m
l
4
(
lm

)
!

It can be seen



m

l


g
(
,)

A
Y
(
,)


l
m
l
m
l

0
m


l
2
l

1
(
l

m
)!
(
l

m
)!
m
m

im

m
*
Y

(

1
)
P
(
x
)
e

(

1
)
Y
l
,

m
l
lm
4
(
l

m
)!
(
l

m
)!
47

The normalization and orthogonality conditions are










2
dd
s
i
n
Y
(
,
)
Y
(
,
)



l
'
m
'
l
m
l
'
lm
'
m
*
00
The complete relation is








m

l
*
lmlm
l

0
m


l
Y
(
'
,
'
)
Y
(
,
)

(

'
)
(cos

cos
'
)


1
l
0
,Y
0
0
4

l
1

3
i
Y


s
i
n

e
11
8


3
Y

c
o
s

1
0
4


48
Note that, for m=0,
2
l
1
Y
(

,

)

P

)
l
0
l(cos
4

which recovers the ordinary Legendre functions.
Since the spherical harmonics form a complete orthogonal set of function
on the surface of the unit sphere, then an arbitrary function
g(,)
can be expanded in spherical harmonics



m

l


g
(
,)

A
Y
(
,)


l
m
l
m
l

0
m


l
Where the coefficients




*
A

Y
(,)
g
(,)
d
sin
d
lm
lm
49
A point of interest to us is the form for   0


2
l

1
[
g
(,)
]

 A


0
l
0
l

0 4
with




2
l

1
A
 
d

P
(
c
o
s
)
g
(
,)
l
0
l
4

The general sol. for a Boundary-value problem in m 0
spherical coordinates can be written





m

l
l

(
l

1
)


l
m
l
ml
m
l

0
m


l
V
(
r
,
,
)

[
A
r

B
r
]
Y
(
,
)
If V is specified on a spherical surface, the coefficient can
be determined by this BC.
50
6 Laplace’s Equation in Cylindrical Coordinates; Bessel Function
In cylindrical coordinates (
, ,z)
2

V
0



V
1

V
1

V

V
  
0






z
2
2
The separation of variables
We get three ordinary differential equations:
2
2 2
2
2

 

V
(
,,
z
)

R
(
)
Q
(
)
Z
(
z
)
2
dZ
2

kZ
0
2
d
z
2
dQ
2

vQ
0
2
d

2
2
d
R
1
dR
2v
 
(
k

)
R

0
2
2
d d
 

51
The solutions of the first two eqs. are elementary

k
z
Z
()
z e
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
2
2
d
R
1
d
R v
 
(
1

)
R

0
2
2
x
d
x x
d
x
This is the Bessel’s Eq., and the solutions are called Bessel functions of order 
J
(
x
)
,
it
is
a
power
series
solution
given
by
v
j
j

(

1
)
(

1
)
2
j

v
2
j
J
(
x
)

(
x
/
2
)
(
x
/
2
) J
(
x
)

(
x
/
2
)
(
x
/
2
)

v


v
j
!

(
j

v

1
)
j

0
!

(
j

v

1
) 52
j

0j

v
(1)
Bessel
functions
of
the
first
kind
These
solutions
are
called
Bessel
functions
of
the
first
order


.The
series
conv
for
all
finite
values
of
x.
If

is
not
an
integer,
J
(
x
),
J
(
x
)
form
a
pair
of
linea
v

v
independ
solutions.
(2)
Bessel
function
of
the
second
kind
(Neumann
function)
m
If


m
(integer),
we
have
J
(
x
)

(

1
)
J
(
x
).
We
need
to
find
anothe
(

m
)
m
linearly
indenenden
tsolution.
J
(
xvJ
)
c
o
s
(
x
)
v

v
N
(
x
)

v
s
i
n
v



This
is
Neumann
function
(Besse
functi
of
seco
the
kin
N
(
x
)
and
J
(
x
)
are
linearly
indepe
t
even
if
is
not
an
int
v
v
(
3
)
The
Bessel
function
of
the
third
kind,
called
Hankel
functi
are
def
as
:
1
H
(
x
)

J
(
x
)

iN
(
x
)
v
v
v
2
H
(
x
)

J
(
x
)

iN
(
x
)
v
v
v
The
Hankel
functions
form
a
fundament
l
set
of
solution
to
the
Bess
eq.
53
Fourier-Bessel series
Bessel function has an infinite number of roots,
J
(
x
)

0
,n

1
, 2
,3

v
vn


x
is
the
nth
root
of
J
(x)
.

n
v
We
consider
only
Bessel
functions
of
first
the
kind.
Func
J
(
x
/
a
),
v
vn
for
fixed
v

0
,
n

1
,
2
,....,
form
an
orthogo
set
on
intev
0


the
a
.
The
normalizat
ion
integral
:


a

d

J
(
x
)
J
(
x
)

[
J
(
x
)]


a a
2
a
2
vvn
vvn
2
v

1
vn
n
n
 
0
Assumin
that
set
of
Bessel
functi
is
com
can
we
the
ex
an
ar
function
f
(
)
on
interva
0


a
in
a
Four
the
Bes
ser
:

f
(

)

A
J
(
x)

a

v
nv v
n
n

1
Where the coefficients:



2a
A

(
)
J
(
x
)
d
f
v
n
v
v
n
2
2
a
a
J
(
x
)
0
v

1
v
n
54
Modified Bessel functions

2
2
Note
:
if
the
sparation
constant
k
is
replace
by
k
, the
eq.
for
R(
)
be
2
2


d
R
1
dR
v




1

R

0
2
2


dx
x
dx
x


The solutions of this equation are called modified Bessel functions; They are
just Bessel functions of pure imaginary argument. The usual choices of linearly
independent solutions are denoted by


v
v

1
(
1
)
I
(
x
)

i
J
(
ix
),
K
(
x
)

i
H
(
ix
)
v
v
v
v
2
55
7
Boundary Value Problems in Cylindrical Coordinates
Consider the specific Boundary - value problem in Fig.
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=VWe
want to find the potential at any point inside the
cylinder.

 
V
(
z

L
)

V
(
,
),
V
(
z

0
)

V
(

a
)

0

 

V
(
,,
z
)

R
(
)
Q
(
)
Z
(
z
)
  
In order that V be single valued and vanish at z=0
kz

kz
Q
(
)

A
s
i
n
(
m
)c

B
o
s
(
m
)Z
(
z
)

{
c
e

c
e

sin
k
1 2}
where v=m is an integer and k is a constant to be determined.
56
The radial factor is
  
R
(
)

CJ
(
k
)

DN
(
k
)
m
m
If V is finite at D should be equal to 0.


2

ln(
x
/
2
)

0
.
5772
v

0





x

1
,
N
(
x
)


 

v
(
v
)
v

 (
2
/
x
) v

0







The
requiremen
t that
potential
the
vanish
as


a
means
that
k
can
on
only
take
special
those
values
:
x
mn
J
(
k
a
)

0

k

,n

1
,
2
,........
m
mn
mn
a
where
x
are
the
roots
of
J
(
x
)

0
.
mn
m
mn
57
Combining all these conditions, we find that the general form of the solution is





 
V
(
,
,
z
)

J
(
k
)
sinh(
k
z
)[
A
sin(
m
)

B
cos
m
)]


m
mn
mn
mn
mn
m

0
n

1
At z=L, we are given the potential as V There we have





 
V
(
,
)

J
(
k
)
sinh(
k
L
)[
A
sin(
m
)

B
cos
m
)]


m
mn
mn
mn
mn


m

0
n

1

 
This
is
a
Fourier
series
in
and
a
Bessel
Fourier
in
.


From
the
orthogonal
properties
of
sin
(
m
)
,
cos(
m
)
(0


2
)
and
J
(
k
)
(0


a),
m
mn
2


sin(
m

)
sin(
m
'
)
d




mm
'


0

a
2
a
2



J
(
k

)
J
(
k

)
d


J
(
k
a
)
mmn
'
mmn
m

1 mn
n
'n


2
0

58
2

a
2
cos
ech
(
k
L
)
mn
A

d
d
V
(
,
)
J
(
k
)
sin
m
mn
m
mn
2
2


a
J
(
k
a
)
m

1
mn
0
0



















2a
2
cos
ech
(
k
L
)
mn
B

d
d
V
(
,
)
J
(
k
)
cos
m
mn
m
mn
2
2
a
J
(
k
a
)
mn
'
mn
0
0


For
m

0,
should
we
use
B
/
2
in
serie
The
resu
is
app
the
e
0
n
for
a
finite
interval
in
,
0


a.
If
a


,
seri
the
goe
ov
in
an
in


For
example
we
want
know
poten
the
for
z

0
wh
V(


)

0
to
and
the
whole
plane
z

0
be
to
V(
,
).
Then
gen
sol
sh
b

  



k
z

 m
m
m

0
0
V
(
,
,
z
)

d
k
e
J
(
k
)
[
A
(
k
)
s
i
n
(
m
)

B
(
k
)
c
o
s
(
m
)
]
m

  
The coefficients are determined by


V
(
,
)

d
k
J
(
k
)
[
A
s
i
n
(
m
)

B
c
o
s
(
m
)
]


m
m
m
m

0
0
59
The variation in  is just a Fourier series. Consequently the
coefficients are separately specified by the integral relations:







2
 

A
(
k
'
)
sin
m


1
m
V
(
,
)
d

J
(
k
')
'

 dk


m
B
(
k
'
)
cos
m
0
0



m
Since


1
x
J
(
kx
)
J
(
k
'x
)
dx
(
k

k
')
m
m

k
0



Multiplyi
g
both
sides
by
J
(
k
)
and
integ
g
over
,
fin
we
th
m
coefficien
ts
are
determi
by
integr
over
who
area
of
pla
the

0
:
z









2

A
s
i
n
m

k
m


d
d
V
(
,
)
J
(
k
)



m
B
c
o
s
m
0
0

m
As
usual,
for
m

0,
must
we
use
B
(
k
)
/
2
in
ser
th
0
60