Physics 712
Chapter 3 Problems
5. [15] A semi-infinite cylinder of radius a has potential  = 0 on the lateral surface, and 
= V on the surface at z = 0. Write the potential as an infinite series. Assuming the
potential does not diverge as z  , which coefficients must vanish? Find the potential
everywhere, and numerically at  = 0 and z = a.
Since the potential vanishes at  = a, it can be written as a superposition of Bessel
functions times eim, that is

   , , z  

  A  z J
m  n 1
mn
m
 xmn   im

e
 a 
where xmn is the n’th root of Jm. It must satisfy Laplace’s equation in the interior, which tells us
 2
2 
1 
1 2
 xmn   im





  2    2  2 z 2  Amn  z  J m  a  e


m  n 1 

2
2
2
2



 
1  m 
x 
   eim  2 
 2
 2  Amn  z  J m  mn 
2
    z 
 a 
m  n 1
 
0  2 



x 
J m  mn  eim


 a 
m  n 1


2
 xmn
d2 



 Amn  z 
2
dz 2 
 a
Since J m  xmn  a  eim forms an independent set of functions on  and , the only way this can
be achieved is if the coefficients all vanish, which implies
2
 xmn
d2 



 Amn  z   0 , or
2
dz 2 
 a
2
xmn
d2
A
z

Amn  z  .


mn
dz 2
a2
This equation has two linearly independent solutions, so
x z
 x z
Amn  z    mn exp  mn    mn exp   mn  .
 a 
 a 
But we only want solutions that do not blow up at infinity, so we demand  mn  0 , and then have
   , , z  


 
m  n 1
mn
x 
 x z
J m  mn  eim exp   mn  e  xmn z a .
a 
 a 

It remains to find the coefficients  mn . Evaluating this expression at z = 0, we must have
V


 
m  n 1
mn
x 
J m  mn  eim .
 a 
We can use the orthonogonality of the functions J m  xmn  a  eim to then find:
 mn 

2
a
a
2V  m 0
1
2
 xmn  
x 
 im
e
d
VJ
d
J 0  0n   d 



 2 2

m


2
2



0
0
0
a J1  x0 n 
2 a J m 1  xmn 
 a 
 a 
x0 n
2V  m 0
J 0  y  y dy ,
x02n J12  x0 n  0
where in the last step we made the substitution   ay x0 n . You can get Maple to do the
integrals numerically, or if you want to be cleverer, you can use the recursion relations
J m 1  x  
m
d
Jm  x 
Jm  x
x
dx
to rewrite the final integral as

x0 n
0
J 0  y  y dy  
x0 n
0
x0 n d
x0 n
1

d
 y J1  y   dy J1  y   y dy  0 dy  yJ1  y   dy  yJ1  y  0  x0 n J1  x0 n  .


We therefore find  0 n  2V  x0 n J1  x0 n   . The first several terms can be
found in the table at right. The potential is
n
 0n V
1
1.601974697
2 -1.064799259

2V
 x0 n  
 x0 n z 
3
0.851399193
J0 
   , , z   
 exp  
.
 a 
 a 
n 1 x0 n J1  x0 n 
4 -0.729645240
5
0.648523614
Substituting in z = a and using the fact that J 0  0   1 , we have
6 -0.589542829

7
0.544180196
2Ve  x0 n
  0,  , a   
 0.1405064065V .
8 -0.507893631
n 1 x0 n J1  x0 n 
9
0.478012498
We let Maple do the sum for us. The numerical value was found adding seven terms.
> add(2/exp(x)/x/BesselJ(1,x),x=evalf(BesselJZeros(0,1..7)));