Physics 712
Chapter 2 Problems
3. [15] Consider a square of side a with  = 0 on three sides and  = V on the surface y =
a in two dimensions. Our goal is to compute the potential everywhere, and particularly

  14 a, 12 a  . Write the potential in the form   x, y    n 1 An  y  sin  n x a  . What is
the form of the functions An  y  ? By matching appropriate boundary conditions,
determine any unknown coefficients, and find   x, y  as an infinite sum. Sum it
numerically to find   14 a, 12 a  . Compare your results with the results of problem 1.7.
We want the Laplacian to vanish, so this implies

 2
   nx 
 2n2
0   2   x, y     2 An  y   2 An  y   sin 
.
a
n 1  y
  a 
Since the sine functions are an independent set of functions, the only way this can vanish is if the
expression in square brackets vanishes. We therefore have
d2
 2n2

A
y
An  y  .


n
dy 2
a2
This has general solution
An  y    n e ny a   n e  ny a .
However, this function must also vanish at y = 0, so this implies  n   n , and our function is

  nx 
  ny 
  x, y    2 n sin 
 sinh 

 a 
 a 
n 1
We now start working on the constants  n . We note that if we set y = a, we must have

  nx 
V   2 n sin 
 sinh  n 
 a 
n 1
If we multiply both sides of this equation by sin  mx a  and integrate over x, we can use the
fact that the functions sin  mx a  are orthogonal to find

a
  mx 
  nx    mx 
V
dx

sin
2 n  sin 



 sin 
 dx sinh  n  ,
0
0
 a 
 a   a 
n 1
a
a

Va
  mx 

  2 n 12 a nm sinh  n  ,
cos 

m
 a  m 0 n 1
V 
m
1   1    m sinh  m 


m
The expression in square brackets vanishes for m even and is 2 for m odd. Substituting this back
into our expression for the potential, we have
  x, y  

4V
  nx 
  ny 
 sinh 

a 
 a 
  n sinh  n  sin 
n odd
We have been asked to compute   14 a, 12 a  , which is therefore
  14 a, 12 a  

2sin  14  n 
4V
1
1
sin

n
sinh

n
V
,







4
2
1
n odd  n sinh   n 
n odd  n cosh  2  n 

where at the last step we used the double angle formula sinh  2   2sinh  cosh  to simplify a
bit. We now let Maple do the sum numerically for us:
> add(evalf(sin(Pi*(2*n-1)/4)/cosh(Pi*(n-1/2))/Pi/(n-1/2)),
n=1..6);
We find   14 a, 12 a   0.1820283319V , adding just six terms. In problem 1.7, our best estimate
was   14 a, 12 a   0.182027585V , which is correct to about six digits, so I was a bit optimistic
there when I claimed seven digits.