Required math: calculus
Required physics: electrostatics
Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problems 3.47.
Here is another example of solving Laplace’s equation in rectangular coordinates. We have a rectangular pipe extending in the direction with boundaries at , and . The potential on each face is held constant and satisfies
The problem is to find everywhere inside the pipe. We can use the separation of variables technique to arrive at a solution of form:
Because the problem is symmetric in we must have , which means that and we can write
where we’ve absorbed the constants and into and .
The boundary condition at means that . At first glance we can’t do much with the boundary conditions at the other three faces. Let’s leave this solution for the moment and try a different approach.
In the original derivation of the separation of variables solution, we had the equations
We made the implicit assumption that the constants were non-zero, requiring only that their sum is zero so as to satisfy the full Laplace’s equation:
In this problem, though, we can use a trick to get the solution. Since the problem doesn’t depend on , we can consider the two equations
The general solution of this pair of equations is linear in each of and separately in order for the second derivatives to be zero. Thus we get
and the overall solution is thus
Applying the boundary conditions on we get
This must be true for all so either or . If we choose the latter case, then everywhere, which isn’t much use, so we’ll take .
At the other boundary, we have
Again, this must be true for all , so and . Thus a solution that satisfies the boundary conditions is
Clearly this doesn’t satisfy the boundary conditions, but we can now return to our earlier solutions to the separation of variables problem to see how to fix this. If we find a solution that satisfies the new boundary conditions
then we can add this solution to the solution we just found. This new solution will satisfy all four boundary conditions, since the sums of the potentials at each of the four boundaries come out to what was originally specified. We can now use the usual technique of building an infinite series of solutions and using the orthogonality of the sine functions to work out the coefficients. That is, returning to our first solution, we have, after satisfying the boundary condition at :
If we now require this solution to be zero at , then we must have
for . We can then construct the sum of these terms and combine it with the other solution we found to give
To satisfy the boundary conditions we must have
for . We now multiply both sides of this equation by and integrate from 0 to , using the orthogonality of the sine, that is
The final form of the potential is