In this section the method of separation of variables will be applied to a problem in polar coordinates. The selected problem turns out to have two eigenfunctions for each eigenvalue other than the lowest.
The problem is to find the ideal flow in a unit circle if the normal
(radial) velocity on the perimeter is known.
We will try to find a solution of this problem in the form
The reason to take the as the eigenfunctions and not the is because separation of variables needs homogeneous boundary conditions. The direction has an inhomogeneous boundary condition at .
This follows the same procedures as in the first example. We
substitute a single term
homogeneous partial differential equation
Now which ordinary differential equation gives us the Sturm-Liouville problem, and thus the eigenvalues? Not the one for ; has an inhomogeneous boundary condition on the perimeter . Eigenvalue problems must be homogeneous; they simply don't work if anything is inhomogeneous.
We are in luck with however. The unknown has ``periodic'' boundary conditions in the -direction. If increases by an amount , returns to exactly the same values as before: it is a ``periodic function'' of . Periodic boundary conditions are homogeneous: the zero solution satisfies them. After all, zero remains zero however many times you go around the circle.
The Sturm-Liouville problem for is:
Pretend that we do not know the solution of this Sturm-Liouville problem!
Write the characteristic equation of the ordinary differential equation:
We will be lazy and try to do the cases of positive and negative
at the same time. For positive , the cleaned-up
Lets write down the boundary conditions first:
These two equations are a bit less simple than the ones we saw so far.
Rather than directly trying to solve them and make mistakes, this time
let us write out the augmented matrix of the system of equations for
If is negative, which is always greater than one for nonzero .
For the found eigenvalues, the system of equations for and becomes:
We had this situation before with eigenvector in the case of double
eigenvalues, where an eigenvalue gave rise two linearly independent
eigenvectors. Basically we have the same situation here: each
eigenvalue is double. Similar to the case of eigenvectors of
symmetric matrices, here we want two linearly independent, and more
specifically, orthogonal eigenfunctions. A suitable pair is
We can now tabulate the complete set of eigenvalues and eigenfunctions
We will again expand all variables in the problem in a Fourier series.
Let's start with the function giving the outflow through
Since is supposedly known, we should again be able to find
its Fourier coefficients using orthogonality. The formulae
are as before.
Since I hate typing big formulae, allow me to write the Fourier series
for much more compactly as
Next, let's write the unknown
as a compact Fourier
We put this into partial differential equation
We get the following ordinary differential equation for :
Fortunately, we have seen this one before: it is the Euler equation.
You solved that one by changing to the logarithm of the independent
variable, in other words, by rewriting the equation in terms of
The ordinary differential equation becomes in terms of :
Now both as well as are infinite when . But that is in the middle of our flow region, and the flow is obviously not infinite there. So from the `boundary condition' at that the flow is not singular, we conclude that all the -coefficients must be zero. Since , all coefficients are of the form , including the one for .
Hence our solution can be more precisely written
Next we expand the boundary condition
at in a Fourier series:
For , we see immediately that can be anything, but we need
for a solution to exist! According to the orthogonality
relationship for , this requires:
For nonzero :
Let's summarize our results, and write the eigenfunctions out in terms of the individual sines and cosines.
Required for a solution is that: