7.38, §3 Eigenfunctions

If we substitute a trial solution into the homogeneous P.D.E. ,we get:

which separates into

Make sure that all r terms are at the same side of the equation!

Now which ODE gives us the Sturm-Liouville problem, and thus the eigenvalues? Not the one for R(r); u has an inhomogeneous boundary condition on the perimeter r=1. 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:

Note that for a second order ODE, we need two boundary conditions. So we wrote down that both , as well as its derivative are exactly the same at and .

Pretend that we do not know the solution of this Sturm-Liouville problem! Write the characteristic equation of the ODE:

Lets look at all possibilities:

Case :

Since k₁ = k₂ = 0:

Boundary conditions:

That can only be true if B=0. Then the second boundary condition is

hence . No undetermined constants in eigenfunctions! Simplest is to choose A=1:

Case :

We will be lazy and try to do the cases of positive and negative at the same time. For positive , the cleaned-up solution is

This also applies for negative , except that the square roots are then imaginary.

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 A and B:

Any nontrivial solution must be nonunique (since zero is also a solution). So the determinant of the matrix must be zero, which is:

or

A cosine is only equal to 1 when its argument is an integer multiple of . Hence the only possible eigenvalues are

If is negative, which is always greater than one for nonzero .

For the found eigenvalues, the system of equations for A and B becomes:

Hence we can find neither A or B; there are two undetermined constants in the solution:

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

Total:

We can tabulate the complete set of eigenvalues and eigenfunctions now as: