7.38 U, §3 Eigenfunctions

Let's start trying to get rid of one variable first. We might try a solution of the form

where the R_n would be the eigenfunctions and the the corresponding Fourier coefficients. Unfortunately, if we try to substitute a single term of the form into the homogeneous PDE, we are not able to take all r terms to the same side of the equation and and t terms to the other side. So we do not get a Sturm-Liouville problem for R_n.

Try again, this time

If we substitute into the homogeneous PDE we get:

This, fortunately, can be separated:

So we have a Sturm-Liouville problem for :

with boundary conditions that are periodic of period . This problem was already fully solved in 7.38. It was the standard Fourier series for a function of period . In particular, the eigenfunctions were , , and , .

Like we did in 7.38, in order to cut down on writing, we will indicate those eigenfunctions compactly as , where and .

So we can concisely write

Now, if you put this into the PDE, you will see that you get rid of the coordinate as usual, but that still leaves you with r and t. So instead of ODE in t, you get PDE involving both r and t derivatives. That is not good enough.

We must go one step further: in addition we need to expand each Fourier coefficient uⁱ_n(r,t) in a generalized Fourier series in r:

Now, if you put a single term of the form into the homogeneous PDE, you get

Since , this is separable:

So we get a Sturm-Liouville problem for Rⁱ_n with eigenvalue

with again the same homogeneous boundary conditions as u:

We need to find all solutions to this problem.

Unfortunately, the ODE above is not a constant coefficient one, so we cannot write a characteristic equation. However, we have seen the special case that before, 7.38. It was a Euler equation. We found in 7.38 that the only solutions that are regular at r=0 were found to be A_n rⁿ. But over here, the only one of that form that also satisfies the boundary condition Rⁱ_n'=0 at r=a is the case n=0. So, for , we only get a single eigenfunction

R₀₀ = 1

For the case , the trick is to define a stretched r coordinate as

This equation can be found in any mathematical handbook in the section on Bessel functions. It says there that solutions are the Bessel functions of the first kind J_n and of the second kind Y_n:

Now we need to apply the boundary conditions. Now if you look up the graphs for the functions Y_n, or their power series around the origin, you will see that they are all singular at r=0. So, regularity at r=0 requires B_n=0.

The boundary condition at the perimeter is

Since is nonzero, nontrivial solutions only occur if

Now if you look up the graphs of the various functions J₀, J₁, , you will see that they are all oscillatory functions, like decaying sines, and have an infinity of maxima and minima where the derivative is zero.

Each of the extremal points gives you a value of , so you will get an infinite of values , , ,, , . There is no simple formula for these values, but you can read them off from the graph. Better still, you can find them in tables for low values of n and m. (Schaum's gives a table containing both the zeros of the Bessel functions and the zeros of their derivatives.)

So the r-eigenvalues and eigenfunctions are:

where m is the counter over the nonzero stationary points of J_n. To include the special case , we can simply add ,Rⁱ₀₀ = J₀(0)=1 to the list above.

In case of negative , the Bessel function J_n of imaginary argument becomes a modified Bessel function I_n of real argument, and looking at the graph of those, you see that there are no solutions.