Eigenvalues and Eigenfunctions of a singular Sturm-Liouville operator using Bessel functions

Question

I’m trying to find the eigenvalues and eigenvectors of the Singular Sturm-Liouville operator:

• $$Lu=xu''+u'$$

  $$u(1)=0$$

  $$u(0) \text{ is finite}$$

  $$0 < x < 1$$

My approach to solving this problem:

I’m using Bessel functions as my solution:

$$R'' + 1/r \times R' + ((λ^2) - (\frac{v}{r})^2 )R = 0$$

solution is

$$R(r)=c_1 J_v (λr)+c_2 Y_v (λr)$$

where

$$J_v (λr) \text{ are finite at r=0}$$

$$Y_v (λr) \text{ are singular as r→0}$$

When $r=0$,EXCLUDE $Y_v$ from the solution

So, using the above:

Let $u(x) = c_1 J_v (λx) + c_2 Y_v (λx)$

But, I excluded $c_2 Y_v (λx)$ because my problem is singular.

So, my solution is $u(x)=cJ_v (λx)$

How am I doing?

Answer 1

A small contribution: Using Frobenius's method you can conclude that

$$ y(x;\lambda)=\sum_{n=0}^\infty \frac{\lambda^n x^n}{(n!)^2} $$ with $$ \lambda\neq 0 \mbox{ such that } \sum_{n=0}^\infty \frac{\lambda^n}{(n!)^2}=0 $$

I add a graphic to see the eigenvalues, that is the zeros of the function: