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

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:

