# 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?

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Not sure. I think the equation you have is not a Bessel equation. Lu=xu′′+u′=λu,u(1)=0 and u(0) finite. How do you transform this equation in the second one? – Pocho la pantera Sep 1 '13 at 1:01
@Pocho la pantera I thought it was a Bessel because I divided my equation by x and it fit that form. Did I make a mistake? – user92520 Sep 1 '13 at 4:26
@Stefan4024 thank you for the edits. – user92520 Sep 1 '13 at 4:29
@user92520: I think you made a mistake: starting w/ $xu'' + u'=\lambda u$ & dividing by $x$ yields $u'' + \frac{1}{x} u' - \frac{\lambda}{x} u = 0$. This does not fit the Bessel eqn. form b/c of the $1/x-$dependence of the zeroth order term; Bessel would require a $1/x^2-$dependence. As is usually the case, everyone would benefit from a bit more background on your problem (e.g., was it given to you like that? Did you derive the ODE yourself? etc.), if of course you have the time. Thanks! – automaton 3 Sep 1 '13 at 10:25
The problem was given like that. I've been solving regular S-L problems without a problem. But when I looked for an ansatz for this problem, I noticed that my problem seemed to match Bessel. – user92520 Sep 1 '13 at 19:33

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