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In chapter 3 of the paper by Barry Simon about Sturm Oscillation and Comparison theorems, he assumes for simplicity that the potential of the equation is bounded, i.e. that $V$ is a real valued bounded function and the equation of interest is $$ - u'' + Vu = Eu $$ for $E$ a real number. Then he goes on and proofs the aforementioned theorems by Sturm in this case.

My question is the following. Suppose that $V$ is not necessarily bounded but has a singularity at 0. Explicitly, suppose that $V = \frac{1}{x^2}W$ where $W$ is a positive exponentially decreasing function.

I am pretty sure that the theorems can be extended to such a case, but I can't figure out how. The first problem is that for this kind of equation, the solutions behave badly at 0: One solution behaves like $\sqrt{x}$ and the other like $\log(x) \sqrt{x}$. Thus, if one tries to naivly redo the proof of Simon, one fails, where he assumes that there exists solutions $u(x)$ such that $$ u(0) =0, \text{ and } u'(0)=1 $$ which is impossible for the potential I am looking at. If there is an extension to such a singular case, what conditions do we have to ask for a solution $u(x)$?

Any help or references would be very appreciated.

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    Restrict to finite subinterval where the potential is bounded. How far does that take you? – 2017-01-20
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    @TrialAndError I came up with an answer. Thank you for your comment. – 2017-01-22

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After reading in this book in chapter 9 about Sturm-Liouville operators, I was able to adapt the proof to my case.

First of all note that we need a self adjoint extension of the differential operator. Since the operator is Limit Cirlce at 0 and limit point at infinity, by the Weyl alternative we have to choose a function $f$ in the domain of the differential operator. The domain is then all functions which have vanishing Wronskian with $f$ at 0. See again chapter 9.

Since this problem is motivated by a phisical application, I choose the fucntion which behaves like $\sqrt{x}$, since the other one would be "non physical".

With this self-adjoint extension, one can redo/adapt the proof done by Simon.

In conclusion I had to understand the choice of self-adjoint extension of the operator, which is not discussed in the Simon's paper.