I was wondering if the following holds:

If you have an ODE $$-y''(x) + q(x) y(x) = \lambda y(x)$$ on a finite interval $(a,b)$ and you know that this equation is limit-circle or limit-point at the end-points.

If you now add a nice smooth + bounded -potential $V \in C^{\infty}(\mathbb{R})$ to your current potential, so that you end up with the ODE

$$-y''(x) + (q(x)+V(x)) y(x) = \lambda y(x),$$

is it still clear that your differential equation is limit-circle or limit-point at the endpoints?

I mean, somehow I feel that this statement should hold, as it is somehow natural to assume that a nice potential should keep the nice properties of the operator, but I could not find a reference for this.

  • $\begingroup$ @TobiasHurth: $V$ bounded is definitely sufficient; I had $V\in C^{\infty}(a,b)$ (open!) in mind. I in fact wasn't sure about the exact situation the OP had in mind as both $(a,b)$ and $\mathbb R$ are mentioned as intervals. $\endgroup$ – Christian Remling Oct 21 '14 at 18:02
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    $\begingroup$ @TobiasHurth: Yes, I think such an argument should also work. (One inconvenience is that one doesn't really have precise information on what the solutions look like.) $\endgroup$ – Christian Remling Oct 21 '14 at 18:12

The LP/LC classification is certainly not affected by bounded perturbations. One easy argument is to observe that a bounded $V$ is also bounded as an operator, and the deficiency indices are stable under bounded perturbations.

  • $\begingroup$ just to understand the relationship: the deficiency indices are (2,2) iff we have just limit circle, (1,1) iff we have limit circle and limit point and (0,0) iff we have limit points at both ends, is this correct? $\endgroup$ – Fabiano Oct 22 '14 at 9:22
  • $\begingroup$ @Fabiano: Yes, this is right; also, LP/LC is a local issue, so I can cut the interval into two halves and then apply the argument (this shows that there is no swapping of LP/LC between endpoints in the $(1,1)$ case). $\endgroup$ – Christian Remling Oct 22 '14 at 17:58

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