Consider the following Sturm–Liouville (SL) eigenvalue problem defined in $(-\infty,0]$ or $[0,\infty)$ or $(-\infty,+\infty)$ $$(py')'-qy=-\lambda^2wy,$$ in which $p(x)=x^2$, $w(x)=1$, and $q(x)=(x/2+a)^2+a$ with parameter $a>0$. It has a regular singularity $x=0$. We basically hope for something like homogeneous Dirichlet b.c.

It is solved by making the substitution $y(x)=e^{x/2}x^{-\frac{1}{2}+\sqrt{(a+\frac{1}{2})^2-\lambda^2}}u(x)$, leading to Kummer's equation with two independent solutions (1st & 2nd kind) $$xu''(x)+(\gamma-x)u'(x)-\alpha u(x)=0,$$ in which $\alpha=\sqrt{(a+\frac{1}{2})^2-\lambda^2}-a+\frac{1}{2}$ and $\gamma=1+2\sqrt{(a+\frac{1}{2})^2-\lambda^2}$.
Let's then follow the ubiquitous argument.

Requiring nondivergence at $x=0$, the 2nd kind solution is dropped. Requiring nondivergence at infinity, the 1st kind is reduced to a polynomial when $-\alpha$ is a non-negative integer and eigenvalue $\lambda^2$ is attained. Seen from this condition for $\alpha$, we only have a bounded and finite sequence of eigenvalues. Therefore, one probably also expects an additional continuous spectrum. But I'm not sure whether it starts from the largest eigenvalue.


Is it possible to know the limit circle/point classification of this ODE near $0,\pm\infty$? Is the above solution useful for this purpose? How should one proceed? I also found page 13 of this paper has a limit point/circle classification of the Kummer equation I used. Not sure if relevant.

  • $\begingroup$ The procedure you describe seems designed to detect the eigenvalues, but the spectrum need not consist of eigenvalues only, so there's nothing special about having only finitely many (or none at all) of these. $\endgroup$ – Christian Remling Jan 5 at 22:50
  • $\begingroup$ @ChristianRemling Sorry, I don't quite get what you mean. It appears to me that the problem satisfies the condition for a singular SL to have an unbounded infinite sequence of discrete eigenvalues. There must be something simple that I missed. Could you please clarify a bit? Thanks. $\endgroup$ – xiaohuamao Jan 5 at 22:59
  • $\begingroup$ What condition exactly are you referring to? There is no simple general condition that would give you this conclusion, and many SL problems have no eigenvalues at all (for example the free equation $-y''=\lambda y$). $\endgroup$ – Christian Remling Jan 5 at 23:35
  • $\begingroup$ @ChristianRemling What I mentioned is Atkinson's result as summarized in the 1st section of this paper, but probably I misunderstood it? I'm not familiar with this. Let me see if I understand correctly. You mean there's no general condition or conclusion on singular SL spectrum, right? In my case, should it be finite discrete eigenvalues plus a continuous spectrum? $\endgroup$ – xiaohuamao Jan 6 at 3:33
  • $\begingroup$ @ChristianRemling Thanks for your patience. A closely related issue that confuses me. What if we consider this equation on $(-\infty,+\infty)$ with homogeneous Dirichlet b.c.? The solution shown with $e^{x/2}$ doesn't seem to allow this at $+\infty$. Did I do anything wrong? $\endgroup$ – xiaohuamao Jan 6 at 3:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.