There are some known criteria for the Sturm-Liouville Problem

\begin{equation} \tag{1} \frac {\mathrm {d} }{\mathrm {d} x}\left[p(x){\frac {\mathrm {d} y}{\mathrm {d} x}}\right]+q(x)y=-\lambda w(x)y \end{equation}

to have a spectrum discrete and bounded below (BD) in the singular case: If singular endpoints are Limit-Circle and Non-Oscillating (LCNO), results from (Niessen, H.-D. and Zettl, A. (1992). Singular Sturm-Liouville Problems: The Friedrichs extension and comparison of eigenvalues.) apply. There also is the famous criterion by Molchanov and generalizations for $w, p \neq 1$ in (Kwong and Zettl - Discreteness conditions for the spectrum of ordinary differential operators). Various other generalizations exist.

As I understand it, the Molchanov criterion and its generalizations are not applicable to the case $q = 0$ or $q$ constant. Inherently, $q$ must run to infinity at the endpoint in a certain way (details depending on $p$ and $w$). (See related questions For which type of potentials a Schrödinger operator will have discrete spectrum?, Harmonic oscillator discrete spectrum)

My questions:

Is there a theory and criteria for (1) to be BD for the case $q = 0$ if endpoints are in the Limit-Point (LP) case? I specifically appreciate pointers to literature. Or do I overlook some tweak, allowing to apply Molchanov-style criteria to the case $q = 0$?

What I read and reasoned so far:

As far as I can assess, transformations to normal form are not suitable to work around this particular issue. Also, according to Zettl (Zettl, A. (2010). Sturm-Liouville Theory), it is questionable whether such a transformation preserves the desired properties of the spectrum (discussion in Chapter 10.13).

I looked into various books and papers and googled around, but I almost exclusively found Molchanov-style criteria, most usually setted for the half-line $[a > -\infty, \infty)$. It appears that Schrödinger operators are the main interest in this field and the Molchanov criterion is a good fit for them. I am more interested in self-adjoint operators similar to the one yielding Hermite polynomials:

\begin{equation} \tag{2} q = 0, \quad w = p = e^{-{\frac {x^{2}}{2}}} \end{equation}

on the real line with two singular LP endpoints. I used to perceive Hermite Polynomials as an important standard example and it puzzles me that it seems to be so hard to find criteria for their spectrum. I mean criteria that are not totally specific to equation (1) with values (2), but also hold if some other (normal-like?) probability distribution is used in (2), e.g. with adjusted variance, skewness, kurtosis etc. (not expecting the solutions to be (orthogonal) polynomials anymore, but still to have BD spectrum). (Edit: It occurred to me that the case of (2) with adjusted variance is known as the generalized Hermite polynomials.) E.g. I skimmed through some books by L. L. Littlejohn and A.M. Krall about orthogonal polynomials in context of singular SL theory, e.g. (L. L. Littlejohn and A.M. Krall, Orthogonal polynomials and singular Sturm-Liouville Systems, I) but they do not seem to discuss the discreteness of the spectrum (sorry if I overlooked it, pointers welcome).

Secondary question: What about the full line vs. the Half line?

It appears that in his original work, Molchanov considered the full line (A.M. Molchanov, Conditions for the discreteness of the spectrum of self-adjoint second-order differential equations.) however, I cannot read Russian and only got this impression from the math terms. Also (Inge Brinck, Self-Adjointness and spectra of Sturm-Liouville operators) assumes the full line, but (Kwong and Zettl - Discreteness conditions for the spectrum of ordinary differential operators) and (D. Hinton, Molchanov’s discrete spectra criterion for a weighted operator) assume the half line. This gives me the impression that the distinction between full line and half line is not essential, but I did not find it discussed anywhere. A pointer to literature that specifically discusses in what sense results are transferable between the half line and full line case would be welcome. (Niessen, H.-D. and Zettl, A. (1992). Singular Sturm-Liouville Problems: The Friedrichs extension and comparison of eigenvalues.) discuss it a bit, but not in context of criteria for BD in the LP case (or I don't understand how to conclude it).

Thanks in advance, also for hints, comments and improvements!

  • 1
    $\begingroup$ Whole line is the same question as half line because the whole line essential spectrum is the union of the two half line essential spectra (so I guess what I'm saying is that a criterion for half line problems will settle the whole line case too). $\endgroup$ Commented Apr 4, 2020 at 18:22
  • $\begingroup$ Just by chance, do you know a reference for the whole line's essential spectrum being the union of the two half line essential spectra? This is not obvious to me as the two half line problems involve a boundary condition at the split point the full line problem does not have. But this boundary condition is not relevant for the essential spectrum I guess... $\endgroup$
    – stewori
    Commented Apr 5, 2020 at 14:29
  • 1
    $\begingroup$ The argument is that cutting the whole line operator into two half lines is a rank $2$ (so in particular compact) perturbation of the resolvent, so the essential spectra agree by Weyl's theorem. Weidmann's Springer Lecture Notes discuss this for sure, but it must be in many other books (for example, somewhere in Teschl's books, which are available on his homepage). $\endgroup$ Commented Apr 5, 2020 at 17:21
  • $\begingroup$ For the record, I found it in Weidmann's Springer Lecture Notes, but he merely states it in a remark and refers the reader to Naimark. Finally I found it stated in (M.A. Naimark, Linear Differential Operators), §24, Theorem 1. Thanks again for all your help! $\endgroup$
    – stewori
    Commented Apr 9, 2020 at 17:47
  • $\begingroup$ No problem, I enjoyed your question! $\endgroup$ Commented Apr 10, 2020 at 16:52

1 Answer 1


The case $q=0$ actually has a complete answer, and the reason this works is that we can solve the equation for $\lambda =0$ explicitly then, by $u=1$ and $v=\int_0^x \frac{dt}{p(t)}$. The spectrum is purely discrete if and only if ($w\in L^1$ and) $$ \lim_{x\to\infty}\int_0^x wv^2\, dt \int_x^{\infty} w\, dt =0 . \quad\quad\quad\quad (1) $$ (This holds in your example, if I managed the asymptotics of the error functions correctly.)

I obtained this by rewriting the SL equation as a canonical system $Jy'=-\lambda Hy$, and the coefficient matrix $H$ we obtain here is given by $$ H = w\begin{pmatrix} 1 & v \\ v & v^2 \end{pmatrix} . $$ There is a precise criterion for purely discrete spectrum for (general) canonical systems, and this gave me (1).

This will probably all sound quite cryptic if you haven't seen these things before, and there seem to be too many small details to explain them all here. You could take a look at my papers 1, 2 on my homepage for full background information.

Also, if I hadn't had that available, I would have tried to analyze the Prufer equation $$ \varphi' = \frac{1}{p}\cos^2\varphi + (\lambda w+q)\sin^2\varphi $$ for the phase of the solution vector $(y,py')$. Purely discrete spectrum is equivalent to $\varphi(x)$ staying bounded in $x$ for all $\lambda$. This might also be good enough to get (1).

  • $\begingroup$ Wow, that's a stunningly beautiful criterion. Thanks for this work! Do I understand correctly that for the full line, (1) must additionally hold with every $\infty$ replaced by $-\infty$? You defined $u=1$ but did not use it any more. Was it a typo? $\endgroup$
    – stewori
    Commented Apr 5, 2020 at 14:21
  • $\begingroup$ Forget the question about $u$, I see, it's just another solution for $\lambda = 0$. Read it too quickly... $\endgroup$
    – stewori
    Commented Apr 5, 2020 at 16:29
  • $\begingroup$ @stewori: Yes, that's correct (what you say about the whole line problem). The other solution $u$ does make another appearance, though I don't call it that, the general form of $H$ is $H=w \bigl( \begin{smallmatrix} u^2 & uv \\ uv & v^2 \end{smallmatrix} \bigr)$. $\endgroup$ Commented Apr 5, 2020 at 17:17
  • $\begingroup$ Right :) I had just found the example in Section 5 of your paper 1 which explains the construction. Then I wrote "forget the question about $u$". The example is so hidden at the end... I had overlooked it when I skimmed through the paper at first. $\endgroup$
    – stewori
    Commented Apr 6, 2020 at 19:32

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