Consider the following singular Sturm-Liouville problem: $$ -(r^{N - 1}h')' - r^{N - 1}c(r)h = \lambda r^{N - 3} h \text{ in } (0, 1), \qquad h(1) = \alpha $$ where

  • $N \in \mathbb N$, $N \geq 3$;
  • $c(r) \in L^\infty(0, 1)$;
  • $\alpha > 0$.

We begin by defining the functional setting:

  • $\displaystyle L_N^2 \:= \left\{v:(0, 1) \to \mathbb R \ : \ \int_0^1 r^{N - 1} v^2 \ dr < + \infty \right\}$;

  • $\displaystyle H^1_N := \left\{v \in L_N^2 \ : \ v' \in L_N^2 \right\}$ with the norm $\displaystyle \|v\|_N^2 = \int_0^1 r^{N - 1} (v^2 + |v'|^2) \ dr$;

  • $\displaystyle H_{0, N}^1 := \left\{v \in H_N^1 \ : \ v(1) = 0 \right\}$;

  • $\displaystyle \mathcal L_N := \left\{ v:(0, 1) \to \mathbb R \ : \ \int_0^1 r^{N - 3} v^2 \ dr < + \infty \right\}$;

  • $\mathcal H_N := H_N^1 \cap \mathcal L_N$;

  • $\mathcal H_{0, N} := H_{0, N}^1 \cap \mathcal L_N$;

  • $K_N := \mathcal H_{0, N} + \{\widetilde h\}$ where $\widetilde h \in \mathcal H_N$ is such that $\widetilde h(1) = \alpha$

Observe that $K_N$, being an affine space, has a Hilbert manifold structure with tangent space $T_vK_N = \mathcal H_{0, N}$ at every $v \in K_N$. Then we can define a weak solution of as $h \in K_N$ such that \begin{equation} \int_0^1 r^{N - 1}(h' v' - c(r)hv) \ dr = \lambda \int_0^1r^{N - 3} h v \ dr \quad \forall v \in \mathcal H_{0, N}. \end{equation}

Question: What can one say about existence of (weak?) solutions $(h, \lambda)$ of this problem such that the eigenvalue $\lambda < 0$ and the eigenfunction $h$ is defined in the whole interval $(0, 1)$? Any hope that there will be a solution $h$ for every $\lambda < 0$?

Remark: As noted by Igor Khavkine in the comments, the usage of "eigen-" is imprecise here. I keep it nonetheless.

Thanks in advance

  • $\begingroup$ The ODE is singular only at $r=0$, not at $r=1$. Pick any additional initial condition $h'(1)=\beta$, and you have your solution $h(r)$ on $(0,1]$ by standard existence/uniqueness for ODEs. Did you mean to impose boundedness or other regularity on $h(r)$ as $r\to 0$? $\endgroup$ Nov 2, 2022 at 12:04
  • $\begingroup$ @IgorKhavkine, thanks for your reply. Do you recommend any references for these facts? I have a very poor background on ODEs. Regarding the regularity, I have proved that once we have the existence of a weak solution, it holds $h \in C[0, 1] \cap C^1(0, 1]$ with $h = O(t^\theta)$ where $\theta$ is a positive exponent depending on $\lambda$. $\endgroup$ Nov 2, 2022 at 13:13
  • $\begingroup$ But are you really sure that this holds? In their paper arxiv.org/pdf/1805.04321.pdf Amadori and Gladiali prove that the spectrum of this equation is discrete, with finitely many negative eigenvalues, under the condition $h(1) = 0$. What changes? $\endgroup$ Nov 2, 2022 at 13:16
  • 1
    $\begingroup$ Amadori-Gladiali impose that $h(r)$ belongs to a weighted Lebesgue or Sobolev space on $(0,1)$. Perhaps you meant to do the same, but didn't explicitly write it in the question. Indeed, even if a unique solution exists for any $\lambda$ and initial data $\alpha$, $\beta$, there is no guarantee that it will have finite norm near the singular point $r=0$. Indeed, imposing $\|h\|<\infty$ puts a restriction on the $(\lambda,\alpha,\beta)$ data. In A-G, under $\alpha=0$ and $h(r)\ne 0$, they obtained the discrete $\lambda$ spectrum. Your situation is analogous, but not entirely. ... $\endgroup$ Nov 2, 2022 at 16:49
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    $\begingroup$ For instance, the inhomogenous initial condition $h(1) = \alpha \ne 0$ means that you are no longer dealing with an eigenvalue problem, simply because the solutions of your equation no longer form a vector space. So referring to it as an eigenvalue problem is ambiguous. Perhaps you should rephrase your question, to make it unambiguous and make it clear in what way you want to stay close to or depart from the setting of A-G. $\endgroup$ Nov 2, 2022 at 16:53

1 Answer 1


Write your differential equation as $Ph = \lambda h$, the differential operator is unbounded and symmetric on the Hilbert space $\mathcal{L}_N$, with an appropriate domain. As analyzed by Amadori-Gladiali in the reference you pointed out, with the boundary condition $h(1)=0$ it is also self-adjoint and has a discrete spectrum, meaning that the resolvent $(P-\lambda)^{-1}$ exists as a bounded operator for all $\lambda \not\in \sigma(P)$ not belonging to the spectrum.

Now comes a standard idea. Choosing some sufficiently regular $\tilde{h}$ satisfying $\tilde{h}(1) = \alpha$ and setting $h=\tilde{h}+f$, your homogeneous equation with inhomogenous boundary conditions becomes an inhomogeneous equation with homogeneous boundary conditions $(P-\lambda)f = -(P-\lambda)\tilde{h}$ (when $\tilde{h}$ is sufficiently regular, the right-hand side is in $\mathcal{L}_N$). The solution $f = (P-\lambda)^{-1} [-(P-\lambda)\tilde{h}]$ will be in $\mathcal{L}_N$ and unique as long as $\lambda\not\in\sigma(P)$. Since the spectrum only has eigenvalues, it's a matter simple linear algebra to figure out what happens when $\lambda \in \sigma(P)$.

  • $\begingroup$ I hadn't noticed when the OP was originally edited. Hope this is still helpful. $\endgroup$ Nov 24, 2022 at 15:44

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