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I am considering the following eigenvalue problem in $\Omega=\mathbb{R}^n_+$ $$-\operatorname{div}(a(x)\nabla \varphi) = \lambda a(x)w(x)\varphi$$ where the weights $a>0$ and $w\in L^{\infty}$ (and is a radial function such that $w>0$), $\lambda>0$ and $\varphi$ lives in the appropriate weighted Sobolev space. I am interested in any results involving the dimension of the second Eigenspace associated with the second smallest eigenvalue to this equation.

I already have shown a weighted Sobolev embedding type estimate, $$\int_{\Omega} a(x) |u|^{2^*}dx \leq C\int_{\Omega} a(x)|\nabla u|^2 dx$$ for functions $u\in C^{\infty}_c(\mathbb{R}).$ The function $w$ is an extremizer for the above inequality and satisfies, $$\int_{\Omega} a(x) |w|^{2^*}dx = C\int_{\Omega} a(x)|\nabla w|^2 dx$$ and $\frac{\partial w}{\partial x_n} =0$ on $\partial \Omega.$

Usually, if $a$ was radial I would use Separation of Variables and Sturm-Liouville Theory to arrive at some conclusion but this will not work here. I also tried looking at some papers related to weighted elliptic eigenvalue problems, but most of the papers were concerned with the first eigenvalue of this equation, so I am not sure if I can use the techniques developed in those articles.

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    $\begingroup$ Does w have an arbitrary sign? $\endgroup$
    – username
    Commented Oct 10, 2021 at 14:06
  • $\begingroup$ @username It's positive actually, let me add that to the question. Thanks! $\endgroup$
    – Student
    Commented Oct 10, 2021 at 14:21
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    $\begingroup$ I would be suprised if there was such a result. If you simply consider the Laplacian on a bounded domain, it is difficult to predict the multiplicity of the eigenvalues, without precise informations on the geometry. Naturally, the multiplicity of the first one is known, by Krein-Rutman. But after that you know that that the dimension of the eigenspace is finite and that's about it. $\endgroup$
    – username
    Commented Oct 10, 2021 at 17:30
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    $\begingroup$ The most common reason eigenspaces are finite is that your operator's inverse is compact. This does not follow from the information you provided, but the information provided is kind of vague: you do not give a boundary condition along $\partial \Omega$, or the assumptions on $w$ and $a$. The point is that you explicitly state that the domain is unbounded, which is potentially dangerous for having eigenfunctions/values at all. $\endgroup$
    – user378654
    Commented Oct 11, 2021 at 2:22
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    $\begingroup$ To complement the comment above, assuming that you have some natural boundary conditions on $\partial\Omega$, Neumann or Dirichlet, and that you have setup your problem in the right space, such a Beppo-Levi space (see e.g. [Deny and Lions] (eudml.org/doc/73718) in French), you have a good chance of showing compactness. $\endgroup$
    – username
    Commented Oct 11, 2021 at 7:37

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