Consider a bounded Euclidean domain $\Omega \subset \mathbb{R}^n$ (for simplicity, let's say, $\Omega$ has smooth boundary and is simply connected). Let $p \in \Omega$ be a point, and call $\Omega_n = \Omega \setminus B(p, \frac{1}{n})$. Let $\Delta$ (respectively, $\Delta_n$) denote the Dirichlet Laplacian on $\Omega$ (respectively, $\Omega_n$). I am interested in when the eigenvalues and eigenfunctions of $-\Delta_n$ converge to the eigenvalues and eigenfunctions of $-\Delta$, and in what sense (norm etc.) the eigenfunction convergence happens. The same question can be asked about the Neumann Laplacian.

The chief reason for my interest is that I am studying problems in homogenization and inverse problems (for example, papers of Vogelius, Friedman, Ammari et al). In these problems the setting is typically more complex, and problems involve an increasing number of perforations with decreasing "size". I am trying to understand a "toy case" thoroughly.

Edit: The main difficulty in the question seems to be for the case of repeated eigenvalues.

  • $\begingroup$ The corresponding quadratic forms are all equal, but the domains increase. As long as the union of the Sobolev spaces $H_0^1(\Omega_n)$ is dense in $H_0^1(\Omega)$, the eigenvalues converge (by the min-max characterisation). Convergence of eigenfunctions is trickier if the eigenvalues are degenerate. $\endgroup$ Mar 1, 2020 at 19:05
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    $\begingroup$ If I remember correctly, Jerison and Kenig developed "domain deformation" techniques in the 1990s. [D. Jerison and N. Nadirashvili, The “hot spots” conjecture for domains with two axes of symmetry, J. Amer. Math. Soc. 13 (2000) 741–772] might be a good source of references. If I find time tomorrow, I will have a look at it. $\endgroup$ Mar 1, 2020 at 19:18
  • $\begingroup$ @MateuszKwaśnicki By "degenerate" I guess you mean non-repeated eigenvalues, correct? If the eigenvalues are non-repeated, and the union of $H^1_0(\Omega_n)$ is dense in $H^1_0(\Omega)$, then the eigenfunctions attaining the min-max should also converge. Is that correct? $\endgroup$
    – user153038
    Mar 2, 2020 at 6:40

1 Answer 1


The only thing you need to show is that the convergence of the resolvant operator is sequentially compact. Then, general results about sequential compactness give you the result that the eigenvalues converges to the eigenvalues, in the sense that for any given open neighborhood of any eigenvalue in $\mathbb C$ of $T$ of multiplicity $d$, the limit compact operator, for $n$ large enough, $T_n$ (the resolvant compact operator) will have exactly $d$ eigenvalues in that neighborhood.

Proving compactness just follows from Rellich-Kondrachov in this case.


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