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If $\Omega$ is a bounded domain in $\mathbb{R}^n$, let $\lambda(\Omega)$ be an eigenvalue of the problem $$ \begin{cases} -\Delta u=\lambda u & \mbox{in }\Omega\\ u=0 & \mbox{on }\partial\Omega. \end{cases}$$

If $A_k=\{x \in \mathbb{R}^n, \frac{1}{k} < |x|< 1 + \frac{1}{k}\}$ and $B=\{x \in \mathbb{R}^n, |x|<1\}$. Is it known whether we have the limit identity $$\lim_{k\to +\infty}\lambda(A_k)=\lambda(B)?$$

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In my opinion it should work at least for the first eigenvalue $\lambda_1$ when $n\ge 3$ at least if we consider $$ A_k = \{ x\in\mathbb R^n\mid \frac1k\le |x|\le 1+\frac1k \}. $$ Since it is clear that $\lambda_1(B)\le\lambda_1(A_k)$ it suffices to prove that $\lim_k \lambda_1(A_k)\le \lambda_1(B)$.

Letting $E_k(f) = \int_{A_k}|\nabla f|^2\,dx$ be the energy associated with the operator $\Delta$ on $A_k$, by the min-max principle we know that $$ \lambda_1(A_k) = \min\left\{ \frac{E_k(f)}{\|f\|_{L^2(A_k)}^2} \mid f \in H^1_0(A_k), f\neq 0 \right\}. $$ Similarly for $\lambda_1(B)$ using the energy $E(f)=\int_B |\nabla f|^2\,dx$.

Consider the eigenfunction $\phi$ of the problem on the ball $B$ and a piece-wise smooth cut-off function $\chi_k$ such that $$ 0\le\chi_k\le1, \quad \chi|_{B\setminus A_{2k}}=0, \quad \chi|_{A_{k}}=1, \quad |\nabla \chi_k|\le k \text{ a.e.}. $$ Then, $\phi_k := \chi_k \phi\in H^1_0(A_{2k})\subset H^1_0(B)$. Exploiting the cut-off function is simple to see that $$ \|\phi-\phi_k\|_{L^2(B)}^2 \lesssim \frac1{k^n}, \quad E(\phi-\phi_k)\lesssim \frac 1 {k^{n-2}}. $$ Thus, if $n\ge 3$, $\phi_k\rightarrow \phi$ in $H^1_0(B)$. Then the claim follows by $$ \lim_k\lambda_1(A_{2k})\le \lim_{k} \frac{E_{2k}(\phi_k)}{\|\phi_k\|_{L^2(A_{2k})}^2} = \lim_{k} \frac{E(\phi_k)}{\|\phi_k\|_{L^2(B)}^2} = \frac{E(\phi)}{\|\phi\|_{L^2(B)}^2} = \lambda_1(B). $$

Maybe this idea can be generalized to higher eigenvalues by exploiting the correct min-max principle.

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  • $\begingroup$ Dear Dario, I found very interesting your argument, you characterized the first eigenvalue $\lambda_1$, right? I think that $E_k(f)=\int_{A_k}|\nabla\,f|^2dx$ and $E(f)=\int_{B}|\nabla\,f|^2dx$ is the correct. Another question, why $\lambda(B)\leq \lambda(A_k)$ and $H^1_0(A_{2k}) \subset H^1_0(B)$ if $A_k=\{x \in \mathbb{R}^n, \frac{1}{k} < |x|< 1 + \frac{1}{k}\}$ and $B=\{x \in \mathbb{R}^n, |x|<1\}$? $\endgroup$ – de Araujo Jul 21 '15 at 15:44
  • $\begingroup$ You are right, sorry but I misread the question and thought it to be only about the first eigenvalue. I edited the answer accordingly. $\endgroup$ – Dario Prandi Jul 21 '15 at 15:47
  • $\begingroup$ Indeed I misread the question also regarding the sets... what I said works for the sets $A_k=\{\frac1k \le |x|\le 1 -\frac1k\}$. I edited the answer to specify this, I'll look into the real question as soon as I have a moment! $\endgroup$ – Dario Prandi Jul 21 '15 at 15:55
  • $\begingroup$ In the set $A_k$ modified his argument is good. Thank you for their collaboration. I try to use your argument for the other eigenvalues. $\endgroup$ – de Araujo Jul 21 '15 at 16:09
  • $\begingroup$ Sorry Dario, in your argument $\|\phi-\phi_k\|_{L^2(B)}^2 \lesssim \frac1{k^n},$ I only get $$\|\phi-\phi_k\|_{L^2(B)}^2 \lesssim \frac1{k^n} + (1-(\frac{1}{2k})^n) + ((1-\frac{1}{2k})^n-(\frac{1}{k})^n) + ((\frac{1}{k})^n-(\frac{1}{2k})^n)$$ and for $E(\phi - \phi_k)$ I do not get the result. $\endgroup$ – de Araujo Jul 21 '15 at 19:10
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This is basically a comment on Dario's answer. I'm going to compare the Dirichlet problem on $B$ with the one on $B_0\equiv B\setminus\{ 0\}$ (though I'm not going to justify formally that this is what the limit in the OP will give), and Dario's argument will show that these are identical. In particular, all eigenvalues agree.

Recall that we may obtain the Dirichlet Laplacian on an open set $\Omega\subset\mathbb R^n$ from its quadratic form $$ Q(f) = \int_{\Omega} |\nabla f|^2\, dx , \quad\quad f\in H_0^1(\Omega) ; $$ in other words, $H_0^1$ is the domain of $(-\Delta_D)^{1/2}$, and $\Delta_D$ denotes the self-adjoint operator on $L^2(\Omega)$ we are interested in here (the Dirichlet Laplacian). See for example here for background.

Now in the situation we are interested in and for $n\ge 3$, we have that $$ H_0^1(B)=H_0^1(B_0) \quad\quad\quad\quad (1) $$ (as subspaces of $L^2(B)$, say).

(1) is established by the argument given by Dario. Since $H_0^1(\Omega)$ may be obtained as the closure of $C_0^{\infty}(\Omega)$ under the Sobolev norm $(Q(f)+\|f\|^2)^{1/2}$, it suffices to show that any $f\in C_0^{\infty}(B)$ can be approximated by smooth functions whose support is separated from $0$. This we do as in Dario's answer by taking $f_n(x)=\chi_n(|x|) f(x)$, with $0\le \chi_n\le 1$, $\chi_n=0$ near $0$, $\chi_n(r)=1$ for $r\ge 1/n$ and $\chi'_n\lesssim n$.

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  • $\begingroup$ Yes, your argument togheter Dario's argument are good and complement each, but I do not could verify the inequalities $$ \|\phi-\phi_k\|_{L^2(B)}^2 \lesssim \frac1{k^n}, \quad E(\phi-\phi_k)\lesssim \frac 1 {k^{n-2}}. $$ as can be seen in the previous discussion. $\endgroup$ – de Araujo Jul 22 '15 at 14:50
  • $\begingroup$ @deAraujo: (In my version:) Use that $f,\nabla f$ are bounded and $\chi'_k$, $(1-\chi_k)$ are non-zero only on the ball $|x|<1/k$, which has volume $\sim k^{-n}$. $\endgroup$ – Christian Remling Jul 22 '15 at 21:09
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It seems that limits $\mu=\lim_{k\to\infty}\lambda(A_k)$ of eigenvalues of the domains $A_k$ should be eigenvalues for the problem $$ \begin{cases} -\Delta u=\mu u & \mbox{in }B\setminus\{0\}\\ u=0 & \mbox{on }\partial B\cup\{0\}. \end{cases} $$ In other words, the limit of the sets $A_k$ in a spectral sense should be the punctured ball, not the ball.

If you take the normalized positive first eigenfunction of the Dirichlet Laplacian on $A_k$, they all vanish at the inner boundary $\partial B(0,1/k)$, so the limit function will certainly vanish at the origin. The limit function will be strictly positive elsewhere, so it will not solve the PDE at the origin. However, the first eigenfunction in the ball $B$ (normalized and chosen positive) is strictly positive and smooth at the origin. Therefore $\lambda_1(B)<\lim_{k\to\infty}\lambda_1(A_k)$, where the subscript refers to the first eigenvalue. I guess that the same inequality holds for all eigenvalues, perhaps with the exception of eigenfunctions of the ball that vanish at the origin.

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  • $\begingroup$ You don't say very explicitly what you mean by the Dirichlet problem on $B\setminus \{ 0\}$, but if you adopt the standard definition, then this is not correct. Maybe my answer can shed more light on this. $\endgroup$ – Christian Remling Jul 21 '15 at 21:18
  • $\begingroup$ @ChristianRemling, my answer was indeed vague and was meant to be an idea instead of a complete answer. I didn't have the time to give proper proofs of my claims back when I gave the answer. And it seems that I don't have them now, either, since my intuition seems to have failed me (to some extent at least). In the Dirichlet problem I mean boundary values in the strong sense for functions $u\in W^{1,2}(B\setminus0)\cap C(\bar B)$. It seems to me that the first eigenfunctions on $A_k$ should converge pointwise and the limit should not be the first eigenfunction on $B$. $\endgroup$ – Joonas Ilmavirta Jul 22 '15 at 8:23
  • $\begingroup$ Yes, if you do this, then (equivalent description) you are keeping precisely those eigenfunctions of $-\Delta_D$ on $B$ that happen to be zero at the origin. This is not the eigenvalue problem of a self-adjoint operator, though (for example, completeness fails, by construction). More to the point, I don't think this is what the limit in the OP delivers, though I have to admit that I haven't thought about this point carefully (the DP on $B\setminus \{ 0\}$ feels right to me as the limiting problem). $\endgroup$ – Christian Remling Jul 22 '15 at 21:15
  • $\begingroup$ @ChristianRemling, I feel that the limiting problem forces all functions to vanish at the origin, but I have nothing rigorous to back up my feeling at this point. It just seems to me that a "limiting eigenfunction" should be a limit of eigenfunctions on $A_k$ (corresponding to a converging eigenvalue). I would like to see an example where this limit function does not vanish (or blow up) at the origin if that can indeed be the case. The correct limiting problem may also depend on dimension (through removability of the origin for the Sobolev space); the answers by you and Dario assume $n\geq3$. $\endgroup$ – Joonas Ilmavirta Jul 23 '15 at 6:52
  • $\begingroup$ @JoonasIllmavirta: Dario does address the OP. The answer depends on the dimension for sure: introducing a Dirichlet BC at an interior point in one dimension will drive up (some) eigenvalues. $\endgroup$ – Christian Remling Jul 23 '15 at 17:08

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