I am trying to solve the simple Eigenvalue problem

$-\Delta{u}=\lambda u, \; u = 0$ on $\partial{\Omega}$

in Sobolev spaces on a smooth exterior (and thus unbounded) domain $\Omega \subset \mathbb{R}^{n}$. Unfortunately, the Rellich-Kondrachov compactness theorem cannot be used on $\Omega$.

I read something online about using weighted Sobolev Spaces, but didn´t find any concrete references and now tried to do it on my own.

I am trying to prove the compactness of the embedding of $H^{1}_{0}(\Omega,\omega)$ in $L^{2}(\Omega,\omega)$, where $H^{1}_{0}(\Omega,\omega)$ is the completion of $C^{\infty}_{c}(\Omega)$ in regards to the norm $||u||^{2}:=\int_{\Omega}|u|^{2}\omega + |Du|^{2}$, where $\omega(x)=1/|x|^{2+\delta}$ for some small $\delta > 0$.

My idea: Let $(u_{n}) \subset H^{1}_{0}(\Omega,\omega)$ be a bounded set. Using the Hardy inequality i get $\int_{\Omega/B_{R}(0)}|u_{n}|^{2}\omega \le 1/|R|^{\delta} \cdot C \cdot \int_{\Omega/B_{R}(0)}|Du_{n}|^{2}$, which can be made $< \epsilon$ for sufficiently large $R$.

On $B_{R}(0)\cap \Omega$ I use the classic Rellich-Kondrachov theorem to find a finite number of functions in $L^{2}(\Omega\cap B_{R}(0),\omega)$ which cover the $u_{n}$ up to $\epsilon$. Thus, the $u_{n}$ are totally bounded.

Does this make sense?

  • $\begingroup$ Do you assume that $0\in \Omega$ so that $L^2(\Omega\cap B_R)=L^2(\Omega\cap B_R,w)$ with equivalent norms? Otherwise I don't see how you use Rellich-Kondrachov for the bounded part. The rest seems fine to me. $\endgroup$ – MaoWao Feb 11 at 15:19
  • $\begingroup$ Thanks! I will adjust $\omega$ slightly to make it work in the general case of $0 \notin \Omega$. One more question: Since the dual $(H^{1}_{0}(\Omega,\omega))´$ is way smaller than the weighted space itself, I am having trouble applying the methods one usually uses with compact operators. Could you briefly explain how I use the spectral theorem in this case? $\endgroup$ – Falc14 Feb 12 at 10:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.