# The first eigenvalue of Laplacian

It's well known that for any bounded domain $$\Sigma$$, there exist unique $$\lambda_\Sigma >0$$ and nonnegative $$\varphi\in H_0^1(\Sigma)$$ such that \label{Eq-La-Bel-op} \left\{ \begin{aligned} &-\Delta\varphi=\lambda_\Sigma\varphi \quad\mbox{in }\Sigma, \\ &\varphi=0 \quad\quad\quad\quad~\mbox{on } \partial\Sigma,\\ \end{aligned} \right. where $$\Delta$$ is the Laplace operator and $$\lambda_\Sigma$$ is the first eigenvalue of it.

Question: For any $$\varepsilon>0$$ small, does there eixist some smooth domain $$\Omega$$ such that $$\Sigma\subsetneq\Omega$$ and $$\lambda_{\Sigma}\leq\lambda_{\Omega}+\varepsilon$$?

• Are you sure this is what you need? Because there is a trivial answer; just take $\Omega$ equal to a translate of $\Sigma$. – Giuseppe Negro Nov 14 '18 at 14:11
• A translate of $\Sigma$ is not a superset of $\Sigma$. – Kusma Nov 14 '18 at 14:15
• How about $\Omega$ is not a planar domain? e.g., $\Omega$ is a subdomain of unit sphere. (There is a result: If $\Omega$ is a planar domain, then $\lambda_{(\alpha\Omega)}=\frac{\lambda_{\Omega}}{\alpha^2}$) – xiaobiaoJia Nov 14 '18 at 14:17
• OH sorry, that symbol made me think that $\Sigma \not\subset \Omega$. – Giuseppe Negro Nov 14 '18 at 14:49
• Just scaling your domain only satisfied your condition of $\Sigma\subsetneq \Omega$ if it is star shaped. – Kusma Nov 14 '18 at 14:49

1. if $$\Sigma$$ is convex and $$\Omega_n$$ are convex sets tending to $$\Sigma$$ in Hausdorff distance, then $$\lambda_k(\Omega_n)\to \lambda_k(\Sigma)$$.
2. If $$\Omega_n$$ are uniformly Lipschitz and converging to $$\Sigma$$ in Hausdorff distance, then $$\lambda_k(\Omega_n)\to \lambda_k(\Sigma)$$
3. If $$\Omega_n\subset \mathbb{R}^2$$ such that the number of connected components of the complement is uniformly bounded, and converging to $$\Sigma$$ in Hausdorff distance, then $$\lambda_k(\Omega_n)\to \lambda_k(\Sigma)$$.
Depending on the regularity and connectedness properties of your set $$\Sigma$$ (and the dimension), this could answer your problems. I am not aware of any necessary conditions for continuity, but of course there might be some.