# Identical muliplicity for eigenvalues of Laplacian

Suppose $\Omega\subset\mathbb R^2$ is a bounded simply connected domain. Consider the Laplacian eigenvalue problem with Dirichlet boundary condition :

\begin{cases} \hfill \Delta u=\lambda u \hfill & \Omega \\ \hfill u=0 & \partial\Omega. \\ \end{cases} $\textbf Q1:$ Is that the case that all eigenvalues are of the same multiplicity?(It seems very unlikely, but what are some examples?)

$\textbf Q2:$If the answer to the first question is negative, is there a characterization of domains that satisfy the property in $(\textbf Q1)$?

$\textbf {Edit 1:}$ This link sounds interestinhg.

Think of a simple example where solutions are $$\phi_{n,m}(x,y)=\sin(n x)\sin(m y),\;\;\; \Omega = [0,\pi]\times[0,\pi].$$ The associated eigenvalues are $-(n^{2}+m^{2})$. The base eigenvalue is $\lambda=-2$, and this eigenspace has multiplicity $1$. The next is $\lambda=-5$, which is of multiplicity $2$.
It's common to have multiplicity $1$ for the base eigenvalue because the base eigenfunction typically cannot vanish inside.
Based on the comments, it's worth adding: \begin{align} \mbox{Multiplicity 2 (\lambda=-3):}&\; 1^2+2^2 = 2^2+1^2 \\ \mbox{Multiplicity 3 (\lambda=-50):}&\; 1^2+7^2 = 7^2+1^2 =5^2+5^2 \\ \mbox{Multiplicity 4 (\lambda=-65):}&\; 1^2+8^2 = 8^2+1^2 = 4^2+7^2 = 7^2+4^2 \end{align}
• @BigM I will guess that for Q1, it is generic that all eigenvalues are of multiplicity one, since somehow higher multplicities comes from some symmetry of $\Omega$. For example, if we consider $\Omega = [0,1] \times [0,r]$ for some $r\neq 1$, then all eigenvalues are simple. – user99914 Nov 19 '15 at 22:52
• @BigM : $1^{2}+7^{2}=7^{2}+1^{2}=5^{2}+5^{2}$. So in this problem we have multiplicity $1,2,3$. Are there others? Yes, $8^2+1^2 = 1^2+8^2=4^2+7^2=7^2+4^2$. So, multiplicity $1,2,3,4$. – DisintegratingByParts Nov 19 '15 at 23:01
• @TrialAndError : Note that everything are off by a $\pi^2$ now. – user99914 Nov 19 '15 at 23:12
• @JohnMa : I put the $\pi$ into the domain instead now. That's easier to fix. :) – DisintegratingByParts Nov 19 '15 at 23:15