1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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: $$ \begin{align} \mbox{Multiplicity 2 ($\lambda=-3$):}&\; 1^2+2^2 = 2^2+1^2 \\ \mbox{Multiplicity 3 ($\lambda=-50$):}&\; 1^2+7^2 = 7^2+1^2 =5^2+5^2 \\ \mbox{Multiplicity 4 ($\lambda=-65$):}&\; 1^2+8^2 = 8^2+1^2 = 4^2+7^2 = 7^2+4^2 \end{align} $$

$\endgroup$
  • 2
    $\begingroup$ @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. $\endgroup$ – user99914 Nov 19 '15 at 22:52
  • 2
    $\begingroup$ @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 $1,2,3,4$. $\endgroup$ – DisintegratingByParts Nov 19 '15 at 23:01
  • 1
    $\begingroup$ @TrialAndError : Note that everything are off by a $\pi^2$ now. $\endgroup$ – user99914 Nov 19 '15 at 23:12
  • 1
    $\begingroup$ @JohnMa : I put the $\pi$ into the domain instead now. That's easier to fix. :) $\endgroup$ – DisintegratingByParts Nov 19 '15 at 23:15
  • 1
    $\begingroup$ @BigM : That's for a manifold without boundary and of genus 0, which is not what we have here. Very interesting reference. $\endgroup$ – DisintegratingByParts Nov 20 '15 at 0:22

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.