Helmholtz solutions on compact domains

30 Views Asked by At

Consider two compact domains $\Omega_1$ and $\Omega_2$ in $\mathbb{R}^n$, such that $\partial \Omega_1 \cap \partial\Omega_2$ is a real analytic hypersurface. Suppose I have an eigenfunction $\varphi$ of the Dirichlet Laplacian on $\Omega_1$. If I understand this correctly, by using the Schwarz reflection principle, I can continue this eigenfunction $\varphi$ across $\partial \Omega_1 \cap \partial\Omega_2$ into $\Omega_2$, which then gives another uniquely defined Dirichlet eigenfunction on $\Omega_2$ by the unique continuation principle. My question is, if the volumes of $\Omega_1$ and $\Omega_2$ are comparable, are the $L^2$ norms of $\varphi|_{\Omega_1}$ and $\varphi|_{\Omega_2}$ comparable?

1

There are 1 best solutions below

0
On BEST ANSWER

No, you don't get an eigenfunction on $\Omega_2$ like that.

Let's take $\partial \Omega_1 \cap \partial\Omega_2$ to be a hyperplane for simplicity. Still, there are two problems:

  1. The extension is only defined on the mirror image of $\Omega_1$. It does not extend any further in general.
  2. If $\Omega_2 $ is properly contained in the mirror image of $\Omega_1$, then we do have an eigenfunction but the Dirichlet boundary condition need not hold on $\partial\Omega_2$.

So, the only case when you do get a Dirichlet eigenfunction is when $\Omega_2$ is precisely the mirror image of $\Omega_1$. Of course, then their volumes are equal and so are the $L^2$ norms of $\varphi$.