# Laplace eigenfunction on a polygonal domain symmetric about an axis

Consider a polygon $$\Omega \subseteq \mathbb{R}^2$$, and let us consider the usual Laplacian operator $$\Delta = \partial_x^2 + \partial_y^2$$, with Dirichlet boundary conditions. My question comes from trying to understand certain things in Chavel's book "Eigenvalues in Riemannian geometry". Suppose $$\phi_2$$ is a Dirichlet eigenfunction corresponding to the second eigenvalue $$\lambda_2$$ of $$\Omega$$. Now, let us reflect $$\Omega$$ about one of its sides, resulting in a polygon $$\Omega^*$$ that is symmetric about some axis. The extension of $$\phi_2$$ via reflection, which we call $$\phi_2^*$$, is clearly still a Dirichlet eigenfunction on $$\Omega^*$$.

My question: is there any way to understand what Dirichlet eigenvalue of $$\Omega^*$$ the extension $$\phi^*_2$$ corresponds to? My guess is that it should one of the low eigenvalues, but cannot prove this rigorously. Does it help to know whether $$\lambda_2$$ is non-repeated or not? Any help will be appreciated, thanks!

Remark: I asked this on MSE, but not getting any comments there, and I am now thinking that this question is probably more suited to MO. Please excuse the repetition.

• The reflection gives an isometric $C_2$-action on $\Omega^*$, so that the Dirichlet spectrum of its Laplacian decomposes into even and odd eigenfunctions. It's easy to see that restricting to $\Omega$ identifies odd eigenfunctions with Dirichlet eigenfunctions on $\Omega$, while even eigenfunctions correspond to von Neumann boundary conditions on the special side and Dirichlet boundary elsewhere. So you would have to bound the spectrum of the latter operator via that of the former. Dec 12, 2020 at 8:47

There is no uniform bound, independent of the domain. Consider a rectangle with side lengths $$\pi, \pi/K$$, with $$K\gg 1$$. The eigenvalues are $$m^2+n^2K^2$$, $$m,n\ge 1$$. In particular, $$\lambda_2$$ is obtained for $$m=2$$, $$n=1$$.
If we now reflect about the long side, then the eigenvalues of the doubled rectangle are $$m^2+n^2K^2/4$$, and $$2^2+K^2$$ is no longer a low eigenvalue (it is about the $$K$$th eigenvalue).