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](https://math.stackexchange.com/questions/3944824/laplace-eigenfunction-on-a-polygonal-domain-symmetric-about-an-axis), but not getting any comments there, and I am now thinking that this question is probably more suited to MO. Please excuse the repetition.



Laplace eigenfunction on a polygonal domain symmetric about an axis
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).


