There is a theorem which states that if $f$ is an eigenfunction of the Dirichlet problem of the Laplacian: $$-\Delta f =\lambda f$$ $$f|_{\partial \Omega }=0$$ Then if $f$ has a single nodal domain (meaning, the set {$x\in \Omega|f(x)\neq 0$} has a single connected component), then the corresponding eigenvalue $\lambda$ is the first eigenvalue of the Laplacian, and it is a simple eigenvalue.
I was wondering if there is a similar result for the case where instead of the Dirichlet problem we're looking at the Neumann problem (where the normal derivative of $f$ vanishes at the boundary), and we replace the assumption that $f$ has a single nodal domain with the assumption that the derivative of $f$ does not vanish in $\Omega$ (only at the boundary). Can I say anything 'interesting' about the eigenfunction/eigenvalue in this case? The proof I know of the theorem above (which uses orthogonality of the eigenfunctions) cannot be applied for this case since $f$ does not have a constant sign in $\Omega$.
I know that in the Neumann case, the first eigenvalue must be $0$, so unless $f$ is constant (and we assume it's not, since it's derivative does not vanish) then the eigenvalue $\lambda$ will definitely not be the first eigenvalue. But maybe it has to be the second? Or maybe it still needs to be simple or something? If anyone knows of a relevant result (or an interesting counter example) - please let me know. I'm looking for anything which can be said about $f$ or $\lambda$ based on only the given assumptions. I'm interested mainly in results for manifolds and metric graphs, but feel free to share anything related.
Thanks a lot in advance.
This wouldn't need to be the second eigenfunction, nor simple. On a cylinder $\mathbb{S}^1 \times [0,b]$ for small $b$, the eigenfunction $\cos(\pi y/b)$ has eigenvalue $-\frac{4\pi^2}{b^2}$, so it appears arbitrarily far down the spectrum as $b \to 0$. On the 3D version $\mathbb{S}^1 \times [0,b]^2$, the eigenfunctions $\cos(\pi y/b)$ and $\cos(\pi z/b)$ have the same eigenvalue.