# Estimating laplace-beltrami spectra for a graph surface in $R^3$

Consider a surface $\Gamma$ in $R^3$. The surface $\Gamma$ is a graph, i.e. $\Gamma = (x,y, h(x,y))$, for $x \in R^2$ and some smooth function $h$, where $h$ and all its derivatives are periodic on [0,1]^2.

I'm trying to estimate the eigenfunctions and eigenvalues of the Laplace-Beltrami operator for this surface. Of course, the laplace beltrami operator can be very easily be expressed in local coordinates on $[0,1]^2$, and so the problem is effectively approximating the spectra of this second order elliptic operator on $L^2[0,1]^2$ with periodic boundary conditions. This is what I am doing, however I still cannot obtain any useful estimates.

Is anyone aware of any references, or has had any experience with this problem?

• Are you interested in the low lying eigenvalues or the asymptotics of the eigenvalues? The second one should follow like Weyl's law .. – Helge Mar 12 '11 at 18:18
• Hi, thanks for your answer. It's mainly the low lying eigenvalues. Even a spectral gap result would be a great start. – RadonNikodym Mar 13 '11 at 9:33
• Are you not interested also in the spectrum of the non-compact surface $\Gamma$. Periodic BC gives the (discrete) spectrum of the (non flat) torus $\Gamma/Z^2$. – BS. Mar 13 '11 at 13:16
• @BS: There is more than one operator involved for the non-compact case. It's still "estimating eigenvalues". – Helge Mar 14 '11 at 3:41
• I'm interested in the discrete spectrum and the corresponding eigenfunctions. My ultimate aim is to estimate the H1 inner product of a function ($x\cdot k$, where $k$ is a constant vector) against the eigenfunctions. – RadonNikodym Mar 14 '11 at 12:23

Some suggestions:

1. To show there is a spectral gap, one can use the following approach: The variational characterization of the first eigenvalue $$E_1 = \inf_{\|\psi\| = 1} \langle \psi, H \psi \rangle$$ can be used to obtain an upper bound on the first eigenvalue. Then Temple's inequality (I think it's in Reed-Simon IV) can be used to obtain a lower bound on the second eigenvalue. Note in order to show the existence of a spectral gap, one needs to study all the operators $H(k)$ of the Floquet-Bloch decomposition.

2. But I would guess that at least as long as $h$ is "sufficiently small", there is no spectral gap (without having done any of the computations). Denote by $E_1(k)$ and $E_2(k)$ the first and second eigenvalue of $H_0(k)$. Here $H_0$ denotes the usual Laplacian and the Floquet--Bloch decomposition is done with respect to $[0,1]^2$. I believe that one has that $$\sup_{k} E_1(k) > \inf_{k} E_2(k).$$ Then using that your operator will be a small perturbation of the Laplacian (in an appropriate sense), one should obtain that there is no spectral gap.

2b. (added in edit) It is clear that for fixed $k$, one has $E_1(k) < E_2(k)$. This follows from the first eigenfunction being positive.

3. There is what is called "Bohr-Sommerfeld conjecture", which is related to higher eigenvalues.

• The first eigenvalue will always be zero, with corresponding eigenfunction being constant. By spectral gap I mean the gap between first and second eigenvalues. Numerically I have computed the spectrum of my operator for various $h$ and have observed that a spectral gap exists and appears to be inversely proportional to the surface area of my surface (over [0,1]^2), though I can't prove this. Temple's inequality seems to be a useful tool for this type of problem. Thanks for telling me about it. I admit I know nothing about Floquet-Bloch decompositions, I'm reading about this now. Thanks. – RadonNikodym Mar 14 '11 at 12:38
• The first eigenvalue of $H_0(k)$ is not always $0$ nor is the eigenfunction constant. Look up Floquet-Bloch decomposition. You need to consider multiple operators. – Helge Mar 14 '11 at 19:20
• I am reading about the Floquet-Bloch decomposition and I am realising that rather that do all the estimates above, I simply need to compute the Hessian of the first bloch eigenvalue. I suspect this might be just as hard as what I was doing previously, but I will attempt it. – RadonNikodym Mar 15 '11 at 10:43