Estimating laplace-beltrami spectra for a graph surface in $R^3$ - MathOverflow most recent 30 from http://mathoverflow.net2013-06-19T11:07:32Zhttp://mathoverflow.net/feeds/question/58242http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/58242/estimating-laplace-beltrami-spectra-for-a-graph-surface-in-r3Estimating laplace-beltrami spectra for a graph surface in $R^3$RadonNikodym2011-03-12T08:19:50Z2011-07-31T00:12:02Z
<p>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.</p>
<p>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.</p>
<p>Is anyone aware of any references, or has had any experience with this problem?</p>
http://mathoverflow.net/questions/58242/estimating-laplace-beltrami-spectra-for-a-graph-surface-in-r3/58381#58381Answer by Helge for Estimating laplace-beltrami spectra for a graph surface in $R^3$Helge2011-03-14T03:41:06Z2011-03-14T19:39:25Z<p>Some suggestions:</p>
<ol>
<li><p>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.</p></li>
<li><p>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.</p>
<p>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.</p></li>
<li><p>There is what is called "Bohr-Sommerfeld conjecture", which is related to higher eigenvalues.</p></li>
</ol>