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I asked this on mathoverflow, and was suggested to ask it here.

Someone mentioned, in passing, to me that $u \mapsto \nabla \cdot ( c^2 \nabla u)$ is a Laplace-Beltrami operator. Does anyone have some insight into this? From my understanding, the Laplace-Beltrami operator generalizes the Laplacian to Riemannian manifolds, by taking the trace of the Hessian. I don't really see the connection to that and the above operator, unless c=1.

This operator comes from the wave equation, where $\partial^2_t u -\nabla \cdot ( c^2 \nabla u) = f$. There may or not be some smoothness conditions on $c$, and all these are functions on subsets of $\mathbb{R}^n$.

Thanks for any help, -Nick

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You may benefit from reading about the Laplace-Beltrami operator here, paying special attention to the form in local coordinates $\triangle_g f = (\det g)^{-1/2} \partial_i (\det g)^{1/2} g^{ij} \partial_j f$. In particular, think about the case where your Riemannian metric is conformal to the Euclidean one. –  Willie Wong Feb 18 '11 at 10:21

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The Laplace-Beltrami operator, as far as I am aware, is `unique'. Perhaps you mean that this operator is a Laplacian---that is certainly true. The idea is that it is still a divergence, and thus (with some assumptions on $c$) hopefully elliptic.

The most common generalisation of the Laplacian in this direction is the $p$-Laplacian, on which you can learn more here: http://en.wikipedia.org/wiki/P-Laplacian. A simple google search for "p laplacian" also turns up plenty of elementary material.

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in the original question the OP stated that $c$ is some function on $\mathbb{R}^n$, and not a function of $u$ and its derivatives. So I don't think the p-laplacian is quite what he is looking for. –  Willie Wong Feb 16 '11 at 23:12
    
@Willie: maybe not (since he didn't reply, we don't know). but the p-Laplacian is the first thing I thought of when I read the post, and figured it was possibly what he heard from "someone". –  Glen Wheeler Feb 18 '11 at 8:25

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