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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](http://en.wikipedia.org/wiki/Laplace%E2%80%93Beltrami_operator), 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.2011-02-18

2 Answers 2