8

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

doraemonpaul
  • 16,178
  • 3
  • 31
  • 75
nick maxwell
  • 231
  • 3
    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

2 Answers2

2

Willie's hint in his comment is a bit misleading; your operator is equivalent to the Laplace-Beltrami operator with respect to some (conformally equivalent) metric only if $c$ is constant.

More generally, an elliptic operator $Lf = \operatorname{div} M \nabla f$ with $M$ self-adjoint and positive-definite is equivalent to the Laplacian of some new metric if and only if $\det M$ is constant. For more information see my answer to this related question: Generalized Laplace--Beltrami operators

Community
  • 1
user7530
  • 49,280
2

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.

Glen Wheeler
  • 2,925
  • 1
    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 08:25