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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-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

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This seems more suitable for math.stackexchange.com. –  Deane Yang Feb 15 '11 at 21:50
    
Or, you may try yourself to solve this exercise using the explicit general form of a LB operator in terms of the metric, see e.g. the Wikipedia article –  Piero D'Ancona Feb 15 '11 at 22:48
    
OK, I asked math.stackexchange.com. Is this a silly question? It's not a homework exercise. I don't know much differential geometry, so I don't know how to thing about these things. –  nick maxwell Feb 16 '11 at 4:31
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It seems to me that it is a Laplace-Beltrami, up to a factor. For a metric $g:=c^{4/(n-2)}$, one has $\Delta_g=c^{-2n/(n-2)}{\rm div}(c^2\nabla)$. There remains a problem with the case $n=2$, presumably due to the rigidity of complex variable. A metric in the conformal class of $dx^2$ corresponds to a conformal transform, that is to a holomorphic change of variable. But this preserves the class of harmonic functions. Therefore a L-B operator of the form $a{\rm div}(b\nabla)$ has to be the usual Laplacian. –  Denis Serre Feb 16 '11 at 9:12
    
But what happens if $c^{2}$ is not analytic? Then it seems an argument based upon conformal transformation may not apply. –  drbobmeister Feb 16 '11 at 19:44

1 Answer 1

  1. As it was mentioned by @Denis Serre, for dimensions $\ne 2$, your operator is proportional to the Beltrami-Laplace operator. The proof is essentially in the comment of Denis Serre; one can also obtain it following suggestion of @Piero D'Ancona and writing this operator in local coordinates (instead of covariant derivative the usual derivatives appear by the price the determinant of the metric appears twice as factors).

  2. For nontrivial $c$, the operator in NOT the Beltrami-Laplace of any metric. I will explain it in the case your metric and $c$ are generic and dimension is $>2$, but it seems that the proof works without this assumption.

Proof of (2). Consider the symbol of your operator. It is a well-defined (2,0)-tensor and it is equal to $c^2 \cdot g^{ij}$. Would you operator be the Beltrami-Laplace of some $g'$, its symbol will be covariantly constant, and in the case of generic $g$ this would imply that the metric $g'$ is a constant factor of $\frac{1}{c^2}g_{ij}$ (I assume $n>2$). Thus, we have only one choice for $g'$. Calculating the Beltrami-Laplace for $g'$, we see that it does not coincide with your operator (if $c\ne \mathrm{const}$).

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