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I already asked this question at stackexchange with no response - so I'll try here.

I'm reading a paper on discrete differential geometry: Meyer et.al.

They define the Laplace-Beltrami operator at a point $P$ by $$\vec{K}(p) = 2k_H(P)\vec{n}(P)$$ where $\vec{n}(p)$ is the normal vector at $p$ and $k_H(p)$ the mean curvature. Then in Section 3.2 they give an error bound for the discrete Laplace-Beltrami operator obtained in the previous section.

At one step they write $||\vec{K}(x) - \vec{K}(x_i)||^2 \leq C_i^2||x-x_i||^2$ where $C_i$ is the Lipschitz constant of the Laplace-Beltrami operator.

Does anyone have a reference that proves that the Laplace-Beltrami operator is Lipschitz (at least for surfaces in $R^3$)?

Thank you.

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First, a comment on terminology. What you denote by $\vec K(p)$ seems to me to just be the vector version of the mean curvature (that's what I'm going to call it). This is a vector, not an operator. The "Laplace-Beltrami operator" is a generalization of the Laplacian on a Riemannian manifold (see http://en.wikipedia.org/wiki/Laplace%E2%80%93Beltrami_operator); it takes as input a function on the manifold and returns another one (and is thus called an operator). This is at least related to the mean curvature, in that probably applying the Laplace beltrami operator to the embedding $S\to\mathbb R^3$ would give the function $\vec K(p)$, but it's wrong to call $\vec K(p)$ the Laplace-Beltrami operator.

Now to answer your question. Let's call the surface $S\subseteq\mathbb R^3$. A function $K:S\to\mathbb R^3$ (like the mean curvature $\vec K(p)$) is called Lipschitz iff $C:=\sup_{x,y}\|K(x)-K(y)\|/\|x-y\|$ is finite. So, your equation is basically the definition of the mean curvature vector being Lipschitz.

For any given surface, the mean curvature may or may not be Lipschitz. It amounts to the existence of some higher derivatives of the embedding. In particular, if your surface is parameterized by a thrice-differentiable function (whose derivative is nonzero), then $K$ is Lipschitz. If not, the probably not, and there might not be a $C$ for which your inequality holds for all $x,y\in S$. For instance, on this surface, $K$ is not even continuous, let alone Lipschitz, even though the surface doesn't have any "corners".

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Thank you. May I ask if you have a reference to this? In particular, if $S^{n} \subseteq \mathbb{R}^{n+1}$ is a hypersurface - when is the mean curvature vector Lipschitz and when is it not? –  Magnus Botnan May 25 '11 at 14:32
    
On <a href="en.wikipedia.org/wiki/Mean_curvature">wikipedia</…; there is an explicit formula for the mean curvature vector for a surface of the form $z=S(x,y)$ (any surface can be put in this form at least locally once it is properly rotated). The formula only involves second derivatives of $S$, so if $S$ is thrice continuously differentiable, the mean curvature is continuously differentiable, and is thus Lipschitz (well, possibly only locally Lipschitz if your surface is non-compact). –  John Pardon May 25 '11 at 15:20
    
Ah, right, this was just about using the Lipschitz condition. Thank you for your answers and clarifying the inconvenient notation. –  Magnus Botnan May 25 '11 at 15:27
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