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Given a domain $\Omega \subset \mathbb{R}^d$ which is convex and smooth and $| \Omega|=1$, it is well known that the metric converges exponentially fast to that of the sphere under volume preserving MCF. I would like to know the following:

Question: Does the rate of convergence depend on the domain $\Omega$? If so, in what way? If I know in particular that $\frac{d}{dt} \|\kappa - \bar \kappa\|_{L^2(\partial \Omega)}^2 \leq -C \|\kappa - \bar \kappa\|_{L^2(\partial \Omega)}^2$ , can I say that $C$ is independent of $\Omega$? If not, can I see in which explicit way it does depend on $\Omega$? I suppose this is equivalent to asking if there is a particular $\Omega$ so that the convergence is slowest.

Thanks.

edited Apr 4 2012 at 11:32

Willie Wong
6,987●20●43

asked Mar 1 2012 at 18:25

Dorian
1,687●3●18

For the linearization of MCF, you will get a function $f\colon \mathbb S^2\to \mathbb R$ which has slowest convergence to zero (probably something like the restriction of $x_1^2-\oint_{\mathbb S^2}x_1^2$ to $\mathbb S^2$). It will be easy to find the constant for this $f$ and this should be the constant for MCF. The domain $\Omega$ might have a faster convergence if it has some symmetry which the function $f$ does not have. – Anton Petrunin Mar 1 2012 at 19:01

So you mean for fixed volume $|\Omega|=1$, this function should give the optimal constant? It doesn't get much worse if you blow up the perimeter or something? – Dorian Mar 1 2012 at 21:05

I am not sure, but it is easy to check. Also, I have no proof, it is only my feeling. – Anton Petrunin Mar 2 2012 at 4:51

Well when you do the calculation, one has: $d/dt \|\kappa - \bar \kappa\|_{L^2}^2 = \int_{\partial \Omega} (\kappa - \bar \kappa) \Delta \kappa = - \int_{\partial \Omega} |\nabla \kappa|^2$ since it seems that $\partial_t \kappa = \nabla \kappa$ for mean curvature flow with an area form missing from my calculations. Combining this with an elliptic estimate and Sobolev inequality, one obtains the exponential convergence. However this constant is precisely some sort of Sobolev constant for the manifold it would seem, and I'm not sure if that is the same as the constant for the convergence rate. – Dorian Mar 2 2012 at 5:41

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Looks to me like the convergence rate can be estimated using the first nonzero eigenvalue of the Laplacian (which as you say can be estimated using the Sobolev constant). This constant certainly changes as the domain changes, but there is probably a bound on the eigenvalue in terms of the mean curvature of the hypersurface. I suggest digging around for this using google or Mathscinet. – Deane Yang Mar 2 2012 at 11:13

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Rbega · Answer 1 · 2012-04-04 16:13:41Z UTC

This is answered in a paper of N. Sesum

See here for Mathscinet review.

Basically since knowing something about the rate only matters at large scales you can assume the flow is near a sphere (by Huisken's result). Then the rate just depends on spectral properties of the Laplacian on the sphere. In particular, the rate is at worst the gap between (if I recall correctly) the first non-trivial eigenvalue and the second non-trivial eigenvalue. This is sharp as one can deform the sphere a small amount in the normal direction by an amount given by the eigenfunction associated to the second non-trivial eigenvalue.

What is the constant in the rate of exponential convergence for mean curvature flow?

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