MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


share|cite|improve this question
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 '12 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 '12 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 '12 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 '12 at 5:41
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 '12 at 11:13

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.

share|cite|improve this answer
This is for regular mean curvature flow, not volume preserving mean curvature flow. – Dorian Apr 7 '12 at 18:18
well since since the rescaled mean curvature flow (which is what Sesum studies) and the volume normalized mean curvature flow agree with each other to high order near a sphere (once you normalize correctly for the radius of the sphere) the results should be the same. In any case, you should be linearizing the flow near the sphere and not trying to look at the flow itself (since that is likely hopeless). – Rbega Apr 9 '12 at 16:39

Here is an explicit example, for the Ricci flow, with which one can compare general computations. For the MCF in dimension 1, an analogue is the Galaktionov--Angenent oval (a.k.a. paper clip). The King-Rosenau (a.k.a. sausage model) solution to the backward Ricci flow on $S^{2}$ can be viewed as the two-point compactification of $g_{\tau}(x)=\frac{\operatorname{s}% (\tau)(dx^{2}+d\theta^{2})}{\operatorname{c}(x)+\operatorname{c}(\tau)}$, $\tau\in(0,\infty)$, on the cylinder $\mathbb{R\times(R}/4\pi\mathbb{Z})$, where $\operatorname{s}=\sinh$ and $\operatorname{c}=\cosh$. Its area is $\operatorname{A}_{\tau}=8\pi\tau$ and its scalar curvature is $R_{g_{\tau} }(x)=\frac{\operatorname{c}(\tau)\operatorname{c}(x)+1}{\operatorname{s} (\tau)\left( \operatorname{c}(x)+\operatorname{c}(\tau)\right) }$, with average $r_{g_{\tau}}=\tau^{-1}$; so it satisfies $\frac{\partial} {\partial\tau}g_{\tau}=R_{g_{\tau}}g_{\tau}$. Using a table of integrals of rational functions of hyperbolic functions, I think we obtain $\frac{1}{4\pi }\int(R_{g_{\tau}}-r_{g_{\tau}})^{2}d\mu_{g_{\tau}}=\frac{\operatorname{c} (\tau)}{\operatorname{s}(\tau)}+\frac{\tau}{\operatorname{s}^{2}(\tau)} -\frac{2}{\tau}=\frac{2}{45}\tau^{3}+\operatorname{O}(\tau^{5})$ as $\tau\rightarrow0^{+}$. Given $a>0$, $h_{\tau}=ag\left( a^{-1}\tau\right) $ is also a solution. Then $\frac{1}{4\pi}\int(R_{h_{\tau}}-r_{h_{\tau}}% )^{2}d\mu_{h_{\tau}}=\frac{a^{-1}}{4\pi}\int(R_{g_{a^{-1}\tau}}-r_{g_{a^{-1} \tau}})^{2}d\mu_{g_{a^{-1}\tau}}=a^{-4}\frac{2}{45}\tau^{3}+\operatorname{O} (\tau^{5})$. For the corresponding solutions to the normalized backward Ricci flow, dilation by $a$ corresponds to time translation by $\ln a$. Let $\bar {g}_{\tau}=\tau^{-1}g_{\tau}$, so that $\overline{\operatorname{A}}_{\tau }=8\pi$ and $\bar{r}_{\tau}=1$; let $\bar{\tau}=\ln\tau$. Then we have the eternal solution $\frac{\partial}{\partial\bar{\tau}}\bar{g}_{\bar{\tau} }=(R_{\bar{g}_{\bar{\tau}}}-1)\bar{g}_{\bar{\tau}}$ and $$ \frac{1}{4\pi}\int(R_{\bar{g}_{\bar{\tau}}}-1)^{2}d\mu_{\bar{g}_{\bar{\tau}} }=\frac{e^{\bar{\tau}}\operatorname{c}(e^{\bar{\tau}})}{\operatorname{s} (e^{\bar{\tau}})}+\frac{e^{2\bar{\tau}}}{\operatorname{s}^{2}(e^{\bar{\tau}} )}-2=\frac{2}{45}e^{4\bar{\tau}}+\operatorname{O}(e^{6\bar{\tau}} )\quad\;\text{as }\bar{\tau}\rightarrow-\infty. $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.