Is the closure of the orbits of the mean curvature flow compact for a finite time? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T05:07:54Z http://mathoverflow.net/feeds/question/115353 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115353/is-the-closure-of-the-orbits-of-the-mean-curvature-flow-compact-for-a-finite-ti Is the closure of the orbits of the mean curvature flow compact for a finite time? Hengyu Zhou 2012-12-04T03:41:05Z 2012-12-07T23:50:32Z <p>Question: Suppose $N$ is an $n$ dimensional immersed submanifold in a complete, non-compact manifold $Ｍ$， with bounded geometry. Suppose the mean curvature flow $F:Ｎ\times [0,q)\rightarrow M$ satisfies $$\begin{split} \frac{d}{dt}F&amp;=\vec{H};\ F(p,0)&amp;=F_{0}; \end{split}$$ If $F(p,t)$ exists for $t\in [0,q)$, is there any possible the closure of $F(N\times[0,t))$ unbounded in $M$. In other words, can the mean curvature flow go to infinity for a finite time? I didn't find the contra_example. I hope the answer is yes. In Euclidean space, definitely this is ture? Is any one know any related reference about this question? You are very appreciated if you can offer some opinions or information. </p> http://mathoverflow.net/questions/115353/is-the-closure-of-the-orbits-of-the-mean-curvature-flow-compact-for-a-finite-ti/115357#115357 Answer by Anton Petrunin for Is the closure of the orbits of the mean curvature flow compact for a finite time? Anton Petrunin 2012-12-04T05:02:07Z 2012-12-07T23:47:56Z <p>I guess you assume that $F_0$ is bounded. In this case the answer is "yes".</p> <p>Fix a point $p$. Fix small $\varepsilon>0$ so that principle curvatures of $\varepsilon$ spheres in $M$ are uniformly bounded. (It is possible since $M$ has bounded geometry.)</p> <p>Denote by $S_r$ the sphere with center $p$ and radius $r$ and let $\Sigma_r$ be the inward $\varepsilon$-equidistant to $S_r$. </p> <p>Note that for $r>2\cdot\varepsilon$ there is a fixed upper bound for principle curvatures of $\Sigma_r$. Therefore $\ell(t)=\max_{x\in F_t}\{\mathop{\rm dist}_px\}$ grows at most linearly. </p> <p><strong>P.S.</strong> I used that $M$ has bounded geometry in an essential way, but I am not sure if it is necessary.</p> http://mathoverflow.net/questions/115353/is-the-closure-of-the-orbits-of-the-mean-curvature-flow-compact-for-a-finite-ti/115661#115661 Answer by Hengyu Zhou for Is the closure of the orbits of the mean curvature flow compact for a finite time? Hengyu Zhou 2012-12-07T01:13:21Z 2012-12-07T01:13:21Z <p>Thanks Anton. I got you. Your proof is right. </p>