I'm studying by myself Mean Curvature Flow and I'm reading the paper "Interior estimates for hypersurfaces moving by mean curvature" by Klaus Ecker and Gerhard Huisken, specifically, I'm reading the following theorem:

$\textbf{Theorem 2.1}$ Let $R > 0$ and $\textbf{x}_0 \in R^{n+1}$ be arbitrary and define $\varphi(\textbf{x},t)= R^2 - |\textbf{x} - \textbf{x}_0| - 2nt$. If $\varphi_+$ denotes the positive part of $\varphi$ we have the estimate

$$v(\textbf{x},t) \varphi_+(\textbf{x},t) \leq \sup_{M_0} v \varphi_+$$

as long as $v(x,t)$ is defined everywhere on the support $\varphi_+$.

The authors denote $\textbf{x} := F(p,t)$ for a solution of the MCF, $\textbf{x}_0 := F(p,0) = M_0$ and define a gradient function $v := \left( \langle \nu, \omega \rangle \right)^{-1}$, where $\nu$ is the unit normal vector of the hypersurface and $\omega$ is some fixed vector such that $\langle \nu, \omega \rangle > 0$ and $|\omega| = 1$. They defined in the proof a function $\eta := \left( R^2 - r \right)^2$, where $r := |\textbf{x}| - 2nt$ (they assumed w.l.o.g. that $\textbf{x}_0 = 0$) and they got

$$\left( \frac{d}{dt} - \triangle \right) v^2 \eta \leq -12 v \nabla v \cdot \nabla \eta + \eta^{-1} \nabla \eta \cdot \nabla (v^2 \eta)$$

They stated

If we replace $\eta$ by $(\varphi_+)^2$ this computation remains valid on the support of $\varphi_+$ as long as $v$ is defined. The weak parabolic maximum principle then implies the result.

I would like to know what is this weak parabolic maximum principle.

This is what I thought about my question:

Initially, I thought that could be the weak parabolic maximum principle on $\mathbb{R}^n$ which is common to see in PDE courses, but the problem is that I'm working with a manifold which receive a local treatment, then I thought that the weak parabolic maximum principle on $\mathbb{R}^n$ could be extended to a manifold, but I couldn't extend the result.

I found a maximum principle applied on MCF in this lecture notes (it's the theorem $2.2.1$ on page $17$), but I couldn't see how this helps me to conclude the inequality of the theorem $2.1$ of the paper.

  • 1
    $\begingroup$ In my answer I gave a short proof of the parabolic maximum principle for subsolutions of a Dirichlet--Cauchy problem. This is far from being as general as possible. Lieberman has more maximum principles, which, as a rule of thumb, always carry over to manifolds whenever the manifold in question is compact. For more maximum principles, see Protter--Weinberger or Volume 2 of Chow et at.'s Ricci flow saga. $\endgroup$
    – Ryan Unger
    Oct 22, 2018 at 23:33
  • $\begingroup$ Firstly, I would like to thank you for the patience to answer my question with details and indicate some reference for parabolic PDEs. Secondly, I have a doubt about why this maximum principle is valid for manifolds. I read the lemma 2.3 of Lieberman's book (indeed the chapter 1) and he assumed that $\Omega$ is a domain seen as a subset of $\mathbb{R}^{n+1}$. I think when you said "maximum principles, which, as a rule of thumb, always carry over to manifolds whenever the manifold in question is compact." you mean that the compactness of $M$ imply that $M$ is closed and bounded, then $\endgroup$
    – George
    Oct 25, 2018 at 3:28
  • $\begingroup$ $M$ is a domain, but $M$ is not a subset of $\mathbb{R}^{n+1}$ at first. Are you assuming w.l.o.g. that $M \subset \mathbb{R}^{n+1}$ by some embedding's theorem like Whitney's theorem? $\endgroup$
    – George
    Oct 25, 2018 at 3:29
  • $\begingroup$ Lieberman works on spacetime domains in $\Bbb R^{n+1}$, sure. My claim was not that you can quote his result, my claim is that the proof works for the more general case. What you need, exactly, is for $\Omega$ to have compact closure in $M\times [0,\infty)$. If $M$ is compact and you're only interested in finite times, then you get this for free, of course. $\endgroup$
    – Ryan Unger
    Oct 25, 2018 at 3:43
  • $\begingroup$ Although it's not hard to extend the maximum principle for all time by just taking $M\times [0,T]$ and letting $T\to\infty$...compactness of the spatial part is the crucial thing. Though in Chow's book there are maximum principles for complete noncompact manifolds. $\endgroup$
    – Ryan Unger
    Oct 25, 2018 at 3:45

1 Answer 1


Firstly, you have the wrong inequality (there is a small typo in the paper). Young's inequality is typically written for nonnegative numbers, but for any $a,b\in\Bbb R$ we have \begin{align*} -ab&\le |ab|\\ &\le \frac{1}{2}a^2+\frac{1}{2}b^2. \end{align*} In the context of the Ecker--Huisken, \begin{align*} -6v \nabla v\cdot\nabla\eta&\le |6v\nabla v\cdot\nabla\eta|\\ &\le 6|\nabla v|^2\eta+6|\nabla |\mathbf x|^2|^2v^2. \end{align*} When rearranged, $$ -6|\nabla v|^2\eta-6|\nabla |\mathbf x|^2|^2v^2 \le 6v \nabla v\cdot\nabla\eta.$$ We therefore have $$(\partial_t-\Delta)v^2\eta\le \eta^{-1}\nabla\eta\cdot\nabla(v^2\eta).$$

Here is the kind of theorem you need now.

Weak Maximum Principle. Let $M$ be a smooth manifold and consider a linear parabolic operator $$Lu=\partial_tu-a^{ij}(X)\nabla_{ij}u+b^i(X)\nabla_iu$$ on a bounded domain $\Omega\subset M\times[0,\infty)$. If $u\in C^{1,2}(\Omega)\cap C^0(\overline\Omega)$, $Lu\le 0$ in $\Omega$ and if $u\le 0$ on $\mathcal P\Omega$ (parabolic boundary), then $u\le 0$ in $\Omega$.

Proof. The same proof as Lieberman Lemma 2.3 works here. Here is the idea: Let $w=e^{-t}u$. Then $w$ and $u$ have the same sign and $$\partial_tw=e^{-t}(\partial_tu-u).$$ Substituting $$\partial_tu\le a\cdot\nabla^2u+b\cdot \nabla u$$ gives $$\partial_tw\le a\cdot\nabla^2 w+b\cdot\nabla w-w.$$ Let $X=(x,t)$ be a point where $w$ assumes a positive maximum. By hypothesis, $X\in \overline\Omega\setminus\mathcal P\Omega$. Then $a\cdot\nabla^2w(X)\le 0$ (here we use the fact that $a$ is a symmetric, positive definite matrix), $\nabla w(X)=0$, and $\partial_tw(X)\ge 0$. This leads to a contradiction if $X\in \Omega$. Therefore $X\in \partial\Omega\setminus\mathcal P\Omega$. This is a bit more technical, but morally if $w$ is positive somewhere on this set then for some time slice of $\Omega$ close to the "top," there will be a point $X^*$ in the slice which is a spatial maximum and such that $\partial_t w(X^*)\ge 0$. In this case we get a contradiction as well. See Lieberman for details.

Now we define $u=(v\varphi_+)^2-(v\varphi_+)^2(0).$ We consider $\Omega$ to be the set of spacetime points $X$ satisfying $\varphi(X)>0$. Then $\Omega$ is actually a cone and $\mathcal P\Omega$ is just $\{\varphi(X)=0\}\cup \{\Omega\cap \{t=0\}\}$. Clearly $u\le 0$ here, so the WMP gives $$u\le 0\quad\text{on }\Omega.$$ Therefore, $$v\varphi_+(X)\le \sup_{x\in \Omega\cap \{t=0\}}v\varphi_+(x,0)\quad \text{for }X\in\Omega.$$ Using the fact that $\varphi_+=0$ outside of $\Omega$, this gives the desired inequality.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .