Weak parabolic maximum principle on Riemannian manifolds

Question

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.

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.