Compare of a PDE and ODE

Question

On closed smooth convex manifold $M^n\subset \mathbb R^{n+1}$, $H(x,t)$ is mean curvature , $A$ is second fundamental form, satisfy $$ \partial_t H =\Delta H + |A|^2H - |A|^2 $$ Consider ODE $$ \frac{d}{dt} \varphi=\frac{\varphi^3 - \varphi^2}{n} ~~~~~~~~ \\ \varphi(0)=H_\min(\cdot , 0) >1 $$ How to prove $H(\cdot, t )\ge \varphi(t)$ for any $t\ge 0$ ? I try to use maximum principle , but fail.

Answer 1

Since $|A|^2 = \sum_i^n \kappa_i^2$ and $H = \sum_i^n \kappa_i$ we have $H^2 \le n |A|^2$. Thus we have $$|A|^2H - |A|^2 = |A|^2 (H-1) \ge \frac 1 n \left(H^3 - H^2\right)$$ whenever $H-1 \ge 0.$

Thus the comparison principle for parabolic PDE gives $H \ge \varphi$ for $\varphi$ any solution of $\varphi'(t) = \frac 1 n \left( \varphi^3 - \varphi^2 \right)$ with $\varphi(0) \le \inf H(\cdot,0)$, so long as $\varphi - 1 \ge 0.$

Since this inequality is true by assumption at the initial time, we now just need to show that it is preserved by the ODE. As $F(\varphi) = \frac 1 n \left( \varphi^3 - \varphi^2 \right)$ is locally Lipschitz near $1$ and $\varphi(t) = 1$ is a solution, the uniqueness theorem tells us that a solution that is initially $\ge 1$ cannot fail to be so at a later time.