Ran across this problem for a poisson equation. Looks like some sort ofaximum principle for the gradient. Can't quite wrap my head around it.
Let $\Omega$ be an open subset of $\mathbb{R}^n$. Suppose $u \in C^2(\overline{\Omega})$ is a solution of the equation $\Delta u = u^3$ with the property that $|\nabla u| \leq 1$ on $\partial \Omega$. Show $|\nabla u| \leq 1$ on all of $\Omega$.
Thank you for any help.
Edit: I will suppose $u\in C^3$.
It suffices to show that $v=|\nabla u|^2$ is subharmonic, or $\Delta v\ge 0$, since such functions satisfy the one-sided maximum principle $\sup_\Omega v\le \sup_{\partial\Omega}v$. We have $$ \Delta v=\partial_i\partial_i(u_ju_j)=2\partial_i(u_ju_{ij})=2u_{ij}u_{ij}+2u_ju_{iij}. $$ To deal with the last term, we differentiate the Poisson equation: $\Delta u_j=3u^2u_j$. Substituting this in, we get $$ \Delta v=2u_{ij}u_{ij}+6u^2u_ju_j=2|D^2u|^2+6u^2|\nabla u|^2, $$ which is clearly nonnegative.