If $-\Delta u(x)+\partial _{x_1}u=0$ prove that $\sup_{\bar \Omega }u=\sup_{\partial \Omega }u$.

Question

Let $\Omega $ a smooth bounded open subset of $\mathbb R^d$ ($d\geq 2$) and let $u\in \mathcal C^2(\bar \Omega )$ the solution of the equation $$-\Delta u(x)+\partial _{x_1}u(x)=0\ \ \ in\ \ \Omega. $$

Prove that $$\sup_{\overline \Omega }u=\sup_{\partial \Omega }u.$$

My first idea was to construct $v$ s.t. $\Delta v=0$, and get $$\sup_{\overline{\Omega }}u=\sup_{\overline{\Omega }}v\underset{v\ harmonic}{=}\sup_{\Omega }v=\sup_{\Omega }u,$$ but I don't find such $v$. Maybe, there is an other way ?

My second idea was to remark that $\Delta u=div(\nabla u)$, and thus, using divergence theorem, $$\int_\Omega \partial _{x_1}u=\int_\Omega \Delta u=\int_\Omega div(\nabla u)=\int_{\partial \Omega }\partial u\cdot \nu,$$ but it still can't conclude.

Answer 1

Suppose that there is $x_0\in\Omega$ such that $\sup_{\bar{\Omega}}u=u(x_0)$. Since $u$ attains the maximum at $x_0\in\Omega$, then $\nabla u(x_0)=0$ and $H=\left(\frac{\partial^2u}{\partial x_i\partial x_j}(x_0)\right)$ is semi-negative definite. One can assume that $H$ is negative definite since otherwise one can use the perturbation $v=u+\epsilon e^{\alpha x_1}$ and send $\epsilon\to0$. Thus $$\partial_{x_1}u(x_0)=0,-\Delta u(x_0)=-\sum_{i=1}^n\frac{\partial^2u}{\partial x_i^2}(x_0)>0$$ which is contradictory to $-\Delta u(x_0)+\partial_{x_1}u(x_0)=0$. So $$\sup_{\overline \Omega }u=\sup_{\partial \Omega }u.$$

Answer 2

Assume $u(x_{0})=max \ u$, $x_{0} \in int \ \Omega$. Then you have $\partial_{x_1} u \leq 0 $. Plugging that into the equation you have $-\Delta u\leq 0 $ In other words, the hessian statisfies $Hu\geq0$ as $\Delta u=tr(Hu)$. But if we had a maximizum, $Hu<0$ would be true. A contradiction.