A PDE Exercise using the Maximum Principle

I have an exercise from the PDE class:

Suppose u is a smooth solution to \left\{ \begin{aligned} Lu:= \Delta u + cu = \vert \nabla u \vert^2 \quad \text{in}\ \Omega, \\ u = 0 \quad \text{on}\ \partial \Omega, \end{aligned} \right. where $$\Omega \subset \mathbb{R}^n$$ is a open bounded region and $$c \le 0$$ is a constant. Show that $$u\equiv 0$$ in $$\Omega$$.

The professor talked about a version of Maximum Principle like this:

Let $$u \in C^2(\Omega) \cap C(\overline{\Omega})$$ satisfies $$Lu \ge 0$$ and $$c(x) \le 0$$ in $$\Omega$$. Then the nonnegative maximum of $$u$$ in $$\overline{\Omega}$$ can be attained only on $$\partial \Omega$$, unless $$u$$ is a constant.

My attempt:

First deal with the case of $$c < 0$$.

If $$u$$ is not constant on $$\Omega$$, by the Maximum Principle we must have $$u < 0$$ in $$\Omega$$. Take $$M = \inf_{x \in \Omega} u$$ and $$M<0$$. Since $$u$$ is smooth and $$\Omega$$ is bounded, we can assume $$u$$ attains the minimum at $$x_0 \in \Omega$$. Then by the equation we have $$\Delta u(x_0) = \vert \nabla u(x_0) \vert^2 - cu(x_0) = -cM < 0.$$ This implies that the Hessian $$D^2u$$ at $$x_0$$ must have a negative eigenvalue, so $$u(x_0)$$ cannot be the minimum inside $$\Omega$$.

However, the above argument does not stand in the case of $$c=0$$. Any hint will be appreciated!

For $$c=0$$, we have \left\{ \begin{aligned} \Delta u = \vert \nabla u \vert^2 \quad \text{in}\ \Omega, \\ u = 0 \quad \text{on}\ \partial \Omega, \end{aligned} \right. we will prove that $$u\geq 0$$ in $$\Omega$$.

Construct auxiliary function $$w=u+\epsilon(e^{Md}-e^{Mx_1}),$$ where $$d$$ is the maximum distance of the point in $$\Omega$$ with the original point，and $$M=\sup_{\Omega}\{\vert\nabla u\vert\}$$. Let $$\overrightarrow{A}=\nabla u$$.

Then, we have \left\{ \begin{aligned} \Delta w-\nabla u\cdot\nabla w = -\epsilon M^2e^{Mx_1}+\epsilon M A_1e^{Mx_1}\quad \text{in}\ \Omega, \\ w = \epsilon(e^{Md}-e^{Mx_1})\quad \text{on}\ \partial \Omega, \end{aligned} \right.

Suppose, $$\min w=w(x_0),x_0\ \text{in}\ \Omega$$, then $$\nabla w(x_0)=0,$$ and $$$$\Delta w\geq 0> \epsilon Me^{Mx_1}(A_1-M)$$$$ which is a contradiction.

Thus, $$w>0$$ and as $$\epsilon\rightarrow 0$$ we get $$u\geq 0$$.

Let $$v=-u$$, and we have \left\{ \begin{aligned} -\Delta u = \vert \nabla u \vert^2 \quad \text{in}\ \Omega, \\ u = 0 \quad \text{on}\ \partial \Omega, \end{aligned} \right. With the same auxiliary function we can derive that $$v\geq 0$$, and so $$u\leq 0$$.

Hence, $$u\equiv 0$$.

*The problem is a direct application of Strong maximum principle in Evans's book (P348, second edition).

• Thanks for the answer! A typo: the equation of $w$ in $\Omega$ should be $\Delta w - \nabla u \cdot \nabla w = \epsilon M e^{Mx_1}(A_1-M)$. (The $M$ was missed in the exponent. ) Jun 23, 2020 at 8:59