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!
 A: For $c=0$, we have
\begin{equation}
\left\{ 
\begin{aligned}
\Delta u  = \vert \nabla u \vert^2 \quad \text{in}\ \Omega, \\
u = 0 \quad \text{on}\ \partial \Omega, 
\end{aligned}
\right.
\end{equation}
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
\begin{equation}
\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.
\end{equation}
Suppose, $\min w=w(x_0),x_0\ \text{in}\ \Omega$,
then $\nabla w(x_0)=0,$ and
\begin{equation}
\Delta w\geq 0> \epsilon Me^{Mx_1}(A_1-M)
\end{equation}
which is a contradiction.
Thus, $w>0$ and as $\epsilon\rightarrow 0$ we get $u\geq 0$.
Let $v=-u$, and we have
\begin{equation}\left\{ 
\begin{aligned}
-\Delta u  = \vert \nabla u \vert^2 \quad \text{in}\ \Omega, \\
u = 0 \quad \text{on}\ \partial \Omega, 
\end{aligned}
\right.
\end{equation}
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).
