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!


1 Answer 1


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).

  • $\begingroup$ 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. ) $\endgroup$
    – Zhixuan Li
    Jun 23, 2020 at 8:59

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