MHB Why can't $u$ achieve its maximum in the interior of $\Omega$?

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In case users haven't read the announcement, Ackbach has stepped down as POTW director, and I'll be taking his place. Here is this week's POTW.

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Let $\Omega$ be a bounded domain in $\Bbb R^2$. Suppose $u$ is a nonconstant, nonnegative solution of the PDE $\Delta u = mu$ in $\Omega$ where $m : \Omega \to (0,\infty)$ is continuous. Prove that $u$ cannot achieve its maximum in the interior of $\Omega$.
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This week's problem was solved correctly by Janssens. You can read his solution below.

Spoiler

Suppose $\mathbf{x}_0$ is an interior point of $\Omega$ such that $u_0 := u(\mathbf{x}_0) \ge u(\mathbf{x})$ for all $\mathbf{x} \in \Omega$. If $u_0 = 0$ then the nonnegativity of $u$ implies that $u$ is constantly zero in $\Omega$, which is a contradiction. So, we may assume that $u_0 > 0$ strictly. By the (single-variable) second derivative test, we have $D_i^2 u(\mathbf{x}_0) \le 0$ where $i = 1,2$ indicates the variable. This implies that
$$
\Delta u(\mathbf{x}_0) = D_1^2 u(\mathbf{x}_0) + D_2^2 u(\mathbf{x}_0) \le 0.
$$
On the other hand,
$$
\Delta u(\mathbf{x}_0) = m(\mathbf{x}_0) u_0 > 0,
$$
a contradiction. This shows that $\mathbf{x}_0$ does not exist, so $u$ does not achieve its maximum in the interior of $\Omega$.