I'm studying PDE and at the moment I'm reading L. Evans' book.

The strong maximum principle states that; if $u\in C^2(U)\cup C(\bar U)$ is harmonic in $U$, where $U$ is connected and if there exists $x_0$ such that $u(x_0)=\max _\bar U u$, then $u$ is constant within $U$.

A little later Evans states that if $U$ is connected and $u$ is a solution of the pde $\Delta u=0$ in $U$ and $u=g$ on $\partial U$, where $g\geq 0$. Then $u$ is positive in $U$ everywhere if $g$ is positive somewhere on $\partial U$.

Why is this true? I can't see how this follows immediately from the strong maximum principle. I'm guessing it is trivial and I'm just over thinking it. Can somebody help me?

  • $\begingroup$ It's actually easy to see using brownian motion on the domain $U$ but I imagine you want a "pde"s like answer. $\endgroup$ – muaddib Jun 4 '15 at 11:34

Suppose $u$ has a point where it is negative. Consider the associated problem where $\tilde{g} = -g$ on the boundary. Then $\tilde{u} = -u$ solves Laplace's equation with those new boundary conditions. So $\tilde{u}$ is everywhere non-positive on the boundary but has point in the interior of $U$ where it is positive. That violates the maximum principle.

  • $\begingroup$ What if we changed the problem so that we have $\Delta u = g$ in $U$ and $u = 0$ on $\partial U$, where $g \ge 0$? $\endgroup$ – nukeguy Feb 17 '17 at 15:30
  • 1
    $\begingroup$ So $u\ge 0$, right? But the statement says $u\gt 0$ in $U$. $\endgroup$ – Shivering Soldier Sep 30 '19 at 11:01

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