3
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.