Prove that if $\phi>0$ on $\partial\Omega$ then $u(x)\geq0$

Question

Let $\Omega\subset R^n$ be a bounded regular domain. Consider a non linear boundary value problem with $u\in C^1(\Omega)$

$\left\{\begin{matrix} -\Delta u +\kappa_{(u>0)}=0\\ u=\phi \text{ on }\partial\Omega \end{matrix}\right.$

Where
$\kappa_{(u>0)}(x)=$$\left\{\begin{matrix} 1 \text{ if }u(x)>0 \\ 0 \text{ if } u(x)\leq0 \end{matrix}\right.$

Prove that if $\phi>0$ on $\partial\Omega$ then $u(x)\geq0$.

My Attempt:
Suppose $\phi>0$ on $\partial\Omega$ and to the contrary assume that $\exists x_0\in\Omega$ such that $u(x_0)<0$.
Then since $u$ is continuous there is an open ball $B(x_0,r)\subset\Omega$ such that $\forall x\in B(x_0,r), $ $u(x)<0$.
Then for this Ball $\kappa(x)=0$.
Thus $\Delta u=0$ which is the Laplace equation.

Therefore the minimum for $u$ on the ball should occur at the boundary. (Property of a Laplace function)

After this I don't see how to proceed. If I can say that at the boundary of the ball ($B(x_0,r)$) u must be positive then there is a contradiction. But we only know the values of $u$ on $\partial\Omega$

May be there is a completely a different way of doing this problem... Appreciate your help.

Answer 1

Since $u$ is continuous on $\overline\Omega$, we assume that the minimum of $u$ occurs at $x_0$. If $u(x_0)<0$, then $x_0 \in \Omega$ since $\phi >0$. Then $u<0$ in an open neighborhood of $x_0$. Thus $u$ satisfies $\Delta u =0$. The maximum principle implies that $u$ is constant in that neighborhood.

The above argument can be used to show that $u(x) = u(x_0)$ on the connected component $U$ of $\{ u <0\}$ in $\Omega$ containing $x_0$ (Note that $\phi >0$ so $\{ u<0\}$ is actually in $\Omega$). Since at the boundary of $U$ one must have $u=0$ (See the remark). This is impossible since $u$ is continuous. Thus $u(x_0)<0$ is impossible.

Remark Let $x\in \partial U$. Then $x\notin U$ since $U$ is open. Thus $U(x) \ge 0$. Next, since $x\in \partial U$, there is a sequence $x_n$ in $U$ converging to $x$. Thus $$ u(x) = \lim_n\to \infty u(x_n), $$ which implies that $u(x) \le 0$ since $u(x_n) <0$ for all $n$. Thus $u(x) = 0$.

Answer 2

If we assume that there exists $x_0\in \Omega$ such that $u(x_0) < 0$, then using the continuity of $u$ we can define the following open set $\tilde{\Omega} = \{x\in \Omega: u(x) < 0\}$. Since $u = \phi > 0$ on $\partial\Omega$ then we know that $\overline{\tilde{\Omega}}\subset \Omega$. Well $\overline{\tilde{\Omega}} = \{x\in\Omega: u(x)\leq 0\}$. Since $u$ must satisfy our original PDE, then we have that $-\Delta u + \kappa_{\{u > 0\}} = 0$ must hold on $\tilde{\Omega}$. But $u < 0$ on $\tilde{\Omega}$ so the original PDE simplifies to $\Delta u = 0$ on $\tilde{\Omega}$. Now on $\partial\tilde{\Omega}$ we have that $u = 0$. We can now define a new PDE that $u$ must satisfy, namely $$\Delta u = 0\hspace{8mm} \text{on}\hspace{4mm}\tilde{\Omega} $$ $$u = 0\hspace{4mm}\text{on} \hspace{4mm}\partial\tilde{\Omega}$$ We apply a consequence of the weak maximum principle to see that $|u|$ on $\overline{\tilde{\Omega}}$ is maximized on $\partial \tilde{\Omega}$. Hence $u = 0$ on $\tilde{\Omega}$ which contradicts the definition of the set. Hence $\tilde{\Omega} = \emptyset$ and we conclude that $u \geq 0$