We are studying about the maximum principle this semester and the following is an exercise which has been assigned to us.
$Let\;u\in\;\mathcal C^2(D)\cap \mathcal C(\bar
D)\;where\;D=\{(x,y)\in \mathbb R^2 :\;x^2+y^2\lt
1\}.\;If\;u\;satisfies\;\\u_{xx}+u_{yy}=u^3\;\;on\;D\;and\;u=0\;on\;the\;boundary\;of\;D,\;prove\;that\;u\;is\;identically\;zero\;on\;D.$
Well,I tried to solve this by separating cases for the sign of $u$.
CASE 1 $\;\;\;\;u\le 0\;\forall (x,y)\in D\;$
Then $\Delta u \le 0\;\;$ and by the weak maximum principle for superharmonic functions we conclude $\min_{\bar D} u = min_{boundary\;of\;D} u=0$ $(1)$
CASE2$\;\;\;\;u\ge 0\;\forall (x,y)\in D$
Then $\Delta u\ge 0\;\;$ and by the weak maximum principle for subharmonic functions we conclude $\max_{\bar D} u = max_{boundary\;of\;D} u=0$ $(2)$
Now from $(1),(2)$ follows that $u$ is identically zero on $D$.
My question is if the above thought is correct.I have doubts about the separation because I am not 100% sure that $u$ preseves its sign on D.Could somebody fix this problem for me or give me some hints for the exercise?
I would appreciate any help! Thanks in advance
Best Answer
By continuity, there exists $x_0 \in D$ such that $u(x_0)=\max_{x\in\overline{D}}u(x)$.
If $x_0 \in \partial D$, then $\max u=0$. Hence $u\leq 0$. And you can apply what you did.
If $x_0 \in D$, then $\Delta u(x_0) \leq 0$, so that $u^3(x_0)\leq 0$, which givess $u(x_0)\leq 0$. Hence $u\leq 0$ and you can apply what you did.