# Maximum Principle fot the pde : $\Delta u=u^3$

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

• Another way to do it is to rewrite the PDE as $u^4 + (\nabla u)^2 = \nabla\cdot(u\nabla u)$ and integrate this over $D$. The right hand side integral is $0$ by the divergence theorem since $u=0$ on $\partial D$. Since the left hand side is strictly positive it must be identical to zero in $D$. – Winther Nov 17 '16 at 18:24
• @Winther this is something I couldn't think of. I'll try this one too. Thank you for the idea! – kaithkolesidou Nov 17 '16 at 18:28

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.
• Just to make clear that I completely understood your answer, by doing this I derive CASE 1, right? So for the complete answer I should do the same for $x_1\;\in D\;$ such that $u(x_1)=\min_{\bar D} u\;$. Did I miss something? However, thank you a lot for your time and help – kaithkolesidou Nov 17 '16 at 18:21
• If $u(x_0)\leq 0$, then $u(x)\leq 0$ for all $x\in \overline{D}$. – anonymus Nov 17 '16 at 18:33
• With your method, we just consider the cases of maximum and minimum separately. One implies $u\le 0$ and the other one implies $u\ge 0$, and so $u=0$. We even don't need the maximum principle for the superharmonic functions and the subharmonic functions. – Sam Wong Oct 11 '18 at 7:49