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Let $\Omega$ be a bounded domain. Suppose $u \in C^2(\Omega) \cap C(\bar{\Omega})$ which satisfies the boundary value problem : \begin{cases} -\Delta u=u-u^3 &\text{ in } \Omega \\ u=0 & \text{ on } \partial\Omega \end{cases} Show that $\sup_{\Omega} |u| \le 1$. Can $u$ take both the values $\pm1$ in $\Omega$ ?

My attempt: I am trying this problem by a result on apriori estimates : Let $\mathscr{L}:=-\sum_{i,j=1}^n a_{ij}(x) \partial_{ij}+\sum_{i=1}^n b_i(x)\partial_i+c(x)$ be a uniformly elliptic operator, with ellipticity constant $\Lambda$, continuous coefficients and we set $\Theta$ as follows, $$\sum_{i,j=1}^n max|a_{ij}|+\sum_{i=1}^n max|b_i| \le \Theta$$ . Let $\Omega$ be a bounded domain and let $f \in C(\Omega), g \in C(\partial\Omega)$. uppose $u \in C^2(\Omega) \cap C(\bar{\Omega})$ which satisfies the boundary value problem : \begin{cases} \mathscr{L}u=f &\text{ in } \Omega \\ u=g & \text{ on } \partial\Omega \end{cases} with $c(x) \ge 0$ in $\Omega$. Then we have $\sup_{\Omega}|u| \le C\sup_{\Omega}|f|+\sup_{\partial\Omega}|g|$ where $C>0$ is a constant depends only on $\Lambda, \Theta,|\Omega|$

So I take, $f=u-u^3,g=0,\mathscr{L}:=-\Delta$ then, $\Theta \ge n$. And letting $\sup_{\Omega}|u|=t$ we get that $$t \le C \sup_{\Omega}|u-u^3| \le Ct \sup_{\Omega}|1-u^2|$$ As $t :=\sup_{\Omega}|u| \ge 0$ we get that $C \sup_{\Omega}|1-u^2| \ge 1$. I don't know how to proceed from here.

I am completely new to PDEs. Thanks in advance for help!

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  • $\begingroup$ Weak maximum principle. See what happens at the maximum of $u$ $\endgroup$ Sep 13, 2020 at 20:54

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P.S: too long for a comment hence posting here.

Proceeding with ArcticChar's hint, in order to get a subsolution for $\mathscr L := -\Delta$, we must have $u^3-u \ge 0$ in $\Omega$. Now by weak maximum principle, $u \le 0$, hence $u^2-1 \le 0$, thus $\sup_{\Omega}|u| \le 1$. In this case $u$ cannot take any strictly positive value in $\Omega$ (again by weak maximum principle) hence cannot assume 1 anywhere in the interior.

For the other case, we proceed by starting with a supersolution...

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  • $\begingroup$ @ArcticChar in this argument we are assuming that $\mathscr{L}$ preserves its sign throughout in $\Omega$, why is that? $\endgroup$
    – Brozovic
    Sep 14, 2020 at 8:22
  • $\begingroup$ It's not quite clear how you got to $u\le 0$. $\endgroup$ Sep 14, 2020 at 9:12
  • $\begingroup$ @ArcticChar by weak maximum principle (while considering a subsolution).. BTW this is what I could come up with. It'd be great if you kindly post a complete answer. $\endgroup$
    – Brozovic
    Sep 14, 2020 at 9:32

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