# Apriori estimate on this boundary value problem

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!

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

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.
• @ArcticChar in this argument we are assuming that $\mathscr{L}$ preserves its sign throughout in $\Omega$, why is that? Sep 14, 2020 at 8:22
• It's not quite clear how you got to $u\le 0$. Sep 14, 2020 at 9:12