Proof $|u|\leq 1$ for pde $\Delta u=u^3-u$

Question

Let $\Omega \subseteq \mathbb{R}$ be open and bounded and let $u\in C^2(\Omega)\cap C(\overline{\Omega})$ be solution of

\begin{align} \Delta u=u^3-u\quad &\text{in } \Omega\\ u=0 \quad &\text{at } \partial \Omega \end{align}

Show that $|u|\leq 1$ in $\Omega$.

Now to proof that $u\geq 1$ ist not possible is simple: Assuming $u\geq 1$ means that $\Delta u \geq 0$ and therefore all the prerequisites for the weak maximum principle are met. Hence the maximum has to be 0, which contradicts our assumption. The same kind of argument can be done to proof that $u\leq -1$ is not possible.

However I'm having trouble with the possible change of the signs on the rhs. I tried to split $\Omega$ but had troubles because I couldn't handle the possible new boundary.

Would be glad if someone could help me.

Answer 1

Suppose that there exists $x_0\in \Omega$ such that $$u(x_0)<-1.\tag{1}$$ Define $\Omega_2=\{x\in \Omega\mid u(x)<-1\}$. Then, $$-\Delta u=-(u^3-u)>0\quad\text{in}\quad \Omega_2.$$ Thus, the weak maximum principle implies that $$-1>u(x_0)\geq\min_{\overline{\Omega_2}}u=\min_{\partial \Omega_2}u=-1,$$ which is a contradiction. Therefore, there is no $x_0\in\Omega$ satisfying $(1)$. In other words: $$u(x)\geq -1,\quad\forall\ x\in \Omega.$$

Answer 2

Define $$ s^+=s, \text{ if }s\ge0, s^+=0 \text{ if }s<0. $$ Clearly $(u-1)^+=0$ on $\partial\Omega$. Multiplying both sides by $(u-1)^+$ and integrating, one has $$ -\int_{\Omega}|\nabla (u-1)^+|^2dx=\int_{\Omega}u(u+1)((u-1)^+)^2dx\ge0. $$ Thus $$ (u-1)^+=0 $$ which implies $u<1$. Similarly $u>-1$. So one must have $|u|\le1$.

Answer 3

You can just argue locally. If $\max_\Omega u>1$, lets take $x_0$ such that the maximum is attained(which exists by continuity of $u$). Then $u(x_0) > 1$ which implies $\Delta u(x_0) > 0$, but this contradicts the fact that $u(x_0)$ is maximum. The same if $\min_\Omega u < -1$.