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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.

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3 Answers 3

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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 $\min_\Omega u < -1$.

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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 $|u|\le1$.

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  • $\begingroup$ The negation of $|u|>1$ everywhere is $|u|\leq 1$ in at least one point, which is not what you need to prove. $\endgroup$
    – themaker
    Commented Jun 12, 2019 at 21:27
  • $\begingroup$ @themaker, I made a mistake. Thank you for pointing. $\endgroup$
    – xpaul
    Commented Jun 12, 2019 at 21:31
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    $\begingroup$ @themaker, the problem is fixed now. $\endgroup$
    – xpaul
    Commented Jun 12, 2019 at 23:08
  • $\begingroup$ Ok, i changed my vote. There is still another factor $(u - 1)$ missing in the last integral (which doesn't change the proof). $\endgroup$
    – themaker
    Commented Jun 14, 2019 at 18:56
  • $\begingroup$ @themaker, thank you again for pointing. $\endgroup$
    – xpaul
    Commented Jun 14, 2019 at 19:45
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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: $$u(x)\geq -1,\quad\forall\ x\in \Omega.$$

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