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I've been given the following BVP: \begin{align*} -\Delta u = u- u^3,\: x\in \Omega \end{align*}\begin{align} u = 0,\: x\in \partial \Omega \end{align} where $\Omega\subset \mathbb{R}^N$ is bounded.

I am supposed to show that $-1< u(x)< 1$ for all $x\in\mathbb{R}^N$.

I have experimented with sub/sup solutions, but this yields something different, and I suspect it is (very) wrong.

Any thoughts/hints?

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Homework, voting to close. – Michael Renardy Feb 5 at 15:34
You can just use plain vanilla maximum principles. Assume $u\geq1$ or $u\leq-1$ and see what happens. Also, are you sure you have to consider all $x\in\mathbb{R}^N$? In any case, I think your question is more suitable for MSE. – timur Feb 5 at 15:45

1 Answer

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Multiply your equation times the function $v(x) = \max(u-1,0)$. After you integrate by parts the Laplacian you will get an equality with opposite signs in each side that gives you a contradiction unless $v \equiv 0$.

Then do the same thing with $\min(u+1,0)$.

There are other ways to do it, including using sub/super-solutions.

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