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Let us consider that $u\in C^2(\Omega)\cap C^0(\Omega)$ and satisfying the following equation . $\Delta u=u^3-u , x\in\Omega$ and

$u=0 $ in $\partial \Omega$

$\Omega \subset \mathbb R^n$ and bounded .

I need hints to find out what possible value $u$ can take ? Thank you for ur hints in advance .

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    @timur : i got this from a book on partial differential equation by Gregory T. von Nessi. I am still not getting how do i solve it :( .2012-07-20

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The first thing to observe is that the right hand side $u^3-u$ is positive if $u>1$and negative if $u<-1$. Now suppose that at some point $u$ becomes larger than $1$. This would mean the maximum of $u$ is larger than $1$. Let $x\in\Omega$ be such a maximum point. It is obvious that $x$ must be an interior point, since $u=0$ at the boundary. Let us look at $\Delta u$ at $x$. Of course, $\Delta u\leq0$ at $x$. But we know that $u^3-u>0$ at $x$, leading to the conclusion that at $x$, the equation $\Delta u = u^3 - u$ cannot be satisfied, hence $u$ is not a solution to our equation.

I leave the consideration of a minimum with $u<-1$ to you.

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We must of course assume $u$ is continuous on the closure of $\Omega$. Since this is bounded and $u = 0$ on $\partial \Omega$, if $u > 0$ somewhere it must achieve a maximum in $\Omega$. Now $u$ is subharmonic on any part of the domain where $u > 1$, so the maximum principle says ...