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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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If you can prove a uniqueness result, then $u = 0$ is the only solution. Not sure if that helps you. –  user12014 Jul 19 '12 at 6:29
    
In fact, you can prove the following with the help of the inverse function theorem: if $f \in C^3$ with $f(0) = f'(0) = 0$, then $\Delta u + u + f(u) = g$ has a unique solution for small enough $g$. In your case, $g = 0$ is small enough and $f(u) = -u^3$ satisfies the conditions. –  user12014 Jul 19 '12 at 6:31
    
@PZZ : the suggested answer to me is $-1 \le u\le 1$ –  Theorem Jul 19 '12 at 6:45
    
@PZZ : I think somehow i should be able to use maximum principle here. –  Theorem Jul 19 '12 at 6:46
    
Is $\Omega$ assumed to be bounded? Otherwise e.g. in $n=1$, $\Omega = (0,\infty)$, $u(x) = \tanh(x/\sqrt{2})$ is a solution. –  Robert Israel Jul 19 '12 at 6:46

2 Answers 2

up vote 2 down vote accepted

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 ...

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