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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.2012-07-19
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    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.2012-07-19
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    @PZZ : the suggested answer to me is $-1 \le u\le 1$2012-07-19
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    @PZZ : I think somehow i should be able to use maximum principle here.2012-07-19
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    Is $\Omega$ assumed to be bounded? Otherwise e.g. in $n=1$, $\Omega = (0,\infty)$, $u(x) = \tanh(x/\sqrt{2})$ is a solution.2012-07-19
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    @RobertIsrael : yes sir, its bounded .2012-07-19
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    @PZZ: e.g. $u(x) = \sqrt{8/5} \text{JacobiSN}(x/\sqrt{5},2)$ is a solution in the case $n=1$ with $u(0)=0$ and $u(b)=0$ for $b \approx 3.769452386$, so even if $\Omega$ is bounded your assertion is incorrect.2012-07-19
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    @RobertIsrael : sir , I am not even able to solve in the case $n=1$ also . how did u proceed ? give me some hint.2012-07-19
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    @RobertIsrael, Theorem Right. When I was thinking about the equation in question was still $\Delta u - u - u^3 = 0$, in which case my assertion is (I believe) still correct. I naively thought the change of sign on $u$ would not matter, but of course it does, because the linearization is now no longer positive definite.2012-07-19
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    According to Maple the general solution of $u'' = u^3 - u$ is $$u \left( x \right) =\frac{k\sqrt {2}}{\sqrt{k^2+1}}{\it JacobiSN} \left( {\frac {x}{\sqrt { {k}^{2}+1}}},k \right) $$2012-07-19
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    @PZZ : Since $u'' = u^3 - u$ is autonomous, you get a separable first-order DE for $u$ and $v = u'$: $\dfrac{dv}{du} = \dfrac{u^3-u}{v}$. Solve that to get $u' = \pm \sqrt{C + u^4/2 - u^2}$. Integration gives you elliptic functions...2012-07-19
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    @RobertIsrael : Sir , the answer was suggest to me by my friend. I tried to get taylor expansion but i didn't get it . Can you help me .2012-07-19
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    The Jacobi functions give you periodic solutions, which stay in the interval $[-1,1]$ (look at a phase portrait). But for $C > 1/2$, so that $C + u^4/2 - u^2 > 0$ for all $u$, you get unbounded solutions.2012-07-19
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    I should say, the Jacobi functions for *real* $k$ give you periodic solutions. For example, the solution with $u(0)=0$, $u'(0)=1$ corresponds to $k = (1 \pm i)/\sqrt{2}$. It goes to $\infty$ as $x \to \int_0^\infty \frac{du}{\sqrt{u^4/2-u^2+1}} \approx 2.854209407$2012-07-19
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    This is a straightforward application of the maximum principle. Where did you get this exercise? (Looks very familiar)2012-07-20
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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 ...