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Let $\Omega\in\mathbf{R}^N$ an unbounded domain and $u\in C^2(\Omega)\cap C(\overline\Omega)$, $u>0$ such that $$\Delta u + f(u)=0, \ \ \ \mbox{em} \ \ \Omega,$$ where $f$ is a bounded lipschitz continuous function. Then $u$ is bounded.

I don't know where I can find this result, and I believe that this assumptions implies $\nabla u$ is bounded too. Someone can help me?

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1 Answer 1

up vote 1 down vote accepted

This appears to be a counterexample: $\Omega=\{(x,y)\in \mathbb R^2: x>1\}$; $u(x,y)=x$; $f\equiv 0$.

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Yes... you give the counter example, thanks. But, you know that this result is true to the first derivate? $|\nabla u|$ is bounded in this situation? –  José Carlos Dec 15 '12 at 20:37
    
@JoséCarlos On the same domain, $u(x,y)=x^2$ and $f\equiv 2$ give an example with unbounded gradient. –  user53153 Dec 15 '12 at 20:51
    
Pavel thank you for the counter examples! Im going to search these estimates at least in bounded domains. –  José Carlos Dec 17 '12 at 1:08
    
@JoséCarlos I think the key term for that is "Alexandrov-Bakelman-Pucci". –  user53153 Dec 17 '12 at 1:37
    
Thank you very much! –  José Carlos Dec 18 '12 at 1:04

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