# Someone know a specific maximum principle…? [Solved]

I need a maximum principle in unbounded domains: if $u$ is a solution, bounded in $\Omega$, satisfying

$$\Delta u+c(x)u=0, \ \ in \ \Omega,$$ $c\in L^\infty$, $$u\leq0 \ \ in \ \Omega$$ $$u(x_0)=0, \ \ x_0\in\Omega$$ Then $$u\equiv0 \ \ in \ \Omega$$ Someone know where I can find this statement?

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First you say $u$ is positive, then you say $u \le 0$. So which is it? If you mean $u \ge 0$ and $u \le 0$, well, then you shouldn't be surprised that $u \equiv 0$. – Robert Israel Nov 15 '12 at 7:26
Im sorry, $u$ is only nonpositive. – José Carlos Nov 15 '12 at 15:02

## 1 Answer

You can apply the Hopf maximum principle to the operator $Lu = \Delta u - c^{-}u$, where $c^-$ denotes the negative part of $c$: The function $u$ satisfies (since $u\le 0$) $$\Delta u - c^- u = -c^+u \ge 0$$ The Hopf maximum principle now asserts that $u$ cannot take on an interior maximum in any ball $B_R$ unless $u$ is constant there. In particular, you can take any ball around $x_0$ and see that $u=0$ on $B_R(x_0)$. But this holds for arbitrary $R>0$, so you're done.

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Did you write $c=c^+-c^-$? But this is positive and the function $c$ could change the sign. I didn't understand this part. – José Carlos Nov 15 '12 at 21:24
Yes $c = c^+ - c^-$. And $c^- = \frac{|c|-c}2 \ge 0$ is the negative part of $c$. I think you'll have to explain in a bit more detail what exactly it is you didn't understand for me to be able to clarify things. – Sam Nov 15 '12 at 21:34
What is your definition of positive and negative part @SamL.? As far i know, $c^+=max(c,0)$, $c^-=min(c,0)$ and $c=c^++c^-$ – Tomás Nov 16 '12 at 2:02
I think your definition of $c^-$ is $c^-=max(-c,0)$. Am i right? – Tomás Nov 16 '12 at 2:08
Where is using the limitation of the function c? – José Carlos Nov 16 '12 at 4:30