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Let $\Omega \subset \mathbb R^d$ a smooth bounded open set ($d\geq 2$) and $f\in \mathcal C(\mathbb R)$. Let $u\in \mathcal C^2(\overline{\Omega })$ be a solution of $$-\Delta u(x)+f(u(x))=0\text{ in }\Omega .$$ Assume that $|f(t)|\leq |t|$ for $t\in \mathbb R$. Prove that $$\int_{B(x,r)}|\nabla u|^2\leq \frac{C}{r^2}\int_{B(x,2r)}|u|^2,$$ for all $x\in \Omega $ and $r>0$ with $B(x,2r)\subset \subset \Omega $ for some positive constant $C$ independant of $u$, $x$ and $r$.

Attempts.

Let $\xi\in \mathcal C^\infty _0(\mathbb R^n)$ a cut-off function s.t. $\xi(x)=1$ on $B(x,r)$, $\xi(x)=0$ on $B(x,2r)^c$, $\xi(x)\in [0,1]$ when $x\in B(x,2r)\backslash B(x,r)$ and $|\nabla \xi|\leq \frac{C}{r}.$ By several calculations, I arrive at $$\int_{B(x,2r)}|\nabla (u\xi)|^2=\int_{B(x,2r)} u^2|\nabla \xi|^2-\int_{B(x,2r)} u\xi^2 f(u),$$ but I can't arrive at $$\int_{B(x,2r)}|\nabla (u\xi)|^2\leq \int_{B(x,2r)} u^2|\nabla \xi|^2.$$ I tried to prove that $ \int_{B(x,2r)} u\xi^2 f(u)\geq 0$ using the fact that $|f(u(x))|\leq |u(x)|$, but it didn't work. Any idea ?

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    $\begingroup$ Another very similar problem: math.stackexchange.com/q/2087724/9464. Where is this exercise from? $\endgroup$ – Jack Jun 4 '17 at 12:35
  • $\begingroup$ @Jack: I've seen this link already, and no it's not the same thing ! Where I'm stuck is not comparable to the problem on the link. This exercise come from an exercise of one of my course. $\endgroup$ – user386627 Jun 4 '17 at 13:00
  • $\begingroup$ Relax. I'm not saying that your question is a duplicate at all. No. :-) It is just a comment. People who are interested in your question might want to look at that one as well because the ways they are phrased are almost identical. I'm just curious about what books/references the exercise is from. $\endgroup$ – Jack Jun 4 '17 at 13:36
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    $\begingroup$ @Jack: No problem :) For the book, may be the evans ? (my teacher is a fan of this book). $\endgroup$ – user386627 Jun 4 '17 at 13:52
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Hint

Using brute force :

$$\int_{B(x,2r)}u^2|\nabla \xi|^2-\int_{B(x,2r)}u\xi^2 f(u)\leq \int_{B(x,2r)}u^2|\nabla \xi|^2+\int_{B(x,2r)}|u|\xi^2 |f(u)| $$$$\leq \int_{B(x,2r)}u^2|\nabla \xi|^2+\int_{B(x,2r)}|u|^2\leq\left(1+\frac{C}{r^2}\right)\int_{B(x,2r)}|u|^2.$$ Using the fact that $\Omega $ is bounded, you can conclude.

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  • $\begingroup$ Since $r^2+C\leq \text{diam}(\Omega )^2+C$ by setting $D=\text{diam}(\Omega )^2+C$ I conclude. Thanks a lot. $\endgroup$ – user386627 Jun 4 '17 at 13:01

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