1
$\begingroup$

I conjecture the following.

Let $\Omega=\mathbb{R}^3-\overline{B_1(0)}$. Define $$E_{\Omega}(u)=\frac{1}{2}\int_{\Omega}|\nabla u|^2dx-\frac{1}{6}\int_{\Omega}|u|^6dx.$$ $E_{\mathbb{R}^3}$ is defined similarly: $$E_{\mathbb{R}^3}(u)=\frac{1}{2}\int_{\\mathbb{R}^3}|\nabla u|^2dx-\frac{1}{6}\int_{\mathbb{R}^3}|u|^6dx.$$

Consider the exterior problem $$ \Delta u+|u|^4u=0,~~ ,~~~u|_{\partial\Omega}=0$$

It's well-known that if $\Omega=\mathbb{R}^3$, then the problem has a unique radial positive solution given by $$W(x)=\frac{1}{(1+\frac{|x|^2}{3})^{1/2}}.$$

Conjecture: If $\Omega=\mathbb{R}^3-B_1(0)$, then the problem admits a unique nontrivial nonnegative radial solution $u$. Moreover, $E_{\Omega}(u)=E_{\mathbb{R}^3}(W)$.

I appreciate very much if anynone can prove this or can tell me the existed source of answer or give counterexamples.

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ You want $u$ to be zero (and not equal to, say, $1$) along the unit sphere? $\endgroup$
    – Deane Yang
    Jan 5, 2013 at 20:37
  • $\begingroup$ Yes, I want zero boundary condition. $\endgroup$
    – user30263
    Jan 5, 2013 at 22:33

1 Answer 1

Reset to default
3
$\begingroup$

Try taking a Kelvin transform of the PDE. It will send it to a new PDE on $|x| \le 1$. IN this case, since hte PDE is critical, you should get the same PDE. Some care will need to be taken at the origin. If you are looking for a "fast decay" solution of the exterior problem, ie. one for which $ |x| |u(x)| $ is bounded then the Kelvin transform will be a classical solution on the unit ball and yet I think this is impossible. So I think the only fast decay solution is $u=0$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Craig, thanks for telling me this technique. I also realized that the answer should be straight forward from the known result. Assume $u$ is a solution to the problem (with certain regularity, for example, $C^1$), then extend $u$ by zero to the whole space, then we must have $u=0$ by the characterization of solutions to the cauchy problem $\Delta u+u^5=0$ in $\mathbb{R}^3$. $\endgroup$
    – user30263
    Feb 1, 2013 at 20:31
  • $\begingroup$ I don't think that works. Assume $u$ is positive solution on exteror of ball. By Hopft you will have non zero gradient on |x|=1. Let $v$ denote extension to full space (zero inside ball). CHeck carefully that $-\Delta v$ has a non zero surface measure supported on $|x|=1$. So $ v$ satisfies something along the liens of $ -\Delta v = v^5 + \sigma $ in R^N where sigma the surface measure. $\endgroup$
    – Craig
    Feb 3, 2013 at 23:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.