# A 'conjecture' on critical elliptic pde

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.

• You want $u$ to be zero (and not equal to, say, $1$) along the unit sphere? – Deane Yang Jan 5 '13 at 20:37
• Yes, I want zero boundary condition. – user30263 Jan 5 '13 at 22:33

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 $u=0$.
• 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 $\mathbb{R}^3$. – user30263 Feb 1 '13 at 20:31
• 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. – Craig Feb 3 '13 at 23:47