Question

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.

Answer 1

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 $u=0$.