I'm trying to find the solution of $$ \int_{\mathbb{R}^n}|\nabla u(x)|^2dx$$ on $$ A_R = \{ u \in Lip(\mathbb{R}^n ) : u(x) \ge 1, |x| \le 1 \wedge u(x) = 0, |x| \ge R\} $$ for $n\ge 1$ and $R> 1$ minimizing the integral.
I was able to prove the solution needs to be radial and computet the EL equation (where $\lambda=\int\dots \int r^{n-1}\prod_{k=1}^{n-2}sin^{n-1-k}d\theta d\phi_1\dots d\phi_{n-2}$ comes from the change of variables from cartesian to spherical) $$div(\lambda \varphi'(r)r^{n-1})=0$$
Now this tells me that $$ \lambda\varphi'(r)r^{n-1}=\kappa\iff$$$$ \varphi(r)=\frac{\kappa}{\lambda(2-n)}r^{2-n}+\rho\wedge =\frac{\kappa}{\lambda}ln(r)+\rho\wedge= \frac{\kappa}{\lambda}r+\rho.$$
But now I don't know how to include the conditions on the boundary and morover most of this functions are not continuous in 0 as they should.
What am I doing wrong? Any help is appreciated.
The idea is to use the solution from EL only between 1 and R.
We have the solutions $$ \lambda\phi'(r)r^{n-1}=\kappa\iff$$$$ \phi(r)=\frac{\kappa}{\lambda(2-n)}r^{2-n}+\rho\wedge \frac{\kappa}{\lambda}ln(r)+\rho\wedge \frac{\kappa}{\lambda}r+\rho$$ respectively for $n>2,\; n=2,\; n=1$.\ We now need to impose the "boundary" conditions remembering that we only require this function in the set $\{x|1\le|x|\le R\}$ where they are all well defined and we get the following three systems of equations for different n.
$$n=1\Rightarrow \begin{cases}\phi(1)=\frac{\kappa}{\lambda}+\rho=1\wedge \phi(R)=\frac{R\kappa}{\lambda}+\rho=0\end{cases}\Rightarrow \phi(r)=\frac{1}{1-R}\left(r-R\right)$$ $$n=2\Rightarrow \begin{cases} \phi(1)=\frac{\kappa}{\lambda}ln(1)+\rho=1\\ \phi(R)=\frac{R\kappa }{\lambda}ln(R)+\rho=0 \end{cases}\Rightarrow \phi(r)=\frac{ln\left(r\right)}{ln\left(\frac{1}{R}\right)}+1 ;$$ $$ n>2\Rightarrow\begin{cases}\phi(1)=\frac{\kappa}{\lambda(2-n)}+\rho=1 \quad \phi(R)=\frac{R^{2-n}\kappa }{\lambda(2-n)}+\rho=0 \end{cases}\Rightarrow \phi(r)=\frac{1}{1-R^{2-n}}\left(r^{2-n}-R^{2-n}\right) ;$$ Anyway the general solution will be $$ \varphi(r)=\begin{cases} 1&\Leftarrow 0\le r\le 1\\ \phi(r)&\Leftarrow 1\le r\le R\\ 0&\Leftarrow r\le R; \end{cases}$$.