Weak solutions of Dirichlet problem for Laplace equation on a punctured disk Let $\Omega = \{ x \in \mathbb{R}^n: 0<|x|<1 \}$ and consider the Dirichlet problem
\begin{align}
\Delta u &= 0 \\
u(0) &= 1 \\
u &= 0 ~~~\text{if} ~~|x|=1
\end{align}
This problem does not admit a continuous solution in $\Omega$, but is it possible to find a weak solution? 
If so, does a sequence of such weak functions exist that minimizes the Dirichlet integral?
 A: A weak solution of Laplace's equation is also a classical solution, per Weyl's lemma. So you don't get anything new. If $u$ is a  bounded harmonic function in $\Omega$ and $u(x)=0$ when $|x|=1 $, then $u$ is identically zero (the singularity at $0$ is removable). There is no way to satisfy the condition $u(0)= 1$. 
You also mention  minimizing the Dirichlet integral under these boundary conditions. The thing is, its infimum is $0$ which is not attained. 
Let $$\phi(r) = \begin{cases} \log  1/r,\quad n=2 \\r^{2-n} - 1,\quad n>2\end{cases}$$
For a small $\epsilon>0$ define $u(x)=1$ when $|x|<\epsilon$, and $u_\epsilon(x)=\phi(|x|)/\phi(\epsilon )$ for $\epsilon\le |x|<1$. A computation shows that the Dirichlet integral of $u_\epsilon$ tends to zero as $\epsilon\to 0$. On the other hand, a function with zero Dirichlet integral would have to be constant, violating the boundary condition. 
The usual way out of the conundrum is to relax the boundary condition, allowing $u$ to ignore it on a polar set. Then $u\equiv0$ is the solution of such generalized Dirichlet problem.
