Laplace equation in a domain with holes

Question

Suppose $u_r(x_1,x_2)$ in $B_1(0) \setminus B_r(0)$ satisfy \begin{align*} \begin{cases} \Delta u_r &=0\\ u_r(|x|=r)&=-\log r \\ u_r(|x|=1)&=0 \end{cases} \end{align*} with $0<r<1$.

Then one can show that $u_r=\log (1/|x|)$ converges to $u^*=\log (1/|x|)$, as $r$ goes to $0$, in $B_1(0) \setminus \{0\}$.

But if we have finitely many holes and consider the domain $B_1(0) \setminus \bigcup B_{r_i}(x_i)$

\begin{align*} \begin{cases} \Delta u_r &=0 \quad\text{ in } B_1(0) \setminus \bigcup B_{r_i}(x_i)\\ u_r(|x-x_i|=r_i)&=-\log r_i \\ u_r(|x|=1)&=0 \end{cases} \end{align*} with all $\overline{B_{r_i}(x_i)} \subset B_1(0)$, how to show that the solution still converges and is locally bounded in $B_1(0)\setminus \{\bigcup x_i\}$, as all $r_i$ go to 0 ?

Answer 1

Let $v$ be the solution of $$ \left\{ \enspace \begin{aligned} &\Delta v =0 && \text{in $B_1(0)$,} \\ &v = -\,\sum\nolimits_i \log (1/|x-x_i|) && \text{on $\partial B_1(0)$.} \end{aligned} \right. $$ Then $u_r$ converges locally uniformly in $\overline B_1(0)\setminus \bigcup_i\{x_i\}$ as $r=\{r_i\}\to 0$ to $$ u = v + \sum\nolimits_i \log (1/|x-x_i|). $$ Indeed, $u(x) = \log (1/|x-x_j|) + O(1)$ as $x\to x_j$ for all $j$. Therefore, $(u-u_r)\bigr|_{\,K}=O(1)$ as $r\to0$ for any compact $K\subset \overline B_1(0)\setminus \bigcup_i\{x_i\}$. Now, if $u_r \to \tilde u$ locally uniformly in $\overline B_1(0)\setminus \bigcup_i\{x_i\}$ as $r\to0$ along a subsequence, $u-\tilde u$ is harmonic in $B_1(0)$ (the singularities at the $x_i$ are removable) and vanishes on $\partial B_1(0)$. Hence, $u\equiv\tilde u$.