, gives $u_i=0$. Thus,

$$\tilde u_{i,n} \to 0,\quad \text{in}\; C^1(B_{L}(0)),$$ for any $L>0$, which implies that $u_n=o(1)$ on $\partial
B_{Ls_{n.i}}(z_{i,n})$.

By assumption,

$$Q_{{\delta}_n} L_{{\delta}_n} u_n = 0,\quad\text{in}\; {\Omega}\setminus
\cup_{i=1}^k B_{L s_{n,i}}(z_{i,n}).$$

On the other hand, by Lemma \[al1\], for $j=1,\cdots,m$, we have

$$\left(P_{{\delta}_n,Z_n} -\kappa_j-\frac{2\pi q(x)}{|\ln{\varepsilon}_n|} \right)_+
=0, \quad x\in {\Omega}_j\setminus  B_{L s_{n,j}}(z_{j,n}).$$ Thus, we find that

$$-\Delta u_n  =0,\quad \text{in}~{\Omega}\setminus  \cup_{i=1}^m  B_{L
s_{n,i}}(z_{i,n}).$$ However, $u_n=0$ on $\partial{\Omega}$ and $u_n=o(1)$ on $\partial
B_{Ls_{n,i}}(z_{i,n})$, $i=1,\cdots,m$. So we have

$$u_n=o(1).$$ This is a contradiction.

\[p32\]

$Q_{\delta}L_{\delta}$ is one to one and onto from $E_{{\delta},Z}$ to $F_{{\delta},Z}$.

Suppose that $Q_{\delta}L_{\delta}u=0$. Then, by Lemma \[l31\], $u=0$. Thus, $Q_{\delta}L_{\delta}$ is one to one.

Next, we prove that $Q_{\delta}L_{\delta}$ is an onto map from $E_{{\delta},Z}$ to $F_{{\delta},Z}$.

Denote

$$\tilde E= \Bigl\{ u:  u\in H_0^1(\Omega), \; \int_{\Omega}
D\frac{\partial P_{{\delta},Z,j}}{\partial z_{j,h} } Du=0,\;
j=1,\cdots,m, h=1,  2\Bigr\}.$$ Note that $E_{{\delta},Z}=\tilde E\cap W^{2,p}(\Omega)$.

For any $\tilde h\in F_{{\delta},Z}$, by the Riesz representation theorem, there is a unique $u\in H_0^1({\Omega})$, such that

$$\label{3.8}
{\delta}^2\int_{\Omega}Du D\varphi =\int_{\Omega}\tilde h \varphi,\quad \forall\;
\varphi\in   H_0^1({\Omega}).$$

On the other hand, from $\tilde h\in F_{{\delta},Z}$, we find that $u\in
\tilde E$. Moreover, by the $L^p$-estimate, we deduce that $u\in
W^{2,p}(\Omega)$. As a result, $u\in E_{{\delta},Z}$. Thus, we see 