${\delta}$ is small then

$$\left(P_{{\delta},Z}
-\kappa_j-\frac{2\pi q(x)}{|\ln{\varepsilon}|} \right)_+=0, \quad x\in
{\Omega}_j\setminus  B_{Ls_{{\delta},j}}(z_j),j=1,\cdots,m.$$

Let

$$M=  E_{{\delta},Z}\cap\Bigl\{ \|\omega\|_\infty\le
{\delta}|\ln{\delta}|^{\frac{p-1}{2}}\Big\}.$$ Then $M$ is complete under $L^\infty$ norm and $G_{\delta}$ is a map from $ E_{{\delta},Z}$ to $ E_{{\delta},Z}$. We will show that $G_{\delta}$ is a contraction map from $M$ to $M$.

Step 1. $G_{\delta}$ is a map from $M$ to $M$.

For any $\omega\in M$, similar to Lemma \[al1\], it is easy to prove that for large $L>0$, ${\delta}$ small

$$\label{3.14}
 \left(P_{{\delta},Z}+\omega
-\kappa_j-\frac{2\pi q(x)}{|\ln{\varepsilon}|}\right)_+=0, \quad \text{in}\;
{\Omega}_j\setminus  B_{Ls_{{\delta},j}}(z_j).$$

Note also that for any $u\in L^\infty({\Omega})$,

$$Q_{\delta}u= u\quad \text{in}\; {\Omega}\setminus  \cup_{j=1}^m
B_{Ls_{{\delta},j}}(z_j).$$ Therefore, using Lemma \[al1\], and , we find that for any $\omega\in M$,

$$Q_{\delta}l_{\delta}+ Q_{\delta}R_{\delta}(\omega)= l_{\delta}+ R_{\delta}(\omega)
=0, \quad \text{in}\; {\Omega}\setminus  \cup_{j=1}^m
B_{Ls_{{\delta},j}}(z_j).$$ So, we can apply Lemma \[l31\] to obtain

$$\| (Q_{\delta}L_{\delta})^{-1} \bigl(
 Q_{\delta}l_{\delta}+ Q_{\delta}R_{\delta}(\omega)\bigr)\|_\infty
\le \frac{C|\ln{\delta}|^{\frac{(p-1)^2}{p}}}{{\delta}^{\frac{2}{p}}} \| Q_{\delta}l_{\delta}+  Q_{\delta}R_{\delta}(\omega)\|_p.$$

Thus, for any $\omega\in M$, we have

$$\label{3.15}
\begin{split}
\| G_{\delta}(\omega)\|_\infty =& \| (Q_{\delta}L_{\delta})^{-1} Q_{\delta}\bigl(
  l_{\delta}+  R_{\delta}(\omega)\bigr)\|_\infty\\
\le & \frac{C|\ln{\delta}|^{\frac{(p-1)^2}{p}}}{{\delta}^{\frac{2}{p}}}\|Q_{\delta}\bigl( l_{\delta}+ R_{\delta}(\omega)\bigr)\|_p.
\end{split}$$

It follows from – that the constant $b_{j,h}$, corresponding to $u\in L^\infty({\Omega})$, satisfies

$$|b_{j,h}| \le C |\ln{\delta}|^{p+1}\sum_{i,\,\bar h}
 \int_{\Omega}\Bigl|\frac{\partial P_{{\delta},Z,i}}
{\partial z_{i,\bar h}}\Bigr| |u|.$$

Since

$$l_{\delta}+ R_{\delta}(\omega)=0, \quad \text{in}\; {\Omega}\setminus  \cup_{j=1}^m
B_{Ls_{{\delta},j}}(z_j),$$ we find that the constant $b_{j,h}$, 