is an unique $\omega_{\delta}\in M$, such that $\omega_{\delta}= G_{\delta}\omega_{\delta}$. Moreover, it follows from that

$$\|\omega_{\delta}\|_\infty\le {\delta}|\ln{\delta}|^{\frac {p-1}2}.$$

Proof of The main results
=========================

In this section, we will choose $Z$, such that $ \sum_{j=1}^m
P_{{\delta},Z,j}+\omega_{\delta}$, where $\omega_{\delta}$ is the map obtained in Proposition \[p33\], is a solution of .

Define $$I(u)=\frac{{\delta}^2}{2}\int_\Omega
|Du|^2-\sum_{j=1}^m\frac{1}{p+1}\int_\Omega\chi_{{\Omega}_j}\left(u-\kappa_j-\frac{2\pi
q(x)}{|\ln{\varepsilon}|}\right)_+^{p+1}$$ and $$\label{K}
 K(Z)= I\left( P_{{\delta},Z} +\omega_{\delta}\right).$$ It is well known that if $Z$ is a critical point of $K(Z)$, then $\sum_{j=1}^m P_{{\delta},Z,j} +\omega_{\delta}$ is a solution of .

In the following, we will prove that $K(Z)$ has a critical point.

\[l42\]

We have

$$K(Z)= I\left(\sum_{j=1}^m P_{{\delta},Z,j}
\right)+O\left(
 \frac{{\varepsilon}^3}{|\ln{\varepsilon}|^p}\right).$$

Recall that

$$P_{{\delta},Z}=\sum_{j=1}^m P_{{\delta},Z,j}.$$ We have

$$\begin{split}
K(Z)=& I\bigl( P_{{\delta},Z}\bigr) +\int_{\Omega}{\delta}^2
DP_{{\delta},Z}D\omega_{\delta}+\frac{{\delta}^2}{2}\int_\Omega |D\omega_{\delta}|^2\\
&-\frac1{p+1} \sum_{j=1}^m\int_{\Omega}\chi_{{\Omega}_j}\Biggl[ \biggl(
P_{{\delta},Z} +\omega_{\delta}-\kappa_j-\frac{2\pi q(x)}{|\ln{\varepsilon}|}
\biggr)_+^{p+1}- \biggl( P_{{\delta},Z}-\kappa_j-\frac{2\pi
q(x)}{|\ln{\varepsilon}|} \biggr)_+^{p+1}\Biggr].
\end{split}$$

Using Proposition \[p33\] and , we find

$$\begin{split}
&  \int_{{\Omega}_j} \Biggl[ \biggl( P_{{\delta},Z}
+\omega_{\delta}-\kappa_j-\frac{2\pi q(x)}{|\ln{\varepsilon}|} \biggr)_+^{p+1}-
\biggl(
P_{{\delta},Z}-\kappa_j-\frac{2\pi q(x)}{|\ln{\varepsilon}|} \biggr)_+^{p+1}\Biggr]\\
= &\int_{ B_{Ls_{{\delta},j}}(z_j)} \Biggl[ \biggl( P_{{\delta},Z}
+\omega_{\delta}-\kappa_j-\frac{2\pi q(x)}{|\ln{\varepsilon}|} \biggr)_+^{p+1}-
\biggl(
P_{{\delta},Z}-\kappa_j-\frac{2\pi q(x)}{|\ln{\varepsilon}|} \biggr)_+^{p+1}\Biggr]\\
=&O\left(\frac{s_{{\delta},j}^2\|\omega_{\delta}\|_\infty}{|\ln{\varepsilon}|^{p}}\right)
= O\Bigl(
 \frac{{\varepsilon}^3}{|\ln{\varepsilon}|^{p}}\Bigr).
\end{split}$$

On the other hand,

$$\begin{split}
&{\delta}^2 \int_{\Omega}DP_{{\delta},Z}D\omega_{\delta}=
 \sum_{j=1}^m\int_{\Omega}\left(W_{{\delta},z_j,a_{{\delta},j}}-a_{{\delta},j}\right)_+^{p} \omega_{\delta}\\
=&\sum_{j=1}^m
 \int_{\cup_{k=1}^m B_{s_{{\delta},k}}(z_k)}
(W_{{\delta},z_j,a_{{\delta},j}}-a_{{\delta},j})_+^{p} \omega_{\delta}\\
=& O\Bigl(
 \frac{{\varepsilon}^3}{|\ln{\varepsilon}|^{p}}\Bigr).
\end{split}$$ Finally, we estimate ${\delta}^2\int_{\Omega}|D\omega_{\delta}|^2$.\
Note that $$\begin{split}
-{\delta}^2\Delta\omega_{\delta}=&\sum_{j=1}^m\chi_{{\Omega}_j}\left(P_{{\delta},Z}+\omega_{\delta}-\kappa_j-\frac{2\pi
q(x)}{|\ln{\varepsilon}|}\right)_+^{p}
-\sum_{j=1}^m\left(W_{{\delta},z_j,a_{{\delta},j}}-a_{{\delta},j}\right)_+^{p}\\
&+\sum_{j=1}^m\sum_{\bar h=1}^2b_{j,\bar
h}\left(-{\delta}^2\Delta\frac{\partial P_{{\delta},Z,j}}{\partial z_{j,\bar
h}}\right),
\end{split}$$ Hence, 