\documentclass[reqno]{amsart}
\usepackage{hyperref}
\usepackage{amssymb}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 46, pp. 1--17.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2016/46\hfil Uniform estimate and strong convergence]
{Uniform estimate and strong convergence of minimizers of a $p$-energy
functional with penalization}

\author[B. Wang, Y. Cai \hfil EJDE-2016/46\hfilneg]
{Bei Wang, Yuze Cai}

\address{Bei Wang \newline
School of mathematics and information technology,
Jiangsu Second Normal University,
Nanjing, Jiangsu 210013, China}
\email{jsjywang@126.com}

\address{Yuze Cai \newline
Department of Basic Science,
Shazhou Professional Institute of Technology,
Zhangjiagang, Jiangsu 215600, China}
\email{caibcd@163.com}

\thanks{Submitted January 20, 2015. Published February 10, 2016.}
\subjclass[2010]{35B25, 35J70, 49K20, 58G18}
\keywords{p-energy functional; p-energy minimizer;
$W^{1,p}$ uniform estimates}

\begin{abstract}
 This article concerns the asymptotic behavior of minimizers
 of a $p$-energy functional with penalization as a parameter
 $\varepsilon$ approaches zero. By establishing  $W^{1,p}$ uniform
 estimates, we obtain $W^{1,p}$ convergence of the
 minimizer to a p-harmonic map.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\allowdisplaybreaks

\section{Introduction}

Let $G \subset \mathbb{R}^2$ be a bounded and simply connected domain with
smooth boundary $\partial G$, and $B_1=\{x \in \mathbb{R}^2;x_1^2+x_2^2<1\}$.
Denote $S^1=\{x \in \mathbb{R}^3;x_1^2+x_2^2=1,x_3=0\}$ and
 $S^2=\{x \in \mathbb{R}^3;x_1^2+x_2^2+x_3^2=1\}$. Sometimes we write the vector value
function $u=(u_1,u_2,u_3)$ as $(u',u_3)$. Let $g=(g',0)$ be a smooth
map from $\partial G$ into $S^1$ satisfying $d=\deg(g',\partial G)
\neq 0$. Without loss of generality, we may assume $d>0$. Consider
the energy functional
$$
E_\varepsilon(u,G)= \frac{1}{p}\int_G|\nabla u|^p dx
+\frac{1}{2\varepsilon^p}\int_G u_3^2 dx,\quad p>2
$$
with a small parameter $\varepsilon >0$. From the direct method in
the calculus of variations it is easy to see that the functional
achieves its minimum in the function class $W_g^{1,p}(G,S^2)$.
Obviously, the minimizer $u_{\varepsilon}$ on $W_g^{1,p}(G,S^2)$
is a weak solution of
$$
-\operatorname{div}(|\nabla u|^{p-2}\nabla u) =u|\nabla u|^p+
\frac{1}{\varepsilon^p}(uu_3^2-u_3e_3),\quad \text{on } G,
$$
where $e_3=(0,0,1)$. Namely, for any $\psi \in W_0^{1,p}(G,\mathbb{R}^3)$,
$u_{\varepsilon}$ satisfies
\begin{equation} \label{1.1}
\int_G|\nabla u|^{p-2}\nabla u\nabla \psi dx=\int_Gu\psi |\nabla
u|^p dx+\frac{1}{\varepsilon^p}\int_G\psi (uu_3^2-u_3e_3)dx.
\end{equation}
Without loss of generality, we assume $u_3\geq 0$, otherwise we
may consider $|u_3|$ in view of the expression of the functional.

When $p=2$, the functional $E_{\varepsilon}(u,G)$ was introduced
in the study of some simplified model of high-energy physics,
which controls the statics of planner ferromagnets and
antiferromagnets (see \cite{KP,PS}). The asymptotic
behavior of minimizers of $E_{\varepsilon}(u,G)$ has been
considered by Fengbo Hang and Fanghua Lin in \cite{HL-01}. When
the term $\frac{u_3^2}{2\varepsilon^2}$ replaced by
$\frac{(1-|u|^2)^2}{4\varepsilon^2}$ and $S^2$ replaced by $\mathbb{R}^2$,
the problem becomes the simplified model of the Ginzburg-Landau
theory for superconductors and was well studied in many papers
such as \cite{BBH-93, BBH-94,Lin,Stru}.
These works enunciate that the study of minimizers of the
functional with some penalization terms is connected tightly with
the study of harmonic maps with $S^1$-value. When $p>2$, it also
shows an enlightenment, namely, the properties (such as the
partial regularity, the properties of singularities) of p-harmonic
maps can be seen via studying the asymptotic properties of
minimizers of some p-energy functional with penalization (cf.
\cite{ABGS-09,ABGS-11,Lei-04,Lei,Lei-07,Lei-10,Wang}).

In this article, as in \cite{BBH-93,BBH-94,HL-01}, we concern with the
asymptotic behavior of minimizers of functional $E_\varepsilon(u,G)$
on $W_g^{1,p}(G,S^2)$ where $p>2$ as $\varepsilon \to 0$.

\begin{theorem}[{\cite[Theorem 1.1]{Lei-08}}]  \label{thm1.1}
Assume $u_{\varepsilon}$ is a minimizer of $E_{\varepsilon}(u,G)$
on \\ $W_g^{1,p}(G,S^2)$. Then all the zeros of
$|u'_{\varepsilon}|$ are included in
finite, disintersected discs $B(x_j^{\varepsilon},h\varepsilon),
j=1,2,\dots ,N_1$ where $N_1$ and $h>0$ do not depend on $\varepsilon \in
(0,1)$.
\end{theorem}

As $\varepsilon \to 0$, there exists a subsequence
$x_i^{\varepsilon_k}$ of the center $x_i^{\varepsilon}$ and
$a_i \in \overline{G}$ such that $x_i^{\varepsilon_k} \to a_i$,
$i=1,2,\dots ,N_1$. Perhaps there may be at least two subsequences
converging to the same point, we denote by $a_1,a_2,\dots ,a_N$,
$N\leq N_1$, the collection of distinct points in
$\{a_i\}_{i=1}^{N_1}$. Although the relationship between $N$ and
$d$ is unknown, the integer $N$ is independent of $\varepsilon \in
(0,1)$. By virtue of Theorem \ref{thm1.1}, we see that all the zeros of
$|u'_\varepsilon|$ converge to $a_1,a_2,\dots ,a_N$ as $\varepsilon$
tends to $0$. In addition, (2.3) in \cite{Lei-08} shows
\begin{equation} \label{1.2}
|u'_\varepsilon| \geq 1/2 \quad\text{on }K,
\end{equation}
where $K$ is an arbitrary compact subset of $G \setminus\cup_{i=1}^N\{a_i\}$.


\begin{theorem}[{\cite[Theorem 1.2]{Lei-08}}] \label{thm1.2*}
Assume $u_{\varepsilon}$ is a minimizer of $E_{\varepsilon}(u,G)$ on \\
$W_g^{1,p}(G,S^2)$. $K$ is an arbitrary compact subset of
$\overline{G}\setminus \cup_{j=1}^N\{a_j\}$. Then there exists a
subsequence $u_{\varepsilon_k}$ of $u_{\varepsilon}$ such that as
$k \to \infty$,
$$
u_{\varepsilon_k} \to u_p=(u'_p,0),\quad\text{weakly in }
W^{1,p}(K,\mathbb{R}^3),
$$
where $u'_p$ is a map of the least p-energy
$\int_K|\nabla u|^p dx$ in $W^{1,p}(K,\partial B_1)$.
\end{theorem}

We shall give the uniform $L_{\rm loc}^p$ estimate of
$\nabla u_{\varepsilon}$ in $\S3$. Recalling the case that the parameter
$p$ equals to the dimension $2$, we know it is available to
estimate the upper bound and the lower bound of
$\int|\nabla u_{\varepsilon}|^2 dx$ since we can use the property of conformal
transformation of $\int|\nabla u_{\varepsilon}|^2 dx$ (the idea of
which can be seen in \cite{BBH-94,HL-96,HL-01,Hong}).
In fact, when scaling $x=y\varepsilon$ in
$E_{\varepsilon}(u,G)$, there is a coefficient
$\varepsilon^{\lambda}$ appearing in the scaled energy functional.
when $p=2$, it can be derived that the exponent $\lambda$ of
$\varepsilon$ is zero. Therefore, the estimate of the upper bound
$$
E_\varepsilon(u_\varepsilon,G) \leq C_1 \ln\frac{1}{\varepsilon}+C
$$
and the lower bound
$$
\frac{1}{2}\int_{G\setminus\cup_{i=1}^dB(a_i,h\varepsilon)}
|\nabla u'_\varepsilon|^2dx\geq C_2\ln\frac{1}{\varepsilon}-C
$$
can be obtained, where $C_1=C_2=\pi d$ (cf. \cite[\S4]{HL-01}). The
uniform estimate is deduced at once. When $p>2$, the property of
conformal transformation of $\int|\nabla u_{\varepsilon}|^p dx$ is
invalid. Therefore, $\lambda\neq 0$. It is impossible to derive such
results as the case $p=2$ if the idea of estimating the upper and
the lower bounds of $\int|\nabla u_{\varepsilon}|^p dx$ is
adopted. In fact, the upper bound
$$
E_\varepsilon(u_\varepsilon,G) \leq C_3 \varepsilon^{2-p}+C
$$
and the lower bound
$$
\frac{1}{p}\int_{G\setminus\cup_{i=1}^NB(a_i,h\varepsilon)}
|\nabla u'_\varepsilon|^pdx\geq C_4 \varepsilon^{2-p}-C,
$$
are also obtained. However, the relationship between $C_3$ and
$C_4$ is not clear except that $C_4$ may be smaller. In
\cite{Lei-08}, a comparison method was used to obtain a uniform
estimate where the average functions come into plays.

Here, we use the iteration technique introduced in \cite{Lei-06}
to obtain the uniform $L^p$ estimate of $\nabla u_\varepsilon$.
In fact, the term $\int_K|\nabla u_{\varepsilon}|^p dx$ of the functional
$E_{\varepsilon}(u_{\varepsilon},K)$ can be divided into three
terms, $\int_K|\nabla |u'_{\varepsilon}||^p dx$, $\int_K|\nabla
u_3|^p dx$ and $\int_K|u'_\varepsilon|^p|\nabla
\frac{u'_{\varepsilon}}{|u'_{\varepsilon}|}|^p dx$. We will prove
that $\int_K|\nabla |u'_{\varepsilon}||^p dx+\int_K|\nabla u_3|^p
dx +\frac{1}{\varepsilon^p} \int_K u_{\varepsilon 3}^2 dx$ may be
bounded by $O(\varepsilon^{\lambda})$ with $\lambda > 0$ as
$\varepsilon \to 0$. Using this estimate we will prove
$$
\int_K|\nabla u_{\varepsilon}|^p dx\leq C +O(\varepsilon^{\lambda}).
$$

Based on the Theorem \ref{thm1.2*}, we will prove in \S3 that the
p-harmonic map $u_p$ is a map of least p-energy $\int_K|\nabla
u|^p dx$, and the convergence is also in strong $W_{\rm loc}^{1,p}$
sense.

\begin{theorem} \label{thm1.2}
Assume  $u_{\varepsilon}$ is a minimizer of $E_{\varepsilon}(u,G)$
on 
$W_g^{1,p}(G,S^2)$. $K$ is an arbitrary compact subset of
$\overline{G}\setminus \cup_{j=1}^N\{a_j\}$. Then there exists a
subsequence $u_{\varepsilon_k}$ of $u_{\varepsilon}$ such that as
$k \to \infty$,
$$
u_{\varepsilon_k} \to u_p=(u'_p,0),\quad in \quad W^{1,p}(K,\mathbb{R}^3),
$$
where $u'_p$ is the map in Theorem \ref{thm1.2*}.
\end{theorem}


\section{Uniform estimate}

The following inverse H\"older inequality will be applied later.

\begin{proposition} \label{prop2.1}
Assume that $p>1$, and $u_{\varepsilon}$ is a minimizer of
$E_{\varepsilon}(u,G)$ on  \\ 
$W_g^{1,p}(G,S^2)$. Then there exist
constants $t, R_0 \in (0,1/2)$ and $C>0$ which is independent of
$\varepsilon$, such that for any $B_R \subset G$ $(2R<R_0)$, we
have
$$
\Big(\int_{B_R}|\nabla u_{\varepsilon}|^qdx\Big)^{1/q} 
\leq C\Big(\int_{B_{2R}}(|\nabla u_{\varepsilon}|^2+1) ^{p/2}dx\Big)^{1/p},
\quad \forall q \in [p,p+2t).
$$
\end{proposition}

The above proposition is a corollary from \cite[Theorem 4.1]{GG},
 with a rescaling.

\begin{theorem} \label{thm2.2}
Let $R>0$ be a small constant such that $B(x,2R) \Subset
G\setminus \cup_{j=1}^N\{a_j\}$. There exist constant
$\varepsilon_0>0$ and $C_j>0$, and $R_j=2R-\frac{jR}{[p]+1}$ such
that for $j=2,3,\dots ,[p]$,
\begin{equation} \label{2.1}
E_{\varepsilon}(u_{\varepsilon},B_j)\leq C_j\varepsilon^{j-p}
\end{equation}
where $\varepsilon \in (0,\varepsilon_0), B_j=B(x,R_j)$, and $[p]$
is the integer part of $p$.
\end{theorem}

For $j=2$, the inequality \eqref{2.1} is follows from
\cite[Proposition 2.1]{Lei-08}. Suppose that \eqref{2.1} holds
for all $j\leq m$. Then we have, in particular,
\begin{equation} \label{2.2}
E_{\varepsilon}(u_{\varepsilon},B_m)\leq C_m\varepsilon^{m-p}.
\end{equation}
If $m=[p]$, then we are done. Suppose $m<[p]$, we want to prove
\eqref{2.1} for $j=m+1$.

Applying \eqref{1.2} we have $\frac{1}{2}\leq
|u'_{\varepsilon}(y)|\leq 1$, for all $y \in B(x,2R)$. Using the
integral mean value theorem we know that there exists $r \in
[R_{m+1/2},R_m]$ such that
$$
E_{\varepsilon}(u_{\varepsilon},B_m\setminus
B_{m+1/2})=C_0(r)\int_{\partial B(x,r)} [\frac{1}{p}|\nabla
u_{\varepsilon}|^p +\frac{1}{4\varepsilon^p}u_{\varepsilon 3}^2]
d\xi,
$$
and applying \eqref{2.2}, we  see that
\begin{equation} \label{2.3}
\int_{\partial B(x,r)}|\nabla u_{\varepsilon}|^pd\xi
+\frac{1}{\varepsilon^p} \int_{\partial B(x,r)}u_{\varepsilon
3}^2d\xi \leq C_0^{-1}(r)C_m\varepsilon^{m-p}.
\end{equation}


We denote  $B=B(x,r)$, and introduce two propositions.

\begin{proposition} \label{prop2.3}
If $\rho_1$ is a minimizer of the functional
$$
E(\rho,B)=\frac{1}{p} \int_B(|\nabla \rho|^2+1)^{p/2} dx
+\frac{1}{2\varepsilon^p} \int_B(1-\rho)^2 dx,
$$
on $W_{|u'_{\varepsilon}|}^{1,p}(B,\mathbb{R}^+\cup\{0\})$. Then
$E(\rho_1,B)\leq C\varepsilon^{m-p+1}$.
\end{proposition}

\begin{proof} Obviously, the minimizer $\rho_1$
exists and satisfies
\begin{gather} \label{2.4}
-\operatorname{div}(v^{(p-2)/2}\nabla \rho) =\frac{1}{\varepsilon^p}(1-\rho)
\quad\text{on } B, \\
 \label{2.5}
\rho|_{\partial B}=|u'_{\varepsilon}|,
\end{gather}
where $v=|\nabla \rho|^2+1$. Since $1/2\leq |u'_{\varepsilon}|\leq
1$, it follows from the maximum principle that on $\overline{B}$,
\begin{equation} \label{2.6}
\frac{1}{2}\leq \rho_1\leq 1.
\end{equation}
Applying \eqref{2.2} and noting $(1-|u'|)^2 \leq u_3^2$, we see easily
that
\begin{equation} \label{2.7}
E(\rho_1,B)\leq E(|u'_{\varepsilon}|,B) \leq
CE_{\varepsilon}(u_{\varepsilon},B) \leq C\varepsilon^{m-p}.
\end{equation}

Multiplying \eqref{2.4} by $\partial_\nu \rho$, where $\rho$ denotes
$\rho_1$, and integrating over $B$, we have
\begin{equation} \label{2.8}
\begin{aligned}
&-\int_{\partial B}v^{(p-2)/2}(\partial_\nu \rho)^2d\xi
+\int_Bv^{(p-2)/2}\nabla \rho \nabla (\partial_\nu \rho)dx\\
&=\frac{1}{\varepsilon^p} \int_B(1-\rho)(\partial_\nu \rho)dx,
\end{aligned}
\end{equation}
where $\nu$ denotes the unit outside norm vector on $\partial B$.
Using \eqref{2.7} we obtain
\begin{equation} \label{2.9}
\begin{aligned}
\big|\int_Bv^{(p-2)/2}\nabla \rho \cdot \nabla
(\partial_\nu \rho)dx\big|
&\leq C\int_Bv^{(p-2)/2}|\nabla \rho|^2dx
+\frac{1}{p}\big|\int_B \nu \cdot \nabla (v^{p/2})dx\big|\\
& \leq C\varepsilon^{m-p}+\frac{1}{p}\int_{\partial B}v^{p/2}d\xi.
\end{aligned}
\end{equation}
Combining \eqref{2.3}, \eqref{2.5} and \eqref{2.7} we also have
$$
\big|\frac{1}{\varepsilon^p} \int_B(1-\rho)(\partial_\nu \rho)dx\big|
\leq \frac{1}{2\varepsilon^p} |\int_B(1-\rho)^2div\nu dx
-\int_{\partial B}(1-\rho)^2d\xi| \leq C\varepsilon^{m-p}.
$$
Substituting this result and \eqref{2.9} into \eqref{2.8} yields
\begin{equation} \label{2.10}
\big|\int_{\partial B}v^{(p-2)/2}(\partial_\nu \rho)^2d\xi\big|
\leq C\varepsilon^{m-p} +\frac{1}{p}\int_{\partial B}v^{p/2}d\xi.
\end{equation}
Applying \eqref{2.3}, \eqref{2.5}, \eqref{2.10} and the Young inequality, we obtain
that for any $\delta \in (0,1)$,
\begin{align*}
\int_{\partial B} v^{p/2}d\xi
&=\int_{\partial B}v^{(p-2)/2}[1+(\partial_\nu \rho)^2
+(\partial_\tau \rho)^2]d\xi\\
&\leq \int_{\partial B}v^{(p-2)/2}d\xi
+\int_{\partial B}v^{(p-2)/2} (\partial_\nu
\rho)^2d\xi\\
&\quad +\Big(\int_{\partial B}v^{p/2}d\xi\Big)^{(p-2)/p}
\Big(\int_{\partial B}
(\tau \cdot \nabla |u_{\varepsilon}|)^pd\xi\Big)^{2/p}\\
&\leq C(\delta)\varepsilon^{m-p}+(\frac{1}{p}
+2\delta)\int_{\partial B}v^{p/2}d\xi,
\end{align*}
where $\tau$ denotes the unit tangent vector on $\partial B$.
Therefore, it follows by choosing $\delta>0$ sufficiently small that
\begin{equation} \label{2.11}
\int_{\partial B}v^{p/2}d\xi \leq C\varepsilon^{m-p}.
\end{equation}

We multiply both sides of \eqref{2.4} by $(1-\rho)$ and integrate over
$B$. Then
$$
\int_Bv^{(p-2)/2} |\nabla \rho|^2dx+\frac{1}{\varepsilon^p}
\int_B(1-\rho)^2 dx=-\int_{\partial B} v^{(p-2)/2}(\nu \cdot
\nabla \rho)(1-\rho)d\xi,
$$
whose left hand side is proportional to $E(\rho_1,B)$. Thus
$$
E(\rho_1,B)\leq C\big| \int_{\partial B}v^{(p-2)/2}(\nu \cdot \nabla
\rho)(1-\rho)d\xi\big|.
$$
Applying Holder's inequality and \eqref{2.3}, \eqref{2.5}, \eqref{2.6} and \eqref{2.11},
we obtain
\begin{equation} \label{2.12}
\begin{aligned}
E(\rho_1,B)
&\leq C|\int_{\partial B} v^{p/2}d\xi|^{(p-1)/p} \Big|\int_{\partial
B}(1-\rho^2)^2d\xi\Big|^{1/p}\\
&\leq C\varepsilon^{(m-p)(p-1)/p} \Big|\int_{\partial B}
u_{\varepsilon 3}^2 d\xi\Big|^{1/p} \leq C\varepsilon^{m-p+1}.
\end{aligned}
\end{equation}
The proof is complete.
\end{proof}

\begin{proposition} \label{prop2.4}
Denote $h=|u'_\varepsilon|$. Then there is $t \in (0,1/2)$ such
that for any $\delta \in (0,1/2)$,
\begin{equation} \label{2.13}
\begin{aligned}
& \frac{1}{p}\int_B|\nabla
h|^pdx+\frac{1}{p}\int_B|\nabla
u_3|^pdx+\frac{1}{4\varepsilon^p}\int_B(1-h^2)^2dx\\
& \leq C\varepsilon^{m-p+1} +\delta\int_B|\nabla
u_{\varepsilon}|^p dx
+C\Big(\int_{B(x,2r)} |\nabla
u_{\varepsilon}|^pdx+1\Big)\\
&\quad\times \Big[\int_B(1-h^2)^2dx\Big]^{t/(p+t)}.
\end{aligned}
\end{equation}
\end{proposition}

\begin{proof}
Let $U=(\sqrt{2\rho_1-\rho_1^2}
w,1-\rho_1)$ on $B$; $U=u_{\varepsilon}$ on $G\setminus B$, where
$w=w_\varepsilon=\frac{u'_{\varepsilon}}{|u'_{\varepsilon}|}$.
Then $U \in W_g^{1,p}(G,S^2)$. Since $u_{\varepsilon}$ is a
minimizer of $E_{\varepsilon}(u,G)$, we have
$$
E_{\varepsilon}(u_{\varepsilon},G) \leq E_{\varepsilon}(U,G)
=E_{\varepsilon}(U,B) +E_{\varepsilon}(u_{\varepsilon},G\setminus B),
$$
which means $E_{\varepsilon}(u_{\varepsilon},B) \leq
E_{\varepsilon}(U,B)$. Using \eqref{2.12} it is not difficult to see
that for any $\delta>0$,
\begin{align*}
\int_B|\nabla \rho_1|^2|\nabla w|^{p-2} dx
&\leq (\int_B|\nabla \rho_1|^pdx)^{2/p}
(\int_B|\nabla w|^pdx)^{\frac{p-2}{p}}\\
&\leq \delta\int_B|\nabla u_{\varepsilon}|^pdx
+C\varepsilon^{m+1-p}.
\end{align*}
By using \eqref{2.6} and the mean value theorem,
\begin{align*}
& \int_B(\frac{(1-\rho_1)^2}{2\rho_1-\rho_1^2}|\nabla \rho_1|^2
+(2\rho_1-\rho_1^2)|\nabla w|^2)^{p/2}dx
-\int_B((2\rho_1-\rho_1^2)|\nabla w|^2)^{p/2}dx\\
&\leq C\int_B(|\nabla \rho_1|^p +|\nabla
\rho_1|^2|\nabla w|^{p-2})dx,
\end{align*}
and noting $2\rho-\rho^2-1=-(1-\rho)^2 \leq 0$, we have
\begin{align*}
 E_{\varepsilon}(u_{\varepsilon},B)
&\leq E_{\varepsilon}(U,B)\\
&\leq \frac{1}{p} \int_B((2\rho_1-\rho_1^2)|\nabla
w|^2)^{p/2}dx +C\int_B(|\nabla \rho_1|^p +|\nabla \rho_1|^2|\nabla
w|^{p-2})dx \\
&\quad +\frac{1}{4\varepsilon^p} \int_B(1-\rho_1)^2dx\\
&\leq \frac{1}{p} \int_B|\nabla
w|^pdx+\delta\int_B|\nabla u_{\varepsilon}|^p
dx+C\varepsilon^{m+1-p}+CE(\rho_1,B).
\end{align*}
From this result and \eqref{2.12}, we  deduce
\begin{equation} \label{2.14}
E_{\varepsilon}(u_{\varepsilon},B) \leq \frac{1}{p} \int_B|\nabla
w|^pdx+C\varepsilon^{m+1-p} +\delta\int_B|\nabla
u_{\varepsilon}|^pdx.
\end{equation}
By Jensen's inequality and \eqref{2.14}, we obtain
\begin{equation} \label{2.15}
\begin{aligned}
&\frac{1}{p}\int_B|\nabla h|^pdx
+\frac{1}{p}\int_B(h^p-1)|\nabla w|^pdx
+\frac{1}{p}\int_B|\nabla u_3|^pdx\\
&+\frac{1}{4\varepsilon^p}\int_B(1-h^2)^2dx\\
&\leq E_{\varepsilon}(u_{\varepsilon},B)
-\frac{1}{p}\int_B|\nabla w|^p dx\\
&\leq C\varepsilon^{m-p+1}+\delta\int_B|\nabla u_{\varepsilon}|^p dx.
\end{aligned}
\end{equation}
Since $h \geq 1/2$ and Proposition \ref{prop2.1},  there
exists a $t \in (0,1/2)$ such that
\begin{equation} \label{2.16}
\begin{aligned}
&\frac{1}{p} \int_B(1-h^p)|\nabla w_\varepsilon|^p dx\\
&\leq \frac{2^p}{p}\int_B(1-h^p)|\nabla u_{\varepsilon}|^pdx\\
&\leq C\Big(\int_B |\nabla
u_{\varepsilon}|^{p+t}dx\Big)^{p/(p+t)}
\Big(\int_B(1-h^p)^{(p+t)/t}dx\Big)^{t/(p+t)}\\
&\leq C\Big(\int_{B(x,2r)} |\nabla
u_{\varepsilon}|^pdx+1\Big) \Big(\int_B(1-h^2)^2dx\Big)^{t/(p+t)}.
\end{aligned}
\end{equation}
Combining this with \eqref{2.15} we complete the proof.
\end{proof}

\subsection*{Proof of Theorem \ref{thm2.2}} \quad

\noindent\textbf{Step 1.} Since $|u'_{\varepsilon}|\geq 1/2$, there exists
$\phi \in W^{1,p}(B(x,3R)$,$[0,2\pi))$ such that
$w=\frac{u'_\varepsilon}{|u'_\varepsilon|}=(\cos \phi,\sin\phi)$.
Obviously, $|\nabla w|^2=|\nabla \phi|^2$. Substituting this into
\eqref{1.1} with the test function $(\psi,0)$ yields
\begin{align*}
&\int_{B(x,3R)}|\nabla u|^{p-2}(w\nabla h+h\nabla w)\nabla \psi dx\\
&=\int_{B(x,3R)}hw|\nabla u|^p \psi dx
+\frac{1}{\varepsilon^p}\int_{B(x,3R)}hw\psi (1-h^2) dx
\end{align*}
where $\psi \in W_0^{1,p}(G,\mathbb{R}^2)$.
Let $e^{i\phi}=\cos \phi +i \sin \phi$. Then
\begin{align*}
&\int_{B_{3R}(x)}he^{i\phi}|\nabla u|^p \psi
dx+\frac{1}{\varepsilon^p} \int_{B_{3R}(x)}h\psi e^{i
\phi}(1-h^2)dx\\
&=\int_{B_{3R}(x)}|\nabla u|^{p-2}(e^{i \phi}\nabla h
+h ie^{i\phi}\nabla \phi)\nabla \psi dx.
\end{align*}
Taking $\psi=e^{-i\phi} \zeta$, where $\zeta \in
W_0^{1,p}(B(x,3R),\mathbb{R}^2)$, we obtain
\begin{gather}
\label{2.17}
\begin{aligned}
&\frac{1}{\varepsilon^p} \int_{B(x,3R)}h(1-h^2)\zeta dx\\
&=\int_{B(x,3R)}|\nabla u|^{p-2}(\nabla h \nabla \zeta +h(|\nabla
\phi|^2-|\nabla u|^2)\zeta)dx.
\end{aligned} \\
\label{2.18}
0=\int_{B(x,3R)}|\nabla u|^{p-2} (h\nabla \phi \nabla \zeta
 -\zeta\nabla h\nabla \phi)dx.
\end{gather}
Taking $\zeta=h\xi$ in \eqref{2.18}, where $\xi \in
W_0^{1,p}(B(x,3R),\mathbb{R}^2)$, we have
\begin{equation} \label{2.19}
0=\int_{B(x,3R)}|\nabla u|^{p-2}h^2\nabla \phi\nabla \xi dx.
\end{equation}


Assume $\rho$ is an arbitrary constant in $(0,3R/2)$. Let $\zeta
\in W_0^{1,p}(B(x,2\rho),[0,1])$, and $\zeta=1$ on $B(x,\rho)$.
Taking $\xi=\phi \zeta^2$ in \eqref{2.19} and using the Young
inequality, for any $\eta \in (0,1)$ we obtain
$$
\int_{B(x,2\rho)}|\nabla u|^{p-2}h^2|\nabla \phi|^2\zeta^2 dx
\leq C\int_{B(x,2\rho)}|\nabla u|^{p-2}h^2 (\eta|\nabla
\phi|^2\zeta^2+C(\eta))dx.
$$
Choosing $\eta$ sufficiently small and noticing $\zeta=1$ on
$B(x,\rho)$, we obtain
\begin{equation} \label{2.20}
\int_{B(x,\rho)}|\nabla u|^{p-2}h^2|\nabla \phi|^2 dx\leq
C\Big(\int_{B(x,2\rho)}|\nabla u|^pdx\Big)^{1-2/p}.
\end{equation}
Applying \eqref{2.20} with $\rho=r$ we obtain
\begin{equation} \label{2.21}
\begin{aligned}
\int_B|\nabla u|^p
&\leq \int_B|\nabla u|^{p-2}(h^2|\nabla \phi|^2
 +|\nabla h|^2+|\nabla u_3|^2)dx\\
&\leq C\Big(\int_{B(x,2r)}|\nabla u|^pdx\Big)^{1-2/p}\\
&\quad +\Big(\int_B(|\nabla h|^p+|\nabla u_3|^p)dx\Big)^{2/p}
\Big(\int_B|\nabla u|^pdx\Big)^{(p-2)/p}\\
&\leq C\Big(\int_{B(x,2r)}|\nabla u|^pdx\Big)^{1-2/p} +\delta
\int_B|\nabla u|^p dx\\
&\quad +C(\delta)\int_B(|\nabla h|^p+|\nabla u_3|^p)dx.
\end{aligned}
\end{equation}
Substituting \eqref{2.13} into \eqref{2.21} and choosing $\delta>0$
sufficiently small we derive
\begin{equation} \label{2.22}
\begin{aligned}
\int_B|\nabla u|^p dx
&\leq C\Big(\int_{B(x,2r)}|\nabla u|^pdx\Big)^{1-2/p}
+C\varepsilon^{m-p+1}\\
&\quad +C\Big(\int_{B(x,2r)} |\nabla
u_{\varepsilon}|^pdx+1\Big) \Big[\int_B(1-h^2)^2dx\Big]^{t/(p+t)}.
\end{aligned}
\end{equation}
From \eqref{2.2} it follows that
\begin{equation} \label{2.23}
\int_B|\nabla u|^p dx\leq C(\varepsilon^{m-p})^{1-2/p}
+C\varepsilon^{m-p+1}
+C\varepsilon^{m-p+\frac{mt}{p+t}}=I_1+I_2+I_3.
\end{equation}
\smallskip

\noindent\textbf{Step 2.}
When $m \leq p/2$, then $m+1-p\leq (m-p)(1-2/p)$.
Therefore $I_1 \leq I_2$. Let $k_0 \in N$ be the minimum with the
property $m+1\leq (1+\frac{t}{p+t})^{k_0}m$.

In the following we shall improve the exponent
$m-p+\frac{t}{p+t}m$ of $\varepsilon$ in $I_3$ to $m-p+1$. Assume
$\zeta \in C_0^{\infty}(B(x,2R),[0,1])$ satisfying $\zeta=1$ on
$B_{m+1/2}$ and $|\nabla \zeta| \leq C$. Taking the test function
as $h \zeta(1-h)$ in \eqref{2.17}, we have
\begin{align*}
&\frac{1}{\varepsilon^p} \int_Bh^2(1-h^2)\zeta(1-h)dx
+\int_B|\nabla u|^{p-2}|\nabla h|^2h\zeta dx+\int_Bh^2|\nabla
u|^p(1-h)\zeta dx\\
&\leq \int_B|\nabla u|^{p-2}\nabla h\nabla \zeta
h(1-h) dx+\int_B|\nabla u|^p\zeta (1-h) \leq C\int_B|\nabla u|^pdx
\end{align*}
Noting $\zeta=1$ on $B_{m+1/2}$, applying $h \geq 1/2$ and \eqref{2.22}, we
obtain
$$
\frac{1}{\varepsilon^p} \int_{B_{m+1/2}}(1-h^2)^2 dx\leq
\frac{C}{\varepsilon^p} \int_Bh^2(1-h^2)(1-h)\zeta dx \leq
C(1+\varepsilon^{m-p+\frac{t}{p+t}m}),
$$
which implies
\begin{equation} \label{2.24}
\int_{B_{m+1/2}}(1-h^2)^2 dx\leq
C\varepsilon^{m(1+\frac{t}{p+t})}, \quad \varepsilon \in
(0,\varepsilon_0).
\end{equation}

On the other hand, similar to the derivation of \eqref{2.14}, for
$B_{m+1/2}$ we still conclude that for
any $\delta>0$,
$$
E_{\varepsilon}(u_{\varepsilon},B_{m+1/2})\leq
\frac{1}{p}\int_{B_{m+1/2}} |\nabla w|^pdx+C\varepsilon^{m-p+1}
+\delta\int_{B_{m+1/2}}|\nabla u_{\varepsilon}|^pdx.
$$
Therefore, \eqref{2.15} can be written as
\begin{equation} \label{2.25}
\begin{aligned}
&\frac{1}{p}\int_{B_{m+1/2}}|\nabla h|^pdx
+\frac{1}{p}\int_{B_{m+1/2}}|\nabla u_3|^pdx
+\frac{1}{4\varepsilon^p}
\int_{B_{m+1/2}}(1-h^2)^2dx\\
&\leq C\varepsilon^{m-p+1}+\frac{1}{p}
\int_{B_{m+1/2}} (1-h^p)|\nabla w|^pdx+\delta\int_{B_{m+1/2}}
|\nabla u_{\varepsilon}|^pdx.
\end{aligned}
\end{equation}
To estimate the second term of the right hand side of \eqref{2.25}, we
apply \eqref{2.23} and \eqref{2.24} to obtain
$$
\frac{1}{p} \int_{B_{m+1/2}}(1-h^p)|\nabla w|^p dx\leq
C\varepsilon^{(m+\frac{t}{p+t}m)\frac{t}{p+t}+m +\frac{t}{p+t}m-p}
=C\varepsilon^{m(1+\frac{t}{p+t})^2-p}
$$
by the same way as for \eqref{2.16}. Substituting this
into \eqref{2.25} yields
$$
\frac{1}{p}\int_{B_{m+1/2}}(|\nabla h|^p+|\nabla u_3|^p) dx\leq
C(\varepsilon^{m-p+1}+\varepsilon^{m(1+\frac{t}{p+t})^2-p})
+\delta\int_{B_{m+1/2}}|\nabla u_{\varepsilon}|^pdx.
$$
Using this instead of \eqref{2.13} and by the same argument of Step 1
we can improve \eqref{2.23} as
$$
\int_{B_{m+1/2}}|\nabla u_{\varepsilon}|^p dx\leq
C+C(\varepsilon^{m-p+1}+\varepsilon^{m(1+ \frac{t}{p+t})^2-p})
 \leq C\varepsilon^{m(1+\frac{t}{p+t})^2-p}.
$$
Now, we use this inequality replacing \eqref{2.23} to discuss,
thus \eqref{2.24} can be written as
$$
\int_{B_{m+3/4}}(1-h^2)^2 dx\leq
C\varepsilon^{m(1+\frac{t}{p+t})^2}, \quad \varepsilon \in
(0,\varepsilon_0).
$$
As a result, it is also follows that, as the derivation of \eqref{2.16}
and \eqref{2.23},
\begin{gather*}
\frac{1}{p} \int_{B_{m+3/4}}(1-h^p)|\nabla w|^p dx\leq
C\varepsilon^{m(1+\frac{t}{p+t})^3-p},
\\
\int_{B_{m+3/4}}|\nabla u_{\varepsilon}|^p dx\leq
C+C(\varepsilon^{m-p+1}+\varepsilon^{m(1+ \frac{t}{p+t})^3-p})\leq
C\varepsilon^{m(1+\frac{t}{p+t})^3-p}.
\end{gather*}

If we do in this way, and noting the definition of $k_0$, we can
derive by $k_0$ steps that
$$
\int_{B_{m+1-1/{2^{k_0-1}}}}|\nabla u_{\varepsilon}|^pdx \leq
C+C(\varepsilon^{m-p+1}+\varepsilon^{m
(1+\frac{t}{p+t})^{k_0}-p}).
$$
Thus
$$
\int_{B_{m+1}}|\nabla u_{\varepsilon}|^p dx\leq
\int_{B_{m+1-1/{2^{k_0-1}}}}|\nabla u_{\varepsilon}|^p dx\leq
C(\varepsilon^{m-p+1}+1).
$$
This is \eqref{2.2} for $j=m+1$.
\smallskip

\noindent\textbf{Step 3.}
When $m>p/2$, $(m-p)(1-2/p)<m+1-p$. Let $k \geq 1$
be an integer such that $(m-p)(1-2/p)^k \leq
m+1-p<(m-p)(1-2/p)^{k+1}$. Now, $I_1 \geq I_2$ in \eqref{2.23}.
Thus,
$$
\int_B|\nabla u|^p dx\leq C(\varepsilon^{m-p})^{1-2/p}
+C\varepsilon^{m-p+\frac{mt}{(p+t)}}.
$$
Similar to Step 2, we may improve the exponent
$m-p+\frac{mt}{p+t}$ of $\varepsilon$ in $I_3$ to $(m-p)(1-2/p)$
since we may find $k_0>0$ such that
$m(1+\frac{t}{p+t})^{k_0}-p>(m-p)(1-2/p)$. Namely, there is a
constant $r_1 \in (R_{m+1},r)$ such that
$$
\int_{B(x,r_1)}|\nabla u_{\varepsilon}|^p dx\leq
C\varepsilon^{(m-p)(1-2/p)}.
$$
Therefore, as the derivation of \eqref{2.24},
$$
\int_{B(x,2r_1/3)}(1-h^2)^2dx \leq C\varepsilon^{(m-p)(1-2/p)+p}.
$$
Substituting these into \eqref{2.22} we have
\begin{align*}
&\int_{B(x,r_1/2)}|\nabla u_{\varepsilon}|^pdx\\
&\leq C\varepsilon^{m+1-p}+ C\Big[\int_{B(x,r)}|\nabla
u_{\varepsilon}|^pdx\Big]^{1-2/p}\\
&\quad +C\Big(\int_{B(x,r)}|\nabla u_{\varepsilon}|^pdx+1\Big)
\Big[\int_{B(x,r)}(1-h^2)^2dx\Big]^{\frac{t}{p+t}}\\
&\leq C\varepsilon^{m+1-p}+
C\varepsilon^{(m-p)(1-2/p)^2}+C\varepsilon^
{(m-p)(1-2/p)+[(m-p)(1-2/p)+p]\frac{t}{p+t}}.
\end{align*}
Noting $(m-p)(1-2/p)^2<m+1-p$, we can see that
$$
\int_{B(x,r_1/2)}|\nabla u_{\varepsilon}|^p dx\leq
C\varepsilon^{(m-p)(1-2/p)^2}+C\varepsilon^
{(m-p)(1-2/p)+[(m-p)(1-2/p)+p]\frac{t}{p+t}}.
$$
Using the idea of Step 2, we can improve the exponent
$(m-p)(1-2/p)+[(m-p)(1-2/p)+p]\frac{t}{p+t}$ of $\varepsilon$ to
$(m-p)(1-2/p)^2$. Namely, there is a constant 
$r_2 \in (R_{m+1},r_1/2)$ such that
$$
\int_{B(x,r_2)}|\nabla u_{\varepsilon}|^p dx\leq
C\varepsilon^{(m-p)(1-2/p)^2}.
$$
Suppose that for some $l \leq k-1$,
$$
\int_{B(x,r_{l-1})}|\nabla u_{\varepsilon}|^p dx\leq
C\varepsilon^{(m-p)(1-2/p)^l}
$$
holds, where $R_{m+1}<r_{l+1}<r_l/2$ for $l=2,3,\cdots,k-1$.
Therefore, as the derivation of \eqref{2.24},
$$
\int_{B(x,r_{l-1})}(1-h^2)^2dx 
\leq C\varepsilon^{(m-p)(1-2/p)^l+p}.
$$
Substituting these inequalities into \eqref{2.22} yields
\begin{align*}
&\int_{B(x,r_l)}|\nabla u_{\varepsilon}|^pdx \\
&\leq C\varepsilon^{m+1-p}+ C\varepsilon^{(m-p)(1-2/p)^{l+1}}
+C\varepsilon^ {(m-p)(1-2/p)^l+[(m-p)(1-2/p)^l+p]\frac{t}{p+t}}\\
&\leq C\varepsilon^{(m-p)(1-2/p)^{l+1}}+C\varepsilon^
{(m-p)(1-2/p)^l+[(m-p)(1-2/p)^l+p]\frac{t}{p+t}}
\end{align*}
Similar to Step 2, we may improve again the exponent
$(m-p)(1-2/p)^l+[(m-p)(1-2/p)^l+p]\frac{t}{p+t}$ of $\varepsilon$
to $(m-p)(1-2/p)^{l+1}$. Namely, it can be seen that
$$
\int_{B(x,r_l)}|\nabla u_{\varepsilon}|^p dx\leq
C\varepsilon^{(m-p)(1-2/p)^{l+1}}.
$$
From this result it follows that for $l=k-1$,
$$
\int_{B(x,r_{k-1})}|\nabla u_{\varepsilon}|^p \leq
C\varepsilon^{(m-p)(1-2/p)^k}.
$$
Therefore, as the derivation of \eqref{2.24},
$$
\int_{B(x,r_{l-1})}(1-h^2)^2dx \leq
C\varepsilon^{(m-p)(1-2/p)^k+p}.
$$
Combining these with \eqref{2.22} we obtain
\begin{align*}
&\int_{B(x,\frac{r_{k-1}}{2})}|\nabla
u_{\varepsilon}|^pdx \\
&\leq C\varepsilon^{m+1-p}+
C\varepsilon^{(m-p)(1-2/p)^{k+1}}+C\varepsilon^
{(m-p)(1-2/p)^k+[(m-p)(1-2/p)^k+p]\frac{t}{p+t}}\\
&\leq C\varepsilon^{m+1-p}+
C\varepsilon^{(m-p)(1-2/p)^k+[(m-p)(1-2/p)^k+p]\frac{t}{p+t}}.
\end{align*}
As in Step 2 and noting the definition of $k$, we may
also improve the exponent of $\varepsilon$ to $m+1-p$ finally.
Namely, we have
$$
\int_{B(x,r_{k-1}/2)}|\nabla u_{\varepsilon}|^p \leq
C\varepsilon^{m+1-p}.
$$
This is \eqref{2.2} for $j=m+1$ and proof of
Theorem \ref{thm2.2} is complete.

\begin{theorem} \label{thm2.5}
For an arbitrary compact subset $K$ of $G\setminus
\{a_1,a_2,\dots ,a_N\}$. There exists a constant $C>0$ which does not
depend on $\varepsilon \in (0,1)$ such that
$E_{\varepsilon}(u_{\varepsilon},K)\leq C$.
\end{theorem}

\begin{proof} 
It is sufficient to prove that
$E_{\varepsilon}(u_{\varepsilon},B(x,R))\leq C$, where $B(x,R)$ is
the disc in $G\setminus \{a_1,a_2,\dots ,a_N\}$. 
Theorem \ref{thm2.2} shows that
\begin{equation} \label{2.26}
E_{\varepsilon}(u_{\varepsilon},B_{[p]})\leq C\varepsilon^{[p]-p}.
\end{equation}
Using this and the integral mean value theorem,
there exists a constant $r \in [R_{[p]+1/2},R_{[p]}]$ such that
\begin{equation} \label{2.27}
\int_{\partial B(x,r)}|\nabla u_{\varepsilon}|^pd\xi
+\frac{1}{\varepsilon^p} \int_{\partial B(x,r)}u_{\varepsilon
3}^2d\xi \leq C(r)\varepsilon^{[p]-p}.
\end{equation}
Consider the functional
$$
E(\rho,B)=\frac{1}{p} \int_B(|\nabla \rho|^2+1)^{p/2}dx
+\frac{1}{2\varepsilon^p} \int_B(1-\rho)^2dx,
$$
where $B=B(x,r)$. It is easy to prove that the minimizer $\rho_2$
of $E(\rho,B)$ on $W_{|u'_{\varepsilon}|}^{1,p} (B,\mathbb{R}^+\cup\{0\})$
exists. Similar to the proof of proposition \ref{prop2.3}, by \eqref{2.26} and
\eqref{2.27} we can derive
\begin{equation} \label{2.28}
E(\rho_2,B)\leq C\varepsilon^{[p]-p+1}.
\end{equation}
From this it follows that for any $\delta>0$,
$$
\int_B|\nabla \rho_2|^2|\nabla w|^{p-2} dx\leq \delta\int_B|\nabla
u_{\varepsilon}|^p dx+C\varepsilon^{[p]+1-p}.
$$
Since $u_{\varepsilon}$ is a minimizer of $E_{\varepsilon}(u,G)$,
we have
\begin{equation} \label{2.29}
\begin{aligned}
E_{\varepsilon}(u_{\varepsilon},B)
&\leq E_{\varepsilon}((\rho_2 w,\sqrt{1-\rho_2^2}),B)\\
&\leq \frac{1}{p} \int_B(\rho_2^2|\nabla
w|^2)^{p/2}dx +C\int_B(|\nabla \rho_2|^p +|\nabla \rho_2|^2|\nabla
w|^{p-2})dx\\
&\quad +\frac{1}{4\varepsilon^p} \int_B(1-\rho_2^2)^2dx.
\end{aligned}
\end{equation}
Therefore,
$$
E_{\varepsilon}(u_{\varepsilon},B) 
\leq \frac{1}{p}\int_B|\nabla w|^pdx 
+C\varepsilon^{[p]+1-p}+\delta\int_B|\nabla u_{\varepsilon}|^p dx.
$$
Combining this with Jensen's inequality yields
\begin{equation} \label{2.30}
\begin{aligned}
&\frac{1}{p}\int_B|\nabla
h|^pdx+\frac{1}{p}\int_B|\nabla
u_3|^pdx +\frac{1}{4\varepsilon^p}\int_B(1-h^2)^2 \\ 
&\leq E_{\varepsilon}(u_{\varepsilon},B)
-\frac{1}{p}\int_B|\nabla
w|^pdx+\frac{1}{p}\int_B(1-h^p)|\nabla w|^pdx\\
&\leq C\varepsilon^{[p]+1-p}+\delta \int_B|\nabla
u_{\varepsilon}|^p dx+\frac{1}{p}\int_B(1-h^p)|\nabla
w|^pdx.
\end{aligned}
\end{equation}
To estimate the third term of the right hand side, we proceed in
the same way of the proof of Proposition \ref{prop2.4}, and use
$\frac{1}{\varepsilon^p} \int_B(1-h^2)^2dx\leq
C\varepsilon^{[p]-p}$ which is implied by \eqref{2.26}. As a
result, there exists $t \in (0,1/2)$ such that
$$
\frac{1}{p}\int_B(1-h^p)|\nabla w|^p dx\leq C
\varepsilon^{[p]+[p]t/(p+t)-p}.
$$
Substituting this into \eqref{2.30} yields
\begin{align*}
&\frac{1}{p}\int_B(|\nabla h|^p+|\nabla u_3|^p)dx+
\frac{1}{4\varepsilon^p}\int_B(1-h^2)^2 dx \\
&\leq C(\varepsilon^{[p]+1-p}+\varepsilon^{[p]+\frac{[p]t}{p+t}-p})
+\delta\int_B|\nabla u_{\varepsilon}|^pdx.
\end{align*}
This and \eqref{2.21} imply that
\begin{equation} \label{2.31}
\int_B|\nabla u_{\varepsilon}|^p dx\leq
C\varepsilon^{[p]-p+1}+C\varepsilon^{[p]-p+\frac{t}{p+t}m}
+C\varepsilon^{([p]-p)(1-2/p)}+C,
\end{equation}
as long as we choose $\delta>0$ sufficiently small. Discussing in
the same way to Step 2 and Step 3, we may improve the exponent of
$\varepsilon$ in the second and the third terms of the right hand
side of \eqref{2.31} step by step such that the improved exponent is not
smaller than $[p]-p+1$, thus for some $B_{[p]+1} \subset B$, there
exists $C$ independent of $\varepsilon \in (0,\varepsilon_0)$ with
$\varepsilon_0$ sufficiently small such that
$$
\int_{B_{[p]+1}}|\nabla u_{\varepsilon}|^p dx\leq
C+C\varepsilon^{[p]+1-p}\leq C.
$$
The proof is complete.
\end{proof}


\section{Proof of Theorem \ref{thm1.2}}

\textbf{Step 1.} Suppose $B(x_0,2\sigma) \subset [G\setminus
\cup_{j=1}^N \{a_j\}]$, where the constant $\sigma$ may be
sufficiently small but independent of $\varepsilon$. Since theorem
2.5 implies $E_{\varepsilon}(u_{\varepsilon},B(x_0,2\sigma)
\setminus B(x_0,\sigma)) \leq C$, there is a constant $r \in
(\sigma, 2\sigma)$ such that
$$
\int_{\partial B(x_0,r)}|\nabla u_{\varepsilon}|^pd\xi
+\frac{1}{\varepsilon^p}\int _{\partial B(x_0,r)}u_{\varepsilon
3}^2d\xi \leq C(r).
$$
Thus, we can find a subsequence $u_{\varepsilon_k}$ of
$u_{\varepsilon}$ such that $u_{\varepsilon_k} \to u_p=(u'_p,0)$
in $C(\partial B(x_0,r),\mathbb{R}^3)$, where $u'_p$ is the $S^1$-valued
harmonic map, which leads to
\begin{equation} \label{3.2}
\frac{u'_{\varepsilon_k}}{|u'_{\varepsilon_k}|} \to u'_p,
\quad\text{in} \quad C(\partial B(x_0,r)).
\end{equation}
\smallskip

\noindent\textbf{Step 2.} 
Denote $B=B(x_0,r)$. It is easy to see the existence
of the solution $w_\varepsilon$ of
\begin{equation} \label{3.3}
\min\{\int_B|\nabla u|^p dx: u \in
W_{\frac{u'_{\varepsilon}}{|u'_{\varepsilon}|}}^{1,p}(B,\partial
B_1)\}.
\end{equation}
Theorem \ref{thm2.5} and $|u'_{\varepsilon}|\geq 1/2$ on $B$ imply
$2^{-p}\int_B|\nabla
\frac{u'_{\varepsilon}}{|u'_{\varepsilon}|}|^p dx\leq
\int_B|\nabla u_{\varepsilon}|^p dx\leq C$, and hence
\begin{equation} \label{3.4}
\int_B|\nabla w_\varepsilon|^p dx\leq \int_B|\nabla
\frac{u'_{\varepsilon}}{|u'_{\varepsilon}|}|^pdx\leq C.
\end{equation}
From this and \eqref{2.28} it follows that $\int_B|\nabla
\rho_2|^2|\nabla w_\varepsilon|^{p-2} dx\leq
C\varepsilon^{2([p]+1-p)/p}$, where $\rho_2$ is the minimizer of
$E(\rho,B)$ on $W_{|u'_{\varepsilon}|}^{1,p} (B,\mathbb{R}^+\cup\{0\})$.
Substituting this result into \eqref{2.29} and using \eqref{2.28},
we obtain
\begin{equation} \label{3.5}
\int_B|\nabla u_{\varepsilon}|^pdx \leq
C\varepsilon^{2([p]+1-p)/p}+\int_B|\nabla w_\varepsilon|^pdx.
\end{equation}
\smallskip

\noindent\textbf{Step 3.} Let $w_\varepsilon^{\tau}$ be a solution of
\begin{equation} \label{3.6}
\min\big\{\int_B(|\nabla w|^2+\tau)^{p/2} dx: w \in
W_{\frac{u'_{\varepsilon}}{|u'_{\varepsilon}|}}^{1,p} (B,\partial
B_1)\big\},\quad \tau \in (0,1).
\end{equation}
Clearly, $w_\varepsilon^{\tau}$ also solves
\begin{equation} \label{3.7}
-\mathop{\rm div}({v_\varepsilon^{\tau}}^{(p-2)/2}\nabla w)
=w|\nabla w|^2{v_\varepsilon^{\tau}}^{(p-2)/2}, \quad
v_\varepsilon^{\tau}=|\nabla w|^2+\tau.
\end{equation}
Noticing $\frac{u'_{\varepsilon}}{|u'_{\varepsilon}|} \in
W_{\frac{u'_{\varepsilon}}{|u'_{\varepsilon}|}}^{1,p} (B,\partial
B_1)$, we have
\begin{equation} \label{3.8}
\begin{aligned}
\int_B|\nabla w_\varepsilon^{\tau}|^p dx 
&\leq \int_B(|\nabla w_\varepsilon^{\tau}|^2+\tau)^{p/2} dx\\
&\leq \int_B(|\nabla \frac{u'_{\varepsilon}}{|u'_{\varepsilon}|}|^2
 +\tau)^{p/2} dx \\
&\leq \int_B(|\nabla
\frac{u'_{\varepsilon}}{|u'_{\varepsilon}|}|^2+1)^{p/2}dx \leq C
\end{aligned}
\end{equation}
by using \eqref{3.4}, where $C$ is a constant which is independent
of $\varepsilon, \tau$. Then there exist 
$w^* \in W_{\frac{u'_{\varepsilon}}{|u'_{\varepsilon}|}}^{1,p} (B,\partial
B_1)$ and a subsequence of $w_\varepsilon^{\tau}$ denoted still by
itself such that
\begin{equation} \label{3.9}
\lim_{\tau \to 0}w_\varepsilon^{\tau} = w^*\quad\text{weakly in } W^{1,p}(B,{R}^2).
\end{equation}
Noting the weak lower semi-continuity of $\int_B|\nabla w|^p$, we
have
\begin{equation} \label{3.10}
\int_B|\nabla w^*|^p dx\leq \liminf_{\tau \to 0}
\int_B|\nabla w_\varepsilon^{\tau}|^p dx \leq
\limsup_{\tau \to 0} \int_B|\nabla w_\varepsilon^{\tau}|^p
dx.
\end{equation}
The fact that $w_\varepsilon^{\tau}$ solves \eqref{3.6} implies
$$
\limsup_{\tau \to 0} \int_B(|\nabla
w_\varepsilon^{\tau}|^2+\tau)^{p/2} dx\leq \lim_{\tau \to 0}
\int_B(|\nabla w_\varepsilon|^2+\tau)^{p/2}dx =\int_B|\nabla
w_\varepsilon|^pdx,
$$
where $w_\varepsilon$ is a solution of \eqref{3.3}. This and \eqref{3.10} lead
to
\begin{equation} \label{3.11}
\int_B|\nabla w^*|^p dx
\leq \liminf_{\tau \to 0} \int_B|\nabla w_\varepsilon^{\tau}|^p dx
\leq \limsup_{\tau \to 0} \int_B|\nabla w_\varepsilon^{\tau}|^p dx
\leq \int_B|\nabla w_\varepsilon|^p dx.
\end{equation}
Since $w^* \in
W_{\frac{u'_{\varepsilon}}{|u'_{\varepsilon}|}}^{1,p} (B,\partial
B_1)$, we know $w^*$ also solves \eqref{3.3}, namely
\begin{equation} \label{3.12}
\int_B|\nabla w_\varepsilon|^p dx=\int_B|\nabla w^*|^p dx.
\end{equation}
Combining this with \eqref{3.11} yields 
$\lim_{\tau \to 0}\int_B|\nabla
w_\varepsilon^{\tau}|^p dx=\int_B|\nabla w^*|^pdx$, which and
\eqref{3.9} imply that as $\tau \to 0$,
\begin{equation} \label{3.13}
\nabla w_\varepsilon^{\tau} \to \nabla w^* \quad \text{in }
L^p(B,{R}^2).
\end{equation}
\smallskip

\noindent\textbf{Step 4.} By the same argument as  in Step 3, we obtain the
following conclusion: Let $u^{\tau}$ be a solution of
\begin{equation} \label{3.14}
\min\{\int_B(|\nabla u|^2+\tau)^{p/2} dx: u \in
W_{u'_p}^{1,p}(B,\partial B_1)\},\quad \tau \in (0,1).
\end{equation}
Then $u^{\tau}$ satisfies
\begin{equation} \label{3.15}
\int_B|\nabla u^{\tau}|^p dx\leq C,
\end{equation}
where $C$ is which is independent of $\tau$, and $u^{\tau}$ solves
\begin{equation} \label{3.16}
-\mathop{\rm div}[(v^{\tau})^{(p-2)/2}\nabla u] =u|\nabla
u|^2v^{(p-2)/2}, \quad v^{\tau}=|\nabla u|^2+\tau.
\end{equation}
As $\tau \to 0$, there exists a subsequence of $u^{\tau}$ denoted
by itself such that
\begin{equation} \label{3.17}
\nabla u^{\tau} \to \nabla u^* \quad\text{in } L^p(B,{R}^2),
\end{equation}
where $u^*$ is a minimizer of $\int_B|\nabla u|^p dx$ in
$W_{u'_p}^{1,p}(B,\partial B_1)$. It is well-known that $u^*$ is a
map of the least p-energy, and also a p-harmonic map.
\smallskip

\noindent\textbf{Step 5.} 
From \cite[Lemma 1, Page 65]{BZ}, we can write
\begin{gather*}
w_\varepsilon^{\tau}=(\cos \phi_\varepsilon^{\tau},\sin
\phi_\varepsilon^{\tau}),\quad
u^{\tau}=(\cos \psi^{\tau},\sin \psi^{\tau}),\\
w_\varepsilon=(\cos \phi_\varepsilon^*,\sin
\phi_\varepsilon^*),\quad
u^*=(\cos \psi^*,\sin \psi^*),\\
\frac{u'_\varepsilon}{|u'_\varepsilon|}|_{\partial B} =(\cos
\phi_\varepsilon,\sin \phi_\varepsilon),\quad u'_p|_{\partial
B}=(\cos \psi,\sin \psi),
\end{gather*}
where $\phi_\varepsilon^{\tau}, \psi^{\tau}, \phi_{\varepsilon}^*,
\psi^*$ belong to $W^{1,p}(B,R)$, $\phi^*, \psi$ belong to
$W^{1,p}(\partial B,R)$, and they are all single-valued functions
since their degrees around $\partial B$ are zero. Therefore,
\begin{equation} \label{3.18}
\phi_\varepsilon^{\tau}|_{\partial B}=\phi_\varepsilon,
\quad\psi^{\tau}|_{\partial B}=\psi,
\end{equation}
and $|\nabla w_\varepsilon^{\tau}|=|\nabla \phi_\varepsilon^{\tau}|$, 
$|\nabla u^{\tau}|=|\nabla \psi^{\tau}|$. 
$|\nabla w_\varepsilon|=|\nabla \phi_\varepsilon^*|$, 
$|\nabla u^*|=|\nabla \psi^*|$. Moreover, by
\eqref{3.7} and \eqref{3.16}, we obtain that both
$\phi_\varepsilon^{\tau}$ and $\psi^{\tau}$ satisfy 
$-\operatorname{div}[(|\nabla \Phi|^2+\tau)^{(p-2)/2}\nabla \Phi]=0$. Thus,
\begin{equation} \label{3.19}
-\operatorname{div}[(|\nabla \phi_\varepsilon^{\tau}|^2+\tau)^{(p-2)/2} \nabla
\phi_\varepsilon^{\tau} -(|\nabla
\psi^{\tau}|^2+\tau)^{(p-2)/2}\nabla \psi^{\tau}]=0.
\end{equation}
Multiplying both sides of \eqref{3.19} by 
$\phi_\varepsilon^{\tau} -\psi^{\tau}$ and integrating over $B$, we obtain
\begin{equation} \label{3.20}
\begin{aligned}
&-\int_{\partial
B}({v_\varepsilon^{\tau}}^{(p-2)/2}\phi_{\nu}
-v^{(p-2)/2}\psi_{\nu})(\phi-\psi)d\xi\\
&+\int_B({v_\varepsilon^{\tau}}^{(p-2)/2}\nabla \phi
-v^{(p-2)/2}\nabla \psi)\nabla (\phi-\psi)dx=0,
\end{aligned}
\end{equation}
where $\nu$ denotes the unit outside-norm vector of $\partial B$.

Let $w=w_\varepsilon^{\tau}$ be a solution of \eqref{3.6}.
Integrating both sides of \eqref{3.7} over $B$, we have
$$
-\int_{\partial B}{v_\varepsilon^{\tau}}^{(p-2)/2}w_{\nu}d\xi
=\int_Bw|\nabla w|^2{v_\varepsilon^{\tau}}^{(p-2)/2}dx,
$$
this and \eqref{3.8} imply
\begin{equation} \label{3.21}
\big|\int_{\partial B}{v_\varepsilon^{\tau}}^{(p-2)/2}\phi_{\nu}d\xi\big|
=\big|\int_{\partial B}{v_\varepsilon^{\tau}}^{(p-2)/2}w_{\nu}d\xi\big|
\leq \int_B {v_\varepsilon^{\tau}}^{p/2} dx\leq C.
\end{equation}
An analogous discussion shows that for the solution $u=u^{\tau}$
of \eqref{3.14} which is equipped with \eqref{3.15}, we may also obtain
\begin{equation} \label{3.22}
\big|\int_{\partial B}v^{(p-2)/2}\psi_{\nu}d\xi\big| 
=\big|\int_{\partial B}v^{(p-2)/2}u_{\nu}d\xi\big| 
\leq \int_B|\nabla u|^p dx\leq C.
\end{equation}
Combining \eqref{3.18} with \eqref{3.20}-\eqref{3.22}, we derive
$$
\int_B({v_\varepsilon^{\tau}}^{(p-2)/2}\nabla \phi
-v^{(p-2)/2}\nabla \psi)\nabla (\phi-\psi) dx\leq C\sup_{\partial
B}|\phi_\varepsilon^{\tau}-\psi^{\tau}| =C\sup_{\partial
B}|\phi_\varepsilon-\psi|,
$$
where $C$ is independent of $\varepsilon, \tau$. Letting $\tau \to
0$ and applying \eqref{3.13} and \eqref{3.17}, we obtain
$$
\big|\int_B(|\nabla \phi_\varepsilon^*|^{(p-2)/2}\nabla
\phi_\varepsilon^* -|\nabla \psi^*|^{(p-2)/2}\nabla \psi^*)\nabla
(\phi_\varepsilon^*-\psi^*)dx\big| 
\leq C\sup_{\partial B}|\phi_\varepsilon-\psi|,
$$
which implies $\int_B|\nabla \phi_\varepsilon^*-\nabla \psi^*|^p
dx\leq C\sup_{\partial B}|\phi_\varepsilon-\psi|$. Letting
$\varepsilon \to 0$ and using \eqref{3.2}, we obtain 
$\int_B|\nabla \phi_\varepsilon^*|^p dx\to \int_B|\nabla \psi^*|^pdx$.
 That is,
\begin{equation} \label{3.23}
\int_B|\nabla w_\varepsilon|^pdx \to \int_B|\nabla u^*|^pdx.
\end{equation}
\smallskip

\noindent\textbf{Step 6.} 
Since $\int_B|\nabla u|^pdx$ is weak lower
semi-continuous, from Theorem \ref{thm1.2*} we deduce 
$\int_B|\nabla u_p|^pdx \leq \liminf_{\varepsilon_k \to 0} \int_B|\nabla
u_{\varepsilon_k}|^pdx$. Combining this result with \eqref{3.5},
\eqref{3.12} and \eqref{3.23}, we obtain
\begin{align*}
\int_B|\nabla u_p|^p dx
&\leq \liminf_{\varepsilon_k \to 0}\int_B|\nabla
u_{\varepsilon_k}|^p dx\leq \limsup_{\varepsilon_k \to
0}\int_B|\nabla u_{\varepsilon_k}|^pdx\\
&\leq \lim_{\varepsilon_k \to 0}\int_B|\nabla
w_\varepsilon|^pdx =\int_B|\nabla u^*|^pdx.
\end{align*}
Recalling the definition of $u^*$ in Step 4, and noticing $u'_p
\in W_{u'_p}^{1,p}(B,\partial B_1)$, we know that $u'_p$ is also a
minimizer of $\int_B|\nabla u|^p$, and
$$
\lim_{\varepsilon_k \to 0}\int_B|\nabla
u_{\varepsilon_k}|^pdx=\int_B|\nabla u_p|^pdx=\int_B|\nabla
u^*|^pdx.
$$
This result and Theorem \ref{thm1.2*} imply $\nabla u_{\varepsilon_k} \to
\nabla u_p$ in $L^p(B,{R}^3)$. when $\varepsilon_k \to 0$.
Combining this with the fact $u_{\varepsilon_k} \to u_p$ in
$L^p(B,{R}^3)$, which is implied by Theorem \ref{thm1.2*}, we obtain
$$
u_{\varepsilon_k} \to u_p, \quad\text{in } W^{1,p}(B,{R}^3)
$$
as $\varepsilon_k \to 0$. Then it is not difficult to complete the
proof of this theorem.


\subsection*{Acknowledgements} 
The authors wish to express their appreciation to
the anonymous referees for their suggestions that greatly improved
this article.


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\end{document}