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\markboth{\hfil Asymptotic behavior of regularizable minimizers
\hfil EJDE--2001/15}
{EJDE--2001/15\hfil Yutian Lei \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 2001}(2001), No. 15, pp. 1--28. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu \quad ftp ejde.math.unt.edu (login: ftp)}
 \vspace{\bigskipamount} \\
 %
 Asymptotic behavior of regularizable minimizers
 of a Ginzburg-Landau functional in higher dimensions
 %
\thanks{ {\em Mathematics Subject Classifications:} 35J70.
\hfil\break\indent
{\em Key words:} Ginzburg-Landau functional, module and zeroes of
regularizable minimizers.
\hfil\break\indent
\copyright 2001 Southwest Texas State University. \hfil\break\indent
Submitted December 13, 2000. Published February 23, 2001.} }
\date{}
%
\author{Yutian Lei }
\maketitle

\begin{abstract}
 We study the asymptotic behavior of the regularizable minimizers 
 of a Ginzburg-Landau type functional. We also dicuss the location 
 of the zeroes of the minimizers.
\end{abstract}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}

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\@addtoreset{equation}{section}
\catcode`@=12

\section{Introduction }
Let $G \subset \mathbb{R}^n$ ($n\geq 2$) be a bounded and simply connected
domain with smooth boundary $\partial G$. Let $g$ be a smooth map
from $\partial G$ into $S^{n-1}$ satisfying $d=\deg(g,\partial G)
\neq 0$. Consider the Ginzburg-Landau-type functional
$$
E_\varepsilon(u,G)=
\frac{1}{p}\int_G|\nabla u|^p
+\frac{1}{4\varepsilon^p}\int_G(1-|u|^2)^2 ,\quad (p>1)
$$
with a small parameter $\varepsilon >0$. It is
known that this functional achieves its minimum on $$ W_p=\{v \in
W^{1,p}(G,\mathbb{R}^n) :v|_{\partial G}=g\} $$ at a function
$u_{\varepsilon}$. We are concerned with the asymptotic behavior of
$u_{\varepsilon}$ and the location of the zeroes of $u_{\varepsilon}$
as $\varepsilon \to 0$.

The functional $E_{\varepsilon}(u,G)$ was introduced in
the study of the Ginzburg-Landau vortices by F. Bethue,
H. Brezis and F. Helein [1] in the case $p=n=2$.
Similar models are also used
in many other theories of phase transition.
The minimizer $u_{\varepsilon}$ of
$E_{\varepsilon}(u,G)$ represents a complex order parameter.
The zeroes of $u_{\varepsilon}$ and the module $|u_{\varepsilon}|
$ both have physics senses, for example,
in superconductivity $|u_{\varepsilon}|^2$
is proportional to the density of supercoducting electrons,
and the zeroes of $u_{\varepsilon}$
are the vortices, which were introduced
in the type-II superconductors.

In the case $1<p<n$, it is easily seen that $W_g^{1,p}(G,S^{n-1})
\neq \emptyset$. It is not difficult to prove that the existence of
solution $u_p$ for the minimization problem
$$
\min\{\int_G|\nabla u|^p : u \in W_g^{1,p}(G,S^{n-1})\}
$$
by taking the minimizing sequence. This solution
is called a map of the least p-energy with boundary value
$g$. Using the variational methods,
we can proved that the solution
$u_p$ is also p-harmonic map on $G$ with the boundary data $g$,
namely, it is a weak solution of the following equation
$$
-\mathop{\rm div}(|\nabla u|^{p-2}\nabla u)
=u|\nabla u|^p.
$$
As $\varepsilon \to 0$, there exists a
subsequence $u_{\varepsilon_k}$ of
$u_{\varepsilon}$, the minimizer of $E_{\varepsilon}(u,G)$, such that
$$
u_{\varepsilon_k} \to u_p, \quad\mbox{in } W^{1,p}(G,\mathbb{R}^n).
$$

In the case $p>n$, $W_g^{1,p}(G,S^{n-1})=\emptyset$. Thus there is no
 map of least p-energy on $G$
with the boundary value $g$. It seems to be very
difficult to study the convergence for minimizers of
$E_{\varepsilon}(u,G)$ in $W_p$. Some results on the asymptotic behavior
of the radial minimizers of $E_{\varepsilon}(u,G)$ were presented in [7].

When $p=n$, this problem was introduced in [1] (the open problem 17).
M. C. Hong studied the asymptotic behavior for the
regularizable minimizers of $E_{\varepsilon}(u,G)$
in $W_n$ [6]. He proved
that there exist $\{a_1,a_2,\dots,a_J\}\subset \overline{G}$, $J\in N$
and a subsequence $u_{\varepsilon_k}$ of the regularizable
minimizers $u_{\varepsilon}$ such that
$$
u_{\varepsilon_k} \stackrel{w}{\to} u_n,
\quad\mbox{in } W_{{\rm loc}}^{1,n}(G\setminus
\{a_1,a_2,\dots,a_J\},\mathbb{R}^n)
\eqno{(1.1)}
$$
as $\varepsilon_k \to 0$, where $u_n$ is an n-harmonic map.

In this paper we shall discuss the
asymptotic behavior for  the regularizable
minimizers of $E_{\varepsilon}(u,G)$ on $W_n$ in the case $p=n$.
Without loss of generality, we may assume $d>0$.
Recalling a minimizer of $E_{\varepsilon}(u,G)$ on $W_n$
be called the regularizable minimizer, if it is the limit of the
minimizer of the regularized functional
$$
E_{\varepsilon}^{\tau}(u,G)
=\frac{1}{p}\int_G(|\nabla u|^2+\tau)^{p/2}
+\frac{1}{4\varepsilon^p}\int_G(1-|u|^2)^2
,\quad (\tau \in (0,1))
$$
on $W_n$ in $W^{1,p}$. It is not difficult to prove that the regularizable
minimizer is also a minimizer of $E_{\varepsilon}(u,G)$.
In order to find the zeroes of the minimizers,
we should first locate the singularities of the
n-harmonic map $u_n$.

\begin{theorem} \label{th 1.1}
If $a_j \in \overline{G}, j=1,2,\dots,J$ are the singularities of
$n$-harmonic map $u_n$, then $J=d$, the degree $\deg(u_n, a_j)=1$,
and $\{a_j\}_{j=1}^d \subset G$. Moreover, for every $j$,
there exists at least
one zero of the regularizable minimizer $u_{\varepsilon}$ near to $a_j$.
\end{theorem}

Because the module of the minimizer has the physics sense,
we have also studied its asymptotic behavior.

\begin{theorem} \label{th 1.2}
Let $u_{\varepsilon}$ be a regularizable
minimizer of $E_{\varepsilon}(u,G)$,
$\rho=|u_{\varepsilon}|$, then
there exists a constant $C$ independent of $\varepsilon$ such that
$$
\int_G|\nabla \rho|^n \leq C, \quad\mbox{and} \quad
\frac{1}{\varepsilon^n}\int_G(1-\rho^2)
\leq C(1+|\ln \varepsilon|).
$$
For any given $\eta>0$, denote $G_{\eta}=G\setminus
\cup_{j=1}^dB(a_j,\eta)$, then
as $\varepsilon \to 0$,
$$
\frac{1}{\varepsilon^n}
\int_{G_{\eta}}(1-\rho^2)^2
\to 0,
$$
$$
\rho \to 1,\quad\mbox{ in } C_{{\rm loc}}(G_{\eta},R).
$$
\end{theorem}

At last, we develop the conclusion of (1.1) into following

\begin{theorem} \label{th 1.3}
There exists a subsequence
$u_{\varepsilon_k}$ of $u_{\varepsilon}$
such that as $\varepsilon \to 0$,
$$
u_{\varepsilon_k} \to u_n,
\quad\mbox{ in } W_{{\rm loc}}^{1,n}(G\setminus
\cup_{j=1}^d\{a_j\},\mathbb{R}^n).
$$
\end{theorem}

We shall prove Theorems 1.2 and 1.3 in $\S5$ and $\S7$
respectively, and the proof of Theorem 1.1 will be given in $\S6$.

\section{Basic properties of the regularizable minimizers }

First we recall the minimizer of the regularized functional
$$
E_{\varepsilon}^{\tau}(u,G)
=\frac{1}{n}\int_G
(|\nabla u|^2+\tau)^{n/2}
+\frac{1}{4\varepsilon^n}
\int_G(1-|u|^2)^2, \quad \tau \in (0,1)
$$
on $W_n$, denoted by $u_{\varepsilon}^{\tau}$.
As $\tau \to 0$, there exists a
subsequence $u_{\varepsilon}^{\tau_k}$
of $u_{\varepsilon}^{\tau}$ such that
$$
\lim_{\tau_k \to 0}u_{\varepsilon}^{\tau_k}=u_{\varepsilon},
\quad\mbox{in } W^{1,n}(G,\mathbb{R}^n),
\eqno{(2.1)}
$$
and the limit $u_{\varepsilon}$ is one
minimizer of $E_{\varepsilon}(u,G)$ on $W_n$, which
is named the regularizable minimizer. It is not difficult to prove
that $u_{\varepsilon}^{\tau}$ solves the problem
$$\displaylines{
\hfill -\mathop{\rm div}[(|\nabla u|^2+\tau)^{(n-2)/2}\nabla u]
=\frac{1}{\varepsilon^n}u(1-|u|^2),\quad\mbox{on }G, \hfill
\llap{(2.2)}\cr
u|_{\partial G}=g
}$$
and satisfies the maximum principle:
$|u_{\varepsilon}^{\tau}|\leq 1$ on
$\overline{G}$. Moreover

\begin{proposition}[{Theorem 2.2 in [6]}] \label{prop2.1}
For any $\delta >0$,
there exists a constant $C$ independent of $\varepsilon$ such that
$$
\overline{\lim}_{\tau \to 0}
|\nabla u_{\varepsilon}^{\tau}|\leq
C\varepsilon^{-1},\quad\mbox{on }G^{\delta \varepsilon},
\eqno{(2.3)}
$$
where  $G^{\delta \varepsilon}=\{x \in G:
\mathop{\rm dist}(x,\partial G)\geq \delta \varepsilon \}$.
\end{proposition}

In this section we shall present some basic properties of the
regularizable minimizer $u_{\varepsilon}$.
Clearly it is a weak solution of the
equation
$$
-\mathop{\rm div}(|\nabla u|^{n-2}\nabla u)=
\frac{1}{\varepsilon^n}u(1-|u|^2),\quad\mbox{on }G,
\eqno{(2.4)}$$
and it is known that $|u_{\varepsilon}|\leq 1$ a.e.
on $\overline{G}$ [6]. We also have

\begin{proposition} \label{prop2.2}  For any $\delta>0$,
there exists a constant $C$ independent of $\varepsilon$ such that
$$
\|\nabla u_{\varepsilon}\|_{L^{\infty}
(B(x,\delta \varepsilon/8,\mathbb{R}^n)} \leq C\varepsilon^{-1},
\quad\mbox{if }x \in G^{\delta \varepsilon}.
$$
\end{proposition}

\paragraph{Proof.} Let $y=x\varepsilon^{-1}$ in (2.4) and denote
$v(y)=u(x)$,
$G_{\varepsilon}=\{y=x\varepsilon^{-1} :x \in G\},
\quad G^{\delta}=\{y
\in G_{\varepsilon}:\mathop{\rm dist}(y,\partial G_{\varepsilon})
>\delta\}$.
Since that $u$ is a weak
solution of (2.4), we have
$$
\int_{G_{\varepsilon}}|\nabla v|^{n-2}\nabla v\nabla \phi
=\int_{G_{\varepsilon}}v (1-|v|^2)\phi,
\quad \phi \in W_0^{1,n}(G_{\varepsilon},\mathbb{R}^n).
$$
Taking $\phi=v \zeta^n, \zeta \in C_0^{\infty}
(G_{\varepsilon},R)$, we obtain
$$
\int_{G_{\varepsilon}}|\nabla v|^n\zeta^n
\leq n\int_{G_{\varepsilon}}|\nabla v|^{n-1}\zeta^{n-1}
|\nabla \zeta||v|
+\int_{G_{\varepsilon}}|v|^2(1-|v|^2)\zeta^n.
$$
Setting $y \in G^{\delta}, B(y,\delta/2) \subset G_{\varepsilon}$,
and $\zeta=1\quad\mbox{in } B(y,\delta/4),
\zeta=0 \quad\mbox{in } G_{\varepsilon}
\setminus B(y,\delta/2), |\nabla \zeta| \leq  C(\delta)$,
we have
$$
\int_{B(y,\delta/2)}|\nabla v|^n\zeta^n
\leq C(\delta)\int_{B(y,\delta/2)}
|\nabla v|^{n-1}\zeta^{n-1}+C(\delta).
$$
Using Holder inequality we can derive
$\int_{B(y,\delta/4)}|\nabla v|^n \leq C(\delta)$.
Combining this with the theorem of [9] yields
$$
\|\nabla v \|_{L^{\infty}(B(y,\delta/8))}^n
\leq C(\delta)\int_{B(y,\delta/4)}(1+|\nabla v|)^n
\leq C(\delta)
$$
which implies
$$
\|\nabla u\|_{L^{\infty}(B(x,\varepsilon \delta
/8))} \leq C(\delta)\varepsilon^{-1}.
$$

\begin{proposition}[{Lemma 2.1 in [6]}] \label{prop2.3}
There exists a constant $C$ independent of
$\varepsilon$ such that for $\varepsilon \in
(0,1)$,
$$
E_{\varepsilon}(u_{\varepsilon},G)\leq
d \frac{(n-1)^{n/2}}{n}|S^{n-1}||\ln \varepsilon|+C.
\eqno{(2.5)}
$$
\end{proposition}

\begin{proposition} \label{prop2.4}
There exists a constant $C$ independent of $\varepsilon$ such that
$$
\frac{1}{\varepsilon^n}
\int_G(1-|u_{e}|^2)^2 \leq C. \eqno{(2.6)}
$$
\end{proposition}

\paragraph{Proof.}
By (3.6) in [6],
$$
\int_G|\nabla u_{\varepsilon}|^n
\geq d(n-1)^{n/2}|S^{n-1}||\ln \varepsilon|-C.
$$
Applying Proposition 2.3 we may obtain (2.6).

\section{A class of bad balls}

Fix $\rho >0$. For the regularizable minimizer $u_{\varepsilon}$,
from Theorem 2.2 in [6] we know
$$
|u_{\varepsilon}| \geq \frac{1}{2},
\quad\mbox{on }G\setminus G^{\rho \varepsilon},
\eqno{(3.1)}
$$
where $G^{\rho \varepsilon}=\{x \in G :
\mathop{\rm dist}(x,\partial G)\geq \rho \varepsilon \}$.
Thus there exists no zero of $u_{\varepsilon}$
on $G\setminus G^{\rho \varepsilon}$.

\begin{proposition} \label{prop3.1}
Let $u_{\varepsilon}$ be a regularizable
minimizer of $E_{\varepsilon}(u,G)$, There exist
positive constants $\lambda, \mu$
which are independent of $\varepsilon \in (0,1)$ such that if
$$
\frac{1}{\varepsilon^n}
\int_{G^{\rho \varepsilon}\cap B^{2l\varepsilon}}
(1-|u_{\varepsilon}|^2)^2 \leq \mu,
\eqno{(3.2)}
$$
where $B^{2l\varepsilon}$ is some ball of
radius $2l\varepsilon$ with $l\geq \lambda$, then
$$
|u_{\varepsilon}| \geq \frac{1}{2},
\quad \forall x \in G^{\rho \varepsilon}\cap B^{l\varepsilon}.
\eqno{(3.3)}
$$
\end{proposition}

\paragraph{Proof.}
First it is known that there exists a constant $\beta>0$ such that
for any $x \in G^{\rho \varepsilon}$ and $0<r \leq 1$,
$$
|G^{\rho \varepsilon} \cap B(x,r)| \geq \beta r^n.
$$
Next we take
$$
\lambda=\min(\frac{1}{4C},
\frac{1}{8} \rho), \quad
\mu=\frac{\beta {\lambda}^n}{16}
$$
where $C$ is the constant in Proposition 2.2.

Suppose  that there is a point $x_0 \in
G^{\rho \varepsilon} \cap B^{l\varepsilon}$ such that
$|u_\varepsilon(x_0)| < 1/2$,
then applying Proposition 2.2 we have
$$
|u_\varepsilon(x)-u_\varepsilon(x_0)|
\leq C \varepsilon^{-1}|x-x_0|=\frac{1}{4},
\quad x \in B(x_0,\lambda \varepsilon)
\cap G^{\rho \varepsilon}.
$$
Hence
$$
(1-|u_\varepsilon(x)|^2)^2 > \frac{1}{16}, \quad
\forall x \in B(x_0,\lambda \varepsilon)\cap G^{\rho \varepsilon},
$$
$$\int_{B(x_0,\lambda \varepsilon)
\cap G^{\rho \varepsilon}}(1-|u_\varepsilon|^2)^2
> \frac{1}{16}
|G^{\rho \varepsilon} \cap B(x_0,\lambda \varepsilon)|
\geq \beta \frac{1}{16}(\lambda \varepsilon)^n
=\mu \varepsilon^n.
\eqno{(3.4)}
$$
Since $x_0 \in B^{l\varepsilon} \cap G^{\rho \varepsilon}$, we have
$(B(x_0,\lambda \varepsilon) \cap G^{\rho \varepsilon})
\subset (B^{2l\varepsilon} \cap G^{\rho \varepsilon})$,
thus (3.4) implies
$$
\int_{B^{2l\varepsilon} \cap
G^{\rho \varepsilon}}(1-|u_\varepsilon|^2)^2
> \mu \varepsilon^n
$$
which contradicts (3.2) and thus the proposition is proved.

To find the zeroes of the regularizable minimizer $u_{\varepsilon}$
based on Proposition 3.1,
we may take (3.2) as the ruler to distinguish
the ball of radius $\lambda \varepsilon$ which contain the zeroes.

Let $\lambda,\mu$ be  constants
in Proposition 3.1.
If
$$
\frac{1}{\varepsilon^n}
\int_{G^{\rho \varepsilon}\cap
B(x^{\varepsilon},2\lambda \varepsilon)}
(1-|u_{\varepsilon}|^2)^2 \leq \mu,
$$
then $B(x^{\varepsilon},\lambda \varepsilon)$ is
called good ball. Otherwise $B(x^{\varepsilon},\lambda \varepsilon)$
is called bad ball. From Proposition 3.1 we are led to
$$
|u_{\varepsilon}| \geq \frac{1}{2},
\quad\mbox{on }G^{\rho \varepsilon}\setminus \cup_{
x^{\varepsilon} \in \Lambda} B(x^{\varepsilon},\lambda \varepsilon),
\eqno{(3.5)}
$$
where $\Lambda$ is the set of the centres of all bad balls.
(3.5) and (3.1) imply that the zeroes of $u_{\varepsilon}$
are contained in these bad balls.

Now suppose that $\{B(x_i^{\varepsilon},\lambda \varepsilon),
i \in I\}$ is a family of balls satisfying
\begin{description}
\item{(i)} $x_i^{\varepsilon} \in G^{\rho \varepsilon},i \in I$
\item{(ii)} $G^{\rho \varepsilon} \subset
\cup_{i \in I}B(x_i^{\varepsilon},\lambda \varepsilon)$
\item{(iii)} $$B(x_i^{\varepsilon},\lambda \varepsilon /4) \cap
B(x_j^{\varepsilon},\lambda \varepsilon /4)=\emptyset,i \neq j\,.
\eqno{(3.6)}$$
\end{description}
Let
$J_\varepsilon=\{i \in I : B(x_i^{\varepsilon},
\lambda \varepsilon)\mbox{ is a bad ball}\}$.

\begin{proposition} \label{prop3.2}
There exists a positive integer
$N$ which is independent of $\varepsilon$
such that the number of bad balls
$\mathop{\rm card}J_\varepsilon \leq N$.
\end{proposition}

\paragraph{Proof.}
Since (3.6) implies that every point in $G^{\rho \varepsilon}$
can be covered by finite,
say m (independent of $\varepsilon$) balls,
from (2.6) and the definition of bad
balls,we have
\begin{eqnarray*}
\mu \varepsilon^n \mathop{\rm card}  J_\varepsilon
&\leq& \sum_{i \in J_\varepsilon}
\int_{B(x_i^{\varepsilon},2\lambda \varepsilon)
                 \cap G^{\rho \varepsilon}}(1-|u_\varepsilon|^2)^2\\
&\leq& m\int_{\cup_{i \in J_\varepsilon}
B(x_i^{\varepsilon},2\lambda \varepsilon)
\cap G^{\rho \varepsilon}}(1-|u_\varepsilon|^2)^2\\
&\leq& m\int_G(1-|u_\varepsilon|^2)^2
\leq mC\varepsilon^n
\end{eqnarray*}
and hence $\mathop{\rm card} J_\varepsilon\leq \frac{mC}{\mu} \leq N$.

Similar to the argument of Theorem IV.1 in [1],
we have

\begin{proposition} \label{prop3.3}
There exist a subset $J \subset J_{\varepsilon}$ and a
constant $h\geq \lambda$
such that
$$\displaylines{
\cup_{i \in J_{\varepsilon}}B(x_i^{\varepsilon},\lambda \varepsilon)
\subset \cup_{i \in J}B(x_j^{\varepsilon},h \varepsilon), \cr
\hfill |x_i^{\varepsilon}-x_j^{\varepsilon}|>
8h\varepsilon,\quad i,j \in J,\quad i \neq j.
\hfill\llap{(3.7)}
}$$
\end{proposition}

\paragraph{Proof.}
If there are two points $x_1, x_2$ such that (3.7) is not true with
$h=\lambda$, we take $h_1=9\lambda$ and
$J_1=J_{\varepsilon}\setminus\{1\}$. In this case,
if (3.7) holds we are done.
Otherwise we continue to choose a pair points
$x_3, x_4$ which does not satisfy (3.7) and take $h_2=9h_1$ and
$J_2=J_{\varepsilon}\setminus\{1,3\}$.
After at most $N$ steps we may conclude
this proposition.

Applying Proposition 3.3 we may modify the family of bad balls such
that the new one, denoted by
$\{B(x_i^{\varepsilon},h\varepsilon) :i \in J\}$, satisfies
$$\displaylines{
\cup_{i \in J_\varepsilon}B(x_i^{\varepsilon},\lambda \varepsilon)
\subset \cup_{i \in J}B(x_i^{\varepsilon},h \varepsilon),
\cr
\hfill \lambda \leq h; \quad \mathop{\rm card}  J
\leq \mathop{\rm card}  J_\varepsilon,
\hfill\llap{(3.8)} \cr
|x_i^{\varepsilon}-x_j^{\varepsilon}|>8h
\varepsilon,i,j \in J,i \neq j.
}$$
The last condition implies that every
two balls in the new family do not intersect.

As $\varepsilon \to 0$,
there exist a subsequence
$x_i^{\varepsilon_k}$ of $x_i^{\varepsilon}$ and
$a_i \in \overline{G}$ such that
$$
x_i^{\varepsilon_k} \to a_i,\quad i=1,2,\dots,N_1=\mathop{\rm card} J.
$$
Perhaps there may be at least two subsequences converge to the same
point, we denote by
$$
a_1,a_2,\dots,a_{N_2},\quad N_2 \leq N_1
$$
the collection of distinct points in $\{a_i\}_1^{N_1}$.

To prove $a_j \overline{\in}\partial G$,
it is convenient to enlarge a little
$G$. Assume $G'\subset \mathbb{R}^n$ is a bounded,
simply connected domain with
smooth boundary such that
$\overline{G} \subset G'$, and take a smooth map $\bar
{g}:(G'\setminus G) \to S^{n-1}$
such that $\bar{g}=g$ on $\partial G$.
We extend the definition domain of
every element in $\{u:G\to \mathbb{R}^n :
u|_{\partial G}=g\}$ to $G'$ such that
$u=\overline{g}$ on $G'\setminus G $.
In particular, the regularizable minimizer
$u_{\varepsilon}$ can be defined  on $G'$.

Fix a small constant $\sigma >0$ such that
$$\displaylines{
\overline{B(a_j,\sigma)} \subset G',\quad j=1,2,\dots,N_2;\cr
4\sigma < |a_j -a_i|,\quad i \neq j;\quad 4\sigma <
\mathop{\rm dist}(G,\partial G').
}$$
Writing $\Lambda_j=\{i \in J :x_i^{\varepsilon_k}
\to a_j\},j=1,2,\dots,N_2$,
we have
$$\displaylines{
\cup _{i \in \Lambda_j} \overline{B(x_i^{\varepsilon_k},
h \varepsilon_k)} \subset B(a_j,\sigma),
\quad j=1,2,\dots,N_2
 \cr
\cup _{j \in J}B(x_j^{\varepsilon_k},h \varepsilon_k)
\subset \cup _{j=1}^{N_2}B(a_j,\sigma /4)
\cr
B(x_i^{\varepsilon_k},h \varepsilon_k)
\cap B(x_j^{\varepsilon_k},h \varepsilon_k)=\emptyset,\quad
i,j \in J,i \neq j
}$$
as long as $\varepsilon_k$ is small enough.
Let $u_{\varepsilon}$ is the regularizable
minimizer of $E_{\varepsilon}(u,G)$ and denote
$d_i^k=deg({u_{\varepsilon_k}},\partial
B(x_i^{\varepsilon_k},h \varepsilon_k)),
l_j^k=deg({u_{\varepsilon_k}},\partial B(a_j,\sigma))$,
thus
$$l_j^k=\sum_{i \in \Lambda_j}d_i^k,\quad
d=\sum_{j=1}^{N_2} l_j^k.
\eqno{(3.9)}
$$

To prove that the degrees $d_i^k$ and $l_j^k$
are independent of $\varepsilon_k$, we
recall a proposition stated in [6] (Lemma 3.3) or [2] (Theorem 8.2).

\begin{proposition} \label{prop3.4}
Let $\phi :S^{n-1} \to S^{n-1}$
be a $C^0$-map with $\deg \phi=d$. Then
$$
\int_{S^{n-1}}|\nabla _{\tau}\phi |^{n-1}dx
\geq |d|(n-1)^{(n-1)/2}|S^{n-1}|.
$$
\end{proposition}

\begin{proposition} \label{prop3.5}
There exists a constant $C$ which is independent of $\varepsilon_k$
such that
$$
|d_i^k| \leq C,i \in J;\quad |l_j^k| \leq C,j=1,2,\dots,N_2.
$$
\end{proposition}

\paragraph{Proof.}
Since $u=u_{\varepsilon}$ is a weak solution of (2.4),
applying the theory of the
local regularity in [9], we know
$u \in C(\partial B(x_i^{\varepsilon_k},h \varepsilon_k))$.
Since (3.5) implies $|u|\geq 1/2$ on
$\partial B(x_i^{\varepsilon_k},h \varepsilon_k)$,
thus $\phi=\frac{u}{|u|} \in C(\partial
B(x_i^{\varepsilon_k},h \varepsilon_k),S^{n-1})$.
From Proposition 3.4, we have
$$
|d_i^k| \leq |S^{n-1}|^{-1}(n-1)^{(1-n)/2}
\int_{\partial B(x_i^{\varepsilon_k},h \varepsilon_k)
}|(\frac{u}{|u|})_{\tau}|^{n-1}.
$$
Since $|u|\geq \frac{1}{2}$
on $G'\setminus G^{\rho \varepsilon}$, there is no
zero of $u_{\varepsilon}$ in it. Thus
$$
\deg(u_{\varepsilon_k},\partial
B(x_i^{\varepsilon_k},h \varepsilon_k))
=\deg(u_{\varepsilon_k},\partial (B(x_i^{\varepsilon_k},
h \varepsilon_k)\cap G^{\rho \varepsilon_k}))
$$
and
$$
|d_i^k| \leq |S^{n-1}|^{-1}(n-1)^{(1-n)/2}
\int_{\partial[B(x_i^{\varepsilon_k},h \varepsilon_k)
\cap G^{\rho \varepsilon}]}|(\frac{u}{|u|})_{\tau}|^{n-1}.
\eqno{(3.10)}
$$
Substituting (2.3) and the fact
$|u_{\varepsilon_k}| \geq \frac{1}{2}$
on $\partial [B(x_i^{\varepsilon_k},h \varepsilon_k)\cap
G^{\rho \varepsilon}]$ into (3.10), we obtain
$$
|d_i^k| \leq C\varepsilon_k^{1-n}|S^{n-1}|^{-1}(n-1)^{(1-n)/2}
(h \varepsilon_k)^{n-1} \leq C,
$$
where $C$ is a constant which is independent of $\varepsilon_k$.
Combining this with (3.9) we can complete the proof of the proposition.

Proposition 3.5 implies that there
exist a number $k_j$ which is independent of
$\varepsilon_k$ and a subsequence of $l_j^k$ denoted itself such that
$$
l_j^k \to k_j,\quad as \quad k \to \infty.
$$
Since $l_j^k, k_j \in N, \{l_j^k\}$ must be constant sequence for any
fixed $j$, namely $l_j^k=k_j$.
The same reason shows $d_i^k$ can be writen as
$d_i$ which is  also a number independent of $\varepsilon_k$ later.

\section{An estimate for the lower bound }

Write $\Omega '=G'\setminus \cup_{j=1}^{N_2}B(a_j,\sigma)$.
Fixing $j \in \{1,2,\dots,N_2\}$ and taking $i_0 \in \Lambda_j$, we have
$x_{i_0} \to a_j$ as  $\varepsilon \to 0$. Thus
$$
\cup_{i \in \Lambda_j}\overline{B(x_i^{\varepsilon},
h \varepsilon)}\subset B(x_{i_0},\sigma /4)
\subset B(a_j,\sigma)
\eqno{(4.1)}
$$
holds with $\varepsilon$ small enough.

Denote
$\Omega_j=B(a_j,\sigma)\setminus\cup_{i \in \Lambda_j}
B(x_i^{\varepsilon},h \varepsilon),
\Omega_{j \sigma}=B(x_{i_0},\sigma /4)
\setminus\cup_{i \in \Lambda_j}B(x_i^{\varepsilon},h \varepsilon)$.
To estimate the lower bound of $\|\nabla u_{\varepsilon}\|
_{L^n(\Omega_j)}$, the following proposition is necessary that was given
by Theorem 3.9 in [6].

\begin{proposition} \label{prop4.1}
Let $A_{s,t}(x_i)=(B(x_i,s)\setminus B(x_i,t))\cap G$
with $\varepsilon \leq t<s \leq
R$. Assume that $u \in W_g^{1,n}(G,\mathbb{R}^n)$
and $\frac{1}{2} \leq |u| \leq 1$ on
$A_{s,t}(x_i)$. If there is a constant $C$ such that
$$
\frac{1}{\varepsilon^n}
\int_{A_{s,t}(x_i)}(1-|u|^2)^2 \leq C.
$$
Then for $\varepsilon < \varepsilon_0$ there holds
$$
\int_{A_{s,t}(x_i)}|\nabla u|^n
\geq |d_i|^{n/(n-1)}(n-1)^{n/2}|S^{n-1}|
\ln \frac{s}{t}-C,
$$
where $C$ is a constant which is independent of
$\varepsilon$ and $d_i$ is the degree of u
on each $\partial (B(x_i,r)\cap G), t\leq r\leq s$.
\end{proposition}

\begin{proposition} \label{prop4.2}
Assume $Card\Lambda_j=N$. Then
$$
\int_{\Omega_j}|\nabla {u_{\varepsilon}}|^n
\geq \int_{\Omega_{j,\sigma}}|\nabla
{u_{\varepsilon}}|^n \geq
 (n-1)^{n/2}|S^{n-1}||k_j|
 \ln\frac{\sigma}{\varepsilon}-C
\eqno{(4.2)}$$
where $C$ is a constant which is independent of $\varepsilon$.
\end{proposition}

\paragraph{Proof.}
We give the proof following that in [6]
(see Theorem 3.10), and the idea comes from [8].
Suppose $x_1,x_2,\dots,x_N$ converge to $a_j$,
and $d_{i,R}(i=1,2,\dots,N)$ is
the degree of $u_{\varepsilon}$ around $\partial B(x_i,R)$.
Let $R_{\varepsilon}^{\sigma}$ denote the set of
all numbers $R \in [\varepsilon, \sigma]$
such that $\partial B(x_i,R)\cap
B(x_j,\varepsilon)=\emptyset$ for all
$i \neq j$ and such that for some
collection $J_R \subset \{1,2,\dots,N\}$, satisfying $J_R \subset
J_{R'}$ if $R' \leq R$, the family
$\{B(x_i,R)\}_{i \in J_R}$ is disjoint
and
$$
\cup_{i=1}^N B(x_i,\varepsilon)
\subset \cup_{i \in J_{R'}} B(x_i,R')
\subset \cup_{i \in J_{R}} B(x_i,R),\quad R' \leq R.
$$
Note that $R_{\varepsilon}^{\sigma}$ is the
union of closed intervals $[R_0^l,\mathbb{R}^l],
1\leq l\leq L$, whose right endpoints
correspond to a number $R=\mathbb{R}^l$ such
that $\partial B(x_i,R) \cap \overline{B(x_j,R)} \neq \emptyset$
for some pair $i \neq j \in J_R$
and whose left endpoints correspond to a
number $R_0^l$ such that $\overline
{B(x_i,\mathbb{R}^{l-1})}\setminus \cup_{j \in
J_0}B(x_j,R_0^l) \neq \emptyset$ for $i \overline{\in}J_{R_0^l}$.
$J_R=J^l$ is a constant for $R \in [R_0^l,\mathbb{R}^l]$ and $J^{l+1} \subset
J^l, J^{l+1} \neq J^l$.
Thus $L \leq N$. Moreover, there exists a constant
$M=M(h)>0$ such that
$$
R_0^l \leq M \varepsilon,\quad \mathbb{R}^L
\geq \sigma /M,\quad R_0^{l+1}\leq MR^l
\eqno{(4.3)}
$$
for all $l=1,2,\dots,L-1$. Finally, observe that for all $R \in
R_{\varepsilon}^{\sigma}$ and $J \in J_R$,
$$
|k_j|=|\sum_{i \in J_R}d_{i,R}|\leq
\sum_{i \in J_R}|d_{i,R}|^{n/(n-1)}.
\eqno{(4.4)}
$$
Applying (4.3)(4.4) and proposition 4.1 we have
\begin{eqnarray*}
\int_{\Omega_{j,\sigma}}|\nabla {u_{\varepsilon}}|^n
&\geq& \sum_{l=1}^L \sum_{i \in J^l}
|\int_{A_{\mathbb{R}^l,R_0^l}(x_i)}
\nabla u_{\varepsilon}|^n\\
&\geq& \sum_{l=1}^L \sum_{i \in J^l}
|S^{n-1}|(n-1)^{n/2} |d_{i,\mathbb{R}^l}|\ln
(\mathbb{R}^l/R_0^l)-C\\
&\geq& |S^{n-1}|(n-1)^{n/2} |k_j|\sum_{l}(\ln
\mathbb{R}^l-\ln R_0^l)-C\\
&\geq& (n-1)^{n/2}|S^{n-1}||k_j|
\ln\frac{\sigma}{\varepsilon}-C.
\end{eqnarray*}
This and (4.1) imply that (4.2) holds.

\paragraph{Remark} In fact the following results
$$
\int_{\Omega_j}|\nabla
\frac{u_{\varepsilon}}{|u_{\varepsilon}|}|^n
\geq (n-1)^{n/2}|S^{n-1}||k_j|^{n/(n-1)}\ln
{\frac{\sigma}{\varepsilon}},
$$
and
$$
\int_{\Omega_j}(1-|u_{\varepsilon}|^n)
|\nabla \frac{u_{\varepsilon}}
{|u_{\varepsilon}|}|^n\leq C
$$
had been presented in the proof of Theorem 3.9 in [6],
where $C$ which is independent of $\varepsilon$. Noticing
$$
\int_{\Omega_j}|u_{\varepsilon}|^n
|\nabla \frac{u_{\varepsilon}}{|u_{\varepsilon}|}|^n
=\int_{\Omega_j}|\nabla
\frac{u_{\varepsilon}}{|u_{\varepsilon}|}|^n
-\int_{\Omega_j}(1-|u_{\varepsilon}|^n)
|\nabla \frac{u_{\varepsilon}}{|u_{\varepsilon}|}|^n,
$$
we have
$$
\int_{\Omega_j}|u_{\varepsilon}|^n
|\nabla \frac{u_{\varepsilon}}{|u_{\varepsilon}|}|^n
\geq (n-1)^{n/2}|k_j|^{n/(n-1)}|S^{n-1}|\ln
\frac{\sigma}{\varepsilon}-C.
$$

\begin{theorem} \label{th4.3}
There exists a constant $C$ which is independent of
$\varepsilon, \sigma \in (0,1)$ such that
$$
 \int_{\cup_{j=1}^{N_2}\Omega_j}|
 \nabla {u_{\varepsilon}}|^n \geq
 (n-1)^{n/2}|S^{n-1}|d\ln\frac{\sigma}{\varepsilon}-C,
\eqno{(4.5)}
$$
$$
\frac{1}{n}\int_{G_{\sigma}}
|\nabla {u_{\varepsilon}}|^n
+\frac{1}{4\varepsilon^n}\int_G
(1-|{u_{\varepsilon}}|^2)^2 \leq
\frac{1}{n}(n-1)^{n/2}
|S^{n-1}|d\ln\frac{1}{\sigma}+C
\eqno{(4.6)}
$$
where $G_{\sigma}=G\setminus \cup_{j=1}^{N_2}B(a_j,\sigma)$.
\end{theorem}

\paragraph{Proof.}
From (4.2) and Proposition 2.3 we have
$$
(n-1)^{n/2}|S^{n-1}|(\sum_{j=1}^{N_2}|k_j|)
\ln\frac{\sigma}{\varepsilon} \leq
(n-1)^{n/2}|S^{n-1}|d\ln\frac{1}{\varepsilon}+C
$$
or $(\sum_{j=1}^{N_2}|k_j|-d)\ln\frac{1}{\varepsilon} \leq C$.
It is seen as $\varepsilon$ small enough
$$
\sum_{j=1}^{N_2}|k_j| \leq d=\sum_{j=1}^{N_2}k_j
$$
which implies
$$
k_j \geq 0.
\eqno{(4.7)}
$$
This and (3.9) imply
$$\sum_{j=1}^{N_2}|k_j|=\sum_{j=1}^{N_2}k_j=d.
\eqno{(4.8)}
$$
Substituting (4.8) into (4.2) yields (4.5), and (4.6) may be concluded
from (4.5) and Proposition 2.3.

From (4.6) and the fact $|u_{\varepsilon}| \leq 1$
a.e. on $G$, we may conclude that
there exists a subsequence $u_{\varepsilon_k}$
of $u_{\varepsilon}$ such that
$$
u_{\varepsilon_k} \stackrel{w}{\to} u_*,
\quad W^{1,n}(G_{\sigma},\mathbb{R}^n)
\eqno{(4.9)}
$$
as $\varepsilon_k \to 0$.
Compare (4.9) with (1.1) we known
$u_*=u_n$ on $G_{\sigma}$, and
$$
\{a_j\}_{j=1}^{N_2}=\{a_j\}_{j=1}^J.
\eqno{(4.10)}
$$
These points were called the singularities of $u_n$.

To show these singularities
$a_j \overline{\in}{\partial G}$, the following
conclussion is necessary.

\begin{proposition} \label{prop4.4}
Assume $a \in \partial G$ and $\sigma \in (0,R)$
with a small constant $R$. If
$$
u\in W^{1,n}(A_{R,\sigma}(a),S^{n-1})\cap C^0, \quad u=\overline{g}
$$
on $(G'\setminus G)\cap B(a,R)$
and $\deg(u,\partial B(a,R))=1$, then there
exists a constant $C$ which is independent of $\sigma$ such that
$$
\int_{A_{R,\sigma}(a)}|\nabla u|^n
\geq 2^{\frac{1}{n}}(n-1)^{n/2}|S^{n-1}|
\ln\frac{1}{\sigma}-C\,.
\eqno{(4.11)}
$$
\end{proposition}

\paragraph{Proof.}
Similar to the proof of Lemma VI.1 in [1], we may write $G$ as
the half space
$$\{(x_1,x_2,\dots,x_n) :x_n >0\}$$
locally and $a$ as $0$ by a conformal change.

Denote $S_t=\partial B(0,t), t \in (\sigma,R)$.
Noticing that $\overline{g}$ is
smooth on $G'\setminus G$, we have
$$
\sup_{\overline{G'}\setminus G}|\overline{g}_{\tau}| \leq C_1.
$$
Taking $t$ sufficiently small such that
$$
t \leq (n-1)^{1/2}\frac{(2^{n-1}-1)^{1/(n-1)}}{2C_1},
$$
then
$$\int_{S_t^{-}}|\bar{g}_{\tau}|^{n-1}
\leq |S_t^-|C_1^{n-1} \leq |S^{n-1}|t^{n-1}C_1^{n-1}
\leq (n-1)^{(n-1)/2}|S^{n-1}|(1-2^{1-n})
\eqno{(4.12)}
$$
with $R<1$ small enough, where $S_t^{-}=S_t \cap \{x_n <0\}$.
On the other hand we can be led to
$$
(n-1)^{(n-1)/2}|S^{n-1}| \leq \int_{S_t}|u_{\tau}|^{n-1}
=\int_{S_t^{+}}|u_{\tau}|^{n-1}
+\int_{S_t^{-}}|\bar{g}_{\tau}|^{n-1}
$$
from Proposition 3.4. Here $S_t^+=S_t\setminus S_t^-$.
Combining this with (4.12) yields
\begin{eqnarray}
\int_{S_t^{+}}|u_{\tau}|^n
&\geq& |S_t^+|^{-1/(n-1)}(\int_{S_t^+}
|u_{\tau}|^{n-1})^{n/(n-1)}\\
&\geq& 2^{\frac{1}{n}}|S^{n-1}|(n-1)^{n/2}t^{-1}.
\end{eqnarray}
Integrating this over $(\sigma,R)$, we obtain
$$
\int_{A_{R,\sigma}}|\nabla u|^n
\geq 2^{\frac{1}{n}}|S^{n-1}|(n-1)^{n/2}
\ln\frac{R}{\sigma}
$$
which implies (4.11).
To prove $k_j=1$ for any $j$, we suppose $R>2\sigma$
is a small constant such that
$$
\overline{B(a_j,R)} \subset G';\quad B(a_j,R)\cap B(a_i,R)=\emptyset,
i \neq j.
\eqno{(4.13)}
$$
Denote
$\Pi=\{v \in W^{1,n}(\Omega',S^{n-1})\cap C^0:
\deg(v,\partial B(a_j,r))=k_j,r \in (\sigma,R),
j=1,2,\dots,N_2\}$.

\begin{proposition} \label{prop4.5}
For any $v \in \Pi$, if $k_j\geq 0,
j=1,2,\dots,N_2$, then
there exists a constant $C=C(R)$ which is independent of $\sigma$
such that
$$
\int_{\Omega'}|\nabla v|^n
\geq (n-1)^{n/2}|S^{n-1}|(\sum_{j=1}^{N_2}k_j^{\frac
{n}{n-1}})\ln\frac{1}{\sigma}-C.
\eqno{(4.14)}
$$
\end{proposition}

\paragraph{Proof.}
Write $A_{R,\sigma}(a_j)=B(a_j,R)\setminus B(a_j,\sigma)$,
thus $\cup_{j=1}^{N_2}A_{R,\sigma}(a_j) \subset \Omega'$.
From Proposition 3.4 we have
\begin{eqnarray*}
k_j=|k_j| &\leq& (n-1)^{(1-n)/2}|S^{n-1}|^{-1}
\int_{S^{n-1}}|v_{\tau}|^{n-1}\\
&\leq& (n-1)^{(1-n)/2}|S^{n-1}|^{(n-1)/n}
(\int_{S^{n-1}}|v_{\tau}|^n)^{(n-1)/n}
\end{eqnarray*}
namely
$$
\int_{S^{n-1}}|v_{\tau}|^n
\geq (n-1)^{n/2}|S^{n-1}|k_j^{n/(n-1)}.
$$
On the other hand, we may obtain
\begin{eqnarray*}
\int_{\Omega'}|\nabla v|^n
&\geq& \sum_{j=1}^{N_2}
\int_{A_{R,\sigma}(a_j)}|\nabla v|^n\\
&\geq& \sum_{j=1}^{N_2}
\int_{\sigma}^R \int_{S^{n-1}}r^{-n}|\nabla _{\tau}v|
     ^nr^{n-1}d\zeta dr\\
&\geq& (n-1)^{n/2}|S^{n-1}|\sum_{j=1}^{N_2}k_j^{n/(n-1)}
     \int_{\sigma}^Rr^{-1}dr\\
&=&(n-1)^{n/2}|S^{n-1}|(\sum_{j=1}^{N_2}k_j^
{n/(n-1)})\ln\frac{R}{\sigma}
\end{eqnarray*}
which implies (4.14).

\section{The proof of Theorem 1.2 }

Let $u_{\varepsilon}$be a regularizable
minimizer of  $E_{\varepsilon}(u,G)$.
Proposition 2.4 has given one estimate of convergence rate of
$|u_{\varepsilon}|$. Moreover, we also have

\begin{theorem} \label{th5.1}
There exists a constant $C$ which is independent
of $\varepsilon \in (0,1)$ such that
\begin{equation} \label{5.1}
\frac{1}{\varepsilon^n}
\int_G(1-|{u_{\varepsilon}}|^2)
\leq C(1+\ln\frac{1}{\varepsilon}).
\end{equation}
\end{theorem}

\paragraph{Proof.}
The minimizer $u=u_{\varepsilon}^{\tau}$ of the regularized functional
$E_{\varepsilon}^{\tau}(u,G)$ solves (2.2).
Taking the inner product of the both
sides of (2.2) with $u$ and integrating over $G$ we have
\begin{eqnarray}
\frac{1}{\varepsilon^n}\int_G|u|^2(1-|u|^2)
&=&-\int_Gdiv(v^{(n-2)/2}\nabla u)u \nonumber \\
&=&\int_Gv^{(n-2)/2}|\nabla u|^2
-\int_{\partial G}v^{(n-2)/2}uu_n \\ %\eqno(5.22)
&\leq& \int_Gv^{(n-2)/2}|\nabla u|^2
+C\int_{\partial G}v^{n/2}+C \nonumber
\end{eqnarray}
where $n$ denotes the unit outward normal to $\partial G$ and $u_n$ the
derivative with respect to $n$.

To estimate $\int_{\partial G}v^{n/2}$,
we choose a smooth vector field $\nu$
such that $\nu |_{\partial G}=n$.
Multiplying (2.2) by $(\nu\cdot \nabla u)$ and
integrating over $G$, we obtain
\begin{eqnarray*}
\frac{1}{\varepsilon^n}
\int_Gu(1-|u|^2)(\nu \cdot \nabla u)
&=&-\int_Gdiv(v^{(n-2)/2}\nabla u)(\nu \cdot \nabla u)\\
&=&\int_Gv^{(n-2)/2}\nabla u \cdot (\nu \cdot \nabla u)
  -\int_{\partial G}v^{(n-2)/2}|u_n|^2.
\end{eqnarray*}
Combining this with
\begin{eqnarray*}
\frac{1}{\varepsilon^n}
\int_Gu(1-|u|^2)(\nu \cdot \nabla u)
&=&\frac{1}{2\varepsilon^n}
\int_G(1-|u|^2)(\nu \cdot \nabla (|u|^2))\\
&=&-\frac{1}{4\varepsilon^n}
\int_G(1-|u|^2)^2\mathop{\rm div}\nu
\end{eqnarray*}
and
\begin{eqnarray*}
\lefteqn{ \int_Gv^{(n-2)/2}\nabla u
\cdot \nabla(\nu \cdot \nabla u) }\\
&=&\int_Gv^{(n-2)/2}|\nabla u|^2
\mathop{\rm div}\nu+\frac{1}{n}\int_G\nu \cdot
\nabla (v^{n/2})\\
&=&\int_Gv^{(n-2)/2}|\nabla u|^2
\mathop{\rm div}\nu+\frac{1}{n}\int_{\partial G}v^{n/2}
-\frac{1}{n}\int_Gv^{n/2}\mathop{\rm div}\nu
\end{eqnarray*}
we obtain
$$
\int_{\partial G}v^{(n-2)/2}|u_n|^2 \leq
\frac{C}{4\varepsilon^n}\int_G(1-|u|^2)^2
+C\int_Gv^{n/2}
+\frac{1}{n}\int_{\partial G}v^{n/2}.
$$
Thus
\begin{eqnarray*}
\int_{\partial G}v^{n/2}
&=&\int_{\partial G}
v^{(n-2)/2}(|u_n|^2+|g_t|^2+\tau)\\
&\leq& C\int_{\partial G}
v^{(n-2)/2}+\frac{1}{n}
\int_{\partial G}v^{n/2}
+CE_{\varepsilon}^{\tau}(u_{\varepsilon}^{\tau},G).
\end{eqnarray*}
Substituting this into (5.2) yields
$$
\frac{1}{\varepsilon^n}\int_G
|u|^2(1-|u|^2) \leq CE_{\varepsilon}^{\tau}(u_{\varepsilon}^{\tau},
G).
$$
Let $\tau \to 0$, applying (2.1) and Proposition 2.3  we have
$$
\frac{1}{\varepsilon^n}\int_G
|u_{\varepsilon}|^2(1-|u_{\varepsilon}|^2)
\leq CE_{\varepsilon}(u_{\varepsilon},G)
\leq C(1+|\ln\varepsilon|)
$$
which and (2.6) imply (5.1).

\begin{theorem} \label{th5.2}
Denote $\rho=|u_{\varepsilon}|$.
There exists a constant $C$ which is independent of
$\varepsilon \in (0,1)$ such that
$$
\|\nabla \rho \|_{L^n(G)} \leq C.
\eqno{(5.3)}
$$
\end{theorem}

\paragraph{Proof.}
Denote $u=u_{\varepsilon}$.
From the Remark in $\S4$ we know
$$
\int_{\Omega_j}|u|^n|\nabla
\frac{u}{|u|}|^n dx
\geq (n-1)^{n/2}|k_j|^{\frac{n}{n-1}}|S^{n-1}|
 \ln{\frac{\sigma}{\varepsilon}} -C.
$$
Thus we may modify (4.5) as
$$
 \int_{\cup_{j=1}^{N_2}\Omega_j}\rho^n|\nabla
 \frac{u}{|u|}|^n \geq
 (n-1)^{n/2}|S^{n-1}|d\ln\frac{\sigma}{\varepsilon}-C.
$$
Combining this with
$$
 \int_{\cup_{j=1}^{N_2}\Omega_j}|\nabla u|^n \geq
 \int_{\cup_{j=1}^{N_2}
 \Omega_j}\rho^n|\nabla \frac{u}{|u|}|^n
+ \int_{\cup_{j=1}^{N_2}\Omega_j}|\nabla \rho|^n-C
$$
and Proposition 2.3, we derive
$$ \int_{\cup_{j=1}^{N_2}\Omega_j}|\nabla \rho|^n \leq C.
\eqno{(5.4)}
$$
On the other hand, from (2.1) and Proposition 2.1 we are led to
$$
\int_{G^{\rho \varepsilon}
\cap B(x_i,h \varepsilon)}|\nabla {u_{\varepsilon}}|^n
=\lim_{\tau_k \to 0}
\int_{G^{\rho \varepsilon}
\cap B(x_i,h \varepsilon)}|\nabla {u_{\varepsilon}}^{\tau_k}|^n
\leq C(\lambda \varepsilon)^n (C/{\varepsilon})^n \leq C,
$$
for $i \in \Lambda_j$.
Summarizing for $i$ and using (5.4) we can obtain (5.3).

\begin{theorem} \label{th5.3} For the $\sigma >0$ in Theorem 4.4,
then as $\varepsilon \to 0$,
$$
\frac{1}{\varepsilon^n}
\int_{G_{3\sigma}}(1-\rho^2)^2
\to 0,
\eqno{(5.5)}
$$
where $G_{3\sigma}=G\setminus \cup_{j=1}^{N_2}B(a_j,3\sigma)$.
\end{theorem}

\paragraph{Proof.}
The regularizable minimizer $u_{\varepsilon}$ satisfies
$$
\int_{G_{\sigma}}|\nabla u|^{n-2}\nabla u \nabla \phi
=\frac{1}{\varepsilon^n}
\int_{G_{\sigma}}u \phi (1-|u|^2),
\eqno{(5.6)}
$$
where $\phi \in W_0^{1,n}(G_{\sigma},\mathbb{R}^n)$
since $u_{\varepsilon}$ is a
weak solution of (2.4).
Denoting $u=u_{\varepsilon}^{\tau}=\rho w,
\rho=|u|,w=\frac{u}{|u|}$ in $G_{\sigma}$
and taking $\phi =\rho w \zeta, \zeta \in W_0^{1,n}(G_{\sigma},
\mathbb{R}^n)$, we have
$$
\int_{G_{\sigma}}|\nabla
u|^{n-2}(w\nabla \rho+\rho \nabla w)
(\rho \zeta \nabla w+\rho w \nabla \zeta +w\zeta \nabla \rho)
=\frac{1}{\varepsilon^n}
\int_{G_{\sigma}}\rho^2 \zeta (1-\rho^2).
\eqno{(5.7)}
$$
Substituting $2w\nabla w=\nabla (|w|^2)=0$ into (5.7), we obtain
$$
\int_{G_{\sigma}}|\nabla u|^{n-2}
(\rho \nabla \rho \nabla \zeta
+|\nabla u|^2 \zeta)
=\frac{1}{\varepsilon^n}
\int_{G_{\sigma}}\rho^2 \zeta (1-\rho^2).
\eqno{(5.8)}
$$
Set $S=\{x \in G_{\sigma} :\rho(x)>1
-\varepsilon^{\beta}\}$ for some fixed
$\beta \in (0,n/2)$ and $\overline{\rho}=\max(\rho,
1-\varepsilon^{\beta})$, thus $\rho=\overline{\rho}$ on $S$.
In (5.8) taking $\zeta=(1-\overline{\rho})\psi$,
where $\psi \in C^{\infty}(G_{\sigma},R), \psi=0$ on
$G_{\sigma} \setminus G_{2\sigma}, 0<\psi<1$ on $G_{2\sigma}
\setminus G_{3\sigma}, \psi=1$ on $G_{3\sigma}$, we have
\setcounter{equation}{8}
\begin{eqnarray}
\lefteqn{\int_{G_{\sigma}}|\nabla u|^{n-2}
\rho \nabla \rho \cdot \nabla \bar{\rho}\psi
+\frac{1}{\varepsilon^n}
\int_{G_{\sigma}}l^2(1-\rho^2)(1-\bar{\rho})\psi }\\
&=&\int_{G_{\sigma}}|\nabla u|^{n-2}\rho
\nabla \rho \nabla \psi(1-\bar{\rho})
+\int_{G_{\sigma}}|\nabla u|^n\psi(1-\overline{\rho}) \nonumber
\end{eqnarray}
Noticing $1/2 \leq l\leq 1$ in $G_{\sigma}$ and  applying
(4.6) we obtain
$$
\frac{1}{\varepsilon^n}
\int_{G_{3\sigma}}(1-\overline{\rho})(1-\rho^2)
+\int_{S \cap G_{3\sigma}}
|\nabla u|^{n-2}|\nabla \rho|^2
\leq C \varepsilon^{\beta}.
\eqno{(5.10)}
$$
On the other hand, (2.6) implies
$$
\varepsilon^{2\beta}|G_{\sigma}\setminus S|
\leq \int_{G_{\sigma}\setminus S}
(1-l^2)^2 \leq C\varepsilon^n,
$$
namely
$|G_{\sigma}\setminus S|\leq C\varepsilon^{n-2\beta}$.
Then there exists a small constant $\varepsilon_0>0$ such that
$$
G_{3\sigma} \subset S \cup E
$$
as $\varepsilon \in (0,\varepsilon_0)$ where $E$
is a set,
the measure of which converges to zero. Thus
$$
\lim_{\varepsilon \to 0}
\int_{G_{3\sigma}}(1-\rho^2)(1-\overline{\rho})
=\lim_{\varepsilon \to 0}
\int_{G_{3\sigma}}(1+\rho)(1-\rho)^2.
$$
By (5.10),
\begin{eqnarray*}
\lefteqn{ \lim_{\varepsilon \to 0}
\frac{1}{\varepsilon^n}
\int_{G_{3\sigma}}(1+\rho)^2(1-\rho)^2}\\
&\leq& \lim_{\varepsilon \to 0}
\frac{2}{\varepsilon^n}
\int_{G_{3\sigma}}(1-\overline{\rho})(1-\rho^2)
=0
\end{eqnarray*}
This is our conclusion.

\begin{theorem} \label{th5.4}
Assume $B(x,2\sigma) \subset G_{\sigma}$ satisfies
$$
\frac{1}{\varepsilon^n}
\int_{B(x,\sigma)}(1-|u_{\varepsilon}|^2)^2 \to 0,
\mbox{ as }\varepsilon \to 0,
\eqno{(5.11)}
$$
then
$|u_{\varepsilon}| \to 1$ in $C(B(x,\sigma),R)$.
\end{theorem}

\paragraph{Proof.}
Since $B(x,2\sigma) \subset G_{\sigma}$,
there exists $\varepsilon_0$ sufficiently small so
that $B(x,\sigma) \subset
G^{2\delta \varepsilon_0}$. We always
assume $\varepsilon <\varepsilon_0$. For $x_0
\in B(x,\sigma)$, set
$\alpha =|u_{\varepsilon}(x_0)|$. Proposition 2.2 implies
$$
|u_{\varepsilon}(x)-u_{\varepsilon}(x_0)|
<C\varepsilon^{-1}\tau \varepsilon,\quad\mbox{if } x \in B(x_0,\tau \varepsilon),
$$
where $\tau=(1-\alpha)(NC)^{-1}, C$
is the constant in Proposition 2.2 and $N$ is
a large number such that $\tau < \delta$. Thus $B(x_0,\tau \varepsilon)
\subset B(x,\sigma)$ and
$$
|u_{\varepsilon}(x)| \leq \alpha +C\tau,
\quad\mbox{if } x \in B(x_0, \tau \varepsilon),
$$
$$
\int_{B(x_0,\tau \varepsilon)}(1-|u_{\varepsilon}(x)|^2)^2
\geq (1-1/N)^2(1-\alpha)^{n+2}\pi \varepsilon^n(NC)^{-n}.
$$
Combining this with (5.11) we obtain $(1-\alpha)^{n+2} = o(1)$
as $\varepsilon \to 0$.
Thus it is not difficult to complete the proof of Theorem.

\section{The proof of Theorem 1.1 }

It is known that the singularities of $u_n$ are in $\overline{G}$
from the discussion in $\S3$.
Since $\deg(g,\partial G)>0$, we can see that
the zeroes of $u_{\varepsilon}$ are also in $G$ .
Moreover, the zeroes are contained in finite bad balls, i.e.
$B(x_i^{\varepsilon},h \varepsilon), i \in J$.
As $\varepsilon \to 0,
B(x_i^{\varepsilon}, h \varepsilon) \to a_j,
i\in \Lambda_j$. This implies that the zeroes of
$u_{\varepsilon}$ distribute near
these singularities of $u_n$ as
$\varepsilon \to 0$. Thus it is necessary to
describe these singularities $\{a_j\}, j=1,2,\dots,N_2$.

\begin{proposition} \label{prop6.1}
$k_j=\deg(u_n,a_j)$.
\end{proposition}

\paragraph{Proof.}
Denote
$\Omega'=G'\setminus \cup_{j=1}^{N_2}B(a_j,\sigma)$.
Combining (4.6) and
$$
\int_{G'\setminus G}|\nabla {u_{\varepsilon}}|^n
=\int_{G'\setminus G}|\nabla \bar{g}|^n \leq C,
$$
we have
$$
\int_{\Omega'}|\nabla
{u_{\varepsilon}}|^n \leq C+(n-1)^{n/2}|S^{n-1}|d|\ln \sigma|,
\eqno{(6.1)}
$$
where $C$ is a constant which is independent of $\varepsilon$.
For $R$ in (4.13), from (6.1) we have
$$
\int_{A_{R,\sigma}(a_j)}
|\nabla {u_{\varepsilon}}|^n \leq C.
$$
Then we know that there exists a
constant $r \in (\sigma,R)$ such that
$$
\int_{\partial B(a_j,r)}
|\nabla {u_{\varepsilon}}|^n \leq C(r)
$$
by using integral mean value theorem. Thus there exists a subsequence
$u_{\varepsilon_k}$ of $u_{\varepsilon}$ such that
$$
{u_{\varepsilon_k}} \to u_n,\quad \mbox{in } C(\partial B(a_j,r))
$$
as $\varepsilon_k \to 0$, which implies
$$
k_j=\deg({u_{\varepsilon}},\partial B(a_j,\sigma))=\deg(u_n,a_j).
$$

\begin{proposition} \label{prop6.2}
$k_j=0$ or $k_j=1$.
\end{proposition}

\paragraph{Proof.}
From the regularity results on n-harmonic maps (see [3][5] or [9]),
we know $u_n \in C^0(G_{\sigma},\mathbb{R}^n)$. Set
$$
w=\left\{\begin{array}{ll} \bar{g} &\mbox{on }G'\setminus G,\\
u_n &\mbox{on } G_{\sigma}, \end{array}\right.
$$
then $w \in \Pi$.
Using Proposition 4.5 and (4.7) we have
$$
\int_{\Omega'}|\nabla w|^n
\geq (n-1)^{n/2}|S^{n-1}|(\sum_{j=1}^{N_2}k_j^{\frac
{n}{n-1}})\ln\frac{1}{\sigma}-C(R).
\eqno{(6.2)}
$$
On the other hand, (6.1) and (4.9) imply
$$
{u_{\varepsilon_k}} \stackrel{w}{\to} w,
\quad\mbox{in } W^{1,n}(\Omega',\mathbb{R}^n).
$$
Noting this and the weak lower semicontinuity of
$\int_{\Omega'}|\nabla u|^n$, applying (6.1) we have
$$
\int_{\Omega'}|\nabla w|^n
\leq \underline{\lim}_{\varepsilon_k \to 0}
\int_{\Omega'}|\nabla u_{\varepsilon_k}|^n
\leq (n-1)^{n/2}|S^{n-1}|d\ln\frac{1}{\sigma}+C.
\eqno{(6.3)}
$$
Combining this with (6.2), we obtain
$$
(\sum_{j=1}^{N_2}k_j^{\frac{n}{n-1}}-d)\ln
\frac{1}{\sigma} \leq C
\quad \mbox{or} \quad
\sum_{j=1}^{N_2}k_j^{\frac{n}{n-1}} \leq d=\sum_{j=1}^{N_2}k_j
$$
for $\sigma$ small enough. Thus
$(k_j^{1/(n-1)}-1)k_j \leq 0$
which implies that the Proposition holds.

\begin{proposition} \label{prop6.3}
$k_j >0$, $j=1,2,\dots,N_2$.
\end{proposition}

\paragraph{Proof.}
Suppose $k_1=0$ and $k_2,k_3,\dots,k_{N_2}>0$.
Similar to the proof of Theorem 4.3 we have
$$
 \int_{\cup_{j=2}^{N_2}\Omega_j}
 |\nabla {u_{\varepsilon}}|^n \geq
 (n-1)^{n/2}|S^{n-1}|d\ln\frac{\sigma}{\varepsilon}-C.
$$
By this we can rewrite (4.6) as
$$
\int_{G\setminus \cup_{j=2}^{N_2}B(a_j,\sigma)}
|\nabla {u_{\varepsilon}}|^n
+\frac{1}{4\varepsilon^n}\int_G
(1-|{u_{\varepsilon}}|^2)^2 \leq C(\sigma).
$$
Thus similar to the proof of Theorem 5.3 we may modify (5.5) as
$$
\frac{1}{\varepsilon^n}
\int_{G\setminus \cup_{j=2}^{N_2}B(a_j,3\sigma)}
(1-|{u_{\varepsilon}}|^2)^2 \to 0
\eqno{(6.4)}
$$
as $\varepsilon \to 0$. Noticing
$$
G \cap B(a_1,\sigma) \subset G \cap B(a_1,R) \subset
G\setminus \cup_{j=2}^{N_2}B(a_j,R) \subset G\setminus
\cup_{j=2}^{N_2}B(a_j,3\sigma)
$$
we have
$$
\frac{1}{\varepsilon^n}
\int_{G \cap B(a_1,\sigma)}
(1-|{u_{\varepsilon}}|^2)^2 \to 0.
\eqno{(6.5)}
$$
On the other hand, the definition of $a_1$ implies that there exists at
least one bad ball $B(x_0^{\varepsilon},h \varepsilon)$ such that
$$
G \cap B(x_0^{\varepsilon},h \varepsilon)\subset G \cap B(a_1,\sigma).
$$
Applying the definition of bad ball we obtain
$$
\frac{1}{\varepsilon^n}
\int_{G \cap B(a_1,\sigma)}
(1-|{u_{\varepsilon}}|^2)^2
\geq \frac{1}{\varepsilon^n}
\int_{G \cap B(x_0^{\varepsilon},h \varepsilon)}
(1-|{u_{\varepsilon}}|^2)^2
\geq \mu >0
$$
which is contrary to (6.5).
This contradiction shows $k_1 >0$.

\paragraph{Remark} We may conclude
$k_j=1,j=1,2,\dots,N_2$ from Proposition 6.2
and Proposition 6.3. Noticing
$d=\sum_{j=1}^{N_2}k_j$, we obtain
$$
N_2=d,\quad 1=k_j=\sum_{i \in \Lambda_j}d_i.
$$
Thus on one hand, although the number of
the singularities of $n-$ harmonic maps
is indefinite (see Theorem A and Theorem C in [3]),
we can say that for this
$n-$ harmonic map $u_n$, the limit of the
regularizable minimizer $u_{\varepsilon_k}$ in $W^{1,n}$
as $k \to \infty$, the number of its
singularities is just the degree $d$ by applying (4.10).
On the other hand, there exists at least one $i_0
\in \Lambda_j$ such that $d_{i_0} \neq 0$.
Then we know that there exists
at least one  zero of $u_{\varepsilon}$
in $B(x_{i_0}^{\varepsilon},h \varepsilon)$
by using Kronecker's theorem.

\begin{theorem} \label{th6.4}
$a_j \in G,\quad j=1,2,\dots,d$.
\end{theorem}

\paragraph{Proof.}
Suppose $a_1 \in \partial G$, $a_2,a_3,\dots,a_d \in G$. Set
$$
\Omega_{\sigma}=(G'\setminus B(a_1,R))-\cup_{j=2}^dB(a_j,\sigma),
\quad
w=\left\{ \begin{array}{ll}
u_n &\mbox{on } G_{\sigma},\\
\bar{g} &\mbox{on }G'\setminus G.\end{array}\right.
$$
Using Proposition 4.5 on $\Omega_{\sigma}$ we have
$$
\int_{\Omega_{\sigma}}|\nabla w|^n
\geq (n-1)^{n/2}|S^{n-1}|(d-1)
\ln\frac{1}{\sigma}-C(R).
\eqno{(6.6)}
$$
Taking $u=w,a=a_1$ in Proposition 4.4 we have
$$
\int_{A_{R,\sigma}(a_1)}|\nabla w|^n
\geq 2^{\frac{1}{n}}(n-1)^{n/2}|S^{n-1}|
\ln\frac{1}{\sigma}-C.
$$
Combining this with (6.6) yields
$$
\int_{\Omega'}|\nabla w|^n
\geq (d+2^{\frac{1}{n}}-1)(n-1)^{n/2}|S^{n-1}|
\ln\frac{1}{\sigma}-C.
$$
Compare this to (6.3) we obtain
$$
(d+2^{\frac{1}{n}}-1-d)\ln\frac{1}{\sigma}\leq C
$$
where $C$ is a constant which is independent of $\sigma$.
It is impossible as $\sigma$ small enough, so $a_1 \in
G$.

\section{The proof of Theorem 1.3 }

\begin{theorem} \label{th 7.1}
Let $u_{\varepsilon}$ be the regularizable
minimizer of $E_{\varepsilon}(u,G)$.
Then there exists a subsequence
$u_{\varepsilon_k}$ of $u_{\varepsilon}$
such that
$$
u_{\varepsilon_k} \to u_n,
\quad\mbox{ in } W_{{\rm loc}}^{1,n}(G\setminus \cup
_{j=1}^d \{a_j\},\mathbb{R}^n).
$$
\end{theorem}

\paragraph{Proof.}
Step 1: Suppose the ball $B(x_0,2\sigma) \subset G\setminus \cup_{j=1}^d
\{a_j\}$, where the constant $\sigma$ may be sufficiently small but
independent of $\varepsilon$. Since (4.6) implies
$$
E_{\varepsilon}(u_{\varepsilon},B(x_0,2\sigma)
\setminus B(x_0,\sigma)) \leq C,
$$
we know there is a constant $r \in (\sigma, 2\sigma)$ such that
$$
\int_{\partial B(x_0,r)}|\nabla u_{\varepsilon}|^n
+\frac{1}{\varepsilon^n}\int
_{\partial B(x_0,r)}(1-|u_{\varepsilon}|^2)^2 \leq C(r),
\eqno{(7.1)}
$$
by applying the integral mean value theorem. Thus, there exists a
subsequence $u_{\varepsilon_k}$ of $u_{\varepsilon}$ such that
$$
u_{\varepsilon_k} \to u_n,
\quad\mbox{in } C(\partial B(x_0,r),\mathbb{R}^n),
$$
which leads to
$$
\frac{u_{\varepsilon_k}}{|u_{\varepsilon_k}|}
\to u_n, \quad\mbox{in } C(\partial B(x_0,r),\mathbb{R}^n).
\eqno{(7.2)}
$$

Step 2: Denote $\rho=|u_{\varepsilon}|$
on $B=B(x_0,r)$. It is not difficult to
prove that the minimizer $w$ of the problem
$$
\min\{\int_B|\nabla u|^n : u \in
W_{\frac{u_{\varepsilon}}{|u_{\varepsilon}|}}^{1,n}(B,S^{n-1})\}
\eqno{(7.3)}
$$
exists. Noting $u_{\varepsilon}$ be a
minimizer of $E_{\varepsilon}(u,G)$, we have
$$
E_{\varepsilon}(u_{\varepsilon},B)
\leq
\frac{1}{n}\int_B|\nabla (\rho w)|^n
+\frac{1}{4\varepsilon^n}\int_B(1-\rho^2)^2.
$$
Obviously (4.6) and $|u_{\varepsilon}|\geq 1/2$ on $B$ imply
$$
\frac{1}{2^n}\int_B|\nabla
\frac{u_{\varepsilon}}{|u_{\varepsilon}|}|^n
\leq \int_B|\nabla u_{\varepsilon}|^n \leq C,
$$
thus
$$
\int_B|\nabla w|^n
\leq \int_B|\nabla
\frac{u_{\varepsilon}}{|u_{\varepsilon}|}|^n\leq C.
\eqno{(7.4)}
$$
Applying this we may claim that
$$
\int_B|\nabla u_{\varepsilon}|^n
\leq C\varepsilon^{\lambda}+\int_B|\nabla w|^n,
\eqno{(7.5)}
$$
for some $\lambda >0$. Its proof can be seen in $\S8$.

Step 3: Let $w^{\tau}$ is a solution of
$$
\min\{\int_B(|\nabla w|^2+\tau)^{n/2} : w \in
W_{\frac{u_{\varepsilon}}{|u_{\varepsilon}|}}^{1,n}
(B,S^{n-1})\},\quad \tau \in (0,1).
\eqno{(7.6)}
$$
It is easy to see that $w^{\tau}$ solves
$$
-\mathop{\rm div}(v_{\varepsilon}^{(n-2)/2}\nabla w)
=w|\nabla w|^2v_{\varepsilon}^{(n-2)/2},
\quad v_{\varepsilon}=|\nabla w|^2+\tau.
\eqno{(7.7)}
$$
as $\tau \to 0$. Noticing
$\frac{u_{\varepsilon}}{|u_{\varepsilon}|}
\in W_{\frac{u_{\varepsilon}}{|u_{\varepsilon}|}}^{1,n}
(B,S^{n-1})$ we have
\setcounter{equation}{7}
\begin{eqnarray}
\int_B|\nabla w^{\tau}|^n
&\leq& \int_B(|\nabla w^{\tau}|^2+\tau)^{n/2}\\
&\leq& \int_B(|\nabla
\frac{u_{\varepsilon}}{|u_{\varepsilon}|}|^2+\tau)^{n/2}
\leq \int_B(|\nabla
\frac{u_{\varepsilon}}{|u_{\varepsilon}|}|^2+1)^{n/2}
\leq C \nonumber
\end{eqnarray}
by using (7.4),
where $C$ is a constant which is independent of $\varepsilon, \tau$.
Then there exist $w^*
\in W_{\frac{u_{\varepsilon}}{|u_{\varepsilon}|}}^{1,n}
(B,S^{n-1})$ and a subsequence of
$w^{\tau}$  such that
\begin{equation}
w^{\tau} \stackrel{w}{\longrightarrow} w^*,\quad\mbox{in }
W^{1,n}(B,\mathbb{R}^n).
\end{equation}
Noting the weak lower semicontinuity of $\int_B|\nabla w|^n$, we have
\begin{eqnarray}
\int_B|\nabla w^*|^n
&\leq& \underline{\lim}_{\tau \to 0}
\int_B|\nabla w^{\tau}|^n \\
&\leq& \overline{\lim}_{\tau \to 0}
\int_B|\nabla w^{\tau}|^n
\leq \overline{\lim}_{\tau \to 0}
\int_B(|\nabla w^{\tau}|^2+\tau)^{n/2}. \nonumber
\end{eqnarray}
The fact that $w^{\tau}$ solves (7.6) implies
$$
\overline{\lim}_{\tau \to 0}
\int_B(|\nabla w^{\tau}|^2+\tau)^{n/2}
\leq \lim_{\tau \to 0}
\int_B(|\nabla w_*|^2+\tau)^{n/2}
=\int_B|\nabla w_*|^n,
$$
where $w_*$ is a solution of (7.3). This and (7.10) lead to
$$
\int_B|\nabla w^*|^n
\leq \underline{\lim}_{\tau \to 0}
\int_B|\nabla w^{\tau}|^n
\leq \overline{\lim}_{\tau \to 0}
\int_B|\nabla w^{\tau}|^n
\leq \int_B|\nabla w_*|^n.
\eqno{(7.11)}
$$
Since $w^* \in
W_{\frac{u_{\varepsilon}}{|u_{\varepsilon}|}}^{1,n}
(B,S^{n-1})$, we know $w^*$ also
solves (7.3), namely
$$
\int_B|\nabla w_*|^n=\int_B|\nabla w^*|^n.
$$
Combining this with (7.11) yields
$$
\lim_{\tau \to 0}\int_B|\nabla w^{\tau}|^n
=\int_B|\nabla w^*|^n,
$$
which and (7.9) imply
$$
\nabla w^{\tau} \to \nabla w^*, \quad \mbox{in } L^n(B,\mathbb{R}^n).
\eqno{(7.12)}
$$

Step 4: Similar to the discussion of Step 3, we may derive the
following conclusion: Let $u^{\tau}$ be a solution of
$$
\min\{\int_B(|\nabla u|^2+\tau)^{n/2} : u \in
W_{u_n}^{1,n}(B,S^{n-1})\},\quad \tau \in (0,1).
\eqno{(7.13)}
$$
Then $u^{\tau}$ satisfies
$$
\int_B|\nabla u^{\tau}|^n \leq C,
\eqno{(7.14)}
$$
where $C$ is which is independent of $\tau$, and $u^{\tau}$ solves
$$
-\mathop{\rm div}(v^{(n-2)/2}\nabla u)=u|\nabla u|^2v^{(n-2)/2},
\quad v=|\nabla u|^2+\tau.
\eqno{(7.15)}
$$
As $\tau \to 0$, there exists a
subsequence of $u^{\tau}$ denoted itself
such that
$$
\nabla u^{\tau} \to \nabla u^*, \quad\mbox{in } L^n(B,\mathbb{R}^n),
\eqno{(7.16)}
$$
where  $u^*$ is a minimizer of $\int_B|\nabla u|^n$ in
$W_{u_n}^{1,n}(B,S^{n-1})$. It is well-known that $u^*$ is a
map of the least
n-energy, and also an n-harmonic map.

Fix $R>2\sigma$ such that $B(x_0,R) \subset G\setminus \cup_{j=1}^d
\{a_j\}$. Applying the regularity results on the map of the least
n-energy (for
example, Theorem 3.1 in [5]),
we have
$$
\sup_{B(x_0,r)}|\nabla u^*|^n
\leq \sup_{\overline{B(x_0,R)}}|\nabla u^*|^n:=C_0.
\eqno{(7.17)}
$$
It is obvious that $C_0$ is a constant which is independent of $r$.

Step 5: From (7.7) subtracts (7.15). Then
$$
-\mathop{\rm div}(v_{\varepsilon}^{(n-2)/2}\nabla w-v^{(n-2)/2}\nabla u)
=w|\nabla w|^2v_{\varepsilon}^{(n-2)/2}-u|\nabla u|^2v^{(n-2)/2}.
\eqno{(7.18)}
$$
Multiplying both sides of (7.18) by $w-u$ and integrating over $B$ we
obtain
\begin{eqnarray*}
\lefteqn{-\int_{\partial B}(v_{\varepsilon}^{(n-2)/2}w_{\nu}
-v^{(n-2)/2}u_{\nu})(w-u)}\\
\lefteqn{+\int_B(v_{\varepsilon}^{(n-2)/2}\nabla w
-v^{(n-2)/2}\nabla u)\nabla (w-u)}\\
&=&\int_B(w|\nabla w|^2v_{\varepsilon}^{(n-2)/2}
-u|\nabla u|^2v^{(n-2)/2})(w-u),
\end{eqnarray*}
where $\nu$ denotes the unit outside-norm vector of $\partial B$. Thus
\setcounter{equation}{18}
\begin{eqnarray}
\lefteqn{|\int_B(v_{\varepsilon}^{(n-2)/2}\nabla w
-v^{(n-2)/2}\nabla u)\nabla (w-u)| }\nonumber\\
&\leq& |\int_{\partial B}(v_{\varepsilon}^{(n-2)/2}w_{\nu}
-v^{(n-2)/2}u_{\nu})(w-u)|\\
&&+|\int_B(w|\nabla u|^2v^{(n-2)/2}
-u|\nabla u|^2v^{(n-2)/2})(w-u)| \nonumber\\
&&+|\int_B(w|\nabla w|^2v_{\varepsilon}^{(n-2)/2}
-w|\nabla u|^2v^{(n-2)/2})(w-u)| \nonumber\\
&=&I_1+I_2+I_3. \nonumber
\end{eqnarray}
First we give an estimate for $I_1$. Let $w=w^{\tau}$ is a solution  of
(7.6). Integrating both sides of (7.7) over $B$, we have
$$
-\int_{\partial B}v_{\varepsilon}^{(n-2)/2}w_{\nu}
=\int_Bw|\nabla w|^2v_{\varepsilon}^{(n-2)/2},
$$
which and (7.8) imply
$$
|\int_{\partial B}v_{\varepsilon}^{(n-2)/2}w_{\nu}|
\leq \int_B v_{\varepsilon}^{n/2} \leq C.
\eqno{(7.20)}
$$
An analogous discussion shows that for the solution $u=u^{\tau}$  of
(7.13) which equips with (7.14), we may also obtain
$$
|\int_{\partial B}v^{(n-2)/2}u_{\nu}|
\leq \int_B|\nabla u|^n \leq C.
\eqno{(7.21)}
$$
Applying (7.20)(7.21) we derive
\setcounter{equation}{21}
\begin{eqnarray}
I_1 &\leq& \sup_{\partial B}|w-u|
(|\int_{\partial B}v_{\varepsilon}^{(n-2)/2}w_{\nu}|
+|\int_{\partial B}v^{(n-2)/2}u_{\nu}|)\\
&\leq& C\sup_{\partial B}|w-u|
=C\sup_{\partial B}|\frac{u_{\varepsilon}}
{|u_{\varepsilon}|}-u_n|, \nonumber
\end{eqnarray}
where $C$ is independent of $\varepsilon, \tau$.
For the estimate of $I_3$, we have
\begin{eqnarray}
I_3 &\leq& \int_B|u-w|||\nabla u|^2v^{(n-2)/2}-|\nabla
w|^2v_{\varepsilon}^{(n-2)/2}|\\
&\leq& 2\int_B||\nabla u|^2v^{(n-2)/2}
-|\nabla w|^2v_{\varepsilon}^{(n-2)/2}|. \nonumber
\end{eqnarray}
For estimating $I_2$, we multiply both sides of (7.15) by $(u-w)$ and
integrate over $B$, then
\begin{eqnarray*}
\lefteqn{-\int_{\partial B}v^{(n-2)/2}u_{\nu}(u-w)
+\int_Bv^{(n-2)/2} \nabla u \nabla (u-w)}\\
&=&\int_B|\nabla u|^2v^{(n-2)/2}u(u-w)
=\int_B|\nabla u|^2v^{(n-2)/2}(1-uw).
\end{eqnarray*}
Thus, we have
\begin{eqnarray*}
I_2 &\leq& \int_B|\nabla u|^2v^{(n-2)/2}|u-w|^2
=2\int_B|\nabla u|^2v^{(n-2)/2}(1-uw)\\
&\leq& 2|\int_{\partial B}v^{(n-2)/2}u_{\nu}(u-w)|
+2|\int_Bv^{(n-2)/2}\nabla u \nabla (u-w)|.
\end{eqnarray*}
Noting (7.21) we may derive
$$
I_2 \leq C\sup_{\partial B}|
\frac{u_{\varepsilon}}{|u_{\varepsilon}|}-u_n|
+2|\int_Bv^{(n-2)/2}\nabla u \nabla (u-w)|.
\eqno{(7.24)}
$$

Step 6: Substituting (7.22)-(7.24) into (7.19) yields
\begin{eqnarray*}
\lefteqn{|\int_B(v_{\varepsilon}^{(n-2)/2}\nabla w
-v^{(n-2)/2}\nabla u)\nabla (w-u)|}\\
&\leq& C\sup_{\partial B}|\frac
{u_{\varepsilon}}{|u_{\varepsilon}|}-u_n|
+2|\int_Bv^{(n-2)/2}\nabla u \nabla (u-w)| \\
&& +2\int_B|v_{\varepsilon}^{(n-2)/2}|\nabla w|^2
-v^{(n-2)/2}|\nabla u|^2|.
\end{eqnarray*}
Letting $\tau \to 0$ and applying (7.12)(7.16) we obtain
\begin{eqnarray*}
\lefteqn{|\int_B(|\nabla w^*|^{(n-2)/2}\nabla w^*
-|\nabla u^*|^{(n-2)/2}\nabla u^*)\nabla (w^*-u^*)|}\\
&\leq& C\sup_{\partial B}|\frac
{u_{\varepsilon}}{|u_{\varepsilon}|}-u_n|
+2|\int_B|\nabla u^*|^{n-1} \nabla (u^*-w^*)|
 +2\int_B||\nabla w^*|^n -|\nabla u^*|^n|.
\end{eqnarray*}
Using Lemma 1.2 in [4], we have
$$
2^{n-1}\int_B|\nabla w^*-\nabla u^*|^n \leq
|\int_B(|\nabla w^*|^{(n-2)/2}\nabla w^*
-|\nabla u^*|^{(n-2)/2}\nabla u^*)\nabla (w^*-u^*)|.
$$
Thus
$$
(2^{n-1}-2)\int_B|\nabla w^*-\nabla u^*|^n
\leq C\sup_{\partial B}|\frac
{u_{\varepsilon}}{|u_{\varepsilon}|}-u_n|
+2|\int_B|\nabla u^*|^{n-1} \nabla (u^*-w^*)|.
$$
Denote $\psi (\varepsilon)=\int_B
|\nabla w^*-\nabla u^*|^n$ and  let $\varepsilon \to 0$, then
$$
(2^{n-1}-2)\psi(\varepsilon)
\leq o(1)+2(C_0|B|)^{(n-1)/n}(\psi(\varepsilon))^{1/n}
\eqno{(7.25)}
$$
holds by using (7.2), where $C_0$ is the constant in (7.17).

We claim that for some small
constant $\sigma >0$, the following holds:
$$
\psi(\varepsilon) \to 0,
\quad\mbox{ as } \varepsilon \to 0.
\eqno{(7.26)}
$$
Suppose (7.26) is not true,
then there exists $\tau>0$, for any $\varepsilon_0>0$,
such that as $\varepsilon<\varepsilon_0$
we have $\psi(\varepsilon) \geq 2\tau >\tau$ or
$$
(\psi(\varepsilon))^{(n-1)/n}>\tau^{(n-1)/n},
\quad \forall \varepsilon<\varepsilon_0.
\eqno{(7.27)}
$$
Taking $\sigma$ small enough so that
$$
2(C_0|B(x_0,r)|)^{(n-1)/n}=(2^{n-2}-1)\tau^{(n-1)/n},
$$
we obtain from (7.25) \setcounter{equation}{27}
\begin{eqnarray}
\lefteqn{(\psi(\varepsilon))^{1/n}[(\psi(\varepsilon))^{(n-1)/n}
-\frac{2(C_0|B|)^{(n-1)/n}}{2^{n-1}-2}]}\\
&=&(\psi(\varepsilon))^{1/n}[(\psi(\varepsilon))^{(n-1)/n}
-\frac{1}{2}\tau^{(n-1)/n}] = o(1). \nonumber
\end{eqnarray}
Substituting (7.27) into (7.28) we derive
$(\psi(\varepsilon))^{1/n} = o(1)$,
which is contrary to (7.27).

Step 7: Noting the weak lower semicontinuity of the functional
$\int_B|\nabla u|^n$, from (4.9) we are led to
$$
\int_B|\nabla u_n|^n
\leq \underline{\lim}_{\varepsilon_k \to 0}
\int_B|\nabla
u_{\varepsilon_k}|^n.
$$
Combining this with (7.5) and (7.26) we obtain
\begin{eqnarray*}
\int_B|\nabla u_n|^n
\leq \underline{\lim}_{\varepsilon_k
\to 0}\int_B|\nabla
u_{\varepsilon_k}|^n
&\leq& \overline{\lim}_{\varepsilon_k
\to 0}\int_B|\nabla
u_{\varepsilon_k}|^n \\
&\leq& \lim_{\varepsilon_k \to 0}\int_B|\nabla w^*|^n
=\int_B|\nabla u^*|^n.
\end{eqnarray*}
Recalling the definition of $u^*$ in Step 4, and noticing $u_n \in
W_{u_n}^{1,n}(B,S^{n-1})$, we know that $u_n$ is also a minimizer of
$\int_B|\nabla u|^n$ and
$$
\lim_{\varepsilon_k \to 0}\int_B|\nabla
u_{\varepsilon_k}|^n=\int_B|\nabla
u_n|^n=\int_B|\nabla
u^*|^n,
\eqno{(7.29)}$$
which and (4.9) imply
$$
\nabla u_{\varepsilon_k} \to \nabla u_n,
\quad\mbox{ in } L^n(B,\mathbb{R}^n).
$$
Combining this with the fact
$$
u_{\varepsilon_k} \to u_n, \quad\mbox{ in } L^n(B,\mathbb{R}^n),
$$
which can be deduced from (4.6), we derive
$$
u_{\varepsilon_k} \to u_n, \quad\mbox{ in } W^{1,n}(B,\mathbb{R}^n).
$$
Then it is not difficult to complete the proof of this theorem.

\section{The proof of (7.5) }

To prove (7.5), we will introduce a comparison function first.
Consider the functional
$$
E(\rho,B)=\frac{1}{n}\int_B(|\nabla \rho|^2+1)^{n/2}
+\frac{1}{2\varepsilon^n}\int_B(1-\rho)^2.
$$
It is easy to prove that the minimizer $\rho_1$ of $E(\rho,B)$
on $W_{|u_{\varepsilon}|}^{1,n}(B,R^+)$ exists and satisfies
$$
-div(v^{(n-2)/2}\nabla \rho)
=\frac{1}{\varepsilon^n}(1-\rho)
\quad on \quad B,
\eqno{(8.2)}
$$
$$
\rho|_{\partial B}=|u_{\varepsilon}|,
\eqno{(8.3)}
$$
where $v=|\nabla \rho|^2+1$. Since $1/2 \leq |u_{\varepsilon}|
\leq 1$ on $B$, it follows from the maximum principle that
$$
1/2 \leq |u_{\varepsilon}| \leq \rho_1
\leq 1
\eqno{(8.4)}
$$
on $\overline{B}$.

Applying (4.6) we see easily that
$$
E(\rho_1,B) \leq E(|u_{\varepsilon}|,B)
\leq CE_{\varepsilon}(u_{\varepsilon},B) \leq C.
\eqno{(8.5)}
$$

Multiplying (8.2) by $(\nu \cdot \nabla \rho)$, where
$\rho=\rho_1$, and integrating over $B$, we obtain
$$
-\int_{\partial B}v^{(n-2)/2}(\nu \cdot \nabla \rho)^2
+\int_Bv^{(n-2)/2}\nabla \rho \cdot \nabla (\nu \cdot
\nabla \rho)
=\frac{1}{\varepsilon^n}\int_B(1-\rho)(\nu \cdot \nabla \rho),
\eqno{(8.6)}
$$
where $\nu$ denotes the unit outside norm vector on $\partial B$.
Using (8.5) we have
$$\begin{array}{ll}
&~~|\int_Bv^{(n-2)/2}\nabla \rho
\nabla(\nu \cdot \nabla \rho)|
\leq C\int_Bv^{(n-2)/2}|\nabla \rho|^2
+\frac{1}{2}|\int_Bv^{(n-2)/2}\nu \cdot \nabla v|\\[3mm]
&\leq C+\frac{1}{n}|\int_B\nu \cdot \nabla(v^{n/2})|
\leq C+\frac{1}{n}\int_B|div(\nu v^{n/2})-v^{n/2}div \nu|\\[3mm]
&C+\frac{1}{n}\int_{\partial B}v^{n/2}.
\end{array}
\eqno{(8.7)}
$$
Combining (8.3)(7.1) and (8.5) we also have
$$\begin{array}{ll}
|\frac{1}{\varepsilon^n}\int_B(1-\rho)(\nu \cdot \nabla \rho)|
\leq \frac{1}{2\varepsilon^n}
|\int_B(1-\rho)^2div\nu-\int_{\partial B}(1-\rho)^2|\\[3mm]
\leq \frac{1}{2\varepsilon^n}
\int_B(1-\rho)^2|div\nu|
+\frac{1}{2\varepsilon^n}\int_{\partial B}(1-\rho)^2 \leq C.
\end{array}
$$
Substituting this and (8.7) into (8.6) yields
$$
|\int_{\partial B}v^{(n-2)/2}
(\nu \cdot \nabla \rho)^2|
\leq C+\frac{1}{n}\int_{\partial B}v^{n/2}.
\eqno{(8.8)}
$$
Applying (8.3)(7.1) and (8.8), we obtain for any $\delta
\in (0,1)$,
$$\begin{array}{ll}
&~~\int_{\partial B}v^{n/2}=
\int_{\partial B}v^{(n-2)/2}[1+
(\tau \cdot \nabla \rho)^2
+(\nu \cdot \nabla \rho)^2]\\[3mm]
&=\int_{\partial B}v^{(n-2)/2}[1+
(\tau \cdot \nabla |u_{\varepsilon}|)^2
+(\nu \cdot \nabla \rho)^2]\\[3mm]
&\leq \int_{\partial B}v^{(n-2)/2}
+\int_{\partial B}v^{(n-2)/2}(\nu \cdot \nabla \rho)^2\\[3mm]
&~~+(\int_{\partial B}v^{n-2})^{(n-2)/n}
(\int_{\partial B}(\tau \cdot \nabla |u_{\varepsilon}|)^n)^{2/n}\\[3mm]
&\leq C(\delta)+(\frac{1}{n}+2\delta)\int_{\partial B}v^{n/2},
\end{array}
$$
where $\tau$ denotes the unit tangent vector on $\partial B$.
Hence it follows by choosing $\delta >0$ so small that
$$
\int_{\partial B}v^{n/2} \leq C.
\eqno{(8.9)}
$$

Now we multiply both sides of (8.2) by $(1-\rho)$ and integrate over
$B$. Then
$$
\int_Bv^{(n-2)/2}|\nabla \rho|^2
+\frac{1}{\varepsilon^n}\int_B(1-\rho)^2
=-\int_{\partial B}v^{(n-2)/2}
(\nu \cdot \nabla \rho)(1-\rho).
$$
From this, using (7.1)(8.3)(8.4) and (8.9) we obtain
$$\begin{array}{ll}
&~~E(\rho_1,B) \leq
C|(\nu \cdot \nabla \rho)(1-\rho)|\\[3mm]
&\leq C|\int_{\partial B}v^{n/2}|^{(n-1)/n}
|\int_{\partial B}(1-\rho)^2|^{1/n}\\[3mm]
&\leq C|\int_{\partial B}(1-|u_{\varepsilon}|)^2|^{1/n}
\leq C\varepsilon
\end{array}
\eqno{(8.10)}
$$

Since $u_{\varepsilon}$ is a minimizer of $E_{\varepsilon}(u,B)$, we have
$$\begin{array}{ll}
&~~E_{\varepsilon}(u_{\varepsilon},B)
\leq E_{\varepsilon}(\rho_1 w,B)\\[3mm]
&=\frac{1}{n}\int_B(|\nabla \rho_1|^2
+\rho_1^2|\nabla w|^2)^{n/2}
+\frac{1}{4\varepsilon^n}\int_B(1-\rho_1^2)^2,
\end{array}
\eqno{(8.11)}
$$
where $w$ is a solution of (7.3). On on hand,
$$\begin{array}{ll}
&~~\int_B(|\nabla \rho_1|^2
+\rho_1^2|\nabla w|^2)^{n/2}dx
-\int_B(\rho_1^2|\nabla w|^2)^{n/2}dx\\[3mm]
&=\frac{n}{2}\int_B\int_0^1[(|\nabla \rho_1|^2
+\rho_1^2|\nabla w|^2)^{(n-2)/2}s
+(\rho_1^2|\nabla w|^2)^{(n-2)/2}(1-s)]ds|\nabla \rho_1|^2dx\\[3mm]
&\leq C\int_B(|\nabla \rho_1|^n
+|\nabla \rho_1|^2|\nabla w|^{(n-2)/2})dx.
\end{array}
\eqno{(8.12)}
$$
On the other hand, by using (8.10) and (7.4) we have
$$
\int_B|\nabla \rho_1|^2|\nabla w|^{(n-2)/2}
\leq (\int_B|\nabla \rho_1|^{4n/(n+2)})^{(n+2)/2n}
(\int_B|\nabla w|^n)^{(n-2)/2n} \leq C\varepsilon^{\lambda}.
\eqno{(8.13)}
$$
Combining (8.11)-(8.13), we can derive
$$
E_{\varepsilon}(u_{\varepsilon},B)
\leq \frac{1}{n}\int_B\rho_1^n|\nabla w|^n
+C\varepsilon^{\lambda},
$$
where $\lambda$ is a constant only depending on $n$.
Thus (7.5) can be seen by (8.4).

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\noindent{\sc Yutian Lei} \\
Mathematics  Department, SuZhou University\\
1 Shizi street \\
Suzhou, Jiangsu, 215006\\
P. R. China \\
e-mail: lythxl@163.com
\end{document}