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\markboth{\hfil $C^{1,\alpha}$ convergence of minimizers \hfil EJDE--2000/14}
{EJDE--2000/14\hfil Yutian Lei \& Zhuoqun Wu \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol.~{\bf 2000}(2000), No.~14, pp.~1--20. \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)}
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  $C^{1,\alpha}$ convergence of minimizers of a Ginzburg-Landau functional
\thanks{ {\em 1991 Mathematics Subject Classifications:} 35J70.
\hfil\break\indent
{\em Key words and phrases:} Ginzburg-Landau functional, regularizable minimizer.
\hfil\break\indent
\copyright 2000 Southwest Texas State University  and University of
North Texas. \hfil\break\indent
Submitted September 2, 1999. Published February 21, 2000.} }
\date{}

% 
\author{Yutian Lei \& Zhuoqun Wu}
\maketitle

\begin{abstract}
In this article we study the minimizers of the functional
$$
E_\varepsilon(u,G)=\frac{1}{p}\int_G|\nabla u|^p+\frac{1}{4\varepsilon^p}
\int_G(1-|u|^2)^2,
$$
on the class $W_g=\{v \in W^{1,p}(G,{\mathbb R}^2);v|_{\partial G}=g\}$,
where $g:\partial G \to S^1$ is a smooth map with Brouwer degree zero, and
$p$ is greater than 2.
In particular, we show that the minimizer converges to the $p$-harmonic map
in $C_{\mbox{\scriptsize\rm loc}}^{1,\alpha}(G,{\mathbb R}^2)$ as
$\varepsilon$ approaches zero.
\end{abstract}

\renewcommand{\theequation}{\thesection.\arabic{equation}}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}

\section{Introduction }

Let $G \subset {\mathbb R}^2$ be a bounded and simply connected domain
with smooth boundary $\partial G$ and $g$ be a smooth map from 
$\partial G$ into $S^1=\{x \in {\mathbb R}^2;|x|=1\}$.
Consider the Ginzburg-Landau-type functional
$$
E_\varepsilon(u,G)=\frac{1}{p}\int_G|\nabla u|^p
+\frac{1}{4\varepsilon^p}\int_G(1-|u|^2)^2
$$
with a small parameter $\varepsilon>0$. This functional has been studied
in [1] for $p=2$, $d=\mathop{\rm deg}(g,\partial G)=0$,
and in [2] for $p=2$, $d=\mathop{\rm deg}(g,\partial G) \neq 0$.
Here $d=\mathop{\rm deg}(g,\partial G)$ denotes the Brouwer degree of the
map $g$. For other related papers, we refer to [3]--[11].

In this paper we are concerned with the case $p>2$,
$d=\mathop{\rm deg}(g,\partial G)=0$. It is easy to see that
 the functional $E_\varepsilon(u,G)$ achieves its minimum on
$$
W_g=\{v \in W^{1,p}(G,{\mathbb R}^2):v|_{\partial G}=g\}
$$
at a function $u_\varepsilon$ and that
\begin{equation}
\lim_{\varepsilon \to 0}u_\varepsilon= u_p\quad
\mbox{in }W^{1,p}(G,{\mathbb R}^2) \label{(1.1)}
\end{equation}
where $u_p$ is a $p$-harmonic map from $G$ into $S^1$ with boundary value
$g$ [9]. Recall that $u \in W^{1,p}(G,S^1)$ is said to
be a $p$-harmonic map on $G$, if $u$ is a weak solution of the
equation
$$
-\mathop{\rm div}(|\nabla u|^{p-2}\nabla u)=u|\nabla u|^p.
$$
Under the condition $d=0$, there exists exactly one
$p$-harmonic map on $G$ with the given boundary value $g$.
However, there may be several minimizers of the functional.
Let $\tilde{u}_\varepsilon$ be a minimizer that
can be obtained as the limit of a subsequence
of the minimizers $u_\varepsilon^{\tau}$ of the regularized functionals
$$
E_\varepsilon^{\tau}(u,G)=
\frac{1}{p}\int_G(|\nabla u|^2+\tau)^{p/2} +\frac{1}{4\varepsilon^p}
\int_G(1-|u|^2)^2,\quad (\tau>0)
$$
on $W_g$ as $\tau_k \to 0$, namely
\begin{equation}
\lim_{\tau_k \to 0}u_\varepsilon^{\tau_k}=\tilde{u}_\varepsilon
\quad \mbox{in }W^{1,p}(G,{\mathbb R}^2). \label{(1.2)}
\end{equation}
$\tilde{u}_\varepsilon$ is called the regularizable minimizer
of $E_\varepsilon(u,G)$.
Our main result reads as follows.

\begin{theorem} \label{th1.1}
 Assume that $p>2$, $d=\mathop{\rm deg}(g,\partial G)=0$.
Let $\tilde{u}_\varepsilon$
be a regularizable minimizer of $E_\varepsilon(u,G)$.
Then for some $\alpha \in (0,1)$ we have
$$
\lim_{\varepsilon \to 0}\tilde{u}_\varepsilon
= u_p\quad \mbox{in } C_{\mbox{\scriptsize\rm loc}}^{1,\alpha}(G,{\mathbb R}^2).
$$
\end{theorem}

We shall prove a series of preliminary propositions
in Sections 2, 3, and 4. Then we complete the proof of the
main theorem in $\S5$. In $\S6$, we indicate how to extend  our result
to the higher dimensional case.

\section{Convergence of $|u_\varepsilon^{\tau}|$ } \setcounter{equation}{0}

We start our argument with the following proposition.

\begin{proposition} \label{prop2.1}
\begin{equation}
\lim_{\varepsilon,\tau \to 0}|u_\varepsilon^{\tau}|=1
\quad \mbox{in }C(\overline{G},{\mathbb R}^2). \label{(2.1)}
\end{equation}
\end{proposition}

\paragraph{Proof.} For $\tau \in (0,1)$, we have
\begin{eqnarray*}
E_\varepsilon^{\tau}(u_\varepsilon,G)
&\leq& E_\varepsilon^{\tau}(u_p,G)
=\frac{1}{p}\int_G(|\nabla u_p|^2+\tau)^{p/2}\\
& \leq& \frac{1}{p}\int_G(|\nabla u_p|^2+1)^{p/2}=C\,.
\end{eqnarray*}
Hence
\begin{eqnarray}
&\int_G|\nabla u_\varepsilon^{\tau}|^p
\leq \int_G(|\nabla u_\varepsilon^{\tau}|^2+\tau)^{p/2}
\leq C,& \label{(2.2)} \\
&\int_G(1-|u_\varepsilon^{\tau}|^2)^2 \leq C\varepsilon^p\,.& \label{(2.3)}
\end{eqnarray}
Here and below, we denote by $C$ a universal constant
which may take different values on different occasions.
If necessary, we indicate explicitly its dependence.

>From (2.3) it follows that there exists a subsequence $u_{\varepsilon_k}
^{\tau_k}$ of $u_\varepsilon^{\tau}$ with
$\varepsilon_k \to 0,\tau_k \to 0$
as $k \to \infty$, such that
\begin{equation}
\lim_{k \to \infty}|u_{\varepsilon_k}^{\tau_k}|
=1,\quad \mbox{ a. e. in }G\,.\label{(2.4)}
\end{equation}
Inequality (2.2) combined with $|u_\varepsilon^{\tau}| \leq 1$ (which follows from
the maximum principle) means that $\|u_\varepsilon^{\tau}\|_{W^{1,p}
(G,{\mathbb R}^2)} \leq C$ which implies that there exist a function
$u_* \in W^{1,p}(G,{\mathbb R}^2)$ and a subsequence of $u_{\varepsilon_k}^{\tau_k}$,
supposed to be $u_{\varepsilon_k}^{\tau_k}$
itself without loss of generality, such that
\begin{equation}
\lim_{k \to \infty}u_{\varepsilon_k}^{\tau_k}=u_*
\quad \mbox{in }C(\overline{G},{\mathbb R}^2)\,. \label{(2.5)}
\end{equation}
Combining (2.5) with (2.4) yields $|u_*|=1$ in $G$ and hence
$$
\lim_{k \to \infty}|u_{\varepsilon_k}^{\tau_k}|=1
\quad \mbox{in }C(\overline{G},{\mathbb R}^2)\,.
$$
Since any subsequence of $|u_\varepsilon^{\tau}|$ contains a uniformly
convergent subsequence and the limit is the same number 1, we may assert
(2.1) which completes the proof.

Next, we prove some related facts about the
asymptotic behaviour of $|u_\varepsilon^{\tau}|$, although
Proposition~\ref{prop2.1} is enough for proving the next steps.

\begin{proposition} \label{prop2.2}
For all $q \in (1,p)$, there exist constants $C$,
$\lambda>0$, independent of $\varepsilon$ such that
\begin{equation}
\int_G|\nabla |u_\varepsilon^{\tau}||^q
\leq C\varepsilon^{\lambda} \label{(2.6)}
\end{equation}
for $\tau \in (0,1)$ and $\varepsilon \in (0,\eta)$ for some small $\eta>0$.
\end{proposition}

\paragraph{Proof.} As a minimizer of $E_\varepsilon^{\tau}(u,G)$,
$u=u_\varepsilon^{\tau}$ satisfies
the corresponding Euler equation
\begin{eqnarray}
&-\mathop{\rm div}(v^{(p-2)/2}\nabla u)
=\frac{1}{\varepsilon^p}u(1-|u|^2) \quad \mbox{in }G &\label{(2.7)}\\
&u|_{\partial G}=g\,,&
\end{eqnarray}
where $v=|\nabla u|^2+\tau$.
Set $u=h(\cos \phi,\sin \phi)$ and $h=|u|$. Then
\begin{eqnarray}
&-\mathop{\rm div}(v^{(p-2)/2}h^2 \nabla \phi)=0 & \\
&-\mathop{\rm div}(v^{(p-2)/2}\nabla h)+h|\nabla \phi|^2v^{(p-2)/2}=
\frac{1}{\varepsilon^p}h(1-h^2)\,. &\label{(2.8)}
\end{eqnarray}
Fix $\beta \in (0,p/2)$ and set
$$
S=\{x \in G;|h(x)|>1-\varepsilon^{\beta}\},
\tilde{h}=\max(h,1-\varepsilon^{\beta}).
$$
Multiplying (2.8) with $h(1-\tilde{h})$,
integrating over $G$ and noticing that
$\tilde{h}|_{\partial G}=1$, we have
$$ \displaylines{
-\int_Gv^{(p-2)/2}h \nabla h \nabla \tilde{h}
+\int_Gv^{(p-2)/2}|\nabla h|^2(1-\tilde{h})
+\int_Gv^{(p-2)/2}h^2|\nabla \phi|^2(1-\tilde{h}) \cr
=\frac{1}{\varepsilon^p}
\int_Gh^2(1-h^2)(1-\tilde{h}) \cr}
$$
and thus we obtain
\begin{equation}
\int_Gv^{(p-2)/2}h \nabla h \nabla \tilde{h}
\leq C\varepsilon^{\beta} \label{(2.9)}
\end{equation}
by using (2.2) and the facts $|\nabla u|^2=|\nabla h|^2+h^2|\nabla \phi|^2$ and
$h=|u| \leq 1$. Since $\tilde{h}=1-\varepsilon^{\beta}$ on $G \setminus S$,
$\tilde{h}=h$ on $S$
and $h>1/2$ for $\varepsilon>0$ small enough, (2.9) implies
$$
\int_Sv^{(p-2)/2}|\nabla h|^2 \leq C\varepsilon^{\beta}
$$
and hence
\begin{equation}
\int_S|\nabla h|^p \leq C\varepsilon^{\beta}. \label{(2.10)}
\end{equation}
On the other hand, from the definition of $S$ and (2.3), we have
$$
C\mathop{\rm meas}(G \setminus S)\varepsilon^{2\beta}
\leq \int_{G \setminus S}(1-|u|^2)^2 \leq C\varepsilon^p,
$$
namely
$$
\mathop{\rm meas}(G \setminus S) \leq C\varepsilon^{p-2\beta}
$$
and hence using (2.2) again we see that for any $q \in (1,p)$
$$
\int_{G \setminus S}|\nabla h|^q \leq \mathop{\rm meas}(G-S)^{1-q/p}
(\int_G|\nabla h|^p)^{q/p}
\leq C\varepsilon^{(p-2\beta)(1-q/p)}
$$
which and (2.10) imply the conclusion of Proposition~\ref{prop2.2}.

\begin{proposition} \label{prop2.3} There exists a constant
$C$ independent of $\varepsilon,\tau\in (0,1)$, such that
\begin{equation}
\frac{1}{\varepsilon^p}
\int_G(1-|u_\varepsilon^{\tau}|^2) \leq C\,.\label{(2.11)}
\end{equation}
\end{proposition}

\paragraph{Proof.}
First we take the inner product of the both sides of (2.7)
with $u$ and then integrate over $G$
$$
-\int_G\mathop{\rm div}(v^{(p-2)/2}\nabla u)u
=\frac{1}{\varepsilon^p}\int_G|u|^2(1-|u|^2).
$$
Integrating by parts and using (2.2) and the Holder inequality we
obtain
\begin{eqnarray}
\frac{1}{\varepsilon^p}\int_G|u|^2(1-|u|^2)
&\leq&\int_Gv^{(p-2)/2}|\nabla u|^2
+\int_{\partial G}v^{(p-2)/2}|u_n||u| \nonumber\\
&\leq& C+\int_{\partial G}v^{(p-2)/2}|u_n| \label{(2.12)}\\
&\leq& C+C\int_{\partial G}v^{(p-2)/2}
+C\int_{\partial G}v^{(p-2)/2}|u_n|^2 \nonumber\\
&\leq& C+C\int_{\partial G}v^{p/2} \nonumber
\end{eqnarray}
where $n$ denotes the unit outward normal to $\partial G$ and $u_n$ the
derivative with respect to $n$.

To estimate $\int_{\partial G}v^{p/2}$, we choose a smooth vector field
$\nu=(\nu_1,\nu_2)$ such that $\nu|_{\partial G}=n$,
take the inner product of the both
sides of (2.7) with $\nu \cdot \nabla u$ and integrate over $G$.
Then we have
$$
-\int_G\mathop{\rm div}(v^{(p-2)/2}\nabla u)(\nu \cdot \nabla u)
=\frac{1}{2\varepsilon^p}
\int_G(1-|u|^2)(\nu \cdot \nabla |u|^2).
$$
Integrating by parts and noticing $|u|_{\partial G}=|g|=1$ and
$$
\int_G(1-|u|^2)(\nu \cdot \nabla |u|^2)
=-\frac{1}{2}\int_G\nabla (1-|u|^2)^2 \cdot \nu
=\frac{1}{2}\int_G(1-|u|^2)^2\mathop{\rm div} \nu
$$
yield
\begin{equation}
-\int_{\partial G}v^{(p-2)/2}|u_n|^2
+\int_Gv^{(p-2)/2}\nabla u \cdot \nabla (\nu \cdot \nabla u)
=\frac{1}{4\varepsilon^p}
\int_G(1-|u|^2)^2\mathop{\rm div} \nu\,.\label{(2.13)}
\end{equation}
>From the smoothness of $\nu$ and (2.2), (2.3) we have
\begin{equation}
\frac{1}{\varepsilon^p}\int_G (1-|u|^2)^2|\mathop{\rm div} \nu| \leq C
\label{(2.14)}
\end{equation}
\begin{eqnarray}
\int_Gv^{(p-2)/2} \nabla u \nabla (\nu \cdot \nabla u)
&\leq& C\int_Gv^{(p-2)/2}|\nabla u|^2
+\frac{1}{2}\int_G
v^{(p-2)/2} \nu \cdot \nabla v \nonumber\\
&\leq& C+\frac{1}{p} \int_G\nu \cdot \nabla (v^{p/2})  \label{(2.15)} \\
&\leq& C+\frac{1}{p}\int_G\mathop{\rm div}(\nu v^{p/2})
-\frac{1}{p}\int_Gv^{p/2}\mathop{\rm div}\nu \nonumber\\
&\leq& C+\frac{1}{p}\int_{\partial G}v^{p/2} \nonumber
\end{eqnarray}
and
\begin{eqnarray}
\int_{\partial G}v^{p/2}
&=&\int_{\partial G}v^{(p-2)/2}(|u_n|^2+|g_t|^2+\tau) \nonumber\\
&\leq& \int_{\partial G}v^{(p-2)/2}
|u_n|^2+C\int_{\partial G}v^{(p-2)/2}\label{(2.16)}
\end{eqnarray}
where $g_t$ denotes the derivative of $g$ with respect to the tangent
vector $t$ to $\partial G$. Combining (2.13)-(2.16) we obtain
$$
\int_{\partial G}v^{p/2}
\leq C\int_{\partial G}v^{(p-2)/2}+C
+\frac{1}{p}\int_{\partial G}v^{p/2}
$$
and derive
\begin{equation}
\int_{\partial G}v^{p/2} \leq C \label{(2.17)}
\end{equation}
by using the Young inequality. Substituting (2.17) into (2.12) yields
$$
\frac{1}{\varepsilon^p}\int_G|u|^2(1-|u|^2) \leq C
$$
which together with (2.3) implies (2.11).

Using Proposition~\ref{prop2.2} and Proposition~\ref{prop2.3},
we may obtain the following result which
is similar to but stronger than the result in Proposition~\ref{prop2.1}.

\begin{proposition} \label{prop2.4}
Uniformly for $\tau \in (0,1)$,
$$
\lim_{\varepsilon \to 0}|u_\varepsilon^{\tau}|
=1\quad \mbox{in }C(\overline{G},{\mathbb R}^2)\,.
\label{(2.18)}
$$
\end{proposition}

\paragraph{Proof.} From (2.6) and (2.11), we have
$$
\int_G|\nabla |u_\varepsilon^{\tau}||^{(p+2)/2}
\leq C\varepsilon^{\lambda},\quad ~
\forall \varepsilon \in (0,\eta),\tau \in (0,1)
$$
$$
\int_G(1-|u_\varepsilon^{\tau}|)^{(p+2)/2}
\leq \int_G(1-|u_\varepsilon^{\tau}|)
\leq \int_G(1-|u_\varepsilon^{\tau}|^2)
\leq C\varepsilon^p,\quad ~\forall \varepsilon,\tau \in (0,1)
$$
Thus
$$
\|1-|u_\varepsilon^{\tau}|\|_{W^{1,(p+2)/2}(G,{\mathbb R}^2)}
\leq C\varepsilon^{\lambda},
\quad ~\forall \varepsilon \in (0,\eta),\tau \in (0,1)
$$
and hence by the embedding inequality, we obtain
$$
\|1-|u_\varepsilon^{\tau}|\|_{C(G,{\mathbb R}^2)} \leq C\varepsilon^{\lambda}
$$
which is a conclusion stronger than (2.18).

\section{Estimate for $\|\nabla u_\varepsilon^{\tau}\|_
{L_{\mbox{\scriptsize\rm loc}}^l}$ }  \setcounter{equation}{0}

The main goal of this section is to establish
uniform estimates for $\|\nabla u_\varepsilon^{\tau}\|_{L_{loc}^l}$.

\begin{proposition} \label{prop3.1} There exists a constant $C$
independent of $\varepsilon,\tau \in (0,\eta)$ for small $\eta>0$, such that
\begin{equation}
\|\nabla u_\varepsilon^{\tau}\|_{L^l(K,{\mathbb R}^2)} \leq C=C(K,l) \label{(3.1)}
\end{equation}
where $K \subset G$ is an arbitrary compact subset and $l>1$.
\end{proposition}

\paragraph{Proof.} Differentiate (2.7) with respect to $x_j$
$$
-(v^{(p-2)/2}u_{x_i})_{x_ix_j}
=\frac{1}{\varepsilon^p}(u(1-|u|^2))_{x_j}\,.
$$
Here and in the sequel, double indices indicate summation.

Let $\zeta \in C_0^{\infty}(G,R)$ be a function such that $\zeta=1$ on
$K$, $\zeta =0$ on $G \setminus \overline{G}_1,
0 \leq \zeta \leq 1, |\nabla \zeta| \leq C$ on $G$,
where $K \subset G_1$ and $G_1 \subset \subset G$ be a sub-domain.
Taking the the inner product of the both sides of (2.7) with
$u_{x_j}\zeta^2$ and integrating over $G$, we obtain
$$
\int_G(v^{(p-2)/2}u_{x_i})_{x_j}(\zeta^2 u_{x_j})_{x_i}
=\frac{1}{\varepsilon^p}
\int_G(1-|u|^2)\zeta^2|\nabla u|^2
-\frac{1}{2\varepsilon^p}
\int_G\zeta^2(|u|^2)_{x_j}^2.
$$
Summing up for $j=1,2$ and computing the term of the left side yield
\begin{eqnarray} \lefteqn{
\int_G \zeta^2v^{(p-2)/2} \sum_{j=1}^2|\nabla u_{x_j}|^2
+\frac{p-2}{4} \int_G\zeta^2v^{(p-4)/2}|\nabla v|^2 }\nonumber \\
&\leq& \frac{1}{\varepsilon^p} \int_G(1-|u|^2)\zeta^2|\nabla  u|^2
+2|\int_G(v^{(p-2)/2}u_{x_i})_{x_j}u_{x_j}\zeta\zeta_{x_i}|\,.\label{(3.2)}
\end{eqnarray}
Applying Proposition~\ref{prop2.1}
and the Young inequality, we derive from (2.7)
that for any $\delta \in (0,1)$
\begin{eqnarray}\lefteqn{
\frac{1}{\varepsilon^p}
\int_G(1-|u|^2)\zeta^2|\nabla u|^2 }\nonumber\\
&\leq& \int_G|u|^{-1}|\nabla u|^2\zeta^2
|\mathop{\rm div}(v^{(p-2)/2}\nabla u)| \label{(3.3)}\\
&\leq& C\int_G\zeta^2v(v^{(p-2)/2}|\Delta u|
+\frac{p-2}{2}v^{(p-3)/2}|\nabla v|) \nonumber \\
&\leq& C(\delta)\int_G\zeta^2v^{(p+2)/2}
+\delta \int_G \zeta^2|\Delta u|^2v^{(p-2)/2}
+\delta\int_G\zeta^2v^{(p-4)/2}|\nabla v|^2 \nonumber\\
&\leq& C(\delta)\int_G\zeta^2v^{(p+2)/2}
+\delta \int_G \zeta^2
\sum_{j=2}^2|\nabla u_{x_j}|^2v^{(p-2)/2}
+\delta\int_G\zeta^2v^{(p-4)/2}|\nabla v|^2 \,,\nonumber
\end{eqnarray}
where $\varepsilon,\tau \in (0,\eta)$ with $\eta>0$ small enough.
Since
\begin{eqnarray*}
|(v^{(p-2)/2}u_{x_i})_{x_j}u_{x_j}\zeta\zeta_{x_i}|
&=&|\frac{1}{2}v^{(p-2)/2}v_{x_i}
+\frac{p-2}{2}v^{(p-4)/2}v_{x_j}
u_{x_i}u_{x_j}\zeta\zeta_{x_i}|\\
&\leq& Cv^{(p-2)/2}\zeta|\nabla \zeta||\nabla v|\,,
\end{eqnarray*}
 using the Young inequality again, for $\delta \in (0,1)$ we
 have
\begin{eqnarray}
|\int_G(v^{(p-2)/2}u_{x_i})_{x_j}u_{x_j}\zeta\zeta_{x_i}|
&\leq& C\int_Gv^{(p-2)/2}\zeta|\nabla \zeta||\nabla v| \label{(3.4)}\\
&\leq& C(\int_Gv^{(p-4)/2}|\nabla v|^2\zeta^2)^{1/2}
(\int_Gv^{p/2}|\nabla \zeta|^2)^{1/2} \nonumber\\
&\leq& \delta\int_Gv^{(p-4)/2}|\nabla v|^2\zeta^2
+C(\delta)\int_Gv^{p/2}|\nabla \zeta|^2\,. \nonumber
\end{eqnarray}
Substituting (3.3) and (3.4) into (3.2) and choosing $\delta$ small enough,yield
\begin{eqnarray}
\lefteqn{ \int_G\zeta^2v^{(p-2)/2}\sum_{j=1}^2|\nabla u_{x_j}|^2
+\int_G\zeta^2v^{(p-4)/2}|\nabla v|^2 }\nonumber\\
&\leq& C\int_G\zeta^2v^{(p+2)/2}
+C\int_Gv^{p/2}|\nabla \zeta|^2.\label{(3.5)}
\end{eqnarray}
Hence, by applying (2.2) to the last term, we obtain
\begin{equation}
\int_G\zeta^2v^{(p-4)/2}|\nabla v|^2
\leq C\int_G\zeta v^{(p+2)/2}+C. \label{(3.6)}
\end{equation}

Now we estimate $\int_G\zeta v^{(p+2)/2}$. To do this, we take
$\phi=\zeta^{2/q}v^{(p+2)/2q}$ in the interpolation inequality
$$
\|\phi\|_{L^q}\leq C\|\nabla \phi\|_{L^1}^{\alpha}
\|\phi\|_{L^1}^{1-\alpha},
\quad ~q \in (1,2),\alpha=2(1-1/q).
$$
Noticing that
$$
|\nabla \phi| \leq C\zeta^{2/q-1}|\nabla \zeta|v^{(p+2)/2q}
+C\zeta^{2/q}v^{(p+2)/2q-1}|\nabla v|,
$$
we have
\begin{eqnarray}
\int_G\zeta^2v^{(p+2)/2}
&\leq& C(\int_G\zeta^{2/q}v^{(p+2)/2q})^{q(1-\alpha)} \label{(3.7)}\\
&&\times(\int_G\zeta^{2/q-1}|\nabla \zeta|v^{(p+2)/2q}
+\int_G\zeta^{2/q}v^{(p+2)/2q-1}|\nabla v|)^{q\alpha}\,.\nonumber
\end{eqnarray}
Since $p>2$, we can choose $q \in (1+2/p,2)$ and hence
$\frac{p+2}{2q}<\frac{p}{2}$.
Thus using (2.2) again,we derive that
$$
\int_G\zeta^{2/q}v^{(p+2)/2q}\mbox{ and }
\int_G\zeta^{2/q-1}|\nabla \zeta|v^{(p+2)/2q}
$$
are bounded by
$$
C\int_Gv^{(p+2)/2q} \leq C(\int_Gv^{p/2})^{(p+2)/pq}\leq C\,.
$$
Substituting these inequalities into (3.7) gives
\begin{eqnarray}
\int_G\zeta^2v^{(p+2)/2} &\leq&
C(1+\int_G\zeta^{2/q} v^{(p+2)/2q-1}|\nabla v|)^{q\alpha} \label{(3.8)}\\
&\leq& C[1+(\int_G\zeta^2v^{(p-4)/2}|\nabla v|^2)^{1/2}
(\int_G\zeta^{4/q-2}v^{(p+2)/q-p/2})^{1/2}]^{q\alpha} \nonumber\\
&\leq& C+C(\int_G\zeta^2v^{(p-4)/2}|\nabla v|^2)^{q\alpha/2}
(\int_G\zeta^{4/q-2}v^{(p+2)/q-p/2})^{q\alpha/2}\,. \nonumber
\end{eqnarray}
Here we have used the inequality
$$
(a+b)^{\lambda} \leq C(a^{\lambda}+b^{\lambda}),\quad (a,b \geq 0).
$$
Since $q \in (1+\frac{2}{p},2)$, we have
$\frac{q\alpha}{2}<1,\frac{p+2}{q}-\frac{p}{2}<\frac{p}{2}$.
Thus, using the Holder inequality and (2.2), we obtain
$$
\int_G\zeta^{4/q-2}v^{(p+2)/q-p/2}
\leq C(\int_Gv^{p/2})^{2(p+2)/pq-1} \leq C\,.
$$
Hence from (3.8),we have for any $\delta \in (0,1)$
\begin{eqnarray*}
\int_G\zeta^2v^{(p+2)/2}
&\leq& C+C(\int_G\zeta^2v^{(p-4)/2}
|\nabla v|^2)^{q\alpha/2}\\
&\leq& C(\delta)+\delta\int_G
\zeta^2v^{(p-4)/2}|\nabla v|^2
\end{eqnarray*}
since $\frac{q\alpha}{2}<1$.
Combining the last inequality with (3.6) we
derive
$$
\int_G\zeta^2v^{(p-4)/2}|\nabla v|^2 \leq C
$$
or
$$
\int_G\zeta^2|\nabla w|^2 \leq C
$$
where $w=v^{p/4}$. Since (2.2) implies
$\int_G\zeta^2|w|^2 \leq C$,
we have $\zeta w \in W^{1,2}(G,R)$, and thus the embedding inequality
gives
$$
\int_G(\zeta w)^l \leq C(l)
$$
for any $l>1$, which implies (3.1) since $\zeta = 1$ on $K$.


\section{Estimate for $\|\nabla
u_\varepsilon^{\tau}\|_{L_{\mbox{\scriptsize\rm loc}}^{\infty}}$  }
 \setcounter{equation}{0}

By means of the Moser iteration, from the estimate (3.1) we can further
prove

\begin{proposition} \label{prop4.1}  There exists a constant $C$
 independent of $\varepsilon,\tau\in (0.\eta)$ for small $\eta>0$ such that
\begin{equation}
\|\nabla u_\varepsilon^{\tau}\|_{L^{\infty}(K,{\mathbb R}^2)} \leq C=C(K)
\label{(4.1)}
\end{equation}
where $K \subset G$ is an arbitrary compact subset.
\end{proposition}

\paragraph{Proof.}
Given any $x_0 \in G$. Let $r$ be small enough and positive so that $B(x_0,
2r) \subset G$. Denote $Q_m=B(x_0,r_m),r_m=r+\frac{r}{2^m}$. Choose
$\zeta_m \in C_0^{\infty}(Q_m,R)$ such that
$\zeta_m=1$ on $Q_{m+1},|\nabla\zeta_m| \leq Cr^{-1}2^m,(m=1,2,\dots)$.
Differentiate both sides of (2.7) with respect
$x_j$, multiply by $\zeta_m^2v^bu_{x_j}$ for
$b \geq 1$ and integrate over $Q_m$.  Then
\begin{eqnarray*}
\lefteqn{ \int_{Q_m}(v^{(p-2)/2}u_{x_i})_{x_j}(\zeta_m^2v^bu_{x_j})_{x_i}
}\\
&=&\frac{1}{\varepsilon^p}
\int_{Q_m}(1-|u|^2)\zeta_m^2v^b|\nabla u|^2
-\frac{1}{2\varepsilon^p}\int_{Q_m}\zeta_m^2v^b(|u|^2)_{x_j}^2\,.
\end{eqnarray*}
Similar to the derivation of (3.2) we can obtain
\begin{eqnarray}\lefteqn{
\int_{Q_m}\zeta_m^2v^{(p+2b-2)/2}\sum_j|\nabla u_{x_j}|^2
+\frac{p+2b-2}{4}
\int_{Q_m}\zeta_m^2v^{(p+2b-4)/2}|\nabla v|^2 } \nonumber \\
&\leq& \frac{1}{\varepsilon^p}
\int_{Q_m}(1-|u|^2)\zeta_m^2v^{b+1}
+|\int_{Q_m}(v^{(p-2)/2}u_{x_i})_{x_j}v^bu_{x_j}\zeta_m\zeta_{mx_i}|\,.
\label{(4.2)} \end{eqnarray}
Also for any $\delta \in (0,1)$, we have
\begin{eqnarray}\lefteqn{
\frac{1}{\varepsilon^p}\int_{Q_m}(1-|u|^2)\zeta_m^2v^{b+1} }\nonumber\\
&\leq& \int_{Q_m}|u|^{-1}
\zeta_m^2v^{b+1}|\mathop{\rm div}(v^{(p-2)/2}\nabla u)| \nonumber\\
&\leq& C\int_{Q_m}\zeta_m^2v^{(p+2b)/2}|\Delta u|+\frac{C(p+2b-2)}{2}
\int_{Q_m}\zeta_m^2v^{(p+2b-2)/2}|\nabla v| \label{(4.3)}\\
&\leq& C(\delta)\int_{Q_m}\zeta_m^2v^{(p+2b+2)/2}
+\delta\int_{Q_m}\zeta_m^2v^{(p+2b-2)/2}|\Delta u|^2 \nonumber\\
&&+\frac{C(\delta)(p+2b-2)}{2}
\int_{Q_m}\zeta_m^2v^{(p+2b+2)/2} \nonumber\\
&&+\frac{\delta(p+2b-2)}{2}
\int_{Q_m}\zeta_m^2v^{(p+2b-4)/2}|\nabla v|^2\nonumber
\end{eqnarray}
and
\begin{eqnarray}\lefteqn{
\bigg|\int_{Q_m}(v^{(p-2)/2}
u_{x_i})_{x_j}v^bu_{x_j}\zeta_m\zeta_{mx_i}\bigg| }\nonumber \\
&\leq& C\int_{Q_m}v^{(p+2b-2)/2}
|\nabla v|\zeta_m|\nabla \zeta_m| \label{(4.4)}\\
&\leq& \delta\int_{Q_m}v^{(p+2b-1)/2}|\nabla v|^2\zeta_m^2
+C(\delta)\int_{Q_m}v^{(p+2b)/2}|\nabla \zeta_m|^2\,,\nonumber
\end{eqnarray}
where the constants $C$ and $C(\delta)$
are independent of $b,m,\varepsilon,\tau$.
Combining (4.2) with (4.3)(4.4) and
choosing $\delta$ small enough yield
\begin{equation}
\int_{Q_m}\zeta_m^2v^{(p+2b-4)/2}|\nabla v|^2
\leq C\int_{Q_m}v^{(p+2b)/2}|\zeta_m|^2
+C\int_{Q_m}\zeta_m^2v^{(p+2b+2)/2}. \label{(4.5)}
\end{equation}
To estimate $\int_{Q_m}\zeta_m^2v^{(p+2b+2)/2}$, we take
$$
\phi =\zeta_m^{2/q}v^{(p+2b+2)/2q}
$$
in the interpolation inequality (3.6) and observe that
$$
|\nabla \phi| \leq \frac{2}{q}
\zeta_m^{2/q-1}|\nabla \zeta_m|v^{(p+2b+2)/2q}
+\frac{p+2b+2}{2q}\zeta_m^{2/q}v^{(p+2b+2)/2q-1}|\nabla v|.
$$
Then we obtain
\begin{eqnarray} \lefteqn{
\int_{Q_m}\zeta_m^2v^{(p+2b+2)/2} }\nonumber\\
&\leq& C(\int_{Q_m}\zeta_m^{2/q}
v^{(p+2b+2)/2q})^{q(1-\alpha)}
 \bigg(\frac{2}{q}\int_{Q_m}\zeta_m^{2/q-1}|\nabla
\zeta_m|v^{(p+2b+2)/2q}    \nonumber\\
&&+\frac{p+2b+2}{2q}
\int_{Q_m}\zeta_m^{2/q}v^{(p+2b+2)/2q-1}|\nabla
v|)^{q\alpha} \nonumber\\
&\leq& C(\int_{Q_m}\zeta_m^{2/q}
v^{(p+2b+2)/2q})^{q(1-\alpha)}
(\frac{2}{q})^{q\alpha} (\int_{Q_m}\zeta_m^{2/q-1}|\nabla
\zeta_m|v^{(p+2b+2)/2q}\bigg)^{q\alpha}     \label{(4.6)}\\
&&+(\frac{p+2b+2}{2q})^{q\alpha}
(\int_{Q_m}\zeta_m^{2/q}v^{(p+2b+2)/2q-1}|\nabla v|)^{q\alpha}\,.
\nonumber
\end{eqnarray}
Now we estimate all integrals on the right-hand side of (4.6).
Choose $r$ small enough so that $\mathop{\rm meas}(Q_m) \leq 1$.
In computing we need to notice that $q\in (1+\frac{2}{p},2)$,
which implies $q>1+\frac{2}{p+2b}$ or
$\frac{p+2b}{2q}<\frac{p+2b}{2}$. We have
\begin{eqnarray*}
\lefteqn{\int_{Q_m}\zeta_m^{2/q}v^{(p+2b+2)/2q} }\\
&\leq& \int_{Q_m}v^{(p+2b+2)/2q} \\
&\leq& (\mathop{\rm meas}(Q_m))^{1-(p+2b+2)/(q(p+2b))}
(\int_{Q_m}v^{(p+2b)/2})^{(p+2b+2)/(q(p+2b))}\\
&\leq& (\int_{Q_m}v^{(p+2b)/2})^{(p+2b+2)/(q(p+2b))},
\end{eqnarray*}
\begin{eqnarray*}
\int_{Q_m}\zeta_m^{2/q-1}|\nabla \zeta_m|v^{(p+2b+2)/2q}
&\leq& \frac{2^m}{r} \int_{Q_m}v^{(p+2b+2)/2q}\\
&\leq& \frac{2^m}{r}(\int_{Q_m}v^{(p+2b)/2})^{(p+2b+2)/(q(p+2b))}
\end{eqnarray*}
and
\begin{eqnarray*}
\lefteqn{\int_{Q_m}\zeta_m^{2/q}v^{(p+2b+2)/2q-1}|\nabla v|}\\
&\leq& (\int_{Q_m}\zeta_m^2v^{(p+2b-4)/2}|\nabla v|^2)^{1/2}
(\int_{Q_m}\zeta_m^{4/q-2}v^{(p+2b+2)/q-(p+2b)/2})^{1/2}\\
&\leq& (\int_{Q_m}\zeta_m^2v^{(p+2b-4)/2}|\nabla v|^2)^{1/2}
(\int_{Q_m}v^{(p+2b)/2})^{(p+2b+2)/(q(p+2b))-1/2}\,.
\end{eqnarray*}
Combining these inequalities with (4.5) and (4.6) yields
\begin{equation}
I_1 \leq C[(\frac{2^m}{r})^2I_2
+(\frac{2^m}{r})^{q\alpha}I_2^{1+2/(p+2b)}
+(\frac{p+2b+2}{2q})^{q\alpha}
I_1^{q\alpha/2}I_2^{1+2/(p+2b)-q\alpha/2}]\,,
\label{(4.7)}
\end{equation}
where
$$
I_1=\int_{Q_m}\zeta_m^2v^{(p+2b-4)/2}|\nabla v|^2, \mbox{ and }
I_2=\int_{Q_m}v^{(p+2b)/2}.
$$
Let
$$
p+2b=s^m,\quad w=v^{(p+2b)/4}=v^{s^m/4}
$$
with $s>2$ to be determined later. Then (4.7) becomes
$$
I_1 \leq C[(\frac{2^m}{r}
I_2+(\frac{2^m}{r})^{q\alpha}I_2^{1+2/{S^m}}
+(\frac{s^m+2}{2q})^{q\alpha}
I_1^{q\alpha/2}I_2^{1+2/{s^m}-q\alpha/2}]\,.
$$
The Young inequality applied to the last term on the right-hand side
yields
\begin{eqnarray*}
\lefteqn{C(\frac{s^m+2}{2q})^{q\alpha}
I_1^{q\alpha/2}I_2^{1+2/(s^m)-q\alpha/2} }\\
&\leq& \delta I_1+C(\delta)[(\frac{s^m+2}{2q})^{q\alpha}
I_2^{1+2/(s^m)-q\alpha/2}]^{2/(2-q\alpha)}\\
&=&\delta I_1+C(\delta)(\frac{s^m+2}{2q})^{2q\alpha/(2-q\alpha)}
I_2^{2(1+2/(s^m)-q\alpha/2)/(2-q\alpha)}\,.
\end{eqnarray*}
Thus we obtain
\begin{eqnarray}
I_1 &\leq& C(\delta)[(\frac{2^m}{r})^2I_2
+(\frac{2^m}{r})^{q\alpha}I_2^{1+2/(s^m)}  \label{(4.8)}\\
&&+(\frac{s^m+2}{2q})^{2q\alpha/(2-q\alpha)}
I_2^{2(1+2/(s^m)-q\alpha/2)/(2-q\alpha)}]\,. \nonumber
\end{eqnarray}
By the embedding theorem, we have for any $s>1$
\begin{eqnarray*}
\int_{Q_m}(\zeta_mw)^{2s}
&\leq& C(s)[\int_{Q_m}(\zeta_mw)^2
+\int_{Q_m}|\nabla(\zeta_mw)|^2]^s\\
&\leq& C(s)[\int_{Q_m}(\zeta_mw)^2
+\int_{Q_m}|\nabla \zeta_m|^2w^2
+\int_{Q_m}\zeta_m^2|\nabla w|^2]^s\\
&\leq& C(s)[(1+(\frac{2^m}{r})^2)I_2
+(\frac{s^m}{4})^2I_1]^s\,,
\end{eqnarray*}
which by using (4.8), turns out to be
\begin{eqnarray}
\int_{Q_m}(\zeta_mw)^{2s}
&\leq& C(s)[(1+(\frac{2^m}{r})^2
+(\frac{s^m}{4})^2(\frac{2^m}{r})^2)I_2
+(\frac{s^m}{4})^2
(\frac{2^m}{r})^{q\alpha}I_2^{1+\frac{2}{s^m}} \label{(4.9)}\\
&&+(\frac{s^m}{4})^2
(\frac{s^m+2}{2q})^{\frac{2q\alpha}{2-q\alpha}}
I_2^{(1+\frac{2}{s^m}-\frac{q\alpha}{2})\frac{2}{2-q\alpha}}]^s\,.\nonumber
\end{eqnarray}

If there exists a subsequence of positive integers
$\{m_i\}$ with $m_i \to \infty$ such that
$$
I_2=\int_{Q_{m_i}}v^{s^{m_i}/2} < 1\,,
$$
then as $m_i \to \infty$,
\begin{equation}
\|v\|_{L^{\infty}(Q_{\infty},R)} \leq C(r)\,.\label{(4.10)}
\end{equation}
Otherwise, there must be a positive integer $m_0$ such that
$$
I_2=\int_{Q_m}v^{s^m/2} \geq 1\,,\ \forall m \geq m_0\,.
$$
Since
$$
(1+\frac{2}{s^m}-\frac{q\alpha}{2})\frac{2}{2-q\alpha}
=1+\frac{2}{s^m}\frac{1}{2-q}>1+\frac{2}{s^m}>1\,,
$$
the exponent of the last term in (4.9) is higher than those of the
other terms. Now comparing the coefficients of the terms in (4.9), we
have
$$
(\frac{s^m}{r})^2\geq 1\,, \quad
(\frac{2^m}{r})^2 \geq (\frac{2^m}{r})^{q\alpha}
$$
and, if we choose $s>2q(\frac{2}{r})^{\frac{2(q-1)}{2-q}}$, $r \leq 1$, then
$$
(\frac{s^m+2}{2q})^{\frac{2q\alpha}{2-q\alpha}}
=(\frac{s^m+2}{2q})^{\frac{2(q-1)}{2-q}}
\geq (\frac{2^m}{r})^2.
$$
Therefore, the coefficient of the last term in (4.9) is greater than those
of the other terms. Hence we have
$$
\int_{Q_m}(\zeta_mw)^{2s}\leq C[(\frac{s^m}{4})^2
(\frac{s^m+2}{2q})^{\frac{2(q-1)}{2-q}}
I_2^{1+\frac{2}{s^m}\frac{1}{2-q}}]^s
$$
or
\begin{equation}
\int_{Q_{m+1}}v^{s^{m+1}/2}
\leq (C_0C_1^m)^s(\int_{Q_m}v^{s^m/2})^{(1+C_2 /S^m)s}
\label{(4.11)}
\end{equation}
with some constant $C_0>0$, $C_2=\frac{2}{2-q}$,
$C_1=s^{(2+\frac{2(q-1)}{2-q})s}$.
Using an iteration proposition which  will be stated and proved
later, we also reach estimate (4.10). Thus the proof of
Proposition~\ref{prop4.1} is complete.

\begin{proposition} \label{prop4.2}
Let $Q_m(m=1,2,\dots) \subset G$ be a sequence of bounded,
open subsets such that $Q_{m+1} \subset Q_m$. If for any $l \geq 1$,
$v \in L^l(Q_1,R)$ and there exist constants
$\lambda, C_0, C_1, C_2>0$, $s>1$, $\lambda s \geq 1$, such that
$$
\int_{Q_{m+1}}|v|^{\lambda s^{m+1}}\,dx
\leq (C_0C_1^m)^s(\int_{Q_m}
|v|^{\lambda s^m}\,dx)^{(1+C_2/s^m)s},
\label{(4.12)}
$$
for $m=1,2,\dots$, then
$$
\|v\|_{L^{\infty}(Q_{\infty},R)} \leq
C_0^{A_1}C_1^{A_2}(\int_{Q_{n_0}}
|v|^{\lambda s^{n_0}}\,dx)^{A_3/(\lambda s^{n_0})},
\label{(4.13)}
$$
where $A_1, A_2, A_3$ are constants depending only on $s$, $C_2$, and $n_0$
is an arbitrary nonnegative integer.
\end{proposition}

\paragraph{Proof.} From (4.12) by iteration, we obtain
\begin{equation}
\int_{Q_{m+1}}|v|^{\lambda s^{m+1}}\,dx
\leq C_0^{X_m}C_1^{Y_m}(\int_{Q_{n_0}}
|v|^{\lambda s^{n_0}}\,dx)^{s^{m-n_0+1}Z_m}
\label{(4.14)}
\end{equation}
where
\begin{eqnarray*}
X_m&=&s+s^2\lambda_m+\dots+s^{m-n_0+1}
\lambda_m\lambda_{m-1}\dots\lambda_{n_0+1}\\
Y_m&=&ms+(m-1)s^2\lambda_m+\dots+{n_0}s^{m-n_0+1}
\lambda_m\lambda_{m-1}\dots\lambda_{n_0+1}\\
Z_m&=&\lambda_{n_0}\dots\lambda_{m-1}\lambda_m
\end{eqnarray*}
with $\lambda_m=1+C_2/s^m$. Since $\lambda_j \geq 1$
for $j=n_0-1,\dots,m-1,m,\dots$,
$Z_m$ is an increasing sequence.
Noticing that $\ln(1+x) \leq x$ for $x>0$,
we have
\begin{eqnarray*}
lnZ_m &=&ln\lambda_{n_0}+\dots+ln\lambda_{m-1}+ln\lambda_m\\
&\leq& C_2[(\frac{1}{s})^{n_0}+\dots+
(\frac{1}{s})^{m-1}+(\frac{1}{s})^m]\\
&\leq& C_2\frac{(1/s)^{n_0}}{1-1/s}=\gamma
\end{eqnarray*}
or $Z_m \leq e^{\gamma}$. Hence
$\lim_{m \to \infty}Z_m=A_3$ exists. Clearly, we also have
\begin{eqnarray*}
X_m &\leq& e^{\gamma}[s+s^2+\dots+s^{m-n_0+1}]\\
Y_m &\leq& e^{\gamma}[ms+(m-1)s^2+\dots+n_0s^{m-n_0+1}]\,.
\end{eqnarray*}
 From which it is easily seen that the following two limits exist: \\
$\lim_{m\to\infty}s^{-(m+1)}X_m=A_1$
and $\lim_{m\to \infty}s^{-(m+1)}Y_m=A_2$.
Taking the $1/\lambda s^{m+1}$ power on the both sides of
(4.14), letting $m \to \infty$, and noticing that
$$
\|v\|_{L^{\infty}(Q_{\infty},R)}
=\lim_{m \to \infty}
(\int_{Q_{m+1}}|v|^{\lambda s^{m+1}}\,dx)^{1/\lambda s^{m+1}},
$$
we obtain (4.13).

\section{Completion of the proof } \setcounter{equation}{0}

Once the locally uniform estimate  $\|\nabla
u_\varepsilon^{\tau}\|_{L^{\infty}(K)}$
is established, it is not difficult to
prove the following proposition.

\begin{proposition} \label{prop5.1} Let $\psi_\varepsilon^{\tau}
=\frac{1}{\varepsilon^p}(1-|u_\varepsilon^{\tau}|^2)$.
Then there exists a constant $C$
independent of $\varepsilon,\tau \in (0,\eta)$
with $\eta>0$ small enough, such
that
$$
\|\psi_\varepsilon^{\tau}\|_{L^{\infty}(K,R)} \leq C=C(K),
\label{(5.1)}
$$
where $K \subset G$ is an arbitrary compact subset.
\end{proposition}

\paragraph{Proof.}
Take the inner product of both sides of (2.7) with $u$,
$$
-\mathop{\rm div}(v^{(p-2)/2}\nabla u)u
=\frac{1}{\varepsilon^p}|u|^2(1-|u|^2)=|u|^2\psi
$$
where $u=u_\varepsilon^{\tau}, \psi=\psi_\varepsilon^{\tau}$. This and
$$\displaylines{
\nabla \psi=-\frac{2}{\varepsilon^p}u \cdot \nabla u \cr
-\mathop{\rm div}(v^{(p-2)/2}\nabla u)u
=-\mathop{\rm div}(v^{(p-2)/2}u \cdot \nabla u)
+v^{(p-2)/2}|\nabla u|^2 \cr}
$$
give
$$
|u|^2\psi=v^{(p-2)/2}|\nabla u|^2+\frac{\varepsilon^p}{2}
\mathop{\rm div}(v^{(p-2)/2}\nabla \psi).
$$
Using Proposition~\ref{prop2.1} we  obtain
$$
\frac{1}{2}\psi
\leq v^{(p-2)/2}|\nabla u|^2
+\frac{\varepsilon^p}{2}\mathop{\rm div}(v^{(p-2)/2}\nabla \psi),
~\forall \varepsilon,\tau \in (0,\eta).
\label{(5.2)}
$$
Since at the point where $\psi$ achieves its maximum,
$\nabla \psi=0$, $\Delta \psi \leq 0$, and
$$
\mathop{\rm div}(v^{(p-2)/2}\nabla \psi)
=v^{(p-2)/2}\Delta \psi
+\frac{p-2}{2}v^{(p-4)/2}\nabla v \nabla \psi
\leq 0\,,
$$
we derive (5.1) from (5.2) by using Proposition~\ref{prop4.1}.

To complete the proof of Theorem 1.1, we apply a theorem in [12]
(Page 244 Line 19--23).
Now according to Proposition~\ref{prop5.1}
the right hand side of (2.7) is bounded
on every compact subset $K \subset G$ uniformly in $\varepsilon, \tau \in
(0,\eta)$. Thus applying the theorem in [12] (Page 244) yields

\begin{equation}
\|u_\varepsilon^{\tau}\|_{C^{1,\beta}(K)} \leq C=C(K)
\label{(5.3)}
\end{equation}
for some $\beta \in (0,1)$, where the constant $C$
does not depend on $\varepsilon,\tau \in (0,\eta)$. From this
it follows that there exist a function $u_*$ and a subsequence
$u_{\varepsilon_k}^{\tau_k}(\varepsilon_k,
\tau_k \to 0$, as $k \to \infty$) of
$u_\varepsilon^{\tau}$, such that
$$
\lim_{k \to \infty}u_{\varepsilon_k}^{\tau_k}=u_*,
\quad \mbox{ in }C^{1,\alpha}(K,{\mathbb R}^2),\alpha \in (0,\beta)\,.
$$
By an argument similar to that in the proof of (1.1) and (1.2), we obtain
$$
\lim_{\varepsilon,\tau \to 0}
u_\varepsilon^{\tau}=u_p,\quad\mbox{in }W^{1,p}(K,{\mathbb R}^2).
$$
Certainly $u_*=u_p$. From the fact that any subsequence of
$u_\varepsilon^{\tau}$ contains a subsequence
convergent in $C^{1,\alpha}(K,{\mathbb R}^2)$
and the limit is the same function $u_p$, we may assert
\begin{equation}
\lim_{\varepsilon,\tau \to 0}
u_\varepsilon^{\tau}=u_p,\quad \mbox{ in }C^{1,\alpha}(K,{\mathbb R}^2).
\label{(5.4)}
\end{equation}
On the other hand, for any $\varepsilon \in (0,\eta)$,
as a regularizable minimizer
of $E_\varepsilon(u,G)$, $\tilde{u}_\varepsilon$
is the limit of some subsequence
$u_\varepsilon^{\tau_k}$ of $u_\varepsilon^{\tau}$
in $W^{1,p}(G,{\mathbb R}^2)$. For large $k$,
$u_\varepsilon^{\tau_k}$ satisfies (5.3)
and hence it contains a subsequence,
for simplicity we suppose it
is $u_\varepsilon^{\tau_k}$ itself, such that
$$
\lim_{k \to \infty}u_\varepsilon^{\tau_k}
=w,\quad \mbox{ in }C^{1,\alpha}(K,{\mathbb R}^2)
$$
where the function $w$ must be $\tilde{u}_\varepsilon$.
Combining this with (5.4) we
finally obtain
$$
\lim_{\varepsilon \to 0}\tilde{u}_\varepsilon
=u_p,\quad \mbox{ in }C^{1,\alpha}(K,{\mathbb R}^2)
$$
and complete the proof of Theorem 1.1.

\paragraph{Remark} Using Proposition~\ref{prop2.4}
instead of Proposition~\ref{prop2.1} we may
also prove our theorem. In this way,
we may obtain (3.1), (4.1) and (5.1)
for $\varepsilon \in (0,\eta),\tau \in (0,1)$
instead of those for $\varepsilon,\tau \in (0,\eta)$.
The remainder of the proof is just the same as above.


\section{Extension of the argument} \setcounter{equation}{0}

Our argument can be extended to the higher dimensional case. Let $n>2$
$G \subset {\mathbb R}^n$ be a bounded and simply connected domain with smooth
boundary $\partial G$, and
$g:\partial G \to S^{n-1}=\{x \in {\mathbb R}^n;|x|=1\}$
be a smooth map with $d=\mathop{\rm deg}(g,\partial
G)=0$. Consider the functional
$$
E_\varepsilon(u,G)=\frac{1}{p}\int_G|\nabla u|^p
+\frac{1}{4\varepsilon^p}
\int_G(1-|u|^2)^2,\quad (\varepsilon>0)
$$
and its regularized functional
$$
E_\varepsilon^{\tau}(u,G)
=\frac{1}{p}\int_G(|\nabla u|^2+\tau)^{p/2}
+\frac{1}{4\varepsilon^p}
\int_G(1-|u|^2)^2,\quad (\varepsilon,\tau>0)
$$
on
$$
W_g=\{v \in W^{1,p}(G,{\mathbb R}^n);v|_{\partial G}=g\}.
$$

Similar to the case $n=2$, we may prove that if $p>1$, then
$E_\varepsilon(u,G)$ and $E_\varepsilon^{\tau}(u,G)$
achieve their minimum on $W_g$ by
some $u_\varepsilon$ and $u_\varepsilon^{\tau}$;
$u_\varepsilon$ and $u_\varepsilon^{\tau}$ satisfy
$$
-\mathop{\rm div}(|\nabla u|^{p-2} \nabla u)
=\frac{1}{\varepsilon^p}u(1-|u|^2),\quad \mbox{in }G
$$
and
$$
-\mathop{\rm div}(v^{(p-2)/2} \nabla u)
=\frac{1}{\varepsilon^p}u(1-|u|^2),\quad \mbox{in }G
\label{(6.1)}
$$
respectively where $v=|\nabla u|^2+\tau$,  and
$$
|u_\varepsilon|,|u_\varepsilon^{\tau}| \leq 1,\quad\mbox{ in }G.
$$

It can also be proved that if $p>1$, then there exists a subsequence
$u_{\varepsilon_k}$ of $u_\varepsilon$
with $\varepsilon_k \to 0$ such that
$$
\lim_{\varepsilon_k \to 0}
u_{\varepsilon_k}=u_p,\quad \mbox{in }W^{1,p}(G,{\mathbb R}^n)
$$
where $u_p$ is a $p$-harmonic map with
boundary value $g$. However, differ to
the case $n=2$, here we can only prove the convergence for a subsequence
because of the lack of uniqueness result for $p$-harmonic map with given
boundary value.

Similarly, for some subsequence we have
$u_\varepsilon^{\tau_k}(\tau_k \to 0)$
of $u_\varepsilon^{\tau}$
$$
\lim_{\tau_k \to 0}u_\varepsilon^{\tau_k}
=\tilde{u}_\varepsilon,\quad \quad{in } W^{1,p}(G,{\mathbb R}^n)
$$
and the limit $\tilde{u}_\varepsilon$ is a
minimizer of $E_\varepsilon(u,G)$, called
regularizable minimizer.
The main result is the following

\begin{theorem} \label{th6.1} Assume that $p>2n-2$ and
$d=\mathop{\rm deg}(g,\partial G)=0$. Let
$\tilde{u}_\varepsilon$ be a regularizable
minimizer of $E_\varepsilon(u,G)$. Then there
exists a subsequence $\tilde{u}_{\varepsilon_k}$
of $\tilde{u}_\varepsilon$ with $\varepsilon_k \to 0$
such that for some $\alpha \in (0,1)$
$$
\lim_{\varepsilon_k \to 0}
\tilde{u}_{\varepsilon_k}=u_p,\quad \mbox{in }
C_{\mbox{\scriptsize\rm loc}}^{1,\alpha}(G,{\mathbb R}^n).
$$
\end{theorem}

The proof is similar to the case $n=2$. First we have
$$
\lim_{\varepsilon,\tau \to 0}
|u_\varepsilon^{\tau}|=1\quad \mbox{in }C(\overline{G},{\mathbb R}^n)
$$
and also
$$
\lim_{\varepsilon \to 0}
|u_\varepsilon^{\tau}|=1\quad \mbox{in }C(\overline{G},{\mathbb R}^n)
$$
uniformly for $\tau \in (0,1)$.

Next we prove
$$
\|\nabla u_\varepsilon^{\tau}\|_{L^l(K,{\mathbb R}^n)} \leq C=C(K,l)
\label{(6.2)}
$$
where $K \subset G$ is an arbitrary compact subset and $l>1$.

For this purpose, we proceed as in section 3: first differentiate (6.1)
with respect to $x_j$, take the inner product of the
both sides with $\zeta^2v^bu_{x_j}(b \geq 0)$, where $\zeta \in
C_{0}^{\infty}(G,R)$ with $0 \leq \zeta \leq 1$,
and integrate over $G$. Then as in (3.5) and in (4.5), we obtain
\begin{equation}
\int_G\zeta^2v^{(p+2b-4)/2}|\nabla v|^2
\leq C\int_G\zeta^2v^{(p+2b+2)/2}
+C\int_Gv^{(p+2b)/2}|\nabla \zeta|^2\,.
\label{(6.3)}
\end{equation}
Using the interpolation inequality
$$
\|\phi\|_{L^q} \leq C\|\nabla \phi\|
_{L^1}^{\alpha}\|\phi\|_{L^1}^{1-\alpha},
\quad q \in (1,n/(n-1)),\ \alpha=n(q-1)/q
$$
for $\phi =\zeta^{2/q}v^{(p+2b+2)/2q}$ to estimate the last term of
(6.3) yields
\begin{eqnarray*}
\lefteqn{\int_G\zeta^2v^{(p+2b-4)/2}|\nabla v|^2 }\\
&\leq& C\int_G|\nabla \zeta|^2v^{(p+2b)/2}\\
&& +C(\int_G\zeta^{2/q}v^{(p+2b+2)/2q})^{q(1-\beta)}
(\int_G \zeta^{2/q-1}
|\nabla \zeta|v^{(p+2b+2)/2q})^{q\beta}\\
&& +C(\int_G\zeta^{2/q}v^{(p+2b+2)/2q})^{\lambda_1}
(\int_G \zeta^{4/q-2}v^{(p+2b+2)/q-(p+2b)/2})^{\lambda_2}\,,
\end{eqnarray*}
where the constants $\lambda_1$, $\lambda_2>0$ depend on
$p$, $q$, $b$, $\alpha$ only. Set
$w=v^{(p+2b)/4}$. Since $p>2n-2$, we may choose $q \in (1-2/p,n/(n-1))$
such that
$$
\frac{p+2b+2}{2q}<\frac{p+2b}{2} \mbox{ and } \frac{p+2b+2}{q}-\frac{p+2b}{2}
<\frac{p+2b}{2}\,.
$$
Then use the Holder inequality to obtain
$$
\int_G\zeta^2|\nabla w|^2 \leq C\int_G|\nabla
\zeta|^2w^2+C(\int_Gw^2)^{\lambda}
[1+(\int_G|\nabla \zeta|^{\lambda})^{\lambda}]\,,
\label{(6.4)}
$$
where the constants $C$ and $\lambda>0$ are independent of
$\varepsilon$ and $\tau$.

Now we choose $\zeta$ such that $\zeta=1$ on $G_1$, where $G_1,
G_0$ are  sub-domains of $G$ satisfying $K \subset G_1 \subset \subset G_0
\subset \subset G$, $\zeta=0$ on
$G \setminus \overline{G}_0$, $|\nabla \zeta| \leq C$ on $G$ and $b=0$. From
$$
E_\varepsilon^{\tau}(u_\varepsilon^{\tau},G)
\leq E_\varepsilon^{\tau}(u_p,G) \leq C
$$
we have
$$
\int_G \zeta^2|\nabla w|^2 \leq C
$$
and hence $\|\zeta w\|_{L^2(G,{\mathbb R}^n)} \leq C$. By the embedding
theorem,
\begin{equation}
\|\zeta w\|_{L^r(G,{\mathbb R}^n)}=(\int_G(\zeta w)^r)^{1/r}
\leq C\|\zeta w\|_{L^2(G,{\mathbb R}^n)} \leq C\,,
\label{(6.5)}
\end{equation}
where $r \leq \frac{2n}{n-2}$.
Clearly $r=2+\frac{8}{np} \leq \frac{2n}{n-2}$.
Choosing $r=2+\frac{8}{np}$ in (6.5) and noticing
that $\zeta=1$ on $G_1$ we see that $\nabla u \in L^{s_1}(G_1)$and
\begin{equation}
\int_{G_1}|\nabla u|^{s_1} \leq C\,, \label{(6.6)}
\end{equation}
where $s_1=p+\frac{4}{n}$.
In the present case $n>2$, we can not
derive (6.2) directly by using the embedding theorem once. To prove (6.2)
we choose $G_2$, a sub-domain of $G_1$,
 such that $K \subset G_2 \subset \subset G_1$ and
$\zeta=1$ on $G_2,\zeta=0$ on $G \setminus
\overline{G}_1,|\nabla \zeta| \leq C$ on $G$. Set
$b=\frac{2}{n},w=v^{(p+4/n)/4}$. Then from (6.6)
$$
\int_{G_1}w^2 =\int_{G_1}v^{(p+4/n)/2}
=\int_{G_1}|\nabla u|^{s_1} \leq C\,,
$$
and from (6.4),
$$
\int_{G_1}\zeta^2 |\nabla w|^2 \leq C
$$
Thus $\|\zeta w\|_{L^2(G,{\mathbb R}^n)} \leq C$. Applying the embedding theorem to
$\zeta w$, we obtain
$$
\int_{G_2}|\nabla u|^{s_2} \leq C\,
$$
where
$$
s_2=s_1+\frac{4(n+2)}{n^2}=p+\frac{4}{n}+\frac{4(n+2)}{n^2}
=p+\frac{8}{n}+\frac{8}{n^2}
$$
For a given $l>1$, proceeding inductively, we  find an $s_i$ for some
$i$ such that $s_i>l$ and
$$
\int_{G_i}|\nabla u|^{s_i} \leq C\,,
$$
where $G_i$ is a sub-domain of $G_{i-1}$
such that $K \subset G_i \subset \subset
G_{i-1} \subset \subset \dots\subset \subset G$. Thus (6.2) is proved.

The remainder of the proof is just the
same as in sections 4 and 5. However, we are not
able to establish the convergence for the
full $\tilde{u}_\varepsilon$ because of the
lack of uniqueness result for the
$p$-harmonic map with the given boundary value.


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\end{thebibliography} \medskip

\noindent
{\sc Yutian Lei} \\
{\sc Zhuoqun Wu} (e-mail: wzq@mail.jlu.edu.cn )\\
Institute of Mathematics,  Jilin University \\
130023 Changchun,  P. R. China

\end{document}