\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small {\em Electronic Journal of
Differential Equations}, Vol. 2003(2003), No. 35, pp. 1--12.
\newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu
or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu  (login: ftp)}
\thanks{\copyright 2003 Southwest Texas State University.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE--2003/35\hfil Radial minimizer of a variant \dots ]
{Radial minimizer of a variant of the p-Ginzburg-Landau functional}

\author[Yutian Lei\hfil EJDE--2003/35\hfilneg]
{Yutian Lei}

\address{Yutian Lei \hfill\break
Department of Mathematics,
Nanjing Normal University,
Nanjing, Jiangsu 210097, China}
\email{lythxl@163.com}

\date{}
\thanks{Submitted December 30, 2002. Published April 3, 2003.}
\subjclass[2000]{35J70, 49K20}
\keywords{Radial minimizer, variant of p-Ginzburg-Landau functional}


\begin{abstract}
 We study the asymptotic behavior of the radial minimizer
 of a variant of the p-Ginzburg-Landau functional when $p \geq n$.
 The location of the zeros and the uniqueness of the radial minimizer
 are derived. We also prove the $W^{1,p}$ convergence of
 the radial minimizer for this functional.
\end{abstract}

\maketitle

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

\section{Introduction}

Let $n \geq 2$, $B=\{x \in R^n;|x|<1\}$. Consider the minimizers
of the variant for the p-Ginzburg-Landau-type functional
$$
E_\varepsilon(u,B)=
\frac{1}{p}\int_B|\nabla u|^p
+\frac{1}{4\varepsilon^p}\int_B|u|^2(1-|u|^2)^2
,\quad (p \geq n)
$$
on the class functions
$$
W=\big\{u(x)=f(r)\frac{x}{|x|} \in W^{1,p}(B,R^n);
f(1)=1,  r=|x|\big\}.
$$
By the direct method in the calculus of variations we see that the
minimizer $u_{\varepsilon}$ exists. It will be called the {\it
radial minimizer}.

When $p=n=2$, the asymptotic behavior of the minimizer
$u_{\varepsilon}$ of $E_\varepsilon(u,B)$ in the class $H_g^1$
were studied in [5]. In this paper, we will study the asymptotic
behavior of the radial minimizer $u_\varepsilon$. We will prove
the following theorems.

\begin{theorem} \label{thm1.1}
Let $u_\varepsilon$ be a radial minimizer of $E_\varepsilon(u,B)$. Then
for any $\eta \in (0,1/2)$, there exists a constant $h=h(\eta)$
independent of $\varepsilon \in (0,1)$ such that $
Z_\varepsilon=\{x \in B; |u_\varepsilon(x)|<1-\eta\} \subset
B(0,h\varepsilon)$. For any given $\varepsilon\in
(0,\varepsilon_0)$, the radial minimizers $u _{\varepsilon}$ of
$E_{\varepsilon}(u,B)$ are unique on $W$.
\end{theorem}

\begin{theorem} \label{thm1.2}
Let $u_\varepsilon$ be a radial minimizer of
$E_\varepsilon(u,B)$. Then as $\varepsilon \to 0$,
$$
u_\varepsilon \to \frac{x}{|x|},\quad \mbox{in }
W_{\rm loc}^{1,p}(\overline{B}\setminus \{0\},R^n).
$$
\end{theorem}


Some basic properties of minimizers are given in $\S2$. The proof
of Theorem 1.1 is presented in $\S3$. The proof of Theorem 1.2. is
based uniform estimates proved in $\S4$.

\section{ Preliminaries } %2

Let
\[
V=\big\{f \in W_{\rm loc}^{1,p}(0,1];r^{\frac{n-1}{p}}f_r\in L^p(0,1),
r^{(n-1-p)/p}f\in L^p(0,1), f(1)=1\big\}\,.
\]
Then $V=\{f(r);u(x)=f(r)\frac{x}{|x|} \in W\}$. As stated in [6,
Proposition 2.1], we have

\begin{proposition} \label{prop2.1}
The set $V$ defined  above is a subset of $\{f \in
C[0,1];f(0)=0\}$.
\end{proposition}

\begin{proposition} \label{prop2.2}
The minimizer $u_{\varepsilon}\in W$ is a weak radial solution of
\[
-\mathop{\rm div}(|\nabla u|^{p-2}\nabla
u)=\frac{1}{\varepsilon^p}u(1-|u|^2)|u|^2
-\frac{1}{2\varepsilon^p}u(1-|u|^2)^2, \quad on \quad B,
\tag{2.1}
\]
\end{proposition}

\begin{proof}
Denote $u_{\varepsilon}$ by $u$. For any $t \in [0,1)$ and
$\phi=f(r)\frac{x}{|x|} \in W_0^{1,p}
(B,R^n)$, we have $u+t\phi \in W$ as long
as $t$ is small sufficiently. Since $u$ is a minimizer
we obtain
$\frac{dE_{\varepsilon}(u+t\phi,B)}{dt}|_{t=0}=0$,
namely,
\[
0=\int_B|\nabla u|^{p-2} \nabla u \nabla \phi dx
-\frac{1}{\varepsilon^p}\int_Bu\phi (1-|u|^2)|u|^2 dx
+\frac{1}{2\varepsilon^p}\int_Bu\phi (1-|u|^2)^2 dx.
\tag{2.2}
\]
\end{proof}

\begin{proposition} \label{prop2.3}
Let $u_\varepsilon \in W$ satisfying (2.2). Then
$|u_{\varepsilon}| \leq 1$ a.e. on $\overline{B}$.
\end{proposition}

\begin{proof}
Let $u=u_\varepsilon$ in (2.2) and set $\phi=u(|u|^2-1)_+$,
where for a positive constant $k$, $(|u|^2-1)_+=\min(k,\max(0,|u|^2-1))$.
Then
\begin{align*}
& \int_B|\nabla u|^p(|u|^2-1)_++2\int_B|\nabla u|^{p-2}(u\nabla u)^2 \\
& +\frac{1}{\varepsilon^p} \int_B |u|^4(|u|^2-1)_+^2
+\frac{1}{2\varepsilon^p} \int_B |u|^2(|u|^2-1)_+(|u|^2-1)^2 =0
\end{align*}
from which it follows that
$$
\frac{1}{\varepsilon^p} \int_B |u|^4(|u|^2-1)_+^2 =0.
$$
Thus $|u|=0$ or $(|u|^2-1)_+=0$  a.e. on $B$. Using proposition
\ref{prop2.1} we know that $|u|=|u_\varepsilon| \leq 1$ a.e. on $B$.
\end{proof}

By the same argument as in \cite[Proposition 2.5]{l3}, we obtain
the following statement.

\begin{proposition} \label{prop2.4}
Assume $u_{\varepsilon}$ is a weak radial solution of (2.1).
Then there exist positive constants $C_1,\rho$ which are both
independent of $\varepsilon$ such that
\begin{gather}
\|\nabla u_{\varepsilon}(x)\|_{L(B(x,\rho\varepsilon/8))}\leq
C_1\varepsilon^{-1}, \quad if \quad x \in B(0,1-\rho\varepsilon),
\tag{2.3} \\
|u_{\varepsilon}(x)|\geq \frac{29}{30}, \quad if \quad x \in
\overline{B}\setminus B(0,1-2\rho\varepsilon). \tag{2.4}
\end{gather}
\end{proposition}


\begin{proposition} \label{prop2.5}
Let $u_\varepsilon$ be a radial minimizer of $E_\varepsilon(u,B)$.
Then there exists a constant $C$ independent of $\varepsilon \in
(0,1)$ such that
\begin{gather}
E_\varepsilon(u_\varepsilon,B) \leq C\varepsilon^{n-p}+C;\quad
\mbox{for } p>n, \tag{2.5} \\
E_\varepsilon(u_\varepsilon,B) \leq C|\ln\varepsilon|+C, \quad
\mbox{for } p=n. \tag{2.6}
\end{gather}
\end{proposition}

\begin{proof}
Let
$$
I(\varepsilon,R)=\min\big\{\int_{B(0,R)} [\frac{1}{p}|\nabla u|^p
+\frac{1}{\varepsilon^p}(1-|u|^2)^2]; u \in W_R\big\},
$$
where
$W_R=\{u(x)=f(r)\frac{x}{|x|}
\in W^{1,p}(B(0,R),R^n); r=|x|, f(R)=1\}$.
Then
\[ \begin{aligned}
 I(\varepsilon,1)&=E_{\varepsilon}(u_{\varepsilon},B)\\
&=\frac{1}{p}\int_B |\nabla u_{\varepsilon}|^pdx
+\frac{1}{4\varepsilon^p} \int_B
(1-|u_{\varepsilon}|^2)^2|u_\varepsilon|^2 dx\\
&=\varepsilon^{n-p}[\frac{1}{p} \int_{B(0,\varepsilon^{-1})}
|\nabla u_{\varepsilon}|^pdy +\frac{1}{4}
\int_{B(0,\varepsilon^{-1})}
(1-|u_{\varepsilon}|^2)^2|u_\varepsilon|^2dy\\
&=\varepsilon^{n-p}I(1,\varepsilon^{-1}).
\end{aligned} \tag{2.7}
\]
Let $u_1$ be a solution of $I(1,1)$ and define
$$
u_2=\begin{cases} u_1,&\mbox{if }0<|x|<1\\
  \frac{x}{|x|}, &\mbox{if } 1 \leq |x|\leq \varepsilon^{-1}.
\end{cases}
$$
Thus $u_2 \in W_{\varepsilon^{-1}}$, and
\[ \begin{aligned}
I(1,\varepsilon^{-1}) &\leq \frac{1}{p}
\int_{B(0,\varepsilon^{-1})}|\nabla u_2|^p +\frac{1}{4}
\int_{B(0,\varepsilon^{-1})}(1-|u_2|^2)^2|u_2|^2 \\
&=\frac{1}{p}\int_B|\nabla u_1|^p
+\frac{1}{4}\int_B(1-|u_1|^2)^2|u_1|^2 +\frac{1}{p}
\int_{B(0,\varepsilon^{-1})\setminus B}|\nabla
\frac{x}{|x|}|^p\\
&=I(1,1)+\frac{(n-1)^{p/2}|S^{n-1}|}{p}
\int_1^{\varepsilon^{-1}}r^{n-p-1}dr
\end{aligned}
\]
Hence
$$
I(1,\varepsilon^{-1})\leq
I(1,1)+\frac{(n-1)^{p/2}|S^{n-1}|}{p(p-n)} (1-\varepsilon^{p-n})
\leq C,\quad \mbox{for } p>n;
$$
$$
I(1,\varepsilon^{-1})\leq I(1,1)+\frac{(n-1)^{p/2}|S^{n-1}|}{p}
|\ln\varepsilon|,\quad \mbox{for } \quad p=n.
$$
Substituting this into (2.7) yields (2.5) and (2.6). \end{proof}

\section{ Proof of Theorem \ref{thm1.1}} %sec. 3

\begin{proposition} \label{prop3.1}
Let $u_\varepsilon$ be a radial minimizer of $E_\varepsilon(u,B)$.
Then there exists a positive constant
$\varepsilon_0$ such that as $\varepsilon \in (0,\varepsilon_0)$,
\[
\frac{1}{\varepsilon^n}
\int_B|u_\varepsilon|^2(1-|u_\varepsilon|^2)^2 \leq
C, \tag{3.1}
\]
where $C$ is independent of $\varepsilon$.
\end{proposition}

\begin{proof}
When $p>n$, the conclusion follows from multiplying (2.5) by
$\varepsilon^{p-2}$. When $p=n$, the proof is similar to the proof
in \cite[Theorem 1]{s1}. Thus we can obtain this proposition by using (2.6).
\end{proof}

\begin{proposition} \label{prop3.2}
Let $u_\varepsilon$ be a radial minimizer of $E_\varepsilon(u,B)$.
Assume $p>n$. Then for any $\eta \in (0,1/2)$, there exist
positive constants $\lambda, \mu$ independent of $\varepsilon \in
(0,1)$ such that if
\[
\frac{1}{\varepsilon^p}\int_{B \cap
B^{2l\varepsilon}}|u_\varepsilon|^2(1-|u_\varepsilon|^2)^2 \leq
\mu, \tag{3.2}
\]
where $B^{2l\varepsilon}$ is some ball of
radius $2l\varepsilon$ with $l \geq \lambda$, then
$$
|u_\varepsilon(x)| \in [0,1-\eta]\cup [1-\eta/2,1], \quad \forall
x \in B \cap B^{l\varepsilon}.
$$
\end{proposition}

\begin{proof}
First we observe that there exists a constant $\beta>0$ such that
for any $x \in B$ and $0<\rho \leq 1$,
$|B \cap B(x,\rho)| \geq \beta \rho^2$.

From Proposition \ref{prop2.3} and (2.5) it follows that
$\|u_\varepsilon\|_{W^{1,p}(B)}\leq C\varepsilon^{\frac{2-p}{2}}$.
By embedding theorem we know that there exists a positive constant
$C_0$ which is independent of $\varepsilon$, such that for any $x,
x_0\in B$,
$$
|u_{\varepsilon}(x)-u_{\varepsilon}(x_0)| \leq
C_0\varepsilon^{\frac{2-p}{p}}|x-x_0|^{1-\frac{2}{p}}.
$$
To obtain the conclusion, we choose
\[
\lambda=\frac{\eta}{4C_0},\quad
\mu=\frac{\beta}{16}\eta^2(1-\eta)^2\lambda^n.
\tag{3.3}
\]


Suppose  that there is a point $x_0 \in B \cap B^{l\varepsilon}$
such that $1-\eta<|u_\varepsilon(x_0)| < 1-\eta/2$. Then
\[
|u_\varepsilon(x)-u_\varepsilon(x_0)| \leq
C_0\varepsilon^{\frac{2-p}{p}} |x-x_0|^{1-\frac{2}{p}}
\leq C_0\lambda=\frac{\eta}{4},
 \quad \forall x \in B(x_0,\lambda \varepsilon)
\]
Hence
$(1-|u_\varepsilon(x)|^2)^2 > (\frac{\eta}{4})^2$,
for all $x \in B(x_0,\lambda \varepsilon)$, and
\[
\int_{B(x_0,\lambda \varepsilon) \cap
B}|u_\varepsilon|^2(1-|u_\varepsilon|^2)^2
> \frac{\eta^2}{16}(1-\eta)^2
|B \cap B(x_0,\lambda \varepsilon)|
\geq \beta \frac{\eta^2}{16}(1-\eta)^2(\lambda \varepsilon)^n
=\mu \varepsilon^n
\tag{3.4}
\]
Since $x_0 \in B^{l\varepsilon} \cap B$, and
$(B(x_0,\lambda \varepsilon) \cap B)
\subset (B^{2l\varepsilon} \cap B)$, (3.4) implies
$$
\int_{B^{2l\varepsilon} \cap
B}|u_\varepsilon|^2(1-|u_\varepsilon|^2)^2> \mu \varepsilon^n
$$
which contradicts (3.2) and thus proposition \ref{prop3.2} is proved.
\end{proof}

Let $u_\varepsilon$ be a radial minimizer of $E_\varepsilon(u,B)$,
$p>n$ . Given $\eta \in (0,1/2)$. Let $\lambda,\mu$ be  constants
in Proposition \ref{prop3.2} corresponding to $\eta$. If
\[
\frac{1}{\varepsilon^n} \int_{B(x^{\varepsilon},2\lambda
\varepsilon) \cap B}|u_\varepsilon|^2(1-|u_\varepsilon|^2)^2 \leq
\mu \tag{3.5}
\]
then $B(x^{\varepsilon},\lambda \varepsilon)$
is called good ball.
Otherwise $B(x^{\varepsilon},\lambda\varepsilon)$
is called bad ball.

Now suppose that $\{B(x_i^{\varepsilon},\lambda \varepsilon),
i \in I\}$ is a family of balls
satisfying
\[ \begin{aligned}
&(i):\quad x_i^{\varepsilon} \in B,i \in I;\\
&(ii):\quad B \subset\cup_{i \in I}B(x_i^{\varepsilon},\lambda \varepsilon)\\
&(iii):\quad B(x_i^{\varepsilon},\lambda \varepsilon /4) \cap
B(x_j^{\varepsilon},\lambda \varepsilon /4)=\emptyset,i \neq j
\end{aligned}
\tag{3.6}
\]
Denote $J_\varepsilon=\{i \in I; B(x_i^{\varepsilon},
\lambda \varepsilon) \mbox{ is a bad ball}\}$.

\begin{proposition} \label{prop3.3}
Assume $p>n$, there exists a positive integer $N$ independent of
$\varepsilon \in (0,1)$, such that the number of bad balls satisfies
$\mathop{\rm Card} J_\varepsilon \leq N$.
\end{proposition}

\begin{proof}
Since (3.6) implies that every point in $B$ can be covered by finite,
say m (independent of $\varepsilon$) balls,
from Proposition \ref{prop3.1} and the definition of bad
balls, we have
\begin{align*}
\mu \varepsilon^n CardJ_\varepsilon
&\leq \sum_{i \in J_\varepsilon}
\int_{B(x_i^{\varepsilon},2\lambda \varepsilon)
  \cap B}|u_\varepsilon|^2(1-|u_\varepsilon|^2)^2\\
&\leq m\int_{\cup_{i \in J_\varepsilon}
B(x_i^{\varepsilon},2\lambda \varepsilon)
\cap B}|u_\varepsilon|^2(1-|u_\varepsilon|^2)^2\\
&\leq m\int_B|u_\varepsilon|^2(1-|u_\varepsilon|^2)^2
\leq mC\varepsilon^n
\end{align*}
and hence $\mathop{\rm Card}J_\varepsilon
\leq \frac{mC}{\mu} \leq N$.
\end{proof}
Similar to the argument in \cite[Theorem IV.1]{b1},
we have the following statement.

\begin{proposition} \label{prop3.4}
Assume $p>n$, there exist a subset $J \subset J_{\varepsilon}$ and a
constant $h\in[\lambda,\lambda 9^N]$ such that
\[
\cup_{i \in J_{\varepsilon}}B(x_i^{\varepsilon},\lambda \varepsilon)
\subset \cup_{i \in J}B(x_j^{\varepsilon},h \varepsilon), \quad
\hfill |x_i^{\varepsilon}-x_j^{\varepsilon}|>
8h\varepsilon,\quad i,j \in J,\quad i \neq j.
\tag{3.7}
\]
\end{proposition}

Applying proposition \ref{prop3.4}, 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
\begin{gather*}
\cup_{i \in J_\varepsilon}B(x_i^{\varepsilon},\lambda \varepsilon)
\subset \cup_{i \in J}B(x_i^{\varepsilon},h\varepsilon), \\
\lambda \leq h; \quad \mathop{\rm Card}J \leq \mathop{\rm Card}J_\varepsilon \\
|x_i^{\varepsilon}-x_j^{\varepsilon}|>8h\varepsilon,i,j \in J,i \neq j\,.
\end{gather*}
The last condition implies that every two balls in the new family
are not intersected.
Now we prove our main result of this section.

\begin{theorem} \label{thm3.5}
Let $u_\varepsilon$ be a radial minimizer of $ E_\varepsilon(u,B)$.
Assume $p\geq n$. Then for any $\eta \in (0,1/2)$, there exists a
constant $h=h(\eta)$ independent of $\varepsilon \in (0,1)$ such
that $ Z_\varepsilon=\{x \in B; |u_\varepsilon(x)|<1-\eta\}
\subset B(0,h\varepsilon)$. In particular the zeroes of
$u_\varepsilon$ are contained in $B(0,h\varepsilon)$.
\end{theorem}

\begin{proof}
When $p>n$. Denote $Y_\varepsilon=\{x\in B;1-\eta\leq
|u_\varepsilon(x)|\leq 1-\eta/2\}$. Suppose there exists a point
$x_0 \in Y_\varepsilon$ such that $x_0
\overline{\in}B(0,h\varepsilon)$. Then all points on the circle
$S_0=\{x \in B;~|x|=|x_0|\}$ satisfy $|u_\varepsilon(x)|<1-\eta$
and hence by virtue of Proposition \ref{prop3.3} all points on $S_0$ are
contained in bad balls. However, since $|x_0| \geq
h\varepsilon,S_0$ can not be covered by a single bad ball. $S_0$
can be covered by at least two bad balls. However this is
impossible. This means $Y_\varepsilon \subset B(0,h\varepsilon)$.

Furthermore, for any given $y_0$ satisfying
$|u_\varepsilon(y_0)|=f(r_0)< 1-\eta$, where $|y_0|=r_0$, we claim
$y_0 \in B(0,h\varepsilon)$. In fact, From $f(r_0) < 1-\eta$,
$f(1)=1>1-\eta/2$, and the continuity of $f$, it follows that
there exists $\xi \in (r_0,1)$ such that $1-\eta<f(\xi)<1-\eta/2$,
so $\xi \in Y_\varepsilon \subset (0,h\varepsilon)$ which implies
$r_0 \in (0,h\varepsilon)$.

When $p=n$, The space $W^{1,n}(B)$ does not embed into
$C^{\alpha}(\overline{B})$. Hence in the proof of Proposition
\ref{prop3.2} we can not derive the similar conclusion in
$\overline{B}$ globally. Now, by virtue of Proposition 2.4, we may
do argument {\it on $B(0,1-\rho\varepsilon)$} instead of {\it on
$B$} in the proof of Proposition \ref{prop3.2} by using (2.3) and
it is also true that we may take
$$
\frac{1}{\varepsilon^n} \int_{B(x^{\varepsilon},2\lambda
\varepsilon) \cap
B(0,1-\rho\varepsilon)}|u_\varepsilon|^2(1-|u_\varepsilon|^2)^2
\leq \mu
$$
as a ruler to distinguish the bad balls in
$B(0,1-\rho\varepsilon)$. Similarly, we also obtain that the set
$\{x \in B(0,1-\rho\varepsilon);1-\eta\leq |u_\varepsilon(x)|\leq
1-\eta/2\}$ must be covered by finite disintersected bad balls for
any $\eta \in (0,1/2)$. Moreover, it follows that the set $\{x \in
B(0,1-\rho\varepsilon);|u_\varepsilon(x)|\leq 1-\eta\} \subset
B(0,h\varepsilon)$ by the same argument above. Noting (2.4), we
can see that the theorem holds.
\end{proof}

By Proposition \ref{prop2.4}, Proposition \ref{prop3.2} and
Theorem \ref{thm3.5} we can see that
\[
|u_\varepsilon(x)|\geq \min(\frac{29}{30},1-2\eta),\quad \forall x
\in \overline{B} \setminus B(0,h\varepsilon). \tag{3.8}
\]


\begin{theorem}  \label{thm3.6}
For any given $\varepsilon\in (0,\varepsilon_0)$, the radial
minimizers $u _{\varepsilon}$ of $E_{\varepsilon}(u,B)$ are unique
on $W$.
\end{theorem}

\begin{proof}
Fix $\varepsilon \in (0,1)$. Suppose
$u_1(x)=f_1(r)\frac{x}{|x|}$ and
$u_2(x)=f_2(r)\frac{x}{|x|}$ are both radial
minimizers of $E_{\varepsilon}(u,B)$ on $W$, then they are both
weak radial solutions of (2.1). Namely, they satisfy
$$
\int_B|\nabla u|^{p-2}\nabla u \nabla
\phi+\frac{1}{2\varepsilon^p}\int_B[(1+3|u|^4)-4|u|^2]\phi=0
$$
Taking $\phi=u_1-u_2=(f_1-f_2)\frac{x}{|x|}$, we have
\begin{align*}
&\int_B(|\nabla u_1|^{p-2}\nabla u_1
-|\nabla u_2|^{p-2}\nabla u_2)\nabla (u_1-u_2)dx\\
& +\frac{1}{2\varepsilon^p}\int_B
(f_1-f_2)^2[1+3(f_1^4+f_1^3f_2+f_1^2f_2^2+f_1f_2^3+f_2^4)\\
&-4(f_1^2+f_2^2+f_1f_2)]dx=0
\end{align*}
Letting $\eta$ in (3.8) be sufficiently small such that
$$
1 \geq f_1,\quad f_2 \geq \frac{29}{30},\quad\mbox{on } B\setminus
B(0,h(\eta)\varepsilon)
$$
for any given
$\varepsilon \in (0,1)$. Hence
$$
\int_B(|\nabla u_1|^{p-2}\nabla u_1
-|\nabla u_2|^{p-2}\nabla u_2)\nabla (u_1-u_2)dx
\leq \frac{C}{\varepsilon^p}\int_{B(0,h\varepsilon)}
(f_1-f_2)^2dx.
$$
Applying (2.11) of \cite{t1}, we can see that there exists a positive
constant $\gamma$ independent of $\varepsilon$ and $h$ such that
\[
\gamma\int_B|\nabla(u_1-u_2)|^2dx \leq
\frac{1}{\varepsilon^p}\int_{B(0,h\varepsilon)}
(f_1-f_2)^2dx, \tag{3.9}
\]
which implies
\[
\int_B|\nabla(f_1-f_2)|^2dx \leq
\frac{1}{\gamma\varepsilon^p}\int_{B(0,h\varepsilon)}
(f_1-f_2)^2dx. \tag{3.10}
\]
When $n>2$. Applying \cite[Theorem 2.1]{l1}, we have
$\|f\|_{\frac{2n}{n-2}} \leq \beta \|\nabla f\|_2$, where
$\beta=\frac{2(n-1)}{n-2}$. Taking $f=f_1-f_2$ and applying
(3.10), we obtain $f(|x|)=0$ as $x \in \partial B$ and
$$
\Big[\int_B|f|^{\frac{2n}{n-2}}dx\Big]^{\frac{n-2}{n}} \leq
\beta^2 \int_B|\nabla f|^2dx \leq \beta^2 \gamma^{-1}
\int_G|f|^2dx \varepsilon^{-p},
$$
where $G=B(0,h\varepsilon)$. Using Holder inequality, we derive
$$
 \int_G|f|^2dx \leq |G|^{1-\frac{n-2}{n}}
[\int_G|f|^{\frac{2n}{n-2}}dx]^{\frac{n-2}{n}}
\leq |B|^{1-\frac{n-2}{n}}h^2
\varepsilon^{2-p}\frac{\beta^2}{\gamma}
\int_G|f|^2dx.
$$
Hence for any given $\varepsilon \in (0,1)$,
\[
\int_G|f|^2dx \leq C(\beta,|B|,\gamma,\varepsilon)
h^2\int_G|f|^2dx. \tag{3.11}
\]
Denote $F(\eta)=\int_{B(0,h(\eta)\varepsilon)}|f|^2dx$, then
$F(\eta)\geq 0$ and (3.11) implies that
\[
F(\eta)(1-C(\beta,|B|,\gamma,\varepsilon) h^2) \leq 0.
\tag{3.12}
\]
On the other hand, since $C(\beta,|B|,\gamma,\varepsilon)$ is
independent of $\eta$, we may take $\eta$ so small that
$h=h(\eta)\leq \lambda 9^N=9^N\frac{\eta}{2C_0}$ (which is implied
by (3.3)) satisfies
$$
0<1-C(\beta,|B|,\gamma,\varepsilon)h^2
$$
for the fixed $\varepsilon \in (0,1)$, which and (3.12) imply that
$F(\eta)=0$. Namely $f=0$ a.e. on $G$, or
$$
f_1=f_2, \quad a.e. \quad on \quad B(0,h\varepsilon).
$$
Substituting this into (3.9), we know that $u_1-u_2=C$ a.e. on
$B$. Noticing the continuity of $u_1,u_2$ which is implied by
Proposition \ref{prop2.1}, and $u_1=u_2=x$ on $\partial B$, we can see at
last that
$$
u_1=u_2,\quad \mbox{on }\overline{B}.
$$
When $n=2$, applying \cite[Theorem 2.1]{l1}, we have
$\|f\|_{6} \leq \beta \|\nabla f\|_{2/3}$, where $\beta$ does not
depend on $\eta$. By the similar argument above, we may see the
same conclusion.
\end{proof}

\section{Proof of Theorem \ref{thm1.2}} %sec. 4

Let $u_\varepsilon(x)=f_\varepsilon(r) \frac{x}{|x|}$ be a radial
minimizer of $E_\varepsilon(u,B_1)$, namely $f_\varepsilon$ be a
minimizer of $E_\varepsilon(f)$ in $V$. From Proposition \ref{prop2.5}, we
have
\[
E_\varepsilon(f_\varepsilon) \leq C \varepsilon^{n-p},\quad\mbox{for }
 p>n; \quad E_\varepsilon(f_\varepsilon) \leq
C|\ln\varepsilon|,\quad \mbox{for } p=n
 \tag{4.1}
\]
for some constant $C$ independent of $\varepsilon \in (0,1)$.
In this section we further prove that for any given $R \in (0,1)$,
there exists a constant $C(R)$ such that
\[
E_\varepsilon(f_\varepsilon;R) \leq C(R) \tag{4.2}
\]
for $\varepsilon \in (0,\varepsilon_0)$ with $\varepsilon_0>0$
sufficiently small, where
$$
E_\varepsilon(f;R) =\frac{1}{p}\int_{R}^1
(f_r^2+(n-1)r^{-2}f^2)^{p/2}r^{n-1}\,dr +\frac{1}{4\varepsilon^p}
\int_{R}^1f^2(1-f^2)^2r^{n-1}\,dr.
$$

\begin{proposition} \label{prop4.1}
Assume $p>n$. Given $T \in (0,1)$. There exist constants $T_j \in
[\frac{(j-1)T}{N+1}, \frac{jT}{N+1}],(N=[p])$ and $C_j$, such that
\[
E_{\varepsilon}(f_{\varepsilon}; T_j) \leq C_j\varepsilon^{j-p}
\tag{4.3}
\]
for $j=n,n+1,\dots,N$, where $\varepsilon \in (0,\varepsilon_0)$
with $\varepsilon_0$ sufficiently small.
\end{proposition}

\begin{proof}
For $j=n$, the inequality (4.3) can be obtained by (4.1) easily.
Suppose that (4.3) holds for all $j\leq m$. Then we have, in
particular,
\[
E_{\varepsilon}(f_{\varepsilon}; T_m) \leq C_m\varepsilon^{m-p}.
\tag{4.4}
\]
If $m=N$ then we are done. Suppose $m<N$, we want to prove (4.3)
for $j=m+1$.

 From (4.4) and integral mean value theorem,
we can see that there exists
$T_{m+1} \in [\frac{mT}{N+1}, \frac{(m+1)T}{N+1}] $ such that
\[
\frac{1}{\varepsilon^p}(1-f_\varepsilon^2)^2|_{r=T_{m+1}} \leq
\frac{C}{f_\varepsilon^2(T_{m+1})}E_{\varepsilon}(u_{\varepsilon},\partial
B(0,T_{m+1})) \leq C_m\varepsilon^{m-p} \tag{4.5}
\]
It is used that $f_\varepsilon(T_{m+1}) \geq \frac{29}{30}$ by
virtue of (3.8) as long as $\varepsilon_0$ and $\eta$ sufficiently
small. Consider the minimizer $\rho_1$ of the functional
$$
E(\rho,T_{m+1})=\frac{1}{p}\int_{T_{m+1}}^1 (\rho_r^2+1)^{p/2}dr
+\frac{1}{2\varepsilon^p} \int_{T_{m+1}}^1(1-\rho)^2dr
$$
It is easy to prove that the minimizer $\rho_{\varepsilon}$ of
$E(\rho,T_{m+1})$ on $W_{f_\varepsilon}^{1,p}((T_{m+1},1), R^+)$
exists and satisfies
\begin{gather}
-\varepsilon^p(v^{(p-2)/2}\rho_r)_r=1-\rho, \quad in~~(T_{m+1},1),
\tag{4.6}\\
\rho|_{r=T_{m+1}}=f_\varepsilon,~~ \rho|_{r=1}=f_\varepsilon(1)=1,
\tag{4.7}
\end{gather}
where $v=\rho_r^2+1$. Since $f_\varepsilon \leq 1$, it follows
from the maximum principle
\[
\rho_{\varepsilon} \leq 1. \tag{4.8}
\]
Applying (4.1) we see easily that
\[
E(\rho_{\varepsilon};T_{m+1}) \leq E(f_\varepsilon;T_{m+1}) \leq
CE_\varepsilon(f_\varepsilon;T_{m+1}) \leq C\varepsilon^{m-p}.
\tag{4.9}
\]
Now choosing a smooth function $0 \leq \zeta(r) \leq 1$ in (0,1]
such that $\zeta=1$ on $(0,T_{m+1}),\zeta=0$ near $r=1$ and
$|\zeta_r| \leq C(T_{m+1})$, multiplying (4.6) by $\zeta \rho_r
(\rho=\rho_{\varepsilon})$ and integrating over $(T_{m+1},1)$ we
obtain
\[
v^{(p-2)/2}\rho_r^2|_{r=T_{m+1}} +\int_{T_{m+1}}^1
v^{(p-2)/2}\rho_r(\zeta_r\rho_r+\zeta\rho_{rr})\,dr
=\frac{1}{\varepsilon^p}\int_{T_{m+1}}^1 (1-\rho)\zeta\rho_r\,dr.
\tag{4.10}
\]
Using (4.9) we have
\[\begin{aligned}
&\big|\int_{T_{m+1}}^1
v^{(p-2)/2}\rho_r(\zeta_r\rho_r+\zeta\rho_{rr})\,dr\big|\\
&\leq \int_{T_{m+1}}^1v^{(p-2)/2}|\zeta_r|\rho_r^2\,dr
+\frac{1}{p}\big| \int_{T_{m+1}}^1(v^{p/2}\zeta)_r\,dr
-\int_{T_{m+1}}^1v^{p/2}\zeta_r\,dr\big|\\
&\leq C\int_{T_{m+1}}^1v^{p/2} +\frac{1}{p}v^{p/2}
\big|_{r=T_{m+1}}+\frac{C}{p}
\int_{T_{m+1}}^1v^{p/2}dr\\
&\leq C\varepsilon^{m-p}+\frac{1}{p}v^{p/2}\big|_{r=T_{m+1}}
\end{aligned}
\tag{4.11}
\]
and using (4.5),(4.7) and (4.9) we have
\[\begin{aligned}
&\big|\frac{1}{\varepsilon^p}\int_{T_{m+1}}^1
(1-\rho)\zeta\rho_r\,dr\big|\\
&=\frac{1}{2\varepsilon^p}\big| \int_{T_{m+1}}^1((1-\rho)^2\zeta)_r \,dr
-\int_{T_{m+1}}^1(1-\rho)^2\zeta_r \,dr\big|\\
&\leq \big| \frac{1}{2\varepsilon^p}(1-\rho)^2\big|_{r=T_{m+1}}
+\frac{C}{2\varepsilon^p} \int_{T_{m+1}}^1(1-\rho)^2 \,dr\big| \leq
C\varepsilon^{m-p}.
\end{aligned} \tag{4.12}
\]
Combining (4.10) with (4.11), (4.12) yields
$$
v^{(p-2)/2}\rho_r^2|_{r=T_{m+1}} \leq C \varepsilon^{m-p}
+\frac{1}{p}v^{p/2}|_{r=T_{m+1}}.
$$
Hence for any $\delta \in (0,1)$,
\[\begin{aligned}
 v^{p/2}|_{r=T_{m+1}}
&=v^{(p-2)/2}(\rho_r^2+1)|_{r=T_{m+1}}\\
&=v^{(p-2)/2}\rho_r^2|_{r=T_{m+1}}
+v^{(p-2)/2}|_{r=T_{m+1}}\\
&\leq C\varepsilon^{m-p} +\frac{1}{p}v^{p/2}|_{r=T_{m+1}}
+v^{(p-2)/2}|_{r=T_{m+1}}\\
&=C \varepsilon^{m-p}+(\frac{1}{p} +\delta)v^{p/2}|_{r=T_{m+1}}
+C(\delta)
\end{aligned}
\]
from which it follows by choosing $\delta>0$ small enough that
\[
v^{p/2}\big|_{r=T_{m+1}} \leq C \varepsilon^{m-p}. \tag{4.13}
\]
Now we multiply both sides of (4.6) by $\rho-1$ and integrate.
Then
\[
-\varepsilon^p\int_{T_{m+1}}^1 [v^{(p-2)/2}\rho_r(\rho-1)]_r\,dr
+\varepsilon^p\int_{T_{m+1}}^1 v^{(p-2)/2}\rho_r^2\,dr
+\int_{T_{m+1}}^1 (\rho-1)^2\,dr=0.
\]
 From this, using(4.5), (4.7) and (4.13), we obtain
\[\begin{aligned}
E(\rho_{\varepsilon};T_{m+1})&\leq C |\int_{T_{m+1}}^1
[v^{(p-2)/2}\rho_r(\rho-1)]_r\,dr|\\
&=Cv^{(p-2)/2}|\rho_r||\rho-1|_{r=T_{m+1}}
\leq Cv^{(p-1)/2}|\rho-1|_{r=T_{m+1}}\\
&\leq (C \varepsilon^{m-p})^{(p-1)/p} (C\varepsilon^m)^{1/2} \leq
C\varepsilon^{m-p+1}.
\end{aligned}
\tag{4.14}
\]
Define
$$w_\varepsilon=\begin{cases} f_\varepsilon &\mbox{for } r \in (0,T_{m+1})\\
\rho_{\varepsilon} &\mbox{for } r \in [T_{m+1},1]
\end{cases}
$$
Since $f_\varepsilon$ is a minimizer of $E_\varepsilon(f)$, we have
$E_\varepsilon(f_\varepsilon) \leq E_\varepsilon(w_\varepsilon)$.
Thus, it follows that
$$
E_\varepsilon(f_\varepsilon;T_{m+1}) \leq
\frac{1}{p}\int_{T_{m+1}}^1(\rho_r^2+(n-1)r^{-2}\rho^2)^{p/2}r^{n-1}\,dr
 +\frac{1}{4\varepsilon^p}\int_{T_{m+1}}^1
\rho^2(1-\rho^2)^2r^{n-1}\,dr
$$
by virtue of $\Gamma \leq \varepsilon<T_{m+1}$ since $\varepsilon$
is sufficiently small. Noticing that
\begin{align*}
&\int_{T_{m+1}}^1 (\rho_r^2+(n-1)r^{-2}\rho^2)^{p/2}r^{n-1}dr
-\int_{T_{m+1}}^1((n-1)r^2\rho^2)^{p/2}r^{n-1}dr\\
&=\frac{p}{2} \int_{T_{m+1}}^1\int_0^1[\rho_r^2
+(n-1)r^{-2}\rho^2)s\\
&\quad +(n-1)r^{-2}\rho^2(1-s)]^{(p-2)/2}ds
\rho_r^2r^{n-1}dr\\
& \leq C\int_{T_{m+1}}^1 (\rho_r^2+(n-1)r^{-2}\rho^2)^{(p-2)
/2}\rho_r^2r^{n-1}dr\\
&\quad +C\int_{T_{m+1}}^1((n-1)r^{-2}\rho^2)
^{(p-2)/2}\rho_r^2r^{n-1}dr\\
& \leq C\int_{T_{m+1}}^1(\rho_r^p+\rho_r^2)dr
\end{align*}
and using (4.8) we obtain
\begin{align*}
&E_\varepsilon(f_\varepsilon;T_{m+1})\\
&\leq \frac{1}{p}\int_{T_{m+1}}^1
((n-1)r^{-2}\rho^2)^{p/2}r^{n-1}\,dr+
C\int_{T_{m+1}}^1(\rho_r^p+\rho_r^2)dr +\frac{C}{4\varepsilon^p}
\int_{T_{m+1}}^1(1-\rho^2)^2dr\\
&\leq \frac{1}{p}\int_{T_{m+1}}^1 ((n-1)r^{-2})^{p/2}r^{n-1}\,dr
+CE(\rho_\varepsilon;T_{m+1}).
\end{align*}
Combining this with (4.14) yields (4.3) for $j=m+1$. It is just
(4.3) for $j=m+1$.
\end{proof}

\begin{proposition} \label{prop4.2}
Assume  $p\geq n$. Given $T\in (0,1)$. There exist constants
$T_{N+1}\in (0,T]$ and $C>0$ such that
\begin{gather*}
E_{\varepsilon}(u_{\varepsilon};T_{N+1}) -
(n-1)^{p/2}\frac{|S^{n-1}|}{p}\int_{T_{N+1}}^1r^{n-p-1}dr
\leq C\varepsilon^{N+1-p},(p>n);\\
E_{\varepsilon}(u_{\varepsilon};T_{N+1}) -
(n-1)^{p/2}\frac{|S^{n-1}|}{p} \int_{T_{N+1}}^1r^{n-p-1}dr
\leq C\varepsilon|\ln\varepsilon|,(p=n),
\end{gather*}
where $N=[p]$.
\end{proposition}

\begin{proof}
From (4.1) and (4.3) we can see
$E_{\varepsilon}(u_{\varepsilon};T_N) \leq CF(\varepsilon)$, where
$F(\varepsilon)=|\ln\varepsilon|$ as $p=n$, and $F(\varepsilon)=
\varepsilon^{N-p}$ as $p>n$. Hence by using integral mean value
theorem we know that there exists $T_{N+1}\in (0,T]$ such that
\[
\frac{1}{p}\int_{\partial B(0,T_{N+1})}|\nabla u_\varepsilon|^p
dx+\frac{1}{4\varepsilon^p} \int_{\partial
B(0,T_{N+1})}|u_\varepsilon|^2(1-|u_\varepsilon|^2)^2 dx \leq
CF(\varepsilon). \tag{4.15}
\]
Note that $\rho_2$ is a minimizer of the functional
$$
E(\rho,T_{N+1})=\frac{1}{p} \int_{T_{N+1}}^1(\rho_r^2+1)^{p/2}dr
+\frac{1}{2\varepsilon^p} \int_{T_{N+1}}^1(1-\rho)^2dr
$$
on $W_{f_{\varepsilon}}^{1,p}((T_{N+1},1), R^+\cup \{0\})$. It is
not difficult to prove by maximum principle that
\[
\rho_2 \leq 1. \tag{4.16}
\]
As in the derivation of (4.14), from (4.3) and (4.15)
it can be proved that
\[
E(\rho_2,T_{N+1}) \leq C\varepsilon F(\varepsilon).
\tag{4.17}
\]
Using that $u_{\varepsilon}$ is a minimizer and
$\rho_2\frac{x}{|x|}\in W_2$, we also have
\[ \begin{aligned}
E_{\varepsilon}(f_{\varepsilon};T_{N+1})
&\leq E_{\varepsilon}(\rho_2;T_{N+1})\\
&\leq \frac{1}{p} \int_{T_{N+1}}^1
[\rho_{2r}^2+\rho_2^2(n-1)r^{-2}]^{p/2}r^{n-1}dr
+\frac{1}{2\varepsilon^p} \int_{T_{N+1}}^1\rho^2(1-\rho_2)^2dr.
\end{aligned}
\tag{4.18}
\]
On the other hand,
\[ \begin{aligned}
& \int_{T_{N+1}}^1
[\rho_r^2+(n-1)r^{-2}\rho^2]^{p/2}r^{n-1}dr -\int_{T_{N+1}}^1
[(n-1)r^{-2}\rho^2]^{p/2}r^{n-1}dr\\
&=\frac{p}{2}\int_{T_{N+1}}^1 \int_0^1
[\rho_r^2+(n-1)r^{-2}\rho^2]^{(p-2)/2}s
 +(n-1)r^{-2}\rho^2(1-s)ds\rho_r^2r^{n-1}dr\\
&\leq C\int_{T_{N+1}}^1
[\rho_r^2+(n-1)r^{-2}\rho^2]^{(p-2)/2}\rho_r^2r^{n-1}dr\\
&\quad+C\int_{T_{N+1}}^1[(n-1)r^{-2}\rho^2]
^{(p-2)/2}\rho_r^2r^{n-1}dr\\
&\leq C\int_{T_{N+1}}^1[\rho_r^p+\rho_r^2]dr.
\end{aligned}
\]
Substituting this into (4.18), we have
\[\begin{aligned}
&E_{\varepsilon}(f_{\varepsilon};T_{N+1})\\
&\leq \frac{1}{p} \int_{T_{N+1}}^1 (n-1)^{p/2}\rho_2^pr^{n-p-1}dr
+C\int_{T_{N+1}}^1(\rho_{2r}^p+\rho_{2r}^2)dr
+\frac{C}{\varepsilon^p}
\int_{T_{N+1}}^1(1-\rho_2)^2dr\\
&\leq \frac{1}{p}\int_{T_{N+1}}^1
(n-1)^{p/2}\rho_2^pr^{n-p-1}dr+C\varepsilon F(\varepsilon)\\
&\leq \frac{1}{p} (n-1)^{p/2}\int_{T_{N+1}}^1r^{n-p-1}dr
+C\varepsilon F(\varepsilon),
\end{aligned}
\]
using (4.16) and (4.17). This completes the proof.
\end{proof}

\begin{theorem} \label{thm4.3}
Let $u_\varepsilon=f_\varepsilon(r)\frac{x}{|x|}$ be a radial
minimizer of $E_\varepsilon(u,B_1)$. Then
$$
\lim_{\varepsilon \rightarrow 0}u_\varepsilon= \frac{x}{|x|},
\quad \mbox{in } W^{1,p}(K,R^n)
$$
for any compact subset $K \subset \overline{B_1} \setminus \{0\}$.
\end{theorem}

\begin{proof}
Without loss of generality, we may assume $K=\overline{B_1}
\setminus B(0,T_{N+1})$. From Proposition \ref{prop4.2}, we have
\[
E_\varepsilon(u_\varepsilon,K) =|S^{n-1}|
E_\varepsilon(f_\varepsilon;T_{N+1}) \leq C\,,
\tag{4.19}
\]
where $C$ is independent of $\varepsilon$. This and
$|u_\varepsilon| \leq 1$ imply the existence of  a subsequence
$u_{\varepsilon_k}$
of $u_\varepsilon$ and a function $u_* \in W^{1,p}(K,R^n)$, such
that
\[ \begin{gathered}
\lim_{\varepsilon_k \rightarrow 0} u_{\varepsilon_k}=u_*, \quad
\mbox{weakly in }W^{1,p}(K,R^n),\\
\lim_{\varepsilon_k \rightarrow 0} u_{\varepsilon_k}=u_*, \quad
\mbox{in }L^{q}(K,R), \quad\forall q>0, \\
\lim_{\varepsilon_k \rightarrow 0} f_{\varepsilon_k}(r)=|u_*|,
\quad \mbox{in } C^{\alpha}([T_{N+1},1],R), \quad\alpha >1-1/p.
\end{gathered}\tag{4.20}
\]
Inequality (4.19) implies $|u_*|\in\{0,1\}$. Using also (4.20) and
$f_{\varepsilon_k}(1)=1$ we see that $|u_*|=1$ or
$u_*=\frac{x}{|x|}$. Hence, noticing that any subsequence of
$u_\varepsilon$ has a convergent subsequence and the limit is
always $x/|x|$, we can assert
\begin{gather}
\lim_{\varepsilon \rightarrow 0}u_\varepsilon=\frac{x}{|x|},
 \quad \mbox{weakly in }W^{1,p}(K,R^n). \tag{4.21}\\
\lim_{\varepsilon \rightarrow 0} u_{\varepsilon}=u_*, \quad
\mbox{in } L^{q}(K,R), \quad\forall q>0. \tag{4.22}
\end{gather}
 From this and the weakly lower semicontinuity of $\int_K|\nabla
u|^p$, using Proposition \ref{prop4.2}, it follows that
\[\begin{aligned}
\int_K|\nabla \frac{x}{|x|}|^p &\leq
\liminf_{\varepsilon_k \rightarrow 0} \int_K|\nabla
u_\varepsilon|^p \leq \limsup_{\varepsilon_k \rightarrow 0}
\int_K|\nabla u_\varepsilon|^p\\
&\leq |S^{n-1}| \int_{T_{N+1}}^1 ((n-1)r^{-2})^{p/2}r^{n-1}\,dr
\end{aligned}
\]
and hence
$$
\lim_{\varepsilon \rightarrow 0} \int_K|\nabla u_\varepsilon|^p
=\int_K|\nabla \frac{x}{|x|}|^p
$$
since
$$
\int_K|\nabla \frac{x}{|x|}|^p =|S^{n-1}| \int_{T_{N+1}}^1
((n-1)r^{-2})^{p/2}r^{n-1}\,dr.
$$
Combining this with (4.21)(4.22) completes the proof.
\end{proof}


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\end{document}