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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 05, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/05\hfil Asymptotic profile]
{Asymptotic profile of a radially symmetric solution
with transition layers for an unbalanced bistable equation}
\author[H. Matsuzawa\hfil EJDE-2006/05\hfilneg]
{Hiroshi Matsuzawa}

 \address{Hiroshi Matsuzawa \hfill\break
Numazu National College of Technology \\
 Ooka 3600, Numazu-city, Shizuoka 410-8501, Japan}
\email{hmatsu@numazu-ct.ac.jp}

\date{}
\thanks{Submitted August 31, 2005. Published January 11, 2006.}
\subjclass[2000]{35B40, 35J25, 35J55, 35J50, 35K57}
\keywords{Transition layer; Allen-Cahn equation; bistable equation; unbalanced}

\begin{abstract}
 In this article, we consider the semilinear elliptic problem
 $$
 -\varepsilon^{2}\Delta u=h(|x|)^2(u-a(|x|))(1-u^2)
 $$
 in $B_1(0)$ with the Neumann boundary condition.
 The function $a$ is a $C^1$ function satisfying $|a(x)|< 1$
 for $x\in [0,1]$ and $a'(0)=0$. In particular we consider
 the case $a(r)=0$ on some interval $I\subset [0,1]$.
 The function $h$ is a positive $C^1$ function satisfying $h'(0)=0$.
 We investigate an asymptotic profile of the global minimizer
 corresponding to the energy functional as $\varepsilon\to 0$.
 We use the variational procedure used in \cite{DS} with a few
 modifications prompted by the presence of the function $h$.
\end{abstract}

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

\section{Introduction and Statement of Main Results}

In this article, we consider the  boundary value problem
\begin{equation} \label{Pe}
\begin{gathered}
-\varepsilon^2\Delta u=h(|x|)^2(u-a(|x|))(1-u^2) \quad\text{in }B_1(0) \\
\frac{\partial u}{\partial \nu}=0 \quad\text{on } \partial B_1(0)
\end{gathered}
\end{equation}
where $\varepsilon$ is a small positive parameter, $B_1(0)$ is a unit
ball in $\mathbb{R}^N$ centered at the origin, and the function $a$
is a $C^1$ function on $[0,1]$ satisfying $-1<a(|x|)<1$ and
$a'(0)=0$. The function $h$ is a positive $C^1$ function on
$[0,1]$ satisfying $h'(0)=0$. We set $r=|x|$.

Problem \eqref{Pe} appears in various models such as population
genetics, chemical reactor theory and phase transition phenomena.
See \cite{AMP} and the references therein. If the function $h$
satisfies $h(r)\equiv 1$ and the function $a$ satisfies
$a(r)\not\equiv 0$, then this problem \eqref{Pe} has been studied
in \cite {AMP}, \cite{DS} and \cite{NM}. In this case, it is shown
that there exist radially symmetric solutions with transition
layers near the set $\{x\in B_1(0)|a(|x|)=0\}$. If the set
$\{r\in\mathbb{R}|a(r)=0\}$ contains an interval $I$, then the
problem to decide the configuration of transition layer on $I$ is
more delicate.

When $N=1$, if the function $h$ satisfies $h(r)\not\equiv 1$ and
the function $a$ satisfies $a(r)\equiv 0$, then problem \eqref{Pe}
has been studied in \cite{N} and \cite{NT}. In this case, it is
shown that there exist stable solutions with transition layers
near prescribed local minimum points of $h$.

In this paper, we consider the case where the function $a$
satisfies $a(r)\not\equiv 0$ with $a(r)=0$ on some interval
$I\subset (0,1)$. We show the minimum point of the function
$r^{N-1}h(r)$ on $I$ has very important role to decide the
configuration of transition layer on $I$ in this case.

We note that in \cite{DS}, Dancer and Shusen Yan considered a
problem  similar to ours. They assume that $N\ge 2$, $h\equiv 1$
and the nonlinear term is $u(u-a|x|)(1-u)$ satisfying $a(r)=1/2$
on $I=[l_1,l_2]$ and $a(r)<1/2$ for $l_1-r>0$ small and $a(r)>1/2$
for $r-l_2>0$ small, then a global minimizer of the corresponding
functional has a transition layer near the $l_1$, that is, the
minimum point of $r^{N-1}$ on $I$ (see \cite[Theorem 1.3]{DS}). In
this sense, we can say that our results are natural extension of
the results in \cite{DS}. We are going to follow throughout the
variational procedure used in \cite{DS} with a few modifications
prompted by the presence of the function $h$.

Here we state the energy functional, corresponding to \eqref{Pe},
\[
J_{\varepsilon}(u)=\int_{B_1(0)}\frac{\varepsilon^2}{2}|\nabla u|^2-F(|x|,u)
dx,
\]
where $F(|x|,u)=\int_{-1}^u f(|x|,s)ds$ and
$f(|x|,u)=h(|x|)^2(u-a(|x|))(1-u^2)$.
It is easy to see that the following minimization problem has a minimizer
\begin{equation}\label{mini}
\inf\{J_{\varepsilon}(u)|u\in H^1(B_1(0))\}.
\end{equation}
Let $A_-=\{x\in B_1(0)|a(|x|)<0\}$ and $A_+=\{x\in B_1(0)|a(|x|)>0\}$.

In this paper, we will analyze the profile of the minimizer of \eqref{mini},
and prove the following results.

\begin{theorem} \label{thm1.1}
Let $u_{\varepsilon}$ be a global minimizer of \eqref{mini}.
Then $u_{\varepsilon}$ is radially symmetric and
\[
u_{\varepsilon}\to \begin{cases}
 1, &\mbox{uniformly on each compact subset of } A_-, \\
-1, &\mbox{uniformly on each compact subset of } A_+,
\end{cases}
\]
as $\varepsilon\to 0$. In particular $u_{\varepsilon}$ converges
uniformly near the boundary of $B_1(0)$, that is, if $a(r)<0$ on
$[r_0,1]$ for some $r_0>0$, $u_{\varepsilon}\to 1$ uniformly on
$\overline{B_1(0)}\backslash B_{r_0}(0)$ and if $a(r)>0$ on
$[r_0,1]$ for some $r_0>0$, $u_{\varepsilon}\to -1$ uniformly on
$\overline{B_1(0)}\backslash B_{r_0}(0)$. Moreover, for any
$0<r_1\le r_2<1$ with $a(r_i)=0$, $i=1,2$, $a(r)\ne 0$ for
$r_1-r>0$  small and for $r-r_2>0$ small, $a(r)=0$ if
$r\in [r_1,r_2]$, we have:
\begin{itemize}
\item[(i)] If $a(r)<0$ for $r_1-r>0$ small and $a(r)>0$ for $r-r_2>0$,
then for any small $\eta>0$ and for any small $\theta>0$, there exists
 a positive number $\varepsilon_0$ which has the following properties:
\begin{itemize}
\item[(a)] For all $\varepsilon\in (0, \varepsilon_0]$, there exist
$t_{\varepsilon,1}<t_{\varepsilon,2}$ such that
\begin{gather*}
u_{\varepsilon}(r)>1-\eta \quad \text{for }
r\in [r_1-\theta,t_{\varepsilon,1}), \\
u_{\varepsilon}(t_{\varepsilon,1})=1-\eta,  \\
u_{\varepsilon}(t_{\varepsilon, 2})=-1+\eta,  \\
u_{\varepsilon}(r)<-1+\eta, \quad\text{for }
 r\in (t_{\varepsilon,2},r_2+\theta].
\end{gather*}
\item[(b)] The function $u_{\varepsilon}(r)$ is decreasing on the interval
 $(t_{\varepsilon,1},t_{\varepsilon, 2})$
\item[(c)] The inequality
$0<R_1\le\frac{t_{\varepsilon,2}-t_{\varepsilon,1}}{\varepsilon}\le R_2$
holds, where $R_1$ and $R_2$ are two constants independent of $\varepsilon>0$.
\item[(d)] If $t_{\varepsilon_j,1}$, $t_{\varepsilon_j,2}\to \overline{t}$
for some positive sequence $\{\varepsilon_j\}$ converging to zero as
$j\to\infty$, then $\overline{t}$ satisfies
$h(\overline{t})\overline{t}^{N-1}=\min_{s\in [r_1, r_2]}h(s) s^{N-1}$.
\end{itemize}
\item[(ii)]
 If $a(r)>0$ for $r_1-r>0$ small and $a(r)<0$ for $r-r_2>0$, then for each
small $\eta>0$ and for each small $\theta>0$, there exists a positive
number $\varepsilon_0$ which has the following properties:
For each $\varepsilon\in (0, \varepsilon_0]$, there exist
$t_{\varepsilon,1}<t_{\varepsilon,2}$ such that
\begin{itemize}
\item[(a)]
\begin{gather*}
u_{\varepsilon}(r)<-1+\eta \quad\text{for }
 r\in [r_1-\theta,t_{\varepsilon,1}), \\
u_{\varepsilon}(t_{\varepsilon,1})=-1+\eta,  \\
u_{\varepsilon}(t_{\varepsilon, 2})=1-\eta,  \\
u_{\varepsilon}(r)>1-\eta, \quad\text{for }
 r\in (t_{\varepsilon,2},r_2+\theta].
\end{gather*}
\item[(b)] The function $u_{\varepsilon}(r)$ is increasing
 in $(t_{\varepsilon,1},t_{\varepsilon, 2})$.
\item[(c)] The inequality
 $0<R_1\le\frac{t_{\varepsilon,2}-t_{\varepsilon,1}}{\varepsilon}\le R_2$
 holds, where $R_1$ and $R_2$ are two constants independent of
 $\varepsilon>0$.
\item[(d)] If $t_{\varepsilon_j,1}$, $t_{\varepsilon_j,2}\to \overline{t}$
for some positive sequence $\{\varepsilon_j\}$ converging to zero
as $j\to\infty$, then $\overline{t}$ satisfies
$h(\overline{t})\overline{t}^{N-1}=\min_{s\in [r_1, r_2]}h(s) s^{N-1}$.
\end{itemize}
\end{itemize}
\end{theorem}

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig1}
\end{center}
\caption{Profile of the global minimizer $u_{\varepsilon}$}
\end{figure}

\subsection*{Remarks.}
\begin{itemize}
\item Note that results from (a) to (c) both in cases (i) and (ii)
are not related to the presence of the function $h$.
The effect of presence of function $h$ appears in the result (d) in (i)
and (ii).

\item If $\min_{s\in [r_1,r_2]}s^{N-1}h(s)$ is attained at a unique point
$\overline{t}$, we can show $t_{\varepsilon,1}$,
$t_{\varepsilon,2}\to \overline{t}$ as $\varepsilon\to 0$ without
taking subsequences.

\item If the function $r^{N-1}h(r)$ is constant on $[r_1, r_2]$, it is
a very difficult problem to know the location of the point
$\overline{t}\in [r_1,r_2]$.
\end{itemize}


This paper is organized as follows: In section 2, we present some
preliminary results. In section 3, we prove the main theorem.

\section{Preliminary Results}

Let $D$ is a bounded domain in $\mathbb{R}^N$. Let
$\overline{f}(x,t)$ be a function defined on
$\overline{D}\times\mathbb{R}$ which is bounded on
$\overline{D}\times[-1,1]$. Suppose $\overline{f}$ is continuous
on $t\in\mathbb{R}$ for each $x\in\overline{D}$ and is measurable
in $D$ for each $t\in\mathbb{R}$. We also assume
\begin{equation}\label{comp1}
\begin{gathered}
\overline{f}(x,t)>0 \quad\text{for } x\in\overline{D},\; t<-1;\\
\overline{f}(x,t)<0 \quad\text{for } x\in\overline{D},\; t>1.
\end{gathered}
\end{equation}
Consider the  minimization problem
\begin{equation}\label{comp2}
\inf\left\{\overline{J}_{\varepsilon}(u,D)
:=\int_{D}\frac{\varepsilon^2}{2}|\nabla u|^2-\overline{F}(x, u)dx:
 u-\eta\in H^1_0(D)\right\},
\end{equation}
where $\eta\in H^1(D)$ with $-1\le \eta\le 1$ on $D$ and
\[
\overline{F}(x,t)=\int_{-1}^t \overline{f}(x,s)ds.
\]
We can prove next two lemmas by methods similar to \cite{DS}. For
the readers convenience, we prove these lemmas in this section.

\begin{lemma}\label{complem1}
Suppose that $\overline{f}(x,t)$ satisfies \eqref{comp1}.
Let $u_{\varepsilon}$ be a minimizer of \eqref{comp2}.
Then $-1\le u_{\varepsilon}\le 1$ on $D$.
\end{lemma}

\begin{proof}
We prove $-1\le u_{\varepsilon}$ on $D$. Let $M=\{x:u_{\varepsilon}(x)<-1\}$.
Define $\tilde{u}_{\varepsilon}$ by
\[
\tilde{u}_{\varepsilon}(x)=\begin{cases}
u_{\varepsilon}(x) & \text{if } x\in D\backslash M \\
-1   &\text{if } x\in M.
\end{cases}
\]
Since $u_{\varepsilon}(x)=\eta\ge -1$ on $\partial D$, we see that
$M$ is compactly contained in $D$. Thus $\tilde{u}-\eta\in H^1_0(D)$.
If the measure $m(M)$ of $M$ is positive, we have
$\overline{J}_{\varepsilon}(\tilde{u}_{\varepsilon}, D)
<\overline{J}_{\varepsilon}(u_{\varepsilon},D)$.
Because $u_{\varepsilon}$ is a minimizer, we see $m(M)=0$,
where $m(A)$ denotes the Lebesgue measure of the set $A$.
Thus $u_{\varepsilon}\ge -1$. Similarly we can prove that
$u_{\varepsilon}\le 1$.
\end{proof}

\begin{lemma}\label{complem2}
Suppose that $\overline{f}_1(x,t)$ and $\overline{f}_2(x,t)$
both satisfy \eqref{comp1} and the same regularity assumption on
$\overline{f}$. Assume that $\eta_i\in H^1(D)$ satisfy $-1\le\eta_i\le 1$
 on $D$ for $i=1,2$. Let $u_{\varepsilon,i}$ be a corresponding
minimizer of {\rm \eqref{comp2}}, where
$\overline{f}=\overline{f}_i$  and $\eta=\eta_i$, $i=1,2$. Suppose
that $\overline{f}_1(x,t)\ge\overline{f}_2(x,t)$ for all
$(x,t)\in\overline{D}\times [-1,1]$ and $1\ge \eta_1\ge \eta_2\ge
-1$. Then $u_{\varepsilon,1}\ge u_{\varepsilon, 2}$.
\end{lemma}

\begin{proof}
Let $M=\{x\in D:u_{\varepsilon, 2}>u_{\varepsilon,1}\}$.
Define $\varphi_{\varepsilon}=(u_{\varepsilon,2}-u_{\varepsilon,1})^+$.
Since $\eta_1\ge \eta_2$, we have $\varphi_{\varepsilon}\in H^1_0(D)$.
Set $\overline{F}_i(x,u)=\int_{-1}^u \overline{f}_i(x,s)ds$.
Since $u_{\varepsilon,i}$ is a minimizer of
\[
J_{\varepsilon,i}(u):=\int_{D}\frac{\varepsilon^2}{2}|\nabla u|^2
-\overline{F}_i(x,u)dx
\]
and $\varphi_{\varepsilon}=0$ for $x\in D\backslash M$, we have
\begin{align*}
0 &\le J_{\varepsilon, 1}(u_{\varepsilon, 1}
  +\varphi_{\varepsilon})-J_{\varepsilon, 1}(u_{\varepsilon, 1})  \\
& =  \int_{M}\frac{\varepsilon^2}{2}(|\nabla (u_{\varepsilon, 1}
  +\varphi_{\varepsilon})|^2-|\nabla u_{\varepsilon,1}|^2)dx
  -\int_{M}\int_{u_{\varepsilon,1}}^{u_{\varepsilon,1}
  +\varphi_{\varepsilon}}\overline{f}_1(x,s)ds \\
& \le  \int_{M}\frac{\varepsilon^2}{2}(|\nabla (u_{\varepsilon, 1}
  +\varphi_{\varepsilon})|^2-|\nabla u_{\varepsilon,1}|^2)dx
  -\int_{M}\int_{u_{\varepsilon,1}}^{u_{\varepsilon,1}
  +\varphi_{\varepsilon}}\overline{f}_2(x,s)ds \\
& =  J_{\varepsilon,2}(u_{\varepsilon,2})-J_{\varepsilon,2}(u_{\varepsilon,2}
  -\varphi_{\varepsilon})\le 0.
\end{align*}
This implies that $u_{\varepsilon, 1}+\varphi_{\varepsilon}$ is also a
 minimizer of $J_{\varepsilon,1}(u)$. Let $L>0$ be large enough such
 that $\overline{f}_1(x,t)+Lt$ is strictly increasing for $x\in\overline{D}$,
 $t\in[-1,1]$. From
\[
-\varepsilon^2\Delta (u_{\varepsilon,1}+\varphi_{\varepsilon})
=\overline{f}_1(u_{\varepsilon,1}+\varphi_{\varepsilon}),
\]
we obtain
\[
 -\varepsilon^2\Delta\varphi_{\varepsilon}
 =\overline{f}_1(u_{\varepsilon,1}+\varphi_{\varepsilon})
 -\overline{f}_1(u_{\varepsilon,1}).
\]
Thus
\[
-\varepsilon^2\Delta\varphi_{\varepsilon}+L\varphi_{\varepsilon}
=\overline{f}_1(u_{\varepsilon,1}+\varphi_{\varepsilon})
 +L(u_{\varepsilon,1}+\varphi_{\varepsilon})
 -(\overline{f}_1(u_{\varepsilon,1})+Lu_{\varepsilon,1})>0
\]
in $D$. Fix $z_0\in M$. Let $x_0\in\partial M$ such that
$|x_0-z_0|={\rm dist}(z_0,\partial M)$. Using the Strong maximum principle
and Hopf's lemma in $B_{{\rm dist}(z_0, \partial M)}(z_0)$, we obtain
that $\frac{\partial\varphi_{\varepsilon}}{\partial\nu}(x_0)<0$,
where $\nu=(x_0-z_0)/|x_0-z_0|$. But $\varphi_{\varepsilon}(x)=0$
for $x\notin M$. Thus,
$\frac{\partial\varphi_{\varepsilon}}{\partial\nu}(x_0)=0$. This is a
contradiction. Thus we obtain $M=\emptyset$.
\end{proof}

\section{Proof of Main Theorem}

To prove Theorem \ref{thm1.1}, the following proposition is used as
the first step.

\begin{proposition} \label{prop3.1}
Let $u_{\varepsilon}$ be a global minimizer of the problem
\eqref{mini}. Then $u_{\varepsilon}$ satisfies
\[
u_{\varepsilon}\to \begin{cases}
1 & \mbox{uniformly on each compact subset of } A_- \\
-1 & \mbox{uniformly on each compact subset of } A_+
\end{cases}
\]
as $\varepsilon\to 0$.
\end{proposition}

\begin{proof}
Let $x_0\in A_-$. Choose $\delta>0$ small so that
$B_{\delta}(x_0)\subset\subset A$. Take
$b\in (\max_{z\in\overline{B_{\delta}(x_0)}}a(z), 1/2)$. Define
$f_{x_0,\delta,b}(t)=(\min_{z\in B_{\delta}(x_0)} h(|z|)^2)(t-b)(1-t^2)$.
Then for $x\in \overline{B_{\delta}(x_0)}$, $t\in [-1, 1]$,
we have $f(|x|,t)\ge f_{x_0,\delta,b}(t)$. Let $u_{\varepsilon,x_0,\delta,b}$
be the minimizer of
\[
\inf\Big\{\int_{B_{\delta}(x_0)}\frac{\varepsilon^2}{2}|\nabla u|^2-F_{x_0,
\delta,b}(u)dx: u+1\in H^1_0(B_{\delta}(x_0))\Big\},
\]
where $F_{x_0,\delta,b}(t)=\int_{-1}^t f_{x_0,\delta,b}(s)ds$.
It follows from Lemmas \ref{complem1} and \ref{complem2} that
\[
u_{\varepsilon, x_0, \delta,b}(x)\le u_{\varepsilon}(x)\le 1, \quad
\text{for } x\in B_{\delta}(x_0).
\]
Since $\int_{-1}^1 f_{x_0,\delta,b}(s)ds>0$, it follows from \cite{CP,CS}
that $u_{\varepsilon,x_0,\delta,b}(x)\to 1$ as $\varepsilon\to 0$
uniformly in $B_{\delta/2}(x_0)$, thus $u_{\varepsilon}(x)\to 1$
as $\varepsilon \to 0$ uniformly in $B_{\delta/2}(x_0)$.
\end{proof}

To prove the rest of Theorem \ref{thm1.1}, we need the following proposition
and lemma.

\begin{proposition}\label{rad}
Let $u$ be a local minimizer of the  problem
\[
\inf\Big\{\int_{B_1(0)}\frac{1}{2}|\nabla u|^2-G(|x|,u)dx:
u\in H^1(B_1(0))\Big\}.
\]
Here $G(r,t)=\int_{-1}^t g(r,s)ds$, $g(r,t)$ is $C^1$ in $t\in\mathbb{R}$
for each $r\ge 0$,  $g(r,t)$ and $g_t(r,t)$ are measurable
on $[0,+\infty)$ for each $t\in\mathbb{R}$, $g(r,t)<0$ if $t<-1$ or $t>1$
and $|g(r,t)|+|g_t(r,t)|$ is bounded on $[0, k]\times [-2,2]$ for
any $k>0$. Then $u$ is radial, i.e., $u(x)=u(|x|)$.
\end{proposition}

The proof of the above proposition can be found in
\cite[Proposition 2.6]{DS}.


\begin{lemma}\label{seed3}
Let $0<\eta<1$ be any fixed constant and $w$ satisfies
\begin{gather*}
-w_{zz}=w(1-w^2) \quad \text{on } \mathbb{R}, \\
w(0)=-1+\eta\quad \mbox{(resp. $w(0)=1-\eta$)},  \\
w(z)\le -1+\eta\quad\mbox{(resp. $w(z)\ge 1-\eta$)} \quad
\text{for }z\le 0, \\
w \mbox{ is bounded on } \mathbb{R}.
\end{gather*}
Then $w$ is a unique solution of
\begin{gather*}
-w_{zz}=w(1-w^2) \quad\text{on } \mathbb{R}, \\
w(0)=-1+\eta\quad\mbox{(resp. $w(0)=1-\eta$)},  \\
w'(z)>0\quad \mbox{(resp. $w'(z)<0$)}\quad  z\in\mathbb{R}, \\
w(z)\to\pm 1\quad\mbox{(resp. $w(z)\to \mp 1$)} \quad\text{as }
 z\to\pm\infty.
\end{gather*}
\end{lemma}

The proof of the above lemma  can be found in \cite{Ma}.
Now we prove the rest of Theorem \ref{thm1.1}.

\begin{proof}[Proof of Theorem \ref{thm1.1}]
For the sake of simplicity, we prove for the case where $a(r)<0$ on
$[0,r_1)$, $a(r)=0$ on $[r_1, r_2]$ and $a(r)>0$ on $(r_2,1]$
for some $0<r_1<r_2<1$ (see Figure 1 in Section 1).

\subsection*{Part 1.}
First we show that $u_{\varepsilon}$ converges uniformly near the
boundary of $B_1(0)$, that is, $u_{\varepsilon}\to -1$ uniformly on
$\overline{B_1(0)}\backslash B_{r_2+\tau}(0)$ for any small $\tau>0$.
We note that we have $u_{\varepsilon}\to -1$ uniformly on
$\overline{B_{1-\tau}(0)}\backslash B_{r_2+\tau}(0)$ as $\varepsilon\to 0$.
Now we claim that
$u_{\varepsilon}(r)\le u_{\varepsilon}(1-\tau)=:T_{\varepsilon}$
for $r\in [1-\tau, 1]$. We define the function $\tilde{u}_{\varepsilon}$
by
\[
\tilde{u}_{\varepsilon}(r)=\begin{cases}
u_{\varepsilon}(r) & \mbox{if $r\in [0, 1-\tau]$} \\
u_{\varepsilon}(r) & \mbox{if $u_{\varepsilon}(r)<T_{\varepsilon}$
and $r\in [1-\tau, 1]$}, \\
T_{\varepsilon} & \mbox{if $u_{\varepsilon}(r)\ge T_{\varepsilon}$
and $r\in [1-\tau, 1]$}.
\end{cases}
\]
We note that $\tilde{u}_{\varepsilon}\in H^1(B_1(0))$ and
$-F(r,T_{\varepsilon})\le -F(r,t)$ for $\varepsilon>0$ and $|r-1|$
small and $t\ge T_{\varepsilon}$. Hence we obtain
$J_{\varepsilon}(\tilde{u}_{\varepsilon})<J_{\varepsilon}(u_{\varepsilon})$
and we have a contradiction if we assume that the measure of the set
$\{r\in [0,1]|u_{\varepsilon}(r)>T_{\varepsilon}\ {\rm and}\
r\in [1-\tau, 1]\}$ is positive. Hence
$-1<u_{\varepsilon}(r)\le T_{\varepsilon}$ and $u_{\varepsilon}\to -1$
uniformly on $\overline{B_1(0)}\backslash B_{r_2+\tau}(0)$.

\subsection*{Part 2.}
We remark that, by Proposition \ref{prop3.1}, $u_{\varepsilon}$ is radially
symmetric and we note that for any $t_2>t_1$, $u_{\varepsilon}$ is a
minimizer of the following problem
\[
\inf\{J_{\varepsilon}(u, B_{t_2}(0)\overline{\backslash B_{t_1}(0)}):
u-u_{\varepsilon}\in H^1_0(B_{t_2}(0)\overline{\backslash B_{t_1}(0)})\},
\]
where
\[
J_{\varepsilon}(u,M)=\int_M\frac{\varepsilon^2}{2}|\nabla u|^2-F(|x|,u)dx
\]
for any open set $M$. Let $m_{\varepsilon,t_1,t_2}$ be the minimum value
of this minimization problem.

In this part we show that $u_{\varepsilon}$ has exactly one layer near
the interval $[r_1, r_2]$.
\smallskip

\noindent{\bf Step 2.1.} First we estimate the energy of transition layer.
Let $\eta>0$ and $\theta>0$ be small numbers. Since $u_{\varepsilon}\to 1$
uniformly on $[0, r_1-\theta]$ and $u_{\varepsilon}\to -1$
uniformly on $[r_2+\theta,1-\theta]$, we can find
$\overline{r}_{\varepsilon}\in (r_1-\theta, r_2+\theta)$ such that
$u_{\varepsilon}(r)\ge 1-\eta$ if $r\in [0, \overline{r}_{\varepsilon}]$,
 $u_{\varepsilon}(r)<1-\eta$ for $r-\overline{r}_{\varepsilon}>0$ small.
Let $\tilde{r}_{\varepsilon}>\overline{r}_{\varepsilon}$ be such
that $u_{\varepsilon}(r)\le\eta$ if $r\in [\tilde{r}_{\varepsilon},1-\theta]$,
 $u_{\varepsilon}(r)>\eta$ for $\tilde{r}_{\varepsilon}-r>0$ small.
We may assume that $\overline{r}_{\varepsilon}\to \overline{r}\in [r_1, r_2]$
and $\tilde{r}_{\varepsilon}\to\tilde{r}\in[r_1,r_2]$

We employ the so-called blow-up argument. Let $v_{\varepsilon}(t)=u_{\varepsilon}(\varepsilon t+\overline{r}_{\varepsilon})$. Then
\[
-v_{\varepsilon}''-\varepsilon\frac{N-1}{\varepsilon
t+\overline{r}_{\varepsilon}}v_{\varepsilon}'=f(\varepsilon
t+\overline{r}_{\varepsilon},v_{\varepsilon}),
\]
$-1\le v_{\varepsilon}\le 1$ and $v_{\varepsilon}(0)=1-\eta$. Since
$\overline{r}_{\varepsilon}\to\overline{r}\in [r_1, r_2]$, it is
easy to see that $v_{\varepsilon}\to v$ in $C^1_{\rm loc}(\mathbb{R})$ and
\[
-v''=h(\overline{r})^2(v-v^3),\quad  t\in\mathbb{R}.
\]
and $v(t)\ge 1-\eta$ for $t\le 0$. If we set $v(t)=V(h(\overline{r})t)$,
the function $V(t)$ satisfies
\begin{equation}
\begin{gathered}
-V''=V-V^3 \quad \text{on } \mathbb{R}, \\
V(0)=1-\eta,  \\
V'(t)\ge 1-\eta \quad t\le 0.
\end{gathered}
\end{equation}
Hence by Lemma \ref{seed3}, the function $V$ is a unique solution for
\begin{equation}\label{seed}
\begin{gathered}
-V''=V-V^3 \quad\text{on } \mathbb{R}, \\
V(0)=1-\eta,  \\
V'(t)<0 \quad t\le 0. \\
V(t)\to \pm 1 \quad\text{as } t\to\mp\infty.
\end{gathered}
\end{equation}
Thus, we can find an $R>0$ large, such that $v(R)=\eta$. Since
$v_{\varepsilon}\to v$ in $C^1_{\rm loc}(\mathbb{R})$, we can find
an $R_{\varepsilon}\in (R-1, R+1)$, such that
$v_{\varepsilon}'(r)<0$ if $r\in [0, R_{\varepsilon}]$ and
$v_{\varepsilon}(R_{\varepsilon})=-1+\eta$. Hence
$u_{\varepsilon}'(r)<0$ if $r\in[\overline{r}_{\varepsilon},
\overline{r}_{\varepsilon}+\varepsilon R_{\varepsilon}]$ and
$u_{\varepsilon}(\overline{r}_{\varepsilon}+\varepsilon
R_{\varepsilon})=-1+\eta$. Then we have
\begin{equation}\label{1lay}
\begin{aligned}
&  J_{\varepsilon}(u_{\varepsilon}, B_{\overline{r}_{\varepsilon}
+\varepsilon R_{\varepsilon}}(0)\backslash
\overline{B_{\overline{r}_{\varepsilon}}(0)})  \\
&=\omega_{N-1}(\overline{r}_{\varepsilon}^{N-1}+o_{\varepsilon}(1))
 \int_{\overline{r}_{\varepsilon}}^{\overline{r}_{\varepsilon}
 +\varepsilon R_{\varepsilon}}\left(\frac{\varepsilon^2}{2}|u_{\varepsilon}'|^2
  -F(t, u_{\varepsilon})\right)dt \\
&=\omega_{N-1}(\overline{r}_{\varepsilon}^{N-1}+o_{\varepsilon}(1))
  \varepsilon\int_0^{R_{\varepsilon}}\left(\frac{1}{2}|v_{\varepsilon}'|^2
  -F(\varepsilon t+\overline{r}_{\varepsilon}, v_{\varepsilon})\right)dt  \\
&=\omega_{N-1}(\overline{r}_{\varepsilon}^{N-1}+o_{\varepsilon}(1))
(\beta_{h(\overline{r})}+O(\eta)+o_{\varepsilon}(1))\varepsilon,
\end{aligned}
\end{equation}
where $\omega_{N-1}$ is the area of the unit sphere in $\mathbb{R}^N$,
$o_{\varepsilon}(1)\to 0$ as $\varepsilon\to 0$, $\beta_{h(s)}$ is the
positive value defined by
\begin{align*}
\beta_{h(s)}
&=\int_{-\infty}^{+\infty}
\Big(\frac{1}{2}|w_{h(s)}'(t)|^2+h(s)^2\frac{(w_{h(s)}^2-1)^2}{4}\Big)dt \\
&=h(s)\int_{-\infty}^{+\infty}\frac{1}{2}|V'(t)|^2+\frac{(V(t)^2-1)^2}{4}dt \\
&=h(s)\beta_1
\end{align*}
and $w_{h(s)}(t)=V(h(s)t)$ for $s\in[0,1]$. We note that although the
function $V$ depends on $\eta$, the value
\[
\beta_1=\int_{-\infty}^{+\infty}\frac{1}{2}|V'(t)|^2+\frac{(V(t)^2-1)^2}{4}dt
\]
is independent of $\eta$. \smallskip


\noindent{\bf Step 2.2.} We claim $u_{\varepsilon}$ has exactly one layer
near the interval $[r_1, r_2]$. To show $u_{\varepsilon}$ has exactly one
layer near the interval $[r_1, r_2]$, it sufficient to prove the following
claim

\noindent{\bf Claim.}
$\tilde{r}_{\varepsilon}=\overline{r}_{\varepsilon}
+\varepsilon R_{\varepsilon}$.

Suppose that the claim is not true. Then we can find a
$t_{\varepsilon}>\overline{r}_{\varepsilon}+R_{\varepsilon}\varepsilon$
such that $u_{\varepsilon}(r)<-1+\eta$ if $r\in
(\overline{r}_{\varepsilon}+R_{\varepsilon}\varepsilon,t_{\varepsilon})$,
$u_{\varepsilon}(t_{\varepsilon})=-1+\eta$. Thus we can use the
blow-up argument again at $t_{\varepsilon}$ to deduce that there
is a $\tilde{t}_{\varepsilon}=t_{\varepsilon}+\varepsilon
\tilde{R}_{\varepsilon}$ with $u_{\varepsilon}'(r)>0$ if $r\in
(t_{\varepsilon}, \tilde{t}_{\varepsilon})$,
$u_{\varepsilon}(\tilde{t}_{\varepsilon})=1-\eta$. We may assume
that $t_{\varepsilon},\tilde{t}_{\varepsilon}\to\overline{t}$ as
$\varepsilon\to 0$ for some $\overline{t}\in [r_2, r_3]$. Moreover
\begin{equation}\label{2lay}
J_{\varepsilon}(u_{\varepsilon}, B_{\tilde{t}_{\varepsilon}}(0)\backslash \overline{B_{t_{\varepsilon}}(0)})=\omega_{N-1}(t_{\varepsilon}^{N-1}+o_{\varepsilon}(1))(\beta_{h(\overline{t})}+O(\eta))\varepsilon+o_{\varepsilon}(1)
\end{equation}
Now we claim $\tilde{t}_{\varepsilon}\ge r_1$.
Suppose $\tilde{t}_{\varepsilon}<r_1$.
Let $F_a(t)=\int_{-1}^t (v-a)(1-v^2)dv$. Then for any $t>0$ small
and $s\in[-1+t,1-t]$,
\begin{equation} \label{cont}
\begin{aligned}
& F_a(1-t)-F_a(s)  \\
&=F_0(1-t)-F_0(s)+F_a(1-t)-F_0(1-t)-F_a(s)+F_0(s)   \\
&=\big[\frac{(v^2-1)^2}{4}\big]_{s}^{1-t}-a\int_{s}^{1-t}(1-v^2)dv
\end{aligned}
\end{equation}
Thus it follows from \eqref{cont} that if $a<0$, then
\begin{equation}\label{cont2}
F_a(1-t)-F_a(s)>0
\end{equation}
for $s\in [-1+t,1-t]$. Define
\[
\overline{u}_{\varepsilon}(r):=\begin{cases}
1-\eta & r\in [\overline{r}_{\varepsilon},\overline{r}_{\varepsilon}
+R_{\varepsilon}\varepsilon]\cup[t_{\varepsilon},\tilde{t}_{\varepsilon}], \\
-u_{\varepsilon}(r) & r\in [\overline{r}_{\varepsilon}
+R_{\varepsilon}\varepsilon, t_{\varepsilon}].
\end{cases}
\]
By the assumption that $\tilde{t}_{\varepsilon}<r_1$ and using
(\ref{cont2}), we see $F(r,u_{\varepsilon})<F(r,\overline{u}_{\varepsilon})$
if $r\in[\overline{r}_{\varepsilon}, \tilde{t}_{\varepsilon}]$.
Hence, we obtain
\[
 J_{\varepsilon}(\overline{u}_{\varepsilon}, B_{\tilde{t}_{\varepsilon}}(0)
\backslash \overline{B_{\overline{r}_{\varepsilon}}(0)})
<J_{\varepsilon}(u_{\varepsilon}, B_{\tilde{t}_{\varepsilon}}(0)\backslash
\overline{B_{\overline{r}_{\varepsilon}}(0)}).
\]
Thus we obtain a contradiction. Therefore we have that
$\tilde{t}_{\varepsilon}\ge r_1$.

Since $a(r)\ge 0$ for $r\in [r_1,1]$, we see $F(r,t)\le F(r, -1)=0$
if $r\in [r_1, 1]$. Since $u_{\varepsilon}(r)\in (-1, -1+\eta)$ for
$r\in[\overline{r}_{\varepsilon}+R_{\varepsilon}\varepsilon, t_{\varepsilon}]$,
 we have
\begin{equation}\label{lower}
\begin{aligned}
     m_{\varepsilon,\overline{r}_{\varepsilon},\tilde{r}_{\varepsilon}}
&= J_{\varepsilon}(\overline{u}_{\varepsilon}, B_{\overline{r}_{\varepsilon}
+\varepsilon R_{\varepsilon}}(0)\backslash
\overline{B_{\overline{r}_{\varepsilon}}(0)})
+J_{\varepsilon}(\overline{u}_{\varepsilon}, B_{\tilde{t}_{\varepsilon}}(0)
\backslash \overline{B_{t_{\varepsilon}}(0)})   \\
&\quad +J_{\varepsilon}(\overline{u}_{\varepsilon}, B_{t_{\varepsilon}}(0)
\backslash \overline{B_{\overline{r}_{\varepsilon}
+\varepsilon R_{\varepsilon}}(0)})+J_{\varepsilon}(\overline{u}_{\varepsilon},
 B_{\tilde{r}_{\varepsilon}}(0)\backslash
\overline{B_{\tilde{t}_{\varepsilon}}(0)})  \\
&\ge \omega_{N-1}(\overline{r}_{\varepsilon}^{N-1}\beta_{h(\overline{r})}\varepsilon+t_{\varepsilon}^{N-1}\beta_{h(\overline{t})}\varepsilon)+O(\eta\varepsilon)+o(\varepsilon)  \\
&\quad +\inf\Big\{-\int_{B_{t_{\varepsilon}}(0)\backslash
B_{\overline{r}_{\varepsilon}+\varepsilon R_{\varepsilon}}(0)}F(r, w)
:-1\le w\le\-1+\eta \Big\} \\
&\quad +\inf\Big\{-\int_{B_{\tilde{r}_{\varepsilon}}(0)\backslash
B_{\tilde{t}_{\varepsilon}}(0)}F(r, w):-1 \le w\le 1 \Big\}  \\
&\ge \omega_{N-1}(\overline{r}_{\varepsilon}^{N-1}\beta_{h(\overline{r})}
\varepsilon+t_{\varepsilon}^{N-1}\beta_{h(\overline{t})}\varepsilon)
+O(\eta\varepsilon)+o(\varepsilon)
\end{aligned}
\end{equation}
Now we give an upper bound for
$m_{\varepsilon,\overline{r}_{\varepsilon}, \tilde{r}_{\varepsilon}}$.
Let $R>0$ be such that $V(h(\overline{r})R)=\eta$, where $V$ is a unique
solution to (\ref{seed}). Define $\overline{u}_{\varepsilon}$ by
\begin{equation}\label{testfunction}
\overline{u}_{\varepsilon}(r):=\begin{cases}
V(h(\overline{r})\frac{r-\overline{r}_{\varepsilon}}{\varepsilon})
& r\in [\overline{r}_{\varepsilon},\overline{r}_{\varepsilon}+\varepsilon R] \\
 -1+\eta-\frac{\eta}{\varepsilon}(r-\overline{r}_{\varepsilon}-\varepsilon R)
& r\in [\overline{r}_{\varepsilon}+\varepsilon R, \overline{r}_{\varepsilon}
 +\varepsilon R+\varepsilon] \\
-1  & r\in[\overline{r}_{\varepsilon}+\varepsilon R+\varepsilon,
  \tilde{r}_{\varepsilon}-\varepsilon] \\
-1+\frac{\eta}{\varepsilon}(r-\tilde{r}_{\varepsilon}+\varepsilon)
& r\in[\tilde{r}_{\varepsilon}-\varepsilon,\tilde{r}_{\varepsilon}]
\end{cases}
\end{equation}
Now we note that $|F(r,t)|=O(\eta)$ for
$r\in [\overline{r}_{\varepsilon},\tilde{r}_{\varepsilon}]$ and
$-1\le t\le -1+\eta$ . Then we have
\begin{equation} \label{upper}
\begin{aligned}
m_{\varepsilon,\overline{r}_{\varepsilon},\tilde{r}_{\varepsilon}}
&\le J_{\varepsilon}(\overline{u}_{\varepsilon},
B_{\tilde{r}_{\varepsilon}}(0)\backslash
\overline{B_{\overline{r}_{\varepsilon}}(0)})  \\
&\le J_{\varepsilon}(\overline{u}_{\varepsilon}, B_{\overline{r}_{\varepsilon}
+R\varepsilon}(0)\backslash \overline{B_{\overline{r}_{\varepsilon}}(0)})
+J_{\varepsilon}(\overline{u}_{\varepsilon}, B_{\tilde{r}_{\varepsilon}}(0)
\backslash \overline{B_{\tilde{r}_{\varepsilon}-\varepsilon}(0)}) \\
&\quad +J_{\varepsilon}(\overline{u}_{\varepsilon}, B_{\tilde{r}_{\varepsilon}
-\varepsilon}(0)\backslash \overline{B_{\overline{r}_{\varepsilon}
+\varepsilon R}(0)})  \\
&\le \omega_{N-1}\overline{r}_{\varepsilon}^{N-1}(\beta_{h(\overline{r})}
+O(\eta))\varepsilon+o(\varepsilon)+O(\varepsilon\eta)+o(\varepsilon)  \\
&= \omega_{N-1}\overline{r}_{\varepsilon}^{N-1}\beta_{h(\overline{r})}
+O(\eta\varepsilon)+o(\varepsilon)
\end{aligned}
\end{equation}
By (\ref{lower}) and (\ref{upper}), we have
\[
\omega_{N-1}(\overline{r}_{\varepsilon}^{N-1}\beta_{h(\overline{r})}
+t_{\varepsilon}^{N-1}\beta_{h(\overline{t})})\varepsilon\le
\omega_{N-1}\overline{r}_{\varepsilon}^{N-1}\beta_{h(\overline{r})}
\varepsilon+O(\varepsilon\eta)+o(\varepsilon)
\]
This is a contradiction. So we can conclude
$\tilde{r}_{\varepsilon}=\overline{r}_{\varepsilon}
+\varepsilon R_{\varepsilon}$.

\subsection*{Part 3.}
It remains to prove that if $\overline{r}_{\varepsilon_j}\to\overline{r}$
for some positive sequence $\{\varepsilon_j\}$ converging to zero as
$j\to\infty$ then $\overline{r}$ satisfies
\[
\overline{r}^{N-1}h(\overline{r})=\min_{s\in [r_1, r_2]}s^{N-1}h(s).
\]
{\bf Step 3.1.} First we note that from Part 1, the function
$u_{\varepsilon}$ satisfies $-1\le u_{\varepsilon}\le -1+\eta$ for
$r\in [\overline{r}_{\varepsilon}+\varepsilon R_{\varepsilon},1]$
in this case.

\noindent{\bf Step 3.2.} Set $H(s)=s^{N-1}h(s)$. Assume that the result is
not true. Then there exists a subsequence of
$\{\overline{r}_{\varepsilon}\}$ (denoted by
$\overline{r}_{\varepsilon}$) such that
$\overline{r}_{\varepsilon}\to r'\in [r_1, r_2]$ and
$H(r')>\min_{s\in [r_1, r_2]}H(s)$. Then we can find a point
$\overline{t}\in (r_1, r_2)$ such that $H(r')>H(\overline{t})$.

Now we give a lower estimate for $J_{\varepsilon}(u_{\varepsilon})$.
We have
\begin{equation} \label{lower21}
J_{\varepsilon}(u_{\varepsilon})
=J_{\varepsilon}(u_{\varepsilon},
 B_{\overline{r}_{\varepsilon}}(0))+J_{\varepsilon}(u_{\varepsilon},
 B_{\overline{r}_{\varepsilon}+\varepsilon R_{\varepsilon}}(0)
 \backslash \overline{B_{\overline{r}_{\varepsilon}}(0)})
+J_{\varepsilon}(u_{\varepsilon}, B_1(0)\backslash
\overline{B_{\overline{r}_{\varepsilon}+R_{\varepsilon}\varepsilon}(0)}).
\end{equation}
First we note that $1-\eta\le u_{\varepsilon}(r)\le 1$  for
$r\le\overline{r}_{\varepsilon}$ and for sufficiently small $\eta>0$,
$-F(r,u)\ge -F(r,1)$ ($u\in [1-\eta, 1]$). We also remark that since
$a(r)<0$ for $r<r_1$ and $a(r)=0$ for $r_1\le r\le r_2$ and $a(r)>0$
for $r>r_2$, we have $-F(r,1)<0$ for $r<r_1$ and $-F(r,1)=0$ for
$r_1\le r\le r_2$ and $-F(r,1)>0$ for $r>r_2$. Hence we have
$-\int_{r_1}^{\overline{r}_{\varepsilon}}r^{N-1}F(r,1)dr\ge 0$ and we
obtain the  estimate
\begin{equation} \label{lower22}
\begin{aligned}
J_{\varepsilon}(u_{\varepsilon}, B_{\overline{r}_{\varepsilon}}(0))
&\ge -\int_0^{\overline{r}_{\varepsilon}}r^{N-1}F(r,u_{\varepsilon})dr  \\
&\ge -\int_0^{\overline{r}_{\varepsilon}}r^{N-1}F(r,1)dr  \\
&= -\int_0^{r_1}r^{N-1}F(r,1)dr-\int_{r_1}^{\overline{r}_{\varepsilon}}r^{N-1}F(r,1)dr  \\
&\ge -\int_0^{r_1}r^{N-1}F(r,1)dr=:A.
\end{aligned}
\end{equation}
Using methods similar to those in the proof of (\ref{1lay}), we obtain
\begin{equation}\label{lower23}
J_{\varepsilon}(u_{\varepsilon},
B_{\overline{r}_{\varepsilon}+R_{\varepsilon}\varepsilon}(0)\backslash
\overline{B_{\overline{r}_{\varepsilon}}(0)})\ge
\omega_{N-1}H(r')\beta_1\varepsilon+O(\eta\varepsilon)+o(\varepsilon).
\end{equation}
Since $-1\le u_{\varepsilon}(r)\le -1+\eta$ for
$r\ge \overline{r}_{\varepsilon}+\varepsilon R_{\varepsilon}$ and for
sufficiently small $\eta>0$, $-F(r,u)\ge-F(r,-1)=0$ ($u\in [-1,-1+\eta]$),
we obtain the  estimate
\begin{equation} \label{lower24}
\begin{aligned}
J_{\varepsilon}(u_{\varepsilon}, B_1(0)\backslash
B_{\overline{r}_{\varepsilon}+R_{\varepsilon}\varepsilon}(0))
&\ge -\int_{\overline{r}_{\varepsilon}+\varepsilon R_{\varepsilon}}^1
  r^{N-1}F(r,u_{\varepsilon})dr  \\
&\ge -\int_{\overline{r}_{\varepsilon}+\varepsilon R_{\varepsilon}}^1
 r^{N-1}F(r,-1)dr=0.
\end{aligned}
\end{equation}
Thus we obtain
\begin{equation}\label{lower25}
J(u_{\varepsilon})\ge
A+\omega_{N-1}H(r')\beta_1\varepsilon+O(\eta\varepsilon)+o(\varepsilon).
\end{equation}
Next we give an upper bound for $J_{\varepsilon}(u_{\varepsilon})$.
Consider the  function
\[
\overline{w}_{\varepsilon}(r):=\begin{cases}
1 & r\in [0,\overline{t}-\varepsilon] \\
1-\frac{\eta}{\varepsilon}(r-\overline{t}+\varepsilon)
 & r\in[\overline{t}-\varepsilon,\overline{t}] \\
V\big(h(\overline{t})\frac{r-\overline{t}}{\varepsilon}\big)
 & r\in [\overline{t}, \overline{t}+\varepsilon R'] \\
-1-\frac{\eta}{\varepsilon}(r-\overline{t}-\varepsilon R'-\varepsilon)
 & r\in [\overline{t}+\varepsilon R', \overline{t}+\varepsilon R'
 +\varepsilon ] \\
-1 & r\in [\overline{t}+\varepsilon R'+\varepsilon, 1],
\end{cases}
\]
where $R'>0$ is the number satisfying
$V(h(\overline{t})R')=-1+\eta$. Then
\begin{equation}\label{upper21}
J_{\varepsilon}(u_{\varepsilon})\le J_{\varepsilon}
(\overline{w}_{\varepsilon})\le A+\omega_{N-1}H(\overline{t})
\beta_1\varepsilon+O(\eta\varepsilon)+o(\varepsilon).
\end{equation}
By (\ref{lower25}) and (\ref{upper21}) we have a contradiction.
 The proof of Theorem \ref{thm1.1} is complete.
The more complicate case, can be shown by a similar method
(see Remark below).
\end{proof}

\subsection*{Remark}
We briefly show the more complicate case, that is, when $a$ is the
function as in Figure 2. More precisely we set $I_1:=[r_1, r_2]$
and $I_2:=[r_3, r_4]$ and we assume $a>0$ on $[0, r_1)\cup(r_4, 1]$
and $a<0$ on $(r_3, r_4)$.

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.8\textwidth]{fig2}
\end{center}
\caption{Special case of coefficient $a(t)$}
\end{figure}

Let $\eta>0$ and $\theta>0$ be small numbers. As in Part 1, we can
find pairs of numbers
$(\overline{r}_{1, \varepsilon}, \overline{r}_{2,\varepsilon})$ and
$(R_{1,\varepsilon}, R_{\varepsilon,2})$ satisfying
$\overline{r}_{1,\varepsilon}\in(r_1-\theta, r_2+\theta)$,
$\overline{r}_{2,\varepsilon}\in(r_3-\theta, r_4+\theta)$,
$\sup_{\varepsilon}|R_{1,\varepsilon}|<\infty$,
$\sup_{\varepsilon}|R_{2,\varepsilon}|<\infty$ and
\begin{gather*}
u_{\varepsilon}(r)<-1+\eta \quad\text{for } 0<r<\overline{r}_{1,\varepsilon} \\
u_{\varepsilon}(\overline{r}_{1,\varepsilon})=-1+\eta  \\
u_{\varepsilon}(\overline{r}_{1,\varepsilon}+\varepsilon R_{1,\varepsilon})
  =1-\eta  \\
u_{\varepsilon}(r)>1-\eta \quad\text{for } \overline{r}_{1,\varepsilon}
  +\varepsilon R_{1,\varepsilon}<r<\overline{r}_{2,\varepsilon} \\
u_{\varepsilon}(\overline{r}_{2,\varepsilon})=1-\eta  \\
u_{\varepsilon}(\overline{r}_{2,\varepsilon}+\varepsilon R_{2,\varepsilon})
  =-1+\eta  \\
u_{\varepsilon}(r)<-1+\eta \quad\text{for } \overline{r}_{2,\varepsilon}
  +\varepsilon R_{2,\varepsilon}<r<1
\end{gather*}
We assume that $\overline{r}_{1,\varepsilon_j}\to\overline{r}_1\in I_1$
and that $\overline{r}_{2, \varepsilon_j}\to\overline{r}_2\in I_2$ for some
sequence $\{\varepsilon_j\}$ which converges to $0$ as $j\to\infty$.
In this case it is easy to show that the energy of global minimizer
$J(u_{\varepsilon})$ is estimated as follows
\begin{equation}\label{sevest1}
J_{\varepsilon_j}(u_{\varepsilon_j})\ge J_{\varepsilon_j}(u_{\varepsilon_j}, B_{r_2-\varepsilon}(0))+\varepsilon_j \omega_{N-1}H(\overline{r}_2)\beta_1+B+O(\varepsilon_j\eta)+o(\varepsilon_j),
\end{equation}
where $B=-\int_{r_2}^{r_3}r^{N-1}F(r,1)dr$.

Let us assume the result does not hold. Then
$H(\overline{r}_1)>\min_{s\in I_1}H(s)$ or $H(\overline{r}_2)>\min_{s\in I_2}$
hold. We assume $H(\overline{r}_1)=\min_{s\in I_1}$ and
$H(\overline{r}_2)>\min_{s\in I_2}H(s)$. We also assume $r_1=\overline{r}_1$.
We note that if $H(\overline{r}_1)>\min_{s\in I_1}H(s)$ or
$\overline{r}_1\in {\rm int}I_1$, the proof is more easy.

Let we take $\tilde{r}_2\in {\rm int}I_2$ such that
$H(\overline{r}_2)>H(\tilde{r}_2)>\min_{s\in I_2}H(s)$ and consider the
 function
\[
\tilde{u}_{\varepsilon}(r):=\begin{cases}
u_{\varepsilon}(r) \quad\text{on }[0, r_2-\varepsilon) \\
1+\frac{\eta}{\varepsilon}(r-r_2) \quad\text{on }[r_2-\varepsilon, r_2]\\
1 \quad\text{on }[r_2, \tilde{r}_2-\varepsilon] \\
1-\frac{\eta}{\varepsilon}(r-\tilde{r}_2+\varepsilon)
 \quad\text{on }[\tilde{r}_2-\varepsilon, \tilde{r}_2] \\
V\left(h(\tilde{r}_2)\frac{r-\tilde{r}_2}{\varepsilon} \right)
 \quad\text{on }[\tilde{r}_2, \tilde{r}_2+\varepsilon R''] \\
-1-\frac{\eta}{\varepsilon}(r-\tilde{r}_2-\varepsilon R''-\varepsilon)
 \quad\text{on }[\tilde{r}_2+\varepsilon R'', \tilde{r}_2+\varepsilon R''
 +\varepsilon] \\
-1 \quad\text{on }[\tilde{r}_2+\varepsilon R''+\varepsilon, 1],
\end{cases}
\]
where $V$ is the unique solution of (\ref{seed}) and $R''$ is the unique
value such that $V(h(r_1)R'')=-1+\eta$.

Since $u_{\varepsilon}$ is global minimizer, we can estimate the energy
of $J_{\varepsilon}(\tilde{u}_{\varepsilon})$ as follows
\begin{equation}\label{sevest2}
J_{\varepsilon}(u_{\varepsilon})\le J_{\varepsilon}(\tilde{u}_{\varepsilon})
\le J_{\varepsilon}(u_{\varepsilon}, B_{r_2-\varepsilon}(0))
+ \varepsilon\omega_{N-1}H(\tilde{r}_2)\beta_1+B+O(\varepsilon\eta)
+o(\varepsilon).
\end{equation}
Then we have a contradiction from (\ref{sevest1}) and (\ref{sevest2})
by taking $\varepsilon=\varepsilon_j$ and sufficiently large $j$.

\subsection*{Acknowledgments}
The author would like to thank Professor Kazuhiro Kurata for
his valuable advice and help, also to the anonymous referee for the
numerous and useful comments.

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