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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 15, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/15\hfil Entire solutions]
{Entire solutions for a class of $p$-Laplace equations in
$\mathbb{R}^2$}

\author[Z. Zhou\hfil EJDE-2010/15\hfilneg]
{Zheng Zhou}

\address{Zheng Zhou \newline
College of Mathematics and Econometrics,
Hunan  University, Changsha, China}
\email{zzzzhhhoou@yahoo.com.cn}

\thanks{Submitted September 15, 2009. Published January 21, 2010.}
\subjclass[2000]{35J60, 35B05, 35B40} 
\keywords{Entire solution; $p$-Laplace Allen-Cahn equation; \hfill\break\indent 
Variational methods}

\begin{abstract}
 We study the entire solutions of
 the $p$-Laplace equation
 \[
 -\mathop{\rm div}(|\nabla u|^{p-2}\nabla u)+a(x,y)W'(u(x,y))=0, \quad
 (x,y)\in {\mathbb{R}}^2
 \]
 where $a(x,y)$ is a periodic in $x$ and $y$, positive function. Here
 $W:\mathbb{R}\to\mathbb{R}$ is a two well potential. Via variational
 methods,  we show that there is layered solution which is heteroclinic
 in $x$ and periodic in $y$ direction.
\end{abstract}

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

\section{Introduction}

In this paper we consider the $p$-Laplacian Allen-Cahn
equation
\begin{equation}\label{eq1.1}
\begin{gathered}
 -\mathop{\rm div}(|\nabla u|^{p-2}\nabla u)+a(x,y)W'(u(x,y))=0,
\quad (x,y)\in {\mathbb{R}}^2\\
\lim_{x\to \pm\infty}u(x,y)=\pm\sigma \quad
\text{uniformly  w.r.t. }y\in \mathbb{R}.
\end{gathered}
\end{equation}
where we assume $2<p<\infty$ and
\begin{itemize}

\item[(H1)] $a(x,y)$ is H\"older
continuous on ${\mathbb{R}}^2$, positive and
\begin{itemize}
 \item[(i)] $a(x+1,y)=a(x,y)=a(x,y+1)$.
 \item[(ii)] $a(x,y)=a(x,-y)$.
\end{itemize}
\item[(H2)] $W\in C^2(\mathbb{R})$ satisfies
\begin{itemize}
 \item[(i)] $0=W(\pm\sigma)<W(s)$ for any
 $s\in{\mathbb{R}}\setminus\{\pm\sigma\}$, and
 $W(s)=O(|s\mp\sigma|^{p})$ as
 $s\to\pm\sigma$;
 \item[(ii)] there exists $R_0>\sigma$ such that $W(s)>W(R_0)$ for any
 $|s|>R_0$.
\end{itemize}
\end{itemize}
For example, here we may take
$W(t)=\frac{p-1}{p}|\sigma^2-t^2|^{p}$. This is similar with case
$p=2$, where the typical examples of $W$ are given by
$W(t)=\frac{1}{4}\prod_{i=1}^{k}(t-z_{i})^2$, where
$z_{i}$, $i=1,2,\ldots k<\infty$ are zeros of $W(t)$. The case $p=2$
can be viewed as stationary Allen-Cahn equation introduced in 1979
by Allen and Cahn. We recall that the Allen-Cahn equation is
a model for phase transitions in binary metallic alloys which
corresponds to taking a constant function $a$ and the double well
potential $W(t)$. The function $u$ in these models is considered as
an order parameter describing pointwise the state of the material.
The global minima of $W$ represent energetically favorite pure
phases and different values of $u$ depict mixed configurations.

In 1978, De Giorgi \cite{De} formulated the following question.
Assume $N>1$ and consider a solution $u\in C^2({\mathbb{R}}^{N})$
of the scalar Ginzburg-Laudau equation:
\begin{equation} \label{eq1.2}
\Delta u = u(u^2-1)
\end{equation}
satisfying $|u(x)|\leq1$, $\frac{\partial u}{\partial x_N}>0$  for
every $x=(x', x_N)\in {\mathbb{R}}^{N}$ and
$\lim\limits_{x_N\to\pm\infty} u(x',x_N)=\pm1$. Then the level
sets of $u(x)$ must be hyperplanes; i.e., there exists
$g\in C^2(\mathbb{R})$ such that $u(x)=g(ax'-x_n) $ for some fixed
$a\in{\mathbb{R}}^{N-1}$. This conjecture was first proved for $N=2$
by Ghoussoub and Gui in \cite{GG} and for $N=3$ by Ambrosio
and Cabr\'{e} in \cite{AC}. For $4\leq N\leq8$ and assuming an
additional limiting condition on $u$, the conjecture has been proved
by Savin in \cite{Sa} .

Alessio, Jeanjean and Montecchiari \cite{AJM} studied the equation
$-\triangle u+a(x)W'(u)=0$ and obtained the existence of layered
solutions based on the crucial condition that there is some discrete
structure of the solutions to the corresponding ODE.

In \cite{AJc}, when $a(x,y)>0$ is periodic in $x$ and $y$,  the
authors got the existence of infinite multibump type solutions,
where $a(x,y)=a(x,-y)$ takes an important role \cite{AJc}(see also
\cite{AJc,R,Ra,Rab,RS1,RS2}).

Inherited from the above results, I wonder under what condition
p-Laplace type equation \eqref{eq1.1} would have two dimensional
layered solutions periodical in $y$. Adapting the renormalized
variational introduced in \cite{AJM,AJc} (see also
\cite{Ra,Rab}) to the p-Laplace case, we prove

\begin{theorem}\label{th1.1}
Assume {\rm (H1)--(H2)}. Then there exists entire
solution for \eqref{eq1.1}, which behaves heteroclinic in $x$ and
periodic in $y$ direction.
\end{theorem}


\section{The periodic problem}

 To prove Theorem \ref{th1.1}, we first consider the equation
\begin{equation}\label{eq2.1}
\begin{gathered}
 -\mathop{\rm div}(|\nabla u|^{p-2}\nabla u)+a(x,y)W'(u(x,y))=0,
\quad (x,y)\in {\mathbb{R}}^2\\
u(x,y)=u(x,y+1)\\
\lim_{x\to\pm\infty}u(x,y)=\pm\sigma \quad
\text{uniformly  w.r.t. } y\in \mathbb{R}.
\end{gathered}
\end{equation}
The main feature of this problem is that it has mixed boundary
conditions, requiring the solution to be periodic in the $y$
variable and of the heteroclinic type in the $x$ variable.

Letting $S_0={\mathbb{R}}\times[0,1]$, we look for minima of the
Euler-Lagrange functional
\[
I(u)=\int_{S_0}\frac{1}{p}|\nabla u(x,y)|^p+a(x,y)W(u(x,y))\,dx\,dy
\]
on the class
$$
\Gamma=\{u\in
W_{\rm loc}^{1,p}(S_0):\|u(x,\cdot)\mp\sigma\|_{L^p(0,1)}\to0.\;
x\to\pm\infty\}
$$
 where
$\|u(x_1,\cdot)-u(x_2,\cdot)\|_{L^p(0,1)}^p=\int_0^1
|u(x_1,y)-u(x_2,y)|^pdy$.
 Setting
\begin{gather*}
\Gamma_p=\{u\in\Gamma:u(x,0)=u(x,1)\text{for a.e. }
x\in{\mathbb{R}}\} \\
c_p=\inf_{\Gamma_p} I\quad \text{and}\quad
 {\mathcal {K}}_p=\{u\in\Gamma_p : I(u)=c_p\}
 \end{gather*}

Then we use the reversibility assumption (H1)-(ii) to show that
the minima $c$ on $\Gamma$ equals minima $c_p$ on $\Gamma_p$, and so
solutions of \eqref{eq2.1}.

Note the assumptions on $a$ and $W$ are sufficient to prove that $I$
is lower semicontinuous with respect to the weak convergence in
$W_{\rm loc}^{1,p}(S_0)$; i.e., if $u_n\to u$ weakly in
$W_{\rm loc}^{1,p}(\Omega)$ for any $\Omega$ relatively compact in
$S_0$, then $I(u)\leq\liminf_{n\to\infty}I(u_n)$. Moreover we have

\begin{lemma} \label{lem2.1}
If $(u_n)\subset W_{\rm loc}^{1,p}(S_0)$ is such that $u_n\to u$
weakly in $W_{\rm loc}^{1,p}(S_0)$ and $I(u_n)\to I(u)$, then
$I(u)\leq\liminf_{n\to\infty}u_n$ and
\begin{gather*}
\int_{S_0}a(x,y)W(u_n)\,dx\,dy\to\int_{S_0}a(x,y)W(u)\,dx\,dy\\
\int_{S_0}|\nabla u_n|^p\,dx\,dy\to\int_{S_0}|\nabla u|^p\,dx\,dy
\end{gather*}
\end{lemma}

\begin{proof}
Since $u_n\to u$ weakly in $W_{\rm loc}^{1,p}(S_0)$,
$\|\nabla u\|_{L^p(S_0)}\leq\liminf_{n\to\infty} \|\nabla
u_n\|_{L^p(S_0)}$ by the lower semicontinuous of the norm. By
compact embedding theorem, we have $u_n\to u$ in
$L_{\rm loc}^p(S_0)$, using pointwise convergence and Fatou lemma, we
have
$\int_{S_0}a(x,y)W(u)\,dx\,dy\leq\liminf_{n\to\infty}
\int_{S_0}a(x,y)W(u_n)\,dx\,dy$, then
\begin{align*}
\int_{S_0}a(x,y)W(u)\,dx\,dy&\leq \limsup_{n\to\infty}
\int_{S_0}a(x,y)W(u_n)\,dx\,dy\\
&= \limsup_{n\to\infty}\Big[I(u_n)-\int_{S_0}\frac{1}{p}
|\nabla u_n|^p\,dx\,dy\Big]\\
&= I(u)-\liminf_{n\to\infty}\int_{S_0}\frac{1}{p}
|\nabla u_n|^p\,dx\,dy\\
&\leq \int_{S_0}a(x,y)W(u)\,dx\,dy.
\end{align*}
Thus, $\int_{S_0}a(x,y)W(u_n)\,dx\,dy\to\int_{S_0}a(x,y)W(u)\,dx\,dy$,
and since $I(u_n)\to I(u)$, we have
$\int_{S_0}|\nabla u_n|^p\,dx\,dy\to\int_{S_0}|\nabla u|^p\,dx\,dy$.
\end{proof}

By Fubini's Theorem, if $u\in W_{\rm loc}^{1,p}(S_0)$, then
$u(x,\cdot)\in W^{1,p}(0,1)$, and for all $x_1,x_2\in{\mathbb{R}}$,
we have
\begin{align*}
\int_0^1|u(x_1,y)-u(x_2,y)|^pdy
&= \int_0^1|\int_{x_1}^{x_2} \partial_xu(x,y)dx|^pdy\\
&\leq |x_1-x_2|^{p-1}\int_0^1
\int_{x_1}^{x_2}|\partial_xu(x,y)dx|^p\,dx\,dy\\
&\leq pI(u)|x_1-x_2|^{p-1}.
\end{align*}
If $I(u)<+\infty$, the function $x\to u(x,\cdot)$ is H\"older
continuous from a dense subset of $\mathbb{R}$ with values in
$L^p(0,1)$ and so it can be extended to a continuous function on
$\mathbb{R}$. Thus, any function $u\in
W_{\rm loc}^{1,p}(S_0)\cap\{I<+\infty\}$ defines a continuous trajectory
in $L^p(0,1)$ verifying
\begin{equation}
\begin{aligned}
\mathrm{d}(u(x_1,\cdot),u(x_2,\cdot))^p
&= \int_0^1|u(x_1,y)-u(x_2,y)|^pdy \\
&\leq  pI(u)|x_1-x_2|^{p-1}, \forall x_1,x_2\in
\mathbb{R}.
\end{aligned}\label{eq2.2}
\end{equation}

\begin{lemma}\label{lem2.2}
For all $r>0$, there exists $\mu_r>0$, such that if
$u\in W_{\rm loc}^{1,p}(S_0)$ satisfies
$\min\|u(x,\cdot)\pm\sigma\|_{W^{1,p}(0,1)}\geq r$ for a.e.
$x\in (x_1,x_2)$, then
\begin{equation}\label{eq2.3}
\begin{aligned}
&\int_{x_1}^{x_2}\Big[\int_0^1\frac{1}{p}|\nabla
u|^p+a(x,y)W(u(x,y))dy\Big]dx\\
&\geq \frac{1}{p(x_2-x_1)^{p-1}}\mathrm{d}(u(x_1,\cdot),u(x_2,\cdot))^p
+\frac{p-1}{p}\mu_r^{\frac{p}{p-1}}(x_2-x_1)\\
&\geq \mu_r\mathrm{d}(u(x_1,\cdot),u(x_2,\cdot))
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
We define  the functional
\[
F(u(x,\cdot))=\int_0^1\frac{1}{p}|\partial_yu(x,y)|^p+
\underline{a}W(u(x,y))dy
\]
on $W^{1,p}(0,1)$, where
$\underline{a}=\min_{{\mathbb{R}}^2}a(x,y)>0$. To prove the lemma,
we first to claim that:
\par
 For any $r>0$,  there exists $\mu_r>0$, such that if $q(y)\in
W^{1,p}(0,1)$ is such that $\min\|q(y)\pm\sigma\|_{W^{1,p}(0,1)}\geq
r$, then$ F(q(y))\geq\frac{p-1}{p}\mu_r^{\frac{p}{p-1}}$. Namely, if
$q_n(\cdot)\in W^{1,p}(0,1)$ and $F(q_n)\to0$, then
$\min\|q_n\pm\sigma\|_{W^{1,p}(0,1)}\to0$.

Assume by contradiction that if $F(q_n)\to0$ and
$\min\|q_n\pm\sigma\|_{L^\infty(0,1)}\geq\varepsilon_0>0$. Then
there exists a sequence $(y_n^1)\subset[0,1]$ such that
$\min|q_n(y_n^1)\pm\sigma|\geq\varepsilon_0$. Since
$\int_0^1\underline{a}W(q_n)dy\to0$ there exists a sequence
$(y_n^2)\subset[0,1]$ such that
$|q_n(y_n^2)\pm\sigma|<\frac{\varepsilon_0}{2}$. Then
\begin{align*}
\frac{\varepsilon_0}{2}
&\leq |q_n(y_n^2)-q_n(y_n^1)|\\
&\leq |\int_{y_n^1}^{y_n^2}|\dot{q}_n(t)|dt~|\\
&\leq |y_n^2-y_n^1|^{1-\frac{1}{p}}\Big[\int_0^1|\dot{q}_n(t)|^pdt\Big]
^{1/p}\\
&\leq p^{\frac{1}{p}}(F(q_n))^{1/p}\to0.
\end{align*}
It is a contradiction.

Since $\min\|q_n\pm\sigma\|_{L^\infty(0,1)}\to0$ as $F(q_n)\to0$,
then $\int_0^1|\dot{q}_n(y)|^pdy\to0$, and it follows that
$\|q_n-\sigma\|_{W^{1,p}(0,1)}\to0$ as $F(q_n)\to0$.

Observe that if $(x_1,x_2)\subset\mathbb{R}$ and $u\in
W_{\rm loc}^{1,p}(S_0)$ are such that
$F(u(x,\cdot))\geq\frac{p-1}{p}\mu_r^{\frac{p}{p-1}}$ for a.e.
$x\in(x_1,x_2)$, by H\"older's and Yung's inequalities we
have
\begin{align*}
& \int_{x_1}^{x_2}\Big[\int_0^1\frac{1}{p}|\nabla
u|^p+a(x,y)W(u(x,y))dy\Big]dx\\
&\geq \int_{x_1}^{x_2}\int_0^1\frac{1}{p}|\partial_xu|^p\,dy\,dx+
\int_{x_1}^{x_2}\int_0^1\frac{1}{p}|\partial_yu|^p+\underline{a}W(u)\,dy\,dx\\
&= \frac{1}{p}\int_0^1\int_{x_1}^{x_2}|\partial_xu|^p\,dx\,dy+
\int_{x_1}^{x_2}F(u(x,\cdot))dx\\
&\geq \frac{1}{p(x_2-x_1)^{p-1}}\mathrm{d}(u(x_1,\cdot),u(x_2,\cdot))^p
+\frac{p-1}{p}\mu_r^{\frac{p}{p-1}}(x_2-x_1)\\
&\geq \mu_r\mathrm{d}(u(x_1,\cdot),u(x_2,\cdot)).
\end{align*}
The proof is complete.
\end{proof}

As a direct consequence of Lemma \ref{lem2.2}, we have the following
result.

\begin{lemma}\label{lem2.3}
If $u\in W_{\rm loc}^{1,p}(S_0)\cap \{I<+\infty\}$, then
$\mathrm{d}\big(u(x,\cdot),\pm\sigma\big)\to0$ as $x\to\pm\infty$.
\end{lemma}

\begin{proof}
Note that since
\[
I(u)=\int_{S_0}\frac{1}{p}|\nabla
u|^p+a(x,y)W(u(x,y))\,dx\,dy<+\infty,
\]
$W(u(x,y))\to0$ as $|x|\to+\infty$. Then by Lemma \ref{lem2.2},
$\liminf_{x\to+\infty}\mathrm{d}\big(u(x,\cdot),\sigma\big)=0$.
Next we show that $\limsup_{x\to+\infty} \mathrm{d}\big(u(x,\cdot),
\sigma\big)=0$ by contradiction. We assume that there exists
$r\in(0,\sigma/4)$ such that
$\limsup_{x\to+\infty}\mathrm{d}(u(x,\cdot),\sigma)>2r$,
by \eqref{eq2.2} there exists infinite
intervals $(p_i,s_i),i\in \mathbb{N}$ such that
$\mathrm{d}\big(u(p_i,\cdot),\sigma\big)=r$,
$\mathrm{d}\big(u(s_i,\cdot),\sigma\big)=2r$ and
$r\leq \mathrm{d}\big(u(x,\cdot),\sigma\big)\leq2r$ for
 $x\in\cup_i(p_i,s_i)$,
$i\in \mathbb{N}$ by Lemma \ref{lem2.2}, this implies
$I(u)=+\infty$, it's a contradiction. Similarly, we can prove that
$\lim_{x\to-\infty}\mathrm{d}\big(u(x,\cdot),-\sigma\big)=0$.
\end{proof}

Now we consider the functional on the class
$$\Gamma=\{u\in W_{\rm loc}^{1,p}(S_0):I(u)<+\infty,
~{\rm d}\big(u(x,\cdot),\pm\sigma\big)\to0{\rm
~as~x\to\pm\infty}\}
$$
Let
\begin{equation}\label{eq2.4}
c=\inf_\Gamma I\quad\text{and}\quad
{\mathcal {K}}= \{u\in\Gamma:I(u)=c\}
\end{equation}
 We will show that $\mathcal {K}$ is not
empty, and we start noting that the trajectory in $\Gamma$ with
action close to the minima has some concentration properties.

For any $\delta>0$, we set
\begin{equation}\label{eq2.5}
\lambda_\delta=\frac{1}{p}\delta^p+\max_{\mathbb{R}^2}a(x,y)\cdot
\max_{|s\pm\sigma|\leq p^{1/p}\delta}W(s).
\end{equation}

\begin{lemma}\label{lem2.4}
There exists $\bar{\delta}_0\in(0,\sigma/2)$ such that for any
$\delta\in(0,\bar{\delta}_0)$ there exists $\rho_\delta>0$ and
$l_\delta>0$, for which, if $u\in\Gamma$ and $I(u)\leq
c+\lambda_\delta$, then
\begin{itemize}
\item[(i)] $\min\|u(x,\cdot)\pm\sigma\|_{W^{1,p}(0,1)}\geq\delta$
for a.e.
$x\in(s,p)$ then $p-s\leq l_\delta$.

\item[(ii)] if $\|u(x_-,\cdot)+\sigma\|_{W^{1,p}(0,1)}\leq\delta$, then
$\mathrm{d}(u(x_-,\cdot),-\sigma)\leq\rho_\delta$ for any $x\leq x_-$,
and if $\|u(x_+,\cdot)-\sigma\|_{W^{1,p}(0,1)}\leq\delta$, then
$\mathrm{d}(u(,\cdot),\sigma)\leq\rho_\delta$ for any $x\geq x_+$.
\end{itemize}
\end{lemma}

\begin{proof}
By Lemma \ref{lem2.2}, as in this case, there exists $\mu_\delta>0$
such that
\[
\int_s^p\int_0^1\frac{1}{p}|\nabla
u|^p+a(x,y)W(u)\,dx\,dy\geq \mu_\delta(p-s).
\]
Since $I(u)\leq c+\lambda_\delta$ there exists $l_\delta<+\infty$
such that $p-s<l_\delta$.

To prove (ii), we first do some preparation, $\mu_{r_\delta}\geq
\frac{p-1}{p}\lambda_\delta$,
$\rho_\delta=\max\{\delta,r_\delta\}
+3(\frac{p-1}{p\mu_{r_\delta}})^\frac{p-1}{p}\lambda_\delta$. Let
$\bar{\delta}_0\in(0,\sigma/2)$ be such that $\rho_\delta<\sigma/2$
for all $\delta\in(0,\bar{\delta}_0)$. Let
$\delta\in(0,\bar{\delta}_0),~u\in\Gamma, I(u)\leq+\infty$ and
$x_-\in\mathbb{R}$ be such that
$\|u(x_-,\cdot)+\sigma\|_{W^{1,p}(0,1)}\leq\delta$.
 Define
\[
u_-(x,y)=\begin{cases}
 -1&\text{if } x<x_--1,\\
 x-x_-+(x-x_-+1)u(x_-,y)& \text{if } x_--1\leq x_-,\\
 u(x,y)& \text{if } x\geq x_-.
\end{cases}
\]
and note that $u_-\in\Gamma$ and $I(u_-)\geq c$, then
$\|u_-+\sigma\|_{W^{1,p}(0,1)}=|x-x_-+1|\cdot\|u(x_-,\cdot)+
\sigma\|_{W^{1,p}(0,1)}\leq\delta$ when $x_--1\leq x\leq x_-$.
Recall that $\|q\|_{L^\infty(0,1)}\leq
p^{1/p}\|q\|_{W^{1,p}(0,1)}$
for any $q\in W^{1,p}(0,1)$, then
$\|u_-+\sigma\|_{L^\infty(0,1)}\leq
p^{1/p}\|u_-+\sigma\|_{W^{1,p}(0,1)}\leq p^{1/p}\delta$,
by definition \eqref{eq2.5} of $\lambda_\delta$, we have
$$
\int_{x_--1}^{x_-}[\int_0^1\frac{1}{p}|\nabla u_-|^p+
a(x,y)W(u_-)dy]dx\leq \lambda_\delta.
$$
Since
\begin{align*}
I(u_-)&= I(u)-\int_{-\infty}^{x_-}\int_0^1\frac{1}{p}|\nabla u|^p+
a(x,y)W(u)\,dy\,dx\\
&\quad +\int_{x_--1}^{x_-}\int_0^1\frac{1}{p}|\nabla u_-|^p+
a(x,y)W(u_-)\,dy\,dx
\end{align*}
we obtain
\begin{equation} \label{eq2.6}
\int_{-\infty}^{x_-}\int_0^1\frac{1}{p}|\nabla u|^p+
a(x,y)W(u)\,dy\,dx\leq 2\lambda_\delta.
\end{equation}
Now, assume by contradiction that there exists $x_1<x_-$ such that
$\mathrm{d}(u(x_1,\cdot),-\sigma)\geq\rho_\delta$, by \eqref{eq2.2}
there exists $x_2\in(x_1,x_-)$ such that
$\mathrm{d}(u(x,\cdot),-\sigma)\geq\max\{\delta,r_\delta\}$ for
$x\in(x_1,x_2)$ and
$\mathrm{d}(u(x_1,\cdot),u(x_1,\cdot))\geq\rho_\delta
-\max\{\delta,r_\delta\}$.
By Lemma \ref{lem2.2}, we have
\[
\int_{-\infty}^{x_-}\int_0^1\frac{1}{p}|\nabla u|^p+ a(x,y)W(u)\,dy\,dx
\geq(\frac{p\mu_{r_\delta}}{p-1})^\frac{p-1}{p}\big(\rho_\delta-
\max\{\delta,r_\delta\}\big)\geq3\lambda_\delta
\]
which contradicts \eqref{eq2.6}.
Thus $\mathrm{d}(u(x,\cdot),-\sigma)\leq\rho_\delta$ for any $x\leq x_-$.
Analogously, we can prove if
$\|u(x_+,\cdot)-\sigma\|_{W^{1,p}(0,1)}\leq\delta$,
then $\mathrm{d}(u(x,\cdot),\sigma)\leq\rho_\delta$ as $x\geq x_+$.
\end{proof}

To exploit the compactness of $I$ on $\Gamma$, we set the function
$X:W_{\rm loc}^{1,p}(S_0)\to{\mathbb{R}}\cup\{+\infty\}$ given by
$$
X(u)=\sup\{x:\mathrm{d}(u(x,\cdot),\sigma)\}\geq\sigma/2.
$$
Setting $\chi(s)=\min|s\pm\sigma|$, by$(H_3)$, there exist
$0<w_1<w_2$ such that
\begin{equation}\label{eq2.7}
  w_1\chi^p(s)\leq W(s)\leq
w_2\chi^p(s){\rm~ when~} \chi(s)\leq\sigma/2.
\end{equation}
Now, we can get the compactness of the minimizing sequence of $I$ in
$\Gamma$.

\begin{lemma}\label{lem2.5}
If $(u_n)\subset\Gamma$ is such that $I(u_n)\to c$ and
$X(u_n)\to X_0\in \mathbb{R}$, then there exists $u_0\in\mathcal {K}$
such that, along a sequence, $u_n\to u_0$ weakly in ${W^{1,p}(S_0)}$.
\end{lemma}

\begin{proof}
We now show that $(u_n)$ is bounded in
$W_{\rm loc}^{1,p}(S_0)$, i.e., $(u_n)$ is bounded in
$L_{\rm loc}^p(S_0),~(\nabla u_n)$ is bounded in $L_{\rm loc}^p(S_0)$.
Since $I(u_n)\to c$and $\int_{S_0}|\nabla u_n|^p\,dx\,dy\leq pI(u_n)$,
we have that $(\nabla u_n)$ is bounded in $L_{\rm loc}^p(S_0)$.
If we can prove that $u_n(x,\cdot)$ is bounded in $L^p(0,1)$ for a.e.
$x\in\mathbb{R}$, then $(u_n)$ is bounded in $L_{\rm loc}^p(S_0)$.

Let $B_r=\{q\in L^p(0,1)/\|q\|_{L^p(0,1)}\leq r\}$, we assume by
contradiction that for any $R>2\sigma$, there exists
$\bar{x}\in\mathbb{R}$ such that $u(\bar{x},\cdot)\notin B_R$ for
$u\in\Gamma\cap\{I(u)\leq c+\lambda\},\lambda>0$, such that
$\|u(\bar{x},\cdot)\|_{L^p(0,1)}\geq R$, then
$\mathrm{d}(u(\bar{x},\cdot),\sigma)\geq\|u(\bar{x},\cdot)\|_{L^p(0,1)}
-\|\sigma\|_{L^p(0,1)}\geq R-\sigma$.
Since $\mathrm{d}(u(x,\cdot),\pm\sigma)\to0$ as $x\to\pm\infty$,
by continuity there exists $x_1>\bar{x}$ such that
$\mathrm{d}(u(x_1,\cdot),\sigma)\leq\sigma/2$ and
$\mathrm{d}(u(x,\cdot),\sigma)\geq\sigma/2$ for $x\in(\bar{x},x_1)$.
Using Lemma 2.2, we get
$$
c+\lambda\geq I(u)\geq \mu_{\sigma/2}\mathrm{d}(u(x_1,\cdot),
u(\bar{x},\cdot))
\geq \mu_{\sigma/2}(R-3\sigma/2).
$$
which is a contradiction for $R$
large enough. We conclude that $(u_n)$ is bounded in
$W_{\rm loc}^{1,p}(S_0)$, thus there exists
$u_0\in W_{\rm loc}^{1,p}(S_0)$
such that up to a sequence, $u_n\to u_0$ weakly in
$W_{\rm loc}^{1,p}(S_0)$. We shall prove that $u_0\in\Gamma$; i.e.,
$\mathrm{d}(u_0(x,\cdot),\pm\sigma)\to0$ as $x\to\pm\infty$. First we
claim that:
\begin{quote}
For any small $\varepsilon>0$, there exists
$\lambda(\varepsilon)\in(0,\lambda_{\bar{\delta}})$ and
$l(\varepsilon)>l_{\bar{\delta}}$ such that if
$u\in\Gamma\cap\{I(u)\leq c+\lambda(\varepsilon)\}$ then
\begin{equation}\label{eq2.8}
\int_{|x-X(u)|\geq l(\varepsilon)}\int_0^1
W(u(x,y))\,dy\,dx\leq\varepsilon.
\end{equation}
\end{quote}
Indeed, let $\delta<\bar{\delta}$ be such that
$3\lambda_\delta\leq\underline{a}w_1\varepsilon$ where
$\underline{a}=\min_{{\mathbb{R}}^2}a(x,y)$. Given any
$u\in\Gamma\cap\{I(u)\leq c+\lambda_\delta\}$, by Lemma
\ref{lem2.4}, there exists $x_-\in(X(u)-l_\delta,X(u))$ and
$x_+\in(X(u),X(u)+l_\delta)$ such that
$\|u(x_-,\cdot)+\sigma)\|_{W^{1,p}(0,1)}\leq\delta$ and
$\|u(x_+,\cdot)-\sigma\|_{W^{1,p}(0,1)}\leq\delta$. We define the
function
\[
\tilde{u}(x,y)=\begin{cases}
 -\sigma& \text{if }  x<x_--1,\\
 \sigma(x-x_-)+(x-x_-+1)u(x_-,y)& \text{if }x_--1\leq x_-,\\
 u(x,y)&\text{if }x_-\leq x\leq x_+,\\
(x_+-x+1)u(x_+,y)+\sigma(x-x_+)&\text{if }x_+\leq x<x_++1,\\
\sigma &\text{if } x>x_++1
\end{cases}
\]
which belongs to $\Gamma$, and $I(\tilde{u})\geq c$,
\begin{align*}
&\int_{|x-X(u)|\geq l_\delta}\int_0^1\frac{1}{p}|\nabla
       u|^p+a(x,y)W(u)\,dy\,dx\\
&\leq  I_{-\infty}^{x_-}(u)+I_{x_+}^{+\infty}(u)\\
&=I(u)-I(\tilde{u})+I_{x_--1}^{x_-}(\tilde{u})
+I_{x_+}^{x_++1}(\tilde{u}) \\
&\leq 3\lambda_\delta
\end{align*}
then \eqref{eq2.8} follows setting $l(\varepsilon)=l_{\bar{\delta}}$
and $\lambda(\varepsilon)=\lambda_\delta$.

 From \eqref{eq2.8} it is easy to see that $u(x,y)\to \sigma$
as $x\to +\infty$. Combining \eqref{eq2.8} and \eqref{eq2.7} we obtain
\[
\int_{|x-X(u)|\geq l(\varepsilon)}\int_0^1
w_1|u(x,y)-\sigma|^p\,dx\,dy
\leq \int_{|x-X(u)|\geq l(\varepsilon)}\int_0^1W(u(x,y))\,dy\,dx
\leq\varepsilon;
\]
i.e., $\mathrm{d}\big(u(x,\cdot),\sigma\big)\to0$ as $x\to+\infty$.
Analogously, we can get that $\mathrm{d}\big(u(x,\cdot),-\sigma\big)\to0$
as $x\to-\infty$, it follows
that $u_0\in\Gamma$.
\end{proof}

As a consequence, we get the following existence result.

\begin{proposition}\label{p2.1}
$\mathcal {K}\neq\emptyset$ and any $u\in\mathcal {K}$ satisfies
$u\in C^{1,\alpha}({\mathbb{R}}^2)$ is a solution of
$-\mathop{\rm div}(|\nabla u|^{p-2}\nabla u)+a(x,y)W'(u(x,y))=0$
on $S_0$ with $\partial_yu(x,0)=\partial_yu(x,1)=0$ for all
$x\in\mathbb{R}$, and
$\|u\|_{L^\infty}(S_0)\leq R_0$. Finally, $u(x,y)\to\pm\sigma$ as
$x\to\pm\infty$ uniformly in $y\in[0,1]$.
\end{proposition}

\begin{proof}
By Lemma \ref{lem2.5}, the set $\mathcal {K}$ is not empty. By
$(H_2)$, $\|u\|_{L^\infty(S_0)}\leq R_0$. Indeed,
$\tilde{u}=\max\{-R_0,\min\{R_0,u\}\}$ is a fortiori minimizer. Let
$\eta\in C_0^\infty(S_0)$ and $\tau\in\mathbb{R}$, then
$u+\tau\eta\in\Gamma$ and since $u\in\mathcal {K}$,
$I(u+\tau\eta)$ is
a $C^1$ function of $\tau$ with a local minima at $\tau=0$.
Therefore,
\[
I'(u)\eta=\int_{S_0}|\nabla u|^{p-2}\nabla u\nabla\eta+aW'(u)\eta
\,dx\,dy=0
\]
for all such $\eta$, namely $u$ is a weak solution of the equation
$-\mathop{\rm div}(|\nabla u|^{p-2}\nabla u)+a(x,y)W'(u(x,y))=0$
on $S_0$.
Standard regularity arguments show that $u\in C^{1,\alpha}(S_0)$ for
some $\alpha\in(0,1)$ and satisfies the Neumann boundary condition
(see \cite{GT}\cite{L2}\cite{To}). Since
$\|u\|_{L^\infty(S_0)}\leq R_0$, there exists $C>0$ such that
$\|u\|_{C^{1,\alpha}(S_0)}\leq C$, which guarantees that $u$
satisfies the boundary conditions.
Indeed, assume by contradiction that $u$ does not verify
$u(x,y)\to-\sigma$ as $x\to-\infty$ uniformly with respect to
$y\in[0,1]$. Then there exists $\delta>0$ and a sequence
$(x_n,y_n)\in S_0$ with $x_n\to-\infty$ and
$|u(x_n,y_n)+\sigma|\geq2\delta$ for all $n\in\mathbb{N}$. The
$C^{1,\alpha}$ estimate of $u$ implies that there exists $\rho>0$
such that $|u(x,y)+\sigma|\geq\delta$ for $\forall\,(x,y)\in
B_\rho(x_n,y_n),n\in\mathbb{N}$. Along a subsequence
$x_n\to-\infty,~y_n\to y_0\in[0,1],~|u(x,y)+\sigma|\geq\delta$ for
$(x,y)\in B_{\rho/2}(x_n,y_0)$, which contradicts with the fact that
$\mathrm{d}(u(x,\cdot),-\sigma)\to0$ as $x\to-\infty$ since
$u\in\Gamma$. The other case is similar.
\end{proof}

We shall explore the reversibility condition of (H1)-(ii), and we
will prove that the minimizer on $\Gamma$ is in fact a solution of
\eqref{eq2.1}.

\begin{lemma} \label{lem2.6}
$c_p=c$.
\end{lemma}

\begin{proof}
Since $\Gamma_p\subset\Gamma$, $c_p\geq c$. Assume by contradiction
that $c_p>c$, then there exists $u\in\Gamma$ such that $I(u)<c_p$.
Writing
\begin{align*}
I(u)&= \int_{\mathbb{R}}\Big[\int_0^{1/2}\frac{1}{p}|\nabla
u|^p+aW(u)dy\Big]dx+\int_{\mathbb{R}}
\Big[\int_{1/2}^1\frac{1}{p}|\nabla u|^p+aW(u)dy\Big]dx\\
&= I_1+I_2
\end{align*}
it follows that $\min\{I_1,I_2\}<\frac{c_p}{2}$. Suppose for example
$I_1<c_p/2$, define
\[
v(x,y)=\begin{cases}
 u(x,y)&\text{if } x\in\mathbb{R}\text{ and }0\leq y\leq\frac{1}{2},\\
 u(x,1-y)&\text{if } x\in\mathbb{R}\text{ and }\frac{1}{2}\leq y\leq1.
\end{cases}
\]
Then $v\in\Gamma_p$, by condition (H1)-(ii), $I(v)=2I_1<c_p$,
this is a contradiction.
\end{proof}

We shall prove that any $u\in\mathcal {K}$ is periodic in $y$.

\begin{lemma} \label{lem2.7}
If $u\in\mathcal {K}$ then $u(x,0)=u(x,1)$ for all $x\in\mathbb{R}$.
\end{lemma}

\begin{proof}
Suppose $u\in\mathcal {K}$ and $v$ as above, then $v(x,y)=u(x,y)$
for $y\in[0,1/2]$. By $(H_1)$-(ii), $I(u)=c=c_p=I(v)$, so
$v\in\mathcal{K}$. Then $u$ and $v$ are solutions of
\begin{equation}\label{eq2.12}
\begin{gathered}
 -\mathop{\rm div}(|\nabla u|^{p-2}\nabla u)+aW'(u(x,y))=0,
\quad \text{on } S_0,\\
\partial_yu(x,0)=\partial_yu(x,1)=0\quad \text{for all }
x\in\mathbb{R}.
\end{gathered}
\end{equation}
Since $u=v$ for $y\in[0,1/2]$, by the principle of unique
continuation (see \cite{BI}), we have $u=v$ in
$\mathbb{R}\times[0,1]$. i.e. $u(x,0)=u(x,1)$.
\end{proof}

\begin{remark} \rm
It is an open problem for the principle for p-harmonic functions in
case $n\geq3$ and $p\neq2$. When $p=\infty$, the principle of unique
continuation does not hold.
\end{remark}

\begin{proof}[Proof of Theorem \ref{th1.1}]
We now extend $u$ periodically in $y$ direction to the entire space
${\mathbb{R}}^2$, and write it as $U(x,y)$. As a consequence of the
above lemmas and proposition \ref{p2.1}, $U(x,y)$ is an entire
solution of \eqref{eq1.1}, which is heteroclinic in $x$ and
1-periodic in $y$ direction.
\end{proof}

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\end{document}