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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 137, pp. 1--14.\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-2010/137\hfil Heteroclinic solutions]
{Heteroclinic solutions to an asymptotically autonomous
second-order equation}

\author[G. S. Spradlin\hfil EJDE-2010/137\hfilneg]
{Gregory S. Spradlin} 

\address{Gregory S. Spradlin\hfill\break
Department of Mathematics\\
Embry-Riddle University\\
Daytona Beach, Florida 32114-3900, USA}
\email{spradlig@erau.edu}

\thanks{Submitted May 21, 2009. Published September 23, 2010.}
\subjclass[2000]{34B40, 34C37}
\keywords{Heteroclinic; non-autonomous equation;
 bounded solution; \hfill\break\indent variational methods}

\begin{abstract}
 We study the differential equation $\ddot{x}(t) = a(t)V'(x(t))$,
 where $V$ is a double-well potential with minima at
 $x = \pm 1$ and $a(t) \to l > 0$ as $|t| \to \infty$.
 It is proven that under certain additional assumptions on $a$,
 there exists a heteroclinic solution $x$ to the differential
 equation with $x(t) \to -1$ as $t \to -\infty$ and
 $x(t) \to 1$ as $t \to \infty$.  The assumptions allow $l-a(t)$
 to change sign for arbitrarily large values of $|t|$, and do not
 restrict the decay rate of $|l-a(t)|$ as $|t| \to \infty$.
\end{abstract}

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

\section{Introduction}

    Consider the autonomous second-order differential equation
\begin{gather}\label{e1.00001}
\ddot{x}(t)= lV'(x(t)), \\
\label{e1.00002}
x(t) \to -1  \text{ as } t \to -\infty, \quad
x(t) \to 1 \text{ as } t \to \infty.
\end{gather}
where $l > 0$, $V \in  C^2(\mathbb{R}, [0, \infty))$, $V(-1) =
V(1) = 0$, and $V > 0$ on $(-1,1)$. The presence of $l$ seems
superfluous at this point; however, we will use it later. It is
easy to show that \eqref{e1.00001}-\eqref{e1.00002} has a
solution: multiply both sides of \eqref{e1.00001} by $\dot{x}(t)$
and integrate, and conclude that $\frac{1}{2}\dot{x}(t)^2 -
lV(x(t))$ is constant.  Assuming that $V(x) \leq c(1 \pm x)^2$ for
some $c > 0$ in a neighborhood of $-1$ and $1$ respectively, then
setting the constant equal to zero, we find that
\eqref{e1.00001}-\eqref{e1.00002} has a solution, which solves the
first-order equation $\dot{x}(t)= \sqrt{2lV(x(t))}$.  That
solution is unique if we impose the condition $x(0) = 0$.  From
now on, we will refer to the unique solution of
\eqref{e1.00001}-\eqref{e1.00002} with $x(0) = 0$ as $\omega$.

The function   $\omega$ can also be characterized as the unique
(modulo translation) minimizer of the functional
%
\begin{equation}\label{e1.00003}
F_l(u) = \int_{-\infty}^\infty \frac{1}{2}\dot{u}(t)^2 - lV(u(t))\,dt
\end{equation}
%
over the affine space
%
\begin{equation}\label{e1.00004}
W = \{u \in W^{1,2}_{\rm loc}(\mathbb{R}) : u + 1 \in
W^{1,2}((-\infty,0]),\; u - 1 \in W^{1,2}([0,\infty))\}.
\end{equation}
%
An interesting problem is to replace $l$ by a nonconstant,
positive coefficient function $a(t)$ and find conditions on
$a$ under which
%
\begin{equation}\label{e1.00005}
\ddot{x}(t)= a(t)V'(x(t))
\end{equation}
%
with \eqref{e1.00002}
has solutions.  We must assume something: note that if $a$ is
continuous and increasing, then if $x$ solves
\eqref{e1.00001}-\eqref{e1.00002}, then
$\frac{1}{2}\dot{x}(t)^2 - a(t)V(x(t)) \to 0$ as
$|t| \to \infty$, but
%
\begin{equation}\label{e1.00006}
\begin{aligned}
\frac{d}{dt}(\frac{1}{2}\dot{x}(t)^2 - a(t)V(x(t)))
&=  \ddot{x}(t)\dot{x}(t) - a(t)V'(x(t))\dot{x}(t) - \dot{a}(t)V(x(t))\\
& = - \dot{a}(t)V(x(t)) <0.
\end{aligned}
\end{equation}
%
This is impossible.

There are many results concerning equations like \eqref{e1.00005}
in which the analogue of $a(t)$ is periodic, and homoclinic,
heteroclinic, and multitransition solutions of the equations are
found.  See \cite{CR1}, \cite{R}.  There seems to be fewer results
for the case
\begin{itemize}
\item[(A1)] $a(t) \to l > 0$ as $|t| \to \infty$
\end{itemize}
In \cite[Chapter 2, Thm.~2.2]{B}, a solution is found for when $0
< a(t) \leq l$ for all $t \in \mathbb{R}$.  In \cite{CT}
(Section~5, Example~1) a solution is found when the coefficient
$a(t)$ is definitively increasing with respect to $|t|$.   In
\cite{G}, a solution is found in the case $l \leq a(t) \leq L$ and
$L$ is suitably bounded from above.  This result is a specific
case of the result proven in this paper and is described more
precisely later.  In \cite{GS}, a solution is found when $a(t)$ is
increasing on $[t_0, \infty)$ and decreasing on $(-\infty, t_0]$
for some $t_0 > 0$ and $l - a(t)$ decays to zero slowly enough as
$|t| \to \infty$.  In this paper, we find conditions on $a$ which
allow $l -a(t)$ to change sign for arbitrarily large $|t|$ and do
not require any assumptions on the monotonicity of $a$ or the
decay rate of $l - a(t)$ as $|t| \to \infty$. In more related
work, in \cite{CR2} heteroclinic orbits to a nonautonomous
differential equation are found that connect stationary points of
different energy levels.  In \cite{BSTT}, heteroclinic solutions
connecting nonconsecutive equilibria of a triple-well potential
are found for a fourth-degree ordinary differential equation.


Let $V$ satisfy
\begin{itemize}
\item[(V1)] $V \in C^2(\mathbb{R},\mathbb{R})$;
\item[(V2)] $V(x) \geq 0$ for all $x \in \mathbb{R}$;
\item[(V3)] $V(-1)= V(1) = 0$;
\item[(V4)] $V > 0$ on $(-1, 1)$;
\item[(V5)] $V''(-1) > 0$, $V''(1) > 0$.
\end{itemize}
Let
\begin{equation}\label{e1.01}
 \xi_- = \min\{x : x > -1, V'(x) = 0\},\quad
 \xi_+ = \max\{x : x < 1, V'(x) = 0\}.
\end{equation}
%
Note that $\xi_-$ and $\xi_+$ are well-defined by (V3)-(V5).
Define
%
\begin{equation}\label{e1.02}
\nu = \min\Big(\int_{-1}^{\xi_-} \sqrt{V(x)}\,dx,\int_{\xi_+}^1
\sqrt{V(x)}\,dx\Big) > 0.
\end{equation}
%
Let $a:\mathbb{R} \to \mathbb{R}$ be a measurable function
satisfying (A1) and
\begin{itemize}
\item[(A2)] $0 < \underline{l} \leq a(t) \leq L \equiv l
+ 4\nu\sqrt{l \underline{l}} /\int_{-1}^1 \sqrt{V(x)}\,dx$ for all
$t \in \mathbb{R}$
\end{itemize}
We will prove the following result.

\begin{theorem} \label{thm1.03}
Let $V$ and $a$ satisfy {\rm (V1)-(V5), (A1)-(A2)}.  Then
\eqref{e1.00005}, \eqref{e1.00002} has a solution taking values in
$(-1,1)$.
\end{theorem}

Note: if $V$ is even and $V > 0$ on $(-1,0)$, then $L = l +
2\sqrt{l \underline{l}}$ in (A2).  If $\underline{l} = l$, we
obtain the result of \cite{G}.  Due to a dearth of
counterexamples, it is not known whether the upper bound on $a$ in
(A2) is really necessary.


This paper is organized as follows: Section~2 lays out the
variational methods used in the proof and an outline of the proof.
Section~3 contains the proofs of some subordinate propositions and
lemmas, with the most involved proposition concerning the
convergence of Palais-Smale sequences of the functional associated
with \eqref{e1.00005}.  Section~4 wraps up the proof of
Theorem~\ref{thm1.03}.


\section{Variational Method and Outline of Proof}

Define the functional $F: W^{1,2}_{\rm loc}(\mathbb{R}) \to [0,
\infty]$ by
%
\begin{equation} \label{e2.04}
F(x) = \int_{-\infty}^\infty \frac{1}{2} \dot{x}(t)^2 + a(t)V(x(t))\,dt.
\end{equation}
%
By (V1)-(V3), $F(x) < \infty$ for all $x \in W$.
$F:W \to \mathbb{R}^+$ is Fr\'echet differentiable with
%
\begin{equation}\label{e2.06}
F'(x)u = \int_{-\infty}^\infty \dot{x}(t)\dot{u}(t) + V'(x(t))u(t)\,dt
\end{equation}
%
for all $x \in W$, $u \in W^{1,2}(\mathbb{R})$.  Critical points
of $F: W \to \mathbb{R}^+$ are solutions
of \eqref{e1.00005}, \eqref{e1.00002}.  We will show via a minimax
argument that $F$ has at least one critical point.

Define
%
\begin{equation}\label{e2.09}
{\mathcal B} = F_l(\omega) > 0,
\end{equation}
%
where $F_l$ is as in \eqref{e1.00003}.
A \emph{Palais-Smale sequence} for $F$ is a sequence
$(x_n) \subset W$ with $F(x_n)$ convergent and
$\|F'(x_n)\| \to 0$ as $n \to \infty$, where $\|F'(x)\|$ is
defined by the operator norm
%
\begin{equation}\label{e2.10}
\|F'(x)\| = \sup\{F'(x)u : u \in W^{1,2}(\mathbb{R}),\;
 \|u\|_{W^{1,2}(\mathbb{R})} = 1\}.
\end{equation}
%
We will use the usual norm on $W^{1,2}(\mathbb{R})$,
%
\begin{equation}\label{e2.11}
\|u\|_{W^{1,2}(\mathbb{R})}
= \Big(\int_{-\infty}^\infty \dot{u}(t)^2 + u(t)^2\,dt\Big)^{1/2}.
\end{equation}
%
The $W^{1,2}(\mathbb{R})$-norm will be denoted simply
by $\|\cdot\|$ for the rest of this article.
We will prove the following proposition.

\begin{proposition}\label{p2.12}
Let $(x_n) \subset W$ with $F'(x_n) \to 0$ and
%
\begin{equation}\label{e2.13}
F(x_n) \to b \in [0,\mathcal{B}) \cup (\mathcal{B}, \mathcal{B}
+ 2\nu\sqrt{2\underline{l}}).
\end{equation}
%
Then, there exists $\bar{x} \in W$ solving \eqref{e1.00005},
\eqref{e1.00002} and a subsequence of
$(x_n)$ (also called $(x_n)$) with $\|x_n - \bar{x}\| \to 0$ as
$n \to \infty$.
\end{proposition}

It is interesting that the conclusion of Proposition~\ref{p2.12}
fails precisely when $b = \mathcal{B}$.  To verify this, define
the translation operator $\tau$ by $\tau_a u(t) = u(t-a)$ for
any $u:\mathbb{R}\to\mathbb{R}$ and $a, t \in \mathbb{R}$.
Then the Palais-Smale sequence
$(x_n) = (\tau_n \omega)$ satisfies $F(x_n) \to \mathcal{B}$ and
$F'(x_n) \to 0$ as $n \to \infty$,
but $x_n \to -1$ pointwise.

    We use a minimax argument similar to that in \cite{G}.  Define
%
\begin{equation}\label{e2.14}
\Gamma = \{\gamma \in C(\mathbb{R},W) : \|\tau_t \omega - \gamma(t)\| \to 0 \text{ as } |t| \to \infty\}
\end{equation}
%
and
%
\begin{equation}\label{e2.15}
c = \inf_{\gamma \in \Gamma} \sup_{t \in \mathbb{R}} F(\gamma(t)).
\end{equation}
%
Clearly $c \geq \mathcal{B}$.  We will show in Section~4 that $c <
\mathcal{B} + 2\nu\sqrt{2\underline{l}}$.  There are two cases to
consider: $c = \mathcal{B}$ and $c > \mathcal{B}$.  If $c >
\mathcal{B}$, then a standard deformation argument shows that
there exists a Palais-Smale sequence $(x_n)$ with $F(x_n) \to c$
and $F'(x_n) \to 0$ as $n \to \infty$.  Applying
Proposition~\ref{p2.12}, there exists a solution $\bar{x}$ of
\eqref{e1.00005}, \eqref{e1.00002} and a subsequence of $(x_n)$
(also denoted $(x_n)$) with $\|x_n - \bar{x}\| \to 0$ as $n \to
\infty$. If $c = \mathcal{B}$, then for every $n \in \mathbb{N}$,
there exists $\gamma_n \in \Gamma$ with $\sup_{t\in \mathbb{R}}
F(\gamma_n(t)) < \mathcal{B} + 1/n$.  Choose $t_n \in \mathbb{R}$
with $\gamma_n(t_n)(0) = 0$ and let $x_n = \gamma_n(t_n)$.  Since
$(F(x_n))$ is bounded, we will show there exists a subsequence
(also called $(x_n)$) and $x \in W^{1,2}_{\rm loc}(\mathbb{R})$
such that $(x_n)$ converges to $x$ locally uniformly and weakly in
$W^{1,2}([-T,T])$ for all $T > 0$.  We will show in Section~4 that
in fact $x \in W$ and $F(x) \leq \mathcal{B}$.  If $x$ is a
critical point of $F$, then Theorem~\ref{thm1.03} is proven.
Otherwise, let $\mathcal{W}(y)$ denote the gradient of $F$ at $y$
for all $y \in W$; that is, for all $y \in W$ and $\varphi \in
W^{1,2}(\mathbb{R})$,
%
\begin{equation}\label{e2.16}
(\mathcal{W}(y),\varphi)_{W^{1,2}(\mathbb{R})} = F'(y)\varphi,
\end{equation}
%
where $(\cdot ,\cdot )$ is the standard inner product on
$W^{1,2}(\mathbb{R})$,
%
\begin{equation}\label{e2.17}
(f,g)_{W^{1,2}(\mathbb{R})} = \int_{-\infty}^\infty \dot{f}(t)\dot{g}(t) + f(t)g(t)\,dt.
\end{equation}
%
Let $\eta$ denote the solution of the gradient vector flow induced by the initial value problem:
%
\begin{equation}\label{e2.18}
\frac{d}{ds}\eta(s,u) = -\mathcal{W}(\eta(s,u));\ \eta(0,u) = u.
\end{equation}
%
We will show in Section~4 that $\eta$ is well-defined
on $[0,\infty) \times W$.

Recall that we have $x \in W$ with $F(x) \leq \mathcal{B}$ and
$F'(x) \neq 0$.  Since $F$ is nonnegative, there exists a sequence
$(s_n) \subset \mathbb{R}^+$ with $F'(\eta(s_n,x)) \to 0$ as
$n \to \infty$.  By Proposition~\ref{p2.12}, there exists $\bar{x}$
satisfying \eqref{e1.00005}, \eqref{e1.00002}.

\section{Palais-Smale Sequences}

In this section, we prove Proposition~\ref{p2.12} and some
subsidiary lemmas and propositions leading up to it.  Although the
full strength of Proposition~\ref{p2.12} is not necessary to prove
Theorem~\ref{thm1.03}, the strong convergence of Palais-Smale
sequences that it implies is interesting and may be useful for
other problems. From now on we assume that
%
\begin{equation}\label{e3.01}
V(x) > 0 \quad\text{ for all } |x| > 1, \quad\text{and}\quad
     {\lim}_{|x|\to\infty} V(x) = \infty.
\end{equation}
%
We may make this assumption because the solution we will find to
\eqref{e1.00005} takes values in $(-1,1)$.

\begin{lemma}\label{l3.01}
If $x \in W^{1,2}_{\rm loc}(\mathbb{R})$ with $F(x) < \infty$,
then $x(t) \to -1$ or $x(t) \to 1$ as $t \to -\infty$, and $x(t)
\to 1$ or $x(t) \to -1$ as $t \to \infty$.  In fact, $x + 1 \in
W^{1,2}((-\infty,0])$ or $x - 1 \in W^{1,2}((-\infty,0])$, and $x
+ 1 \in W^{1,2}([0,\infty))$ or $x - 1 \in W^{1,2}([0,\infty))$.
\end{lemma}

\begin{proof}
Suppose the lemma is false.  Then there exist $x \in W^{1,2}_{\rm
loc}(\mathbb{R})$ with $F(x) < \infty$, $\delta > 0$ and a
sequence $(t_n)$ with $|t_n| \to \infty$ as $n \to \infty$
%
\begin{equation}\label{e3.02}
x_n(t) \in (-\infty, -1-\delta) \cup (-1+\delta, 1-\delta) \cup (1+\delta, \infty).
\end{equation}
%
Let
%
\begin{equation}\label{e3.04}
d = \inf\{V(x) : x \in (-\infty, -1-\delta/2) \cup (-1+\delta/2,
1-\delta/2) \cup (1+\delta/2, \infty)\} > 0.
\end{equation}
%
Assume without loss of generality that $t_n \to \infty$, and
taking  a subsequence if necessary, that $t_{n+1} \geq t_n + 1$
for all $n$.  If $x(t) \in (-\infty, -1-\delta/2) \cup
(-1+\delta/2, 1-\delta/2) \cup (1+\delta/2, \infty)$ for all $t
\in [t_n, t_n + 1]$, then $\int_{t_n}^{t_n + 1} V(x(t))\,dt \geq
\delta$.  Otherwise, there exists $t^* \in [t_n,t_{n+1}]$ with
$|x(t_n) - x(t^*)| \geq \delta/2$, and by the Cauchy-Schwarz
inequality,
%
\begin{equation}\label{e3.05}
\begin{aligned}
\delta/2 &\leq |x(t_n) - x(t^*)|
 \leq \int_{t_n}^{t^*} |\dot{x}(t)|\,dt \\
&\leq   \sqrt{t^* - t_n}
\Big(\int_{t_n}^{t^*} \dot{x}(t)^2\,dt\Big)^{1/2}
\leq \Big(\int_{t_n}^{t^*} \dot{x}(t)^2\,dt\Big)^{1/2}, \\
\end{aligned}
\end{equation}
%
\begin{equation}\label{e3.06}
\int_{t_n}^{t_n + 1} \dot{x}(t)^2\,dt \geq \int_{t_n}^{t^*} \dot{x}(t)^2\,dt \geq \delta^2/4.
\end{equation}
%
Either way,
\begin{equation}\label{e3.07}
\int_{t_n}^{t_n + 1} \frac{1}{2}\dot{x}(t)^2 + a(t)V(x(t))\,dt \geq
        \min(\delta^2/8, d\underline{l}),
\end{equation}
and
\begin{equation}\label{e3.08}
F(x) \geq \sum_{n=1}^\infty
    \int_{t_n}^{t_n + 1} \frac{1}{2}\dot{x}(t)^2 + a(t)V(x(t))\,dt
\geq \sum_{n=1}^\infty \min(\delta^2/8, d\underline{l}) = \infty,
\end{equation}
%
which is a contradiction.  So $x(t) \to -1$ or $x(t) \to 1$ as $t
\to \infty$.  Similarly, $x(t) \to -1$ or $x(t) \to 1$ as
$t \to -\infty$.

By (V5), there exists $\epsilon > 0$ with $V(x) \geq \epsilon(x+1)^2$
for all $x \in (-1-\epsilon, -1+\epsilon)$ and
$V(x) \geq \epsilon(x-1)^2$ for all
$x \in (1-\epsilon, 1+\epsilon)$.  So if $x(t) \to 1$ as
$t \to \infty$, there exists
$T > 0$ such that
%
\begin{equation}\label{e3.09}
\int_T^\infty (x(t)-1)^2\,dt \leq  \int_T^\infty V(x(t))/\epsilon\,dt
\leq   \frac{1}{\epsilon\underline{l}}\int_T^\infty a(t)V(x(t))\,dt
\leq   \frac{F(x)}{\epsilon\underline{l}} < \infty
\end{equation}
%
and $x-1 \in W^{1,2}([0,\infty))$.  Similar arguments apply to the
cases $x(t) \to -1$ as $t \to \infty$, $x(t) \to 1$ as $t \to -\infty$,
and $x(t) \to -1$ as $t \to -\infty$.
\end{proof}

Next we show that Palais-Smale sequences are bounded in $W^{1,2}_{\rm
loc}(\mathbb{R})$.

\begin{lemma}\label{l3.091}
Let $A, T > 0$.  There exists $B > 0$ such that if $x \in
W^{1,2}_{\rm loc}(\mathbb{R})$ with $F(x) \leq A$, then
$\|x\|_{W^{1,2}([-T,T])} \leq B$.
\end{lemma}

\begin{proof}
Clearly $\int_{-T}^T \dot{x}(t)^2\,dt \leq 2A$, so it suffices to
find an upper bound on $|x|$ over $[-T,T]$.  Let $C > 0$ such that
$V(x) > C$ for all $|x| \geq A/2T$.  Since
$\int_{-T}^T V(x(t))\,dx \leq A$,
there exists $t^* \in [T,T]$ with $V(t^*) \leq A/2T$ and
$|x(t^*)| \leq C$.  For any $s \in [-T,T]$,
%
\begin{equation}\label{e3.092}
\begin{aligned}
|x(s)| &\leq |x(t^*)| + |\int_{t^*}^s \dot{x}(t)\,dt| \\
&\leq |x(t^*)| + \sqrt{|s-t^*|}\Big|\int_{t^*}^s \dot{x}(t)^2\,dt\Big|^{1/2}\\
 &\leq C + \sqrt{2T}\cdot\sqrt{2A}.
\end{aligned}
\end{equation}
\end{proof}

    For $\Omega \subset \mathbb{R}$, define
\begin{equation}\label{e3.10}
F_\Omega (x) = \int_\Omega \frac{1}{2}\dot{x}(t)^2 + a(t)V(x(t))\,dt.
\end{equation}
Then we have the following lemma.

\begin{lemma}\label{l3.11}
If $x_0, x_1 \in (-1,1)$, $t_0 < t_1$, and $x \in W^{1,2}([t_0,t_1])$
with $x(t_0) = x_0$ and $x(t_1) = x_1$, then
%
\begin{equation}\label{e3.12}
F_{[t_0,t_1]}(x) \geq \sqrt{2 \underline{l}}
 \big|\int_{x_0}^{x_1} \sqrt{V(x)}\,dx \big|.
\end{equation}
\end{lemma}

\begin{proof} Let $\omega_{\underline{l}}$ denote the unique
solution in $W$ of the differential equation
%
\begin{equation}\label{e3.13}
\ddot{x}(t) = \underline{l}V'(x(t))
\end{equation}
%
satisfying $\omega_{\underline{l}}(0) = 0$.
Then $\omega_{\underline{l}}$ minimizes the functional
%
\begin{equation}\label{e3.14}
F_{\underline{l}}(u) = \int_{-\infty}^\infty \frac{1}{2} \dot{u}(t)^2 + \underline{l}V(u(t))\,dt
\end{equation}
%
over $W$.  By the argument following \eqref{e1.00002},
%
\begin{equation}\label{e3.15}
\dot{\omega}_{\underline{l}}(t)
= \sqrt{2\underline{l}V(\omega_{\underline{l}}(t))}.
\end{equation}

Let $x_0, x_1 \in (-1,1), t_0 < t_1$, and $x \in W^{1,2}([t_0,t_1])$
with $x(t_0) = x_0$ and $x(t_1) = x_1$.  Assume $x_0 < x_1$. Now
%
\begin{equation}\label{e3.16}
\int_{t_0}^{t_1} \frac{1}{2}\dot{x}(t)^2 + \underline{l}V(x(t))\,dt \geq
            \int_{t_0}^{t_1} \frac{1}{2}\dot{\omega}_{\underline{l}}(t)^2 +
                        \underline{l}V(\omega_{\underline{l}}(t))\,dt;
\end{equation}
%
otherwise, we could replace
$\omega_{\underline{l}}|_{[\omega^{-1}_{\underline{l}}(x_0),
\omega^{-1}_{\underline{l}}(x_1)]}$
by $x|_{[t_0, t_1]}$ to obtain $\tilde{\omega} \in W$ with
$F_{\underline{l}}(\tilde{\omega}) < F_{\underline{l}}(\omega_{\underline{l}})$, contradicting
the optimality of $\omega_{\underline{l}}$.
$\tilde{\omega}$ is defined by
%
\begin{equation}\label{e3.17}
\begin{aligned}
&\tilde{\omega}(t) \\
&= \begin{cases}
 \omega_{\underline{l}}(t),
 &t \leq \omega^{-1}_{\underline{l}}(x_0); \\
x(t - \omega^{-1}_{\underline{l}}(x_0)+ t_0),
 & \omega^{-1}_{\underline{l}}(x_0) \leq t
\leq \omega^{-1}_{\underline{l}}(x_0)+t_1 - t_0;\\
\omega_{\underline{l}} (t+(\omega^{-1}_{\underline{l}}(x_1)
-\omega^{-1}_{\underline{l}}(x_0))- (t_1 - t_0)),
 &t \geq \omega^{-1}_{\underline{l}}(x_0)+t_1 - t_0.
\end{cases}
\end{aligned}
\end{equation}
%
Now by \eqref{e3.15}-\eqref{e3.16},
%
\begin{equation}\label{e3.18}
F_{[t_0,t_1]}(x) \geq \int_{t_0}^{t_1} \dot{\omega}_{\underline{l}}(t)^2 =
    \int_{t_0}^{t_1}\dot{\omega}_{\underline{l}}(t)
                \sqrt{2\underline{l}V(\omega_{\underline{l}}(t))}\,dt =
                    \int_{x_0}^{x_1} \sqrt{2\underline{l}V(x(t))}\,dt.
\end{equation}

    For the case $x_0 > x_1$, define $x_R$, the reversal of $x$ on
$[t_0,t_1]$, by $x_R(t)=x(t_0 + t_1 - t)$.  Then $x_R(t_0) = x_1$
and $x_R(t_1) = x_0$ so by the first case,
%
\begin{equation}\label{e3.19}
\begin{aligned}
F_{[t_0,t_1]}(x)
&\geq \int_{t_0}^{t_1} \frac{1}{2} \dot{x}(t)^2
 + \underline{l}V(x(t))\,dt
=\int_{t_0}^{t_1} \frac{1}{2} \dot{x_R}(t)^2
 + \underline{l}V(x_R(t))\,dt \\
&\geq \int_{x_1}^{x_0} \sqrt{2\underline{l}V(x_R(t))}\,dt
 = \big| \int_{x_0}^{x_1} \sqrt{2\underline{l} V(x(t))}\,dt\big|.
\end{aligned}
\end{equation}
\end{proof}

Recall that $\xi_-$ and $\xi_+$ from \eqref{e1.01}, and assume from
now on that
%
\begin{equation}\label{e3.20}
\begin{gathered}
    V(x) = V(-1 + (-1-x)) \quad\text{ for all } x \in [-1-(\xi_- + 1),
  -1], \\
    V(x) = V(1 - (x-1)) \quad \text{ for all } x \in [1,1+(1-\xi_+)].
\end{gathered}
\end{equation}
%
Again, we may assume this because our solution of
\eqref{e1.00005}, \eqref{e1.00002} will take values in $(-1,1)$.
To prove Proposition~\ref{p2.12}, we will use the following result.

\begin{proposition}\label{p3.205}
If $(x_n) \subset W$ with $F'(x_n) \to 0$,
%
\begin{equation}\label{e3.206}
F(x_n) \to b < 2\mathcal{B} + \sqrt{2 \underline{l}}\int_{-1}^1\sqrt{V(x)}\,dx,
\end{equation}
and
$x_n \to \bar{x} \in W$ locally uniformly and weakly in
$W^{1,2}([-T,T])$ for all $T > 0$ as $n \to \infty$, then
$\bar{x}$ solves \eqref{e1.00005} and $\|x_n - \bar{x}\| \to 0$ as
$n \to \infty$.
\end{proposition}

\begin{proof}
Let $(x_n)$ and $\bar{x}$ be as in the Proposition statement.
To prove $\bar{x}$ solves \eqref{e1.00005}, let
$\varphi \in C^\infty_0$.  Then
%
\begin{equation}\label{e3.21}
\begin{aligned}
0 &= \lim_{n \to \infty} F'(x_n)\varphi
 = \lim_{n \to \infty} \int_{-\infty}^\infty \dot{x}_n(t)
 \dot{\varphi}(t) +  V'(x_n(t))\varphi(t)\,dt \\
&=  \int_{-\infty}^\infty \dot{\bar{x}}(t)\dot{\varphi}(t) +
                V'(\bar{x}(t))\varphi(t)\,dt = F'(\bar{x})\varphi,
\end{aligned}
\end{equation}
%
and $\bar{x}$ is a weak solution of \eqref{e1.00005}.  Next we
show that for any $T > 0$, $\|x_n - \bar{x}\|_{W^{1,2}([-T,T])} \to 0$
as $n \to \infty$.  Let $T>0$.  Since $x_n \to \bar{x}$ uniformly
on $[-T,T]$, $\int_{-T}^T (x_n(t) - \bar{x}(t))^2\,dt \to 0$ as
$n \to \infty$.  We must therefore show that
$\int_{-T}^T (\dot{x}_n(t) - \dot{\bar{x}}(t))^2\,dt \to 0$ as
$n \to \infty$.  Since $\dot{x}_n \to \dot{\bar{x}}$ weakly in
$L^2([-T,T])$,
%
\begin{equation}\label{e3.22}
\begin{aligned}
&\limsup_{n \to \infty}\int_{-T}^T (\dot{x}_n(t)-\bar{x}(t))^2\,dt \\
&= \limsup_{n \to \infty} \int_{-T}^T \dot{x}_n(t)^2 -
            2\int_{-T}^T \dot{x}_n(t)\dot{\bar{x}}(t)\,dt +
                        \int_{-T}^T \dot{\bar{x}}(t)^2\,dt  \\
&= \limsup_{n\to \infty} \int_{-T}^T \dot{x}_n(t)^2
 - \dot{\bar{x}}(t)^2\,dt,
\end{aligned}
\end{equation}
%
and it suffices to prove
$\lim_{n \to \infty}\int_{-T}^T \dot{x}_n(t)^2 - \dot{\bar{x}}(t)^2\,dt
= 0$.
Define $(u_n) \subset W^{1,2}(\mathbb{R})$ by
%
\begin{equation}\label{e3.23}
u_n(t)= \begin{cases}
    0 & t \leq -T-1 \\
(x_n(-T)-\bar{x}(-T))(t+T+1) & -T-1 \leq t \leq -T\\
 x_n(t)-\bar{x}(t) & -T \leq t \leq T\\
(x_n(T)-\bar{x}(T))(-t + T + 1) &T \leq t \leq T+1\\
    0 &t \geq T+1
\end{cases}
\end{equation}
%
Clearly, $(u_n)$ is bounded in $W^{1,2}(\mathbb{R})$.
Since $u_n \to 0$ uniformly on $[-T-1,T+1]$,
%
\begin{equation}\label{e3.24}
\begin{aligned}
0 &= \lim_{n\to \infty} F'(x_n)u_n + F'(\bar{x})u_n  \\
  &= \lim_{n \to \infty} (x_n,u_n)_{W^{1,2}([-T-1,T+1])} +
           (\bar{x},u_n)_{W^{1,2}([-T-1,T+1])}  \\
  &\quad- \int_{-T-1}^{T+1} a(t)V'(x_n(t))u_n(t)\,dt -
               \int_{-T-1}^{T+1} a(t)V'(\bar{x}(t))u_n(t)\,dt  \\
  &= \lim_{n \to \infty} (x_n,u_n)_{W^{1,2}([-T-1,T+1])} +
           (\bar{x},u_n)_{W^{1,2}([-T-1,T+1])}.
\end{aligned}
\end{equation}
%
Since $\|u_n\|_{W^{1,2}([-T-1,-T])} \to 0$ and
        $\|u_n\|_{W^{1,2}([T,T+1])} \to 0$ as $n \to \infty$,
%
\begin{equation}\label{e3.25}
\begin{aligned}
0 &= \lim_{n \to \infty} (x_n,u_n)_{W^{1,2}([-T,T])} +
                         (\bar{x},u_n)_{W^{1,2}([-T,T])}  \\
&= \lim_{n \to \infty}\int_{-T}^T
           \dot{x}_n(t)(\dot{x}_n(t)-\dot{\bar{x}}(t)) +
                     x_n(t)(x_n(t)-\bar{x}(t))  \\
&\quad + \dot{\bar{x}}(t)(\dot{x}_n(t)-\dot{\bar{x}}(t)) +
                     \bar{x}(t)(x_n(t)-\bar{x}(t))\,dt  \\
&= \lim_{n\to\infty} \int_{-T}^T \dot{x}^2_n(t)
  - \dot{\bar{x}}(t)^2 + x_n(t)^2 - \bar{x}(t)^2\,dt \\
&= \lim_{n\to\infty} \int_{-T}^T \dot{x}^2_n(t)
 - \dot{\bar{x}}(t)^2\,dt.
\end{aligned}
\end{equation}
%
Therefore, $\|x_n - \bar{x}\|_{W^{1,2}([-T,T])} \to 0$ as
$n \to \infty$.

    Suppose $\|x_n - \bar{x}\| \not\to 0$ as $n \to \infty$.
Then there exist $\delta > 0$ and a sequence $(T_n)$ with
$T_n \to \infty$ and
%
\begin{equation}\label{e3.26}
    \|x_n - \bar{x}\|_{\mathbb{R} \setminus [-T_n, T_n]}^2
\geq 4\delta^2
\end{equation}
%
for all $n$.  Along a subsequence, either
%
\begin{equation}\label{e3.261}
\|x_n - \bar{x}\|_{W^{1,2}((-\infty, -T_n])}^2 \geq 2\delta^2\quad
\text{or}\quad
\|x_n - \bar{x}\|_{W^{1,2}([T_n,\infty))}^2 \geq 2\delta^2.
\end{equation}
%
Let us assume the former; the latter case is similar.
Since $1 + \bar{x} \in W^{1,2}((-\infty,0])$,
%
\begin{equation}\label{e3.262}
\|x_n + 1\|_{W^{1,2}((-\infty, -T_n])} \geq \delta
\end{equation}
%
for large $n$.  There are two cases to consider:
%\label{e3.27}
\begin{itemize}
\item[Case I:] For all $\epsilon > 0$, there exists $M > 0$ such that
 $|1+x_n(t)|<\epsilon$ for all $n$ and $t \leq -M$.
\item[Case II:] There exists $d \in (0,1)$ and a sequence
$(t_n) \subset \mathbb{R}$ with
 $t_n \to -\infty$ and $|1+ x_n(t_n)| > d$ for all $n$.
\end{itemize}

Case I: let $\xi^* \in (-1, \xi_-)$ and $c \in (0,1)$ such that
%
\begin{equation}\label{e3.28}
V'(x)x \geq c(1+x)^2
\end{equation}
%
for all $x \in [-1-(\xi^* + 1), \xi^*]$.  This is possible by
(V3)-(V5), \eqref{e3.20}, and the definition of $\xi_-$. Let
$M > 0$ be large enough so that
%
\begin{equation}\label{e3.281}
|1 + x_n(t)| < \min\big(1 + \xi^*, \frac{c\delta^2}{8(1+\sqrt{b})}\big)
\end{equation}
%
for all $n \in \mathbb{N}$, $t \leq -M$.
Define $(u_n) \subset W^{1,2}(\mathbb{R})$ by
%
\begin{equation}\label{e3.29}
\begin{aligned}
u_n(t) =
\begin{cases}
    1+x_n(t) &t \leq -M \\
    (1+x_n(-M))(1-M-t) & -M \leq t \leq -M+1\\
    0 & t\geq -M+1
\end{cases}
\end{aligned}
\end{equation}
%
We will show $(u_n)$ is uniformly bounded in $W^{1,2}(\mathbb{R})$.
Let $K > 0$ so
%
\begin{equation}\label{e3.30}
|V'(x)| \leq K \quad\text{and}\quad (x+1)^2 \leq KV(x)
\end{equation}
%
for all $x \in [-1-(\xi^* + 1),\xi^*]$.  This is possible by
(V1)-(V5),\eqref{e3.20}, and the definition of $\xi_-$.  For large
$n$,
%
\begin{equation}\label{e3.31}
\begin{aligned}
\|u_n\|^2
 &= \int_{-\infty}^{-M} \dot{x}_n(t)^2 + (1+x_n(t))^2\,dt
            + (1 + x_n(-M))^2 + \frac{1}{2}(1+x_n(-M))  \\
 &\leq (2 + \frac{K}{\underline{l}}) \int_{-\infty}^{-M}
              \frac{1}{2}\dot{x}_n(t)^2 + a(t)V(x_n(t))\,dt  \\
 &\quad + (1 + \bar{x}(-M))^2 + \frac{1}{2}(1+\bar{x}(-M))+ 1  \\
 &\leq (2 + \frac{K}{\underline{l}})F(x_n) +
   (1 + \bar{x}(-M))^2 + \frac{1}{2}(1+\bar{x}(-M))+ 1  \\
            &\leq (2 + \frac{K}{\underline{l}})(2b) +
                (1 + \bar{x}(-M))^2 + \frac{1}{2}(1+\bar{x}(-M))+ 1.
\end{aligned}
\end{equation}
%
Since $F'(x_n) \to 0$, $F'(x_n)u_n \to 0$ as $n \to \infty$.
But for large $n$,
%
\begin{equation}\label{e3.32}
\begin{aligned}
&F'(x_n)u_n \\
&= \int_{-\infty}^{-M} \dot{x}_n(t)^2 +V'(x_n(t))(1+x_n(t))\,dt
   +\int_{-M}^{-M+1}(1+x_n(-M))\dot{x}_n(t)\,dt \\
&\quad + \int_{-M}^{-M+1}(1+x_n(-M))(1-M-t)\,dt \\
&\geq \int_{-\infty}^{-M} \dot{x}_n(t)^2 +c(1+x_n(t))^2\,dt\\
&\quad -|1+x_n(-M)|\Big(\int_{-M}^{-M+1}\dot{x}_n(t)^2\,dt\Big)^{1/2}
  -\frac{1}{2}|1+x_n(-M)| \\
&\geq c\|1+x_n\|^2_{W^{1,2}(-\infty,-M])} -
                |1+x_n(-M)|(\sqrt{2F(x_n)} +1) \\
&\geq c\delta^2 - (1 + 2\sqrt{b})|1+x_n(-M)|
\geq \frac{1}{2}c\delta^2
\end{aligned}
\end{equation}
%
by \eqref{e3.281}. This is impossible.

Case II: by the arguments of Lemma~\ref{l3.11},
%
\begin{equation}\label{e3.34}
F(x) \geq \sqrt{2\underline{l}}\int_{-1}^1 \sqrt{V(x)}\,dx
\end{equation}
%
for all $x \in W$, including $\bar{x}$.  Let $d$ and $(t_n)$ be as
in Case I.  Let
$M > 0$ be large enough so that $|1+\bar{x}(t)| < d/2$ for all
$t \leq -M$, and
%
\begin{equation}\label{e3.35}
F_{[-M,M]}(\bar{x})
> \sqrt{2\underline{l}}\int_{-1}^1 \sqrt{V(x)}\,dx -
    \frac{1}{10}(2\mathcal{B}+  \sqrt{2\underline{l}}\int_{-1}^1
 \sqrt{V(x)}\,dx - b).
\end{equation}
%
Define $\alpha_n \leq \beta_n < -M$ by
%
\begin{equation}\label{e3.36}
\beta_n = \max\{t < -M : |1+x_n(t)| = d\},\quad
\alpha_n = \min\{t : |1+x_n(t)| = d\}
\end{equation}
%
Since $x_n \to \bar{x}$ locally uniformly and $|1+\bar{x}(t)| <
d/2 < 1$ for all $t \leq -M$, $\beta_n \to -\infty$ as $n \to
\infty$.  Define $v_n = \tau_{-\beta_n}x_n$.  By Fatou's Lemma,
the weak lower semicontinuity of $\int_{-\infty}^\infty
\dot{x}(t)^2\,dt$, and Lemma~\ref{l3.091}, there exists $\bar{v}
\in W^{1,2}(\mathbb{R})$ with $F(\bar{v}) < \infty$ and $v_n \to
\bar{v}$ locally uniformly and weakly in $W^{1,2}([-T,T])$ for all
$T>0$.  By the arguments of \eqref{e3.21}, $F'_l(\bar{v}) = 0$. By
the definition of $\beta_n$, $\bar{v}(t) \leq -1 + d < 0$ for all
$t > 0$. Therefore, by the arguments of Lemma~\ref{l3.01} applied
to $F_l$ instead of $F$, $\bar{v}(t) \to -1$ as $t \to \infty$.
By the arguments following \eqref{e1.00002}, $\dot{\bar{v}}(t) =
-\sqrt{2lV(\bar{v}(t))}$ for all $t \in \mathbb{R}$.  Let
$\omega_R$ denote the reversal of $\omega$: $\omega_R(t) =
\omega(-t)$ for all $t$.  Clearly $\bar{v} = \tau_\lambda
\omega_R$ for some $\lambda \in \mathbb{R}$.  By the arguments of
\eqref{e3.22}-\eqref{e3.25},
%
\begin{equation}\label{e3.37}
\|x_n - \tau_{\lambda+\beta_n}\omega_R\|_{W^{1,2}
([\beta_n - T,\beta_n + T])} \to 0
\end{equation}
%
as $n \to \infty$
for all $T > 0$.  This implies $\beta_n - \alpha_n \to \infty$
as $n \to \infty$.  For all $n$ and all $t < \alpha_n$,
 $x_n(t) < -1 + d/2 < 0$.  Therefore, arguments similar to those
above show that there exists
$\lambda_2 \in \mathbb{R}$ with
%
\begin{equation}\label{e3.38}
\|x_n - \tau_{\lambda_2+\alpha_n}\omega\|_{W^{1,2}([\alpha_n - T,
\alpha_n + T])} \to 0
\end{equation}
%
for all $T > 0$ as $n \to \infty$. For $\Omega \subset \mathbb{R}$,
define
%
\begin{equation}\label{e3.385}
{F_l}_\Omega (x) = \int_\Omega \frac{1}{2}\dot{x}(t)^2 + lV(x(t))\,dt
\end{equation}
%
Still assuming that $M$ is large enough so that \eqref{e3.35} holds,
assume also that $M$ is large enough that
%
\begin{equation}\label{e3.39}
\begin{gathered}
{F_l}_{[-M,M]}(\tau_\lambda \omega_R)
>\mathcal{B} - \frac{1}{10}(2\mathcal{B}
 + \sqrt{2\underline{l}}\int_{-1}^1 \sqrt{V(x)}\,dx - b),\\
{F_l}_{[-M,M]}(\tau_{\lambda_2} \omega)
 >\mathcal{B} - \frac{1}{10}(2\mathcal{B}
+\sqrt{2\underline{l}}\int_{-1}^1 \sqrt{V(x)}\,dx - b),
\end{gathered}
\end{equation}
%
Then for large $n$, \eqref{e3.35}, \eqref{e3.37}-\eqref{e3.39},
$\alpha_n \to -\infty$, $\beta_n \to -\infty$, and $a(t) \to l$ as
$|t| \to \infty$ imply
%
\begin{equation}\label{e3.40}
\begin{aligned}
F(x_n) &\geq F_{[\alpha_n - M,\alpha_n +M]}(x_n) +
             F_{[\beta_n - M,\beta_n +M]}(x_n) +
             F_{[-M,M]}(x_n)  \\
 &\geq (\mathcal{B} - \frac{1}{5}(2\mathcal{B} +
                 \sqrt{2\underline{l}}\int_{-1}^1 \sqrt{V(x)}\,dx - b)\\
 &\quad +    (\mathcal{B} - \frac{1}{5}(2\mathcal{B} +
                 \sqrt{2\underline{l}}\int_{-1}^1 \sqrt{V(x)}\,dx - b)\\
 &\quad+(\sqrt{2\underline{l}}\int_{-1}^1 \sqrt{V(x)}\,dx -
            \frac{1}{5}(2\mathcal{B} +
                 \sqrt{2\underline{l}}\int_{-1}^1 \sqrt{V(x)}\,dx - b))\\
 &\quad- \frac{1}{5}(2\mathcal{B} +
              \sqrt{2\underline{l}}\int_{-1}^1 \sqrt{V(x)}\,dx - b)\\
  &= 2\mathcal{B} + \sqrt{2\underline{l}}\int_{-1}^1 \sqrt{V(x)}\,dx -
              \frac{4}{5}(2\mathcal{B} +
               \sqrt{2\underline{l}}\int_{-1}^1 \sqrt{V(x)}\,dx - b)\\
& \equiv b^+ > b.
\end{aligned}
\end{equation}
%
This is impossible. Proposition~\ref{p3.205} is proven.
\end{proof}


\begin{proof}[Proof of Proposition~\ref{p2.12}]
 There are two cases:
$b < \mathcal{B}$ and $b > \mathcal{B}$.  The case $b <
\mathcal{B}$ is easier.  Let $(x_n) \subset W$ with $F(x_n) \to b
< \mathcal{B}$ and $F'(x_n) \to 0$ as $n \to \infty$.  By
Lemma~\ref{l3.091}, $(x_n)$ converges locally uniformly and weakly
in $W^{1,2}([-T,T])$ for all $T > 0$ to some function $\bar{x} \in
W^{1,2}_{\rm loc}(\mathbb{R})$.  By Fatou's Lemma and the weak
lower semicontinuity of $\int_{-\infty}^\infty \dot{x}(t)^2\,dt$,
$F(\bar{x}) < \infty$.  By Proposition~\ref{p3.205}, it suffices
to show $\bar{x} \in W$.  Suppose $\bar{x} \not\in W$.  Then by
Lemma~\ref{l3.01}, $\bar{x}(t) \to 1 $ as $t \to -\infty$ or
$\bar{x}(t) \to -1$ as $t \to \infty$.  Suppose $\bar{x}(t) \to
-1$ as $t \to \infty$ (the proof for $\bar{x}(t) \to 1$ as $t \to
-\infty$ is similar). Define
%
\begin{equation}\label{e3.41}
\mathcal{B}_\epsilon = \int_{\omega^{-1}(-1+\epsilon)}
^{\omega^{-1}(1-\epsilon)}
                \frac{1}{2}\dot{\omega}^2(t) + lV(\omega(t))\,dt
\end{equation}
%
for $\epsilon > 0$.  Let $\epsilon> 0$ be small enough that
%
\begin{equation}\label{e3.42}
\Big( \frac{l-\epsilon}{l} \Big)\mathcal{B}_\epsilon > b.
\end{equation}
%
Let $T > 0$ be large enough so that $a \geq l - \epsilon$ on $[T,\infty)$ and
$\bar{x}(T) < -1 + \epsilon$.  Let $n$ be large enough that
$x_n(T) < -1 + \epsilon$.  Let
$T<\alpha<\beta$ with $x_n(\alpha) = -1+\epsilon$, $x_n(\beta) = 1-\epsilon$.  By arguments similar to those of Lemma~\ref{l3.11},
%
\begin{equation}\label{e3.43}
\begin{aligned}
F(x_n) &\geq F_{[\alpha,\beta]}(x_n)
  =  \int_\alpha^\beta \frac{1}{2}\dot{x}_n(t)^2 + a(t) V(x_n(t))\,dt \\
 &\geq \int_\alpha^\beta \frac{1}{2}\dot{x}_n(t)^2
   + (l-\epsilon) V(x_n(t))\,dt  \\
 &\geq \frac{l-\epsilon}{l}\int_\alpha^\beta \frac{1}{2}\dot{x}_n(t)^2
  +  lV(x_n(t))\,dt  \\
 &\geq \frac{l-\epsilon}{l}\mathcal{B}_\epsilon \equiv b^+ > b.
\end{aligned}
\end{equation}
%
This is a contradiction.

Now suppose $b \in (\mathcal{B},
\mathcal{B}+2\nu\sqrt{l\underline{l}})$.  As before, along a
subsequence, $(x_n)$ converges locally uniformly and weakly in
$W^{1,2}([-T,T])$ for all $T > 0$ to a function $\bar{x} \in
W^{1,2}_{\rm loc}(\mathbb{R})$ with $F(\bar{x}) \leq b$.  We must
show $\bar{x} \in W$; then applying Proposition~\ref{p3.205}
proves Theorem~\ref{thm1.03}.  Suppose $\bar{x}(t) \not\to 1$ as
$t \to \infty$ (the proof for $\bar{x}(t) \not\to -1$ as $t \to
-\infty$ is similar).  By Lemma~\ref{l3.01}, $\bar{x}(t) \to -1$
as $t \to \infty$.  Let $t_n = \max\{t : x_n(t) = 0\}$.  Then $t_n
\to \infty$ as $n \to \infty$.  By the arguments following
\eqref{e3.36} and the arguments of \eqref{e3.22}-\eqref{e3.25},
%
\begin{equation}\label{e3.44}
\|x_n - \tau_{t_n}\omega\|_{W^{1,2}([t_n -M,t_n + M])} \to 0
\end{equation}
%
as $n \to \infty$ for all $M > 0$.  Let $-1 < e_1 < \xi_i < \xi_-$
with
%
\begin{equation}\label{e3.45}
\sqrt{2\underline{l}}\int_{e_1}^{\xi_1}\sqrt{V(x)}\,dx >
    \nu\sqrt{2\underline{l}} -
      \frac{1}{5}(\mathcal{B} + 2\nu\sqrt{2\underline{l}}   -b).
\end{equation}
%
Let $c \in (0,1)$ with
%
\begin{equation}\label{e3.46}
V'(x)(1+x) \geq cV(x)
\end{equation}
%
for all $x \in [-1, \xi_1]$.  Let $K > 0$ be large enough that
%
\begin{equation}\label{e3.47}
|V'(x)| < K
\end{equation}
%
for all $x \in [-1,1]$.  Let $M > 0$ be large enough that
%
\begin{gather}\label{e3.48}
{F_l}_{[-M,M]}(\omega) > \mathcal{B} -
\frac{1}{6}(\mathcal{B}+2\nu\sqrt{2\underline{l}} - b),\\
\label{e3.49} 1+ \omega(-M) <
\min(\frac{c(b-\mathcal{B})}{16(K+2\sqrt{b})}, 1 + e_1).
\end{gather}
%
By \eqref{e3.44} and the fact that $a(t) \to l$ as $t \to \infty$,
%
\begin{equation}\label{e3.50}
F_{[t_n - M,t_n+M]}(x_n) < \mathcal{B} + \frac{1}{2}(b-\mathcal{B})
\end{equation}
%
for large $n$, so
%
\begin{equation}\label{e3.51}
F_{(-\infty, t_n-M]}(x_n) + F_{[t_n + M,\infty)}(x_n) >
                    \frac{1}{3}(b-\mathcal{B}).
\end{equation}
%
Assume $F_{(-\infty, t_n - M]} > (b-\mathcal{B})/6$ (the case
$F_{[t_n + M,\infty)} > (b-\mathcal{B})/6$ is similar).
There are two possible cases: along a subsequence,
\begin{itemize} %\label{e3.52}
\item[Case I:] $|1+x_n(\alpha_n)| \geq 1 + \xi_1$ for
$\alpha_n < t_n -M$,
\item[Case II:] $|1+x_n(t)| < 1 + \xi_1$ for all $t < t_n - M$.
\end{itemize}

For Case I, assume $1+ x_n(\alpha_n) \geq 1 + \xi_1$ (the case $1
+ x_n(\alpha_n) \leq -(1+\xi_1)$ is similar due to \eqref{e3.20}).
For large $n$, by Lemma~\ref{l3.11}, \eqref{e3.45}, \eqref{e3.49},
\eqref{e3.44}, \eqref{e3.48}, (A1), and $t_n \to \infty$,
%
\begin{equation}\label{e3.53}
\begin{aligned}
F(x_n) &\geq F_{(-\infty, \alpha_n]}(x_n) +
             F_{[\alpha_n,t_n - M]}(x_n) +
             F_{[t_n-M, t_n+M]}(x_n)  \\
 &\geq 2(\nu\sqrt{2\underline{l}}
  - \frac{1}{5}(\mathcal{B} + 2\nu\sqrt{2\underline{l}} - b)) +
    (\mathcal{B} -\frac{1}{5}(\mathcal{B} + 2\nu\sqrt{2\underline{l}}
    - b)) \\
 &= \mathcal{B} + 2\nu\sqrt{2\underline{l}} -
                \frac{3}{5}(\mathcal{B}
+ 2\nu\sqrt{2\underline{l}} - b) \equiv  b^+ > b.
\end{aligned}
\end{equation}
%
This is impossible.

For Case II, define $(u_n) \subset W^{1,2}(\mathbb{R})$ by
%
\begin{equation}\label{e3.54}
u_n(t) =
\begin{cases}
    1+x_n(t) & t \leq t_n - M \\
    (1+x_n(t_n-M))(t_n-M+1-t) & t_n - M \leq t \leq t_n -M + 1\\
  0 & t \geq t_n -M +1.
\end{cases}
\end{equation}
%
The sequence $(u_n)$ is uniformly bounded in $W^{1,2}(\mathbb{R})$,
as in \eqref{e3.31}.  So $F'(x_n)u_n \to 0$.  But for large $n$,
%
\begin{equation}\label{e3.55}
\begin{aligned}
&F'(x_n)u_n\\
&= \int_{-\infty}^{t_n-M} \dot{x}_n(t)^2
   + a(t)V'(x_n(t))(1+x_n(t))\,dt \\
&\quad -(1+x_n(t_n-M))\int_{t_n-M}^{t_n-M+1}\dot{x}_n(t)\,dt \\
&\quad + (1+x_n(t_n - M))\int_{t_n-M}^{t_n-M+1}V'(x_n(t))(t_n
-M+1-t)    \,dt  \\
&\geq  \int_{-\infty}^{t_n-M} \dot{x}_n(t)^2 + ca(t)V(x_n(t))\,dt \\
&\quad- |1+x_n(t_n - M)|\Big(\int_{t_n-M}^{t_n-M+1}
\dot{x}_n(t)^2\,dt\Big)^{1/2} -  K|1+x_n(t-M)| \\
&\geq c\int_{-\infty}^{t_n-M}\frac{1}{2}\dot{x}_n(t)^2
  +a(t)V(x_n(t))\,dt
  -(K + 2\sqrt{b})|1+x_n(t_n-M)|  \\
&= cF_{(-\infty, t_n-M]}(x_n) -
                        (K + 2\sqrt{b})|1+x_n(t_n-M)|  \\
&\geq \frac{1}{6}c(b-\mathcal{B}) - \frac{1}{12}c(b-\mathcal{B}) =
\frac{1}{12}c(b-\mathcal{B}) > 0
\end{aligned}
\end{equation}
%
by \eqref{e3.49}.  This is impossible.  Case~II is proven.
Proposition~\ref{p2.12} is proven.
\end{proof}

\section{Completion of Proof}

    In this section we tie up some loose ends from Section~2.
It was asserted that $c < \mathcal{B} + 2\nu\sqrt{2\underline{l}}$,
where $c$ is
    from \eqref{e2.15}.  Define $\gamma_0 \in \Gamma$ by
$\gamma_0(t) = \tau_t(\omega)$.  We will show
$\sup_{t\in \mathbb{R}} F(\omega_0(t)) < \mathcal{B}$.  Since
$F(\gamma_0(t)) \to \mathcal{B}$ as $|t| \to \infty$, and
$F(\gamma_0(t))$ is continuous in $t$, it suffices to prove that
$F(\gamma_0(t)) < \mathcal{B} + 2\nu\sqrt{2\underline{l}}$ for
all $t \in \mathbb{R}$.  We will prove this for $t = 0$;
the proof is similar for other $t$.  After \eqref{e1.00002},
it is proven that
%
\begin{equation}\label{e4.01}
V(\omega(t)) = \frac{\dot{\omega}(t)}{\sqrt{2l}}\sqrt{V(\omega(t))}
\end{equation}
%
for all $t$.  Since $a(t) \to l$ as $|t| \to \infty$, and
$\omega(t) \in (-1, 1)$ for all $t$, (A2) gives us
%
\begin{equation}\label{e4.02}
\begin{aligned}
F(\gamma_0(0))
&= F(\omega)
 = \int_{-\infty}^\infty \frac{1}{2}\dot{\omega}(t)^2
   + a(t)V(\omega(t))\,dt  \\
&< \int_{-\infty}^\infty \frac{1}{2}\dot{\omega}(t)^2 +
                        LV(\omega(t))\,dt  \\
&= \int_{-\infty}^\infty \frac{1}{2}\dot{\omega}(t)^2 +
     lV(\omega(t))\,dt + (L-l)\int_{-\infty}^\infty
      V(\omega(t))\,dt  \\
&= \mathcal{B} + \frac{4\nu\sqrt{l\underline{l}}}
                        {\int_{-1}^1 \sqrt{V(x)}\,dx}
     \int_{-\infty}^\infty   \frac{1}{\sqrt{2l}}\dot{\omega}(t)
     \sqrt{V(\omega(t))}\,dt =
      \mathcal{B} + 2\nu\sqrt{2\underline{l}}.
\end{aligned}
\end{equation}
%
 We must prove that the gradient vector flow from \eqref{e2.18}
is well-defined on $\mathbb{R}^+ \times W$.  Since $F$ is $C^2$,
it suffices to show that for all $A > 0$, there exists $B > 0$
 such that if $x \in W$ with $F(x) \leq A$, $\|F'(x)\| \leq B$:
 By (V5), it is possible to extend $V$ from
$[-1-(\xi_-+1),1+(1-\xi_+)]$ (see \eqref{e3.20}) to $\mathbb{R}$
such that there exists $K > 0$ with $V'(x)^2 \leq KV(x)$ for all
real $x$.  Let $x \in W$ with $F(x) \leq A$ and
    $u \in W^{1,2}(\mathbb{R})$ with
$\|u\|_{W^{1,2}(\mathbb{R})} = 1$.  Then
%
\begin{equation}\label{e4.021}
\begin{aligned}
F'(x)u
&= \int_{-\infty}^\infty \dot{x}(t)\dot{u}(t) + a(t)V'(x(t))u(t)\,dt \\
&\leq \Big(\int_{-\infty}^\infty \dot{x}(t)^2\,dt\Big)^{1/2}
      \Big(\int_{-\infty}^\infty u(t)^2\,dt\Big)^{1/2} \\
&\quad + L\Big(\int_{-\infty}^\infty V'(x(t))^2\,dt\Big)^{1/2}
      \Big(\int_{-\infty}^\infty u(t)^2\,dt\Big)^{1/2} \\
&\leq \sqrt{2A} + L\Big(\int_{-\infty}^\infty KV(x(t))\,dt\Big)^{1/2}  \\
&\leq \sqrt{2A} + L\sqrt{K/\underline{l}}
      \Big(\int_{-\infty}^\infty a(t) V(x(t))\,dt\Big){1/2}  \\
 &\leq \sqrt{2A} + L\sqrt{KA/\underline{l}}.
\end{aligned}
\end{equation}
%
Here is the ``standard deformation argument'' alluded to after
\eqref{e2.15}: suppose $c > \mathcal{B}$, and suppose there does
not exist a Palais-Smale sequence $(x_n) \subset W$ with $F(x_n)
\to c$ and $F'(x_n) \to 0$.  Then there exist $\epsilon, \delta >
0$ such that $\|F'(x_n)\| > \delta$ for all $x \in W$ with $F(x)
\in [c-\epsilon, c+\epsilon]$.  Let $\gamma \in \Gamma$ with
$\sup_{t \in \mathbb{R}} F(\gamma(t)) < c + \epsilon$.  Let $T >0$
be large enough so that $F(\gamma(t)) < c\ (> \mathcal{B})$ for
$|t| \geq T$. Let $\varphi\in C(\mathbb{R}, [0,1])$ with $\varphi
= 0$ on $(-\infty, -T-1] \cup [T+1,\infty)$ and $\varphi = 1$ on
$[-T,T]$. Define $\gamma_2 \in \Gamma$ by $\gamma_2(t) =
\eta(\frac{2\varphi(t)\epsilon}{\delta^2},\gamma(t))$, where
$\eta$ is the gradient vector flow from \eqref{e2.18}.  Since
$\frac{d}{ds}F(\eta(s,u)) = -\|F'(\eta(s,u)\|^2$ for all $u \in
W$, $s \in \mathbb{R}^+$, $F(\gamma_2(t)) < c$ for all $t \in
\mathbb{R}$. $F(\gamma_2(t)) \to \mathcal{B}$ as $|t| \to \infty$,
so $\sup_{t\in \mathbb{R}} F(\gamma_2(t)) < c$, contradicting the
definition of $c$.

    In the $c = \mathcal{B}$ case after \eqref{e2.15}, we have
$(x_n) \subset W$ with $F(x_n) \to b \leq \mathcal{B}$ as $n \to \infty$
and $x_n(0) = 0$ for all $n$.  Since $F(x_n)$ is bounded, there
exists $\bar{x} \in W^{1,2}_{\rm loc}(\mathbb{R})$ and a
subsequence of $(x_n)$ (also denoted $(x_n)$) such that $x_n \to
\bar{x}$ locally uniformly and weakly in $W^{1,2}([-T,T])$ for all
$T > 0$.   As before, $F(\bar{x}) \leq b \leq \mathcal{B}$.  We must
prove $\bar{x} \in W$.  Suppose otherwise.  By Lemma~\ref{l3.01},
$\bar{x}(t) \to 1$ or $-1$ as $t \to \infty$ and $\bar{x}(t) \to
1$ or $-1$ as $t \to -\infty$.  Suppose $\bar{x}(t) \to -1$ as $t
\to \infty$ (the proof for $\bar{x}(t) \to 1$ as $t \to -\infty$
is similar).  Let $\mathcal{B}_\epsilon$ be as in \eqref{e3.41}
and let $\epsilon > 0$ be small enough that
%
\begin{equation}\label{e4.03}
\frac{l-\epsilon}{l}\mathcal{B}_\epsilon >
                    \mathcal{B} - F_{[-1,1]}(\bar{x})/2.
\end{equation}
%
Let $T > 1$ be large enough so $a \geq l - \epsilon$ on $[t,
\infty)$ and $\bar{x}(T) < -1 + \epsilon$.  Then, as in
\eqref{e3.43}, for large $n$,
%
\begin{equation}
F(x_n) \geq F_{[-1,1]}(x_n) + F_{[T,\infty)}(x_n) \geq
  F_{[-1,1]}(\bar{x})/2 + \frac{l - \epsilon}{l}\mathcal{B}_\epsilon > \mathcal{B}.
\end{equation}
%
This is impossible.

The final step in the proof is to show that a solution of
\eqref{e1.00001} in $W$ takes values in $(-1,1)$.  Suppose $x \in
W$ and solves \eqref{e1.00001}.  If $x(t) > 1$ for some real $t$,
then let $t_{\rm max} \in \mathbb{R}$ with $x(t_{\rm max}) =
\max_{t \in \mathbb{R}} x(t)$.
    $\ddot{x}(t_{\rm max}) \leq 0$, but $V(x(t_{\rm max})) > 0$.  This is impossible.  Similarly, $x(t) \leq 1$ for all real $t$.  Now suppose $x(t^*) = 1$.
Then $x$ satisfies the Cauchy problem \eqref{e1.00001}, $x(t^*) =
1$, $\dot{x}(t^*) = 0$, so by (V1), $x \equiv 1$.  This is a
contradiction.  Similarly, $x(t) > -1$ for all real $t$.


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\end{document}