\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 % $$\label{e1.00003} F_l(u) = \int_{-\infty}^\infty \frac{1}{2}\dot{u}(t)^2 - lV(u(t))\,dt$$ % over the affine space % $$\label{e1.00004} W = \{u \in W^{1,2}_{\rm loc}(\mathbb{R}) : u + 1 \in W^{1,2}((-\infty,0]),\; u - 1 \in W^{1,2}([0,\infty))\}.$$ % An interesting problem is to replace $l$ by a nonconstant, positive coefficient function $a(t)$ and find conditions on $a$ under which % $$\label{e1.00005} \ddot{x}(t)= a(t)V'(x(t))$$ % 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 % \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} % 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 $$\label{e1.01} \xi_- = \min\{x : x > -1, V'(x) = 0\},\quad \xi_+ = \max\{x : x < 1, V'(x) = 0\}.$$ % Note that $\xi_-$ and $\xi_+$ are well-defined by (V3)-(V5). Define % $$\label{e1.02} \nu = \min\Big(\int_{-1}^{\xi_-} \sqrt{V(x)}\,dx,\int_{\xi_+}^1 \sqrt{V(x)}\,dx\Big) > 0.$$ % 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 % $$\label{e2.04} F(x) = \int_{-\infty}^\infty \frac{1}{2} \dot{x}(t)^2 + a(t)V(x(t))\,dt.$$ % By (V1)-(V3), $F(x) < \infty$ for all $x \in W$. $F:W \to \mathbb{R}^+$ is Fr\'echet differentiable with % $$\label{e2.06} F'(x)u = \int_{-\infty}^\infty \dot{x}(t)\dot{u}(t) + V'(x(t))u(t)\,dt$$ % 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 % $$\label{e2.09} {\mathcal B} = F_l(\omega) > 0,$$ % 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 % $$\label{e2.10} \|F'(x)\| = \sup\{F'(x)u : u \in W^{1,2}(\mathbb{R}),\; \|u\|_{W^{1,2}(\mathbb{R})} = 1\}.$$ % We will use the usual norm on $W^{1,2}(\mathbb{R})$, % $$\label{e2.11} \|u\|_{W^{1,2}(\mathbb{R})} = \Big(\int_{-\infty}^\infty \dot{u}(t)^2 + u(t)^2\,dt\Big)^{1/2}.$$ % 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 % $$\label{e2.13} F(x_n) \to b \in [0,\mathcal{B}) \cup (\mathcal{B}, \mathcal{B} + 2\nu\sqrt{2\underline{l}}).$$ % 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 % $$\label{e2.14} \Gamma = \{\gamma \in C(\mathbb{R},W) : \|\tau_t \omega - \gamma(t)\| \to 0 \text{ as } |t| \to \infty\}$$ % and % $$\label{e2.15} c = \inf_{\gamma \in \Gamma} \sup_{t \in \mathbb{R}} F(\gamma(t)).$$ % 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 $\varphi \in W^{1,2}(\mathbb{R})$, % $$\label{e2.16} (\mathcal{W}(y),\varphi)_{W^{1,2}(\mathbb{R})} = F'(y)\varphi,$$ % where $(\cdot ,\cdot )$ is the standard inner product on $W^{1,2}(\mathbb{R})$, % $$\label{e2.17} (f,g)_{W^{1,2}(\mathbb{R})} = \int_{-\infty}^\infty \dot{f}(t)\dot{g}(t) + f(t)g(t)\,dt.$$ % Let $\eta$ denote the solution of the gradient vector flow induced by the initial value problem: % $$\label{e2.18} \frac{d}{ds}\eta(s,u) = -\mathcal{W}(\eta(s,u));\ \eta(0,u) = u.$$ % 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 % $$\label{e3.01} V(x) > 0 \quad\text{ for all } |x| > 1, \quad\text{and}\quad {\lim}_{|x|\to\infty} V(x) = \infty.$$ % 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 $n \to \infty$ % $$\label{e3.02} x_n(t) \in (-\infty, -1-\delta) \cup (-1+\delta, 1-\delta) \cup (1+\delta, \infty).$$ % Let % $$\label{e3.04} d = \inf\{V(x) : x \in (-\infty, -1-\delta/2) \cup (-1+\delta/2, 1-\delta/2) \cup (1+\delta/2, \infty)\} > 0.$$ % 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, % \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} % $$\label{e3.06} \int_{t_n}^{t_n + 1} \dot{x}(t)^2\,dt \geq \int_{t_n}^{t^*} \dot{x}(t)^2\,dt \geq \delta^2/4.$$ % Either way, $$\label{e3.07} \int_{t_n}^{t_n + 1} \frac{1}{2}\dot{x}(t)^2 + a(t)V(x(t))\,dt \geq \min(\delta^2/8, d\underline{l}),$$ and $$\label{e3.08} F(x) \geq \sum_{n=1}^\infty \int_{t_n}^{t_n + 1} \frac{1}{2}\dot{x}(t)^2 + a(t)V(x(t))\,dt \geq \sum_{n=1}^\infty \min(\delta^2/8, d\underline{l}) = \infty,$$ % 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 % $$\label{e3.09} \int_T^\infty (x(t)-1)^2\,dt \leq \int_T^\infty V(x(t))/\epsilon\,dt \leq \frac{1}{\epsilon\underline{l}}\int_T^\infty a(t)V(x(t))\,dt \leq \frac{F(x)}{\epsilon\underline{l}} < \infty$$ % 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 $s \in [-T,T]$, % \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{proof} For $\Omega \subset \mathbb{R}$, define $$\label{e3.10} F_\Omega (x) = \int_\Omega \frac{1}{2}\dot{x}(t)^2 + a(t)V(x(t))\,dt.$$ 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 % $$\label{e3.12} F_{[t_0,t_1]}(x) \geq \sqrt{2 \underline{l}} \big|\int_{x_0}^{x_1} \sqrt{V(x)}\,dx \big|.$$ \end{lemma} \begin{proof} Let $\omega_{\underline{l}}$ denote the unique solution in $W$ of the differential equation % $$\label{e3.13} \ddot{x}(t) = \underline{l}V'(x(t))$$ % satisfying $\omega_{\underline{l}}(0) = 0$. Then $\omega_{\underline{l}}$ minimizes the functional % $$\label{e3.14} F_{\underline{l}}(u) = \int_{-\infty}^\infty \frac{1}{2} \dot{u}(t)^2 + \underline{l}V(u(t))\,dt$$ % over $W$. By the argument following \eqref{e1.00002}, % $$\label{e3.15} \dot{\omega}_{\underline{l}}(t) = \sqrt{2\underline{l}V(\omega_{\underline{l}}(t))}.$$ 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 % $$\label{e3.16} \int_{t_0}^{t_1} \frac{1}{2}\dot{x}(t)^2 + \underline{l}V(x(t))\,dt \geq \int_{t_0}^{t_1} \frac{1}{2}\dot{\omega}_{\underline{l}}(t)^2 + \underline{l}V(\omega_{\underline{l}}(t))\,dt;$$ % 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 % \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} % Now by \eqref{e3.15}-\eqref{e3.16}, % $$\label{e3.18} F_{[t_0,t_1]}(x) \geq \int_{t_0}^{t_1} \dot{\omega}_{\underline{l}}(t)^2 = \int_{t_0}^{t_1}\dot{\omega}_{\underline{l}}(t) \sqrt{2\underline{l}V(\omega_{\underline{l}}(t))}\,dt = \int_{x_0}^{x_1} \sqrt{2\underline{l}V(x(t))}\,dt.$$ 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, % \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{proof} Recall that $\xi_-$ and $\xi_+$ from \eqref{e1.01}, and assume from now on that % $$\label{e3.20} \begin{gathered} V(x) = V(-1 + (-1-x)) \quad\text{ for all } x \in [-1-(\xi_- + 1), -1], \\ V(x) = V(1 - (x-1)) \quad \text{ for all } x \in [1,1+(1-\xi_+)]. \end{gathered}$$ % 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 $F'(x_n) \to 0$, % $$\label{e3.206} F(x_n) \to b < 2\mathcal{B} + \sqrt{2 \underline{l}}\int_{-1}^1\sqrt{V(x)}\,dx,$$ 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 % \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} % 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 $L^2([-T,T])$, % \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} % 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 % $$\label{e3.23} u_n(t)= \begin{cases} 0 & t \leq -T-1 \\ (x_n(-T)-\bar{x}(-T))(t+T+1) & -T-1 \leq t \leq -T\\ x_n(t)-\bar{x}(t) & -T \leq t \leq T\\ (x_n(T)-\bar{x}(T))(-t + T + 1) &T \leq t \leq T+1\\ 0 &t \geq T+1 \end{cases}$$ % Clearly, $(u_n)$ is bounded in $W^{1,2}(\mathbb{R})$. Since $u_n \to 0$ uniformly on $[-T-1,T+1]$, % \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} % Since $\|u_n\|_{W^{1,2}([-T-1,-T])} \to 0$ and $\|u_n\|_{W^{1,2}([T,T+1])} \to 0$ as $n \to \infty$, % \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} % 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 % $$\label{e3.26} \|x_n - \bar{x}\|_{\mathbb{R} \setminus [-T_n, T_n]}^2 \geq 4\delta^2$$ % for all $n$. Along a subsequence, either % $$\label{e3.261} \|x_n - \bar{x}\|_{W^{1,2}((-\infty, -T_n])}^2 \geq 2\delta^2\quad \text{or}\quad \|x_n - \bar{x}\|_{W^{1,2}([T_n,\infty))}^2 \geq 2\delta^2.$$ % Let us assume the former; the latter case is similar. Since $1 + \bar{x} \in W^{1,2}((-\infty,0])$, % $$\label{e3.262} \|x_n + 1\|_{W^{1,2}((-\infty, -T_n])} \geq \delta$$ % 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 % $$\label{e3.28} V'(x)x \geq c(1+x)^2$$ % 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 % $$\label{e3.281} |1 + x_n(t)| < \min\big(1 + \xi^*, \frac{c\delta^2}{8(1+\sqrt{b})}\big)$$ % for all $n \in \mathbb{N}$, $t \leq -M$. Define $(u_n) \subset W^{1,2}(\mathbb{R})$ by % \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} % We will show $(u_n)$ is uniformly bounded in $W^{1,2}(\mathbb{R})$. Let $K > 0$ so % $$\label{e3.30} |V'(x)| \leq K \quad\text{and}\quad (x+1)^2 \leq KV(x)$$ % for all $x \in [-1-(\xi^* + 1),\xi^*]$. This is possible by (V1)-(V5),\eqref{e3.20}, and the definition of $\xi_-$. For large $n$, % \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} % Since $F'(x_n) \to 0$, $F'(x_n)u_n \to 0$ as $n \to \infty$. But for large $n$, % \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} % by \eqref{e3.281}. This is impossible. Case II: by the arguments of Lemma~\ref{l3.11}, % $$\label{e3.34} F(x) \geq \sqrt{2\underline{l}}\int_{-1}^1 \sqrt{V(x)}\,dx$$ % 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 % $$\label{e3.35} F_{[-M,M]}(\bar{x}) > \sqrt{2\underline{l}}\int_{-1}^1 \sqrt{V(x)}\,dx - \frac{1}{10}(2\mathcal{B}+ \sqrt{2\underline{l}}\int_{-1}^1 \sqrt{V(x)}\,dx - b).$$ % Define $\alpha_n \leq \beta_n < -M$ by % $$\label{e3.36} \beta_n = \max\{t < -M : |1+x_n(t)| = d\},\quad \alpha_n = \min\{t : |1+x_n(t)| = d\}$$ % 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}, % $$\label{e3.37} \|x_n - \tau_{\lambda+\beta_n}\omega_R\|_{W^{1,2} ([\beta_n - T,\beta_n + T])} \to 0$$ % 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 % $$\label{e3.38} \|x_n - \tau_{\lambda_2+\alpha_n}\omega\|_{W^{1,2}([\alpha_n - T, \alpha_n + T])} \to 0$$ % for all $T > 0$ as $n \to \infty$. For $\Omega \subset \mathbb{R}$, define % $$\label{e3.385} {F_l}_\Omega (x) = \int_\Omega \frac{1}{2}\dot{x}(t)^2 + lV(x(t))\,dt$$ % Still assuming that $M$ is large enough so that \eqref{e3.35} holds, assume also that $M$ is large enough that % $$\label{e3.39} \begin{gathered} {F_l}_{[-M,M]}(\tau_\lambda \omega_R) >\mathcal{B} - \frac{1}{10}(2\mathcal{B} + \sqrt{2\underline{l}}\int_{-1}^1 \sqrt{V(x)}\,dx - b),\\ {F_l}_{[-M,M]}(\tau_{\lambda_2} \omega) >\mathcal{B} - \frac{1}{10}(2\mathcal{B} +\sqrt{2\underline{l}}\int_{-1}^1 \sqrt{V(x)}\,dx - b), \end{gathered}$$ % 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 % \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} % 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 % $$\label{e3.41} \mathcal{B}_\epsilon = \int_{\omega^{-1}(-1+\epsilon)} ^{\omega^{-1}(1-\epsilon)} \frac{1}{2}\dot{\omega}^2(t) + lV(\omega(t))\,dt$$ % for $\epsilon > 0$. Let $\epsilon> 0$ be small enough that % $$\label{e3.42} \Big( \frac{l-\epsilon}{l} \Big)\mathcal{B}_\epsilon > b.$$ % 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}, % \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} % 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}, % $$\label{e3.44} \|x_n - \tau_{t_n}\omega\|_{W^{1,2}([t_n -M,t_n + M])} \to 0$$ % as $n \to \infty$ for all $M > 0$. Let $-1 < e_1 < \xi_i < \xi_-$ with % $$\label{e3.45} \sqrt{2\underline{l}}\int_{e_1}^{\xi_1}\sqrt{V(x)}\,dx > \nu\sqrt{2\underline{l}} - \frac{1}{5}(\mathcal{B} + 2\nu\sqrt{2\underline{l}} -b).$$ % Let $c \in (0,1)$ with % $$\label{e3.46} V'(x)(1+x) \geq cV(x)$$ % for all $x \in [-1, \xi_1]$. Let $K > 0$ be large enough that % $$\label{e3.47} |V'(x)| < K$$ % 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 $t \to \infty$, % $$\label{e3.50} F_{[t_n - M,t_n+M]}(x_n) < \mathcal{B} + \frac{1}{2}(b-\mathcal{B})$$ % for large $n$, so % $$\label{e3.51} F_{(-\infty, t_n-M]}(x_n) + F_{[t_n + M,\infty)}(x_n) > \frac{1}{3}(b-\mathcal{B}).$$ % 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 $t_n \to \infty$, % \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} % This is impossible. For Case II, define $(u_n) \subset W^{1,2}(\mathbb{R})$ by % $$\label{e3.54} u_n(t) = \begin{cases} 1+x_n(t) & t \leq t_n - M \\ (1+x_n(t_n-M))(t_n-M+1-t) & t_n - M \leq t \leq t_n -M + 1\\ 0 & t \geq t_n -M +1. \end{cases}$$ % 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 $n$, % \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} % 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 % $$\label{e4.01} V(\omega(t)) = \frac{\dot{\omega}(t)}{\sqrt{2l}}\sqrt{V(\omega(t))}$$ % for all $t$. Since $a(t) \to l$ as $|t| \to \infty$, and $\omega(t) \in (-1, 1)$ for all $t$, (A2) gives us % \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} % 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 % \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} % 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 % $$\label{e4.03} \frac{l-\epsilon}{l}\mathcal{B}_\epsilon > \mathcal{B} - F_{[-1,1]}(\bar{x})/2.$$ % 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 $n$, % $$F(x_n) \geq F_{[-1,1]}(x_n) + F_{[T,\infty)}(x_n) \geq F_{[-1,1]}(\bar{x})/2 + \frac{l - \epsilon}{l}\mathcal{B}_\epsilon > \mathcal{B}.$$ % 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. 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