\documentstyle[twoside]{article}
\pagestyle{myheadings}
\markboth{\hfil HOMOCLINIC ORBITS \hfil EJDE--1994/01}%
{EJDE--1994/01\hfil P. Korman \& A.C. Lazer\hfil}
\begin{document}
\ifx\Box\undefined \newcommand{\Box}{\diamondsuit}\fi
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc Electronic Journal of Differential Equations}\newline
Vol. {\bf 1994}(1994), No. 01, pp. 1-10. Published February 15, 1994.\newline
ISSN 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp (login: ftp) 147.26.103.110 or 129.120.3.113 }
\vspace{\bigskipamount} \\
Homoclinic Orbits for a Class of Symmetric Hamiltonian Systems
\thanks{ {\em 1991 Mathematics Subject Classifications:}
34B15, 34A34.\newline\indent
{\em Key words and phrases:} Homoclinic orbits, mountain-pass lemma.
\newline\indent
\copyright 1994 Southwest Texas State University and University of
North Texas.\newline\indent
Submitted: October 22, 1993.\newline\indent
Supported in part by the Taft Faculty Grant at
the University of Cincinnati (P. K.)\newline\indent
Supported in part by NSF under Grant DMS-91023 (A. C. L.)} }
\date{}
\author{Philip Korman\\ and \\ Alan C. Lazer}
\newtheorem{thm}{Theorem}[section]
\newtheorem{lma}{Lemma}[section]
\newtheorem{cor}{Corollary}
\newtheorem{ex}{Example}
\renewcommand{\theequation}{\arabic{section}.\arabic{equation}}
\newcommand{\beq}{\begin{equation}}
\newcommand{\eeq}{\end{equation}}
\newcommand{\s}{\;\;\;}
\newcommand{\ol}{\overline}
\maketitle
\begin{abstract}
We study existence of homoclinic orbits for a class of Hamiltonian systems
that are symmetric with respect to independent variable (time). For the
scalar case we prove existence and uniqueness of a positive homoclinic
solution. For the system case we prove existence of symmetric homoclinic
orbits. We use variational approach.
\end{abstract}
\section{Introduction}
\setcounter{equation}{0}
Recently variational techniques have been applied to obtain existence of
homoclinic orbits of the Hamiltonian systems
\beq
u'' - L(t)u + V_u(t,u) = 0 \, ,
\label{1.1}
\eeq
see e.g. P.H. Rabinowitz \cite{4} and W. Omana and M. Willem \cite{3}. Here
$L(t)$ is a positive definite $n\times n$ matrix, $V$ is assumed to be
superquadratic at infinity and subquadratic at zero in $u\,$, and the
solution $u(t) \in H^1(R,R^n)$ is homoclinic at zero, i.e.,
$\lim_{t\rightarrow\pm\infty} u(t) = 0 \,$. The technical difficulty in
applying the mountain pass lemma on the infinite interval
$t\in(-\infty,\infty)\,$, is in verifying the Palais-Smale or ``compactness''
condition. In \cite{3} a new compact embedding theorem was used to verify
the Palais-Smale condition. However, one had to assume that the smallest
eigenvalue of $L(t)$ tends to $\infty$ as $|t|\rightarrow\infty\,$, which is
a rather restrictive and not very natural condition, as it excludes e.g. the
case of constant $L \,$. In this paper we show that this assumption
is not necessary in case $L(t)$ and $V(t,u)$ are even in $t\,$. For the
scalar case we also show existence and uniqueness of a positive homoclinic
orbit.
We begin by considering the equation (\ref{1.1}) on a finite interval
$(-T,T)$ together with the boundary conditions
\beq
u(-T) = u(T) = 0 \, .
\label{1.2}
\eeq
Using the mountain pass lemma we show that the problem (\ref{1.1}-{1.2}) has
a nontrivial solution. Moreover, using variational approach, we derive
uniform in $T$ estimate of $H^1$ norm of the solution. This is the crucial
step, which allows us to obtain homoclinic orbits by letting
$T\rightarrow\infty \,$.
\section{Positive homoclinics for a scalar equation}
\setcounter{equation}{0}
In this section we will prove existence and uniqueness of positive
homoclinics for a model problem with a cubic nonlinearity. Namely, we are
looking for a positive solution of
\beq
u'' - a(x)u + b(x)u^3 = 0 \,, \s -\infty < x < \infty \, ,
\label{2.1}
\eeq
\beq
u(-\infty) = u(\infty) = u'(-\infty) = u'(\infty) = 0 \, .
\label{2.2}
\eeq
We assume that the functions $a(x), b(x) \in C^1(-\infty,\infty)$ are
strictly positive on $(-\infty,\infty)\,$, i.e. $a(x) \geq a_0 > 0$ and $b(x)
\geq b_0 > 0$ and moreover we assume that $a(x)$ and $b(x)$ are even with
respect to some real number $c \,$. Without loss of generality we will
assume that $c=0\,$, i.e. $a(x)$ and $b(x)$ are even functions. We assume
additionally that $xa'(x) > 0$ and $xb'(x) < 0$ for all $x \neq 0 \,$.
We shall obtain the solution of (\ref{2.1}-\ref{2.2}) as the limit as $T
\rightarrow \infty$ of the solutions of
\beq
u'' - a(x)u + b(x)u^3 = 0 \s \mbox{for} \s x\in(-T,T), \s u(-T) = u(T)
= 0 \, .
\label{2.3}
\eeq
We shall need the following lemma from P. Korman and T. Ouyang \cite{2}.
(Except for the last assertion, this lemma is also included in B. Gidas,
W.-M. Ni and L. Nirenberg \cite{1}).
\begin{lma}
Consider the problem
\beq
u'' + f(x,u) = 0 \s \mbox{for} \s x\in(-T,T), \s u(-T) = u(T) = 0 \, .
\label{2.4}
\eeq
Assume that the function $f\in C^1([-T,T] \times R_+)$ is such that
\beq
f(-x,u) = f(x,u) \s \mbox{for} \s x\in(-T,T) \s \mbox{and} \s u>0 \, ,
\label{2.5}
\eeq
\beq
xf_x(x,u) < 0 \s \mbox{for} \s x\in(-T,T) \backslash \{0\} \s \mbox{and} \s
u>0 \, .
\label{2.6}
\eeq
Then any positive solution of (\ref{2.4}) is an even function with $u'(x)
< 0$ on $(0,T)\,$. Moreover any two positive solutions of (\ref{2.4})
cannot intersect on $(-T,T)$ (and hence they are strictly ordered on
$(-T,T)$).
\end{lma}
Clearly, under our conditions, the Lemma 2.1 applies to the problem
(\ref{2.3}).
We are looking for solution of (\ref{2.3}) which is strictly positive on
$(-T,T)$, and so we can work with an equivalent problem
\beq
\s u'' - a(x)u + b(x) (u^+)^3 = 0 \s \mbox{for $x\in(-T,T)$ , $u(-T) =
u(T) = 0 \, ,$}
\label{2.7}
\eeq
where $u^+ = \max(u,0)$.
\begin{lma}
The problem (\ref{2.3}) has under our conditions a unique positive solution
for any $T\geq 1 \,$. Moreover, for this solution we have an estimate
\beq
\int^T_{-T} \left(u'^2 + u^2 \right) \, dx \, \leq c \s \mbox{uniformly
in} \s T\geq 1 \, .
\label{2.8}
\eeq
\end{lma}
\noindent
{\bf Proof.} \hspace{.15in} We shall work with the problem (\ref{2.7})
using the space $H^1_0[-T,T]$ of absolutely continuous functions, which
vanish at $\pm T\,$, with the norm $\|u\|^2 = \int^T_{-T}(u'^2 + u^2)\, dx
\,$. We consider a functional $f: H^1_0 [-T,T] \rightarrow R\,$, defined by
\[
f(u) = \int^T_{-T} \left[ \frac{u'^2}{2} + a(x) \frac{u^2}{2} - b(x)
\frac{(u^+)^4}{4} \right] \, dx \, .
\]
Clearly $f(0) = 0\,$, and it is standard to verify that $f(u)$ satisfies the
Palais-Smale condition, and that $f(u)$ has a strict local minimum at
$u=0\,$. Starting with any function $u_0(x) \in H^1_0 [-1,1]\,$, such that
$u^+_0 \not\equiv 0\,$, we define $u^* \in H^1_0[-T,T]$ as follows: $u^*(x)
= \lambda u_0(x)$ for $x\in [-1,1]$, and $u^*(x) = 0$ for
$x\in[-T,T]\backslash [-1,1]\,$. Then $f(u^*) < 0$ for $\lambda$
sufficiently large, and we conclude by the well-known mountain pass theorem
that $f(u)$ has a nontrivial critical point $u(x) \in H^1_0[-T,T]\,$, which
is then easily seen to be a strictly positive solution of (\ref{2.3}).
The variational approach also allows us to derive the estimate (\ref{2.8}).
Indeed, if we define the set of paths
\[
\Gamma_T = \{ g(\tau):[0,1] \rightarrow H^1_0[-T,T] \; | \; g(0) = 0, \s g(1)
= u^* \} \, ,
\]
then the solution of $u(x)$ of (\ref{2.3}) is the point where
\beq
\inf_{g\in\Gamma_T} \; \max_{\tau\in[0,1]} \; f(g(\tau)) \equiv c_T \, ,
\label{2.9}
\eeq
is achieved. Let now $T_1 > T\, $. Then $\Gamma_T \subset \Gamma_{T_1}$,
since any function in $H^1_0[-T,T]$ can be regarded as belonging to
$H^1_0[-T_1,T_1]\,$, if one extends it by zero in $[-T_1,T_1] \backslash
[-T,T]\, $. Hence for $T_1$ the set of competing paths in (\ref{2.9}) is
greater than that for $T\,$, which implies that
\beq
c_{T_1} \leq c_T \leq c_1 \s (T_1 > T \geq 1) \, .
\label{2.10}
\eeq
So that for the positive solution of (\ref{2.3}),
\beq
\int^T_{-T} \left( \frac{u'^2}{2} + a(x) \frac{u^2}{2} - b(x) \frac{u^4}{4}
\right) \, dx \, \leq c_1 \s \mbox{uniformly in} \s T \geq 1 \, .
\label{2.11}
\eeq
Multiply (\ref{2.3}) by $u$ and integrate,
\beq
\frac{1}{4} \int^T_{-T} \left( u'^2 + a(x)u^2 - b(x) u^4 \right) \, dx
\, = 0 \, .
\label{2.12}
\eeq
Subtracting (\ref{2.12}) from (\ref{2.11}), we establish the estimate
(\ref{2.8}).
Turning to the uniqueness, we notice that any two different solutions $u(x)$
and $v(x)$ of (\ref{2.3}) would have to be ordered by Lemma 2.1, i.e. $u(x)
< v(x)$ for all $x\in (-T,T)\,$, which leads to a contradiction by an
application of Sturm comparison theorem.
\begin{thm}
The problem (\ref{2.1}-\ref{2.2}) has under our conditions exactly one
positive solution. Moreover this solution is an even function with $u'(x)
< 0$ for $x>0\, $.
\end{thm}
\noindent
{\bf Proof.} \hspace{.15in} Take a sequence $T_n \rightarrow \infty\,$, and
consider the problem (\ref{2.3}) on the interval $(-T_n,T_n)\,$, i.e.
consider
\beq
\s u'' - a(x)u + b(x)u^3 = 0 \s \mbox{on} \s (-T_n,T_n), \s u(-T_n) =
u(T_n) = 0 \, .
\label{2.3u}
\eeq
By Lemma 2.2 the problem (\ref{2.3u}) has a unique positive solution
$u_n(x)\,$, and
\beq
\int^{T_n}_{-T_n} \left( u'^2_n + u^2_n \right) \, dx \, \leq c \s
\mbox{uniformly in} \s n \, .
\label{2.13}
\eeq
By Lemma 2.1, $u_n(x)$ takes its maximum at $x=0\,$, which implies that
$u''_n(0) \leq 0\,$, and then from the equation (\ref{2.3u}),
\beq
u_n(0) \geq \sqrt{\frac{a(0)}{b(0)}} \, .
\label{2.14}
\eeq
Writing
\[
u_n(x_1) - u_n(x_2) = \int^{x_2}_{x_1} u'_n \,dx\, \leq \sqrt{x_2 - x_1}
\left( \int^{x_2}_{x_1} u'^2_n \, dx \, \right)^{1/2} \, ,
\]
we conclude that the sequence $\{u_n(x)\}$ is equicontinuous and uniformly
bounded on every interval $[-T_n,T_n]\,$. Hence it has a uniformly
convergent subsequence on every $[-T_n,T_n]\,$.
So let $\{u^1_{n_k}\}$ be a subsequence of $\{u_n\}$ that converges on
$[-T_1,T_1]\,$. Consider this subsequence on $[-T_2,T_2]$ and select a
further subsequence $\{u^2_{n_k}\}$ of $\{u^1_{n_k}\}$ that converges
uniformly on $[-T_2,T_2]\,$. Repeat this procedure for all $n\,$, and then
take a diagonal sequence $\{u_{n_k}\}\,$, which consists of $u^1_{n_1},
u^2_{n_2}, u^3_{n_3}, \ldots\,$. Since the diagonal sequence is a
subsequence of $\{u^p_{n_k}\}$ for any $p\geq 1\,$, it follows that it
converges uniformly on any bounded interval to a function $u(x)\,$.
Expressing $u''_{n_k}$ from the equation (\ref{2.3u}), we conclude that the
sequence $\{u''_{n_k}\}\,$, and then also $\{u'_{n_k}\}\,$, converge
uniformly on bounded intervals. Writing
\[
u_{n_k}(x) = \int^x_a (x-\xi) u''_{n_k} (\xi)\,d\xi \s \mbox{with} \s
a = -T_{n_k} - 1 \, ,
\]
we conclude that $u(x) \in C^2(-\infty,\infty)\,$, and that $u''_{n_k}
\rightarrow u''$ uniformly on bounded intervals. Hence, we can pass to
the limit in the equation (\ref{2.3u}), and we conclude that $u(x)$ solves
(\ref{2.1}). By (2.15) $u(x)$ is not identically zero.
Writing
\[
u^2(x) = \int^x_0 2uu' \, dx \, + u'^2(0) \, ,
\]
we conclude that the limits of $u(x)$ as $x\rightarrow\pm\infty$ exist
$(uu' \in L^1(-\infty,\infty))\,$. The only possibility is $u(\pm\infty) =
0\,$. Next we notice from (\ref{2.1}) that
\beq
|u''(x)| < c \s \mbox{for all real $x$ and some $c>0$ \, .}
\label{2.15}
\eeq
We claim that $u'(\pm\infty) = 0\,$. If not, there is an $\varepsilon > 0$
and a sequence of $x_n \rightarrow \infty$, such that
\[
|u'(x_n)| \geq \varepsilon \s \mbox{for all} \s n \, .
\]
By (\ref{2.15}) we can find $a$ $\delta\,$, such that
\[
|u'(x)| \geq \frac{\varepsilon}{2} \s \mbox{for all $x\in(x_n-\delta,
x_n+\delta)$ and all $n$} \, .
\]
This implies that $\int^{T_{n_k}}_{-T_{n_k}} u'^2(x)\, dx \,$ becomes large
with $k\,$, say, $\int^{T_{n_k}}_{-T_{n_k}} u'^2(x)\, dx \, \geq 2c$ for some
$k\,$, where $c$ is the constant from (\ref{2.13}). Then by fixing $j>k$ so
large that $u_{n_j}(x)$ and $u'_{n_j}(x)$ are uniformly close to $u(x)$ and
$u'(x)$ respectively on the interval $(-T_{n_k},T_{n_k})$ we get, using
(\ref{2.13}),
\[
\int^{T_{n_k}}_{-T_{n_k}} u'^2\, dx \, \simeq \int^{T_{n_k}}_{-T_{n_k}}
u'^2_{n_j} \, dx \, \leq \int^{T_{n_j}}_{-T_{n_j}} u'^2_{n_j} \, dx \, \leq c
\, ,
\]
which is a contradiction. So that $u(x)$ is a solution of our problem
(\ref{2.1}-\ref{2.2}).
Since by Lemma 2.1 the functions $u_{n_k}(x)$ are even, with the only
maximum at $x=0\,$, the same is true for their limit $u(x)\,$. That $u'(x) <
0$ for $x>0$ is easily seen by differentiating (\ref{2.1}) (a similar
argument can be found in \cite{2}).
Turning to uniqueness, let $v(x)$ be another positive solution of (2.1), (2.2)
(which is also an even function with the only maximum at $x=0\,$, as follows
by an easy modification of the proof of lemma 1 in [2]). Since
\[
\int^\infty_{-\infty} b(x) uv(u^2 - v^2) \, dx \, = 0 \, ,
\]
it follows that the solutions $u(x)$ and $v(x)$ cannot be ordered, and so
have to intersect. Two cases are possible: either $u(x)$ and $v(x)$ have at
least two positive points of intersection, or only one positive point of
intersection. Assume first $\xi_1 > 0$ is the smallest positive point of
intersection and $\xi_2 > \xi_1$ the next one, and $u(x) < v(x)$ on
$(\xi_1,\xi_2)\,$. Multiply the equation (\ref{2.1}) by $u'$ and integrate
from $\xi_1$ to $\xi_2\,$. Denoting by $x=x_1(u)$ the inverse function of
$u(x)$ on $(\xi_1,\xi_2)\,$, we obtain denoting $f(x,u) = -a(x)u +
b(x)u^3\,$, and $u_1 = u(\xi_1) = v(\xi_1)\,$, $u_2 = u(\xi_2) = v(\xi_2)\,$,
\beq
\frac{1}{2} u'^2(\xi_2) - \frac{1}{2} u'^2(\xi_1) + \int^{u_2}_{u_1}
f(x_1(u),u)\,du = 0 \, .
\label{2.16}
\eeq
Doing the same for $v(x)\,$, and denoting its inverse on $(\xi_1,\xi_2)$ by
$x = x_2(v)\,$, we obtain
\beq
\frac{1}{2} v'^2(\xi_2) - \frac{1}{2} v'^2(\xi_1) + \int^{u_2}_{u_1}
f(x_2(v),v)\, dv = 0 \, .
\label{2.17}
\eeq
Subtracting (\ref{2.17}) from (\ref{2.16}),
\begin{eqnarray}
\frac{1}{2}\left(u'^2(\xi_2) - v'^2(\xi_2)\right) + \frac{1}{2}
\left(v'^2(\xi_1) - u'^2(\xi_1)\right) \\
+ \int^{u_1}_{u_2} \left[f(x_2(u),u) - f(x_1(u),u)\right]\, du = 0 \, .
\nonumber
\label{2.18}
\end{eqnarray}
Notice, $u_2 < u_1$ and $x_2(u) > x_1(u)$ for all $u\in(u_2,u_1)\,$. Keeping
in mind that $f(x,u)$ is decreasing in $x\,$, and using uniqueness theorem for
initial value problems, we conclude that all three terms on the left in
(2.19) are negative. This is a contradiction, which rules out the case
of two positive intersection points. If $\xi_1$ is the only intersection
point, we integrate from $\xi_1$ to $\infty\,$, obtaining a similar
contradiction. Uniqueness of solution follows, completing the proof of the
theorem.
\section{Homoclinic orbits for a class of Hamiltonian systems}
\setcounter{equation}{0}
We are looking for nontrivial solutions $u\in H^1(R^n,R)$ of the system
\beq
u'' - L(t)u + \nabla V(u) = 0
\label{3.1}
\eeq
\beq
u(\pm\infty) = u'(\pm\infty) = 0 \, .
\label{3.2}
\eeq
We assume that $V \in C^1(R^n,R)\,$, and the following conditions
\begin{equation}
\begin{array}{l}
L(t) \s \mbox{is a positive definite matrix with entries of class $C^1(R)$,} \\
\mbox{and $L(-t) = L(t)$ for all $t$}\, , \end{array}
\label{3.3}
\end{equation}
\beq
(L'(t)\xi,\xi) \geq 0 \s \mbox{for all $\xi\in R^n$ and $t\geq 0$} \, ,
\label{3.4}
\eeq
\beq
\; \; 0 < \gamma V(\xi) \leq (\nabla V(\xi),\xi) \s \mbox{for some constant
$\gamma > 2$ and all $\xi \in R^n$} \, .
\label{3.6}
\eeq
\begin{thm}
Under assumptions (3.3-3.5) the problem (\ref{3.1}-\ref{3.2}) has
a nontrivial solution $u(t)\,$, with $u(-t) = u(t)$ for all $t\,$.
\end{thm}
We postpone the proof of the theorem, and present two examples, which show
that our result on scalar equations cannot be expected to carry over to
systems.
\vspace{.15in}\noindent
{\bf Example 1} \hspace{.15in} On some interval $(a,b)$ consider a system
\begin{equation}
\begin{array}{rcl}
u'' - 2u + u(u^2+v^2) = 0 &\mbox{ for }& x\in(a,b), \s u(a) = u(b) = 0\\
v'' - v + v(u^2+v^2) = 0 &\mbox{ for }& x\in(a,b), \s v(a) = v(b) = 0\,.
\end{array}\label{3.7} \end{equation}
This system has no positive solution (i.e. solution with $u>0$ and $v>0$ on
$(a,b)$). Indeed, we can regard the first equation in (3.6) as a
linear equation of the form $u'' + c(x)u = 0$, and the second one as $v'' +
d(x)v = 0\,$. Since $d(x) > c(x)$, the claim follows by the Sturm's
comparison theorem. This system is of type (\ref{3.1}) with $V = \frac{1}{4}
\left(u^4 + v^4\right) + \frac{1}{2} u^2 v^2\,$.
\vspace{.15in}
\noindent
{\bf Example 2} \hspace{.15in} The problem
\begin{equation}\begin{array}{rcl}
u'' - u + u(u^2+v^2) = 0 & \mbox{for} & x\in(a,b), \s u(a) = u(b) = 0 \\
v'' - v + v(u^2+v^2) = 0 & \mbox{for} & x\in(a,b), \s v(a) = v(b) = 0 \, ,
\end{array} \label{3.8}\end{equation}
has infinitely many positive solutions, all of the form $u = \alpha v\,$,
where $\alpha$ is an arbitrary positive constant. Indeed, regarding
$u^2 + v^2$ as a known function, we see that $u$ and $v$ are positive
solutions of the same linear equation, and so have to be multiples of one
another. Setting $u = \alpha v\,$, we find $u$ to be the unique (in view of
Lemma 2.2) positive solution of
\beq
u'' - u + u^3 \left( 1 + \frac{1}{\alpha^2} \right) = 0 \s \mbox{for} \s
x\in (a,b), \s u(a) = u(b) = 0 \, ,
\label{3.9}
\eeq
while $v$ is the unique positive solution of
\beq
v'' - v + v^3(\alpha^2 + 1) = 0 \s \mbox{for} \s x\in (a,b), \s v(a) =
v(b) = 0 \, .
%\label{3.10}
\eeq
Setting $u = \alpha v$ in (\ref{3.9}), we obtain (3.9), so that the pair
$(u,v)$ is indeed a solution of (3.7).
\vspace{.25in}
{\bf Proof of the Theorem 3.1.} We begin by showing that our condition (3.5)
implies that
\beq
\frac{V(\xi)}{|\xi|^2} \rightarrow 0 \s \mbox{as} \s |\xi| \rightarrow 0 \, .
\label{3.10}
\eeq
Indeed, write (3.5) at $r\xi$ with some constant $r>0$,
\[
r(\nabla V(r\xi),\xi) \geq \gamma V(r\xi)
\]
or
\[
\frac{d}{dr} V(r\xi) - \frac{\gamma}{r} V(r\xi) \geq 0 \, .
\]
Multiplying by $r^{-\gamma}$ and integrating over $(\epsilon, 1)\, ,$
$0 < \epsilon < 1\,$,
\[
V(\xi) - \frac{V(\epsilon \xi)}{\epsilon^\gamma} \geq 0 \, .
\]
Let now $|\xi| = 1$ and set $\eta = \epsilon \xi \, , \, |\eta| = \epsilon
\, $. Then
\[
\frac{|V(\eta)|}{|\eta|^\gamma} \leq c \, ,
\]
and (3.10) follows.
As in the previous section, we approximate the solution of
(\ref{3.1}-\ref{3.2}) by the problem
\beq
u'' - L(t)u + \nabla V(u) = 0 \; \mbox{for} \; t\in (-T,T), \;
u(-T) = u(T) = 0 \, ,
\label{3.11}
\eeq
\beq
u(-t) = u(t) \s \mbox{for all real $t$} \, .
\label{3.12}
\eeq
The key step is to show the existence of $\delta > 0\,$, such that any
nontrivial solution of (3.11), (3.12) satisfies
\beq
|u(0)| > \delta \s \mbox{independently of $T>0$} \, .
\label{3.13}
\eeq
To prove (\ref{3.13}) we introduce the ``energy'' function for the solution
$u(t)$ of (\ref{3.11}),
\[
E(t) = \frac{1}{2} |u'(t)|^2 - \frac{1}{2}(L(t)u,u) + V(u(t)) \, .
\]
Differentiating $E(t)\,$, and using the equation (\ref{3.11}) and the
condition (\ref{3.4}), we express
\[
E'(t) = -\frac{1}{2}(L'(t)u,u) \leq 0 \s \mbox{for all} \s 0\leq t \leq T
\, ,
\]
and hence
\[
E(0) \geq E(T) = \frac{|u'(T)|^2}{2} \geq 0 \, .
\]
Since $u(t)$ is even, $u'(0) = 0\,$, and then
\[
E(0) = V(u(0)) - \frac{1}{2} (L(0)u(0), u(0)) \geq 0 \, ,
\]
or
\[
V(u(0)) \geq \frac{1}{2} (L(0)u(0),u(0)) \geq c|u(0)|^2 \, ,
\]
\beq
\frac{V(u(0))}{|u(0)|^2} \geq c \, .
\label{3.14}
\eeq
Comparing (\ref{3.14}) with (\ref{3.10}), we conclude the estimate
(\ref{3.13}).
The rest of the proof is the same as that of the Theorem 2.1, except that we
use (3.13) instead of (2.15), so we only sketch it. We take a sequence
$\{T_k\} \rightarrow \infty$ as $k \rightarrow \infty\,$, $0 < T_k < T_{k+1}$
for all $k\,$. By $E_k$ we denote the subspace of $H^1_0(-T_k,T_k)\,$,
consisting of even functions. By taking zero extensions, we see that $E_k
\subset E_{k+1}\,$. We consider the functionals $f_k: E_k \rightarrow R\,$,
defined by
\[
f_k(u) = \int^{T_k}_{-T_k} \left[ \frac{1}{2} |u'|^2 + \frac{1}{2} (L(t)u,u)
- V(u) \right] \, dt \,.
\]
The mountain pass theorem applies to $f_k(u)\,$, producing $u_k \in E_k\,$,
which is an even nontrivial solution of
\beq
u''_k - L(t)u_k + \nabla V(u_k) = 0 \s \mbox{for} \s t\in(-T_k,T_k)
\label{3.15}
\eeq
\beq
u_k(-T) = u_k(T) = 0 \, .
\label{3.16}
\eeq
The critical values $c_k = f_k(u_k) > 0$ are non-increasing in $k\,$. And
$u_k(0) > \delta$ uniformly in $k\,$, in view of (\ref{3.13}).
We show next that the $H^1$ norm of the solution of (\ref{3.15}-\ref{3.16})
is bounded uniformly in $k\,$. Multiply the equation (\ref{3.15}) by $u_k$
and integrate,
\beq
\int^{T_k}_{-T_k} \left[ |u'_k|^2 + (L(t)u_k,u_k) - (\nabla V(u_k),u_k)
\right] \, dt \, = 0 \, .
\label{3.17}
\eeq
On the other hand by the definition of $c_k$ and (\ref{3.6}),
\begin{eqnarray}
\int^{T_k}_{-T_k} \left[ \frac{1}{2} |u'_k|^2 + \frac{1}{2} (L(t)u_k,u_k)
\right] \, dt \, = \int^{T_k}_{-T_k} V(u_k) \, dt \, + c_k \\
\leq \frac{1}{\gamma} \int^{T_k}_{-T_k} (\nabla V(u_k),u_k)\, dt \, + c_k
\,. \nonumber
\label{3.18}
\end{eqnarray}
Using (\ref{3.17}) in (3.18), we obtain
\beq
\left(\frac{1}{2} - \frac{1}{\gamma}\right) \int^{T_k}_{-T_k} \left[
|u'_k|^2 + (L(t)u_k,u_k) \right] \, dt \, \leq c_k \leq c_1 \, .
\label{3.19}
\eeq
Since $\frac{1}{2} - \frac{1}{\gamma} > 0\,$, and
\[
(L(t)u_k,u_k) \geq (L(0)u_k,u_k) \geq c|u_k|^2 \, ,
\]
we conclude from (\ref{3.19}) that
\[
\int^{T_k}_{-T_k} \left(|u'_k|^2 + |u_k|^2 \right) \, dt \, \leq c \s
\mbox{uniformly in $k$} \, .
\]
The rest of the proof is the same as in the Theorem 2.1.
\vspace{.15in}
\noindent
{\bf Remark.} We remark that we can prove a similar result with $V$
depending on $t\,$, provided the condition (3.5) is uniform in $t\,$,
and (3.4) is replaced by
\[
(L'(t)\xi, \xi) - V_t(t,\xi) \geq 0 \; \; \mbox{for all} \; \; \xi \in
R^n \; \; \mbox{and} \; \; t \geq 0 \, .
\]
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{\sc Philip Korman\newline
Institute for Dynamics and Department of
Mathematical Sciences\newline
University of Cincinnati\newline
Cincinnati, OH 45221-0025} \newline
E-mail address: korman@ucbeh.san.uc.edu \par
{\sc Alan C. Lazer\newline
Department of Mathematics \& Computer Science\newline
University of Miami\newline
Coral Gables, FL 33124}\newline
E-mail address: lazer@mthvax.cs.miami.edu
\end{document}