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\markboth{\hfil Interfering solutions \hfil EJDE--2001/47}
{EJDE--2001/47\hfil Gregory S. Spradlin \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 2001}(2001), No. 47, pp. 1--10. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu  (login: ftp)}
 \vspace{\bigskipamount} \\
 %
  Interfering solutions of a nonhomogeneous Hamiltonian system
 %
\thanks{ {\em Mathematics Subject Classifications:} 35A15.
\hfil\break\indent
{\em Key words:} Variational methods, minimax argument, 
nonhomogeneous linearity, \hfil\break\indent
Hamiltonian system, Nehari manifold.
\hfil\break\indent
\copyright 2001 Southwest Texas State University. \hfil\break\indent
Submitted April 30, 2001. Published June 21, 2001.} }
\date{}
%
\author{Gregory S. Spradlin}
\maketitle

\begin{abstract} 
 A Hamiltonian system is studied which has a term approaching a constant at an
 exponential rate at infinity .  A minimax argument is used to show that the
 equation has a positive homoclinic solution.  The proof employs the interaction
 between translated solutions of the corresponding homogeneous equation. 
 What distinguishes this result from its few predecessors is that the equation 
 has a nonhomogeneous nonlinearity.
\end{abstract}


\newtheorem{theorem}{Theorem}[section]
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\newtheorem{proposition}[theorem]{Proposition}

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\section{ Introduction}

	This paper is inspired by a remarkable result of Bahri 
and Li \cite{b1}, which is a proof of a result of Bahri and 
Lions \cite{b2} employing a minimax method.  The papers treat an 
elliptic partial differential equation of the form 
$-\Delta u + u = b(x)u^p$ on ${\mathbb{R}^N}$, 
with $u^p$ a 
superquadratic, subcritical nonlinearity ($1 < p$, and
$p < (N+2)/(N-2)$ for $N \geq 3$), and $b(x)\to b_\infty > 0$
as $|x| \to \infty$.  One searches for positive solutions $u$ 
that decay to zero as $|x| \to \infty$.  This nonautonomous 
problem on the noncompact domain ${\mathbb{R}^N}$ is difficult to solve
without assuming symmetry on $b(x)$.  
Bahri and Li found decaying 
positive solutions under the assumption that $b$ is positive and 
continuous, and $(b(x) - b_\infty)_-$
(the negative part of $b(x) - b_\infty$) decays exponentially at 
a fast enough rate as $|x| \to \infty$ (to be precise, 
$b(x) - b_\infty = O(\exp(-(2 + \delta))$ for some $\delta > 0$).
 Surprisingly, unlike 
in other, perturbative results, $b(x) - b_\infty$ may be 
``large'' in just about any other sense, such as $L^q$ norm, 
$1 \leq q \leq \infty$.  

	The proof in \cite{b1} is variational.  A minimax class is formed
using sums of translates of a solution of
 the corresponding autonomous 
problem $-\Delta u + u = b_\infty u^p$, and exploiting the 
``interference'' between ``tails'' of that solution.  This 
idea contrasts with many ``multibump'' results, 
in which a 
multibump solution is one that resembles a sum of translates 
of solutions of an equation \cite{c2,c3,s1}.  In most of these results, the
interference between bumps is a problem that must be managed
by separating the translates by a great distance.  There seems 
to have been little work done
in exploiting interference 
of solutions in either multibump or simple existence results 
in elliptic PDE.  A notable exception is a singularly 
perturbed elliptic equation studied by Wei and Xiaosong \cite{w1}.  
They employed interference between bumps to ``hold them together'' and 
counteract the lack of compactness in the problem and find 
multibumps.  A related result is \cite{d1}.  

Articles \cite{b1,w1,d1} all involve an equation with homogeneous, 
or power linearity.  The new paper \cite{a1} proves a result similar 
to \cite{b1} for a problem with nonhomogeneous nonlinearity, but 
with a symmetry condition on the coefficient function.  We seek to 
avoid any symmetry assumptions.  We begin 
here by studying an ordinary differential equation, with the 
aim of extending it later to the PDE case.  


	Here is the theorem:

\begin{theorem} \label{thm1.0}
Let $f$ and $b$ satisfy
\begin{description}
\item{$(f_1)$} $f \in C([0,\infty), [0, \infty))$ 
\item{$(f_2)$} $f(0) = 0$, $f(q) > 0$ for $q > 0$
\item{$(f_3)$} there exists $\mu > 2$ such that
$f(q)q/\mu \geq F(q) \equiv \int_0^q f(s)\,ds$
for all $q \geq 0$, and
\item{$(f_4)$} $f(q)/q$ is an increasing 
function of $q$ for $q > 0$,  and %

\item{$(b_1)$} $b  \in C(\mathbb{R}, (0, \infty))$
\item{$(b_2)$} $b(t) \to b_\infty > 0$ as $|t| \to \infty$, and 
\item{$(b_3)$} there exist $\delta > 2\mu/(\mu - 2)$ and $A > 0$ 
such that $b(t)-b_\infty > -Ae^{\delta |t|}$ for all $t \in \mathbb{R}$.
\end{description}
Then the Hamiltonian system
\begin{equation}
 -u''  + u = b(t)f(u)  \label{e1.1}
\end{equation}
has a positive solution homoclinic to zero, that is, a solution
$u$ with $u(t) \to 0$ and $u'(t) \to 0$ as $|t| \to \infty$.
\end{theorem}


Hypotheses $(f_1)$-$(f_3)$ imply 
that $F$ is ``superquadratic,'' that is, 
for small $q$, $F(q) = o(q^2)$ and for large $q$, $F(q) > O(q^2)$.
Condition $(f_4)$ is due to Nehari 
and has many helpful
consequences, as we will see.  $(f_1)-(f_4)$ are 
all satisfied in the canonical case $f(q) = q^p$, $p>1$.


This paper is organized as follows: Section~2 contains 
the variational framework of the proof and the beginning
of the proof.  Section~3 contains the conclusion, which
exploits the fact that $g$ decays exponentially, as 
do homoclinic solutions of the autonomous equation 
associated with (\ref{e1.1}).

\section{The variational argument}

	Let $E \equiv W^{1,2}(\mathbb{R})$, with the 
standard inner product and norm.  
Extend $f$ to $0$ on the negative 
reals, and define the $C^2$ functional $I:E \to \mathbb{R}$
by 
$$ I(u) = { 1 \over 2}\|u\|^2 - {\int_\mathbb{R}} b(t)F(u(t))\,dt.
			$$
By regularity theory, and the maximum principle, all 
nonzero critical points of $I$ are 
positive homoclinic solutions
to (\ref{e1.1}), and vice versa. 
>From now on, without loss of generality assume
$b_\infty = 1$.  Then the functional 
corresponding to the 
autonomous equation $-u'' + u = f(u)$
is
%
$$ I_0(u) = { 1 \over 2}\|u\|^2 - {\int_\mathbb{R}} F(u(t))\,dt.
			$$
%
An analysis of the equation $-u'' + u = f(u)$
in the $(u, u')$ phase plane shows that the equation
has a unique nonzero homoclinic solution, modulo 
translation, which is positive.  Let us denote 
by $\omega$ the positive 
solution satisfying $\omega(0) = \max \omega$.
$\omega$ is even in $t$ and decays 
exponentially.
$I_0$ has a unique nonzero critical 
value, $c_0 = I_0(\omega)$. $c_0$ is the ``mountain pass''
value for $I_0$.  That is, defining the set of paths
%
$$ \Phi_0 = \{h \in C([0,1], E) \mid
		h(0) = 0, I_0(h(1)) < 0\}, 
$$					% e2.2
%
$c_0$ is the minimax value
%
$$	c_0 = \inf_{h \in \Phi_0} 
		\max_{\theta \in [0,1]}
		    I_0 (h(1)) > 0.    % e2.3
$$
Define $c$, the mountain pass value for $I$, by 
defining the set of paths
%
$$ \Phi = \{h \in C([0,1], E) \mid
		h(0) = 0, I(h(1)) < 0\}, 
$$					
and
%
$$	c = \inf_{h \in \Phi} 
		\max_{\theta \in [0,1]}
		    I (h(\theta)) > 0. $$ %e.25
%

	We will use a concentration compactness type
result to describe Palais-Smale sequences of $I$.  
Recall that a Palais-Smale sequence of $I$ is a 
sequence $(u_m) \subset E$ with $I'(u_m) \to 0$ and
$(I(u_m))$ convergent.  Define the translation 
operator $\tau$ as follows: for $t_0 \in \mathbb{R}$ and
$u: \mathbb{R} \to \mathbb{R}$, let $\tau_{t_0} u$ be 
$u$ translated $t_0$ units to the right, that is,
$\tau_{t_0} u(t) = u(t - t_0)$ for all $t \in \mathbb{R}$.
The proposition below states that a Palais-Smale 
sequence ``splits'' into the sum of a critical point
of $I$ and critical points of $I_0$:


\begin{lemma} \label{lm2.6}
If $(u_m) \subset E$ with
$I'(u_m) \to 0$ and $I(u_m) \to a > 0$, there exist
$v \in E$, $k \geq 0$, and sequences 
$(t^i_m)_{\overset {1 \leq i \leq k} \to {m \geq 1}} 
\subset \mathbb{R}$, such that, along a subsequence 
(also denoted $(u_m)$), 
\begin{description}
\item{(i)} $\|u_m - (v + \sum_{i=1}^k \tau_{t^i_m}\omega)\| \to 0$
			
\item{(ii)} $|t^i_m| \to \infty$  as $m \to \infty$ for $i= 1, \ldots, k$ 
			
\item{(iii)} $t^{i+1}_m - t^i_m \to \infty$  as 
	$m \to \infty$ for $i = 1, \ldots, k-1$
			
\item{(iv)} $kc_0 + I(v) = a$ 
			
\end{description}
\end{lemma}

	A proof can be found in \cite{c2}.  By standard 
deformation arguments, there exists \cite{r1} a Palais-Smale 
sequence $(u_m)$ for $I$ along which $I$ converges to 
$c$.  Suppose $c < c_0$.  Then by applying Lemma~\ref{lm2.6}, 
and the fact that $I$ has no negative 
critical values, (iv) implies that the 
``$k$'' value is zero, so along a 
subsequence, $(u_m)$ converges to a critical point 
$v$ of $I$.  Since $I(v) = c > 0$, $v$ is nontrivial.

	By ($b_2$), $c \leq c_0$.  So from now on, we assume
%
$$		c = c_0.	$$ % e2.7
%
By ($f_4$), $I$ has the following property (as 
does $I_0$):  For any $u \in E \setminus \{0\}$, the 
mapping $s \mapsto I(su)$ is $0$ at $s = 0$, increases
for small positive $s$, attains a maximum, then 
decreases to $-\infty$ (see \cite{c2} for proof).  
Define the Nehari manifold $\mathcal{S}$
for $I$ by
%
$$	\mathcal{S} = \{u \in E \mid u \neq 0,\ I'(u)u=0\}.
				$$ % \label{e2.8}
%
Note that a nonzero function $u$ is in $\mathcal{S}$ 
if and only if $I(u) = \sup_{s > 0} I(su)$.  Then
%
$$		c = \inf_\mathcal{S} I.
			$$ %\label{e2.9}
%
Define the ``location'' function 
${\cal L}:E\setminus \{0\} \to \mathbb{R}$ by 
%
$$	{\int_\mathbb{R}} u^2 \tan^{-1}(t - {\cal L}(u))\,dt = 0.
				$$ % \label{e2.10}
%
By the 
Implicit Function Theorem,
${\cal L}$ is a well defined, continuous function.  
${\cal L}(\omega) = 0$, and 
${\cal L}(\tau_t u) = {\cal L}(u) + t$ for any 
$u \in E \setminus \{0\}$ and $t \in \mathbb{R}$.  
Define
\begin{equation}
\beta = \inf\{I(u) \mid u \in \mathcal{S},\ 
	           {\cal L}(u) = 0\}. \label{e2.11}
\end{equation} 
%
Clearly $\beta \geq c = c_0$.  If $\beta = c_0$, then 
$c_0$ is a critical value of $I$:  suppose 
$\beta = c_0$.  Take $(u_m) \subset \mathcal{S}$ with 
${\cal L}(u_m) = 0$ and $I(u_m) \to c_0$.  
Along a subsequence, $I'|_\mathcal{S}(u_m) \to 0$. 
By $(f_4)$, $I'(u_m) \to 0$ (see \cite{f1} for similar
minimax arguments on a Nehari manifold). Apply 
Lemma~\ref{lm2.6}.  (iv) shows, again, that either 
$k = 0$, or $k = 1$ with $v = 0$.  The latter alternative 
is impossible because ${\cal L}(u_m) = 0$.  
Therefore $(u_m)$ converges (along a subsequence) to a
critical point of $I$.  Thus we assume from now on that
\begin{equation}
	\beta > c_0.		\label{e2.12}
\end{equation}
%
Define $\mathcal{P}: E \setminus \{0\} \to \mathcal{S}$ to be
radial projection onto $\mathcal{S}$, that is,
%
$$	\mathcal{P}(u) = tu;\ t>0,\ tu \in \mathcal{S}.	
			$$ %	\label{e2.13} 
%
For $R > 0$ define the minimax class
%
$$  \Gamma_R = \{\gamma \in C([0,1],\mathcal{S}) \mid
		  \gamma(0) = \mathcal{P}(\tau_{-R}\omega),
	          \gamma(1) = \mathcal{P}(\tau_{R}\omega)\}
		$$    %\label{e2.14}
%
and the minimax value
%
$$	c[R] = \inf_{\gamma \in \Gamma_R}
		 \max_{\theta \in [0,1]}
			I(\gamma(\theta)).
			$$ %	\label{e2.15}
%
Theorem~\ref{thm1.0} will follow from the following 
proposition:

\begin{proposition} \label{prop2.16} 
There exists $R_0 = R_0(f,g)$ with 
the following property: if $R \geq R_0$, then 
\begin{description}
\item{(i)} $I(\mathcal{P}(\tau_{-R}\omega)) < \beta$ and 
$I(\mathcal{P}(\tau_{R}\omega)) < \beta$
			
\item{(ii)}  $c[R] \geq \beta$	
			
\item{(i)}  $c[R] < 2c_0$ 
\end{description}
\end{proposition}

	To prove Theorem~\ref{thm1.0} from 
Proposition~\ref{prop2.16}, let $R \geq R_0$.  
By (i)-(ii), there exists a sequence 
$(u_m) \subset \mathcal{S}$ with $I(u_m) \to c[R]$ and 
$I'|_\mathcal{S} (u_m) \to 0$, and 
$I'(u_m) \to 0$.
Apply Lemma~\ref{lm2.6} to $(u_m)$.
Since $c_0 < \beta < c[R] < 2c_0$, 
Lemma~\ref{lm2.6}(iv)  implies that 
$c[R]$ or $c[R] - c_0$ is a critical value of 
$I$. Since $0 < c[R] - c_0 < c_0$, 
assumption $(f_4)$
implies 
that $c[R] - c_0$ is not a critical value of $I$.  
Therefore $c[R]$ is a positive critical value of $I$.
Maximum principle arguments show that
(\ref{e1.1}) has a positive homoclinic solution. 
Theorem~\ref{thm1.0} is proven.


\section{Interfering Tails}


This section contains the proof of Proposition~\ref{prop2.16}.  
Parts (i) and (ii) are easy.  Using $(b_2)$, it is 
straightforward to show that 
$I(\mathcal{P}(\tau_{-R}\omega)) \to c_0$ and
$I(\mathcal{P}(\tau_{R}\omega)) \to c_0$ as $R \to \infty$, 
and we assumed in (\ref{e2.12}) that $c_0 < \beta$.  (ii) holds 
for all $R > 0$, not necessarily large: let 
$\gamma \in \Gamma_R$.  Since 
${\cal L}(\mathcal{P}(\tau_{-R}\omega)) = -R$, 
${\cal L}(\mathcal{P}(\tau_{R}\omega)) = R$, and ${\cal L}$ is continuous, 
 there exists 
$\theta^* \in [0,1]$ with ${\cal L}(\gamma(\theta^*)) = 0$,
so by the definition of $\beta$, 
$I(\gamma(\theta^*)) \geq \beta$.  The remainder of this
section is devoted to the proof of (iii), which is 
more difficult.

	We adopt a construction similar to that 
of \cite{b1}.  For $R > 0$, define $\gamma_R: [0,1] \to E \setminus \{0\}$ 
by
%
$$ \gamma_R(\theta) =
	\max((1-\theta)\tau_{-R}\omega, \theta\tau_R \omega)
					$$ %\label{e3.0}
%
and ${ { \hat \gamma} }_R \in \Gamma_R$ by
%
$$ { { \hat \gamma} }_R(\theta) =
	\mathcal{P}(\gamma_R(\theta))
$$ %	\label{e3.1}
We will show that for large enough $R$,
\begin{equation}
\max_{\theta \in [0,1]}
	  I({ { \hat \gamma} }_R(\theta)) < 2c_0,	\label{e3.2}
\end{equation} 
proving Proposition~\ref{prop2.16}(iii).

	To avoid the complications in working with 
the manifold $\mathcal{S}$, we have the following lemma.  
``For large enough $R$'' means there exists 
$R_0 = R_0(f,g)$ such that if $R \geq R_0$, etc., 
as in Proposition~\ref{prop2.16}.


\begin{lemma} \label{lm3.3} 
There exists $T = T(f,g)$  such that for 
large enough $R$, and all $\theta \in [0,1]$,
%
$$    I(T\gamma_R(\theta)) < 0.   $$
\end{lemma}

\paragraph{Proof:} let $T > 0$ be large enough so that 
$I_0(T \omega) < - c_0 - 2$.  Then for all $\theta \in [0,1]$,
$I_0(\theta T \omega) + I_0((1- \theta)T \omega) < -2$.  
Let ${ { \tilde \omega }} \in E$ have compact support in $\mathbb{R}$ and be close
enough to $\omega$ that 
$I_0(\theta T { \tilde \omega }) + I_0((1- \theta)T{ \tilde \omega }) < -1$
for all $\theta \in [0,1]$. 
$I$ is Lipschitz on bounded subsets of $E$ (see \cite{r1} for a proof
in a similar setting), 
so for large enough $R$, and all $\theta \in [0,1]$, 
$I(T\gamma_R(\theta)) <  -1 + 1/2 < 0$.


	By Lemma~\ref{lm3.3}, in order to prove (\ref{e3.2}),
it suffices to show that for large enough $R$, 
all $\theta \in [0,1]$ and $s \in [0,T]$,
\begin{equation}
  I(s\gamma_R(\theta)) < 2c_0.  \label{e3.4}
\end{equation}
We will treat separately the case where 
$\theta$ is close to $0$ or $1$:



\begin{lemma} \label{lm3.5} 
 For large enough $R$, all $s \in [0,T]$
and all $\theta \in [0, 1/4] \cup [3/4, 1]$,
\begin{equation}
    I(s\gamma_R(\theta)) < 2c_0.   \label{e3.6}
\end{equation}
\end{lemma}

\paragraph{Proof:} without loss of generality let $\theta \in [0, 1/4]$.
If $I_0(s\theta \omega) < I_0(\omega /2)$, then 
$I_0(s\theta \omega) + I_0((1-\theta)s\omega) < c_0 + I_0(\omega/2)$.
If $I_0(s\theta\omega) \geq I_0(\omega/2)$, then $s\theta \geq 1/2$,
so $(1-\theta)s = ({1 \over \theta} - 1)(s\theta) \geq 3({ 1 \over 2})
\geq 3/2$, so $I_0((1-\theta)s\omega) \leq 
I_0({3 \over 2}\omega)$,
$I_0(s\theta\omega) + I_0((1 - \theta)s\omega) < 
c_0 + I_0({3 \over 2}\omega)$.  Assume that $R$ is large enough so that 
for all $s \in [0,T]$,
%
$$  |I(s\gamma_R(\theta) - 
          (I_0(s(1-\theta)\omega) + I_0(s\theta\omega))| <
	{ 1 \over 2}\min(c_0 - I_0(\omega/2), c_0 - I_0(3\omega/2)).
	$$ %\label{e3.7}
%
Then the triangle inequality gives (\ref{e3.6}).


Now we must prove (\ref{e3.4}) for $\theta \in [1/4, 3/4]$.  
For all 
$R>0$ and $s \geq 0$, 
$I_0(s\tau_{-R}\omega) \leq c_0$ and 
$I_0(s\tau_{R}\omega) \leq c_0$.  So it suffices to show
that for large enough $R$, $s \in [0,T]$, and
$\theta \in (1/4, 3/4)$,
%
\begin{eqnarray}
\lefteqn{ 2c_0 - I(s\gamma_R(\theta)) } \nonumber\\
&\geq&  \Big[(I_0((1-\theta)s \tau_{-R}\omega) + 
       I_0(\theta s\tau_{R}\omega) ) -
      (I((1-\theta)s\tau_{-R}\omega) +
            I(\theta s\tau_{R}\omega))\Big] \nonumber \\
&&+ \big[
     (I((1-\theta)s \tau_{-R}\omega) + I(\theta s\tau_{R}\omega)) -
     I(s\gamma_R(\theta)) \big] \label{e3.8} \\
&\equiv& A(R, \theta, s) + B(R, \theta, s) >0\,.  \nonumber
\end{eqnarray} 
%
We begin with $B(R, \theta, s)$.  To 
analyze $I(s\gamma_R(\theta))$, we have the following
lemma:

\begin{lemma} \label{lm3.9} 
 For large enough $R$, and 
$\theta \in (1/4, 3/4)$, there exists 
a unique $t \in \mathbb{R}$ with 
$(1- \theta)\tau_{-R}\omega(t) = \theta\tau_R\omega(t)$.
Calling this value of $t$, $t_{R,\theta}$, 
%
$$	t_{R,\theta}/R \to 0 \mbox{ as } R \to \infty.	
$$ %\label{e3.10}
\end{lemma}

\paragraph{Proof:}
  let $\epsilon \in (0,1)$.  Let $R$ be large enough
so that $\omega(t + 2R) < \omega(t)/4$ for all 
$t \geq -(1-\epsilon)R$.  This is possible because $\omega$ 
decays exponentially.  Now 
$(1-\theta)\tau_{-R}\omega \equiv (1 -\theta)\omega(\cdot + R)$
is decreasing on $[-\epsilon R, \epsilon R]$ and
$\theta\tau_{R}\omega$ is increasing on $[-\epsilon R, \epsilon R]$.
Thus, to prove the existence and uniqueness of 
$t_{R,\theta}$, it now suffices to prove that for 
large $R$ and $\theta \in (1/4, 3/4)$,
%
\begin{description}
\item{(i)}  $(1-\theta)\tau_{-R}\omega > \theta\tau_R\omega$
		on $(-\infty, -\epsilon R]$ , and
			
\item{(ii)} $\theta\tau_{R}\omega > (1- \theta)\tau_{-R}\omega$
		on $[\epsilon R, \infty)$.
\end{description}				
%
By the symmetry of the problem, the proof of (i) and 
(ii) are practically the same, so we prove (ii). 
Let $t \geq \epsilon R$.  Then $t - R \geq -(1-\epsilon)R$, so
$\omega(T+R) < \omega(t-R)/4$, and
%
$$  (1-\theta)\tau_{-R}\omega(t) \equiv
	(1- \theta)\omega(t+R) <
		{1 \over 4}(1-\theta)\omega(t-R) <
			{1 \over 4}\omega(t-R) <
				\theta\tau_R \omega(t).
					$$ %\label{e3.12}
%
For $U \subset \mathbb{R}$, define
$\|u\|_U = \|u\|_{W^{1,2}(U)}$.    For large $R$
and $s \in [0,T]$,
%
\begin{equation}\begin{aligned}
	B(R,\theta,s) &=
	  I((1-\theta)s\tau_{-R}\omega) + 
          I(\theta s\tau_{R}\omega) -
	  I(s\gamma_R (\theta)) \\
&= { 1 \over 2}(1-\theta)^2 s^2 
	   \|\tau_{-R}\omega\|^2_{[t_{R,\theta},\infty)} +
           { 1 \over 2}\theta^2 s^2 
	   \|\tau_{R}\omega\|^2_{(-\infty, t_{R,\theta}]} \\
&  - \int_{t_{R,\theta}}^\infty
		b(t)F(s(1-\theta)\omega)\tau_{-R}\omega
	  - \int_{-\infty}^{t_{R,\theta}}
		b(t)F(s\theta\omega)\tau_{R}\omega.
\end{aligned} \label{e3.13}
\end{equation}
Assume that $R$ is large enough so that
$|t_{R,\theta}| < R/2$ and that for all 
$\theta \in (1/4, 3/4)$, $s \in [0,T]$
 and $t \geq R/2$, 
\begin{equation}
  b(t)F(s(1-\theta)\omega(t)) \leq
	{ 1 \over 4} s^2 (1-\theta)^2\omega(t)^2. \label{e3.14}
\end{equation}	
%
Since $t_{R,\theta} > -R/2$, $t + R > R/2$, 
so (\ref{e3.14}) gives, for 
all $t \geq t_{R,\theta}$,
\begin{equation}
  b(t)F(s(1-\theta)\tau_{-R}\omega(t)) \leq
	{ 1 \over 4} s^2 (1-\theta)^2\tau_{-R}\omega(t)^2. \label{e3.15}
\end{equation} 
% 
Similarly, for all $t \leq t_{R,\theta}$,
\begin{equation}
  b(t)F(s\theta\tau_{R}\omega(t)) \leq
	{ 1 \over 4} s^2 \theta^2\tau_{R}\omega(t)^2, \label{e3.16}
\end{equation} 	
%
so (\ref{e3.13}), (\ref{e3.15}), (\ref{e3.16}) and the Sobolev inequality
$\|u\|_{L^\infty(0,\infty)} \leq \|u\|_{W^{1,2}(0,\infty)}$
give
\begin{equation}\begin{aligned} 
     B(R,\theta,s) &\geq
	{ 1 \over 4} s^2 \big[
	    (1-\theta)^2
             \|\tau_{-R}\omega\|^2_{[t_{R,\theta},\infty)}	
	+ \theta^2
             \|\tau_{R}\omega\|^2_{(-\infty, t_{R,\theta}]}
			\big] \\
	&\geq s^2(\omega(R + t_{R,\theta})^2 +
                \omega(R - t_{R,\theta})^2)/16. 
\end{aligned} \label{e3.17}
\end{equation}
%
Since $\delta > 2\mu/(\mu-2)$ ($(b_3)$), we may choose
$$ 
d \in ({2 \over \delta}, 1 - {2 \over \mu}).
$$
Since $(1-d)\mu > 2$ and $d\delta > 2$,
we may choose $\epsilon_1 \in (0,1)$ and small enough so that
\begin{equation}
 2(1+\epsilon_1)^2 < \min((1-\epsilon_1)\mu(1-d),\ \delta d).  
					\label{e3.19}
\end{equation}
%
By the maximum
principle, and the superlinear 
growth of $f$ near $0$, 
there exists $l > 0$ such that
%
$$	\omega(t) > le^{-(1 + \epsilon_1)|t|}
			$$ %\label{e3.20}
%
for all $t \in \mathbb{R}$.  Assume $R$ is large enough 
(Lemma~\ref{lm3.9}) so 
$$|t_{R,\theta}| < \epsilon_1 R\,.$$
Then 
%
$$   \tau_{-R}\omega(t_{R,\theta}) =
	\omega(R + t_{R,\theta}) \geq
		\omega((1 + \epsilon_1)R) >
			le^{-(1 + \epsilon_1)^2 R},
			$$	%\label{e3.22}
%
so by (\ref{e3.17}),
\begin{equation}
 B(R,\theta, s) \geq s^2 l^2 e^{-2(1+\epsilon_1) R}/8. \label{e3.23}
\end{equation} 

	Next, let us estimate $A(R,\theta,s)$ from
(\ref{e3.8}), still assuming $\theta \in (1/4, 3/4)$ and
$s \in [0,T]$.  
>From now on, $C$ will denote a large positive
constant depending only on $f$, $b$, and the 
already chosen $T$.  The value of 
$C$ may change from line to line.  By $(f_3)$, for all 
$q \in [0, T\max \omega]$,
%
$$	F(q) \leq Cq^\mu. $$
%
By the form 
of $A(R,\theta,s)$, all of the ``$\|\ \|^2$'' terms
in $A(R,\theta,s)$
cancel
out.  Since $f \geq 0$, it is easy to show, by 
comparing $\omega$ to a solution $v$ of 
$-v'' + v = 0$, that $\omega(t) \leq Ce^{-|t|}$ 
for all $t \in \mathbb{R}$.  Therefore,
%
$$\begin{aligned}
	A(R,\theta,s) &= 
		{\int_\mathbb{R}} (b(t)-1)F((1-\theta)s\tau_{-R}\omega) +
		{\int_\mathbb{R}} (b(t)-1)F(\theta s\tau_{R}\omega) \\
&\geq -C\big[{\int_\mathbb{R}} e^{-\delta |t|}
		       F((1-\theta)s\omega(t+R))\,dt +
		{\int_\mathbb{R}} e^{-\delta |t|}
		      F(\theta s\omega(t-R))\,dt \big] \\
&\geq -Cs^\mu \big[{\int_\mathbb{R}} e^{-\delta |t|}
		       e^{-\mu|t-R|}\,dt +
                    {\int_\mathbb{R}} e^{-\delta |t|}
		       e^{-\mu|t+R|}\,dt \big]\\
&\geq -Cs^2{\int_\mathbb{R}} e^{-\delta |t|}
		       e^{-\mu|t-R|}\,dt.
\end{aligned} %\label{e3.25}
$$
The integral can be estimated by
%
$$\begin{aligned} 
     {\int_\mathbb{R}} e^{-\delta |t|}e^{-\mu|t-R|}\,dt
&\leq
     \int_{-\infty}^{dR}e^{-\mu|t-R|}\,dt +
     \int_{dR}^\infty e^{-\delta |t|}\,dt \\
&= {	{  e^{\mu(t-R)}  } \over  {\mu}
         }\big|_{t = -\infty}^{t=dR} +
	   {
	     {e^{-\delta t}}
	 	  \over
	     {-\delta}
	   }\big|_{t = dR}^{t=\infty} \leq
		C\big[e^{-\mu(1-d)R} +
		  e^{-d\delta R}\big].
\end{aligned} \label{e3.26}
$$
By (\ref{e3.23}) and (\ref{e3.26}), for large $R$, $s \in [0,T]$, and
$\theta \in (1/4, 3/4)$,
%%
$$\begin{gathered} 
A(R,\theta,s) \geq -Cs^2[e^{  -\mu(1-d)(1-\epsilon_1)R  } +
   e^{  -d\delta R  } ] \quad \hbox{ and } \\
B(R, \theta, s) \geq l^2 s^2 e^{  -2(1 + \epsilon_1)^2 R  }/8.
\end{gathered} 
$$
By (\ref{e3.19}), $A(R,\theta,s) + B(R,\theta,s) > 0$ for 
large $R$.  By (\ref{e3.8}), Lemma~\ref{lm3.5} is proven, from which follow
Proposition~\ref{prop2.16} and Theorem~\ref{thm1.0}.


The proof of Theorem~\ref{thm1.0} was indirect, and we can not say much 
about the positive homoclinic 
solution of (\ref{e1.1}).  We do know that if $c < c_0$, then $I$ has a critical 
point at critical level $c$.  If the alternative 
$c = c_0$ holds, then if 
$\beta = c_0$ (see (\ref{e2.11})), $c$ is a critical level of $I$.  If 
$c = c_0$ and $\beta > c_0$, then for large enough $R$, 
$c[R]\  (\in (c_0, 2c_0))$ is a critical level of $I$
(assuming otherwise, then a deformation argument yields
a contradiction to the definition of $c[R]$).

	The proof of Theorem~\ref{thm1.0} suggests
that, under additional conditions on $b$, (\ref{e1.1}) may have ``two-bump''
solutions, with parts resembling translations of $\omega$ to the left 
and to the right of zero. 
 

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\noindent\textsc{Gregory S. Spradlin}\\
 Department of Computing and Mathematics\\
 Embry-Riddle Aeronautical University \\
 Daytona Beach, Florida 32114-3900 USA \\
 e-mail: spradlig@db.erau.edu 


\end{document}
