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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 94, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/94\hfil Existence of solutions]
{Existence of solutions for second-order differential
equations and systems on infinite intervals}

\author[T. Moussaoui, R. Precup\hfil EJDE-2009/94\hfilneg]
{Toufik Moussaoui, Radu Precup}  % in alphabetical order

\address{Toufik Moussaoui \newline
Department of Mathematics, E.N.S.\\
P.O. Box 92, 16050 Kouba, Algiers, Algeria}
\email{moussaoui@ens-kouba.dz}

\address{Radu Precup \newline
Department of Applied Mathematics \\
Babe\c{s}-Bolyai University\\
400084 Cluj, Romania}
\email{r.precup@math.ubbcluj.ro}

\thanks{Submitted October 7, 2008. Published August 6, 2009.}
\subjclass[2000]{34B40}
\keywords{Boundary value problem; fixed point theorem}

\begin{abstract}
 We study the existence of nontrivial solutions to the
 boundary-value problem
 \begin{gather*}
 -u''+cu'+\lambda u  =  f(x,u),\quad -\infty <x<+\infty , \\
 u(-\infty )=u(+\infty )  =0
 \end{gather*}
 and to the system
 \begin{gather*}
 -u''+c_1u'+\lambda _1u  =  f(x,u,v),\quad
 -\infty <x<+\infty , \\
 -v''+c_2v'+\lambda _2v  =  g(x,u,v),\quad
 -\infty <x<+\infty , \\
 u(-\infty )=u(+\infty )  =0,   \quad
 v(-\infty )=v(+\infty )  =0,
 \end{gather*}
 where $c,c_1,c_2,\lambda ,\lambda _1,\lambda _2$ are real
 positive constants and the nonlinearities $f$ and $g$ satisfy
 suitable conditions.
 The proofs are based on fixed point theorems.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]

\section{Introduction}

In this article, we consider the existence of nontrivial solutions
to the boundary-value problem
\begin{equation}
\begin{gathered}
-u''+cu'+\lambda u  =  f(x,u),\quad -\infty <x<+\infty , \\
u(-\infty )=u(+\infty )  =0
\end{gathered} \label{problem1}
\end{equation}
and to the system
\begin{equation}
\begin{gathered}
-u''+c_1u'+\lambda _1u  =  f(x,u,v),\quad -\infty <x<+\infty . \\
-v''+c_2v'+\lambda _2v  =  g(x,u,v),\quad -\infty <x<+\infty . \\
u(-\infty )=u(+\infty )  =0,   \quad
v(-\infty )=v(+\infty )  =0.
\end{gathered} \label{problem2}
\end{equation}
Here $c,c_1,c_2,\lambda ,\lambda _1,\lambda _2$ are positive
numbers.

By a solution of Problem \eqref{problem1} we mean a function
$u \in C^1( {\mathbb{R}},\mathbb{R}) $ satisfying
\eqref{problem1} and by a solution of Problem \eqref{problem2}
we mean a vector-valued function
$(u,v)\in C^1({\mathbb{R}},\mathbb{R}^2)
:=C^1({\mathbb{R}},\mathbb{R})\times C^1({\mathbb{R}},
\mathbb{R})$ satisfying \eqref{problem2}.

Some of the ideas used in this paper are motivated by Djebali and
Moussaoui \cite{DjeMou1,DjeMou2} and Djebali and Mebarki \cite{DjeMeb1}.

In \cite{DjeMou1}, the authors study the boundary value problem
\eqref{problem1}. According to the behavior of the nonlinear source term,
existence results of bounded solutions, positive solutions, classical as
well as weak solutions are provided. They mainly use fixed point arguments.

Under some relations upon the real parameters and coefficients, the
authors in \cite{DjeMou2} present some existence and nonexistence
results for Problem \eqref{problem1}. They use a variational method
and fixed point arguments.

In \cite{DjeMeb1}, the authors study existence of positive nontrivial
solutions for a boundary value problem on the positive half-line arising
from epidemiology. They mainly use the fixed point theorem of cone expansion
and compression of functional type.

In this paper, we study existence of nontrivial solutions for both
 Problems \eqref{problem1} and \eqref{problem2}. Our arguments are based on fixed point
theory. So, let us recall for the sake of completeness,
Krasnosel'sk'ii-Zabreiko's and Schauder's fixed point theorems:

\begin{theorem}[\cite{KZ1}] \label{KZ}
Let $(E,\| .\| )$ be a Banach space,
and $T:E\to E$ be completely continuous. Assume that $A:E\to E$
is a bounded linear operator such that $1$ is not an eigenvalue of $A$
and
\[
\lim_{\| u\| \to \infty }\frac{\| Tu-Au\| }{\| u\|}=0.
\]
Then $T$ has a fixed point in $E$.
\end{theorem}

\begin{theorem}[\cite{Ze}] \label{thm1.2}
Let $E$ be a Banach space and $K\subset E$ a nonempty,
bounded, closed and convex subset of $E$. Let $T:K\to K$ be a
completely continuous operator. Then $T$ has a fixed point in $K$.
\end{theorem}

In what follows, $C_{0}({\mathbb{R}},{\mathbb{R}})$ stands for the Banach
space of continuous functions defined on the real line and vanishing at
infinity, endowed with the sup-norm
\[
\| u\| _{0}=\sup_{{x\in {\mathbb{R}}}}| u(x)| .
\]
Recall that $L^1({\mathbb{R}},{\mathbb{R}})$ is the Banach space of
integrable functions on ${\mathbb{R}}$ endowed with norm
\[
|u|_1=\int_{-\infty }^{+\infty }|u(s)|\,ds.
\]
In the sequel, we put
\[
k=\sqrt{c^2+4\lambda },\quad
k_1=\sqrt{c_1^2+4\lambda _1},\quad
k_2=\sqrt{c_2^2+4\lambda _2}.
\]

This paper is organized as follows. In Section 2, we state the main result
concerning the existence of solutions for the boundary value problem (\ref
{problem1}) (Theorem \ref{thm1}) by using the fixed point theorem of
Krasnosel'sk'ii-Zabreiko. In Section 3, we generalize the result obtained in
Section 2 to systems of differential equations by using the same technique
(Theorem \ref{thm2}). In Section 4, we present two other results concerning
Problem \eqref{problem2} by using Schauder's fixed point theorem
(Theorem \ref{thm3} and Theorem \ref{thm4}). Finally, in the last section, we give some examples
to illustrate our results.

\section{Existence result for a generalized Fisher-like equation}

In this section, we study the  boundary-value problem
\begin{equation}
\begin{gathered}
-u''+cu'+\lambda u  = p(x)f(u),\quad -\infty <x<+\infty , \\
u(-\infty )=u(+\infty )  =0
\end{gathered}  \label{nonl1}
\end{equation}
where $p\in L^1({\mathbb{R}},{\mathbb{R}}_{+})$ and
$f:{\mathbb{R}}\to {\mathbb{R}}$ is a continuous function with
$f(0)\neq 0$.
Our main result in this section is the following theorem.

\begin{theorem} \label{thm1}
Assume that
\begin{equation}
\lim_{u\to \infty }\frac{f(u)}{u}=f_{\infty }\quad \text{with }
|f_{\infty }|<\frac{k}{|p|_1}.  \label{H1}
\end{equation}
Then Problem \eqref{nonl1} has at least one nontrivial solution
$u\in C_{0}({\mathbb{R}},{\mathbb{R}})$.
\end{theorem}

\begin{proof}
It is clear that Problem \eqref{nonl1} is equivalent to the integral
equation
\[
u(x)=\int_{-\infty }^{+\infty }G(x,s)p(s)f(u(s))\,ds
\]
with the Green function
\[
G(x,s)=\frac{1}{\rho _1-\rho _2}
\begin{cases}
e^{\rho_1(x-s)} & \text{if }  x\leq s \\
e^{\rho _2(x-s)} & \text{if }  x\geq s
\end{cases}
\]
and characteristic roots
\[
\rho_1=\frac{c+\sqrt{c^2+4\lambda }}{2},\quad
\rho_2=\frac{c-\sqrt{c^2+4\lambda }}{2}\,.
\]
Notice that
\[
0<G(x,s)\leq \frac{1}{k}\quad \text{for all }x,s\in {\mathbb{R}}.
\]
Define the mapping $T: C_{0}({\mathbb{R}},{\mathbb{R}})
\to C_{0}({\mathbb{R}},{\mathbb{R}})$ by
\[
Tu(x)=\int_{-\infty }^{+\infty }G(x,s)p(s)f(u(s))\,ds.
\]
In view of Krasnosel'sk'ii-Zabreiko's fixed point theorem, we look
for fixed points of the operator $T$ in the Banach space
$C_{0}({\mathbb{R}},{\mathbb{R}})$. The proof is split into four
steps.

 \textbf{Claim 1:} The mapping $T$ is well defined; indeed, for
any $u\in C_{0}({\mathbb{R}},{\mathbb{R}})$, by Assumption
(\ref{H1}), we obtain the following estimates:
\[
|Tu(x)| \leq \int_{-\infty }^{+\infty }G(x,s)|p(s)f(u(s))|\,ds
\leq \frac{1}{k}|p|_1\max_{|y|\leq \| u\| _{0}}|f(y)|.
\]
The convergence of the integral defining $(Tu)(x)$ is then
established. In addition for any $s\in {\mathbb{R}}$,
$G(\pm \infty ,s)=0$,
and then, taking the limit in $(Tu)(x)$, we obtain
$Tu(\pm \infty )=0$. Therefore, the mapping
$T:C_{0}({\mathbb{R}},{\mathbb{R}})\to C_{0}({\mathbb{R}},{\mathbb{R}})$
is well defined.

\textbf{Claim 2:} The operator $T$ is continuous. Let
$(u_{n})_{n}\subset C_{0}({\mathbb{R}},{\mathbb{R}})$
be a sequence which converges uniformly to $u_{0}$ on each compact
subinterval of ${\mathbb{R}}$. For some fixed $a>0$, we will prove
the uniform convergence of $(Tu_{n})_{n}$ to
$Tu_{0}$ on the interval $[-a,a]$. Let $\varepsilon >0$ and
choose some $b>a$ large enough. By the uniform convergence of the
sequence $(u_{n})_{n}$ on $[-b,b]$, there exists an integer
$N=N(\varepsilon ,b)$ satisfying
\[
I_1: =\sup_{x\in {\mathbb{R}}
}\int_{-b}^{+b}G(x,s)|p(s)||f(u_{n}(s))
-f(u_{0}(s))|\,ds<\frac{\varepsilon }{2}\quad \text{for all }n\geq N.
\]
For $x\in [-a,a]$, we have that
$|(Tu_{n})(x)-(Tu_{0})(x)|\leq I_1+I_2+I_{3}$, where
\[
I_2: =\sup_{x\in {\mathbb{R}}}\int_{{\mathbb{R}}
-[-b,+b]}G(x,s)|p(s)f(u_{0}(s))|\,ds\leq \frac{\varepsilon }{4}
\]
(by the Cauchy Convergence Criterion and $\lim_{|s|\to
+\infty }p(s)f(u_{0}(s))=0$),
\[
I_{3}: =\sup_{x\in {\mathbb{R}}}\,\int_{{\mathbb{R}}
-[-b,+b]}G(x,s)|p(s)f(u_{n}(s))|\,ds\leq \frac{\varepsilon }{4}
\]
(by the Lebesgue Dominated Convergence Theorem). This proves the uniform
convergence of the sequence $(Tu_{n})_{n}$ to $Tu_{0}$ on $[-a,a]$.


\textbf{Claim 3:} For any $M>0$, the set $\{Tu,\,\| u\| _{0}\leq M\}$
is relatively compact in $C_{0}({\mathbb{R}},{\mathbb{R}})$. By
the Ascoli-Arzela Theorem, it is sufficient to prove that all the functions
of this set are equicontinuous on every subinterval $[-a,a]$
and that there exists a function $\gamma \in C_{0}({\mathbb{R}},{
\mathbb{R}})$ such that for any $x\in {{\mathbb{R}}}$, $|Tu(x)|\leq
\gamma (x)$. Let $x_1,x_2\in [-a,a]$. We have successively
the estimates
\begin{align*}
|Tu(x_2)-Tu(x_1)| &\leq \int_{-\infty }^{+\infty
}|G_1(x_2,s)-G_1(x_1,s)||p(s)f(u(s))|\,ds \\
&\leq \max_{y\in \lbrack -M,M]}|f(y)|\int_{-\infty }^{+\infty
}|G_1(x_2,s)-G_1(x_1,s)||p(s)|\,ds.
\end{align*}
By the continuity of the Green function $G$, the last term tends to $0$, as $
x_2$ tends to $x_1$, whence comes the equicontinuity of the functions
from $\{Tu;\| u\| _{0}\leq M\}$. Analogously we have
\begin{align*}
|Tu(x)| &\leq \int_{-\infty }^{+\infty }G(x,s)|p(s)f(u(s))|\,ds \\
&\leq \max_{y\in \lbrack -M,M]}|f(y)|\int_{-\infty }^{+\infty
}G(x,s)|p(s)|\,ds: =\gamma (x).
\end{align*}
Clearly, $\gamma \in C_{0}({\mathbb{R}},{\mathbb{R}})$.

Now, we consider the  boundary-value problem
\begin{equation}
\begin{gathered}
-u''+cu'+\lambda u  =  f_{\infty }p(x)u(s),\quad
-\infty <x<+\infty \\
u(-\infty )=u(+\infty )  =0,
\end{gathered}  \label{lin1}
\end{equation}
and  define operator $A$ by
\[
Au(x)=f_{\infty }\int_{-\infty }^{+\infty }G(x,s)p(s)u(s)\,ds.
\]
Obviously, $A$ is a bounded linear operator. Furthermore,
any fixed point of $A$ is a solution of Problem (\ref{lin1}),
and conversely.

We now claim that $1$ is not an eigenvalue of $A$.
In fact, if $f_{\infty}=0$, then Problem (\ref{lin1}) has no
nontrivial solutions. Let $f_{\infty}\neq 0$ and assume that
Problem (\ref{lin1}) has a nontrivial solution $u$.
Then
\begin{align*}
|Au(x)| &\leq |f_{\infty }|\int_{-\infty }^{+\infty }G(x,s)|p(s)u(s)|\,ds \\
&\leq \frac{1}{k}|f_{\infty }||p|_1\| u\| _{0} \\
&< \| u\| _{0}.
\end{align*}
Hence $\| Au\| _{0}<\| u\| _{0}$. This contradiction means that
Problem (\ref{lin1}) has no nontrivial solution. Thus $1$ is not an
eigenvalue of $A$.

Finally, we prove that
\[
\lim_{\| u\| _{0}\to \infty }\frac{\| Tu-Au\| _{0}}{
\| u\| _{0}}=0.
\]
According to $\lim_{u\to \infty }f(u)/u=f_{\infty }$, for
any $\varepsilon >0$, there exists $R>0$ such that
\[
|f(u)-f_{\infty }u|<\varepsilon |u|,\quad |u|>R.
\]
Set $R^{\ast }=\max_{|u|\leq R}|f(u)|$ and select $M>0$ such that $R^{\ast
}+|f_{\infty }|R<\varepsilon M$. Denote
\[
I_1=\{x\in {\mathbb{R}}:|u(x)|\leq R\},\quad
I_2=\{x\in {\mathbb{R}}:|u(x)|>R\}.
\]
Thus for any $u\in C_{0}({\mathbb{R}},{\mathbb{R}})$ with
$\| u\|_{0}>M$, when $x\in I_1$, we have
\[
|f(u(x))-f_{\infty }u(x)|\leq |f(u(x))|+|f_{\infty }||u(x)|
\leq R^{\ast}+|f_{\infty }|R<\varepsilon M<\varepsilon \| u\| _{0}.
\]
Similarly, we conclude that for any
$u\in C_{0}({\mathbb{R}},{\mathbb{R}})$
with $\| u\| _{0}>M$, when $x\in I_2$, we also have that
\[
|f(u(x))-f_{\infty }u(x)|<\varepsilon \| u\| _{0}.
\]
We conclude that for any $u\in C_{0}({\mathbb{R}},{\mathbb{R}})$ with
$\|u\| _{0}>M$, we have
\[
|f(u(x))-f_{\infty }u(x)|<\varepsilon \| u\| _{0}.
\]
Then for any $u\in C_{0}({\mathbb{R}},{\mathbb{R}})$ with
$\| u\|_{0}>M$, one has
\begin{align*}
|f(u(x))|
&\leq |f(u(x))-f_{\infty }u(x)|+|f_{\infty }u(x)| \\
&\leq \varepsilon \| u\| _{0}+|f_{\infty }|\| u\| _{0} \\
&\leq (f_{\infty }|+\varepsilon )\| u\| _{0}.
\end{align*}
Hence we obtain
\begin{align*}
|Tu(x)| &= \Big|\int_{-\infty }^{+\infty }G(x,s)p(s)f(u(s))\,ds-f_{\infty
}\int_{-\infty }^{+\infty }G(x,s)p(s)u(s)\,ds\Big| \\
&\leq \frac{1}{k}\int_{-\infty }^{+\infty }|p(s)||f(u(s))-f_{\infty
}u(s)|\,ds \\
&< \frac{1}{k}|p|_1\varepsilon \| u\| _{0}.
\end{align*}
Then we have
\[
\lim_{\| u\| _{0}\to \infty }\frac{\| Tu-Au\| _{0}}{
\| u\| _{0}}=0.
\]
Theorem \ref{KZ} now guarantees that Problem \eqref{nonl1} has a
nontrivial solution. This completes the proof.
\end{proof}

\section{Existence result for a generalized Fisher-like system}

In this section, we study the system
\begin{equation}
\begin{gathered}
-u''+c_1u'+\lambda _1u  =  p(x)f(u,v),\quad
-\infty <x<+\infty , \\
-v''+c_2v'+\lambda _2v  =  q(x)g(u,v),\quad
-\infty <x<+\infty , \\
u(-\infty )=u(+\infty )  =0,   \quad
v(-\infty )=v(+\infty )  =0.
\end{gathered}  \label{nonl2}
\end{equation}
where $p,q\in L^1({\mathbb{R}},{\mathbb{R}}_{+})$,
$f,g:{\mathbb{R}}\times {\mathbb{R}}\to {\mathbb{R}}$
are continuous functions with $f(0,0)\neq 0$ or $g(0,0)\neq 0$.

Our main theorem in this section reads as follows.

\begin{theorem}\label{thm2}
Assume that
\begin{gather}
\lim_{u+v\to \infty }\frac{f(u,v)}{u+v}=f_{\infty }\quad \text{with }
|f_{\infty }|<\frac{k_1}{\alpha |p|_1},  \label{H2}
\\
\lim_{u+v\to \infty }\frac{g(u,v)}{u+v}=g_{\infty }\quad \text{with }
|g_{\infty }|<\frac{k_2}{\beta |q|_1},  \label{H3}
\end{gather}
for some positive real numbers $\alpha $ and $\beta $ satisfying
$\frac{1}{\alpha }+\frac{1}{\beta }=1$. Then Problem
\eqref{nonl2} has at least one nontrivial solution
$(u,v)\in C_{0}({\mathbb{R}},{\mathbb{R}}^2)$.
\end{theorem}

\begin{proof}
Let the Banach space $C_{0}({\mathbb{R}},{\mathbb{R}}^2)
:=C_{0}({\mathbb{R}},{\mathbb{R}})\times C_{0}({\mathbb{R}
},{\mathbb{R}})$ be endowed with the norm $\| (.,.)\| $ given
by
\[
\| (u,v)\| =\| u\| _{0}+\| v\| _{0}.
\]
It is clear that Problem \eqref{nonl2} is equivalent to the integral
equation:
\begin{align*}
&\Big(u(x),v(x)\Big) \\
&= \Big(\int_{-\infty }^{+\infty
}G_1(x,s)p(s)f(u(s),v(s))\,ds,\int_{-\infty }^{+\infty
}G_2(x,s)q(s)g(u(s),v(s))\,ds\Big)
\end{align*}
with Green functions
\[
G_1(x,s)=\frac{1}{r_1-r_1'}
\begin{cases}
e^{r_1(x-s)} & \text{if }  x\leq s \\
e^{r_1'(x-s)} & \text{if }  x\geq s
\end{cases}
\]
and
\[
G_2(x,s)=\frac{1}{r_2-r_2'}
\begin{cases}
e^{r_2(x-s)} & \text{ if }  x\leq s \\
e^{r_2'(x-s)} & \text{ if }  x\geq s
\end{cases}
\]
and characteristic roots
\begin{gather*}
r_1=\frac{c_1+\sqrt{c_1^2+4\lambda _1}}{2}\,,\quad
r_1'=\frac{c_1-\sqrt{c_1^2+4\lambda _1}}{2}\,, \\
r_2=\frac{c_2+\sqrt{c_2^2+4\lambda _2}}{2}\,,\quad
r_2'=\frac{c_2-\sqrt{c_2^2+4\lambda _2}}{2}\,.
\end{gather*}

Define the mapping $T: C_{0}({\mathbb{R}},{\mathbb{R}}
^2)\to C_{0}({\mathbb{R}},{\mathbb{R}}^2)$
by
$T=(T_1,T_1)$
where
\begin{gather*}
T_1(u,v)(x)=\int_{-\infty }^{+\infty }G_1(x,s)p(s)f(u(s),v(s))\,ds\,,\\
T_2(u,v)(x)=\int_{-\infty }^{+\infty }G_2(x,s)q(s)g(u(s),v(s))\,ds.
\end{gather*}
In view of Krasnosel'sk'ii-Zabreiko's fixed point theorem, we look
for fixed points for the operator $T$ in the Banach space
$C_{0}({\mathbb{R}},{\mathbb{R}}^2)$. The proof is split in four
steps.

 \textbf{Claim 1:} The mapping $T=(T_1,T_2)$ is well defined;
indeed, for any $(u,v)\in C_{0}({\mathbb{R}},{\mathbb{R}}^2)$,
we get, by Assumptions (\ref{H2}), (\ref{H3}), the following estimate
\begin{align*}
|T_1(u,v)(x)| &\leq \int_{-\infty }^{+\infty
}G_1(x,s)|p(s)f(u(s),v(s))|\,ds \\
&\leq \frac{1}{k_1}|p|_1\max_{|y|\leq \| u\| _{0},|z|\leq
\| v\| _{0}}|f(y,z)|.
\end{align*}
In the same way, we find that
\[
|T_2(u,v)(x)|\leq \frac{1}{k_2}|q|_1\max_{|y|\leq \| u\|
_{0},|z|\leq \| v\| _{0}}|g(y,z)|.
\]
The convergence of the integrals defining $T_1(u,v)(x)$ and $T_2(u,v)(x)$
is then established. In addition, for any $s\in {\mathbb{R}}$,
$G_1(\pm \infty ,s)=0,G_2(\pm \infty ,s)=0$, and then, taking
the limit in $T_1(u,v)(x)$ and $T_2(u,v)(x)$, we get
\[
T(u,v)(\pm \infty )=\Big(T_1(u,v)(\pm \infty ),T_2(u,v)(\pm \infty )\Big)
=(0,0).
\]
Therefore, the mapping $T:C_{0}({\mathbb{R}},{\mathbb{R}}^2)
\to C_{0}({\mathbb{R}},{\mathbb{R}}^2)$ is well
defined.

\textbf{Claim 2:} The operator $T$ is continuous. It suffices to prove
that
both $T_1$ and $T_1$ are continuous.
Let $\big((u_{n},v_{n})\big)_{n}\subset C_{0}({\mathbb{R}},{\mathbb{R}}
^2)$ be a sequence which converges uniformly to $(u_{0},v_{0})$ on
each compact subinterval of ${\mathbb{R}}$. For some fixed $a>0$, we will
prove the uniform convergence of $(T(u_{n},v_{n}))_{n}$ to $T(u_{0},v_{0})$
on $[-a,a]$. Let $\varepsilon >0$ and choose some $b>a$ large
enough. By the uniform convergence of the sequence
$\big((u_{n},v_{n})\big)_{n}$ on $[-b,b]$, there exists an integer
$N=N(\varepsilon ,b)$
satisfying
\begin{align*}
I_1: &= \sup_{x\in {\mathbb{R}}
}\int_{-b}^{+b}G_1(x,s)|p(s)f(u_{n}(s),v_{n}(s))-p(s)f(u_{0}(s),v_{0}(s))|
\,ds \\
&< \frac{\varepsilon }{2}\quad \text{for  }n\geq N.
\end{align*}
For $x\in [-a,a]$, we have that
$|T_1(u_{n},v_{n})(x)-T_1(u_{0},v_{0})(x)|\leq I_1+I_2+I_{3}$ with
\[
I_2: =\sup_{x\in {\mathbb{R}}}\,\int_{{\mathbb{R}}
-[-b,+b]}G_1(x,s)|p(s)f(u_{0}(s),v_{0}(s))|\,ds
\leq \frac{\varepsilon }{4}
\]
(by the Cauchy Convergence Criterion and $\lim_{|s|\to
+\infty }p(s)f(u_{0}(s),v_{0}(s))=0$) and
\[
I_{3}: =\sup_{x\in {\mathbb{R}}}\,\int_{{\mathbb{R}}
-[-b,+b]}G_1(x,s)|p(s)f(u_{n}(s),v_{n}(s))|\,ds
\leq \frac{\varepsilon }{4}
\]
(by the Lebesgue Dominated Convergence Theorem). This proves the uniform
convergence of $(T_1(u_{n},v_{n}))_{n}$ to $T_1(u_{0},v_{0})$ on
$[-a,a]$.
Similarly, one can prove the uniform convergence of
$(T_2(u_{n},v_{n}))_{n} $ to $T_2(u_{0},v_{0})$ on $[-a,a]$.


\textbf{Claim 3:} For any $M>0$, the set
$\{T(u,v);\,\| (u,v)\| \leq M\}$ is relatively compact in
$C_{0}({\mathbb{R}},{\mathbb{R}}^2)$. By the Ascoli-Arzela Theorem,
it is sufficient to prove that
the functions of this set are equicontinuous on every subinterval
$[-a,a]$ and that there exist functions
$\gamma _1,\gamma _2\in C_{0}({\mathbb{R}},{\mathbb{R}})$ such that
for any $x\in {{\mathbb{R}}}$, $|T_1(u,v)(x)|\leq \gamma _1(x)$ and
$|T_2(u,v)(x)|\leq \gamma _2(x)$. Let $x_1,x_2\in [-a,a]$. Then
\begin{align*}
&|T_1(u,v)(x_2)-T_1(u,v)(x_1)| \\
&\leq \int_{-\infty }^{+\infty
}|G_1(x_2,s)-G_1(x_1,s)||p(s)f(u(s),v(s))|\,ds \\
&\leq \max_{y+z\in \lbrack -M,M]}|f(y,z)|\int_{-\infty }^{+\infty
}|G_1(x_2,s)-G_1(x_1,s)||p(s)|\,ds.
\end{align*}
By the continuity of the Green function $G_1$, the last term tends to $0$,
as $x_2$ tends to $x_1$.
Similarly,
\begin{align*}
&|T_2(u,v)(x_2)-T_2(u,v)(x_1)| \\
&\leq \max_{y+z\in \lbrack -M,M]}|g(y,z)|\int_{-\infty }^{+\infty
}|G_2(x_2,s)-G_2(x_1,s)||q(s)|\,ds.
\end{align*}
By the continuity of the Green function $G_2$, the last term tends to
$0$, as $x_2$ tends to $x_1$, whence comes the equicontinuity of
$\{T(u,v);\| (u,v)\| \leq M\}$. Now, we check analogously the second
statement:
\begin{align*}
|T_1(u,v)(x)|
&\leq \int_{-\infty }^{+\infty
}G_1(x,s)|p(s)f(u(s),v(s))|\,ds \\
&\leq \max_{y+z\in \lbrack -M,M]}|f(y,z)|\int_{-\infty }^{+\infty
}G_1(x,s)|p(s)|\,ds: =\gamma _1(x)
\end{align*}
and $\gamma _1\in C_{0}({\mathbb{R}},{\mathbb{R}})$.
Also
\[
|T_2(u,v)(x)|\leq \max_{y+z\in \lbrack -M,M]}|g(y,z)|\int_{-\infty
}^{+\infty }G_2(x,s)|q(s)|\,ds: =\gamma _2(x)
\]
and $\gamma _2\in C_{0}({\mathbb{R}},{\mathbb{R}})$.

Now we consider the  boundary-value problem
\begin{equation}
\begin{gathered}
-u''+c_1u'+\lambda _1u  =  p(x)(u(x)+v(x)),
\quad -\infty <x<+\infty , \\
-v''+c_2v'+\lambda _2v  =  q(x)(u(x)+v(x)),
\quad -\infty <x<+\infty , \\
u(-\infty )=u(+\infty )  =0,   \quad
v(-\infty )=v(+\infty )  =0.
\end{gathered}  \label{lin2}
\end{equation}
Define the operator $A=(A_1,A_2)$ by
\begin{gather*}
A_1(u,v)(x)=f_{\infty }\int_{-\infty }^{+\infty
}G_1(x,s)p(s)(u(s)+v(x))\,ds,
\\
A_2(u,v)(x)=g_{\infty }\int_{-\infty }^{+\infty
}G_2(x,s)q(s)(u(s)+v(x))\,ds.
\end{gather*}
Obviously, $A$ is a bounded linear operator. Furthermore, any fixed
point of $A$ is a solution of the Problem (\ref{lin2}), and conversely.

We now assert that $1$ is not an eigenvalue of $A$. In fact,
if $f_{\infty}=0$ and $g_{\infty }=0$, then the Problem (\ref{lin2})
has no nontrivial solutions. If $f_{\infty }\neq 0$ or
$g_{\infty }\neq 0$, suppose that the
Problem (\ref{lin2}) has a nontrivial solution $(u,v)$. Then
\begin{align*}
|A_1(u,v)(x)|
&\leq |f_{\infty }|\int_{-\infty }^{+\infty }G_1(x,s)|p(s)(u(s)+v(s))|\,ds \\
&\leq \frac{1}{k_1}|f_{\infty }||p|_1(\| u\| _{0}+\| v\|
_{0}) \\
&= \frac{1}{k_1}|f_{\infty }||p|_1\| (u,v)\| \\
&< \frac{1}{\alpha }\| (u,v)\|
\end{align*}
and
\begin{align*}
|A_2(u,v)(x)|
&\leq |g_{\infty }|\int_{-\infty }^{+\infty
 }G_2(x,s)|q(s)(u(s)+v(s))|\,ds \\
&\leq \frac{1}{k_2}|g_{\infty }||q|_1(\| u\| _{0}+\| v\|
_{0}) \\
&= \frac{1}{k_2}|g_{\infty }||q|_1\| (u,v)\| \\
&< \frac{1}{\beta }\| (u,v)\| .
\end{align*}
Hence
\[
\| A(u,v)\| = \| A_1(u,v)\| _{0}+\| A_2(u,v)\| _{0}
< (\frac{1}{\alpha }+\frac{1}{\beta })\| (u,v)\|
= \| (u,v)\| .
\]
This contradiction proves that Problem (\ref{lin2}) has no nontrivial
solution. Thus, $1$ is not an eigenvalue of $A$.

Finally, we prove that
\[
\lim_{\| (u,v)\| \to \infty }\frac{\| T(u,v)-A(u,v)\| }{
\| (u,v)\| }=0
\]
which is equivalent to
\[
\lim_{\| u\| _{0}+\| v\| _{0}\to \infty }\frac{\|
T_1(u,v)-A_1(u,v)\| _{0}}{\| u\| _{0}+\| v\| _{0}}
+\lim_{\| u\| _{0}+\| v\| _{0}\to \infty }\frac{\|
T_2(u,v)-A_2(u,v)\| _{0}}{\| u\| _{0}+\| v\| _{0}}
=0.
\]

According to
$\lim_{u+v\to \infty }\frac{f(u,v)}{u+v}=f_{\infty }$ and
$\lim_{u+v\to \infty }\frac{g(u,v)}{u+v}=g_{\infty }$,
for any $\varepsilon >0$, there exists $R>0$ such that
\begin{gather*}
|f(u,v)-f_{\infty }(u+v)|<\varepsilon |u+v|,\quad \text{for}\;|u+v|>R,
\\
|g(u,v)-g_{\infty }(u+v)|<\varepsilon |u+v|,\quad \text{for}\;|u+v|>R.
\end{gather*}
Set $R^{\ast }=\max \Big\{\max_{|u+v|\leq
R}|f(u,v)|,\max_{|u+v|\leq R}|g(u,v)|\Big\}$ and select $M>0$ such
that $R^{\ast }+\max \{|f_{\infty }|,|g_{\infty }|\}R<\varepsilon M$.
Denote
\[
I_1=\{x\in {\mathbb{R}}:|u(x)+v(x)|\leq R\},\quad
I_2=\{x\in {\mathbb{R}}:|u(x)+v(x)|>R\}.
\]
Thus for any $(u,v)\in C_{0}({\mathbb{R}},{\mathbb{R}}^2)$ with
$\|(u,v)\| >M$, when $x\in I_1$, we have
\begin{align*}
|f(u(x),v(x))-f_{\infty }(u(x)+v(x))|
&\leq |f((u(x),v(x))|+|f_{\infty }||u(x)+v(x)| \\
&\leq R^{\ast }+|f_{\infty }|R\\
&<\varepsilon M<\varepsilon \| (u,v)\| .
\end{align*}
Similarly, we conclude that for any
$(u,v)\in C_{0}({\mathbb{R}},{\mathbb{R}}^2)$ with
$\| (u,v)\| >M$, when $x\in I_2$, we also have that
\[
|f(u(x),v(x))-f_{\infty }(u(x)+v(x)|<\varepsilon \| (u,v)\| .
\]
We conclude that for any $(u,v)\in C_{0}({\mathbb{R}},{\mathbb{R}}^2)$
with $\| (u,v)\| >M$, one has
\[
|f(u(x),v(x))-f_{\infty }(u(x)+v(x))|<\varepsilon \| (u,v)\| .
\]
Then for any $(u,v)\in C_{0}({\mathbb{R}},{\mathbb{R}}^2)$ with
$\|(u,v)\| >M$, we have
\begin{align*}
|f(u(x),v(x))|
&\leq |f(u(x),v(x))-f_{\infty }(u(x)+v(x))|+|f_{\infty
}(u(x)+v(x))| \\
&\leq \varepsilon \| (u,v)\| +|f_{\infty }|\| (u,v)\| \\
&\leq (|f_{\infty }|+\varepsilon )\| (u,v)\| .
\end{align*}
In the same way, we find that for any $(u,v)\in C_{0}({\mathbb{R}},{\mathbb{R
}}^2)$ with $\| (u,v)\| >M$, we have
\[
|g(u(x),v(x))|\leq (|g_{\infty }|+\varepsilon )\| (u,v)\| .
\]
Hence we obtain
\begin{align*}
&|T_1(u,v)(x)-A_1(u,v)(x)|\\
&= |\int_{-\infty }^{+\infty }G_1(x,s)p(s)f(u(s),v(s))\,ds-f_{\infty
}\int_{-\infty }^{+\infty }G_1(x,s)p(s)(u(s)+v(s))\,ds| \\
&\leq \frac{1}{k_1}\int_{-\infty }^{+\infty
}|p(s)||f(u(s),v(s)))-f_{\infty }(u(s)+v(s))|\,ds \\
&< \frac{1}{k_1}|p|_1\varepsilon \| (u,v)\| \\
&= \frac{1}{k_1}|p|_1\varepsilon (\| u\| _{0}+\| v\| _{0}).
\end{align*}
Similarly,
\[
|T_2(u,v)(x)-A_2(u,v)(x)|<\frac{1}{k_2}|q|_1\varepsilon (\|
u\| _{0}+\| v\| _{0}).
\]
Then we have
\begin{gather*}
\lim_{\| u\| _{0}+\| v\| _{0}\to \infty }\frac{\|
T_1(u,v)-A_1(u,v)\| _{0}}{\| u\| _{0}+\| v\| _{0}} = 0,
\\
\lim_{\| u\| _{0}+\| v\| _{0}\to \infty }\frac{\|
T_2(u,v)-A_2(u,v)\| _{0}}{\| u\| _{0}+\| v\| _{0}} = 0
\end{gather*}
and hence
\[
\lim_{\| u+v\| \to \infty }\frac{\| T(u,v)-A(u,v)\| }{\| u+v\| }=0.
\]
Theorem \ref{KZ} now guarantees that Problem \eqref{nonl2} has at
least one nontrivial solution. This completes the proof.
\end{proof}

\section{Further results}

In this section, we study the system
\begin{equation}
\begin{gathered}
-u''+c_1u'+\lambda _1u  =  f(x,u,v),\quad -\infty <x<+\infty , \\
-v''+c_2v'+\lambda _2v  =  g(x,u,v),\quad -\infty <x<+\infty , \\
u(-\infty )=u(+\infty ) =0,   \quad
v(-\infty )=v(+\infty ) =0.
\end{gathered}  \label{nonl3}
\end{equation}
where $f,g:{\mathbb{R}}\times {\mathbb{R}}^2\to {\mathbb{R}}$ are
continuous functions with $f(x,0,0)\not\equiv 0$ or
$g(x,0,0)\not\equiv 0$.

The main existence result of this section is as follows.

\begin{theorem}\label{thm3}
Assume that there exist two functions
$\varphi ,\psi :{\mathbb{R}}_{+}\times {\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$ continuous
and nondecreasing with respect to their two variables and there
exist two positive continuous functions
$p,q:{\mathbb{R}}\to {\mathbb{R}}_{+}$ and $M_{0}>0$ such that
\begin{equation}
\begin{gathered}
|f(x,u,v)|\leq p(x)\varphi (|u|,|v|),\quad\text{for }
(x,u,v)\in {{\mathbb{R}}}^{3}, \\
|g(x,u,v)|\leq q(x)\psi (|u|,|v|),\quad \text{for }
(x,u,v)\in {{\mathbb{R}}}^{3}, \\
\max \Big\{\frac{1}{k_1}|p|_1\varphi (M_{0},M_{0}),\frac{1}{k_2}
|q|_1\psi (M_{0},M_{0})\Big\}\leq M_{0}.
\end{gathered}  \label{S}
\end{equation}
Then Problem \eqref{nonl3} admits at least one nontrivial solution
$(u,v)\in C_{0}({\mathbb{R}},{\mathbb{R}}^2)$.
\end{theorem}

\begin{proof}
Define the mapping $T: \,C_{0}({\mathbb{R}},{\mathbb{R}}
^2)\to \,C_{0}({\mathbb{R}},{\mathbb{R}}^2)$
by
$T=(T_1,T_1)$
where
\begin{gather*}
T_1(u,v)(x)=\int_{-\infty }^{+\infty }G_1(x,s)f(s,u(s),v(s))\,ds,\\
T_2(u,v)(x)=\int_{-\infty }^{+\infty }G_2(x,s)g(s,u(s),v(s))\,ds.
\end{gather*}
In view of Schauder's fixed point theorem, we look for fixed points of
$T$ in the Banach space $C_{0}({\mathbb{R}},{\mathbb{R}}^2)$. The
proof is split into four steps.

 \textbf{Claim 1:} The mapping $T$ is well defined. Indeed, for
any $(u,v)\in C_{0}({\mathbb{R}},{\mathbb{R}}^2)$, we get, by
Assumptions \eqref{S}, the  estimate
\begin{align*}
|T_1(u,v)(x)| &\leq \int_{-\infty }^{+\infty
}G_1(x,s)|f(s,u(s),v(s))|\,ds \\
&\leq \int_{-\infty }^{+\infty }G_1(x,s)p(s)\varphi (|u(s)|,|v(s)|)\,ds \\
&\leq \varphi (\| u\| ,\| v\| )\int_{-\infty }^{+\infty
}G_1(x,s)p(s)\,ds,\;\forall \,x\in {{\mathbb{R}}} \\
&\leq \frac{1}{k_1}|p|_1\varphi (\| u\| ,\| v\| ).
\end{align*}
In the same way, one can prove that
\[
|T_2(u,v)(x)\leq \frac{1}{k_2}|q|_1\psi (\| u\| ,\| v\| ).
\]
The convergence of the integrals defining $T(u,v)(x)$ is then established.
In addition for any $s\in {\mathbb{R}}$, $G_1(\pm \infty ,s)=0,\;G_1(\pm
\infty ,s)=0$, and then, taking the limit in $
T(u,v)(x)=(T_1(u,v)(x),T_2(u,v)(x))$, we obtain $T(u,v)(\pm \infty )=0$.
Therefore, the mapping $T:C_{0}({\mathbb{R}},{\mathbb{R}}^2)
\to C_{0}({\mathbb{R}},{\mathbb{R}}^2)$ is well
defined.

\textbf{Claim 2:} As in Section 3, one can prove easily that the
operator $T=(T_1,T_2)$ is continuous.

\textbf{Claim 3:} As in Section 3, one can prove easily that for
any $M>0$, the set $\{T(u,v)=(T_1(u,v),T_2(u,v));\,\| (u,v)\| \leq M\}$
is relatively compact in $C_{0}({\mathbb{R}},{\mathbb{R}}^2)$.

\textbf{Claim 4:} There exists a nonempty closed bounded convex $K$ such
that $T$ maps $K$ into itself. Let
\[
K=\left\{ (u,v)\in C_{0}({\mathbb{R}},{\mathbb{R}}^2):\|
u\| _{0}\leq M_{0},\| v\| _{0}\leq M_{0}\right\} .
\]
From assumption \eqref{S}, we know that
\[
\frac{1}{k_1}|p|_1\varphi (M_{0},M_{0})\leq 1,\quad
\frac{1}{k_2}|q|_1\psi (M_{0},M_{0})\leq 1.
\]
If $\| u\| \leq M_{0}$ and $\| v\| \leq M_{0}$, then
\begin{align*}
\| T_1(u,v)\| _{0} &\leq \sup_{x\in {{\mathbb{R}}}}\int_{-\infty
}^{+\infty }G_1(x,s)p(s)\varphi (|u(s)|,|v(s)|)\,ds \\
&\leq \frac{1}{k_1}|p|_1\varphi (M_{0},M_{0})\\
&\leq M_{0}.
\end{align*}
Similarly,
$\| T_2(u,v)\| _{0}\leq M_{0}$.
Therefore, the operator $T$ maps $K$ into itself.
The proof of Theorem \ref{thm2} then follows from Schauder's
fixed point theorem.
\end{proof}

 Using of Schauder's theorem, one can also prove the
existence of a positive solution under some integral conditions on the
nonlinear terms:

\begin{theorem}\label{thm4}
Suppose that the functions $f$ and $g$ are positive with
$f(x,0,0)\not\equiv 0$ or $g(x,0,0)\not\equiv 0$,
and satisfy the following two mean growth assumptions:
\begin{gather*}
|f(x,u,v)|\leq \varphi (x,|u|,|v|),\\
|g(x,u,v)|\leq \psi (x,|u|,|v|)
\end{gather*}
where $\varphi ,\psi :{\mathbb{R}}\times {\mathbb{R}}_{+}\times {\mathbb{R}
}_{+}\to {\mathbb{R}}_{+}$ are continuous, nondecreasing with
respect to their two last arguments and verify
\begin{equation}
\int_{-\infty }^{+\infty }\varphi (x,M_{0},M_{0})\,dx\leq k_1M_{0},\quad
\int_{-\infty }^{+\infty }\psi (x,M_{0},M_{0})\,dx\leq k_2M_{0}  \label{ST}
\end{equation}
for some constant $M_{0}>0$. Then Problem \eqref{nonl3} has a
positive solution $(u,v)\in C_{0}({\mathbb{R}},{\mathbb{R}}_{+}\times {
\mathbb{R}}_{+})$.
\end{theorem}

\begin{proof}
To prove Theorem \ref{thm4}, we proceed as in Theorem \ref{thm3} by taking
the closed convex subset $K$ of
$C_{0}({\mathbb{R}},{\mathbb{R}}^2)$ defined by:
\[
K=\{(u,v)\in C_{0}({\mathbb{R}},{\mathbb{R}}^2):0\leq
u(x)\leq M_{0},\;0\leq v(x)\leq M_{0}\text{ on }{\mathbb{R}}\}.
\]
Using Assumption (\ref{ST}) and the fact that the mapping $\varphi $
and $\psi $ are nondecreasing with respect to their two last arguments,
we find that $T$ maps $K$ into itself. Indeed, we derive the estimates:
\begin{align*}
0&\leq T_1(u,v)(x) \leq \int_{-\infty }^{+\infty }G_1(x,s)\varphi
(s,|u(y)|,|v(y)|)\,ds \\
&\leq \frac{1}{k_1}\int_{-\infty }^{+\infty }\varphi (s,M_{0},M_{0})\,ds
\leq M_{0}
\end{align*}
and
\[
0\leq T_2(u,v)(x)\leq M_{0}.
\]
In addition, the mapping $T$ is continuous as can easily be seen and
one can check that $T(K)$ is relatively compact.
Then the claim of Theorem \ref{thm4} follows.
\end{proof}

\section{Examples}

In this section, we give some examples to illustrate our results.

(1) Consider the boundary-value problem
\begin{equation} \label{e5.1}
\begin{gathered}
-u''+u'+2u  =  \frac{1}{\pi (x^2+1)}[2u+1+\lg (1+|u|)], \\
u(-\infty )=u(+\infty )  =  0.
\end{gathered}
\end{equation}
Here $p(x)=\frac{1}{\pi (x^2+1)}$ and
$f(u)=2u+1+\lg (1+|u|)$.
Notice that $k=3$, $|p|_1=1$ and $f_{\infty }=2$.
Thus, by Theorem \ref{thm1}, Problem \eqref{e5.1} has at
least one nontrivial
solution $u\in C_{0}(R,R)$.

(2) Consider the boundary-value system
\begin{equation} \label{e5.2}
\begin{gathered}
-u''+u'+2u  =  \frac{1}{2\pi (x^2+1)}\big[2(u+v)+1+\lg (1+|u+v|)\big], \\
-v''+\sqrt{5}v'+v  =  \frac{1}{2\sqrt{\pi }}
e^{-x^2}\big[2(u+v)+1+\sqrt{1+|u+v|}\big], \\
u(-\infty )=u(+\infty )  =  0, \quad
v(-\infty )=v(+\infty )  =  0.
\end{gathered}
\end{equation}
Set $p(x)=\frac{1}{2\pi (x^2+1)}$,
$q(x)=\frac{1}{2\sqrt{\pi }}e^{-x^2}$,
$f(u,v)=2(u+v)+1+\lg (1+|u+v|)$, and
$g(u,v)=2(u+v)+1+ \sqrt{1+|u+v|}$.
Notice that $k_1=k_2=3$, $\alpha =\beta=2$,
$|p|_1=|q|_1=\frac{1}{2}$, and $f_{\infty }=g_{\infty }=2$.
Thus, by Theorem \ref{thm2}, Problem \eqref{e5.3} has at least one nontrivial
solution $(u,v)\in C_{0}(R,R^2)$.


(3) Consider the boundary-value system
\begin{equation} \label{e5.3}
\begin{gathered}
-u''+u'+2u  =  \frac{1}{\pi (x^2+1)}\big(
\left\vert v\right\vert ^{\mu }+1\big), \\
-v''+\sqrt{5}v'+v  =  \frac{1}{\sqrt{\pi }}\exp
^{-x^2}\big(\left\vert u\right\vert ^{\nu }+1\big), \\
u(-\infty )=u(+\infty )  =  0, \quad
v(-\infty )=v(+\infty )  =  0.
\end{gathered}
\end{equation}
where $\mu $ and $\nu $ are real numbers such that
$0<\mu <1$ and $0<\nu <1$.  Set
$p(x)=\frac{1}{\pi (x^2+1)}$,
$q(x)=\frac{1}{\sqrt{\pi }}e^{-x^2}$,
$\varphi (y,z)=z^{\mu }+1$, and
$\psi (y,z)=y^{\nu }+1$. Notice that
$k_1=k_2=3$, $|p|_1=|q|_1=1$  and if we choose any
$M_{0}$ large enough, then condition \eqref{S} is satisfied.
Thus, by Theorem \ref{thm3}, Problem \eqref{e5.3} has at
least one nontrivial solution $(u,v)\in C_{0}(R,R^2)$.


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\bibitem{DjeMou2} S. Djebali and T. Moussaoui;
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\bibitem{DjeMeb1} S. Djebali and K. Mebarki;
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\bibitem{KZ1} M. A. Krasnoselskii, P. P. Zabreiko;
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\bibitem{Ze} E. Zeidler;
\emph{Nonlinear Functional Analysis,} T1,
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\end{thebibliography}

\end{document}