\documentclass[reqno]{amsart}
%\usepackage[notref,notcite]{showkeys}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 39, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2016/39\hfil Existence of non-oscillatory solutions]
{Existence of non-oscillatory solutions to
first-order neutral differential equations}

\author[T. Candan \hfil EJDE-2016/39\hfilneg]
{Tuncay Candan}

\address{Tuncay Candan  \newline
Department of Mathematics,
Faculty of Arts and Sciences,
Ni\u{g}de University,
Ni\u{g}de 51200, Turkey}
\email{tcandan@nigde.edu.tr}

\thanks{Submitted October 14, 2015. Published January 27, 2016.}
\subjclass[2010]{34K11, 34C10} 
\keywords{Neutral equations; fixed point; non-oscillatory solution}

\begin{abstract}
 This article presents sufficient conditions for the existence
 of non-oscillatory solutions to first-order differential equations
 having both delay and advance terms, known as mixed equations.
 Our main tool is the Banach contraction principle.
\end{abstract}

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

\section{Introduction}

In this article, we consider a first-order  neutral
differential equation
\begin{equation}\label{e:1}
\begin{aligned}
&\frac{d}{dt}[x(t)+P_1(t)x(t-\tau_1)+P_2(t)x(t+\tau_2)]\\
&+Q_1(t)x(t-\sigma_1)-Q_2(t)x(t+\sigma_2)=0,
\end{aligned}
\end{equation}
where $P_i\in C([t_0,\infty),\mathbb{R})$, 
$Q_i\in C([t_0,\infty),[0,\infty))$,
$\tau_i>0$ and $\sigma_i\geq   0$ for $i=1,2$. We give some new criteria
for the existence of  non-oscillatory  solutions of \eqref{e:1}.

Recently,  the existence of non-oscillatory solutions of
first-order neutral functional differential  equations has been
investigated by many authors.  Yu and Wang \cite{Yu} showed
that the equation
\[
\frac{d}{dt}\left[x(t)+px(t-c)\right]+Q(t)x(t-\sigma)=0,\quad t\geq
t_0
\]
 has a non-oscillatory solution for $p\geq 0$. Later, in $1993$, Chen et  al
\cite{Che} studied the same equation and they extended the results to the
case $p\in\mathbb{R}\backslash\{-1\}$. Zhang et al \cite{Zha}
investigated  the existence of non-oscillatory solutions of the
first-order neutral delay differential equation with variable
coefficients
\[
\frac{d}{dt}[x(t)+P(t)x(t-\tau)]+Q_1(t)x(t-\sigma_1)-Q_2(t)x(t-\sigma_2)=0,\quad
t\geq t_0 \,.
\]
They obtained sufficient conditions for the existence of
non-oscillatory solutions depending on the four different ranges of
$P(t)$. In \cite{Doroc}, existence of non-oscillatory solutions of
first-order neutral differential equations
\[
\frac{d}{dt}[x(t)-a(t) x(t-\tau)]=p(t)f(x(t-\sigma))
\]
was studied.

On the other hand, there has been  research activities about
the oscillatory behavior of first and higher order neutral
differential equations with advanced terms. For  instance, in
\cite{Aga0} and \cite{Tun0}, n-th order neutral differential
equations with advanced term of the form
\[
[x(t)+ax(t-\tau)+bx(t+\tau)]^{(n)}+\delta
\left(q(t)x(t-g)+p(t)x(t+h)\right)=0
\]
and
\[
[x(t)+\lambda ax(t-\tau)+\mu
bx(t+\tau)]^{(n)}+\delta\Big(
\int_c^dq(t,\xi)x(t-\xi)d\xi+\int_c^dp(t,\xi)x(t+\xi)d\xi\Big)=0,
\]
were studied, respectively.

This article was motivated by the above studies. To the best of our
knowledge, this current  paper is the only paper regarding to the
existence of non-oscillatory  solutions of neutral differential
equation with advanced term. Some other papers  for the existence
of non-oscillatory solutions of first, second  and higher order
neutral functional differential and difference equations; see
\cite{Kule, Zho, Tun1, Tun2, Tun3, Tia} and the references
contained therein. We refer the reader to the books \cite{Lad,
Gyo, Bai, Erb, Aga1, Aga2} on the subject of neutral differential
equations.


Let $m=\max\{\tau_1,\sigma_1\}$. By a solution of  \eqref{e:1} we
mean a function $x\in C([t_1-m,\infty),\mathbb{R})$, for some
$t_1\geq t_0$, such that
$x(t)+P_1(t)x(t-\tau_1)+P_2(t)x(t+\tau_2)$ is  continuously
differentiable on $[t_1,\infty)$ and \eqref{e:1}  is satisfied for
$t\geq t_1$.

As it is customary, a solution of \eqref{e:1} is said to be
oscillatory if it has arbitrarily large zeros. Otherwise the
solution is called non-oscillatory.

The following theorem will be used to prove the theorems.

\begin{theorem}[Banach's Contraction Mapping Principle]
A contraction mapping on a complete metric space has exactly one
fixed point.
\end{theorem}

\section{Main Results}

To show that an operator $S$  satisfies the
conditions for the contraction mapping principle, we  consider
different cases for the ranges of the coefficients $P_1(t)$ and
$P_2(t)$.

\begin{theorem}\label{t:1}
 Assume that $ 0\leq P_1(t)\leq p_1<1$, $0\leq P_2(t)\leq p_2<1-p_1$ and
\begin{equation}\label{e:2}
\int_{t_0}^{\infty}Q_1(s) ds<\infty,\quad
\int_{t_0}^{\infty}Q_2(s) ds<\infty,
\end{equation}
then \eqref{e:1} has a bounded non-oscillatory solution.
\end{theorem}

\begin{proof} Because of \eqref{e:2}, we can choose a
$t_1>t_0$,
\begin{equation}\label{e:3}
t_1\geq t_0+\max\{\tau_1,\sigma_1\}
\end{equation}
sufficiently large such that
\begin{gather}\label{e:4}
\int_{t}^{\infty}Q_1(s) ds\leq \frac{M_2-\alpha}{M_2},\quad t\geq t_1, \\
\label{e:5}
\int_{t}^{\infty}Q_2(s) ds\leq \frac{\alpha-(p_1+p_2)M_2-M_1}{M_2},
\quad t\geq t_1,
\end{gather}
 where $M_1$ and $M_2$ are positive constants such that
\[
(p_1+p_2)M_2+M_1<M_2\quad\text{and}\quad 
\alpha\in\big((p_1+p_2)M_2+M_1, M_2\big).
\]
Let $\Lambda$ be the set of all continuous and bounded functions
on $[t_0,\infty)$ with the supremum norm. Set
\[
\Omega=\{x\in \Lambda:M_1\leq x(t)\leq M_2,\; t\geq t_0 \}.
\]
 It is clear that  $\Omega$ is a bounded,
closed and convex subset of $\Lambda$. Define  an operator
$S:\Omega\to\Lambda$ as follows:
\[
(Sx)(t)= \begin{cases}
\alpha-P_1(t)x(t-\tau_1)-P_2(t)x(t+\tau_2)\\
+\int_{t}^{\infty}[Q_1(s)x(s-\sigma_1)-Q_2(s)x(s+\sigma_2)]ds, & t\geq  t_1,\\[4pt]
 (Sx)(t_1),& t_0\leq t\leq t_1.
\end{cases}
\]
Obviously, $Sx$ is continuous. For $t\geq t_1$ and $x\in \Omega$,
from \eqref{e:4} and  \eqref{e:5}, respectively, it follows that
\[
(Sx)(t)\leq \alpha+\int_{t}^{\infty}Q_1(s)x(s-\sigma_1) ds
\leq \alpha+M_2\int_{t}^{\infty}Q_1(s) ds \leq M_2
\]
and
\begin{align*}
(Sx)(t)&\geq
\alpha-P_1(t)x(t-\tau_1)-P_2(t)x(t+\tau_2)
 -\int_{t}^{\infty}Q_2(s)x(s+\sigma_2)ds\\
&\geq \alpha- p_1M_2- p_2M_2-M_2\int_{t}^{\infty}Q_2(s) ds
\geq M_1.
\end{align*}
This means that $S\Omega\subset \Omega$. To apply the
contraction mapping principle, the remaining is to show that $S$
is a contraction mapping on $\Omega$.
 Thus, if $x_1, x_2\in \Omega$ and $t\geq t_1$,
\begin{align*}
&|(Sx_1)(t)-(Sx_2)(t)|\\
&\leq P_1(t)|x_1(t-\tau_1)-x_2(t-\tau_1)|+P_2(t)|x_1(t+\tau_2)-x_2(t+\tau_2)|\\
&\quad +\int_{t}^{\infty} \left(Q_1(s)|x_1(s-\sigma_1)-x_2(s-\sigma_1)|
+Q_2(s)|x_1(s+\sigma_2)-x_2(s+\sigma_2)|\right) ds
\end{align*}
or
\begin{align*}
&|(Sx_1)(t)-(Sx_2)(t)|\\
&\leq \|x_1-x_2\|\Big(p_1+p_2+\int_{t}^{\infty}\left(Q_1(s)
+Q_2(s)\right) ds\Big)\\
&\leq \Big(p_1+p_2+\frac{M_2-\alpha}{M_2}
 +\frac{\alpha-(p_1+p_2)M_2-M_1}{M_2}\Big)\|x_1-x_2\|\\
&=\lambda_1\|x_1-x_2\|,
\end{align*}
where $\lambda_1=(1-\frac{M_1}{M_2})$.
 This implies  that
\[
\|Sx_1-Sx_2\|\leq \lambda_1\|x_1-x_2\|,
\]
where the supremum norm is used. Since  $\lambda_1<1$, $S$ is a
contraction mapping on $\Omega$. Thus $S$ has a unique fixed
point which is a positive and bounded solution of \eqref{e:1}.
This completes the proof.
\end{proof}


\begin{theorem}\label{t:2}
Assume that $ 0\leq P_1(t)\leq p_1<1$, $p_1-1<p_2\leq P_2(t)\leq 0$ and
\eqref{e:2} hold, then \eqref{e:1} has a
bounded non-oscillatory solution.
\end{theorem}

\begin{proof}
Because of \eqref{e:2},  we can choose a $t_1>t_0$
sufficiently large satisfying  \eqref{e:3} such that
\begin{gather}\label{e:6}
\int_{t}^{\infty}Q_1(s) ds\leq \frac{(1+p_2)N_2-\alpha}{N_2},\quad
t\geq t_1, \\
\label{e:7}
\int_{t}^{\infty}Q_2(s) ds\leq \frac{\alpha-p_1N_2-N_1}{N_2}, \quad
t\geq t_1,
\end{gather}
where $N_1$ and $N_2$ are positive constants such that
\[
N_1+p_1N_2<(1+p_2)N_2\quad\text{and}\quad
\alpha \in (N_1+p_1N_2,(1+p_2)N_2).
\]
Let $\Lambda$ be the set of all continuous and bounded functions
on $[t_0,\infty)$ with the supremum norm. Set
\[
\Omega=\{x\in \Lambda: N_1\leq x(t)\leq N_2,\; t\geq t_0 \}.
\]
It is clear that  $\Omega$ is a bounded,
closed and convex subset of $\Lambda$. Define  an operator
$S:\Omega\to\Lambda$ as follows:
\[
(Sx)(t)=\begin{cases}
\alpha-P_1(t)x(t-\tau_1)-P_2(t)x(t+\tau_2)\\
+\int_{t}^{\infty}\left[Q_1(s)x(s-\sigma_1)-Q_2(s)x(s+\sigma_2)\right]ds,& t\geq  t_1,\\[4pt]
(Sx)(t_1),& t_0\leq t\leq t_1.
\end{cases}
\]
Obviously, $Sx$ is continuous. For $t\geq t_1$ and $x\in \Omega$,
from \eqref{e:6} and  \eqref{e:7}, respectively, it follows that
\begin{gather*}
(Sx)(t)\leq \alpha-p_2N_2+N_2\int_{t}^{\infty}Q_1(s) ds\leq N_2, \\
(Sx)(t)\geq\alpha- p_1N_2-N_2\int_{t}^{\infty}Q_2(s) ds\geq N_1.
\end{gather*}
This proves that $S\Omega\subset \Omega$. To apply the
contraction mapping principle, it remains to show that $S$
is a contraction mapping on $\Omega$.
 Thus, if $x_1, x_2\in \Omega$ and $t\geq t_1$,
\begin{align*}
|(Sx_1)(t)-(Sx_2)(t)|
&\leq \|x_1-x_2\|\Big(p_1-p_2+\int_{t}^{\infty}\left(Q_1(s)
+Q_2(s)\right) ds\Big)\\
&\leq \lambda_2\|x_1-x_2\|,
\end{align*}
where $\lambda_2=(1-\frac{N_1}{N_2})$.
 This implies
\[
\|Sx_1-Sx_2\|\leq \lambda_2\|x_1-x_2\|,
\]
where  the supremum norm is used. Since  $\lambda_2<1$, $S$ is a
contraction mapping on $\Omega$. Thus $S$ has a unique fixed
point which is a positive and bounded solution of \eqref{e:1}.
This completes the proof.
\end{proof}

\begin{theorem}\label{t:3}
Assume that  $1<p_1\leq P_1(t)\leq p_{1_0}<\infty$,
$0\leq P_2(t)\leq p_2<p_1-1$  and \eqref{e:2}
hold, then \eqref{e:1} has a bounded non-oscillatory solution.
\end{theorem}

\begin{proof}
In view of \eqref{e:2}, we can choose a $t_1>t_0$,
\begin{equation}\label{e:8}
t_1+\tau_1\geq t_0+\sigma_1,
\end{equation}
sufficiently large such that
\begin{gather}\label{e:9}
\int_{t}^{\infty}Q_1(s) ds\leq \frac{p_1M_4-\alpha}{M_4},\quad t\geq t_1, \\
\label{e:10}
\int_{t}^{\infty}Q_2(s) ds\leq
\frac{\alpha-p_{1_0}M_3-(1+p_2)M_4}{M_4},\quad t \geq  t_1,
\end{gather}
where $M_3$ and $M_4$ are positive constants such that
\[
p_{1_0}M_3+(1+p_2)M_4 <p_1M_4\quad\text{and}\quad
\alpha\in\big( p_{1_0}M_3+(1+p_2)M_4 , p_1M_4\big)\,.
\]
Let $\Lambda$ be the set of all continuous and bounded functions
on $[t_0,\infty)$ with the supremum norm. Set
\[
\Omega=\{x\in \Lambda:M_3\leq x(t)\leq M_4,\; t\geq t_0 \}.
\]
 It is clear that  $\Omega$ is a bounded,
closed and convex subset of $\Lambda$. Define a mapping
$S:\Omega\to \Lambda$ as follows:
\[
(Sx)(t)=\begin{cases}
\frac{1}{P_1(t+\tau_1)}\{\alpha-x(t+\tau_1)-P_2(t+\tau_1)x(t+\tau_1+\tau_2)\\
+\int_{t+\tau_1}^{\infty}\left[Q_1(s)x(s-\sigma_1)-Q_2(s)x(s+\sigma_2)\right]ds\},
& t\geq t_1,\\[4pt]
 (Sx)(t_1), &  t_0\leq t\leq t_1.
\end{cases}
\]
Clearly, $Sx$ is continuous. For $t\geq t_1$ and $x\in \Omega$,
from \eqref{e:9} and  \eqref{e:10}, respectively, it follows that
\[
(Sx)(t)
\leq \frac{1}{P_1(t+\tau_1)}\Big(\alpha+M_4\int_{t}^{\infty}Q_1(s)ds\Big)
\leq \frac{1}{p_1}\Big(\alpha+M_4\int_{t}^{\infty}Q_1(s) ds\Big)\leq M_4
\]
and
\begin{align*}
(Sx)(t)
&\geq \frac{1}{P_1(t+\tau_1)}\Big(\alpha-(1+p_2)M_4-M_4\int_{t}^{\infty}Q_2(s)
ds\Big)\\
&\geq \frac{1}{p_{1_0}}\Big(\alpha-(1+p_2)M_4-M_4\int_{t}^{\infty}Q_2(s)
ds\Big)\geq M_3.
\end{align*}
This means that $S\Omega\subset \Omega$. To apply the
contraction mapping principle it remains to show that $S$
is a contraction mapping on $\Omega$.
 Thus, if $x_1, x_2\in \Omega$ and $t\geq t_1$,
\begin{align*}
|(Sx_1)(t)-(Sx_2)(t)|
&\leq \frac{1}{p_1} \|x_1-x_2\|\Big(1+p_2+\int_{t}^{\infty}
\left(Q_1(s) +Q_2(s)\right) ds\Big)\\
&\leq \lambda_3\|x_1-x_2\|,
\end{align*}
where $\lambda_3=(1-\frac{p_{1_0}M_3}{p_1M_4})$.
 This implies
\[
\|Sx_1-Sx_2\|\leq \lambda_3\|x_1-x_2\|,
\]
where the supremum norm is used. Since  $\lambda_3<1$, $S$ is a
contraction mapping on $\Omega$. Thus $S$ has a unique fixed
point which is a positive and bounded solution of \eqref{e:1}.
This completes the proof.
\end{proof}

\begin{theorem}\label{t:4}
Assume that  $1<p_1\leq P_1(t)\leq p_{1_0}<\infty$,
 $1-p_1< p_2\leq P_2(t)\leq 0$  and  \eqref{e:2}
hold, then \eqref{e:1} has a bounded non-oscillatory solution.
\end{theorem}

\begin{proof}
In view of \eqref{e:2},  we can choose a $t_1>t_0$ sufficiently
large satisfying  \eqref{e:8} such that
\begin{gather}\label{e:11}
\int_{t}^{\infty}Q_1(s) ds\leq \frac{(p_1+p_2)N_4-\alpha}{N_4},\quad
t\geq t_1, \\
\label{e:12}
\int_{t}^{\infty}Q_2(s) ds\leq
\frac{\alpha-p_{1_0}N_3-N_4}{N_4},\quad t \geq  t_1,
\end{gather}
where $N_3$ and $N_4$ are positive constants such that
\[
p_{1_0}N_3+N_4<(p_1+p_2)N_4\quad\text{and}\quad
\alpha \in\big(p_{1_0}N_3+N_4, (p_1+p_2)N_4\big)\, .
\]
Let $\Lambda$ be the set of all continuous and bounded functions
on $[t_0,\infty)$ with the supremum norm. Set
\[
\Omega=\{x\in \Lambda:N_3\leq x(t)\leq N_4,\; t\geq t_0 \}.
\]
 It is clear that  $\Omega$ is a bounded,
closed and convex subset of $\Lambda$. Define a mapping
$S:\Omega\to \Lambda$ as follows:
\[
(Sx)(t)= \begin{cases}
\frac{1}{P_1(t+\tau_1)}\{\alpha-x(t+\tau_1)-P_2(t+\tau_1)x(t+\tau_1+\tau_2)\\
+\int_{t+\tau_1}^{\infty}\left[Q_1(s)x(s-\sigma_1)-Q_2(s)x(s+\sigma_2)\right]ds\},
&t\geq t_1,\\[4pt]
(Sx)(t_1), &  t_0\leq t\leq t_1.
\end{cases}
\]
Clearly, $Sx$ is continuous. For $t\geq t_1$ and $x\in \Omega$,
from \eqref{e:11} and  \eqref{e:12}, respectively, it follows that
\begin{align*}
(Sx)(t)
&\leq \frac{1}{P_1(t+\tau_1)}\Big(\alpha-p_2N_4+N_4\int_{t}^{\infty}Q_1(s)
ds\Big)\\
&\leq \frac{1}{p_1}\Big(\alpha-p_2N_4+N_4\int_{t}^{\infty}Q_1(s)
ds\Big)\leq N_4
\end{align*}
and
\begin{align*}
(Sx)(t)
&\geq \frac{1}{P_1(t+\tau_1)}\Big(\alpha-N_4-N_4\int_{t}^{\infty}Q_2(s)
ds\Big)\\
&\geq \frac{1}{p_{1_0}}\Big(\alpha-N_4-N_4\int_{t}^{\infty}Q_2(s)
ds\Big)\geq N_3.
\end{align*}
This proves that $S\Omega\subset \Omega$. To apply the
contraction mapping principle it remains to show that $S$
is a contraction mapping on $\Omega$.
 Thus, if $x_1, x_2\in \Omega$ and $t\geq t_1$,
\begin{align*}
|(Sx_1)(t)-(Sx_2)(t)|
&\leq \frac{1}{p_1} \|x_1-x_2\|\Big(1-p_2+\int_{t}^{\infty}\left(Q_1(s)
+Q_2(s)\right) ds\Big)\\
&\leq \lambda_4\|x_1-x_2\|,
\end{align*}
where $\lambda_4=(1-\frac{p_{1_0}N_3}{p_1N_4})$.
 This implies
\[
\|Sx_1-Sx_2\|\leq \lambda_4\|x_1-x_2\|,
\]
where the supremum norm is used. Since  $\lambda_4<1$, $S$ is a
contraction mapping on $\Omega$. Thus $S$ has a unique fixed
point which is a positive and bounded solution of \eqref{e:1}.
This completes the proof.
\end{proof}


\begin{theorem}\label{t:5}
Assume that   $-1<p_1\leq P_1(t)\leq 0$, $0\leq P_2(t)\leq p_2<1+p_1$ and
\eqref{e:2} hold, then \eqref{e:1} has a bounded non-oscillatory solution.
\end{theorem}

\begin{proof} Because of \eqref{e:2},  we can choose a $t_1>t_0$
sufficiently large satisfying  \eqref{e:3} such that
\begin{equation}\label{e:13}
\int_{t}^{\infty}Q_1(s) ds\leq \frac{(1+p_1)M_6-\alpha}{M_6},\quad
t\geq t_1,
\end{equation}
and
\begin{equation}\label{e:14}
\int_{t}^{\infty}Q_2(s) ds\leq \frac{\alpha-p_2M_6-M_5}{M_6}, \quad
t\geq t_1\,,
\end{equation}
 where $M_5$ and $M_6$ are positive constants such that
\[
M_5+p_2M_6<(1+p_1)M_6\quad\text{and}\quad
\alpha \in ( M_5+p_2M_6, (1+p_1)M_6)\,.
\]
Let $\Lambda$ be the set of all continuous and bounded functions
on $[t_0,\infty)$ with the supremum norm. Set
\[
\Omega=\{x\in \Lambda:M_5\leq x(t)\leq M_6,\; t\geq t_0 \}.
\]
 It is clear that  $\Omega$ is a bounded,
closed and convex subset of $\Lambda$. Define an operator
$S:\Omega\to\Lambda$ as follows:
\[
(Sx)(t)=\begin{cases}
\alpha-P_1(t)x(t-\tau_1)-P_2(t)x(t+\tau_2)\\
+\int_{t}^{\infty}\left[Q_1(s)x(s-\sigma_1)-Q_2(s)x(s+\sigma_2)\right]ds,
& t\geq  t_1,\\[4pt]
(Sx)(t_1), & t_0\leq t\leq t_1.
\end{cases}
\]
Obviously, $Sx$ is continuous. For $t\geq t_1$ and $x\in \Omega$,
from \eqref{e:13} and  \eqref{e:14}, respectively, it follows that
\begin{gather*}
(Sx)(t)\leq  \alpha-p_1M_6+M_6\int_{t}^{\infty}Q_1(s) ds\leq M_6, \\
(Sx)(t)\geq\alpha- p_2M_6-M_6\int_{t}^{\infty}Q_2(s) ds\geq M_5.
\end{gather*}
This proves that $S\Omega\subset \Omega$. To apply the
contraction mapping principle it remains  to show that $S$
is a contraction mapping on $\Omega$.
 Thus, if $x_1, x_2\in \Omega$, $t\geq t_1$,
\begin{align*}
|(Sx_1)(t)-(Sx_2)(t)|
&\leq \|x_1-x_2\|\Big(-p_1+p_2+\int_{t}^{\infty}\left(Q_1(s)
+Q_2(s)\right) ds\Big)\\
&\leq  \lambda_5\|x_1-x_2\|,
\end{align*}
where $\lambda_5=(1-\frac{M_5}{M_6})$.
 This implies
\[
\|Sx_1-Sx_2\|\leq \lambda_5\|x_1-x_2\|,
\]
where the supremum norm is used. Since  $\lambda_5<1$, $S$ is a
contraction mapping on $\Omega$. Thus $S$ has a unique fixed
point which is a positive and bounded solution of \eqref{e:1}.
This completes the proof.
\end{proof}


\begin{theorem}\label{t:6}
 Assume that   $-1<p_1\leq P_1(t)\leq 0$, $-1-p_1<p_2\leq P_2(t)\leq 0 $ and
 \eqref{e:2} hold, then \eqref{e:1} has
a bounded non-oscillatory solution.
\end{theorem}

\begin{proof}
Because of \eqref{e:2},  we can choose a $t_1>t_0$
sufficiently large satisfying  \eqref{e:3} such that
\begin{equation}\label{e:15}
\int_{t}^{\infty}Q_1(s) ds\leq
\frac{(1+p_1+p_2)N_6-\alpha}{N_6},\quad t\geq t_1,
\end{equation}
and
\begin{equation}\label{e:16}
\int_{t}^{\infty}Q_2(s) ds\leq \frac{\alpha-N_5}{N_6}, \quad t\geq
t_1,
\end{equation}
 where $N_5$ and $N_6$ are positive constants such that
\[
N_5<(1+p_1+p_2)N_6\quad\text{and}\quad
\alpha \in( N_5,(1+p_1+p_2)N_6) .
\]
Let $\Lambda$ be the set of  continuous and bounded functions
on $[t_0,\infty)$ with the supremum norm. Set
\[
\Omega=\{x\in \Lambda:N_5\leq x(t)\leq N_6,\; t\geq t_0 \}.
\]
 It is clear that  $\Omega$ is a bounded,
closed and convex subset of $\Lambda$. Define  an operator
$S:\Omega\to\Lambda$ as follows:
\[
(Sx)(t)=\begin{cases}
\alpha-P_1(t)x(t-\tau_1)-P_2(t)x(t+\tau_2)\\
+\int_{t}^{\infty}\left[Q_1(s)x(s-\sigma_1)-Q_2(s)x(s+\sigma_2)\right]ds,\
& t\geq  t_1,\\[4pt]
(Sx)(t_1),& t_0\leq t\leq t_1.
\end{cases}
\]
Obviously, $Sx$ is continuous. For $t\geq t_1$ and $x\in \Omega$,
from \eqref{e:15} and  \eqref{e:16}, respectively, it follows that
\begin{gather*}
(Sx)(t) \leq  \alpha-p_1N_6-p_2N_6+N_6\int_{t}^{\infty}Q_1(s)
ds\leq N_6, \\
(Sx)(t)\geq\alpha-N_6\int_{t}^{\infty}Q_2(s) ds\geq N_5.
\end{gather*}
This proves that $S\Omega\subset \Omega$. To apply the
contraction mapping principle it remains  to show that $S$
is a contraction mapping on $\Omega$.
 Thus, if $x_1, x_2\in \Omega$ and $t\geq t_1$,
\begin{align*}
|(Sx_1)(t)-(Sx_2)(t)|
&\leq \|x_1-x_2\|\Big(-p_1-p_2+\int_{t}^{\infty}\left(Q_1(s)
+Q_2(s)\right) ds\Big)\\
&\leq  \lambda_6\|x_1-x_2\|,
\end{align*}
where $\lambda_6=(1-\frac{N_5}{N_6})$.
 This implies
\[
\|Sx_1-Sx_2\|\leq \lambda_6\|x_1-x_2\|,
\]
where  the supremum norm is used. Since  $\lambda_6<1$, $S$ is a
contraction mapping on $\Omega$. Thus $S$ has a unique fixed
point which is a positive and bounded solution of \eqref{e:1}.
This completes the proof.
\end{proof}

\begin{theorem}\label{t:7}
 Assume that  $-\infty<p_{1_0}\leq P_1(t)\leq p_1 <-1$,
$0\leq P_2(t)\leq p_2<-p_1-1$  and  \eqref{e:2} hold, then
\eqref{e:1} has a bounded non-oscillatory solution.
\end{theorem}

\begin{proof}
In view of \eqref{e:2},  we can choose a $t_1>t_0$ sufficiently
large satisfying  \eqref{e:8} such that
\begin{equation}\label{e:17}
\int_{t}^{\infty}Q_1(s) ds\leq \frac{p_{1_0}M_7+\alpha}{M_8},\quad
t\geq t_1,
\end{equation}
and
\begin{equation}\label{e:18}
\int_{t}^{\infty}Q_2(s)
ds\leq \frac{(-p_1-1-p_2)M_8-\alpha}{M_8},\quad t\geq t_1,
\end{equation}
where $M_7$ and $M_8$ are positive constants such that
\[
-p_{1_0}M_7<(-p_1-1-p_2)M_8\quad\text{and}\quad 
\alpha \in ( -p_{1_0}M_7, (-p_1-1-p_2)M_8)\,.
\]
Let $\Lambda$ be the set of all continuous and bounded functions
on $[t_0,\infty)$ with the supremum norm. Set
\[
\Omega=\{x\in \Lambda:M_7\leq x(t)\leq M_8,\; t\geq t_0 \}.
\]
 It is clear that  $\Omega$ is a bounded,
closed and convex subset of $\Lambda$. Define a mapping
$S:\Omega\to \Lambda$ as follows:
\[
(Sx)(t)=\begin{cases}
\frac{-1}{P_1(t+\tau_1)}\{\alpha+x(t+\tau_1)+P_2(t+\tau_1)x(t+\tau_1+\tau_2)\\
-\int_{t+\tau_1}^{\infty}\left[Q_1(s)x(s-\sigma_1)-Q_2(s)x(s+\sigma_2)\right]ds\},
& t\geq t_1\\[4pt]
(Sx)(t_1), &  t_0\leq t\leq t_1.
\end{cases}
\]
Clearly, $Sx$ is continuous. For $t\geq t_1$ and $x\in \Omega$,
from \eqref{e:18} and  \eqref{e:17}, respectively, it follows that
\[
(Sx)(t)\leq \frac{-1}{p_1}\left(\alpha+M_8+p_2M_8+M_8\int_{t}^{\infty}Q_2(s)
ds\right)\leq M_8
\]
and
\[
(Sx)(t)
\geq \frac{-1}{p_{1_0}}\left(\alpha-M_8\int_{t}^{\infty}Q_1(s)ds\right)
\geq M_7.
\]
This implies that $S\Omega\subset \Omega$. To apply the
contraction mapping principle it remains to show that $S$
is a contraction mapping on $\Omega$.
 Thus, if $x_1, x_2\in \Omega$ and $t\geq t_1$,
\begin{align*}
|(Sx_1)(t)-(Sx_2)(t)|
&\leq \frac{-1}{p_1}
\|x_1-x_2\|\Big(1+p_2+\int_{t}^{\infty}\left(Q_1(s)
+Q_2(s)\right) ds\Big)\\
&\leq \lambda_7\|x_1-x_2\|,
\end{align*}
where $\lambda_7=(1-\frac{p_{1_0}M_7}{p_1M_8})$.
 This implies
\[
\|Sx_1-Sx_2\|\leq \lambda_7\|x_1-x_2\|,
\]
where the supremum norm is used. Since  $\lambda_7<1$, $S$ is a
contraction mapping on $\Omega$. Thus $S$ has a unique fixed
point which is a positive and bounded solution of \eqref{e:1}.
This completes the proof.
\end{proof}

\begin{theorem}\label{t:8}
 Assume that  $-\infty<p_{1_0}\leq P_1(t)\leq p_1 <-1$, $p_1+1<p_2\leq P_2(t)\leq 0$
and  \eqref{e:2} hold, then
\eqref{e:1} has a bounded non-oscillatory solution.
\end{theorem}

\begin{proof}
In view of \eqref{e:2},  we can choose a $t_1>t_0$ sufficiently
large satisfying  \eqref{e:8} such that
\begin{equation}\label{e:19}
\int_{t}^{\infty}Q_1(s) ds\leq
\frac{p_{1_0}N_7+p_2N_8+\alpha}{N_8},\quad t\geq t_1,
\end{equation}
and
\begin{equation}\label{e:20}
\int_{t}^{\infty}Q_2(s) ds\leq \frac{(-p_1-1)N_8-\alpha}{N_8},\quad
t\geq t_1,
\end{equation}
 where $N_7$ and $N_8$ are positive constants such that
\[
-p_{1_0}N_7-p_2N_8<(-p_1-1)N_8\quad\text{and}\quad
\alpha \in ( -p_{1_0}N_7-p_2N_8,(-p_1-1)N_8).
\]
Let $\Lambda$ be the set of  continuous and bounded functions
on $[t_0,\infty)$ with the supremum norm. Set
\[
\Omega=\{x\in \Lambda:N_7\leq x(t)\leq N_8,\; t\geq t_0 \}.
\]
 It is clear that  $\Omega$ is a bounded,
closed and convex subset of $\Lambda$. Define a mapping
$S:\Omega\to \Lambda$ as follows:
\[
(Sx)(t)=\begin{cases}
\frac{-1}{P_1(t+\tau_1)}\{\alpha+x(t+\tau_1)+P_2(t+\tau_1)x(t+\tau_1+\tau_2)\\
-\int_{t+\tau_1}^{\infty}\left[Q_1(s)x(s-\sigma_1)-Q_2(s)x(s+\sigma_2)\right]ds\},
& t\geq t_1,\\
(Sx)(t_1),&  t_0\leq t\leq t_1.
\end{cases}
\]
Clearly, $Sx$ is continuous. For $t\geq t_1$ and $x\in \Omega$,
from \eqref{e:20} and  \eqref{e:19}, respectively, it follows that
\[
(Sx)(t)\leq \frac{-1}{p_1}\Big(\alpha+N_8+N_8\int_{t}^{\infty}Q_2(s)
ds\Big)\leq N_8
\]
and
\[
(Sx)(t)
\geq \frac{-1}{p_{1_0}}\Big(\alpha+p_2N_8-N_8\int_{t}^{\infty}Q_1(s)ds\Big)
\geq N_7.
\]
These prove that $S\Omega\subset \Omega$. To apply the
contraction mapping principle it remains  to show that $S$
is a contraction mapping on $\Omega$.
 Thus, if $x_1, x_2\in \Omega$, $t\geq t_1$,
\begin{align*}
|(Sx_1)(t)-(Sx_2)(t)|
&\leq \frac{-1}{p_1} \|x_1-x_2\|\Big(1-p_2+\int_{t}^{\infty}\left(Q_1(s)
+Q_2(s)\right) ds\Big)\\
&\leq \lambda_8\|x_1-x_2\|,
\end{align*}
where $\lambda_8=(1-\frac{p_{1_0}N_7}{p_1N_8})$.
 This implies
\[
\|Sx_1-Sx_2\|\leq \lambda_8\|x_1-x_2\|,
\]
where the supremum norm is used. Since  $\lambda_8<1$, $S$ is a
contraction mapping on $\Omega$. Thus $S$ has a unique fixed
point which is a positive and bounded solution of \eqref{e:1}.
This completes the proof.
\end{proof}

\begin{example} \label{ex1} \rm
Consider the equation
\begin{equation}\label{e:21}
\begin{aligned}
&\Big[x(t)-\frac{1}{2}x(t-2\pi)
+\big[\frac{1}{2}-\exp(-\frac{t}{2})\big]x(t+5\pi)\Big]'\\
&+\frac{1}{2}\exp(-\frac{t}{2})
x(t-4\pi)-\exp(-\frac{t}{2})  x(t+\frac{5\pi}{2})=0, \quad
t>-2\ln(1/2)
\end{aligned}
\end{equation}
 and note that 
\[
P_1(t)=-\frac{1}{2},\quad P_2(t)=\frac{1}{2}-\exp(-\frac{t}{2}),\quad
Q_1(t)=\frac{1}{2}\exp(-\frac{t}{2}),\quad
Q_2(t)=\exp(-\frac{t}{2}).
\]
 A straightforward verification yields that the conditions of Theorem~\ref{t:5}
 are valid. We note that $x(t)=2+\sin t$ is a non-oscillatory solution
of \eqref{e:21}.
\end{example}

\begin{example} \label{ex2} \rm
Consider the equation
\begin{equation}\label{e:22}
\begin{aligned}
&\Big[x(t)-\frac{1}{\exp(1)}\big[\frac{3}{4}-\exp(-t)\big]x(t-1)
-\exp(1/4)\big[\frac{1}{4}+\exp(-t)\big]x(t+\frac{1}{4})\Big]'\\
&+ \exp(-t-1)x(t-1)-\exp(-t+\frac{1}{4})  x(t+\frac{1}{4})=0, \quad t\geq \frac{3}{2}
\end{aligned}
\end{equation}
and note that
\begin{gather*}
P_1(t)=-\frac{1}{\exp(1)}\big[\frac{3}{4}-\exp(-t)\big],\quad
P_2(t)=-\exp(\frac{1}{4})\big[\frac{1}{4}+\exp(-t)\big],\\
Q_1(t)=\exp(-t-1), \quad Q_2(t)=\exp(-t+\frac{1}{4}) .
\end{gather*}
 It is easy to verify that the conditions of Theorem~\ref{t:6}
 are valid. We note that $x(t)=1+\exp(-t)$ is a non-oscillatory
solution of \eqref{e:22}.
\end{example}


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\end{document}