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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 163, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or
 http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/163\hfil Non-oscillatory behaviour]
{Non-oscillatory behaviour of  higher order functional
differential equations \\ of neutral type}

\author[R. N. Rath, N. Misra, P. P. Mishra, L. N. Padhy \hfil EJDE-2007/163\hfilneg]
{Radhanath  Rath,  Niyati Misra, \\
 Prayag Prasad Mishra, Laxmi Narayan Padhy} 

\address{Radhanath Rath \newline
Department of Mathematics,
Khallikote Autonomous College,
Berhampur, 760001 Orissa, India}
\email{radhanathmath@yahoo.co.in}

\address{Niyati Misra \newline
Department of Mathematics,
Berhampur University,
Berhampur, 760007 Orissa, India}
\email{niyatimath@yahoo.co.in}


\address{Prayag Prasad Mishra \newline
Department of Mathematics,
Silicin Institute of Technology,
Bhubaneswar, Orissa, India}
\email{prayag@silicon.ac.in}


\address{Laxmi Narayan Padhy \newline
Department Of Computer Science and Engineering, K.I.S.T,
Bhubaneswar Orissa, India}
\email{ln\_padhy\_2006@yahoo.co.in}


\thanks{Submitted September 24, 2007. Published November 30, 2007.}
\subjclass[2000]{34C10, 34C15, 34K40}
\keywords{Oscillatory solution; nonoscillatory solution;
asymptotic behaviour}

\begin{abstract}
In this paper, we obtain sufficient conditions so that the
 neutral functional differential equation
\begin{equation*}
\big[r(t) [y(t)-p(t)y(\tau (t))]'\big]^{(n-1)} +
q(t) G(y(h(t))) = f(t)
\end{equation*}
has a bounded and positive solution.
Here $n\geq 2$; $q,\tau, h$ are continuous functions  
with $q(t) \geq 0$; $h(t)$ and $\tau(t)$
are  increasing functions which are less than $t$,
and approach infinity as $t \to \infty$.
In our work, $r(t) \equiv 1$ is admissible, and neither we assume 
that $G$ is non-decreasing, that $xG(x) > 0$ for $x \neq 0$, nor that
$G$ is Lipschitzian. Hence the results of this paper  generalize
many results in \cite{d1} and \cite{p1}--\cite{r3}.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}

\section{Introduction}

In this paper we find sufficient conditions for the neutral delay
differential equation (NDDE in short), of order $n\geq 2$,
\begin{equation}
\big[r(t) [y(t)-p(t)y(\tau (t))]'\big]^{(n-1)} +
q(t) G(y(h(t))) = f(t)
\label{eE}
\end{equation}
to have a bounded positive solution which does not tend to zero as
$t \to \infty$. Here $q, h ,\tau\in C ([0, \infty),R)$ such that
$q(t) \geq 0, h(t)$ and $\tau(t)$ are increasing functions which
are  less thatn or equal to $t$, and approach $\infty$ as
$t \to \infty$, $r \in C^{(n-1)} ([0, \infty), (0, \infty))$, 
$p \in C^{(n)} ([0, \infty), \mathbb{R})$,
$G \in C(\mathbb{R},\mathbb{R})$.

We need some of the following assumptions in the sequel.
\begin{itemize}
\item[(H1)] There exists a bounded function $F(t)$ such that
$F^{(n-1)} (t) = f(t)$.

\item[(H2)] $\int^\infty_{t_0}t^{n-2} q(t) dt < \infty$.

\item[(H3)] $\int^\infty_{t_0} {dt \over r(t)} = \infty$.

\item[(H4)]$\int^\infty_{t_0} {dt \over r(t)} < \infty$.

\item[(H5)] $\int^\infty_{t_0} ({1\over r(t)} \int^\infty_t
(s-t)^{n-2}q(s) ds)dt < \infty$.
\end{itemize}

\begin{remark} \label{rmk1} \rm
Since $r(t) > 0$, it follows that
\begin{itemize}
\item[(i)] either (H3) or (H4) holds exclusively.

\item[(ii)] If (H3) holds then (H5) implies (H2) but not
conversely.

\item[(iii)] If (H4) holds then (H2) implies (H5) but not conversely.
\end{itemize}
\end{remark}


 The study of
oscillation and non-oscillation properties of neutral delay
differential equations has attracted the attention of many authors all
over the world during the last two decades.In \cite{d1,p1,p2,r1,r2,r3}
the authors have proved
the existence  of a bounded positive solution of neutral delay
differential equations
\begin{gather}
(y(t) - p(t) y(t - \beta))' + q(t) G (y(t -
\delta))= f(t) ,\label{e1}\\
(y(t) - p(t) y(t - \beta))'' + q(t) G (y(t -
\delta))= f(t), \label{e2},\\
(y(t) - p(t) y(t - \beta))^{(n)} + q(t) G (y(t -
\delta))= f(t), \label{e3}
\end{gather}
where $\beta$ and $\delta$ are constants.
For that purpose the authors  assume the following hypothesis.
\begin{itemize}
\item[(H6)] There exists a function $F(t)$ such that $F(t)
\to 0$ as $t \to \infty$ and $F^{n} (t) = f(t)$.

\item[(H7)] $\big| \int^\infty_{t_0}t^{n-1} f(t) dt\big|< \infty$.

\item[(H8)] $G$ is Lipschitzian in every interval of the form
$[a,b]$, with $0 < a < b$.

\item[(H9)] $x G(x) > 0$ for $x \neq 0$, and $G$ is non-decreasing.

\end{itemize}
It is obvious  that (H6) $\Leftrightarrow$ (H7) and
(H1) is weaker than both (H6) and (H7).In \eqref{eE} if we put $r(t)=1$
 ,
$\tau(t)=t-\beta$, $h(t)=t-\delta$ then it reduces to \eqref{e3}.
We find almost no result with the NDDE \eqref{eE} in the literature.
For example if $\tau(t)=t/2$ and $h(t)=t/3$ then  the existing results
fail to answer any thing. Since we formulate our results with (H1) and
 do
not assume either (H8) or (H9), therefore our work extends,
improves and generalizes some of the results of
 \cite{d1,p1,p2,r1,r2,r3}.
While studying the existence of a positive
solution of neutral delay differential equation \eqref{e3}
for $n \geq 2$, the authors take $p(t)$ in different ranges.
But some how we find  no result when  $p(t)
\equiv -1$, in these papers. However, in this work
we consider $p(t)$ in different ranges including $p(t) = \pm 1$.
Our results hold good when  $n$ is  both   odd or even but $\geq 2$.

Let $T_y > 0$ and $T_0 = \min \{h(T_y), \tau(T_y)\}$. Suppose
$\phi \in C ([T_0 T_y], R)$. By a solution of $\eqref{eE}$, we
mean a real valued continuous function
 $y \in C^{(n)} ([T_0, \infty), R)$ such that
$y(t) = \phi(t)$ for $T_0 \leq t \leq T_y$ and
$y(t) - p(t) y(t - \tau)$ is differentiable,
$r(t) (y(t) - p(t) y(t-\tau))'$ is $(n-1)$ times further
differentiable and then for $t\geq T_y$ the neutral equation \eqref{eE}
 is satisfied.Such a solution is said
to be oscillatory if it has arbitrarily large zeros, otherwise it
is called non-oscillatory.

So far as existence and uniqueness of solutions of \eqref{eE} are
concerned one may refer \cite{g1}, but in this work we assume the
existence of solutions of \eqref{eE} and study its non-oscillatory
 behaviour.

\section{Main Results}

In this section we assume that there exists positive real numbers $p,c,$ and $d$ such that $p(t)$  satisfies one of the following
conditions.
\begin{itemize}
\item[(A1)] $0 \leq p(t) \leq p < 1$.

\item[(A2)] $-1 < -p \leq p(t) \leq 0$.

\item[(A3)] $-d < p(t) \leq -c < -1$.

\item[(A4)] $1 < c \leq p(t) < d$.
\end{itemize}

For our work we need the following Lemma from [3].

\begin{lemma}[Krasnoselskiis Fixed point Theorem \cite{e1}]
 \label{lem1}
Let $X$ be a Banach space. Let $\Omega$ be a bounded closed convex
subset of $X$ and let $S_1, S_2$ be maps of $\Omega$ into $X$ such
that $S_1 x+ S_2 y \in \Omega$ for every pair $x, y \in \Omega$.
If $S_1$ is a contraction and $S_2$ is completely continuous, then
the equation
\begin{equation*}
S_1 x + S_2x = x
\end{equation*}
has a solution in $\Omega$.
\end{lemma}

\begin{theorem}\label{thm2.2}
Let {\rm (A1), (H1), (H4), (H5)} hold. Then there exists a
bounded solution of $\eqref{eE}$ which is bounded below by a positive
constant i.e there exists a solution of \eqref{eE} which neither
oscillates nor tends to zero as $t\to\infty$.
\end{theorem}

\begin{proof}
Since $G \in C(\mathbb{R},\mathbb{R})$, then let
\begin{equation}\label{f2}
\mu = \max \{G(x) : {3\over 5}(1 -p) \leq x \leq 1\}.
\end{equation}
From (H1) , we find $\alpha >0$ and  $t_1 > t_0>0$ such that
\begin{equation}\label{f1}
|F(t)| < \alpha \quad \text{for }  t \geq t_1.
\end{equation}
Then using (H4) we find $t_2>t_1$ such that $t \geq t_2$ implies
\begin{equation}\label{f}
 \int^\infty_t {1 \over r(s)} ds  < {1 - p \over 10\,\alpha}.
\end{equation}
 From \eqref{f1} and \eqref{f} it follows that for $t>t_3>t_2$
\begin{equation}\label{f3}
 \int^\infty_t {|F(s)| \over r(s)} ds  < {1 - p \over 10}.
\end{equation}
 From (H5)  we find $t_4>t_3$ such that $t > t_4$ implies
\begin{equation}\label{f4}
\frac{\mu}{(n-2)!} \int^\infty_t {1 \over r(s)} \int^\infty_s
 (u-s)^{n-2}q(u)
du\, ds < {1-p \over 10}.
\end{equation}
Let $T >t_4$ and $T_0 = \min \{ \tau(T), h (T)\}$.
Then for $t \geq T$, \eqref{f3} and \eqref{f4} hold.
Let $X = C([T_0, \infty), R )$ be the set of all
continuous functions with norm
$\|x\| = \sup_{t \geq T_0} |x(t)| < \infty$.
Clearly $X$ is a Banach space. Let
\begin{equation}\label{f5}
S = \big\{u \in BC ([T_0, \infty),R ): {3 \over
5}(1-p) \leq u(t) \leq 1\big\},
\end{equation}
with the supremum norm $\|u \|= \sup \{ |u(t)|: t \geq T_0 \}$.
Clearly $S$ is a closed, bounded and convex subset of
$C([T_0, \infty), R)$. Define two maps $A$ and $B: S \to X$ as follows.
For $x \in S$,
\begin{equation} \label{f6}
Ax(t)=\begin{cases}Ax(T), & t \in [T_0, T];\\
p(t) x(\tau(t)) + {4(1 -p)\over 5} ,& t \geq T,
\end{cases}
\end{equation}
and
\begin{equation}\label{f7}
Bx (t)=\begin{cases}Bx(T),  & t \in [T_0, T];\\
\frac{(-1)^{n-1}}{(n-2)!}\int^\infty_t
{1\over r(s)}\int^\infty_s (u-s)^{n-2}q(u) G(x(h(u)))du\,ds \\
-\int^\infty_t {F(s)\over r(s)}ds, & t \geq T.
\end{cases}
\end{equation}
First we show that if $x, y \in S$ then $Ax + By \in S$.
In fact, for every $x, y \in S$ and $t \geq T$, we get
\begin{equation*}
\begin{split}
(Ax)(t) + (By) (t)
& \leq p(t) x(\tau(t))+ {4(1-p)\over 5} - \int^\infty_t {F(s)
\over r(s)}ds\\
& \quad +\frac{(-1)^{n-1}}{(n-2)!}\int^\infty_t
{1\over r(s)}\int^\infty_s (u-s)^{n-2}q(u) G(y(h(u)))du\,ds\\
& \leq  p + {4(1 - p)\over 5} + {1 - p \over 10}+{1 - p \over 10}
  \leq 1.
\end{split}
\end{equation*}
On the other hand for $t \geq T$,
\begin{align*}
(Ax)(t) + (By) (t) & \geq {4(1-p)\over 5} -
\int^\infty_t {F(s)\over r(s)}ds\\
& \quad +\frac{(-1)^{n-1}}{(n-2)!}
  \int^\infty_t {1
\over r(s)} \int^\infty_s (u-s)^{n-2}q(u) G(y(h(u)))du \,ds\\
& \geq {4(1-p)\over 5} -
\alpha\int^\infty_t {1\over r(s)}ds\\
&\quad -\frac{\mu}{(n-2)!} \int^\infty_t {1\over
r(s)}\int^\infty_s(u-s)^{n-2}q(u) du\,ds\\
& \geq {4(1-p)\over 5} - {1-p \over 10} - {1-p \over 10} =
{3\over 5}(1-p).
\end{align*}
Hence
\begin{equation*}
{3\over 5}(1-p) \leq (Ax)(t) + (By)(t) \leq 1
\end{equation*}
for $t \geq T$. So that $Ax + By\in S$ for all $x, y \in S$.

Next we show that $A$ is a contraction in $S$. In fact,  for $x, y
\in S$ and $t \geq T$, we have
\begin{align*}
|(Ax)(t) - (Ay)(t)| &\leq |p(t) \{x(\tau(t)) - y(\tau(t))\}|\\
& \leq|p(t)| |x(\tau(t)) - y(\tau(t))|\\
& \leq p\parallel x -y \parallel.
\end{align*}
Since $0 < p < 1$, we conclude that $A$ is a contraction mapping on
$S$.

We now show that $B$ is completely continuous. First, we shall
show that $B$ is continuous. Let $x_k = x_k(t) \in S$ for $k=1,2,\dots
 $.
 be such that
$\sup_{t\geq T}|x_k(t)-x(t)|\to 0$ as $k \to \infty$.Because $S$ is
 closed,
$x = x(t) \in S$. For $t \geq T$, we have
\begin{align*}
& |(B  x_k)(t) - (Bx)(t)| \\
& \leq \frac{1}{(n-2)!}\int^\infty_t {1\over r(s)}
\int^\infty_s (u-s)^{n-2}q(u)
|G(x(h(u)))- G(x_k(h(u)))|du \, ds.
\end{align*}
Since for all $t\geq T , x_k(t),k=1,2...$, tend uniformly to $x(t)$ as
$t\to\infty$ and $G$ is continuous, therefore
$|G(x(h(u))) -G (x_k(h(u)))|\to 0$ as $k \to \infty$.
We conclude that $\lim_{k \to \infty} |(B x_k)(t) - (Bx)(t)| = 0$.
This means that $B$ is continuous.

Next, we show that $BS$ is relatively compact. It suffices to show
that the family of functions $\{Bx : x \in S\}$ is uniformly
bounded and equicontinuous on $[T_0, \infty)$. The uniform
boundedness is obvious. For the equicontinuity, according to
Levitan's result we only  need to show that, for any given $\epsilon >
0, [T_0, \infty)$ can be decomposed into finite subintervals in
such a way that on each subinterval all functions of the family
have change of amplitude less than $\epsilon$. From (H5) and (H4),
 it follows that for any $\epsilon > 0$,we can find  $T^\ast \geq T$
large enough so that
\begin{equation*}
\frac{\mu}{(n-2)!} \int^\infty_{T^\ast} {1 \over r(s)}
\int^\infty_s (u-s)^{n-2}q(u) du\, ds < {\epsilon \over 4},
\end{equation*}
and
\begin{equation*}
\alpha \int^\infty_{T^\ast} {ds \over r(s)} < {\epsilon \over 4}.
\end{equation*}
Then for $x \in S$ and $t_2> t_1 \geq T^*$,
\begin{align*}
&|(Bx)(t_2) - (Bx)(t_1)|\\
&= \Big| \frac{(-1)^n}{(n-2)!}\int^\infty_{t_2} {1 \over r(s)}
\int^\infty_s(u-s)^{n-2} q(u) G(x(h(u)))du\,ds
  - \int^\infty_{t_2} {F(s)\over r(s)} ds\\
& \quad - \frac{(-1)^n}{(n-2)!}\int^\infty_{t_1} {1 \over r(s) }
\int^\infty_s (u-s)^{n-2}q(u)G(x(h(u)))du\,ds
+ \int^\infty_{t_1} {F(s)\over r(s)}ds   \Big|  \\
& \leq \frac{\mu}{(n-2)!} \int^\infty_{t_1} {1 \over r(s)}
\int^\infty_s(u-s)^{n-2} q(u)du\,ds
 + \alpha \int^\infty_{t_1} {ds\over r(s)}\\
 & \quad + \frac{\mu}{(n-2)!} \int^\infty_{t_2} {1\over r(s)}
 \int^\infty_s (u-s)^{n-2}q(u) du\,ds
+ \alpha \int^\infty_{t_2} {ds \over r(s)}\\
&  < 4{\epsilon \over 4}=\epsilon.
\end{align*}
For $x \in S$ and $T \leq t_1 < t_2 \leq T^\ast$,
\begin{align*}
&|(Bx)(t_2) - (Bx)(t_1)|\\
& \leq \frac{\mu}{(n-2)!} \int^{t_2}_{t_1} {1\over r(s)}
\int^\infty_s (u-s)^{n-2}q(u) \,du\,ds
+ \alpha \int^{t_2}_{t_1} {1\over r(s)} ds \\
&\leq \max_{T \leq s \leq T^\ast}
\Big[{\mu \over (n-2)!r(s)} \int^\infty_s (u-s)^{n-2}q(u)\,du
+ {\alpha \over r(s)}\Big](t_2 - t_1).
\end{align*}
Thus there exists a $\delta > 0$ such that
\begin{equation*}
|(Bx)(t_2) - (Bx)(t_1)| < \epsilon \quad \text{if }  0 < |t_2 - t_1| <
\delta.
\end{equation*}
For any $x \in S, T_0 \leq t_1 < t_2 \leq T$, it is easy to see
that
\begin{equation*}
| (Bx) (t_2) - (Bx)(t_1)| = 0 < \epsilon.
\end{equation*}
Therefore, $\{Bx : x \in S\}$ is uniformly bounded and
equicontinuous on $[T_0, \infty)$ and  hence $BS$ is relatively
compact. By Lemma \ref{lem1}, there is an $x_0 \in S$ such that $Ax_0 +
Bx_0 = x_0$. It is easy to see that $x_0(t)$ is  the required non
oscillatory solution of the equation $\eqref{eE}$, which is bounded
 below
by  the positive constant ${3(1-p)\over 4}$.
\end{proof}

\begin{corollary}\label{cor2.3}
Let {\rm (A1), (H1), (H2), (H4)} hold. Then there exists a
bounded solution of $\eqref{eE}$ which is bounded below by a positive
constant.
\end{corollary}

\begin{proof}
By remark \ref{rmk1}(iii) (H2) and (H4) imply (H5). Hence
the proof follows from the proof of the above theorem,.
\end{proof}

\begin{theorem}\label{thm2.4}
Let {\rm (A1), (H3), (H5)} hold. Suppose there exists
$\alpha > 0$ such that for large $t$
\begin{equation}\label{f8}
r(t) > {1 \over \alpha},
\end{equation}
and
\begin{equation}\label{f9}
\big|\int^\infty_0 F(t) dt\big| < \infty \quad \text{with}\quad
 F^{n-1} (t) = f(t).
\end{equation}
Then there exists a bounded solution of $\eqref{eE}$ which is bounded
below by a positive constant.
\end{theorem}

\begin{proof}
Using \eqref{f8}  and \eqref{f9} we can get \eqref{f3}. Rest of the
proof is similar to that of the Theorem \ref{thm2.2}.
\end{proof}

\begin{corollary}\label{cor2.5}
Let {\rm (A1), (H5)}, \eqref{f8}, \eqref{f9} hold. Then there exists a
bounded solution of $\eqref{eE}$ which is bounded below by a positive
constant.
\end{corollary}

\begin{proof}
By Remark \ref{rmk1}(i) we have either (H3) holds or (H4)
holds. If (H3) holds then we proceed as in the proof of Theorem
\ref{thm2.4}. On the other hand if (H4) holds then from
\eqref{f8}  and \eqref{f9} we  get \eqref{f3}  and then proceed
as in the proof of  Theorem \ref{thm2.2} to  get the desired  result.
\end{proof}

\begin{remark} \label{rmk3}\rm
If in (H5) we take $r(t)\equiv 1$ then it
reduces to
\begin{equation}\label{f10}
\int^\infty_{t_0} \int^\infty_t (u-t)^{n-2} q(u) du < \infty.
\end{equation}
The above condition is required for our next result which follows
from Corollary \ref{cor2.5} when $r(t) \equiv 1$.
\end{remark}

\begin{corollary}\label{cor2.6}
Inequality \eqref{f10} is a sufficient condition for the $n$th order
 NDDE
\begin{equation}\label{f11}
\big(y(t) - p(t) y(\tau(t))\big)^{n}+ q(t)
G(y(h(t))) = f(t)
\end{equation}
to have a solution bounded below by a positive constant under the
assumptions (A1), \eqref{f8} and \eqref{f9}.
\end{corollary}

\begin{remark} \label{rmk4} \rm
We claim that the condition
\begin{equation}\label{f12}
\int^\infty_{t_0} u^{n-1} q(u) ds < \infty
\end{equation}
implies \eqref{f10}.
It is clear that \eqref{f12} is equivalent to
$\int^\infty_s (u - s)^{n-1} q(u) du < \infty$. Let
$$
K(s) = \int^\infty_s (u -s)^{n-1} q(u) du.
$$
This implies
$\lim_{s \to \infty}K(s) = 0$.
 Differentiating, we get
$$
K' (s) = - (n-1)\int^\infty_s (u-s)^{n-2}q(u) du.
$$
 From this integrating between s=t and s=T ,we obtain
 $$
\int^T_t K' (s) ds = -(n-1)\int^T_t \int^\infty_s (u-s)^{n-2}q(u)
 \,du\,ds.
$$
Hence
 $$
-(n-1)\int^T_t \int^\infty_s (u-s)^{n-2}q(u) \,du\, ds = K(T) - K(t).
$$
In the limit as $T \to \infty$, we obtain
 $$
\int^\infty_t \int^\infty_s (u-s)^{n-2} q(u) \,du\,ds
={ K(t)\over (n-1)} = {1 \over (n-1)}\int^\infty_t (u-t)^{n-1}q(u) du <
 \infty.
$$
Hence the claim holds.
\end{remark}

\begin{remark} \label{rmk5} \rm
Corollary \ref{cor2.6} improves   \cite[Theorem 1]{d1}
and \cite[Theorem 4.3]{p1}  because the authors assumed  $G$
to satisfy (H9) and to be Lipschizian.

It may be noted in view of the Remark\ref{rmk4} that the
condition  \eqref{f10} used in Coprollary\ref{cor2.6} is weaker
than the condition
\eqref{f12} used in \cite{d1,p1}.
\end{remark}

\begin{theorem} \label{thm2.7}
Let {\rm (A2), (H1), (H4), (H5)} hold. Then there exists a
bounded solution of $\eqref{eE}$ which is bounded below by a positive
constant.
\end{theorem}

\begin{proof}
We proceed as in the proof of the Theorem \ref{thm2.2} with the
following changes
\begin{equation*}
\mu = \max \{|G(x)|: {1-p \over 10} \leq x \leq 1 \}.
\end{equation*}
By {\rm (H1), (H4), (H5)}, we find $T$ such that for $t\geq T$
\begin{equation*}
\frac{\mu}{(n-2)!} \int^\infty_t {1\over r(s)} \int^\infty_s (u-s)^{n-2}q(u) du\,ds
< {1 - p \over 10},
\end{equation*}
and
\begin{equation*}
\int_t^\infty {|F(t)| \over r(t)} dt < \alpha \int_t^\infty
{dt \over r(t)} < {1-p \over 10}.
\end{equation*}
Let $S = \{y \in X : {1-p \over 10} \leq y(t) \leq 1, t \geq T_0\}$.
\begin{equation*}
(Ay)(t) = \begin{cases}
{7p+3 \over 10} + p(t) y(t - \tau)-\int^\infty_t {F(s)\over r(s)} ds,
&\text{for }  t \geq T;\\
Ay(T), & \text{for } T_0 \leq t \leq T.
\end{cases}
\end{equation*}

\begin{equation*}
(By)(t) = \begin{cases}
By(T), \quad  \text{for }  T_0 \leq t \leq T;\\
\frac{(-1)^{n-1}}{(n-2)!}\int^\infty_t {1\over r(s)}
\int^\infty_s (u-s)^{n-2}q(u)G(y(h(u)))du\,ds
\quad   \text{for }  t \geq T.
\end{cases}
\end{equation*}

Then as  in  Theorem\ref{thm2.2} we prove (i) $Ax + By \in S$
(ii) $A$ is a contraction, and finally
(iii) $B$ is completely continuous. Then by Lemma \ref{lem1} there
is a fixed point $x_0$ in
$S$ such that $Ax_0 + Bx_0 = x_0$ which is the required solution
 bounded
below by a positive constant.
\end{proof}

\begin{remark}\label{rmkn}\rm
The above theorem substantially improves   \cite[Theorem 3.1]{r3}
where the authors obtained a positive bounded solution of \eqref{eE}
with assumptions (A2), (H2), (H4), (H6), (H8),  (H9).
It may be noted that (H6) implies (H1) and (H2) with (H4)implies (H5).
Further we did not require (H8) and (H9).
\end{remark}

\begin{theorem}\label{thm2.8}
Let {\rm (A2), (H3), (H5)}, \eqref{f8} and \eqref{f9} hold. Then there
exists a bounded solution of $\eqref{eE}$ which  is bounded below by a
positive constant.
\end{theorem}
The proof of the above Theorem is similar to that of Theorem
\ref{thm2.7}.

\begin{definition}\label{d1}
For any $t>t_0$, define
$$
\tau _{-1}(t) =\{s\text{ is a real number } : s  \geq t \text{ and }
 \tau(s)=t \}.
$$
\end{definition}

\begin{remark}\label{rk2} \rm
The function $\tau_{-1}$ defined above is the inverse function of $\tau
 (t)$.
Since $\tau (t)$ is increasing it is one-one.Clearly  $\tau _{-1}(\tau
 (t))=t$
for $t>\tau_{-1}(t_0)$.
\end{remark}

\begin{theorem}\label{thm2.9}
Let {\rm (A3), (H1), (H4), (H5)} hold. Then there exists  a
bounded  solution of $\eqref{eE}$ which is bounded below by a positive
constant.
\end{theorem}

\begin{proof}
 If necessary increment $d$ such that $d>1+\frac{2}{c}$.
 Choose  positive numbers $\epsilon < \frac{c-1}{2}$,
$h=(c-1)-\epsilon$ and  $H = d-1+\frac{2\epsilon}{c}$.
Then $H > h > 0$. Set
$$
\mu = \max \{|G(x)|: h \leq x \leq H\}.
$$
 From (H4) and (H5) one can find $T > 0$ such that $t \geq T$
implies
\begin{equation*}
\int^\infty_{\tau_{-1}(t)} {F(s) \over r(s)} < {\epsilon  \over 2},
\end{equation*}
and
\begin{equation*}
 \frac{\mu}{(n-2)!} \int^\infty_{\tau_{-1}(t)} {1 \over r(s)}
\int^\infty_s (u-s)^{n-2}q(u)du\,ds < {\epsilon \over 2}.
\end{equation*}
Define
\begin{equation*}
S = \{ y(t) \in X : h \leq y(t) \leq H, t \geq T_0\}.
\end{equation*}
Define
\begin{equation*}
Ax(t) = \begin{cases} Ax (T), & \text{if } t \in [T_0, T];\\
{x(\tau_{-1}(t))\over p(\tau_{-1}(t))}- {cd - 1\over p(\tau_{-1}(t))}
+ {1\over p(\tau_{-1}(t))} \int^\infty_{\tau_{-1}(t)} {F(s)\over r(s)}
 ds ,
& \text{if } t \geq T.
\end{cases}
\end{equation*}

\begin{equation*}
Bx(t) = \begin{cases} Bx(T) ,& \text{if }  t \in [T_0, T];\\
\frac{(-1)^n}{(n-2)!\, p(\tau_{-1}(t))}
\int^\infty_{\tau_{-1}(t)} {1\over r(s)}
\int^\infty_s (u-s)^{n-2}q(u) G (y(h(u)))du\,ds,
&\text{if } t \geq T.
\end{cases}
\end{equation*}
We show that if $x,y \in S$, then $Ax+By \in S$. For $t\geq T$ we
 obtain
\begin{equation*} \begin{split}
Ax+By=
&\frac{-1}{p(\tau_{-1}(t))}\Big[-x(\tau_{-1}(t))-\int^\infty_{\tau_{-1}(t)}
 {F(s)\over r(s)} ds+(cd-1)\\
& +\frac{(-1)^{n-1}}{(n-2)!}\int^\infty_{\tau_{-1}(t)} {1\over r(s)}
(\int^\infty_s (u-s)^{n-2} q(u)G (y(h(u)))du)ds\Big]\\
&\leq
 \frac{1}{c}\big[-h+\frac{\epsilon}{2}+\frac{\epsilon}{2}+(cd-1)\big] \\
& =\frac{1}{c}\big[2\epsilon+ c(d-1)\big]=(d-1)+\frac{2\epsilon}{c}\\
& \leq H.
\end{split}
\end{equation*}
Further,
\begin{equation*}
\begin{split}
Ax+By=&\frac{-1}{p(\tau_{-1}(t))}\Big[-x(\tau_{-1}(t))-\int^\infty_{\tau_{-1}(t)}
 {F(s)\over r(s)} ds+(cd-1)\\
& +\frac{(-1)^{n-1}}{(n-2)!}\int^\infty_{\tau_{-1}(t)} {1\over r(s)}
\int^\infty_s (u-s)^{n-2} q(u)G (y(h(u)))du\,ds\Big]\\
& \geq \frac{1}{d} \big[-H -\frac{\epsilon}{2} +
 (cd-1)-\frac{\epsilon}{2}\big]\\
& =\frac{1}{d}\big[d(c-1)-\epsilon\frac{(c+2)}{c}\big]\\ &
 >c-1-\epsilon = h.
\end{split}
\end{equation*}
Thus $Ax+By \in S$.
Next we show that $A$ is a contraction in $S$.In fact for $x,y \in S$
and $t \geq T$ we have
$$
\| Ax-Ay\|\leq |\frac{1}{p(\tau_{-1}(t))}|
 |x(\tau_{-1}(t))-y(\tau_{-1}(t))|
\leq \frac{1}{c}\|x-y\|.
$$
Hence $A$ is a contraction because, $0<\frac{1}{c}<1$.Next we prove $B$
is completely continuous as in the proof of Theoren\ref{thm2.2}.
Then by Lemma \ref{lem1} there is a fixed point $x_0$ in
$S$ such that
 $$
Ax_0 + Bx_0 = x_0.
$$
Writing $x_0=y(t)$ and multiplying both sides of the above equation
by $p(\tau _{-1}(t))$ we obtain,
  \begin{equation*}
  \begin{split}
   p(\tau_{-1}(t))y(t)
&=y(\tau_{-1}(t))+\int^\infty_{\tau_{-1}(t)} {F(s)\over r(s)} ds
 -(cd-1)\\
  &\quad +\frac{(-1)^n}{(n-2)!} \int^\infty_{\tau_{-1}(t)} {1\over
 r(s)}
\int^\infty_s (u-s)^{n-2} q(u)G (y(h(u)))du\,ds.
\end{split}
\end{equation*}
  Then we replace $t$ by $\tau(t)$ ,use the fact that
  $\tau_{-1}(\tau(t))=t$ and finally  with some rearrangement of  terms, obtain
 \begin{equation*}
  \begin{split}
  y(t)-p(t)y(\tau(t))=&-\int_t^\infty\frac{F(s)}{r(s)}+(cd-1)\\
  &  +\frac{(-1)^{n-1}}{(n-2)!}\int_t^\infty \frac{1}{r(s)}
\int_s^{\infty}(u-s)^{n-2}q(u)G(y(h(u)))du \,ds .
  \end{split}
  \end{equation*}
 First differentiating the above equation once and then multiplying
 bothsides by $r(t)$ and finally differentiating $n-1$ times ,we see
 that,
$x_0$ is  the required  solution of \eqref{eE}, which is bounded
below by a positive constant.

\end{proof}
\begin{theorem}\label{thm2.10}
Let {\rm (A3), (H3), (H5)}, \eqref{f8}, \eqref{f9} hold. Then there
exists a bounded solution of $\eqref{eE}$ which is bounded below by a
positive constant.
\end{theorem}

The proof of the above theorem is similar to that of the above theorem.

\begin{corollary}\label{cor2.11}
Let {\rm (A3), (H5)}, \eqref{f8}, \eqref{f9} hold. Then there exists a
bounded solution of $\eqref{eE}$ which is bounded below by a positive
constant.
\end{corollary}

\begin{proof}
In view of Remark \ref{rmk1} (i) the proof follows lines similar to
those in Theorem \ref{thm2.9} and \ref{thm2.10}.

The results for the range (A4) are similar to those under
condition (A3). Hence we skip all proofs except the following
one.
\end{proof}

\begin{theorem}\label{thm2.12}
Let {\rm (A4), (H1), (H4), (H5)} hold. Then there exists a bounded
solution of $\eqref{eE}$ which is bounded below by a positive constant.
\end{theorem}

\begin{proof}
We proceed as in the proof of the Theorem\ref{thm2.9} with the
following changes. Choose
\begin{equation*}
\mu = \max \{|G(x)|: {c-1 \over d} \leq x \leq 2 \}.
\end{equation*}
\begin{equation*}
S = \{y \in X : {c-1 \over d} \leq y \leq 2\}.
\end{equation*}
From (H1), (H4) and (H5) we can find $T > 0$ such that $t \geq T$
implies
\begin{equation*}
\int^\infty_{\tau_{-1}(t)} {|F(s)| \over r(s)} < {c-1  \over 2},
\end{equation*}
and
\begin{equation*}
 \frac{\mu}{(n-2)!} \int^\infty_{\tau_{-1}(t)} {1 \over r(s)}
 \int^\infty_s
(u-s)^{n-2}q(u)du\,ds < {c-1 \over 2}.
\end{equation*}
Define
\begin{equation*}
Ax(t) = \begin{cases} Ax (T), & \text{if }  t \in [T_0, T];\\
{x(\tau_{-1}(t))\over p(\tau_{-1}(t))}-
{2c - 2\over p(\tau_{-1}(t))}
+ {1\over p(\tau_{-1}(t))} \int^\infty_{\tau_{-1}(t)} {F(s)\over r(s)}
 ds ,
& \text{if }  t \geq T.
\end{cases}
\end{equation*}
\begin{equation*}
Bx(t) = \begin{cases}
Bx(T) ,& \text{if }   t \in [T_0, T];\\
\frac{(-1)^n}{(n-2)! \, p(\tau_{-1}(t))}
\int^\infty_{\tau_{-1}(t)} {1\over r(s)}\\
\times \int^\infty_s (u-s)^{n-2}q(u) G (y(h(u)))du\,ds,
& \text{if }  t \geq T.
\end{cases}
\end{equation*}
For the rest of the proof we may refer the proofs  of the
Theorems \ref{thm2.2} and \ref{thm2.9}.
\end{proof}

\section{Positive solution for $p(t) = \pm 1$ }

In this section we find sufficient condition for the NDDE
\begin{equation}\label{g1}
(r(t) (y(t) + y(\tau(t)))' )^{n-1} + q(t)
G(y(h(t))) = f(t),
\end{equation}
or
\begin{equation}\label{g2}
(r(t) (y(t) - y(\tau(t)))' )^{n-1} + q(t)
G(y(h(t))) = f(t),\end{equation}
to have a bounded positive solution.

The results with NDDE \eqref{g1} are rare in the literature. We dont
find such a result in \cite{d1} or \cite{p1,p2,r1,r2,r3}. To achieve
our result we need the following Lemma.

\begin{lemma}[Schauder's Fixed Point Theorem \cite{e1}] \label{lem3.1}
 Let  $\Omega$ be a closed
convex and nonempty subset of a Banach space $X$. Let $B : \Omega
\to \Omega$ be a continuous mapping such that $B \Omega$
is a relatively compact subset of $X$. Then $B$ has at least one
fixed point in $\Omega$. That is there exists an $x \in \Omega$
such that $Bx = x$.
\end{lemma}

%\begin{definition}\label{d2}\rm
 For $t \geq t_0$, define $\tau_{-1}^{0}(t)=t$,
$\tau_{-1}^{1}(t)=\tau_{-1}(t),$\,
 $\tau_{-1}^{2}(t)=\tau_{-1}(\tau_{-1}(t))$.
  For any positive integer $i >2$, we define
$$
\tau_{-1}^{i}(t)=\tau_{-1}(\tau_{-1}^{i-1}(t)).
$$
%\end{definition}


\begin{theorem}\label{thm3.2}
Suppose {\rm (H1), (H4), (H5)} hold. Then there exists a
solution of \eqref{g1} which is bounded below by a positive
constant, that is, it neither oscillates nor tends to zero as $t$
tends to $\infty$.
\end{theorem}

\begin{proof}
We proceed as in the proof of Theorem  \ref{thm2.2} with the
following changes. Let
\begin{equation*}
\mu = \max \{|G(x)|: 1 \leq x \leq 5\}.
\end{equation*}
From (H1), (H4) and (H5) there exists $T > 0$ such that for
$t \geq T$ implies
\begin{equation}\label{new0}
\frac{\mu}{(n-2)!}\big|\int^\infty_t {1 \over r(s)}
 \int^\infty_s (u-s)^{n-2}q(u) du\,ds\big|< 1,
\end{equation}
and
\begin{equation}\label{new1}
\big|\int^\infty_t {F(s) \over r(s)} ds\big| < 1.
\end{equation}
For any continuous function  $g(t)$, it is clear that
\begin{equation}\label{new2}
 \sum^\infty_{l = 1} \int^{\tau_{-1}^{2l}(t)}_{\tau_{-1}^{2l-1}(t)}
 g(s)ds
<\int^\infty_t g(s)ds.
 \end{equation}
 Hence using the above inequality in \eqref{new0} and \eqref{new1},
we conclude that for $t\geq T$
 \begin{equation} \label{new3}
  \frac{\mu}{(n-2)!}\sum^\infty_{l = 1}
\int^{\tau_{-1}^{2l}(t)}_{\tau_{-1}^{2l-1}(t)} {1
\over r(s)} \int^\infty_s (u-s)^{n-2}q(u)du \,ds <1,
\end{equation}
and
\begin{equation}\label{new4}
\sum^\infty_{l = 1} \int^{\tau_{-1}^{2l}(t)}_{\tau_{-1}^{2l-1}(t)}
 {|F(s)|
\over r(s)} ds <1.
\end{equation}
Set
$S = \{y \in X : 1 \leq y(t) \leq 5,\; t \geq T_0 \}$.
Next we define the mapping $B : S \to X$ as
\begin{equation*}
By(t) = \begin{cases}
By(T), &   T_0 \leq t \leq T;\\
3-\sum^\infty_{l = 1} \int^{\tau_{-1}^{2l}(t)}_{\tau_{-1}^{2l-1}(t)}
 {F(s)
\over r(s)} ds +\frac{(-1)^{n-1}}{(n-2)!}
 \sum^\infty_{l = 1} \int^{\tau_{-1}^{2l}(t)}_{\tau_{-1}^{2l-1}(t)}\big({1\over r(s)}\\
\times  \int^\infty_s (u-s)^{n-2}q(u)
 G(y(h(u))\big)du \big)ds,  & t \geq T.
\end{cases}
\end{equation*}
Then using \eqref{new3} and \eqref{new4} we find that $By<5$ and
 $By>1$.
Hence $By \in S$ for $y \in S$.Next we show
$BS$ is relatively compact as in the proof of Theorem 2.2.
Then by Lemma 3.1 there is a fixed
point $y_0$ in $S$ such that $By_0 = y_0$.Hence for $t\geq T$, we
 obtain
\begin{align*}
 y_0(t)=& 3+\frac{(-1)^{n-1}}{(n-2)!}\sum^\infty_{l = 1}
\int^{\tau_{-1}^{2l}(t)}_{\tau_{-1}^{2l-1}(t)} {1
\over r(s)} \int^\infty_s (u-s)^{n-2}q(u) G(y_0(h(u)))du \,ds\\
&-\sum^\infty_{l = 1} \int^{\tau_{-1}^{2l}(t)}_{\tau_{-1}^{2l-1}(t)}
{F(s) \over r(s)} ds.
\end{align*}
In the above  we replace $t$ by $\tau(t)$ and note that
$\tau_{-1}^{m}(\tau(t))=\tau_{-1}^{m-1}(t)$ and
$\tau_{-1}^0(t)=t$. Then It follows for $t\geq T$ that
\begin{align*}
 y_0(\tau(t))=& 3+\frac{(-1)^{n-1}}{(n-2)!}\sum^\infty_{l = 1}
 \int^{\tau_{-1}^{2l-1}(t)}_{\tau_{-1}^{2l-2}(t)} {1
\over r(s)} \int^\infty_s (u-s)^{n-2}q(u) G(y_0(h(u)))du \,ds\\
& -\sum^\infty_{l = 1} \int^{\tau_{-1}^{2l-1}(t)}_{\tau_{-1}^{2l-2}(t)}
 {F(s)
\over r(s)}\, ds.
\end{align*}
Then for $t\geq T$ we have
\begin{align*}
y_0(t)+y_0(\tau(t))=& 6+\frac{(-1)^{n-1}}{(n-2)!}\int_t^\infty {1
\over r(s)} \int^\infty_s (u-s)^{n-2}q(u) G(y_0(h(u)))du \,ds\\
&-\int_t^\infty {F(s)
\over r(s)} ds.
\end{align*}
Differentiating the above equation first and then multiplying
 by $r(t)$ to both sides and after that  differentiating again for $n-1$ times,
we see that $y_0$ is the required  solution of \eqref{g1} which is
bounded below by a positive constant.Hence this solution neither
oscillates nor tends to zero as $t\to\infty$.
 Hence the theorem is proved.
\end{proof}

\begin{corollary}\label{cor3.3}
If {\rm (H1), (H2), (H4)} hold, then there exists a positive solution
of \eqref{g1} which is bounded below by a positive constant.
\end{corollary}

\begin{proof}
The proof follows from Remark \ref{rmk1} and the above Theorem.
\end{proof}

\begin{theorem}\label{thm3.4}
Let {\rm (H3), (H5)}, \eqref{f8} and \eqref{f9} hold. Then there
exists a positive solution of \eqref{g1} which is bounded below by a
positive constant that is, it neither oscillates nor tends to zero as
$t$ tends to $\infty$.
\end{theorem}

The proof of the above theorem is similar to that of
Theorem \ref{thm3.2}.

\begin{theorem}\label{thm3.5}
Suppose {\rm (H1)} hold. Assume for $t \geq t_0$
\begin{equation}\label{g3}
\sum^\infty_{i = 1} \int^\infty_{\tau_{-1}^{i}(t)} {1 \over r(s)}
\int^\infty_s (u-s)^{n-2}q(u) du\,ds < \infty,
\end{equation}
 and
\begin{equation}\label{g4}
\sum^\infty_ {i = 1} \int^\infty_{\tau_{-1}^{i}(t)}  {1\over r(s)} ds <
 \infty.
\end{equation}
Then \eqref{g2} has a solution bounded below by a positive
constant.
\end{theorem}

\begin{proof}
We proceed as in the proof of Theorem \ref{thm3.2} with the following
changes. Let
\begin{equation*}
\mu = \max \{|G(x)|: 1 \leq x \leq 5\}.
\end{equation*}
Then from (H1), \eqref{g3} and \eqref{g4}, there exists $T> 0$
such that for $t \geq T$
\begin{equation*}
\frac{\mu}{(n-2)!} \sum^\infty_{i = 1} \int^\infty_{\tau_{-1}^{i}(t)}\frac{1}{r(s)}
\int^\infty_s (u-s)^{n-2}q(u) du\, ds < 1,
\end{equation*}
and
\begin{equation*}
\sum^\infty_{i=1} \int^\infty_{\tau_{-1}^{i}(t)} {F(s) \over r(s)} ds
 <1.
\end{equation*}
Let
$ S = \{y \in X : 1 \leq y \leq 5 ,\, t \geq T_0\}$.
Then define
\begin{equation*}
By(t) = \begin{cases}
By (T) ,& \text{for }   t \in [T_0, T];\\
3 + \frac{(-1)^n}{(n-2)!}\sum^{\infty}_{i = 1}
\int^\infty_{\tau_{-1}^{i}(t)} {1 \over
r(s)} \int^\infty_s (u-s)^{n-2}\\
\times q(u) G(y(h(u)))du\,ds
+ \sum^\infty_{i =1} \int^\infty_{\tau_{-1}^{i}(t)} {F(s)\over r(s)} ds
 ,&
\text{for }  t \geq T.
\end{cases}
\end{equation*}
Then as in Theorem \ref{thm2.2} we prove (i)
$By \in S$ for $y \in S$, and (ii) $BS$ is relatively compact.
Then by lemma \ref{lem3.1} there
exists a fixed point $y_0 \in S$ such that $By_0 = y_0$, Putting
$y_0 = y(t)$, we get
\begin{align*}
y(t) = 3 &- \frac{(-1)^{n-1}}{(n-2)!} \sum^\infty_{i =1}
\int^\infty_{\tau_{-1}^{i}(t)} {1
\over r(s)} \int^\infty_s (u-s)^{n-2}q(u) G(y(h(u)))du\,ds\\
&+ \sum^\infty_{i = 1} \int^\infty_{\tau_{-1}^{i}(t)} {F(s)\over
 r(s)}ds.
\end{align*}
Then replacing $t$ by $\tau(t)$ in the above and using
$\tau_{-1}^i(\tau(t))=\tau_{-1}^{i-1}(t)$, we may obtain
$y(\tau(t))$. Consequently for $t \geq T$, we find
\begin{align*}
y(t) - y(\tau(t)) =
&\frac{(-1)^{n-1}}{(n-2)!} \int^\infty_t {1 \over r(s)} \int^\infty_s
(u-s)^{n-2}q(u) G(y(h(u)))du \,ds\\
&-\int^\infty_t {F(s)\over r(s)}ds.
\end{align*}
We may differentiate the above and then multiply by $r(t)$ and  then
again differentiate $n-1$ times to arrive at \eqref{g2}. This solution
 is
bounded below by a positive constant.
\end{proof}

\begin{remark} \label{rmkn1} \rm
It is not difficult to verify that the above theorem still holds,
if we replace \eqref{g4} and (H1) by the following  assumption
\begin{equation}\label{g5}
\sum^\infty_{i = 1} \int^\infty_{\tau_{-1}^{i}(t)} {1 \over r(s)}
\int^\infty_s (u-s)^{n-2}f(u) du\,ds < \infty.
\end{equation}
Of course, in that case we have  to modify the definition of
the mapping $B$ as follows.
\begin{equation*}
By(t) = \begin{cases}
By (T), & \text{for }  t \in [T_0, T];\\
3 -\frac{(-1)^{n-1}}{(n-2)!}\sum^{\infty}_{i = 1}
\int^\infty_{\tau_{-1}^{i}(t)} {1 \over
r(s)} \int^\infty_s (u-s)^{n-2}\\
\times q(u) G(y(h(u)))du\,ds\\
+ \frac{(-1)^{n-1}}{(n-2)!}
\sum^{\infty}_{i = 1} \int^\infty_{\tau_{-1}^{i}(t)} {1 \over
r(s)} \int^\infty_s (u-s)^{n-2}f(u) du\,ds ,
&\text{for }  t \geq T.
\end{cases}
\end{equation*}
If we put $r(t)=1$ in \eqref{g3} and \eqref{g5} then we obtain
\begin{equation}\label{g6}
\sum^\infty_{i = 1} \int^\infty_{\tau_{-1}^{i}(t)}
\int^\infty_s (u-s)^{n-2}q(u) du\,ds < \infty,
\end{equation}
and
\begin{equation}\label{g7}
\sum^\infty_{i = 1} \int^\infty_{\tau_{-1}^{i}(t)}
\int^\infty_s (u-s)^{n-2}f(u) du\,ds < \infty.
\end{equation}
\end{remark}
Then from the above theorem the following result follows directly.

\begin{corollary}\label{cor3.6}
If \eqref{g6} and \eqref{g7} hold for $t>t_0$,  then the NDDE
\begin{equation}\label{g8}
(y(t) -  y(t - \tau))^{(n)} + q(t) G (y(t -
\sigma))= f(t)
\end{equation}
has a solution, bounded below by a positive constant.
\end{corollary}

The above corollary improves and generalizes  \cite[Theorem 3.1]{p2}
and \cite[Theorem 2.5]{r2}, because in these papers, the authors assume
the following additional conditions that we don't require.
\begin{itemize}
\item[(i)] $n$ is  odd.
\item[(ii)] $G$ is non-decreasing and $xG(x)>0$ for $x \neq 0$.
\end{itemize}

Before we close this article we present an interesting example
which illustrates most of the results of this paper.

\begin{example} \label{ex1}\rm
Consider NDDE
\begin{equation}\label{g10}
(r(t) (y(t) - py(t/2))')^{n-1} + \frac{1}{t^{n+2}}
G(y(t/3)) = 0 \quad \text{for } t \geq t_0.
\end{equation}
In this example suppose that  $p$ is any constant and $r(t)\equiv 1$
or $r(t)\equiv \frac{1}{t^2}$.
If we compare this equation \eqref{g10} with NDDE \eqref{eE} then
$\tau(t)=\frac{t}{2}$, $h(t)=\frac{t}{3}$ and $q(t)=\frac{1}{t^{n+2}}$.
It is not difficult to verify that  $q(t)$ satisfies
{(H2), (H5)} and \eqref{g3}. Suppose that $G(u) = 1- u^3$ and it is
 decreasing.
Clearly the NDDE \eqref{g10} has a positive solution $y(t) \equiv 1$.
Hence this example illustrates all the results of this paper.
However since $G$ is decreasing and $\tau(t)$ is not of the form $t-k$,
the existing results of
\cite{d1,p1,p2,r1,r2,r3} are not applicable to  \eqref{g10}.
\end{example}

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\end{document}