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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 151, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/151\hfil Existence of solutions?]
{Existence of solutions to n-th order neutral dynamic equations
on time scales}

\author[Q. Li, Z. Zhang\hfil EJDE-2010/151\hfilneg]
{Qiaoluan Li, Zhenguo Zhang}  % in alphabetical order

\address{Qiaoluan Li \newline
College of Mathematics and Information Science,
Hebei Normal University, \newline
Shijiazhuang, 050016, China}
\email{qll71125@163.com}

\address{Zhenguo Zhang \newline
College of Mathematics and Information Science,
Hebei Normal University, \newline
Shijiazhuang, 050016, China. \newline
Information College,  Zhejiang Ocean University,
Zhoushan, 316000, China}
\email{zhangzhg@mail.hebtu.edu.cn}

\thanks{Submitted June 6, 2010. Published October 21, 2010.}
\subjclass[2000]{34K40, 34N99, 39A10}
\keywords{Time scales; dynamic equations; non-oscillatory solution.}

\begin{abstract}
 In this article,  we study n-th order neutral nonlinear dynamic
 equation on time scales. We obtain sufficient conditions for
 the existence of non-oscillatory solutions by using fixed point
 theory.
\end{abstract}

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

\section{Introduction}

This article concerns the n-th order neutral dynamic equation
\begin{equation}
(x(t)+p(t)x(\tau(t)))^{\Delta^n}+f_1(t,x(\tau_1(t)))
-f_2(t,x(\tau_2(t)))=0,\label{e1}
\end{equation}
for $t\geq t_0$, where $t\in \mathbb{T}$, $n\in \mathbb{N}$.
We assume $p\in C_{rd}(\mathbb{T}$,
$\mathbb{R}),\tau,\tau_i\in C_{rd}(\mathbb{T},\mathbb{T})$,
$\tau$ is strictly increasing, $\tau(t)<t$,
$\tau(t)\to\infty$, $\tau_i(t)\to \infty$,
$f_i\in C_{rd}(\mathbb{T}\times \mathbb{R}, \mathbb{R})$,
$f_1(t,u)f_2(t,u)>0$, and $f_i$ is non-decreasing in $u$.
In the sequel, without loss of generality, we assume that
$f_i(t, u)>0$, $i=1,2$.

In 1988, Stephan Hilger \cite{h1} introduced the theory of time scales
as a means of unifying discrete and continuous calculus. Several
authors have expounded on various aspects of this new theory, see
\cite{b1,e3,z2} and references therein. Recently, much attention is
concerned with questions of existence of non-oscillatory solutions
for dynamic equations on time scales. For significant works along
this line, see \cite{e2,h2,l1,s1}.
 Many results have been obtained for
first and second order dynamic equations, however, few results are
available for higher order dynamic equations. Motivated by these
works, we investigate the existence of non-oscillatory solutions of
\eqref{e1}.

 In Section 2, we present some preliminary material that we will
need to show the existence of solutions of \eqref{e1}. We present our
main results in Section 3.

\section{Preliminaries}

 We assume the reader is familiar with the notation and basic
results for dynamic equations on time scales. For a review of this
topic we direct the reader to the monographs \cite{b2,b3}.

We recall $x$ is a solution of \eqref{e1} provided that
$x(t)+p(t)x(\tau(t))$ is n times differentiable, and $x$ satisfies
\eqref{e1}. A solution $x$ of \eqref{e1} is called  non-oscillatory
 if $x$ is of one sign when $t\geq T$.

We define a sequence of functions $g_k(s,t)$, $k=1,2,\dots$ as
 follows.
\begin{equation}
\begin{gathered}
 g_0(s,t)\equiv 1,\quad s,t\in \mathbb{T}^{\kappa},\\
 g_{k+1}(s,t)=\int_{t}^{s}g_k(\sigma(u),t)\Delta u,\quad
 s,t\in \mathbb{T}^{\kappa}.
\end{gathered}\label{e2}
\end{equation}
For $g_{k}(s,t)$, we have the following Lemma.


\begin{lemma}[\cite{z1}] \label{lem1}
 Assume $s$ is fixed, and let
$g_k^{\Delta}(s, t)$ be the derivative of $g_{k}(s,t)$ with
respect to $t$. Then
\begin{equation}
g_k^{\Delta}(s, t)=-g_{k-1}(s, t), \quad k\in \mathbb{N},\; t\in
\mathbb{T}^{\kappa}.\label{e3}
\end{equation}
\end{lemma}

\begin{lemma}[\cite{e1}] \label{lem2}
Let $X$ be a Banach space, $\Omega$
be a bounded closed convex subset of $X$ and let $A, B$ be maps from
 $\Omega$ to $X$ such that $Ax+By\in \Omega$ for every pair
$x, y\in \Omega$. If  $A$ is a contraction and $B$ is completely
continuous, then the equation $Ax+Bx=x$ has a solution in
 $\Omega$.
\end{lemma}

\begin{lemma}[\cite{e1}] \label{lem3}
Let $X$ be a locally convex linear space, $S$ be a compact
convex subset of $X$, and  $T: S\to S$ be a continuous mapping
with $T(S)$ compact. Then $T$ has a fixed point in $S$.
\end{lemma}

\section{Main Results}

\begin{theorem} \label{thm1}
 Assume that $0< p(t)\leq p<1$, and
 there exists $ b>0$ such that
\begin{equation}
\int_{t_0}^{\infty}g_{n-1}(\sigma(s), 0)f_i(s, b)\Delta
s<\infty,\quad i=1, 2.\label{e4}
\end{equation}
Then \eqref{e1} has a bounded non-oscillatory solution which
is bounded away from zero.
\end{theorem}

\begin{proof}
Let $BC$ be the set of bounded functions on $[t_0,\infty)$
 with sup norm  $\|x\|=\sup_{t\geq t_0}|x(t)|$,
$t\in \mathbb{T}$.
Let $\Omega\subset BC$, $\Omega=\{x\in BC, 0<M_1\leq x(t)\leq M_2<b,
t\geq t_0,t\in \mathbb{T}\}$,
where $M_1<(1-p)M_2$, then $\Omega$ is a closed bounded and
convex subset of $BC$.

Choose $\alpha$ such that $pM_2+M_1<\alpha<M_2$,
and $c=\min\{M_2-\alpha, \alpha-pM_2-M_1\}$. We choose
$t_1>t_0$, such that
$ \tau(t)\geq t_0$, $\tau_i(t)\geq t_0$, $i=1,2$ $t\geq t_1$ and
$\int_{t_1}^{\infty}g_{n-1}(\sigma(s), 0)f_i(s, b)\Delta
s\leq c$, $i=1,2$.  Define a mapping $\Gamma$ on $\Omega$ as
follows.
$$
(\Gamma x)(t)=(\Gamma_1 x)(t)+(\Gamma_2 x)(t),
$$
where
\begin{gather*}
(\Gamma_1 x)(t)=\begin{cases}
 \alpha-p(t)x(\tau(t)),& t\geq t_1,\; t\in\mathbb{T},\\
 (\Gamma_1 x)(t_1), & t_{0}\leq t\leq t_1,\; t\in\mathbb{T}.
\end{cases}
\\
(\Gamma_2 x)(t)=\begin{cases}
 (-1)^{n-1}\int_{t}^{\infty}g_{n-1}(\sigma(s), t)\big[f_1(s,
 x(\tau_1(s)))\\
-f_2(s, x(\tau_2(s)))\big]\Delta s, & t\geq t_1,\\[4pt]
 (\Gamma_2 x)(t_1), & t_{0}\leq t\leq t_1.
\end{cases}
\end{gather*}
For any $x, y \in \Omega$, $t\ge t_0$, $t\in \mathbb{T}$, we have
\begin{gather*}
(\Gamma_1 x)(t)+(\Gamma_2 y)(t)\leq \alpha+c\leq M_2,\\
(\Gamma_1 x)(t)+(\Gamma_2 y)(t)\geq \alpha-pM_2-c\geq M_1.
\end{gather*}
Hence for $t\geq t_0$, $t\in \mathbb{T}$,
$\Gamma_1x+\Gamma_2y\in \Omega$.
Clearly, $\Gamma_1$ is a contraction mapping on $\Omega$ and
$\Gamma_2$ is continuous. We shall show that $\Gamma_2$ is
completely continuous.
In fact, for any $x\in \Omega$, for $t_0\leq t\leq t_1$,
$(\Gamma_2x)(t)=(\Gamma_2x)(t_1)$, and for $t\geq t_1$, we have
 \begin{align*}
|(\Gamma_2x)(t)|&\leq  \int_{t}^{\infty}g_{n-1}(\sigma(s),
t)|f_1(s, x(\tau_1(s)))-f_2(s, x(\tau_2(s)))|\Delta s\\
&\leq  \int_{t}^{\infty}g_{n-1}(\sigma(s), t)f_1(s,
x(\tau_1(s)))\Delta s\\
&\leq  \int_{t}^{\infty}g_{n-1}(\sigma(s), 0)f_1(s, b)\Delta
s\leq c.
\end{align*}
Hence $\Gamma_2\Omega$ is uniformly bounded. For
$\varepsilon>0$, there exists a $T$, such that for $t\geq T$,
$$
\int_{t}^{\infty}g_{n-1}(\sigma(s), 0)f_i(s, b)\Delta
s<\frac{\varepsilon}{2}.
$$
For $t, t'>T$, we have
 $$
|(\Gamma_2x)(t)-(\Gamma_2x)(t')|\leq 2\int_{T}^{\infty}
g_{n-1}(\sigma(s), 0)f_i(s, b)\Delta s< \varepsilon.
$$
For $t, t'\in [t_1, T]$, we have
\begin{align*}
&|(\Gamma_2x)(t)-(\Gamma_2x)(t')|\\
&=  |\int_{t}^{\infty}g_{n-1}(\sigma(s), t)[f_1(s, x(\tau_1(s)))-
f_2(s, x(\tau_2(s)))]\Delta s\\
&\quad -\int_{t'}^{\infty}g_{n-1}(\sigma(s), t')[f_1(s,
x(\tau_1(s)))- f_2(s, x(\tau_2(s)))]\Delta s|\\
&\leq  |\int_{t}^{t'}g_{n-1}(\sigma(s), t)[f_1(s,
x(\tau_1(s)))- f_2(s, x(\tau_2(s)))]\Delta s|\\
&\quad + \int_{t'}^{T}|g_{n-1}(\sigma(s), t)-g_{n-1}(\sigma(s),
t')\|f_1(s, x(\tau_1(s)))- f_2(s, x(\tau_2(s)))|\Delta s\\
&\quad + \int_{T}^{\infty}|g_{n-1}(\sigma(s), t)-g_{n-1}(\sigma(s),
t')\|f_1(s, x(\tau_1(s)))- f_2(s, x(\tau_2(s)))|\Delta s.
\end{align*}
There exists a $\delta$, so that when $|t-t'|<\delta$,
$|(\Gamma_2x)(t)-(\Gamma_2x)(t')|<\varepsilon$, which shows that
the family $\Gamma_2\Omega$ is equicontinuous, $\Gamma_2$ is
completely continuous.

By Lemma 2, there exists a fixed point $x\in \Omega$, such that
$\Gamma x=x$. It is easily to see that $x$ is a bounded
non-oscillatory solution which is bounded away
from zero.
\end{proof}

\begin{theorem} \label{thm2}
 Assume that $1<p_1\leq p(t)\leq p_2$, and
\eqref{e4}  holds. Then \eqref{e1} has a bounded non-oscillatory
solution which is bounded away from zero.
\end{theorem}

\begin{proof} We choose $t_1>t_0$ such that
$$
T_0=\min\{\tau(t_1),\inf_{t\geq t_1}(\tau_1(t)),
 \inf_{t\geq t_1}(\tau_2(t))\}\geq t_0.
$$
Let $BC$ be the set of bounded functions on $[t_0,\infty)$
with supremum norm
$\|x\|=\sup_{t\geq t_0}|x(t)|$, $t\in \mathbb{T}$.
Define a set $ \Omega \subset BC$ as follows:
\begin{align*}
\Omega=\Big\{&x\in BC, x^{\Delta}(t)\leq 0,
0<M_1\leq x(t)\leq p_{1}M_1<b, t\geq t_1,\\
&x(t)=x(t_1),  \quad T_{0}\leq t\leq t_1.\Big\}
\end{align*}
Then $\Omega$ is a closed bounded and convex subset of $BC$.
Let $c=\min\{\alpha-M_1, p_1M_1-\alpha\}$, where
$M_1<\alpha<p_1M_1 $. We choose $t_2\geq t_1$, such that for
$t\geq t_2$,
$$
\int_{t}^{\infty}g_{n-1}(\sigma(s), 0)f_i(s, b)\Delta s\leq  c.
$$
For $x\in \Omega$, define
$$
\psi(t)=\begin{cases}
 \sum_{i=1}^{\infty}\frac{(-1)^{i-1}x(\tau^{-i}(t))}{H_{i}
 (\tau^{-i}(t))},& t\geq t_2,\\
 \psi(t_2), & T_{0}\leq t\leq t_2,
\end{cases}
$$
where $\tau^{0}(t)=t$, $\tau^{i}(t)=\tau(\tau^{i-1}(t))$,
$\tau^{-i}(t)=\tau^{-1}(\tau^{-(i-1)}(t))$, $H_{0}(t)=1$,
$H_{i}(t)= \prod_{j=0}^{i-1}p(\tau^{j}(t))$, $i=1,2,\dots$.
   From $M_1\leq x(t)\leq p_{1}M_1$, we have
$$
0<\psi (t)\leq p_1M_1,\quad t\geq t_2,\; t\in \mathbb{T}.
$$
Define a mapping  $\Gamma$ on $\Omega$ as follows
$$
(\Gamma x)(t)=\begin{cases}
 \alpha+(-1)^{n-1}\int_{t}^{\infty}g_{n-1}(\sigma(s), t)\\
 \times \sum_{i=1}^{2}(-1)^{i+1}f_i(s, \psi(\tau_{i}(s)))\Delta s,
& t\geq t_2,\\[4pt]
 (\Gamma x)(t_2),& T_{0}\leq t\leq t_2.
\end{cases}
$$
Then $\Gamma$ satisfies the following conditions:\\
(a) $\Gamma \Omega\subseteq \Omega$.
 In fact, for any $x\in \Omega$, $(\Gamma x)(t)\geq \alpha-c\geq
 M_1$,  $(\Gamma x)(t)\leq \alpha+c\leq  p_1M_1$.\\
(b) $\Gamma$ is continuous which is easy to show.\\
(c) Similar to Theorem 1, $\Gamma $ is equicontinuous.

 By Lemma 3, there exists
$x\in \Omega$, such that $x=\Gamma x$; i.e.,
$$
x(t)=\alpha+(-1)^{n-1}\int_{t}^{\infty}
g_{n-1}(\sigma(s), t)[f_1(s,
\psi(\tau_{1}(s)))-f_2(s,\psi(\tau_{2}(s)))]\Delta s.
$$
Since $\psi(t)+p(t)\psi(\tau(t))=x(t)$, we obtain
\begin{align*}
&\psi(t)+p(t)\psi(\tau(t))\\
&=\alpha+(-1)^{n-1}\int_{t}^{\infty}
g_{n-1}(\sigma(s), t)[f_1(s,
\psi(\tau_{1}(s)))-f_2(s,\psi(\tau_{2}(s)))]\Delta s.
\end{align*}
So $\psi(t)$ satisfies \eqref{e1} for $t\geq t_0$,
$t\in \mathbb{T}$, and
$\frac{p_{1}-1}{p_{1}p_{2}}x(\tau^{-1}(t))\leq \psi(t)\leq
 x(t)$. The proof is complete.
\end{proof}


\begin{theorem} \label{thm3}
Assume that $-1<p_1\leq p(t)\leq 0$, and \eqref{e4}  holds.
Then \eqref{e1} has a bounded non-oscillatory solution which is
bounded away from zero.
\end{theorem}

\begin{proof}
Let $BC$ be the set of bounded functions on $[t_0,\infty)$
with sup norm  $\|x\|=\sup_{t\geq t_0}|x(t)|$.
We choose $M_1, M_2<b$ such that $0< M_1<\alpha<(1+p_1)M_2$.
Let $\Omega=\{x\in BC, M_1\leq x(t)\leq M_2< b, t\geq t_0\}$.
Then $\Omega$ is a closed bounded and convex subset of $BC$.
Let $c=\min\{\alpha-M_1, (1+p_1)M_2-\alpha\}$.
We choose $t_1\geq t_0$, such that
$\tau(t)\geq t_0, \tau_i(t)\geq t_0$, for $t\geq t_1$ and
$\int_{t_1}^{\infty}g_{n-1}(\sigma(s), 0)f_i(s,b)\Delta s\leq c$,
$i=1,2$.
Define a mapping $\Gamma$ on $\Omega$ as follows:
$$
(\Gamma x)(t)=(\Gamma_1 x)(t)+(\Gamma_2  x)(t),
$$
where
\begin{gather*}
(\Gamma_1 x)(t)=\begin{cases}
 \alpha-p(t)x(\tau(t)),& t\geq t_1,\\
 (\Gamma_1 x)(t_1), & t_{0}\leq t\leq t_1,
\end{cases}
\\
(\Gamma_2 x)(t)=\begin{cases}
 (-1)^{n-1}\int_{t}^{\infty}g_{n-1}(\sigma(s), t)
\big[f_1(s, x(\tau_1(s)))\\
-f_2(s, x(\tau_2(s)))\big]\Delta s, & t\geq t_1, \\[4pt]
 (\Gamma_2 x)(t_1), & t_{0}\leq t\leq t_1.
\end{cases}
\end{gather*}
For  $x, y \in \Omega$, $t\ge t_0$, we have
\begin{gather*}
(\Gamma_1 x)(t)+(\Gamma_2 y)(t)\leq \alpha-p_1M_2+c\leq M_2,\\
(\Gamma_1 x)(t)+(\Gamma_2 y)(t)\geq \alpha-c\geq M_1.
\end{gather*}
Hence for $t\geq t_0$, $\Gamma_1x+\Gamma_2y\in \Omega$.
Clearly, $\Gamma_1$ is a contraction mapping on $\Omega$
and $\Gamma_2$ is continuous. Similar to Theorem 1, we can
prove that $\Gamma_2$ is completely continuous.
So that there exists $x\in \Omega$ such that
$x=\Gamma x$. The proof is complete.
\end{proof}

\begin{theorem} \label{thm4}
Assume that $p_1\leq p(t)\leq p_2<-1$ and
\eqref{e4} holds. Then \eqref{e1} has a bounded non-oscillatory
solution which is bounded away from zero.
\end{theorem}

\begin{proof} Let $BC$ be the  bounded functions on $[t_0,\infty)$.
We choose $ 0<M_1<M_2<b$, such that $-p_1M_1<\alpha
<(-p_2-1)M_2$.
Let $\Omega=\{x\in BC, M_1\leq x(t)\leq M_2,t\geq t_0\}$,
$c=\min\{\frac{(\alpha+M_1p_1)p_2}{p_1},(-p_2-1)M_2-\alpha\}$.
Choose $t_1\geq t_0$ such that for $ t\geq t_1$,
$$
\tau^{-1}(\tau_i(t))\geq t_0,\quad
\int_{\tau^{-1}(t)}^{\infty}g_{n-1}(\sigma(s), 0)f_i(s,b)\Delta
s\leq c,\quad i=1,2.
$$
 Define two maps $\Gamma_1,\Gamma_2$ on $\Omega$ as follows:
\begin{gather*}
(\Gamma_1 x)(t)=\begin{cases}
 -\frac{\alpha}{p(\tau^{-1}(t))}-\frac{x(\tau^{-1}(t))}{p(\tau^{-1}(t))}
 ,& t\geq t_1,\\
 (\Gamma_1 x)(t_1), & t_{0}\leq t\leq t_1.
\end{cases}
\\
(\Gamma_2 x)(t)=\begin{cases}
 \frac{(-1)^{n-1}}{p(\tau^{-1}(t))}
 \int_{\tau^{-1}(t)}^{\infty}g_{n-1}(\sigma(s), t)
 \big[f_1(s, x(\tau_{1}(s)))\\
-f_2(s, x(\tau_{2}(s)))\big]\Delta s,
& t\geq t_1,\\[4pt]
(\Gamma_2 x)(t_1),& t_{0}\leq t\leq t_1.
\end{cases}
\end{gather*}
For  $x, y \in \Omega$,  $(\Gamma_1 x)(t)+(\Gamma_2 y)(t)\geq
\frac{-\alpha}{p_1}+\frac{c}{p_2}\geq M_1$,
 $(\Gamma_1 x)(t)+(\Gamma_2 y)(t)\leq
\frac{-\alpha}{p_2}-\frac{M_2}{p_2}-\frac{c}{p_2}\leq M_2$. So
$(\Gamma_1 x)(t)+(\Gamma_2 y)(t) \in \Omega$. Since $p_1\leq
p(t)\leq p_2\leq -1$, we get $\Gamma_1$ is contraction.
We shall prove that $\Gamma_2$ is completely continuous.
In fact, for all $x\in \Omega$, $t_0\leq t\leq t_1$,
$(\Gamma_2x)(t)=(\Gamma_2x)(t_1)$.
For $t\geq t_1$,
$$
|(\Gamma_2x)(t)|\leq
-\frac{1}{p_2}\int_{\tau^{-1}(t)}^{\infty}g_{n-1}(\sigma(s),
0)f_i(s, x(\tau_i(s)))\Delta s\leq -\frac{c}{p_2},
$$
so $\Gamma_2\Omega$ is uniformly bounded.
By the conditions, for all $\varepsilon>0$, there exists a
$T>t_1$ such that
$$
\int_{\tau^{-1}(T)}^{\infty}g_{n-1}(\sigma(s), 0)f_i(s,
b)\Delta s<\frac{-p_2\varepsilon}{2}.
$$
For all $x\in \Omega$, $t,t'\geq T$, we have
$$
|(\Gamma_2x)(t)-(\Gamma_2x)(t')|\leq -\frac{2}{p_2}
 \int_{\tau^{-1}(T)}^{\infty}g_{n-1}(\sigma(s), 0)f_i(s,
b)\Delta s<\varepsilon.
$$
Since $\tau^{-1}(t),\frac{1}{p(\tau^{-1}(t))}$ are continuous on
$[t_1,T]$, they are uniformly continuous on $[t_1, T]$. Let
$|g_{n-1}(\sigma(t),0)f_i(t, b)|\leq M$,
when $t\in [t_1, T]$.
Hence for each $\varepsilon>0$, there exists a $\delta>0$ such
that for $ t, t'\in [t_1, T]$, $|t-t'|<\delta$, we have
\begin{gather*}
|\frac{1}{p(\tau^{-1}(t))}-\frac{1}{p(\tau^{-1}(t'))}|
<\frac{\varepsilon}{3c},\quad
|\tau^{-1}(t)-\tau^{-1}(t')|<\frac{-p_2 \varepsilon}{3M},\\
\int_{\tau^{-1}(t_1)}^{\infty}|g_{n-1}(\sigma(s),
t)-g_{n-1}(\sigma(s), t')|f_i(s, b)\Delta
s<\frac{|p_2|\varepsilon}{3}.
\end{gather*}
For all $x\in\Omega$, when
$t,t'\in [t_1,T]$ and $|t-t'|<\delta$, we have
\begin{align*}
&|(\Gamma_2x)(t)-(\Gamma_2x)(t')|\\
&= |\frac{1}{p(\tau^{-1}(t))}\int_{\tau^{-1}(t)}^{\infty}g_{n-1}
 (\sigma(s), t)[f_1(s, x(\tau_1(s)))-f_2(s, x(\tau_2(s)))]\Delta s\\
&\quad -\frac{1}{p(\tau^{-1}(t'))}\int_{\tau^{-1}(t')}^{\infty}g_{n-1}
 (\sigma(s), t')[f_1(s, x(\tau_1(s)))-f_2(s, x(\tau_2(s)))]\Delta s|\\
&\leq  |\frac{1}{p(\tau^{-1}(t))}-\frac{1}{p(\tau^{-1}(t'))}|
\int_{\tau^{-1}(t)}^{\infty}g_{n-1}(\sigma(s), t)|f_1(s,
x(\tau_1(s)))-f_2(s, x(\tau_2(s)))|\Delta s\\
&\quad +
|\frac{1}{p(\tau^{-1}(t'))}\|\int_{\tau^{-1}(t)}^{\tau^{-1}(t')}
g_{n-1}(\sigma(s), t)[f_1(s, x(\tau_1(s)))-f_2(s,
x(\tau_2(s)))]\Delta s|\\
&\quad +|\frac{1}{p(\tau^{-1}(t'))}|\int_{\tau^{-1}(t')}^{\infty}
|g_{n-1}(\sigma(s), t)-g_{n-1}(\sigma(s),t')|\\
&\quad\times |\sum_{i=1}^{2}(-1)^{i+1}f_i(s, x(\tau_i(s)))|\Delta s\\
&< \frac{\varepsilon
c}{3c}+\frac{M}{|p_2|}\cdot\frac{|p_2|\varepsilon}{3M}
+\frac{|p_2|\varepsilon}{|p_2|3}=\varepsilon,
\end{align*}
which shows that the family $\Gamma_2\Omega$ is equicontinuous, so
$\Gamma_2$ is completely continuous.
By Lemma 2, there exists a fixed point $x\in \Omega$ such that
$\Gamma x=x$. It is easily to see that $x$ is a bounded
non-oscillatory solution which is bounded away from zero.
\end{proof}

\subsection*{Example}
On the time scale $\mathbb{T} =\{q^{n}: n\in \mathbb{N}_0,\,q>1\}$,
consider the dynamic equation
\begin{equation}
\begin{aligned}
&(x(t)-\frac{1}{\sqrt{q}}x(\rho(t)))^{\Delta^4}
+2\frac{(\sqrt{q}-1)(q+1)^2(q^2+1)(q^2+q+1)}
{q^{10}t^3(t+q^2)^2}x^2(\rho^2(t))\\
&- \frac{(\sqrt{q}-1)(q+1)^2(q^2+1)(q^2+q+1)}
{q^{10}t^3(t+q^2)^3}x^2(\rho^3(t))=0,
\end{aligned}\label{e5}
\end{equation}
where $\rho$ is the backward operator,
$\rho^2(t)=\rho(\rho(t)), \rho^3(t)=\rho(\rho^2(t))$.
In this equation, $n=4$, $p(t)=-\frac{1}{\sqrt{q}}$,
$\tau(t)=\rho(t)=\frac{t}{q}$,
$\tau_1(t)=\rho^2(t)$, $\tau_2(t)=\rho^3(t)$,
\begin{gather*}
f_1(t,b)=2\frac{(\sqrt{q}-1)(q+1)^2(q^2+1)(q^2+q+1)}
{q^{10}t^3(t+q^2)^2}b^2, \\
f_2(b)=\frac{(\sqrt{q}-1)(q+1)^2(q^2+1)(q^2+q+1)}
{q^{10}t^3(t+q^2)^3}b^2.
\end{gather*}
By the definition of  $g_k(s, t)$,
\begin{align*}
g_{4-1}(\sigma(s), 0)\cdot f_1(s, b)
& \leq  s^3 2\frac{(\sqrt{q}-1)(q+1)^2(q^2+1)(q^2+q+1)}
{q^{10}s^3(s+q^2)^2}b^2\\
&\leq  2\frac{(\sqrt{q}-1)(q+1)^2(q^2+1)(q^2+q+1)}
{q^{10}s^2}b^2,
\end{align*}
\begin{align*}
g_{4-1}(\sigma(s), 0)\cdot f_2(s, b)
& \leq  s^3 \frac{(\sqrt{q}-1)(q+1)^2(q^2+1)(q^2+q+1)}
{q^{10}s^3(s+q^2)^3}b^2\\
&\leq  \frac{(\sqrt{q}-1)(q+1)^2(q^2+1)(q^2+q+1)} {q^{10}s^3}b^2,
\end{align*}
and
\begin{gather*}
\int_{t_0}^{\infty}\frac{(\sqrt{q}-1)(q+1)^2(q^2+1)(q^2+q+1)}
{q^{10}s^2}b^2\Delta s<\infty,\\
\int_{t_0}^{\infty} \frac{(\sqrt{q}-1)(q+1)^2(q^2+1)(q^2+q+1)}
{q^{10}s^3}b^2\Delta s<\infty.
\end{gather*}
It is obviously that  \eqref{e5} satisfies all conditions of
Theorem 3. Hence  \eqref{e5} has a bounded non-oscillatory solution
which is bounded away from zero. In fact $x(t)=1+\frac{1}{t} $ is
a solution of  \eqref{e5}.

\subsection*{Acknowledgements}
This research was supported by grants L2009Z02 from the Main
Foundation of Hebei Normal University, and L2006B01 from the
Doctoral Foundation of Hebei Normal University.
The authors would like to thank the anonymous referee for his or
her careful reading and the comments on improving
the presentation of this article.

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