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\begin{document}

\title{Delay Dynamic Equations with Stability}

\author[Anderson]{Douglas R. Anderson}
\address{Department of Mathematics, Concordia College, Moorhead, MN 56562}
\email{andersod@cord.edu}

\author[Krueger]{Robert J. Krueger}
\address{Department of Mathematics, Concordia University, St. Paul, MN 55104}
\email{rkrueger@csp.edu}

\author[Peterson]{Allan C. Peterson}\thanks{This research was supported by NSF grant 0354281.}
\address{Department of Mathematics, University of Nebraska-Lincoln, Lincoln, NE 68588} \email{apeterso@math.unl.edu}


\date{7/21/05}
\begin{abstract}
We first give conditions which guarantee that every solution of a first order linear delay dynamic equation for isolated 
time scales vanishes at infinity. Several interesting examples are given. In the last half of the paper, we give 
conditions under which the trivial solution of a nonlinear delay dynamic equation is asymptotically stable, for
arbitrary time scales.
\end{abstract}

\maketitle\thispagestyle{empty}

\section{Preliminaries}

The unification and extension of continuous calculus, discrete calculus, $q$-calculus, and indeed arbitrary 
real-number calculus to time-scale calculus, where a time scale is simply any nonempty closed set of real numbers, 
were first accomplished by Hilger in \cite{h}. Since then, time-scale calculus has made steady inroads in explaining
the interconnections that exist among the various calculuses, and in extending our understanding to a new, more general 
and overarching theory. The purpose of this work is to illustrate this new understanding by extending some continuous
and discrete delay equations to certain time scales.  Examples will include specific cases in differential equations,
difference equations, $q$-difference equations, and harmonic-number equations.

The definitions that follow here will serve as a short primer on the time-scale calculus; they can be
found in \cite{bp1} and \cite{bp2} and the references therein.

\begin{definition} Define the forward (backward) jump operator $\sigma (t)$ at $t$ for $t < \sup \mathbb{T}$ (respectively $\rho(t)$ at $t$ for
$t > \inf \mathbb{T}$) by
$$\sigma (t) = \inf \{ \tau > t : \tau \in \mathbb{T} \}, \ \ \ (\rho (t) = \sup \{ \tau < t : \tau \in \mathbb{T} \},) \ 
\mbox{for all} \ t \in \mathbb{T}.$$
Also define $\sigma(\sup\T)=\sup\T$, if $\sup\T<\infty$, and $\rho(\inf\T)=\inf\T$, if $\inf\T>-\infty$. 
Define the graininess function $\mu:\mathbb{T}\rightarrow \mathbb{R}$ by $\mu(t)=\sigma(t)-t.$
\end{definition}

Throughout this work the assumption is made that $\T$ is unbounded above and has the topology that it inherits from the standard topology on the real numbers $\R$.  Also assume throughout that $a < b$ are points in $\mathbb{T}$ and define the time scale interval $[a,b]_{\T} = \{ t \in \T: a \le t \le b \}$.  The jump operators $\sigma$ and $\rho$ allow the classification of points in a time scale in the following way:
If $\sigma (t) >t$ then call the point $t$ right-scattered; while if $\rho(t) < t$ then we say $t$ is left-scattered.
If $\sigma(t) = t$ then call the point $t$ right-dense; while if $t>\inf\T$ and $\rho(t) = t$ then we say $t$ is left-dense.
 
We next define the so-called delta derivative. The novice could skip this definition and look at the results stated in Theorem \ref{props}. In particular in part $(2)$ of Theorem \ref{props} we see what the delta derivative is at right--scattered points and in part $(3)$ of Theorem \ref{props} we see that at right--dense points the derivative is similar to the definition given in calculus.

\begin{definition}
Fix $t \in \T$ and let $y: \mathbb{T} \rightarrow \R$. Define $y^{\Delta}(t)$ to be the number (if it exists) 
with the property that given $\epsilon > 0$ there is a
neighbourhood $U$ of $t$ such that, for all $s \in U$,
$$| [y(\sigma(t)) - y(s)] - y^{\Delta}(t) [\sigma(t) - s] | \leq \epsilon |\sigma (t) - s |.$$
Call $y^{\Delta}(t)$ the (delta) derivative of $y(t)$ at $t$.
\end{definition}

\begin{definition}
If $F^{\Delta}(t) = f(t)$ then define the (Cauchy) delta integral by
$$ \int_a^t f(s) \Delta s = F(t) - F(a). $$
\end{definition}
The following theorem is due to Hilger \cite{h}.
\begin{theorem}\label{props}
Assume that $f:\mathbb{T} \rightarrow \mathbb{R}$ and let $t \in \T$.
\begin{enumerate}
\item If $f$ is differentiable at $t$, then $f$ is continuous at $t$.
\item If $f$ is continuous at $t$ and $t$ is right-scattered, then $f$ is differentiable at $t$ with
$$f^{\Delta} (t) = \frac{f(\sigma(t)) - f(t)}{\sigma(t) - t}.$$
\item If $f$ is differentiable and $t$ is right-dense, then
$$f^{\Delta} (t) =\lim_{s \rightarrow t} \frac{f(t) - f(s)}{t - s}.$$
\item If $f$ is differentiable at $t$, then $f(\sigma(t)) = f(t) + \mu(t) f^{\Delta} (t)$.
\end{enumerate}
\end{theorem}
Next we define the important concept of right-dense continuity. An important fact concerning right-dense continuity is that every right-dense continuous function has a delta antiderivative \cite[Theorem 1.74]{bp1}. This implies that the delta definite integral of any right-dense continuous function exists.

\begin{definition}
We say that $f:\T\rightarrow\R$ is right-dense continuous (and write $f \in C_{rd}(\T; \R)$) provided $f$ is continuous at every right-dense point $t \in \mathbb{T}$, and
$\lim_{s \rightarrow t^-} f(s)$ exists and is finite at every left-dense point $t \in \mathbb{T}$.
\end{definition}

We say $p$ is regressive provided $1+\mu(t)p(t) \neq 0, \forall t \in \T$. Let
$$\cR:=\{p \in C_{rd}(\T; \R):1+\mu(t)p(t)\neq 0, t \in \T\}.$$
Also, $p \in \cR^+$ iff $1+\mu(t)p(t)>0, \: \forall t \in \T$.
Then if $p \in \cR, t_0 \in \T$, one can define the generalized exponential function $e_{p}(t,t_0)$ to be the unique solution of the initial value problem
$$ x^\Delta = p(t)x, \quad x(t_0)=1. $$
We will use many of the properties of this generalized exponential function $e_{p}(t,t_0)$ listed in Theorem \ref{expthm}.

\begin{theorem}[Theorem 2.36, \cite{bp1}]\label{expthm}
If $p,q \in \cR$ and $s,t \in \T$, then
\begin{enumerate}
	\item $e_{0} (t,s)\equiv 1$ and $e_{p} (t,t)\equiv 1$;
	\item $e_{p} (\sigma(t),s) = (1+\mu(t)p(t))e_{p}(t,s)$;
	\item $\frac{1}{e_{p}(t,s)} = e_{\ominus p}(t,s)$, where $\ominus p:=\frac{-p}{1+\mu p}$;
	\item $e_{p}(t,s)= \frac{1}{e_{p}(s,t)}=e_{\ominus p}(s,t)$;
	\item $e_{p}(t,s) e_{p}(s,r)=e_{p}(t,r)$;
	\item $e_{p}(t,s) e_{q}(t,s) = e_{p \oplus q}(t,s)$, where $p\oplus q:=p+q+\mu pq$;
	\item $\frac{e_{p}(t,s)}{e_{q}(t,s)}=e_{p \ominus q}(t,s)$.
\end{enumerate}
\end{theorem}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                                           %
%          Delay Equation Intro             %
%                                           %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Introduction to a Delay Dynamic Equation}

Since we are interested in the asymptotic properties of solutions we assume as mentioned earlier that our time scale 
$\T$ is unbounded above. Consider the delay dynamic equation
\begin{equation} \label{eq1}
x^\Delta (t) = - a(t)x(\delta(t))\delta ^\Delta (t), \quad t \in [t_0,\infty)_\T,
\end{equation}
where the delay function $\delta: [t_0,\infty)_\T \ra [\delta(t_0),\infty)_\T$ is strictly increasing and delta differentiable with $\delta(t)<t$ for $t \in (t_0,\infty)_\T$.

For example, if $\T=[-m,\infty)$, and $\delta(t):= t-m$ where $m>0$, then \eqref{eq1} becomes the well-studied delay differential equation
$$ x'(t)=-a(t)x(t-m). $$

If $\T=\{-N,-N+1, \cdots, 0, 1, 2, \cdots \}$, and $\delta(t):= t-m$ where $m$ is a positive integer, then \eqref{eq1} becomes 
$$ \Delta x(t)=-a(t)x(t-m), $$
where $\Delta$ is the forward difference operator defined by $\Delta x(t)=x(t+1)-x(t)$.

If $\T=q^{\N_0}\cup \{q^{-1}, q^{-2}, \cdots, q^{-m}\}$ where $q^{\N_0}:=\{1, q, q^2,\ldots\}$, $q>1$, and $\delta(t):= \frac{1}{q^m} t$ where $m \in \N$, then \eqref{eq1} becomes the delay quantum equation
$$ D_q x(t)=-\frac{1}{q^m} a(t)x \lt(\frac{1}{q^m} t \rt), $$
where 
$$ D_q x(t):= \frac{x(qt)-x(t)}{(q-1)t} $$
is the so-called quantum derivative studied in Kac and Cheung \cite{kc}.

We need the following three lemmas to prove our main result.

\begin{lemma}[Chain Rule]\label{chain}
 Assume $\T$ is an isolated time scale.  Assume $g(\sigma(t))=\sigma(g(t))$ for $t \in \T$. If $g:\T \ra \T$ and $h:\T \ra \R$, then
\begin{equation*}
\lt(\int_{t_0}^{g(t)} h(s) \Delta s \rt)^\Delta = h(g(t))g^\Delta(t).
\end{equation*}
\end{lemma}

\begin{proof}
Since $t$ is right-scattered,
\begin{eqnarray*}
\lt(\int_{t_0}^{g(t)} h(s) \Delta s \rt)^\Delta &=& \frac{1}{\mu(t)} \lt(\int_{t_0}^{g(\sigma(t))} h(s) \Delta s - \int_{t_0}^{g(t)} h(s) \Delta s \rt)\\
&=& \frac{1}{\mu(t)} \int_{g(t)}^{g(\sigma(t))} h(s) \Delta s\\
&=& \frac{1}{\mu(t)} \int_{g(t)}^{\sigma(g(t))} h(s) \Delta s\\
&=& \frac{1}{\mu(t)} h(g(t))(\sigma(g(t))-g(t))\\
&=& h(g(t)) \frac{g(\sigma(t))-g(t)}{\mu(t)}\\
&=& h(g(t))g^\Delta(t).
\end{eqnarray*}
\end{proof}

\begin{lemma}
Assume $\T$ is an isolated time scale and the delay $\delta$ satisfies $\delta \circ\sigma = \sigma \circ \delta$. Then the delay equation \eqref{eq1} is equivalent to the delay equation
\begin{equation} \label{eq2}
x^\Delta (t) = - a(\delta^{-1}(t))x(t) + \left(\int^{t}_{\delta(t)}  a(\delta^{-1}(s))x(s)\Delta s \right)^\Delta.
\end{equation}
\end{lemma}

\begin{proof}
Assume $x$ is a solution of \eqref{eq2}.  Then using the chain rule [Lemma \ref{chain}],
\begin{eqnarray*}
x^\Delta (t) &=& - a(\delta^{-1}(t))x(t)+ \left(\int^{t}_{\delta(t)}  a(\delta^{-1}(s))x(s)\Delta s \right)^\Delta \\
 &=& - a(\delta^{-1}(t))x(t)+a(\delta^{-1}(t))x(t)- a(t)x(\delta(t))\delta ^\Delta (t)\\
 &=& - a(t)x(\delta(t))\delta ^\Delta (t).
\end{eqnarray*}
Hence $x$ is a solution of \eqref{eq1}.  Reversing the above steps, we obtain the desired result.
\end{proof}

\begin{lemma}\label{sollem}
If $x$ is a solution of \eqref{eq1} with initial function $\psi$, then
\begin{eqnarray} \label{sol}
x(t)&=& \nonumber e_{-a(\delta^{-1})} (t,t_0)\psi(t_0)\\
&+& \nonumber \int^{t}_{\delta(t)} a(\delta^{-1}(s))x(s)\Delta s - e_{-a(\delta^{-1})} (t,t_0) \int^{t_0}_{\delta(t_0)} a(\delta^{-1}(s))\psi(s)\Delta s \\ 
&-&\int^{t}_{t_0} \frac{a(\delta^{-1}(\tau))}{1-\mu(\tau) a(\delta^{-1}(\tau))} e_{-a(\delta^{-1})}(t,\tau) \left(\int^{\tau}_{\delta(\tau)} a(\delta^{-1}(s))x(s)\Delta s \right) \Delta \tau.
\end{eqnarray}
\end{lemma}

\begin{proof}
We use the variation of constants \cite[page 77]{bp1} on \eqref{eq2}, to obtain
$$x(t) = e_{-a(\delta^{-1})} (t,t_0)x(t_0)+ \int^{t}_{t_0} e_{-a(\delta^{-1})}  (t,\sigma(\tau))\left(\int^{\tau}_{\delta(\tau)} a(\delta^{-1}(s))x(s)\Delta s \right)^{\Delta_\tau} \Delta \tau.$$
Using integration by parts \cite[page 28]{bp1},
\begin{eqnarray*}
x(t) &=&  e_{-a(\delta^{-1})} (t,t_0)x(t_0)+ e_{-a(\delta^{-1})}  (t,\tau)\int^{\tau}_{\delta(\tau)} a(\delta^{-1}(s))x(s)\Delta s \lt. \rt|^{t}_{t_0}\\
&-& \int^{t}_{t_0} e^{\Delta_{\tau}} _{-a(\delta^{-1})}  (t,\tau)\left(\int^{\tau}_{\delta(\tau)} a(\delta^{-1}(s))x(s)\Delta s \right) \Delta \tau.
\end{eqnarray*}
It follows from Theorem \ref{expthm} that
\begin{eqnarray*}
x(t)&=& e_{-a(\delta^{-1})} (t,t_0)x(t_0)\\
&+& \int^{t}_{\delta(t)} a(\delta^{-1}(s))x(s)\Delta s - e_{-a(\delta^{-1})} (t,t_0) \int^{t_0}_{\delta(t_0)} a(\delta^{-1}(s))x(s)\Delta s\\
&-&\int^{t}_{t_0} e^{\Delta_{\tau}}_{\ominus(-a(\delta^{-1}))}(\tau,t) \left(\int^{\tau}_{\delta(\tau)} a(\delta^{-1}(s))x(s)\Delta s \right) \Delta \tau\\
&=& e_{-a(\delta^{-1})} (t,t_0)x(t_0)\\
&+& \int^{t}_{\delta(t)} a(\delta^{-1}(s))x(s)\Delta s - e_{-a(\delta^{-1})} (t,t_0) \int^{t_0}_{\delta(t_0)} a(\delta^{-1}(s))x(s)\Delta s\\
&-&\int^{t}_{t_0} \ominus (-a(\delta^{-1})) (e_{\ominus(-a(\delta^{-1}))}(\tau,t) \left(\int^{\tau}_{\delta(\tau)} a(\delta^{-1}(s))x(s)\Delta s \right) \Delta \tau.
\end{eqnarray*}
Finally, using Theorem \ref{expthm} once again and $x(t)=\psi(t)$ for $t \in [\delta^{-1}(t_0),t_0]$,
\begin{eqnarray*}
x(t)&=& e_{-a(\delta^{-1})} (t,t_0)\psi(t_0)\\
&+&\int^{t}_{\delta(t)} a(\delta^{-1}(s))x(s)\Delta s - e_{-a(\delta^{-1})} (t,t_0) \int^{t_0}_{\delta(t_0)} a(\delta^{-1}(s))\psi(s)\Delta s \\ 
&-&\int^{t}_{t_0} \frac{a(\delta^{-1}(\tau))}{1-\mu(\tau) a(\delta^{-1}(\tau))} e_{-a(\delta^{-1})}(t,\tau) \left(\int^{\tau}_{\delta(\tau)} a(\delta^{-1}(s))x(s)\Delta s \right) \Delta \tau.
\end{eqnarray*}
\end{proof}

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%                                           %
%         Asymptotic Properties             %
%                                           %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Asymptotic Properties of the Delay Equation}

Let $\psi:[\delta(t_0),t_0]_\T \ra \R$ be a bounded initial function and say $x(t):=x(t,t_0,\psi)$ is a solution of 
\eqref{eq1} if $x(t) = \psi (t)$ on $[\delta(t_0),t_0]_\T$ and $x(t)$ satisfies \eqref{eq1} on $[t_0,\infty)_\T$. 

Let $\left\| \phi \right\| = \sup \left|\phi(t)\right|$ for $t \in [\delta(t_0),\infty)_\T$, and
define the Banach space $B=\lt\{ \phi \in C(\T):\phi \mbox{ is bounded on } [\delta(t_0),\infty)_\T \mbox{ with } \phi(t) \ra 0 \mbox{ as } t \ra \infty \rt\}$, with
$$ S:=\lt\{\phi \in B: \phi(t)=\psi(t) \quad \forall t \in [\delta(t_0),t_0]_\T \rt\}. $$

In the following we assume 
\begin{equation} \label{ezero}
e_{-a(\delta^{-1})} (t,t_0) \ra 0 \mbox{ as } t \ra \infty,
\end{equation}
and take $D:[t_0,\infty)_{\T}\rightarrow\R$ to be the function
\begin{eqnarray} \label{Dt}
D(t)&:=& \nonumber \int^{t}_{t_0} \lt( \lt| \frac{a(\delta^{-1}(\tau))}{1-\mu(\tau) a(\delta^{-1}(\tau))} \rt| \lt| e_{-a(\delta^{-1})}(t,\tau) \rt| \int^{\tau}_{\delta(\tau)} \lt| a(\delta^{-1}(s)) \rt| \Delta s \rt) \Delta \tau\\
&& \; + \int^{t}_{\delta(t)} \lt| a(\delta^{-1}(s)) \rt| \Delta s.
\end{eqnarray}
To enable the use of the contraction mapping theorem, we in fact assume there exists $\alpha \in (0,1)$ such that
\begin{equation}\label{alphaterm}
D(t)\leq \alpha, \quad t \in [t_0,\infty)_{\T}.
\end{equation}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                                           %
%              Main Theorem                 %
%                                           %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\begin{theorem}\label{mainthm}
Assume \eqref{ezero} and \eqref{alphaterm} hold.  Then every solution of \eqref{eq1} goes to zero at infinity.
\end{theorem}

\begin{proof}
Fix $\psi:[\delta(t_0),t_0] \ra \R$ and define $P:S \ra S$ by $(P \phi)(t):=\psi(t)$ for $t \leq t_0$ and for $t \geq t_0$,
\begin{eqnarray*}
(P\phi)(t) &=& \psi(t_0) e_{-a(\delta^{-1})}(t,t_0) + \int^{t}_{\delta(t)} a(\delta^{-1}(s)) \phi(s) \Delta s\\
&-& e_{-a(\delta^{-1})}(t,t_0) \int^{t_0}_{\delta(t_0)} a(\delta^{-1}(s)) \psi(s) \Delta s\\
&-& \int^{t}_{t_0} \left( \frac{a(\delta^{-1}(\tau))}{1-\mu(\tau) a(\delta^{-1}(\tau))} e_{-a(\delta^{-1})}(t,\tau) \int^{\tau}_{\delta(\tau)} a(\delta^{-1}(s))\phi(s)\Delta s \right) \Delta \tau.
\end{eqnarray*}

Then by Lemma \ref{sollem}, it suffices to show that $P$ has a fixed point.  We will use the contraction mapping theorem to show $P$ has a fixed point.

By rd-continuity of $\delta$ at $t_0$ and the discrete nature of all $t>t_0$, $P\phi$ is continuous.  By \eqref{ezero}, there exists $Q>0$ such that 

\begin{equation} \label{theQ}
e_{-a(\delta^{-1})}(t,t_0) \leq Q, \forall t \geq t_0.
\end{equation}

Let $\phi \in S, \left\|\phi\rt\| \leq K$ for some $K>0$, and $\psi$ be an initial function with $\left|\psi\rt| <\gamma$ for $\gamma>0$. Then, using \eqref{alphaterm}, 

\begin{eqnarray*}
\lt\|(P\phi)\rt\| &=& \gamma Q + \int^{t}_{\delta(t)} \lt| a(\delta^{-1}(s))\rt| \lt\|\phi\rt\| \Delta s\\
&+& Q \int^{t_0}_{\delta(t_0)} \lt| a(\delta^{-1}(s)) \rt| \lt\|\phi\rt\| \Delta s\\
&+& \int^{t}_{t_0} \lt( \lt| \frac{a(\delta^{-1}(\tau))}{1-\mu(\tau) a(\delta^{-1}(\tau))} \rt| \lt| e_{-a(\delta^{-1})}(t,\tau) \rt| \int^{\tau}_{\delta(\tau)} \lt| a(\delta^{-1}(s)) \rt| \Delta s \rt) \Delta \tau \ \lt\| \phi \rt\| \\
&\leq& \gamma Q +  \lt\|\phi\rt\| \alpha + Q \int^{t_0}_{\delta(t_0)} \lt| a(\delta^{-1}(s)) \rt| \lt\|\phi\rt\| \Delta s\\
&\leq& \gamma Q + K \alpha + Q K \int^{t_0}_{\delta(t_0)} \lt| a(\delta^{-1}(s)) \rt| \Delta s\\
&\leq& \gamma Q + K \alpha + Q K \alpha.
\end{eqnarray*}

Therefore, $P\phi$ is bounded.  To show that $(P\phi)(t) \ra 0$ as $t \ra \infty$, note that the first and third terms on the right-hand side of $(P\phi)(t)$ go to zero by \eqref{ezero}.  From  \eqref{Dt} and \eqref{alphaterm} and the fact that $\phi(t) \ra 0$ as $t \ra \infty$, we have that 
$$ \lt|\phi(t)\rt| \int^{t}_{\delta(t)} \lt| a(\delta^{-1}(s)) \rt| \Delta s \leq \lt|\phi(t)\rt|\alpha \ra 0, \quad t \ra \infty. $$
Let $\epsilon>0$ be given and choose $t^* \in \T$ so that 
$$ \alpha \lt\|\phi\rt\| \lt| e_{-a(\delta^{-1})}(t,T) \rt| < \frac{\epsilon}{2}, \quad \forall t > t^*, $$
for some large $t^*>T$.  For the same $T$ it is possible to make
$$ \alpha \lt\|\phi\rt\|_{[\delta(T),\infty)_{\T}} < \frac{\epsilon}{2}, $$
where $\left\| \phi \right\|_{[\delta(T),\infty)_{\T}} = \sup \{\left|\phi(t)\right|$, $t \in [\delta(T),\infty)_\T\}.$  By \eqref{sol} and \eqref{ezero}, for $t \geq T$,

\begin{eqnarray*}
&& \int^{t}_{t_0} \lt( \lt| \frac{a(\delta^{-1}(\tau))}{1-\mu(\tau) a(\delta^{-1}(\tau))} \rt| \lt| e_{-a(\delta^{-1})}(t,\tau) \rt| \int^{\tau}_{\delta(\tau)} \lt| a(\delta^{-1}(s)) \phi(s) \rt| \Delta s \rt) \Delta \tau\\
 &=& \lt( \int^{T}_{t_0}+ \int^t_T \rt) \lt( \lt| \frac{a(\delta^{-1}(\tau))}{1-\mu(\tau) a(\delta^{-1}(\tau))} \rt| \lt| e_{-a(\delta^{-1})}(t,\tau) \rt| \int^{\tau}_{\delta(\tau)} \lt| a(\delta^{-1}(s)) \phi(s) \rt| \Delta s \rt) \Delta \tau\\
 &=& \int^{T}_{t_0} \lt( \lt| \frac{a(\delta^{-1}(\tau))}{1-\mu(\tau) a(\delta^{-1}(\tau))} \rt| \lt| e_{-a(\delta^{-1})}(t,T) e_{-a(\delta^{-1})}(T,\tau) \rt| \int^{\tau}_{\delta(\tau)} \lt| a(\delta^{-1}(s)) \phi(s) \rt| \Delta s \rt) \Delta \tau\\
 & & \; +\int^t_T \lt( \lt| \frac{a(\delta^{-1}(\tau))}{1-\mu(\tau) a(\delta^{-1}(\tau))} \rt| \lt| e_{-a(\delta^{-1})}(t,\tau) \rt| \int^{\tau}_{\delta(\tau)} \lt| a(\delta^{-1}(s)) \phi(s) \rt| \Delta s \rt) \Delta \tau\\
 &\leq& \lt| e_{-a(\delta^{-1})}(t,T) \rt| \lt\|\phi\rt\|  \int^{T}_{t_0} \lt( \lt| \frac{a(\delta^{-1}(\tau))}{1-\mu(\tau) a(\delta^{-1}(\tau))} \rt| \lt|  e_{-a(\delta^{-1})}(T,\tau) \rt| \int^{\tau}_{\delta(\tau)} \lt| a(\delta^{-1}(s)) \rt| \Delta s \rt) \Delta \tau\\
 & & \; +\left\| \phi \right\|_{[\delta(T),\infty)_{\T}} \int^t_T \lt( \lt| \frac{a(\delta^{-1}(\tau))}{1-\mu(\tau) a(\delta^{-1}(\tau))} \rt| \lt| e_{-a(\delta^{-1})}(t,\tau) \rt| \int^{\tau}_{\delta(\tau)} \lt| a(\delta^{-1}(s)) \rt| \Delta s \rt) \Delta \tau\\
 &\leq& \alpha \lt| e_{-a(\delta^{-1})}(t,T) \rt| \lt\|\phi\rt\| + \alpha \left\| \phi \right\|_{[\delta(T),\infty)_{\T}}\\
 &<& \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.
\end{eqnarray*}

Hence $(P\phi)(t) \ra 0$ as $t \ra \infty$ and therefore, $P$ maps $S$ into $S$.  It remains to show that $P$ is a contraction under the sup norm.  Let $x,y \in S$.  Then

\begin{eqnarray*}
&&\lt|(Px)(t)-(Py)(t)\rt|\\
&\leq& \int^{t}_{\delta(t)} \lt| a(\delta^{-1}(s))\rt| \lt|(x(s)-y(s))\rt| \Delta s\\
&+& \int^{t}_{t_0} \lt( \lt| \frac{a(\delta^{-1}(\tau))}{1-\mu(\tau) a(\delta^{-1}(\tau))} \rt| \lt| e_{-a(\delta^{-1})}(t,\tau) \rt| \int^{\tau}_{\delta(\tau)} \lt| a(\delta^{-1}(s)) \rt|  \lt| x(s)-y(s) \rt|\Delta s \rt) \Delta \tau \\
&\leq&  \lt\|x-y\rt\| \lt[ \int^{t}_{\delta(t)} \lt| a(\delta^{-1}(s))\rt|  \Delta s \rt.\\
&+& \lt. \int^{t}_{t_0} \lt( \lt| \frac{a(\delta^{-1}(\tau))}{1-\mu(\tau) a(\delta^{-1}(\tau))} \rt| \lt| e_{-a(\delta^{-1})}(t,\tau) \rt| \int^{\tau}_{\delta(\tau)} \lt| a(\delta^{-1}(s)) \rt|  \Delta s \rt) \Delta \tau \rt]\\
&\leq& \alpha \lt\|x-y\rt\|.
\end{eqnarray*}
Therefore, by the contraction mapping principle \cite[page 300]{kp}, $P$ has a unique fixed point in $S$.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%     Examples     %%     Examples     %%     Examples     %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{example}
For any real number $q>1$ and positive integer $m$, define $$\T=\{q^{-m}, q^{-m+1}, \cdots, q^{-1}, 1, q, q^2, \cdots\}.$$ 
We show if $0<c<\frac{q^m}{2m(q-1)}$, then for any initial function $\psi(t), t \in [q^{-m},1]_{\T}$, the solution of the delay initial value problem
\begin{eqnarray}\label{delayIVP}
D_q x(t)&=&  -\frac{1}{q^m} \frac{c}{t} x\lt(\frac{1}{q^m}t\rt),\quad t \in [1,\infty]_{\T}, \\
x(t)&=& \nonumber \psi(t), \quad t \in [q^{-m},1]_{\T}
\end{eqnarray} 
goes to zero as $t \ra \infty$.
\end{example}

\noindent To obtain equation \eqref{delayIVP} from \eqref{eq1}, take $a(t)=\frac{c}{t}$ and $\delta(t)=q^{-m}t$ which implies $a(\delta^{-1}(t))=\frac{c}{q^m t}$ and $\delta^{\Delta}(t)=q^{-m}$.  To use Theorem \ref{mainthm}, we verify that conditions \eqref{ezero} and \eqref{alphaterm} hold.  Note that 
$$e_{-a(\delta^{-1})}(t,1) = \prod_{s \in [1,t)_\T} \lt[1-s(q-1)a(q^ms)\rt]=(1-q^{-m}c(q-1))^n$$
for $t=q^n$.

If $c \in \lt(0,\frac{q^m}{2m(q-1)}\rt)$, then $c \in \lt(0,\frac{2q^m}{q-1}\rt)$ so that $1-q^{-m}c(q-1) \in (-1,1)$ and
$$\lim_{t \ra \infty} e_{-a(\delta^{-1})}(t,1) =\lim_{n \ra \infty} (1-q^{-m}c(q-1))^n = 0.$$
Thus, \eqref{ezero} is satisfied. Now consider $D(t)$ as defined in \eqref{Dt}. We seek $\alpha \in(0,1)$ such that $D(t) \leq \alpha, \forall t \in [1,\infty)_{\T}$.  Here we have $t_0=1, \mu(t)=(q-1)t$, and
$$e_{-a(\delta^{-1})}(t,\tau) =(1-q^{-m}c(q-1))^{n-k}$$ for $t=q^n, \tau = q^k$ with $k<n$.  For the second integral in $D(t)$, note that
$$\int_u^{qu} f(\zeta) \Delta \zeta = (qu-u)f(u),$$ whence
\begin{eqnarray*}
\int^{t}_{\delta(t)} a(\delta^{-1}(s))\Delta s &=& \lt(\int_{q^{-m}t}^{q^{-m+1}t} + \int_{q^{-m+1}t}^{q^{-m+2}t} +\cdots+ \int_{q^{-1}t}^{t} \rt) \lt(\frac{c}{q^m s}\rt) \Delta s\\
&=& \lt(\frac{c}{q^m (q^{-m}t)}q^{-m}t + w \rt)(q-1)\\
&=& \frac{mc}{q^m}(q-1),
\end{eqnarray*}
which is independent of $t$. It follows that 
\begin{eqnarray*}
D(t)&=& \frac{mc}{q^m}(q-1) +\frac{mc}{q^m}(q-1)\int_1^t \lt(\frac{c}{q^m \tau} \cdot \frac{1}{1-\frac{(q-1)\tau c}{q^m \tau}} e_{-a(\delta^{-1})}(t,\tau)\rt)\Delta \tau\\
&=& \frac{mc}{q^m}(q-1) +\frac{mc}{q^m}(q-1) \sum_{k=0}^{n-1} \frac{c}{q^m q^k - q^k(q-1)c} (1-q^{-m}c(q-1))^{n-k}(q-1)q^k\\
&=& \frac{mc}{q^m}(q-1) +\frac{mc}{q^m} \sum_{k=0}^{n-1} \frac{c(q-1)^2}{q^m -(q-1)c} (1-q^{-m}c(q-1))^{n-k}\\
&=& \frac{mc}{q^m}(q-1) +\frac{mc^2q^{-m}(q-1)^2}{q^m-c(q-1)} \frac{1-q^{-m}c(q-1)}{-q^{-m} (q-1)c} \lt((1-q^{-m}c(q-1))^n -1\rt)\\
&=&\frac{mc(q-1)}{q^m} +\frac{mcq^{-m}(q-1)}{1-cq^{-m}(q-1)} (1-q^{-m}c(q-1)) \lt(1-(1-q^{-m}c(q-1))^n \rt).
\end{eqnarray*}
Therefore, 
$$ D(t) =\frac{mc(q-1)}{q^m}\lt[2-(1-q^{-m}c(q-1))^n\rt]<\frac{2mc(q-1)}{q^m}, \;\forall t=q^n \in [1,\infty)_\T. $$
Therefore, since $0<c<\frac{q^m}{2m(q-1)}$, defining $\alpha:= \frac{2mc(q-1)}{q^m}$, \eqref{alphaterm} is satisfied by
$$ D(t)<\alpha<1, \;\forall t \in [1,\infty)_\T. $$
Conditions \eqref{ezero} and \eqref{alphaterm} are met, so that by Theorem \ref{mainthm}, \eqref{eq1} has a solution that goes to zero as $t \ra \infty$.

\begin{example}
Consider the time scale of harmonic numbers 
$$ \T=\{H_{-m}, H_{-m+1}, \cdots, H_0, H_1, \cdots\} $$ 
for some $m\in \N$, where $H_0:=0$, $H_n:=\sum_{j=1}^n \frac{1}{j}$ and $H_{-n}:=-H_n$ for $n \in \N$.  We will show that if
$$ c \in \left(0,\frac{H_m}{2m}\right),$$
then for any initial function $\psi(t), t \in [H_{-m},0]_\T$, the solution of the delay initial value problem
\begin{eqnarray}\label{delayIVP2}
\Delta_n x(H_n)&=&  -\frac{(n-m+1)c}{H_m} x(H_{n-m}) \Delta_n H_{n-m},\quad n \in \N_0\\
x(H_n)&=& \nonumber \psi(H_n), \quad n=0,-1,\cdots, -m
\end{eqnarray} 
goes to zero as $t \ra \infty$. 
\end{example}

\noindent To get \eqref{delayIVP2} from equation \eqref{eq1}, take 
$$ a(t)=a(H_n)=\frac{(n-m+1)c}{H_m},\quad \delta(t)=\delta(H_n)=H_{n-m}. $$
It follows that 
$$e_{-a(\delta^{-1})}(H_n,0) =\lt(1-\frac{c}{H_m}\rt)^n \mbox{ for } n \in \N_0.$$
If we restrict $c \in (0,2H_m)$, $$\lim_{t \ra \infty} e_{-a(\delta^{-1})}(t,0) =\lim_{n \ra \infty} \lt(1-\frac{c}{H_m}\rt)^n =0,$$ satisfying \eqref{ezero}.  Simplifying \eqref{Dt},

\begin{eqnarray*}
D(t)&=&\int_{H_{n-m}}^{H_n} \frac{(s+1)c}{H_m} \Delta s\\
&+& \int_0^{H_n} \left( \frac{(\tau+1)c}{H_m} \frac{1}{1-\frac{(\tau+1)c}{(\tau+1)H_m}}\left(1-\frac{c}{H_m}\right)^{n-\tau} \int_{H_{\tau-m}}^{H_{\tau}} \frac{(s+1)c}{H_m} \Delta s\right) \Delta \tau\\
&=& \frac{cm}{H_m} +\frac{c^2m}{H_m} \cdot \frac{1}{H_m-c}\sum_{\tau=0}^{n-1} \left(1-\frac{c}{H_m}\right)^{n-\tau}\\
&=& \frac{cm}{H_m} +\frac{c^2m}{H_m(H_m-c)} \left(\frac{H_m-c}{H_m}\right) \left(\frac{-H_m}{c}\right) \left[\left(1-\frac{c}{H_m}\right)^{n}-1\right]\\
&=& \frac{cm}{H_m} \left[2-\left(1-\frac{c}{H_m}\right)^{n}\right]<\frac{2cm}{H_m}
\end{eqnarray*}
for all $t=H_n \in [0,\infty)_{\T}$.  By choosing $c \in (0,\frac{H_m}{2m}), \: D(t)<\alpha := \frac{2cm}{H_m}<1, \: \forall t=H_n \in [0,\infty)_{\T}$, satisfying \eqref{alphaterm}.  Thus by Theorem \ref{mainthm}, for any given initial function $\psi$, \eqref{eq1} has a unique solution that goes to zero as $t=H_n$ goes to infinity.

\begin{example}
Let $\T=\{-mh, \cdots, -h, 0, h, 2h, \cdots\}$ where $m \in \N$, then \eqref{eq1} becomes 
\begin{equation} \label{ex2}
\Delta_h x(t)=-c x(t-hm),\quad t \in h\N,
\end{equation}
where $\Delta_h x(t):=\frac{x(t+h)-x(t)}{h}$. Our results give that if 
$$ 0<c<\frac{1}{2mh}, $$ 
then all solutions go to zero as $t \ra \infty$.
\end{example}

\noindent In this case, $a(t)=c, \delta(t)=t-mh$, and $\delta^{-1}(t)=t+mh$.  It can be shown that
$$ e_{-a(\delta^{-1})}(t,0)=(1-ch)^{\frac{t}{h}}.$$
Note for $0<ch<2$, condition \eqref{ezero} is satisfied because
$$ e_{-a(\delta^{-1})}(t,0) \ra 0$$
as $t \ra \infty$.  It also can be shown that
$$ D(t) = mhc+mhc(1-|1-ch|^{\frac{t}{h}}) \leq 2mhc $$
if $0<ch<1$.  Hence \eqref{alphaterm} holds if $0<c<\frac{1}{2mh}$.  Therefore, we use Theorem \ref{mainthm} to conclude that all solutions of \eqref{ex2} go to zero as $t \ra \infty$.

It can be shown that if $c \leq 0$, then there is $\lambda \geq 1$ such that $x(t)=\lambda ^{ht}$ is a solution of \eqref{ex2}.  However, $x(t)$ does not approach zero as $t \ra \infty$.  Hence our lower estimate for $c$ is sharp.

If $m=1$, then it can be shown that all solutions of \eqref{ex2} go to zero if $0<c<\frac{1}{h}$. If $m=2$, our example shows that if $0<c<\frac{1}{4h}$ all solutions go to zero as $t \ra \infty$.  It can be shown that if $c = \frac{1}{2h}$, then there is a solution that does not go to zero as $t \ra \infty$.

Next we give an example where $\T$ is the real interval $[-m,\infty), m>0$.  Delay differential equations have been studied extensively; for example, see \cite{LLZ}.

\begin{example}
Let $\T=[-m,\infty)$. For the delay differential equation
\begin{equation}\label{ex3}
x'(t)=-c x(t-m), \quad t \in [0,\infty),
\end{equation}
if $0<c<\frac{1}{2m}$, then all solutions $x(t)$ approach zero as $t \ra \infty$.
\end{example}

\noindent Note that equation \eqref{eq1} reduces to equation \eqref{ex3} if $a(t)=c$, $\delta(t)=t-m$, and $\T=[-m,\infty)$. It can be shown that
$$ e_{-a(\delta^{-1})}(t,\tau)=e^{-c(t-\tau)}, \quad D(t)=cm(2-e^{-ct}). $$
Note that $c>0$ implies $e_{-a(\delta^{-1})}(t,0) \ra 0$ as $t \ra \infty$ and $D(t)\leq 2cm<1$ if $c<\frac{1}{2m}$.  Our result follows from Theorem \ref{mainthm}.  Also note that if $c=\frac{\pi}{2m}$, then $x(t)=\sin\lt(\frac{\pi}{2m} t\rt)$ is a solution that does not approach zero as $t \ra \infty$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                              %
%    Asymptotic Stability      %
%                              %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Asymptotic Stability of a Nonlinear Delay Dynamic Equation}
In this section we consider, on arbitrary time scales, the nonlinear delay dynamic equation
\begin{equation}\label{feq}
  x^{\Delta}(t)=-\int_{\delta(t)}^t \left(\sum_{i=1}^n f_i(t,x(s))\right)\Delta s,\quad t \in [t_0,\infty)_\T,
\end{equation}
where $f_i(t,x)$ for each fixed $t \in \T$ is continuous with respect to $x$.  In addition, we always suppose
\begin{eqnarray*}
\mbox{(H1)} && xf_i(t,x)\geq 0 \mbox{ and } \sum_{i=1}^n f_i(t,x)=0 \iff x=0,\: t \in [t_0,\infty)_\T,\\
\mbox{(H2)} && \delta:\T\! \ra\! \T \mbox{ is continuous and nondecreasing, with } \delta(t) \!\le\! t \mbox{ and } \lim_{t \ra \infty} \delta(t)\!=\!\infty.
\end{eqnarray*}

The initial condition associated with \eqref{feq} takes the form 
\begin{equation}\label{delaysol}
x(t)=\psi(t),\quad t \in [\delta(t_0),t_0], \quad \psi \mbox{ is rd-continuous on } [\delta(t_0),t_0].
\end{equation}

Equation \eqref{feq} is studied extensively in \cite{kuang} in the case when $\T = \R$; indeed many of our techniques 
in this section are motivated by those in \cite{kuang}.  See also a related discussion in \cite{hk}. 

%%%%%%%%%
% Lemma %
%%%%%%%%%

\begin{lemma}\label{bddosc}
Assume $x$ is a global solution for \eqref{feq}, \eqref{delaysol}.  Then either $x$ is bounded or $x$ is oscillatory.
\end{lemma}

\begin{proof}
If $x$ is nonoscillatory, then there exists $T>0$ such that for $t>T$, $x$ does not change sign. Without loss of 
generality, we suppose $x(t)>0$ for $t>T$.  By (H1), $\sum_{i=1}^n f_i(t,x(t))>0$ for $t>T$, which together with 
\eqref{feq} yields that $x^\Delta (t)<0$ for large $t$. Therefore, $x$ is strictly decreasing on each interval, 
so by continuity, $x$ is bounded.
\end{proof}

%%%%%%%%%%%%%%%
% Theorem 4.2 %
%%%%%%%%%%%%%%%

\begin{theorem}\label{thm4.3}
Assume there exists $M>0$ such that for $|c| \leq M$,
\begin{enumerate}
	\item[(i)] $\int_{\delta(t)}^{\sigma(t)} \sum_{i=1}^n \frac{1}{c} f_i(\tau,c)(\tau-\delta(\tau))\Delta \tau \leq \xi<1, \: c\neq 0$, for $t \in [t_1,\infty)_\T$;
	\item[(ii)] $\int_{t_0}^{\infty} \sum_{i=1}^n \left|f_i(\tau,c)\right|(\tau-\delta(\tau))\Delta \tau=\infty$;
	\item[(iii)] $\left|\sum_{i=1}^nf_i(\tau,c)\right|\leq \sum_{i=1}^na_i(\tau)|c|$ where the $a_i$ are rd-continuous and nonnegative for $\tau \geq t_0$;
	\item[(iv)] $p(t):=\sum_{i=1}^n a_i(t)(t-\delta(t)) \in \cR^+$;
	\item[(v)] $\left|\sum_{i=1}^nf_i(\tau,x)\right|\leq \sum_{i=1}^n f_i(\tau,y)$ for $|x|\leq y\leq M$, $\tau\in[\delta(t_0),\infty)_{\T}$.
\end{enumerate}
Then for any $0 < \epsilon \leq M$, there is a $\eta(\epsilon)>0$ such that for any rd-continuous initial function $\psi$ with $\lt\|\psi\rt\|_{[\delta(t_0),t_0]_\T}<\eta(\epsilon)$, the solution $x$ of \eqref{feq}, \eqref{delaysol} satisfies
$$|x(t)|<\epsilon \: \forall t \in [\delta(t_0),\infty)_\T, \quad \lim_{t \ra \infty} x(t)= 0.$$
In other words, the trivial solution of \eqref{feq} is asymptotically stable. 
\end{theorem}

\begin{proof}
Let $\epsilon\in(0,M]$ and $t_1\in[t_0,\infty)_{\T}$ be given. Define $\eta(\epsilon):=\frac{\epsilon}{e_p(t_1,t_0)}$ and
take initial function $\psi$ with $\lt\|\psi\rt\|_{[\delta(t_0),t_0]_\T}<\eta(\epsilon)$. Integrating \eqref{feq} from 
$t_0$ to $t \in [t_0,t_1]_\T$, we get 
$$ x(t)=x(t_0)-\int_{t_0}^t \lt( \int_{\delta(\tau)}^{\tau} \sum_{i=1}^n f_i(\tau,x(s)) \Delta s \rt)\Delta \tau. $$
Taking absolute values,
$$ |x(t)|\leq |x(t_0)|+\int_{t_0}^t \lt( \int_{\delta(\tau)}^{\tau} \left|\sum_{i=1}^n f_i(\tau,x(s))\right| \Delta s \rt)\Delta \tau.$$
By property (iii),
\begin{equation}\label{roy} 
|x(t)|\leq |x(t_0)|+\int_{t_0}^t \lt( \int_{\delta(\tau)}^{\tau} \sum_{i=1}^n a_i(\tau)|x(s)| \Delta s \rt)\Delta \tau.
\end{equation}
Let $\tilde{x}(t):=\max \{|x(s)|: s \in [\delta(t_0),t]\}$.  Then rewriting \eqref{roy}, we have
$$ \tilde{x}(t)\leq \tilde{x}(t_0)+\int_{t_0}^t \tilde{x}(\tau) \sum_{i=1}^n a_i(\tau) (\tau-\delta(\tau))\Delta \tau. $$
By Gronwall's inequality \cite[page 257]{bp1} and property (iv),
$$ \tilde{x}(t)\leq \tilde{x}(t_0)e_p(t,t_0),\quad \forall t \in [t_0,\infty)_\T. $$
Therefore by the definition of $\eta(\epsilon)$ and the choice of $psi$, it follows that $\tilde{x}(t_0)<\eta(\epsilon)$ 
and $|x(t)|\leq \tilde{x}(t)<\epsilon$ for $t \in [t_0,t_1]_\T$.  Suppose there exists $t^*\geq t_1$ such that 
$x(t^*)=\epsilon$ and $x^{\Delta}(t^*)\geq 0$.  Note the case when $x(t^*)=-\epsilon$ and $x^{\Delta}(t^*)\leq 0$ is 
similar.  By \eqref{feq}, $\exists \bar{t} \in [\delta(t^*),t^*)_\T$ such that $x(\bar{t})\leq 0$.  Integrating 
\eqref{feq} from $\bar{t}$ to $t^*$, we obtain
$$ \epsilon - x(\bar{t})= -\int_{\bar{t}}^{t^*} \lt( \int_{\delta(\tau)}^{\tau} \sum_{i=1}^n f_i(\tau,x(s)) \Delta s \rt)\Delta \tau. $$
Since $x(\bar{t})\leq 0$, 
$$ \epsilon \leq \int_{\delta(t^*)}^{t^*} \lt( \int_{\delta(\tau)}^{\tau} \left|\sum_{i=1}^n f_i(\tau,x(s))\right| \Delta s \rt)\Delta \tau. $$
By (H1) and (v),
$$ \epsilon \leq \int_{\delta(t^*)}^{t^*} \lt( \int_{\delta(\tau)}^{\tau} \sum_{i=1}^n |f_i(\tau,\epsilon)| \Delta s \rt)\Delta \tau, $$
so that
$$ 1\leq \frac{1}{\epsilon} \int_{\delta(t^*)}^{t^*} (\tau-\delta(\tau)) \sum_{i=1}^n f_i(\tau,\epsilon)\Delta \tau,$$ which is a contradiction of (i).  Therefore no such $t^*$ exists.  Now suppose $\exists t^* \in [t_1,\infty)_\T$ such that $t^*$ is right scattered, $|x(t)|\leq \epsilon, \: \forall t \in [\delta(t_0),t^*)_\T, x(t^*) \in (-\epsilon, \epsilon)$, but $|x(\sigma(t^*))|\geq \epsilon.$ Without loss of generality, assume
\begin{equation}\label{zig}
x(\sigma(t^*)) \geq \epsilon.
\end{equation}
By \eqref{feq},
$$ \int_{\delta(t^*)}^{t^*} \lt( \sum_{i=1}^n f_i(t^*,x(s)) \rt)\Delta s<0.$$
Therefore, $\exists \bar{t} \in [\delta(t^*),t^*)_\T$ such that $x(\bar{t})\leq 0$, by (H1) and \eqref{feq}. Integrate 
\eqref{feq} from $\bar{t}$ to $\sigma(t^*)$ and use \eqref{zig} to see that
$$ \epsilon \leq x(\sigma(t^*))\leq x(\sigma(t^*))-x(\bar{t})
   =-\int_{\bar{t}}^{\sigma(t^*)} \lt( \int_{\delta(\tau)}^{\tau} \sum_{i=1}^n f_i(\tau,x(s)) \Delta s \rt)\Delta \tau. $$
Thus, comparing the extremities,
\begin{eqnarray*}
  \epsilon &\leq& \int_{\delta(t^*)}^{\sigma(t^*)} \lt( \int_{\delta(\tau)}^{\tau} \left|\sum_{i=1}^n f_i(\tau,x(s))\right| \Delta s \rt)\Delta \tau \\
           &\leq& \int_{\delta(t^*)}^{\sigma(t^*)} \sum_{i=1}^n f_i(\tau,\epsilon) (\tau-\delta(\tau)) \Delta \tau;
\end{eqnarray*}
division by $\epsilon$ yields
$$ 1\leq \frac{1}{\epsilon} \int_{\delta(t^*)}^{\sigma(t^*)} \sum_{i=1}^n f_i(\tau,\epsilon) (\tau-\delta(\tau)) \Delta \tau, $$ 
which also contradicts (i) and provides the desired result.

Now we show the limit of $x(t)$ goes to zero as $t \ra \infty$.

%%%%%%%%%%%%%
%  Case 1   %
%%%%%%%%%%%%%
  
\noindent \underline{Case 1}: Let $x$ be a nonoscillatory function.  Assume $\xi<1$ but there exists a solution $x$ of \eqref{feq} such that $$\lim_{t \ra \infty} x(t)\neq 0.$$
Since $x$ is nonoscillatory, there exists $T_1>t_0$ such that $x(t)x(T_1)>0, \: \forall t \in [T_1,\infty)_\T$.  Without loss of generality, assume $x(t)>0, \: \forall t \in [T_1,\infty)_\T$.  From \eqref{feq} and (H1), $x^{\Delta}(t)<0$. Hence there exists a constant $x^*>0$ such that
$$\lim_{t \ra \infty} x(t)=x^*$$ and there exists $T^* \in \T$ such that $\forall t \in [T_1,\infty)_\T, x(t) \in \lt[x^*, \frac{3x^*}{2}\rt]$.  Integrate \eqref{feq} from $\delta^{-1}(T^*)$ to $t$ to obtain
\begin{equation}\label{con1}
x(\delta^{-1}(T^*))-x(t) = \int_{\delta^{-1}(T^*)}^{t} \lt( \int_{\delta(\tau)}^{\tau} \sum_{i=1}^n f_i(\tau,x(s)) \Delta s \rt)\Delta \tau.
\end{equation}

For $t \geq \delta^{-1}(T^*),$
\begin{equation}\label{con2}
x(\delta^{-1}(T^*))-x(t)\leq \frac{x^*}{2},
\end{equation}
but
\begin{eqnarray*}
\int_{\delta^{-1}(T^*)}^{t} \lt( \int_{\delta(\tau)}^{\tau} \sum_{i=1}^n f_i(\tau,x(s)) \Delta s \rt)\Delta \tau
 &\geq& \int_{\delta^{-1}(T^*)}^{t} \lt( \int_{\delta(\tau)}^{\tau} \sum_{i=1}^n f_i(\tau,x^*) \Delta s \rt)\Delta \tau\\
 &=&\int_{\delta^{-1}(T^*)}^{t} \lt(  \sum_{i=1}^n f_i(\tau,x(s)) (\tau - \delta(\tau))\rt) \Delta \tau,
\end{eqnarray*}
which goes to $\infty$ as $t \ra \infty$ by (H1) and properties (ii) and (v). This contradicts \eqref{con1} and \eqref{con2}, so when $x$ is nonoscillatory, the limit of $x(t)$ goes to zero as $t \ra \infty$.

%%%%%%%%%%%%%
%  Case 2   %
%%%%%%%%%%%%%

\noindent \underline{Case 2}: Now assume $x$ is oscillatory. Pick $\epsilon \in (0,M]$ and $T_2>t_0$ such that
$$\int_{\delta(t)}^{t} \lt(  \sum_{i=1}^n \frac{1}{c} f_i(\tau,c)(\tau - \delta(\tau))\rt) \Delta \tau<\xi$$
for $ 0 <|c|\leq \epsilon$ and $ \forall t \in [T_2,\infty)_\T$.  Assume 
\begin{equation*}
  \bar{x}:=\limsup_{t \ra \infty} |x(t)| \neq 0.
\end{equation*}
Without loss of generality, we will assume
\begin{equation*}
  \limsup_{t \ra \infty} x(t)=\bar{x} \neq 0.
\end{equation*}
Since the solution is oscillatory, there exists a sequence $\lt\{t_j\rt\}_{j=1}^{\infty}$ in $\T$ such that
$\lim_{j \ra \infty} t_j = \infty$, $|x(\sigma(t_j))|\ra \bar{x}\leq \epsilon$ as $j \ra \infty$, and 
$x^\Delta(t_j)\geq 0$.  Let $\beta:=\frac{\bar{x}}{2}(1-\xi)$ and $T_3>T_2$ such that $|x(t)|<\beta + \bar{x}$ 
for $t \in [T_3,\infty)_\T.$  Further assume the $t_j$ were chosen such that $\delta(t_j)>T_3$ and 
$x(\sigma(t_j))>\bar{x}-\beta$. From \eqref{feq} and the choice of $t_j$, we see that 
$$ 0\leq x^{\Delta}(t_j) = -\int_{\delta(t_j)}^{t_j} \lt(  \sum_{i=1}^n f_i(t_j,x(s))\rt)\Delta s. $$
By (H1), there exists $t^* \in [\delta(t_j),t_j]_\T$ such that $x(t^*)\leq 0.$ Then
\begin{eqnarray*}
\bar{x} - \beta &<& x(\sigma(t_j)) \leq x(\sigma(t_j))-x(t^*)=-\int_{t^*}^{\sigma(t_j)} \lt( \int_{\delta(\tau)}^{\tau} \sum_{i=1}^n f_i(\tau,x(s)) \Delta s \rt)\Delta \tau\\
\bar{x} - \beta &<& \int_{t^*}^{\sigma(t_j)} \lt( \int_{\delta(\tau)}^{\tau} \left|\sum_{i=1}^n f_i(\tau,x(s))\right| \Delta s \rt)\Delta \tau\\
&\leq&  \int_{t^*}^{\sigma(t_j)} \sum_{i=1}^n f_i(\tau,\beta+\bar{x})(\tau-\delta(\tau))\Delta \tau\\
&\leq&  \int_{\delta(t_j)}^{\sigma(t_j)} \sum_{i=1}^n f_i(\tau,\beta+\bar{x})(\tau-\delta(\tau))\Delta \tau\\
&<&(\beta +\bar{x})\xi
\end{eqnarray*}
by (i). But then $\beta > \lt(\frac{1-\xi}{1+\xi}\rt)\bar{x}$, a contradiction of our selection of $\beta$.  
Therefore $\bar{x}=0$, in other words, the solution goes to zero as $t \ra \infty.$
\end{proof}

%%%%%%%%%%%%%%%%%%
% Theorem %
%%%%%%%%%%%%%%%%%%

\begin{theorem}
Assume there exists $M>0$ and $t_1\geq t_0$ such that for $|c| \geq M$,
\begin{enumerate}
  \item[(i)] $\int_{\delta(t)}^{\sigma(t)} \sum_{i=1}^n \frac{1}{c} f_i(\tau,c)(\tau-\delta(\tau))\Delta \tau \leq \xi<1, \:$ for $t \in [t_1,\infty)_\T$;
  \item[(ii)] $\left|\sum_{i=1}^n f_i(\tau,c)\right| \leq \sum_{i=1}^n a_i(\tau)|c|$ where the $a_i$ are rd-continuous and nonnegative for $\tau \geq t_0$.
  \item[(iii)] $\left|\sum_{i=1}^n f_i(t,c)\right| \leq \sum_{i=1}^n f_i(t,y)$ for $|c|\leq y$, $t\in[\delta(t_0),\infty)_{\T}$.
\end{enumerate}
Then every solution $x$ of \eqref{feq}, \eqref{delaysol} for bounded initial function $\psi$ is bounded and satisfies
$$ \limsup_{t\rightarrow\infty}|x(t)|\leq M. $$ 
\end{theorem}

\begin{proof}
Appealing to assumption (ii), $x^{\Delta}(t)$ exists and is finite for each $t\in[t_0,\infty)_{\T}$, so that solutions of 
\eqref{feq}, \eqref{delaysol} are global. Suppose $x$ is an unbounded solution of \eqref{feq}, \eqref{delaysol} with 
bounded initial function $\psi$. By Lemma \ref{bddosc}, $x$ is also oscillatory. As in Case 2 of the proof of Theorem
\ref{thm4.3}, without loss of generality there exists a sequence $\lt\{t_j\rt\}_{j=1}^{\infty}$ in $\T$ such that
$\lim_{j \ra \infty} t_j = \infty$, $M \leq x(\sigma(t_j)) < x(\sigma(t_{j+1}))$ with $x(\sigma(t_j))\ra\infty$ as 
$j \ra \infty$, and there exist corresponding $t^*_j\in[\delta(t_j),t_j]_{\T}$ satisfying $x(t^*_j)\leq 0$. Then
$$ x(\sigma(t_j)) \leq x(\sigma(t_j))-x(t^*_j) 
    = -\int_{t^*_j}^{\sigma(t_j)} \lt( \int_{\delta(\tau)}^{\tau} \sum_{i=1}^n f_i(\tau,x(s)) \Delta s \rt)\Delta \tau, $$
so that by (iii) we have
\begin{eqnarray*}
 x(\sigma(t_j)) &\leq& \int_{t^*_j}^{\sigma(t_j)} \lt( \int_{\delta(\tau)}^{\tau} \left|\sum_{i=1}^n f_i(\tau,x(s))\right| \Delta s \rt)\Delta \tau\\
  &\leq& \int_{\delta(t_j)}^{\sigma(t_j)} \int_{\delta(\tau)}^{\tau} \sum_{i=1}^n f_i(\tau,x(\sigma(t_j)))\Delta s \Delta \tau.
\end{eqnarray*}
But then 
$$ \xi < 1 \leq \int_{\delta(t_j)}^{\sigma(t_j)} \frac{\tau-\delta(\tau)}{x(\sigma(t_j))} \sum_{i=1}^n f_i(\tau,x(\sigma(t_j))) \Delta \tau, $$ 
a contradiction of (i).  Therefore $x$ must be bounded.  In the same way, if
$$ \limsup_{t\rightarrow\infty} |x(t)| = \bar{M} > M, $$
there exists a sequence $\{\bar{t}_j\}_{j=1}^{\infty}$ as before, which would likewise lead to a contradiction, whereby
the conclusion of the theorem holds.
\end{proof}

%%%%%%%%%%%
% Example %
%%%%%%%%%%%

\begin{example}
Let $\T=\Z$, the set of integers, the delay function $\delta(t):=t-\lambda$ for some positive integer
$\lambda$, and $f_1(t,x(s)):=kx(s)$ for some constant $k$. Then the five conditions of Theorem \ref{thm4.3} are met
if $0<k<\frac{\xi}{\lambda(\lambda+1)}$ for any $\xi\in(0,1)$, whereby the trivial solution of \eqref{feq} is
asymptotically stable.
\end{example}
\noindent For $\T=\Z$, $n=1$ and $f_1$ as above, \eqref{feq} becomes the difference equation
$$ x(t+1) = x(t) - \sum_{s=t-\lambda}^{t-1}kx(s), \quad t\in\Z. $$
Fix $\xi\in(0,1)$, $\lambda\in\N$, and $0<k<\frac{\xi}{\lambda(\lambda+1)}$. To check (i), note that
$$ \sum_{\tau=t-\lambda}^t \frac 1c kc(\tau-(\tau-\lambda))=k\lambda(\lambda+1)\leq \xi<1, \quad t\in\Z. $$
In (ii), $\sum_{\tau=t_0}^{\infty}k|c|\lambda=\infty$. For (iii), $|f_1(\tau,c)|=k|c|$, so we take $a_1(\tau)\equiv k>0$,
which is rd-continuous. Since $p(t):=a_1(t)(t-\delta(t))=k\lambda>0$, $p\in\mathcal{R}^+$, and (iv) holds. Finally,
(v) is met as $|f_1(\tau,x)|=k|x|\leq ky=f_1(\tau,y)$ for $|x|\leq y \leq M$ with $\tau\in[t_0,\infty)_{\T}$.

%%%%%%%%%%%
% Example %
%%%%%%%%%%%

\begin{example}
Equation \eqref{eq1} is a special case of \eqref{feq} if $\T$ is an isolated time scale.
\end{example}
\noindent If $\T$ is an isolated time scale, then the delay dynamic equation \eqref{feq} with 
$f_1(t,x)=a(t)\delta^{\Delta}(t)x$ and $f_i(t,x)=0, 2 \leq i \leq n$ becomes
$$ x^\Delta (t) = - \int_{\tau(t)}^t a(t)\delta ^\Delta (t) x(s) \Delta_s g(t,s), $$
where $\tau(t) \leq \delta(t)$ are delay functions and
$$ g(t,s)=\begin{cases} 0, & \tau(t)\leq s \leq \delta(t)\\ 1, & s > \delta(t). \end{cases} $$
Then (the Dirac delta function)
$$ g^{\Delta_s}(t,s)=\begin{cases} \frac{1}{\mu(\tau)}, & s = \delta(t)\\ 0, & \mbox{otherwise.} \end{cases} $$
Using the above we obtain
\begin{eqnarray*}
x^\Delta (t) &=& - \int_{\tau(t)}^t a(t)\delta ^\Delta (t) x(s) \Delta_s g(t,s)\\
 &=& - a(t)\delta ^\Delta (t) \int_{\tau(t)}^t  x(s)  g^{\Delta_s}(t,s) \Delta s\\
 &=& - a(t)\delta ^\Delta (t) \int_{\delta(t)}^{\sigma(\delta(t))} x(s) g^{\Delta_s}(t,s) \Delta s\\
 &=&- a(t)\delta ^\Delta(t) x(\delta(t))\frac{1}{\mu(\delta(t))} \mu(\delta(t))\\
 &=&- a(t)\delta ^\Delta(t) x(\delta(t)).
\end{eqnarray*}

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\end{document}