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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 12, 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 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/12\hfil Boundedness and exponential stability]
{Boundedness and exponential stability of solutions to dynamic
equations on time scales}

\author[A.-L. Liu\hfil EJDE-2006/12\hfilneg]
{Ai-Lian Liu} 

\address{Ai-Lian Liu \newline
Department of Statistics and Mathematics,
Shandong Economics University, \newline
Jinan 250014, China}
\email{ailianliu2002@163.com}

\thanks{Submitted January 16, 2006. Published January 8, 2007.}
\thanks{Supported by grants 10371135 and 10571183 from NNSF of China,
 and by grant 05M12  \hfill\break\indent from  Zhongshan University
Advanced Research Centre}
\subjclass[2000]{34K11, 34K40, 39A12}
\keywords{Time scale; Lyapunov-type function on time scales;
 boundedness; \hfill\break\indent exponential stability}

\begin{abstract}
 Making use of the generalized time scales exponential function,
 we give a new definition for the exponential stability of solutions
 for dynamic equations on time scales. Employing Lyapunov-type
 functions on time scales, we investigate the boundedness and the
 exponential stability of solutions  to first-order dynamic equations
 on time scales, and some sufficient conditions are obtained.
 Some examples are given at the end of this paper.
\end{abstract}

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

\section{Introduction}

In this paper, we consider the boundedness and exponential
stability of solutions to the first-order dynamic equations
\begin{equation}
x^\Delta =f(t,x)\quad t\geq t_0,\;t \in \mathbb{T} , \label{e1.1}
\end{equation}
 subject to the initial condition
\begin{equation}
x(t_0)=x_0\quad t_0\in \mathbb{T},\;x_0\in \mathbb{R}^n,
\label{e1.2}
\end{equation}
 where $f:\mathbb{T} \times \mathbb{R}^n\to \mathbb{R}^n$ is a
so called rd-continuous function and $t$ is from a
so called  ``time scale'' $\mathbb{T}$.

If $\mathbb{T}=\mathbb{R}$, then $x^\Delta=x'$ and \eqref{e1.1},
\eqref{e1.2} is the following initial value-problem for ordinary
differential equations,
\begin{gather}
x'=f(t,x), \label{e1.3} \\
x(t_0)=x_0. \label{e1.4}
\end{gather}
If $\mathbb{T}=\mathbb{Z}$, then $x^\Delta=\Delta x$
(the forward difference calculus),
 and \eqref{e1.1}, \eqref{e1.2} corresponds to the
initial value-problem  for the O$\Delta$E
\begin{gather}
x(n+1)-x(n)=f(n,x(n)) , \label{e1.5} \\
x(n_0)=x_0.  \label{e1.6}
\end{gather}

Recently Raffoul~\cite{you} used Lyapunov-type function to
formulate some sufficient conditions that ensure all solutions of
\eqref{e1.3}, \eqref{e1.4} are uniformly bounded. In \cite{ac}, by
a kind of suitable and easy-to-calculate Lyapunov type I function
on time scales, Peterson and Tisdell formulated some appropriate
inequalities on these functions that guarantee solutions to
\eqref{e1.1}, \eqref{e1.2} are uniformly bounded. Other results on
boundedness can be found, for example, in \cite{ph} and \cite{ht}.

The investigation of stability analysis of nonlinear systems has
produced a vast body of important results. In \cite{mu}, Muhammad
made use of non-negative definite Lyapunov functions to study the
exponential stability of the zero solution of nonlinear discrete
system \eqref{e1.5}, \eqref{e1.6}, and in \cite{nm}, \cite{sy} they
gave sufficient conditions for the exponential stability of a
class of nonlinear time-varying differential equations.

 In this paper, we first define the boundedness and the exponential
stability of solutions  on time scales, then making use of the
Lyapunov-type function on time scales, we get  sufficient
 conditions that
guarantee the boundedness and exponential stability of zero
solution to \eqref{e1.1}, \eqref{e1.2}, which generalize the
results in \cite{mu,nm,sy}. Some examples are also
presented at the end of this paper.

Throughout this paper, the following notation will be used:
$\mathbb{R}^n$ is the $n$-dimen\-sional
 Euclidean vector space; $\mathbb{R}^+$ is the set of all non-negative
 real numbers; $\|x\|$ is
 the Euclidean norm of a vector $x\in\mathbb{R}^n$.

\section{Preliminaries}

The theory of dynamic equations on time scales (or more generally,
measure chains) was introduced in Stefan Hilger's PhD thesis in
1988. The theory presents a structure where, once a result is
established for general time scales, special cases include a
result for differential equations (obtained by taking the time
scale to be the real numbers) and a result for difference
equations (obtained by taking the time scales to be the integers).
A great deal of work has been done since 1988, unifying the theory
of differential equations and the theory of difference equations
by establishing the corresponding results in time scale setting.

\begin{definition}\label{def2.1} \rm
 A \emph{time scale} $\mathbb{T}$ is an
arbitrary nonempty closed subset of the set $\mathbb{R}$, with the
topology and ordering inherited from $ \mathbb{R}$. We
assume throughout the paper that $\mathbb{T}$ is unbounded above.
\end{definition}

\begin{definition}\label{def2.2} \rm
 For $t\in \mathbb{T}$ \emph{the forward jump
operator} $\sigma(t):\mathbb{T}\to \mathbb{T}$ is defined by
$$
\sigma(t):=\inf\{s\in {\mathbb{T}}:s>t\}.
$$
The jump operator $\sigma$ gives the classifications of points on
time scales. A point $t\in \mathbb{T}$ is called
\emph{right dense}
if $\sigma(t)=t$, and \emph{right scattered} if $\sigma(t)>t$.
\emph{The graininess function} $\mu(t):\mathbb{T}\to \mathbb{R}$ is
defined by $\mu(t)=\sigma(t)-t$.
\end{definition}

\begin{definition}\label{def2.3} \rm
 Fix $t\in \mathbb{T}$ and let $x:\mathbb{T}\to \mathbb{R}
^n$, define the \emph{delta-derivative} $x^\Delta(t)$ of $x$
at $t \in \mathbb{T}$ to be the vector (if it exists) with the property
that given any $\epsilon>0$, there is a neighborhood
 $U\subset\mathbb{T}$  of   $t$ such that
$$
|[x_i (\sigma(t))-x_i(s)]-x_i^\Delta (t)[\sigma (t)-s]|
\leq \epsilon |\sigma(t)-s|
$$
for all $s\in U$ and $i=1,2,\dots,n.$
 At this
time we say $x(t)$ is (delta) differentiable. In the case of
$\mathbb{T}=\mathbb{R}$, $x^\Delta (t)= x^\prime (t)$.
When $\mathbb{T}=\mathbb{Z}$, $x^\Delta(t)$ is the standard forward
difference operator $x(n+1)-x(n)$.
\end{definition}

\begin{definition}\label{def2.4} \rm
 For $f:\mathbb{T}\to \mathbb{R}^n$ and $F: \mathbb{T}\to \mathbb{R}^n$, if
$F^\Delta(t)=f(t)$ for all  $t\in \mathbb{T}$, then $F$ is said to be an
\emph{antiderivative} of $f$. And define the \emph{cauchy
integral} by the formula
$$
\int_a^b f(\tau) \Delta \tau = F(b)-F(a)\quad \mbox{for }a,b\in \mathbb{T}.
$$
\end{definition}

\begin{definition}\label{def2.5} \rm
 A function $f:\mathbb{T}\to \mathbb{R}$ is called
\emph{right-dense continuous} provided it is continuous at
right dense points of $\mathbb{T}$ and its left sided limit exists
(finite) at left dense points of $\mathbb{T}$. The set of all right-dense
continuous function on $\mathbb{T}$ is denoted by
$$
C_{rd}=C_{rd}(\mathbb{T})= C_{rd}(\mathbb{T},\mathbb{R}).
$$
Consequences of these definitions and properties can be found in
\cite{bp1,bp2,hilger}.
\end{definition}

\begin{definition}\label{def2.6} \rm
 A function $p: \mathbb{T}\to \mathbb{R}$ is
\emph{regressive} provided that
$$
1+\mu (t)p(t)\neq 0\quad \mbox{for all }t\in \mathbb{T}.
$$
The set of all regressive and right-dense continuous function is
denoted by $\mathcal{R}=\mathcal{R}(\mathbb{T})=\mathcal{R}(\mathbb{T},
\mathbb{R})$. The set $\mathcal{R} ^+$ of all \emph{positively
regressive} function is
$$
\mathcal{R} ^+ =\mathcal{R} ^+
(\mathbb{T},\mathbb{R})=\{p  \in \mathcal{R} :
  1+\mu (t) p(t) > 0  \mbox{ for all }t\in {\mathbb{T}} \}.
$$
\end{definition}

\begin{definition}\label{def2.7} \rm
 For  $p(t)\in \mathcal{R}$, we define the
\emph{generalized exponential function} as
 $$
e_p(t,t_0)=\exp(\int_{t_0}  ^t\frac{Log(1+\mu(\tau)p(\tau))}
{\mu(\tau)}\Delta\tau) \quad \mbox{for }t_0,t\in {\mathbb{T}}.
$$
 \end{definition}

\begin{remark}\label{rem2.8} \rm
 Consider the dynamic initial-value problem
\begin{equation}
x^\Delta=p(t)x,\quad x(t_0)=x_0, \label{e2.1}
\end{equation}
where $t_0\in\mathbb{T}$ and
$p(t)\in\mathcal{R}$. The exponential function $x(t)=x_0 e_p(t,t_0)$ is
 the unique solution to system \eqref{e2.1}.
 \end{remark}

\begin{theorem}\label{thm2.9}
\begin{itemize}
\item[(i)] If $p\in \mathcal{R}^+$, then $ e_p(t,t_0)>0$;
\item[(ii)] $e_p(\sigma(t),t_0)=(1+\mu(t)p(t))e_p(t,t_0)$;
\item[(iii)] $e_{\ominus
p}(t,t_0)=\frac{1}{e_p(t,t_0)}$, where
$$
\ominus p= \frac{-p}{1+p\mu(t)};
$$
 \item[(iv)]  If $p,q\in
\mathcal{R}$, then $e_p(t,t_0)e_q(t,t_0)=e_{p\oplus q}(t,t_0)$;

\item[(v)] If $p$ is a positive constant, then
$\lim_{t\to \infty }e_p(t,t_0)=\infty,
\lim_{t\to \infty }e_{\ominus p}(t,t_0)=0$.
\end{itemize}
\end{theorem}

Other relevant theorems can be found in \cite{bp1},\cite{bp2}. In
the following discussions, we assume conditions are imposed on
system \eqref{e1.1},\eqref{e1.2} such that the existence of
solutions is guaranteed when
$t\in \mathbb{T}_{t_0}^+=\{t\in\mathbb{T}:t\geq t_0\}$.

\section{Boundedness of Solutions}

\begin{definition}\label{def3.1} \rm
 We say a solution  $x(t)$  of  system
\eqref{e1.1},\eqref{e1.2} is \textit{bounded} if there exists a
constant $C(t_0,x_0)$ (that may depend on $t_0$ and $x_0$) such
that
$$\|x(t)\|\leq C(t_0,x_0)\quad\quad\quad \mbox{for}\quad
 t\in \mathbb{T}_{t_0}^+.
$$
We say that solutions of \eqref{e1.1},\eqref{e1.2} are
\textit{uniformly bounded} if $C$ is independent of $t_0$.
\end{definition}

 Assume $V:\mathbb{T}_{t_0}^+\times \mathbb{R}^n\ \to \mathbb{R}^+$
is delta differentiable in variable $t$ and continuously
differentiable in variable $x$, and $x(t)$ is any solution of
dynamic system \eqref{e1.1}, \eqref{e1.2}, then from
\cite{barna,chainrule} we know the delta derivative along $x(t)$ for
$V(t,x)$ is the following
\begin{align*}
V^\Delta (t,x)=V^\Delta (t,x(t))
&=V_t^\Delta(t,x(\sigma(t)))+
\int_0 ^1 V_x'(t,x(t)+h\mu(t)x^\Delta(t))dh\,x^\Delta(t)\\
&=V_t^\Delta(t,x(\sigma(t)))+\int_0^1
V_x'(t,x(t)+h\mu(t)f(t,x))dh\,f(t,x),
\end{align*}
where $V_t^\Delta$ is considered as the delta derivative in the
first variable $t$ and $V_x'$ is taken as the normal derivative
in variable $x$.
 Then we call $V(t,x)$ a \textit{Lyapunov-type function} on time scales.

To calculate the derivative is not an easy work generally, but if
$V(t,x)$ is explicitly
 independent of $t$, i.e. $V:\mathbb{R}^n\ \to \mathbb{R}^+$ and
$V(x)=V_1(x_1)+\dots +V_n(x_n)$, this is the type I Lyapunov
function introduced in \cite{ac} and $(V\circ x)^\Delta(t)$ is
easy to handle at this time.

 In this section we want to point out that the results in \cite{ac}
are held to be true
theoretically for general  Lyapunov-type function on time scales.

\begin{theorem}\label{thm3.2}
 Assume $D$ is an open and convex set in $\mathbb{R}^n$. Suppose
there exists a Lyapunov-type function $V:\mathbb{T}_{t_0}^+\times
D\to \mathbb{R}^+$ that satisfies
 \begin{gather}
 \lambda_1\|x\|^p\leq V(t,x)\leq \lambda_2\|x\|^q, \label{e3.1}
\\
 V^\Delta(t,x)\leq \frac{-\lambda_3\|x\|^r+L}{1+M\mu(t)}, \label{e3.2}
\\
 V(t,x)-V^{r/q}(t,x)\leq \gamma,   \label{e3.3}
 \end{gather}
where $\lambda_1,\lambda_2,\lambda_3, p,q,r$ are positive
constants, $L$ and $\gamma$ are
 nonnegative constants, and $M= \lambda_3/\lambda_2^{r/q}$.
Then all solutions of
 \eqref{e1.1}, \eqref{e1.2} that stay in $D$ are uniformly bounded.
 \end{theorem}

 \begin{proof} Note that $M= \lambda_3/\lambda_2^{r/q}$, so
  $M\in \mathcal{R}^+$ and $e_M(t,t_0)$ is well defined and is positive.
From the derivative formula   of products and condition \eqref{e3.2},
 \begin{align*}
(V(t,x)e_M(t,t_0))^\Delta
&= V^\Delta(t,x)e_M(\sigma(t),t_0)+ V(t,x)e_M ^\Delta(t,t_0)\\
&\leq \frac{-\lambda_3\|x\|^r+L}{1+M\mu(t)}(1+M\mu(t))e_M(t,t_0)+M V(t,x)e_M(t,t_0)\\
&= (-\lambda_3\|x\|^r+L+M V(t,x))e_M(t,t_0).
\end{align*}
 From \eqref{e3.1}, we have $ \|x\|^q\geq V(t,x)/\lambda_2$,
consequently $ -\|x\|^r\leq (V(t,x)/\lambda_2)^{r/q}$. So by
\eqref{e3.3},
 \begin{align*}
(V(t,x)e_M(t,t_0))^\Delta
&\leq [-(\lambda_3/\lambda_2^{r/q})V^{r/q}(t,x)
              + MV(t,x)+L]e_M(t,t_0)\\
&= [M(V(t,x)-V^{r/q}(t,x))+L]e_M(t,t_0)\\
&\leq (M\gamma +L)e_M(t,t_0).
\end{align*}
Integrating the above inequality from $t_0$ to $t$
$(t\in \mathbb{T}_{t_0}^+)$, we obtain
 \begin{align*}
V(t,x)e_M(t,t_0)
&\leq  V(t_0,x_0)+\frac{M\gamma+L}{M}(e_M(t,t_0)-e_M(t_0,t_0))\\
&\leq V(t_0,x_0)+\frac{M\gamma+L}{M}e_M(t,t_0)\\
&\leq \lambda_2\|x_0\|^q+\frac{M\gamma+L}{M}e_M(t,t_0).
\end{align*}
Hence
\begin{align*}
V(t,x)
&\leq \lambda_2\|x_0\|^q e_{\ominus M}(t,t_0)+\frac{M\gamma+L}{M}\\
&\leq \lambda_2\|x_0\|^q +\frac{M\gamma+L}{M}.
\end{align*}
 From \eqref{e3.1}, we have $\lambda_1\|x\|^p\leq V(t,x)$, which
implies
$$
\|x(t)\|\leq \big(\frac{1}{\lambda_1}\big)^{1/p}
\big(\lambda_2\|x_0\|^q +\frac{M\gamma+L}{M}\big)^{1/p}
\quad\mbox{for all } t\in\mathbb{T}_{t_0}^+.
$$
This completes the proof.
\end{proof}

\begin{theorem}\label{thm3.3}
 Assume $D\subset \mathbb{R}^n$ is  open and convex, and there
exists a Lyapunov-type function $V:\mathbb{T}_{t_0}^+\times D\to
\mathbb{R}^+ $ that satisfies
 \begin{gather}
 \lambda_1(t)\|x\|^p\leq V(t,x)\leq \lambda_2(t)\|x\|^q,  \label{e3.4}
\\
V^\Delta(t,x)\leq \frac{-\lambda_3(t)\|x\|^r+L}{1+M\mu(t)}, \label{e3.5}
\\
V(t,x)-V^{r/q}(t,x)\leq \gamma,   \label{e3.6}
 \end{gather}
for some positive constants $p,q,r$ and positive functions
$\lambda_1(t),\lambda_2(t),\lambda_3(t)$, where $\lambda_1(t)$ is
nondecreasing,  $L$ and $\gamma$ are
 nonnegative constants, and
$$
M=\inf_{t\in \mathbb{T}_{t_0 }^+} \lambda_3(t)/\lambda_2^{r/q}(t)>0.
$$
 Then all solutions of
 \eqref{e1.1}, \eqref{e1.2} that stay in $D$ are bounded.
 \end{theorem}

\begin{proof}
 For $M=\inf_{t\in \mathbb{T}_{t_0 }^+}\lambda_3(t)/\lambda_2^{r/q}(t)>0$,
by calculating $(V(t,x)e_M(t,t_0))^\Delta $ and then by the similar
argument as in Theorem \ref{thm3.2}, we obtain
\begin{align*}
V(t,x)
&\leq \lambda_2(t_0)\|x_0\|^q e_{\ominus M}(t,t_0)+\frac{M\gamma+L}{M}\\
&\leq \lambda_2(t_0)\|x_0\|^q +\frac{M\gamma+L}{M}.
\end{align*}
Using condition \eqref{e3.4}, we arrive at
$$
\|x\|\leq \big(\frac{V(t,x)}{\lambda_1(t)}\big)^{1/p}
\leq \big(\frac{V(t,x)}{\lambda_1(t_0)}\big)^{1/p}.
$$
Combining the above two inequalities, we get
$$
\|x(t)\|\leq \big(\frac{1}{\lambda_1(t_0)}\big)^{1/p}
\big(\lambda_2(t_0)\|x_0\|^q
+\frac{M\gamma+L}{M}\big)^{1/p}
\quad\mbox{for all } t\in\mathbb{T}_{t_0}^+,
$$
which concludes the proof.
\end{proof}

\section{Exponential Stability of Solutions}

For the dynamic equation \eqref{e1.1}, we assume $f(t,0)=0$ for all
$t\in \mathbb{T}$, so $x(t)=0$ is the trivial solution of \eqref{e1.1}.

 There are different  definitions for the
 exponential stability of the zero solution according
 to different authors (\cite{chr,jeff}). P\"{o}tzsche \cite{chr}
 gave the definition by the  regular exponential function
 $e^{-p(t-t_0)}$(constant $p>0$); and Dacunha \cite{jeff} defined the
 exponential stability in terms of $e_{-p}(t,t_0)$
(constant $p>0, -p\in\mathcal{R}^+$).
For constant $p>0, -p \in\mathcal{R}^+$, taking into consideration
the following relationship
 $$
e_{-p}(t,t_0)\leq e^{-p(t-t_0)}\leq e_{\ominus p}(t, t_0)
 \quad t,t_0\in\mathbb{T},t\geq t_0,
$$
 here we introduce a new  definition, which is more general than the present,
 by  the use of the generalized time scales exponential function
$e_{\ominus p}(t,t_0)$.

\begin{definition} \rm
The zero solution to system \eqref{e1.1} is \textit{exponentially
stable} if any solution $x(t)$ of \eqref{e1.1}, \eqref{e1.2}
satisfies
$$
\|x(t)\|\leq \beta (\|x_0\|,t_0)e_{\ominus p}^\alpha (t,t_0)\quad
    t\in \mathbb{T}_{t_0}^+,
$$
where $\beta: \mathbb{R}^+\times \mathbb{T}\to \mathbb{R}^+$ is a nonnegative
function, and $\alpha, p$ are positive constants. If $\beta
(\|x_0\|,t_0)$ does not depend on $t_0$, the zero solution is
called \textit{uniformly exponentially stable}.
\end{definition}

 For the rest of this article, to shorten expressions, instead of saying
the zero solution is stable, we say that
 the system \eqref{e1.1},\eqref{e1.2} is stable.

\begin{lemma}\label{lem4.2}
For any time scale $\mathbb{T}$, assume the graininess function $\mu(t)$
is bounded above, i.e. there exists a constant $ B_{\mathbb{T}}$ (that
may depend on $\mathbb{T}$) such that $\mu(t)\leq B_{\mathbb{T}}$,
then for any
positive constants $M,\delta$ satisfying $M<\delta$, we have
$$
t\to \int_{t_0}^t e_{M\ominus\delta}(\tau , t_0)\Delta\tau \quad
 \mbox{is   bounded   above}.
$$
\end{lemma}

\begin{proof} From the properties of generalized exponential function
(Theorem \ref{thm2.9}),
$$
\int_{t_0}^t e_{M\ominus\delta}(\tau , t_0)\Delta\tau=
    \int_{t_0}^t \frac{e_M(\tau , t_0)}{e_\delta (\tau,t_0)}\Delta\tau
    =\frac{1}{M}\int_{t_0}^t \frac{e_M^\Delta(\tau , t_0)}{e_\delta (\tau,t_0)}\Delta\tau.$$
By the integration by parts formula \cite[Theorem 1.77]{bp1},
we have
\begin{align*}
\int_{t_0}^t \! e_{M\ominus\delta}(\tau , t_0)\Delta\tau
&=
 \frac{1}{M}\Big(e_{M\ominus\delta}(t , t_0)-1-\int_{t_0}^t
    e_M(\sigma(\tau),t_0)\frac{-e_\delta^\Delta(\tau, t_0)}{e_\delta(\tau,t_0)
           e_\delta (\sigma(\tau),t_0)}\Delta\tau\Big)\\
&= \frac{1}{M}\Big(e_{M\ominus\delta}(t , t_0)-1+
              \delta \int_{t_0}^t\frac{e_M(\sigma(\tau),t_0)}{e_\delta
              (\sigma(\tau),t_0)} \Delta\tau  \Big)\\
&= \frac{1}{M}\Big(e_{M\ominus\delta}(t , t_0)-1+\delta
            \int_{t_0}^t\frac{(1+\mu(\tau)M)e_M(\tau,t_0)}{e_\delta
              (\sigma(\tau),t_0)} \Delta\tau \Big).
\end{align*}
Due to  the assumption, we have
\[
\int_{t_0}^t e_{M\ominus\delta}(\tau , t_0)\Delta\tau
\leq
 \frac{1}{M}\Big(e_{M\ominus\delta}(t , t_0)-1+\delta (1+B_{\mathbb{T}} M)
            \int_{t_0}^t\frac{e_M(\tau,t_0)}{e_\delta
              (\sigma(\tau),t_0)} \Delta\tau \Big).
\]
From the formula in \cite[Theorem 2.38]{bp1},
$$
\int_{t_0}^t\frac{e_M(\tau,t_0)}{e_\delta (\sigma(\tau),t_0)} \Delta\tau=
 \frac{1}{M-\delta}\int_{t_0}^te_{M\ominus\delta}^\Delta(\tau ,
t_0)\Delta\tau.
$$
Hence
\begin{align*}
\int_{t_0}^t e_{M\ominus\delta}(\tau , t_0)\Delta\tau
&\leq
  \frac{1}{M}\Big(  1-1+\frac{\delta(1+B_{\mathbb{T}}M)}{M-\delta}
  (e_{M\ominus\delta}(t , t_0)-1)\Big)\\
&=  \frac{1}{M}\Big(\frac{\delta(1+B_{\mathbb{T}}M)}{\delta-M}-
  \frac{\delta(1+B_{\mathbb{T}}M)}{\delta-M} e_{M\ominus\delta}(t , t_0) \Big)\\
&\leq  \frac{\delta(1+B_{\mathbb{T}}M)}{M(\delta-M)}
=  \mbox{Constant}.
\end{align*}
That completes the proof.
\end{proof}

The results obtained in this section are under the assumption that
Lemma \ref{lem4.2} holds to be true, i.e., throughout this section
we assume $\sup_{t\in\mathbb{T}_{t_0}^+}\mu(t)<\infty$ with its
bound dependent on the time scale $\mathbb{T}$.

\begin{theorem}\label{thm4.3}
 Assume $D\subset \mathbb{R}^n$ is an open and convex set containing
the origin and there exists a Lyapunov-type function
$V:\mathbb{T}_{t_0}^+\times D\to \mathbb{R}^+$ which satisfies
\begin{gather}
\lambda_1\|x\|^p\leq V(t,x) \leq\lambda_2\|x\|^q, \label{e4.1}
\\
V^{\Delta}(t,x)\leq \frac{-\lambda_3\|x\|^r+K
e_{\ominus\delta}(t,t_0)}{1+\mu(t)M},   \label{e4.2}
\end{gather}
where $\lambda_1, \lambda_2, \lambda_3, K, p,q,r,\delta$ are
positive numbers and the following two conditions hold for all
$(t,x)\in \mathbb{T}_{t_0}^+\times D$
\begin{equation}
\delta > \lambda_3/(\lambda_2)^{r/q} =M,   \label{e4.3}
\end{equation}
and there  exists  a $\gamma \geq 0$ such that
\begin{equation}
 V(t,x)-V^{r/q}(t,x)\leq\gamma e_{\ominus\delta}(t,t_0).
 \label{e4.4}
\end{equation}
 Then system \eqref{e1.1}, \eqref{e1.2} is uniformly exponentially stable.
 \end{theorem}

\begin{proof} Let $x(t)$ be a solution of \eqref{e1.1}, \eqref{e1.2} and let
 $$
Q(t,x)=V(t,x)e_M(t,t_0).
$$
 Then
$$
Q^\Delta (t,x)=V^\Delta (t,x)e_M(\sigma(t),t_0)+V(t,x)e_M^\Delta (t,t_0).
$$
 Taking \eqref{e4.2} into account, for all $t\in \mathbb{T}_{t_0}^+, x\in D$,
we have
 $$
Q^\Delta (t,x)\leq \frac{-\lambda_3\|x\|^r+Ke_{\ominus\delta}(t,t_0)}{
 1+\mu(t)M}(1+\mu(t)M)e_M(t,t_0)+V(t,x)M e_M(t,t_0).
$$
 By \eqref{e4.1}, we have $\|x\|^q\geq V(t,x)/\lambda_2$, or
equivalently
 $-\|x\|^r\leq -(V(t,x)/\lambda_2)^{r/q}$.
 Therefore,
 $$
Q^\Delta (t,x)\leq \Big( -V^{r/q}(t,x)\lambda_3/\lambda_2^{r/q}
+Ke_{\ominus\delta}(t,t_0) +MV(t,x)\Big)e_M(t,t_0).
$$
 Since
$\lambda_3/\lambda_2^{r/q}=M$,
 we have
 $$
Q^\Delta (t,x)\leq M(V(t,x)-V^{r/q}(t,x))e_M(t,t_0)
+Ke_{M\ominus \delta}(t,t_0).
$$
 Using \eqref{e4.4}, we obtain
 $$
Q^\Delta (t,x)\leq  (M\gamma+K)e_{M\ominus \delta}(t,t_0).
$$
 Integrating both sides of the above inequality from $t_0$ to $t$, we obtain
\[
Q(t,x)-Q(t_0,x_0)\leq \int_{t_0}^t (M\gamma+K)
e_{M\ominus \delta}(\tau ,t_0)\Delta \tau
=  (M\gamma+K)\int_{t_0}^te_{M\ominus \delta}(\tau ,t_0)\Delta \tau.
\]
 From Lemma \ref{lem4.2}, we have
\[
Q(t,x)-Q(t_0,x_0)\leq  \frac{(M\gamma+K)}{M}
\frac{\delta(1+B_{\mathbb{T}} M)}{\delta-M}.
\]
Since $Q(t_0,x_0)=V(t_0,x_0)\leq \lambda_2\|x_0\|^q$, we have
\[
Q(t,x)\leq \lambda_2\|x_0\|^q+\frac{(M\gamma+K)}{M}
\frac{\delta(1+B_{\mathbb{T}} M)}{\delta-M}=: \beta(\|x_0\|).
\]
We have $Q(t,x)\leq \beta(\|x_0\|)$. On the other hand, from
\eqref{e4.1}  it follows that
$$
\|x\|\leq \Big(\frac{V(t,x)}{\lambda_1}\Big)^{1/p}.
$$
Substituting $V(t,x)=Q(t,x)e_{\ominus M}(t,t_0)$ in the last
inequality, we obtain
\[
\|x(t)\|\leq \Big( \frac{Q(t,x)e_{\ominus
M}(t,t_0)}{\lambda_1}\Big)^{1/p}
\leq \Big(\frac{\beta(\|x_0\|)}{\lambda_1}\Big)^{1/p}
e_{\ominus M}^{1/p}(t,t_0).
\]
This inequality shows that system \eqref{e1.1}, \eqref{e1.2} is
uniformly exponentially stable. Therefore
the  proof is complete.
\end{proof}

\begin{theorem}\label{thm4.4}
  Assume $D\subset \mathbb{R}^n$ is an open and convex set
containing the origin and there exists a Lyapunov-type function
$V:\mathbb{T}_{t_0}^+\times D\to \mathbb{R}^+$ which satisfies
\begin{gather}
\lambda_1(t)\|x\|^p\leq V(t,x) \leq\lambda_2(t)\|x\|^q, \label{e4.5}
\\
V^{\Delta}(t,x)\leq \frac{-\lambda_3(t)\|x\|^r+K
e_{\ominus\delta}(t,t_0)}{1+\mu(t)M}, \label{e4.6}
\\
\delta > \inf_{t\in \mathbb{T}_{t_0}^+}\lambda_3(t)/(\lambda_2(t))^{r/q}=M>0,
 \label{e4.7}
\\
\exists  \gamma \geq 0,\mbox{ such that }
 V(t,x)-V^{r/q}(t,x)\leq\gamma e_{\ominus\delta}(t,t_0),  \label{e4.8}
\end{gather}
 where $\lambda_1(t), \lambda_2(t), \lambda_3(t)$ are positive functions,
 $\lambda_1(t)$  is nondecreasing for all $t\in \mathbb{T}_{t_0}^+$,
and $K, p,q,r,\delta$ are positive constants.
 Then system \eqref{e1.1}, \eqref{e1.2} is exponentially stable.
 \end{theorem}

 \begin{proof} We consider the function
$$
Q(t,x)=V(t,x)e_M(t,t_0).
$$
 By a similar argument used in  the proof of Theorem \ref{thm4.3},
we arrive at
 $$
Q^\Delta (t,x)\leq\Big(-\lambda_3(t)\|x\|^r+Ke_{\ominus\delta}(t,t_0)
\Big)e_M(t,t_0)  +V(t,x)M e_M(t,t_0).
$$
 Taking condition \eqref{e4.5} into account and  by the assumption
$\lambda_2(t)>0$ for all  $t\in \mathbb{T}_{t_0}^+$, we have
 $\|x\|^q\geq V(t,x)/\lambda_2(t)$, so equivalently
 $-\|x\|^r\leq -(V(t,x)/\lambda_2(t))^{r/q}$.
  Therefore,
$$
Q^\Delta (t,x)\leq \Big( -V^{r/q}(t,x)\lambda_3(t)/\lambda_2(t)^{r/q}
+Ke_{\ominus\delta}(t,t_0)  +MV(t,x)\Big)e_M(t,t_0).
$$
 Since $\lambda_3(t)/\lambda_2(t)^{r/q}\geq M$,
 by condition \eqref{e4.8}, we obtain
\begin{align*}
Q^\Delta (t,x)
& \leq  M(V(t,x)-V^{r/q}(t,x))e_M(t,t_0)+K e_{M\ominus \delta}(t,t_0)\\
&\leq  (M\gamma+K)e_{M\ominus \delta}(t,t_0).
\end{align*}
Thus integrating both sides of the above inequality from $t_0$ to
$t$ and applying Lemma \ref{lem4.2},
\begin{align*}
Q(t,x)-Q(t_0,x_0)
&\leq \int_{t_0}^t (M\gamma+K)e_{M\ominus \delta}(\tau ,t_0)\Delta \tau\\
&=  \frac{(M\gamma+K)}{M} \frac{\delta(1+B_{\mathbb{T}} M)}{\delta-M}.
\end{align*}
Since $Q(t_0,x_0)=V(t_0,x_0)\leq \lambda_2(t_0)\|x_0\|^q$, we have
\begin{align*}
Q(t,x) \leq
\lambda_2(t_0)\|x_0\|^q+\frac{(M\gamma+K)}{M}
\frac{\delta(1+B_{\mathbb{T}} M)}{\delta-M}
=:\beta(\|x_0\|,t_0).
\end{align*}
 Furthermore, from
\eqref{e4.5}  it follows that
$$
\|x\|\leq \Big(\frac{V(t,x)}{\lambda_1(t)}\Big)^{1/p}.
$$
Since $\lambda_1(t)$ is non-decreasing, hence
$\lambda_1(t)\geq\lambda_1(t_0)$. So
$$
\|x\|\leq \Big(\frac{V(t,x)}{\lambda_1(t_0)}\Big)^{1/p}.
$$
Substituting $V(t,x)=Q(t,x)e_{\ominus M}(t,t_0)$ into the last
inequality, we obtain
\[
\|x(t)\|\leq  \Big( \frac{Q(t,x)e_{\ominus
M}(t,t_0)}{\lambda_1(t_0)}\Big)^{1/p}
\leq \Big(\frac{\beta(\|x_0\|,t_0)}{\lambda_1(t_0)}\Big)^{1/p}
e_{\ominus M}^{1/p}(t,t_0).
\]
This inequality shows that system \eqref{e1.1}, \eqref{e1.2} is
exponential stable.
\end{proof}

\begin{theorem} \label{thm4.5}
Assume $D\subset \mathbb{R}^n$ is an open and convex set containing
the origin, and there exists a Lyapunov-type function
 $V:\mathbb{T}_{t_0}^+\times D\to \mathbb{R}^+$ such that
\begin{gather*}
\lambda_1(t)\|x\|^p\leq V(t,x)\leq \Phi (\|x\|),\\
V^{\Delta}(t,x)\leq \frac{\Psi(\|x\|)+L e_{\ominus \delta }(t,t_0)}{1+\mu(t)},\\
\Psi(\Phi ^{-1}(V(t,x)))+V(t,x)\leq \gamma e_{\ominus \delta }(t,t_0),
\end{gather*}
where $\Phi:[0, \infty)\to[0,\infty)$,
$\Psi:[0, \infty)\to(-\infty,0]$,
$\lambda_1 (t): \mathbb{T}_{t_0}^+\to (0,+\infty)$,
$\Psi$ is nonincreasing,  $\lambda_1(t), \Phi$ is nondecreasing,
and $\Phi^{-1}$ exists, $L$ and $\gamma$ are nonnegative
 constants, $\delta >1$.
Then system \eqref{e1.1}, \eqref{e1.2} is uniformly exponentially
 stable.
 \end{theorem}

 \begin{proof}
Let $x(t)$ be a solution of system \eqref{e1.1}, \eqref{e1.2},
then
  \begin{align*}
    [V(t, x)e_1(t, t_0)]^{\Delta}
&= V^{\Delta}(t, x)e_1(\sigma(t),t_0)+V(t,x)e_1^{\Delta}(t, t_0)\\
&\leq  \frac{\Psi(\|x\|)+L e_{\ominus \delta }(t,t_0)}{1+\mu(t)}
(1+\mu(t))e_1(t,t_0)     +V(t,x)e_1(t,t_0)\\
&\leq  (\Psi(\Phi ^{-1}(V(t,x)))+Le_{\ominus \delta }(t,t_0) )e_{1 }(t,t_0)+
     V(t, x)e_1(t,t_0)\\
&\leq  (\gamma +L) e_{1\ominus \delta }(t,t_0)).
\end{align*}
  Integrating both sides from $t_0$ to $t$, we obtain
  \begin{align*}
  V(t, x(t))e_1(t, t_0)
&\leq  V(t_0,x_0)+(\gamma +L)\int_{t_0}^t
       e_{1\ominus \delta }(\tau,t_0)\Delta\tau\\
&\leq  V(t_0,x_0)+  (\gamma +L) \frac{\delta(1+B_{\mathbb{T}})}{\delta -1}
   =:\beta(\|x_0\|, t_0),
  \end{align*}
  so that
  $V(t, x(t))\leq \beta(\|x_0\|, t_0)e_{\ominus 1 }(t,t_0)$.
 From the assumption, we have
  $$
\|x\|\leq \Big(\frac{1}{\lambda_1(t)}\beta(\|x_0\|, t_0)
  e_{\ominus 1 }(t,t_0)\Big)^{1/p}
  \leq \Big(\frac{1}{\lambda_1(t_0)}\beta(\|x_0\|, t_0)\Big)^{1/p}
  e_{\ominus 1 }^{1/p}(t,t_0),$$
 which completes the proof.
 \end{proof}

\begin{remark} \rm
 In Theorems \ref{thm4.4} and \ref{thm4.5}, we can replace the
nondecreasing assumption of $\lambda_1(t)$ by the following
assumption:
There exists $a >0$  such that $a<M$  and
$$
 \lambda_1(t)\geq e_{\ominus a}(t,t_0),\quad
\mbox{for all } t\in \mathbb{T}_{t_0}^+,
$$
where $M= \inf_{t\in \mathbb{T}_{t_0}^+}\lambda_3(t)/(\lambda_2(t))^{r/q}$.
Take $M=1$ in Theorem \ref{thm4.5}.
\end{remark}

The following theorem does not require an upper bound on the
Lyapunov function $V(t,x)$.

 \begin{theorem} \label{thm4.7}
 Assume $D\subset \mathbb{R}^n$ is an open and convex set containing
the origin. Let $V:\mathbb{T}_{t_0}^+\times D\to \mathbb{R}^+$ be a
 given Lyapunov-type function satisfying
\begin{gather}
\lambda_1\|x\|^p\leq V(t,x),  \label{e4.9}
\\
V^{\Delta}(t,x)\leq \frac{-\lambda_2V(t,x)+K
e_{\ominus\delta}(t,t_0)} {1+\mu(t)\varepsilon},  \label{e4.10}
\end{gather}
for some positive constants
$\lambda_1, \lambda_2, p,K, \delta , \varepsilon $ with
 $\varepsilon \leq\lambda_2, \varepsilon <\delta$.
 Then system \eqref{e1.1}, \eqref{e1.2} is exponentially stable.
 \end{theorem}

 \begin{proof} Let
$$
Q(t,x)=V(t,x)e_{\varepsilon}(t,t_0).
$$
 By an argument similar to the one used in  the proof of the two theorems
above,  and taking into consideration
 conditions \eqref{e4.9}, \eqref{e4.10}, we arrive at
 \begin{align*}
 Q^\Delta (t,x)
&=  V^\Delta (t,x)e_{\varepsilon}(\sigma(t),t_0)+
 V(t,x)e_{\varepsilon}^\Delta (t,t_0)\\
 &\leq   \frac{-\lambda_2V(t,x)+Ke_{\ominus\delta}(t,t_0)}{
 1+\mu(t)\varepsilon }(1+\mu(t)\varepsilon)e_{\varepsilon}(t,t_0)+
\varepsilon V(t,x) e_{\varepsilon}(t,t_0) \\
 &= \left(-\lambda_2V(t,x)+Ke_{\ominus\delta}(t,t_0)\right)e_{\varepsilon}(t,t_0)+
 \varepsilon V(t,x) e_{\varepsilon}(t,t_0)\\
  &=  \left(-\lambda_2+\varepsilon\right)V(t,x)e_{\varepsilon}(t,t_0)+
  Ke_{\varepsilon\ominus\delta}(t,t_0)\\
 &\leq  Ke_{\varepsilon\ominus\delta}(t,t_0).
\end{align*}
 So by Lemma \ref{lem4.2},
\begin{align*}
V(t,x)e_{\varepsilon}(t,t_0)
&\leq V(t_0,x_0) + K \int_{t_0}^t e_{\varepsilon\ominus\delta}
(\tau,t_0)\Delta\tau\\
&\leq  V(t_0,x_0) +K\frac{\delta(1+B_{\mathbb{T}}\varepsilon)}
{\varepsilon(\delta-\varepsilon)}
=:\beta(\|x_0\|,t_0).
\end{align*}
Hence
$V(t,x)\leq   \beta(\|x_0\|,t_0)e_{\ominus\varepsilon}(t,t_0)$.
 From \eqref{e4.9}, we get
\[
\|x\| \leq \Big(\frac{V(t,x)}{\lambda_1}\Big)^{1/p}
  \leq \Big(\frac{\beta(\|x_0\|,t_0)}{\lambda_1}\Big)^{1/p}
 e_{\ominus\varepsilon}^{1/p}(t,t_0).
\]
This completes the proof.
\end{proof}

Now we will present some examples to illustrate the theory developed above.

\begin{example} \label{exa4.1} \rm
 Consider the dynamic equation
\begin{equation}
x^\Delta (t)=a\, x +R \,x^{1/3}\,
e_{\ominus\delta}^{1/3}\,(t,t_0), \label{e4.11}
\end{equation}
 where $a<0, a\in \mathcal{R}^+$,  $R>0$ and $\delta>0$.
If there exist positive constants $\lambda_3, K$ such that
\begin{equation}
\begin{gathered}
\delta>\lambda_3, \\
(1+\mu(t)\lambda_3)(2a+\mu(t)a^2+\frac{4}{3}R+\frac{4}{3}\mu(t)aR+\frac{1}{3}\mu(t)R^2)
\leq-\lambda_3,\\
(1+\mu(t)\lambda_3)(\frac{2}{3}R+\frac{2}{3}\mu(t)a
R+\frac{2}{3}\mu(t)R^2)\leq K,
\end{gathered} \label{e4.12}
\end{equation}
then system \eqref{e4.11} is uniformly exponentially stable.
\end{example}

To see this, let $V(t,x)=x^2$. By calculating $V^{\Delta}(t,x)$
along the solutions of \eqref{e4.11}, we obtain
\begin{align*}
V^\Delta (t,x)
&= 2x f(t,x)+\mu(t)(f(t,x))^2\\
 &= 2x\left(ax+Rx^{1/3}e_{\ominus\delta}^{1/3}(t,t_0)\right)+
    \mu(t)\left(ax+Rx^{1/3}e_{\ominus\delta}^{1/3}(t,t_0)\right)^2\\
 &=  \left(2a+\mu(t)a^2\right)x^2+\left(2Re_{\ominus\delta}^{1/3}(t,t_0)
  +2\mu(t)a   Re_{\ominus\delta}^{1/3}(t,t_0)\right)
 x^{4/3}\\
&\quad +\mu(t)R^2e_{\ominus\delta}^{2/3}(t,t_0)x^{2/3}.
\end{align*}
Using  the Young's inequality
($wz<\frac{w^e}{e}+\frac{z^f}{f}$
 with $\frac{1}{e}+\frac{1}{f}=1$), we have
 \begin{align*}
x^{4/3}e_{\ominus\delta}^{1/3}(t,t_0)&\leq
\Big(\frac{(x^{4/3})^{3/2}}{3/2}+
\frac{(e_{\ominus\delta}^{1/3}(t,t_0))^3}{3}\Big)
= \frac{2}{3}x^2+\frac{1}{3}e_{\ominus\delta}(t,t_0),\\
x^{2/3}e_{\ominus\delta}^{2/3}(t,t_0)&\leq
\Big(\frac{(x^{2/3})^{3}}{3}+
\frac{(e_{\ominus\delta}^{2/3}(t,t_0))^{3/2}}{3/2}\Big)
= \frac{1}{3}x^2+\frac{2}{3}e_{\ominus\delta}(t,t_0).
\end{align*}
Thus
\begin{align*}
V^\Delta (t,x)
&\leq
\left(2a+\mu(t)a^2\right)x^2+\left(2R+2\mu(t)a
   R\right)\Big(\frac{2}{3}x^2+\frac{1}{3}e_{\ominus\delta}(t,t_0)\Big)\\
&\quad  +\mu(t)R^2\Big( \frac{1}{3}x^2+\frac{2}{3}e_{\ominus\delta}(t,t_0)\Big)\\
&\leq  \Big(2a+\mu(t)a^2 +\frac{4}{3}R+\frac{4}{3}\mu(t)aR +
 \frac{1}{3}\mu(t)R^2 \Big)x^2 \\
&\quad +\Big(\frac{2}{3}R+\frac{2}{3}\mu(t)a R+\frac{2}{3}\mu(t)R^2 \Big)
 e_{\ominus\delta}(t,t_0)\\
&= \frac{1}{1+\mu(t)\lambda_3} \Big\{
\left(1+\mu(t)\lambda_3\right)\big(2a+\mu(t)a^2 +
\frac{4}{3}R+\frac{4}{3}\mu(t)aR +
\frac{1}{3}\mu(t)R^2 \big)x^2\\
& \quad +\left(1+\mu(t)\lambda_3\right)
\Big(\frac{2}{3}R+ \frac{2}{3}\mu(t)a
R+\frac{2}{3}\mu(t)R^2 \Big)
e_{\ominus\delta}(t,t_0)  \Big\}.
\end{align*}
Under the above assumptions, one can check that conditions
\eqref{e4.1}--\eqref{e4.4} of  Theorem \ref{thm4.3}
  are satisfied. Hence system
\eqref{e4.11} is uniformly exponentially stable.

In fact, if there exist constants $\lambda_3>0,\ K>0$, such that
\begin{equation}
\begin{gathered}
\delta>\lambda_3 ,\\
(1+\lfloor\mu\rfloor\lambda_3) (2a+\lceil\mu\rceil
a^2+\frac{4}{3}R+\frac{4}{3}\lfloor\mu\rfloor aR
+\frac{1}{3}\lceil\mu\rceil R^2)
\leq-\lambda_3 ,\\
(1+\lceil\mu\rceil\lambda_3)(\frac{2}{3}R+\frac{2}{3}
\lfloor\mu\rfloor a R+
\frac{2}{3}\lceil\mu\rceil R^2)\leq K,
\end{gathered} \label{e4.13}
\end{equation}
here $\lceil\mu\rceil=\sup_{t\in\mathbb{T}}\mu(t)$,
$\lfloor\mu\rfloor=\inf_{t\in\mathbb{T}}\mu(t) $, then
\eqref{e4.12} will hold evidently.

\noindent\textbf{Case 1:}
If $\mathbb{T}=\mathbb{R}$, then
$\mu(t)=\lceil\mu\rceil=\lfloor\mu\rfloor=0$ and the conditions in
\eqref{e4.13} reduce to that positive constants $\lambda_3, K$ need to
exist such that
$$
\delta>\lambda_3\leq -(2a +\frac{4}{3}R),\quad
\frac{2}{3}R\leq K,
$$
then system \eqref{e4.11} is uniformly exponentially stable.

\noindent\textbf{Case 2:}
If $\mathbb{T}=h\mathbb{Z}$, then
$\mu(t)=\lceil\mu\rceil=\lfloor\mu\rfloor=h$. The conditions in
\eqref{e4.13} reduce to that there exist $\lambda_3>0, K>0$ such that
\begin{gather*}
\delta>\lambda_3, \\
2a+ha^2+\frac{4}{3}R+\frac{4}{3}ha R+\frac{1}{3}h R^2
\leq-\lambda_3/(1+h\lambda_3),\\
\frac{2}{3}R+\frac{2}{3}h  a R+\frac{2}{3}hR^2\leq
K/(1+h\lambda_3),
\end{gather*}
then system \eqref{e4.11} is uniformly exponentially stable.

\noindent\textbf{Case 3:} When
$\mathbb{T}=\bigcup_{k=0}^\infty[k(l+h), k(l+h)+l)]$, here $l, h$
are positive constants,  this kind of time scales could exactly
describe many phenomena which are common in nature, such as the
life span of certain species and the change of electric circuit
with time progressing etc. \cite{bp1,RMD}. At this time,
$\lfloor\mu\rfloor=0$, $\lceil\mu\rceil=h$. If there exist
constants $\lambda_3 >0$, $K>0$, such that
\begin{gather*}
\lambda_3<\delta,\\
2a +h a^2+\frac{4}{3}R+\frac{1}{3}R^2 h\leq -\lambda_3,\\
(1+h\lambda_3)(\frac{2}{3}R+\frac{2}{3}R^2h)\leq K,
\end{gather*}
then \eqref{e4.13} will hold. So system \eqref{e4.11} is uniformly
exponentially stable.

\begin{example} \label{exa4.2} \rm
Consider the  system
\begin{gather}
x_1^{\Delta}=-a x_1+ax_2, \label{e4.14}\\
x_2^{\Delta}=-ax_1-ax_2,  \label{e4.15}\\
(x_1(t_0),x_2(t_0))=(c,d), \label{e4.16}
\end{gather}
for certain constants $a>0$, $-a\in\mathcal{R}^+$, and $c,d$ are
any constants. If there is a constant $\lambda_2>0$ such that for
all $t\in \mathbb{T}_{t_0}^+$
\begin{equation}
\lambda_2/(1+\lambda_2 \mu(t))\leq 2a(1-a\mu(t)), \label{e4.17}
\end{equation}
then system \eqref{e4.14}--\eqref{e4.16} is exponentially stable.
\end{example}

\begin{proof} We will show that, under the above assumptions, the
conditions of Theorem \ref{thm4.7} are satisfied. Choose $D=\mathbb{R}^2$,
$V(t,x)=\|x\|^2=x_1^2+x_2^2$, so \eqref{e4.9} holds. From \cite{ac}, we
have
\begin{align*}
V^{\Delta}(t, x)
&=  2x f(t,x) +\mu(t)\|f(t,x)\|^2\\
&= -2a(1-a\mu(t))\|x\|^2\\
&\leq \frac{-\lambda_2V(t,x)}{1+\lambda_2 \mu(t)}\\
&\leq \frac{-\lambda_2V(t,x)+Ke_{\ominus\delta}(t,t_0)}{1+\lambda_2 \mu(t)}\\
&= \frac{-\lambda_2V(t,x)+Ke_{\ominus\delta}(t,t_0)}{1+\varepsilon
\mu(t)}.
\end{align*}
Here we let $\varepsilon=\lambda_2$, $\delta>\varepsilon>0$, and
$K$ be arbitrary positive constant, so $(4.10)$ holds. Therefore,
system \eqref{e4.14}--\eqref{e4.16} is exponentially stable.

In fact, if there exists a constant $\lambda_2>0$, such that
\begin{equation}
2a(1-a\lceil\mu\rceil)(1+\lambda_2\lfloor\mu\rfloor)\geq \lambda_2,
\label{e4.18}
\end{equation}
then condition \eqref{e4.17} will hold.

\noindent\textbf{Case 1:} If $\mathbb{T}=\mathbb{R}$, then
$\mu(t)=\lceil\mu\rceil=\lfloor\mu\rfloor=0$ and \eqref{e4.18} will
hold to be true if there exists a constant $\lambda_2$ satisfying
$$
0<\lambda_2\leq 2a.
$$
 So system \eqref{e4.14}--\eqref{e4.16} is exponentially stable.

\noindent\textbf{Case 2:}
If $\mathbb{T}=h\mathbb{Z}$, then
$\mu(t)=\lceil\mu\rceil=\lfloor\mu\rfloor=h$. If we can find a
constant $\lambda_2>0$ such that
$$
\frac{\lambda_2}{1+h\lambda_2}\leq 2a(1-ah),
$$
then condition \eqref{e4.18} would hold. So system
\eqref{e4.14}--\eqref{e4.16} is exponentially stable.

\noindent\textbf{Case 3:}
If $\mathbb{T}=\bigcup_{k=0}^\infty[k(l+h), k(l+h)+l]$
(as in the above example), then $\lfloor\mu\rfloor=0$,
$\lceil\mu\rceil=h$. The condition \eqref{e4.18} reduces to that a
constant $\lambda_2$ exists such that
$$
0<\lambda_2\leq 2a(1-ah),
$$
so that system \eqref{e4.14}--\eqref{e4.16} is exponentially stable.
\end{proof}

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\end{document}
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