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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 110, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/110\hfil
 Stability of quasi-linear differential equation]
{Stability of quasi-linear differential equations
with transition conditions}

\author[W. Feng, Y. Liu\hfil EJDE-2008/110\hfilneg]
{Weizhen Feng, Yubin Liu}  % in alphabetical order

\address{Weizhen Feng\newline
Department of Mathematics, South China Normal University,
Guangzhou 510631, China}
\email{fengweizhen1@yahoo.cn}

\address{Yubin Liu \newline
Department of Mathematics,
Huizhou University, Huizhou 516007, China}
\email{gdliuyubin@163.com}

\thanks{Submitted January 14, 2008. Published August 15, 2008.}
\thanks{Supported by grants 011471 from the Natural Science Foundation of
Guangdong Province, \hfill\break\indent
 and 0120 from the Natural Science
Foundation of Guangdong Higher Education Bureau.}

\subjclass[2000]{35J20, 35J25}
\keywords{Differential equations on time scales;
 transition conditions; \hfill\break\indent
 impulsive differential equations; stability}

\begin{abstract}
 This paper investigates the  stability of quasi-linear
 differential equations on certain time scales with transition
 condition (DETC). We establish Sufficient conditions for
 stability and illustrate our results with examples.
\end{abstract}

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

\section{Introduction}

 The study of quasi-linear systems is valuable because it is an
 important transition from linear systems to nonlinear systems.
 B. Liu \cite{l1} introduced the stability of a class of quasi-linear
 impulsive hybrid systems by using the methods of Lyapunov functions
 and the linearization technique. Y. Liu \cite{l2} gained sufficient
 conditions of the exponential stability of a class of quasi-linear
 switched systems by using the Cauchy matrices of its
 corresponding linear systems.

 Differential equations  on certain time scales with transition
 conditions (DETS) are in some sense more general than dynamic
 equations on time scales. Akhmet \cite{a1} investigated DETS on the basic
 of reduction to the impulsive differential equations (IDE). A
 special transformation ($\psi$-substitution) was used in \cite{a1}
 which allowed the reduced IDE to inherit all similar properties of
 the corresponding DETC.

 In this paper, we make an attempt to investigate  the
 stability of the DETC
 \begin{equation} \label{e1.1}
\begin{gathered}
y'=A(t)y+f(t,y)+A_iY(t_{2i+1}),\quad t\in[t_{2i+1},t_{2i+2}],\;t\geq t^0,\\
 y(t_{2i+1})=B_iy(t_{2i})+J_i(y(t_{2i}))+y(t_{2i}),\quad t_{2i}\geq t^0,\\
 y(t^0)=y_0,\quad t^0\in T_0,
\end{gathered}
\end{equation}
where the time scale
$T_0=\bigcup_{i=0}^{+\infty}[t_{2i-1},t_{2i}],t_i<t_{i+1}$,
$t_{-1}<0<t_0, \lim_{i\to+\infty}t_i=+\infty$, and the derivative is
one sided at the boundary points of
$T_0$,
\[
Y(t_{2i+1})=\begin{cases}
y_0, & t_{2i+1}<t^0<t_{2i+2},\\
y(t_{2i+1}),& t_{2i+1}\geq t^0\,.
\end{cases}
\]
Note that \eqref{e1.1} is more general than
\begin{equation}
y^{\Delta}=A(t)y+f(t,y),\quad t\in T_0,\label{e1.2}
\end{equation}
since this equation can be written as
 \begin{equation}
\begin{gathered}
y'=A(t)y+f(t,y),\quad t\in[t_{2i+1},t_{2i+2}],\; t\geq t^0,\\
y(t_{2i+1})=[A(t_{2i})y(t_{2i})+f(t_{2i},y(t_{2i}))](t_{2i+1}-t_{2i})
+y(t_{2i}), \\
y(t^0)=y_0,\quad t^0\in T_0.
\end{gathered} \label{e1.3}
\end{equation}

\section{Preliminaries}

Let $\mathbb{R}^n$ denote the $n$-dimensional real space and
$\|A\|$ the norm of an $n\times n$ matrix $A$ induced by the
Euclidean vector norm, i.e.
$\|A\|=[\lambda_{\rm max}(A^TA)]^{\frac{1}{2}}$.
Let $\mathbb{R}^+=[0,+\infty)$,
$N_+=\{1,2,\dots\}$ and $N=\{0\}\cup N_+$.

 In the following discussion, we suppose that:
\begin{itemize}
\item[(A1)]  $A(t)\in C(T_0,\mathbb{R}^{n\times n})$;
$B_i,A_i\in \mathbb{R}^{n\times n}$;
$f(t,y)\in C(T_0\times \mathbb{R}^n,\mathbb{R}^n)$;
$J_i(y)\in C(\mathbb{R}^n,\mathbb{R}^n)$; $f(t,0)\equiv0$
and $J_i(0)\equiv0$ for $t\in T_0$ and $i\in N$.
\end{itemize}
 The following conditions will be used in some theorems:
\begin{itemize}
\item[(H2)] There exist $\delta^*>0$, $\sigma_i>0$ ($i\in N$) such that
$\|B_iy+J_i(y)+y\|\leq\sigma_i\|y\|$ whenever $\|y\|\leq \delta^*$;

\item[(H3)] $0<\lambda_1\leq\inf\{t_{2i}-t_{2i-1}\}\leq\sup_{i\in
N}\{t_{2i}-t_{2i-1}\}=\lambda<+\infty$,
\[
0<\lambda_3\leq\inf\{t_{2i+1}-t_{2i}\}
\leq\sup_{i\in N}\{t_{2i+1}-t_{2i}\}=\lambda_2<+\infty;
\]

\item[(H4)] $\|f(t,y)\|\leq\|F(t)\|\cdot\|y\|$, $(t,y)\in T_0\times
\mathbb{R}^n$;

\item[(H5)] for all $\varepsilon\geq0$ there exists $\delta(\varepsilon)>0$,
such that $\|f(t,y)\|\leq\varepsilon\|y\|$ for all $t\in T_0$
whenever $\|y\|\leq\delta$.

\end{itemize}
 Without loss of generality, we assume $t^0\in[t_{2m-1},t_{2m}]$
and $t^0\geq 0$. Denote
$d(T_0[t^0,t])=t-t_{2p+1}+\sum_{i=m+1}^p(t_{2i}-t_{2i-1})+(t_{2m}-t^0)$
for $t\in[t_{2p+1},t_{2p+2}]$, and
$T_0{[a,+\infty)}=T_0\cap[a,+\infty)$.

A function $y: [t^0,+\infty)\cap T_0\to \mathbb{R}^n$ is said to
be a solution of \eqref{e1.1} if
\begin{itemize}
\item[(i)]  $y(t^0)=y_0$;
\item[(ii)] $y'=A(t)y+f(t,y)+A_iY(t_{2i+1})$ if
$t\in[t_{2i+1},t_{2i+2}]$ and $t\geq t^0, i\in N$;
\item[(iii)] $ y(t_{2i+1})=B_iy(t_{2i})+J_i(y(t_{2i}))+y(t_{2i})$,
$t_{2i}\geq t^0$, $i\in N$.
\end{itemize}

 Denote the solution of \eqref{e1.1} as $y(t,t^0,y_0)$.

\begin{definition}[\cite{a1}] \label{def2}\rm
The $\psi$-substitution on the set
$T_0'=T_0\setminus\bigcup_{i=0}^{+\infty}\{t_{2i-1}\} $
is defined as
\begin{equation}
\psi(t)=t-\sum_{0\leq t_{2k}<t}\delta_k,\quad t\in T_0',\; t\geq 0,\label{e2.1}
\end{equation}
where $\delta_k=t_{2k+1}-t_{2k}$.
\end{definition}

\begin{definition} \label{def3} \rm
(I) The zero solution of \eqref{e1.1} is said to be
uniformly stable, if for all $\varepsilon>0$, there
exists $\delta(\varepsilon)>0$, such that for any
$t^0\in T_0,\|y_0\|<\delta$ implies $\|y(t,t^0,y_0)\|<\varepsilon$ for all
$t\in T_0{[t^0,+\infty)}$;

 (II) The zero solution of \eqref{e1.1} is said to be
uniformly attractive, if
there exists $\delta_0>0$, such for all $\varepsilon>0$,
there exists $T(\varepsilon)>0$
such that for any $t^0\in T_0,\|y_0\|<\delta_0$ implies
$\|y(t,t^0,y_0)\|<\varepsilon$ for all
$t\in T_0{[t^0+T,+\infty)}$;

(III) The zero solution of \eqref{e1.1} is said to be uniformly
asymptotically stable if the zero solution of \eqref{e1.1} is uniformly
stable and uniformly attractive;

(IV) The zero solution of \eqref{e1.1} is said to be exponentially stable,
if there exists $\alpha>0$ such that for all $\varepsilon>0$,
there exists $\delta(\varepsilon)>0$,
such that for any $t^0\in T_0,\|y_0\|<\delta$ implies
$\|y(t,t^0,y_0)\|<\varepsilon e^{-\alpha(t-t^0)}$ for all $t\in
T_0{[t^0,+\infty)}$.
\end{definition}

 Let $x(s^+)=\lim_{h\to0^+}x(s+h), x(s)=x(s^-)=\lim_{h\to0^-}x(s+h)$.

\begin{lemma}[\cite{a1}] \label{lem1}
$\psi(t)$ is a one-to-one map, $\psi(0)=0$, $\psi(T_0')=R$.
The inverse transformation is
\begin{equation}
\psi^{-1}(s)=s+\sum_{0\leq s_k<s}\delta_k,\quad s\geq0,\;
s_i=\psi(t_{2i}),\;i\in N.
\label{e2.2}
\end{equation}
\end{lemma}

\begin{lemma}[\cite{a1}] \label{lem2}
 $\psi'(t)=1$ if $t\in T_0'$. $\frac{d}{ds}(\psi^{-1}(s))=1$
if $s\neq s_i$, $i\in N$.
\end{lemma}

\begin{lemma}\label{lem3}
 If $y(t)$ is the solution of  \eqref{e1.1}, then
$x(s)=y(\psi^{-1}(s))$ is the solution of
\begin{equation} \label{e2.3}
\begin{gathered}
x'=\widetilde{A}(s)x+\widetilde{f}(s,x)+A_i\widetilde{X}(s_i^+),\quad
  s\in(s_i,s_{i+1}], s\geq s_0,\\
x(s_i^+)=B_ix(s_i)+J_i(x(s_i))+x(s_i),\quad s_i\geq s_0,\\
x(s^0)=y_0.
\end{gathered}
\end{equation}
and vice versa, where
$\widetilde{A}(s)=A(\psi^{-1}(s))$,
$\widetilde{f}(s,x)=f(\psi^{-1}(s),x)$, $s^0=\psi(t_0)$,
\[
\widetilde{X}(s_i^+)=\begin{cases}
y_0, & s_i<s^0<s_{i+1},\\
x(s_i^+), & s_i\geq s^0,
\end{cases}
\]
$x(s_i^+)=y(t_{2i+1})$.
\end{lemma}

\begin{lemma} \label{lem4}
The zero solution of  \eqref{e1.1} is uniformly stable if and only if the
zero solution of  \eqref{e2.3} is uniformly stable.
\end{lemma}

\begin{lemma} \label{lem5}
Suppose that {\rm (H3)} holds, then the zero solution of  \eqref{e1.1}
 is uniformly asymptotically stable if the
zero solution of  \eqref{e2.3} is uniformly asymptotically stable.
\end{lemma}

\begin{proof}
 If the zero solution of \eqref{e2.3} is uniformly attractive, then there
exists a $\delta_0>0$ such that for all $\varepsilon>0$, there exists
$T_1(\varepsilon)>0$, such that $\|y_0\|<\delta,t^0\in T_0$ implies
\begin{equation}
\|x(s,s^0,y_0)\|<\varepsilon\quad  \text{for all }  s\geq s^0+T_1.\label{e2.4}
\end{equation}
Hence, for the  $\varepsilon$ above, there exists a
$T=T_1+\frac{T_1}{\lambda_1}\lambda_2>0$, such that
$t\in T_0[t^0+T,+\infty)$ implies $s=\psi(t)\geq s^0+T_1$; that is,
\[
\|y(t,t^0,y_0)\|=\|x(s,s^0,y_0)\|<\varepsilon\quad  \text{for all }
 t\in T_0[t^0+T,+\infty)\setminus\{t_{2i+1}\},s=\psi(t),%\label{e2.5}
\]
and
\begin{equation}
\|y(t_{2i+1},t^0,y_0)\|=\|x(s_i^+,s^0,y_0)\|
=\lim_{h\to0^+}\|x(s_i+h,s^0,y_0)\|\leq \varepsilon. \label{e2.6}
\end{equation}
Therefore, the zero solution of \eqref{e1.1} is uniformly attractive.

 By the above conclusion and Lemma \ref{lem4}, we can get that the zero
solution of \eqref{e1.1} is uniformly asymptotically stable.
\end{proof}

\begin{lemma} \label{lem6}
Suppose that {\rm (H3)} holds, then the zero solution of  \eqref{e1.1}
 is exponentially stable if the
zero solution of  \eqref{e2.3} is exponentially stable.
\end{lemma}

\begin{proof} If the zero solution of \eqref{e2.3} is exponentially stable,
then there exists $\alpha>0$ such that for all
$\varepsilon>0$ there exists $\delta(\varepsilon)>0$,
such that $\|y_0\|<\delta$ implies
\begin{equation}
\|x(s,s^0,y_0)\|\leq \varepsilon e^{-\alpha(s-s^0)}\quad \text{for }
 s\geq s^0.\label{e2.7}
\end{equation}
For $t\geq t^0, t\in[t_{2(m+p)-1},t_{2(m+p)}]$, by (H3), we have
\begin{equation}
d(T_0[t^0,t])\geq p\lambda_1
=\frac{p\lambda_1}{(p+1)\lambda+p\lambda_2}[(p+1)\lambda+p\lambda_2]\geq
l(t-t^0),\label{e2.8}
\end{equation}
where
$l=\frac{p\lambda_1}{(p+1)\lambda+p\lambda_2}$.
Hence, for the above $\varepsilon$, and $\|y_0\|<\delta$, we
have
\begin{equation}
\|y(t,t^0,y_0)\|\leq \varepsilon e^{-\alpha
d(T_0[t^0,t])}\leq \varepsilon e^{-\alpha l(t-t^0)}, \ t\in
T_0[t^0,+\infty);\label{e2.9}
\end{equation}
that is, the zero solution of \eqref{e1.1} is exponentially stable.
\end{proof}

 Denote $Y(t,\tau)$, $X(s,\gamma)$ as the Cauchy matrices of
$y'=A(t)y$ and $x'=\widetilde{A}(s)x$ respectively. It is
easy to verify that $Y(t,\tau)=X(s,\gamma)$ where $t=\psi^{-1}(s),
\tau=\psi^{-1}(\gamma)$ when $\gamma\neq s_i, t=\psi^{-1}(s)$ and
$\tau=t_{2i+1}$ when $\gamma=s_i$.
 Consider
\begin{equation} \begin{gathered}
y'=A(t)y,\quad t\in[t_{2i+1},t_{2i+2}],\; t\geq t^0,\\
 y(t_{2i+1})=B_iy(t_{2i})+y(t_{2i}),\quad t_{2i}\geq t^0,\\
 y(t^0)=y_0
\end{gathered} \label{e2.10}
\end{equation}
and
\begin{equation} \begin{gathered}
x'=\widetilde{A}(s)x,\quad s\neq s_i,\; s\geq s^0,\\
x(s_i^+)=(B_i+I)x(s_i),\quad s_i\geq s^0,\\
x(s^0)=y_0.
\end{gathered} \label{e2.11}
\end{equation}
It is easy to verify that the solution of \eqref{e2.10}, for
$t\in[t_{2p+1},t_{2p+2}]$, is
\begin{equation}
y(t)=Y(t,t_{2p+1})(B_p+I)[\prod_{k=m+1}^p Y(t_{2k},t_{2k-1})(B_{k-1}+I)]
Y(t_{2m},t^0)y_0, \label{e2.12}
\end{equation}
and the solution of \eqref{e2.11}, for
$s\in(s_p,s_{p+1}]$, is
\begin{equation}
x(s)=X(s,s_p^+)(B_p+I)[\prod_{k=m+1}^p X(s_k,s_{k-1})(B_{k-1}+I)]
X(s_m,s^0)y_0. \label{e2.13}
\end{equation}
Denote
\begin{gather*}
E(t,t^0)=Y(t,t_{2p+1})(B_p+I)[\prod_{k=m+1}^p Y(t_{2k},t_{2k-1})(B_{k-1}+I)]
Y(t_{2m},t^0),\\
E_1(s,s^0)=X(s,s_p^+)(B_p+I)[\prod_{k=m+1}^p X(s_k,s_{k-1})(B_{k-1}+I)]
X(s_m,s^0).
\end{gather*}

\section{Stability}

\begin{lemma}\label{lem7}
 The zero solution of \eqref{e2.10} is uniformly stable if and only if
$E(t,t^0)$ is uniformly bounded on $T_0$ with respect to $t^0$.
\end{lemma}

The above lemma follows from \eqref{e2.12}; so we omit the proof.

\begin{theorem}\label{thm1}
  Assume that
\begin{itemize}
\item[(i)]  the zero solution of  \eqref{e2.10} is uniformly stable;

\item[(ii)]  (H4) is satisfied and there exists a $\beta\geq0$ such
that for all $\tau\in T_0$,\\
$\int_{T_0[\tau,+\infty)}\|F(t)\|dt\leq\beta$;

\item[(iii)]  $A_i=0, J_i(y)\equiv0$ for all $y\in \mathbb{R}^n$, $i\in N$.
\end{itemize}
Then the zero solution of {\em \eqref{e1.1}} is uniformly stable.
\end{theorem}

\begin{proof}  By (i), Lemma \ref{lem3} and Lemma \ref{lem7}, we
can get that there exists an $M>0$ such that $\|E_1(t,\tau)\|\leq M$
for $t,\tau\in T_0$, $t\geq\tau$.

 We are going to prove that the zero solution of \eqref{e2.3} is uniformly
stable. It is important that under (iii), \eqref{e2.3} can be re-written
\begin{equation} \begin{gathered}
x'=\widetilde{A}(s)x+\widetilde{f}(s,x(s)),\quad s\in (s_i,s_{i+1}],\;
  s\geq s^0,\\
x(s_i^+)=B_ix(s_i)+x(s_i),\quad s_i\geq s^0,\\
x(s^0)=y_0.\end{gathered} \label{e3.1}
\end{equation}
 For $s\in(s^0,s_m]$, the solution of \eqref{e3.1} is
\begin{gather*}
x(s)=X(s,s^{0+})y_0+\int_{s^0}^sX(s,\tau)\widetilde{f}
(\tau,x(\tau))d\tau,\label{e3.2} \\
x(s_m^+)=(B_m+I)X(s_m,s^{0+})y_0+\int_{s^0}^{s_m}(B_m+I)X(s_m,\tau)
\widetilde{f}(\tau,x(\tau))d\tau\,.\label{e3.3}
\end{gather*}
For $s\in(s_m,s_{m+1}]$,
\begin{align*}
x(s)
&=X(s,s_m^+)x(s_m^+)+\int_{s_m}^sX(s,\tau)\widetilde{f}(\tau,x(\tau))d\tau\\
&=X(s,s_m^+)(B_m+I)X(s_m,s^{0+})y_0\\
&\quad +\int_{s^0}^{s_m}X(s,s_m^+)(B_m+I)X(s_m,\tau)\widetilde{f}
 (\tau,x(\tau))d\tau
 +\int_{s_m}^sX(s_m,\tau)\widetilde{f}(\tau,x(\tau))d\tau\\
&=E_1(s,s^{0+})y_0+\int_{s^0}^{s_m}E_1(s,\tau)\widetilde{f}(\tau,x(\tau))d\tau
+\int_{s_m}^sE_1(s,\tau)\widetilde{f}(\tau,x(\tau))d\tau\\
&=E_1(s,s^{0+})y_0+\int_{s^0}^sE_1(s,\tau)\widetilde{f}
(\tau,x(\tau))d\tau\,.
\end{align*} %\label{e3.4}
 By \eqref{e3.1}, the above equality,  and mathematical induction,
we conclude easily that
\begin{equation}
x(s)=E_1(s,s^{0+})y_0+\int_{s^0}^sE_1(s,\tau)\widetilde{f}(\tau,x(\tau))
d\tau,\quad s\geq s^0.\label{e3.5}
\end{equation}
 Therefore,
\begin{equation}
\|x(s)\|\leq M\|y_0\|+M\int_{s^0}^s\|F(\tau)\|\cdot\|x(\tau)\|d\tau.
\label{e3.6}
\end{equation}
 By the Bellman inequality,
\begin{equation}
\|x(s)\|\leq M\|y_0\|\exp(\int_{s^0}^s\|F(\tau)\|d\tau)
\leq M\|y_0\|e^\beta,\label{e3.7}
\end{equation}
which leads to that the solution $y(t)$ of \eqref{e1.1} satisfies
$$
\|y(t)\|\leq Me^\beta\|y_0\|.%\label{e3.8}
$$
which yields that the zero solution of \eqref{e1.1} is uniformly
stable. The proof is complete.
\end{proof}

\begin{theorem}\label{thm2}
 Assume that {\rm (H2), (H3), (H5)} hold and
\begin{itemize}
\item[(i)]  there exist $M,\alpha>0$ such that
$\sup_{t\in[t_{2i-1},t_{2i}]}\|Y(t,t_{2i-1})\| \leq M$,\\
$ \sup_{t\in[t_{2i-1},t_{2i}]}\|A(t)\|\leq\alpha$, $i\in N$;

\item[(ii)]  there exist $\mu,\sigma>0$ such that $\|A_i\|\leq\mu$,
$\sigma_i\leq\sigma$ for $i\in N$ and $M(1+\lambda\mu)\sigma\leq q<1$.
\end{itemize}
Then the zero solution of \eqref{e1.1} is uniformly asymptotically stable.
\end{theorem}

\begin{proof} From (H5)  there exists $\delta_1>0$
such that
\begin{equation}
\|\widetilde{f}(t,x)\|\leq\|x\|,\quad  \text{whenever }  \|x\|<\delta_1.
\label{e3.9}
\end{equation}
We claim that there exists $\delta_2>0$ such that the solution
of \eqref{e2.3} satisfies
\begin{equation}
\|x(s)\|<\delta_1, s\in(s^0,s_m]\quad \text{if }
 \|x(s^{0+})\|<\delta_2,\label{e3.10}
\end{equation}
and
\begin{equation}
\|x(s)\|<\delta_1, s\in(s_k,s_{k+1}]\quad \text{whenever }
 \|x(s^+_k)\|<\delta_2,k\geq m,\label{e3.11}
\end{equation}
where $\delta_2(1+\lambda\mu)\exp[(1+\alpha)\lambda]\leq\delta_1$.
Note that $x(s^{0+})=x(s^0)$ if $s^0\in(s_i,s_{i+1})$, and
$x(s^{0+})=(B_i+I)x(s^0)$ if $s^0=s_i, i\in N$.

Otherwise, if \eqref{e3.10} is not true, there must exists
$\tau_0\in(s^0,s_m]$ such that $\|x(\tau_0)\|=\delta_1$ while
$\|x(s)\|<\delta_1$ for all $s\in(s^0,\tau_0]$. Then,
$\|\widetilde{f}(s,x(s))\|\leq\|x(s)\|$ for all $s\in(s^0,\tau_0]$.

 For any $s\in(s^0,\tau_0]$, it is true that
$$
x(s)=x(s^{0+})+\int_{s^0}^s[\widetilde{A}(r)x(r)+\widetilde{f}(r,x(r))
+A_{m-1}x(s^{0+})]dr,
$$
which implies
\begin{equation}
\|x(s)\|\leq\|x(s^{0+})\|(1+\lambda\mu)+\int_{s^0}^s(1+\alpha)\|x(r)\|dr.
\label{e3.12}
\end{equation}
By the Bellman inequality,  for $s\in(s^0,\tau_0]$,
\begin{equation}
\|x(s)\|\leq\|x(s^{0+})\|(1+\lambda\mu)\exp[(1+\alpha)\lambda];\label{e3.13}
\end{equation}
that is,
$$
\|x(s)\|\leq\delta_2(1+\lambda\mu)\exp[(1+\alpha)\lambda]<\delta_1,
$$
which contradicts the fact that $\|x(\tau_0)\|=\delta_1$. Hence,
\eqref{e3.10} is true. Similarly, we can get that \eqref{e3.11} is true.
Thus, if $\|x(s^+_k)\|<\delta_2$, then
\begin{equation}
\|x(s)\|\leq\|x(s^+_k)\|(1+\lambda\mu)\exp[(1+\alpha)\lambda],\quad
s\in(s_k,s_{k+1}],\; k\geq m.\label{e3.14}
\end{equation}
For a given $\varepsilon_0$: $0<\varepsilon_0<\min\{1-q,\frac{1-q}{\sigma}\}$,
we choose $\varepsilon_1$: $0<\varepsilon_1<1$ such that
$\varepsilon_1M\lambda(1+\lambda\mu)\exp[(1+\alpha)\lambda]\leq\varepsilon_0$.
By (H5)  there exists a $\delta_3$: $0<\delta_3<\min\{\delta_1,\delta^*\}$
such that
$\|\widetilde{f}(s,x)\|\leq\varepsilon_1\|x\|$ whenever
$\|x\|<\delta_3$.

 By a proof similar to the above discussion, we can obtain
$0<\delta_4<\delta_2$ such that
$\|x(s)\|\leq\delta_3,s\in(s^0,s_m]$, and \eqref{e3.13} holds for
$s\in(s^0,s_m]$ if $\|x(s^{0+})\|\leq\delta_4$. We can also get
that, for $k\geq m$, $\|x(s)\|\leq\delta_3,s\in(s_k,s_{k+1}]$ and
\eqref{e3.14} holds for $s\in(s_k,s_{k+1}]$ whenever
$\|x(s^+_k)\|\leq\delta_4$.

 Denote $\delta_5:\
0<\delta_5<\min\{\delta_4,\frac{\delta_4}{\sigma}\}$. Let
$\|y_0\|<\delta_5$. Then $\|x(s^{0+})\|<\delta_4$.
 From \eqref{e2.3} we get that for $s\in(s^0,s_m]$,
\begin{align*}
x(s)&=X(s,s^{0+})x(s^{0+})+\int_{s^0}^sX(s,\tau)[\widetilde{f}
(\tau,x(\tau))+A_{m-1}x(s^{0+})] d\tau\\
&=[X(s,s^{0+})+\int_{s^0}^sX(s,\tau)d\tau\cdot
A_{m-1}]x(s^{0+})+\Delta_s,
\end{align*}
where $\Delta_s=\int_{s^0}^sX(s,\tau)\widetilde{f}(\tau,x(\tau))d\tau$.

 That $\|y_0\|<\delta_5$ and that \eqref{e3.13} holds for
$s\in(s^0,s_m]$ leads us to
$$
\Delta_s\leq M\varepsilon_1\int_{s^0}^s\|x(\tau)\|d\tau
\leq \|x(s^{0+})\|\cdot
M\varepsilon_1(1+\mu\lambda)\exp[(1+\alpha)\lambda]\lambda
\leq\varepsilon_0\|x(s^{0+})\|
$$
for $s\in(s^0,s_m]$.
 Hence, for $s\in(s^0,s_m]$,
\begin{equation}
\|x(s)\|\leq(M+\varepsilon_0+M\mu\lambda)\|x(s^{0+})\|.\label{e3.15}
\end{equation}
Therefore,
\begin{equation}
\begin{aligned}
\|x(s_m^+)\|&=\|B_mx(s_m)+J_m(x(s_m))+x(s_m)\|\\
&\leq\sigma(M+\varepsilon_0+M\mu\lambda)\|x(s^{0+})\|\\
&\leq(q+\sigma\varepsilon_0)\|x(s^{0+})\|<\delta_4.
\end{aligned}\label{e3.16}
\end{equation}
 From \eqref{e2.3} we  obtain that for $s\in(s_m,s_{m+1}]$,
\begin{equation}
x(s)=[X(s,s^+_m)+\int_{s_m}^sX(s,\tau)d\tau\cdot A_m]x(s^+_m)
+\Delta_s,\label{e3.17}
\end{equation}
$\Delta_s=\int_{s_m}^sX(s,\tau)\widetilde{f}(\tau,x(\tau))d\tau$,
$s\in(s_m,s_{m+1}]$.

As  for \eqref{e3.15}, we can get that, for $s\in(s_m,s_{m+1}]$,
\begin{gather}
\|x(s)\|\leq(M+\varepsilon_0+M\mu\lambda)\|x(s^+_m)\|,\label{e3.18}\\
\|x(s_{m+1}^+)\|\leq(q+\sigma\varepsilon_0)\|x(s^+_m)\|<\delta_4.
\label{e3.19}
\end{gather}
 By  mathematical induction,  for
$s\in(s_k,s_{k+1}]$, we obtain
\begin{gather}
\|x(s_k^+)\|<\delta_4,\quad  \|x(s)\|\leq(M+\varepsilon_0+M\mu\lambda)
\|x(s^+_k)\|,\\
\|x(s_{k+1}^+)\|\leq(q+\sigma\varepsilon_0)\|x(s^+_k)\|<\delta_4, k\geq m.
\label{e3.20}
\end{gather}
Hence,
\begin{equation}
\|x(s_k^+)\|\leq(q+\sigma\varepsilon_0)^{k-m+1}\|x(s^{0+})\|, k\geq m.
\label{e3.21}
\end{equation}
 Since for each $\varepsilon>0$, there exists $N_1\in N_+$ such that
$n\geq N_1$ implies $(q+\sigma\varepsilon_0)^n<\varepsilon$, there exists
$T_1=N_1\lambda$ such that for any $\|y_0\|<\delta_5$, $s^0\in T_0,
s_k\geq s^0+T_1$,
$$
\|x(s_k^+)\|<\delta_5\varepsilon,\quad \text{and}\quad
\|x(s)\|\leq(M+\varepsilon_0+M\mu\lambda)\delta_5\varepsilon,\quad
s\in(s_k,s_{k+1}];
$$
that is, the zero solution of \eqref{e2.3} is uniformly attractive.

 Form \eqref{e3.16},\eqref{e3.21}, we  conclude that
\begin{equation}
\|x(s_k^+)\|\leq\|x(s^{0+})\|.\label{e3.22}
\end{equation}
 Therefore, for each $\varepsilon>0$, there
exists $0<\delta<\min\{\delta_5,\varepsilon\}$, such that
$\|y_0\|<\delta$ implies
$$
\|x(s_k^+)\|\leq(1+\sigma)\varepsilon,k\geq m,
$$
and
\begin{equation}
\|x(s)\|\leq(1+\sigma)(M+\varepsilon_0+M\mu\lambda)\varepsilon,\quad
s\in[s^0,+\infty)\cup T_0;\label{e3.23}
\end{equation}
that is, the zero solution of \eqref{e2.3} is uniformly stable.

 Summing up the above discussion, we can get that the zero
solution of \eqref{e2.3} is uniformly asymptotically stable. Hence, by
Lemma \ref{lem5}, we get that the zero solution of \eqref{e1.1} is uniformly
asymptotically stable.
\end{proof}

\begin{theorem} \label{thm3}
Assume that {\rm (H2), (H3), (H5)}  hold and for $k\in N$,
\begin{itemize}
\item[(i)] There exist $M,\alpha>0$ such that
 $\|Y(t,s)\|\leq Me^{-\alpha(t-s)}$, $t_{2k-1}\leq s\leq t\leq t_{2k}$;

 \item[(ii)] there exist $\sigma,\beta>0$ such that $\sigma_k\leq\sigma$,
 $\beta<\alpha$, and \\
$\sigma_kM(1+\frac{1}{\alpha})e^{-(\alpha-\beta)(t_{2k}-t_{2k-1})}\leq1$;

\item[(iii)]  $\|A_k\|\leq e^{-\alpha\lambda}$.
\end{itemize}
Then the zero solution of \eqref{e1.1} is exponentially
 stable.
 \end{theorem}

\begin{proof} At first, we investigate the solution
  of \eqref{e2.3},
\begin{equation}
x(s)=\begin{cases}
 X(s,s_p^+)x(s_p^+)+\int_{s_p}^sX(s,\tau)\widetilde{f}(\tau,x(\tau))d\tau\\
 +x(s_p^+)\int_{s_p}^sX(s,\tau)d\tau\cdot A_p,
 & s\in(s_p,s_{p+1}],\; p\geq m,\\[3pt]
 X(s,s^{0+})x(s^{0+})+\int_{s^0}^sX(s,\tau)\widetilde{f}(\tau,x(\tau))d\tau\\
 +x(s^{0+})\int_{s^0}^sX(s,\tau)d\tau\cdot A_{m-1},
 & s\in(s^0,s_m]\,.
 \end{cases} \label{e3.24}
\end{equation}

\noindent\textbf{Case I: $M>1$.}
  Let $M'=\max\{1,\sigma\}, \varepsilon_0>0$ such that
 $\varepsilon_0M' M^2(1+\frac{1}{\alpha})<\beta$. (H5)
 yields that there exists a $\delta(\varepsilon_0)>0$ which ensure
 $\|\widetilde{f}(s,x)\|\leq\varepsilon_0\|x\|$ whenever
 $\|x\|<\delta$ and $s\in T_0$.

 For
 $\|y_0\|<\frac{\delta}{2MM^{\prime 2}(1+\frac{1}{\alpha})}$ (which
 ensure $
 \|x(s^{0+})\|<\frac{\delta}{2MM'(1+\frac{1}{\alpha})}$), there exists
$T_1: s^0+T_1< s_m$, such that $\|x(s)\|<\delta$ for $s\in( s^0,s^0+T_1]$.
Hence, for $s\in( s^0,s^0+T_1]$,
\begin{equation}  \label{e3.25}
\begin{aligned}
\|x(s)\|
&\leq \|X(s,s^0)\|\cdot\|x(s^{0+})\|+\int_{s^0}^s\|X(s,\tau)\|\cdot
 \|\widetilde{f}(\tau,x(\tau))\|d\tau\\
&\quad +\|A_{m-1}\|\cdot\|x(s^{0+})\|\int_{s^0}^s\|X(s,\tau)\|d\tau\\
&\leq Me^{-\alpha(s-s^0)}\|x(s^{0+})\|
  +\int_{s^0}^sMe^{-\alpha(s-\tau )}\varepsilon_0\|x(\tau)\| d\tau\\
&\quad +e^{-\alpha\lambda}\|x(s^{0+})\|\int_{s^0}^sMe^{-\alpha(s-\tau )}d\tau\\
&\leq M(1+\frac{1-e^{-\alpha\lambda}}{\alpha})e^{-\alpha(s-s^0)}\|x(s^{0+})\|
 +M\varepsilon_0\int_{s^0}^se^{-\alpha(s-\tau )}\|x(\tau)\|d\tau\\
 &\leq M(1+\frac{1}{\alpha})e^{-\beta(s-s^0)}\|x(s^{0+})\|
 +M' M^2\varepsilon_0(1+\frac{1}{\alpha})\int_{s^0}^s
 e^{-\beta(s-\tau )}\|x(\tau)\|d\tau\,;
\end{aligned}
\end{equation}
that is,
\begin{equation}
\|x(s)\|e^{\beta s}\leq M(1+\frac{1}{\alpha})e^{\beta s^0}\|x(s^{0+})\|
 +M' M^2\varepsilon_0(1+\frac{1}{\alpha})\int_{s^0}^s
 e^{\beta\tau}\|x(\tau)\|d\tau.\label{e3.26}
\end{equation}
 By the Bellman inequality,  for $s\in(s^0,s^0+T_1]$,
\begin{equation}
\|x(s)\|\leq M(1+\frac{1}{\alpha})e^{-[\beta-\varepsilon_0M'
 M^2(1+\frac{1}{\alpha})]
(s-s^0)}\|x(s^{0+})\|,\label{e3.27}
\end{equation}
which leads to
\begin{equation}
\|x(s)\|<\frac{\delta}{2} \quad  \text{for } s\in(s^0,s^0+T_1].
\label{e3.28}
\end{equation}
 We claim that
\begin{equation}
\|x(s)\|<\delta \quad  \text{for } s\in(s^0,s_m].\label{e3.29}
\end{equation}
 In fact, if \eqref{e3.29} is not true, there exists
$s'\in(s^0+T_1,s_m]$, such that $\|x(s')\|=\delta$ and
$\|x(s)\|<\delta$ for $s\in(s^0,s')$. As above,  we can prove
that \eqref{e3.27} holds for $s\in(s^0,s']$; that is,
$\|x(s)\|<\frac{\delta}{2}$ for $s\in(s^0,s')$. Hence,
$\|x(s')\|=\lim_{s\to s^{\prime-}}\|x(s)\|\leq \frac{\delta}{2}<\delta$,
which is a contradiction to that
$\|x(s')\|=\delta$. \eqref{e3.29} leads us to that \eqref{e3.25} is true for
$s\in(s^0,s_m]$.

 Also \eqref{e3.29} yields that \eqref{e3.27} is true for $s\in(s^0,s_m]$.
Therefore, $\|x(s_m)\|\leq\frac{\delta}{2M'}$, and
\begin{equation}
\|x(s_m^+)\|=\|B_mx(s_m)+J_m(x(s_m))+x(s_m)\|
\leq \sigma_m\|x(s_m)\|\leq\frac{\delta}{2}.\label{e3.30}
\end{equation}
 So, there exists $T_2: s_m+T_2<s_{m+1}$, such that $\|x(s)\|<\delta$ for
$s\in(s_m,s_m+T_2]$. Since \eqref{e3.25} holds for $s\in(s^0,s_m]$,
for $s\in(s_m,s_m+T_2]$,
\begin{align*}
\|x(s)\|&\leq \|X(s,s_m)\|\cdot\|x(s^+_m)\|+\int_{s_m}^s\|X(s,\tau)\|\cdot
  \|\widetilde{f}(\tau,x(\tau))\|d\tau\\
&\quad +\|A_m\|\cdot\|x(s^+_m)\|\int_{s_m}^s\|X(s,\tau)\|d\tau\\
&\leq[Me^{-\alpha(s-s_m)}+e^{-\alpha\lambda}M\int_{s_m}^s
 e^{-\alpha(s-\tau)}d\tau]\|x(s^+_m)\|\\
&\quad +\varepsilon_0M\int_{s_m}^se^{-\alpha(s-\tau )}\|x(\tau)\|d\tau\\
&\leq\sigma_mM(1+\frac{1}{\alpha})e^{-\alpha(s-s_m)}[M(1+\frac{1}{\alpha})
 e^{-\alpha(s_m-s^0)}  \|x(s^{0+})\| \\
&\quad +\varepsilon_0M\int_{s^0}^{s_m}e^{-\alpha(s_m-\tau)}\|x(\tau)\|d\tau]
 +M\varepsilon_0\int_{s_m}^se^{-\alpha(s-\tau )}\|x(\tau)\|d\tau\\
&=\sigma_mM^2(1+\frac{1}{\alpha})^2e^{-\alpha(s-s^0)}\|x(s^{0+})\|\\
&\quad +\sigma_mM^2(1+\frac{1}{\alpha})\varepsilon_0\int_{s^0}^{s_m}
 e^{-\alpha(s-\tau)}\|x(\tau)\|d\tau
 +M\varepsilon_0\int_{s_m}^se^{-\alpha(s-\tau )}\|x(\tau)\|d\tau.
 \end{align*}
 Hence,  for $s\in(s_m,s_m+T_2]$,
\begin{equation}
\|x(s)\|\leq M(1+\frac{1}{\alpha})e^{-\beta(s-s^0)}\|x(s^{0+})\|
 +M' M^2\varepsilon_0(1+\frac{1}{\alpha})\int_{s^0}^s
 e^{-\beta(s-\tau )}\|x(\tau)\|d\tau. \label{e3.31}
\end{equation}
  Hence,  for $s\in(s_m,s_m+T_2]$, \eqref{e3.27} holds. So,
$\|x(s)\|<\delta$ for $s\in(s_m,s_m+T_2]$. Similarly, we can prove
that $\|x(s)\|<\delta$ for $s\in(s_m,s_{m+1}]$ and \eqref{e3.27} holds
for $s\in(s_m,s_{m+1}]$.

 Suppose that for $\|y_0\|<\frac{\delta}{2MM^{\prime2}(1+\frac{1}{\alpha})}$,
\eqref{e3.27} holds for $s\in(s^0,s_k]$, $k\geq m$. Then by \eqref{e3.27},
we have
$\|x(s_k)\|<\frac{\delta}{2M'}$. Hence,
$\|x(s_k^+)\|\leq\sigma_k\|x(s_k)\|<\delta/2$. There must be
an $H_1>0: s_k+H_1<s_{k+1}$, such that $\|x(s)\|<\delta$ for
$s\in(s_k,s_k+H_1]$. Therefore, for $s\in(s_k,s_k+H_1]$,
\begin{align*}
&\|x(s)\| \\
&\leq M(1+\frac{1}{\alpha})e^{-\alpha(s-s_k)}\|x(s^+_k)\|
 +M\varepsilon_0\int_{s_k}^se^{-\alpha(s-\tau )}\|x(\tau)\|d\tau\\
&\leq M(1+\frac{1}{\alpha})e^{-\alpha(s-s_k)}\sigma_k[M(1+\frac{1}{\alpha})
 e^{-\alpha(s_k-s_{k-1})} \|x(s^+_{k-1})\|\\
&\quad +M\varepsilon_0\int_{s_{k-1}}^{s_k}e^{-\alpha(s_k-\tau
 )}\|x(\tau)\|d\tau]
 +M\varepsilon_0\int_{s_k}^se^{-\alpha(s-\tau )}\|x(\tau)\|d\tau
\\
 &\leq \sigma_kM^2(1+\frac{1}{\alpha})^2e^{-\alpha(s-s_{k-1})}\|x(s^+_{k-1})\|\\
&\quad +\sigma_kM^2(1+\frac{1}{\alpha})\varepsilon_0\int_{s_{k-1}}^{s_k}e^{-\alpha(s-\tau
 )}\|x(\tau)\|d\tau
 +M\varepsilon_0\int_{s_k}^se^{-\alpha(s-\tau )}\|x(\tau)\|d\tau\\
&\leq M(1+\frac{1}{\alpha})e^{-\beta(s-s_{k-1})}\|x(s^+_{k-1})\|
 +M' M^2(1+\frac{1}{\alpha})\varepsilon_0\int_{s_{k-1}}^s
 e^{-\beta(s-\tau)}\|x(\tau)\|d\tau
\\
&\leq M(1+\frac{1}{\alpha})e^{-\beta(s-s_{k-1})}[M(1+\frac{1}{\alpha})
 e^{-\alpha(s_{k-1}-s_{k-2})}\|x(s^+_{k-2})\|
\\
&\quad +M\varepsilon_0\int_{s_{k-2}}^{s_{k-1}}e^{-\alpha(s_{k-1}-\tau
 )}\|x(\tau)\|d\tau]+M' M^2(1+\frac{1}{\alpha})\varepsilon_0\int_{s_{k-1}}^s
 e^{-\beta(s-\tau)}\|x(\tau)\|d\tau\\
&\leq \sigma_{k-1}M^2(1+\frac{1}{\alpha})^2e^{-\beta(s-s_{k-1})
  -\alpha(s_{k-1}-s_{k-2})}\|x(s^+_{k-2})\|\\
&\quad +M'M^2(1+\frac{1}{\alpha})\varepsilon_0\int_{s_{k-2}}^{s_{k-1}}
 e^{-\beta(s-\tau )}\|x(\tau)\|d\tau\\
&\quad +M' M^2(1+\frac{1}{\alpha})\varepsilon_0\int_{s_{k-1}}^s
 e^{-\beta(s-\tau)}\|x(\tau)\|d\tau
\\
 &\leq M(1+\frac{1}{\alpha})e^{(\alpha-\beta)(s_{k-1}-s_{k-2})-\beta(s-s_{k-1})-\alpha(s_{k-1}-s_{k-2})}
\|x(s^+_{k-2})\|
\\
&\quad +M' M^2(1+\frac{1}{\alpha})\varepsilon_0\int_{s_{k-2}}^s
  e^{-\beta(s-\tau)}\|x(\tau)\|d\tau\\
&\leq M(1+\frac{1}{\alpha})e^{-\beta(s-s_{k-2})}\|x(s^+_{k-2})\|
+M' M^2(1+\frac{1}{\alpha})\varepsilon_0\int_{s_{k-2}}^s
 e^{-\beta(s-\tau)}\|x(\tau)\|d\tau\\
&\leq\dots
\leq M(1+\frac{1}{\alpha})e^{-\beta(s-s^0)}\|x(s^{0+})\|+M'
M^2(1+\frac{1}{\alpha})\varepsilon_0\int_{s^0}^s
 e^{-\beta(s-\tau)}\|x(\tau)\|d\tau
\end{align*}
So, for $s\in(s_k,s_k+H_1]$, \eqref{e3.27} holds, which implies that
$\|x(s)\|<\delta$ for $s\in(s^0,s_k+H_1]$. Similarly, we can prove
that $\|x(s)\|<\delta$ for $s\in(s^0,s_{k+1}]$ and \eqref{e3.31} holds for
$s\in(s^0,s_{k+1}]$. Therefore \eqref{e3.27} holds for $s\in(s^0,s_{k+1}]$.
 By mathematical induction, we can conclude that
$\|y_0\|<\frac{\delta}{2MM^{\prime2}(1+\frac{1}{\alpha})}$ leads to
\eqref{e3.27} holds for $s\in(s^0,+\infty)$.

\noindent\textbf{Case II: $M\leq1$.}
 Let $0<\varepsilon_0<\frac{\beta}{M'(1+\frac{1}{\alpha})}$. We can
get
$$
\|x(s)\|\leq(1+\frac{1}{\alpha})e^{-(\beta-\varepsilon_0
M'(1+\frac{1}{\alpha}))(s-s^0)}\|x(s^{0+})\|.%\label{e3.32}
$$
 Summing up the above discussion, we can conclude that no matter
if $M>1$ or $M\leq1$, there exists a $\gamma(M,M',\alpha)>0$,
such that
$$
\|x(s)\|\leq M'(1+\frac{1}{\alpha})e^{-\gamma(s-s^0)}\|y_0\|,\ s\geq s^0.
$$
Therefore, the zero solution of \eqref{e2.3} is exponentially stable. By
Lemma \ref{lem6}, we can conclude that the zero solution of \eqref{e1.1} is
exponentially stable.
\end{proof}

\section{Examples}

As a first example consider the equation
\begin{equation} \label{e4.1}
\begin{gathered}
y'=Ay+f(t,y),\quad t\in[2k+1,2k+2]\\
y(2k+1)=By(2k)+y(2k),\quad k\in N
\end{gathered}
\end{equation}
where $A,B\in \mathbb{R}^{n\times n}$, $\|f(t,y)\|\leq F(t)\cdot\|y\|$,
$\int_0^{+\infty}F(t)dt<+\infty$.

 (i) If $e^{\|A\|}\cdot\|B+I\|\leq1$, then the zero solution
of \eqref{e4.1} is uniformly stable.

 (ii) If
\begin{equation}
\min\{\mathop{\rm Re}(\lambda_j): \lambda_j
\text{ is an eigenvalue of } A\}=-\alpha<0 \label{e4.2}
\end{equation}
and $e^{-\alpha}\|B+I\|\leq 1$, then the zero solution of \eqref{e4.1} is
uniformly stable.

As a second example consider the equation
\begin{equation} \label{e4.3}
\begin{gathered}
y'=Ay+f(t,y)+CY(2k+1),\quad t\in[2k+1,2k+2]\\
y(2k+1)=By(2k)+y(2k),\quad k\in N
\end{gathered}
\end{equation}
 where $A,B\in \mathbb{R}^{n\times n}$, $f(t,y)$ satisfies (H5)
(for example, $f(t,y)=(y_1^2,\dots,y_n^2)^T)$.

(i) If $e^{\|A\|}\cdot(1+\|C\|)\|B+I\|\leq q<1$, then the zero solution
of \eqref{e4.3} is uniformly stable.

(ii) If \eqref{e4.2} holds,
$e^{-\alpha}(1+\|C\|)\|B+I\|\leq q<1$, then the zero solution of
\eqref{e4.3} is uniformly stable.

(iii) If \eqref{e4.2} holds, $e^{-\alpha}(1+\frac{1}{\alpha})\|B+I\|\leq q<1$
and $\|C\|\leq e^{-\alpha}$, then the zero solution of
\eqref{e4.3} is exponentially stable.

\begin{thebibliography}{9}

\bibitem{a1} M. U. Akhmet, M. Turan.
\emph{The differential equations on time scales through impulsive
differential equations}. Nonlinear Analysis, 65(2006), 2043-2060.

\bibitem{d1} Jefferey J. Dacunha.
\emph{Stability for time varying linear
dynamic systems on time scales}. Journal of Computational and Applied
Mathematics, 176(2005), 381-410.

\bibitem{l1} Bin Liu, Xinzhi Liu, and Xiaoxin Liao.
\emph{Stability and robustness of
quasi-linear impulsive hybrid systems}, J. Math. Anal. Appl,
283(2003), 416-430.

\bibitem{l2} Yubin Liu, Weizhen Feng.
\emph{Stability of linear impulsive systmes and
their nonlinear perturbed systems under arbitrary switched}.
Journal of South China Normal University(Natural Science Edition), No. 1
2007, 8-14.

\end{thebibliography}

\end{document}