Linear impulsive Volterra integro-dynamic system on time scales

Abstract

This paper deals with the asymptotic stability and boundedness of the solution of a time-varying impulsive Volterra integro-dynamic system on time scales in which the coefficient matrix is not necessarily stable. We generalize to a time scale some known properties concerning the asymptotic behavior and boundedness from the continuous case.

1 Introduction

Impulsive differential systems represent a natural framework for mathematical modelling of several processes in the applied sciences [1–4]. Basic qualitative and quantitative results about impulsive Volterra integro differential equations were studied in the literature (see [5–8]). Volterra-type equations (integral and integro-dynamic) on time scales were studied in [9–15]. In [16] the authors presented a theory for linear impulsive dynamic systems on time scales and recently in [17] various results concerning the asymptotic stability and boundedness of Volterra integro-dynamic equations on time scales were developed. Motivated by these papers, we generalize these results to impulsive integro-dynamic systems on time scales.

2 Preliminaries

In this paper we assume that the reader is familiar with the basic calculus of time scales. Let ${\mathbb{R}}^n$ be the space of n-dimensional column vectors $x=col(x_1,x_2,\dots ,x_n)$ with a norm $\parallel \cdot \parallel$. Also, with the same symbol $\parallel \cdot \parallel$ we denote the corresponding matrix norm in the space $M_n(\mathbb{R})$ of $n\times n$ matrices. If $A\in M_n(\mathbb{R})$, then we denote by $A^T$ its conjugate transpose. We recall that $\parallel A\parallel :=sup\{\parallel Ax\parallel ;\parallel x\parallel \le 1\}$ and the following inequality $\parallel Ax\parallel \le \parallel A\parallel \parallel x\parallel$ holds for all $A\in M_n(\mathbb{R})$ and $x\in {\mathbb{R}}^n$. A time scale $\mathbb{T}$ is a nonempty closed subset of ℝ. The set of all rd-continuous functions $f:\mathbb{T}\to {\mathbb{R}}^n$ will be denoted by $C_{\mathrm{rd}}(\mathbb{T},{\mathbb{R}}^n)$.

The notations $[a,b]$, $[a,b)$, and so on, denote time scale intervals such as $[a,b]:=\{t\in \mathbb{T};a\le t\le b\}$, where $a,b\in \mathbb{T}$. Also, for any $\tau \in \mathbb{T}$, let ${\mathbb{T}}_{\tau }:=[\tau ,\mathrm{\infty })\cap \mathbb{T}$ and ${\mathbb{T}}_0:=[0,\mathrm{\infty })\cap \mathbb{T}$.

We denote by ℛ (respectively ${\mathcal{R}}^{+}$) the set of all regressive (respectively positively regressive) functions from $\mathbb{T}$ to ℝ. The space of all rd-continuous and regressive functions from $\mathbb{T}$ to ℝ is denoted by $C_{\mathrm{rd}}\mathcal{R}(\mathbb{T},\mathbb{R})$. Also,

$$C_{\mathrm{rd}}^{+}\mathcal{R}(\mathbb{T},\mathbb{R}):=\{p\in C_{\mathrm{rd}}\mathcal{R}(\mathbb{T},\mathbb{R});1+\mu (t)p(t)>0\text{ for all }t\in \mathbb{T}\}.$$

We denote by $C_{\mathrm{rd}}^1(\mathbb{T},{\mathbb{R}}^{{}_n})$ the set of all functions $f:\mathbb{T}\to {\mathbb{R}}^n$ that are differentiable on $\mathbb{T}$ with its delta-derivative $f^{\mathrm{\Delta }}(t)\in C_{\mathrm{rd}}(\mathbb{T},{\mathbb{R}}^n)$. The set of rd-continuous (respectively rd-continuous and regressive) functions $A:\mathbb{T}\to M_n(\mathbb{R})$ is denoted by $C_{\mathrm{rd}}(\mathbb{T},M_n(\mathbb{R}))$ (respectively by $C_{\mathrm{rd}}\mathcal{R}(\mathbb{T},M_n(\mathbb{R}))$). We recall that a matrix-valued function A is said to be regressive if $I+\mu (t)A(t)$ is invertible for all $t\in \mathbb{T}$, where I is the $n\times n$ identity matrix. For a comprehensive review on time scales, we refer the reader to [18] and [19].

Lemma 2.1 ([[18], Theorem 2.38])

Let $p,q\in C_{\mathrm{rd}}\mathcal{R}(\mathbb{T},\mathbb{R})$. Then $e_{p\ominus q}^{\mathrm{\Delta }}(\cdot ,t_0)=(p-q)\frac{e_p(\cdot ,t_0)}{e_q^{\sigma }(\cdot ,t_0)}$.

Lemma 2.2 ([[18], Theorem 6.2])

Let $\alpha \in \mathbb{R}$ with $\alpha \in C_{\mathrm{rd}}^{+}\mathcal{R}(\mathbb{T},\mathbb{R})$. Then

$$e_{\alpha }(t,s)\ge 1+\alpha (t-s)\mathit{\text{for all }}t\ge s.$$

Theorem 2.3 ([[13], Theorem 7])

Let $a,b\in \mathbb{T}$ with $b>a$ and assume that $f:\mathbb{T}\times \mathbb{T}\to \mathbb{R}$ is integrable on $\{(t,s)\in \mathbb{T}\times \mathbb{T}:b>t>s\ge a\}$. Then

$${\int }_a^b{\int }_a^{\eta }f(\eta ,\xi )\mathrm{\Delta }\xi \mathrm{\Delta }\eta ={\int }_a^b{\int }_{\sigma (\xi )}^{b}f(\eta ,\xi )\mathrm{\Delta }\eta \mathrm{\Delta }\xi .$$

It is easy to verify that the above result holds for $f\in C_{\mathrm{rd}}(\mathbb{T}\times \mathbb{T},{\mathbb{R}}^n)$.

Lemma 2.4 ([[16], Lemma 2.1])

Let $t_0\in {\mathbb{T}}_0$, $y\in C_{\mathrm{rd}}\mathcal{R}({\mathbb{T}}_0,\mathbb{R})$, $p\in C_{\mathrm{rd}}^{+}\mathcal{R}({\mathbb{T}}_0,\mathbb{R})$ and $c,b_k\in {\mathbb{R}}_{+}$, $k=1,2,\dots$ . Then

$$y(t)\le c+{\int }_{t_0}^{t}p(s)y(s)\mathrm{\Delta }s+\sum _{t_0<t_k<t}{b}_{k}y(t_k),t\in {\mathbb{T}}_0$$

implies

$$y(t)\le c\prod _{t_0<t_k<t}(1+b_k)e_p(t,\tau ),t\ge t_0.$$

Consider the Volterra time-varying impulsive integro-dynamic system

$$\{\begin{array}{c}{x}^{\mathrm{\Delta }}(t)=A(t)x(t)+{\int }_{t_0}^{t}K(t,s)x(s)\mathrm{\Delta }s+F(t),t\in {\mathbb{T}}_0\mathrm{\setminus }\{t_k\},\hfill \\ x(t_k^{+})=(I+C_k)x(t_k),t=t_k,k=1,2,\dots ,\hfill \\ x(t_0)=x_0,\hfill \end{array}$$

(1)

where A (not necessarily stable) is an $n\times n$ matrix function and F is an n-vector function, which is piecewise continuous on ${\mathbb{T}}_0$, K is an $n\times n$ matrix function, which is piecewise continuous on $\mathrm{\Omega }:=\{(t,s)\in {\mathbb{T}}_0\times {\mathbb{T}}_0:t_0\le s\le t<\mathrm{\infty }\}$, $C_k\in M_n({\mathbb{R}}_{+})$, $0\le t_0<t_1<t_2<\cdots <t_k<\cdots$ , with ${lim}_{k\to \mathrm{\infty }}{t}_k=\mathrm{\infty }$ and the impulsive points $t_k$ are right dense. Note that $x(t_k^{-})$ represents the left limit of $x(t)$ at $t=t_k$ and $x(t_k^{+})$ represents the right limit of $x(t)$ at $t=t_k$.

The rest of the paper is organized as follows. In Section 3, we investigate the asymptotic behavior of solutions of system (1), which generalizes the continuous version ($\mathbb{T}=\mathbb{R}$) of [[8], Theorem 2.5]. In Section 4 we discuss the uniform boundedness of solutions of (1) by constructing a Lyapunov functional. Further results for boundedness, uniform boundedness and stability of solutions will also be developed.

3 Asymptotic stability

Our first result in this section is to present a system equivalent to (1) which involves an arbitrary function.

Theorem 3.1 Let $L(t,s)$ be an $n\times n$ continuously differentiable matrix function with respect to s on $t_{k-1}<s\le t_k<t$ with $L(t,t_k^{+})={(I+C_k)}^{-1}L(t,t_k)$ for each $k=1,2,\dots$ . Then (1) is equivalent to the system

$$\{\begin{array}{c}{y}^{\mathrm{\Delta }}(t)=B(t)y(t)+{\int }_{t_0}^{t}G(t,s)y(s)\mathrm{\Delta }s+H(t),t\in {\mathbb{T}}_0\mathrm{\setminus }\{t_k\},\hfill \\ y(t_k^{+})=(I+C_k)y(t_k),t=t_k,k=1,2,\dots ,\hfill \\ y(t_0)=y_0,\hfill \end{array}$$

(2)

where

$$\begin{array}{r}B(t)=A(t)-L(t,t),\\ H(t)=F(t)+L(t,t_0)x_0+{\int }_{t_0}^{t}L(t,\sigma (s))F(s)\mathrm{\Delta }s,\end{array}$$

(3)

and

$$\begin{array}{rl}G(t,s)=& K(t,s)+{\mathrm{\Delta }}_{s}L(t,s)+L(t,\sigma (s))A(s)\\ +{\int }_{\sigma (s)}^{t}L(t,\sigma (\tau ))K(\tau ,s)\mathrm{\Delta }\tau ,s,t\ne t_k.\end{array}$$

(4)

Proof Let $x(t)$ be any solution of (1) on ${\mathbb{T}}_0$. If we take $p(s)=L(t,s)x(s)$, then for $t_{k-1}<s\le t_k<t$, we have

$$p^{\mathrm{\Delta }}(s)={\mathrm{\Delta }}_{s}L(t,s)x(s)+L(t,\sigma (s))x^{\mathrm{\Delta }}(s)$$

and by (1) it follows that

$$\begin{array}{rcl}{p}^{\mathrm{\Delta }}(s)& =& {\mathrm{\Delta }}_{s}L(t,s)x(s)+L(t,\sigma (s))A(s)x(s)\\ +L(t,\sigma (s)){\int }_{t_0}^{s}K(s,\tau )x(\tau )\mathrm{\Delta }\tau +L(t,\sigma (s))F(s).\end{array}$$

Integration from $t_0$ to t yields

$$\begin{array}{r}p(t)-p(t_0)-\sum _{t_0<t_k<t}\mathrm{\Delta }p(t_k)\\ ={\int }_{t_0}^t{\mathrm{\Delta }}_{s}L(t,s)x(s)\mathrm{\Delta }s+{\int }_{t_0}^{t}L(t,\sigma (s))A(s)x(s)\mathrm{\Delta }s\\ +{\int }_{t_0}^{t}L(t,\sigma (s)){\int }_{t_0}^{s}K(s,\tau )x(\tau )\mathrm{\Delta }\tau \mathrm{\Delta }s\\ +{\int }_{t_0}^{t}L(t,\sigma (s))F(s)\mathrm{\Delta }s.\end{array}$$

Using Theorem 2.3, we obtain

$$\begin{array}{rcl}p(t)-p(t_0)-\sum _{t_0<t_k<t}\mathrm{\Delta }p(t_k)& =& {\int }_{t_0}^t{\mathrm{\Delta }}_{s}L(t,s)x(s)\mathrm{\Delta }s+{\int }_{t_0}^{t}L(t,\sigma (s))A(s)x(s)\mathrm{\Delta }s\\ +{\int }_{t_0}^t({\int }_{\sigma (\tau )}^{t}L(t,\sigma (s))K(s,\tau )\mathrm{\Delta }s)x(\tau )\mathrm{\Delta }\tau \\ +{\int }_{t_0}^{t}L(t,\sigma (s))F(s)\mathrm{\Delta }s.\end{array}$$

By a change of variables in the double integral term, we have

$$\begin{array}{rcl}p(t)-p(t_0)-\sum _{t_0<t_k<t}\mathrm{\Delta }p(t_k)& =& {\int }_{t_0}^t[{\mathrm{\Delta }}_{s}L(t,s)+L(t,\sigma (s))A(s)\\ +{\int }_{\sigma (s)}^{t}L(t,\sigma (u))K(u,s)\mathrm{\Delta }u]x(s)\mathrm{\Delta }s\\ +{\int }_{t_0}^{t}L(t,\sigma (s))F(s)\mathrm{\Delta }s.\end{array}$$

Using (3) and (4), we obtain

$$(A(t)-B(t))x(t)={\int }_{t_0}^t(G(t,s)-K(t,s))x(s)\mathrm{\Delta }s+H(t)-F(t)+\sum _{t_0<t_k<t}\mathrm{\Delta }p(t_k).$$

From (1), we have

$$x^{\mathrm{\Delta }}(t)=B(t)x(t)+{\int }_{t_0}^{t}G(t,s)x(s)\mathrm{\Delta }s+H(t)+\sum _{t_0<t_k<t}\mathrm{\Delta }p(t_k).$$

For $t_0<s\le t_k<t$, we obtain

$$\begin{array}{rcl}\mathrm{\Delta }p(t_k)& =& L(t,t_k^{+})x\left(t_k^{+}\right)-L(t,t_k)x(t_k)\\ =& [L(t,t_k^{+})(I+C_k)-L(t,t_k)]x(t_k)=0.\end{array}$$

Hence, $x(t)$ is a solution of (2).

Conversely, let $y(t)$ be any solution of (2) on ${\mathbb{T}}_0$. We shall show that it satisfies (1). Consider

$$Z(t)=y^{\mathrm{\Delta }}(t)-F(t)-A(t)y(t)-{\int }_{t_0}^{t}K(t,s)y(s)\mathrm{\Delta }s.$$

Then by (2) and (3) we have

$$\begin{array}{rcl}Z(t)& =& -L(t,t)y(t)+L(t,t_0)x_0+{\int }_{t_0}^{t}G(t,s)y(s)\mathrm{\Delta }s\\ +{\int }_{t_0}^{t}L(t,\sigma (s))F(s)\mathrm{\Delta }s-{\int }_{t_0}^{t}K(t,s)y(s)\mathrm{\Delta }s.\end{array}$$

Using (4), we obtain

$$\begin{array}{rcl}Z(t)& =& -L(t,t)y(t)+L(t,t_0)x_0+{\int }_{t_0}^{t}L(t,\sigma (s))F(s)\mathrm{\Delta }s\\ -{\int }_{t_0}^{t}K(t,s)y(s)\mathrm{\Delta }s+{\int }_{t_0}^t[K(t,s)+{\mathrm{\Delta }}_{s}L(t,s)+L(t,\sigma (s))A(s)\\ +{\int }_{\sigma (s)}^{t}L(t,\sigma (\tau ))K(\tau ,s)\mathrm{\Delta }\tau ]y(s)\mathrm{\Delta }s.\end{array}$$

Again by Theorem 2.3, we have

$$\begin{array}{rl}Z(t)=& -L(t,t)y(t)+{\int }_{t_0}^t[{\mathrm{\Delta }}_{s}L(t,s)+L(t,\sigma (s))A(s)]y(s)\mathrm{\Delta }s\\ +{\int }_{t_0}^{t}L(t,\sigma (s))[{\int }_{t_0}^{s}K(s,\tau )y(\tau )\mathrm{\Delta }\tau ]\mathrm{\Delta }s\\ +L(t,t_0)x_0+{\int }_{t_0}^{t}L(t,\sigma (s))F(s)\mathrm{\Delta }s.\end{array}$$

(5)

Now, by setting $q(s)=L(t,s)y(s)$, then for $t_{k-1}<s<t_k<t$, we get

$$q^{\mathrm{\Delta }}(s)={\mathrm{\Delta }}_{s}L(t,s)y(s)+L(t,\sigma (s))y^{\mathrm{\Delta }}(s).$$

(6)

Integrating (6) from $t_0$ to t yields

$$q(t)-q(t_0)-\sum _{t_0<t_k<t}\mathrm{\Delta }q(t_k)={\int }_{t_0}^t[{\mathrm{\Delta }}_{s}L(t,s)y(s)+L(t,\sigma (s))y^{\mathrm{\Delta }}(s)]\mathrm{\Delta }s$$

and therefore, we have

$$L(t,t)y(t)-L(t,t_0)x_0-\sum _{t_0<t_k<t}\mathrm{\Delta }q(t_k)={\int }_{t_0}^t[{\mathrm{\Delta }}_{s}L(t,s)y(s)+L(t,\sigma (s))y^{\mathrm{\Delta }}(s)]\mathrm{\Delta }s.$$

(7)

Since $\mathrm{\Delta }q(t_k)=0$, substituting (7) in (5), we obtain

$$\begin{array}{rcl}Z(t)& =& -{\int }_{t_0}^{t}L(t,\sigma (s))y^{\mathrm{\Delta }}(s)\mathrm{\Delta }s+{\int }_{t_0}^{t}L(t,\sigma (s))A(s)y(s)\mathrm{\Delta }s\\ +{\int }_{t_0}^{t}L(t,\sigma (s))[{\int }_{t_0}^{s}K(s,\tau )y(\tau )\mathrm{\Delta }\tau ]\mathrm{\Delta }s+{\int }_{t_0}^{t}L(t,\sigma (s))F(s)\mathrm{\Delta }s\\ =& -{\int }_{t_0}^{t}L(t,\sigma (s))Z(s)\mathrm{\Delta }s,\end{array}$$

which implies $Z(t)\equiv 0$, by the unique solution of Volterra integral equations [12] and the fact that $\mathrm{\Delta }Z(t_k)=0$. Hence $y(t)$ is a solution of (1). □

For our next result we assume that matrix B commutes with its integral, so B commutes with its matrix exponential, that is, $B(t)e_B(t,s)=e_B(t,s)B(t)$, [20].

Theorem 3.2 Let $B\in C(\mathbb{T},M_n(\mathbb{R}))$ and $M,\alpha >0$. Assume that matrix B commutes with its integral. If

$$\parallel e_B(t,s)\parallel \le M{e}_{\alpha }(s,t),t,s\in \mathrm{\Omega },$$

(8)

then every solution $x(t)$ of (1) satisfies

$$\begin{array}{rl}\parallel x(t)\parallel \le & M\parallel x_0\parallel e_{\alpha }(t_0,t)+M{\int }_{t_0}^t{e}_{\alpha }(\sigma (s),t)\parallel H(s)\parallel \mathrm{\Delta }s\\ +M{\int }_{t_0}^t\left[{\int }_{\sigma (s)}^t{e}_{\alpha }(\sigma (\tau ),t)\parallel G(\tau ,s)\parallel \mathrm{\Delta }\tau \right]\parallel x(s)\parallel \mathrm{\Delta }s\\ +M{e}_{\alpha }(t_0,t)\sum _{t_0<t_k<t}\parallel {\beta }_k\parallel \parallel x(t_k)\parallel ,\end{array}$$

(9)

where ${\beta }_k=[e_B(t_0,t_k^{+})(I+C_k)-e_B(t_0,t_k)]$.

Proof Let $x(t)$ be the solution of (2) and define $q(t)=e_B(t_0,t)x(t)$. Then

$$q^{\mathrm{\Delta }}(t)=-B(t)e_B(t_0,\sigma (t))x(t)+e_B(t_0,\sigma (t))x^{\mathrm{\Delta }}(t).$$

Substituting for $x^{\mathrm{\Delta }}(t)$ from (2) and integrating from $t_0$ to t, we obtain

$$\begin{array}{rcl}q(t)-q(t_0)-\sum _{t_0<t_k<t}\mathrm{\Delta }q(t_k)& =& {\int }_{t_0}^t{e}_B(t_0,\sigma (s))H(s)\mathrm{\Delta }s\\ +{\int }_{t_0}^t{e}_B(t_0,\sigma (s))[{\int }_{t_0}^{s}G(s,\tau )x(\tau )\mathrm{\Delta }\tau ]\mathrm{\Delta }s.\end{array}$$

Using Theorem 2.3 and applying the semigroup property of exponential functions [[18], Theorem 2.36], we obtain

$$\begin{array}{rl}x(t)=& e_B(t,t_0)x_0+{\int }_{t_0}^t{e}_B(t,\sigma (s))H(s)\mathrm{\Delta }s+e_B(t,t_0)\sum _{t_0<t_k<t}\mathrm{\Delta }q(t_k)\\ +{\int }_{t_0}^t[{\int }_{\sigma (s)}^t{e}_B(t,\sigma (\tau ))G(\tau ,s)\mathrm{\Delta }\tau ]x(s)\mathrm{\Delta }s.\end{array}$$

(10)

For $t_0<s<t_k<t$, we have

$$\begin{array}{rcl}\mathrm{\Delta }q(t_k)& =& e_B(t_0,t_k^{+})x\left(t_k^{+}\right)-e_B(t_0,t_k)x(t_k)\\ =& [e_B(t_0,t_k^{+})(I+C_k)-e_B(t_0,t_k)]x(t_k)\\ =& {\beta }_{k}x(t_k).\end{array}$$

Hence, using (8) and applying the norm on (10), we obtain (9), which completes the proof. □

In the next theorem we present sufficient conditions for asymptotic stability.

Theorem 3.3 Let $L(t,s)$ be an $n\times n$ continuously differentiable matrix function with respect to s on Ω such that

• (a) the assumptions of Theorem  3.2 hold,

• (b) $\parallel L(t,s)\parallel \le \frac{L_0{e}_{\gamma }(s,t)}{(1+\mu (t)\alpha )(1+\mu (t)\gamma )}$,

• (c) ${sup}_{t_0\le s\le t<\mathrm{\infty }}{\int }_{\sigma (s)}^t{e}_{\alpha }(\sigma (\tau ),t)\parallel G(\tau ,s)\parallel \mathrm{\Delta }\tau \le {\alpha }_0$,

• (d) $F(t)\equiv 0$ and

• (e) ${\prod }_{t_0<t_k<t}[1+M^2(d_k+1)]\le e_{\lambda }(t_k,0)$, where $d_k=\parallel (I+C_k)\parallel$: $d_k\to 0$ as $k\to \mathrm{\infty }$,

where $L_0,\gamma >\alpha$, ${\alpha }_0$, λ are positive real constants.

If $\alpha \ominus M{\alpha }_0\ominus \lambda >0$, then every solution $x(t)$ of (1) tends to zero exponentially as $t\to +\mathrm{\infty }$.

Proof In view of Theorem 3.1 and the fact that $L(t,s)$ satisfies (a), it is enough to show that every solution of (2) tends to zero as $t\to +\mathrm{\infty }$. From (a) and using (9), we obtain

$$\begin{array}{rl}{e}_{\alpha }(t,0)\parallel x(t)\parallel \le & M\parallel x_0\parallel e_{\alpha }(t_0,0)+M{\int }_{t_0}^t{e}_{\alpha }(\sigma (s),0)\parallel H(s)\parallel \mathrm{\Delta }s\\ +M{\int }_{t_0}^t\left[{\int }_{\sigma (s)}^t{e}_{\alpha }(\sigma (\tau ),0)\parallel G(\tau ,s)\parallel \mathrm{\Delta }\tau \right]\parallel x(s)\parallel \mathrm{\Delta }s\\ +M{e}_{\alpha }(t_0,0)\sum _{t_0<t_k<t}\parallel {\beta }_k\parallel \parallel x(t_k)\parallel .\end{array}$$

(11)

Since

$${\int }_{t_0}^t{e}_{\alpha }(\sigma (s),0)\parallel H(s)\parallel \mathrm{\Delta }s\le L_0\parallel x_0\parallel e_{\gamma }(t_0,0){\int }_{t_0}^t\frac{e_{\alpha }(\sigma (s),0)e_{\gamma }(0,s)}{(1+\mu (s)\alpha )(1+\mu (s)\gamma )}\mathrm{\Delta }s,$$

then by Lemma 2.1 and the fact that $\gamma >\alpha$, we obtain

$${\int }_{t_0}^t{e}_{\alpha }(\sigma (s),0)\parallel H(s)\parallel \mathrm{\Delta }s\le \frac{L_0\parallel x_0\parallel e_{\alpha }(t_0,0)}{\gamma -\alpha }.$$

Using (11), (b), (c) and (d), we have

$$\begin{array}{rcl}{e}_{\alpha }(t,0)\parallel x(t)\parallel & \le & M\parallel x_0\parallel e_{\alpha }(t_0,0)+M{L}_0\parallel x_0\parallel \frac{e_{\alpha }(t_0,0)}{\gamma -\alpha }+M{\int }_{t_0}^t{\alpha }_0{e}_{\alpha }(s,0)\parallel x(s)\parallel \mathrm{\Delta }s\\ +M{e}_{\alpha }(t_0,0)\sum _{t_0<t_k<t}\parallel {\beta }_k\parallel \parallel x(t_k)\parallel .\end{array}$$

From Theorem 3.2, we have

$$\parallel {\beta }_k\parallel \le \parallel e_B(t_0,t_k^{+})\parallel \parallel (I+C_k)\parallel +\parallel e_B(t_0,t_k)\parallel \le M{e}_{\alpha }(t_k,t_0)(1+d_k),$$

(12)

which implies

$$\begin{array}{rl}{e}_{\alpha }(t,0)\parallel x(t)\parallel \le & M\parallel x_0\parallel (1+\frac{L_0}{\gamma -\alpha })e_{\alpha }(t_0,0)+M{\int }_{t_0}^t{\alpha }_0{e}_{\alpha }(s,0)\parallel x(s)\parallel \mathrm{\Delta }s\\ +\sum _{t_0<t_k<t}{M}^2(1+d_k)e_{\alpha }(t_k,0)\parallel x(t_k)\parallel .\end{array}$$

(13)

Lemma 2.4 yields that

$$e_{\alpha }(t,0)\parallel x(t)\parallel \le M\parallel x_0\parallel (1+\frac{L_0}{\gamma -\alpha })e_{\alpha }(t_0,0)\prod _{t_0<t_k<t}[1+M^2(1+d_k)]e_{M{\alpha }_0}(t,t_0).$$

Using [[18], Theorem 2.36], (e) and the fact that $t_0<t_k<t$, we obtain

$$\parallel x(t)\parallel \le M\parallel x_0\parallel (1+\frac{L_0}{\gamma -\alpha })e_{\alpha \ominus M{\alpha }_0}(t_0,0)e_{\alpha \ominus M{\alpha }_0\ominus \lambda }(0,t),$$

where $\alpha \ominus M{\alpha }_0\ominus \lambda =\frac{\alpha -M{\alpha }_0-\lambda (1+\mu (t)M{\alpha }_0)}{(1+\mu (t)M{\alpha }_0)(1+\mu (t)\lambda )}$ [[18], Exercise 2.28]. By Lemma 2.2, we have $e_{\alpha \ominus M{\alpha }_0\ominus \lambda }(0,t)\le \frac{1}{1+(\alpha \ominus M{\alpha }_0\ominus \lambda )t}$, so we obtain

$$\parallel x(t)\parallel \le \frac{M\parallel x_0\parallel (1+\frac{L_0}{\gamma -\alpha })e_{\alpha \ominus M{\alpha }_0}(t_0,0)}{1+(\alpha \ominus M{\alpha }_0\ominus \lambda )t}.$$

Hence, in view of (e) and the fact $\alpha \ominus M{\alpha }_0\ominus \lambda >0$, we obtain the required result. □

Example 3.4 Let us consider the Volterra integro-dynamic equation

$$\{\begin{array}{c}{x}^{\mathrm{\Delta }}(t)=\ominus 2x(t)+{\int }_0^t{e}_{\ominus 2}(t,s)x(s)\mathrm{\Delta }s,\hfill \\ x(t_k^{+})=\frac{1}{4k}x(t_k),\hfill \\ x(0)=1,\hfill \end{array}$$

(14)

where $A(t)=\ominus 2$, $K(t,s)=e_{\ominus 2}(t,s)$, $1+C_k=\frac{1}{4k}$ and the impulsive points $t_k=2k$. Now consider $L(t,s)=0$ so then $B(t)=\ominus 2$. The matrix function $G(t,s)$ given in (4) becomes

$$G(t,s)=e_{\ominus 2}(t,s).$$

(15)

In the following, we check the assumptions of Theorem 3.3 when $\mathbb{T}=\mathbb{R}$.

Let $\mathbb{T}=\mathbb{R}$. Then we have

$$|e_B(t,s)|=|e_{-2}(t,s)|=e^{2(s-t)}\le M{e}^{2(s-t)},M=2,$$

and

$$0=|L(t,s)|<L_0{e}^{3(s-t)},L_0=1.$$

Here the constants are $\alpha =2$ and $\gamma =3$. From (15) it follows that

$$G(t,s)=e^{-2(t-s)}.$$

(16)

Then from (16) we obtain that $G(t,s)$ is a positive function, and

$$\begin{array}{rcl}{\int }_s^t{e}^{2(\tau -t)}|G(\tau ,s)|d\tau & =& {\int }_s^t{e}^{2(\tau -t)}{e}^{-2(\tau -s)}d\tau \\ =& e^{2(s-t)}(t-s)\\ \le & \frac{(t-s)}{1+2(t-s)}\\ <& \frac{1}{2},\end{array}$$

from which it follows that

$$\underset{0\le s\le t<\mathrm{\infty }}{sup}{\int }_s^t{e}^{\frac{1}{2}(\tau -t)}|G(\tau ,s)|d\tau \le \frac{1}{2}$$

and

$$\prod _{t_0<t_k<t}(5+\frac{1}{k})\le e^{\frac{19k}{10}}.$$

Since ${\alpha }_0=\frac{1}{2}$ and $\lambda =\frac{19}{20}$, then we have that $\alpha -M{\alpha }_0-\lambda >0$. Therefore, since all the assumptions of Theorem 3.3 hold for system (14), it follows that the solution of (14) tends to zero exponentially as $t\to +\mathrm{\infty }$.

If $\mathbb{T}=\mathbb{N}$, then all points are right scattered and there is no impulse condition. So, from [[17], Example 3.7] it follows that the solution of (14) tends to zero exponentially as $t\to +\mathrm{\infty }$.

Theorem 3.5 Let $L\in C(\mathrm{\Omega },M_n(\mathbb{R}))$ such that ${\mathrm{\Delta }}_{s}L(t,s)\in C(\mathrm{\Omega },M_n(\mathbb{R}))$ for $(t,s)\in \mathrm{\Omega }$ and

• (i) assumptions (a), (b), (d) and (e) of Theorem  3.3 hold,

• (ii) $\parallel {\mathrm{\Delta }}_{s}L(t,s)\parallel \le N_0{e}_{\delta }(s,t)$ and $\parallel K(t,s)\parallel \le K_0{e}_{\theta }(s,t)$,

• (iii) $\parallel A(t)\parallel \le A_0$ for $t_0\le t<\mathrm{\infty }$,

• (iv) ${sup}_{t_0\le s\le t<\mathrm{\infty }}{\int }_{\sigma (s)}^t[(K_0+N_0)(1+\mu (\tau )\alpha )+\frac{A_0{L}_0+(\tau -\sigma (s))L_0{K}_0}{\mu (\tau )\alpha }]\mathrm{\Delta }\tau \le {\alpha }_0^{\star }$ for some ${\alpha }_0^{\star }>0$,

where $A_0$, $N_0$, $K_0$, δ and θ are positive real numbers such that $\gamma >\delta >\alpha$, $\theta >\alpha$.

If $\alpha \ominus M{\alpha }_0^{\star }\ominus \lambda >0$, then every solution $x(t)$ of (1) tends to zero exponentially as $t\to +\mathrm{\infty }$.

Proof From (4), we obtain

$$\begin{array}{rcl}\parallel G(t,s)\parallel & \le & \parallel K(t,s)\parallel +\parallel {\mathrm{\Delta }}_{s}L(t,s)\parallel +\parallel L(t,\sigma (s))\parallel \parallel A(s)\parallel \\ +{\int }_{\sigma (s)}^t\parallel L(t,\sigma (u))\parallel \parallel K(u,s)\parallel \mathrm{\Delta }u,\end{array}$$

which implies

$$\begin{array}{rl}\parallel G(t,s)\parallel \le & K_0{e}_{\theta }(s,t)+N_0{e}_{\delta }(s,t)+\frac{L_0{e}_{\gamma }(s,t)}{(1+\mu (t)\alpha )(1+\mu (t)\gamma )}{A}_0\\ +{\int }_{\sigma (s)}^t\frac{L_0{K}_0{e}_{\gamma }(u,t)e_{\theta }(s,u)}{(1+\mu (t)\alpha )(1+\mu (t)\gamma )}\mathrm{\Delta }u.\end{array}$$

(17)

Since $\gamma >\delta >\alpha$, $\theta >\alpha$, then from (i), (ii) and (iii), (17) becomes

$$\begin{array}{rl}\parallel G(t,s)\parallel \le & K_0{e}_{\alpha }(s,t)+N_0{e}_{\alpha }(s,t)\\ +\frac{L_0{e}_{\alpha }(s,t)}{(1+\mu (t)\alpha )(1+\mu (t)\gamma )}{A}_0+\frac{(t-\sigma (s))L_0{K}_0{e}_{\alpha }(s,t)}{(1+\mu (t)\alpha )(1+\mu (t)\gamma )}\end{array}$$

(18)

and

$$e_{\alpha }(\sigma (t),0)\parallel G(t,s)\parallel \le [(K_0+N_0)(1+\mu (t)\alpha )+\frac{A_0{L}_0+(t-\sigma (s))L_0{K}_0}{\mu (t)\alpha }]e_{\alpha }(s,0).$$

Integrating the above inequality and using (iv), we obtain

$${\int }_{\sigma (s)}^t{e}_{\alpha }(\sigma (\tau ),0)\parallel G(\tau ,s)\parallel \mathrm{\Delta }\tau \le {\alpha }_0^{\star }{e}_{\alpha }(s,0).$$

(19)

Substituting (19) in (11), we obtain

$$\begin{array}{rcl}{e}_{\alpha }(t,0)\parallel x(t)\parallel & \le & M\parallel x_0\parallel (1+\frac{L_0}{\gamma -\alpha })e_{\alpha }(t_0,0)+M{\int }_{t_0}^t{\alpha }_0^{\star }{e}_{\alpha }(s,0)\parallel x(s)\parallel \mathrm{\Delta }s\\ +\sum _{t_0<t_k<t}{M}^2(1+d_k)e_{\alpha }(t_k,0)\parallel x(t_k)\parallel .\end{array}$$

Lemma 2.4 yields that

$$\begin{array}{rcl}{e}_{\alpha }(t,0)\parallel x(t)\parallel & \le & M\parallel x_0\parallel (1+\frac{L_0}{\gamma -\alpha })e_{\alpha }(t_0,0)\\ \times \prod _{t_0<t_k<t}[1+M^2(1+d_k)]e_{M{\alpha }_0^{\star }}(t,t_0).\end{array}$$

Using [[18], Theorem 2.36], (e) and the fact that $t_0<t_k<t$, we obtain

$$\parallel x(t)\parallel \le M\parallel x_0\parallel (1+\frac{L_0}{\gamma -\alpha })e_{\alpha \ominus M{\alpha }_0^{\star }}(t_0,0)e_{\alpha \ominus M{\alpha }_0^{\star }\ominus \lambda }(0,t).$$

Then by Lemma 2.2, we have

$$\parallel x(t)\parallel \le \frac{M\parallel x_0\parallel (1+\frac{L_0}{\gamma -\alpha })e_{\alpha \ominus M{\alpha }_0^{\star }}(t_0,0)}{1+(\alpha \ominus M{\alpha }_0^{\star }\ominus \lambda )t}.$$

Hence, in view of (i) and $\alpha \ominus M{\alpha }_0^{\star }\ominus \lambda >0$, we obtain the required result. □

Corollary 3.6 Let $L(t,s)$ be an $n\times n$ continuously differentiable matrix function with respect to s on $t_{k-1}<s\le t_k<t$ with $L(t,t_k^{+})={(I+C_k)}^{-1}L(t,t_k)$ for each $k=1,2,\dots$ . Then (1) is equivalent to the impulsive dynamic system

$$\{\begin{array}{c}{y}^{\mathrm{\Delta }}(t)=B(t)y(t)+H(t),t\in {\mathbb{T}}_0\mathrm{\setminus }\{t_k\},\hfill \\ y(t_k^{+})=(I+C_k)y(t_k),t=t_k,k=1,2,\dots ,\hfill \\ y(t_0)=y_0,\hfill \end{array}$$

(20)

where

$$\begin{array}{r}B(t)=A(t)-L(t,t),\\ H(t)=F(t)+L(t,t_0)x_0+{\int }_s^{t}L(t,\sigma (s))F(s)\mathrm{\Delta }s,\end{array}$$

(21)

and

$$\begin{array}{r}K(t,s)+{\mathrm{\Delta }}_{s}L(t,s)+L(t,\sigma (s))A(s)\\ +{\int }_{\sigma (s)}^{t}L(t,\sigma (u))K(u,s)\mathrm{\Delta }u=0,s,t\ne t_k.\end{array}$$

(22)

Proof The proof follows an argument similar to that in Theorem 3.1 with $G(t,s)=0$. □

4 Boundedness

In the first result of this section, we give sufficient conditions to insure that (1) has bounded solutions. Our results apply to (1) whether $A(t)$ is stable, identically zero, or completely unstable, and do not require $A(t)$ to be constant nor $K(t,s)$ to be a convolution kernel. Let $C(t)$ and $D(t,s)$ be continuous $n\times n$ matrices, $t_0\le s\le t<\mathrm{\infty }$. Let $s\in [t_0,\mathrm{\infty })$ and assume that $C(t)$ is an $n\times n$ regressive matrix. The unique matrix solution of the initial valued problem

$$Y^{\mathrm{\Delta }}=C(t)Y,Y\left(t_k^{+}\right)=(I+C_k)Y(t_k),Y(s)=I,$$

(23)

is called the impulsive transition matrix (at s) and it is denoted by $S_C(t,s)$ (see [[16], Corollary 3.1]). Also, if $H(t,s)$ is an $n\times n$ regressive matrix satisfying

$$\{\begin{array}{c}{\mathrm{\Delta }}_{t}H(t,s)=C(t)H(t,s)+D(t,s),\hfill \\ H(t_k^{+},s)=(I+C_k)H(t_k,s),\hfill \\ H(s,s)=A(s)-C(s),\hfill \end{array}$$

(24)

then

$$H(t,s)=S_C(t,s)[A(s)-C(s)]+{\int }_s^t{S}_C(t,\sigma (\tau ))D(\tau ,s)\mathrm{\Delta }\tau .$$

(25)

Theorem 4.1 Let $e_C(t,s)$ be the solution of (23), and suppose that there are positive constants N, J and M such that

• (i) $\parallel S_C(t,t_0)\parallel \le N$,

• (ii) ${\int }_{t_0}^t\parallel S_C(t,s)[A(s)-C(s)]+{\int }_s^t{S}_C(t,\sigma (\tau ))K(\tau ,s)\mathrm{\Delta }\tau \parallel \mathrm{\Delta }s\le J<1$,

• (iii) $\parallel {\int }_{t_0}^t{S}_C(t,\sigma (u))[F(u)-G(t)x(t)]\mathrm{\Delta }u\parallel \le M$.

Then all the solutions of (1) are uniformly bounded, and the zero solution of the corresponding homogenous equation of (1) is uniformly stable with the initial condition $x(t_0)=0$.

Proof Consider the following functional

$$V(t,x(\cdot ))=x(t)-{\int }_{t_0}^{t}H(t,s)x(s)\mathrm{\Delta }s.$$

(26)

The derivative of $V(t,x(\cdot ))$ along a solution $x(t)=x(t,t_0,x_0)$ of (1) satisfies

$$V^{\mathrm{\Delta }}(t,x(\cdot ))=x^{\mathrm{\Delta }}(t)-{\mathrm{\Delta }}_t{\int }_{t_0}^{t}H(t,s)x(s)\mathrm{\Delta }s.$$

From [[18], Theorem 1.117], we obtain

$$\begin{array}{rcl}{V}^{\mathrm{\Delta }}(t,x(\cdot ))& =& x^{\mathrm{\Delta }}(t)-H(\sigma (t),t)x(t)-{\int }_{t_0}^t{\mathrm{\Delta }}_{t}H(t,s)x(s)\mathrm{\Delta }s\\ =& A(t)x(t)-H(\sigma (t),t)x(t)+{\int }_{t_0}^{t}K(t,s)x(s)\mathrm{\Delta }s\\ -{\int }_{t_0}^t{\mathrm{\Delta }}_{t}H(t,s)x(s)\mathrm{\Delta }s+F(t)\end{array}$$

or

$$\begin{array}{rcl}{V}^{\mathrm{\Delta }}(t,x(\cdot ))& =& [A(t)-H(\sigma (t),t)]x(t)+F(t)\\ +{\int }_{t_0}^t[K(t,s)-{\mathrm{\Delta }}_{t}H(t,s)]x(s)\mathrm{\Delta }s.\end{array}$$

(27)

By using (25), [[18], Theorems 1.75] and [[16], Theorem 3.4], we have the following expression:

$$\begin{array}{rcl}H(\sigma (t),t)& =& S_C(\sigma (t),t)[A(t)-C(t)]+{\int }_t^{\sigma (t)}{S}_C(\sigma (t),\sigma (\tau ))D(\tau ,t)\mathrm{\Delta }\tau \\ =& (I+\mu (t)C(t))S_C(t,t)[A(t)-C(t)]+\mu (t)S_C(\sigma (t),\sigma (t))D(t,t)\\ =& (I+\mu (t)C(t))[A(t)-C(t)]+\mu (t)D(t,t)\\ =& [A(t)-C(t)]+\mu (t)[C(t)A(t)-C^2(t)+D(t,t)],\end{array}$$

which implies that

$$H(\sigma (t),t)=[A(t)-C(t)]+G(t),$$

(28)

where $G(t)=\mu (t)[C(t)A(t)-C^2(t)+D(t,t)]$. Substituting (28) in (27) yields

$$V^{\mathrm{\Delta }}(t,x(\cdot ))=C(t)x(t)-G(t)x(t)+{\int }_{t_0}^t[K(t,s)-{\mathrm{\Delta }}_{t}H(t,s)]x(s)\mathrm{\Delta }s+F(t).$$

From (24) and (26), we have

$$V^{\mathrm{\Delta }}(t,x(\cdot ))=C(t)V(t,x(\cdot ))+{\int }_{t_0}^t[K(t,s)-D(t,s)]x(s)\mathrm{\Delta }s+F(t)-G(t)x(t),$$

and it is easy to see that

$$V(t_k^{+},x(\cdot ))=(I+C_k)V(t_k,x(\cdot )).$$

Thus

$$V(t,x(\cdot ))=S_C(t,t_0)x_0+{\int }_{t_0}^t{S}_C(t,\sigma (u))g(u,x(\cdot ))\mathrm{\Delta }u,$$

(29)

where

$$g(t,x(\cdot ))={\int }_{t_0}^t[K(t,s)-D(t,s)]x(s)\mathrm{\Delta }s+F(t)-G(t)x(t).$$

Let $D(t,s)=K(t,s)$. Then by (25), (ii) is precisely ${\int }_{t_0}^t\parallel H(t,s)\parallel \mathrm{\Delta }s\le J<1$. By (29) and (i)-(iii),

$$\begin{array}{rcl}\left|V(t,x(\cdot ))\right|& =& \parallel S_C(t,t_0)x_0+{\int }_{t_0}^t{S}_C(t,\sigma (u))[F(u)-G(t)x(t)]\mathrm{\Delta }u\parallel \\ \le & \parallel S_C(t,t_0)\parallel \parallel x_0\parallel +\parallel {\int }_{t_0}^t{S}_C(t,\sigma (u))[F(u)-G(t)x(t)]\mathrm{\Delta }u\parallel \\ \le & N\parallel x_0\parallel +M.\end{array}$$

If $\parallel x_0\parallel <B_1$ for some constant, and if $Q=N{B}_1+M$, then by (26) we obtain

$$\parallel x(t)\parallel -{\int }_{t_0}^t\parallel H(t,s)\parallel \parallel x(s)\parallel \mathrm{\Delta }s\le \parallel V(t,x(\cdot ))\parallel \le Q.$$

(30)

Now, either there exists $B_2>0$ such that $\parallel x(t)\parallel <B_2$ for all $t\ge t_0$, and thus $x(t)$ is uniformly bounded, or there exists a monotone sequence $\{t_n\}$ tending to infinity such that $\parallel x(t_n)\parallel ={max}_{t_0\le t\le t_n}\parallel x(t)\parallel$ and $\parallel x(t_n)\parallel \to \mathrm{\infty }$ as $t_n\to \mathrm{\infty }$, and by (ii) and (30) we have

$$\parallel x(t_n)\parallel (1-J)\le \parallel x(t_n)\parallel -{\int }_{t_0}^{t_n}\parallel H(t_n,s)\parallel \parallel x(s)\parallel \mathrm{\Delta }s\le Q,$$

a contradiction. This completes the proof. □

In the second part of this section, we consider system (1) with $F(t)$ bounded and suppose that

$$C(t,s)=-{\int }_t^{\mathrm{\infty }}K(u,s)\mathrm{\Delta }u$$

(31)

is defined and continuous on Ω. The matrix $E(t)$ on $[t_0,\mathrm{\infty })$ is defined by

$$E(t)=A(t)-C(\sigma (t),t).$$

(32)

Then (1) is equivalent to the system

$$\{\begin{array}{c}{x}^{\mathrm{\Delta }}(t)=E(t)x(t)+{\mathrm{\Delta }}_t{\int }_{t_0}^{t}C(t,s)x(s)\mathrm{\Delta }s+F(t),t\in {\mathbb{T}}_0\mathrm{\setminus }\{t_k\},\hfill \\ x(t_k^{+})=(I+C_k)x(t_k),t=t_k,k=1,2,\dots ,\hfill \\ x(t_0)=x_0.\hfill \end{array}$$

(33)

Theorem 4.2 Let $E\in C(\mathbb{T},M_n(\mathbb{R}))$ and $M,\alpha >0$. Assume that $E(t)$ commutes with its integral. If

$$\parallel e_E(t,s)\parallel \le M{e}_{\alpha }(s,t),t,s\in \mathrm{\Omega },$$

(34)

then every solution $x(t)$ of (1) with $x(t_0)=x_0$ satisfies

$$\begin{array}{rl}\parallel x(t)\parallel \le & M\parallel x_0\parallel e_{\alpha }(t_0,t)+M{\int }_{t_0}^t{e}_{\alpha }(\sigma (s),t)\parallel F(s)\parallel \mathrm{\Delta }s\\ +M{\int }_{t_0}^t\parallel E(u)\parallel e_{\alpha }(\sigma (u),t)\left[{\int }_{t_0}^u\parallel C(u,s)\parallel \parallel x(s)\parallel \mathrm{\Delta }s\right]\mathrm{\Delta }u\\ +{\int }_{t_0}^t\parallel C(t,s)\parallel \parallel x(s)\parallel \mathrm{\Delta }s+M{e}_{\alpha }(t_0,t)\sum _{t_0<t_k<t}\parallel {\beta }_k\parallel \parallel x(t_k)\parallel ,\end{array}$$

(35)

where ${\beta }_k=[e_E(t_0,t_k^{+})(I+C_k)-e_E(t_0,t_k)]$.

Proof Let $x(t)$ be the solution of (1) and define $q(t)=e_E(t_0,t)x(t)$. Then

$$q^{\mathrm{\Delta }}(t)=-E(t)e_E(t_0,\sigma (t))x(t)+e_E(t_0,\sigma (t))x^{\mathrm{\Delta }}(t).$$

Substituting for $x^{\mathrm{\Delta }}(t)$ from (33) and integrating from $t_0$ to t, yields

$$\begin{array}{rcl}q(t)-q(t_0)-\sum _{t_0<t_k<t}\mathrm{\Delta }q(t_k)& =& {\int }_{t_0}^t{e}_E(t_0,\sigma (s))F(s)\mathrm{\Delta }s\\ +{\int }_{t_0}^t{e}_E(t_0,\sigma (u))[{\mathrm{\Delta }}_u{\int }_{t_0}^{u}C(u,s)x(s)\mathrm{\Delta }s]\mathrm{\Delta }u.\end{array}$$

Applying the integration by parts on the second term of the right-hand side [[18], Theorem 1.77] and the semigroup property of exponential functions [[18], Theorem 2.36], we obtain

$$\begin{array}{rl}x(t)=& e_E(t,t_0)x_0+{\int }_{t_0}^t{e}_E(t,\sigma (s))F(s)\mathrm{\Delta }s+{\int }_{t_0}^{t}C(t,s)x(s)\mathrm{\Delta }s\\ +{\int }_{t_0}^{t}E(u)e_E(t,\sigma (s))[{\int }_{t_0}^{u}C(u,s)x(s)\mathrm{\Delta }s]\mathrm{\Delta }u+e_E(t,t_0)\sum _{t_0<t_k<t}\mathrm{\Delta }q(t_k).\end{array}$$

(36)

For $t_0<s<t_k<t$, we have

$$\begin{array}{rcl}\mathrm{\Delta }q(t_k)& =& e_E(t_0,t_k^{+})x\left(t_k^{+}\right)-e_E(t_0,t_k)x(t_k)\\ =& [e_E(t_0,t_k^{+})(I+C_k)-e_E(t_0,t_k)]x(t_k)\\ =& {\beta }_{k}x(t_k).\end{array}$$

Hence, using (34) and applying the norm on (36), we obtain (35), which completes the proof. □

Assume that the hypotheses of Theorem 4.2 hold for next results.

Theorem 4.3 Let $x(t)$ be a solution of (1). If $\parallel E(t)\parallel \le d$ on $[t_0,\mathrm{\infty })$ for some $d>0$, $F(t)$ is bounded and ${sup}_{t_0\le t<\mathrm{\infty }}{\int }_{t_0}^t\parallel C(t,s)\parallel \mathrm{\Delta }s+\frac{1}{d}{\sum }_{t_0<t_k<t}\parallel {\beta }_k\parallel \le \beta$, with β sufficiently small, then $x(t)$ is bounded.

Proof For the given $t_0$ and bounded $F(t)$, there is $C_1>0$ with

$$M\parallel x_0\parallel e_{\alpha }(t_0,t)+M\underset{t_0\le t<\mathrm{\infty }}{sup}{\int }_{t_0}^t{e}_{\alpha }(\sigma (s),t)\parallel F(s)\parallel \mathrm{\Delta }s<C_1.$$

(37)

Substituting (37) in (35), we obtain

$$\begin{array}{rcl}\parallel x(t)\parallel & \le & C_1+Md{\int }_{t_0}^t{e}_{\alpha }(\sigma (u),t)\left[{\int }_{t_0}^u\parallel C(u,s)\parallel \parallel x(s)\parallel \mathrm{\Delta }s\right]\mathrm{\Delta }u\\ +{\int }_{t_0}^t\parallel C(t,s)\parallel \parallel x(s)\parallel \mathrm{\Delta }s+M{e}_{\alpha }(t_0,t)\sum _{t_0<t_k<t}\parallel {\beta }_k\parallel \parallel x(t_k)\parallel ,\\ \le & C_1+\frac{Md}{\alpha }\beta \underset{t_0\le s<\mathrm{\infty }}{sup}\parallel x(s)\parallel +\beta \underset{t_0\le s<\mathrm{\infty }}{sup}\parallel x(s)\parallel \\ =& C_1+\beta [1+\frac{Md}{\alpha }]\underset{t_0\le s<\mathrm{\infty }}{sup}\parallel x(s)\parallel .\end{array}$$

Let β be chosen so that $\beta [1+\frac{Md}{\alpha }]=m<1$. Then

$$\parallel x(t)\parallel \le C_1+m\underset{t_0\le s<t}{sup}\parallel x(s)\parallel .$$

Let $C_2>\parallel x_0\parallel$ and $C_1+m{C}_2<C_2$. If $\parallel x(t)\parallel$ is not bounded, then there exists a first $t_1>t_0$ with $\parallel x(t_1)\parallel =C_2$, and then

$$C_2=\parallel x(t_1)\parallel \le C_1+m{C}_2<C_2,$$

a contradiction. This completes the proof. □

Theorem 4.4 If $F(t)=0$ in (1), $\parallel E(t)\parallel \le d$ on $[t_0,\mathrm{\infty })$ for some $d>0$, and ${\int }_{t_0}^t\parallel C(t,s)\parallel \mathrm{\Delta }s+\frac{1}{d}{\sum }_{t_0<t_k<t}\parallel {\beta }_k\parallel \le \beta$ for β sufficiently small, then the zero solution of (1) with the initial condition $x(t_0)=0$ is uniformly stable.

Proof Let $ϵ>0$ be given. We wish to find $\delta >0$ such that $t_0\ge 0$, $\parallel x_0\parallel <\delta$, and $t\ge t_0$ implies $\parallel x(t,x_0)\parallel <ϵ$. Let $\delta <ϵ$ with δ yet to be determined. If $\parallel x_0\parallel <\delta$, then $M\parallel x_0\parallel \le M\delta$. From (35) with $F(t)=0$,

$$\begin{array}{rcl}\parallel x(t)\parallel & \le & M\delta +\frac{Md}{\alpha }\beta \underset{t_0\le s<t}{sup}\parallel x(s)\parallel +\beta \underset{t_0\le s<t}{sup}\parallel x(s)\parallel \\ =& M\delta +\beta [1+\frac{Md}{\alpha }]\underset{t_0\le s<t}{sup}\parallel x(s)\parallel .\end{array}$$

First take β so that $\beta [1+\frac{Md}{\alpha }]\le \frac{3}{4}$ and δ so that $M\delta +\frac{3}{4}ϵ<ϵ$. If $\parallel x_0\parallel <\delta$ and if there exists $t_1>t_0$ with $\parallel x(t_1)\parallel =ϵ$, we have

$$ϵ=\parallel x(t_1)\parallel <M\delta +\frac{3}{4}ϵ<ϵ,$$

a contradiction. Thus the zero solution is uniformly stable. The proof is complete. □

Example 4.5 Let us consider the following system:

$$\{\begin{array}{c}{x}^{\mathrm{\Delta }}(t)=-\frac{1}{{(\sigma (t))}^2}+{\int }_0^t(\frac{1}{t^2\sigma (t)}+\frac{1}{t{(\sigma (t))}^2})x(s)\mathrm{\Delta }s,\hfill \\ x(t_k^{+})=(1+C_k)x(t_k),\hfill \\ x(0)=0,\hfill \end{array}$$

(38)

where $A(t)=-\frac{1}{{(\sigma (t))}^2}$ and $K(t,s)=(\frac{1}{t^2\sigma (t)}+\frac{1}{t{(\sigma (t))}^2})$. It is easy to check that

$${\left(\frac{1}{u^2}\right)}^{\mathrm{\Delta }}=-\frac{1}{u^2\sigma (u)}-\frac{1}{u{(\sigma (u))}^2}.$$

(39)

By using (31) and (39), we obtain

$$\begin{array}{rcl}C(t,s)& =& {\int }_t^{\mathrm{\infty }}(-\frac{1}{u^2\sigma (u)}-\frac{1}{u{(\sigma (u))}^2})\mathrm{\Delta }u\\ =& {\int }_t^{\mathrm{\infty }}{\left(\frac{1}{u^2}\right)}^{\mathrm{\Delta }}\mathrm{\Delta }u\\ =& \frac{1}{t^2}.\end{array}$$

This implies that $E(t)=0$ and

$${\int }_0^t\frac{1}{t^2}\mathrm{\Delta }s=\frac{1}{t}.$$

(40)

Finally, by taking the supremum over t in (40), over ${[0,\mathrm{\infty })}_{\mathbb{T}}$, we obtain

$$sup{\int }_0^t\frac{1}{t^2}\mathrm{\Delta }s=0.$$

Since ${\beta }_k=C_k$, so we can choose $C_k$ such that $\frac{1}{d}{\sum }_{t_0<t_k<t}\parallel {\beta }_k\parallel \le \beta$ for β sufficiently small. It follows that all the assumptions of Theorem 4.3 are satisfied, hence all the solutions of (38) are bounded. Moreover, Theorem 4.4 yields that the zero solution of (38) is uniformly stable on an arbitrary time scale.