Exponential Stability of Volterra Integro Dynamic Equations on Time Scales

Abstract

In this paper, we give new sufficient conditions for boundedness and exponential stability of solutions for nonlinear Volterra integro dynamic equations from above on unbounded time scales using first Lyapunovs method. To prove this result we reduce the n-dimensional problem to the corresponding scalar one using the concept of matrix measure and a new simpler proof of Coppel’s inequality on the time scales. There is an example that illustrates the conditions of the theorem.

1. Introduction

The theory of time scales was introduced by Hilger in his Ph.D. thesis “Ein Maßkettenkalkül mit Anwendung auf Zentrumsmannigfaltigkeiten” [1] in 1988. The main idea of time scale calculus is to unify and extend continuous and discrete analysis. For an introduction to time scales, we refer to the monographs by Bohner and Peterson [2,3].

The concept of the logarithmic norm of a matrix was introduced independently in 1958 by Dahlquist [4] and Lozinskii [5] when they estimated the error of discretization in numerical analysis of differential equations. Russo and Wirth [6] introduced the concept of matrix measure to time scales, thus extending the concept of logarithmic norm to time scales. Often, the use of the concept of matrix measures allows the solution of the n-dimensional problem reduced to the solution of the corresponding scalar problem and, as a result, we investigate a new and more complete solution to the given one.

Let $\mathbb{T}$ be an unbounded from above time scale with graininess $\mu :\mathbb{T}\to [0,+\infty )$. In this paper, we consider nonlinear Volterra integro dynamic equation

$$x^{\Delta }\left(t\right)=A\left(t\right)x\left(t\right)+f(t,x\left(t\right))+{\int }_s^{t}K(t,r,x\left(r\right))\Delta r,x\left(s\right)=x_s,$$

(1)

where $s,t\in \mathbb{T}$, $x,x_s\in {\mathbb{R}}^n$. The right hand side of (1) is rd-continuous with respect to $t\in \mathbb{T}$, locally Lipschitz continuous with respect to $x\in {\mathbb{R}}^n$. Matrix valued function $A:\mathbb{T}\to {\mathbb{R}}^n\times {\mathbb{R}}^n$ is rd-continuos and regressive, i.e., $A\in \mathcal{R}$. Vector valued functions $f:\mathbb{T}\times {\mathbb{R}}^n\to {\mathbb{R}}^n$ and $K:\mathbb{T}\times \mathbb{T}\times {\mathbb{R}}^n\to {\mathbb{R}}^n$ satisfy conditions

$$|K(t,r,x)-K(t,r,x^{\prime })|\le L(t,r)|x-x^{\prime }|,$$

$$|f(t,x)-f(t,x^{\prime })|\le \epsilon \left(t\right)|x-x^{\prime }|,$$

where functions $L:\mathbb{T}\times \mathbb{T}\to \mathbb{R}$ and $\epsilon :\mathbb{T}\to \mathbb{R}$ are non-negative and rd-continuous.

We are interested to give the sufficient conditions for boundedness and exponential stability of solutions of (1) by using a novel matrix measure approach on the time scales. Let us note that the nonlinear Equation (1) contains an additionally Volterra type integral term, which takes into account the behaviour of the solution in the past.

To study the stability of a solution in the sense of Lyapunov, one usually uses the first or second Lyapunov method. According to the second Lyapunov method, a Lyapunov function with defined properties must be found. There is no general method for constructing the corresponding Lyapunov function, and often finding the Lyapunov function is very complicated.

According to the first Lyapunov method, the stability of the solution is studied based on the properties of the right-hand side of the equation. With the help of an appropriate Lyapunov transformation, the linear part of the equation is transformed into a simpler form. For example, if $\mathbb{T}=\mathbb{R}$ and the matrix A is constant or periodic, the linear part of the equation is transformed into Jordan normal form. In the case of exponential stability, the Jordan matrix measure is negative, although the norm of the matrix itself is positive. The matrix measure can be found without difficulty even using numerical methods. In addition, we note that the measure of the matrix changes a little for sufficiently small $C^1$ type perturbations of the dynamical system. In the above case, an interesting and at the same time difficult problem arises—finding an appropriate Lyapunov-type transformation for dynamic equations on a time scale.

In their monograph Adivar and Raffoul [7] study and summarize their own and other research [8,9,10,11] on the boundedness and exponential stability of solutions to the first order dynamic equations and also to Volterra integro dynamic equations on the time scale using the Lyapunov-type function method. Qiang et al. [12,13] papers are also dedicated to exponential stability for dynamic equations on time scales, based on the Lyapunov’s second method. Compared with the traditional exponential stability results of perturbed systems, the time derivatives of related Lyapunov functions are not required to be negative definite for all time. Du and Thien [14] proved the preservation of exponential stability under small enough Lipschitz perturbations for the first order dynamic equations on the time scales. The scalar Volterra integro dynamic equation on a time scales was studied by Ostaszewska et al. [15]. Recently Harisha et al. [16] consider Voltera integro dynamic Sylvester matrix system on time scales.

Also worth noticing is monograph by Georgiev [17] on functional dynamic equations on time scales and the references cited therein. We can also note the works of other authors related to the Volterra integro dynamic equations [18,19,20,21,22] and using matrix measure.

As far as we know, no one has studied exponential stability of nonlinear Volterra integro dynamic equations on time scales based on matrix measure on n-dimensional space.

2. Preliminaries

In this section, we recall some basic definitions and modify some results that will be useful for the further development of this work.

2.1. Time Scales

A time scale is an arbitrary nonempty closed subset of the real numbers $\mathbb{R}$ with the topology inherited from the standard one in $\mathbb{R}$. We denote a time scale by the symbol $\mathbb{T}$ and assume that a time scale is unbounded above, because we are interested in the behaviour of the functions near ∞.

Since a time scale may or may not be connected, we need the concept of the jump operators to describe the structure of the time scale under consideration and also to define the derivative. The forward and the backward jump operators $\sigma ,\rho :\mathbb{T}\to \mathbb{T}$ and the graininess $\mu :\mathbb{T}\to [0,+\infty )$ are defined, respectively, by $\sigma \left(t\right)=inf\{s\in \mathbb{T}\mid s>t\}$, $\rho \left(t\right)=sup\{s\in \mathbb{T}\mid s<t\}$, $\mu \left(t\right)=\sigma \left(t\right)-t.$

The jump operators allow the classification of points in a time scale $\mathbb{T}$. If $\sigma \left(t\right)>t$, then the point $t\in \mathbb{T}$ is called right scattered, while if $\rho \left(t\right)<t$, then the point $t\in \mathbb{T}$ is called left scattered. If $\sigma \left(t\right)=t$, then $t\in \mathbb{T}$ is called right dense, while if $\rho \left(t\right)=t$, then $t\in \mathbb{T}$ is called left dense.

We say that function $f:\mathbb{T}\to {\mathbb{R}}^{n\times n}$ is rd-continuous provided f is continuous at each right dense point of $\mathbb{T}$ and has a finite left sided limit at each left dense point of $\mathbb{T}$. Every rd-continuous function on a compact interval is bounded. The function $f:\mathbb{T}\to {\mathbb{R}}^{n\times n}$ is regressive with respect to $\mathbb{T}$ if the mapping

$$I+\mu \left(t\right)f\left(t\right):{\mathbb{R}}^{n\times n}\to {\mathbb{R}}^{n\times n}$$

is invertible for all $t\in \mathbb{T}$, where $I\in {\mathbb{R}}^{n\times n}$ is an identity matrix. The set of all regressive and rd-continuous functions is denoted by $\mathcal{R}=\mathcal{R}(\mathbb{T},{\mathbb{R}}^{n\times n})$.

A function $p:\mathbb{T}\to \mathbb{R}$ is called positively regressive if

$$1+\mu \left(t\right)p\left(t\right)>0$$

for all $t\in \mathbb{T}$. The set of all positively regressive and rd-continuous functions is denoted by ${\mathcal{R}}^{+}={\mathcal{R}}^{+}(\mathbb{T},\mathbb{R})$ and is an Abelian or regressive group together with the circle addition ⊕. If $p,q\in {\mathcal{R}}^{+}$, then circle addition ⊕ defined by $(p\oplus q)\left(t\right)=p\left(t\right)+q\left(t\right)+\mu \left(t\right)p\left(t\right)q\left(t\right)$, the inverse element is given by $(\ominus p)\left(t\right)=-{(1+\mu \left(t\right)p\left(t\right))}^{-1}p\left(t\right)$ and circle subtraction ⊖ defined by $(p\ominus q)\left(t\right)=(p\oplus (\ominus q\left)\right)\left(t\right)$.

Assume that $|·|$ is the Euclidean vector norm on ${\mathbb{R}}^n$ with induced matrix norm $\parallel ·\parallel$. We say that function $f:\mathbb{T}\to {\mathbb{R}}^{n\times n}$ is differentiable at $t\in \mathbb{T}$ if there exists $f^{\Delta }\left(t\right)\in {\mathbb{R}}^{n\times n}$ such that for each $\epsilon >0$ there exists a neighbourhood U of t with

$$∥f\left(\sigma \left(t\right)\right)-f\left(s\right)-f^{\Delta }\left(t\right)(\sigma \left(t\right)-s)∥\le \epsilon \left|\sigma \left(t\right)-s\right|\mathrm{for}\mathrm{all}s\in U.$$

In this case $f^{\Delta }\left(t\right)$ is said to be the delta derivative of f at $t\in \mathbb{T}$.

A function $F:\mathbb{T}\to {\mathbb{R}}^{n\times n}$ is called an antiderivative of $f:\mathbb{T}\to {\mathbb{R}}^{n\times n}$ provided $F^{\Delta }\left(t\right)=f\left(t\right)$ holds for all $t\in \mathbb{T}$. In this case, we define the (Cauchy) delta integral of f by

$${\int }_s^{t}f\left(r\right)\Delta r=F\left(t\right)-F\left(s\right)\mathrm{for}\mathrm{all}s,t\in \mathbb{T}.$$

2.2. Exponential Functions on Time Scales

Let $A\in \mathcal{R}$ and let us consider the (Cauchy) initial value problem for linear dynamic equation

$$x^{\Delta }\left(t\right)=A\left(t\right)x\left(t\right),x\left(s\right)=x_s,t,s\in \mathbb{T}$$

and the corresponding matrix initial value problem

$$X^{\Delta }\left(t\right)=A\left(t\right)X\left(t\right),X\left(s\right)=I,t,s\in \mathbb{T}.$$

(2)

According to the Existence and Uniqueness Theorem [2,3], the Cauchy problem has a unique continuous solution

$$x(t,s)=e_A(t,s)x_s,$$

where $e_A(·,s):\mathbb{T}\to {\mathbb{R}}^{n\times n}$, $e_A(s,s)=I$ is continuous solution of matrix initial value problem (2). Since the right-hand side of (2) is rd-continuous, than the solution $e_A(·,s)$ is rd-differentiable i.e., differentiable with rd-continuous derivative. The unique solution of (2) is called the matrix exponential function.

In the following theorem, we use some fundamental properties of the matrix exponential function on time scales.

Theorem 1.

If $A\in \mathcal{R}$, $t,s,r\in \mathbb{T}$ and $e_A(·,s):\mathbb{T}\to {\mathbb{R}}^{n\times n}$ is matrix exponential function, then

(i) 

$e_0(t,s)=I$ and $e_A(t,t)=I$,

(ii) 

$e_A(\sigma \left(t\right),s)=(I+\mu \left(t\right)A\left(t\right))e_A(t,s)$,

(iii) 

semigroup property $e_A(t,s)e_A(s,r)=e_A(t,r)$,

(iv) 

If the point $t\in \mathbb{T}$ is right dense, then for every $\epsilon >0$ there is $\delta >0$ such that

$$\parallel e_A(t+h,s)-e_A(t,s)-hA\left(t\right)e_A(t,s)\parallel \le \epsilon h$$

for all $t,t+h\in \mathbb{T}$ and $0<h<\delta$.

Proof. 

Refer to [2,3] for the proof. □

Let $p\in \mathcal{R}$. The Cauchy problem for homogeneous linear equation on time scales

$$x^{\Delta }\left(t\right)=p\left(t\right)x\left(t\right),x\left(s\right)=1,t,s\in \mathbb{T}$$

(3)

has a unique continuous solution $e_p(·,s):\mathbb{T}\to \mathbb{R}$. The unique solution of (3) is called the generalized exponential function.

We also use some additional properties of the generalized exponential function on time scales.

Theorem 2.

If $p,q\in \mathcal{R}$ and $t,s\in \mathbb{T}$, then

(v) 

${\int }_s^{t}p\left(r\right)e_p(t,\sigma \left(r\right))\Delta r=e_p(t,s)-1$,

(vi) 

$e_p(t,s)e_q(t,s)=e_{p\oplus q}(t,s)$,

(vii) 

If $p\in {\mathcal{R}}^{+}$, then $e_p(t,s)>0$,

(viii) 

If $p\in {\mathcal{R}}^{+}$ and $p\left(t\right)\le q\left(t\right)$ for $t\in \mathbb{T}$, then $e_p(t,s)\le e_q(t,s)$ for $s,t\in \mathbb{T}$ and $t\ge s$.

Proof. 

Look up [2,3] for the proof of (v)–(vii) and [23] for the proof of (viii). □

We can prove some more useful relations.

Corollary 1.

Suppose $p\in {\mathcal{R}}^{+}$, $p\left(t\right)\le 0$ for $s,t\in \mathbb{T}$. Then

(ix) 

we have by Theorem (1) (i) and Theorem (2) (viii)

$$0<e_p(t,s)\le 1$$

for $t\ge s$;

(x) 

we have by Theorem (2) (v)

$$0\le {\int }_s^t|p\left(r\right)|e_p(t,\sigma \left(r\right))\Delta r=-{\int }_s^{t}p\left(r\right)e_p(t,\sigma \left(r\right))\Delta r=1-e_p(t,s)<1$$

for $t\ge s$.

2.3. Matrix Measure and Coppel’s Inequality on Time Scales

To introduce the matrix measure and to prove the Coppel’s inequality, we used the properties of a convex function.

Definition 1.

A function $\phi :(a,b)\to \mathbb{R}$ is called convex if for each pair of points $t_1,t_2\in (a,b)$ and each λ with $0\le \lambda \le 1$ the condition

$$\phi (\lambda t_1+(1-\lambda )t_2)\le \lambda \phi \left(t_1\right)+(1-\lambda )\phi \left(t_2\right)$$

is satisfied.

Theorem 3.

Assume that $\phi :(a,b)\to \mathbb{R}$ is a convex function. Then φ has finite left and right hand derivative at each point $t\in (a,b)$.

Proof. 

Refer to [24] for the proof. □

Corollary 2.

Consider the function $\phi :\mathbb{R}\to \mathbb{R}$, where $\phi \left(h\right)=\parallel (I+hA)J\parallel$ and $A,J\in {\mathbb{R}}^{n\times n}$. Then function φ is convex. Indeed

$$\begin{array}{c}\hfill \phi (\lambda h_1+(1-\lambda )h_2)=\parallel \lambda (I+h_{1}A)J+(1-\lambda )(I+h_{2}A)J\parallel \\ \hfill \le \lambda \parallel (I+h_{1}A)J\parallel +(1-\lambda )\parallel (I+h_{2}A)J\parallel =\lambda \phi \left(h_1\right)+(1-\lambda )\phi \left(h_2\right).\end{array}$$

It follows that there exists finite right hand derivative at point $h=0$

$$\underset{h\to +0}{lim}{\displaystyle \frac{\phi \left(h\right)-\phi \left(0\right)}{h}}=\underset{h\to +0}{lim}{\displaystyle \frac{\parallel (I+hA)J\parallel -\parallel J\parallel }{h}}.$$

If $J=I$, then there also exists a finite limit

$$\underset{h\to +0}{lim}{\displaystyle \frac{\parallel I+hA\parallel -1}{h}}.$$

Finally we have

$$\underset{h\to +0}{lim}{\displaystyle \frac{\parallel (I+hA)J\parallel -\parallel J\parallel }{h}}\le \underset{h\to +0}{lim}{\displaystyle \frac{\parallel I+hA\parallel -1}{h}}\parallel J\parallel .$$

Definition 2

([6]). Assume $A:\mathbb{T}\to {\mathbb{R}}^{n\times n}$ is matrix valued function. The matrix measure $m\left(A\right):\mathbb{T}\to \mathbb{R}$ is defined for $t\in \mathbb{T}$ as:

$$m\left(A\left(t\right)\right)=\left\{\begin{array}{cc}{\displaystyle \frac{\parallel I+\mu \left(t\right)A\left(t\right)\parallel -1}{\mu \left(t\right)}},\hfill & if \mu \left(t\right)>0,\hfill \\ \underset{h\to +0}{lim}{\displaystyle \frac{\parallel I+hA\left(t\right)\parallel -1}{h}},\hfill & if \mu \left(t\right)=0.\hfill \end{array}\right.$$

Let us note that if the matrix valued function A is rd-continuous, then matrix measure $m\left(A\right)$ is also rd-continuous as composition of continuous and rd-continuous functions [2,3].

Lemma 1

([6]). $m\left(A\right)\in {\mathcal{R}}^{+}$ if and only if $I+\mu \left(t\right)A\left(t\right)\ne 0$.

Proof. 

The result is trivial if $\mu \left(t\right)=0$. We should only consider the case when $\mu \left(t\right)>0$. We have

$$1+\mu \left(t\right)m\left(A\left(t\right)\right)=1+\mu \left(t\right){\displaystyle \frac{\parallel I+\mu \left(A\right(t\left)\right)\parallel -1}{\mu \left(t\right)}}=\parallel I+\mu \left(t\right)A\left(t\right)\parallel .$$

□

Now we prove Coppel’s inequality [6,25] by giving a new simple proof on the time scales.

Theorem 4.

Let $\mathbb{T}$ be an unbounded from above time scale with graininess $\mu :\mathbb{T}\to [0,+\infty )$ and suppose that $m\left(A\right):\mathbb{T}\to \mathbb{R}$ is a matrix measure of rd-continuous matrix valued function A. If matrix measure is positively regressive $m\left(A\right)\in {\mathcal{R}}^{+}$, then

$$\parallel e_A(t,s)\parallel \le e_{m\left(A\right)}(t,s)fors,t\in \mathbb{T}andt\in {\mathbb{T}}_s=\mathbb{T}\cap [s,+\infty ).$$

(4)

Proof. 

The function $e_A(·,s):\mathbb{T}\to {\mathbb{R}}^{n\times n}$, $e_A(s,s)=I$ is a continuous solution of matrix initial value problem (2) for each fixed $s\in \mathbb{T}$. Therefore, $\parallel e_A(·,s)\parallel :\mathbb{T}\to \mathbb{R}$ is also a continuous function. Next, we prove that $\parallel e_A(·,s)\parallel$ is rd-differentiable i.e., differentiable with rd-continuous right hand derivative.

We prove the inequality (4) by considering two cases: $\mu \left(t\right)>0$ and $\mu \left(t\right)=0$. Let $\mu \left(t\right)>0$ and $t\in \mathbb{T}$ is right scattered. Then $e_A(·,s)$ is differentiable at t with

$$\begin{array}{c}\hfill \parallel e_A{(t,s)\parallel }^{\Delta }={\displaystyle \frac{\parallel e_A(\sigma \left(t\right),s)\parallel - \parallel e_A(t,s)\parallel }{\mu \left(t\right)}}\\ \hfill ={\displaystyle \frac{\parallel (I+\mu \left(t\right)A\left(t\right))e_A(t,s)\parallel - \parallel e_A(t,s)\parallel }{\mu \left(t\right)}}\\ \hfill \le {\displaystyle \frac{\parallel I+\mu \left(t\right)A\left(t\right)\parallel \parallel e_A(t,s)\parallel - \parallel e_A(t,s)\parallel }{\mu \left(t\right)}}\\ \hfill ={\displaystyle \frac{\parallel I+\mu \left(t\right)A\left(t\right)\parallel -1}{\mu \left(t\right)}}\parallel e_A(t,s)\parallel =m\left(A\left(t\right)\right)\parallel e_A(t,s)\parallel .\end{array}$$

Since $\mu$ and A are rd-continuous, then from the expression for $\parallel e_A{(·,s)\parallel }^{\Delta }$ it follows that it is also rd-continuous.

Let $\mu \left(t\right)=0$ and $t\in \mathbb{T}$ is right dense. According to the Corollary 2 and Definition 2 of matrix measure we have

$$\underset{h\to +0}{lim}{\displaystyle \frac{\parallel (I+hA\left(t\right))e_A(t,s)\parallel - \parallel e_A(t,s)\parallel }{h}}\le m\left(A\left(t\right)\right)\parallel e_A(t,s)\parallel .$$

Let’s look at the expression

$$\left|{\displaystyle \frac{\parallel e_A(t+h,s)\parallel - \parallel e_A(t,s)\parallel }{h}}-{\displaystyle \frac{\parallel (I+hA\left(t\right))e_A(t,s)\parallel - \parallel e_A(t,s)\parallel }{h}}\right|$$

$$=\left|{\displaystyle \frac{\parallel e_A(t+h,s)\parallel - \parallel (I+hA\left(t\right))e_A(t,s)\parallel }{h}}\right|$$

$$\le {\displaystyle \frac{\parallel e_A(t+h,s)-(I+hA\left(t\right))e_A(t,s)\parallel }{h}}.$$

The right-hand side of the expression tends to 0 as $h\to +0$ according to Theorem 1 (iv). The second term on the left hand side converges as $h\to +0$. It follows that there exists right hand derivative

$$\parallel e_A{(t,s)\parallel }_{+}^{\Delta }=\underset{h\to +0}{lim}{\displaystyle \frac{\parallel e_A(t+h,s)\parallel - \parallel e_A(t,s)\parallel }{h}}$$

$$=\underset{h\to +0}{lim}{\displaystyle \frac{\parallel (I+hA\left(t\right))e_A(t,s)\parallel - \parallel e_A(t,s)\parallel }{h}}\le m\left(A\left(t\right)\right)\parallel e_A(t,s)\parallel .$$

From the last relation it follows that $\parallel e_A{(·,s)\parallel }_{+}^{\Delta }$ is also rd-continuous. We get that the function $\parallel e_A(·,s)\parallel$ is rd-differentiable i.e., differentiable with rd-continuous right hand derivative.

Let us consider continuous and rd-differentiable difference with right hand derivative $\eta :{\mathbb{T}}_s\to \mathbb{R}$

$$\eta \left(t\right)=\parallel e_A(t,s)\parallel -e_{m\left(A\right)}(t,s),\eta \left(s\right)=0,s,t\in \mathbb{T},\mathrm{and}t\ge s.$$

We get

$${\eta }_{+}^{\Delta }\left(t\right)\le m\left(A\left(t\right)\right)\parallel e_A(t,s)\parallel -m\left(A\left(t\right)\right)e_{m\left(A\right)}(t,s)=m\left(A\left(t\right)\right)\eta \left(t\right).$$

Since $m\left(A\right)\in {\mathcal{R}}^{+}$, we have $e_{m\left(A\right)}(t,s)>0$ for $s,t\in \mathbb{T}$. Therefore for $t\ge s$

$${\displaystyle \frac{{\eta }_{+}^{\Delta }\left(t\right)e_{m\left(A\right)}(t,s)-m\left(A\left(t\right)\right)\eta \left(t\right)e_{m\left(A\right)}(t,s)}{e_{m\left(A\right)}(t,s)e_{m\left(A\right)}(\sigma \left(t\right),s)}}\le 0$$

or

$${\zeta }_{+}^{\Delta }\left(t\right)={\left({\displaystyle \frac{\eta \left(t\right)}{e_{m\left(A\right)}(t,s)}}\right)}_{+}^{\Delta }\le 0.$$

Given that $\eta \left(s\right)=0$. Than $\zeta \left(s\right)=0$. A continuous function $\zeta$ is nonincreasing if its right hand derivative is not positive [25,26]. According to the Comparison Theorem [23,27], it follows that $\zeta \left(t\right)\le 0$ and $\eta \left(t\right)\le 0$ for $t\ge s$. From here the inequality (4) follows. □

3. Volterra Integro Dynamic Equation

Let us consider the Cauchy problem for a nonlinear Volterra integro dynamic Equation (1) on time scales ${\mathbb{T}}_s=\mathbb{T}\cap [s,+\infty )$.

3.1. Boundedness

First, we provide sufficient conditions for the existence of a bounded solution to the nonlinear Volterra integro dynamic Equation (1) on time scales ${\mathbb{T}}_s$ using Coppel’s inequality. Thus, we generalize the scalar linear case proved by [15] to the general n-dimensional case.

Theorem 5.

Let $m\left(A\right)\in {\mathcal{R}}^{+}$ and $m\left(A\right(t\left)\right)\le 0$ for $s,t\in \mathbb{T}$ and $t\in {\mathbb{T}}_s$. Moreover, assume that there exist non-negative rd-continuous function $l:{\mathbb{T}}_s\to \mathbb{R}$ and non-negative constant $N\ge 0$ such that

$$\epsilon \left(t\right)+{\int }_s^{t}L(t,r)\Delta r\le l\left(t\right)$$

(5)

and

$$|f(t,0)|+{\int }_s^t|K(t,r,0)|\Delta r\le Nl\left(t\right).$$

(6)

If

$${\int }_s^t{e}_{m\left(A\right)}(t,\sigma \left(r\right))l\left(r\right)\Delta r\le c<1,$$

(7)

where constant $c\in (0,1)$ is independent of $t\in {\mathbb{T}}_s$, then all solutions of (1) are bounded for $s,t\in \mathbb{T}$ and $t\in {\mathbb{T}}_s$.

Note that if $l\left(t\right)\le c|m(A\left(t\right))|$ for $t\in {\mathbb{T}}_s$, then

$${\int }_s^t{e}_{m\left(A\right)}(t,\sigma \left(r\right))l\left(r\right)\Delta r\le c{\int }_s^t{e}_{m\left(A\right)}(t,\sigma \left(r\right))|m\left(A\left(r\right)\right)|\Delta r\le c<1.$$

Corollary (1) (x) implies the convergence of the given integral. So, the inequality (7) holds for $s,t\in \mathbb{T}$ and $t\in {\mathbb{T}}_s$.

Proof. 

Let $A\in \mathcal{R}$. Using the Variation of Constants formula [27], we obtain the following Volterra type integral equation

$$\begin{array}{c}\hfill x\left(t\right)=e_A(t,s)x_s+{\int }_s^t{e}_A(t,\sigma \left(r\right))f(r,x\left(r\right))\Delta r\\ \hfill +{\int }_s^t{e}_A(t,\sigma \left(r\right)){\int }_s^{r}K(r,\tau ,x\left(\tau \right))\Delta \tau \Delta r.\end{array}$$

(8)

Note that every solution of (1) is also a solution of (8) and vice versa.

To find sufficient conditions for the existence of bounded solutions of (8), we consider the linear space of all continuous functions $x:{\mathbb{T}}_s\to {\mathbb{R}}^n$, such that

$$\underset{t\in {\mathbb{T}}_s}{sup}|x\left(t\right)|<\infty$$

and denote this space by $C({\mathbb{T}}_s,{\mathbb{R}}^n)$. The space $C({\mathbb{T}}_s,{\mathbb{R}}^n)$ endowed with a supremum norm

$${\parallel x\parallel }_{{\mathbb{R}}^n}=\underset{t\in {\mathbb{T}}_s}{sup}|x\left(t\right)|$$

is Banach space. Let us consider the operator $F:C({\mathbb{T}}_s,{\mathbb{R}}^n)\to C({\mathbb{T}}_s,{\mathbb{R}}^n)$ defined by

$$\begin{array}{ccc}\hfill \left[Fx\right]\left(t\right)& =& e_A(t,s)x_s+{\int }_s^t{e}_A(t,\sigma \left(r\right))\left(f(r,x(r\left)\right)-f(r,0)\right)\Delta r\hfill \\ \hfill & +& {\int }_s^t{e}_A(t,\sigma \left(r\right)){\int }_s^r\left(K(r,\tau ,x(\tau \left)\right)-K(r,\tau ,0)\right)\Delta \tau \Delta r\hfill \\ & +& {\int }_s^t{e}_A(t,\sigma \left(r\right))\left(f(r,0)+{\int }_s^{r}K(r,\tau ,0)\Delta \tau \right)\Delta r.\hfill \end{array}$$

(9)

Let us note that the fixed point of operator F will be the solution of (8). Thus, we want to prove that there exists a unique $x_{*}\in C({\mathbb{T}}_s,{\mathbb{R}}^n)$, such that $F{x}_{*}=x_{*}$. We apply Banach’s fixed point theorem, further developed by [28,29], to prove the existence of a solution to the Volterra integral equations on an unbounded above time scales. Taking norms in (9) and using the Coppel’s inequality (4), where $m\left(A\right(t\left)\right)\le 0$ and the inequalities (5)–(7), we get that

$$\begin{array}{ccc}\hfill {\parallel Fx\parallel }_{{\mathbb{R}}^n}& \le & |x_s|+\underset{t\in {\mathbb{T}}_s}{sup}\left({\int }_s^t{e}_{m\left(A\right)}(t,\sigma \left(r\right))l\left(r\right)\left({\parallel x\parallel }_{{\mathbb{R}}^n}+N\right)\Delta r\right)\hfill \\ & \le & |x_s{|+c\parallel x\parallel }_{{\mathbb{R}}^n}+cN\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \parallel Fx-F{x}^{\prime }{\parallel }_{{\mathbb{R}}^n}& \le & \underset{t\in {\mathbb{T}}_s}{sup}\left({\int }_s^t{e}_{m\left(A\right)}(t,\sigma \left(r\right))l\left(r\right){\parallel x-x^{\prime }\parallel }_{{\mathbb{R}}^n}\Delta r\right)\hfill \\ & \le & c\parallel x-x^{\prime }{\parallel }_{{\mathbb{R}}^n}.\hfill \end{array}$$

Note that $c\in (0,1)$, $\parallel Fx-F{x}^{\prime }{\parallel }_{{\mathbb{R}}^n}\le c{\parallel x-x^{\prime }\parallel }_{{\mathbb{R}}^n}$, ${\parallel Fx\parallel }_{{\mathbb{R}}^n}\le {\parallel x\parallel }_{{\mathbb{R}}^n}$ if ${\parallel x\parallel }_{{\mathbb{R}}^n}\le {\displaystyle \frac{|x_s|+cN}{1-c}}$. Hence, we see that the operator F is a contraction operator with contraction constant $c<1$ and $F:C({\mathbb{T}}_s,{\mathbb{R}}^n)\to C({\mathbb{T}}_s,{\mathbb{R}}^n)$. The Banach fixed point theorem is applicable to prove the existence of a unique fixed point $F{x}_{*}=x_{*}$. □

Example 1.

Consider the nonlinear Volterra integro dynamic equation

$$x^{\Delta }\left(t\right)=A\left(t\right)x\left(t\right)+{\int }_0^{t}K(t,r,x\left(r\right))\Delta r,$$

(10)

where $0\in \mathbb{T}$, $t\in \mathbb{T}\cap [0,+\infty )$ and $x\left(0\right)=x_0$.

Assume that

$$A\left(t\right)=\left(\begin{array}{cc}a\left(t\right)& -a\left(t\right)\\ a\left(t\right)& a\left(t\right)\end{array}\right)$$

and

$$K(t,r,x)={\displaystyle \frac{\nu }{2t^3+1}}\left(\begin{array}{c}(r+\sigma \left(r\right))\sqrt{|x_2{|}^2+{(sint)}^2}\\ (r+\sigma \left(r\right))\sqrt{|x_1{|}^2+1-cos2t}\end{array}\right),$$

where $1+a\left(t\right)\mu \left(t\right)>0$, $a\left(t\right)<0$ and $\nu >0$. Then

$$a\left(t\right)\le m\left(A\right(t\left)\right)\le a\left(t\right)(1+a(t\left)\mu \right(t\left)\right),$$

$$l\left(t\right)={\int }_0^{t}L(t,r)\Delta r={\int }_0^t{\displaystyle \frac{\nu (r+\sigma (r\left)\right)}{2t^3+1}}\Delta r={\displaystyle \frac{\nu t^2}{2t^3+1}}\le {\displaystyle \frac{\nu }{3}},$$

$${\int }_0^t|K(t,r,x)-K(t,r,x^{\prime })|\Delta r\le {\displaystyle \frac{\nu t^2}{2t^3+1}}\parallel x-x^{\prime }{\parallel }_{{\mathbb{R}}^n}=l\left(t\right){\parallel x-x^{\prime }\parallel }_{{\mathbb{R}}^n},$$

$${\int }_0^t|K(t,r,0)|\Delta r\le {\displaystyle \frac{\nu t^2}{2t^3+1}}\sqrt{3}|sint|=l\left(t\right)\sqrt{3}|sint|\le {\displaystyle \frac{\nu }{\sqrt{3}}}.$$

If

$$l\left(t\right)={\displaystyle \frac{\nu t^2}{2t^3+1}}\le c|m\left(A\left(t\right)\right)|\le c|a\left(t\right)(1+a\left(t\right)\mu \left(t\right))|,$$

where $c<1$, then from Theorem (5) it follows that all solutions of Equation (10) are bounded.

At the same time, in our example, it should be noted that in the case $\mathbb{T}=\mathbb{R}$ ($\mu \left(t\right)=0$), the estimate of the measure of the matrix A is $m\left(A\right(t\left)\right)=a\left(t\right)$. If $a\left(t\right)<0$, then the given estimate of the measure of the matrix A is both a necessary and sufficient condition for the exponential stability of the linear part of the given equation.

3.2. Exponential Stability

Definition 3

([7,15]). Equation (1) is said to be exponentially stable on ${\mathbb{T}}_s=\mathbb{T}\cap [s,+\infty )$ if there exists a positive constant d with $-d\in {\mathcal{R}}^{+}$ and a constant $C\in {\mathbb{R}}_{+}$ such that for any solution $x:{\mathbb{T}}_s\to {\mathbb{R}}^n$, $x\left(s\right)=x_s$, $x_s\in {\mathbb{R}}^n$, $s,t\in \mathbb{T}$ of (1) the following inequality holds

$$|x\left(t\right)|\le C(|x_s|,s)e_{-d}(t,s)$$

for all $t\in {\mathbb{T}}_s$.

Equation (1) is said to be uniformly exponentially stable on ${\mathbb{T}}_s$ if it is exponentially stable and C and d are independent of s.

Assume that $f(t,0)=0$ and $K(t,r,0)=0$. Then Equation (1) and accordingly, Equation (8) have the trivial solution. In addition, we assume that all solutions of (1) and accordingly, (8) are bounded.

Theorem 6.

Let $m\left(A\right)\in {\mathcal{R}}^{+}$ and $m\left(A\right(t\left)\right)<0$. Moreover, assume that the following conditions are satisfied:

(i) 

there exist positive constants α and γ such that $-\alpha \in {\mathcal{R}}^{+}$, $-\gamma \in {\mathcal{R}}^{+}$ and

$$m\left(A\right(t\left)\right)\le -\alpha <-\gamma <0,$$

(ii) 

there exists positive constant $\lambda >\alpha$ such that $-\lambda \in {\mathcal{R}}^{+}$ and constant $\mathsf{\Lambda }\ge 0$ such that

$$L(r,\tau )\le \mathsf{\Lambda }{e}_{(-\lambda )\oplus (-\gamma )}(\sigma \left(r\right),s),r\ge \tau \ge s,$$

(iii) 

there exist constant ${\epsilon }_0\ge 0$ such that

$$\epsilon \left(r\right)\le {\epsilon }_0{e}_{(-\alpha )\oplus (-\gamma )}(\sigma \left(r\right),s),r\ge s.$$

Then Volterra integro dynamic Equation (1) is uniformly exponentially stable on ${\mathbb{T}}_s$.

Proof. 

Let $x:{\mathbb{T}}_s\to {\mathbb{R}}^n$ be a solution of (8) and

$$M=\underset{t\in {\mathbb{T}}_s}{sup}|x\left(t\right)|.$$

Therefore, using Coppel’s inequality, we obtain that

$$\begin{array}{ccc}\hfill |x(t)|& \le & |x_s|e_{-\alpha }(t,s)+M{\int }_s^t{e}_{-\alpha }(t,\sigma \left(r\right))\epsilon \left(r\right)\Delta r\hfill \\ & +& M{\int }_s^t{e}_{-\alpha }(t,\sigma \left(r\right)){\int }_s^{r}L(r,\tau )\Delta \tau \Delta r\hfill \end{array}$$

for arbitrary $t\in {\mathbb{T}}_s$. Let us note that

$$e_{-\gamma }(\sigma \left(r\right),s)\le e_{-\gamma }(r,s)\le e_{-\gamma }(r,\tau )\le e_{-\gamma }(r,\sigma \left(\tau \right)).$$

Applying the conditions (ii) and (iii) of the Theorem 6, we get

$$\begin{array}{ccc}\hfill |x(t)|& \le & |x_s|e_{-\alpha }(t,s)+{\displaystyle \frac{M{\epsilon }_0{e}_{-\alpha }(t,s)}{\gamma }}{\int }_s^t\gamma e_{-\gamma }(r,s)\Delta r\hfill \\ & +& {\displaystyle \frac{M\mathsf{\Lambda }{e}_{-\alpha }(t,s)}{\gamma }}{\int }_s^t{e}_{(-\lambda )\ominus (-\alpha )}(\sigma \left(r\right),s){\int }_s^r\gamma e_{-\gamma }(r,\sigma \left(\tau \right))\Delta \tau \Delta r\hfill \\ & =& |x_s|e_{-\alpha }(t,s)+{\displaystyle \frac{M{\epsilon }_0{e}_{-\alpha }(t,s)}{\gamma }}(1-e_{-\gamma }(t,s))\hfill \\ & +& {\displaystyle \frac{M\mathsf{\Lambda }{e}_{-\alpha }(t,s)}{\gamma }}{\int }_s^t{e}_{(-\lambda )\ominus (-\alpha )}(\sigma \left(r\right),s)(1-e_{-\gamma }(r,s))\Delta r\hfill \\ & \le & |x_s|e_{-\alpha }(t,s)+{\displaystyle \frac{M{e}_{-\alpha }(t,s)}{\gamma }}\left({\epsilon }_0+\mathsf{\Lambda }{\int }_s^t{e}_{(-\lambda )\ominus (-\alpha )}(\sigma \left(r\right),s)\Delta r\right).\hfill \end{array}$$

Since $\lambda >\alpha >0$, we get that $\left(\right(-\lambda )\ominus (-\alpha \left)\right)<0$ and $|(-\lambda )\ominus (-\alpha )|>\lambda -\alpha$. Furthermore, we note that $e_{(-\lambda )\ominus (-\alpha )}(\sigma \left(r\right),s)\le e_{(-\lambda )\ominus (-\alpha )}(r,s)$ because the exponential rate of the corresponding exponential function is negative and therefore the exponential function is decreasing with respect to the first variable. It follows that

$$\begin{array}{ccc}\hfill |x(t)|& \le & e_{-\alpha }(t,s)\left(|x_s|+{\displaystyle \frac{M}{\gamma }}\left({\epsilon }_0+\mathsf{\Lambda }{\displaystyle \frac{\left(1-e_{(-\lambda )\ominus (-\alpha )}(t,s)\right)}{\gamma |(-\lambda )\ominus (-\alpha )|}}\right)\right)\hfill \\ & \le & e_{-\alpha }(t,s)\left(|x_s|+{\displaystyle \frac{M}{\gamma }}\left({\epsilon }_0+{\displaystyle \frac{\mathsf{\Lambda }}{\lambda -\alpha }}\right)\right)\mathrm{for}\mathrm{all}t\in {\mathbb{T}}_s.\hfill \end{array}$$

The corresponding constants C and d in the solution estimate do not depend on the initial moment $s\in \mathbb{T}$. According to the definition of uniform exponential stability, the nonlinear Volterra integro dynamic Equation (1) is uniformly exponentially stable on ${\mathbb{T}}_s$. □

4. Conclusions

In this article, we studied boundedness and exponential stability of nonlinear n-dimensional Volterra integro dynamic equation on unbounded above time scales. To achieve this, we reduce the n-dimensional equation to a corresponding scalar equation using the definition of matrix measure and a simplified proof of Coppel’s inequality on the time scales. The aim of this article was achieved by proving the relationship between boundedness and exponential stability and matrix measure on time scales.

It should be noted that the matrix measure can be negative (the matrix norm is always non-negative). This can be a significant advantage in research when using the first Lapunov method. By modifying the classical Lyapunov transformation or applying an analogous transformation to the equations on a time scale, one would expect to obtain the best possible estimate of the matrix measure (similar to the case when $\mathbb{T}=\mathbb{R}$ and the corresponding matrix has constant or periodic coefficients). In addition, we note that the matrix measure changes a little under the $C^1$ perturbation of the matrix in a finite interval.