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An extended Halanay inequality with unbounded coefficient functions on time scales

Journal of Inequalities and Applications20162016:316

https://doi.org/10.1186/s13660-016-1259-x

  • Received: 29 July 2016
  • Accepted: 22 November 2016
  • Published:

Abstract

In this paper, we obtain an extended Halanay inequality with unbounded coefficient functions on time scales, which extends an earlier result in Wen et al. (J. Math. Anal. Appl. 347:169-178, 2008). Two illustrative examples are also given.

Keywords

  • time scales
  • delay dynamic equation
  • coefficient functions
  • inequality

1 Introduction and preliminaries

As is well known, Halanay-type differential inequalities have been very useful in the stability analysis of time-delay systems and these have led to some interesting new stability conditions (see [14] and the references therein).

In [3], Halanay proved the following inequality.

Lemma 1.1

Halanay’s inequality

If
$$ x'(t)\leq-\alpha x(t)+\beta\sup_{s\in[t-\tau, t]}x(s), \quad\textit{for } t\geq t_{0}, \tau>0, $$
(1.1)
and \(\alpha>\beta>0\), then there exist \(\gamma>0\) and \(K>0\) such that
$$ x(t)\leq Ke^{-\gamma(t-t_{0})}, \quad\textit{for } t\geq t_{0}. $$
(1.2)

In [5], Baker and Tang obtained the following Halanay-type inequality with unbounded coefficient functions.

Lemma 1.2

see [5]

Let \(x(t)>0, t\in(-\infty ,+\infty)\), and
$$\begin{aligned} &\frac{dx(t)}{dt}\leq-a(t)x(t)+b(t)\sup_{t-\tau(t)\leq s\leq t}{x(s)}, \quad t>t_{0}, \end{aligned}$$
(1.3)
$$\begin{aligned} &x(t)=\bigl|\varphi(t)\bigr|,\quad t\leq t_{0} , \end{aligned}$$
(1.4)
where \(\varphi(t)\) is bounded and continuous for \(t\leq t_{0}\), and \(a(t)\geq0, b(t)\geq0\) for \(t\in[t_{0},\infty),\tau(t)\geq0\) and \(t-\tau(t)\rightarrow\infty\) as \(t\rightarrow\infty\). If there exists \(\sigma>0\) such that
$$\begin{aligned} -a(t)+b(t)\leq- \sigma< 0 \quad\textit{for } t\geq t_{0}, \end{aligned}$$
(1.5)
then
$$\begin{aligned} &(\mathrm{i}) \quad x(t)\leq\|\varphi\|^{(-\infty,t_{0}]}, \quad t\geq t_{0} \quad\textit{and} \quad (\mathrm{ii})\quad x(t)\rightarrow0 \quad \textit{as } t\rightarrow\infty, \end{aligned}$$
(1.6)
where \(\|\varphi\|^{(-\infty,t_{0}]}= \sup_{t\in(-\infty,t_{0}]}|\varphi(t)|<\infty\).

In [1], Wen et al. obtained an extension of Lemma 1.2.

In this paper, we extend the main results of [5] to time scale. As an application, we consider the stability of the following delay dynamic equation:
$$\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} x^{\triangle}(t)=-a(t)x^{\sigma}(t)+b(t)x(t-\tau(t))+c(t),\quad t\in [t_{0},+\infty)_{{\mathbb {T}}},\\ x(s)=|\varphi(s)| \quad \mbox{for }s\in(-\infty,t_{0}]_{\mathbb{T}}, \end{array}\displaystyle \right . \end{aligned}$$
(1.7)
where \(\varphi(s)\) is bounded rd-continuous for \(s\in(-\infty,t_{0}]_{\mathbb{T}}\) and \(\tau(t)\), \(a(t)\), \(b(t)\), \(c(t)\) are nonnegative, rd-continuous functions for \(t\in[t_{0},\infty)_{\mathbb{T}}\) and \(c(t)\) is bounded. We prove that the zero solution of the delay difference equation
$$\begin{aligned} \triangle x(n)=-2(n+1)x(n+1)+\frac{n^{2}}{2n+1}x(n-2),\quad n\geq0, \end{aligned}$$
(1.8)
is stable.

For completeness, we introduce the following concepts related to the notions of time scales. We refer to [6] for additional details concerning the calculus on time scales.

Definition 1.1

see [6]

A function \(h: \mathbb{T}\rightarrow\mathbb{R}\) is said to be regressive provided \(1+\mu(t)h(t)\neq0\) for all \(t\in\mathbb{T}^{\kappa}\), where \(\mu(t)=\sigma(t)-t\). The set of all regressive rd-continuous functions \(\varphi: \mathbb{T}\rightarrow\mathbb{R}\) is denoted by \(\mathfrak{R}\) while the set \(\mathfrak{R}^{+}\) is given by \(\mathfrak{R}^{+}=\{{\varphi\in}\mathfrak{R}:1+\mu(t)\varphi (t)>0\mbox{ for all }t\in\mathbb{T}\}\). If \(\varphi\in\mathfrak{R}\), the exponential function is defined by
$$ e_{\varphi}(t,s)=\exp \biggl( \int_{s}^{t}\xi_{\mu(r)}\bigl(\varphi (r) \bigr)\triangle r \biggr), \quad\mbox{for } t\in\mathbb{T}, s\in\mathbb {T}^{\kappa}, $$
(1.9)
where \(\xi_{\mu(s)}\) is the cylinder transformation given by
$$\begin{aligned} \xi_{\mu(r)}\bigl(\varphi(r)\bigr):= \textstyle\begin{cases} \frac{1}{\mu(r)}\operatorname{Log} (1+\mu(r)\varphi(r) ), & \mu(r)> 0,\\ \varphi(r), & \mu(r)=0, \end{cases}\displaystyle \end{aligned}$$
and some properties of the exponential function are given in the following lemma.

Lemma 1.3

see [7]

Let \(\varphi\in\mathfrak{R}\), Then
  1. (i)

    \(e_{0}(s,t)\equiv1, e_{\varphi}(t,t)\equiv1\) and \(e_{\varphi}(\sigma(t),s) = (1+\mu(t)\varphi(t) )e_{\varphi}(t,s)\);

     
  2. (ii)

    \(\frac{1}{e_{\varphi}(t,s)} =e_{\ominus\varphi}(t,s)\), where \(\ominus\varphi(t)=-\frac {\varphi(t)}{1+\mu(t)\varphi(t)}\);

     
  3. (iii)

    \((\frac{1}{e_{\varphi}(t,s)} )^{\triangle} =-\frac{\varphi(t)}{e_{\varphi}(\sigma(t),s)}\);

     
  4. (iv)

    \([e_{\varphi}(c,t)]^{\Delta}=-\varphi(t)e_{\varphi}(c,\sigma (t))\), where \(c\in\mathbb{T}\);

     
  5. (v)

    \(e_{p}(t,s)=\frac{1}{e_{p}(s,t)}=e_{\ominus p}(s,t)\).

     

Lemma 1.4

see [8]

For a nonnegative φ with \(-\varphi\in\mathfrak{R}^{+}\), we have the inequalities
$$\begin{aligned} 1- \int_{s}^{t}\varphi(u)\Delta u\leq e_{-\varphi}(t,s)\leq\exp \biggl\{ - \int_{s}^{t}\varphi(u)\Delta u \biggr\} \quad \textit{for all }t\geq s. \end{aligned}$$
(1.10)
If φ is rd-continuous and nonnegative, then
$$\begin{aligned} 1+ \int_{s}^{t}\varphi(u)\Delta u\leq e_{\varphi}(t,s)\leq\exp \biggl\{ \int_{s}^{t}\varphi(u)\Delta u \biggr\} \quad \textit{for all }t\geq s. \end{aligned}$$
(1.11)

Remark 1.1

If \(\varphi\in\mathfrak{R}^{+}\) and \(\varphi(r)>0\) for all \(r\in [s,t]_{\mathbb{T}}\), then
$$\begin{aligned} e_{\varphi}(t,r)\leq e_{\varphi}(t,s) \quad\mbox{and} \quad e_{\varphi}(a,b)< 1 \quad\mbox{for } s\leq a< b\leq t. \end{aligned}$$
(1.12)

Proof

By \(\varphi(r)>0\), \(\varphi\in\mathfrak{R}^{+}\) and Lemma 1.3(iv) we have \([e_{\varphi}(c,t)]^{\Delta}=-\varphi(t)e_{\varphi }(c,\sigma(t))<0\), so
$$e_{\varphi}(t,r)\leq e_{\varphi}(t,s). $$
Since \(a< b\), from the above result, we have
$$e_{\varphi}(a,b)< e_{\varphi}(a,a)=1. $$
 □

2 Main results

Throughout this paper, we assume that the following conditions hold:
(H1): 
Let \(x(t)\) be a nonnegative right-dense function satisfying
$$\left \{ \textstyle\begin{array}{@{}l} x^{\triangle}(t)\leq-a(t)x(t)+b(t)\sup_{t-\tau(t)\leq s\leq t}{x(s)}+c(t)\\ \hphantom{x^{\triangle}(t)\leq}{}+d(t)\int^{\infty}_{0}K(t,s)x(t-s)\Delta s,\quad t\in[t_{0},\infty),\\ x(t)=|\varphi(t)|,\quad t\in(-\infty,t_{0}], \end{array}\displaystyle \right . $$
where \(\varphi(t)\) is bounded rd-continuous for \(t\in (-\infty,t_{0}]_{\mathbb{T}}\) and \(\sup_{t\leq t_{0}}{|\varphi(t)|}=M\).
(H2): 

\(a(t)\), \(b(t)\), \(c(t)\), \(\tau(t)\) are nonnegative, rd-continuous functions for \(t\in[t_{0},\infty)_{\mathbb{T}}\) and \(c(t)\) is bounded, such that \(\sup_{t\geq t_{0}}{c(t)}=\overline {c}\), \(\lim_{t\rightarrow\infty}(t-\tau(t))=+\infty\).

(H3): 

There exists \(\delta>0\) such that \(a(t)-b(t)-d(t)\int_{0}^{\infty}K(t,s)\triangle s>\delta>0\), for \(t\in [t_{0},\infty)_{\mathbb{T}}\), where the delay kernel \(K(t,s)\) is a nonnegative, rd-continuous for \((t,s)\in\mathbb{ T}\times[0,\infty)\) and satisfies \(\forall t\in\mathbb{T},\int_{0}^{\infty}K(t,s)\triangle s<\infty\).

Theorem 2.1

Assume that (H1)-(H3) and \(-a(t)\in\mathfrak{R}^{+}\) hold, then we have
  1. (i)
    $$ x(t)\leq \frac{\overline{c}}{\delta}+M ,\quad t\in[t_{0},+\infty ). $$
    (2.1)
    If we assume further that \(d(t)=0\) in (H1), (H3) and there exists \(0<\kappa<1\) such that
    $$ \kappa a(t)-b(t)>0 \quad\textit{for } t\in[t_{0},+ \infty)_{{\mathbb {T}}}, $$
    (2.2)
    then we have
     
  2. (ii)
    for any given \(\epsilon>0\), there exists \(\widetilde {t}=\widetilde {t}(M,\epsilon)>t_{0}\), such that
    $$ x(t)\leq \frac{\overline{c}}{\delta}+\epsilon,\quad t\in[\widetilde{t}, \infty). $$
    (2.3)
     

Proof

We now consider the following two cases successively.

Case 1. \(\overline{c}>0\).

Proof of Theorem 2.1 (i).

For any \(\varepsilon>1\), we have from (H1)
$$\begin{aligned} \forall t\leq t_{0},\quad x(t)=\bigl|\varphi(t)\bigr|\leq\sup _{t\leq t_{0}}{\bigl|\varphi(t)\bigr|}=M< \frac{\overline{c}}{\delta}+\varepsilon M, \end{aligned}$$
(2.4)
from this we shall deduce that
$$\begin{aligned} \forall t\geq t_{0},\quad x(t)< \frac{\overline{c}}{\delta}+ \varepsilon M. \end{aligned}$$
(2.5)

To prove (2.5), let \(t_{1} =\sup\) \(\{t| x(s)\leq\frac {\overline {c}}{\delta}+\varepsilon M, s\in[t_{0},t]_{\mathbb{T}}\}>t_{0}\), we will show \(t_{1}=\infty\).

Suppose \(t_{1}<\infty\). Clearly we have \(x(t_{1})\leq\frac{\overline {c}}{\delta}+\varepsilon M\).

In fact, suppose that \(x(t_{1})\leq\frac{\overline{c}}{\delta}+\varepsilon M\) fails, then we have \(x(t_{1})>\frac{\overline{c}}{\delta}+\varepsilon M\).

If \(t_{1}\) is left-dense, there is \(\{t_{n}\}\) satisfying: \(t_{n}< t_{1}, t_{n}\rightarrow t_{1}\) (\(n\rightarrow\infty\)), and \(x(t_{n})\leq\frac{\overline {c}}{\delta}+\varepsilon M\), we have \(x(t_{1})=\lim_{n\rightarrow\infty}x(t_{n})\leq\frac{\overline{c}}{\delta }+\varepsilon M\), which contradicts \(x(t_{1})>\frac{\overline {c}}{\delta }+\varepsilon M\).

If \(t_{1}\) is left-scattered, \(\rho(t_{1})< t_{1}\) and \(x(\rho (t_{1}))\leq\frac{\overline{c}}{\delta}+\varepsilon M\); \(x(t_{1})>\frac {\overline{c}}{\delta}+\varepsilon M\), then we have sup \(\{t| x(s)<\frac{\overline{c}}{\delta}+\varepsilon M, s\in[t_{0},t]\} =\rho (t_{1})<t_{1}\), which contradicts the definition of \(t_{1}\).

Therefore we can suppose \(t_{1}<\infty, x(t_{1})\leq\frac{\overline {c}}{\delta}+\varepsilon M\). We will discuss two cases:

Case 1.1. Suppose \(x(t_{1})=\frac{\overline{c}}{\delta }+\varepsilon M, t_{1}>t_{0}\),
$$ \forall t\in[t_{0},t_{1})_{{\mathbb {T}}},\quad x(t)\leq\frac{\overline{c}}{\delta }+\varepsilon M,\qquad x(t_{1})=\frac{\overline{c}}{\delta}+ \varepsilon M. $$
(2.6)
Clearly we have \(x^{\triangle}(t_{1})\geq0\). In fact, suppose that \(x^{\triangle}(t_{1})\geq0\) fails, then we have \(x^{\triangle}(t_{1})<0\).

If \(t_{1}\) is right-dense, \(\forall s>t_{1}\), from \(x^{\triangle}(t_{1})= {\lim_{s\rightarrow t_{1}^{+}}} \frac {x(t_{1})-x(s)}{t_{1}-s}<0\), we get \(x(s)< x(t_{1})=\frac{\overline {c}}{\delta}+\varepsilon M\), which contradicts the definition of \(t_{1}\).

If \(t_{1}\) is right-scattered, from \(x^{\triangle}(t_{1})=\frac{x(\sigma(t_{1}))-x(t_{1})}{\mu (t_{1})}<0\), we get \(x(\sigma(t_{1}))< x(t_{1})=\frac{\overline{c}}{\delta }+\varepsilon M\), which contradicts the definition of \(t_{1}\).

We have from (2.6), (H1), and (H3)
$$\begin{aligned} x^{\triangle}(t_{1})&\stackrel{( \mathrm{H}_{1})}{\leq} -a(t_{1})x(t_{1})+b(t_{1}) \sup_{t_{1}-\tau(t_{1})\leq s\leq t_{1}}{x(s)}+c(t_{1})+d(t_{1}) \int_{0}^{\infty }K(t_{1},s)x(t_{1}-s) \triangle s \\ &\stackrel{(2.6)}{=}-a(t_{1}) \biggl(\frac{\overline {c}}{\delta}+ \varepsilon M\biggr)+b(t_{1})\sup_{t_{1}-\tau(t_{1})\leq s\leq t_{1}}{x(s)}+c(t_{1})+d(t_{1}) \int_{0}^{\infty }K(t_{1},s)x(t_{1}-s) \triangle s \\ &\stackrel{(2.6)}{\leq}- \biggl(a(t_{1})-b(t_{1})+d(t_{1}) \int _{0}^{\infty}K(t_{1},s)\triangle s \biggr) \biggl(\frac{\overline{c}}{\delta }+\varepsilon M\biggr)+\overline{c} \\ &\overset{(\mathrm{H}_{3})}{< } -\delta\biggl(\frac{\overline{c}}{\delta }+\varepsilon M \biggr)+\overline{c}=-\delta\varepsilon M< 0, \end{aligned}$$
(2.7)
which contradicts \(x^{\triangle}(t_{1})\geq0\).
Case 1.2. Suppose \(x(t_{1})<\frac{\overline{c}}{\delta}+\varepsilon M\). In this case, \(t_{1}\) must be right-scattered, for otherwise if \(t_{1}\) is right-dense, there exists \(\epsilon_{1}\) sufficiently small so that \(x(t)<\frac{\overline{c}}{\delta}+\varepsilon M\), for \(t\in[t_{1},t_{1}+\epsilon_{1}]_{{\mathbb {T}}}\). Therefore, \(x(t)\leq\frac{\overline{c}}{\delta}+\varepsilon M\), for \(t\in[t_{0}, t_{1}+\epsilon_{1}]_{\mathbb{T}}\). This contradicts the definition of \(t_{1}\). Hence, since \(t_{1}\) is right-scattered, we have
$$ x\bigl(\sigma(t_{1})\bigr)>\frac{\overline{c}}{\delta}+ \varepsilon M\quad \mbox{and}\quad x(t)\leq\frac{\overline{c}}{\delta}+\varepsilon M\quad \mbox{for all }t\leq t_{1}< \sigma(t_{1}). $$
(2.8)
We have from (2.8) and (H1)
$$\begin{aligned} \frac{x(\sigma(t_{1}))-x(t_{1})}{\mu(t_{1})}={}&x^{\triangle}(t_{1}) \\ \stackrel{(\mathrm{H}_{1})}{\leq}{}& {-}a(t_{1})x(t_{1})+b(t_{1}) \sup_{t_{1}-\tau(t_{1})\leq s\leq t_{1}}{x(s)}+c(t_{1}) \\ &{}+d(t_{1}) \int _{0}^{\infty}K(t_{1},s)x(t_{1}-s) \triangle s \\ \stackrel{(2.8)}{< }{}& {-}a(t_{1})x(t_{1})+ \biggl(b(t_{1})+d(t_{1}) \int_{0}^{\infty }K(t_{1},s)\triangle s \biggr) \biggl(\frac{\overline{c}}{\delta }+\varepsilon M\biggr)+\overline{c}. \end{aligned}$$
(2.9)
By (2.8), (2.9), (H3), and \(1-\mu(t)a(t)>0, t\in\mathbb{T} \), we get
$$\begin{aligned} \frac{\overline{c}}{\delta}+\varepsilon M< {}&x\bigl(\sigma(t_{1}) \bigr) \\ \stackrel{(2.9)}{< }{}& \bigl(1-\mu(t_{1})a(t_{1}) \bigr)x(t_{1}) +\mu(t_{1}) \biggl( b(t_{1})+d(t_{1}) \int_{0}^{\infty }K(t_{1},s)\triangle s \biggr) \biggl(\frac{\overline{c}}{\delta}+\varepsilon M\biggr) \\ &{}+\mu (t_{1}) \overline {c} \\ \stackrel{(2.8)}{< }{}& \biggl(1-\mu(t_{1})a(t_{1})+ \mu(t_{1})b(t_{1}) +\mu(t_{1})d(t_{1}) \int_{0}^{\infty}K(t_{1},s)\triangle s\biggr) \biggl(\frac {\overline{c}}{\delta}+\varepsilon M\biggr) \\ &{}+\mu(t_{1})\overline {c} \\ \stackrel{(\mathrm{H}_{3})}{\leq}{}& \bigl(1-\delta \mu(t_{1}) \bigr) \biggl(\frac {\overline {c}}{\delta}+\varepsilon M\biggr)+ \mu(t_{1})\overline{c}=\frac{\overline {c}}{\delta}+\varepsilon M-\delta \varepsilon M\mu(t_{1}), \end{aligned}$$
(2.10)
which leads to a contradiction.

Hence the inequality (2.5) must hold.

Since \(\varepsilon>1\) is arbitrary, we let \(\varepsilon\rightarrow 1^{+}\) and obtain
$$\begin{aligned} \forall t\geq t_{0},\quad x(t)\leq\frac{\overline{c}}{\delta}+M. \end{aligned}$$
(2.11)

Proof of Theorem 2.1 (ii).

If \(M=0\), it is evident from (2.1) that (2.3) holds. Now we assume \(M>0\). Let \(\limsup_{t\rightarrow\infty}x(t)=\alpha\), then \(0\leq\alpha\leq \frac{\overline{c}}{\delta}+M\). Now we prove that \(\alpha\leq\frac{\overline{c}}{\delta}\).

Suppose this is not true, i.e. \(\alpha>\frac{\overline {c}}{\delta}\), then we can choose \(\varepsilon_{2}>0\) such that \(\alpha=\frac{\overline{c}}{\delta}+\varepsilon_{2}\).

Since \(\tau(t)\geq0\), and \(\lim_{t\rightarrow\infty}(t-\tau (t))=+\infty\), we have \(\limsup_{t\rightarrow\infty}\sup_{t-\tau (t)\leq s\leq t}{x(s)}=\alpha\).

Clearly, there exists a sufficiently large \(T>0\) and T is fixed, such that
$$\begin{aligned} &\lambda:=\kappa+(1-\kappa)\exp(-\delta T)< 1. \end{aligned}$$
(2.12)
Taking \(\theta:0<\theta<\frac{1-\lambda}{1+\lambda}\varepsilon_{2}\), using the properties of the superior limits we see that there exists a sufficiently large \(t^{*}>t_{0}\), such that
$$ \left \{ \textstyle\begin{array}{@{}l} x(t^{*})>\alpha-\theta,\\ x(t)< \alpha+\theta,\quad t\in[t^{*}-T,t^{*}]\\ \sup_{t-\tau(t)\leq s\leq t}{x(s)}\leq\alpha+\theta,\quad t\in[t^{*}-T, t^{*}]. \end{array}\displaystyle \right . $$
(2.13)
On the other hand, it follows from (H1) and (H3) that
$$\begin{aligned} x^{\triangle}(t)&\overset{(\mathrm{H}_{1})}{\leq} -a(t)x(t)+b(t)\sup_{t-\tau(t)\leq s\leq t}{x(s)}+c(t) \\ &\overset{(\mathrm{H}_{3})}{\leq}-a(t)x(t)+b(t)\sup _{t-\tau(t)\leq s\leq t}{x(s)}+\frac{a(t)-b(t)}{\delta}\bar{c} \\ &=-a(t) \biggl(x(t)-\frac{\overline{c}}{\delta}\biggr)+b(t)\sup_{t-\tau (t)\leq s\leq t}{ \biggl(x(s)-\frac{\overline{c}}{\delta}\biggr)}. \end{aligned}$$
(2.14)
Denote \(y(t)=x(t)-\frac{\overline{c}}{\delta}\), and (2.13) implies that
$$ \left \{ \textstyle\begin{array}{@{}l} y(t^{*})=x(t^{*}) -\frac{\overline{c}}{\delta}>\alpha-\theta- \frac{\bar{c}}{\delta}=\varepsilon_{2}-\theta,\\ y(t)=x(t)-\frac{\overline{c}}{\delta}\leq\alpha+\theta-\frac {\overline {c}}{\delta}=\varepsilon_{2}+\theta,\quad t\in[t^{*}-T, t^{*}]\\ \sup_{t-\tau(t)\leq s\leq t}{y(s)}=\sup_{t-\tau(t)\leq s\leq t} (x(s)-\frac{\overline{c}}{\delta} )\leq\alpha +\theta -\frac{\overline{c}}{\delta}=\varepsilon_{2}+\theta,\\ \quad t\in[t^{*}-T, t^{*}]. \end{array}\displaystyle \right . $$
(2.15)
By (2.2), (2.14), (2.15), and \(y(t)=x(t)-\frac {\overline{c}}{\delta}\), we have
$$\begin{aligned} y^{\triangle}(t)&\stackrel{(2.14)}{\leq} -a(t)y(t)+b(t)\sup_{t-\tau(t)\leq s\leq t}{y(s)} \\ &\stackrel{(2.15)}{\leq} -a(t)y(t)+(\varepsilon _{2}+ \theta )b(t) \\ &\stackrel{(2.2)}{< } -a(t)y(t)+\kappa(\varepsilon_{2}+ \theta)a(t), \end{aligned}$$
(2.16)
which implies
$$\begin{aligned} \bigl(y(t)-\kappa(\varepsilon_{2}+\theta) \bigr)^{\triangle}\leq -a(t) \bigl(y(t)-\kappa(\varepsilon_{2}+ \theta) \bigr); \end{aligned}$$
(2.17)
then we have
$$\begin{aligned} \biggl(\frac{y(t)-\kappa(\varepsilon_{2}+\theta)}{e_{-a}(t,t_{0})} \biggr)^{\triangle}= \frac{ (y(t)-\kappa(\varepsilon_{2}+\theta) ) ^{\triangle}+a(t) (y(t)-\kappa(\varepsilon_{2}+\theta) )}{e_{-a}(\sigma(t),t_{0})}\overset{(2.17)}{\leq}0, \end{aligned}$$
(2.18)
where we used the property of the exponential function: if \(p\in \mathfrak{R}^{+}\) and \(t_{0}\in\mathbb{T}\), then \(e_{p}(t,t_{0})>0\) for all \(t\in\mathbb{T}\).
Integrating both sides of (2.18) from \(t^{*}-T\) to \(t^{*}\) and by (1.11) we obtain
$$\begin{aligned} (1-\kappa) (\varepsilon_{2}+\theta)-2\theta&= \varepsilon_{2}-\theta -\kappa (\varepsilon_{2}+\theta) \\ &< y\bigl(t^{*}\bigr)-\kappa(\varepsilon_{2}+\theta) \\ &\stackrel{(\mathrm{Lemma}\ 1.3(\mathrm{v}))}{\leq} e_{-a}\bigl(t^{*},t^{*}-T\bigr) \bigl[y \bigl(t^{*}-T\bigr)-\kappa(\varepsilon_{2}+\theta ) \bigr] \\ &\stackrel{(1.11),\ (2.15)}{\leq} \bigl(\varepsilon _{2}+\theta -\kappa(\varepsilon_{2}+\theta) \bigr)\exp \biggl(- \int _{t^{*}-T}^{t^{*}}a(u)\Delta u \biggr) \\ &< (1-\kappa) (\varepsilon_{2}+\theta)\exp(-\delta T), \end{aligned}$$
(2.19)
where we used \(a(t)\geq a(t)-b(t)\geq\delta>0\) in the last step.
By (2.19), we have
$$\theta\geq\frac{1-[\kappa+(1-\kappa)\exp(-\delta T)]}{1+[\kappa+(1-\kappa)\exp(-\delta T)]}\varepsilon_{2}=\frac{1-\lambda}{1+\lambda} \varepsilon_{2}. $$
This contradicts the choice of θ, so we get \(\alpha\leq\frac {\overline{c}}{\delta}\). From the definition of the superior limits we obtain (2.3).

Case 2. \(\overline{c}=0\).

If only we replace in the proof of Case 1 by \(\overline{c}+\epsilon_{3}\) for any given \(\epsilon_{3}>0\), then let \(\epsilon_{3}\rightarrow0^{+}\), we find that (2.1) and (2.3) hold.

A combination of Cases 1 and 2 completes the proof of Theorem 2.1. □

Remark 2.1

When \(M=0\), from (2.7), (H3) must have the form that there exists \(\delta>0\) such that
$$\begin{aligned} a(t)-b(t)-d(t) \int_{0}^{\infty}K(t,s)\triangle s>\delta>0,\quad \mbox{for }t\in [t_{0},\infty)_{\mathbb{T}}. \end{aligned}$$
When \(M>0\), (H3) may have the form that there exists \(\delta>0\) such that
$$\begin{aligned} a(t)-b(t)-d(t) \int_{0}^{\infty}K(t,s)\triangle s\geq\delta>0,\quad \mbox{for } t\in[t_{0},\infty)_{\mathbb{T}}. \end{aligned}$$
Similarly, in [1] when \(G=0\), (2.10) must have the form that
$$ \alpha(t)+\beta(t)< -\sigma< 0 \quad\mbox{for } t\geq t_{0}; $$
when \(G>0\), (2.10) may have the form that
$$ \alpha(t)+\beta(t)\leq-\sigma< 0 \quad\mbox{for } t\geq t_{0}. $$

Theorem 2.1 can be regarded as the extension of the main theorem of [5], Theorem 2.3 of [1].

3 Applications and examples

Consider the delay dynamic equation
$$\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} x^{\triangle}(t)=-a(t)x^{\sigma}(t)+b(t)x(t-\tau(t))+c(t)+d(t)\int ^{\infty}_{0}K(t,s)x(t-s)\triangle s,\\ \quad t\in[t_{0},+\infty)_{{\mathbb {T}}},\\ x(t)=|\varphi(t)|\quad \mbox{for }t\in(-\infty,t_{0}]_{\mathbb{T}}, \end{array}\displaystyle \right . \end{aligned}$$
(3.1)
where \(\varphi(t)\) is bounded rd-continuous for \(s\in(-\infty,t_{0}]_{\mathbb{T}}\) and \(\tau(t)\), \(a(t)\), \(b(t)\), \(c(t)\), \(d(t)\) are nonnegative, rd-continuous functions for \(t\in[t_{0},\infty)_{\mathbb{T}}\) and \(c(t)\) is bounded,
$$\sup_{t\leq t_{0}}{\bigl|\varphi(t)\bigr|}=M,\qquad \sup_{t\geq t_{0}}{c(t)}= \overline{c},\qquad \lim_{t\rightarrow\infty}\bigl(t-\tau (t)\bigr)=+\infty. $$
Assume there exists \(\delta>0\) such that
$$ a(t)-b(t)-d(t) \int_{0}^{\infty} K(t,s)\triangle s>\delta>0\quad \mbox{for } t\in[t_{0},\infty)_{\mathbb{T}}, $$
(3.2)
where the delay kernel \(K(t,s)\) is a nonnegative, rd-continuous for \((t,s)\in[0,\infty)_{\mathbb{T}}\times[0,\infty)_{{\mathbb {T}}}\).
From (3.1), we have
$$\begin{aligned} x(t)={}&x(t_{0})e_{\ominus a}(t,t_{0}) \\ &{}+ \int^{t}_{t_{0}}e_{\ominus a}(t,s) \biggl[b(s)x \bigl(s-\tau(s)\bigr)+c(s)+d(s) \int^{\infty}_{0}K(s,v)x(s-v)\triangle v \biggr] \triangle s. \end{aligned}$$
(3.3)
Let the functions \(y(t)\) be defined as follows: \(y(t)=|x(t)|\) for \(t\in(-\infty,t_{0}]_{{\mathbb {T}}}\), and
$$\begin{aligned} y(t)={}&\bigl|x(t_{0})\bigr|e_{\ominus a}(t,t_{0})\\ &{}+ \int^{t}_{t_{0}}e_{\ominus a}(t,s) \biggl[b(s)\bigl|x \bigl(s-\tau(s)\bigr)\bigr|+c(s) +d(s) \int^{\infty}_{0}K(s,v)\bigl|x(s-v)\bigr|\triangle v \biggr] \triangle s, \end{aligned}$$
for \(t>t_{0}\). Then we have \(|x(t)|\leq y(t)\), for all \(t\in(-\infty ,+\infty)_{{\mathbb {T}}}\).
By [9], Theorem 5.37, we get
$$\begin{aligned} y^{\triangle}(t)={}&\ominus a(t) \biggl\{ \bigl|x(t_{0})\bigr|e_{\ominus a}(t,t_{0})+ \int ^{t}_{t_{0}}e_{\ominus a}(t,s) \biggl[b(s)\bigl|x \bigl(s-\tau(s)\bigr)\bigr|+c(s) \\ &{}+d(s) \int^{\infty}_{0}K(s,v)\bigl|x(s-v)\bigr|\triangle v \biggr] \triangle s \biggr\} \\ &{}+e_{\ominus a}\bigl(\sigma(t),t\bigr) \biggl\{ b(t)\bigl|x \bigl(t-\tau(t)\bigr)\bigr|+c(t)+d(t) \int ^{\infty}_{0}K(t,v)\bigl|x(t-v)\bigr|\triangle v \biggr\} \\ \leq{}& \frac{1}{1+\mu(t)a(t)} \biggl\{ -a(t)y(t)+b(t)\sup_{t-\tau (t)\leq \theta\leq t}y( \theta)+c(t) \\ &{}+d(t) \int^{\infty}_{0}K(t,v)y(t-v)\triangle v \biggr\} ,\quad t\in[t_{0},\infty)_{\mathbb{T}}. \end{aligned}$$
(3.4)

Example 1

Let \({\mathbb {T}}={\mathbb {R}}^{+}\), then system (H1) is expressed as
$$ \left \{ \textstyle\begin{array}{@{}l} x'(t)\leq-a(t)x(t)+b(t)\sup_{t-\tau(t)\leq s\leq t}{x(s)}+c(t)\\ \hphantom{x'(t)\leq}{}+d(t)\int_{0}^{\infty} K(t,s)x(t-s)\,ds,\quad t\geq0,\\ x(t)=|\varphi(t)|,\quad t\leq0, \end{array}\displaystyle \right . $$
(3.5)
where \(\varphi(t)\) is bounded continuous for \(t\in(-\infty,0]\) and \(\sup_{t\leq0}{|\varphi(t)|}=M\).
We choose some explicit nonnegative, continuous functions for \(a(t), b(t), c(t), d(t), \tau(t), K(t,s)\). Let
$$\begin{aligned} &a(t)=\frac{(t+1)^{2}}{t+2},\qquad b(t)=\frac{t^{2}+t}{2e(t+2)},\qquad c(t)= \biggl( \frac{t+2}{t+1} \biggr)^{t},\qquad d(t) =\frac{t}{e(2-e^{-t^{2}})\sqrt{\pi}}, \\ &K(t,s)=(2-\cos2ts)e^{-s^{2}},\quad (t,s)\in[0,\infty)\times[0,\infty), \tau (t)< t \mbox{ and }\lim_{t\rightarrow\infty}\bigl(t-\tau(t)\bigr)=+\infty. \end{aligned}$$
Obviously, \(a(t), b(t), d(t)\) are unbounded for \(t\geq0\) and \(\sup_{t\geq0}{c(t)}=\overline{c}=e\).
(1) \(\forall t\in[0,\infty), g(t):=\int_{0}^{\infty}K(t,s)\,ds =\int_{0}^{\infty}(2-\cos2ts)e^{-s^{2}}\,ds\), then since \(\forall(t,s)\in[0,\infty)\times[0,\infty)\),
$$\begin{aligned} &\bigl|K(t,s)\bigr|\leq3e^{-s^{2}}\quad \mbox{and}\quad \biggl|\frac{\partial K(t,s)}{\partial t}\biggr|=\bigl|2se^{-s^{2}} \sin2ts\bigr|\leq 2se^{-s^{2}}, \\ & \int_{0}^{\infty}e^{-s^{2}}\,ds= \frac{\sqrt{\pi}}{2}=g(0)\quad \mbox{and}\quad \int_{0}^{\infty}2se^{-s^{2}}\,ds=1, \end{aligned}$$
we have \(g(t)=\int_{0}^{\infty}K(t,s)\,ds\) is convergent for \(t\in [0,\infty)\) and \(\int_{0}^{\infty}K_{t}(t,s)\,ds\) is uniformly convergent for \(t\in[0,\infty)\).
So
$$\begin{aligned} g'(t)= \int_{0}^{\infty}K_{t}(t,s)\,ds = \int_{0}^{\infty}2se^{-s^{2}}\sin2ts\,ds =-2tg(t)+2 \sqrt{\pi}t. \end{aligned}$$
Rearrange terms and obtain
$$ \frac{(g(t)-\sqrt{\pi})'}{g(t)-\sqrt{\pi}}=-2t. $$
(3.6)
Solving (3.6) for \(g(t)\), we have
$$ g(t)= \int_{0}^{\infty} K(t,s)\,ds=\frac{\sqrt{\pi }}{2} \bigl(2-e^{-t^{2}}\bigr)< \sqrt{\pi}. $$
(2) There exists \(\delta=\frac{1}{2}>0\), such that
$$\begin{aligned} a(t)-b(t) =\frac{(t+1)^{2}}{t+2}-\frac{t^{2}+t}{2e(t+2)}\geq\frac {1}{2}=\delta>0, \quad t\in[0,\infty). \end{aligned}$$
By (i) of Theorem 2.1, we have \(|x(t)|\leq\frac{\overline{c}}{\delta }+M=2e+M, t\geq0\).
Take \(\kappa=\frac{1}{2}\in(0,1)\), it is easy to see that
$$\kappa a(t)-b(t)=\frac{1}{e}\cdot\frac{(t+1)^{2}}{t+2}\geq \frac{1}{4}>0,\quad t\in[0,\infty). $$
By (ii) of Theorem 2.1, for any given \(\epsilon>0\), there exists \(\widetilde{t}=\widetilde{t}(M,\epsilon)>0\), such that \(|x(t)|\leq\frac{\overline{c}}{\delta}+\epsilon=2e+\epsilon,t\geq \widetilde{t}>0\).

Taking \(c(t)\equiv0\), we have \(|x(t)|\leq\epsilon\), \(t\geq \widetilde{t}>0\). So the zero solution of the system (3.1) is stable.

Example 2

Consider the delay dynamic equation
$$ \left \{ \textstyle\begin{array}{@{}l} x^{\triangle}(t)=-a(t)x^{\sigma}(t)+b(t)x(t-\tau(t))+c(t),\quad t\in [t_{0},+\infty)_{{\mathbb {T}}},\\ x(t)=|\varphi(t)|\quad \mbox{for }t\in(-\infty,t_{0}]_{\mathbb{T}}, \end{array}\displaystyle \right . $$
(3.7)
where \(\varphi(t)\) is bounded, rd-continuous for \(t\leq t_{0}\) and \(\sup_{t\leq t_{0}}{|\varphi(t)|}=M\), \(a(t), b(t), c(t), \tau(t)\) are nonnegative, rd-continuous functions for \(t\in[t_{0},\infty)_{\mathbb {T}}\) and \(\sup_{t\geq t_{0}}{c(t)}=\overline{c}\).
If there exists \(\delta>0\) such that
$$\begin{aligned} a(t)-b(t)\geq\delta>0 \quad\mbox{for } t\geq t_{0}. \end{aligned}$$
(3.8)
Similar to Example 1, we get
$$y^{\triangle}(t)\leq \frac{1}{1+\mu(t)a(t)} \Bigl\{ -a(t)y(t)+b(t)\sup _{t-\tau(t)\leq s\leq t}y(s)+c(t) \Bigr\} ,\quad t\in[t_{0},+ \infty)_{{\mathbb {T}}}. $$
In particular, we take \(\mathbb{T}=\mathbb{N}\), (3.7) reduces to
$$\begin{aligned} \triangle x(n)=-a(n)x(n+1)+b(n)x(n-2)+c(n),\quad n\geq2. \end{aligned}$$
Let \(a(n)=2(n+1), b(n)=\frac{n^{2}}{2n+1},c(n)=\frac{5n}{\sqrt [n]{n!}}, \tau(n)=2\).
Obviously, \(a(n), b(n)\) are unbounded for \(n\in\mathbb{N}\) and \(\sup_{n\in\mathbb{N}}{c(n)}= {\lim_{n\rightarrow\infty }}\frac{5n}{\sqrt[n]{n!}}=\overline{c}=5e\).
  1. (1)

    \(\forall n\geq2, \frac{a(n)-b(n)}{1+a(t)}=\frac{2(n+1)-\frac {n^{2}}{2n+1}}{2n+3}=\frac{3n^{2}+6n+2}{(2n+1)(2n+3)}\geq\frac {2}{3}=\delta\).

     
  2. (2)
    Take \(\kappa=0.9\in(0,1)\), it is easy to see that
    $$\frac{\kappa a(n)-b(n)}{1+a(n)}>0. $$
     

By (ii) of Theorem 2.1, for any given \(\epsilon>0\), there exists \(\widetilde{t}=\widetilde{t}(M,\epsilon)>0\), such that \(|x(t)|\leq\frac{\overline{c}}{\delta}+\epsilon=\frac {15e}{2}+\epsilon ,t\geq\widetilde{t}>0\).

Taking \(c(n)\equiv0\), we have \(|x(t)|\leq\epsilon\), \(t\geq \widetilde{t}>0\). So the zero solution of the system (3.7) is stable.

Declarations

Acknowledgements

The fourth author was supported by the National Natural Science Foundation of China (No. 11271380) and the Guangdong Province Key Laboratory of Computational Science.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
School of Mathematics and Computation Science, Lingnan Normal University, Zhanjiang, 524048, China
(2)
Department of Mathematics, Guangdong University of Petrochemical Technology, Maoming, 525000, China
(3)
School of Mathematics, Sun Yat-Sen University, Guangzhou, 510275, China

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