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 [1–4] 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 c̅ 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.