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.
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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
and \(\alpha>\beta>0\), then there exist \(\gamma>0\) and \(K>0\) such that
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
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
then
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:
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
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
where \(\xi_{\mu(s)}\) is the cylinder transformation given by
and some properties of the exponential function are given in the following lemma.
Lemma 1.3
see [7]
Let \(\varphi\in\mathfrak{R}\), Then
-
(i)
\(e_{0}(s,t)\equiv1, e_{\varphi}(t,t)\equiv1\) and \(e_{\varphi}(\sigma(t),s) = (1+\mu(t)\varphi(t) )e_{\varphi}(t,s)\);
-
(ii)
\(\frac{1}{e_{\varphi}(t,s)} =e_{\ominus\varphi}(t,s)\), where \(\ominus\varphi(t)=-\frac {\varphi(t)}{1+\mu(t)\varphi(t)}\);
-
(iii)
\((\frac{1}{e_{\varphi}(t,s)} )^{\triangle} =-\frac{\varphi(t)}{e_{\varphi}(\sigma(t),s)}\);
-
(iv)
\([e_{\varphi}(c,t)]^{\Delta}=-\varphi(t)e_{\varphi}(c,\sigma (t))\), where \(c\in\mathbb{T}\);
-
(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
If φ is rd-continuous and nonnegative, then
Remark 1.1
If \(\varphi\in\mathfrak{R}^{+}\) and \(\varphi(r)>0\) for all \(r\in [s,t]_{\mathbb{T}}\), then
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
Since \(a< b\), from the above result, we have
□
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
-
(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
-
(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)
from this we shall deduce that
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 \(x(t_{1})=\frac{\overline{c}}{\delta }+\varepsilon M, t_{1}>t_{0}\),
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)
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
We have from (2.8) and (H1)
By (2.8), (2.9), (H3), and \(1-\mu(t)a(t)>0, t\in\mathbb{T} \), we get
which leads to a contradiction.
Hence the inequality (2.5) must hold.
Since \(\varepsilon>1\) is arbitrary, we let \(\varepsilon\rightarrow 1^{+}\) and obtain
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
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
On the other hand, it follows from (H1) and (H3) that
Denote \(y(t)=x(t)-\frac{\overline{c}}{\delta}\), and (2.13) implies that
By (2.2), (2.14), (2.15), and \(y(t)=x(t)-\frac {\overline{c}}{\delta}\), we have
which implies
then we have
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
where we used \(a(t)\geq a(t)-b(t)\geq\delta>0\) in the last step.
By (2.19), we have
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
When \(M>0\), (H3) may have the form that there exists \(\delta>0\) such that
Similarly, in [1] when \(G=0\), (2.10) must have the form that
when \(G>0\), (2.10) may have the form that
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
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,
Assume there exists \(\delta>0\) such that
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
Let the functions \(y(t)\) be defined as follows: \(y(t)=|x(t)|\) for \(t\in(-\infty,t_{0}]_{{\mathbb {T}}}\), and
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
Example 1
Let \({\mathbb {T}}={\mathbb {R}}^{+}\), then system (H1) is expressed as
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
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 \(\forall(t,s)\in[0,\infty)\times[0,\infty)\),
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
Rearrange terms and obtain
Solving (3.6) for \(g(t)\), we have
(2) There exists \(\delta=\frac{1}{2}>0\), such that
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
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
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
Similar to Example 1, we get
In particular, we take \(\mathbb{T}=\mathbb{N}\), (3.7) reduces to
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)
\(\forall n\geq2, \frac{a(n)-b(n)}{1+a(t)}=\frac{2(n+1)-\frac {n^{2}}{2n+1}}{2n+3}=\frac{3n^{2}+6n+2}{(2n+1)(2n+3)}\geq\frac {2}{3}=\delta\).
-
(2)
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.
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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.
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Ou, B., Lin, Q., Du, F. et al. An extended Halanay inequality with unbounded coefficient functions on time scales. J Inequal Appl 2016, 316 (2016). https://doi.org/10.1186/s13660-016-1259-x
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DOI: https://doi.org/10.1186/s13660-016-1259-x