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Oscillation and nonoscillation theorems of neutral dynamic equations on time scales

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Abstract

We present the oscillation criteria for the following neutral dynamic equation on time scales:

$$ \bigl(y(t)-C(t)y(t-\zeta )\bigr)^{\Delta }+P(t)y(t-\eta )-Q(t)y(t-\delta )=0, \quad t\in {\mathbb{T}}, $$

where \(C, P, Q\in C_{\mathit{rd}}([t_{0},\infty ),{\mathbb{R}}^{+})\), \({\mathbb{R}} ^{+}=[0,\infty )\), \(\gamma , \eta , \delta \in {\mathbb{T}}\) and \(\gamma >0\), \(\eta >\delta \geq 0\). New conditions for the existence of nonoscillatory solutions of the given equation are also obtained.

Introduction

In the past two decades, there has been shown a growing interest in the study of oscillation and stability of delay dynamic equations on time scales. Several excellent monographs [1,2,3,4,5] on the topic indeed reflect its popularity. Some recent results on oscillation and existence of nonoscillatory solutions for dynamic equations can be found in the articles [6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23] and the references cited therein.

Motivated by aforementioned work, in this paper, we consider the following neutral dynamic equation on time scales:

$$ \bigl(y(t)-C(t)y(t-\zeta )\bigr)^{\Delta }+P(t)y(t-\eta )-Q(t)y(t-\delta )=0, \quad t\in {\mathbb{T}}, $$
(1)

where \(C, P, Q\in C_{\mathit{rd}}([t_{0},\infty ),{\mathbb{R}}^{+})\), \({\mathbb{R}} ^{+}= [0, \infty )\), \(C_{\mathit{rd}}\) denotes the class of right-dense continuous functions, \(\zeta , \eta , \delta \in {\mathbb{T}}\) and \(\zeta >0\), \(\eta >\delta \geq 0\). Some conditions for oscillation of Eq. (1) are obtained. We also discuss the existence of nonoscillatory solutions for Eq. (1).

A time scale is an arbitrary nonempty closed subset of the real numbers. We denote the time scale by the symbol \({\mathbb{T}}\). For \(t\in {\mathbb{T}}\) we define the forward jump operator \(\sigma :{\mathbb{T}} \rightarrow {\mathbb{T}}\) by \(\sigma (t):=\inf \{s\in {\mathbb{T}}: s > t \}\). Let \(C_{\mathit{rd}}({\mathbb{T}}, {\mathbb{R}})\) denote the space of functions which are right-dense continuous on \({\mathbb{T}}\). In addition, we define the interval \([t_{0},\infty )\) in \({\mathbb{T}}\) by \([t_{0},\infty ):=\{t \in {\mathbb{T}}: t_{0}\leq t<\infty \}\).

Definition 1.1

For \(h\geq 0\), we define the cylinder transformation \(\xi _{h}\) by

$$ \xi _{h}(z)=\textstyle\begin{cases} \frac{\log (1+hz)}{h}, & \text{if } h\neq 0, \\ z, &\text{if } h=0. \end{cases} $$

Definition 1.2

A solution of (1) is said to be oscillatory if it is neither eventually positive nor eventually negative, otherwise it is nonoscillatory.

Lemma 1.3

If \(f: {\mathbb{T}}\rightarrow {\mathbb{R}}\) is differentiable and \(f^{\Delta }\geq 0\), then f is nondecreasing on \({\mathbb{T}}\).

Lemma 1.4

If \(f: {\mathbb{T}}\rightarrow {\mathbb{R}}\) is differentiable at t, then f is continuous at t.

Oscillation

In this section, we derive the main results for oscillation of Eq. (1). For that, we assume the following conditions:

\((c_{1})\) :

\(0\leq C(t)+ \int ^{t-\delta }_{t-\eta }Q(s+\delta )\Delta s\leq 1 \);

\((c_{2})\) :

\(\bar{R}(t)=P(t)-Q(t-\eta +\delta )\geq 0\) and \(\liminf_{t\rightarrow \infty } \int _{t-\eta }^{t}\bar{R}(s)\Delta s>\gamma >0\).

The following lemmas are useful in proving the main results of this section.

Lemma 2.1

Assume that the conditions \((c_{1})\) and \((c_{2})\) are satisfied. Let \(y(t)\) be an eventually positive solution of (1) such that

$$ u(t)=y(t)-C(t)y(t-\zeta )- \int ^{t-\delta }_{t-\eta }Q(s+\delta )y(s) \Delta s. $$
(2)

Then eventually

$$ u^{\Delta }(t)\leq 0, \qquad u(t)>0. $$

Proof

Since \(y(t)\) is an eventually positive solution of (1), there exists \(t_{1}\geq t_{0}\) such that \(y(t-m)>0\) for \(t\geq t_{1}\), where \(m=\max \{\zeta , \eta , \delta \}\). In view of (1) and (2), we get

$$\begin{aligned} u^{\Delta }(t) =&\bigl(y(t)-C(t)y(t-\zeta )\bigr)^{\Delta }- \biggl( \int ^{t- \delta }_{t-\eta }Q(s+\delta )y(s)\Delta s \biggr)^{\Delta } \\ =&-P(t)y(t-\eta )+Q(t)y(t-\delta )-Q(t)y(t-\delta )+Q(t-\eta +\delta )y(t-\eta ) \\ =&-\bigl(P(t)-Q(t-\eta +\delta )\bigr)y(t-\eta ) \\ =&-\bar{R}(t)y(t-\eta ) \\ \leq & 0, \end{aligned}$$

which implies that \(u(t)\) is decreasing. Next, we shall show that \(u(t)>0\). If \(u(t)\rightarrow -\infty \) as \(t\rightarrow \infty \), then \(y(t)\) must be unbounded. Therefore there exists \(\{t_{n}'\}\) with \(t_{n}'\geq t_{2}\), \(t_{2}=t_{1}+m\) such that

$$ \lim_{n\rightarrow \infty }t_{n}'=\infty , \qquad \lim _{n\rightarrow \infty }y\bigl(t_{n}'\bigr)=\infty , $$

and \(y(t_{n}')=\max_{t_{2}\leq t\leq t_{n}'}y(t)\). Hence, we have

$$\begin{aligned} u\bigl(t_{n}'\bigr) =&y\bigl(t_{n}' \bigr)-C\bigl(t_{n}'\bigr)y\bigl(t_{n}'- \zeta \bigr)- \int ^{t_{n}'-\delta } _{t_{n}'-\eta }Q(s+\delta )y(s)\Delta s \\ \geq &y\bigl(t_{n}'\bigr) \biggl(1-C \bigl(t_{n}'\bigr)- \int ^{t_{n}'-\delta }_{t_{n}'- \eta }Q(s+\delta )\Delta s \biggr)\geq 0. \end{aligned}$$

In consequence, we get

$$ \lim_{t\to \infty }u(t)=\lim_{n\to \infty }u \bigl(t_{n}'\bigr)\geq 0, $$

which is a contradiction. Hence \(\lim_{t\rightarrow \infty }u(t)=l\) exists. As before, if \(y(t)\) is unbounded, then \(l\geq 0\). Now we consider the case when \(y(t)\) is bounded. Let \(\bar{l}= \limsup_{t\rightarrow \infty } y(t)=\lim_{t'\rightarrow \infty }y(t')\). Then

$$\begin{aligned} y\bigl(t'\bigr)-u\bigl(t'\bigr) =&C \bigl(t'\bigr)y\bigl(t'-\zeta \bigr)+ \int ^{t'-\delta }_{t'-\eta }Q(s+ \delta )y(s)\Delta s \\ \leq & y(\xi _{t'}) \biggl(C\bigl(t'\bigr)+ \int ^{t'-\delta }_{t'-\eta }Q(s+ \delta )\Delta s \biggr), \end{aligned}$$

where \(y(\xi _{t'})=\max \{\{y(s):s\in (t'-\eta ,t'-\delta )\}, y(t'- \zeta )\}\). Hence, it follows that \(\xi _{t'}\to \infty \) as \(t'\to \infty \) and \(\limsup_{t'\to \infty }y(\xi _{t'})\leq \bar{l}\). Thus, we get

$$ y\bigl(t'\bigr)-u\bigl(t'\bigr) \leq y( \xi _{t'}), $$
(3)

which, on taking superior limit, leads to \(\bar{l}-l\leq \bar{l}\). Therefore \(l\geq 0\). Hence \(u(t)>0\) eventually. The proof is complete. □

Lemma 2.2

Suppose that the conditions \((c_{1})\) and \((c_{2})\) hold and that \(y(t)\) is an eventually positive solution of (1) satisfying (2). Then the set \(\varLambda = \{ \lambda >0: u^{\Delta }(t)+ \lambda \bar{R}u(t)\leq 0, \textit{eventually} \} \) is nonempty and there exists an upper bound of Λ which is independent of solution \(y(t)\).

Proof

From the given assumptions, there exists a \(t_{1}\geq t_{0}\), such that \(y(t-m)>0\) for \(t\geq t_{1}\), where \(m=\{\zeta ,\eta ,\delta \}\). It follows from (2) that \(u(t)\leq y(t)\) for \(t\geq t_{1}\). Then

$$ u^{\Delta }(t)=-\bar{R}(t)y(t-\eta )\leq -\bar{R}(t)u(t-\eta ) \leq - \bar{R}(t)u(t), \quad t\geq t_{1}+m, $$
(4)

that is, \(\lambda =1\in \varLambda \). Therefore Λ is nonempty.

Let

$$ 3k=\liminf_{t\to \infty } \int _{t-\eta }^{t} \bar{R}(s)\Delta s. $$

By \((c_{2})\), we have \(k>0\), and there exists a \(t_{2}>t_{1}+m\) such that

$$ \int _{t-\eta }^{t} \bar{R}(s)\Delta s>2k:=\gamma , \quad t\geq t_{2}. $$

Therefore, for any \(t\geq t_{2}\), there exists \(t^{*}>t>t^{*}-\eta \) such that

$$ \int _{t}^{t^{*}}\bar{R}(s)\Delta s> k,\qquad \int _{t^{*}-\eta }^{t} \bar{R}(s)\Delta s> k. $$

Integrating (4) from t to \(t^{*}\) and noting that \(u^{\Delta }(t)\leq 0\), \(u(t)>0\) for \(t\geq t_{2}\), we find that

$$ u(t)-u\bigl(t^{*}\bigr)\leq - \int _{t}^{t^{*}}\bar{R}(s)u(s-\eta )\Delta s, $$

which implies that

$$ u(t)\geq \int _{t}^{t^{*}}\bar{R}(s)u(s-\eta )\Delta s\geq u \bigl(t^{*}- \eta \bigr) \int _{t}^{t^{*}}\bar{R}(s)\Delta s>k u \bigl(t^{*}-\eta \bigr). $$

Next, integrating (4) from \(t^{*}-\eta \) to t, we get

$$ u\bigl(t^{*}-\eta \bigr)>ku(t-\eta ). $$

Hence

$$ u(t)>k^{2}u(t-\eta ), \quad t\geq t_{2}. $$
(5)

Let us define

$$ \liminf_{t\rightarrow \infty } y(t-\eta )=I. $$
(6)

Since \(y(t-m)>0\), (6) implies that \(I\geq 0\). On the other hand, there exists a sequence \(\{t'_{n}\}\) such that \(t_{n}'\geq t_{2}\) and \(t'_{n}\rightarrow \infty \) as \(n\rightarrow \infty \) and

$$ \liminf_{t\rightarrow \infty } \int _{t-\eta }^{t}\bar{R}(s)\Delta s= \lim _{n\rightarrow \infty } \int _{t'_{n}-\eta }^{t'_{n}}\bar{R}(s) \Delta s. $$
(7)

From (4), we have

$$ y(\xi _{n}-\eta ) \int _{t'_{n}-\eta }^{t'_{n}}\bar{R}(s)\Delta s= \int _{t'_{n}-\eta }^{t'_{n}}\bar{R}(s) y(s-\eta )\Delta s=-u \bigl(t'_{n}\bigr)+u\bigl(t'_{n}- \eta \bigr), $$
(8)

where \(\xi _{n}\in [t_{n}'-\eta ,t_{n}']\), and \(\xi _{n}\to \infty \) as \(n\to \infty \). Hence, we can find an increasing subsequence in \(\{\xi _{n}\}\) and so, without loss of generality, we may assume that the sequence numbers \(\{\xi _{n}\}\) is also increasing. Let

$$ F(t)=\inf \bigl\{ y(s-\eta ):s\geq t\bigr\} , \quad t\geq t_{2}. $$

Then we have

$$ \lim_{t\rightarrow \infty } F(t)=\liminf_{t\rightarrow \infty } y(t- \eta ). $$

Since \(\{\xi _{n}\}\) is an increasing sequence of numbers, we get

$$ \bigl\{ y\bigl(\xi _{n}'-\eta \bigr):n'\geq n\bigr\} \subset \bigl\{ y(s-\eta ):s\geq \xi _{n}\bigr\} . $$

Therefore

$$ F(\xi _{n})=\inf \bigl\{ y(s-\eta ):s\geq \xi _{n}\bigr\} \leq \inf \bigl\{ y\bigl(\xi _{n}'- \eta \bigr):n'\geq n\bigr\} , $$

which implies that

$$ \liminf_{t\rightarrow \infty } y(t-\eta )=\lim_{n\rightarrow \infty }F( \xi _{n})\leq \lim_{n\rightarrow \infty }\inf y(\xi _{n}-\eta ), $$

that is,

$$ \liminf_{t\rightarrow \infty } y(t-\eta )\leq \lim _{n\rightarrow \infty }\inf y(\xi _{n}-\eta ). $$
(9)

On the other hand, \(\lim_{t\rightarrow \infty }u(t)\) exists and is a finite number. Therefore, it follows from (7)–(9) that

$$\begin{aligned} I \biggl(\liminf_{t\rightarrow \infty } \int _{t^{-}\eta }^{t}\bar{R}(s) \Delta s \biggr) =& \Bigl( \liminf_{t\rightarrow \infty } y(t-\eta ) \Bigr) \biggl(\liminf _{t\rightarrow \infty } \int _{t^{-}\eta }^{t}\bar{R}(s) \Delta s \biggr) \\ \leq & \Bigl(\lim_{n\rightarrow \infty }\inf y(\xi _{n}-\eta ) \Bigr) \biggl(\lim_{n\rightarrow \infty } \int _{t'_{n}-\eta }^{t'_{n}}\bar{R}(s) \Delta s \biggr) \\ =& \Bigl(\lim_{n\rightarrow \infty }\inf y(\xi _{n}-\eta ) \Bigr) \biggl(\lim_{n\rightarrow \infty }\inf \int _{t'_{n}-\eta }^{t'_{n}} \bar{R}(s)\Delta s \biggr) \\ \leq &\lim_{n\rightarrow \infty }\inf \bigl(y(\xi _{n}-\eta )\bigr) \int _{t'_{n}-\eta }^{t'_{n}}\bar{R}(s)\Delta s \\ =&\lim_{n\rightarrow \infty }\inf \int _{t'_{n}-\eta }^{t'_{n}} \bar{R}(s)y(s-\eta )\Delta s) \\ =&-\lim_{n\rightarrow \infty }u\bigl(t'_{n}\bigr)+\lim _{n\rightarrow \infty }u\bigl(t'_{n}- \eta \bigr) \\ =& 0, \end{aligned}$$

that is,

$$ I \biggl(\liminf_{t\rightarrow \infty } \int _{t^{-}\eta }^{t}\bar{R}(s) \Delta s \biggr)\leq 0. $$
(10)

From condition \((c_{2})\), (10) and the fact that \(I\geq 0\), we deduce that \(I=0\). Thus, we obtain

$$ \liminf_{t\rightarrow \infty } y(t-\eta )=0. $$

Hence there exists a sequence \(\{s_{n}\}\) with \(s_{n}\geq t_{2}+2m\), such that \(y(s_{n})\rightarrow 0\) as \(n \rightarrow \infty \) and \(y(s_{n}-\eta )=\min_{t_{2}\leq s\leq s_{n}-\eta }y(s)\) for \(n=1,2,\ldots \) . Then, from (4) for \(n=1,2,\ldots \) , we have

$$\begin{aligned} u(s_{n})-u(s_{n}-\eta ) =&- \int _{s_{n}-\eta }^{s_{n}}\bar{R}(s)y(s- \eta )\Delta s \\ \leq &-y(s_{n}-\eta ) \int _{s_{n}-\eta }^{s_{n}}\bar{R}(s)\Delta s \\ < &-2k y(s_{n}-\eta ). \end{aligned}$$

Hence

$$ u(s_{n}-\eta )>2k y(s_{n}-\eta ), \quad n=1,2,\ldots . $$
(11)

Also, from (4), (5) and (11), for \(n=1,2,\ldots \) , we have

$$\begin{aligned} u^{\Delta }(s_{n})=-\bar{R}(s_{n})y(s_{n}- \eta ) >-\frac{1}{2k} \bar{R}(s_{n})u(s_{n}-\eta ) \geq -\frac{1}{2k^{3}}\bar{R}(s_{n})u(s _{n}), \end{aligned}$$

which implies that

$$ u^{\Delta }(s_{n})+\frac{1}{2k^{3}} \bar{R}(s_{n})u(s_{n})>0, \quad n=1,2,\ldots . $$
(12)

Now we may assert that \(\frac{1}{2k^{3}}\bar{\in } \varLambda \). In fact, if \(\frac{1}{2k^{3}}\in \varLambda \), then there exists some \(T'\) by the definition of Λ such that, for all \(t\geq T'\), the following inequality holds true:

$$ u^{\Delta }(t)+\frac{1}{2k^{3}}\bar{R}(t)u(t)\leq 0. $$
(13)

On the other hand, in view of the fact that \(s_{n}\rightarrow 0\) as \(n \rightarrow \infty \), from \(\{s_{n}\}\) we find some \(s_{n}'\) such that \(s_{n}'\geq T'\). Then it follows from (12) that

$$ u^{\Delta }\bigl(s_{n}'\bigr)+\frac{1}{2k^{3}} \bar{R}\bigl(s_{n}'\bigr)u\bigl(s_{n}' \bigr)>0, $$

which contradicts (13). Therefore, \(\frac{1}{2k^{3}}\) is an upper bound of Λ which is independent of solution \(y(t)\). The proof is complete. □

Theorem 2.3

Assume that the conditions \((c_{1})\) and \((c_{2})\) are satisfied. In addition it is assumed that there exist \(T \geq t_{1}+m\) and \(\lambda >0\) such that

$$ \begin{aligned}[b] &\inf_{t\geq T,\lambda >0} \biggl\{ \frac{1}{\lambda }\exp \biggl(- \int ^{t}_{t-\eta }\xi _{\mu }\bigl( -\lambda \bar{R}(s)\bigr)\Delta s \biggr) +C(t- \eta )\exp \biggl(- \int ^{t}_{t-\zeta }\xi _{\mu }\bigl( -\lambda \bar{R}(s)\bigr) \Delta s \biggr) \\ &\quad {}+ \int ^{t-\delta }_{t-\eta }Q(s+\delta -\eta )\exp \biggl(- \int ^{t} _{s}\xi _{\mu }\bigl( -\lambda \bar{R}(u)\bigr)\Delta u \biggr)\Delta s \biggr\} > 1. \end{aligned} $$
(14)

Then every solution of Eq. (1) is oscillatory.

Proof

On the contrary, let \(y(t)\) be a nonoscillatory solution of Eq. (1). Without loss of generality, it can be assumed that \(y(t)\) is an eventually positive solution. Moreover, let \(u(t)\) be the same as defined in (2) and the set Λ as given in Lemma 2.2. Then, by Lemma 2.2, we see that there exists a \(t_{2}\geq t_{0}\) such that

$$ u^{\Delta }(t)\leq 0, \qquad u(t)>0, \quad \text{for }t\geq t_{2}. $$

From condition (14), there exists a constant \(\alpha >1\) such that

$$ \begin{aligned}[b] &\inf_{t\geq T,\lambda >0} \biggl\{ \frac{1}{\lambda }\exp \biggl(- \int ^{t}_{t-\eta }\xi _{\mu }\bigl( -\lambda \bar{R}(s)\bigr)\Delta s \biggr) +C(t- \eta )\exp \biggl(- \int ^{t}_{t-\zeta }\xi _{\mu }\bigl( -\lambda \bar{R}(s)\bigr) \Delta s \biggr) \\ &\quad {}+ \int ^{t-\delta }_{t-\eta }Q(s+\delta -\eta )\exp \biggl(- \int ^{t} _{s}\xi _{\mu }\bigl( -\lambda \bar{R}(u)\bigr)\Delta u \biggr)\Delta s \biggr\} \geq \alpha > 1. \end{aligned} $$
(15)

Let \(\lambda _{0}\in \varLambda \). Then we shall show that \(\alpha \lambda _{0} \in \varLambda \). In fact, \(\lambda _{0}\in \varLambda \) implies that

$$ u^{\Delta }(t)+\lambda _{0}\bar{R}(t)u(t)\leq 0. $$
(16)

Define

$$ w(t)=u(t)\exp \biggl(- \int ^{t}_{t_{0}}\xi _{\mu }\bigl( -\lambda _{0}\bar{R}(s)\bigr) \Delta s \biggr) $$
(17)

and note that \(w(t)\) is well defined. Let us introduce

$$ v(t)=\exp \biggl( \int ^{t}_{t_{0}}\xi _{\mu }\bigl( -\lambda _{0}\bar{R}(s)\bigr) \Delta s \biggr) $$

and note that

$$\begin{aligned} w^{\Delta }(t) =& \biggl(\frac{u(t)}{v(t)} \biggr)^{\Delta } \\ =&\frac{u^{\Delta }(t)v(t)-u(t)v^{\Delta }(t)}{v(t)v(\sigma (t))} \\ \leq &\frac{-\lambda _{0}\bar{R}(t)u(t)v(t)-u(t)[-\lambda _{0}\bar{R}(t)v(t)]}{v(t)v( \sigma (t))} \\ =&\frac{-\lambda _{0}\bar{R}(t)u(t)v(t)+u(t)\lambda _{0}\bar{R}(t)v(t)}{v(t)v( \sigma (t))} \\ =&0. \end{aligned}$$

Hence, \(w(t)\) is nonincreasing. From (2), we get \(u^{\Delta }(t)=- \bar{R}(t)y(t-\eta )\), which together with (16) yields \(y(t-\eta ) \geq \lambda _{0}u(t)\). Therefore

$$\begin{aligned} u^{\Delta }(t) =&-\bar{R}(t)y(t-\eta ) \\ =&-\bar{R}(t) \biggl[u(t-\eta )+C(t-\eta )y(t-\eta -\zeta )+ \int ^{t-\eta -\delta }_{t-2\eta }Q(s+\delta )y(s)\Delta s \biggr] \\ \leq &-\bar{R}(t) \biggl[u(t-\eta )+\lambda _{0}C(t-\eta )u(t-\zeta )+ \lambda _{0} \int ^{t-\eta -\delta }_{t-2\eta }Q(s+\delta )u(s+\eta ) \Delta s \biggr] \\ =&-\bar{R}(t) \biggl[u(t-\eta )+\lambda _{0}C(t-\eta )u(t-\zeta )+ \lambda _{0} \int ^{t-\delta }_{t-\eta }Q(s+\delta -\eta )u(s)\Delta s \biggr] \\ =&-\bar{R}(t) \biggl[w(t-\eta )\exp \biggl( \int ^{t-\eta }_{t_{0}}\xi _{\mu }\bigl( -\lambda _{0}\bar{R}(s)\bigr)\Delta s \biggr) \\ &{}+\lambda _{0}C(t-\eta )w(t-\zeta )\exp \biggl( \int ^{t-\zeta }_{t_{0}} \xi _{\mu }\bigl(-\lambda _{0}\bar{R}(s)\bigr)\Delta s \biggr) \\ &{}+\lambda _{0} \int ^{t-\delta }_{t-\eta }Q(s+\delta -\eta )w(s)\exp \biggl( \int ^{s}_{t_{0}}\xi _{\mu }\bigl( -\lambda _{0}\bar{R}(u)\bigr)\Delta u \biggr) \Delta s \biggr] \\ \leq &-\bar{R}(t) \biggl[w(t)\exp \biggl( \int ^{t-\eta }_{t_{0}}\xi _{ \mu }\bigl( -\lambda _{0}\bar{R}(s)\bigr)\Delta s \biggr)\\ &{}+\lambda _{0}C(t-\eta )w(t) \exp \biggl( \int ^{t-\zeta }_{t_{0}}\xi _{\mu }\bigl( -\lambda _{0}\bar{R}(s)\bigr) \Delta s \biggr) \\ &{}+\lambda _{0} \int ^{t-\delta }_{t-\eta }Q(s+\delta -\eta )w(s)\exp \biggl( \int ^{s}_{t_{0}}\xi _{\mu }\bigl( -\lambda _{0}\bar{R}(u)\bigr)\Delta u \biggr) \Delta s \biggr] \\ =&-\bar{R}(t) \biggl[u(t)\exp \biggl(- \int ^{t}_{t-\eta }\xi _{\mu }\bigl( - \lambda _{0}\bar{R}(s)\bigr)\Delta s \biggr)\\ &{}+\lambda _{0}C(t-\eta )u(t) \exp \biggl(- \int ^{t}_{t-\zeta }\xi _{\mu }\bigl( -\lambda _{0}\bar{R}(s)\bigr) \Delta s \biggr) \\ &{}+\lambda _{0} \int ^{t-\delta }_{t-\eta }Q(s+\delta -\eta )u(s)\exp \biggl(- \int ^{t}_{s}\xi _{\mu }\bigl( -\lambda _{0}\bar{R}(u)\bigr)\Delta u \biggr) \Delta s \biggr] \\ \leq &-\bar{R}(t) \biggl[\exp \biggl(- \int ^{t}_{t-\eta }\xi _{\mu }\bigl( - \lambda _{0}\bar{R}(s)\bigr)\Delta s \biggr)+\lambda _{0}C(t-\eta ) \exp \biggl(- \int ^{t}_{t-\zeta }\xi _{\mu }\bigl( -\lambda _{0}\bar{R}(s)\bigr)\Delta s \biggr) \\ &{}+\lambda _{0} \int ^{t-\delta }_{t-\eta }Q(s+\delta -\eta )\exp \biggl(- \int ^{t}_{s}\xi _{\mu }\bigl( -\lambda _{0}\bar{R}(u)\bigr)\Delta u \biggr)\Delta s \biggr]u(t) \\ \leq &-\inf_{t\geq T} \biggl[\exp \biggl(- \int ^{t}_{t-\eta }\xi _{ \mu }\bigl( -\lambda _{0}\bar{R}(s)\bigr)\Delta s \biggr)+\lambda _{0}C(t-\eta ) \exp \biggl(- \int ^{t}_{t-\zeta }\xi _{\mu }\bigl( -\lambda _{0}\bar{R}(s)\bigr) \Delta s \biggr) \\ &{}+\lambda _{0} \int ^{t-\delta }_{t-\eta }Q(s+\delta -\eta )\exp \biggl(- \int ^{t}_{s}\xi _{\mu }\bigl( -\lambda _{0}\bar{R}(u)\bigr)\Delta u \biggr)\Delta s \biggr]\bar{R}(t)u(t) \\ \leq &-\alpha \lambda _{0}\bar{R}(t)u(t). \end{aligned}$$

Thus, \(\alpha \lambda _{0}\in \varLambda \). Repeating this procedure, one finds that \(\alpha ^{m}\lambda _{0}\in \varLambda \) for any integer m, which contradicts the boundedness of Λ. The proof is complete. □

Corollary 2.4

Assume that \(P(t)\geq 0\), \(\liminf_{t\rightarrow \infty }\int _{t- \eta }^{t}P(s)\Delta s>0\) and there exist T and \(\lambda >0\) such that

$$ \inf_{t\geq T,\lambda >0} \biggl\{ \frac{1}{\lambda }\exp \biggl(- \int ^{t}_{t-\eta }\xi _{\mu }\bigl( -\lambda \bar{R}(s)\bigr)\Delta s \biggr) \biggr\} > 1. $$

Then every solution of the equation

$$ y^{\Delta }(t)+P(t)y(t-\eta )=0 $$

is oscillatory.

Nonoscillation

Here we derive some results for the existence of a positive solution of (1).

Lemma 3.1

Assume that

  1. (i)

    \(\bar{R}(t)=P(t)-Q(t-\eta -\delta )\geq 0\);

  2. (ii)

    the inequality

    $$ C(t)z(t-\zeta )+ \int ^{t-\delta }_{t-\eta }Q(s+\delta )z(s)\Delta s + \int _{t-\eta }^{\infty }\bar{R}(s+\eta )z(s)\Delta s\leq z(t), \quad \textit{for } t\geq t_{1}, $$
    (18)

    has a continuous positive solution \(Z(t)\): \([t_{1}-m, \infty )\rightarrow (0,\infty )\) with \(\lim_{t\rightarrow \infty }Z(t)=0\).

Then the equation

$$ C(t)y(t-\zeta )+ \int ^{t-\delta }_{t-\eta }Q(s+\delta )y(s)\Delta s + \int _{t-\eta }^{\infty }\bar{R}(s+\eta )y(s)\Delta s= y(t), \quad \textit{for } t\geq t_{1}, $$
(19)

has a continuous positive solution \(y(t)\) with \(0< y(t)\leq Z(t)\) for \(t\geq t_{1}\).

Proof

Take \(T>t_{1}\) large enough so that \(z(t)>Z(t)\) for \(t\in [t_{1}-m,T)\). Define a set

$$ \varOmega =\bigl\{ {\omega \in C_{\mathit{rd}} \bigl([t_{1}-m, \infty \bigr),{\mathbb{R}}^{+}) : 0 \leq \omega (t)\leq Z(t),t\geq t_{1}-m \bigr\} } $$

and introduce an operator S on Ω as follows:

$$ (S\omega ) (t)=\textstyle\begin{cases} C(t)\omega (t-\zeta )+\int ^{t-\delta }_{t-\eta }Q(s+\delta )\omega (s) \Delta s + \int _{t-\eta }^{\infty }\bar{R}(s+\eta )\omega (s)\Delta s, & t\in ( T ,\infty ), \\ (S\omega )(T)+z(t)-Z(T), & t\in [t_{1}-m,T ]. \end{cases} $$

It is clear that \(S \varOmega \subset \varOmega \), and \(\omega _{1}, \omega _{2}\in \varOmega \) with \(\omega _{1}\leq \omega _{2}\) implies \(S\omega _{1}\leq S \omega _{2}\).

Define a sequence on Ω as

$$ z _{0}(t)=Z (t ), \qquad z_{k} (t)=S z _{k -1}(t), \quad k =1,2,\ldots . $$

It is not difficult to prove that

$$ 0\leq z_{k}(t)\leq z_{k-1}(t)\leq \cdots \leq z_{1}(t )\leq z(t), \quad t\in [t_{1}-m,\infty ). $$

Therefore, the sequence \(\{z _{k}(t)\}\) has a limiting function \(y(t)\) with \(\lim_{t\rightarrow \infty }z _{k} (t)=y(t)\) for \(t\in [t_{1}-m,\infty )\) and \(y(t)\) satisfies (19) by Lebesgue’s convergence theorem. It is easy to see that \(y(t)>0\) for \(t\in [t_{1}-m,T ]\) and hence \(y(t)>0\) for all \(t\in [t_{1}-m, \infty )\) with \(0< y(t)\leq Z(t)\). The proof is complete. □

Theorem 3.2

Assume that

  1. (i)

    \(\bar{R}(t)=P(t)-Q(t-\eta -\delta )\geq 0\);

  2. (ii)

    there exist \(T\geq t_{1}+m\) and \(\lambda ^{*}>0\) such that

    $$ \begin{aligned}[b] &\sup_{t\geq T} \biggl\{ \frac{1}{\lambda ^{*}}\exp \biggl(- \int ^{t}_{t- \eta }\xi _{\mu }\bigl( -\lambda ^{*}\bar{R}(u)\bigr)\Delta u \biggr) +C(t-\eta ) \exp \biggl(- \int ^{t}_{t-\zeta }\xi _{\mu }\bigl( -\lambda ^{*}\bar{R}(s)\bigr) \Delta s \biggr) \\ &\quad {}+ \int ^{t-\delta }_{t-\eta }Q(s+\delta -\eta )\exp \biggl(- \int ^{t} _{s}\xi _{\mu }\bigl( -\lambda ^{*}\bar{R}(u)\bigr)\Delta u \biggr)\Delta s \biggr\} \leq 1. \end{aligned} $$
    (20)

Then Eq. (1) has a positive solution \(y(t)\) with \(\lim_{t\rightarrow \infty }y(t)=0\).

Proof

Set

$$ z(t)=\exp \biggl( \int ^{t+\eta }_{t_{1}}\xi _{\mu }\bigl( -\lambda ^{*} \bar{R}(s)\bigr)\Delta s \biggr). $$
(21)

Obviously \(z(t)\) is well defined, positive and continuous. From the condition (20), for \(t\geq T\geq T-\eta \), we have

$$ \begin{aligned}[b] &\frac{1}{\lambda ^{*}} \biggl\{ \exp \biggl(- \int ^{t+\eta }_{t}\xi _{ \mu }\bigl( -\lambda ^{*}\bar{R}(u)\bigr)\Delta u \biggr) +C(t)\exp \biggl(- \int ^{t+\eta }_{t+\eta -\zeta }\xi _{\mu }\bigl( -\lambda ^{*}\bar{R}(s)\bigr) \Delta s \biggr) \\ &\quad {}+ \int ^{t-\delta +\eta }_{t}Q(s+\delta -\eta )\exp \biggl(- \int ^{t+ \eta }_{s}\xi _{\mu }\bigl( -\lambda ^{*}\bar{R}(u)\bigr)\Delta u \biggr)\Delta s \biggr\} \leq 1. \end{aligned} $$
(22)

Substituting (21) into (22), we get

$$ \frac{1}{\lambda ^{*}}\frac{z(t-\eta )}{z(t)}+C(t) \frac{z(t-\zeta )}{z(t)} + \int ^{t-\delta +\eta }_{t}Q(s+\delta - \eta ) \frac{z(s-\eta )}{z(t)} \Delta s\leq 1. $$
(23)

From (21), it is easy to see that \(z^{\Delta }(t)=-\lambda ^{*} \bar{R}(t+\eta )z(t)\), and hence we have

$$ \int ^{\infty }_{t-\eta }\bar{R}(s+\eta )z(s)\Delta s=- \frac{1}{\lambda ^{*}} \int ^{\infty }_{t-\eta }z^{\Delta }(s)\Delta s= \frac{z(t-\eta )}{ \lambda ^{*}}. $$
(24)

Combining (23) and (24), we obtain

$$ \int ^{\infty }_{t-\eta }\bar{R}(s+\eta )z(s)\Delta s+C(t)z(t- \zeta )+ \int ^{t+\eta -\delta }_{t}Q(s+\delta -\eta ) z(s-\eta )\leq z(t). $$

Thus the desired conclusion follows by Lemma 3.1. The proof is complete. □

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Acknowledgements

This project was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, Saudi Arabia, under grant No. (DF-070-130-1441). The author, therefore, acknowledge with thanks DSR technical and financial support.

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Funding

This project was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, under grant No. (DF-070-130-1441).

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All authors, YZ, BA and AA, contributed equally to each part of this work. All authors read and approved the final manuscript.

Correspondence to Yong Zhou.

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MSC

  • 34A99
  • 34C10
  • 34N05
  • 39A11

Keywords

  • Oscillation
  • Nonoscillation
  • Dynamic equations
  • Time scales