Oscillation and nonoscillation theorems of neutral dynamic equations on time scales

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.

1 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.

2 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.

3 Nonoscillation

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

Lemma 3.1

Assume that

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

• (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

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

• (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. □