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Advances in Difference Equations

, 2019:16

# The Kamenev type interval oscillation criteria of mixed nonlinear impulsive differential equations under variable delay effects

• Xiaoliang Zhou
• Changdong Liu
• Ruyun Chen
Open Access
Research
• 169 Downloads

## Abstract

In this paper, a class of mixed nonlinear impulsive differential equations is studied. When the delay $$\sigma(t)$$ is variable, each given interval is divided into two parts on which the quotients of $$x(t-\sigma(t))$$ and $$x(t)$$ are estimated. Then, by introducing binary auxiliary functions and using the Riccati transformation, several Kamenev type interval oscillation criteria are established. The well-known results obtained by Liu and Xu (Appl. Math. Comput. 215:283–291, 2009) for $$\sigma(t)=0$$ and by Guo et al. (Abstr. Appl. Anal. 2012:351709, 2012) for $$\sigma(t)=\sigma_{0}$$ ($$\sigma_{0}\geq0$$) are developed. Moreover, an example illustrating the effectiveness and non-emptiness of our results is also given.

## Keywords

Interval oscillation Impulsive differential equation Variable delay Interval delay function

## 1 Introduction

We consider the following mixed nonlinear impulsive differential equations with variable delay:
\begin{aligned} & \bigl(r(t)\varPhi_{\alpha} \bigl(x'(t)\bigr) \bigr)'+p_{0}(t) \varPhi_{\alpha}\bigl(x(t)\bigr) +\sum_{i=1}^{n}p_{i}(t) \varPhi_{\beta_{i}}\bigl(x\bigl(t-\sigma(t)\bigr)\bigr)=f(t),\quad t\geq t_{0}, t\neq\tau_{k}, \\ &x\bigl(\tau^{+}_{k}\bigr)=a_{k}x( \tau_{k}), \qquad x'\bigl(\tau^{+}_{k} \bigr)=b_{k}x'(\tau_{k}), \quad k=1,2,\ldots, \end{aligned}
(1)
where $$\varPhi_{*}(s)=|s|^{*-1}s$$, $$\{\tau_{k}\}$$ denotes the impulse moments, $$0\leq t_{0}<\tau_{1}<\tau_{2}<\cdots<\tau_{k}<\cdots$$ and $$\lim_{k\rightarrow\infty}\tau_{k}=\infty$$, $$\{a_{k}\}$$ and $$\{b_{k}\}$$ are real constant sequences and $$b_{k}\geq a_{k}>0$$ for $$k=1,2,\ldots$$ , $$\sigma(t)\in C([t_{0},\infty))$$ and there exists a nonnegative constant $$\sigma_{0}$$ such that $$0\leq\sigma(t)\leq\sigma_{0}$$ for all $$t\geq t_{0}$$, $$r(t)\in C^{1}([t_{0},\infty),(0,\infty))$$ is nondecreasing.
For some particular cases of (1), many authors have devoted work to the interval oscillation problem (see [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]). Particularly, when $$\alpha=1$$, $$a_{k}=b_{k}=1$$ and $$\sigma(t)=0$$, (1) reduces to the mixed type Emden–Fowler equation
$$\bigl(r(t)x'(t) \bigr)'+p_{0}(t)x(t)+ \sum_{i=1}^{n}p_{i}(t) \varPhi_{\beta _{i}}\bigl(x(t)\bigr)=f(t),$$
(2)
which was given much attention due to the effect of modeling the growth of bacteria population with competitive species. For example, in [14] and [15], the authors established interval oscillation theorems for (2) which improved the well-known criteria of [16] and [17]. For additional studies of Emden–Fowler differential equations, see [18, 19, 20].
In [1], the authors considered (2) with impulse effects,
\begin{aligned} &\bigl(r(t)x'(t) \bigr)'+p_{0}(t)x(t)+\sum_{i=1}^{n}p_{i}(t) \varPhi_{\beta _{i}}\bigl(x(t)\bigr)=f(t),\quad t\geq t_{0}, t\neq \tau_{k}, \\ &x\bigl(\tau^{+}_{k}\bigr)=a_{k}x( \tau_{k}),\qquad x'\bigl(\tau^{+}_{k} \bigr)=b_{k}x'(\tau_{k}),\quad k=1,2,\ldots, \end{aligned}
(3)
and established some interval oscillation results which extended those of [14, 15] and [21].
When $$\sigma(t)=0$$, (1) becomes the following impulse equations without delay:
\begin{aligned} & \bigl(r(t)\varPhi_{\alpha} \bigl(x'(t)\bigr) \bigr)'+p_{0}(t) \varPhi_{\alpha}\bigl(x(t)\bigr) +\sum_{i=1}^{n}p_{i}(t) \varPhi_{\beta_{i}}\bigl(x(t)\bigr)=f(t), \quad t\geq t_{0}, t\neq \tau_{k}, \\ &x\bigl(\tau^{+}_{k}\bigr)=a_{k}x( \tau_{k}),\qquad x'\bigl(\tau^{+}_{k} \bigr)=b_{k}x'(\tau_{k}),\quad k=1,2,\ldots. \end{aligned}
(4)
In [22], Özbekler and Zafer investigated (4). They considered the coefficients $$p_{i}(t)$$ ($$i=1,2,\ldots,n$$) satisfying two cases: (i) $$p_{i}(t)\geq0$$ for $$i=1,2,\ldots,n$$ and (ii) $$p_{i}(t)\geq0$$ for $$i=1,2,\ldots,m$$; $$p_{i}(t)$$ are allowed to be negative for $$i=m+1,\ldots,n$$ and obtained several interval oscillation results which recovered the early ones in [8] and [14].

When $$\sigma(t)$$ is a nonnegative constant, i.e., $$\sigma(t)=\sigma_{0}$$ ($$\sigma_{0}\geq0$$), by idea of [23], Guo et al. [2] studied (1) and developed the results of [1, 22, 24].

Recently, in [25], the authors studied (1) with the assumption of delay $$\sigma(t)$$ being variable. They used Riccati transformation and univariate ω functions to obtain some generalized interval oscillation results.

In this paper, we continue the discussion of the interval oscillation of (1). Unlike the methods of [22, 25], we introduce a binary auxiliary function, divide each given interval into two parts and then estimate the quotients of $$x(t-\sigma(t))$$ and $$x(t)$$. Due to the considered delay being variable, the results obtained here are the development of some well-known ones, such as in [1] and [2]. Moreover, we also give an example to illustrate the effectiveness and non-emptiness of our results.

## 2 Main results

First, we define a functional space $$C_{-}(I,\mathbb{R})$$ as follows:
\begin{aligned} C_{-}(I,\mathbb{R}) :=&\bigl\{ y: I\rightarrow\mathbb{R} \mid I \mbox{ is a real interval}, y \mbox{ is continuous on } I\setminus\{t_{i}\} \mbox{ and} \\ & y\bigl(t_{i}^{-}\bigr)=y(t_{i}), i \in\mathbb{N}\bigr\} . \end{aligned}
In the following, we always assume:
1. (A)

the exponents satisfy $$\beta_{1}>\cdots>\beta_{m}>\alpha>\beta_{m+1}>\cdots>\beta_{n}>0$$; $$f(t), p_{i}(t)\in C_{-}([t_{0},\infty),\mathbb{R})$$, $$i=0,1,\ldots,n$$; $$\tau_{k+1}-\tau_{k}>\sigma_{0}$$ for all $$k=1,2,\ldots$$ .

Let $$k(s)=\max\{i:t_{0}<\tau_{i}<s\}$$. For any given intervals $$[c_{j},d_{j}]$$ ($$j=1,2$$), we suppose that $$k(c_{j})< k(d_{j})$$ ($$j=1,2$$), then there exist impulse moments $$\tau_{k(c_{j})+1},\ldots ,\tau_{k(d_{j})}$$ in $$[c_{j},d_{j}]$$ ($$j=1,2$$) and we have the following cases to consider.
($$S_{1}$$)

$$\tau_{k(c_{j})}+\sigma_{0}< c_{j}$$ and $$\tau _{k(d_{j})}+\sigma_{0}< d_{j}$$;

($$S_{2}$$)

$$\tau_{k(c_{j})}+\sigma_{0}< c_{j}$$ and $$\tau _{k(d_{j})}+\sigma_{0}>d_{j}$$;

($$S_{3}$$)

$$\tau_{k(c_{j})}+\sigma_{0}>c_{j}$$ and $$\tau _{k(d_{j})}+\sigma_{0}>d_{j}$$;

($$S_{4}$$)

$$\tau_{k(c_{j})}+\sigma_{0}>c_{j}$$ and $$\tau _{k(d_{j})}+\sigma_{0}< d_{j}$$.

We further assume that there exist points $$\delta_{j}\in(c_{j},d_{j}) \setminus\{\tau_{k}\}$$ ($$j=1,2$$) which divide intervals $$[c_{j},d_{j}]$$ into two parts $$[c_{j},\delta _{j}]$$ and $$[\delta_{j},d_{j}]$$. In view of whether or not there are impulsive moments of $$x(t)$$ in $$[c_{j},\delta_{j}]$$ and $$[\delta_{j},d_{j}]$$, we should consider the following cases.
($$\bar{S}_{1}$$)

$$k(c_{j})< k(\delta_{j})< k(d_{j})$$;

($$\bar{S}_{2}$$)

$$k(c_{j})=k(\delta_{j})< k(d_{j})$$;

($$\bar{S}_{3}$$)

$$k(c_{j})< k(\delta_{j})=k(d_{j})$$.

We define a interval delay function ([12]):
$$D_{k}(t)=t-\tau_{k}-\sigma(t), \quad t\in( \tau_{k},\tau_{k+1}], k=1,2,\ldots,$$
and we assume there is a point $$t_{k}\in(\tau_{k},\tau_{k+1}]$$ such that $$D_{k}(t_{k})=0$$, $$D_{k}(t)<0$$ for $$t\in(\tau_{k},t_{k})$$ and $$D_{k}(t)>0$$ for $$t\in(t_{k},\tau_{k+1}]$$.
Moreover, for the relationship of the division point $$\delta_{j}$$ and the zero point $$t_{k(\delta_{j})}$$ of $$D_{k(\delta_{j})}$$ on $$[\tau_{k(\delta _{j})},\tau_{k(\delta_{j})+1}]$$ we should have
($$\bar{\bar{S}}_{1}$$)

$$t_{k(\delta_{j})}<\delta_{j}$$;

($$\bar{\bar{S}}_{2}$$)

$$t_{k(\delta_{j})}>\delta_{j}$$; or

($$\bar{\bar{S}}_{3}$$)

$$t_{k(\delta_{j})}=\delta_{j}$$.

We only consider the case of combination of ($$S_{1}$$) with ($$\bar{S}_{1}$$) and ($$\bar{\bar{S}}_{1}$$). For the other cases, the discussion will be omitted here.

### Lemma 2.1

Assume that, for any$$T\geq t_{0}$$, there exist$$T< c_{1}-\sigma _{0}< c_{1}<\delta_{1}<d_{1}$$and
$$f(t)\leq0, \qquad p_{i}(t)\geq0,\quad t \in[c_{1}-\sigma_{0},d_{1}]\setminus \{\tau _{k}\}, i=0,1,2,\ldots,n,$$
(5)
and$$t_{k}$$is a zero point of$$D_{k}(t_{k})$$in$$(\tau_{k},\tau_{k+1}]$$. If$$x(t)$$is a positive solution of (1), then the ratio$${x(t-\sigma(t))}/{x(t)}$$will be estimated as follows:
1. (a)

$$\frac{x(t-\sigma(t))}{x(t)}>\frac{t-\tau_{k}-\sigma (t)}{t-\tau_{k}}$$, $$t\in(t_{k},\tau_{k+1}]$$for$$k=k(c_{1})+1,\ldots,k(d_{1})-1$$, $$k\neq k(\delta_{1})$$;

2. (b)

$$\frac{x(t-\sigma(t))}{x(t)}>\frac{t-\tau _{k}}{b_{k}(t+\sigma(t)-\tau_{k})}$$, $$t\in(\tau_{k},t_{k}]$$for$$k=k(c_{1})+1,\ldots,k(d_{1})$$;

3. (c)

$$\frac{x(t-\sigma(t))}{x(t)}>\frac{t-\tau_{k(\delta _{1})}-\sigma(t)}{t-\tau_{k(\delta_{1})}}$$, $$t\in(t_{k(\delta_{1})},\delta_{1}]$$;

4. (d)

$$\frac{x(t-\sigma(t))}{x(t)}>\frac{t-\tau_{k(d_{1})}-\sigma (t)}{t-\tau_{k(d_{1})}}$$, $$t\in(t_{k(d_{1})},d_{1}]$$;

5. (e)

$$\frac{x(t-\sigma(t))}{x(t)}>\frac{t-\tau_{k(\delta _{1})}-\sigma(t)}{t-\tau_{k(\delta_{1})}}$$, $$t\in(\delta_{1},\tau_{k(\delta_{1})+1}]$$;

6. (f)

$$\frac{x(t-\sigma(t))}{x(t)}>\frac{t-\tau_{k(c_{1})}-\sigma (t)}{t-\tau_{k(c_{1})}}$$, $$t\in[c_{1},\tau_{k(c_{1})+1}]$$.

### Proof

From (1), (5) and (A), we obtain, for $$t\in[c_{1},d_{1}]\setminus \{\tau_{k}\}$$,
$$\bigl(r(t)\varPhi_{\alpha}\bigl(x'(t)\bigr) \bigr)'=f(t)-p_{0}(t)\varPhi_{\alpha }\bigl(x(t)\bigr)- \sum_{i=1}^{n}p_{i}(t) \varPhi_{\beta_{i}}\bigl(x\bigl(t-\sigma(t)\bigr)\bigr)\leq0.$$
Hence $$r(t)\varPhi_{\alpha}(x'(t))$$ is nonincreasing on the interval $$[c_{1},d_{1}]\setminus\{\tau_{k}\}$$. Next, we give the proof of case (a) only. For the other cases, the proof is similar and will be omitted.
If $$t_{k}< t\leq\tau_{k+1}$$, from $$D_{k}(t)>0$$, we know $$(t-\sigma (t),t)\subset(\tau_{k},\tau_{k+1}]$$. Thus there is no impulse moment in $$(t-\sigma(t),t)$$. Therefore, for any $$s\in(t-\sigma (t),t)$$, there exists a $$\xi_{k}\in(\tau_{k},s)$$ such that $$x(s)-x(\tau_{k}^{+})=x'(\xi_{k})(s-\tau_{k})$$. Since $$x(\tau _{k}^{+})>0$$, $$r(s)$$ is nondecreasing, $$\varPhi_{\alpha}(\cdot)$$ is an increasing function and $$r(t)\varPhi_{\alpha }(x'(t))$$ is nonincreasing on $$(\tau_{k},\tau_{k+1}]$$, we have
\begin{aligned} \varPhi_{\alpha}\bigl(x(s)\bigr) \geq&\frac{r(\xi_{k})}{r(s)}\varPhi_{\alpha} \bigl(x(s)\bigr)>\frac {r(\xi_{k})}{r(s)} \varPhi_{\alpha}\bigl(x'( \xi_{k}) (s-\tau_{k})\bigr)=\frac{r(\xi_{k})\varPhi_{\alpha }(x'(\xi_{k}))}{r(s)}(s- \tau_{k})^{\alpha} \\ \geq&\frac{r(s)\varPhi_{\alpha}(x'(s))}{r(s)}(s-\tau_{k})^{\alpha}=\varPhi _{\alpha}\bigl(x'(s) (s-\tau_{k})\bigr). \end{aligned}
Therefore,
$$\frac{x'(s)}{x(s)}< \frac{1}{s-\tau_{k}}.$$
Integrating both sides of the above inequality from $$t-\sigma(t)$$ to t, we obtain
$$\frac{x(t-\sigma(t))}{x(t)}>\frac{t-\tau_{k}-\sigma(t)}{t-\tau_{k}},\quad t\in(t_{k}, \tau_{k+1}].$$
The proof is completed. □

### Lemma 2.2

Assume that, for any$$T\geq t_{0}$$, there exist$$T< c_{2}-\sigma _{0}< c_{2}<\delta_{2}<d_{2}$$and
$$f(t)\geq0,\qquad p_{i}(t)\geq0, \quad t \in[c_{2}-\sigma_{0},d_{2}]\setminus \{\tau _{k}\}, \quad i=0,1,2,\ldots,n,$$
(6)
and$$t_{k}$$is a zero point of$$D_{k}(t_{k})$$in$$(\tau_{k},\tau_{k+1}]$$. If$$x(t)$$is a negative solution of (1), then estimations (a)(f) in Lemma 2.1are correct with the replacement of$$c_{1}$$, $$d_{1}$$and$$\delta_{1}$$by$$c_{2}$$, $$d_{2}$$and$$\delta _{2}$$, respectively.

The proof of Lemma 2.2 is similar to that of Lemma 2.1 and will be omitted.

### Lemma 2.3

Assume that for any$$T\geq t_{0}$$there exists$$T< c_{1}-\sigma _{0}< c_{1}< d_{1}$$and (5) holds. If$$x(t)$$is a positive solution of (1) and$$u(t)$$is defined by
$$u(t):=\frac{r(t)\varPhi_{\alpha}(x'(t))}{\varPhi_{\alpha}(x(t))},\quad t\in[c_{1},d_{1}],$$
(7)
then we have the following estimations of$$u(t)$$:
1. (g)

$$u(\tau_{k+1})\leq\frac{\widetilde{r}}{(\tau_{k+1}-\tau _{k})^{\alpha}}$$, $$\tau_{k+1}\in[c_{1},d_{1}]$$, $$k=k(c_{1})+1,\ldots,k(d_{1})-1$$, $$k\neq k(\delta_{1})$$;

2. (h)

$$u(\tau_{k(c_{1})+1})\leq\frac{\widetilde{r}}{(\tau _{k(c_{1})+1}-c_{1})^{\alpha}}$$, $$\tau_{k(c_{1})+1}\in[c_{1},d_{1}]$$;

3. (i)

$$u(\tau_{k(\delta_{1})+1})\leq\frac{\widetilde{r}}{(\tau _{k(\delta_{1})+1}-\delta_{1})^{\alpha}}$$, $$\tau_{k(\delta_{1})+1}\in[c_{1},d_{1}]$$,

where$$\widetilde{r}=\max_{t\in[c_{1},d_{1}]\cup[c_{2},d_{2}]}\{r(t)\}$$.

### Proof

For $$t\in(\tau_{k},\tau_{k+1}]\subset[c_{1},d_{1}]$$, $$k=k(c_{1})+1,\ldots,k(d_{1})-1$$, there exists $$\varsigma_{k}\in(\tau _{k},t)$$ such that
$$x(t)-x\bigl(\tau^{+}_{k}\bigr)=x'(\varsigma_{k}) (t-\tau_{k}).$$
In view of $$x(\tau^{+}_{k})>0$$ and the monotone properties of $$\varPhi_{\alpha}(\cdot)$$, $$r(t)\varPhi_{\alpha}(x'(t))$$ and $$r(t)$$, we obtain
$$\varPhi_{\alpha}\bigl(x(t)\bigr)>\varPhi_{\alpha}\bigl(x'( \varsigma_{k})\bigr)\varPhi_{\alpha}(t-\tau_{k}) \geq \frac{r(t)}{r(\varsigma_{k})}\varPhi_{\alpha}\bigl(x'(t)\bigr) (t-\tau _{k-1})^{\alpha}.$$
That is,
$$\frac{r(t)\varPhi_{\alpha}(x'(t))}{\varPhi_{\alpha}(x(t))}< \frac{r(\varsigma _{k})}{(t-\tau_{k})^{\alpha}}.$$
Letting $$t\rightarrow\tau^{-}_{k+1}$$, we obtain conclusion (g). Using a similar analysis on $$(c_{1},\tau_{k(c_{1})+1}]$$ and $$(\delta _{1},\tau_{k(\delta_{1})+1}]$$, we can get (h) and (i). The proof is completed. □

### Lemma 2.4

Assume that, for any$$T\geq t_{0}$$, there exist$$c_{2},d_{2},\delta _{2}\notin\{\tau_{k}\}$$such that$$T< c_{2}-\sigma_{0}< c_{2}<\delta _{2}<d_{2}$$and (6) hold. If$$x(t)$$is a negative solution of (1) and$$u(t)$$is defined by
$$u(t):=\frac{r(t)\varPhi_{\alpha}(x'(t))}{\varPhi_{\alpha}(x(t))},\quad t\in[c_{2},d_{2}],$$
(8)
then the estimations (g)(i) in Lemma 2.3are correct with the replacement of$$c_{1}$$, $$d_{1}$$and$$\delta_{1}$$by$$c_{2}$$, $$d_{2}$$and$$\delta_{2}$$, respectively.

The proof of Lemma 2.4 is similar to that of Lemma 2.3 and will be omitted.

### Lemma 2.5

(cf. Lemma 1.1 in [22])

Let$$\{\beta_{1},\ldots,\beta _{n}\}$$be the n-tuple satisfying (A). Then there exists an n-tuple$$(\eta_{1},\eta_{2},\ldots,\eta_{n})$$satisfying
\begin{aligned} (\mathrm{i})&\quad \sum _{i=1}^{n}\beta_{i}\eta_{i}= \alpha,\quad \textit{and} \\ (\mathrm{ii})&\quad \sum_{i=1}^{n} \eta_{i}=\lambda< 1, \quad 0< \eta_{i}< 1. \end{aligned}
(9)

In the following we will establish Kamenev type interval oscillation criteria for (1) by the idea of Philos [26]. For the research of Kamenev/Philos-type oscillation criteria for differential equations, see [27, 28, 29, 30, 31].

Let $$E=\{(t,s):t_{0}\leq s\leq t\}$$, $$H_{1},H_{2}\in C^{1}(E,\mathbb{R})$$. Then a pair of functions $$H_{1}$$, $$H_{2}$$ is said to belong to a function set $$\mathscr{H}$$, denoted by $$(H_{1},H_{2})\in\mathscr{H}$$, if there exist $$h_{1},h_{2}\in L_{\mathrm{loc}}(E,\mathbb{R})$$ satisfying the following conditions:
($$C_{1}$$)

$$H_{1}(t,t)=H_{2}(t,t)=0$$, $$H_{1}(t,s)>0$$, $$H_{2}(t,s)>0$$ for $$t>s$$;

($$C_{2}$$)

$$\frac{\partial}{\partial t}H_{1}(t,s)=h_{1}(t,s)H_{1}(t,s)$$, $$\frac{\partial}{\partial s}H_{2}(t,s)=h_{2}(t,s)H_{2}(t,s)$$.

For convenience in the expression below, we also use the following notation:
$$\int_{[c,d]}:= \int_{c}^{\tau_{k(c)+1}}+ \sum_{k=k(c)+1}^{k(d)-1} \biggl( \int_{\tau_{k}}^{t_{k}} + \int_{t_{k}}^{\tau_{k+1}} \biggr) + \int_{\tau_{k(d)}}^{t_{k(d)}}+ \int_{t_{k(d)}}^{d}.$$

### Lemma 2.6

Assume that the conditions of Lemma 2.1hold. Let$$x(t)$$be a positive solution of (1) and$$u(t)$$be defined by (7). Then, for any$$(H_{1},H_{2})\in\mathscr{H}$$, we have
\begin{aligned}& \int_{[c_{1},\delta_{1}]}\psi(t)H_{1}(t,c_{1}) \frac{x^{\alpha}(t-\sigma (t))}{x^{\alpha}(t)} \,\mathrm{d}t \\& \qquad {} + \int_{c_{1}}^{\delta_{1}}H_{1}(t,c_{1}) \biggl[p_{0}(t)-\frac {r(t)}{(1+\alpha)^{1+\alpha}} \bigl\vert h_{1}(t,c_{1}) \bigr\vert ^{1+\alpha} \biggr]\,\mathrm{d}t \\& \quad \leq \sum_{i=k(c_{1})+1}^{k(\delta_{1})} \frac{b_{i}^{\alpha }-a_{i}^{\alpha}}{a_{i}^{\alpha}} H_{1}(\tau_{i},c_{1})u( \tau_{i})-H_{1}(\delta_{1},c_{1})u( \delta_{1}) \end{aligned}
(10)
and
\begin{aligned}& \int_{[\delta_{1},d_{1}]}\psi(t)H_{2}(d_{1},t) \frac{x^{\alpha }(t-\sigma(t))}{x^{\alpha}(t)}\,\mathrm{d}t \\& \qquad {} + \int_{\delta_{1}}^{d_{1}}H_{2}(d_{1},t) \biggl[p_{0}(t) -\frac{r(t)}{(1+\alpha)^{1+\alpha}} \bigl\vert h_{2}(d_{1},t) \bigr\vert ^{1+\alpha} \biggr]\,\mathrm{d}t \\& \quad \leq \sum_{i=k(\delta_{1})+1}^{k(d_{1})} \frac{b_{i}^{\alpha }-a_{i}^{\alpha}}{a_{i}^{\alpha}} H_{2}(d_{1},\tau_{i})u( \tau_{i})+H_{2}(d_{1},\delta_{1})u( \delta_{1}), \end{aligned}
(11)
where$$\psi(t)=\eta_{0}^{-\eta_{0}}|f(t)|^{\eta_{0}}\prod_{i=1}^{n}\eta _{i}^{-\eta_{i}}(p_{i}(t))^{\eta_{i}}$$with$$\eta_{0}=1-\sum_{i=1}^{n}\eta_{i}$$and$$\eta_{1},\eta_{2},\ldots,\eta_{n}$$are positive constants satisfying conditions of Lemma 2.5.

### Proof

Differentiating $$u(t)$$ and in view of (1), we obtain, for $$t\neq\tau_{k}$$,
\begin{aligned} u'(t) =&- \Biggl[\sum_{i=1}^{n}p_{i}(t) \varPhi_{\beta_{i}-\alpha}\bigl(x\bigl(t-\sigma(t)\bigr)\bigr) +\frac{|f(t)|}{\varPhi_{\alpha}(x(t-\sigma(t)))} \Biggr]\frac{\varPhi_{\alpha }(x(t-\sigma(t)))}{\varPhi_{\alpha}(x(t))} \\ &{}-\frac{\alpha}{r^{1/\alpha}(t)} \bigl\vert u(t) \bigr\vert ^{1+{1}/{\alpha}}-p_{0}(t). \end{aligned}
(12)
Let
\begin{aligned} v_{0}=\eta_{0}^{-1}\frac{|f(t)|}{\varPhi_{\alpha}(x(t-\sigma(t)))},\qquad v_{i}=\eta_{i}^{-1}p_{i}(t) \varPhi_{\beta_{i}-\alpha}\bigl(x\bigl(t-\sigma(t)\bigr)\bigr),\quad i=1,2,\ldots,n, \end{aligned}
where $$\eta_{1},\eta_{2},\ldots,\eta_{n}$$ are chosen to satisfy conditions of Lemma 2.5 with $$\eta_{0}=1-\sum_{i=1}^{n}\eta_{i}$$ for given $$\beta_{1},\ldots,\beta _{n}$$ and α. Employing the arithmetic–geometric mean inequality (see [32])
$$\sum_{i=0}^{n}\eta_{i}v_{i} \geq\prod_{i=0}^{n}v_{i}^{\eta_{i}},$$
we have, from (12),
$$u'(t)\leq-\psi(t)\frac{x^{\alpha}(t-\sigma(t))}{x^{\alpha}(t)} - \frac{\alpha}{r^{1/\alpha}(t)} \bigl\vert u(t) \bigr\vert ^{1+{1}/{\alpha}}-p_{0}(t),$$
(13)
where
$$\psi(t)=\eta_{0}^{-\eta_{0}} \bigl\vert f(t) \bigr\vert ^{\eta_{0}}\prod_{i=1}^{n}\eta _{i}^{-\eta_{i}}\bigl(p_{i}(t)\bigr)^{\eta_{i}}.$$
Multiplying both sides of (13) by $$H_{1}(t,c_{1})$$ and integrating it from $$c_{1}$$ to $$\delta_{1}$$, we have
\begin{aligned} \int_{[c_{1},\delta_{1}]}H_{1}(t,c_{1})u'(t) \,\mathrm{d}t \leq&- \int _{[c_{1},\delta_{1}]}\psi(t)H_{1}(t,c_{1}) \frac{x^{\alpha}(t-\sigma(t))}{x^{\alpha}(t)}\,\mathrm{d}t \\ &{}-\alpha \int_{[c_{1},\delta_{1}]}H_{1}(t,c_{1}) \frac {|u(t)|^{1+{1}/{\alpha}}}{r^{1/\alpha}(t)}\,\mathrm{d}t \\ &{}- \int_{c_{1}}^{\delta_{1}}H_{1}(t,c_{1})p_{0}(t) \,\mathrm{d}t. \end{aligned}
(14)
Noticing that the impulse moments $$\tau_{k(c_{1})+1},\tau_{k(c_{1})+2}, \ldots,\tau_{k(\delta_{1})}$$ are in $$[c_{1},\delta_{1}]$$ and using the integration by parts formula on the left-hand side of the above inequality, we obtain
\begin{aligned} \int_{[c_{1},\delta_{1}]}H_{1}(t,c_{1})u'(t) \,\mathrm{d}t =&\sum_{i=k(c_{1})+1}^{k(\delta_{1})} \biggl(1- \frac{b_{i}^{\alpha }}{a_{i}^{\alpha}} \biggr) H_{1}(\tau_{i},c_{1})u( \tau_{i}) +H_{1}(\delta_{1},c_{1})u( \delta_{1}) \\ &{}- \int_{[c_{1},\delta_{1}]}H_{1}(t,c_{1})h_{1}(t,c_{1})u(t) \,\mathrm{d}t. \end{aligned}
(15)
Substituting (15) into (14), we obtain
\begin{aligned}& \int_{[c_{1},\delta_{1}]}\psi(t)H_{1}(t,c_{1}) \frac{x^{\alpha}(t-\sigma (t))}{x^{\alpha}(t)}\,\mathrm{d}t \\& \quad \leq \sum_{i=k(c_{1})+1}^{k(\delta_{1})} \biggl( \frac{b_{i}^{\alpha }}{a_{i}^{\alpha}}-1 \biggr) H_{1}(\tau_{i},c_{1})u( \tau_{i})-H_{1}(\delta_{1},c_{1})u( \delta _{1}) \\& \qquad {}- \int_{c_{1}}^{\delta_{1}}p_{0}(t)H_{1}(t,c_{1}) \,\mathrm{d}t + \int_{[c_{1},\delta_{1}]}H_{1}(t,c_{1})V\bigl(u(t)\bigr) \,\mathrm{d}t, \end{aligned}
where $$V(u(t))= [|h_{1}(t,c_{1})||u(t)| -\frac{\alpha}{r^{1/\alpha}(t)}|u(t)|^{1+1/\alpha} ]$$. We easily see that
$$V\bigl(u(t)\bigr)\leq\sup_{u\in\mathbb{R}}V\bigl(u(t)\bigr)= \frac{r(t)}{(1+\alpha )^{1+\alpha}} \bigl\vert h_{1}(t,c_{1}) \bigr\vert ^{1+\alpha}.$$
Thus, we obtain (10).

Multiplying both sides of (13) by $$H_{2}(d_{1},t)$$ and using a similar analysis to the above, we can obtain (11). The proof is completed. □

### Lemma 2.7

Assume that the conditions of Lemma 2.2hold. Let$$x(t)$$be a negative solution of (1) and$$u(t)$$be defined by (8). Then for any$$(H_{1},H_{2})\in\mathscr{H}$$we have
\begin{aligned}& \int_{[c_{2},\delta_{2}]}\psi(t)H_{1}(t,c_{2}) \frac{x^{\alpha}(t-\sigma (t))}{x^{\alpha}(t)}\,\mathrm{d}t \\& \qquad {} + \int_{c_{2}}^{\delta_{2}}H_{1}(t,c_{2}) \biggl[p_{0}(t) -\frac{r(t)}{(1+\alpha)^{1+\alpha}} \bigl\vert h_{1}(t,c_{2}) \bigr\vert ^{1+\alpha} \biggr]\,\mathrm{d}t \\& \quad \leq \sum_{i=k(c_{2})+1}^{k(\delta_{2})} \frac{b_{i}^{\alpha}-a_{i}^{\alpha }}{a_{i}^{\alpha}} H_{1}(\tau_{i},c_{2})u( \tau_{i})-H_{1}(\delta_{2},c_{2})u( \delta_{2}) \end{aligned}
(16)
and
\begin{aligned}& \int_{[\delta_{2},d_{2}]}\psi(t)H_{2}(d_{2},t) \frac{x^{\alpha }(t-\sigma(t))}{x^{\alpha}(t)}\,\mathrm{d}t \\& \qquad {} + \int_{\delta_{2}}^{d_{2}}H_{2}(d_{2},t) \biggl[p_{0}(t) -\frac{r(t)}{(1+\alpha)^{1+\alpha}} \bigl\vert h_{2}(d_{2},t) \bigr\vert ^{1+\alpha} \biggr]\,\mathrm{d}t \\& \quad \leq \sum_{i=k(\delta_{2})+1}^{k(d_{2})} \frac{b_{i}^{\alpha}-a_{i}^{\alpha }}{a_{i}^{\alpha}} H_{2}(d_{1},\tau_{i})u( \tau_{i})-H_{2}(d_{2},\delta_{2})u( \delta_{2}), \end{aligned}
(17)
where$$\psi(t)$$is defined as in Lemma 2.6.

The proof of Lemma 2.7 is similar to that of Lemma 2.6 and will be omitted.

For two constants $$\nu_{1},\nu_{2}\notin\{\tau_{k}\}$$ with $$\nu_{1}< \nu_{2}$$ and $$k(\nu_{1})< k(\nu_{2})$$, using function $$\varphi\in C([\nu_{1},\nu_{2}],\mathbb {R})$$ and function $$\phi\in C_{-}([\nu_{1},\nu_{2}],\mathbb{R})$$, we define a functional $$Q:C([\nu_{1},\nu_{2}],\mathbb{R})\rightarrow\mathbb{R}$$ by
$$Q_{\nu_{1}}^{\nu_{2}}[\varphi]:= \frac{\widetilde{r}(b^{\alpha}_{k(\nu_{1})+1}-a^{\alpha}_{k(\nu_{1})+1})\varphi (\tau_{k(\nu_{1})+1})}{ a^{\alpha}_{k(\nu_{1})+1}(\tau_{k(\nu_{1})+1}-\nu_{1})^{\alpha}} + \sum_{k=k(\nu_{1})+2}^{k(\nu_{2})}\frac{\widetilde{r}(b^{\alpha}_{k}-a^{\alpha}_{k})\varphi(\tau_{k})}{ a^{\alpha}_{k}(\tau_{k}-\tau_{k-1})^{\alpha}},$$
(18)
where $$\sum_{m}^{n}=0$$ if $$m>n$$, and a functional $$L:C_{-}([\nu_{1},\nu_{2}],\mathbb{R})\rightarrow\mathbb{R}$$ by
\begin{aligned} L_{\nu_{1}}^{\nu_{2}}[\phi] :=& \int_{\nu_{1}}^{\tau_{k(\nu_{1})+1}}\phi(t)\frac{(t-\tau_{k(\nu_{1})} -\sigma(t))^{\alpha}}{(t-\tau_{k(\nu_{1})})^{\alpha}}\,\mathrm{d}t \\ &{}+\sum_{k=k(\nu_{1})+1}^{k(\nu_{2})-1} \biggl[ \int_{\tau_{k}}^{t_{k}} \phi(t)\frac{(t-\tau_{k})^{\alpha}}{b^{\alpha}_{k}(t-\tau_{k}+\sigma (t))^{\alpha}}\, \mathrm{d}t + \int_{t_{k}}^{\tau_{k+1}}\phi(t)\frac{(t-\tau_{k}-\sigma(t))^{\alpha}}{(t-\tau_{k})^{\alpha}}\,\mathrm{d}t \biggr] \\ &{}+ \int_{\tau_{k(\nu_{2})}}^{t_{k(\nu_{2})}}\phi(t)\frac{(t-\tau_{k(\nu _{2})})^{\alpha}}{b^{\alpha}_{k(\nu_{2})}(t-\tau_{k(\nu_{2})} +\sigma(t))^{\alpha}}\,\mathrm{d}t \\ &{}+ \int_{t_{k(\nu_{2})}}^{\nu_{2}}\phi(t)\frac{(t-\tau_{k(\nu_{2})}-\sigma (t))^{\alpha}}{(t-\tau_{k(\nu_{2})})^{\alpha}}\, \mathrm{d}t, \end{aligned}
(19)
where $$t_{k}$$ are zero points of $$D_{k}(t)$$ on $$[\tau_{k},\tau_{k+1}]$$ for $$k=k(\nu_{1})+1,\ldots,k(\nu_{2})$$.
For convenience in the expression below, we define, for $$j=1,2$$,
$$\varPi_{c_{j}}^{\delta_{j}}\bigl[H_{1}(t,c_{j}) \bigr]:=L_{c_{j}}^{\delta_{j}}\bigl[\psi (t)H_{1}(t,c_{j}) \bigr]+ \int_{c_{j}}^{\delta_{j}}H_{1}(t,c_{j}) \biggl[p_{0}(t) -\frac{r(t)}{(1+\alpha)^{1+\alpha}} \bigl\vert h_{1}(t,c_{j}) \bigr\vert ^{1+\alpha} \biggr]\,\mathrm{d}t$$
and
$$\varPi_{\delta_{j}}^{d_{j}}\bigl[H_{2}(d_{j},t) \bigr]:=L_{\delta_{j}}^{d_{j}}\bigl[\psi (t)H_{2}(d_{j},t) \bigr]+ \int_{\delta_{j}}^{d_{j}}H_{2}(d_{j},t) \biggl[p_{0}(t) -\frac{r(t)}{(1+\alpha)^{1+\alpha}} \bigl\vert h_{2}(d_{j},t) \bigr\vert ^{1+\alpha} \biggr]\,\mathrm{d}t,$$
where $$\psi(t)=\eta_{0}^{-\eta_{0}}|f(t)|^{\eta_{0}}\prod_{i=1}^{n}\eta _{i}^{-\eta_{i}}(p_{i}(t))^{\eta_{i}}$$.

### Theorem 2.1

Assume that, for any$$T\geq t_{0}$$, there exist$$T< c_{1}-\sigma_{0}< c_{1}< d_{1}\leq c_{2}-\sigma_{0}< c_{2}< d_{2}$$and (5) and (6) hold. If there exists a pair of$$(H_{1},H_{2})\in\mathscr{H}$$such that
$$\frac{\varPi_{c_{j}}^{\delta_{j}}[H_{1}(t,c_{j})]}{H_{1}(\delta_{j},c_{j})} +\frac{\varPi_{\delta_{j}}^{d_{j}}[H_{2}(d_{j},t)]}{H_{2}(d_{j},\delta _{j})}>\frac{Q_{c_{j}}^{\delta_{j}}[H_{1}(\cdot,c_{j})]}{H_{1}(\delta _{j},c_{j})} + \frac{Q_{\delta_{j}}^{d_{j}}[H_{2}(d_{j},\cdot)]}{H_{2}(d_{j},\delta_{j})},\quad j=1,2,$$
(20)
then (1) is oscillatory.

### Proof

Assume, to the contrary, that $$x(t)$$ is a nonoscillatory solution of (1). If $$x(t)$$ is a positive solution, we choose the interval $$[c_{1}, d_{1}]$$ to consider.

From Lemma 2.6, we obtain (10) and (11). Applying the estimation (a)–(f) into the left side, meanwhile (g)–(i) into the right side, of (10) and (11), we get
$$\varPi_{c_{1}}^{\delta_{1}}\bigl[H_{1}(t,c_{1}) \bigr]\leq Q_{c_{1}}^{\delta _{1}}\bigl[H_{1}( \cdot,c_{1})\bigr]-H_{1}(\delta_{1},c_{1})u( \delta_{1})$$
(21)
and
$$\varPi_{\delta_{1}}^{d_{1}}\bigl[H_{2}(d_{1},t) \bigr]\leq Q_{\delta _{1}}^{d_{1}}\bigl[H_{2}(d_{1}, \cdot)\bigr]+H_{2}(d_{1},\delta_{1})u( \delta_{1}).$$
(22)
Dividing (21) and (22) by $$H_{1}(\delta_{1},c_{1})$$ and $$H_{2}(d_{1},\delta_{1})$$, respectively, and adding them, we get
$$\frac{\varPi_{c_{1}}^{\delta_{1}}[H_{1}(t,c_{1})]}{H_{1}(\delta_{1},c_{1})} +\frac{\varPi_{\delta_{1}}^{d_{1}}[H_{2}(d_{1},t)]}{H_{2}(d_{1},\delta _{1})}\leq\frac{Q_{c_{1}}^{\delta_{1}}[H_{1}(\cdot,c_{1})]}{H_{1}(\delta _{1},c_{1})} + \frac{Q_{\delta_{1}}^{d_{1}}[H_{2}(d_{1},\cdot)]}{H_{2}(d_{1},\delta_{1})},$$
which contradicts (20) for $$j=1$$.

If $$x(t)$$ is a negative solution of (1), we choose interval $$[c_{2},d_{2}]$$ and can get a contradiction to (20) for $$j=2$$. The details will be omitted.

The proof is complete. □

### Remark 2.1

When $$\sigma(t)=0$$, i.e., the delay disappears and $$\alpha=1$$ in (1), Theorem 2.1 reduces to Theorem 2.2 in [1].

### Remark 2.2

When $$\sigma(t)=\sigma_{0}$$, i.e., the delay is constant, Theorem 2.1 reduces to Theorem 2.8 in [2].

In Eq. (19), zero points $$t_{k}$$ of $$D_{k}(t)$$ appear at upper limit (or lower limit) of integrals. However, these zero points are generally not easy to solve from $$D_{k}(t)=0$$, which will lead to difficult in the calculation of (19). To overcome this difficulty we need to re-estimate $$x(t-\sigma(t))/x(t)$$ on $$(t_{k},\tau _{k+1}]$$, $$(\tau_{k},t_{k})$$, $$(t_{k(d_{j})},d_{j})$$ and $$(\tau _{k(d_{j})},t_{k(d_{j})})$$ in Lemma 2.1 and Lemma 2.2.

If $$x(t)$$ is a positive solution of (1), from (a) in Lemma 2.1, we have, for $$k=k(c_{1})+1,\ldots,k(d_{1})-1$$, $$k\neq k(\delta_{1})$$,
$$\frac{x(t-\sigma(t))}{x(t)}>\frac{t-\tau_{k}-\sigma(t)}{t-\tau _{k}}>\frac{t-\tau_{k}-\sigma(t)}{t},\quad t \in(t_{k},\tau_{k+1}].$$
(23)
From (b) in Lemma 2.1, we have, for $$k=k(c_{1})+1,\ldots,k(d_{1})$$,
$$\frac{x(t-\sigma(t))}{x(t)}>\frac{t-\tau_{k}}{b_{k}(t+\sigma(t)-\tau _{k})}>0>\frac{t-\tau_{k}-\sigma(t)}{t}, \quad t \in(\tau_{k},t_{k}].$$
(24)
Combining (23) with (24), we obtain estimation of $$x^{\alpha}(t-\sigma(t))/x^{\alpha}(t)$$ on $$(\tau_{k},\tau_{k+1}]$$ for $$k=k(c_{1})+1,\ldots,k(d_{1})-1$$, $$k\neq k(\delta_{1})$$,
$$(\bar{\mathrm{a}})\quad \frac{x^{\alpha}(t-\sigma(t))}{x^{\alpha}(t)}>\varPhi_{\alpha } \biggl( \frac{t-\tau_{k}-\sigma(t)}{t} \biggr), \quad t\in(\tau_{k},\tau_{k+1}].$$
Similarly, from (b) and (c) in Lemma 2.1, we have
$$(\bar{\mathrm{b}})\quad \frac{x^{\alpha}(t-\sigma(t))}{x^{\alpha}(t)}>\varPhi_{\alpha } \biggl( \frac{t-\tau_{k(\delta_{1})}-\sigma(t)}{t} \biggr),\quad t\in(\tau _{k(\delta_{1})},t_{k(\delta_{1})}] \cup(t_{k(\delta_{1})},\delta_{1}],$$
from (b) and (d) in Lemma 2.1, we have
$$(\bar{\mathrm{c}})\quad \frac{x^{\alpha}(t-\sigma(t))}{x^{\alpha}(t)}>\varPhi_{\alpha } \biggl( \frac{t-\tau_{k(d_{1})}-\sigma(t)}{t} \biggr),\quad t\in(\tau _{k(d_{1})},t_{k(d_{1})}] \cup(t_{k(d_{1})},d_{1}],$$
and from (e) and (f) in Lemma 2.1, we have
$$(\bar{\mathrm{d}})\quad \frac{x^{\alpha}(t-\sigma(t))}{x^{\alpha}(t)} >\varPhi_{\alpha} \biggl( \frac{t-\tau_{k(\delta_{1})}-\sigma(t)}{t-\tau _{k(\delta_{1})}} \biggr),\quad t\in[\delta_{1},\tau_{k(\delta_{1})+1}],$$
and
$$(\bar{\mathrm{e}})\quad \frac{x^{\alpha}(t-\sigma(t))}{x^{\alpha}(t)} >\varPhi_{\alpha} \biggl( \frac{t-\tau_{k(c_{1})}-\sigma(t)}{t-\tau _{k(c_{1})}} \biggr),\quad t\in[c_{1},\tau_{k(c_{1})+1}].$$

If $$x(t)$$ is a negative solution of (1), from Lemma 2.2, we can get similar estimations to the above for $$t\in[c_{2},d_{2}]$$.

For convenience, we define functional $$\widetilde{L}: C_{-}([c_{j},d_{j}],\mathbb{R})\rightarrow\mathbb{R}$$, for $$j=1,2$$, by
\begin{aligned} \widetilde{L}_{c_{j}}^{{\delta_{j}}}[\phi] :=& \int_{c_{j}}^{\tau_{k(c_{j})+1}} \phi(t)\varPhi_{\alpha} \biggl( \frac{t-\tau_{k(c_{j})}-\sigma(t)}{t-\tau _{k(c_{j})}} \biggr)\,\mathrm{d}t \\ &{}+\sum_{k=k(c_{j})+1}^{k(\delta_{j})-1} \int_{\tau_{k}}^{\tau_{k+1}} \phi(t)\varPhi_{\alpha} \biggl( \frac{t-\tau_{k}-\sigma(t)}{t} \biggr)\,\mathrm {d}t \\ &{}+ \int_{\tau_{k(\delta_{j})}}^{\delta_{j}}\phi(t)\varPhi_{\alpha} \biggl( \frac{t-\tau_{k(\delta_{j})}-\sigma(t)}{t} \biggr)\,\mathrm{d}t \end{aligned}
and
\begin{aligned} \widetilde{L}_{\delta_{j}}^{d_{j}}[\phi] :=& \int_{\delta_{j}}^{\tau _{k(\delta_{j})+1}} \phi(t)\varPhi_{\alpha} \biggl( \frac{t-\tau_{k(\delta_{j})}-\sigma(t)}{t-\tau _{k(\delta_{j})}} \biggr)\,\mathrm{d}t \\ &{}+\sum_{k=k(\delta_{j})+1}^{k(d_{j})-1} \int_{\tau_{k}}^{\tau_{k+1}} \phi(t)\varPhi_{\alpha} \biggl( \frac{t-\tau_{k}-\sigma(t)}{t} \biggr)\,\mathrm {d}t \\ &{}+ \int_{\tau_{k(\delta_{j})}}^{d_{j}}\phi(t)\varPhi_{\alpha} \biggl( \frac {t-\tau_{k(\delta_{j})}-\sigma(t)}{t} \biggr)\,\mathrm{d}t. \end{aligned}
Further, we define, for $$j=1,2$$,
$$\widetilde{\varPi}_{c_{j}}^{\delta_{j}}\bigl[H_{1}(t,c_{j}) \bigr]:=\widetilde {L}_{c_{j}}^{\delta_{j}}\bigl[\psi(t)H_{1}(t,c_{j}) \bigr]+ \int_{c_{j}}^{\delta _{j}}H_{1}(t,c_{j}) \biggl[p_{0}(t) -\frac{r(t)}{(1+\alpha)^{1+\alpha}} \bigl\vert h_{1}(t,c_{j}) \bigr\vert ^{1+\alpha} \biggr]\,\mathrm{d}t$$
and
$$\widetilde{\varPi}_{c_{j}}^{\delta_{j}}\bigl[H_{2}(d_{j},t) \bigr]:=\widetilde {L}_{c_{j}}^{\delta_{j}}\bigl[\psi(t)H_{2}(d_{j},t) \bigr]+ \int_{c_{j}}^{\delta _{j}}H_{2}(d_{j},t) \biggl[p_{0}(t) -\frac{r(t)}{(1+\alpha)^{1+\alpha}} \bigl\vert h_{2}(d_{j},t) \bigr\vert ^{1+\alpha} \biggr]\,\mathrm{d}t,$$
where $$\psi(t)=\eta_{0}^{-\eta_{0}}|f(t)|^{\eta_{0}}\prod_{i=1}^{n}\eta _{i}^{-\eta_{i}}(p_{i}(t))^{\eta_{i}}$$.

Using similar proof method to that of Theorem 2.1 and applying estimations (ā)–(ē), we can obtain following theorem.

### Theorem 2.2

Assume that, for any$$T\geq t_{0}$$, there exist$$T< c_{1}-\sigma _{0}< c_{1}< d_{1}\leq c_{2}-\sigma_{0}< c_{2}< d_{2}$$and (5) and (6) hold. If there exists a pair of$$(H_{1},H_{2})\in\mathscr{H}$$such that
$$\frac{\widetilde{\varPi}_{c_{j}}^{\delta_{j}}[H_{1}(t,c_{j})]}{H_{1}(\delta_{j},c_{j})} +\frac{\widetilde{\varPi}_{\delta _{j}}^{d_{j}}[H_{2}(d_{j},t)]}{H_{2}(d_{j},\delta_{j})} >\frac{Q_{c_{j}}^{\delta_{j}}[H_{1}(\cdot,c_{j})]}{H_{1}(\delta_{j},c_{j})} + \frac{Q_{\delta_{j}}^{d_{j}}[H_{2}(d_{j},\cdot)]}{H_{2}(d_{j},\delta_{j})},\quad j=1,2,$$
(25)
then (1) is oscillatory.

## 3 Example

In this section, we give an example to illustrate the effectiveness and non-emptiness of our results.

### Example 3.1

Consider the following equation:
\begin{aligned} &x''(t)+ \mu_{1} p_{1}(t)\varPhi_{\frac{5}{2}} \bigl(x\bigl(t-\sigma(t) \bigr) \bigr) +\mu_{2} p_{2}(t)\varPhi_{\frac{1}{2}} \bigl(x \bigl(t-\sigma(t)\bigr) \bigr)=f(t),\quad t\neq \tau_{k}, \\ &x\bigl(\tau^{+}_{k}\bigr)=a_{k}x( \tau_{k}), \qquad x'\bigl(\tau^{+}_{k} \bigr)=b_{k}x'(\tau _{k}),\quad k=1,2,\ldots, \end{aligned}
(26)
where $$\varPhi_{*}(s)=|s|^{*-1}s$$, $$\sigma(t)=\frac{1}{3}\sin^{2}(\pi t)$$, $$\mu_{1}$$, $$\mu_{2}$$ are positive constants and $$\tau_{k}$$: $$\tau_{n,1}=8n+\frac{3}{2}$$, $$\tau_{n,2}=8n+\frac{5}{2}$$, $$\tau_{n,3}=8n+\frac{11}{2}$$, $$\tau_{n,4}=8n+\frac{13}{2}$$, $$n\in \mathbb{N}$$.
Let
$$p_{1}(t)=p_{2}(t)= \textstyle\begin{cases} (t-8n),& t\in[8n,8n+3] , \\ 3,& t\in[8n+3,8n+5] , \\ (8n+8-t),& t\in[8n+5,8n+8] , \end{cases}$$
and
$$f(t)= \textstyle\begin{cases} (t-8n)(t-8n-4)^{3},& t\in[8n,8n+4] , \\ (t-8n-4)^{3}(8n+8-t),& t\in[8n+4,8n+8] . \end{cases}$$

For any $$t_{0}>0$$, we choose n large enough such that $$t_{0}<8n$$ and let $$[c_{1},d_{1}]=[8n+1,8n+3]$$, $$[c_{2},d_{2}]=[8n+5,8n+7]$$, $$\delta_{1} =8n+2$$ and $$\delta_{2}=8n+6$$. We see that there has a zero point of $$D_{k}(t)$$ on each interval of $$[c_{1},\delta_{1}]$$, $$[\delta_{1}, d_{1}]$$, $$[c_{2},\delta_{2}]$$ and $$[\delta_{2}, d_{1}]$$. By approximate calculation, we get $$t_{1}\approx8n+1.709$$, $$t_{2}\approx8n+2.710$$, $$t_{3}\approx8n+5.709$$ and $$t_{4}\approx8n+6.710$$. Moreover, from conditions $$\alpha=1$$, $$\beta_{1}=5/2$$ and $$\beta_{2}=1/2$$, we can choose $$\eta_{1}=1/3$$, $$\eta_{1}=1/3$$ and $$\eta_{0}=1-\eta_{1}-\eta_{2}=1/3$$. So, the conditions of Lemma 2.5 are satisfied.

Letting $$H_{1}(t,s)=H_{2}(t,s)=(t-s)^{2}$$ and $$h_{1}(t,s)=-h_{2}(t,s)=\frac{2}{t-s}$$. By simple calculation, we have, for $$t\in[c_{1},\delta_{1}]$$,
\begin{aligned} & \int_{c_{1}}^{\delta_{1}}H_{1}(t,c_{1}) \biggl[p_{0}(t)-\frac {r(t)}{(1+\alpha)^{1+\alpha}} \bigl\vert h_{1}(t,c_{1}) \bigr\vert ^{1+\alpha} \biggr]\,\mathrm {d}t \\ &\quad = \int_{8n+1}^{8n+2}(t-8n-1)^{2} \biggl(0- \frac{2^{2}}{{2^{2}}(t-8n-1)^{2}} \biggr)\,\mathrm{d}t=-1. \end{aligned}
Let
\begin{aligned} \phi_{1}(t) :=&\psi(t)H_{1}(t,c_{1})= \eta_{0}^{-\eta_{0}} \bigl\vert f(t) \bigr\vert ^{\eta _{0}} \prod_{i=1}^{2}\eta_{i}^{-\eta_{i}} \bigl(p_{i}(t)\bigr)^{\eta _{i}}(t-c_{1})^{2} \\ =&3\sqrt[3]{\mu_{1}\mu_{2}}(t-8n) (t-8n-4) (t-8n-1)^{2}. \end{aligned}
Then
\begin{aligned} L_{c_{1}}^{\delta_{1}}\bigl[\phi_{1}(t)\bigr] =& \int_{8n+1}^{8n+\frac{3}{2}}\phi_{1}(t) \frac{t-8n+\frac{3}{2}-\frac{1}{3}\sin^{2}(\pi t)}{t-8n+\frac {3}{2}}\,\mathrm{d}t \\ &{}+ \int_{8n+\frac{3}{2}}^{t_{1}}\phi_{1}(t) \frac{t-8n-\frac{3}{2}}{b_{n,1}(t-8n-\frac{3}{2}+\frac{1}{3}\sin^{2}(\pi t))}\,\mathrm{d}t \\ &{} + \int_{t_{1}}^{8n+2}\phi_{1}(t) \frac{t-8n-\frac{3}{2}-\frac{1}{3}\sin ^{2}(\pi t)}{t-8n-\frac{3}{2}}\,\mathrm{d}t \\ =&3\sqrt[3]{\mu_{1}\mu_{2}}\biggl( \int_{1}^{\frac{3}{2}}\frac {t(4-t)(t-1)^{2}(t+\frac{3}{2}-\frac{1}{3}\sin^{2}(\pi t))}{t+\frac {3}{2}}\,\mathrm{d}t \\ &{}+ \int_{\frac{3}{2}}^{1.709}\frac{t(4-t)(t-1)^{2}(t-\frac {3}{2})}{b_{n,1}(t-\frac{3}{2}+\frac{1}{3}\sin^{2}(\pi t))}\,\mathrm{d}t \\ &{}+ \int_{1.709}^{2}\frac{t(4-t)(t-1)^{2}(t-\frac{3}{2}-\frac {1}{3}\sin^{2}(\pi t))}{t-\frac{3}{2}}\,\mathrm{d}t\biggr) \\ \approx& 3\sqrt[3]{\mu_{1}\mu_{2}} \biggl(2.373+ \frac{0.2551}{b_{n,1}} \biggr). \end{aligned}
Therefore,
$$\varPi_{c_{1}}^{\delta_{1}}\bigl[H_{1}(t,c_{1}) \bigr]=3\sqrt[3]{\mu_{1}\mu_{2}} \biggl(2.373+ \frac{0.2551}{b_{n,1}} \biggr)-1.$$
Similarly, for $$t\in[\delta_{1},d_{1}]$$, we have
\begin{aligned}& \phi_{2}(t):=\psi(t)H_{2}(d_{1},t)=3\sqrt[3]{ \mu_{1}\mu _{2}}(t-8n) (8n+4-t) (8n+3-t)^{2}, \\& \int_{\delta_{1}}^{d_{1}}H_{2}(d_{1},t) \biggl[p_{0}(t)-\frac {r(t)}{(1+\alpha)^{1+\alpha}} \bigl\vert h_{2}(d_{1},t) \bigr\vert ^{1+\alpha} \biggr]\,\mathrm{d}t=-1, \end{aligned}
and
\begin{aligned} L_{\delta_{1}}^{d_{1}}\bigl[\phi_{2}(t)\bigr] =&3\sqrt[3]{ \mu_{1}\mu_{2}}\biggl( \int_{2}^{\frac{5}{2}}\frac {t(4-t)(3-t)^{2}(t-\frac{3}{2}-\frac{1}{3}\sin^{2}(\pi t))}{t-\frac {3}{2}}\,\mathrm{d}t \\ &{}+ \int_{\frac{5}{2}}^{2.71}\frac{t(4-t)(3-t)^{2}(t-\frac {5}{2})}{b_{n,2}(t-\frac{5}{2}+\frac{1}{3}\sin^{2}(\pi t))}\,\mathrm{d}t \\ &{}+ \int_{2.71}^{3}\frac{t(4-t)(3-t)^{2}(t-\frac{5}{2}-\frac {1}{3}\sin^{2}(\pi t))}{t-\frac{5}{2}}\,\mathrm{d}t\biggr) \\ \approx& 3\sqrt[3]{\mu_{1}\mu_{2}} \biggl(2.964+ \frac{0.078}{b_{n,2}} \biggr). \end{aligned}
Therefore,
$$\varPi_{\delta_{1}}^{d_{1}}\bigl[H_{2}(d_{1},t) \bigr]=3\sqrt[3]{\mu_{1}\mu_{2}} \biggl(2.964+ \frac{0.078}{b_{n,2}} \biggr)-1.$$
Since
$$H_{1}(\delta_{1},c_{1})=(\delta_{1}-c_{1})^{2}=1, \qquad H_{2}(d_{1},\delta _{1})=(d_{1}- \delta_{1})^{2}=1,$$
the left-hand side of inequality (20) is
$$\frac{\varPi_{c_{1}}^{\delta_{1}}[H_{1}(t,c_{1})]}{H_{1}(\delta_{1},c_{1})} +\frac{\varPi_{\delta_{1}}^{d_{1}}[H_{2}(d_{1},t)]}{H_{2}(d_{1},\delta _{1})}\approx 3\sqrt[3]{\mu_{1} \mu_{2}} \biggl(5.337+\frac{0.255}{b_{n,1}}+\frac {0.078}{b_{n,2}} \biggr)-2.$$
Because $$\widetilde{r}_{1}=\widetilde{r}_{2}=1$$, $$\tau_{k(c_{1})+1}=\tau_{k(\delta_{1})}=\tau_{n,1}=8n+\frac{3}{2}\in (c_{1},\delta_{1})$$ and $$\tau_{k(\delta_{1})+1}=\tau_{k(d_{1})}=\tau_{n,2}=8n+\frac{5}{2}\in (\delta_{1},d_{1})$$, it is easy to see that the right-hand side of inequality (20) for $$j=1$$ is
$$\frac{Q_{c_{1}}^{\delta_{1}}[H_{1}(\cdot,c_{1})]}{H_{1}(\delta_{1},c_{1})} +\frac{Q_{\delta_{1}}^{d_{1}}[H_{2}(d_{1},\cdot)]}{H_{2}(d_{1},\delta_{1})} =\frac{b_{n,1}-a_{n,1}}{4a_{n,1}}+\frac{b_{n,2}-a_{n,2}}{4a_{n,2}}.$$
Thus (20) is satisfied with $$j=1$$ if
$$3\sqrt[3]{\mu_{1}\mu_{2}} \biggl(5.337+\frac{0.255}{b_{n,1}}+ \frac {0.078}{b_{n,2}} \biggr) >2+\frac{b_{n,1}-a_{n,1}}{4a_{n,1}}+\frac{b_{n,2}-a_{n,2}}{4a_{n,2}}.$$
When $$j=2$$, with the same argument as above we see that the left-hand side of inequality (20) is
\begin{aligned}& \frac{\varPi_{c_{2}}^{\delta_{2}}[H_{1}(t,c_{2})]}{H_{1}(\delta_{2},c_{2})} +\frac{\varPi_{\delta_{2}}^{d_{2}}[H_{2}(d_{2},t)]}{H_{2}(d_{2},\delta_{2})} \\& \quad = 3\sqrt[3]{\mu_{1}\mu_{2}} \biggl( \int_{1}^{\frac{3}{2}}\frac{t(4-t)(t-1)^{2}(t+\frac{3}{2}-\frac {1}{3}\sin^{2}(\pi t))}{t+\frac{3}{2}}\,\mathrm{d}t \\& \qquad {} + \int_{\frac{3}{2}}^{1.709}\frac{t(4-t)(t-1)^{2}(t-\frac {3}{2})}{b_{n,3}(t-\frac{3}{2}+\frac{1}{3}\sin^{2}(\pi t))}\,\mathrm{d}t \\& \qquad {} + \int_{1.709}^{2}\frac{t(4-t)(t-1)^{2}(t-\frac{3}{2}-\frac{1}{3}\sin ^{2}(\pi t))}{t-\frac{3}{2}}\,\mathrm{d}t \\& \qquad {} + \int_{2}^{\frac{5}{2}}\frac{t(4-t)(3-t)^{2}(t-\frac{3}{2}-\frac {1}{3}\sin^{2}(\pi t))}{t-\frac{3}{2}}\,\mathrm{d}t \\& \qquad {} + \int_{\frac{5}{2}}^{2.71}\frac{t(4-t)(3-t)^{2}(t-\frac {5}{2})}{b_{n,4}(t-\frac{5}{2}+\frac{1}{3}\sin^{2}(\pi t))}\,\mathrm{d}t \\& \qquad {} + \int_{2.71}^{3}\frac{t(4-t)(3-t)^{2}(t-\frac{5}{2}-\frac {1}{3}\sin^{2}(\pi t))}{t-\frac{5}{2}}\,\mathrm{d}t \biggr)-2 \\& \quad \approx 3\sqrt[3]{\mu_{1}\mu_{2}} \biggl(5.337+ \frac {0.255}{b_{n,3}}+\frac{0.078}{b_{n,4}} \biggr)-2, \end{aligned}
and the right-hand side of inequality (20) is
$$\frac{Q_{c_{2}}^{\delta_{2}}[H_{1}(\cdot,c_{2})]}{H_{2}(\delta_{2},c_{2})} +\frac{Q_{\delta_{2}}^{d_{2}}[H_{2}(d_{2},\cdot)]}{H_{2}(d_{2},\delta_{2})} =\frac{b_{n,3}-a_{n,3}}{4a_{n,3}}+ \frac{b_{n,4}-a_{n,4}}{4a_{n,4}}.$$
Therefore, (20) is satisfied for $$j=2$$, if
$$3\sqrt[3]{\mu_{1}\mu_{2}} \biggl(5.337+\frac{0.255}{b_{n,1}}+ \frac {0.078}{b_{n,2}} \biggr) >2+\frac{b_{n,3}-a_{n,3}}{4a_{n,3}}+\frac{b_{n,4}-a_{n,4}}{4a_{n,4}}.$$
Hence, by Theorem 2.1, Eq. (26) is oscillatory, if
\begin{aligned} \textstyle\begin{cases} 3\sqrt[3]{\mu_{1}\mu_{2}} (5.337+\frac{0.255}{b_{n,1}}+\frac {0.078}{b_{n,2}} ) >2+\frac{b_{n,1}-a_{n,1}}{4a_{n,1}}+\frac{b_{n,2}-a_{n,2}}{4a_{n,2}}, \\ 3\sqrt[3]{\mu_{1}\mu_{2}} (5.337+\frac{0.255}{b_{n,1}}+\frac {0.078}{b_{n,2}} ) >2+\frac{b_{n,3}-a_{n,3}}{4a_{n,3}}+\frac{b_{n,4}-a_{n,4}}{4a_{n,4}}. \end{cases}\displaystyle \end{aligned}
(27)

## Notes

### Acknowledgements

The authors sincerely thank the editors and reviewers for their valuable suggestions and useful comments that have led to the present improved version of the original manuscript.

Not applicable.

### Authors’ contributions

Each of the authors, XZ, CL and RC contributed to each part of this study equally and read and approved the final version of the manuscript.

### Funding

This work has been supported by the NNSF of China [Grant No. 11561019], the Science and Technology Planning Projects of Guangdong Province [Grant No. 2017A030303085] and of Zhanjiang City [Grant No. 2016A01001], the Key Subject Program of Lingnan Normal University [Grant No. 1171518004] and the Special Funds for the Cultivation of Guangdong College Students’ Scientific and Technological Innovation [Grant No. pdjh0304].

### Competing interests

The authors declare that they have no competing interests.

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## Authors and Affiliations

• Xiaoliang Zhou
• 1
• Changdong Liu
• 2
• Ruyun Chen
• 2
1. 1.Department of MathematicsLingnan Normal UniversityZhanjiangP.R. China
2. 2.Department of MathematicsGuangdong Ocean UniversityZhanjiangP.R. China