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

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.

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:

• (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:

• (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 $k=k(c_{1})+1,\ldots,k(d_{1})-1$, $k\neq k(\delta_{1})$;

• (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 $k=k(c_{1})+1,\ldots,k(d_{1})$;

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

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

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

• (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 2.1 are 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 $u(t)$:

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

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

• (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 2.3 are 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 2.1 hold. 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 2.2 hold. 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)