Oscillation criteria of third-order nonlinear dynamic equations with nonpositive neutral coefficients on time scales

Abstract

In this paper, we establish oscillation criteria of third-order nonlinear dynamic equations with nonpositive neutral coefficients on time scales by a generalized Riccati transformation and employing functions in some function classes. Two examples are presented to show the significance of the results.

1 Introduction

In this paper, we consider third-order nonlinear dynamic equations with nonpositive neutral coefficients of the form

$$ \bigl(r_{1}(t) \bigl( \bigl(r_{2}(t) \bigl(z^{\varDelta }(t) \bigr)^{\gamma _{2}} \bigr)^{\varDelta } \bigr)^{\gamma_{1}} \bigr)^{\varDelta }+f \bigl(t,x \bigl(h(t) \bigr) \bigr)=0, $$

(1)

where $z(t)=x(t)-p(t)x(g(t))$, on a time scale $\mathbb{T}$ satisfying $\inf\mathbb{T}=t_{0}$ and $\sup\mathbb{T}=\infty$. Throughout this paper we assume that:

• (C1) $r_{1}, r_{2}\in C_{\mathrm{rd}}(\mathbb{T}, (0,\infty))$ such that

  $$\begin{aligned} \int^{\infty}_{t_{0}}\frac{1}{r_{1}^{1/\gamma_{1}}(t)}\varDelta t=\infty, \qquad \int^{\infty}_{t_{0}}\frac{1}{r_{2}^{1/\gamma_{2}}(t)}\varDelta t= \infty; \end{aligned}$$

• (C2) γ, $\gamma_{1}$, $\gamma_{2}$ are all quotients of odd positive integers, and $\gamma=\gamma_{1}\cdot\gamma_{2}$;

• (C3) $p\in C_{\mathrm{rd}}(\mathbb{T}, [0,\infty))$ and there exists a constant $p_{0}$ with $0\leq p_{0}<1$ such that

  $$\begin{aligned} \lim_{t \rightarrow\infty}p(t)=p_{0}; \end{aligned}$$

• (C4) $g\in C_{\mathrm{rd}}(\mathbb{T}, \mathbb{T})$, $g(t)\leq t$, $\lim_{t \rightarrow\infty}g(t)=\infty$, and there exists a sequence $\{c_{k}\}_{k\geq0}$ such that $\lim_{k \rightarrow\infty}c_{k}=\infty$ and $g(c_{k+1})=c_{k}$;

• (C5) $h\in C_{\mathrm{rd}}(\mathbb{T}, \mathbb{T})$, and for any $t\in \mathbb{T}$,

  $$\begin{aligned} h(t)\geq \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \sigma(t),& 0< \gamma< 1,\\ t, & \gamma\geq1; \end{array}\displaystyle \right . \end{aligned}$$

• (C6) $f\in C(\mathbb{T}\times\mathbb{R}, \mathbb{R})$ and there exists a function $q\in C_{\mathrm{rd}}(\mathbb{T}, (0, \infty))$ such that $uf(t,u)\geq q(t)u^{\gamma+1}$;

• (C7) When $0<\gamma<1$, it always satisfies

  $$\begin{aligned} \int^{\infty}_{t_{0}}q(t)\varDelta t< \infty. \end{aligned}$$

Definition 1.1

A solution x of (1) is said to have a generalized zero at $t^{*}\in\mathbb{T}$ if $x(t^{*})x(\sigma(t^{*}))\leq0$, and it is said to be nonoscillatory on $\mathbb{T}$ if there exists $t_{0}\in\mathbb{T}$ such that $x(t)x(\sigma(t))>0$ for all $t>t_{0}$. Otherwise, it is oscillatory. Equation (1) is said to be oscillatory if all solutions of (1) are oscillatory.

In 1988, the theory of time scales was introduced by Hilger in his Ph.D. thesis [1] to unify continuous and discrete analysis; see also [2]. Since then, the theory had received a lot of attention. The details of time scales can be found in [3–6] and are omitted here.

There has been many achievements of the study of oscillation of nonlinear dynamic equations on time scales in the last few years; see [7–16] and the references therein. Hassan [8], Erbe et al. [7], and Zhang and Wang [16] gave some oscillation criteria successively for the third-order nonlinear delay dynamic equation

$$\begin{aligned} \bigl(a(t) \bigl[ \bigl(r(t)x^{\varDelta }(t) \bigr)^{\varDelta } \bigr]^{\gamma } \bigr)^{\varDelta }+f \bigl(t,x \bigl(\tau (t) \bigr) \bigr)=0. \end{aligned}$$

Saker et al. [13] studied the oscillation of the second-order damped dynamic equation

$$\begin{aligned} \bigl(a(t)x^{\varDelta }(t) \bigr)^{\varDelta }+p(t)x^{\varDelta ^{\sigma }}(t)+q(t) \bigl(f\circ x^{\sigma} \bigr)=0. \end{aligned}$$

Qiu and Wang [10] considered second-order nonlinear dynamic equation

$$\begin{aligned} \bigl(p(t)\psi \bigl(x(t) \bigr)k\circ x^{\varDelta }(t) \bigr)^{\varDelta }+f \bigl(t, x \bigl(\sigma(t) \bigr) \bigr)=0. \end{aligned}$$

Employing a generalized Riccati transformation

$$\begin{aligned} u(t)=A(t)\frac{p(t)\psi(x(t))k\circ x^{\varDelta }(t)}{x(t)}+B(t), \end{aligned}$$

the authors established some Kamenev-type oscillation criteria. Şenel [14] investigated the oscillation of the second-order nonlinear dynamic equation of the form

$$ \bigl(r(t) \bigl(x^{\varDelta }(t) \bigr)^{\gamma} \bigr)^{\varDelta }+p(t) \bigl(x^{\varDelta }(t) \bigr)^{\gamma}+f \bigl(t, x \bigl(g(t) \bigr) \bigr)=0. $$

(2)

Qiu and Wang [11] corrected some mistakes in [14] and established correct oscillation criteria for (2). Yu and Wang [15] considered the third-order nonlinear dynamic equation

$$ \biggl(\frac{1}{a_{2}(t)} \biggl( \biggl(\frac{1}{a_{1}(t)} \bigl(x^{\varDelta }(t) \bigr)^{\alpha_{1}} \biggr)^{\varDelta } \biggr)^{\alpha_{2}} \biggr)^{\varDelta }+q(t)f \bigl(x(t) \bigr)=0 $$

(3)

under the condition $\alpha_{1}\alpha_{2}=1$, and they established some sufficient conditions which guarantee that every solution x of (3) oscillates or converges to zero on a time scale $\mathbb{T}$. Li et al. [9] studied the second-order neutral delay differential equation

$$\begin{aligned} \bigl(r(t) \bigl(z'(t) \bigr)^{\alpha} \bigr)'+q(t)f \bigl(x \bigl(\sigma(t) \bigr) \bigr)=0,\quad t\geq t_{0}>0, \end{aligned}$$

where $z(t)=x(t)-p(t)x(\tau(t))$ and $\alpha>0$ is the ratio of two odd integers. Qiu [12] obtained some significant results for the existence of nonoscillatory solutions to the third-order nonlinear neutral dynamic equation of the form

$$\begin{aligned} \bigl(r_{1}(t) \bigl(r_{2}(t) \bigl(x(t)+p(t)x \bigl(g(t) \bigr) \bigr)^{\varDelta } \bigr)^{\varDelta } \bigr)^{\varDelta }+f \bigl(t,x \bigl(h(t) \bigr) \bigr)=0, \end{aligned}$$

where $\lim_{t \rightarrow\infty}p(t)=p_{0}\in(-1,1)$.

In this paper, motivated by [9, 10, 12, 14, 15], we will establish oscillation criteria of (1), which are more general than (3), by a generalized Riccati transformation, and give two examples to show the significance of the results.

For the sake of simplicity, we denote $(a,b)\cap\mathbb{T}=(a,b)_{\mathbb{T}}$ throughout the paper, where $a, b\in\mathbb{R}$, and $[a,b]_{\mathbb{T}}$, $\left .[a,b) \right ._{\mathbb{T}}$, $\left .(a,b] \right ._{\mathbb{T}}$ are similar notations.

2 Preliminary results

To establish the oscillation criteria of (1), we give six lemmas in this section.

Lemma 2.1

Suppose that $x(t)$ is an eventually positive solution of (1), and there exists a constant $a\geq0$ such that $\lim_{t\rightarrow \infty}z(t)=a$. Then we have

$$\begin{aligned} \lim_{t\rightarrow\infty}x(t)=\frac{a}{1-p_{0}}. \end{aligned}$$

Proof

Suppose that $x(t)$ is an eventually positive solution of (1). In view of (C3) and (C5), there exist $T\in \left .[t_{0},\infty) \right ._{\mathbb{T}}$ and $p_{0}< p_{1}<1$ such that $x(t)>0$, $x(g(t))>0$, and $p(t)\leq p_{1}$ for $t\in [T,\infty)_{\mathbb{T}}$. We claim that $x(t)$ is bounded on $[T,\infty)_{\mathbb{T}}$. Assume not; then there exists $\{t_{n}\}\in[T,\infty)_{\mathbb{T}}$ with $t_{n}\rightarrow \infty$ as $n\rightarrow\infty$ such that

$$\begin{aligned} x(t_{n})=\max_{t\in[T,t_{n}]_{\mathbb{T}}}x(t) \quad\mbox{and}\quad \lim _{n\rightarrow\infty}x(t_{n})=\infty. \end{aligned}$$

Noting that $g(t)\leq t$, we have

$$\begin{aligned} z(t_{n})=x(t_{n})-p(t_{n})x \bigl(g(t_{n}) \bigr)\geq (1-p_{1})x(t_{n}) \rightarrow\infty \end{aligned}$$

as $n\rightarrow\infty$, which contradicts the fact that $\lim_{t\rightarrow\infty}z(t)=a$. Therefore, $x(t)$ is bounded. Then assume that

$$\begin{aligned} \limsup_{t\rightarrow\infty}x(t)=\overline{x} \quad\mbox{and}\quad \liminf _{t\rightarrow\infty}x(t)=\underline{x}. \end{aligned}$$

Since $0\leq p_{0}<1$, we have

$$\begin{aligned} a\geq\overline{x}-p_{0}\overline{x} \quad\mbox{and}\quad a\leq \underline{x}-p_{0}\underline{x}, \end{aligned}$$

which implies that $\overline{x}\leq\underline{x}$. So $\overline{x}=\underline{x}$, and we see that $\lim_{t\rightarrow \infty}x(t)$ exists and $\lim_{t\rightarrow \infty}x(t)=a/(1-p_{0})$. The proof is complete. □

Lemma 2.2

Assume that $x(t)$ is an eventually positive solution of (1), then there exists a sufficiently large $T\in \left .[t_{0},\infty) \right ._{\mathbb{T}}$ such that, for $t\in \left .[T,\infty) \right ._{\mathbb{T}}$, we have

$$\begin{aligned} \bigl(r_{2}(t) \bigl(z^{\varDelta }(t) \bigr)^{\gamma_{2}} \bigr)^{\varDelta }>0 \end{aligned}$$

and

$$\begin{aligned} z^{\varDelta }(t)>0 \quad\textit{or}\quad z^{\varDelta }(t)< 0. \end{aligned}$$

Proof

Suppose that $x(t)$ is an eventually positive solution of (1). From (C3) and (C5), there exist $t_{1}\in \left .[t_{0},\infty) \right ._{\mathbb{T}}$ and $p_{0}< p_{1}<1$ such that $x(t)>0$, $x(g(t))>0$, $x(h(t))>0$, and $p(t)\leq p_{1}$ for $t\in [t_{1},\infty)_{\mathbb{T}}$. By (1) and (C6), it follows that, for $t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$,

$$\begin{aligned} \bigl(r_{1}(t) \bigl( \bigl(r_{2}(t) \bigl(z^{\varDelta }(t) \bigr)^{\gamma _{2}} \bigr)^{\varDelta } \bigr)^{\gamma_{1}} \bigr)^{\varDelta } =-f \bigl(t,x \bigl(h(t) \bigr) \bigr)< 0. \end{aligned}$$

(4)

Hence, $r_{1}(t) ( (r_{2}(t) (z^{\varDelta }(t) )^{\gamma _{2}} )^{\varDelta } )^{\gamma_{1}}$ is strictly decreasing on $\left .[t_{1},\infty) \right ._{\mathbb{T}}$. We claim that

$$\begin{aligned} r_{1}(t) \bigl( \bigl(r_{2}(t) \bigl(z^{\varDelta }(t) \bigr)^{\gamma _{2}} \bigr)^{\varDelta } \bigr)^{\gamma_{1}}>0,\quad t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}. \end{aligned}$$

(5)

Assume not; then there exists $t_{2}\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$ such that

$$\begin{aligned} r_{1}(t) \bigl( \bigl(r_{2}(t) \bigl(z^{\varDelta }(t) \bigr)^{\gamma _{2}} \bigr)^{\varDelta } \bigr)^{\gamma_{1}}< 0 \end{aligned}$$

for $t\in \left .[t_{2},\infty) \right ._{\mathbb{T}}$. So there exists a constant $c<0$ and we have $t_{3}\in \left .[t_{2},\infty) \right ._{\mathbb{T}}$ such that $r_{1}(t) ( (r_{2}(t) (z^{\varDelta }(t) )^{\gamma _{2}} )^{\varDelta } )^{\gamma_{1}}\leq c$ for $t\in \left .[t_{3},\infty) \right ._{\mathbb{T}}$, which means that

$$\begin{aligned} \bigl(r_{2}(t) \bigl(z^{\varDelta }(t) \bigr)^{\gamma_{2}} \bigr)^{\varDelta }\leq \biggl(\frac{c}{r_{1}(t)} \biggr)^{1/\gamma_{1}},\quad t\in \left .[t_{3},\infty) \right ._{\mathbb{T}}. \end{aligned}$$

(6)

Substituting s for t, and integrating (6) from $t_{3}$ to $t\in \left .[\sigma(t_{3}),\infty) \right ._{\mathbb{T}}$, we obtain

$$\begin{aligned} r_{2}(t) \bigl(z^{\varDelta }(t) \bigr)^{\gamma_{2}}\leq r_{2}(t_{3}) \bigl(z^{\varDelta }(t_{3}) \bigr)^{\gamma_{2}}+c^{1/\gamma _{1}}\int^{t}_{t_{3}} \frac{\varDelta s}{r_{1}^{1/\gamma_{1}}(s)}. \end{aligned}$$

Letting $t\rightarrow\infty$, by (C1) we have $r_{2}(t) (z^{\varDelta }(t) )^{\gamma_{2}}\rightarrow -\infty$. Then there exists $t_{4}\in \left .[t_{3},\infty) \right ._{\mathbb{T}}$ such that $r_{2}(t) (z^{\varDelta }(t) )^{\gamma_{2}}\leq r_{2}(t_{4}) (z^{\varDelta }(t_{4}) )^{\gamma_{2}}<0$ for $t\in \left .[t_{4},\infty) \right ._{\mathbb{T}}$, which implies that

$$\begin{aligned} z^{\varDelta }(t)\leq r_{2}^{1/\gamma_{2}}(t_{4})z^{\varDelta }(t_{4}) \cdot \frac{1}{r_{2}^{1/\gamma_{2}}(t)}. \end{aligned}$$

(7)

Substituting s for t, and integrating (7) from $t_{4}$ to $t\in \left .[\sigma(t_{4}),\infty) \right ._{\mathbb{T}}$, we obtain

$$\begin{aligned} z(t)-z(t_{4})\leq r_{2}^{1/\gamma_{2}}(t_{4})z^{\varDelta }(t_{4}) \int_{t_{4}}^{t} \frac{\varDelta s}{r_{2}^{1/\gamma_{2}}(s)}. \end{aligned}$$

Letting $t\rightarrow\infty$, by (C1) we have $z(t)\rightarrow -\infty$. Then there exists $t_{5}\in \left .[t_{4},\infty) \right ._{\mathbb{T}}$ such that $z(t)<0$ or

$$\begin{aligned} x(t)< p(t)x \bigl(g(t) \bigr)\leq p_{1}x \bigl(g(t) \bigr), \quad t\in \left .[t_{5},\infty) \right ._{\mathbb{T}}. \end{aligned}$$

By (C4), we can choose some positive integer $k_{0}$ such that $c_{k}\in \left .[t_{5},\infty) \right ._{\mathbb{T}}$ for all $k\geq k_{0}$. Then for any $k\geq k_{0}+1$, we have

$$\begin{aligned} x(c_{k})&< p_{1}x \bigl(g(c_{k}) \bigr)=p_{1}x(c_{k-1})< p_{1}^{2}x \bigl(g(c_{k-1}) \bigr) =p_{1}^{2}x(c_{k-2})< \cdots \\ &< p_{1}^{k-k_{0}}x \bigl(g(c_{k_{0}+1}) \bigr)=p_{1}^{k-k_{0}}x(c_{k_{0}}). \end{aligned}$$

The inequality above implies that $\lim_{k\rightarrow \infty}x(c_{k})=0$. It follows that

$$\begin{aligned} \lim_{k\rightarrow\infty}z(c_{k})=0, \end{aligned}$$

and this contradicts $\lim_{t\rightarrow\infty}z(t)=-\infty$. So (5) holds, which implies that

$$\begin{aligned} \bigl(r_{2}(t) \bigl(z^{\varDelta }(t) \bigr)^{\gamma_{2}} \bigr)^{\varDelta }>0,\quad t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}. \end{aligned}$$

Therefore, $r_{2}(t) (z^{\varDelta }(t) )^{\gamma_{2}}$ is strictly increasing on $\left .[t_{1},\infty) \right ._{\mathbb{T}}$. It follows that $r_{2}(t) (z^{\varDelta }(t) )^{\gamma_{2}}$ is eventually positive or $r_{2}(t) (z^{\varDelta }(t) )^{\gamma_{2}}<0$ on $\left .[t_{1},\infty) \right ._{\mathbb{T}}$. Lemma 2.2 is proved. □

Lemma 2.3

Assume that $x(t)$ is an eventually positive solution of (1), then $z(t)$ is eventually positive or $\lim_{t\rightarrow \infty}x(t)=0$.

Proof

Suppose that $x(t)$ is an eventually positive solution of (1), by Lemma 2.2 there exists $t_{1}\in \left .[t_{0},\infty) \right ._{\mathbb{T}}$ such that $z^{\varDelta }(t)>0$ or $z^{\varDelta }(t)<0$, $t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$.

(i) $z^{\varDelta }(t)>0$, $t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$. Then it follows that $z(t)$ is eventually positive or eventually negative. If $z(t)$ is eventually positive, the lemma is proved. If $z(t)$ is eventually negative, we see that $\lim_{t\rightarrow\infty}z(t)$ exists. Assume that $\lim_{t\rightarrow\infty}z(t)<0$. Similarly as in the proof of Lemma 2.2, we will have the contradiction. Hence, $\lim_{t\rightarrow\infty}z(t)=0$. Then it follows that $\lim_{t\rightarrow\infty}x(t)=0$ by Lemma 2.1.

(ii) $z^{\varDelta }(t)<0$, $t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$. Similarly, we see that $z(t)$ is eventually positive or eventually negative. Assume that $z(t)$ is eventually negative, there exists a constant $c<0$ and we have $t_{2}\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$ such that $z(t)< c$, $t\in \left .[t_{2},\infty) \right ._{\mathbb{T}}$. It will cause a similar contradiction as in the proof of Lemma 2.2. Hence, $z(t)$ is eventually positive and the lemma is proved.

The proof is complete. □

Lemma 2.4

For $0<\gamma<1$, assume that $x(t)$ is an eventually positive solution of (1), and $z(t)$, $z^{\varDelta }(t)$ are both eventually positive. Then there exists $t_{1}\in \left .[t_{0},\infty) \right ._{\mathbb{T}}$ such that

$$\begin{aligned} \biggl(\frac{z^{\varDelta }(t)}{z^{\sigma}(t)} \biggr)^{1-\gamma}\geq \alpha(t)= \biggl( \frac{\delta(t)}{r_{2}(t)} \biggr)^{(1-\gamma)/\gamma _{2}} \biggl(\int_{t}^{\infty}q(s) \varDelta s \biggr)^{(1-\gamma)/\gamma}, \quad t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}, \end{aligned}$$

where

$$\begin{aligned} \delta(t)=\int_{t_{1}}^{t}\frac{\varDelta s}{r_{1}^{1/\gamma_{1}}(s)}. \end{aligned}$$

Proof

Suppose that $x(t)$ is an eventually positive solution of (1), and $z(t)$, $z^{\varDelta }(t)$ are both eventually positive, then there exists $t_{1}\in \left .[t_{0},\infty) \right ._{\mathbb{T}}$ such that $x(t)>0$, $x(g(t))>0$, $x(h(t))>0$, $z(t)>0$, and $z^{\varDelta }(t)>0$ for $t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$. By Lemma 2.2 we have

$$\begin{aligned} \bigl(r_{2}(t) \bigl(z^{\varDelta }(t) \bigr)^{\gamma_{2}} \bigr)^{\varDelta }>0,\quad t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}. \end{aligned}$$

By $z^{\varDelta }(t)>0$ and $z(t)=x(t)-p(t)x(g(t))\leq x(t)$, it follows that, for $t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$,

$$\begin{aligned} &\bigl(r_{1}(t) \bigl( \bigl(r_{2}(t) \bigl(z^{\varDelta }(t) \bigr)^{\gamma _{2}} \bigr)^{\varDelta } \bigr)^{\gamma_{1}} \bigr)^{\varDelta } \\ &\quad=-f \bigl(t,x \bigl(h(t) \bigr) \bigr) \leq-q(t)x^{\gamma} \bigl(h(t) \bigr)\leq-q(t)z^{\gamma} \bigl(h(t) \bigr)\leq -q(t)z^{\gamma} \bigl(\sigma(t) \bigr)< 0. \end{aligned}$$

(8)

Substituting s for t, and integrating (8) from $t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$ to ∞, we obtain

$$\begin{aligned} r_{1}(t) \bigl( \bigl(r_{2}(t) \bigl(z^{\varDelta }(t) \bigr)^{\gamma _{2}} \bigr)^{\varDelta } \bigr)^{\gamma_{1}}\geq \int _{t}^{\infty}q(s)z^{\gamma} \bigl(\sigma(s) \bigr) \varDelta s\geq z^{\gamma} \bigl(\sigma(t) \bigr)\int_{t}^{\infty}q(s) \varDelta s. \end{aligned}$$

As $r_{1}(t) ( (r_{2}(t) (z^{\varDelta }(t) )^{\gamma _{2}} )^{\varDelta } )^{\gamma_{1}}$ is strictly decreasing on $\left .[t_{1},\infty) \right ._{\mathbb{T}}$, we have, for $t\in \left .[\sigma(t_{1}),\infty) \right ._{\mathbb{T}}$,

$$\begin{aligned} r_{2}(t) \bigl(z^{\varDelta }(t) \bigr)^{\gamma_{2}} =&r_{2}(t_{1}) \bigl(z^{\varDelta }(t_{1}) \bigr)^{\gamma_{2}} +\int _{t_{1}}^{t}\frac{r_{1}^{1/\gamma_{1}}(s) (r_{2}(s) (z^{\varDelta }(s) )^{\gamma_{2}} )^{\varDelta }}{r_{1}^{1/\gamma _{1}}(s)}\varDelta s \\ \geq& r_{1}^{1/\gamma_{1}}(t) \bigl(r_{2}(t) \bigl(z^{\varDelta }(t) \bigr)^{\gamma _{2}} \bigr)^{\varDelta }\int _{t_{1}}^{t}\frac{1}{r_{1}^{1/\gamma _{1}}(s)}\varDelta s \\ \geq& \delta(t) \biggl(z^{\gamma} \bigl(\sigma(t) \bigr)\int _{t}^{\infty}q(s)\varDelta s \biggr)^{1/\gamma_{1}}= \delta(t)z^{\gamma_{2}} \bigl(\sigma(t) \bigr) \biggl(\int_{t}^{\infty}q(s) \varDelta s \biggr)^{1/\gamma_{1}}. \end{aligned}$$

Hence, when $0<\gamma<1$, we have

$$\begin{aligned} \frac{z^{\varDelta }(t)}{z^{\sigma}(t)}\geq \biggl(\frac{\delta(t)}{r_{2}(t)} \biggr)^{1/\gamma_{2}} \biggl( \int_{t}^{\infty}q(s)\varDelta s \biggr)^{1/\gamma}, \quad t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}, \end{aligned}$$

which implies that

$$\begin{aligned} \biggl(\frac{z^{\varDelta }(t)}{z^{\sigma}(t)} \biggr)^{1-\gamma}\geq \alpha(t), \quad t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}. \end{aligned}$$

Lemma 2.4 is proved. □

Lemma 2.5

For $\gamma\geq1$, assume that $x(t)$ is an eventually positive solution of (1), and $z^{\varDelta }(t)$ is eventually negative. If it satisfies

$$ \int_{t_{0}}^{\infty}q(t)\varDelta t=\infty, $$

(9)

then $\lim_{t\rightarrow\infty}x(t)=0$.

Proof

Suppose that $x(t)$ is an eventually positive solution of (1) and $z^{\varDelta }(t)$ is eventually negative. By the proof of Lemma 2.3, we see that $z(t)$ is eventually positive. Then there exists $t_{1}\in \left .[t_{0},\infty) \right ._{\mathbb{T}}$ such that $x(t)>0$, $x(g(t))>0$, $x(h(t))>0$, $z(t)>0$, and $z^{\varDelta }(t)<0$ for $t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$. By Lemma 2.2 we have

$$\begin{aligned} \bigl(r_{2}(t) \bigl(z^{\varDelta }(t) \bigr)^{\gamma_{2}} \bigr)^{\varDelta }>0,\quad t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}. \end{aligned}$$

By $z^{\varDelta }(t)<0$, we claim that there exists $b\geq0$ such that $\lim_{t\rightarrow\infty}z(t)=b$. Assume not; then there exists $t_{2}\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$ such that $z(t)<0$ for $t\in \left .[t_{2},\infty) \right ._{\mathbb{T}}$. It will cause a similar contradiction as in the proof of Lemma 2.2. Then assuming $b>0$, by (8) and $z(\sigma(t)), z(g(t))>b$, we obtain

$$\begin{aligned} \bigl(r_{1}(t) \bigl( \bigl(r_{2}(t) \bigl(z^{\varDelta }(t) \bigr)^{\gamma _{2}} \bigr)^{\varDelta } \bigr)^{\gamma_{1}} \bigr)^{\varDelta }\leq -q(t)z^{\gamma} \bigl( \sigma(t) \bigr)< -b^{\gamma}q(t). \end{aligned}$$

(10)

Letting $v(t)=r_{1}(t) ( (r_{2}(t) (z^{\varDelta }(t) )^{\gamma _{2}} )^{\varDelta } )^{\gamma_{1}}$, $t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$, we have $v(t)>0$, and

$$\begin{aligned} v^{\varDelta }(t)< -b^{\gamma}q(t),\quad t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}. \end{aligned}$$

(11)

Substituting s for t, and integrating (11) from $t_{1}$ to $t\in \left .[\sigma(t_{1}),\infty) \right ._{\mathbb{T}}$, we obtain

$$\begin{aligned} v(t)< v(t_{1})-b^{\gamma}\int_{t_{1}}^{t}q(s) \varDelta s. \end{aligned}$$

By (9), there exists a sufficiently large $t_{2}\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$ such that $v(t)<0$, $t\in \left .[t_{2},\infty) \right ._{\mathbb{T}}$, which contradicts $v(t)>0$. So $b=0$, and Lemma 2.5 is proved. □

Lemma 2.6

Assume that $x(t)$ is an eventually positive solution of (1), and there exists $t_{1}\in \left .[t_{0},\infty) \right ._{\mathbb{T}}$ such that $x(t)>0$, $x(g(t))>0$, $x(h(t))>0$, $z(t)>0$, and $z^{\varDelta }(t)>0$ for $t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$. For $t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$, define

$$ u(t)=A(t)\frac{r_{1}(t) ( (r_{2}(t) (z^{\varDelta }(t) )^{\gamma_{2}} )^{\varDelta } )^{\gamma_{1}}}{z^{\gamma}(t)}+B(t), $$

(12)

where $A \in C^{1}_{\mathrm{rd}}(\mathbb{T}, (0, \infty))$, $B \in C^{1}_{\mathrm{rd}}(\mathbb{T},\mathbb{R})$. Then $u(t)$ satisfies

$$\begin{aligned} u^{\varDelta }(t)+A(t)q(t)-B^{\varDelta }(t)- \varPhi _{0}(t)\leq0, \end{aligned}$$

(13)

where

$$\begin{aligned} \varPhi _{0}(t)= \left \{ \textstyle\begin{array}{@{}l@{\quad}l} A^{\varDelta }(t) (\frac{u(t)-B(t)}{A(t)} )^{\sigma}-\gamma A(t)\alpha(t) (\frac{\delta(t)}{r_{2}(t)} )^{\gamma_{1}} [ (\frac{u(t)-B(t)}{A(t)} )^{\sigma} ]^{2},& 0< \gamma< 1,\\ A^{\varDelta }(t) (\frac{u(t)-B(t)}{A(t)} )^{\sigma}-\gamma A(t) (\frac{\delta(t)}{r_{2}(t)} )^{1/\gamma_{2}} [ (\frac{u(t)-B(t)}{A(t)} )^{\sigma} ]^{(1+\gamma)/\gamma},& \gamma\geq1. \end{array}\displaystyle \right . \end{aligned}$$

Proof

Since $x(t)$ is an eventually positive solution of (1), and there exists $t_{1}\in \left .[t_{0},\infty) \right ._{\mathbb{T}}$ such that $x(t)>0$, $x(g(t))>0$, $x(h(t))>0$, $z(t)>0$, and $z^{\varDelta }(t)>0$ for $t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$, Lemmas 2.2 and 2.4 hold. Let $u(t)$ be defined by (12). Then, differentiating (12) and using (1), it follows that

$$\begin{aligned} u^{\varDelta }(t) =&\frac{A(t)}{z^{\gamma}(t)} \bigl(r_{1}(t) \bigl( \bigl(r_{2}(t) \bigl(z^{\varDelta }(t) \bigr)^{\gamma_{2}} \bigr)^{\varDelta } \bigr)^{\gamma_{1}} \bigr)^{\varDelta } \\ &{} + \biggl(\frac{A(t)}{z^{\gamma}(t)} \biggr)^{\varDelta } \bigl(r_{1}(t) \bigl( \bigl(r_{2}(t) \bigl(z^{\varDelta }(t) \bigr)^{\gamma _{2}} \bigr)^{\varDelta } \bigr)^{\gamma_{1}} \bigr)^{\sigma}+B^{\varDelta }(t) \\ =&-\frac{A(t)}{z^{\gamma}(t)}\cdot f \bigl(t,x \bigl(h(t) \bigr) \bigr)+B^{\varDelta }(t) \\ &{} +\frac{A^{\varDelta }(t)z^{\gamma}(t)-A(t)(z^{\gamma}(t))^{\varDelta }}{z^{\gamma}(t)z^{\gamma}(\sigma(t))} \bigl(r_{1}(t) \bigl( \bigl(r_{2}(t) \bigl(z^{\varDelta }(t) \bigr)^{\gamma_{2}} \bigr)^{\varDelta } \bigr)^{\gamma_{1}} \bigr)^{\sigma}. \end{aligned}$$

Using the fact that

$$\begin{aligned} f \bigl(t,x \bigl(h(t) \bigr) \bigr)\geq q(t)x^{\gamma} \bigl(h(t) \bigr) \geq q(t)z^{\gamma} \bigl(h(t) \bigr)\geq q(t)z^{\gamma}(t), \end{aligned}$$

we obtain

$$\begin{aligned} u^{\varDelta }(t) \leq& -A(t)q(t)+B^{\varDelta }(t)+A^{\varDelta }(t) \biggl(\frac{u(t)-B(t)}{A(t)} \biggr)^{\sigma} \\ &{} -A(t)\frac{(z^{\gamma}(t))^{\varDelta }}{z^{\gamma}(t)} \biggl(\frac {u(t)-B(t)}{A(t)} \biggr)^{\sigma}. \end{aligned}$$

(14)

When $0<\gamma<1$, using the Pötzsche chain rule (see [5]), we have

$$\begin{aligned} \bigl(z^{\gamma}(t) \bigr)^{\varDelta }=\gamma \int_{0}^{1} \bigl(z(t)+h\mu(t)z^{\varDelta }(t) \bigr)^{\gamma-1}\,dh \cdot z^{\varDelta }(t)\geq\gamma \bigl(z^{\sigma}(t) \bigr)^{\gamma-1}z^{\varDelta }(t), \end{aligned}$$

and it follows that

$$\begin{aligned} \frac{(z^{\gamma}(t))^{\varDelta }}{z^{\gamma}(t)}\geq\frac{\gamma (z^{\sigma}(t))^{\gamma-1}z^{\varDelta }(t)}{z^{\gamma}(t)}=\gamma \frac{z^{\varDelta }(t)}{z^{\sigma}(t)} \biggl( \frac{z^{\sigma }(t)}{z(t)} \biggr)^{\gamma}. \end{aligned}$$

By Lemmas 2.2 and 2.4, for $t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$, we obtain

$$\begin{aligned} \frac{z^{\varDelta }(t)}{z^{\sigma}(t)} =&\frac{1}{r_{2}^{\gamma _{1}}(t)}\frac{r_{2}^{\gamma_{1}}(t) (z^{\varDelta }(t) )^{\gamma }}{(z^{\sigma}(t))^{\gamma}} \biggl( \frac{z^{\varDelta }(t)}{z^{\sigma }(t)} \biggr)^{1-\gamma} \\ \geq&\alpha(t) \biggl(\frac{\delta(t) }{r_{2}(t)} \biggr)^{\gamma_{1}} \frac{r_{1}(t) ( (r_{2}(t) (z^{\varDelta }(t) )^{\gamma_{2}} )^{\varDelta } )^{\gamma _{1}}}{(z^{\gamma}(t))^{\sigma}} \\ \geq&\alpha(t) \biggl(\frac{\delta(t) }{r_{2}(t)} \biggr)^{\gamma_{1}} \frac{ (r_{1}(t) ( (r_{2}(t) (z^{\varDelta }(t) )^{\gamma_{2}} )^{\varDelta } )^{\gamma_{1}} )^{\sigma}}{(z^{\gamma}(t))^{\sigma}} \\ =&\alpha(t) \biggl(\frac{\delta(t) }{r_{2}(t)} \biggr)^{\gamma_{1}} \biggl( \frac{u(t)-B(t)}{A(t)} \biggr)^{\sigma} \end{aligned}$$

and

$$\begin{aligned} \frac{z^{\sigma}(t)}{z(t)}\geq1. \end{aligned}$$

So (14) becomes

$$\begin{aligned} u^{\varDelta }(t) \leq& -A(t)q(t)+B^{\varDelta }(t)+A^{\varDelta }(t) \biggl(\frac{u(t)-B(t)}{A(t)} \biggr)^{\sigma} \\ &{} -\gamma A(t)\alpha(t) \biggl(\frac{\delta(t)}{r_{2}(t)} \biggr)^{\gamma_{1}} \biggl[ \biggl(\frac{u(t)-B(t)}{A(t)} \biggr)^{\sigma} \biggr]^{2}. \end{aligned}$$

(15)

When $\gamma\geq1$, we have

$$\begin{aligned} \bigl(z^{\gamma}(t) \bigr)^{\varDelta }=\gamma \int_{0}^{1} \bigl(z(t)+h\mu(t)z^{\varDelta }(t) \bigr)^{\gamma-1}\,dh \cdot z^{\varDelta }(t)\geq\gamma z^{\gamma-1}(t)z^{\varDelta }(t), \end{aligned}$$

and it follows that

$$\begin{aligned} \frac{(z^{\gamma}(t))^{\varDelta }}{z^{\gamma}(t)}\geq\frac{\gamma z^{\gamma-1}(t)z^{\varDelta }(t)}{z^{\gamma}(t)}= \frac{\gamma z^{\varDelta }(t)}{z(t)}. \end{aligned}$$

By Lemmas 2.2 and 2.4, for $t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$, we obtain

$$\begin{aligned} \biggl(\frac{z^{\varDelta }(t)}{z(t)} \biggr)^{\gamma} =&\frac {1}{r_{2}^{\gamma_{1}}(t)} \frac{r_{2}^{\gamma_{1}}(t) (z^{\varDelta }(t) )^{\gamma}}{z^{\gamma}(t)} \\ \geq& \biggl(\frac{\delta(t)}{r_{2}(t)} \biggr)^{\gamma_{1}}\frac{r_{1}(t) ( (r_{2}(t) (z^{\varDelta }(t) )^{\gamma_{2}} )^{\varDelta } )^{\gamma_{1}}}{z^{\gamma}(t)} \\ \geq& \biggl(\frac{\delta(t)}{r_{2}(t)} \biggr)^{\gamma_{1}}\frac{ (r_{1}(t) ( (r_{2}(t) (z^{\varDelta }(t) )^{\gamma _{2}} )^{\varDelta } )^{\gamma_{1}} )^{\sigma}}{(z^{\gamma }(t))^{\sigma}} \\ =& \biggl(\frac{\delta(t)}{r_{2}(t)} \biggr)^{\gamma_{1}} \biggl(\frac {u(t)-B(t)}{A(t)} \biggr)^{\sigma}, \end{aligned}$$

which implies that

$$\begin{aligned} \frac{z^{\varDelta }(t)}{z(t)}\geq \biggl(\frac{\delta(t)}{r_{2}(t)} \biggr)^{1/\gamma_{2}} \biggl[ \biggl(\frac {u(t)-B(t)}{A(t)} \biggr)^{\sigma} \biggr]^{1/\gamma}. \end{aligned}$$

So (14) becomes

$$\begin{aligned} u^{\varDelta }(t) \leq& -A(t)q(t)+B^{\varDelta }(t)+A^{\varDelta }(t) \biggl(\frac{u(t)-B(t)}{A(t)} \biggr)^{\sigma} \\ & {} -\gamma A(t) \biggl(\frac{\delta(t)}{r_{2}(t)} \biggr)^{1/\gamma_{2}} \biggl[ \biggl(\frac{u(t)-B(t)}{A(t)} \biggr)^{\sigma} \biggr]^{(1+\gamma)/\gamma}. \end{aligned}$$

(16)

By (15) and (16), (13) holds. Lemma 2.6 is proved. □

3 Main results

In this section, we establish oscillation criteria of (1) by a generalized Riccati transformation. Firstly, we give some definitions as follows.

Let $D_{0}=\{s\in\mathbb{T}: s\geq0\}$ and $D=\{(t,s)\in\mathbb{T}^{2}: t\geq s\geq0\}$. For any function $f(t,s)$: $\mathbb{T}^{2}\rightarrow\mathbb{R}$, denote by $f^{\varDelta }_{2}$ the partial derivative of f with respect to s. Define

$$\begin{aligned}& (\mathscr{A}, \mathscr{B})= \bigl\{ (A, B): A(s) \in C^{1}_{\mathrm{rd}} \bigl(D_{0}, (0, \infty) \bigr), B(s) \in C^{1}_{\mathrm{rd}}(D_{0}, \mathbb{R}), s\in D_{0} \bigr\} ; \\& \mathscr{H}= \bigl\{ H(t,s)\in C^{1} \bigl(D, [0, \infty )\bigr): H(t,t)=0, H(t,s)>0, H^{\varDelta }_{2}(t,s)\leq0, t>s\geq0 \bigr\} . \end{aligned}$$

These function classes will be used in the sequel. Now, we give our first theorem.

Theorem 3.1

Assume that there exist $(A,B)\in(\mathscr{A}, \mathscr{B})$ and $H\in\mathscr{H}$ such that, for any $t_{1}\in \left .[t_{0},\infty) \right ._{\mathbb{T}}$,

$$\begin{aligned} \limsup_{t\rightarrow\infty}\frac{1}{H(t,t_{1})}\int ^{t}_{t_{1}} \bigl[H(t,s) \bigl(A(s)q(s)-B^{\varDelta }(s) \bigr) -H_{2}^{\varDelta }(t,s)B^{\sigma}(s)- \varPhi _{1}(s) \bigr]\varDelta s=\infty, \end{aligned}$$

(17)

where

$$\begin{aligned} \varPhi _{1}(s)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} (\frac{r_{2}(s)}{\delta(s)} )^{\gamma_{1}}\frac{ (H_{2}^{\varDelta }(t,s)A^{\sigma}(s)+H(t,s)A^{\varDelta }(s) )^{2}}{4\gamma H(t,s)A(s)\alpha(s)},& 0< \gamma< 1,\\ (\frac{r_{2}(s)}{\delta(s)} )^{\gamma_{1}}\frac{1}{ (H(t,s)A(s) )^{\gamma}} (\frac{H_{2}^{\varDelta }(t,s)A^{\sigma}(s) +H(t,s)A^{\varDelta }(s)}{1+\gamma} )^{1+\gamma},& \gamma\geq 1. \end{array}\displaystyle \right . \end{aligned}$$

Then (1) is oscillatory or $\lim_{t\rightarrow\infty}x(t)$ exists.

Proof

Assume that (1) is not oscillatory. Without loss of generality, we may suppose that $x(t)$ is an eventually positive solution of (1). Then by Lemma 2.3, we have $z(t)$ is eventually positive or $\lim_{t\rightarrow\infty}x(t)=0$.

If $\lim_{t\rightarrow\infty}x(t)=0$, the theorem is proved. While $z(t)$ is eventually positive, it follows that there exists $T\in \left .[t_{0},\infty) \right ._{\mathbb{T}}$ such that $z(t)>0$ for $t\in \left .[T,\infty) \right ._{\mathbb{T}}$. By Lemma 2.2, there exists $t_{1}\in \left .[T,\infty) \right ._{\mathbb{T}}$ such that either $z^{\varDelta }(t)>0$ or $z^{\varDelta }(t)<0$ holds for $t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$. Assume that $z^{\varDelta }(t)>0$, $t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$. Let $u(t)$ be defined by (12). Then by Lemma 2.6, (13) holds.

Multiplying (13), where t is replaced by s, by H, and integrating it with respect to s from $t_{1}$ to t with $t\in \left .[\sigma(t_{1}), \infty) \right ._{\mathbb{T}}$, we obtain

$$\begin{aligned} &\int^{t}_{t_{1}}H(t,s) \bigl(A(s)q(s)-B^{\varDelta }(s) \bigr)\varDelta s \\ &\quad\leq-\int^{t}_{t_{1}}H(t,s)u^{\varDelta }(s) \varDelta s+\int^{t}_{t_{1}}H(t,s)\varPhi _{0}(s) \varDelta s. \end{aligned}$$

Noting that $H(t,t)=0$, by the integration by parts formula we have

$$\begin{aligned} & \int^{t}_{t_{1}}H(t,s) \bigl(A(s)q(s)-B^{\varDelta }(s) \bigr)\varDelta s \\ &\quad\leq H(t,t_{1})u(t_{1})+\int^{t}_{t_{1}} \bigl(H_{2}^{\varDelta }(t,s)u^{\sigma }(s)+H(t,s) \varPhi _{0}(s) \bigr) \varDelta s \\ &\quad=H(t,t_{1})u(t_{1})+\int^{t}_{t_{1}}H_{2}^{\varDelta }(t,s)B^{\sigma}(s) \varDelta s \\ &\qquad{} +\int^{t}_{t_{1}} \biggl(H_{2}^{\varDelta }(t,s)A^{\sigma}(s) \biggl(\frac {u(s)-B(s)}{A(s)} \biggr)^{\sigma}+H(t,s)\varPhi _{0}(s) \biggr) \varDelta s. \end{aligned}$$

(18)

When $0<\gamma<1$, we have

$$\begin{aligned} & H_{2}^{\varDelta }(t,s)A^{\sigma}(s) \biggl( \frac {u(s)-B(s)}{A(s)} \biggr)^{\sigma}+H(t,s)\varPhi _{0}(s) \\ &\quad= \bigl(H_{2}^{\varDelta }(t,s)A^{\sigma}(s)+H(t,s)A^{\varDelta }(s) \bigr) \biggl(\frac{u(s)-B(s)}{A(s)} \biggr)^{\sigma} \\ &\qquad{} -\gamma H(t,s)A(s)\alpha(s) \biggl(\frac{\delta(s)}{r_{2}(s)} \biggr)^{\gamma _{1}} \biggl[ \biggl(\frac{u(s)-B(s)}{A(s)} \biggr)^{\sigma} \biggr]^{2} \\ &\quad= \biggl(\frac{r_{2}(s)}{\delta(s)} \biggr)^{\gamma_{1}}\frac{ (H_{2}^{\varDelta }(t,s)A^{\sigma}(s)+H(t,s)A^{\varDelta }(s) )^{2}}{4\gamma H(t,s)A(s)\alpha(s)} \\ &\qquad{} -\gamma H(t,s)A(s)\alpha(s) \biggl(\frac{\delta(s)}{r_{2}(s)} \biggr)^{\gamma _{1}} \biggl[ \biggl(\frac{u(s)-B(s)}{A(s)} \biggr)^{\sigma} \\ &\qquad{} - \biggl(\frac{r_{2}(s)}{\delta(s)} \biggr)^{\gamma_{1}} \frac{H_{2}^{\varDelta }(t,s)A^{\sigma}(s)+H(t,s)A^{\varDelta }(s)}{2\gamma H(t,s)A(s)\alpha(s)} \biggr]^{2} \\ &\quad\leq \biggl(\frac{r_{2}(s)}{\delta(s)} \biggr)^{\gamma_{1}}\frac{ (H_{2}^{\varDelta }(t,s)A^{\sigma}(s)+H(t,s)A^{\varDelta }(s) )^{2}}{4\gamma H(t,s)A(s)\alpha(s)}. \end{aligned}$$

When $\gamma\geq1$, we have

$$\begin{aligned} & H_{2}^{\varDelta }(t,s)A^{\sigma}(s) \biggl( \frac {u(s)-B(s)}{A(s)} \biggr)^{\sigma}+H(t,s)\varPhi _{0}(s) \\ &\quad= \bigl(H_{2}^{\varDelta }(t,s)A^{\sigma}(s)+H(t,s)A^{\varDelta }(s) \bigr) \biggl(\frac{u(s)-B(s)}{A(s)} \biggr)^{\sigma} \\ &\qquad{} -\gamma H(t,s)A(s) \biggl(\frac{\delta(s)}{r_{2}(s)} \biggr)^{1/\gamma_{2}} \biggl[ \biggl(\frac{u(s)-B(s)}{A(s)} \biggr)^{\sigma} \biggr]^{(1+\gamma)/\gamma}. \end{aligned}$$

Using the inequality

$$\begin{aligned} \lambda ab^{\lambda-1}-a^{\lambda}\leq(\lambda-1)b^{\lambda}, \end{aligned}$$

let $\lambda=\frac{1+\gamma}{\gamma}$, and

$$\begin{aligned}& a^{\lambda}=a^{(1+\gamma)/\gamma}=\gamma H(t,s)A(s) \biggl(\frac{\delta(s)}{r_{2}(s)} \biggr)^{1/\gamma_{2}} \biggl[ \biggl(\frac{u(s)-B(s)}{A(s)} \biggr)^{\sigma} \biggr]^{(1+\gamma)/\gamma}, \\& b^{\lambda-1}=b^{1/\gamma}=\frac{\gamma}{1+\gamma} \biggl(\frac {r_{2}(s)}{\delta(s)} \biggr)^{\gamma_{1}/(1+\gamma)} \frac{H_{2}^{\varDelta }(t,s)A^{\sigma}(s)+H(t,s)A^{\varDelta }(s)}{(\gamma H(t,s)A(s))^{\gamma/(1+\gamma)}}, \end{aligned}$$

then we have

$$\begin{aligned} &H_{2}^{\varDelta }(t,s)A^{\sigma}(s) \biggl( \frac{u(s)-B(s)}{A(s)} \biggr)^{\sigma}+H(t,s)\varPhi _{0}(s) \\ &\quad\leq \biggl(\frac{r_{2}(s)}{\delta(s)} \biggr)^{\gamma_{1}}\frac {1}{(H(t,s)A(s))^{\gamma}} \biggl( \frac{H_{2}^{\varDelta }(t,s)A^{\sigma }(s)+H(t,s)A^{\varDelta }(s)}{1+\gamma} \biggr)^{1+\gamma}. \end{aligned}$$

Therefore, for all $\gamma>0$, by (18) we have

$$\begin{aligned} &\int^{t}_{t_{1}}H(t,s) \bigl(A(s)q(s)-B^{\varDelta }(s) \bigr)\varDelta s \\ &\quad\leq H(t,t_{1})u(t_{1})+\int^{t}_{t_{1}}H_{2}^{\varDelta }(t,s)B^{\sigma}(s) \varDelta s+\int^{t}_{t_{1}}\varPhi _{1}(s)\varDelta s, \end{aligned}$$

which implies that

$$\begin{aligned} \int^{t}_{t_{1}} \bigl[H(t,s) \bigl(A(s)q(s)-B^{\varDelta }(s) \bigr)-H_{2}^{\varDelta }(t,s)B^{\sigma}(s)- \varPhi _{1}(s) \bigr]\varDelta s \leq H(t,t_{1})u(t_{1}). \end{aligned}$$

Hence,

$$\begin{aligned} \frac{1}{H(t,t_{1})}\int^{t}_{t_{1}} \bigl[H(t,s) \bigl(A(s)q(s)-B^{\varDelta }(s) \bigr) -H_{2}^{\varDelta }(t,s)B^{\sigma}(s)- \varPhi _{1}(s) \bigr]\varDelta s \leq u(t_{1})< \infty, \end{aligned}$$

which contradicts (17). So $z^{\varDelta }(t)<0$, $t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$, and it is clear that $\lim_{t\rightarrow\infty}z(t)$ exists. By Lemma 2.1 we see that $\lim_{t\rightarrow\infty}x(t)$ exists. The proof is completed. □

When $\gamma\geq1$, if (9) holds, we have the following corollary on the basis of Lemma 2.5 and Theorem 3.1.

Corollary 3.2

When $\gamma\geq1$, assume that (9) holds and there exist $(A,B)\in(\mathscr{A}, \mathscr{B})$ and $H\in\mathscr{H}$ such that, for any $t_{1}\in \left .[t_{0},\infty) \right ._{\mathbb{T}}$,

$$\begin{aligned} \limsup_{t\rightarrow\infty}\frac{1}{H(t,t_{1})}\int ^{t}_{t_{1}} \bigl[H(t,s) \bigl(A(s)q(s)-B^{\varDelta }(s) \bigr) -H_{2}^{\varDelta }(t,s)B^{\sigma}(s)- \varPhi _{1}(s) \bigr]\varDelta s=\infty, \end{aligned}$$

(19)

where

$$\begin{aligned} \varPhi _{1}(s)= \biggl(\frac{r_{2}(s)}{\delta(s)} \biggr)^{\gamma_{1}} \frac {1}{ (H(t,s)A(s) )^{\gamma}} \biggl(\frac{H_{2}^{\varDelta }(t,s)A^{\sigma}(s) +H(t,s)A^{\varDelta }(s)}{1+\gamma} \biggr)^{1+\gamma}. \end{aligned}$$

Then (1) is oscillatory or $\lim_{t\rightarrow\infty}x(t)=0$.

Remark 3.3

In Corollary 3.2, letting $(A,B)=(1,0)$, we can simplify (19) as

$$\begin{aligned} \limsup_{t\rightarrow\infty}\frac{1}{H(t,t_{1})}\int^{t}_{t_{1}} \biggl[H(t,s)q(s) - \biggl(\frac{r_{2}(s)}{\delta(s)} \biggr)^{\gamma_{1}}\frac{1}{H^{\gamma }(t,s)} \biggl( \frac{H_{2}^{\varDelta }(t,s)}{1+\gamma} \biggr)^{1+\gamma} \biggr]\varDelta s=\infty. \end{aligned}$$

When $B=0$, (12) is simplified as

$$ u(t)=A(t)\frac{r_{1}(t) ( (r_{2}(t) (z^{\varDelta }(t) )^{\gamma_{2}} )^{\varDelta } )^{\gamma_{1}}}{z^{\gamma}(t)},\quad t\in \left .[t_{1}, \infty) \right ._{\mathbb{T}}. $$

(20)

Now we have the following theorem.

Theorem 3.4

Assume that there exists $A\in C^{1}_{\mathrm{rd}}(D_{0},(0, \infty))$ such that, for any $t_{1}\in \left .[t_{0},\infty) \right ._{\mathbb{T}}$,

$$\begin{aligned} \limsup_{t\rightarrow\infty}\int^{t}_{t_{1}} \bigl[A(s)q(s)-\varPhi _{2}(s) \bigr]\varDelta s=\infty, \end{aligned}$$

(21)

where

$$\begin{aligned} \varPhi _{2}(s)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} (\frac{r_{2}(s)}{\delta(s)} )^{\gamma_{1}}\frac{ (A^{\varDelta }(s) )^{2}}{4\gamma A(s)\alpha(s)},& 0< \gamma< 1,\\ (\frac{r_{2}(s)}{\delta(s)} )^{\gamma_{1}}\frac{1}{A^{\gamma }(s)} (\frac{A^{\varDelta }(s)}{1+\gamma} )^{1+\gamma},& \gamma\geq1. \end{array}\displaystyle \right . \end{aligned}$$

Then (1) is oscillatory or $\lim_{t\rightarrow\infty}x(t)$ exists.

Proof

Assume that (1) is not oscillatory. Without loss of generality, we may suppose that $x(t)$ is an eventually positive solution of (1). Similarly as in the proof of Theorem 3.1, we have $z(t)$ is eventually positive or $\lim_{t\rightarrow\infty}x(t)=0$.

If $\lim_{t\rightarrow\infty}x(t)=0$, the theorem is proved. If $z(t)$ is eventually positive, there exists $t_{1}\in \left .[t_{0},\infty) \right ._{\mathbb{T}}$ such that $z(t)>0$, and either $z^{\varDelta }(t)>0$ or $z^{\varDelta }(t)<0$ holds for $t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$ by Lemma 2.2. Assume that $z^{\varDelta }(t)>0$, $t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$. Let $u(t)$ be defined by (20). Then by Lemma 2.6, we have

$$\begin{aligned} u^{\varDelta }(t)+A(t)q(t)-\varPhi _{0}(t)\leq0, \end{aligned}$$

where $\varPhi _{0}(t)$ is simplified as

$$\begin{aligned} \varPhi _{0}(t)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} A^{\varDelta }(t) (\frac{u(t)}{A(t)} )^{\sigma}-\gamma A(t)\alpha(t) (\frac{\delta(t)}{r_{2}(t)} )^{\gamma_{1}} [ (\frac{u(t)}{A(t)} )^{\sigma} ]^{2},& 0< \gamma< 1,\\ A^{\varDelta }(t) (\frac{u(t)}{A(t)} )^{\sigma}-\gamma A(t) (\frac{\delta(t)}{r_{2}(t)} )^{1/\gamma_{2}} [ (\frac{u(t)}{A(t)} )^{\sigma} ]^{(1+\gamma)/\gamma},& \gamma\geq1. \end{array}\displaystyle \right . \end{aligned}$$

When $0<\gamma<1$, we have

$$\begin{aligned} u^{\varDelta }(t) \leq& -A(t)q(t)+A^{\varDelta }(t) \biggl(\frac{u(t)}{A(t)} \biggr)^{\sigma}-\gamma A(t)\alpha(t) \biggl(\frac{\delta(t)}{r_{2}(t)} \biggr)^{\gamma_{1}} \biggl[ \biggl(\frac{u(t)}{A(t)} \biggr)^{\sigma} \biggr]^{2} \\ =&-A(t)q(t)+ \biggl(\frac{r_{2}(t)}{\delta(t)} \biggr)^{\gamma_{1}}\frac { (A^{\varDelta }(t) )^{2}}{4\gamma A(t)\alpha(t)} \\ &{} -\gamma A(t)\alpha(t) \biggl(\frac{\delta(t)}{r_{2}(t)} \biggr)^{\gamma_{1}} \biggl[ \biggl(\frac{u(t)}{A(t)} \biggr)^{\sigma}- \biggl( \frac{r_{2}(t)}{\delta (t)} \biggr)^{\gamma_{1}} \frac{A^{\varDelta }(t)}{2\gamma A(t)\alpha(t)} \biggr]^{2} \\ \leq& -A(t)q(t)+ \biggl(\frac{r_{2}(t)}{\delta(t)} \biggr)^{\gamma_{1}} \frac { (A^{\varDelta }(t) )^{2}}{4\gamma A(t)\alpha(t)}. \end{aligned}$$

When $\gamma\geq1$, we have

$$\begin{aligned} u^{\varDelta }(t)\leq -A(t)q(t)+A^{\varDelta }(t) \biggl(\frac{u(t)}{A(t)} \biggr)^{\sigma}-\gamma A(t) \biggl(\frac{\delta(t)}{r_{2}(t)} \biggr)^{1/\gamma_{2}} \biggl[ \biggl(\frac{u(t)}{A(t)} \biggr)^{\sigma} \biggr]^{(1+\gamma)/\gamma}. \end{aligned}$$

Using the inequality

$$\begin{aligned} \lambda ab^{\lambda-1}-a^{\lambda}\leq(\lambda-1)b^{\lambda}, \end{aligned}$$

let $\lambda=\frac{1+\gamma}{\gamma}$, and

$$\begin{aligned}& a^{\lambda}=a^{(1+\gamma)/\gamma}=\gamma A(t) \biggl(\frac{\delta(t)}{r_{2}(t)} \biggr)^{1/\gamma_{2}} \biggl[ \biggl(\frac{u(t)}{A(t)} \biggr)^{\sigma} \biggr]^{(1+\gamma)/\gamma}, \\& b^{\lambda-1}=b^{1/\gamma}=\frac{\gamma}{1+\gamma} \biggl(\frac {r_{2}(t)}{\delta(t)} \biggr)^{\gamma_{1}/(1+\gamma)}\frac{A^{\varDelta }(t)}{(\gamma A(t))^{\gamma/(1+\gamma)}}, \end{aligned}$$

then we have

$$\begin{aligned} u^{\varDelta }(t)\leq -A(t)q(t)+ \biggl(\frac{r_{2}(t)}{\delta(t)} \biggr)^{\gamma_{1}}\frac {1}{A^{\gamma}(t)} \biggl(\frac{A^{\varDelta }(t)}{1+\gamma} \biggr)^{1+\gamma}. \end{aligned}$$

Therefore, for all $\gamma>0$, we always have

$$\begin{aligned} u^{\varDelta }(t)\leq-A(t)q(t)+\varPhi _{2}(t), \end{aligned}$$

which implies that

$$\begin{aligned} A(t)q(t)-\varPhi _{2}(t)\leq-u^{\varDelta }(t). \end{aligned}$$

(22)

Letting t be replaced by s, and integrating (22) with respect to s from $t_{1}$ to $t\in \left .[\sigma(t_{1}),\infty) \right ._{\mathbb{T}}$, we obtain

$$\begin{aligned} \int^{t}_{t_{1}} \bigl[A(s)q(s)- \varPhi _{2}(s) \bigr]\varDelta s\leq -\int^{t}_{t_{1}}u^{\varDelta }(s) \varDelta s= u(t_{1})-u(t)< u(t_{1})< \infty, \end{aligned}$$

which is a contradiction of (21). So $z^{\varDelta }(t)<0$, $t\in \left .[t_{1},\infty) \right ._{\mathbb{T}}$, and as before, $\lim_{t\rightarrow\infty}z(t)$ and $\lim_{t\rightarrow\infty}x(t)$ exist. The proof is completed. □

When $\gamma\geq1$, if (9) holds, from Lemma 2.5 and Theorem 3.4 we have the following result.

Corollary 3.5

When $\gamma\geq1$, assume that (9) holds and there exists $A \in C^{1}_{\mathrm{rd}}(D_{0}, (0, \infty))$ such that, for any $t_{1}\in \left .[t_{0},\infty) \right ._{\mathbb{T}}$,

$$\begin{aligned} \limsup_{t\rightarrow\infty}\int^{t}_{t_{1}} \biggl[A(s)q(s) - \biggl(\frac{r_{2}(s)}{\delta(s)} \biggr)^{\gamma_{1}} \frac{1}{A^{\gamma }(s)} \biggl(\frac{A^{\varDelta }(s)}{1+\gamma} \biggr)^{1+\gamma} \biggr]\varDelta s=\infty. \end{aligned}$$

(23)

Then (1) is oscillatory or $\lim_{t\rightarrow\infty}x(t)=0$.

Remark 3.6

It is not difficult to satisfy the conditions in Corollary 3.5. Indeed, letting $A=1$, by (9) we have (23). The condition (23) can be deleted in Corollary 3.5. Therefore, when $\gamma\geq1$, assume that (9) holds, then it follows that (1) is oscillatory or $\lim_{t\rightarrow\infty}x(t)=0$.

Remark 3.7

Take $r_{1}(t)=1/a_{2}(t)$, $r_{2}(t)=1/a_{1}(t)$, $\gamma_{1}=\alpha_{2}$, $\gamma_{2}=\alpha_{1}$, $\gamma=1$, $p(t)=0$, $h(t)=t$, and $f(t,x)=q(t)f_{0}(x)$, where $f_{0}$ is equivalent to f in Yu and Wang [15]. It is obvious that the conclusions in this paper extend the ones in [15]. Meanwhile, the proofs and results above may provide some enlightenment to the study of oscillation of higher-order nonlinear dynamic equations with nonpositive neutral coefficients on time scales.

4 Examples

In this section, the application of our oscillation criteria will be shown in two examples. Now we give the first example to demonstrate Theorem 3.1 (or Corollary 3.2).

Example 4.1

Let $\mathbb{T}=\bigcup_{n=1}^{\infty}[2n-1,2n]$. Consider the equation

$$\begin{aligned} \biggl(t \biggl( \biggl(\frac{1}{t} \biggl( \biggl(x(t)- \frac {t-1}{2t}x(t-2) \biggr)^{\varDelta } \biggr)^{1/3} \biggr)^{\varDelta } \biggr)^{5} \biggr)^{\varDelta } +\frac{2+\sin t}{t}x^{5/3} \bigl(h(t) \bigr)=0, \end{aligned}$$

(24)

where $r_{1}(t)=t$, $r_{2}(t)=1/t$, $p(t)=(t-1)/2t$, $g(t)=t-2$, $\gamma_{1}=5$, $\gamma_{2}=1/3$, $\gamma=5/3$, $h(t)\geq t$, and $t_{0}=1$. By (C3) we have $p_{0}=1/2$, and by (C6) we take $q(t)=1/t$. Since

$$\begin{aligned} \int^{\infty}_{t_{0}}\frac{1}{r_{1}^{1/\gamma_{1}}(t)}\varDelta t=\int ^{\infty}_{1}\frac{1}{t^{1/5}}\varDelta t=\infty, \qquad \int^{\infty}_{t_{0}}\frac{1}{r_{2}^{1/\gamma_{2}}(t)}\varDelta t=\int ^{\infty}_{1}t^{3}\varDelta t=\infty \end{aligned}$$

and

$$\begin{aligned} \int^{\infty}_{t_{0}}q(t)\varDelta t=\int ^{\infty}_{1}\frac{1}{t}\varDelta t=\infty, \end{aligned}$$

it is obvious that the coefficients of (24) satisfy (C1)-(C6) and (9). Letting $H(t,s)=(t-s)^{2}$ and $(A,B)=(s,0)$, we have

$$\begin{aligned} \delta(t)=\int_{t_{1}}^{t}\frac{\varDelta s}{r_{1}^{1/\gamma_{1}}(s)} =\int _{t_{1}}^{t}\frac{\varDelta s}{s^{1/5}}=O \bigl(t^{4/5} \bigr) \end{aligned}$$

and

$$\begin{aligned} \varPhi _{1}(s) = & \biggl(\frac{r_{2}(s)}{\delta(s)} \biggr)^{\gamma_{1}} \frac {1}{ (H(t,s)A(s) )^{\gamma}} \biggl(\frac{H_{2}^{\varDelta }(t,s)A^{\sigma}(s) +H(t,s)A^{\varDelta }(s)}{1+\gamma} \biggr)^{1+\gamma} \\ =& \biggl(\frac{s^{-1}}{O(s^{4/5})} \biggr)^{5}\frac{1}{ ((t-s)^{2}s )^{5/3}} \biggl( \frac{O(s)\cdot O(s)+(t-s)^{2}}{8/3} \biggr)^{8/3}=O \bigl(s^{-26/3} \bigr). \end{aligned}$$

Hence,

$$\begin{aligned} &\limsup_{t\rightarrow\infty}\frac{1}{H(t,t_{1})}\int^{t}_{t_{1}} \bigl[H(t,s) \bigl(A(s)q(s)-B^{\varDelta }(s) \bigr)-H_{2}^{\varDelta }(t,s)B^{\sigma }(s)- \varPhi _{1}(s) \bigr]\varDelta s \\ &\quad=\limsup_{t\rightarrow\infty}\frac{1}{(t-t_{1})^{2}}\int ^{t}_{t_{1}} \bigl[(t-s)^{2}-O \bigl(s^{-26/3} \bigr) \bigr]\varDelta s=\infty. \end{aligned}$$

That is, (19) holds. By Theorem 3.1 (or Corollary 3.2) we see that (24) is oscillatory or $\lim_{t\rightarrow\infty}x(t)=0$.

The second example illustrates Theorem 3.4.

Example 4.2

Let $\mathbb{T}=\bigcup_{n=0}^{\infty}[3^{n},2\cdot3^{n}]$. Consider the equation

$$\begin{aligned} \biggl(\frac{1}{t^{2}} \biggl( \biggl(\sqrt{t} \biggl( \biggl(x(t)-\frac {1}{t}x \biggl(\frac{t}{3} \biggr) \biggr)^{\varDelta } \biggr)^{5/3} \biggr)^{\varDelta } \biggr)^{1/5} \biggr)^{\varDelta } +\frac{2+t^{2}}{t^{2} (1+t^{2} )}x^{1/3} \bigl(h(t) \bigr)=0, \end{aligned}$$

(25)

where $r_{1}(t)=1/t^{2}$, $r_{2}(t)=\sqrt{t}$, $p(t)=1/t$, $g(t)=t/3$, $\gamma_{1}=1/5$, $\gamma_{2}=5/3$, $\gamma=1/3$, $h(t)\geq\sigma(t)$, and $t_{0}=1$. By (C3) we have $p_{0}=0$, and by (C6) we take $q(t)=1/t^{2}$. Since

$$\begin{aligned} \int^{\infty}_{t_{0}}\frac{1}{r_{1}^{1/\gamma_{1}}(t)}\varDelta t=\int ^{\infty}_{1}t^{10}\varDelta t=\infty,\qquad \int ^{\infty}_{t_{0}}\frac{1}{r_{2}^{1/\gamma_{2}}(t)}\varDelta t=\int ^{\infty}_{1}\frac{1}{t^{3/10}}\varDelta t=\infty \end{aligned}$$

and

$$\begin{aligned} \int^{\infty}_{t_{0}}q(t)\varDelta t=\int ^{\infty}_{1}\frac{1}{t^{2}}\varDelta t< \infty, \end{aligned}$$

it is obvious that the coefficients of (25) satisfy (C1)-(C7). Then, letting $(A,B)=(s^{2},0)$, we obtain

$$\begin{aligned}& \delta(t)=\int_{t_{1}}^{t}\frac{\varDelta s}{r_{1}^{1/\gamma_{1}}(s)} =\int _{t_{1}}^{t}s^{10}\varDelta s=O \bigl(t^{11} \bigr), \\& \begin{aligned}[b] \alpha(t)&= \biggl(\frac{\delta(t)}{r_{2}(t)} \biggr)^{(1-\gamma)/\gamma _{2}} \biggl(2\int_{t}^{\infty}q(s)\varDelta s \biggr)^{(1-\gamma)/\gamma} \\ &= \biggl(\frac{O(t^{11})}{\sqrt{t}} \biggr)^{2/5} \bigl(O \bigl(t^{-1} \bigr) \bigr)^{2}=O \bigl(t^{11/5} \bigr), \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \varPhi _{2}(s) =& \biggl(\frac{r_{2}(s)}{\delta(s)} \biggr)^{\gamma_{1}} \frac { (A^{\varDelta }(s) )^{2}}{4\gamma A(s)\alpha(s)} \\ =& \biggl(\frac{\sqrt{s}}{O(s^{11})} \biggr)^{1/5}\frac{O(s^{2})}{4/3\cdot s^{2}\cdot O(s^{11/5})}=O \bigl(s^{-43/10} \bigr). \end{aligned}$$

Therefore,

$$\begin{aligned} \limsup_{t\rightarrow\infty}\int^{t}_{t_{1}} \bigl[A(s)q(s)-\varPhi _{2}(s) \bigr]\varDelta s=\limsup_{t\rightarrow\infty} \int^{t}_{t_{1}} \bigl[1-O \bigl(s^{-43/10} \bigr) \bigr]\varDelta s=\infty. \end{aligned}$$

That is, (21) holds. By Theorem 3.4 we see that (25) is oscillatory or $\lim_{t\rightarrow\infty}x(t)$ exists.