Theory and Modern Applications

# Nonoscillation for higher-order nonlinear delay dynamic equations on time scales

## Abstract

In this paper, we investigate the nonoscillation of the higher-order nonlinear delay dynamic equation

\begin{aligned} &\bigl(a_{n-1}(t) \bigl(a_{n-2}(t) \bigl(\cdots \bigl(a_{1}(t)x^{\Delta}(t)\bigr)^{\Delta}\cdots \bigr)^{\Delta}\bigr)^{\Delta}\bigr)^{\Delta} +u(t)g\bigl(x\bigl( \delta(t)\bigr)\bigr)=R(t) \\ &\quad\mbox{for } t\in [t_{0}, \infty)_{\mathbb{T}}, \end{aligned}

where $$\mathbb{T}$$ is a scale with $$\sup\mathbb{T}=\infty$$, $$t_{0}\in\mathbb{T}$$, and $$[t_{0},\infty)_{\mathbb{T}}= \{t\in\mathbb{T}:t\geq t_{0}\}$$. We obtain some sufficient conditions for all solutions of this equation to be nonoscillatory.

## 1 Introduction

A time scale $$\mathbb{ T}$$ is an arbitrary nonempty closed subset of the real numbers. Thus, the set $$\mathbb{R}$$ of all real numbers, the set $$\mathbb{N}$$ of all natural numbers, and the set $$\mathbb{Z}$$ of all integers are examples of time scales. On a time scale $$\mathbb{ T}$$, the forward jump operator, the backward jump operator, and the graininess function are defined as

$$\sigma(t)=\inf\{s\in\mathbb{ T}:s>t\}, \qquad \rho(t)=\sup\{s\in \mathbb{ T}:s< t \}, \quad\mbox{and}\quad \mu(t)=\sigma(t)-t,$$

respectively.

In this paper, we investigate the nonoscillation of the higher-order nonlinear delay dynamic equation

\begin{aligned} &\bigl(a_{n-1}(t) \bigl(a_{n-2}(t) \bigl(\cdots \bigl(a_{1}(t)x^{\Delta}(t)\bigr)^{\Delta}\cdots \bigr)^{\Delta}\bigr)^{\Delta}\bigr)^{\Delta} +u(t)g\bigl(x\bigl( \delta(t)\bigr)\bigr)=R(t) \\ & \quad\mbox{for } t\in [t_{0}, \infty)_{\mathbb{T}}, \end{aligned}
(1.1)

where $$t_{0}\in{\mathbb { T}}$$, the time scale interval $$[t_{0},\infty)_{\mathbb{ T}}\equiv\{t\in\mathbb{ T}:t\geq t_{0}\}$$, $$a_{i}\in C_{rd}([t_{0},\infty)_{\mathbb{ T}}, (0,\infty))$$ ($$1\leq i\leq n-1$$), $$u,R\in C_{rd}([t_{0},\infty)_{\mathbb{ T}}, {\mathbb{ R}})$$, $$\delta\in C_{rd}([t_{0},\infty)_{\mathbb{T}},\mathbb{T})$$ is surjective with $$\delta(t)\leq t$$ and $$\delta(t)\rightarrow\infty$$ as $$t\rightarrow\infty$$, and $$g\in C([t_{0},\infty)_{\mathbb{ T}}\times{\mathbb{R}}, {\mathbb{ R}})$$. Our goal is to obtain sufficient conditions for all solutions of (1.1) to be nonoscillatory.

We define

$$R_{i}\bigl(t,x(t)\bigr)= \left \{ \textstyle\begin{array}{@{}l@{\quad}l} x(t) & \mbox{if } i=0,\\ a_{i}(t)R^{\triangle}_{i-1}(t,x(t)) & \mbox{if } 1\leq i\leq n-1. \end{array}\displaystyle \right .$$
(1.2)

Then (1.1) reduces to the equation

$$R_{n-1}^{\triangle}\bigl(t,x(t)\bigr)+u(t)g\bigl(x\bigl(\delta(t) \bigr)\bigr)=R(t).$$
(1.3)

We can suppose the $$\sup{\mathbb{ T}}=\infty$$ since we are interested in the oscillatory behavior of solutions near infinity. By a solution of (1.1) we mean a nontrivial real-valued function $$x\in C_{rd}([T_{x},\infty)_{\mathbb{ T}},{\mathbb{ R}})$$, $$T_{x}\geq t_{0}$$, such that $$R_{n-1}(t,x(t))\in C^{1}_{rd}([T_{x},\infty)_{\mathbb{ T}},{\mathbb{ R}})$$ and satisfies (1.1) on $$[T_{x},\infty)$$. Since we are working on a time scale, the notion of oscillation takes the form of what is known as a generalized zero of a function. We say that $$x(t)$$ has a generalized zero at a point T if $$x(T)x(\sigma(T))\leq0$$. A function is said to be oscillatory if it has arbitrarily large generalized zeros and nonoscillatory otherwise.

In order to create a theory that can unify discrete and continuous analysis, the theory of time scale was initiated by Hilgerâ€™s landmark paper [1], which has received a lot of attention. There exist a variety of interesting time scales, and they give rise to many applications (see [2]). We refer the reader to [3, 4] for further results on time-scale calculus. In the thousands of papers in the literature, finding sufficient conditions for all solutions of an equation to be oscillatory have been a major focus of study (see [5â€“28]), but finding necessary and sufficient conditions for the existence of a nonoscillatory bounded solution of an equation are more rare (see [29]).

Zhu and Wang [21] studied the existence of nonoscillatory solutions to neutral dynamic equation

$$\bigl[x(t)+p(t)x\bigl(g (t)\bigr)\bigr]^{\Delta} +f\bigl(t,x\bigl(h(t) \bigr)\bigr)=0.$$

Karpuz and Ã–calan [22] studied the asymptotic behavior of delay dynamic equations of the form

$$\bigl[x(t)+A(t)x\bigl(\alpha(t)\bigr)\bigr]^{\triangle} +B(t)F\bigl(x\bigl( \beta(t)\bigr)\bigr)-C(t)G\bigl(x\bigl(\gamma (t)\bigr)\bigr)=\varphi(t).$$

Wu et al. [25] investigated the oscillation of the higher-order dynamic equation

$$\bigl\{ r_{n}(t)\bigl[\bigl(r_{n-1}(t) \bigl(\cdots \bigl(r_{1}(t)x(t)^{\Delta}\bigr)^{\Delta}\cdots \bigr)^{\Delta}\bigr)^{\Delta}\bigr]^{\gamma}\bigr\} ^{\Delta} +F\bigl(t,x\bigl(\tau(t)\bigr)\bigr)=0.$$

Sun et al. [26] obtained some necessary and sufficient conditions for the existence of nonoscillatory solution for the higher-order equation

$$\bigl\{ a(t) \bigl[\bigl(x(t)-p(t)x\bigl(\tau(t)\bigr)\bigr)^{\Delta^{m}} \bigr]^{\alpha} \bigr\} ^{\Delta}+f\bigl(t,x\bigl(\delta(t)\bigr) \bigr)=0.$$

## 2 Auxiliary results

We state the following conditions, which are needed in the sequel.

(H1):

There exist constants $$\alpha,\beta\geq0$$ and $$\gamma\geq0$$ such that $$|g(u)|\leq \alpha|u|^{\gamma}+\beta$$.

(H2):

$$\int_{t_{0}}^{\infty}\frac{\Delta s_{1}}{a_{1}(s_{1})}\int_{t_{0}}^{s_{1}}\frac{\Delta s_{2}}{a_{2}(s_{2})}\cdots\int_{t_{0}}^{s_{n-2}}\frac{\Delta s_{n-1}}{a_{n-1}(s_{n-1})} \int_{t_{0}}^{s_{n-1}}|R(s_{n})|\Delta s_{n}<\infty$$.

(H3):

$$\int_{t_{0}}^{\infty}\frac{\Delta s_{1}}{a_{1}(s_{1})}\int_{t_{0}}^{s_{1}}\frac{\Delta s_{2}}{a_{2}(s_{2})}\cdots\int_{t_{0}}^{s_{n-2}}\frac{\Delta s_{n-1}}{a_{n-1}(s_{n-1})} \int_{t_{0}}^{s_{n-1}}|u(s_{n})|\Delta s_{n}<\infty$$.

We shall employ the following lemma.

### Lemma 2.1

Let $$\mathbb{R}_{+}\equiv[0,\infty)$$ and $$H=\{(t,s_{1},s_{2},\ldots,s_{n-1}):0\leq s_{n-1}\leq s_{n-2}\leq\cdots\leq s_{1}\leq t<\infty\}$$. Suppose that $${r}\in C_{rd}([t_{0},\infty)_{\mathbb{T}},\mathbb{R}_{+})$$, $$h\in C_{rd}(H,\mathbb{R}_{+})$$, and that $$p\in C(\mathbb{R}_{+},\mathbb{R}_{+})$$ is nondecreasing with $$p(r)>0$$ for $$r>0$$. If there exists a constant $$c>0$$ such that

$${r}(t)\leq c+ \int_{t_{0}}^{t}\Delta s_{1} \int_{t_{0}}^{s_{1}}\Delta s_{2} \int_{t_{0}}^{s_{2}}\cdots \int_{t_{0}}^{s_{n-1}} h(s_{1},s_{2}, \ldots,s_{n})p\bigl({r}(s_{n})\bigr)\Delta s_{n},$$
(2.1)

then

$${r}(t)\leq P^{-1} \biggl(P(c)+ \int_{t_{0}}^{t}\Delta s_{1} \int _{t_{0}}^{s_{1}}\Delta s_{2} \int_{t_{0}}^{s_{2}}\cdots \int_{t_{0}}^{s_{n-1}} h(s_{1},s_{2}, \ldots,s_{n})\Delta s_{n} \biggr),$$

where

$$P(w)= \int_{w_{0}}^{w}\frac{ds}{p(s)},\quad w_{0},w>0,$$

$$P^{-1}$$ is the inverse function of P, and

$$P(c)+ \int_{t_{0}}^{t}\Delta s_{1} \int_{t_{0}}^{s_{1}}\Delta s_{2} \int_{t_{0}}^{s_{2}}\cdots \int_{t_{0}}^{s_{n-1}} h(s_{1},s_{2}, \ldots,s_{n})\Delta s_{n}\in \operatorname{Dom}\bigl(P^{-1} \bigr).$$
(2.2)

### Proof

Let $$z(t)$$ denote the right side of inequality (2.1). Then $$z(t_{0})=c$$, $${r}(t)\leq z(t)$$, and

\begin{aligned} z^{\Delta} (t) =& \int_{t_{0}}^{t}\Delta s_{2} \int_{t_{0}}^{s_{2}}\cdots \int_{t_{0}}^{s_{n-1}}h(t,s_{2}, \ldots,s_{n})p\bigl({r}(s_{n})\bigr)\Delta s_{n} \\ \leq&p\bigl(z(t)\bigr) \int_{t_{0}}^{t}\Delta s_{2} \int_{t_{0}}^{s_{2}}\cdots \int_{t_{0}}^{s_{n-1}}h(t,s_{2}, \ldots,s_{n})\Delta s_{n}. \end{aligned}

Since $$z^{\Delta} (t)\geq0$$ and p is nondecreasing, we obtain

$$\frac{z^{\Delta} (t)}{p(z(t))}\leq \int_{t_{0}}^{t}\Delta s_{2} \int _{t_{0}}^{s_{2}}\cdots \int_{t_{0}}^{s_{n-1}}h(t,s_{2},\ldots ,s_{n})\Delta s_{n}.$$
(2.3)

Noting that

$$P^{\Delta}\bigl(z(t)\bigr)=z^{\Delta}(t) \int_{0}^{1}\frac{dh}{p[hz(\sigma(t)) +(1-h)z(t)]}\leq\frac{z^{\Delta} (t)}{p(z(t))},$$

we have

$$P\bigl(z(t)\bigr)\leq P\bigl(z(t_{0})\bigr)+ \int_{t_{0}}^{t}\Delta s_{1} \int _{t_{0}}^{s_{1}}\Delta s_{2} \int_{t_{0}}^{s_{2}}\cdots \int_{t_{0}}^{s_{n-1}} h(s_{1},s_{2}, \ldots,s_{n})\Delta s_{n}.$$

Since $$P(w)$$ is increasing, we have

$$z(t)\leq P^{-1} \biggl(P(c)+ \int_{t_{0}}^{t}\Delta s_{1} \int _{t_{0}}^{s_{1}}\Delta s_{2} \int_{t_{0}}^{s_{2}}\cdots \int_{t_{0}}^{s_{n-1}} h(s_{1},s_{2}, \ldots,s_{n})\Delta s_{n} \biggr).$$

The proof is complete.â€ƒâ–¡

Notice that taking $$p(\nu)=\nu^{\xi}$$ and $$\xi>1$$ in LemmaÂ 2.1, we have

$$P\bigl(z(t)\bigr)-P\bigl(z(t_{0})\bigr)=\frac{1}{1-\xi} \bigl[z^{1-\xi}(t)-z^{1-\xi}(t_{0})\bigr].$$

So

$$\frac{1}{1-\xi}z^{1-\xi}(t)\leq\frac{1}{1-\xi}z^{1-\xi}(t_{0})+ \int _{t_{0}}^{t}\Delta s_{1} \int_{t_{0}}^{s_{1}}\Delta s_{2} \int_{t_{0}}^{s_{2}}\cdots \int_{t_{0}}^{s_{n-1}} h(s_{1},s_{2}, \ldots,s_{n})\Delta s_{n},$$

that is,

$$z^{1-\xi}(t)\geq z^{1-\xi}(t_{0})+(1-\xi) \int_{t_{0}}^{t}\Delta s_{1} \int_{t_{0}}^{s_{1}}\Delta s_{2} \int_{t_{0}}^{s_{2}}\cdots \int_{t_{0}}^{s_{n-1}} h(s_{1},s_{2}, \ldots,s_{n})\Delta s_{n}.$$

We have

$${r}(t)\leq \biggl[c^{1-\xi}-(\xi-1) \int_{t_{0}}^{t}\Delta s_{1} \int _{t_{0}}^{s_{1}}\Delta s_{2} \int_{t_{0}}^{s_{2}}\cdots \int_{t_{0}}^{s_{n-1}} h(s_{1},s_{2}, \ldots,s_{n})\Delta s_{n} \biggr]^{\frac{-1}{\xi-1}},$$

provided that

$$\int_{t_{0}}^{t}\Delta s_{1} \int_{t_{0}}^{s_{1}}\Delta s_{2} \int_{t_{0}}^{s_{2}}\cdots \int_{t_{0}}^{s_{n-1}} h(s_{1},s_{2}, \ldots,s_{n})\Delta s_{n}< \frac{c^{1-\xi}}{\xi-1}.$$
(2.4)

## 3 Main results

Now, we state and prove our main results.

### Theorem 3.1

Assume that conditions (H1)-(H3) hold and for some $$k\geq0$$,

$$\int_{t_{0}}^{\infty}\frac{\Delta s_{1}}{a_{1}(s_{1})} \int _{t_{0}}^{s_{1}}\frac{\Delta s_{2}}{a_{2}(s_{2})}\cdots \int _{t_{0}}^{s_{n-2}}\frac{\Delta s_{n-1}}{a_{n-1}(s_{n-1})} \int_{t_{0}}^{s_{n-1}}\bigl|u(s_{n})\bigr| \delta^{k\gamma}(s_{n})\Delta s_{n}< \infty.$$
(3.1)

If $$x(t)$$ is an oscillatory solution of (1.1) such that

$$\bigl|x(t)\bigr|=O\bigl(t^{k}\bigr),\quad t\rightarrow\infty,$$
(3.2)

then $$x(t)\rightarrow0$$ as $$t\rightarrow\infty$$.

### Proof

We will show $$\limsup_{t\longrightarrow\infty}x(t)=0$$ and $$\liminf_{t\longrightarrow\infty}x(t)=0$$. Suppose that $$\limsup_{t\longrightarrow\infty}x(t) =L>0$$. Then for any $$t_{1}\geq t_{0}$$, there exists $$t_{2}\geq t_{1}$$ such that $$x(t_{2})>\frac{L}{2}$$. In view of conditions (H2), (H3), (3.1), and (3.2), there exist $$T_{0}\geq t_{0}$$ and $$K>0$$ such that $$|x(t)|\leq Kt^{k}$$ ($$t\geq T_{0}$$) and

\begin{aligned} &\int_{T_{0}}^{\infty}\frac{\Delta s_{1}}{a_{1}(s_{1})} \int_{T_{0}}^{s_{1}}\frac{\Delta s_{2}}{a_{2}(s_{2})}\cdots \int_{T_{0}}^{s_{n-2}}\frac{\Delta s_{n-1}}{a_{n-1}(s_{n-1})} \\ &\quad{}\times \int_{T_{0}}^{s_{n-1}}\bigl\{ \bigl|R(s_{n})\bigr|+\bigl|u(s_{n})\bigr| \bigl[\alpha K^{\gamma}\delta^{k\gamma}(s_{n})+\beta\bigr] \bigr\} \Delta s_{n}< \frac{L}{4}. \end{aligned}
(3.3)

Since $$x(t)$$ is an oscillatory solution of (1.1), every $$R_{i}{(t,x(t))}$$ is oscillatory for $$i=1,2,\ldots,n-1$$. Choose $$T_{0}< T_{1}\leq T_{2}\leq\cdots\leq T_{n-1}$$ such that

$$R_{n-i}\bigl(T_{i},x(T_{i})\bigr)R_{n-i}\bigl( \sigma(T_{i}),x\bigl(\sigma(T_{i})\bigr)\bigr)\leq0,\quad i=1,2, \ldots,n-1,$$
(3.4)

and

$$R_{n-i}\bigl(T_{i},x(T_{i})\bigr)\leq0,\quad i=1,2,\ldots,n-1.$$
(3.5)

Integrating (1.1) from $$T_{i}$$ to t, $$i=1,2,\ldots,n-1$$, successively $$n-1$$ times with $$t>T_{n-1}$$, we obtain

\begin{aligned} a_{1}x^{\Delta}(t) =&a_{1}(T_{n-1})x^{\Delta}(T_{n-1})+ \int _{T_{n-1}}^{t}\frac{R_{2}(T_{n-2},x(T_{n-2}))}{a_{2}(s_{n-2})}\Delta s_{n-2} \\ &{}+ \int_{T_{n-1}}^{t}\frac{\Delta s_{n-2}}{a_{2}(s_{n-2})} \int _{T_{n-2}}^{s_{n-2}}\frac {R_{3}(T_{n-3},x(T_{n-3}))}{a_{3}(s_{n-3})}\Delta s_{n-3} \\ &{}+\cdots \\ &{}+ \int_{T_{n-1}}^{t}\frac{\Delta s_{n-2}}{a_{2}(s_{n-2})} \int_{T_{n-2}}^{s_{n-2}}\frac{\Delta s_{n-3}}{a_{3}(s_{n-3})}\cdots \int_{T_{2}}^{s_{2}} \frac{R_{n-1}(T_{1},x(T_{1}))}{a_{n-1}(s_{1})}\Delta s_{1} \\ &{}+ \int_{T_{n-1}}^{t}\frac{\Delta s_{n-2}}{a_{2}(s_{n-2})} \int _{T_{n-2}}^{s_{n-2}}\frac{\Delta s_{n-3}}{a_{3}(s_{n-3})}\cdots \int _{T_{1}}^{s_{1}}\bigl[R(s)-u(s)g\bigl(x\bigl(\delta(s) \bigr)\bigr)\bigr]\Delta s \\ \leq& \int_{T_{n-1}}^{t}\frac{\Delta s_{n-2}}{a_{2}(s_{n-2})} \int_{T_{n-2}}^{s_{n-2}}\frac{\Delta s_{n-3}}{a_{3}(s_{n-3})}\cdots \int_{T_{1}}^{s_{1}}\bigl[R(s)-u(s)g\bigl(x\bigl(\delta(s) \bigr)\bigr)\bigr]\Delta s. \end{aligned}
(3.6)

Choose $$T_{n}>T_{n-1}$$ so that

$$x(T_{n})x\bigl(\sigma(T_{n})\bigr)\leq0 \quad\mbox{and} \quad x(T_{n})\leq0.$$

Take $$T_{n+1}\geq T_{n}$$ such that

$$x(T_{n+1})\geq\frac{L}{2} \quad\mbox{and}\quad x(t)>0 ,\quad t \in(T_{n},T_{n+1}).$$

Note that such $$T_{n+1}$$ exists since $$\limsup_{t\longrightarrow\infty}x(t)>\frac{L}{2}$$. Dividing (3.6) by $$a_{1}(t)$$ and integrating once more from $$T_{n}$$ to $$T_{n+1}$$, we have

\begin{aligned} \frac{L}{2} \leq& x(T_{n+1})\leq \int_{T_{n}}^{T_{n+1}}\frac{\Delta s_{n-1}}{a_{1}(s_{n-1})} \int_{T_{n-1}}^{s_{n-1}}\frac{\Delta s_{n-2}}{a_{2}(s_{n-2})} \int_{T_{n-2}}^{s_{n-2}}\frac{\Delta s_{n-3}}{a_{3}(s_{n-3})} \\ &{}\cdots \int_{T_{1}}^{s_{1}}\bigl[R(s)-u(s)g\bigl(x\bigl(\delta(s) \bigr)\bigr)\bigr] \Delta s. \end{aligned}
(3.7)

It follows from (H1) that

\begin{aligned} \frac{L}{2} \leq& \int_{T_{n}}^{T_{n+1}}\frac{\Delta s_{n-1}}{a_{1}(s_{n-1})} \int_{T_{n-1}}^{s_{n-1}}\frac{\Delta s_{n-2}}{a_{2}(s_{n-2})}\cdots \int _{T_{1}}^{s_{1}}\bigl[\bigl|R(s)\bigr|+\bigl|u(s)\bigr|\bigl|g\bigl(x\bigl( \delta(s)\bigr)\bigr)\bigr|\bigr]\Delta s \\ \leq& \int_{T_{n}}^{T_{n+1}}\frac{\Delta s_{n-1}}{a_{1}(s_{n-1})} \int _{T_{n-1}}^{s_{n-1}}\frac{\Delta s_{n-2}}{a_{2}(s_{n-2})}\\ &{}\cdots \int _{T_{1}}^{s_{1}}\bigl\{ \bigl|R(s)\bigr|+\bigl|u(s)\bigr|\bigl[\alpha\bigl|x \bigl(\delta(s)\bigr)\bigr|^{\gamma}+\beta\bigr]\bigr\} \Delta s \\ \leq& \int_{T_{n}}^{T_{n+1}}\frac{\Delta s_{n-1}}{a_{1}(s_{n-1})} \int_{T_{n-1}}^{s_{n-1}}\frac{\Delta s_{n-2}}{a_{2}(s_{n-2})}\cdots \int_{T_{1}}^{s_{1}}\bigl\{ \bigl|R(s)\bigr|+\bigl|u(s)\bigr|\bigl[\alpha K^{\gamma}\delta^{k\gamma}(s)+\beta\bigr]\bigr\} \Delta s . \end{aligned}

In view of (3.3), we have a contradiction.

In a similar fashion, we can show that $$\liminf_{t\longrightarrow\infty }x(t)=0$$. The proof is complete.â€ƒâ–¡

### Theorem 3.2

Assume that conditions (H1)-(H3) hold with $$\gamma\geq1$$. Then every oscillatory solution of (1.1) is bounded.

### Proof

Let $$x(t)$$ be an oscillatory solution of (1.1), and $$d>0$$ be a constant.

If $$\gamma>1$$, then it follows from conditions (H2) and (H3) that there exists $$T^{*}\geq t_{0}$$ such that

$$\int_{T^{*}}^{\infty}\frac{\Delta s_{1}}{a_{1}(s_{1})} \int _{T^{*}}^{s_{1}}\frac{\Delta s_{2}}{a_{2}(s_{2})}\cdots \int _{T^{*}}^{s_{n-2}}\frac{\Delta s_{n-1}}{a_{n-1}(s_{n-1})} \int_{T^{*}}^{s_{n-1}}\bigl[\bigl|R(s_{n})\bigr|+ \beta\bigl|u(s_{n})\bigr|\bigr]\Delta s_{n}< d$$
(3.8)

and

$$\int_{T^{*}}^{\infty}\frac{\Delta s_{1}}{a_{1}(s_{1})} \int_{T^{*}}^{s_{1}}\frac{\Delta s_{2}}{a_{2}(s_{2})}\cdots \int_{T^{*}}^{s_{n-2}}\frac{\Delta s_{n-1}}{a_{n-1}(s_{n-1})} \int_{T^{*}}^{s_{n-1}}\alpha\bigl|u(s_{n})\bigr|\Delta s_{n}< \frac{d^{1-\gamma}}{2(\gamma-1)} .$$
(3.9)

We will show that eventually for any interval on which $$x(t)$$ is positive, we have that $$x(t)$$ is bounded by a constant independent of $$x(t)$$. Choose $$T^{*}< T_{1}\leq T_{2}\leq\cdots\leq T_{n-1}\leq T_{n}$$ so that (3.4)-(3.5) are satisfied, $$\delta(t)>T_{n-1}$$ for $$t\geq T_{n}$$, and $$x(\delta(T_{n}))x(\delta(\sigma(T_{n})))\leq0$$ with $$x(\delta(T_{n}))\leq0$$. As in the proof of TheoremÂ 3.1, using (3.8), we have

\begin{aligned} x\bigl(\delta(t)\bigr) \leq& \int_{T_{1}}^{\delta(t)}\frac{\Delta s_{1}}{a_{1}(s_{1})} \int_{T_{1}}^{s_{1}}\frac{\Delta s_{2}}{a_{2}(s_{2})} \\ &{}\cdots \int_{T_{1}}^{s_{n-2}}\frac{\Delta s_{n-1}}{a_{n-1}(s_{n-1})} \int_{T_{1}}^{s_{n-1}}\bigl\{ \bigl|R(s_{n})\bigr|+\bigl|u(s_{n})\bigr| \bigl[\alpha\bigl|x\bigl(\delta (s_{n})\bigr)\bigr|^{\gamma}+\beta\bigr] \bigr\} \Delta s_{n} \\ =& \int_{T_{1}}^{\delta(t)}\frac{\Delta s_{1}}{a_{1}(s_{1})} \int _{T_{1}}^{s_{1}}\frac{\Delta s_{2}}{a_{2}(s_{2})}\cdots \int _{T_{1}}^{s_{n-2}}\frac{\Delta s_{n-1}}{a_{n-1}(s_{n-1})} \int_{T_{1}}^{s_{n-1}}\bigl[\bigl|R(s_{n})\bigr|+ \beta\bigl|u(s_{n})\bigr|\bigr]\Delta s_{n} \\ &{}+ \int_{T_{1}}^{\delta(t)}\frac{\Delta s_{1}}{a_{1}(s_{1})} \int _{T_{1}}^{s_{1}}\frac{\Delta s_{2}}{a_{2}(s_{2})}\cdots \int _{T_{1}}^{s_{n-2}}\frac{\Delta s_{n-1}}{a_{n-1}(s_{n-1})} \int_{T_{1}}^{s_{n-1}}\bigl|u(s_{n})\bigr|\alpha\bigl|x\bigl( \delta(s_{n})\bigr)\bigr|^{\gamma}\Delta s_{n} \\ \leq& d+ \int_{T_{1}}^{\delta(t)}\frac{\Delta s_{1}}{a_{1}(s_{1})} \int _{T_{1}}^{s_{1}}\frac{\Delta s_{2}}{a_{2}(s_{2})} \\ &{}\cdots \int _{T_{1}}^{s_{n-2}}\frac{\Delta s_{n-1}}{a_{n-1}(s_{n-1})} \int_{T_{1}}^{s_{n-1}}\bigl|u(s_{n})\bigr|\alpha\bigl|x\bigl( \delta(s_{n})\bigr)\bigr|^{\gamma}\Delta s_{n}. \end{aligned}
(3.10)

We can apply LemmaÂ 2.1 with $$c=d$$, $$h(s_{1},s_{2},\ldots,s_{n})=\frac{\alpha |u(s_{n})|}{a_{1}(s_{1})a_{2}(s_{2})\cdots a_{n-1}(s_{n-1})}$$, $$\xi=\gamma$$, and $$p(s)=s^{\gamma}$$. From condition (3.9) we have

\begin{aligned} &d^{1-\gamma}-(\gamma-1)\alpha \int_{T_{1}}^{\delta(t)}\frac{\Delta s_{1}}{a_{1}(s_{1})} \int_{T_{1}}^{s_{1}}\frac{\Delta s_{2}}{a_{2}(s_{2})}\cdots \int_{T_{1}}^{s_{n-2}}\frac{\Delta s_{n-1}}{a_{n-1}(s_{n-1})} \int_{T_{1}}^{s_{n-1}}\bigl|u(s_{n})\bigr|\Delta s_{n} \\ &\quad>d^{1-\gamma}-(\gamma-1)\frac{d^{1-\gamma}}{2(\gamma-1)} =\frac{d^{1-\gamma}}{2}>0. \end{aligned}

Thus, (2.4) holds. It follows from LemmaÂ 2.1 that

\begin{aligned} &x\bigl(\delta(t)\bigr) \\ &\quad\leq \biggl[d^{1-\gamma}-(\gamma-1)\alpha \int_{T_{1}}^{\delta(t)}\frac {\Delta s_{1}}{a_{1}(s_{1})} \int_{T_{1}}^{s_{1}}\frac{\Delta s_{2}}{a_{2}(s_{2})}\cdots \int_{T_{1}}^{s_{n-2}}\frac{\Delta s_{n-1}}{a_{n-1}(s_{n-1})} \int_{T_{1}}^{s_{n-1}}\bigl|u(s_{n})\bigr|\Delta s_{n} \biggr]^{\frac{-1}{\gamma -1}} \\ &\quad\leq\frac{2^{\frac{1}{\gamma-1}}}{d}. \end{aligned}

So $$x(\delta(t))$$ is bounded. A similar argument holds for intervals where $$x(t)$$ is negative.

If $$\gamma=1$$, then choose $$\hat{T}\geq t_{0}$$ so that (3.8) holds with $$T^{*}$$ replaced by TÌ‚ and

$$\int_{\hat{T}}^{\infty}\frac{\Delta s_{1}}{a_{1}(s_{1})} \int_{\hat{T}}^{s_{1}}\frac{\Delta s_{2}}{a_{2}(s_{2})}\cdots \int_{\hat{T}}^{s_{n-2}}\frac{\Delta s_{n-1}}{a_{n-1}(s_{n-1})} \int_{\hat{T}}^{s_{n-1}}\bigl|u(s_{n})\bigr|\Delta s_{n}< \frac{1}{\alpha +1}.$$

Choose $$T^{*}< T'_{1}\leq T'_{2}\leq\cdots\leq T'_{n-1}\leq T'_{n}$$ so that $$R_{n-i}(T'_{i},x(T'_{i}))R_{n-i}(\sigma(T'_{i}),x(\sigma(T'_{i})))\leq 0$$ with $$R_{n-i}(T'_{i},x(T'_{i}))\geq0$$ for $$1\leq i\leq n$$ and $$\delta(t)>T'_{n-1}$$ for $$t\geq T'_{n}$$. As in the proof of TheoremÂ 3.1, using (3.8), we have

\begin{aligned} x\bigl(\delta(t)\bigr) \geq-d- \int_{T'_{1}}^{\delta(t)}\frac{\Delta s_{1}}{a_{1}(s_{1})} \int_{T'_{1}}^{s_{1}}\frac{\Delta s_{2}}{a_{2}(s_{2})}\cdots \int_{T'_{1}}^{s_{n-2}}\frac{\Delta s_{n-1}}{a_{n-1}(s_{n-1})} \int_{T'_{1}}^{s_{n-1}}\bigl|u(s_{n})\bigr|\alpha\bigl|x\bigl( \delta(s_{n})\bigr)\bigr|\Delta s_{n}. \end{aligned}

Combining (3.10) with this inequality, we obtain

\begin{aligned} \bigl|x\bigl(\delta(t)\bigr)\bigr| \leq& d+ \int_{L}^{\delta(t)}\frac{\Delta s_{1}}{a_{1}(s_{1})} \int_{L}^{s_{1}}\frac{\Delta s_{2}}{a_{2}(s_{2})} \\ &{}\cdots \int_{L}^{s_{n-2}}\frac{\Delta s_{n-1}}{a_{n-1}(s_{n-1})} \int_{L}^{s_{n-1}}\bigl|u(s_{n})\bigr|\alpha\bigl|x\bigl( \delta(s_{n})\bigr)\bigr|\Delta s_{n}, \end{aligned}
(3.11)

where $$L=\min\{T_{1},T'_{1}\}$$. Denoting by $$z(t)$$ the right side of inequality (3.11), we see that $$|x(\delta(t))|\leq z(t)$$, $$z(\delta (t))\leq z(t)$$, and

\begin{aligned} z(t) =&d+ \int_{L}^{\delta(t)}\frac{\Delta s_{1}}{a_{1}(s_{1})} \int_{L}^{s_{1}}\frac{\Delta s_{2}}{a_{2}(s_{2})}\cdots \int_{L}^{s_{n-2}}\frac{\Delta s_{n-1}}{a_{n-1}(s_{n-1})} \int_{L}^{s_{n-1}}\bigl|u(s_{n})\bigr|\alpha\bigl|x\bigl( \delta(s_{n})\bigr)\bigr|\Delta s_{n} \\ \leq& d+ \int_{L}^{\delta(t)}\frac{\Delta s_{1}}{a_{1}(s_{1})} \int_{L}^{s_{1}}\frac{\Delta s_{2}}{a_{2}(s_{2})}\cdots \int_{L}^{s_{n-2}}\frac{\Delta s_{n-1}}{a_{n-1}(s_{n-1})} \int_{L}^{s_{n-1}}\bigl|u(s_{n})\bigr|\alpha z(s_{n})\Delta s_{n} \\ \leq&d+z(t) \int_{L}^{\delta(t)}\frac{\Delta s_{1}}{a_{1}(s_{1})} \int_{L}^{s_{1}}\frac{\Delta s_{2}}{a_{2}(s_{2})}\cdots \int_{L}^{s_{n-2}}\frac{\Delta s_{n-1}}{a_{n-1}(s_{n-1})} \int_{L}^{s_{n-1}}\bigl|u(s_{n})\bigr|\alpha\Delta s_{n} \\ \leq&d+\frac{\alpha}{\alpha+1}z(t), \end{aligned}

which implies $$x(\delta(t))\leq d(\alpha+1)$$. The proof is complete.â€ƒâ–¡

After seeing the proof of TheoremÂ 3.2, the proof of the following TheoremÂ 3.3 becomes obvious.

### Theorem 3.3

Assume that conditions (H1)-(H3) hold with $$\gamma\geq1$$. If (3.1) holds, then every oscillatory solution of (1.1) converges to zero as $$t\rightarrow\infty$$.

In a similar fashion as before, we can show the following theorem.

### Theorem 3.4

Assume that conditions (H1)-(H3) hold with $$0<\gamma< 1$$. If (3.1) holds, then every oscillatory solution of (1.1) is bounded and converges to zero as $$t\rightarrow\infty$$.

### Proof

Notice that taking $$p(\nu)=\nu^{\xi}$$ and $$0<\xi<1$$ in LemmaÂ 2.1, we have

$$P\bigl(z(t)\bigr)-P\bigl(z(t_{0})\bigr)=\frac{1}{1-\xi} \bigl[z^{1-\xi}(t)-z^{1-\xi}(t_{0})\bigr].$$

So

$$\frac{1}{1-\xi}z^{1-\xi}(t)\leq\frac{1}{1-\xi}z^{1-\xi}(t_{0})+ \int _{t_{0}}^{t}\Delta s_{1} \int_{t_{0}}^{s_{1}}\Delta s_{2} \int_{t_{0}}^{s_{2}}\cdots \int_{t_{0}}^{s_{n-1}} h(s_{1},s_{2}, \ldots,s_{n})\Delta s_{n},$$

that is,

$$z(t)\leq \biggl[z^{1-\xi}(t_{0})+(1-\xi) \int_{t_{0}}^{t}\Delta s_{1} \int _{t_{0}}^{s_{1}}\Delta s_{2} \int_{t_{0}}^{s_{2}}\cdots \int_{t_{0}}^{s_{n-1}} h(s_{1},s_{2}, \ldots,s_{n})\Delta s_{n} \biggr]^{\frac{1}{1-\xi}}.$$

We have

$${r}(t)\leq \biggl[z^{1-\xi}(t_{0})+(1-\xi) \int_{t_{0}}^{t}\Delta s_{1} \int_{t_{0}}^{s_{1}}\Delta s_{2} \int_{t_{0}}^{s_{2}}\cdots \int_{t_{0}}^{s_{n-1}} h(s_{1},s_{2}, \ldots,s_{n})\Delta s_{n}\biggr]^{\frac{1}{1-\xi}}.$$

Further, the proof is similar to that of TheoremÂ 3.2, so we have

\begin{aligned} x\bigl(\delta(t)\bigr) \leq& \biggl[d^{1-\gamma}+(1-\gamma)\alpha \int_{t_{0}}^{{\delta(t)}}\frac {\Delta s_{1}}{a_{1}(s_{1})} \int_{t_{0}}^{s_{1}}\frac{\Delta s_{2}}{a_{2}(s_{2})} \\ &{}\cdots \int_{t_{0}}^{s_{n-2}}\frac{\Delta s_{n-1}}{a_{n-1}(s_{n-1})} \int_{t_{0}}^{s_{n-1}}\bigl|u(s_{n})\bigr|\Delta s_{n}\biggr]^{\frac{1}{1-\gamma}}. \end{aligned}

So we can conclude that every oscillatory solution of (1.1) is bounded, and by TheoremÂ 3.1 $$x(t)$$ converges to zero as $$t\rightarrow\infty$$. The proof is complete.â€ƒâ–¡

### Theorem 3.5

Assume that conditions (H1)-(H3) hold with $$g(0)=0$$. If there exists $$N>0$$ such that for all large T, either

$$\liminf_{t\longrightarrow\infty} \int_{T}^{t}\bigl[R(s)-N\bigl|u(s)\bigr|\bigr]\Delta s>0$$
(3.12)

or

$$\limsup_{t\longrightarrow\infty} \int_{T}^{t}\bigl[R(s)+N\bigl|u(s)\bigr|\bigr]\Delta s< 0,$$
(3.13)

then all solutions of (1.1) are nonoscillatory.

### Proof

For contradiction, let $$x(t)$$ be an oscillatory solution of (1.1). By TheoremÂ 3.3 and TheoremÂ 3.4, $$x(t)$$ converges to 0 as $$t\rightarrow\infty$$. Hence, there exists $$T_{0}\geq t_{0}$$ such that $$|g(x(\delta(t)))|\leq N$$ for $$t\geq T_{0}$$. From (1.3) we have

$$R(t)-N\bigl|u(t)\bigr|\leq R^{ \Delta}_{n-1}\bigl(t,x(t)\bigr)\leq R(t)+N\bigl|u(t)\bigr|.$$
(3.14)

If (3.12) holds, then we choose $$T\geq T_{0}$$ such that $$\delta(t)\geq T_{0}$$ for $$t\geq T$$,

$$R_{n-2}\bigl(T,x(T)\bigr)R_{n-2}\bigl(\sigma(T),x\bigl( \sigma(T)\bigr)\bigr)\leq 0 ,\qquad R_{n-2}\bigl(T,x(T)\bigr) \geq0,$$
(3.15)

and integrating the left inequality in (3.14) from T to t, we obtain

$$R_{n-2}\bigl(T,x(T)\bigr)+ \int^{t}_{T}\bigl[R(s)-N\bigl|u(s)\bigr|\bigr]\Delta s\leq R_{n-2}\bigl(t,x(t)\bigr).$$

This is a contradiction since if $$x(t)$$ is oscillatory, then $$R_{n-2}(t,x(t))$$ is also oscillatory.

If (3.13) holds, then we choose T so that the second inequality in (3.15) is reversed. This completes the proof of the theorem.â€ƒâ–¡

## 4 Example

In this section, we give an example to illustrate our main results.

### Lemma 4.1

[23, 24]

Assume that $$s,t\in {\mathbb{ T}}$$ and $$g\in C_{rd}({\mathbb{ T}}\times{\mathbb{ T}},{\mathbb{ R}})$$. Then

$$\int_{s}^{t} \biggl[ \int_{\eta}^{t} g(\eta,\zeta)\Delta\zeta \biggr]\Delta \eta = \int_{s}^{t} \biggl[ \int_{s}^{\sigma(\zeta)}g(\eta ,\zeta)\Delta\eta \biggr]\Delta \zeta.$$

### Example 4.1

Let $$\mathbb{ T}=\{q^{n}:n\in\mathbb{ Z}\}\cup\{0\}$$ with $$q>1$$. Consider the higher-order dynamic equation

$$\bigl(t\bigl(t\bigl(\cdots\bigl(t^{2+\frac{1}{\gamma}}x^{\Delta} \bigr)^{\Delta}\cdots\bigr)^{\Delta }\bigr)^{\Delta} \bigr)^{\Delta}+\frac{1}{ t^{1+k\gamma+\frac{1}{\gamma}}} \biggl|x \biggl(\frac{t}{q} \biggr) \biggr|^{\gamma}\operatorname{sgn} x \biggl(\frac{t}{q} \biggr)= \frac{1}{t^{1+\frac{1}{\gamma}}},$$
(4.1)

where $$t\in[q,\infty)_{\mathbb{ T}}$$, $$\gamma >0$$, $$k\geq0$$, $$a_{1}(t)=t^{2+\frac{1}{\gamma}}$$, $$a_{i}(t)=t$$ ($$2\leq i\leq n-1$$), $$u(t)=\frac{1}{t^{1+k\gamma}+\frac{1}{\gamma}}$$, $$\delta(t)=\frac{t}{q}$$, $$R(t)=\frac{1}{t^{1+\frac{1}{\gamma}}}$$, and $$g(u)=|u|^{\gamma}\operatorname{sgn}(u)$$.

It is easy to verify that $$R(t)$$ and $$u(t)$$ satisfy the condition (3.12). We will use the following inequality: if $$s>t\geq q$$, then

\begin{aligned} \int_{t}^{s}\frac{1}{\tau}\Delta\tau &= \int_{t}^{qt}\frac{1}{\tau}\Delta\tau+ \int_{qt}^{q^{2}t}\frac{1}{\tau }\Delta\tau+\cdots+ \int_{q^{n-1}t}^{q^{n}t=s}\frac{1}{\tau}\Delta\tau \\ &=\frac{qt-t}{t}+\frac{q^{2}t-qt}{qt}+\cdots+\frac {q^{n}t-q^{n-1}t}{q^{n-1}t} \\ &=n(q-1)\leq q^{n+1}\leq q^{n}t=s. \end{aligned}

Applying LemmaÂ 4.1 and the last inequality, we have

\begin{aligned} & \int_{q}^{\infty}\frac{\Delta s_{1}}{a_{1}(s_{1})} \int_{q}^{s_{1}}\frac{\Delta s_{2}}{a_{2}(s_{2})}\cdots \int_{q}^{s_{n-2}}\frac{\Delta s_{n-1}}{a_{n-1}(s_{n-1})} \int_{q}^{s_{n-1}}\bigl|u(s_{n})\bigr|\Delta s_{n} \\ &\quad\leq \int_{q}^{\infty}\frac{\Delta s_{1}}{a_{1}(s_{1})} \int_{q}^{s_{1}}\frac{\Delta s_{2}}{a_{2}(s_{2})}\cdots \int_{q}^{s_{n-2}}\frac{\Delta s_{n-1}}{a_{n-1}(s_{n-1})} \int_{q}^{s_{n-1}}\bigl|R(s_{n})\bigr|\Delta s_{n} \\ &\quad= \int_{q}^{\infty}\frac{\Delta s_{1}}{s_{1}^{2+\frac{1}{\gamma}}} \int_{q}^{s_{1}}\frac{\Delta s_{2}}{s_{2}}\cdots \int_{q}^{s_{n-2}}\frac{\Delta s_{n-1}}{s_{n-1}} \int_{q}^{s_{n-1}}\frac{1}{s_{n}^{1+\frac{1}{\gamma}}}\Delta s_{n} \\ &\quad= \int_{q}^{\infty}\frac{\Delta s_{1}}{s_{1}^{2+\frac{1}{\gamma}}} \int_{q}^{s_{1}}\frac{\Delta s_{2}}{s_{2}}\cdots{ \int_{q}^{s_{n-3}}\frac{\Delta s_{n-2}}{s_{n-2}} \int_{q}^{s_{n-2}}\Delta s_{n-1} \int_{q}^{s_{n-1}}\frac{1}{s_{n-1}}\frac{1}{s_{n}^{1+\frac{1}{\gamma }}} \Delta s_{n}} \\ &\quad\leq \int_{q}^{\infty}\frac{\Delta s_{1}}{s_{1}^{2+\frac{1}{\gamma}}} \int_{q}^{s_{1}}\frac{\Delta s_{2}}{s_{2}}\cdots{ \int_{q}^{s_{n-3}}\frac{\Delta s_{n-2}}{s_{n-2}}} \int_{q}^{s_{n-2}}\Delta s_{n-1} \int_{q}^{\sigma(s_{n-1})}\frac{1}{s_{n-1}}\frac{1}{s_{n}^{1+\frac {1}{\gamma}}} \Delta s_{n} \\ &\quad= \int_{q}^{\infty}\frac{\Delta s_{1}}{s_{1}^{2+\frac{1}{\gamma}}} \int_{q}^{s_{1}}\frac{\Delta s_{2}}{s_{2}}\cdots{ \int_{q}^{s_{n-3}}\frac{\Delta s_{n-2}}{s_{n-2}}} \int_{q}^{s_{n-2}}\frac{\Delta s_{n}}{s_{n}^{1+\frac{1}{\gamma}}} \int_{s_{n}}^{s_{n-2}}\frac{1}{s_{n-1}}\Delta s_{n-1} \\ &\quad\leq \int_{q}^{\infty}\frac{\Delta s_{1}}{s_{1}^{2+\frac{1}{\gamma}}} \int_{q}^{s_{1}}\frac{\Delta s_{2}}{s_{2}}\cdots \int_{q}^{s_{n-3}}\frac{\Delta s_{n-2}}{s_{n-2}} \int _{q}^{s_{n-2}}\frac{s_{n-2}}{s_{n}^{1+\frac{1}{\gamma}}} \Delta s_{n} \\ &\quad= \int_{q}^{\infty}\frac{\Delta s_{1}}{s_{1}^{2+\frac{1}{\gamma}}} \int_{q}^{s_{1}}\frac{\Delta s_{2}}{s_{2}}\cdots{ \int_{q}^{s_{n-4}}\frac{\Delta s_{n-3}}{s_{n-3}}} \int_{q}^{s_{n-3}}\Delta s_{n-2} \int_{q}^{s_{n-2}}\frac{1}{s_{n}^{1+\frac{1}{\gamma}}} \Delta s_{n} \\ &\quad\leq \int_{q}^{\infty}\frac{\Delta s_{1}}{s_{1}^{2+\frac{1}{\gamma}}} \int_{q}^{s_{1}}\frac{\Delta s_{2}}{s_{2}}\cdots{ \int_{q}^{s_{n-4}}\frac{\Delta s_{n-3}}{s_{n-3}}} \int_{q}^{s_{n-3}}\Delta s_{n-2} \int_{q}^{\sigma(s_{n-2})}\frac{1}{s_{n}^{1+\frac{1}{\gamma}}} \Delta s_{n} \\ &\quad= \int_{q}^{\infty}\frac{\Delta s_{1}}{s_{1}^{2+\frac{1}{\gamma}}} \int_{q}^{s_{1}}\frac{\Delta s_{2}}{s_{2}}\cdots{ \int_{q}^{s_{n-4}}\frac{\Delta s_{n-3}}{s_{n-3}}} \int_{q}^{s_{n-3}}\Delta s_{n} \int_{s_{n}}^{s_{n-3}}\frac{1}{s_{n}^{1+\frac{1}{\gamma}}} \Delta s_{n-2} \\ &\quad\leq \int_{q}^{\infty}\frac{\Delta s_{1}}{s_{1}^{2+\frac{1}{\gamma }}} \int_{q}^{s_{1}}\frac{\Delta s_{2}}{s_{2}}\cdots \int _{q}^{s_{n-3}}\frac{s_{n-3}}{s_{n}^{1+\frac{1}{\gamma}}} \Delta s_{n} \\ &\quad\cdots \\ &\quad\leq \int_{q}^{\infty}\frac{\Delta s_{1}}{s_{1}^{2+\frac{1}{\gamma}}} \int_{q}^{s_{1}}\frac {s_{1}}{s_{n}^{1+\frac{1}{\gamma}}} \Delta s_{n}\leq \int_{q}^{\infty}\frac{\Delta s_{1}}{s_{1}^{1+\frac{1}{\gamma}}} \int_{q}^{\sigma (s_{1})}\frac{1}{s_{n}^{1+\frac{1}{\gamma}}} \Delta s_{n} \\ &\quad= \int_{q}^{\infty}\frac{\Delta s_{n}}{s_{n}^{1+\frac{1}{\gamma}}} \int_{s_{n}}^{\infty}\frac {1}{s_{1}^{1+\frac{1}{\gamma}}} \Delta s_{1} \\ &\quad\leq \int_{q}^{\infty}\frac{\Delta s_{n}}{s_{n}^{1+\frac{1}{\gamma}}} \int_{q}^{\infty}\frac {1}{s_{1}^{1+\frac{1}{\gamma}}} \Delta s_{1}= \biggl(\frac{q-1}{q^{\frac{1}{\gamma}}-1} \biggr)^{2}< \infty. \end{aligned}

Thus, conditions (H1)-(H3) and (3.1) hold. Then it follows from TheoremÂ 3.5 that every solution $$x(t)$$ of (4.1) is nonoscillatory.

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## Acknowledgements

This project is supported by NNSF of China (11461003).

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Correspondence to Taixiang Sun.

### Competing interests

The authors declare that they have no competing interests.

### Authorsâ€™ contributions

All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.

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Tao, C., Sun, T. & He, Q. Nonoscillation for higher-order nonlinear delay dynamic equations on time scales. Adv Differ Equ 2016, 58 (2016). https://doi.org/10.1186/s13662-016-0786-6