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 Hilger’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 [5–28]), 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 Öcalan [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. □

Notice that taking $p(\nu)=\nu^{\xi}$ and $\xi>1$ in Lemma 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. □

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 Theorem 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 Lemma 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 Lemma 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 T̂ 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 Theorem 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. □

After seeing the proof of Theorem 3.2, the proof of the following Theorem 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 Lemma 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 Theorem 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 Theorem 3.1 $x(t)$ converges to zero as $t\rightarrow\infty$. The proof is complete. □

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 Theorem 3.3 and Theorem 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. □

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 Lemma 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 Theorem 3.5 that every solution $x(t)$ of (4.1) is nonoscillatory.