# Existence of the nonoscillatory solutions of higher order neutral dynamic equations on time scales

## Abstract

In this paper, we investigate the existence of the nonoscillatory solutions of the following higher order neutral dynamic equation: $$\{r_{n}(t)[(r_{n-1}(t)(\cdots(r_{1}(t)(x(t)-q(t)x(\tau(t)))^{\Delta })^{\Delta}\cdots)^{\Delta})^{\Delta}]^{\gamma}\}^{\Delta} +f(t,x(\delta (t)))=0$$ for $$t\in \left .[t_{0},\infty) \right ._{\mathbb{T}}$$, and obtain some necessary and sufficient conditions for the existence of nonoscillatory bounded solutions for this equation.

## Introduction

A time scale $$\mathbb{T}$$ is an arbitrary nonempty closed subset of the real numbers. Thus the real numbers $$\mathbb {R}$$, the integers $$\mathbb{Z}$$ and the natural numbers $$\mathbb{N}$$ 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

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

respectively.

In this paper, we investigate the existence of the nonoscillatory solutions of the following higher order neutral dynamic equation:

\begin{aligned}& \bigl\{ r_{n}(t)\bigl[\bigl(r_{n-1}(t) \bigl(\cdots \bigl(r_{1}(t) \bigl(x(t)-q(t)x\bigl(\tau(t)\bigr)\bigr)^{\Delta } \bigr)^{\Delta}\cdots\bigr)^{\Delta}\bigr)^{\Delta}\bigr]^{\gamma}\bigr\} ^{\Delta} \\& \quad{} +f\bigl(t,x\bigl(\delta(t)\bigr)\bigr)=0 \quad \mbox{for } t\in \bigl.[t_{0},\infty)\bigr._{\mathbb{ T}}, \end{aligned}
(1.1)

where γ is the quotient of odd positive integers, $$t_{0}\in {\mathbb{T}}$$, the time scale interval $$\left .[t_{0},\infty) \right ._{\mathbb{ T}}=\{t\in\mathbb{T}:t\geq t_{0}\}$$, $$r_{1}\in C_{\mathrm{rd}}( \left .[t_{0},\infty) \right ._{\mathbb{ T}}, [L,\infty))$$ for some constant $$L>0$$ and $$r_{k}\in C_{\mathrm{rd}}( \left .[t_{0},\infty) \right ._{\mathbb{ T}}, (0,\infty))$$ ($$2\leq k\leq n$$), $$q\in C_{\mathrm{rd}}( \left .[t_{0},\infty) \right ._{\mathbb{ T}}, {\mathbb{ R}})$$, $$\tau, \delta\in C({\mathbb{ T}},{\mathbb { T}})$$ with $$\lim_{t\rightarrow\infty}\tau(t)=\lim_{t\rightarrow\infty}\delta (t)=\infty$$ and $$f\in C( \left .[t_{0},\infty) \right ._{\mathbb{ T}}\times{\mathbb{ R}}, {\mathbb{ R}})$$ satisfying the following conditions:

1. (i)

$$uf(t,u)>0$$ for any $$t\in \left .[t_{0},\infty) \right ._{\mathbb{ T}}$$ and $$u\neq0$$;

2. (ii)

$$f(t,u)$$ is nondecreasing in u for any $$t\in \left .[t_{0},\infty) \right ._{\mathbb{ T}}$$.

In the sequel, write

$$W_{k}\bigl(t,x(t)\bigr)= \left \{ \textstyle\begin{array}{l@{\quad}l} x(t) & \mbox{if } k=0 , \\ r_{k}(t)W^{\triangle}_{k-1}(t,x(t)) & \mbox{if } 1\leq k\leq n-1, \\ r_{n}(t)[W^{\triangle}_{n-1}(t,x(t))]^{\gamma}& \mbox{if } k= n \end{array}\displaystyle \right .$$

and

$$z(t)=x(t)-q(t)x\bigl(\tau(t)\bigr).$$
(1.2)

Then Eq. (1.1) reduces to the equation

$$W_{n}^{\triangle}\bigl(t,z(t)\bigr)+f\bigl(t,x\bigl(\delta(t)\bigr) \bigr)=0.$$
(1.3)

We can suppose that $$\sup{\mathbb{ T}}=\infty$$ since we are interested in the oscillatory behavior of solutions near infinity. We say a nontrivial real-valued function $$x\in C_{\mathrm{rd}}( \left .[T_{x},\infty) \right ._{\mathbb{ T}},{\mathbb{ R}})$$ ($$T_{x}\geq t_{0}$$) to be a solution of Eq. (1.1) if $$W_{n}(t,z(t))\in C^{1}_{\mathrm{rd}}( \left .[T_{x},\infty) \right ._{\mathbb{ T}},{\mathbb{ R}})$$ and satisfies Eq. (1.1) on $$[T_{x},\infty)$$. The solutions vanishing in some neighborhood of infinity will be excluded from our consideration. A solution x of Eq. (1.1) is said to be oscillatory if it is neither eventually positive nor eventually negative, otherwise it is called nonoscillatory.

The theory of time scale was initiated by Hilger’s landmark paper  in order to create a theory that can unify discrete and continuous analysis, which has received a lot of attention. There exists a variety of interesting time scales, and they give rise to many applications (see ). 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 has been a major focus of study. Necessary and sufficient conditions for the existence of a nonoscillatory bounded solution are more rare because it is much more difficult to find necessary and sufficient conditions for a solution of higher order equations.

In a number of papers, there has been much research activity concerning the oscillation and nonoscillation of solutions of various equations on time scales even today, many similar equations on time scales can be found in .

In  Zhu and Wang studied the existence of nonoscillatory solutions to the 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 et al.  studied the asymptotic behavior of delay dynamic equations having the following form:

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

In  Wu et al. investigated the oscillation of the following higher order dynamic equation:

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

## Auxiliary results

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

(H1):

$$\int^{\infty}_{t_{0}}\frac{1}{r_{k}(s)}\triangle s=\int^{\infty}_{t_{0}}[\frac{1}{r_{n}(s)}]^{\frac{1}{\gamma}}\triangle s=\infty$$ ($$1\leq k\leq n-1$$).

(H2):

There exist constants $$\alpha_{1},\beta_{1}\in[0,1)$$ with $$\alpha_{1}+\beta_{1}<1$$ such that $$-\alpha_{1}\leq q(t)\leq\beta_{1}$$ for all $$t\in \left .[t_{0},\infty) \right ._{\mathbb{ T}}$$.

(H3):

There exist constants $$\alpha_{2},\beta_{2}\in(1,\infty)$$ such that $$-\alpha_{2}\leq q(t)\leq-\beta_{2}$$ for all $$t\in \left .[t_{0},\infty) \right ._{\mathbb{ T}}$$.

(H4):

There exist constants $$\alpha_{3},\beta_{3}\in(1,\infty)$$ such that $$\alpha_{3}\leq q(t)\leq\beta_{3}$$ for all $$t\in \left .[t_{0},\infty) \right ._{\mathbb{ T}}$$.

Let $$\mathit{BC}_{\mathrm{rd}}( \left .[t_{0},\infty) \right ._{\mathbb{ T}},{\mathbb{R}})$$ be the Banach space of all bounded rd-continuous functions on $$\left .[t_{0},\infty) \right ._{\mathbb{ T}}$$ with sup norm $$\|x\|=\sup_{t\geq t_{0}}|x(t)|$$. Let $$X\subset \mathit{BC}_{\mathrm{rd}}( \left .[t_{0},\infty) \right ._{\mathbb{ T}},{\mathbb{R}})$$, we say that X is equi-continuous on $$[a,b]_{\mathbb{ T}}$$ if for any given $$\varepsilon>0$$, there exists $$\delta>0$$ such that for any $$x\in X$$ and $$u,v\in [a,b]_{\mathbb{ T}}$$ with $$|u-v|<\delta$$, $$|x(u)-x(v)|<\varepsilon$$. X is said to be uniformly Cauchy if for any given $$\varepsilon>0$$, there exists $$t_{1}>t_{0}$$ such that for any $$x\in X$$, $$|x(u)-x(v)|<\varepsilon$$ for all $$u,v\in \left .[t_{1},\infty) \right ._{\mathbb{ T}}$$. $$U:X\rightarrow \mathit{BC}_{\mathrm{rd}}( \left .[t_{0},\infty) \right ._{\mathbb{ T}},{\mathbb{R}})$$ is called completely continuous if it is continuous and maps bounded sets into relatively compact sets.

### Lemma 2.1



Let $$m\in\mathbf{N}$$. Then

1. (1)

$$\liminf_{t\rightarrow\infty}W_{m}(t,x(t))>0$$ implies $$\lim_{t\rightarrow\infty}W_{i}(t,x(t))=\infty$$ for all $$0\leq i\leq m-1$$;

2. (2)

$$\limsup_{t\rightarrow\infty}W_{m}(t,x(t))<0$$ implies $$\lim_{t\rightarrow\infty}W_{i}(t,x(t))=-\infty$$ for all $$0\leq i\leq m-1$$.

### Lemma 2.2

Let $$x(t)\in{C_{\mathrm{rd}}( \left .[t_{0},\infty) \right ._{\mathbb{ T}},(0,\infty))}$$ be bounded for $$t\in \left .[t_{0},\infty) \right ._{\mathbb{ T}}$$. If $$W_{n}^{\Delta}(t,x(t))<0$$ on $$\left .[t_{1},+\infty) \right ._{\mathbb{ T}}$$ for some $$t_{1}\geq t_{0}$$, then

$$(-1)^{n-k}W_{k}\bigl(t,x(t)\bigr)>0 \quad \textit{for } t\geq t_{1}, k=0,1,2,\ldots,n$$
(2.1)

and

$$\lim_{t\rightarrow\infty}W_{k}\bigl(t,x(t)\bigr)=0, \quad k=0,1,2, \ldots,n.$$
(2.2)

### Proof

Since $$W_{n}^{\Delta}(t,x(t))<0$$, $$W_{n}(t,x(t))$$ is strictly decreasing on $$\left .[t_{1},\infty) \right ._{\mathbb { T}}$$. We claim that $$W_{n}(t,x(t))>0$$ for all $$t\in \left .[t_{1},\infty) \right ._{\mathbb{ T}}$$. If not, there exists $$t_{2}\in \left .[t_{1},\infty) \right ._{\mathbb{ T}}$$ such that, for all $$t\geq t_{2}$$,

$$W_{n}\bigl(t,x(t)\bigr)\leq W_{n}\bigl(t_{2},x(t_{2}) \bigr)< 0.$$

Then by Lemma 2.1 we get $$\lim_{t\rightarrow\infty}x(t)=-\infty$$, which contradicts the fact that $$x(t)$$ is bounded. Then $$W_{n}(t,x(t))>0$$ for all $$t\in \left .[t_{1},\infty) \right ._{\mathbb{ T}}$$, and $$W_{n}^{\Delta}(t,x(t))<0$$. Let

$$\lim_{t\rightarrow\infty}W_{n}\bigl(t,x(t)\bigr)=L_{1} \geq0\quad \mbox{and}\quad W_{n}\bigl(t,x(t)\bigr)\geq L_{1},\quad t\in \bigl.[t_{1},\infty)\bigr._{\mathbb{ T}}.$$

If $$L_{1}>0$$, then we get from Lemma 2.1 that $$\lim_{t\rightarrow\infty}x(t)=\infty$$, which is a contradiction to the boundedness of $$x(t)$$. Therefore $$L_{1}=0$$, i.e.,

$$\lim_{t\rightarrow\infty}W_{n}\bigl(t,x(t)\bigr)=0.$$
(2.3)

Because of $$W_{n}(t,x(t))>0$$, we know $$W_{n-1}(t,x(t))$$ is strictly increasing on $$\left .[t_{1},\infty) \right ._{\mathbb{ T}}$$, this implies that exactly one of the following is true:

(a1):

$$W_{n-1}(t,x(t))<0$$ for $$t\geq t_{1}$$;

(b1):

there exists $$t_{3}\geq t_{1}$$ such that $$W_{n-1}(t,x(t))\geq W_{n-1}(t_{3},x(t_{3}))>0$$ for $$t\geq t_{3}$$.

If (b1) holds, by Lemma 2.1, we obtain

$$\lim_{t\rightarrow\infty}W_{n-2}\bigl(t,x(t)\bigr)=\lim _{t\rightarrow\infty }W_{n-3}\bigl(t,x(t)\bigr)=\cdots=\lim _{t\rightarrow\infty}x(t)=\infty,$$

which contradicts the fact that $$x(t)$$ is bounded. Then $$W_{n-1}(t,x(t))<0$$ for all $$t\in \left .[t_{1},\infty) \right ._{\mathbb{ T}}$$. Thus, we conclude

$$\lim_{t\rightarrow\infty}W_{n-1}\bigl(t,x(t)\bigr)=L_{2} \leq0 \quad \mbox{and}\quad W_{n-1}\bigl(t,x(t)\bigr)\leq L_{2},\quad t\in \bigl.[t_{1},\infty)\bigr._{\mathbb{ T}}.$$

If $$L_{2}<0$$, then by Lemma 2.1 we get $$\lim_{t\rightarrow\infty}x(t)=-\infty$$, which is a contradiction to the boundedness of $$x(t)$$. Therefore $$L_{2}=0$$, i.e.,

$$\lim_{t\rightarrow\infty}W_{n-1}\bigl(t,x(t)\bigr)=0.$$
(2.4)

Since $$W^{\Delta}_{n-2}(t,x(t))=\frac{S_{n-1}(t,x(t))}{r_{n-1}(t)}<0$$, we see $$W_{n-2}(t,x(t))$$ is strictly decreasing, which implies that exactly one of the following is true:

(a2):

$$W_{n-2}(t,x(t))>0$$ for $$t\geq t_{1}$$;

(b2):

there exists $$t_{4}\geq t_{1}$$ such that $$W_{n-2}(t,x(t))\leq W_{n-2}(t_{4},x(t_{4}))<0$$ for $$t\geq t_{4}$$.

If (b2) holds, by Lemma 2.1, we obtain

$$\lim_{t\rightarrow\infty}W_{n-3}\bigl(t,x(t)\bigr)=\lim _{t\rightarrow\infty }W{n-4}\bigl(t,x(t)\bigr)=\cdots=\lim_{t\rightarrow\infty}x(t)=- \infty,$$

which contradicts the fact that $$x(t)$$ is bounded. Then $$W_{n-2}(t,x(t))>0$$ for all $$t\in \left .[t_{1},\infty) \right ._{\mathbb{ T}}$$.

Therefore we can repeat the above argument and show that Lemma 2.2 holds. The proof is completed. □

### Lemma 2.3



Suppose that $$X\subset \mathit{BC}_{\mathrm{rd}}( \left .[t_{0},\infty) \right ._{\mathbb{ T}},{\mathbb{R}})$$ is bounded and uniformly Cauchy. Further, suppose that X is equi-continuous on $$[t_{0},t_{1}]_{\mathbb{ T}}$$ for any $$t_{1}\in \left .[t_{0},\infty) \right ._{\mathbb{ T}}$$. Then X is relatively compact.

### Lemma 2.4



Suppose that X is a Banach space and Ω is a bounded, convex and closed subset of X. Further, suppose that there exist two operators U and $$V:\Omega\rightarrow X$$ such that

1. (i)

$$Ux +Vy\in\Omega$$ for all $$x,y\in\Omega$$;

2. (ii)

U is a contraction mapping;

3. (iii)

V is completely continuous.

Then $$U +V$$ has a fixed point in Ω.

## Main results

Now, we state and prove our main results.

### Theorem 3.1

Assume that (H1) and (H2) hold. Then Eq. (1.1) has a nonoscillatory bounded solution $$x(t)$$ with $$\lim\inf_{t\rightarrow\infty} |x(t)|>0$$ if and only if there exists some constant $$M\neq0$$ such that

\begin{aligned}& \int_{t_{0}}^{\infty}\frac{1}{r_{1}(u_{1})}\int _{u_{1}}^{\infty}\frac {1}{r_{2}(u_{2})}\cdots\int _{u_{n-2}}^{\infty}\frac {1}{r_{n-1}(u_{n-1})}\int_{u_{n-1}}^{\infty}A(u_{n})\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1} \\& \quad < \infty, \end{aligned}
(3.1)

where $$A(u_{n})= [\frac{1}{r_{n}(u_{n})}\int_{u_{n}}^{\infty}f(s,|M|)\Delta s ]^{\frac{1}{\gamma}}$$.

### Proof

Necessity. Assume that Eq. (1.1) has a nonoscillatory bounded solution $$x(t)$$ on $$\left .[t_{0},\infty) \right ._{\mathbb{ T}}$$ with $$\lim\inf_{t\rightarrow\infty} |x(t)|>0$$. Without loss of generality, we assume that there is a constant $$M>0$$ and some $$t_{1}\geq t_{0}$$ such that $$x(t)>M$$ and $$x(\delta(t))>M$$ for $$t\geq t_{1}$$. By assumption that $$x(t)$$ is bounded and condition (H2), we see that $$z(t)$$ is bounded and $$W_{n}^{\Delta}(t,z(t))=-f(t,x(\delta(t)))<0$$. Thus, by Lemma 2.2 we know that there exists $$t_{2}\geq t_{1}$$ such that

$$(-1)^{n-k}W_{k}\bigl(t,z(t)\bigr)>0 \quad \mbox{for } t\geq t_{2} \mbox{ and } 0\leq k\leq n.$$
(3.2)

Integrating Eq. (1.3) from t ($$\geq t_{2}$$) to ∞, we have

$$\int_{t}^{\infty}W_{n}^{\Delta}\bigl(s,z(s)\bigr)\Delta s=-\int_{t}^{\infty}f \bigl(s,x\bigl(\delta(s)\bigr)\bigr)\Delta s.$$

That is,

\begin{aligned} r_{n}(t)\bigl[W_{n-1}^{\Delta}\bigl(t,z(t)\bigr) \bigr]^{\gamma} =&W_{n}\bigl(t,z(t)\bigr)= \int_{t}^{\infty}f\bigl(s,x\bigl(\delta(s)\bigr)\bigr)\Delta s \\ \geq&\int_{t}^{\infty}f(s,M)\Delta s \quad ( t\geq t_{2}). \end{aligned}
(3.3)

From (3.2) and (3.3), we have that for $$t\geq t_{2}$$,

\begin{aligned}& \int_{t}^{\infty}\frac{1}{r_{1}(u_{1})}\int _{u_{1}}^{\infty}\frac{1}{r_{2}(u_{2})}\cdots\int _{u_{n-2}}^{\infty}\frac {1}{r_{n-1}(u_{n-1})}\int_{u_{n-1}}^{\infty}A(u_{n})\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1} \\& \quad \leq \int_{t}^{\infty}\frac{1}{r_{1}(u_{1})}\int _{u_{1}}^{\infty}\frac{1}{r_{2}(u_{2})}\cdots \int _{u_{n-2}}^{\infty}\frac{1}{r_{n-1}(u_{n-1})} \\& \qquad {}\times\int_{u_{n-1}}^{\infty}W_{n-1}^{\Delta}\bigl(u_{n},z(u_{n})\bigr) \Delta u_{n} \Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1} \\& \quad = \int_{t}^{\infty}\frac{1}{r_{1}(u_{1})}\int _{u_{1}}^{\infty}\frac {1}{r_{2}(u_{2})}\cdots\int _{u_{n-2}}^{\infty}\frac{1}{r_{n-1}(u_{n-1})} (-1)^{n-(n-1)} \\& \qquad {}\times W_{n-1} \bigl(u_{n-1},z(u_{n-1})\bigr)\Delta u_{n-1}\cdots \Delta u_{2}\Delta u_{1} \\& \quad = \int_{t}^{\infty}\frac{1}{r_{1}(u_{1})}\int _{u_{1}}^{\infty}\frac {1}{r_{2}(u_{2})}\cdots\int _{u_{n-2}}^{\infty}(-1)^{n-(n-1)}W_{n-2}^{\Delta}\bigl(u_{n-1},z(u_{n-1})\bigr)\Delta u_{n-1}\cdots \Delta u_{2}\Delta u_{1} \\& \quad = \int_{t}^{\infty}\frac{1}{r_{1}(u_{1})}\int _{u_{1}}^{\infty}\frac {1}{r_{2}(u_{2})}\cdots\int _{u_{n-3}}^{\infty}\frac {1}{r_{n-2}(u_{n-2})}(-1)^{n-(n-2)} \\& \qquad {}\times W_{n-2} \bigl(u_{n-2},z(u_{n-2})\bigr)\Delta u_{n-2}\cdots \Delta u_{2}\Delta u_{1} \\& \qquad \cdots \\& \quad = \int_{t}^{\infty}\frac {1}{r_{1}(u_{1})}(-1)^{n-1}W_{1} \bigl(u_{1},z(u_{1})\bigr)\Delta u_{1} \\& \quad = \int_{t}^{\infty}(-1)^{n-1}z^{\Delta}(u_{1}) \Delta u_{1} \\& \quad = (-1)^{n-1}z(u_{1})|_{t}^{\infty}< \infty. \end{aligned}

By using the boundedness of $$z(t)$$, we see that (3.1) holds.

Sufficiency. Suppose that there exists some constant $$M>0$$ such that

$$\int_{t_{0}}^{\infty}\frac{1}{r_{1}(u_{1})}\int _{u_{1}}^{\infty}\frac {1}{r_{2}(u_{2})}\cdots\int _{u_{n-2}}^{\infty}\frac {1}{r_{n-1}(u_{n-1})}\int_{u_{n-1}}^{\infty}A(u_{n})\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1}< \infty.$$

Then we may choose $$t_{1}\geq t_{0}$$ such that

\begin{aligned}& \int_{t}^{\infty}\frac{1}{r_{1}(u_{1})}\int _{u_{1}}^{\infty}\frac {1}{r_{2}(u_{2})}\cdots\int _{u_{n-2}}^{\infty}\frac {1}{r_{n-1}(u_{n-1})}\int_{u_{n-1}}^{\infty}A(u_{n})\Delta u_{n}\cdots\Delta u_{2}\Delta u_{1} \\& \quad < \frac{(1-\beta_{1})(1-\beta_{1}-\alpha_{1})}{2}M \end{aligned}

and $$\min\{\delta(t),\tau(t)\}\geq t_{0}$$ for $$t\geq t_{1}$$. Let

$$\Omega=\bigl\{ x\in \mathit{BC}_{\mathrm{rd}}\bigl(\bigl.[t_{0},\infty )\bigr._{\mathbb{ T}},{\mathbb{ R}}\bigr):\beta _{1}(1- \beta_{1}-\alpha_{1})M\leq x (t)\leq M \mbox{ for } t\geq t_{0}\bigr\} .$$

It is easy to verify that Ω is a bounded, convex and closed subset of $$\mathit{BC}_{\mathrm{rd}}( \left .[t_{0},\infty) \right ._{\mathbb{ T}},{\mathbb{ R}})$$.

Now we define two operators U and $$V:\Omega\rightarrow \mathit{BC}_{\mathrm{rd}}( \left .[t_{0},\infty) \right ._{\mathbb{ T}},{\mathbb{ R}})$$ as follows:

$$(Ux) (t)=q\bigl(t^{*}\bigr)x\bigl(\tau\bigl(t^{*}\bigr)\bigr)$$

and

\begin{aligned} (Vx) (t) =& \frac{(1-\beta_{1})(1+\alpha_{1}+\beta_{1})}{2}M \\ &{}+(-1)^{n}\int_{t^{\ast }}^{\infty}\frac{1}{r_{1}(u_{1})}\int_{u_{1}}^{\infty}\frac{1}{r_{2}(u_{2})} \cdots \int_{u_{n-2}}^{\infty}\frac{1}{r_{n-1}(u_{n-1})} \\ &{}\times\int _{u_{n-1}}^{\infty}A(u_{n},x)\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1}, \end{aligned}

where $$t^{*}=\max\{t,t_{1}\}$$, $$A(u_{n},x)= [\frac{1}{r_{n}(u_{n})}\int_{u_{n}}^{\infty}f(s,x(\delta(s)))\Delta s ]^{\frac{1}{\gamma}}$$. Now we show that U and V satisfy the conditions in Lemma 2.4.

(1) We will prove that $$Ux+Vy\in\Omega$$ for any $$x,y\in \Omega$$. In fact, for any $$x,y\in\Omega$$ and $$t\geq t_{0}$$, $$x(t),y(t)\in[\beta_{1}(1-\beta_{1}-\alpha_{1})M, M]$$ and

\begin{aligned} (Ux) (t)+(Vy) (t) =& \frac{(1-\beta_{1})(1+\alpha_{1}+\beta_{1})}{2}M+q\bigl(t^{*}\bigr)x\bigl(\tau \bigl(t^{*} \bigr)\bigr) \\ &{}+(-1)^{n}\int_{t^{\ast}}^{\infty}\frac{1}{r_{1}(u_{1})}\int_{u_{1}}^{\infty}\frac{1}{r_{2}(u_{2})} \cdots \int_{u_{n-2}}^{\infty}\frac{1}{r_{n-1}(u_{n-1})} \\ &{}\times\int _{u_{n-1}}^{\infty}A(u_{n},x)\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1} \\ \leq& \frac{(1-\beta_{1})(1+\alpha_{1}+\beta_{1})}{2}M+\beta_{1}M+\frac {(1-\beta_{1})(1-\beta_{1}-\alpha_{1})}{2}M \\ =&M \end{aligned}

and

\begin{aligned} (Ux) (t)+(Vy) (t) =&\frac{(1-\beta_{1})(1+\alpha_{1}+\beta _{1})}{2}M+q\bigl(t^{*}\bigr)x\bigl(\tau\bigl(t^{*} \bigr)\bigr) \\ &{}+(-1)^{n}\int_{t^{\ast}}^{\infty}\frac{1}{r_{1}(u_{1})}\int_{u_{1}}^{\infty}\frac{1}{r_{2}(u_{2})} \cdots \int_{u_{n-2}}^{\infty}\frac{1}{r_{n-1}(u_{n-1})} \\ &{}\times\int _{u_{n-1}}^{\infty}A(u_{n},x)\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1} \\ \geq&\frac{(1-\beta_{1})(1+\alpha_{1}+\beta_{1})}{2}M-\alpha_{1}M-\frac {(1-\beta_{1})(1-\beta_{1}-\alpha_{1})}{2}M \\ =&\beta_{1}(1-\beta_{1}-\alpha_{1})M, \end{aligned}

which implies that $$Ux+Vy\in\Omega$$ for any $$x,y\in\Omega$$.

(2) We will show that U is a contraction mapping. Indeed, for any $$x,y\in\Omega$$ and $$t\geq t_{0}$$, we have

\begin{aligned} \bigl\vert (Ux) (t)-(Uy) (t)\bigr\vert =&\bigl\vert q\bigl(t^{*}\bigr)x \bigl(\tau\bigl(t^{*}\bigr)\bigr)-q\bigl(t^{*}\bigr)y\bigl(\tau \bigl(t^{*}\bigr)\bigr) \bigr\vert \\ \leq& \max\{\alpha_{1},\beta_{1}\}\|x-y\|. \end{aligned}

Therefore, we conclude

$$\|Ux-Uy\|\leq\max\{\alpha_{1},\beta_{1}\}\|x-y\|,$$

then U is a contraction mapping.

(3) We will show that V is a completely continuous mapping.

(i) By the proof of (1), we see that $$\beta_{1}(1-\beta_{1}-\alpha_{1})M\leq(Vx)(t)\leq M$$ for $$t\in \left .[t_{0},\infty) \right ._{\mathbb{ T}}$$. That is, $$V\Omega\subset\Omega$$.

(ii) We consider the continuity of V. Let $$x_{n}\in\Omega$$ and $$\|x_{n}-x\|\rightarrow0$$ as $$n\rightarrow\infty$$, then $$x\in \Omega$$ and $$|x_{n}(t)-x(t)|\rightarrow0$$ as $$n\rightarrow\infty$$ for any $$t\in \left .[t_{0},\infty) \right ._{\mathbb{ T}}$$. Consequently, for any $$s\in \left .[t_{1},\infty) \right ._{\mathbb{ T}}$$, we have

\begin{aligned}& \lim_{n\rightarrow\infty}\biggl\vert \frac{1}{r_{1}(u_{1})}\int _{u_{1}}^{\infty}\frac{1}{r_{2}(u_{2})}\cdots\int _{u_{n-2}}^{\infty}\frac {1}{r_{n-1}(u_{n-1})} \\& \quad {}\times \int_{u_{n-1}}^{\infty}\bigl[A(u_{n},x_{n})-A(u_{n},x) \bigr] \Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\biggr\vert =0. \end{aligned}

Since

\begin{aligned}& \biggl\vert \frac{1}{r_{1}(u_{1})}\int_{u_{1}}^{\infty}\frac {1}{r_{2}(u_{2})}\cdots\int_{u_{n-2}}^{\infty}\frac {1}{r_{n-1}(u_{n-1})}\int_{u_{n-1}}^{\infty}\bigl[A(u_{n},x_{n})-A(u_{n},x)\bigr]\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\biggr\vert \\& \quad \leq 2\frac{1}{r_{1}(u_{1})}\int_{u_{1}}^{\infty}\frac {1}{r_{2}(u_{2})}\cdots\int_{u_{n-2}}^{\infty}\frac {1}{r_{n-1}(u_{n-1})}\int_{u_{n-1}}^{\infty}A(u_{n}) \Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2} \end{aligned}

and for any $$t\in \left .[t_{0},\infty) \right ._{\mathbb{ T}}$$,

\begin{aligned}& \bigl\vert (Vx_{n}) (t)-(Vx) (t)\bigr\vert \\& \quad \leq \int_{t_{1}}^{\infty}\frac{1}{r_{1}(u_{1})}\int _{u_{1}}^{\infty}\frac {1}{r_{2}(u_{2})}\cdots\int _{u_{n-2}}^{\infty}\frac {1}{r_{n-1}(u_{n-1})} \\& \qquad {}\times \int_{u_{n-1}}^{\infty}\bigl\vert A(u_{n},x_{n})-A(u_{n},x) \bigr\vert \Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1}, \end{aligned}

we have

\begin{aligned}& \|Vx_{n}-Vx\| \\& \quad \leq \int_{t_{1}}^{\infty}\frac{1}{r_{1}(u_{1})}\int _{u_{1}}^{\infty}\frac {1}{r_{2}(u_{2})}\cdots\int _{u_{n-2}}^{\infty}\frac {1}{r_{n-1}(u_{n-1})} \\& \qquad {}\times \int_{u_{n-1}}^{\infty}\bigl\vert A(u_{n},x_{n})-A(u_{n},x)\bigr\vert \Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1}. \end{aligned}

By Chapter 5 in , we see that the Lebesgue dominated convergence theorem satisfies the integral on time scales. Then

$$\lim_{n\rightarrow\infty}\|Vx_{n}-Vx\|=0,$$

which implies that V is continuous on Ω.

(iii) We show that VΩ is uniformly Cauchy. In fact, for any $$\varepsilon>0$$, let us take $$t_{2}>t_{1}$$ so that

$$\int_{t_{2}}^{\infty}\frac{1}{r_{1}(u_{1})}\int _{u_{1}}^{\infty}\frac {1}{r_{2}(u_{2})}\cdots\int _{u_{n-2}}^{\infty}\frac {1}{r_{n-1}(u_{n-1})}\int_{u_{n-1}}^{\infty}A(u_{n})\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1}< \varepsilon.$$

Then, for any $$x\in\Omega$$ and $$u,v\in \left .[t_{2},\infty) \right ._{\mathbb{ T}}$$, we have

$$\bigl\vert (Vx) (u)-(Vx) (v)\bigr\vert < 2\varepsilon,$$

which implies that VΩ is uniformly Cauchy.

(iv) We show that VΩ is equi-continuous on $$[t_{0},t_{2}]_{\mathbb{ T}}$$ for any $$t_{2}\in \left .[t_{0},\infty) \right ._{\mathbb{ T}}$$. Without loss of generality, we assume $$t_{2}\geq t_{1}$$. Note $$L\leq r_{1}(t)$$. For any $$\varepsilon>0$$, choose $$\delta=L\varepsilon/(1+\int_{t_{0}}^{\infty}\frac{1}{r_{2}(u_{2})}\cdots \int_{u_{n-2}}^{\infty}\frac{1}{r_{n-1}(u_{n-1})} \int_{u_{n-1}}^{\infty}A(u_{n})\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2})$$, then when $$u,v\in [t_{0},t_{2}]$$ with $$|u-v|<\delta$$, for any $$x\in\Omega$$, we have

\begin{aligned}& \bigl\vert (Vx) (u)-(Vx) (v)\bigr\vert \\& \quad = \biggl\vert \int_{u^{\ast}}^{\infty}\frac{1}{r_{1}(u_{1})}\int_{u_{1}}^{\infty}\frac{1}{r_{2}(u_{2})} \cdots\int_{u_{n-2}}^{\infty}\frac {1}{r_{n-1}(u_{n-1})}\int _{u_{n-1}}^{\infty}A(u_{n})\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1} \\& \qquad {} - \int_{v^{\ast}}^{\infty}\frac{1}{r_{1}(u_{1})}\int _{u_{1}}^{\infty}\frac {1}{r_{2}(u_{2})}\cdots\int _{u_{n-2}}^{\infty}\frac {1}{r_{n-1}(u_{n-1})}\int_{u_{n-1}}^{\infty}A(u_{n})\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1}\biggr\vert \\& \quad = \biggl\vert \int_{u^{\ast}}^{v^{\ast}} \frac{1}{r_{1}(u_{1})}\int_{u_{1}}^{\infty}\frac{1}{r_{2}(u_{2})} \cdots\int_{u_{n-2}}^{\infty}\frac {1}{r_{n-1}(u_{n-1})}\int _{u_{n-1}}^{\infty}A(u_{n})\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1}\biggr\vert \\& \quad \leq \biggl\vert \frac{1}{L}\int_{u^{\ast}}^{v^{\ast}} \int_{t_{0}}^{\infty}\frac {1}{r_{2}(u_{2})}\cdots\int _{u_{n-2}}^{\infty}\frac {1}{r_{n-1}(u_{n-1})}\int_{u_{n-1}}^{\infty}A(u_{n})\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1}\biggr\vert \\& \quad \leq \frac{1}{L}\bigl\vert u^{\ast}-v^{\ast}\bigr\vert \int_{t_{0}}^{\infty}\frac {1}{r_{2}(u_{2})}\cdots\int _{u_{n-2}}^{\infty}\frac {1}{r_{n-1}(u_{n-1})}\int_{u_{n-1}}^{\infty}A(u_{n})\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2} \\& \quad < \varepsilon, \end{aligned}

which implies that VΩ is equi-continuous on $$[t_{0},t_{2}]_{\mathbb{ T}}$$ for any $$t_{2}\in \left .[t_{0},\infty) \right ._{\mathbb{ T}}$$.

We see from Lemma 2.3 that V is a completely continuous mapping. By Lemma 2.4 it follows that there exists $$x\in\Omega$$ such that $$(U +V)x =x$$, which is the desired bounded solution of Eq. (1.1) with $$\lim\inf_{t\rightarrow\infty} |x(t)|>0$$. The proof is completed. □

### Theorem 3.2

Assume that (H1) and (H3) hold, and that τ has the inverse $$\tau^{-1}\in C({\mathbb{ T}},{\mathbb{ T}})$$. Then Eq. (1.1) has a nonoscillatory bounded solution $$x(t)$$ with $$\lim\inf_{t\rightarrow\infty} |x(t)|>0$$ if and only if there exists some constant $$M\neq0$$ such that (3.1) holds.

### Proof

The proof of necessity is similar to that of Theorem 3.1.

Sufficiency. Suppose that there exists some constant $$M>0$$ such that

$$\int_{t_{0}}^{\infty}\frac{1}{r_{1}(u_{1})}\int _{u_{1}}^{\infty}\frac {1}{r_{2}(u_{2})}\cdots\int _{u_{n-2}}^{\infty}\frac {1}{r_{n-1}(u_{n-1})}\int_{u_{n-1}}^{\infty}A(u_{n})\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1}< \infty.$$

Then we may choose $$t_{1}\geq t_{0}$$ such that

\begin{aligned}& \int_{t}^{\infty}\frac{1}{r_{1}(u_{1})}\int _{u_{1}}^{\infty}\frac {1}{r_{2}(u_{2})}\cdots\int _{u_{n-2}}^{\infty}\frac {1}{r_{n-1}(u_{n-1})}\int_{u_{n-1}}^{\infty}A(u_{n})\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1} \\& \quad < \frac{\beta_{2}(\beta_{2}-1)}{2(\beta_{2}+1)}M \end{aligned}

and $$\min\{\delta(\tau^{-1}(t)),\tau^{-1}(t)\}\geq t_{0}$$ for $$t\geq t_{1}$$. Let

$$\Omega=\biggl\{ x\in \mathit{BC}_{\mathrm{rd}}\bigl(\bigl.[t_{0},\infty )\bigr._{\mathbb{ T}},{\mathbb{R}}\bigr):\frac {(\beta_{2}-1)}{\alpha_{2}(\beta_{2}+1)}M\leq x (t)\leq M \mbox{ for } t\geq t_{0}\biggr\} .$$

It is easy to verify that Ω is a bounded, convex and closed subset of $$\mathit{BC}_{\mathrm{rd}}( \left .[t_{0},\infty) \right ._{\mathbb{ T}},{\mathbb{R}})$$.

Now we define two operators U and $$V:\Omega\rightarrow \mathit{BC}_{\mathrm{rd}}( \left .[t_{0},\infty) \right ._{\mathbb{ T}},{\mathbb{ R}})$$ as follows:

$$(Ux) (t)=\frac{x(\tau^{-1}(t^{*}))}{q(\tau^{-1}(t^{*}))}+\frac{(\beta _{2}+3)\beta_{2}M}{-2(\beta_{2}+1)q(\tau^{-1}(t^{*}))}$$

and

\begin{aligned} (Vx) (t) =&\frac{1}{q(\tau^{-1}(t^{*}))}(-1)^{n}\int_{\tau^{-1}(t^{\ast })}^{\infty}\frac{1}{r_{1}(u_{1})}\int_{u_{1}}^{\infty}\frac {1}{r_{2}(u_{2})} \cdots \int_{u_{n-2}}^{\infty}\frac{1}{r_{n-1}(u_{n-1})} \\ &{}\times\int _{u_{n-1}}^{\infty}A(u_{n},x)\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1}, \end{aligned}

where $$t^{*}=\max\{t,t_{1}\}$$ for any $$t\in \left .[t_{0},\infty) \right ._{\mathbb{ T}}$$. In order to prove the theorem, we will show that U and V satisfy the conditions in Lemma 2.4.

We first show that $$Ux+Vy\in\Omega$$ for any $$x,y\in\Omega$$. In fact, for any $$x,y\in\Omega$$ and $$t\geq t_{0}$$, $$x(t),y(t)\in [\frac{\beta_{2}-1}{\alpha_{2}(\beta_{2}+1)}M, M]$$ and

\begin{aligned} (Ux) (t)+(Vy) (t) =& \frac{x(\tau^{-1}(t^{\ast}))}{q(\tau^{-1}(t^{\ast }))}+\frac{1}{-q(\tau^{-1}(t^{\ast}))} \biggl[ \frac{(\beta_{2}+3)\beta_{2}}{2(\beta_{2}+1)}M \\ & {}+(-1)^{n+1}\int_{\tau^{-1}(t^{\ast})}^{\infty}\frac{1}{r_{1}(u_{1})} \int_{u_{1}}^{\infty}\frac{1}{r_{2}(u_{2})} \cdots \int_{u_{n-2}}^{\infty}\frac{1}{r_{n-1}(u_{n-1})} \\ & {}\times\int _{u_{n-1}}^{\infty}A(u_{n},x)\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1} \biggr] \\ \leq& \frac{1}{\beta_{2}} \biggl[\frac{(\beta_{2}+3)\beta_{2}}{2(\beta _{2}+1)}M+\frac{\beta_{2}(\beta_{2}-1)}{2(\beta_{2}+1)}M \biggr] \\ =& M \end{aligned}

and

\begin{aligned} (Ux) (t)+(Vy) (t) =& \frac{x(\tau^{-1}(t^{\ast}))}{q(\tau^{-1}(t^{\ast }))}+\frac{1}{-q(\tau^{-1}(t^{\ast}))} \biggl[ \frac{(\beta_{2}+3)\beta_{2}}{2(\beta_{2}+1)}M \\ & {}+(-1)^{n+1}\int_{\tau^{-1}(t^{\ast})}^{\infty}\frac{1}{r_{1}(u_{1})} \int_{u_{1}}^{\infty}\frac{1}{r_{2}(u_{2})} \cdots \int_{u_{n-2}}^{\infty}\frac{1}{r_{n-1}(u_{n-1})} \\ & {}\times\int _{u_{n-1}}^{\infty}A(u_{n},x)\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1} \biggr] \\ \geq& \frac{1}{\alpha_{2}} \biggl[\frac{(\beta_{2}+3)\beta_{2}}{2(\beta _{2}+1)}M-M-\frac{\beta_{2}(\beta_{2}-1)}{2(\beta_{2}+1)}M \biggr] \\ =& \frac{\beta_{2}-1}{\alpha_{2}(\beta_{2}+1)}M, \end{aligned}

and $$|(Vx)(t)|\leq \frac{\beta_{2}(\beta_{2}-1)}{2(\beta_{2}+1)}M$$, which means that $$Ux+Vy\in\Omega$$ for any $$x,y\in\Omega$$ and VΩ is uniformly bounded.

Now we show that VΩ is equi-continuous on $$[t_{0},t_{2}]_{\mathbb{ T}}$$ for any $$t_{2}\in \left .[t_{0},\infty) \right ._{\mathbb{ T}}$$. Without loss of generality, we assume $$t_{2}\geq t_{1}$$. Since $$1/q(\tau^{-1}(t))$$, $$\tau^{-1}(t)$$ are continuous on $$[t_{0},t_{2}]_{\mathbb{ T}}$$, so they are uniformly continuous on $$[t_{0},t_{2}]_{\mathbb{ T}}$$. For any $$\varepsilon>0$$, choose $$\delta>0$$ such that when $$u,v\in[t_{0},t_{2}]_{\mathbb{ T}}$$ with $$|u-v|<\delta$$, we have

$$\biggl\vert \frac{1}{q(\tau^{-1}(u))}-\frac{1}{q(\tau^{-1}(v))}\biggr\vert < \frac{\varepsilon}{1+A}$$

and

$$\bigl\vert \tau^{-1}(u)-\tau^{-1}(v)\bigr\vert < \frac{L\varepsilon}{1+B},$$

where

\begin{aligned} \begin{aligned} &A=\int_{t_{0}}^{\infty}\frac{1}{r_{1}(u_{1})}\int _{u_{1}}^{\infty}\frac{1}{r_{2}(u_{2})}\cdots \int _{u_{n-2}}^{\infty}\frac{1}{r_{n-1}(u_{n-1})}\int_{u_{n-1}}^{\infty}A(u_{n},x)\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1}, \\ &B=\int_{t_{0}}^{\infty}\frac{1}{r_{2}(u_{2})}\cdots \int _{u_{n-2}}^{\infty}\frac{1}{r_{n-1}(u_{n-1})}\int_{u_{n-1}}^{\infty}A(u_{n},x)\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}. \end{aligned} \end{aligned}

Then, we obtain that for any $$x\in\Omega$$,

\begin{aligned}& \bigl\vert (Vx) (u)-(Vx) (v)\bigr\vert \\& \quad = \biggl\vert \frac{1}{q(\tau^{-1}(u^{*}))}\int_{\tau^{-1}(u^{\ast})}^{\infty}\frac{1}{r_{1}(u_{1})}\int_{u_{1}}^{\infty}\frac{1}{r_{2}(u_{2})} \cdots \int_{u_{n-1}}^{\infty}A(u_{n},x)\Delta u_{n}\cdots\Delta u_{2}\Delta u_{1} \\& \qquad {} -\frac{1}{q(\tau^{-1}(v^{*}))}\int_{\tau^{-1}(v^{\ast})}^{\infty}\frac{1}{r_{1}(u_{1})}\int_{u_{1}}^{\infty}\frac{1}{r_{2}(u_{2})} \cdots \int_{u_{n-1}}^{\infty}A(u_{n},x)\Delta u_{n}\cdots\Delta u_{2}\Delta u_{1}\biggr\vert \\& \quad \leq \biggl\vert \biggl[\frac{1}{q(\tau^{-1}(u^{*}))}-\frac{1}{q(\tau ^{-1}(v^{*}))}\biggr] \\& \qquad {}\times\int _{\tau^{-1}(u^{\ast})}^{\infty}\frac {1}{r_{1}(u_{1})}\int _{u_{1}}^{\infty}\frac{1}{r_{2}(u_{2})}\cdots\int _{u_{n-1}}^{\infty}A(u_{n},x)\Delta u_{n}\cdots\Delta u_{2}\Delta u_{1}\biggr\vert \\& \qquad {} +\biggl\vert \frac{1}{q(\tau^{-1}(v^{*}))}\int_{\tau^{-1}(u^{\ast})}^{\tau ^{-1}(v^{\ast}) } \frac{1}{r_{1}(u_{1})}\int_{u_{1}}^{\infty}\frac {1}{r_{2}(u_{2})} \cdots\int_{u_{n-1}}^{\infty}A(u_{n},x)\Delta u_{n}\cdots\Delta u_{2}\Delta u_{1}\biggr\vert \\& \quad \leq \biggl\vert \biggl[\frac{1}{q(\tau^{-1}(u^{*}))}-\frac{1}{q(\tau ^{-1}(v^{*}))}\biggr] \\& \qquad {}\times\int _{t_{0}}^{\infty}\frac{1}{r_{1}(u_{1})}\int_{u_{1}}^{\infty}\frac{1}{r_{2}(u_{2})}\cdots\int_{u_{n-1}}^{\infty}A(u_{n},x)\Delta u_{n}\cdots\Delta u_{2}\Delta u_{1}\biggr\vert \\& \qquad {} +\biggl\vert \frac{1}{q(\tau^{-1}(v^{*}))}\frac{1}{L}\int _{\tau^{-1}(u^{\ast })}^{\tau^{-1}(v^{\ast}) } \int_{t_{0}}^{\infty}\frac {1}{r_{2}(u_{2})}\cdots\int_{u_{n-1}}^{\infty}A(u_{n},x)\Delta u_{n}\cdots\Delta u_{2}\Delta u_{1}\biggr\vert \\& \quad \leq \varepsilon+\frac{1}{L}\bigl\vert \tau^{-1} \bigl(u^{*}\bigr)-\tau^{-1}\bigl(v^{*}\bigr)\bigr\vert \int _{t_{0}}^{\infty}\frac{1}{r_{2}(u_{2})}\cdots\int _{u_{n-1}}^{\infty}A(u_{n},x)\Delta u_{n}\cdots\Delta u_{2} \\& \quad < 2\varepsilon, \end{aligned}

which implies that VΩ is equi-continuous on $$[t_{0},t_{2}]_{\mathbb{ T}}$$ for any $$t_{2}\in \left .[t_{0},\infty) \right ._{\mathbb{ T}}$$. The rest of the proof is similar to that of Theorem 3.1. The proof is completed. □

### Theorem 3.3

Assume that (H1) and (H4) hold, and that τ has the inverse $$\tau^{-1}\in C({\mathbb{ T}},{\mathbb{ T}})$$. Then Eq. (1.1) has a nonoscillatory bounded solution $$x(t)$$ with $$\lim\inf_{t\rightarrow\infty} |x(t)|>0$$ if and only if there exists some constant $$M\neq0$$ such that (3.1) holds.

### Proof

The proof of necessity is similar to that of Theorem 3.1.

Sufficiency. Suppose that there exists some constant $$M>0$$ such that

$$\int_{t_{0}}^{\infty}\frac{1}{r_{1}(u_{1})}\int _{u_{1}}^{\infty}\frac {1}{r_{2}(u_{2})}\cdots\int _{u_{n-2}}^{\infty}\frac {1}{r_{n-1}(u_{n-1})}\int_{u_{n-1}}^{\infty}A(u_{n})\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1}< \infty.$$

Then we may choose $$t_{1}\geq t_{0}$$ such that

$$\int_{\tau^{-1}(t)}^{\infty}\frac{1}{r_{1}(u_{1})}\int _{u_{1}}^{\infty}\frac{1}{r_{2}(u_{2})}\cdots\int _{u_{n-2}}^{\infty}\frac {1}{r_{n-1}(u_{n-1})}\int_{u_{n-1}}^{\infty}A(u_{n})\Delta u_{n}\cdots\Delta u_{2}\Delta u_{1}< \frac{\alpha_{3}-1}{3}M,$$

and $$\min\{\delta(\tau^{-1}(t)),\tau^{-1}(t)\}\geq t_{0}$$ for $$t\geq t_{1}$$. Let

$$\Omega=\biggl\{ x\in \mathit{BC}_{\mathrm{rd}}\bigl(\bigl.[t_{0},\infty )\bigr._{\mathbb{ T}},{\mathbb{R}}\bigr):\frac {\alpha_{3}-1}{3(\beta_{3}-1)}M\leq x (t)\leq M \mbox{ for } t\geq t_{0}\biggr\} .$$

It is easy to verify that Ω is a bounded, convex and closed subset of $$\mathit{BC}_{\mathrm{rd}}( \left .[t_{0},\infty) \right ._{\mathbb{ T}},{\mathbb{ R}})$$.

Now we define two operators U and $$V:\Omega\rightarrow \mathit{BC}_{\mathrm{rd}}( \left .[t_{0},\infty) \right ._{\mathbb{ T}},{\mathbb{R}})$$ as follows:

$$(Ux) (t)=\frac{x(\tau^{-1}(t^{*}))}{q(\tau^{-1}(t^{*}))}+\frac{2(\alpha _{3}-1)M}{3q(\tau^{-1}(t^{*}))}$$

and

\begin{aligned} (Vx) (t) =& \frac{1}{q(\tau^{-1}(t^{*}))}(-1)^{n}\int_{\tau^{-1}(t^{\ast})}^{\infty}\frac{1}{r_{1}(u_{1})}\int_{u_{1}}^{\infty}\frac{1}{r_{2}(u_{2})} \cdots \int_{u_{n-2}}^{\infty}\frac{1}{r_{n-1}(u_{n-1})} \\ &{}\times\int _{u_{n-1}}^{\infty}A(u_{n},x)\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1}, \end{aligned}

where $$t^{*}=\max\{t,t_{1}\}$$ for any $$t\in \left .[t_{0},\infty) \right ._{\mathbb{ T}}$$. Now we show that U and V satisfy the conditions in Lemma 2.4.

We will show that $$Ux+Vy\in\Omega$$ for any $$x,y\in\Omega$$. In fact, for any $$x,y\in\Omega$$ and $$t\geq t_{0}$$, $$x(t), y(t)\in [(\alpha_{3}-1)M/3(\beta_{3}-1), M]$$ and

\begin{aligned} (Ux) (t)+(Vy) (t) =&\frac{x(\tau^{-1}(t^{\ast}))}{q(\tau^{-1}(t^{\ast }))}+\frac{1}{q(\tau^{-1}(t^{\ast}))} \biggl[ \frac{2\alpha_{3}}{3}M \\ &{}+(-1)^{n+1}\int_{\tau^{-1}(t^{\ast})}^{\infty}\frac{1}{r_{1}(u_{1})}\int_{u_{1}}^{\infty}\frac{1}{r_{2}(u_{2})} \cdots \int_{u_{n-2}}^{\infty}\frac{1}{r_{n-1}(u_{n-1})} \\ &{}\times\int _{u_{n-1}}^{\infty}A(u_{n},x)\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1} \biggr] \\ \leq&\frac{1}{\alpha_{3}} \biggl[\frac{2(\alpha_{3}-1)}{3}M+M+\frac{\alpha _{3}-1}{3}M \biggr] \\ =&M \end{aligned}

and

\begin{aligned} (Ux) (t)+(Vy) (t) =&\frac{x(\tau^{-1}(t^{\ast}))}{q(\tau^{-1}(t^{\ast }))}+\frac{1}{q(\tau^{-1}(t^{\ast}))} \biggl[ \frac{2(\alpha_{3}-1)}{3}M \\ &{}+(-1)^{n+1}\int_{\tau^{-1}(t^{\ast})}^{\infty}\frac{1}{r_{1}(u_{1})}\int_{u_{1}}^{\infty}\frac{1}{r_{2}(u_{2})} \cdots \int_{u_{n-2}}^{\infty}\frac{1}{r_{n-1}(u_{n-1})} \\ &{}\times\int _{u_{n-1}}^{\infty}A(u_{n},x)\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1} \biggr] \\ \geq&\frac{1}{\beta_{3}} \biggl[\frac{2(\alpha_{3}-1)}{3}M+\frac{\alpha _{3}-1}{3(\beta_{3}-1)}M- \frac{\alpha_{3}-1}{3}M \biggr] \\ =&\frac{\alpha_{3}-1}{3(\beta_{3}-1)}M, \end{aligned}

and $$|(Vx)(t)|\leq(\alpha_{3}-1)M/3$$, which implies that $$Ux+Vy\in \Omega$$ for any $$x,y\in\Omega$$ and VΩ is uniformly bounded. The rest of the proof is similar to that of Theorem 3.2. The proof is completed. □

## 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 $$g\in C_{\mathrm{rd}}({\mathbb{ T}}\times{\mathbb{ T}},{\mathbb{ R}})$$, 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 following higher order dynamic equation:

\begin{aligned}& \bigl\{ t^{\gamma}\bigl[\bigl(t\bigl(\cdots\bigl(t \bigl(x(t)-q_{k}(t)x(qt) \bigr)^{\Delta}\bigr)^{\Delta}\cdots \bigr)^{\Delta}\bigr)^{\Delta}\bigr]^{\gamma}\bigr\} ^{\Delta} \\& \quad {}+ \frac{q^{n\gamma +1}-1}{(q-1)q^{n\gamma+1}t^{n\gamma+2}}x^{2r+1}\bigl(q^{3+r}t\bigr)=0, \end{aligned}
(4.1)

where $$t\in \left .[q,\infty) \right ._{\mathbb{ T}}$$, γ is the quotient of odd positive integers and r is a positive integer, $$q_{k}(t)=-2[(-1)^{k}k^{2}+(-1)^{\log_{q} t}]/5$$ ($$k\in\{1,2,3\}$$), $$r_{n}(t)=t^{\gamma}$$, $$r_{k}(t)=t$$ ($$1\leq k \leq n-1$$), $$\tau(t)=qt$$, $$\delta(t)=q^{3+r}t$$ and $$f(t,u)=\frac{q^{n\gamma+1}-1}{(q-1)q^{n\gamma+1}t^{n\gamma+2}}u^{2r+1}$$.

It is easy to verify that $$q_{k}(t)$$ satisfies the condition $$(H_{k+1})$$. On the other hand, we have

\begin{aligned}& \int_{q}^{\infty}\biggl[\frac{1}{r_{n}(t)} \biggr]^{\frac{1}{\gamma}}\Delta t=\int_{q}^{\infty}\biggl[\frac{1}{t^{\gamma}} \biggr]^{\frac{1}{\gamma}}\Delta t=\infty, \\& \int_{q}^{\infty}\frac{1}{r_{k}(t)}\Delta t=\int _{q}^{\infty}\frac{1}{t}\Delta t=\infty,\quad 1 \leq k\leq n-1. \end{aligned}

If $$s>t\geq q$$ and $$s=q^{n}t$$, then we have the following inequality:

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

Combining Lemma 4.1 with the above inequality, we see that for any $$M>0$$,

\begin{aligned}& \int_{q}^{\infty}\frac{1}{u_{1}}\int _{u_{1}}^{\infty}\frac{1}{u_{2}}\cdots\int _{u_{n-2}}^{\infty}\frac{1}{u_{n-1}}\int _{u_{n-1}}^{\infty}\biggl[\frac {1}{u_{n}^{\gamma}}\int _{u_{n}}^{\infty}f(s,M)\Delta s \biggr]^{\frac{1}{\gamma}} \Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1} \\& \quad = M^{\frac{2r+1}{\gamma}}\int_{q}^{\infty}\frac{1}{u_{1}}\int_{u_{1}}^{\infty}\frac{1}{u_{2}} \cdots\int_{u_{n-2}}^{\infty}\frac{1}{u_{n-1}} \\& \qquad {}\times\int _{u_{n-1}}^{\infty}\biggl[\frac {1}{u_{n}^{\gamma}}\int _{u_{n}}^{\infty}\biggl(\frac{-1}{s^{n\gamma+1}} \biggr)^{\Delta}\Delta s \biggr]^{\frac{1}{\gamma}}\Delta u_{n}\Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1} \\& \quad = M^{\frac{2r+1}{\gamma}}\int_{q}^{\infty}\frac{1}{u_{1}}\int_{u_{1}}^{\infty}\frac{1}{u_{2}} \cdots\int_{u_{n-2}}^{\infty}\frac{1}{u_{n-1}}\int _{u_{n-1}}^{\infty}\frac {1}{u_{n}^{n+1+\frac{1}{\gamma}}} \Delta u_{n} \Delta u_{n-1}\cdots\Delta u_{2}\Delta u_{1} \\ & \quad = M^{\frac{2r+1}{\gamma}}\int_{q}^{\infty}\frac{1}{u_{1}}\int_{u_{1}}^{\infty}\frac{1}{u_{2}} \cdots\int_{u_{n-3}}^{\infty}\frac{1}{u_{n-2}} \\ & \qquad {}\times\biggl[\int _{u_{n-2}}^{\infty}\int_{u_{n-1}}^{\infty}\frac{1}{u_{n-1}}\frac{1}{u_{n}^{n+1+\frac{1}{\gamma}}} \Delta u_{n}\Delta u_{n-1} \biggr]\Delta u_{n-2}\cdots\Delta u_{2} \Delta u_{1} \\ & \quad = M^{\frac{2r+1}{\gamma}}\int_{q}^{\infty}\frac{1}{u_{1}}\int_{u_{1}}^{\infty}\frac{1}{u_{2}} \cdots\int_{u_{n-3}}^{\infty}\frac{1}{u_{n-2}} \\ & \qquad {}\times\biggl[\int _{u_{n-2}}^{\infty}\frac {1}{u_{n}^{n+1+\frac{1}{\gamma}}}\int _{u_{n-2}}^{\sigma(u_{n})}\frac{1}{u_{n-1}} \Delta u_{n-1} \Delta u_{n} \biggr]\Delta u_{n-2}\cdots\Delta u_{2}\Delta u_{1} \\ & \quad \leq M^{\frac{2r+1}{\gamma}}\int_{q}^{\infty}\frac{1}{u_{1}}\int_{u_{1}}^{\infty}\frac{1}{u_{2}} \cdots\int_{u_{n-3}}^{\infty}\frac{1}{u_{n-2}} \biggl[\int _{u_{n-2}}^{\infty}\frac {1}{u_{n}^{n+\frac{1}{\gamma}}} \Delta u_{n} \biggr]\Delta u_{n-2}\cdots\Delta u_{2}\Delta u_{1} \\ & \quad = M^{\frac{2r+1}{\gamma}}\int_{q}^{\infty}\frac{1}{u_{1}}\int_{u_{1}}^{\infty}\frac{1}{u_{2}} \cdots \biggl[\int_{u_{n-3}}^{\infty}\int _{u_{n-2}}^{\infty}\frac{1}{u_{n-2}}\frac {1}{u_{n}^{n+\frac{1}{\gamma}}} \Delta u_{n}\Delta u_{n-2} \biggr]\cdots\Delta u_{2}\Delta u_{1} \\ & \quad = M^{\frac{2r+1}{\gamma}}\int_{q}^{\infty}\frac{1}{u_{1}}\int_{u_{1}}^{\infty}\frac{1}{u_{2}} \cdots \biggl[\int_{u_{n-3}}^{\infty}\frac{1}{u_{n}^{n+\frac{1}{\gamma}}}\int _{u_{n-3}}^{\sigma(u_{n})}\frac{1}{u_{n-2}} \Delta u_{n-2} \Delta u_{n} \biggr]\cdots\Delta u_{2}\Delta u_{1} \\ & \quad \leq M^{\frac{2r+1}{\gamma}}\int_{q}^{\infty}\frac{1}{u_{1}}\int_{u_{1}}^{\infty}\frac{1}{u_{2}} \cdots \biggl[\int_{u_{n-3}}^{\infty}\frac{1}{u_{n}^{n-1+\frac{1}{\gamma}}} \Delta u_{n} \biggr]\cdots\Delta u_{2}\Delta u_{1} \\ & \qquad \cdots \\ & \quad = M^{\frac{2r+1}{\gamma}}\int_{q}^{\infty} \frac{1}{u_{1}} \biggl[\int_{u_{1}}^{\infty}\int _{u_{2}}^{\infty}\frac{1}{u_{2}}\frac {1}{u_{n}^{4+\frac{1}{\gamma}}}\Delta u_{n}\Delta u_{2} \biggr]\Delta u_{1} \\ & \quad = M^{\frac{2r+1}{\gamma}}\int_{q}^{\infty} \frac{1}{u_{1}} \biggl[\int_{u_{1}}^{\infty}\int _{u_{1}}^{\sigma(u_{n})}\frac{1}{u_{2}}\frac {1}{u_{n}^{4+\frac{1}{\gamma}}}\Delta u_{2}\Delta u_{n} \biggr]\Delta u_{1} \\ & \quad \leq M^{\frac{2r+1}{\gamma}}\int_{q}^{\infty}\int _{u_{1}}^{\infty}\frac {1}{u_{1}}\frac{1}{u_{n}^{3+\frac{1}{\gamma}}} \Delta u_{n}\Delta u_{1} \\ & \quad \leq M^{\frac{2r+1}{\gamma}}\int_{q}^{\infty} \frac{1}{u_{n}^{2+\frac {1}{\gamma}}} \Delta u_{n} \\ & \quad = M^{\frac{2r+1}{\gamma}}\frac{q-1}{q^{1+\frac{1}{\gamma}}-1} \\ & \quad < \infty. \end{aligned}

Thus conditions (H1) and (3.1) hold. By Theorem 3.1, Theorem 3.2 and Theorem 3.3, we see that Eq. (4.1) has a nonoscillatory bounded solution $$x(t)$$ with $$\lim\inf_{t\rightarrow\infty} |x(t)|>0$$.

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## Acknowledgements

This project is supported by NNSF of China (11461003) and NSF of Guangxi (2012GXNSFDA276040).

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Correspondence to Taixiang Sun.

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

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