Existence of nonoscillatory solutions tending to zero of third-order neutral dynamic equations on time scales

Abstract

We study the existence of nonoscillatory solutions tending to zero of a class of third-order nonlinear neutral dynamic equations on time scales by employing Krasnoselskii’s fixed point theorem. Two examples are given to illustrate the significance of the conclusions.

1 Introduction

In this paper, we consider the existence of nonoscillatory solutions tending to zero of a class of third-order nonlinear neutral dynamic equations

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

(1.1)

on a time scale $\mathbb{T}$ satisfying $\sup \mathbb{T}=\infty$, where $t\in [t_{0},\infty )_{\mathbb{T}}=[t_{0},\infty )\cap {\mathbb{T}}$ with $t_{0}\in \mathbb{T}$. The following conditions are assumed to hold throughout this paper:

• (C1) $r_{1}, r_{2}\in \mathrm{ C}_{\mathrm{rd}}([t_{0},\infty )_{\mathbb{T}}, (0, \infty ))$;

• (C2) $p\in \mathrm{ C}_{\mathrm{rd}}([t_{0},\infty )_{\mathbb{T}}, \mathbb{R})$ and $\lim_{t \rightarrow \infty }p(t)=p_{0}$, where $|p_{0}|<1$;

• (C3) $g,h\in \mathrm{ C}_{\mathrm{rd}}([t_{0},\infty )_{\mathbb{T}}, \mathbb{T})$, $g(t)\leq t$, and $\lim_{t \rightarrow \infty }g(t)=\lim_{t \rightarrow \infty }h(t)= \infty$; if $p_{0}\in (-1,0]$, then there exists a sequence $\{c_{k}\}_{k\geq 0}$ such that $\lim_{k \rightarrow \infty }c_{k}=\infty$ and $g(c_{k+1})=c_{k}$;

• (C4) $f\in \mathrm{ C}([t_{0},\infty )_{\mathbb{T}}\times \mathbb{R}, \mathbb{R})$, $f(t,x)$ is nondecreasing in x, and $xf(t,x)>0$ for $x\neq 0$.

The details of the theory of time scales can be found in [1–4, 8, 9] and hence they are omitted here. In recent years, the existence of nonoscillatory solutions of neutral dynamic equations on time scales has been studied successively in [6, 7, 11, 13–17]. Zhu and Wang [17] were concerned with a first-order neutral dynamic equation

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

Afterward, Deng and Wang [6] and Gao and Wang [7] investigated a second-order neutral dynamic equation

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

under the different assumptions $\int _{t_{0}}^{\infty }1/r(t)\Delta t=\infty$ and $\int _{t_{0}}^{\infty }1/r(t)\Delta t<\infty$, respectively. Furthermore, Qiu [11] studied (1.1) with $\int _{t_{0}}^{\infty }1/r_{1}(t)\Delta t=\int _{t_{0}}^{\infty }1/r_{2}(t) \Delta t=\infty$, whereas other cases of the convergence and divergence of $\int _{t_{0}}^{\infty }1/r_{1}(t)\Delta t$ and $\int _{t_{0}}^{\infty }1/r_{2}(t)\Delta t$ were considered in [14–16]. Similar sufficient conditions for the existence of nonoscillatory solutions tending to zero of neutral dynamic equations have been presented. However, it is not easy to find a necessary condition for equations to have a nonoscillatory solution tending to zero asymptotically.

Mojsej and Tartal’ová [10] studied the asymptotic behavior of nonoscillatory solutions to a third-order differential equation

$$ \biggl(\frac{1}{p(t)} \biggl(\frac{1}{r(t)}x'(t) \biggr)' \biggr)'+q(t)f\bigl(x(t)\bigr)=0. $$

They stated some necessary and sufficient conditions ensuring the existence of nonoscillatory solutions tending to zero. Motivated by [10], Qiu [12] studied the existence of nonoscillatory solutions tending to zero of (1.1) under the conditions $0\leq p_{0}<1$ and $g(t)\geq t$. The conclusions extend and improve the results reported in the papers [11, 14–16].

The purpose of this paper is to further discuss the same problem of (1.1) with $|p_{0}|<1$ and $g(t)\leq t$. The existence of nonoscillatory solutions tending to zero of (1.1) is established by employing Krasnoselskii’s fixed point theorem. Finally, two examples are presented to show the versatility of the conclusions.

2 Auxiliary results

Let $\mathrm{ BC}[T_{0},\infty )_{\mathbb{T}}$ denote the Banach space of all bounded continuous functions mapping $[T_{0},\infty )_{\mathbb{T}}$ into $\mathbb{R}$ with the norm $\|x\|=\sup_{t\in [T_{0},\infty )_{\mathbb{T}}}|x(t)|$. For the sake of convenience, we define

$$ z(t)=x(t)+p(t)x\bigl(g(t)\bigr), $$

(2.1)

and state the following lemmas which will be used in the sequel.

Lemma 2.1

(see [5, Krasnoselskii’s fixed point theorem])

LetXbe a Banach space andΩbe a bounded, convex, and closed subset ofX. If there exist two operators$U,V:\varOmega \rightarrow X$such that$Ux+Vy\in \varOmega$for all$x,y\in \varOmega$, whereUis a contraction mapping andVis completely continuous, then$U+V$has a fixed point inΩ.

Lemma 2.2

Suppose thatxis an eventually positive solution of (1.1) and there exists a constant$a\geq 0$such that$\lim_{t\rightarrow \infty }z(t)=a$. Then

$$ \lim_{t\rightarrow \infty }x(t)=\frac{a}{1+p_{0}}. $$

The proof is similar to those of [6, Lemma 2.3], [7, Theorem 1], and [17, Theorem 7], and thus is omitted.

3 Main results

In this section, our existence criteria for eventually positive solutions tending to zero as $t\rightarrow \infty$ of (1.1) are established by employing Krasnoselskii’s fixed point theorem.

Theorem 3.1

Assume that

$$ H_{1}(t_{0})< \infty\quad \textit{and}\quad \int _{t_{0}}^{\infty } \int _{t_{0}}^{s} \frac{f(u,2H_{1}(h(u)))}{r_{1}(s)}\Delta u\Delta s< \infty , $$

(3.1)

where

$$ H_{1}(t)= \int _{t}^{\infty }\frac{\Delta v}{r_{2}(v)}, $$

which satisfies$\lim_{t\rightarrow \infty }H_{1}(g(t))/H_{1}(t)=1$. Then (1.1) has an eventually positive solutionxwith$\lim_{t\rightarrow \infty }x(t)=0$, where$r_{2}z^{\Delta }$and$r_{1} (r_{2}z^{\Delta } )^{\Delta }$are both eventually negative.

Proof

Suppose that (3.1) holds. There will be two cases to be considered.

Case (i). $0\leq p_{0}<1$. Take $p_{1}$ such that $p_{0}< p_{1}<(1+4p_{0})/5<1$. When $p_{0}>0$, choose a sufficiently large $T_{0}\in [t_{0},\infty )_{\mathbb{T}}$ such that

$$\begin{aligned} &p(t)>0,\qquad \frac{5p_{1}-1}{4}\leq p(t)\leq p_{1}< 1, \qquad p(t) \frac{H_{1}(g(t))}{H_{1}(t)}\geq \frac{5p_{1}-1}{4},\quad t\in [T_{0}, \infty )_{\mathbb{T}}, \\ &\int _{T_{0}}^{\infty } \int _{T_{0}}^{s} \frac{f(u,2H_{1}(h(u)))}{r_{1}(s)}\Delta u\Delta s\leq \frac{1-p_{1}}{4}. \end{aligned}$$

(3.2)

When $p_{0}=0$, choose $p_{1}$ such that $|p(t)|\leq p_{1}\leq 1/13$ for $t\in [T_{0},\infty )_{\mathbb{T}}$. In view of (C3), there exists a $T_{1}\in (T_{0},\infty )_{\mathbb{T}}$ such that $g(t)\geq T_{0}$ and $h(t)\geq T_{0}$ for $t\in [T_{1},\infty )_{\mathbb{T}}$.

Define

$$ \varOmega _{1}=\bigl\{ x\in \mathrm{ BC}[T_{0},\infty )_{\mathbb{T}}:H_{1}(t)\leq x(t) \leq 2H_{1}(t)\bigr\} . $$

It is easy to prove that $\varOmega _{1}$ is a bounded, convex, and closed subset of $\mathrm{ BC}[T_{0},\infty )_{\mathbb{T}}$. Define $U_{1}$ and $V_{1}$: $\varOmega _{1}\rightarrow \mathrm{ BC}[T_{0},\infty )_{\mathbb{T}}$ as follows:

$$\begin{aligned} \begin{aligned} &(U_{1}x) (t)= \textstyle\begin{cases} (U_{1}x)(T_{1}),& t\in [T_{0},T_{1})_{ \mathbb{T}}, \\ 3p_{1}H_{1}(t)/2-p(t)x(g(t)), &t\in [T_{1},\infty )_{\mathbb{T}}, \end{cases}\displaystyle \\ &(V_{1}x) (t)= \textstyle\begin{cases} (V_{1}x)(T_{1}), & t\in [T_{0},T_{1})_{ \mathbb{T}}, \\ 3H_{1}(t)/2& \\ \quad{}+\int _{t}^{\infty }\int _{T_{1}}^{v}\int _{T_{1}}^{s}f(u,x(h(u)))/(r_{1}(s)r_{2}(v)) \Delta u\Delta s\Delta v, & t\in [T_{1},\infty )_{\mathbb{T}}. \end{cases}\displaystyle \end{aligned} \end{aligned}$$

(3.3)

We can prove that $U_{1}$ and $V_{1}$ satisfy all conditions in Lemma 2.1. The proof is expatiatory but similar to those of [6, Theorem 2.5], [7, Theorem 2], [12, Theorem 3.1], and [17, Theorem 8]; so we omit it here. By virtue of Lemma 2.1, there exists an $x\in \varOmega _{1}$ such that $(U_{1}+V_{1})x=x$. Hence, for $t\in [T_{1},\infty )_{\mathbb{T}}$, we have

$$ x(t)=\frac{3(1+p_{1})}{2}H_{1}(t)-p(t)x\bigl(g(t)\bigr)+ \int _{t}^{\infty } \int _{T_{1}}^{v} \int _{T_{1}}^{s} \frac{f(u,x(h(u)))}{r_{1}(s)r_{2}(v)}\Delta u\Delta s \Delta v. $$

Since

$$ \int _{t}^{\infty } \int _{T_{1}}^{v} \int _{T_{1}}^{s} \frac{f(u,x(h(u)))}{r_{1}(s)r_{2}(v)}\Delta u\Delta s \Delta v\leq H_{1}(t) \int _{T_{1}}^{\infty } \int _{T_{1}}^{s} \frac{f(u,2H_{1}(h(u)))}{r_{1}(s)}\Delta u\Delta s $$

and

$$ \lim_{t\rightarrow \infty }H_{1}(t) \int _{T_{1}}^{\infty } \int _{T_{1}}^{s} \frac{f(u,2H_{1}(h(u)))}{r_{1}(s)}\Delta u\Delta s=0, $$

we arrive at $\lim_{t\rightarrow \infty }z(t)=0$, which implies that $\lim_{t\rightarrow \infty }x(t)=0$ with the help of Lemma 2.2. For $t\in [T_{1},\infty )_{\mathbb{T}}$, we obtain

$$ r_{2}(t)z^{\Delta }(t)=-\frac{3(1+p_{1})}{2}- \int _{T_{1}}^{t} \int _{T_{1}}^{s} \frac{f(u,x(h(u)))}{r_{1}(s)}\Delta u\Delta s< 0 $$

and

$$ r_{1}(t) \bigl(r_{2}(t)z^{\Delta }(t) \bigr)^{\Delta }=- \int _{T_{1}}^{t}f\bigl(u,x\bigl(h(u)\bigr)\bigr) \Delta u< 0. $$

Case (ii). $-1< p_{0}<0$. Take $p_{1}$ satisfying $-p_{0}< p_{1}<(1-4p_{0})/5<1$. Choose a sufficiently large $T_{0}\in [t_{0},\infty )_{\mathbb{T}}$ such that (3.2) holds and

$$ p(t)< 0, \qquad \frac{5p_{1}-1}{4}\leq -p(t)\leq p_{1}< 1,\qquad -p(t) \frac{H_{1}(g(t))}{H_{1}(t)}\geq \frac{5p_{1}-1}{4}, \quad t\in [T_{0}, \infty )_{\mathbb{T}}. $$

Proceeding as in the proof of Case (i), define $V_{1}$ as in (3.3) and $U'_{1}$ on $\varOmega _{1}$ as follows:

$$ \bigl(U'_{1}x\bigr) (t)= \textstyle\begin{cases} (U'_{1}x)(T_{1}), & t \in [T_{0},T_{1})_{\mathbb{T}}, \\ -3p_{1}H_{1}(t)/2-p(t)x(g(t)), & t\in [T_{1},\infty )_{\mathbb{T}}. \end{cases} $$

Similarly, there exists an $x\in \varOmega _{1}$ such that $(U'_{1}+V_{1})x=x$. For $t\in [T_{1},\infty )_{\mathbb{T}}$, we have

$$ x(t)=\frac{3(1-p_{1})}{2}H_{1}(t)-p(t)x\bigl(g(t)\bigr)+ \int _{t}^{\infty } \int _{T_{1}}^{v} \int _{T_{1}}^{s} \frac{f(u,x(h(u)))}{r_{1}(s)r_{2}(v)}\Delta u\Delta s \Delta v $$

and we arrive at the same conclusions as in Case (i). This completes the proof. □

Theorem 3.2

Assume that

$$ H_{1}(t_{0})=\infty \quad\textit{or}\quad \int _{t_{0}}^{\infty } \int _{t_{0}}^{v} \frac{1}{r_{1}(s)r_{2}(v)}\Delta s\Delta v= \infty. $$

Then (1.1) has no eventually positive solutionsxsatisfying that$r_{2}z^{\Delta }$and$r_{1} (r_{2}z^{\Delta } )^{\Delta }$are both eventually negative.

Proof

Suppose that x is an eventually positive solution of (1.1), and there exists a $T_{0}\in [t_{0},\infty )_{\mathbb{T}}$ such that

$$ r_{2}(t)z^{\Delta }(t)< 0, \qquad r_{1}(t) \bigl(r_{2}(t)z^{\Delta }(t) \bigr)^{\Delta }< 0, \quad t\in [T_{0},\infty )_{\mathbb{T}}. $$

From (C3), there exists a $T_{1}\in (T_{0},\infty )_{\mathbb{T}}$ such that $h(t)\geq T_{0}$ for $t\in [T_{1},\infty )_{\mathbb{T}}$. Integrating (1.1) from $T_{1}$ to s, $s\in [\sigma (T_{1}),\infty )_{\mathbb{T}}$, by (C4) we obtain

$$ r_{1}(s) \bigl(r_{2}(s)z^{\Delta }(s) \bigr)^{\Delta }-r_{1}(T_{1}) \bigl(r_{2}(T_{1})z^{\Delta }(T_{1}) \bigr)^{\Delta } =- \int _{T_{1}}^{s}f\bigl(u,x\bigl(h(u)\bigr)\bigr) \Delta u< 0, $$

which yields

$$ \bigl(r_{2}(s)z^{\Delta }(s) \bigr)^{\Delta }< \frac{r_{1}(T_{1}) (r_{2}(T_{1})z^{\Delta }(T_{1}) )^{\Delta }}{r_{1}(s)}. $$

(3.4)

Integrating (3.4) from $T_{1}$ to v, $v\in [\sigma (T_{1}),\infty )_{\mathbb{T}}$, we get

$$ r_{2}(v)z^{\Delta }(v)-r_{2}(T_{1})z^{\Delta }(T_{1}) < r_{1}(T_{1}) \bigl(r_{2}(T_{1})z^{\Delta }(T_{1}) \bigr)^{\Delta } \int _{T_{1}}^{v} \frac{1}{r_{1}(s)}\Delta s $$

or

$$ z^{\Delta }(v)< \frac{r_{2}(T_{1})z^{\Delta }(T_{1})}{r_{2}(v)}+ \frac{r_{1}(T_{1}) (r_{2}(T_{1})z^{\Delta }(T_{1}) )^{\Delta }}{r_{2}(v)} \int _{T_{1}}^{v}\frac{1}{r_{1}(s)}\Delta s. $$

(3.5)

Integrating (3.5) from $T_{1}$ to t, $t\in [\sigma (T_{1}),\infty )_{\mathbb{T}}$, we obtain

$$\begin{aligned} z(t)< {}&z(T_{1})+r_{2}(T_{1})z^{\Delta }(T_{1}) \int _{T_{1}}^{t} \frac{1}{r_{2}(v)}\Delta v \\ &{} +r_{1}(T_{1}) \bigl(r_{2}(T_{1})z^{\Delta }(T_{1}) \bigr)^{\Delta } \int _{T_{1}}^{t} \int _{T_{1}}^{v}\frac{1}{r_{1}(s)r_{2}(v)}\Delta s \Delta v. \end{aligned}$$

Letting $t\rightarrow \infty$, we have $z(t)\rightarrow -\infty$. From (2.1), it follows that $p_{0}\in (-1,0]$, and then there exist a $T_{2}\in [T_{1},\infty )_{\mathbb{T}}$ and a $p_{1}$ with $-p_{0}< p_{1}<1$ such that $z(t)<0$ or

$$ x(t)< -p(t)x\bigl(g(t)\bigr)\leq p_{1}x\bigl(g(t)\bigr),\quad t\in [T_{2},\infty )_{ \mathbb{T}}. $$

By (C3), choose some positive integer $k_{0}$ such that $c_{k}\in [T_{2},\infty )_{\mathbb{T}}$ for all $k\geq k_{0}$. Then, for any $k\geq k_{0}+1$, we have

$$ x(c_{k})< p_{1}x(c_{k-1})< p_{1}^{2}x(c_{k-2})< \cdots < p_{1}^{k-k_{0}}x(c_{k_{0}}). $$

This inequality implies that $\lim_{k\rightarrow \infty }x(c_{k})=0$. It follows from (2.1) that $\lim_{k\rightarrow \infty }z(c_{k})=0$ which contradicts $z(t)\rightarrow -\infty$ as $t\rightarrow \infty$. The proof is complete. □

Theorem 3.3

Assume that

$$ H_{2}(t_{0})< \infty \quad \textit{and}\quad \int _{t_{0}}^{\infty }f\bigl(t,2H_{2}\bigl(h(t) \bigr)\bigr) \Delta t< \infty , $$

(3.6)

where

$$ H_{2}(t)= \int _{t}^{\infty } \int _{v}^{\infty } \frac{1}{r_{1}(s)r_{2}(v)}\Delta s\Delta v, $$

which satisfies$\lim_{t\rightarrow \infty }H_{2}(g(t))/H_{2}(t)=1$. Then (1.1) has an eventually positive solutionxwith$\lim_{t\rightarrow \infty }x(t)=0$, where$r_{2}z^{\Delta }$is eventually negative and$r_{1} (r_{2}z^{\Delta } )^{\Delta }$is eventually positive.

Proof

Suppose that (3.6) holds. There are two cases to be considered.

Case (i). $0\leq p_{0}<1$. Take $p_{1}$ as in Case (i) of Theorem 3.1. When $p_{0}>0$, choose a sufficiently large $T_{0}\in [t_{0},\infty )_{\mathbb{T}}$ such that

$$\begin{aligned} &p(t)>0,\qquad \frac{5p_{1}-1}{4}\leq p(t)\leq p_{1}< 1,\qquad p(t) \frac{H_{2}(g(t))}{H_{2}(t)}\geq \frac{5p_{1}-1}{4},\quad t\in [T_{0}, \infty )_{\mathbb{T}}, \\ &\int _{T_{0}}^{\infty }f\bigl(t,2H_{2}\bigl(h(t) \bigr)\bigr)\Delta t\leq \frac{1-p_{1}}{4}. \end{aligned}$$

When $p_{0}=0$, choose $p_{1}$ such that $|p(t)|\leq p_{1}\leq 1/13$ for $t\in [T_{0},\infty )_{\mathbb{T}}$. By virtue of (C3), there exists a $T_{1}\in (T_{0},\infty )_{\mathbb{T}}$ such that $g(t)\geq T_{0}$ and $h(t)\geq T_{0}$ for $t\in [T_{1},\infty )_{\mathbb{T}}$.

Define

$$ \varOmega _{2}=\bigl\{ x\in \mathrm{ BC}[T_{0},\infty )_{\mathbb{T}}:H_{2}(t)\leq x(t) \leq 2H_{2}(t)\bigr\} $$

(3.7)

and $U_{2}$, $V_{2}$: $\varOmega _{2}\rightarrow \mathrm{ BC}[T_{0},\infty )_{\mathbb{T}}$ as follows:

$$\begin{aligned} \begin{aligned} &(U_{2}x) (t)= \textstyle\begin{cases} (U_{2}x)(T_{1}), & t \in [T_{0},T_{1})_{\mathbb{T}}, \\ 3p_{1}H_{2}(t)/2-p(t)x(g(t)), & t\in [T_{1},\infty )_{\mathbb{T}}, \end{cases}\displaystyle \\ & (V_{2}x) (t)= \textstyle\begin{cases} (V_{2}x)(T_{1}), & t\in [T_{0},T_{1})_{ \mathbb{T}}, \\ 3H_{2}(t)/2& \\ \quad{}+\int _{t}^{\infty }\int _{v}^{\infty }\int _{s}^{\infty }f(u,x(h(u)))/(r_{1}(s)r_{2}(v)) \Delta u\Delta s\Delta v, & t\in [T_{1},\infty )_{\mathbb{T}}. \end{cases}\displaystyle \end{aligned} \end{aligned}$$

(3.8)

The remainder of the proof is similar to that of Theorem 3.1 and so is omitted. By Lemma 2.1, there exists an $x\in \varOmega _{2}$ such that $(U_{2}+V_{2})x=x$. For $t\in [T_{1},\infty )_{\mathbb{T}}$, we have

$$ x(t)=\frac{3(1+p_{1})}{2}H_{2}(t)-p(t)x\bigl(g(t)\bigr)+ \int _{t}^{\infty } \int _{v}^{\infty } \int _{s}^{\infty } \frac{f(u,x(h(u)))}{r_{1}(s)r_{2}(v)}\Delta u\Delta s \Delta v. $$

Since

$$ \int _{t}^{\infty } \int _{v}^{\infty } \int _{s}^{\infty } \frac{f(u,x(h(u)))}{r_{1}(s)r_{2}(v)}\Delta u\Delta s \Delta v\leq H_{2}(t) \int _{T_{1}}^{\infty }f\bigl(u,2H_{2}\bigl(h(u) \bigr)\bigr)\Delta u $$

and

$$ \lim_{t\rightarrow \infty }H_{2}(t) \int _{T_{1}}^{\infty }f\bigl(u,2H_{2}\bigl(h(u) \bigr)\bigr) \Delta u=0, $$

we get $\lim_{t\rightarrow \infty }z(t)=0$, which implies that $\lim_{t\rightarrow \infty }x(t)=0$ due to Lemma 2.2. For $t\in [T_{1},\infty )_{\mathbb{T}}$, we obtain

$$ r_{2}(t)z^{\Delta }(t)=-\frac{3(1+p_{1})}{2} \int _{t}^{\infty } \frac{1}{r_{1}(s)}\Delta s- \int _{t}^{\infty } \int _{s}^{\infty } \frac{f(u,x(h(u)))}{r_{1}(s)}\Delta u\Delta s< 0 $$

and

$$ r_{1}(t) \bigl(r_{2}(t)z^{\Delta }(t) \bigr)^{\Delta }= \frac{3(1+p_{1})}{2}+ \int _{t}^{\infty }f\bigl(u,x\bigl(h(u)\bigr)\bigr)\Delta u>0. $$

Case (ii). $-1< p_{0}<0$. Introduce $\mathrm{ BC}[T_{0},\infty )_{\mathbb{T}}$ and its subset $\varOmega _{2}$ as in (3.7). Define $V_{2}$ as in (3.8) and $U'_{2}$ on $\varOmega _{2}$ as follows:

$$ \bigl(U'_{2}x\bigr) (t)= \textstyle\begin{cases} (U'_{2}x)(T_{1}),& t \in [T_{0},T_{1})_{\mathbb{T}}, \\ -3p_{1}H_{2}(t)/2-p(t)x(g(t)), & t\in [T_{1},\infty )_{\mathbb{T}}. \end{cases} $$

The following proof is similar to Case (i) and we omit it here. There exists an $x\in \varOmega _{2}$ such that $(U'_{2}+V_{2})x=x$. For $t\in [T_{1},\infty )_{\mathbb{T}}$, we have

$$ x(t)=\frac{3(1-p_{1})}{2}H_{2}(t)-p(t)x\bigl(g(t)\bigr)+ \int _{t}^{\infty } \int _{v}^{\infty } \int _{s}^{\infty } \frac{f(u,x(h(u)))}{r_{1}(s)r_{2}(v)}\Delta u\Delta s \Delta v $$

and obtain the similar results as in Case (i). This completes the proof. □

4 Examples

In this section, two examples are presented to show the applications of our results. The first example is given to illustrate Theorem 3.1.

Example 4.1

Let $\mathbb{T}=\bigcup_{n=1}^{\infty }[3n-2,3n]$. For $t\in [4,\infty )_{\mathbb{T}}$, consider

$$ \biggl(t^{4} \biggl(t^{2} \biggl(x(t)- \frac{t-1}{2t}x(t-3) \biggr)^{ \Delta } \biggr)^{\Delta } \biggr)^{\Delta } +tx^{3}(t)+ \frac{x(t)}{t^{2}}=0. $$

(4.1)

Here, $r_{1}(t)=t^{4}$, $r_{2}(t)=t^{2}$, $p(t)=-(t-1)/(2t)$, $g(t)=t-3$, $h(t)=t$, and $f(t,x)=tx^{3}+x/t^{2}$. It is obvious that the coefficients of (4.1) satisfy (C1)–(C4). Since

$$ \int _{t_{0}}^{\infty }\frac{\Delta t}{r_{2}(t)}= \int _{4}^{\infty } \frac{\Delta t}{t^{2}}< 1 $$

and

$$ H_{1}(t)= \int _{t}^{\infty }\frac{\Delta v}{v^{2}}< 1, $$

we obtain

$$ f\bigl(u,2H_{1}\bigl(h(u)\bigr)\bigr)=u\cdot \bigl(2H_{1}(u) \bigr)^{3}+\frac{2H_{1}(u)}{u^{2}}< 8u+ \frac{2}{u^{2}}< 9u $$

and

$$\begin{aligned} \int _{t_{0}}^{\infty } \int _{t_{0}}^{s} \frac{f(u,2H_{1}(h(u)))}{r_{1}(s)}\Delta u\Delta s&< 9 \int _{4}^{ \infty } \int _{4}^{s}\frac{u}{s^{4}}\Delta u\Delta s \\ & < 9 \int _{4}^{\infty } \int _{4}^{s}\frac{1}{s^{3}}\Delta u\Delta s< 9 \int _{4}^{\infty }\frac{1}{s^{2}}\Delta s< \infty. \end{aligned}$$

By Theorem 3.1, we see that (4.1) has an eventually positive solution x satisfying $\lim_{t\rightarrow \infty }x(t)=0$, where $r_{2}z^{\Delta }$ and $r_{1} (r_{2}z^{\Delta } )^{\Delta }$ are both eventually negative.

Now, we give the second example to demonstrate Theorems 3.2 and 3.3.

Example 4.2

Let $\mathbb{T}=\bigcup_{n=1}^{\infty }[2^{n}-1,2^{n}]$. For $t\in [3,\infty )_{\mathbb{T}}$, consider

$$ \biggl(t^{3} \biggl(t \biggl( \biggl( \frac{3}{2}+\frac{1}{t} \biggr)x(t) \biggr)^{\Delta } \biggr)^{\Delta } \biggr)^{\Delta } + \frac{1}{t^{2}}x \biggl( \frac{t}{2} \biggr)=0. $$

(4.2)

Here, $r_{1}(t)=t^{3}$, $r_{2}(t)=t$, $p(t)=1/2+1/t$, $g(t)=t$, $h(t)=t/2$, and $f(t,x)=x/t^{2}$. It is obvious that the coefficients of (4.2) satisfy (C1)–(C4). Since

$$ \int _{t_{0}}^{\infty }\frac{\Delta t}{r_{2}(t)}= \int _{3}^{\infty } \frac{\Delta t}{t}=\infty , $$

in terms of Theorem 3.2, we deduce that (4.2) has no eventually positive solutions x satisfying $\lim_{t\rightarrow \infty }x(t)=0$, where $r_{2}z^{\Delta }$ and $r_{1} (r_{2}z^{\Delta } )^{\Delta }$ are both eventually negative. However, we have

$$ \int _{t_{0}}^{\infty } \int _{v}^{\infty }\frac{1}{r_{1}(s)r_{2}(v)} \Delta s\Delta v= \int _{3}^{\infty } \int _{v}^{\infty } \frac{1}{s^{3}v}\Delta s\Delta v< \infty $$

and

$$ H_{2}(t)= \int _{t}^{\infty } \int _{v}^{\infty }\frac{1}{s^{3}v} \Delta s\Delta v. $$

Furthermore, there exists a constant $M>0$ such that

$$ f\bigl(t,2H_{2}\bigl(h(t)\bigr)\bigr)=\frac{2}{t^{2}}H_{2} \biggl(\frac{t}{2} \biggr)\leq \frac{2M}{t^{2}} $$

and

$$ \int _{t_{0}}^{\infty }f\bigl(t,2H_{2}\bigl(h(t) \bigr)\bigr)\Delta t\leq 2M \int _{3}^{ \infty }\frac{\Delta t}{t^{2}}< \infty. $$

By virtue of Theorem 3.3, we conclude that (4.2) has an eventually positive solution x satisfying $\lim_{t\rightarrow \infty }x(t)=0$, where $r_{2}z^{\Delta }$ is eventually negative and $r_{1} (r_{2}z^{\Delta } )^{\Delta }$ is eventually positive.