On nonoscillatory solutions tending to zero of third-order nonlinear dynamic equations on time scales

Abstract

By Kranoselskii’s fixed point theorem, we state some sufficient conditions of the existence of nonoscillatory solutions tending to zero as $t\rightarrow\infty$ of a class of third-order nonlinear dynamic equations on time scales. Two examples are presented to show the significance of the results.

1 Introduction

In 1988, Hilger introduced the theory of time scales in his Ph.D. thesis [1]; see also [2]. We refer the reader to [3, 4] for details on time scales. We remark that some researchers have studied the existence of nonoscillatory solutions of several kinds of nonlinear dynamic equations on time scales, which can be found in [5–10].

Zhu and Wang [10] studied first-order nonlinear neutral dynamic equations

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

on a time scale $\mathbb{T}$. Then Gao and Wang [7], Deng and Wang [6] discussed second-order nonlinear neutral dynamic equations

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

under different conditions successively. Inspired by [6, 7], Qiu [9] considered third-order nonlinear neutral dynamic equations

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

(1)

under the condition $g(t)\leq t$, and established the existence of nonoscillatory solutions of equation (1). However, the conditions ensuring the existence of the nonoscillatory solutions tending to zero as $t\rightarrow\infty$ of equation (1) were exceptional, and as a result, the applications were within limits (see [9], Theorems 3.5 and 3.6).

In 2011, Mojsej and Tartal’ová [8] studied the asymptotic behavior of nonoscillatory solutions of third-order nonlinear differential equations with quasiderivatives of the form

$$\begin{aligned} \biggl(\frac{1}{p(t)} \biggl(\frac{1}{r(t)}x'(t) \biggr)' \biggr)'+q(t)f\bigl(x(t)\bigr)=0,\quad t\geq a, \end{aligned}$$

(2)

and stated the necessary and sufficient conditions ensuring the existence of nonoscillatory solutions tending to zero as $t\rightarrow\infty$.

In this paper, we investigate the existence of nonoscillatory solutions tending to zero as $t\rightarrow\infty$ of equation (1) on a time scale $\mathbb{T}$, which satisfies $\inf \mathbb{T}=t_{0}$ and $\sup\mathbb{T}=\infty$. We state the following conditions, which hold throughout this paper:

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

• (C2) $p\in C_{rd}(\mathbb{T}, [0,\infty))$ and there exists a constant $p_{0}$ with $0\leq p_{0}<1$ such that $\lim_{t \rightarrow\infty}p(t)=p_{0}$.

• (C3) $g,h\in C_{rd}(\mathbb{T}, \mathbb{T})$, $g(t)\geq t$, and $\lim_{t \rightarrow\infty}h(t)=\infty$.

• (C4) $f\in C(\mathbb{T}\times\mathbb{R}, \mathbb{R})$ and $xf(t,x)>0$ for $t\in\mathbb{T}$ and $x\neq0$.

In the sequel, there are two cases of the function f to be considered:

• (C4A) There exist $q\in C(\mathbb{T}, (0,\infty))$ and $f_{0}\in C(\mathbb{R}, \mathbb{R})$ such that $xf(t,x)\leq xq(t)f_{0}(x)$.

• (C4B) $f(t,x)$ is nondecreasing in x.

Motivated by [8, 9], we will establish the existence of nonoscillatory solutions tending to a zero of equation (1) by employing Kranoselskii’s fixed point theorem and, finally, two examples are given to show the versatility of the conclusions.

Definition 1.1

A solution x of equation (1) is said to be eventually positive (or eventually negative) if there exists $c\in\mathbb{T}$ such that $x(t)>0$ (or $x(t)<0$) for all $t\geq c$ in $\mathbb{T}$. x is said to be nonoscillatory if it is either eventually positive or eventually negative; otherwise, it is oscillatory.

2 Auxiliary results

Let $BC(\left.[T_{0},\infty)\right._{\mathbb{T}},\mathbb{R})$ denote the Banach space of all bounded continuous functions on $\left.[T_{0},\infty)\right._{\mathbb{T}}$ with the norm

$$\begin{aligned} \|x\|=\sup_{t\in \left.[T_{0},\infty)\right._{\mathbb{T}}}\bigl|x(t)\bigr|. \end{aligned}$$

Definition 2.1

Let $X \subseteq BC\left.[T_{0},\infty)\right._{\mathbb{T}}$, we say that X is uniformly Cauchy, if for any given $\epsilon>0$, there exists a $T_{1}\in \left.[T_{0},\infty)\right._{\mathbb{T}}$ such that $|x(t_{1})-x(t_{2})|<\epsilon$ for any $x\in X$ and $t_{1},t_{2}\in \left.[T_{1},\infty)\right._{\mathbb{T}}$.

Definition 2.2

X is said to be equi-continuous on $[a,b]_{\mathbb{T}}$, if for any given $\epsilon>0$, there exists $\delta>0$ such that $|x(t_{1})-x(t_{2})|<\epsilon$ for any $x\in X$ and $t_{1},t_{2}\in [a,b]_{\mathbb{T}}$ with $|t_{1}-t_{2}|<\delta$.

Lemma 2.3

([10], Lemma 4)

Suppose that $X \subseteq BC\left.[T_{0},\infty)\right._{\mathbb{T}}$ 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

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

Suppose that X is a Banach space and Ω is a bounded, convex, and closed subset of X. Suppose further that there exist two operators $U,S:\varOmega \rightarrow X$ such that

• (1) $Ux+Sy\in \varOmega$ for all $x,y\in \varOmega$;

• (2) U is a contraction mapping;

• (3) S is completely continuous.

Then $U+S$ has a fixed point in Ω.

Without loss of generality, we shall consider the case that eventually positive solutions of equation (1) in the following. It is because if $x(t)$ is an eventually negative solution of equation (1), then $y(t)=-x(t)$ will satisfy

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

Note that $\overline{f}(t,u):=-f(t,-u)$ satisfies (C4), and (C4A) or (C4B) similarly as $f(t,u)$.

Define

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

(3)

then we will have the following lemma.

Lemma 2.5

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

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

Proof

Suppose that $x(t)$ is an eventually positive solution of equation (1). In view of (C3), there exists $t_{1}\in \left.[t_{0},\infty)\right._{\mathbb{T}}$ such that $x(t)>0$, $x(g(t))>0$ for $t\in \left.[t_{1},\infty)\right._{\mathbb{T}}$. We claim that $x(t)$ is bounded on $\left.[t_{1},\infty)\right._{\mathbb{T}}$. Assume not; then we have

$$\begin{aligned} z(t)=x(t)+p(t)x\bigl(g(t)\bigr)\geq x(t)\rightarrow\infty \end{aligned}$$

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

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

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

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

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

3 Main results

In this section, we state and prove our existence criteria for eventually positive solutions tending to zero as $t\rightarrow \infty$ of equation (1).

Theorem 3.1

Assume that the function f satisfies (C4A) and there exists $M_{1}>0$ such that

$$\begin{aligned} \int_{t_{0}}^{\infty}\frac{1}{r_{2}(t)}\varDelta t=M_{1}< \infty \end{aligned}$$

(4)

and

$$\begin{aligned} \int_{t_{0}}^{\infty} \int_{t_{0}}^{s}\frac{q(u)}{r_{1}(s)}\varDelta u\varDelta s< \infty. \end{aligned}$$

(5)

Define $H_{1}(t)=\int_{t}^{\infty}\frac{1}{r_{2}(v)}\varDelta v$, which satisfies $\lim_{t\rightarrow \infty}\frac{H_{1}(g(t))}{H_{1}(t)}=\eta_{1}\in(0,1]$. If there exists $L>0$ such that

$$\begin{aligned} \bigl|f(t,x_{1})-f(t,x_{2})\bigr|\leq L\cdot q(t)|x_{1}-x_{2}|,\quad x_{1},x_{2} \in[0,2M_{1}], \end{aligned}$$

(6)

then equation (1) has an eventually positive solution $x(t)$ with $\lim_{t\rightarrow\infty}x(t)=0$, and there exists $T_{1}\in \left.[t_{0},\infty)\right._{\mathbb{T}}$ such that

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

Proof

From (C2), for $0\leq p_{0}<1$, choose $p_{1}$ such that $p_{0}< p_{1}<(1+4p_{0})/5<1$. By (4) and (5), there exists $T_{0}\in \left.[t_{0},\infty)\right._{\mathbb{T}}$ such that

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

(7)

and

$$\begin{aligned} \int_{T_{0}}^{\infty} \int_{T_{0}}^{s}\frac{q(u)}{r_{1}(s)}\varDelta u\varDelta s\leq \min\biggl\{ \frac{1-p_{1}\eta_{1}}{4K},1 \biggr\} , \end{aligned}$$

(8)

where $K=\max\{|f_{0}(x)|:x\in[0,2M_{1}]\}$. From (C3), there always exists $T_{1}\in(T_{0},\infty)_{\mathbb{T}}$ such that

$$\begin{aligned} h(t)\geq T_{0},\quad t\in \left.[T_{1},\infty)\right._{\mathbb{T}}. \end{aligned}$$

Define

$$\begin{aligned} \varOmega _{1}=\bigl\{ x(t)\in BC\left.[T_{0}, \infty)\right._{\mathbb{T}}:H_{1}(t)\leq x(t)\leq2H_{1}(t)\bigr\} . \end{aligned}$$

(9)

It is clear that $\varOmega _{1}$ is a bounded, convex, and closed subset of $BC\left.[T_{0},\infty)\right._{\mathbb{T}}$. For any $x\in \varOmega _{1}$, by (C4A) we have

$$\begin{aligned} 0< f\bigl(t,x\bigl(h(t)\bigr)\bigr)\leq q(t)f_{0}\bigl(x\bigl(h(t) \bigr)\bigr),\quad t\in \left.[T_{1},\infty)\right._{\mathbb{T}}. \end{aligned}$$

Define the operators $U_{1}$ and $S_{1}$: $\varOmega _{1}\rightarrow BC\left.[T_{0},\infty)\right._{\mathbb{T}}$ as

$$\begin{aligned} (U_{1}x) (t)=\frac{3}{2}p_{1}\eta_{1}H_{1}(t)-p(t)x \bigl(g(t)\bigr),\quad t\in \left.[T_{0},\infty)\right._{\mathbb{T}}, \end{aligned}$$

and

$$\begin{aligned} (S_{1}x) (t)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{3}{2}H_{1}(t)+\int_{t}^{\infty}\int_{T_{1}}^{v}\int _{T_{1}}^{s}\frac{f(u,x(h(u)))}{r_{1}(s)r_{2}(v)}\varDelta u\varDelta s\varDelta v,& t\in \left.[T_{1},\infty)\right._{\mathbb{T}},\\ (S_{1}x)(T_{1}),& t\in [T_{0},T_{1}]_{\mathbb{T}}. \end{array}\displaystyle \right . \end{aligned}$$

(10)

Then we prove that $U_{1}$ and $S_{1}$ satisfy all the conditions in Lemma 2.4.

(1) For any $x,y\in \varOmega _{1}$ and $t\in \left.[T_{1},\infty)\right._{\mathbb{T}}$, by (7)-(9) we obtain

$$\begin{aligned} &(U_{1}x) (t)+(S_{1}y) (t)\\ &\quad\leq\frac{3(1+p_{1}\eta_{1})}{2}H_{1}(t)-\frac{5p_{1}-1}{4}\eta _{1}H_{1}(t)+ \int_{t}^{\infty} \int_{T_{1}}^{v} \int_{T_{1}}^{s}\frac {q(u)f_{0}(y(h(u)))}{r_{1}(s)r_{2}(v)}\varDelta u\varDelta s\varDelta v\\ &\quad\leq\frac{3(1+p_{1}\eta_{1})}{2}H_{1}(t)-\frac{5p_{1}-1}{4}\eta _{1}H_{1}(t)+KH_{1}(t) \int_{T_{1}}^{\infty} \int_{T_{1}}^{s}\frac {q(u)}{r_{1}(s)}\varDelta u\varDelta s\\ &\quad\leq\frac{3(1+p_{1}\eta_{1})}{2}H_{1}(t)-\frac{5p_{1}-1}{4}\eta _{1}H_{1}(t)+\frac{1-p_{1}\eta_{1}}{4}H_{1}(t)\\ &\quad=\frac{7+\eta_{1}}{4}H_{1}(t)\leq2H_{1}(t) \end{aligned}$$

and

$$\begin{aligned} &(U_{1}x) (t)+(S_{1}y) (t)\\ &\quad=\frac{3(1+p_{1}\eta_{1})}{2}H_{1}(t)-p(t)x\bigl(g(t)\bigr)+ \int_{t}^{\infty} \int_{T_{1}}^{v} \int_{T_{1}}^{s}\frac {f(u,y(h(u)))}{r_{1}(s)r_{2}(v)}\varDelta u\varDelta s\varDelta v\\ &\quad\geq\frac{3(1+p_{1}\eta_{1})}{2}H_{1}(t)-2p_{1} \eta_{1}H_{1}(t)= \frac{3-p_{1}\eta_{1}}{2}H_{1}(t)>H_{1}(t). \end{aligned}$$

Similarly, $H_{1}(t)\leq(U_{1}x)(t)+(S_{1}y)(t)\leq2H_{1}(t)$ also holds for any $x,y\in \varOmega _{1}$ and $t\in [T_{0},T_{1}]_{\mathbb{T}}$. It follows that $U_{1}x+S_{1}y\in \varOmega _{1}$ for any $x,y\in \varOmega _{1}$.

(2) For any $x,y\in \varOmega _{1}$ and $t\in \left.[T_{0},\infty)\right._{\mathbb{T}}$, we always have

$$\begin{aligned} \bigl|(U_{1}x) (t)-(U_{1}y) (t)\bigr|=\bigl|p(t) \bigl(x\bigl(g(t) \bigr)-y\bigl(g(t)\bigr) \bigr)\bigr|\leq p_{1}\sup_{t\in \left.[T_{0},\infty )\right._{\mathbb{T}}}\bigl|x(t)-y(t)\bigr|. \end{aligned}$$

It follows that $\|U_{1}x-U_{1}y\|\leq p_{1}\|x-y\|$ for any $x,y\in \varOmega _{1}$. That is, $U_{1}$ is a contraction mapping.

(3) For $t\in \left.[T_{0},\infty)\right._{\mathbb{T}}$, we have

$$\begin{aligned} (S_{1}x) (t)\leq\frac{3}{2}H_{1}(t)+ \frac{1-p_{1}\eta_{1}}{4}H_{1}(t)=\frac{7-p_{1}\eta _{1}}{4}H_{1}(t)< 2H_{1}(t) \end{aligned}$$

and

$$\begin{aligned} (S_{1}x) (t)>\frac{3}{2}H_{1}(t)>H_{1}(t). \end{aligned}$$

It is easy to see that $S_{1}$ maps $\varOmega _{1}$ into $\varOmega _{1}$.

For any $x\in \varOmega _{1}$ and $t\in \left.[T_{0},\infty)\right._{\mathbb{T}}$, let $x_{n}\in \varOmega _{1}$ and $\|x_{n}-x\|\rightarrow0$ as $n\rightarrow\infty$. For $t\in \left.[T_{1},\infty)\right._{\mathbb{T}}$, by (4), (6), and (8), we have

$$\begin{aligned} &\bigl\vert (S_{1}x_{n}) (t)-(S_{1}x) (t) \bigr\vert \\ &\quad\leq \int_{t}^{\infty} \int_{T_{1}}^{v} \int_{T_{1}}^{s}\frac {|f(u,x_{n}(h(u)))-f(u,x(h(u)))|}{r_{1}(s)r_{2}(v)}\varDelta u\varDelta s\varDelta v \\ &\quad\leq L \int_{T_{1}}^{\infty}\frac{1}{r_{2}(v)}\varDelta v\cdot \int_{T_{1}}^{\infty} \int_{T_{1}}^{s}\frac {q(u)|x_{n}(h(u))-x(h(u))|}{r_{1}(s)}\varDelta u\varDelta s \\ &\quad\leq LM_{1}\cdot\sup_{t\in \left.[T_{0},\infty)\right._{\mathbb{T}}}\bigl|x_{n}(t)-x(t)\bigr|. \end{aligned}$$

For $t\in[T_{0},T_{1}]_{\mathbb{T}}$, we also have the result above. By Lebesgue’s dominated convergence theorem (see Chapter 5 in [11]), it follows that

$$\begin{aligned} \|S_{1}x_{n}-S_{1}x\|\rightarrow0,\quad n \rightarrow\infty. \end{aligned}$$

Therefore, $S_{1}$ is continuous.

It is obvious that $S_{1}\varOmega _{1}$ is bounded. On the other hand, by (4), for any $\epsilon>0$ there exists $T_{2}\in \left.[T_{1},\infty)\right._{\mathbb{T}}$ such that

$$\begin{aligned} H_{1}(T_{2})= \int_{T_{2}}^{\infty}\frac{1}{r_{2}(t)}\varDelta t< \frac{\epsilon}{2K+3}. \end{aligned}$$

For $t_{1},t_{2}\in[T_{0},T_{1}]_{\mathbb{T}}$, we always have $|(S_{1}x)(t_{1})-(S_{1}x)(t_{2})|=0$. Then, for any $x\in \varOmega _{1}$ and $t_{1},t_{2}\in \left.[T_{2},\infty)\right._{\mathbb{T}}$, we have

$$\begin{aligned} &\bigl\vert (S_{1}x) (t_{1})-(S_{1}x) (t_{2})\bigr\vert \\ &\quad\leq\biggl\vert \int_{t_{1}}^{\infty} \int_{T_{1}}^{v} \int_{T_{1}}^{s}\frac {f(u,x(h(u)))}{r_{1}(s)r_{2}(v)}\varDelta u\varDelta s\varDelta v- \int_{t_{2}}^{\infty} \int_{T_{1}}^{v} \int_{T_{1}}^{s}\frac {f(u,x(h(u)))}{r_{1}(s)r_{2}(v)}\varDelta u\varDelta s\varDelta v\biggr\vert \\ &\qquad{} +\frac{3}{2}\bigl|H_{1}(t_{1})-H_{1}(t_{2})\bigr| \\ &\quad\leq2 \int_{T_{2}}^{\infty} \int_{T_{1}}^{v} \int_{T_{1}}^{s}\frac {f(u,x(h(u)))}{r_{1}(s)r_{2}(v)}\varDelta u\varDelta s\varDelta v+3H_{1}(T_{2}) \\ &\quad\leq2KH_{1}(T_{2}) \int_{T_{1}}^{\infty} \int_{T_{1}}^{s}\frac {q(u)}{r_{1}(s)}\varDelta u\varDelta s+3H_{1}(T_{2}) \\ &\quad\leq(2K+3)H_{1}(T_{2})< \epsilon. \end{aligned}$$

It is clear that $S_{1}\varOmega _{1}$ is uniformly Cauchy.

For $x\in \varOmega _{1}$ and $t_{1},t_{2}\in [\min\{T_{1}-1,T_{0}\},T_{2}+1]_{\mathbb{T}}$, we have

$$\begin{aligned} &\bigl\vert (S_{1}x) (t_{1})-(S_{1}x) (t_{2})\bigr\vert \\ &\quad\leq\biggl\vert \int_{t_{1}}^{t_{2}} \int_{T_{1}}^{v} \int_{T_{1}}^{s}\frac {f(u,x(h(u)))}{r_{1}(s)r_{2}(v)}\varDelta u\varDelta s\varDelta v\biggr\vert +\frac{3}{2}\biggl\vert \int_{t_{1}}^{t_{2}}\frac {1}{r_{2}(t)}\varDelta t\biggr\vert \\ &\quad\leq K \int_{T_{1}}^{\infty} \int_{T_{1}}^{s}\frac{q(u)}{r_{1}(s)}\varDelta u\varDelta s\biggl\vert \int_{t_{1}}^{t_{2}}\frac{1}{r_{2}(t)}\varDelta t\biggr\vert + \frac{3}{2}\biggl\vert \int_{t_{1}}^{t_{2}}\frac{1}{r_{2}(t)}\varDelta t\biggr\vert \\ &\quad\leq\biggl(K+\frac{3}{2} \biggr)\bigl|H_{1}(t_{1})-H_{1}(t_{2})\bigr|. \end{aligned}$$

Therefore, there exists $0<\delta<1$ such that $|(S_{1}x)(t_{1})-(S_{1}x)(t_{2})|<\epsilon$ for any $t_{1},t_{2}\in [T_{0}, T_{2}+1]_{\mathbb{T}}$ with $|t_{1}-t_{2}|<\delta$. We conclude that $S_{1}\varOmega _{1}$ is equi-continuous.

According to Lemma 2.3, it follows that $S_{1}\varOmega _{1}$ is relatively compact and $S_{1}$ is completely continuous. Then by Lemma 2.4, there exists $x\in \varOmega _{1}$ such that $(U_{1}+S_{1})x=x$. It means that $x(t)$ is an eventually positive solution of equation (1), and for $t\in \left.[T_{1},\infty)\right._{\mathbb{T}}$, we have

$$\begin{aligned} x(t)=\frac{3(1+p_{1}\eta_{1})}{2}H_{1}(t)-p(t)x\bigl(g(t)\bigr)+ \int_{t}^{\infty } \int_{T_{1}}^{v} \int_{T_{1}}^{s}\frac {f(u,x(h(u)))}{r_{1}(s)r_{2}(v)}\varDelta u\varDelta s\varDelta v. \end{aligned}$$

As

$$\begin{aligned} \int_{t}^{\infty} \int_{T_{1}}^{v} \int_{T_{1}}^{s}\frac {f(u,x(h(u)))}{r_{1}(s)r_{2}(v)}\varDelta u\varDelta s\varDelta v\leq KH_{1}(t) \int_{T_{1}}^{\infty} \int_{T_{1}}^{s}\frac {q(u)}{r_{1}(s)}\varDelta u\varDelta s \end{aligned}$$

for $t\in \left.[T_{1},\infty)\right._{\mathbb{T}}$, and

$$\begin{aligned} \lim_{t\rightarrow \infty}KH_{1}(t) \int_{T_{1}}^{\infty} \int_{T_{1}}^{s}\frac {q(u)}{r_{1}(s)}\varDelta u\varDelta s=0, \end{aligned}$$

it follows that $\lim_{t\rightarrow\infty}z(t)=0$. By Lemma 2.5, we obtain $\lim_{t\rightarrow\infty}x(t)=0$. Furthermore, for $t\in \left.[T_{1},\infty)\right._{\mathbb{T}}$, we have

$$\begin{aligned} r_{2}(t)z^{\varDelta }(t)=-\frac{3(1+p_{1}\eta_{1})}{2}- \int_{T_{1}}^{t} \int_{T_{1}}^{s}\frac{f(u,x(h(u)))}{r_{1}(s)}\varDelta u\varDelta s< 0 \end{aligned}$$

and

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

The proof is complete. □

Theorem 3.2

Assume that

$$\begin{aligned} \int_{t_{0}}^{\infty}\frac{1}{r_{2}(t)}\varDelta t=\infty \quad\textit{or}\quad \int_{t_{0}}^{\infty} \int_{t_{0}}^{v}\frac{1}{r_{1}(s)r_{2}(v)}\varDelta s\varDelta v= \infty. \end{aligned}$$

Then equation (1) has no eventually positive solution $x(t)$, for which $r_{2}(t)z^{\varDelta }(t)$ and $r_{1}(t) (r_{2}(t)z^{\varDelta }(t) )^{\varDelta }$ are both eventually negative.

Proof

Suppose that $x(t)$ is an eventually positive solution of equation (1), and there exists $T_{0}\in \left.[t_{0},\infty)\right._{\mathbb{T}}$ such that

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

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

$$\begin{aligned} r_{1}(s) \bigl(r_{2}(s)z^{\varDelta }(s) \bigr)^{\varDelta }-r_{1}(T_{1}) \bigl(r_{2}(T_{1})z^{\varDelta }(T_{1}) \bigr)^{\varDelta } =- \int_{T_{1}}^{s}f\bigl(u,x\bigl(h(u)\bigr)\bigr)\varDelta u< 0, \end{aligned}$$

which implies that

$$\begin{aligned} \bigl(r_{2}(s)z^{\varDelta }(s) \bigr)^{\varDelta }< \frac{r_{1}(T_{1}) (r_{2}(T_{1})z^{\varDelta }(T_{1}) )^{\varDelta }}{r_{1}(s)}. \end{aligned}$$

(11)

Integrating (11) from $T_{1}$ to $v\in \left.[\sigma(T_{1}),\infty)\right._{\mathbb{T}}$, we obtain

$$\begin{aligned} r_{2}(v)z^{\varDelta }(v)-r_{2}(T_{1})z^{\varDelta }(T_{1}) < r_{1}(T_{1}) \bigl(r_{2}(T_{1})z^{\varDelta }(T_{1}) \bigr)^{\varDelta } \int_{T_{1}}^{v}\frac{1}{r_{1}(s)}\varDelta s \end{aligned}$$

or

$$\begin{aligned} z^{\varDelta }(v)< \frac{r_{2}(T_{1})z^{\varDelta }(T_{1})}{r_{2}(v)}+ \frac{r_{1}(T_{1}) (r_{2}(T_{1})z^{\varDelta }(T_{1}) )^{\varDelta }}{r_{2}(v)} \int_{T_{1}}^{v}\frac{1}{r_{1}(s)}\varDelta s. \end{aligned}$$

(12)

Integrating (12) from $T_{1}$ to $t\in \left.[\sigma(T_{1}),\infty)\right._{\mathbb{T}}$, we obtain

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

Letting $t\rightarrow\infty$, we have $z(t)\rightarrow-\infty$. It is a contradiction because $z(t)=x(t)+p(t)x(g(t))$ is eventually positive. The proof is complete. □

By Theorems 3.1 and 3.2, we have the following corollary.

Corollary 3.3

Assume that

$$\begin{aligned} \int_{t_{0}}^{\infty} \int_{t_{0}}^{s}\frac{q(u)}{r_{1}(s)}\varDelta u\varDelta s< \infty. \end{aligned}$$

If the function f satisfies (C4A) and there exists $L>0$ such that

$$\begin{aligned} \bigl|f(t,x_{1})-f(t,x_{2})\bigr|\leq L\cdot q(t)|x_{1}-x_{2}|,\quad x_{1},x_{2}\in\biggl[0,2 \int_{t_{0}}^{\infty}\frac{1}{r_{2}(t)}\varDelta t \biggr], \end{aligned}$$

then $\int_{t_{0}}^{\infty}\frac{1}{r_{2}(t)}\varDelta t<\infty$ is a necessary and sufficient condition for equation (1) to have an eventually positive solution $x(t)$ satisfying the requirement that $\lim_{t\rightarrow\infty}x(t)=0$ and $r_{2}(t)z^{\varDelta }(t)$, $r_{1}(t) (r_{2}(t)z^{\varDelta }(t) )^{\varDelta }$ are both eventually negative.

Theorem 3.4

Assume that the function f satisfies (C4A) and there exists $M_{2}>0$ such that

$$\begin{aligned} \int_{t_{0}}^{\infty} \int_{v}^{\infty}\frac{1}{r_{1}(s)r_{2}(v)}\varDelta s\varDelta v=M_{2}< \infty \end{aligned}$$

(13)

and

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

(14)

Define $H_{2}(t)=\int_{t}^{\infty}\int_{v}^{\infty}\frac {1}{r_{1}(s)r_{2}(v)}\varDelta s\varDelta v$, which satisfies $\lim_{t\rightarrow \infty}\frac{H_{2}(g(t))}{H_{2}(t)}=\eta_{2}\in(0,1]$. If there exists $L>0$ such that

$$\begin{aligned} \bigl|f(t,x_{1})-f(t,x_{2})\bigr|\leq L\cdot q(t)|x_{1}-x_{2}|, \quad x_{1},x_{2} \in[0,2M_{2}], \end{aligned}$$

(15)

then equation (1) has an eventually positive solution $x(t)$ with $\lim_{t\rightarrow\infty}x(t)=0$, and there exists $T_{1}\in \left.[t_{0},\infty)\right._{\mathbb{T}}$ such that

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

Proof

From (C2), for $0\leq p_{0}<1$, choose $p_{1}$ such that $p_{0}< p_{1}<(1+4p_{0})/5<1$. By (13) and (14), there exists $T_{0}\in \left.[t_{0},\infty)\right._{\mathbb{T}}$ such that

$$\begin{aligned} \frac{5p_{1}-1}{4}\leq p(t)\leq p_{1}< 1,\qquad p(t) \frac{H_{2}(g(t))}{H_{2}(t)}\geq\frac{5p_{1}-1}{4}\eta_{2},\quad t \in \left.[T_{0},\infty)\right._{\mathbb{T}} , \end{aligned}$$

(16)

and

$$\begin{aligned} \int_{T_{0}}^{\infty}q(t)\varDelta t\leq\min\biggl\{ \frac{1-p_{1}\eta_{2}}{4K},1 \biggr\} , \end{aligned}$$

where $K=\max\{|f_{0}(x)|:x\in[0,2M_{2}]\}$. Similarly, there always exists $T_{1}\in(T_{0},\infty)_{\mathbb{T}}$ such that $h(t)\geq T_{0}$ for $t\in \left.[T_{1},\infty)\right._{\mathbb{T}}$.

Define

$$\begin{aligned} \varOmega _{2}=\bigl\{ x(t)\in BC\left.[T_{0},\infty)\right._{\mathbb{T}}:H_{2}(t) \leq x(t)\leq2H_{2}(t)\bigr\} . \end{aligned}$$

Then $\varOmega _{2}$ is also a bounded, convex, and closed subset of $BC\left.[T_{0},\infty)\right._{\mathbb{T}}$. For any $x\in \varOmega _{2}$, by (C4A) we have

$$\begin{aligned} 0< f\bigl(t,x\bigl(h(t)\bigr)\bigr)\leq q(t)f_{0}\bigl(x\bigl(h(t) \bigr)\bigr),\quad t\in \left.[T_{1},\infty)\right._{\mathbb{T}}. \end{aligned}$$

Define the operators $U_{2}$ and $S_{2}: \varOmega _{2}\rightarrow BC\left.[T_{0},\infty)\right._{\mathbb{T}}$ as

$$\begin{aligned} (U_{2}x) (t)=\frac{3}{2}p_{1}\eta_{2}H_{2}(t)-p(t)x \bigl(g(t)\bigr),\quad t\in \left.[T_{0},\infty)\right._{\mathbb{T}}, \end{aligned}$$

and

$$\begin{aligned} (S_{2}x) (t)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{3}{2}H_{2}(t)+\int_{t}^{\infty}\int_{v}^{\infty}\int _{s}^{\infty }\frac{f(u,x(h(u)))}{r_{1}(s)r_{2}(v)}\varDelta u\varDelta s\varDelta v,& t\in \left.[T_{1},\infty)\right._{\mathbb{T}},\\ (S_{2}x)(T_{1}),& t\in [T_{0},T_{1}]_{\mathbb{T}}. \end{array}\displaystyle \right . \end{aligned}$$

Similarly to the proof of Theorem 3.1, we can conclude that $U_{2}$ and $S_{2}$ satisfy all the conditions in Lemma 2.4. Therefore, there exists $x\in \varOmega _{2}$ such that $(U_{2}+S_{2})x=x$, which means that $x(t)$ is an eventually positive solution of equation (1). It follows that for $t\in \left.[T_{1},\infty)\right._{\mathbb{T}}$, we have

$$\begin{aligned} x(t)=\frac{3(1+p_{1}\eta_{2})}{2}H_{2}(t)-p(t)x\bigl(g(t)\bigr)+ \int_{t}^{\infty } \int_{v}^{\infty} \int_{s}^{\infty}\frac {f(u,x(h(u)))}{r_{1}(s)r_{2}(v)}\varDelta u\varDelta s\varDelta v. \end{aligned}$$

As

$$\begin{aligned} \int_{t}^{\infty} \int_{v}^{\infty} \int_{s}^{\infty}\frac {f(u,x(h(u)))}{r_{1}(s)r_{2}(v)}\varDelta u\varDelta s\varDelta v\leq KH_{2}(t) \int_{T_{1}}^{\infty}q(u)\varDelta u \end{aligned}$$

for $t\in \left.[T_{1},\infty)\right._{\mathbb{T}}$, and

$$\begin{aligned} \lim_{t\rightarrow\infty}KH_{2}(t) \int_{T_{1}}^{\infty}q(u)\varDelta u=0, \end{aligned}$$

we have $\lim_{t\rightarrow\infty}z(t)=0$, which implies that $\lim_{t\rightarrow\infty}x(t)=0$ by Lemma 2.5. For $t\in \left.[T_{1},\infty)\right._{\mathbb{T}}$, we obtain

$$\begin{aligned} r_{2}(t)z^{\varDelta }(t)=-\frac{3(1+p_{1}\eta_{2})}{2} \int_{t}^{\infty }\frac{1}{r_{1}(s)}\varDelta s- \int_{t}^{\infty} \int_{s}^{\infty}\frac{f(u,x(h(u)))}{r_{1}(s)}\varDelta u\varDelta s< 0 \end{aligned}$$

and

$$\begin{aligned} r_{1}(t) \bigl(r_{2}(t)z^{\varDelta }(t) \bigr)^{\varDelta }=\frac{3(1+p_{1}\eta _{2})}{2}+ \int_{t}^{\infty}f\bigl(u,x\bigl(h(u)\bigr)\bigr)\varDelta u>0. \end{aligned}$$

The proof is complete. □

Remark 3.5

Actually, (6) in Theorem 3.1 and (15) in Theorem 3.4 will hold especially when $f(t,x)=q(t)f_{0}(x)$ and $f_{0}(x)$ satisfies the Lipschitz condition on $[0,2M_{1}]$ and $[0,2M_{2}]$, respectively.

Remark 3.6

Letting $\mathbb{T}=[a,\infty)$, $r_{1}(t)=\frac{1}{p(t)}$, $r_{2}(t)=\frac{1}{r(t)}$, $p(t)=0$, and the function f satisfying (C4A), equation (1) is simplified as (2). Therefore, Theorems 3.1, 3.2, 3.4, and Corollary 3.3 complement and extend the results in Mojsej and Tartal’ová [8].

In the sequel, we change the condition of the function f from (C4A) to (C4B), and get some similar conclusions as follows.

Theorem 3.7

Assume that the function f satisfies (C4B) and

$$\begin{aligned} \int_{t_{0}}^{\infty}\frac{1}{r_{2}(t)}\varDelta t< \infty \end{aligned}$$

(17)

and

$$\begin{aligned} \int_{t_{0}}^{\infty} \int_{t_{0}}^{s}\frac {f(u,2H_{1}(h(u)))}{r_{1}(s)}\varDelta u\varDelta s< \infty, \end{aligned}$$

(18)

where $H_{1}(t)=\int_{t}^{\infty}\frac{1}{r_{2}(v)}\varDelta v$, which satisfies $\lim_{t\rightarrow \infty}\frac{H_{1}(g(t))}{H_{1}(t)}=\eta_{1}\in(0,1]$. Then equation (1) has an eventually positive solution $x(t)$ with $\lim_{t\rightarrow\infty}x(t)=0$, and there exists $T_{1}\in \left.[t_{0},\infty)\right._{\mathbb{T}}$ such that

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

Proof

From (C2), for $0\leq p_{0}<1$, choose $p_{1}$ such that $p_{0}< p_{1}<(1+4p_{0})/5<1$. By (17) and (18), there exists $T_{0}\in \left.[t_{0},\infty)\right._{\mathbb{T}}$ such that (7) holds and

$$\begin{aligned} \int_{T_{0}}^{\infty} \int_{T_{0}}^{s}\frac {f(u,2H_{1}(h(u)))}{r_{1}(s)}\varDelta u\varDelta s\leq \min\biggl\{ \frac{1-p_{1}\eta_{1}}{4},1 \biggr\} . \end{aligned}$$

Similarly, there always exists $T_{1}\in (T_{0},\infty)_{\mathbb{T}}$ such that $h(t)\geq T_{0}$ for $t\in \left.[T_{1},\infty)\right._{\mathbb{T}}$.

Define $\varOmega _{1}$ as in Theorem 3.1. For any $x\in \varOmega _{1}$, by (C4B) we have

$$\begin{aligned} 0< f\bigl(t,x\bigl(h(t)\bigr)\bigr)\leq f\bigl(t,2H_{1}\bigl(h(t) \bigr)\bigr), \quad t\in \left.[T_{1},\infty)\right._{\mathbb{T}}. \end{aligned}$$

In addition, define $U_{1}$ and $S_{1}$ as in Theorem 3.1. It is obvious that $U_{1}$ and $S_{1}$ also satisfy all the conditions in Lemma 2.4. Hence, there exists $x\in \varOmega _{1}$ such that $(U_{1}+S_{1})x=x$. That is, $x(t)$ is an eventually positive solution of equation (1). For $t\in \left.[T_{1},\infty)\right._{\mathbb{T}}$, it follows that

$$\begin{aligned} x(t)=\frac{3(1+p_{1}\eta_{1})}{2}H_{1}(t)-p(t)x\bigl(g(t)\bigr)+ \int_{t}^{\infty } \int_{T_{1}}^{v} \int_{T_{1}}^{s}\frac {f(u,x(h(u)))}{r_{1}(s)r_{2}(v)}\varDelta u\varDelta s\varDelta v. \end{aligned}$$

As

$$\begin{aligned} \int_{t}^{\infty} \int_{T_{1}}^{v} \int_{T_{1}}^{s}\frac {f(u,x(h(u)))}{r_{1}(s)r_{2}(v)}\varDelta u\varDelta s\varDelta v\leq H_{1}(t) \int_{T_{1}}^{\infty} \int_{T_{1}}^{s}\frac {f(u,2H_{1}(h(u)))}{r_{1}(s)}\varDelta u\varDelta s \end{aligned}$$

for $t\in \left.[T_{1},\infty)\right._{\mathbb{T}}$, and

$$\begin{aligned} \lim_{t\rightarrow \infty}H_{1}(t) \int_{T_{1}}^{\infty} \int_{T_{1}}^{s}\frac {f(u,2H_{1}(h(u)))}{r_{1}(s)}\varDelta u\varDelta s=0, \end{aligned}$$

it follows that $\lim_{t\rightarrow\infty}z(t)=0$. By Lemma 2.5, we obtain $\lim_{t\rightarrow\infty}x(t)=0$. Furthermore, for $t\in \left.[T_{1},\infty)\right._{\mathbb{T}}$, it satisfies

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

The proof is complete. □

Theorem 3.8

Assume that the function f satisfies (C4B) and

$$\begin{aligned} \int_{t_{0}}^{\infty} \int_{v}^{\infty}\frac{1}{r_{1}(s)r_{2}(v)}\varDelta s\varDelta v< \infty \end{aligned}$$

(19)

and

$$\begin{aligned} \int_{t_{0}}^{\infty}f\bigl(t,2H_{2}\bigl(h(t) \bigr)\bigr)\varDelta t< \infty, \end{aligned}$$

(20)

where $H_{2}(t)=\int_{t}^{\infty}\int_{v}^{\infty}\frac {1}{r_{1}(s)r_{2}(v)}\varDelta s\varDelta v$, which satisfies $\lim_{t\rightarrow \infty}\frac{H_{2}(g(t))}{H_{2}(t)}=\eta_{2}\in(0,1]$. Then equation (1) has an eventually positive solution $x(t)$ with $\lim_{t\rightarrow\infty}x(t)=0$, and there exists $T_{1}\in \left.[t_{0},\infty)\right._{\mathbb{T}}$ such that

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

Proof

From (C2), for $0\leq p_{0}<1$, choose $p_{1}$ such that $p_{0}< p_{1}<(1+4p_{0})/5<1$. By (19) and (20), there exists $T_{0}\in \left.[t_{0},\infty)\right._{\mathbb{T}}$ such that (16) holds and

$$\begin{aligned} \int_{T_{0}}^{\infty}f\bigl(t,2H_{2}\bigl(h(t) \bigr)\bigr)\varDelta t\leq\min\biggl\{ \frac{1-p_{1}\eta_{2}}{4},1 \biggr\} . \end{aligned}$$

From (C3), there always exists $T_{1}\in (T_{0},\infty)_{\mathbb{T}}$ such that $h(t)\geq T_{0}$ for $t\in \left.[T_{1},\infty)\right._{\mathbb{T}}$.

Define $\varOmega _{2}$ as in Theorem 3.4. For any $x\in \varOmega _{1}$, by (C4B) we have

$$\begin{aligned} 0< f\bigl(t,x\bigl(h(t)\bigr)\bigr)\leq f\bigl(t,2H_{2}\bigl(h(t) \bigr)\bigr),\quad t\in \left.[T_{1},\infty)\right._{\mathbb{T}}. \end{aligned}$$

In addition, define the operators $U_{2}$ and $S_{2}$ as in Theorem 3.4. It is obvious that $U_{2}$ and $S_{2}$ also satisfy all the conditions in Lemma 2.4. Then there exists $x\in \varOmega _{2}$ such that $(U_{2}+S_{2})x=x$, which means that $x(t)$ is an eventually positive solution of equation (1). For $t\in \left.[T_{1},\infty)\right._{\mathbb{T}}$, it follows that

$$\begin{aligned} x(t)=\frac{3(1+p_{1}\eta_{2})}{2}H_{2}(t)-p(t)x\bigl(g(t)\bigr)+ \int_{t}^{\infty } \int_{v}^{\infty} \int_{s}^{\infty}\frac {f(u,x(h(u)))}{r_{1}(s)r_{2}(v)}\varDelta u\varDelta s\varDelta v. \end{aligned}$$

As

$$\begin{aligned} \int_{t}^{\infty} \int_{v}^{\infty} \int_{s}^{\infty}\frac {f(u,x(h(u)))}{r_{1}(s)r_{2}(v)}\varDelta u\varDelta s\varDelta v\leq H_{2}(t) \int_{T_{1}}^{\infty}f\bigl(u,2H_{2}\bigl(h(u) \bigr)\bigr)\varDelta u \end{aligned}$$

for $t\in \left.[T_{1},\infty)\right._{\mathbb{T}}$, and

$$\begin{aligned} \lim_{t\rightarrow \infty}H_{2}(t) \int_{T_{1}}^{\infty}f\bigl(u,2H_{2}\bigl(h(u) \bigr)\bigr)\varDelta u=0, \end{aligned}$$

we have $\lim_{t\rightarrow\infty}z(t)=0$, which implies that $\lim_{t\rightarrow\infty}x(t)=0$ by Lemma 2.5. We also obtain

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

The proof is complete. □

Remark 3.9

Indeed, if the function f satisfies both of (C4A) and (C4B), compared to Theorems 3.1 and 3.4, Theorems 3.7 and 3.8 are more convenient to employ, respectively.

4 Examples

In this section, we give two examples to show the application of our results. The first example is presented to illustrate Theorems 3.1 and 3.7.

Example 4.1

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

$$ \biggl(t^{5} \biggl(t^{2} \biggl(x(t)+ \frac{t-1}{2t}x(t+2) \biggr)^{\varDelta } \biggr)^{\varDelta } \biggr)^{\varDelta } +t^{2}x(t)+\frac{x^{3}(t)}{t}=0, $$

(21)

where $r_{1}(t)=t^{5}$, $r_{2}(t)=t^{2}$, $p(t)=\frac{t-1}{2t}$, $g(t)=t+2$, $h(t)=t$, $f(t,x)=t^{2}x+\frac{x^{3}}{t}$, $t_{0}=1$.

Take $q(t)=t^{2}$ and $f_{0}(x)=x+x^{3}$. It is obvious that the coefficients of equation (21) satisfy (C1)-(C4), (C4A), and (C4B). Since

$$\begin{aligned}& \int_{t_{0}}^{\infty}\frac{1}{r_{2}(t)}\varDelta t= \int_{1}^{\infty}\frac{1}{t^{2}}\varDelta t=M_{1}< 2,\\& H_{1}(t)= \int_{t}^{\infty}\frac{1}{r_{2}(v)}\varDelta v= \int_{t}^{\infty}\frac{1}{v^{2}}\varDelta v< 2,\\& \lim_{t\rightarrow \infty}\frac{H_{1}(g(t))}{H_{1}(t)}=\lim_{t\rightarrow \infty} \frac{\int_{t+2}^{\infty}\frac{1}{v^{2}}\varDelta v}{\int_{t}^{\infty}\frac{1}{v^{2}}\varDelta v}=\lim_{t\rightarrow \infty}\frac{t^{2}}{(t+2)^{2}}=1,\\& \int_{t_{0}}^{\infty} \int_{t_{0}}^{s}\frac{q(u)}{r_{1}(s)}\varDelta u\varDelta s= \int_{1}^{\infty} \int_{1}^{s}\frac{u^{2}}{s^{5}}\varDelta u\varDelta s< \int_{1}^{\infty} \int_{1}^{s}\frac{1}{s^{3}}\varDelta u\varDelta s< \int_{1}^{\infty}\frac{1}{s^{2}}\varDelta s< \infty, \end{aligned}$$

and, for any $x_{1},x_{2}\in[0,4]$,

$$\begin{aligned} \bigl|f(t,x_{1})-f(t,x_{2})\bigr| =&\biggl\vert t^{2}(x_{1}-x_{2})+\frac {1}{t} \bigl(x_{1}^{3}-x_{2}^{3}\bigr)\biggr\vert \\ \leq& t^{2}\bigl(|x_{1}-x_{2}|+\bigl|x_{1}^{3}-x_{2}^{3}\bigr| \bigr)\leq\bigl(1+x_{1}^{2}+x_{1}x_{2}+x_{2}^{2} \bigr)\cdot t^{2}|x_{1}-x_{2}| \\ \leq&49\cdot q(t)|x_{1}-x_{2}|, \end{aligned}$$

by Theorem 3.1 we see that equation (21) has an eventually positive solution $x(t)$ with $\lim_{t\rightarrow \infty}x(t)=0$, and $r_{2}(t)z^{\varDelta }(t)$, $r_{1}(t) (r_{2}(t)z^{\varDelta }(t) )^{\varDelta }$ are both eventually negative.

On the other hand, since $H_{1}(t)<2$, we obtain

$$\begin{aligned} f\bigl(u,2H_{1}\bigl(h(u)\bigr)\bigr)=u^{2} \cdot2H_{1}(u)+\frac{8H_{1}^{3}(u)}{u}\leq u^{2} \bigl(2H_{1}(u)+8H_{1}^{3}(u) \bigr)< 68u^{2} \end{aligned}$$

and

$$\begin{aligned} \int_{t_{0}}^{\infty} \int_{t_{0}}^{s}\frac {f(u,2H_{1}(h(u)))}{r_{1}(s)}\varDelta u\varDelta s < &68 \int_{1}^{\infty} \int_{1}^{s}\frac{u^{2}}{s^{5}}\varDelta u\varDelta s \\ < &68 \int_{1}^{\infty} \int_{1}^{s}\frac{1}{s^{3}}\varDelta u\varDelta s< 68 \int_{1}^{\infty}\frac{1}{s^{2}}\varDelta s< \infty. \end{aligned}$$

By Theorem 3.7 we have the same conclusion.

Then the second example demonstrates Theorems 3.4 and 3.8.

Example 4.2

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

$$ \biggl(t^{2} \biggl(t \biggl( \biggl(1+ \frac{1}{t} \biggr)x(t) \biggr)^{\varDelta } \biggr)^{\varDelta } \biggr)^{\varDelta } +\frac{x (\frac{t}{2} )}{t^{2}}=0, $$

(22)

where $r_{1}(t)=t^{2}$, $r_{2}(t)=t$, $p(t)=\frac{1}{t}$, $g(t)=t$, $h(t)=\frac{t}{2}$, $f(t,x)=\frac{x}{t^{2}}$, $t_{0}=1$.

Take $q(t)=\frac{1}{t^{2}}$ and $f_{0}(x)=x$. It is obvious that the coefficients of equation (22) satisfy (C1)-(C4), (C4A), and (C4B). Since

$$\begin{aligned}& \begin{aligned}[b] \int_{t_{0}}^{\infty} \int_{v}^{\infty}\frac{1}{r_{1}(s)r_{2}(v)}\varDelta s\varDelta v&= \int_{1}^{\infty} \int_{v}^{\infty}\frac{1}{s^{2}v}\varDelta s\varDelta v\\ &= \int_{1}^{\infty} \int_{1}^{s}\frac{1}{s^{2}v}\varDelta v\varDelta s=M_{2}< \infty, \end{aligned}\\& H_{2}(t)= \int_{t}^{\infty} \int_{v}^{\infty}\frac {1}{r_{1}(s)r_{2}(v)}\varDelta s\varDelta v= \int_{t}^{\infty} \int_{v}^{\infty}\frac{1}{s^{2}v}\varDelta s\varDelta v\leq M_{2},\\& \lim_{t\rightarrow \infty}\frac{H_{2}(g(t))}{H_{2}(t)}=\lim_{t\rightarrow \infty} \frac{H_{2}(t)}{H_{2}(t)}=1,\\& \int_{t_{0}}^{\infty}q(t)\varDelta t= \int_{1}^{\infty}\frac{1}{t^{2}}\varDelta t< \infty, \end{aligned}$$

and for any $x_{1},x_{2}\in[0,2M_{2}]$,

$$\begin{aligned} \bigl|f(t,x_{1})-f(t,x_{2})\bigr|=\frac{|x_{1}-x_{2}|}{t^{2}}=1\cdot q(t)|x_{1}-x_{2}|, \end{aligned}$$

by Theorem 3.4 we see that equation (22) has an eventually positive solution $x(t)$ with $\lim_{t\rightarrow \infty}x(t)=0$, $r_{2}(t)z^{\varDelta }(t)$ is eventually negative and $r_{1}(t) (r_{2}(t)z^{\varDelta }(t) )^{\varDelta }$ is eventually positive.

On the other hand, since $H_{2}(t)\leq M_{2}$, we obtain

$$\begin{aligned} f\bigl(t,2H_{2}\bigl(h(t)\bigr)\bigr)=\frac{2H_{2}(\frac {t}{2})}{t^{2}}\leq \frac{2M_{2}}{t^{2}} \end{aligned}$$

and

$$\begin{aligned} \int_{t_{0}}^{\infty}f\bigl(t,2H_{2}\bigl(h(t) \bigr)\bigr)\varDelta t\leq2M_{2} \int_{1}^{\infty}\frac{1}{t^{2}}\varDelta t< \infty. \end{aligned}$$

By Theorem 3.8 we have the same conclusion.