Oscillation criteria for a class of higher odd order neutral difference equations with continuous variable

Abstract

In this paper, we are mainly concerned with oscillatory behavior of solutions for a class of higher odd order nonlinear neutral difference equations with continuous variable. By converting the above difference equations to the corresponding differential equations and inequalities, the oscillatory criteria are obtained. In addition, examples are given to illustrate the obtained criteria, respectively.

1 Introduction

Difference equations have attracted a great deal of attention of researchers in mathematics, biology, physics, and economy. This is specially due to the applications in various problems of biology, physics, economy. Among the topics studied for oscillation of the solutions has been investigated intensively. Please see [1–18].

In this paper, we deal with the nonlinear neutral difference equation with continuous variable of the form

$$ \Delta^{m}_{\tau}\bigl( x(t)-px(t-r)\bigr)+f \bigl(t,x\bigl(g(t)\bigr)\bigr)=0, $$

(1.1)

where $m\geq3$, $p\geq0$, τ and r are positive constants, $\Delta_{\tau}x(t)=x(t+\tau)-x(t)$, $0< g(t)< t$, $g\in C^{1}([t_{0},\infty), R _{+})$, $g'(t)>0$, and $f \in C([t_{0},\infty)\times R, R)$. Throughout this paper we assume that

$$ g(t+\tau)\geq g(t)+\tau \quad\mbox{for } t\geq t_{0} $$

(1.2)

and

$$ {f(t, u)}/{u}\ge q(t)>0 \quad\mbox{for } u\neq0 \mbox{ and some } q\in C(R, R_{+}). $$

(1.3)

Let $t'_{0}=\min\{g(t_{0}), t_{0}-r\}$ and $I_{0}=[t'_{0}, t_{0}]$. A function x is called the solution of (1.1) with $x(t)=\varphi(t)$ for $t\in I_{0}$ and $\varphi\in {C(I_{0}, R)}$ if it satisfies (1.1) for $t\ge t_{0}$.

A solution x is said to be oscillatory if it is neither eventually positive nor eventually negative; it is called nonoscillatory if it is not oscillatory.

The organization of this paper is as follows. We will give the main results in Section 2 and leave the proofs to Section 5. Three demonstrated examples will be presented in Section 3. In Section 4, some lemmas will be given to prove the main results.

2 Statement of the main results

For later convenience, let

$$ \bar{q}(t)=\alpha\min_{t \leq s\leq t+m\tau }\bigl\{ q(s)\bigr\} \Bigl(\min_{g(t)\le s\le g(t)+m\tau}\bigl\{ \bigl(g^{-1}(s) \bigr)'\bigr\} \Bigr)^{m}, $$

(2.1)

where $0<\alpha<1$. Throughout this paper, the function $\bar{q}$ will play an important role in the oscillatory criteria for (1.1). Let

$$ \beta_{1}=\inf_{t \geq T} \biggl\{ \frac{(g^{-1}(t)-t)^{m-1}\bar {q}(g^{-1}(t))}{(m-1)!\tau^{m}} \biggr\} $$

(2.2)

and

$$ \beta_{2}=\inf_{t \geq T} \biggl\{ \frac{(g^{-1}(t)-t)^{m-1}\bar {q}(t)}{(m-1)!\tau^{m}} \biggr\} , $$

(2.3)

where $T \geq t_{0}$ is sufficiently large.

Theorem 2.1

Assume that (1.1) with $0 < p <1$ satisfies

$$ r\beta_{1}\sum_{i=1}^{n} i p^{i} \geq1 $$

(2.4)

and

$$ 0 \leq\liminf_{t \to\infty} \int_{t-r}^{t} \bigl(g^{-1}(s)-s\bigr)^{m-1}\bar {q}\bigl(g^{-1}(s) \bigr)\,ds \leq\frac{(m-1)!\tau^{m}(1-p)e^{-1}}{p-p^{n+1}} $$

(2.5)

for some integer $n \geq1$. Also assume that $\bar{q}(t)$ given by (2.1) is nonincreasing. Then, for every bounded solution $x(t)$ of (1.1), either $x(t)$ is oscillatory or $\liminf_{t \to\infty } (|x(t)|-p|x(t-r)| )<0$.

Corrollary 2.2

The conclusion of Theorem  2.1 still holds if (2.5) is replaced by

$$ 0 \leq\liminf_{t \to\infty} \int_{t-r}^{t} \bigl(g^{-1}(s)-s\bigr)^{m-1}\bar {q}\bigl(g^{-1}(s) \bigr)\,ds \leq\frac{(m-1)!\tau^{m}}{ep}. $$

(2.6)

Corrollary 2.3

Assume that (1.1) with $0 < p <1$ satisfies

$$ r\beta_{2}\sum_{i=1}^{n} i p^{i} \geq1 $$

(2.7)

and

$$ 0\leq\liminf_{t \to\infty} \int_{t-r}^{t} \bigl(g^{-1}(s)-s\bigr)^{m-1}\bar {q}(s)\,ds \leq\frac{(m-1)!\tau^{m}(1-p)e^{-1}}{p-p^{n+1}} $$

(2.8)

for some integer $n \geq1$. Also assume that $\bar{q}(t)$ given by (2.1) is nondecreasing. Then the conclusion of Theorem  2.1 holds.

Corrollary 2.4

The conclusion of Corollary  2.3 still holds if (2.8) is replaced by

$$ 0 \leq\liminf_{t \to\infty} \int_{t-r}^{t} \bigl(g^{-1}(s)-s\bigr)^{m-1}\bar {q}(s)\,ds \leq\frac{(m-1)!\tau^{m}}{ep}. $$

(2.9)

Corrollary 2.5

Assume $0 < p <1$ and $r=k\tau$. Under the assumptions of either Theorem  2.1 or Corollary  2.2 or Corollary  2.3 or Corollary  2.4, every bounded solution $x(t)$ of (1.1) is oscillatory.

The following results are for the bounded solutions of (1.1) with $p>1$.

Theorem 2.6

Assume that $p>1$, $r=k\tau$, $k \in N$, $r \geq t+m\tau-g(t)$, and

$$ r\beta_{2}\sum_{i=1}^{n} \frac{(i-1)}{p^{i}} \geq1 $$

(2.10)

for some integer $n \geq2$. Also assume that $\bar{q}(t)$ given by (2.1) is nondecreasing. Then every bounded solution $x(t)$ of (1.1) is oscillatory.

Corrollary 2.7

Assume that $p >1$, $r=k\tau$, $k \in N$, $r \geq t+m\tau-g(t)$, and

$$ r\beta_{1} \sum_{i=1}^{n} \frac{(i-1)}{p^{i}} \geq1 $$

(2.11)

for some integer $n \geq2$. Also assume that $\bar{q}(t)$ given by (2.1) is nonincreasing. Then every bounded solution $x(t)$ of (1.1) is oscillatory.

3 Examples

Three examples will be given in this section to demonstrate the applications of the obtained results. From (2.2) and (2.3) it is clear that both $\beta_{1}$ and $\beta_{2}$ are nondecreasing functions of T. The following examples show that $\beta_{1}$ and $\beta_{2}$ may be independent of T or increasing functions of T.

Example 1

Consider the difference equation

$$ \Delta^{m}_{1} \biggl(x(t)-\frac{1}{2}x(t-1) \biggr)+ \biggl((m-1)!+\frac{1}{t} \biggr)x (t- 1) =0 $$

(3.1)

for $t >0$, where m is an odd positive integer $m \geq3$. Viewing (3.1) as (1.1), we have $\tau=1$, $0< p=1/2<1$, $r=1$, $q(t)=(m-1)!+1/t$ and $g(t)=t-1$. Then, according to (2.1),

$$\bar{q}(t)=\alpha \biggl((m-1)!+ \frac{1}{t+m} \biggr). $$

So

$$\beta_{1}=\inf_{t \geq T} \biggl\{ \frac{(t+1-t)^{m-1}\cdot\alpha ((m-1)!+\frac{1}{t+m+1} )}{(m-1)!\cdot1^{m}} \biggr\} =\alpha $$

with $T \geq3$. Since

$$\beta_{1} \sum_{i=1}^{3}irp^{i}= \alpha\cdot \biggl(\frac{1}{2}+2 \times \frac{1}{4}+3 \times \frac{1}{8} \biggr)=\frac{11\alpha}{8} \geq1 $$

holds for $\alpha\in[8/11, 1)$ and

$$0 \leq \liminf_{t \to\infty} \int_{t-1}^{t}(s+1-s)^{m-1} \cdot\alpha \biggl((m-1)!+\frac{1}{s+m+1} \biggr)\,ds = \alpha\cdot(m-1)! \leq \frac{2\cdot(m-1)!}{e} $$

holds for any $\alpha\in(0, 2/e]$, (2.4) and (2.6) are satisfied for $n=3$ and $\alpha\in[8/11, 2/e]$. Since $r=1=\tau$, by Corollaries 2.2 and 2.5, every bounded solution $x(t)$ of (3.1) is oscillatory.

Example 2

Consider the difference equation

$$ \Delta^{m}_{\frac{\pi}{m}} \bigl(x(t)-2x(t-4\pi) \bigr)+8x(t-\pi)+ \frac{8\sigma}{1+t^{2}}x^{3}(t-\pi) =0, $$

(3.2)

for $t >0$, where m is an odd positive integer with $m \geq3$ and σ is a positive real number. Regarding (3.2) as (1.1), we have $\tau=\pi/m$, $p=2$, $r=4\pi$, $g(t)=t-\pi$ and $q(t)=8$. Then, for some $\alpha\in(0,1)$, $\bar{q}=8\alpha$ by (2.1). Moreover, $r \geq t+m\tau-g(t)$ and $r=k\tau$ are satisfied. In addition,

$$\beta_{2}=\inf_{t \geq T} \biggl\{ \frac{8\alpha\cdot(t+\pi -t)^{m-1}}{(m-1)!(\frac{\pi}{m})^{m}} \biggr\} =\frac{8m^{m}\alpha}{\pi(m-1)!}, $$

where $T \geq12\pi$. So (2.10) is satisfied since

$$\beta_{2}\sum_{i=1}^{3} \frac{4 \pi(i-1)}{p^{i}}=\frac{8m^{m}\alpha}{\pi (m-1)!}\times4\pi\times \biggl(\frac{1}{2^{2}}+ \frac{2}{2^{3}} \biggr)=\alpha\frac{16m^{m}}{(m-1)!} \geq1 $$

holds for $\alpha\in[(m-1)!/(16m^{m}), 1)$. By Theorem 2.6, every bounded solution $x(t)$ of (3.2) is oscillatory.

Example 3

Consider the difference equation

$$ \Delta^{m}_{\frac{\pi}{m}} \bigl(x(t)-2x(t-2\pi) \bigr)+e^{-\frac{\sigma}{t}}x(t- \pi)=0, $$

(3.3)

for $t >0$, where m is an odd positive integer with $m \geq3$ and σ is a positive constant. Regarding (3.3) as (1.1), we have $\tau=\pi/m$, $p=2$, $r=2\pi$, $g(t)=t-\pi$, and $q(t)=e^{-\frac{\sigma}{t}}$. Then, for some $\alpha\in (0,1)$, $\bar{q}=\alpha e^{-\frac{\sigma}{t}}$ by (2.1). Moreover, $r \geq t+m\tau-g(t)$ and $r=k\tau$ are satisfied. In addition,

$$\beta_{2}=\inf_{t \geq T} \biggl\{ \frac{\alpha e^{-\frac{\sigma}{t}} \cdot (t+\pi-t)^{m-1}}{(m-1)!(\frac{\pi}{m})^{m}} \biggr\} =\frac{m^{m}\alpha }{e^{\sigma/T}\pi(m-1)!}\to\frac{m^{m}\alpha}{\pi(m-1)!}, $$

as $T \to\infty$. So (2.10) is satisfied when T is large enough since

$$\beta_{2}\sum_{i=1}^{3} \frac{2 \pi(i-1)}{p^{i}} \to \frac{\alpha m^{m}}{\pi(m-1)!}\times2\pi\times \biggl( \frac{1}{2^{2}}+\frac {2}{2^{3}} \biggr)=\alpha\frac{m^{m}}{(m-1)!} > 1 $$

as $T \to\infty$ for $\alpha\in ((m-1)!/(m^{m}), 1 )$. By Theorem 2.6, every bounded solution $x(t)$ of (3.3) is oscillatory.

4 Related lemmas

To prove the main results, we need to prove the following lemmas first. The first lemma is about a function $x(t)$ satisfying the differential inequality

$$ x'(t)+q(t)x\bigl(\tau(t)\bigr) \leq0, $$

(4.1)

where $q, \tau\in C([t_{0},\infty), R_{+})$, $\tau(t) \leq t$, and $\lim_{t \to\infty}\tau(t)=\infty$. Let

$$\eta=\liminf_{t \to\infty} \int_{\tau(t)}^{t}q(s)\,ds. $$

Lemma 4.1

Assume that τ is nondecreasing, $0 \leq\eta\leq e^{-1}$, and $x(t)$ is an eventually positive function satisfying (4.1). Set

$$r=\liminf_{t \to\infty} \frac{x(t)}{x(\tau(t))}. $$

Then r satisfies

$$\frac{1-\eta-\sqrt{1-2\eta-\eta^{2}}}{2}\leq r \leq1. $$

The above lemma can be found in [6], p.18.

Lemma 4.2

Let $0 \leq p <1$. Assume that $x(t)$ is a bounded and eventually positive (negative) solution of (1.1) with $z(t)=x(t)-px(t-r)$ and $\liminf_{t \to\infty} z(t) \geq0$ ($\limsup_{t \to\infty}z(t) \leq0$). Let

$$y(t)=\int_{t}^{t+\tau}dt_{1}\int _{t_{1}}^{t_{1}+\tau}dt_{2} \cdots \int _{t_{m-1}}^{t_{m-1}+\tau} z(\theta)\,d\theta. $$

Then $y(t)>0$ (<0), $(-1)^{k}y^{(k)}(t)>0$ (<0) for $1 \leq k \leq m$ eventually. Moreover,

$$ \Delta_{\tau}^{m}y(t)+\bar{q}(t)\sum _{i=0}^{n}p^{i}y\bigl(g(t)-ir\bigr)< 0\ (>0) $$

(4.2)

holds for any fixed natural number n and for all large enough t.

Proof

Suppose $x(t)$ is a bounded and eventually positive solution. Notice that $g(t)< t$ and $g'(t)>0$ for all $t \geq t_{0}$. So there exists a $t_{1} > t_{0}$ such that $x(g(t))>0$ for all $t \geq t_{1}$. From (1.1) it follows that

$$\Delta_{\tau}^{m}z(t)+f\bigl(t, x\bigl(g(t)\bigr)\bigr)=0. $$

By (1.3), we have $f(t, x(g(t))) \geq q(t)x(g(t)) >0$ for $t \geq t_{1}$. Therefore,

$$ y^{(m)}(t) +q(t)x\bigl(g(t)\bigr) \leq0 $$

(4.3)

for $t \geq t_{1}$. According to $q(t)x(g(t))> 0$, $y^{(m)}(t)<0$ for all $t \geq t_{1}$. Thus, $y^{(m-1)}(t)$ is decreasing so either $y^{(m-1)}(t)>0$ for all $t \geq t_{1}$ or $y^{(m-1)}(t) \leq y^{(m-1)}(t_{2}) <0$ for some $t_{2} >t_{1}$ and for all $t \geq t_{2}$. If the latter holds, then

$$y^{(m-k)}(t) \to-\infty, \quad k=2, 3, \ldots, m, $$

as $t \to\infty$, a contradiction to the boundedness of x and z. Therefore we have $y^{(m-1)}(t)>0$ for all $t \geq t_{1}$. Thus, $y^{(m-2)}(t)$ is increasing so either $y^{(m-2)}(t)<0$ for all $t \geq t_{1}$ or $y^{(m-2)}(t) \geq y^{(m-2)}(t_{3})>0$ for some $t_{3} \geq t_{1}$ and all $t \geq t_{3}$. If the latter holds, then

$$y^{(m-k)}(t) \to\infty,\quad k=3, 4, \ldots, m, $$

as $t \to\infty$, a contradiction again to the boundedness of x and z. Hence, we must have $y^{(m-2)}(t)<0$ for all $t \geq t_{1}$. Repeating the above process, we obtain $(-1)^{k}y^{(k)}(t)>0$ for $1 \leq k \leq m$ and all $t \geq t_{1}$. Therefore, $y(t)$ is decreasing so either $y(t)>0$ for all $t \geq t_{1}$ or there is a $t_{4} \geq t_{1}$ such that $y(t) \leq y(t_{4}) <0$ for $t \geq t_{4}$. Suppose the latter case holds. Then

$$\begin{aligned}& \int_{t}^{t+\tau}dt_{1}\int _{t_{1}}^{t_{1}+\tau}dt_{2}\cdots\int _{t_{m-1}}^{t_{m-1}+\tau} x(\theta)\,d\theta \\& \quad= y(t)+p \int_{t}^{t+\tau}dt_{1}\int _{t_{1}}^{t_{1}+\tau}dt_{2}\cdots\int _{t_{m-1}}^{t_{m-1}+\tau} x(\theta-r)\,d\theta \\& \quad\leq y(t_{4})+p \int_{t}^{t+\tau}dt_{1} \int_{t_{1}}^{t_{1}+\tau} dt_{2}\cdots\int _{t_{m-1}}^{t_{m-1}+\tau} x(\theta-r)\,d\theta \\& \quad \cdots \\& \quad \leq y(t_{4})\sum_{i=0}^{s-1} p^{i}+ p^{s} \int_{t}^{t+\tau}dt_{1} \int_{t_{1}}^{t_{1}+\tau}dt_{2} \cdots \int _{t_{m-1}}^{t_{m-1}+\tau} x(\theta-s r)\,d\theta \\& \quad \leq \frac{y(t_{4})(1-p^{s})}{1-p}+p^{s}M \tau^{m} \end{aligned}$$

for $t \geq t_{4}+sr$, where $M=\sup_{t \geq t_{0}}{x(t)}$ and s is any positive integer. Let $s \to\infty$ so $t \to\infty$ as well, $p^{s}M \tau^{m}$ then is arbitrarily small due to $0\leq p <1$. Thus,

$$\int_{t}^{t+\tau}dt_{1}\int _{t_{1}}^{t_{1}+\tau}dt_{2}\cdots\int _{t_{m-1}}^{t_{m-1}+\tau} x(\theta)\,d\theta< 0, $$

which contradicts the assumption that $x(t)$ is eventually positive. Therefore, we must have $y(t)>0$ for all $t \geq t_{1}$.

From (4.3) it follows that

$$\Delta_{\tau}^{m} z(t) +q(t)z\bigl(g(t)\bigr)+p q(t)x \bigl(g(t)-r\bigr) \leq0. $$

According to the definition of $z(t)$, the above inequality becomes

$$\Delta_{\tau}^{m} z(t) +q(t)z\bigl(g(t)\bigr)+p q(t)z \bigl(g(t)-r\bigr)+p^{2} q(t) x\bigl(g(t)-2r\bigr) \leq0. $$

Proceeding in the same way as the above, we have

$$\Delta_{\tau}^{m} z(t) +q(t)\sum_{i=0}^{n}p^{i}z \bigl(g(t)-ir\bigr)+p^{n+1} q(t) x\bigl(g(t)-(n+1)r\bigr) \leq0. $$

Since $q(t)p^{n+1}x(g(t)-(n+1)r)>0$ when t is large enough, the above inequality implies that

$$\Delta_{\tau}^{m} z(t) +q(t)\sum_{i=0}^{n}p^{i}z \bigl(g(t)-ir\bigr) < 0. $$

In order to integrate the above inequality, we need to show that $z(t)$ is positive. If $p=0$, then $z(t)=x(t)>0$ holds eventually. Now suppose $0< p<1$. Since $y^{(m)}(t)=\Delta_{\tau}^{m}z(t)<0$ for $t \geq t_{1}$,

$$\Delta_{\tau}^{m-1}z\bigl(t+(h+1)\tau\bigr)- \Delta_{\tau}^{m-1}z(t+h\tau)=\Delta _{\tau}^{m}z(t+h \tau)< 0 $$

so $\Delta_{\tau}^{m-1}z(t+h\tau)$ is decreasing as h increases. By the boundedness of $x(t)$ we know that $\lim_{h \to \infty}\Delta_{\tau}^{m-1}z(t+h\tau)$ exists. If $\lim_{h \to\infty} \Delta_{\tau}^{m-1}z(t+h\tau)=S(t)\neq0$, then

$$\Delta_{\tau}^{m-2}z\bigl(t+(h+1)\tau\bigr) \to-\infty \mbox{ or } \infty $$

as $h \to\infty$, a contradiction to the boundedness of $\Delta_{\tau}^{m-2}z(t)$. Thus, for each $t \geq t_{1}$, $\Delta_{\tau}^{m-1}z(t+h\tau)$ is decreasing and tends to 0 as $h \to\infty$. Repeating the same procedure, we see that $\Delta _{\tau}z(t+h\tau)$ is increasing as h increases and $\Delta_{\tau}z(t+h\tau) \to0$ as $h \to\infty$; $z(t+h\tau)$ is decreasing as h increases so $\lim_{h \to\infty}z(t+h\tau)$ exists for each $t \geq t_{1}$. By assumption, $\liminf_{t\to\infty}z(t) \geq0$. Then $z(t+h\tau)$ is decreasing and $\lim_{h \to\infty} z(t+h\tau) \geq 0$ so $z(t+h\tau)>0$ for all $t \geq t_{1}$ and $h \geq1$. Integrating $q(t)z(g(t)-ir)$, by the assumptions on g and q, we obtain

$$\begin{aligned}& \int_{t}^{t+\tau} ds_{1} \int _{s_{1}}^{s_{1}+\tau} ds_{2} \cdots \int _{s_{m-1}}^{s_{m-1}+\tau} z\bigl(g(\theta)-ir\bigr)q(\theta)\,d\theta \\& \quad\geq\min_{t \leq s \leq t+m\tau}\bigl\{ q(s)\bigr\} \int_{t}^{t+\tau} ds_{1} \int_{s_{1}}^{s_{1}+\tau} ds_{2} \cdots \int_{s_{m-1}}^{s_{m-1}+\tau} z\bigl(g(\theta)-ir\bigr)\,d \theta \\& \quad \geq\min_{t \leq s \leq t+m\tau}\bigl\{ q(s)\bigr\} \int_{g(t)}^{g(t+\tau)} \bigl(g^{-1}(s_{1})\bigr)'\,ds_{1}\int _{s_{1}}^{g(g^{-1}(s_{1})+\tau)} \bigl(g^{-1}(s_{2}) \bigr)' \,ds_{2} \cdots \\& \quad\quad{}\times \int_{s_{m-1}}^{g(g^{-1}(s_{m-1})+\tau)} z(\theta-ir) \bigl(g^{-1}(\theta )\bigr)'\,d\theta \\& \quad\geq\min_{t \leq s \leq t+m\tau}\bigl\{ q(s)\bigr\} \Bigl(\min _{g(t)\leq s \leq g(t)+m\tau}\bigl(g^{-1}(s)\bigr)' \Bigr)^{m} \int_{g(t)}^{g(t)+\tau}ds_{1} \int_{s_{1}}^{s_{1}+\tau} ds_{2} \cdots \\& \quad\quad{}\times\int_{s_{m-1}}^{s_{m-1}+\tau} z(\theta-ir)\,d\theta \\& \quad\geq\min_{ t \leq s \leq t+m\tau}\bigl\{ q(s)\bigr\} \Bigl(\min _{g(t)\leq s \leq g(t)+m\tau}\bigl(g^{-1}(s)\bigr)' \Bigr)^{m} y\bigl(g(t)-ir\bigr) \\& \quad \geq\bar{q}(t)y\bigl(g(t)-ir\bigr). \end{aligned}$$

Therefore,

$$\Delta_{\tau}^{m}y(t)+\bar{q}(t)\sum _{i=0}^{n}p^{i}y\bigl(g(t)-ir\bigr)< 0 $$

holds for any fixed natural number n and for all large enough t. If $x(t)$ is a bounded and eventually negative solution, then the above proof with obvious changes shows the conclusion within brackets. □

Lemma 4.3

Let $0 \leq p <1$ and $r=k\tau$. Assume that $x(t)$ is a bounded and eventually positive (negative) solution of (1.1). Let

$$\begin{aligned}& z(t) = x(t)-px(t-r), \\& y(t) = \int_{t}^{t+\tau}dt_{1}\int _{t_{1}}^{t_{1}+\tau}dt_{2} \cdots \int _{t_{m-1}}^{t_{m-1}+\tau}z(\theta)\,d\theta. \end{aligned}$$

Then the conclusion of Lemma  4.2 holds.

Proof

The proof is the same as that of Lemma 4.2 until $\lim_{h\to\infty}z(t+h\tau)$ exists for each $t \geq t_{1}$. Suppose there is a $t' > t_{1}$ such that $\lim_{h \to \infty}z(t'+h\tau)=\delta<0$. Then $z(t'+h\tau) \leq\delta/2 <0$ for $h \geq h_{1}>0$ so $z(t'+hr+h_{1}\tau) = z(t'+(kh+h_{1})\tau) \leq\delta/2$ for $h\geq0$. Thus, for $h>0$,

$$\begin{aligned} x\bigl(t'+hr+h_{1}\tau\bigr) \leq& \delta/2 + px \bigl(t'+(h-1)r+h_{1}\tau\bigr) \\ \leq& \delta\bigl(1 + p + \cdots+ p^{h-1}\bigr)/2 +p^{h}x \bigl(t'+h_{1}\tau\bigr). \end{aligned}$$

This implies $x(t'+k(h+h_{1})\tau) <0$ for large h, a contradiction to the assumption that x is eventually positive. Therefore $\lim_{h \to\infty}z(t+h\tau) \geq0$ for $t \geq t_{1}$. Since $z(t+h\tau)$ is decreasing as h increases, $z(t)>0$ for all $t \geq t_{1}$. The rest of the proof of Lemma 4.2 is still valid here. □

Lemma 4.4

Under the assumptions of Lemma  4.2 or Lemma  4.3, let

$$v(t)=\int_{t}^{t+\tau}dt_{1}\int _{t_{1}}^{t_{1}+\tau}dt_{2}\cdots \int _{t_{m-1}}^{t_{m-1}+\tau}y(\theta) \,d\theta. $$

Then $v(t)>0$ (<0), $(-1)^{k}v^{(k)}(t)>0$ (<0) for $1 \leq k \leq m$ eventually. Moreover,

$$ v^{(m)}(t)+ \frac{1}{\tau^{m}}\bar{q}(t)\sum _{i=0}^{n}p^{i}v\bigl(g(t)-ir\bigr)< 0\ (>0) $$

(4.4)

holds for any fixed natural number n and for all large enough t.

Proof

By the definition of $v(t)$, $v(t)$ has the same sign as $y(t)$ for all $t \geq t_{1}$. Furthermore, we have

$$v'(t)=\int_{t}^{t+\tau}dt_{1} \int_{t_{1}}^{t_{1}+\tau} dt_{2} \cdots\int _{t_{m-1}}^{t_{m-1}+\tau}y'(\theta)\, d\theta. $$

Then $v'(t)$ has the same sign as $y'(t)$. Similarly, $v^{(j)}(t)$ has the same sign as $y^{(j)}(t)$ for all $j=1, 2, \ldots, m$. Notice also that $v^{(m)}(t)=\Delta_{\tau}^{m}y(t)$. If $y'(t)<0$, then

$$\begin{aligned} v\bigl(g(t)-ir\bigr) =&\int_{g(t)}^{g(t)+\tau}dt_{1} \int_{t_{1}}^{t_{1}+\tau} dt_{2} \cdots\int _{t_{m-1}}^{t_{m-1}+\tau}y(\theta-ir)\,d\theta \\ \leq& \int_{g(t)}^{g(t)+\tau}dt_{1}\int _{t_{1}}^{t_{1}+\tau} dt_{2} \cdots\int _{t_{m-1}}^{t_{m-1}+\tau}y(t_{m-1}-ir)\,d\theta \\ \leq& \tau\int_{g(t)}^{g(t)+\tau}dt_{1}\int _{t_{1}}^{t_{1}+\tau} dt_{2} \cdots\int _{t_{m-2}}^{t_{m-2}+\tau}y(t_{m-1}-ir)\,dt_{m-1} \\ \cdots& \\ \leq&\tau^{m-1}\int_{g(t)}^{g(t)+\tau}y(t_{1}-ir)\,dt_{1} \\ \leq& \tau^{m}y\bigl(g(t)-ir\bigr). \end{aligned}$$

Hence, from (4.2) it follows that

$$v^{(m)}(t)+\frac{1}{\tau^{m}}\bar{q}(t)\sum _{i=0}^{n}p^{i}v\bigl(g(t)-ir\bigr)< 0 $$

holds for any fixed natural number n and for all large enough t. If $y'(t)>0$, then $v(g(t)-ir) \geq\tau^{m}y(g(t)-ir)$ so

$$v^{(m)}(t)+\frac{1}{\tau^{m}}\bar{q}(t)\sum _{i=0}^{n}p^{i}v\bigl(g(t)-ir\bigr)>0. $$

 □

Lemma 4.5

Under the assumptions of Lemma  4.4, for each $t \geq t_{1}$ there is a $\theta\in(g(t), t)$ such that

$$ \bigl|v'\bigl(g(t)\bigr)\bigr| > \frac{(t-g(t))^{m-1}}{(m-1)!}\bigl|v^{(m)}( \theta)\bigr|. $$

(4.5)

Proof

Under the assumptions of Lemma 4.4, we know that $(-1)^{j}v^{(j)}(t)$ for $j=1, 2, \ldots,m$ have the same sign. According to Taylor’s formula, we have

$$\begin{aligned} v'\bigl(g(t)\bigr) =&v'(t)+v''(t) \bigl(g(t)-t\bigr)+\frac{1}{2}v^{(3)}(t) \bigl(g(t)-t \bigr)^{2}+\cdots \\ &{} +\frac{1}{(m-1)!}v^{(m)}(\theta) \bigl(g(t)-t\bigr)^{m-1} \end{aligned}$$

for some $\theta\in(g(t), t)$ and (4.5) follows immediately. □

The next lemmas are for the bounded solutions of (1.1) with $p >1$.

Lemma 4.6

Let $p>1$ and $r=k\tau$, $k\in N$. Assume that $x(t)$ is a bounded and eventually positive (negative) solution of (1.1). Let

$$\begin{aligned}& z(t) = x(t)-px(t-r), \\& y(t) = \int_{t}^{t+\tau}dt_{1}\int _{t_{1}}^{t_{1}+\tau}dt_{2}\cdots \int _{t_{m-1}}^{t_{m-1}+\tau} z(\theta)\,d\theta. \end{aligned}$$

Then $y(t)<0$ (>0), $(-1)^{k}y^{(k)}(t)>0$ (<0) for $1 \leq k \leq m$ eventually. Moreover,

$$ \Delta_{\tau}^{m}y(t)-\bar{q}(t)\sum _{i=1}^{n}\frac{1}{p^{i}}y\bigl(g(t)+ir\bigr)< 0\ (>0) $$

(4.6)

holds for any fixed integer $n \geq1$ and for all large enough t.

Proof

Suppose $x(t)$ is a bounded and eventually positive solution. Since $g(t)< t$ and $g'(t)>0$, from the assumptions, there exists a $t_{1} > t_{0}$ such that $x(g(t))>0$ for all $t \geq t_{1}$. Notice also that

$$\Delta^{m}_{\tau}z(t)+f\bigl(t, x\bigl(g(t)\bigr)\bigr)=0. $$

According to (1.3), we have $f(t, x(g(t))) \geq q(t)x(g(t)) >0$ for $t \geq t_{1}$. Therefore

$$ \Delta^{m}_{\tau}z(t) +q(t)x\bigl(g(t)\bigr) \leq0 $$

(4.7)

for $t \geq t_{1}$. By the definition of $y(t)$, $y^{(m)}(t)=\Delta_{\tau}^{m} z(t)$. Thus, from (4.7) it follows that

$$ y^{(m)}(t) +q(t)x\bigl(g(t)\bigr) \leq0 $$

(4.8)

for $t \geq t_{1}$. Due to $q(t)x(g(t))> 0$, $y^{(m)}(t)<0$ for all $t \geq t_{1}$. From the proof of Lemma 4.2 we know that $(-1)^{k}y^{(k)}(t)>0$ holds for $1 \leq k \leq m$ and all $t \geq t_{1}$. Thus, $y(t)$ is decreasing. We now prove that $y(t)<0$ for all $t \geq t_{1}$. Since $y^{(m)}(t)=\Delta_{\tau}^{m}z(t)$ for all $t \geq t_{1}$, from the proof of Lemma 4.2 we know that $z(t+h\tau)$ is decreasing for each fixed $t \geq t_{1}$ as h increases. Next we show that $z(t)<0$, so that $y(t)<0$ for some $t_{2} \geq t_{1}$ and all $t \geq t_{2}$. Suppose there is a $t' >t_{1}$ such that $z(t'+h\tau)>0$ for all $h \geq1$. Under $r=k\tau$, we then have $z(t'+hr)>0$ for all $h \geq1$ so $x(t'+hr) > p^{h}x(t')$ for all $h \geq1$. So $x(t'+hr) \to\infty$ as $h \to\infty$, a contradiction to the boundedness of x. Therefore, for each $t \in[t_{1}, t_{1}+\tau]$, $z(t+h\tau)$ is decreasing as h increases and there is an integer $H(t)>0$ such that $z(t+h\tau) < z(t+H(t)\tau)<0$ for all $h > H(t)$. Since $z(t)$ is continuous for each $t' \in[t_{1}, t_{1}+\tau]$, there is an open interval $I(t')$ such that $z(t+h\tau)< z(t+H(t')\tau)<0$ hold for all $t \in I(t')$ and $h> H(t')$. Since $[t_{1}, t_{1}+\tau]$ is compact and $\{I(t'): t' \in[t_{1}, t_{1}+\tau]\}$ is an open cover of $[t_{1}, t_{1}+\tau]$, there is a finite subset of $\{I(t'): t' \in[t_{1}, t_{1}+\tau]\}$ covering $[t_{1}, t_{1}+\tau]$. Therefore, there is a $K>0$ such that

$$z(t+h\tau) \leq z(t+K\tau) < 0 $$

for all $t \in[t_{1}, t_{1}+\tau]$ and all $h \geq K$. Hence, there is a $t_{3}>t_{1}$ such that $z(t)<0$, so that $y(t)<0$ for all $t \geq t_{3}$.

From (4.7), we have

$$\Delta^{m}_{\tau}z(t)-\frac{q(t)}{p}z\bigl(g(t)+r\bigr)+ \frac {q(t)}{p}x\bigl(g(t)+r\bigr)\leq0. $$

According to the definition of $z(t)$, it follows from the above inequality that

$$\Delta^{m}_{\tau}z(t)-\frac{q(t)}{p}z\bigl(g(t)+r\bigr)+ \frac{q(t)}{p} \biggl(-\frac{1}{p}z\bigl(g(t)+2r\bigr)+ \frac{1}{p}x\bigl(g(t)+2r\bigr) \biggr)\leq0. $$

Repeating the above procedure, we obtain

$$\Delta^{m}_{\tau}z(t)-q(t)\sum_{i=1}^{n} \frac {1}{p^{i}}z\bigl(g(t)+ir\bigr)+q(t)\frac{1}{p^{n}} x\bigl(g(t)+ nr \bigr)\leq0. $$

Since $q(t)x(g(t)+nr)>0$ for sufficiently large t, we have

$$\Delta^{m}_{\tau}z(t)-q(t)\sum_{i=1}^{n} \frac{1}{p^{i}}z\bigl(g(t)+ir\bigr)< 0. $$

Integrating $q(t)z(g(t)+ir)$, by the assumptions on p and g, we obtain

$$\begin{aligned}& \int_{t}^{t+\tau} ds_{1} \int _{s_{1}}^{s_{1}+\tau} ds_{2} \cdots\int _{s_{m-1}}^{s_{m-1}+\tau} z\bigl(g(\theta)+ir\bigr)q(\theta)\,d\theta \\& \quad\leq \min_{t \leq l \leq t+m\tau}\bigl\{ q(l)\bigr\} \int_{t}^{t+\tau} ds_{1} \int_{s_{1}}^{s_{1}+\tau} ds_{2} \cdots \int_{s_{m-1}}^{s_{m-1}+\tau} z\bigl(g(\theta)+ir\bigr)\,d \theta \\& \quad\leq \min_{t \leq l \leq t+m\tau}\bigl\{ q(l)\bigr\} \int_{g(t)}^{g(t+\tau)} \bigl(g^{-1}(s_{1})\bigr)'\,ds_{1}\int _{s_{1}}^{g(g^{-1}(s_{1})+\tau)} \bigl(g^{-1}(s_{2}) \bigr)' \,ds_{2} \cdots \\& \quad\quad{}\times \int_{s_{m-1}}^{g(g^{-1}(s_{m-1})+\tau)} z(\theta+ir) \bigl(g^{-1}(\theta )\bigr)'\,d\theta \\& \quad\leq \min_{t \leq l \leq t+m\tau}\bigl\{ q(l)\bigr\} \Bigl(\min _{g(t)\leq s \leq g(t)+m\tau}\bigl(g^{-1}(s)\bigr)' \Bigr)^{m} \int_{g(t)}^{g(t)+\tau}ds_{1} \int_{s_{1}}^{s_{1}+\tau} ds_{2} \cdots \\& \quad\quad{}\times \int_{s_{m-1}}^{s_{m-1}+\tau} z(\theta+ir)\,d\theta \\& \quad\leq \min_{t \leq l \leq t+m\tau}\bigl\{ q(l)\bigr\} \Bigl(\min _{g(t)\leq s \leq g(t)+m\tau}\bigl(g^{-1}(s)\bigr)' \Bigr)^{m} y\bigl(g(t)+ir\bigr) \\& \quad\leq \bar{q}(t)y\bigl(g(t)+ir\bigr). \end{aligned}$$

Therefore,

$$\Delta_{\tau}^{m}y(t)-\bar{q}(t)\sum _{i=1}^{n}\frac{1}{p^{i}}y\bigl(g(t)+ir\bigr)< 0 $$

holds for any fixed integer $n \geq1$ and for all large enough t. If $x(t)$ is a bounded and eventually negative solution, then the conclusion within brackets follows from the above proof with minor modification. □

Lemma 4.7

Under the assumptions of Lemma  4.6, let

$$v(t)=\int_{t}^{t+\tau}dt_{1}\int _{t_{1}}^{t_{1}+\tau}dt_{2}\cdots \int _{t_{m-1}}^{t_{m-1}+\tau}y(\theta)\, d\theta. $$

Then $v(t)<0$ (>0), $(-1)^{k}v^{(k)}(t)>0$ (<0) for $1 \leq k \leq m$ eventually. Moreover,

$$ v^{(m)}(t)- \frac{1}{\tau^{m}}\bar{q}(t)\sum _{i=1}^{n}\frac{1}{p^{i}}v\bigl(g(t)-m\tau+ir \bigr)< 0\ (>0) $$

(4.9)

holds for any fixed integer $n \geq1$ and for all large enough t.

Proof

By the definition of $v(t)$, $v(t)$ has the same sign as $y(t)$. Further, we have

$$v'(t)=\int_{t}^{t+\tau}dt_{1} \int_{t_{1}}^{t_{1}+\tau} dt_{2} \cdots\int _{t_{m-1}}^{t_{m-1}+\tau}y'(\theta) \,d\theta. $$

Then $v'(t)$ has the same sign as $y'(t)$. Similarly, $(-1)^{k} v^{(k)}(t)$ for $1 \leq k \leq m$ and $(-1)^{j}y^{(j)}(t)$ for $1 \leq j \leq m$ all have the same sign. Note also that $v^{(m)}(t)=\Delta_{\tau}^{m}y(t)$. If $y'(t)<0$, then

$$\begin{aligned} v\bigl(g(t)+ir\bigr) =&\int_{g(t)}^{g(t)+\tau}dt_{1} \int_{t_{1}}^{t_{1}+\tau} dt_{2}\cdots\int _{t_{m-1}}^{t_{m-1}+\tau}y(\theta+ir)\,d\theta \\ \geq& \int_{g(t)}^{g(t)+\tau}dt_{1}\int _{t_{1}}^{t_{1}+\tau} dt_{2} \cdots\int _{t_{m-1}}^{t_{m-1}+\tau}y(t_{m-1}+\tau+ir)\,d\theta \\ \geq& \tau\int_{g(t)}^{g(t)+\tau}dt_{1}\int _{t_{1}}^{t_{1}+\tau }dt_{2} \cdots\int _{t_{m-2}}^{t_{m-2}+\tau} y(t_{m-1}+\tau+ir)\,dt_{m-1} \\ \cdots& \\ \geq&\tau^{m-1}\int_{g(t)}^{g(t)+\tau}y \bigl(t_{1}+(m-1)\tau+ir\bigr)\,dt_{1} \\ \geq& \tau^{m}y\bigl(g(t)+m\tau+ir\bigr). \end{aligned}$$

Hence, from (4.6) we have

$$v^{(m)}(t)-\frac{1}{\tau^{m}}\bar{q}(t)\sum _{i=1}^{n}\frac {1}{p^{i}}v\bigl(g(t)-m\tau+ir \bigr)< 0 $$

for any fixed integer $n \geq1$ and for all large enough t. If $y'(t)>0$, then $0 < v(g(t)+ir) \leq\tau^{m} y(g(t)+m\tau+ir)$ so

$$v^{(m)}(t)-\frac{1}{\tau^{m}}\bar{q}(t)\sum _{i=1}^{n}\frac {1}{p^{i}}v\bigl(g(t)-m\tau+ir \bigr)>0. $$

 □

Lemma 4.8

Assume that $x(t)$ is an eventually positive (negative) and bounded solution of (1.1). Let $z(t)$ and $v(t)$ be defined as in Lemma  4.6 and Lemma  4.7. Then, under the assumptions of Lemma  4.6, for any given $t \geq t_{1}$, there is a $\theta\in(g(t), t)$ such that

$$ \bigl|v'\bigl(g(t)\bigr)\bigr| > \frac{(t-g(t))^{m-1}}{(m-1)!}\bigl|v^{(m)}( \theta)\bigr|. $$

(4.10)

Proof

The proof of Lemma 4.5 is still valid for Lemma 4.8. □

5 Proofs of the main results

Here, the proofs of the main results will be presented.

Proof of Theorem 2.1

Suppose the conclusion is not true. Let $x(t)$ be an eventually positive and bounded solution of (1.1) with $\liminf_{t \to \infty}(x(t)-px(t-r)) \geq0$. Let $y(t)$ be defined as in Lemma 4.2 and $v(t)$ be defined as in Lemma 4.4. By Lemma 4.4, we know that $v(t)>0$, $(-1)^{k}v^{(k)}(t)> 0$ for $1 \leq k \leq m$ and (4.4), i.e.,

$$v^{(m)}(t)+\frac{1}{\tau^{m}}\bar{q}(t) \sum _{i=0}^{n}p^{i} v\bigl(g(t)-ir\bigr) < 0 $$

holds for any fixed natural number n and for all large enough t. By Lemma 4.5, we know that

$$v'\bigl(g(t)\bigr)\frac{(m-1)!}{(t-g(t))^{m-1}} < v^{(m)}(\theta) $$

for some $\theta\in(g(t), t)$. Since $\bar{q}(\theta) \geq\bar{q}(t)$ by assumption and

$$v\bigl(g(\theta)-ir\bigr) \geq v\bigl(g(t)-ir\bigr), $$

from (4.4) with t replaced by θ, it follows that

$$v'\bigl(g(t)\bigr)\frac{(m-1)!}{(t-g(t))^{m-1}}+\frac{1}{\tau^{m}}\bar{q}(t) \sum_{i=0}^{n}p^{i} v \bigl(g(t)-ir\bigr) < 0 $$

i.e.,

$$ v'\bigl(g(t)\bigr)+\frac{(t-g(t))^{m-1}}{(m-1)! \tau^{m}} \bar{q}(t) \sum _{i=0}^{n}p^{i} v\bigl(g(t)-ir \bigr) < 0. $$

(5.1)

With the replacement of t by $g^{-1}(t)$, (5.1) yields

$$ v'(t)+\frac{(g^{-1}(t)-t)^{m-1}}{(m-1)! \tau^{m}} \bar{q} \bigl(g^{-1}(t)\bigr) \sum_{i=0}^{n}p^{i} v(t-ir) < 0. $$

(5.2)

Assume that $(-1)^{k}v^{(k)}(t)>0$ and (5.2) hold for $0 \leq k \leq m$ and $t \geq t_{1} \geq t_{0}$. Without loss of generality, we may assume $T \geq t_{1}+nr$. Let

$$w(t)=\frac{-v'(t)}{v(t)}. $$

Note that $w(t)>0$ and $v(t)=v(T)\exp\int_{T}^{t}-w(\theta )\,d\theta$ for all $t \geq T \geq t_{1}+nr$. From (5.2) it follows that

$$ w(t) > \frac{(g^{-1}(t)-t)^{m-1}\bar{q}(g^{-1}(t))}{(m-1)! \tau^{m}}\sum_{i=0}^{n}p^{i} \exp\int_{t-ir}^{t}w(s)\,ds, $$

(5.3)

i.e.,

$$ w(t) > \frac{Q_{1}(t)}{(m-1)! \tau^{m}} \sum_{i=0}^{n}p^{i} \exp \int_{t-ir}^{t} w(s)\,ds $$

(5.4)

for all $t \geq T x$, where $Q_{1}(t)= (g^{-1}(t)-t)^{m-1}\bar{q} x(g^{-1}(t))>0$.

Let $w_{0}(t)\equiv0$ for $t\geq T x-nr$ and let

$$w_{k+1}(t)=\frac{Q_{1}(t)}{(m-1)! \tau^{m}} \sum_{i=0}^{n}p^{i} \exp\int_{t-ir}^{t} w_{k}(s)\,ds $$

for each $k \in N$ and $t \geq T x+nkr$. Let

$$\alpha_{1k}=\inf_{t \geq T +(k-1)nr} \bigl\{ w_{k}(t) \bigr\} ,\quad k \in\bar{N}. $$

Then

$$\alpha_{1k+1} \geq\inf_{t \geq T } \Biggl\{ \frac{Q_{1}(t)}{(m-1)!\tau ^{m}}\sum_{i=0}^{n}p^{i}e^{ir\alpha_{1k}} \Biggr\} =\beta_{1}\sum_{i=0}^{n}p^{i}e^{ir\alpha_{1k}}. $$

Now, (5.4), (2.4), and the definition of $\{\alpha _{1k}\}$ imply that $\{\alpha_{1k}\}$ is an increasing sequence. Suppose

$$\lim_{k \to\infty} \alpha_{1k}=\rho_{1} < \infty. $$

So $\rho_{1} \geq\beta_{1}\sum_{i=0}^{n}p^{i}e^{ir\rho_{1}}$. Let

$$F_{1}(x)=\beta_{1}\sum_{i=0}^{n}p^{i}e^{irx}-x. $$

Then $F'_{1}(x)=\beta_{1}\sum_{i=1}^{n}irp^{i}e^{irx}-1$ and $F''_{1}(x)>0$, so $F'_{1}(x)$ is increasing. Since $F'_{1}(0)=\beta_{1} \sum_{i=0}^{n}irp^{i}-1 \geq0$ by (2.4), then $F'_{1}(x)>0$ for $x >0$. Hence $F_{1}(x)$ is increasing. Thus, from $F_{1}(0)=\beta_{1}\sum_{i=0}^{n}p^{i}>0$ we have $F_{1}(x)>0$ for all $x \geq0$. This shows that no positive number $\rho_{1}$ satisfies $\rho_{1} \geq\beta_{1} \sum_{i=0}^{n}p^{i}e^{ir\rho_{1}}$. Therefore, we must have $\alpha_{1k} \to\infty$ as $k \to\infty$. Note that $w(t) \geq w_{k+1}(t) \geq\alpha_{1k+1}$ for $t \geq T_{3}+nkr$. Thus $w(t) \to\infty$ as $t \to\infty$. Notice also that

$$w(t) \geq w_{k+1}(t) \geq\alpha_{1k+1}\quad \mbox{for } t \geq T +nkr. $$

Thus $w(t) \to\infty$ as $t \to\infty$, which implies

$$ \frac{v(t)}{v(t+r)}=\exp\int_{t}^{t+r}w(s)\,ds \to\infty \quad\mbox{as } t \to\infty. $$

(5.5)

On the other hand, since $v'(t)<0$ and $v(t)>0$, (5.2) yields (by dropping the $i=0$ term)

$$\begin{aligned} v'(t) < & -\frac{(g^{-1}(t)-t)^{m-1}}{(m-1)!\tau^{m}}\bar {q} \bigl(g^{-1}(t)\bigr)\sum_{i=1}^{n}p^{i}v(t-ir) \\ < & -\frac{(g^{-1}(t)-t)^{m-1}}{(m-1)!\tau^{m}}\cdot\frac {p-p^{n+1}}{1-p}\cdot\bar{q}\bigl(g^{-1}(t) \bigr)v(t-r). \end{aligned}$$

(5.6)

By (2.5) and Lemma 4.1,

$$\liminf_{t \to\infty} \frac{v(t)}{v(t-r)} \in(0,1]. $$

Thus $v(t+r)/v(t)$ has a positive lower bound so $v(t)/v(t+r)$ has a positive upper bound. This contradicts (5.5). Assume that $x(t)$ is an eventually negative and bounded solution of (1.1) with $\limsup_{t \to\infty} (x(t)-px(t-r)) \leq0$. Then the above proof with a minor modification also leads to a contradiction. Therefore, the conclusion of the theorem holds. □

Proof of Corollary 2.2

The proof is the same as that of Theorem 2.1 except (5.6). The conclusion still holds if (5.6) is replaced by

$$v'(t) < -\frac{(g^{-1}(t)-t)^{m-1}}{\tau^{m}(m-1)!} \bar {q}\bigl(g^{-1}(t)\bigr)p v(t-r). $$

 □

Proof of Corollary 2.3

The proof of Theorem 2.1 is still valid after the replacement of $\bar{q}(\theta) \geq\bar{q}(t)$ by $\bar {q}(\theta) \geq\bar{q}(g(t))$. □

The proof of Corollary 2.4 is similar to that of Corollary 2.2.

Proof of Corollary 2.5

The proof of Theorem 2.1 is still valid after the replacement of Lemma 4.2 by Lemma 4.3. □

Proof of Corollary 2.6

Suppose the conclusion is not true. Without loss of generality, assume that (1.1) has an eventually positive and bounded solution $x(t)$. Let $y(t)$ be defined as in Lemma 4.6 and $v(t)$ be defined as in Lemma 4.7. By Lemma 4.7, we know that $v(t)<0$, $(-1)^{k}v^{(k)}(t)> 0$ for $1 \leq k \leq m$, and (4.9), i.e.,

$$v^{(m)}(t)-\frac{1}{\tau^{m}}\bar{q}(t) \sum _{i=1}^{n}\frac{1}{p^{i}} v\bigl(g(t)-m\tau+ir \bigr) < 0 $$

holds for any fixed integer $n \geq1$ and for all large enough t. By Lemma 4.8, we know that

$$v'\bigl(g(t)\bigr)\frac{(m-1)!}{(t-g(t))^{m-1}} < v^{(m)}(\theta) $$

for some $\theta\in(g(t), t)$. Since $\bar{q}(\theta) \geq\bar {q}(g(t))$ and $v(g(\theta)-ir) \leq v(g(g(t))-ir)$, with the replacement of t by θ, (4.9) yields

$$v'\bigl(g(t)\bigr)\frac{(m-1)!}{(t-g(t))^{m-1}}-\frac{1}{\tau^{m}}\bar{q} \bigl(g(t)\bigr) \sum_{i=1}^{n} \frac{1}{p^{i}} v\bigl(g\bigl(g(t)\bigr)-m\tau+ir\bigr) \leq0 $$

i.e.,

$$ v'\bigl(g(t)\bigr)-\frac{(t-g(t))^{m-1}}{(m-1)! \tau^{m}} \bar{q} \bigl(g(t)\bigr) \sum_{i=1}^{n} \frac{1}{p^{i}} v\bigl(g\bigl(g(t)\bigr)-m\tau+ir\bigr) \leq0. $$

(5.7)

With the replacement of t by $g^{-1}(t)$, (5.7) becomes

$$ v'(t)-\frac{(g^{-1}(t)-t)^{m-1}}{(m-1)! \tau^{m}} \bar{q}(t) \sum _{i=1}^{n}\frac{1}{p^{i}} v\bigl(g(t)-m\tau+ir \bigr) \leq0. $$

(5.8)

Assume that $v(t)$, $(-1)^{k}v^{(k)}(t)>0$ ($1 \leq k \leq m$) and (5.8) hold for $t \geq t_{1} \geq t_{0}$ and, without loss of generality, that $T \geq t_{1}+nr$. Let

$$w(t)=\frac{v'(t)}{v(t)}. $$

Note that $w(t)>0$ and $v(t)=v(t')\exp\int_{t'}^{t}w(\theta )\,d\theta$ for all $t, t' \geq T$. From (5.8), we have

$$ w(t) \geq\frac{(g^{-1}(t)-t)^{m-1}\bar{q}(t)}{(m-1)! \tau^{m}}\sum_{i=1}^{n} \frac{1}{p^{i}} \exp\int_{t}^{g(t)-m\tau+ir}w(s)\,ds, $$

(5.9)

i.e.,

$$ w(t) \geq\frac{Q_{m2}(t)}{(m-1)! \tau^{m}} \sum_{i=1}^{n} \frac{1}{p^{i}} \exp\int_{t}^{g(t)-m\tau+ir} w(s)\,ds $$

(5.10)

for all $t\geq T$, where $Q_{2}(t)= (g^{-1}(t)-t)^{m-1}\bar{q}(t) >0$.

Let $w_{0}(t)\equiv0$ for $t\geq T$. For each $k \in\bar{N}$ and $t \geq T$, let

$$w_{k+1}(t)=\frac{Q_{2}(t)}{(m-1)! \tau^{m}} \sum_{i=1}^{n} \frac{1}{p^{i}} \exp\int_{t}^{g(t)-m\tau+ir} w_{k}(s)\,ds $$

and

$$\alpha_{2k}=\inf_{t \geq T}\bigl\{ w_{k}(t) \bigr\} ,\quad k \in\bar{N}. $$

So $w(t) \geq w_{k+1}(t) \geq w_{k}(t) \geq\alpha_{2k}$ for all $k \in N$ and $t \geq T$. By assumption, we therefore have

$$\begin{aligned} \alpha_{2k+1} \geq&\inf_{t \geq T} \Biggl\{ Q_{2}(t)\cdot\frac {1}{(m-1)!\tau^{m}}\sum_{i=1}^{n} \frac{1}{p^{i}}e^{[g(t)-t-m\tau +ir]\alpha_{2k}} \Biggr\} \\ \geq&\beta_{2}\sum_{i=1}^{n} \frac{1}{p^{i}}e^{(i-1)r \alpha_{2k}}. \end{aligned}$$

Note that $\{\alpha_{2k}\}$ is a bounded nondecreasing sequence and suppose that

$$ \lim_{k \to\infty} \alpha_{2k}= \rho_{2}< \infty. $$

(5.11)

So $\rho_{2} \geq\beta_{2}\sum_{i=1}^{n}(e^{(i-1)r\rho_{2}}/p^{i})$. Let

$$F_{2}(x)=\beta_{2}\sum_{i=1}^{n} \frac{e^{(i-1)rx}}{p^{i}}-x. $$

Then $F_{2}'(x)=\beta_{2}\sum_{i=2}^{n}((i-1)re^{(i-1)rx}/p^{i})-1$ and $F''_{2}(x)>0$, so $F'_{2}(x)$ is increasing. Since $F'_{2}(0)=\beta_{2} \sum_{i=1}^{n}((i-1)r/p^{i})-1 \geq0$ by (2.10), $F'_{2}(x)>0$ for $x >0$. Hence $F_{2}(x)$ is increasing. Thus, from $F_{2}(0)=\beta_{2}\sum_{i=1}^{n}1/p^{i}>0$ we have $F_{2}(x)>0$ for all $x \geq0$. This shows that no positive number $\rho_{2}$ satisfies $\rho_{2} \geq\beta_{2} \sum_{i=1}^{n}(e^{(i-1)r\rho_{2}}/p^{i})$. This contradiction shows that the conclusion holds. □

Proof of Corollary 2.7

The proof of Theorem 2.6 is still valid after the replacement of $\bar{q}(\theta) \geq\bar{q}(g(t))$ by $\bar {q}(\theta) \geq\bar{q}(t)$. □