The distribution of generalized zeros of oscillatory solutions for second-order nonlinear neutral delay difference equations

Abstract

In this paper, the distributions of generalized zeros of oscillatory solutions for second-order nonlinear neutral delay difference equations are studied. By means of inequality techniques, specific function sequences and non-increasing solutions for corresponding first-order difference inequality, some new estimates for the distribution of the zeros of oscillatory solutions are presented, which extend and improve some well-known results.

1 Introduction

In recent years, the study of oscillation of differential equations has become more and more perfect, including various sufficient conditions, necessary conditions, the existence of non-oscillatory solutions, and even the zeros distribution of oscillatory solutions.

In 2017, Li et al. [1] studied the distribution of zeros of oscillatory solutions for second-order nonlinear neutral delay differential equation

$$\bigl(a(t)z'(t)\bigr)'+q(t)f\bigl(x(t-\sigma)\bigr)=0, \quad t\geqslant t_{0}, $$

and obtained a sufficient condition for oscillation of differential equation.

However, most of references about oscillation of difference equations are concerned with sufficient or necessary conditions for oscillation; see [2,3,4,5,6,7,8]. We will also naturally ask some questions of difference equations: Are there any bounds for the distance between adjacent generalized zeros of oscillatory solutions when equations show oscillation? And how do we estimate these bounds? Therefore, we obtain the oscillation criteria of difference equations by studying the distribution of zeros.

The distribution of generalized zeros of oscillation solutions for first-order dynamic equations and second-order non-neutral dynamic equations on time scale can be found in [9,10,11]. However, most oscillatory results for second-order neutral dynamic equations are sufficient conditions for oscillation; see [12,13,14,15,16,17,18,19]. To the best of our knowledge, there is no paper on the generalized zero distribution of oscillation solutions for second-order neutral dynamic equations on time scale.

Motivated by the above papers, we consider the second-order neutral difference equation of the following form:

$$ \Delta\bigl(a(t)\Delta z(t)\bigr)+q(t)f\bigl(x(t-\sigma)\bigr)=0, \quad t\in[ t_{0},\infty )_{\mathbb{Z}}, $$

(1.1)

where Δ denotes the forward difference operator $\Delta x(t)=x(t+1)-x(t)$, $z(t)=x(t)+p(t)x(t-\tau)$, $\mathbb{Z}$ represents the set of all integers and

$$\sum_{s=t_{0}}^{\infty}\frac{1}{a(s)}=\infty. $$

Throughout this paper, we assume that the following hypotheses are satisfied:

$(H_{1})$ :

$a(t),q(t),p(t)\in(0,\infty)$, where $t\in[ t_{0},\infty )_{\mathbb{Z}}$.

$(H_{2})$ :

$\tau, \sigma\in\mathbb{R^{+}}$, where $\mathbb{R^{+}}$ represents the set of all positive real numbers, and $\sigma>\tau$.

$(H_{3})$ :

There exists a positive constant k such that $\frac {f(u)}{u}\geqslant k$ for all $u\neq0$.

$(H_{4})$ :

There exists a function $H(t)$ which satisfies $H(t)\geqslant\frac{p(t-\sigma)q(t)}{q(t-\tau)}$ and $\Delta H(t)\leqslant0$, $t\geqslant t_{1}$ for some $t_{1}\geqslant t_{0}+\sigma$, where $t\in\mathbb{Z}$.

In this paper, we relate the distance between adjacent generalized zeros of an oscillation solution of (1.1) to a positivity problem of certain solution for a first-order delay difference inequality

$$ \Delta x(t)+P(t)x(t-r_{1})\leqslant0, \quad t\in[ t_{0},\infty)_{\mathbb{Z}}, $$

(1.2)

where $P(t)\in[ 0,1)$ which define by (2.1), $r_{1}$ is a constant satisfying $r_{1}\geqslant2$.

2 Preliminaries

In order to prove our main results, we establish some fundamental results in this section.

For convenience, we define a sequence $\{F_{n}(t)\}\in[ 0,1)$ by

$$ \begin{aligned} &F_{0}(t)=P(t):= \frac{kq(t)}{1+H(t+1)}\sum_{s=T_{0}}^{t-1} \frac {1}{a(s-\sigma)} ,\quad t\in[ t_{0},\infty)_{\mathbb{Z}}, \\ &F_{n}(t)=F_{n-1}(t)\sum_{s=t-r}^{t-1}F_{n-1}(s) \prod_{\zeta =s-r}^{t}\frac{1}{1-F_{n-1}(\zeta)}, \quad t\in[ t_{0}+2nr,\infty)_{\mathbb {Z}}, n=1,2,\ldots, \end{aligned} $$

(2.1)

where $T_{0}$ satisfies $x(t)>0$, $t\geqslant T_{0}$ when $x(t)$ is eventually positive solution.

If $t_{n}$ is a generalized zero of solution of (1.1), then it satisfies $x([t_{n}])\cdot x([t_{n}]+1)\leqslant0$. Let $d_{s}(x)$ be the least upper bound of the distance between adjacent generalized zeros of a solution $x(t)$ of Eq. (1.1) on $[s,\infty)$.

Lemma 2.1

Assume that $x(t)$ is an eventually positive solution of (1.1), and $(H_{1})\sim(H_{3})$ hold. Then $z(t)$ satisfies $z(t)>0$, $\Delta z(t)>0$, $\Delta(a(t)\Delta z(t))<0$.

Proof

If $x(t)$ is an eventually positive solution of Eq. (1.1), then there exists a $t_{1}>t_{0}$ such that $x(t)>0$, $x(t-\tau)>0$ and $x(t-\sigma)>0$ for all $t\geqslant t_{1}$. Thus $z(t)=x(t)+p(t)x(t-\tau)>0$. From (1.1) and condition $(H_{3})$, we obtain

$$ \begin{aligned} \Delta\bigl(a(t)\Delta z(t)\bigr)=-q(t) \frac{f(x(t-\sigma))}{x(t-\sigma )}x(t-\sigma)\leqslant-kq(t)x(t-\sigma)< 0,\quad t\geqslant t_{1}, \end{aligned} $$

(2.2)

so we can conclude $a(t)\Delta z(t)$, $t\geqslant t_{1}$ is decreasing. It can be seen that there exists a $t_{2}>t_{1}$ such that $\Delta z(t)>0$ or $\Delta z(t)<0$ for $t\geqslant t_{2}$. Now, we prove $\Delta z(t)>0$, $t\geqslant t_{2}$. If not, assume that $\Delta z(t)<0$, $t\geqslant t_{2}$, then also $a(t)\Delta z(t)<-c<0$ and summing up it from $t_{2}$ to $t-1$, we have

$$ \begin{aligned} z(t)-z(t_{2})< -c\sum _{s=t_{2}}^{t-1}\frac{1}{a(s)}. \end{aligned} $$

(2.3)

Taking limits of both sides for the above inequality, we have $\lim_{t\rightarrow\infty}z(t)=-\infty$, which is a contradiction. The proof is completed. □

In the following lemmas, let $r=[r_{1}]:=\max\{a|a\leqslant r_{1},a\in \mathbb{Z}\}$, where $r_{1}$ is the delay argument of (1.2). And δ is a constant satisfying $|\delta|\leqslant r$.

Lemma 2.2

Let n be a positive integer such that

$$ \begin{aligned} \sum_{s=t-r}^{t-1}F_{n}(s)\,ds \geqslant1,\quad t\in\bigl[ t_{0}+(2n+1)r,\infty \bigr)_{\mathbb{Z}}, n=1,2, \ldots. \end{aligned} $$

(2.4)

If $x(t)$ is a non-increasing function on $[T_{1}-\delta,T]_{\mathbb {Z}}$ which satisfies (1.2) on $[T_{1},T]_{\mathbb{Z}}$, then $x(t)$ cannot be positive on $[T_{1},T]_{\mathbb{Z}}$, where $T>T_{1}+(3n+1)r+(n+1)-\delta$, $T_{1}\geqslant t_{0}+(2n+1)r$.

Proof

Without loss of generality, we assume that $x(t)$ is positive on $[T_{1},T]_{\mathbb{Z}}$. Summing up (1.2) from $t-r$ to $t-1$, we have

$$x(t)-x(t-r)+\sum_{s=t-r}^{t-1}P(s)x(s-r_{1}) \leqslant0,\quad t\in [T_{1}+r,T]_{\mathbb{Z}}. $$

Multiplying this inequality by $P(t)$ and using (1.2), we get

$$ \begin{aligned} \Delta x(t)+P(t)x(t)+P(t)\sum _{s=t-r}^{t-1}P(s)x(s-r_{1})\leqslant0,\quad t \in[T_{1}+r+1-\delta,T]_{\mathbb{Z}}, \end{aligned} $$

(2.5)

so

$$\prod_{\zeta=t_{0}}^{t}\frac{1}{1-P(\zeta)} \Biggl( \Delta x(t)+P(t)x(t)+P(t)\sum_{s=t-r}^{t-1}P(s)x(s-r_{1}) \Biggr)\leqslant0,\quad t\in[T_{1}+r+1-\delta,T]_{\mathbb{Z}}. $$

Using $\Delta x(t)=x(t+1)-x(t)$, we get

$$ \begin{aligned}[b] &\prod_{\zeta=t_{0}}^{t} \frac{1}{1-P(\zeta)}x(t+1)-\prod_{\zeta =t_{0}}^{t-1} \frac{1}{1-P(\zeta)}x(t) +\prod_{\zeta=t_{0}}^{t} \frac{1}{1-P(\zeta)}P(t)\sum_{s=t-r}^{t-1}P(s)x(s-r_{1}) \leqslant0, \\ &\quad t\in[T_{1}+r+1-\delta,T]_{\mathbb{Z}}. \end{aligned} $$

(2.6)

Let $y_{1}(t):=x(t)\prod_{\zeta=t_{0}}^{t-1}\frac{1}{1-P(\zeta)}$. Then $y_{1}(t)>0$ on $[T_{1},T]_{\mathbb{Z}}$ and

$$ \begin{aligned} \Delta y_{1}(t)=\prod _{\zeta=t_{0}}^{t}\frac{1}{1-P(\zeta )}x(t+1)-\prod _{\zeta=t_{0}}^{t-1}\frac{1}{1-P(\zeta)}x(t). \end{aligned} $$

(2.7)

Using (2.7) in (2.6), we have

$$\Delta y_{1}(t)+P(t)\sum_{s=t-r}^{t-1}P(s) \Biggl(x(s-r_{1})\prod_{\zeta=t_{0}}^{t} \frac{1}{1-P(\zeta)} \Biggr)\leqslant0, \quad t\in [T_{1}+r+1- \delta,T]_{\mathbb{Z}}, $$

i.e.

$$ \begin{aligned}[b] &\Delta y_{1}(t)+P(t)\sum _{s=t-r}^{t-1} \Biggl(P(s)x(s-r_{1}) \prod_{\zeta=t_{0}}^{s-r-1}\frac{1}{1-P(\zeta)}\prod _{\zeta=s-r} ^{t} \frac{1}{1-P(\zeta)} \Biggr) \\ &\quad \leqslant0,\quad t\in[T_{1}+r+1-\delta,T]_{\mathbb{Z}}. \end{aligned} $$

(2.8)

From the definition of $y_{1}(t)$ and $\Delta x(t)\leqslant0$, $t\in [T_{1}+r+1-\delta,T]_{\mathbb{Z}}$ we obtain

$$\begin{aligned} \Delta y_{1}(t) =&\Delta x(t)\prod_{\zeta=t_{0}}^{t} \frac {1}{1-P(\zeta)}+x(t)\Delta\Biggl(\prod_{\zeta=t_{0}}^{t-1} \frac{1}{ 1-P(\zeta)}\Biggr) \\ =&\Delta x(t)\prod_{\zeta=t_{0}}^{t} \frac{1}{1-P(\zeta)}+x(t)\prod_{\zeta=t_{0}}^{t} \frac{1}{1-P(\zeta)}-x(t) \prod_{\zeta=t_{0}}^{t-1} \frac{1}{1-P(\zeta)} \\ =&\prod_{\zeta=t_{0}}^{t}\frac{1}{1-P(\zeta)}\bigl( \Delta x(t)+x(t)-x(t) \bigl(1-P(t)\bigr)\bigr) \\ =&\prod_{\zeta=t_{0}}^{t}\frac{1}{1-P(\zeta)}\bigl( \Delta x(t)+P(t)x(t)\bigr) \\ \leqslant&\prod_{\zeta=t_{0}}^{t}\frac{1}{1-P(\zeta)} \bigl(\Delta x(t)+P(t)x(t-r_{1})\bigr). \end{aligned}$$

Since $\Delta x(t)+P(t)x(t-r_{1})\leqslant0$, we can conclude $\Delta y_{1}(t)\leqslant0$, $t\in[T_{1}+r+1-\delta,T]_{\mathbb{Z}}$, and from (2.8), we have

$$\Delta y_{1}(t)+F_{1}(t)y_{1}(t-r)\leqslant0, \quad t\in[T_{1}+3r+1-\delta ,T]_{\mathbb{Z}}. $$

Repeating the above procedure to this inequality, we get

$$ \begin{aligned} \Delta y_{1}(t)+F_{1}(t)y_{1}(t)+F_{1}(t) \sum_{s=t-r}^{t-1}F_{1}(s)y_{1}(s-r) \leqslant0,\quad t\in[T_{1}+4r+1-\delta ,T]_{\mathbb{Z}}. \end{aligned} $$

(2.9)

Let $y_{2}(t):=y_{1}(t)\prod_{\zeta=t_{0}+2r}^{t-1}\frac {1}{1-F_{1}(\zeta)}$. It follows from (2.9) that

$$\Delta y_{2}(t)+F_{1}(t)\sum_{s=t-r}^{t-1}F_{1}(s)y_{2}(s-r) \prod_{\zeta=s-r}^{t}\frac{1}{1-F_{1}(s)}\leqslant0, \quad t\in [T_{1}+4r+1-\delta,T]_{\mathbb{Z}}, $$

where $\Delta y_{2}(t)\leqslant0$ for $t\in[T_{1}+4r+2-\delta ,T]_{\mathbb{Z}}$ and hence

$$\Delta y_{2}(t)+F_{2}(t)y_{2}(t-r)\leqslant0, \quad t\in[T_{1}+6r+2-\delta ,T]_{\mathbb{Z}}. $$

Repeating this argument n times, we obtain

$$ \begin{aligned} \Delta y_{n}(t)+F_{n}(t)y_{n}(t-r) \leqslant0,\quad t\in[T_{1}+3nr+n-\delta ,T]_{\mathbb{Z}}, \end{aligned} $$

(2.10)

where $\Delta y_{n}(t)\leqslant0$ for $t\in[T_{1}+(3n-2)r+n-\delta ,T]_{\mathbb{Z}}$. Now, summing up (2.10) from $t-r$ to $t-1\in [T_{1}+(3n+1)r+n-\delta,T]_{\mathbb{Z}}$, we have

$$y_{n}(t)-y_{n}(t-r)+\sum_{s=t-r}^{t-1}F_{n}(s)y_{n}(s-r) \leqslant0. $$

Since $y(t)$ is decreasing, we obtain

$$y_{n}(t)+y_{n}(t-r) \Biggl[\sum _{s=t-r}^{t-1}F_{n}(s)-1 \Biggr]\leqslant0, \quad t\in\bigl[T_{1}+(3n+1)r+(n+1)-\delta,T\bigr]_{\mathbb{Z}}, $$

which is a contradiction with hypothesis (2.4). The proof of Lemma 2.2 is complete. □

Lemma 2.3

Assume that $\sum_{s=t-r}^{t-2}P(s)\geqslant\beta$ for $0<\beta<1$ and there exist $T_{2}\geqslant t_{0}+r$, $T\geqslant T_{2}+(1+n)r-\delta$, $n=1,2,\ldots$ and a function $x(t)$ satisfying inequality (1.2) on $[T_{2},T]_{\mathbb{Z}}$ with $\Delta x(t)\leqslant0$ for $t\in [T_{2}-\delta,T]_{\mathbb{Z}}$. If $x(t)$ is positive on $[T_{2},T]_{\mathbb{Z}}$, then

$$ \begin{aligned} \frac{x(t-r)}{x(t)}\geqslant f_{n}(\beta)>0,\quad t\in\bigl[T_{2}+(1+n)r-\delta ,T \bigr]_{\mathbb{Z}}, \end{aligned} $$

(2.11)

for some integer $n\geqslant0$, where $f_{n}(\beta)$ is defined by

$$f_{0}(\beta)=1, \qquad f_{1}(\beta)=\frac{1}{1-\beta}, \qquad f_{n+2}(\beta)=\frac {r-\beta f_{n}(\beta)}{r-\beta f_{n}(\beta)-\beta+\frac{\beta f_{n}(\beta)}{f_{n+1}(\beta)} }. $$

Proof

Since $x(t)$ is non-increasing on $[T_{2}-\delta ,T]_{\mathbb{Z}}$, we find

$$ \begin{aligned} \frac{x(t-r)}{x(t)}\geqslant f_{0}(\beta)=1,\quad t\in[T_{2}+r-\delta ,T]_{\mathbb{Z}}. \end{aligned} $$

(2.12)

Summing inequality (1.2) from $t-r+1$ to $t-1$, where $t\in [T_{2}+2r-\delta,T]_{\mathbb{Z}}$, we obtain

$$ \begin{aligned} x(t-r)\geqslant x(t)+\sum _{s=t-r+1}^{t-1}P(s)x(s-r_{1})\geqslant x(t)+ \beta x(t-r). \end{aligned} $$

(2.13)

Therefore

$$\frac{x(t-r)}{x(t)}\geqslant\frac{1}{1-\beta}=f_{1}(\beta)>0,\quad t \in [T_{2}+2r-\delta,T]_{\mathbb{Z}}. $$

On the other hand, dividing inequality (1.2) by $x(t)$,

$$\frac{\Delta x(t)}{x(t)}=-P(t)\frac{x(t-r_{1})}{x(t)}, $$

because of $\Delta x(t)<0$,

$$\frac{x(t+1)}{x(t)}\leqslant1-P(t)\frac{x(t-r)}{x(t)}. $$

Multiplying from $s-r$ to $t-r-1$ where $s\in[t-r+1,t-1]_{\mathbb {Z}}$, we find

$$\frac{x(s-r)}{x(t-r)}\geqslant\prod_{u=s-r}^{t-r-1} \frac {1}{1-P(u)\frac{x(u-r)}{x(u)}},\quad t\in[T_{2}+3r-\delta,T]_{\mathbb{Z}}. $$

According to (2.12), this yields

$$ \begin{aligned} \frac{x(s-r)}{x(t-r)}\geqslant\prod _{u=s-r}^{t-r-1}\frac {1}{1-f_{0}(\beta)P(u)},\quad t \in[T_{2}+3r-\delta,T]_{\mathbb{Z}}. \end{aligned} $$

(2.14)

We can easily obtain

$$ \begin{aligned}[b] &\Delta \Biggl(-\frac{1}{f_{0}(\beta)} \prod_{u=t-r}^{s-1}\bigl(1-f_{0}( \beta)P(u)\bigr) \Biggr) \\ &\quad =-\frac{1}{f_{0}(\beta)}\prod_{u=t-r}^{s} \bigl(1-f_{0}(\beta)P(u)\bigr)+\frac {1}{f_{0}(\beta)}\prod _{u=t-r}^{s-1}\bigl(1-f_{0}(\beta) P(u)\bigr) \\ &\quad =-\frac{1}{f_{0}(\beta)}\bigl(1-f_{0}(\beta)P(s)-1\bigr)\prod _{u=t-r}^{s-1}\bigl(1-f_{0}(\beta)P(u) \bigr) \\ &\quad =P(s)\prod_{u=t-r}^{s-1} \bigl(1-f_{0}(\beta)P(u)\bigr). \end{aligned} $$

(2.15)

Combining (2.14), (2.15) with (2.13), and because of the fact

$$\frac{n}{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}}\leqslant \sqrt[n]{a_{1}a_{2}\cdots a_{n}}\leqslant \frac{a_{1}+a_{2}+\cdots+a_{n}}{ n} ,\quad a_{i}>0, i=1,2, \ldots,n, $$

we have

$$\begin{aligned} x(t-r)-x(t) \geqslant& x(t-r)\sum_{s=t-r+1}^{t-1}P(s) \frac {x(s-r)}{x(t-r)} \\ \geqslant& x(t-r) \sum_{s=t-r+1}^{t-1}P(s) \Biggl(\prod_{u=s-r}^{t-r-1}\frac {1}{1-f_{0}(\beta)P(u)} \Biggr) \\ \geqslant& x(t-r)\sum_{s=t-r+1}^{t-1}P(s) \Biggl(\prod_{u=s-r}^{s-1}\frac {1}{1-f_{0}(\beta)P(u)}\prod _{u=t-r}^{s-1}{1-f_{0} (\beta)P(u)} \Biggr) \\ \geqslant& x(t-r)\frac{r}{r-\beta f_{0}(\beta)}\sum_{s=t-r+1}^{t-1}P(s) \Biggl(\prod_{u=t-r}^{s-1}\bigl(1-f_{0}( \beta )P(u)\bigr) \Biggr) \\ =& x(t-r)\frac{r}{r-\beta f_{0}(\beta)}\sum_{s=t-r+1}^{t-1} \Delta \Biggl(-\frac{1}{f_{0}(\beta)}\prod_{u=t-r}^{s-1} \bigl(1-f_{0}(\beta )P(u)\bigr) \Biggr) \\ =& x(t-r)\frac{r}{r-\beta f_{0}(\beta)}\Biggl(-\frac{1}{f_{0}(\beta )}\prod _{u=t-r}^{t-1}\bigl(1-f_{0}(\beta)P(u)\bigr) \\ &{} +\frac{1}{f_{0}(\beta)}\prod_{u=t-r}^{t-r} \bigl(1-f_{0}(\beta)P(u)\bigr)\Biggr) \\ \geqslant& x(t-r)\frac{r}{r-\beta f_{0}(\beta)}\frac{1}{f_{0}(\beta )}\bigl(1-f_{0}( \beta)P(t-r)\bigr) \biggl(1-\frac{r-1-\beta f_{0}(\beta)}{r-1} \biggr) \\ \geqslant& x(t-r)\frac{r}{r-\beta f_{0}(\beta)}\frac{1}{f_{0}(\beta )}\biggl(1- \frac{f_{0}(\beta)}{f_{1}(\beta)}\biggr) \biggl(1-\frac{r-1-\beta f_{0}(\beta)}{r-1} \biggr) \\ \geqslant& x(t-r)\frac{\beta}{r-\beta f_{0}(\beta)}\biggl(1-\frac {f_{0}(\beta)}{f_{1}(\beta)}\biggr). \end{aligned}$$

Thus

$$\frac{x(t-r)}{x(t)}\geqslant\frac{r-\beta f_{0}(\beta)}{r-\beta f_{0}(\beta)-\beta+\frac{\beta f_{0}(\beta)}{f_{1}(\beta)} }=f_{2}(\beta)>1,\quad t \in[T_{2}+3r-\delta,T]_{\mathbb{Z}}. $$

Repeating this argument, it follows by induction that

$$\frac{x(t-r)}{x(t)}\geqslant f_{n}(\beta)>0, \quad t\in \bigl[T_{2}+(n+1)r-\delta ,T\bigr]_{\mathbb{Z}}. $$

The proof is complete. □

Remark

It can easily be seen that either $f_{n}(\beta)$ satisfies $\lim_{t\rightarrow\infty}f_{n}(\beta)=1$ or $f_{n}(\beta)$ is nondecreasing and $\lim_{t\rightarrow\infty} f_{n}(\beta)=\infty$ or $f_{n}(\beta)\rightarrow\infty$ after finite number of terms or $f_{n}(\beta)$ is negative.

Lemma 2.4

Assume that $\sum_{s=t-r}^{t-2}P(s)\geqslant\beta$, $t\geqslant t_{0}$ holds for some $0<\beta<1$ and there exists a function $x(t)$ satisfying inequality (1.2) on $[T_{2},T+Nr+1]_{\mathbb{Z}}$ for some positive integer N such that $\Delta x(t)\leqslant0$ on $[T_{2}-\delta ,T_{2}+Nr+1]_{\mathbb{Z}}$ where $T_{2}\geqslant t_{0}+r$. If $x(t)$ is positive on $[T_{2},T_{2}+Nr+1]_{\mathbb{Z}}$, then

$$ \begin{aligned} 0< \frac{x(t-r)}{x(t)}\leqslant g_{m}(\beta), \quad t\in\bigl[T_{2}+2r-\delta ,T_{2}+(N-m)r+1\bigr]_{\mathbb{Z}}, \end{aligned} $$

(2.16)

where m is a positive integer, $N\geqslant m+2-\frac{\delta}{r}$, and $g_{m}(\beta)$ is defined by

$$g_{1}(\beta)=\frac{2(1-\beta)}{\beta^{2}}, \qquad g_{m+1}(\beta)= \frac {2(1-\beta-\frac{1}{g_{m}(\beta)})}{\beta^{2}},\quad m=1,2,\ldots. $$

Proof

From $\sum_{s=t-r}^{t-1}P(s)\geqslant\beta$, $t\geqslant t_{0}$, we see that $\sum_{s=t}^{t+r-1}P(s)\geqslant\beta$ for $t\geqslant T_{2}$. Summing both sides of (1.2) from t to $t+r-1$, we obtain

$$ \begin{aligned} x(t)-x(t+r)\geqslant\sum _{s=t}^{t+r-1}P(s)x(s-r_{1}),\quad t\in \bigl[T_{2}+r,T_{2}+(N-1)r+1\bigr]_{\mathbb{Z}}. \end{aligned} $$

(2.17)

Since $T_{2}+r\leqslant t\leqslant s\leqslant t+r-1$, it follows $T_{2}\leqslant t-r\leqslant s-r\leqslant t-1$. Again, summing (1.2) from $s-r$ to t yields

$$x(s-r)-x(t)\geqslant\sum_{u=s-r}^{t-1}P(u)x(u-r_{1}). $$

It is clear that $x(u-r_{1})$ is non-increasing on $[s-r,t+1]_{\mathbb {Z}}\subseteq[T_{2}+r-\delta,T_{2}+(N-1)r+1]_{\mathbb{Z}}$. Thus,

$$\begin{aligned} x(s-r) \geqslant& x(t)+\sum_{u=s-r}^{t-1}P(u)x(u-r_{1}) \\ \geqslant& x(t)+x(t-r)\sum_{u=s-r}^{t-1}P(u) \\ =&x(t)+x(t-r) \Biggl[\sum_{u=s-r}^{s}P(u)- \sum_{u=t}^{s}P(u) \Biggr] \\ \geqslant& x(t)+x(t-r) \Biggl[\beta-\sum_{u=t}^{s}P(u) \Biggr]. \end{aligned}$$

In view of the last inequality and (2.17), we obtain

$$\begin{aligned} x(t) \geqslant& x(t+r)+\sum _{s=t}^{t+r-1}P(s)x(s-r_{1}) \\ \geqslant& x(t+r)+\sum_{s=t}^{t+r-1}P(s) \Biggl[x(t)+x(t-r) \Biggl(\beta-\sum_{u=t}^{s}P(u) \Biggr) \Biggr] \\ \geqslant& x(t+r)+\beta x(t)+\beta^{2}x(t-r)-x(t-r)\sum _{s=t}^{t+r-1}P(s) \Biggl(\sum _{u=t}^{s}P(u) \Biggr), \end{aligned}$$

(2.18)

for all $t\in[T_{2}+2r-\delta,T_{2}+(N-1)r+1]_{\mathbb{Z}}$. As is well known, we have the identity

$$\sum_{s=t}^{t+r-1}\sum _{u=t}^{s} \bigl(P(s)P(u) \bigr)=\sum _{u=t}^{t+r-1}\sum_{s=u}^{t+r-1} \bigl(P(u)P(s) \bigr) =\sum_{s=t}^{t+r-1}\sum _{u= s}^{t+r-1} \bigl(P(s)P(u) \bigr). $$

Consequently,

$$\sum_{s=t}^{t+r-1}\sum _{u=t}^{s} \bigl(P(s)P(u) \bigr)>\frac {1}{2} \sum_{s=t}^{t+r-1} \sum _{u= t}^{t+r-1} \bigl(P(s)P(u) \bigr)=\frac{1}{2} \Biggl(\sum_{s=t}^{t+r-1}P(s) \Biggr)^{2}\geqslant\frac{\beta^{2}}{2}. $$

Substituting into (2.18),

$$ \begin{aligned} x(t)\geqslant x(t+r)+\beta x(t)+ \frac{\beta^{2}}{2}x(t-r),\quad t\in \bigl[T_{2}+2r- \delta,T_{2}+(N-1)r+1\bigr]_{\mathbb{Z}}. \end{aligned} $$

(2.19)

Since $x(t+r)>0$ on $[T_{2}+2r-\delta,T_{2}+(N-1)r]_{\mathbb{Z}}$, we get

$$ \begin{aligned} \frac{x(t-r)}{x(t)}< \frac{2(1-\beta)}{\beta^{2}}=g_{1}(\beta),\quad t\in \bigl[T_{2}+2r- \delta,T_{2}+(N-1)r+1\bigr]_{\mathbb{Z}}. \end{aligned} $$

(2.20)

On the other hand, when $t\in[T_{2}+2r-\delta,T_{2}+(N-2)r+1]_{\mathbb {Z}}$, we have $T_{2}+2r-\delta\leqslant t\leqslant t+r\leqslant T_{2}+(N-1)r$. So (2.20) leads to

$${x(t+r)}>\frac{1}{g_{1}(\beta)}{x(t)},\quad t\in\bigl[T_{1}+2r-\delta ,T_{1}+(N-2)r+1\bigr]_{\mathbb{Z}}. $$

Since $x(t)$ is non-increasing on $[T_{2}-\delta,T+Nr+1]$, it follows that

$${x(t+r)}>\frac{1}{g_{1}(\beta)}{x(t)}\geqslant\frac{1}{g_{1}(\beta )}{x(t)}, \quad t\in \bigl[T_{2}+2r-\delta,T_{2}+(N-2)r+1\bigr]_{\mathbb{Z}}. $$

From this inequality and (2.19), we obtain

$$x(t)\geqslant\frac{1}{g_{1}(\beta)}{x(t)}+\beta x(t)+\frac{\beta ^{2}}{2}x(t-r), \quad t\in\bigl[T_{2}+2r-\delta,T_{2}+(N-2)r+1 \bigr]_{\mathbb{Z}}. $$

Rearranging,

$$\frac{x(t-r)}{x(t)}< \frac{2(1-\beta-\frac{1}{g_{1}(\beta)})}{\beta ^{2}}=g_{2}(\beta),\quad t\in \bigl[T_{2}+2r-\delta,T_{2}+(N-2)r+1\bigr]_{\mathbb{Z}}. $$

Repeating the above procedure, we get

$$\frac{x(t-r)}{x(t)}< \frac{2(1-\beta-\frac{1}{g_{m-1}(\beta)})}{\beta ^{2}}=g_{m}(\beta),\quad t\in \bigl[T_{2}+2r-\delta,T_{2}+(N-m)r+1\bigr]_{\mathbb{Z}}. $$

The proof of Lemma 2.4 is complete. □

Remark

Wu and Xu [18] proved that $g_{m}(\beta)$ is decreasing. They found also that $g_{m+1}(\beta)>\frac{1-\beta}{\beta_{2}}$ for $m=1,2,\ldots$ . So when $0<\beta\leqslant\sqrt{2}-1$, there exists a function $g(\beta)=\frac{2(1-\beta-\frac{1}{g(\beta)})}{\beta_{2}}$ such that $\lim_{m\rightarrow\infty}g_{m}(\beta)=g(\beta)$.

Lemma 2.5

Assume that $\sum_{s=t-r}^{t-2}P(s)\geqslant\beta$ holds for some $\beta>\sqrt{2}-1$ and $x(t)$ is a function satisfying inequality (1.2) on $[T_{2},T]_{\mathbb{Z}}$ with $\Delta x(t)\leqslant0$ for $[T_{2}-\delta,T]_{\mathbb{Z}}$, $T\geqslant T_{2}+(k_{\beta}+1)r-\delta$, $T_{2}\geqslant t_{0}+r$ and $k_{\beta}$ is defined by

$$ \begin{aligned} &k_{\beta}= \textstyle\begin{cases} 1, \quad \beta\geqslant1,\\ \min\{\alpha,\gamma\},\quad \sqrt{2}-1< \beta< 1, \end{cases}\displaystyle \\ &\alpha=\min_{n\geqslant1,m\geqslant1}\bigl\{ n+m|f_{n}(\beta)\geqslant g_{m}(\beta)\bigr\} , \\ &\gamma=1+\min_{n\geqslant1}\bigl\{ n|f_{n}(\beta)< 0 \textit{ or } f_{n+1}(\beta)=\infty\bigr\} . \end{aligned} $$

(2.21)

Then $x(t)$ is positive on $[T_{2},T]_{\mathbb{Z}}$.

Proof

Suppose, for the sake of contradiction, that $x(t)$ is positive on $[T_{2},T]$. We consider two cases:

Case 1: $\beta\geqslant1$. In this case $k_{\beta}=1$ and $T\geqslant T_{2}+2r-\delta$. Since $\Delta x(t)\leqslant0$ on $[T_{2}-\delta ,T]_{\mathbb{Z}}$, we obtain

$$x(t)\geqslant x(T_{2}+r-\delta), \quad t\in[T_{2}- \delta,T_{2}+r-\delta ]_{\mathbb{Z}}. $$

Summing both sides of (1.2) from $T_{2}+r-\delta$ to $T_{2}+2r-\delta-1$ and using the above inequality, we obtain

$$\begin{aligned} x(T_{2}+2r-\delta) \leqslant& x(T_{2}+r- \delta)-\sum_{s=T_{2}+r-\delta}^{T_{2}+2r-\delta-1}P(s)x(s-r) \\ \leqslant& x(T_{2}+r-\delta)-x(T_{2}+r-\delta)\sum _{s=T_{2}+r-\delta }^{T_{2}+2r-\delta-1}P(s) \\ =&x(T_{2}+r-\delta) \Biggl[1-\sum_{s=T_{2}+r-\delta}^{T_{2}+2r-\delta -1}P(s) \Biggr]< 0, \end{aligned}$$

which is a contradiction.

Case 2: $\sqrt{2}-1<\beta<1$. If $k_{\beta}=n^{\ast}+m^{\ast}$, then

$$ \begin{aligned} f_{n^{\ast}}(\beta)\geqslant g_{m^{\ast}}(\beta). \end{aligned} $$

(2.22)

From Lemma 2.3, it follows that

$$ \begin{aligned} \frac{x(t-r)}{x(t)}\geqslant f_{n^{\ast}}(\beta),\quad t\in\bigl[T_{2}+\bigl({n^{\ast}}+1\bigr)r-\delta,T\bigr]_{\mathbb{Z}}. \end{aligned} $$

(2.23)

On the other hand, by Lemma 2.4 we find

$$ \begin{aligned} \frac{x(t-r)}{x(t)}< g_{m^{\ast}}( \beta),\quad t\in\bigl[T_{2}+2r-\delta ,T_{2}+ \bigl(N-{m^{\ast}}r\bigr)+1\bigr]_{\mathbb{Z}}. \end{aligned} $$

(2.24)

So, when $t=T_{2}+({n^{\ast}}+1)r-\delta$ in (2.23) and (2.24), it follows that

$$f_{n^{\ast}}(\beta)\leqslant\frac{x(T_{2}+{n^{\ast}}r-\delta )}{x(T_{2}+({n^{\ast}}+1)r-\delta)}< g_{m^{\ast}}(\beta), $$

which contradicts (2.22). If

$$k_{\beta}=1+\min_{n\geqslant1}\bigl\{ n|f_{n+1}( \beta)< 0 \text{ or } f_{n+1}(\beta)=\infty\bigr\} , $$

then Lemma 2.3 implies a contradiction and the proof is complete. □

3 Main results

In this section, we obtain sufficient oscillation conditions for Eq. (1.1) about the distribution of generalized zeros.

Theorem 3.1

Let $(H_{1})$–$(H_{4})$ and (2.4) establish for some positive integer n with $r_{1}=\sigma-\tau$. Then the equation (1.1) oscillates and $d_{\tilde{t}}(x)\leqslant2\sigma +3n(\sigma-\tau)$, where $\tilde{t}=t_{1}+(2n+1)(\sigma-\tau)$.

Proof

If Eq. (1.1) has a non-oscillatory solution $x(t)$, and $-x(t)$ is also the solution of Eq. (1.1), so we only consider the situation of the solution of (1.1) is eventually positive. We assume $x(t)>0$ on $[T_{0},T]_{\mathbb{Z}}$ for some integer $T_{0}\geqslant\tilde{t}$ where $T>T_{0}+2\sigma+3n(\sigma-\tau)$. Since $z(t)=x(t)+p(t)x(t-\tau)$ for $t\in[T_{0}+\tau,T]$, $z(t)>0$ on $[T_{0}+\tau,T]_{\mathbb{Z}}$. From inequality (2.2), we have

$$ \begin{aligned}[b] &\Delta\bigl(a(t)\Delta z(t)\bigr) \\ &\quad \leqslant-kq(t)z(t-\sigma)+kq(t)p(t-\sigma )x(t-\tau-\sigma), \quad t \in[T_{0}+\sigma+\tau,T]_{\mathbb{Z}}. \end{aligned} $$

(3.1)

Also from (2.2), we obtain $x(t-\tau-\sigma)\leqslant-\frac {\Delta(a(t-\tau)\Delta z(t-\tau))}{kq(t-\tau)}$, $t\in[T_{0}+\sigma+\tau,T]_{\mathbb{Z}}$. Then inequality (3.1) can be rewritten as

$$\begin{aligned}& \Delta\bigl(a(t)\Delta z(t)\bigr) \\& \quad \leqslant-kq(t)z(t-\sigma)-kq(t)p(t-\sigma )\frac{\Delta(a(t-\tau)\Delta z(t-\tau))}{kq(t-\tau)} \\& \quad =-kq(t)z(t-\sigma)-p(t-\sigma)\frac{q(t)}{q(t-\tau)}\Delta \bigl(a(t-\tau) \Delta z(t-\tau)\bigr),\quad t\in[T_{0}+\sigma,T]_{\mathbb{Z}}, \end{aligned}$$

i.e.

$$\begin{aligned}& \Delta\bigl(a(t)\Delta z(t)\bigr)+p(t-\sigma)\frac{q(t)}{q(t-\tau)}\Delta \bigl(a(t- \tau)\Delta z(t-\tau)\bigr)+kq(t)z(t-\sigma) \\& \quad \leqslant0,\quad t\in [T_{0}+\sigma,T]_{\mathbb{Z}}. \end{aligned}$$

We can conclude from condition $(H_{4})$ and $\Delta(a(t)\Delta z(t))<0$,

$$ \begin{aligned}[b] &\Delta\bigl(a(t)\Delta z(t) \bigr)+H(t)\Delta\bigl(a(t-\tau)\Delta z(t-\tau )\bigr)+kq(t)z(t-\sigma) \\ &\quad \leqslant0,\quad t\in[T_{0}+\sigma,T]_{\mathbb{Z}}. \end{aligned} $$

(3.2)

Let

$$ \begin{aligned} w(t)=a(t)\Delta z(t)+H(t) \bigl(a(t- \tau)\Delta z(t-\tau)\bigr), \quad t\in [T_{0}+\sigma+ \tau,T]_{\mathbb{Z}}. \end{aligned} $$

(3.3)

So

$$ \begin{aligned} w(t)\leqslant\bigl(1+H(t)\bigr) \bigl(a(t-\tau)\Delta z(t-\tau)\bigr). \end{aligned} $$

(3.4)

Differentiating both sides of (3.3), and because of (3.2), and $\Delta(a(t)\Delta z(t))<0$, we obtain

$$ \begin{aligned}[b] \Delta w(t)&\leqslant\Delta H(t) \bigl(a(t-\tau+1)\Delta z(t-\tau +1)\bigr)-kq(t)z(t-\sigma) \\ &< \Delta H(t) \bigl(a(t-\tau)\Delta z(t-\tau)\bigr)-kq(t)z(t-\sigma),\quad t\in [T_{0}+\sigma,T]_{\mathbb{Z}}. \end{aligned} $$

(3.5)

From (3.4), we get

$$\Delta z(t-\sigma)=\Delta z(t+\tau-\sigma-\tau)\geqslant\frac {w(t+\tau-\sigma)}{a(t-\sigma)(1+H(t+\tau-\sigma))},\quad t\in [T_{0}+2\sigma,T]_{\mathbb{Z}}. $$

Summing up the above form from $T_{0}$ to $t-1$, we have

$$z(t-\sigma)-z(T_{0}-\sigma)\geqslant\sum _{s=T_{0}}^{t-1}\frac {w(s+\tau-\sigma)}{1+H(s+\tau-\sigma)}\frac{1}{ a(s-\sigma)}, $$

therefore

$$ \begin{aligned} z(t-\sigma)\geqslant\sum _{s=T_{0}}^{t-1}\frac{w(s+\tau-\sigma )}{1+H(s+\tau-\sigma)}\frac{1}{a(s-\sigma)}. \end{aligned} $$

(3.6)

Let

$$ \begin{aligned} y(t)=\frac{w(t)}{1+H(t)}>0,\quad t \in[T_{0}+\sigma+\tau,T]_{\mathbb{Z}}. \end{aligned} $$

(3.7)

Then

$$ \begin{aligned} \Delta y(t)=\frac{\Delta w(t)(1+H(t))-w(t)\Delta (1+H(t))}{(1+H(t+1))(1+H(t))},\quad t \in[T_{0}+\sigma+\tau,T]_{\mathbb{Z}}. \end{aligned} $$

(3.8)

Adding (3.3) and (3.5) to (3.8), we have

$$ \begin{aligned} \Delta y(t)+\frac{kq(t)z(t-\sigma)}{1+H(t+1)}< 0,\quad t\in[T_{0}+\sigma +\tau,T]_{\mathbb{Z}}. \end{aligned} $$

(3.9)

From (3.6), (3.7) and the decreasing of $y(t)$, we get

$$z(t-\sigma)\geqslant\sum_{s=T_{0}}^{t-1} \frac{1}{a(s-\sigma )}y(s+\tau-\sigma)\geqslant y(t+\tau-\sigma)\sum _{s=T_{0}}^{t-1}\frac{1}{a(s-\sigma)},\quad t \in[T_{0}+\sigma+\tau,T]_{\mathbb{Z}}. $$

Substituting the above inequality into (3.9), we obtain

$$ \begin{aligned} \Delta y(t)+\frac{kq(t)y(t+\tau-\sigma)}{1+H(t+1)}\sum _{s=T_{0}}^{t-1}\frac{1}{a(s-\sigma)}< 0,\quad t \in[T_{0}+\sigma+\tau ,T]_{\mathbb{Z}}. \end{aligned} $$

(3.10)

Set $r_{1}=\sigma-\tau$, $T_{1}=T_{0}+\sigma+\tau$ and $P(t)=\frac {kq(t)}{1+H(t+1)}\sum_{s=T_{0}}^{t-1}\frac{1}{a(s-\sigma)}$, we conclude

$$\Delta y(t)+P(t)y(t-r_{1})< 0, \quad t\in[T_{1},T]_{\mathbb{Z}}. $$

What is more, (2.1) holds and $y(t)$ is decreasing. Then we can conclude from Lemma 2.2 that $y(t)$ cannot be positive on $[T_{1},T]_{\mathbb{Z}}$ when $r_{1}=\sigma-\tau$, where $T>T_{1}+3n(\sigma-\tau)$. This is a contradiction with (3.7). The proof is completed. □

Assume the following condition holds:

($H_{5}$):

$\sum_{s=t-r}^{t-1}\frac{q(s)}{1+H(s+1)}\sum_{v=T_{0}}^{s-1}\frac{1}{a(v-\sigma)}\geqslant\beta$, $t\geqslant t_{2}$ for some $t_{2}\geqslant t_{0}+2\sigma-\tau$.

Then we can obtain some further conclusions by means of Theorem 3.1.

Corollary 3.1

Suppose conditions $(H_{1})$–$(H_{5})$ hold and a sequence $\{\beta _{n}\}$ is defined by

$$\beta_{0}=\beta>0,\qquad \beta_{n}={\beta_{0}}^{n+1}, \quad n=0,1,2,\ldots. $$

If there is some positive constant $n_{0}\in\mathbb{N}$ such that $1\leqslant\beta< r$, then Eq. (1.1) is oscillatory and $d_{\tilde{t}}\leqslant2\sigma+3n_{0}(\sigma-\tau)$, where $\tilde{t}=t_{1}+(2n_{0}+1)(\sigma-\tau)$.

Proof

According to condition $(H_{5})$, we have $\sum_{s=t-r}^{t-1}F_{0}(s)\geqslant\beta$. In addition, from the iterative sequence $\{F_{n}(t)\}$, we get

$$\begin{aligned} \sum_{s=t-r}^{t-1}F_{1}(s) =&\sum _{s=t-r}^{t-1}F_{0}(s)\sum _{v=s-r}^{s-1}F_{0}(s) \prod _{\zeta=v-r}^{s-1}\frac{1}{1-F_{0}(\zeta)} \\ \geqslant&\sum_{s=t-r}^{t-1}F_{0}(s) \sum_{v=s-r}^{s-1}F_{0}(s)\prod _{\zeta=v-r}^{v-1} \frac{1}{ 1-F_{0}(\zeta)} \prod _{\zeta=v}^{s-1}\frac{1}{1-F_{0}(\zeta)} \\ \geqslant&\frac{r}{r-\beta}\sum_{s=t-r}^{t-1}F_{0}(s) \sum_{v=s-r}^{s-1}F_{0}(s) \\ =&\frac{r}{r-\beta}{\beta}^{2}\geqslant\beta^{2}. \end{aligned}$$

In the same way, continuing the calculation n times, we obtain $\sum_{s=t-r}^{t-1}F_{n}(s)\geqslant\beta_{n}$ for $n=2,3,\ldots$ . Therefore, by mathematical induction, we have

$$\sum_{s=t-r}^{t-1}F_{n}(s)\geqslant \beta_{n},\quad \text{for all } n\in \mathbb{N}. $$

Let $n=n_{0}$. According to Theorem 3.1, the proof is completed. □

Theorem 3.2

Let $(H_{1})$–$(H_{5})$ hold. Then Eq. (1.1) oscillates and $d_{t_{2}}(x)\leqslant2\sigma+k_{\beta}(\sigma-\tau)$, where $k_{\beta}$ is defined by (2.21).

Proof

As usual, we assume (1.1) has a solution $x(t)>0$ on $[T_{0},T]_{\mathbb{Z}}$ where $T>T_{0}+2\sigma+k_{\beta}(\sigma -\tau)$ and $T_{0}\geqslant t_{2}$. Proceed as in the proof of Theorem 3.1, when $T_{1}=T_{0}+2\sigma$. It follows that

$$\Delta y(t)+p(t)y(t+\tau-\sigma)< 0,\quad t\in[T_{1},T]_{\mathbb{Z}}, $$

where

$$y(t)>0, \quad t\in\bigl[T_{1}-2(\sigma-\tau),T\bigr]_{\mathbb{Z}}. $$

Also from (3.10) we obtain

$$\Delta y(t)< 0,\quad t\in\bigl[T_{1}-(\sigma-\tau),T \bigr]_{\mathbb{Z}}. $$

Since $(H_{5})$ holds, we conclude from Lemma 2.5 with $\delta =\sigma-\tau$ that $y(t)$ cannot be positive on $[T_{1},T]_{\mathbb{Z}}$, where $T>T_{1}+k_{\beta}(\sigma-\tau)$. This contradiction completes the proof. □

4 Example

In this section, we will present an example to illustrate main results.

Example 4.1

We consider the delay difference equation

$$ \begin{aligned} \Delta\bigl(\Delta\bigl(x(t)+x(t-1) \bigr)\bigr)+\frac{7}{48(t-2)}x(t-2)=0,\quad t\geqslant1. \end{aligned} $$

(4.1)

Compared with (1.1), we denote $a(t)=1$ where $c>t-1$, $p(t)=1$, $q(t)=\frac{7}{48(t-2)}$, $\tau=1$, $\sigma=3$, $r=\sigma-\tau=2$, $f(u)=u$ for $u\neq0$. It is easy to verify that conditions $(H_{1})$ and $(H_{2})$. Since $f(u)=u$, we get $\frac{f(u)}{u}=1$. $(H_{3})$ holds. Now take $H(t)\equiv1$ which satisfies $(H_{4})$. According to (2.1), we obtain $F_{0}(t)=\frac{kq(t)}{1+H(t+1)}\sum_{s=T_{0}}^{t-1}\frac {1}{a(s-\sigma)}=\frac{7}{2\times48(t-2)} \sum_{s=2}^{t-1}1=\frac{7}{24}$, $t\geqslant1$ and $\frac {1}{1-F_{0}(\zeta)}=\frac{24}{17}$, so

$$\begin{aligned} F_{1}(t) =&F_{0}(t)\sum_{s=t-r}^{t-1}F_{0}(s) \prod_{\zeta =s-r}^{t}\frac{1}{1-F_{0}(\zeta)} \\ =&\frac{7}{24}\frac{7}{24} \biggl(\frac{24}{17} \biggr)^{5}+\frac {7}{24}\frac{7}{24} \biggl( \frac{24}{17} \biggr)^{4} \approx0.815 , \quad t\geqslant6. \end{aligned}$$

Therefore, $\sum_{s=t-2}^{t-1}F_{1}(s)\,ds\geqslant1$ for all $t\geqslant6$. Here, all conditions of Theorem 3.1 are satisfied with $n=1$, then we derive that (4.1) shows oscillatory and $d_{\tilde {t}}(x)\leqslant2\sigma+3n(\sigma-\tau)=12$, where $\tilde{t}=t_{1}+(2n+1)(\sigma-\tau)=t_{1}+6$ and $t_{1}\geqslant t_{0}+\sigma=4$.

5 Conclusion

In this paper, two theorems on the distribution of oscillation zeros for second-order nonlinear neutral delay difference equations are obtained by means of inequality techniques, specific function sequences and non-increasing solutions for corresponding first-order difference inequality. Comparing with the corresponding differential equation, it is more complex to deal with the lower bound of summation. Function $\frac{1}{a(t)}\prod_{s=1}^{t-1} (1+a(s))$ is invariant after derivation in difference equation, which is equivalent to $e^{x}$ in differential equation. That is the difficulty we address and the innovation of this paper. We study a second-order equation under the canonical form, and it is also of great significance for the study of non-canonical forms. Moreover, this paper can be extended to the dynamic equation on time scale.