Oscillation Results Using Linearization of Quasi-Linear Second Order Delay Difference Equations

Abstract

In this paper, the authors investigate the oscillatory behavior of quasilinear second order delay difference equations of the form

$$\begin{aligned} \Delta (b(n)(\Delta u(n))^{\alpha })+p(n)u^{\beta }(n-\sigma )=0. \end{aligned}$$

By obtaining new monotonic properties of the nonoscillatory solutions and using them to linearize the equation leads to new oscillation criteria. The criteria obtained improve existing ones in the literature. Two examples are included to show the importance of the main results.

1 Introduction

In this paper, we investigate the oscillatory and asymptotic behavior of solutions of the second order quasilinear delay difference equation

where $n_0$ is a positive integer. We assume that the following conditions hold throughout this paper without further mention:

(C$_1)$:

$\{b(n)\}$ and $\{p(n)\}$ are positive real sequences;

(C$_2)$:

$\alpha$ and $\beta$ are ratios of odd positive integers;

(C$_3)$:

$\sigma$ is a positive integer;

(C$_4)$:

$\displaystyle {B(n)=\sum _{s=n_0}^{n-1}b^{-1/\alpha }(s)\rightarrow \infty \ \text{ as } n\rightarrow \infty .}$

By a solution of (E), we mean a real sequence $\{u(n)\}$ defied for $n\ge n_0-\sigma$ satisfying equation (E) for all $n\ge n_0$. A nontrivial solution of (E) is called oscillatory if it is neither eventually negative nor eventually positive, and it is called nonoscillatory otherwise. Equation (E) is called oscillatory if all its solutions are oscillatory.

The problem of investigating oscillation criteria for various types of difference equations has been a very active research area over the past several decades. A large number of papers and monographs have been devoted to this problem; for a few examples, see [1,2,3, 7,8,9, 11,12,17] and the references contained therein. Koplatadze [11] obtained some very nice oscillatory criteria for the equation

$$\begin{aligned} \Delta ^2u(n)+p(n)u(n-\sigma )=0 \end{aligned}$$

(1.1)

based on the following monotonic properties of positive solutions:

$$\begin{aligned} u(n) \ \ \text {is increasing and} \ \frac{u(n)}{n} \ \ \text {is decreasing.} \end{aligned}$$

(1.2)

The main aim of the paper is to establish some new comparison theorems for investigation of oscillatory behavior of solutions of (E). First, we will linearize equation (E). Then we will deduce the oscillation of (E) from that of its linearized forms. To achieve this, we obtain some results on the monotonic properties of nonoscillatory solutions of (E) that are new even for (1.1) and improves those in (1.2). We will demonstrate the usefulness of our main results via some examples. The technique of proof used here is based in part on some recent papers of Baculíková [4,5,6] on the oscillation of solutions of differential equations.

2 Auxiliary Results

We begin with some useful lemmas concerning monotonic properties of nonoscillatory solutions of (E).

Lemma 2.1

Let $\{u(n)\}$ be a positive solution of (E). Then:

(P$_1)$:

$\{u(n)\}$ is eventually increasing and $\{b(n)(\Delta u(n))\}$ is eventually decreasing;

(P$_2)$:

$\{\frac{u(n)}{B(n)}\}$ is eventually decreasing.

Moreover, if

$$\begin{aligned} \sum _{n=n_0}^{\infty }p(n)B^{\beta }(n-\sigma )=\infty , \end{aligned}$$

(2.1)

then

(P$_3)$:

$\displaystyle {\lim _{n\rightarrow \infty }\frac{u(n)}{B(n)}=0.}$

Proof

Let $\{u(n)\}$ be a positive solution of (E). Then $\Delta (b(n)(\Delta u(n))^{\alpha })<0$, and there is an integer $n_1\ge n_0$ that $b(n)(\Delta u(n))^{\alpha }$ has a constant sign for all $n\ge n_1$. We claim that $b(n)(\Delta u(n))^{\alpha }>0$ eventually. To show this, assume that $b(n)(\Delta u(n))^{\alpha }<0$ for $n\ge n_2$ for some $n_2 \ge n_1$. Then there exists a constant $M>0$ such that $b(n)(\Delta u(n))^{\alpha }<-M<0$ for $n \ge n_2$. Summing the last inequality from $n_2$ to $n-1$ and using (C$_4$), we have

$$\begin{aligned} u(n)\le u(n_2) - MB(n)\rightarrow -\infty \ \ \text {as} \ \ n\rightarrow \infty \end{aligned}$$

which is a contradiction and proves our claim. Employing the monotonicity of $b^{1/\alpha }(n)\Delta u(n)$, we obtain

$$\begin{aligned} u(n)\ge \sum _{s=n_2}^{n-1}\frac{b^{1/\alpha }(s)\Delta u(s)}{b^{1/\alpha }(s)}\ge B(n)b^{1/\alpha }(n)\Delta u(n), \end{aligned}$$

(2.2)

which implies $\Delta \left( \frac{u(n)}{B(n)}\right) <0$. Since $\frac{u(n)}{B(n)}$ is positive and decreasing, there exists $c\in {{\mathbb {R}}}$ such that

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{u(n)}{B(n)} = c\ge 0. \end{aligned}$$

If $c>0$, then, $u(n)\ge cB(n)$ for $n\ge n_3 \ge n_2$. Using this in (E) and then summing from $n_3$ to $n-1$, we obtain

$$\begin{aligned} b(n_3)(\Delta u(n_3))^{\alpha }\ge c^{\beta }\sum _{s=n_3}^{n-1}p(s)B^{\beta }(s-\sigma ), \end{aligned}$$

which as $n\rightarrow \infty$ contradicts (2.1). Thus, $c=0$, that is $\lim _{n\rightarrow \infty }\frac{u(n)}{B(n)}=0$, which competes the proof of the lemma. $\square$

Remark 2.2

In the case of equation (1.1) where $b(n) \equiv 1$ and $\beta = 1$, the three properties of nonoscillatory solutions described in Lemma 2.1 become:

(P$_1)$:

$\{u(n)\}$ is eventually increasing and $\{\Delta u(n)\}$ is eventually decreasing;

(P$_2)$:

$\{\frac{u(n)}{n-1}\}$ is eventually decreasing;

(P$_3)$:

$\lim _{n\rightarrow \infty }\frac{u(n)}{B(n)}=0$.

Lemma 2.3

Let $\{u(n)\}$ be an eventually increasing solution of (E). Then, $u^{\beta -\alpha }(n)\ge \eta (n)$, where $\eta (n)$ is given by

$$\begin{aligned} \eta (n)= {\left\{ \begin{array}{ll} 1, &{} \text {if} \ \alpha =\beta , \\ a_1, &{} \text {if} \ \alpha <\beta , \\ a_2B^{\beta -\alpha }(n), &{} \text {if} \ \alpha >\beta , \end{array}\right. } \end{aligned}$$

and $a_1$ and $a_2$ are positive constants.

Proof

Since u(n) is a positive increasing solution of (E), there exists a constant $M>0$ such that $u(n)\ge M$ for all $n\ge n_1$ for some $n_1 \ge n_0$. From $(P_2)$, we see that $\frac{u(n)}{B(n)}$ is decreasing and so

$$\begin{aligned} \frac{u(n)}{B(n)}\le \frac{u(n_1)}{B(n_1)} = M_1. \end{aligned}$$

Thus,

$$\begin{aligned} u^{\beta -\alpha }(n)\ge \eta (n)= {\left\{ \begin{array}{ll} 1, &{} \text {for} \ \alpha =\beta , \\ a_1, &{} \text {for} \ \alpha <\beta , \\ a_2B^{\beta -\alpha }(n), &{} \text {for} \ \alpha >\beta , \end{array}\right. } \end{aligned}$$

where $a_1=M^{\beta -\alpha }$ and $a_2=M_{1}^{\beta - \alpha }$. This proves the lemma. $\square$

Lemma 2.4

Let (2.1) hold and assune there exists a constant $\delta \in [0,1)$ such that

$$\begin{aligned} p(n)\eta (n-\sigma )B^{\alpha }(n-\sigma )B(n)b^{1/\alpha }(n)\ge \alpha \delta , \ n\ge n_0. \end{aligned}$$

(2.3)

If $\{u(n)\}$ is a positive solution of (E), then

$$\begin{aligned} u(n)\ge \frac{B(n)b^{1/\alpha }(n)\Delta u(n)}{(1-\delta )} \ \text{ eventually } \end{aligned}$$

(2.4)

and

$$\begin{aligned} \frac{u(n)}{B^{\delta _1}(n)} \ \text{ is } \text{ eventually } \text{ increasing, } \text{ where } \ \delta _1=\delta ^{1/\alpha }. \end{aligned}$$

(2.5)

Proof

In view of (2.1) and from Lemma 2.1, we see that (P$_1$) holds and

$$\begin{aligned}&\Delta (b(n)(\Delta u(n))^{\alpha }B^{\alpha \delta }(n))=\Delta (b(n)(\Delta u(n))^{\alpha })B^{\alpha \delta }(n+1)\nonumber \\&+b(n)(\Delta u(n))^{\alpha }\Delta B^{\alpha \delta }(n). \end{aligned}$$

(2.6)

By the Mean-value Theorem,

$$\begin{aligned} B^{\alpha \delta }(n+1)-B^{\alpha \delta }(n) \le {\left\{ \begin{array}{ll} \alpha \delta B^{\alpha \delta -1}(n+1)\Delta B(n), &{} \text{ if } \ \alpha \delta >1,\\ \alpha \delta B^{\alpha \delta -1}(n)\Delta B(n), &{} \text{ if } \ \alpha \delta <1, \end{array}\right. } \end{aligned}$$

where $B(n)<B(n+1)$. Since $\Delta B(n)= b^{-1/\alpha }(n)$, we have

$$\begin{aligned} \Delta B^{\alpha \delta }(n) \le \alpha \delta \frac{B^{\alpha \delta }(n+1)}{B(n)}b^{-1/\alpha }(n), \end{aligned}$$

and using this in (2.6), we obtain

$$\begin{aligned}&\Delta (b(n)(\Delta u(n))^{\alpha }B^{\alpha \delta }(n)) \le -p(n)\eta (n-\sigma )u^{\alpha }(n-\sigma )B^{\alpha \delta }(n+1)\nonumber \\&\quad + \alpha \delta b(n)(\Delta u(n))^{\alpha }\frac{b^{-1/\alpha }(n)B^{\alpha \delta }(n+1)}{B(n)}. \end{aligned}$$

(2.7)

Since $b^{1/\alpha }(n)(\Delta u(n))$ is decreasing, from (2.2)

$$\begin{aligned} u^{\alpha }(n-\sigma )\ge B^{\alpha }(n-\sigma )b(n)(\Delta u(n))^{\alpha }. \end{aligned}$$

(2.8)

Combining (2.7) and (2.8), and then using (2.3), we obtain

$$\begin{aligned}&\Delta (b(n)(\Delta u(n))^{\alpha }B^{\alpha \delta }(n)) \\&\quad \le -b(n)(\Delta u(n))^{\alpha } B^{\alpha \delta }(n+1) \left[ p(n)\eta (n-\sigma )B^{\alpha }(n-\sigma ) -\frac{\alpha \delta }{b^{1/\alpha }(n)B(n)}\right] < 0. \end{aligned}$$

Hence, $\{b(n)(\Delta u(n))^{\alpha }B^{\alpha \delta }(n)\}$ is decreasing and thus there exists an integer $n_1\ge n_0$ such that

$$\begin{aligned} u(n)&\ge b^{1/\alpha }(n)B^{\delta }(n)\Delta u(n)\sum _{s=n_1}^{n-1}\frac{B^{-\delta }(s)}{b^{1/\alpha }(s)}\\&\ge b^{1/\alpha }(n)B^{\delta }(n)\Delta u(n)\sum _{s=n_1}^{n-1}\int _{B(s)}^{B(s+1)}\frac{dx}{x^{\delta }}\\&= b^{1/\alpha }(n)B^{\delta }(n)\Delta u(n)\left( \frac{B^{1-\delta }(n)}{1-\delta }\right) , \end{aligned}$$

which proves (2.4).

To prove (2.5) first note that (P$_3$) implies

$$\begin{aligned} \lim _{n\rightarrow \infty } b^{1/\alpha }(n)\Delta u(n)=0. \end{aligned}$$

(2.9)

Therefore, a summation of (E) yields

$$\begin{aligned} b^{1/\alpha }(n)\Delta u(n)\ge \left[ \sum _{s=n}^{\infty }p(s)\eta (s-\sigma )u^{\alpha }(s-\sigma )\right] ^{1/\alpha }. \end{aligned}$$

(2.10)

Using (2.3) and the facts that $\frac{u(n)}{B(n)}$ is decreasing and u(n) is increasing, it follows from (2.10) that

$$\begin{aligned} b^{1/\alpha }(n)\Delta u(n)\ge & {} u(n)\left[ \sum _{s=n}^{\infty }\frac{p(s)\eta (s-\sigma )B^{\alpha }(s-\sigma )}{B^{\alpha }(s)}\right] ^{1/\alpha }\nonumber \\\ge & {} u(n)\left[ \sum _{s=n}^{\infty }\frac{\alpha \delta }{B^{\alpha +1}(s)b^{1/\alpha }(s)}\right] ^{1/\alpha }\nonumber \\\ge & {} \frac{\delta _1u(n)}{B(n)}. \end{aligned}$$

(2.11)

Now

$$\begin{aligned} \Delta \left( \frac{u(n)}{B^{\delta _1}(n)}\right) = \frac{B^{\delta _1}(n)\Delta u(n) - u(n)\Delta B^{\delta _1}(n)}{B^{\delta _1}(n)B^{\delta _1}(n+1)}. \end{aligned}$$

(2.12)

By the Mean-Value Theorem,

$$\begin{aligned} B^{\delta _1}(n+1)-B^{\delta _1}(n)\le \delta _1B^{\delta _1-1}(n)\Delta B(n), \end{aligned}$$

since $\delta _1=\delta ^{1/\alpha }<1$ and $\Delta B(n)=b^{-1/\alpha }(n)$. Using this in (2.12), we obtain

$$\begin{aligned} \Delta \left( \frac{u(n)}{B^{\delta _1}(n)}\right) \ge \frac{B(n)b^{1/\alpha }(n)\Delta u(n)-\delta _1u(n)}{B(n)B^{\delta _1}(n+1)b^{1/\alpha }(n)}\ge 0 \end{aligned}$$

in view of (2.11). This proves (2.5) and completes the proof of the lemma. $\square$

Remark 2.5

The monotone increasing property of $\{\frac{u(n)}{B^{\delta _1}(n)}\}$ obtained in Lemma 2.4 improves that for $\{u(n)\}$. This is new even for equation (1.1) for which it takes the form $\frac{u(n)}{(n-1)^{\delta _1}}$.

The following lemma taken from [16] will also be needed in the proofs of our results.

Lemma 2.6

([16, Lemma 1]) Let F(n, u) be a continuous function defined on $\mathbb {N}_0 \times {\mathbb {R}}$ that is nondecreasing in u with ${{\,\mathrm{sgn}\,}}F(n,u) = {{\,\mathrm{sgn}\,}}u$, and let $\alpha$ and $\sigma$ be as above. If the difference inequality

$$\begin{aligned} \Delta ((\Delta x_n)^{\alpha }) + F(n,x(n-\sigma ) \le 0 \end{aligned}$$

has an eventually positive solution, then so does the difference equation

$$\begin{aligned} \Delta ((\Delta y_n)^{\alpha }) + F(n,y(n-\sigma )) = 0. \end{aligned}$$

3 Comparison Results

In this section, we present new comparison principles that significantly simply the examination of (E).

Theorem 3.1

Let conditions (2.1) and (2.3) hold. Then Eq. (E) is oscillatory provided that the equation

$$\begin{aligned} \Delta w(n)+\frac{p(n)\eta (n-\sigma )B^{\alpha }(n-\sigma )}{(1-\delta )^{\alpha }}w(n-\sigma )=0 \end{aligned}$$

(3.1)

is oscillatory.

Proof

Assume to the contrary that $\{u(n)\}$ is a nonoscillatory solution of (E), say $u(n) > 0$ for $n \ge n_1$ for some $n_1 \ge n_0$. Then using (2.4) in (E), we obtain

$$\begin{aligned} \Delta (b(n)(\Delta u(n))^{\alpha })+ \frac{p(n)\eta (n-\sigma )B^{\alpha }(n-\sigma )b(n-\sigma )(\Delta u(n-\sigma ))^{\alpha }}{(1-\delta )^{\alpha }} \le 0. \end{aligned}$$

Letting $w(n)=b(n)(\Delta u(n))^{\alpha }$, we see that $\{w(n)\}$ is a positive solution of the inequality

$$\begin{aligned} \Delta w(n)+\frac{p(n)\eta (n-\sigma )B^{\alpha }(n-\sigma )}{(1-\delta )^{\alpha }}w(n-\sigma ) \le 0. \end{aligned}$$

By Lemma 2.6, the corresponding difference equation (3.1) also has a positive solution. This contradiction completes the proof of the theorem. $\square$

Theorem 3.2

Let $\alpha >1$ and conditions (2.1) and (2.3) hold. Then Eq. (E) is oscillatory provided that

$$\begin{aligned} \Delta (b^{1/\alpha }(n)\Delta u(n))+\frac{(1-\delta )^{\alpha -1}}{\alpha }B^{\alpha -1}(n-\sigma )p(n)\eta (n-\sigma )u(n-\sigma )=0 \end{aligned}$$

(3.2)

is oscillatory.

Proof

Assume to the contrary that $\{u(n)\}$ is a positive solution of (E), say $u(n) > 0$ for $n \ge n_1 \ge n_0$. It is easy to see that by the Mean-Value Theorem,

$$\begin{aligned} \Delta (b(n)(\Delta u(n))^{\alpha })\ge \alpha (b^{1/\alpha }(n)\Delta u(n))^{\alpha -1}\Delta (b^{1/\alpha }(n)\Delta u(n)), \end{aligned}$$

or

$$\begin{aligned} -p(n)\eta (n-\sigma )u^{\alpha }(n-\sigma )\ge \alpha (b^{1/\alpha }(n)\Delta u(n))^{\alpha -1}\Delta (b^{1/\alpha }(n)\Delta u(n)), \end{aligned}$$

which implies

$$\begin{aligned} \Delta (b^{1/\alpha }(n)\Delta u(n))+\frac{1}{\alpha }(b^{1/\alpha }(n)\Delta u(n))^{1-\alpha }p(n)\eta (n-\sigma )u^{\alpha }(n-\sigma )\le 0.\nonumber \\ \end{aligned}$$

(3.3)

Using (2.4) in (3.3) and taking into account that $b^{1/\alpha }(n)\Delta u(n)$ is decreasing, we have

$$\begin{aligned} \Delta (b^{1/\alpha }(n)\Delta u(n))+\frac{(1-\delta )^{\alpha -1}}{\alpha }B^{\alpha -1}(n-\sigma )p(n)\eta (n-\sigma )u(n-\sigma )\le 0. \end{aligned}$$

But by Lemma 2.6, the corresponding Eq. (3.2) has a positive solution, and so this contradiction completes the proof. $\square$

Before stating our next theorem, first note that since B(n) is increasing, there exists a constant $\lambda \ge 1$ such that

$$\begin{aligned} \frac{B(n)}{B(n-\sigma )} \ge \lambda . \end{aligned}$$

(3.4)

Theorem 3.3

Let $0<\alpha <1$ and conditions (2.1) and (2.3) hold. Then Eq. (E) is oscillatory provided the equation

$$\begin{aligned} \Delta (b^{1/\alpha }(n)\Delta u(n))+\frac{\delta ^{\frac{1-\alpha }{\alpha }}\lambda ^{(1-\alpha )}}{\alpha (1-\delta _1)^{\frac{1-\alpha }{\alpha }}}B^{\alpha -1}(n+1)p(n)\eta (n-\sigma )u(n-\sigma )=0\nonumber \\ \end{aligned}$$

(3.5)

is oscillatory.

Proof

Let $\{u(n)\}$ be a nonoscillatory solution of (E) with $u(n) > 0$ for $n \ge n_1 \ge n_0$. From (2.9) and (E), we have

$$\begin{aligned} b^{1/\alpha }(n)\Delta u(n)=\left[ \sum _{s=n}^{\infty }p(s)u^{\beta }(s-\sigma )\right] ^{\frac{1}{\alpha }}. \end{aligned}$$

Hence,

$$\begin{aligned} \Delta (b^{1/\alpha }(n)\Delta u(n))=\Delta \left( \left[ \sum _{s=n}^{\infty }p(s)u^{\beta }(s-\sigma )\right] ^{\frac{1}{\alpha }}\right) . \end{aligned}$$

By the Mean-Value Theorem,

$$\begin{aligned}&\Delta (b^{1/\alpha }(n)\Delta u(n))+\frac{1}{\alpha }\left[ \sum _{s=n+1}^{\infty }p(s)\eta (s-\sigma )u^{\alpha }(s-\sigma )\right] ^{\frac{1}{\alpha }-1}\\&\quad p(n)\eta (n-\sigma )u^{\alpha }(n-\sigma )\le 0. \end{aligned}$$

Since $\left\{ \frac{u(n)}{B^{\delta _1}(n)}\right\}$ is increasing,

$$\begin{aligned}&\Delta (b^{1/\alpha }(n)\Delta u(n)) +\frac{1}{\alpha }\frac{u^{1-\alpha }(n-\sigma )}{B^{\delta _1(1-\alpha )}(n-\sigma )} \left[ \sum _{s=n+1}^{\infty }p(s)\eta (s-\sigma )B^{\alpha \delta _1}(s-\sigma )\right] ^{\frac{1-\alpha }{\alpha }}\\&\quad p(n)\eta (n-\sigma )u^{\alpha }(n-\sigma )\le 0. \end{aligned}$$

Therefore, $\{u(n)\}$ satisfies the linear difference inequality

$$\begin{aligned}&\Delta (b^{1/\alpha }(n)\Delta u(n))+\frac{1}{\alpha }\frac{p(n)\eta (n-\sigma )}{B^{\delta _1(1-\alpha )}(n-\sigma )}\nonumber \\&\quad \left[ \sum _{s=n+1}^{\infty }p(s)\eta (s-\sigma )B^{\alpha \delta _1}(s-\sigma )\right] ^{\frac{1-\alpha }{\alpha }}u(n-\sigma )\le 0. \end{aligned}$$

(3.6)

Since $\delta _1 <1$, from (3.4) we obtain

$$\begin{aligned} B^{\alpha (\delta _1-1)}(n-\sigma ) \ge \lambda ^{\alpha (1-\delta _1)} B^{\alpha (\delta _1-1)}(n) \end{aligned}$$

Using this and (2.3), we have

$$\begin{aligned} \sum _{s=n+1}^{\infty } p(s)\eta (s-\sigma )B^{\alpha \delta _1}(s-\sigma )&\ge \alpha \delta \sum _{s=n+1}^{\infty }\frac{B^{\alpha (\delta _1-1)}(s-\sigma )}{B(s)b^{1/\alpha }(s)} \\&\ge \alpha \delta \lambda ^{\alpha (1-\delta _1)} \sum _{s=n+1}^{\infty }\frac{B^{\alpha (\delta _1-1) -1}(s)}{b^{1/\alpha }(s)} \\&= \alpha \delta \lambda ^{\alpha (1-\delta _1)} \sum _{s=n+1}^{\infty } B^{\alpha (\delta _1-1) -1}(s) \Delta B(s)\\&\ge \alpha \delta \lambda ^{\alpha (1-\delta _1)} \sum _{s=n+1}^{\infty } \int _{B(s)}^{B(s+1)} \frac{du}{u^{\alpha (1-\delta _1)+1}}\\&=\frac{\delta \lambda ^{\alpha (1-\delta _1)}}{(1-\delta _1)} B^{\alpha (\delta _1-1)}(n+1). \end{aligned}$$

Substituting this into (3.6) gives

$$\begin{aligned}&\Delta (b^{1/\alpha }(n)\Delta u(n))+\frac{\delta ^{\frac{1-\alpha }{\alpha }}\lambda ^{(1-\delta _1)(1-\alpha )}}{\alpha (1-\delta _1)^{\frac{1-\alpha }{\alpha }}}\\&\frac{B^{(\delta _1-1)(1-\alpha )}(n+1)}{B^{\delta _1(1-\alpha )}(n-\sigma )}p(n)\eta (n-\sigma )u(n-\sigma ) \le 0 \end{aligned}$$

which in view of (3.4) yields that $\{u(n)\}$ is a positive solution of the difference inequality

$$\begin{aligned} \Delta (b^{1/\alpha }(n)\Delta u(n))+\frac{\delta ^{\frac{1-\alpha }{\alpha }}\lambda ^{1-\alpha }}{\alpha (1-\delta _1)^{\frac{1-\alpha }{\alpha }}} B^{(\alpha -1)}(n+1)p(n)\eta (n-\sigma )u(n-\sigma )\le 0. \end{aligned}$$

By Lemma 2.6, the corresponding difference equation (3.5) has also a positive solution, so the proof is complete. $\square$

Remark 3.4

The comparison results presented in Theorems 3.1–3.3 reduce the examination of oscillatory properties of (E) to those of linear equations.

4 Oscillation Criteria

In this section, we apply the results from the previous section to establish new oscillation criteria for Eq. (E).

Theorem 4.1

Let (2.1) and (2.3) hold. If

$$\begin{aligned} \lim _{n\rightarrow \infty }\inf \sum _{s=n-\sigma }^{n-1}p(s)\eta (s-\sigma )B^{\alpha }(s-\sigma )>(1-\delta )^{\alpha }\left( \frac{\sigma }{\sigma +1}\right) ^{\sigma +1}, \end{aligned}$$

(4.1)

then (E) is oscillatory.

Proof

In view of (4.1) and Theorem 7.6.1 of [10], it is easy to see that Eq. (3.1) is oscillatory. Therefore, by Theorem 3.1, Eq. (E) is oscillatory. $\square$

Theorem 4.2

Let $\alpha >1$ and conditions (2.1) and (2.3) hold. If

$$\begin{aligned}&\limsup _{n\rightarrow \infty }\left\{ \frac{1}{B(n-\sigma )}\sum _{s=n_1}^{n-\sigma -1}p(s) \eta (s-\sigma )B(s)B^{\alpha }(s-\sigma )\right. \nonumber \\&\qquad \left. + \sum _{s=n-\sigma }^{n-1}p(s) \eta (s-\sigma )B^{\alpha }(s-\sigma )\right. \nonumber \\&\qquad \left. +B^{1-\delta _1}(n-\sigma )\sum _{s=n}^{\infty }p(s)\eta (s-\sigma )B^{\alpha +\delta _1-1}(s-\sigma )\right\} >\frac{1}{k} \end{aligned}$$

(4.2)

for some $n_1 \ge n_0 + \sigma$, where $k=\frac{(1-\delta )^{\alpha -1}}{\alpha }$, then equation (E) is oscillatory.

Proof

Assume that (E) is not oscillatory. By Theorem 3.2, Eq. (3.2) is also nonoscillatory and we may assume that it possess an eventually positive solution $\{u(n)\}$ with $u(n) > 0$ for $n \ge n_1 \ge n_0 + \sigma$ such that (4.2) holds. Summing (3.2) yields

$$\begin{aligned} \Delta u(n)\ge \frac{k}{b^{1/\alpha }(n)}\sum _{s=n}^{\infty }p(s)\eta (s-\sigma )B^{\alpha -1}(s-\sigma )u(s-\sigma ). \end{aligned}$$

Summing once more gives

$$\begin{aligned} u(n)\ge & {} k\sum _{s=n_1}^{n-1}\frac{1}{b^{1/\alpha }(s)}\sum _{t=s}^{\infty }p(t)\eta (t-\sigma )B^{\alpha -1}(t-\sigma )u(t-\sigma )\\= & {} k\sum _{s=n_1}^{n-1}\frac{1}{b^{1/\alpha }(s)}\sum _{t=s}^{n-1}p(t)\eta (t-\sigma )B^{\alpha -1}(t-\sigma )u(t-\sigma )\\&+ k\sum _{s=n_1}^{n-1}\frac{1}{b^{1/\alpha }(s)}\sum _{t=n}^{\infty }p(t)\eta (t-\sigma )B^{\alpha -1}(t-\sigma )u(t-\sigma ). \end{aligned}$$

Using summation by parts,

$$\begin{aligned} u(n)&\ge k\sum _{s=n_1}^{n-1}p(s)\eta (s-\sigma )B(s+1)B^{\alpha -1}(s-\sigma )u(s-\sigma )\\&\quad +k B(n)\sum _{t=n}^{\infty }p(t)\eta (t-\sigma )B^{\alpha -1}(t-\sigma )u(t-\sigma ). \end{aligned}$$

Hence,

$$\begin{aligned} u(n-\sigma )&\ge k\sum _{s=n_1}^{n-\sigma -1}p(s)\eta (s-\sigma )B(s+1)B^{\alpha -1}(s-\sigma )u(s-\sigma )\\&\quad + k B(n-\sigma )\sum _{s=n-\sigma }^{n-1}p(s)\eta (s-\sigma )B^{\alpha -1}(s-\sigma )u(s-\sigma )\\&\quad + k B(n-\sigma )\sum _{s=n}^{\infty }p(s)\eta (s-\sigma )B^{\alpha -1}(s-\sigma )u(s-\sigma ). \end{aligned}$$

In view of the fact that $\frac{u(n)}{B(n)}$ is decreasing and $\frac{u(n)}{B^{\delta _1}(n)}$ is increasing (see Lemmas 2.1 and 2.4), the previous inequality gives

$$\begin{aligned} u(n-\sigma )&\ge k\frac{u(n-\sigma )}{B(n-\sigma )} \sum _{s=n_1}^{n-\sigma -1}p(s)\eta (s-\sigma )B(s+1)B^{\alpha }(s-\sigma )\\&\quad + k u(n-\sigma ) \sum _{s=n-\sigma }^{n-1}p(s)\eta (s-\sigma )B^{\alpha }(s-\sigma )\\&\quad + k \frac{B(n-\sigma )u(n-\sigma )}{B^{\delta _1}(n-\sigma )} \sum _{s=n}^{\infty }p(s)\eta (s-\sigma )B^{\alpha +\delta _1 -1}(s-\sigma ). \end{aligned}$$

Simplifying, we obtain

$$\begin{aligned}&\left\{ \frac{1}{B(n-\sigma )} \sum _{s=n_1}^{n-\sigma -1}p(s)\eta (s-\sigma )B(s+1)B^{\alpha }(s-\sigma )+ \sum _{s=n-\sigma }^{n-1}p(s)\eta (s-\sigma )B^{\alpha }(s-\sigma )\right. \\&\quad \left. + B^{1-\delta _1}(n-\sigma )\sum _{s=n}^{\infty }p(s)\eta (s-\sigma )B^{\alpha +\delta _1-1}(s-\sigma )\right\} \le \frac{1}{k}. \end{aligned}$$

This is a contradiction and proves the theorem. $\square$

For our next and final result we set

$$\begin{aligned} L=\frac{\delta ^{\frac{1-\alpha }{\alpha }}\lambda ^{1-\alpha }}{\alpha (1-\delta _1)^{\frac{1-\alpha }{\alpha }}}. \end{aligned}$$

Theorem 4.3

Let $0<\alpha <1$ and conditions (2.1) and (2.4) hold. If

$$\begin{aligned}&\limsup _{n\rightarrow \infty }\left\{ \frac{1}{B(n-\sigma )}\sum _{s=n_1}^{n-\sigma -1}B^{\sigma }(s+1)B(s-\sigma )p(s)\eta (s-\sigma )\right. \nonumber \\&\quad \left. + \sum _{s=n-\sigma }^{n-1}B^{\alpha -1}(s+1)B(s-\sigma )p(s)\eta (s-\sigma )\right. \nonumber \\&\quad \left. +B^{1-\delta _1}(n-\sigma )\sum _{s=n}^{\infty }B^{\alpha -1}(s+1)B^{\delta _1}(s-\sigma )p(s)\eta (s-\sigma )\right\} >\frac{1}{L},\nonumber \\ \end{aligned}$$

(4.3)

then (E) is oscillatory.

Proof

Assume that equation (E) is not oscillatory. By Theorem 3.3, Eq. (3.5) is also nonoscillatory. Without loss of generality, we may assume that it possesses an eventually positive solution $\{u(n)\}$ for $n \ge n_1 \ge n_0$. Summing (3.5) gives

$$\begin{aligned} \Delta u(n)\ge \frac{L}{b^{1/\alpha }(n)}\sum _{s=n}^{\infty }B^{\alpha -1}(s+1)p(s)\eta (s-\sigma )u(s-\sigma ). \end{aligned}$$

Then,

$$\begin{aligned} u(n)&\ge L\sum _{s=n_1}^{n-1}\frac{1}{b^{1/\alpha }(s)}\sum _{t=s}^{\infty }B^{\alpha -1}(t+1)p(t)\eta (t-\sigma )u(t-\sigma )\\&= L\sum _{s=n_1}^{n-1}\frac{1}{b^{1/\alpha }(s)}\sum _{t=s}^{n-1}B^{\alpha -1}(t+1)p(t)\eta (t-\sigma )u(t-\sigma )\\&\quad + L\sum _{s=n_1}^{n-1}\frac{1}{b^{1/\alpha }(s)}\sum _{t=n}^{\infty }B^{\alpha -1}(t-\sigma )p(t)\eta (t-\sigma )u(t-\sigma ). \end{aligned}$$

Thus,

$$\begin{aligned} u(n)&\ge L\sum _{s=n_1}^{n-1}B^{\alpha }(s+1)p(s)\eta (s-\sigma )u(s-\sigma )\\&\quad + L B(n)\sum _{s=n}^{\infty }B^{\alpha -1}(s+1)p(s)\eta (s-\sigma )u(s-\sigma ), \end{aligned}$$

and so

$$\begin{aligned} u(n-\sigma )&\ge L \sum _{s=n_1}^{n-\sigma -1}B^{\alpha }(s+1)p(s)\eta (s-\sigma )u(s-\sigma )\\&\quad + L B(n-\sigma )\sum _{s=n-\sigma }^{n-1}B^{\alpha -1}(s+1)p(s)\eta (s-\sigma )u(s-\sigma )\\&\quad + L B(n-\sigma )\sum _{s=n}^{\infty }B^{\alpha -1}(s+1)p(s)\eta (s-\sigma )u(s-\sigma ). \end{aligned}$$

Since $\frac{u(n)}{B(n)}$ is decreasing and $\frac{u(n)}{B^{\delta _1}(n)}$ is increasing, the last inequality implies that

$$\begin{aligned} u(n-\sigma )&\ge L\frac{u(n-\sigma )}{B(n-\sigma )} \sum _{s=n_1}^{n-\sigma -1}B^{\alpha }(s+1)B(s-\sigma )p(s)\eta (s-\sigma )\\&\quad + L u(n-\sigma ) \sum _{s=n-\sigma }^{n-1}B^{\alpha -1}(s+1)p(s)\eta (s-\sigma )B(s-\sigma )\\&\quad + L \frac{B(n-\sigma )u(n-\sigma )}{B^{\delta _1}(n-\sigma )} \sum _{s=n}^{\infty }B^{\alpha -1}(s+1)B^{\delta _1}(s-\sigma )p(s)\eta (s-\sigma ). \end{aligned}$$

Hence,

$$\begin{aligned}&\left\{ \frac{1}{B(n-\sigma )} \sum _{s=n_1}^{n-\sigma -1}B^{\alpha }(s+1)B(s-\sigma )p(s)\eta (s-\sigma )\right. \\&\qquad \left. + \sum _{s=n-\sigma }^{n-1}B^{\alpha -1}(s+1)B(s-\sigma )p(s)\eta (s-\sigma )\right. \\&\quad \quad \left. + B^{1-\delta _1}(n-\sigma ) \sum _{s=n}^{\infty }B^{\alpha -1}(s+1)B^{\delta _1}(s-\sigma )p(s)\eta (s-\sigma )\right\} \le \frac{1}{L}. \end{aligned}$$

This contradicts (4.3), and completes the proof of the theorem. $\square$

5 Examples

In this section, we illustrate the oscillation criteria obtained in the previous section with examples of Euler type difference equations.

Example 5.1

Consider the second order delay difference equation

$$\begin{aligned} \Delta ((\Delta u(n))^{3})+\frac{a}{n^4}u^{3}(n-2)=0, \ n\ge 3, \end{aligned}$$

(5.1)

with $a>0$. Here we have $b(n)=1$, $\sigma =2$, and $\alpha =\beta =3$. A simple calculation shows that $B(n)=n-3$, and by taking $\lambda =1$ and $\delta =\frac{1}{8}$, we have $\delta _1=\frac{1}{2}$, $k=\frac{49}{192}$, and $\eta (n-\sigma )=1$. Condition (2.1) becomes

$$\begin{aligned} \sum _{n=3}^{\infty }\frac{a}{n^4}(n-5)^3=\infty \end{aligned}$$

and (2.3) is satisfied if $a\ge 32$. Condition (4.2) becomes

$$\begin{aligned}&\limsup _{n\rightarrow \infty }\left\{ \frac{1}{(n-3)}\sum _{s=6}^{n-3} \frac{a}{s^4}(s-1)(s-3)^3 + \sum _{s=n-2}^{n-1}a\frac{(s-3)^3}{s^4} \right. \\&\quad \left. + (n-3)^{\frac{1}{2}}\sum _{s=n}^{\infty } \frac{a}{s^4} (s-3)^{\frac{5}{2}} \right\} =3a > \frac{192}{49}, \end{aligned}$$

and hence (4.2) is satisfied for $n_1 = 6$ if $a>\frac{64}{49}$. Therefore, by Theorem 4.2, Eq. (5.1) is oscillatory if $a\ge 32$.

Example 5.2

Consider the second order delay difference equation

$$\begin{aligned} \Delta (n^{\frac{2}{9}}(\Delta u(n))^{\frac{1}{3}})+\frac{a}{n^{\frac{10}{9}}}u^{\frac{1}{3}}(n-1)=0, \ n\ge 1, \end{aligned}$$

(5.2)

where $a>0$. We have $b(n)=n^{2/9}$, $\sigma =1$, and $\alpha =\beta =1/3$. Then, with $\lambda =1$ and $\delta =\frac{1}{2}$, we have $\delta _1=\frac{1}{8}$, $\eta (n-\sigma )=1$, $L=\frac{48}{49}$, and $B(n)\approx 3n^{1/3}$. Condition (2.1) becomes

$$\begin{aligned} \sum _{n=1}^{\infty }\frac{27a}{n^{\frac{1}{9}}}=\infty , \end{aligned}$$

and (2.3) is satisfied for $a\ge \frac{1}{18}(2/27)^{\frac{1}{9}}$. Condition (4.3) reduces to

$$\begin{aligned}&\limsup _{n\rightarrow \infty } \left\{ \frac{1}{3(n-1)^{\frac{1}{3}}} \sum _{s=n_1}^{n-2} 9 (s+1)^{\frac{1}{3}} (s-1)^{\frac{1}{3}} \frac{a}{s^{\frac{10}{9}}} \right. \\&\quad + \sum _{s=n-1}^{n-1}[3(s+1)^{\frac{1}{3}}]^{-\frac{2}{3}} 3(s-1)^{\frac{1}{3}} \frac{a}{s^{\frac{10}{9}}}\\&\quad \left. + [3(n-1)^{\frac{1}{3}}]^{\frac{7}{8}} \sum _{s=n}^{\infty } [3(s+1)^{\frac{1}{3}}]^{-\frac{2}{3}} [3(s-1)^{\frac{1}{3}}]^{\frac{1}{8}} \frac{a}{s^{\frac{10}{9}}}\right\} = \frac{7a}{72} > \frac{49}{48}. \end{aligned}$$

Therefore, by Theorem 4.3, Eq. (5.2) is oscillatory if $a> \frac{21}{2}$.