Oscillatory behavior of solutions of certain fractional difference equations

Abstract

In this paper, we consider the oscillation behavior of solutions of the following fractional difference equation:

$$ \Delta \bigl( c ( t ) \Delta \bigl( a ( t ) \Delta \bigl( r ( t ) \Delta^{\alpha }x ( t ) \bigr) \bigr) \bigr) +q ( t ) G ( t ) =0, $$

where $t\in \mathbf{N}_{t_{0}+1-\alpha }$, $G ( t ) = \sum_{s=t_{0}}^{t-1+\alpha } ( t-s-1 ) ^{-\alpha }x ( s )$, and $\Delta^{\alpha }$ denotes a Riemann–Liouville fractional difference operator of order $0<\alpha \leq 1$. By using the generalized Riccati transformation technique, we obtain some oscillation criteria. Finally we give an example.

1 Introduction and preliminaries

Fractional differential (or difference) equations are a more general form of differential equations with integer order. And there is an increasing interest in the study of them due to some important contributions [1, 2].

Many authors have been focused on various equations like ordinary and partial differential equations [3,4,5,6], difference equations [7,8,9], dynamic equations on time scales [10,11,12,13,14], and fractional differential (difference) equations [15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31] obtaining some oscillation criteria. Recently, oscillation studies have become a very hot topic. That is why, we consider the following fractional difference equation:

$$ \Delta \bigl( c ( t ) \Delta \bigl( a ( t ) \Delta \bigl( r ( t ) \Delta^{\alpha }x ( t ) \bigr) \bigr) \bigr) +q ( t ) G ( t ) =0, $$

(1)

where $t\in \mathbf{N} _{t_{0}+1-\alpha }$, $G ( t ) = \sum_{s=t_{0}}^{t-1+\alpha } ( t-s-1 ) ^{ ( -\alpha ) }x ( s )$, $c ( t )$, $a ( t )$, $r ( t )$, and $q ( t )$ are positive sequences, and $\Delta^{\alpha }$ denotes the Riemann–Liouville fractional difference operator of order $0<\alpha \leq 1$.

By a solution of Eq. (1), we mean a real-valued sequence $x ( t )$ satisfying Eq. (1) for $t\in \mathbf{N} _{t_{0}}$. A solution $x ( t )$ of Eq. (1) is called oscillatory if it is neither eventually positive nor eventually negative, otherwise it is called non-oscillatory. Equation (1) is called oscillatory if all its solutions are oscillatory.

Definition 1

([32])

Let $v>0$. The vth fractional sum f is defined by

$$ \Delta^{-v}f ( t ) =\frac{1}{\varGamma ( v ) }\sum_{s=a} ^{t-v} ( t-s-1 ) ^{v-1}f ( s ) , $$

(2)

where f is defined for $s\equiv a \mathbf{mod} ( 1 )$, $\Delta^{-v}f$ is defined for $t\equiv ( a+v ) \mathbf{mod} ( 1 )$, and $t^{ ( v ) }=\frac{\varGamma ( t+1 ) }{\varGamma ( t-v+1 ) }$. The fractional sum $\Delta^{-v}f$ maps functions defined on $\mathbf{N} _{a}$ to functions defined on $\mathbf{N} _{a+v}$, where $\mathbf{N} _{t}= \{ t,t+1,t+2,\ldots \}$.

Definition 2

([32])

Let $v>0$ and $m-1<\mu <m$, where m denotes a positive integer, $m= \lceil \mu \rceil$. Set $v=m-\mu$. The μth fractional difference is defined as

$$ \Delta^{\mu }f ( t ) =\Delta^{m-v}f ( t ) =\Delta^{m} \Delta^{-v}f ( t ) , $$

(3)

where $\lceil \mu \rceil$ is the ceiling function of μ.

Lemma 1

([33])

Assume that A and B are nonnegative real numbers. Then

$$ \lambda AB^{\lambda -1}-A^{\lambda }\leq ( \lambda -1 ) B ^{\lambda } $$

(4)

for all $\lambda >1$.

2 Main results

Throughout this paper, we denote

$$ \phi ( t ) =\sum_{s=t_{1}}^{t-1} \frac{1}{c ( s ) };\quad\quad \vartheta ( t ) =\sum_{s=t_{2}}^{t-1} \frac{\phi ( s ) }{a ( s ) };\quad\quad \delta ( t ) =\sum_{s=t_{3}} ^{t-1}\frac{\vartheta ( s ) }{r ( s ) }. $$

For simplification, we consider

$$ \Delta \gamma_{+} ( s ) =\max \bigl\{ 0,\Delta \gamma ( s ) \bigr\} $$

and

$$ \Delta \beta_{+} ( s ) =\max \bigl\{ 0,\Delta \beta ( s ) \bigr\} . $$

Lemma 2

([28])

Let $x ( t )$ be a solution of Eq. (1), and let

$$ G ( t ) =\sum_{s=t_{0}}^{t-1+\alpha } ( t-s-1 ) ^{ ( -\alpha ) }x ( s ) , $$

(5)

then

$$ \Delta \bigl( G ( t ) \bigr) =\varGamma ( 1-\alpha ) \Delta^{\alpha }x ( t ) . $$

(6)

Lemma 3

Assume that $x ( t )$ is an eventually positive solution of Eq. (1). If

$$ \sum_{s=t_{0}}^{\infty }\frac{1}{c ( s ) }=\sum _{s=t_{0}} ^{\infty }\frac{1}{a ( s ) }=\sum _{s=t_{0}}^{\infty }\frac{1}{r ( s ) }=\infty , $$

(7)

then we have two possible cases for $t\in [ t_{1},\infty )$, $t_{1}>t_{0}$ is sufficiently large:

• Case 1 $\Delta^{\alpha }x ( t ) >0$, $\Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) >0$, $\Delta ( a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) ) >0$ or

• Case 2 $\Delta^{\alpha }x ( t ) >0$, $\Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) <0$, $\Delta ( a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) ) >0$.

Proof

From the hypothesis, there exists $t_{1}$ such that $x ( t ) >0$ on $[ t_{1},\infty )$, so that $G ( t ) >0$ on $[ t_{1},\infty )$, and from Eq. (1), we have

$$ \Delta \bigl( c ( t ) \Delta \bigl( a ( t ) \Delta \bigl( r ( t ) \Delta^{\alpha }x ( t ) \bigr) \bigr) \bigr) =-q ( t ) G ( t ) < 0. $$

(8)

Then $c ( t ) \Delta ( a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) )$ is an eventually non-increasing sequence on $[ t_{1},\infty )$. We know that $\Delta^{\alpha }x ( t )$, $\Delta ( r ( t ) \Delta^{\alpha }x ( t ) )$, and $\Delta ( a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) )$ are eventually of one sign. For $t_{2}>t_{1}$ is sufficiently large, we claim that $\Delta ( a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) ) >0$ on $[ t_{2}, \infty )$. Otherwise, assume that there exists sufficiently large $t_{3}>t_{2}$ such that $\Delta ( a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) ) <0$ on $[ t_{3},\infty )$. For $[ t_{3},\infty )$ and there exists a constant $l_{1}>0$, we have

$$ \Delta \bigl( a ( t ) \Delta \bigl( r ( t ) \Delta^{\alpha }x ( t ) \bigr) \bigr) \leq -\frac{l_{1}}{c ( t ) }< 0. $$

Hence, there exist a constant $l_{2}>0$ and sufficiently large $t_{4}>t_{3}$ such that

$$ \Delta \bigl( r ( t ) \Delta^{\alpha }x ( t ) \bigr) \leq - \frac{l_{2}}{a ( t ) }< 0. $$

(9)

Then there exist a constant $l_{3}>0$ and sufficiently large $t_{5}>t_{4}$ such that

$$ \Delta^{\alpha }x ( t ) \leq -\frac{l_{3}}{r ( t ) }, $$

that is,

$$ \Delta G ( t ) \leq -\frac{\varGamma ( 1-\alpha ) l _{3}}{r ( t ) }< 0. $$

By (7), we obtain $\lim_{t\rightarrow \infty }G ( t ) =-\infty$. This is a contradiction. If $\Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) <0$, then $\Delta^{\alpha }x ( t ) >0$ due to $\sum_{s=t_{0}}^{\infty }\frac{1}{r ( s ) }=\infty$. If $\Delta ( r ( t ) \Delta^{ \alpha }x ( t ) ) >0$, then $\Delta^{\alpha }x ( t ) >0$ due to $\Delta ( a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) ) >0$. So, the proof is complete. □

Lemma 4

Assume that $x ( t )$ is an eventually positive solution of Eq. (1), which satisfies Case 1 of Lemma 3. Then

$$ a ( t ) \Delta \bigl( r ( t ) \Delta^{\alpha }x ( t ) \bigr) \geq c ( t ) \Delta \bigl( a ( t ) \Delta \bigl( r ( t ) \Delta^{\alpha }x ( t ) \bigr) \bigr) \sum_{s=t_{0}}^{t-1}\frac{1}{c ( s ) }. $$

If there exists a positive sequence ϕ such that, for $t\in [ t_{1},\infty )$,

$$ \frac{\phi ( t ) }{c ( t ) \sum_{s=t_{0}}^{t-1}\frac{1}{c ( s ) }}-\Delta \phi ( t ) \leq 0, $$

where $t_{1}$ is sufficiently large, then $a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) / \phi ( t )$ is a non-increasing sequence on $[ t_{1}, \infty )$ and

$$ r ( t ) \Delta^{\alpha }x ( t ) \geq \Delta \bigl( r ( t ) \Delta^{\alpha }x ( t ) \bigr) \frac{a ( t ) }{\phi ( t ) }\sum _{s=t_{1}}^{t-1}\frac{\phi ( s ) }{a ( s ) }. $$

Furthermore, if there exists a positive sequence ϑ and $t_{2}>t_{1}$ is sufficiently large such that, for $t\in [ t_{2}, \infty )$,

$$ \frac{\vartheta ( t ) }{\frac{a ( t ) }{\phi ( t ) }\sum_{s=t_{2}}^{t-1}\frac{\phi ( s ) }{a ( s ) }}-\Delta \vartheta ( t ) \leq 0, $$

then $r ( t ) \Delta^{\alpha }x ( t ) /\vartheta ( t )$ is a non-increasing sequence on $[ t_{2}, \infty )$ and

$$ G ( t ) \geq \Delta G ( t ) \frac{r ( t ) }{ \vartheta ( t ) }\sum_{s=t_{2}}^{t-1} \frac{\vartheta ( s ) }{r ( s ) }. $$

Suppose also that there exists a positive sequence δ and $t_{3}>t_{2}$ is sufficiently large such that, for $t\in [ t_{3}, \infty )$,

$$ \frac{\delta ( t ) }{\frac{r ( t ) }{\vartheta ( t ) }\sum_{s=t_{2}}^{t-1}\frac{\vartheta ( s ) }{r ( s ) }}-\Delta \delta ( t ) \leq 0. $$

Then $G ( t ) /\delta ( t )$ is a non-increasing sequence on $[ t_{3},\infty )$.

Proof

Assume that x is an eventually positive solution of Eq. (1). Then we have that $\Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) >0$ and $\Delta ( c ( t ) \Delta ( a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) ) ) <0$ on $[ t_{0},\infty )$. So,

$$\begin{aligned} a ( t ) \Delta \bigl( r ( t ) \Delta^{\alpha }x ( t ) \bigr) =&a ( t_{0} ) \Delta \bigl( r ( t_{0} ) \Delta^{\alpha }x ( t_{0} ) \bigr) \\ &{}+\sum_{s=t_{0}}^{t-1}\frac{c ( s ) \Delta ( a ( s ) \Delta ( r ( s ) \Delta^{\alpha }x ( s ) ) ) }{c ( s ) } \\ \geq & c ( t ) \Delta \bigl( a ( t ) \Delta \bigl( r ( t ) \Delta^{\alpha }x ( t ) \bigr) \bigr) \sum_{s=t_{0}}^{t-1} \frac{1}{c ( s ) }, \end{aligned}$$

and then

$$\begin{aligned}& \Delta \biggl( \frac{a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) }{\phi ( t ) } \biggr) \\& \quad = \frac{\Delta ( a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) ) \phi ( t ) -a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) \Delta \phi ( t ) }{\phi ( t ) \phi ( t+1 ) } \\& \quad \leq \frac{\Delta ( a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) ) }{\phi ( t ) \phi ( t+1 ) } \biggl( \frac{\phi ( t ) }{c ( t ) \sum_{s=t_{1}}^{t-1}\frac{1}{c ( s ) }}-\Delta \phi ( t ) \biggr) \leq 0. \end{aligned}$$

Hence, $a ( t ) \Delta ( r ( t ) \Delta^{ \alpha }x ( t ) ) /\phi ( t )$ is a non-increasing sequence on $[ t_{1},\infty )$ where $t_{1}>t_{0}$ is sufficiently large. Then we have

$$\begin{aligned} r ( t ) \Delta^{\alpha }x ( t ) =&r ( t_{1} ) \Delta^{\alpha }x ( t_{1} ) +\sum_{s=t_{1}}^{t-1} \frac{a ( s ) \Delta ( r ( s ) \Delta^{\alpha }x ( s ) ) }{\phi ( s ) }\frac{\phi ( s ) }{a ( s ) } \\ \geq &\frac{a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) }{\phi ( t ) } \sum_{s=t_{1}}^{t-1} \frac{\phi ( s ) }{a ( s ) } \end{aligned}$$

and

$$\begin{aligned} \Delta \biggl( \frac{r ( t ) \Delta^{\alpha }x ( t ) }{\vartheta ( t ) } \biggr) =&\frac{\Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) \vartheta ( t ) -r ( t ) \Delta^{\alpha }x ( t ) \Delta \vartheta ( t ) }{\vartheta ( t ) \vartheta ( t+1 ) } \\ \leq &\frac{r ( t ) \Delta^{\alpha }x ( t ) }{ \vartheta ( t ) \vartheta ( t+1 ) } \biggl( \frac{ \vartheta ( t ) }{\frac{a ( t ) }{\phi ( t ) }\sum_{s=t_{1}}^{t-1}\frac{\phi ( s ) }{a ( s ) }}-\Delta \vartheta ( t ) \biggr) \leq 0. \end{aligned}$$

So $r ( t ) \Delta^{\alpha }x ( t ) /\vartheta ( t )$ is a non-increasing sequence on $[ t_{2}, \infty )$ where $t_{2}>t_{1}$ is sufficiently large. Then we have

$$\begin{aligned} G ( t ) =&G ( t_{2} ) +\varGamma ( 1-\alpha ) \sum _{s=t_{2}}^{t-1}\frac{r ( s ) \Delta^{\alpha }x ( s ) }{\vartheta ( s ) }\frac{ \vartheta ( s ) }{r ( s ) } \\ \geq &\frac{r ( t ) \varGamma ( 1-\alpha ) \Delta^{\alpha }x ( t ) }{\vartheta ( t ) } \sum_{s=t_{2}}^{t-1} \frac{\vartheta ( s ) }{r ( s ) } \\ =&\Delta G ( t ) \frac{r ( t ) }{\vartheta ( t ) }\sum_{s=t_{2}}^{t-1} \frac{\vartheta ( s ) }{r ( s ) }, \end{aligned}$$

and then

$$\begin{aligned} \Delta \biggl( \frac{G ( t ) }{\delta ( t ) } \biggr) =&\frac{ ( \Delta G ( t ) ) \delta ( t ) -G ( t ) \Delta \delta ( t ) }{\delta ( t ) \delta ( t+1 ) } \\ \leq &\frac{G ( t ) }{\delta ( t ) \delta ( t+1 ) } \biggl( \frac{\delta ( t ) }{\frac{r ( t ) }{\vartheta ( t ) }\sum_{s=t_{2}}^{t-1}\frac{\vartheta ( s ) }{r ( s ) }}-\Delta \delta ( t ) \biggr) \leq 0. \end{aligned}$$

Then $G ( t ) /\delta ( t )$ is a non-increasing sequence on $[ t_{3},\infty )$ where $t_{3}>t_{2}$ is sufficiently large. So the proof is complete. □

Theorem 1

Assume that (7) holds and there exists a positive sequence γ such that, for all sufficiently large t,

$$ \lim_{t\rightarrow \infty }\sup \sum_{s=t_{3}}^{t-1} \Biggl( \frac{ \varGamma ( 1-\alpha ) \gamma ( s ) q ( s ) }{\vartheta ( s ) \phi ( s+1 ) }\sum_{u=t_{2}} ^{s-1} \frac{\vartheta ( u ) }{r ( u ) }\sum_{u=t _{1}}^{s-1} \frac{\phi ( u ) }{a ( u ) }-\frac{c ( s ) ( \Delta \gamma_{+} ( s ) ) ^{2}}{4 \gamma ( s ) } \Biggr) =\infty . $$

(10)

If there exist positive sequences β, λ such that, for all sufficiently large t,

$$ \frac{\lambda ( t ) }{r ( t ) \sum_{s=t_{1}}^{t-1}\frac{1}{r ( s ) }}-\Delta \lambda ( t ) \leq 0 $$

(11)

and

$$ \lim_{t\rightarrow \infty }\sup \sum_{\zeta =t_{2}}^{t-1} \Biggl( \frac{ \beta ( \zeta ) \lambda ( \zeta ) }{\lambda ( \zeta +1 ) a ( \zeta ) }\sum_{s=\zeta }^{\infty } \Biggl( \frac{1}{c ( s ) }\sum_{v=s}^{\infty }q ( v ) \Biggr) -\frac{r ( \zeta ) ( \Delta \beta_{+} ( \zeta ) ) ^{2}}{4\varGamma ( 1-\alpha ) \beta ( \zeta ) } \Biggr) =\infty . $$

(12)

Then every solution of Eq. (1) is oscillatory.

Proof

Suppose to the contrary that $x(t)$ is a non-oscillatory solution of Eq. (1). Then, without loss of generality, we may assume that there is a solution $x ( t )$ of Eq. (1) such that $x ( t ) >0$ on $[ t_{0},\infty )$, where $t_{0}$ is sufficiently large. From Lemma 3, $x ( t )$ satisfies Case 1 or Case 2. Firstly, let Case 1 hold. Then we define the following function:

$$ \omega ( t ) =\gamma ( t ) \frac{c ( t ) \Delta ( a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) ) }{a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) }. $$

For $t\in [ t_{0},\infty )$, we have

$$\begin{aligned} \Delta \omega ( t ) =&\Delta \gamma ( t ) \frac{ \omega ( t+1 ) }{\gamma ( t+1 ) }+\gamma ( t ) \Delta \biggl( \frac{c ( t ) \Delta ( a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) ) }{a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) } \biggr) \\ =&\Delta \gamma ( t ) \frac{\omega ( t+1 ) }{ \gamma ( t+1 ) }-\gamma ( t ) \frac{q ( t ) G ( t ) }{a ( t+1 ) \Delta ( r ( t+1 ) \Delta^{\alpha }x ( t+1 ) ) } \\ &{}-\gamma ( t ) \frac{c ( t ) \Delta ( a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) ) \Delta ( a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) ) }{a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) a ( t+1 ) \Delta ( r ( t+1 ) \Delta^{\alpha }x ( t+1 ) ) }. \end{aligned}$$

Since $a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) /\phi ( t )$ is a non-increasing sequence on $[ t_{1},\infty )$, we have

$$ \frac{a ( t+1 ) \Delta ( r ( t+1 ) \Delta^{ \alpha }x ( t+1 ) ) }{\phi ( t+1 ) }\leq \frac{a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) }{\phi ( t ) }. $$

From Lemma 4, we obtain

$$\begin{aligned}& \frac{G ( t ) }{a ( t+1 ) \Delta ( r ( t+1 ) \Delta^{\alpha }x ( t+1 ) ) } \\& \quad = \frac{1}{a ( t+1 ) }\frac{G ( t ) }{\Delta G ( t ) }\frac{\Delta G ( t ) }{\Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) }\frac{\Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) }{\Delta ( r ( t+1 ) \Delta^{\alpha }x ( t+1 ) ) } \\& \quad \geq \frac{1}{a ( t+1 ) } \Biggl( \frac{r ( t ) }{ \vartheta ( t ) }\sum _{s=t_{2}}^{t-1}\frac{\vartheta ( s ) }{r ( s ) } \Biggr) \Biggl( \frac{\varGamma ( 1- \alpha ) }{r ( t ) }\frac{a ( t ) }{\phi ( t ) }\sum_{s=t_{1}}^{t-1} \frac{\phi ( s ) }{a ( s ) } \Biggr) \frac{\phi ( t ) a ( t+1 ) }{\phi ( t+1 ) a ( t ) } \\& \quad = \frac{\varGamma ( 1-\alpha ) }{\vartheta ( t ) \phi ( t+1 ) }\sum_{s=t_{2}}^{t-1} \frac{\vartheta ( s ) }{r ( s ) } \Biggl( \sum_{s=t_{1}}^{t-1} \frac{\phi ( s ) }{a ( s ) } \Biggr) \end{aligned}$$

and

$$\begin{aligned} \Delta \omega ( t ) \leq &\Delta \gamma_{+} ( t ) \frac{\omega ( t+1 ) }{\gamma ( t+1 ) }-\gamma ( t ) q ( t ) \frac{\varGamma ( 1-\alpha ) }{\vartheta ( t ) \phi ( t+1 ) }\sum _{s=t_{2}} ^{t-1}\frac{\vartheta ( s ) }{r ( s ) } \Biggl( \sum _{s=t_{1}}^{t-1}\frac{\phi ( s ) }{a ( s ) } \Biggr) \\ &{}-\frac{\gamma ( t ) }{c ( t ) }\frac{\omega^{2} ( t+1 ) }{\gamma^{2} ( t+1 ) }. \end{aligned}$$

Setting $\lambda =2$, $A= ( \frac{\gamma ( t ) }{c ( t ) } ) ^{1/2}\frac{\omega ( t+1 ) }{\phi ( t+1 ) }$, and $B=\frac{1}{2} ( \frac{c ( t ) }{ \gamma ( t ) } ) ^{1/2}\Delta \gamma_{+} ( t )$ using Lemma 1, we obtain

$$ \Delta \omega ( t ) \leq -\gamma ( t ) q ( t ) \frac{\varGamma ( 1-\alpha ) }{\vartheta ( t ) \phi ( t+1 ) }\sum _{s=t_{2}}^{t-1}\frac{\vartheta ( s ) }{r ( s ) } \Biggl( \sum _{s=t_{1}}^{t-1}\frac{ \phi ( s ) }{a ( s ) } \Biggr) + \frac{c ( t ) }{4\gamma ( t ) } \bigl( \Delta \gamma_{+} ( t ) \bigr) ^{2}. $$

Summing both sides of the above inequality from $t_{3}$ to $t-1$, we get

$$\begin{aligned}& \sum_{s=t_{3}}^{t-1} \Biggl( \frac{\varGamma ( 1-\alpha ) \gamma ( s ) q ( s ) }{\vartheta ( s ) \phi ( s+1 ) } \sum_{u=t_{2}}^{s-1} \frac{\vartheta ( u ) }{r ( u ) } \Biggl( \sum_{u=t_{1}}^{s-1} \frac{\phi ( u ) }{a ( u ) } \Biggr) - \frac{c ( s ) ( \Delta \gamma_{+} ( s ) ) ^{2}}{4\gamma ( s ) } \Biggr) \\& \quad \leq \omega ( t_{3} ) -\omega ( t ) \leq \omega ( t_{3} ) . \end{aligned}$$

This contradicts (10). Now we consider Case 2. Then we define the following function:

$$ \omega_{2} ( t ) =\beta ( t ) \frac{r ( t ) \Delta^{\alpha }x ( t ) }{G ( t ) }. $$

Then

$$\begin{aligned} \Delta \omega_{2} ( t ) =&\Delta \beta ( t ) \frac{ \omega ( t+1 ) }{\beta ( t+1 ) }+ \beta ( t ) \Delta \biggl( \frac{r ( t ) \Delta^{\alpha }x ( t ) }{G ( t ) } \biggr) \\ =&\Delta \beta ( t ) \frac{\omega ( t+1 ) }{ \beta ( t+1 ) }+\beta ( t ) \biggl( \frac{\Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) G ( t ) -r ( t ) \Delta^{\alpha }x ( t ) \Delta G ( t ) }{G ( t ) G ( t+1 ) } \biggr) \\ =&\Delta \beta ( t ) \frac{\omega ( t+1 ) }{ \beta ( t+1 ) }+\beta ( t ) \frac{\Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) }{G ( t+1 ) }-\beta ( t ) \frac{r ( t ) \Delta^{ \alpha }x ( t ) \Delta G ( t ) }{G ( t ) G ( t+1 ) }. \end{aligned}$$

Hence we have

$$\begin{aligned} G ( t ) =&G ( t_{1} ) +\varGamma ( 1-\alpha ) \sum _{s=t_{1}}^{t-1}\frac{r ( s ) \Delta^{\alpha }x ( s ) }{r ( s ) } \\ \geq &\varGamma ( 1-\alpha ) r ( t ) \Delta^{ \alpha }x ( t ) \sum _{s=t_{1}}^{t-1}\frac{1}{r ( s ) }. \end{aligned}$$

That is,

$$ \frac{G ( t ) }{r ( t ) \sum_{s=t_{1}}^{t-1}\frac{1}{r ( s ) }}\geq \varGamma ( 1-\alpha ) \Delta^{\alpha }x ( t ) = \Delta G ( t ) $$

and

$$\begin{aligned} \Delta \biggl( \frac{G ( t ) }{\lambda ( t ) } \biggr) =&\frac{\Delta G ( t ) \lambda ( t ) -G ( t ) \Delta \lambda ( t ) }{\lambda ( t ) \lambda ( t+1 ) } \\ \leq &\frac{G ( t ) }{\lambda ( t ) \lambda ( t+1 ) } \biggl( \frac{\lambda ( t ) }{r ( t ) \sum_{s=t_{1}}^{t-1}\frac{1}{r ( s ) }}-\Delta \lambda ( t ) \biggr) \leq 0. \end{aligned}$$

Thus we have $G ( t ) /\lambda ( t )$ is eventually non-increasing and

$$ \frac{G ( t ) }{G ( t+1 ) }\geq \frac{\lambda ( t ) }{\lambda ( t+1 ) }. $$

(13)

Using the fact that $r ( t ) \Delta^{\alpha }x ( t )$ is strictly decreasing, we have

$$ r ( t ) \Delta^{\alpha }x ( t ) \geq r ( t+1 ) \Delta^{\alpha }x ( t+1 ) $$

and $\Delta G ( t ) >0$, then $G ( t+1 ) >G ( t )$, it follows that

$$\begin{aligned} \Delta \omega_{2} ( t ) \leq &\Delta \beta_{+} ( t ) \frac{\omega ( t+1 ) }{\beta ( t+1 ) }+\beta ( t ) \frac{\Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) }{G ( t+1 ) } \\ &{}-\frac{\varGamma ( 1-\alpha ) \beta ( t ) }{r ( t ) }\frac{\omega_{2}^{2} ( t+1 ) }{\beta^{2} ( t+1 ) }. \end{aligned}$$

From 8, we have

$$\begin{aligned}& c ( u ) \Delta \bigl( a ( u ) \Delta \bigl( r ( u ) \Delta^{\alpha }x ( u ) \bigr) \bigr) -c ( t ) \Delta \bigl( a ( t ) \Delta \bigl( r ( t ) \Delta^{\alpha }x ( t ) \bigr) \bigr) \\ & \quad =-\sum_{s=t}^{u-1}q ( s ) G ( s ) \end{aligned}$$

for $\Delta G ( t ) >0$, and letting $u\rightarrow \infty$, we get

$$ -c ( t ) \Delta \bigl( a ( t ) \Delta \bigl( r ( t ) \Delta^{\alpha }x ( t ) \bigr) \bigr) \leq -G ( t ) \sum_{s=t}^{\infty }q ( s ) $$

or

$$ \Delta \bigl( a ( t ) \Delta \bigl( r ( t ) \Delta^{\alpha }x ( t ) \bigr) \bigr) \geq \frac{G ( t ) }{c ( t ) }\sum_{s=t}^{\infty }q ( s ) . $$

And so

$$ a ( u ) \Delta \bigl( r ( u ) \Delta^{\alpha }x ( u ) \bigr) -a ( t ) \Delta \bigl( r ( t ) \Delta^{\alpha }x ( t ) \bigr) \geq G ( t ) \sum _{s=t} ^{u-1} \Biggl( \frac{1}{c ( s ) }\sum _{v=s}^{\infty }q ( v ) \Biggr) . $$

Letting $u\rightarrow \infty$, we have

$$ \Delta \bigl( r ( t ) \Delta^{\alpha }x ( t ) \bigr) \leq -G ( t ) \frac{1}{a ( t ) }\sum_{s=t}^{\infty } \Biggl( \frac{1}{c ( s ) }\sum_{v=s}^{\infty }q ( v ) \Biggr) $$

due to $\lim_{u\rightarrow \infty }a ( u ) \Delta ( r ( u ) \Delta^{\alpha }x ( u ) ) =k<0$. Then, by (13), we obtain

$$\begin{aligned} \frac{\Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) }{G ( t+1 ) } \leq& -\frac{G ( t ) }{G ( t+1 ) }\frac{1}{a ( t ) }\sum _{s=t}^{\infty } \Biggl( \frac{1}{c ( s ) }\sum _{v=s}^{\infty }q ( v ) \Biggr) \\ \leq &-\frac{\lambda ( t ) }{\lambda ( t+1 ) }\frac{1}{a ( t ) }\sum _{s=t}^{\infty } \Biggl( \frac{1}{c ( s ) }\sum _{v=s}^{\infty }q ( v ) \Biggr) . \end{aligned}$$

So,

$$\begin{aligned} \begin{aligned} \Delta \omega_{2} ( t ) &\leq \Delta \beta_{+} ( t ) \frac{\omega_{2} ( t+1 ) }{\beta ( t+1 ) }-\beta ( t ) \frac{\lambda ( t ) }{\lambda ( t+1 ) } \frac{1}{a ( t ) }\sum_{s=t}^{\infty } \Biggl( \frac{1}{c ( s ) }\sum_{v=s}^{\infty }q ( v ) \Biggr) \\ &{}-\frac{\varGamma ( 1-\alpha ) \beta ( t ) }{r ( t ) }\frac{\omega_{2}^{2} ( t+1 ) }{\beta^{2} ( t+1 ) }.\end{aligned} \end{aligned}$$

Setting $\lambda =2$, $A= ( \frac{\varGamma ( 1-\alpha ) \beta ( t ) }{r ( t ) } ) ^{1/2}\frac{\omega_{2} ( t+1 ) }{\beta ( t+1 ) }$, and $B=\frac{1}{2} ( \frac{r ( t ) }{\varGamma ( 1-\alpha ) \beta ( t ) } ) ^{1/2}\Delta \beta_{+} ( t )$ using Lemma 1, we obtain

$$ \Delta \omega_{2} ( t ) \leq -\beta ( t ) \frac{ \lambda ( t ) }{\lambda ( t+1 ) } \frac{1}{a ( t ) }\sum_{s=t}^{\infty } \Biggl( \frac{1}{c ( s ) }\sum_{v=s} ^{\infty }q ( v ) \Biggr) +\frac{r ( t ) ( \Delta \beta_{+} ( t ) ) ^{2}}{4\varGamma ( 1-\alpha ) \beta ( t ) }. $$

Summing both sides of the above inequality from $t_{2}$ to $t-1$, we have

$$\begin{aligned}& \sum_{\zeta =t_{2}}^{t-1} \Biggl( \beta ( \zeta ) \frac{ \lambda ( \zeta ) }{\lambda ( \zeta +1 ) }\frac{1}{a ( \zeta ) }\sum_{s=\zeta }^{\infty } \Biggl( \frac{1}{c ( s ) }\sum_{v=s}^{\infty }q ( v ) \Biggr) -\frac{r ( \zeta ) ( \Delta \beta_{+} ( \zeta ) ) ^{2}}{4\varGamma ( 1-\alpha ) \beta ( \zeta ) } \Biggr) \\& \quad \leq \omega_{2} ( t_{2} ) -\omega_{2} ( t ) \leq \omega_{2} ( t_{2} ) < \infty , \end{aligned}$$

which contradicts (12). So, the proof is complete. □

Theorem 2

Let (7) hold. Assume that there exists a positive sequence γ such that, for all sufficiently large t,

$$ \lim_{t\rightarrow \infty }\sup \sum_{s=t_{3}}^{t-1} \Biggl( \gamma ( s ) q ( s ) \frac{\varGamma ( 1-\alpha ) }{\vartheta ( s+1 ) }\sum _{u=t_{2}}^{s-1}\frac{\vartheta ( u ) }{r ( u ) }-\frac{a ( s ) \vartheta ( s+1 ) ( \Delta \gamma_{+} ( s ) ) ^{2}}{4\gamma ( s ) \vartheta ( s ) \sum_{u=t_{0}} ^{s-1}\frac{1}{c ( u ) }} \Biggr) =\infty . $$

(14)

If there exist positive sequences β, λ such that (11) and (12) hold, then Eq. (1) is oscillatory.

Proof

Suppose to the contrary that $x(t)$ is a non-oscillatory solution of (1). Then, without loss of generality, we may assume that there is a solution $x ( t )$ of Eq. (1) such that $x ( t ) >0$ on $[ t_{0},\infty )$ where $t_{0}$ is sufficiently large. From Lemma 3, $x ( t )$ satisfies Case 1 or Case 2. Firstly, let Case 1 hold. Then we define the following function:

$$ \pi ( t ) =\gamma ( t ) \frac{c ( t ) \Delta ( a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) ) }{r ( t ) \Delta^{\alpha }x ( t ) }. $$

For $t\in [ t_{0},\infty )$, we have

$$\begin{aligned} \Delta \pi ( t ) =&\Delta \gamma ( t ) \frac{ \pi ( t+1 ) }{\gamma ( t+1 ) }+\gamma ( t ) \Delta \biggl( \frac{c ( t ) \Delta ( a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) ) }{r ( t ) \Delta^{\alpha }x ( t ) } \biggr) \\ =&\Delta \gamma ( t ) \frac{\pi ( t+1 ) }{ \gamma ( t+1 ) }-\gamma ( t ) \frac{q ( t ) G ( t ) }{r ( t+1 ) \Delta^{\alpha }x ( t+1 ) } \\ &{}-\gamma ( t ) \frac{c ( t ) \Delta ( a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) }{r ( t ) \Delta^{\alpha }x ( t ) r ( t+1 ) \Delta^{\alpha }x ( t+1 ) }. \end{aligned}$$

From Lemma 4, we obtain

$$\begin{aligned}& \Delta \bigl( r ( t ) \Delta^{\alpha }x ( t ) \bigr) \geq \frac{\sum_{s=t_{0}}^{t-1}\frac{1}{c ( s ) }}{a ( t ) }c ( t ) \Delta \bigl( a ( t ) \Delta \bigl( r ( t ) \Delta^{\alpha }x ( t ) \bigr) \bigr) , \\& 1\leq \frac{r ( t+1 ) \Delta^{\alpha }x ( t+1 ) }{r ( t ) \Delta^{\alpha }x ( t ) }\leq \frac{\vartheta ( t+1 ) }{\vartheta ( t ) }, \\& \frac{\vartheta ( t ) }{\vartheta ( t+1 ) }\leq \frac{r ( t+1 ) \Delta^{\alpha }x ( t+1 ) }{r ( t ) \Delta^{\alpha }x ( t ) } \end{aligned}$$

or

$$ \frac{r ( t+1 ) \vartheta ( t ) }{r ( t ) \vartheta ( t+1 ) }\leq \frac{\Delta G ( t ) }{ \Delta G ( t+1 ) } $$

and

$$\begin{aligned} \frac{G ( t ) }{r ( t+1 ) \Delta^{\alpha }x ( t+1 ) } =&\frac{\varGamma ( 1-\alpha ) }{r ( t+1 ) }\frac{G ( t ) }{\Delta G ( t ) } \frac{ \Delta G ( t ) }{\Delta G ( t+1 ) } \\ \geq &\frac{\varGamma ( 1-\alpha ) }{r ( t+1 ) } \Biggl( \frac{r ( t ) }{\vartheta ( t ) }\sum _{s=t _{2}}^{t-1}\frac{\vartheta ( s ) }{r ( s ) } \Biggr) \frac{r ( t+1 ) \vartheta ( t ) }{r ( t ) \vartheta ( t+1 ) } \\ =&\frac{\varGamma ( 1-\alpha ) }{\vartheta ( t+1 ) }\sum_{s=t_{2}}^{t-1} \frac{\vartheta ( s ) }{r ( s ) }. \end{aligned}$$

Hence,

$$\begin{aligned} \Delta \pi ( t ) \leq &\Delta \gamma_{+} ( t ) \frac{ \pi ( t+1 ) }{\gamma ( t+1 ) }- \gamma ( t ) q ( t ) \frac{\varGamma ( 1-\alpha ) }{ \vartheta ( t+1 ) }\sum_{s=t_{2}}^{t-1} \frac{\vartheta ( s ) }{r ( s ) } \\ &{}-\frac{\gamma ( t ) \vartheta ( t ) }{\vartheta ( t+1 ) }\frac{\sum_{s=t_{0}}^{t-1}\frac{1}{c ( s ) }}{a ( t ) }\frac{\pi^{2} ( t+1 ) }{\gamma^{2} ( t+1 ) }. \end{aligned}$$

In Lemma 1, choosing $\lambda =2$, $A= ( \frac{\gamma ( t ) \vartheta ( t ) }{\vartheta ( t+1 ) }\frac{ \sum_{s=t_{1}}^{t-1}\frac{1}{c ( s ) }}{a ( t ) } ) ^{1/2}\frac{\pi ( t+1 ) }{\gamma ( t+1 ) }$, and $B=\frac{1}{2} ( \frac{a ( t ) \vartheta ( t+1 ) }{\gamma ( t ) \vartheta ( t ) \sum_{s=t_{0}}^{t-1}\frac{1}{c ( s ) }} ) ^{1/2} \Delta \gamma_{+} ( t )$, we obtain

$$ \Delta \pi ( t ) \leq -\gamma ( t ) q ( t ) \frac{\varGamma ( 1-\alpha ) }{\vartheta ( t+1 ) } \sum _{s=t_{2}}^{t-1}\frac{\vartheta ( s ) }{r ( s ) }+\frac{a ( t ) \vartheta ( t+1 ) ( \Delta \gamma_{+} ( t ) ) ^{2}}{4\gamma ( t ) \vartheta ( t ) \sum_{s=t_{0}}^{t-1}\frac{1}{c ( s ) }}. $$

Summing both sides of the above inequality from $t_{3}$ to $t-1$, we have

$$\begin{aligned}& \sum_{s=t_{3}}^{t-1} \Biggl( \gamma ( s ) q ( s ) \frac{ \varGamma ( 1-\alpha ) }{\vartheta ( s+1 ) } \sum_{u=t_{2}}^{s-1} \frac{\vartheta ( u ) }{r ( u ) }-\frac{a ( s ) \vartheta ( s+1 ) ( \Delta \gamma_{+} ( s ) ) ^{2}}{4\gamma ( s ) \vartheta ( s ) \sum_{u=t_{0}}^{s-1}\frac{1}{c ( u ) }} \Biggr) \\& \quad \leq \pi ( t_{1} ) -\pi ( t ) \\& \quad \leq \pi ( t_{2} ) < \infty , \end{aligned}$$

which contradicts (14). And the proof of Case 2 is the same as that of Theorem 1 and hence is omitted. This completes the proof. □

Theorem 3

Let (7) hold. Assume that there exists a positive sequence γ such that, for all sufficiently large t,

$$ \lim_{t\rightarrow \infty }\sup \sum_{s=t_{2}}^{t-1} \biggl( \gamma ( s ) q ( s ) \frac{\delta ( s ) }{ \delta ( s+1 ) }-\frac{r ( s ) \phi ( s ) ( \Delta \gamma_{+} ( s ) ) ^{2}}{4\gamma ( s ) \sum_{s=t_{1}}^{u-1}\frac{\phi ( u ) }{a ( u ) }\sum_{u=t_{0}}^{s-1}\frac{1}{c ( u ) }} \biggr) = \infty . $$

(15)

If there exist positive sequences β, λ such that (11) and (12) hold, then Eq. (1) is oscillatory.

Proof

Suppose to the contrary that $x(t)$ is a non-oscillatory solution of (1). Then, without loss of generality, we may assume that there is a solution $x ( t )$ of Eq. (1) such that $x ( t ) >0$ on $[ t_{0},\infty )$, where $t_{0}$ is sufficiently large. From Lemma 3, $x ( t )$ satisfies Case 1 or Case 2. Firstly, let Case 1 hold. Then we define the following function:

$$ \nu ( t ) =\gamma ( t ) \frac{c ( t ) \Delta ( a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) ) }{G ( t ) }. $$

For $t\in [ t_{0},\infty )$, we get

$$\begin{aligned} \Delta \nu ( t ) =&\Delta \gamma ( t ) \frac{ \nu ( t+1 ) }{\gamma ( t+1 ) }+\gamma ( t ) \Delta \biggl( \frac{c ( t ) \Delta ( a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) ) }{G ( t ) } \biggr) \\ =&\Delta \gamma ( t ) \frac{\nu ( t+1 ) }{ \alpha ( t+1 ) }-\gamma ( t ) \frac{q ( t ) G ( t ) }{G ( t+1 ) } \\ &{}-\gamma ( t ) \frac{c ( t ) \Delta ( a ( t ) \Delta ( r ( t ) \Delta^{\alpha }x ( t ) ) ) \Delta G ( t ) }{G ( t ) G ( t+1 ) }. \end{aligned}$$

From Lemma 4, we have

$$ \Delta G ( t ) \geq \frac{1}{r ( t ) } \Biggl( \frac{a ( t ) }{\phi ( t ) }\sum _{s=t_{1}}^{t-1}\frac{ \phi ( s ) }{a ( s ) } \Biggr) \frac{\sum_{s=t_{0}} ^{t-1}\frac{1}{c ( s ) }}{a ( t ) }c ( t ) \Delta \bigl( a ( t ) \Delta \bigl( r ( t ) \Delta^{\alpha }x ( t ) \bigr) \bigr) $$

and

$$ \frac{G ( t ) }{G ( t+1 ) }\geq \frac{\delta ( t ) }{\delta ( t+1 ) }. $$

Thus we obtain

$$\begin{aligned} \Delta \nu ( t ) \leq &\Delta \gamma_{+} ( t ) \frac{ \nu ( t+1 ) }{\gamma ( t+1 ) }- \gamma ( t ) p ( t ) \frac{\delta ( t ) }{\delta ( t+1 ) } \\ &{}-\frac{\gamma ( t ) }{r ( t ) \phi ( t ) }\sum_{s=t_{1}}^{t-1} \frac{\phi ( s ) }{a ( s ) } \sum_{s=t_{0}}^{t-1} \frac{1}{c ( s ) }\frac{\nu^{2} ( t+1 ) }{\gamma^{2} ( t+1 ) }. \end{aligned}$$

Then, setting $\lambda =2$,

$$\begin{aligned}& A= \Biggl( \frac{\gamma ( t ) }{r ( t ) \phi ( t ) }\sum_{s=t_{1}}^{t-1} \frac{ \phi ( s ) }{a ( s ) }\sum_{s=t_{0}}^{t-1} \frac{1}{c ( s ) } \Biggr) ^{1/2}\frac{\nu ( t+1 ) }{\gamma ( t+1 ) },\quad \text{and} \\& B=\frac{1}{2} \biggl( \frac{r ( t ) \phi ( t ) }{\gamma ( t ) \sum_{s=t_{1}}^{t-1}\frac{ \phi ( s ) }{a ( s ) }\sum_{s=t_{0}}^{t-1}\frac{1}{c ( s ) }} \biggr) ^{1/2}\Delta \gamma_{+} ( t ) \end{aligned}$$

using Lemma 1, we obtain

$$ \Delta \nu ( t ) \leq -\gamma ( t ) q ( t ) \frac{\delta ( t ) }{\delta ( t+1 ) }+ \frac{r ( t ) \phi ( t ) ( \Delta \gamma_{+} ( t ) ) ^{2}}{4\gamma ( t ) \sum_{s=t_{1}}^{t-1}\frac{ \phi ( s ) }{a ( s ) }\sum_{s=t_{0}}^{t-1}\frac{1}{c ( s ) }}. $$

Summing both sides of the above inequality from $t_{2}$ to $t-1$, we have

$$\begin{aligned} \begin{aligned} \sum_{s=t_{2}}^{t-1} \biggl( \gamma ( s ) q ( s ) \frac{ \delta ( s ) }{\delta ( s+1 ) }-\frac{r ( s ) \phi ( s ) ( \Delta \gamma_{+} ( s ) ) ^{2}}{4\gamma ( s ) \sum_{s=t_{1}}^{u-1}\frac{ \phi ( u ) }{a ( u ) }\sum_{u=t_{0}}^{s-1}\frac{1}{c ( u ) }} \biggr) & \leq \nu ( t_{2} ) -\nu ( t ) \\ & \leq \nu ( t_{2} ) < \infty ,\end{aligned} \end{aligned}$$

which contradicts (15). The proof of Case 2 is the same as that of Theorem 1 and hence is omitted. This completes the proof. □

3 Applications

Example 1

Consider the following fractional difference equation for $t\geq 2$:

$$ \Delta^{3+\alpha }x ( t ) +t^{-2} \Biggl( \sum _{s=t_{0}}^{t-1+ \alpha } ( t-s-1 ) ^{ ( -\alpha ) }x ( s ) \Biggr) =0. $$

(16)

This corresponds to Eq. (1) with $\alpha \in ( 0,1 ]$, $t_{0}=2$, $c ( t ) =a ( t ) =r ( t ) =1$, and $q ( t ) =t^{-2}$. Then $\phi ( t ) =\lambda ( t ) =t-t_{1}$, $\vartheta ( t ) =\sum_{s=t_{2}} ^{t-1} ( s-t_{1} )$, $\gamma ( t ) =\beta ( t ) =t$. For $k\in ( 0,1 )$, it can be written $kt \leq \phi ( t ) \leq t$, $k^{2}t^{2}/2\leq \vartheta ( t ) \leq t^{2}/2$, $k^{3}t^{3}/3\leq \sum_{s=t_{3}}^{t-1}k^{2}s ^{2}\leq t^{3}/3$. So,

$$\begin{aligned}& \lim_{t\rightarrow \infty }\sup \sum_{s=t_{3}}^{t-1} \Biggl( \frac{ \varGamma ( 1-\alpha ) \gamma ( s ) q ( s ) }{\vartheta ( s ) \phi ( s+1 ) }\sum_{u=t_{2}} ^{s-1} \frac{\vartheta ( u ) }{r ( u ) }\sum_{u=t _{1}}^{s-1} \frac{\phi ( u ) }{a ( u ) }-\frac{c ( s ) ( \Delta \gamma_{+} ( s ) ) ^{2}}{4 \gamma ( s ) } \Biggr) \\& \quad \geq \lim_{t\rightarrow \infty }\sup \sum_{s=t_{3}}^{t-1} \biggl( \frac{ \varGamma ( 1-\alpha ) k^{5}s^{2}}{6 ( s+1 ) }- \frac{1}{4s} \biggr) =\infty \end{aligned}$$

and

$$\begin{aligned}& \lim_{t\rightarrow \infty }\sup \sum_{\zeta =t_{2}}^{t-1} \Biggl( \frac{ \beta ( \zeta ) \lambda ( \zeta ) }{\lambda ( \zeta +1 ) a ( \zeta ) }\sum_{s=\zeta }^{\infty } \Biggl( \frac{1}{c ( s ) }\sum_{v=s}^{\infty }q ( v ) \Biggr) -\frac{r ( \zeta ) ( \Delta \beta_{+} ( \zeta ) ) ^{2}}{4\varGamma ( 1-\alpha ) \beta ( \zeta ) } \Biggr) \\& \quad \geq \lim_{t\rightarrow \infty }\sup \sum_{\zeta =t_{2}}^{t-1} \Biggl( \frac{\zeta^{2}}{ ( \zeta +1 ) }\sum_{s=\zeta }^{ \infty } \Biggl( \sum_{v=s}^{\infty }v^{-2} \Biggr) -\frac{1}{4\varGamma ( 1-\alpha ) \zeta } \Biggr) \\& \quad =\infty . \end{aligned}$$

Thus, (16) is oscillatory from Theorem 1.