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Oscillation theorems for nonlinear fractional difference equations

Boundary Value Problems20182018:178

https://doi.org/10.1186/s13661-018-1098-4

• Accepted: 15 November 2018
• Published:

Abstract

In this study, we discuss some theorems related to the oscillatory behavior of nonlinear fractional difference equations equipped with well-known fractional Riemann–Liouville difference operator. Then we give an example for the illustration of the results obtained.

Keywords

• Fractional difference operator
• Riemann–Liouville
• Oscillation
• Oscillation theory
• Fractional difference equations
• Oscillation criteria

1 Introduction and preliminaries

Fractional calculus has proved to be valuable tools in describing and solving a large number of problems in various fields of sciences and engineering [1, 2]. Their treatment from the viewpoint of difference equations can additionally open up new perspectives. Thus we decided to study fractional difference equations.

Up to now, many authors have investigated the oscillatory behaviors of solutions of various equations, including differential equations, fractional differential equations, difference equations, fractional difference equations, partial differential equations, fractional partial differential equations and dynamic equations on time scales . Motivated by this work, we are concerned with the following equations:
$$\Delta \bigl( r ( t ) \Delta^{\alpha}x ( t ) \bigr) +q ( t ) f \bigl( G ( t ) \bigr) =0,$$
(1)
where $$t\in \mathbf{N} _{t_{0}+1-\alpha}$$, $$G ( t ) =\sum_{s=t_{0}}^{t-1+\alpha} ( t-s-1 ) ^{ ( -\alpha ) }x ( s )$$, $$q ( t )$$ and $$r ( t )$$ is positive sequences, $$f:\mathbf{R} \rightarrow \mathbf{R}$$ is continuous function satisfies $$xf ( x ) >0$$ for $$x\neq0$$ and $$\Delta^{\alpha}$$ denotes the Riemann–Liouville fractional difference operator of order $$0<\alpha\leq1$$. Throughout the study, we consider
$$R ( t ) =\sum_{s=t_{0}}^{t-1} \frac{1}{r ( t ) }$$
and $$\lim_{t\rightarrow\infty}R ( t ) =\infty$$.

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 for every positive integers $$T_{0}>t_{0}$$ there exists $$t\geq T_{0}$$ such that $$x ( t ) x ( t+1 ) \leq0$$, otherwise it is called non-oscillatory. Equation (1) is called oscillatory if all its solutions are oscillatory.

Definition 1

()

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 $$s\equiv a\mod ( 1 )$$, $$\Delta^{-v}f$$ is defined for $$t\equiv ( a+v ) \mod ( 1 )$$ 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

()

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.

2 Main result

Lemma 1

()

Let $$x ( t )$$ be a solution of Eq. (1) and let
$$G ( t ) =\sum_{s=t_{0}}^{t-1+\alpha} ( t-s-1 ) ^{ ( -\alpha ) }x ( s ) ,$$
(4)
then
$$\Delta \bigl( G ( t ) \bigr) =\varGamma ( 1-\alpha ) \Delta^{\alpha}x ( t ) .$$
(5)

Theorem 1

Suppose that
$$\sum_{s=t_{0}}^{\infty}q ( s ) = \infty$$
(6)
and
$$\lim_{t\rightarrow\infty}\inf f ( t ) >0.$$
(7)
Then every solution of (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 assume that $$x ( t )$$ is eventually positive solution of (1) on $$[ t_{0},\infty )$$, then $$G ( t ) >0$$ on $$[ t_{0},\infty )$$. From (1), we have
$$\Delta \bigl( r ( t ) \Delta^{\alpha}x ( t ) \bigr) =-q ( t ) f \bigl( G ( t ) \bigr) < 0.$$
(8)
That is, $$r ( t ) \Delta^{\alpha}x ( t )$$ is an eventually nonincreasing sequence on $$[ t_{0},\infty )$$. We claim that $$r ( t ) \Delta^{\alpha}x ( t ) >0$$ on $$[ t_{1},\infty )$$, where $$t_{1}$$ is sufficiently large. Otherwise, assume that there exists a $$t_{2}>t_{1}$$ such that $$r ( t ) \Delta ^{\alpha}x ( t ) <0$$ on $$[ t_{2},\infty )$$. Then we have
$$r ( t ) \Delta^{\alpha}x ( t ) \leq r ( t_{2} ) \Delta^{\alpha}x ( t_{2} ) =c< 0,$$
or
$$\Delta^{\alpha}x ( t ) \leq\frac{c}{r ( t ) }.$$
Then we get
$$\Delta \bigl( G ( t ) \bigr) \leq\varGamma ( 1-\alpha ) \frac{c}{r ( t ) }.$$
(9)
Summing both sides of (9) from $$t_{2}$$ to $$t-1$$, we have
$$G ( t ) \leq G ( t_{2} ) +\varGamma ( 1-\alpha ) \sum _{s=t_{2}}^{t-1}\frac{c}{r ( s ) }.$$
Letting $$t\rightarrow\infty$$, we get $$\lim_{t\rightarrow\infty}G ( t ) =-\infty$$, which contradicts the fact that $$G ( t ) >0$$ on $$[ t_{0},\infty )$$. Hence we obtain $$r ( t ) \Delta^{\alpha}x ( t ) >0$$ on $$[ t_{1},\infty )$$. Hence we obtain $$\Delta^{\alpha}x ( t ) >0$$ and $$\Delta ( r ( t ) \Delta^{\alpha}x ( t ) ) <0$$ on $$[ t_{1},\infty )$$. Now we consider that
$$\lim_{t\rightarrow\infty}G ( t ) =k.$$
Then $$k>0$$ is finite or infinite.

Case 1 $$k>0$$ is finite.

Since f is a continuous function, we get
$$\lim_{t\rightarrow\infty}f \bigl( G ( t ) \bigr) =f ( k ) >0.$$
This implies we have for sufficiently large $$t_{3}>t_{2}$$ and $$t\geq t_{3}$$
$$f \bigl( G ( t ) \bigr) >\frac{1}{2}f ( k ).$$
(10)
Substituting (10) in (8), we get
$$\Delta \bigl( r ( t ) \Delta^{\alpha}x ( t ) \bigr) \leq-q ( t ) \frac{1}{2}f ( k )$$
or
$$\Delta \bigl( r ( t ) \Delta^{\alpha}x ( t ) \bigr) +\frac{1}{2}f ( k ) q ( t ) \leq0.$$
(11)
Then summing both sides of the last inequality from $$t_{3}$$ to $$t-1$$, we have
$$r ( t ) \Delta^{\alpha}x ( t ) -r ( t_{3} ) \Delta^{\alpha}x ( t_{3} ) +\frac{1}{2}f ( k ) \sum _{s=t_{3}}^{t-1}q ( t ) \leq0.$$
Hence we obtain for $$t\geq t_{3}$$
$$\frac{1}{2}f ( k ) \sum_{s=t_{3}}^{t-1}q ( t ) \leq r ( t_{3} ) \Delta^{\alpha}x ( t_{3} ),$$

Case 2 $$k=\infty$$.

From the condition (7), we have
$$\lim_{t\rightarrow\infty}\inf f \bigl( G ( t ) \bigr) >0.$$
Then we can choose a positive constant c and sufficiently large $$t_{4}>t_{3}$$ such that for $$t\geq t_{4}$$
$$f \bigl( G ( t ) \bigr) >c.$$
(12)
Substituting (12) in (8),
$$\Delta \bigl( r ( t ) \Delta^{\alpha}x ( t ) \bigr) \leq-cq ( t ) .$$
This implies
$$\Delta \bigl( r ( t ) \Delta^{\alpha}x ( t ) \bigr) +cq ( t ) \leq0.$$
Summing the last inequality from $$t_{4}$$ to $$t-1$$, we get
$$r ( t ) \Delta^{\alpha}x ( t ) -r ( t_{4} ) \Delta^{\alpha}x ( t_{4} ) +c\sum_{s=t_{4}}^{\infty}q ( s ) \leq0.$$
Thus
$$c\sum_{s=t_{4}}^{\infty}q ( s ) \leq r ( t_{4} ) \Delta^{\alpha}x ( t_{4} ),$$
which contradicts (6). Then the proof is complete. □

Theorem 2

Assume that
$$\sum_{s=t_{0}}^{\infty}R ( s ) q ( s ) =\infty .$$
(13)
Then every bounded solution of (1) is oscillatory.

Proof

Proceeding as in the proof of Theorem 1 with the assumption that $$x ( t )$$ is a bounded non-oscillatory solution of (1), from (11), we have for $$t\geq t_{0}$$ where $$t_{0}$$ is sufficiently large
$$R ( t ) \Delta \bigl( r ( t ) \Delta^{\alpha}x ( t ) \bigr) + \frac{1}{2}f ( k ) R ( t ) q ( t ) \leq0.$$
(14)
$$R ( t ) \Delta \bigl( r ( t ) \Delta^{\alpha}x ( t ) \bigr) \geq\Delta \bigl( R ( t ) r ( t ) \Delta^{\alpha}x ( t ) \bigr) - \bigl( \Delta R ( t ) \bigr) r ( t ) \Delta^{\alpha}x ( t ) .$$
(15)
From (14) and (15),
$$\Delta \bigl( R ( t ) r ( t ) \Delta^{\alpha}x ( t ) \bigr) - \frac{\Delta G ( t ) }{\varGamma ( 1-\alpha ) }+\frac{1}{2}f ( k ) R ( t ) q ( t ) \leq0$$
and summing both sides of the last inequality from $$t_{0}$$ to $$t-1$$, we get
\begin{aligned} &R ( t ) r ( t ) \Delta^{\alpha}x ( t ) -R ( t_{0} ) r ( t_{0} ) \Delta^{\alpha}x ( t_{0} ) -\frac{1}{\varGamma ( 1-\alpha ) } \bigl( G ( t ) -G ( t_{0} ) \bigr) \\ &\qquad {}+\frac{1}{2}f ( k ) \sum_{s=t_{0}}^{t-1}R ( s ) q ( s ) \leq0. \end{aligned}
That is,
$$\frac{1}{2}f ( k ) \sum_{s=t_{0}}^{t-1}R ( s ) q ( s ) \leq\frac{G ( t ) -G ( t_{0} ) }{\varGamma ( 1-\alpha ) }+R ( t_{0} ) r ( t_{0} ) \Delta^{\alpha}x ( t_{0} ) .$$
Since $$x ( t )$$ is bounded, we can choose a positive constant c such that
$$\sum_{s=t_{0}}^{t-1}R ( s ) q ( s ) \leq c,$$
which is a contradiction to the assumption of the theorem. □

Theorem 3

Assume that (6), f is non-decreasing and there is a non-negative constant M such that
$$\lim_{t\rightarrow0}\sup\frac{t}{f ( t ) }=M$$
(16)
and there exists a $$r_{1} ( t )$$ positive subsequence of $$r ( t )$$ such that $$r_{1} ( t ) \leq1$$. Then the fractional difference $$\Delta^{\alpha}x ( t )$$ of every solution $$x ( t )$$ of (1) oscillates.

Proof

Suppose to the contrary that Eq. (1) has a solution $$x(t)$$ such that its fractional difference $$\Delta^{\alpha}x ( t )$$ is non-oscillatory. Firstly we assume that $$\Delta^{\alpha}x ( t )$$ is eventually negative. Then there exists a positive integer $$t_{0}$$ such that $$\Delta^{\alpha}x ( t ) <0$$ and $$G ( t )$$ is decreasing on $$[ t_{0},\infty )$$. This implies that $$x ( t )$$ is also non-oscillatory. Then we consider the following function for $$t\geq t_{1}\geq t_{0}$$:
$$\omega ( t ) =\frac{r ( t ) \Delta^{\alpha}x ( t ) }{f ( G ( t ) ) }.$$
Thus
\begin{aligned} \Delta\omega ( t ) =&\frac{\Delta ( r ( t ) \Delta^{\alpha}x ( t ) ) f ( G ( t ) ) -r ( t ) \Delta^{\alpha}x ( t ) \Delta f ( G ( t ) ) }{f ( G ( t ) ) f ( G ( t+1 ) ) } \\ =&\frac{ \{ r ( t+1 ) \Delta^{\alpha}x ( t+1 ) -r ( t ) \Delta^{\alpha}x ( t ) \} f ( G ( t ) ) }{f ( G ( t ) ) f ( G ( t+1 ) ) } \\ &{}-\frac{r ( t ) \Delta^{\alpha}x ( t ) \{ f ( G ( t+1 ) -f ( G ( t ) ) ) \}}{f ( G ( t ) ) f ( G ( t+1 ) ) } \\ =&\frac{r ( t+1 ) \Delta^{\alpha}x ( t+1 ) f ( G ( t ) ) -r ( t ) \Delta^{\alpha}x ( t ) f ( G ( t+1 ) ) }{f ( G ( t ) ) f ( G ( t+1 ) ) } \\ =&\frac{r ( t+1 ) \Delta^{\alpha}x ( t+1 ) }{f ( G ( t+1 ) ) }-\frac{r ( t ) \Delta^{\alpha }x ( t ) }{f ( G ( t ) ) } \\ =&\frac{r ( t+1 ) \Delta^{\alpha}x ( t+1 ) }{f ( G ( t+1 ) ) }-\frac{ \{ r ( t+1 ) \Delta^{\alpha}x ( t+1 ) -\Delta ( r ( t ) \Delta^{\alpha}x ( t ) ) \} }{f ( G ( t ) ) } \\ =&\frac{\Delta ( r ( t ) \Delta^{\alpha}x ( t ) ) }{f ( G ( t ) ) }+\frac{r ( t+1 ) \Delta^{\alpha}x ( t+1 ) \{ f ( G ( t ) ) -f ( G ( t+1 ) ) \} }{f ( G ( t ) ) f ( G ( t+1 ) ) } \\ \leq&\frac{\Delta ( r ( t ) \Delta^{\alpha}x ( t ) ) }{f ( G ( t ) ) }=-q ( t ). \end{aligned}
That is,
$$\Delta\omega ( t ) \leq-q ( t ) .$$
Summing both sides of last inequality from $$t_{1}$$ to $$t-1$$, we have
$$\omega ( t ) -\omega ( t_{1} ) \leq-\sum _{s=t_{1}}^{t-1}q ( s ) .$$
That is,
$$\lim_{t\rightarrow\infty}\omega ( t ) =-\infty,$$
(17)
thus $$f ( G ( t ) ) >0$$ and hence
$$G ( t )>0.$$
(18)
From (17), for sufficiently large $$t_{2}>t_{1}$$ and $$t\geq t_{2}$$
$$\omega ( t ) \leq- ( M+1 ).$$
Then we obtain
$$\omega ( t ) =\frac{r ( t ) \Delta^{\alpha}x ( t ) }{f ( G ( t ) ) }\leq- ( M+1 )$$
or
$$r ( t ) \Delta^{\alpha}x ( t ) + ( M+1 ) f \bigl( G ( t ) \bigr) \leq0.$$
(19)
Set $$\lim_{t\rightarrow\infty}G ( t ) =L$$. Then $$L\geq0$$. We claim that $$L=0$$. If $$L>0$$, then $$\lim_{t\rightarrow\infty}f ( G ( t ) ) =f ( L ) >0$$ by the continuity of f. For sufficiently large $$t_{3}>t_{2}$$ and $$t\geq t_{3}$$ we get
$$f \bigl( G ( t ) \bigr) >\frac{1}{2}f ( L ).$$
Substituting the last inequality in (19), we have
$$\Delta^{\alpha}x ( t ) +\frac{ ( M+1 ) f ( L ) }{2r ( t ) }\leq0;$$
in other words
$$\Delta G ( t ) +\frac{\varGamma ( 1-\alpha ) ( M+1 ) f ( L ) }{2r ( t ) }\leq0.$$
Summing both sides of the last inequality from $$t_{3}$$ to $$t-1$$, we obtain
$$G ( t ) -G ( t_{3} ) +\frac{1}{2}\varGamma ( 1-\alpha ) ( M+1 ) f ( L ) \sum_{s=t_{3}}^{t-1} \frac{1}{r ( s ) }\leq0.$$
This implies $$\lim_{t\rightarrow\infty}G ( t ) =-\infty$$, which contradicts (18). Thus $$\lim_{t\rightarrow\infty}G ( t ) =0$$. By the assumption (16), we get
$$\lim_{t\rightarrow\infty}\sup\frac{G ( t ) }{f ( G ( t ) ) }\leq M.$$
Then for sufficiently large $$t_{4}>t_{3}$$ and $$t\geq t_{4}$$ we obtain
$$\frac{G ( t ) }{f ( G ( t ) ) }< M+1.$$
That is,
$$G ( t ) < ( M+1 ) f \bigl( G ( t ) \bigr) .$$
From (19), we obtain
$$r ( t ) \Delta^{\alpha}x ( t ) +G ( t ) < 0.$$
Then we consider subsequence $$r_{1} ( t )$$,
$$\Delta^{\alpha}x ( t ) +G ( t ) \leq r_{1} ( t ) \Delta^{\alpha}x ( t ) +G ( t ) < 0,$$
and we have
$$\frac{\Delta G ( t ) }{\varGamma ( 1-\alpha ) }+G ( t ) < 0$$
or
$$0< G ( t+1 ) -G ( t ) +\varGamma ( 1-\alpha ) G ( t ) < 0,$$
which contradicts (18). The case that $$\Delta^{\alpha}x ( t )$$ is eventually positive can be proved in a similar manner and so the proof is complete. □

3 Applications

Example

Consider the following fractional difference equation:
$$\footnotesize \Delta \bigl( t\Delta^{\alpha}x ( t ) \bigr) +t^{7} \Biggl( \sum_{s=t_{0}}^{t-1+\alpha} ( t-s-1 ) ^{ ( -\alpha ) }x ( s ) \Biggr) \exp \Biggl( \sum _{s=t_{0}}^{t-1+\alpha} ( t-s-1 ) ^{ ( -\alpha ) }x ( s ) \Biggr) =0.$$
(20)
This corresponds to Eq. (1) with $$\alpha\in ( 0,1 ]$$, $$r ( t ) =t$$, $$q ( t ) =t^{7}$$ and $$f ( x ) =xe^{x}$$. Then we have $$\lim_{t\rightarrow\infty}R ( t ) =\infty$$, $$xf ( x ) >0$$,
$$\sum_{s=t_{0}}^{\infty}q ( s ) =\sum _{s=t_{0}}^{\infty}t^{7}=\infty,$$
and
$$\lim_{t\rightarrow\infty}\inf f ( t ) >0.$$
Thus, (20) is oscillatory from Theorem 1.

Declarations

Acknowledgements

The author is grateful to the scholars who provided the literature sources.

Not applicable.

Authors’ contributions

HA contributed to the work totally, and he read and approved the final version of the manuscript.

Competing interests

The author declares that they have no competing interests. 