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R E S E A R C H

Open Access

Oscillatory behavior of solutions of certain

fractional difference equations

Hakan Adiguzel

1*

*Correspondence:

[email protected] 1Department of Mechatronics

Engineering, Istanbul Gelisim University, Istanbul, Turkey

Abstract

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

(

c(t)

(

a(t)

(

r(t)

αx(t)

)))

+q(t)G(t) = 0,

wheretNt0+1–α,G(t) =

t–1+α

s=t0 (ts– 1)

αx(s), and

αdenotes a Riemann–Liouville

fractional difference operator of order 0 <

α

≤1. By using the generalized Riccati transformation technique, we obtain some oscillation criteria. Finally we give an example.

Keywords: Oscillation; Oscillation criteria; Fractional difference operator; Riemann–Liouville; Fractional difference equations; Riccati technique; Hardy inequalities

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 differen-tial equations [3–6], difference equations [7–9], dynamic equations on time scales [10–14], and fractional differential (difference) equations [15–31] obtaining some oscillation crite-ria. Recently, oscillation studies have become a very hot topic. That is why, we consider the following fractional difference equation:

c(t)a(t)r(t)αx(t)+q(t)G(t) = 0, (1)

wheretNt0+1–α,G(t) =

t–1+α

s=t0 (ts– 1)

(–α)x(s),c(t),a(t),r(t), andq(t) are positive

se-quences, andαdenotes the Riemann–Liouville fractional difference operator of order 0 <α≤1.

By a solution of Eq. (1), we mean a real-valued sequencex(t) satisfying Eq. (1) fortNt0. A solutionx(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.

(2)

Definition 1([32]) Letv> 0. Thevth fractional sumf is defined by

vf(t) = 1

Γ(v)

tv

s=a

(ts– 1)v–1f(s), (2)

wherefis defined forsamod(1),vf is defined fort≡(a+v)mod(1), andt(v)= Γ(t+1)

Γ(tv+1). The fractional sum vf maps functions defined on N

a to functions defined on Na+v,

where Nt={t,t+ 1,t+ 2, . . .}.

Definition 2([32]) Letv> 0 andm– 1 <μ<m, wheremdenotes a positive integer,m=

μ. Setv=mμ. Theμth fractional difference is defined as

μf(t) =mvf(t) =mvf(t), (3)

whereμis the ceiling function ofμ.

Lemma 1([33]) Assume that A and B are nonnegative real numbers.Then

λABλ–1Aλ(λ– 1)Bλ (4)

for allλ> 1.

2 Main results

Throughout this paper, we denote

φ(t) =

t–1

s=t1

1

c(s); ϑ(t) =

t–1

s=t2 φ(s)

a(s); δ(t) =

t–1

s=t3 ϑ(s)

r(s).

For simplification, we consider

γ+(s) =max

0,γ(s)

and

β+(s) =max0,β(s).

Lemma 2([28]) Let x(t)be a solution of Eq. (1),and let

G(t) =

t–1+α

s=t0

(ts– 1)(–α)x(s), (5)

then

(3)

Lemma 3 Assume that x(t)is an eventually positive solution of Eq. (1).If

s=t0

1

c(s)=

s=t0

1

a(s)=

s=t0

1

r(s)=∞, (7)

then we have two possible cases for t∈[t1,∞),t1>t0is sufficiently large: Case 1 αx(t) > 0,(r(t)αx(t)) > 0,(a(t)(r(t)αx(t))) > 0or

Case 2 αx(t) > 0,(r(t)αx(t)) < 0,(a(t)(r(t)αx(t))) > 0.

Proof From the hypothesis, there existst1such thatx(t) > 0 on [t1,∞), so thatG(t) > 0 on [t1,∞), and from Eq. (1), we have

c(t)a(t)r(t)αx(t)= –q(t)G(t) < 0. (8)

Then c(t)(a(t)(r(t)αx(t))) is an eventually non-increasing sequence on [t1,). We

know thatαx(t),(r(t)αx(t)), and(a(t)(r(t)αx(t))) are eventually of one sign. For t2>t1is sufficiently large, we claim that(a(t)(r(t)αx(t))) > 0 on [t2,∞). Otherwise, assume that there exists sufficiently larget3>t2 such that(a(t)(r(t)αx(t))) < 0 on [t3,∞). For [t3,∞) and there exists a constantl1> 0, we have

a(t)r(t)αx(t)≤– l1

c(t)< 0.

Hence, there exist a constantl2> 0 and sufficiently larget4>t3such that

r(t)αx(t)≤– l2

a(t)< 0. (9)

Then there exist a constantl3> 0 and sufficiently larget5>t4such that

αx(t)≤– l3

r(t),

that is,

G(t)≤–Γ(1 –α)l3

r(t) < 0.

By (7), we obtain limt→∞G(t) = –∞. This is a contradiction. If (r(t)αx(t)) < 0,

then αx(t) > 0 due to s=t0 r1(s) =∞. If (r(t)αx(t)) > 0, then αx(t) > 0 due to

(a(t)(r(t)αx(t))) > 0. So, the proof is complete.

Lemma 4 Assume that x(t)is an eventually positive solution of Eq. (1),which satisfies Case1of Lemma3.Then

a(t)r(t)αx(t)≥c(t)a(t)r(t)αx(t)

t–1

s=t0

1

(4)

If there exists a positive sequenceφsuch that,for t∈[t1,∞),

φ(t)

c(t)ts–1=t0 c(1s)φ(t)≤0,

where t1is sufficiently large,then a(t)(r(t)αx(t))/φ(t)is a non-increasing sequence on [t1,∞)and

r(t)αx(t)≥r(t)αx(t)a(t)

φ(t)

t–1

s=t1 φ(s)

a(s).

Furthermore,if there exists a positive sequenceϑand t2>t1is sufficiently large such that,

for t∈[t2,∞),

ϑ(t)

a(t)

φ(t) t–1

s=t2 φ(s)

a(s)

ϑ(t)≤0,

then r(t)αx(t)/ϑ(t)is a non-increasing sequence on[t2,)and

G(t)≥G(t)r(t)

ϑ(t)

t–1

s=t2 ϑ(s)

r(s).

Suppose also that there exists a positive sequenceδand t3>t2is sufficiently large such that,

for t∈[t3,∞),

δ(t)

r(t)

ϑ(t) t–1

s=t2 ϑ(s)

r(s)

δ(t)≤0.

Then G(t)/δ(t)is a non-increasing sequence on[t3,∞).

Proof Assume that x is an eventually positive solution of Eq. (1). Then we have that

(r(t)αx(t)) > 0 and(c(t)(a(t)(r(t)αx(t)))) < 0 on [t0,). So,

a(t)r(t)αx(t)=a(t0)r(t0)αx(t0)

+

t–1

s=t0

c(s)(a(s)(r(s)αx(s))) c(s)

c(t)a(t)r(t)αx(t)

t–1

s=t0

1

c(s),

and then

a(t)(r(t)αx(t)) φ(t)

=(a(t)(r(t)

αx(t)))φ(t) –a(t)(r(t)αx(t))φ(t)

φ(t)φ(t+ 1)

(a(t)(r(t)αx(t))) φ(t)φ(t+ 1)

φ(t)

(5)

Hence,a(t)(r(t)αx(t))/φ(t) is a non-increasing sequence on [t1,) wheret

1>t0is suf-ficiently large. Then we have

r(t)αx(t) =r(t1)αx(t1) +

So the proof is complete.

Theorem 1 Assume that(7)holds and there exists a positive sequenceγ such that,for all sufficiently large t,

lim

If there exist positive sequencesβ,λsuch that,for all sufficiently large t,

λ(t)

(6)

and

Proof Suppose to the contrary thatx(t) is a non-oscillatory solution of Eq. (1). Then, with-out loss of generality, we may assume that there is a solutionx(t) of Eq. (1) such thatx(t) > 0 on [t0,∞), wheret0 is sufficiently large. From Lemma3,x(t) satisfies Case 1 or Case 2. Firstly, let Case 1 hold. Then we define the following function:

ω(t) =γ(t)c(t)(a(t)(r(t)

From Lemma4, we obtain

(7)

Settingλ= 2,A= (γc((tt)))1/2ω(t+1)

φ(t+1), andB= 1 2(

c(t)

γ(t))1/2γ+(t) using Lemma1, we obtain

ω(t)≤–γ(t)q(t) Γ(1 –α)

ϑ(t)φ(t+ 1)

t–1

s=t2 ϑ(s)

r(s) t–1

s=t1 φ(s)

a(s)

+ c(t) 4γ(t)

γ+(t)2.

Summing both sides of the above inequality fromt3tot– 1, we get

t–1

s=t3

Γ(1 –α)γ(s)q(s)

ϑ(s)φ(s+ 1)

s–1

u=t2 ϑ(u)

r(u) s–1

u=t1 φ(u)

a(u)

c(s)(γ+(s)) 2

4γ(s)

ω(t3) –ω(t)≤ω(t3).

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

ω2(t) =β(t)r(t)

αx(t)

G(t) .

Then

ω2(t) =β(t)ω(t+ 1)

β(t+ 1)+β(t)

r(t)αx(t)

G(t)

=β(t)ω(t+ 1)

β(t+ 1)+β(t)

(r(t)αx(t))G(t) –r(t)αx(t)G(t) G(t)G(t+ 1)

=β(t)ω(t+ 1)

β(t+ 1)+β(t)

(r(t)αx(t)) G(t+ 1) –β(t)

r(t)αx(t)G(t) G(t)G(t+ 1) .

Hence we have

G(t) =G(t1) +Γ(1 –α)

t–1

s=t1

r(s)αx(s)

r(s)

Γ(1 –α)r(t)αx(t)

t–1

s=t1

1

r(s).

That is,

G(t)

r(t)ts–1=t1 r(1s)Γ(1 –α)

α

x(t) =G(t)

and

G(t)

λ(t) =

G(t)λ(t) –G(t)λ(t)

λ(t)λ(t+ 1)

G(t) λ(t)λ(t+ 1)

λ(t)

r(t)ts–1=t1 r(1s)λ(t) ≤0.

Thus we haveG(t)/λ(t) is eventually non-increasing and

G(t)

G(t+ 1)≥

λ(t)

(8)
(9)

Settingλ= 2,A= (Γ(1–r(αt))β(t))1/2ω2(t+1)

Summing both sides of the above inequality fromt2tot– 1, we have

t–1

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

Theorem 2 Let(7)hold.Assume that there exists a positive sequenceγ such that,for all sufficiently large t,

lim

Proof Suppose to the contrary thatx(t) is a non-oscillatory solution of (1). Then, without loss of generality, we may assume that there is a solutionx(t) of Eq. (1) such thatx(t) > 0 on [t0,∞) wheret0is sufficiently large. From Lemma3,x(t) satisfies Case 1 or Case 2. Firstly, let Case 1 hold. Then we define the following function:

π(t) =γ(t)c(t)(a(t)(r(t)

From Lemma4, we obtain

(10)

or

Summing both sides of the above inequality fromt3tot– 1, we have

t–1

which contradicts (14). And the proof of Case 2 is the same as that of Theorem1and hence

is omitted. This completes the proof.

Theorem 3 Let(7)hold.Assume that there exists a positive sequenceγ such that,for all sufficiently large t,

(11)

Proof Suppose to the contrary thatx(t) is a non-oscillatory solution of (1). Then, without loss of generality, we may assume that there is a solutionx(t) of Eq. (1) such thatx(t) > 0 on [t0,∞), wheret0is sufficiently large. From Lemma3,x(t) satisfies Case 1 or Case 2. Firstly, let Case 1 hold. Then we define the following function:

ν(t) =γ(t)c(t)(a(t)(r(t)

αx(t)))

G(t) .

Fort∈[t0,∞), we get

ν(t) =γ(t)ν(t+ 1)

γ(t+ 1)+γ(t)

c(t)(a(t)(r(t)αx(t))) G(t)

=γ(t)ν(t+ 1)

α(t+ 1)–γ(t)

q(t)G(t)

G(t+ 1)

γ(t)c(t)(a(t)(r(t)

αx(t)))G(t)

G(t)G(t+ 1) .

From Lemma4, we have

G(t)≥ 1

r(t)

a(t)

φ(t)

t–1

s=t1 φ(s)

a(s)

t–1

s=t0

1

c(s)

a(t) c(t)

a(t)r(t)αx(t)

and

G(t)

G(t+ 1)≥

δ(t)

δ(t+ 1).

Thus we obtain

ν(t)≤γ+(t)

ν(t+ 1)

γ(t+ 1)–γ(t)p(t)

δ(t)

δ(t+ 1)

γ(t)

r(t)φ(t)

t–1

s=t1 φ(s)

a(s)

t–1

s=t0

1

c(s)

ν2(t+ 1)

γ2(t+ 1).

Then, settingλ= 2,

A=

γ(t)

r(t)φ(t)

t–1

s=t1 φ(s)

a(s)

t–1

s=t0

1

c(s) 1/2

ν(t+ 1)

γ(t+ 1), and

B=1 2

r(t)φ(t)

γ(t)ts–1=t1 φa((ss))st–1=t0 c(1s) 1/2

γ+(t)

using Lemma1, we obtain

ν(t)≤–γ(t)q(t) δ(t)

δ(t+ 1)+

(12)

Summing both sides of the above inequality fromt2tot– 1, we have

which contradicts (15). The proof of Case 2 is the same as that of Theorem1and hence is

omitted. This completes the proof.

3 Applications

Example1 Consider the following fractional difference equation fort≥2:

3+αx(t) +t–2

Thus, (16) is oscillatory from Theorem1.

Acknowledgements

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

Funding

Not applicable.

Availability of data and materials

Not applicable.

Competing interests

The author declares that they have no competing interests.

Authors’ contributions

(13)

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Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

Received: 2 October 2018 Accepted: 26 November 2018 References

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References

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