# The Asymptotic Behavior of Solutions for a Class of Nonlinear Fractional Difference Equations with Damping Term

Authors: Zhihong Bai; Run Xu
Journal: Discrete Dynamics in Nature and Society
 (2018)

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## Abstract

Based on generalized Riccati transformation and some inequalities, some oscillation results are established for a class of nonlinear fractional difference equations with damping term. An example is given to illustrate the validity of the established results.


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## Body

## 1. Introduction

During the past two decades, fractional calculus arose in the field of viscoelasticity, electrochemistry, physics, control, porous media, electromagnetism, and so forth; see [1–11] and the references therein. Fractional differential equations and difference equations can be used to describe some complex systems more accurately, and we may get the corresponding equations from those phenomena. However, as we all know, it is usually difficult to find the exact solutions to fractional differential or difference equations. In recent years, the study on qualitative properties of solutions of fractional differential equations, such as the existence, uniqueness, boundedness, oscillation, and other asymptotic behaviors, attracted much attention and some excellent results were obtained; we refer the reader to see [12–29] and the references cited therein.In [12], Marian et al. studied the oscillation of fractional nonlinear difference equations of the form(1)Δαxt+f1t,xt+α=vt+f2t,xt+α,t∈N0,0<α≤1.Sagayaraj et al. [13] and Selvam et al. [14] investigated the oscillation of the following nonlinear fractional difference equations:(2)ΔptΔαxtγ+qtf∑s=t0t-1+αt-s-1-αxs=0,t∈Nt0+1-α,ΔctΔαxtγ+ptΔαxtγ+qt∑s=t0t-1+αt-s-1-αxsγ=0,t∈Nt0+1-α,where 0<α≤1 and γ>0 is a quotient of odd positive integers.In [15], Sagayaraj et al. studied oscillatory behavior of the following fractional difference equations:(3)ΔptΔrtΔαxtη+Ft,∑s=t0t-1+αt-s-1-αxs=0,t∈Nt0+1-α,where 0<α≤1 and η>0 is a quotient of odd positive integers.In [16], Li investigated the oscillation of forced fractional difference equations with damping term of the form(4)1+ptΔΔαxt+ptΔαxt+ft,xt=gt,t∈N0,with initial condition Δα-1xtt=0=x0, where 0<α<1.In [24], Secer and Adiguzel established the oscillation results for a class of nonlinear fractional difference equations of the form(5)ΔatΔrtΔαxtγ1γ2+qtf∑s=t0t-1+αt-s-1-αxs=0,t∈Nt0+1-α,where 0<α≤1 and γ1 and γ2 are the quotients of two odd positive numbers.Motivated by the idea in [24], in this paper, we are concerned with the oscillation of a class of nonlinear fractional difference equations with damping term of the form(6)ΔctΔrtΔαxtγ+ptΔrtΔαxtγ+qtf∑s=t0t-1+αt-s-1-αxs=0,t∈Nt0,where γ≥1 is a quotient of two odd positive integers,  0<α≤1 is a constant, Δα denotes the Riemann-Liouville fractional difference operator of order α, and Nt0={t0,t0+1,t0+2,…}.By a solution of (6), we mean a real-valued sequence x(t) satisfying (6) for t∈Nt0. A nontrivial solution x(t) of (6) is called oscillatory if it is neither eventually positive nor eventually negative; otherwise, it is called nonoscillatory. Equation (6) is called oscillatory if all of its solutions are oscillatory.Throughout this paper, we assume that the following conditions hold:(A)
c(t), r(t), p(t), and q(t) are positive sequences, c(t)>p(t).(B)
f is a monotone decreasing function satisfying xf(x)>0; f(x)/xγ≥L>0 for x≠0.For convenience, in the rest of this paper, we set(7)Gt=∑s=t0t-1+αt-s-1-αxs,δt,t1=∑s=t1t-11uscs1/γ.


## 2. Preliminaries and Lemmas

In this section, the definitions of the Riemann-Liouville fractional sum and difference are given; then some basic lemmas are presented, which will be used in the following proof.Definition 1 (see [28]).
Theυth fractional sum of f, for υ>0, is defined by(8)Δ-υft=1Γυ∑s=at-υt-s-1υ-1fs,where f is defined for s=amod(1), Δ-υf is defined for t=(a+υ)mod(1), and t(υ)=Γ(t+1)/Γ(t+1-υ). The fractional sum Δ-υ is a map from Na to Na+υ, where Nt={t,t+1,t+2,…}.Definition 2 (see [28]).
Letμ>0 and m-1<μ<m, where m denotes a positive integer; m=⌈μ⌉. Set υ=m-μ; then μth fractional difference is defined as(9)Δμft=Δm-υft=ΔmΔ-υft.Lemma 3 (see [14]).
Letx(t) be a solution of (6), G(t)=∑s=t0t-1+αt-s-1(-α)x(s), and then(10)ΔGt=Γ1-αΔαxt.Lemma 4 (see [29]).
The product and quotient rules of the difference operatorΔ are as follows:(11)Δxtyt=xt+1Δyt+Δxt·yt(12)=Δxt·yt+1+xtΔyt,(13)Δxtyt=Δxt·yt-xtΔytytyt+1(14)=Δxt·yt+1-xt+1Δytytyt+1,where Δx(t)=x(t+1)-x(t).Lemma 5.
Ifγ≥1 is a quotient of two odd positive integers, then the following two inequalities are established:(15)ifGt+1>Gt>0,thenΔGγt≥ΔGtγ,(16)ifGt+1<Gt<0,thenΔGγt≤ΔGtγ.Proof.
Using the inequality (see [30])(17)x-yp≤xp-ypwherex≥0,y≥0,p≥1,we have the following results.
IfG(t+1)>G(t)>0, let x=G(t+1), y=G(t), and p=γ; then x>y>0; it follows from (17) that ΔGγ(t) = Gγ(t+1)-Gγ(t) ≥ (G(t+1)-G(t))γ = (ΔG(t))γ, so ΔGγ(t)≥(ΔG(t))γ.
IfG(t+1)<G(t)<0, then -G(t+1)>-G(t)>0; let x=-G(t+1), y=-G(t), and p=γ; then x>y>0; by (17), we obtain(18)-Gt+1γ--Gtγ≥-Gt+1+Gtγ.Since γ≥1 is a quotient of two odd positive integers, then(19)-Gt+1γ--Gtγ=-Gγt+1+Gγt=-ΔGγt,-Gt+1+Gtγ=-ΔGtγ=-ΔGtγ.Substituting (19) into (18), we have -ΔGγ(t)≥-(ΔG(t))γ, which means ΔGγ(t)≤(ΔG(t))γ.Lemma 6.
Leta>0, b,X∈R; then bX-aX2≤b2/4a.Proof.
(20)
b
X
-
a
X
2
=
b
2
4
a
-
a
X
-
b
2
a
2
≤
b
2
4
a
.We define the following sequence:(21)ut=∏s=t0t-1cscs-ps;then(22)ut>0,ut+1=ctct-ptut,Δut=ut+1-ut=ptct-ptut.


## 3. Main Results

Lemma 7.
Assume thatx(t) is an eventually positive solution of (6) and(23)∑s=t0∞1uscs1/γ=∞,(24)∑s=t0∞1rs=∞,(25)∑ξ=t0∞1rξ∑τ=ξ∞1uτcτ∑s=τ∞us+1qs1/γ=∞;then, there exists a sufficiently large T∈Nt0 such that ΔrtΔαxt>0 on [T,∞) and one of the following two conditions holds: (i) Δαx(t)>0 on [T,∞) and (ii) Δαx(t)<0 on [T,∞) and limt→∞⁡G(t)=0.Proof.
Sincex(t) is an eventually positive solution of (6), then there exists a sufficiently large t1 such that x(t)>0,t∈[t1,∞). So G(t)>0,t∈[t1,∞). Noting assumption (B), from (6) we obtain(26)ΔctΔrtΔαxtγ+ptΔrtΔαxtγ=-qtfGt≤-LqtGγt.Therefore, it follows from the definition of u(t) and the product rule (11) of the difference operator Δ that(27)ΔutctΔrtΔαxtγ=ut+1ΔctΔrtΔαxtγ+Δut·ctΔrtΔαxtγ=ut+1ΔctΔrtΔαxtγ+ptct-ptct-ptctut+1ctΔrtΔαxtγ=ut+1ΔctΔrtΔαxtγ+ptΔrtΔαxtγ≤-Lut+1qtGγt<0.Then, u(t)c(t)[Δ(r(t)Δαx(t))]γ is strictly decreasing on [t1,∞), and thus Δ(r(t)Δαx(t)) is eventually of one sign. For t2>t1 is sufficiently large, we claim that Δ(r(t)Δαx(t))>0 on [t2,∞). Otherwise, assume that there exists a sufficiently large t3>t2 such that Δ(r(t3)Δαx(t3))<0; then, for t∈(t3,∞), we get(28)utctΔrtΔαxtγ<ut3ct3Δrt3Δαxt3γ=K<0,that is,(29)ΔrtΔαxt<Kutct1/γ<0,t∈t3,∞;So we can get Δ(r(t)Δαx(t))<0 on [t3,∞). From these terms, for t∈[t3,∞), we have(30)rtΔαxt-rt3Δαxt3=∑s=t3t-1uscs1/γΔrsΔαxsuscs1/γ≤ut3ct31/γΔrt3Δαxt3∑s=t3t-11uscs1/γ.By (23), we obtain limt→∞⁡r(t)Δαx(t)=-∞, which means, for some sufficiently large t4>t3, Δαx(t)<0 on [t4,∞). By Lemma 3, we have(31)Gt-Gt4=∑s=t4t-1ΔGs=Γ1-α∑s=t4t-1Δαxs=Γ1-α∑s=t4t-1rsΔαxsrs≤Γ1-αrt4Δαxt4∑s=t4t-11rs.By (24), we obtain limt→∞⁡G(t)=-∞, which contradicts G(t)>0,t∈[t1,∞). Therefore, ΔrtΔαxt>0,t∈[t2,∞). Thus, Δαx(t) is eventually of one sign. There are two possibilities: (i) Δαx(t)>0 on [T,∞) and (ii) Δαx(t)<0 on [T,∞), where T is sufficiently large.
Now, we assume thatΔαx(t)<0,t∈[t5,∞), where t5>t4 is sufficiently large. Then, by Lemma 3, we have ΔG(t)=Γ(1-α)Δαx(t)<0,t∈[t5,∞). Since G(t)>0,t∈[t1,∞), we have limt→∞⁡G(t)=β≥0. We claim that β=0. Otherwise, assume that β>0. Then, G(t)≥β,t∈[t5,∞). By (27), we have(32)ΔutctΔrtΔαxtγ≤-Lut+1qtGγt≤-Lut+1qtβγ.Substituting t with s in (32), a summation for (32) with respect to s from t to ∞ yields(33)∑s=t∞ΔuscsΔrsΔαxsγ≤-Lβγ∑s=t∞us+1qs,which implies(34)-utctΔrtΔαxtγ≤-lims→∞⁡uscsΔrsΔαxsγ-Lβγ∑s=t∞us+1qs<-Lβγ∑s=t∞us+1qs;therefore,(35)ΔrtΔαxt>L1/γβ1utct∑s=t∞us+1qs1/γ.Substituting t with τ in (35), a summation for (35) with respect to τ from t to ∞ yields(36)∑τ=t∞ΔrτΔαxτ>L1/γβ∑τ=t∞1uτcτ∑s=τ∞us+1qs1/γ;namely,(37)-rtΔαxt>-limτ→∞⁡rτΔαxτ+L1γβ∑τ=t∞1uτcτ∑s=τ∞us+1qs1γ>L1γβ∑τ=t∞1uτcτ∑s=τ∞us+1qs1γ;therefore,(38)Δαxt<-L1γβ1rt∑τ=t∞1uτcτ∑s=τ∞us+1qs1γ;that is,(39)ΔGt<-Γ1-αL1/γβ1rt∑τ=t∞1uτcτ∑s=τ∞us+1qs1/γ.Substituting t with ξ in (39), a summation for (39) with respect to ξ from t5 to t-1 yields(40)∑ξ=t5t-1ΔGξ<-Γ1-αL1γβ∑ξ=t5t-11rξ∑τ=ξ∞1uτcτ∑s=τ∞us+1qs1γ;then(41)Gt-Gt5<-Γ1-αL1/γβ∑ξ=t5t-11rξ∑τ=ξ∞1uτcτ∑s=τ∞us+1qs1/γ.By (25), it follows from (41) that limt→∞⁡G(t)=-∞, which contradicts G(t)>0,t∈[t1,∞). Then we get that β=0, which is limt→∞⁡G(t)=0. This completes the proof of Lemma 7.By the same proof as above, ifx(t) is an eventually negative solution of (6), we can obtain Δ(u(t)c(t)[Δ(r(t)Δαx(t))]γ)>0 and Lemma 8 holds.Lemma 8.
Assume thatx(t) is an eventually negative solution of (6) and (23)–(25) hold. Then, there exists a sufficiently large T∈Nt0 such that Δ(r(t)Δαx(t))<0 on [T,∞) and one of the following two conditions holds: (i) Δαx(t)<0 on [T,∞) and (ii) Δαx(t)>0 on [T,∞) and limt→∞⁡G(t)=0.Lemma 9.
Assume thatx(t) is an eventually positive solution of (6) such that Δ(r(t)Δαx(t))>0, Δαx(t)>0 on [t1,∞), where t1≥t0 is sufficiently large. Then(42)ΔGt≥Γ1-αδt,t1utct1/γΔrtΔαxtrt.Proof.
Assume thatxt is an eventually positive solution of (6); then we obtain that u(t)c(t)[Δ(r(t)Δαx(t))]γ is strictly decreasing on [t1,∞); by (27) we have,(43)rtΔαxt≥rtΔαxt-rt1Δαxt1=∑s=t1t-1uscs1/γΔrsΔαxsuscs1/γ≥utct1/γΔrtΔαxt∑s=t1t-11uscs1/γ=δt,t1utct1/γΔrtΔαxt.By Lemma 3, we obtain(44)ΔGt≥Γ1-αδt,t1utct1/γΔrtΔαxtrt.Then the proof is complete.With the same proof as that in Lemma9, we can obtain the following.Lemma 10.
Assume thatx(t) is an eventually negative solution of (6) such that Δ(r(t)Δαx(t))<0, Δαx(t)<0 on [t1,∞), where t1≥t0 is sufficiently large. Then(45)ΔGt≤Γ1-αδt,t1utct1/γΔrtΔαxtrt<Γ1-αδt,t1ut+1ct+11/γΔrt+1Δαxt+1rt.Theorem 11.
Assume that (23)–(25) hold. If(46)lim⁡supt→∞⁡∑s=Tt-1Lusqs-Δus24usus+1Rs=∞,where T is sufficiently large, u(t) is defined as in (21), and R(t)=Γ(1-α)δ(t,t1)/r(t)γ, then (6) is oscillatory or satisfies limt→∞⁡G(t)=0.Proof.
Suppose to the contrary that (6) has a nonoscillatory solution x(t),  t∈Nt0; then x(t) is either eventually positive or eventually negative.
In the case whenx(t) is eventually positive, we assume that x(t)>0 on [t1,∞), where t1∈Nt0 is sufficiently large; then G(t)>0. By Lemma 7, we obtain Δ(r(t)Δαx(t))>0 on [t2,∞), where t2>t1 is sufficiently large, and either Δαx(t)>0 on [t2,∞) or limt→∞⁡G(t)=0.
IfΔαx(t)>0 on [t2,∞), then the conclusion of Lemma 9 holds.
Define the generalized Riccati function as follows:(47)wt=utctΔrtΔαxtγGγt,t∈t2,∞.It is clear that w(t)>0. By the product rule (12) and the quotient rule (13), for t∈[t2,∞), we have(48)Δwt=Δut·wt+1ut+1+utΔctΔrtΔαxtγGγt=Δut·wt+1ut+1+utΔctΔrtΔαxtγ·Gγt-ctΔrtΔαxtγΔGγtGγtGγt+1=Δut·wt+1ut+1+ut-ptΔrtΔαxtγ-qtfGtGγt+1-utctΔrtΔαxtγΔGγtGγtGγt+1=Δutut+1wt+1-utptΔrtΔαxtγGγt+1-utqtfGtGγt+1-utctΔrtΔαxtγΔGγtGγtGγt+1.From Δ(r(t)Δαx(t))>0 and Lemma 3, we have(49)utptΔrtΔαxtγGγt+1>0,and ΔGt>0; then G(t+1)>G(t)>0; it follows from (B) that(50)fGtGγt+1>fGt+1Gγt+1≥L,and ΔGγ(t)≥(ΔG(t))γ by (15).
Using Lemma9 and the fact that u(t)c(t)[Δ(r(t)Δαx(t))]γ is strictly decreasing, we have(51)utctΔrtΔαxtγΔGγtGγtGγt+1≥utctΔrtΔαxtγΔGtγGγtGγt+1≥utctΔrtΔαxtγΓ1-αδt,t1/rtγutctΔrtΔαxtγGγtGγt+1>utctΔrtΔαxtγ2RtG2γt+1>ut+1ct+1Δrt+1Δαxt+1γ2RtG2γt+1=Rtw2t+1.Now substituting (49), (50), and (51) into (48), we obtain(52)Δwt<Δutut+1wt+1-Lutqt-Rtw2t+1.Taking a=R(t)>0, b=Δu(t)/u(t+1), and X=w(t+1) in (52), using Lemma 6, we obtain(53)Δwt<-Lutqt+Δut24Rtut+12.Substituting t with s in (53), a summation for (53) with respect to s from t2 to t-1 yields(54)∑s=t2t-1Lusqs-Δus24Rsus+12<-∑s=t2t-1Δws=wt2-wt<wt2<∞.Since u(s+1)>u(s)>0, it is clear that(55)Δus24usus+1Rs>Δus24us+12Rs;therefore,(56)∑s=t2t-1Lusqs-Δus24us+12Rs>∑s=t2t-1Lusqs-Δus24usus+1Rs,(57)lim⁡supt→∞⁡∑s=t2t-1Lusqs-Δus24us+12Rs≥lim⁡supt→∞⁡∑s=t2t-1Lusqs-Δus24usus+1Rs.From (46) and (57), we obtain that(58)lim⁡supt→∞⁡∑s=t2t-1Lusqs-Δus24us+12Rs=∞.Taking lim⁡sup in (54) as t→∞, we have(59)lim⁡supt→∞⁡∑s=t2t-1Lusqs-Δus24Rsus+12≤wt2<∞,which contradicts (58).
IfΔαx(t)<0 on [t2,∞), then, from Lemma 7, we get that limt→∞⁡G(t)=0.
In the case whenx(t) is eventually negative, we assume that x(t)<0 on [t1,∞), where t1∈Nt0 is sufficiently large; then G(t)<0. By Lemma 8, we obtain ΔrtΔαxt<0 on [t2,∞), where t2>t1 is sufficiently large, and either Δαx(t)<0 on [t2,∞) or limt→∞⁡G(t)=0.
IfΔαx(t)<0 on [t2,∞), then the conclusion of Lemma 10 holds.
Definew(t) as in (47); since ΔrtΔαxt<0, G(t)<0, then w(t)>0.
Using the product rule (12) and the quotient rule (14), for t∈[t2,∞), we get(60)Δwt=Δutut+1wt+1-utptΔrtΔαxtγGγt-utqtfGtGγt-utct+1Δrt+1Δαxt+1γΔGγtGγtGγt+1.By Δ(r(t)Δαx(t))<0 and G(t)<0, we have(61)utptΔrtΔαxtγGγt>0,and ΔG(t)<0 by Lemma 3; then G(t+1)<G(t)<0; from (B) and (16), we obtain that(62)fGtGγt≥L,(63)ΔGγt≤ΔGtγ.Proceeding the proof of Lemma 7, we have(64)ΔutctΔrtΔαxtγ≥-Lut+1qtGγt>0,which implies u(t)c(t)[Δ(r(t)Δαx(t))]γ is strictly increasing. By virtue of Lemma 10, ΔGγ(t)≤(ΔG(t))γ, and(65)utct+1Δrt+1Δαxt+1γGγtGγt+1<0,we have(66)utct+1Δrt+1Δαxt+1γΔGγtGγtGγt+1≥utct+1Δrt+1Δαxt+1γΔGtγGγtGγt+1>utut+1ct+1Δrt+1Δαxt+1γ2Γ1-αδt,t1/rtγGγtGγt+1=utut+1ct+1Δrt+1Δαxt+1γ2RtGγtGγt+1=utG2γt+1Rtut+1GγtGγt+1w2t+1>utRtut+1w2t+1,where Gγ(t+1)<Gγ(t)<0 and Gγ(t+1)/Gγ(t)>1.
Combining (60), (61), (62), and (66), we can obtain(67)Δwt<Δutut+1wt+1-Lutqt-utRtut+1w2t+1.Taking(68)a=utRtut+1>0,b=Δutut+1,X=wt+1,using Lemma 6, we have(69)Δwt<-Lutqt+Δut24utut+1Rt.Substituting t with s in (69), a summation for (69) with respect to s from t2 to t-1 yields(70)∑s=t2t-1Lusqs-Δus24usus+1Rs<∑s=t2t-1-Δws=wt2-wt<wt2<∞.Taking lim⁡sup in (70) as t→∞, we have(71)lim⁡supt→∞⁡∑s=t2t-1Lusqs-Δus24usus+1Rs≤wt2<∞,which contradicts (46).
IfΔαx(t)>0 on [t2,∞), then, from Lemma 8, we get that limt→∞⁡G(t)=0.
The proof of Theorem11 is complete.Theorem 12.
Assume that (23)–(25) hold and there exists a positive sequence H(t,s) such that(72)Ht,t=0fort≥t0,Ht,s>0fort>s≥t0,Δ2Ht,s=Ht,s+1-Ht,s<0fort≥s≥t0.
If(73)lim⁡supt→∞⁡1Ht,t0∑s=t0t-1LusqsHt,s-h2t,sus+14Ht,susRs=∞,where ht,s=Δ2Ht,s+Ht,sΔus/us+1 and ut and Rt are the same as in Theorem 11, then (6) is oscillatory or satisfies limt→∞⁡Gt=0.Proof.
Suppose on the contrary thatx(t) is a nonoscillatory solution of (6); then x(t) is either eventually positive or eventually negative.
In the case whenx(t) is eventually positive, we assume that x(t)>0 on [t1,∞), where t1∈Nt0 is sufficiently large. According to the proof of Theorem 11, if Δαx(t)>0 on [t2,∞), then (52) holds.
Substitutingt with s in (52), multiplying both sides by H(t,s), and then summing with respect to s from t2 to t-1 yield(74)∑s=t2t-1LusqsHt,s<-∑s=t2t-1Ht,sΔws+∑s=t2t-1Ht,sΔusus+1ws+1-∑s=t2t-1Ht,sRsw2s+1.Using summation by parts formula, we obtain(75)-∑s=t2t-1Ht,sΔws=-Ht,swss=t2t+∑s=t2t-1ws+1Δ2Ht,s=Ht,t2wt2+∑s=t2t-1ws+1Δ2Ht,s.Therefore,(76)∑s=t2t-1LusqsHt,s<Ht,t2wt2+∑s=t2t-1Δ2Ht,s+Ht,sΔusus+1ws+1-Ht,sRsw2s+1=Ht,t2wt2+∑s=t2t-1ht,sws+1-Ht,sRsw2s+1.Taking a=H(t,s)R(s)>0,  b=h(t,s), and X=w(t+1) in (76), using Lemma 6, we obtain(77)∑s=t2t-1LusqsHt,s<Ht,t2wt2+∑s=t2t-1h2t,s4Ht,sRs,which means(78)∑s=t2t-1LusqsHt,s-h2t,s4Ht,sRs<Ht,t2wt2<Ht,t0wt2,for t>t2>t1>t0. Then(79)∑s=t0t-1LusqsHt,s-h2t,s4Ht,sRs=∑s=t0t2-1LusqsHt,s-h2t,s4Ht,sRs+∑s=t2t-1LusqsHt,s-h2t,s4Ht,sRs<∑s=t0t2-1LusqsHt,s-h2t,s4Ht,sRs+Ht,t0wt2<Ht,t0∑s=t0t2-1Lusqs+Ht,t0wt2,which means(80)1Ht,t0∑s=t0t-1LusqsHt,s-h2t,s4Ht,sRs<∑s=t0t2-1Lusqs+wt2.Since u(s+1)>u(s)>0, it is obvious that(81)h2t,sus+14Ht,susRs>h2t,s4Ht,sRs;therefore,(82)1Ht,t0∑s=t0t-1LusqsHt,s-h2t,s4Ht,sRs>1Ht,t0∑s=t0t-1LusqsHt,s-h2t,sus+14Ht,susRs,(83)lim⁡supt→∞⁡1Ht,t0∑s=t0t-1LusqsHt,s-h2t,s4Ht,sRs≥lim⁡supt→∞⁡1Ht,t0∑s=t0t-1LusqsHt,s-h2t,sus+14Ht,susRs.By (73) and (83), we have(84)lim⁡supt→∞⁡1Ht,t0∑s=t0t-1LusqsHt,s-h2t,s4Ht,sRs=∞.Taking lim⁡sup in (80) as t→∞, we obtain(85)lim⁡supt→∞⁡1Ht,t0∑s=t0t-1LusqsHt,s-h2t,s4Ht,sRs≤∑s=t0t2-1Lusqs+wt2<∞,which contradicts (84).
IfΔαx(t)<0 on [t2,∞), then, from Lemma 7, we have limt→∞⁡G(t)=0.
In the case whenx(t) is eventually negative, it can be proved similarly; here we omit it.
The proof of Theorem12 is complete.


## 4. Applications

In this section, an example is shown to illustrate the validity of the established results above.Example 13.
Consider the following fractional difference equation:(86)Δt2Δt-1Δαxt3+tΔt-1Δαxt3+t2∑s=2t-1+αt-s-1-αxs-1=0,t∈N2,where 0<α<1 and N2={2,3,4,…}. Comparing with (6), we have(87)ct=t2,rt=t-1,pt=t,qt=t2,fx=x-1,γ=3,t0=2,fxxγ=1x4>ε=L>0,x≠0,where ε is a certain positive number.
It is clear that (A) and (B) hold. Moreover, it follows from (21) that(88)ut=∏s=t0t-1cscs-ps=∏s=2t-1s2s2-s=t-1,Δut=ut+1-ut=1.Furthermore,(89)∑s=t0∞1uscs1/γ=∑s=2∞1s-1s21/3>∑s=2∞1s=∞,∑s=t0∞1rs=∑s=2∞s=∞,∑ξ=t0∞1rξ∑τ=ξ∞1uτcτ∑s=τ∞us+1qs1/γ=∑ξ=2∞ξ∑τ=ξ∞1τ-1τ2∑s=τ∞s31/3=∞,which means (23), (24), and (25) hold.
For a sufficiently larget1, we have(90)δt,t1=∑s=t1t-11uscs1/γ=∑s=t1t-11s-1s21/3,Rt=Γ1-αδt,t1rtγ=Γ1-αt∑s=t1t-11s-1s21/33.For t2>t1,(91)∑s=t2t-1Lusqs-Δus24usus+1Rs=∑s=t2t-1εs-1s2-14s-1sΓ1-αs∑ξ=t1s-11/ξ-1ξ21/33=∑s=t2t-1εs-1s2-14s-1s4Γ1-α∑ξ=t1s-11/ξ-1ξ21/33>∑s=t2t-1εs-1s2-14Γ1-α3s-1s41/t1-1t12=∑s=t2t-1εs-1s2-t1-1t124Γ1-α3s-1s4.Hence,(92)lim⁡supt→∞⁡∑s=t2t-1Lusqs-Δus24usus+1Rs≥lim⁡supt→∞⁡∑s=t2t-1εs-1s2-t1-1t124Γ1-α3s-1s4=∞,which implies that condition (46) is satisfied. Therefore, (86) is oscillatory or satisfies limt→∞⁡G(t)=0 by virtue of Theorem 11.



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*Source: 5232147-2018-01-29.xml*