# Asymptotic Stability Results for Nonlinear

Authors: Fulai Chen; Zhigang Liu
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We present some results for the asymptotic
stability of solutions for nonlinear fractional difference equations
involvingRiemann-Liouville-likedifference operator. The
results are obtained by using Krasnoselskii's fixed point theorem
and discrete Arzela-Ascoli's theorem. Three examples are also
provided to illustrate our main results.


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## Body

## 1. Introduction

In this paper we consider the asymptotic stability of solutions for nonlinear fractional difference equations:(1.1)Δαx(t)=f(t+α,x(t+α)),t∈N0,0<α≤1,Δα-1x(t)|t=0=x0,
where Δα is aRiemann-Liouville-likediscrete fractional difference, f:[0,+∞)×R→R is continuous with respect to t and x, Na={a,a+1,a+2,…}.Fractional differential equations have received increasing attention during recent years since these equations have been proved to be valuable tools in the modeling of many phenomena in various fields of science and engineering. Most of the present works were focused on fractional differential equations, see [1–12] and the references therein. However, very little progress has been made to develop the theory of the analogous fractional finite difference equation [13–19].Due to the lack of geometry interpretation of the fractional derivatives, it is difficult to find a valid tool to analyze the stability of fractional difference equations. In the case that it is difficult to employ Liapunov’s direct method, fixed point theorems are usually considered in stability [20–25]. Motivated by this idea, in this paper, we discuss asymptotic stability of nonlinear fractional difference equations by using Krasnoselskii’s fixed point theorem and discrete Arzela-Ascoli’s theorem. Different from our previous work [18], in this paper, the sufficient conditions of attractivity are irrelevant to the initial value x0.


## 2. Preliminaries

In this section, we introduce preliminary facts of discrete fractional calculus. For more details, see [14].Definition 2.1 (see [14]).
Letν>0. The ν-th fractional sum x is defined by
(2.1)Δ-νf(t)=1Γ(ν)∑s=at-ν(t-s-1)(ν-1)f(s),
where f is defined for s=a mod (1) and Δ-νf is defined for t=(a+ν) mod (1), and t(ν)=Γ(t+1)/Γ(t-ν+1). The fractional sum Δ-ν maps functions defined on Na to functions defined on Na+ν.Definition 2.2 (see [14]).
Letμ>0 and m-1<μ<m, where m denotes a positive integer, m=⌈μ⌉, ⌈·⌉ ceiling of number. Set ν=m-μ. The μ-th fractional difference is defined as
(2.2)Δμf(t)=Δm-νf(t)=Δm(Δ-νf(t)).Theorem 2.3 (see [15]).
Letf be a real-value function defined on Na and μ,ν>0, then the following equalities hold:(i)
Δ-ν[Δ-μf(t)]=Δ-(μ+ν)f(t)=Δ-μ[Δ-νf(t)];(ii)
Δ-νΔf(t)=ΔΔ-νf(t)-(t-a)(ν-1)Γ(ν)f(a).Lemma 2.4 (see [15]).
Letμ≠1 and assume μ+ν+1 is not a nonpositive integer, then
(2.3)Δ-νt(μ)=Γ(μ+1)Γ(μ+ν+1)t(μ+ν).Lemma 2.5 (see [15]).
Assume that the following factorial functions are well defined:(i)
If0<α<1, then t(αγ)≥(t(γ))α;(ii)
t(β+γ)=(t-γ)(β)t(γ).Lemma 2.6 (see [13]).
Letμ>0 be noninteger, m=⌈μ⌉, ⌈·⌉, ν=m-μ, thus one has(2.4)∑s=a+νt-μ(t-s-1)(μ-1)=(t-a-ν)(μ)μ.Lemma 2.7.
The equivalent fractional Taylor’s difference formula of (1.1) is
(2.5)x(t)=x0Γ(α)t(α-1)+1Γ(α)∑s=0t-α(t-s-1)(α-1)f(s+α,x(s+α)),t∈Nα.Proof.
Apply theΔ-α operator to each side of the first formula of (1.1) to obtain
(2.6)Δ-αΔαx(t)=Δ-αf(t+α,x(t+α)),t∈Nα.
Apply Theorem2.3 to the left-hand side of (2.6) to obtain
(2.7)Δ-αΔαx(t)=Δ-αΔΔ-(1-α)x(t)=ΔΔ-αΔ-(1-α)x(t)-t(α-1)Γ(α)x(α-1)=x(t)-x0Γ(α)t(α-1).
So, applying Definition2.1 to the right-hand side of (2.6), for t∈Nα we obtain (2.5). The recursive iteration to this Taylor’s difference formula implies that (2.5) represents the unique solution of the IVP (1.1). This completes the proof.Lemma 2.8 (see [4, (1.5.15)]).
The quotient expansion of two gamma functions at infinityis(2.8)Γ(z+a)Γ(z+b)=za-b[1+O(1z)],(|arg(z+a)|<π,|z|⟶∞).Corollary 2.9.
One has(2.9)t(-β)>(t+α)(-β)forα,β,t>0.Proof.
According to Lemma2.8,
(2.10)t(-β)(t+α)(-β)=Γ(t+1)Γ(t+β+1)⋅Γ(t+α+β+1)Γ(t+α+1)=Γ(t+1)Γ(t+α+1)⋅Γ(t+α+β+1)Γ(t+β+1)=t-α[1+O(1t)]⋅(t+β)α[1+O(1t+β)]=(1+βt)α[1+O(1t)][1+O(1t+β)]>1.
Then, t(-β)>(t+α)(-β) for α,β,t>0. This completes the proof.Definition 2.10.
The solutionx=φ(t) of the IVP (1.1) is said to be(i)
stable if for anyɛ>0 and t0∈R+, there exists a δ=δ(t0,ɛ)>0 such that(2.11)|x(t,x0,t0)-φ(t)|<ɛ
for |x0-φ(t0)|≤δ(t0,ɛ) and all t≥t0;(ii)
attractive if there existsη(t0)>0 such that ∥x0∥≤η implies(2.12)lim⁡t→∞x(t,x0,t0)=0;(iii)
asymptotically stable if it is stable and attractive.
The spaceln0∞ is the set of real sequences defined on the set of positive integers where any individual sequence is bounded with respect to the usual supremum norm. It is well known that under the supremum norm ln0∞ is a Banach space [26].Definition 2.11 (see [27]).
A setΩ of sequences in ln0∞ is uniformly Cauchy (or equi-Cauchy), if for every ɛ>0, there exists an integer N such that |x(i)-x(j)|<ɛ, whenever i,j>N for any x={x(n)} in Ω.Theorem 2.12 (see [27, (discrete Arzela-Ascoli’s theorem)]).
A bounded, uniformly Cauchy subsetΩ of ln0∞ is relatively compact.Theorem 2.13 (see [20, (Krasnoselskii’s fixed point theorem)]).
LetS be a nonempty, closed, convex, and bounded subset of the Banach space X and let A:X→X and B:S→X be two operators such that(a)
A  is a contraction with constantL<1,(b)
B  is continuous,BS resides in a compact subset of X,(c)
[x=Ax+By,y∈S]⇒x∈S.Then the operator equation Ax+Bx=x has a solution in S.


## 3. Main Results

Letlα∞ be the set ofall real sequencesx={x(t)}t=α∞ with norm ∥x∥=supt∈Nα|x(t)|, then lα∞ is a Banach space.Define the operator(3.1)Px(t)=x0Γ(α)t(α-1)+1Γ(α)∑s=0t-α(t-s-1)(α-1)f(s+α,x(s+α)),Ax(t)=x0Γ(α)t(α-1),Bx(t)=1Γ(α)∑s=0t-α(t-s-1)(α-1)f(s+α,x(s+α)),t∈Nα.
Obviously, Px=Ax+Bx, the operator A is a contraction with the constant 0, which implies that condition (a) of Theorem 2.13 holds, and x(t) is a solution of (1.1) if it is a fixed point of the operator P.Lemma 3.1.
Assume that the following condition is satisfied:
(H1) there exist constants β1∈(α,1) and L1≥0 such that
(3.2)|f(t,x(t))|≤L1t(-β1)fort∈Nα.
Then the operator B is continuous and BS1 is a compact subset of R for t∈Nα+n1, where
(3.3)S1={x(t):|x(t)|≤t(-γ1)fort∈Nα+n1},γ1=(-1/2)(α-β1), and n1∈N satisfies that
(3.4)|x0|Γ(α)(α+n1+γ1)((1/2)(α+β1)-1)+L1Γ(1-β1)Γ(1+α-β1)(α+n1+γ1)(-γ1)≤1.Proof.
Fort∈Nα, apply Lemma 2.8 and γ1>0,
(3.5)t(-γ1)=Γ(t+1)Γ(t+γ1+1)=t-γ1[1+O(1t)],
and we have that t(-γ1)→0 as t→∞, then there exists a n1∈N such that inequality (3.4) holds, which implies that the set S1 exists.
We firstly show thatB maps S1 in S1.
It is easy to know thatS1 is a closed, bounded, and convex subset of R.
Apply condition(H1), Lemma 2.5, Corollary 2.9 and (3.4), for t∈Nα+n1, we have
(3.6)|Bx(t)|≤1Γ(α)∑s=0t-α(t-s-1)(α-1)|f(s+α,x(s+α))|≤1Γ(α)∑s=0t-α(t-s-1)(α-1)L1(s+α)(-β1)=L1Δ-α(t+α)(-β1)=L1Γ(1-β1)Γ(1+α-β1)(t+α)(α-β1)<L1Γ(1-β1)Γ(1+α-β1)t(α-β1)=L1Γ(1-β1)Γ(1+α-β1)(t+γ1)(-γ1)t(-γ1)≤L1Γ(1-β1)Γ(1+α-β1)(α+n1+γ1)(-γ1)t(-γ1)≤t(-γ1),
which implies that BS1⊂S1 for t∈Nα+n1.
Nextly, we show thatB is continuous on S1.
Letɛ>0 be given then there exist T1∈N and T1≥n1 such that t∈Nα+T1 implies that
(3.7)L1Γ(1-β1)Γ(1+α-β1)t(α-β1)<ɛ2.
Let{xn} be a sequence such that xn→x. For t∈{α+n1,α+n1+1,…,α+T1-1}, applying the continuity of f and Lemma 2.6, we have
(3.8)|Bxn(t)-Bx(t)|≤1Γ(α)∑s=0t-α(t-s-1)(α-1)|f(s+α,xn(s+α))-f(s+α,x(s+α))|≤max⁡s∈{0,1,…,T1-1}|f(s+α,xn(s+α))-f(s+α,x(s+α))|×1Γ(α)∑s=0t-α(t-s-1)(α-1)=t(α)Γ(α+1)max⁡s∈{0,1,…,T1-1}|f(s+α,xn(s+α))-f(s+α,x(s+α))|≤(α+T1-1)(α)Γ(α+1)max⁡s∈{0,1,…,T1-1}|f(s+α,xn(s+α))-f(s+α,x(s+α))|⟶0asn⟶∞.
Fort∈Nα+T1,
(3.9)|Bxn(t)-Bx(t)|≤1Γ(α)∑s=0t-α(t-s-1)(α-1)[|f(s+α,xn(s+α))|+|f(s+α,x(s+α))|]≤2L1Γ(α)∑s=0t-α(t-s-1)(α-1)(s+α)(-β1)=2L1Δ-α(t+α)(-β1)=2L1Γ(1-β1)Γ(1+α-β1)(t+α)(α-β1)<2L1Γ(1-β1)Γ(1+α-β1)t(α-β1)<ɛ.
Thus, for allt∈Nα+n1, we have
(3.10)|Bxn(t)-Bx(t)|⟶0asn→∞.
which implies that B is continuous.
Lastly, we show thatBS1 is relatively compact.
Lett1,t2∈Nα+T1 and t2>t1, thus we have
(3.11)|Bx(t2)-Bx(t1)|=|1Γ(α)∑s=0t2-α(t2-s-1)(α-1)f(s+α,x(s+α))-1Γ(α)∑s=0t1-α(t1-s-1)(α-1)f(s+α,x(s+α))|≤1Γ(α)∑s=0t2-α(t2-s-1)(α-1)|f(s+α,x(s+α))|+1Γ(α)∑s=0t1-α(t1-s-1)(α-1)|f(s+α,x(s+α))|≤L1Γ(1-β1)Γ(1+α-β1)(t2+α)(α-β1)+L1Γ(1-β1)Γ(1+α-β1)(t1+α)(α-β1)<L1Γ(1-β1)Γ(1+α-β1)t2(α-β1)+L1Γ(1-β1)Γ(1+α-β1)t1(α-β1)<ɛ.
Thus, {Bx:x∈S1} is a bounded and uniformly Cauchy subset by Definition 2.11, and BS1 is relatively compact by means of Theorem 2.12. This completes the proof.Lemma 3.2.
Assume that condition(H1) holds, then a solution of (1.1) is in S1 for t∈Nα+n1.Proof.
Notice if thatx(t) is a fixed point of P, then it is a solution of (1.1). To prove this, it remains to show that, for fixed y∈S1, x=Ax+By⇒x∈S1 holds.
Ifx=Ax+By, applying condition (H1) and (3.4), for t∈Nα+n1, we have
(3.12)|x(t)|≤|Ax(t)|+|By(t)|≤|x0|Γ(α)t(α-1)+1Γ(α)∑s=0t-α(t-s-1)(α-1)|f(s+α,y(s+α))|≤|x0|Γ(α)t(α-1)+L1Γ(1-β1)Γ(1+α-β1)(t+α)(α-β1)<|x0|Γ(α)t(α-1)+L1Γ(1-β1)Γ(1+α-β1)t(α-β1)=[|x0|Γ(α)(t+γ1)((1/2)(α+β1)-1)+L1Γ(1-β1)Γ(1+α-β1)(t+γ1)(-γ1)]t(-γ1)≤[|x0|Γ(α)(α+n1+γ1)((1/2)(α+β1)-1)+L1Γ(1-β1)Γ(1+α-β1)(α+n1+γ1)(-γ1)]t(-γ1)≤t(-γ1).
Thus, x(t)∈S1 for t∈Nα+n1. According to Theorem 2.13 and Lemma 3.1, there exists a x∈S1 such that x=Ax+Bx, that is, P has a fixed point in S1 which is a solution of (1.1) for t∈Nα+n1. This completes the proof.Theorem 3.3.
Assume that condition(H1) holds, then the solutions of (1.1) is attractive.Proof.
By Lemma3.2, the solutions of (1.1) exist and are in S1. All functions x(t) in S1 tend to 0 as t→∞. Then the solutions of (1.1) tend to zero as t→∞. This completes the proof.Theorem 3.4.
Assume that the following condition is satisfied:
(H2) there exist constants β2∈(α,1) and L2≥0 such that
(3.13)|f(t,x(t))-f(t,y(t))|≤L2t(-β2)‖x-y‖fort∈Nα.
Then the solutions of (1.1) are stable provided that
(3.14)c:=L2Γ(1+α)Γ(1-β2)Γ(1+α-β2)Γ(1+β2)<1.Proof.
Letx(t) be a solution of (1.1), and let x̃(t) be a solution of (1.1) satisfying the initial value condition x̃(0)=x̃0. For t∈Nα, applying condition (H2), we have
(3.15)|x(t)-x̃(t)|≤t(α-1)Γ(α)|x0-x̃0|+1Γ(α)∑s=0t-α(t-s-1)(α-1)×|f(s+α,x(s+α))-f(s+α,x̃(s+α))|≤t(α-1)Γ(α)|x0-x̃0|+L2Γ(α)∑s=0t-α(t-s-1)(α-1)(s+α)(-β2)‖x-x̃‖=t(α-1)Γ(α)|x0-x̃0|+L2Δ-α(t+α)(-β2)‖x-x̃‖=t(α-1)Γ(α)|x0-x̃0|+L2Γ(1-β2)Γ(1+α-β2)(t+α)(α-β2)‖x-x̃‖≤α(α-1)Γ(α)|x0-x̃0|+L2Γ(1-β2)Γ(1+α-β2)α(α-β2)‖x-x̃‖=α|x0-x̃0|+L2Γ(1+α)Γ(1-β2)Γ(1+α-β2)Γ(1+β2)‖x-x̃‖=α|x0-x̃0|+c‖x-x̃‖,
which implies that
(3.16)‖x-x̃‖≤α1-c|x0-x̃0|.
For any givenɛ>0, let δ=((1-c)/α)ɛ, |x0-x̃0|<δ follows that ∥x-x̃∥<ɛ, which yields that the solutions of (1.1) are stable. This completes the proof.Theorem 3.5.
Assume that conditions(H1) and (H2) hold, then the solutions of (1.1) are asymptotically stable provided that (3.14) holds.
Theorem3.5 is the simple consequence of Theorems 3.3 and 3.4.Theorem 3.6.
Assume that the following condition is satisfied:
(H3) there exist constants β3∈(α,(1/2)(1+α)), γ2=(1/2)(1-α), and L3≥0 such that
(3.17)|f(t,x(t))|≤L3(t+γ2)(-β3)|x(t)|fort∈Nα.
Then the solutions of (1.1) is attractive.Proof.
Set(3.18)S2={x(t):|x(t)|≤t(-γ2)fort∈Nα+n2},
where n2∈N satisfies that
(3.19)|x0|Γ(α)(α+n2+γ2)(-γ2)+L3Γ(1-β3-γ2)Γ(1+α-β3-γ2)(α+n2+γ2)(α-β3)≤1.
We first prove condition(c) of Theorem 2.13, that is, for fixed y∈S2 and for all x∈R, x=Ax+By⇒x∈S2 holds.
Ifx=Ax+By, applying condition (H3) and (3.19), for t∈Nα+n2, we have
(3.20)|x(t)|≤|Ax(t)|+|By(t)|≤|x0|Γ(α)t(α-1)+1Γ(α)∑s=0t-α(t-s-1)(α-1)|f(s+α,y(s+α))|≤|x0|Γ(α)t(α-1)+1Γ(α)∑s=0t-α(t-s-1)(α-1)L3(s+α+γ2)(-β3)|y(s+α)|≤|x0|Γ(α)t(α-1)+L3Γ(α)∑s=0t-α(t-s-1)(α-1)(s+α+γ2)(-β3)(s+α)(-γ2)≤|x0|Γ(α)t(α-1)+L3Γ(α)∑s=0t-α(t-s-1)(α-1)(s+α)(-β3-γ2)≤|x0|Γ(α)t(α-1)+L3Γ(1-β3-γ2)Γ(1+α-β3-γ2)(t+α)(α-β3-γ2)<|x0|Γ(α)t(α-1)+L3Γ(1-β3-γ2)Γ(1+α-β3-γ2)t(α-β3-γ2)=[|x0|Γ(α)(t+γ2)(-γ2)+L3Γ(1-β3-γ2)Γ(1+α-β3-γ2)(t+γ2)(α-β3)]t(-γ2)≤[|x0|Γ(α)(α+n2+γ2)(-γ2)+L3Γ(1-β3-γ2)Γ(1+α-β3-γ2)(α+n2+γ2)(α-β3)]t(-γ2)≤t(-γ2).
Thus, condition (c) of Theorem 2.13 holds.
The proof of condition(b) of Theorem 2.13 is similar to that of Lemma 3.1, and we omit it. Therefore, P has a fixed point in S2 by using Theorem 2.13, that is, the IVP (1.1) has a solution in S2. Moreover, all functions in S2 tend to 0 as t→∞, then the solution of (1.1) tends to zero as t→∞, which shows that the zero solution of (1.1) is attractive. This completes the proof.Theorem 3.7.
Assume that conditions(H2) and (H3) hold, then the solutions of (1.1) are asymptotically stable provided that (3.14) holds.Theorem 3.8.
Assume that the following condition is satisfied:
(H4) there exist constants η∈(0,1),β4∈(α,(2+αη)/(2+η)), and L4≥0 such that
(3.21)|f(t,x(t))|≤L4(t+1)(-β4)|x(t)|ηfort∈Nα.
Then the solutions of (1.1) is attractive.Proof.
Set(3.22)S3={x(t):|x(t)|≤t(-γ3)fort∈Nα+n3},
where γ3=(1/2)(β4-α), and n3∈N satisfies that
(3.23)|x0|Γ(α)(α+n3+γ3)(α-1+γ3)+L4Γ(1-β4-γ3η)Γ(1+α-β4-γ3η)(α+n3+γ3)-γ3≤1.
Here we only prove that condition(c) of Theorem 2.13 holds, and the remaining part of the proof is similar to that of Theorem 3.6.
Sinceη∈(0,1),β4∈(α,(2+αη)/(2+η)), and γ3=(1/2)(β4-α), then γ3,γ3η,α+γ3∈(0,1),β4+γ3η∈(α,1).
Ifx=Ax+By, applying condition (H4), Lemma 2.5 and (3.23), for t∈Nα+n3, we have
(3.24)|x(t)|≤|Ax(t)|+|By(t)|≤|x0|Γ(α)t(α-1)+1Γ(α)∑s=0t-α(t-s-1)(α-1)|f(s+α,y(s+α))|≤|x0|Γ(α)t(α-1)+1Γ(α)∑s=0t-α(t-s-1)(α-1)L4(s+α+1)(-β4)|y(s+α)|η≤|x0|Γ(α)t(α-1)+L4Γ(α)∑s=0t-α(t-s-1)(α-1)(s+α+γ3η)(-β4)[(s+α)(-γ3)]η≤|x0|Γ(α)t(α-1)+L4Γ(α)∑s=0t-α(t-s-1)(α-1)(s+α+γ3η)(-β4)(s+α)(-γ3η)=|x0|Γ(α)t(α-1)+L4Γ(α)∑s=0t-α(t-s-1)(α-1)(s+α)(-β4-γ3η)≤|x0|Γ(α)t(α-1)+L4Γ(1-β4-γ3η)Γ(1+α-β4-γ3η)(t+α)(α-β4-γ3η)<|x0|Γ(α)t(α-1)+L4Γ(1-β4-γ3η)Γ(1+α-β4-γ3η)t(α-β4-γ3η)≤|x0|Γ(α)t(α-1)+L4Γ(1-β4-γ3η)Γ(1+α-β4-γ3η)t(α-β4)=[|x0|Γ(α)(t+γ3)(α-1+γ3)+L4Γ(1-β4-γ3η)Γ(1+α-β4-γ3η)(t+γ3)(-γ3)]t(-γ3)≤[|x0|Γ(α)(α+n3+γ3)(α-1+γ3)+L4Γ(1-β4-γ3η)Γ(1+α-β4-γ3η)(α+n3+γ3)(-γ3)]t(-γ3)≤t(-γ3).
Thus, condition (c) of Theorem 2.13 holds. This completes the proof.


## 4. Examples

Example 4.1.
Consider(4.1)Δ0.5x(t)=0.2(t+0.5)(-0.75)sin⁡(x(t+0.5)),t∈N0,Δ-0.5x(t)|t=0=x0,
where f(t,x(t))=0.2t(-0.75)sin(x(t)), t∈N0.5.
Since(4.2)|f(t,x(t))|=|0.2t(-0.75)sin⁡(x(t))|≤0.2t(-0.75),thisimplies that condition (H1) holds.
In addition,(4.3)|f(t,x(t))-f(t,y(t))|≤0.2t(-0.75)‖x-y‖.
Thus, condition (H2) is satisfied.
Moreover, fromL2=0.2, α=0.5, and β2=0.75, we have
(4.4)c=L2Γ(1+α)Γ(1-β2)Γ(1+α-β2)Γ(1+β2)=0.2Γ(1.5)Γ(0.25)Γ(1.25)Γ(1.75)≈0.7716<1,
which implies that inequality (3.14) holds.
Thus the solutions of (4.1) are asymptotically stable by Theorem 3.5.Example 4.2.
Consider(4.5)Δ0.5x(t)=0.2(t+1.5)(-0.6)x(t+0.5),t∈N0,Δ-0.5x(t)|t=0=x0,
where f(t,x(t))=0.2(t+1)(-0.6)x(t), t∈N0.5.
Sinceβ3=0.6,α=0.5, we have that β3∈(α,(1/2)(1+α)), γ2=0.25 and
(4.6)|f(t,x(t))|=|0.2(t+1)(-0.6)x(t)|≤0.2(t+0.25)(-0.6)|x(t)|,
which implies that condition (H3) is satisfied.
Meanwhile,(4.7)|f(t,x(t))-f(t,y(t))|≤0.2(t+1)(-0.6)‖x-y‖≤0.2t(-0.6)‖x-y‖,
which implies that condition (H2) is satisfied.
FromL2=0.2, α=0.5, and β2=0.6, we have
(4.8)c=L2Γ(1+α)Γ(1-β2)Γ(1+α-β2)Γ(1+β2)=0.2Γ(1.5)Γ(0.4)Γ(0.9)Γ(1.6)≈0.4120<1,
which implies that inequality (3.14) holds.
Thus the solutions of (4.5) are asymptotically stable by Theorem 3.7.Example 4.3.
Consider(4.9)Δ0.5x(t)=(t+1.5)(-0.6)x1/3(t+0.5),t∈N0,Δ-0.5x(t)|t=0=x0,
where f(t,x(t))=(t+1)(-0.6)x1/3(t), t∈N0.5.
Sinceα=0.5,β4=0.6,η=1/3, we have that η∈(0,1),β4∈(α,(2+αη)/(2+η)) and
(4.10)|f(t,x(t))|≤(t+1)(-0.6)|x(t)|1/3,
then condition (H4) is satisfied.
The solutions of (4.9) are attractive by Theorem 3.8.



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*Source: 879657-2012-03-08.xml*