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Journal of Applied Mathematics
Volume 2012 (2012), Article ID 879657, 14 pages
http://dx.doi.org/10.1155/2012/879657
Research Article

Asymptotic Stability Results for Nonlinear Fractional Difference Equations

Department of Mathematics, Xiangnan University, Chenzhou 423000, China

Received 1 August 2011; Revised 27 December 2011; Accepted 2 January 2012

Academic Editor: Michela Redivo-Zaglia

Copyright © 2012 Fulai Chen and Zhigang Liu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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.

1. Introduction

In this paper we consider the asymptotic stability of solutions for nonlinear fractional difference equations:Δ𝛼𝑥(𝑡)=𝑓(𝑡+𝛼,𝑥(𝑡+𝛼)),𝑡𝑁0Δ,0<𝛼1,𝛼1𝑥(𝑡)|𝑡=0=𝑥0,(1.1) where Δ𝛼 is aRiemann-Liouville-likediscrete fractional difference, 𝑓[0,+)×𝑅𝑅 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 [112] and the references therein. However, very little progress has been made to develop the theory of the analogous fractional finite difference equation [1319].

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 [2025]. 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]). Let 𝜈>0. The 𝜈-th fractional sum 𝑥 is defined by Δ𝜈1𝑓(𝑡)=Γ(𝜈)𝑡𝜈𝑠=𝑎(𝑡𝑠1)(𝜈1)𝑓(𝑠),(2.1) where 𝑓 is defined for 𝑠=𝑎 mod (1) and Δ𝜈𝑓 is defined for 𝑡=(𝑎+𝜈) mod (1), and 𝑡(𝜈)=Γ(𝑡+1)/Γ(𝑡𝜈+1). The fractional sum Δ𝜈 maps functions defined on 𝑁𝑎 to functions defined on 𝑁𝑎+𝜈.

Definition 2.2 (see [14]). Let 𝜇>0 and 𝑚1<𝜇<𝑚, where 𝑚 denotes a positive integer, 𝑚=𝜇, ceiling of number. Set 𝜈=𝑚𝜇. The 𝜇-th fractional difference is defined as Δ𝜇𝑓(𝑡)=Δ𝑚𝜈𝑓(𝑡)=Δ𝑚(Δ𝜈𝑓(𝑡)).(2.2)

Theorem 2.3 (see [15]). Let 𝑓 be a real-value function defined on 𝑁𝑎 and 𝜇,𝜈>0, then the following equalities hold:(i)Δ𝜈[Δ𝜇𝑓(𝑡)]=Δ(𝜇+𝜈)𝑓(𝑡)=Δ𝜇[Δ𝜈𝑓(𝑡)]; (ii)Δ𝜈Δ𝑓(𝑡)=ΔΔ𝜈𝑓(𝑡)(𝑡𝑎)(𝜈1)Γ(𝜈)𝑓(𝑎).

Lemma 2.4 (see [15]). Let 𝜇1 and assume 𝜇+𝜈+1 is not a nonpositive integer, then Δ𝜈𝑡(𝜇)=Γ(𝜇+1)𝑡Γ(𝜇+𝜈+1)(𝜇+𝜈).(2.3)

Lemma 2.5 (see [15]). Assume that the following factorial functions are well defined:(i)If 0<𝛼<1, then 𝑡(𝛼𝛾)(𝑡(𝛾))𝛼;(ii)𝑡(𝛽+𝛾)=(𝑡𝛾)(𝛽)𝑡(𝛾).

Lemma 2.6 (see [13]). Let 𝜇>0 be noninteger, 𝑚=𝜇, , 𝜈=𝑚𝜇, thus one has𝑡𝜇𝑠=𝑎+𝜈(𝑡𝑠1)(𝜇1)=(𝑡𝑎𝜈)(𝜇)𝜇.(2.4)

Lemma 2.7. The equivalent fractional Taylor’s difference formula of (1.1) is 𝑥𝑥(𝑡)=0𝑡Γ(𝛼)(𝛼1)+1Γ(𝛼)𝑡𝛼𝑠=0(𝑡𝑠1)(𝛼1)𝑓(𝑠+𝛼,𝑥(𝑠+𝛼)),𝑡𝑁𝛼.(2.5)

Proof. Apply the Δ𝛼 operator to each side of the first formula of (1.1) to obtain Δ𝛼Δ𝛼𝑥(𝑡)=Δ𝛼𝑓(𝑡+𝛼,𝑥(𝑡+𝛼)),𝑡𝑁𝛼.(2.6)
Apply Theorem 2.3 to the left-hand side of (2.6) to obtain Δ𝛼Δ𝛼𝑥(𝑡)=Δ𝛼ΔΔ(1𝛼)𝑥(𝑡)=ΔΔ𝛼Δ(1𝛼)𝑡𝑥(𝑡)(𝛼1)𝑥Γ(𝛼)𝑥(𝛼1)=𝑥(𝑡)0𝑡Γ(𝛼)(𝛼1).(2.7)
So, applying Definition 2.1 to the right-hand side of (2.6), for 𝑡𝑁𝛼 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 infinityisΓ(𝑧+𝑎)Γ(𝑧+𝑏)=𝑧𝑎𝑏11+𝑂𝑧,||||.arg(𝑧+𝑎)<𝜋,|𝑧|(2.8)

Corollary 2.9. One has 𝑡(𝛽)>(𝑡+𝛼)(𝛽)for𝛼,𝛽,𝑡>0.(2.9)

Proof. According to Lemma 2.8, 𝑡(𝛽)(𝑡+𝛼)(𝛽)=Γ(𝑡+1)Γ(𝑡+𝛽+1)Γ(𝑡+𝛼+𝛽+1)=Γ(𝑡+𝛼+1)Γ(𝑡+1)ΓΓ(𝑡+𝛼+1)(𝑡+𝛼+𝛽+1)Γ(𝑡+𝛽+1)=𝑡𝛼11+𝑂𝑡(𝑡+𝛽)𝛼11+𝑂=𝛽𝑡+𝛽1+𝑡𝛼11+𝑂𝑡11+𝑂𝑡+𝛽>1.(2.10) Then, 𝑡(𝛽)>(𝑡+𝛼)(𝛽) for 𝛼,𝛽,𝑡>0. This completes the proof.

Definition 2.10. The solution 𝑥=𝜑(𝑡) of the IVP (1.1) is said to be(i) stable if for any 𝜀>0 and 𝑡0𝑅+, there exists a 𝛿=𝛿(𝑡0,𝜀)>0 such that||𝑥𝑡,𝑥0,𝑡0||𝜑(𝑡)<𝜀(2.11) for |𝑥0𝜑(𝑡0)|𝛿(𝑡0,𝜀) and all 𝑡𝑡0;(ii) attractive if there exists 𝜂(𝑡0)>0 such that 𝑥0𝜂 implieslim𝑡𝑥𝑡,𝑥0,𝑡0=0;(2.12)(iii) asymptotically stable if it is stable and attractive.
The space 𝑙𝑛0 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 set Ω of sequences in 𝑙𝑛0 is uniformly Cauchy (or equi-Cauchy), if for every 𝜀>0, there exists an integer 𝑁 such that |𝑥(𝑖)𝑥(𝑗)|<𝜀, 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 subset Ω of 𝑙𝑛0 is relatively compact.

Theorem 2.13 (see [20, (Krasnoselskii’s fixed point theorem)]). Let 𝑆 be a nonempty, closed, convex, and bounded subset of the Banach space 𝑋 and let 𝐴𝑋𝑋 and 𝐵𝑆𝑋 be two operators such that(a)A is a contraction with constant 𝐿<1,(b)B is continuous, 𝐵𝑆 resides in a compact subset of 𝑋,(c)[𝑥=𝐴𝑥+𝐵𝑦,𝑦𝑆]𝑥𝑆.Then the operator equation 𝐴𝑥+𝐵𝑥=𝑥 has a solution in 𝑆.

3. Main Results

Let 𝑙𝛼 be the set ofall real sequences𝑥={𝑥(𝑡)}𝑡=𝛼 with norm 𝑥=sup𝑡𝑁𝛼|𝑥(𝑡)|, then 𝑙𝛼 is a Banach space.

Define the operator𝑥𝑃𝑥(𝑡)=0𝑡Γ(𝛼)(𝛼1)+1Γ(𝛼)𝑡𝛼𝑠=0(𝑡𝑠1)(𝛼1)𝑥𝑓(𝑠+𝛼,𝑥(𝑠+𝛼)),𝐴𝑥(𝑡)=0𝑡Γ(𝛼)(𝛼1),1𝐵𝑥(𝑡)=Γ(𝛼)𝑡𝛼𝑠=0(𝑡𝑠1)(𝛼1)𝑓(𝑠+𝛼,𝑥(𝑠+𝛼)),𝑡𝑁𝛼.(3.1) 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 𝐿10 such that ||||𝑓(𝑡,𝑥(𝑡))𝐿1𝑡(𝛽1)for𝑡𝑁𝛼.(3.2) Then the operator 𝐵 is continuous and 𝐵𝑆1 is a compact subset of 𝑅 for 𝑡𝑁𝛼+𝑛1, where 𝑆1=||||𝑥(𝑡)𝑥(𝑡)𝑡(𝛾1)for𝑡𝑁𝛼+𝑛1,(3.3)𝛾1=(1/2)(𝛼𝛽1), and 𝑛1𝑁 satisfies that ||𝑥0||Γ(𝛼)𝛼+𝑛1+𝛾1((1/2)(𝛼+𝛽1)1)+𝐿1Γ1𝛽1Γ1+𝛼𝛽1𝛼+𝑛1+𝛾1(𝛾1)1.(3.4)

Proof. For 𝑡𝑁𝛼, apply Lemma 2.8 and 𝛾1>0, 𝑡(𝛾1)=Γ(𝑡+1)Γ𝑡+𝛾1+1=𝑡𝛾111+𝑂𝑡,(3.5) and we have that 𝑡(𝛾1)0 as 𝑡, then there exists a 𝑛1𝑁 such that inequality (3.4) holds, which implies that the set 𝑆1 exists.
We firstly show that 𝐵 maps 𝑆1 in 𝑆1.
It is easy to know that 𝑆1 is a closed, bounded, and convex subset of 𝑅.
Apply condition (𝐻1), Lemma 2.5, Corollary 2.9 and (3.4), for 𝑡𝑁𝛼+𝑛1, we have ||||1𝐵𝑥(𝑡)Γ(𝛼)𝑡𝛼𝑠=0(𝑡𝑠1)(𝛼1)||||1𝑓(𝑠+𝛼,𝑥(𝑠+𝛼))Γ(𝛼)𝑡𝛼𝑠=0(𝑡𝑠1)(𝛼1)𝐿1(𝑠+𝛼)(𝛽1)=𝐿1Δ𝛼(𝑡+𝛼)(𝛽1)=𝐿1Γ1𝛽1Γ1+𝛼𝛽1(𝑡+𝛼)(𝛼𝛽1)<𝐿1Γ1𝛽1Γ1+𝛼𝛽1𝑡(𝛼𝛽1)=𝐿1Γ1𝛽1Γ1+𝛼𝛽1𝑡+𝛾1(𝛾1)𝑡(𝛾1)𝐿1Γ1𝛽1Γ1+𝛼𝛽1𝛼+𝑛1+𝛾1(𝛾1)𝑡(𝛾1)𝑡(𝛾1),(3.6) which implies that 𝐵𝑆1𝑆1 for 𝑡𝑁𝛼+𝑛1.
Nextly, we show that 𝐵 is continuous on 𝑆1.
Let 𝜀>0 be given then there exist 𝑇1𝑁 and 𝑇1𝑛1 such that 𝑡𝑁𝛼+𝑇1 implies that 𝐿1Γ1𝛽1Γ1+𝛼𝛽1𝑡(𝛼𝛽1)<𝜀2.(3.7)
Let {𝑥𝑛} be a sequence such that 𝑥𝑛𝑥. For 𝑡{𝛼+𝑛1,𝛼+𝑛1+1,,𝛼+𝑇11}, applying the continuity of 𝑓 and Lemma 2.6, we have ||𝐵𝑥𝑛||1(𝑡)𝐵𝑥(𝑡)Γ(𝛼)𝑡𝛼𝑠=0(𝑡𝑠1)(𝛼1)||𝑓𝑠+𝛼,𝑥𝑛||(𝑠+𝛼)𝑓(𝑠+𝛼,𝑥(𝑠+𝛼))max𝑠0,1,,𝑇11||𝑓𝑠+𝛼,𝑥𝑛||×1(𝑠+𝛼)𝑓(𝑠+𝛼,𝑥(𝑠+𝛼))Γ(𝛼)𝑡𝛼𝑠=0(𝑡𝑠1)(𝛼1)=𝑡(𝛼)Γ(𝛼+1)max𝑠0,1,,𝑇11||𝑓𝑠+𝛼,𝑥𝑛||(𝑠+𝛼)𝑓(𝑠+𝛼,𝑥(𝑠+𝛼))𝛼+𝑇11(𝛼)Γ(𝛼+1)max𝑠0,1,,𝑇11||𝑓𝑠+𝛼,𝑥𝑛||(𝑠+𝛼)𝑓(𝑠+𝛼,𝑥(𝑠+𝛼))0as𝑛.(3.8)
For 𝑡𝑁𝛼+𝑇1, ||𝐵𝑥𝑛||1(𝑡)𝐵𝑥(𝑡)Γ(𝛼)𝑡𝛼𝑠=0(𝑡𝑠1)(𝛼1)||𝑓𝑠+𝛼,𝑥𝑛||+||||(𝑠+𝛼)𝑓(𝑠+𝛼,𝑥(𝑠+𝛼))2𝐿1Γ(𝛼)𝑡𝛼𝑠=0(𝑡𝑠1)(𝛼1)(𝑠+𝛼)(𝛽1)=2𝐿1Δ𝛼(𝑡+𝛼)(𝛽1)=2𝐿1Γ1𝛽1Γ1+𝛼𝛽1(𝑡+𝛼)(𝛼𝛽1)<2𝐿1Γ1𝛽1Γ1+𝛼𝛽1𝑡(𝛼𝛽1)<𝜀.(3.9)
Thus, for all 𝑡𝑁𝛼+𝑛1, we have ||𝐵𝑥𝑛||(𝑡)𝐵𝑥(𝑡)0as𝑛.(3.10) which implies that 𝐵 is continuous.
Lastly, we show that 𝐵𝑆1 is relatively compact.
Let 𝑡1,𝑡2𝑁𝛼+𝑇1 and 𝑡2>𝑡1, thus we have ||𝑡𝐵𝑥2𝑡𝐵𝑥1||=|||||1Γ(𝛼)𝑡2𝛼𝑠=0𝑡2𝑠1(𝛼1)1𝑓(𝑠+𝛼,𝑥(𝑠+𝛼))Γ(𝛼)𝑡1𝛼𝑠=0𝑡1𝑠1(𝛼1)|||||1𝑓(𝑠+𝛼,𝑥(𝑠+𝛼))Γ(𝛼)𝑡2𝛼𝑠=0𝑡2𝑠1(𝛼1)||||+1𝑓(𝑠+𝛼,𝑥(𝑠+𝛼))Γ(𝛼)𝑡1𝛼𝑠=0𝑡1𝑠1(𝛼1)||||𝐿𝑓(𝑠+𝛼,𝑥(𝑠+𝛼))1Γ1𝛽1Γ1+𝛼𝛽1𝑡2+𝛼(𝛼𝛽1)+𝐿1Γ1𝛽1Γ1+𝛼𝛽1𝑡1+𝛼(𝛼𝛽1)<𝐿1Γ1𝛽1Γ1+𝛼𝛽1𝑡2(𝛼𝛽1)+𝐿1Γ1𝛽1Γ1+𝛼𝛽1𝑡1(𝛼𝛽1)<𝜀.(3.11) Thus, {𝐵𝑥𝑥𝑆1} 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 𝑡𝑁𝛼+𝑛1.

Proof. Notice if that 𝑥(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 𝑦𝑆1, 𝑥=𝐴𝑥+𝐵𝑦𝑥𝑆1 holds.
If 𝑥=𝐴𝑥+𝐵𝑦, applying condition (𝐻1) and (3.4), for 𝑡𝑁𝛼+𝑛1, we have ||||||||+||||||𝑥𝑥(𝑡)𝐴𝑥(𝑡)𝐵𝑦(𝑡)0||𝑡Γ(𝛼)(𝛼1)+1Γ(𝛼)𝑡𝛼𝑠=0(𝑡𝑠1)(𝛼1)||||||𝑥𝑓(𝑠+𝛼,𝑦(𝑠+𝛼))0||𝑡Γ(𝛼)(𝛼1)+𝐿1Γ1𝛽1Γ1+𝛼𝛽1(𝑡+𝛼)(𝛼𝛽1)<||𝑥0||Γ𝑡(𝛼)(𝛼1)+𝐿1Γ1𝛽1Γ1+𝛼𝛽1𝑡(𝛼𝛽1)=||𝑥0||Γ(𝛼)𝑡+𝛾1((1/2)(𝛼+𝛽1)1)+𝐿1Γ1𝛽1Γ1+𝛼𝛽1𝑡+𝛾1(𝛾1)𝑡(𝛾1)||𝑥0||Γ(𝛼)𝛼+𝑛1+𝛾1((1/2)(𝛼+𝛽1)1)+𝐿1Γ1𝛽1Γ1+𝛼𝛽1𝛼+𝑛1+𝛾1(𝛾1)𝑡(𝛾1)𝑡(𝛾1).(3.12) Thus, 𝑥(𝑡)𝑆1 for 𝑡𝑁𝛼+𝑛1. According to Theorem 2.13 and Lemma 3.1, there exists a 𝑥𝑆1 such that 𝑥=𝐴𝑥+𝐵𝑥, that is, 𝑃 has a fixed point in 𝑆1 which is a solution of (1.1) for 𝑡𝑁𝛼+𝑛1. This completes the proof.

Theorem 3.3. Assume that condition (𝐻1) holds, then the solutions of (1.1) is attractive.

Proof. By Lemma 3.2, 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 𝐿20 such that ||||𝑓(𝑡,𝑥(𝑡))𝑓(𝑡,𝑦(𝑡))𝐿2𝑡(𝛽2)𝑥𝑦for𝑡𝑁𝛼.(3.13) Then the solutions of (1.1) are stable provided that 𝐿𝑐=2Γ(1+𝛼)Γ1𝛽2Γ1+𝛼𝛽2Γ1+𝛽2<1.(3.14)

Proof. Let 𝑥(𝑡) be a solution of (1.1), and let ̃𝑥(𝑡) be a solution of (1.1) satisfying the initial value condition ̃𝑥(0)=̃𝑥0. For 𝑡𝑁𝛼, applying condition (𝐻2), we have ||||𝑡𝑥(𝑡)̃𝑥(𝑡)(𝛼1)||𝑥Γ(𝛼)0̃𝑥0||+1Γ(𝛼)𝑡𝛼𝑠=0(𝑡𝑠1)(𝛼1)×||||𝑡𝑓(𝑠+𝛼,𝑥(𝑠+𝛼))𝑓(𝑠+𝛼,̃𝑥(𝑠+𝛼))(𝛼1)||𝑥Γ(𝛼)0̃𝑥0||+𝐿2Γ(𝛼)𝑡𝛼𝑠=0(𝑡𝑠1)(𝛼1)(𝑠+𝛼)(𝛽2)=𝑡𝑥̃𝑥(𝛼1)Γ||𝑥(𝛼)0̃𝑥0||+𝐿2Δ𝛼(𝑡+𝛼)(𝛽2)=𝑡𝑥̃𝑥(𝛼1)||𝑥Γ(𝛼)0̃𝑥0||+𝐿2Γ1𝛽2Γ1+𝛼𝛽2(𝑡+𝛼)(𝛼𝛽2)𝛼𝑥̃𝑥(𝛼1)||𝑥Γ(𝛼)0̃𝑥0||+𝐿2Γ1𝛽2Γ1+𝛼𝛽2𝛼(𝛼𝛽2)||𝑥𝑥̃𝑥=𝛼0̃𝑥0||+𝐿2Γ(1+𝛼)Γ1𝛽2Γ1+𝛼𝛽2Γ1+𝛽2||𝑥𝑥̃𝑥=𝛼0̃𝑥0||+𝑐𝑥̃𝑥,(3.15) which implies that 𝛼𝑥̃𝑥||𝑥1𝑐0̃𝑥0||.(3.16)
For any given 𝜀>0, let 𝛿=((1𝑐)/𝛼)𝜀, |𝑥0̃𝑥0|<𝛿 follows that 𝑥̃𝑥<𝜀, 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.
Theorem 3.5 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)(1𝛼), and 𝐿30 such that ||||𝑓(𝑡,𝑥(𝑡))𝐿3𝑡+𝛾2(𝛽3)||||𝑥(𝑡)for𝑡N𝛼.(3.17) Then the solutions of (1.1) is attractive.

Proof. Set 𝑆2=||||𝑥(𝑡)𝑥(𝑡)𝑡(𝛾2)for𝑡𝑁𝛼+𝑛2,(3.18) where 𝑛2𝑁 satisfies that ||𝑥0||Γ(𝛼)𝛼+𝑛2+𝛾2(𝛾2)+𝐿3Γ1𝛽3𝛾2Γ1+𝛼𝛽3𝛾2𝛼+𝑛2+𝛾2(𝛼𝛽3)1.(3.19)
We first prove condition (c) of Theorem 2.13, that is, for fixed 𝑦𝑆2 and for all 𝑥𝑅, 𝑥=𝐴𝑥+𝐵𝑦𝑥𝑆2 holds.
If 𝑥=𝐴𝑥+𝐵𝑦, applying condition (𝐻3) and (3.19), for 𝑡𝑁𝛼+𝑛2, we have ||||||||+||||||𝑥𝑥(𝑡)𝐴𝑥(𝑡)𝐵𝑦(𝑡)0||𝑡Γ(𝛼)(𝛼1)+1Γ(𝛼)𝑡𝛼𝑠=0(𝑡𝑠1)(𝛼1)||||||𝑥𝑓(𝑠+𝛼,𝑦(𝑠+𝛼))0||𝑡Γ(𝛼)(𝛼1)+1Γ(𝛼)𝑡𝛼𝑠=0(𝑡𝑠1)(𝛼1)𝐿3𝑠+𝛼+𝛾2(𝛽3)||||||𝑥𝑦(𝑠+𝛼)0||Γ𝑡(𝛼)(𝛼1)+𝐿3Γ(𝛼)𝑡𝛼𝑠=0(𝑡𝑠1)(𝛼1)𝑠+𝛼+𝛾2(𝛽3)(𝑠+𝛼)(𝛾2)||𝑥0||𝑡Γ(𝛼)(𝛼1)+𝐿3Γ(𝛼)𝑡𝛼𝑠=0(𝑡𝑠1)(𝛼1)(𝑠+𝛼)(𝛽3𝛾2)||𝑥0||𝑡Γ(𝛼)(𝛼1)+𝐿3Γ1𝛽3𝛾2Γ1+𝛼𝛽3𝛾2(𝑡+𝛼)(𝛼𝛽3𝛾2)<||𝑥0||𝑡Γ(𝛼)(𝛼1)+𝐿3Γ1𝛽3𝛾2Γ1+𝛼𝛽3𝛾2𝑡(𝛼𝛽3𝛾2)=||𝑥0||Γ(𝛼)𝑡+𝛾2(𝛾2)+𝐿3Γ1𝛽3𝛾2Γ1+𝛼𝛽3𝛾2𝑡+𝛾2(𝛼𝛽3)𝑡(𝛾2)||𝑥0||Γ(𝛼)𝛼+𝑛2+𝛾2(𝛾2)+𝐿3Γ1𝛽3𝛾2Γ1+𝛼𝛽3𝛾2𝛼+𝑛2+𝛾2(𝛼𝛽3)𝑡(𝛾2)𝑡(𝛾2).(3.20) 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 𝐿40 such that ||||𝑓(𝑡,𝑥(𝑡))𝐿4(𝑡+1)(𝛽4)||||𝑥(𝑡)𝜂for𝑡𝑁𝛼.(3.21) Then the solutions of (1.1) is attractive.

Proof. Set 𝑆3=||||𝑥(𝑡)𝑥(𝑡)𝑡(𝛾3)for𝑡𝑁𝛼+𝑛3,(3.22) where 𝛾3=(1/2)(𝛽4𝛼), and 𝑛3𝑁 satisfies that ||𝑥0||Γ(𝛼)𝛼+𝑛3+𝛾3(𝛼1+𝛾3)+𝐿4Γ1𝛽4𝛾3𝜂Γ1+𝛼𝛽4𝛾3𝜂𝛼+𝑛3+𝛾3𝛾31.(3.23)
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.
Since 𝜂(0,1),𝛽4(𝛼,(2+𝛼𝜂)/(2+𝜂)), and 𝛾3=(1/2)(𝛽4𝛼), then 𝛾3,𝛾3𝜂,𝛼+𝛾3(0,1),𝛽4+𝛾3𝜂(𝛼,1).
If 𝑥=𝐴𝑥+𝐵𝑦, applying condition (𝐻4), Lemma 2.5 and (3.23), for 𝑡𝑁𝛼+𝑛3, we have ||||||||+||||||𝑥𝑥(𝑡)𝐴𝑥(𝑡)𝐵𝑦(𝑡)0||𝑡Γ(𝛼)(𝛼1)+1Γ(𝛼)𝑡𝛼𝑠=0(𝑡𝑠1)(𝛼1)||||||𝑥𝑓(𝑠+𝛼,𝑦(𝑠+𝛼))0||𝑡Γ(𝛼)(𝛼1)+1Γ(𝛼)𝑡𝛼𝑠=0(𝑡𝑠1)(𝛼1)𝐿4(𝑠+𝛼+1)(𝛽4)||||𝑦(𝑠+𝛼)𝜂||𝑥0||𝑡Γ(𝛼)(𝛼1)+𝐿4Γ(𝛼)𝑡𝛼𝑠=0(𝑡𝑠1)(𝛼1)𝑠+𝛼+𝛾3𝜂(𝛽4)(𝑠+𝛼)(𝛾3)𝜂||𝑥0||Γ𝑡(𝛼)(𝛼1)+𝐿4Γ(𝛼)𝑡𝛼𝑠=0(𝑡𝑠1)(𝛼1)𝑠+𝛼+𝛾3𝜂(𝛽4)(𝑠+𝛼)(𝛾3𝜂)=||𝑥0||𝑡Γ(𝛼)(𝛼1)+𝐿4Γ(𝛼)𝑡𝛼𝑠=0(𝑡𝑠1)(𝛼1)(𝑠+𝛼)(𝛽4𝛾3𝜂)||𝑥0||𝑡Γ(𝛼)(𝛼1)+𝐿4Γ1𝛽4𝛾3𝜂Γ1+𝛼𝛽4𝛾3𝜂(𝑡+𝛼)(𝛼𝛽4𝛾3𝜂)<||𝑥0||𝑡Γ(𝛼)(𝛼1)+𝐿4Γ1𝛽4𝛾3𝜂Γ1+𝛼𝛽4𝛾3𝜂𝑡(𝛼𝛽4𝛾3𝜂)||𝑥0||𝑡Γ(𝛼)(𝛼1)+𝐿4Γ1𝛽4𝛾3𝜂Γ1+𝛼𝛽4𝛾3𝜂𝑡(𝛼𝛽4)=||𝑥0||Γ(𝛼)𝑡+𝛾3(𝛼1+𝛾3)+𝐿4Γ1𝛽4𝛾3𝜂Γ1+𝛼𝛽4𝛾3𝜂𝑡+𝛾3(𝛾3)𝑡(𝛾3)||𝑥0||Γ(𝛼)𝛼+𝑛3+𝛾3(𝛼1+𝛾3)+𝐿4Γ1𝛽4𝛾3𝜂Γ1+𝛼𝛽4𝛾3𝜂𝛼+𝑛3+𝛾3(𝛾3)𝑡(𝛾3)𝑡(𝛾3).(3.24) Thus, condition (c) of Theorem 2.13 holds. This completes the proof.

4. Examples

Example 4.1. Consider Δ0.5𝑥(𝑡)=0.2(𝑡+0.5)(0.75)sin(𝑥(𝑡+0.5)),𝑡𝑁0,Δ0.5𝑥(𝑡)|𝑡=0=𝑥0,(4.1) where 𝑓(𝑡,𝑥(𝑡))=0.2𝑡(0.75)sin(𝑥(𝑡)), 𝑡𝑁0.5.
Since ||||=||𝑓(𝑡,𝑥(𝑡))0.2𝑡(0.75)||sin(𝑥(𝑡))0.2𝑡(0.75),(4.2)thisimplies that condition (𝐻1) holds.
In addition, ||||𝑓(𝑡,𝑥(𝑡))𝑓(𝑡,𝑦(𝑡))0.2𝑡(0.75)𝑥𝑦.(4.3) Thus, condition (𝐻2) is satisfied.
Moreover, from 𝐿2=0.2, 𝛼=0.5, and 𝛽2=0.75, we have 𝐿𝑐=2Γ(1+𝛼)Γ1𝛽2Γ1+𝛼𝛽2Γ1+𝛽2=0.2Γ(1.5)Γ(0.25)Γ(1.25)Γ(1.75)0.7716<1,(4.4) which implies that inequality (3.14) holds.
Thus the solutions of (4.1) are asymptotically stable by Theorem 3.5.

Example 4.2. Consider Δ0.5𝑥(𝑡)=0.2(𝑡+1.5)(0.6)𝑥(𝑡+0.5),𝑡𝑁0,Δ0.5𝑥(𝑡)|𝑡=0=𝑥0,(4.5) where 𝑓(𝑡,𝑥(𝑡))=0.2(𝑡+1)(0.6)𝑥(𝑡), 𝑡𝑁0.5.
Since 𝛽3=0.6,𝛼=0.5, we have that 𝛽3(𝛼,(1/2)(1+𝛼)), 𝛾2=0.25 and ||||=||𝑓(𝑡,𝑥(𝑡))0.2(𝑡+1)(0.6)||𝑥(𝑡)0.2(𝑡+0.25)(0.6)||||,𝑥(𝑡)(4.6) which implies that condition (𝐻3) is satisfied.
Meanwhile, ||||𝑓(𝑡,𝑥(𝑡))𝑓(𝑡,𝑦(𝑡))0.2(𝑡+1)(0.6)𝑥𝑦0.2𝑡(0.6)𝑥𝑦,(4.7) which implies that condition (𝐻2) is satisfied.
From 𝐿2=0.2, 𝛼=0.5, and 𝛽2=0.6, we have 𝐿𝑐=2Γ(1+𝛼)Γ1𝛽2Γ1+𝛼𝛽2Γ1+𝛽2=0.2Γ(1.5)Γ(0.4)Γ(0.9)Γ(1.6)0.4120<1,(4.8) which implies that inequality (3.14) holds.
Thus the solutions of (4.5) are asymptotically stable by Theorem 3.7.

Example 4.3. Consider Δ0.5𝑥(𝑡)=(𝑡+1.5)(0.6)𝑥1/3(𝑡+0.5),𝑡𝑁0,Δ0.5𝑥(𝑡)|𝑡=0=𝑥0,(4.9) where 𝑓(𝑡,𝑥(𝑡))=(𝑡+1)(0.6)𝑥1/3(𝑡), 𝑡𝑁0.5.
Since 𝛼=0.5,𝛽4=0.6,𝜂=1/3, we have that 𝜂(0,1),𝛽4(𝛼,(2+𝛼𝜂)/(2+𝜂)) and ||||𝑓(𝑡,𝑥(𝑡))(𝑡+1)(0.6)||||𝑥(𝑡)1/3,(4.10) then condition (𝐻4) is satisfied.
The solutions of (4.9) are attractive by Theorem 3.8.

Acknowledgments

Thisresearch was supported by the NSF of Hunan Province (10JJ6007, 2011FJ3013), the Scientific Research Foundation of Hunan Provincial Education Department, and the Construct Program of the Key Discipline in Hunan Province.

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