Abstract

A class of nonautonomous two-species competitive system with stage structure and impulse is considered. By using the continuation theorem of coincidence degree theory, we derive a set of easily verifiable sufficient conditions that guarantee the existence of at least a positive periodic solution, and, by constructing a suitable Lyapunov functional, the uniqueness and global attractivity of the positive periodic solution are presented. Finally, an illustrative example is given to demonstrate the correctness of the obtained results.

1. Introduction

In recent years, with the increasing applications of theory of differential equations in mathematical ecology, various mathematical models have been proposed in the study of population [125]. But most of the previous results focused on the dynamical behaviors (including the stability, attractiveness, persistence, and periodicity of solution) of the systems which have fixed parameters and there is no impulse. Considering that harvest of many populations are not continuous and the periodic environmental factor, it is reasonable to investigate the systems with periodic coefficients and impulse. Impulsive differential systems display a combination of characteristics of both the continuous-time and discrete-time systems [2630]. In 2006, Chen [1] studied the following non-autonomous almost periodic competitive two-species model with stage structure in one species:̇𝑥1(𝑡)=𝑎1(𝑡)𝑥1(𝑡)+𝑏1(𝑡)𝑥2(𝑡),̇𝑥2(𝑡)=𝑎2(𝑡)𝑥1(𝑡)𝑏2(𝑡)𝑥2(𝑡)𝑐(𝑡)𝑥22(𝑡)𝛽1(𝑡)𝑥2(𝑡)𝑥2(𝑡)𝑥3(𝑡),̇𝑥3(𝑡)=𝑥3(𝑡)𝑑(𝑡)𝑒(𝑡)𝑥3(𝑡)𝛽2(𝑡)𝑥2,(𝑡)(1.1) where 𝑥1(𝑡) and 𝑥2(𝑡) are immature and mature population densities of one species, respectively; 𝑥3(𝑡) represents the population density of another species; 𝑎𝑖(𝑡),𝑏𝑖(𝑡),𝛽𝑖(𝑡)(𝑖=1,2),𝑐(𝑡),𝑑(𝑡),𝑒(𝑡) are all continuous, almost periodic functions. The competition is between 𝑥2(𝑡) and 𝑥3(𝑡). Chen [1] obtained sufficient conditions for the existence of a unique, globally attractive, strictly positive almost periodic solution for system (1.1).

Considering that the harvest is an annual harvest pulse, to describe a system more accurately, we should consider the impulsive differential equation. Motivated by this point of view, we revised system (1.1) into the following form:̇𝑥1(𝑡)=𝑎1(𝑡)𝑥1(𝑡)+𝑏1(𝑡)𝑥2(𝑡),𝑡𝑡𝑘,̇𝑥2(𝑡)=𝑎2(𝑡)𝑥2(𝑡)𝑏2(𝑡)𝑥2(𝑡)𝑐(𝑡)𝑥22(𝑡)𝛽1(𝑡)𝑥2(𝑡)𝑥3(𝑡),𝑡𝑡𝑘,̇𝑥3(𝑡)=𝑥3(𝑡)𝑑(𝑡)𝑒(𝑡)𝑥3(𝑡)𝛽2(𝑡)𝑥2(𝑡),𝑡𝑡𝑘,Δ𝑥𝑖𝑡𝑘=𝑥𝑖𝑡+𝑘𝑥𝑖𝑡𝑘=𝛾𝑖𝑘𝑥𝑖(𝑡𝑘),𝑖=1,2,3,𝑘=1,2,,𝑞,(1.2) where Δ𝑥𝑖(𝑡𝑘)=𝑥𝑖(𝑡+𝑘)𝑥𝑖(𝑡𝑘) are the impulses at moments 𝑡𝑘 and 𝑡1<𝑡2< is a strictly increasing sequence such that lim𝑘𝑡𝑘=+;𝑥1(𝑡) and 𝑥2(𝑡) are immature and mature population densities of one species, respectively, and 𝑥3(𝑡) represents the population density of another species. The competition is between 𝑥2(𝑡) and 𝑥3(𝑡).

Throughout the paper, we always assume the following. (H1)𝑎𝑖(𝑡),𝑏𝑖(𝑡),𝛽𝑖(𝑡)(𝑖=1,2),𝑐(𝑡),𝑑(𝑡),𝑒(𝑡) are all continuous 𝜔 periodic; that is, 𝑎𝑖(𝑡+𝜔)=𝑎𝑖(𝑡), 𝑏𝑖(𝑡+𝜔)=𝑏𝑖(𝑡), 𝛽𝑖(𝑡+𝜔)=𝛽𝑖(𝑡)(𝑖=1,2), 𝑐(𝑡+𝜔)=𝑐(𝑡), 𝑑(𝑡+𝜔)=𝑑(𝑡), 𝑒(𝑡+𝜔)=𝑒(𝑡) for any 𝑡𝑅.(H2)𝑎𝑖(𝑡),𝑏𝑖(𝑡),𝛽𝑖(𝑡)(𝑖=1,2),𝑐(𝑡),𝑑(𝑡),𝑒(𝑡) are all positive.(H3)0<𝛾𝑖𝑘<1,𝑖=1,2,3 for all 𝑘𝑁, and there exists a positive integer 𝑞 such that 𝑡𝑘+𝑞=𝑡𝑘+𝜔,𝛾𝑖(𝑘+𝑞)=𝛾𝑖𝑘,𝑖=1,2,3.

The principle object of this paper is by using Mawhin’s continuation theorem of coincidence degree theory and by constructing the Lyapunov functions to investigate the stability and existence of periodic solutions of (1.2). To the best of my knowledge, it is the first time to deal with the existence and stability of periodic solutions of (1.2).

The organization of the paper is as follows. In Section 2, we introduce some notations and definitions and state some preliminary results needed in later sections. We then establish, in Section 3, some simple criteria for the existence of positive periodic solutions of system (1.2) by using the continuation theorem of coincidence degree theory proposed by Gaines and Mawhin [31]. The uniqueness and global attractivity of the positive periodic solution are presented in Section 4. In Section 5, an illustrative example is given to demonstrate the correctness of the obtained results.

2. Preliminaries

We will introduce some notations and definitions and state some preliminary results. Consider the impulsive systeṁ𝑥(𝑡)=𝑓(𝑡,𝑥),𝑡𝑡𝑘,𝑘=1,2,,Δ𝑥(𝑡)𝑡=𝑡𝑘=𝐼𝑘𝑥𝑡𝑘,(2.1) where 𝑥𝑅𝑛,𝑓𝑅×𝑅𝑛𝑅𝑛 is continuous and 𝑓(𝑡+𝜔,𝑥)=𝑓(𝑡,𝑥);𝐼𝑘𝑅𝑛𝑅𝑛 are continuous, and there exists a positive integer 𝑞 such that 𝑡𝑘+𝑞=𝑡𝑘+𝜔,𝐼𝑘+𝑞(𝑥)=𝐼𝑘(𝑥) with 𝑡𝑘𝑅,𝑡𝑘+1>𝑡𝑘,lim𝑘=,Δ𝑥(𝑡)𝑡=𝑡𝑘=𝑥(𝑡+𝑘)𝑥(𝑡𝑘). For 𝑡𝑘0(𝑘=1,2,),[0,𝜔]{𝑡𝑘}={𝑡1,𝑡2,,𝑡𝑞}. As we know, {𝑡𝑘} are called points of jump.

Let us recall some definitions. For the Canchy problem,[]̇𝑥(𝑡)=𝑓(𝑡,𝑥),𝑡0,𝜔,𝑡𝑡𝑘,Δ𝑥(𝑡)𝑡=𝑡𝑘=𝐼𝑘𝑥𝑡𝑘,𝑥(0)=𝑥0.(2.2)

Definition 2.1. A map 𝑥[0,𝜔]𝑅𝑛 is said to be a solution of (2.2), if it satisfied the following conditions:(i)𝑥(𝑡) is a piecewise continuous map with first-class discontinuity points in 𝑡𝑘[0,𝜔], and at each discontinuity point it is continuous on the left;(ii)𝑥(𝑡) satisfies (2.2).

Definition 2.2. A map 𝑥[0,𝜔]𝑅𝑛 is said to be an 𝜔 periodic solution of (2.1), if (i)𝑥(𝑡) satisfies (i) and (ii) of Definition 2.1 in the interval [0,𝜔] and(ii)𝑥(𝑡) satisfies 𝑥(𝑡+𝜔0)=𝑥(𝑡0),𝑡𝑅.

Obviously, if 𝑥(𝑡) is a solution of (2.2) defined on [0,𝜔], such that 𝑥(0)=𝑥(𝜔), then, by the periodicity of (2.2) in 𝑡, the function 𝑥(𝑡) defined by 𝑥[]𝑡(𝑡)=𝑥(𝑡𝑗𝜔),𝑡𝑗𝜔,(𝑗+1)𝜔𝑘,𝑥(𝑡)isleftcontinuousat𝑡=𝑡𝑘(2.3) is a 𝜔 periodic solution of (2.1).

For system (1.2), seeking the periodic solutions is equivalent to seeking solutions of the following boundary value problem:̇𝑥1(𝑡)=𝑎1(𝑡)𝑥1(𝑡)+𝑏1(𝑡)𝑥2(𝑡),𝑡𝑡𝑘[],𝑡0,𝜔,𝑘=1,2,,𝑞,̇𝑥2(𝑡)=𝑎2(𝑡)𝑥2(𝑡)𝑏2(𝑡)𝑥2(𝑡)𝑐(𝑡)𝑥22(𝑡)𝛽1(𝑡)𝑥2(𝑡)𝑥3(𝑡),𝑡𝑡𝑘[],𝑡0,𝜔,𝑘=1,2,,𝑞,̇𝑥3(𝑡)=𝑥3(𝑡)𝑑(𝑡)𝑒(𝑡)𝑥3(𝑡)𝛽2(𝑡)𝑥2(𝑡),𝑡𝑡𝑘[],𝑡0,𝜔,𝑘=1,2,,𝑞,Δ𝑥𝑖𝑡𝑘=𝑥𝑖𝑡+𝑘𝑥𝑖𝑡𝑘=𝛾𝑖𝑘𝑥𝑖𝑡𝑘,𝑖=1,2,3,𝑥𝑖(0)=𝑥𝑖(𝜔),𝑘=1,2,,𝑞.(2.4)

3. Existence of Positive Periodic Solutions

In this section, based on the Mawhin’s continuation theorem, we shall study the existence of at least one periodic solution of (1.1). To do so, we shall make some preparations.

Let 𝑋,𝑌 be normed vector spaces; 𝐿Dom𝐿𝑋𝑌 is a linear mapping; 𝑁𝑋𝑌 is a continuous mapping. The mapping 𝐿 will be called a Fredholm mapping of index zero if dimKer𝐿=codimIm𝐿<+ and Im𝐿 is closed in 𝑌. If 𝐿 is a Fredholm mapping of index zero and there exist continuous projectors 𝑃𝑋𝑋 and 𝑄𝑌𝑌 such that Im𝑃=Ker𝐿,Im𝐿=Ker𝑄=Im(𝐼𝑄), it follows that 𝐿Dom𝐿Ker𝑃(𝐼𝑃)𝑋Im𝐿 is invertible. We denote the inverse of that map by 𝐾𝑃. If Ω is an open bounded subset of 𝑋, the mapping 𝑁 will be called 𝐿-compact on Ω if 𝑄𝑁(Ω) is bounded and 𝐾𝑃(𝐼𝑄)𝑁Ω𝑋 is compact. Since Im𝑄 is isomorphic to Ker𝐿, there exist isomorphisms 𝐽Im𝑄Ker𝐿.

Now we introduce Mawhin’s continuation theorem [31] as follows.

Lemma 3.1 (Continuation Theorem [31]). Let 𝐿 be a Fredholm mapping of index zero, and let 𝑁 be 𝐿-compact on Ω. Suppose(a)for each 𝜆(0,1), every solution 𝑥 of 𝐿𝑥=𝜆𝑁𝑥 is such that 𝑥𝜕Ω.(b)𝑄𝑁𝑥0 for each 𝑥Ker𝐿𝜕Ω, and deg{JQN,Ω𝜕Ker𝐿,0}0.Then the equation 𝐿𝑥=𝑁𝑥 has at least one solution lying in Dom𝐿Ω.

For convenience and simplicity in the following discussion, we always use the notations below throughout the paper: 1𝑓=𝜔𝜔0𝑓(𝑡)𝑑𝑡,𝑓𝐿=min[]𝑡0,𝜔𝑓(𝑡),𝑓𝑀=max[]𝑡0,𝜔𝑓(𝑡),||𝑓||=1𝜔𝜔0||||𝑓(𝑡)𝑑𝑡,(3.1) where 𝑓(𝑡) is a 𝜔 continuous periodic function. For any nonnegative integer 𝑝, let 𝐶(𝑝)[0,𝜔;𝑡1,𝑡2,,𝑡𝑞]={𝑥[0,𝜔]𝑅𝑚𝑥(𝑝)(𝑡) exist for 𝑡𝑡1,,𝑡𝑞;𝑥(𝑝)(𝑡+0), and let 𝑥(𝑝)(𝑡0) exist at 𝑡1,𝑡2,,𝑡𝑞, and 𝑥(𝑗)(𝑡𝑘)=𝑥(𝑗)(𝑡𝑘0),𝑘=1,,𝑚,𝑗=0,1,2,,𝑝} with the norm 𝑥𝑝=max{sup𝑡[0,𝜔]𝑥(𝑗)(𝑡)}𝑝𝑗=1, where is any norm of 𝑅𝑚. It is easy to see that 𝐶(𝑝)[0,𝜔;𝑡1,𝑡2,,𝑡𝑞] is a Banach space.

Now we are now in a position to state and prove the existence of periodic solutions of (2.4).

Theorem 3.2. In addition to (H1),(H2),(H3), assume further that the following hold: H4𝑃min1,𝑃2,𝑃3H>0,5𝑎1𝜔>𝑞𝑘=1ln1𝛾1𝑘,H6𝑑𝜔+𝑞𝑘=1ln1𝛾3𝑘>𝛽2𝜔𝑒𝐵4,𝛽1𝛽2𝑐𝑒,(3.2) where 𝑃1=||𝑎2𝑏2||𝜔+𝑞𝑘=1ln1𝛾2𝑘𝑐𝜔𝑒𝐵6,𝑃2=||𝑎2𝑏2||𝜔+𝑞𝑘=1ln1𝛾2𝑘𝛽1𝜔𝑒𝐵29,𝑃3=||𝑎2𝑏2||𝜔+𝑞𝑘=1ln1𝛾2𝑘𝑐𝜔,(3.3) and 𝐵4,𝐵6,𝐵29 are defined by (3.27), (3.32), and (3.61), respectively. Then the system (1.2) has at least a 𝜔 periodic solution.

Proof. According to the discussion above in Section 2, we need only to prove that the boundary value problem (2.4) has a solution. Since solutions of (2.4) remained positive for all 𝑡0, we let 𝑢1𝑥(𝑡)=ln1(𝑡),𝑢2𝑥(𝑡)=ln2(𝑡),𝑢3𝑥(𝑡)=ln3(𝑡),(3.4) then system (2.4) can be translated to ̇𝑢1(𝑡)=𝑎1(𝑡)+𝑏1(𝑡)𝑒(𝑢2(𝑡)𝑢1(𝑡)),𝑡𝑡𝑘[],𝑡0,𝜔,𝑘=1,2,,𝑞,̇𝑢2(𝑡)=𝑎2(𝑡)𝑏2(𝑡)𝑐(𝑡)𝑒𝑢2(𝑡)𝛽1(𝑡)𝑒𝑢3(𝑡),𝑡𝑡𝑘[],𝑡0,𝜔,𝑘=1,2,,𝑞,̇𝑢3(𝑡)=𝑑(𝑡)𝑒(𝑡)𝑒𝑢3(𝑡)𝛽2(𝑡)𝑒𝑢2(𝑡),𝑡𝑡𝑘[],𝑡0,𝜔,𝑘=1,2,,𝑞,Δ𝑢𝑖𝑡𝑘=ln1𝛾𝑖𝑘,𝑖=1,2,3,𝑢𝑖(0)=𝑢𝑖(𝜔).(3.5) It is easy to see that if system (3.5) has one 𝜔 periodic solution (𝑢1(𝑡),𝑢2(𝑡),𝑢3(𝑡))𝑇, then (𝑥1(𝑡),𝑥2(𝑡),𝑥3(𝑡))𝑇=(𝑒𝑢1(𝑡),𝑒𝑢2(𝑡),𝑒𝑢3(𝑡))𝑇 is a positive solution of system (1.2). Therefore, to complete the proof, it suffices to show that system (3.5) has at least one 𝜔 periodic solution.
In order to use the continuation theorem of coincidence degree theory, we take𝑋=𝑢𝐶0,𝜔;𝑡1,𝑡2,,𝑡𝑞,𝑌=𝑋×𝑅3×(𝑞+1).(3.6) Then 𝑋 is a Banach space with norm 0, and 𝑌 is also a Banch space with norm 𝑧=𝑥0+𝑦,𝑥𝑋,𝑌𝑅3𝑞.
Let the following hold:𝑢dom𝐿=𝑥=1,𝑢2,𝑢3𝑇[]𝐶0,𝜔;𝑡1,𝑡2,,𝑡𝑞,𝑥𝐿Dom𝐿𝑋𝑌,𝑥𝑡,Δ𝑥𝑘𝑞𝑘=1,𝑁𝑋𝑌,𝑁𝑢=𝑎1(𝑡)+𝑏1𝑢(𝑡)exp2(𝑡)𝑢1𝑎(𝑡)2(𝑡)𝑏2(𝑡)𝑐(𝑡)𝑒𝑢2(𝑡)𝛽1(𝑡)𝑒𝑢3(𝑡)𝑑(𝑡)𝑒(𝑡)𝑒𝑢3(𝑡)𝛽2(𝑡)𝑒𝑢2(𝑡),ln1𝛾11ln1𝛾21ln1𝛾31,ln1𝛾12ln1𝛾22ln1𝛾32,,ln1𝛾1𝑞ln1𝛾2𝑞ln1𝛾3𝑞,0(3.7) Obviously, Ker𝐿=𝑢𝑢(𝑡)=𝑅3[],,𝑡0,𝜔Im𝐿=𝑧=𝑓,𝑎1,𝑎2,,𝑎𝑞,𝑑𝑌𝜔0𝑓(𝑠)𝑑𝑠+𝑞𝑘=1𝑎𝑘+𝑑=0=𝑋×𝑅3×𝑞×{0},dimKer𝐿=3=codimIm𝐿.(3.8) So, Im𝐿 is closed in 𝑌; 𝐿 is a Fredholm mapping of index zero. Define two projectors 1𝑃𝑥=𝜔𝜔0𝑥(𝑡)𝑑𝑡,𝑄𝑧=𝑄𝑓,𝑎1,𝑎2,,𝑎𝑞=1,𝑑𝜔𝜔0𝑓(𝑠)𝑑𝑠+𝑞𝑘=1𝑎𝑘.+𝑑,,0,0,,0(3.9) It is easy to show that 𝑃 and 𝑄 are continuous and satisfy Im𝑃=Ker𝐿,Im𝐿=Ker𝑄=Im(𝐼𝑄).
Further, by direct computation, we can find that the inverse 𝐾𝑃 of 𝐿, 𝐾𝑃Im𝐿Ker𝑃Dom𝐿 has the following form:𝐾𝑃(𝑧)=𝑡0𝑓(𝑠)𝑑𝑠+𝑡𝑘<𝑡𝑎𝑘1𝜔𝜔0𝑡0𝑓(𝑠)𝑑𝑠𝑑𝑡𝑞𝑘=1𝑎𝑘+1𝜔𝑞𝑘=1𝑎𝑘𝑡𝑘.(3.10) Moreover, it is easy to check that 1𝑄𝑁𝑢=𝜔𝑡0𝐹11(𝑠)𝑑𝑠+𝜔𝑞𝑘=1ln1𝛾1𝑘1𝜔𝑡0𝐹21(𝑠)𝑑𝑠+𝜔𝑞𝑘=1ln1𝛾2𝑘1𝜔𝑡0𝐹31(𝑠)𝑑𝑠+𝜔𝑞𝑘=1ln1𝛾3𝑘,𝐾,0,0,,0𝑃(𝐼𝑄)𝑁𝑢=𝑡0𝐹1(𝑠)𝑑𝑠+𝑡>𝑡𝑘ln1𝛾1𝑘𝑡0𝐹2(𝑠)𝑑𝑠+𝑡>𝑡𝑘ln1𝛾2𝑘𝑡0𝐹3(𝑠)𝑑𝑠+𝑡>𝑡𝑘ln1𝛾3𝑘1𝜔𝜔0𝑡0𝐹1(𝑠)𝑑𝑠𝑑𝑡𝑞𝑘=1ln1𝛾1𝑘+1𝜔𝑞𝑘=1ln1𝛾1𝑘1𝜔𝜔0𝑡0𝐹2(𝑠)𝑑𝑠𝑑𝑡𝑞𝑘=1ln1𝛾2𝑘+1𝜔𝑞𝑘=1ln1𝛾2𝑘1𝜔𝜔0𝑡0𝐹3(𝑠)𝑑𝑠𝑑𝑡𝑞𝑘=1ln1𝛾3𝑘+1𝜔𝑞𝑘=1ln1𝛾3𝑘𝑡𝜔12𝜔0𝐹1(𝑠)𝑑𝑠+𝑞𝑘=1ln1𝛾1𝑘𝜔0𝐹2(𝑠)𝑑𝑠+𝑞𝑘=1ln1𝛾2𝑘𝜔0𝐹3(𝑠)𝑑𝑠+𝑞𝑘=1ln1𝛾3𝑘,(3.11) where 𝐹1(𝑠)=𝑎1(𝑠)+𝑏1(𝑠)𝑒(𝑢2(𝑠)𝑢1(𝑠)),𝐹2(𝑠)=𝑎2(𝑠)𝑏2(𝑠)𝑐(𝑠)𝑒𝑢2(𝑠)𝛽1(𝑠)𝑒𝑢3(𝑠),𝐹2(𝑠)=𝑑(𝑠)𝑒(𝑠)𝑒𝑢3(𝑠)𝛽2(𝑠)𝑒𝑢2(𝑠).(3.12) Obviously, 𝑄𝑁 and 𝐾𝑃(𝐼𝑄)𝑁 are continuous. Using the Ascoli-Arzela theorem, it is not difficult to show that 𝐾𝑃(𝐼𝑄)𝑁(Ω) is compact for any open bounded set Ω𝑋. Moreover, 𝑄𝑁(Ω) is bounded. Thus, 𝑁 is 𝐿-compact on Ω with any open bounded set Ω𝑋.
Now we are at the point to search for an appropriate open, bounded subset Ω for the application of the continuation theorem. Corresponding to the operator equation 𝐿𝑢=𝜆𝑁𝑢,𝜆(0,1), we havė𝑢1(𝑡)=𝜆𝑎1(𝑡)+𝑏1(𝑡)𝑒(𝑢2(𝑡)𝑢1(𝑡)),𝑡𝑡𝑘[],𝑡0,𝜔,𝑘=1,2,,𝑞,̇𝑢2𝑎(𝑡)=𝜆2(𝑡)𝑏2(𝑡)𝑐(𝑡)𝑒𝑢2(𝑡)𝛽1(𝑡)𝑒𝑢3(𝑡),𝑡𝑡𝑘[],𝑡0,𝜔,𝑘=1,2,,𝑞,̇𝑢3𝑑(𝑡)=𝜆1(𝑡)𝑒(𝑡)𝑒𝑢3(𝑡)𝛽2(𝑡)𝑒𝑢2(𝑡),𝑡𝑡𝑘[],𝑡0,𝜔,𝑘=1,2,,𝑞,Δ𝑢𝑖𝑡𝑘=𝜆ln1𝛾𝑖𝑘,𝑖=1,2,3,𝑢𝑖(0)=𝑢𝑖(𝜔).(3.13) Suppose that 𝑢(𝑡)=(𝑢1(𝑡),𝑢2(𝑡),𝑢3(𝑡))𝑇𝑋 is an arbitrary solution of system (3.13) for a certain 𝜆(0,1), integrating both sides of (3.13) over the interval [0,𝜔] with respect to 𝑡, we obtain 𝜔0𝑏1(𝑡)𝑒(𝑢2(𝑡)𝑢1(𝑡))𝑑𝑡+𝑞𝑘=1ln1𝛾1𝑘=𝜔0𝑎1(𝑡)𝑑𝑡,𝜔0𝑐(𝑡)𝑒𝑢2(𝑡)+𝛽1(𝑡)𝑒𝑢3(𝑡)𝑑𝑡=𝜔0𝑎2(𝑡)𝑏2(𝑡)𝑑𝑡+𝑞𝑘=1ln1𝛾2𝑘,𝜔0𝑒(𝑡)𝑒𝑢3(𝑡)+𝛽2(𝑡)𝑒𝑢2(𝑡)𝑑𝑡=𝜔0𝑑(𝑡)𝑑𝑡+𝑞𝑘=1ln1𝛾3𝑘.(3.14) From (3.13) and (3.14), we obtain 𝜔0||̇𝑢1||(𝑡)𝑑𝑡<𝜔0𝑎1(𝑡)𝑑𝑡+𝜔0𝑏1(𝑡)𝑒(𝑢2(𝑡)𝑢1(𝑡))𝑑𝑡=2𝑎1𝜔𝑞𝑘=1ln1𝛾1𝑘,(3.15)𝜔0||̇𝑢2||(𝑡)𝑑𝑡<𝜔0𝑎2(𝑡)+𝑏2(𝑡)𝑑𝑡+𝜔0𝑐(𝑡)𝑒𝑢2(𝑡)+𝛽1(𝑡)𝑒𝑢3(𝑡))𝑑𝑡=2𝑎2𝜔+𝑞𝑘=1ln1𝛾2𝑘,(3.16)𝜔0||̇𝑢3||(𝑡)𝑑𝑡<𝜔0𝑑(𝑡)𝑑𝑡+𝜔0𝑒(𝑡)𝑒𝑢3(𝑡)+𝛽2(𝑡)𝑒𝑢2(𝑡)𝑑𝑡=2𝑑𝜔+𝑞𝑘=1ln1𝛾3𝑘.(3.17) Let the following hold: 𝑢𝑖𝜉𝑖=min[]𝑡0,𝜔𝑢𝑖(𝑡),𝑢𝑖𝜂𝑖=max𝑡[0,𝜔]𝑢𝑖(𝑡),𝑖=1,2,3.(3.18) From the second and the third equations of (3.14), we can obtain ||𝑎2𝑏2||𝜔+𝑞𝑘=1ln1𝛾2𝑘>𝜔0𝑐(𝑡)𝑒𝑢2(𝑡)𝑑𝑡𝜔0𝑐(𝑡)𝑒𝑢2(𝜉2)𝑑𝑡,𝑑𝜔+𝑞𝑘=1ln1𝛾3𝑘>𝜔0𝑒(𝑡)𝑒𝑢3(𝑡)𝑑𝑡𝜔0𝑒(𝑡)𝑒𝑢3(𝜉3)𝑑𝑡=𝑒𝜔𝑒𝑢3(𝜉3),(3.19) then 𝑢2𝜉2<ln||𝑎2𝑏2||𝜔+𝑞𝑘=1ln1𝛾2𝑘,𝑢𝑐𝜔(3.20)3𝜉3<ln𝑑𝜔+𝑞𝑘=1ln1𝛾3𝑘.𝑒𝜔(3.21) Thus 𝑢2(𝑡)=𝑢2𝜉2+𝑡𝜉2̇𝑢2(𝑡)𝑑𝑡𝑢2𝜉2+𝜔0||̇𝑢2(||𝑡)𝑑𝑡<ln||𝑎2𝑏2||𝜔+𝑞𝑘=1ln1𝛾2𝑘𝑐𝜔+2𝑎2𝜔+𝑞𝑘=1ln1𝛾2𝑘=𝐵1.(3.22)
In the following, we will consider four cases.
Case 1 (if 𝑢1(𝑡)>0,𝑢2(𝑡)>0). From the first equation of (3.14), we have𝑎1𝜔<𝜔0𝑏1(𝑡)𝑒𝑢2(𝑡)𝑑𝑡+𝑞𝑘=1ln1𝛾1𝑘𝑏1𝜔𝑒𝑢2(𝜂2)+𝑞𝑘=1ln1𝛾3𝑘,𝑎1𝜔>𝜔0𝑏1(𝑡)𝑒𝑢1(𝑡)𝑑𝑡+𝑞𝑘=1ln1𝛾1𝑘𝜔0𝑏1(𝑡)𝑒𝑢1(𝜂1)𝑑𝑡+𝑞𝑘=1ln1𝛾1𝑘,(3.23) that is, 𝑢2𝜂2>ln𝑎1𝜔𝑞𝑘=1ln1𝛾1𝑘𝑏1𝜔,𝑢1𝜂1>ln𝑏1𝜔𝑎1𝜔𝑞𝑘=1ln1𝛾1𝑘.(3.24) Then 𝑢2(𝑡)=𝑢2𝜂2𝜂2𝑡̇𝑢2(𝑡)𝑑𝑡𝑢2𝜂2𝜔0||̇𝑢2||(𝑡)𝑑𝑡ln𝑎1𝜔𝑞𝑘=1ln1𝛾1𝑘𝑏1𝜔2𝑎2𝜔𝑞𝑘=1ln1𝛾2𝑘=𝐵2,𝑢(3.25)1(𝑡)=𝑢1𝜂1𝜂1𝑡̇𝑢1(𝑡)𝑑𝑡𝑢1𝜂1𝜔0||̇𝑢1(||𝑡)𝑑𝑡ln𝑏1𝜔𝑎1𝜔𝑞𝑘=1ln1𝛾1𝑘2𝑎1𝜔𝑞𝑘=1ln1𝛾1𝑘=𝐵3.(3.26) Thus, from (3.22) and (3.25), we obtain ||𝑢2||||𝐵(𝑡)max1||,||𝐵2||=𝐵4.(3.27) By the first and the third equations of (3.14), we get 𝜔0𝑏1(𝑡)𝑒𝐵4𝑢1(𝜉1)𝑑𝑡+𝑞𝑘=1ln1𝛾1𝑘>𝑎1𝜔,𝜔0𝑒(𝑡)𝑒𝑢3(𝜂3)+𝛽2(𝑡)𝑒𝐵4𝑑𝑡>𝑑𝜔+𝑞𝑘=1ln1𝛾3𝑘,(3.28) then 𝑢1𝜉1<ln𝑏1𝜔𝑒𝐵4𝑎1𝜔𝑞𝑘=1ln1𝛾1𝑘𝑢,(3.29)3𝜂3>ln𝑑𝜔+𝑞𝑘=1ln1𝛾3𝑘𝛽2𝜔𝑒𝐵4.𝑒𝜔(3.30) From (3.15), (3.17), (3.21), and (3.30), we have 𝑢1(𝑡)=𝑢1𝜉1+𝑡𝜉1̇𝑢1(𝑡)𝑑𝑡𝑢1𝜉1+𝜔0||̇𝑢1(||𝑡)𝑑𝑡<ln𝑏1𝜔𝑒𝐵4𝑎1𝜔𝑞𝑘=1ln1𝛾1𝑘+2𝑎1𝜔+𝑞𝑘=1ln1𝛾1𝑘=𝐵5,𝑢(3.31)3(𝑡)=𝑢3𝜉3+𝑡𝜉3̇𝑢3(𝑡)𝑑𝑡𝑢3𝜉3+𝜔0||̇𝑢3(||𝑡)𝑑𝑡<ln𝑑𝜔+𝑞𝑘=1ln1𝛾3𝑘𝑒𝜔+2𝑑𝜔+𝑞𝑘=1ln1𝛾3𝑘=𝐵6,𝑢(3.32)3(𝑡)=𝑢3𝜂3𝜂3𝑡̇𝑢3(𝑡)𝑑𝑡𝑢3𝜂3𝜔0||̇𝑢3||(𝑡)𝑑𝑡>ln𝑑𝜔+𝑞𝑘=1ln1𝛾3𝑘𝛽2𝜔𝑒𝐵4𝑒𝜔2𝑑𝜔+𝑞𝑘=1ln1𝛾3𝑘=𝐵7.(3.33) Thus, ||𝑢1||(𝑡)𝐵5,||𝑢3||||𝐵(𝑡)max6||,||𝐵7||=𝐵8.(3.34)Case 2 (if 𝑢1(𝑡)>0,𝑢2(𝑡)<0). By the first equation of (3.14), we have𝑎1𝜔<𝜔0𝑏1(𝑡)𝑒𝑢2(𝑡)𝑑𝑡+𝑞𝑘=1ln1𝛾1𝑘𝑏1𝜔𝑒𝑢2(𝜂2)+𝑞𝑘=1ln1𝛾3𝑘,(3.35) namely, 𝑢2𝜂2>ln𝑎1𝜔𝑞𝑘=1ln1𝛾1𝑘𝑏1𝜔.(3.36) Then 𝑢2(𝑡)=𝑢2𝜂2𝜂2𝑡̇𝑢2(𝑡)𝑑𝑡𝑢2𝜂2𝜔0||̇𝑢2||(𝑡)𝑑𝑡ln𝑎1𝜔𝑞𝑘=1ln1𝛾1𝑘𝑏12𝑎2𝜔𝑞𝑘=1ln1𝛾2𝑘=𝐵2.(3.37) From (3.22) and (3.37), we obtain ||𝑢2||||𝐵(𝑡)max1||,||𝐵2||=𝐵3.(3.38) By the first equation of (3.14), we also have 𝜔0𝑏1(𝑡)𝑒𝐵3𝑒𝑢1(𝑡)𝑑𝑡+𝑞𝑘=1ln1𝛾1𝑘>𝑎1𝜔,𝜔0𝑏1(𝑡)𝑒𝐵3𝑒𝑢1(𝑡)𝑑𝑡+𝑞𝑘=1ln1𝛾1𝑘<𝑎1𝜔,(3.39) Then 𝑏1𝜔𝑒𝐵3𝑒𝑢1(𝜉1)+𝑞𝑘=1ln1𝛾1𝑘>𝑎1𝜔,𝑏1𝜔𝑒𝐵3𝑒𝑢1(𝜂1)+𝑞𝑘=1ln1𝛾1𝑘<𝑎1𝜔.(3.40) that is, 𝑢1𝜉1<ln𝑏1𝜔𝑒𝐵3𝑎1𝜔𝑞𝑘=1ln1𝛾1𝑘=𝐵8,𝑢1𝜂1<ln𝑏1𝜔𝑒𝐵3𝑎1𝜔𝑞𝑘=1ln1𝛾1𝑘=𝐵9.(3.41) Thus, 𝑢1(𝑡)=𝑢1𝜉1+𝑡𝜉1̇𝑢1(𝑡)𝑑𝑡𝑢1𝜉1+𝜔0||̇𝑢1(||𝑡)𝑑𝑡<𝐵8+2𝑎1𝜔𝑞𝑘=1ln1𝛾1𝑘=𝐵10,𝑢1(𝑡)=𝑢1𝜂1+𝜂1𝑡̇𝑢1(𝑡)𝑑𝑡𝑢1𝜂1+𝜔0||̇𝑢1||(𝑡)𝑑𝑡>𝐵92𝑎1𝜔𝑞𝑘=1ln1𝛾1𝑘=𝐵11.(3.42) From (3.42), we have ||𝑢1||𝐵(𝑡)max10||,||𝐵11||=𝐵12.(3.43) By the second equation of (3.14), we have 𝜔0𝑐(𝑡)𝑒𝐵3𝑑𝑡+𝜔0𝛽1(𝑡)𝑒𝑢3(𝑡)𝑑𝑡>||𝑎2𝑏2||𝜔+𝑞𝑘=1ln1𝛾2𝑘,𝜔0𝑐(𝑡)𝑒𝐵3𝑑𝑡+𝜔0𝛽1(𝑡)𝑒𝑢3(𝑡)𝑑𝑡<||𝑎2𝑏2||𝜔+𝑞𝑘=1ln1𝛾2𝑘.(3.44) Then 𝑐𝜔𝑒𝐵3+𝛽1𝜔𝑒𝑢3(𝜂3)>||𝑎2𝑏2||𝜔+𝑞𝑘=1ln1𝛾2𝑘,𝑐𝜔𝑒𝐵3+𝛽1𝜔𝑒𝑢3(𝜉3)<||𝑎2𝑏2||𝜔+𝑞𝑘=1ln1𝛾2𝑘=𝐵13,(3.45) that is, 𝑢3𝜂3>ln||𝑎2𝑏2||𝜔+𝑞𝑘=1ln1𝛾2𝑘𝑐𝜔𝑒𝐵6𝛽1𝜔,𝑢3𝜉3<ln||𝑎2𝑏2||𝜔+𝑞𝑘=1ln1𝛾2𝑘𝑐𝜔𝑒𝐵3𝛽1𝜔=𝐵14.(3.46) Therefore, we get 𝑢3(𝑡)=𝑢3𝜉3+𝑡𝜉3̇𝑢3(𝑡)𝑑𝑡𝑢3𝜉3+𝜔0||̇𝑢3(||𝑡)𝑑𝑡<𝐵14+2𝑑1𝜔𝑞𝑘=1ln1𝛾3𝑘=𝐵15,𝑢3(𝑡)=𝑢3𝜂3𝜂3𝑡̇𝑢3(𝑡)𝑑𝑡𝑢3𝜂3𝜔0||̇𝑢3||(𝑡)𝑑𝑡>𝐵142𝑑1𝜔𝑞𝑘=1ln1𝛾3𝑘=𝐵16.(3.47) Hence, we have ||𝑢3||||𝐵(𝑡)max15||,||𝐵16||=𝐵17.(3.48)Case 3 (if 𝑢1(𝑡)<0,𝑢2(𝑡)>0). By the first equation of (3.14), we have𝜔0𝑏1(𝑡)𝑒𝑢1(𝑡)𝑑𝑡+𝑞𝑘=1ln1𝛾1𝑘<𝜔0𝑎1(𝑡)𝑑𝑡,𝜔0𝑏1(𝑡)𝑒(𝑢2(𝜂2)𝑢1(𝑡))𝑑𝑡+𝑞𝑘=1ln1𝛾1𝑘>𝜔0𝑎1(𝑡)𝑑𝑡.(3.49) Then 𝑏1𝜔𝑒𝑢1(𝜂1)+𝑞𝑘=1ln1𝛾1𝑘<𝑎1𝜔,𝑏1𝜔𝑒𝐵19𝑒𝑢2(𝜂2)+𝑞𝑘=1ln1𝛾1𝑘>𝑎1𝜔.(3.50) namely, 𝑢1𝜂1>ln𝑏1𝜔𝑎1𝑞𝑘=1ln1𝛾1𝑘=𝐵18,𝑢2𝜂2>ln𝑎1𝜔𝑞𝑘=1ln1𝛾1𝑘𝑏1𝜔𝑒𝐵19=𝐵20.(3.51) Therefore, 𝑢1(𝑡)=𝑢1𝜂1𝜂1𝑡̇𝑢1(𝑡)𝑑𝑡𝑢1𝜂1𝜔0||̇𝑢1||(𝑡)𝑑𝑡>𝐵182𝑎1𝜔𝑞𝑘=1ln1𝛾1𝑘=𝐵19,𝑢2(𝑡)=𝑢2𝜉2+𝑡𝜉2̇𝑢2(𝑡)𝑑𝑡𝑢2𝜉2+𝜔0||̇𝑢2(||𝑡)𝑑𝑡<ln||𝑎2𝑏2||𝜔+𝑞𝑘=1ln1𝛾2𝑘𝑐𝜔+2𝑎2𝜔+𝑞𝑘=1ln1𝛾2𝑘=𝐵21,𝑢2(𝑡)=𝑢2𝜂2𝜂2𝑡̇𝑢2(𝑡)𝑑𝑡𝑢2𝜂1𝜔0||̇𝑢2||(𝑡)𝑑𝑡>𝐵202𝑎2𝜔𝑞𝑘=1ln1𝛾2𝑘=𝐵22.(3.52) So 𝐵19<𝑢1||𝑢(𝑡)<0,2||||𝐵(𝑡)max21||,||𝐵22||=𝐵23.(3.53) By the third equation of (3.14), we obtain 𝜔0𝑒(𝑡)𝑒𝑢3(𝜂3)𝑑𝑡+𝜔0𝛽2(𝑡)𝑒𝐵23𝑑𝑡>||𝑎2𝑏2||𝜔+𝑞𝑘=1ln1𝛾2𝑘,(3.54) that is, 𝑢3𝜂3>ln||𝑎2𝑏2||𝜔+𝑞𝑘=1ln1𝛾2𝑘𝛽2𝜔𝑒𝐵23𝑒𝜔=𝐵24.(3.55) Thus, ||𝑢3||||𝐵(𝑡)max23||,||𝐵24||=𝐵25.(3.56)Case 4 (if 𝑢1(𝑡)<0,𝑢2(𝑡)<0). By the second equation of (3.14), we have𝜔0𝑐(𝑡)𝑑𝑡+𝜔0𝛽1(𝑡)𝑒𝑢3(𝑡)𝑑𝑡>||𝑎2𝑏2||𝜔+𝑞𝑘=1ln1𝛾2𝑘.(3.57) Then 𝑐𝜔+𝛽1𝜔𝑒𝑢3(𝜂3)>||𝑎2𝑏2||𝜔+𝑞𝑘=1ln1𝛾2𝑘,(3.58) that is, 𝑢3𝜂3ln|𝑎2𝑏2|𝜔+𝑞𝑘=1ln1𝛾2𝑘𝑐𝜔𝛽1𝜔=𝐵26.(3.59) Therefore, 𝑢3(𝑡)=𝑢3𝜉3+𝑡𝜉3̇𝑢3(𝑡)𝑑𝑡𝑢3𝜉3+𝜔0||̇𝑢3(||𝑡)𝑑𝑡<ln𝑑𝜔+𝑞𝑘=1ln1𝛾3𝑘𝑒𝜔+2𝑑𝜔+𝑞𝑘=1ln1𝛾3𝑘=𝐵27,𝑢3(𝑡)=𝑢3𝜂3𝜂3𝑡̇𝑢3(𝑡)𝑑𝑡𝑢3𝜂3𝜔0||̇𝑢3(||𝑡)𝑑𝑡>𝐵262𝑑𝜔𝑞𝑘=1ln1𝛾3𝑘=𝐵28.(3.60) Thus, ||𝑢3||||𝐵(𝑡)max27||,||𝐵28||=𝐵29.(3.61) By the second equation of (3.14), we obtain 𝜔0𝑐(𝑡)𝑒𝑢2(𝜂2)𝑑𝑡+𝜔0𝛽1(𝑡)𝑒𝐵29𝑑𝑡>||𝑎2𝑏2||𝜔+𝑞𝑘=1ln1𝛾2𝑘,(3.62) that is, 𝑐𝑒𝑢2(𝜂2)+𝛽1𝜔𝑒𝐵29>||𝑎2𝑏2||𝜔+𝑞𝑘=1ln1𝛾2𝑘.(3.63) Thus, 𝑢2𝜂2>ln||𝑎2𝑏2||𝜔+𝑞𝑘=1ln1𝛾2𝑘𝛽1𝜔𝑒𝐵29.𝑐𝜔(3.64) Then, from (3.16) and (3.20), we get 𝑢2(𝑡)=𝑢2𝜉2+𝑡𝜉2̇𝑢2(𝑡)𝑑𝑡𝑢2𝜉2+𝜔0||̇𝑢2(||𝑡)𝑑𝑡<ln||𝑎2𝑏2||𝜔+𝑞𝑘=1ln1𝛾2𝑘𝑐𝜔+2𝑎2𝜔+𝑞𝑘=1ln1𝛾2𝑘=𝐵31,𝑢2(𝑡)=𝑢2𝜂2𝜂2𝑡̇𝑢2(𝑡)𝑑𝑡𝑢2𝜂2𝜔0||̇𝑢2||(𝑡)𝑑𝑡>𝐵302𝑎2𝜔𝜔𝑞𝑘=1ln1𝛾2𝑘=𝐵32.(3.65) Thus, ||𝑢2||||𝐵(𝑡)max31||,||𝐵32||=𝐵33.(3.66) By the first equation of (3.14), we have 𝜔0𝑏1(𝑡)𝑒𝑢2(𝑡)𝑢1(𝜂1)𝑑𝑡+𝑞𝑘=1ln1𝛾1𝑘<𝜔0𝑎1(𝑡)𝑑𝑡.(3.67) Then 𝑏1𝜔𝑒𝐵33𝑒𝑢1(𝜂1)+𝑞𝑘=1ln1𝛾1𝑘<𝑎1𝜔.(3.68) Thus, 𝑢1𝜂1>ln𝑏1𝜔𝑒𝐵32𝑎1𝜔𝑞𝑘=1ln1𝛾1𝑘=𝐵34.(3.69) Hence, we have 𝐵34<𝑢1(𝑡)<0.(3.70) Based on the discussion above, we can easily obtain 𝑢1𝐵(𝑡)max5,𝐵12,||𝐵19||,||𝐵34||,𝑢2(𝐵𝑡)max3,𝐵4,𝐵23,𝐵33,𝑢3𝐵(𝑡)max8,𝐵17,𝐵25,𝐵29.(3.71) Obviously, 𝐵𝑖(𝑖=1,2,,34) are independent of 𝜆(0,1). Similar to the proof of Theorem 2.1 of [17], we can easily find a sufficiently large 𝑀>0 so that we denote the set 𝑢Ω=𝑢(𝑡)=1(𝑡),𝑢2(𝑡),𝑢3(𝑡)𝑇𝑡𝑥𝑢<𝑀,𝑢+𝑘Ω,𝑘=1,2,,𝑞.(3.72) It is clear that Ω satisfies the requirement (a) in Lemma 3.1.
When(𝑢1(𝑡),𝑢2(𝑡),𝑢3(𝑡))𝑇𝜕ΩKer𝐿=𝜕Ω𝑅3 and 𝑢={(𝑢1,𝑢2,𝑢3)𝑇} is a constant vector in 𝑅3 with 𝑢=(𝑢1(𝑡),𝑢2(𝑡),𝑢3(𝑡))𝑇=𝑀, then we have𝑄𝑁𝑢=𝑎1+𝑏1𝑒𝑢2𝑢1+1𝜔𝑞𝑘=1ln1𝛾1𝑘𝑎2𝑏2𝑐𝑒𝑢2𝛽1𝑒𝑢3+1𝜔𝑞𝑘=1ln1𝛾2𝑘𝑑𝑒𝑒𝑢3𝛽2𝑒𝑢2+1𝜔𝑞𝑘=1ln1𝛾3𝑘,0,,00.(3.73) Letting 𝐽Im𝑄Ker𝐿,(𝑟,0,,0,0)𝑟 and, by direct calculation, we get 𝑢deg𝐽𝑄𝑁1,𝑢2,𝑢3𝑇;𝜕Ωker𝐿;0=signdet𝑏1𝑒𝑢2𝑢1𝑏1𝑒𝑢2𝑢100𝑐𝑒𝑢2𝛽1𝑒𝑢30𝛽2𝑒𝑢2𝑒𝑒𝑢3=sign𝑏1𝛽1𝛽2𝑏1𝑐𝑒𝑒2𝑢2𝑢1+𝑢30.(3.74) This proves that condition (b) in Lemma 3.1 is satisfied. By now, we have proved that Ω verifies all requirements of Lemma 3.1, then it follows that 𝐿𝑢=𝑁𝑢 has at least one solution (𝑢1(𝑡),𝑢2(𝑡),𝑢3(𝑡))𝑇 in Dom𝐿Ω; that is, to say, (3.5) has at least one 𝜔 periodic solution in Dom𝐿Ω. Then we know that ((𝑥1(𝑡),𝑥2(𝑡),𝑥3(𝑡))𝑇=(𝑒𝑢1(𝑡),𝑒𝑥𝑢2(𝑡),𝑒𝑢3(𝑡))𝑇 is an 𝜔 periodic solution of system (2.4) with strictly positive components. This completes the proof.

4. Uniqueness and Global Attractivity of Periodic Solutions

Under the hypotheses (H1),(H2),(H3), we consider the following ordinary differential equation without impulsive:̇𝑧1(𝑡)=𝑧1(𝑡)𝑎1(𝑡)+𝑏1(𝑡)0<𝑡𝑘<𝑡1𝛾2𝑘𝑧2(𝑡)0<𝑡𝑘<𝑡1𝛾1𝑘𝑧1,(𝑡)̇𝑧2(𝑡)=𝑧2(𝑎𝑡)2(𝑡)𝑏2(𝑡)𝑐(𝑡)0<𝑡𝑘<𝑡1𝛾2𝑘𝑧2(𝑡)𝛽1(𝑡)0<𝑡𝑘<𝑡1𝛾3𝑘𝑧3(,𝑡)̇𝑧3(𝑡)=𝑧3(𝑡)𝑑(𝑡)𝑒(𝑡)0<𝑡𝑘<𝑡1𝛾3𝑘𝑧3(𝑡)𝛽2(𝑡)0<𝑡𝑘<𝑡1𝛾2𝑘𝑧2,(𝑡)(4.1) with the initial conditions 𝑧𝑖(0)>0,𝑖=1,2,3.

The following lemmas will be helpful in the proofs of our results. The proof of the following Lemma 4.1 is similar to that of Theorem 1 in [18], and it will be omitted.

Lemma 4.1. Assume that (H1),(H2),(H3) hold, then one has the following.(i)If 𝑧(𝑡)=(𝑧1(𝑡),𝑧2(𝑡),𝑧3(𝑡))𝑇 is a solution of (4.1) on [0,+), then 𝑥𝑖(𝑡)=0<𝑡𝑘<𝑡(1𝛾𝑖𝑘)𝑧𝑖(𝑡)(i=1,2,3) is a solution of (2.4) on [0,+).(ii)If 𝑥(𝑡)=(𝑥1(𝑡),𝑥2(𝑡),𝑥3(𝑡))𝑇 is a solution of (2.4) on [0,+), then 𝑧𝑖(𝑡)=0<𝑡𝑘<𝑡(1𝛾𝑖𝑘)1𝑥𝑖(𝑡)(𝑖=1,2,3) is a solution of (4.1) on [0,+).

Lemma 4.2. Let 𝑧(𝑡)=(𝑧1(𝑡),𝑧2(𝑡),𝑧3(𝑡))𝑇denote any positive solution of system (4.1) with initial conditions 𝑧𝑖(0)>0,𝑖=1,2,3. Assume that the following condition holds, H7𝑎𝑀2>𝑏𝐿2,𝑑𝑀>𝑒𝐿.(4.2) Then there exists a 𝑇3>0 such that 0<𝑧𝑖(𝑡)𝑀𝑖,(𝑖=1,2,3),for𝑡𝑇3,(4.3) where 𝑀1>𝑀1=0<𝑡𝑘<𝑡1𝛾2𝑘𝑀2𝑎𝐿10<𝑡𝑘<𝑡1𝛾1𝑘,𝑀2>𝑀2=𝑎𝑀2𝑏𝐿2𝑐𝐿0<𝑡𝑘<𝑡1𝛾2𝑘,𝑀3>𝑀3=𝑑𝑀𝑒𝐿𝑒𝐿0<𝑡𝑘<𝑡1𝛾3𝑘.(4.4)

Proof. From the second equation of (4.1), we can obtain ̇𝑧2(𝑡)𝑧2𝑎(𝑡)2(𝑡)𝑏2(𝑡)𝑐(𝑡)0<𝑡𝑘<𝑡1𝛾2𝑘𝑧2(𝑡)𝑧2𝑎(𝑡)𝑀2𝑏𝐿2𝑐𝐿0<𝑡𝑘<𝑡1𝛾2𝑘𝑧2.(𝑡)(4.5) By (4.5), we can derive the following.(A1)If 𝑧2(0)𝑀2, then 𝑧2(𝑡)𝑀2,𝑡0.(A2)If 𝑧2(0)>𝑀2, let 𝛼1=𝑀2[𝑎𝑀2𝑏𝐿2𝑐𝐿0<𝑡𝑘<𝑡(1𝛾2𝑘)𝑀2],(𝛼1>0). Then there exists 𝜀1>0 such that 𝑡[0,𝜀1), then 𝑧2(𝑡)>𝑀2, and also we have ̇𝑧2(𝑡)<𝛼1<0.(4.6) From what has been discussed above, we can easily conclude that, if 𝑧2(0)>𝑀2, then 𝑧2(𝑡) is strictly monotone decreasing with speed at least 𝛼1. Therefore, there exists a 𝑇1>0 such that 𝑡>𝑇1, then 𝑧2(𝑡)𝑀2.
From the third equation of (4.1), we can obtaiṅ𝑧3(𝑡)𝑧3(𝑡)𝑑(𝑡)𝑒(𝑡)0<𝑡𝑘<𝑡1𝛾3𝑘𝑧3(𝑡)𝑧3𝑑(𝑡)𝑀𝑒𝐿0<𝑡𝑘<𝑡1𝛾3𝑘𝑧3.(𝑡)(4.7) By (4.7), we can derive the following(B1)If 𝑧3(0)𝑀3, then 𝑧3(𝑡)𝑀3,𝑡0.(B2)If 𝑧3(0)>𝑀3, let 𝛼2=𝑀3[𝑑𝑀𝑒𝐿0<𝑡𝑘<𝑡(1𝛾3𝑘)𝑀3],(𝛼2>0). Then there exists 𝜀2>0 such that 𝑡[0,𝜀2), then 𝑧3(𝑡)>𝑀3, and also we have ̇𝑧3(𝑡)<𝛼2<0.(4.8) From what has been discussed above, we can easily conclude that, if 𝑧3(0)>𝑀3, then 𝑧3(𝑡) is strictly monotone decreasing with speed at least 𝛼2. Therefore, there exists a 𝑇2>0 such that 𝑡>𝑇2, then 𝑧3(𝑡)𝑀3.
From the first equation of (4.1), we can obtaiṅ𝑧1(𝑡)𝑧1(𝑡)𝑎1(𝑡)+𝑏1(𝑡)0<𝑡𝑘<𝑡1𝛾2𝑘𝑀20<𝑡𝑘<𝑡1𝛾1𝑘𝑧1(𝑡)=𝑎1(𝑡)𝑧1(𝑡)+0<𝑡𝑘<𝑡1𝛾2𝑘𝑀20<𝑡𝑘<𝑡1𝛾1𝑘𝑎𝐿1𝑧1(𝑡)+0<𝑡𝑘<𝑡1𝛾2𝑘𝑀20<𝑡𝑘<𝑡1𝛾1𝑘.(4.9) Then we have 𝑧1(𝑡)𝑀1,for𝑡𝑇1.(4.10) Set 𝑇3=max{𝑇1,𝑇2}, then we have 0<𝑧𝑖(𝑡)𝑀𝑖,(𝑖=1,2,3),for𝑡𝑇3.(4.11) The proof is complete.

Lemma 4.3. Let (H1),(H2),(H3) hold. Assume that the following condition holds. H8𝑎𝐿2>𝑏𝑀2+𝛽𝑀10<𝑡𝑘<𝑡1𝛾3𝑘,𝑑𝐿>𝑒𝑀𝛽𝑀20<𝑡𝑘<𝑡1𝛾2𝑘.(4.12) Then there exists positive constants 𝑇>0 and 𝑚𝑖(𝑖=1,2,3) such that, for all 𝑡>𝑇, 𝑚𝑖<𝑧𝑖(𝑡),(𝑖=1,2,3),for𝑡𝑇,(4.13) in which 𝑚1<𝑚1=𝑏𝐿10<𝑡𝑘<𝑡1𝛾2𝑘𝑚2𝑎𝑀10<𝑡𝑘<𝑡1𝛾1𝑘,𝑚2<𝑚2=𝑎𝐿2𝑏𝑀2𝛽𝑀10<𝑡𝑘<𝑡1𝛾3𝑘𝑐𝑀0<𝑡𝑘<𝑡1𝛾2𝑘,𝑚3<𝑚3=𝑑𝐿𝑒𝑀𝛽𝑀20<𝑡𝑘<𝑡1𝛾2𝑘𝑀2𝑒𝑀0<𝑡𝑘<𝑡1𝛾3𝑘.(4.14)

Proof. By the second equation of (4.1), It is easy to obtain that, for 𝑡𝑇3, ̇𝑧2(𝑡)𝑧2𝑎(𝑡)𝐿2𝑏𝑀2𝑐𝑀0<𝑡𝑘<𝑡1𝛾2𝑘𝑧2(𝑡)𝛽𝑀10<𝑡𝑘<𝑡1𝛾3𝑘𝑀3,(4.15) where 𝑇3 is defined in Lemma 4.1. (C1)If 𝑧2(𝑇3)𝑚2, then 𝑧2(𝑡)𝑚2,𝑡𝑇3. (C2)If 𝑧2(𝑇3)<𝑚2 and let 𝜇1=𝑧2𝑇3𝑎𝐿2𝑏𝑀2𝑐𝑀0<𝑡𝑘<𝑡1𝛾2𝑘𝑚2,(4.16) then there exists 𝜀3>0 such that 𝑡[𝑇3,𝑇3+𝜀3), then 𝑧2(𝑡)>𝑚2, and also we have ̇𝑧2(𝑡)>𝜇1>0.(4.17) Then we know that if 𝑧2(𝑇3)<𝑚2, 𝑧2(𝑡) will strictly monotonically increase with speed 𝜇2. Thus, there exists 𝑇4>𝑇3 such that if 𝑡𝑇4, then 𝑧2(𝑡)𝑚2.
By the third equation of (4.1), It is easy to obtain that for 𝑡𝑇3,̇𝑧3(𝑡)𝑧3𝑑(𝑡)𝐿𝑒𝑀0<𝑡𝑘<𝑡1𝛾3𝑘𝑧3(𝑡)𝛽𝑀20<𝑡𝑘<𝑡1𝛾2𝑘𝑀2,(4.18) where 𝑇3 is defined in Lemma 4.2. (D1)If 𝑧2(𝑇3)𝑚3, then 𝑧3(𝑡)𝑚3,𝑡𝑇3.(D2)If 𝑧2(𝑇3)<𝑚3, and let 𝜇2=𝑧3𝑇3𝑑𝐿𝑒𝑀0<𝑡𝑘<𝑡1𝛾3𝑘𝑚3𝛽𝑀20<𝑡𝑘<𝑡1𝛾2𝑘𝑀2,(4.19) then there exists 𝜀4>0 such that 𝑡[𝑇3,𝑇3+𝜀4), then 𝑧3(𝑡)>𝑚3, and also we have ̇𝑧3(𝑡)>𝜇2>0.(4.20) Then we know that if 𝑧3(𝑇3)<𝑚3, 𝑧3(𝑡) will strictly monotonically increase with speed 𝜇2. Thus, there exists 𝑇5>𝑇3 such that, if 𝑡𝑇5, then 𝑧3(𝑡)𝑚3.
Finally, by the third equation of (4.1), we obtaiṅ𝑧1(𝑡)𝑧1(𝑡)𝑎𝑀1+𝑏𝐿10<𝑡𝑘<𝑡1𝛾2𝑘𝑚20<𝑡𝑘<𝑡1𝛾1𝑘𝑧1(𝑡)=𝑎𝑀1𝑧1(𝑡)+𝑏𝐿10<𝑡𝑘<𝑡1𝛾2𝑘𝑚20<𝑡𝑘<𝑡1𝛾1𝑘𝑧1.(𝑡)(4.21) Thus, we have 𝑧1(𝑡)𝑚1,(4.22) for 𝑡𝑇4. Set 𝑇=max{𝑇4,𝑇5}, then we have 𝑧𝑖(𝑡)>𝑚𝑖,(𝑖=1,2,3),for𝑡𝑇.(4.23) In the sequel, we formulate the uniqueness and global attractivity of the 𝜔 periodic solution 𝑥(𝑡) in Theorem 4.4. It is immediate that if 𝑥(𝑡) is global attractivity, then 𝑥(𝑡) is in fact unique.

Theorem 4.4. In addition to (H1)(H8), assume further (H9)lim𝑡inf𝐵𝑖(𝑡)>0, where 𝐵1(𝑡)=𝑐(𝑡)𝛽2𝑏(𝑡)𝑀𝑚10<𝑡𝑘<𝑡1𝛾1𝑘0<𝑡𝑘<𝑡1𝛾2𝑘,𝐵2(𝑡)=𝑒(𝑡)𝛽1(𝑡)0<𝑡𝑘<𝑡1𝛾3𝑘.(4.24) Then system (2.4) has a unique positive 𝜔 periodic solution 𝑥(𝑡)=(𝑥1(𝑡),𝑥2(𝑡),𝑥3(𝑡))𝑇 which is global attractivity.

Proof. According to the conclusion of Theorem 3.2, we only need to show that the positive periodic solution of (2.4) is global asymptotical stable. Let 𝑥(𝑡)=(𝑥1(𝑡),𝑥2(𝑡),𝑥3(𝑡))𝑇 be a positive 𝜔 periodic solution of system (2.4) let 𝑥(𝑡)=(𝑥1(𝑡),𝑥2(𝑡),𝑥3(𝑡))𝑇 be any positive solution of system (2.4). Then 𝑧(𝑡)=(𝑧1(𝑡),𝑧2(𝑡),𝑧3(𝑡))𝑇,(𝑧1(𝑡)=0<𝑡𝑘<𝑡(1𝛾1𝑘)𝑥1(𝑡),𝑧2(𝑡)=0<𝑡𝑘<𝑡(1𝛾2𝑘)𝑥2(𝑡),𝑧3(𝑡)=0<𝑡𝑘<𝑡𝛾(13𝑘)𝑥3(𝑡)) is the positive 𝜔 periodic solution of (4.1), and 𝑧(𝑡) is the positive solution of (4.1). It follows from Lemma 4.2 and 4.3 that there exists positive constants 𝑇>0,𝑀𝑖 and 𝑚𝑖 (defined by Lemmas 4.2 and 4.3, resp.) such that, for all 𝑡>𝑇, 𝑚𝑖<𝑧𝑖(𝑡)𝑀𝑖,𝑚𝑖<𝑧𝑖(𝑡)𝑀𝑖,𝑖=1,2,3.(4.25) Define ||𝑉(𝑡)=ln𝑧1(𝑡)ln𝑧1||+||(𝑡)ln𝑧2(𝑡)ln𝑧2||+||(𝑡)ln𝑧3(𝑡)ln𝑧3||.(𝑡)(4.26) Calculating the upper-right derivative of 𝑉(𝑡) along the solution of (4.1), it follows for 𝑡𝑇that 𝐷+𝑉(𝑡)=3𝑖=1𝑧𝑖(𝑡)𝑧𝑖𝑧(𝑡)𝑖(𝑡)𝑧𝑖𝑧(𝑡)sgn𝑖(𝑡)𝑧𝑖𝑧(𝑡)=sgn1(𝑡)𝑧1𝑏(𝑡)1(𝑡)0<𝑡𝑘<𝑡1𝛾2𝑘0<𝑡𝑘<𝑡1𝛾1𝑘𝑧2(𝑡)𝑧1𝑧(𝑡)2(𝑡)𝑧1𝑧(𝑡)+sgn2(𝑡)𝑧2(𝑡)𝑐(𝑡)0<𝑡𝑘<𝑡1𝛾2𝑘𝑧2(𝑡)𝑧2(𝑡)𝛽1(𝑡)0<𝑡𝑘<𝑡1𝛾3𝑘𝑧3(𝑡)𝑧3(𝑧𝑡)+sgn3(𝑡)𝑧3(𝑡)𝑒(𝑡)0<𝑡𝑘<𝑡1𝛾3𝑘𝑧3(𝑡)𝑧3(𝑡)𝛽2(𝑡)0<𝑡𝑘<𝑡1𝛾2𝑘𝑧2(𝑡)𝑧2(𝑡)𝑐(𝑡)0<𝑡𝑘<𝑡1𝛾2𝑘||𝑧2(𝑡)𝑧2||(𝑡)+𝛽1(𝑡)0<𝑡𝑘<𝑡1𝛾3𝑘||𝑧3(𝑡)𝑧3||(𝑡)𝑒(𝑡)0<𝑡𝑘<𝑡1𝛾3𝑘||𝑧3(𝑡)𝑧3||(𝑡)+𝛽2(𝑡)0<𝑡𝑘<𝑡1𝛾2𝑘||𝑧2(𝑡)𝑧2||(𝑡)+𝐷1(𝑡),(4.27) where 𝐷1𝑏(𝑡)=1(𝑡)0<𝑡𝑘<𝑡1𝛾2𝑘0<𝑡𝑘<𝑡1𝛾1𝑘𝑧2(𝑡)𝑧1𝑧(𝑡)2(𝑡)𝑧1(𝑡),𝑧1(𝑡)>𝑧1𝑏(𝑡),1(𝑡)0<𝑡𝑘<𝑡1𝛾2𝑘0<𝑡𝑘<𝑡1𝛾1𝑘𝑧2(𝑡)𝑧1𝑧(𝑡)2(𝑡)𝑧1(𝑡),𝑧1(𝑡)<𝑧1(𝑡).(4.28) In the sequel, we will estimate 𝐷1(𝑡) under the following two cases.(i)If 𝑧1(𝑡)𝑧1(𝑡), then𝐷1𝑏(𝑡)1(𝑡)0<𝑡𝑘<𝑡1𝛾2𝑘𝑧1(𝑡)0<𝑡𝑘<𝑡1𝛾1𝑘𝑧2(𝑡)𝑧2𝑏(𝑡)𝑀0<𝑡𝑘<𝑡1𝛾2𝑘𝑚10<𝑡𝑘<𝑡1𝛾1𝑘||𝑧2(𝑡)𝑧2||.(𝑡)(4.29)(ii)If 𝑧1(𝑡)<𝑧1(𝑡), then𝐷1𝑏(𝑡)1(𝑡)0<𝑡𝑘<𝑡1𝛾2𝑘𝑧1(𝑡)0<𝑡𝑘<𝑡1𝛾1𝑘𝑧2(𝑡)𝑧2𝑏(𝑡)𝑀0<𝑡𝑘<𝑡1𝛾2𝑘𝑚10<𝑡𝑘<𝑡1𝛾1𝑘||𝑧2(𝑡)𝑧2||.(𝑡)(4.30) Combining the conclusions of (4.29) and (4.30), we obtain 𝐷1𝑏(𝑡)𝑀0<𝑡𝑘<𝑡1𝛾2𝑘𝑚10<𝑡𝑘<𝑡1𝛾1𝑘||𝑧2(𝑡)𝑧2||(𝑡).(4.31) It follows from (4.31) that 𝐷+𝑉(𝑡)𝑐(𝑡)0<𝑡𝑘<𝑡1𝛾2𝑘||𝑧2(𝑡)𝑧2||(𝑡)+𝛽1(𝑡)0<𝑡𝑘<𝑡1𝛾3𝑘||𝑧3(𝑡)𝑧3||(𝑡)𝑒(𝑡)0<𝑡𝑘<𝑡1𝛾3𝑘||𝑧3(𝑡)𝑧3||(𝑡)+𝛽2(𝑡)0<𝑡𝑘<𝑡1𝛾2𝑘||𝑧2(𝑡)𝑧2||+𝑏(𝑡)𝑀0<𝑡𝑘<𝑡1𝛾2𝑘𝑚10<𝑡𝑘<𝑡1𝛾1𝑘||𝑧2(𝑡)𝑧2||=𝑏(𝑡)𝑀𝑚10<𝑡𝑘<𝑡1𝛾1𝑘𝑐(𝑡)+𝛽2(𝑡)0<𝑡𝑘<𝑡1𝛾2𝑘||𝑧2(𝑡)𝑧2||+𝛽(𝑡)1(𝑡)𝑒(𝑡)0<𝑡𝑘<𝑡1𝛾3𝑘||𝑧3(𝑡)𝑧3||𝐵(𝑡)1||𝑧(𝑡)2(𝑡)𝑧2||(𝑡)+𝐵2||𝑧(𝑡)3(𝑡)𝑧3||,(𝑡)(4.32) where 𝐵1(𝑡) and 𝐵2(𝑡) are defined in Theorem 4.4. By hypothesis (H8), there exist constants 𝛼𝑖,(𝑖=2,3) and 𝑇>𝑇 such that 𝐵𝑖(𝑡)𝛼𝑖>0,(𝑖=2,3),for𝑡𝑇.(4.33) Integrating both sides of (4.32) on interval [𝑇,𝑡] yields 𝑉(𝑡)+3𝑖=2𝑡𝑇𝐵𝑖||𝑧(𝑡)𝑖(𝑡)𝑧𝑖||𝑇(𝑡)𝑑𝑠𝑉.(4.34) It follows from (4.33) and (4.34) that 3𝑖=2𝑡𝑇𝐵𝑖||𝑧(𝑡)𝑖(𝑡)𝑧𝑖||𝑇(𝑡)𝑑𝑠𝑉<,for𝑡𝑇.(4.35) Since 𝑧𝑖(𝑡) and 𝑧𝑖(𝑡) (𝑖=2,3) are bounded for 𝑡𝑇, so |𝑧𝑖(𝑡)𝑧𝑖(𝑡)|(𝑖=2,3) are uniformly continuous on [𝑇,). By Barbalat’s Lemma [32], we have lim𝑡||𝑧𝑖(𝑡)𝑧𝑖||(𝑡)=lim𝑡0<𝑡𝑘<𝑡1𝛾𝑖𝑘1||𝑥𝑖(𝑡)𝑥𝑖||(𝑡)=0,(𝑖=2,3).(4.36) Thus, lim𝑡||𝑥𝑖(𝑡)𝑥𝑖||(𝑡)=0,(𝑖=2,3).(4.37) By (4.37) and the first equation of (2.4), one can easily obtain that lim𝑡||𝑥1(𝑡)𝑥1||(𝑡)=0.(4.38) By Theorems 7.4 and 8.2 in [33], we know that the positive periodic solution 𝑥(𝑡)=(𝑥1(𝑡),𝑥2(𝑡),𝑥3(𝑡))𝑇 of (2.4) is uniformly asymptotically stable. The proof of Theorem 4.4 is complete.

5. An Example

As an application of our main results, we consider the following system:̇𝑥1(𝑡)=2𝑥1(𝑡)+𝑥2(𝑡),𝑡𝑡𝑘,̇𝑥2(𝑡)=(4+cos𝑡)𝑥2(𝑡)(2+cos𝑡)𝑥2(𝑡)(1sin𝑡)𝑥221(𝑡)2𝑒200𝜋+1+sin𝑡𝑥2(𝑡)𝑥3(𝑡),𝑡𝑡𝑘,̇𝑥3(𝑡)=𝑥3(𝑡)50+sin𝑡50𝑒200𝜋𝑥+1+sin𝑡3(𝑡)49𝑒50𝑥3𝜋cos𝑡2(𝑡),𝑡𝑡𝑘,Δ𝑥1𝑡𝑘=𝑥1𝑡+𝑘𝑥1𝑡𝑘1=2𝑥1𝑡𝑘,𝑘=1,2,,Δ𝑥2𝑡𝑘=𝑥2𝑡+𝑘𝑥2𝑡𝑘1=3𝑥2𝑡𝑘,𝑘=1,2,,Δ𝑥3𝑡𝑘=𝑥3𝑡+𝑘𝑥3𝑡𝑘1=4𝑥3𝑡𝑘,𝑘=1,2,,(5.1) in which 𝑡𝑘+2=𝑡𝑘+2𝜋, [0,2𝜋]{𝑡𝑘}={𝑡1,𝑡2}, 𝑎1(𝑡)=2, 𝑏1(𝑡)=1, 𝑎2(𝑡)=4+cos𝑡, 𝑏2(𝑡)=2+cos𝑡, 𝛽1(𝑡)=(1/2𝑒200𝜋+1)+sin𝑡, 𝛽2(𝑡)=(49/𝑒503𝜋)cos𝑡, 𝑐(𝑡)=1sin𝑡, 𝑑(𝑡)=50+sin𝑡, and 𝑒(𝑡)=50𝑒200𝜋+1+sin𝑡. By direct computation, we can obtain 𝑎1=2,𝑎2=4,𝑐=1,|𝑎2𝑏2|=2,𝑑=50,𝛽1=12𝑒200𝜋+1,𝛽2=49𝑒50,3𝜋𝑒=50𝑒200𝜋+1,𝑏1𝐵=1,1=ln4𝜋+2ln(2/3)22𝜋+16𝜋+2ln3𝐵50.07,2=ln4𝜋2ln(1/2)22𝜋16𝜋2ln3𝐵47.07,6=𝐵27=ln100𝜋2ln(3/4)2𝜋50𝑒200𝜋3+200𝜋+2ln4𝐵0.5754,26=ln2𝜋+2ln(2/3)𝜋/𝑒200𝜋+1𝐵629.55,28=𝐵263200𝜋2ln42.9362.(5.2) Then 𝐵450.07,𝐵292.9362. It is easy to check that (5.1) satisfies all the conditions of Theorems 3.2 and 4.4; hence, (5.1) has a positive 2𝜋 periodic solution which is global attractivity.

Acknowledgments

The authors would like to thank the referees for their helpful comments and valuable suggestions regarding this paper. This work is supported by the National Natural Science Foundation of China (no. 10771215), the Scientific Research Fund of Hunan Provincial Education Department (no. 10C0560), the Doctoral Foundation of Guizhou College of Finance and Economics (2010) and the Science and technology Program of Hunan Province (no. 2010FJ6021).