# Existence of Multiple Positive Periodic Solutions to Two Species Parasitical Model with Impulsive Effects and Harvesting Terms

Authors: Zhouhong Li
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

By applying Mawhin’s continuation theorem of coincidence degree theory and some skills of inequalities, we establish the existence of four positive solutions for two species parasitical system with impulsive effects and harvesting terms. Finally, an example is given to illustrate the effectiveness of our results.


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## Body

## 1. Introduction

In recent years, the existence of periodic solutions in biological models has been widely studied. Models with harvesting terms are often considered. Generally, the model with harvesting terms is described as follows:(1)x˙1=x1f(x1,x2)-h,x˙2=x2g(x1,x2)-k,
where x1 and x2 are functions of two species, respectively; h and k are harvesting terms standing for the harvests (see [1, 2]). Because of the effect of changing environment such as the weather, season, and food, the number of species population periodically varies with the time. The rate of change usually is not a constant. Motivated by this, we consider the periodic nonautonomous population models. For example, two species parasitical system with harvesting terms is as follows [3]:
(2)x˙=x(t)(a1(t)-b1(t)x(t))-h1(t),y˙=y(t)(a2(t)-b2(t)y(t)+c(t)x(t))-h2(t),
where x(t) and y(t) denote the densities of the host and the parasites, respectively; ai(t),bi(t),c(t), and hi(t)(i=1,2) are all positive continuous functions and denote the intrinsic growth rate, death rate, obtaining nutriment rate from the host, and harvesting rate, respectively. In the model (2), the parasitical influence on its host is negligible. As we know, in population dynamics, many evolutionary processes experience short-time rapid chance after undergoing relatively long sooth variation. Examples include stocking of species and annual immigration. Incorporating these phenomena gives us impulsive differential equations. For the theory of impulsive differential equations, we refer the reader to [4, 5].However, to the best of our knowledge, there are few results on the existence of multiple periodic solutions for the delay parasitical with impulsive effects in the literatures. This motivates us to consider the existence of multiple periodic solutions for following parasitical with impulsive effects and harvesting terms nonautonomous model:(3)x˙1(t)=x1(t)(a1(t)-b1(t)x1(t-τ11(t)))-h1(t),t≠tk,x˙2(t)=x2(t)(a2(t)-b2(t)x2(t-τ22(t))+c(t)x1(t-τ21(t)))-h2(t),t≠tk,Δxi(tk)=xi(tk+)-xi(tk-)=gikxi(tk),t=tk,
where gik∈(-1,+∞)(i=1,2;k∈ℕ={1,2}).    {tk}k∈ℕ is a strictly increasing sequence with t1>0 and limt→∞tk=∞.  xi(t)(i=1,2) is the ith species population density.   ai(t)>0(i=1,2) denotes the intrinsic growth rate; bi(t)>0 and hi(t)>0(i=1,2) stand for death rate, obtaining nutriment rate from the host, and harvesting rate, respectively. c(t)>0 represents obtaining nutriment rate from the host; τ21(t)≥0 stands for the time-lag in the process of transformation from the 1th species to the 2th species. τii(t)≥0(i=1,2) represents the time-lag in the process of intraspecific competition.In addition, the effects of a periodically varying environment are important for evolutionary theory as the selective forces on systems in a fluctuating environment differ from those in a stable environment. Therefore, the assumptions of periodicity of the parameters are a way of incorporating the periodicity of the environment (e.g, seasonal effects of weather, food supplies, mating habits, etc.), which leads us to assume thatai(t),bi(t),c(t),τii(t),τ21(t), and hi(t)(i=1,2) are all continuous ω-periodic functions. For impulsive effects, we further assume that there exists a q∈ℕ such that gi(k+q)=gik(i=1,2) and tk+q=tk+ω.Since a very basic and important problem in the study of a population growth model with a periodic environment is the global existence and stability of a positive periodic solution, which plays a similar role as a globally stable equilibrium does in an autonomous model, also, on the existence of positive periodic solutions to system (3), few results are found in the literatures. This motivates us to investigate the existence of a positive periodic or multiple positive periodic solutions for system (3). In fact, it is more likely for some biological species to take on multiple periodic change regulations and have multiple local stable periodic phenomena. Therefore, it is essential for us to investigate the existence of multiple positive periodic solutions for population models. Our main aim of this paper is by using Mawhin’s continuation theorem of coincidence degree theory to establish the existence of four positive periodic solutions for system (3). For the work concerning the multiple existence of periodic solutions of periodic population models which was done using coincidence degree theory, we refer the reader to [6–8].This paper is organized as follows. In Section2, by using the continuation theorem of coincidence degree theory and the skills of inequalities, we establish the existence of at least four positive periodic solutions of system (3). An example is presented in the last section to illustrate the effectiveness of our results.


## 2. Existence of at Least Four Positive Periodic Solutions

We first summarize a few concepts from the book by Gaines and Mawhin [9].LetX and Z be real normed vector spaces. Let L:DomL⊂X→Z be a linear mapping and N:X×[0,1]→Z a continuous mapping. The mapping L will be called a Fredholm mapping of index zero if dimKerL = codimImL<∞, and ImL is closed in Z. If L is a Fredholm mapping of index zero, then there exist continuous projectors P:X→X and Q:Z→Z such that ImP=KerL and KerQ=ImL=Im(I-Q), and X=KerL⊕KerP,Z=ImL⊕ImQ. It follows that L|DomL∩KerP:(I-P)X→ImL is invertible, and its inverse is denoted by KP. If Ω is a bounded open subset of X, the mapping N is called L-compact on Ω-×[0,1], and if QN(Ω-×[0,1]) is bounded, KP(I-Q)N:Ω-×[0,1]→X is compact. Because ImQ is isomorphic to KerL, there exists an isomorphism J:ImQ→KerL.Lemma 1 (see [9]).
LetL be a Fredholm mapping of index zero, and let N be L-compact on Ω-×[0,1]. Assume (a)
for eachλ∈(0,1), every solution x of Lx=λN(x,λ) is such that x∉∂Ω∩DomL;(b)
QN(x,0)x≠0 for each x∈∂Ω∩KerL;(c)
deg(JQN(x,0),Ω∩KerL,0)≠0.
ThenLx=N(x,1) has at least one solution in Ω¯∩DomL.LetT be a given positive constant, and a finite number of points of the sequence {τk} lies in the interval [0,T]. Let PC([0,T],ℝn) be the set of functions x:[0,T]→ℝn which are piecewise continuous in [0,T] and have points of discontinuous τk∈[0,T], where they are continuous from the left. In the set PC([0,T],ℝn) introduce the norm ∥x∥=sup|x(t)|:t∈[0,T] with which PC([0,T],ℝn) becomes a Banach space with the uniform convergence topology.Definition 2.
The setℱ⊂PC([0,T],ℝn) is said to be quasiequicontinuous in [0,T], if for any ϵ>0, there exists δ>0 such that if x∈ℱ;k∈ℤ;t1,t2∈(τk-1,τk]∩[0,T] and |t1-t2|<δ, then |x(t1)-x(t2)|<ϵ.The following result called compactness criterion gives a necessary and sufficient condition for relative compactness inPC([0,T],ℝn).Lemma 3 (see [4]).
The setℱ⊂PC([0,T],ℝn) is relatively compact if and only if(a)
ℱ is bounded, namely, ∥x∥≤c, for each x∈ℱ and some c>0;(b)
ℱ is quasiequicontinuous in [0,T].For the sake of convenience, we denotef-=(1/ω)∫0ω‍f(t)dt; here f(t) is a continuous ω-periodic function. Let PCω denote the space of ω-periodic functions ψ:ℝ→ℝ which are continuous for t≠tk, are continuous from the left for t∈ℝ, and have possible discontinuities of the first kind at points t=tk; that is, the limit from the right of tk exists but may be different from the value at tk. We also denote PCω1={ψ∈PCω:ψ˙∈PCω}. Obviously, ψ is continuous if ψ∈PCω1.For simplicity, we need to introduce some notations as follows:(4)gi=12∑k=1q‍ln(1+gik),Gi=12∑k=1q‍[ln(1+gik)+|ln(1+gik)|],ρi=ea¯iω+Gi,l1±=(a-1+2g1/ω)ρ1±[(a-1+2g1/ω)ρ1]2-4b-1h-1ρ12b-1,l2±=(±[(c-l1++a2+2g2ω)ρ2]2-4b-2h-2ρ2)(c-l1++a2+2g2ω)ρ2±[(c-l1++a2+2g2ω)ρ2]2-4b-2h-2ρ2)(2b-2)-1,Ai±=(a-i+2gi/ω)±(a-i+2gi/ω)2-4b-ih-iρi2b-i,Bi±=Ai±ρi,Hi=li-ρi,
where i=1,2.Throughout this paper, we need the following assumptions:(T1)
(a-i+2gi/ω)ρi-1/2>2b-ih-i(i=1,2);(T2)
ρi>1(i=1,2).The following results will play an important role in the proof of our main result.Lemma 4 (see [5]).
Supposeψ∈PCω1, then(5)0≤sups∈[0,ω]ψ(s)-infs∈[0,ω]ψ(s)≤12[∫0ω‍|ψ′(s)|ds+∑k=1q‍|Δψ(tk)|].Lemma 5.
Letx>0,y>0,z>0, and x>2yz; for the functions f(x,y,z)=(x+x2-4yz)/2z and g(x,y,z)=(x-x2-4yz)/2z, the following assertions hold. (1)
f(x,y,z) and g(x,y,z) are monotonically increasing and monotonically decreasing on the variable x∈(0,∞), respectively.(2)
f(x,y,z) and g(x,y,z) are monotonically decreasing and monotonically increasing on the variable y∈(0,∞), respectively.(3)
f(x,y,z) and g(x,y,z) are monotonically decreasing and monotonically increasing on the variable z∈(0,∞), respectively.Proof.
In fact, for allx>0,y>0, and z>0, we have(6)∂f∂x=x+x2-4yz2zx2-4yz>0,∂g∂x=x2-4yz-x2zx2-4yz<0,∂f∂y=-1x2-4yz<0,∂g∂y=1x2-4yz>0,∂f∂z=-(x+x2-4yz)24z2x2-4yz<0,∂g∂z=(x-x2-4yz)24z2x2-4yz>0.
By the relationship of the derivative and the monotonicity, the above assertions obviously hold. The proof of Lemma 5 is complete.Lemma 6.
Assume that(T1) and (T2) hold, then one has the following inequalities: (1)
lnli+>lnAi+>lnAi->lnli->lnHi(i=1,2);(2)
lnli+>lnBi+>lnBi->lnli->lnHi(i=1,2);(3)
min{lnAi+,lnBi+}>max{lnAi-,lnBi-}(i=1,2).Proof.
Since(7)(a-1+2g1ω)ρ1>a-1+2g1ω>a-1+2g1/ωρ1>2b-1h-1>0,(c-l2++2a2+2g2ω)ρ2>a-2+2g2ω>2g2/ω+a-2ρ2>2b-2h-2>0,b-i>0,h-i>0,i=1,2.
Applying Lemma 5, we have
(8)0<H1=l1-ρ1<l1-=g((a-1+2g1ω)ρ1,b-1,h-1ρ1)<g(a-1+2g1ω,b-1,h-1ρ1)=A1-<A1+=f(a-1+2g1ω,b-1,h-1ρ1)<f((a-1+2g1ω)ρ1,b-1,h-1ρ1)=l1+,0<H2=l2-ρ2<l2-=g((c-l2++a-2+2g2ω)ρ2,b-2,h-2ρ2)<g(2g2ω+a-2,b-2,h-2ρ2)=A2-<A2+=f(2g2ω+a-2,b-2,h-2ρ2)<f((c-l1++a-2+2g2ω)ρ2,b-2,h-2ρ2)=l2+.
Thus, we havelnHi<lnli-<lnAi-<lnAi+<lnli+,i=1,2. The proof of (1) of Lemma 6 is complete. For (2) of Lemma 6, we similarly have
(9)0<H1=l1-ρ1<l1-=g((a-1+2g1ω)ρ1,b-1,h-1ρ1)<g(a-1+2g1ω,b-1,h-1ρ1)=B1-<B1+=f(a-1+2g1ω,b-1,h-1ρ1)<f((a-1+2g1ω)ρ1,b-1,h-1ρ1)=l1+,0<H2=l2-ρ2<l2-=g((a-2+2g2ω)ρ2,b-2,h-2ρ2)<g((2g2ω+a-2)ρ2,b-2,h-2ρ2)=B2-<B2+=f((2g2ω+a-2)ρ2,b-2,h-2ρ2)<f((c-l1++a-2+2g2ω)ρ2,b-2,h-2ρ2)=l2+,
which imply that lnHi<lnli-<lnBi-<lnBi+<lnli+(i=1,2); that is, the inequality (2) of Lemma 6 holds. Let us now prove the inequality (3) of Lemma 6. In fact, since
(10)max{lnAi-,lnBi-}=lnAi-+lnBi-+|lnAi--lnBi-|2(i=1,2),min{lnAi+,lnBi+}=lnAi++lnBi++|lnAi+-lnBi+|2(i=1,2),
we have
(11)min{lnAi+,lnBi+}-max{lnAi-,lnBi-}=(lnAi++lnBi+-lnAi--lnBi--|lnAi--lnBi-|-|lnAi+-lnBi+|)(2)-1.
IfAi+≥Bi+, then Ai-≤Bi-, and
(12)min{lnAi+,lnBi+}-max{lnAi-,lnBi-}=lnBi+-lnBi->0.
IfAi+<Bi+, then Ai->Bi-, and
(13)min{lnAi+,lnBi+}-max{lnAi-,lnBi-}=lnAi+-lnAi->0.
Hence, we drive min{lnAi+,lnBi+}>max{lnAi-,lnBi-}. From the above, we have completed the proof of Lemma 6.Theorem 7.
Assume that(T1) and (T2) hold. Then system (3) has at least four positive ω-periodic solutions.Proof.
By making the substitution(14)xi(t)=eui(t),i=1,2,
system (3) can be reformulated as
(15)u˙1(t)=a1(t)-b1(t)eu1(t-τ11(t))-h1(t)e-u1(t),t≠tk,u˙2(t)=a2(t)-b2(t)eu2(t-τ22(t))+c(t)eu1(t-τ21(t))-h2(t)e-u2(t),t≠tk,Δui(tk)=ui(tk+)-ui(tk-)=ln(1+gik),i=1,2,t=tk.
Let(16)X={u=(u1,u2)T:ui∈PCω1,i=1,2},Z=X×ℝ2q,
and define
(17)∥u∥0=∑i=12‍supt∈[0,ω]|ui(t)|,u∈X,∥z∥1=∥u∥0+∑j=1q‍|ξj|,z=(u,ξ1,ξ2,…,ξq)∈Z.
Then both (X,∥·∥0) and (Z,∥·∥1) are Banach spaces. Let
(18)DomL={u∈X:x˙∈X},L:DomL⟶Z,L(u=(u1,u2)T)=((u˙1u˙2),{(Δu1(tk)Δu2(tk))}k=1q),N:X⟶Z,N(u,λ)=((F1(t,λ)F2(t,λ)),{(ln(1+g1k)ln(1+g2k))}k=1q),
where
(19)F1(t,λ)=a1(t)-b1(t)eu1(t-τ11(t))-h1(t)e-u1(t),F2(t,λ)=a2(t)-b2(t)eu2(t-τ22(t))+λc(t)eu1(t-τ21(t))-h2(t)e-u2(t),P:X→X,Q:Z→Z,P(u=(u1,u2)T)=(u-1,u-2)T,u∈X,Q((u1u2),{(m1km2k)}k=1q)=((u-1+1ω∑k=1q‍m1ku-2+1ω∑k=1q‍m2k),{(00)}k=1q).
Obviously,
(20)KerL={u=(u1,u2)T∈X:(u1(t),u2(t)(t))KerL==(c1,c2)∈ℝ2,t∈ℝ},ImL={((u1u2),{(m1km2k)}k=1q)∈Z:ωu-1+∑k=1qm1k=0ωu-2+∑k=1qm2k=0}.
Since ImL is closed in Z,ImP=KerL,KerQ=ImL, and dim KerL=2=codim ImL; we know that L is a Fredholm mapping of index zero. Furthermore, the generalized inverse (to L) KP:ImL→KerP⋂‍DomL is given by(21)KP(z=((u1u2),{(m1km2k)}k=1q))=(∫0t‍u1(s)ds+∑0<tk<t‍m1k-1ω∑k=1q‍m1k-1ω∫0ω‍∫0t‍u1(s)dsdt∫0t‍u2(s)ds+∑0<tk<t‍m2k-1ω∑k=1q‍m2k-1ω∫0ω‍∫0t‍u2(s)dsdt).Then
(22)QN(u,λ)=((1ω∫0ω‍F1(s,λ)ds+2g1ω1ω∫0ω‍F2(s,λ)ds+2g2ω),{(00)}k=1q),KP(I-Q)N(u,λ)=(∫0t‍F1(s,λ)ds+∑0<tk<t‍ln(1+g1k)∫0t‍F1(s,λ)ds+∑0<tk<t‍ln(1+g1k))+(12-tω)(∫0ω‍F1(s,λ)ds+2g1∫0ω‍F2(s,λ)ds+2g2)-1ω(∫0ω‍∫0t‍F1(s,λ)dsdt+2g1∫0ω‍∫0t‍F2(s,λ)dsdt+2g2),
where Fi(s,λ)(i=1,2) are very similar to that of Fi(t,λ)(i=1,2). By the Lebesgue convergence theorem, QN and KP(I-Q)N are continuous. Moreover, because of periodicity, it follows from Lemma 3 that QN(Ω¯),KP(I-Q)N(Ω¯) are relatively compact for any open bounded set Ω⊂X. Thus, N is L-compact on Ω¯ for any open bounded set Ω⊂X.
In order to use Lemma1, we have to find at least four appropriate open bounded subsets of X. Considering the operator equation Lu=λN(u,λ),λ∈(0,1), we have
(23)u˙1(t)=λ(a1(t)-b1(t)eu1(t-τ11(t))-h1(t)e-u1(t)),u˙2(t)=λ(a2(t)-b2(t)eu2(t-τ22(t))+λc(t)eu1(t-τ21(t))-h2(t)e-u2(t)),Δuj(tk)=λln(1+gjk),j=1,2.
Assume that u∈X is a ω-periodic solution of system (23) for some λ∈(0,1). Integrating (23) from 0 to ω, we have
(24)-2g1=∫0ω‍{a1(t)-b1(t)eu1(t-τ11(t))-h1(t)e-u1(t)}dt,-2g2=∫0ω‍{a2(t)-b2(t)eu2(t-τ22(t))-2g2=∫0ω‍+λc(t)eu1(t-τ21(t))-h2(t)e-u2(t)}dt.
Then
(25)∫0ω‍{b1(t)eu1(t-τ11(t))+h1(t)e-u1(t)}dt=a-1ω+2g1,∫0ω‍{b2(t)eu2(t-τ22(t))-λc(t)eu1(t-τ21(t))+h2(t)e-u2(t)}dt=a-2ω+2g2.
Furthermore, note that u=(u1,u2)T∈X. There exists ξi,ηi∈[0,ω],i=1,2, such that ui(ξi)=supt∈[0,ω]ui(t),ui(ηi)=inft∈[0,ω]ui(t).
On the one hand, according to the first equation of (23) and (25), we have
(26)∫0ω‍|u˙1(t)|dt<2a-1ω+2g1.In the light of the first equation of (25) and (26)  and Lemma 4, we get
(27)ωb-1eu1(ξ1)-(a-1ω+G1)+ωh-1e-u1(ξ1)≤ωb-1eu1(ξ1)-(1/2)[∫0ω‍|u˙1(s)|ds+∑k=1q|Δu1(tk)|]+ωh-1e-u1(ξ1)<ωb-1eu1(η1)+ωh-1e-u1(ξ1)≤∫0ω‍{b1(t)eu1(t-τ11(t))+h1(t)e-u1(t)}dt=a-1ω+2g1,
namely,
(28)b-1e2u1(ξ1)-(a-1+2g1ω)ea-1ω+G1eu1(ξ1)+h-1ea-1ω+G1<0,
which implies that
(29)lnl1-<u1(ξ1)<lnl1+.
Applying Lemma (23), we drive
(30)lnH1=lnl1--(a-1ω+G1)<u1(η)≤u1(ξ1)<lnl1+.
According to (23) and (25), we get
(31)∫0ω‍|u˙2(t)|dt<2a-2ω+2g2.
By (25) and (31), we have
(32)ωb-2eu2(ξ2)-(a-2ω+G2)-ωc-l1++ωh-2e-u2(ξ2)≤ωb-2eu2(ξ2)-(1/2)[∫0ω‍|u˙2(s)|ds+∑k=1q|Δu2(tk)|]-ωc-l1++ωh-2e-u2(ξ2)<ωb-2eu2(η2)-ωc-l1++ωh-2e-u2(ξ2)<∫0ω‍{b2(t)eu2(t-τ22(t))-λc(t)eu1(t-τ21(t))+h2(t)e-u2(t)}dt=a-2ω+2g2,
that is,
(33)b-2e2u2(ξ2)-(c-l1++a2+2g2ω)ea-2ω+G2eu2(ξ2)+h-2ea-2ω+G2<0,
which implies that
(34)lnl2-<u2(ξ2)<lnl2+.
Applying Lemma 4, we drive
(35)lnH2=lnl2--(a-2ω+G2)<u2(η)≤u2(ξ2)<lnl2+.
On the other hand, in the light of the first equation of (25) and (26), we have
(36)a-1ω+2g1=∫0ω‍{b1(t)eu1(t-τ11(t))+h1(t)e-u1(t)}dt≤b-1ωeu1(ξ1)+h-1ωe-u1(η1)<b-1ωeu1(ξ1)+h-1ωe-u1(ξ1)+(1/2)[∫0ω‍|u˙1(s)|ds+∑k=1q|Δu1(tk)|]<b-1ωeu1(ξ1)+h-1ωe-u1(ξ1)+a1ω+G1,
namely,
(37)b-1e2u1(ξ1)-(a-1+2g1ω)eu1(ξ1)+h-1ea-1ω+G1>0,
which implies that
(38)lnA1+<u1(ξ1)oru1(ξ1)<lnA1-,a-1ω+2g1=∫0ω‍{b1(t)eu1(t-τ11(t))+h1(t)e-u1(t)}dt≤ω(b-1eu1(ξ1)+h-1e-u1(η1))<ω(b-1eu1(η1)+(1/2)[∫0ω‍|u˙1(s)|ds+∑k=1q|Δu1(tk)|]+h-1e-u1(η1))<b-1ωeu1(η1)+a-1ω+G1+h-1ωe-u1(η1),
that is,
(39)b-1e2u1(η1)-(a-1+2g1ω)e-(a-1ω+G1)eu1(η1)+h-1e-(a-1ω+G1)>0,
which implies that
(40)lnB1+<u1(η1)oru1(η1)<lnB1-.
By (25) and (31), we have
(41)2g2<a-2ω+2g2=∫0ω‍{b2(t)eu2(t-τ22(t))-λc(t)eu1(t-τ21(t))+h2(t)e-u2(t)}dt<b-2ωeu2(ξ2)+h-2ωe-u2(η2)<b-2ωeu2(ξ2)+h-2ωe-u2(ξ2)+(1/2)[∫0ω‍|u˙2(s)|ds+∑k=1q|Δu2(tk)|]<b-2ωeu2(ξ2)+h-2ωe-u2(ξ2)+a-2ω+G2,
namely,
(42)b-2e2u2(ξ2)-(a-2+2g2ω)eu2(ξ2)+h-2ea-2ω+G2>0,
which implies that
(43)lnA2+<u2(ξ2)oru2(ξ2)<lnA2-,2g2<a-2ω+2g2=∫0ω‍{b2(t)eu2(t-τ22(t))-λc(t)eu1(t-τ21(t))+h2(t)e-u2(t)}dt<b-2ωeu2(ξ2)+h-2ωe-u2(η2)<b-2ωeu2(η2)+(1/2)[∫0ω‍|u˙2(s)|ds+∑k=1q|Δu2(tk)|]+h-2ωe-u2(η2)<b-2ωeu2(η2)+a-2ω+G2+h-2ωe-u2(η2),
that is,
(44)b-2e2u2(η2)-(2g2ω+a-2)e-(a-2ω+G2)eu2(η2)+h-2e-(a-2ω+G2)>0,
which implies that
(45)lnB2+<u2(η2)oru2(η2)<lnB2-.
In view of (30)–(45) and Lemma 6, we have
(46)min{lnA1+,lnB1+}<u1(η1)<u1(ξ1)<lnl1+,min{lnA2+,lnB2+}<u2(η2)<u2(ξ2)<lnl2+
or
(47)lnH1<u1(η1)<u1(ξ1)<max{lnA1-,lnB1-},lnH2<u2(η2)<u2(ξ2)<max{lnA2-,lnB2-},
which imply that, for all t∈R,
(48)min{lnA1+,lnB1+}<u1(t)<lnl1+orlnH1-<u1(t)<max{lnA1-,lnB1-},min{lnA2+,lnB2+}<u2(t)<lnl2+orlnH2-<u2(t)<max{lnA2-,lnB2-}.
Clearly,lnl1±,lnl2±,lnA1±,lnA2±,lnB1±, and lnB2± are independent of λ. Now let
(49)Ω1={u=(u1,u2)T∈X:min{lnA1+,lnB1+}<u1(t)<lnl1+,min{lnA2+,lnB2+}<u2(t)<lnl2+(u1,u2)T},Ω2={u=(u1,u2)T∈X:min{lnA1+,lnB1+}<u1(t)lnH2<u2(t)<max{lnA2-,lnB2-}(u1,u2)T},Ω3={u=(u1,u2)T∈X:lnH1<u1(t)<max{lnA1-,lnB1-},min{lnA2+,lnB2+}<u2(t)<lnl2+(u1,u2)T(u1,u2)T},Ω4={u=(u1,u2)T∈X:lnH1<u1(t)<max{lnA1-,lnB1-},lnH2<u2(t)<max{lnA2-,lnB2-}(u1,u2)T}.
Obviously, the number of the above sets is four. We denote these sets as Ωk,k=1,2,3, and 4.   Ωk are bounded open subsets of X,Ωi∩Ωj=ϕ,i≠j. Thus Ωk(k=1,2,3, and 4) satisfies the requirement (a) in Lemma 1.
Now we show that(b) of Lemma 1 holds; that is, we prove when u∈∂Ωk∩KerL=∂Ωk∩R2,QN(u,0)≠(0,0,)T,k=1,2,3, and 4. If it is not true, then when u∈∂Ωk∩KerL=∂Ωk∩R2,i=1,2,3, and 4, constant vector u=(u1,u2)T with u∈∂Ωk(k=1,2,3, and 4) satisfies
(50)-2g1=∫0ω‍{a1(t)-b1(t)eu1(t-τ11(t))-h1(t)e-u1(t)}dt,-2g2=∫0ω‍{a2(t)-b2(t)eu2(t-τ22(t))-2g2=∫0ω‍+λc(t)eu1(t-τ21(t))-h2(t)e-u2(t)}dt,
that is,
(51)-2g1ω=a-1-b-1eu1-h-1e-u1,-2g2ω=a-2-b-2eu2-h-2e-u2.
Similar to the process of (25)–(48), we obtain
(52)lnHi<ui-<max{lnAi-,lnBi-}<min{lnAi+,lnBi+}<ui+<lnli+(1≤i≤2).
Then u belongs to one of Ωk∩R2,k=1,2,3, and 4. This contradicts the fact that u∈∂Ωk∩R2,k=1,2,3, and 4. This proves that (b) in Lemma 1 holds.
Finally, in order to show that(c) in Lemma 1 holds, we only prove that for u∈∂Ωk∩KerL=∂Ωk∩R2,k=1,2,3, and 4, then it holds that deg{JQN(u,0),Ωk∩KerL,(0,0)T}≠0. To this end, we define the mapping ϕ:DomL×[0,1]→X by
(53)ϕ(u,μ)=μQN(u,0)+(1-μ)G(u),
here μ∈[0,1] is a parameter, and G(u) is defined by (54)G(u)=((1ω∫0ω‍(a1(s)-b1(s)eu1(s)-h1(s)e-u1(s))ds+2g1ω1ω∫0ω‍(a2(s)-b2(s)eu2(s)-h2(s)e-u2(s))ds+2g2ω),{(00)}k=1q(1ω∫0ω‍(a1(s)-b1(s)eu1(s)-h1(s)e-u1(s))ds+2g1ω1ω∫0ω‍(a2(s)-b2(s)eu2(s)-h2(s)e-u2(s))ds+2g2ω)).We show that for u∈∂Ωk∩KerL=∂Ωk∩R2,k=1,2,3, and 4,μ∈[0,1], then it holds that ϕ(u,μ)≠(0,0,)T. Otherwise, parameter μ and constant vector u=(u1,u2)T∈R2 satisfy ϕ(u,μ)=(0,0)T, that is,
(55)0=μ∫0ω‍(a1(s)-b1(s)eu1-h1(s)e-u1+2g1ω)ds+(1-μ)∫0ω‍(a1(s)-b1(s)eu1-h1(s)e-u1+2g1ω)ds,0=μ∫0ω‍(a2(s)-b2(s)eu2-h2(s)e-u2+2g2ω)ds+(1-μ)∫0ω‍(a2(s)-b2(s)eu2-h2(s)e-u2+2g2ω)ds,
that is,
(56)a-1-b-1eu1-h-1e-u1=-2g1ω,a-2-b-2eu2-h-2e-u2=-2g2ω.
Following the argument of (25)–(48), we obtain
(57)lnHi<ui-<max{lnAi-,lnBi-}<min{lnAi+,lnBi+}<ui+<lnli+(1=1,2).
Equation (57) gives that u belongs to one of Ωk∩R2,k=1,2,3, and 4. This contradicts the fact that u∈∂Ωk∩R2,k=1,2,3, and 4. This proves ϕ(u,μ)≠(0,0)T holds. Note that the system of algebraic equations,
(58)a-1-b-1ex-h-1e-x=-2g1ω,a-2-b-2ey-h-2e-y=-2g2ω,
has four distinct solutions since (T1) and (T2) hold:
(59)(x1*,y1*)=(lnx+,lny+),(x2*,y2*)=(lnx+,lny-),(x3*,y3*)=(lnx-,lny+),(x4*,y4*)=(lnx-,lny-),
where
(60)x±=a-1+2g1/ω±(a-1+2g1/ω)2-4b-1h-12b-1,y±=a-2+2g2/ω±(a-2+2g2/ω)2-4b-2h-22b-2.
Similar to the proof of Lemma 6, it is easy to verify that
(61)lnH1-<lnx-<max{lnB1-,lnA1-}<min{lnA1+,lnB1+}<lnx+<lnl1+,lnH2-<lny-<max{lnB2-,lnA2-}<min{lnA2+,lnB2+}<lny+<lnl2+.
Therefore(62)(x1*,y1*)∈Ω1,(x2*,y2*)∈Ω2,(x3*,y3*)∈Ω3,(x4*,y4*)∈Ω4,
uniquely belongs to the corresponding Ωk. Since KerL=Im Q, we can take J=I. A direct computation gives, for k=1,2,3, and 4,
(63)deg{JQN(u,0),Ωk∩KerL,(0,0)T}=sign|-b-1x*+h-1x*00-b-2y*+h-2y*|=sign[(-b-1x*2+h-1)(-b-2y*2+h-2)].
Since (a-1+2g1/ω)-b-1x*-h-1/x*=0,(a-2+2g2/ω)-b-2y*-h-1/y*=0.
Hence,(64)deg{JQN(u,0),Ω1∩KerL,(0,0)T}=sign{(-b-1(x1*)2+h-1)(-b-2(y1*)2+h-2)}=1,deg{JQN(u,0),Ω2∩KerL,(0,0)T}=sign{(-b-1(x2*)2+h-1)(-b-2(y2*)2+h-2)}=-1,deg{JQN(u,0),Ω3∩KerL,(0,0)T}=sign{(-b-1(x3*)2+h-1)(-b-2(y3*)2+h-2)}=1,deg{JQN(u,0),Ω4∩KerL,(0,0)T}=sign{(-b-1(x4*)2+h-1)(-b-2(y4*)2+h-2)}=-1.
So far, we have proved thatΩk(k=1,2,3, and 4) satisfies all the assumptions in Lemma 1. Hence, system (15) has at least four different ω-periodic solutions. Thus by Lemma 1, system (3) has at least four different positive ω-periodic solutions. This completes the proof of Theorem 7.


## 3. An Example

Example 8.
Consider the following time-delay parasitical system with impulsive effectiveness and harvesting terms:(65)x˙1(t)=x1(t)(3+sint2-2+sint10e2πx1(t-|sint|))-4+cost20e2π,x˙2(t)=x2(t)×(2+cost25-5+cost10eπx2(t-|sin2t|)+2+sin2t8x1(t-|cos2t|))-2+cost25eπ,Δx1(π2)=e-π-1,Δx1(3π2)=eπ-1,Δx2(π2)=e-π2-1,Δx2(3π2)=e3π/2-1.
In this case, a1(t)=(3+sint)/2,  b1(t)=(2+sint)/10e2π,  h1(t)=(4+cost)/20e2π,   τ11(t)=|sint|,  a2(t)=(2+cost)/25,  b2(t)=(5+cost)/10,  h2(t)=(2+cost)/25eπ,  c(t)=(2+sin2t)/8,  τ22(t)=|cos2t|, and τ21(t)=|sin2t|. It is not difficult to calculate ρ1>e4π>1,  ρ2=e2π>1. Since
(66)l1±=(a-1+2g1/ω)ρ1±[(a-1+2g1/ω)ρ1]2-4b-1h-1ρ12b-1=(15e4π±3225e8π-16)4e-2π,l2±=(±[(c-l1++a-2+(2g_2)/ω)ρ2]2-4b-2h-2ρ2(c-l1++a-2+2g2/ω)ρ2±[(c-l1++a-2+2g2/ω)ρ2]2-4b-2h-2ρ2)(2b-2)-1=15e2π±225e4π-6420e-π.
Then
(67)a-1+2g1/ωρ1=12e-2π>25e-2π=2b-1h-1,2g2/ω+a-2ρ2=2125e-π>25e-π=2b-2h-2.
Therefore, all conditions of Theorem7 are satisfied. By Theorem 7, system (65) has at least four positive 2π-periodic solutions.



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*Source: 198927-2013-03-19.xml*