# Periodic Solutions of a Nonautonomous Plant-Hare Model with Impulses

Authors: Haihui Wu; Yan Zhou
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

A plant-hare model subjected by the effect of impulses is studied in this paper. Sufficient conditions are obtained for the existence of at least one positive periodic solution.


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## Body

## 1. Introduction

Classical predator-prey model has been well studied (e.g., see [1–8] and the references cited therein). To explore the impact of plant toxicity on the dynamics of plant-hare interactions, Gao and Xia [9] consider a nonautonomous plant-herbivore dynamical system with a toxin-determined functional response:
(1)N˙(t)=r(t)N(t)[1-N(t)K]-C(N(t))P(t),P˙(t)=B(t)C(N(t))P(t)-d(t)P(t),C(N(t))=f(N(t))[1-f(N(t))4G],f(N(t))=eδN(t)1+heδN(t),
where N(t) denotes the density of plant at time t, P(t) denotes the herbivore biomass at time t, r(t) is the plant intrinsic growth rate at time t, d(t) is the per capita rate of herbivore death unrelated to plant toxicity at time t, B(t) is the conversion rate at time t, e is the encounter rate per unit plant, δ is the fraction of food items encountered that the herbivore ingests, K is the carrying capacity of plant, G measures the toxicity level, and h is the time for handing one unit of plant. To explore the impact of environmental factors (e.g., seasonal effects of weather, food supplies, mating habits, harvesting, etc.), the assumption of periodicity of parameters is more realistic and important. To this reason, they assumed that r(t), d(t), and B(t) are continuously positive periodic functions with period ω and e, δ, K, G, h are five positive real constants.However, birth of many species is an annual birth pulse, for having more accurate description of the system, we need to consider using the impulsive differential equations. To see how impulses affect the differential equations, for examples, one can refer to [10–17]. Motivated by the above-mentioned works, in this paper, we consider the above system with impulses:
(2)N˙(t)=N(t)[4GeδP(t)+(4Gh-1)e2δ2N(t)P(t)4G(1+heδN(t))2r(t)(1-N(t)K)-4GeδP(t)+(4Gh-1)e2δ2N(t)P(t)4G(1+heδN(t))2],P˙(t)=P(t)[(4G(1+heδN(t))2)-1(4GeδB(t)N(t)+(4Gh-1)e2δ2×B(t)N2(t))(4G(1+heδN(t))2)-1-d(t)],t≠tk;ΔN(tk)=N(tk+)-N(tk-)=c1kN(tk),ΔP(tk)=P(tk+)-P(tk-)=c2kP(tk),t=tk,k=1,2,…,
where the assumptions on r, d, B, e, δ, K, G, and h are the same as before, cjk∈(-1,∞)(j=1,2,k∈ℕ=1,2,…), {tk}∈ℕ is a strictly increasing sequence with t1>0, and limk→∞tk=∞. We further assume that there exists a q∈ℕ such that cj(k+q)=cjk(j=1,2) and tk+q=tk+ω for k∈ℕ.Without loss of generality, we will assumetk≠0 for k=1,2,…, and [0,ω]∩{tk}={t1,t2,…,tm}; hence q=m.


## 2. Preliminaries

In this section, we cite some definitions and lemmas.LetPCω 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}.For the convenience, we list the following definitions and lemmas.Definition 1 (see [10]).
The set𝔽⊂PCω is said to be quasi-equicontinuous in [0,ω] if for any ϵ>0 there exists a δ>0 such that if x∈𝔽; k∈ℤ; τ1,τ2∈(tk-1,tk)∩[0,ω] and |τ1-τ2|<δ, then
(3)|x(τ1)-x(τ2)|<ϵ.Lemma 2 (see [10]).
The set𝔽⊂PCω is relatively compact if and only if (1)
𝔽 is bounded, that is, ∥x∥≤M, for each x∈𝔽, and some M>0;(2)
𝔽 is quasi-equicontinuous in [0,ω].Lemma 3 (see [11]).
Assume thatψ∈PCω1, then the following inequality holds:
(4)sups∈[0,ω]ψ(s)-infs∈[0,ω]ψ(s)≤12[∫0ω‍|ψ˙(s)|ds+∑k=1m‍|Δψ(tk)|].
Before starting the main result, for the sake of convenience, one denotes
(5)f¯=1ω∫0ω‍f(t)dtf∈PCω,cj=∑k=1m‍ln(1+cjk),j=1,2,C1=∑k=1m‍|ln(1+c1k)|+c1,C2=∑k=1m‍|ln(1+c2k)|-c2.


## 3. Existence of Positive Periodic Solutions

In order to obtain the existence of positive periodic solutions of (2), for convenience, we will summarize in the following a few concepts and results from [18] that will be basic for this section.LetX, Y be normed vector spaces, let L:DomL⊂X→Y be a linear mapping, and N:X→Y a continuous mapping. The mapping L is 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, there exist continuous projectors P:X→X and Q:Y→Y such that ImP=KerL, KerQ=ImL=Im(I-Q). It follows that L∣domL∩KerP:(I-P)X→ImL is invertible. We denote the inverse of that map by Kp. If Ω is an open bounded subset of X, then the mapping N will be called L-compact on Ω¯ if QN(Ω¯) is bounded and Kp(I-Q)N:Ω¯→X is compact. Since ImQ is isomorphic to KerL, there exists an isomorphism J:ImQ→KerL.Lemma 4 (see [18]).
LetΩ⊂X be an open and bounded set. Let L be a Fredholm mapping of index zero and let N be L-compact on Ω¯. Assume(a)
for eachλ∈(0,1), x∈∂Ω∩DomL, Lx≠λNx;(b)
for eachx∈∂Ω∩KerL, QNx≠0;(c)
deg{JQN,Ω∩KerL,0}≠0. Then Lx=Nx has at least one solution in Ω¯∩DomL.Iff is a continuous ω-periodic function, then we set
(6)f¯=1ω∫0ω‍f(t)dt.
The following assumptions are valid throughout this paper:(A¯1)
(4Gd¯h2/(4Gh-1))<B¯<2h(d¯-(c2/ω)),(A¯2)
K(1+(c1/r¯ω))-eH1>0, H1=max{|lnK+2r¯ω|,|ln(2(d¯-(c2/ω)-(B¯/2h))/eδB¯)-2r¯ω|}.For convenience, we introduce two numbers as follows:(7)u+=(4GeδB¯-8Gheδ(d¯-(c2/ω)))+Δ2[4Gd¯h2e2δ2-(4Gh-1)e2δ2B¯],u-=(4GeδB¯-8Gheδ(d¯-(c2/ω)))-Δ2[4Gd¯h2e2δ2-(4Gh-1)e2δ2B¯],
where Δ=(4GeδB¯-8Gheδ(d¯-(c2/ω)))2-16Gd¯[4Gd¯h2e2δ2-(4Gh-1)e2δ2B¯].Theorem 5.
In addition to (A¯1), (A¯2), suppose that (A¯3)
12h<G<1h.Then system (2) has at least one positive ω-periodic solution.Remark 6.
If the impulsive operators disappear, thenc1=c2=0. Then Theorem 5 reduces to the main results in Gao and Xia [9]. This implies that our result generalizes the previous one. It shows that the impulses do affect the system indeed.Proof.
Making the change of variables(8)N(t)=exp(u1(t)),P(t)=exp(u2(t)).
Then, system (2) can be rewritten as
(9)u˙1(t)=r(t)-r(t)Kexp(u1(t))-4Geδexp(u2(t))+(4Gh-1)e2δ2exp(u1(t)+u2(t))4G(1+heδexp(u1(t)))2≡f1(t),u˙2(t)=-d(t)+(e2δ2B(t)exp(2u1(t))e2δ2B(t)exp(2u1(t))4GeδB(t)exp(u1(t))+(4Gh-1)×e2δ2B(t)exp(2u1(t)))×(4G(1+heδexp(u1(t)))2)-1≡f2(t),t≠tk,Δu1(tk)=ln(1+c1k),Δu2(tk)=ln(1+c2k),t=tk.
Take
(10)X={x=(u1,u2)T:uj∈PCω,j=1,2,x(t+ω)=x(t)},Y=X×ℝ2q
and define
(11)∥x∥0=∑j=12‍supt∈[0,ω]|uj(t)|,x=(u1,u2)∈X,∥y∥1=∥x∥0+∑j=1q‍∥ξj∥,y=[x,ξ1,…,ξq]∈Y.
Both (X,∥·∥) and (Y,∥·∥1) are Banach spaces.
Define(12)DomL={x∈X:x˙∈X},L:DomL⟶Y,L((u1,u2)T)=((u˙1u˙2),(Δu1(tk)Δu2(tk))k=1q);N:X→Y,
(13)N(u1u2)=((f1(t)f2(t)),(ln(1+c1k)ln(1+c2k))k=1q);P:X→X, P((u1,u2)T)=(u1¯,u2¯)T; Q:Y→Y,
(14)Q((u1u2),{(mknk)}k=1q)=((u1¯+1ω∑k=1qmku2¯+1ω∑k=1qnk),{(00)}k=1q).
It is not difficult to show that KerL={x∣x∈X,(u1,u2)T∈X:(u1(t),u2(t))T=(c1,c2)T∈ℝ2,t∈ℝ(u1(t),u2(t))T}.(15)ImL={[(u1u2),{(mknk)}k=1q]∈Y:ωu1¯+∑k=1qmk=0ωu2¯+∑k=1qnk=0}.
Since ImL is closed in Y, P and Q are continuous projectors such that
(16)ImP=KerL,KerQ=ImL=Im(I-Q),dimKerL=codimImL=2.
It follows that L is a Fredholm mapping of index zero. Furthermore, the generalized inverse (to L) Kp:ImL→DomL∩KerP exists, which is given by
(17)Kp[(u1u2),{(mknk)}k=1q]=(∫0tu1(s)ds+∑0<tk<tmk-1ω∑k=1qmk-∫0tu1(s)ds¯∫0t‍u2(s)ds+∑0<tk<tnk-1ω∑k=1qnk-∫0tu2(s)ds¯).
Then QN:X→Y and Kp(I-Q)N:X→X are defined by
(18)QN(u1u2)=((f1¯+c1ωf2¯+c2ω),{(00)}k=1q),Kp(I-Q)N(u1u2)=(∫0tf1(s)ds+∑0<tk<tln(1+c1k)∫0t‍f2(s)ds+∑0<tk<tln(1+c2k))+(12-tω)(∫0ωf1(t)dt+c1∫0ωf2(t)dt+c2)-1ω(∫0ω∫0tf1(s)dsdt+c1∫0ω∫0tf2(s)dsdt+c2).
Clearly, QN and Kp(I-Q)N are continuous. By using the Arzela-Ascoli theorem (see [10]), it is not difficult to prove that Kp(I-Q)N(Ω¯)¯ is compact for any open bounded set Ω⊂X. Moreover, QN(Ω¯) is bounded. Therefore, N is L-compact on Ω¯ with any open bounded set Ω⊂X.
Now, we reach the position to search for an appropriate open, bounded subsetΩ for the application of the continuation theorem.
Corresponding to the operator equationLx=λNx, λ∈(0,1), we have
(19)u˙1(t)=λ[r(t)-r(t)Kexp(u1(t))-((4G(1+heδexp(u1(t)))2)-1(e2δ2exp(u1(t)+u2(t))4Geδexp(u2(t))+(4Gh-1)×e2δ2exp(u1(t)+u2(t)))×(4G(1+heδexp(u1(t)))2)-1)r(t)K],u˙2(t)=λ[(4G(1+heδexp(u1(t)))2)-1)-d(t)+((4G(1+heδexp(u1(t)))2)-1(e2δ2B(t)exp(2u1(t))4GeδB(t)exp(u1(t))+(4Gh-1)×e2δ2B(t)exp(2u1(t)))×(4G(1+heδexp(u1(t)))2)-1)],t≠tk,Δu1(tk)=λln(1+c1k),Δu2(tk)=λln(1+c2k),t=tk.
Suppose x=(u1(t),u2(t))T∈X is a solution of (19) for a certain λ∈(0,1). Integrating the first equation of (19) over the interval [0,ω], we obtain
(20)∫0ω‍r(t)Kexp(u1(t))dt+∫0ω‍(×(4G(1+heδexp(u1(t)))2)-1(e2δ24Geδexp(u2(t))+(4Gh-1)×e2δ2exp(u1(t)+u2(t)))×(4G(1+heδexp(u1(t)))2)-1)dt=r¯ω+c1.
Similarly, integrating the second equation of (19) over the interval [0,ω], we obtain
(21)∫0ω‍((4G(1+heδexp(u1(t)))2)-1(e2δ2B(t)exp(2u1(t))4GeδB(t)exp(u1(t))+(4Gh-1)×e2δ2B(t)exp(2u1(t)))×(4G(1+heδexp(u1(t)))2)-1)dt=d¯ω-c2.
It follows from the first equation of (19) and (20) and (A¯2) that
(22)∫0ω‍|u˙1(t)|dt=λ∫0ω‍|[r(t)-r(t)Kexp(u1(t))-((4G(1+heδexp(u1(t)))2)-1(e2δ24Geδexp(u2(t))+(4Gh-1)×e2δ2exp(u1(t)+u2(t)))×(4G(1+heδexp(u1(t)))2)-1)r(t)K]|dt<∫0ω‍r(t)dt+∫0ω‍r(t)Kexp(u1(t))dt+∫0ω‍((4G(1+heδexp(u1(t)))2)-1(e2δ24Geδexp(u2(t))+(4Gh-1)×e2δ2exp(u1(t)+u2(t)))×(4G(1+heδexp(u1(t)))2)-1)dt=∫0ω‍r(t)dt+r¯ω+c1=2r¯ω+c1.
That is,
(23)∫0ω‍|u˙1(t)|dt<2r¯ω+c1.
Similarly, it follows from the second equation of (19) and (21) and (A¯2) that
(24)∫0ω‍|u˙2(t)|dt<2d¯ω-c2.
Since (u1(t),u2(t))T∈X, there exists ξi,ηi∈[0,ω] such that
(25)ui(ξi)=inft∈[0,ω]ui(t),ui(ηi)=supt∈[0,ω]ui(t),i=1,2.
From (20), we see that
(26)r¯ω+c1≥∫0ω‍r(t)Kexp(u1(t))dt,
which implies
(27)r¯+c1ω≥r¯Kexp(u1(ξ1)).
So
(28)u1(ξ1)≤lnK(1+c1r¯ω):=Δ11.
This, combined with (23), gives
(29)u1(t)≤u1(ξ1)+∫0ω‍|u˙1(t)|dt<Δ11+2r¯ω:=H11.
Similarly, it follows from (21) that
(30)d¯ω-c2≤∫0ω‍4GeδB(t)exp(u1(t))8Gheδexp(u1(t))dt+∫0ω‍4Ghe2δ2B(t)exp(2u1(t))8Gheδexp(u1(t))dt,
which implies
(31)d¯-c2ω≤B¯2h+eδB¯2exp(u1(η1)).
It follows from (A¯1) that
(32)u1(η1)≥ln22(d¯-(B¯/2h)-(c2/ω))eδB¯:=Δ12.
This, combined with (23), gives
(33)u1(t)≥u1(η1)-∫0ω‍|u˙1(t)|dt>Δ12-2r¯ω:=H12.
It follows from (29) and (33) that
(34)maxt∈[0,ω]u1(t)<max{|H11|,|H12|}:=H1.
On the other hand, it follows from (21) and (34) that
(35)r¯ω+c1≤r¯ωKexp(H1)+∫0ω‍4Geδexp(u2(t))4Gdt+∫0ω‍4Ghe2δ2exp(u1(t)+u2(t))8Gheδexp(u1(t))dt,
which implies
(36)r¯+c1ω≤r¯Kexp(H1)+32eδexp(u2(η2)).
It follows from (A¯2) that
(37)u2(η2)≥ln2r¯(1+(c1/r¯ω)-(eH1/K))3eδ:=Δ22.
This, combined with (25), gives
(38)u2(t)≥u2(η2)-∫0ω‍|u˙2(t)|dt>Δ22-2d¯ω:=H22.
Similarly, it follows from (21) and (34) that
(39)r¯ω+c1≥∫0ω‍4Geδexp(u2(ξ2))4G(1+heδexp(H1))2dt,
which implies
(40)r¯+c1ω≥eδexp(u2(ξ2))(1+heδexp(H1))2.
So
(41)u2(ξ2)≤lnr¯(1+(c1/r¯ω))(1+heδexp(H1))2eδ:=Δ21.
This, combined with (25), gives
(42)u2(t)≤u2(ξ2)+∫0ω‍|u˙2(t)|dt<Δ21+2d¯ω:=H21.
It follows from (38) and (42) that
(43)maxt∈[0,ω]u2(t)<max{|H21|,|H22|}:=H2.
Now, let us consider QNx with x=(u1,u2)T∈ℝ2. Note that
(44)QN(u1,u2)=[4GeδB¯exp(u1)+(4Gh-1)e2δ2B¯exp(2u1)4G(1+heδexp(u1))2r¯-r¯Kexp(u1)-4Geδexp(u2)+(4Gh-1)e2δ2exp(u1+u2)4G(1+heδexp(u1))2+c1ω,-d¯+4GeδB¯exp(u1)+(4Gh-1)e2δ2B¯exp(2u1)4G(1+heδexp(u1))2+c2ω].
It follows from (A¯1), (A¯2), and (A¯3) that u+<0, which implies that the equation QN(u1,u2)=0 has only one solution
(45)u~=(lnu-,ln4G(r¯-(r¯/K)u-)(1+heδu-)24Geδ+(4Gh-1)e2δ2u-).
Choose C>0 such that
(46)C>|ln4G(r¯-(r¯/K)u-)(1+heδu-)24Geδ+(4Gh-1)e2δ2u-|.
Set H=∥(H1,H2)T∥+C; then ∥x∥<H. Let
(47)Ω={x(t)=(u1(t),u2(t))T∈X:∥x(t)∥<H}.
It is clear that Ω verifies the requirement (a) in Lemma 4. When x∈KerL∩∂Ω, x is a constant with ∥x∥=H. Then QNx≠0 for x∈KerL∩∂Ω. Simple computation shows that deg{JQN,Ω∩KerL,0}≠0. Here, J is taken as the identity mapping since ImQ=KerL.By now, we have proved thatΩ verifies all the requirements in Lemma 4. Hence, (2) has at least one ω-periodic solution in Ω¯.



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