# Periodicity and Permanence of a Discrete Impulsive Lotka-Volterra Predator-Prey Model Concerning Integrated Pest Management

Authors: Chang Tan; Jun Cao
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

By piecewise Euler method, a discrete Lotka-Volterra predator-prey model with impulsive effect at fixed moment is proposed and investigated. By using Floquets theorem, we show that a globally asymptotically stable pest-eradication periodic solution exists when the impulsive period is less than some critical value. Further, we prove that the discrete system is permanence if the impulsive period is larger than some critical value. Finally, some numerical experiments are given.


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## Body

## 1. Introduction

Impulsive equations are found in almost every domain of applied science, such as population dynamics, ecology, biological systems, and optimal control. In recent years, the theory of impulsive differential equations has been an object of active research (see [1–4] and reference therein) since it is much richer than the corresponding theory of differential equations without impulsive effects.It is well known that continuous-time dynamic systems play an important role in control theory, population dynamics, and so on. But in applications of continuous-time dynamic systems to some practical problems, such as computer simulation, experimental, or computational purposes, it is usual to formulate a discrete-time system which is a version of the continuous-time system. In some sense, the discrete time model inherits the dynamical characteristics of the continuous-time systems. We refer to [4–16] for related discussions of the importance and the need for discrete-time analogs to reflect the dynamics of their continuous-time counterparts. Nevertheless, the discrete-time version can but not always preserve the dynamics of its initial version because the theory of difference equations is a lot richer than the corresponding theory of differential equations as pointed out in [17, 18]. Therefore, it is important to study the dynamics of its initial version alone.Due to the above facts, we construct the following discrete impulsive Lotka-Volterra predator-prey model concerning integrated pest management by piecewise Euler method:(1)xkT+l+1=xkT+lexp{r-axkT+l-bykT+l},ykT+l+1=ykT+lexp{-d+cxkT+l},xkT+0=(1-p1)xkT,ykT+0=(1-p2)ykT+μ,l=0,1,2,…,T-1,dddddddddddddddddd.dddddk=0,1,2,…,
where r>0 is the intrinsic growth rate of pest, a>0 is the coefficient of intraspecific competition, b>0 is the per-capita rate of predation of the predator, d>0 is the death rate of predator, c>0 denotes the product of the per-capita rate of predation and the rate of conversing pest into predator, and T is the period of the impulsive effect. 0≤p1<1(0≤p2<1) represents the fraction of pest (predator) which dies due to the pesticide, and μ>0 is the release amount of predator at kT, k∈ℤ+. That is, we can use a combination of biological (periodic releasing natural enemies) and chemical (spraying pesticide) tactics that eradicates the pest to extinction and show the efficiency of integrated pest management strategy.System (1) can be regarded as a discrete analogy of the following impulsive Lotka-Volterra predator-prey model concerning integrated pest management:
(2)dx1dt=x1{r-ax1-bx2},dx2dt=x2{-d+cx1},Δx1=(1-p1)x1,Δx2=(1-p2)x2+μ,
where Δxi(t)=xi(t+)-xi(t), i=1,2. Liu et al. [19] discussed the dynamical behavior of model (2).Recently, the studies of discrete impulsive model have received great attention from more scholar (see [5, 15, 16, 20–22]). The main difficulty for dynamical analysis of such equations comes from impulsive effect on the equations since the corresponding theory for impulsive difference equations have not yet been fully developed. The discrete impulsive model (1) gives a new form of describing the impulsive moment. In some papers, authors use (kT)+ to denote impulsive moment (see [20]). It is obvious that describing the impulsive moment of model (1) is easily realized at computer. In addition, some authors use kT-1 to denote impulsive moment (see [16, 21]). Compared with it, model (1) is a better analogue of the continuous-time dynamic system.The main aim of this paper is to construct the discrete impulsive model (1) and discuss the dynamical behaviors of the discrete impulsive model (1). We investigate the globally asymptotical stability of pest-eradication periodic solution system (1) and the permanence of system (1).


## 2. Global Qualitative Analysis for Model (1)

Before our main results, we will give some lemmas which will be useful for our main results. First, we present the Floquent theory for the linearT-periodic difference equation
(3)un+1=Anun,
where un∈Rs, n=0,1,2,…. An is a real s×s matrix whose entries are functions of n satisfying An=An+T for a positive integer T and is nonsingular; that is, det An≠0 for all n=0,1,2,… . As usual M=AT-1AT-2⋯A0 denotes the monodromy of (3).Lemma 1 (see [5, 6]).
LetB be an s×s nonsingular matrix and let T be any positive integer. Then there exists some s×s matrix C, such that CT=B.Lemma 2 (see [5, 6]).
If we writeM=DT, then there exists a change of variable un=Snyn, with Sn+T=Sn for all n=0,1,2,…, such that (3) becomes
(4)yn+1=Dyn.Lemma 3 (see [5]).
LetakT+l be a solution of the following impulsive inequality:
(5)akT+l+1≥akT+lexp(-d),akT+0=(1-p2)akT+μ,l=0,1,2,…,T-1,a0+0=a0,k=0,1,2,…
and let bkT+l be a solution of the following impulsive inequality
(6)bkT+l+1=bkT+lexp(-d),bkT+0=(1-p2)bkT+μ,l=0,1,2,…,T-1,b0+0=a0,k=0,1,2,….
Then akT+l≥bkT+l, l=0,1,2,…,T-1.Proof.
First, we know thata0+0=a0=b0+0. By induction, we assume that akT+l≥bkT+l, k=0,1,2,…, l=1,2,…,T-1.
There are two cases.
Case  1. Considering  l≠0, k=0,1,2,… or k=0, l=0,1,2,…,T-1, we have
(7)akT+l+1≥akT+lexp(-d)≥bkT+lexp(-d)=bkT+l+1.Case  2. Considering l=0,  k=1,2,…, we have
(8)akT+1≥akT+0exp(-d)=((1-p2)akT+μ)exp(-d)≥((1-p2)bkT+μ)exp(-d)=bkT+1.
Therefore, for both cases we haveakT+l≥bkT+l. The proof is complete.Lemma 4.
Suppose(xkT+l,ykT+l) is a solution of (1) subject to x0>0, y0>0; then xkT+l>0,ykT+l>0 for all kT+l≥0, k=0,1,2,…, l=0,1,2,…,T-1.


### 2.1. Global Stability of the Pest-Eradication Periodic Solution

Consider the following system:(9)ykT+l+1=ykT+lexp(-d),ykT+0=(1-p2)ykT+μ,l=0,1,2,…,T-1,y0+0=y0>0,k=0,1,2,….Clearly,ykT+l~=(μ/(1-(1-p2)exp(-dT)))exp(-dl),  l=0,1,2,…,T-1, with y0+0=(μ/(1-(1-p2)exp(-dT))), is a positive periodic solution of (9). Since the solution of (9) is ykT+l=(1-p2)k(y0+0-(μ/(1-(1-p2)exp(-dT))))exp(-(kT+l)d)+ykT+l~, l=0,1,2,…,T-1, k=0,1,2,3,…, we have the following lemma.Lemma 5.
Equation (9) has a positive periodic solution ykT+l, k=0,1,2,…, l=0,1,2,…,T-1, and, for every solution ykT+l of (9), we have ykT+l→ykT+l~ as k→∞.Therefore, system (1) has a pest-eradication periodic solution:
(10)(0,ykT+l~)=(0,μ1-(1-p2)exp(-dT)exp(-dl))
and y0+0=μ/(1-(1-p2)exp(-dT)).Now we give the conditions which assure the globally stability of the pest-eradication periodic solution(0,ykT+l~).Theorem 6.
Let(xkT+l,ykT+l) be any solution of (1); then (0,ykT+l~) is globally asymptotically stable provided
(11)rT-bμ(1-exp(-dT))(1-(1-p2)exp(-dT))(1-exp(-d))<ln(11-p1).Proof.
Firstly, we proved the local stability. The local stability of periodic solution(0,ykT+l~) may be determined by considering the behavior of small amplitude perturbations of the solution. Defining xkT+l=ukT+l, ykT+l=vkT+l+ykT+l~, there may be written
(12)(ukT+l+1vkT+l+1)=(ukT+lexp(r-aukT+l-b(vkT+l+ykT+l~))(vkT+l+ykT+l~)exp(-d+cukT+l)).
Equation (12) can be expanded in a Taylor series: after neglecting higher-order terms, the linearized equations read as
(13)ukT+l+1=exp{r-bykT+l~}ukT+l,vkT+l+1=ykT+l~cexp(-d)ukT+l+exp(-d)vkT+l,ukT+0=(1-p1)ukT,vkT+0=(1-p2)vkT,l=0,1,2,…,T-1,dddddddddddd.d.dddk=0,1,2,….
Hence
(14)AkT+l=(exp(r-bykT+l~)0ykT+l~cexp(-d)exp(-dT));
the fundamental solution matrix is(15)M=AkT+0AkT+1AkT+2⋯AkT+T-1(1-p1001-p2)=(exp(rT-bμ1-(1-p2)exp(-dT)1-exp(-dT)1-exp(-d))0*exp(-dT))(1-p1001-p2)=((1-p1)exp(rT-bμ1-(1-p2)exp(-dT)1-exp(-dT)1-exp(-d))0*(1-p2)exp(-dT)).
There is no need to calculate the exact form of(*) as it is not required in the analysis that follows.
The stability of the periodic solution(0,ykT+l~) is determined by the eigenvalues of M.
Letλ1,λ2 be eigenvalues of matrix M. Then
(16)λ1=(1-p1)×exp(rT-bμ1-(1-p2)exp(-dT)1-exp(-dT)1-exp(-d)),λ2=(1-p2)exp(-dT),
according to Lemmas 1 and 2, (0,ykT+l~) is locally stable if |λ1|<1 and |λ2|<1. Obviously, λ2<1. So, (0,ykT+l~) is locally stable if |λ1|<1. That is,
(17)rT-bμ(1-exp(-dT))(1-(1-p2)exp(-dT))(1-exp(-d))<ln(11-p1).
In the following we prove the global attractivity. Chooseε>0 such that
(18)δ≜(1-p1)×exp(bμ(1-exp(-dT))(1-(1-p2)exp(-dT))(1-exp(-d))rT+bεTffdfd.ff-bμ(1-exp(-dT))(1-(1-p2)exp(-dT))(1-exp(-d)))<1.
Noting thatykT+l+1≥ykT+lexp(-d), consider the following impulsive equation:
(19)zkT+l+1=zkT+lexp{-d},zkT+0=(1-p2)zkT+μ,l=0,1,2,…,T-1,ddddddddddddddddddddddddk=0,1,2,…,z0+0=y0≥0.
By Lemmas3 and 5, we have
(20)ykT+l≥zkT+l>ykT+l~-ε,
for k large enough, so
(21)xkT+l+1≤xkT+lexp(r-bykT+l)≤xkT+lexp(r-b(ykT+l~-ε)),
which leads to
(22)xkT+0=(1-p1)xkT≤(1-p1)xkT-1exp(r-b(ykT-1~-ε))≤(1-p1)xkT-2exp(r-b(ykT-2~-ε))×exp(r-b(ykT-1~-ε))⋮≤(1-p1)x(k-1)T+0×exp(y(k-1)T-1~Tr+bTεddddd-b(y(k-1)T+0~+y(k-1)T-1~+⋯+ykT-1~))≤(1-p1)x(k-1)T+0×exp(bμ(1-exp(-dT))(1-(1-p2)exp(-dT))(1-exp(-d))Tr+bTεddddddddd-bμ(1-exp(-dT))(1-(1-p2)exp(-dT))(1-exp(-d))).
Therefore,xkT+0≤x0+0δk and xkT+0→0 as k→+∞. Moreover, 0<xkT+l≤xkT+0exp(rT+bTε), so we have xkT+l→0 as k→∞.
Next, we prove thatykT+l→ykT+l~ as k→∞ if limk→∞xkT+l=0. For 0<ε1<d/c, there exists an K0>0 such that 0<xkT+l<ε1 for all l=0,1,2,…,T-1, and k>K0; then from (1) we have
(23)ykT+lexp(-d)<ykT+l+1<ykT+lexp(-d+cε1).
By Lemmas3 and 5, we obtain
(24)ykT+l1≤ykT+l≤ykT+l2,ykT+l1⟶ykT+l~,ykT+l2⟶ykT+l2~ask⟶∞,
where ykT+l1 and ykT+l2 are solutions of
(25)ykT+l+11=ykT+l1exp{-d},ykT+01=(1-p2)ykT1+μ,l=0,1,2,…,T-1,dddddddddddddddd.dddddddddk=0,1,2,…,y0+01=y0≥0,(26)ykT+l+12=ykT+l2exp{-d+cε1},ykT+02=(1-p2)ykT2+μ,l=0,1,2,…,T-1,dddddddddddddddddddddddddk=0,1,2,…,y0+02=y0≥0,
respectively, and ykT+l2~=(μ/(1-(1-p2)exp((-d+cε1)T)))exp((-d+cε1)l),  l=0,1,2,…,T-1.
Therefore, for anyε2>0 there exists a K^>0 such that
(27)ykT+l~-ε2<ykT+l<ykT+l2~+ε2.
Letε1→0; we have
(28)ykT+l2~⟶ykT+l~ask⟶∞,
which implies ykT+l→ykT+l~ as k→∞. This completes the proof.


### 2.2. Permanence

Now we investigate the permanence of system (1). For convenience, let n=kT+l. Firstly, we give the following definition.Definition 7.
System (1) is said to be permanent if there are constants m,M>0 (independent of initial value) and a finite time T0 such that for all solutions (xn,yn) with all initial values (x0+0>0,y0+0>0),
(29)m<xn<M,m<yn<M
hold for all n>T0. Here T0 may depend on the initial values (x0+0,y0+0).Lemma 8.
There exist two positive constantsM and N, such that for every solution (xn,yn) of system (1), we have
(30)xn<Mforn>N.Proof.
To prove (30), we have two cases.
Case I. n≠kT+0, k=0,1,2,3,…. For any x0>0, we have
(31)xn+1=xnexp{r-axn-byn}≤xnexp{r-axn}≤exp{r-1}a≜M;
here we used
(32)maxx≥0{xexp{b-ax}}=exp{b-1}a.
Case II.
n
=
k
T
+
0, k=0,1,2,3,…, we have
(33)xkT+0=(1-p1)xkT=(1-p1)xkT-1exp{r-axkT-1-bykT-1}≤(1-p1)xkT-1exp{r-axkT-1}≤(1-p1)M<M.
This completes the proof.In the following, without loss of generality, we assumeM>d/c is large enough.Lemma 9.
There exists a constantM1 such that, for every solution (xn,yn) of (1), we have
(34)limsupn→∞yn<M1.Proof.
We first prove that there exists aK>0 such that
(35)liminfn→∞yn<K.
There exist K, ε0, M2>μ/p2>0 such that
(36)ε0<dc;Mexp(r-bK)<ε0;K>μeB-1+p2,μ(1-p2)exp(AT)-1<K+1,(37)1AlnM22(K+1)>3,
where A=-d+cM, B=d-cε0. Now we claim that (35) holds. Otherwise, there would exist N1∈ℕ and ε>0, such that yn>K+ε for n>N1. By Lemma 8, there exists an N2>N1 such that
(38)xn<M,forn>N2.
Then if n>N2(39)xn+1≤xnexp(r-b(K+ε))≤Mexp(r-bK)<ε0.
There are two cases as follows.(1)
Ifn=kT+0>N2+1,
(40)yn+1=ykT+1≤ykT+0exp(-d+cxkT+0)≤((1-p2)ykT+μ)exp(-d+cε0)≤((1-p2)yn+μ)exp(-B).We claim thatyn+1<yn. Otherwise, yn+1≥yn, so
(41)((1-p2)yn+μ)exp(-B)≥yn;hence(42)yn≤μeB-(1-p2)<K,which is a contradiction.(2)
Ifn≠kT+0>N2+1,
(43)yn+1≤ynexp(-d+cxkT+0)≤ynexp(-d+cε0)≤ynexp(-B)<yn.
Therefore, for both cases, we haveyn+1<yn, so limn→∞yn=0, which is a contradiction.
ForM2 given by (37), in the following we will prove that (34) holds.
Otherwise,(44)limsupn→∞yn>M2.
By (35) and (44), there exist
(45)n2>n1>N2,
such that
(46)yn1≤K+1,yn2≥M2,K+1<yn<M2forn∈(n1,n2).(i)
Ifn1,n2 are at the same region ∈[kT,(k+1)T), then
(47)yn2≤yn2-1exp(A)≤yn2-2exp(2A)⋮≤yn1exp((n2-n1)A)so(48)n2-n1≥1AlnM2K+1>3.(ii)
Ifn1,n2 are not at the same region, we assume n2∈[kT,(k+1)T) and n1∈[k~T,(k~+1)T). Then
(49)yn2≤yn2-1exp(A)≤yn2-2exp(2A)⋮≤(1-p2)ykTexp((n2-kT)A)+μexp((n2-kT)A)⋮≤(1-p2)2y(k-1)Texp[(n2-(k-1)T)A]+μ(1-p2)exp[(n2-(k-1)T)A]+μexp((n2-kT)A)⋮≤(1-p2)k-k~y(k~+1)Texp[(n2-(k~+1)T)A]+μ(1-p2)k-k~-1exp[(n2-(k~+1)T)A]+μ(1-p2)k-k~-2exp[(n2-(k~+2))TA]+⋯+μ(1-p2)2exp[(n2-(k-2)T)A]+μ(1-p2)exp[(n2-(k-1)T)A]+μexp((n2-kT)A)≤(1-p2)k-k~yn1exp((n2-n1)A)+μexp((n2-kT)A)[1-(1-p2)exp[(k-k~)AT]]1-(1-p2)exp(AT)≤yn1exp((n2-n1)A)+μ(1-p2)exp[(n2-k~T)A-exp((n2-kT)A)](1-p2)exp(AT)-1≤2(K+1)exp[(n2-n1)A].
So(50)n2-n1≥1AlnM22(K+1)>3.
Therefore, the set{n∣n1<n<n2} is nonempty. So if n2-1≠kT+0>N2+1, then
(51)xn2-1=xn2-2exp(r-axn2-2-byn2-2)≤xn2-2exp(r-byn2-2)≤Mexp(r-b(K+1))≤ε0.
Ifn2-1=kT+0>N2+1, then
(52)xn2-1=xkT+0=(1-p1)xkT≤(1-p1)xkT-1exp(r-b(K+1))≤Mexp(r-b(K+1))≤ε0.
There are three case as follows.(1)
Ifn2=kT+1, k=1,2,3,… and M2>μ/p2,
(53)yn2=ykT+1=ykT+0exp(-d+cxkT+0)=((1-p2)ykT+μ)exp(-d+cxkT+0)≤((1-p2)ykT+μ)exp(-d+cε0)≤((1-p2)M2+μ)exp(-d+cε0)≤(1-p2)M2+μ≤M2.(2)
Ifn2=kT+0,k=1,2,3,… and M2>μ/p2,
(54)yn2=ykT+0=((1-p2)ykT+μ)=((1-p2)ykT-1exp(-d+cxkT-1)+μ)≤((1-p2)ykT-1exp(-d+cε0)+μ)≤((1-p2)M2exp(-d+cε0)+μ)≤M2.(3)
Other cases are as follows:(55)yn2=yn2-1exp(-d+cxn2-1)=yn2-1exp(-d+cxn2-1)≤M2exp(-d+cε0)≤M2,which is a contradiction. This completes the proof.Theorem 10.
Equation (1) is permanence provided that
(56)rT-bμ(1-exp(-dT))(1-(1-p2)exp(-dT))(1-exp(-d))>ln(11-p1)
holds true.Proof.
Suppose(xn,yn) is a solution of (1) with (x0>0,y0>0). By Lemmas 8 and 9, we have proved there exists a constant M>0 such that (xn<M,yn<M) for n large enough.
From (20), we know yn>y~n-ε for all n large enough and some ε>0, so yn≥(μ/(1-(1-p2)exp(-dT)))exp(-dT)-ε≜m2 for n large enough.
Thus, we only need to findm1>0 such that xn≥m1 for n large enough.
We will do it in the following two steps.
Step I. From condition (56), let 0<m3<d/c, ε1>0 be small enough such that η≜(1-p1)exp(T(r-am3-bε1)-(μ/(1-(1-p2)exp((-d+cm3)T)))(b(1-exp((-d+cm3)T))/(1-exp(-d+cm3))))>1; we will prove that xn<m3 cannot hold for all n>0. Otherwise,
(57)yn+1≤ynexp(-d+cm3),ykT+0=(1-p2)ykT+μ.
So we haveyn≤zn and zn→z~n,  n→∞, where zn is the solution of
(58)zn+1=znexp(-d+cm3),zkT+0=(1-p2)zkT+μ,z0+0=y0+0,
and z~n=(μ/(1-(1-p2)exp((-d+cm3)T)))exp((-d+cm3)l),  l=0,1,2,…,T-1, with y0+0=μ/(1-(1-p2)exp((-d+cm3)T)), is a positive periodic solution of (58).
Therefore, there exists aN1>0 such that
(59)yn≤zn<z~n+ε1,(60)xn+1≥xnexp(r-am3-b(z~n+ε1)),xkT+0=(1-p1)xkT
for n>N1. Therefore there exists a K1, such that if k>K1(61)xkT+T≥xkT+T-1exp(r-am3-b(zkT+T-1~+ε1))≥xkT+T-1exp(r-am3-bε1)exp(-bzkT+T-1~)≥xkT+T-2exp(2(r-am3-bε1))×exp(-b(zkT+T-1~+zkT+T-2~))≥⋮≥xkT+0exp(T(r-am3-bε1))×exp(-b(zkT+T-1~+zkT+T-2~+⋯+zkT+0)~))≥(1-p1)xkT×exp(b(1-exp((-d+cm3)T))1-exp(-d+cm3)T(r-am3-bε1).dddddd-μ1-(1-p2)exp((-d+cm3)T)d.dddd.×b(1-exp((-d+cm3)T))1-exp(-d+cm3))=xkTη,
then x(k+K1)T≥xK1Tηk→∞ as k→∞, which is a contradiction to the boundedness of xkT+l. Hence there exists a n1>0 such that xn1≥m3.
Step II. If xn≥m3 for all n≥n1, then our aim is achieved. Otherwise, xn<m3 for some n>n1. Set n*=minn>n1{n∣xn<m3}. We have xn≥m3 for n∈[n1,n*).
Assume thatn*∈[kT,(k+1)T). It is easy to see that xn*<m3, xn*-1≥m3.
There are two possible cases forn∈[n*,(k+1)T). Here, for M large enough, we have η1=r-am3-bM<0 and η2=r-aM-bM<0.
Case (a). There exists a n′′∈[n*,(k+1)T), such that xn′′≥m3. Let n^=minn>n*{n∣xn≥m3}; then xn<m3 for n∈[n*,n^) and xn^≥m3.
Forn∈[n*,n^-1),
(62)xn+1≥xnexp(η1)⋮≥xn*exp((n+1-n*)η1).
Ifn*≠kT+0, then
(63)xn*=xn*-1exp(r-axn*-1-byn*-1)≥m3exp(η2),xn+1≥xn*exp((n+1-n*)η1)≥m3exp(η2)exp(η1)≥(1-p1)m3exp(η2)exp(η1).
Ifn*=kT+0, then kT+0>n1.
Case (1). Consider kT=n1,
(64)xn*=(1-p1)xkT≥(1-p1)m3≥(1-p1)m3exp(η2).
Case (2). Consider kT>n1,
(65)xn*=(1-p1)xkT≥(1-p1)xkT-1exp(r-axkT-1-bykT-1)≥(1-p1)m3exp(η2),xn+1≥xn*exp((n+1-n*)η1)≥(1-p1)m3exp(η2)exp(η1)≥(1-p1)m3exp(η2)exp(η1).
Letm1=(1-p1)m3exp(η2)exp(η1). So we have xn≥m1 for n∈[n*,n^-1). For n≥n^ the same arguments can be continued since xn^≥m3.
Case (b). There does not exist a n′′∈[n*,(k+1)T), such that xn′′≥m3; namely, xn<m3 for n∈[n*,(k+1)T).
Choosen2,n3∈ℤ+ such that
(66)n2T>1-d+cm3lnε1′M+μ,exp(η2)(1-p1)n2+1exp(η1(n2+1)T)η1n3>1.
LetT′=(n2+n3)T. We claim that there must be a n~∈[(k+1)T,(k+1)T+T′), such that xn~≥m3. Otherwise, xn<m3, n∈[(k+1)T,(k+1)T+T′). Consider (58) with z(k+1)T+0=y(k+1)T+0; we have
(67)zmT+l=(z(k+1)T+0-μ1-(1-p1)exp((-d+cm3)T))×exp((-d+cm3)(mT+l-(k+1)T))+z~mT+l
for mT+l∈(nT,(n+1)T) and k+1≤n≤k+1+n2+n3. Then for m∈(k+1+n2,k+1+n2+n3),
(68)|zmT+l-z~mT+l|<(M+μ)exp((-d+cm3)(mT+l-(k+1)T))≤(M+μ)exp((-d+cm3)(mT-(k+1)T))≤(M+μ)exp((-d+cm3)(n2T))<ε1′,ymT+l≤zmT+l≤z~mT+l+ε1′,
which implies that (60) holds for m∈(k+1+n2,k+1+n2+n3).
So as in Step I, we have(69)x(k+1+n2+n3)T≥x(k+1+n2)Tηn3.
Sincexn<m3,n∈[n*,(k+1)T), then xn<m3 for all n∈[n*,((k+1)+n2)T].
So,(70)xn+1≥xnexp(η1),(71)x((k+1)+n2)T≥x((k+1)+n2)T-1exp(η1)≥⋮≥xn*(1-p1)n2exp(η1((k+1+n2)T)-n*).
Ifn*≠kT+0, then
(72)x((k+1)+n2)T=xn*-1exp[r-axn*-1-byn*-1]×(1-p1)n2exp(η1((k+1+n2)T)-n*)≥m3exp(η2)(1-p1)n2exp(η1((k+1+n2)T)-n*)≥m3exp(η2)(1-p1)n2exp(η1(n2+1)T).
Ifn*=kT+0, then kT+0>n1(73)x((k+1)+n2)T=xn*(1-p1)n2+1exp(η1((k+1+n2)T)-n*)=xn*-1exp[r-axn*-1-byn*-1]×(1-p1)n2+1exp(η1((k+1+n2)T)-n*)≥m3exp(η2)(1-p1)n2+1exp(η1((k+1+n2)T)-n*)≥m3exp(η2)(1-p1)n2+1exp(η1(n2+1)T).
So,(74)x(k+1)+n2+n3≥m3exp(η2)(1-p1)n2+1×exp(η1(n2+1)T)η1n3>m3,
which is a contradiction.
Letn¯=minn>n*{n∣xn≥m3}. Then xn¯≥m3, xn¯-1<m3. For n∈[n*,n¯-1], (70) holds. Suppose n¯∈[(k+1)T+(k′-1)T,(k+1)T+k′T)], k′≤n2+n3.
We have(75)xn≥xn1exp(η1)⋮≥xn*(1-p1)k′exp((n-n*)η1).
Ifn*≠kT+0, then
(76)xn≥m3exp(η2)(1-p1)k′exp((n-n*)η1)≥m3exp(η2)(1-p1)k′exp((n2+n3+1)Tη1).
Ifn*=kT+0, then kT+0>n1(77)xn≥m3exp(η2)(1-p1)k′+1exp((n-n*)η1)≥m3exp(η2)(1-p1)k′+1exp((n2+n3+1)Tη1).
Let m1′=m3exp(η2)(1-p1)n2+n3+1exp((n2+n3+1)η1T)<m1. So xn>m1′ for [n*,n¯-1]. For n>n¯-1, the same arguments can be continued since xn¯≥m3.
The proof is completed.Remark 11.
Let(78)f(T)=rT-bμ(1-exp(-dT))(1-(1-p2)exp(-dT))(1-exp(-d))-ln(11-p1).
Since f(0)=-ln(1/(1-p1))<0, f(T)→∞ as T→∞ and
(79)f′′(T)=bμp2d21-exp(-d)exp(-dT)(1+(1-p2)exp(-dT))(1-(1-p2)exp(-dT))3>0,
so f(T)=0 has a unique positive root, denoted by Tmax. From Theorem 6 we know that the pest-eradication periodic solution is globally stable when T<Tmax. From Theorem 10, system (1) is permanent if T>Tmax.


## 2.1. Global Stability of the Pest-Eradication Periodic Solution

Consider the following system:(9)ykT+l+1=ykT+lexp(-d),ykT+0=(1-p2)ykT+μ,l=0,1,2,…,T-1,y0+0=y0>0,k=0,1,2,….Clearly,ykT+l~=(μ/(1-(1-p2)exp(-dT)))exp(-dl),  l=0,1,2,…,T-1, with y0+0=(μ/(1-(1-p2)exp(-dT))), is a positive periodic solution of (9). Since the solution of (9) is ykT+l=(1-p2)k(y0+0-(μ/(1-(1-p2)exp(-dT))))exp(-(kT+l)d)+ykT+l~, l=0,1,2,…,T-1, k=0,1,2,3,…, we have the following lemma.Lemma 5.
Equation (9) has a positive periodic solution ykT+l, k=0,1,2,…, l=0,1,2,…,T-1, and, for every solution ykT+l of (9), we have ykT+l→ykT+l~ as k→∞.Therefore, system (1) has a pest-eradication periodic solution:
(10)(0,ykT+l~)=(0,μ1-(1-p2)exp(-dT)exp(-dl))
and y0+0=μ/(1-(1-p2)exp(-dT)).Now we give the conditions which assure the globally stability of the pest-eradication periodic solution(0,ykT+l~).Theorem 6.
Let(xkT+l,ykT+l) be any solution of (1); then (0,ykT+l~) is globally asymptotically stable provided
(11)rT-bμ(1-exp(-dT))(1-(1-p2)exp(-dT))(1-exp(-d))<ln(11-p1).Proof.
Firstly, we proved the local stability. The local stability of periodic solution(0,ykT+l~) may be determined by considering the behavior of small amplitude perturbations of the solution. Defining xkT+l=ukT+l, ykT+l=vkT+l+ykT+l~, there may be written
(12)(ukT+l+1vkT+l+1)=(ukT+lexp(r-aukT+l-b(vkT+l+ykT+l~))(vkT+l+ykT+l~)exp(-d+cukT+l)).
Equation (12) can be expanded in a Taylor series: after neglecting higher-order terms, the linearized equations read as
(13)ukT+l+1=exp{r-bykT+l~}ukT+l,vkT+l+1=ykT+l~cexp(-d)ukT+l+exp(-d)vkT+l,ukT+0=(1-p1)ukT,vkT+0=(1-p2)vkT,l=0,1,2,…,T-1,dddddddddddd.d.dddk=0,1,2,….
Hence
(14)AkT+l=(exp(r-bykT+l~)0ykT+l~cexp(-d)exp(-dT));
the fundamental solution matrix is(15)M=AkT+0AkT+1AkT+2⋯AkT+T-1(1-p1001-p2)=(exp(rT-bμ1-(1-p2)exp(-dT)1-exp(-dT)1-exp(-d))0*exp(-dT))(1-p1001-p2)=((1-p1)exp(rT-bμ1-(1-p2)exp(-dT)1-exp(-dT)1-exp(-d))0*(1-p2)exp(-dT)).
There is no need to calculate the exact form of(*) as it is not required in the analysis that follows.
The stability of the periodic solution(0,ykT+l~) is determined by the eigenvalues of M.
Letλ1,λ2 be eigenvalues of matrix M. Then
(16)λ1=(1-p1)×exp(rT-bμ1-(1-p2)exp(-dT)1-exp(-dT)1-exp(-d)),λ2=(1-p2)exp(-dT),
according to Lemmas 1 and 2, (0,ykT+l~) is locally stable if |λ1|<1 and |λ2|<1. Obviously, λ2<1. So, (0,ykT+l~) is locally stable if |λ1|<1. That is,
(17)rT-bμ(1-exp(-dT))(1-(1-p2)exp(-dT))(1-exp(-d))<ln(11-p1).
In the following we prove the global attractivity. Chooseε>0 such that
(18)δ≜(1-p1)×exp(bμ(1-exp(-dT))(1-(1-p2)exp(-dT))(1-exp(-d))rT+bεTffdfd.ff-bμ(1-exp(-dT))(1-(1-p2)exp(-dT))(1-exp(-d)))<1.
Noting thatykT+l+1≥ykT+lexp(-d), consider the following impulsive equation:
(19)zkT+l+1=zkT+lexp{-d},zkT+0=(1-p2)zkT+μ,l=0,1,2,…,T-1,ddddddddddddddddddddddddk=0,1,2,…,z0+0=y0≥0.
By Lemmas3 and 5, we have
(20)ykT+l≥zkT+l>ykT+l~-ε,
for k large enough, so
(21)xkT+l+1≤xkT+lexp(r-bykT+l)≤xkT+lexp(r-b(ykT+l~-ε)),
which leads to
(22)xkT+0=(1-p1)xkT≤(1-p1)xkT-1exp(r-b(ykT-1~-ε))≤(1-p1)xkT-2exp(r-b(ykT-2~-ε))×exp(r-b(ykT-1~-ε))⋮≤(1-p1)x(k-1)T+0×exp(y(k-1)T-1~Tr+bTεddddd-b(y(k-1)T+0~+y(k-1)T-1~+⋯+ykT-1~))≤(1-p1)x(k-1)T+0×exp(bμ(1-exp(-dT))(1-(1-p2)exp(-dT))(1-exp(-d))Tr+bTεddddddddd-bμ(1-exp(-dT))(1-(1-p2)exp(-dT))(1-exp(-d))).
Therefore,xkT+0≤x0+0δk and xkT+0→0 as k→+∞. Moreover, 0<xkT+l≤xkT+0exp(rT+bTε), so we have xkT+l→0 as k→∞.
Next, we prove thatykT+l→ykT+l~ as k→∞ if limk→∞xkT+l=0. For 0<ε1<d/c, there exists an K0>0 such that 0<xkT+l<ε1 for all l=0,1,2,…,T-1, and k>K0; then from (1) we have
(23)ykT+lexp(-d)<ykT+l+1<ykT+lexp(-d+cε1).
By Lemmas3 and 5, we obtain
(24)ykT+l1≤ykT+l≤ykT+l2,ykT+l1⟶ykT+l~,ykT+l2⟶ykT+l2~ask⟶∞,
where ykT+l1 and ykT+l2 are solutions of
(25)ykT+l+11=ykT+l1exp{-d},ykT+01=(1-p2)ykT1+μ,l=0,1,2,…,T-1,dddddddddddddddd.dddddddddk=0,1,2,…,y0+01=y0≥0,(26)ykT+l+12=ykT+l2exp{-d+cε1},ykT+02=(1-p2)ykT2+μ,l=0,1,2,…,T-1,dddddddddddddddddddddddddk=0,1,2,…,y0+02=y0≥0,
respectively, and ykT+l2~=(μ/(1-(1-p2)exp((-d+cε1)T)))exp((-d+cε1)l),  l=0,1,2,…,T-1.
Therefore, for anyε2>0 there exists a K^>0 such that
(27)ykT+l~-ε2<ykT+l<ykT+l2~+ε2.
Letε1→0; we have
(28)ykT+l2~⟶ykT+l~ask⟶∞,
which implies ykT+l→ykT+l~ as k→∞. This completes the proof.


## 2.2. Permanence

Now we investigate the permanence of system (1). For convenience, let n=kT+l. Firstly, we give the following definition.Definition 7.
System (1) is said to be permanent if there are constants m,M>0 (independent of initial value) and a finite time T0 such that for all solutions (xn,yn) with all initial values (x0+0>0,y0+0>0),
(29)m<xn<M,m<yn<M
hold for all n>T0. Here T0 may depend on the initial values (x0+0,y0+0).Lemma 8.
There exist two positive constantsM and N, such that for every solution (xn,yn) of system (1), we have
(30)xn<Mforn>N.Proof.
To prove (30), we have two cases.
Case I. n≠kT+0, k=0,1,2,3,…. For any x0>0, we have
(31)xn+1=xnexp{r-axn-byn}≤xnexp{r-axn}≤exp{r-1}a≜M;
here we used
(32)maxx≥0{xexp{b-ax}}=exp{b-1}a.
Case II.
n
=
k
T
+
0, k=0,1,2,3,…, we have
(33)xkT+0=(1-p1)xkT=(1-p1)xkT-1exp{r-axkT-1-bykT-1}≤(1-p1)xkT-1exp{r-axkT-1}≤(1-p1)M<M.
This completes the proof.In the following, without loss of generality, we assumeM>d/c is large enough.Lemma 9.
There exists a constantM1 such that, for every solution (xn,yn) of (1), we have
(34)limsupn→∞yn<M1.Proof.
We first prove that there exists aK>0 such that
(35)liminfn→∞yn<K.
There exist K, ε0, M2>μ/p2>0 such that
(36)ε0<dc;Mexp(r-bK)<ε0;K>μeB-1+p2,μ(1-p2)exp(AT)-1<K+1,(37)1AlnM22(K+1)>3,
where A=-d+cM, B=d-cε0. Now we claim that (35) holds. Otherwise, there would exist N1∈ℕ and ε>0, such that yn>K+ε for n>N1. By Lemma 8, there exists an N2>N1 such that
(38)xn<M,forn>N2.
Then if n>N2(39)xn+1≤xnexp(r-b(K+ε))≤Mexp(r-bK)<ε0.
There are two cases as follows.(1)
Ifn=kT+0>N2+1,
(40)yn+1=ykT+1≤ykT+0exp(-d+cxkT+0)≤((1-p2)ykT+μ)exp(-d+cε0)≤((1-p2)yn+μ)exp(-B).We claim thatyn+1<yn. Otherwise, yn+1≥yn, so
(41)((1-p2)yn+μ)exp(-B)≥yn;hence(42)yn≤μeB-(1-p2)<K,which is a contradiction.(2)
Ifn≠kT+0>N2+1,
(43)yn+1≤ynexp(-d+cxkT+0)≤ynexp(-d+cε0)≤ynexp(-B)<yn.
Therefore, for both cases, we haveyn+1<yn, so limn→∞yn=0, which is a contradiction.
ForM2 given by (37), in the following we will prove that (34) holds.
Otherwise,(44)limsupn→∞yn>M2.
By (35) and (44), there exist
(45)n2>n1>N2,
such that
(46)yn1≤K+1,yn2≥M2,K+1<yn<M2forn∈(n1,n2).(i)
Ifn1,n2 are at the same region ∈[kT,(k+1)T), then
(47)yn2≤yn2-1exp(A)≤yn2-2exp(2A)⋮≤yn1exp((n2-n1)A)so(48)n2-n1≥1AlnM2K+1>3.(ii)
Ifn1,n2 are not at the same region, we assume n2∈[kT,(k+1)T) and n1∈[k~T,(k~+1)T). Then
(49)yn2≤yn2-1exp(A)≤yn2-2exp(2A)⋮≤(1-p2)ykTexp((n2-kT)A)+μexp((n2-kT)A)⋮≤(1-p2)2y(k-1)Texp[(n2-(k-1)T)A]+μ(1-p2)exp[(n2-(k-1)T)A]+μexp((n2-kT)A)⋮≤(1-p2)k-k~y(k~+1)Texp[(n2-(k~+1)T)A]+μ(1-p2)k-k~-1exp[(n2-(k~+1)T)A]+μ(1-p2)k-k~-2exp[(n2-(k~+2))TA]+⋯+μ(1-p2)2exp[(n2-(k-2)T)A]+μ(1-p2)exp[(n2-(k-1)T)A]+μexp((n2-kT)A)≤(1-p2)k-k~yn1exp((n2-n1)A)+μexp((n2-kT)A)[1-(1-p2)exp[(k-k~)AT]]1-(1-p2)exp(AT)≤yn1exp((n2-n1)A)+μ(1-p2)exp[(n2-k~T)A-exp((n2-kT)A)](1-p2)exp(AT)-1≤2(K+1)exp[(n2-n1)A].
So(50)n2-n1≥1AlnM22(K+1)>3.
Therefore, the set{n∣n1<n<n2} is nonempty. So if n2-1≠kT+0>N2+1, then
(51)xn2-1=xn2-2exp(r-axn2-2-byn2-2)≤xn2-2exp(r-byn2-2)≤Mexp(r-b(K+1))≤ε0.
Ifn2-1=kT+0>N2+1, then
(52)xn2-1=xkT+0=(1-p1)xkT≤(1-p1)xkT-1exp(r-b(K+1))≤Mexp(r-b(K+1))≤ε0.
There are three case as follows.(1)
Ifn2=kT+1, k=1,2,3,… and M2>μ/p2,
(53)yn2=ykT+1=ykT+0exp(-d+cxkT+0)=((1-p2)ykT+μ)exp(-d+cxkT+0)≤((1-p2)ykT+μ)exp(-d+cε0)≤((1-p2)M2+μ)exp(-d+cε0)≤(1-p2)M2+μ≤M2.(2)
Ifn2=kT+0,k=1,2,3,… and M2>μ/p2,
(54)yn2=ykT+0=((1-p2)ykT+μ)=((1-p2)ykT-1exp(-d+cxkT-1)+μ)≤((1-p2)ykT-1exp(-d+cε0)+μ)≤((1-p2)M2exp(-d+cε0)+μ)≤M2.(3)
Other cases are as follows:(55)yn2=yn2-1exp(-d+cxn2-1)=yn2-1exp(-d+cxn2-1)≤M2exp(-d+cε0)≤M2,which is a contradiction. This completes the proof.Theorem 10.
Equation (1) is permanence provided that
(56)rT-bμ(1-exp(-dT))(1-(1-p2)exp(-dT))(1-exp(-d))>ln(11-p1)
holds true.Proof.
Suppose(xn,yn) is a solution of (1) with (x0>0,y0>0). By Lemmas 8 and 9, we have proved there exists a constant M>0 such that (xn<M,yn<M) for n large enough.
From (20), we know yn>y~n-ε for all n large enough and some ε>0, so yn≥(μ/(1-(1-p2)exp(-dT)))exp(-dT)-ε≜m2 for n large enough.
Thus, we only need to findm1>0 such that xn≥m1 for n large enough.
We will do it in the following two steps.
Step I. From condition (56), let 0<m3<d/c, ε1>0 be small enough such that η≜(1-p1)exp(T(r-am3-bε1)-(μ/(1-(1-p2)exp((-d+cm3)T)))(b(1-exp((-d+cm3)T))/(1-exp(-d+cm3))))>1; we will prove that xn<m3 cannot hold for all n>0. Otherwise,
(57)yn+1≤ynexp(-d+cm3),ykT+0=(1-p2)ykT+μ.
So we haveyn≤zn and zn→z~n,  n→∞, where zn is the solution of
(58)zn+1=znexp(-d+cm3),zkT+0=(1-p2)zkT+μ,z0+0=y0+0,
and z~n=(μ/(1-(1-p2)exp((-d+cm3)T)))exp((-d+cm3)l),  l=0,1,2,…,T-1, with y0+0=μ/(1-(1-p2)exp((-d+cm3)T)), is a positive periodic solution of (58).
Therefore, there exists aN1>0 such that
(59)yn≤zn<z~n+ε1,(60)xn+1≥xnexp(r-am3-b(z~n+ε1)),xkT+0=(1-p1)xkT
for n>N1. Therefore there exists a K1, such that if k>K1(61)xkT+T≥xkT+T-1exp(r-am3-b(zkT+T-1~+ε1))≥xkT+T-1exp(r-am3-bε1)exp(-bzkT+T-1~)≥xkT+T-2exp(2(r-am3-bε1))×exp(-b(zkT+T-1~+zkT+T-2~))≥⋮≥xkT+0exp(T(r-am3-bε1))×exp(-b(zkT+T-1~+zkT+T-2~+⋯+zkT+0)~))≥(1-p1)xkT×exp(b(1-exp((-d+cm3)T))1-exp(-d+cm3)T(r-am3-bε1).dddddd-μ1-(1-p2)exp((-d+cm3)T)d.dddd.×b(1-exp((-d+cm3)T))1-exp(-d+cm3))=xkTη,
then x(k+K1)T≥xK1Tηk→∞ as k→∞, which is a contradiction to the boundedness of xkT+l. Hence there exists a n1>0 such that xn1≥m3.
Step II. If xn≥m3 for all n≥n1, then our aim is achieved. Otherwise, xn<m3 for some n>n1. Set n*=minn>n1{n∣xn<m3}. We have xn≥m3 for n∈[n1,n*).
Assume thatn*∈[kT,(k+1)T). It is easy to see that xn*<m3, xn*-1≥m3.
There are two possible cases forn∈[n*,(k+1)T). Here, for M large enough, we have η1=r-am3-bM<0 and η2=r-aM-bM<0.
Case (a). There exists a n′′∈[n*,(k+1)T), such that xn′′≥m3. Let n^=minn>n*{n∣xn≥m3}; then xn<m3 for n∈[n*,n^) and xn^≥m3.
Forn∈[n*,n^-1),
(62)xn+1≥xnexp(η1)⋮≥xn*exp((n+1-n*)η1).
Ifn*≠kT+0, then
(63)xn*=xn*-1exp(r-axn*-1-byn*-1)≥m3exp(η2),xn+1≥xn*exp((n+1-n*)η1)≥m3exp(η2)exp(η1)≥(1-p1)m3exp(η2)exp(η1).
Ifn*=kT+0, then kT+0>n1.
Case (1). Consider kT=n1,
(64)xn*=(1-p1)xkT≥(1-p1)m3≥(1-p1)m3exp(η2).
Case (2). Consider kT>n1,
(65)xn*=(1-p1)xkT≥(1-p1)xkT-1exp(r-axkT-1-bykT-1)≥(1-p1)m3exp(η2),xn+1≥xn*exp((n+1-n*)η1)≥(1-p1)m3exp(η2)exp(η1)≥(1-p1)m3exp(η2)exp(η1).
Letm1=(1-p1)m3exp(η2)exp(η1). So we have xn≥m1 for n∈[n*,n^-1). For n≥n^ the same arguments can be continued since xn^≥m3.
Case (b). There does not exist a n′′∈[n*,(k+1)T), such that xn′′≥m3; namely, xn<m3 for n∈[n*,(k+1)T).
Choosen2,n3∈ℤ+ such that
(66)n2T>1-d+cm3lnε1′M+μ,exp(η2)(1-p1)n2+1exp(η1(n2+1)T)η1n3>1.
LetT′=(n2+n3)T. We claim that there must be a n~∈[(k+1)T,(k+1)T+T′), such that xn~≥m3. Otherwise, xn<m3, n∈[(k+1)T,(k+1)T+T′). Consider (58) with z(k+1)T+0=y(k+1)T+0; we have
(67)zmT+l=(z(k+1)T+0-μ1-(1-p1)exp((-d+cm3)T))×exp((-d+cm3)(mT+l-(k+1)T))+z~mT+l
for mT+l∈(nT,(n+1)T) and k+1≤n≤k+1+n2+n3. Then for m∈(k+1+n2,k+1+n2+n3),
(68)|zmT+l-z~mT+l|<(M+μ)exp((-d+cm3)(mT+l-(k+1)T))≤(M+μ)exp((-d+cm3)(mT-(k+1)T))≤(M+μ)exp((-d+cm3)(n2T))<ε1′,ymT+l≤zmT+l≤z~mT+l+ε1′,
which implies that (60) holds for m∈(k+1+n2,k+1+n2+n3).
So as in Step I, we have(69)x(k+1+n2+n3)T≥x(k+1+n2)Tηn3.
Sincexn<m3,n∈[n*,(k+1)T), then xn<m3 for all n∈[n*,((k+1)+n2)T].
So,(70)xn+1≥xnexp(η1),(71)x((k+1)+n2)T≥x((k+1)+n2)T-1exp(η1)≥⋮≥xn*(1-p1)n2exp(η1((k+1+n2)T)-n*).
Ifn*≠kT+0, then
(72)x((k+1)+n2)T=xn*-1exp[r-axn*-1-byn*-1]×(1-p1)n2exp(η1((k+1+n2)T)-n*)≥m3exp(η2)(1-p1)n2exp(η1((k+1+n2)T)-n*)≥m3exp(η2)(1-p1)n2exp(η1(n2+1)T).
Ifn*=kT+0, then kT+0>n1(73)x((k+1)+n2)T=xn*(1-p1)n2+1exp(η1((k+1+n2)T)-n*)=xn*-1exp[r-axn*-1-byn*-1]×(1-p1)n2+1exp(η1((k+1+n2)T)-n*)≥m3exp(η2)(1-p1)n2+1exp(η1((k+1+n2)T)-n*)≥m3exp(η2)(1-p1)n2+1exp(η1(n2+1)T).
So,(74)x(k+1)+n2+n3≥m3exp(η2)(1-p1)n2+1×exp(η1(n2+1)T)η1n3>m3,
which is a contradiction.
Letn¯=minn>n*{n∣xn≥m3}. Then xn¯≥m3, xn¯-1<m3. For n∈[n*,n¯-1], (70) holds. Suppose n¯∈[(k+1)T+(k′-1)T,(k+1)T+k′T)], k′≤n2+n3.
We have(75)xn≥xn1exp(η1)⋮≥xn*(1-p1)k′exp((n-n*)η1).
Ifn*≠kT+0, then
(76)xn≥m3exp(η2)(1-p1)k′exp((n-n*)η1)≥m3exp(η2)(1-p1)k′exp((n2+n3+1)Tη1).
Ifn*=kT+0, then kT+0>n1(77)xn≥m3exp(η2)(1-p1)k′+1exp((n-n*)η1)≥m3exp(η2)(1-p1)k′+1exp((n2+n3+1)Tη1).
Let m1′=m3exp(η2)(1-p1)n2+n3+1exp((n2+n3+1)η1T)<m1. So xn>m1′ for [n*,n¯-1]. For n>n¯-1, the same arguments can be continued since xn¯≥m3.
The proof is completed.Remark 11.
Let(78)f(T)=rT-bμ(1-exp(-dT))(1-(1-p2)exp(-dT))(1-exp(-d))-ln(11-p1).
Since f(0)=-ln(1/(1-p1))<0, f(T)→∞ as T→∞ and
(79)f′′(T)=bμp2d21-exp(-d)exp(-dT)(1+(1-p2)exp(-dT))(1-(1-p2)exp(-dT))3>0,
so f(T)=0 has a unique positive root, denoted by Tmax. From Theorem 6 we know that the pest-eradication periodic solution is globally stable when T<Tmax. From Theorem 10, system (1) is permanent if T>Tmax.


## 3. Numerical Experiments

In Figure1, we choose parameters of system (1) as d=0.4, a=0.4, b=0.9, c=0.6, r=0.2, p1=0.8, p2=0.4, μ=0.2, T=10, and initial value x0=1, y0=1. It is easy to verify that condition (11) holds, and, by Theorem 6, the pest-eradication periodic solution of (1) is the global stability.Dynamical behavior of the pest-predator system of (1).
(a)
Time-series of the pest population
(b)
Time-series of the predator populationIn Figure2, we choose parameters of system (1) as d=0.4, a=0.4, b=1, c=0.6, r=0.3, p1=0.2, p2=0.01, μ=0.2, T=10, and initial value x0=1, y0=1. It is easy to verify that condition (56) holds, and, by Theorem 10, (1) is permanence.Dynamical behavior of the pest-predator system of (1).
(a)
Time-series of the pest population
(b)
Time-series of the predator population


## 4. Conclusion

In this paper, by piecewise Euler method, we construct a discrete impulsive Lotka-Volterra predator-prey model concerning integrated pest management. The discrete impulsive model gives a new form of describing the impulsive moment. On the other hand, model (1) is a better analogue of the continuous-time impulsive dynamic system. By using Floquets theorem, we show that a globally asymptotically stable pest-eradication periodic solution exists when the impulsive period is less than some critical value and the discrete system is permanence if the impulsive period is larger than some critical value. By (78), the impulsive period critical value Tmax can be obtained. Since Tmax is a direct function with respect to p1, p2, and μ, in order to obtain the object of integrated pest management, we can determine the impulsive period T according to effect of the chemical pesticides on the populations and cost of the releasing natural enemies.



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*Source: 767526-2013-12-18.xml*