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2019 Guangzhou Workshop

Survival analysis of a stochastic predator–prey model with prey refuge and fear effect

&
Pages 871-892
Received 14 Jan 2020
Accepted 10 Nov 2020
Published online: 03 Dec 2020

In this paper, we propose and investigate a stochastic Holling type-II predator–prey model with prey refuge and fear effect. We first prove the existence and uniqueness of the global positive solution. Then we perform the survival analysis of the model, including the existence of a unique ergodic stationary distribution and the extinction of the model. Numerical simulations are carried out to validate our analytical results. Our findings indicate that the white noise is adverse to the growth of predator and prey populations, and the increase of fear effect will lead to the decrease of predator density, but with no obvious effect on prey density.

This article is part of the following collections:
Mathematical Modeling and Analysis of Populations and Infectious Diseases

1. Introduction

Population models, as an important part of ecology, have been extensively studied and explored for their rich dynamic behaviour with the aim to provide a theoretical guidance for the protection, development and utilization of biological resources [2,5]. Among the most important population models, predator–prey models play an important role in understanding the interaction of different species in volatile natural environments and have been widely investigated [1,4,9,12,30,33].

As is well known, it is common for predators to hunt preys in nature. However, not all preys are captured by the predators because the preys usually have refuges where they can evade the predation [10,18,27]. This indicates that a prey refuge can protect and prevent the extinction of preys, and thus, it plays an important role in the interaction between predator and prey populations. Recently, the research of the predator–prey models with prey refuges has attracted many mathematicians' attention and some interesting results have been obtained [19,23,25,32].

Notice that in most previous predator–prey models, only the direct killing of preys in the presence of predators is considered due to its obviousness in nature. But in reality, the preys may change their behaviour when the predator they faced is powerful [13,26,28]. Some studies have shown that all preys respond to the risk of predation and show all sorts of anti-predator responses, including habitat changes, foraging, vigilance and different physiological changes. Such anti-predator behaviour can reduce the long-term cost of reproduction and increase the probability of adult survival [4]. In addition, frightened preys often forage less, so their birth rate decline and they use some survival mechanisms like starvation [3].

Considering the effect of prey refuge and fear effect, in a recent paper [37], Zhang et al. proposed a predator–prey model which takes the following form: (1) dxdt=αx1+Kybx2β(1m)xy1+a(1m)x,dydt=γy+cβ(1m)xy1+a(1m)x,(1) where x(t) and y(t) denote, respectively, the densities of prey and predator population at time t, α is the intrinsic growth rate of prey, b is the strength of intraspecific competition of prey, γ is the death rate of the predator, c is the conversion coefficient, K refers to the level of fear and β/a is the maximum number of prey eaten per predator per unit of time. All these parameters are assumed to be positive. The term βx/(1+ax) represents the Holling type-II functional response, and (1m)x is the quantity of preys available to the predators, where m[0,1) is the protection rate of the prey refuge for prey. For this model, the authors performed a detailed stability and bifurcation analysis. We refer the readers to Ref. [37] for more details.

In fact, a real ecosystem is inevitably affected by environmental noise. Therefore, the shifting environmental effects can not be neglected, which indicates that deterministic prey–predator models have limitations in predicting dynamics accurately, while stochastic models can make it [38]. Many researchers introduced random disturbances into deterministic systems to reveal the influence of environmental noise [14,16,17,22,29,31,34–36,39–41]. Ripa et al. [24] examined the impact of environmental noise on populations and presented a general theory of environmental noise in ecological food webs. Mao et al. [21] proved that the explosion of population dynamics can be suppressed by environmental Brownian noise and obtained that even a sufficiently small noise perturbation can suppress explosions in population dynamics. Hence, it is of great significance to further incorporate environmental randomness into the deterministic model (1). Applying the technique used in Ref. [6] to include stochastic effects, we can obtain the stochastic version of model (1) as follows (see the appendix for specific proof): (2) dx=αx1+Kybx2β(1m)xy1+a(1m)xdt+σ1xdB1(t),dy=γy+cβ(1m)xy1+a(1m)xdt+σ2ydB2(t),(2) where B1(t) and B2(t) are two independent standard Brownian motions defined in a complete probability space (Ω,F,{Ft}t0,P) with a filtration {Ft}t0 satisfying the usual conditions, and σi2,i=1,2 are the intensities of the white noises. In this paper, we will devote ourselves to the investigation on the dynamics of the stochastic model (2).

The organization of this paper is as follows. In the next section, we present some necessary notations and preliminary results of model (2). In Section 3, we prove the existence of a unique ergodic stationary distribution of the model, and in Section 4, we perform the extinction analysis of the model. Some numerical simulations are carried out in Section 5 to illustrate our theoretical results. We close the paper with conclusion in Section 6.

2. Preliminaries

In this section, we present some notations and basic results about model (2), which are the basis for further investigation on the dynamics of model (2).

Consider the following n-dimensional stochastic differential equation: (3) dx(t)=f(x(t),t)dt+g(x(t),t)dB(t)for tt0.(3) Denote by C2,1(Rn×[t0,);R+) the family of all nonnegative functions V(x,t) defined on Rn×[t0,) such that they are continuously twice differentiable in x and once in t. We introduce the differential operator defined in Ref. [20] L=t+i=1nfi(x,t)xi+12i,j=1n[gT(x,t)g(x,t)]ij2xixj.If L acts on a function VC2,1(Rn×[t0,);R+), then LV(x,t)=Vt(x,t)+Vx(x,t)f(x,t)+12trace[gT(x,t)Vxx(x,t)g(x,t)],where Vt=V/t,Vx=(V/x1,,V/xn) and Vxx=(2V/xixj)n×n. From Itô's formula, if x(t)Rn, then dV(x(t),t)=LV(x(t),t)dt+Vx(x(t),t)g(x(t),t)dB(t).The following theorem is about the existence and uniqueness of the global positive solution of model (2).

Theorem 2.1

For any given initial value (x(0),y(0))R+2, there exists a unique solution (x(t),y(t)) of model (2) on t0 and the solution will remain in R+2 with probability one, that is to say, (x(t),y(t))R+2 for all t0 almost surely (a.s.).

Proof.

Since the coefficients of model (2) satisfy the local Lipschitz condition, then there exists a unique local solution (x(t),y(t)) of model (2) on t[0,τe), a.s., where τe denotes the explosion time. To show that this solution is global, we only need to prove τe=, a.s. Let k0>0 be sufficiently large such that both x(0) and y(0) lie within the interval [1/k0,k0]. For each integer kk0, define the stopping time τk=inf{t[0,τe):min{x(t),y(t)}1k, or max{x(t),y(t)}k},where τk is increasing as k. Set τ=limkτk, which implies ττe, a.s. Hence to complete the proof, we only need to show that τ=, a.s. We prove this by contradiction. If τ<, then there exist a pair of constants ϵ(0,1) and T>0 such that P(τ<T)>ϵ.Therefore, there exists k1k0 such that for all kk1, P(Ωk)ϵ,where Ωk={τkT}. Then define a C2-function V: R+2R+ by V=xγ2cβ(1m)γ2cβ(1m)ln2cβ(1m)xγ+1c(y1lny).From Itô's formula, (4) dV(x,y)=LV(x,y)dt+σ1xγσ12cβ(1m)dB1(t)+σ2ycσ2cdB2(t),(4) where LV(x,y)=αx1+Kybx2β(1m)xy1+a(1m)xαγ2cβ(1m)(1+Ky)+bγx2cβ(1m)+γ2cy1+a(1m)x+γσ124cβ(1m)γcy+β(1m)xy1+a(1m)x+γcβ(1m)x1+a(1m)x+σ222cαxbx2+bγx2cβ(1m)+γ2cy+γσ124cβ(1m)γcy+γc+σ222cbx2+α+bγ2cβ(1m)xγ2cy+γσ124cβ(1m)+γc+σ222cγσ124cβ(1m)+γc+σ222c+J~:=J.Here, J~=sup(x,y)(0,){bx2+α+bγ2cβ(1m)xγ2cy}.Integrating and taking the expectation of both sides of (4) from 0 to τkT, we obtain (5) EVxτkT,yτkTVx0,y0+JEτkTVx0,y0+JT.(5) Noting that for every ωΩk, x(τk,ω) or y(τk,ω) equals either k or 1/k. Hence Vxτk,yτkkγ2cβ(1m)γ2cβ(1m)ln2cβ(1m)kγ1kγ2cβ(1m)γ2cβ(1m)ln2cβ(1m)γk1c(k1lnk)1c1k1+lnk.From (5), we have Vx0,y0+JTEIΩkVxτk,yτkϵ{kγ2cβ(1m)γ2cβ(1m)ln2cβ(1m)kγ1kγ2cβ(1m)γ2cβ(1m)ln2cβ(1m)γk1c(k1lnk)1c1k1+lnk}.This leads to a contradiction as we let k, >Vx0,y0+JT=.So we have that τ=, a.s.

Next, we present a lemma which gives a condition for the existence of a unique ergodic stationary distribution of system (3). Let X(t) be a homogeneous Markov process in Rn described by the following stochastic differential equation: dXt=bXdt+r=1kgrXdBrt,and the diffusion matrix is A(x)=(aij(x)),aij(x)=r=1kgri(x)grj(x).

Lemma 2.2

[7]

The Markov process X(t) has a unique ergodic stationary distribution μ(), if there exists a bounded domain DRn with regular boundary Γ and

  • there is a positive number M such that i,j=1naijxξiξjMξ2,xD, ξRn.

  • There exists a nonnegative C2 function V such that LV is negative for any RnD. Then PlimT1T0TfXtdt=Rnfxμdx=1for all xRn, where f() is a function integrable with respect to the measure μ.

3. Stationary distribution

In this section, we establish sufficient conditions for the existence of a unique ergodic stationary distribution. First, we give the following assumptions:

(H1)

γ>12max{σ12,σ22};

(H2)

λ=2αcβ(1m)b+2aα(1m)2αcβ(1m)3b[1+a(1m)(2α/b)]24c2α23bc2ασ122bγσ222>0, where c2 is a positive constant satisfying that (6) c2>3cβa(1m)2b1+a(1m)2αb3+cβ(1m)α1+a(1m)2αb2.(6)

Theorem 3.1

Let (x(t),y(t)) be the solution of model (2) with any given initial value (x(0),y(0))R+2. If assumptions (H1) and (H2) are satisfied, then model (2) admits a unique stationary distribution μ() and it has the ergodic property.

Proof.

By the condition (B.2) in Lemma 2.2, we need to construct a C2-continuous function V:R+2R+ and a bounded closed Uϵ such that LV1 for (x,y)R+2Uϵ. To this end, let σmax=max{σ1,σ2} and choose a constant θ>0 such that ρ=γ(1/2)σmax2(θ+1)>0, and define a C2-function V~:R+2R as follows: V~=Mc1x+c2xαblnx+c2β(1m)αbγylny+1θ+2x+ycθ+2,where c1=(cβ(1m))/(3α[1+a(1m)(2α/b)]2)+(2/3)c2 and c2 is defined in (6) (obviously, condition (6) implies that c2>c1); M=(2/λ)max{2,R1} and R1 is to be determined later. We can easily check that lim infϵ0,(x,y)R+2UϵV~(x,y)=,where Uϵ:=(ϵ,1/ϵ)×(ϵ,1/ϵ), and ϵ>0 is a sufficiently small number. Notice that V~(x,y) has a global minimum point (x0,y0) in the interior of R+2. Then, we can define a nonnegative C2-function V:R+2R+ by (7) V(x,y)=V~(x,y)V~(x0,y0)=MV1(x,y)+V2(x,y),(7) where V1(x,y)=c1x+c2xαblnx+c2β(1m)αbγylny,V2(x,y)=1θ+2x+ycθ+2V~(x0,y0).Applying Itô's formula to V1(x,y), we have LV1=c1αx1+Ky+c1bx2+c1β(1m)xy1+a(1m)x+c2αx1+Kyc2bx2c2β(1m)xy1+a(1m)xc2α2b(1+Ky)+c2αx+c2αβ(1m)yb[1+a(1m)x]+c2ασ122bc2β(1m)αby+c2cβ2(1m)2αxybγ[1+a(1m)x]+γcβ(1m)x1+a(1m)x+σ222α(c2c1)x1+Ky+c1bx2+c1β(1m)xyc2bx2+c2αx+c2αβ(1m)by+c2ασ122bc2αβ(1m)by+c2cβ2(1m)2αxybγ+γcβ(1m)x1+a(1m)x+σ222c2αxc1αx+c1bx2c2bx2+c2αxcβ(1m)x1+a(1m)x+[c1β(1m)+c2cβ2(1m)2αbγ]xy+c2ασ122b+γ+σ222=c2αx2bαx+c1αxbαx12αbcβ(1m)x1+a(1m)x+[c1β(1m)+c2cβ2(1m)2αbγ]xy+2c1α2b+c2ασ122b+γ+σ222=cβ(1m)(2α/b)1+a(1m)(2α/b)+2c1α2b+c2ασ122b+γ+σ222+{cβ(1m)x1+a(1m)x+cβ(1m)(2α/b)1+a(1m)(2α/b)+c2αx2bαx+c1αxbαx12αb}+c1β(1m)+c2cβ2(1m)2αbγxy=cβ(1m)(2α/b)1+a(1m)(2α/b)+2c1α2b+c2ασ122b+γ+σ222+[c1β(1m)+c2cβ2(1m)2αbγ]xy+F(x),where F(x)=cβ(1m)x1+a(1m)x+cβ(1m)(2α/b)1+a(1m)(2α/b)+c2αx2bαx+c1αxbαx12αb.We can compute that F(x)=cβ(1m)[1+a(1m)x]2+2c2α2c2bx+2c1bxc1αand F(x)=2cβa(1m)2[1+a(1m)x]32c2b+2c1b.Therefore, F(x)|x=2α/b=cβ(1m)1+a(1m)(2α/b)22c2α+3c1α=0,due to the equality c1=(cβ(1m))/(3α[1+a(1m)(2α/b)]2)+(2/3)c2. Moreover, we can compute that F(x)|x=2α/b=2cβa(1m)21+a(1m)(2α/b)32c2b+2c1b<0,where we have used inequality (6). Hence, we have that for all xR+, F(x)F2αb=0.Consequently, (8) LV1cβ(1m)(2α/b)1+a(1m)(2α/b)+2c1α2b+c2ασ122b+γ+σ222+[c1β(1m)+c2cβ2(1m)2αbγ]xy=cβ(1m)(2α/b)1+a(1m)(2α/b)+2αcβ(1m)3b1+a(1m)(2α/b)2+4c2α23b+c2ασ122b+γ+σ222+c1β(1m)+c2cβ2(1m)2αbγxy:=λ+c1β(1m)+c2cβ2(1m)2αbγxy,(8) where λ is a positive number defined in assumption (H2).

Similarly, we can compute that (9) LV2=x+ycθ+1αx1+Kybx2β(1m)xy1+a(1m)xγcy+β(1m)xy1+a(1m)x+12(θ+1)x+ycθσ12x2+σ22yc2x+ycθ+1(α+γ)xbx2γxγcy+12(θ+1)x+ycθσ12x2+σ22yc2x+ycθ+1Gγx+yc+12σmax2(θ+1)x+ycθ+2=Gx+ycθ+1γx+ycθ+2+12σmax2(θ+1)x+ycθ+2=Gx+ycθ+1γ12σmax2(θ+1)x+ycθ+2Fρ2x+ycθ+2Fρ2xθ+2+ycθ+2,(9) where G=sup(x,y)R+2{(α+γ)xbx2}<,F=sup(x,y)R+2{Gx+ycθ+1ρ2x+ycθ+2}<.From (7)–(9), we have that (10) LV=L(V1+V2)M{λ+c1β(1m)+c2cβ2(1m)2αbγxy}ρ2xθ+2+ycθ+2+F.(10) Next, we prove LV1 on R+2Uϵ. Take ϵ>0 sufficiently small such that the following inequalities hold: (11) 0<ϵ<λbγ4[c1β(1m)bγ+c2cβ2(1m)2α],(11) (12) 0<ϵ<ρbγ4Mcθ+2[c1β(1m)bγ+c2cβ2(1m)2α],(12) (13) 0<ϵ<ρbγ4M[c1β(1m)bγ+c2cβ2(1m)2α],(13) (14) Mλρ4ϵθ+2+R21,(14) (15) Mλρ4cθ+2ϵθ+2+R31,(15) where R2 and R3 are constants given by (17) and (18). Denote Uϵ1=x,yR+2:0<x<ϵ,Uϵ2=x,yR+2:0<y<ϵ,Uϵ3=x,yR+2:x>1ϵ,Uϵ4=x,yR+2:y>1ϵ.Obviously, R+2Uϵ=Uϵ1Uϵ2Uϵ3Uϵ4. Thus we only need to validate LV1 in each domain Uϵi, where i = 1, 2, 3, 4.

Case 1. On the domain Uϵ1, we have xyϵ(1+yθ+2). It then follows from (10) that LVMλ4+Mλ4+Mϵc1β(1m)+c2cβ2(1m)2αbγ+Mϵc1β(1m)+c2cβ2(1m)2αbγρ4cθ+2yθ+2ρ4xθ+2+Mλ2+R1,where (16) R1=sup(x,y)R+2{ρ4xθ+2+yθ+2cθ+2}+F=F.(16) By M=(2/λ)max{2,R1}, one can see that Mλ/41. Hence we have that for all (x,y)Uϵ1, LV(x,y)Mλ4ρ4xθ+2Mλ41,where (11) and (12) have been used.

Case 2. On the domain Uϵ2, we have xyϵ(1+xθ+2). It then follows from (10) that LVMλ4+Mλ4+Mϵc1β(1m)+c2cβ2(1m)2αbγ+Mϵc1β(1m)+c2cβ2(1m)2αbγρ4xθ+2ρ4cθ+2yθ+2+Mλ2+R1.In view of (11) and (13), we get LV(x,y)Mλ4ρ4cθ+2yθ+2Mλ41.

Case 3. On Uϵ3, we have from (14) that for all (x,y)Uϵ3, LVMλρ4xθ+2+Mc1β(1m)+c2cβ2(1m)2αbγxyρ4xθ+2ρ2cθ+2yθ+2+FMλρ4xθ+2+R2Mλρ4ϵθ+2+R21,where (17) R2=sup(x,y)R+2{Mc1β(1m)+c2cβ2(1m)2αbγxyρ4xθ+2ρ2cθ+2yθ+2+F}.(17)

Case 4. On Uϵ4, we have from (15) that for all (x,y)Uϵ4, LVMλρ4cθ+2yθ+2+Mc1β(1m)+c2cβ2(1m)2αbγxyρ2xθ+2ρ4cθ+2yθ+2+FMλρ4cθ+21ϵθ+2+R31,where (18) R3=sup(x,y)R+2{Mc1β(1m)+c2cβ2(1m)2αbγxyρ2xθ+2ρ4cθ+2yθ+2+F}.(18) To sum up, we can take a sufficiently small ϵ such that LV(x,y)1forall(x,y)R+2Uϵ, meaning that condition (B.2) in Lemma 2.2 holds.

In addition, we find M0=min(x,y)Uϵ{σ12x2,σ22y2} such that i,j=12aijxξiξj=σ12x2ξ12+σ22y2ξ22M0ξ2,xUϵ, ξ=(ξ1,ξ2)R2,which implies that condition (B.1) in Lemma 2.2 also holds. Therefore, thanks to Lemma 2.2, model (2) has a unique stationary distribution μ() and it is ergodic. This completes the proof.

4. Extinction

In this section, we derive sufficient criteria for the extinction of model (2) in two scenarios, one is both the predator and the prey go extinct, the other is the predator goes to extinction but the prey persists.

Theorem 4.1

Let (x(t),y(t)) be the solution of model (2) with any positive initial value (x(0),y(0))R+2. We have

  1. if α<σ12/2, then both x(t) and y(t) will be extinct, a.s., that is, limtx(t)=0,limty(t)=0, a.s.

  2. if α>σ12/2, cβ(1m)((α(σ12/2))/(b))γ(σ22/2)<0, then y(t) will go to extinction, a.s., i.e. limty(t)=0, a.s. Moreover, the distribution of x(t) converges weakly to the measure which has the following density: π(x)=Cσ12x(2α/σ122)e(2b/σ12)x,x(0,),where C=(σ12(σ12/2b)1(2α/σ12))/(Γ((2α/σ12)1)) is a constant such that 0π(x)dx=1.

Proof.

(i) Making use of Itô's formula on lnx(t), we get dlnx(t)=α1+Ky(t)bx(t)β(1m)y(t)1+a(1m)x(t)σ122dt+σ1dB1(t)ασ122dt+σ1dB1(t),from which we have lnx(t)t lnx(0)t+ασ122+σ1B1(t)t.Noticing that limtB1(t)/t=0, a.s. and α<σ12/2, we have lim suptlnx(t)tασ122<0,which means that (19) limtx(t)=0,a.s.(19) Similarly, by using Itô's formula to lny(t), we obtain dlny(t)=γ+cβ(1m)x(t)1+a(1m)x(t)σ222dt+σ2dB2(t)cβ(1m)x(t)γσ222dt+σ2dB2(t).Then, (20) lny(t)tlny(0)t+cβ(1m)t0tx(s)dsγσ222+σ2B2(t)t.(20) It follows from (19) and limtB2(t)/t=0, a.s., we get lim suptlny(t)tγσ222<0,which implies that limty(t)=0, a.s.

(ii) For any positive initial value (x(0),y(0))R+2, we have dx(t)x(t)(αbx(t))dt+σ1x(t)dB1(t),then by Lemma A.1 in Ref. [11] and the comparison theorem, we obtain (21) lim supt1t0tx(s)dsα(σ12/2)b a.s.(21) Combining (20) and (21), it follows that if α>σ12/2, cβ(1m)((α(σ12/2))/b)γ(σ22/2)<0, then we have that lim suptlny(t)tcβ(1m)α(σ12/2)bγσ222<0,that is, limty(t)=0, a.s.

On the other hand, consider the following one-dimensional stochastic differential equation: (22) dX(t)=X(t)(αbX(t))dt+σ1X(t)dB1(t),(22) with the initial value X(0)=x(0)>0. We know from Ref. [15] that model (22) has the ergodic property and the invariant density is given by π(x)=Cσ12x(2α/σ12)2e(2b/σ12)x,x(0,),where C=(σ12(σ12/2b)1(2α/σ12))/(Γ((2α/σ12)1)) is a constant satisfying that 0π(x)dx=1.

Since limty(t)=0, a.s., for any small ϵ, there are T and a set ΩϵΩ such that P(Ωϵ)>1ϵ, (β(1m)xy)/(1+a(1m)x)β(1m)xyϵx and α/(1+Ky)α/(1+Kϵ) for tT and ωΩϵ. Then, we have x(t)α1+Kϵbx(t)ϵx(t)dt+σ1x(t)dB1(t)dx(t)x(t)(αbx(t))dt+σ1x(t)dB1(t),that is to say, when ϵ0, we obtain that the distribution of the process x(t) converges weakly to the measure with the density of π(x). This completes the proof.

5. Numerical simulations

In this section, we will perform some numerical simulations using MATLAB R2016b to illustrate the effect of white noise, fear effect and a prey refuge on the dynamics of model (2). The numerical scheme obtained through Milstein's higher order method [8] applied on stochastic model (2) under consideration is given by xj+1=xj+xjα1+Kyjbxjβ(1m)yj1+a(1m)xjΔt+σ1ζjΔt+12σ12Δt(ζj21),yj+1=yj+yjcβ(1m)xj1+a(1m)xjγΔt+σ2ξjΔt+12σ22Δt(ξj21),where ζj and ξj are two independent Gaussian random variables N(0,1) for j=1,2,,n. All numerical simulations reported here are carried out with the choice of time stepping Δt=0.01. In this section, we always take the following parameter values: (23) α=0.6,b=0.3,β=0.3,c=0.8,a=0.3,γ=0.1,(23) and assume that the initial value of model (2) is (x(0),y(0))=(0.6,0.5).

First, in order to investigate the impact of white noise on the dynamics of model (2), we choose all parameter values in (23) and m=0.1,K=0.3,σ1=σ2=0.01. By calculating, we get from (6) that c2>0.148. Now we take c2=0.1481, then we compute λ=0.0118>0. Thus Assumptions (H1) and (H2) are satisfied. By Theorem 3.1, there exists a unique ergodic stationary distribution for the solutions of model (2). Figure 1 confirms this.

Figure 1. (a) and (c): The asymptotic behaviour of the solutions to stochastic model (2) around the positive equilibrium of model (1) with initial value (x(0),y(0))=(0.6,0.5); (b) and (d): The density function diagrams of x(t) and y(t), respectively. The parameters are taken as (23) and m = 0.1, K = 0.3, σ1=σ2=0.01.

To obtain deep insights of the influences of white noise on population dynamics, we keep the model parameter values the same as in (23) but let σ1=1.1, which implies that condition α<σ12/2 holds. Then we get that both prey and predator populations are extinction by Theorem 4.1, shown as in Figure 2. Now we take σ1=0.1,σ2=0.9 and the other parameters remain unchanged, we can easily check that the condition α>σ12/2,cβ(1m)((α(σ12/2)/b)γσ222<0 are satisfied. By Theorem 4.1, we know that the prey population is persistent and the predator population will be extinct (see Figure 3). These show that the random disturbance will change the dynamics when the environment noise is large enough.

Figure 2. Numerical simulation for model (1) and model (2) with initial value (x(0),y(0))=(0.6,0.5). The parameters are taken as (23) and m = 0.1, K = 0.3, σ1=1.1, σ2=0.01.

Figure 3. Numerical simulation for model (1) and model (2) with initial value (x(0),y(0))=(0.6,0.5). The parameters are taken as (23) and m = 0.1, K = 0.3, σ1=0.1, σ2=0.9.

Next, in order to illustrate the influence of fear effect to model (2) through numerical simulation, we choose different values of K, say K = 0, K = 0.1 and K = 1, σ1=σ2=0.01. For the remaining parameter values, we keep them the same as in (23). We can compute that these parameters satisfy Assumptions (H1) and (H2) and therefore model (2) has a stationary distribution, which means that both prey and predator populations are persistent. Moreover, from Figure 4 we find that the increase of fear effect will reduce the density of predator but with no obvious effect on the density of prey.

Figure 4. Numerical simulation for model (1) and model (2) with initial value (x(0),y(0))=(0.6,0.5) and different K, respectively. The parameters are taken as (23) and m = 0.1, σ1=σ2=0.01.

Finally, we numerically illustrate the impact of a prey refuge to model (2) and choose different values of m, say m = 0, m = 0.2, m = 0.5 and m = 0.7, σ1=σ2=0.01. For the remaining parameter values, we keep them the same as in (23). We can easily compute that these parameters satisfy assumptions (H1) and (H2). So we obtain that there is a unique ergodic stationary distribution for model (2), which means that both prey and predator populations are persistent. From Figure 5, we can see that with the increase of the amount of refuges m, the density of the prey increases while the predator density decreases. This means that the increase of the amount of refuges m is favourable to prey population but it is adverse to the persistence of predator population. Moreover, we can also see that large K may lead to the predator density decreasing fast. When we choose σ1=1.1,σ2=0.01 and the other parameters are the same as in Figure 5, we simply calculate that condition α<σ12/2 holds. Then we obtain that both prey and predator populations are extinction by Theorem 4.1, shown in Figure 6. The results again show that the large noise is very destructive to the persistence of the population.

Figure 5. The solutions of model (2) with the initial value (x(0),y(0))=(0.6,0.5) and different K,m, respectively. The parameters are taken as (23) and σ1=σ2=0.01.

Figure 6. The solutions of model (2) with the initial value (x(0),y(0))=(0.6,0.5) and different K,m, respectively. The parameters are taken as (23) and σ1=1.1,σ2=0.01.

6. Conclusion

This paper focuses on a stochastic Holling type-II prey–predator model with a prey refuge and fear effect. Mathematically, the sufficient criteria for the existence of a unique ergodic stationary distribution and the extinction of the model have been obtained. Ecologically, we get the following conclusions:

  1. Random disturbance may change the dynamic behaviour of the population, especially when the noise is large, it may lead to the extinction of the prey and predator populations.

  2. By simulations, we find that the increase of fear effect K will reduce the density of predator, but the effect on the density of prey is not obvious. In addition, one can see that the increase of the amount of refuges m is favourable to prey population but it is adverse to the persistence of predator population.

There are some interesting themes worthy of further research. On the one hand, we can incorporate some other environmental noises into model (2), such as coloured noise, Poisson noise and so on. On the other hand, to make the model (2) more realistic, we can further consider the factors such as the influence of impulsive perturbations and delay. We leave these for future investigations.

Disclosure statement

No potential conflict of interest was reported by the author(s).

Additional information

Funding

Research is supported by the National Natural Science Foundation of China (Grant Nos. 11671260 and 12071293).

References

Appendix 1: Construction of the stochastic model (2)

By the methods used in Ref. [6] about stochastic effects, we first consider a discrete time Markov chain. For x(t) and y(t) in model (2), given Δt>0, we present a process X(Δt)(t)=(x(Δt)(t),y(Δt)(t))T,t=0, Δt, 2Δt,with initial value X(Δt)(0)=(x(Δt)(0),y(Δt)(0))TR+2. Let normal distribution random variable sequence {Ri(Δt)(l)}l=0 satisfies (A1) E[Ri(Δt)(l)]=0,E[Ri(Δt)(l)]2=σi2Δt,E[Ri(Δt)(l)]4=o(Δt),(A1) where i=1,2, l=0,1,2, and σi2 express the intensities of stochastic effects. On each period [lΔt,(l+1)Δt), we assume that X(Δt)(t) grows according to model (2) and is also affected by the random (x(Δt)R1(Δt)(l),y(Δt)R2(Δt)(l))T. Specifically, for l=0,1,, we get x(Δt)((l+1)Δt)=x(Δt)(lΔt)+x(Δt)(lΔt)R1(Δt)(l)+x(Δt)(lΔt)α1+Ky(Δt)(lΔt)bx(Δt)(lΔt)β(1m)y(Δt)(lΔt)1+a(1m)x(Δt)(lΔt)Δt,y(Δt)((l+1)Δt)=y(Δt)(lΔt)+y(Δt)(lΔt)R2(Δt)(l)+γy(Δt)(lΔt)+cβ(1m)x(Δt)(lΔt)y(Δt)(lΔt)1+a(1m)x(Δt)(lΔt)Δt.Next we demonstrate that X(Δt)(t) converges to a diffusion process as Δt0. And we need to determine the drift coefficient and diffusion coefficient of the diffusion process. Let P(Δt)(x¯,dz) denote the transition probabilities of the homogeneous Markov chain {X(Δt)(lΔt)}, that is P(Δt)(x¯,B)=Prob{X(Δt)((l+1)Δt)B|X(Δt)=x¯}for all x¯=(x,y)R+2 and all Borel sets BR+2. From (A1) 1Δt(z1x)P(Δt)(x¯,dz)=xα1+kybxβ(1m)y1+a(1m)x+xΔtER1(Δt)(0)=xα1+kybxβ(1m)y1+a(1m)x,1Δt(z2y)P(Δt)(x¯,dz)=γy+cβ(1m)xy1+a(1m)x+yΔtER2(Δt)(0)=γy+cβ(1m)xy1+a(1m)x.To determine the diffusion coefficients, we consider the following moments: gij(Δt)(x)=1Δt(zixi)(zjxj)P(Δt)(x,dz),i,j=1,2.By (A1) g11(Δt)(x)σ12x2=1ΔtE{Δtxα1+Kybxβ(1m)y1+a(1m)x+R1(Δt)(0)x}2σ12x2=|Δtxα1+Kybxβ(1m)y1+a(1m)x2+2x2α1+Kybxβ(1m)y1+a(1m)xER1(Δt)(0)+1Δtx2ER1(Δt)(0)2σ12x2|=Δtxα1+Kybxβ(1m)y1+a(1m)x2,g22(Δt)(x)σ22y2=1ΔtE{Δtγy+cβ(1m)xy1+a(1m)x+R2(Δt)(0)y}2σ22y2=|Δtγy+cβ(1m)xy1+a(1m)x2+2y2γ+cβ(1m)x1+a(1m)xER2(Δt)(0)+1Δty2ER2(Δt)(0)2σ22y2|=Δtγy+cβ(1m)xy1+a(1m)x2.Therefore, limΔt0+sup||x||Lg11(Δt)(x)σ12x2=0,limΔt0+sup||x||Lg22(Δt)(x)σ22y2=0for all 0<L<. Similarly, limΔt0+sup||x||Lg12(Δt)(x)=0.In addition, from (A1), we obtain that for all 0<L< limΔt0+sup||x||L1Δt||zx||3P(Δt)(x,dz)=0.Finally, extend the definition of X(Δt)(t) to all t0 by setting X(Δt)(t)=X(Δt)(lΔt) for all t[lΔt,(l+1)Δt). On the basis of Theorem 7.1, Lemma 8.2 in Ref. [6], we can verify that as t0, X(Δt)(t) converges weakly to the solution of the stochastic system as follows: dx=αx1+Kybx2β(1m)xy1+a(1m)xdt+σ1xdB1(t),dy=γy+cβ(1m)xy1+a(1m)xdt+σ2ydB2(t),where B1(t) and B2(t) are two independent Brownian motions with the intensities represented by two positive constants σ1 and σ2.

 

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