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Research Article

A stochastic predator–prey model with two competitive preys and Ornstein–Uhlenbeck process

Article: 2193211 | Received 06 May 2022, Accepted 18 Nov 2022, Published online: 22 Mar 2023

Abstract

In this paper, a stochastic predator–prey model with two competitive preys and Ornstein–Uhlenbeck process is formulated and analysed, which is used to obtain a better understanding of the population dynamics. At first, we validate that the stochastic system has a unique global solution with any initial value. Then we analyse the stochastic dynamics of the model in detail, including pth moment boundedness, asymptotic pathwise estimation in turn. After that, we obtain sufficient conditions for the existence of a stationary distribution of the system by adopting stochastic Lyapunov function methods. In addition, under some mild conditions, we derive the specific form of covariance matrix in the probability density near the quasi-positive equilibrium of the stochastic system. Finally, numerical illustrative examples are depicted to confirm our theoretical findings.

1. Introduction

Predator–prey interaction with multiple prey is common in nature, for example, Pelican birds in Salton sea eat Tilapia Oreochromis mossambicus with Avian botulism [Citation3]. In order to study the evolution law of predator–prey interaction with multiple preys, a lot of predator–prey models with multiple preys were established and analysed during the past few decades [Citation2, Citation14, Citation16, Citation23, Citation29]. The well-known predator–prey model with competitive interaction between two preys can be expressed as follows: (1) {dx1(t)dt=x1(t)[a1b11x1(t)b12x2(t)b13x3(t)],dx2(t)dt=x2(t)[a2b21x1(t)b22x2(t)b23x3(t)],dx3(t)dt=x3(t)[a3+b31x1(t)+b32x2(t)b33x3(t)],(1) where ai and bij (i,j=1,2,3) are all positive constants, x1(t) and x2(t) represent the population densities of two competitive species and x3(t) stands for the population density of predator species. The biological significance of the parameters is as follows: a1 and a2 denote the intrinsic growth rates of the two competitive preys and a3 represents the natural death rate of the predator population. The coefficients b11, b22 and b33 denote the intra-specific competitive coefficients of two prey species x1, x2 and the predator species x3, respectively. b13 and b23 represent the capturing rate of the predator at time interval, b12 and b21 are the inter-specific competitive coefficients of two prey species x1, x2 and b31, b32 stand for the rate of conversion of nutrients of the predator species. All the parameters in the model (Equation1) are assumed to be positive constants.

On the other hand, in the natural world, the growth of population is always disturbed by environmental noises [Citation4, Citation5], such as random changes in epidemics, droughts, floods, temperatures, etc. These changes sometimes are so strong that they can change the behaviours of population significantly [Citation32]. May [Citation24] pointed out that due to environmental fluctuations, the birth rates, death rates, competition coefficients, transmission coefficients and other parameters involved in the system should exhibit random fluctuation to some extent. Therefore, the parameters a1, a2 and a3 should be regarded as random variables involved in randomness. Accordingly, many scholars have introduced white noise in the deterministic population systems to study how the noise affects the dynamics of stochastic population models [Citation9, Citation10, Citation12, Citation13, Citation15, Citation19, Citation21, Citation25] and this way is common.

However, except using the common way to simulate the random disturbances, we can also assume that the parameters involved in the system satisfy the well-known Ornstein–Uhlenbeck process. The Ornstein–Uhlenbeck process is an Itô process which can be described by the following stochastic differential equation: da(t)=θ[μa(t)]dt+σdB(t),where θ, μ, σ are all positive constants and θ is a mean reversion constant that denotes the strength of the process attracted by the mean [Citation26], μ is the mean value, σ represents the degree of volatility around the mean value μ caused by shocks, B(t) denotes the standard Brownian motion.

By the work of Allen [Citation1], the stochastic mean-reverting process should incorporate the influences of environmental perturbations into the parameters of a system. So in this paper, by considering the second way of introducing randomness in model (Equation1), we assume that a1(t), a2(t) and a3(t) are affected by three Ornstein–Uhlenbeck processes, i.e. (2) da1(t)=α1[a¯1a1(t)]dt+σ1dB1(t),(2) (3) da2(t)=α2[a¯2a2(t)]dt+σ2dB2(t),(3) (4) da3(t)=α3[a¯3a3(t)]dt+σ3dB3(t),(4) where αi, a¯i, σi are all positive constants, and αi denotes the speed of reversion; σi are the intensities of the volatility of process ai; a¯i represent the mean reversion levels of ai(t), i = 1, 2, 3. {B1(t)}t0, {B2(t)}t0 and {B3(t)}t0 are mutually independent standard Brownian motions defined on a complete probability space (Ω,F,{Ft}t0,P) with a filtration {Ft}t0 satisfying the usual conditions [Citation22].

By performing stochastic integral operation to (Equation2)–(Equation4), respectively, we have ai(t)=a¯i+(ai0a¯i)eαit+σi0teαi(ts)dBi(s),i=1,2,3,where ai(0)=ai0 are the initial values of Ornstein–Uhlenbeck processes ai(t), i = 1, 2, 3. It is easy to see that the expected values and the variances of ai(t) over an interval [0,t] are as follows E[ai(t)]=a¯i+(ai0a¯i)eαit,Var[ai(t)]=σi22αi(1e2αit),i=1,2,3.According to the property of Brownian motion, we obtain that σi0teαi(ts)dBi(s) follows the distribution N(0,σi2(1e2αit)/2αi) and ai follows the normal distribution N(a¯i+(ai0a¯i)eαit,σi2(1e2αit)/2αi). Apparently, the Ornstein–Uhlenbeck processes ai(t) follow the distribution N(ai0,0) as t tends to zero, i = 1, 2, 3, which can be seen as a both mathematically and biologically reasonable method to simulate the random impacts of key parameters in a population system. Keeping in mind of the above facts and combining with (Equation2)–(Equation4), we get the stochastic form corresponding to system (Equation1) which is presented as follows: (5) {dx1(t)=x1(t)[a1(t)b11x1(t)b12x2(t)b13x3(t)]dt,dx2(t)=x2(t)[a2(t)b21x1(t)b22x2(t)b23x3(t)]dt,dx3(t)=x3(t)[a3(t)+b31x1(t)+b32x2(t)b33x3(t)]dt,da1(t)=α1[a¯1a1(t)]dt+σ1dB1(t),da2(t)=α2[a¯2a2(t)]dt+σ2dB2(t),da3(t)=α3[a¯3a3(t)]dt+σ3dB3(t).(5) Here the parameters involved in system (Equation5) are the same as those in system (Equation1).

The rest of the paper is arranged as follows. In Section 2, the existence and uniqueness of the global solution of the stochastic system (Equation5) is proved which is necessary to study the dynamics of a population model. In Section 3, the boundedness of the pth moment of solutions to system (Equation5) is shown. In Section 4, the asymptotic pathwise estimation of the solutions of system (Equation5) is investigated. In Section 5, sufficient criteria for the existence of a stationary distribution of positive solutions to system (Equation5) are derived. In Section 6, we gain the exact form of covariance matrix in the probability density around the origin point. In Section 7, through the numerical simulations of several examples, we validate the effectiveness of our theoretical findings. Finally, we present a brief conclusion to end the paper.

Next, we present the following notations which will be used in the whole paper: R+d={x=(x1,,xd)Rd:xi>0,1id}andR¯+d={x=(x1,,xd)Rd:xi0,1id}.xy=min{x,y}foranyx,yR.Let C2(Rd;R¯+) be the set of all nonnegative functions V(x) defined on Rd which means it contains all continuously twice differentiable in x. We use the notation IA to denote the indicator function of the set A. If G is a matrix, we use the notation GT to represent its transpose. If G is an invertible matrix, its inverse matrix is denoted by G1. If G is a square matrix, we use the notation |G| to denote its determinant. Let G(k) be the kth sequential principal submatrix of the matrix G. In addition, for any two d-dimensional symmetric matrices H and J, we define H0:Hisapositivedefinitematrix,H0:Hisapositivesemidefinitematrix,HJ:HJisatleastapositivesemidefinitematrix.In this sense, H0 if J0 and HJ.

Throughout this paper, the following hypothesis always holds: (A)b12b21b33+b12b23b31+b13b21b32b11b22b33b11b23b32b13b22b31<0,a¯2b12b33+a¯2b13b32+a¯3b12b23a¯1b22b33a¯1b23b32a¯3b13b22<0,a¯1b21b33+a¯1b23b31+a¯3b13b21a¯2b11b33a¯2b13b31a¯3b11b23<0,a¯1b21b32+a¯2b12b31+a¯3b11b22a¯1b22b31a¯2b11b32a¯3b12b21<0.

Remark 1.1

The Assumption (A) is used to ensure the positive equilibrium E¯(x¯1,x¯2,x¯3) of system (Equation1) exists.

Now we will present basic theories on stochastic differential equations (for more details, we refer the reader to [Citation22]).

Let X(t) be a d-dimensional stochastic process on t0 given by the following stochastic differential equation: dX(t)=f(X(t))dt+g(X(t))dB(t),where fL1(R+;Rd), gL2(R+;Rd×m) and B(t)={(Bt1,Bt2,,Btd)T}t0 is a d-dimensional Brownian motion defined on the complete probability space (Ω,F,{Ft}t0,P). We define the differential operator L as follows: L=i=1dfi(X(t))Xi+12i,j=1d(gT(X(t))g(X(t)))ij2XiXj.If L acts on a function VC2(Rd;R), then LV(X(t))=VX(X(t))f(X(t))+12trace[gT(X(t))VXX(X(t))g(X(t))],where VX=(V/X1,,V/Xd) and VXX=(2V/(XiXj))d×d. If X(t)Rd, then by means of Itô's formula, we gain dV(X(t))=LV(X(t))dt+VX(X(t))g(X(t))dB(t).

2. Existence and uniqueness of the global solution

To investigate the dynamical behaviour of system (Equation5), we should first ensure whether the solution is global or not. To this end, we present the following theorem which is related to the existence and uniqueness of the global solution to system (Equation5).

Theorem 2.1

Given initial value (x1(0),x2(0),x3(0),a1(0),a2(0),a3(0))TR+3×R3, there is a unique global solution (x1(t),x2(t),x3(t),a1(t),a2(t),a3(t))T to system (Equation5) which will remain in R+3×R3 almost surely (a.s.).

Proof.

It is noticed that all the coefficients of system (Equation5) are locally Lipschitz continuous, then for any initial value (x1(0),x2(0),x3(0),a1(0),a2(0),a3(0))TR+3×R3, there exists a unique local solution (x1(t),x2(t),x3(t),a1(t),a2(t),a3(t))TR+3×R3 of system (Equation5) on the interval [0,τe) a.s., where τe represents the explosion time [Citation22]. Next, let n01 be large enough such that x1(0), x2(0), x3(0), ea1(0), ea2(0) and ea3(0) all lie within the interval [1/n0,n0]. For each integer nn0, define the stopping time as below [Citation22] τn=inf{t[0,τe):min{x1(t),x2(t),x3(t),ea1(t),ea2(t),ea3(t)}1normax{x1(t),x2(t),x3(t),ea1(t),ea2(t),ea3(t)}n}.It is clear that τn is increasing as n. Denote by τ=limnτn, then ττe a.s. If τ= a.s. is true, then τe= a.s. and (x1(t),x2(t),x3(t),a1(t),a2(t),a3(t))TR+3×R3 a.s. In other words, to complete the proof, τ= a.s. needs to be proved. If this assertion is not correct, then there are two constants T>0 and ϵ(0,1) such that P{τT}>ϵ.Hence, there exists an integer n1n0 such that (6) P{τnT}ϵforallnn1.(6) Define a C2-function V:R+3×R3R¯+ by V(x1,x2,x3,a1,a2,a3)=c1(x11lnx1)+c2(x21lnx2)+(x31lnx3)+a14+a24+a34,where c1 and c2 are positive constants to be determined later. The nonnegativity of this function can be derived from u1lnu0,u>0.Applying Itô's formula [Citation22] to V, we obtain (7) dV(x1,x2,x3,a1,a2,a3)=LV(x1,x2,x3,a1,a2,a3)dt+4σ1a13dB1(t)+4σ2a23dB2(t)+4σ3a33dB3(t),(7) where LV:R+3×R3R is defined by LV=c1x1(a1b11x1b12x2b13x3)+c2x2(a2b21x1b22x2b23x3)+x3(a3+b31x1+b32x2b33x3)c1a1+c1b11x1+c1b12x2+c1b13x3c2a2+c2b21x1+c2b22x2+c2b23x3+a3b31x1b32x2+b33x3+4α1a¯1a13+6σ12a12+4α2a¯2a23+6σ22a22+4α3a¯3a33+6σ32a324α1a144α2a244α3a34c1a1x1+c2a2x2a3x3c1b11x12c2b22x22b33x32+(b31c1b13)x1x3+(b32c2b23)x2x3c1a1+a3c2a2+(c1b11+c2b21)x1+(c1b12+c2b22)x2+(c1b13+c2b23+b33)x3+4α1a¯1a13+6σ12a12+4α2a¯2a23+6σ22a22+4α3a¯3a33+6σ32a324α1a144α2a244α3a34c1(23x132+13|a1|3)+c2(23x232+13|a2|3)+(23x332+13|a3|3)c1b11x12c2b22x22b33x32+(c1b11+c2b21)x1+(c1b12+c2b22)x2+(c1b13+c2b23+b33)x3+c1|a1|+c2|a2|+|a3|+4α1a¯1a13+6σ12a12+4α2a¯2a23+6σ22a22+4α3a¯3a33+6σ32a324α1a144α2a244α3a34+(b31c1b13)x1x3+(b32c2b23)x2x3supx1(0,){c1b11x12+2c13x132+(c1b11+c2b21)x1}+supx2(0,){c2b22x22+2c23x232+(c1b12+c2b22)x2}+supx3(0,){b33x32+23x332+(c1b13+c2b23+b33)x3}+supa1(,){4α1a14+4α1a¯1a13+c13|a1|3+6σ12a12+c13c1|a1|}+supa2(,){4α2a24+4α2a¯2a23+c23|a2|3+6σ22a22+c2|a2|}+supa3(,){4α3a34+4α3a¯3a33+13|a3|3+6σ32a32+|a3|}+(b31c1b13)x1x3+(b32c2b23)x2x3,where in the second inequality, we have used the Young inequality a1x123x132+13|a1|3,a2x223x232+13|a2|3anda3x323x332+13|a3|3.Choose c1=b31/b13, c2=b32/b23 such that b31c1b13=0 and b32c2b23=0, then we have LVsupx1(0,){c1b11x12+2c13x132+(c1b11+c2b21)x1}+supx2(0,){c2b22x22+2c23x232+(c1b12+c2b22)x2}+supx3(0,){b33x32+23x332+(c1b13+c2b23+b33)x3}+supa1(,){4α1a14+4α1a¯1a13+c13|a1|3+6σ12a12+c13c1|a1|}+supa2(,){4α2a24+4α2a¯2a23+c23|a2|3+6σ22a22+c2|a2|}+supa3(,){4α3a34+4α3a¯3a33+13|a3|3+6σ32a32+|a3|}:=K1.Here K1 is a positive constant which is independent of the variables x1, x2, x3, a1, a2 and a3. The rest of the proof is similar to that of Zhang and Yuan [Citation31] and so it is omitted here. The proof is confirmed.

3. The pth moment boundedness

In this section, the boundedness of the pth moment of solutions to system (Equation5) is shown by establishing some Lyapunov functions.

Theorem 3.1

Given initial value (x1(0),x2(0),x3(0),a1(0),a2(0),a3(0))TR+3×R3 and any integer p3, there exists a positive constant M(p) such that the solution (x1(t),x2(t),x3(t),a1(t),a2(t),a3(t))T of system (Equation5) has the property lim suptE[(b31b13x1(t)+b32b23x2(t)+x3(t))pp+a12p(t)2p+a22p(t)2p+a32p(t)2p]M(p).

Proof.

For any integer p3, define V(x,a)=(b31b13x1+b32b23x2+x3)pp+a12p2p+a22p2p+a32p2p=V1(x)+V2(a),x=(x1,x2,x3)TR+3,a=(a1,a2,a3)TR3,where V1(x):=(b31b13x1+b32b23x2+x3)pp,V2(a):=a12p2p+a22p2p+a32p2p.Applying Itô's formula [Citation22] to V1(x), we have LV1=(b31b13x1+b32b23x2+x3)p1{b31b13x1(a1b11x1b12x2b13x3)+b32b23x2(a2b21x1b22x2b23x3)+b31b13x3(a3+b31x1+b32x2b33x3)}=(b31b13x1+b32b23x2+x3)p1(b11b31b13x12b22b32b23x22b33x32)+(b31b13x1+b32b23x2+x3)p1(b31b13a1x1+b32b23a2x2a3x3).Let A=(b11b31b13000b22b32b23000b33),then A is positive definite. Let λ0 be the minimum eigenvalue of the matrix A, i.e.λ0=min{b11b31/b13,b22b32/b23,b33}, then we get (x1x2x3)A(x1x2x3)λ0(x12+x22+x32).In addition, we have (b31b13x1+b32b23x2+x3)2(b312b132+b322b232+1)(x12+x22+x32).Hence (8) LV1λ0(b31b13x1+b32b23x2+x3)p1(x12+x22+x32)+(b31b13x1+b32b23x2+x3)p1(b31b13a1x1+b32b23a2x2a3x3)λ0b132b232b132b232+b132b322+b232b312(b31b13x1+b32b23x2+x3)p+1+(b31b13x1+b32b23x2+x3)p1(b31b13a1x1+b32b23a2x2b32b23a3x3).(8) Similarly, applying Itô's formula [Citation22] to V2(r), we obtain (9) LV2=α1a12p1(a¯1a1)+α2a22p1(a¯2a2)+α3a32p1(a¯3a3)+2p12(σ12a12p2+σ22a22p2+σ32a32p2)α12a12pα22a22pα32a32p+K2,(9) where K2:=sup(a1,a2,a3)TR3{α12a12pα22a22pα32a32p+α1a¯1a12p1+α2a¯2a22p1+α3a¯3a32p1+2p12σ12a12p2+2p122p12σ22a22p2+2p12σ32a32p2}<.From (Equation8) and (Equation9), it follows that (10) LVλ0b132b232b132b232+b132b322+b232b312(b31b13x1+b32b23x2+x3)p+1+(b31b13x1+b32b23x2+x3)p1(b31b13a1x1+b32b23a2x2a3x3)α12a12pα22a22pα32a32p+K2=λ0b132b232pp+1pb132b232+b132b322+b232b312V1p+1p+pV1b31b13a1x1+b32b23a2x2a3x3b31b13x1+b32b23x2+x3α12a12pα22a22pα32a32p+K2λ0b132b232pp+1pb132b232+b132b322+b232b312V1p+1p+p(|a1|+|a2|+|a3|)V1α12a12pα22a22pα32a32p+K2λ0b132b232pp+1pb132b232+b132b322+b232b312V1p+1p+p(|a1|p+32p+32+p+12p+32V11+22p+1)+p(|a2|p+32p+32+p+12p+32V11+22p+1)+p(|a3|p+32p+32+p+12p+32V11+22p+1)α12a12pα22a22pα32a32p+K2λ0b132b232pp+1p2(b132b232+b132b322+b232b312)V1p+1pα14a12pα24a22pα34a32p+K3,(10) where in the second inequality, we have used the Young inequality |a1|V1|a1|p+32p+32+p+12p+32V11+22p+1,|a2|V1|a2|p+32p+32+p+12p+32V11+22p+1and|a3|V1|a3|p+32p+32+p+12p+32V11+22p+1and K3:=sup(x1,x2,x3,a1,a2,a3)TR+3×R3{λ0b132b232pp+1p2(b132b232+b132b322+b232b312)V1p+1pα14a12pα24a22pα34a32p+3p(p+12)p+32V11+22p+1+22p+1pp+32|a1|p+32+pp+32|a2|p+32+pp+32|a3|p+32+K2}<.Hence, for any 0<k<(p/2)min{α1,α2,α3}, by means of (Equation10), we obtain (11) L(ektV(x(t),a(t)))=ekt(kV(x(t),a(t))+LV(x(t),a(t)))ekt(λ0b132b232pp+1p2(b132b232+b132b322+b232b312)V1p+1pα14a12pα24a22pα34a32p+kV1+k2pa12p+k2pa22pk2pa32p+K3λ0b132b232pp+1p2(b132b232+b132b322+b232b312)V1p+1p)H(p)ekt,(11) where H(p):=sup(x1,x2,x3,a1,a2,a3)TR+3×R3{λ0b132b232pp+1p2(b132b232+b132b322+b232b312)V1p+1pα14a12pα24a22pα34a32p+kV1+k2pa12p+k2pk2pa22p+k2pa32p+K3λ0b132b232pp+1p2(b132b232+b132b322+b232b312)}<.Furthermore, according to (Equation11), we have (12) d[ektV(x,a)]=ektkV(x,a)dt+ektdV(x,a)H(p)ektdt+σ1ekta12p1dB1(t)+σ2ekta22p1dB2(t)+σ3ekta32p1dB3(t).(12) By integrating (Equation12) from 0 to t and then taking expectations on both sides, one has EV(x(t),a(t))ektV(x(0),a(0))+H(p)k(1ekt).Letting t leads to that lim suptEV(x(t),a(t))H(p)k:=M(p).The proof is confirmed.

Remark 3.1

In view of Theorem 3.1, it is easy to obtain that there exists a positive constant S(p) such that lim suptE(x1p(t))S(p),lim suptE(x2p(t))S(p),lim suptE(x3p(t))S(p),lim suptE(a12p(t))S(p),lim suptE(a22p(t))S(p),lim suptE(a32p(t))S(p).

4. Asymptotic pathwise estimation

In this section, we will verify pathwise properties of the solution (x1(t),x2(t),x3(t),a1(t),a2(t),a3(t))T of system (Equation5).

Theorem 4.1

Given initial value (x1(0),x2(0),x3(0),a1(0),a2(0),a3(0))TR+3×R3, the solution (x1(t),x2(t),x3(t),a1(t),a2(t),a3(t))T of system (Equation5) has the property lim suptln(b31b13x1(t)+b32b23x2(t)+x3(t))lnt1pa.s.,foranyintegerp3.In addition lim suptlnx1(t)t0,lim suptlnx2(t)t0andlim suptlnx3(t)t0a.s.

Proof.

According to Theorem 3.1, one can see that there is a positive constant M1(p) such that lim suptE((b31b13x1(t)+b32b23x2(t)+x3(t))pp)M1(p)foranyintegerp3,which together with the continuity of x1(t), x2(t) and x3(t) implies that there is a positive constant M2(p) such that (13) E((b31b13x1(t)+b32b23x2(t)+x3(t))pp)M2(p)forallt[0,).(13) For sufficiently small δ>0, k=1,2,, in view of (Equation13), we get E[supkδt(k+1)δ(b31b13x1(t)+b32b23x2(t)+x3(t))pp]E[(b31b13x1(kδ)+b32b23x2(kδ)+x3(kδ))pp]+IM2(p)+I,where I:=E[supkδt(k+1)δ|kδt(λ0b132b232pp+1p2(b132b232+b132b322+b232b312)V1p+1p(s)α14a12p(s)α24a22p(s)α34a32p(s)+K3λ0b132b232pp+1p2(b132b232+b132b322+b232b312))ds|]E[supkδt(k+1)δ(kδtλ0b132b232pp+1p2(b132b232+b132b322+b232b312)V1p+1p(s)ds+α14kδt|a12p(s)|ds+α24kδt|a22p(s)|ds+kδt|K3|ds+α34kδt|a32p(s)|dsλ0b132b232pp+1p2(b132b232+b132b322+b232b312)]λ0b132b232pp+1p2(b132b232+b132b322+b232b312)kδ(k+1)δEV1p+1p(s)ds+α14kδ(k+1)δE|a12p(s)|ds+α24kδ(k+1)δE|a22p(s)|ds+α34kδ(k+1)δE|a32p(s)|ds+kδ(k+1)δ|K3|dsλ0b132b232pp+1p2(b132b232+b132b322+b232b312)M(p)+α1δ4S(p)+α2δ4S(p)+α3δ4S(p)+|K3|δ:=M3(p).So E[supkδt(k+1)δ(b31b13x1(t)+b32b23x2(t)+x3(t))pp]E[(b31b13x1(kδ)+b32b23x2(kδ)+x3(kδ))pp]+M3(p)M2(p)+M3(p):=M4(p).Let ϵx>0 be arbitrary. By Chebyshev's inequality [Citation22], we obtain P{supkδt(k+1)δ(b31b13x1(t)+b32b23x2(t)+x3(t))pp>(kδ)1+ϵx}E[supkδt(k+1)δ(b31b13x1(t)+b32b23x2(t)+x3(t))pp](kδ)1+ϵxM4(p)(kδ)1+ϵx,k=1,2,By Borel–Cantelli Lemma [Citation22], one obtains that for almost all ωΩ, (14) supkδt(k+1)δ(b31b13x1(t)+b32b23x2(t)+x3(t))pp(kδ)1+ϵx,(14) holds for all but finitely many k. As a result, there exists a k0(ω) for almost all ωΩ, for which (Equation14) holds whenever kk0. Thus, for almost all ωΩ, if kk0 and kδt(k+1)δ, ln((b31b13x1(t)+b32b23x2(t)+x3(t))pp)lnt(1+ϵx)ln(kδ)ln(kδ)=1+ϵxa.s.Therefore lim suptln((b31b13x1(t)+b32b23x2(t)+x3(t))pp)lnt1+ϵxa.s.Letting ϵx0+ gives that lim suptln(b31b13x1(t)+b32b23x2(t)+x3(t))plnt1a.s.and hence lim suptln(b31b13x1(t)+b32b23x2(t)+x3(t))lnt1pa.s.Since the integer p is arbitrary and greater than 0, it can be seen that lim suptlnx1(t)t0,lim suptlnx2(t)t0andlim suptlnx3(t)t0a.s.This completes the proof.

5. Existence of a stationary distribution

In this section, sufficient conditions which are used to ensure the existence of a stationary distribution of the stochastic system (Equation5) are presented.

Let X(t) be a homogeneous Markov process defined in Rd which is described by the following stochastic differential equation: (15) dX(t)=b(X(t))dt+r=1kσr(X(t))dBr(t).(15)

Lemma 5.1

[Citation11]

Assume that the vectors b(X), σ1(X),,σk(X) (tt0,XRd) are continuous functions with respect to X and satisfy the following conditions:

(A1) There is a constant B such that (16) |b(X1)b(X2)|+r=1k|σr(X1)σr(X2)|B|X1X2|,|b(X)|+r=1k|σr(X)|B(1+|X|).(16) (A2) There exists a nonnegative C2-function V(X) in Rd such that LV(X)1 outside some compact set. Then system (Equation15) admits at least one stationary distribution on Rd.

Remark 5.1

We can use the global existence of solutions of (Equation15) to replace the conditions (Equation16) in Lemma 5.1 and the conclusion can be found in Remark 5 of Xu et al. [Citation30].

Theorem 5.1

Let Assumption (A) hold. If the following conditions hold b11>σ12,b22>σ22,b33>σ32,4b13b23b31b32(b11σ12)(b22σ22)>(b12b23b31+b13b21b32)2,ω<min{b31b13(b11σ12)ϵ2(b12b31b13+b21b32b23)x¯12,+b32b23(b22σ22)12ϵ(b12b31b13+b21b32b23)x¯22,(b33σ32)x¯32},and ϵ>0 is a sufficiently small constant satisfying b12b23b31+b13b21b322b13b32(b22σ22)<ϵ<2b23b31(b11σ12)b12b23b31+b13b21b32,then system (Equation5) has at least one stationary distribution π() on R+3×R3, where ω=b31σ1/(2b13α1)+b32σ2/(2b23α2)+σ3/(2α3) and (x¯1,x¯2,x¯3) is the positive equilibrium of system (Equation1).

Proof.

By Theorem 2.1, we have proved that there is a global solution to system (Equation5). Hence, in order to complete the proof of Theorem 5.1, we only need to prove the condition (A2). That is to say, we need to construct a nonnegative C2-function V(x1,x2,x3,a1,a2,a3) and a compact set DR+3×R3 such that LV1 for all (x1,x2,x3,a1,a2,a3)(R+3×R3)D.

Define a nonnegative C2-function V:R+3×R3R as follows: V(x1,x2,x3,a1,a2,a3)=b31b13(x1x¯1x¯1lnx1x¯1)+b32b23(x2x¯2x¯2lnx2x¯2)+(x3x¯3x¯3lnx3x¯3)+b312b13α1σ1(a1a¯1)2+b322b23α2σ2(a2a¯2)2+12α3σ3(a3a¯3)2,where (x¯1,x¯2,x¯3) is the positive equilibrium of system (Equation1) which satisfies the following algebraic equations (17) {a¯1b11x¯1b12x¯2b13x¯3=0,a¯2b21x¯1b22x¯2b23x¯3=0,a¯3+b31x¯1+b32x¯2b33x¯3=0.(17) For convenience, denote by V1(x1)=x1x¯1x¯1lnx1x¯1,V2(x2)=x2x¯2x¯2lnx2x¯2,V3(x3)=x3x¯3x¯3lnx3x¯3,V4(a1)=b312b13α1σ1(a1a¯1)2,V5(a2)=b322b23α2σ2(a2a¯2)2,V6(a3)=12α3σ3(a3a¯3)2.Applying Itô's formula [Citation22] to V1 and using (Equation17), we have (18) LV1=(x1x¯1)(a¯1b11x1b12x2b13x3)+(x1x¯1)(a1a¯1)=(x1x¯1)(b11x¯1+b12x¯2+b13x¯3b11x1b12x2b13x3)+(x1x¯1)(a1a¯1)=b11(x1x¯1)2b12(x1x¯1)(x2x¯2)b13(x1x¯1)(x3x¯3)+(x1x¯1)(a1a¯1)(b11σ12)(x1x¯1)2b12(x1x¯1)(x2x¯2)b13(x1x¯1)(x3x¯3)+(a1a¯1)22σ1,(18) where in the above inequality, we have used the Young inequality (x1x¯1)(a1a¯1)σ12(x1x¯1)2+12σ1(a1a¯1)2and σ1(0,2b11) is a sufficiently small constant to ensure the coefficient in front of (x1x¯1)2 is negative.

Similarly, we obtain (19) LV2=(x2x¯2)(a¯2b21x1b22x2b23x3)+(x2x¯2)(a2a¯2)=(x2x¯2)(b21x¯1+b22x¯2+b23x¯3b21x1b22x2b23x3)+(x2x¯2)(a2a¯2)=b22(x2x¯2)2b21(x1x¯1)(x2x¯2)b23(x2x¯2)(x3x¯3)+(x2x¯2)(a2a¯2)(b22σ22)(x2x¯2)2b21(x1x¯1)(x2x¯2)b23(x2x¯2)(x3x¯3)+(a2a¯2)22σ2,(19) (20) LV3=(x3x¯3)(a¯3+b31x1+b32x2b33x3)+(x3x¯3)(a3+a¯3)=(x3x¯3)(b31x¯1b32x¯2+b33x¯3+b31x1+b32x2b33x3)+(x3x¯3)(a3+a¯3)=b33(x3x¯3)2+b31(x1x¯1)(x3x¯3)+b32(x2x¯2)(x3x¯3)+(x3x¯3)(a3+a¯3)(b33σ32)(x3x¯3)2+b31(x1x¯1)(x3x¯3)+b32(x2x¯2)(x3x¯3)+(a3a¯3)22σ3,(20) where σ2(0,2b22), σ3(0,2b33) are sufficiently small constants to ensure the coefficients in front of (x2x¯2)2, (x3x¯3)2 are negative, respectively.

Moreover, we have (21) LV4=b31b13σ1(a1a¯1)(a¯1a1)+b31σ12b13α1=b31b13σ1(a1a¯1)2+b31σ12b13α1,(21) (22) LV5=b32b23σ2(a2a¯2)(a¯2a2)+b32σ22b23α2=b32b23σ2(a2a¯2)2+b32σ22b23α2,(22) and (23) LV6=1σ3(a3a¯3)(a¯3a3)+σ32α3=1σ3(a3a¯3)2+σ32α3.(23) Then according to (Equation18)–(Equation23), we have LV(b33σ32)(x3x¯3)2b31b13(b11σ12)(x1x¯1)2b32b23(b22σ22)(x2x¯2)2b312b13σ1(a1a¯1)2b322b23σ2(a2a¯2)212σ3(a3a¯3)2+b31σ12b13α1+b32σ22b23α2+σ32α3(b12b31b13+b21b32b23)(x1x¯1)(x2x¯2)(b33σ32)(x3x¯3)2b31b13(b11σ12)(x1x¯1)2b32b23(b22σ22)(x2x¯2)2b312b13σ1(a1a¯1)2b322b23σ2(a2a¯2)212σ3(a3a¯3)2+b31σ12b13α1+b32σ22b23α2+σ32α3+ϵ2(b12b31b13+b21b32b23)(x1x¯1)2+12ϵ(b12b31b13+b21b32b23)(x2x¯2)2=[b31b13(b11σ12)ϵ2(b12b31b13+b21b32b23)](x1x¯1)2[b32b23(b22σ22)12ϵ(b12b31b13+b21b32b23)](x2x¯2)2(b33σ32)(x3x¯3)2b312b13σ1(a1a¯1)2b322b23σ2(a2a¯2)212σ3(a3a¯3)2+b31σ12b13α1+b32σ22b23α2+σ32α3,where in the second inequality we have used the Young inequality (x1x¯1)(x2x¯2)ϵ2(x1x¯1)2+12ϵ(x2x¯2)2and b12b23b31+b13b21b322b13b32(b22σ22)<ϵ<2b23b31(b11σ12)b12b23b31+b13b21b32such that b31b13(b11σ12)ϵ2(b12b31b13+b21b32b23)>0andb32b23(b22σ22)12ϵ(b12b31b13+b21b32b23)>0.Let ω=b31σ12b13α1+b32σ22b23α2+σ32α3,and note that if the condition ω<min{[b31b13(b11σ12)ϵ2(b12b31b13+b21b32b23)]x¯12,[b32b23(b22σ22)12ϵ(b12b31b13+b21b32b23)]x¯22,(b33σ32)x¯32}holds, then the ellipsoid [b31b13(b11σ12)ϵ2(b12b31b13+b21b32b23)](x1x¯1)2[b32b23(b22σ22)12ϵ(b12b31b13+b21b32b23)](x2x¯2)2(b33σ32)(x3x¯3)2b312b13σ1(a1a¯1)2b322b23σ2(a2a¯2)212σ3(a3a¯3)2+ω=0lies entirely in R+3×R3. Thus, we can take Dϵ to be a neighbourhood of the ellipsoid with D¯ϵR+3×R3, where Dϵ:=(ϵ,1/ϵ)×(ϵ,1/ϵ)×(ϵ,1/ϵ)×(1/ϵ,1/ϵ)×(1/ϵ,1/ϵ)×(1/ϵ,1/ϵ) and D¯ϵ denotes the compact closure of Dϵ. So for (x1,x2,x3,a1,a2,a3)T(R+3×R3)Dϵ, LV1, which implies that the condition (A2) in Lemma 5.1 holds. According to Lemma 5.1, we obtain that there is a solution (x1(t),x2(t),x3(t),a1(t),a2(t),a3(t))T of system (Equation5) which is a stationary Markov process. This completes the proof.

6. Local density function for system (5)

In this section, we will get the specific form of covariance matrix in the probability density around the quasi-positive equilibrium of the stochastic system (Equation5).

6.1. Equivalent transformation of system (5)

Firstly, we define a quasi-positive equilibrium E=(x1,x2,x3,a1,a2,a3)T involved in stochasticity by the equations (24) {x1(a1b11x1b12x2b13x3)=0,x2(a2b21x1b22x2b23x3)=0,x3(a3+b31x1+b32x2b33x3)=0,α1(a¯1a1)=0,α2(a¯2a2)=0,α3(a¯3a3)=0.(24) By solving Equation (Equation24), we obtain that if Assumption (A) holds, then x1=x¯1>0,x2=x¯2>0,x3=x¯3>0,a1=a¯1,a2=a¯2,a3=a¯3,where (x¯1,x¯2,x¯3) is the positive equilibrium of system (Equation1).

Let (u1,u2,u3,u4,u5,u6)T=(x1x1,x2x2,x3x3,a1a1,a2a2,a3a3)T. In view of Itô's integral and system (Equation5), the corresponding linearised system of (Equation5) around E can be expressed as follows (25) {du1=(q11u1q12u2q13u3+q14u4)dt,du2=(q21u1q22u2q23u3+q25u5)dt,du3=(q31u1+q32u2q33u3q36u6)dt,du4=α1u4dt+σ1dB1(t),du5=α2u5dt+σ2dB2(t),du6=α3u6dt+σ3dB3(t),(25) where q11=2b11x1+b12x2+b13x3a1=b11x1>0, q12=b12x1>0, q13=b13x1>0, q14=x1>0, q21=b21x2>0, q22=b21x1+2b22x2+b23x3a¯2=b22x2>0, q23=b23x2>0, q25=x2>0, q31=b31x3>0, q32=b32x3>0, q33=2b33x3+a¯3b31x1b32x2=b33x3>0, q36=x3>0.

6.2. Local density function for system (5)

In this subsection, we will obtain the specific form of covariance matrix in the probability density around the quasi-positive equilibrium of the stochastic system (Equation5). To this end, we need to introduce a definition and a lemma.

Definition 6.1

[Citation20]

The characteristic polynomial of the square matrix An is defined as φAn(λ)=λn+a1λn1++an1λ+an, then An is called a Hurwitz matrix if and only if An has all negative real-part eigenvalues, i.e. Hp=|a1a3a5a2p11a2a4a2p20a1a3a2p301a2a2p4000ap|>0,p=1,2,,n,where the complementary definition aj=0, j>n.

Lemma 6.1

[Citation7]

For the three-dimensional real algebraic equation H02+A0Σ0+Σ0A0T=0, where H0=diag(1,0,0), and A0=(τ1τ2τ3100010).If Σ0 is a three-dimensional symmetric matrix, τ1>0,τ3>0andτ1τ2τ3>0,then Σ0 is a positive definite matrix, where Σ0=(τ22(τ1τ2τ3)012(τ1τ2τ3)012(τ1τ2τ3)012(τ1τ2τ3)0τ12τ3(τ1τ2τ3)).Here A0 is called the standard R1 matrix.

Theorem 6.1

Let (u1(t),u2(t),u3(t),u4(t),u5(t),u6(t))T be a solution of system (Equation25) with the initial value (u1(0),u2(0),u3(0),u4(0),u5(0),u6(0))TR6. If Assumption (A) holds, then there is a local density function of a multivariate normal distribution Φ(u1,u2,u3,u4,u5,u6) around the origin point (0,0,0,0,0,0)T, which takes the form Φ(u1,u2,u3,u4,u5,u6)=(2π)3|Σ|12e12(u1,u2,u3,u4,u5,u6)Σ1(u1,u2,u3,u4,u5,u6)T,where Σ=(σij)6×6 denotes the covariance matrix of (u1,u2,u3,u4,u5,u6)T which is positive definite and takes the following form

  1. If ω1=0, ω2=0, ω3=0, then Σ=ρ~12(J6J5J2J1)1Σ~01[(J6J5J2J1)1]T+ρ~22(P6P5P2P1)1Σ~02[(P6P5P2P1)1]T+ρ~32(Q6Q5Q2Q1)1Σ~03[(Q6Q5Q2Q1)1]T.

  2. If ω1=0, ω20, ω30, then Σ=ρ~12(J6J5J2J1)1Σ~01[(J6J5J2J1)1]T+ρ22(P4P3P2P1)1Σ02[(P4P3P2P1)1]T+ρ32(Q4Q3Q2Q1)1Σ03[(Q4Q3Q2Q1)1]T.

  3. If ω10, ω2=0, ω30, then Σ=ρ12(J4J3J2J1)1Σ01[(J4J3J2J1)1]T+ρ~22(P6P5P2P1)1Σ~02[(P6P5P2P1)1]T+ρ32(Q4Q3Q2Q1)1Σ03[(Q4Q3Q2Q1)1]T.

  4. If ω10, ω20, ω3=0, then Σ=ρ12(J4J3J2J1)1Σ01[(J4J3J2J1)1]T+ρ22(P4P3P2P1)1Σ02[(P4P3P2P1)1]T+ρ~32(Q6Q5Q2Q1)1Σ~03[(Q6Q5Q2Q1)1]T.

  5. If ω1=0, ω2=0, ω30, then Σ=ρ~12(J6J5J2J1)1Σ~01[(J6J5J2J1)1]T+ρ~22(P6P5P2P1)1Σ~02[(P6P5P2P1)1]T+ρ32(Q4Q3Q2Q1)1Σ03[(Q4Q3Q2Q1)1]T.

  6. If ω1=0, ω20, ω3=0, then Σ=ρ~12(J6J5J2J1)1Σ~01[(J6J5J2J1)1]T+ρ22(P4P3P2P1)1Σ02[(P4P3P2P1)1]T+ρ~32(Q6Q5Q2Q1)1Σ~03[(Q6Q5Q2Q1)1]T.

  7. If ω10, ω2=0, ω3=0, then Σ=ρ12(J4J3J2J1)1Σ01[(J4J3J2J1)1]T+ρ~22(P6P5P2P1)1Σ~02[(P6P5P2P1)1]T+ρ~32(Q6Q5Q2Q1)1Σ~03[(Q6Q5Q2Q1)1]T.

  8. If ω10, ω20, ω30, then Σ=ρ12(J4J3J2J1)1Σ01[(J4J3J2J1)1]T+ρ22(P4P3P2P1)1Σ02[(P4P3P2P1)1]T+ρ32(Q4Q3Q2Q1)1Σ03[(Q4Q3Q2Q1)1]T,

here ω1=q21q22q31q212q32q21q31q33q23q312q312,ω2=q11q12q32q122q31q13q322q12q32q33q322,ω3=q13q22q23+q132q21q12q232q11q13q23q232,ρ1=σ1q14q31ω1,ρ~1=σ1q14q31,ρ2=σ2q25q32ω2,ρ~2=σ2q25q32,ρ3=σ3q23q36ω3,ρ~3=σ3q23q36,τ1=q11+q22+q33,τ2=q11q22+q11q33+q13q31+q22q33+q23q32q12q21,τ3=q11q22q33+q11q23q32+q13q22q31q13q21q32q12q21q33q12q23q31,p1=q11+q33+q21q32q31,p2=q11q33+q12q21+q13q31+q11q21q32q31,p3=(α1+q21q32q31q22)[q32(q11q22+q21q32q31)q12q31],r1=q22+q33+q12q31q32,r2=q22(q33+q12q31q32)+q23q32q12q21,r3=(α2+q12q31q32q11)[q31(q22q11+q12q31q32)q21q32],l1=q22+q33+q13q21q23,l2=q13q31+q22q33+q23q32+q13q21q33q23,l3=(α3+q13q21q23q11)[q21(q11q33q13q21q23)q23q31],a3(25)=q25(q32ω1+(q21q32q22q31)2q312),a3(26)=q36ω1(q22+q33)q21q36(q32ω1+(q21q32q22q31)2q312)q31,a5(23)=q12q21q11(q33+q21q32q31)q13q31,a5(24)=q11q32+q21q322q31q22q32q12q31,c3(25)=q14(q31ω2+(q12q31q11q32)2q322),c3(26)=q36ω2(q11+q33)q12q36q32(q31ω2+(q12q31q11q32)2q322),c5(23)=q12q21q22(q33+q12q31q32)q23q32,c5(24)=q31(q22q11+q12q31q32)q21q32,d3(25)=q14((q11q23q13q21)2q232ω3q21),d3(26)=q25[q13(q11q23q13q21)2q233+ω3(q11+q22q13q21q23)],d5(23)=(q13q31+q22q33+q23q32+q13q21q33q23),d5(24)=q21(q11q33q13q21q23)q23q31,Σ01=2(α13+α12τ1+α1τ2+τ3)(τ1τ2τ3),Σ02=2(α23+α22τ1+α2τ2+τ3)(τ1τ2τ3),Σ03=2(α33+α32τ1+α3τ2+τ3)(τ1τ2τ3),and J1=(000100100000001000010000000010000001),J2=(10000001000000100000q21q31100000010000001),J3=(1000000ω1q31ω1(q22+q33)ω1q32+(q21q32q22q31)2q3120000ω1q21q32q22q31q3100000100000010000001),J4=(q14q31ω1τ1τ2τ3a3(25)a3(26)010000001000000100000010000001),J5=(1000000q31q33q21q32q31q3200001000000100000010000001),J6=(q14q31(q11+q33+q21q32q31)a5(23)a5(24)q25q32q33q36010000001000000100000010000001),P1=(000010010000001000100000000100000001),P2=(10000001000000100000q12q32100000010000001),P3=(1000000ω2q32ω2(q11+q33)ω2q31+(q12q31q11q32)2q3220000ω2q12q31q11q32q3200000100000010000001),P4=(q25q32ω2τ1τ2τ3c3(25)c3(26)010000001000000100000010000001),P5=(1000000q32q33q12q31q32q3100001000000100000010000001),P6=(q25q32(q22+q33+q12q31q32)c5(23)c5(24)q14q31q33q36010000001000000100000010000001),Q1=(000001001000010000100000000100000010),Q2=(10000001000000100000q13q23100000010000001),Q3=(1000000ω3q23ω3(q11+q22)(q13q21q11q23)2q232ω3q210000ω3q13q21q11q23q2300000100000010000001),Q4=(q23q36ω3τ1τ2τ3d3(25)d3(26)010000001000000100000010000001),Q5=(1000000q23q22q13q21q23q2100001000000100000010000001)Q6=(q23q36(q22+q33+q13q21q23)d5(23)d5(24)q14q21q22q25010000001000000100000010000001),Σ01=(α12(τ1τ2τ3)+τ2(α1τ2+τ3)Σ010α1τ2+τ3Σ010α1τ2+τ3Σ010α1τ2+τ3Σ010α1+τ1Σ010α1+τ1Σ010000000000α1+τ1Σ0100000α1τ1(α1+τ1)+τ1τ2τ3α1τ3Σ0100000000),Σ~01=(α1p1+p22p1(α1p1+p2+α12)012p1(α1p1+p2+α12)012p1(α1p1+p2+α12)012p1(α1p1+p2+α12)0α1+p12α1p1p2(α1p1+p2+α12)000000000000000000000000000),Σ02=(α22(τ1τ2τ3)+τ2(α2τ2+τ3)Σ020α2τ2+τ3Σ020α2τ2+τ3Σ020α2τ2+τ3Σ020α2+τ1Σ020α2+τ1Σ020000000000α2+τ1Σ0200000α2τ1(α2+τ1)+τ1τ2τ3α2τ3Σ0200000000),Σ~02=(α2r1+r22r1(α2r1+r2+α22)0012r1(α2r1+r2+α22)12r1(α2r1+r2+α22)000000012r1(α2r1+r2+α22)0000000α2+r12α2r1r2(α2r1+r2+α22)000000000000000),Σ03=(α32(τ1τ2τ3)+τ2(α3τ2+τ3)Σ030α3τ2+τ3Σ030α3τ2+τ3Σ030α3τ2+τ3Σ030α3+τ1Σ030α3+τ1Σ030000000000α3+τ1Σ0300000α3τ1(α3+τ1)+τ1τ2τ3α3τ3Σ0300000000),Σ~03=(α3l1+l22l1(α3l1+l2+α32)0012l1(α3l1+l2+α32)12l1(α3l1+l2+α32)000000012l1(α3l1+l2+α32)0000000α3+l12α3l1l2(α3l1+l2+α32)000000000000000).

Proof.

Let X=(u1,u2,u3,u4,u5,u6)T, B(t)=(0,0,0,B1(t),B2(t),B3(t))T, A=(q11q12q13q1400q21q22q230q250q31q32q3300q36000α1000000α2000000α3),H=(000000000000000000000σ1000000σ2000000σ3).With these notations, system (Equation25) can be written in the following equivalent form: (26) dX(t)=AX(t)dt+HdB(t).(26) On the basis of Gardiner [Citation6], system (Equation26) has a unique density function Φ(u1,u2,u3,u4,u5,u6,t), which can be determined by the following six-dimensional Fokker–Planck equation (27) tΦ(X(t),t)+X(t)[AX(t)Φ(X(t),t)]σ1222u42Φ(X(t),t)σ2222u52Φ(X(t),t)σ3222u62Φ(X(t),t)=0.(27) Next, we will give the accurate expression of the density function by solving Equation (Equation27). Note that Φ(X(t),t)/t=0 under a stationary case, then (Equation27) becomes u1[(q11u1q12u2q13u3+q14u4)Φ]+u2[(q21u1q22u2q23u3+q25u5)Φ]+u3[(q31u1+q32u2q33u3q36u6)Φ]+u4(α1u4Φ)+u5(α2u5Φ)+u6(α3u6Φ)σ1222u42Φσ2222u52Φσ3222u62Φ=0.Additionally, because the diffusion matrix H of system (Equation26) is a constant matrix, then by the work of Gardiner [Citation6], we get (28) H2+AΣ+ΣAT=0.(28) Because of the mutual independence of Brownian motions B1(t), B2(t) and B3(t), in view of the finite independent superposition principle, system (Equation28) can be equivalent to the sum of solutions of the following three sub-equations Hi2+AΣi+ΣiAT=0,i=1,2,3,where H12=diag(0,0,0,σ12,0,0), H22=diag(0,0,0,0,σ22,0), H32=diag(0,0,0,0,0,σ32), Σi (i=1,2,3) are their corresponding solutions. One can easily get that Σ=Σ1+Σ2+Σ3 and H2=H12+H22+H32.

Let A(3):=(q11q12q13q21q22q23q31q32q33).In order to prove the matrix A(3) is a Hurwitz matrix, in view of Definition 6.1, we only need to show that all the eigenvalues of A(3) have negative real parts. To this end, define the characteristic equation of A(3) by φA(3)(λ)=λ3+τ1λ2+τ2λ+τ3=0,where τ1=q11+q22+q33>0,τ2=q11q22+q11q33+q13q31+q22q33+q23q32q12q21>0,τ3=q11q22q33+q11q23q32+q13q22q31q13q21q32q12q21q33q12q23q31>0.By direct computation, one has τ1τ2τ3=(q11+q22+q33)(q11q22+q11q33+q13q31+q22q33+q23q32q12q21)(q11q22q33+q11q23q32+q13q22q31q13q21q32q12q21q33q12q23q31)2=x1x2x3(b12b23b31+b13b21b32+2b11b22b33)+(x1)2x2(b112b22b11b12b21)+x1(x2)2(b11b222b12b21b22)+(x1)2x3(b112b33+b11b13b31)+x1(x3)2(b13b31b33+b11b332)+(x2)2x3(b222b33+b22b23b32)+x2(x3)2(b22b332+b23b32b33)>0.By means of Lemma 6.1, we obtain that the matrix A(3) has all negative real-part eigenvalues.

Next, we will accomplish the proof of Theorem 6.1 in three steps.

Step 1. Consider the algebraic equation (29) H12+AΣ1+Σ1AT=0,(29) where H12=diag(0,0,0,σ12,0,0).

Define A1=J1AJ11, where the ordering matrix J1 is given by J1=(000100100000001000010000000010000001).Direct calculation leads to that A1=(α100000q14q11q13q12000q31q33q320q360q21q23q22q2500000α2000000α3).Define A2=J2A1J21, where the elimination matrix J2 takes the form J2=(10000001000000100000q21q31100000010000001),then we obtain A2=(α100000q14q11q12q21q31q13q12000q31q33q21q32q31q320q3600ω1q21q32q31q22q25q21q36q310000α2000000α3),where ω1=(q21q22q31q212q32q21q31q33q23q312)/q312. In view of the value of ω1, we will divide the following discussion into two parts.

Case 1. If ω10, consider the standard transformation matrix J3=(1000000ω1q31ω1(q22+q33)ω1q32+(q21q32q22q31)2q3120000ω1q21q32q22q31q3100000100000010000001),such that A3=J3A2J31=(α100q14q31ω1(q11+q22+q33)a3(23)010001000000000a3(24)a3(25)a3(26)0q25(q21q32q22q31)q31a3(36)0q25q21q36q310α2000α3),where a3(23)=q12q21q11q22q11q33q13q31q22q33q23q32,a3(24)=q13q21q32+q12q21q33+q12q23q31q11q22q33q11q23q32q13q22q31,a3(25)=q25(q32ω1+(q21q32q22q31)2q312),a3(26)=q36ω1(q22+q33)q21q36(q32ω1+(q21q32q22q31)2q312)q31,a3(36)=q36ω1q21q36(q21q32q22q31)q312.Moreover, define A4=J4A3J41, where the standard transformation matrix J4 takes the form J4=(q14q31ω1τ1τ2τ3a3(25)a3(26)010000001000000100000010000001).Direct calculation gives that A4=((τ1+α1)(τ1α1+τ2)(τ2α1+τ3)100010001000000τ3α1a4(15)a4(16)0000q25(q21q32q22q31)q31a4(36)0q25q21q36q310α2000α3),where a4(15)=q25{(α1α2)(q32ω1+(q21q32q22q31)2q312)τ3τ2(q21q32q22q31)q31},a4(16)=q21q36τ3q31+(α1α3τ2)×[q36ω1(q22+q33)q21q36(q32ω1+(q21q32q22q31)2q312)q31],a4(36)=q36ω1(q22+q33)q21q36(q32ω1+(q21q32q22q31)2q312)q31.In addition, an equivalent algebraic equation of (6.6) can be described by (J4J3J2J1)H12(J4J3J2J1)T+A4(J4J3J2J1)Σ1(J4J3J2J1)T+(J4J3J2J1)Σ1(J4J3J2J1)TA4T=0.Let Σ01=1ρ12(J4J3J2J1)Σ1(J4J3J2J1)T,G0=diag(1,0,0,0,0,0),where ρ1=σ1q14q31ω1, then we have (30) G02+A4Σ01+Σ01A4T=0.(30) By solving Equation (Equation30), we obtain Σ01=(α12(τ1τ2τ3)+τ2(α1τ2+τ3)Σ010α1τ2+τ3Σ010α1τ2+τ3Σ010α1τ2+τ3Σ010α1+τ1Σ010α1+τ1Σ010000000000α1+τ1Σ0100000α1τ1(α1+τ1)+τ1τ2τ3α1τ3Σ0100000000),where Σ01=2(α13+α12τ1+α1τ2+τ3)(τ1τ2τ3).

Noticeably, the matrix Σ01 is positive semi-definite and its submatrix Σ01(4)=(α12(τ1τ2τ3)+τ2(α1τ2+τ3)Σ0100α1τ2+τ3Σ01α1τ2+τ3Σ0100α1+τ1Σ01α1τ2+τ3Σ0100α1+τ1Σ01α1+τ1Σ0100α1τ1(α1+τ1)+τ1τ2τ3α1τ3Σ01)is positive definite. Thus, the matrix Σ1=ρ12(J4J3J2J1)1Σ01[(J4J3J2J1)1]T is also positive semi-definite and there is a positive constant η1 such that Σ1η1(100000010000001000000100000000000000).Case 2. If ω1=0, consider the standardized transformation matrix J5=(1000000q31q33q21q32q31q3200001000000100000010000001),such that A5=J5A2J51=(α100q14q31(q11+q33+q21q32q31)a5(23)010000000000000a5(24)q25q32q33q3600q36q21q32q31q22q25q21q36q310α2000α3),where a5(23)=q12q21q11(q33+q21q32q31)q13q31,a5(24)=q11q32+q21q322q31q22q32q12q31.Then continue to do a matrix similarity transformation to A5, choose the matrix J6=(q14q31(q11+q33+q21q32q31)a5(23)a5(24)q25q32q33q36010000001000000100000010000001),such that A6=J6A5J61=((p1+α1)(p1α1+p2)p2α1100010000000000p3a6(15)a6(16)00000q36q21q32q31q22q25q21q36q310α2000α3),where p1=q11+q33+q21q32q31,p2=a5(23),p3=(α1+q21q32q31q22)a5(24),a6(15)=q25(q11q32q12q31q22q32+q21q322q31)+q25q32(α1α2),a6(16)=q13q31q36+q33q36(α1α3+q11)+q21q32q36q31(q22q21q32q31).Furthermore, an equivalent algebraic equation of (6.6) can be described by (J6J5J2J1)H12(J6J5J2J1)T+A6(J6J5J2J1)Σ1(J6J5J2J1)T+(J6J5J2J1)Σ1(J6J5J2J1)TA6T=0.Denote by Σ~01=1ρ~12(J6J5J2J1)Σ1(J6J5J2J1)T,ρ~1=σ1q14q31,then (31) G02+A6Σ~01+Σ~01A6T=0.(31) By solving Equation (Equation31), we obtain Σ~01=(α1p1+p22p1(α1p1+p2+α12)012p1(α1p1+p2+α12)012p1(α1p1+p2+α12)012p1(α1p1+p2+α12)0α1+p12α1p1p2(α1p1+p2+α12)000000000000000000000000000).It is obvious that the matrix Σ~01 is positive semi-definite and its submatrix Σ~01(4)=(α1p1+p22p1(α1p1+p2+α12)0012p1(α1p1+p2+α12)12p1(α1p1+p2+α12)00012p1(α1p1+p2+α12)000α1+p12α1p1p2(α1p1+p2+α12)000)is also positive semi-definite. So the matrix Σ1=ρ~12(J6J5J2J1)1Σ~01[(J6J5J2J1)1]T is positive semi-definite and there is a positive constant η~1 such that Σ1η~1(100000000000000000000100000000000000).Step 2. Consider the algebraic equation (32) H22+AΣ2+Σ2AT=0,(32) where H22=diag(0,0,0,0,σ22,0).

Let C1=P1AP11, where the ordering matrix P1 takes the form P1=(000010010000001000100000000100000001).Then C1=(α200000q25q22q23q21000q32q33q310q360q12q13q11q1400000α1000000α3).Let C2=P2C1P21, where the elimination matrix P2 takes the form P2=(10000001000000100000q12q32100000010000001),then we get C2=(α200000q25q22q12q21q32q23q21000q32q33q12q31q32q310q3600ω2q12q31q32q11q14q12q36q320000α1000000α3),where ω2=(q11q12q32q122q31q13q322q12q32q33)/q322. By means of the value of ω2, we will divide the following discussion into two parts.

Case 1. If ω20, consider the standardized transformation matrix P3=(1000000ω2q32ω2(q11+q33)ω2q31+(q12q31q11q32)2q3220000ω2q12q31q11q32q3200000100000010000001),such that C3=P3C2P31=(α2000ω2q25q32(q11+q22+q33)c3(23)c3(24)010000100000000000c3(25)c3(26)q14(q12q31q11q32)q32c3(36)q14q12q36q32α100α3),where c3(23)=q12q21q11q22q11q33q22q33q23q32+q31ω2+q12q31q32(q33q11)+q122q312q322,c3(24)=ω2(q22q31q21q32)+q12q31q11q32q322×(q12q22q31+q22q32q33+q23q322q12q21q32),c3(25)=q14(q31ω2+(q12q31q11q32)2q322),c3(26)=q36ω2(q11+q33)q12q36q32(q31ω2+(q12q31q11q32)2q322),c3(36)=q36ω2q12q36(q12q31q11q32)q322.Then continue to do a matrix similarity transformation to C3, choose the matrix P4=(q25q32ω2τ1τ2τ3c3(25)c3(26)010000001000000100000010000001),such that C4=P4C3P41=((τ1+α2)(τ1α2+τ2)(τ2α2+τ3)100010001000000τ3α2c4(15)c4(16)0000q14(q12q31q11q32)q32c4(36)0q14q12q36q320α1000α3),where c4(15)=q14{(α2α1)[q112q12q31(q11q33)q32]τ3+τ2(q11q32q12q31)q32},c4(16)=q36{(α3α2)[q12q32(q332q13q31)+q11q13+q13q33]τ2(q13+q12q33q32)+q12τ3q32},c4(36)=q36(q12q33+q13q32)q32.Moreover, an equivalent algebraic equation of (Equation32) can be described by (P4P3P2P1)H22(P4P3P2P1)T+C4(P4P3P2P1)Σ2(P4P3P2P1)T+(P4P3P2P1)Σ2(P4P3P2P1)TC4T=0.Let Σ02=1ρ22(P4P3P2P1)Σ2(P4P3P2P1)T,where ρ2=σ2q25q32ω2, then we have (33) G02+C4Σ02+Σ02C4T=0.(33) By solving Equation (Equation33), we have Σ02=(α22(τ1τ2τ3)+τ2(α2τ2+τ3)Σ020α2τ2+τ3Σ020α2τ2+τ3Σ020α2τ2+τ3Σ020α2+τ1Σ020α2+τ1Σ020000000000α2+τ1Σ0200000α2τ1(α2+τ1)+τ1τ2τ3α2τ3Σ0200000000),where Σ02=2(α23+α22τ1+α2τ2+τ3)(τ1τ2τ3).

It is easy to see that the matrix Σ02 is positive semi-definite and its submatrix Σ02(4)=(α22(τ1τ2τ3)+τ2(α2τ2+τ3)Σ0200α2τ2+τ3Σ02α2τ2+τ3Σ0200α2+τ1Σ02α2τ2+τ3Σ0200α2+τ1Σ02α2+τ1Σ0200α2τ1(α2+τ1)+τ1τ2τ3α2τ3Σ02)is positive definite. Accordingly, the matrix Σ2=ρ22(P4P3P2P1)1Σ02[(P4P3P2P1)1]T is also positive semi-definite and there is a positive constant η2 such that Σ2η2(100000010000001000000000000010000000).Case 2. If ω2=0, consider the standardized transformation matrix P5=(1000000q32q33q12q31q32q3100001000000100000010000001),such that C5=P5C2P51=(α200q25q32(q22+q33+q12q31q32)c5(23)010000000000000c5(24)q14q31q33q3600q36q12q31q32q11q14q12q36q320α1000α3),where c5(23)=q12q21q22(q33+q12q31q32)q23q32,c5(24)=q31(q22q11+q12q31q32)q21q32.Then continue to do a matrix similarity transformation to C5, choose the matrix P6=(q25q32(q22+q33+q12q31q32)c5(23)c5(24)q14q31q33q36010000001000000100000010000001),such that C6=P6C5P61=((r1+α2)(r1α2+r2)r2α2r3c6(15)c6(16)10000001000q36000q12q31q32q11q14q12q36q320000α1000000α3),where r1=q22+q33+q12q31q32,r2=c5(23),r3=(α2+q12q31q32q11)c5(24),c6(15)=q14q31(q22q11+q12q31q32)q14q21q32+q14q31(α2α1),c6(16)=q33q36(α2α3)+q22q33q36+q23q32q36+q11q12q31q36q32q122q312q36q322.Furthermore, an equivalent algebraic equation of (6.6) can be described by (P6P5P2P1)H22(P6P5P2P1)T+C6(P6P5P2P1)Σ2(P6P5P2P1)T+(P6P5P2P1)Σ2(P6P5P2P1)TC6T=0.Let Σ~02=1ρ~22(P6P5P2P1)Σ2(P6P5P2P1)T,ρ~2=σ2q25q32,then (34) G02+C6Σ~02+Σ~02C6T=0.(34) By solving Equation (Equation34), we obtain Σ~02=(α2r1+r22r1(α2r1+r2+α22)0012r1(α2r1+r2+α22)12r1(α2r1+r2+α22)000000012r1(α2r1+r2+α22)0000000α2+r12α2r1r2(α2r1+r2+α22)000000000000000).Obviously, the matrix Σ~02 is positive semi-definite and its submatrix Σ~02(4)=(α2r1+r22r1(α2r1+r2+α22)012r1(α2r1+r2+α22)0012r1(α2r1+r2+α22)0012r1(α2r1+r2+α22)0α2+r12α2r1r2(α2r1+r2+α22)00000)is also positive semi-definite. So the matrix Σ2=ρ~22(P6P5P2P1)1Σ~02[(P6P5P2P1)1]T is positive semi-definite and there is a positive constant η~2 such that Σ2η~2(000000010000000000000000000010000000).Step 3. Consider the algebraic equation (29) H32+AΣ3+Σ3AT=0,(29) where H32=diag(0,0,0,0,0,σ32).

Let D1=Q1AQ11, where the ordering matrix Q1 takes the form Q1=(000001001000010000100000000100000010).By a simple computation, we have D1=(α300000q36q33q32q31000q23q22q210q250q13q12q11q1400000α1000000α2).Choose the elimination matrix Q2=(10000001000000100000q13q23100000010000001),such that D2=Q2D1Q21=(α300000q36q33q32+q13q31q23q31000q23q22q13q21q23q210q2500ω3q13q21q23q11q14q13q25q230000α1000000α2),where ω3=(q13q22q23+q132q21q12q232q11q13q23)/q232. According to the value of ω3, we will divide the following discussion into two parts.

Case 1. If ω30, consider the standardized transformation matrix Q3=(1000000ω3q23ω3(q11+q22)(q13q21q11q23)2q232ω3q210000ω3q13q21q11q23q2300000100000010000001),such that D3=Q3D2Q31=(α3000q23q36ω3(q11+q22+q33)d3(23)d3(24)010000100000000000d3(25)d3(26)q14(q13q21q11q23)q23d3(36)q14q13q25q23α100α2),where d3(23)=q132q212q232+q13q21(q22q21)q23q11q22q11q33q13q31q21ω3q22q33q23q32,d3(24)=q13q21q33q23(q22+q13q11)+q13q21q32ω3(q21q33+q23q31)q11q23q32q11q13q31q11q22q33,d3(25)=q14((q11q23q13q21)2q232ω3q21),d3(26)=q25[q13(q11q23q13q21)2q233+ω3(q11+q22q13q21q23)],d3(36)=q25[q13(q11q23q13q21)q232+ω3].Then continue to do a matrix similarity transformation to D3, choose the matrix Q4=(q23q36ω3τ1τ2τ3d3(25)d3(26)010000001000000100000010000001),such that D4=Q4D3Q41=((τ1+α3)(τ1α3+τ2)(τ2α3+τ3)τ3α310000100001000000000d4(15)d4(16)00q14(q13q21q11q23)q23d4(36)q14q13q25q23α100α2),where d4(15)=q14{(α3α1)[q12q21q112q13q21(q11+q22)q23]τ3+τ2q11q23q13q21q23},d4(16)=q25{(α2α3)[q13q23(q222q12q21)q11q12q12q22]+q13q23τ3+τ2q23(q12q23q13q22)},d4(36)=q25(q13q22q12q23)q23.Besides, an equivalent algebraic equation of (6.6) takes the following form (Q4Q3Q2Q1)H32(Q4Q3Q2Q1)T+D4(Q4Q3Q2Q1)Σ3+Σ3(Q4Q3Q2Q1)TD4T=0.Let Σ03=1ρ32(Q4Q3Q2Q1)Σ3(Q4Q3Q2Q1)T,where ρ3=σ3q23q36ω3, we derive (35) G02+D4Σ03+Σ03D4T=0.(35) By solving Equation (Equation35), we have Σ03=(α32(τ1τ2τ3)+τ2(α3τ2+τ3)Σ030α3τ2+τ3Σ030α3τ2+τ3Σ030α3τ2+τ3Σ030α3+τ1Σ030α3+τ1Σ030000000000α3+τ1Σ0300000α3τ1(α3+τ1)+τ1τ2τ3α3τ3Σ0300000000),where Σ03=2(α33+α32τ1+α3τ2+τ3)(τ1τ2τ3).

It is easy to see that the matrix Σ03 is positive semi-definite and its submatrix Σ03(4)=(α32(τ1τ2τ3)+τ2(α3τ2+τ3)Σ0300α3τ2+τ3Σ03α3τ2+τ3Σ0300α3+τ1Σ03α3τ2+τ3Σ0300α3+τ1Σ03α3+τ1Σ0300α3τ1(α3+τ1)+τ1τ2τ3α3τ3Σ03)is positive definite. Therefore, the matrix Σ3=ρ32(Q4Q3Q2Q1)1Σ03[(Q4Q3Q2Q1)1]T is also positive semi-definite and there is a positive constant η3 such that Σ3η3(100000010000001000000000000000000001).Case 2. If ω3=0, consider the standardized transformation matrix Q5=(1000000q23q22q13q21q23q2100001000000100000010000001),such that D5=Q5D2Q51=(α300q23q36(q22+q33+q13q21q23)d5(23)010000000000000d5(24)q14q21q22q2500q25q13q21q23q11q14q13q25q230α1000α2),where d5(23)=(q13q31+q22q33+q23q32+q13q21q33q23),d5(24)=q21(q11q33q13q21q23)q23q31.Then continue to do a matrix similarity transformation to D5, choose the matrix Q6=(q23q36(q22+q33+q13q21q23)d5(23)d5(24)q14q21q22q25010000001000000100000010000001),such that D6=Q6D5Q61=((l1+α3)(l1α3+l2)l2α3l3d6(15)d6(16)10000001000q25000q13q21q23q11q14q13q25q230000α1000000α2),where l1=q22+q33+q13q21q23,l2=q13q31+q22q33+q23q32+q13q21q33q23,l3=(α3+q13q21q23q11)[q21(q11q33q13q21q23)q23q31],d6(15)=q14[q21(q11q33q13q21q23)q23q31]+q14q21(α1α3),d6(16)=q22q25(α2α3)q25(q22q33+q23q32)q13q25q23(q11q21q13q212q23).Furthermore, an equivalent algebraic equation of (6.6) can be described by (Q6Q5Q2Q1)H32(Q6Q5Q2Q1)T+D6(Q6Q5Q2Q1)Σ3(Q6Q5Q2Q1)T+(Q6Q5Q2Q1)Σ3(Q6Q5Q2Q1)TD6T=0.Denote by Σ~03=1ρ~32(Q6Q5Q2Q1)Σ3(Q6Q5Q2Q1)T,ρ~3=σ3q23q36,then (36) G02+D6Σ~03+Σ~03D6T=0.(36) By solving Equation (Equation36), we obtain Σ~03=(α3l1+l22l1(α3l1+l2+α32)0012l1(α3l1+l2+α32)12l1(α3l1+l2+α32)000000012l1(α3l1+l2+α32)0000000α3+l12α3l1l2(α3l1+l2+α32)000000000000000).It is clear that the matrix Σ~03 is positive semi-definite and its submatrix Σ~03(4)=(α3l1+l22l1(α3l1+l2+α32)012l1(α3l1+l2+α32)0012l1(α3l1+l2+α32)0012l1(α3l1+l2+α32)0α3+l12α3l1l2(α3l1+l2+α32)00000)is also positive semi-definite. Hence the matrix Σ3=ρ~32(Q6Q5Q2Q1)1Σ~03[(Q6Q5Q2Q1)1]T is positive semi-definite and there is a positive constant η~3 such that Σ3η~3(000000000000001000000000000000000001).Now we prove the matrix Σ=Σ1+Σ2+Σ3 in Equation (Equation28) is positive definite. If ω1=0, ω2=0, ω3=0, then the covariance matrix Σ=Σ1+Σ2+Σ3η~1(100000000000000000000100000000000000)+η~2(000000010000000000000000000010000000)+η~3(000000000000001000000000000000000001)(η~1η~2η~3)(100000010000001000000100000010000001)is positive definite. Similarly, we can show that under the other seven cases, the covariance matrix Σ is also positive definite. This completes the proof.

7. Numerical simulations

In this section, we will present some numerical simulations to show the effectiveness of our theoretical findings. By employing the Milstein higher-order method introduced in [Citation8], we obtain the corresponding discretization equations of system (Equation5) which are given by: (37) {x1j+1=x1j+x1j(a1jb11x1jb12x2jb13x3j)Δt,x2j+1=x2j+x2j(a2jb21x1jb22x2jb23x3j)Δt,x3j+1=x3j+x3j(a3j+b31x1j+b32x2jb33x3j)Δt,a1j+1=a1j+α1(a¯1a1j)Δt+σ1Δtε1,j,a2j+1=a2j+α2(a¯2a2j)Δt+σ2Δtε2,j,a3j+1=a3j+α3(a¯3a3j)Δt+σ3Δtε3,j,(37) where (x1j,x2j,x3j,a1j,a2j,a3j)T is the corresponding value of the jth iteration of the Equation (Equation37). Δt is the time increment which is positive, σi2 are the noise intensities, εij are mutually independent Gaussian random variables which follow the distribution N(0,1) for i=1,2,3;j=1,,n. We choose real parameter values from the previous references, and all the values of biological parameters are given in Table .

Table 1. List of parameter values of system (Equation5).

Example 7.1

To show the existence of a stationary distribution numerically, we choose σ1=σ2=σ3=0.001 and the other parameter values are given in Table . By simple computation, we obtain b12b21b33+b12b23b31+b13b21b32b11b22b33b11b23b32b13b22b31=6.2334<0,a¯2b12b33+a¯2b13b32+a¯3b12b23a¯1b22b33a¯1b23b32a¯3b13b22=4.3433<0,a¯1b21b33+a¯1b23b31+a¯3b13b21a¯2b11b33a¯2b13b31a¯3b11b23=1.6<0,a¯1b21b32+a¯2b12b31+a¯3b11b22a¯1b22b31a¯2b11b32a¯3b12b21=0.52855<0,b11=1>0.0005=σ12,b22=2>0.0005=σ22,b33=3>0.0005=σ32,4b13b23b31b32(b11σ12)(b22σ22)=0.0255808032>6.5536×104=(b12b23b31+b13b21b32)2,ω=b31σ12b13α1+b32σ22b23α2+σ32α3=0.0075<0.02156952448=min{b31b13(b11σ12)ϵ2(b12b31b13+b21b32b23)x¯12,b32b23(b22σ22)12ϵ(b12b31b13+b21b32b23)x¯22,(b33σ32)x¯32},and b12b23b31+b13b21b322b13b32(b22σ22)=0.16004001<0.2=ϵ=0.2<6.246875=2b23b31(b11σ12)b12b23b31+b13b21b32.

That is to say, the conditions of Theorem 5.1 hold. According to Theorem 5.1, we get that system (Equation5) admits a stationary distribution π() which indicates that all the species are coexistent a.s. in the long time. See Figure .

Figure 1. The left column displays the trajectory of the species x1, x2 and x3 of system (Equation5) and their corresponding deterministic model (Equation1) with noise intensities σ1=σ2=σ3=0.001. The right column shows the frequency histograms and marginal density functions of x1, x2 and x3 in system (Equation5).

Figure 1. The left column displays the trajectory of the species x1, x2 and x3 of system (Equation5(5) {dx1(t)=x1(t)[a1(t)−b11x1(t)−b12x2(t)−b13x3(t)]dt,dx2(t)=x2(t)[a2(t)−b21x1(t)−b22x2(t)−b23x3(t)]dt,dx3(t)=x3(t)[−a3(t)+b31x1(t)+b32x2(t)−b33x3(t)]dt,da1(t)=α1[a¯1−a1(t)]dt+σ1dB1(t),da2(t)=α2[a¯2−a2(t)]dt+σ2dB2(t),da3(t)=α3[a¯3−a3(t)]dt+σ3dB3(t).(5) ) and their corresponding deterministic model (Equation1(1) {dx1(t)dt=x1(t)[a1−b11x1(t)−b12x2(t)−b13x3(t)],dx2(t)dt=x2(t)[a2−b21x1(t)−b22x2(t)−b23x3(t)],dx3(t)dt=x3(t)[−a3+b31x1(t)+b32x2(t)−b33x3(t)],(1) ) with noise intensities σ1=σ2=σ3=0.001. The right column shows the frequency histograms and marginal density functions of x1, x2 and x3 in system (Equation5(5) {dx1(t)=x1(t)[a1(t)−b11x1(t)−b12x2(t)−b13x3(t)]dt,dx2(t)=x2(t)[a2(t)−b21x1(t)−b22x2(t)−b23x3(t)]dt,dx3(t)=x3(t)[−a3(t)+b31x1(t)+b32x2(t)−b33x3(t)]dt,da1(t)=α1[a¯1−a1(t)]dt+σ1dB1(t),da2(t)=α2[a¯2−a2(t)]dt+σ2dB2(t),da3(t)=α3[a¯3−a3(t)]dt+σ3dB3(t).(5) ).

Example 7.2

To illustrate the existence of the probability density near the origin point (0,0,0,0,0,0)T, we choose σ1=σ2=σ3=0.001 and the other parameter values are presented in Table . Direct calculation gives that (x1,x2,x3,a¯1,a¯2,a¯3)T=(0.6968,0.2567,0.0848,0.8,0.6,0.05)T and b12b21b33+b12b23b31+b13b21b32b11b22b33b11b23b32b13b22b31=6.2334<0,a¯2b12b33+a¯2b13b32+a¯3b12b23a¯1b22b33a¯1b23b32a¯3b13b22=4.3433<0,a¯1b21b33+a¯1b23b31+a¯3b13b21a¯2b11b33a¯2b13b31a¯3b11b23=1.6<0,a¯1b21b32+a¯2b12b31+a¯3b11b22a¯1b22b31a¯2b11b32a¯3b12b21=0.52855<0.In other words, the conditions of Theorem 6.1 hold. Accordingly, system (Equation5) has a Gaussian probability density near the origin point (0,0,0,0,0,0)T. Hence, in view of Theorem 6.1, we calculate the specific form of the covariance matrix Σ, Σ=(0.000151035300.036922458290.000006409560.036922458299.523407716930.001643331960.000006409560.001643331960.000002142530.000007845830.000000038120.000000171730.000022077410.005738864260.000001025270.000000230560.000000074200.000000463060.000007845830.000022077410.000000230560.000000038120.005738864260.000000074200.000000171730.000001025270.000000463060.000013779000000.000013656880000.00002734324),and the corresponding probability density Φ(u1,u2,u3,u4,u5,u6) takes the form Φ(u1,u2,u3,u4,u5,u6)=(2π)3|Σ|12e12(u1,u2,u3,u4,u5,u6)Σ1(u1,u2,u3,u4,u5,u6)T=8390771386.35635060275e12(u1,u2,u3,u4,u5,u6)Σ1(u1,u2,u3,u4,u5,u6)T.Figure  shows this.

Figure 2. Numerical simulations for: (i) the frequency histogram fitting density curves of x1, x2, x3, a1, a2 and a3 of system (Equation5) with 200,000 iteration points, respectively. (ii) The marginal probability densities of x1, x2, x3, a1, a2 and a3 of system (Equation5). All of the parameter values are the same as in Figure .

Figure 2. Numerical simulations for: (i) the frequency histogram fitting density curves of x1, x2, x3, a1, a2 and a3 of system (Equation5(5) {dx1(t)=x1(t)[a1(t)−b11x1(t)−b12x2(t)−b13x3(t)]dt,dx2(t)=x2(t)[a2(t)−b21x1(t)−b22x2(t)−b23x3(t)]dt,dx3(t)=x3(t)[−a3(t)+b31x1(t)+b32x2(t)−b33x3(t)]dt,da1(t)=α1[a¯1−a1(t)]dt+σ1dB1(t),da2(t)=α2[a¯2−a2(t)]dt+σ2dB2(t),da3(t)=α3[a¯3−a3(t)]dt+σ3dB3(t).(5) ) with 200,000 iteration points, respectively. (ii) The marginal probability densities of x1, x2, x3, a1, a2 and a3 of system (Equation5(5) {dx1(t)=x1(t)[a1(t)−b11x1(t)−b12x2(t)−b13x3(t)]dt,dx2(t)=x2(t)[a2(t)−b21x1(t)−b22x2(t)−b23x3(t)]dt,dx3(t)=x3(t)[−a3(t)+b31x1(t)+b32x2(t)−b33x3(t)]dt,da1(t)=α1[a¯1−a1(t)]dt+σ1dB1(t),da2(t)=α2[a¯2−a2(t)]dt+σ2dB2(t),da3(t)=α3[a¯3−a3(t)]dt+σ3dB3(t).(5) ). All of the parameter values are the same as in Figure 1.

8. Conclusion

Predator–prey system is an important component of a real ecosystem. In fact, predator–prey interaction with multiple preys is common and more complicated in the nature. In this work, we concentrate on a stochastic predator–prey model with two competitive preys. In general, existing literature always simulated environmental fluctuations with white noise or coloured noise. While in this paper, we employ another way to introduce stochastic perturbations to simulate random interferences in the environment, i.e. the Ornstein–Uhlenbeck process. To ensure the mathematical and biological rationality of the stochastic system (Equation5), we first show that the solution of the system is global. Then we study the dynamics of system (Equation5) in detail. More precisely, analytical findings of this paper are presented as below:

  • Given initial value (x1(0),x2(0),x3(0),a1(0),a2(0),a3(0))TR+3×R3 and any integer p3, there exists a positive constant M(p) such that the solution (x1(t),x2(t),x3(t),a1(t),a2(t),a3(t))T of system (Equation5) has the property lim suptE[(b31b13x1(t)+b32b23x2(t)+x3(t))pp+a12p(t)2p+a22p(t)2p+a32p(t)2p]M(p).

  • Given initial value (x1(0),x2(0),x3(0),a1(0),a2(0),a3(0))TR+3×R3, the solution (x1(t),x2(t),x3(t),a1(t),a2(t),a3(t))T of system (Equation5) has the property lim suptln(b31b13x1(t)+b32b23x2(t)+x3(t))lnt1pa.s.,foranyintegerp3.In addition lim suptlnx1(t)t0,lim suptlnx2(t)t0andlim suptlnx3(t)t0a.s.

  • Let Assumption (A) hold. If the following conditions hold b11>σ12,b22>σ22,b33>σ32,4b13b23b31b32(b11σ12)(b22σ22)>(b12b23b31+b13b21b32)2,ω<min{b31b13(b11σ12)ϵ2(b12b31b13+b21b32b23)x¯12,b32b23(b22σ22)12ϵ(b12b31b13+b21b32b23)x¯22,(b33σ32)x¯32}, and ϵ>0 is a sufficiently small constant satisfying b12b23b31+b13b21b322b13b32(b22σ22)<ϵ<2b23b31(b11σ12)b12b23b31+b13b21b32,then system (Equation5) has at least one stationary distribution π() on R+3×R3, where ω=b31σ1/(2b13α1)+b32σ2/(2b23α2)+σ3/(2α3) and (x¯1,x¯2,x¯3) is the positive equilibrium of system (Equation1).

  • Let (u1(t),u2(t),u3(t),u4(t),u5(t),u6(t))T be a solution of system (Equation25) with the initial value (u1(0),u2(0),u3(0),u4(0),u5(0),u6(0))TR6. If Assumption (A) holds, then there is a local density function of a multivariate normal distribution Φ(u1,u2,u3,u4,u5,u6) near the origin point (0,0,0,0,0,0)T, which takes the form Φ(u1,u2,u3,u4,u5,u6)=(2π)3|Σ|12e12(u1,u2,u3,u4,u5,u6)Σ1(u1,u2,u3,u4,u5,u6)T,where the exact form of covariance matrix Σ can be obtained in the proof process of Theorem 6.1.

However, some other topics of interest are worthy of further consideration. For example, in this paper, we only suppose that the parameters ai (i=1,2,3) satisfy the Ornstein–Uhlenbeck process, as a matter of fact, if we suppose that other parameters involved in system (Equation5) satisfy the Ornstein–Uhlenbeck process, what kind of dynamic behaviour should system (Equation5) exhibit? Secondly, as far as we know, most of key parameters in population systems are usually affected by coloured noise [Citation27]. Hence, the corresponding generalised stochastic system (Equation5) with discrete Markov switching should be studied. Finally, in this paper, we do not consider the effects of spatial diffusion factor, if we take into account the factor, i.e. we formulate a stochastic partial differential equation model, then the system (Equation5) will show more abundant dynamic behaviours. These matters will be the subject of our future work.

 

Acknowledgments

The author read and approved the final manuscript.

Availability of data and materials

Data sharing is not applicable to this article as no datasets are generated or analysed during the current study.

Disclosure statement

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

Additional information

Funding

The author thanks the National Natural Science Foundation of P.R. China [grant number 12001090] and the Jilin Provincial Science and Technology Development Plan Project [grant number YDZJ202201ZYTS633].

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