DISCRETE AND CONTINUOUS doi:10.3934/dcds.2013.33.2495DYNAMICAL SYSTEMSVolume 33, Number 6, June 2013 pp. 2495–2522

POPULATION DYNAMICAL BEHAVIOR OF

LOTKA-VOLTERRA COOPERATIVE SYSTEMS WITH RANDOM

PERTURBATIONS

Meng Liu

School of Mathematical Science

Huaiyin Normal University, Huaian 223300, China

Department of MathematicsHarbin Institute of Technology, Weihai 264209, China

Ke Wang

Department of Mathematics

Harbin Institute of Technology, Weihai 264209, China

(Communicated by Hinke Osinga)

Abstract. This paper is concerned with two n-species stochastic coopera-

tive systems. One is autonomous, the other is non-autonomous. For the first

system, we prove that for each species, there is a constant which can be repre-sented by the coefficients of the system. If the constant is negative, then the

corresponding species will go to extinction with probability 1; If the constant

is positive, then the corresponding species will be persistent with probability1. For the second system, sufficient conditions for stochastic permanence and

global attractivity are established. In addition, the upper- and lower-growthrates of the positive solution are investigated. Our results reveal that, firstly,

the stochastic noise of one population is unfavorable for the persistence of all

species; secondly, a population could be persistent even the growth rate of thispopulation is less than the half of the intensity of the white noise.

1. Introduction. Mutualism is an important ecological interaction in nature. Leaf-cutter ants and the fungi they cultivate are a classic example of such a relationship,but others abound. Some plants produce fruit attractive to birds, so that the plantsseeds are dispersed more effectively while the birds have a food source ([4]). As weknow, several different cooperative models have been developed to describe thistype of ecological interaction (see e.g. [4, 9, 15]).

Among various types of cooperative models, we should specially mention thefollowing Lotka-Volterra system:

dxidt

= xi[ri −n∑j=1

aijxj ], i = 1, ..., n (1)

or in matrix form

dx(t)

dt= diag(x1(t), ..., xn(t))[(r − Ax(t)],

2010 Mathematics Subject Classification. Primary: 34F05, 60H10; Secondary: 92B05, 60J27.Key words and phrases. Cooperative system, random perturbations, persistence, extinction,

global attractivity.

2495

2496 MENG LIU AND KE WANG

where xi(t) is the population size of the ith species at time t, aii > 0, aij < 0,i 6= j, i, j = 1, 2, ..., n, and x = (x1, ..., xn)T , r = (r1, ..., rn)T , A = (aij)n×n. Herewe choose the Lotka-Volterra system (1) not other models on the grounds that, onthe one hand, system (1) is the most basic and important; on the other hand, somebiologists have argued that in many cases, system (1) can describe the reality well(see e.g. [4]). In fact, owing to its theoretical and practical significance, system (1)has been extensively studied, and we here mention [1, 2, 3, 9, 10, 12, 15, 17, 19, 26,27, 31, 33, 35, 36] among others.

On the other hand, population systems in nature are often subject to environ-mental noise. It is therefore important to reveal how the noise affects the populationsystems. As a matter of fact, stochastic cooperative system has recently been stud-ied by many authors, see e.g. [5, 8, 14, 22, 28, 29, 32, 34]. Here Mao, Sabais,and Renshaw [29] revealed that the environmental noise may suppress a potentialpopulation explosion while Luo and Mao [28] showed that the stochastic model canproduce many desired properties for population dynamics, for example, stochasti-cally ultimate boundedness and the moment average boundedness in time. Amongvarious types of stochastic cooperative models, we should specially mention thefollowing system:

dxi = xi[ri −n∑j=1

aijxj ]dt+ σixidBi(t), i = 1, ..., n, (2)

where (B1(t), ..., Bn(t))T is an n-dimensional Brownian motion defined on a com-plete probability space (Ω,F ,P). For a symmetric n × n matrix Q = (qij)n×n,define λ+max(Q) = supx∈Rn

+,|x|=1 xTQx, where Rn+ = a = (a1, a2, ..., an) : ai >

0, 1 ≤ i ≤ n. For system (2), Cheng [8] established the following result.

Theorem 1.1. Suppose that there exist positive numbers p1, ..., pn such thatλ+max(−PA− ATP ) ≤ 0, where P = diag(p1, ..., pn). Moreover suppose that

σiσj > ri + rj , 1 ≤ i, j ≤ n. (3)

Then for any given initial value x(0) ∈ Rn+, the solution x(t) = (x1(t), ..., xn(t))T

of Eq. (2) obeys limt→+∞

xi(t) = 0 almost surely (a.s.), 1 ≤ i ≤ n.

However, some interesting questions are that

(Q1) Theorem 1.1 tells us that if the noise is sufficiently large (i.e. (3) holds),then all the populations modeled by (2) will go to extinction a.s. Then itis interesting and important to investigate what happens if the noise is notsufficiently large (i.e. (3) is not satisfied). Moreover, when the species will bepersistent? Can we find the critical value between extinction and persistence?

(Q2) It is well known that in the study of stochastic population systems, stochasticpermanence, which indicates that the species will survive forever, is one ofthe most important and interesting topics. However, stochastic permanenceof system (2) was not investigated in Cheng [8]. Then the second interestingquestion is that when system (2) will be stochastically permanent?

(Q3) Global attractivity also is an important topic. However, in [8], this questionwas not studied. Then the third interesting question is that when the solutionof system (2) will be globally attractive?

The main aims of this paper are to investigate the above questions one by one. InSection 2, we will give some preliminaries. In Section 3, we shall study (Q1). We

POPULATION DYNAMICAL BEHAVIOR 2497

will prove that for each population represented by (2), there exists a constant whichcan be represented by the coefficients of (2). We shall see that if the constant isnegative, then the corresponding population will go to extinction with probability1; if the constant is positive, then the corresponding population will be persistent

with probability 1. In addition, we shall estimate limt→+∞

t−1∫ t0xi(s)ds.

In Section 4, we shall discuss (Q2). Since the natural growth of many populationsvary with t in real situation, for example, due to the seasonality, then from thissection we shall consider the following non-autonomous model:

dxi = xi[ri(t)−n∑j=1

aij(t)xj ]dt+ σi(t)xidBi(t), i = 1, ..., n, (4)

where ri(t), aij(t) and σi(t) are continuous and bounded functions on t ≥ 0 andaii(t) > 0, aij(t) < 0 for i, j = 1, ..., n, j 6= i. It is easy to see that Eq.(2) is aspecial case of Eq.(4). In this section, we will establish a sufficient condition forstochastic permanence of system (4). In Section 5, the growth rate of the positivesolution of Eq. (4) will be estimated.

In Section 6 we shall investigate (Q3) by establishing a sufficient condition forglobal attractivity of Eq.(4). In Section 7, we will introduce some figures to illustrateour main results. We shall give the conclusions and the future directions of theresearch in the last section.

2. Preliminaries. Throughout this paper, let (Ω,F ,P) be a complete probabil-ity space with a filtration Ftt∈R+

satisfying the usual conditions (i.e., it is rightcontinuous and increasing while F0 contains all P-null sets). Let (B1(t), B2(t), ..,Bn(t))T denote an n-dimensional Brownian motion defined on this probabilityspace. If Q is a vector or matrix, its transpose is denoted by QT . If Q is amatrix, its trace norm is denoted by |Q| =

√trace(QTQ) whilst its operator norm

is denoted by ||Q|| = sup|Ax| : |x| = 1.Let Lk([0,+∞);Rn) be the family of all Rn valued measurable Ft-adapted

processes f(t) satisfying∫ T0|f(t)|kdt <∞ a.s. for every T > 0.

Lemma 2.1. (Ito’s formula, see e.g. [30], P.38) Let x(t) be an n−dimensional Itoprocess on t ≥ 0 with the stochastic differential

dx(t) = f(t)dt+ g(t)dB(t),

where f ∈ L1([0,+∞);Rn) and g ∈ L2([0,+∞);Rn×m). Let V ∈ C2,1(Rn ×[0,+∞);R). Then V (x(t), t) is again an Ito process with the stochastic differen-tial

dV (x(t), t) =

[Vt(x(t), t) + Vx(x(t), t)f(t)

+0.5trace(gT (t)Vxx(x(t), t)g(t))

]dt+ Vx(x(t), t)g(t)dB(t), a.s.

Lemma 2.2. (Young’s inequality, see e.g. [30], P.52) Suppose that m1, m2 ∈R, α, β > 0, α+ β = 1, then |m1|α|m2|β ≤ α|m1|+ β|m2|.

Lemma 2.3. (Chebyshev’s inequality, see e.g. [30], P.5) If c > 0, p > 0 andX ∈ Lp, where Lp = Lp(Ω;Rn) is the family of Rn-valued random variables X withE|X|p <∞, then Pω : |X(ω)| ≥ c ≤ c−pE|X|p.

2498 MENG LIU AND KE WANG

Lemma 2.4. (Discrete Holder’s inequality, see e.g. [30], P.53) For p > 1,∣∣∣∣ k∑i=1

ci

∣∣∣∣p ≤ kp−1 k∑i=1

|ci|p.

Lemma 2.5. (Burkholder-Davis-Gundy’s inequality, see e.g. [30], P.40) Let g ∈L2([0,+∞);Rn×m). Define, for t ≥ 0

x(t) =

∫ t

0

g(s)dB(s), X(t) =

∫ t

0

|g(s)|2ds.

Then

E

(sup

0≤s≤t|x(s)|2

)≤ 4E|X(t)|.

Lemma 2.6. (Borel-Cantelli lemma, see e.g. [30], P.7) If Uk ⊂ F and∑∞k=1 P(Uk) <∞, then

P(

lim supk→∞

Uk

)= 0.

Lemma 2.7. (Moment inequality, see e.g. [30], P.69) Let p ≥ 2. Let

g ∈ M [0, T ];Rn×m such that E∫ T0|g(s)|pds < ∞, where M [0, T ];Rn×m de-

notes the family of all n × m-matrix-valued measurable Ft-adapted processes

f = fij(t)0≤t≤T such that E∫ T0|f(s)|2ds <∞. Then

E

∣∣∣∣ ∫ T

0

g(s)dB(s)

∣∣∣∣p ≤ (p(p− 1)

2

) p2

Tp−22 E

∫ T

0

|g(s)|pds.

3. Persistence and extinction of Eq. (2). For simplicity, we introduce somenotations.A = det(A).Ak: k-order leading principal minor of the determinant A.(Ak)ij : The complement minor of the element aij in Ak.

A(i)k : k-order determinant obtained by removing the ith row and the ith column of

Ak.Aik: k-order determinant obtained by changing the ith column of determinant Akto (r1, r2, ..., rk)T .

Aik: k-order determinant obtained by changing the ith column of Ak to (σ21/2, ...,

σ2k/2)T .

Bijk : k-order determinant obtained by changing the ith column of Ak to (r1, r2, ...,rk)T and the jth column to (σ2

1/2, ..., σ2k/2)T .

C(i)k (respectively, C

(i)k ): k-order determinant obtained by removing the ith row and

the ith column of Ck (respectively, Ck), where Ck = Akk (respectively, Ck = Akk).Without loss of generality, in this section, we assume that r1/σ

21 > r2/σ

22 >

... > rn/σ2n, which means that the order of persistent ability of populations (from

strongest to weakest) is x1, x2, ..., xn. We also assume that(H1): Ak > 0, Aik > 0 for i = 1, 2, ..., k; k = 1, 2, ..., n, which means that Eq. (2)has a positive equilibrium before the introduction of the stochastic disturbance.(H2): Aik > 0, Bmkk > 0 for k = 1, 2, ..., n; i = 1, 2, ..., k; m = 1, 2, ..., k − 1.(H3): There exist positive numbers p1, ..., pn such that

λ+max(−PA− ATP ) < 0.

POPULATION DYNAMICAL BEHAVIOR 2499

Lemma 3.1. Under (H3), the solution of (2) obeys

lim supt→+∞

ln |x(t)|/ ln t ≤ 1 a.s. (5)

The proof is rather standard and hence is omitted.

Lemma 3.2. Suppose that x(t) ∈ C[Ω× [0,+∞), R+] and limt→+∞

δ(t) = 0.

(I) If there exist two positive constants T1 and λ0 such that

lnx(t) ≤ λt+ δ(t)t− λ0∫ t

0

x(s)ds+

n∑i=1

αiBi(t) (6)

for all t ≥ T1, where Bi(t), 1 ≤ i ≤ n, are independent standard Brownian motionsand αi, 1 ≤ i ≤ n, are constants, then

lim supt→+∞

⟨x(t)

⟩≤ λ/λ0 a.s., if λ ≥ 0;

limt→+∞

x(t) = 0 a.s., if λ < 0,

where 〈f(t)〉 :=1

t

∫ t

0

f(s)ds;

(II)If there exist three positive constants T1, λ and λ0 such that

lnx(t) ≥ λt− λ0∫ t

0

x(s)ds+

n∑i=1

αiBi(t)

for all t ≥ T1, where Bi(t), 1 ≤ i ≤ n, are independent standard Brownian motionsand αi, 1 ≤ i ≤ n, are constants, then lim inf

t→+∞

⟨x(t)

⟩≥ λ/λ0 a.s.

Proof. (I) Suppose that λ > 0. Set g(t) =

∫ t

0

x(s)ds for all t ≥ T1, then dg(t)/dt =

x(t) ≥ 0 for all t ≥ T1. Substituting g(t) into Eq. (6) and noting that δ(t) < ε forsufficiently large t > T2, we get

lndg(t)

dt≤ (λ+ ε)t− λ0g +

n∑i=1

αiBi(t), t ≥ T,

where T = minT1, T2. Consequently

exp

λ0g(t)

dg(t)

dt≤ exp

n∑i=1

αiBi(t)

exp

(λ+ ε)t

, t ≥ T.

Integrating this inequality from T to t and then using the mean value theorem forintegral, one can see that

λ−10

[expλ0g(t) − expλ0g(T )

]≤∫ t

T

exp

n∑i=1

αiBi(s)

exp

(λ+ ε)s

ds

= [(λ+ ε)]−1 exp

n∑i=1

αiBi(τ)

[exp(λ+ ε)t − exp(λ+ ε)T

],

2500 MENG LIU AND KE WANG

where τ ∈ [T, t]. Rewriting this inequality, we obtain that

exp

λ0g(t)

≤ exp

λ0g(T )

+

[λ+ ε

]−1λ0 exp

(λ+ ε)t+

n∑i=1

αiBi(τ)

−λ0

[λ+ ε

]−1exp

n∑i=1

αiBi(τ)

exp

(λ+ ε)T

.

That is to say

g(t) ≤ λ−10 ln

λ0[λ+ ε]−1 exp

n∑i=1

αiBi(τ) exp(λ+ ε)t

+ expλ0g(T )+ λ0[λ+ ε]−1 expn∑i=1

αiBi(τ) exp(λ+ ε)T.

Therefore⟨x(t)

⟩≤ λ−10 ln

exp

n∑i=1

αiBi(τ)/t ×λ0[λ+ ε]−1 exp(λ+ ε)t

+ expλ0g(T )−n∑i=1

αiBi(τ) − λ0[λ+ ε]−1 exp(λ+ ε)T1

t.

Making use oflim

t→+∞Bi(t)/t = 0 (7)

and the basic inequality (a+ b+ c)p ≤ [3(a ∨ b ∨ c)]p, we obtain

lim supt→+∞

⟨x(t)

⟩≤ lim sup

t→+∞λ−10

t−1 ln

3λ0[λ+ ε]−1 exp(λ+ ε)t

= (λ+ ε)/λ0.

Then the desired assertion follows from the arbitrariness of ε.If λ = 0, then for ∀ ε > 0, ∃ T such that

lnx(t) ≤ εt− λ0∫ t

0

x(s)ds+

n∑i=1

γiBi(t)

for all t > T . Therefore lim supt→+∞

⟨x(t)

⟩≤ ε/λ0. Making use of the arbitrariness of ε

leads to the required assertion.If λ < 0, then for ∀ ε > 0, ∃ T such that

lnx(t) ≤ λt/2− λ0∫ t

0

x(s)ds+

n∑i=1

βiBi(t) ≤ λt/2 +

n∑i=1

βiBi(t)

for all t > T . Then we get

x(t) ≤ exp

λt/2 +

n∑i=1

βiBi(t)

= exp

t

[λ/2 +

n∑i=1

βiBi(t)/t

].

Applying (8) and the fact that λ/2 < 0, we have limt→+∞

x(t) = 0.

The proof of (II) is similar to (I). This completes the proof.

Lemma 3.3. Let πk = Ck/Ck. If (H1) and (H2) hold, then we have

(−1)i+j(Ak)ij > 0, (8)

πk > 2rk/σ2k, (9)

POPULATION DYNAMICAL BEHAVIOR 2501

C(i)k /C

(i)k < Ck/Ck (10)

and πk is a decreasing sequence of numbers with increasing of k, 1 ≤ k ≤ n.Proof. Let (aij)kk be a k-order matrix corresponding to the leading principal minorAk. Under (H1), it is easy to see that (aij)kk is an M -matrix (see, e.g. [30], P. 68).Thus (−1)i+j(Ak)ij ≥ 0 (see, e.g. [7]). An application of the Laplace Theorem ondeterminant Ak leads to

aj1(−1)i+1(Ak)i1 + ...+ ajj(−1)i+j(Ak)ij + ...+ ajk(−1)i+k(Ak)ik = 0

for i 6= j. Then the required assertion (8) follows from ajj > 0, ajm < 0 (m 6= j)and (−1)i+m(Ak)im ≥ 0.

Now let us turn to (9). Compute that

σ2kCk/2 −rkCk = (−1)k−1

∣∣∣∣∣∣∣∣∣∣σ2k/2 rk 0 0 ... 0σ21/2 r1 a11 a12 ... a1(k−1)σ22/2 r2 a21 a22 ... a2(k−1)... ... ... ... ... ...σ2k/2 rk ak1 ak2 ... ak(k−1)

∣∣∣∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣∣∣∣0 ak1 ak2 ... ak(k−1) 0r1 a11 a12 ... a1(k−1) σ2

1/2r2 a21 a22 ... a2(k−1) σ2

2/2... ... ... ... ... ...rk ak1 ak1 ... ak(k−1) σ2

k/2

∣∣∣∣∣∣∣∣∣∣= (−1)3ak1B

1kk + (−1)5ak2B

2kk + ...+ (−1)2k−1ak(k−1)B

(k−1)kk

= −ak1B1kk − ak2B2k

k − ...− ak(k−1)B(k−1)kk

Since akm < 0, Bmkk > 0 (m = 1, ..., k − 1), we obtain (9).As for inequality (10), let

F1 =

a11 ... a1(i−1) a1(i+1) ... a1(k−1)... ... ... ... ... ...

a(i−1)1 ... a(i−1)(i−1) a(i−1)(i+1) ... a(i−1)(k−1)a(i+1)1 ... a(i+1)(i−1) a(i+1)(i+1) ... a(i+1)(k−1)... ... ... ... ... ...ak1 ... ak(i−1) ak(i+1) ... ak(k−1)0 ... 0 0 ... 0... ... ... ... ... ...0 ... 0 0 ... 0

F2 =

σ21/2 r1 0 0 ... 0... ... ... ... ... ...

σ2i−1/2 ri−1 0 0 ... 0σ2i+1/2 ri+1 0 0 ... 0... ... ... ... ... ...σ2k/2 rk 0 0 ... 0σ21/2 r1 a11 a12 ... a1(k−1)... ... ... ... ... ...σ2k/2 rk ak1 ak2 ... ak(k−1)

According to the Laplace Theorem, expanding determinant F = |F1;F2| by its first

(k−1) rows, we obtain F = (−1)k(k−1)+1+k−1[C(i)k Ck−C(i)

k Ck]. On the other hand,

2502 MENG LIU AND KE WANG

make an elementary transformation to determinant F , one can see that F = |F3;F4|,where

F3 =

a11 ... a1(i−1) a1(i+1) ... a1(k−1)... ... ... ... ... ...

a(i−1)1 ... a(i−1)(i−1) a(i−1)(i+1) ... a(i−1)(k−1)a(i+1)1 ... a(i+1)(i−1) a(i+1)(i+1) ... a(i+1)(k−1)... ... ... ... ... ...

a(k−1)1 ... a(k−1)(i−1) a(k−1)(i+1) ... a(k−1)(k−1)ak1 ... ak(i−1) ak(i+1) ... ak(k−1)0 ... 0 0 ... 0... ... ... ... ... ...0 ... 0 0 ... 0

F4 =

0 0 −a11 ... −a1(k−1)... ... ... ... ...

0 0 −a(i−1)1 ... −a(i−1)(k−1)0 0 −a(i+1)1 ... −a(i+1)(k−1)... ... ... ... ...0 0 −a(k−1)1 ... −a(k−1)(k−1)

σ2k/2 rk 0 ... 0σ21/2 r1 a11 ... a1(k−1)... ... ... ... ...σ2k/2 rk ak1 ... ak(k−1)

Expanding determinant F by its first (k − 2) rows yields

F = (−1)(k−1)(k−2)+1+2+k−1A(i)k−1[rkCk − σ2

kCk/2].

Consequently

C(i)k Ck − C(i)

k Ck = A(i)k−1[rkCk − σ2

kCk/2].

By virtue of (9) and A(i)k−1 = (Ak−1)ii = (−1)i+i(Ak−1)ii > 0, we then obtain (10).

In order to prove the last assertion, set

G =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣

a11 a12 ... a1k σ21/2 r1 0 0 ... 0

... ... ... ... ... ... ... ... ... ...ak1 ak2 ... akk σ2

k/2 rk 0 0 ... 0a(k+1)1 a(k+1)2 ... a(k+1)k σ2

k+1/2 rk+1 0 0 ... 00 0 ... 0 σ2

1/2 r1 a11 a12 ... a1(k−1)... ... ... ... ... ... ... ... ... ...0 0 ... 0 σ2

k/2 rk ak1 ak2 ... ak(k−1)

∣∣∣∣∣∣∣∣∣∣∣∣∣∣Using the Laplace Theorem gives

G = (−1)k−1CkCk+1 − (−1)k−1CkCk+1. (11)

On the other hand, it is easy to get that

G =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣

a11 ... a1k 0 0 −a11 −a12 ... −a1(k−1)... ... ... ... ... ... ... ... ...ak1 ... akk 0 0 −ak1 −ak2 ... −ak(k−1)

a(k+1)1 ... a(k+1)k σ2k+1/2 rk+1 0 0 ... 0

0 ... 0 σ21/2 r1 a11 a12 ... a1(k−1)

... ... ... ... ... ... ... ... ...0 ... 0 σ2

k/2 rk ak1 ak2 ... ak(k−1)

∣∣∣∣∣∣∣∣∣∣∣∣∣∣.

POPULATION DYNAMICAL BEHAVIOR 2503

Making use of the Laplace Theorem again one can observe that

G = (−1)k−1Ak[σ2k+1Ck/2− rk+1Ck].

This, together with (11), mean that

CkCk+1 − CkCk+1 = Ak[σ2k+1Ck/2− rk+1Ck].

Note that Ak > 0, rk/σ2k > rk+1/σ

2k+1, it then follows from (9) that

CkCk+1 > CkCk+1, (12)

which is the desired assertion.

Now we are in the position to give our main results of this section. To this end,we quote some classical concepts (see e.g. [11, 23, 24, 25]).

Definition 3.4. • x(t) is said to be extinctive if limt→+∞

x(t) = 0;

• x(t) is said to be nonpersistent on average if limt→+∞

⟨x(t)

⟩= 0;

• x(t) is said to be weakly persistent on average if 〈x〉∗ > 0, wheref∗ := lim sup

t→+∞f(t);

• x(t) is said to be strongly persistent on average if 〈x〉∗ > 0, wheref∗ := lim inf

t→+∞f(t).

Theorem 3.5. Suppose that (H1)-(H3) hold, then for xk (1 ≤ k ≤ n) representedby model (2), we have:

(A) If πk − 1 > 0, then xk is weakly persistent on average a.s.;(B) If πk − 1 = 0, then xk is nonpersistent on average a.s.;(C) If πk − 1 < 0, then xk goes to extinction a.s..

Proof. (A). Applying Ito’s formula to (2) gives

t−1 ln

(xi(t)/xi(0)

)= bi −

n∑j=1

aij⟨xi(t)

⟩+ σiBi(t)/t, i = 1, ..., n (13)

where bi = ri − σ2i /2. Note that akm < 0 and

⟨xm(t)

⟩≥ 0 (k < m ≤ n), then

1

tln

(x1(t)

x1(0)

)≥ b1 − a11

⟨x1(t)

⟩− ...− a1k

⟨xk(t)

⟩+ σ1B1(t)/t,

1

tln

(x2(t)

x2(0)

)≥ b2 − a21

⟨x2(t)

⟩− ...− a2k

⟨xk(t)

⟩+ σ2B2(t)/t,

..............,

1

tln

(xk(t)

xk(0)

)≥ bk − ak1

⟨x1(t)

⟩− ...− akk

⟨xk(t)

⟩+ σkBk(t)/t.

(14)

2504 MENG LIU AND KE WANG

Let h1, h2,..., hk−1 satisfy the equations

a11h1 + a21h2 + ...+ a(k−1)1hk−1 = ak1,

a12h1 + a22h2 + ...+ a(k−1)2hk−1 = ak2,

..............,

a1(k−1)h1 + a2(k−1)h2 + ...+ a(k−1)(k−1)hk−1 = ak(k−1).

Solving the above equations, we can obtain that hi = (−1)k+i+1(Ak)ik/Ak−1. By(8), one can derive that hi < 0 for i = 1, 2, ..., k − 1.

Now, multiplying both sides of each inequality of (14) by −h1,−h2,...,−hk−1 and1, respectively, and then adding these k inequalities, we get

ln(xk(t)/xk(0)

)t

≥ h1t

ln

(x1(t)

x1(0)

)+h2t

ln

(x2(t)

x2(0)

)+ ...+

hk−1t

ln

(xk−1(t)

xk−1(0)

)+Ck − CkAk−1

− AkAk−1

⟨xk(t)

⟩−k−1∑i=1

hiσiBi(t)/t+ σkBk(t)/t.

Taking the superior limit and then using the properties of superior limit and inferiorlimit, one can obtain that

0 ≥[

lnxk(t)

t

]∗≥ Ck − Ck

Ak−1− AkAk−1

⟨xk⟩∗.

In the proof, we have used the fact that Ck > 0, Ck > 0, Ak > 0, Ak−1 > 0 andlim

t→+∞W (t)/t = 0, where W (t) is a standard Brownian motion. In other words, if

1 < πk = Ck/Ck, then⟨xk⟩∗> 0.

(B). We will prove (B) and (C) together. To begin with, let us show that

limt→+∞

xn(t) = 0 if 1 > πn and 〈xn⟩∗

= 0 if 1 = πn. In fact, by (13), for suffi-

ciently large t,

1

tln

(x1(t)

x1(0)

)≤ b1 + ε1 − a11

⟨x1(t)

⟩− a12

⟨x2⟩∗ − ...− a1n⟨xn⟩∗ +

σ1B1(t)

t,

1

tln

(x2(t)

x2(0)

)≤ b2 + ε2 − a21

⟨x1⟩∗ − a22⟨x2(t)

⟩− ...− a2n

⟨xn⟩∗

+σ2B2(t)

t,

..............,

1

tln

(xn(t)

xn(0)

)≤ bn + εn − an1

⟨x1⟩∗ − an2⟨x2⟩∗ − ...− ann⟨xn(t)

⟩+σnBn(t)

t.

That is to say

1

tln

(xi(t)

xi(0)

)≤ λi − aii

⟨xi(t)

⟩+σiBi(t)

t, i = 1, ..., n (15)

where

λi = bi + εi − ai1⟨x1⟩∗ − ...− ai(i−1)⟨xi−1⟩∗ − ai(i+1)

⟨xi+1

⟩∗ − ...− ain⟨xn⟩∗.

POPULATION DYNAMICAL BEHAVIOR 2505

Let E1 denote the set ⟨xi⟩∗> 0 for all 1 ≤ i ≤ n− 1, and let E2 be the set

there are i1, i2, ..., im such that⟨xi1⟩∗

= 0,⟨xi2⟩∗

= 0, ...,⟨xim⟩∗

= 0.

Then PE1+ PE2 = 1. We shall divide the following proof into several steps.Step 1. Suppose that PE1 > 0. Then we get E3 := λi > 0 for all 1 ≤ i ≤n− 1 ⊃ E1 (In fact, if there is an i such that λi ≤ 0, then it follows from (15) and

Lemma 3.2 that⟨xi⟩∗

= 0). For ∀ω1 ∈ E3, applying Lemma 3.2 to the first n − 1inequalities of (15) leads to⟨

xi(ω1)⟩∗ ≤ λi(ω1)

aii, i = 1, ..., n− 1.

It then follows from the arbitrariness of εi that for ∀ω1 ∈ E3,

a11⟨x1⟩∗

+ ...+ a1(n−1)⟨xn−1

⟩∗ ≤ b1 − a1n⟨xn⟩∗,a21⟨x1⟩∗

+ ...+ a2(n−1)⟨xn−1

⟩∗ ≤ b2 − a2n⟨xn⟩∗,..............,

a(n−1)1⟨x1⟩∗

+ ...+ a(n−1)(n−1)⟨xn−1

⟩∗ ≤ bn−1 − a(n−1)n⟨xn⟩∗.(16)

In these inequalities, for the sake of convenience, we have omitted ω1. Solving theabove inequalities, we obtain⟨

xi⟩∗ ≤ Ain−1 − Ain−1

An−1− (−1)n−1−i(An)ni

An−1

⟨xn⟩∗, 1 ≤ i ≤ n− 1. (17)

Now, we are going to prove P⟨xn⟩∗

= 0

∣∣∣∣E1

= 1, where P

·∣∣∣∣ · denotes the

conditional probability. Otherwise, P⟨xn⟩∗> 0

∣∣∣∣E1

> 0. That is to say,

P⟨

xn⟩∗> 0

⋂E1

> 0.

On the other hand, it follows from Lemma 3.2 that

λn > 0

⊃⟨xn⟩∗

> 0

.

Then for ∀ω2 ∈⟨xn⟩∗> 0

∩ E1, applying Lemma 3.2 to the last inequality of

(15) gives⟨xn(ω2)

⟩∗ ≤ λnann

=bn + εn − an1

⟨x1(ω2)

⟩∗ − ...− an(n−1)⟨xn−1(ω2)⟩∗

ann.

It then follows from the arbitrariness of εn that⟨xn(ω2)

⟩∗ ≤ bn − an1⟨x1(ω2)

⟩∗ − ...− an(n−1)⟨xn−1(ω2)⟩∗

ann.

Note that ω2 ∈ E1 ⊂ E3, which implies that (17) is valid. Substituting (17) intothe above inequality, we obtain

Cn − CnAn−1

− A

An−1

⟨xn(ω2)

⟩∗ ≥ 0.

2506 MENG LIU AND KE WANG

In other words,

P

1 < Cn/Cn

≥ P

⟨xn⟩∗> 0

⋂E1

> 0.

Thus Cn/Cn > 1. This contradicts 1 ≥ πn. Consequently, P⟨xn⟩∗> 0

∣∣∣∣E1

= 0.

Furthermore, if 1 > πn, for ∀ω ∈ E1, substituting (17) into the last inequality of(15), one can derive that

1

tln

(xn(t, ω)

xn(0)

)≤ bn −

n−1∑i=1

ani

[Ain−1 − Ain−1

An−1− (−1)n+i+1(An)ni

An−1

⟨xn(ω)

⟩∗]−ann

⟨xn(t, ω)

⟩+ σnBn(t)/t

=Cn − CnAn−1

− A− annAn−1An−1

⟨xn(ω)

⟩∗ − ann⟨xn(t, ω)⟩

+σnBn(t)

t

=Cn − CnAn−1

− A

An−1

⟨xn(ω)

⟩∗+ δ(t, ω) + σnBn(t)/t

≤ Cn − CnAn−1

− A

An−1

⟨xn(t, ω)

⟩+ δ(t, ω) + σnBn(t)/t,

where δ(t, ω) = ann

(⟨xn(ω)

⟩∗ − ⟨xn(t, ω)⟩). Since 1 > πn =

Cn

Cn, then we have⟨

xn(ω)⟩∗

= 0, which is to say, δ(t, ω) → 0. Then using Lemma 3.2 again leads to

P

limt→+∞

xn(t) = 0

∣∣∣∣E1

= 1.

Step 2. Suppose that PE2 > 0. Then it corresponds to removing the ijth(j = 1, 2, ...,m) rows and ijth (j = 1, 2, ...,m) columns of the coefficient determinantAn−1 of (16). Using the same arguments given in Sept 1 it should be clear that

1

tln

(xn(t)

xn(0)

)≤ C

(i1,...,im)n − C(i1,...,im)

n

A(i1,...,im)n−1

− A(i1,...,im)

A(i1,...,im)n−1

⟨xn(t)

⟩+ ε(t) +

σnBn(t)

t,

where limt→+∞

ε(t) = 0. Thus, from Lemma 3.2, we get that if C(i1,...,im)n ≤ C(i1,...,im)

n ,

then P⟨xn⟩∗

= 0

∣∣∣∣E2

= 1. Especially, if C

(i1,...,im)n < C

(i1,...,im)n , then

P

limt→+∞

xn(t) = 0

∣∣∣∣E2

= 1. Making use of (10), one can see that if Cn ≤ Cn, then

P⟨xn⟩∗

= 0

∣∣∣∣E2

= 1. In particular, if Cn < Cn, P

lim

t→+∞xn(t) = 0

∣∣∣∣E2

= 1.

Step 3. If Cn ≤ Cn, then it follows from Steps 1 and 2 that

P⟨xn⟩∗

= 0

= P

⟨xn⟩∗

= 0

∣∣∣∣E1

· PE1

+ P

⟨xn⟩∗

= 0

∣∣∣∣E2

· PE2

= P

E1

+ P

E2

= 1.

POPULATION DYNAMICAL BEHAVIOR 2507

Particularly, if Cn < Cn,

P

limt→+∞

xn(t) = 0

= P

lim

t→+∞xn(t) = 0

∣∣∣∣E1

· PE1

+P

limt→+∞

xn(t) = 0

∣∣∣∣E2

· PE2

= P

E1

+ P

E2

= 1.

In the above, we have shown that if 1 > πn, then limt→+∞

xn = 0; if 1 = πn, then⟨xn⟩∗

= 0. Now let us prove that xk goes to extinction if 1 > πk, 1 ≤ k ≤ n − 1.When k ≤ n − 1, if 1 > πk, then it follows from (12) that 1 > πn. From Step3, we obtain lim

t→+∞xn = 0. So model (2) becomes an (n − 1)–dimensional system.

Consequently, similar to Steps 1-3, we can show that⟨xn−1

⟩∗= 0 provided 1 = πn−1

and limt→+∞

xn−1(t) = 0 provided 1 > πn−1. In turn, we can also get the rest of the

consequences. Therefore, all the conclusions of this theorem have been proved.Remark 1. (I) It is useful to point out that some similar conclusions of Theorem

3.5 can be obtained if Bmkk < 0.(II) Let us consider model (1), which can be regarded as Eq. (2) in the absence of

random disturbances. On the one hand, Akk =

k∑i=1

σ2i

2(−1)i+k(Ak)ik. Making

use of (8), we get Akk > 0. Consequently,

Ck − Ck = Akk − Akk < Akk = Ck.

On the other hand, for model (1), similar to Theorem 3.5, we have:

(i) If Ck > 0, then xk represented by (1) obeys⟨xk⟩∗> 0;

(ii) If Ck < 0, then xk goes to extinction.By comparing (i) with result (A) in Theorem 3.5, we can see that the sto-chastic noise of xi is unfavorable for the persistence of all species xj , i, j =1, 2, ..., n. From a biological point of view, this is reasonable. Note that sys-tem (2) describes a population model, in which each individual enhances thegrowth of others, since the stochastic noise of xi is unfavorable for the persis-tence of xi, then xj will get less supports. That is to say, the stochastic noiseof xi is unfavorable for the persistence of xj , i, j = 1, 2, ..., n.

(III) Let us compare Theorem 3.5 with Theorem 1.1. It is easy to see that ifσ2i /2 > ri for 1 ≤ i ≤ n (which is weaker than condition (3)), then it follows

from (9) that π1 < 1. In other words, all the species will go to extinction.That is to say, our conditions of Theorem 3.5 are much weaker than Theorem1.1 in some cases.

From Theorem 3.5, we can see that 1 − πk is the critical value between weakpersistence on average and extinction for xk. Moreover, it follows from the lastassertion of Lemma 3.3 that if πk+1 < 1 < πk, then xi (k + 1 ≤ i ≤ n) goes toextinction and xj (1 ≤ j ≤ k) is weakly persistent on average almost surely. Nowlet us give another persistence result.

(H4): For all 1 ≤ i ≤ n,

n∑j=1

βijbj > 0, where(βij)n×n =

(A)−1

representing the

inverse matrix of A, bj = rj − 0.5σ2j .

2508 MENG LIU AND KE WANG

Theorem 3.6. Under (H1-H4), if πn > 1, then for system (2),

limt→+∞

t−1∫ t0x(s)ds = µ > 0 a.s., where µ =

(A

)−1(b1, ..., bn

)T.

Proof. Since πn > 1, which implies that⟨xi⟩∗> 0 for all 1 ≤ i ≤ n, then it follows

from (15) and Lemma 3.2 that

a11⟨x1⟩∗

+ a12⟨x2⟩∗

+ ...+ a1n⟨xn⟩∗ ≤ b1,

..............,

an1⟨x1⟩∗

+ an2⟨x2⟩∗...+ ann

⟨xn⟩∗ ≤ bn.

That is, in the matrix form,(⟨x1⟩∗,⟨x2⟩∗, ...,

⟨xn⟩∗)T ≤ (A)−1(b1, ..., bn)T . (18)

On the other hand, left multiplication of (13) with the inverse matrix of A,(A)−1 (

ln[x1(t)/x1(0)]

t, ...,

ln[xn(t)/xn(0)]

t

)T= (A)−1

(b1, ..., bn

)T−I(⟨x1(t)

⟩, ...,

⟨xn(t)

⟩)T+ (A)−1

(σ1B1(t)/t, ..., σnBn(t)/t

)T,

where I is the unit matrix. In other words, we have shown that

n∑j=1

βij ln

[xj(t)/xj(0)

]/t =

n∑j=1

βijbj −⟨xi(t)

⟩+

n∑j=1

βijσjBj(t)/t (19)

for all 1 ≤ i ≤ n. Note that βij = (−1)i+j(An)ij/A, then (8) indicates βij > 0. It

then follows from

[lnxj(t)

t

]∗≤ 0 and (19) that

βii ln[xi(t)/xi(0)]/t ≥n∑j=1

βijbj −⟨xi(t)

⟩+

n∑j=1

βijσjBj(t)/t

for all 1 ≤ i ≤ n. Making use of (H4) and the second part of Lemma 3.2, we can

obtain⟨xi⟩∗ ≥

n∑j=1

βijbj for all 1 ≤ i ≤ n. That is, in matrix form,

(⟨x1⟩∗,⟨x2⟩∗, ...,

⟨xn⟩∗

)T≥ (A)−1

(b1, ..., bn

)T. (20)

Then the desired assertion follows from (18) and (20).

Remark 2. In Theorem 3.6, for the sake of convenience, we suppose that πn > 1.It is useful to point out that if πk > 1 > πk+1, then xj (k + 1 ≤ j ≤ n) will go toextinction, thus Eq.(2) becomes an k−dimensional system. In this case, a similar

result of Theorem 3.6 can be obtained under (H1-H3) and

k∑j=1

βijbj > 0 for all

1 ≤ i ≤ k.

POPULATION DYNAMICAL BEHAVIOR 2509

4. Stochastic permanence of Eq. (4). In the previous section, we have studiedthe persistence and extinction for autonomous system (2). From now on, let us moveto the non-autonomous system (4). To begin with, in this section, let us consider thestochastic permanence of (4). To this end, define fu = supt≥0 f(t), f l = inft≥0 f(t).

Definition 4.1. (see e.g. [16, 20, 21, 22, 24, 25]) Eq. (4) is said to be stochasticallypermanent if for any ε ∈ (0, 1), there are positive constants δ = δ(ε) and χ = χ(ε)such that for any initial value x(0) ∈ Rn+, the solution obeys

P∗|x(t)| =

√√√√ n∑i=1

x2i (t) ≥ δ≥ 1− ε, P∗

|x(t)| ≤ χ

≥ 1− ε.

Lemma 4.2. For any given initial value x(0) ∈ Rn+, if there exist positive numbers

p1, ..., pn such that supt≥0 λ+max(−PA(t)−AT (t)P ) < 0, then model (4) has a unique

solution x(t) on t ≥ 0 and the solution will remain in Rn+ with probability 1, where

A(t) = (aij(t))n×n.

The proof is similar to that in Cheng [8] and hence is omitted. And from nowon, we suppose that for system (4), the following assumption always hold.(H3’): There exist positive numbers p1, ..., pn such that

supt≥0

[λ+max(−PA(t)− AT (t)P )

]< 0.

We also need the following assumption.(H5): For all 1 ≤ i ≤ n, alii >

∑nj=1,j 6=i(−aij)u, alii >

∑nj=1,j 6=i(−aji)u.

Lemma 4.3. Suppose that x(t) is an arbitrary positive solution of (4). If (H5)hold, then for all p > 1, i = 1, 2..., n, there exists a positive constant L = L(p) suchthat

E

[xpi (t)

]≤ L(p); t ≥ 0. (21)

Proof. Define V (x, y) =

n∑i=1

etxpi for x ∈ Rn+, where p > 1. Applying Ito’s formula

results in

dV (x, y) = etn∑i=1

xpi dt+ etn∑i=1

pxp−1i dxi + 0.5p(p− 1)xp−2i (dxi)

2

= LV (x)dt+ etp

n∑i=1

σi(t)xpi dBi(t).

2510 MENG LIU AND KE WANG

where

LV (x) = etn∑i=1

p

[1/p+ ri(t) + 0.5(p− 1)σ2

i (t)

]xpi

−aii(t)xp+1i −

n∑j=1,j 6=i

aij(t)xpi xj

≤ et

n∑i=1

p

[1/p+ ri(t) + 0.5(p− 1)σ2

i (t)

]xpi

−aii(t)xp+1i −

n∑j=1,j 6=i

aij(t)

[pxp+1

i

p+ 1+xp+1j

p+ 1

]= etp

n∑i=1

[1/p+ ri(t) + 0.5(p− 1)σ2

i (t)

]xpi

−[aii(t) +

n∑j=1,j 6=i

(aij(t)p

p+ 1+aji(t)

p+ 1

)]xp+1i

≤ etp

n∑i=1

[1/p+ rui + 0.5(p− 1)(σ2

i )u]xpi

−[alii −

n∑j=1,j 6=i

((−aij)upp+ 1

+(−aji)u

p+ 1

)]xp+1i

≤ et

n∑i=1

Ki(p).

The first inequality follows from Young’s inequality and the last inequality follows

from (H5)

(note that it follows from (H5) that for all p > 1 and 1 ≤ i ≤ n,

alii >

n∑j=1,j 6=i

((−aij)upp+ 1

+(−aji)u

p+ 1

)). In other words, we have shown that

etE

[ n∑i=1

xpi (t)

]≤

n∑i=1

xpi (0) + E

∫ t

0

esn∑i=1

Ki(p)ds =

n∑i=1

xpi (0) +

n∑i=1

Ki(p)(et − 1).

Consequently

lim supt→+∞

E

[ n∑i=1

xpi (t)

]≤

n∑i=1

Ki(p) =: L1(p).

Thus there exists a T > 0 such that

n∑i=1

E

(xpi (t)

)≤ 1.5L1(p) for all t ≥ T. At the

same time, it follows from the continuity of

n∑i=1

E

(xpi (t)

)that there exists L(p) > 0

such that∑ni=1E

(xpi (t)

)≤ L(p) for t ≤ T. Let L(p) = max1.5L1(p), L(p), then

for all t ≥ 0, we get

n∑i=1

E

(xpi (t)

)≤ L(p).

(H6): min1≤i≤n

bli > 0, where bi(t) = ri(t)− 0.5σ2i (t), 1 ≤ i ≤ n.

Theorem 4.4. If (H3’), (H5) and (H6) hold, then (4) is stochastically permanent.

POPULATION DYNAMICAL BEHAVIOR 2511

Proof. To begin with, let us demonstrate that for arbitrarily given ε > 0, thereexists a constant δ > 0 such that P∗|x(t)| ≥ δ ≥ 1− ε. Define

V1(x) =1∑ni=1 xi

for x ∈ Rn+. Applying Ito’s formula leads to

dV1(x) =

− V 2

1 (x)

n∑i=1

xi

[ri(t)−

n∑j=1

aij(t)xj

]+ V 3

1 (x)

n∑i=1

σ2i (t)x2i

dt

−V 21 (x)

n∑i=1

σi(t)xidBi(t).

It follows from (H6) that we can choose a positive constant θ such that

min1≤i≤n

bli > 0.5θ max1≤i≤n

(σ2i )u. (22)

Define

V2(x) = (1 + V1(x))θ.

Then by Ito’s formula

dV2(x) = θ(1 + V1(x))θ−1dV1(x) + 12θ(θ − 1)(1 + V1(x))θ−2d(V1(x))2

= θ(1 + V1(x))θ−2− (1 + V1(x))V 2

1 (x)

n∑i=1

xi

[ri(t)−

n∑j=1

aij(t)xj

]+V 3

1 (x)(1 + V1(x))

n∑i=1

σ2i (t)x2i +

θ − 1

2V 41 (x)

n∑i=1

σ2i (t)x2i

dt

−θ(1 + V1(x))θ−1V 21 (x)

n∑i=1

σi(t)xidBi(t).

(23)Now let κ be sufficiently small to guarantee that

0 < κ/θ < min1≤i≤n

bli − 0.5θ max1≤i≤n

(σ2i )u.

Define

V3(x(t)) = eκtV2(x(t)).

Then it follows from Ito’s formula that

dV3(x) = κeκtV2(x)dt+ eκtdV2(x)

= LV3(x)dt− θeκt(1 + V1(x))θ−1V 21 (x)

n∑i=1

σi(t)xidBi(t),

2512 MENG LIU AND KE WANG

where

LV3(x) = θeκt(1 + V1(x))θ−2κ(1 + V1(x))2/θ − (1 + V1(x))V 2

1 (x)

n∑i=1

xi[ri(t)

−n∑j=1

aij(t)xj ] + V 31 (x)(1 + V1(x))

n∑i=1

σ2i (t)x2i +

θ − 1

2V 41 (x)

n∑i=1

σ2i (t)x2i

≤ θeκt(1 + V1(x))θ−2

κ(1 + V1(x))2/θ − (1 + V1(x))V 2

1 (x)

n∑i=1

xi[ri(t)− aii(t)xi]

+V 31 (x)

n∑i=1

σ2i (t)x2i +

θ + 1

2V 41 (x)

n∑i=1

σ2i (t)x2i

≤ θeκt(1 + V1(x))θ−2

×κ(1 + V1(x))2/θ − V 3

1 (x)n∑i=1

ri(t)xi + (1 + V1(x))V 21 (x)[ max

1≤i≤nauii]

n∑i=1

x2i

+V 31 (x)[ max

1≤i≤n(σ2i )u]

n∑i=1

x2i +θ + 1

2V 41 (x)

n∑i=1

σ2i (t)x2i

≤ θeκt(1 + V1(x))θ−2

×κ(1 + V1(x))2/θ − V 4

1 (x)

n∑i=1

ri(t)x2i +

θ + 1

2V 41 (x)

n∑i=1

σ2i (t)x2i

+(1 + V1(x)) max1≤i≤n

auii + V1(x) max1≤i≤n

(σ2i )u

= θeκt(1 + V1(x))θ−2V 21 (x)κ/θ − V 4

1 (x)

n∑i=1

[bi(t)− 0.5θσ2

i (t)

]x2i

+V1(x)

[max1≤i≤n

auii + max1≤i≤n

(σ2i )u + 2κ/θ

]+ max

1≤i≤nauii + κ/θ

≤ θeκt(1 + V1(x))θ−2

− V 2

1 (x)

[min

1≤i≤nbli − 0.5θ max

1≤i≤n(σ2i )u − κ/θ

]+V1(x)

[max1≤i≤n

auii + max1≤i≤n

(σ2i )u + 2κ/θ

]+

[max1≤i≤n

auii + κ/θ

]= eκtJ(x),

where

J(x) = θ(1 + V1(x))θ−2− V 2

1 (x)

[min

1≤i≤nbli − 0.5θ max

1≤i≤n(σ2i )u − κ/θ

]+V1(x)

[max1≤i≤n

auii + max1≤i≤n

(σ2i )u + 2κ/θ

]+

[max1≤i≤n

auii + κ/θ

].

Note that J(x) is upper bounded in Rn+, namely M1 := supx∈Rn+J(x) < +∞. Then

dV3(x(t)) ≤M1eκtdt− θ eκt(1 + V1(x))θ−1V 2

1 (x)

n∑i=1

σi(t)xidBi(t).

Integrating and then taking expectations, we have

E

[V3(x(t))

]= E

[eκt(1 + V1(x(t)))θ

]≤(

1 + V1(x(0))

)θ+M1e

κt/κ.

In other words, we have shown that

lim supt→+∞

E

[V θ1 (x(t))

]≤ lim sup

t→+∞E

[(1 + V1(x(t)))θ

]≤M1/κ.

POPULATION DYNAMICAL BEHAVIOR 2513

For x(t) ∈ Rn+, note that( n∑i=1

xi(t)

)θ≤(n max

1≤i≤nxi(t)

)θ= nθ

(max1≤i≤n

x2i (t)

)0.5θ

≤ nθ|x(t)|θ.

That is to say

lim supt→+∞

E

[|x(t)|−θ

]≤ nθM1/κ =: M. (24)

Now for any ε > 0, let δ = ε1/θ/M1/θ, in view of Chebyshev’s inequality, we get

P|x(t)| < δ

= P

1/|x(t)|θ > 1/δθ

≤E

[1/|x(t)|θ

]1/δθ

= δθE

[1/|x(t)|θ

].

Consequently, P∗|x(t)| < δ

≤ δθM = ε. Thus P∗

|x(t)| ≥ δ

≥ 1− ε.

Next we need to show that for every ε > 0, there exists a positive constant χ

such that P∗|x(t)| ≤ χ

≥ 1−ε. This assertion follows from (21) and Chebyshev’s

inequality.

5. Asymptotic property of Eq. (4).

Theorem 5.1. Suppose that (H3’), (H5) and (H6) hold, then the solution of Eq.(4)has the property that

lim inft→+∞

ln |x(t)|ln t

≥ −1/θ, (25)

where θ satisfies inequality (22).

Proof. By (24), there exists a positive constant M2 such that

E

[(1 + V1(x(t))

)θ]≤M2, t ≥ 0. (26)

At the same time, it follows from (23) that

dV2(x) = θ(1 + V1(x))θ−2− (1 + V1(x))V 2

1 (x)n∑i=1

xi

[ri(t)−

n∑j=1

aij(t)xj

]+V 3

1 (x)

n∑i=1

σ2i (t)x2i +

θ + 1

2V 41 (x)

n∑i=1

σ2i (t)x2i

dt

−θ(1 + V1(x))θ−1V 21 (x)

n∑i=1

σi(t)xidBi(t).

≤ θ(1 + V1(x))θ−2− V 2

1 (x)

[min

1≤i≤nbli − 0.5θ max

1≤i≤n(σ2i )u]

+V1(x)

[max1≤i≤n

auii + max1≤i≤n

(σ2i )u]

+

[max1≤i≤n

auii

]dt

−θ(1 + V1(x))θ−1V 21 (x)

n∑i=1

σi(t)xidBi(t)

≤ θM3(1 + V1(x))θ−2V 21 (x) + V1(x) + 1

dt

−θ(1 + V1(x))θ−1V 21 (x)

n∑i=1

σi(t)xidBi(t),

2514 MENG LIU AND KE WANG

whereM3 = max

min

1≤i≤n

[bli−0.5θ(σ2

i )u], max1≤i≤n

[auii+(σ2

i )u],[

max1≤i≤n

auii]. Let µ > 0

be sufficiently small for

θM3µ+ 6nθ√

max1≤i≤n

(σ2)uµ0.5 < 0.5. (27)

Let k = 1, 2..., then we have

E

(sup

(k−1)µ≤t≤kµ(1 + V1(x(t))θ

)≤ E

(1 + V1(x((k − 1)µ))

)θ+E

(sup

(k−1)µ≤t≤kµ

∣∣∣∣ ∫ t

(k−1)µθM3

(1 + V1(x(s))

)θ−2×V 21 (x(s)) + V1(x(s)) + 1

ds

∣∣∣∣)+E

(sup

(k−1)µ≤t≤kµ

∣∣∣∣ ∫ t

(k−1)µθ

(1 + V1(x(s))

)θ−1V 21 (x(s))

n∑i=1

σi(s)xidBi(s)

∣∣∣∣).(28)

Compute that

E

(sup

(k−1)µ≤t≤kµ

∣∣∣∣ ∫ t

(k−1)µM3

(1 + V1(x(s))

)θ−2V 21 (x(s)) + V1(x(s)) + 1

ds

∣∣∣∣)≤ E

(sup

(k−1)µ≤t≤kµ

∣∣∣∣ ∫ t

(k−1)µM3

(1 + V1(x(s))

)θ−2(1 + V1(x(s))

)2

ds

∣∣∣∣)≤ E

(∫ kµ

(k−1)µ

∣∣∣∣M3

(1 + V1(x(s))

)θ∣∣∣∣ds)≤M3µE

(sup

(k−1)µ≤t≤kµ

(1 + V1(x(t))

)θ).

(29)By the Burkholder-Davis-Gundy inequality, we derive that

E

(sup

(k−1)µ≤t≤kµ

∣∣∣∣ ∫ t

(k−1)µθ

(1 + V1(x(s))

)θ−1V 21 (x(s))

n∑i=1

σi(s)xidBi(s)

∣∣∣∣)≤

n∑i=1

E

(sup

(k−1)µ≤t≤kµ

∣∣∣∣ ∫ t

(k−1)µθ

(1 + V1(x(s))

)θ−1V 21 (x(s))σi(s)xidBi(s)

∣∣∣∣)≤ 6

n∑i=1

E

(∫ kµ

(k−1)µθ2(

1 + V1(x(s))

)2θ−2

V 41 (x(s))σ2

i (s)x2i ds

)0.5

= 6

n∑i=1

E

(∫ kµ

(k−1)µθ2(

1 + V1(x(s))

)2θV 21 (x(s))(

1 + V1(x(s))

)2

σ2i (s)x2i

(∑ni=1 xi)

2ds

)0.5

≤ 6θ√

max1≤i≤n

(σ2)un∑i=1

E

(∫ kµ

(k−1)µ

(1 + V1(x(s))

)2θ

ds

)0.5

≤ 6θ√

max1≤i≤n

(σ2)uµ0.5n∑i=1

E

(sup

(k−1)µ≤t≤kµ

(1 + V1(x(t))

)2θ)0.5

≤ 6nθ√

max1≤i≤n

(σ2)uµ0.5E

(sup

(k−1)µ≤t≤kµ

(1 + V1(x(t))

)θ)(30)

POPULATION DYNAMICAL BEHAVIOR 2515

Substituting (29) and (30) into (28) results in

E

(sup

(k−1)µ≤t≤kµ(1 + V1(x(t))θ

)≤ E

(1 + V1

(x((k − 1)µ)

))θ+[θM3µ+ 6nθ

√max1≤i≤n

(σ2)uµ0.5]E

(sup

(k−1)µ≤t≤kµ

(1 + V1(x(t))

)θ) (31)

It then follows from (26) and (27) that

E

(sup

(k−1)µ≤t≤kµ(1 + V1(x(t))θ

)≤ 2M2.

Let ε > 0 be arbitrary. Then by Chebyshev’s inequality, we obtain

Pω : sup

(k−1)µ≤t≤kµ(1 + V1(x(t))θ > (kµ)1+ε

≤ 2M2

(kµ)1+ε, k = 1, 2, ...

An application of the Borel-Cantelli lemma leads to that for almost all ω ∈ Ω, thereexists a random integer k0 = k0(ω) such that for k ≥ k0 and (k − 1)µ ≤ t ≤ kµ,

ln(1 + V1(x(t))θ

ln t≤ (1 + ε) ln(kµ)

ln((k − 1)µ).

In other words, we have shown that

lim supt→+∞

ln(1 + V1(x(t))θ

ln t≤ 1 + ε.

Letting ε→ 0 yields

lim supt→+∞

ln(1 + V1(x(t))θ

ln t≤ 1.

Recalling the definition of V1(x), we have

lim supt→+∞

ln(|x(t)|−θ)ln t

≤ 1.

That is to say

lim supt→+∞

ln |x(t)|ln t

≥ −1/θ,

which is the required assertion (25).

Remark 3. It is useful to point out that if (H3’) holds, then the solution of Eq.(4) satisfies (5). The proof is similar to that in Cheng [8] and hence is omitted.

6. Global attractivity of Eq. (4). In this section, let us study the global at-tractivity of (4).

Definition 6.1. (see e.g. [21]) Let x(t) and y(t) be two arbitrary solutions of (4)with initial values x(0) ∈ Rn+ and y(0) ∈ Rn+ respectively. If for all 1 ≤ i ≤ n,

limt→+∞

|xi(t)− yi(t)| = 0 a.s., then we say (4) is globally attractive.

2516 MENG LIU AND KE WANG

Lemma 6.2. (see e.g. Karatzas [18]) Suppose that an n-dimensional stochasticprocess X(t) on t ≥ 0 satisfies the condition

E|X(t)−X(s)|λ1 ≤ c|t− s|1+λ2 , 0 ≤ s, t <∞,

for some positive constants λ1, λ2 and c. Then there exists a continuous modificationX(t) of X(t) which has the property that for every γ ∈ (0, λ2/λ1) there is a positiverandom variable h(ω) such that

Pω : sup

0<|t−s|<h(ω), 0≤s,t<∞

|X(t, ω)−X(t, ω)||t− s|γ

≤ 2

1− 2−γ

= 1.

In other words, almost every sample path of X(t) is locally but uniformly Holdercontinuous with exponent γ.

Lemma 6.3. Let x(t) be a positive solution of (4), if (H5) holds, then almost everysample path of x(t) is uniformly continuous.

Proof. Eq. (4) is equivalent to the following stochastic integral equation

xi(t) = xi(0) +

∫ t

0

xi(s)[ri(s)−n∑j=1

aij(s)xj(s)]ds+

∫ t

0

σi(s)xi(s)dBi(s).

Compute that

E

∣∣∣∣xi(ri(t)− n∑j=1

aij(t)xj)

∣∣∣∣p = E

[∣∣xi∣∣p∣∣∣∣ri(t)− n∑j=1

aij(t)xj

∣∣∣∣p]≤ 0.5E|xi|2p + 0.5E

∣∣∣∣ri(t)− n∑j=1

aij(t)xj

∣∣∣∣2p≤ 0.5

L(2p) + (n+ 1)2p−1

[(|ri|u)2p + (auii)

2pL(2p) + L(2p)∑

j=1,j 6=i

(|aij |u)2p]

=: K1(p),

where the second inequality follows from the discrete Holder’s inequality. Moreover,making use of the moment inequality for stochastic integrals we can observe thatfor 0 ≤ t1 ≤ t2 and p > 2,

E

∣∣∣∣ ∫ t2

t1

σi(s)x(s)dBi(s)

∣∣∣∣p≤(|σi|u

)p[p(p− 1)

2

]p/2(t2 − t1)(p−2)/2

∫ t2

t1

E|xi(s)|pds

≤(|σi|u

)p[p(p− 1)

2

]p/2(t2 − t1)p/2L(p)

POPULATION DYNAMICAL BEHAVIOR 2517

Thus for 0 < t1 < t2 <∞, t2 − t1 ≤ 1, 1/p+ 1/q = 1, p > 2, we obtain

E

(|xi(t2)− xi(t1)|p

)= E

∣∣∣∣ ∫ t2

t1

xi(s)

[ri(s)−

n∑j=1

aij(s)xj(s)

]ds+

∫ t2

t1

σi(s)xi(s)dBi(s)

∣∣∣∣p≤ 2p−1E

∣∣∣∣ ∫ t2

t1

xi(s)

[ri(s)−

n∑j=1

aij(s)xj(s)

]ds

∣∣∣∣p+2p−1E

∣∣∣∣ ∫ t2

t1

σi(s)xi(s)dBi(s)

∣∣∣∣p≤ 2p−1(t2 − t1)p/q

∫ t2

t1

E

∣∣∣∣xi(s)[ri(s)− n∑j=1

aij(s)xj(s)

]∣∣∣∣pds+2p−1

(|σi|u

)p[p(p− 1)

2

]p/2(t2 − t1)p/2L(p)

≤ 2p−1(t2 − t1)p/q+1K1(p) + 2p−1(|σi|u

)p[p(p− 1)

2

]p/2(t2 − t1)p/2L(p)

≤ 2p−1(t2 − t1)p/2[(t2 − t1)p/2 +

(p(p− 1)

2

)p/2]K2(p)

≤ 2p−1(t2 − t1)p/2[1 +

(p(p− 1)

2

)p/2]K2(p),

where K2(p) = max

K1(p),

(|σi|u

)p. Then it follows from Lemma 6.2 that al-

most every sample path of xi(t) is locally but uniformly Holder continuous with

exponent γ ∈(

0,p− 2

2p

)and therefore almost every sample path of xi(t) is uni-

formly continuous on t ≥ 0.

Lemma 6.4. (see e.g. [6]) Let f be a non-negative function defined on [0,+∞)such that f is integrable and is uniformly continuous. Then lim

t→+∞f(t) = 0.

Theorem 6.5. If (H3’) and (H5) hold, then Eq. (4) is globally attractive.

Proof. Let x(t) and y(t) be two solutions of (4) with initial values x(0) ∈ Rn+ and

y(0) ∈ Rn+ respectively. Define V (t) =∑ni=1

∣∣∣∣ lnxi(t) − ln yi(t)

∣∣∣∣. A calculation of

2518 MENG LIU AND KE WANG

the right differential d+V (t), and then by Ito’s formula,

d+V (t) =

n∑i=1

sgn

(xi(t)− yi(t)

)d

(lnxi(t)− ln yi(t)

)= −

n∑i=1

sgn

(xi(t)− yi(t)

) n∑j=1

aij(t)

(xj(t)− yj(t)

)dt

≤n∑i=1

[− aii(t)

∣∣xi(t)− yi(t)∣∣+

n∑j=1,j 6=i

(−aij(t))∣∣xj(t)− yj(t)∣∣]dt

=

n∑i=1

[− aii(t) +

n∑j=1,j 6=i

(−aji(t))]∣∣xi(t)− yi(t)∣∣dt

≤n∑i=1

[− alii +

n∑j=1,j 6=i

(−aji)u]∣∣xi(t)− yi(t)∣∣dt

That is to say

V (t) ≤ V (0)−∫ t

0

n∑i=1

[alii −

n∑j=1,j 6=i

(−aji)u]|xi(s)− yi(s)|ds.

Consequently

V (t) +

∫ t

0

n∑i=1

[alii −

n∑j=1,j 6=i

(−aji)u]|xi(s)− yi(s)|ds ≤ V (0) <∞.

It then follow from V (t) ≥ 0 and (H5) that |xi(t) − yi(t)| is integrable. Then thedesired assertion follows from Lemmas 6.3 and 6.4.

7. Example and numerical simulations. Consider modeldx1 = x1[r1 − a11x1 − a12x2]dt+ σ1x1dB1(t),

dx2 = x2[r2 − a21x1 − a22x2]dt+ σ2x2dB2(t).

(32)

Suppose that r1 > 0, r2 > 0, a11 > 0, a22 > 0, a12 < 0, a21 < 0, det(A) =a11a22−a12a21 > 0, then (H1) and (H3) hold. Moreover, without loss of generality,suppose that r1/σ

21 > r2/σ

22 , then B12

2 > 0 and (H2) holds and

π1 =C1

C1

=r1σ21/2

>

∣∣∣∣ a11 r1a21 r2

∣∣∣∣ / ∣∣∣∣ a11 σ21/2

a21 σ22/2

∣∣∣∣ =C2

C2

= π2.

According to Theorem 3.5, we have(i) If 1 > π1, then both x1 and x2 go to extinction a.s.;(ii) If π1 > 1 > π2, then x1 is weakly persistent on average a.s. and x2 goes toextinction a.s.;(iii) If π2 > 1, then both x1 and x2 are weakly persistent on average a.s. Moreover, ifπ2 > 1, then a11b2 > a21b1 and b1 > 0. It then follows from det(A) > 0 that a22b1 >

a12b2. Clearly, the inverse matrix of A is (a11a22−a12a21)−1(

a22 −a12−a21 a11

).Thus

(H4) holds. By Theorem 3.6,

limt→+∞

t−1∫ t

0

x1(s)ds =

∣∣∣∣ b1 a12b2 a22

∣∣∣∣ / ∣∣∣∣ a11 a12a21 a22

∣∣∣∣ ;

POPULATION DYNAMICAL BEHAVIOR 2519

limt→+∞

t−1∫ t

0

x2(s)ds =

∣∣∣∣ a11 b1a21 b2

∣∣∣∣ / ∣∣∣∣ a11 a12a21 a22

∣∣∣∣ .Now let us use Milstein’s method (see e.g. Higham [13]) to illustrate the analyt-

ical findings. Consider the discretization equation:

x(k+1)1 = x

(k)1 + x

(k)1

[r1 − a11x(k)1 − a12x(k)2

]∆t

+σ1x(k)1

√∆tξ(k) +

σ21

2 x(k)1 ((ξ(k))2∆t−∆t);

x(k+1)2 = x

(k)2 + x

(k)2

[r2 − a21x(k)1 − a22x(k)2

]∆t

+σ2x(k)2

√∆tη(k)k +

σ22

2 x(k)2 ((η(k))2∆t−∆t),

where ξ(k) and η(k),k = 1, 2, ...,K, are the Gaussian random variables.In Fig.1, we choose r1 = 0.055, r2 = 0.045, a11 = a22 = 0.01, a21 = a12 =

−0.005. The only difference between the conditions of Fig.1(a), Fig.1(b) andFig.1(c) is that the values of σ2

1 and σ22 are different. In Fig.1(a), we choose

σ21/2 = σ2

2/2 = 0.06, then π1 < 1. By (i) in this section, both x1 and x2 goto extinction. Fig.1(a) confirms these. Then we choose σ2

1/2 = σ22/2 = 0.05 in

Fig.1(b), it is easy to see that π2 < 1 < π1. Making use of (ii) leads to that x1 isweakly persistent on average and x2 goes to extinction. See Fig.1(b). Finally, inFig.1(c), we choose σ2

1/2 = 0.05, σ22/2 = 0.046, then 1 < π2. Using (iii) leads to

limt→+∞

t−1∫ t

0

x1(s)ds = 0.06; limt→+∞

t−1∫ t

0

x2(s)ds = 0.02.

Fig.1(c) reveals an interesting result: although r2 < 0.5σ22 , x2 does not go to extinc-

tion due to the action of x1. However, it is well-known that for stochastic logisticequation dx = x(r − nx) + σxdB(t), if r < 0.5σ2, then x goes to extinction a.s.

In Fig.2, we choose r1 = 0.55, r2 = 0.24, a11 = a22 = 0.95, a21 = a12 =−0.05, σ2

1/2 = 0.05, σ22/2 = 0.08, x1(0) = 0.9, x2(0) = 0.8, y1(0) = 0.2, y2(0) =

0.15, then (H3), (H5) and (H6) hold. In view of Theorem 4.4, system (32) isstochastically permanent. Fig.2(a) confirms this. At the same time, by virtue ofTheorem 6.5, system (32) is globally attractive. This can be seen from Fig.2(b).

8. Conclusions and remarks. This paper is concerned with two n-dimensionalstochastic Lotka-Volterra cooperative systems. One is autonomous (i.e., system(2)), the other is non-autonomous (i.e., system (4)). For model (2), we showedthat, under (H1)-(H3), for all 1 ≤ k ≤ n,(A) If πk < 1, then xk, xk+1,..., xn go to extinction a.s..(B) If πk = 1, then xk is nonpersistent on average a.s..(C) If πk > 1, then x1, x2,..., xk are weakly persistent on average a.s..(D) Moreover, if (H4) holds and πn > 1, then all the populations, x1, x2,..., xn, are

strongly persistent on average and limt→+∞

t−1∫ t0xi(s)ds = a positive constant,

i = 1, ..., n.

For model (4), we established some sufficient conditions for stochastic permanenceand global attractivity. Moreover, we investigated the growth rates of the positivesolution of (4). We showed that under some conditions,

lim supt→+∞

ln |x(t)|ln t

≤ 1, lim inft→+∞

ln |x(t)|ln t

≥ −1/θ, a.s.

2520 MENG LIU AND KE WANG

0 500 1000 1500 2000 2500 30000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

Time

Fig.1(a)

x1

x2

(a)

0 500 1000 1500 2000 2500 3000 3500 40000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time

Fig.1(b)

x1

x2

(b)

0 1 2 3 4 5

x 104

0

0.2

0.4

0.6

0.8

1

Time

Fig.1(c)

x1

x2

<x1>

<x2>

(c)

Figure 1. Solutions of system (32) for r1 = 0.055, r2 =0.045, a11 = a22 = 0.01, a21 = a12 = −0.005, x1(0) =0.65, x2(0) = 0.35, step size ∆t = 0.001. The horizontal axisin this and following figures represents the time t. (a) is withσ21/2 = σ2

2/2 = 0.06; (b) is with σ21/2 = σ2

2/2 = 0.05. (c) is withσ21/2 = 0.05, σ2

2/2 = 0.046.

0 100 200 300 400 5000

0.2

0.4

0.6

0.8

1

1.2

Time

Fig.2(a)

sqrt(x12+x

22)

(a)

0 10 20 30 40 500

0.2

0.4

0.6

0.8

1

Time

Fig.2(b)

x1

y1

x2

y2

(b)

Figure 2. Solutions of system (32) for r1 = 0.55, r2 = 0.24, a11 =a22 = 0.95, a21 = a12 = −0.05, σ2

1/2 = 0.05, σ22/2 = 0.08, x1(0) =

0.9, x2(0) = 0.8, y1(0) = 0.2, y2(0) = 0.15, step size ∆t = 0.001.

POPULATION DYNAMICAL BEHAVIOR 2521

Some interesting topics deserve further investigation. To begin, in lieu of themodel considered, we could study some more realistic and complex models, forexample, time delay. Also it is interesting to study the n-dimensional stochasticfood chain model or competitive system. Another problem of interest is to considerthe persistence and extinction of (4). In fact, we also attempted to study thisquestion. Unfortunately, we only obtain some persistence and extinction results,but can not obtain the critical value at present stage, and we leave this for futurework.

Acknowledgments. The authors thank the editor and referees for their very im-portant and helpful comments and suggestions. The authors also thank Dr. H.Qiuand Dr. X.Zou for helping us to improve the English exposition. This research issupported by the NSFC of PR China (Nos.11126219, 11171081 and 11171056).

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Received March 2011; revised October 2012.

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