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Dynamics of a stochastic non-autonomous predator-prey system with Beddington-DeAngelis functional response

Abstract

A stochastic non-autonomous predator-prey system with Beddington-DeAngelis functional response is proposed, the existence of a global positive solution and stochastically ultimate boundedness are derived. Sufficient conditions for extinction, non-persistence in the mean, weak persistence in the mean and strong persistence in the mean are established. The global attractiveness of the solution is also considered. Finally, numerical simulations are carried out to support our findings.

MSC:92B05, 34F05, 60H10, 93E03.

1 Introduction

In population dynamics, the relationship between predator and prey plays an important role due to its universal existence. There are many significant functional responses in order to model various different situations. In fact, most of the functional responses are prey-dependent; however, the functional response should also be predator-dependent, especially when predators have to search for food. Beddington [1] and DeAngelis et al. [2] in 1975 first introduced the Beddington-DeAngelis type predator-prey model taking the form

{ d u d t = r 1 u g 1 u 2 f 1 u v β + γ u + δ v , d v d t = r 2 v g 2 v 2 + f 2 u v β + γ u + δ v ,
(1)

where u and v denote the population densities of prey and predator. Although the Beddington-DeAngelis functional response is similar to the Holling type-II functional, it can reflect mutual interference among predators. That is to say, this kind of functional response is affected by both predator and prey. In the last years, some experts have studied the system [37]. In the following, we introduce a non-autonomous predator-prey model with Beddington-DeAngelis functional response:

{ d x d t = x ( a 1 ( t ) b 1 ( t ) x c 1 ( t ) y m 1 ( t ) + m 2 ( t ) x + m 3 ( t ) y ) , d y d t = y ( a 2 ( t ) b 2 ( t ) y + c 2 ( t ) x m 1 ( t ) + m 2 ( t ) x + m 3 ( t ) y ) ,
(2)

where x(t) and y(t) represent the population density of prey and predator at time t, respectively. a 1 (t) denotes the intrinsic growth rate of prey. b 1 (t) and b 2 (t) stand for the density-dependent coefficients of prey and predator, respectively. a 2 (t) is the death rate of predator. c 1 (t) is the capturing rate of predator, c 2 (t) represents the rate of conversion of nutrients into the reproduction of predator. For the other coefficients’ biological representation, we refer the reader to [1] and [2].

On the other hand, population systems are often subject to environmental noise; therefore, it is important to study how the noise affects the population systems. In fact, stochastic population systems have been studied recently by many authors [817]. However, there are not many papers considering non-autonomous stochastic systems [1823]. In this paper, considering the effect of environmental noise, we introduce stochastic perturbation into the intrinsic growth rate of prey and the death rate of predator in system (2) and assume the parameters a 1 (t) and a 2 (t) are disturbed to a 1 (t)+α(t)d B 1 (t), a 2 (t)+β(t)d B 2 (t), respectively. Then corresponding to the deterministic system (2), we obtain the following stochastic system:

{ d x = x ( a 1 ( t ) b 1 ( t ) x c 1 ( t ) y m 1 ( t ) + m 2 ( t ) x + m 3 ( t ) y ) d t + α ( t ) x d B 1 ( t ) , d y = y ( a 2 ( t ) b 2 ( t ) y + c 2 ( t ) x m 1 ( t ) + m 2 ( t ) x + m 3 ( t ) y ) d t + β ( t ) y d B 2 ( t ) .
(3)

We assume all the coefficients are continuous bounded nonnegative functions on R + =[0,+).

If f(t) is a continuous bounded function on R + , define

f u = sup t R + f(t), f l = inf t R + f(t).

Throughout this paper, suppose that a 2 l >0, m i l >0 (i=1,2,3), b i l >0 (i=1,2).

Definition 1 (1) The population x(t) is said to be non-persistent in the mean if x =0, where f(t)= 1 t 0 t f(s)ds, f = lim sup t + f(t), f = lim inf t + f(t).

  1. (2)

    The population x(t) is said to be weakly persistent in the mean if x >0.

  2. (3)

    The population x(t) is said to be strongly persistent in the mean if x >0.

Throughout this paper, unless otherwise specified, let (Ω,F,P) be a complete probability space with a filtration { F t } t R satisfying the usual conditions (i.e., it is right continuous and increasing and F 0 contains all P-null sets), here B ˙ i (t) (i=1,2) is a standard Brownian motion defined on this probability space. In addition, R + 2 denotes {(x,y) R 2 :x>0,y>0}.

Here we give the following auxiliary statements which are introduced in [24]. Consider the d-dimensional stochastic differential equation

dx(t)=f ( x ( t ) , t ) dt+g ( x ( t ) , t ) dB(t),t t 0 .

Denote by C 2 , 1 ( R d ×[ t 0 ,); R + ) the family of all nonnegative functions V(x,t) defined on R d ×[ t 0 ,) such that they are continuously twice differentiable in x and once in t. The differential operator L of the above equation is defined by the formula

L= t + i = 1 d f i (x,t) x i + 1 2 i , j = 1 d [ g T ( x , t ) g ( x , t ) ] i j 2 x i x i .

If L acts on a function V C 2 , 1 ( R d ×[ t 0 ,); R + ), then

LV(x,t)= V t (x,t)+ V x (x,t)f(x,t)+ 1 2 trace [ g T ( x , t ) V x x ( x , t ) g ( x , t ) ] .

2 Global positive solution and stochastic boundedness

For a model of population dynamics, the first thing considered is whether the solution is globally existent and nonnegative, hence in this section we will show it. In order for a stochastic differential equation to have a unique global solution for any given initial value, the coefficients of the equation are generally required to satisfy the linear growth condition and the local Lipschitz condition [24]. The coefficients of model (3) do not satisfy the linear growth condition, though they are locally Lipschitz continuous, the solution of system (3) may explode at a finite time (cf. [24]). In the following, using the Lyapunov analysis method, we can prove that system (3) has a global positive solution.

Theorem 1 For any initial value ( x 0 , y 0 ) R + 2 , there is a unique solution (x(t),y(t)) of system (3) on t0, and the solution will remain in R + 2 with probability 1.

Proof Since the coefficients of model (3) are locally Lipschitz continuous, for any given initial value ( x 0 , y 0 ) R + 2 , there is a unique local solution (x(t),y(t)) on t[0, τ e ), where τ e is the explosion time [24]. To show the solution is global, we need to show that τ e =.

Let n 0 >0 be sufficiently large for x 0 and y 0 lying within the interval [1/ n 0 , n 0 ]. For each integer n> n 0 , we define the stopping times

τ n =inf { t [ 0 , τ e ) : x ( t ) ( 1 / n , n )  or  y ( t ) ( 1 / n , n ) } ,

where, throughout this paper, we set inf= ( denotes the empty set). Obviously, τ n is increasing as n. Let τ = lim n τ n , hence, τ τ e a.s. Now, we only need to show τ =. If this statement is false, there is a pair of constants T>0 and ε(0,1) such that P{ τ T}>ε. Consequently, there exists an integer n 1 n 0 such that

P{ τ n T}>ε,n n 1 .
(4)

Define a C 2 -function V: R + 2 R + by

V(x,y)=(x1logx)+(y1logy).

The nonnegativity of this function can be seen from u1logu0, u>0. Applying Itô’s formula, we get

d V = ( x 1 ) ( a 1 ( t ) b 1 ( t ) x c 1 ( t ) y m 1 ( t ) + m 2 ( t ) x + m 3 ( t ) y ) d t + α 2 ( t ) 2 d t + ( y 1 ) ( a 2 ( t ) b 2 ( t ) y + c 2 ( t ) x m 1 ( t ) + m 2 ( t ) x + m 3 ( t ) y ) d t + β 2 ( t ) 2 d t + ( x 1 ) α ( t ) d B 1 ( t ) + ( y 1 ) β ( t ) d B 2 ( t ) ,
(5)

then

L V = a 1 ( t ) x b 1 ( t ) x 2 c 1 ( t ) x y m 1 ( t ) + m 2 ( t ) x + m 3 ( t ) y a 1 ( t ) + b 1 ( t ) x + c 1 ( t ) y m 1 ( t ) + m 2 ( t ) x + m 3 ( t ) y + α 2 ( t ) 2 a 2 ( t ) y b 2 ( t ) y 2 + c 2 ( t ) x y m 1 ( t ) + m 2 ( t ) x + m 3 ( t ) y + a 2 ( t ) + b 2 ( t ) y c 2 ( t ) x m 1 ( t ) + m 2 ( t ) x + m 3 ( t ) y + β 2 ( t ) 2 a 1 ( t ) x b 1 ( t ) x 2 a 1 ( t ) + b 1 ( t ) x + c 1 ( t ) y m 1 ( t ) + m 2 ( t ) x + m 3 ( t ) y + α 2 ( t ) 2 a 2 ( t ) y b 2 ( t ) y 2 + c 2 ( t ) x y m 1 ( t ) + m 2 ( t ) x + m 3 ( t ) y + a 2 ( t ) + b 2 ( t ) y + β 2 ( t ) 2 a 1 u x b 1 l x 2 a 1 l + b 1 u x + c 1 u m 3 l + ( α u ) 2 2 a 2 l y b 2 l y 2 + c 2 u m 2 l y + a 2 u + b 2 u y + ( β u ) 2 2 ( a 1 u + b 1 u ) x b 1 l x 2 a 1 l + c 1 u m 3 l + ( α u ) 2 2 + ( b 2 u + c 2 u m 2 l a 2 l ) y b 2 l y 2 + a 2 u + ( β u ) 2 2 K ,

where K is a positive number. Substituting this inequality into Eq. (5), we see that

dV ( x ( t ) , y ( t ) ) Kdt+(x1)α(t)d B 1 (t)+(y1)β(t)d B 2 (t),

which implies that

0 τ n T d V ( x ( t ) , y ( t ) ) 0 τ n T K d t + 0 τ n T α ( s ) ( x ( s ) 1 ) d B 1 ( s ) + 0 τ n T β ( s ) ( y ( s ) 1 ) d B 2 ( s ) ,

where τ n T=min{ τ n ,T}. Taking the expectations of the above inequality leads to

EV ( x ( τ n T ) , y ( τ n T ) ) V( x 0 , y 0 )+KE( τ n T)V( x 0 , y 0 )+KT.
(6)

Let Ω n ={ τ n T} for n n 1 , then by (4) we have P( Ω n )ε. Note that for every ω Ω n , there is at least one of x( τ n ,ω) and y( τ n ,ω) equaling either n or 1 n ; therefore, V(x( τ n ,ω),y( τ n ,ω)) is no less than min{(n1logn),( 1 n 1log 1 n )}. It then follows from (6) that

V ( x 0 , y 0 ) + K T E [ 1 Ω n ( ω ) V ( x ( τ n ) , y ( τ n ) ) ] ε min { ( n 1 log n ) , ( 1 n 1 log 1 n ) } ,

where 1 Ω n ( ω ) is the indicator function of Ω n , letting n, we have that

>V( x 0 , y 0 )+KT=

is a contradiction, then we must have τ =. Therefore, the solution of system (3) will not explode at a finite time with probability 1. This completes the proof. □

Definition 2 The solution X(t)=(x(t),y(t)) of system (3) is said to be stochastically ultimately bounded if for any ε(0,1), there is a positive constant χ(χ(ε)) such that for any initial value ( x 0 , y 0 ) R + 2 , the solution of (3) has the property that

lim sup t P { | X ( t ) | > χ } <ε.

Lemma 1 Let (x(t),y(t)) be a solution of system (3) with initial value ( x 0 , y 0 ) R + 2 , for all p>1, there exist K 1 (p) and K 2 (p) such that

E [ x p ( t ) ] K 1 (p),E [ y p ( t ) ] K 2 (p),t[0,+),

where K 1 (p)=max{ K 1 ¯ (p), ( a 1 u + 1 2 p ( α u ) 2 b 1 l ) p + ε ˜ }, K 2 (p)=max{ K 2 ¯ (p), ( c 2 u m 2 l + 1 2 p ( β u ) 2 b 2 l ) p + ε ˜ }, K i ¯ (p) (i=1,2) and ε ˜ are both positive constants.

Proof Define V 1 (x)= x p for x R + , where p>1. Applying Itô’s formula leads to

d V 1 ( x ) = p x p 1 d x + 1 2 p ( p 1 ) x p 2 ( d x ) 2 = p x p [ a 1 ( t ) b 1 ( t ) x c 1 ( t ) y m 1 ( t ) + m 2 ( t ) x + m 3 ( t ) y + 1 2 ( p 1 ) α 2 ( t ) ] d t + p x p α ( t ) d B 1 ( t ) p x p [ a 1 u + 1 2 p ( α u ) 2 b 1 l x ] d t + p x p α ( t ) d B 1 ( t ) .

Similarly, we have

d y p = p y p [ a 2 ( t ) b 2 ( t ) y + c 2 ( t ) x m 1 ( t ) + m 2 ( t ) x + m 3 ( t ) y + 1 2 ( p 1 ) β 2 ( t ) ] d t + p y p β ( t ) d B 2 ( t ) p y p [ c 2 u m 2 l + 1 2 p ( β u ) 2 b 2 l y ] d t + p y p β ( t ) d B 2 ( t ) .

Taking expectation, we have

d E [ x p ( t ) ] d t p { [ a 1 u + 1 2 p ( α u ) 2 ] E [ x p ( t ) ] b 1 l E [ x p + 1 ( t ) ] } p { [ a 1 u + 1 2 p ( α u ) 2 ] E [ x p ( t ) ] b 1 l [ E [ x p ( t ) ] ] 1 + 1 p } E [ x p ( t ) ] p { [ a 1 u + 1 2 p ( α u ) 2 ] b 1 l [ E [ x p ( t ) ] ] 1 p }

and

d E [ y p ( t ) ] d t p { [ c 2 u m 2 l + 1 2 p ( β u ) 2 ] E [ y p ( t ) ] b 2 l E [ y p + 1 ( t ) ] } p { [ c 2 u m 2 l + 1 2 p ( β u ) 2 ] E [ y p ( t ) ] b 2 l [ E [ y p ( t ) ] ] 1 + 1 p } E [ y p ( t ) ] p { [ c 2 u m 2 l + 1 2 p ( β u ) 2 ] b 2 l [ E [ y p ( t ) ] ] 1 p } .

Therefore, by the comparison theorem, we get

lim sup t E [ x p ( t ) ] ( a 1 u + 1 2 p ( α u ) 2 b 1 l ) p , lim sup t E [ y p ( t ) ] ( c 2 u m 2 l + 1 2 p ( β u ) 2 b 2 l ) p .

Thus, for a given constant ε ˜ >0, there exists a T>0 such that for all t>T,

E [ x p ( t ) ] ( a 1 u + 1 2 p ( α u ) 2 b 1 l ) p + ε ˜ ,E [ y p ( t ) ] ( c 2 u m 2 l + 1 2 p ( β u ) 2 b 2 l ) p + ε ˜ .

Together with the continuity of E[ x p (t)] and E[ y p (t)], there exist K 1 ¯ (p)>0, K 2 ¯ (p)>0 such that E[ x p (t)] K 1 ¯ (p), E[ y p (t)] K 2 ¯ (p) for tT. Let

K 1 ( p ) = max { K 1 ¯ ( p ) , ( a 1 u + 1 2 p ( α u ) 2 b 1 l ) p + ε ˜ } , K 2 ( p ) = max { K 2 ¯ ( p ) , ( c 2 u m 2 l + 1 2 p ( β u ) 2 b 2 l ) p + ε ˜ } ,

then for all t R + ,

E [ x p ( t ) ] K 1 (p),E [ y p ( t ) ] K 2 (p).

 □

Theorem 2 The solutions of system (3) with initial value ( x 0 , y 0 ) R + 2 are stochastically ultimately bounded.

Proof If X=(x,y) R 2 , its norm here is denoted by |X|= ( x 2 + y 2 ) 1 2 , then

| X ( t ) | p 2 p 2 ( | x ( t ) | p + | y ( t ) | p )

by Lemma 1, E | X ( t ) | p K(p), t(0,+). K(p) is dependent of ( x 0 , y 0 ) and defined by K(p)= 2 p 2 ( K 1 (p)+ K 2 (p)). By virtue of Chebyshev’s inequality, the above result is straightforward. □

3 Persistence in the mean and extinction

Lemma 2 The solutions of system (3) with initial value ( x 0 , y 0 ) R + 2 have the following properties:

lim sup t ln x ( t ) t 0, lim sup t ln y ( t ) t 0,a.s.

Proof It follows from system (3) that

{ d x x ( a 1 ( t ) b 1 ( t ) x ) d t + α ( t ) x d B 1 ( t ) , d y y ( c 2 ( t ) m 2 ( t ) b 2 ( t ) y ) d t + β ( t ) y d B 2 ( t ) .
(7)

Set

{ d x ¯ = x ¯ ( a 1 ( t ) b 1 ( t ) x ¯ ) d t + α ( t ) x ¯ d B 1 ( t ) , d y ¯ = y ¯ ( c 2 ( t ) m 2 ( t ) b 2 ( t ) y ¯ ) d t + β ( t ) y ¯ d B 2 ( t ) ,
(8)

where ( x ¯ (t), y ¯ (t)) is a solution of system (8) with initial value x 0 >0 and y 0 >0. By the comparison theorem for stochastic differential equations, it is easy to have

x(t) x ¯ (t),y(t) y ¯ (t),a.s. t[0,+).
(9)

By Lemma 3.4 in [20], it is easy to get the following result:

dx(t)=x(t) [ ( b ( t ) a 11 ( t ) x ( t ) ) d t + σ ( t ) d B ( t ) ]

here b(t), a 11 (t) and σ(t) are all nonnegative functions defined on R + . If a 11 l >0, then lim sup t ln ( | x ( t ) | ) ln t 1, a.s.

Note that b i l >0 (i=1,2), then it follows from Eqs. (8) and (9),

lim sup t ln x ( t ) ln t lim sup t ln x ¯ ( t ) ln t 1, lim sup t ln y ( t ) ln t lim sup t ln y ¯ ( t ) ln t 1,a.s.

In addition,

lim sup t ln x ( t ) t = lim sup t ln x ( t ) ln t lim sup t ln t t lim sup t ln t t =0;

therefore, it leads to lim sup t ln x ( t ) t 0, a.s. Similarly, we can have lim sup t ln y ( t ) t 0, a.s. □

Lemma 3 [25]

Suppose that x(t)C[Ω× R + , R + 0 ], where R + 0 :={a|a>0,aR}.

  1. (I)

    If there are positive constants λ 0 , T and λ0 such that

    lnx(t)λt λ 0 0 t x(s)ds+ i = 1 n β i B i (t)

for tT, where β i is a constant, 1in, then x λ/ λ 0 , a.s. (i.e., almost surely).

  1. (II)

    If there are positive constants λ 0 , T and λ0 such that

    lnx(t)λt λ 0 0 t x(s)ds+ i = 1 n β i B i (t)

for tT, where β i is a constant, 1in, then x λ/ λ 0 , a.s.

In the following, we give the result about weak persistence in the mean and extinction of the prey and predator population. Applying Itô’s formula to Eq. (3) leads to

(10)
(11)

Let r 1 (t)= a 1 (t) α 2 ( t ) 2 , r 2 (t)= a 2 (t) β 2 ( t ) 2 , then r 2 <0. For the prey population x(t) of system (3), we have

Theorem 3 (i) If r 1 <0, then the prey population x(t) will go to extinction a.s.

  1. (ii)

    If r 1 =0, then the prey population x(t) will be non-persistent in the mean a.s.

  2. (iii)

    If r 1 >0, then the prey population x(t) will be weakly persistent in the mean a.s.

  3. (iv)

    If r 1 c 1 ( t ) m 3 ( t ) >0, then the prey population x(t) will be strongly persistent in the mean a.s.

Proof (i) It follows from (10) that

ln x ( t ) ln x 0 = 0 t [ r 1 ( s ) b 1 ( s ) x ( s ) c 1 ( s ) y ( s ) m 1 ( s ) + m 2 ( s ) x ( s ) + m 3 ( s ) y ( s ) ] d s + 0 t α ( s ) d B 1 ( s ) ,
(12)

lnx(t)ln x 0 0 t r 1 (s)ds+ 0 t α(s)d B 1 (s), set M 1 (t)= 0 t α(s)d B 1 (s), it is a martingale whose quadratic variation is M 1 , M 1 t = 0 t α 2 (s)ds ( α u ) 2 t. Making use of the strong law of large numbers for martingale yields

lim t + M 1 ( t ) t =0,a.s.
(13)

then

ln x ( t ) ln x 0 t 1 t 0 t r 1 (s)ds+ M 1 ( t ) t .
(14)

Taking superior limit on both sides of inequality (14) leads to lim sup t + ln x ( t ) t r 1 <0, we can see that lim t + x(t)=0.

  1. (ii)

    By Eq. (12), we have

    ln x ( t ) ln x 0 t r 1 b 1 l x ( t ) + M 1 ( t ) t .
    (15)

It follows from the property of superior limit and (13) that for arbitrary ε>0, there exists T>0 such that r 1 r 1 + ε 2 and M 1 ( t ) t ε 2 for all tT. Substituting these inequalities into (15) yields

lnx(t)ln x 0 ( r 1 + ε 2 ) t b 1 l 0 t x(s)ds+ ε 2 t ( r 1 + ε ) t b 1 l 0 t x(s)ds,

when r 1 =0, then

ln x ( t ) x 0 εt b 1 l x 0 0 t x ( s ) x 0 ds.

By b 1 l >0 and Lemma 3, we have x ( t ) ε b 1 l , by virtue of the arbitrariness of ε, x ( t ) 0. Since the solution of system (3) is nonnegative, it is easy to have x ( t ) =0, that is to say, the prey population x(t) is non-persistent in the mean a.s.

  1. (iii)

    We only need to show that there exists a constant μ 1 >0 such that for any solution (x(t),y(t)) of system (3) with initial value ( x 0 , y 0 ) R + 2 , x ( t ) μ 1 >0 a.s. Otherwise, for arbitrary ε 1 >0, there exists a solution ( x ˜ (t), y ˜ (t)) with positive initial value x 0 >0 and y 0 >0 such that P{ x ˜ ( t ) < ε 1 }>0.

Let ε 1 be sufficiently small so that

r 1 b 1 u ε 1 >0, r 2 + c 2 u m 1 l ε 1 <0.
(16)

It follows from Eq. (11) that

ln y ˜ ( t ) ln y 0 t r 2 b 2 l y ˜ ( t ) + c 2 u m 1 l x ˜ ( t ) + M 2 ( t ) t
(17)

here M 2 (t)= 0 t β(s)d B 2 (s), it also has

lim t M 2 ( t ) t =0.
(18)

By virtue of (17), it leads to [ t 1 ln y ˜ ( t ) ] r 2 + c 2 u m 1 l ε 1 <0, thus

lim t y ˜ (t)=0.
(19)

On the other hand, it follows from Eq. (12) that

ln x ˜ ( t ) ln x 0 t r 1 b 1 u x ˜ ( t ) c 1 u m 1 l y ˜ ( t ) + M 1 ( t ) t .

Taking the superior limit to the above inequality and making use of (13), (16) and (19), we have [ t 1 ln x ˜ ( t ) ] r 1 b 1 u ε 1 >0, that is to say, we have shown P{ [ t 1 ln x ˜ ( t ) ] >0}>0, this is a contradiction to Lemma 2. Therefore, x ( t ) >0, the prey population x(t) will be weakly persistent in the mean a.s.

  1. (iv)

    By Eq. (10), we get

    dlnx ( r 1 ( t ) b 1 ( t ) x c 1 ( t ) m 3 ( t ) ) dt+α(t)d B 1 (t).

It is easy to have

ln x ( t ) ln x 0 t r 1 c 1 ( t ) m 3 ( t ) b 1 u x ( t ) + M 1 ( t ) t .

If r 1 c 1 ( t ) m 3 ( t ) >0, there exists sufficiently small ε>0 such that r 1 c 1 ( t ) m 3 ( t ) ε>0. It follows from the property of superior limit, interior limit and (13) that for positive number ε, there exists a T>0 such that

r 1 > r 1 ε 3 , c 1 ( t ) m 3 ( t ) < c 1 ( t ) m 3 ( t ) + ε 3 , M 1 ( t ) t > ε 3

for all t>T. Then

ln x ( t ) ln x 0 t r 1 ε 3 c 1 ( t ) m 3 ( t ) ε 3 b 1 u x ( t ) ε 3 ( r 1 c 1 ( t ) m 3 ( t ) ε ) b 1 u x ( t ) .

By virtue of Lemma 3 and the arbitrariness of ε, we have

x ( t ) r 1 c 1 ( t ) m 3 ( t ) b 1 u >0.

In other words, the prey population x(t) is strongly persistent in the mean a.s. □

Remark 1 The results of Theorem 3 illustrate that r 1 is the threshold between weak persistence in the mean and extinction. Note r 1 (t)= a 1 (t) α 2 ( t ) 2 , if α 2 ( t ) 2 > a 1 (t), then the prey population will be extinct, no matter whether there are predators. However, the prey population will survive when not considering environmental noise. This indicates that when the density of environmental noise is larger than the intrinsic growth rate of prey, it will cause the extinction of prey population. Therefore, it is more suitable to take into account stochastic perturbation in the systems. Here we can also find that the condition (iv) implies condition (iii), that is to say, the prey population must be weakly persistent in the mean when it is strongly persistent in the mean.

For the predator population, we have the following result.

Theorem 4 (i) If ( b 1 ) r 2 + ( c 2 ( t ) m 1 ( t ) ) r 1 <0, then the predator population y(t) will go to extinction a.s.

  1. (ii)

    If ( b 1 ) r 2 + ( c 2 ( t ) m 1 ( t ) ) r 1 =0, then the predator population y(t) will be non-persistent in the mean a.s.

  2. (iii)

    If r 2 + c 2 ( t ) x ¯ ( t ) m 1 ( t ) + m 2 ( t ) x ¯ ( t ) + m 3 ( t ) y ¯ ( t ) >0, then the predator population y(t) will be weakly persistent in the mean a.s. where ( x ¯ (t), y ¯ (t)) is the solution of Eq. (8) with initial value ( x 0 , y 0 ) R + 2 .

  3. (iv)

    If r 2 + c 2 ( t ) m 2 ( t ) >0, then the predator population y(t) has a superior bound in time average, that is, y ( t ) r 2 + c 2 ( t ) m 2 ( t ) b 2 l .

Proof (i) If r 1 0, then it follows from Theorem 3 that x ( t ) =0. By Eq. (11),

ln y ( t ) ln y 0 t r 2 + c 2 u m 1 l x ( t ) + M 2 ( t ) t ;

therefore, [ t 1 ln y ( t ) ] r 2 <0, then lim t y(t)=0.

Now, if r 1 >0, it follows from the property of superior limit, interior limit and (13) that for sufficiently small ε, there exists a T>0 such that

ln x ( t ) ln x 0 t r 1 ( b 1 ) x ( t ) + M 1 ( t ) t r 1 + ε 2 ( b 1 ) x ( t ) + ε 2

for all t>T. Applying Lemma 3 and the arbitrariness of ε yield

x ( t ) r 1 ( b 1 ) .
(20)

Substituting the above inequality into (11) gives

[ t 1 ln y ( t ) ] r 2 + c 2 ( t ) m 1 ( t ) x ( t ) r 2 + ( c 2 ( t ) m 1 ( t ) ) x ( t ) r 2 + ( c 2 ( t ) m 1 ( t ) ) r 1 ( b 1 ) ,
(21)

then

[ t 1 ln y ( t ) ] ( b 1 ) r 2 + ( c 2 ( t ) m 1 ( t ) ) r 1 ( b 1 ) <0,

which means lim t y(t)=0 a.s.

  1. (ii)

    In the case (i), we have shown that if r 1 0, then lim t y(t)=0, therefore, y ( t ) =0. Now, we will prove that y ( t ) =0 is still valid when r 1 >0. Otherwise, if y ( t ) >0, then it follows from Lemma 2 that [ t 1 ln y ( t ) ] =0. Making use of (21), one can see that

    0= [ t 1 ln y ( t ) ] r 2 + ( c 2 ( t ) m 1 ( t ) ) x ( t ) .
    (22)

On the other hand, for arbitrary ε>0, there exists a T ¯ >0 such that

r 2 < r 2 + ε 3 , c 2 ( t ) m 1 ( t ) x ( t ) < ( c 2 ( t ) m 1 ( t ) ) x ( t ) + ε 3 , M 2 ( t ) t < ε 3

for all t> T ¯ . Substituting these inequalities into (11) yields

ln y ( t ) ln y 0 t r 2 + c 2 ( t ) m 1 ( t ) x ( t ) b 2 ( t ) y ( t ) + M 2 ( t ) t r 2 + ε 3 + ( c 2 ( t ) m 1 ( t ) ) x ( t ) + ε 3 ( b 2 ) y ( t ) + ε 3 r 2 + ε + ( c 2 ( t ) m 1 ( t ) ) x ( t ) ( b 2 ) y ( t ) .

Then an application of Lemma 3 and (22) results in y ( t ) r 2 + ε + ( c 2 ( t ) m 1 ( t ) ) x ( t ) ( b 2 ) . By virtue of (20) and the arbitrariness of ε,

y ( t ) r 2 ( b 1 ) + ( c 2 ( t ) m 1 ( t ) ) r 1 ( b 1 ) ( b 2 ) =0,

which is a contradiction to our assumption, therefore, y ( t ) =0 a.s.

  1. (iii)

    In the following, we need to show that y ( t ) >0 a.s. Otherwise, for arbitrary ε 2 >0, there exists a solution ( x ˆ (t), y ˆ (t)) of system (3) with positive initial value ( x 0 , y 0 ) R + 2 such that P{ y ˆ ( t ) < ε 2 }>0. Let ε 2 be sufficiently small so that

    r 2 + c 2 ( t ) x ¯ ( t ) m 1 ( t ) + m 2 ( t ) x ¯ ( t ) + m 3 ( t ) y ¯ ( t ) > ( b 2 u + 2 c 1 u c 2 u b 1 l ( m 1 l ) 2 ) ε 2 .
    (23)

It follows from (11) that

ln y ˆ ( t ) ln y 0 t = r 2 + c 2 ( t ) x ¯ ( t ) m 1 ( t ) + m 2 ( t ) x ¯ ( t ) + m 3 ( t ) y ¯ ( t ) b 2 ( t ) y ˆ ( t ) + M 2 ( t ) t + c 2 ( t ) x ˆ ( t ) m 1 ( t ) + m 2 ( t ) x ˆ ( t ) + m 3 ( t ) y ˆ ( t ) c 2 ( t ) x ¯ ( t ) m 1 ( t ) + m 2 ( t ) x ¯ ( t ) + m 3 ( t ) y ¯ ( t ) .

Here ( x ¯ (t), y ¯ (t)) is the solution of model (8) with initial value ( x 0 , y 0 ) R + 2 and x ˆ (t) x ¯ (t), y ˆ (t) y ¯ (t), a.s. for t[0,+).

Because of

c 2 ( t ) x ˆ ( t ) m 1 ( t ) + m 2 ( t ) x ˆ ( t ) + m 3 ( t ) y ˆ ( t ) c 2 ( t ) x ¯ ( t ) m 1 ( t ) + m 2 ( t ) x ¯ ( t ) + m 3 ( t ) y ¯ ( t ) = c 2 ( t ) m 3 ( t ) x ¯ ( y ¯ y ˆ ) c 2 ( t ) m 1 ( t ) ( x ¯ x ˆ ) c 2 ( t ) m 3 ( t ) y ¯ ( x ¯ x ˆ ) ( m 1 ( t ) + m 2 ( t ) x ˆ ( t ) + m 3 ( t ) y ˆ ( t ) ) ( m 1 ( t ) + m 2 ( t ) x ¯ ( t ) + m 3 ( t ) y ¯ ( t ) ) c 2 ( t ) m 3 ( t ) x ¯ ( y ¯ y ˆ ) ( m 1 ( t ) + m 2 ( t ) x ˆ ( t ) + m 3 ( t ) y ˆ ( t ) ) ( m 1 ( t ) + m 2 ( t ) x ¯ ( t ) + m 3 ( t ) y ¯ ( t ) ) c 2 ( t ) m 1 ( t ) m 1 2 ( t ) ( x ¯ x ˆ ) c 2 ( t ) m 3 ( t ) y ¯ m 1 ( t ) m 3 ( t ) y ¯ ( x ¯ x ˆ ) 2 c 2 ( t ) m 1 ( t ) ( x ¯ x ˆ )

then

ln y ˆ ( t ) ln y 0 t r 2 + c 2 ( t ) x ¯ ( t ) m 1 ( t ) + m 2 ( t ) x ¯ ( t ) + m 3 ( t ) y ¯ ( t ) b 2 ( t ) y ˆ ( t ) + M 2 ( t ) t 2 c 2 ( t ) m 1 ( t ) ( x ¯ ( t ) x ˆ ( t ) ) r 2 + c 2 ( t ) x ¯ ( t ) m 1 ( t ) + m 2 ( t ) x ¯ ( t ) + m 3 ( t ) y ¯ ( t ) b 2 ( t ) y ˆ ( t ) + M 2 ( t ) t 2 c 2 u m 1 l ( x ¯ ( t ) x ˆ ( t ) ) .
(24)

Consider the Lyapunov function V 2 (t)=|ln x ¯ (t)ln x ˆ (t)|, then V 2 (t) is a positive function on R + , by Itô’s formula, (8) and (10), we have

d + V 2 ( t ) = [ b 1 ( t ) ( x ¯ ( t ) x ˆ ( t ) ) + c 1 ( t ) y ˆ ( t ) m 1 ( t ) + m 2 ( t ) x ˆ ( t ) + m 3 ( t ) y ˆ ( t ) ] d t [ c 1 u m 1 l y ˆ ( t ) b 1 l ( x ¯ ( t ) x ˆ ( t ) ) ] d t .

Integrating from 0 to t and dividing by t on both sides of the inequality yield

V 2 ( t ) V 2 ( 0 ) t c 1 u m 1 l y ˆ ( t ) b 1 l x ¯ ( t ) x ˆ ( t ) .

Owing to V 2 ( t ) t 0, it leads to

b 1 l x ¯ ( t ) x ˆ ( t ) c 1 u m 1 l y ˆ ( t ) + V 2 ( 0 ) t ,

here V 2 (0)=0, then x ¯ (t) x ˆ (t) c 1 u m 1 l b 1 l y ˆ (t). Substituting the above inequality into (24), we have

ln y ˆ ( t ) ln y 0 t r 2 + c 2 ( t ) x ¯ ( t ) m 1 ( t ) + m 2 ( t ) x ¯ ( t ) + m 3 ( t ) y ¯ ( t ) b 2 ( t ) y ˆ ( t ) + M 2 ( t ) t 2 c 1 u c 2 u b 1 l ( m 1 l ) 2 y ˆ ( t ) .

Taking superior limit of the above inequality, we get

[ t 1 ln y ˆ ( t ) ] r 2 + c 2 ( t ) x ¯ ( t ) m 1 ( t ) + m 2 ( t ) x ¯ ( t ) + m 3 ( t ) y ¯ ( t ) ( b 2 u + 2 c 1 u c 2 u b 1 l ( m 1 l ) 2 ) ε 2 >0,

which contradicts Lemma 2, then y ( t ) >0 a.s., that is to say, the predator population y(t) is weakly persistent in the mean a.s.

  1. (iv)

    It follows from Eq. (11) that

    dlny ( r 2 ( t ) b 2 ( t ) y + c 2 ( t ) m 2 ( t ) ) dt+β(t)d B 2 (t),

that is to say,

ln y ( t ) ln y 0 t r 2 + c 2 ( t ) m 2 ( t ) b 2 l y ( t ) + M 2 ( t ) t .

The following proof is similar to the proof of (iv) in Theorem 3, here we omit it. □

Remark 2 From the proof of Theorem 4, we can observe that if r 1 <0, then ( b 1 ) r 2 + ( c 2 ( t ) m 1 ( t ) ) r 1 <0 is straightforward. It shows that if the prey population goes to extinction, the predator population will also go to extinction, which is consistent with the reality. In the other case, r 1 >0, ( b 1 ) r 2 + ( c 2 ( t ) m 1 ( t ) ) r 1 <0, which means the prey population will survive, but the predator population will go to extinction. Notice r 2 (t)= a 2 (t) β 2 ( t ) 2 , this phenomenon may be caused by the large death rate of predator a 2 (t) or the noise density  β 2 (t).

4 Global attractiveness of the solution

In this section, we give sufficient conditions of global attractiveness.

Definition 3 System (3) is said to be globally attractive if

lim t | x 1 ( t ) x 2 ( t ) | = lim t | y 1 ( t ) y 2 ( t ) | =0a.s.,

where ( x 1 (t), y 1 (t)) and ( x 2 (t), y 2 (t)) are two arbitrary positive solutions of system (3) with initial values ( x 10 , y 10 ) R + 2 and ( x 20 , y 20 ) R + 2 .

To give the result of global attractiveness, we show some lemmas first.

Lemma 4 [26]

Suppose that an n-dimensional stochastic process X(t) on t0 satisfies the condition

E | X ( t ) X ( s ) | α 1 c | t s | 1 + α 2 ,0s,t<

for some positive constants α 1 , α 2 and c. Then there exists a continuous modification X ˜ (t) of X(t) which has the property that for every ϑ(0, α 2 / α 1 ), there is a positive random variable h(ω) such that

P { ω : sup 0 < | t s | < h ( ω ) , 0 s , t < | X ˜ ( t , ω ) X ( t , ω ) | | t s | ϑ 2 1 2 ϑ } =1.

In other words, almost every sample path of X ˜ (t) is locally but uniformly Hölder continuous with an exponent ϑ.

Lemma 5 Let (x(t),y(t)) be a solution of system (3) on t0 with initial value ( x 0 , y 0 ) R + 2 , then almost every sample path of (x(t),y(t)) is uniformly continuous.

Proof Equation (3) is equivalent to the following stochastic integral equation:

x ( t ) = x 0 + 0 t x ( s ) [ a 1 ( s ) b 1 ( s ) x ( s ) c 1 ( s ) y ( s ) m 1 ( s ) + m 2 ( s ) x ( s ) + m 3 ( s ) y ( s ) ] d s + 0 t α ( s ) x ( s ) d B 1 ( s ) .

Let f(s)=x(s)[ a 1 (s) b 1 (s)x(s) c 1 ( s ) y ( s ) m 1 ( s ) + m 2 ( s ) x ( s ) + m 3 ( s ) y ( s ) ], g(s)=α(s)x(s), notice that

Moreover, in view of the moment inequality for stochastic integrals in [[24], p.39], one can obtain that for 0 t 1 t 2 and p>2,

E | t 1 t 2 g ( s ) d B 1 ( s ) | p [ p ( p 1 ) 2 ] p 2 ( t 2 t 1 ) ( p 2 ) 2 t 1 t 2 E | g ( s ) | p d s [ p ( p 1 ) 2 ] p 2 ( t 2 t 1 ) p 2 G ( p ) .

Then for 0< t 1 < t 2 <, t 2 t 1 1, 1 p + 1 q =1, we have

E | x ( t 2 ) x ( t 1 ) | p = E | t 1 t 2 f ( s ) d s + t 1 t 2 g ( s ) d B 1 ( s ) | p 2 p 1 E | t 1 t 2 f ( s ) d s | p + 2 p 1 E | t 1 t 2 g ( s ) d B 1 ( s ) | p 2 p 1 ( t 1 t 2 1 q d s ) p q E ( t 1 t 2 | f ( s ) | p d s ) + 2 p 1 [ p ( p 1 ) 2 ] p 2 ( t 2 t 1 ) p 2 G ( p ) 2 p 1 ( t 2 t 1 ) p q F ( p ) ( t 2 t 1 ) + 2 p 1 [ p ( p 1 ) 2 ] p 2 ( t 2 t 1 ) p 2 G ( p ) 2 p 1 ( t 2 t 1 ) p F ( p ) + 2 p 1 [ p ( p 1 ) 2 ] p 2 ( t 2 t 1 ) p 2 G ( p ) 2 p 1 ( t 2 t 1 ) p 2 { ( t 2 t 1 ) p 2 + [ p ( p 1 ) 2 ] p 2 } M ( p ) 2 p 1 ( t 2 t 1 ) p 2 { 1 + [ p ( p 1 ) 2 ] p 2 } M ( p ) ,

where M(p)=F(p)+G(p), it follows from Lemma 4 that almost every sample path of x(t) is locally but uniformly Hölder continuous with an exponent ϑ for ϑ(0, p 2 2 p ) and therefore almost every sample path of x(t) is uniformly continuous on t R + . In the same way, we can demonstrate that almost every sample path of y(t) is uniformly continuous. □

Lemma 6 [27]

Let f be a nonnegative function defined on R + such that f is integrable and is uniformly continuous, then lim t f(t)=0.

Theorem 5 System (3) is globally attractive if

b 1 l 2 c 2 u m 1 l c 1 u m 2 u m 1 l m 3 l >0, b 2 l 2 c 1 u m 1 l >0.
(25)

Proof For ( x 1 (t), y 1 (t)) and ( x 2 (t), y 2 (t)), any two positive solutions of system (3) with initial values ( x 10 , y 10 ) R + 2 and ( x 20 , y 20 ) R + 2 define

V(t)= | ln x 1 ( t ) ln x 2 ( t ) | + | ln y 1 ( t ) ln y 2 ( t ) | ,

then V(t) is a continuous positive function on t0. A direct calculation of the right differential d + V(t) of V(t), by (10) and (11) and Itô’s formula,

d + V ( t ) = sgn ( x 1 x 2 ) { [ d x 1 x 1 ( d x 1 ) 2 2 x 1 2 ] [ d x 2 x 2 ( d x 2 ) 2 2 x 2 2 ] } + sgn ( y 1 y 2 ) { [ d y 1 y 1 ( d y 1 ) 2 2 y 1 2 ] [ d y 2 y 2 ( d y 2 ) 2 2 y 2 2 ] } = { b 1 ( t ) sgn ( x 1 x 2 ) ( x 1 x 2 ) b 2 ( t ) sgn ( y 1 y 2 ) ( y 1 y 2 ) sgn ( x 1 x 2 ) c 1 ( t ) m 1 ( t ) ( y 1 y 2 ) + c 1 ( t ) m 2 ( t ) [ x 2 ( y 1 y 2 ) y 2 ( x 1 x 2 ) ] ( m 1 ( t ) + m 2 ( t ) x 1 + m 3 ( t ) y 1 ) ( m 1 ( t ) + m 2 ( t ) x 2 + m 3 ( t ) y 2 ) + sgn ( y 1 y 2 ) c 2 ( t ) m 1 ( t ) ( x 1 x 2 ) c 2 ( t ) m 3 ( t ) [ x 2 ( y 1 y 2 ) y 2 ( x 1 x 2 ) ] ( m 1 ( t ) + m 2 ( t ) x 1 + m 3 ( t ) y 1 ) ( m 1 ( t ) + m 2 ( t ) x 2 + m 3 ( t ) y 2 ) } d t { b 1 ( t ) | x 1 x 2 | b 2 ( t ) | y 1 y 2 | + c 1 ( t ) m 1 ( t ) | y 1 y 2 | + c 1 ( t ) m 2 ( t ) x 2 | y 1 y 2 | + c 1 ( t ) m 2 ( t ) y 2 | x 1 x 2 | ( m 1 ( t ) + m 2 ( t ) x 1 + m 3 ( t ) y 1 ) ( m 1 ( t ) + m 2 ( t ) x 2 + m 3 ( t ) y 2 ) + c 2 ( t ) m 1 ( t ) | x 1 x 2 | c 2 ( t ) m 3 ( t ) x 2 | y 1 y 2 | + c 2 ( t ) m 3 ( t ) y 2 | x 1 x 2 | ( m 1 ( t ) + m 2 ( t ) x 1 + m 3 ( t ) y 1 ) ( m 1 ( t ) + m 2 ( t ) x 2 + m 3 ( t ) y 2 ) } d t { b 1 ( t ) | x 1 x 2 | b 2 ( t ) | y 1 y 2 | + c 1 ( t ) m 1 ( t ) | y 1 y 2 | + c 1 ( t ) m 1 ( t ) | y 1 y 2 | + c 1 ( t ) m 2 ( t ) m 1 ( t ) m 3 ( t ) | x 1 x 2 | + c 2 ( t ) m 1 ( t ) | x 1 x 2 | + c 2 ( t ) m 1 ( t ) | x 1 x 2 | } d t { ( b 1 ( t ) 2 c 2 ( t ) m 1 ( t ) c 1 ( t ) m 2 ( t ) m 1 ( t ) m 3 ( t ) ) | x 1 x 2 | ( b 2 ( t ) 2 c 1 ( t ) m 1 ( t ) ) | y 1 y 2 | } d t { ( b 1 l 2 c 2 u m 1 l c 1 u m 2 u m 1 l m 3 l ) | x 1 x 2 | ( b 2 l 2 c 1 u m 1 l ) | y 1 y 2 | } d t .

Integrating both sides leads to

V(t)V(0) 0 t ( b 1 l 2 c 2 u m 1 l c 1 u m 2 u m 1 l m 3 l ) | x 1 ( s ) x 2 ( s ) | + ( b 2 l 2 c 1 u m 1 l ) | y 1 ( s ) y 2 ( s ) | ds.

Consequently,

V(t)+ 0 t ( b 1 l 2 c 2 u m 1 l c 1 u m 2 u m 1 l m 3 l ) | x 1 ( s ) x 2 ( s ) | + ( b 2 l 2 c 1 u m 1 l ) | y 1 ( s ) y 2 ( s ) | dsV(0)<.

It then follows from V(t)0 and (25) that

| x 1 ( t ) x 2 ( t ) | L 1 [0,+), | y 1 ( t ) y 2 ( t ) | L 1 [0,+).

Then from Lemmas 5 and 6, we get the desired assertion. □

5 Numerical simulation

In this section, to support the main results in our paper, some simulation figures are introduced. Here we use the Milstein method mentioned in Higham [28] to simulate stochastic equations, considering the following discrete equations:

ξ k and η k (k=1,2,,n) are the Gaussian random variables N(0,1). Let a 1 (t)=0.4+0.01sint, b 1 (t)=0.7+0.01sint, c 1 (t)=0.3+0.02sint, c 2 (t)=0.1+0.02sint, a 2 (t)=0.2+0.05sint, b 2 (t)=0.2+0.01sint, m 1 (t)=0.2+0.04sint, m 2 (t)=0.5+0.05sint, m 3 (t)=0.3+0.05sint with initial value ( x 0 , y 0 )=(0.5,0.6). In Figure 1, we choose α 2 ( t ) 2 =0.5+0.02sint, β 2 ( t ) 2 =0.3+0.02sint, then it is easy to have r 1 =0.1<0. In view of Theorems 3 and 4, both prey population x and predator population y go to extinction. Figure 1 confirms this. In Figure 2, we choose α 2 ( t ) 2 =0.2+0.02sint, β 2 ( t ) 2 =0.4+0.02sint, the other parameters are the same. At this time, we get r 1 =0.2>0 and ( b 1 ) r 2 + ( c 2 ( t ) m 1 ( t ) ) r 1 =0.314<0. By virtue of Theorems 3 and 4, the prey population x is weakly persistent in the mean and the predator population y goes to extinction, which is shown in Figure 2.

Figure 1
figure1

Both populations go to extinction. Solutions of system (3) for α 2 (t)=1+0.04sint, β 2 (t)=0.6+0.04sint, with step size Δt=0.05>0. Prey and predator populations are represented by red line and blue line with a star, respectively.

Figure 2
figure2

Prey is weakly persistent in the mean, predator goes to extinction. Solutions of system (3) for α 2 (t)=0.4+0.04sint, β 2 (t)=0.8+0.04sint, with step size Δt=0.05>0. Prey and predator populations are represented by red line and blue line with a star, respectively.

In Figure 3, we choose c 1 (t)=0.1+0.02sint, c 2 (t)=0.2+0.02sint, m 1 =0.1+0.04sint, α 2 ( t ) 2 =0.02+0.02sint, β 2 ( t ) 2 =0.02+0.02sint, the other parameters are the same; then r 1 =0.38>0, but r 2 + c 2 ( t ) x ¯ ( t ) m 1 ( t ) + m 2 ( t ) x ¯ ( t ) + m 3 ( t ) y ¯ ( t ) >0 is difficult to verify at present. Figure 3 illustrates that the situation that both prey population x and predator population y are weakly persistent in the mean exists.

Figure 3
figure3

Both populations are weakly persistent in the mean. Solutions of system (3) for α 2 (t)=0.04+0.04sint, β 2 (t)=0.04+0.04sint, with step size Δt=0.1>0. Prey and predator populations are represented by red line and blue dotted line, respectively.

6 Conclusions

Owing to theoretical and practical importance, the predator-prey system with Beddington-DeAngelis functional response has received great attention and has been studied extensively, but for stochastic non-autonomous case, there is none. Here, we consider a stochastic non-autonomous predator-prey system with Beddington-DeAngelis functional response. Firstly, we show that the solution of system (3) is globally positive and stochastically ultimately bounded. Sufficient conditions for extinction, non-persistence in the mean, weak persistence in the mean and strong persistence in the mean are obtained. The threshold between weak persistence and extinction for prey population is established. We also show that the solution of system (3) is globally attractive under some sufficient conditions. These results are useful to estimate the risk of extinction of species in the system. Besides, global attractiveness means that all the species in the community can coexist.

There are still some interesting questions deserved to be study. For example, here the condition of the weak persistence in the mean for predator is only a sufficient condition, which is not so ideal. Maybe we can give the threshold between weak persistence in the mean and extinction for predator in the future. Moreover, we can consider colored noise in the models owing to sudden environmental changes caused by seasons or other reasons.

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Acknowledgements

This work was supported by National Science Foundation of China (Nos. 11071275 and 11228104) and by the Special Fund for Basic Scientific Research of Central Colleges (CCNU10B01005).

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SL proposed the model and completed the main part of the manuscript. XZ checked all the results and polished the language. All authors read and approved the final manuscript.

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Li, S., Zhang, X. Dynamics of a stochastic non-autonomous predator-prey system with Beddington-DeAngelis functional response. Adv Differ Equ 2013, 19 (2013). https://doi.org/10.1186/1687-1847-2013-19

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Keywords

  • predator-prey
  • non-autonomous
  • stochastic
  • Beddington-DeAngelis
  • persistence
  • extinction