The interaction between predator and prey has long been one of the themes of mathematical biology because of the ubiquity and importance of predation. Due to the imbalance of species in the ecological environment, some species are endangered and many of them have become extinct. Therefore, protecting the diversity of ecological species has become one of the main topics in today’s society. This will inspire more scholars to devote themselves to this research.

It is well known that, in the ecosystem, many factors affect the dynamics of ecological models. One of the key elements is called “functional response”, which represents the consumption per unit of time. Holling-II functional response is one of the most important functional responses [1] . In the past few decades, the deterministic predator model has attracted much attention. For example, Li and Gao [2] introduced the following predator-prey system with Holling II functional response:

{ d x = x ( r 1 − b 1 x − c 1 y 1 + x ) d t , d y = y ( − r 2 − b 2 y + c 2 x 1 + x ) d t , (1.1)

where x ( t ) , y ( t ) stand for prey and predator densities at time t, respectively. Parameters r 1 , r 2 , b 1 , b 2 , c 1 and c 2 are positive constants; r 1 and r 2 stand for intrinsic growth rates of prey x ( t ) and predator y ( t ) respectively. Parameters b 1 and b 2 describe the strength of competition among individuals of species x ( t ) or y ( t ) . Parameters c 1 and c 2 represent the capture rate of predators and the reproduction rate of converting nutrients into predators, respectively.

However, in order to better describe the phenomenon in population dynamics, Hassell initially proposed a nonlinear function of the interaction size of species. He found that as the population grew, the interference became stronger. Therefore, he introduced the concept of mutual interference constant m ∈ ( 0 , 1 ] (see e.g. [3] [4] [5] ). The deterministic predator-prey model with mutual interference and Holling-II function response can be expressed as

{ d x ( t ) = x ( t ) ( r 1 − b 1 x ( t ) − c 1 y m ( t ) 1 + x ) d t , d y ( t ) = ( − r 2 y ( t ) − b 2 y 2 ( t ) + c 2 x ( t ) 1 + x y m ( t ) ) d t , (1.2)

In recent years, system (1.2) and its various extension forms have been extensively studied by scholars (e.g. [6] [7] ).

The growth of species in nature is often restricted by the environment. Because of environmental fluctuations, the parameters involved in the population model are not constant, and they may fluctuate around some average values. Based on this factor, more and more people begin to pay attention to the random population system [8] [9] [10] . We assume that environmental fluctuations mainly affect the internal growth rate r 1 ( t ) and the mortality r 2 ( t ) of predators, that is,

r 1 ( t ) → r 1 ( t ) + σ 1 ( t ) B ′ 1 ( t ) , − r 2 ( t ) → − r 2 ( t ) + σ 2 ( t ) B ′ 2 ( t ) .

where B 1 and B 2 are independent Brownian motions, σ i 2 ( t ) ( i = 1 , 2 ) denotes the intensity of white noise. That is, we consider the following stochastic non-autonomous predator-prey system with Holling-II functional response and mutual interference:

{ d x ( t ) = x ( t ) ( r 1 − b 1 x ( t ) − c 1 y m ( t ) 1 + x ) d t + σ 1 x ( t ) d B 1 ( t ) , d y ( t ) = ( − r 2 y ( t ) − b 2 y 2 ( t ) + c 2 x ( t ) 1 + x y m ( t ) ) d t + σ 2 ( t ) y ( t ) d B 2 ( t ) , (1.3)

On the other hand, the life of species often has some big abrupt changes, such as drought, earthquake, typhoon and other big natural disasters, as well as interference from human activities, such as large-scale hunting, policy protection, etc., which will bring great changes to the number and density of species in a short period of time. Therefore, the interference of impulses to the model needs to be considered, and the following model can be established:

{ d x ( t ) = x ( t ) ( r 1 ( t ) − b 1 ( t ) x ( t ) − c 1 ( t ) 1 + x ( t ) y m ( t ) ) d t + σ 1 ( t ) x ( t ) d B 1 ( t ) , d y ( t ) = ( − r 2 ( t ) y ( t ) − b 2 ( t ) y 2 ( t ) + c 2 ( t ) 1 + x ( t ) x ( t ) y m ( t ) ) d t + σ 2 ( t ) y ( t ) d B 2 ( t ) , } t ≠ t k x ( t k + ) − x ( t k ) = α k x ( t k ) y ( t k + ) − y ( t k ) = β k y ( t k ) } t = t k , k = 1 , 2 , 3 , ⋯ . (1.4)

where r i ( t ) , b i ( t ) , c i ( t ) and σ i 2 ( t ) ( i = 1 , 2 ) are positive and continuous T-periodic functions; and the time sequences satisfies 0 < t 1 < t 2 < t 3 < ⋯ , and lim k → ∞ t k = + ∞ . In addition, Parameters α k , β k represent the impulsive effects, and α k > 0 , β k > 0 denote the planting of the species, and if α k < 0 , β k < 0 , then they represent the harvest of the species. From a biological point of view, we are only looking at the positive solution of this equation. Therefore, it’s a natural constraint that

1 + α k > 0 , 1 + β k > 0 , k = 1 , 2 , 3 , ⋯ .

For the periodicity, we assume that there exists a positive integer p such that t k + p = t k + T , α k + p = α k , β k + p = β k , k ∈ Z . Without loss of generality, we assume [ 0 , T ) ∩ { t k , k ∈ Z } = { t 1 , t 2 , t 3 ⋯ , t p } .

From the biological point of view, the population density will change with the changes of some factors, such as rainfall, drought, plague, which is random. All the possible outcomes from a set Ω with typical element w ∈ Ω . A filtration { F t } t ≥ 0 is the smallest σ-algebra σ ( Ω ) , which contains Ω . Throughout this paper, let ( Ω , { F t } t ≥ 0 , ℙ ) is a complete probability space with a filtration { F t } satisfying the usual normal conditions. (i.e., it is increasing and right continuous while F 0 contains all ℙ -null sets), and we define f u = sup t → ∞ f ( t ) , f l = inf t → ∞ f ( t ) .

The main purpose of this paper is to study the existence and uniqueness of global positive periodic solutions as well as the permanence and extinction of species of system (1.4).

The rest of this paper is organized as follows. In the second part, the existence and uniqueness of global positive solution are proved. In the third part, we get that, under certain conditions, system (1.4) exists T-periodic solution. In the fourth part, under some conditions, we discuss the stochastic extinction and permanence of species. In the fifth part, we use numerical simulation to illustrate our results. Finally we draw a conclusion to conclude this paper.

Definition 2.1 ( [11] ). Consider the following impulsive stochastic differential equation (ISDE)

{ d x ( t ) = f ( t , x ( t ) ) d t + g ( t , x ( t ) ) d B ( t ) , t ≠ t k , t > 0 , x ( t k + ) − x ( t k ) = α k x ( t k ) , t = t k , k = 1 , 2 , 3 , ⋯ . (2.1)

with the initial value

x ( 0 ) = x 0 ∈ ℝ n .

A stochastic process x ( t ) = ( x 1 ( t ) , x 2 ( t ) , ⋯ , x n ( t ) ) T , t ∈ [ 0 , + ∞ ) is said to be a solution of ISDE (2.1), if x ( t ) satisfies

1) x ( t ) is F t adapted and is continuous on ( 0 , t 1 ) and each interval ( t k , t k + 1 ) , k ∈ ℕ and f ( t , x ( t ) ) ∈ L 1 ( ℝ + , ℝ n ) , g ( t , x ( t ) ) ∈ L 2 ( ℝ + , ℝ n ) ;

2) x ( t ) obeys the equivalent integral equation of (2.1) for almost every t ∈ ℝ + \ t k and satisfies the impulsive conditions at each t ∈ ℝ + , k ∈ ℕ a.s.;

3) For each t k , k ∈ ℕ , x ( t k + ) = lim t → t k + x ( t ) and x ( t k − ) = lim t → t k − x ( t ) exist and x ( t k − ) = x ( t k ) with probability one.

As to the existence and uniqueness of global positive solution of system (1.4), we have the following result.

Theorem 2.1 For any initial value ( x 0 , y 0 ) ∈ R + 2 , system (1.4) has a unique global positive solution ( x ( t ) , y ( t ) ) for t ≥ 0 and the solution remains in ℝ + with probability one.

Proof. First, we construct the following stochastic differential equation(SDE) without impulses:

{ d y 1 ( t ) = y 1 ( t ) ( r 1 ( t ) + 1 T ∑ j = 1 p ln ( 1 + α j ) − b 1 ( t ) A 1 ( t ) y 1 ( t )                       − c 1 ( t ) 1 + A 1 ( t ) y 1 ( t ) A 2 m ( t ) y 2 m ( t ) ) d t + σ 1 ( t ) y 1 ( t ) d B 1 ( t ) , d y 2 ( t ) = y 2 ( t ) ( − r 2 ( t ) + 1 T ∑ j = 1 p ln ( 1 + β j ) − b 2 ( t ) A 2 ( t ) y 2 ( t )                       + c 2 ( t ) 1 + A 1 ( t ) y 1 ( t ) A 1 ( t ) A 2 m − 1 ( t ) y 1 ( t ) y 2 m − 1 ( t ) ) d t + σ 2 ( t ) y 2 ( t ) d B 2 ( t ) , (2.2)

with the initial value ( y 1 ( 0 ) , y 2 ( 0 ) = ( x 0 , y 0 ) ) . According to the classic theory of SDE without impulse, SDE (2.2) has a unique global positive solution y ( t ) = ( y 1 ( t ) , y 2 ( t ) ) (more details see [11] ). Let x ( t ) = A 1 ( t ) y 1 ( t ) , y ( t ) = A 2 ( t ) y 2 ( t ) , then we claim that ( x ( t ) , y ( t ) ) is the solution of the system (1.4).

In fact, it is easy to check that x ( t ) and y ( t ) are continuous on ( 0 , t 1 ) and ( t k , t k + 1 ) ⊂ [ 0 , + ∞ ) , k ∈ N ,

and for t ≠ t k , we have

d x ( t ) = A ′ 1 ( t ) y 1 ( t ) d t + A 1 ( t ) d y 1 ( t )                 = A 1 ( t ) y 1 ( t ) [ r 1 ( t ) − b 1 ( t ) A 1 ( t ) y 1 ( t ) − c 1 ( t ) 1 + A 1 ( t ) y 1 ( t ) A 2 m y 2 m ] d t                       + σ 1 ( t ) A 1 ( t ) y 1 ( t ) d w 1 ( t ) .

Similarly, we have

d y ( t ) = y ( t ) [ − r 2 ( t ) − b 2 ( t ) y ( t ) + c 2 ( t ) 1 + x ( t ) x ( t ) y m − 1 ( t ) ] d t − σ 2 ( t ) y ( t ) d w 2 ( t ) .

And for every k ∈ N ,

x ( t k + ) = lim t → t k + A 1 ( t ) y 1 ( t ) = ( ∏ j = 1 p ( 1 + α j ) ) − t k T ∏ 0 ≤ t j ≤ t k ( 1 + α j ) y 1 ( t k + ) = ( 1 + α j ) x ( t k ) ,

x ( t k − ) = lim t → t k − A 1 ( t ) y 1 ( t ) = A 1 ( t k ) y 1 ( t k − ) = x ( t k ) .

In the same way, we have

y ( t k + ) = ( 1 + β j ) y ( t k ) , y ( t k − ) = y ( t k ) .

This completes the proof.

In this section, we give the existence of the positive periodic solution of the stochastic system (1.4) with impulses. For convenience of readers, we first give the definition of the periodic solution of the impulsive stochastic differential equation in the sense of distribution and the results of the existence of periodic solutions (see [12] [13] ).

Definition 3.1 ( [13] ). A stochastic process ξ ( t ) = ξ ( t , ω ) is said to be periodic with period T, if for every finite sequence of numbers t 1 , t 2 , ⋯ , t n , the joint distribution of random variables ξ ( t 1 + h ) , ξ ( t 2 + h ) , ⋯ , ξ ( t n + h ) is independent of h, where h = k T ( k = ± 1 , ± 2 , ⋯ ).

Consider the following periodic stochastic differential equation without impulse:

d X ( t ) = f ( t , X ( t ) ) d t + g ( t , X ( t ) ) d B ( t ) , t ≥ 0 , (3.1)

where g ( t , X ( t ) ) n × l is a n × l matrix function, f ( t , X ( t ) ) and the matrix g ( t , X ( t ) ) n × l are T-periodic in t. Then, Itô’s formula can be applied to F ( t , X ) where X satisfies (3.1). This yields the stochastic differential for F of the form

d F ( t , X ( t ) ) = ( ∂ F ( t , X ) ∂ t + f ( t , X ) ∂ F ( t , X ) ∂ x + 1 2 g 2 ( t , X ) ∂ 2 F ( t , X ) ∂ x 2 ) d t                                         + g ( t , X ) ∂ F ( t , X ) ∂ x d W ( t ) . (3.2)

Lemma 3.1 ( [14] ). Assume that system (3.1) has a global solution, and there exists a T-periodic function V ( x , t ) such that the following conditions hold:

1) L V ( t , x ) ≤ − 1 on the outside of some compact set, where

L V ( t , x ) = ∂ V ∂ t + ∂ V ∂ x f + 1 2 ∂ 2 V ∂ x 2 g 2 .

2) inf | x | > R → ∞ , as R → ∞ .

Then (3.1) has a T-periodic solution.

According to Lemma 3.1, we can obtain the main result in this section. Firstly, we can translate system (1.4) into the following two cases:

Case I. When m = 1 , we have

{ d x ( t ) = x ( t ) ( r 1 ( t ) − b 1 ( t ) x ( t ) − c 1 ( t ) 1 + x ( t ) y ( t ) ) d t + σ 1 ( t ) x ( t ) d B 1 ( t ) d y ( t ) = y ( t ) ( − r 2 ( t ) − b 2 ( t ) y ( t ) + c 2 ( t ) 1 + x ( t ) x ( t ) ) d t + σ 2 ( t ) y ( t ) d B 2 ( t ) } t ≠ t k x ( t k + ) − x ( t k ) = α k x ( t k ) y ( t k + ) − y ( t k ) = β k y ( t k ) } t = t k , k = 1 , 2 , 3 , ⋯ (3.3)

Theorem 3.1 Assume that the following assumption hold

(H1): c 1 u A 2 u > c 1 ( t ) A 2 ( t ) ,

(H2): λ = 1 T ∫ 0 T c 2 l r 1 u + b 1 u A 1 u ( r 1 ( t ) − σ 1 2 ( t ) 2 ) − ( r 2 ( t ) + σ 2 2 ( t ) 2 ) d t           + 1 T ∑ j = 1 p ln ( 1 + α j ) + 1 T ∑ j = 1 p ln ( 1 + β j ) > 0.

Then system (3.2) has a positive T-periodic solution.

Proof. We only need to prove the existence of a periodic solution of the equivalent system (3.3) without impulses as follows:

{ d y 1 ( t ) = y 1 ( t ) ( r 1 ( t ) + 1 T ∑ j = 1 p ln ( 1 + α j ) − b 1 ( t ) A 1 ( t ) y 1 ( t )                           − c 1 ( t ) 1 + A 1 ( t ) y 1 ( t ) A 2 ( t ) y 2 ( t ) ) d t + σ 1 ( t ) y 1 ( t ) d B 1 ( t ) , d y 2 ( t ) = y 2 ( t ) ( − r 2 ( t ) + 1 T ∑ j = 1 p ln ( 1 + β j ) − b 2 ( t ) A 2 ( t ) y 2 ( t )                           + c 2 ( t ) 1 + A 1 ( t ) y 1 ( t ) A 1 ( t ) y 1 ( t ) ) d t + σ 2 ( t ) y 2 ( t ) d B 2 ( t ) , (3.4)

The global existence of the solution has been ensured by Theorem 2.1. Then, we only have to verify the conditions of Lemma 3.1.

Define a C²-function V ( t , x , y ) : ℝ + 2 → ℝ + as follows:

V ( x , y , t ) = M [ − ln y + c 2 l r 1 u + b 1 u A 1 u ( − ln ( x 1 + x ) + c 1 u A 2 u r 2 l y ) ] + ( x + p y ) θ + 1 θ + 1 + M W ( t ) = V 1 ( x , y ) + V 2 ( x , y ) + V 3 ( t ) (3.5)

where p = ( c 1 l c 2 u ) , θ is a constant such that 0 < θ < min { 1 , 2 r 2 l ( σ 1 u ) 2 } . Parameter M > 0 will be given later and W ( t ) satisfies

W ′ ( t ) = − λ + c 2 l r 1 u + b 1 u A 1 u ( r 1 ( t ) − σ 1 2 ( t ) 2 ) − ( r 2 ( t ) + σ 2 2 ( t ) 2 ) . (3.6)

It is easy to check that W ( t ) is a T-periodic function. Indeed

W ( t + T ) − W ( t ) = ∫ t t + T W ′ ( s ) d s = − ∫ 0 T [ c 2 l r 1 u + b 1 u A 1 u ( r 1 ( t ) − σ 1 2 ( t ) 2 ) − ( r 2 ( t ) + σ 2 2 ( t ) 2 ) ] d s         + ∫ T t + T [ c 2 l r 1 u + b 1 u A 1 u ( r 1 ( t ) − σ 1 2 ( t ) 2 ) − ( r 2 ( t ) + σ 2 2 ( t ) 2 ) ] d s = 0.

According to the periodicity of r i ( t ) , σ i ( t ) , i = 1 , 2 , W ( t ) is a T-periodic function. To verify condition (2) of Lemma 3.1, we only need to show that

inf ( t , x , y ) ∈ [ 0 , + ∞ ) × ( R 2 \ U k ) V ( t , x , y ) → ∞ m as k → ∞ .

Here, U k = ( 1 k , k ) × ( 1 k , k ) . All the coefficients of the quadratic term in V ( t , x , y ) are positive, thus, condition (2) of Lemma 3.1 is satisfied.

Next we prove condition (1) of Lemma 3.1. By the Itô’s formula (3.2), we have

L ( − ln y ) = r 2 ( t ) − 1 T ∑ j = 1 p ln ( 1 + β j ) + b 2 ( t ) A 2 ( t ) y − c 2 ( t ) x A 1 ( t ) 1 + x A 1 ( t ) + σ 2 2 ( t ) 2 , (3.7)

L ( − ln ( x 1 + x ) ) = − 1 1 + x [ r 1 ( t ) + 1 T ∑ j = 1 p ln ( 1 + α j ) − b 1 ( t ) A 1 ( t ) x − c 1 ( t ) y A 2 ( t ) 1 + x A 1 ( t ) ]     + σ 1 2 ( t ) ( 1 + x − x 2 ) 2 ( 1 + x ) 2 ≤ − [ r 1 ( t ) + 1 T ∑ j = 1 p ln ( 1 + α j ) − σ 1 2 ( t ) 2 ] + x 1 + x ( r 1 ( t ) + b 1 ( t ) A 1 ( t ) )       − σ 1 2 ( t ) x 2 2 + c 1 ( t ) y A 2 ( t ) 1 + x A 1 ( t ) 1 1 + A 1 ( t ) x

≤ − [ r 1 ( t ) + 1 T ∑ j = 1 p ln ( 1 + α j ) − σ 1 2 ( t ) 2 ] + c 1 ( t ) y A 2 ( t ) + x 1 + x [ r 1 u + b 1 u A 1 u + 1 T ∑ j = 1 p ln ( 1 + α j ) ] . (3.8)

Using (3.6), (H1) and (3.7), we have

L V 1 ( x , y ) ≤ M [ r 2 ( t ) − 1 T ∑ j = 1 p ln ( 1 + β j ) + b 2 ( t ) A 2 ( t ) y + σ 2 2 ( t ) 2     − c 2 l r 1 u + b 1 u A 1 u ( r 1 ( t ) + 1 T ∑ j = 1 p ln ( 1 + α j ) − σ 1 2 ( t ) 2 ) ]     + M q ( 1 T ∑ j = 1 p ln ( 1 + β j ) y + c 1 ( t ) x y 1 + x )     + M c 2 l r 1 u + b 1 u A 1 u 1 T ∑ j = 1 p ln ( 1 + α j ) (3.9)

where q = c 2 l r 1 u + b 1 u A 1 u × c 1 u A 2 u r 2 l . From (3.5) and (3.8), we have

L V 1 + L V 3 ≤ M [ r 2 ( t ) − 1 T ∑ j = 1 p ln ( 1 + β j ) + σ 2 2 ( t ) 2 − c 2 l r 1 u + b 1 u A 1 u ( r 1 ( t ) + 1 T ∑ j = 1 p ln ( 1 + α j ) − σ 1 2 ( t ) 2 ) ]     + M q ( 1 T ∑ j = 1 p ln ( 1 + β j ) y + c 1 ( t ) x y 1 + x ) + M c 2 l r 1 u + b 1 u A 1 u [ r 1 + 1 T ∑ j = 1 p ln ( 1 + α j ) − σ 1 2 ( t ) 2 ]     − M λ − M [ r 2 − 1 T ∑ j = 1 p ln ( 1 + β j ) + σ 2 2 ( t ) 2 ] + M c 2 l r 1 u + b 1 u A 1 u 1 T ∑ j = 1 p ln ( 1 + α j ) ≤ − M λ + M q c 1 u x y 1 + x + C (3.10)

Here, C = M q 1 T ∑ j = 1 p ln ( 1 + β j ) y + M c 2 l r 1 u + b 1 u A 1 u 1 T ∑ j = 1 p ln ( 1 + α j ) , and

L V 2 ( x , y ) = ( x + p y ) θ [ r 1 ( t ) x + 1 T ∑ j = 1 p ln ( 1 + α j ) x − b 1 ( t ) A 1 ( t ) x 2 + 1 T ∑ j = 1 p ln ( 1 + β j ) y       − p 2 b 2 ( t ) y 2 − p r 2 ( t ) y + − c 1 ( t ) + p c 2 ( t ) 1 + x A 1 ( t ) x y A 1 ( t ) A 2 ( t ) ]       + θ 2 ( x + p y ) θ − 1 [ σ 1 2 ( t ) x 2 + p 2 σ 2 2 ( t ) y 2 ]

≤ ( x + p y ) θ r 1 u x − b 1 l A 1 l x 2 + θ − p 1 + θ r 2 l y 1 + θ − p 2 + θ b 2 l y 2 + θ + θ 2 p 1 + θ ( σ 2 u ) 2 y 1 + θ     + ( x + p y ) θ [ 1 T ∑ j = 1 p ln ( 1 + α j ) x + 1 T ∑ j = 1 p ln ( 1 + β j ) y ] + θ 2 ( σ 1 u ) 2 x 1 + θ ≤ − b 1 l A 1 l x 2 + θ 2 − p 1 + θ 2 ( r 2 l − θ 2 ( σ 2 u ) 2 ) y 1 + θ + C 1 (3.11)

where, C 1 = sup ( x , y ) ∈ R + 2 { − b 1 l A 1 l x 2 + θ 2 − p 1 + θ 2 ( r 2 l − θ 2 ( σ 2 u ) 2 ) y 1 + θ + ( x + p y ) θ r 1 u x     + θ 2 ( σ 1 u ) 2 x 1 + θ + ( x + p y ) θ [ 1 T ∑ j = 1 p ln ( 1 + α j ) x + 1 T ∑ j = 1 p ln ( 1 + β j ) y ] } ,

and in (3.11), we use ( x + p y ) θ ≤ x θ , ( x + p y ) θ ≤ p θ y θ in the second inequality. Then we have

L V ( t , x , y ) = L V 1 + L V 2 + L V 3 ≤ − M λ + M q c 1 u x y 1 + x − b 1 l A 1 l x 2 + θ 2 − p 1 + θ 2 ( r 2 l − θ 2 ( σ 2 u ) 2 ) y 1 + θ + C 2 . (3.12)

where C 2 = C + C 1 .

Let M = 2 λ max { 2 , sup ( x , y ) ∈ R + 2 { − b 1 l A 1 l x 2 + θ 4 − p 1 + θ 4 ( r 2 l − θ 2 ( σ 2 u ) 2 ) y 1 + θ + C 2 } } .

To confirm the condition (1) of Lemma 3.2, we choose a sufficiently small constant ε such that:

0 < ε ≤ { λ 4 q , p 1 + θ 4 M q ( r 2 l − θ 2 ( σ 2 u ) 2 ) , b 1 l A 1 l 4 M q } , (3.13)

− M λ − b 1 l A 1 l 4 ε 2 + θ + C 3 ≤ − 1 , (3.14)

− M λ − p 1 + θ 4 ε 2 + θ ( r 2 l − θ 2 ( σ 2 u ) 2 ) + C 4 ≤ − 1 . (3.15)

Define a bounded closed set as follows:

D = { ( x , y ) ∈ R + 2 : ε ≤ x ≤ 1 ε , ε ≤ y ≤ 1 ε } .

Denote

D ε 1 = { ( x , y ) ∈ R + 2 | 0 < x ≤ ε } , D ε 2 = { ( x , y ) ∈ R + 2 | 0 < y ≤ ε } ,

D ε 3 = { ( x , y ) ∈ R + 2 | x ≥ 1 ε } , D ε 4 = { ( x , y ) ∈ R + 2 | y ≥ 1 ε } .

Clearly, D ε C = D ε 1 ∪ D ε 2 ∪ D ε 3 ∪ D ε 4 .

Now we prove L V ( t , x , y ) ≤ − 1 on each domain.

Case 1. If ( t , x , y ) ∈ [ 0 , + ∞ ) × D ε 1 , then it is easy to verify that x y 1 + x ≤ x y ≤ ε y ≤ ε ( 1 + y 1 + θ ) , and

L V ( t , x , y ) ≤ − M λ 4 − b 1 l A 1 l x 2 + θ 4 + [ − M λ 4 + M q ε ] + [ − p 1 + θ 4 ( r 2 l − θ 2 ( σ 2 u ) 2 ) + M q ε ] y 1 + θ     + [ − M λ 2 + sup ( x , y ) ∈ R + 2 { − b 1 l A 1 l x 2 + θ 4 − p 1 + θ 4 ( r 2 l − θ 2 ( σ 2 u ) 2 ) y 1 + θ + C 2 } ] .

Combining with the definition of M λ 4 ≥ 1 and (3.13), we have

L V ( t , x , y ) ≤ − M λ 4 − b 1 l A 1 l x 2 + θ 4 ≤ − M λ 4 ≤ − 1 ,

Thus, L V ( t , x , y ) ≤ − 1 for al ( x , y ) ∈ D ε 1 .

Case 2. Similarly, for any ( x , y ) ∈ D ε 2 , owing to x y 1 + x ≤ x y ≤ ε x ≤ ε ( 1 + x 2 + θ ) , we have that

L V ( t , x , y ) ≤ − M λ 4 + [ − M λ 4 + M q ε ] − p 1 + θ 4 ( r 2 l − θ 2 ( σ 2 u ) 2 ) y 1 + θ + [ − b 1 l A 1 l 4 + ε M q ] x 2 + θ     + [ − M λ 2 + { sup ( x , y ) ∈ R + 2 − b 1 l A 1 l 4 x 2 + θ − p 1 + θ 4 ( r 2 l − θ 2 ( σ 2 u ) 2 ) y 1 + θ + C 2 } ] .

Together with (3.13), we can also get

L V ( t , x , y ) ≤ − M λ 4 − p 1 + θ 4 ( r 2 l − θ 2 ( σ 2 u ) 2 ) y 1 + θ ≤ − M λ 4 ≤ − 1 .

Thus, L V ( t , x , y ) ≤ − 1 for all ( x , y ) ∈ D ε 2 .

Case 3. For any ( x , y ) ∈ D ε 3 , since x y 1 + x ≤ y , it is easy to have

L V ( t , x , y ) ≤ − M λ − b 1 l A 1 l 4 ε 2 + θ + [ − b 1 l A 1 l 4 x 2 + θ − p 1 + θ 2 ( r 2 l − θ 2 ( σ 2 u ) 2 ) y 1 + θ + M q c 1 u x y 1 + x + C 2 ] ≤ − M λ − b 1 l A 1 l 4 ε 2 + θ + C 3 .

where C 3 = sup ( x , y ) ∈ R + 2 { − b 1 l A 1 l 4 x 2 + θ − p 1 + θ 2 ( r 2 l − θ 2 ( σ 2 u ) 2 ) y 1 + θ + M q c 1 u x y 1 + x + C 2 } , and we get L V ( t , x , y ) ≤ − 1 in this domain from (3.14).

Case 4. Similarly, for any ( x , y ) ∈ D ε 4 , we have

L V ( t , x , y ) ≤ − M λ − p 1 + θ 4 ε 2 + θ ( r 2 l − θ ( σ 2 u ) 2 2 ) − p 1 + θ 4 ( r 2 l − θ ( σ 2 u ) 2 2 ) y 1 + θ     − b 1 l A 1 l 2 x 2 + θ + M q c 1 u x y 1 + x + C 2 ≤ − M λ − p 1 + θ 4 ε 2 + θ ( r 2 l − θ 2 ( σ 2 u ) 2 ) + C 4 .

where C 4 = sup ( x , y ) ∈ R + 2 { − b 1 l A 1 l 2 x 2 + θ − p 1 + θ 4 ( r 2 l − θ 2 ( σ 2 u ) 2 ) y 1 + θ + M q c 1 u x y 1 + x + C 2 } . It is clear that L V ( t , x , y ) ≤ − 1 in this domain.

Therefore

L V ( t , x , y ) ≤ − 1 , for all ( t , x , y ) ∈ [ 0 , + ∞ ) × D C .

That is to say, Condition (1) of Lemma 3.1 is verified. Thus system (3.2) has a T-periodic solution. This completes the proof.

Now we’re going to talk about another case.

Case II. If 0 < m < 1 , then we have

{ d x ( t ) = x ( t ) ( r 1 ( t ) − b 1 ( t ) x ( t ) − c 1 ( t ) 1 + x ( t ) y m ( t ) ) d t + σ 1 ( t ) x ( t ) d B 1 ( t ) , d y ( t ) = y ( t ) ( − r 2 ( t ) − b 2 ( t ) y ( t ) + c 2 ( t ) 1 + x ( t ) x ( t ) y m − 1 ( t ) ) d t + σ 2 ( t ) y ( t ) d B 2 ( t ) , } t ≠ t k x ( t k + ) − x ( t k ) = α k x ( t k ) y ( t k + ) − y ( t k ) = β k y ( t k ) } t = t k , k = 1 , 2 , 3 , ⋯ . (3.16)

Theorem 3.3 Assume that the following assumption holds

λ 1 = 1 T ∫ 0 T ( r 1 ( t ) − σ 1 2 ( t ) 2 ) d t − 1 M T ∫ 0 T ( r 2 ( t ) + σ 2 2 ( t ) 2 ) d t             + 1 T ∑ j = 1 p ln ( 1 + α j ) − 1 T ∑ j = 1 p ln ( 1 + β j ) > 0

Then system (3.16) has a positive T-periodic solution.

Proof. We only need to prove the existence of a periodic solution of the equivalent system (2.2) without impulses. The global existence of the solution has been ensured by Theorem 2.1. Then, we only have to verify the conditions of Lemma 3.1.

Define a C²-function V 4 ( t , x , y ) : ℝ + 2 → ℝ +

V 4 ( t , x , y ) = M 1 [ − b 1 l ( A 2 1 − m ) l c 2 u y 1 − m 1 − m − ln x + H A 2 u b 2 l y ] − ln y + c 2 u c 1 l y + A 1 u x + M 1 W 1 ( t ) = V 5 ( x , y ) + V 6 ( t ) . (3.17)

Here, H > 0 will be given in (3.18), and W 1 ( t ) is T-periodic. Same as case I, so the condition (2) of Lemma 3.1 is satisfied.

Next, by the Itô’s formula, we show that the condition (2) of the Lemma 3.1.

L ( − b 1 l A 2 1 − m c 2 u y 1 − m 1 − m ) ≤ b 1 l ( A 2 1 − m ) l c 2 u y 1 − m ( r 2 ( t ) − 1 T ∑ j = 1 p ln ( 1 + β j ) + m σ 2 2 ( t ) 2 )                                                         + b 1 l b 2 l c 2 u ( A 2 2 − m ) l y 2 − m − b 1 l A 1 l x 1 + A 1 u x .

L ( − ln x ) ≤ − ( r 1 ( t ) + 1 T ∑ j = 1 p ln ( 1 + α j ) − σ 1 2 ( t ) 2 ) + b 1 u A 1 u x + c 1 u ( A 2 m ) u y m .

Let V ^ = − b 1 l ( A 2 1 − m ) l c 2 u y 1 − m 1 − m − ln x , then

L ( V ^ ) ≤ b 1 l ( A 2 1 − m ) l c 2 u y 1 − m ( r 2 ( t ) − 1 T ∑ j = 1 p ln ( 1 + β j ) + m σ 2 2 ( t ) 2 )     − ( r 1 ( t ) + 1 T ∑ j = 1 p ln ( 1 + α j ) − σ 1 2 ( t ) 2 )     + b 1 l b 2 l c 2 u ( A 2 2 − m ) l y 2 − m + b 1 u A 1 u x + c 1 u ( A 2 m ) u y m .

By Young inequality, there exists a positive constant H such that

L ( V ^ ) ≤ H ( A 2 2 ) l y 2 + b 1 u A 1 u x − ( r 1 ( t ) + 1 T ∑ j = 1 p ln ( 1 + α j ) − σ 1 2 ( t ) 2 ) . (3.18)

Then

L ( V ^ + H A 2 u b 2 l y ) ≤ b 1 u A 1 u x − ( r 1 ( t ) + 1 T ∑ j = 1 p ln ( 1 + α j ) − σ 1 2 ( t ) 2 )                                               + H A 2 u b 2 l 1 T ∑ j = 1 p ln ( 1 + β j ) y + H b 2 l c 2 u A 1 u ( A 2 m ) u x y m .

Therefore,

L V 5 ( x , y ) ≤ M 1 [ b 1 u A 1 u x − ( r 1 ( t ) + 1 T ∑ j = 1 p ln ( 1 + α j ) − σ 1 2 ( t ) 2 ) + H A 2 u b 2 l T ∑ j = 1 p ln ( 1 + β j ) y ]         + H b 2 l c 2 u A 1 u ( A 2 m ) u x y m + ( r 2 ( t ) − 1 T ∑ j = 1 p ln ( 1 + β j ) + σ 2 2 ( t ) 2 ) − b 2 u A 2 u y         − b 1 l A 1 l x 2 + A 1 u x [ r 1 u + 1 T ∑ j = 1 p ln ( 1 + α j ) ] − c 1 l A 2 l c 2 u y [ r 2 l + b 2 l A 2 l y ] .

L V 4 ( t , x , y ) = L V 5 ( x , y ) + L V 6 ( t ) ≤ M 1 [ b 1 u A 1 u x + H A 2 u b 2 l 1 T ∑ j = 1 p ln ( 1 + β j ) y + H b 2 l c 2 u A 1 u ( A 2 m ) u x y m ] − b 1 l A 1 l x 2     + b 2 u A 2 u y + A 1 u x [ r 1 u + 1 T ∑ j = 1 p ln ( 1 + α j ) ] − c 1 l A 2 l c 2 u y [ r 2 l + b 2 l A 2 l y ] − M 1 λ 1 ≤ M 1 [ H b 2 l c 2 u A 1 u ( A 2 u ) m x y m − λ 1 ] − b 1 l A 1 l x 2 2 − c 1 l ( A 2 l ) 2 2 c 2 u b 2 l y 2 + Q .

where

Q = sup ( x , y ) ∈ R + 2 { M 1 b 1 u A 1 u x + b 2 u A 2 u y + A 1 u x [ r 1 u + 1 T ∑ j = 1 p ln ( 1 + α j ) ] − b 1 l A 1 l x 2 2             − c 1 l ( A 2 l ) 2 2 c 2 u b 2 l y 2 − c 1 l r 2 l A 2 l c 2 u y + M 1 H A 2 u b 2 l 1 T ∑ j = 1 p ln ( 1 + β j ) y }

Let M 1 = 2 + Q λ 1 , then − λ 1 M 1 + Q = − 2 . We choose a sufficiently small constant δ such that

0 < δ ≤ { λ 1 4 , c 1 l ( A 2 l ) 2 b 2 l 4 c 2 u M 1 , b 1 l A 1 l 4 M 1 } , (3.19)

− M 1 λ 1 − b 1 l A 1 l 4 δ 2 + Q 1 ≤ − 1 , (3.20)

− M 1 λ 1 − c 1 l ( A 2 l ) 2 b 2 l 4 δ 2 c 2 u + Q 2 ≤ − 1 . (3.21)

Here

Q 1 = − b 1 l A 1 l x 2 4 − c 1 l ( A 2 l ) 2 2 c 2 u b 2 l y 2 + M 1 H b 2 l c 2 u A 1 u ( A 2 u ) m x y m + Q ,

Q 2 = − b 1 l A 1 l x 2 2 − c 1 l ( A 2 l ) 2 4 c 2 u b 2 l y 2 + M 1 H b 2 l c 2 u A 1 u ( A 2 u ) m x y m + Q .

The following proof is similar to the proof of m = 1 and is omitted. This completes the proof.

In Section 3, we showed that under certain conditions, the system (1.4) has a periodic solution. Because in system (1.4), when 0 < m < 1 the predator birth rate is of the form c 2 x ( t ) 1 + x ( t ) y m − 1 ( t ) . Therefore, the predator birth rate goes to infinity when y → 0 , provided the prey population exists. So in this section, we will show that if the noise is sufficiently large, the solutions to the associated stochastic model will become extinct with probability one when m = 1 .

Definition 4.1 [15] . Let x ( t ) be a solution to system (1.4).

1) If lim t → ∞ x ( t ) = 0 a.s., then species x ( t ) is said to be extinct;

2) If lim t → ∞ 1 t ∫ 0 t   x ( s ) d s > 0 a.s., then species x ( t ) is said to be persistent in the mean.

Theorem 4.1. For any initial value ( x 0 , y 0 ) ∈ R + 2 , the solution X ( t ) = ( x ( t ) , y ( t ) ) of (3.2) obeys

lim t → ∞ sup ln x ( t ) t ≤ r 1 ( t ) − σ 1 2 ( t ) 2 + 1 T ∑ j = 1 p ln ( 1 + α j ) < 0 ,

lim t → ∞ sup ln y ( t ) t ≤ − r 2 ( t ) + c 2 ( t ) − σ 2 2 ( t ) 2 + 1 T ∑ j = 1 p ln ( 1 + β j ) < 0 .

that is, the solution of (3.2) is extinct exponentially with probability.

Proof. For system (3.2), using the Itô’s formula, we have

d ln x ≤ ( r 1 ( t ) + 1 T ∑ j = 1 p ln ( 1 + α j ) − σ 1 2 ( t ) 2 ) d t + σ 1 ( t ) d B 1 ( t ) ,

and

d ln y ≤ ( − r 2 ( t ) + 1 T ∑ j = 1 p ln ( 1 + β j ) + c 2 ( t ) − σ 2 2 ( t ) 2 ) d t + σ 2 ( t ) d B 2 ( t ) .

Integrate both sides from 0 to t, then we get

ln x ( t ) − ln x ( 0 ) t ≤ 1 t ∫ 0 t ( r 1 ( s ) + 1 T ∑ j = 1 p ln ( 1 + α j ) − σ 1 2 ( s ) 2 ) d s + 1 t ∫ 0 t σ 1 ( t ) d B 1 ( t ) ,

ln y ( t ) − ln y ( 0 ) t ≤ 1 t ∫ 0 t ( − r 2 ( s ) + 1 T ∑ j = 1 p ln ( 1 + β j ) + c 2 ( s ) − σ 2 2 ( s ) 2 ) d s + 1 t ∫ 0 t σ 2 ( t ) d B 2 ( t ) .

Let P i ( t ) = ∫ 0 t   σ i ( t ) d B i ( t ) , i = 1 , 2 , then P i ( t ) is a local martingale, and ∫ 0 t ( σ i 2 ) u d s ≤ ( σ i 2 ) u t .

By the strong law of martingale, we have

lim t → ∞ P i ( t ) t = 0     a . s . .

Hence we can derive

lim t → ∞ sup ln x ( t ) t ≤ r 1 ( t ) − σ 1 2 ( t ) 2 + 1 T ∑ j = 1 p ln ( 1 + α j ) < 0     a . s .

lim t → ∞ sup ln y ( t ) t ≤ − r 2 ( t ) + 1 T ∑ j = 1 p ln ( 1 + β j ) + c 2 ( t ) − σ 2 2 ( t ) 2 < 0     a . s .

Thus, lim t → ∞ x ( t ) = 0 , lim t → ∞ y ( t ) = 0 a.s..

Theorem 4.2. Assume

∫ 0 t ( r 1 ( s ) − σ 1 2 ( s ) 2 ) d s + 1 T ∑ j = 1 p ln ( 1 + α j ) > 0 ,

∫ 0 t ( − r 2 ( s ) + c 2 ( s ) − σ 2 2 ( s ) 2 ) d s + 1 T ∑ j = 1 p ln ( 1 + β j ) < 0 .

then the predators of system (3.2) will eventually extinction, and prey populations go to persistent.

Proof. By Theorem 4.1, we have

when ∫ 0 t ( − r 2 ( s ) + c 2 ( s ) − σ 2 2 ( s ) 2 ) d s + 1 T ∑ j = 1 p ln ( 1 + β j ) < 0 , then lim t → ∞ y ( t ) = 0 a.s.

So predators will eventually extinction. Next, we prove the persistence of prey quantity.

For system (3.2), by Itô’s formula (3.2), we have

d ln x = ( r 1 ( t ) + 1 T ∑ j = 1 p ln ( 1 + α j ) − b 1 ( t ) A 1 ( t ) x − c 1 ( t ) y A 2 ( t ) 1 + x A 1 ( t ) − σ 1 2 ( t ) 2 ) d t     + σ 1 ( t ) d B 1 ( t ) .

Integrate both sides of this equation from 0 to t and divide by t, then

ln x ( t ) − ln x ( 0 ) t = 1 t ∫ 0 t ( r 1 ( s ) + 1 T ∑ j = 1 p ln ( 1 + α j ) − σ 1 2 ( s ) 2 ) d s − 1 t ∫ 0 t   b 1 ( s ) A 1 ( s ) x d s     − 1 t ∫ 0 t c 1 ( s ) y A 2 ( s ) 1 + x A 1 ( s ) d s + 1 t ∫ 0 t σ 1 ( t ) d B 1 ( t ) . (4.1)

According to the periodicity of r 1 ( t ) , σ 1 2 ( t ) , we get

1 t ∫ 0 t ( r 1 ( s ) − σ 1 2 ( s ) 2 ) d s = 1 T ∫ 0 t ( r 1 ( s ) − σ 1 2 ( s ) 2 ) d s .

By Theorem 4.1, we have

lim t → ∞ ∫ 0 t σ 1 ( t ) d B 1 ( t ) t = 0     a . s .

Suppose y ( t ) is bounded on t, and lim t → ∞ y ( t ) = 0 a.s. , then

lim t → ∞ 1 t ∫ 0 t c 1 ( s ) y ( s ) A 2 ( s ) 1 + x A 1 ( s ) d s = 0 .

Then, Equation (4.1) can be changed to

0 = 1 T ∫ 0 t ( r 1 ( s ) − σ 1 2 ( s ) 2 ) d s + 1 T ∑ j = 1 p ln ( 1 + α j ) − lim t → ∞ 1 t ∫ 0 t   b 1 ( s ) A 1 ( s ) x d s .

So

b 1 u lim t → ∞ 1 t ∫ 0 t   x ( s ) d s ≥ 1 T ∫ 0 t ( r 1 ( s ) − σ 1 2 ( s ) 2 ) d s + 1 T ∑ j = 1 p ln ( 1 + α j ) > 0 .

which shows the number of prey is persistent. The proof is completed.

In this section, by using the Higham [16] simulation method, we illustrate our main results. The red and blue lines represent predators and prey, respectively.

Example 1. Let r 1 ( t ) = 0.7 , m = 1 , b 1 ( t ) = 0.55 , r 2 ( t ) = 0.09 , b 2 ( t ) = 0.15 , c 1 ( t ) = c 2 ( t ) = 0.35 , σ 1 = 0.03 , σ 2 = 0.05 , p = 1 , α = β = 0.01 , T = 2 π , and initial value ( x ( 0 ) , y ( 0 ) ) = ( 1 , 0.7 ) . Thus, λ ≈ 0.21 > 0 , ∫ 0 2π ( r 1 ( t ) − σ 1 2 ( t ) 2 ) d t + 1 T ∑ j = 1 p ln ( 1 + α j ) ≈ 4.4 > 0 . From Theorems 3.1, we know that system (1.4) has a periodic solution. We use Figure 1 to illustrate this result (see Figure 1).

Example 2. When we let σ 1 = 0.82 , σ 2 = 0.85 , and the other parameters are the same as Example 1. we have ∫ 0 2π ( r 1 ( t ) − σ 1 2 ( t ) 2 ) d t + 1 T ∑ j = 1 p ln ( 1 + α j ) ≈ 3.42 > 0 , ∫ 0 T ( − r 2 ( t ) + c 2 ( t ) − σ 2 2 ( t ) 2 ) d s + 1 T ∑ j = 1 p ln ( 1 + β j ) ≈ − 4.23 < 0 , then Theorem 3.1 and 4.2 tell us that big white noise intensity can cause the predators extinction. We use Figure 2 to explain these results (see Figure 2).

Example 3. Set r 1 ( t ) = 0.5 , b 1 ( t ) = 0.45 , r 2 ( t ) = 0.5 , b 2 ( t ) = 0.4 , c 1 ( t ) = 0.45 , p = 1 , c 2 ( t ) = 0.35 , σ 1 ( t ) = 1.2 , σ 2 ( t ) = 0.85 , α = β = 0.3 ,

m = 1 , T = 2 π , Thus, we have ∫ 0 2π ( r 1 ( t ) − σ 1 2 ( t ) 2 ) d t + 1 T ∑ j = 1 p ln ( 1 + α j ) ≈ − 0.01 > 0 , ∫ 0 T ( − r 2 ( t ) + c 2 ( t ) − σ 2 2 ( t ) 2 ) d s + 1 T ∑ j = 1 p ln ( 1 + β j ) ≈ − 0.1 < 0 . By Theorems 4.1, we can determine that the two species will become extinct (see Figure 3).

In this paper, we propose a predator-prey model with impulsive disturbance in a random environment. We prove that system (1.4) has a globally unique positive solution and a positive periodic solution. In this article, our results are as follows:

a) The effects of white noise and impulse interference on the model are considered.

b) If the impulses are large enough such that the conditions of λ , λ 1 hold, then system (1.4) has a positive T-periodic solution.

c) The main results reveal that large white noise will force the population to become extinct while the population may be persistent under a relatively small white noise.

This work was supported by the National Natural Science Foundation of China (11861027) and Natural Science Foundation of Guangxi (2016 GXNSFAA380194).

The authors declare no conflicts of interest regarding the publication of this paper.

Yang, Y.F., Shao, Y.F. and Li, M.W. (2019) Periodic Solution for a Stochastic Predator-Prey Model with Impulses and Holling-II Functional Response. Journal of Applied Mathematics and Physics, 7, 2212-2230. https://doi.org/10.4236/jamp.2019.710152