DSpace at VNU: On the dynamics of a stochastic ratio-dependent predator–prey model with a specific functional response

DSpace at VNU: On the dynamics of a stochastic ratio-dependent predator–prey model with a specific functional response t...

O R I G I NA L R E S E A R C H

On the dynamics of a stochastic ratio-dependent

predator–prey model with a specific functional response

Yan Zhang · Shujing Gao · Kuangang Fan ·

Yanfei Dai

Received: 14 May 2014

© Korean Society for Computational and Applied Mathematics 2014

Abstract In this paper, a new stochastic two-species predator–prey model which

is ratio-dependent and a specific functional response is considered in, is proposed The existence of a global positive solution to the model for any positive initial value

is shown Stochastically ultimate boundedness and uniform continuity are derived Moreover, under some sufficient conditions, the stochastic permanence and extinction are established for the model At last, numerical simulations are carried out to support our results

Keywords Itô’s formula· Stochastically permanent · Extinction · White noise

1 Introduction

Predator–prey relationship can be important in regulating the number of prey and predators And the dynamic relationship between predators and their preys has long been and will continue to be one of the dominant themes in both ecology and mathe-matical ecology due to its universal existence and importance [1] In the past decades, more and more mathematical models for predator–prey behavior are carried out, see [2 4] and the references cited therein

Y Zhang (B) · S Gao · Y Dai

Key Laboratory of Jiangxi Province for Numerical Simulation and Emulation Techniques,

Gannan Normal University, Ganzhou 341000, People’s Republic of China

e-mail: zhyan8401@163.com

S Gao

e-mail: gaosjmath@126.com

K Fan

School of Mechanical and Electrical Engineering, Jiangxi University of Science

and Technology, Ganzhou 341000, People’s Republic of China

e-mail: kuangangfriend@163.com

Y Zhang et al.

When investigating biological phenomena, there are many factors which affect dynamical properties of biological and mathematical models One of the familiar nonlinear factors is functional response [5] There are many significant functional responses in order to model various different situations and evidences show that when predators have to search or compete for food, a more suitable functional response depending on the densities of both prey and predator, which is called a ratio-dependent functional response, should be introduced in realistic models [6,7] In the past decades, many authors proposed different forms of ratio-dependent functional responses to model this process and the three classical predator-dependent functions are Crowley– Martin type [8], Beddington–DeAngelis type by Beddington [9] and DeAngelis et al [10], as well as Hassell–Varley type [11] In this paper, we consider the following model with a specific functional response:

⎧

⎪

⎪

˙x(t) = x(t)



1+ α1x (t) + α2 y (t) + α3 x (t)y(t)



,

˙y(t) = y(t)



1+ α1x (t) + α2 y (t) + α3 x (t)y(t)



.

(1.1)

where x (t), y(t) stand for the population densities of prey and predator at time t, respectively These parameter are defined as follows: r is the growth rate of the prey, a measures the strength of competition among individuals of species x, ω is the capturing rate of predator, f is the rate of conversion of nutrients into the production of predator,

h , b have the similar meaning to r, a, respectively. 1+α1x (t)+α ωx(t)2y (t)+α3x (t)y(t) is the radio-dependent functional response, whereα1, α2, α3 ≥ 0 are constants It is very important to note that this functional response becomes a linear mass-action function response (or Holling type I functional response) ifα1 = α2 = α3 = 0, the Holling type-II functional response ifα2 = α3 = 0, the modified Holling type-II functional response proposed in [12,13] whenα3= 0, and Crowley–Martin functional response presented in [6,8,14] ifα3 = α1α2.

As a matter of fact, population systems in the real world are often inevitably subject

to environmental noises Many researchers pointed out the fact that due to environ-mental noise, the birth rate, carrying capacity and other parameters involved in the model exhibit random fluctuation to a greater or lesser extent Therefore, more and more interest is focused on stochastic systems and many research has been done in this field , see e.g., [15–20] In this paper, taking into the effect of randomly fluctuating environment, we incorporate white noise in each equations of the system (1.1) and we assume that fluctuations in environment will manifest themselves mainly as fluctua-tions in the growth rate and capturing rate of the prey population and the growth rate and conversion rate of the predator population, therefore, the corresponding stochastic system to Eq (1.1) can be described as follows:

d x = x(t)



1+ α1x (t) + α2 y (t) + α3 x (t)y(t)



dt + x(t)



1+ α1x (t) + α2 y (t) + α3 x (t)y(t)



d B1(t),

d y = y(t)



1+ α1x (t) + α2 y (t) + α3 x (t)y(t)



+ y(t)



1+ α1x (t) + α2 y (t) + α3 x (t)y(t)



d B2(t). (1.2)

where r, a, h, b, ω, f, α i are positive parameters,i = 1, 2, 3 σ2

i and δ2

i represent

the intensities of the white noises, i = 1, 2 B1(t), B2(t) are independent Brownian motions defined on a complete probability space(, F, P) with a filtration {F t}t ∈R+

satisfying the usual conditions(i.e it is right continuous and increasing while F0

con-tains all P −null sets) We denote by R2

+the positive cone in R2, and also denote by

X (t) = (x(t), y(t)) and |X(t)| = (x2(t) + y2(t))1

As far as we know, there is no work has been done on the stochastic model (1.2), and the aim of this paper is to analyze the dynamical properties of the model and show the impact of environmental noise on the population system (1.2)

The rest of the paper is arranged as follows In this paper, we firstly show that the system (1.2) has a unique global solution for any positive initial value in Sect.2.1 The stochastic boundedness of the positive solution is discussed in Sect.2.2 Moreover, we show that the solution is uniformly continuous in Sect.2.3 In Sect.3, by constructing

a suitable Lyapunov function, we establish the sufficient conditions on the stochastic permanence and extinction of the model At last, in Sect.4, a numerical simulation which verifies our qualitative results is given and the paper is ended with a discussion

2 Properties of the solution

2.1 Existence, uniqueness and global positive solution

In this section, as x (t), y(t) in system (1.2) stand for the population densities of prey

and predator at time t, respectively, we will only interested in the positive solutions

of system (1.2) 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 local Lipschitz condition [21] However, the coefficients of system (1.2) neither satisfy the linear growth condition, nor local Lipschitz continuous In the following, by making the change of variables, existence, uniqueness of the positive solution will be shown

Theorem 1 For any initial value x0 > 0, y0 > 0, there is a unique positive local

solution (x(t), y(t)) for t ∈ [0, τ e ) of system (1.2) a.s.

Proof Let u (t) = lnx(t), v(t) = lny(t), then we obtain the following equations

du (t) =



1+ α1e u (t) + α2e v(t) + α3e u (t) e v(t)

− 0.5



1+ α1e u (t) + α2e v(t) + α3e u (t) e v(t)

2

dt

Y Zhang et al.

+



1+ α1e u (t) + α2e v(t) + α3e u (t) e v(t)



d B1(t),

d v(t) =



1+ α1e u (t) + α2e v(t) + α3e u (t) e v(t)

− 0.5



1+ α1e u (t) + α2e v(t) + α3e u (t) e v(t)

2

dt

+



δ1+ δ2 e u (t)

1+ α1e u (t) + α2e v(t) + α3e u (t) e v(t)



d B2(t). (2.1)

for t ≥ 0 with initial value u0 = lnx0,v0 = lny0 Obviously, the coefficients of model (2.1) satisfy the local Lipschitz condition Therefore, there exists a unique local

solution u(t), v(t) on t ∈ [0, τ e ), where τ eis the explosion time So, by Itô’s formula,

it is easy to obtain that x (t) = e u (t) , y(t) = e v(t)is the unique positive local solution

to system (1.2) with initial value x0 > 0, y0 > 0.  Next, we will show the unique positive solution of model (2.1) is global, i.e.,

τ e= ∞

Theorem 2 For any given value (x(0), y(0)) = X0 ∈ R+2 , there is a unique solution

(x(t), y(t)) to Eq (1.2) on t ≥ 0 and the solution will remain in R2

+with probability

1, namely (x(t), y(t)) in R2

+for all t ≥ 0 almost surely

Proof Our Proof is motivated by the works of Mao et al [22] and Liu and Wang [23] Let k0 > 0 be sufficiently large for X0 lying within the interval [1/k0, k0]

For each integer k > k0, define the stopping times τ k = inf{t ∈ [0, τ e] :

x (t)∈(1/k, k), or y(t)∈(1/k, k)}, where throughout this paper we set in f ∅ = ∞.

Obviously,τ k is increasing as k → ∞ Let τ∞= lim

k→+∞τ k, thenτ∞≤ τ e If we can showτ∞= ∞ a.s., then we can obtain that τ e = ∞a.s and (x(t), y(t)) ∈ R2

+a.s for

all t ≥ 0 In other words, to complete the proof all of we need to show is that τ∞= ∞ a.s

If this statement is false, then there exist constants T > 0 and ε ∈ (0, 1) such that

P {τ∞≤ T } > ε Hence, there exists an integer k1≥ k0such that

P {τ k ≤ T } ≥ ε for all k ≥ k1 (2.2)

Define a C2-function V : R2

+→ R+by V (x, y) = (x − 1 − lnx) + (y − 1 − lny) Then the nonnegativity of the function V (x, y) can easily be seen from

u − 1 − lnu ≥ 0 on u > 0

If(x(t), y(t)) ∈ R+2, in view of I t ˆo’s formula, we obtain that

d V (x, y) = V x d x + 0.5V x x (dx)2+ V y d y + 0.5V yy (dy)2

=



(1 − 1/x)x



1+ α1x (t) + α2y (t) + α3x (t)y(t)



+ (1 − 1/y)y



1+ α1x (t) + α2y (t) + α3x (t)y(t)



dt + 0.5



σ1+ σ2y (t)

1+ α1x (t) + α2y (t) + α3x (t)y(t)

dt + 0.5



δ1+ δ2x (t)

1+ α1x (t) + α2y (t) + α3x (t)y(t)

dt + (1 − 1/x)x



σ1+ σ2y (t)

1+ α1x (t) + α2y (t) + α3x (t)y(t)



d B1(t)

+ (1 − 1/y)y



δ1+ δ2x (t)

1+ α1x (t) + α2y (t) + α3x (t)y(t)



d B2(t)

≤



(r +a)x(t)−ax2(t)+ ω

α2+f ω

α2)2

dt

+ 0.5



δ1+δ2

α1



σ1+ σ2y (t)

1+α1x (t)+α2y (t)+α3x (t)y(t)



d B1(t)

+ (y(t) − 1)



δ1+ δ2x (t)

1+ α1x (t) + α2y (t) + α3x (t)y(t)



d B2(t)

Thus, there exists a positive number G such that

d V (x, y) ≤ Gdt + (x(t) − 1)



1+ α1x (t) + α2 y (t) + α3 x (t)y(t)



d B1(t)

+ (y(t) − 1)



1+ α1x (t) + α2 y (t) + α3 x (t)y(t)



d B2(t)

Taking integral for each side of the above inequality from 0 toτ k T yields that

τ k T

0

d V (x, y) ≤ τ k T

0

Gdt

0



(x(t)−1)



σ1+ σ2y (t)

1+α1x (t)+α2y (t)+α3x (t)y(t)



d B1(t)

+ (y(t)−1)



δ1+ δ2x (t)

1+α1x (t)+α2y (t)+α3x (t)y(t)



d B2(t)

whereτ k T = min{τ k , T } Whence taking expectation, we have that

E V (x(τ k T ), y(τ k T )) ≤ V (x(0), y(0)) + G1 E (τ k T ) ≤ V (x(0), y(0))

Set  k = {τ k ≤ T } for k ≥ k1 and by (2.2), P( k ) ≥ ε Note that for every

ω ∈  k , there is x (τ k , ω) or y(τ k , ω) equals either k or 1/k, and hence

V (x(τ k , ω), y(τ k , ω)) ≥ min{k − 1 − lnk, 1/k − 1 + lnk}

Y Zhang et al.

It then follows from (2.3) that

V (x(0), y(0)) + G1 T ≥ E[1  k (ω)V (x(ω), y(ω))]

≥ ε min{k − 1 − lnk, 1/k − 1 + lnk}

where 1 kis the indicator function of k Letting k→ ∞ leads to the contradiction

∞ > V (x(0), y(0)) + G1T = ∞

So we must haveτ∞= ∞ a.s This completes the proof of Theorem2 

2.2 Stochastic boundedness

Definition 1 (See [24]) The solution X (t) = (x(t), y(t)) of Eq (1.2) is said to be stochastically ultimately bounded, if for anyε ∈ (0, 1), there is a positive constant

δ = δ(ε), such that for any given initial value X0 ∈ R2+, the solution X(t) to (1.2) has the property that

lim sup

t→∞ P {|X(t)| > δ} < ε.

Theorem 3 The solutions of model (1.2) are stochastically ultimately bounded for

any initial value X0= (x0, y0) ∈ R2+.

Proof Define Lyapunov functions V1= e t x p and V2 = e t y prespectively, for(x, y) ∈

R+

2 and p > 0 Then by the Itô formula, we compute

d (e t x p ) = e t x p dt + e t

px p−1d x + 0.5p(p − 1)x p−2(dx)2

= e t

x p (t)dt + pe t

x p (t)



1+α1x (t)+α2 y (t)+α3 x (t)y(t)



dt + 0.5p(p − 1)e t

x p (t)



1+ α1x (t) + α2 y (t) + α3 x (t)y(t)

2

dt + pe t

x p (t)



1+ α1x (t) + α2 y (t) + α3 x (t)y(t)



d B1(t)

and

d (e t

y p ) = e t

y p (t)dt + pe t

y p (t)



1+α1x (t)+α2 y (t)+α3 x (t)y(t)



dt + 0.5p(p − 1)e t

y p (t)



1+ α1x (t) + α2 y (t) + α3 x (t)y(t)

2

dt + pe t

y p (t)



1+ α1x (t) + α2 y (t) + α3 x (t)y(t)



d B2(t)

L V1= e t

x p (t)dt + pe t

x p (t)



1+ α1x (t) + α2 y (t) + α3 x (t)y(t)



+ 0.5p(p − 1)e t

x p (t)



1+ α1x (t) + α2 y (t) + α3 x (t)y(t)

2

≤ e t

x p (t) 1+ pr + 0.5p(p − 1)



σ1+σ2

α2

2

− apx



≤ G2(p)et

(2.4) and

L V2= e t

y p (t)dt + pe t

y p (t)



1+ α1x (t) + α2 y (t) + α3 x (t)y(t)



+ 0.5p(p − 1)e t

y p (t)



1+ α1x (t) + α2 y (t) + α3 x (t)y(t)

2

≤ e t

y p (t)



1+ ph + p f ω

α1 + 0.5p(p − 1)



δ1+ δ2

α1

2

− pby



≤ G3(p)et

(2.5) where

G2(p) =



1+ pr + 0.5p(p − 1)σ1+σ2

α2

2p+1

a p (p + 1) p+1 ,

G3(p) =



1+ ph + p f ω

α1 + 0.5p(p − 1)δ1+ δ2

α1

2p+1

b p (p + 1) p+1

Thus, taking integral and expectations on both sides of (2.4) and (2.5), respectively,

we have the following equalities

lim sup

t→∞ E x

p ≤ G2(p) < +∞, lim sup

t→∞ E y

p ≤ G3(p) < +∞

On the other hand, for X (t) = (x(t), y(t)) ∈ R+2, note that

|X(t)| p≤2 max{x2(t), y2(t)}

p

2

≤ 2p (x p (t) + y p (t)).

Consequently,

lim sup

t→∞ E |X(t)| p ≤ G4(p) < +∞

Y Zhang et al.

where G4(p) = 2 p (G2(p) + G3(p)) By the Chebyshev inequality, the proof is

2.3 Uniform continuity

In this section, we continue to show the positive solution X (t) = (x(t), y(t)) is

uniformly H¨older continuous Main tools are to use appropriate Lyapunov functions

and fundamental inequalities

Lemma 1 (See [25,26]) Suppose that a n − dimensional stochastic process X(t) on

t ≥ 0 satisfies the condition

E |X(t) − X(s)| α1 ≤ c|t − s|1+α2, 0 ≤ s, t < ∞,

for some positive constants α1, α2 and c Then there exists a continuous modification

˜X(t) of X(t) which has the property that for every υ ∈ 0,α2

α1



, there is a positive random variable ψ(ω) such that

P



0<|t−s|<ψ(ω),0≤s,t<∞

| ˜X(t, ω) − X(t, ω)|

1− 2−υ



= 1.

In other words, almost every sample path of ˜ X (t) is locally but uniformly H ¨older

continuous with exponent υ.

Theorem 4 Let (x(t), y(t)) be a solution of system (1.2) on t ≥ 0 with initial value (x0, y0) ∈ R2

+, then almost every sample path of (x(t), y(t)) is uniformly continuous.

Proof The first equation of system (1.2) is equivalent to the following stochastic integral equation

x (t) = x0+ t

0

x (s)



1+ α1x (s) + α2 y (s) + α3 x (s)y(s)



ds

0

1+ α1x (s) + α2 y (s) + α3 x (s)y(s)



d B1(s),

Let

f1(s) = x(s)



1+ α1x (s) + α2 y (s) + α3 x (s)y(s)



,

f2(s) = x(s)σ1+ σ2 y (s)

1+ α1x (s) + α2 y (s) + α3 x (s)y(s)



then we have

E | f1(t)|p = E

1+ α1x (s) + α2 y (s) + α3 x (s)y(s)



p

= E



|x| p

r − ax −1+ α1x (s) + α2 ωy y (s) + α3 x (s)y(s)p

≤ 1

2E |x| 2 p+1

2E |r + ax + ωy| 2 p

≤ 1

2E |x| 2 p+1

23

2 p−1[r 2 p + a 2 p

E |x| 2 p + ω 2 p

E |y| 2 p]

≤ 1

2G2(2p) +32 p−1

2 [r 2 p + a 2 p

G2(2p) + ω2 p

G3(2p)]

 F1(p)

E | f2(t)|p = E

1+ α1x (s) + α2 y (s) + α3 x (s)y(s)



p

≤



σ1+σ2

α2

p

E |x| p≤



σ1+σ2

α2

p

G2(p)

 F2(p)

In view of the moment inequality for stochastic integrals, we can have that for

0≤ t1≤ t2and p > 2,

E

 t2

t1

f2(s)d B1(s)

p≤



p (p − 1)

2

p

2

(t2 − t1)p−22

t2

t1

E | f2(s)|p

ds

≤



p (p − 1)

2

p

2

(t2 − t1)p2F2(p)

Then for 0< t1 < t2 < ∞, t2 − t1≤ 1, 1

p+ 1

q = 1, we have

E |x(t2)−x(t1)|p = E

 t2

t1

f1(s)ds + t2

t1

f2(s)d B1(s)

p

≤ 2p−1E

 t t2 1

f1(s)ds

p+ 2p−1E

 t t2 1

f2(s)d B1(s)

p

≤ 2p−1(t2 − t1)p q E

t2

t1

| f1(s)|p ds

+ 2p−1

p (p − 1)

2

p

(t2 − t1)p F2(p)

≤ 2p−1(t2 −t1)p F1(p)(t2−t1)+2p−1

p (p−1)

2

p

(t2 −t1)p F2(p)

= 2p−1(t2 − t1)p F1(p) + 2p−1

p (p − 1)

2

p

(t2 − t1)p F2(p)

Y Zhang et al.

= 2p−1(t2 − t1)p



(t2 − t1)p F1(p) +



p (p − 1)

2

p

F2(p)



≤ 2p−1(t2 − t1)p2



1+



p (p − 1)

2

p

F (p)

where F (p) = max{F1(p), F2(p)} From Lemma1, we can obtain that almost every

sample path of x(t) is locally but uniformly H ¨older-continuous with exponent υ for

everyυ ∈0, p−2

2 p



and thus almost every sample path of x (t) is uniformly continuous

on t ∈ R+ In the same way, we can prove that almost every sample path of y(t) is also uniformly continuous on t ∈ R+ That is, almost every sample path of(x(t), y(t)) to

3 The long behavior of system ( 1.2 )

In this section, we will investigate the long time behavior of system (1.2) such as

permanence and extinction to discuss how the solution varies in R+

2 in more detail Firstly, let us impose the following hypothesis which will be useful in the following

(H1) ω

α3+ 2



max σ1, α1 σ2 , δ1, α2 δ2

2

< min r−α2 ω , h



;

(H2) r − 0.5σ2

1 < 0, h + f ω

α1 − 0.5δ

2

1< 0.

3.1 Stochastic permanence

Now we are in the position to show the stochastic permanence which is defined below

Definition 2 (See [24]) The solution X (t) = (x(t), y(t)) of Eq (1.2) is said to be stochastically permanent, if for anyε ∈ (0, 1), there exists a pair of positive constants

δ = δ(ε) and χ = χ(ε) such that for any initial value X(0) = (x(0), y(0)) ∈ R+2, the

solution X (t) to (1.2) has the properties that

lim inf

t→∞ P {|X(t)| ≥ δ} ≥ 1 − ε, lim inf

t→∞ P {|X(t)| ≤ χ} ≥ 1 − ε.

Then we have the following theorem:

Theorem 5 Suppose that assumption (H1) holds, then system (1.2) is stochastically

permanent.

Proof Firstly, we will prove that for any initial value X (0) = (x(0), y(0)) ∈ R2+, the

solution X (t) = (x(t), y(t)) satisfies that

lim sup

t→∞ E

 1

|X(t)| θ



≤ M,

3 The long... functions

and fundamental inequalities

Lemma (See [25,26]) Suppose that a n − dimensional stochastic process X(t) on< /i>

t ≥ satisfies the condition...