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Global Attractivity of a Holling-Tanner Model with Beddington-DeAngelis Functional Response:

with or without Prey Refuge

Baoguo Chen,

Abstract—A Holling-Tanner system with Beddington- DeAngelis functional response and prey refuge takes the form

x(t) = x(t)(1 − x(t)) − (1 − m)x(t)y(t) a1+ b(1 − m)x(t) + c1y(t), y(t) = y(t)h

δ − βy(t) (1 − m)x(t)

i

is investigated in this paper, where a1, b, c1, δ, and β are all positive constants, m is a nonnegative constant which satisfies 0 ≤ m < 1. For the system without prey refuge, i.e., m = 0 case, by developing the new analysis technique, we show that c12 is enough to ensure the global attractivity of the positive equilibrium of the system, such a result seems amazing since it is independent of the parameter a1, b, δ and β. Consequently, we can draw the conclusion that for the most of the parameters, system admits a unique globally attractive positive equilibrium.

For 0 < c1 < 2, we also investigate the stability property of the positive equilibrium. Two examples together with their numerical simulations show the feasibility of the main results.

For the system with prey refuge, we show that there exists a m, such that for all m > m, the system always admits a unique positive equilibrium, which means that enough large prey refuge can improve the coexistence of the species. Refuge plays important role on the persistent property of the system.

Index Terms—Beddington-DeAngelis functional response, Holling-Tanner, Global attractivity, Prey refuge.

I. INTRODUCTION

L

U and Liu [1] proposed the following Holling-Tanner model with Beddington-DeAngelis functional response

dx

dt = rx 1 − x

k − αx(t)y(t) a+ bx(t) + cy(t), dy

dt = y



s 1 − hy(t) x(t)



 ,

(1.1)

where r, k, α, a, b, c, s and h are all positive constants. By nondimensionalize system (1.1) with the following scaling

w= rt, ˜x(ω) = x(t)

k , y(ω) =˜ αy(t) rk , δ= s

r, β=sh

α, a1= a

k, c1= cr α.

(1.2)

System (1.1) changes to the system

x(t) = x(t)(1 − x(t)) − x(t)y(t) a1+ bx(t) + c1y(t), y(t) = y(t)h

δ −βy(t) x(t)

i.

(1.3)

Corresponding author. B. Chen is with the Research Institute of Science Technology and Society, Fuzhou University, Fuzhou, Fujian, 350116, China.

E-mails: [email protected](B. G. Chen).

By simple computation, one could easily see that system (1.3) admits a unique positive equilibrium E(x, y), where

x = −∆ +p∆2+ 4(bβ + c1δ)a1β 2(bβ + c1δ) , y = δx

β ,

(1.4)

and∆ = a1β − bβ − c1δ+ δ.

For the rest of the paper, let’s set

M2def

= δ

β, (1.5)

m1

def= a1+ c1M2− M2

a1+ c1M2

, (1.6)

m2 def= δ

βm1. (1.7)

Concerned with the stability property of this positive equilibrium, by constructing Lyapunov function, Lu and Liu[1] obtained the following result.

Theorem A. If

a1+ c1M2− M2>0 (1.8) and

1 − b a1

− 1 2a1

− δ

2βm1

>0 (1.9) hold, then the positive equilibrium E(x, y) is globally asymptotically stable.

At first sight, condition (1.9) is very simple, however, one could see that (1.9) requires the following three inequalities hold.

b < a1, 1 < 2a1, δ <2βm1. (1.10) Now let’s consider the following example.

Example 1.1.

dx

dt = x 1 − x − xy

0.4 + 0.3x + 3y, dy

dt = y 1 − y x, x(0) > 0, y(0) > 0.

(1.11)

Here, we take a1= 0.4, δ = 2, β = 1, b = 1, c1= 3, and so b= 1 > 0.4 = a1, 1 > 2a1= 0.8,

δ= 2 > 118 = 2 ×1116 = 2βm1.

(1.12)

That is, none of the inequalities in (1.10) holds, let along the condition (1.9), however, numeric simulation (Fig. 1)

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Fig. 1. Dynamic behaviors of system (1.11) with the ini- tial condition (x(0), y(0)) = (1, 1.5), (1.5, 1.5), (1.5, 0.5), (0.1, 0.2), (1.5, 0.2) and (0.5, 1.5), respectively.

shows that system (1.11) admits a unique globally attractive positive equilibrium E1(x1, y1), where x1= 0.7396379247, y1≈0.7396379247.

We found that if we use the method of constructing the Lyapunov function, then (1.9) is necessary. But for the system itself, this condition may not be necessary. Motivated by the above problem, we will revisit the stability property of the positive equilibrium of system (1.3) again. By using a new method which is very different to that of [1], we will establish the following results:

Theorem 1.1. Assume that c12 holds, then the positive equilibrium E(x, y) of (1.3) is globally asymptotically stable.

Theorem 1.2. Assume that 0 < c1 < 2 and (1.8) hold, assume further that

a1(a1+ b)β2+ δ(2a1c1+ bc1−2b)β + δ2c1(c1−1) > 0 (1.13) and

bβ > c1δ(or bβ + δ < c1δ) (1.14) holds, then the positive equilibrium E(x, y) of (1.3) is globally asymptotically stable.

Remark 1.1. In Example 1.1, since c1 = 3, condition of Theorem 1.1 holds, and consequently, the positive equilibri- um of the system (1.11) is globally attractive.

Remark 1.2. Theorem 1.1 shows that for almost all of the parameters (only require c1≥2, and without any restriction on other parameters) of the system (1.3), two species could be coexist in a stable state, this seems very interesting, c1 can be seen as the most important parameters in the system.

Remark 1.3. One could easily see that if a1 large enough, b large enough (or small enough, in this case, requires 1 < c1 <2), inequalities (1.13)-(1.14) hold, and it follows from Theorem 1.2 that two species could be coexist in a stable state.

On the other hand, the existence of refuges have the

important effects on the coexistence of predators and prey, research on the dynamic behaviors of predator-prey system incorporating a prey refuge become a poplar topic during the last decade, see [5]-[14]. Chen and Chen[11] showed that Gause type predator prey system incorporating prey refuge on prey species could admits more than one positive equilib- rium, they also gave sufficient conditions which guarantee the existence and uniqueness of limit cycle; Ma, Chen and Wu[6]

considered a Lotka-Volterra predator-prey model incorpo- rating a prey refuge and predator mutual interference, they showed that the system admits a unique positive equilibrium, which is globally asymptotically stable. Also, Chen, Chen and Xie[14] investigated the dynamic behaviors of the Leslie- Gower predator prey model with prey refuge. They showed that the system admits the unique positive equilibrium, which is globally asymptotically stable, however, the prey refuge has different influence to the final density of both prey and predator species. To the best of our knowledge, to this day, still no scholars incorporate the prey refuge to system (1.3) and studied the influence of the prey refuge. The success of Chen and Chen[11], Ma, Chen and Wu[6] and Chen, Chen and Xie[14] motivated us to propose the following system:

x(t) = x(t)(1 − x(t)) − (1 − m)x(t)y(t) a1+ b(1 − m)x(t) + c1y(t), y(t) = y(t)h

δ − βy(t) (1 − m)x(t)

i

(1.15) where a1, b, c1, δ, and β are all positive constants, m is a nonnegative constant which satisfies0 ≤ m < 1, m describe the prey refuge, and mx is the number of prey species stay in the prey refuge.

By simple computation, one could easily see that system (1.3) admits a unique positive equilibrium E(x, y), where

xm = −∆m+p∆2m+ 4(bβ + c1δ)(1 − m)a1β 2(bβ + c1δ)(1 − m) , ym = δ(1 − m)xm

β ,

(1.16) and∆m= a1β − bβ(1 − m) − c1δ(1 − m) + δ(1 − m)2. As for as system (1.15) is concerned, one interesting problem is whether the system admits a unique positive equilibrium which is globally asymptotically stable if the refuge is enough large, which means that the coexistence of the two species. We will give an affirm answer to this problem, indeed, we have the following result:

Theorem 1.3. Assume that m >maxn

1 − ra1β

δ ,1 − c1,1 − a1

b

o (1.17)

holds, then the positive equilibrium E(x, y) of (1.15) is globally asymptotically stable.

The paper is arranged as follows: In Section 2, some useful Lemmas are established and then we prove Theorem 1.1 and 1.2 in Section 3, then we prove Theorem 1.3 in Section 4. In Section 5, two examples together with their numeric simulations are presented to illustrate the feasibility of Theorem 1.2. We end this paper by a briefly discussion.

For more works on Leslie-Gower predator-prey model, one could refer to [1-36] and the references cited therein.

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II. LEMMAS

Now we state and prove several useful Lemmas.

Lemma 2.1.[4] If a >0, b > 0 and ˙x ≥ x(b − ax), when t ≥0 and x(0) > 0, we have

lim inf

t→+∞x(t) ≥ b a.

If a >0, b > 0 and ˙x ≤ x(b−ax), when t ≥ 0 and x(0) > 0, we have

lim sup

t→+∞ x(t) ≤ b a.

By using Lemma 2.1, similar to the proof of Lemma 2 and Theorem 3 in [1], we can obtain the following Lemma 2.2 and 2.3.

Lemma 2.2. Let(x(t), y(t)) be any positive solution of the system (1.3), then

lim sup

t→+∞ x(t) ≤ 1, lim sup

t→+∞ y(t) ≤ δ β

def= M2.

Lemma 2.3. Let(x(t), y(t)) be any positive solution of the system (1.3), assume that a1+ c1M2− M2>0 holds, then

lim inf

t→+∞x(t) ≥ m1, lim inf

t→+∞y(t) ≥ m2,

where m1and m2are defined by (1.6) and (1.7), respectively.

Lemma 2.4. Assume that one of the following assumption holds

(1) c11;

(2)0 < c1<1, a1+ c1M2− M2>0, then system

dx dt = x

1 − x − B

a1+ bx + c1B

 (2.1)

admits a unique positive equilibrium x(B), which is glob- ally attractive, where B ∈ (m2 − ε, M2 + ε) is some positive constant, and ε > 0 is enough small such that

−a1+ (βδ + ε)(1 − c1) < 0 holds.

Proof. The positive equilibrium of system (2.1) satisfies the equation

1 − x − B

a1+ bx + c1B = 0, (2.2) which is equivalent to

bx2+ (Bc1+ a1− b)x + B(1 − c1) − a1= 0. (2.3) Obviously, under the assumption of Lemma 2.4, B(1 − c1)−

a1<0, and so, system (2.3) has a unique positive solution

x(B) = −∆1+q

21−4b B(1 − c1) − a1

2b . (2.4)

where∆1= Bc1+ a1− b.

Set

F(x) = 1 − x − B a1+ bx + c1B, since

F(0) = 1 − B

a1+ c1B ≥1 −

δ β+ ε

a1+ c1(βδ + ε) >0

and F(x) = 0, from the continuity of the function F (x), it follows that

F(x) > 0 for all x ∈ (0, x) and

F(x) < 0 for all x ∈ (x,+∞),

and so apply Theorem 2.1 in [3] to system (2.1), one could see that x is globally stable, i. e., lim

t→+∞x(t) = x. This ends the proof of Lemma 2.4.

Lemma 2.5. Let x(B) be defined by (2.4), assume that the conditions of Lemma 2.3 and (1.13) hold, then x(B), B ∈ [m2, M2] is a strictly decreasing function of B.

Proof. Since x(B) is the positive solution of (2.3). Let’s consider the function

G(x, B)

= b(x)2+ (Bc1+ a1− b)x+ B(1 − c1) − a1, (2.5) where x∈[m1,1], B ∈ (m2, M2].

It then follows from (1.13) that

∂G

∂x = Bc1+ 2bx+ a1− b

≥ 2bm1+ m2c1+ a1− b

= A1β2+ A2β+ A3

β(a1β+ c1δ) >0,

(2.6)

where

A1 = a1(a1+ b),

A2 = 2a1c1+ b(c1−2)δ, A3 = c1(c1−1)δ2.

(2.7)

and

∂G

∂B = c1x− c1+ 1 ≥ c1m1− c1+ 1

≥ βa1

a1β+ c1δ >0,

(2.8)

Then, it follows from implicit function theorem that dx

dB = −

∂G

∂x

∂G

∂B

<0. (2.9)

Hence, x(B) is the strict decreasing function of B. This ends the proof of Lemma 2.5.

Remark 2.1. (1.13) can be rewrite as follows a1(a1+ b)β2+ δ 2a1c1+ b(c1−2)β

2c1(c1−1) > 0. (2.10) Obviously, if c1≥2, (2.10) holds, and so, the conclusion of Lemma 2.5 holds.

III. PROOF OFTHEOREM1.1AND1.2

Proof of Theorem 1.1. c1 ≥ 2 implies that (1.8) holds, and so, the conclusions of Lemma 2.3 and 2.4 hold. Also, from remark 2.1, Lemma 2.5 holds. Let(x(t), y(t)) be any

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positive solution of system (1.3), let ε > 0 be any positive constant enough small which satisfies

δ

−∆ +q

2−4b (βδ + ε)(1 − c1) − a1

2b −(β + 1)ε > 0,

where

∆ = δ

β + εc1+ a1− b.

It follows from Lemma 2.2 that there exists a T > 0 such that for all t ≥ T ,

x(t) < 1 + εdef= M1(1). (3.1) y(t) < δ

β + εdef= M2(1). (3.2) (3.2) together with the first equation of (1.3) leads to

˙x(t) ≥ x

1−x− M2(1) a1+ bx + c1M2(1)

 for all t ≥ T. (3.3)

Consider the auxiliary equation

˙v(t) = v

1 − v − M2(1) a1+ bv + c1M2(1)

. (3.4)

Since c1≥2, according to Lemma 2.4, (3.4) admits a unique positive equilibrium

v11=−∆2+ q

22−4b M2(1)(1 − c1) − a1



2b , (3.5)

which is globally attractive, where∆2= M2(1)c1+ a1− b.

Hence, by using the differential inequality theory, there exists a T11> T such that

x(t) > v11− εdef= m(1)1 >0 for all t ≥ T11. (3.6) (3.6) together with the second equation of (1.3) leads to

dy dt ≥ y

δ − βy m(1)1



, (3.7)

Applying Lemma 2.1 to (3.7) leads to

lim inf

t→+∞y(t) ≥ δm(1)1

β . (3.8)

That is, for above ε >0, there exists a T12> T11 such that y(t) > δm(1)1

β − εdef= m(1)2 >0 for all t ≥ T12. (3.9) It follows from (3.1),(3.2), (3.6) and (3.9) that for all t ≥ T12,

0 < m(1)1 < x(t) < M1(1), 0 < m(1)2 < y(t) < M2(1).

(3.10)

(3.10) together with the first equation of (1.3) leads to

˙x(t) ≤ x

1 − x − m(1)2 a1+ bx + c1m(1)2

 for all t ≥ T12. (3.11) Consider the auxiliary equation

˙v = v

1 − v − m(1)2 a1+ bv + c1m(1)2

. (3.12)

Since c1 ≥ 2, according to Lemma 2.3, equation (3.12) admits a unique positive equilibrium

v21= −∆3+ q

23−4b m(1)2 (1 − c1) − a1

2b , (3.13)

which is globally attractive, where∆3 = m(1)2 c1+ a1− b.

Hence, by using the differential inequality theory, there exists a T21> T12 such that

x(t) < v21+ε 2

def= M1(2) for all t ≥ T21. (3.14) Since

v21

= −∆3+ q

23−4b m(1)2 (1 − c1) − a1 2b

= −∆3+ q

(m(1)2 c1+ a1+ b)2−4bm(1)2 2b

< −(m(1)2 c1+ a1− b) + q

(m(1)2 c1+ a1+ b)2 2b

= 1,

(3.15)

it follows from (3.1) and (3.15) that

M1(2)< M1(1). (3.16) From (3.16) and the second equation of (1.3), we know that for t ≥ T21,

dy dt ≤ y

δ − βy M1(2)



, (3.17)

Applying Lemma 2.1 to (3.17) leads to

lim sup

t→+∞

y(t) ≤ δ

βM1(2). (3.18) That is, for above ε >0, there exists a T22> T21 such that

y(t) < δ

βM1(2)+ε 2

def= M2(2) for all t ≥ T22. (3.19)

It follows from (3.2), (3.16) and (3.19) that

M2(2)< M2(1). (3.20) Substituting (3.20) into the first equation of system (1.3), we obtain

˙x(t) ≥ x

1 − x − M2(2) a1+ bx + c1M2(2)

 for all t ≥ T22.

Similarly to the analysis of (3.3)-(3.6), there exists a T23>

T22such that x(t) > v22−ε

2

def= m(2)1 >0 for all t ≥ T23, (3.21) where

v22= −∆4+ q

24−4b M2(2)(1 − c1) − a1

2b . (3.22)

here∆4= M2(2)c1+ a1− b.

From (3.5), (3.20) and Lemma 2.5, we have

m(2)1 > m(1)1 . (3.23)

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From (3.23) and the second equation of (1.3), we know that for t ≥ T23,

˙y(t) ≥ y δ − βy

m(2)1

, (3.24)

Applying Lemma 2.1 to (3.24) leads to lim inf

t→+∞y(t) ≥ δ

βm(2)1 . (3.25) That is, for above ε >0, there exists a T24> T23 such that

y(t) > δ

βm(2)1 − ε 2

def= m(2)2 for all t ≥ T24. (3.26)

From (3.8), (3.23) and (3.26) we have

m(2)2 > m(1)2 . (3.27) It follows from (3.16), (3.20), (3.23) and (3.27) that for all t ≥ T24,

0 < m(1)1 < m(2)1 < x(t) < M1(2)< M1(1), 0 < m(1)2 < m(2)2 < y(t) < M2(2)< M2(1).

(3.28)

Repeating the above procedure, we get four sequences Mi(n), m(n)i , i= 1, 2, n = 1, 2, .... such that

M1(n)= vn1

n, m(n)1 = vn2−ε

n, (3.29)

vn1= −∆2n−1+ q

22n−1−4b m(n−1)2 (1 − c1) − a1

2b .

(3.30) vn2= −∆2n+

q

22n−4b M2(n)(1 − c1) − a1

2b . (3.31)

δ

βm(n)1 − ε

n = m(n)2 , δ

βM1(n)

n = M2(n), (3.32) where

2n−1 = m(n−1)2 c1+ a1− b,

2n = M2(n)c1+ a1− b.

Now, we go to show that the sequences Mi(n) is strictly decreasing, and the sequences m(n)i is strictly increasing for i= 1, 2 by induction. Firstly, from (3.28), we have

m(1)i < m(2)i , Mi(2)< Mi(1), i= 1, 2. (3.33) Let us suppose that

m(n−1)i < m(n)i , Mi(n)< Mi(n−1), i= 1, 2. (3.34) It then follows from (3.30) and Lemma 2.5 that

vn1> v(n+1)1. (3.35) From (3.29) we have

M1(n)> M1(n+1). (3.36) By using (3.36), it follows from (3.32) that

M2(n)> M2(n+1). (3.37) It then follows from (3.37), (3.31) and Lemma 2.5 that

v(n+1)2> vn2. (3.38) (3.38) and (3.29) show that

m(n+1)1 > m(n)1 . (3.39)

From the relationship of m(n)1 and m(n)2 , we have

m(n+1)2 > m(n)2 . (3.40) Therefore, we have

0 < m(1)1 < m(2)1 < · · · < m(n)1 < x(t)

< M1(n)< · · · < M1(2)< M1(1),

0 < m(1)2 < m(2)2 < · · · < m(n)2 < y(t)

< M2(n)< · · · < M2(2)< M2(1).

(3.41)

Hence, the limits of Mi(n) and m(n)i , i= 1, 2, n = 1, 2, ...

exist. Denote that

n→+∞lim M1(n) = x, lim

n→+∞m(n)1 = x,

n→+∞lim M2(n) = y, lim

n→+∞m(n)2 = y. (3.42) Then x ≥ x, y ≥ y. Letting n →+∞ in (3.29)-(3.33), we obtain

x = −K1+q

K12−4b y(1 − c1) − a1

2b .

x = −K2+q

K22−4b y(1 − c1) − a1

2b .

δ

βx = y, δ βx= y.

(3.43)

where

K1= yc1+ a1− b, K2= yc1+ a1− b.

(3.43) is equivalent to

bx2+ (c1y+ a1− b)x + y(1 − c1) − a1= 0, bx2+ (c1y+ a1− b)x + y(1 − c1) − a1= 0, δx= βy, δx = βy.

(3.44)

And so,

(x − x) b(x + x) + a1− b

+c1(yx − yx) + (1 − c1)(y − y) = 0. (3.45) Thus,

(x − x) b(x + x) + a1− b +c1δ

β(xx − xx) + (1 − c1)βδ(x − x) = 0. (3.46) and so,

(x − x)

b(x + x) + a1− b −(1 − c1)δ β

= 0. (3.47) From Lemma 2.3 and c1≥2 we have

b(x + x) + a1− b −(1 − c1)βδ

≥ 2bx + a1− b −(1 − c1)δβ

= B1β2+ B2β+ B3

β(a1β+ c1δ)

> 0.

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where

B1 = a1(a1+ b),

B2 = δ(2a1c1+ bc1− a1−2b), B3 = c1δ2(c1−1).

Hence, it follows from (3.47) that x= x.

Also, from (3.43) we have y= y.

Under the assumption of Theorem 1.1, system (1.3) admits a unique positive solution (x, y), hence x = x = x, y= y= y. That is to say,

t→+∞lim x(t) = x, lim

t→+∞y(t) = y. (3.48) This ends the proof of the Theorem 1.1.

Proof of Theorem 1.2. Similarly to the proof of Theorem 1.1, we can finally obtain (3.45)-(3.47). Assume that x 6= x, then from (3.47) we have

x= −x + 1 −a1

b + δ

bβ(1 − c1), (3.49) and

x= −x + 1 −a1

b + δ

bβ(1 − c1). (3.50) Substituting (3.49) and (3.50) to (3.44) leads to

D1x2+ D2x+ D3= 0,

D1x2+ D2x+ D3= 0, (3.51) where

D1 = bβ(bβ − c1δ),

D2 = (bβ − c1δ)(a1β − bβ+ c1δ − δ), D3 = −(bβ − c1δ+ δ)(a1β+ c1δ − δ).

And so, x and x are the positive solution of the equation D1x2+ D2x+ D3= 0. (3.52) From (1.8) and (1.14) one could see that D1>0 and D3<0.

Hence, (3.52) has a unique positive solution, this shows that x= x, the rest of the proof is similar to that of the proof of Theorem 1.1, and we omit the detail here. This ends the proof of Theorem 1.2.

IV. PROOF OFTHEOREM1.3

Concerned with the upper and lower bound of the solutions of system (1.15), we can obtain the following Lemma 4.1 and 4.2.

Lemma 4.1. Let(x(t), y(t)) be any positive solution of the system (1.15), then

lim sup

t→+∞ x(t) ≤ 1def= N1, lim sup

t→+∞ y(t) ≤δ(1 − m) β

def= N2.

Lemma 4.2. Let(x(t), y(t)) be any positive solution of the system (1.15), assume that a1+ c1N2− N2(1 − m) > 0 holds, then

lim inf

t→+∞x(t) ≥ n1, lim inf

t→+∞y(t) ≥ n2, where

n1def

= a1+ c1N2− N2(1 − m) a1+ c1N2

, (4.1)

n2

def= δ(1 − m)

β n1. (4.2)

Remark 4.1 If

m >1 − ra1β

δ , (4.3)

then the inequality a1+c1N2−N2(1−m) > 0 always holds, that is, if the prey refuge is enough large, then the inequality a1+ c1N2− N2(1 − m) > 0 holds.

Lemma 4.3. Assume that

m >1 − c1, (4.4) then system

dx dt = x

1 − x − B(1 − m) a1+ b(1 − m)x + c1B

 (4.5)

admits a unique positive equilibrium xm(B), which is glob- ally attractive, where B ∈(n2− ε, N2+ ε) is some positive constant, and ε >0 is enough small such that ε < m2holds.

Proof. The positive equilibrium of system (4.5) satisfies the equation

1 − x − B(1 − m)

a1+ b(1 − m)x + c1B = 0, (4.6) which is equivalent to

b(1−m)x2+(Bc1+a1−b(1−m))x+B(1−c1−m)−a1= 0.

(4.7) Obviously, under the assumption of Lemma 4.3, B(1 − c1− m) − a1 < 0, and so, system (4.7) has a unique positive solution

xm(B) = −Γ1+q

Γ21−4b(1 − m) B(1 − c1− m) − a1

2b(1 − m) .

(4.8) whereΓ1= Bc1+ a1− b(1 − m).

Similarly to the proof of Lemma 2.4, we could show that

t→+∞lim x(t) = xm(B). This ends the proof of Lemma 4.3.

Lemma 4.4. Let xm(B) be defined by (4.8), assume that the conditions of Lemma 4.3 and (4.4) hold, assume further that

m >1 − a1

b (4.9)

holds, then xm(B), B ∈ [n2, N2] is a strictly decreasing function of B.

Proof. Since x(B) is the positive solution of (4.7). Let’s consider the function

G(x, B)

= b(1 − m)(xm)2+ (Bc1+ a1− b(1 − m))xm

+B(1 − c1− m) − a1= 0,

(4.10)

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where xm∈[n1,1], B ∈ (n2, N2].

It then follows from (4.9) that

∂G

∂xm = Bc1+ 2b(1 − m)xm+ a1+ b(1 − m)

≥ a1− b(1 − m) > 0,

(4.11) and

∂G

∂B = c1xm− c1+ 1 ≥ c1n1− c1+ 1

= βa1(1 − m)

a1β+ c1δ(1 − m) >0,

(4.12)

Then, it follows from implicit function theorem that dxm

dB = −

∂G

∂xm

∂G

∂B

<0. (4.13)

Hence, xm(B) is the strict decreasing function of B. This ends the proof of Lemma 4.4.

Proof of Theorem 1.3. It follows from Lemma 4.1 that there exists a T >0 such that for all t ≥ T ,

x(t) < 1 + εdef= M1(1). (4.14) y(t) <δ(1 − m)

β + εdef= M2(1). (4.15) (4.15) together with the first equation of (1.15) leads to

˙x(t) ≥ x

1 − x − (1 − m)M2(1) a1+ b(1 − m)x + c1M2(1)

. (4.16)

for all t ≥ T .

Consider the auxiliary equation

˙v(t) = v

1 − v − (1 − m)M2(1) a1+ b(1 − m)v + c1M2(1)

. (4.17)

According to Lemma 4.4, (4.17) admits a unique positive equilibrium

v11= −∆2+ q

22−4b(1 − m) M2(1)(1 − c1− m) − a1

2b(1 − m) ,

which is globally attractive, where∆2= M2(1)c1+a1−b(1−

m). Hence, by using the differential inequality theory, there exists a T11> T such that

x(t) > v11− εdef= m(1)1 >0 for all t ≥ T11. (4.18) (4.18) together with the second equation of (1.15) leads to

dy dt ≥ y

δ − βy

(1 − m)m(1)1

, (4.19)

Applying Lemma 2.1 to (4.19) leads to

lim inf

t→+∞y(t) ≥ δ(1 − m)m(1)1

β . (4.20)

That is, for above ε >0, there exists a T12> T11 such that y(t) > δ(1 − m)m(1)1

β − εdef= m(1)2 >0 for all t ≥ T12. (4.21)

Repeating the above procedure, we get four sequences Mi(n), m(n)i , i= 1, 2, n = 1, 2, .... such that

M1(n)= vn1+ ε

n, m(n)1 = vn2− ε

n, (4.22)

vn1= −∆2n−1+q

22n−1−4b(1 − m)L1

2b(1 − m) . (4.23)

vn2=−∆2n+p∆22n−4b(1 − mL2

2b(1 − m) . (4.24)

δ

βm(n)1 − ε

n = m(n)2 , δ

βM1(n)+ ε

n = M2(n), (4.25) where

2n−1 = m(n−1)2 c1+ a1− b(1 − m),

2n = M2(n)c1+ a1− b(1 − m), L1 = m(n−1)2 (1 − c1− m) − a1, L2 = M2(n)(1 − c1− m) − a1.

Similar to the analysis of (3.33)-(3.41), we can show that the sequences Mi(n)is strictly decreasing, and the sequences m(n)i is strictly increasing for i= 1, 2. Hence, the limits of Mi(n) and m(n)i , i= 1, 2, n = 1, 2, ... exist. Denote that

n→+∞lim M1(n) = x, lim

n→+∞m(n)1 = x,

n→+∞lim M2(n) = y, lim

n→+∞m(n)2 = y. (4.26) Similarly to the analysis of (3.42)-(3.48), one could show that under the assumption of Theorem 1.3,

t→+∞lim x(t) = xm, lim

t→+∞y(t) = ym. (4.27) This ends the proof of the Theorem 1.3.

V. NUMERIC SIMULATIONS

In section I, we gave an example to show the feasibility of the Theorem 1.1, now let’s consider the following two examples which illustrate the feasibility of the Theorem 1.2.

Example 5.1 Now let us consider the following system dx

dt = x 1 − x − xy 4 + 2x +12y, dy

dt = y 1 − y x, x(0) > 0, y(0) > 0.

(5.1)

Here, we take a1= 4, δ = β = 1, b = 2, c1= 12, and so, by simple computation, we have

a1(a1+ b)β2+ δ(2a1c1+ bc1−2b)β + δ2c1(c1−1)

= 99 4 >0,

(5.2) bβ= 2 > 1

2 = c1δ (5.3)

and

a1β+ c1δ= 9

2 >1 = δ (5.4)

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Fig. 2. Dynamic behaviors of system (4.1) with the initial condition (x(0), y(0)) = (0.2, 1.5), (1.5, 1.5), (1.5, 0.5), (0.1, 0.3), (1.5, 0.2) and (0.5, 1.5), respectively.

hold, then the positive equilibrium E2(x2, y2) of (4.1) is globally asymptotically stable. Numeric simulation (Fig. 2) shows that system (4.1) admits a unique globally attractive positive equilibrium E2(x2, y2), where x2= y2≈0.8601470509.

Example 5.2 Now let us consider the following system dx

dt = x 1 − x − xy 4 +101x+32y, dy

dt = y 1 −y x, x(0) > 0, y(0) > 0.

(5.5)

Here, we take a1 = 4, δ = β = 1, b = 101, c1 = 32, and so, by simple computation, we have

2a1c1+ bc1−2b = 12 − 1

20 >0, c1−1 = 1

2 >0, (5.6) that is, (1.13) holds. Also,

bβ+ δ = 1 10 <3

2 = c1δ, (5.7) then the positive equilibrium E3(x3, y3) of (5.5) is globally asymptotically stable, where x3 = y3 ≈ 0.8424688318.

Since the numeric simulation is similar to Fig. 2, we omit the detail here.

VI. DISCUSSION

In this paper, we revisit the Holling-Tanner system with Beddington-DeAngelis functional response, which was pro- posed by Lu and Liu[1]. By developing some new analysis technique and using the new method, two set of sufficient conditions which ensure the global attractivity of the positive equilibrium are obtained.

It is nature to conjecture that system (1.3) admits complex dynamic behaviors, since the system contains five parameter- s. However, Theorem 1.1 shows that if c1≥2, then the rest of the parameters have no influence on the dynamic behaviors of the system, and the system always admits a unique positive equilibrium, which is globally attractive. Theorem 1.2 shows that for the case 0 < c1<2, if a1 enough large, b enough

large or enough small, then the system also has a positive equilibrium which is globally attractive.

To summarize: System (1.3) admits a very simple dynamic behaviors for most of the parameters.

On the other hand, we incorporate the prey refuge to system (1.3), this leads to the system (1.15), Theorem 1.3 shows that if the prey refuge is enough large, then two species could be coexist in a stable state.

FUNDING

This work is supported by National Social Science Foundation of China (16BKS132), Humanities and So- cial Science Research Project of Ministry of Education Fund(15YJA710002) and the Natural Science Foundation of Fujian Province (2015J01283).

REFERENCES

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