DSpace at VNU: ON THE DYNAMICS OF PREDATOR-PREY SYSTEMS WITH BEDDINGTON-DEANGELIS FUNCTIONAL RESPONSE

Keywords: Predator-prey system; function response; periodic solution; global stability; Dulac’s criterion.. Introduction The purpose of this paper is to study the permanence of the syste

c World Scientific Publishing Company

DOI: 10.1142/S1793557111000058

ON THE DYNAMICS OF PREDATOR-PREY SYSTEMS WITH BEDDINGTON-DEANGELIS FUNCTIONAL RESPONSE

Faculty of Mathematics, Mechanics and Informatics Hanoi National University, 334 Nguyen Trai, Thanh Xuan,

Hanoi, Vietnam dunh@vnu.edu.vn Tong Thanh Trung Faculty of Economical Mathematics National Economic University (NEU), Hanoi, Vietnam

tongthanhtrung@yahoo.com

Communicated by B.K Dass

Received August 28, 2009 Revised September 23, 2009

This paper studies a predator-prey system with Beddington-DeAngelis functional re-sponse We establish sufficient criteria posed on the coefficients for the permanence of the system, globally asymptotic stability of solutions and the non-existence of periodic orbits.

Keywords: Predator-prey system; function response; periodic solution; global stability; Dulac’s criterion.

AMS Subject Classification: 92D25, 93A30, 93D99

1 Introduction

The purpose of this paper is to study the permanence of the system, the global stability and the non-existence of periodic solutions of the original predator-prey system having the form

(

˙x = ax −b+wx+yf xy − g1x2

˙y = b+wx+yef xy − dy − g2y 2, (1.1) where b, d, e, f, w are positive and g1, g2 are nonnegative The functions x and y stand for the quantity (or density) of the prey and the predator respectively, a

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is the intrinsic growth rate of the prey, d is the mortality rate of the predator,

f is the feeding parameter, e is the conversion efficiency parameter, g1 and g2 are the intraspecies interference parameters, w is the weighting factor, that corre-lates inversely with the prey density at which feeding saturation occurs and b is

a normalization coefficient that relates the densities of the predator and prey to the environment in which they interact The prey-predator system (1.1) was first proposed by DeAngelis [5] in 1975 as a solution of the observed problems in the clas-sic predator-prey theory (with Michaelis-Menten-Holling type functional response) Independently, Beddington [6] offered the same form of a functional response for describing parasite-host interactions

After a change of variables (u = wx

b , v = yb, A = fa, D = d

a, E = waef and h1 = g1 b

aw, h2 = g2 b

a ) and an appropriate time rescaling (t → at), we receive the following model:

(

˙u = u[1 − Av

1+u+v − h1u],

˙v = v[ Eu

By the above simple but crucial change of variables, the system (1.1) is transformed into (1.2) In the system (1.2), the number of parameters are smaller than (1.1) and with a simpler analysis we can understand the system (1.1) through (1.2)

There are some papers where one has considered dynamical properties of special forms of model (1.2) For example, the model (1.2) with h1= h2= 0 was complete mathematical analysis of its dynamics by Dobromir T Dimitrov and Hristo V Kojouharov [2] In that paper, the authors have dealt with the existence of the equilibria and the global dynamics of the system The corresponding model with logistic-growth rate of the prey population (i.e., the case h1 > 0, h2 = 0) was partially analyzed by Robert Stephen Cantrell and Chris Cosner [1] and Tzy-Wei Hwang [3] Some criteria for the permanence and for the predator extinction are derived The global stability is established, provided that the system possesses a positive equilibrium point A ratio-dependent version of (1.1) is also researched by

C Cosner, D L DeAngelis, J S Ault and D B Olson [7]

This paper continues to study further the permanence and stability of the system (1.2) by investigating the case h26= 0 If h26= 0 but h1= 0, we show conditions for the existence of positive equilibria and classify them For the case h1and h2are si-multaneously different from zero, it is difficult to determine the positive equilibrium points because they are the solutions of a third order algebraic equation Although

we can use the Cardano formula to find the solutions, this formula is rather cum-bersome and we cannot use it This difficulty makes a bad effect to analyze directly the stability property of the equilibrium points We can give a sufficient criterion posed on the coefficient for the permanence of the system Further, we show some conditions to ensure the uniqueness of the positive equilibrium point It is known that in that case, if there is no periodic orbit, this (unique) equilibrium point is stable

The paper is organized as follows: In section II, we analyze the steady states of Asian-European J Math 2011.04:35-48 Downloaded from www.worldscientific.com by UNIVERSIDADE FEDERAL DE SAP PAULO on 08/13/13 For personal use only.

the system (1.2) and give a condition to ensure the permanence of solution, global stability and we use the Dulac’s criterion to show that system (1.2) has no periodic solution when it has positive equilibrium In section III, we give some examples to illustrate our main results in section II In the last section, section IV, biological implications and future research directions are outlined

2 Main Results

We find the steady states of the system (1.2) by equating the derivatives on the left-hand sides to zero and solving the resulting algebraic equations It is seen that E1(0, 0); E2(h1

1, 0) are two equilibrium points Apart from that if the system (1.2) has a positive equilibrium point, then it satisfies the following system

(

˙u = 0,

˙v = 0, ⇒

(

1 − Av 1+u+v − h1u = 0, Eu

or equivalently,

v = −h1u 2+ (1 − h1)u + 1

u = h2v

2+ (D + h2)v + D

with u > 0, v > 0

It is easy to classify two trivial equilibrium points E1(0, 0); E2(h11, 0) By simple calculation, the Jacobian matrix of system (1.2) at E1(0, 0) is

1 0

0 −D

 Thus, E1(0, 0) is a saddle point whose stable manifold and unstable manifold are the v-axis and u-axis respectively

For the equilibrium point E2(1

h 1, 0), the Jacobian matrix is

J = −1 h−A1 +1

0 E−D(1+h1 )

h 1 +1

!

If E < (1 + h1)D then det(J) > 0 and trace(J) < 0 which implies E2(h11, 0) to be a stable node point If E > (1 + h1)D then det(J) < 0 Hence, E2(1

h 1, 0) is a saddle point whose stable manifold is u-axis

We now pass to investigate the permanence of the system (1.2) First, we give some definitions which can be referred in [4] and [9] Let R2+ = {(x, y) : x > 0, y > 0}

Definition 2.1 The system (1.2) is called dissipative if there is a bounded set

B ⊂ R2

+ such that, for any (u0, v0) ∈ R2

+, the solution (u(t), v(t)) with the initial condition u(0) = u0, v(0) = v0 satisfies (u(t), v(t)) ∈ B for large enough t

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Definition 2.2 The system (1.2) is said to be (strongly) persistent if lim inf

t→∞ u(t) >

0 and lim inf

t→∞ v(t) > 0 for any solution (u(t), v(t)) starting in any point of intR2

+ Definition 2.3 The system (1.2) is said to be permanent if it is dissipative and persistent

Firstly, we consider the case where h1 = 0 and h2 6= 0, i.e., there is no in-traspecies interference of prey but the environment competition of the predator exists In this case, the system (1.2) has the form

(

˙u = u[1 − Av

1+u+v],

˙v = v[ Eu

Since ˙v = v[ Eu

1+u+v− D − h2v] 6 v[E − D − h2v], it follows that v(t) is bounded

on [0, ∞) when v(0) > 0

Let A 6 1 then 1 − Av

1+u+v > 0 when u > 0 and v > 0 which implies that u(t) is increasing in t and then limt→∞u(t) = ∞

Suppose that A > 1 If the system (2.4) has a positive equilibrium then it satisfies the following system

(

1 − Av 1+u+v = 0, Eu

Let (u∗, v∗) be a positive solution of (2.5) By a simple calculation, we see that

v∗=1 + u∗

A − 1, and u∗ is the solution of the equation

h2Au2− (EA2− 2EA + E − DA2+ DA − 2h2A)u + (h2A + DA2− DA) = 0 (2.6) Since A > 1, h2A + DA2− DA > 0 On the other hand,

∆ = (A − 1)2[(A − 1)2E2− 2(2h2A + DA2− DA)E + D2A2]

This trinomial of second degree ∆ (as a function with variable E) has two solutions E1,2 = A(√h 2 +AD−D±√h 2

A−1 )2 which implies that if A(√h 2 +AD−D−√h 2

A−1 )2 < E < A(√h2 +AD−D+√h 2

A−1 )2then ∆ < 0 In the case where 0 < E 6 A(√h2 +AD−D−√h 2

we see that ∆ > 0 and simultaneously EA2− 2EA + E − DA2+ DA − 2h2A < 0 Thus, when 0 < E < A(√h2 +AD−D+√h 2

At−1 )2, equation (2.6) has also no positive solution, i.e., the system (2.4) has no positive equilibrium point We show that in that case, lim

t→∞u(t) = ∞, i.e., the feeding saturation occurs Indeed, it is easy to see that lim sup

t→∞ u(t) > 0 Suppose that lim inf

t→∞ u(t) = 0 There is a sequence (tn) ↑ ∞ such that lim

n→∞u(tn) = 0 and ˙u(tn) = 0 From the first equation of (2.4) we obtain lim

t n →∞v(tn) = 1

A−1 Therefore, the point (0, 1

A−1) belongs to the ω−limit set of the solution (u(t), v(t)) Since the ω−limit set is invariant, the interval linking the point Asian-European J Math 2011.04:35-48 Downloaded from www.worldscientific.com by UNIVERSIDADE FEDERAL DE SAP PAULO on 08/13/13 For personal use only.

(0, 0) and the point (0, 1

A−1) belongs also to this ω−limit set This is impossible since when v(t) is small, u(t) is increasing By this contradiction it follows that lim inf

t→∞ u(t) > 0 Moreover, if lim sup

t→∞ u(t) < ∞ then lim sup

t→∞

v(t) > 0 By virtue of Bendixson’s theorem (see Theorem 1, page 140, in [8]), there is a periodic orbit in intR2

+ which is contradicting because in a domain surrounded by a periodic orbit, there exists an equilibrium point Thus, lim sup

t→∞ u(t) = ∞ On the other hand, from the boundedless of v(t), there is N such that 1 − 1+u+v(t)Av(t) > 12 for any u > N Hence, u(t) increases when u(t) > N Combining with lim sup

t→∞ u(t) = ∞ we obtain lim

t→∞u(t) = ∞

Let A > 1 and E > A(√h 2 +AD−D+√h 2

A−1 )2 We see that in this case, ∆ > 0 and

EA2−2EA+E −DA2+ DA −2h2A > 0 Therefore, equation (2.6) has two positive solutions

u∗

1= EA

2− 2EA + E − DA2+ DA − 2h2A +√∆

2h2A

u∗2= EA

2− 2EA + E − DA2+ DA − 2h2A −√∆

Hence, the system (2.4) has two positive equilibria, namely E3(u∗

1,uA−1∗+1) and E4(u∗

2,uA−1∗+1) Further, the Jacobian matrix of the system (1.2) at equilibrium E(u∗

i, v∗

i) i = 1, 2 is

Ji=







(A−1)u ∗ i

A(u ∗

i +1) −(A−1)2u∗i

A(u ∗

i +1) E(A+u ∗

i )

A 2 (u ∗

i +1)

E (A−1) 3

u ∗

i −DA 2

(A−1)(u ∗

i +1)−2h 2 A 2

(u ∗

i +1) 2

A 2

(A−1)(u ∗

i +1)







It is easy to show that det(J1) = −√∆u ∗

A 2 (u ∗ +1) < 0, thus, E3is a saddle point Further, det(J2) =

√

∆u∗

A2(u∗+ 1) > 0, trace(J2) =(A

2− A − AE + E + DA2)u∗

2+ DA2

A2(u∗

Therefore, the equilibrium point E4is not stable if (A2− A − AE + E + DA2)u∗

2+

DA2>0 and it is stable if (A2− A − AE + E + DA2)u∗

2+ DA2< 0

Remark 2.1 If h1 = h2 = 0 then the condition E > A(√h2 +AD−D+√h 2

A−1 )2 is equivalent to the condition (3) in [2]

From now on, unless otherwise mention, we assume that h16= 0

Proposition 2.1 Let (u(t), v(t)) be a solution of the system (1.2), starting in a point of R2

+ There hold the following statements:

1) If E 6 (h1+1)D then v(t) convergences exponentially to 0 and lim

t→∞u(t) = h11.

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2) If E > (h1+ 1)D then lim sup

t→∞

u(t) 6 1

h 1 and lim sup

t→∞

v(t) 6E−D(1+h1 )

h 1 D .

By the consequence, the system (1.2) is dissipative on the first quadrant R2

+.

Proof From equation (1.2) we have ˙u 6 u(1 − h1u) which implies that

lim sup t→∞

u(t) 6 1

Therefore, for any ǫ > 0, there is a t1 > 0 such that u(t) < 1

h 1 + ǫ for all t > t1 Hence, Eu(1 + v) 6 Eh11 + ǫ(1 + v) which implies that

Eu

1 + u + v 6

E(1

h 1 + ǫ)

1 + ǫ +h1

1 + v. Substituting this inequality into the second equation of (1.2) we get

˙v(t) < E(

1

h 1 + ǫ)

1 + ǫ +h11+ v(t)− Dv(t) for t > t1 (2.8) Suppose that E 6 (h1+ 1)D We find ǫ > 0 such that E(

1 h1 +ǫ) 1+ǫ+ 1 h1 − D < −ǫ From the relation

˙v(t) < E(

1

h 1 + ǫ)

1 + ǫ + 1

h 1 + v(t)− Dv(t) <E(

1

h 1 + ǫ)

1 + ǫ + 1

h 1

− Dv(t) < −ǫv(t) (2.9) for t > t1, it follows that v(t) convergences exponentially to 0

Given 0 < ε < 1, there exists t2 > 0 such that 1+u(t)+v(t)Av(t) < ε for any t > t2 which implies that

˙u(t) > u(t)(1 − ε − h1u(t)), for any t > t2

Hence,

u(t) > u(t2) exp{(1 − ε)(t − t2)}

1 + u(t2 )h 1

1−ε [exp{(1 − ε)(t − t2)} − 1] for any t > t2.

Thus, lim inf

t→∞ u(t) > 1−εh1 Since ε is arbitrary we see that lim inf

t→∞ u(t) > h11 By using (2.7) we get lim

t→∞u(t) = 1

h 1 Thus 1) is proved

We now assume that E > (h1+ 1)D From (2.8) we see that whenever v(t) > (E−D)(1+εh 1 )

h 1 D − 1 we have ˙v < 0 Therefore, lim sup

t→∞

v(t) 6 (E−D)(1+εh1 )

h 1 D Since ε is arbitrary we get lim sup

t→∞ v(t) 6E−D(h1 +1)

h 1 D Combining with (2.7) we get 2) The proof is complete

Proposition 2.2 If E > (h1+ 1)D then lim inf

t→∞ v(t) > 0 and lim inf

t→∞ u(t) > 0 As

a consequence, if E > (h + 1)D then the system (1.1) is permanent.

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Proof Denote by ω(x, y) the ω−limit set of the solution (u(t), v(t)), starting in (x, y) ∈ R2+ Suppose in the contrary that lim inf

t→∞ v(t) = 0 There are two cases to

be considered:

1) lim t→∞v(t) = 0

Let ε > 0 such that E(1−h1 ε)

1+h 1 − D − h2ε > 0 By 1) of Proposition 2.1, we see that lim

t→∞u(t) = h1

1 Therefore, there exists t3 > 0 such that u(t) > h1

1 − ε and

0 < v(t) < ε for t > t3 Since the function Ex

1+v+x is increasing in x,

˙v(t) v(t) =

Eu

1 + u + v − D − h2v > E(1 − h1 + h11ε)− D − h2ε > 0 ∀ t > t3, which contradicts lim

t→∞v(t) = 0

2) Assume lim sup

t→∞

v(t) > 0

In this case there exists a sequence tn ↑ ∞ such that vn = v(tn) tends to zero and ˙v(tn) = 0 From (2.3) it follows that lim

t→∞u(tn) = E−DD < h11 This property says that the point (E−DD , 0) belongs to the set ω(x, y) Since the set ω(x, y) is invariant, the interval [(0, 0); (0, D

E−D)] ⊂ ω(x, y) That contradicts the fact that (0, 0) is a saddle point Thus, lim inf

t→∞ v(t) > 0

We show that lim inf

t→∞ u(t) > 0 Suppose in the contrary that lim inf

t→∞ u(t) = 0 Then either lim

t→∞u(t) = 0 or lim inf

t→∞ u(t) = 0 but lim sup

t→∞

u(t) > 0

If lim t→∞u(t) = 0 then exists a t4 > 0 such that 1+u(t)+v(t)Eu(t) < D

2 for any t > t4 Therefore, ˙v(t) 6 −D

2v(t), for any t > t4which implies that limt→∞v(t) = 0 This

is a contradiction

If lim inf t→∞ u(t) = 0 and lim sup

t→∞

u(t) > 0 then there exists a sequence tn↑ ∞ such that lim

n→∞u(tn) = 0 and ˙u(tn) = 0 From equation (2.2) it follows that

1 + lim n→∞v(tn) = A lim

If A 6 1, this relation is impossible Let A > 1 From (2.10), lim

n→∞v(tn) = A−11 This means that the point m(0,A−11 ) belongs to the set ω(x, y) Since ω(x, y) is an invariant set, the half straight line [m, ∞) on the vertical axis is a subset of ω(x, y) This contradicts to the fact that our system is dissipative

We study the existence of positive equilibria for the system (1.2) If E 6 (1 + h1)D then system (1.2) has no positive equilibrium because lim

t→∞v(t) = 0 for any initial condition (u(0), v(0)) ∈ R2

+ by Proposition 2.1 We consider the case E > (1 + h1)D Since lim sup

t→∞

u(t) 6 1/h1, for any ε > 0 there is a t5 > 0 such that u(t) 6 mε:= 1/h1+ ε for any t > t5 From the second equation of (1.2) we get

˙v(t) 6 v(t) Emε

1 + mε+ v(t)− D − h2v(t) for any t > t5

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It is easy to see that, for ε small, the equation

Emε

1 + mε+ v − D − h2v = 0 has a unique positive solution, namely v∗

ε By a similar way as in the proof of Proposition2.1we see that

lim sup t→∞

v(t) 6 v∗

ε Hence,

lim sup t→∞

v(t) 6 v∗, where

h1D + h1h2+ h2+p(h1D + h1h2+ h2)2+ 4h1h2(E − D(h1+ 1))

is the unique positive solution of the equation

f (v) := h1h2v2+ (h1D + h1h2+ h2)v + D(h1+ 1) − E = 0 (2.11) Therefore, the positive equilibrium (u, v), if it exists, must satisfy the estimate

Proposition 2.3 Let E > (1 + h1)D Then system (1.2) has at least one positive equilibrium.

Proof By substituting (2.2) into (2.3) we obtained

g(v) = (h1h2E + h22A)v3+ (2h1h2E + h1DE + h2E − 2EAh2+ 2ADh2)v2 + (h1h2E + 2h1DE + h2E + DE − 2ADE + AE2+ AD2− E2)v

+ E(h1D + D − E)

We have

g(v∗) = Ev∗[h1h2v∗2+ (h1D + h1h2+ h2)v∗+ D(h1+ 1) − E]

+ h22Av∗3+ (h1h2E − 2h2AE + 2h2AD)v∗2 + (h1h2E + h1DE + h2E − 2ADE + AE2+ AD2)v∗+ h1DE + DE − E2

= h22Av∗3− 2h2AEv∗2+ 2h2ADv∗2− 2ADEv∗2+ AE2v∗+ AD2v∗

= Av∗[h22v∗2− 2h2Ev∗+ 2h22Dv∗− 2DE + E2+ D2]

= Av∗(h2v∗− E + D)2> 0

Hence, the system (2.1) has at least one solution (u, v) with 0 < v < v∗ It is easy

to check v∗ < E−Dh2 which implies that u > 0 Thus, the system (2.1) has at least one positive solution (u, v) The proof is complete

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M = h1h2E + h2A, N = 2h1h2E + h1DE + h2E − 2EAh2+ 2ADh2,

P = h1h2E + 2h1DE + h2E + DE − 2ADE + AE2+ AD2− E2,

Q = E(h1D + D − E)

Proposition 2.4 Let E > (1+h1)D and one of the following conditions is satisfied

i) P 6 0, ii) P > 0 and N > 0, iii) P > 0, N < 0 and N2− 3MP 6 0, iv) P > 0, N < 0, N2− 3MP > 0 and 9MQ − NP > 0.

Then the positive equilibrium of the system (1.2) is unique.

Proof By substituting (2.2) into (2.1) we obtained g(v) = M v3+ N v2+ P v + Q According to Proposition2.3, the equation

has at least one positive solution Further, from g′(v) = 3M v2+ 2N v + P we get: i) Let P < 0 The equation g′(v) = 0 has a positive root and a negative one v1< 0 < v2 Because g(0) < 0 and g(+∞) = +∞ then v1is the local maximum point and v2 is the local minimum one Hence, there exists a unique positive solution of the system (1.2) If P = 0 then the equation g′(v) = 3M v2+ 2N v has a solution v1 = 0 This means that 0 is an extreme point of the function g(v) Since g(0) < 0, we also conclude that the system (2.13) has only one positive solution

ii) In the case where P > 0 and N > 0, we have g′(v) > 0 for any v > 0 which implies that g(v) is increasing in v on (0, ∞) Since g(0) < 0 and g(+∞) = +∞, equation (2.13) has a unique positive solution

iii) If P > 0, N < 0 and N2− 3MP 6 0 then the function g(v) is monotonous increasing on R Hence, equation (2.13) has unique solution

iv) If P > 0, N < 0, N2− 3MP > 0 and 9MQ − NP < 0 then the function g(v) has the maximum value and the minimum value and the product of these values is positive Hence, g(v) = 0 has a unique solution

Summing up, in these cases, equation (2.13) has a unique positive solution This mean that the positive equilibrium of the system (1.2) is unique

Proposition 2.5 If one of the following conditions is satisfied

a) A < E, b) A = E and h1+ h26= 0,

c) A > E and h = h2= 0, Asian-European J Math 2011.04:35-48 Downloaded from www.worldscientific.com by UNIVERSIDADE FEDERAL DE SAP PAULO on 08/13/13 For personal use only.

d) A > E, h2+ h26= 0 and adding one of three conditions

i) 4h1AD + 4h1h2+ 4h2E − 4h1h2D + AE2+ h1h2E − A2E − h1h2A > 0,

ii) 8h1AD + 4h1h2+ 4h2E − 4h1h2D + 4E 2− 4h1ED − 4AE > 0, iii) 4h1AD + 4h1h2+ 4h2E − 4h1h2D + AE2+ h1h2E − A2E − h1h2A > 0

Then there is no periodic orbit of system (1.2).

Proof We use the Dulac’s criterion (see [8], Theorem 2) to prove this theorem

u α+1 v β+1 and F = (F1, F2) where F1(u, v) = uh1 − Av

1+u+v − h1ui and F2(u, v) = vh1+u+vEu − D − h2vi Consider the Dulac’s function

▽(ϕF (u, v)) =∂u∂ (ϕ(u, v)F1(u, v)) + ∂

∂v(ϕ(u, v)F2(u, v)).

By direct computation we have

▽ (ϕF ) = uα+11vβ+1h− α + Dβ +1 + u + vAαv

(1 + u + v)2 −1 + u + vEβu −(1 + u + v)Euv 2 + h1(α − 1)u + h2(β − 1)vi a) Let A < E, we choose α = β = 0 We see that ▽(ϕF ) < 0

b) If A = E and h1+ h26= 0,we also choose α = β = 0 and obtain ▽(ϕF ) < 0 c) In the case A > E and h1= h2= 0 we can refer in [2]

d) We have

▽(ϕF ) 6uα+11vβ+1h−α+Dβ+m(A − E)4 +Aα + h2(β − 1) + n(A − E)

+−Eβ + h1(α − 1) + p(A − E)

h1(α − 1)(u + v)u (1 + u + v) +

h2(β − 1)(u + v)v (1 + u + v)

i Where the positive numbers m, n and p satisfy the relation m + n + p = 1 and

α 6 1, β 6 1 If A > E, h1+ h26= 0 and either i) or ii) or iii) holds then there exist positive numbers m, n, p such that

(AE + h1h2)m + (4E − 4h1D)n + (4AD + 4h2)p 6 4h1+ 4h1h2+ 4h2E − 4h1h2D

Let α = (mE+4pD)(A−E)−4h1 D

4(E−h 1 D) and β = −4h1 +(4p+mh 1 )(A−E)

4(E−h 1 D) It is easy to verify

−α + Dβ +m(A−E)4 60, Aα + h2(β − 1) + n(A − E) 6 0 and −Eβ + h1(α − 1) + p(A − E) 6 0 Hence ▽(ϕF ) 6 0 and is not identically zero By using Dulac’s criterion ([8], Theorem 2) it follows that there is no periodic orbit for the system (1.2)

Corollary 2.2 If the conditions mentioned in Propositions 2.4 and 2.5 are satis-fied, then the unique positive equilibrium is globally stable.

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