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

February 10, 2011 11:18 WSPC/INSTRUCTION FILE 04˙Dutrung Asian-European Journal of Mathematics Vol 4, No (2011) 35–48 c World Scientific Publishing Company DOI: 10.1142/S1793557111000058 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 ON THE DYNAMICS OF PREDATOR-PREY SYSTEMS WITH BEDDINGTON-DEANGELIS FUNCTIONAL RESPONSE Nguyen Huu Du∗ 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 response 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 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 f xy b+wx+y ef xy b+wx+y − dy x˙ = ax − y˙ = − g1 x2 − g2 y , (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 ∗ Corresponding author 35 February 10, 2011 11:18 WSPC/INSTRUCTION FILE 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 36 04˙Dutrung N H Du & T T Trung 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 correlates 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 classic 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 y f ef d After a change of variables (u = wx b , v = b , A = a , D = a , E = wa and g2 b g1 b h1 = aw , h2 = a ) and an appropriate time rescaling (t → at), we receive the following model: Av 1+u+v − h1 u], Eu v[ 1+u+v − D − h2 v] u˙ = u[1 − v˙ = (1.2) 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 = 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 h2 = If h2 = but h1 = 0, we show conditions for the existence of positive equilibria and classify them For the case h1 and h2 are simultaneously 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 cumbersome 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 February 10, 2011 11:18 WSPC/INSTRUCTION FILE 04˙Dutrung On the Dynamics · · · 37 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 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 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 ( h11 , 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, Av 1+u+v − h1 u = 0, Eu 1+u+v − D − h2 v = 0, 1− (2.1) or equivalently, −h1 u2 + (1 − h1 )u + , h1 u + A − h2 v + (D + h2 )v + D , u= E − D − h2 v v= (2.2) (2.3) with u > 0, v > 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 −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 ( h11 , 0), the Jacobian matrix is J= −1 −A h1 +1 E−D(1+h1 ) h1 +1 If E < (1 + h1 )D then det(J) > and trace(J) < which implies E2 ( h11 , 0) to be a stable node point If E > (1 + h1 )D then det(J) < Hence, E2 ( h11 , 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 February 10, 2011 11:18 WSPC/INSTRUCTION FILE 38 04˙Dutrung N H Du & T T Trung Definition 2.2 The system (1.2) is said to be (strongly) persistent if lim inf u(t) > t→∞ and lim inf v(t) > for any solution (u(t), v(t)) starting in any point of intR2+ t→∞ Definition 2.3 The system (1.2) is said to be permanent if it is dissipative and persistent 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 Firstly, we consider the case where h1 = and h2 = 0, i.e., there is no intraspecies interference of prey but the environment competition of the predator exists In this case, the system (1.2) has the form Av 1+u+v ], Eu −D− v[ 1+u+v u˙ = u[1 − v˙ = (2.4) h2 v] Eu − D − h2 v] v[E − D − h2 v], it follows that v(t) is bounded Since v˙ = v[ 1+u+v on [0, ∞) when v(0) > Av > when u > and v > which implies that u(t) is Let A then − 1+u+v increasing in t and then limt→∞ u(t) = ∞ Suppose that A > If the system (2.4) has a positive equilibrium then it satisfies the following system Av 1+u+v = 0, Eu 1+u+v − D − h2 v 1− (2.5) = Let (u∗ , v ∗ ) be a positive solution of (2.5) By a simple calculation, we see that v∗ = and u∗ is the solution of the equation + u∗ , A−1 h2 Au2 − (EA2 − 2EA + E − DA2 + DA − 2h2 A)u + (h2 A + DA2 − DA) = (2.6) Since A > 1, h2 A + DA2 − DA > On the other hand, ∆ = (A − 1)2 [(A − 1)2 E − 2(2h2 A + DA2 − DA)E + D2 A2 ] This trinomial E) has√ two solutions √ of second√degree ∆ (as a function with variable √ h2 h2 +AD−D− h2 E1,2 = A( h2 +AD−D± ) which implies that if A( ) < E < A−1 A−1 √ √ √ √ h2 h2 A( h2 +AD−D+ ) then ∆ < In the case where < E A( h2 +AD−D− ) A−1 A−1 2 we see that ∆ and simultaneously EA − 2EA + E − DA + DA − 2h A < √ √ h2 has also no positive ) , equation (2.6) Thus, when < E < A( h2 +AD−D+ At−1 solution, i.e., the system (2.4) has no positive equilibrium point We show that in that case, lim u(t) = ∞, i.e., the feeding saturation occurs Indeed, it is easy to see t→∞ that lim sup u(t) > Suppose that lim inf u(t) = There is a sequence (tn ) ↑ ∞ t→∞ t→∞ such that lim u(tn ) = and u(t ˙ n ) = From the first equation of (2.4) we obtain n→∞ lim v(tn ) = tn →∞ A−1 Therefore, the point (0, 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 February 10, 2011 11:18 WSPC/INSTRUCTION FILE 04˙Dutrung On the Dynamics · · · 39 (0, 0) and the point (0, 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 u(t) > Moreover, if lim sup u(t) < ∞ then lim sup v(t) > By virtue of t→∞ t→∞ t→∞ 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 u(t) = ∞ On the other hand, from 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 t→∞ Av(t) the boundedless of v(t), there is N such that − 1+u+v(t) for any u > N Hence, u(t) increases when u(t) > N Combining with lim sup u(t) = ∞ we obtain t→∞ lim u(t) = ∞ t→∞ √ √ h2 Let A > and E A( h2 +AD−D+ ) We see that in this case, ∆ > and A−1 2 EA − 2EA + E − DA + DA − 2h2A > Therefore, equation (2.6) has two positive solutions √ EA2 − 2EA + E − DA2 + DA − 2h2 A + ∆ ∗ u1 = 2h2 A √ 2 EA − 2EA + E − DA + DA − 2h2 A − ∆ ∗ u2 = 2h2 A Hence, the system (2.4) has two positive equilibria, namely E3 (u∗1 , u∗ +1 ) Further, E4 (u∗2 , A−1 E(u∗i , vi∗ ) i = 1, is   Ji =   u∗ +1 A−1 ) and the Jacobian matrix of the system (1.2) at equilibrium (A−1)u∗ i A(u∗ i +1) −  (A−1)2 u∗ i A(u∗ i +1) ∗ ∗ ∗ E(A+u∗ i ) E(A−1) ui −DA (A−1)(ui +1)−2h2 A (ui +1) A2 (u∗ A2 (A−1)(u∗ i +1) i +1) √ ∗    It is easy to show that det(J1 ) = A−2 (u∆u ∗ +1) < 0, thus, E3 is a saddle point Further, √ ∗ ∆u det(J2 ) = ∗ > 0, A (u + 1) (A2 − A − AE + E + DA2 )u∗2 + DA2 trace(J2 ) = A2 (u∗2 + 1) Therefore, the equilibrium point E4 is not stable if (A2 − A − AE + E + DA2 )u∗2 + DA2 and it is stable if (A2 − A − AE + E + DA2 )u∗2 + DA2 < Remark 2.1 If h1 = h2 = then the condition E > A( equivalent to the condition (3) in [2] √ √ h2 +AD−D+ h2 ) A−1 is From now on, unless otherwise mention, we assume that h1 = 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 (h1 +1)D then v(t) convergences exponentially to and lim u(t) = h11 t→∞ February 10, 2011 11:18 WSPC/INSTRUCTION FILE 40 04˙Dutrung N H Du & T T Trung h1 2) If E > (h1 + 1)D then lim sup u(t) t→∞ E−D(1+h1 ) h1 D and lim sup v(t) t→∞ By the consequence, the system (1.2) is dissipative on the first quadrant R2+ Proof From equation (1.2) we have u˙ u(1 − h1 u) which implies that h1 lim sup u(t) t→∞ (2.7) h1 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 Therefore, for any ǫ > 0, there is a t1 > such that u(t) < Hence, Eu(1 + v) h1 E + ǫ for all t > t1 + ǫ (1 + v) which implies that E( h11 + ǫ) Eu 1+u+v 1+ǫ+ h1 +v Substituting this inequality into the second equation of (1.2) we get E( h11 + ǫ) v(t) ˙ < Suppose that E 1+ǫ+ h1 + v(t) − D v(t) (h1 + 1)D We find ǫ > such that for t t1 E( h1 +ǫ) 1+ǫ+ h1 relation E( h11 + ǫ) v(t) ˙ < 1+ǫ+ h1 + v(t) − D v(t) < E( h11 + ǫ) 1+ǫ+ h1 (2.8) − D < −ǫ From the − D v(t) < −ǫv(t) for t t1 , it follows that v(t) convergences exponentially to Av(t) Given < ε < 1, there exists t2 > such that 1+u(t)+v(t) < ε for any t which implies that u(t) ˙ u(t)(1 − ε − h1 u(t)), for any t (2.9) t2 t2 Hence, u(t) Thus, lim inf u(t) t→∞ 1+ u(t2 ) exp{(1 − ε)(t − t2 )} u(t2 )h1 1−ε [exp{(1 − ε)(t − t2 )} 1−ε h1 (2.7) we get lim u(t) = t→∞ − 1] for any t t2 Since ε is arbitrary we see that lim inf u(t) h1 t→∞ h1 By using Thus 1) is proved We now assume that E > (h1 + 1)D From (2.8) we see that whenever v(t) > (E−D)(1+εh1 ) (E−D)(1+εh1 ) − we have v˙ < Therefore, lim sup v(t) Since ε is h1 D h1 D t→∞ arbitrary we get lim sup v(t) t→∞ E−D(h1 +1) h1 D Combining with (2.7) we get 2) The proof is complete Proposition 2.2 If E > (h1 + 1)D then lim inf v(t) > and lim inf u(t) > As t→∞ t→∞ a consequence, if E > (h1 + 1)D then the system (1.1) is permanent February 10, 2011 11:18 WSPC/INSTRUCTION FILE 04˙Dutrung On the Dynamics · · · 41 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 v(t) = There are two cases to t→∞ be considered: 1) lim v(t) = t→∞ ε) Let ε > such that E(1−h − D − h2 ε > By 1) of Proposition 2.1, we see 1+h1 1 that lim u(t) = h1 Therefore, there exists t3 > such that u(t) h1 − ε and t→∞ 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 < v(t) < ε for t t3 Since the function Ex 1+v+x is increasing in x, v(t) ˙ Eu E(1 − h1 ε) − D − h2 ε > = − D − h2 v > v(t) 1+u+v + h1 ∀t t3 , which contradicts lim v(t) = t→∞ 2) Assume lim sup v(t) > t→∞ In this case there exists a sequence tn ↑ ∞ such that = v(tn ) tends to zero D < h11 This property and v(t ˙ n ) = From (2.3) it follows that lim u(tn ) = E−D t→∞ D , 0) belongs to the set ω(x, y) Since the set ω(x, y) is says that the point ( E−D D invariant, the interval [(0, 0); (0, E−D )] ⊂ ω(x, y) That contradicts the fact that (0, 0) is a saddle point Thus, lim inf v(t) > t→∞ We show that lim inf u(t) > Suppose in the contrary that lim inf u(t) = t→∞ t→∞ Then either lim u(t) = or lim inf u(t) = but lim sup u(t) > t→∞ t→∞ t→∞ Eu(t) 1+u(t)+v(t) If lim u(t) = then exists a t4 > such that t→∞ < D for any t t4 −D v(t), for any t t4 which implies that limt→∞ v(t) = This Therefore, v(t) ˙ is a contradiction If lim inf u(t) = and lim sup u(t) > then there exists a sequence tn ↑ ∞ such t→∞ t→∞ that lim u(tn ) = and u(t ˙ n ) = From equation (2.2) it follows that n→∞ + lim v(tn ) = A lim v(tn ) n→∞ If A (2.10) n→∞ 1, this relation is impossible Let A > From (2.10), lim v(tn ) = n→∞ A−1 This means that the point m(0, A−1 ) 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 (1 + h1 )D then system (1.2) has no positive equilibrium because lim v(t) = for any t→∞ initial condition (u(0), v(0)) ∈ R2+ by Proposition 2.1 We consider the case E > (1 + h1 )D Since lim sup u(t) 1/h1 , for any ε > there is a t5 > such that t→∞ u(t) mε := 1/h1 + ε for any t v(t) ˙ v(t) t5 From the second equation of (1.2) we get Emε − D − h2 v(t) + mε + v(t) for any t t5 February 10, 2011 11:18 WSPC/INSTRUCTION FILE 42 04˙Dutrung N H Du & T T Trung It is easy to see that, for ε small, the equation Emε − D − h2 v = + mε + v has a unique positive solution, namely vε∗ By a similar way as in the proof of Proposition 2.1 we see that vε∗ lim sup v(t) t→∞ 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 Hence, v∗ , lim sup v(t) t→∞ where v∗ = h1 D + h1 h2 + h2 + 2(E − D(h1 + 1)) (h1 D + h1 h2 + h2 )2 + 4h1 h2 (E − D(h1 + 1)) is the unique positive solution of the equation f (v) := h1 h2 v + (h1 D + h1 h2 + h2 )v + D(h1 + 1) − E = (2.11) Therefore, the positive equilibrium (u, v), if it exists, must satisfy the estimate 0 (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) = (h1 h2 E + h22 A)v + (2h1 h2 E + h1 DE + h2 E − 2EAh2 + 2ADh2 )v + (h1 h2 E + 2h1 DE + h2 E + DE − 2ADE + AE + AD2 − E )v + E(h1 D + D − E) We have g(v ∗ ) = Ev ∗ [h1 h2 v ∗ + (h1 D + h1 h2 + h2 )v ∗ + D(h1 + 1) − E] + h22 Av ∗ + (h1 h2 E − 2h2 AE + 2h2 AD)v ∗ + (h1 h2 E + h1 DE + h2 E − 2ADE + AE + AD2 )v ∗ + h1 DE + DE − E = h22 Av ∗ − 2h2 AEv ∗ + 2h2 ADv ∗ − 2ADEv ∗ + AE v ∗ + AD2 v ∗ = Av ∗ [h22 v ∗ − 2h2 Ev ∗ + 2h22 Dv ∗ − 2DE + E + D2 ] = Av ∗ (h2 v ∗ − E + D)2 > Hence, the system (2.1) has at least one solution (u, v) with < v < v ∗ It is easy to check v ∗ < E−D which implies that u > Thus, the system (2.1) has at least h2 one positive solution (u, v) The proof is complete February 10, 2011 11:18 WSPC/INSTRUCTION FILE 04˙Dutrung On the Dynamics · · · 43 Denote M = h1 h2 E + h22 A, N = 2h1 h2 E + h1 DE + h2 E − 2EAh2 + 2ADh2 , P = h1 h2 E + 2h1 DE + h2 E + DE − 2ADE + AE + AD2 − E , Q = E(h1 D + D − E) 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 Proposition 2.4 Let E > (1+h1 )D and one of the following conditions is satisfied i) ii) iii) iv) P 0, P > and N 0, P > 0, N < and N − 3M P 0, P > 0, N < 0, N − 3M P > and 9M Q − N P > Then the positive equilibrium of the system (1.2) is unique Proof By substituting (2.2) into (2.1) we obtained g(v) = M v + N v + P v + Q According to Proposition 2.3, the equation g(v) = (2.13) has at least one positive solution Further, from g ′ (v) = 3M v + 2N v + P we get: i) Let P < The equation g ′ (v) = has a positive root and a negative one v1 < < v2 Because g(0) < and g(+∞) = +∞ then v1 is 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 = then the equation g ′ (v) = 3M v + 2N v has a solution v1 = This means that 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 > and N 0, we have g ′ (v) > for any v > which implies that g(v) is increasing in v on (0, ∞) Since g(0) < and g(+∞) = +∞, equation (2.13) has a unique positive solution iii) If P > 0, N < and N − 3M P then the function g(v) is monotonous increasing on R Hence, equation (2.13) has unique solution iv) If P > 0, N < 0, N − 3M P > and 9M Q − N P < then the function g(v) has the maximum value and the minimum value and the product of these values is positive Hence, g(v) = 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 + h2 = 0, c) A > E and h1 = h2 = 0, February 10, 2011 11:18 WSPC/INSTRUCTION FILE 44 04˙Dutrung N H Du & T T Trung d) A > E, h21 + h22 = and adding one of three conditions i) 4h1 AD + 4h1 h2 + 4h2 E − 4h1 h2 D + AE + h1 h2 E − A2 E − h1 h2 A ii) 8h1 AD + 4h1 h2 + 4h2 E − 4h1 h2 D + 4E − 4h1 ED − 4AE 0, iii) 4h1 AD + 4h1 h2 + 4h2 E − 4h1 h2 D + AE + h1 h2 E − A2 E − h1 h2 A 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 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 Av Let ϕ = uα+11vβ+1 and F = (F1 , F2 ) where F1 (u, v) = u − 1+u+v − h1 u and Eu 1+u+v F2 (u, v) = v − D − h2 v Consider the Dulac’s function ▽(ϕF (u, v)) = ∂ ∂ (ϕ(u, v)F1 (u, v)) + (ϕ(u, v)F2 (u, v)) ∂u ∂v By direct computation we have Aαv 1+u+v Auv Eβu Euv + − + h1 (α − 1)u + h2 (β − 1)v − (1 + u + v)2 1+u+v (1 + u + v)2 ▽ (ϕF ) = a) b) c) d) uα+1 v β+1 − α + Dβ + Let A < E, we choose α = β = We see that ▽(ϕF ) < If A = E and h1 + h2 = 0,we also choose α = β = and obtain ▽(ϕF ) < In the case A > E and h1 = h2 = we can refer in [2] We have m(A − E) Aα + h2 (β − 1) + n(A − E) + v −α+Dβ + uα+1 v β+1 1+u+v −Eβ + h1 (α − 1) + p(A − E) h1 (α − 1)(u + v)u h2 (β − 1)(u + v)v + u+ + 1+u+v (1 + u + v) (1 + u + v) ▽(ϕF ) Where the positive numbers m, n and p satisfy the relation m + n + p = and α 1, β If A > E, h1 + h2 = and either i) or ii) or iii) holds then there exist positive numbers m, n, p such that (AE + h1 h2 )m + (4E − 4h1 D)n + (4AD + 4h2 )p (mE+4pD)(A−E)−4h1 D 4(E−h1 D) m(A−E) Dβ + 0, Aα + Let α = and β = 4h1 + 4h1 h2 + 4h2 E − 4h1 h2 D A−E −4h1 +(4p+mh1 )(A−E) 4(E−h1 D) It is easy to verify −α + h2 (β − 1) + n(A − E) and −Eβ + h1 (α − 1) + p(A − E) Hence ▽(ϕF ) 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 satisfied, then the unique positive equilibrium is globally stable February 10, 2011 11:18 WSPC/INSTRUCTION FILE 04˙Dutrung On the Dynamics · · · 45 Remark 2.3 In the case A = E and h1 = h2 = 0, Dobromir T Dimitrov, Hristo V Kojouharov [2] had proved that: If AE − E − AD > then the interior equilibrium (u∗ , v ∗ ) is a global center, which means that all trajectories (except one starting in (u∗ , v ∗ )) are periodic orbits containing (u∗ , v ∗ ) in their interior Examples We can show here some examples to illustrate our results 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 Example 3.1 10v(t) − u(t) , + u(t) + v(t) 8u(t) − 0.6 − 0.5v(t) v(t) ˙ = v(t) + u(t) + v(t) u(t) ˙ = u(t) − (3.1) In this case, A = 10, D = 0.6, E = 8, h1 = 1; h2 = 0.5 and P = 506 > 0, N = −57.20 < 0, N − 3M P = −6595.16 < Therefore, there exists a unique positive equilibrium for the system (3.1) and this equilibrium is globally stable Fig The unique equilibrium point of the system (3.1) is globally stable Example 3.2 We consider the system 3.8v(t) − 0.03u(t) , (1 + u(t) + v(t) 3u(t) − − 0.5v(t) v(t) ˙ = v(t) + u(t) + v(t) u(t) ˙ = u(t) − (3.2) ... 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... w is the weighting factor, that correlates inversely with the prey density at which feeding saturation occurs and b is a normalization coefficient that relates the densities of the predator and... Michaelis-Menten-Holling type functional response) Independently, Beddington [6] offered the same form of a functional response for describing parasite-host interactions y f ef d After a change of variables (u