03964.com

文档资料库 文档搜索专家

文档资料库 文档搜索专家

Nonlinear Analysis: Real World Applications 7 (2006) 1104 – 1118 www.elsevier.com/locate/na

Analysis of a predator–prey model with modi?ed Leslie–Gower and Holling-type II schemes with time delay

A.F. Nindjina , M.A. Aziz-Alaouib,? , M. Cadivelb

a Laboratoire de Mathématiques Appliquées, Université de Cocody, 22 BP 582, Abidjan 22, C?te d’Ivoire, France b Laboratoire de Mathématiques Appliquées, Université du Havre, 25 rue Philippe Lebon, B.P. 540, 76058 Le Havre Cedex, France

Received 17 July 2005; accepted 7 October 2005

Abstract Two-dimensional delayed continuous time dynamical system modeling a predator–prey food chain, and based on a modi?ed version of Holling type-II scheme is investigated. By constructing a Liapunov function, we obtain a suf?cient condition for global stability of the positive equilibrium. We also present some related qualitative results for this system. 2005 Elsevier Ltd. All rights reserved.

Keywords: Time delay; Boundedness; Permanent; Local stability; Global stability; Liapunov functional

1. Introduction The dynamic relationship between predators and their prey has long been and will continue to be one of dominant themes in both ecology and mathematical ecology due to its universal existence and importance. A major trend in theoretical work on prey–predator dynamics has been to derive more realistic models, trying to keep to maximum the unavoidable increase in complexity of their mathematics. In this optic, recently [2], see also [1,5,6] has proposed a ?rst study of two-dimensional system of autonomous differential equation modeling a predator prey system. This model incorporates a modi?ed version of Leslie–Gower functional response as well as that of the Holling-type II. They consider the following model ? c y ? x = a1 ? bx ? 1 x, ?˙ ? x + k1 (1) ? ? ? y = a2 ? c2 y ˙ y x + k2 with the initial conditions x(0) > 0 and y(0) > 0. This two species food chain model describes a prey population x which serves as food for a predator y. The model parameters a1 , a2 , b, c1 , c2 , k1 and k2 are assuming only positive values. These parameters are de?ned as follows: a1 is the growth rate of prey x, b measures the strength of competition among individuals of species x, c1

? Corresponding author. Tel./fax: +1 33 2 32 74 4.

E-mail address: Aziz-Alaoui@univ-lehavre.fr (M.A. Aziz-Alaoui). 1468-1218/$ - see front matter 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.nonrwa.2005.10.003

A.F. Nindjin et al. / Nonlinear Analysis: Real World Applications 7 (2006) 1104 – 1118

1105

is the maximum value of the per capita reduction rate of x due to y, k1 (respectively, k2 ) measures the extent to which environment provides protection to prey x (respectively, to the predator y), a2 describes the growth rate of y, and c2 has a similar meaning to c1 . It was ?rst motivated more by the mathematics analysis interest than by its realism as a model of any particular natural dynamical system. However, there may be situations in which the interaction between species is modelized by systems with such a functional response. It may, for example, be considered as a representation of an insect pest–spider food chain. Furthermore, it is a ?rst step towards a predator–prey model (of Holling–Tanner type) with inverse trophic relation and time delay, that is where the prey eaten by the mature predator can consume the immature predators. Let us mention that the ?rst equation of system (1) is standard. By contrast, the second equation is absolutely not standard. This intactness model contains a modi?ed Leslie–Gower term, the second term on the right-hand side in the second equation of (1). The last depicts the loss in the predator population. The Leslie–Gower formulation is based on the assumption that reduction in a predator population has a reciprocal relationship with per capita availability of its preferred food. Indeed, Leslie introduced a predator prey model where the carrying capacity of the predator environment is proportional to the number of prey. He stresses the fact that there are upper limits to the rates of increase of both prey x and predator y, which are not recognized in the Lotka–Volterra model. In case of continuous time, the considerations lead to the following: dy y = a2 y 1 ? , dt x in which the growth of the predator population is of logistic form, i.e. dy y = a2 y 1 ? . dt C Here, “C” measures the carry capacity set by the environmental resources and is proportional to prey abundance, C = x, where is the conversion factor of prey into predators. The term y/ x of this equation is called the Leslie–Gower term. It measures the loss in the predator population due to the rarity (per capita y/x) of its favorite food. In the case of severe scarcity, y can switch over to other population, but its growth will be limited by the fact that its most favorite food, the prey x, is not available in abundance. The situation can be taken care of by adding a positive constant to the denominator, hence the equation above becomes, dy y = a2 y 1 ? dt x+d and thus, ? dy a2 ? = y ?a2 ? . dt y x+ ? ? d?

that is the second equation of system (1). In this paper, we introduce time delays in model (1), which is a more realistic approach to the understanding of predator–prey dynamics. Time delay plays an important role in many biological dynamical systems, being particularly relevant in ecology, where time delays have been recognized to contribute critically to the stable or unstable outcome of prey densities due to predation. Therefore, it is interesting and important to study the following delayed modi?ed Leslie–Gower and Holling-Type-II schemes: ? c y(t) ? x(t) = a1 ? bx(t) ? 1 x(t), ?˙ ? x(t) + k1 (2) ? ? ? y(t) = a2 ? c2 y(t ? r) ˙ y(t) x(t ? r) + k2 for all t > 0. Here, we incorporate a single discrete delay r > 0 in the negative feedback of the predator’s density.

1106

A.F. Nindjin et al. / Nonlinear Analysis: Real World Applications 7 (2006) 1104 – 1118

Let us denote by R2 the nonnegative quadrant and by int(R2 ) the positive quadrant. For + + following conventional notation: xt ( ) = x( + t). Then the initial conditions for this system take the form x0 ( ) =

1 ( ), y0 ( ) =

2 ( )

∈ [?r, 0], we use the

(3)

for all ∈ [?r, 0], where (

1 ,

2 ) ∈ C([?r, 0], R2 ), x(0) =

1 (0) > 0 and y(0) =

2 (0) > 0. + It is well known that the question of global stability of the positive steady state in a predator–prey system, with a single discrete delay in the predator equation without instantaneous negative feedback, remains a challenge, see [3,5,7]. Our main purpose is to present some results about the global stability analysis on a system with delay containing modi?ed Leslie–Gower and Holling-Type-II terms. This paper is organized as follows. In the next section, we present some preliminary results on the boundedness of solutions for system (2)–(3). Next, we study some equilibria properties for this system and give a permanence result. In Section 5, the analysis of the global stability is made for a boundary equilibrium and suf?cient conditions are provided for the positive equilibrium of both instantaneous system (1) and system with delay (2)–(3) to be globally asymptotically stable. Finally, a discussion which includes local stability results for system (2)–(3) is given. 2. Preliminaries In this section, we present some preliminary results on the boundedness of solutions for system (2)–(3). We consider (x, y) a noncontinuable solution, see [4], of system (2)–(3), de?ned on [?r, A[, where A ∈]0, +∞]. Lemma 1. The positive quadrant int(R2 ) is invariant for system (2). + Proof. We have to show that for all t ∈ [0, A[, x(t) > 0 and y(t) > 0. Suppose that is not true. Then, there exists 0 < T < A such that for all t ∈ [0, T [, x(t) > 0 and y(t) > 0, and either x(T ) = 0 or y(T ) = 0. For all t ∈ [0, T [, we have x(t) = x(0) exp

0 t

a1 ? bx(s) ?

c1 y(s) ds x(s) + k1

(4)

and y(t) = y(0) exp

0 t

a2 ?

c2 y(s ? r) ds . x(s ? r) + k2 0 such that for all t ∈ [?r, T [, c1 y(s) ds x(s) + k1

(5)

As (x, y) is de?ned and continuous on [?r, T ], there is a M x(t) = x(0) exp

0 t

a1 ? bx(s) ?

x(0) exp(?T M)

and y(t) = y(0) exp

0 t

a2 ?

c2 y(s ? r) ds x(s ? r) + k2

y(0) exp(?T M).

Taking the limit, as t → T , we obtain x(T ) x(0) exp(?T M) > 0 and y(T ) y(0) exp(?T M) > 0, which contradicts the fact that either x(T ) = 0 or y(T ) = 0. So, for all t ∈ [0, A[, x(t) > 0 and y(t) > 0.

A.F. Nindjin et al. / Nonlinear Analysis: Real World Applications 7 (2006) 1104 – 1118

1107

Lemma 2. For system (2)–(3), A = +∞ and lim sup x(t) K

t→+∞

(6)

and lim sup y(t) L

t→+∞

(7)

where K = a1 /b and L = a2 /c2 (K + k2 )ea2 r . Proof. From the ?rst equation of system (2)–(3), we have for all t ∈ [0, A[, x(t) < x(t)(a1 ? bx(t)). ˙ A standard comparison argument shows that for all t ∈ [0, A[, x(t) x(t) where x is the solution of the following ? ? ordinary differential equation ˙ x(t) = x(t)(a1 ? bx(t)), ? ? ? x(0) = x(0) > 0. ? As limt→+∞ x(t) = a1 /b, then x and thus x is bounded on [0, A[. Moreover, from Eq. (5), we can de?ne y on all ? ? interval [kr, (k + 1)r], with k ∈ N, and it is easy to see that y is bounded on [0, A[ if A < + ∞. Then A = +∞, see [4, Theorem 2.4]. Now, as for all t 0, x(t) x(t), then ? lim sup x(t)

t→+∞

lim sup x(t) = K. ?

t→+∞

From the predator equation, we have y(t) < a2 y(t), ˙ hence, for t > r, y(t) y(t ? r)ea2 r , which is equivalent, for t > r, to y(t ? r) y(t)e?a2 r . Moreover, for any t > T + r, (8)

> 1, there exists positive T , such that for t > T , x(t) < K. According to (8), we have, for c2 e?a2 r y(t) , K + k2

y(t) < y(t) a2 ? ˙

which implies by the same arguments use for x that, lim sup y(t) L ,

t→+∞

where L = a2 /c2 ( K + k2 )ea2 r . Conclusion of this lemma holds by letting 3. Equilibria

→ 1.

In this section we study some equilibria properties of system (2)–(3). These steady states are determined analytically by setting x = y = 0. They are independent of the delay r. It is easy to verify that this system has three trivial boundary ˙ ˙ equilibria, E0 = (0, 0), E1 = (a1 /b, 0) and E2 (0, a2 k2 /c2 ).

1108

A.F. Nindjin et al. / Nonlinear Analysis: Real World Applications 7 (2006) 1104 – 1118

Proposition 3. System (2)–(3) has a unique interior equilibrium E ? = (x ? , y ? ) (i.e. x ? > 0 and y ? > 0) if the following condition holds a2 k2 a1 k1 < . c2 c1 Proof. From system (2)–(3), such a point satis?es (a1 ? bx ? )(x ? + k1 ) = c1 y ? and y? = a2 (x ? + k2 ) . c2 (11) (10) (9)

? ? ? ? If (9) holds, this system has two solutions (x+ , y+ ) and (x? , y? ) given by ? ? x ? = 1 (?(c1 a2 ? a1 c2 + c2 bk 1 ) ± 1/2 ), ? ± ? 2c2 b ? ? ? ? ? y = a2 (x± + k2 ) , ± c2 ? where = (c1 a2 ? a1 c2 + c2 bk 1 )2 ? 4c2 b(c1 a2 k2 ? c2 a1 k1 ) > 0. Moreover, it is easy to see that, x+ > 0 and ? < 0. x?

Linear analysis of system (2)–(3) shows that point E0 is unstable (it repels in both x and y directions) and point E1 is also unstable (it attracts in the x-direction but repels in the y-direction). For r > 0, the characteristic equation of the linearized system at E2 takes the form P2 ( ) + Q2 ( )e? where P2 ( ) =

2 r

= 0,

?A ,

Q2 ( ) = a2 ( ? A), and A = a1 ? Let us de?ne F2 (y) = |P2 (iy)|2 ? |Q2 (iy)|2 . It is easy to verify that the equation F2 (y) = 0 has one positive root. Therefore, if E2 is unstable for r = 0, it will remain so for all r > 0, and if it is stable for r = 0, there is a positive constant r2 , such that for r > r2 , E2 becomes unstable. It is easy to verify that for r = 0, E2 is asymptotically stable if a2 k2 /c2 > a1 k1 /c1 , stable (but not asymptotically) if a2 k2 /c2 = a1 k1 /c1 and unstable if a2 k2 /c2 < a1 k1 /c1 . 4. Permanence results c 1 a 2 k2 . c2 k 1

De?nition 4. System (2)–(3) is said to be permanent, see [4], if there exist , , 0 < < , independent of the initial condition, such that for all solutions of this system, min lim inf x(t), lim inf y(t)

t→+∞ t→+∞

A.F. Nindjin et al. / Nonlinear Analysis: Real World Applications 7 (2006) 1104 – 1118

1109

and max lim sup x(t), lim sup y(t)

t→+∞ t→+∞

.

Theorem 5. System (2)–(3) is permanent if L< a1 k 1 . c1 = max{K, L} > 0 independent of the initial condition such that . (12)

Proof. By Lemma (2), there is a

max lim sup x(t), lim sup y(t)

t→+∞ t→+∞

We only need to show that there is a > 0, independent of the initial data, such that min lim inf x(t), lim inf y(t)

t→+∞ t→+∞

.

It is easy to see that, for system (2)–(3), for any > 1 and for t large enough, we have y(t) < L. Thus, we obtain x > x a1 ? bx ? ˙ c1 L . k1

By standard comparison arguments, it follows that lim inf x(t)

t→+∞

1 c1 L a1 ? b k1

and letting

t→+∞

→ 1, we obtain 1 c1 L . a1 ? b k1 (13)

lim inf x(t)

Let us denote by N1 = 1/b(a1 ? c1 L/k1 ). If (12) is satis?ed, N1 > 0. From (13) and Lemma 2 and for any > 1, there exists a positive constant, T , such that for t > T , x(t) > N1 / and y(t) < L. Then, for t > T + r, we have y(t) > y(t) a2 ? ˙ c2 y(t ? r) . N1 + k 2 (14)

On the one hand, for t > T + r, these inequalities lead to y(t) > ? ˙

2L y(t), N1 + k 2 2c

which involves, for t > T + r, y(t ? r) < y(t) exp

2L r . N1 + k 2 2c

(15)

On the other hand, from (14) and (15), we have for t > T + r, y(t) > y(t) a2 ? ˙ which yields lim inf y(t)

t→+∞ 2c L a2 (N1 + k2 ) 2 exp ? r =y . c2 N1 + k 2 2c L c2 2 exp r y(t) N1 + k 2 N1 + k 2

1110

A.F. Nindjin et al. / Nonlinear Analysis: Real World Applications 7 (2006) 1104 – 1118

Letting

→ 1, we get a2 (N1 + k2 ) c2 L exp ? r = y1 . c2 N1 + k 2

lim inf y(t)

t→+∞

Let be = min{N1 , y1 } > 0. Then we have shown that system (2)–(3) is permanent. 5. Global stability analysis 5.1. Stability of E2 Theorem 6. If a1 ? c1 y1 < 0, (k1 + K)

then E2 is globally asymptotically stable for system (2)–(3). Proof. As lim inf t→+∞ y(t) y1 and lim sup x(t) K, from the prey’s equation we obtain: ?

t→+∞

1, ?T > 0, ?t > T ,

x(t) < x(t) a1 ? bx ? ˙ As a1 ? c1 y1 < 0, (k1 + K) c 1 y1 (k1 + K)

c 1 y1 (k1 + K)

.

there exists > 1 such that a1 ? < 0.

Then, by standard comparison arguments, it follows that lim supt→+∞ x(t) 0 and thus, limt→+∞ x(t) = 0. The -limit set of every solution with positive initial conditions is then contained in {(0, y), y 0}. Now from (7), we obviously obtain ? {(0, y), 0 y L}. As E0 ∈ , (E0 is unstable is repels in both x and y directions) and as is nonempty closed and invariant set, / therefore = {E2 }. 5.2. Stability of E ? without delay First, we give some suf?cient conditions which insure that the steady state in the instantaneous system, i.e. without time delay, is globally asymptotically stable. Theorem 7. The interior equilibrium E ? is globally asymptotically stable if a1 + c1 < b(k1 + x ? ), a1 a2 < bk 2 (c2 ? a2 ). Proof. The proof is based on constructing a suitable Lyapunov function. We de?ne V (x, y) = (x ? x ? ) ? x ? ln where = k2 c1 /k1 a2 . x + x? (y ? y ? ) ? y ? ln y y? , (16) (17)

A.F. Nindjin et al. / Nonlinear Analysis: Real World Applications 7 (2006) 1104 – 1118

1111

This function is de?ned and continuous on int(R2 ). It is obvious that the function V is zero at the equilibrium E ? + and is positive for all other values of x and y, and thus, E ? is the global minimum of V . The time derivative of V along a solution of system (2)–(3) is given by dV c1 y = (x ? x ? ) a1 ? bx ? dt x + k1 dV = dt + (y ? y ? ) a2 ? c2 y x + k2 .

Centering dV /dt on the positive equilibrium, we get a1 ? bx ? c2 (x ? x ? )2 ? (y ? y ? )2 x + k1 x + k2 a2 c1 (x ? x ? )(y ? y ? ) + ? x + k2 x + k1 a1 ? bx ? c2 = ?b + (x ? x ? )2 ? (y ? y ? )2 x + k1 x + k2 (x ? x ? )2 + (y ? y ? )2 c1 a2 . + + x + k2 x + k1 2 a1 ? bx ? c1 ?b + (x ? x ? )2 + k1 k1 c2 c1 (y ? y ? )2 + ? + x + k2 k1 a1 ? bx ? c1 (x ? x ? )2 ?b + + k1 k1 c1 c 2 k2 (y ? y ? )2 . + x + k2 ? k1 (x + k2 ) a2 ?b + a1 ? bx ? c1 + k1 k1 c 2 k2 a2

(18)

(19)

From (16), we obtain ?b + < 0.

From (6) and (17), there exists > 1 and T > 0, such that K + k2 ? and for t > T , dV dt c1 a1 ? bx ? (x ? x ? )2 + k1 k1 c1 c 2 k2 (y ? y ? )2 . + K + k2 ? k1 (x + k2 ) a2 ?b + < 0,

Thus, dV /dt is negative de?nite provided that (16) and (17) holds true. Finally, E ? is globally asymptotically stable. If in particular, we suppose that environment provides the same protection to both prey and predator (i.e. k1 = k2 ) then Theorem 7 can be simpli?ed as follows. Corollary 8. The interior equilibrium E ? is globally asymptotically stable if k1 = k2 and a1 < b(k1 + x ? ). (21) (20)

1112

A.F. Nindjin et al. / Nonlinear Analysis: Real World Applications 7 (2006) 1104 – 1118

Proof. From (18), we have dV = dt a1 ? bx ? c2 (x ? x ? )2 ? (y ? y ? )2 x + k1 x + k2 c1 (k2 ? k1 )x (x ? x ? )(y ? y ? ) + k1 (x + k2 )(x + k1 ) a1 ? bx ? c2 ?b + (x ? x ? )2 ? (y ? y ? )2 k1 x + k2 c1 (k2 ? k1 )x (x ? x ? )(y ? y ? ), + k1 (x + k2 )(x + k1 ) ?b +

which is negative de?nite provided that (20) and (21) holds true and thus E ? is globally asymptotically stable. 5.3. Stability of E ? with delay In this subsection we shall give a result on the global asymptotic stability of the positive equilibrium for the delayed system. Theorem 9. Assume that parameters of system (2)–(3) satisfy a1 k1 a1 3 a2 k2 . > max , c1 2c2 2 c2 (22)

Then, for b large enough, there exists r0 > 0 such that, for r ∈ [0, r0 ], the interior equilibrium E ? is globally asymptotically stable in R2 . + Proof. First of all, we rewrite Eqs. (2) to center it on its positive equilibrium. By using the following change of variables, X(t) = ln and Y (t) = ln y(t) , y? x(t) x?

the system becomes ? ? c1 y ? ? X(t) = x ? ?b + a1 ? bx (eX(t) ? 1) ? (eY (t) ? 1), ?˙ x(t) + k1 x(t) + k1 (23) c2 y ? a2 x ? ?˙ ? Y (t) = ? (eY (t?r) ? 1) + (eX(t?r) ? 1). x(t ? r) + k2 x(t ? r) + k2 According to the global existence of solutions established in Lemma 2, we can assume that the initial data exists on [?2r, 0] (this can be done by changing initial time). Now, let > 1 be ?xed and let us de?ne the following Liapunov functional V : C([?2r; 0], R2 ) → R, V(

1, 2) =

a2 x ? 0 (e 1 (u) ? 1)2 du 2k2 ?r 0 0 c2 y ? 0 0 2 (s) c2 y ? a2 x ? + e (e 2 (s?r) ? 1)2 + (e 1 (s?r) ? 1)2 ds dv 2k2 ?r v k2 k2 0 c y? c2 a2 x ? 2 + r L (e 2 (s) ? 1)2 + (e 1 (s) ? 1)2 ds. k2 k2 2k2 ?r

1 (0)

(eu ? 1) du +

2 (0)

(eu ? 1) du +

Let us de?ne the continuous and nondecreasing function u : R+ → R+ by u(x) = ex ? x ? 1.

A.F. Nindjin et al. / Nonlinear Analysis: Real World Applications 7 (2006) 1104 – 1118

1113

We have u(0) = 0, u(x) > 0 for x > 0 and u(| (0)|) u(

1 (0)) + u( 2 (0))

V(

1,

2 ),

where = ( 1 , 2 ) and | · | denotes the in?nity norm in R2 . Let v : R+ → R+ be de?ne by v(x) = 2u(x) + (ex ? 1)2 a2 x ? r + 2k2 c2 y ? a2 x ? + k2 k2 c2 y ? r 2 x c2 2 e + r L 4k2 2k2 .

It is clear that v is a continuous and nondecreasing function and satis?es v(0) = 0, v(x) > 0 and V(

1, 2)

v(

),

where · denotes the in?nity norm in C([?2r; 0], R2 ). Now, let (X, Y ) be a solution of (23) and let us compute ˙ V(23) , the time derivative of V along the solutions of (23). First, we start with the function V1 ( We have V1 (Xt , Yt ) = Then, a2 x ? X(t) ˙ ˙ ˙ ((e ? 1)2 ? (eX(t?r) ? 1)2 ). V1|(23) = (eX(t) ? 1)X(t) + (eY (t) ? 1)Y (t) + 2k2 System (23) gives us a1 ? bx ? c1 y ? ˙ (eX(t) ? 1)2 ? (eY (t) ? 1)(eX(t) ? 1) V1|(23) = x ? ?b + x(t) + k1 x(t) + k1 c2 y ? ? (eY (t) ? 1)(eY (t?r) ? 1) x(t ? r) + k2 a2 x ? + (eY (t) ? 1)(eX(t?r) ? 1) x(t ? r) + k2 a2 x ? X(t) + ((e ? 1)2 ? (eX(t?r) ? 1)2 ). 2k2 By using several times the obvious inequalities of type ? c1 y ? (eY (t) ? 1)(eX(t) ? 1) x(t) + k1 c1 y ? x(t) + k1 (eY (t) ? 1)2 (eX(t) ? 1)2 + 2 2

X(t) 0 1, 2) =

1 (0)

(eu ? 1) du +

0

2 (0)

(eu ? 1) du +

0

a2 x ? 2k2

t

0 ?r

(e

1 (u)

? 1)2 du.

(eu ? 1) du +

0

Y (t)

(eu ? 1) du +

a2 x ? 2k2

(eX(u) ? 1)2 du.

t?r

c1 y ? Y (t) ((e ? 1)2 + (eX(t) ? 1)2 ), 2k1 we get ˙ V1|(23) a1 ? bx ? c1 y ? a2 + (eX(t) ? 1)2 + k1 2k2 2k1 c1 y ? a2 x ? (eY (t) ? 1)2 + + 2k1 2k2 c2 y ? (eY (t) ? 1)(eY (t?r) ? 1). ? x(t ? r) + k2 x ? ?b +

1114

A.F. Nindjin et al. / Nonlinear Analysis: Real World Applications 7 (2006) 1104 – 1118

As, eY (t?r) = eY (t) ? we obtain ˙ V1|(23) a1 ? bx ? c1 y ? a2 + (eX(t) ? 1)2 + k1 2k2 2k1 c1 y ? a2 x ? c2 y ? (eY (t) ? 1)2 + + ? 2k1 2k2 x(t ? r) + k2 t c2 y ? ˙ + (eY (t) ? 1) eY (s) Y (s) ds. x(t ? r) + k2 t?r x ? ?b +

t t?r

˙ eY (s) Y (s) ds,

(24)

From (23), we have (eY (t) ? 1) =

t t t?r

˙ eY (s) Y (s) ds

?c2 y ? (eY (t) ? 1)(eY (s?r) ? 1) ds x(s ? r) + k2 t?r t a2 x ? eY (s) (eY (t) ? 1)(eX(s?r) ? 1) ds + x(s ? r) + k2 t?r t a2 x ? c2 y ? + eY (s) ds.(eY (t) ? 1)2 2k2 2k2 t?r c2 y ? t Y (s) Y (s?r) a2 x ? t Y (s) X(s?r) + e (e ? 1)2 ds + e (e ? 1)2 ds 2k2 t?r 2k2 t?r eY (s) L for s large enough, we obtain, for t large enough,

t t?r

and as y ? eY (s) (eY (t) ? 1) +

t t?r

˙ eY (s) Y (s) ds

eY (s)

c2 a2 x ? (eY (t) ? 1)2 + 2k2 2k2 y ? c2 y ? Y (s?r) a2 x ? X(s?r) (e ? 1)2 + (e ? 1)2 ds 2k2 2k2 Lr

and as x(s ? r) ˙ V1|(23)

L for s large enough, (24) becomes, for t large enough, x ? ?b +

a1 ? bx ? c1 y ? a2 + (eX(t) ? 1)2 + k1 2k2 2k1 c1 y ? a2 x ? c2 y ? c2 Lr c2 y ? a2 x ? (eY (t) ? 1)2 + + ? + + 2k1 2k2 x(t ? r) + k2 k2 2k2 2k2 c2 y ? t Y (s) c2 y ? Y (s?r) a2 x ? X(s?r) + e (e ? 1)2 + (e ? 1)2 ds. k2 t?r 2k2 2k2

(25)

The next step consists in computation of the time derivative along the solution of (23), of the term V2 (

1, 2) =

c2 y ? 0 0 e 2k2 ?r v 0 c2 + r L 2k2 ?r

2 (s)

c2 y ? (e k2

c2 y ? (e k2

2 (s)

2 (s?r)

? 1)2 +

? 1)2 +

a2 x ? (e k2

a2 x ? (e k2

1 (s)

1 (s?r)

? 1)2

ds dv

? 1)2 ds.

A.F. Nindjin et al. / Nonlinear Analysis: Real World Applications 7 (2006) 1104 – 1118

1115

We have V2 (Xt , Yt ) =

t c2 y ? Y (s?r) c2 y ? t a2 x ? X(s?r) eY (s) (e ? 1)2 + (e ? 1)2 2k2 t?r v k2 k2 t c2 y ? Y (s) a2 x ? X(s) c2 r L (e ? 1)2 + (e ? 1)2 ds. + 2k2 k2 k2 t?r

ds dv

Then, c2 y ? Y (t) c2 y ? Y (t?r) a2 x ? X(t?r) ˙ V2 |(23) = e (e ? 1)2 + (e ? 1)2 r 2k2 k2 k2 c2 y ? t Y (s) c2 y ? Y (s?r) a2 x ? X(s?r) ? e (e ? 1)2 + (e ? 1)2 ds 2k2 t?r k2 k2 c2 y ? Y (t) a2 x ? X(t) c2 r L (e ? 1)2 + (e ? 1)2 + 2k2 k2 k2 c2 c2 y ? Y (t?r) a2 x ? X(t?r) ? r L (e ? 1)2 + (e ? 1)2 2k2 k2 k2 c2 r c2 y ? Y (t?r) a2 x ? X(t?r) = (e ? 1)2 + (e ? 1)2 (y ? eY (t) ? L) 2k2 k2 k2 c2 y ? t Y (s) c2 y ? Y (s?r) a2 x ? X(s?r) ? e (e ? 1)2 + (e ? 1)2 ds 2k2 t?r k2 k2 c2 y ? Y (t) a2 x ? X(t) c2 r L (e ? 1)2 + (e ? 1)2 . + 2k2 k2 k2 For t large enough, we have y ? eY (t) ? L < 0 and thus, ˙ V2 |(23) ? c2 y ? t Y (s) c2 y ? Y (s?r) a2 x ? X(s?r) e (e ? 1)2 + (e ? 1)2 ds 2k2 t?r k2 k2 c2 y ? Y (t) a2 x ? X(t) c2 r L (e ? 1)2 + (e ? 1)2 . + 2k2 k2 k2

This inequality and (25) lead to, for t large enough, ˙ V |(23) a1 ? bx ? c1 y ? a2 + (eX(t) ? 1)2 + k1 2k2 2k1 c1 y ? a2 x ? c2 y ? c2 Lr c2 y ? a2 x ? + + ? + + 2k1 2k2 x(t ? r) + k2 k2 2k2 2k2 ? ? c2 c2 y Y (t) a2 x X(t) + r L (e ? 1)2 + (e ? 1)2 2k2 k2 k2 x ? ?b + ? + Now, if ? and c1 y ? a2 x ? c2 y ? + ? <0 2k1 2k2 x(t ? r) + k2 (27) a2 x ? c1 y ? bx ? (k1 + x ? ) a1 x ? + + + <0 k1 k1 2k2 2k1 (26)

(eY (t) ? 1)2

bx ? (k1 + x ? ) a1 x ? a2 x ? c1 y ? c 2 a2 x ? r L (eX(t) ? 1)2 + + + + 2 k1 k1 2k2 2k1 2k2 c1 y ? a2 x ? c2 y ? c2 Lr + ? + 2k1 2k2 x(t ? r) + k2 k2 c2 y ? a2 x ? + k2 2k2 (eY (t) ? 1)2 .

1116

A.F. Nindjin et al. / Nonlinear Analysis: Real World Applications 7 (2006) 1104 – 1118

˙ then, for r small enough, we can conclude that V |(23) is negative de?nite. Hence V satis?es all the assumptions of Corollary 5.2 in [4] and the theorem follows. We now study when inequalities (26) and (27) hold. From (10), we have ?bx ? (k1 + x ? ) = c1 (y ? ? a1 (k1 + x ? )/c1 ) and using (11), (26) becomes 3 c1 a2 (x ? + k2 ) a2 x ? ? a1 + <0 2 k1 c2 2k2 which is rewritten as x? As x ? a1 a2 k1 3a2 + c1 k 2 c2 <2 a1 k1 a 2 k2 ?3 . c1 c2 (28)

a1 b,

then the following inequality <b 2 a1 k 1 a 2 k2 ?3 c1 c2 (29)

a 2 k1 3a2 + c1 k 2 c2

implies (28). Now, if the following inequality holds c1 y ? a2 x ? c2 y ? + ? a < 0, 1 2k1 2k2 + k2 b then, for t large enough, (27) holds too. Using (11), (30) is reformulated as c1 y ? c 2 y ? ? a 2 k2 c2 y ? + < a 1 2k1 2k2 + k2 b c1 c2 c2 a2 + < a + ? 1 2k1 2k2 2y + k2 b c1 c2 c2 c2 + < a + ?+k ) 1 2k1 2k2 2(x 2 + k2 b ? ? 2 c1 1 1? ? < c2 ? a + ? ? ?. 1 k1 x + k2 k2 + k2 b As x ? a1 /b, then the last inequality is satis?ed if ? ? 3 1? c1 ? < c2 ? a ? ? 1 k1 k2 + k2 b that is if k2 a1 c 1 k 2 + a1 c2 < b (2c2 k1 ? c1 ). k1 k1 In conclusion, if a1 3 a2 k2 a1 k 1 > max , c1 2c2 2 c2 and for b large enough, there exists a unique interior equilibrium and for r small enough it is globally asymptotically stable. (32) (31) (30)

A.F. Nindjin et al. / Nonlinear Analysis: Real World Applications 7 (2006) 1104 – 1118

1117

6. Discussion It is interesting to discuss the effect of time delay r on the stability of the positive equilibrium of system (2)–(3). We assume that positive equilibrium E ? exists for this system. Linearizing system (2)–(3) at E ? , we obtain ˙ X(t) = A11 X(t) + A12 Y (t), (33) ˙ Y (t) = A21 X(t ? r) + A22 Y (t ? r), where A11 = ?bx ? + A12 = ? A21 = and A22 = ? x? c2 y ? = ?a2 . + k2

r

c1 x ? y ? (x ? + k1 )2

,

c1 x ? , x ? + k1 =

2 a2 c2

c2 (y ? )2 (x ? + k2 )2

The characteristic equation for (33) takes the form P ( ) + Q( )e? in which P( ) = and Q( ) = ?A22 + (A11 A22 ? A12 A21 ). When r = 0, we observe that the jacobian matrix of the linearized system is J= A11 A21 A12 A22 .

2

=0

(34)

? A11

One can verify that the positive equilibrium E ? is stable for r = 0 provided that a1 < bk 1 . Indeed, when (35) holds, then it is easy to verify that T r(J ) = A11 + A22 < 0 and Det(J ) = A11 A22 ? A12 A21 > 0. By denoting F (y) = |P (iy)|2 ? |Q(iy)|2 , we have F (y) = y 4 + (A2 ? A2 )y 2 ? (A11 A22 ? A12 A21 )2 . 11 22 If (35) holds, then = 0 is not a solution of (34). It is also easy to verify that if (35) holds, then equation F (y) = 0 has at least one positive root. By applying standard theorem on the zeros of transcendental equation, see [4, Theorem (35)

1118

A.F. Nindjin et al. / Nonlinear Analysis: Real World Applications 7 (2006) 1104 – 1118

4.1], we see that there is a positive constant r0 (which can be evaluated explicitly), such that for r > r0 , E ? becomes unstable. Then, the global stability of E ? involves restrictions on length of time delay r. Therefore, it is obvious that time delay has a destabilized effect on the positive equilibrium of system (2)–(3). References

[1] M.A. Aziz-Alaoui, Study of Leslie–Gower-type tritrophic population, Chaos Solitons Fractals 14 (8) (2002) 1275–1293. [2] M.A. Aziz-Alaoui, M. Daher Okiye, Boundeness and global stability for a predator–prey model with modi?ed Leslie–Gower and Holling-typeII schemes, Appl. Math. Lett. 16 (2003) 1069–1075. [3] E. Beretta, Y. Kuang, Global analyses in some delayed ratio-depended predator–prey systems, Nonlinear Anal. Theory Methods Appl. 32 (3) (1998) 381–408. [4] Y. Kuang, Delay Differential Equations, with Applications in Population Dynamics, Academic Press, New York, 1993. [5] R.K. Upadhyay, V. Rai, Crisis-limited chaotic dynamics in ecological systems, Chaos Solitons Fractals 12 (2) (2001) 205–218. [6] R.K. Upadhyay, S.R.K. Iyengar, Effect of seasonality on the dynamics of 2 and 3 species prey–predator system, Nonlinear Anal.: Real World Appl. 6 (2005) 509–530. [7] R. Xu, M.A.J. Chaplain, Persistence and global stability in a delayed predator–prey system with Michaelis–Menten type functional response, Appl. Math. Comput. 130 (2002) 441–455.

相关文章:

- 2014年度-理学院邓联望科研立项_图文
- Hopf bifurcation
*analysis*and amplitude control*of*the*modified*Lorenz system,...Ghorai, Dynamic behavior*of**a*delayed*predator*-*prey**model**with*harvesting, ...

更多相关标签:

- Analysis of a Prey-predator Model with Disease in Prey
- Analysis of a Prey-predator Fishery Model with Prey Reserve
- The role of noise in a predator–prey model with Allee effect
- HOPF BIFURCATION ANALYSIS OF A PREDATOR-PREY MODEL WITH TIME-DELAY
- Dynamics of a Stage-Structured Leslie-Gower Predator-Prey Model
- Complex dynamics of prey refuges in a predator-prey model with
- A predator–prey model with infected prey
- Bifurcation analysis of a predator–prey model
- A ratio-dependent predator–prey model with disease in the prey
- Dynamical analysis of a delayed predator-prey system with Holling II functional response and
- Predator MES White Paper v3.0
- Global dynamics of a predator–prey model with time delay and stage
- Bifurcation analysis of an autonomous epidemic predator–prey model with delay
- A ratio-dependent predator–prey model with disease in the prey
- Dynamics of a nonautonomous predator–prey

- 《中华新韵府简表》吟华诗联学苑整理贴出 [建筑/土木]
- Analysis of a predator–prey model with modified Leslie–Gower [演讲/主持]
- 优秀学生干部(共250名) [教学反思/汇报]
- 向日葵杂交制种栽培技术 [农学]
- 向日葵杂交制种技术 [农学]
- 人教版新课标五年级语文下册期末试卷 [语文]
- 本单位公章、合同专用章或政府采购专用章空白印模一份 [表格类模板]
- PS2801A-4-V-F4中文资料 [电子/电路]
- 分析化学选择题 [工学]
- 2014年秋季学期新版新人教版七年级数学上册1.4有理数的乘除法教案4 [初一数学]
- 大连船舶重工ERP采购管理系统研究 [生产/经营管理]