Dynamics of a two-dimensional system of rational difference equations of Leslie--Gower type

Abstract

We investigate global dynamics of the following systems of difference equations

[Equation not available: see fulltext.]

where the parameters [alpha] ^sub 1^, [beta] ^sub 1^, A ^sub 1^, γ ^sub 2^, A ^sub 2^, B ^sub 2^ are positive numbers, and the initial conditions x ^sub 0^ and y ^sub 0^ are arbitrary nonnegative numbers. We show that this system has rich dynamics which depends on the region of parametric space. We show that the basins of attractions of different locally asymptotically stable equilibrium points or non-hyperbolic equilibrium points are separated by the global stable manifolds of either saddle points or non-hyperbolic equilibrium points. We give examples of a globally attractive non-hyperbolic equilibrium point and a semi-stable non-hyperbolic equilibrium point. We also give an example of two local attractors with precisely determined basins of attraction. Finally, in some regions of parameters, we give an explicit formula for the global stable manifold.

Mathematics Subject Classification (2000)

Primary: 39A10, 39A11 Secondary: 37E99, 37D10[PUBLICATION ABSTRACT]

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RESEARCH Open Access

Dynamics of a two-dimensional system of rational difference equations of LeslieGower type

S Kalabui1, MRS Kulenovi2* and E Pilav1

* Correspondence: mailto:[email protected]

Web End [email protected]

2Department of Mathematics, University of Rhode Island, Kingston, RI 02881-0816, USAFull list of author information is available at the end of the article

Abstract

We investigate global dynamics of the following systems of difference equations

xn+1 = 1 + 1xn A1 + yn

yn+1 = 2ynA2 + B2xn + yn

, n = 0, 1, 2, . . .

where the parameters a1, b1, A1, g2, A2, B2 are positive numbers, and the initial conditions x0 and y0 are arbitrary nonnegative numbers. We show that this system has rich dynamics which depends on the region of parametric space. We show that the basins of attractions of different locally asymptotically stable equilibrium pointsor non-hyperbolic equilibrium points are separated by the global stable manifolds of either saddle points or non-hyperbolic equilibrium points. We give examples of a globally attractive non-hyperbolic equilibrium point and a semi-stable non-hyperbolic equilibrium point. We also give an example of two local attractors with precisely determined basins of attraction. Finally, in some regions of parameters, we give an explicit formula for the global stable manifold.

Mathematics Subject Classification (2000)

Primary: 39A10, 39A11 Secondary: 37E99, 37D10

Keywords: Basin of attraction, Competitive map, Global stable manifold, Monotonicity, Period-two solution

1 Introduction

In this paper, we study the global dynamics of the following rational system of difference equations

xn+1 = 1 + 1xn A1 + yn

yn+1 = 2ynA2 + B2xn + yn

, n = 0, 1, 2, . . . (1)

where the parameters a1, b1, A1, g2, A2, B2 are positive numbers and initial conditions x0 and y0 are arbitrary nonnegative numbers.

System (1) was mentioned in [1] as one of three systems of Open Problem 3, which asked for a description of the global dynamics of some rational systems of difference equations. In notation used to label systems of linear fractional difference equations

2011 Kalabui et al; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0

Web End =http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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used in [1], System (1) is referred to as (29, 38). This system is dual to the system where the roles of xn and yn are interchanged, which is labeled as (29, 38) in [1], and so all results proven here extend to the latter system. In this paper, we provide a precise description of the global dynamics of the System (1). We show that System (1) may have between zero and three equilibrium points, which may have different local character. If System (1) has one equilibrium point, then this point is either locally asymptotically stable or saddle point or non-hyperbolic equilibrium point. If System (1) has two equilibrium points, then they are either locally asymptotically stable and non-hyperbolic, or locally asymptotically stable and saddle point. If System (1) has three equilibrium points, then two of equilibrium points are locally asymptotically stable and the third point, which is between these two points in southeast ordering defined below, is a saddle point. The major problem for global dynamics of the System (1) is determining the basins of attraction of different equilibrium points. The difficulty in analyzing the behavior of all solutions of the System (1) lies in the fact that there are many regions of parameters where this system possesses different equilibrium points with different local character and that in several cases, the equilibrium point is non-hyperbolic. However, all these cases can be handled by using recent results from [2].

System (1) is a competitive system, and our results are based on recent results about competitive systems in the plane, see [2,3]. System (1) can be used as a mathematical model for competition in population dynamics. In fact, second equation in (1) is of Leslie-Gower type, and first equation can be considered to be of Leslie-Gower type with stocking which is represented with the term a1, see [4-6].

In the next section, we present some general results about competitive systems in the plane. Section 3 contains some basic facts such as the non-existence of period-two solution of System (1). Section 4 analyzes local stability which is fairly complicated for this system. Finally, Section 5 gives global dynamics for all values of parameters.

2 Preliminaries

A first-order system of difference equations

xn+1 = f (xn, yn)yn+1 = g(xn, yn) , n = 0, 1, 2, . . . (2)

where S 2, (f, g): S S, f, g are continuous functions is competitive if f(x, y) is

non-decreasing in x and non-increasing in y, and g(x, y) is non-increasing in x and non-decreasing in y. If both f and g are non-decreasing in x and y, the System (2) is cooperative. Competitive and cooperative maps are defined similarly. Strongly competitive systems of difference equations or strongly competitive maps are those for which the functions f and g are coordinate-wise strictly monotone.

Competitive and cooperative systems have been investigated by many authors, see [2,3,5-19]. Special attention to discrete competitive and cooperative systems in the plane was given in [2,3,5-7,10,12,17,20]. One of the reasons for paying special attention to two-dimensional discrete competitive and cooperative systems is their applicability and the fact that many examples of mathematical models in biology and economy which involve competition or cooperation are models which involve two species. Another reason is that the theory of two-dimensional discrete competitive and cooperative systems is very well developed, unlike such theory for three and higher

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dimensional systems. Part of the reason for this situation is de Mottoni and Schiaffino theorem given below, which provides relatively simple scenarios for possible behavior of many two-dimensional discrete competitive and cooperative systems. However, this does not mean that one cannot encounter chaos in such systems as has been shown by Smith, see [17].

If v = (u, v) 2, we denote with Ql (v),

{1, 2, 3, 4}, the four quadrants in 2 relative to v, i.e., Q1 (v) = {(x, y) 2: x u, y v}, Q2 (v) = {(x, y) 2: x u, y v},

and so on. Define the South-East partial order se on 2 by (x, y) se (s, t) if and only if x s and y t. Similarly, we define the North-East partial order ne on 2 by (x, y)

ne (s, t) if and only if x s and y t. For A 2 and x 2, define the distance

from x to A as dist(x, A) = inf{||x-y||: y A}. By int A, we denote the interior of a set A.It is easy to show that a map F is competitive if it is non-decreasing with respect to the South-East partial order, that is, if the following holds:

x1 y1

[precedesorcurly]se

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x2y2

F

x1 y1

x2 y2

. (3)

For standard definitions of attracting fixed point, saddle point, stable manifold, and related notions see [11].

We now state three results for competitive maps in the plane. The following definition is from [17].

Definition 1 Let S be a nonempty subset of 2. A competitive map T : S Sis said

to satisfy condition (O+) if for every x, y in S, T(x) ne T (y) implies x ne y, and T is

said to satisfy condition (O-) if for every x, y in S, T(x) ne T (y) implies y ne x.

The following theorem was proved by de Mottoni and Schiaffino [20] for the Poincar map of a periodic competitive Lotka-Volterra system of differential equations. Smith [14,15] generalized the proof to competitive and cooperative maps.

Theorem 1 Let S be a nonempty subset of 2. If T is a competitive map for which (O

+) holds then for all x S, {Tn(x)} is eventually componentwise monotone. If the orbit

of x has compact closure, then it converges to a fixed point of T. If instead (O-) holds, then for all x S, {T2n(x)} is eventually componentwise monotone. If the orbit of x has

compact closure in S, then its omega limit set is either a period-two orbit or a fixed

point.

The following result is from [17], with the domain of the map specialized to be the cartesian product of intervals of real numbers. It gives a sufficient condition for conditions (O+) and (O-).

Theorem 2 Let 2 be the cartesian product of two intervals in . Let T: be a C1 competitive map. If T is injective and det JT (x) >0 for all x then T satisfies (O+). If T is injective and det JT (x) <0 for all x then T satisfies (O-).

The following result is a direct consequence of the Trichotomy Theorem of Dancer and Hess, see [3] and [21] and is helpful for determining the basins of attraction of the equilibrium points.

Corollary 1 If the nonnegative cone of is a generalized quadrant in n, and if T has no fixed points in u1, u2 other than u1 and u2, then the interior of u1, u2

is either a subset of the basin of attraction of u1 or a subset of the basin of attraction of u2.

[precedesorcurly]seF

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Next result is well-known global attractivity result that holds in partially ordered Banach spaces as well, see [21].

Theorem 3 Let T be a monotone map on a closed and bounded rectangular region 2. Suppose that T has a unique fixed point in . Then is a global attractor of T

on .

The following theorems were proved by Kulenovi and Merino [2] for competitive systems in the plane, when one of the eigenvalues of the linearized system at an equilibrium (hyperbolic or non-hyperbolic) is by absolute value smaller than 1 while the other has an arbitrary value. These results are useful for determining basins of attraction of fixed points of competitive maps.

Theorem 4 Let T be a competitive map on a rectangular region 2. Let x R

be a fixed point of T such that : = int (Q1(x) Q3(x))is nonempty (i.e., x is not

the NW or SE vertex of ), and T is strongly competitive on . Suppose that the following statements are true.

a. The map T has a C1 extension to a neighborhood of x.b. The Jacobian JT(x)of T at x has real eigenvalues l, such that 0 <|l| <, where |l| <1, and the eigenspace El associated with l is not a coordinate axis.

Then there exists a curve C through x that is invariant and a subset of the basin

of attraction of x, such that C is tangential to the eigenspace El at x, and C is the graph

of a strictly increasing continuous function of the first coordinate on an interval. Any endpoints of C in the interior of are either fixed points or minimal period-two points.

In the latter case, the set of endpoints of C is a minimal period-two orbit of T.

The situation where the endpoints of C are boundary points of is of interest. The

following result gives a sufficient condition for this case.Theorem 5 For the curve C of Theorem 4 to have endpoints in

that at least one of the following conditions is satisfied.

i. The map T has no fixed points nor periodic points of minimal period-two in .ii. The map T has no fixed points in , det JT(x) > 0, and T(x) = xhas no solutions x .iii. The map T has no points of minimal period-two in , det JT(x) < 0, and T(x) = xhas no solutions x .

The next result is useful for determining basins of attraction of fixed points of competitive maps.

Theorem 6 (A) Assume the hypotheses of Theorem 4, and let C be the curve whose

existence is guaranteed by Theorem 4. If the endpoints of C belong to

, then C sepa

rates into two connected components, namely

W := {x R\C : y C with x[precedesorcurly]sey} and W+ := {x R\C : y C with y [precedesorcurly]sex}, (4) such that the following statements are true.

(i) W- is invariant, and dist (Tn(x), Q2(x)) 0 as n for every x W .

(ii) W

+ is invariant, and dist (Tn(x), Q4(x)) 0 as n for every x W+.

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, it is sufficient

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(B) If, in addition to the hypotheses of part (A), x is an interior point of and T is C2 and strongly competitive in a neighborhood of x, then T has no periodic points in the boundary of (Q1(x) Q3(x)) except for x, and the following statements are true.

(iii) For every x W- there exists n0 N such that Tn(x) int Q2(x)for n n0.

(iv) For every x W

+ there exists n0 N such that Tn(x) int Q4(x)for n n0.

If T is a map on a set and if x is a fixed point of T, the stable set Ws(x) of x is the

set {x R : Tn (x) x} and unstable set Wu(x) of x is the set

{ x R : there exists {xn}0n= R s.t. T(xn) = xn+1, x0 = x, and lim

n

When T is non-invertible, the set Ws(x) may not be connected and made up of infi

nitely many curves, or Wu(x) may not be a manifold. The following result gives a

description of the stable and unstable sets of a saddle point of a competitive map. If the map is a diffeomorphism on , the sets Ws(x) and Wu(x) are the stable and

unstable manifolds of x.

Theorem 7 In addition to the hypotheses of part (B) of Theorem 6, suppose that >1 and that the eigenspace E associated with is not a coordinate axis. If the curve C of

Theorem 4 has endpoints in , then C is the stable set Ws(x) of x, and the unstable

set Wu(x) of x is a curve in that is tangential to E at x and such that it is the

graph of a strictly decreasing function of the first coordinate on an interval. Any endpoints of Wu(x) in are fixed points of T.

The following result gives information on local dynamics near a fixed point of a map when there exists a characteristic vector whose coordinates have negative product and such that the associated eigenvalue is hyperbolic. This is a well-known result, valid in much more general setting that we include it here for completeness. A point (x, y) is a subsolution if T(x, y) se (x, y), and (x, y) is a supersolution if (x, y) se T(x, y). An order interval (a, b), (c, d) is the cartesian product of the two compact intervals [a, c] and [b, d].

Theorem 8 Let T be a competitive map on a rectangular set 2 with an isolated fixed point x R such that R int (Q2(x) Q4(x)) = . Suppose T has a C1

extension to a neighborhood of x. Let v = (v(1), v(2)) 2 be an eigenvector of the Jacobian of T at x, with associated eigenvalue . If v(1)v(2) < 0, then there exists an order interval which is also a relative neighborhood of x such that for every relative neighborhood U of x the following statements are true.

i. If > 1, then U int Q2(x)contains a subsolution and U int Q4(x)contains a

supersolution. In this case for every x I int (Q2(x) Q4(x))there exists N such

that Tn(x) for n N.ii. If < 1, then U int Q2(x)contains a supersolution and U int Q4(x) contains

a subsolution. In this case Tn(x) xfor every x .

3 Some basic facts

In this section, we give some basic facts about the nonexistence of period-two solutions, local injectivity of the map T at the equilibrium point, and boundedness of solutions. See [22] for similar analysis.

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xn = x }

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3.1 Equilibrium points

The equilibrium points (x, y) of System (1) satisfy

x = 1 + 1x A1 + y

, y =

2yA2 + B2x + y

. (5)

Solutions of System (5) are:

(i) y = 0, x =

1 A1 1

, A1 >b1, i.e. E1 =

1 A1 1

, 0

. Thus, the equilibrium point

E1 =

1 A11 , 0

exists if A1 >b1.

(ii) If y = 0, then using System (5), we obtain

y = 2 A2 B2x, x2B2 x(2 + A1 A2 1) + 1 = 0. (6)

Solutions of System (6) are:

x3,2 = 2 + A1 A2 1 D

0

2 A2 A1 + 1 D0

2 , (7)

where D0 = (g2 - A2 + A1 - b1)2 - 4B2a1 which gives a pair of the equilibrium points E2 = (x2, y2) and E3 = (x3, y3).

The criteria for the existence of the three equilibrium points are summarized in Table 1.

3.2 Injectivity

Lemma 1 Assume that (x, y)is an equilibrium of the map T. Then the following holds:

1) If

B2 > A21

1 , A1(B21 A21)2(B21 + (A1A2)1)(A1A21A2 + B21) = 0, (8)

then T

x, A1A21+xA1B21B21A21 = (x, y)for all x 0, where

(x, y) =

x,A1A21 + xA1B21 B21 A21 =

2B2 , y2,3 =

B21 + A21A1B2 ,

A221 + A1A21 + B211 B21 A21

.

That is the line

I =

x, A1A21 + xA1B21

B21 A21

: x 0

is invariant, equilibrium (x, y) Iand for (x, y) the following holds

T(x, y) = (x, y), that is every point of this line is mapped to the equilibrium point (x, y).

1.i) If (B21 A21)2 A21B21 > 0then (x, y) = E3.

1.ii) If (B21 A21)2 A21B21 < 0then (x, y) = E2.

1.iii) If (B21 A21)2 A21B21 = 0then (x, y) = E3 = E2.

2) If

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Table 1 The equilibrium points of System (1)

E1 A1 > 1, A2 < 2 < A1 + A2 1, (A )( A )B < 1 (A A + )4B or

A1 > 1, A2 > 2, 1 (A1A21+2)

2

4B2 , A1 + 2 = A2 + 1

or

2

A1 > 1, A2 = 2, 1 (A1A21+2)

4B2 or

A1 > 1, 1 > (A1A21+2)

2

4B2

E1 E2 E3 A1 > 1, A1 + A2 = 1 + 2, 1 = (A11)(2A2)B2

E1 E3, E2 A1 > 1, A1 + A2 < 1 + 2, A2 < 2, 1 = (A11)(2A2)B2

E1, E2, E3 A1 > 1, A1 + A2 < 1 + 2,

(A11)(2A2)

B2 < 1 < (A1A21+2)

2

4B2

E1, E2 A1 > 1, A2 < 2, 1 < (A11)(2A2)B2

E1 E2 A1 > 1, A2 < 2 < A1 + A2 1, 1 = (A11)(2A2)B2

E1, E2 E3 A1 > 1, A1 + A2 < 1 + 2, 1 = (A1A21+2)

2

4B2

4B2 or

A1 = 1, A1 + A2 < 1 + 2, 1 < (A1A21+2)

2

E2, E3 A1 < 1, A1 + 2 > A2 + 1, 1 < (A1A21+2)

2

4B2

4B2 or

A1 = 1, A1 + A2 < 1 + 2, 1 = (A1A21+2)

2

E2 = E3 A1 < 1 A1 + 2 > A2 + 1, 1 = (A1A21+2)

2

4B2

No equilibrium A1 < 1, A2 < 2 < A1 + A2 + 1, 1 (A1A21+2)

2

4B2 or

2

A1 < 1, A2 2, 1 (A1A21+2)

4B2 or

A1 1, 1 > (A1A21+2)

2

4B2 or

A1 = 1, A1 + A2 > 2 + 1, 1 (A1A21+2)

2

4B2

B2

A21

1 or A1 (B21 A21) 2(B21 + (A1 A2) 1) (A1A2 1A2 + B21) = 0,

then the following holds.

T(x, y) = (x, y) (x, y) = (x, y).

Proof T(x, y) = (x, y) is equivalent to

1 + 1xA1 + y ,2yA2 + B2x + y = (x, y). (9)

Since (x, y) is the equilibrium point of the map T then System (9) is equivalent to

1 + 1xA1 + y ,2yA2 + B2x + y =

1 + 1xA1 + y, 2yA2 + B2x + y

. (10)

System (10) is equivalent to

y1 + y1 yx1 + xy1 + xA11 xA11 = 0 (11)

yA22 yA22 + yxB22 xyB22 = 0. (12)

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Equation 11 implies

y = y1 + xy1 + xA11 xA11

1 + x1

.

and Equation 12 is equivalent to

(x x)

yB21 + yA21 + A1A21 + xA1B21 2 = 0. (13)

We conclude the following: If

yB21 + yA21 + A1A21 + xA1B21

= 0, then x = x

and y = y.On the other hand, if

yB21 + yA21 + A1A21 + xA1B21 = 0, since (x, y) is the

equilibrium of the map T, then

B2 > A21

1 , y =

A1 (A2 + xB2) 1 A21 B21

and

(x, y) =

1 + 1xA1 + y, 2yA2 + B2x + y

.

Using these equations, we have

x = B21 A21A1B2 , y =

1 (A1A2 + 1A2 B21) A21 B21

and

A1 (B21 A21) 2 (B21 + (A1 A2) 1) (A1A2 1A2 + B21) = 0, (14)

which completes the proof of lemma.

3.3 Period-two solutions

In this section, we prove that System (1) has no minimal period-two solutions which will be essential for application of Theorem 4 and Corollary 6.

Lemma 2 System (1) has no minimal period-two solution.

Proof Period-two solution satisfies T2(x, y) = (x, y), that is

T2(x, y) =

1 + 1(1+x1)y+A1

A1 + y

2 y+A2+xB2

, y 22

(y+A2+xB2)((y+A1)A2+B2(1+x1))

y+A1 + y2

= (x, y).

This is equivalent to

(y + A2 + xB2)(x21 11 + (y + A1)(xA1 1)) + xy(y + A1)2

(y + A1)(A1(y + A2 + xB2) + y2) = 0

and

y

y

+ A1

22 y

y + A1 2 +

y A2 xB2 y+ A1

A2 + B2 (1 + x1)

y + A2 + xB2 y+ A1

A2 + B2 (1 + x1) + y

y + A1 2 = 0,

which is equivalent to

y + A2 + xB2 x21 11 +

y + A1 (xA1 1) + xy

y + A1 2 = 0 (15)

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y A2 xB2 y+ A1

A2 + B2 (1 + x1) = 0 (16)

If y = 0, we substitute in (15) to obtain the first fixed point, that is x = 1

A11 i

x = A2B2. Assume

y + A1 22 y

y + A1 2 +

y

y

+ A1

22 y

y + A1 2 +

y A2 xB2 y+ A1

A2 + B2 (1 + x1) = 0.(17)

From (17) we calculate x2. We have

x2 = y

+ A1

A22 +

y2 + A1

y + xB2 + B2

1 + x

y + 1

A2

B221

(18)

y + A1

y

2

xB221 + yB2 (1 + x1) +

2

.

B221

Put (18) into (15), we have that (15) is equivalent to

y + A1 = 0 (19)

or

A1 y + A1 21 22 + y

21 + xB21 A1

y + A1 2

+ y A2 xB2 A221 + B211 + A1 y+ A1

A2 + B2.1 = 0 (20)

If (19) holds, then we obtain a negative solution. Now, assume that (20) holds. We have

x =

A1

y + A1 21 22 y

A1

y + A1 21 2

B2

A2A21 +

yA2 + B21

A1 1

B21 + A21 + y2

y A2

A

221 + B211 + A1

yA2 + B21

A1 1

B21 + A21 + y2 .

(21)

y

+ A1

A2 + B21

B2

+ A2A21 +

Put (21) into (18), we obtain that (18) is equivalent to

y2 + (A1 A2 + 1 + 2) y B21 + 1 (A2 2) + A1 (2 A2) = 0 (22)

or

(A2 + 2)

y2 (A1 + 1) (A1 A2 + 1 2)

(B21 + A1 (A2 + 2) 1 (A2 + 2)) y + (A1 + 1)22 (B21 + A1 (A2 + 2) 1 (A2 + 2)) = 0.

(23)

A21 + (1 A2) A1 + 12

If (22) holds, we obtain the fixed points. So, we assume that (23) holds. Set

: = (A1 + 1)2 (B21 + (A1 1) (A2 + 2))

(B21 + (A1 1) (A2 + 2)) (A1 A2 + 1 2)2 + 42 (A2 + 2) (A1 (A1 A2 + 1) + 12) . (24)

If 0 and A1(A1 - A2 + b1) + b1g2 0 hold, we obtain the real solution of the form

y1 =

( 1 )2 (A2 + 2) (A1 (A1 A2 + 1) + 12)

y2 =

( 1 + )

2 (A2 + 2) (A1 (A1 A2 + 1) + 12)

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where

1 := (A1 + 1) (A1 A2 + 1 2) (B21 + (A1 1) (A2 + 2)) .

Substituting this into (21), we have that the corresponding solutions are

x1 = ( 2

2B2 (A1 + 1) (A1 (A1 A2 + 1) + 12)

x2 = ( 2 +

)2B2 (A1 + 1) (A1 (A1 A2 + 1) + 12)

(A1 + 1)2 + B21 2 + (A1 + A2 1) (A2 (A1 + 1) B21) . (25)

Claim 1 Assume 0. Then we have:

a) If x1 0 then y1 < 0.b) If x2 0 then y2 < 0.

Proof. Since T : [0, )2 [0, )2, T(x1, y1) = (x2, y2) and T(x2, y2) = (x1, y1), it is obvious that if (xi, yi) [0, )2 holds then T(xi, yi) [0, )2 for i = 1, 2. It is enough to show that the assumptions (x1, y1), (x2, y2) [0, )2 and T(x1, y1) = (x2, y2) (x1, y1) lead to a contradiction.

Indeed, if A1(A1 - A2 + b1) + b1g2 > 0 then (x1, y1) se (x2, y2). Since T is strongly competitive map then (x2, y2) = T(x1, y1) <<se T(x2, y2) = (x1, y1) which is impossible since (x1, y1) se (x2, y2).

If A1(A1 - A2 + b1) + b1g2 < 0 then (x2, y2) se (x1, y1) Similarly, we have the same conclusion if A1(A1 - A2 + b1) + b1g2 = 0.

3.4 Boundedness of solutions

Lemma 3 Assume that y0 = 0, x0 +. Then the following statements are true.

(i) If A1 >b1 then yn = 0 xn

1A11, n .

(ii) If A1 <b1 then yn = 0, xn , n .(iii) If A1 = b1, then xn = x0 + 1A1 nand yn = 0, xn .

Assume that y0 0 and (x0, y0) R+2. Then the following statements are true.

(iv) xn+1 1A1 + 1A1 xnfor all n = 0, 1, 2,...

(v) yn g2, n N, yn+1 C

2 A2

1

A11 + , > 0, A1 >b1.

(c) If g2 <A2 then yn 0, n

Proof. Take y0 = 0 and x0 +. Then, we have yn = 0, for all n N, and

(b) xn

Page 10 of 29

)

where

2 := (A1 + 1)

(A1 + 1) 22

nand

(a) xn

1A11, A1 >b1.

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xn+1 = 1

A1 +

1A1 xn. (26)

Solution of Equation 26 is

xn = c

1 A1

n+ 1 A1 1

(27)

A2+B2xn+yn it follows that yn+1 2A2 yn, yn+1 g2, n 0. The proof of Lemma 3 follows from (27).

4 Linearized stability analysis

The map T associated to System (1) is given by

T(x, y) =

1 + 1xA1 + y ,2yA2 + B2x + y

From yn+1 =

2yn

.

The Jacobian matrix of the map T has the form:

JT =

1 A1+y

1+1x(A1+y)2

. (28)

The value of the Jacobian matrix of T at the equilibrium point E = (x, y) is

JT(x, y) =

1A1+y

B22y

(A2+B2x+y)2

2A2+2B2x

(A2+B2x+y)2

x A1+y

B2y

A2+B2 x+y

2A2+2B2 x (A2+B2 x+y)2

. (29)

The determinant of (29) is given by

det JT(x, y) =

1 A1 + y

2A2 + 2B2x (A2 + B2x + y)2

x A1 + y

B2yA2 + B2x + y

.

The trace of (29) is

TrJT(x, y) =

1 A1 + y

+ 2A2 + 2B2x

(A2 + B2x + y)2

.

The characteristic equation has the form

2

1 A1 + y

+ 1(2A2 + 2B2x)(A1 + y)(A2 + B1x + y)2

B1xy(A1 + y)(A2 + B2x + y)= 0.

Theorem 9 Assume that A1 >b1. Then there exists the equilibrium point E1 and:

(i) E1 is locally asymptotically stable if 2 A2 <

B21 A11.

+ 2A2 + 2B2x

(A2 + B2x + y)

(ii) E1 is a saddle point if 2 A2 >

B21A11. The eigenvalues are

1 = 1

A1 , 2 =

(A1 1)2 B21 + A2(A2 1)

.

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The corresponding eigenvectors, respectively, are

v1 = (1, 0), v2 =

1

.

A1(A1 1)

11(A11)A1A2+B21A21

, 1

(iii) E1 is non-hyperbolic if 2 A2 =

B21A11 .The eigenvalues are 1 = 1A1, l2 = 1. The

corresponding eigenvectors are

and (1, 0).

Proof. Evaluating Jacobian (29) at the equilibrium point E1(a1/(A1 - b1), 0),

JT(E1) =

1 A1

1 (A11)2 , 1

1 A1(A11)

0 (A11)2

A2(A11)+B21

. (30)

The determinant of (30) is given by

det JT(x, y) =

12(A1 1)

A1[A2(A1 1) + B21]

.

The trace of (30) is

TrJT(x, y) =

1 A1 +

(A1 1)2 A2(A1 1) + B21

.

The characteristic equation associated to System (1) at E1 has the form

1A1 (A1 1)2A2(A1 1) + B21

= 0. (31)

From Equation 31 we have

1 = 1

A1 , 2 =

(A1 1)2 A2(A1 1) + B21

.

(i) If A1 >b1 and 2 A2 <

B21A11 then l1 < 1 and l2 < 1. Hence, E1 is a sink.

(ii) If A1 >b1 and 2 A2 >

B21A11. Then l1 < 1, and l2 < 1. Hence, E1 is a saddle.

(iii) If A1 >b1 and 2 A2 =

B21A11. Then, using Equation 31, we have that l1 < 1

and l2 < 1.

From (30) we obtain the eigenvectors that correspond to these eigenvalues.

We now perform a similar analysis for the other cases in table. Theorem 10 Assume

A1 > 1, A1 + A2 < 1 + 2, (A1 1) (2 A2)

B2 < 1 <

(A1 A2 1 + 2)2

4B2 .

Then E1, E2, E3 exist and:

(i) Equilibrium E1 is locally asymptotically stable.

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(ii) Equilibrium E3 is a saddle point. The eigenvalues are

1 = y3

A1 + y3 + 2

A1 + 1 + y3 D

22

A1 + y3

and

2 = y3

A1 + y3 + 2

A1 + 1 + y3 + D 22(A1 + y3)

,

and |l1| < 1, |l2| > 1, where

D = y23

A1 + y3

2 22y3

A1 1 2B2x3 + y3 A1 + y3 + 22

A1 1 + y3

2.

The corresponding eigenvectors, respectively, are

v1 =

y3

A1 + y3 + 2

A1 1 + y3 + D, 2B2y3

A1 + y3

A1 + y3 .

(iii) Equilibrium E2 is locally asymptotically stable.

Proof. By Theorem 9 (i) holds.

Equilibrium E3 is a saddle if and only if the following conditions are satisfied

|TrJT(x, y)| > |1 + det JT(x, y)| and Tr2JT(x, y) 4 det JT(x, y) > 0. The first condition is equivalent to

1 A1 + y

+ 2A2 + 2B2x

(A2 + B2x + y)2

v2 =

y3

A1 + y3 + 2

A1 1 + y3

D, 2B2y3

>

1 +

1 (A1 + y)

2A2 + 2B2x (A2 + B2x + y)2

B2xy(A1 + y)(A2 + B2x + y)

which is equivalent to

1(A2 + B2x + y)2 + (A1 + y)(2A2 + 2B2x)

> (A1 + y)(A2 + B2x + y)2 + 12(A2 + B2x) B2xy(A2 + B2x + y). This is equivalent to

(A2 + B2x + y)2(1 A1 y) + 2(A2 + B2x)(A1 + y 1) > B2xy(A2 + B2x + y)

22(1 A1 y) + 2(A2 + B2x)(A1 + y 1) > B22xy

(A1 1 + y)(A2 + B2x 2) > B2xy (1 A1 y)(A2 + B2x 2) < B2xy.

We have to prove that (1 A1 y3)(A2 + B2x3 2) < B2x3y3. Notice that

1 A1 y3 = B2x2 and A2 + B2x3 2 = y3.

Now,

(1 A1 y3)(A2 + B2x3 2) < B2x3y3

is equivalent to B2x2y3 < B2x3y3. This implies x2 < x3 which is true. Condition

Tr2JT(x, y) 4 det JT(x, y) > 0

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is equivalent to

1A1 + y

2A2 + 2B2x (A2 + B2x + y)2 2

+ 4B2xy(A1 + y)(A2 + B2x + y)

> 0

which is clearly satisfied. Hence, E3 is a saddle.

Now, we prove that E2 is locally asymptotically stable. Notice that

|TrJT(x, y)| < 1 + det JT(x, y) < 2

implies x3 > x2 which is true.

The second condition is equivalent to

1 (A1 + y)

2A2 + 2B2x (A2 + B2x + y)2

B2xy(A1 + y)(A2 + B2x + y)

< 1.

This implies the following

12(A2 + B2x) B2xy(A2 + B2x + y) < (A1 + y)(A2 + B2x + y)2.

Now, using Equation 5, we obtain

12(2 y) B2xy2 < (A1 + y) 22

(1y + B2xy) < (A1 1 + y)2

which is true, since the left side is always negative, while the right side is always positive.

Theorem 11 Assume

A1 > 1, A1 + A2 < 1 + 2, 1 = (A1 A2 1 + 2)2

4B2 .

Then E1(a1/(A1 - b1), 0) and E2 = E3 =

2A2+A112B2 , 2A2A1+1 2

exist and

(i) Equilibrium E1 is locally asymptotically stable.(ii) Equilibrium E2 is non-hyperbolic. The eigenvalues are

1 = 1, 2 = A21 A22 + 2A21 21 + 2A22 + 212 22

22(A1 A2 + 1 + 2)

.

The corresponding eigenvectors are

(1/B2, 1),

22(A1 A2 1 + 2)

B2(A1 A2 + 1 + 2)(A1 A2 + 1 + 2)

, 1

.

Proof. By Theorem 9, E1 is locally asymptotically stable. Now, we prove that E2 is non-hyperbolic.

Evaluating Jacobian (29) at the equilibrium point E2 =

2A2+A112B2 , 2A2A1+1 2

,

B2y2

A2+B2 x

2

JT(E2) =

1 A1+y

x A1+y

21 A1+2A2+1

2+A2A1+1

B2(A1+2A2+1)

= B2(2A2A1+1)

22

A2+2+A11

22

. (32)

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The eigenvalues of (32) are

1 = 1, and 2 = A21 A22 + 2A21 21 + 2A22 + 212 22

22(A1 A1 + 1 + 2)

.

Notice that |l2| < 1. Hence, E2 is non-hyperbolic. Theorem 12 Assume

A1 > 1, A2 < 2 < A1 + A2 1,

(A11)(2A2)

B2 < 1 (A1A21+2)

2

4B2

A1 > 1, A2 > 2, 1 (A1A21+2)

2

4B2 , A1 + 2 = A2 + 1

A1 > 1, A2 = 2, 1 (A1A21+2)

2

4B2

A1 > 1, 1 > (A1A21+2)

2

4B2

Then there exists a unique equilibrium E1 (a1/(A1 - b),0) which is locally asymptotically stable.

Proof. Observe that the assumption of Theorem 12 implies that the y coordinates of the equilibrium E2 and E3 are less then zero. By Theorem 9 E1 is locally asymptotically stable.

Theorem 13 Assume

A1 > 1, A2 < 2, 1 < (A1 1) (2 A2)

B2 .

Then then there exist two equilibrium points E1 and E2. E1 is a saddle point. The eigenvalues are

1 = 1A1 , 2 =

(A1 1)2 B21 + A2(A2 1)

.

The corresponding eigenvectors, respectively, are

v1 = (1, 0), v2 =

1

.

A1(A1 1)

11(A11)A1A2+B21A21

, 1

The equilibrium E2 is locally asymptotically stable.

Proof. By Theorem 9 (ii), E1 is a saddle point.

Now, we check the sign of coordinates of the equilibrium point E2. We have that

x2 > 0, since all parameters are positive. Consider y2. Since

(A1 A2 1 + 2)2

4B2

(A1 1) (2 A2)

B2 =

(A1 + A2 1 2)2

4B2 > 0,

we have that (g2 - A2 + A1 - b1)2 - 4a1B2 > 0.

y1 > 0 2 A2 + 1 A1 +

(2 A2 + A1 1)2 41B2 > 0.

This implies

(2 A2 + A1 1)2 41B2 > (A1 1) (2 A2). (33)

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From Equation 33, we see that inequality is always true if A1 - b1 <g2 - A2. If A1 - b1 >g2 - A2, then

(2 A2)2 + 2(2 A2)(A1 1) + (A1 1)2 41B1 > (A1 1)2 2(A1 1)(2 A2)

(2 A2)(A1 1) > 1B2

which is true, since A1 1 >

B212A2. So, in both cases x2 > 0 and y2 > 0.

Notice, that x3 > 0. Now, we check the sign of y3. Assume that y3 > 0. Then, we

have

y2 > 0 (2 A2) (A1 1) > (2 A2 + A1 1)2 41B2.

(2 A2)(A 1 1) < 1B2.

This is a contradiction with the assumption of theorem and so E3 is not in considered domain.

By Theorem 10, E2 is a locally asymptotically stable.

Theorem 14 Assume

A1 > 1, A1 + A2 < 1 + 2, 1 = (A1 1) (2 A2)

B2 .

Then there exist two equilibrium points E1 E3 =

1A11 , 0

and

E2 =

12A2 , 2A2A1+12

, and E1 E3 is non-hyperbolic. The eigenvalues are 1 = 1A1,

l2 = 1. The corresponding eigenvectors are 1(A11)2, 1 and (1. 0) The equilibrium

point E2 is locally asymptotically stable.

Proof. By Theorem 10, E2 is locally asymptotically stable. By Theorem 9 (iii), E1 is non-hyperbolic.

Now, we consider the special case of System (1) when A1 = b1.

In this case, System (1) becomes

xn+1 = 1+A1xnA1+ynyn+1 =2ynA2+B2xn+yn, n = 0, 1, 2, . . . (34)

The map T associated to System (34) is given by

T(x, y) =

1 + A1xA1 + y ,2yA2 + B2x + y

.

The Jacobian matrix of the map T has the form:

JT =

A1 A1+y

1+A1x(A1+y)2

. (35)

The value of the Jacobian matrix of T at the equilibrium point E = (x, y) is

JT(x, y) =

A1A1+y

22y

(A2+B2x+y)2

2A2+2B2x

(A2+B2x+y)2

x A1+y

=

A1A1+y

x A1+y

B2y

A2+B2 x+y

2A2+2B2 x (A2+B2 x+y)2

B2y2

A2+B2 x

2

. (36)

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The characteristic equation of T at (x, y) has the form

2

A1 A1 + y

+ A2 + B2x2

+ A1

A1 + y

A2 + B2x

2

B2xy (A1 + y)2

= 0.

Equilibrium points satisfy the following System

x = 1+A1xA1+y

y = 2yA2+B2 x+y n = 0, 1, . . .

(37)

Notice, if y = 0, then using the first equation of System (37 we obtain a1 = 0 which is

impossible. If y = 0 then, using System (37), we obtain

y = 2 A2 B2x0 = B2x2 x(2 A2) + 1.

and the equilibrium points are:

E3 =

2 A2 +

(2 A2)2 4B21 2B2 ,

2 A2

(2 A2)2 4B212

,

E2 =

2 A2

(2 A2)2 4B21 2B2 ,

2 A2 +

(2 A2)2 4B212

.

We prove the following.

Theorem 15 Assume

A1 = 1.

Then the following statements hold.

(i) If g2 >A2, (g2 - A2)2 - 4B2a1 > 0 then System (34) has two positive equilibrium points

E3 =

2 A2 +

(2 A2)2 4B21

2B2 ,

2 A2

(2 A2)2 4B21

2

and

E2 =

2 A2

(2 A2)2 4B21

2B2 ,

2 A2 +

(2 A2)2 4B21

2

.

E3 is a saddle point. The eigenvalues are

1 = y3(A1 + y3) + 2(2A1 + y3)

F

22(A1 + y3)

, |1| < 1

2 = y3(A1 + y3) + 2(2A1 + y3) +

F

22(A1 + y3)

, 2 > 1,

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where

F = y23(A1 + y3)2 22y3(y3 2B2x3)(A1 + y3) + 22y23.

The corresponding eigenvectors are

v1 = (y3(A1 + y3) + 2y3 + F, 2B2y3(A1 + y3)), v2 = (y3(A1 + y3) + 2y3 F, 2B2y3(A1 + y3)).

The equilibrium E2 is locally asymptotically stable.

(ii) If g2 >A2, (g2 - A2)2 - 4B2a1 > 0 then System (34) has a unique equilibrium point

E =

2A22B2 , 2A22 which is non-hyperbolic. The eigenvalues are l1 = 1 and

2 = 2A1A2A

22+2A12+2A22 22

22(2A1A2+2) . The corresponding eigenvectors are: (-1/B2, 1) and

22 B2(2A1A2+2) , 1

.

(iii) If g2 <A2 and (g2 - A2)2 - 4B2a1 0 or (g2 - A2)2 - 4B2a1 > 0 then System (34) has no equilibrium points.

Proof. (i) First, notice that under these assumptions, E3 and E2 are positive. Now, we prove that E3 is a saddle point.

The equilibrium point E3 is a saddle if and only if the following conditions are satisfied|TrJT(x, y)| > |1 + det JT(x, y)| and Tr2JT(x, y) 4 det JT(x, y) > 0.

The first condition is equivalent to

A1 A1 + y

+ A2 + B2x

2 > 1 +

A1(A2 + B2x)2(A1 + y)

B2xy 2(A1 + y)

,

which is equivalent to

A12 + (A1 + y)(A2 + B2x) > 2(A1 + y) + A1(A1 + B2x) B2x y,

and this is equivalent to

2 A2 < 2B2x.

In the case of equilibrium E3, this condition becomes

2A2 < 2B2x3 2A2 < 2A2+

(2 A2)2 4B21 (2 A2)2 4B21 > 0,

which is true.

The second condition becomes

A1 A1 + y

+ A2 + B2x

2

2 4A1(A2 + B 2x)2(A1 + y)+4 B2xy 2(A1 + y)=

A1A1 + y

A2 + B2x 2

2+4 B2xy 2(A1 + y)

which is greater then zero. Hence, E3 is a saddle.

Now, we prove that E2 is locally asymptotically stable. The equilibrium point E2 is locally asymptotically stable if the following is satisfied

|TrJT(x, y)| < 1 + det JT(x, y) < 2.

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The first condition is equivalent to

A1 A1 + y

+ A2 + B2x

2 < 1 +

A1(A2 + B2x)2(A1 + y)

B2xy 2(A1 + y)

.

This implies

A12 + (A1 + y)(A2 + B2x) < 2(A1 + y) + A1(A2 + B2x) B2xy

which is equivalent to 2 A2 > 2B2x. In the case of the equilibrium point E2, we

have

2 A2 > 2 A2

(2 A2)2 4B21 (2 A2)2 4B21 < 0

which is true.

The second condition is equivalent to

A1(A2 + B2x)2(A1 + y)

B2xy 2(A1 + y)

< 1.

This implies

A1(A2 + B2x) B2xy < 2(A1 + y) A1(A2 2 + B2x) < y(2 + B2x).

Notice that

A22+B2x2 =

A2 2

(2 A2)2 4B212 = y2.

Now, condition A1(A2 2 + B2x) < y(2 + B 2x) becomes A1y2 < y2(2 + B2x2) A1 < 2 + B2x2 which is true. Hence, E2 is locally asympto

tically stable.

(ii) The characteristic equation associated to System (37) at E has the form

2

2A12A1 + 2 A2

(2 A2)2 4B212 =

2 A2 +

+ A2 + 2 22

+ A1

2

(2 A2)2 22(2A1 + 2 A2)

= 0. (38)

Solutions of Equation (38) are l1 = 1 and 2 = 2A1A2A

22+2A12+2A22 22

22(2A1A2+2) .

The corresponding eigenvectors are (-1/B2, 1) and

22 B2(2A1A2+2) , 1

.

If g2 <A2 and (g2 - A2)2 - 4B2a1 0 then x2 < 0 and x3 < 0.

Theorem 16 Assume

A1 < 1, 2 > A2, 2 A2 > 1 A1 and (2 A2 + A1 1)2 4B21 > 0. Then there exist two positive equilibrium points

E2 =

2 A2 + A1 1

(2 A2 + A1 1)2 4B21

2B2 ,

2 A2 A1 + 1 +

(2 A2 + A1 1)2 4B21

2

and

E3 =

2 A2 + A1 1 +

(2 A2 + A1 1)2 4B21

2B2 ,

2 A2 + 1 A1

(2 A2 + A1 1)2 4B21

2

.

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E2 is locally asymptotically stable and E3 is a saddle. The eigenvalues of characteristic equation at E3 are

1 = y3

A1 + y3 + 2

A1 + 1 + y3 D

,

22

A1 + y3

where

D = y23

A1 + y3

2 22y3

A1 1 2B2x3 + y3 A1 + y3 + 22

A1 1 + y3

2.

The corresponding eigenvectors are

v1,2 =

y3

A1 + y3 + 2

A1 + y3 .

Proof. Now, we prove that E2 is a sink. We check the condition

|TrJT(x, y)| < 1 + det JT(x, y) < 2. The first condition is equivalent to1 A1 + y

+ A2 + B2x

2 < 1 +

A1 1 + y3 D, 2B2y3

1(A2 + B2x)2(A1 + y)

B2xy 2(A1 + y)

.

This implies

12 + (A1 + y)(A2 + B2x) < 2(A1 + y) + 1(A2 + B2x) B2xy 2(1 A1 y) + (A2 + B2x)(A1 + y 1) < B2xy(A1 1 + y)(A2 + B2x 2) < B2xy

y(A1 1 + y) > B2xy (A1 1 + y) > B2x.

So, we have to prove

(A1 1 + y2) > B2x2. (39)

Note that

A1 1 + y2 = A1 1 +

2 A2 + 1 A1 +

(2 A2 + A1 1)2 4B21

2

=

A1 1 + 2 A2 +

(2 A2 + A1 1)2 4B21

2B2 B2

= B2x3.

Now, (39) becomes B2x3 > B2x2 x3 > x2 which is true.

The second condition is equivalent to

1(A2 + B2x)

2(A1 + y)

B2xy 2(A1 + y)

< 1.

This implies 1(2 y) B2xy < 2(A1 + y). Using equations of equilibrium points,

we obtain y2(1 + B2x2) > 2(1 A1 y2) and

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1 + B2x2 = 1 +

2 A2 + A1 1

(2 A2 + A1 1)2 4B212

=

2 A2 + A1 + 1

(2 A2 + A1 1)2 4B212 > 0.

On the other hand, we have

(1 A1 y2 = 1 A1

2 A2 + 1 A1 +

(2 A2 + A1 1)2 4B212

(2 A2 + A1 1)2 4B212 < 0

since g2 - A2 >b1 - A1. Hence, E2 is locally asymptotically stable.

Now, we prove that E3 is a saddle.

The equilibrium point E3 is a saddle if and only if the following conditions are satisfied

|TrJT(x, y)| > |1 + det JT(x, y)| and Tr2JT(x, y) 4 det JT(x, y) > 0.

Note that the first condition is equivalent to B2x3 > B2x2 x3 > x2 which is true.

The second condition becomes

1 A1 + y

=

1 A1 + A2 2

+ A2 + B2x2

2 41(A2 + B2x) (A1 + y)2

+4 B2xy 2(A1 + y)

=

1A1 + y

A2 + B2x 2

2+4 B2xy2(A1 + y)> 0.

Hence, E3 is a saddle.

Theorem 17 Assume

A1 < 1, 2 > A2, 2A2 > 1A1 and (2 + A1 A2 1)241B2 = 0. Then there exists a unique equilibrium point

E2 E3 = E =

2A2+A112B2 , 2A2+1A12 which is non-hyperbolic. The eigenvalues are:

1 = 1 and 2 = A21 A22 + 2A21 21 + 2A22 + 212 22

22(A1 A2 + 1 + 2)

.

The corresponding eigenvectors are:

1B2 , 1 , and

22(A1 A2 1 + 2)

B2(A1 A2 + 1 + 2)(A1 A2 + 1 + 2)

, 1

.

Proof. The value of the Jacobian matrix of T at the equilibrium point E = (x, y) is

JT(x, y) =

21 2A2+1+A1

2A2+A11

B2(A1+2A2+1)

B2(2A2+1A1)22

A2+2+A11

22

. (40)

The characteristic equation is given by

2

212 A2 + 1 + A1

+ A2 + 2 + A1 1

22

+ 1(A2 + 2 + A1 1)

2(2 A2 + 1 + A1)

(41)

(2 A2 + 1 A1)(2 A2 + A1 1)

22(A1 + 2 A2 + 1)

= 0.

Kalabui et al. Advances in Difference Equations 2011, 2011:29 http://www.advancesindifferenceequations.com/content/2011/1/29

Solutions of Equation (41) are:

1 = 1 and 2 = A21 A22 + 2A21 21 + 2A22 + 212 22

22(A1 A2 + 1 + 2)

By using (40), we obtain the corresponding eigenvectors.

5 Global behavior

Theorem 18 Table 2 describes the global behavior of System (1).Proof. Throughout the proof of theorem will denote se.

(Ri, i = 1, 4) By Theorem 9, E1 is locally asymptotically stable. Consider M(t) = (0, t) for t g2 - A2. Since M(t) T(M(t)) =

1t+A1 , t(t+A22)t+A2

, we have M(t) T(M(t)) for

t g2 - A2. By induction, TnM(t) Tn+1(M(t))) E1 for all n = 0,1,2,... because M(t) E1 for all t 0. By monotonicity and boundedness, the sequence {Tn(M(t))} has to converge to the unique equilibrium E1. Consider N(u) = (u, 0) for u 0. Lemma 3 implies Tn (N(u)) E1 as n . Take any point (x, y) [0, +)[0, +). Then there exists t*, u* 0, such that M(t*) (x, y) (x, y) N(u*). The monotonicity of the map T implies Tn M(t*)) Tn ((x, y)) Tn (N(u*)). Since Tn M(t*)), Tn (N(u*)) E1 as n

, then Tn ((x, y)) E1. This completes the proof.

(5) The first part of this theorem is proven in Theorem 9. The rest of the proof is similar to the proof of part ((Ri, i = 1, 4)).

(6) By Lemma 3 y0 = 0 implies yn = 0, n N, and xn

shows that x-axis is a subset of the basin of attraction of E1.Furthermore, every solution of (1) enters and stays in the box B(E2) and so we can

restrict our attention to solutions that starts in B(E2). Clearly, the set Q2(E2) B(E2) is

an invariant set with a single equilibrium point E2 and by Theorem 3, every solution that starts there is attracted to E2. In view of Corollary 1, the interior of rectangle E2, E1 is attracted to either E1 or E2, and because E2 is the local attractor, it is attracted to E2. If (x, y) A = B\([[E2, E1]] (Q2(E2) B) {(x, 0) : x 0}), then

there exist the points (xu, yu) A Q4(E2) and (xl, yl) Q2(E2) B such that (xl, yl)

se (x, y) se (xu, yu). Consequently, Tn ((xl, yl)) se Tn ((x, y)) se Tn ((xu, yu)) for all n = 1,2,... and so Tn ((x, y)) E2 as n , which completes the proof.

(7) The first part of this Theorem is proven in Theorem 13. Now, we prove a global result.

JT(E1) =

(A11)2

A1A21A2+B21 and so

B2 2 > 1, A1 > 1 1 < 1.

The eigenvector of T at E1 that corresponds to the eigenvalue l1 < 1 is (1, 0). The rest of the proof is similar to the proof of part (6).

(8, 9) The first part of theorem follows from Theorems 15 and 16. If parameters a1 b1, A1, g2, A2 and B2 do not satisfy the condition (8) of Lemma 1, then the map T defined on the set R = R2+, satisfies all conditions of Theorems 4, 6-8. This implies that

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.

1

A11, n , which

1

A1

1

A1(A11)

0 (A11)2

A1A21A2+B21

(42)

The eigenvalues of JT(E1) are given by 1 = 1A1 and 2 =

A2 < 2, 1 < (A1 1) (2 A2)

Kalabui et al. Advances in Difference Equations 2011, 2011:29 http://www.advancesindifferenceequations.com/content/2011/1/29

Page 23 of 29

Table 2 Global behavior of System (1)

Region Global behavior

R1A1 > 1, A2 < 2 < A1 + A2 1,(A1 1)(2 A2)

B2 < 1

or

R2A1 > 1, A2 > 2, 1

(A1 A2 1 + 2)2 4B2

(A1 A2 1 + 2)2 4B2 ,

A1 + 2 = A2 + 1,

or

R3 A1 > 1, A2 = 2, 1

(A1 A2 1 + 2)2 4B2

or

R4 A1 > 1, 1 > (A1 A2 1 + 2)2 4B2

There exists a unique equilibrium E1, and it is globally asymptotically stable (G.A.S.). The basin of attraction of E1 is

given byB1(E1) = [0, )2

R5 A1 > 1, 2 + 1 A1 + A2, 1 =

(A1 1)(2 A2)

B2

There exists a unique equilibrium E1 = E2 which is non-hyperbolic. Furthermore, this equilibrium is the global attractor. Its basin of attraction is given by

B(E1) = [0, +)2. This is an example of globally attractive non-hyperbolic equilibrium point

R6 A1 > 1, A1 + A2 < 1 + 2, 1 = (A1 1)(2 A2) B2

There exist two equilibrium points E = E1 = E3 which is non-hyperbolic, and E2, which is locally asymptotically stable. Furthermore, the x-axis is the basin of attraction of E1. The

equilibrium point E2 is globally asymptotically stable with the basin of attraction

B(E2) = [0, +) [0, +)

R7 A2 > 1, A2 < 2, 1 < (A1 1)(2 A2) B2

There exist two equilibrium points E1, which is a saddle, and

E2, which is a locally asymptotically stable equilibrium point. Furthermore, the x-axis is the global stable manifold of

Ws

(E1). The equilibrium point E2 is globally asymptotically stable with the basin of attraction

B(E2) = [0, +) [0, +)

R8 A1 < 1, A1 + 2 > A2 + 1, 1 < (A1 A2 1 + 2)24B2There exist two equilibrium points E3, which is a saddle, and

E2, which is locally asymptotically stable.Furthermore, there exists the global stable manifold

Bs

(E3)

that separates the positive quadrant so that all orbits below this manifold are asymptotic to (+, 0), and all orbits above this manifold are asymptotic to the equilibrium point E2. All orbits that starts on

Bs

(E3) are attracted to E3

Kalabui et al. Advances in Difference Equations 2011, 2011:29 http://www.advancesindifferenceequations.com/content/2011/1/29

Page 24 of 29

Table 2 Global behavior of System (1) (Continued) or

R9A1 = 1, A1 + A2 < 1 + 2,

1 < (A1 A2 1 + 2)24B2

R10A1 > 1, A1 + A2 < 1 + 2,

(A1 1)(2 A2)

B2 < 1 <

(A1 A2 1 + 2)2 4B2 ,

There exist three equilibrium points E1, E2, and E3, where E1 and E2 are locally asymptotically stable and E3 is a saddle. There exists the global stable manifold

Ws

(E3) that separates the positive quadrant so that all orbits below this manifold are attracted to the equilibrium point E1, and all orbits above this manifold are attracted to the equilibrium point E2. All

orbits that starts on Ws

(E3) are

attracted to E3. The global unstable manifold Ws

(E3) is the

graph of a continuous strictly decreasing function, and Wu

(E3)

has endpoints E2 and E1

R11 A1 > 1, A1 + A2 < 1 + 2, 1 = (A1 A2 1 + 2)24B2There exist two equilibrium points E = E2 = E3 and E1. E1 is locally asymptotically stable and E is non-hyperbolic. There exists a continuous increasing curve

WE which is a subset of the basin of attraction of E. All orbits that start below this curve are attracted to E1. All orbits that start above this curve are attracted to E

R12 A1 < 1, A1 + 2 > A2 + 1, 1 = (A1 A2 1 + 2)24B2There exists a unique equilibrium point E = E2 = E3

which is non-hyperbolic. There exists a continuous increasing curve W

E which is a subset of basin of attraction of E. All orbits that start below this curve are attracted to (+, 0). All orbits that start above this curve are attracted to E. This is an example of semi-stable non-hyperbolic equilibrium point

R13A1 = 1, A1 + A2 < 1 + 2,

1 = (A1 A2 1 + 2)24B2

R14A1 < 1, A2 < 2 < A1 + A2 + 1,

1

(A1 A2 1 + 2)2 4B2

or

System (1) does not posses an equilibrium point. Its behavior is as follows xn , yn , n

R15A1 < 1, A2 2,

1

(A1 A2 1 + 2)2 4B2

or

(A1 A2 1 + 2)2 4B2

or

or

R16 A1 1, 1 >

R17A1 = 1, A1 + A2 > 2 + 1,

1

(A1 A2 1 + 2)2 4B2

Kalabui et al. Advances in Difference Equations 2011, 2011:29 http://www.advancesindifferenceequations.com/content/2011/1/29

there exists the global stable manifold Ws(E3) that separates the first quadrant into two

invariant regions W-(E3) (above the stable manifold) and W+(E3) (below the stable

manifold) which are connected. Now, we show that each orbit starting in the region

W+(E3) is asymptotic to (,0). Indeed, set T1(x, y) = 1+1xA1+y,

A2+B2x+y. Take x = (x0, y0) W+(E3) (+, -), where (+, -) = {(x, y) : T1(x, y) >x, T2(x, y) <y}. As

is known, see [12], the set (+, -) is invariant. We have

T1(x0, y0) = 1 + 1x0

A1 + y0 > x0, T2(x0, y0) =

which implies the following

(x0, y0)[precedesorcurly]se(T1(x0, y0), T2(x0, y0)) (x0, y0)[precedesorcurly]seT(x0, y0).

By monotonicity, T(x0, y0) se T2 (x0, y0) and by induction, Tn(x0, y0) se Tn+1 (x0, y0). This implies that sequence {xn} is non-decreasing and {yn} is non-increasing. Since, {yn} is bounded from above, hence it must converges. Now limn yn = 0 since otherwise (xn, yn) will converge to another limit which is strictly south-east of E3, which is impossible. By Lemma 3, xn . By Theorems 6-8 for all (x, y) W+(E3), there exists

n0 > 0 such that Tn((x, y)) int(Q4(E3) ), n >n0. We see that for all (x, y) int(Q4

(E3)) ), there exists (xl, yl) W+(E3) (+, -) such that (xl, yl) (x, y). By mono

tonicity Tn ((xl, yl)) Tn ((x, y)) (, 0). This implies Tn ((x, y)) (, 0) as n .

Now, we show that each orbit starting in the region W-(E3) converges to E2. By The

orem 6, for all (x, y) W-(E3), there exists n0 > 0 such that, Tn((x, y)) int(Q2(E3)

), n >n0. Set M(t) = (0, t) By part ((Ri, i = 1, 4)), for t g2 - A2, we have

M(t) [precedesorcurly] T(M(t)) [precedesorcurly] E2.. By using monotonicity, Tn(M(t)) E2 as n . By Corollary 1, the interior of rectangle E2, E3 is attracted to either E2 or E3, and because E2 is local attractor, it is attracted to E2. If (x, y) int(Q2(E3) ), then there exist the points (xr, yr) E2, E3 and t* g2 - A2, such that M(t*) se (x, y) se (xr, yr). Consequently, Tn(M(t*)) se Tn((x, y)) se Tn((xr, yr)) for all n = 1, 2,... and so Tn((x, y)) E2 as n .

Now, assume that parameters a1, b1, A1, g2, A2, and B2 satisfy the condition (8) and inequality 1.i) of Lemma 1. Then the set

I =

x, A1A21 + xA1B21

B21 A21

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T2(x, y) =

2y

2y0

A2 + B2x0 + y0 < y0,

: x 0

is invariant and contains the equilibrium point E3, and T(x, y) = E3 for (x, y) . In view of the uniqueness of global stable manifold, we conclude that Ws(E3) = . Take

any point (x, y) W+(E3). Then there exists the point (xl, yl) such that (xl, yl)

<<se (x, y). Since, the map T is strongly competitive, then E3 = T(xl, yl) <<se T(x, y). This implies T(x, y) int(Q4(E3) ). Similarly, if (x, y) W-(E3), then T(x, y) int

(Q2(E3) ). The rest of the proof is similar to the proof of the first case. This completes the proof.

(10) The first part of the theorem follows from Theorem 10. If parameters a1, b1, A1, g2, A2, and B2 do not satisfy the condition (8) of Lemma 1, then the map T, defined on the set R = R2+,, satisfies all conditions of Theorems 4, 6-8. This implies that there

exists the global stable manifold Ws(E3) that separates the first quadrant into two

Kalabui et al. Advances in Difference Equations 2011, 2011:29 http://www.advancesindifferenceequations.com/content/2011/1/29

invariant regions W+(E3) (below the stable manifold) and W-(E3) (above the stable

manifold) which are connected.Using Theorems 6, 7, and 8, we have that for all (x, y) W+(E3), there exists n0 >0

such that for n > n0, Tn((x, y)) int(Q4(E3) ), and for all (x, y) W-(E3), there

exists n1 >0 such that for all n > n1, Tn((x, y)) int(Q2(E3) ). Now, we show that each orbit starting in the region int(Q4(E3)) converges to E1, and each orbit starting in the region int(Q2(E3)) converges to E2.

Take 0 t (g2 - A2)/B2, U(t) = (t,-A2 - tB2 + g2). It is easy to see that the following holds

U(x) = E = E2 = E3 [precedesorcurly] E1 where x = x2 = x3 and

U(t) T(U(t)) =

(A1 + A2 + 2tB2 + 1 2)2 4B2 (A1 A2 tB2 + 2), 0

Since x2 and x3 are solutions of the equation B2t2 + (-A1 + A2 + b1 - g2) t + a1 = 0 and the following inequality holds A2 + tB2 - g2 <0, we have that U(t) se T(U(t)) for 0 t x2 and x3 t (g 2 - A2)/B2 and T(U(t))) se U(t) for x2 t x3.

By using monotonicity of T, we have that for 0 t < x2, Tn(U(t)) Tn+1(U(t)) E2, and for x2 t < x3, E2 Tn+1(U(t)) Tn(U(t)) E3. This implies Tn(U(t)) E2 as n . Similarly, for x3 t (g2 - A2)/B2, we have E3 Tn(U(t)) Tn+1(U(t)) E1. This implies Tn(U(t)) E1 as n . By using the property of points U(t) and N(u), we have that for each point (x, y) int(Q4(E3) ), there exits x3 < t* <(g 2 - A2)/B2 and u* >0 such that U(t*) (x, y) N(u*). By using monotonicity of T, we have Tn(U(t*))

Tn((x, y)) Tn(N(u*))). This implies Tn((x, y)) E1 as n . Furthermore, for each point (x, y) int(Q2(E3) ), there exist t1 >0 and t2, x2 < t2 < x3 such that M (t1) (x, y) U(t2). By using monotonicity of T, we have Tn(M(t1)) Tn ((x, y)) Tn (U(t2)). This implies Tn((x, y)) E2 as n .

Now, assume that parameters a1, A1, g2, A2, and B2 satisfy the condition (8) and inequality 1.i) of Lemma 1. Then the set

I =

x, A1A21 + xA1B21B21 A21

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.

is invariant and contains the equilibrium point E3 and T(x, y) = E3 for (x, y) I. In view of the uniqueness of global stable manifold, we conclude that Ws(E3) = . Take

any point (x, y) W+(E3), then there exists the point (xl, yl) such that (xl, yl) <<se

(x, y). Since, the map T is strongly competitive, then E3 = T(xl, yl) <<se T(x, y). This implies T(x, y) int(Q4(E3) ). Similarly, if (x, y) W-(E3), then T(x, y) int(Q2

(E3) ). The rest of the proof is similar to the proof of the first case. This completes the proof.

(11) The first part of theorem follows from Theorems 15 and 16. If parameters a1, b1, A1,g2, A2, and B2 do not satisfy the condition (8) of Lemma 1, then the map T, defined on the set R = R2+,, satisfies all conditions of Theorems 4, 6, and 8. This

implies that there exists an invariant curve C, which is a subset of the basin of attrac

tion of the equilibrium point E, which separates the first quadrant into two invariant regions, W+(E) (below the stable manifold) and W-(E) (above the stable manifold)

which are connected.

: x 0

Kalabui et al. Advances in Difference Equations 2011, 2011:29 http://www.advancesindifferenceequations.com/content/2011/1/29

By Theorems 6 and 7 and 8 for all (x, y) W+(E), there exists n0 > 0 such that Tn

((x, y)) int(Q4(E) ) for n >n0. For all (x, y) W-(E), there exists n1 > 0 such that

for all n >n1, Tn((x, y)) int(Q2(E) ). Now, we show that each orbit starting in the region int(Q4 (E)) converges to E1, and each orbit starting in the region int(Q2(E)) converges E.

Now, for 0 t (g2 - A2)/B2, take U(t) = (t,-A2 - tB2 + g2) Since a1 = (A1 - A2 - b1 + g2)2/(4B2), it is easy to see that the following holds

U(x) = E = E2 = E3 [precedesorcurly] E1 where x = x2 = x3 and

U(t) T(U(t)) =

(A1 + A2 + 2tB2 + 1 2)2 4B2 (A1 A2 tB2 + 2), 0

Since A2 + tB2 - g2 < 0, we have U(t)[precedesorcurly]seT(U(t)) [precedesorcurly] for 0 t (g2 - A2)/B2.

By using monotonicity of T, we have that Tn(U(t)) [precedesorcurly] Tn+1(U(t)) [precedesorcurly] E for 0 t < x.

This implies Tn(U(t)) E as n . Similarly, for x t < (2 A2)/B2,

E [precedesorcurly] Tn(U(t)) [precedesorcurly] Tn+1(U(t)) [precedesorcurly] E1. This implies Tn(U(t)) E1 as n . By using the property of the points U(t) and N(u), we have that for each point (x, y) int(Q4(E) ), there exist x < t < (2 A2)/B2 and u* > 0 such that U(t) [precedesorcurly] (x, y) [precedesorcurly] N(u). By

using monotonicity of T, we have that Tn(U(t)) [precedesorcurly] Tn((x, y)) [precedesorcurly] Tn(N(u))). This implies Tn ((x, y)) E1 as n . Furthermore, for each point (x, y) int(Q2(E) ) there exists t1 > 0 such that M(t1) [precedesorcurly] (x, y) [precedesorcurly] E. By using monotonicity of T, we have

Tn(M(t1)) [precedesorcurly] Tn((x, y)) [precedesorcurly] E. This implies Tn ((x, y)) E as n .

Now, assume that parameters a1, b1, A1, g2, A2, and B2 satisfy the condition (8) and inequality 1.i) of Lemma 1. The proof of Theorem is similar to the proof of Theorem in the regions (9) and (10).

(12, 13) The first part of theorem follows from Theorems 15 and 17. If parameters a1, b1, A1 g2, A2, and B2 do not satisfy (8) of Lemma 1, then the map T, defined on the set R = R2+, satisfies all conditions of Theorems 4,6, and 8. This implies that

there exists an invariant curve C, which is a subset of the basin of attraction of the

equilibrium point E, and which separates the first quadrant into two invariant regions,

W+(E) (below the stable manifold) and W-(E) (above the stable manifold) which are

connected.By Theorems 6 and 8 for all (x, y) W+(E), there exists n0 > 0 such that Tn((x, y))

int(Q4(E) ) for n >n0, and for all (x, y) C-(E), there exists n1 > 0 such that Tn((x,

y)) int(Q2(E) ) for all n >n1. Now, we show that each orbit starting in the region int(Q4(E)) is asymptotic to (, 0) and each orbit starting in the region int(Q2(E)) converges to E.

Since a1 = (A1 - A2 b1 + g2)2/(4B2), for 0 t (g2 - A2)/B2, we have U(t) = (t, -A2 -tB2 + g2). It is easy to see

U(x) = E = E2 = E3 where x = x2 = x3 and

U(t) T(U(t)) =

(A1 + A2 + 2tB2 + 1 2)2 4B2 (A1 A2 tB2 + 2), 0

Since A2 + tB2 - g2 < 0, for 0 t (g2 - A2)/B2, we have U(t)[precedesorcurly]seT(U(t)).

By using monotonicity of T, we have E [precedesorcurly] Tn(U(t)) [precedesorcurly] Tn+1(U(t)) [precedesorcurly] E10 t < x. This implies Tn(U(t)) E as n . Similarly, E [precedesorcurly] Tn(U(t)) [precedesorcurly] Tn+1(U(t)) [precedesorcurly] (, 0) for x < t < (2 A2)/B2. This implies Tn(U(t)) (, 0) as n . For each point (x, y)

Page 27 of 29

.

.

Kalabui et al. Advances in Difference Equations 2011, 2011:29 http://www.advancesindifferenceequations.com/content/2011/1/29

int(Q4(E3) ), there exists x < t < (2 A2)/B2 such that

0 t < x. 0 t < x. By monotonicity of T, we have Tn(U(t)) [precedesorcurly] Tn((x, y)) [precedesorcurly] (, 0).

This implies Tn((x, y)) (, 0) as n . Furthermore, for each point (x, y) int(Q2 (E3) ), there exists t1 > 0 such that U(t) [precedesorcurly] (x, y) [precedesorcurly] N(u). By monotonicity of T, we have Tn(M(t1)) [precedesorcurly] Tn((x, y)) [precedesorcurly] E. This implies Tn((x, y)) E as n .

If parameters a1, b1, A1, g2, A2, and B2 satisfy the condition (8) and inequality 1.i) of Lemma 1, then the proof of Theorem is similar to the proof of parts (9) and (10). This completes the proof of Theorem in the regions 12, 13. This is an example of semistable non-hyperbolic equilibrium point.

Ri, i = 14, 17

Assumptions of this theorem imply that equilibrium does not exist.

Set M (t) = (0, t) for t g2 - A2. Since M(t) T(M(t)) =

1t + A1 ,t (t + A2 2) t + A2

,

we have M(t) T(M(t)) for t g2 - A2. By using monotonicity Tn(M(t)) Tn+1(M(t))), for all n = 0, 1, 2,... Set (xn, yn) = Tn(M(t)). This implies that {yn} is non-increasing and

bounded, hence it has to converge. Set limn yn = y. Since {xn} is unbounded and

non-decreasing, we have that xn . By using the second equation of the System (1),

we see that y = 0. Take any point (x, y) [0, )2. Then there exists t*, such that M (t*) (x, y) (, 0). By using monotonicity, Tn(M(t*)) (Tn((x, y)) (, 0) as Since Tn(M(t*)) (, 0) as n , we obtain Tn((x, y)) (, 0) as n , as which completes the proof of theorem.

Author details

1Department of Mathematics, University of Sarajevo, Sarajevo, Bosnia and Herzegovina 2Department of Mathematics, University of Rhode Island, Kingston, RI 02881-0816, USA

Authors contributions

All authors contributed equally to the manuscript and read and approved the final draft.

Competing interests

The authors declare that they have no competing interests.

Received: 26 January 2011 Accepted: 23 August 2011 Published: 23 August 2011

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doi:10.1186/1687-1847-2011-29Cite this article as: Kalabui et al.: Dynamics of a two-dimensional system of rational difference equations of LeslieGower type. Advances in Difference Equations 2011 2011:29.

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