Attractivity of the Recursive Sequence x fi flx F x ?· Attractivity of the Recursive Sequence xn+1…

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  • ISSN 1749-3889 (print), 1749-3897 (online)International Journal of Nonlinear Science

    Vol.7(2009) No.2,pp.201-206

    Attractivity of the Recursive Sequence xn+1 = ( xn1)F (xn)A. M. Ahmed1, , Alaa E. Hamza2

    1 Department of Natural Sciences, Arriyadh Community College,King Saud University, Malaz, P.O Box28095, Riyadh 11437, Saudi Arabia.

    2 Department of Mathematics, Faculty of Science,Cairo University, Giza, (12211), Egypt.(Received 17 February 2008, accepted 13 December 2008)

    Abstract: In this paper, we investigate the global attractivity of the recursive sequence

    xn+1 = ( xn1)F (xn), n = 0, 1, . . .

    where , 0. We show that the unique positive equilibrium point of the equation is a globalattractor with some basin. We apply this result to the rational recursive sequence

    xn+1 = xn1

    + axn + bx2n, n = 0, 1, . . .

    where , , , a, b 0 and prove that the positive equilibrium point of the equation is a globalattractor with a basin that depends on certain conditions posed on the coefficients.

    Key words: difference Equations; asymptotic behavior; attractivity; stabilityAMS Mathematics Subject Classification : 39A10.

    1 Introduction

    The asymptotic stability of the rational recursive sequence

    xn+1 = + xn

    +k

    i=1 ixni, n = 0, 1, . . . (1)

    was investigated when the coefficients , , and i are non-negative (see Kocic, Ladas and Rodrigues [1],and Kocic and Ladas [2-4]). Studying the asymptotic behavior of the rational sequence (1) when some ofthe coefficients are negative was suggested by Kocic and Ladas in [3]. M. T. Aboutaleb et al [5] studied theasymptotic stability of the rational recursive sequence

    xn+1 = xn + xn1

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

    where , and are non-negative with arbitrary initial conditions x1 and x0. An interesting (Lyness-type)special case of equation (2) was investigated in [6]. For the terminology used here, we refer the reader to[3]. Li and Sun [7] extended the above results to the following rational recursive sequence

    xn+1 = xn + xnk

    , n = 0, 1, . . . (3)

    Corresponding author. E-mail address: 1 ahmedelkb@yahoo.comOn leave from Department of Mathematics, Faculty of Science, Al-Azhar University, Nasr City (11884), Cairo, EGYPT.2 E-mail address: hamzaaeg@yahoo.com

    Copyright cWorld Academic Press, World Academic UnionIJNS.2009.04.15/218

  • 202 International Journal of Nonlinear Science,Vol.7(2009),No.2,pp. 201-206

    Also, H. M. El-Owaidy et al [12] investigated the global attractivity of the following rational recursivesequence

    xn+1 = xn1

    + xn, n = 0, 1, . . . (4)

    Some results concerning the difference equation

    xn+1 = ( xn)F (xn1), n = 0, 1, . . . (5)

    For other related results see [8-11]. Section 2 of this paper is devoted to the recursive sequence

    xn+1 = ( xn1)F (xn), n = 0, 1, . . . (6)

    where and are non-negative real numbers and F is non-increasing continuous function. In this sectionwe investigate sufficient conditions for the unique positive equilibrium point to be a global attractor withbasin [0, /] [0, /]. In section 3 we apply the results of section 2 for the rational recursive sequence

    xn+1 = xn1

    + axn + bx2n, n = 0, 1, . . . (7)

    where , , , a, b 0 and prove the global attractivity of the positive equilibrium point of the equationwith basin [0, /] [0, /] when > + b2/2.

    2 The Recursive Sequence xn+1 = ( xn)F (xn1)In this section we study the asymptotic behavior of the difference equation

    xn+1 = ( xn1)F (xn), n = 0, 1, . . . (8)

    where , 0. Here F is a positive non-increasing continuous function.We need the following lemmas in proving the main result.

    Lemma 1 (1) If , > 0, then equation (8) has a unique positive equilibrium point x (0, /).(2) If = 0 and > 0, then equation (8) has a unique equilibrium point x = 0.

    Proof. (1) Since the function

    (x) = x F (x)1 + F (x)

    is a non-decreasing continuous function, (0) < 0 and (/) > 0, then (x) has a unique positive fixedpoint in (0, /), whence equation (8) has a unique positive equilibrium point in this interval.(2) is clear since F is positive.

    Lemma 2 Assume that , > 0 such that F (0) < 1. Let {xn} be a solution of equation (8). If xm [0, /], xm+1 [0, /] for some m 1, then

    C xm+i D, i 4,

    whereC = (1 F (0))F (

    ) and D = F (0). (9)

    Moreoverlim inf

    nxn x lim supnxn.

    IJNS email for contribution: editor@nonlinearscience.org.uk

  • A. M. Ahmed, Alaa E. Hamza: Attractivity of the Recursive Sequence 203

    Proof. We can see that 0 xm+i F (0), i = 2, 3. Then

    C = (1 F (0))F (

    ) xm+4 F (0) = D.

    The result follows by induction.Set

    = lim infnxn and = lim supn

    xn.

    For every > 0, there exists n0 N such that < xn < + n n0. If > x, take = x.There exists n0 N such that x < xn n n0. Hence x > xn+1n n0 + 1 which is a contradiction.Therefore x. Similarly we can see that x .

    In the following theorem we establish sufficient conditions for the positive equilibrium point x to be aglobal attractor with basin [0, /] [0, /]. Set G(x) = ( x)F (x).

    Theorem 3 Assume that , > 0. If F is non-increasing positive continuous function on [0,) such thatF (0) < 1, then the following conditions

    (a)x is the unique fixed point of G2 in [0, /].

    (b)G2(x) > x x (0, x).(c) If and are non-negative numbers in [0, /] such that

    G() x G(), (10)

    then = = x. (11)

    (d) The systemy = G(x) and x = G(y) (12)

    has exactly one solution (x, y) [0, /]2.are equivalent and each of them is a sufficient condition for x to be a global attractor of equation (8) withbasin S = [0, /] [0, /].

    Proof. (a = b) Assume on the contrary that there exists x (0, x) such that G2(x) x. Since G2(0) > 0,then G2 has a fixed point in (0, x) which is a contradiction.(b = c) Assume that , [0, /] such that

    G() x G().

    Since G is non-increasing on [0, /], then G() G2(). Clearly = x, because if < x, thenby (b) G2() > which is impossible.(c = a) Assume towards a contradiction that x0 6= x is another fixed point of G2 in [0, /]. If x0 < x,take = x0 and = G(x0). Then (10) holds but not (11). If x0 > x, take = G(x0) and = x0. Then(10) holds but not (11).(a = d) If the system (12) has a solution (x, y) 6= (x, x) in [0, /]2, then G2 has a fixed point differentfrom x, which contradicts (a).(d = c) Let , [0, /] such that (10) holds. Set

    U1 = G() and L1 = G()

    and for n = 1, 2, . . . , setUn+1 = G(Ln) and Ln+1 = G(Un+1).

    We can see by induction that

    0 < n L2 L1 x U1 U2 Un F (0).

    IJNS homepage:http://www.nonlinearscience.org.uk/

  • 204 International Journal of Nonlinear Science,Vol.7(2009),No.2,pp. 201-206

    Hence, both of {Ln} and {Un} converges to a number say L [0, /] and U [0, /] respectively.Then (L,U) is a solution of the system (12) and L = U = x. Clearly U x L. Therefore = = x.Let {xn} be a solution of equation (8) with initial conditions x1, x0 such that (x1, x0) S. By Lemma(2), xm [C,D] m 4, where C and D are defined in (9). Set

    = lim infnxn and = lim supn

    xn.

    Let > 0 be such that < min{(/) , }. There exists n0 N such that

    < xn < + n n0.

    Hence ( ( + ))F ( + ) < xn+1 < ( ( ))F ( ) n n0 + 1.Then we get the following inequality

    G() x G().

    By (c), = = x.The next theorem presents a detailed description of the semicycles of any solution {xn} of equation

    (8) about the positive equilibrium point x with initial conditions x1 [0,), x0 [0, /]. Also thistheorem establishes the strict oscillation of such solutions.

    Theorem 4 If F is decreasing and continuous function such that F (0) < 1, then every solution {xn} ofequation (8) with initial conditions x1 [0,) and x0 [0, /] ,which are not both equal to x, satisfiesthe following statements

    (1) {xn} cannot have two consecutive terms equal to x.(2) Every semicycle of {xn} has at most two terms.(3) {xn} is strictly oscillatory.Proof. (1) If xk = xk+1 = x for some k N, then xk1 = x and consequently xk1 = xk2 = =x1 = x0 = x which is impossible.(2) Assume that a semicycle C starts with two terms xk1, xk. When C is negative, then 0 xk1, xk < xwhence xk+1 > x. When C is positive, we have either xk1 x and xk > x or xk1 > x and xk x andin both cases xk+1 < x.(3) From (1) and (2), we obtain the strict oscillation of {xn}.

    3 The Recursive Sequence xn+1 = ( xn1)/( + axn + bx2n)In this section we investigate the attractivity of the rational recursive sequence

    xn+1 = xn1

    + axn + bx2n, n = 0, 1, . . . (13)

    where , > 0. Suppose that the following condition holds

    + b2

    2. (14)

    The functionF (x) =

    1 + ax + bx2

    is decreasing, continuous and F (0) < 1. By Lemma (1), equation (13) has a unique positive equilibriumpoint x (0, /).Theorem 5 If condition (14) holds, then x is a global attractor with basin [0, /] [0, /].

    IJNS email for contribution: editor@nonlinearscience.org.uk

  • A. M. Ahmed, Alaa E. Hamza: Attractivity of the Recursive Sequence 205

    Proof. Set G(x) = x+ax+bx2

    . Let , be non-negative numbers in [0, /] such that (10) holds. Then

    ( ) ( ) + b( ) 0. (15)

    If > , then (15) yields

    b < b2

    2,

    which contradicts condition (14). Therefore = = x.

    4 The recursive sequence xn+1 = xn1F (xn)This section is devoted to investigate the attractivity of solutions of the difference equation (8) in case of = 0, that is for the following difference equation

    xn+1 = xn1F (xn), n = 0, 1, . . . (16)

    where F is a continuos non-increasing positive function on some interval [a,), where a > 0. We provethat: If F (a) < 1, then the equilibrium point x = 0 is a global attractor with basin [a, a]2.

    Theorem 6 Suppose that a > 0 such that F (a) < 1.(1)If the initial values x1, x0 [a, 0], then {x4n1}, {x4n} is monotonically increasing to zero while{x4n+1}, {x4n+2} is monotonically decreasing to zero.(2)If the initial values x1 [a, 0] and x0 [0, a], then {x4n1}, {x4n+2}is monotonically increasing tozero while {x4n}, {x4n+1}is monotonically decreasing to zero.(3)If the initial values x1, x0 [0, a], then {x4n+1}, {x4n+2}is monotonically increasing to zero while{x4n1}, {x4n}is monotonically decreasing to zero.(4)If the initial values x1 [0, a] and x0 [a, 0], then {x4n}, {x4n+1}is monotonically increasing tozero while {x4n1}, {x4n+2}is monotonically decreasing to zero .Proof. (1)Let x1, x0 [a, 0],then x1, x2 [0, a] and x3, x4 [a, 0]. By induction we can see that{x4n1}, {x4n} [a, 0] and {x4n+1}, {x4n+2} [0, a], n = 0, 1, ... Since

    x4n+2x4n2

    = F (x4n1)F (x4n+1) < 1,

    thenx4n+2 < x4n2, n = 0, 1, ...

    Similarly we can see that: x4n+5 < x4n+1, x4n+4 > x4n and x4n+3 > x4n1, n = 0, 1, ... and theresult follows.(2)The proof is similar to that of the first case and will be omitted.(3)The proof is similar to that of the first case and will be omitted.(4)The proof is similar to that of the first case and will be omitted.

    As a direct consequence of Theorem 4, we obtain the following results.

    Corollary 7 If F (a) < 1, then the equilibrium point x = 0 is a global attractor for equation (16) withbasin [a, a]2.

    Theorem 8 Assume that the initial conditions x1, x0 [a, a] such that they are not both equal to x = 0.If F (a) < 1, then the following statements are true:(1) {xn} of (16) cannot have two consecutive terms equal to x = 0.(2) Every negative semicycle of {xn} of (16) has at most two terms,while every positive semicycle of {xn}of (16) has at most three terms.

    (3) {xn} of (16) is strictly oscillatory.

    IJNS homepage:http://www.nonlinearscience.org.uk/

  • 206 International Journal of Nonlinear Science,Vol.7(2009),No.2,pp. 201-206

    Proof. (1) If xl = xl+1 = x = 0 for some l N, then xl1 = xl2 = ... = x0 = x1 = x = 0, which is acontradiction.(2) Assume that C is a negative semicycle starts with two terms xl1, xl, then a xl1, xl < 0 whichimplies that

    xl+1 = xl1F (xl) > 0.Now Suppose that C is a positive semicycle starts with two terms xl1, xl,then 0 xl1, xl a which

    implies thatxl+1 = xl1F (xl) 0.

    Note thatxl+1 = 0 if and only if xl1 = 0,

    and so xl > 0,which implies thatxl+2 = xlF (xl+1) < 0.

    (3) From (1) and (2), we get {xn} is strictly oscillatory.

    Acknowledge

    This paper is supported by the Deanship of Scientific Research, King Saud University, The authors wouldlike to thank King Saud University for its academic and financial support.

    References

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    [2] V. L. Kocic and G. Ladas, Global attractivity in a second-order nonlinear difference equations, J. Math.Anal. Appl. 180(1):144-150(1993)

    [3] V. L. Kocic and G. Ladas, Global Behavior of Nonlinear Difference Equations of Higher Order withApplications, Kluwer Academic Publishers, Dordrecht(1993)

    [4] V. L. Kocic and G. Ladas, Permanence and global attractivity in nonlinear difference equations WorldCongress of Nonlinear Analysis, Tampa, Florida, I-IV: 1161-1172(1992)

    [5] M. T. Aboutaleb, M. A. E-Sayed and A. E. Hamza, Stability of the recursive sequence xn+1 = ( xn)/( + xn1), J. M. Anal. Appl. 261:126-133(2001)

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    [9] M. R. S. Kulenovic, G. Ladas and N. R. Prokup, On a rational difference equation, Comp. Math. Appl.,41:671-678(2001)

    [10] X. X. Yan and W. T. Li, The Global attractivity of a higher order delay nonlinear difference equation,Appl. Math. J. Chinese University, Ser. B. (In press)

    [11] H. M. El-Owaidy , A. M. Ahmed and M. S. Mousa, On the the recursive sequences xn+1 =xn1xn ,

    Appl. Math. Comp., 145:747-753(2003)

    [12] H. M. El-Owaidy , A. M. Ahmed and Z.Elsady, Global attractivity of the recursive sequence xn+1 =xn1

    +xn, Appl. Math. Comp.151:827-833(2004)

    IJNS email for contribution: editor@nonlinearscience.org.uk

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