Attractivity of the Recursive Sequence x n+1 = (α βx n 1 )F (x n )

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1 ISSN (print), (online) International Journal of Nonlinear Science Vol.7(2009) No.2,pp Attractivity of the Recursive Sequence (α βx n 1 )F (x n ) A. M. Ahmed 1,, Alaa E. Hamza 2 1 Department of Natural Sciences, Arriyadh Community College,King Saud University, Malaz, P.O Box 28095, 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 (α βx n 1 )F (x n ), n = 0, 1,... where α, β 0. We show that the unique positive equilibrium point of the equation is a global attractor with some basin. We apply this result to the rational recursive sequence α βx n 1 γ + ax n + bx 2, n = 0, 1,... n where α, β, γ, a, b 0 and prove that the positive equilibrium point of the equation is a global attractor with a basin that depends on certain conditions posed on the coefficients. Key words: difference Equations; asymptotic behavior; attractivity; stability AMS Mathematics Subject Classification : 39A10. 1 Introduction The asymptotic stability of the rational recursive sequence α + βx n γ + k i=1 γ ix n i, n = 0, 1,... (1) was investigated when the coefficients α, β, γ and γ 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 of the coefficients are negative was suggested by Kocic and Ladas in [3]. M. T. Aboutaleb et al [5] studied the asymptotic stability of the rational recursive sequence α βx n γ + x n 1, n = 0, 1,... (2) where α, β and γ are non-negative with arbitrary initial conditions x 1 and x 0. 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 α βx n γ + x n k, n = 0, 1,... (3) Corresponding author. address: 1 ahmedelkb@yahoo.com On leave from Department of Mathematics, Faculty of Science, Al-Azhar University, Nasr City (11884), Cairo, EGYPT. 2 address: hamzaaeg@yahoo.com Copyright c World Academic Press, World Academic Union IJNS /218

2 202 International Journal of Nonlinear Science,Vol.7(2009),No.2,pp Also, H. M. El-Owaidy et al [12] investigated the global attractivity of the following rational recursive sequence Some results concerning the difference equation α βx n 1 γ + x n, n = 0, 1,... (4) (α βx n )F (x n 1 ), n = 0, 1,... (5) For other related results see [8-11]. Section 2 of this paper is devoted to the recursive sequence (α βx n 1 )F (x n ), n = 0, 1,... (6) where α and β are non-negative real numbers and F is non-increasing continuous function. In this section we investigate sufficient conditions for the unique positive equilibrium point to be a global attractor with basin [0, α/β] [0, α/β]. In section 3 we apply the results of section 2 for the rational recursive sequence α βx n 1 γ + ax n + bx 2, n = 0, 1,... (7) n where α, β, γ, a, b 0 and prove the global attractivity of the positive equilibrium point of the equation with basin [0, α/β] [0, α/β] when γ > β + bα 2 /β 2. 2 The Recursive Sequence (α βx n )F (x n 1 ) In this section we study the asymptotic behavior of the difference equation (α βx n 1 )F (x n ), n = 0, 1,... (8) where α, β 0. Here F is a positive non-increasing continuous function. We need the following lemmas in proving the main result. Lemma 1 (1) If α, β > 0, then equation (8) has a unique positive equilibrium point x (0, α/β). (2) If α = 0 and β > 0, then equation (8) has a unique equilibrium point x = 0. Proof. (1) Since the function ψ(x) = x αf (x) 1 + βf (x) is a non-decreasing continuous function, ψ(0) < 0 and ψ(α/β) > 0, then ψ(x) has a unique positive fixed point in (0, α/β), whence equation (8) has a unique positive equilibrium point in this interval. (2) is clear since F is positive. Lemma 2 Assume that α, β > 0 such that βf (0) < 1. Let {x n } be a solution of equation (8). If x m [0, α/β], x m+1 [0, α/β] for some m 1, then C x m+i D, i 4, where Moreover C = α(1 βf (0))F ( α ) and D = αf (0). (9) β lim inf x n x lim sup x n. n n IJNS for contribution: editor@nonlinearscience.org.uk

3 A. M. Ahmed, Alaa E. Hamza: Attractivity of the Recursive Sequence 203 Proof. We can see that 0 x m+i αf (0), i = 2, 3. Then The result follows by induction. Set C = α(1 βf (0))F ( α β ) x m+4 αf (0) = D. λ = lim inf x n and Λ = lim sup x n. n n For every ɛ > 0, there exists n 0 N such that λ ɛ < x n < Λ + ɛ n n 0. If λ > x, take ɛ = λ x. There exists n 0 N such that x < x n n n 0. Hence x > x n+1 n n which is a contradiction. Therefore λ x. Similarly we can see that x Λ. In the following theorem we establish sufficient conditions for the positive equilibrium point x to be a global attractor with basin [0, α/β] [0, α/β]. Set G(x) = (α βx)f (x). Theorem 3 Assume that α, β > 0. If F is non-increasing positive continuous function on [0, ) such that βf (0) < 1, then the following conditions (a) x is the unique fixed point of G 2 in [0, α/β]. (b)g 2 (x) > x x (0, x). (c) If λ and Λ are non-negative numbers in [0, α/β] such that then (d) The system G(Λ) λ x Λ G(λ), (10) λ = Λ = x. (11) y = G(x) and x = G(y) (12) has exactly one solution (x, y) [0, α/β] 2. are equivalent and each of them is a sufficient condition for x to be a global attractor of equation (8) with basin S = [0, α/β] [0, α/β]. Proof. (a = b) Assume on the contrary that there exists x (0, x) such that G 2 (x) x. Since G 2 (0) > 0, then G 2 has a fixed point in (0, x) which is a contradiction. (b = c) Assume that λ, Λ [0, α/β] such that G(Λ) λ x Λ G(λ). Since G is non-increasing on [0, α/β], then λ G(Λ) G 2 (λ). Clearly λ = x, because if λ < x, then by (b) G 2 (λ) > λ which is impossible. (c = a) Assume towards a contradiction that x 0 x is another fixed point of G 2 in [0, α/β]. If x 0 < x, take λ = x 0 and Λ = G(x 0 ). Then (10) holds but not (11). If x 0 > x, take λ = G(x 0 ) and Λ = x 0. Then (10) holds but not (11). (a = d) If the system (12) has a solution (x, y) ( x, x) in [0, α/β] 2, then G 2 has a fixed point different from x, which contradicts (a). (d = c) Let λ, Λ [0, α/β] such that (10) holds. Set U 1 = G(λ) and L 1 = G(Λ) and for n = 1, 2,..., set We can see by induction that U n+1 = G(L n ) and L n+1 = G(U n+1 ). 0 < Ł n L 2 L 1 x U 1 U 2 U n αf (0). IJNS homepage:

4 204 International Journal of Nonlinear Science,Vol.7(2009),No.2,pp Hence, both of {L n } and {U n } converges to a number say L [0, α/β] and U [0, α/β] respectively. Then (L, U) is a solution of the system (12) and L = U = x. Clearly U Λ x λ L. Therefore Λ = λ = x. Let {x n } be a solution of equation (8) with initial conditions x 1, x 0 such that (x 1, x 0 ) S. By Lemma (2), x m [C, D] m 4, where C and D are defined in (9). Set λ = lim inf x n and Λ = lim sup x n. n n Let ɛ > 0 be such that ɛ < min{(α/β) Λ, λ}. There exists n 0 N such that λ ɛ < x n < Λ + ɛ n n 0. Hence (α β(λ + ɛ))f (Λ + ɛ) < x n+1 < (α β(λ ɛ))f (λ ɛ) n n Then we get the following inequality G(Λ) λ x Λ G(λ). By (c), λ = Λ = x. The next theorem presents a detailed description of the semicycles of any solution {x n } of equation (8) about the positive equilibrium point x with initial conditions x 1 [0, ), x 0 [0, α/β]. Also this theorem establishes the strict oscillation of such solutions. Theorem 4 If F is decreasing and continuous function such that βf (0) < 1, then every solution {x n } of equation (8) with initial conditions x 1 [0, ) and x 0 [0, α/β],which are not both equal to x, satisfies the following statements (1) {x n } cannot have two consecutive terms equal to x. (2) Every semicycle of {x n } has at most two terms. (3) {x n } is strictly oscillatory. Proof. (1) If x k = x k+1 = x for some k N, then x k 1 = x and consequently x k 1 = x k 2 = = x 1 = x 0 = x which is impossible. (2) Assume that a semicycle C starts with two terms x k 1, x k. When C is negative, then 0 x k 1, x k < x whence x k+1 > x. When C is positive, we have either x k 1 x and x k > x or x k 1 > x and x k x and in both cases x k+1 < x. (3) From (1) and (2), we obtain the strict oscillation of {x n }. 3 The Recursive Sequence (α βx n 1 )/(γ + ax n + bx 2 n) In this section we investigate the attractivity of the rational recursive sequence α βx n 1 γ + ax n + bx 2, n = 0, 1,... (13) n where α, β > 0. Suppose that the following condition holds The function F (x) = γ β + b α2 β 2. (14) 1 γ + ax + bx 2 is decreasing, continuous and βf (0) < 1. By Lemma (1), equation (13) has a unique positive equilibrium point x (0, α/β). Theorem 5 If condition (14) holds, then x is a global attractor with basin [0, α/β] [0, α/β]. IJNS for contribution: editor@nonlinearscience.org.uk

5 A. M. Ahmed, Alaa E. Hamza: Attractivity of the Recursive Sequence 205 Proof. Set G(x) = If Λ > λ, then (15) yields α βx γ+ax+bx 2. Let λ, Λ be non-negative numbers in [0, α/β] such that (10) holds. Then β(λ λ) γ(λ λ) + bλλ(λ λ) 0. (15) γ β bλλ < b α2 β 2, which contradicts condition (14). Therefore λ = Λ = x. 4 The recursive sequence x n 1 F (x n ) This section is devoted to investigate the attractivity of solutions of the difference equation (8) in case of α = 0, that is for the following difference equation x n 1 F (x n ), n = 0, 1,... (16) where F is a continuos non-increasing positive function on some interval [ a, ), where a > 0. We prove that: 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 x 1, x 0 [ a, 0], then {x 4n 1 }, {x 4n } is monotonically increasing to zero while {x 4n+1 }, {x 4n+2 } is monotonically decreasing to zero. (2)If the initial values x 1 [ a, 0] and x 0 [0, a], then {x 4n 1 }, {x 4n+2 }is monotonically increasing to zero while {x 4n }, {x 4n+1 }is monotonically decreasing to zero. (3)If the initial values x 1, x 0 [0, a], then {x 4n+1 }, {x 4n+2 }is monotonically increasing to zero while {x 4n 1 }, {x 4n }is monotonically decreasing to zero. (4)If the initial values x 1 [0, a] and x 0 [ a, 0], then {x 4n }, {x 4n+1 }is monotonically increasing to zero while {x 4n 1 }, {x 4n+2 }is monotonically decreasing to zero. Proof. (1)Let x 1, x 0 [ a, 0],then x 1, x 2 [0, a] and x 3, x 4 [ a, 0]. By induction we can see that {x 4n 1 }, {x 4n } [ a, 0] and {x 4n+1 }, {x 4n+2 } [0, a], n = 0, 1,... Since then x 4n+2 x 4n 2 = F (x 4n 1 )F (x 4n+1 ) < 1, x 4n+2 < x 4n 2, n = 0, 1,... Similarly we can see that: x 4n+5 < x 4n+1, x 4n+4 > x 4n and x 4n+3 > x 4n 1, n = 0, 1,... and the result 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) with basin [ a, a] 2. Theorem 8 Assume that the initial conditions x 1, x 0 [ a, a] such that they are not both equal to x = 0. If F ( a) < 1, then the following statements are true: (1) {x n } of (16) cannot have two consecutive terms equal to x = 0. (2) Every negative semicycle of {x n } of (16) has at most two terms,while every positive semicycle of {x n } of (16) has at most three terms. (3) {x n } of (16) is strictly oscillatory. IJNS homepage:

6 206 International Journal of Nonlinear Science,Vol.7(2009),No.2,pp Proof. (1) If x l = x l+1 = x = 0 for some l N, then x l 1 = x l 2 =... = x 0 = x 1 = x = 0, which is a contradiction. (2) Assume that C is a negative semicycle starts with two terms x l 1, x l, then a x l 1, x l < 0 which implies that x l+1 = x l 1 F (x l ) > 0. Now Suppose that C is a positive semicycle starts with two terms x l 1, x l,then 0 x l 1, x l a which implies that x l+1 = x l 1 F (x l ) 0. Note that and so x l > 0,which implies that x l+1 = 0 if and only if x l 1 = 0, x l+2 = x l F (x l+1 ) < 0. (3) From (1) and (2), we get {x n } is strictly oscillatory. Acknowledge This paper is supported by the Deanship of Scientific Research, King Saud University, The authors would like to thank King Saud University for its academic and financial support. References [1] V. L. Kocic, G. Ladas and I. W. Rodrigues, On Rational Recursive Sequences, J. Math. Anal. Appl. 173(1): (1993) [2] V. L. Kocic and G. Ladas, Global attractivity in a second-order nonlinear difference equations, J. Math. Anal. Appl. 180(1): (1993) [3] V. L. Kocic and G. Ladas, Global Behavior of Nonlinear Difference Equations of Higher Order with Applications, Kluwer Academic Publishers, Dordrecht(1993) [4] V. L. Kocic and G. Ladas, Permanence and global attractivity in nonlinear difference equations World Congress of Nonlinear Analysis, Tampa, Florida, I-IV: (1992) [5] M. T. Aboutaleb, M. A. E-Sayed and A. E. Hamza, Stability of the recursive sequence (α βx n )/(γ + x n 1 ), J. M. Anal. Appl. 261: (2001) [6] J. Feuer, E. J. Janowski, and G. Ladas, Lyness-type equations in the third quardrant, Nonlinear Analysis, Theory, Methods & Applications, 30(2): (1997) [7] W. T. Li and H. R. Sun, Global attractivity in a rational recursive sequence, Dynamic Systems and Applications.(In press) [8] R. Devault, W. Kosmala, G. Ladas and S. W. Schultz, Global behavior of y n+1 = (p + y n k )/(qy n + y n k ), Nonlinear Analysis. 47: (2001) [9] M. R. S. Kulenovic, G. Ladas and N. R. Prokup, On a rational difference equation, Comp. Math. Appl., 41: (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 αx n 1 β±x n, Appl. Math. Comp., 145: (2003) [12] H. M. El-Owaidy, A. M. Ahmed and Z.Elsady, Global attractivity of the recursive sequence, Appl. Math. Comp.151: (2004) α βx n 1 γ+x n IJNS for contribution: editor@nonlinearscience.org.uk