Global Asymptotic Stability of a Nonlinear Recursive Sequence

Transcription

1 International Mathematical Forum, 5, 200, no. 22, Global Asymptotic Stability of a Nonlinear Recursive Sequence Mustafa Bayram Department of Mathematics, Faculty of Arts and Sciences Fatih University, 34500, Buyukcekmece, Istanbul, Turkey mbayram@fatih.edu.tr S. Ebru Daṣ Department of Mathematics, Faculty of Arts and Sciences Yildiz Technical University, 34220, Davutpasa, Istanbul, Turkey eyeni@yildiz.edu.tr Abstract In this work, we investigate the global asymptotic stability of the following nonlinear recursive sequence: x n+ = x nx b n 2 + xb n 3 + xb n + a x n x b n 3 + xb n 2 + xb n + a () where a, b [0, ) and the initial values x 3, x 2, x, x 0 are arbitrary positive real numbers. Mathematics Subject Classification: 39A0 Keywords: Recursive Sequence, Difference equation, Global Stability Introduction Recursive sequences are also called difference equations. Some recursive sequences can be look very simple; but in fact their global behaviors are mostly very complicated. Li and Zhu[] are obtained a sufficient condition to guarantee the global asymptotic stability of the following recursive sequence: x n+ = x nx b n + xb n 2 + a x b n + x n x b n 2 + a (2)

2 084 M. Bayram and S. Ebru Daṣ where a, b [0, ) and the initial values x 2,x,x 0 (0, ). Li[2] use a new method to investigate the qualitative properties of the following rational difference equation: x n+ = x nx n x n 3 + x n + x n + x n 3 + a, n =0,,... (3) x n x n + x n x n 3 + x n x n 3 ++a Li [3] use a new method to investigate the global behavior of the following rational difference equation: x n+ = x n x n 2 x n 3 + x n + x n 2 + x n 3 + a, n =0,,... (4) x n x n 2 + x n x n 3 + x n 2 x n 3 ++a To be motivated by the above studies, in this paper, we consider the following nonlinear difference equation: x n+ = x nx b n 2 + xb n 3 + xb n + a x n x b n 3 + x b n 2 + x b n + a, n =0,, 2,... (5) where a, b [0, ) and the initial values x 3, x 2, x, x 0 are arbitrary positive real numbers. We review some results which will be useful in our investigation. Definition. Let I R and f : I k+ I be a continuously differentiable function. Then for every set of initial conditions x k,x k+,..., x 0 I, the difference equation x n+ = f(x n,x n,..., x n k ), n =0,,... (6) has an unique solution {x n } n= k. A point x I is called an equilibrium point of Eq(6) if x = f( x, x,..., x). Definition.2 Let x be the equilibrium point of the Eq(6). (i) The equilibrium point x of Eq(6) is called locally stable if for every ε>0, there exists δ>0 such that for all x k,x k+,..., x 0 I with x k x + x k+ x x 0 x <δ, we have x n x <ε foralln k (7) (ii) The equilibrium point x of Eq(6) is called a global attractor if for every x k,x k+,..., x 0 I, we have lim n x n = x (8) (iii) The equilibrium point x of Eq(6) is called global asymptotically stable if it is locally stable and a global attractor.

3 Global asymptotic stability 085 Definition.3 Let x be an equilibrium point of Eq (6). A positive semicycle of a solution {x n } n= k of Eq (6) consists of a string of terms {x l,x l+,...,x m } all greater than or equal to x, with l k and m s either l = k or l > k and x l < x and either m = or m < and x m+ < x A negative semicycle of a solution {x n } n= k of Eq (6) consists of a string of terms {x l,x l+,...,x m } all less than x, with l k and m such that either l = k or l > k and x l x and either m = or m < and x m+ x 2 Several Lemmas It is easy to see that the positive equilibrium point x of Eq (5) satisfies x n+ = xb+ +2 x b + a (9) x b+ +2 x b + a from which one can see that Eq (5) has an unique positive equilibrium x =. Lemma 2. A positive solution {x n } n= 3 of Eq (5) is eventually equal to if and only if (x 0 )(x 2 x 3 ) = 0 (0) Proof. Assume that (0) holds. Then, according to Eq (5), it is easy to see that x n = for n. Conversely, assume that Then, we must show that Assume the contrary that for some N, (x 0 )(x 2 x 3 ) 0 () x n for any n (2) x N = and x n for 2 n N (3) Clearly, =x N = x N x b N 3 + x b N 4 + x b N 2 + a x N x b N 4 + xb N 3 + xb N 2 + a (4)

4 086 M. Bayram and S. Ebru Daṣ which implies x N 3 = x N 4 and by (), N 3. Thus, from x N 4 = x N 3 = x N 4x b N 6 + xb N 7 + xb N 5 + a x N 4 x b N 7 + x b N 6 + x b N 5 + a (x N 4 )(x b N 7 (x N 4 +)+x b N 5 + a) = 0 (5) one can obtain that from (x b N 7 (x N 4 +)+x b N 5 + a) 0,x N 4 = which contradicts (3). Lemma 2.2 Let {x n } n= 3 be a positive solution of Eq (5) which is not eventually equal to. Then the following statements are true: (i) (x n+ x n )(x n ) < 0, for n 0 (ii) (x n+ )(x n )(x n 2 x n 3 ) > 0, for n 0 (iii) (x n+ )(x n )(x n 3 ) > 0, for n 3 Proof. From Eq (5), one can see x n+ x n = ( x n)[a +(+x n )x b n 3 + xb n ] x n x b n 3 + x b n 2 + x b n + a, n =0,,... (6) Since [a +(+x n )x b n 3 + xb n ] and [x nx b n 3 + xb n 2 + xb n + a] are positive, (i) holds. From Eq (5), x n+ = (x n )(x n 2 x n 3 )[x b n x b n 3] x n x b n 3 + x b n 2 + x b n + a, n =0,,... (7) Since [x b n xn 3] b and [x n x b n 3 + xb n 2 + xb n + a] are positive, (ii) holds. (iii) is a consequence of inequalities (i) and (ii). Lemma 2.3 If x 3 <x 2 <x <x 0 <, then {x n } n= 3 has a negative semicycle with an infinite number of terms and it monotonically tends to the positive equilibrium point x =. Proof. If x 3 <x 2 <x <x 0 <, from Lemma 2.2.(i) and (ii), for n 3 x 0 <x <... < x n <x n < (8) Clearly,{x n } n= 3 has a negative semicycle with an infinite number of terms. Furthermore, we know that the positive solution is strictly increasing for n 0. So the limit lim x n = L (9) n

5 Global asymptotic stability 087 exist and finite. Taking the limit on both sides of Eq (5), we have L = Lb+ +2L b + a L b+ +2L b + a = (20) We can easily see that {x n } n= 3 tends to the positive equilibrium point x =. Lemma 2.4 Let {x n } n= 3 be a positive solution of Eq (5) which is not eventually less than or equal to. Then, with the possible exception of the first semicycle, the following affirmations hold. (a) Every positive semicycle consists of four, two or one terms; Every negative semicycle consists of three, two or one terms. (b) The positive and negative semicycles of Eq (5) has the form 4 +,, +,, 2 +, 2, +, 3. Theorem 2.5 Let a [0, ) and b>0. Then the positive equilibrium point of Eq (5) is globally asymptotically stable. Proof. We must show that the positive equilibrium point is locally asymptotically stable and global attractor. The linearized equation of Eq (5) is z n+ =0.z n +0.z n +0.z n 2 +0.z n 3 (2) By virtue of [[4],Remark.3.7], x is locally asymptotically stable. Now we must show that every positive solution {x n } n= 3 of Eq (5) converges to as n. That is, lim x n = x (22) n If the solution is nonoscillatory about the positive equilibrium point x of Eq (5), then from Lemma 2. and Lemma 2.2, the solution is either equal to or eventually negative one which has an infinite number of terms and monotonically tends to the positive equilibrium point x of Eq (5), and so Eq (22) holds. Therefore, it suffices to prove that Eq (22) holds for the solution to be strictly oscillatory. Consider now {x n } n= 3 to be strictly oscillatory about the positive equilibrium point x of Eq (5). By virtue of Lemmas 2.2(ii) and Lemmas 2.4, the {x n } n= 3 solution of Eq (5) has the positive and negative semicycles of the form 4 +,, +,, 2 +, 2, +, 3. So we have the following sequences: {x p+5n,x p+5n+,x p+5n+2,x p+5n+3 } +, {x p+5n+4 }, {x p+5n+5 } +, {x p+5n+6 }, {x p+5n+7,x p+5n+8 } +, {x p+5n+9,x p+5n+0 }, {x p+5n+ } +, {x p+5n+2,x p+5n+3,x p+5n+4 } We now have the following assertions:

6 088 M. Bayram and S. Ebru Daṣ (i) x p+5n >x p+5n+ >x p+5n+2 >x p+5n+3, x p+5n+7 >x p+5n+8 x p+5n+0 >x p+5n+9, x p+5n+4 >x p+5n+3 >x p+5n+2 (ii) x p+5n+3 x p+5n+4 >,x p+5n+4 x p+5n+5 <,x p+5n+5 x p+5n+6 >, x p+5n+6 x p+5n+7 <,x p+5n+8 x p+5n+9 >,x p+5n+0 x p+5n+ <, x p+5n+ x p+5n+2 >,x p+5n+4 x p+5n+5 < inequality (i) can be easily seen from Lemma 2.2.(i) for n =0,,... From the observations of x p+5n+4 = x p+5n+3x b p+5n+ + x b p+5n + x b p+5n+2 + a x p+5n+3 x b p+5n + x b p+5n+ + x b p+5n+2 + a > = x p+5n+3 x b p+5n+ + x b p+5n + x b p+5n+2 + a x 2 p+5n+3x b p+5n + x b p+5n+x p+5n+3 + x b p+5n+2x p+5n+3 + ax p+5n+3 x p+5n+3 and x p+5n+5 = x p+5n+4x b p+5n+2 + xb p+5n+ + xb p+5n+3 + a x p+5n+4 x b p+5n+ + x b p+5n+2 + x b p+5n+3 + a < = x p+5n+4 x b p+5n+2 + xb p+5n+ + xb p+5n+3 + a x 2 p+5n+4x b p+5n+ + x b p+5n+2x p+5n+4 + x b p+5n+3x p+5n+4 + ax p+5n+4 x p+5n+4 x p+5n+5 x p+5n+6 >,x p+5n+6 x p+5n+7 <,x p+5n+8 x p+5n+9 >, x p+5n+0 x p+5n+ <,x p+5n+ x p+5n+2 >,x p+5n+4 x p+5n+5 < can be easily shown. From inequality (i) and (ii), x p+5n+5 < x p+5n+4 < x p+5n+3 < x p+5n+9 < x p+5n+8 <x p+5n+7 < x p+5n+4 < x p+5n+3 <x p+5n+2 <x p+5n+ <x p+5n <x p+5n+ < < x p+5n+2 x p+5n+0 <x p+5n+5 < (23) x p+5n+6 From equation (23), we can see that {x p+5n } n=0 is decreasing with lower bound. So the limit lim x p+5n = L (24) n

7 Global asymptotic stability 089 exist and are finite. From equation(23), we obtain lim n x p+5n+5 = lim n x p+5n+ = lim n x p+5n+8 = lim n x p+5n+7 = lim n x p+5n+5 = lim n x p+5n+3 = lim n x p+5n+2 = lim n x p+5n+ = L lim n x p+5n+4 = lim n x p+5n+3 = lim n x p+5n+2 = lim n x p+5n+0 = lim n x p+5n+9 =lim n x p+5n+6 = lim n x p+5n+4 = L, It suffices to verify that L =. For this, x p+5n+5 = x p+5n+4x b p+5n+2 + x b p+5n+ + x b p+5n3 + a x p+5n+4 x b p+5n+3 + x b p+5n+2 + x b p+5n3 + a (25) If we take the limits on both sides of the equation (25), L = + L b + + a L b+ L b L b a L b (26) which imply that L =. References [] X. Li and D. Zhu, Global asymptotic stability of a nonlinear recursive sequence, Compt.Math.Appl., 7 (2004), [2] X. Li, Qualitative properties for a fourth-order rational difference equation, J.Math.Anal.Appl., 3 (2005), [3] X. Li, Global behavior for a fourth-order rational difference equations, J.Math.Anal.Appl., 32 (2005), [4] V. L. Kocic and G. Ladas, Global Behavior of Nonlinear Difference Equations of Higher Order with Applications, Kluwer Academic, Dordrecht, 993. Received: November, 2009