On a difference equation with maximum

On a difference equation with maximum

Applied Mathematics and Computation 181 (2006) 1–5 www.elsevier.com/locate/amc On a difference equation with maximum Xiaofan Yang a,b,* , Xiaofeng L...

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Applied Mathematics and Computation 181 (2006) 1–5 www.elsevier.com/locate/amc

On a difference equation with maximum Xiaofan Yang

a,b,*

, Xiaofeng Liao a, Chuandong Li

a

a

b

College of Computer Science, Chongqing University, Chongqing, 400044, PR China Key Laboratory of Opto-Electronic Technique and System of Education Ministry of China, Chongqing University, Chongqing, 400044, PR China

Abstract This paper studies the asymptotic behavior of positive solutions to the difference equation   1 A xn ¼ max a ; ; n ¼ 0; 1; . . . ; xn1 xn2 where 0 < a < 1, A > 0. We prove that every positive solution of this equation approaches x* = 1 or is eventually periodic with period 4 according as A 6 1 or A > 1. Ó 2006 Elsevier Inc. All rights reserved. Keywords: Difference equation; Global attractivity; Eventual periodicity

1. Introduction Ladas [6] initiated the study of the asymptotic behavior of solutions to the difference equation   A 1 A2 Ap xn ¼ max ; ;...; ; n ¼ 0; 1; . . . xn1 xn2 xnp

ð1:1Þ

Specifically, the following two conjectures were proposed: Conjecture 1. If A1, A2, . . . , Ap are nonnegative real numbers with A1 + A2 +    + Ap > 0, then every positive solution to (1.1) is eventually periodic with period T 2 {2, 3, . . . , 2p}. Conjecture 2. If A1, A2, . . . , Ap are nonnegative real numbers with A1 + A2 +    + Ap > 0, Aq > max{Aj: 1 6 j 6 p, j 5 q}, then every positive solution to (1.1) is eventually periodic with period T = 2q. These conjectures were partially confirmed by Amleh et al. [1], Mishev et al. [7] and Voulov [10,11]. In particular, Amleh et al. [1] considered the difference equation.

*

Corresponding author. Address: College of Computer Science, Chongqing University, Chongqing, 400044, PR China. E-mail address: [email protected] (X. Yang).

0096-3003/$ - see front matter Ó 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2006.01.005

2

X. Yang et al. / Applied Mathematics and Computation 181 (2006) 1–5



1

A



; n ¼ 0; 1; . . . ; ð1:2Þ ; xn1 xn2 where A > 0. They proved that every positive solution to (1.2) is eventually periodic with period T, where T is defined by 8 > < 2 if 0 < A < 1; xn ¼ max

T ¼

> :

3 4

if A ¼ 1; if A > 1.

Recently, Bidwell [2] proved the following result: Theorem 1.1. If A1, A2, . . ., Ap are nonnegative real numbers with A1 + A2 +    + Ap > 0, then (i) every positive solution to (1.1) is eventually periodic, and (ii) the maximum prime period grows exponentially with the system size. As a result of this theorem, Conjecture 1 is half-proved and half-disproved. To our knowledge, Conjecture 2 remains yet to be solved. When some of the coefficients in (1.1) are negative, the asymptotic behavior of solutions was examined by Amleh et al. [1], Bidwell [2] and Szalkai [9]. Also some variants of (1.1) with periodic coefficients were investigated [3–5,8]. Motivated by the aforementioned work and by computer simulations, this paper addresses the asymptotic behavior of positive solutions to the difference equation   1 A xn ¼ max a ; ; n ¼ 0; 1; . . . ; ð1:3Þ xn1 xn2 where 0 < a < 1, A > 0. We prove that every positive solution to this equation approaches x* = 1 or is eventually periodic with period 4 according as A 6 1 or A > 1. 2. The case A = 1 In this section, we consider the difference equation   xn ¼ max 1=xan1 ; 1=xn2 ; n ¼ 0; 1; . . . ;

ð2:1Þ

where 0 < a < 1. Theorem 2.1. Let {xn} be a positive solution to (2.1). Then xn ! 1. Proof. Choose a number B so that 0 < B < 1, Let xn ¼ By n ; n P 2. Then {yn} is a solution to the difference equation y n ¼ minfay n1 ; y n2 g; n ¼ 0; 1; . . . ð2:2Þ To prove xn ! 1, it suffices to prove yn ! 0. Observe that there is a positive integer N such that for all n P 0, y3n+N P 0, y3n+N+1 6 0, y3n+N+2 6 0. By simple computation, we get that, for all n P 0, y 3nþN þ2 ¼ y 3nþN ;

y 3nþN þ4 ¼ ay 3nþN þ3 ;

0 < y 3nþN þ3 6 ay 3nþN .

So, y3n+N ! 0, y3n+N+2 ! 0, y3n+N+1 ! 0. This implies yn ! 0. h 3. The case 0 < A < 1 In this section, we consider the difference equation   xn ¼ max 1=xan1 ; A=xn2 ; n ¼ 0; 1; . . . ; where 0 < a < 1, 0 < A < 1.

ð3:1Þ

X. Yang et al. / Applied Mathematics and Computation 181 (2006) 1–5

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Theorem 3.1. Let {xn} be a positive solution to (3.1). Then xn ! 1. Proof. Let xn ¼ Ay n ; n P 2. Then {yn} is a solution to the difference equation y n ¼ minfay n1 ; 1  y n2 g;

n ¼ 0; 1; . . .

ð3:2Þ

To prove xn! 1, it suffices to prove yn ! 0. If y1 = 0,y2 = 0, then we have yn = 0 for all n P 2. Next, we assume either y1 5 0 or y2 5 0. The following four claims are obviously true. Claim I: Claim II: Claim III: Claim IV:

If If If If

yn1 P 0 and yn2 P 0 for some n, then jynj 6 max{ajyn1j,jyn2j  1}. yn1 6 0 and yn2 6 0 for some n, then jynj 6 ajyn1j. yn1 P 0 and yn2 6 0 for some n, then jynj = ajyn1j. yn1 6 0 and yn2 P 0 for some n, then jynj 6 max{ajyn1j,jyn2j  1}.

In general, we have jy n j 6 maxfajy n1 j; jy n2 j  1g;

ð3:3Þ

n P 0.

A direct consequence of (3.3) is jy n j 6 maxfjy n1 j; jy n2 jg;

ð3:4Þ

n P 0.

From (3.4) and by induction, we get jy n j 6 maxfjy 1 j; jy 2 jg;

ð3:5Þ

n P 0.

Choose a number b so that 1  maxfjy 1 j;jy 1

jy 1 j 6

1 ; 1b

jy 2 j 6

2 jg

6 b < 1. Then

1 . 1b

ð3:6Þ

(3.5) and (3.6) yield jy n j  1 6 bjy n j;

n P 2.

ð3:7Þ

Let c = max{a, b}. Then 0
ð3:8Þ

n P 0.

From (3.8) and by induction, we get jy n j 6 cdn=2e max fjy n2dn=2eþ1 j; jy n2dn=2e jg; This implies yn ! 0.

n P 0.

ð3:9Þ

h

4. The case A > 1 In this section, we consider the difference equation xn ¼ maxf1=xan1 ; A=xn2 g;

n ¼ 0; 1; . . . ;

ð4:1Þ

where A > 1. Theorem 4.1. Let {xn} be a positive solution to (4.1). Then there is an integer N such that xn+2 = xn for all n P N. Proof. Let xn ¼ Ay n þ1=2 , n P 2. Then {yn} is a solution to the difference equation.   1þa ; y n2 ; n ¼ 0; 1; . . . y n ¼ max ay n1  2

ð4:2Þ

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X. Yang et al. / Applied Mathematics and Computation 181 (2006) 1–5

To prove the desired result, it suffices to show that there is an integer N such that yn+2 = yn for all n P N. By computation, we get Claim I: If for some integer N, we have     1 1þa 1þa 1 1þa 1þa yN  ; ay N  yN þ ; ay N þ max 6 y N þ1 6 min ; a 2a 2 a 2a 2 then yn+2 = yn for all n P N. From Claim I, we immediately obtain Claim II: If for some integer N, we have 2



ð1 þ aÞ 1  a2 6 yN 6 ; 2 2ð1 þ a Þ 2ð1 þ a2 Þ

y N þ1 ¼ ay N 

1þa ; 2

then yn+2 = yn for all n P N. From (4.2), we can assume that yN+1 = ayN  (1 + a)/2 holds for some integer N; otherwise the desired ð1þaÞ2 1a2 result is already proven. If  2ð1þa 2 Þ 6 y N 6 2ð1þa2 Þ, the desired result follows from Claim II. Below we examine the two remaining cases. 2

ð1þaÞ 1þa Case 1: y N <  2ð1þa 2 Þ, y N þ1 ¼ ay N  2 . By computation, we get

y N þ2 ¼ y N ;

y N þ3 ¼ y N þ1 . 2

1a 1þa 1þa Observe that y N þ3 < 2ð1þa 2 Þ and y N þ3  y N ¼ ða  1Þy N þ 2 > 2 . We proceed by checking two possibilities. 2

2

ð1þaÞ 1a 1þa Case 1.1:  2ð1þa 2 Þ 6 y N þ3 < 2ð1þa2 Þ. Observe that y N þ4 ¼ ay N þ3  2 , the desired result follows from Claim II. 2

ð1þaÞ Case 1.2: y N þ3 <  2ð1þa 2 Þ. By computation, we can derive

y N þ4 ¼ ay N þ3 

1þa ; 2

y N þ5 ¼ y N þ3 ;

y N þ6 ¼ y N þ4 .

2

2

ð1þaÞ ð1þaÞ 1a Likewise, there are two possibilities:  2ð1þa 2 Þ 6 y N þ6 < 2ð1þa2 Þ or y N þ6 <  2ð1þa2 Þ. Observe that 2

1þa 1þa > ; 2 2 1þa 1þa > ; ¼ ða  1Þy N þ3 þ 2 2

y N þ3  y N ¼ ða  1Þy N þ y N þ6  y N þ3 ...

This process will continue until there is a positive integer M such that 2



ð1 þ aÞ 1  a2 6 y 3MþN < . 2 2ð1 þ a Þ 2ð1 þ a2 Þ

, and the desired result follows from Claim II. Then y 3MþN þ1 ¼ ay 3MþN  1þa 2 1a2 1þa Case 2: y N > 2ð1þa , y ¼ ay 2Þ N þ1 N  2 . By computation, we get y N þ2 ¼ ay N þ1 

1þa . 2 2

ð1þaÞ Observe that y N þ1 <  2ð1þa 2 Þ. The subsequent argument is analogous to that for Case 1.

h

As a direct consequence of Theorem 4.1, we get Theorem 4.2. Let {xn} be a positive solution to (4.1). Then xn is eventually periodic with period 4.

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5. Conclusions and remarks This paper has examined the asymptotic behavior of positive solutions to the difference Eq. (1.3) with 0 < a < 1, A > 0. The method used in this work may provide insight into the asymptotic behavior of positive solutions to the generic difference equation   A 1 A2 Ap xn ¼ max a1 ; a2 ; . . . ; ap ; n ¼ 0; 1; . . . ; ð5:1Þ xn1 xn2 xnp where A1 P 0, A2 P 0, . . . , Ap P 0, 0 < a1 6 1, 0 < a2 6 1, . . . , 0 < ap 6 1, max{a1, a2, . . . , ap} = 1. We close this work by presenting the following conjecture: Conjecture 5.1. Let aq = 1. If Aq > max{Aj: 1 6 j 6 p, j 5 q}, then every positive solution to (5.1) is eventually periodic with period T = 2q. Clearly, this conjecture includes Conjecture 2 given in the first section of this paper as special case. Acknowledgements This work was supported by New Century’s Talents Plan Funds of Education Ministry of China, Doctorate Funds of Education Ministry of China (No. 200506110001), and Natural Science Funds of Chongqing (No. CSTC2005, BB2191). References [1] A.M. Amleh, J. Hoag, G. Ladas, A difference equation with eventually periodic solutions, Computers and Mathematics with Applications 36 (10–12) (1998) 401–404. [2] J.C. Bidwell, On the periodic nature of solutions to the reciprocal delay difference equation with maximum, Ph.D. Dissertation, North Carolina State University, 2005. [3] W.J. Briden, E.A. Grove, C.M. Kent, G. Ladas, Eventually periodic solutions of xn+1 = max{1/xn, An/xn1}, Communications on Applied Nonlinear Analysis 6 (1999) 31–34. [4] W.J. Briden, E.A. Grove, G. Ladas, L.C. McGrath, On the nonautonomous equation xn+1 = max{An/xn,Bn/xn1}, in: Proceedings of the Third International Conference on Difference Equations and Applications, 1997, Taipei, Taiwan, pp. 49–73. [5] E.A. Grove, C. Kent, G. Ladas, M.A. Radin, On xn+1 = max{1/xn,An/xn1} with a period 3 parameter, Fields Institute Communications 29 (2001). 1 [6] G. Ladas, On the recursive sequence xn ¼ maxfxAn1 ; . . . ; xAnkk g, open problems and conjectures, Journal of Difference Equations and Applications 2 (1996) 339–341. [7] D.P. Mishev, W.T. Patula, H.D. Voulov, A reciprocal difference equation with maximum, Computers and Mathematics with Applications 43 (8–9) (2002) 1021–1026. [8] D.P. Mishev, W.T. Patula, H.D. Voulov, Periodic coefficients in a reciprocal difference equation with maximum, PanAmerican Mathematical Journal 13 (3) (2003) 43–57. 1 [9] I. Szalkai, On the periodicity of the sequence xnþ1 ¼ maxfAxn0 ; xAn1 ; . . . ; xAnkk g, Journal of Difference Equations and Applications 5 (1999) 25–29. [10] H.D. Voulov, Periodic solutions to a difference equation with maximum, Proceedings of the American Mathematical Society 131 (7) (2003) 2155–2160. [11] H.D. Voulov, On the periodic nature of the solutions of the reciprocal difference equation with maximum, Journal of Mathematical Analysis and Applications 296 (1) (2004) 32–43.