- Email: [email protected]
On a non-autonomous difference equation
Applied Mathematics and Computation 168 (2005) 380–388 www.elsevier.com/locate/amc
On a non-autonomous difference equationq Xiaofan Yang a
a,*
, David J. Evans b, Graham M. Megson
c
College of Computer Science, Chongqing University, Chongqing 400044, People’s Republic of China b School of Computing and Mathematics, Nottingham Trent University, Room 421a, Newton Building, Burton Street, Nottingham NG1 4BU, UK c Department of Computer Science, School of Systems Engineering, University of Reading, P.O. Box 225, Whiteknights, Reading, Berkshire RG6 6AY, UK
Abstract Abu-Saris and DeVault proposed two open problems about the difference equation xnþ1 ¼ an xn =xn1 ;
n ¼ 0; 1; 2; . . . ;
where an 5 0 for n = 0, 1, 2, . . . , x1 5 0, x0 5 0. In this paper we provide solutions to the two open problems. Ó 2004 Elsevier Inc. All rights reserved. Keywords: Difference equation; Limit cycle; Limit point; Attractivity
1. Introduction Kulenovic and Ladas [2] suggested to investigate the asymptotic behavior of all solutions of the difference equation
q
This work is supported by Chinese National Natural Science Funds Project (No. 60271019) and ChongqingÕs Application-Oriented Fundamentals Research Funds Project (No. 7370). * Corresponding author. E-mail addresses: [email protected], [email protected] (X. Yang). 0096-3003/$ - see front matter Ó 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2004.09.004
X. Yang et al. / Appl. Math. Comput. 168 (2005) 380–388
xnþ1 ¼ an xn =xn1 ;
n ¼ 0; 1; 2; . . . ;
381
ð1:1Þ
where an 5 0 for n = 0, 1, 2, . . . , x1 5 0, x0 5 0. Recently, Abu-Saris and DeVault [1] proposed the following two open problems about Eq. (1.1). Open Problem 1. Suppose that fan g1 n¼0 converges to one. Find a necessary and sufficient condition on {an} so that every solution of Eq. (1.1) converges to a period six cycle. Open Problem 2. Find a necessary and sufficient condition on fan g1 n¼0 so that every solution of Eq. (1.1) converges to zero. In this paper, we also consider Open Problem 3. Find a necessary and sufficient condition on {an} so that every solution of Eq. (1.1) converges to a period six cycle. Our contributions in this paper are to provide solutions to the above three problems.
2. Preliminaries Inductively applying Eq. (1.1), we obtain xnþ1 ¼ an xn =xn1 ¼ an an1 =xn2 ¼ an an1 a1 n3 xn4 =xn3 1 ¼ an an1 a1 n3 an4 xn5 ¼
Finally, we have 8 hn x0 =x1 > > > > > hn =x1 > > > < h =x n 0 xnþ1 ¼ > h x > > n 1 =x0 > > > hn x1 > > : hn x 0
for n 0 ðmod 6Þ; for n 1 ðmod 6Þ; for n 2 ðmod 6Þ;
ð2:1Þ
for n 3 ðmod 6Þ; for n 4 ðmod 6Þ; for n 5 ðmod 6Þ;
1
where fhn gn¼0 is a sequence whose terms are determined by the coefficient se1 quence fan gn¼0 according to the rule: n Y 1 1 1 k a a a a a ¼ axnk ; ð2:2Þ hn ¼ an an1 a1 n6 n7 n3 n4 n9 n10 k¼0
382
X. Yang et al. / Appl. Math. Comput. 168 (2005) 380–388
(x0, x1, x2, x3, x4, x5, . . .) = (1, 1, 0,1, 1, 0, 1, 1, 0, 1, 1, 0, . . .) is a period six sequence. Abu-Saris and DeVault ([1], Lemma 2.1) found that xk ¼ ðkþ1Þp p2ffiffi sin . Clearly, we have 3 3
Lemma 2.1. h6n+3h6n = a6n+3a6n+2, h6n+4h6n+1 = a6n+4a6n+3, h6n+5h6n+2 = a6n+5 a6n+4.
3. A solution to the Open Problem 3 The following theorem solves the Open Problem 3. Theorem 3.1. Eq. (1.1) has a solution converging to a (not necessarily prime) period six cycle if and only if the sequence fh6nþk g1 n¼0 is convergent for 0 6 k 6 5. If so, then every solution of Eq. (1.1) converges to a period six cycle. Proof. Necessity. Suppose that Eq. (1.1) has a solution {xn} converging to a period six cycle. Then the limit limn!1x6n+k exists for 0 6 k 6 5. In view of Eq. (2.1), we have x1 lim x6nþ1 lim h6n ¼
n!1
n!1
x0
lim h6nþ1 ¼ x1 lim x6nþ2 ;
;
n!1
n!1
x0 lim x6nþ4 lim h6nþ2 ¼ x0 lim x6nþ3 ;
n!1
lim h6nþ3 ¼
n!1
lim x6nþ5 lim h6nþ4 ¼ n!1 ; n!1 x1
n!1
;
x1
n!1
ð3:1Þ
lim x6n lim h6nþ5 ¼ n!1 : n!1 x0
So {h6n+k} is convergent for 0 6 k 6 5. Sufficiency. Suppose that {h6n+k} is convergent for 0 6 k 6 5. Let fxn g1 n¼1 be an arbitrary solution of Eq. (1.1). According to Eq. (2.1), we derive x0 lim h6n lim x6n ¼ x0 lim h6nþ5 ;
n!1
n!1
lim h6nþ1 lim x6nþ2 ¼ n!1 ; n!1 x1
lim x6nþ1 ¼
n!1
n!1
x1
;
lim h6nþ2 lim x6nþ3 ¼ n!1 ; n!1 x0
x1 lim h6nþ3 lim x6nþ4 ¼
n!1
So
1 fxn gn¼1
n!1
x0
;
lim x6nþ5 ¼ x1 lim h6nþ4 :
n!1
converges to a period six cycle.
n!1
h
ð3:2Þ
X. Yang et al. / Appl. Math. Comput. 168 (2005) 380–388
383
4. Solution to the Open Problem 1 The following theorem is a solution to the Open Problem 1. Theorem 4.1. Suppose that fan g1 n¼0 converges to one. Then Eq. (1.1) has a solution converging to a period six cycle if and only if the sequence {h6n+k} is convergent to a non-zero value for 0 6 k 6 2. If so, then every solution of Eq. (1.1) converges to a period six cycle. Proof. Necessity. Suppose that Eq. (1.1) has a solution {xn} converging to a period six cycle. By Theorem 3.1, {h6n+k} is convergent for 0 6 k 6 5. The remaining thing is to show that limn!1h6n+k 5 0 for 0 6 k 6 2. Suppose, on the contrary, that there was k in{0, 1, 2} so that limn!1h6n+k = 0. By Lemma 2.1, h6n+k+3h6n+k = a6n+k+3a6n+k+2. Since limn!1(a6n+k+3a6n+k+2) = 1, we could derive limn!hh6n+k+3 = 1. This, however, would contradict that {h6n+k+3} is convergent. Sufficiency. Suppose that {h6n+k} is convergent to a non-zero value for 0 6 k 6 2. According to Lemma 2.1, h6nþ3 h6n ¼ a6nþ3 a6nþ2 ;
h6nþ4 h6nþ1 ¼ a6nþ4 a6nþ3 ;
h6nþ5 h6nþ2 ¼ a6nþ5 a6nþ4 :
Taking the limits on both sides of each of the above equations, we obtain lim h6nþ3 lim h6n ¼ 1;
n!1
n!1
lim h6nþ4 lim h6nþ1 ¼ 1;
n!1
n!1
lim h6nþ5 lim h6nþ2 ¼ 1:
n!1
n!1
So {h6n+k} is convergent to a non-zero value for 3 6 k 6 5. It then follows from Theorem 3.1 that every solution of Eq. (1.1) converges to a period six cycle. h Next we give an easily checked sufficient condition for every solution of Eq. (1.1) to converge to a period six cycle. Theorem 4.2. Suppose that {an} converges to one. If for each 0 6 k 6 2, there is an integer N(k) so that the sequence fa3nþk a3nþkþ1 g1 n¼N ðkÞ is monotonic, then every solution of Eq. (1.1) is convergent to a period six cycle. Proof. Clearly, there is an integer N1 so that an > 0 for n P N1. Let k be an arbitrary integer with 0 6 k 6 2. Here we assume fa3nþk a3nþkþ1 g1 n¼N ðkÞ is monotonically increasing (The case where fa3nþk a3nþkþ1 g1 is monotonin¼NðkÞ cally decreasing can be discussed similarly). Clearly, we have 6ðnN YÞþ5 x N1 N ðkÞ q a6nþkþ6q for n P N ¼ max : h6nþkþ6 ¼ h6N þk ; 2 6 q¼0 ð4:1Þ
384
X. Yang et al. / Appl. Math. Comput. 168 (2005) 380–388
Let q6nþkþ6 ¼
Q6ðnN Þþ5 q¼0
x
q a6nþkþ6q . Then Eq. (4.1) is equivalent to
h6nþkþ6 ¼ h6N þk q6nþkþ6
for n P N :
ð4:2Þ
1
Moreover, the sequence fq6nþk gn¼N satisfies the following recursive relations: q6nþkþ6 ¼
a6nþkþ6 a6nþkþ5 q a6nþkþ3 a6nþkþ2 6nþk
for n P N ;
q6N þk ¼ 1:
ð4:3Þ ð4:4Þ
Clearly, fq6nþk g1 n¼N is monotonically increasing. Inductively applying Eq. (4.3), we derive a6nþk a6nþk1 a6nþkþ3 a6nþkþ2 a6nþk6 a6nþk7 a6N þkþ6 a6N þkþ5 1 a6nþk3 a6nþk4 a6N þkþ9 a6N þkþ8 a6N þkþ3 a6N þkþ2 a6nþkþ6 a6nþkþ5 6 : a6N þkþ3 a6N þkþ2
q6nþkþ6 ¼ a6nþkþ6 a6nþkþ5
ð4:5Þ
Since a6N+k+3 > 0, a6N+k+2 > 0, a6n+k+6 > 0, a6n+k+5 > 0, limn!1a6n+k+6 = 1, and limn!1a6n+k+5 = 1, there is a positive number M so that q6nþkþ6 6
a6nþkþ6 a6nþkþ5 6M a6N þkþ3 a6N þkþ2
for n P N :
ð4:6Þ
1
Thus the sequence fq6nþk gn¼N is convergent to a non-zero value. Taking the limit on both sides of Eq. (4.2), we obtain lim h6nþk ¼ h6N þk lim q6nþk :
n!1
n!1
ð4:7Þ
So the sequence {h6n+k} is also convergent to a non-zero value. According to Theorem 4.1, every solution of Eq. (1.1) is convergent to a period six cycle. h We close this section by checking the prime periods of the limit cycles of Eq. (1.1). Before doing this, it should be noted that a point, a prime period two cycle, or a prime period three cycle can be viewed as a degenerate period six cycle. So, according to Theorem 4.1, Eq. (1.1) with limn!1an = 1 has a solution converging to a point or a prime period two cycle or a prime period three cycle only if {h6n+k} is convergent to a non-zero value for 0 6 k 6 2. Theorem 4.3. Suppose that fan g1 n¼0 converges to one, and {h6n+k} is convergent to a non-zero value for 0 6 k 6 2.
X. Yang et al. / Appl. Math. Comput. 168 (2005) 380–388
385
(1) Eq. (1.1) has a solution converging to a point if and only if limn!1h6nlimn!1h6n+2 = ±limn!1h6n+1. If so, Eq. (1.1) has a unique solution n!1 h6nþ1 {xn}, which is defined by ðx1 ; x0 Þ ¼ limn!1 h6n limn!1 h6nþ2 ; lim , limn!1 h6n that converges to a point. (2) Eq. (1.1) has a solution converging to a prime period two cycle if and only if limn!1h6n limn!1h6n+2 5 ±limn!1h6n+1. If so, Eq. (1.1) has a unique solution {xn}, which is defined by ðx1 ; x0 Þ ¼ ½limn!1 h6n ðlimn!1 h i1 1 ðlimn!1 h6nþ2 Þ2 3 , that converges to a prime h6nþ1 Þ2 limn!1 h6nþ2 3 ; limn!1 h6nþ1 limn!1 h6n period two cycle. (3) Eq. (1.1) has a solution converging to a prime period three cycle if and only if limn!1h6nlimn!1h6n+2 = ±limn!1h6n+1. Moreover, the following conclusions hold. (3a) If limn!1h6nlimn!1h6n+2 = limn!1h6n+1, then Eq. (1.1) has exactly three solutions, which are defined, respectively, by ðx1 ; x0 Þ ¼ lim h6nþ1 ; lim h6nþ2 ; n!1
n!1
ðx1 ; x0 Þ ¼ lim h6nþ1 ; lim h6nþ2 n!1
n!1
and ðx1 ; x0 Þ ¼ lim h6nþ1 ; lim h6nþ2 ; n!1
n!1
that are each convergent to a prime period three cycle. (3b) If limn!1h6nlimn!1h6n+2 = limn!1h6n+1, then Eq. (1.1) has exactly three solutions, which are defined, respectively, by ðx1 ; x0 Þ ¼ lim h6nþ1 ; lim h6nþ2 ; n!1
ðx1 ; x0 Þ ¼
n!1
lim h6nþ1 ; lim h6nþ2
n!1
n!1
and
ðx1 ; x0 Þ ¼ lim h6nþ1 ; lim h6nþ2 ; n!1
n!1
that are each convergent to a prime period three cycle. Proof. From the given conditions and Lemma 2.1, we have lim h6nþ3 ¼
n!1
1 ; lim h6n
n!1
lim h6nþ4 ¼
n!1
1 ; lim h6nþ1
n!1
lim h6nþ5 ¼
n!1
1 : lim h6nþ2
n!1
ð4:8Þ
386
X. Yang et al. / Appl. Math. Comput. 168 (2005) 380–388
Now consider an arbitrary solution {xn} of Eq. (1.1). Substituting Eq. (4.8) into Eq. (3.2), we deduce x0 lim h6n x0 ; lim x6nþ1 ¼ n!1 ; lim x6n ¼ n!1 n!1 lim h6nþ2 x1 n!1
lim h6nþ1 lim x6nþ2 ¼ n!1 ; n!1 x1 x1 lim x6nþ4 ¼ ; n!1 x0 lim h6n n!1
lim h6nþ2 lim x6nþ3 ¼ n!1 ; n!1 x0 x1 lim x6nþ5 ¼ : n!1 lim h6nþ1
ð4:9Þ
n!1
(1) Necessity. Let {xn} be a solution of Eq. (1.1) converging to a point. It follows from Eq. (4.9) that x0 lim h6n lim h6nþ1 lim h6nþ2 x0 x1 x1 ¼ n!1 ¼ n!1 ¼ n!1 ¼ ¼ : lim h6nþ2 x1 x1 x0 x0 lim h6n lim h6nþ1 n!1
n!1
n!1
ð4:10Þ Eq. (4.10) are equivalent to lim h6n lim h6nþ2 ¼ lim h6nþ1 ;
n!1
n!1
n!1
x1 ¼ lim h6n lim h6nþ2 n!1
n!1
and
lim h6nþ1
x0 ¼
n!1
lim h6n
:
n!1
Sufficiency. Suppose that limn!1h6nlimn!1h6n+2 = ± limn!1h6n+1. It is easily verified that the solution {xn} of Eq. (1.1) defined by ðx1 ; x0 Þ ¼ n!1 h6nþ1 limn!1 h6n limn!1 h6nþ2 ; lim converges to a point. limn!1 h6n (2) Necessity. Let {xn} be a solution of Eq. (1.1) converging to a prime period two cycle. It follows from Eq. (4.9) that lim h6nþ1 x0 lim h6n lim h6nþ2 x0 x1 x1 n!1 ¼ n!1 ¼ ; ¼ n!1 ¼ ; lim h6nþ2 x1 x0 lim h6n x1 x0 lim h6nþ1 n!1
n!1
lim h6nþ2 x0 6¼ n!1 : lim h6nþ2 x0
n!1
Eq. (4.11) are equivalent to lim h6n lim h6nþ2 6¼ lim h6nþ1 ; n!1 n!1 n!1 h i13 x1 ¼ lim h6n ðlim h6nþ1 Þ2 lim h6nþ2 ; n!1 n!1 2 31 lim h6nþ1 ð lim h6nþ2 Þ2 3 n!1 5: x0 ¼ 4n!1 lim h6n n!1
n!1
ð4:11Þ
X. Yang et al. / Appl. Math. Comput. 168 (2005) 380–388
387
Sufficiency. Suppose that limn!1h6nlimn!1h6n+2 5 ±limn!1h6n+1. Clearly, Eq. h (1.1) defined 2 i13by ðx1 ; x0 Þ ¼ ð½limn!1 h6n ðlimn!1 h6nþ2 Þ ðlimn!1 h6nþ1 Þ limn!1 h6nþ2 ; limn!1 h6nþ1 Þ converges to a prime perlimn!1 h6n iod two cycle. (3) Necessity. Let {xn} be a solution of Eq. (1.1) converging to a prime period three cycle. It follows from Eq. (4.9) that the
solution
{xn}
of
2
1 3
lim h6nþ2 x0 ¼ n!1 ; lim h6nþ2 x0
x0 lim h6n n!1
x1
n!1
either
x0 6¼ lim h6nþ2
x0 lim h6n n!1
or
x1
n!1
¼
x1 ; x0 lim h6n
lim h6nþ1
n!1
n!1
x1
¼
x1 ; lim h6nþ1
n!1
x1 x1 6¼ : x0 lim h6n lim h6nþ1 n!1
n!1
ð4:12Þ Eq. (4.12) are equivalent to x0 ¼ lim h6nþ2 ; n!1
x1 ¼ lim h6nþ1 ; n!1
lim h6n lim h6nþ2 ¼ lim h6nþ1 ;
n!1
n!1
n!1
lim h6nþ1
: either x1 6¼ lim h6n lim h6nþ2 ; or x0 6¼ n!1 n!1 n!1 lim h6n n!1
ð4:13Þ Here there are two possibilities: Case 1: limn!1h6nlimn!1h6n+2 = limn!1h6n+1. In this case (x1, x0) assumes one of the following three values: (limn!1h6n+1, limn!1h6n+2), (limn!1 h6n+1, limn!1h6n+2), (limn!1h6n+1, limn!1h6n+2). Case 2: limn!1h6nlimn!1h6n+2 = limn!1h6n+1. In this case (x1, x0) assumes one of the following three values: (limn!1h6n+1, limn!1h6n+2), (limn!1h6n+1, limn!1h6n+2), (limn!1h6n+1, limn!1h6n+2). Sufficiency. When limn!1h6nlimn!1h6n+2 = limn!1h6n+1. It is easily verified that each of the three solutions defined by (x1, x0) = (limn!1h6n+1, limn!1 h6n+2), (x1, x0) = (limn!1h6n+1, limn!1h6n+2), and (x1, x0) = (limn!1 h6n+1, limn!1h6n+2), respectively, converges to a prime period three cycle. When limn!1h6nlimn!1h6n+2 = limn!1h6n+1. It is easily verified that each of the three solutions defined by (x1, x0) = (limn!1h6n+1, limn!1h6n+2), (x1, x0) = (limn!1h6n+1, limn!1h6n+2), and (x1, x0) = (limn!1h6n+1, limn!1 h6n+2), respectively, converges to a prime period three cycle. h
5. Solution to the Open Problem 2 The following theorem provides a solution to the Open Problem 2.
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X. Yang et al. / Appl. Math. Comput. 168 (2005) 380–388
Theorem 5.1. Eq. (1.1) has a solution converging to zero if and only if {hn} is convergent to zero. If so, then every solution of Eq. (1.1) approaches to zero. Proof. Necessity. This is a direct consequence of Eq. (3.1). Sufficiency. This is an immediate consequence of Eq. (3.2).
h
References [1] R. Abu-Saris, R. DeVault, On the asymptotic behavior of xn+1 = anxn/xn1, Journal of Mathematical Analysis and Applications 280 (2003) 148–154. [2] M. Kulenovic, G. Ladas, Dynamics of Second Order Rational Difference Equations, Chapman and Hall/CRC Press, 2002.
On a non-autonomous difference equationq Xiaofan Yang a
a,*
, David J. Evans b, Graham M. Megson
c
College of Computer Science, Chongqing University, Chongqing 400044, People’s Republic of China b School of Computing and Mathematics, Nottingham Trent University, Room 421a, Newton Building, Burton Street, Nottingham NG1 4BU, UK c Department of Computer Science, School of Systems Engineering, University of Reading, P.O. Box 225, Whiteknights, Reading, Berkshire RG6 6AY, UK
Abstract Abu-Saris and DeVault proposed two open problems about the difference equation xnþ1 ¼ an xn =xn1 ;
n ¼ 0; 1; 2; . . . ;
where an 5 0 for n = 0, 1, 2, . . . , x1 5 0, x0 5 0. In this paper we provide solutions to the two open problems. Ó 2004 Elsevier Inc. All rights reserved. Keywords: Difference equation; Limit cycle; Limit point; Attractivity
1. Introduction Kulenovic and Ladas [2] suggested to investigate the asymptotic behavior of all solutions of the difference equation
q
This work is supported by Chinese National Natural Science Funds Project (No. 60271019) and ChongqingÕs Application-Oriented Fundamentals Research Funds Project (No. 7370). * Corresponding author. E-mail addresses: [email protected], [email protected] (X. Yang). 0096-3003/$ - see front matter Ó 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2004.09.004
X. Yang et al. / Appl. Math. Comput. 168 (2005) 380–388
xnþ1 ¼ an xn =xn1 ;
n ¼ 0; 1; 2; . . . ;
381
ð1:1Þ
where an 5 0 for n = 0, 1, 2, . . . , x1 5 0, x0 5 0. Recently, Abu-Saris and DeVault [1] proposed the following two open problems about Eq. (1.1). Open Problem 1. Suppose that fan g1 n¼0 converges to one. Find a necessary and sufficient condition on {an} so that every solution of Eq. (1.1) converges to a period six cycle. Open Problem 2. Find a necessary and sufficient condition on fan g1 n¼0 so that every solution of Eq. (1.1) converges to zero. In this paper, we also consider Open Problem 3. Find a necessary and sufficient condition on {an} so that every solution of Eq. (1.1) converges to a period six cycle. Our contributions in this paper are to provide solutions to the above three problems.
2. Preliminaries Inductively applying Eq. (1.1), we obtain xnþ1 ¼ an xn =xn1 ¼ an an1 =xn2 ¼ an an1 a1 n3 xn4 =xn3 1 ¼ an an1 a1 n3 an4 xn5 ¼
Finally, we have 8 hn x0 =x1 > > > > > hn =x1 > > > < h =x n 0 xnþ1 ¼ > h x > > n 1 =x0 > > > hn x1 > > : hn x 0
for n 0 ðmod 6Þ; for n 1 ðmod 6Þ; for n 2 ðmod 6Þ;
ð2:1Þ
for n 3 ðmod 6Þ; for n 4 ðmod 6Þ; for n 5 ðmod 6Þ;
1
where fhn gn¼0 is a sequence whose terms are determined by the coefficient se1 quence fan gn¼0 according to the rule: n Y 1 1 1 k a a a a a ¼ axnk ; ð2:2Þ hn ¼ an an1 a1 n6 n7 n3 n4 n9 n10 k¼0
382
X. Yang et al. / Appl. Math. Comput. 168 (2005) 380–388
(x0, x1, x2, x3, x4, x5, . . .) = (1, 1, 0,1, 1, 0, 1, 1, 0, 1, 1, 0, . . .) is a period six sequence. Abu-Saris and DeVault ([1], Lemma 2.1) found that xk ¼ ðkþ1Þp p2ffiffi sin . Clearly, we have 3 3
Lemma 2.1. h6n+3h6n = a6n+3a6n+2, h6n+4h6n+1 = a6n+4a6n+3, h6n+5h6n+2 = a6n+5 a6n+4.
3. A solution to the Open Problem 3 The following theorem solves the Open Problem 3. Theorem 3.1. Eq. (1.1) has a solution converging to a (not necessarily prime) period six cycle if and only if the sequence fh6nþk g1 n¼0 is convergent for 0 6 k 6 5. If so, then every solution of Eq. (1.1) converges to a period six cycle. Proof. Necessity. Suppose that Eq. (1.1) has a solution {xn} converging to a period six cycle. Then the limit limn!1x6n+k exists for 0 6 k 6 5. In view of Eq. (2.1), we have x1 lim x6nþ1 lim h6n ¼
n!1
n!1
x0
lim h6nþ1 ¼ x1 lim x6nþ2 ;
;
n!1
n!1
x0 lim x6nþ4 lim h6nþ2 ¼ x0 lim x6nþ3 ;
n!1
lim h6nþ3 ¼
n!1
lim x6nþ5 lim h6nþ4 ¼ n!1 ; n!1 x1
n!1
;
x1
n!1
ð3:1Þ
lim x6n lim h6nþ5 ¼ n!1 : n!1 x0
So {h6n+k} is convergent for 0 6 k 6 5. Sufficiency. Suppose that {h6n+k} is convergent for 0 6 k 6 5. Let fxn g1 n¼1 be an arbitrary solution of Eq. (1.1). According to Eq. (2.1), we derive x0 lim h6n lim x6n ¼ x0 lim h6nþ5 ;
n!1
n!1
lim h6nþ1 lim x6nþ2 ¼ n!1 ; n!1 x1
lim x6nþ1 ¼
n!1
n!1
x1
;
lim h6nþ2 lim x6nþ3 ¼ n!1 ; n!1 x0
x1 lim h6nþ3 lim x6nþ4 ¼
n!1
So
1 fxn gn¼1
n!1
x0
;
lim x6nþ5 ¼ x1 lim h6nþ4 :
n!1
converges to a period six cycle.
n!1
h
ð3:2Þ
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383
4. Solution to the Open Problem 1 The following theorem is a solution to the Open Problem 1. Theorem 4.1. Suppose that fan g1 n¼0 converges to one. Then Eq. (1.1) has a solution converging to a period six cycle if and only if the sequence {h6n+k} is convergent to a non-zero value for 0 6 k 6 2. If so, then every solution of Eq. (1.1) converges to a period six cycle. Proof. Necessity. Suppose that Eq. (1.1) has a solution {xn} converging to a period six cycle. By Theorem 3.1, {h6n+k} is convergent for 0 6 k 6 5. The remaining thing is to show that limn!1h6n+k 5 0 for 0 6 k 6 2. Suppose, on the contrary, that there was k in{0, 1, 2} so that limn!1h6n+k = 0. By Lemma 2.1, h6n+k+3h6n+k = a6n+k+3a6n+k+2. Since limn!1(a6n+k+3a6n+k+2) = 1, we could derive limn!hh6n+k+3 = 1. This, however, would contradict that {h6n+k+3} is convergent. Sufficiency. Suppose that {h6n+k} is convergent to a non-zero value for 0 6 k 6 2. According to Lemma 2.1, h6nþ3 h6n ¼ a6nþ3 a6nþ2 ;
h6nþ4 h6nþ1 ¼ a6nþ4 a6nþ3 ;
h6nþ5 h6nþ2 ¼ a6nþ5 a6nþ4 :
Taking the limits on both sides of each of the above equations, we obtain lim h6nþ3 lim h6n ¼ 1;
n!1
n!1
lim h6nþ4 lim h6nþ1 ¼ 1;
n!1
n!1
lim h6nþ5 lim h6nþ2 ¼ 1:
n!1
n!1
So {h6n+k} is convergent to a non-zero value for 3 6 k 6 5. It then follows from Theorem 3.1 that every solution of Eq. (1.1) converges to a period six cycle. h Next we give an easily checked sufficient condition for every solution of Eq. (1.1) to converge to a period six cycle. Theorem 4.2. Suppose that {an} converges to one. If for each 0 6 k 6 2, there is an integer N(k) so that the sequence fa3nþk a3nþkþ1 g1 n¼N ðkÞ is monotonic, then every solution of Eq. (1.1) is convergent to a period six cycle. Proof. Clearly, there is an integer N1 so that an > 0 for n P N1. Let k be an arbitrary integer with 0 6 k 6 2. Here we assume fa3nþk a3nþkþ1 g1 n¼N ðkÞ is monotonically increasing (The case where fa3nþk a3nþkþ1 g1 is monotonin¼NðkÞ cally decreasing can be discussed similarly). Clearly, we have 6ðnN YÞþ5 x N1 N ðkÞ q a6nþkþ6q for n P N ¼ max : h6nþkþ6 ¼ h6N þk ; 2 6 q¼0 ð4:1Þ
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Let q6nþkþ6 ¼
Q6ðnN Þþ5 q¼0
x
q a6nþkþ6q . Then Eq. (4.1) is equivalent to
h6nþkþ6 ¼ h6N þk q6nþkþ6
for n P N :
ð4:2Þ
1
Moreover, the sequence fq6nþk gn¼N satisfies the following recursive relations: q6nþkþ6 ¼
a6nþkþ6 a6nþkþ5 q a6nþkþ3 a6nþkþ2 6nþk
for n P N ;
q6N þk ¼ 1:
ð4:3Þ ð4:4Þ
Clearly, fq6nþk g1 n¼N is monotonically increasing. Inductively applying Eq. (4.3), we derive a6nþk a6nþk1 a6nþkþ3 a6nþkþ2 a6nþk6 a6nþk7 a6N þkþ6 a6N þkþ5 1 a6nþk3 a6nþk4 a6N þkþ9 a6N þkþ8 a6N þkþ3 a6N þkþ2 a6nþkþ6 a6nþkþ5 6 : a6N þkþ3 a6N þkþ2
q6nþkþ6 ¼ a6nþkþ6 a6nþkþ5
ð4:5Þ
Since a6N+k+3 > 0, a6N+k+2 > 0, a6n+k+6 > 0, a6n+k+5 > 0, limn!1a6n+k+6 = 1, and limn!1a6n+k+5 = 1, there is a positive number M so that q6nþkþ6 6
a6nþkþ6 a6nþkþ5 6M a6N þkþ3 a6N þkþ2
for n P N :
ð4:6Þ
1
Thus the sequence fq6nþk gn¼N is convergent to a non-zero value. Taking the limit on both sides of Eq. (4.2), we obtain lim h6nþk ¼ h6N þk lim q6nþk :
n!1
n!1
ð4:7Þ
So the sequence {h6n+k} is also convergent to a non-zero value. According to Theorem 4.1, every solution of Eq. (1.1) is convergent to a period six cycle. h We close this section by checking the prime periods of the limit cycles of Eq. (1.1). Before doing this, it should be noted that a point, a prime period two cycle, or a prime period three cycle can be viewed as a degenerate period six cycle. So, according to Theorem 4.1, Eq. (1.1) with limn!1an = 1 has a solution converging to a point or a prime period two cycle or a prime period three cycle only if {h6n+k} is convergent to a non-zero value for 0 6 k 6 2. Theorem 4.3. Suppose that fan g1 n¼0 converges to one, and {h6n+k} is convergent to a non-zero value for 0 6 k 6 2.
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385
(1) Eq. (1.1) has a solution converging to a point if and only if limn!1h6nlimn!1h6n+2 = ±limn!1h6n+1. If so, Eq. (1.1) has a unique solution n!1 h6nþ1 {xn}, which is defined by ðx1 ; x0 Þ ¼ limn!1 h6n limn!1 h6nþ2 ; lim , limn!1 h6n that converges to a point. (2) Eq. (1.1) has a solution converging to a prime period two cycle if and only if limn!1h6n limn!1h6n+2 5 ±limn!1h6n+1. If so, Eq. (1.1) has a unique solution {xn}, which is defined by ðx1 ; x0 Þ ¼ ½limn!1 h6n ðlimn!1 h i1 1 ðlimn!1 h6nþ2 Þ2 3 , that converges to a prime h6nþ1 Þ2 limn!1 h6nþ2 3 ; limn!1 h6nþ1 limn!1 h6n period two cycle. (3) Eq. (1.1) has a solution converging to a prime period three cycle if and only if limn!1h6nlimn!1h6n+2 = ±limn!1h6n+1. Moreover, the following conclusions hold. (3a) If limn!1h6nlimn!1h6n+2 = limn!1h6n+1, then Eq. (1.1) has exactly three solutions, which are defined, respectively, by ðx1 ; x0 Þ ¼ lim h6nþ1 ; lim h6nþ2 ; n!1
n!1
ðx1 ; x0 Þ ¼ lim h6nþ1 ; lim h6nþ2 n!1
n!1
and ðx1 ; x0 Þ ¼ lim h6nþ1 ; lim h6nþ2 ; n!1
n!1
that are each convergent to a prime period three cycle. (3b) If limn!1h6nlimn!1h6n+2 = limn!1h6n+1, then Eq. (1.1) has exactly three solutions, which are defined, respectively, by ðx1 ; x0 Þ ¼ lim h6nþ1 ; lim h6nþ2 ; n!1
ðx1 ; x0 Þ ¼
n!1
lim h6nþ1 ; lim h6nþ2
n!1
n!1
and
ðx1 ; x0 Þ ¼ lim h6nþ1 ; lim h6nþ2 ; n!1
n!1
that are each convergent to a prime period three cycle. Proof. From the given conditions and Lemma 2.1, we have lim h6nþ3 ¼
n!1
1 ; lim h6n
n!1
lim h6nþ4 ¼
n!1
1 ; lim h6nþ1
n!1
lim h6nþ5 ¼
n!1
1 : lim h6nþ2
n!1
ð4:8Þ
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X. Yang et al. / Appl. Math. Comput. 168 (2005) 380–388
Now consider an arbitrary solution {xn} of Eq. (1.1). Substituting Eq. (4.8) into Eq. (3.2), we deduce x0 lim h6n x0 ; lim x6nþ1 ¼ n!1 ; lim x6n ¼ n!1 n!1 lim h6nþ2 x1 n!1
lim h6nþ1 lim x6nþ2 ¼ n!1 ; n!1 x1 x1 lim x6nþ4 ¼ ; n!1 x0 lim h6n n!1
lim h6nþ2 lim x6nþ3 ¼ n!1 ; n!1 x0 x1 lim x6nþ5 ¼ : n!1 lim h6nþ1
ð4:9Þ
n!1
(1) Necessity. Let {xn} be a solution of Eq. (1.1) converging to a point. It follows from Eq. (4.9) that x0 lim h6n lim h6nþ1 lim h6nþ2 x0 x1 x1 ¼ n!1 ¼ n!1 ¼ n!1 ¼ ¼ : lim h6nþ2 x1 x1 x0 x0 lim h6n lim h6nþ1 n!1
n!1
n!1
ð4:10Þ Eq. (4.10) are equivalent to lim h6n lim h6nþ2 ¼ lim h6nþ1 ;
n!1
n!1
n!1
x1 ¼ lim h6n lim h6nþ2 n!1
n!1
and
lim h6nþ1
x0 ¼
n!1
lim h6n
:
n!1
Sufficiency. Suppose that limn!1h6nlimn!1h6n+2 = ± limn!1h6n+1. It is easily verified that the solution {xn} of Eq. (1.1) defined by ðx1 ; x0 Þ ¼ n!1 h6nþ1 limn!1 h6n limn!1 h6nþ2 ; lim converges to a point. limn!1 h6n (2) Necessity. Let {xn} be a solution of Eq. (1.1) converging to a prime period two cycle. It follows from Eq. (4.9) that lim h6nþ1 x0 lim h6n lim h6nþ2 x0 x1 x1 n!1 ¼ n!1 ¼ ; ¼ n!1 ¼ ; lim h6nþ2 x1 x0 lim h6n x1 x0 lim h6nþ1 n!1
n!1
lim h6nþ2 x0 6¼ n!1 : lim h6nþ2 x0
n!1
Eq. (4.11) are equivalent to lim h6n lim h6nþ2 6¼ lim h6nþ1 ; n!1 n!1 n!1 h i13 x1 ¼ lim h6n ðlim h6nþ1 Þ2 lim h6nþ2 ; n!1 n!1 2 31 lim h6nþ1 ð lim h6nþ2 Þ2 3 n!1 5: x0 ¼ 4n!1 lim h6n n!1
n!1
ð4:11Þ
X. Yang et al. / Appl. Math. Comput. 168 (2005) 380–388
387
Sufficiency. Suppose that limn!1h6nlimn!1h6n+2 5 ±limn!1h6n+1. Clearly, Eq. h (1.1) defined 2 i13by ðx1 ; x0 Þ ¼ ð½limn!1 h6n ðlimn!1 h6nþ2 Þ ðlimn!1 h6nþ1 Þ limn!1 h6nþ2 ; limn!1 h6nþ1 Þ converges to a prime perlimn!1 h6n iod two cycle. (3) Necessity. Let {xn} be a solution of Eq. (1.1) converging to a prime period three cycle. It follows from Eq. (4.9) that the
solution
{xn}
of
2
1 3
lim h6nþ2 x0 ¼ n!1 ; lim h6nþ2 x0
x0 lim h6n n!1
x1
n!1
either
x0 6¼ lim h6nþ2
x0 lim h6n n!1
or
x1
n!1
¼
x1 ; x0 lim h6n
lim h6nþ1
n!1
n!1
x1
¼
x1 ; lim h6nþ1
n!1
x1 x1 6¼ : x0 lim h6n lim h6nþ1 n!1
n!1
ð4:12Þ Eq. (4.12) are equivalent to x0 ¼ lim h6nþ2 ; n!1
x1 ¼ lim h6nþ1 ; n!1
lim h6n lim h6nþ2 ¼ lim h6nþ1 ;
n!1
n!1
n!1
lim h6nþ1
: either x1 6¼ lim h6n lim h6nþ2 ; or x0 6¼ n!1 n!1 n!1 lim h6n n!1
ð4:13Þ Here there are two possibilities: Case 1: limn!1h6nlimn!1h6n+2 = limn!1h6n+1. In this case (x1, x0) assumes one of the following three values: (limn!1h6n+1, limn!1h6n+2), (limn!1 h6n+1, limn!1h6n+2), (limn!1h6n+1, limn!1h6n+2). Case 2: limn!1h6nlimn!1h6n+2 = limn!1h6n+1. In this case (x1, x0) assumes one of the following three values: (limn!1h6n+1, limn!1h6n+2), (limn!1h6n+1, limn!1h6n+2), (limn!1h6n+1, limn!1h6n+2). Sufficiency. When limn!1h6nlimn!1h6n+2 = limn!1h6n+1. It is easily verified that each of the three solutions defined by (x1, x0) = (limn!1h6n+1, limn!1 h6n+2), (x1, x0) = (limn!1h6n+1, limn!1h6n+2), and (x1, x0) = (limn!1 h6n+1, limn!1h6n+2), respectively, converges to a prime period three cycle. When limn!1h6nlimn!1h6n+2 = limn!1h6n+1. It is easily verified that each of the three solutions defined by (x1, x0) = (limn!1h6n+1, limn!1h6n+2), (x1, x0) = (limn!1h6n+1, limn!1h6n+2), and (x1, x0) = (limn!1h6n+1, limn!1 h6n+2), respectively, converges to a prime period three cycle. h
5. Solution to the Open Problem 2 The following theorem provides a solution to the Open Problem 2.
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X. Yang et al. / Appl. Math. Comput. 168 (2005) 380–388
Theorem 5.1. Eq. (1.1) has a solution converging to zero if and only if {hn} is convergent to zero. If so, then every solution of Eq. (1.1) approaches to zero. Proof. Necessity. This is a direct consequence of Eq. (3.1). Sufficiency. This is an immediate consequence of Eq. (3.2).
h
References [1] R. Abu-Saris, R. DeVault, On the asymptotic behavior of xn+1 = anxn/xn1, Journal of Mathematical Analysis and Applications 280 (2003) 148–154. [2] M. Kulenovic, G. Ladas, Dynamics of Second Order Rational Difference Equations, Chapman and Hall/CRC Press, 2002.