Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2009, Article ID 152964, 10 pages
doi:10.1155/2009/152964
Research Article
Global Behavior of the Max-Type Difference
Equation xn1 max{1/xn , An /xn−1 }
Taixiang Sun,1 Bin Qin,2 Hongjian Xi,2 and Caihong Han1
1
2
College of Mathematics and Information Science, Guangxi University, Nanning 530004, China
Department of Mathematics, Guangxi College of Finance and Economics, Nanning 530003, China
Correspondence should be addressed to Taixiang Sun, stx1963@163.com
Received 4 February 2009; Revised 2 March 2009; Accepted 8 March 2009
Recommended by Stevo Stevic
We study global behavior of the following max-type difference equation xn1
max{1/xn , An /xn−1 }, n 0, 1, . . . , where {An }∞
n0 is a sequence of positive real numbers
with 0 ≤ inf An ≤ sup An < 1. The special case when {An }∞
n0 is a periodic sequence with period k
and An ∈ 0, 1 for every n ≥ 0 has been completely investigated by Y. Chen. Here we extend his
results to the general case.
Copyright q 2009 Taixiang Sun et al. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
1. Introduction
In the recent years, there has been a lot of interest in studying the global behavior of, the socalled, max-type difference equations; see, for example, 1–17 see also references therein.
In 1, 3–5, 7, 8, the second order max-type difference equation
xn1 max
1 An
,
,
xn xn−1
n 0, 1, . . .
1.1
has been studied for positive coefficients An , which are periodic with period k. The case k 1
was studied in 1, the case k 2 was studied in 3, the case k 3 was studied in 4, 8,
and the more difficult case k 4 was studied in 7. Chen 5 found that every positive
solution of 1.1 is eventually periodic with period 2 when {An }∞
n0 is a periodic sequence of
positive real numbers with period k ≥ 2 and An ∈ 0, 1 for all n ≥ 0. These results were
also included in the recent monograph 9 along with other related references. In this paper,
we study global behavior of 1.1 when {An }∞
n0 is a sequence of positive real numbers with
0 ≤ infAn ≤ sup An < 1.
2
Abstract and Applied Analysis
2. Main Results
The main results of this paper are established through the following lemmas.
Lemma 2.1. Let {xn }∞
n−1 be a positive solution of 1.1, then
1 xn1 xn ≥ 1 for all n ≥ 0;
2 if xk1 xk > 1 for some k ≥ 1, then xk2 xk1 1.
Proof. 1 is obvious since xn1 ≥ 1/xn for all n ≥ 0.
2 If xk1 xk > 1 for some k ≥ 1, then xk1 xk−1 Ak . Suppose for the sake of
contradiction that xk2 xk1 > 1, then similarly we get xk2 xk Ak1 and
Ak1 Ak xk1 xk−1 xk2 xk ≥ 1.
2.1
This is a contradiction since Ak1 < 1 and Ak < 1. The proof is complete.
Lemma 2.2. Let {xn }∞
n−1 be a positive solution of 1.1 and Pn max{xn , xn−1 } for all n ≥ 1. Then
1 xn1 ≤ Pn and Pn is nonincreasing;
2 xn is bounded, and moreover 1/P1 ≤ xn ≤ P1 for any n ≥ 1.
Proof. By Lemma 2.11 and the assumption An < 1, we obtain that for any n ≥ 1,
xn1 max
An xn
xn−1
,
xn xn−1 xn xn−1
≤ max{xn−1 , xn } Pn .
2.2
Hence
Pn1 max{xn1 , xn } ≤ Pn ,
2.3
xn ≤ P1 .
2.4
which implies that for all n ≥ 1,
Furthermore, it follows that for all n ≥ 1,
xn1 max
1 An
,
xn xn−1
≥
1
1
≥
.
xn P1
2.5
The proof is complete.
Remark 2.3. Note that from the proof of Lemma 2.2 we have that P1 ≥ 1.
Remark 2.4. Various sequences which satisfy inequality in Lemma 2.21, that is, xn1 ≤ Pn
have been studied, for example, in 18–24.
Abstract and Applied Analysis
3
Lemma 2.5. Let {xn }∞
n−1 be a positive solution of 1.1 and limn → ∞ Pn S. Then S lim supn → ∞ xn .
Proof. Since Pn is a subsequence of xn , it follows that
S ≤ lim sup xn .
2.6
n→∞
On the other hand, by xn1 ≤ Pn for all n ≥ 1, we obtain
lim sup xn ≤ lim sup Pn S.
n→∞
2.7
n→∞
The proof is complete.
Remark 2.6. Let {xn }∞
n−1 be a positive solution of 1.1. By Lemma 2.2, we see that if S lim supn → ∞ xn and xN < S for some N > 0, then xN−1 , xN1 ∈ S, ∞. For example, if it were
xN−1 < S, then it would be PN < S, which would imply lim supn → ∞ xn < S.
Lemma 2.7. Suppose that {xn }∞
n−1 is a positive solution of 1.1 and S lim supn → ∞ xn . Write
ωxn x : there exist − 1 ≤ k1 < k2 < · · · < kn < · · · such that lim xkn
n→∞
x .
2.8
Then ωxn {S, 1/S}.
Proof. If ωxn contains only one point, we may assume by taking a subsequence that Ank →
μ< 1. By taking the limit in the following relationship:
xnk 1 max
1 Ank
,
,
xnk xnk −1
2.9
as k → ∞, we obtain
S max
1 μ
,
S S
1
,
S
2.10
which implies that S 1.
If ωxn contains at least two points, let L ∈ ωxn − {S}, then there exists a subsequence xnk
of xn such that
xnk −→ L < S.
2.11
By Remark 2.6, we see that there exists N > 0 such that for every nk > N,
xnk < S,
xnk 1 , xnk −1 ∈ S, ∞,
2.12
4
Abstract and Applied Analysis
from which it follows that
xnk 1 −→ S,
xnk −1 −→ S.
2.13
By taking a subsequence we may assume that Ank → μ< 1. By taking the limit in the
following relationship:
xnk 1
1 Ank
max
,
,
xnk xnk −1
2.14
as k → ∞, we obtain
S max
1 μ
,
L S
1
,
L
2.15
which implies
L
1
.
S
2.16
The proof is complete.
Theorem 2.8. Let {xn }∞
n−1 be a positive solution of 1.1 and S lim supn → ∞ xn . Then one of the
following two statements is true.
1 If there exist infinitely many n such that xn ≥ S and xn1 ≥ S, then {xn }∞
n−1 is eventually
equal to 1.
2 If there exists N such that xN2k < S and xN2k−1 ≥ S for all k ≥ 0, then xN2k → 1/S
and xN2k−1 → S.
Proof. 1 We assume that there exists an infinite sequence n1 < n2 < n3 < · · · < nk < · · · such
that
xnk ≥ S,
xnk 1 ≥ S.
2.17
By taking a subsequence we may assume from Lemma 2.7 that
Ank −→ μ < 1,
xnk −1 −→ l ∈
S,
1
.
S
2.18
By taking the limit in the following relationship:
xnk 1 xnk
Ank xnk
max 1,
,
xnk −1
2.19
Abstract and Applied Analysis
5
as k → ∞, we get
Sμ
.
S max 1,
l
2
2.20
Since Sμ/l ∈ {μ, μS2 } and μ < 1, it follows that S2 1 and ωxn {1}.
In the following, we show that {xn }∞
n−1 is eventually equal to 1. It only needs to prove that
there exists N ≥ 0 such that for all n ≥ N,
An
1
>
.
xn xn−1
2.21
Indeed, if there exist infinitely many nk such that
xnk 1 Ank
,
xnk −1
2.22
by taking a subsequence we may assume that Ank → μ < 1, then it follows that
1
μ
,
1
μ 1,
2.23
which is a contradiction. Therefore there exists N such that for all n ≥ N,
xn1 1
.
xn
2.24
Thus
xn xN ,
for n N 2k,
xn xN1 ,
for n N 2k 1.
2.25
Since xn → 1, we have xN1 xN 1.
2 If S 1, then the result follows from Lemma 2.7. In the following, we assume S /
1.
Suppose for the sake of contradiction that there exists a subsequence xN2ki of xN2k such
that
xN2ki −→ S.
2.26
By taking a subsequence we may assume that
AN2ki −→ μ.
2.27
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Abstract and Applied Analysis
By taking the limit in the following relationship:
xN2ki 1
AN2ki
max
,
,
xN2ki xN2ki −1
1
2.28
as ki → ∞, we get
S max
1 μ
,
,
S S
2.29
which implies
S 1.
2.30
This is a contradiction. The proof is complete.
Corollary 2.9. Let {An }∞
n0 be a periodic sequence of positive real numbers, then every positive
solution of 1.1 is eventually periodic with period 2.
Proof. Let {xn }∞
n−1 be a positive solution of 1.1 and S lim supn → ∞ xn . By Remark 2.6 and
Theorem 2.8, we may assume without loss of generality that x2k < S, x2k−1 ≥ S ≥ 1 for all k ≥
0. Suppose for the sake of contradiction that there exists a sequence m1 < m2 < · · · < mk < · · ·
such that
1 xmk 1 xmk −1 Amk , and xmk 1 xmk > 1;
mk .
2 xn1 xn 1, for n /
Then mk is odd for every k ≥ 1. Let mk 2nk 1, then it follows from Lemma 2.1 that
x2nk 2 x2nk A2nk 1 < 1 x2nk 1 x2nk < x2nk 1 x2nk 2 .
2.31
From this and by 2 it follows that
A2nk1 1
A2nk 1
x2nk < x2nk 2 x2nk 4 · · · x2nk1 < x2nk1 2 .
x2nk 2
x2nk1
2.32
Therefore for every k ≥ 1,
2
2
A2nk 1 < x2n
x2n
< A2nk1 1 ,
k 2
k1
which is a contradiction since {An }∞
n0 is a periodic sequence. The proof is complete.
Remark 2.10. Corollary 2.9 is the main result of 5.
2.33
Abstract and Applied Analysis
7
3. Example
In this section, we give an example for {An }∞
n0 to be no periodic sequence.
Example 3.1. Consider
xn1
1 An
max
,
,
xn xn−1
n 0, 1, . . . ,
3.1
where A2n A2n1 2 − 1/2n 2 − 1/2n1 /16 for any n ≥ 0. Then solution {xn }∞
n−1 of 3.1
with the initial values x−1 1/4 and x0 4 satisfies the following.
1 x2p−1 x2p 1, for any p ≥ 0.
2 x2p−1 < x2p1 A2p
1
< < 2 < x2p2 < x2p , for any p ≥ 0.
x2p−1 2
Proof. By simple computation, we have
A2p
⎧
2
⎪
,
x−1
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
A0 2
⎪
⎪
,
⎪
⎪
⎪
⎨ x−1
2 − 1/2p 2 − 1/2p1
2
> A
2p−2 A2p−6 · · · A2
⎪
16
⎪
⎪
x−1 ,
⎪
⎪
⎪
⎪ A2p−4 A2p−8 · · · A0
⎪
⎪
2
⎪
⎪
⎪
A2p−2 A2p−6 · · · A4 A0
⎪
⎪
⎪
⎩ A2p−4 A2p−8 · · · A2 x−1 ,
if p 0,
if p 1,
if p ≥ 2 is even,
if p ≥ 2 is odd.
It follows from 3.1 and 3.2 that
x1 x−1
x2 x1
x3 x1
x4 x3
x5 x3
x−1
2
max
, A0 max x−1
, A0 A0 ,
x0
x1 A1
A0 A1
max 1,
max 1,
1,
x0
x−1 x0
2
x1
x1
A0 2
max
, A2 max
, A2 max
, A2 A2 ,
x2
x2 x1
x−1
A2 A3
x3 A3
max 1,
1,
max 1,
x2
x2 x1
2
2
x3
x3
x3 x1
max
, A4 max
, A4 max
x−1 , A4
x4
x4 x3
x1 x−1
2
A2
x−1 , A4 A4 ,
max
A0
3.2
8
Abstract and Applied Analysis
x5 A5
A4 A5
x6 x5 max 1,
max 1,
1,
x4
x4 x3
2
x5
x5
x5 x3 x1 x−1 2
x7 x5 max
, A6 max
, A6 max
, A6
x6
x6 x5
x3 x1 x−1
A4 A0 2
, A6 A6 ,
max
A2 x−1
x7 A7
A6 A7
x8 x7 max 1,
max 1,
1.
x6
x6 x5
3.3
By induction, we have from 3.1 and 3.2 that for any p ≥ 1,
x4p1 x4p−1
x4p−1
max
, A4p
x4p
max
⎧
2
⎨ x4p−1
⎩ x4p x4p−1
, A4p
⎫
⎬
⎭
2
max x4p−1
, A4p
⎧
⎫
2
⎨ x4p−1 x4p−3 x4p−5 · · · x1
⎬
max
x−1 , A4p
⎩ x4p−3 x4p−5 x4p−7 · · · x−1
⎭
max
⎧
⎨ A4p−2 A4p−6 · · · A2
⎩
A4p−4 A4p−8 · · · A0
x4p2 x4p1
x4p3 x4p1
x−1
, A4p
⎫
⎬
⎭
A4p ,
A4p A4p1
max 1,
1,
x4p x4p−1
⎫
⎧
2
⎬
⎨ x4p1
x4p1
2
max
, A4p2 max
, A4p2 max x4p1
, A4p2
⎭
⎩ x4p2 x4p1
x4p2
x4p1 A4p1
max 1,
x4p
max
2
⎧
⎨ x4p1 x4p−1 x4p−3 x4p−5 · · · x1 x−1 2
⎩
x4p−1 x4p−3 x4p−5 x4p−7 · · · x1 x−1
, A4p2
⎫
⎬
⎭
⎫
⎧
2
⎬
⎨
A4p A4p−4 · · · A4 A0
, A4p2 A4p2 ,
max
⎭
⎩ A4p−2 A4p−6 · · · A2 x−1
x4p4 x4p3
x4p3 A4p3
max 1,
x4p2
A4p2 A4p3
max 1,
x4p2 x4p1
from which the result follows. The proof is complete.
1.
3.4
Abstract and Applied Analysis
9
Acknowledgments
The project is supported by NNSF of China10861002, NSF of Guangxi 0640205,0728002,
and Innovation Project of Guangxi Graduate Education2008105930701M43.
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