Báo cáo hoa học: " Research Article On Boundedness of Solutions of the Difference Equation xn 1 pxn qxn−1 / 1 xn for q 1 p 1" pot

Stevi´c, “Periodic solutions of the rational difference equation yn y n−3 y n−4 /y n−1 ,” Journal of Di fference Equations and Applications, vol.. Stevi´c, “Periodic character of a class

Volume 2009, Article ID 463169, 11 pages

doi:10.1155/2009/463169

Research Article

On Boundedness of Solutions of the Difference

Hongjian Xi,1, 2Taixiang Sun,1 Weiyong Yu,1 and Jinfeng Zhao1

1 Department of Mathematics, Guangxi University, Nanning, Guangxi 530004, China

2 Department of Mathematics, Guangxi College of Finance and Economics,

Nanning, Guangxi 530003, China

Correspondence should be addressed to Taixiang Sun,stx1963@163.com

Received 4 February 2009; Revised 19 April 2009; Accepted 2 June 2009

Recommended by Agacik Zafer

We study the boundedness of the difference equation xn1  px n  qx n−1 /1  x n , n  0, 1, , where q > 1  p > 1 and the initial values x−1, x0∈ 0, ∞ We show that the solution {x n}∞

n−1of this equation converges tox  q  p − 1 if x n ≥ x or x n ≤ x for all n ≥ −1; otherwise {x n}∞n−1is unbounded Besides, we obtain the set of all initial valuesx−1, x0 ∈ 0, ∞×0, ∞ such that the

positive solutions{x n}∞

n−1of this equation are bounded, which answers the open problem 6.10.12 proposed by Kulenovi´c and Ladas2002

Copyrightq 2009 Hongjian Xi 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 this paper, we study the following difference equation:

x n1 px n  qx n−1

1 x n , n  0, 1, , 1.1

where p, q ∈ 0, ∞ with q > 1  p and the initial values x−1, x0∈ 0, ∞.

The global behavior of1.1 for the case p  q < 1 is certainly folklore It can be found,

for example, in1 see also a precise result in 2

The global stability of1.1 for the case pq  1 follows from the main result in 3 see also Lemma 1 in Stevi´c’s paper4 Some generalizations of Copson’s result can be found, for example, in papers5 8 Some more sophisticated results, such as finding the asymptotic behavior of solutions of 1.1 for the case p  q  1 even when p  0 can be found, for

example, in papers4see also 8 11 Some other properties of 1.1 have been also treated

in4

The case q  1p was treated for the first time by Stevi´c’s in paper 12 The main trick from12 has been later used with a success for many times; see, for example, 13–15 Some existing results for1.1 are summarized as follows16

Theorem A 1 If p  q ≤ 1, then the zero equilibrium of 1.1  is globally asymptotically stable.

2 If q  1, then the equilibrium x  p of 1.1 is globally asymptotically stable.

3 If 1 < q < 1  p, then every positive solution of 1.1 converges to the positive equilibrium

x  p  q − 1.

4 If q  1  p, then every positive solution of 1.1 converges to a period-two solution.

5 If q > 1  p, then 1.1 has unbounded solutions.

In16, Kulenovi´c and Ladas proposed the following open problem

Open problem B (see Open problem 6.10.12of [ 16 ])

Assume that q ∈ 1, ∞.

a Find the set B of all initial conditions x−1, x0 ∈ 0, ∞ × 0, ∞ such that the

solutions{x n}∞

n−1of1.1 are bounded

b Let x−1, x0 ∈ B Investigate the asymptotic behavior of {x n}∞

n−1

In this paper, we will obtain the following results: let p, q ∈ 0, ∞ with q > 1  p, and

let{x n}∞

n−1be a positive solution of1.1 with the initial values x−1, x0 ∈ 0, ∞ × 0, ∞.

If x n ≥ x for all n ≥ −1 or x n ≤ x for all n ≥ −1, then {x n}∞n−1 converges to x  q  p − 1.

Otherwise{x n}∞n−1is unbounded

For closely related results see17–34

2 Some Definitions and Lemmas

In this section, let q > 1  p > 1 and x  q  p − 1 be the positive equilibrium of 1.1 Write

D  0, ∞ × 0, ∞ and define f : D → D by, for all x, y ∈ D,

f

x, y

y, py  qx

1 y



It is easy to see that if{x n}∞

n−1is a solution of1.1, then f n x−1, x0  x n−1 , x n  for any n ≥ 0.

Let

A1 0, x × 0, x, A2 x, ∞ × x, ∞,

A3 0, x × x, ∞, A4 x, ∞ × 0, x,

R0 {x} × 0, x, L0  {x} × x, ∞,

R1 0, x × {x}, L1  x, ∞ × {x}.

2.2

Then D  ∪4

i1 A i  ∪ L0∪ L1∪ R0∪ R1∪ {x, x} The proof ofLemma 2.1is quite similar to that of Lemma 1 in35 and hence is omitted

Lemma 2.1 The following statements are true.

1 f : D → fD is a homeomorphism.

2 fL1  L0and fL0 ⊂ A4.

3 fR1  {x} × p, x and fR0 ⊂ A3.

4 fA3 ⊂ A4and fA4 ⊂ A3.

5 A2∪ L1⊂ fA2 ⊂ A2∪ L1∪ A4and A1∪ R1⊂ fA1 ⊂ A1∪ R1∪ A3.

Lemma 2.2 Let q > 1  p > 1, and let {x n}∞n−1 be a positive solution of1.1.

1 If lim n → ∞ x 2n  a ∈ 0, ∞ and a / p, then lim n → ∞ x 2n1  a  x.

2 If lim n → ∞ x 2n−1  b ∈ 0, ∞ and b / p, then lim n → ∞ x 2n  b  x.

Proof We show only 1 because the proof of 2 follows from 1 by using the change y n 

x n−1and the fact that1 is autonomous Since limn → ∞ x 2n  a ∈ 0, ∞ and a / p, by 1.1

we have

lim

n → ∞ x 2n1 lim

n → ∞

qx 2n − x 2n2

x 2n2 − p 



q − 1a

Also it follows from1.1 that

a  lim

n → ∞ x 2n lim

n → ∞

qx 2n−1 − x 2n1

x 2n1 − p 



q − 12a



q − 1a − pa − p , 2.4

from which we have a  x and lim n → ∞ x 2n1  a  x This completes the proof.

Lemma 2.3 Let q > 1  p > 1, and let {x n}∞n−1 be a positive solution of 1.1 with the initial values

x−1, x0 ∈ A4 If there exists some n ≥ 0 such that x 2n−1 ≥ x 2n1 , then x 2n ≥ x 2n2

Proof Since x−1, x0 ∈ A4, it follows fromLemma 2.1thatx 2n−1 , x 2n  ∈ A4 for any n ≥ 0

Without loss of generality we may assume that n  0, that is, x−1≥ x1 Now we show x0≥ x2.

Suppose for the sake of contradiction that x0< x2, then

x−1≥ x1 px0 qx−1

x0 < x2 px1 qx0

1 x1

By2.5 we have

x0≥ x−1



q − 1

and by2.6 we get



q − 1 − px2

0p2 q − 1 − qx−1



x0 pqx−1> 0. 2.8

Claim 1 If x−1≥ x, then



p2 q − 1 − qx−1

2

− 4q − 1 − ppqx−1≥ 0. 2.9

Proof of Claim 1

Let gx  p2 q − 1 − qx2− 4q − 1 − ppqx x ≥ x, then we have

g x  2q1 qx − p2− q− 4pqq − 1 − p

≥ 2qq − 12

 p2 p1− q p

 2qq − 1

q − p − 1 p2 p

> 0.

2.10

Since x−1≥ x, it follows



p2 q − 1 − qx−1

2

− 4q − 1 − ppqx−1

≥q2 qp − 2q  1 − p22

− 4q − 1 − pqp

q  p − 1

q2− 2q  1 − p22

 2qpq2− 2q  1 − p2

qp2

− 4q2− 2q  1 − p2

pq

q2− 2q  1 − p2− pq2

≥ 0.

2.11

This completes the proof ofClaim 1

By2.8, we have

x0> λ1



1 qx−1− p2− q 1 qx−1− p2− q2− 4pqq − 1 − px−1

2

x0< λ2



1 qx−1− p2− q− 1 qx−1− p2− q2− 4pqq − 1 − px−1

2

Claim 2 We have

λ2≤ x−1



q − 1

Proof of Claim 2

Since

1 qq  p − 1 − p2− q2− 4pqq − 1 − pp  q − 1

 q2− p2− 2q  1 − qp

 2q  p − 1q − 1 − p−1 qq  p − 1− p2− q ,

2.16

we have

λ1



1 qx−1− p2− q 1 qx−1− p2− q2− 4pqq − 1 − px−1

2

q − 1 − p

≥



1 qx − p2− q 1 qx − p2− q2− 4pqq − 1 − px

2

q − 1 − p



1 qq  p − 1− p2− q 1 qq  p − 1 − p2− q2− 4pqq − 1 − pp  q − 1

2

q − 1 − p

≥q  p − 1 x.

2.17

The proof of2.14 is completed

Now we show2.15 Let

hx  pqx − p2−x − pq − 11 qx − p2− qq − 12

q − 1 − px. 2.18

Note that 2pq − 2qq − 1 < 0; it follows that if x ≥ x, then

h x  2pqx − p−q − 1

1 qx − p2− q qq − 1x − p−q − 12

q − 1 − p

≤ 2pqq − 1−

q − 12pq − q − p2 q2− p

q − 1q  pp  1 − q< 0,

2.19

which implies that hx is decreasing for x ≥ x Since x−1≥ x and

hx  pqq − 12−q − 1q − 11 qq  p − 1− p2− q

q − 12

q − 1 − pq  p − 1 0, 2.20

it follows that

hx−1  pqx−1− p2−x−1− pq − 11 qx−1− p2− q

q − 12

q − 1 − px−1≤ hx  0. 2.21

Thus



q − 12

1 qx−1− p2− q2− 4pqq − 1 − px−1



≥ 4p2q2

x−1− p2− 4pqx−1− pq − 11 qx−1− p2− q

q − 12

1 qx−1− p2− q2.

2.22

This implies that



q − 1 1 qx−1− p2− q2− 4pqq − 1 − px−1

≥ 2pqx−1− p−q − 11 qx−1− p2− q. 2.23 Finally we have

x−1

q − 1

x−1− p ≥

4

q − 1 − ppqx−1

2

q − 1 − p 1 qx−1− p2− q 1 qx−1− p2− q2− 4pqq − 1 − px−1





1 qx−1− p2− q− 1 qx−1− p2− q2− 4pqq − 1 − px−1

2

2.24 The proof of2.15 is completed

Note that x0 < x since x−1, x0 ∈ A4 By2.12, 2.13, 2.14, and 2.15, we see x0 <

x−1q − 1/x−1− p, which contradicts to 2.7 The proof ofLemma 2.3is completed

3 Main Results

In this section, we investigate the boundedness of solutions of1.1 Let q > 1  p > 1, and

let{x n}∞n−1be a positive solution of1.1 with the initial values x−1, x0 ∈ 0, ∞ × 0, ∞,

then we see thatx n1 − xx n − x < 0 for some n ≥ −1 or x n ≥ x for all n ≥ −1 or x n ≤ x for all n≥ −1

Theorem 3.1 Let q > 1  p > 1, and let {x n}∞n−1 be a positive solution of 1.1 such that x n ≥ x for

all n ≥ −1 or x n ≤ x for all n ≥ −1, then {x n}∞

n−1 converges to x  q  p − 1.

Proof.

Case 1 0 < x n ≤ x for any n ≥ −1 If 0 < x 2n ≤ q − 1 for some n, then

x 2n1 − x 2n−1 px 2n  qx 2n−1 − x 2n−1 − x 2n−1 x 2n

1 x 2n

If q − 1 < x 2n ≤ x for some n, then

px 2n

x 2n − q  1 ≥

px

x − q  1  x ≥ x 2n−1 , 3.2

which implies that px 2n ≥ x 2n−1 x 2n − q  1 and

x 2n1 − x 2n−1 px 2n  qx 2n−1 − x 2n−1 − x 2n−1 x 2n

1 x 2n ≥ 0. 3.3

Thus x ≥ x 2n1 ≥ x 2n−1 for any n ≥ 0 In similar fashion, we can show x ≥ x 2n2 ≥ x 2nfor any

n ≥ 0 Let lim n → ∞ x 2n1  a and lim n → ∞ x 2n  b, then

a  pb  qa

1 b , b 

pa  qb

which implies a  b  x.

Case 2 x n ≥ x  p  q − 1 for any n ≥ −1 Since fx, y  py  qx/1  y x > p/q is decreasing in y, it follows that for any n ≥ −1,

x n2 px n1  qx n

1 x n1

≤ px  qx n

1 x ≤ x n .

3.5

In similar fashion, we can show that limn → ∞ x 2n1  limn → ∞ x 2n  x This completes the

proof

Lemma 3.2 see 20, Theorem 5 Let I be a set, and let F : I × I → I be a function Fu, v

which decreases in u and increases in v, then for every positive solution {x n}∞n−1 of equation x n1 

Fx n , x n−1 , {x 2n}∞

n0 and {x 2n−1}∞

n0 do exactly one of the following.

1 They are both monotonically increasing.

2 They are both monotonically decreasing.

3 Eventually, one of them is monotonically increasing, and the other is monotonically

decreasing.

Remark 3.3 Using arguments similar to ones in the proof of Lemma 3.2, Stevi´c proved Theorem 2 in25 Beside this, this trick have been used by Stevi´c in 18,28,29

Theorem 3.4 Let q > 1  p > 1, and let {x n}∞n−1 be a positive solution of 1.1 such that x n1−

xx n − x < 0 for some n ≥ −1, then {x n}∞n−1 is unbounded.

Proof We may assume without loss of generality that x0− xx−1− x < 0 and x−1, x0 ∈ A4

the proof for x−1, x0 ∈ A3 is similar FromLemma 2.1 we seex 2n−1 , x 2n  ∈ A4 for all

n ≥ 0.If {x 2n}∞n0is eventually increasing, then it follows fromLemma 2.3that{x 2n−1}∞n0is eventually increasing Thus limn → ∞ x 2n−1  b > x and lim n → ∞ x 2n  a ≤ x, it follows from

Lemma 2.2that b ∞

If{x 2n}∞n0 is not eventually increasing, then there exists some N≥ 0 such that

x 2N ≥ x 2N2 px 2N1  qx 2N

from which we obtain x 2N ≥ px 2N1 /1  x 2N1 − q ≥ p, since x 2N1 ≥ x  p  q − 1 and q > 1 Since fy, x  py  qx/1  y  p  qx − p/1  y x ≥ p, y ≥ p is increasing

in x and is decreasing in y, we have that x 2n ≥ p for any n ≥ N It follows fromLemma 3.2

that{x 2n}∞

n0is eventually decreasing Thus limn → ∞ x 2n  a < x and lim n → ∞ x 2n−1  b ≥ x.

It follows fromLemma 2.2that b ∞ This completes the proof

By Theorems3.1and3.4we have the following

Corollary 3.5 Let q > 1  p > 1, and let {x n}∞n−1 be a positive bounded solution of 1.1, then

x n−1 ≥ x n ≥ x for all n ≥ 0 or x ≥ x n ≥ x n−1 for all n ≥ 0.

Now one can find out the set of all initial valuesx−1, x0 ∈ 0, ∞ × 0, ∞ such that

the positive solutions of1.1 are bounded Let P0 A2, Q0 A1 For any n ≥ 1, let

P n  f−1P n−1 , Q n  f−1Q n−1 . 3.7

It follows fromLemma 2.1that P1  f−1P0 ⊂ P0, Q1  f−1Q0 ⊂ Q0, which implies

P n ⊂ P n−1 , Q n ⊂ Q n−1 3.8

for any n≥ 1

Let S be the set of all initial values x−1, x0 ∈ 0, ∞ × 0, ∞ such that the positive

solutions{x n}∞n−1of1.1 are bounded Then we have the following theorem

Theorem 3.6 S  ∞

n0 Q n ∪ ∞

n0 P n  ⊂ A1∪ A2∪ {x, x}.

Proof Let {x n}∞n−1be a positive solution of1.1 with the initial values x−1, x0 ∈ S.

Ifx−1, x0 ∈ ∞

n0 Q n , then f n x−1, x0  x n−1 , x n  ∈ A1for any n ≥ 0, which implies

x n ≤ x for any n ≥ −1 It follows fromTheorem 3.1that limn → ∞ x n  x.

Ifx−1, x0 ∈∞

n0 P n , then f n x−1, x0  x n−1 , x n  ∈ A2, which implies x n ≥ x for any

n ≥ −1 It follows fromTheorem 3.1that limn → ∞ x n  x.

Now assume that {x n}∞n−1 is a positive solution of 1.1 with the initial values

x−1, x0 ∈ D − S.

If x−1, x0 ∈ A3



A4

L0

L1

R0

R1, then it follows from Lemma 2.1 that

f2 x−1, x0  x1, x2 ∈ {x, y : x − xy − x < 0}, which along withTheorem 3.4implies that{x n} is unbounded

Ifx−1, x0 ∈ A2−∞

n0 P n , then there exists n ≥ 0 such that x−1, x0 ∈ P n − P n1 

f −n A2 − f −n−1 A2 Thus f n x−1, x0  x n−1 , x n  ∈ A2− f−1A2 ByLemma 2.1, we obtain

f n1 x−1, x0 ∈ L1



A4and f n3 x−1, x0  x n2 , x n3  ∈ A4, which along withTheorem 3.4

implies that{x n} is unbounded

Ifx−1, x0 ∈ A1−∞

n1 Q n , then there exists n ≥ 0 such that x−1, x0 ∈ Q n − Q n1  Q n−

f−1Q n  and f n x−1, x0  x n−1 , x n  ∈ A1− f−1A1 Again byLemma 2.1andTheorem 3.4,

we have that{x n} is unbounded This completes the proof

Acknowledgment

Project Supported by NNSF of China10861002 and NSF of Guangxi 0640205, 0728002

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