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Introduction In this paper, we study the convergence of positive solutions of a system of rational dif-ference equations.. Recently there has been published quite a lot of works concerni

xn+1= f (yn − q, xn − s), yn+1= g(xn − t, yn − p)

TAIXIANG SUN AND HONGJIAN XI

Received 20 March 2006; Revised 19 May 2006; Accepted 28 May 2006

We study the global behavior of positive solutions of the system of rational difference equations x n+1 = f (y n − q,x n − s), y n+1 = g(x n − t,y n − p), n =0, 1, 2, , where p, q, s, t ∈ {0, 1, 2, } withs ≥ t and p ≥ q, the initial values x − s,x − s+1, , x0,y − p,y − p+1, , y0∈

(0, +∞) We give sufficient conditions under which every positive solution of this system converges to the unique positive equilibrium

Copyright © 2006 T Sun and H Xi 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 convergence of positive solutions of a system of rational dif-ference equations Recently there has been published quite a lot of works concerning the behavior of positive solutions of systems of rational difference equations [1–7,9,11] Not only these results are valuable in their own right, but also they can provide insight into their differential counterparts

Papaschinopoulos and Schinas [10] studied the oscillatory behavior, the periodicity, and the asymptotic behavior of the positive solutions of systems of rational difference equations

x n+1 = A + x n −1

y n , y n+1 = A + y n −1

x n , n =0, 1, , (1.1)

whereA ∈(0, +∞) and the initial valuesx −1,x0,y −1,y0∈(0, +∞)

Recently, Kulenovi´c and Nurkanovi´c [8] investigated the global asymptotic behavior

of solutions of systems of rational difference equations

x n+1 = a + x n

b + y n, y n+1 = d + y n

e + x n, n =0, 1, , (1.2) wherea, b, d, e ∈(0, +∞) and the initial valuesx0,y0∈(0, +∞)

Hindawi Publishing Corporation

Advances in Di fference Equations

Volume 2006, Article ID 51520, Pages 1 8

DOI 10.1155/ADE/2006/51520

In this paper, we consider the more general equation

x n+1 = f

y n − q,x n − s

, y n+1 = g

x n − t,y n − p

where p, q, s, t ∈ {0, 1, 2, }withs ≥ t and p ≥ q, the initial values x − s,x − s+1, , x0,y − p,

y − p+1, , y0∈(0, +∞) and f satisfies the following hypotheses.

(H1) f (u, v), g(u, v) ∈ C(E × E, (0, + ∞)) with a =inf(u,v) ∈ E × E f (u, v) ∈ E and b =

inf(u,v) ∈ E × E g(u, v) ∈ E, where E ∈ {(0, +∞), [0, +∞)}

(H2) f (u, v) and g(u, v) are decreasing in u and increasing in v.

(H3) Equation

has a unique positive solutionx = x, y = y.

(H4) f (b, x) has only one fixed point in the interval (a, + ∞), denoted byA, and g(a, y)

has only one fixed point in the interval (b, + ∞), denoted byB.

(H5) For everyw ∈ E, f (w, x)/x and g(w, x)/x are nonincreasing in x in (0, + ∞)

2 Main results

Theorem 2.1 Assume that (H1)–(H5) hold and {(x n,y n)} is a positive solution of ( 1.3 ), then there exists a positive integer N such that

f (B, a) ≤ x n ≤ A, g(A, b) ≤ y n ≤ B, for n ≥ N. (2.1) Proof Since a =inf(u,v) ∈ E × E f (u, v) ∈ E and b =inf(u,v) ∈ E × E g(u, v) ∈ E, we have

x = f (y, x) > f (y + 1, x) ≥ a,

Claim 1 g(A, b) < y < B and f (B, a) < x < A.

Proof of Claim 1 If B ≤ y, then it follows from (H2), (H4), and (H5) that

B = g(a, B) > g(x, B) = B g(x, B)

B ≥ B g(x, y)

which is a contradiction Thereforey < B In a similar fashion it is true that x < A.

Sincey < B and x < A, we have that

f (B, a) < f (y, x) = x, g(A, b) < g(x, y) = y, (2.4)

Claim 2 (i) For all n ≥ q + 1, x n+1 ≤ x n − sifx n − s > A and x n+1 ≤ A if x n − s ≤ A.

(ii) For alln ≥ t + 1, y n+1 ≤ y n − pify n − p > B and y n+1 ≤ B if y n − p ≤ B.

Proof of Claim 2 We only prove (i) (the proof of (ii) is similar) Obviously

x n+1 = f

y n − q,x n − s

≤ f

b, x n − s

Ifx n − s ≤ A, then x n+1 ≤ f (b, x n − s)≤ f (b, A) = A.

Ifx n − s > A, then

f

b, x n − s

x n − s ≤ f (b, A)

which impliesx n+1 ≤ f (b, x n − s)≤ x n − s.Claim 2is proven 

Claim 3 (i) There exists a positive integer N1such thatx n ≤ A for all n ≥ N1

(ii) There exists a positive integerN2such thaty n ≤ B for all n ≥ N2

Proof of Claim 3 We only prove (i) (the proof of (ii) is similar) Assume on the contrary

thatClaim 3does not hold Then it follows fromClaim 2that there exists a positive in-tegerR such that x n(s+1)+R ≥ x(n+1)(s+1)+R > A for every n ≥1 Let limn →∞ x n(s+1)+R = A1, thenA1≥ A.

We know fromClaim 2that{ x n }and{ y n }are bounded Letc =limn →∞supy n(s+1)+R − q −1, thenc ≥ b and there exists a sequence n k → ∞such that

lim

k →∞ y n k(s+1)+R − q −1= c. (2.7)

By (1.3) we have that

x n k(s+1)+R = f

y n k(s+1)+R − q −1,x(n k−1)(s+1)+R



from which it follows that

A1= f

c, A1



≤ f

b, A1



= A1f

b, A1



A1 ≤ A1f (b, A)

A = A1. (2.9) This with (H2) and (H4) impliesc = b and A1= A Therefore lim n →∞ y n(s+1)+R − q −1= b.

Since{ x n }and{ y n }are bounded, we may assume (by taking a subsequence) that there exist a sequencel n → ∞andα, β ∈ E such that

lim

k →∞ x l k(s+1)+R − q − t −2= α, lim

k →∞ y l k(s+1)+R − q − p −2= β. (2.10)

By (1.3) we have that

y l k(s+1)+R − q −1= g

x l k(s+1)+R − q − t −2,y l k(s+1)+R − q − p −2



from which it follows that

This is a contradiction.Claim 3is proven 

LetN =max{ N1,N2}+ 2s + 2p, then for all n > N we have that

x n ≤ A, y n ≤ B,

x n = f

y n − q −1,x n − s −1



≥ f (B, a),

y n = g

x n − t −1,y n − p −1 

≥ g(A, b).

(2.13)

Theorem 2.2 Let I =[c, d] and J =[α, β] be intervals of real numbers Assume that f ∈ C(J × I, I) and g ∈ C(I × J, J) satisfy the following properties:

(i) f (u, v) and g(u, v) are decreasing in u and increasing in v;

(ii) if M1,m1∈ I with m1≤ M1 and M2,m2∈ J with m2≤ M2 are a solution of the system

M1= f

m2,M1 

, m1= f

M2,m1 

,

M2= g

m1,M2



, m2= g

M1,m2



then M1= m1and M2= m2.

Then the system

x n+1 = f

y n − q,x n − s



, y n+1 = g

x n − t,y n − p



, n =0, 1, , (2.15)

has a unique equilibrium (S, T) and every solution of ( 2.15 ) with the initial values x − s,x − s+1,

, x0∈ I and y − p,y − p+1, , y0∈ J converges to (S, T).

Proof Let

m0= c, m0= α, M0= d, M0= β, (2.16) and fori =1, 2, ., we define

M i1= f

m i −1

2 ,M i −1 1



, m i1= f

M i −1

2 ,m i −1 1



,

M i

2= g

m i −1

1 ,M i −1 2



, m i

2= g

M i −1

1 ,m i −1 2



It is easy to verify that

m0≤ m1= f

M0,m0

≤ f

m0,M0

= M1≤ M0,

m02≤ m12= g

M10,m02

≤ g

m01,M20

= M21≤ M20. (2.18)

From (i) and (2.18) we obtain

m1= f

M0,m0 

≤ f

M1,m1 

= m2,

m2= f

M1,m1 

≤ f

m1,M1 

= M2,

M2= f

m1,M1 

≤ f

m0,M0 

= M1,

m1= g

M0,m0 

≤ g

M1,m1 

= m2,

m2= g

M1,m1 

≤ g

m1,M1 

= M2,

M2= g

m1,M1 

≤ g

m0,M0 

= M1.

(2.19)

By induction it follows that fori =0, 1, ,

m i1≤ m i+11 ≤ ··· ≤ M1i+1 ≤ M1i,

m i

2≤ m i+1

2 ≤ ··· ≤ M i+1

2 ≤ M i

2.

(2.20)

On the other hand, we have x n ∈[m0,M0] for anyn ≥ − s and y n ∈[m0,M0] for any

n ≥ − p since x − s,x − s+1, , x0∈[m0,M0] andy − p,y − p+1, , y0∈[m0,M0] For anyn ≥0,

we obtain

m1= f

M0,m0 

≤ x n+1 = f

y n − q,x n − s

≤ f

m0,M0 

= M1,

m1= g

M0,m0 

≤ y n+1 = g

x n − t,y n − p

≤ g

m0,M0 

= M1. (2.21)

Letk =max{ s + 1, p + 1 } It follows that for anyn ≥ k,

m2= f

M1,m1 

≤ x n+1 = f

y n − q,x n − s

≤ f

m1,M1 

= M2,

m2= g

M1,m1 

≤ y n+1 = g

x n − t,y n − p

≤ g

m1,M1 

= M2.

(2.22)

By induction, forl =0, 1, , we obtain that for any n ≥ lk,

m l+1

1 ≤ x n+1 ≤ M l+1

2 ≤ y n+1 ≤ M l+1

Let

lim

n →∞ m n

1= m1, lim

n →∞ m n

2= m2, lim

n →∞ M n

1= M1, lim

n →∞ M n

By the continuity of f and g, we have from (2.17) that

M1= f

m2,M1



, M2= g

m1,M2



,

m2= g

M1,m2



, m1= f

M2,m1



Using assumption (ii), it follows from (2.23) that

lim

n →∞ x n = m1= M1= S, lim

n →∞ y n = m2= M2= T. (2.26)

Theorem 2.3 If (H1)–(H5) hold and the system

M1= f

m2,M1



, M2= g

m1,M2



,

m = g

M ,m 

, m = f

M ,m 

with f (B, a) ≤ m1≤ M1≤ A and g(A, b) ≤ m2≤ M2≤ B has the unique solution m1=

M1= x and m2= M2= y, then every solution of ( 1.3 ) converges to the unique positive equi-librium (x, y).

Proof Let {(x n,y n)}is a positive solution of (1.3) ByTheorem 2.1, there exists a positive integerN such that f (B, a) ≤ x n = f (y n − q,x n − s)≤ A and g(A, b) ≤ y n = g(x n − t,y n − p)≤

B for all n ≥ N Since f , g satisfy the conditions (i) and (ii) of Theorem 2.2 in I =

[f (B, a), A] and J =[(A, b), B], it follows that {(x n,y n)}converges to the unique positive

3 Examples

In this section, we will give two applications of the above results

Example 3.1 Consider equation

x n+1 = a + y c + x n − s

n − q, y n+1 = d + y b + x n − p

where p, q, s, t ∈ {0, 1, 2, } with s ≥ t and p ≥ q, the initial values x − s,x − s+1, , x0,

y − p,y − p+1, , y0∈(0, +∞) anda, b, c, d ∈(0, +∞) Ifa > 1 and b > 1, then every positive

solution of (3.1) converges to the unique positive equilibrium

Proof Let E =[0, +∞), it is easy to verify that (H1)–(H5) hold for (3.1) In addition, if

M1= c + M1

a + m2, M2= d + M2

b + m1,

m2= d + m2

b + M1, m1= c + m1

a + M2,

(3.2)

with 0≤ m1≤ M1and 0≤ m2≤ M2, then we have



M1− m1



(a −1)= m1M2− M1m2,



M2− m2



(b −1)= M1m2− m1M2,

(3.3)

from which it follows thatM1= m1andM2= m2 Moreover, it is easy to verify that (3.2) have the unique solution

M1= m1= x = −(a −1)(b −1) +c − d +



(a −1)(b −1) +d − c 2

+ 4c(a −1)(b −1)

M2= m2= y = −(a −1)(b −1) +d − c +



(a −1)(b −1) +c − d 2

+ 4d(a −1)(b −1)

(3.4)

It follows from Theorems2.1and2.3that every positive solution of (3.1) converges to the

Example 3.2 Consider equation

x n+1 = a + x n − s

y n − q, y n+1 = b + y n − p

x n − t, (3.5)

where p, q, s, t ∈ {0, 1, 2, } with s ≥ t and p ≥ q, the initial values x − s,x − s+1, , x0,

y − p,y − p+1, , y0∈(0, +∞) anda, b ∈(0, +∞) Ifa > 1 and b > 1, then every positive

so-lution of (3.5) converges to the unique positive equilibrium

Proof Let E =(0, +∞), it is easy to verify that (H1)–(H5) hold for (3.5) In addition, if

M1= a + M1

m2 , M2= b + M2

m1 ,

m2= b + m2

M1, m1= a + m1

M2,

(3.6)

with 0≤ m1≤ M1and 0≤ m2≤ M2, then (3.6) have the unique solution

M1= m1= x = ab −1

b −1 ,

M2= m2= y = ab −1

a −1 .

(3.7)

It follows from Theorems2.1and2.3that every positive solution of (3.5) converges to the unique positive equilibrium (x, y) =((ab −1)/(b −1), (ab −1)/(a −1)) 

Acknowledgment

The project was supported by NNSF of China (10461001,10361001) and NSF of Guangxi (0447004)

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Taixiang Sun: Department of Mathematics, College of Mathematics and Information Science, Guangxi University, Nanning, Guangxi 530004, China

E-mail address:stx1963@163.com

Hongjian Xi: Department of Mathematics, Guangxi College of Finance and Economics, Nanning, Guangxi 530003, China

E-mail address:xhongjian@263.net