Báo cáo hóa học: "ON THE SYSTEM OF RATIONAL DIFFERENCE EQUATIONS xn+1 = f (xn , yn−k ), yn+1 = f (yn ,xn−k )" potx

We give sufficient conditions under which every positive solution of this equation converges to a positive equilibrium.. If a positive solution of 1.2 is a pair of positive constants x, y,

EQUATIONS xn+1= f (xn, yn − k), yn+1= f (yn, xn − k)

TAIXIANG SUN, HONGJIAN XI, AND LIANG HONG

Received 15 September 2005; Revised 27 October 2005; Accepted 13 November 2005

We study the global asymptotic behavior of the positive solutions of the system of rational difference equations xn+1 = f (x n,y n − k), y n+1 = f (y n,x n − k),n =0, 1, 2, , under

appro-priate assumptions, wherek ∈ {1, 2, }and the initial valuesx − k,x − k+1, ,x0,y − k,y − k+1,

, y0∈(0, +∞) We give sufficient conditions under which every positive solution of this equation converges to a positive equilibrium The main theorem in [1] is included in our result

Copyright © 2006 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

Recently there has been published quite a lot of works concerning the behavior of posi-tive solutions of systems of rational difference equations [2–7] These results are not only valuable in their own right, but they can provide insight into their differential counter-parts

In [1], Camouzis and Papaschinopoulos studied the global asymptotic behavior of the positive solutions of the system of rational difference equations

x n+1 =1 + x n

y n − k,

y n+1 =1 + y n

x n − k,

wherek ∈ {1, 2, }and the initial valuesx − k,x − k+1, ,x0,y − k,y − k+1, , y0∈(0, +∞)

To be motivated by the above studies, in this paper, we consider the more general equation

x n+1 = f (x n,y n − k),

y n+1 = f (y n,x n − k), n =0, 1, 2, , (1.2)

Hindawi Publishing Corporation

Advances in Di fference Equations

Volume 2006, Article ID 16949, Pages 1 7

wherek ∈ {1, 2, }, the initial valuesx − k,x − k+1, ,x0,y − k,y − k+1, , y0∈(0, +∞) and f

satisfies the following hypotheses

(H1) f ∈ C(E × E,(0,+ ∞)) with a =inf(u,v) ∈ E × E f (u,v) ∈ E, where E ∈ {(0, +∞), [0, +∞)}

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

(H3) There exists a decreasing functiong ∈ C((a,+ ∞), (a,+ ∞)) such that

(i) For anyx > a, g(g(x)) = x and x = f (x,g(x));

(ii) limx → a+g(x) =+∞and limx →+∞ g(x) = a.

A pair of sequences of positive real numbers{(x n,y n)} ∞

n =− kthat satisfies (1.2) is a pos-itive solution of (1.2) If a positive solution of (1.2) is a pair of positive constants (x, y),

then (x, y) is called a positive equilibrium of (1.2) In this paper, our main result is the following theorem

Theorem 1.1 Assume that (H1)–(H3) hold Then the following statements are true (i) Every pair of positive constant ( x, y) ∈(a,+ ∞)×(a,+ ∞ ) satisfying the equation

is a positive equilibrium of ( 1.2 ).

(ii) Every positive solution of ( 1.2 ) converges to a positive equilibrium (x, y) of ( 1.2 ) sat-isfying ( 1.3 ) as n → ∞

2 Proof of Theorem 1.1

In this section we will proveTheorem 1.1 To do this we need the following lemma

Lemma 2.1 Let {(x n,y n)} ∞

n =− k be a positive solution of ( 1.2 ) Then there exists a real num-ber L ∈(a,+ ∞ ) with L < g(L) such that x n,y n ∈[L,g(L)] for all n ≥ 1 Furthermore, if

lim supx n = M, liminf x n = m, limsup y n = P, liminf y n = p, then M = g(p) and P =

g(m).

Proof From (H1) and (H2), we have

x i = fx i −1,y i −1− k

> fx i −1,y i −1− k+ 1

≥ a,

y i = fy i −1,x i −1− k

> fy i −1,x i −1− k+ 1

≥ a, for every 1≤ i ≤ k + 1. (2.1)

Since limx → a+g(x) =+∞, there existsL ∈(a,+ ∞) withL < g(L) such that

x i,y i ∈L,g(L) for every 1≤ i ≤ k + 1. (2.2)

It follows from (2.2) and (H3) that

g(L) = fg(L),L≥ x k+2 = fx k+1,y1



≥ fL,g(L)= L, g(L) = fg(L),L≥ y k+2 = fy k+1,x1



Inductively, we have thatx n,y n ∈[L,g(L)] for all n ≥1

Let lim supx n = M, liminf x n = m, limsup y n = P, liminf y n = p, then there exist

se-quencesl n ≥1 ands n ≥1 such that

lim

n →∞ x l n = M, nlim

Without loss of generality, we may assume (by taking a subsequence) that there exist

A,D ∈[m,M] and B,C ∈[p,P] such that

lim

n →∞ x l n −1= A,

lim

n →∞ y l n − k −1= B,

lim

n →∞ y s n −1= C,

lim

n →∞ x s n − k −1= D.

(2.5)

Thus, from (1.2), (H2) and (H3), we have

fM,g(M)= M = f (A,B) ≤ f (M, p),

from which it follows that

By (H3), we obtain

Therefore,M = g(p) By the symmetry, we have also P = g(m).Lemma 2.1is proven 

Proof of Theorem 1.1

(i) Is obvious

(ii) Let{(x n,y n)} ∞

n =− k be a positive solution of (1.2) with the initial conditionsx0,

x −1, ,x − k,y0,y −1, , y − k ∈(0, +∞) ByLemma 2.1, we have that

a < liminf x n = g(P) ≤lim supx n = M < + ∞,

a < liminf y n = g(M) ≤lim supy n = P < + ∞ (2.9)

Without loss of generality, we may assume (by taking a subsequence) that there exists a sequencel n ≥4k such that

lim

n →∞ x l n = M,

lim

n →∞ x l n − j = M j ∈g(P),M, for j ∈ {1, 2, ,3k + 1 }, lim

n →∞ y l n − j = P j ∈g(M),P, forj ∈ {1, 2,···, 3k + 1 }

(2.10)

From (1.2), (2.10) and (H3), we have

fM,g(M)= M = fM1,P k+1

≤ fM1,g(M)≤ fM,g(M), (2.11) from which it follows that

In a similar fashion, we may obtain that

fM,g(M)= M = M1= fM2,P k+2

≤ fM2,g(M)≤ fM,g(M), (2.13) from which it follows that

Inductively, we have that

M j = M,

P k+j = g(M), forj ∈ {1, 2, ,2k + 1 }, (2.15)

from which it follows that

lim

n →∞ x l n − j = M, for j ∈ {0, 1, ,2k + 1 }, lim

n →∞ y l n − j = g(M), for j ∈ { k + 1, ,3k + 1 } (2.16)

In view (2.16), for any 0< ε < M − a, there exists some l s ≥4k such that

M − ε < x l s − j < M + ε, if j ∈ {0, 1, ,2k + 1 },

g(M + ε) < y l s − j < g(M − ε), if j ∈ { k + 1, ,2k + 1 } (2.17)

From (1.2) and (2.17), we have

y l s − k = fy l s − k −1,x l s −2k −1



< fg(M − ε),M − ε= g(M − ε). (2.18) Also (1.2), (2.17) and (2.18) implies

x l s+1= fx l s,y l s − k

> fM − ε,g(M − ε)= M − ε. (2.19) Inductively, it follows that

y l s+n − k < g(M − ε) ∀ n ≥0,

Since lim supx n = M and liminf y n = g(M), we have

lim

Thus limn →∞(x n,y n)=(M,P) with P = g(M).Theorem 1.1is proven 

3 Examples

To illustrate the applicability ofTheorem 1.1, we present the following examples

Example 3.1 Consider equation

x n+1 = p + x n

1 +y n − k,

y n+1 = p + y n

1 +x n − k,

wherek ∈ {1, 2,···}, the initial conditionsx − k,x − k+1, ,x0,y − k,y − k+1, , y0∈(0, +∞) andp ∈(0, +∞) LetE =[0, +∞) and

f (x, y) = p + x

1 +y (x ≥0,y ≥0), g(x) =

p

It is easy to verify that (H1)–(H3) hold for (3.1) It follows fromTheorem 1.1that (i) every pair of positive constant (x, y) ∈(0, +∞)×(0, +∞) satisfying the equation

is a positive equilibrium of (3.1)

(ii) every positive solution of (3.1) converges to a positive equilibrium (x, y) of (3.1) satisfying (3.3) asn → ∞

Example 3.2 Consider equation

x n+1 =1 + x n

y n − k,

y n+1 =1 + y n

x n − k,

wherek ∈{1, 2, }and the initial conditionsx − k,x − k+1, ,x0,y − k,y − k+1, , y0∈(0, +∞) LetE =(0, +∞) and

f (x, y) =1 +x

y (x > 0, y > 0), g(x) =

x

x −1 (x > 1). (3.5)

It is easy to verify that (H1)–(H3) hold for (3.4) It follows fromTheorem 1.1that (i) every pair of positive constant (x, y) ∈(1, +∞)×(1, +∞) satisfying the equation

is a positive equilibrium of (3.4);

(ii) every positive solution of (3.4) converges to a positive equilibrium (x, y) of (3.4) satisfying (3.6) asn → ∞

Example 3.3 Consider equation

x n+1 = p + q + y A + x n

n − k,

y n+1 = p + q + x A + y n

n − k,

where k ∈ {1, 2, }, the initial conditionsx − k,x − k+1, ,x0,y − k,y − k+1, , y0∈(0, +∞),

A ∈(0, +∞) and p,q ∈[0, 1] with p + q =1 LetE =(0, +∞) ifp > 0 and E =[0, +∞) if

p =0 and

thena =inf(x,y) ∈ E × E f (x, y) = p Let g(x) =(pq + px + A)/(x − p) (x > p) It is easy to

verify that (H1)–(H3) hold for (3.7) It follows fromTheorem 1.1that

(i) every pair of positive constant (x, y) ∈(p,+ ∞)×(p,+ ∞) satisfying the equation

is a positive equilibrium of (3.7);

(ii) every positive solution of (3.7) converges to a positive equilibrium (x, y) of (3.7) satisfying (3.9) asn → ∞

Acknowledgments

I would like to thank the reviewers for their constructive comments and suggestions Project Supported by NNSF of China (10361001,10461001) and NSF of Guangxi (0447004)

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Taixiang Sun: Department of Mathematics, 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 530004, China

E-mail address:xhongjian@263.net

Liang Hong: Department of Mathematics, Guangxi University, Nanning, Guangxi 530004, China

E-mail address:stxhql@gxu.edu.cn