Properties of positive solutions of 1.7 In this section, we will study properties of positive solutions of 1.7.. From 2.2, we get the following.. “If ” part is obvious... Global asympt
Trang 1DIFFERENCE EQUATION
HONGJIAN XI AND TAIXIANG SUN
Received 17 January 2006; Revised 6 April 2006; Accepted 12 April 2006
We investigate in this paper the global behavior of the following difference equation:
x n+1 =(P k(x n i0,x n i1, ,x n i2k) +b)/(Q k(x n i0,x n i1, ,x n i2k) +b), n =0, 1, , under
appropriate assumptions, whereb [0,), k1,i0,i1, ,i2 k 0, 1, withi0 < i1 <
< i2 k, the initial conditionsx i2k,x i 2k+1, ,x0 (0,) We prove that unique equilib-riumx =1 of that equation is globally asymptotically stable
Copyright © 2006 H Xi and T Sun 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
For some difference equations, although their forms (or expressions) look very simple,
it is extremely difficult to understand thoroughly the global behaviors of their solutions Accordingly, one is often motivated to investigate the qualitative behaviors of difference equations (e.g., see [2,3,6,9,10])
In [6], Ladas investigated the global asymptotic stability of the following rational dif-ference equation:
(E1)
x n+1 = x n+x n 1 x n 2
x n x n 1+x n 2, n =0, 1, , (1.1) where the initial valuesx 2,x 1,x0 R + (0, +)
In [9], Nesemann utilized the strong negative feedback property of [1] to study the following difference equation:
(E2)
x n+1 = x n 1+x n x n 2
x n x n 1+x n 2, n =0, 1, , (1.2) where the initial valuesx 2,x 1,x0 R +
Hindawi Publishing Corporation
Advances in Di fference Equations
Volume 2006, Article ID 27637, Pages 1 7
DOI 10.1155/ADE/2006/27637
Trang 2In [10], Papaschinopoulos and Schinas investigated the global asymptotic stability of the following nonlinear difference equation:
(E3)
x n+1 =
i Zk j 1,jx n i+x n j x n j+1+ 1
i Zk x n i , n =0, 1, , (1.3) wherek 1, 2, 3, ,j, j1 Zk 0, 1, ,k, and the initial valuesx k,x k+1, , x0 R +
Recently, Li [7,8] studied the global asymptotic stability of the following two nonlinear difference equations:
(E4)
x n+1 = x n 1 x n 2 x n 3+x n 1+x n 2+x n 3+a
x n 1 x n 2+x n 1 x n 3+x n 2 x n 3+ 1 +a, n =0, 1, (1.4)
(E5)
x n+1 = x n x n 1 x n 3+x n+x n 1+x n 3+a
x n x n 1+x n x n 3+x n 1 x n 3+ 1 +a, n =0, 1, , (1.5)
wherea [0,+) and the initial valuesx 3,x 2,x 1,x0 R +
Letk1 andi0,i1, ,i2 k 0, 1, withi0 < i1 < < i2 k LetP0(x n i0)= x n i0and
Q0(x n i0)=1, for any 1 j k, let
P j
x n i0, ,x n i2j
=x n i2j x n i2j 1+ 1
P j 1
x n i0, ,x n i2j 2
+
x n i2j+x n i2j 1
Q j 1
x n i0, ,x n i2j 2
,
Q j
x n i0, ,x n i2j
=x n i2j x n i2j 1+ 1
Q j 1
x n i0, ,x n i2j 2
+
x n i2j+x n i2j 1
P j 1
x n i0, ,x n i2j 2
.
(1.6)
In this paper, we consider the following difference equation:
x n+1 = P k
x n i0,x n i1, ,x n i2k
+b
Q k
x n i0,x n i1, ,x n i2k
+b, n =0, 1, , (1.7)
whereb [0,) and the initial conditionsx i2k,x i2k+1, ,x0 (0,)
It is easy to see that the positive equilibriumx of (1.7) satisfies
x = P k(x,x, ,x) + b
Q k(x,x, ,x) + b
=
x2+ 1
P k 1(x,x, ,x) + 2xQ k 1(x,x, ,x) + b
x2+ 1
Q k 1(x,x, ,x) + 2xP k 1(x,x, ,x) + b .
(1.8)
Trang 3Thus, we have
(x1)
x2+xQ k 1(x,x, ,x) + (x + 1)P k 1(x,x, ,x) + b=0, (1.9) from which one can see that (1.7) has the unique positive equilibriumx =1
Remark 1.1 Let k =1, then (1.7) is (1.4) when (i0,i1,i2)=(1, 2, 3) and is (1.5) when (i0,i1,i2)=(0, 1, 3)
2 Properties of positive solutions of ( 1.7 )
In this section, we will study properties of positive solutions of (1.7) Since
P k
x n i0,x n i1, ,x n i2k
Q k
x n i0,x n i1, ,x n i2k
=x n i2k1
x n i2k 11
P k 1
x n i0, ,x n i2k 2
Q k 1
x n i0, ,x n i2k 2
=x n i2k1
x n i2k 11
x n i2 1
x n i1 1
P0x n i0
Q0x n i0
=x n i0 1
x n i1 1
x n i2k1
,
(2.1)
it follows from (1.7) that for anyn0,
x n+11=
x n i0 1
x n i1 1
x n i2k1
Q k
x n i0,x n i1, ,x n i2k
Definition 2.1 Letx n
n = i2k be a solution of (1.7) anda n
n = i2k a sequence witha n
1, 0, 1for everyn i2 k.a n
n = i2k is called itinerary ofx n
n = i2k ifa n =1 when
x n < 1, a n =0 whenx n =1, anda n =1 whenx n > 1.
From (2.2), we get the following
Proposition 2.2 Letx n
n = i2k be a solution of ( 1.7 ) whose itinerary isa n
n = i2k , then
a n+1 = a n i0a n i1 a n i2k for any n0.
Proposition 2.3 Letx n
n = i2k be a solution of ( 1.7 ), then it follows that x n = 1 for any
n1
i2k
j =0(x j1) = 0.
Proof Let itinerary ofx n
n = i2kbea n
n = i2k, then it follows fromProposition 2.2that
x n =1 for anyn1 a n =0 for anyn1
i2k
j =0a j =0
i2k
j =0(x j1) =0
Proposition 2.4 If gcd( i s+ 1,i2 k+ 1)= 1 for some s 0, 1, ,2k1, then a positive solutionx n
n = i2k of ( 1.7 ) is eventually equal to 1 x p = 1 for some p i2 k
Proof “If ” part is obvious.
“Only if ” part Ifx p =1 for somep i2 k, thena p =0, wherea n
n = i2kis itinerary
ofx n
n = i2k ByProposition 2.2, we havea j(i2k+1)+p = a j(is+1)+p =0 for any j0 Since gcd(i s+ 1,i2 k+ 1)=1, we see that for anyt 0, 1, ,i2 k , there existj t 1, 2, ,i2 k+ 1
Trang 4andm t 0, 1, ,i s+ 1such that
j t
i s+ 1
= m t
i2 k+ 1
Together withProposition 2.2, it follows that
a( is+1)(i2k+1)+t+p =0. (2.4) Again byProposition 2.2, we havea n =0 for anyn(i s+ 1)(i2 k+ 1) +p, which implies
Example 2.5 Consider the equation
x n+1 = x n i0x n i1x n 3+x n i0+x n i1+x n 3+b
x n i0x n i1+x n i0x n 3+x n i1x n 3+ 1 +b, n =0, 1, , (2.5)
whereb [0,+), 0 i0 < i1 < 3, and the initial values x 3,x 2,x 1,x0 R + Letx n
n = 3
be a solution of (2.5) whose itinerary isa n
n = 3, then the following hold
(1) If (i0,i1) (0, 1), (1, 2)andx n
n = 3is not eventually equal to 1, thena n
n = 3
is a periodic sequence of period 7
(2) If (i0,i1)=(0, 2) andx n
n = 3is not eventually equal to 1, thena n
n = 3is a peri-odic sequence of period 6
(3)x n =1 for anyn1
0
j = 3(x j1) =0
(4)x n
n = 3is eventually equal to 1 x p =1 for somep 3
Proof (1) If ( i0,i1)=(0, 1), then fromProposition 2.2, it follows that for anyn0,
a n+4 = a n+3 a n+2 a n = a n+2 a n+1 a n 1 a n+2 a n
= a n+1 a n a n 1 = a n a n 1 a n 3 a n a n 1
= a n 3
(2.6)
If (i0,i1)=(1, 2), then in a similar fashion, it is true thata n+4 = a n 3for anyn0 (2) If (i0,i1)=(0, 2), then fromProposition 2.2, it follows that for anyn0,
a n+3 = a n+2 a n a n 1 = a n+1 a n 1 a n 2 a n a n 1
= a n+1 a n a n 2 = a n a n 2 a n 3 a n a n 2
= a n 3
(2.7)
(3) It follows fromProposition 2.3
(4) It follows fromProposition 2.4since either gcd(i0+ 1, 4)=1 or gcd(i1+ 1, 4)=1
3 Global asymptotic stability of ( 1.7 )
In this section, we will study global asymptotic stability of (1.7) To do this, we need the following lemmas
Trang 5Lemma 3.1 Let ( y0,y1, , y i2k) Ri2k+1
+ (1, 1, ,1)and M =maxy j, 1/y j 0 j i2 k, then
1
M < P k
y i0,y i1, , y i2k
Q k
y i0,y i1, , y i2k
Proof Since ( y0,y1, , y i2k) Ri2k+1
+ (1, 1, ,1)andM =maxy j, 1/y j 0 j i2 k,
we haveM > 1 and either Ma > 1/M or M > a1/M for any a y j, 1/y j 0 j i2 k
It is easy to verify that
P1y i0,y i1,y i2
=y i1y i2+ 1
y i0+
y i1+y i2
<y i1y i2+ 1
M +y i1+y i2
y i0M
= Q1y i0,y i1,y i2
M, P1y i0,y i1,y i2
M =y i1y i2+ 1
y i0+
y i1+y i2
M
>y i1y i2+ 1
+
y i1+y i2
y i0
= Q1y i0,y i1,y i2
.
(3.2)
From that we have
P2y i0,y i1,y i2,y i3,y i4
=y i3y i4+ 1
P1y i0,y i1,y i2
+
y i3+y i4
Q1y i0,y i1,y i2
<y i3y i4+ 1
Q1y i0,y i1,y i2
M +y i3+y i4
P1y i0,y i1,y i2
M
= Q2y i0,y i1,y i2,y i3,y i4
M, P2y i0,y i1,y i2,y i3,y i4
M =y i3y i4+ 1
P1y i0,y i1,y i2
+
y i3+y i4
Q1y i0,y i1,y i2
M
>y i3y i4+ 1
Q1y i0,y i1,y i2
+
y i3+y i4
P1y i0,y i1,y i2
= Q2y i0,y i1,y i2,y i3,y i4
.
(3.3)
By induction, we have that for any 1 j k,
P j
y i0,y i1, , y i2j
< Q j
y i0,y i1, , y i2j
M,
P j
y i0,y i1, , y i2j
M > Q j
y i0,y i1, , y i2j
Thus
1
M < P k
y i0,y i1, , y i2k
Q k
y i0,y i1, , y i2k
Trang 6
Letn be a positive integer and let ρ denote the part-metric onRn
+(see [11]) which is defined by
ρ(x, y) =log min
x i
y i,y i
x i 1 i n
forx =x1, ,x n
, y =y1, , y n
Rn+. (3.6)
It was shown by Thompson [11] that (Rn
+,ρ) is a complete metric space In [4], Krause and Nussbaum proved that the distances indicated by the part-metric and by the Eu-clidean norm are equivalent onRn
+ Lemma 3.2 [5] Let T :Rn
+ Rn
+be a continuous mapping with unique fixed point x
Rn
+ Suppose that there exists some l1 such that for the part-metric ρ,
ρT l x,x
< ρx,x
x = x
Then x
is globally asymptotically stable.
Theorem 3.3 The unique equilibrium x = 1 of (1.7 ) is globally asymptotically stable Proof Letx n
n = i2k be a solution of (1.7) with initial conditionsx i2k,x i2k+1, ,x0
Ri2k+1
+ such thatx n
n = i2kis not eventually equal to 1 since otherwise there is nothing to show Denoted byT :Ri2k+1
+ Ri2k+1
+ the mapping
Tx n i2k,x n i2k+1, ,x n
= x n i2k+1,x n i2k+2, ,x n,P k
x n i0,x n i1, ,x n i2k
+b
Q k
x n i0,x n i1, ,x n i2k
+b
.
(3.8) Then solutionx n
n = i2k of (1.7) is represented by the first component of the solution
y n
n =0of the systemy n+1 = T y nwith initial condition y0 =(x i2k,x i2k+1, ,x0) It fol-lows fromLemma 3.1that for alln0, the following inequalities hold:
min
x n i, 1
x n i 0 i i2 k < x n+1 < max x n i, 1
x n i 0 i i2 k (3.9) Inductively, we obtain that for alln0 and all 1 j i2 k+ 1,
min
x n i, 1
x n i 0 i i2 k < x n+j < max x n i, 1
x n i 0 i i2 k , (3.10) from which it follows that
min
x n i, 1
x n i 0 i i2 k < min x n+i, 1
x n+i 1 i i2 k+ 1 . (3.11)
Trang 7Thus, forx
=(1, 1, ,1) and the part-metric ρ ofRi2k+1
+ , we have
ρT i2k+1
y n
,x
=log min
x n+i, 1
x n+i 1 i i2 k+ 1
<log min
x n i, 1
x n i 0 i i2 k
= ρy n,x
(3.12)
for alln0 It follows fromLemma 3.2that the positive equilibriumx =1 of (1.7) is
Acknowledgments
Project supported by NNSF of China (10461001, 10361001) and NSF of Guangxi (0640205)
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Hongjian Xi: Department of Mathematics, Guangxi College of Finance and Economics, Nanning, Guangxi 530004, China
E-mail address:xhongjian@263.com
Taixiang Sun: Department of Mathematics, College of Mathematics and Information Science, Guangxi University, Nanning, Guangxi 530004, China
E-mail address:stx1963@163.com
... anyn0,a n+4 = a n+3 a n+2 a n = a n+2 a n+1 a n 1 a n+2 a. .. n
= a n+1 a n a n = a n a n 1 a n 3 a n a n... anyn0,
a n+3 = a n+2 a n a n = a n+1 a n 1 a n 2 a n