On the Rational Recursive Sequence x

Volume 2007, Article ID 23618,12pages doi:10.1155/2007/23618

Research Article

On the Rational Recursive Sequence x

(A +

α

x

)/(B +

β

x

)

Received 13 August 2006; Revised 22 January 2007; Accepted 22 January 2007 Recommended by Martin J. Bohner

The main objective of this paper is to study the boundedness character, the periodic char- acter, the convergence, and the global stability of the positive solutions of the difference equationxn+1=(A+^k_i=0αixn−i)/(B+^k_i=0β_ixn−i),n=0, 1, 2,..., whereA,B,αi,β_iand the initial conditionsx₋k,...,x₋1,x0are arbitrary positive real numbers, whilekis a posi- tive integer number.

Copyright © 2007 E. M. E. Zayed and M. A. El-Moneam. 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 prop- erly cited.

1. Introduction

Our goal in this paper is to investigate the boundedness character, the periodic character, the convergence and the global stability of the positive solutions of the difference equation

xn+1=A+^k_i=0αixn−i

B+^k_i=0βixn−i, n=0, 1, 2,..., (1.1) where A,B, αi, βi and the initial conditions x₋k,...,x₋1, x0 are arbitrary positive real numbers, whilekis a positive integer number. The case where any ofA,B,αi,βiis allowed to be zero gives different special cases of (1.1) which are studied by many authors (see, e.g., [1–14]). For the related work, see [15–26]. The study of these equations is challenging and rewarding and is still in its infancy. We believe that the nonlinear rational difference equations are paramount importance in their own right. Furthermore, the results about such equations offer prototypes for the development of the basic theory of the global behavior of nonlinear difference equations. Note that the difference equation (1.1) has

been extensively studied in the special casek=1 in the monograph [14]. So, the results presented in our paper are new.

Definition 1.1. The equilibrium pointxof the difference equation (1.1) is the point that satisfies the conditionx=F(x,x,...,x). That is, the constant sequence {xn}^∞_n=−kwithxn= xfor alln≥ −kis a solution of the difference equation (1.1).

Definition 1.2. Letx∈(0,∞) be an equilibrium point of the difference equation (1.1).

Then, the following hold

(i) The equilibrium pointxof the difference equation (1.1) is called locally stable if for everyε >0 there existsδ >0 such thatx−k,...,x−1,x0∈(0,∞) with|x−k− x|+···+|x₋1−x|+|x0−x|< δ, then|xn−x|< εfor alln≥ −k.

(ii) The equilibrium pointxof the difference equation (1.1) is called locally asymp- totically stable if it is locally stable and if there existsγ >0 such thatx−k,...,x−1, x0∈(0,∞) with|x₋k−x|+···+|x₋1−x|+|x0−x|< γ, thenxn→xasn→ ∞. (iii) The equilibrium pointxof the difference equation (1.1) is called global attractor

if for everyx−k,...,x−1,x0∈(0,∞) one hasxn→xasn→ ∞.

(iv) The equilibrium pointxof the equation (1.1) is called globally asymptotically stable if it is locally stable and global attractor.

(v) The equilibrium pointxof the difference equation (1.1) is called unstable if it is not locally stable.

Definition 1.3. Say that the sequence{xn}^∞_n=−kis bounded and persists if there exist posi- tive constantsmandMsuch that

m≤xn≤M ∀n≥ −k. (1.2)

Definition 1.4. A sequence{xn}^∞_n=−kis said to be periodic with periodpifxn+p=xnfor alln≥ −k. A sequence{xn}^∞_n=−k is said to be periodic with prime period p ifp is the smallest positive integer having this property.

Assume thata=_k

i₌0αi,a=_k

i₌0(−1)ⁱαi,b=_k

i₌0βi, andb=_k

i₌0(−1)ⁱβi. Then the equilibrium pointxof the difference equation (1.1) is the solution of the equation

x=A+ax

B+bx. (1.3)

Consequently, the positive equilibrium pointxof the difference equation (1.1) is given by

x=(a−B) +(a−B)²+ 4Ab

2b . (1.4)

LetF: (0,∞)^k+1→(0,∞) be a continuous function defined by Fu0,u1,...,uk

=A+^k_i=0αiui

B+^k_i=0βiui. (1.5)

Now, we have

yn+1= k j=0

∂F(x,...,x)

∂uj yn−j, (1.6)

and then the linearized equation is yn+1+

k j=0

bjyn−j=0, (1.7)

where

bj=βjx−αj

B+bx . (1.8)

2. The main results

In this section, we establish some results which show that the positive equilibrium point xof the difference equation (1.1) is globally asymptotically stable and every positive so- lution of the difference equation (1.1) is bounded and has prime period two.

Theorem 2.1 (see [4,10,13,17]). Assume thata,b∈Randk∈ {0, 1, 2,...}. Then

|a|+|b|<1 (2.1)

is a sufficient condition for the asymptotic stability of the difference equation

xn+1+axn+bxn−k=0, n=0, 1,.... (2.2) Remark 2.2 (see [13]). Theorem 2.1can be easily extended to a general linear difference equation of the form

xn+k+p1xn+k−1+···+pkxn=0, n=0, 1, 2,..., (2.3) wherep1,p2,...,pk∈Randk∈ {1, 2,...}. Thenequation 2.3is asymptotically stable pro- vided that

k i=1

pi<1. (2.4)

Theorem 2.3. Assume that B >a holds. Let {xn}^∞_n=−k be a solution of the difference equation (1.1) such that for somen0≥0,

eitherxn≥x forn≥n0 (2.5)

orxn≤x forn≥n0. (2.6)

Then{xn}converges toxasn→ ∞, that is,

nlim→∞xn=x. (2.7)

Proof. Assume that (2.5) holds. The case where (2.6) holds is similar and will be omitted.

Then forn≥n0+k, we deduce that xn+1=A+^k_i=0αixn−i

B+^k_i=0βixn−i = k i=0

αixn−i 1 + (A/^k_i=0αixn−i) B+^k_i=0βixn−i

≤ k i=0

αixn−i 1 +A/ax B+bx ⁼

k i=0

αixn−i A+ax axB+bx.

(2.8)

With the aid of (1.3), the last inequality becomes xn+1≤

k i=0

αixn−i/a, (2.9)

and so

xn+1≤max

0≤i≤k

xn−i

forn≥n0+k. (2.10)

Set

yn=max

0_≤i_≤k

xn−i

forn≥n0+k. (2.11)

Then clearly

yn≥xn+1≥x forn≥n0+k. (2.12) Next, we claim that

yn+1≤yn forn≥n0+k. (2.13)

Now, we have yn+1=max

0≤i≤k

xn+1−i

=maxxn+1, max

0≤i≤k

xn−i

≤maxxn+1,yn

=yn. (2.14) From (2.12) and (2.13), it follows that the sequence{yn}is convergent and that

y=_nlim

→∞yn≥x. (2.15)

Furthermore, we get

xn+1≤A+^k_i=0αixn−i

B+bx ^≤

A+a y n

B+bx. (2.16)

From this and by using (2.13) we obtain, xn+i≤A+ay n+i−1

B+bx ^≤ A+ay n

B+bx fori=1,...,k+ 1. (2.17)

Then

yn+k+1= max

1_≤i≤k+1

xn+i

≤A+ay n

B+bx, (2.18)

and by lettingn→ ∞, we obtain

y≤A+a y

B+bx. (2.19)

Consequently, we obtain

y1− a B+bx

≤ A

B+bx. (2.20)

From (1.3) and (2.20), we deduce thaty≤x, and in view of (2.15), we obtainy=x. Thus,

the proof ofTheorem 2.3is completed.

Theorem 2.4. Let{xn}^∞_n=−kbe a positive solution of the difference equation (1.1) andB >1.

Then there exist positive constantsmandMsuch that

m≤xn≤M, n=0, 1,.... (2.21)

Proof. From the difference equation (1.1), we have whenB >1 xn+1≤A

B+1 B

_k

i=0

αixn−i

, n=0, 1,.... (2.22)

Consider the linear difference equation yn+1=A

B+ 1 B

_k

i=0

αiyn−i

, n=0, 1,... (2.23)

with the initial conditionsyi=xi>0,i= −k,...,−1, 0. It follows by induction that

xn≤yn. (2.24)

First of all, assume thatB >a. Then we have A/(B−a) is a particular solution of (2.23) and every solution of the homogeneous equation which is associated with (2.23) tends to zero asn→ ∞. Hence

nlim→∞yn= A

B−a. (2.25)

From this and (2.24), it follows that the sequence{xn}is bounded from above by a posi- tive constantMsay. That is,

xn≤M, n=0, 1,.... (2.26)

Set

m= A

B+bM. (2.27)

Then we have

xn+1=A+^k_i=0αixn−i

B+^k_i=0βixn−i ≥ A

B+bM ⁼m (2.28)

and consequently, we get

m≤xn≤M, n=0, 1,..., (2.29)

which completes the proof ofTheorem 2.4whenB >a. Second, consider the case when B≤a. It suffices to show that {xn}is bounded from above by some positive constant.

For the sake of contradiction, assume that{xn}is unbounded. Then there exists a subse- quence{xnj}such that

limj→∞nj= ∞, lim

j→∞x1+nj= ∞, x1+nj=maxxn:−k≤n≤1 +nj

, (j=0, 1, 2,...). (2.30) From (2.22), we deduce that

k i=0

αix−i+nj≥Bx1+nj−A. (2.31) Taking the limit asj→ ∞of both sides of the last inequality, we obtain

limj→∞

k i=0

αix₋i+nj= ∞. (2.32)

It is easy enough to show thatx₋i+nj≤x1+nj, (i=0, 1, 2,...,k), and then asa=_k

i=0αi, we have

k i=0

αix₋i+nj≤ax 1+nj. (2.33) From the last inequality and the difference equation (1.1), we obtain

0≤ax 1+nj− k i=0

αix−i+nj=aA +^k_i=0αix−i+nj

a−B−_k

i=0βix−i+nj

B+^k_i=0βix₋i+nj

. (2.34)

Consequently, it follows that

k i=0

βix−i+nj ≤a−B. (2.35)

Then for every i=0, 1, 2,...,k for which βi is positive, the subsequence {x₋i+nj} is bounded which implies that the sequence{_k

i=0αix₋i+nj}is also bounded. This contra- dicts (2.32) and the proof ofTheorem 2.4is completed.

Theorem 2.5. Assume thatB >aholds. Then the positive equilibrium pointxof the differ- ence equation (1.1) is globally asymptotically stable.

Proof. The linearized equation (1.7) with (1.8) can be written in the form

yn+1+ k j=0

βjx−αj

B+bx yn−j=0. (2.36)

AsB >a, we get

k j=0

βjx−αj

B+bx ≤

a+ bx

B+bx<1. (2.37)

Thus, byRemark 2.2, we deduce that the equilibrium pointxof the difference equation (1.1) is locally asymptotically stable. It remains to prove that the equilibrium pointx is a global attractor. To this end, setI=limn→∞infxnandS=limn→∞supxn, which by Theorem 2.4are positive numbers. Then, from the difference equation (1.1), we see that

S≤A+aS

B+bI, I≥A+aI

B+bS . (2.38)

Hence

A+ (a−B)I≤bIS≤A+ (a−B)S. (2.39) From which it follows thatI=S. Thus, the proof ofTheorem 2.5is completed.

Theorem 2.6. The necessary and sufficient condition for the difference equation (1.1) to have positive prime period two solutions is that both inequalities

Ab−b²−(a+a)b−b(B+a)< b(B+a)², (2.40)

B+a <0 (2.41)

are valid.

Proof. First, suppose that there exist positive prime period two solutions

...,P,Q,P,Q,... (2.42)

of the difference equation (1.1). We will prove that the condition (2.40) holds. It follows from the difference equation (1.1) that

P=A+α0Q+α1P+α2Q+α3P+···

B+β0Q+β1P+β2Q+β3P+···, Q=A+α0P+α1Q+α2P+α3Q+···

B+β0P+β1Q+β2P+β3Q+···.

(2.43)

Consequently, we obtain

A+α0Q+α1P+α2Q+α3P+··· =BP+β0PQ+β1P²+β2PQ+β3P²+···, (2.44) A+α0P+α1Q+α2P+α3Q+··· =BQ+β0PQ+β1Q²+β2PQ+β3Q²+···. (2.45) By subtracting, we deduce after some reduction that

P+Q= −(B+a)

β1+β3+···, (2.46)

while by adding we obtain

PQ=Aβ1+β3+···

−

α0+α2+···

(B+a)

bβ1+β3+··· , (2.47)

whereB+a <0. Now, it is clear from (2.46) and (2.47) that P and Q are two positive distinct real roots of the quadratic equation

t²−(P+Q)t+PQ=0. (2.48)

Thus, we deduce that −(B+a)

β1+β3+···

2

>4

Aβ1+β3+···

−

α0+α2+···

(B+a) bβ1+β3+···

. (2.49)

From (2.49), we obtain

Ab−b²−(a+a)b−b(B+a)< b(B+a)², (2.50) and hence the condition (2.40) is valid. Conversely, suppose that the condition (2.40) is valid. Then, we deduce immediately from (2.40) that the inequality (2.49) holds. Conse- quently, there exist two positive distinct real numbersPandQsuch that

P= −(B+a) 2β1+β3+···−1

2

T1, (2.51)

Q= −(B+a) 2β1+β3+···+1

2

T1, (2.52)

whereT1>0 which is given by the formula T1=

−(B+a) β1+β3+···

2

−4

Aβ1+β3+···

−α0+α2+···(B+a) bβ1+β3+···

. (2.53) Thus,PandQrepresent two positive distinct real roots of the quadratic equation (2.48).

Now, we are going to prove thatPandQare positive prime period two solutions of the difference equation (1.1). To this end, we assume that

x₋k=P, x₋k+1=Q,..., x₋1=Q, x0=P. (2.54) We wish to show that

x1=Q, x2=P. (2.55)

To this end, we deduce from the difference equation (1.1) that x1=A+α0x0+α1x₋1+···+αkx₋k

B+β0x0+β1x−1+···+βkx−k

=A+Pα0+α2+···

+Qα1+α3+···

B+Pβ0+β2+···

+Qβ1+β3+···.

(2.56)

Dividing the denominator and numerator of (2.56) by−(B+a)/(β1+β3+···) and using (2.51)–(2.53), we obtain

x1

= −2Aβ1+β3+···)/(B+a+1 +K1

α0+α2+···

+1− K1

α1+α3+···

−2Bβ1+β3+···/(B+a) +1 +K1β0+β2+···

+1−

K1β1+β3+···

=

a−2Aβ1+β3+···/(B+a)+aK1

b−2Bβ1+β3+···/(B+a)+bK1

,

(2.57) where

K1=1− Ab−b²−(a+a)b−b(B+a) b(B+a)²

, (2.58)

and from the condition (2.40), we deduce thatK1>0. Multiplying the denominator and numerator of (2.57) by

b−2Bβ1+β3+···

(B+a)

−bK1. (2.59)

We have x1=

a−2Aβ1+β3+···

/(B+a)b−2Bβ1+β3+···

/(B+a)−baK1

b−2Bβ1+β3+···

/(B+a)²−b²K1

+ba−ab −a2Bβ1+β3+···

/(B+a+b2Aβ1+β3+···

/(B+a)K1

b−2Bβ1+β3+···/(B+a²−b²K1

. (2.60) After some reduction, we deduce that

x1= −(B+a) 2β1+β3+···

×

2α1+···

β0+···

−2α0+···

β1+···

−

2β1+···

(B+a(Ab−Ba1+K1

2α1+···

β0+···

−2α0+···

β1+···

−

2β1+···

/(B+a)(Ab−Ba)

=−(B+a)1 +K1

2β1+β3+··· = −(B+a) 2β1+β3+···+1

2

T1=Q.

(2.61) Similarly, we can show that

x2=A+α0x1+α1x0+···+αkx₋(k−1)

B+β0x1+β1x0+···+βkx₋(k₋1) =A+Qα0+α2+···

+Pα1+α3+···

B+Qβ0+β2+···

+Pβ1+β3+··· =P.

(2.62) By using the mathematical induction, we have

xn=P, xn+1=Q ∀n≥ −k. (2.63)

Thus, the difference equation(1.1) has positive prime period two solutions

...,P,Q,P,Q,.... (2.64)

Hence the proof ofTheorem 2.6is completed.

Acknowledgment

The authors would like to express their deep thanks to the referee for his interesting sug- gestions and comments on this paper.

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E. M. E. Zayed: Mathematics Department, Faculty of Science, Zagazig University, Zagazig 44519, Egypt

Current address: Mathematics Department, Faculty of Science, Taif University, P.O. Box 888, El-Taif 5700, Hawai, Saudi Arabia

Email address:[email protected]

M. A. El-Moneam: Mathematics Department, Faculty of Science, Zagazig University, Zagazig 44519, Egypt

Email address:[email protected]