# On the Rational Recursive Sequencexn+1=(α−βxn)/(γ−δxn−xn−k)

Authors: E. M. E. Zayed; A. B. Shamardan; T. A. Nofal
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

We study the global stability, the periodic character, and the boundedness character of the positive solutions of the difference equationxn+1=(α−βxn)/(γ−δxn−xn−k),n=0,1,2,…,k∈{1,2,…}, in the two cases: (i) δ≥0,α>0,γ>β>0; (ii) δ≥0,α=0,γ,β>0, where the coefficients α,β,γ,andδ, and the initial conditions x−k,x−k+1,…,x−1,x0 are real numbers. We show that the positive equilibrium of this equation is a global attractor with a basin that depends on certain conditions posed on the coefficients of this equation.


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## Body

## 1. Introduction

The asymptotic
stability of the rational recursive sequence,(1.1)xn+1=α+βxnγ+∑i=0kγixn−i,n=0,1,2,…, was investigated
when the coefficients α,β,γ,andγi are nonnegative real numbers (see [1–3]). Studying the asymptotic behavior
of the rational sequence (1.1) when some of the coefficients are negative was
suggested in [3]. Recently, Aboutaleb et al. [4] studied the rational recursive
sequence, (1.2)xn+1=α−βxnγ+xn−1,n=0,1,2,…,where α,β,andγ are nonnegative real numbers and obtained
sufficient conditions for the global attractivity of the positive equilibria.
Yan et al. [5] studied recently the rational recursive
sequence, (1.3)xn+1=α−βxnγ−xn−k,n=0,1,2,…, where α≥0,γ,β>0 are real numbers while k≥1 is an integer number, and the initial
conditions x−k,x−k+1,…,x−1,x0 are arbitrary real numbers. They proved that the positive equilibrium x¯ of (1.3) is a global attractor with a basin
that depends on certain conditions of the coefficients. He et al. [6]
studied recently the rational recursive sequence, (1.4)xn+1=a−bxn−kA+xn,n=0,1,2,…, where a≥0,A,b>0 are real numbers while k≥1 is an integer number and the initial
conditions x−k,x−k+1,…,x−1,x0 are arbitrary real numbers. They proved the global attractivity and
periodic character of the positive solution of (1.4). Stević [7] studied
recently the rational recursive sequence, (1.5)xn+1=α+βxnγ−xn−k,n=0,1,2,…, where the parameters α,β,andγ are nonnegative real numbers and k≥1 is an integer number while the initial
conditions x−k,x−k+1,…,x−1,x0 are arbitrary real numbers. Other related results can be found in
[8–19].Our aim in this paper is to study the global
attractivity, the periodicity, and the boundedness of the positive solution of
the following rational recursive sequence:(1.6)xn+1=α−βxnγ−δxn−xn−k,n=0,1,2,…, in the two cases (i) δ≥0,α>0,γ>β>0, (ii) δ≥0,α=0,γ,β>0, where the coefficients α,β,γ,andδ are real numbers and k≥1 is an integer number, while the initial
conditions x−k,x−k+1,…,x−1,x0 are arbitrary real numbers. We will prove that the positive
equilibrium x¯ of (1.6) is a global attractor with a basin
that depends on certain conditions of these coefficients.


## 2. Local Stability and Permanence

We first recall some results which will be useful in
the sequel. LetI be some real interval and let F be a
continuous function defined on Ik+1. Then, for initial conditions x−k,x−k+1,…,x−1,x0∈I, it is easy to see that the difference equation,(2.1)xn+1=F(xn,xn−1,…,xn−k),n=0,1,2,…,k≥1, has a unique solution {xn}.Definition 2.1.
A pointx¯ is called an equilibrium of (2.1), if x¯=F(x¯,…,x¯). That is, xn=x¯ for n≥0 is a solution of (2.1), or equivalently, is a
fixed point of F.Definition 2.2.
An intervalJ⊂I is called an invariant interval of (2.1) if
the initial conditions x−k,x−k+1,…,x−1,x0∈J imply that the solution xn∈J for n>0. That is, every solution of (2.1) with initial conditions in J remains in J.Definition 2.3.
The difference equation (2.1) is said to be
permanent if there exist numbers P and Q with 0<P≤Q<∞ such that for any initial conditions x−k,x−k+1,…,x−1,x0 there exists a positive integer N which depends on the initial
conditions such that P≤xn≤Q, for all n≥N.The linearized equation associated with (2.1) about
the equilibrium x¯ is (2.2)yn+1=∑i=0k∂F(x¯,…,x¯)∂uiyn−i,n=0,1,2,…. Its characteristic equation
is (2.3)λn+1=∑i=0k∂F(x¯,…,x¯)∂uiλn−i,n=0,1,2,….Theorem 2.4 (see [3]).
Assume thatF is a C1-function and let x¯ be an equilibrium of (2.1). Then, the
following statements are true:
(a)
if all the
roots of (2.3) lie in the open unit disk |λ|<1, then the equilibrium x¯ of (2.1) is locally asymptotically stable;(b)
if at least
one root of (2.3) has absolute value greater than one, then the equilibrium x¯ of (2.1) is unstable.Theorem 2.5 (see [3, 8]).
Assume thatp,q∈ℝ,
and k∈{1,2,…}. Then, (2.4)|p|+|q|<1 is a sufficient condition for the
asymptotic stability of the difference equation (2.5)xn+1−pxn+qxn−k=0,n=0,1,2,…. Suppose in addition that one of
the following two cases holds: (i) k is odd and q<0, or (ii)
k is even and pq<0. Then, (2.4) is also a necessary condition for
the asymptotic stability of (2.5) (see [6]).First, we study the rational recursive
sequence(2.6)xn+1=α−βxnγ−δxn−xn−k,n=0,1,2,…, together with the
conditions (2.7)δ≥0,α>0,γ>β>0,k∈{1,2,…}.The unique positive equilibrium pointx¯ of (2.6) is the solution of the
equation (2.8)x¯=α−βx¯γ−(δ+1)x¯, which is given by (2.9)x¯=(γ+β)±T2(δ+1), where (2.10)T=(γ+β)2−4α(δ+1). If (2.7) holds and α=(γ+β)2/4(δ+1), then (2.6) has a unique positive equilibrium x¯0=(γ+β)/2(δ+1). If (2.7) holds and α<(γ+β)2/4(δ+1) then (2.6) has two positive equilibria x¯1,2 given by (2.9).The linearized equation of (2.6) about the equilibrium x¯i(i=0,1,2) is given by (2.11)yn+1+β−δx¯i[γ−(δ+1)x¯i]yn−x¯i[γ−(δ+1)x¯i]yn−k=0. The characteristic equation
associated with (2.6) about x¯0 is (2.12)λk+1+[2βγ−β−δ(γ+β)(δ+1)(γ−β)]λk−γ+β(δ+1)(γ−β)=0. Now, we have the following
results:(a)
if0≤δ<2β/(γ−β),
then (γ+β)/(δ+1)(γ−β)>1 and hence the equilibrium x¯0 of (2.6) is unstable (see Figure 1);(b)
ifδ>2β/(γ−β), then (2.13)|2βγ−β−δ(γ+β)(δ+1)(γ−β)|+|γ+β(δ+1)(γ−β)|=1. Thus, the linearized stability
analysis fails. On the other hand, the characteristic equation associated with
(2.6) about x¯1 is (2.14)λk+1+[2βγ−β−T−δ(γ+β+T)(δ+1)(γ−β−T)]λk−γ+β+T(δ+1)(γ−β−T)=0. Now, we have the following results:Figure 1(a)
if0≤δ<2β/(γ−β),
then it is obvious that (2.15)|γ+β+T(δ+1)(γ−β−T)|≥(γ−βγ+β)[γ+β+T][γ−β−T]>1, hence the equilibrium x¯1 of (2.6) is unstable;(b)
ifδ≥2β/(γ−β),
then it is easy to see that 2β(δ+1)<δ[γ+β+T], and consequently, we have (2.16)|2βγ−β−T−δ[γ+β+T](δ+1)[γ−β−T]|+|γ+β+T(δ+1)[γ−β−T]|=γ−β+Tγ−β−T>1, and hence the equilibrium x¯1 of (2.6) is unstable.For the positive equilibriumx¯2, in view of  conditions (2.7) and α<(γ+β)2/4(δ+1),
we have (2.17)x¯2=γ+β−T2(δ+1)<γ+β2(δ+1)<γδ+1. Hence, if (2.18)0<α≤β(γ−β)δ+1, then (2.19)T≥(γ+β)2−4β(γ−β)>(γ+β)2−(γ+3β)(γ−β)=(γ+β)2−(γ+β)2+4β2=2β. Consequently, we
have (2.20)|β−δx¯2γ−(δ+1)x¯2|+|x¯2γ−(δ+1)x¯2|<β+(δ+1)x¯2γ−(δ+1)x¯2=3β+γ−Tγ−β+T<3β+γ−2βγ−β+2β=1, which by Theorem 2.5 implies that x¯2 is locally asymptotically stable (see Figure 2).Figure 2Lemma 2.6.
Letf(u,v)=(α−βu)/(γ−δu−v) and assume that conditions (2.7) and
(2.18) hold. Then, the following statements are true:
(a)
0<x¯2<α/β,α/β<x¯1<∞;(b)
f(x,x)is a
strictly decreasing function in(−∞,α/β);(c)
letu,v∈(−∞,α/β),then the functionf(u,v)is a strictly
decreasing function inuandastrictly increasing function inv.Proof.
We prove (a) only. The proofs of
(b) and (c) are omitted here. In view of (2.7) and (2.18), we
have (2.21)x¯2=γ+β−T2(δ+1)<γ+β2(δ+1)<γδ+1. From (2.8) and (2.21), we have α−βx¯2>0 and so x¯2<α/β. Also, in view of (2.7) and (2.18), we have (2.22)0<α−βx¯1γ−(δ+1)x¯1=x¯1=γ+β+T2(δ+1)≥γ+β+(γ+β)2−4β(γ−β)2(δ+1)=γ+β+(γ−β)2+4β22(δ+1)>γ+β+(γ−β)22(δ+1)=γδ+1, and so γ−x¯1(δ+1)<0. Consequently, α−βx¯1<0 which implies that x¯1>α/β. The proof is completed.Theorem 2.7.
Assume that the conditions (2.7)
and (2.18) hold. Let {xn} be any solution of (2.6). If xi∈(−∞,α/β],for i=−k,−k+1,…,−1 and if x0∈[0,α/β], then (2.23)0≤xn≤αβ,n=1,2,…. That is the solution {xn} is bounded.Proof.
By part (c) of Lemma2.6, we
have (2.24)0=α−β⋅(α/β)γ−δx0−x−k≤x1=α−βx0γ−δx0−x−k≤α−β⋅0γ−δ⋅0−α/β=αγ−α/β=βαγβ−α. From (2.18), we deduce that γβ−α>β2,
and then we have (2.25)0≤x1≤αβ. Also, we have (2.26)0=α−β⋅(α/β)γ−δx1−x−k+1≤x2=α−βx1γ−δx1−x−k+1≤α−β⋅0γ−δ⋅0−α/β<αβ. Thus, (2.27)0≤x2≤αβ. The result (2.23) now follows by
induction. The proof is completed.


## 3. Global Attractivity

In this section, we will study the global attractivity
of positive solutions of (2.6). We show that the positive equilibrium x¯ of (2.6) is a global attractor with a basin
that depends on certain conditions imposed on the coefficients.Theorem 3.1.
Assume
that the conditions (2.7) and (2.18) hold. Then, the equilibrium point x¯2 of (2.6) is globally asymptotically stable.Proof.
In Section2, we have shown under the assumptions (2.7) and (2.18) that the
equilibrium x¯2 is locally asymptotically stable. It remains to prove that the
equilibrium x¯2 is a global attractor. To this end, set I=limn→∞infxn and S=limn→∞supxn which by Theorem 2.7 exist and are positive numbers. Then, from (2.6) we
deduce that (3.1)S≤α−βSγ−(δ+1)I,I≥α−βIγ−(δ+1)S. Consequently, we have (3.2)−α+(γ+β)S≤(δ+1)IS≤−α+(γ+β)I, from which it follows that I=S. Thus, the the proof of Theorem 3.1 is completed.Lemma 3.2 (see [8]).
Consider the difference
equation(3.3)xn+1=f(xn,xn−k),k≥1,n=0,1,2,…. Let [a,b] be some interval of real numbers, and assume
that f:[a,b]×[a,b]→[a,b] is a continuous function satisfying the
following properties:
(a)
f(u,v)is a nonincreasing function in u, and a
nondecreasing function inv;(b)
if(m,M)∈[a,b]×[a,b]is a solution of the system(3.4)m=f(M,m),M=f(m,M),then,m=M. Then, (3.3) has
a unique equilibrium point x¯ and every solution of (3.3) converges to x¯.Theorem 3.3.
Assume
that the conditions (2.7) and (2.18) hold. Then, the positive equilibrium x¯ of (2.6) is a global attractor with a basin S*=[0,α/β]k+1.Proof.
Foru,v∈[0,α/β], set (3.5)f(u,v)=α−βuγ−δu−v. We claim that f:[0,α/β]×[0,α/β]→[0,α/β]. In fact, if we set a=0,b=α/β,
then (3.6)f(b,a)=α−βbγ−δb−a=α−αγ−δ(α/β)=0=a, and in view of the condition
(2.18), we have (3.7)f(a,b)=α−βaγ−δa−b=αγ−α/β=βαγβ−α<αβ=b. Since f(u,v) is decreasing in u and increasing in v,
it follows that a≤f(u,v)≤b, for all u,v∈[a,b], which implies that our assertion is true. On
the other hand, conditions (a) and (b) of Lemma 3.2 are clearly true. Let {xn} be a solution of (2.6) with the initial conditions (x−k,x−k+1,…,x−1,x0)∈S. By Lemma 3.2, we have limn→∞xn=x¯. The proof is completed.Theorem 3.4.
Assume that conditions (2.7)
and (2.18) hold. Then, the positive equilibrium x¯ of (2.6) is a global attractor with a basin S∗=(−∞,α/β]k×[0,α/β].Proof.
Let{xn} be a solution of (2.6) with the initial conditions x−k,x−k+1,…,x−1,x0∈S*. Then, by Theorem 2.7, we have (3.8)xn∈[0,αβ],n=1,2,…. By Theorem 3.3, we have limn→∞xn+k=x¯ and so limn→∞xn=x¯. The proof is completed.Theorem 3.5.
Assume that conditions (2.7)
hold with 0≤δ<1.
Also, assume that k is an odd positive integer. Then, the necessary and
sufficient condition for (2.6) to have positive solutions of prime period two
is that (3.9)β(γ−β)<α<(γ−β)4[γ+3β−δ(γ−β)].Proof.
First, suppose that there exist distinctive
positive solutions of prime period two,(3.10)…,P,Q,P,Q,…, of the difference equation
(2.6).
Ifk is odd, then xn+1=xn−k. It follows from the difference equation (2.6) that (3.11)P=α−βQγ−δQ−P,Q=α−βPγ−δP−Q. Consequently, we
obtain (3.12)P+Q=γ−β,PQ=α−β(γ−β)1−δ. Thus, we deduce
that (3.13)α>β(γ−β),0≤δ<1. Now it is clear that P,Q are two
positive distinct real roots of the quadratic equation (3.14)t2−(P+Q)t+PQ=0. Therefore, we
have (3.15)(γ−β)2>4[α−β(γ−β)]1−δ. From (3.13) and (3.15) we obtain
condition (3.9). Conversely, suppose that the condition (3.9) is valid.
Then, we deduce that (3.13) and (3.15) hold. Consequently, there exists two
positive distinct real numbers P and Q such that (3.16)P=γ−β2−K2,(3.17)Q=γ−β2+K2, where K>0 is given by (3.18)K=(γ−β)2−4[α−β(γ−β)1−δ],0≤δ<1. Thus, P and Q given by (3.16) and (3.17) represent two
positive distinct real roots of the quadratic equation (3.14). Now, we are going
to prove that P and Q given by (3.16) and (3.17) are positive
solutions of prime period two of the difference equation (2.6). To this end, we
assume that x−k=P,x−k+1=Q,…,x−1=P,x0=Q. We wish to prove that x1=P and x2=Q.
It follows from the difference equation (2.6) and the
formulas (3.16) and (3.17) that (3.19)x1=α−βx0γ−δx0−x−k=α−βQγ−δQ−P=2α−β[γ−β+K]2γ−δ[γ−β+K]−[γ−β−K]=β[2α/β−(γ−β)−K]2γ−(1+δ)(γ−β)+(1−δ)K=(β1−δ){2α/β−(γ−β)−K}{2γ/(1−δ)−((1+δ)/(1−δ))(γ−β)−K}{2γ/(1−δ)−((1+δ)/(1−δ))(γ−β)+K}{2γ/(1−δ)−((1+δ)/(1−δ))(γ−β)−K}. After some reduction, we deduce
that (3.20)x1=(β1−δ)[α(1−δ)+β2][γ−β−K]/β(1−δ)2[α(1−δ)+β2]/(1−δ)2=γ−β−K2=P. Similarly, we can show
that, (3.21)x2=α−βx1γ−δx1−x−k+1=α−βPγ−δP−Q=Q. By using the induction, we
have (3.22)xn=P,xn+1=Q,∀n≥−k. Thus, the difference equation
(2.6) has positive solutions of prime period two. Hence, the proof of Theorem 3.5
is completed.Theorem 3.6.
Assume that conditions (2.7)
hold. If k is even, then (2.6) has no positive solutions of prime period
two.Proof.
Suppose that there exists distinctive positive solutions of prime
period two,(3.23)…,P,Q,P,Q,…, of the difference equation
(2.6).
Ifk is even, then xn=xn−k. It follows from the difference equation (2.6) that (3.24)P=α−βQγ−(δ+1)Q,Q=α−βPγ−(δ+1)P. From which we have (γ−β)(P−Q)=0 and by using (2.7), we deduce that P=Q. This is a contradiction. Thus, the proof of
Theorem 3.6 is completed.


## 4. The Caseα=0

Secondly, we study the rational recursive
sequence(4.1)xn+1=−βxnγ−δxn−xn−k,n=0,1,2,…, where δ≥0,γ,β>0 are real numbers and k∈{1,2,…}. By putting xn=βyn, (4.1) yields (4.2)yn+1=−ynA−δyn−yn−k,n=0,1,2,…, where A=γ/β.
Equation (4.2) has two equilibrium points (4.3)y¯1=0,y¯2=1+A1+δ. The linearized equation
associated with (4.2) about the equilibria y¯i,(i=1,2) is (4.4)zn+1+1−δy¯iA−(δ+1)y¯izn−y¯iA−(δ+1)y¯izn−k=0. The characteristic equation of
(4.4) about the equilibrium y¯2=(1+A)/(1+δ) is (4.5)λk+1+[δA−1δ+1]λk+A+1δ+1=0. Now, we deduce from (4.5) the
following results:(a)
ifδ=0, and since A+1>1, then the equilibrium y¯2 is unstable (see [7]);(b)
ifA>δ>0, and since (A+1)/(δ+1)>1, then the equilibrium y¯2 is unstable;(c)
ifA=δ, then (4.6)|δA−1δ+1|+|δ+1δ+1|=|δ−1|+1. Now, we have the following
results from case (c): (i) if A=δ>1,
then the equilibrium y¯2 is unstable; (ii) if 0<A=δ<1,
then the equilibrium y¯2 is unstable; (iii) if A=δ=1,
then the linearized stability analysis fails;(d)
if1<A<δ, (4.7)|δA−1δ+1|+|A+1δ+1|=δA−1δ+1+A+1δ+1=A(δ+1)δ+1=A>1, and hence the equilibrium y¯2 is unstable;(e)
ifA<δ≤1,(4.8)|δA−1δ+1|+|A+1δ+1|=1−δA+1+Aδ+1≥1+A2(1−δ)≥1, and hence the equilibrium y¯2 is unstable. The
characteristic equation of (4.4) about the equilibrium y¯1=0 is (4.9)λk+1+1Aλk=0. This equation has two
roots (4.10)λ1=0,λ2=−1A. Now, we deduce from (4.10) the
following results:(i)
ifA>1, then the equilibrium y¯1=0 is locally asymptotically stable (see Figure 3);(ii)
if0<A<1, then the equilibrium y¯1=0 is unstable (see Figure 4);(iii)
ifA=1, then the linearized stability analysis fails.Figure 3Figure 4In the following results, we assume thatA≥δ+2, where δ≥0.Lemma 4.1.
Assume that the initial conditionsy−i∈[−1,1], for i=1,2,…,k and y0∈[−1,0]. Then, {y2n−1} is nonnegative and monotonically decreasing to zero, while {y2n} is nonpositive and monotonically increasing to zero.Proof.
Suppose
thaty−i∈[−1,1], for i=1,2,…,k and y0∈[−1,0]. Clearly, 0≤y1≤1 and −1≤y2≤0. By induction, we can see that 0≤y2n−1≤1 and −1≤y2n≤0 for n≥1.
IfA≥δ+2,δ≥0 we have (4.11)y2n−1y2n+1=(A−δy2n−y2n−k)(A−δy2n−1−y2n−k−1)>1, and hence (4.12)y2n−1>y2n+1,n=1,2,…. Similarly, we can show that y2n<y2n+2,n=1,2,…. The proof of Lemma 4.1 is completed.On using arguments similar to that used in Lemma4.1, we
can easily prove the following lemma.Lemma 4.2.
Assume that the initial conditionsy−i∈[−1,1], for i=1,2,…,k and y0∈[0,1]. Then, {y2n−1} is nonpositive and monotonically increasing to zero, while {y2n} is nonnegative and monotonically decreasing to zero.Corollary 4.3.
The equilibrium pointy¯1=0 of (4.1) is a global attractor with a basin S*=[−1,1]k+1.Theorem 4.4.
The
equilibrium pointy¯1=0 of (4.1) is a  global attractor with a basin S*=(−∞,1]k×[(−A+1)/(δ+1),(A−1)/(δ+1)], where δ≥0.Proof.
Assuming that the initial conditionsy−k,y−k+1,…,y−1,y0∈S*. If A≥δ+2, with δ≥0, then we deduce that (4.13)−1≤(1−A)/(1+δ)A−δy0−y−k≤y1=−y0A−δy0−y−k≤(A−1)/(δ+1)(A−1)/(δ+1)=1,−1≤−1A−δy1−y−k+1≤y2=−y1A−δy1−y−k+1≤1A−δ−1≤1. By induction, it follows that yi∈[−1,1] for i≥1. Thus, the proof of Theorem 4.4 follows from Corollary 4.3.Theorem 4.5.
IfA>1, then the equilibrium point y¯1=0 of (4.2) is globally asymptotically stable.Finally, on using arguments similar to that used in
Theorems3.5 and 3.6, we can prove easily the following results.Theorem 4.6.
Assume
thatδandA>1. If k is an odd positive integer, then the
necessary and sufficient condition for (4.2) to have positive solutions of
prime period two is that (see Figure 5) (4.14)(A−1)δ>A+3.Figure 5Theorem 4.7.
If k is an even positive integer,
then (4.2) has no positive solutions of prime period two (see Figure 6).Figure 6



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