# On Global Attractivity of a Class of Nonautonomous Difference Equations

Authors: Wanping Liu; Xiaofan Yang; Jianqiu Cao
Journal: Discrete Dynamics in Nature and Society
 (2010)

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## Abstract

We mainly investigate the global behavior to the family of higher-order nonautonomous recursive equations given byyn=(p+ryn-s)/(q+ϕn(yn-1,yn-2,…,yn-m)+yn-s), n∈ℕ0, with p≥0,r,q>0,s,m∈ℕ and positive initial values, and present some sufficient conditions for the parameters and maps ϕn:(ℝ+)m→ℝ+,n∈ℕ0, under which every positive solution to the equation converges to zero or a unique positive equilibrium. Our main result in the paper extends some related results from the work of Gibbons et al. (2000), Iričanin (2007), and Stević (vol. 33, no. 12, pages 1767–1774, 2002; vol. 6, no. 3, pages 405–414, 2002; vol. 9, no. 4, pages 583–593, 2005). Besides, several examples and open problems are presented in the end.


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## Body

## 1. Introduction

There has been a great interest in studying classes of nonlinear difference equations and systems, particularly those which model real situations in engineering and science, for example, [1–15]. On the other hand, non-autonomous difference equations also have a ubiquitous presence in applications from automatic controlling, ecology, economics, biology, population dynamics and so forth. Thus the main task when dealing them is to know the asymptotical behaviour of their solutions. For some recent advances in this area see [1, 16–24] and the references cited therein.Gibbons et al. [25] discussed the behavior of nonnegative solutions to the rational recursive equation(1.1)xn+1=α+βxn-1γ+xn,n∈ℕ0,
with α,β,γ≥0, and also proposed an open problem, which had been solved by Stević in [4], concerning the particular case α=0,γ=β in (1.1) (see also [26, 27] for the case of some related higher-order difference equations, as well as [28–30]).In [3], Stević studied the behavior of nonnegative solutions of the following second-order difference equation(1.2)xn+1=α+βxn-11+g(xn),n∈ℕ0,
where g:ℝ+⋃{0}→ℝ is a nonnegative increasing mapping. Obviously (1.2) is a generalization of (1.1).Later, Stević [6] extended (1.1) and (1.2) to the following more general equation(1.3)xn+1=α+βxn-kf(xn,…,xn-k+1),n∈ℕ0,
where k∈ℕ,α,β≥0 and f:ℝ+k→ℝ+ is a continuous function nondecreasing in each variable such that f(0,0,…,0)>0; and investigated the oscillatory behavior, the boundedness character and the global stability of nonnegative solutions to the equation.Recently, Iričanin [2] studied the asymptotic behavior of the following class of autonomous difference equations:(1.4)xn=αxn-k1+xn-k+f(xn-1,…,xn-m),n∈ℕ0,
where α>0,k,m∈ℕ and f is a continuous mapping satisfying the condition(1.5)βmin{u1,…,um}≤f(u1,u2,…,um)≤βmax{u1,…,um},
for certain β∈(0,1). In [2] he adopted the approach of frame sequences (a discrete analog of the method of frame curves used in the theory of differential equations), which has been used in the literature for many times, for example, [26–28, 30–38]; and showed that all positive solutions converge to zero if 0<α≤1 and converge to the unique positive equilibrium if α>1.Motivated by the above works, especially [2, 5], our aim in this paper is to study the global attractivity in the following family of non-autonomous difference equations:(1.6)yn=p+ryn-sq+ϕn(yn-1,…,yn-m)+yn-s,n∈ℕ0,
where p≥0,r,q>0,s,m∈ℕ, and ϕn:(ℝ+)m→ℝ+,n∈ℕ0 are mappings satisfying the following condition(1.7)βmin{x1,…,xm}≤ϕn(x1,x2,…,xm)≤βmax{x1,…,xm},
for some fixed β∈(0,+∞).Through careful analysis, we find that the results in [2] also persist if the function f in (1.4) is replaced by variable functions such as ϕn satisfying condition (1.7). If p=0, then (1.6) can be transformed into the following form(1.8)xn=(r/q)xn-s1+Gn(xn-1,…,xn-m)+xn-s,n∈ℕ0,
where Gn(x1,…,xm)=ϕn(qx1,qx2,…,qxm)/q, by setting yn=qxn. Then according to the results in [2], we have that if r≤q, then limn→∞yn=0; and if r>q, then limn→∞yn=qlimn→∞xn=q((r/q-1)/(1+β))=(r-q)/(1+β), for some β∈(0,1). Thus it suffices to consider the case when p>0 in the following.Note that ifp>0, then by relation (1.7), y¯=((q-r)2+4p(1+β)+r-q)/2(1+β) is the unique positive equilibrium of (1.6). And in Section 3, we will prove the following main theorem.Theorem 1.1.
Consider (1.6), where s,m∈ℕ,p,r,q>0 with rq≥p, and ϕn:(ℝ+)m→ℝ+ are functions satisfying the condition
(1.9)βmin{x1,…,xm}≤ϕn(x1,x2,…,xm)≤βmax{x1,…,xm},
for some fixed β∈(0,+∞). If q≥r,β∈(0,+∞) or q<r,β∈(0,β0], where β0=4p/(q-r)2+1, then the unique positive equilibrium y¯ of (1.6) is a global attractor.


## 2. Auxiliary Results

Before proving the main result of this paper, in this section we first confirm two preliminary lemmas.LetΦ:ℝ+⋃{0}×ℝ+→ℝ be the mapping Φ(x,w)=(p+rw)/(q+βx+w), where p,q,r,β>0 and rq≥p, so as Φ is decreasing in the first variable and increasing in the second one. Then (1.6) can be simplified to the following form:(2.1)yn=Φ(ϕn(yn-1,…,yn-m)β,yn-s),n∈ℕ0.Lemma 2.1.
Consider the following higher-order rational difference equation:(2.2)wn=Φ(x,wn-s),n∈ℕ0,
where p,r,q,β∈ℝ+,s∈ℕ, the parameter x≥0 and initial values wk,k∈{-1,…,-s} are arbitrary nonnegative numbers. Then every positive solution (wn)n=-s∞ to (2.2) converges to the unique positive equilibrium point
(2.3)S(x)=12((q-r+βx)2+4p-(q-r+βx)).Proof.
First we show that (2.2) has a unique positive equilibrium. Assume that w¯>0 is an equilibrium point of (2.2), then w¯=(p+rw¯)/(q+βx+w¯) which implies only one positive root
(2.4)w¯=S(x)=12((q-r+βx)2+4p-(q-r+βx)).
If s≥2, then (2.2) can be separated into s analogous first-order difference equations of the form
(2.5)wn(k)=p+rwn-1(k)q+βx+wn-1(k),n∈ℕ0,
with different initial values w-1(k)=w-(k+1), where k∈{0,1,…,s-1}. Note that the equation is Riccati, so it can be solved and the convergence of its solutions can be proved (see, e.g., [39] or a recent comment in [40]).
Let the symbol[] symbolize the greatest integer function and define a sequence P(n)≡n(mods),n∈ℕ0. Obviously, for each positive solution (wn)n=-s∞ to (2.2) we have(2.6)wn=w[n/s](s-P(n)-1),n≥-s.
From the above analysis, it suffices to prove the case whens=1. Suppose that s=1 for (2.2), then for all n∈ℕ0, we have(2.7)wn+1-wn=(w0-w-1)(r(q+βx)-p)n+1(q+βx+wn)(∏i=0n-1(q+βx+wi))2(q+βx+w-1),(2.8)w2k+2-w2k=(r(q+βx)-p)(w2k+1-w2k-1)(q+βx+w2k+1)(q+βx+w2k-1),k∈ℕ0,(2.9)w2k+3-w2k+1=(r(q+βx)-p)(w2k+2-w2k)(q+βx+w2k+2)(q+βx+w2k),k∈ℕ0.Case 1.
Ifr(q+βx)≥p, then by (2.7) (wn)n=-1∞ is either nonincreasing or nondecreasing. On the other hand, we have that
(2.10)min{p,r}max{q+βx,1}≤wn≤max{p,r}min{q+βx,1}
for all n≥0. Therefore, the limit of (wn)n=-1∞ exists, and through simple calculations, we get limn→∞wn=S(x).Case 2.
Ifr(q+βx)<p, then by (2.8) and (2.9) and inductively we have that (w2k) is nonincreasing and (w2k-1) nondecreasing, or (w2k) is nondecreasing and (w2k-1) nonincreasing. Again by (2.10), the limits of (w2k) and (w2k-1) exist, denoted by limk→∞w2k=α and limk→∞w2k-1=γ. From (2.2) we have
(2.11)α=p+rγq+βx+γ,γ=p+rαq+βx+α,
which imply that α=γ=w¯. Hence limn→∞wn=S(x).
The proof of Lemma2.1 is complete.Lemma 2.2.
Suppose that the parameters, in (2.3), satisfy p,r,q,β>0 with t=q-r. Define two sequences (Mk)k=1∞ and (mk)k=1∞ as follows:
(2.12)mk=S(Mk+ɛk),k=1,2,…,Mk=S(mk-1-ɛk-1),k=2,3,…,
where the initial value M1=S(0), and ɛ∈(0,λ),
(2.13)λ=12(1+β)((t+βM1)2+4p(1+β)-(t+βM1)).
If q≥r,β∈(0,+∞) or q<r,β∈(0,β0], where β0=4p/(q-r)2+1, then
(2.14)limk→∞Mk=limk→∞mk.Proof.
By simple calculations, we have(2.15)M2-M1=S(m1-ɛ)-S(0)=12((t+β(m1-ɛ))2+4p-t2+4p-β(m1-ɛ))=12(β2(m1-ɛ)2+2βt(m1-ɛ)(t+β(m1-ɛ))2+4p+t2+4p-β(m1-ɛ))=-β(m1-ɛ)(S(m1-ɛ)+S(0)(t+β(m1-ɛ))2+4p+t2+4p).
Obviously, S(m1-ɛ)+S(0)>0.Claim 1.
m1-ɛ>0.Proof.
Define a functionf(x)=2(S(M1+x)-x). It suffices to prove that f(x)>0 for all x∈(0,λ). The derivative of f(x) is
(2.16)dfdx=β[t+β(M1+x)][t+β(M1+x)]2+4p-β-2<0.
Since f(λ)=0 and f(0)>0, thus f(x)>0 for x∈(0,λ).Therefore, it follows from (2.15) and Claim 1 that
(2.17)M2-M1<0.
Denote
(2.18)Lk=2t+β(Mk+1+Mk+ɛ/k+ɛ/(k+1))[t+β(Mk+1+ɛ/(k+1))]2+4p+[t+β(Mk+ɛ/k)]2+4p,Qk=2t+β(mk+mk-1-ɛ/k-ɛ/(k-1))[t+β(mk-ɛ/k)]2+4p+[t+β(mk-1-ɛ/(k-1))]2+4p.
Simply, we obtain that |Lk|<1 and |Qk|<1.
Observe that(2.19)2(mk+1-mk)=β(1-Lk)[Mk-Mk+1+ɛk(k+1)],k=1,2,…,2(Mk+1-Mk)=β(Qk-1)[mk-mk-1+ɛk(k+1)],k=2,3,….
With (2.17) and (2.19), it follows by induction that (mk)k=1∞,(Mk)k=1∞ are strictly increasing and decreasing, respectively. In addition, Mk>0,k=1,2,…, hence (Mk)k=1∞ possesses a finite limit denoted by φ=limk→∞Mk. From (2.12), we know that the limit of (mk)k=1∞ (denoted by μ=limk→∞mk) also exists. Therefore, taking limits on both sides of (2.12), we have
(2.20)μ=S(φ),φ=S(μ)
which imply that
(2.21)μ2+tμ+βμφ=p,φ2+tφ+βμφ=p,(2.22)(μ-φ)(μ+φ+t)=0.Claim 2.
Ifq≥r,β∈(0,+∞) or q<r,β∈(0,β0], then μ=φ.Proof.
Suppose thatμ≠φ, then it follows from (2.22) that μ=-φ-t. By substituting μ=-φ-t into the second identity of (2.21), we get
(2.23)(1-β)φ2+t(1-β)φ-p=0.(i)
Ifβ=1, then p=0 which is a contradiction to p>0,(ii)
Ifβ∈(0,1), then the unique positive root of (2.23) is
(2.24)φ=t2+4p/(1-β)-t2>M1.
However, φ<M1 since (Mk) is strictly decreasing.(iii)
Ifq=r,β∈(1,+∞), then (2.23) reduces to 0>(1-β)φ2=p>0.(iv)
Ifq≠r,β∈(1,β0), then for (2.23), Δ=t2(1-β)2+4p(1-β)<0 which implies that (2.23) has no real roots.(v)
Forq>r,β=β0, we have Δ=0. So, φ=(r-q)/2<0 which is contradictive to φ≥0.(vi)
Forq>r,β∈(β0,+∞), (2.23) has two negative roots.(vii)
Forq<r,β=β0. Solving (2.23), we get φ=(r-q)/2 implying μ=(r-q)/2. Hence μ=φ, which contradicts the assumption.
Obviously Claim2 follows directly from (i)–(vii).
Applying Claim2 and (2.21), we conclude that(2.25)limk→∞Mk=limk→∞mk=t2+4p(1+β)-t2(1+β).
Hence the lemma is complete.


## 3. Main Results

Obviously, condition (1.7) in Section 1 guarantees the fact that (1.6) possesses a unique equilibrium point y¯=(t2+4p(1+β)-t))/(2(1+β)), where t=q-r.First, we present a proposition concerning the boundedness of all positive solutions to (1.6).Proposition 3.1.
Consider (1.6) with condition (1.7) and p,q,r∈ℝ+, then every positive solution to (1.6) has permanent bounds.Proof.
Let(yn) be a solution to (1.6) with positive initial values. Then, we have
(3.1)yn=p+ryn-sq+ϕn(yn-1,…,yn-m)+yn-s≤max{p,r}min{q+ϕn(yn-1,…,yn-m),1}≤max{p,r}min{q,1}=U,∀n≥0,yn=p+ryn-sq+ϕn(yn-1,…,yn-m)+yn-s≥min{p,r}max{q+ϕn(yn-1,…,yn-m),1}≥min{p,r}max{q+βU,1}=L,∀n≥m.
Thus we have L≤yn≤U, for all n≥m.In the following, we will give the proof of the main result (i.e., Theorem1.1) presented in Section 1.Proof of Theorem1.1.
Letɛ∈ℝ+ be an arbitrary fixed number satisfying 0<ɛ<λ (λ defined by (2.13) in Lemma 2.2). Define two sequences (Mk)k=1∞,(mk)k=1∞ as shown by (2.12) in Lemma 2.2. Let (yn) be any positive solution to (1.6). In the following, we proceed by presenting two claims.Claim 1.
There existsN1∈ℕ, such that m1-ɛ≤yn≤M1+ɛ for all n≥ℕ1.Proof.
From (2.1), we have that
(3.2)yn=Φ(ϕn(yn-1,…,yn-m)β,yn-s)≤Φ(0,yn-s).
Suppose that (xn) is a solution to the following difference equation
(3.3)xn=Φ(0,xn-s),n∈ℕ0,
with initial values x-s=y-s,…,x-1=y-1. From this and in view of the monotonicity of the function f(x)=(p+rx)/(q+x),x∈ℝ+, by induction we can easily get that yn≤xn for n≥-s.
By Lemma2.1, limn→∞xn=M1. Hence, there exists b1∈ℕ such that xn≤M1+ɛ for n≥b1, then(3.4)yn≤M1+ɛ
for all n≥b1.
From (2.1), (1.7), and (3.4), it follows that(3.5)yn≥Φ(max{yn-1,…,yn-m},yn-s)≥Φ(M1+ɛ,yn-s)
for all n≥b1+m.
Suppose that(un) is a solution to the following difference equation:(3.6)un=Φ(M1+ɛ,un-s),n∈ℕ0,
with initial values ub1+m-s=yb1+m-s,…,ub1+m-1=yb1+m-1.
Since the functionh(x)=Φ(M1+ɛ,x) is increasing on the interval (0,+∞), we can easily get by induction that yn≥un for n≥b1+m-s, and by Lemma 2.1, limn→∞un=m1. Hence there exists a natural number N1≥b1 such that un≥m1-ɛ for n≥N1, then m1-ɛ≤yn≤M1+ɛ for n≥N1.
Working inductively, we will eventually reach the following claim.
Claim 2.
For eachk∈ℕ, there exists Nk∈ℕ such that mk-ɛ/k≤yn≤Mk+ɛ/k for all n≥Nk.Proof.
By Claim1, if k=1, we have N1∈ℕ such that m1-ɛ≤yn≤M1+ɛ for all n≥N1. Then by the method of induction, we can assume that for k∈ℕ fixed, there exists Nk∈ℕ such that mk-ɛ/k≤yn≤Mk+ɛ/k for all n≥Nk. Thus, it suffices to show that there exists Nk+1∈ℕ such that mk+1-ɛ/(k+1)≤yn≤Mk+1+ɛ/(k+1) for all n≥Nk+1.
Letz=max{s,m}. Define a sequence (xn(k+1)) as follows(3.7)xn(k+1)=Φ(S-1(Mk+1),xn-s(k+1)),n≥Nk+z,
with xn(k+1)=yn, for n=Nk,…,Nk+z-1.
By reasoning inductively onn≥Nk+z, one has(3.8)yn≤Φ(min{yn-1,…,yn-m},yn-s)≤Φ(S-1(Mk+1),yn-s)≤Φ(S-1(Mk+1),xn-s(k+1))=xn(k+1),∀n≥Nk+z.
By Lemma 2.1, limn→∞xn(k+1)=Mk+1. Therefore, there is bk+1≥Nk such that yn≤Mk+1+ɛ/(k+1)foralln≥bk+1.
Define the other sequence(un(k+1)) as follows:(3.9)un(k+1)=Φ(S-1(mk+1),un-s(k+1)),forn≥bk+1,
where un(k+1)=yn,forn=bk+1,…,bk+1+z-1.
Once more, by induction onn≥bk+1+z,(3.10)yn≥Φ(max{yn-1,…,yn-m},yn-s)≥Φ(Mk+1+ɛk+1,yn-s)≥Φ(S-1(mk+1),un-s(k+1))=un(k+1),∀n≥bk+1+z.
By Lemma 2.1, limn→∞un(k+1)=mk+1. Thus, let Nk+1≥bk+1 be greater enough so as yn≥mk+1-(ɛ/(k+1))=S-1(Mk+2) for all n≥Nk+1.
Therefore, we get that there existsNk+1∈ℕ such that(3.11)mk+1-ɛk+1≤yn≤Mk+1+ɛk+1
for all n≥Nk+1.By Claim 2, we have
(3.12)limk→∞mk=limk→∞(mk-ɛk)≤lim infn→∞yn≤lim supn→∞yn≤limk→∞(Mk+ɛk)=limk→∞Mk.
This plus Lemma 2.2 leads to
(3.13)limn→∞yn=y¯=t2+4p(1+β)-t2(1+β).
The proof is complete.


## 4. Applications and Future Work

Next, several examples are presented.Example 4.1.
Letpn∈(0,+∞) for all n∈ℕ0, and
(4.1)ϕn(x1,x2,…,xm)=β∑i=1mxipnmpn,n∈ℕ0
for some β>0. If rq>p and q≥r,β∈(0,+∞) or q<r,β∈(0,β0], where β0=4p/(q-r)2+1, then by Theorem 1.1 we conclude that every positive solution to the following non-autonomous difference equation:
(4.2)yn=p+ryn-sq+β(∑i=1myn-ipn)/mpn+yn-s,n∈ℕ0,
converges to the unique positive equilibrium point y¯=((q-r)2+4p(1+β)+r-q)/2(1+β).Example4.1 extends Example 2.4 in [2].Example 4.2.
Letϕn(x1,x2,…,xm)=βmax{x1,x2,…,xm} for all n∈ℕ0, then under the conditions of Theorem 1.1, all positive solutions to the recursive equation
(4.3)yn=p+ryn-sq+βmax{yn-1,yn-2,…,yn-m}+yn-s,n∈ℕ0,
converge to the unique positive equilibrium y¯=((q-r)2+4p(1+β)+r-q)/2(1+β).In this paper, the behavior of positive solutions to the case whenrq≥p,q<r,β∈(β0,+∞), where β0=4p/(q-r)2+1, isn't investigated, since we have no further new ideas for the particular case. Through certain calculations, easily we know that the equation S∘S(x)=x has two different positive roots, if q<r,β∈(β0,+∞), which implies limk→∞Mk>limk→∞mk. From this we propose the following open problem.Open Problem 4.
Is there a positive solution(yn) to (1.6) with condition (1.7) when rq≥p,q<r,β∈(β0,+∞), where β0=4p/(q-r)2+1, such that (yn) eventually converges to a periodic solution?Furthermore, the caserq<p for (1.6) is also of extreme value to study.



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