# On the Max-Type Difference Equationxn+1=max{A/xn,xn−3}

Authors: Bratislav D. Iričanin; E. M. Elsayed
Journal: Discrete Dynamics in Nature and Society
 (2010)

---

## Abstract

We show that every well-defined solution of the fourth-order difference equationxn+1=max{A/xn,xn−3},    n∈ℕ0, where parameter A≥0, is eventually periodic with period four.


---

## Body

## 1. Introduction

The study of max-type difference equations attracted recently a considerable attention, see, for example, [1–27], and the references listed therein. This type of difference equations stems from, for example, certain models in automatic control theory (see [28]). In the beginning of the study of these equations experts have been focused on the investigation of the behavior of some particular cases of the following general difference equation of order k∈ℕ:(1.1)xn=max{An(1)xn-1,An(2)xn-2,…,An(k)xn-k},n∈ℕ0,
where k∈ℕ, An(i), i=1,…,k, are real sequences (mostly constant or periodic ones) and where the initial values x-1,…,x-k are different from zero (see, e.g., [2, 3, 6, 7, 9–12, 22–25] and the references cited therein).The study of max-type equations of the following general form(1.2)xn=max{Bn(0),Bn(1)xn-p1r1xn-q1s1,Bn(2)xn-p2r2xn-q2s2,…,Bn(k)xn-pkrkxn-qksk},n∈ℕ0,
where k∈ℕ, pi,qi are natural numbers such that p1<p2<⋯<pk, q1<q2<⋯<qk, ri,si∈ℝ+ and Bn(j), j=0,1,…,k, are sequences of real numbers, was proposed by  Stević in numerous talks, for example, in [13, 14]. For some results in this direction see [1, 4, 15–17, 19–21, 26, 27]. For some nonlinear difference equations related to (1.2) see, for example, [7, 15, 17, 18, 29–38].Definition 1.1.
A sequence(xn)n=-k∞ is said to be eventually periodic with periodp if there is an index n0∈{-k,…,-1,0,1,…} such that xn+p=xn for all n≥n0. Specially, if n0=-k, then the sequence (xn)n=-k∞ is periodic with periodp.Motivated by some ideas due to Stević (e.g., the main lemmas there, Lemmas3.1 and 3.2 are suggested by him), the authors of [26] considered the following second-order max-type difference equation:(1.3)xn+1=max{1xn,Axn-1},n∈ℕ0.
Equation(1.3) is not difficult for handling since, by the change yn=xnxn-1, it is transformed into one of the following first-order difference equations(1.4)yn+1=max{1,Ayn}oryn+1=min{1,Ayn}.
Using these equations, it is easy to see that for the case A=1 every solution of (1.3) is eventually periodic with period two.Recently, in the paper [5] it was showed that every solution of the third-order max-type difference equation(1.5)xn+1=max{Axn,xn-2},n∈ℕ0,
where the initial conditions x-2,x-1,x0are arbitrary nonzero real numbers and A∈ℝ, is eventually periodic with period three. The fact that all solutions of (1.5) are periodic is not a surprising fact (for an explanation see [4]).For some recent papers on difference equations all the solutions of which are periodic see, for example, [7, 39–45] and the references cited therein.Here we show that every well-defined solution of the following fourth-order max-type difference equation(1.6)xn+1=max{Axn,xn-3},n∈ℕ0,
where the parameter A∈ℝ+∪{0}, is eventually periodic with period four.Remark 1.2.
Note that ifA=0, then (1.6) becomes xn+1=xn-3, from which it follows that every solution is periodic with period four. Hence, in the sequel we will consider the case A≠0.In the sequel we will frequently use the following simple lemma, given without a proof (for related results see [4, 5]).Lemma 1.3.
Assume that(xn)n=-3∞ is a solution of (1.6) and there is k0∈ℕ0∪{-3,-2,-1} such that
(1.7)xk0=xk0+4,xk0+1=xk0+5,xk0+2=xk0+6,xk0+3=xk0+7.
Then this solution is eventually periodic with period four.


## 2. Main Results

In this subsection we give a specific form of the solutions of the difference equation (1.6) when the parameter A>0 and in each case we can deduce that every solution of this equation is periodic with period four.Depending on the positivity of four initial values of (1.6), there are the following 16 cases to be considered:(2.1)(i)x-3,x-2,x-1,x0>0,(ii)x-3,x-2,x-1,x0<0,(iii)x0<0,x-3,x-2,x-1>0,(iv)x-1<0,x-3,x-2,x0>0,(v)x-2<0,x-3,x-1,x0>0,(vi)x-3<0,x-2,x-1,x0>0,(vii)x0,x-1<0,x-3,x-2>0,(viii)x0,x-2<0,x-3,x-1>0,(ix)x0,x-3<0,x-1,x-2>0,(x)x-1,x-2<0,x0,x-3>0,(xi)x-1,x-3<0,x0,x-2>0,(xii)x-2,x-3<0,x0,x-1>0,(xiii)x0,x-1,x-2<0,x-3>0,(xiv)x0,x-1,x-3<0,x-2>0,(xv)x0,x-2,x-3<0,x-1>0,(xvi)x-1,x-2,x-3<0,x0>0.First, we prove another auxiliary result.Lemma 2.1.
Assume that the parameterA>0. Then every solution of (1.6) is eventually positive if initial values satisfy one of conditions (i), (iii)–(xvi).Proof.
Ifx0>0 or x-3>0, then
(2.2)x1=max{Ax0,x-3}>0.
From this, (1.6), and by induction it follows that xn>0 for every n∈ℕ0.
Ifx-2>0, then
(2.3)x2=max{Ax1,x-2}>0.
From this, (1.6), and by induction it follows that xn>0 for every n≥2.
Ifx-1>0, then
(2.4)x3=max{Ax1,x-1}>0.
Similar to the previous case, by induction it follows that xn>0 for every n≥3.Now, we can formulate and prove our main results.Theorem 2.2.
Assume that the parameterA>0. Then every solution of (1.6) with positive initial values is eventually periodic with period four.Proof.
From (1.6), we see that
(2.5)x1=max{Ax0,x-3}.
We consider the following two cases.  (a1)
x1=A/x0. In this case A/x0≥x-3, and we see that
(2.6)x2=max{Ax1,x-2}=max{x0,x-2}.Now, there exists two subcases.  (a11)
x2=x0, which occurs when x0≥x-2. We have
(2.7)x3=max{Ax2,x-1}=max{Ax0,x-1}.(a111)
Ifx-1≥A/x0, then x3=x-1,and(2.8)x4=max{Ax3,x0}=max{Ax-1,x0}=x0,x5=max{Ax4,x1}=max{Ax0,Ax0}=Ax0,x6=max{Ax5,x2}=max{x0,x0}=x0.
Hence, x3=x-1, x4=x0, x5=x1, and x6=x2, which implies that (xn)n=-3∞ is an eventually (from x-1) periodic solution with period four. In this case we see that the solution has the following form:
(2.9)(x-3,x-2,x-1,x0,Ax0,x0,x-1,x0,Ax0,x0,…).(a112)
IfA/x0≥x-1, then x3=A/x0,and(2.10)x4=max{Ax3,x0}=max{x0,x0}=x0,x5=max{Ax4,x1}=max{Ax0,Ax0}=Ax0,x6=max{Ax5,x2}=max{x0,x0}=x0,x7=max{Ax6,x3}=max{Ax0,Ax0}=Ax0.
Hence, x4=x0, x5=x1, x6=x2, and x7=x3, which implies that (xn)n=-3∞ is an eventually (from x0) periodic solution with period four (in this case minimal period is two). This solution takes the form
(2.11)(x-3,x-2,x-1,x0,Ax0,x0,Ax0,x0,Ax0,x0,Ax0,…).(a12)
x2=x-2, which occurs when x-2≥x0, and
(2.12)x3=max{Ax2,x-1}=max{Ax-2,x-1}.(a121)
Ifx-1≥A/x-2, then x3=x-1,and
(2.13)x4=max{Ax3,x0}=max{Ax-1,x0}.(a1211)
Ifx0≥A/x-1, then x4=x0,and(2.14)x5=max{Ax4,x1}=max{Ax0,Ax0}=Ax0.
Hence, x2=x-2, x3=x-1, x4=x0, and x5=x1, which implies that (xn)n=-3∞ is an eventually (from x-2) periodic solution with period four. It can be written in the form
(2.15)(xn)n=-3∞=(x-3,x-2,x-1,x0,Ax0,x-2,x-1,x0,Ax0,…).(a1212)
IfA/x-1≥x0, then x4=A/x-1,and(2.16)x5=max{Ax4,x1}=max{x-1,Ax0}=Ax0,x6=max{Ax5,x2}=max{x0,x-2}=x-2,x7=max{Ax6,x3}=max{Ax-2,x-1}=x-1,x8=max{Ax7,x4}=max{Ax-1,Ax-1}=Ax-1.
Hence, x5=x1, x6=x2, x7=x3, and x8=x4, which implies that (xn)n=-3∞ is an eventually (from x1) periodic solution with period four. Moreover, it can be written as follows:
(2.17)(xn)n=-3∞=(x-3,x-2,x-1,x0,Ax0,x-2,x-1,Ax-1,Ax0,x-2,x-1,Ax-1,…).(a122)
IfA/x-2≥x-1, then x3=A/x-2,and(2.18)x4=max{Ax3,x0}=max{x-2,x0}=x-2,x5=max{Ax4,x1}=max{Ax-2,Ax0}=Ax0,x6=max{Ax5,x2}=max{x0,x-2}=x-2,x7=max{Ax6,x3}=max{Ax-2,Ax-2}=Ax-2,x8=max{Ax7,x4}=max{x-2,x-2}=x-2.
As above, the solution is eventually (from x1) periodic with period four and it has the form
(2.19)(x-3,x-2,x-1,x0,Ax0,x-2,Ax-2,x-2,Ax0,x-2,Ax-2,x-2,Ax0,…).(a2)
x1=x-3. In this case x-3≥A/x0, and we see that
(2.20)x2=max{Ax1,x-2}=max{Ax-3,x-2}.There again exist two subcases.  (a21)
x2=A/x-3, which occurs when A/x-3≥x-2. So,
(2.21)x3=max{Ax2,x-1}=max{x-3,x-1}.(a211)
Ifx-3≥x-1, then x3=x-3,and(2.22)x4=max{Ax3,x0}=max{Ax-3,x0}=x0,x5=max{Ax4,x1}=max{Ax0,x-3}=x-3,x6=max{Ax5,x2}=max{Ax-3,Ax-3}=Ax-3,x7=max{Ax6,x3}=max{x-3,x-3}=x-3.
Then we see that the solution is
(2.23)(x-3,x-2,x-1,x0,x-3,Ax-3,x-3,x0,x-3,Ax-3,x-3,x0,x-3,Ax-3,…),
and (xn)n=-3∞ is an eventually (from x0) periodic solution with period four.  (a212)
Ifx-1≥x-3, then x3=x-1,and(2.24)x4=max{Ax3,x0}=max{Ax-1,x0}=x0,x5=max{Ax4,x1}=max{Ax0,x-3}=x-3,x6=max{Ax5,x2}=max{Ax-3,Ax-3}=Ax-3.
So, the solution takes the following form which is an eventually (from x-1) periodic solution with period four:
(2.25)(x-3,x-2,x-1,x0,x-3,Ax-3,x-1,x0,x-3,Ax-3,…).(a22)
x2=x-2, which occurs when x-2≥A/x-3. So,
(2.26)x3=max{Ax2,x-1}=max{Ax-2,x-1}.(a221)
If x-1≥A/x-2, then x3=x-1,and
(2.27)x4=max{Ax3,x0}=max{Ax-1,x0}.(a2211)
Ifx0≥A/x-1, then x4=x0.
Therefore(xn)n=-3∞ is a periodic solution with period four and the solution takes the form
(2.28)(x-3,x-2,x-1,x0,x-3,x-2,x-1,x0,…).(a2212)
IfA/x-1≥x0, then x4=A/x-1,and(2.29)x5=max{Ax4,x1}=max{x-1,x-3}=x-3,x6=max{Ax5,x2}=max{Ax-3,x-2}=x-2,x7=max{Ax6,x3}=max{Ax-2,x-1}=x-1,x8=max{Ax7,x4}=max{Ax-1,Ax-1}=Ax-1.
Therefore, again (xn)n=-3∞ is an eventually (from x1) periodic solution with period four and the solution takes the form
(2.30)(x-3,x-2,x-1,x0,x-3,x-2,x-1,Ax-1,x-3,x-2,x-1,Ax-1,…).(a222)
IfA/x-2≥x-1, then x3=A/x-2,and
(2.31)x4=max{Ax3,x0}=max{x-2,x0}.(a2221)
Ifx-2≥x0, then x4=x-2,and(2.32)x5=max{Ax4,x1}=max{Ax-2,x-3}=x-3,x6=max{Ax5,x2}=max{Ax-3,x-2}=x-2,x7=max{Ax6,x3}=max{Ax-2,Ax-2}=Ax-2,x8=max{Ax7,x4}=max{x-2,x-2}=x-2.
Thus, the solution is in the following form which is eventually (from x1) periodic with period four:
(2.33)(x-3,x-2,x-1,x0,x-3,x-2,Ax-2,x-2,x-3,x-2,Ax-2,x-2,…).(a2222)
Ifx0≥x-2, then x4=x0,and(2.34)x5=max{Ax4,x1}=max{Ax0,x-3}=x-3,x6=max{Ax5,x2}=max{Ax-3,x-2}=x-2,x7=max{Ax6,x3}=max{Ax-2,Ax-2}=Ax-2.
Thus, the solution is of the following form
(2.35)(x-3,x-2,x-1,x0,x-3,x-2,Ax-2,x0,x-3,x-2,Ax-2,…).
Therefore (xn)n=-3∞ is eventually (from x0) periodic with period four. The proof is completed.From Lemma1.3 and Theorem 2.2 we obtain the following result.Theorem 2.3.
Assume the parameterA>0 and that initial values of (1.6) satisfy one of conditions  (i), (iii)–(xvi) in Lemma 2.1. Then every such solution of (1.6) is eventually periodic with period four.Proof.
If initial values of (1.6) satisfy one of conditions (i), (iii)–(xvi) in Lemma 2.1, then by the same lemma it follows that the corresponding solution is eventually positive. This means that there is k1∈ℕ0∪{-3,-2,-1} such that xn>0 for every n≥k1. In particular, we have that xk1,xk1+1,xk1+2,xk1+3>0. Since equation (1.6) is autonomous if (xn)n=-3∞ is a solution of (1.6), then yn=xn+k1+3 is also a solution of (1.6) but such that y-3,y-2,y-1,y0>0. Hence, the problem is reduced to the case when all the initial values are positive. Applying Theorem 2.2, the result follows.In the next theorem we study those solutions of (1.6) such that x-3,x-2,x-1,x0<0. We would like to thank Professor Stević for giving us the elegant proof below which drastically reduced our original proof.Theorem 2.4.
Assume that the parameterA>0 and all the initial values are negative x-3,x-2,x-1,x0<0. Then every solution of (1.6) is eventually periodic with period four.Proof.
Sincex-3<0, x-2<0, x-1<0, x0<0, and A>0, by induction we have xn<0 for each n∈ℕ. By the change xn=-A/zn, (1.6) becomes
(2.36)zn+1=(min{zn,1zn-3})-1=max{1zn,zn-3},
where zn>0 for every n≥-3. Hence by Theorem 2.2 the result follows.Since all the cases are reduced to the case when all initial values are positive, it is of interest to investigate this case in more details. The next theorem describes all eventually constant solutions of (1.6). Our idea stems from [46] (see also [47–50]).Theorem 2.5.
Assume that the parameterA>0. Then all positive solutions to (1.6) which are eventually equal to the positive equilibrium have the following form:
(2.37)(1,c,b,1,1,1,…),forsomeb,c∈(0,1].Proof.
First note that by the changexn=Ayn the problem is reduced to the case A=1. Assume that xn=1 for n≥k and xk-1=a≠1. Then 1=xk=max{1/a,xk-4}, k≥1, which implies xk-4=1 and a>1. On the other hand, we have 1=xk+3=max{1/xk+2,xk-1}=max{1,a}, so that a<1, which is a contradiction. Hence k≤0, that is, (xn)=(d,c,b,1,1,1,…). Since x1=1, we get x-3=1 (case k=1 above). Since 1=x2=max{x1,x-2}=max{1,c} and 1=x3=max{x2,x-1}=max{1,b}, it follows that b,c∈(0,1], as claimed.


## 3. Conclusions and Future Works

We finish this article with some comments which can motivate further works.CaseA<0
The case when the parameterA is negative is not treated in this article. By similar calculations as above, we have managed to show that all the solutions of (1.6) are eventually periodic with period four in many subcases. However, there are too many subcases and calculations without some new ideas, so we decided not to present any result when A<0. We conjecture that every solution of (1.6), when A<0, is eventually periodic with the same period four. What is more interesting is to find a reasonably short proof of the conjecture without using tiresome calculations similar to those in Theorem 2.2.Periodicity
We also want to mention that in [4] we proved that every solution of the equation
(3.1)xn+1=max{Axn-k,xn-l},n∈ℕ0,
where k,l∈ℕ0, is periodic. It is of some interest to find the minimal period of the equation, as well as to get a general result concerning this problem.



---

*Source: 675413-2010-02-21.xml*