# On the Max-Type Equationxn+1=max1/xn,Anxn-1 with a Period-Two Parameter

Authors: İbrahim Yalçınkaya
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## Abstract

We study the behavior of the well-defined solutions of the max type difference equationxn+1=max1/xn,Anxn-1, n=0,1,…, where the initial conditions are arbitrary nonzero real numbers and {An} is a period-two sequence of real numbers with An∈[0,∞).


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## Body

## 1. Introduction and Preliminaries

Recently, the study of max-type difference equations attracted a considerable attention. Although max-type difference equations are relatively simple in form, it is, unfortunately, extremely difficult to understand thoroughly the behavior of their solutions; see, for example [1–39] and the relevant references cited therein. Max-type difference equations stem from certain models in automatic control theory (see [1, 24]). For some papers on periodicity of difference equation, see, for example, [15, 16, 19, 22] and the relevant references cited therein.In [9], Simsek et al. studied the behavior of the solutions of the following max-type difference equation:(1.1)xn+1=max⁡⁡{xn-1,1xn-1},n=0,1,…,
where the initial conditions are nonzero real numbers.In [10], Simsek studied the behavior of the solutions of the following max-type difference equation:(1.2)xn+1=max⁡⁡{xn-2,1xn-2},n=0,1,…,
where the initial conditions are negative real numbers.In [18], Elabbasy and Elsayed studied the behavior of the solutions of (1.2) where the initial conditions are nonzero real numbers.In [20], Elsayed and Stević showed that every well-defined solution of the difference equation(1.3)xn+1=max⁡{Axn,xn-2},n=0,1,…,
where A∈ℝ, is eventually periodic with period three.In [21], Elsayed and Iričanin showed that every positive solution to the following third-order nonautonomous max-type difference equation:(1.4)xn+1=max⁡⁡{Anxn,xn-2},n=0,1,…,
where An is a three-periodic sequence of positive numbers and is periodic with period three.In [29], Yalçinkaya et al. studied the behavior of the solutions of the following max-type difference equation:(1.5)xn+1=max⁡⁡{1xn,Axn-1},n=0,1,…,
where A∈ℝ and initial conditions are nonzero real numbers.In this paper, we study the behavior of the well-defined solutions of the max type difference equation(1.6)xn+1=max⁡⁡{1xn,Anxn-1},n=0,1,…,
where the initial conditions are arbitrary nonzero real numbers and {An} is a period-two sequence of real numbers with An∈[0,∞).We need the following definitions and lemmas.Definition 1.1.
A sequence{xn}n=-k∞ is said to be eventually periodic with period p if there is n0∈{-k,…,-1,0,1,…} such that xn+p=xn for all n≥n0. If n0=-k, then we say that the sequence {xn}n=-k∞ periodic with period p.We make two definitions regarding (1.6).Definition 1.2.
A right semicycle is a string of termsxl,…,xm with l≥1, m≤∞ such that xn=An-1xn-2 for all n=l,…,m. Furthermore, if l>1,xl-1=1/xl-2, and if m<∞,xm+1=1/xm.Definition 1.3.
A left semicycle is a string of termsxl,…,xm with l≥1, m≤∞ such that xn=1/xn-1 for all n=l,…,m. Furthermore, if l>1, xl-1=An-2xn-3, and if m<∞,xm+1=Amxm-1.We give the following lemmas which show us the periodic behavior of the solutions of (1.6).Lemma 1.4.
Assume that{xn}n=-1∞ is a well-defined solution of (1.6). If xn0=xn0+2 and xn0+1=xn0+3 such that n0∈ℕ0∪{-1}, then the solution {xn}n=-1∞ is eventually periodic with period two.Proof.
We prove that(1.7)xn0=xn0+2m,xn0+1=xn0+2m+1,
by induction. For m=1, this is, assumption. Assume that (1.7) holds for all 1≤n≤m0∈ℕ. We may assume that n0 is odd. Then, by the inductive hypothesis, we have
(1.8)xn0+2(m0+1)=max⁡⁡{1xn0+2m0+1,An0+2m0+1xn0+2m0}=max⁡⁡{1xn0+1,A1xn0}=xn0+1=xn0,
from this and the inductive hypothesis, we have
(1.9)xn0+2(m0+1)+1=max⁡{1xn0+2m0+2,An0+2m0+2xn0+2m0+1}=max⁡{1xn0,A0xn0+1}=xn0+3=xn0+1,
which completes the proof (the case n0 is even similar, so it will be omitted).We omit the proof of the following lemma, since it can easily be obtained by induction.Lemma 1.5.
Assume that{xn}n=-1∞ is a well-defined solution of (1.6). If xn0,xn0+1>0 such that n0∈ℕ0∪{-1}, then xn>0 for all n≥n0.Lemma 1.6.
Assume that{xn}n=-1∞ is a well-defined solution of (1.6) and An∈[0,1). If this solution is eventually positive, then it is eventually periodic with period two.Proof.
Assume thatn0∈ℕ0∪{-1} is the smallest index such that xn>0 for all n≥n0. Then, we have
(1.10)xn+1xn=max⁡⁡{1,Anxnxn-1}∀n≥n0+1.
Using this, we have
(1.11)xn0+2xn0+1=max⁡{1,An0+1xn0+1xn0},xn0+3xn0+2=max⁡{1,An0+2xn0+2xn0+1}=max⁡{1,An0+2,An0+2An0+1xn0+1xn0}=max⁡{1,An0+2An0+1xn0+1xn0},xn0+4xn0+3=max⁡{1,An0+3An0+2An0+1xn0+1xn0},⋮
then we get
(1.12)xn0+k+1xn0+k=max⁡⁡{1,xn0+1xn0∏i=1kAn0+i}∀k≥1.
Observe that there exists a positive integer k such that
(1.13)xn0+1xn0∏i=1kAn0+i≤1.
From this directly follows that {xn}n=-1∞ is eventually periodic with period two.Lemma 1.7.
Equation (1.6) has no right semicycle with an infinite terms for the positive initial conditions and 0<A0,A1<1.Proof.
Conversely, assume that (1.6) has a right semicycle with an infinite terms. And, let {an} be periodic sequence of natural numbers with period two such that (an)=(0,1,0,1,…). Without loss of generality, we denote by x1 the first term of right semicycle with an infinite terms. There is at least n0∈ℕ. For all n>n0, we can write
(1.14)xn=max⁡{1A0[|n/2|]anA1[|n/2|]an+1x-1anx0an+1,A0[|(n+1)/2|]an+1A1[|(n+1)/2|]anx-1an+1x0an}=A0[|(n+1)/2|]an+1A1[|(n+1)/2|]anx-1x0,
which implies
(1.15)A0[(n+1)/2]an+1+[n/2]anA1[(n+1)/2]an+[n/2]an+1x-1x0>1∀n>n0.
But this is a contradiction which completes the proof.We omit the proof of the following lemma, since it can easily be obtained similarly.Lemma 1.8.
Equation (1.6) has no right semicycle with an infinite terms for the negative initial conditions and A0,A1>1.


## 2. Main Results

SinceAn is a two periodic, it has the form (A0,A1,A0,A1,…). If A0=A1=0, then (1.6) becomes xn+1=1/xn, from which it follows that every well-defined solution is periodic with period two. Hence, in the sequel, we will consider the case when at least one of A0 and A1 is not zero.


### 2.1. The Case0<A0,A1≤1

Theorem 2.1.
If0<A0,A1≤1 and at least one of the initial conditions is arbitrary positive real number, then every well-defined solution of (1.6) is eventually periodic with period two.Proof.
Firstly, assume thatx-1,x0>0. Then, we have x1=max{1/x0,A0x-1}. There are two cases to be considered.(a)
IfA0x-1x0≤1, then x1=1/x0. Hence,
(2.1)x2=max⁡{1x1,A1x0}=max⁡{x0,A1x0}=x0,x3=max⁡{1x2,A2x1}=max⁡{1x0,A0x0}=1x0,x4=max⁡{1x3,A3x2}=max⁡{x0,A1x0}=x0.
From Lemma 1.4, the result follows.(b)
IfA0x-1x0>1, then x1=A0x-1. We have
(2.2)x2=max⁡{1x1,A1x0}=max⁡{1A0x-1,A1x0}.
There are two subcases to be considered.
(b1)
IfA0A1x-1x0≤1, then x2=1/(A0x-1). Hence,
(2.3)x3=max⁡{1x2,A2x1}=max⁡{A0x-1,A02x-1}=A0x-1,x4=max⁡{1x3,A3x2}=max⁡{1A0x-1,A1A0x-1}=1A0x-1.
From Lemma 1.4, the result follows in this case.(b2)
IfA0A1x-1x0>1, then x2=A1x0. We have
(2.4)x3=max⁡{1x2,A2x1}=max⁡{1A1x0,A02x-1}.
There are two subcases to be considered.
(b21)
IfA02A1x-1x0≤1, then x3=1/(A1x0). We have
(2.5)x4=max⁡{1x3,A3x2}=max⁡{A1x0,A12x0}=A1x0,x5=max⁡{1A1x0,A4x3}=max⁡{1A1x0,A0A1x0}=1A1x0.
From Lemma 1.4, the result follows in this case.(b22)
IfA02A1x-1x0>1, then x3=A02x-1. The result follows Lemma 1.7. Secondly, assume that x0<0<x-1, then we have
(2.6)x1=max⁡{1x0,A0x-1}=A0x-1>0,x2=max⁡{1x1,A1x0}=1x1>0.
From Lemmas 1.5 and 1.6, the result follows (the case x-1<0<x0 is similar, so it will be omitted) which completes the proof.Remark 2.2.
If0<A0,A1≤1 and x-1,x0<0, then every well-defined solution of (1.6) is not periodic.


### 2.2. The CaseA0=0<A1<1 or A1=0<A0<1

Theorem 2.3.
IfA0=0<A1<1 or A1=0<A0<1, then every well-defined solution of (1.6) is eventually periodic with period two.Proof.
First assume thatA1=0<A0<1. Then, we have
(2.7)x1=max⁡{1x0,A0x-1}>0,x2=max⁡{1x1,A1x0}>0.
From Lemmas 1.5 and 1.6, the result follows. The case A0=0<A1<1 is similar, so it will be omitted.


### 2.3. The Other Cases

If at least one ofA0 and A1 greater than one, then we have the well-defined solutions of (1.6), where the positive initial conditions are not periodic. So, there are many cases in which solutions of (1.6) are not periodic. If the solutions of (1.6) are not periodic, then general solution of (1.6) can be obtained for many subcases.Theorem 2.4.
Assume that{xn}n=-1∞ is a well-defined solution of (1.6) for A0>1 and A1=0.(a)
Ifx1=1/x0 and x0,x-1>0 or x-1<0<x0, then
(2.8)xn=(x0A0[|(n-1)/2|])(-1)n.(b)
Ifx1=A0x-1 and x0,x-1>0 or x0<0<x-1, then
(2.9)xn=(1A0[|(n+1)/2|]x-1)(-1)n.Proof.
(a) It can be proved by induction. Letx1=1/x0 and x0,x-1>0. For n=1, (2.8) holds. Assume that (2.8) holds for all 1≤m≤m0. We may assume that m0∈ℕ is even (the case m0 is odd is similar, so it will be omitted). Then, by the inductive hypothesis, we have
(2.10)xm0+1=max⁡{1xm0,Am0xm0-1}=max⁡{A0(m0-2)/2x0,A0m0/2x0}=A0m0/2x0=(x0A0[|m0/2|])(-1)m0+1,
which completes the proof.
(b) Also, this case can be proved similarly.Now, we describe the behavior of solutions of (1.6) for some other cases. We omit the proof of the following theorem, since it can easily be obtained by induction.Theorem 2.5.
Assume that{xn}n=-1∞ is a well-defined solution of (1.6).(a)
IfA0=0,A1>1 and x0,x-1>0(orx-1<0<x0), then(2.11)xn=(A1[|n/2|]x0)(-1)n.(b)
IfA0,A1>1 and x0,x-1>0(orx-1<0<x0,x1=1/x0), then
(2.12)xn=(A0an+1A1an)[|n/2|]x0(-1)n.(c)
IfA0,A1>1 and x0,x-1>0,x1=A0x-1,x2=A1x0, then
(2.13)xn=(A0[|(n+1)/2|]an+1A1[|(n+1)/2|]an)[|n/2|]x-1an+1x0an.There are many different cases. The different cases can be obtained similarly.Theorem 2.6.
IfA0,A1>1 and initial conditions are negative, then every well-defined solution of (1.6) is eventually periodic with period two.Proof.
Assume thatx0,x-1<0. Then,
(2.14)x1=max⁡⁡{1x0,A0x-1}.
There are two cases to be considered.(a)
Ifx1=1/x0, then x2=x0,x3=1/x0, x4=x4=x0. Then, the result follows Lemma 1.4.(b)
Ifx1=A0x-1, then x2=max{1/(A0x-1),A1x0}. There are two subcases.
(b1)
Ifx2=1/(A0x-1), then x3=A0x-1, x4=1/(A0x-1). Then the result follows Lemma 1.4.(b2)
Ifx2=A0x-1, then there will be subcases and from Lemmas 1.4 and 1.8 which completes the proof.


## 2.1. The Case0<A0,A1≤1

Theorem 2.1.
If0<A0,A1≤1 and at least one of the initial conditions is arbitrary positive real number, then every well-defined solution of (1.6) is eventually periodic with period two.Proof.
Firstly, assume thatx-1,x0>0. Then, we have x1=max{1/x0,A0x-1}. There are two cases to be considered.(a)
IfA0x-1x0≤1, then x1=1/x0. Hence,
(2.1)x2=max⁡{1x1,A1x0}=max⁡{x0,A1x0}=x0,x3=max⁡{1x2,A2x1}=max⁡{1x0,A0x0}=1x0,x4=max⁡{1x3,A3x2}=max⁡{x0,A1x0}=x0.
From Lemma 1.4, the result follows.(b)
IfA0x-1x0>1, then x1=A0x-1. We have
(2.2)x2=max⁡{1x1,A1x0}=max⁡{1A0x-1,A1x0}.
There are two subcases to be considered.
(b1)
IfA0A1x-1x0≤1, then x2=1/(A0x-1). Hence,
(2.3)x3=max⁡{1x2,A2x1}=max⁡{A0x-1,A02x-1}=A0x-1,x4=max⁡{1x3,A3x2}=max⁡{1A0x-1,A1A0x-1}=1A0x-1.
From Lemma 1.4, the result follows in this case.(b2)
IfA0A1x-1x0>1, then x2=A1x0. We have
(2.4)x3=max⁡{1x2,A2x1}=max⁡{1A1x0,A02x-1}.
There are two subcases to be considered.
(b21)
IfA02A1x-1x0≤1, then x3=1/(A1x0). We have
(2.5)x4=max⁡{1x3,A3x2}=max⁡{A1x0,A12x0}=A1x0,x5=max⁡{1A1x0,A4x3}=max⁡{1A1x0,A0A1x0}=1A1x0.
From Lemma 1.4, the result follows in this case.(b22)
IfA02A1x-1x0>1, then x3=A02x-1. The result follows Lemma 1.7. Secondly, assume that x0<0<x-1, then we have
(2.6)x1=max⁡{1x0,A0x-1}=A0x-1>0,x2=max⁡{1x1,A1x0}=1x1>0.
From Lemmas 1.5 and 1.6, the result follows (the case x-1<0<x0 is similar, so it will be omitted) which completes the proof.Remark 2.2.
If0<A0,A1≤1 and x-1,x0<0, then every well-defined solution of (1.6) is not periodic.


## 2.2. The CaseA0=0<A1<1 or A1=0<A0<1

Theorem 2.3.
IfA0=0<A1<1 or A1=0<A0<1, then every well-defined solution of (1.6) is eventually periodic with period two.Proof.
First assume thatA1=0<A0<1. Then, we have
(2.7)x1=max⁡{1x0,A0x-1}>0,x2=max⁡{1x1,A1x0}>0.
From Lemmas 1.5 and 1.6, the result follows. The case A0=0<A1<1 is similar, so it will be omitted.


## 2.3. The Other Cases

If at least one ofA0 and A1 greater than one, then we have the well-defined solutions of (1.6), where the positive initial conditions are not periodic. So, there are many cases in which solutions of (1.6) are not periodic. If the solutions of (1.6) are not periodic, then general solution of (1.6) can be obtained for many subcases.Theorem 2.4.
Assume that{xn}n=-1∞ is a well-defined solution of (1.6) for A0>1 and A1=0.(a)
Ifx1=1/x0 and x0,x-1>0 or x-1<0<x0, then
(2.8)xn=(x0A0[|(n-1)/2|])(-1)n.(b)
Ifx1=A0x-1 and x0,x-1>0 or x0<0<x-1, then
(2.9)xn=(1A0[|(n+1)/2|]x-1)(-1)n.Proof.
(a) It can be proved by induction. Letx1=1/x0 and x0,x-1>0. For n=1, (2.8) holds. Assume that (2.8) holds for all 1≤m≤m0. We may assume that m0∈ℕ is even (the case m0 is odd is similar, so it will be omitted). Then, by the inductive hypothesis, we have
(2.10)xm0+1=max⁡{1xm0,Am0xm0-1}=max⁡{A0(m0-2)/2x0,A0m0/2x0}=A0m0/2x0=(x0A0[|m0/2|])(-1)m0+1,
which completes the proof.
(b) Also, this case can be proved similarly.Now, we describe the behavior of solutions of (1.6) for some other cases. We omit the proof of the following theorem, since it can easily be obtained by induction.Theorem 2.5.
Assume that{xn}n=-1∞ is a well-defined solution of (1.6).(a)
IfA0=0,A1>1 and x0,x-1>0(orx-1<0<x0), then(2.11)xn=(A1[|n/2|]x0)(-1)n.(b)
IfA0,A1>1 and x0,x-1>0(orx-1<0<x0,x1=1/x0), then
(2.12)xn=(A0an+1A1an)[|n/2|]x0(-1)n.(c)
IfA0,A1>1 and x0,x-1>0,x1=A0x-1,x2=A1x0, then
(2.13)xn=(A0[|(n+1)/2|]an+1A1[|(n+1)/2|]an)[|n/2|]x-1an+1x0an.There are many different cases. The different cases can be obtained similarly.Theorem 2.6.
IfA0,A1>1 and initial conditions are negative, then every well-defined solution of (1.6) is eventually periodic with period two.Proof.
Assume thatx0,x-1<0. Then,
(2.14)x1=max⁡⁡{1x0,A0x-1}.
There are two cases to be considered.(a)
Ifx1=1/x0, then x2=x0,x3=1/x0, x4=x4=x0. Then, the result follows Lemma 1.4.(b)
Ifx1=A0x-1, then x2=max{1/(A0x-1),A1x0}. There are two subcases.
(b1)
Ifx2=1/(A0x-1), then x3=A0x-1, x4=1/(A0x-1). Then the result follows Lemma 1.4.(b2)
Ifx2=A0x-1, then there will be subcases and from Lemmas 1.4 and 1.8 which completes the proof.



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*Source: 327437-2012-01-03.xml*