# On the Periodicity of a Difference Equation  with Maximum

Authors: Ali Gelisken; Cengiz Cinar; Ibrahim Yalcinkaya
Journal: Discrete Dynamics in Nature and Society
 (2008)

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## Abstract

We investigate the periodic nature of solutions of the max difference equationxn+1=max⁡{xn,A}/(xnxn−1), n=0,1,…, where A is a positive real parameter, and the initial conditions x−1=Ar−1 and x0=Ar0 such that r−1 and r0 are positive rational numbers. The results in this paper answer the Open Problem 6.2 posed by Grove and Ladas (2005).


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## Body

## 1. Introduction

Recently, there
has been a great interest in studying the periodic nature of nonlinear
difference equations. Although difference equations are relatively simple in
form, it is, unfortunately, extremely difficult to understand thoroughly the
periodic behavior of their solutions. The periodic nature of nonlinear
difference equations of the max type has been investigated by many authors.
See, for example, [1–20] (see also the references therein).In [7], the following open problems were posed.Problem 1.1 (Open Problem [7, Page 217, Open Problem 6.1]).
Assume thatA∈(0,∞) and that r1 and r2 are positive rational numbers.
Investigate the periodic nature of the solution of the
difference equation(1.1)xn+1=max⁡{xn,A}xn−1,n=0,1,…, where the initial
conditions x−1=Ar1 and x0=Ar2.Problem 1.2 (Open Problem [7, Page 217, Open Problem 6.2]).
Assume thatA∈(0,∞) and that r1 and r2 are positive rational numbers. Investigate
the periodic nature of the solution of the difference equation (1.2)xn+1=max⁡{xn,A}xnxn−1,n=0,1,…, where the initial
conditions x−1=Ar1 and x0=Ar2.Problem 1.3 (Open Problem [7, Page 217, Open Problem 6.3]).
Assume thatA∈(0,∞) and that r1 and r2 are positive rational numbers. Investigate
the periodic nature of the solution of the difference equation (1.3)xn+1=max⁡{xn2,A}xnxn−1,n=0,1,…, where the initial
conditions x−1=Ar1 and x0=Ar2.Problem 1.4 (Open Problem [7, Page 218, Open Problem 6.4]).
Assume thatA∈(0,∞) and that r1 and r2 are positive rational numbers. Investigate the
periodic nature of the solution of the difference equation (1.4)xn+1=max⁡{xn,A}xn2xn−1,n=0,1,…, where the initial conditions x−1=Ar1 and x0=Ar2.Problem 1.5 (Open Problem [7, Page 218, Open Problem 6.5]).
Assume thatA∈(0,∞),  k and l  are natural numbers, and that  r1 and r2 are positive rational numbers. Investigate the
periodic nature of the solution of the difference equation (1.5)xn+1=max⁡{xnk,A}xnlxn−1,n=0,1,…, where the initial conditions x−1=Ar1 and x0=Ar2.In [6], we solved the open problem 6.1. And, in [20] we solved the open problem 6.4. Now, in this paper we give answer to the open
problem 6.2.


## 2. The CaseA<1

We consider (1.2) where A<1.
It is clear that the change of variables(2.1)xn=Arnforn≥−1reduces (1.2) to the difference equation(2.2)rn+1=min⁡{0,1−rn}−rn−1,n=0,1,…,where the initial conditions are
positive rational numbers.In this section, we consider the behavior of the
solutions of (2.2) (or equivalently of (1.2)) where A<1.
We give the following lemmas which give us explicit solutions of (2.2) for some consecutive terms and show us the
pattern of the behavior of solutions of (2.2) (or equivalently of (1.2)). The proofs of some lemmas and theorems in
this section are similar. So, some will be proved and the proofs of the others
will be omitted.Lemma 2.1.
Suppose that{rn}n=−1∞ is a solution of (2.2). If at least one of the initial conditions of (2.2) is less than or equal to one and max⁡{r−1,r0}=r, then the following statements are true for
some positive integer N.(a)
IfrN=−r, then |rN−1|≤|rN|.(b)
Ifr≤1, then rn=rn+4 for all n≥−1.(c)
Ifr>1,rN=−r, then rN=rN+3 or rN=rN+4.(d)
Ifr>1,rN=rN+3=−r and rN−1<−1, then(2.3)rN≤rN−1,rN+2=1−rN+rN−1.(e)
Ifr>1,rN=rN+4=−r and −1<rN−1,then(2.4)rN−1≤−rN,rN+3=1+rN+rN−1.Proof.
(a) Let1<r. From (2.2),
if r0=r, we have r1=1−r0−r−1 and r2=−r0. Thus, we get r2≤r1 and r3=−r1. If r1≤−1, then we get r4=2−r−1, r5=−r0,
and r4≤−r5. If −1<r1, then we get r4=r0, r5=2−2r0−r−1,  r6=−r0,
and r6<r5.
Similarly, ifr−1=r, we get that r1=−r−1, r0<−r1,
and then r2=−r0,  r3=r−1,  r4=1+r0−r−1,  r5=−r−1, and r5<r4. If this proceeds, we have |rN−1|≤|rN| for rN=−r and 1<r.
Now, letr≤1. If r0=r, from (2.2), we have r1=−r−1 and r2=−r0. So, we obtain r2<r1.
Ifr−1=r, we have r1=−r−1. So, we obtain r0<−r1. If this proceeds, we have |rN−1|≤|rN| if rN=−r and r≤1. So, the proof of (a) is complete.
(b) Letr≤1. Then, we have 0<r−1≤1 and 0<r0≤1.
So, the proof of (b) follows directly from (2.2).
(c) Let1<r and rN=−r. From (2.2),
we get rN+1=−rN−1.
IfrN−1<−1, then rN+2=min⁡{0,1−rN+1}−rN,rN+2=1−rN+rN−1,
and rN+3=min⁡{0,1−rN+2}−rN+1,  rN+3=rN from (a).
If−1<rN−1, then rN+2=−rN,  rN+3=1+rN+rN−1,  rN+4=min⁡{0,−rN−rN−1}−(−rN),
and rN+4=rN from (a). So, the proof of (c) is complete.
(d) Letr>1 and rN=rN+3=−r for some positive integer N. Suppose that rN−1<−1. From (2.2),
we get that rN+1=−rN−1,  rN+2=1−rN+rN−1,
and rN+3=min⁡{0,rN−rN−1}−(−rN−1). From rN=rN+3, we have rN≤rN−1. So, the proof of (d) is complete.
(e) Letr>1 and rN=rN+4=−r for some positive integer N. Suppose that −1<rN−1. From (2.2),
we get that rN+1=−rN−1,  rN+2=−rN,  rN+3=1+rN+rN−1,
and rN+4=min⁡{0,−rN−rN−1}−(−rN). From rN=rN+4, we have rN−1≤−rN. So, the proof of (e) is complete.The proof of the following lemma is similar and will be omitted.Lemma 2.2.
Suppose that{rn}n=−1∞ is a solution of (2.2),
where r−1,r0>1. If r−1+r0=r, then the following statements are true for
some positive integer N.(a)
IfrN=1−r, then |rN−1|≤|rN|.(b)
IfrN=1−r, then rN=rN+3 or rN=rN+4.(c)
IfrN=rN+3=1−r and rN−1<−1, then(2.5)rN≤rN−1,rN+2=1−rN+rN−1.(d)
IfrN=rN+4=1−r and −1<rN−1, then(2.6)rN−1≤−rN,rN+3=1+rN+rN−1.Remark 2.3.
In view of Lemma2.1 for r>1, it is clear that rj=−r and rj=rj+4 for j=1 or j=2. But, it is not clear rj+4=rj+7 or rj+4=rj+8. If rj+4=rj+7, then rj+7=rj+10 or rj+7=rj+11. If rj+4=rj+8, then rj+8=rj+11 or rj+8=rj+12. So, we have rN+3=−r or rN+4=−r, if rN=−r for N>2.
In view of Lemma2.1, for r>1, from Lemma 2.1(c), we obtain
rN+2=1−rN+rN−1,rN+3=rN=−r.Also, from Lemma 2.1(d), we have(2.7)rN+6=2+rN−1,rN+7=rN+3=−r.Now, applying Lemma 2.1(d) and then we get
rN+3=1+rN+rN−1,rN+4=rN=−r.Moreover, from Lemma 2.1(c),
we have
rN+6=2+rN−1,rN+7=rN+4=−r.It shows that the last
corresponding two terms are the same in each two cases. So, we can apply Lemma 2.1(c) or Lemma 2.1(d) for getting the last two terms we need.
Furthermore, it is reality there are infinite number of integers N satisfying Lemma 2.1(b). If we determine exactly the number of integers N,
we can apply Lemmas 2.1(c) and 2.1(d), consecutively. Also, in view of Lemma 2.1,
(2.8)ifrN=−r,rN−1<−1,thenrN=rN+3,ifrN=−r,−1<rN−1,thenrN=rN+4.Remark 2.4.
In view of Lemma2.2, we can get similar results as Remark 2.3. So, we have the following:(2.9)ifrN=1−r,rN−1<−1,thenrN=rN+3,ifrN=1−r,−1<rN−1,thenrN=rN+4.Also, we can apply Lemma 2.2(c) or Lemma 2.2(d) firstly for getting the last two terms we
need.Clearly, there are infinite number of integersN satisfying Lemma 2.1 or Lemma 2.2. We give the following two lemmas about the number of integers N.Lemma 2.5.
Suppose
that{rn}n=−1∞ is a solution of (2.2) satisfying Lemma 2.1.
If max⁡{r−1,r0,1}=k/m and GCD(k,m)=1, then the following statements are true.
(a)
Ifk+m is an even integer, then the number of
integers N satisfying Lemma 2.1(c) is (k−m)/2 and the number of integers N satisfying Lemma 2.1(d) is (k+m)/2 for N<(7k+m)/2.(b)
Ifk+m is an odd integer, then the number of integers N satisfying Lemma 2.1(c) is k−m and the number of integers N satisfying Lemma 2.1(d) is k+m for N<7k+m.Proof.
(a) Assume thatk+m is an even integer. So, both of k and m are even or odd integers. From GCD(k,m)=1,k
and m are odd integers. We have max⁡{r−1,r0,1}=k/m≥1.
Letk/m=1. Then, we have k=m=1,  0<r−1≤1,
and 0<r0≤1. If r−1=r, we get r1=−r from (2.2),
then we get r1=r5=−r from −1<r0, Lemma 2.1(d) and Remark 2.3.
Similarly, If r0=r, we get r2=−r from (2.2),
then we get r2=r6=−r from −1<r1=−r−1, Lemma 2.1(d) and Remark 2.3.
So, the number of  integers N satisfying Lemma 2.1(d) is (1+1)/2=1 for N<(7.1+1)/2=4,
such that N=1 or N=2. There are not any integers N satisfying Lemma 2.1(c).
So, the claim is true for k/m=1.
Letk/m>1 and the number of integers N satisfying Lemma 2.1(c) is ((k−m)/2+1) for N<(7k+m)/2. From Remark 2.3, observe that j is the smallest integer of the integers N satisfying Lemma 2.1(b),
such that j=1 or j=2. This
assumption and Remark 2.3 allow us that Lemma 2.1(c) can be applied consecutively for
iterated ((k−m)/2+1) times such that(2.10)rj+4+3l1=rj+4+3(l1+1)=−r,rj+4+3l1−1<−1forl1=0,1,…,k−m2.
So, we have(2.11)rj+4+3((k−m)/2)−1<−1,rj+3=1+rj+rj−1,rj+6=1−rj+4+rj+3=1+rj+(1−rj)+rj−1,rj+9=1−rj+7+rj+6=1+rj+2(1−rj)+rj−1,⋮rj+4+3((k−m)/2)−1=1+rj+(k−m2)(1−rj)+rj−1.
Thus,(2.12)1+rj+(k−m2)(1−rj)+rj−1=1−km+(k−m2)(1−(−km))+rj−1<−1,rj−1<−2+km−(k−m2)(1+km)≤−km.
This shows thatrj−1<rj. But, it contradicts Lemma 2.1(a). This means Lemma 2.1(c) cannot be applied consecutively for iterated ((k−m)/2+1) times. So, the number of integers N satisfying Lemma 2.1(c) is not more than (k−m)/2 for N<(7k+m)/2.
Similarly, we assume that the number of integersN satisfying Lemma 2.1(d) is ((k+m)/2+1) for N<(7k+m)/2. So, we can apply Lemma 2.1(d) consecutively for iterated ((k+m)/2+1) times such that(2.13)rj+4l2=rj+4(l2+1)=−r,rj+4l2−1>−1forl2=0,1,…,k+m2.
So, we have(2.14)rj+4((k+m)/2)−1>−1,rj+3=1+rj+rj−1,rj+8=1+rj+4+rj+3=2(1+rj)+rj−1,⋮rj+4((k+m)/2)−1=(k+m2)(1+rj)+rj−1.
Thus, we have(2.15)(k+m2)(1+rj)+rj−1=(k+m2)(1+(−km))+rj−1>−1,rj−1>−1−(k+m2)(1−km)≥km.
This means thatrj−1>−rj. But, it contradicts Lemma 2.1(a) So, the number of integers N satisfying Lemma 2.1(c) is not more than (k+m)/2.
We assume that the number of integersN satisfying Lemma 2.1(c) is (k−m)/2−1 for N<(7k+m)/2. We have just had the number of integers N satisfying Lemma 2.1(d) is less than (k+m)/2+1. We apply Lemma 2.1(d) for iterated (k+m)/2 times, and then we get that the first integer N satisfying Lemma 2.1(c) is [j+4((k+m)/2)] such that rj+4((k+m)/2)=rj+4((k+m)/2)+3=−r.
So, we apply Lemma 2.1(c) for iterated ((k−m)/2−2) times, and then we get that the biggest integer N satisfying Lemma 2.1(b) for N<(7k+m)/2 is [j+4((k+m)/2)+3((k−m)/2−2)] such that rj+4((k+m)/2)+3((k−m)/2−2)=−r. From Lemma 2.1(b), rj+4((k+m)/2)+3((k−m)/2−2)+3=−r or rj+4((k+m)/2)+3((k−m)/2−2)+4=−r. But, rj+4((k+m)/2)+3((k−m)/2−2)+4=−r is not possible because  the number of
integers N satisfying Lemma 2.1(d) is not ((k+m)/2+1). So, it must be rj+4((k+m)/2)+3((k+m)/2−2)+3=−r and it contradicts our assumption. Thus, the
number of integers N satisfying Lemma 2.1(c) is (k−m)/2 exactly for N<(7k+m)/2.
Similarly, the number of integersN satisfying Lemma 2.1(d) is not ((k+m)/2−1) and can be showed
for N<(7k+m)/2.
So, the number of integers N satisfying Lemma 2.1(d) is (k+m)/2 exactly for N<(7k+m)/2. The proof is complete.
(b) The proof of (b) is similar and will be omitted.The proof of the following lemma is similar and will be omitted.Lemma 2.6.
Suppose that{rn}n=−1∞ is a solution of (2.2) satisfying Lemma 2.2.
If r=k/m and GCD(k,m)=1, then the following statements are true.
(a)
Ifk is an even integer, then the number of
integers N satisfying Lemma 2.2(a) is (k−2m)/2 and the number of integers N satisfying Lemma 2.2(b) is k/2 for N<(7k−6m)/2.(b)
Ifk is an odd integer, then the number of integers N satisfying Lemma 2.2(a) is k−2m and the number of integers N satisfying Lemma 2.2(b) is k for N<7k−6m.We give the following two lemmas which are generalized
from Lemmas2.1 and 2.5,
and Remark 2.3.
It allows us to more quickly calculate terms in the solution of (2.2).Lemma 2.7.
Suppose that{rn}n=−1∞ is a solution of (2.2) satisfying Lemmas 2.1 and 2.5.
If k+m is an even integer, then the following statements
are true.
(a)
IfrN=rN+4l1=−r, then(2.16)rN+4l1−1=1+rN+4(l1−1)+rN+4(l1−1)−1for l1=1,2,…,(k+m)/2.(b)
IfrN+4((k+m)/2)=rN+4((k+m)/2)+3l2=−r, then(2.17)rN+4((k+m)/2)+3l2−1=1−rN+4((k+m)/2)+3(l2−1)+rN+4((k+m)/2)+3(l2−1)−1for l2=1,2,…,(k−m)/2.Lemma 2.8.
Suppose
that{rn}n=−1∞ is a solution of (2.2) satisfying Lemma 2.1 and 2.5.
If k+m is an odd integer, then the following
statements are true.
(a)
IfrN=rN+4l1=−r, then(2.18)rN+4l1−1=1+rN+4(l1−1)+rN+4(l1−1)−1for l1=1,2,…,k+m.(b)
IfrN+4(k+m)=rN+4(k+m)+3l2=−r, then(2.19)rN+4(k+m)+3l2−1=1−rN+4(k+m)+3(l2−1)+rN+4(k+m)+3(l2−1)−1for l2=1,2,…,k−m.The following two lemmas are generalized from Lemmas2.2, 2.6, and Remark 2.4.Lemma 2.9.
Suppose
that{rn}n=−1∞ is a solution of (2.2) satisfying Lemmas 2.2 and 2.6.
If k is an even integer, then the following
statements are true.
(a)
IfrN=rN+4l1=1−r, then(2.20)rN+4l1−1=1+rN+4(l1−1)+rN+4(l1−1)−1for l1=1,2,…,k/2.(b)
IfrN+4(k/2)=rN+4(k/2)+3l2=1−r, then(2.21)rN+4(k/2)+3l2−1=1−rN+4(k/2)+3(l2−1)+rN+4(k/2)+3(l2−1)−1for l2=1,2,…,(k−2m)/2.Lemma 2.10.
Suppose
that{rn}n=−1∞ is a solution of (2.2) satisfying Lemmas 2.2 and 2.6.
If k is an odd integer, then the following statements are true.
(a)
IfrN=rN+4l1=1−r, then(2.22)rN+4l1−1=1+rN+4(l1−1)+rN+4(l1−1)−1for l1=1,2,…,k.(b)
IfrN+4k=rN+4k+3l2=1−r, then(2.23)rN+4k+3l2−1=1−rN+4k+3(l2−1)+rN+4k+3(l2−1)−1for l2=1,2,…,k−2m.Theorem 2.11.
Suppose
that{xn}n=−1∞ is a solution of (1.2) with the initial conditions x−1=Ar−1 and x0=Ar0,
such that 0<A<1,  r−1 and r0 are positive rational numbers. Let at
least one of r−1 and r0 is less than or equal to one. If max⁡{r−1,r0,1}=k/m and GCD(k,m)=1, then {xn}n=−1∞ is periodic(2.24)withprimeperiod{7k+m2,ifk+misaneveninteger,7k+m,ifk+misanoddinteger.Proof.
Let{rn}n=−1∞ is a solution of (2.2) satisfying Lemmas 2.1 and 2.5.
We assume that k+m is an odd integer. We must show
that(2.25)rn=rn+7k+m∀n≥−1.
From Lemma2.8(a),
we get that(2.26)rN=rN+4(k+m),rN+4(k+m)−1=(k+m)(1+rN)+rN−1.Then, from Lemma 2.8(b),
we get that(2.27)rN+4(k+m)=rN+4(k+m)+3(k−m),rN+4(k+m)+3(k−m)−1=(k+m)(1+rN)+rN−1+(k−m)(1−rN).So, at the end of this  process, we have rN=rN+7k+m and rN−1=rN+7k+m−1.
From rn−1=min⁡{0,1−rn}−rn+1, we get immediately rN=rN+7k+m for
all N≥−1. Also, it is easy to see that rN−1≠rN+4l1−1 and rN−1≠rN+4(k+m)+3l2−1 for l1=1,2,…,k+m and l2=1,2,…,k−m−1. It shows that 7k+m is prime period. So, the proof is complete.
The proof of the case that k+m is an even integer is similar and will be
omitted.The proof of the following theorem is similar and
follows directly from Lemmas2.9 and 2.10.Theorem 2.12.
Suppose
that{xn}n=−1∞ is a solution of (1.2) with the initial conditions x−1=Ar−1 and x0=Ar0 such that 0<A<1,  r−1 and r0 are positive rational numbers. Let 1<r−1,  r0. If r−1+r0=k/m and GCD(k,m)=1, then {xn}n=−1∞ is periodic(2.28)withprimeperiod{7k−6m2ifkisaneveninteger,7k−6mifkisanoddinteger.


## 3. The CaseA>1

We consider (1.2) where A>1.
It is clear that the change of variables(3.1)xn=Arnforn≥−1reduces (1.2) to the difference equation(3.2)rn+1=max⁡{0,1−rn}−rn−1,n=0,1,…,where the initial conditions are
positive rational numbers.In this section, we consider the behavior of the solutions
of (3.2) (or equivalently of (1.2)) where A>1. We omit the proofs of the following results
since they can easily be obtained in a way similar to the proofs of the lemmas
and theorems in the previous section.Lemma 3.1.
Suppose that{rn}n=−1∞ is a solution of (3.2). If max⁡{r−1,r0}=r, then the following statements are true for
some integer N≥−1.(a)
IfrN=r, then |rN−1|≤rN.(b)
IfrN=r, then rN=rN+3 or rN=rN+4.(c)
IfrN=rN+3=r and −1<rN−1, then(3.3)rN−1≤rN,rN+2=1−rN+rN−1.(d)
IfrN=rN+4=r and rN−1<−1, then −rN≤rN−1,rN+3=1+rN+rN−1.Lemma 3.2.
Suppose that{rn}n=−1∞ is a solution of (3.2) satisfying Lemma 3.1.
If max⁡{r−1,r0,1}=k/m and GCD(k,m)=1,then the following statements are true.
(a)
Ifk+m is an even integer, then the number of
integers N satisfying Lemma 3.1(a) is (k+m)/2 and the number of integers N satisfying Lemma 3.1(b) is (k−m)/2 for N<(7k−m)/2.(b)
Ifk+m is an odd integer, then the number of integers N satisfying Lemma 3.1(a) is k+m and the number of integers N satisfying Lemma 3.1(b) is k−m for N<7k−m.The following two lemmas are generalized from Lemmas3.1 and 3.2.Lemma 3.3.
Suppose
that{rn}n=−1∞ is a solution of (3.2) satisfying Lemmas 3.1 and 3.2.
If k+m is an even integer, then the following
statements are true.
(a)
IfrN=rN+4l1=r, then(3.4)rN+4l1−1=1+rN+4(l1−1)+rN+4(l1−1)−1for l1=1,2,…,(k−m)/2.(b)
IfrN+4((k−m)/2)=rN+4((k−m)/2)+3l2=r, then(3.5)rN+4((k−m)/2)+3l2−1=1−rN+4((k−m)/2)+3(l2−1)+rN+4((k−m)/2)+3(l2−1)−1for l2=1,2,…,(k+m)/2.Lemma 3.4.
Suppose
that{rn}n=−1∞ is a solution of (3.2) satisfying Lemmas 3.1 and 3.2.
If k+m is an odd integer, then the following
statements are true.
(a)
IfrN=rN+4l1=r, then(3.6)rN+4l1−1=1+rN+4(l1−1)+rN+4(l1−1)−1for l1=1,2,…,k−m.(b)
IfrN+4(k−m)=rN+4(k−m)+3l2=r, then(3.7)rN+4(k−m)+3l2−1=1−rN+4(k−m)+3(l2−1)+rN+4(k−m)+3(l2−1)−1for l2=1,2,…,k+m.Theorem 3.5.
Suppose that{xn}n=−1∞ is a solution of (1.2) with the initial conditions x−1=Ar−1 and x0=Ar0 such that 1<A,  r−1, and r0 are positive rational numbers. If max⁡{r−1,r0,1}=k/m and GCD(k,m)=1, then {xn}n=−1∞ is periodic(3.8)withprimeperiod{7k−m2ifk+m⁢isaneveninteger,7k−mifk+misanoddinteger.


## 4. Conclusion

In this paper, we have solved the open problem 6.2 which was proposed in [7]. We think that the method used in this work may solve the open problems 6.3 and 6.5 which were proposed in [7].



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*Source: 820629-2008-06-12.xml*