On the Periodicity of a Difference Equation with Maximum

Abstract

We investigate the periodic nature of solutions of the max difference equation 𝑥_𝑛+1=max{𝑥_𝑛,𝐴}/(𝑥_𝑛𝑥_𝑛−1), 𝑛=0,1,…, where 𝐴 is a positive real parameter, and the initial conditions 𝑥₋₁=𝐴^𝑟−1 and 𝑥₀=𝐴^𝑟0 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).

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 that 𝐴∈(0,∞) and that 𝑟₁ and 𝑟₂ are positive rational numbers.
Investigate the periodic nature of the solution of the difference equation 𝑥_𝑛+1=max𝑥_𝑛,𝐴𝑥_𝑛−1,𝑛=0,1,…, (1.1) where the initial conditions 𝑥₋₁=𝐴^𝑟1 and 𝑥₀=𝐴^𝑟2.

Problem 1.2 (Open Problem [7, Page 217, Open Problem 6.2]). Assume that 𝐴∈(0,∞) and that 𝑟₁ and 𝑟₂ are positive rational numbers. Investigate the periodic nature of the solution of the difference equation 𝑥_𝑛+1=max𝑥_𝑛,𝐴𝑥_𝑛𝑥_𝑛−1,𝑛=0,1,…, (1.2) where the initial conditions 𝑥₋₁=𝐴^𝑟1 and 𝑥₀=𝐴^𝑟2.

Problem 1.3 (Open Problem [7, Page 217, Open Problem 6.3]). Assume that 𝐴∈(0,∞) and that 𝑟₁ and 𝑟₂ are positive rational numbers. Investigate the periodic nature of the solution of the difference equation 𝑥_𝑛+1=max𝑥²_𝑛,𝐴𝑥_𝑛𝑥_𝑛−1,𝑛=0,1,…, (1.3) where the initial conditions 𝑥₋₁=𝐴^𝑟1 and 𝑥₀=𝐴^𝑟2.

Problem 1.4 (Open Problem [7, Page 218, Open Problem 6.4]). Assume that 𝐴∈(0,∞) and that 𝑟₁ and 𝑟₂ are positive rational numbers. Investigate the periodic nature of the solution of the difference equation 𝑥_𝑛+1=max𝑥_𝑛,𝐴𝑥²_𝑛𝑥_𝑛−1,𝑛=0,1,…, (1.4) where the initial conditions 𝑥₋₁=𝐴^𝑟1 and 𝑥₀=𝐴^𝑟2.

Problem 1.5 (Open Problem [7, Page 218, Open Problem 6.5]). Assume that 𝐴∈(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 𝑥_𝑛+1=max𝑥^𝑘_𝑛,𝐴𝑥^𝑙_𝑛𝑥_𝑛−1,𝑛=0,1,…, (1.5) where the initial conditions 𝑥₋₁=𝐴^𝑟1 and 𝑥₀=𝐴^𝑟2.

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 Case 𝐴<1

We consider (1.2) where 𝐴<1. It is clear that the change of variables𝑥_𝑛=𝐴^𝑟𝑛for𝑛≥−1 (2.1)reduces (1.2) to the difference equation𝑟_𝑛+1=min0,1−𝑟_𝑛−𝑟_𝑛−1,𝑛=0,1,…, (2.2)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 {𝑟_𝑛}^∞_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)If 𝑟_𝑁=−𝑟, then |𝑟_𝑁−1|≤|𝑟_𝑁|.(b)If 𝑟≤1, then 𝑟_𝑛=𝑟_𝑛+4 for all 𝑛≥−1.(c)If 𝑟>1,𝑟_𝑁=−𝑟, then 𝑟_𝑁=𝑟_𝑁+3 or 𝑟_𝑁=𝑟_𝑁+4.(d)If 𝑟>1,𝑟_𝑁=𝑟_𝑁+3=−𝑟 and 𝑟_𝑁−1<−1, then𝑟_𝑁≤𝑟_𝑁−1,𝑟_𝑁+2=1−𝑟_𝑁+𝑟_𝑁−1. (2.3)(e)If 𝑟>1,𝑟_𝑁=𝑟_𝑁+4=−𝑟 and −1<𝑟_𝑁−1,then𝑟_𝑁−1≤−𝑟_𝑁,𝑟_𝑁+3=1+𝑟_𝑁+𝑟_𝑁−1. (2.4)

Proof.(a) Let 1<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, if 𝑟₋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 |𝑟_𝑁−1|≤|𝑟_𝑁| for 𝑟_𝑁=−𝑟 and 1<𝑟.
Now, let 𝑟≤1. If 𝑟₀=𝑟, from (2.2), we have 𝑟₁=−𝑟₋₁ and 𝑟₂=−𝑟₀. So, we obtain 𝑟₂<𝑟₁.
If 𝑟₋₁=𝑟, we have 𝑟₁=−𝑟₋₁. So, we obtain 𝑟₀<−𝑟₁. If this proceeds, we have |𝑟_𝑁−1|≤|𝑟_𝑁| if 𝑟_𝑁=−𝑟 and 𝑟≤1. So, the proof of (a) is complete.
(b) Let 𝑟≤1. Then, we have 0<𝑟₋₁≤1 and 0<𝑟₀≤1. So, the proof of (b) follows directly from (2.2).
(c) Let 1<𝑟 and 𝑟_𝑁=−𝑟. From (2.2), we get 𝑟_𝑁+1=−𝑟_𝑁−1.
If 𝑟_𝑁−1<−1, then 𝑟_𝑁+2=min{0,1−𝑟_𝑁+1}−𝑟_𝑁,𝑟_𝑁+2=1−𝑟_𝑁+𝑟_𝑁−1, and 𝑟_𝑁+3=min{0,1−𝑟_𝑁+2}−𝑟_𝑁+1, 𝑟_𝑁+3=𝑟_𝑁 from (a).
If −1<𝑟_𝑁−1, then 𝑟_𝑁+2=−𝑟_𝑁, 𝑟_𝑁+3=1+𝑟_𝑁+𝑟_𝑁−1, 𝑟_𝑁+4=min{0,−𝑟_𝑁−𝑟_𝑁−1}−(−𝑟_𝑁), and 𝑟_𝑁+4=𝑟_𝑁 from (a). So, the proof of (c) is complete.
(d) Let 𝑟>1 and 𝑟_𝑁=𝑟_𝑁+3=−𝑟 for some positive integer 𝑁. Suppose that 𝑟_𝑁−1<−1. From (2.2), we get that 𝑟_𝑁+1=−𝑟_𝑁−1, 𝑟_𝑁+2=1−𝑟_𝑁+𝑟_𝑁−1, and 𝑟_𝑁+3=min{0,𝑟_𝑁−𝑟_𝑁−1}−(−𝑟_𝑁−1). From 𝑟_𝑁=𝑟_𝑁+3, we have 𝑟_𝑁≤𝑟_𝑁−1. So, the proof of (d) is complete.
(e) Let 𝑟>1 and 𝑟_𝑁=𝑟_𝑁+4=−𝑟 for some positive integer 𝑁. Suppose that −1<𝑟_𝑁−1. From (2.2), we get that 𝑟_𝑁+1=−𝑟_𝑁−1, 𝑟_𝑁+2=−𝑟_𝑁, 𝑟_𝑁+3=1+𝑟_𝑁+𝑟_𝑁−1, and 𝑟_𝑁+4=min{0,−𝑟_𝑁−𝑟_𝑁−1}−(−𝑟_𝑁). From 𝑟_𝑁=𝑟_𝑁+4, we have 𝑟_𝑁−1≤−𝑟_𝑁. So, the proof of (e) is complete.

The proof of the following lemma is similar and will be omitted.

Lemma 2.2. Suppose that {𝑟_𝑛}^∞_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)If 𝑟_𝑁=1−𝑟, then |𝑟_𝑁−1|≤|𝑟_𝑁|.(b)If 𝑟_𝑁=1−𝑟, then 𝑟_𝑁=𝑟_𝑁+3 or 𝑟_𝑁=𝑟_𝑁+4.(c)If 𝑟_𝑁=𝑟_𝑁+3=1−𝑟 and 𝑟_𝑁−1<−1, then𝑟_𝑁≤𝑟_𝑁−1,𝑟_𝑁+2=1−𝑟_𝑁+𝑟_𝑁−1. (2.5)(d)If 𝑟_𝑁=𝑟_𝑁+4=1−𝑟 and −1<𝑟_𝑁−1, then𝑟_𝑁−1≤−𝑟_𝑁,𝑟_𝑁+3=1+𝑟_𝑁+𝑟_𝑁−1. (2.6)

Remark 2.3. In view of Lemma 2.1 for 𝑟>1, it is clear that 𝑟_𝑗=−𝑟 and 𝑟_𝑗=𝑟_𝑗+4 for 𝑗=1 or 𝑗=2. But, it is not clear 𝑟_𝑗+4=𝑟_𝑗+7 or 𝑟_𝑗+4=𝑟_𝑗+8. If 𝑟_𝑗+4=𝑟_𝑗+7, then 𝑟_𝑗+7=𝑟_𝑗+10 or 𝑟_𝑗+7=𝑟_𝑗+11. If 𝑟_𝑗+4=𝑟_𝑗+8, then 𝑟_𝑗+8=𝑟_𝑗+11 or 𝑟_𝑗+8=𝑟_𝑗+12. So, we have 𝑟_𝑁+3=−𝑟 or 𝑟_𝑁+4=−𝑟, if 𝑟_𝑁=−𝑟 for 𝑁>2.
In view of Lemma 2.1, for 𝑟>1, from Lemma 2.1(c), we obtain 𝑟_𝑁+2=1−𝑟_𝑁+𝑟_𝑁−1,𝑟_𝑁+3=𝑟_𝑁=−𝑟. (1)Also, from Lemma 2.1(d), we have𝑟_𝑁+6=2+𝑟_𝑁−1,𝑟_𝑁+7=𝑟_𝑁+3=−𝑟. (2.7)Now, applying Lemma 2.1(d) and then we get
𝑟_𝑁+3=1+𝑟_𝑁+𝑟_𝑁−1,𝑟_𝑁+4=𝑟_𝑁=−𝑟. (2)Moreover, from Lemma 2.1(c), we have
𝑟_𝑁+6=2+𝑟_𝑁−1,𝑟_𝑁+7=𝑟_𝑁+4=−𝑟. (3)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, if𝑟_𝑁=−𝑟,𝑟_𝑁−1<−1,then𝑟_𝑁=𝑟_𝑁+3,if𝑟_𝑁=−𝑟,−1<𝑟_𝑁−1,then𝑟_𝑁=𝑟_𝑁+4. (2.8)

Remark 2.4. In view of Lemma 2.2, we can get similar results as Remark 2.3. So, we have the following:if𝑟_𝑁=1−𝑟,𝑟_𝑁−1<−1,then𝑟_𝑁=𝑟_𝑁+3,if𝑟_𝑁=1−𝑟,−1<𝑟_𝑁−1,then𝑟_𝑁=𝑟_𝑁+4. (2.9)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 integers 𝑁 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 {𝑟_𝑛}^∞_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)If 𝑘+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)If 𝑘+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 that 𝑘+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.
Let 𝑘/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.
Let 𝑘/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 that𝑟_{𝑗+4+3𝑙1}=𝑟_{𝑗+4+3(𝑙1+1)}=−𝑟,𝑟_{𝑗+4+3𝑙1−1}<−1for𝑙₁=0,1,…,𝑘−𝑚2. (2.10)
So, we have𝑟_{𝑗+4+3((𝑘−𝑚)/2)−1}𝑟<−1,_𝑗+3=1+𝑟_𝑗+𝑟_𝑗−1,𝑟_𝑗+6=1−𝑟_𝑗+4+𝑟_𝑗+3=1+𝑟_𝑗+1−𝑟_𝑗+𝑟_𝑗−1,𝑟_𝑗+9=1−𝑟_𝑗+7+𝑟_𝑗+6=1+𝑟_𝑗+21−𝑟_𝑗+𝑟_𝑗−1,⋮𝑟_{𝑗+4+3((𝑘−𝑚)/2)−1}=1+𝑟_𝑗+𝑘−𝑚21−𝑟_𝑗+𝑟_𝑗−1. (2.11)
Thus,1+𝑟_𝑗+𝑘−𝑚21−𝑟_𝑗+𝑟_𝑗−1𝑘=1−𝑚+𝑘−𝑚2−𝑘1−𝑚+𝑟_𝑗−1𝑟<−1,_𝑗−1𝑘<−2+𝑚−𝑘−𝑚2𝑘1+𝑚𝑘≤−𝑚. (2.12)
This shows that 𝑟_𝑗−1<𝑟_𝑗. 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 integers 𝑁 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 that𝑟_𝑗+4𝑙2=𝑟_{𝑗+4(𝑙2+1)}=−𝑟,𝑟_{𝑗+4𝑙2−1}>−1for𝑙₂=0,1,…,𝑘+𝑚2. (2.13)
So, we have𝑟_{𝑗+4((𝑘+𝑚)/2)−1}𝑟>−1,_𝑗+3=1+𝑟_𝑗+𝑟_𝑗−1,𝑟_𝑗+8=1+𝑟_𝑗+4+𝑟_𝑗+3=21+𝑟_𝑗+𝑟_𝑗−1,⋮𝑟_{𝑗+4((𝑘+𝑚)/2)−1}=𝑘+𝑚21+𝑟_𝑗+𝑟_𝑗−1. (2.14)
Thus, we have𝑘+𝑚21+𝑟_𝑗+𝑟_𝑗−1=𝑘+𝑚2−𝑘1+𝑚+𝑟_𝑗−1𝑟>−1,_𝑗−1>−1−𝑘+𝑚2𝑘1−𝑚≥𝑘𝑚. (2.15)
This means that 𝑟_𝑗−1>−𝑟_𝑗. 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 integers 𝑁 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 𝑟_{𝑗+4((𝑘+𝑚)/2)}=𝑟_{𝑗+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 𝑟_{𝑗+4((𝑘+𝑚)/2)+3((𝑘−𝑚)/2−2)}=−𝑟. From Lemma 2.1(b), 𝑟_{𝑗+4((𝑘+𝑚)/2)+3((𝑘−𝑚)/2−2)+3}=−𝑟 or 𝑟_{𝑗+4((𝑘+𝑚)/2)+3((𝑘−𝑚)/2−2)+4}=−𝑟. But, 𝑟_{𝑗+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 𝑟_{𝑗+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 integers 𝑁 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 {𝑟_𝑛}^∞_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)If 𝑘 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)If 𝑘 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 Lemmas 2.1 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 {𝑟_𝑛}^∞_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)If 𝑟_𝑁=𝑟_𝑁+4𝑙1=−𝑟, then𝑟_{𝑁+4𝑙1−1}=1+𝑟_{𝑁+4(𝑙1−1)}+𝑟_{𝑁+4(𝑙1−1)−1} (2.16)for 𝑙₁=1,2,…,(𝑘+𝑚)/2.(b)If 𝑟_{𝑁+4((𝑘+𝑚)/2)}=𝑟_{𝑁+4((𝑘+𝑚)/2)+3𝑙2}=−𝑟, then𝑟_{𝑁+4((𝑘+𝑚)/2)+3𝑙2−1}=1−𝑟_{𝑁+4((𝑘+𝑚)/2)+3(𝑙2−1)}+𝑟_{𝑁+4((𝑘+𝑚)/2)+3(𝑙2−1)−1} (2.17)for 𝑙₂=1,2,…,(𝑘−𝑚)/2.

Lemma 2.8. Suppose that {𝑟_𝑛}^∞_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)If 𝑟_𝑁=𝑟_𝑁+4𝑙1=−𝑟, then𝑟_{𝑁+4𝑙1−1}=1+𝑟_{𝑁+4(𝑙1−1)}+𝑟_{𝑁+4(𝑙1−1)−1} (2.18)for 𝑙₁=1,2,…,𝑘+𝑚.(b)If 𝑟_{𝑁+4(𝑘+𝑚)}=𝑟_{𝑁+4(𝑘+𝑚)+3𝑙2}=−𝑟, then𝑟_{𝑁+4(𝑘+𝑚)+3𝑙2−1}=1−𝑟_{𝑁+4(𝑘+𝑚)+3(𝑙2−1)}+𝑟_{𝑁+4(𝑘+𝑚)+3(𝑙2−1)−1} (2.19)for 𝑙₂=1,2,…,𝑘−𝑚.

The following two lemmas are generalized from Lemmas 2.2, 2.6, and Remark 2.4.

Lemma 2.9. Suppose that {𝑟_𝑛}^∞_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)If 𝑟_𝑁=𝑟_𝑁+4𝑙1=1−𝑟, then𝑟_{𝑁+4𝑙1−1}=1+𝑟_{𝑁+4(𝑙1−1)}+𝑟_{𝑁+4(𝑙1−1)−1} (2.20)for 𝑙₁=1,2,…,𝑘/2.(b)If 𝑟_{𝑁+4(𝑘/2)}=𝑟_{𝑁+4(𝑘/2)+3𝑙2}=1−𝑟, then𝑟_{𝑁+4(𝑘/2)+3𝑙2−1}=1−𝑟_{𝑁+4(𝑘/2)+3(𝑙2−1)}+𝑟_{𝑁+4(𝑘/2)+3(𝑙2−1)−1} (2.21)for 𝑙₂=1,2,…,(𝑘−2𝑚)/2.

Lemma 2.10. Suppose that {𝑟_𝑛}^∞_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)If 𝑟_𝑁=𝑟_𝑁+4𝑙1=1−𝑟, then𝑟_{𝑁+4𝑙1−1}=1+𝑟_{𝑁+4(𝑙1−1)}+𝑟_{𝑁+4(𝑙1−1)−1} (2.22)for 𝑙₁=1,2,…,𝑘.(b)If 𝑟_𝑁+4𝑘=𝑟_{𝑁+4𝑘+3𝑙2}=1−𝑟, then𝑟_{𝑁+4𝑘+3𝑙2−1}=1−𝑟_{𝑁+4𝑘+3(𝑙2−1)}+𝑟_{𝑁+4𝑘+3(𝑙2−1)−1} (2.23)for 𝑙₂=1,2,…,𝑘−2𝑚.

Theorem 2.11. Suppose that {𝑥_𝑛}^∞_𝑛=−1 is a solution of (1.2) with the initial conditions 𝑥₋₁=𝐴^𝑟−1 and 𝑥₀=𝐴^𝑟0, 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 {𝑥_𝑛}^∞_𝑛=−1 is periodicwithprimeperiod7𝑘+𝑚2,if𝑘+𝑚isaneveninteger,7𝑘+𝑚,if𝑘+𝑚isanoddinteger. (2.24)

Proof.Let {𝑟_𝑛}^∞_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 that𝑟_𝑛=𝑟_{𝑛+7𝑘+𝑚}∀𝑛≥−1. (2.25)
From Lemma 2.8(a), we get that𝑟_𝑁=𝑟_{𝑁+4(𝑘+𝑚)},𝑟_{𝑁+4(𝑘+𝑚)−1}=(𝑘+𝑚)(1+𝑟_𝑁)+𝑟_𝑁−1. (2.26)Then, from Lemma 2.8(b), we get that𝑟_{𝑁+4(𝑘+𝑚)}=𝑟_{𝑁+4(𝑘+𝑚)+3(𝑘−𝑚)},𝑟_{𝑁+4(𝑘+𝑚)+3(𝑘−𝑚)−1}=(𝑘+𝑚)(1+𝑟_𝑁)+𝑟_𝑁−1+(𝑘−𝑚)(1−𝑟_𝑁). (2.27)So, at the end of this process, we have 𝑟_𝑁=𝑟_{𝑁+7𝑘+𝑚} and 𝑟_𝑁−1=𝑟_{𝑁+7𝑘+𝑚−1}. From 𝑟_𝑛−1=min{0,1−𝑟_𝑛}−𝑟_𝑛+1, we get immediately 𝑟_𝑁=𝑟_{𝑁+7𝑘+𝑚} for all 𝑁≥−1. Also, it is easy to see that 𝑟_𝑁−1≠𝑟_{𝑁+4𝑙1−1} and 𝑟_𝑁−1≠𝑟_{𝑁+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 Lemmas 2.9 and 2.10.

Theorem 2.12. Suppose that {𝑥_𝑛}^∞_𝑛=−1 is a solution of (1.2) with the initial conditions 𝑥₋₁=𝐴^𝑟−1 and 𝑥₀=𝐴^𝑟0 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 {𝑥_𝑛}^∞_𝑛=−1 is periodicwithprimeperiod7𝑘−6𝑚2if𝑘isaneveninteger,7𝑘−6𝑚if𝑘isanoddinteger. (2.28)

3. The Case 𝐴>1

We consider (1.2) where 𝐴>1. It is clear that the change of variables𝑥_𝑛=𝐴^𝑟𝑛for𝑛≥−1 (3.1)reduces (1.2) to the difference equation𝑟_𝑛+1=max0,1−𝑟_𝑛−𝑟_𝑛−1,𝑛=0,1,…, (3.2)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 {𝑟_𝑛}^∞_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)If 𝑟_𝑁=𝑟, then |𝑟_𝑁−1|≤𝑟_𝑁.(b)If 𝑟_𝑁=𝑟, then 𝑟_𝑁=𝑟_𝑁+3 or 𝑟_𝑁=𝑟_𝑁+4.(c)If 𝑟_𝑁=𝑟_𝑁+3=𝑟 and −1<𝑟_𝑁−1, then𝑟_𝑁−1≤𝑟_𝑁,𝑟_𝑁+2=1−𝑟_𝑁+𝑟_𝑁−1. (3.3)(d)If 𝑟_𝑁=𝑟_𝑁+4=𝑟 and 𝑟_𝑁−1<−1, then −𝑟_𝑁≤𝑟_𝑁−1,𝑟_𝑁+3=1+𝑟_𝑁+𝑟_𝑁−1. (4)

Lemma 3.2. Suppose that {𝑟_𝑛}^∞_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)If 𝑘+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)If 𝑘+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 Lemmas 3.1 and 3.2.

Lemma 3.3. Suppose that {𝑟_𝑛}^∞_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)If 𝑟_𝑁=𝑟_𝑁+4𝑙1=𝑟, then𝑟_{𝑁+4𝑙1−1}=1+𝑟_{𝑁+4(𝑙1−1)}+𝑟_{𝑁+4(𝑙1−1)−1} (3.4)for 𝑙₁=1,2,…,(𝑘−𝑚)/2.(b)If 𝑟_{𝑁+4((𝑘−𝑚)/2)}=𝑟_{𝑁+4((𝑘−𝑚)/2)+3𝑙2}=𝑟, then𝑟_{𝑁+4((𝑘−𝑚)/2)+3𝑙2−1}=1−𝑟_{𝑁+4((𝑘−𝑚)/2)+3(𝑙2−1)}+𝑟_{𝑁+4((𝑘−𝑚)/2)+3(𝑙2−1)−1} (3.5)for 𝑙₂=1,2,…,(𝑘+𝑚)/2.

Lemma 3.4. Suppose that {𝑟_𝑛}^∞_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)If 𝑟_𝑁=𝑟_𝑁+4𝑙1=𝑟, then𝑟_{𝑁+4𝑙1−1}=1+𝑟_{𝑁+4(𝑙1−1)}+𝑟_{𝑁+4(𝑙1−1)−1} (3.6)for 𝑙₁=1,2,…,𝑘−𝑚.(b)If 𝑟_{𝑁+4(𝑘−𝑚)}=𝑟_{𝑁+4(𝑘−𝑚)+3𝑙2}=𝑟, then𝑟_{𝑁+4(𝑘−𝑚)+3𝑙2−1}=1−𝑟_{𝑁+4(𝑘−𝑚)+3(𝑙2−1)}+𝑟_{𝑁+4(𝑘−𝑚)+3(𝑙2−1)−1} (3.7)for 𝑙₂=1,2,…,𝑘+𝑚.

Theorem 3.5. Suppose that {𝑥_𝑛}^∞_𝑛=−1 is a solution of (1.2) with the initial conditions 𝑥₋₁=𝐴^𝑟−1 and 𝑥₀=𝐴^𝑟0 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 {𝑥_𝑛}^∞_𝑛=−1 is periodicwithprimeperiod7𝑘−𝑚2if𝑘+𝑚isaneveninteger,7𝑘−𝑚if𝑘+𝑚isanoddinteger. (3.8)

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].

Acknowledgment

The authors are grateful to the anonymous referees for their valuable suggestions that improved the quality of this study.