Periodicity in a Class of Systems of Delay Difference Equations

Abstract

We study a system of delay difference equations modeling four-dimensional discrete-time delayed neural networks with no internal decay. Such a discrete-time system can be regarded as the discrete analog of a differential equation with piecewise constant argument. By using semicycle analysis method, it is shown that every bounded solution of this discrete-time system is eventually periodic. The obtained results are new, and they complement previously known results.

1. Introduction

Over the past decades, there has been increasing interest in the potential applications of the dynamics of artificial neural networks in signal and image processing. Among the most popular models in the literature of artificial neural networks is the following well-known Hopfield's model [1, 2]: d𝑢_𝑖(𝑡)d𝑡=𝜇_𝑖𝑢_𝑖(𝑡)+_𝑗≠𝑖𝑇_𝑖𝑗𝑓_𝑗𝑢_𝑗𝑡−𝜏_𝑖𝑗,1≤𝑖≤𝑛, (1.1) where 𝜇_𝑖≥0 is the internal decay rate, 𝜏_𝑖𝑗≥0 is the delay incorporated by Marcus and Westervelt [3] to account for the finite switching speed of amplifiers (neurons), 𝑓_𝑗∶𝐑→𝐑 (the set of all real numbers) is the signal functions, and 𝑇_𝑖𝑗 represents the connection strengths between neurons and if the output from neuron 𝑗 excites (resp., inhibits) neuron 𝑖, then 𝑇_𝑖𝑗>0 (resp., <0), 1≤𝑖,𝑗≤𝑛.

Much work on Hopfield-type neural networks with constant time delays has been carried out. However, from the point of view of electronic implementation of neural networks and control, the time-varying delay case is more suitable for practical neural networks, and we can achieve this by using piecewise constant arguments. Because of the wide application of differential equations with piecewise constant argument in certain biomedical models (see, e.g., [4]), much progress has been made in the study of differential equations with piecewise constant arguments since the pioneering work of Cooke and Wiener [5] and Shah and Wiener [6]. For more details and references on this subject, interested readers may refer to a survey of Cooke and Wiener [7]. Motivated by these discussions, we propose d𝑢_𝑖(𝑡)d𝑡=𝜇_𝑖𝑢_𝑖(𝑡)+_𝑗≠𝑖𝑇_𝑖𝑗𝑓_𝑗𝑢_𝑗𝑡−𝜏_𝑖𝑗,1≤𝑖≤𝑛, (1.2) as a new neural network model, where [⋅] denotes the greatest integer function and 𝑓_𝑗 (1≤𝑗≤𝑛) are the McCulloch-Pitts nonlinear function given by 𝑓(𝑢)=1ifu>𝜎,−1ifu≤𝜎. (1.3) Here, 𝜎∈𝑅 is referred as the threshold. The McCulloch-Pitts nonlinearity reflects the fact that the signal transmission is of digital nature: a neuron is either fully active or completely inactive.

Neural networks are complex and large-scale nonlinear dynamical systems, while the dynamics of the delayed neural networks is even richer and more complicated [8]. In order to obtain a deep and clear understanding of the dynamics of neural networks, the dynamics of delayed neural networks consisting of a few neurons has received increasing attention over the past years [9–14]. The studies on these neural networks of a few neurons are potentially useful because the complexities found in such models often provide promising information for the studies on more complicated neural networks with a large number of neurons [15]. So, we are inspired to consider a special case of (1.2), 𝑥^′₁(𝑡)=𝑓𝑥₂,𝑥([𝑡−𝑘])_′2(𝑡)=𝑓𝑥₃,𝑥([𝑡−𝑘])_′3(𝑡)=−𝑓𝑥₄,𝑥([𝑡−𝑘])_′4(𝑡)=−𝑓𝑥₁([𝑡−𝑘]), (1.4) where 𝑓 is the activation function with McCulloch-Pitts nonlinearity (1.3), 𝑘 is a nonnegative integer and the negative (resp., positive) sign on the right indicates negative (resp., positive) feedback. (1.4) describes the dynamics of four interacting neurons with no internal decays.

To study (1.4), it is convenient to consider the following difference system: 𝑥₁(𝑛+1)=𝑥₁(𝑛)+𝑓𝑥₂(),x𝑛−𝑘₂(𝑛+1)=𝑥₂(𝑛)+𝑓𝑥₃(),𝑥𝑛−𝑘₃(𝑛+1)=𝑥₃(𝑛)−𝑓𝑥₄(),𝑥𝑛−𝑘₄(𝑛+1)=𝑥₄(𝑛)−𝑓𝑥₁(𝑛−𝑘), (1.5) where 𝑘 is a nonnegative integer and 𝑓 satisfies (1.3). In fact, integrate (1.4) from 𝑛 to 𝑡∈[𝑛,𝑛+1) to get 𝑥₁(𝑡)=𝑥₁(𝑛)+^𝑡_𝑛𝑓𝑥₂([𝑠−𝑘])𝑑𝑠=𝑥₁(𝑛)+𝑓𝑥₂()),𝑥𝑛−𝑘(𝑡−𝑛₂(𝑡)=𝑥₂(𝑛)+^𝑡_𝑛𝑓𝑥₃([𝑠−𝑘])𝑑𝑠=𝑥₂(𝑛)+𝑓𝑥₃()(),𝑥𝑛−𝑘𝑡−𝑛₃(𝑡)=𝑥₃(𝑛)−^𝑡_𝑛𝑓𝑥₄([𝑠−𝑘])𝑑𝑠=𝑥₃(𝑛)−𝑓𝑥₄()(),𝑥𝑛−𝑘𝑡−𝑛₄(𝑡)=𝑥₄(𝑛)−^𝑡_𝑛𝑓𝑥₁([𝑠−𝑘])𝑑𝑠=𝑥₄(𝑛)−𝑓𝑥₁(𝑛−𝑘)(𝑡−𝑛). (1.6) Letting 𝑡→𝑛+1, we get a special case of (1.5) with 𝑘=0. Hence, system (1.5) can be regarded as a more general neural network model of four neurons with transmission delay.

Our study is inspired by the work of Chen [16–18], who considered 𝑥()(𝑛)(𝑥(,𝑥()(𝑛)(𝑥(,𝑥()(𝑛)𝑦(),𝑦()(𝑛)(𝑥(,𝑛+1=𝑥−𝑓𝑛−𝑘))𝑛+1=𝑥+𝑓𝑛−𝑘))𝑛+1=𝑥+𝑓𝑛−𝑘𝑛+1=𝑦−𝑓𝑛−𝑘)) (1.7) where 𝑓∶𝐑→𝐑 is given by (1.3).

The main objective of this paper is to study (1.5) with 𝑓 satisfying (1.3). The discontinuity of 𝑓 makes it difficult to apply directly the theory of discrete dynamical systems to system (1.5). However, by using semicycle analysis method, we obtain some interesting results for the periodicity of (1.5). It is shown that every bounded solution of system (1.5) is eventually periodic.

The rest of the paper is organized as follows. In Section 2, we introduce some necessary notations and lemmas that will be used later. In Section 3, we state and prove our main result.

2. Notations and Preliminaries

For the sake of simplicity, we introduce some notations. Let 𝑍 denote the set of all integers. For 𝑎, 𝑏∈𝑍, define 𝑍(𝑎)={𝑎,𝑎+1,…} and 𝑍(𝑎,𝑏)={𝑎,𝑎+1,…,𝑏} whenever 𝑎≤𝑏.

By a solution of (1.5), we mean a sequence {(𝑥₁(𝑛),𝑥₂(𝑛),𝑥₃(𝑛),𝑥₄(𝑛))}^+∞_𝑛=−𝑘 that satisfies (1.5). Clearly, for given (𝑥₁(−𝑘),𝑥₂(−𝑘),𝑥₃(−𝑘),𝑥₄(−𝑘)),(𝑥₁(−𝑘+1),𝑥₂(−𝑘+1), 𝑥₃(−𝑘+1),𝑥₄(−𝑘+1)),…,(𝑥₁(0),𝑥₂(0), 𝑥₃(0), 𝑥₄(0))∈𝐑⁴, system (1.5) has a unique solution {(𝑥₁(𝑛),𝑥₂(𝑛),𝑥₃(𝑛),𝑥₄(𝑛))}^+∞_𝑛=−𝑘 satisfying the initial condition: 𝑥₁(𝑖),𝑥₂(𝑖),𝑥₃(𝑖),𝑥₄(𝑖)=𝑥₁(𝑖),𝑥₂(𝑖),𝑥₃(𝑖),𝑥₄(𝑖)for𝑖∈𝑍(−𝑘,0). (2.1) If all terms of the sequence {𝑣_𝑛}^+∞_𝑛=−𝑘 are integers, we say that it is an integer sequence. If {𝑥_𝑖(𝑛)}^∞_𝑛=−𝑘 for all 𝑖∈𝑍(1,4) are integer sequences and also satisfy (1.5) and (1.3), then {(𝑥₁(𝑛),𝑥₂(𝑛),𝑥₃(𝑛),𝑥₄(𝑛))}^+∞_𝑛=−𝑘 is said to be an integer sequence solution of (1.5) and (1.3).

Let {(𝑥₁(𝑛),𝑥₂(𝑛),𝑥₃(𝑛),𝑥₄(𝑛))}^+∞_𝑛=−𝑘 be a solution of (1.5). Define 𝜏(𝑥)=𝑥−[𝑥]for𝑥≠0,1for𝑥=0. (2.2) Clearly, 𝜏(𝑥)≥0. From (1.5) and (1.3), we have 𝜏(𝑥_𝑖(𝑛)−𝜎)=𝜏(𝑥_𝑖(0)−𝜎), 𝑖∈𝑍(1,4). Denote 𝑥_𝑖(𝑛)=𝑥_𝑖(𝑛)−𝜎−𝜏(𝑥_𝑖(0)−𝜎)+1, 𝑖∈𝑍(1,4). Then 𝑥_𝑖(𝑛)∈𝑍 for all 𝑛∈𝑍(−𝑘), 𝑖∈𝑍(1,4). Moreover, if 𝑥_𝑖(𝑛)>𝜎, then 𝑥_𝑖(𝑛)=𝑥_𝑖(𝑛)−𝜎−𝜏(𝑥_𝑖(0)−𝜎)+1≥𝑥_𝑖(𝑛)−𝜎>0, and if 𝑥_𝑖(𝑛)≤𝜎, then 𝑥_𝑖(𝑛)=𝑥_𝑖(𝑛)−𝜎−𝜏(𝑥_𝑖(0)−𝜎)+1≤−1+1=0, 𝑖∈𝑍(1,4). Hence, {(𝑥₁(𝑛),…,𝑥₄(𝑛))}^+∞_𝑛=−𝑘 is an integer sequence solution of (1.5) and (1.3) with 𝜎=0.

So, without loss of generality, in the paper we always assume that 𝜎=0, and a solution of (1.5) means an integer sequence solution.

A solution {(𝑥₁(𝑛),𝑥₂(𝑛),𝑥₃(𝑛),𝑥₄(𝑛))}^+∞_𝑛=−𝑘 is said to be eventually periodic if there exist 𝑛₀∈{−𝑘,−𝑘+1,−𝑘+2,…} and 𝜔∈{1,2,3,…} such that 𝑥₁𝑛+𝑛₀+𝜔,𝑥₂𝑛+𝑛₀+𝜔,…,𝑥₄𝑛+𝑛₀+𝜔=𝑥₁𝑛+𝑛₀,𝑥₂𝑛+𝑛₀,…,𝑥₄𝑛+𝑛₀, (2.3) for 𝑛=0,1,2,…, and 𝜔 is called a period. The smallest such 𝜔 is called the minimal period of {(𝑥₁(𝑛),𝑥₂(𝑛),𝑥₃(𝑛),𝑥₄(𝑛))}^+∞_𝑛=−𝑘.

Definition 2.1. For a sequence {𝑢_𝑛}^+∞_𝑛=−𝑘, we say that {𝑢_𝑛1,𝑢_𝑛1+1,…,𝑢_{𝑛1+𝑙−1}} with 𝑙≥1 is a positive semicycle with the length 𝑙 if 𝑢_𝑛>0 for 𝑛∈𝑍(𝑛₁,𝑛₁+𝑙−1), and 𝑛₁ and 𝑙 satisfy either𝑛₁=−𝑘or𝑢_𝑛1−1≤0,either𝑙=+∞or𝑢_𝑛1+𝑙≤0. (2.4) We call 𝑙 the length of the positive semicycle. A negative semicycle is defined similarly (replace > by ≤ and vice versa).
To simplify the following arguments, for an integer sequence, we denote 𝐴⁺_𝑟 the positive semicycle {𝑣_𝑛1,…,𝑣_𝑛1+2𝑟} with 𝑣_{𝑛1+𝑟±𝑚}=𝑟−𝑚+1 for 𝑚∈𝑍(0,𝑟) and 𝐴⁻_𝑟 the negative semicycle {𝑣_𝑛1,…,𝑣_𝑛1+2𝑟} with 𝑣_{𝑛1+𝑟±𝑚}=𝑚−𝑟 for 𝑚∈𝑍(0,𝑟).

Definition 2.2. An integer sequence {𝑣_𝑛}^{𝑛1+𝐿−1}_𝑛=𝑛1 (may be infinite terms) is said to be an 𝑟-normal cycle if it possesses one of the representations as follows: {𝐴⁺_𝑟,𝐴⁻_𝑟,…,𝐴⁺_𝑟,𝐴⁻_𝑟}, {𝐴⁺_𝑟,𝐴⁻_𝑟,…,𝐴⁻_𝑟,𝐴⁺_𝑟}, {𝐴⁻_𝑟,𝐴⁺_𝑟,…,𝐴⁻_𝑟,𝐴⁺_𝑟}, and {𝐴⁻_𝑟,𝐴⁺_𝑟,…,𝐴⁺_𝑟,𝐴⁻_𝑟}. In addition, if 𝐿=+∞, then it is said to be a total 𝑟-normal cycle.

Remark 2.3. For a 𝑟-normal cycle {𝑣_𝑛}^{𝑛1+𝐿−1}_𝑛=𝑛1, from Definition 2.2, it is easy to see that there must exist 𝑖∈𝑍(0) such that 𝐿=(2𝑟+1)𝑖. If 𝐿=+∞, that is, {𝑣_𝑛}^{𝑛1+𝐿−1}_𝑛=𝑛1 is a total 𝑟-normal cycle, then 𝑖=+∞.
Throughout this section, we tacitly assume that (𝑆)={(𝑥₁(𝑛),𝑥₂(𝑛),𝑥₃(𝑛),𝑥₄(𝑛))}^+∞_𝑛=−𝑘 is a periodic solution of system (1.5) satisfying the initial values: 𝑥₁(𝑖),𝑥₂(𝑖),𝑥₃(𝑖),𝑥₄(𝑖)∈𝑍⁴for𝑖∈𝑍(−𝑘,0). (2.5) In view of (1.5) and (1.3), it is easy to see that all terms 𝑥₁(𝑛), 𝑥₂(𝑛), 𝑥₃(𝑛), and 𝑥₄(𝑛) of (𝑆) are integers, that is, (𝑆) is an integer sequence solution of (1.5).
For a periodic solution (𝑆) of (1.5), denote 𝑃⁺_𝑥𝑖=sup𝑥_𝑖(𝑛)∶𝑛∈𝑍(−𝑘)−1,𝑃⁻_𝑥𝑖=−inf𝑥_𝑖(𝑛)∶𝑛∈𝑍(−𝑘), (2.6) and let 𝐿⁺_𝑥𝑖 and 𝐿⁻_𝑥𝑖 be the length of the longest positive semicycle of {𝑥_𝑖(𝑛)}^+∞_𝑛=−𝑘 and the length of the longest negative semicycle of {𝑥_𝑖(𝑛)}^+∞_𝑛=−𝑘, respectively, where 𝑖=1,2,…,4. Note that 𝐿⁺_𝑥𝑖,𝐿⁻_𝑥𝑖∈𝑍(1) for 𝑖=1,2,…,4.
The following several lemmas will be useful to the proof of the main result in this paper.

Lemma 2.4. (a)Either 𝐿⁺_𝑥𝑖>𝐿⁺_𝑥𝑗 or 𝐿⁻_𝑥𝑖>𝐿⁺_𝑥𝑗 or min{𝐿⁺_𝑥𝑖,𝐿⁻_𝑥𝑖}≥𝐿⁺_𝑥𝑗 for 𝑖,𝑗=1,2,…,4 and 𝑖≠𝑗.
(b)Either 𝐿⁺_𝑥𝑖>𝐿⁻_𝑥𝑗 or 𝐿⁻_𝑥𝑖>𝐿⁻_𝑥𝑗 or min{𝐿⁺_𝑥𝑖,𝐿⁻_𝑥𝑖}≥𝐿⁻_𝑥𝑗 for 𝑖,𝑗=1,2,…,4 and 𝑖≠𝑗.

Proof. We only prove the conclusion: either 𝐿⁺_𝑥1>𝐿⁺_𝑥2 or 𝐿⁻_𝑥1>𝐿⁺_𝑥2 or min{𝐿⁺_𝑥1,𝐿⁻_𝑥1}≥𝐿⁺_𝑥2. The other conclusions can be proved similarly. Now, we distinguish two cases to finish the proof.
Case 1. 𝐿⁺_𝑥2=2𝑟 for some 𝑟>0. Without loss of generality, we assume that 𝑥₂(0)≤0,𝑥₂(1)>0,𝑥₂(2)>0,…,𝑥₂(2𝑟)>0,𝑥₂(2𝑟+1)≤0,…. Then, from (1.5) and (1.3), we have 𝑥₁(2𝑟+𝑘+1)=𝑥₁(𝑟+𝑘+1)+^2𝑟_𝑖=𝑟+1𝑓𝑥₂(𝑖)=𝑥₁()𝑥𝑟+𝑘+1+𝑟,₁(𝑟+𝑘+1)=𝑥₁(𝑘+1)+^𝑟_𝑖=1𝑓𝑥₂(𝑖)=𝑥₁(𝑘+1)+𝑟. (2.7)(2.8)
(i)If 𝑥₁(𝑟+𝑘+1)≥1, it follows from (2.7) that 𝑥₁(2𝑟+𝑘+1)≥𝑟+1. Since the distance between consecutive elements of {𝑥₁(𝑛)}^∞_𝑛=0 is 1, we have 𝐿⁺_𝑥1≥2𝑟+1>𝐿⁺_𝑥2.
(ii)If 𝑥₁(𝑟+𝑘+1)≤0, it follows from (2.8) that 𝑥₁(𝑘+1)≤−𝑟. Thus, 𝐿⁻_𝑥1≥2𝑟+1>𝐿⁺_𝑥2.
Case 2. 𝐿⁺_𝑥2=2𝑟+1 for some 𝑟>0. Without loss of generality, we assume that 𝑥₂(0)≤0,𝑥₂(1)>0,𝑥₂(2)>0,…,𝑥₂(2𝑟+1)>0, and 𝑥₂(2𝑟+2)≤0,…. Then, from (1.5) and (1.3), we have 𝑥₁(𝑘+2𝑟+2)=𝑥₁(𝑟+𝑘+1)+^2𝑟+1_𝑖=1+𝑟𝑓𝑥₂(𝑖)=x₁()+(),𝑥𝑟+𝑘+1𝑟+1₁(𝑟+𝑘+1)=𝑥₁(𝑘+1)+^𝑟_𝑖=1𝑓𝑥₂(𝑖)=𝑥₁(𝑘+1)+𝑟. (2.9)(2.10)(i)If 𝑥₁(𝑟+𝑘+1)>0, it follows from (2.9) that 𝑥₁(𝑘+2𝑟+2)>𝑟+1. Thus, 𝐿⁺_𝑥1>2𝑟+1=𝐿⁺_𝑥2.(ii)If 𝑥₁(𝑟+𝑘+1)<0, it follows from (2.10) that 𝑥₁(𝑘+1)<−𝑟. Thus, 𝐿⁻_𝑥1>2𝑟+1=𝐿⁺_𝑥2.(iii)If 𝑥₁(𝑟+𝑘+1)=0, it follows from (2.9) and (2.10) that 𝑥₁(𝑘+2𝑟+2)=𝑟+1 and 𝑥₁(𝑘+1)=−𝑟. Thus, min{𝐿⁺_𝑥1,𝐿⁻_𝑥1}≥2𝑟+1=𝐿⁺_𝑥2.
Therefore, we have either 𝐿⁺_𝑥1>𝐿⁺_𝑥2 or 𝐿⁻_𝑥1>𝐿⁺_𝑥2 or min{𝐿⁺_𝑥1,𝐿⁻_𝑥1}≥𝐿⁺_𝑥2. This completes the proof.

Lemma 2.5. 𝐿⁺_𝑥𝑖=𝐿⁻_𝑥𝑗 for 𝑖,𝑗=1,…,4. Moreover, 𝐿⁺_𝑥1 is odd.

Proof. The lemma follows immediately from Lemma 2.4 together with its proof.

Lemma 2.6. 𝑃⁺_𝑥𝑖=𝑃⁻_𝑥𝑗 for 𝑖,𝑗=1,…,4.

Proof. First, we claim that 𝑃⁺_𝑥2≤𝑃⁺_𝑥1. Otherwise, 𝑃⁺_𝑥2>𝑃⁺_𝑥1. Since the distance between consecutive elements of {𝑥₂(𝑛)}^∞_𝑛=0 is 1, it follows easily from the definitions of 𝑃⁺_𝑥2 and 𝐿⁺_𝑥2 that 𝐿⁺_𝑥2≥2𝑃⁺_𝑥2+1. By Lemma 2.5, there exists 𝑟≥0 such that 𝐿⁺_𝑥2=2𝑟+1≥2𝑃⁺_𝑥2+1, which implies that 𝑟≥𝑃⁺_𝑥2. Without loss of generality, we assume that 𝑥₂(0)≤0,𝑥₂(1)>0,𝑥₂(2)>0,…,𝑥₂(2𝑟+1)>0, and 𝑥₂(2𝑟+2)≤0,…. Note that 𝑟≥𝑃⁺_𝑥2 and 𝑃⁺_𝑥2>𝑃⁺_𝑥1. It follows from (2.9) and the definition of 𝑃⁺_𝑥1 that 𝑥₁(𝑘+1+𝑟)=𝑥₁(𝑘+2𝑟+2)−(𝑟+1)≤𝑥₁(𝑘+2𝑟+2)−𝑃⁺_𝑥2−1<𝑥₁(𝑘+2𝑟+2)−𝑃⁺_𝑥1−1≤0. (2.11) Thus, 𝑥₁(𝑘+1+𝑟)≤−1, which, together with (2.10), implies that 𝑥₁(𝑘+1)=𝑥₁(𝑘+1+𝑟)−𝑟≤−(𝑟+1). (2.12) It follows that 𝑃⁻_𝑥1≥𝑟+1 and hence, 𝐿⁻_𝑥1≥2𝑟+3>𝐿⁺_𝑥2, a contradiction to Lemma 2.5. This proves the claim. Repeating the same argument as that in the proof of the above claim, we have 𝑃⁺_𝑥𝑖≤𝑃⁺_𝑥𝑗, 𝑃⁻_𝑥𝑖≤𝑃⁻_𝑥𝑗, and 𝑃⁺_𝑥𝑖≤𝑃⁻_𝑥𝑗 for 𝑖,𝑗=1,…,4 and 𝑖≠𝑗. Thus, we get 𝑃⁺_𝑥𝑖=𝑃⁻_𝑥𝑗 for 𝑖,𝑗=1,…,4. This completes the proof.

Lemma 2.7. 𝐿⁺_𝑥1=2𝑃⁺_𝑥1+1.

Proof. It follows from the definitions of 𝑃⁺_𝑥1 and 𝐿⁺_𝑥1 that 𝐿⁺_𝑥1≥2𝑃⁺_𝑥1+1. We show 𝐿⁺_𝑥1=2𝑃⁺_𝑥1+1 by way of contradiction. Assume that 𝐿⁺_𝑥1>2𝑃⁺_𝑥1+1. By Lemma 2.5, there exists 𝑟≥0 such that 𝐿⁺_𝑥1=2𝑟+1≥2𝑃⁺_𝑥1+1, which implies that 𝑟≥𝑃⁺_𝑥1. Without loss of generality, we assume that 𝑥₁(0)≤0,𝑥₁(1)>0,𝑥₁(2)>0,…,𝑥₁(2𝑟+1)>0, and 𝑥₁(2𝑟+2)≤0,…. Now, we distinguish two cases to finish the proof.
Case 1. 𝑥₄(𝑟+𝑘+1)≥1. For this case, from (1.5) and (1.3), we have 𝑥₄(𝑘+1)=𝑥₄(𝑘+𝑟+1)+^𝑟_𝑖=1𝑓𝑥₁(𝑖)≥1+𝑟>1+𝑃⁺_𝑥1, (2.13) which implies that 𝑃⁺_𝑥4>𝑃⁺_𝑥1, a contradiction to Lemma 2.6.
Case 2. 𝑥₄(𝑟+𝑘+1)≤0. For this case, from (1.5) and (1.3), we have 𝑥₄(𝑘+2𝑟+2)=𝑥₄(𝑟+𝑘+1)−^2𝑟+1_𝑖=1+𝑟𝑓𝑥₁(𝑖(),)≤−1+𝑟 (2.14) which implies that 𝑃⁻_𝑥4≥1+𝑟>𝑃⁺_𝑥1, a contradiction to Lemma 2.6. This completes the proof.

Throughout the remaining part of this paper, we denote 𝑃⁺_𝑥1=𝑝.

Lemma 2.8. (a)𝑥₁(𝑛₁)=𝑝+1 for some 𝑛₁∈𝑍(−𝑘). Moreover, 𝑥₄(𝑛₁+𝑘)=1 and 𝑥₄(𝑛₁+𝑘+𝑝+1)=−𝑝.
(b)𝑥₄(𝑛₂)=𝑝+1 for some 𝑛₂∈𝑍(−𝑘). Moreover, 𝑥₃(𝑛₂+𝑘)=1 and 𝑥₃(𝑛₂+𝑘+𝑝+1)=−𝑝.
(c)𝑥₂(𝑛₃)=𝑝+1 for some 𝑛₃∈𝑍(−𝑘). Moreover, 𝑥₁(𝑛₃+𝑘)=0 and 𝑥₁(𝑛₃+𝑘+𝑝+1)=𝑝+1.
(d)𝑥₃(𝑛₄)=𝑝+1 for some 𝑛₄∈𝑍(−𝑘). Moreover, 𝑥₂(𝑛₄+𝑘)=0 and 𝑥₂(𝑛₄+𝑘+𝑝+1)=𝑝+1.

Proof. We only give the proof of the conclusion (c). The other conclusions can be proved similarly. By the definition of 𝑃⁺_𝑥2, there must exist 𝑛₃∈𝑍(−𝑘) such that 𝑥₂(𝑛₃)=𝑃⁺_𝑥2+1=𝑝+1. Since the distance between consecutive elements of {𝑥₂(𝑛)}^∞_𝑛=0 is 1, it follows that 𝑥₂(𝑛₃±𝑗)>0 for all 𝑗∈𝑍(1,𝑝). We claim that 𝑥₁(𝑛₃+𝑘)=0. Otherwise, 𝑥₁(𝑛₃+𝑘)≤−1 or 𝑥₁(𝑛₃+k)≥1. If 𝑥₁(𝑛₃+𝑘)≥1, it follows from (1.5) and (1.3) that 𝑥₁𝑛₃+𝑘+𝑝+1=𝑥₁𝑛₃+𝑘+^𝑛3+𝑝_𝑗=𝑛3𝑓𝑥₂𝑗=𝑥₁𝑛₃+𝑘+𝑝+1. (2.15) Thus, 𝑥₁(𝑛₃+𝑘+𝑝+1)≥𝑝+2, which implies that 𝑃⁺_𝑥1≥𝑝+2>𝑃⁺_𝑥2, a contradiction to Lemma 2.6. If 𝑥₁(𝑛₃+𝑘)≤−1, it follows from (1.5) and (1.3) that 𝑥₁𝑛₃+𝑘=𝑥₁𝑛₃+𝑘−𝑝+^𝑛3−1_{𝑗=𝑛3−𝑝}𝑓𝑥₂𝑗=𝑥₁𝑛₃+𝑘−𝑝+𝑝. (2.16) Thus, 𝑥₁(𝑛₃+𝑘−𝑝)=𝑥₁(𝑛₃+𝑘)−𝑝≤−1−𝑝, which implies that 𝑃⁻_𝑥1≥𝑝+1>𝑃⁺_𝑥2, a contradiction to Lemma 2.6. This proves the claim. By (1.5) and (1.3), we must have 𝑥₁𝑛₃+𝑘+𝑝+1=𝑥₁𝑛₃+𝑘+^𝑛3+𝑝_𝑗=𝑛3𝑓𝑥₂𝑗=𝑝+1. (2.17) This completes the proof.

Arguing as in the proof of Lemma 2.8, we can get the following result.

Lemma 2.9. (a)𝑥₁(𝑛₁)=−𝑝 for some 𝑛₁∈𝑍(−𝑘). Moreover, 𝑥₄(𝑛₁+𝑘)=0 and 𝑥₄(𝑛₁+𝑘+𝑝+1)=𝑝+1.
(b)𝑥₄(𝑛₂)=−𝑝 for some 𝑛₂∈𝑍(−𝑘). Moreover, 𝑥₃(𝑛₂+𝑘)=0 and 𝑥₃(𝑛₂+𝑘+𝑝+1)=𝑝+1.
(c)𝑥₂(𝑛₃)=−𝑝 for some 𝑛₃∈𝑍(−𝑘). Moreover, 𝑥₁(𝑛₃+𝑘)=1 and 𝑥₁(𝑛₃+𝑘+𝑝+1)=−𝑝.
(d)𝑥₃(𝑛₄)=−𝑝 for some 𝑛₄∈𝑍(−𝑘). Moreover, 𝑥₂(𝑛₄+𝑘)=1 and 𝑥₂(𝑛₄+𝑘+𝑝+1)=−𝑝.

Lemma 2.10. 𝑝≤2𝑘.

Proof. Without loss of generality, we assume that 𝑥₁(0)=0,𝑥₁(1)=1,…,𝑥₁(𝑝)=𝑝,𝑥₁(𝑝+1)=𝑝+1,𝑥₁(𝑝+2)=𝑝,…,𝑥₁(2𝑝)=2, 𝑥₁(2𝑝+1)=1, and 𝑥₁(2𝑝+2)=0. It follows from Lemma 2.8 that 𝑥₄(𝑝+𝑘+1)=1. By (1.5) and (1.3), we have 𝑥₄𝑝+𝑘+1=𝑥₄(𝑘+1)−^𝑝_𝑖=1𝑓𝑥₁(𝑖). (2.18) Thus, 𝑥₄(𝑘+1)=𝑝+1. It follows from Lemma 2.8 that 𝑥₃(𝑝+2𝑘+2)=−𝑝, and hence, by Lemma 2.9, we have 𝑥₂(𝑝+3𝑘+2)=1. From (1.5) and (1.3), we get 𝑥₂𝑝+3𝑘+2=𝑥₂(3𝑘+2)+^𝑝+2𝑘+1_𝑖=2𝑘+2𝑓𝑥₃(𝑖)=𝑥₂(3𝑘+2)−𝑝. (2.19) Thus, 𝑥₂(3𝑘+2)=𝑝+1. Therefore, by Lemma 2.8, we have 𝑥₁(4𝑘+2)=0. Since 𝑥₁(2𝑝+2)=0, it follows that 4𝑘+2≥2𝑝+2, and hence 2𝑘≥𝑝. This completes the proof.

We hereafter use 𝑀_𝑥𝑖 to denote the length of the longest normal cycle of {𝑥_𝑖(𝑛)}^+∞_𝑛=−𝑘, where 𝑖=1,…,4.

Lemma 2.11. 𝑀_𝑥𝑖=+∞, 𝑖=1,…,4.

Proof. We show that 𝑀_𝑥𝑖=+∞ for 𝑖∈𝑍(1,4) by way of contradiction. Without loss of generality, we assume that 𝑀_𝑥1<+∞. We will distinguish several cases to finish the proof.
Case 1. 𝑀_𝑥𝑖=+∞, 𝑖=2,3,4. Without loss of generality, we assume that {𝑥₂(𝑖)}^+∞_𝑛=1 is a total 𝑝-normal cycle and 𝑥₂(1)=1. Then, 𝑥₂(2𝑖(2𝑝+1)+𝑝+1)=𝑝+1 for 𝑖∈𝑍(0), and 𝑥₂(2𝑖(2𝑝+1)+3𝑝+2)=−𝑝 for 𝑖∈𝑍(0). By Lemmas 2.8 and 2.9, we have 𝑥₁2𝑖2𝑝+1+𝑝+1+𝑘+𝑝+1=𝑝+1,𝑥₁2𝑖2𝑝+1+3𝑝+2+𝑘+𝑝+1=−𝑝, (2.20) for 𝑖∈𝑍(0). Therefore, {𝑥₁(𝑛)}^+∞_{𝑛=𝑝+𝑘+2} is a total 𝑝-normal cycle, which is a contradiction to 𝑀_𝑥1<+∞.
Case 2. 𝑀_𝑥𝑖<+∞, 𝑖=2,3,4. Without loss of generality, we assume that 𝑀_𝑥2≥max{𝑀_𝑥1,𝑀_𝑥3,𝑀_𝑥4}. Then, by the definition of 𝑀_𝑥2, there exists 𝐿≥1 such that 𝑀_𝑥2=(2𝑝+1)𝐿 is the length of the longest normal cycle of {𝑥₂(𝑛)}^+∞_𝑛=−𝑘. Without loss of generality, we assume that {𝑥₂(𝑛)}^𝑀𝑥2_𝑛=1 is a 𝑝-normal cycle and 𝑥_{𝑀𝑥2−𝑝}=𝑝+1,𝑥_{𝑀𝑥2−𝑝+1}=𝑝,…,𝑥_{𝑀𝑥2−1}=2,𝑥_𝑀𝑥2=1. To finish the proof of Case 2, we will distinguish two subcases.
Subcase 1. 𝑥₂(1)=1. Then there exists 𝑞≥0 such that 𝐿=2𝑞+1. Thus, 𝑥₂,𝑥2𝑖2𝑝+1+𝑝+1=𝑝+1for𝑖∈𝑍0,𝑞₂2𝑖2𝑝+1+3𝑝+2=−𝑝for𝑖∈𝑍0,𝑞−1. (2.21) By Lemmas 2.8 and 2.9, we have 𝑥₁(2𝑖(2𝑝+1)+(𝑝+1)+(𝑘+𝑝+1))=𝑝+1 for 𝑖∈𝑍(0,𝑞) and 𝑥₁(2𝑖(2𝑝+1)+(3𝑝+2)+(𝑘+𝑝+1))=−𝑝 for 𝑖∈𝑍(0,𝑞−1). It follows from 𝑥₂(𝑝+1)=𝑝+1 that 𝑥₁(𝑝+𝑘+1)=0. This, combined with 𝑥₁𝑝+𝑘+1=𝑥₁(𝑘+1)+^𝑝_𝑖=1𝑓𝑥₂(𝑖)=𝑥₁(𝑘+1)+𝑝, (2.22) gives 𝑥₁(𝑘+1)=−𝑝. By Lemma 2.10, we have 𝑘−𝑝+1≥𝑘−2𝑘+1>−𝑘. Therefore, the total number of 𝑝+1 and −𝑝 contained in {𝑥₁(𝑖)}^{(2𝑞+1)(2𝑝+1)+𝑘+1+𝑝}_{𝑖=𝑘+1−𝑝} is 2𝑞+2. Since 2𝑞+12𝑝+1+𝑘+1+𝑝−𝑘−𝑝+1+1=2𝑞+12𝑝+1+2𝑝+1, (2.23) it follows that {𝑥₁(𝑖)}^{(2𝑞+1)(2𝑝+1)+𝑘+1+𝑝}_{𝑖=𝑘+1−𝑝} is a normal cycle with the length of 𝑀_𝑥2+2𝑝+1, and hence 𝑀_𝑥1>𝑀_𝑥2, which leads to a contradiction.
Subcase 2. 𝑥₂(1)=0. Then, using a similar argument to that in the proof of Subcase 1, we can show that this is also a contradiction.
If 𝑀_𝑥2<+∞, 𝑀_𝑥3=+∞, 𝑀_𝑥4=+∞ or 𝑀_𝑥2<+∞, 𝑀_𝑥3<+∞, 𝑀_𝑥4=+∞, then we have the same contradiction by similar arguments. This completes the proof.

3. Main Result and Proof

Theorem 3.1. For 𝑘∈𝑍(0), every bounded solution of (1.5) is eventually periodic with the minimal period 4𝑟+2 for some 𝑟∈Ω(𝑘), where Ω(𝑘)={𝑟∈𝑍(0)∶(2𝑘+1)/(2𝑟+1)∈𝑍(0)}.

Proof. Let {(𝑥₁(𝑛),𝑥₂(𝑛),𝑥₃(𝑛),𝑥₄(𝑛))}^+∞_𝑛=−𝑘 be a bounded solution of (1.5). Then there exists a positive integer 𝑀₀>0 such that |𝑥₁(𝑛)|+⋯+|𝑥₄(𝑛)|<𝑀₀ for all 𝑛∈𝑍(−𝑘). Denote 𝑥^𝑛_𝑗(𝑖)=𝑥_𝑗(𝑛−𝑖) for 𝑖∈𝑍(0,𝑘) and 𝑗∈𝑍(1,…,4), and 𝑋_𝑛=𝑥^𝑛₁(0),𝑥^𝑛₁(1),…,𝑥^𝑛₁(𝑘),𝑥^𝑛₂(0),𝑥^𝑛₂(1),…,𝑥^𝑛₂(𝑘),…,𝑥^𝑛₄(0),…,𝑥^𝑛₄(𝑘). (3.1) Define 𝑃_𝑖∶𝑍^4𝑘+4→𝑍 by setting 𝑃_𝑖(𝑣₁,𝑣₂,…,𝑣_4𝑘+4)=𝑣_𝑖 for 𝑣_i∈𝑍 and 𝑖∈𝑍(1,4𝑘+4), and define 𝐹=(𝐹₁,𝐹₂,…,𝐹_4𝑘+4) as 𝐹_𝑖(𝑋)=𝑃₁𝑋+𝑓𝑃_2𝑘+2𝑋𝑃for𝑖=1,_𝑖−1𝑃𝑋for2≤𝑖≤k+1,_𝑘+2𝑋+𝑓𝑃_3𝑘+3𝑋𝑃for𝑖=k+2,_𝑖−1𝑃𝑋fork+3≤𝑖≤2k+2,_2𝑘+3𝑋−𝑓𝑃_4𝑘+4𝑋𝑃for𝑖=2k+3,_𝑖−1𝑃𝑋for2k+4≤𝑖≤3k+3,_3𝑘+4𝑋−𝑓𝑃_𝑘+1𝑋𝑃for𝑖=3k+4,_𝑖−1𝑋for3k+5≤𝑖≤4k+4. (3.2) Then, 𝑋_𝑛+1=𝐹(𝑋_𝑛). Since |𝑥₁(𝑛)|+⋯+|𝑥₄(𝑛)|<𝑀₀ for all 𝑛∈𝑍(−𝑘) and 𝑥_𝑗(𝑛)∈𝑍 for all 𝑛∈𝑍(−𝑘) and 𝑗∈𝑍(1,…,4), it follows that {𝑋_𝑛}^+∞_𝑛=−𝑘 is a finite set. Therefore, there exist 𝑛₁>𝑛₂≥1 such that 𝑋_𝑛2=𝑋_𝑛1=𝑋_{𝑛2+𝑛1−𝑛2}. This, combined with 𝑋_𝑛+1=𝐹(𝑋_𝑛), gives 𝑋_𝑛=𝑋_{𝑛+𝑛1−𝑛2} for 𝑛≥𝑛₂+1. Hence, {𝑋_𝑛}^+∞_𝑛=−𝑘 is eventually periodic with the period (𝑛₁−𝑛₂). It follows that {𝑃_𝑗𝑋_𝑛}^+∞_𝑛=−𝑘 are also eventually periodic, 𝑗=1,𝑘+2,2𝑘+3,3𝑘+4. Therefore, {(𝑥₁(𝑛),…,𝑥₄(𝑛))}^+∞_𝑛=−𝑘 is eventually periodic with the minimal period 𝑇∈𝑍(1). By Lemma 2.11, there exists some 𝑟∈𝑍(0) such that 𝑇=4𝑟+2. Without loss of generality, we assume that {𝑥₁(𝑛)}^+∞_𝑛=1 is a total 𝑟-normal cycle, and 𝑥₁(1)=1. Then, 𝑥₁(𝑟+1)=𝑟+1. By Lemma 2.8, we can get that 𝑥₄(𝑟+𝑘+1)=1 and 𝑥₄(2𝑟+2+𝑘)=−𝑟. It follows from Lemma 2.9 that 𝑥₃(2𝑟+2𝑘+2)=0 and 𝑥₃(3𝑟+2𝑘+3)=𝑟+1, and so, again by Lemma 2.8, 𝑥₂(3𝑟+3𝑘+3)=0 and 𝑥₂(4𝑟+3𝑘+4)=𝑟+1. By Lemma 2.8, we have 𝑥₁(4𝑟+4𝑘+4)=0 and 𝑥₁(5𝑟+4𝑘+5)=𝑟+1. From (1.5) and (1.3), we can get 𝑥₁(4𝑟+4𝑘+5)=𝑥₁(4𝑟+4𝑘+4)+𝑓(𝑥₂(4𝑟+3𝑘+4)). Thus, 𝑥₁(4𝑟+4𝑘+5)=0+1=1. Therefore, {𝑥₁(𝑖)}^{4𝑟+4𝑘+4}_𝑖=1 consists of 𝐴⁺_𝑟,𝐴⁻_𝑟,…,𝐴⁺_𝑟,𝐴⁻_𝑟, consecutively, in total of 𝑁 semicycles, in which 𝑁 is a positive even integer. Consequently, there exists an integer 𝑀≥0 such that 4𝑟+4𝑘+4=(2𝑟+1)⋅(2𝑀+2), that is, (2𝑘+1)/(2𝑟+1)=𝑀, which implies that 𝑟∈Ω(𝑘). This completes the proof.

Acknowledgment

This paper is supported by the Fundamental Research Funds for the Central Universities of Hunan University (Grant no. 531107040013).