The Form of the Solutions and Periodicity of Some Systems of Difference Equations

Abstract

This paper is devoted to get the form of the solutions and the periodic nature of the following systems of rational difference equations 𝑥_𝑛+1=𝑥_𝑛−5/(−1+𝑥_𝑛−5𝑦_𝑛−2),𝑦_𝑛+1=𝑦_𝑛−5/(±1±𝑦_𝑛−5𝑥_𝑛−2), where the initial conditions are real numbers.

1. Introduction

Difference equations appear naturally as discrete analogues and as numerical solutions of differential equations. They have many applications in biology, ecology, economy, and physics. So, recently, there has been an increasing interest in the study of qualitative analysis of rational difference equations and systems of difference equations. Although difference equations are very simple in form, it is extremely difficult to understand thoroughly the behaviors of their solutions, see [1–23] and the references cited therein.

Periodic solutions of a difference equations have been investigated by many researchers, and various methods have been proposed for the existence and qualitative properties of the solution.

The periodicity of the positive solutions of the system of rational difference equations 𝑥_𝑛+1=1𝑦_𝑛,𝑦_𝑛+1=𝑦_𝑛𝑥_𝑛−1𝑦_𝑛−1 (1.1) was studied by Çinar in [5].

Elsayed [11] has obtained the solution of the following system of the difference equations: 𝑥_𝑛+1=1𝑦_𝑛−𝑘,𝑦_𝑛+1=𝑦_𝑛−𝑘𝑥_𝑛𝑦_𝑛. (1.2) The behavior of the positive solution of the following system: 𝑥_𝑛+1=𝑥_𝑛−11+𝑥_𝑛−1𝑦_𝑛,𝑦_𝑛+1=𝑦_𝑛−11+𝑦_𝑛−1𝑥_𝑛. (1.3) has been studied by Kurbanli et al. [22].

Özban [24] has investigated the positive solution of the system of rational difference equations as 𝑥_𝑛+1=1𝑦_𝑛−𝑘,𝑦_𝑛+1=𝑦_𝑛𝑥_𝑛−𝑚𝑦_{𝑛−𝑚−𝑘}. (1.4) Özban [25] has investigated the solution of the following system: 𝑥_𝑛+1=𝑎𝑦_𝑛−3,𝑦_𝑛+1=𝑏𝑦_𝑛−3𝑥_𝑛−𝑞𝑦_𝑛−𝑞. (1.5)

In [26] Yalcinkaya investigated the sufficient condition for the global asymptotic stability of the following system of difference equations: 𝑧_𝑛+1=𝑡_𝑛𝑧_𝑛−1+𝑎𝑡_𝑛+𝑧_𝑛−1,𝑡_𝑛+1=𝑧_𝑛𝑡_𝑛−1+𝑎𝑧_𝑛+𝑡_𝑛−1. (1.6) Also, Yalcinkaya [27] has obtained the sufficient conditions for the global asymptotic stability of the system of two nonlinear difference equations as 𝑥_𝑛+1=𝑥_𝑛+𝑦_𝑛−1𝑥_𝑛𝑦_𝑛−1−1,𝑦_𝑛+1=𝑦_𝑛+𝑥_𝑛−1𝑦_𝑛𝑥_𝑛−1.−1 (1.7) Yang et al. [28] has investigated the positive solution of the system following: 𝑥_𝑛=𝑎𝑦_𝑛−𝑝,𝑦_𝑛=𝑏𝑦_𝑛−𝑝𝑥_𝑛−𝑞𝑦_𝑛−𝑞. (1.8) Similar nonlinear systems of rational difference equations were investigated [26–41].

In this paper, we investigate the behavior of the solutions of the difference equations systems as 𝑥_𝑛+1=𝑥_𝑛−5−1+𝑥_𝑛−5𝑦_𝑛−2,𝑦_𝑛+1=𝑦_𝑛−5±1±𝑦_𝑛−5𝑥_𝑛−2, (1.9) where the initial conditions 𝑥_𝑖,𝑦_𝑖 for 𝑖=−5,−4,−3,−2,−1,0 are real numbers.

2. The First System: 𝑥_𝑛+1=𝑥_𝑛−5/(−1+𝑦_𝑛−2𝑥_𝑛−5), 𝑦_𝑛+1=𝑦_𝑛−5/(1+𝑥_𝑛−2𝑦_𝑛−5)

In this section, we investigate the solution of the system of two difference equations as 𝑥_𝑛+1=𝑥_𝑛−5−1+𝑦_𝑛−2𝑥_𝑛−5,𝑦_𝑛+1=𝑦_𝑛−51+𝑥_𝑛−2𝑦_𝑛−5, (2.1) where the initial conditions are arbitrary real numbers with 𝑥₋₅𝑦₋₂, 𝑥₋₄𝑦₋₁, 𝑥₋₃𝑦₀≠1,≠1/2, and 𝑥₋₂𝑦₋₅, 𝑥₋₁𝑦₋₄, 𝑥₀𝑦₋₃≠±1.

The following theorem is devoted to the form of the solutions of system (2.1).

Theorem 2.1. Suppose that {𝑥_𝑛,𝑦_𝑛} are solutions of system (2.1). Also, assume that the initial conditions 𝑥₋₅,𝑥₋₄,𝑥₋₃,𝑥₋₂,𝑥₋₁,𝑥₀,𝑦₋₅,𝑦₋₄,𝑦₋₃,𝑦₋₂,𝑦₋₁𝑎𝑛𝑑𝑦₀are arbitrary real numbers and let 𝑥₋₅=𝑓, 𝑥₋₄=𝑒, 𝑥₋₃=𝑑,𝑥₋₂=𝑐, 𝑥₋₁=𝑏,𝑥₀=𝑎,𝑦₋₅=𝑠,𝑦₋₄=𝑟, 𝑦₋₃=𝑞,  𝑦₋₂=𝑝, 𝑦₋₁=ℎ, 𝑦₀=𝑔. Then for 𝑛=0,1,2,…, one has 𝑥_12𝑛−5=(−1)^𝑛𝑓−1+2𝑝𝑓^𝑛−1+𝑝𝑓^2𝑛,𝑥_12𝑛−4=(−1)^𝑛𝑒(−1+2𝑒ℎ)^𝑛(−1+𝑒ℎ)^2𝑛,𝑥_12𝑛−3=(−1)^𝑛𝑑−1+2𝑑𝑔^𝑛−1+𝑑𝑔^2𝑛,𝑥_12𝑛−2=𝑐(1+𝑠𝑐)^𝑛(1−𝑠𝑐)^𝑛,𝑥_12𝑛−1=𝑏(1+𝑏𝑟)^𝑛(1−𝑏𝑟)^𝑛,𝑥_12𝑛=𝑎1+𝑎𝑞^𝑛1−𝑎𝑞^𝑛,𝑥_12𝑛+1=(−1)^𝑛𝑓−1+2𝑝𝑓^𝑛−1+𝑝𝑓^2𝑛+1,𝑥_12𝑛+2=(−1)^𝑛𝑒(−1+2𝑒ℎ)^𝑛(−1+𝑒ℎ)^2𝑛+1,𝑥_12𝑛+3=(−1)^𝑛𝑑−1+2𝑑𝑔^𝑛−1+𝑑𝑔^2𝑛+1,𝑥_12𝑛+4=−𝑐(1+𝑠𝑐)^𝑛+1(1−𝑠𝑐)^𝑛,𝑥_12𝑛+5=−𝑏(1+𝑏𝑟)^𝑛+1(1−𝑏𝑟)^𝑛,𝑥_12𝑛+6=−𝑎1+𝑎𝑞^𝑛+11−𝑎𝑞^𝑛,𝑦_12𝑛−5=𝑠(1+𝑠𝑐)^𝑛(1−𝑠𝑐)^𝑛,𝑦_12𝑛−4=𝑟(1+𝑏𝑟)^𝑛(1−𝑏𝑟)^𝑛,𝑦_12𝑛−3=𝑞1+𝑎𝑞^𝑛1−𝑎𝑞^𝑛,𝑦_12𝑛−2=(−1)^𝑛𝑝−1+𝑝𝑓^2𝑛−1+2𝑝𝑓^𝑛,𝑦_12𝑛−1=(−1)^𝑛ℎ(−1+𝑒ℎ)^2𝑛(−1+2𝑒ℎ)^𝑛,𝑦_12𝑛=(−1)^𝑛𝑔−1+𝑑𝑔^2𝑛−1+2𝑑𝑔^𝑛,𝑦_12𝑛+1=𝑠(1+𝑠𝑐)^𝑛+1(1−𝑠𝑐)^𝑛,𝑦_12𝑛+2=𝑟(1+𝑏𝑟)^𝑛+1(1−𝑏𝑟)^𝑛,𝑦_12𝑛+3=𝑞1+𝑎𝑞^𝑛+11−𝑎𝑞^𝑛,𝑦_12𝑛+4=(−1)^𝑛𝑝−1+𝑝𝑓^2𝑛+1−1+2𝑝𝑓^𝑛+1,𝑦_12𝑛+5=(−1)^𝑛ℎ(−1+𝑒ℎ)^2𝑛+1(−1+2𝑒ℎ)^𝑛+1,𝑦_12𝑛+6=(−1)^𝑛𝑔−1+𝑑𝑔^2𝑛+1−1+2𝑑𝑔^𝑛+1. (2.2)

Proof. For 𝑛=0, the result holds. Now suppose that 𝑛>0 and that our assumption holds for 𝑛−1, that is, 𝑥_12𝑛−17=(−1)^𝑛−1𝑓−1+2𝑝𝑓^𝑛−1−1+𝑝𝑓^2𝑛−2,𝑥_12𝑛−16=(−1)^𝑛−1𝑒(−1+2𝑒ℎ)^𝑛−1(−1+𝑒ℎ)^2𝑛−2,𝑥_12𝑛−15=(−1)^𝑛−1𝑑−1+2𝑑𝑔^𝑛−1−1+𝑑𝑔^2𝑛−2,𝑥_12𝑛−14=𝑐(1+𝑠𝑐)^𝑛−1(1−𝑠𝑐)^𝑛−1,𝑥_12𝑛−13=𝑏(1+𝑏𝑟)^𝑛−1(1−𝑏𝑟)^𝑛−1,𝑥_12𝑛−12=𝑎1+𝑎𝑞^𝑛−11−𝑎𝑞^𝑛−1,𝑥_12𝑛−11=(−1)^𝑛−1𝑓−1+2𝑝𝑓^𝑛−1−1+𝑝𝑓^2𝑛−1,𝑥_12𝑛−10=(−1)^𝑛−1𝑒(−1+2𝑒ℎ)^𝑛−1(−1+𝑒ℎ)^2𝑛−1,𝑥_12𝑛−9=(−1)^𝑛−1𝑑−1+2𝑑𝑔^𝑛−1−1+𝑑𝑔^2𝑛−1,𝑥_12𝑛−8=−𝑐(1+𝑠𝑐)^𝑛(1−𝑠𝑐)^𝑛−1,𝑥_12𝑛−7=−𝑏(1+𝑏𝑟)^𝑛(1−𝑏𝑟)^𝑛−1,𝑥_12𝑛−6=−𝑎1+𝑎𝑞^𝑛1−𝑎𝑞^𝑛−1,𝑦_12𝑛−17=𝑠(1+𝑠𝑐)^𝑛−1(1−𝑠𝑐)^𝑛−1,𝑦_12𝑛−16=𝑟(1+𝑏𝑟)^𝑛−1(1−𝑏𝑟)^𝑛−1,𝑦_12𝑛−15=𝑞1+𝑎𝑞^𝑛−11−𝑎𝑞^𝑛−1,𝑦_12𝑛−14=(−1)^𝑛−1𝑝−1+𝑝𝑓^2𝑛−2−1+2𝑝𝑓^𝑛−1,𝑦_12𝑛−13=(−1)^𝑛−1ℎ(−1+𝑒ℎ)^2𝑛−2(−1+2𝑒ℎ)^𝑛−1,𝑦_12𝑛−12=(−1)^𝑛−1𝑔−1+𝑑𝑔^2𝑛−2−1+2𝑑𝑔^𝑛−1,𝑦_12𝑛−11=𝑠(1+𝑠𝑐)^𝑛(1−𝑠𝑐)^𝑛−1,𝑦_12𝑛−10=𝑟(1+𝑏𝑟)^𝑛(1−𝑏𝑟)^𝑛−1,𝑦_12𝑛−9=𝑞1+𝑎𝑞^𝑛1−𝑎𝑞^𝑛−1,𝑦_12𝑛−8=(−1)^𝑛−1𝑝−1+𝑝𝑓^2𝑛−1−1+2𝑝𝑓^𝑛,𝑦_12𝑛−7=(−1)^𝑛−1ℎ(−1+𝑒ℎ)^2𝑛−1(−1+2𝑒ℎ)^𝑛,𝑦_12𝑛−6=(−1)^𝑛−1𝑔−1+𝑑𝑔^2𝑛−1−1+2𝑑𝑔^𝑛. (2.3) Now, it follows from (2.1) that 𝑥_12𝑛−5=𝑥_12𝑛−11−1+𝑦_12𝑛−8𝑥_12𝑛−11=(−1)^𝑛−1𝑓−1+2𝑝𝑓^𝑛−1/−1+𝑝𝑓^2𝑛−1−1+(−1)^𝑛−1𝑝−1+𝑝𝑓^2𝑛−1/−1+2𝑝𝑓^𝑛(−1)^𝑛−1𝑓−1+2𝑝𝑓^𝑛−1/−1+𝑝𝑓^2𝑛−1=(−1)^𝑛−1𝑓−1+2𝑝𝑓^𝑛−1−1+𝑝𝑓^2𝑛−1−1+𝑝𝑓/−1+2𝑝𝑓−1+2𝑝𝑓=()−1+2𝑝𝑓−1^𝑛−1𝑓−1+2𝑝𝑓^𝑛−1−1+2𝑝𝑓−1+𝑝𝑓^2𝑛−1=()1−2𝑝𝑓+𝑝𝑓−1^𝑛𝑓−1+2𝑝𝑓^𝑛−1+𝑝𝑓^2𝑛,𝑦_12𝑛−5=𝑦_12𝑛−111+𝑥_12𝑛−8𝑦_12𝑛−11=𝑠/(1+𝑠𝑐)^𝑛(1−𝑠𝑐)^𝑛−11−𝑐(1+𝑠𝑐)^𝑛(1−𝑠𝑐)^𝑛−1𝑠/(1+𝑠𝑐)^𝑛(1−𝑠𝑐)^𝑛−1=𝑠(1+𝑠𝑐)^𝑛(1−𝑠𝑐)^𝑛−1()=𝑠1−𝑠𝑐(1+𝑠𝑐)^𝑛(1−𝑠𝑐)^𝑛,𝑥_12𝑛−4=𝑥_12𝑛−10−1+𝑦_12𝑛−7𝑥_12𝑛−10=(−1)^𝑛−1𝑒(−1+2𝑒ℎ)^𝑛−1/(−1+𝑒ℎ)^2𝑛−1−1+(−1)^𝑛−1ℎ(−1+𝑒ℎ)^2𝑛−1/(−1+2𝑒ℎ)^𝑛(−1)^𝑛−1𝑒(−1+2𝑒ℎ)^𝑛−1/(−1+𝑒ℎ)^2𝑛−1=(−1)^𝑛−1𝑒(−1+2𝑒ℎ)^𝑛−1(−1+𝑒ℎ)^2𝑛−1(−1+ℎ𝑒/(−1+2𝑒ℎ))−1+2𝑒ℎ=()−1+2𝑒ℎ−1^𝑛−1𝑒(−1+2𝑒ℎ)^𝑛(−1+𝑒ℎ)^2𝑛−1()=()1−2𝑒ℎ+ℎ𝑒−1^𝑛𝑒(−1+2𝑒ℎ)^𝑛(−1+𝑒ℎ)^2𝑛,𝑦_12𝑛−4=𝑦_12𝑛−101+𝑥_12𝑛−7𝑦_12𝑛−10=𝑟/(1+𝑏𝑟)^𝑛(1−𝑏𝑟)^𝑛−11−𝑏(1+𝑏𝑟)^𝑛(1−𝑏𝑟)^𝑛−1𝑟/(1+𝑏𝑟)^𝑛(1−𝑏𝑟)^𝑛−1=𝑟(1+𝑏𝑟)^𝑛(1−𝑏𝑟)^𝑛−1()=𝑟1−𝑏𝑟(1+𝑏𝑟)^𝑛(1−𝑏𝑟)^𝑛. (2.4) Similarly, we can prove the other relations.

Lemma 2.2. Let {𝑥_𝑛,𝑦_𝑛} be a positive solution of system (2.1), then {𝑦_𝑛} is bounded and converges to zero.

Proof. It follows from (2.1) that 𝑦_𝑛+1=𝑦_𝑛−51+𝑥_𝑛−2𝑦_𝑛−5≤𝑦_𝑛−5. (2.5) Then, the subsequences {𝑦_6𝑛−5}^∞_𝑛=0, {𝑦_6𝑛−4}^∞_𝑛=0, {𝑦_6𝑛−3}^∞_𝑛=0, {𝑦_6𝑛−2}^∞_𝑛=0, {𝑦_6𝑛−1}^∞_𝑛=0, and {𝑦_6𝑛}^∞_𝑛=0 are decreasing and so are bounded from above by 𝑀=max{𝑦₋₅,𝑦₋₄,𝑦₋₃,𝑦₋₂,𝑦₋₁,𝑦₀}.

Example 2.3. We consider interesting numerical example for the difference system (2.1) with the initial conditions, where𝑥₋₅=0.8, 𝑥₋₄=0.1, 𝑥₋₃=−1.6, 𝑥₋₂=0.3, 𝑥₋₁=0.1, 𝑥₀=−0.7, 𝑦₋₅=1.7, 𝑦₋₄=0.3, 𝑦₋₃=0.4, 𝑦₋₂=−0.2, 𝑦₋₁=0.5, and 𝑦₀=0.6(see Figure 1).

Figure 1 

3. The Second System: 𝑥_𝑛+1=𝑥_𝑛−5/(−1+𝑦_𝑛−2𝑥_𝑛−5), 𝑦_𝑛+1=𝑦_𝑛−5/(−1+𝑥_𝑛−2𝑦_𝑛−5)

In this section, we study the solution of the following system of the difference equations: 𝑥_𝑛+1=𝑥_𝑛−5−1+𝑦_𝑛−2𝑥_𝑛−5,𝑦_𝑛+1=𝑦_𝑛−5−1+𝑥_𝑛−2𝑦_𝑛−5, (3.1) where 𝑛∈ℕ₀ and the initial conditions are arbitrary real numbers such that 𝑥₋₅𝑦₋₂,𝑥₋₄𝑦₋₁,𝑥₋₃𝑦₀, 𝑥₋₂𝑦₋₅, 𝑥₋₁𝑦₋₄, 𝑥₀𝑦₋₃≠1.

Theorem 3.1. Assume that {𝑥_𝑛,𝑦_𝑛} are solutions of system (3.1). Then for 𝑛=0,1,2,…, one has 𝑥_6𝑛−5=𝑓−1+𝑝𝑓^𝑛,𝑥_6𝑛−4=𝑒(−1+𝑒ℎ)^𝑛,𝑥_6𝑛−3=𝑑−1+𝑑𝑔^𝑛,𝑥_6𝑛−2=𝑐(−1+𝑠𝑐)^𝑛,𝑥_6𝑛−1=𝑏(−1+𝑏𝑟)^𝑛,𝑥_6𝑛=𝑎−1+𝑎𝑞^𝑛,𝑦_6𝑛−5=𝑠(−1+𝑠𝑐)^𝑛,𝑦_6𝑛−4=𝑟(−1+𝑏𝑟)^𝑛,𝑦_6𝑛−3=𝑞−1+𝑎𝑞^𝑛,𝑦_6𝑛−2=𝑝−1+𝑝𝑓^𝑛,𝑦_6𝑛−1=ℎ(−1+𝑒ℎ)^𝑛,𝑦_6𝑛=𝑔−1+𝑑𝑔^𝑛. (3.2)

Proof. For 𝑛=0, the result holds. Now suppose that 𝑛>1 and that our assumption holds for 𝑛−1, that is, 𝑥_6𝑛−11=𝑓−1+𝑝𝑓^𝑛−1,𝑥_6𝑛−10=𝑒(−1+𝑒ℎ)^𝑛−1,𝑥_6𝑛−9=𝑑−1+𝑑𝑔^𝑛−1,𝑥_6𝑛−8=𝑐(−1+𝑠𝑐)^𝑛−1,𝑥_6𝑛−7=𝑏(−1+𝑏𝑟)^𝑛−1,𝑥_6𝑛−6=𝑎−1+𝑎𝑞^𝑛−1,𝑦_6𝑛−11=𝑠(−1+𝑠𝑐)^𝑛−1,𝑦_6𝑛−10=𝑟(−1+𝑏𝑟)^𝑛−1,𝑦_6𝑛−9=𝑞−1+𝑎𝑞^𝑛−1,𝑦_6𝑛−8=𝑝−1+𝑝𝑓^𝑛−1,𝑦_6𝑛−7=ℎ(−1+𝑒ℎ)^𝑛−1,𝑦_6𝑛−6=𝑔−1+𝑑𝑔^𝑛−1. (3.3) Now, it follows from (3.1) that 𝑥_6𝑛−5=𝑥_6𝑛−11−1+𝑦_6𝑛−8𝑥_6𝑛−11=𝑓/−1+𝑝𝑓^𝑛−1−1+𝑝−1+𝑝𝑓^𝑛−1×𝑓/−1+𝑝𝑓^𝑛−1=𝑓/−1+𝑝𝑓^𝑛−1=𝑓−1+𝑝𝑓−1+𝑝𝑓^𝑛,𝑦_6𝑛−5=𝑦_6𝑛−11−1+𝑥_6𝑛−8𝑦_6𝑛−11=𝑠/(−1+𝑠𝑐)^𝑛−1−1+𝑐(−1+𝑠𝑐)^𝑛−1𝑠/(−1+𝑠𝑐)^𝑛−1=𝑠/(−1+𝑠𝑐)^𝑛−1()=𝑠−1+𝑐𝑠(−1+𝑠𝑐)^𝑛. (3.4) Also, we see from (3.1) that 𝑥_6𝑛−2=𝑥_6𝑛−8−1+𝑦_6𝑛−5𝑥_6𝑛−8=𝑐(−1+𝑠𝑐)^𝑛−1−1+(𝑠/(−1+𝑠𝑐)^𝑛)𝑐(−1+𝑠𝑐)^𝑛−1=𝑐(−1+𝑠𝑐)^𝑛−1((()−1+𝑠𝑐/−1+𝑠𝑐))−1+𝑠𝑐()=𝑐()−1+𝑠𝑐−1+𝑠𝑐^𝑛()1−𝑠𝑐+𝑠𝑐=𝑐−1+𝑠𝑐^𝑛,𝑦_6𝑛−2=𝑦_6𝑛−8−1+𝑥_6𝑛−5𝑦_6𝑛−8=𝑝−1+𝑝𝑓^𝑛−1−1+𝑝−1+𝑝𝑓^𝑛−1𝑓/−1+𝑝𝑓^𝑛=𝑝−1+𝑝𝑓^𝑛−1−1+𝑝𝑓/−1+𝑝𝑓−1+𝑝𝑓=𝑝−1+𝑝𝑓−1+𝑝𝑓^𝑛1−𝑝𝑓+𝑝𝑓=𝑝−1+𝑝𝑓^𝑛. (3.5) Similarly, we can prove the other relations.

Lemma 3.2. The solutions of system (3.1) has unboundedness solutions except in the following case.

Theorem 3.3. System (3.1) has a periodic solution of period six if and only if 𝑝𝑓=𝑒ℎ=𝑑𝑔=𝑏𝑟=𝑎𝑞=𝑠𝑐=2and it will take the form {𝑥_𝑛}={𝑓,𝑒,𝑑,𝑐,𝑏,𝑎,𝑓,𝑒,…}, {𝑦_𝑛}={𝑠,𝑟,𝑞,𝑝,ℎ,𝑔,𝑠,𝑟,…}.

Proof. First suppose that there exists a prime period-six solution 𝑥_𝑛=𝑓,𝑒,𝑑,𝑐,𝑏,𝑎,𝑓,𝑒,…,𝑦_𝑛=𝑠,𝑟,𝑞,𝑝,ℎ,𝑔,𝑠,𝑟,…, (3.6) of system (3.1). We see from the form of the solution of system (3.1) that 𝑓𝑓=−1+𝑝𝑓^𝑛𝑒,𝑒=(−1+𝑒ℎ)^𝑛𝑑,𝑑=−1+𝑑𝑔^𝑛,𝑐=𝑐(−1+𝑠𝑐)^𝑛,𝑏=𝑏(−1+𝑏𝑟)^𝑛,𝑎=𝑎−1+𝑎𝑞^𝑛,𝑠𝑠=(−1+𝑠𝑐)^𝑛𝑟,𝑟=(−1+𝑏𝑟)^𝑛𝑞,𝑞=−1+𝑎𝑞^𝑛,𝑝=𝑝−1+𝑝𝑓^𝑛,ℎ=ℎ(−1+𝑒ℎ)^𝑛,𝑔=𝑔−1+𝑑𝑔^𝑛. (3.7) Then, we get −1+𝑝𝑓=−1+𝑒ℎ=−1+𝑑𝑔=−1+𝑠𝑐=−1+𝑏𝑟=−1+𝑎𝑞=1. (3.8) Thus, 𝑝𝑓=𝑒ℎ=𝑑𝑔=𝑏𝑟=𝑎𝑞=𝑠𝑐=2. (3.9) Second, assume that 𝑝𝑓=𝑒ℎ=𝑑𝑔=𝑏𝑟=𝑎𝑞=𝑠𝑐=2. Then, we see from the form of the solution of system (3.1) that 𝑥_6𝑛−5=𝑓,𝑥_6𝑛−4=𝑒,𝑥_6𝑛−3𝑥=𝑑,_6𝑛−2=𝑐,𝑥_6𝑛−1=𝑏,𝑥_6𝑛𝑦=𝑎,_6𝑛−5=𝑠,𝑦_6𝑛−4=𝑟,𝑦_6𝑛−3𝑦=𝑞,_6𝑛−2=𝑝,𝑦_6𝑛−1=ℎ,𝑦_6𝑛=𝑔. (3.10) Thus, we have a periodic solution of period six and the proof is complete.

Example 3.4. Figure 2 shows the behavior of the solution of the difference system (3.1) with the initial conditions, where𝑥₋₅=0.18, 𝑥₋₄=−0.41, 𝑥₋₃=.6, 𝑥₋₂=.3, 𝑥₋₁=−0.21, 𝑥₀=.7, 𝑦₋₅=−0.17,𝑦₋₄=1.3,𝑦₋₃=.14, 𝑦₋₂=0.2, 𝑦₋₁=−.15, and 𝑦₀=0.16.

Figure 2 

Example 3.5. If we consider the difference equation system (3.1) with the initial conditions, where𝑥₋₅=5, 𝑥₋₄=−2, 𝑥₋₃=.1, 𝑥₋₂=−7, 𝑥₋₁=4, 𝑥₀=−3,𝑦₋₅=−2/7, 𝑦₋₄=0.5,𝑦₋₃=−2/3, 𝑦₋₂=0.4,  𝑦₋₁=−1, and 𝑦₀=20, then we get the shape of Figure 3.

Figure 3 

4. The Third System: 𝑥_𝑛+1=𝑥_𝑛−5/(−1+𝑦_𝑛−2𝑥_𝑛−5), 𝑦_𝑛+1=𝑦_𝑛−5/(1−𝑥_𝑛−2𝑦_𝑛−5)

In this section, we obtain the form of the solution of the system of two difference equations as 𝑥_𝑛+1=𝑥_𝑛−5−1+𝑦_𝑛−2𝑥_𝑛−5,𝑦_𝑛+1=𝑦_𝑛−51−𝑥_𝑛−2𝑦_𝑛−5, (4.1) where the initial conditions are arbitrary real numbers such that 𝑥₋₅𝑦₋₂, 𝑥₋₄𝑦₋₁, 𝑥₋₃𝑦₀≠±1, and 𝑥₋₂𝑦₋₅, 𝑥₋₁𝑦₋₄, 𝑥₀𝑦₋₃≠1/2,1.

Theorem 4.1. Suppose that {𝑥_𝑛,𝑦_𝑛} are solutions of system (4.1). Then 𝑥_12𝑛−5=(−1)^𝑛𝑓−1+𝑝𝑓^𝑛1+𝑝𝑓^𝑛,𝑥_12𝑛−4=(−1)^𝑛𝑒(−1+𝑒ℎ)^𝑛(1+𝑒ℎ)^𝑛,𝑥_12𝑛−3=(−1)^𝑛𝑑−1+𝑑𝑔^𝑛1+𝑑𝑔^𝑛,𝑥_12𝑛−2=(−1)^𝑛𝑐(1−𝑠𝑐)^2𝑛(−1+2𝑠𝑐)^𝑛,𝑥_12𝑛−1=(−1)^𝑛𝑏(1−𝑏𝑟)^2𝑛(−1+2𝑏𝑟)^𝑛,𝑥_12𝑛=(−1)^𝑛𝑎1−𝑎𝑞^2𝑛−1+2𝑎𝑞^𝑛,𝑥_12𝑛+1=(−1)^𝑛𝑓−1+𝑝𝑓^𝑛+11+𝑝𝑓^𝑛,𝑥_12𝑛+2=(−1)^𝑛𝑒(−1+𝑒ℎ)^𝑛+1(1+𝑒ℎ)^𝑛,𝑥_12𝑛+3=(−1)^𝑛𝑑−1+𝑑𝑔^𝑛+11+𝑑𝑔^𝑛,𝑥_12𝑛+4=(−1)^𝑛𝑐(1−𝑠𝑐)^2𝑛+1(−1+2𝑠𝑐)^𝑛+1,𝑥_12𝑛+5=(−1)^𝑛𝑏(1−𝑏𝑟)^2𝑛+1(−1+2𝑏𝑟)^𝑛+1,𝑥_12𝑛+6=(−1)^𝑛𝑎1−𝑎𝑞^2𝑛+1−1+2𝑎𝑞^𝑛+1,𝑦_12𝑛−5=(−1)^𝑛𝑠(−1+2𝑠𝑐)^𝑛(1−𝑠𝑐)^2𝑛,𝑦_12𝑛−4=(−1)^𝑛𝑟(−1+2𝑏𝑟)^𝑛(1−𝑏𝑟)^2𝑛,𝑦_12𝑛−3=(−1)^𝑛𝑞−1+2𝑎𝑞^𝑛1−𝑎𝑞^2𝑛,𝑦_12𝑛−2=(−1)^𝑛𝑝−1+𝑝𝑓^𝑛1+𝑝𝑓^𝑛,𝑦_12𝑛−1=(−1)^𝑛ℎ(−1+𝑒ℎ)^𝑛(1+𝑒ℎ)^𝑛,𝑦_12𝑛=(−1)^𝑛𝑔−1+𝑑𝑔^𝑛1+𝑑𝑔^𝑛,𝑦_12𝑛+1=(−1)^𝑛𝑠(−1+2𝑠𝑐)^𝑛(1−𝑠𝑐)^2𝑛+1,𝑦_12𝑛+2=(−1)^𝑛𝑟(−1+2𝑏𝑟)^𝑛(1−𝑏𝑟)^2𝑛+1,𝑦_12𝑛+3=(−1)^𝑛𝑞−1+2𝑎𝑞^𝑛1−𝑎𝑞^2𝑛+1,𝑦_12𝑛+4=(−1)^𝑛+1𝑝−1+𝑝𝑓^𝑛+11+𝑝𝑓^𝑛,𝑦_12𝑛+5=(−1)^𝑛+1ℎ(−1+𝑒ℎ)^𝑛+1(1+𝑒ℎ)^𝑛,𝑦_12𝑛+6=(−1)^𝑛+1𝑔−1+𝑑𝑔^𝑛+11+𝑑𝑔^𝑛, (4.2) where𝑥₋₅=𝑓, 𝑥₋₄=𝑒, 𝑥₋₃=𝑑, 𝑥₋₂=𝑐, 𝑥₋₁=𝑏,  𝑥₀=𝑎, 𝑦₋₅=𝑠, 𝑦₋₄=𝑟,𝑦₋₃=𝑞, 𝑦₋₂=𝑝, 𝑦₋₁=ℎ, 𝑦₀=𝑔.

Proof. As the proof of Theorem 2.1, and so it will be omitted.

Example 4.2. Figure 4 shows the behavior of the solutions of the system (4.1) with the initial conditions𝑥₋₅=0.05, 𝑥₋₄=−.42, 𝑥₋₃=.11, 𝑥₋₂=0.07, 𝑥₋₁=−0.4,𝑥₀=−3, 𝑦₋₅=−1.7,  𝑦₋₄=0.12,𝑦₋₃=−1.2, 𝑦₋₂=0.2, 𝑦₋₁=−1, and 𝑦₀=0.13.

Figure 4 

5. The Fourth System: 𝑥_𝑛+1=𝑥_𝑛−5/(−1+𝑦_𝑛−2𝑥_𝑛−5), 𝑦_𝑛+1=𝑦_𝑛−5/(−1−𝑥_𝑛−2𝑦_𝑛−5)

We get, in this section, the solution of the following system of the difference equations: 𝑥_𝑛+1=𝑥_𝑛−5−1+𝑦_𝑛−2𝑥_𝑛−5,𝑦_𝑛+1=𝑦_𝑛−5−1−𝑥_𝑛−2𝑦_𝑛−5, (5.1) where 𝑛∈ℕ₀ and the initial conditions are arbitrary real numbers.

Theorem 5.1. Let {𝑥_𝑛,𝑦_𝑛}^+∞_𝑛=−5 be solutions of system (5.1). Then for 𝑛=0,1,2,…one has𝑥_6𝑛−5=𝑓^𝑛−1_𝑖=01−(2𝑖)𝑝𝑓−1+(2𝑖+1)𝑝𝑓,𝑥_6𝑛−4=𝑒^𝑛−1_𝑖=0(1−(2𝑖)𝑒ℎ)(()),𝑥−1+2𝑖+1𝑒ℎ_6𝑛−3=𝑑^𝑛−1_𝑖=01−(2𝑖)𝑑𝑔−1+(2𝑖+1)𝑑𝑔,𝑥_6𝑛−2=𝑐^𝑛−1_𝑖=0(−1−(2𝑖+1)𝑠𝑐)(()),𝑥1+2𝑖+2𝑠𝑐_6𝑛−1=𝑏^𝑛−1_𝑖=0(−1−(2𝑖+1)𝑏𝑟)(1+(2𝑖+2)𝑏𝑟),𝑥_6𝑛=𝑎^𝑛−1_𝑖=0−1−(2𝑖+1)𝑎𝑞(),𝑦1+2𝑖+2𝑎𝑞_6𝑛−5=𝑠^𝑛−1_𝑖=0(1+(2𝑖)𝑠𝑐)(−1−(2𝑖+1)𝑠𝑐),𝑦_6𝑛−4=𝑟^𝑛−1_𝑖=0(1+(2𝑖)𝑏𝑟)(()),𝑦−1−2𝑖+1𝑏𝑟_6𝑛−3=𝑞^𝑛−1_𝑖=01+(2𝑖)𝑎𝑞−1−(2𝑖+1)𝑎𝑞,𝑦_6𝑛−2=𝑝^𝑛−1_𝑖=0−1+(2𝑖+1)𝑝𝑓(),𝑦1−2𝑖+2𝑝𝑓_6𝑛−1=ℎ^𝑛−1_𝑖=0(−1+(2𝑖+1)𝑒ℎ)(1−(2𝑖+2)𝑒ℎ),𝑦_6𝑛=𝑔^𝑛−1_𝑖=0−1+(2𝑖+1)𝑑𝑔(),1−2𝑖+2𝑑𝑔 (5.2) where𝑥₋₅=𝑓, 𝑥₋₄=𝑒, 𝑥₋₃=𝑑, 𝑥₋₂=𝑐,  𝑥₋₁=𝑏, 𝑥₀=𝑎, 𝑦₋₅=𝑠, 𝑦₋₄=𝑟, 𝑦₋₃=𝑞, 𝑦₋₂=𝑝, 𝑦₋₁=ℎ, 𝑦₀=𝑔.

Proof. For 𝑛=0, the result holds. Now suppose that 𝑛>1 and that our assumption holds for 𝑛−1, that is, 𝑥_6𝑛−11=𝑓^𝑛−2_𝑖=01−(2𝑖)𝑝𝑓−1+(2𝑖+1)𝑝𝑓,𝑥_6𝑛−10=𝑒^𝑛−2_𝑖=0(1−(2𝑖)𝑒ℎ)(()),𝑥−1+2𝑖+1𝑒ℎ_6𝑛−9=𝑑^𝑛−2_𝑖=01−(2𝑖)𝑑𝑔−1+(2𝑖+1)𝑑𝑔,𝑥_6𝑛−8=𝑐^𝑛−2_𝑖=0(−1−(2𝑖+1)𝑠𝑐)(()),𝑥1+2𝑖+2𝑠𝑐_6𝑛−7=𝑏^𝑛−2_𝑖=0(−1−(2𝑖+1)𝑏𝑟)(1+(2𝑖+2)𝑏𝑟),𝑥_6𝑛−6=𝑎^𝑛−2_𝑖=0−1−(2𝑖+1)𝑎𝑞(),𝑦1+2𝑖+2𝑎𝑞_6𝑛−11=𝑠^𝑛−2_𝑖=0(1+(2𝑖)𝑠𝑐)(−1−(2𝑖+1)𝑠𝑐),𝑦_6𝑛−10=𝑟^𝑛−2_𝑖=0(1+(2𝑖)𝑏𝑟)(()),𝑦−1−2𝑖+1𝑏𝑟_6𝑛−9=𝑞^𝑛−2_𝑖=01+(2𝑖)𝑎𝑞−1−(2𝑖+1)𝑎𝑞,𝑦_6𝑛−8=𝑝^𝑛−2_𝑖=0−1+(2𝑖+1)𝑝𝑓(),𝑦1−2𝑖+2𝑝𝑓_6𝑛−7=ℎ^𝑛−2_𝑖=0(−1+(2𝑖+1)𝑒ℎ)(1−(2𝑖+2)𝑒ℎ),𝑦_6𝑛−6=𝑔^𝑛−2_𝑖=0−1+(2𝑖+1)𝑑𝑔().1−2𝑖+2𝑑𝑔 (5.3) It follows from (3.1) that 𝑥_6𝑛−5=𝑥_6𝑛−11−1+𝑦_6𝑛−8𝑥_6𝑛−11=𝑓∏^𝑛−2_𝑖=01−(2𝑖)𝑝𝑓/−1+(2𝑖+1)𝑝𝑓−1+𝑝∏^𝑛−2_𝑖=0−1+(2𝑖+1)𝑝𝑓/1−(2𝑖+2)𝑝𝑓𝑓∏^𝑛−2_𝑖=0()/()=𝑓∏1−2𝑖𝑝𝑓−1+2𝑖+1𝑝𝑓_{𝑛−2𝑖=0}1−(2𝑖)𝑝𝑓/−1+(2𝑖+1)𝑝𝑓−1+𝑝𝑓∏^𝑛−2_𝑖=0()/()=𝑓∏1−2𝑖𝑝𝑓1−2𝑖+2𝑝𝑓^{𝑛−2𝑖=0}1−(2𝑖)𝑝𝑓/−1+(2𝑖+1)𝑝𝑓−1+𝑝𝑓/1−(2𝑛−2)𝑝𝑓=𝑓^𝑛−2_𝑖=01−(2𝑖)𝑝𝑓()()−1+2𝑖+1𝑝𝑓1−2𝑛−2𝑝𝑓−1+(2𝑛−2)𝑝𝑓+𝑝𝑓=𝑓^𝑛−1_𝑖=01−(2𝑖)𝑝𝑓(),𝑦−1+2𝑖+1𝑝𝑓_6𝑛−5=𝑦_6𝑛−11−1−𝑥_6𝑛−8𝑦_6𝑛−11=∏^𝑛−2_𝑖=0((1+(2𝑖)𝑠𝑐)/(−1−(2𝑖+1)𝑠𝑐))−1−𝑐∏^𝑛−2_𝑖=0((−1−(2𝑖+1)𝑠𝑐)/(1+(2𝑖+2)𝑠𝑐))𝑠∏^𝑛−2_𝑖=0())/(()=∏((1+2𝑖𝑠𝑐−1−2𝑖+1𝑠𝑐))_{𝑛−2𝑖=0}((1+(2𝑖)𝑠𝑐)/(−1−(2𝑖+1)𝑠𝑐))−1−𝑠𝑐∏^𝑛−2_𝑖=0())/(()=∏((1+2𝑖𝑠𝑐1+2𝑖+2𝑠𝑐))^{𝑛−2𝑖=0}((1+(2𝑖)𝑠𝑐)/(−1−(2𝑖+1)𝑠𝑐))((()=−1−𝑠𝑐/1+2𝑛−2𝑠𝑐))_{𝑛−2𝑖=0}(1+(2𝑖)𝑠𝑐)(())(())−1−2𝑖+1𝑠𝑐1+2𝑛−2𝑠𝑐(()).−1−2𝑛−2𝑠𝑐−𝑠𝑐 (5.4) Then, we see that 𝑦_6𝑛−5=𝑠^𝑛−1_𝑖=0(1+(2𝑖)𝑠𝑐)(()).−1−2𝑖+1𝑠𝑐 (5.5) Similarly, we can prove the other relations. This completes the proof.

Lemma 5.2. If 𝑥₋₅,𝑥₋₄,𝑥₋₃,𝑥₋₂,𝑥₋₁,𝑥₀,𝑦₋₅,𝑦₋₄,𝑦₋₃,𝑦₋₂,𝑦₋₁, and 𝑦₀ are arbitrary real numbers and {𝑥_𝑛,𝑦_𝑛} are solutions of system (5.1), then the following statements are true.(i)If 𝑥₋₅=0, 𝑦₋₂≠0, then we have 𝑥_6𝑛−5=0and𝑦_6𝑛−2=(−1)^𝑛𝑦₋₂.(ii)If 𝑥₋₄=0, 𝑦₋₁≠0, then we have 𝑥_6𝑛−4=0 and 𝑦_6𝑛−1=(−1)^𝑛𝑦₋₁.(iii)If 𝑥₋₃=0, 𝑦₀≠0, then we have 𝑥_6𝑛−3=0 and 𝑦_6𝑛=(−1)^𝑛𝑦₀.(iv)If 𝑥₋₂=0, 𝑦₋₅≠0, then we have 𝑥_6𝑛−2=0 and 𝑦_6𝑛−5=(−1)^𝑛𝑦₋₅.(v)If 𝑥₋₁=0, 𝑦₋₄≠0, then we have 𝑥_6𝑛−1=0 and 𝑦_6𝑛−4=(−1)^𝑛𝑦₋₄.(vi)If 𝑥₀=0, 𝑦₋₃≠0, then we have 𝑥_6𝑛=0 and 𝑦_6𝑛−3=(−1)^𝑛𝑦₋₃.(vii)If 𝑦₋₅=0,𝑥₋₂≠0, then we have 𝑦_6𝑛−5=0 and 𝑥_6𝑛−2=(−1)^𝑛𝑥₋₂.(viii)If 𝑦₋₄=0,𝑥₋₁≠0, then we have 𝑦_6𝑛−4=0 and 𝑥_6𝑛−1=(−1)^𝑛𝑥₋₁.(ix)If 𝑦₋₃=0, 𝑥₀≠0, then we have 𝑦_6𝑛−3=0 and 𝑥_6𝑛=(−1)^𝑛𝑥₀.(x)If 𝑦₋₂=0, 𝑥₋₅≠0, then we have 𝑦_6𝑛−2=0 and 𝑥_6𝑛−5=(−1)^𝑛𝑥₋₅.(xi)If 𝑦₋₁=0, 𝑥₋₄≠0, then we have 𝑦_6𝑛−1=0 and 𝑥_6𝑛−4=(−1)^𝑛𝑥₋₄.(xii)If 𝑦₀=0, 𝑥₋₃≠0, then we have 𝑦_6𝑛=0 and 𝑥_6𝑛−3=(−1)^𝑛𝑥₋₃.

Proof. The proof follows from the form of the solution of system (5.1).

Example 5.3. If we take the system of difference equations (5.1) with the initial conditions𝑥₋₅=0.05,𝑥₋₄=−0.42,𝑥₋₃=.101,𝑥₋₂=0.07,𝑥₋₁=−0.4,𝑥₀=−3,𝑦₋₅=−0.7,𝑦₋₄=.12,𝑦₋₃=−1.2, 𝑦₋₂=0.2, 𝑦₋₁=−0.11, and 𝑦₀=0.13, we get the following shape of the solution, see Figure 5.

Figure 5 

Acknowledgments

This paper was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. The authors, therefore, acknowledge with thanks the DSR technical and financial support.