# Qualitative Behavior of Solutions of Tenth-Order Recursive Sequence Equation

Authors: E. M. Elsayed; B. S. Alofi; Abdul Qadeer Khan
Journal: Mathematical Problems in Engineering
 (2022)

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## Abstract

Most nonlinear difference equations have exact solutions that are not always possible to obtain theoretically. As a result, a large number of researchers investigate several qualitative aspects of difference equations in order to predict their lengthy behavior. The goal of our research is to obtain the solutions of a tenth-order difference equationUn+1=Un−9Un−5Un−1/Un−7Un−3±1±Un−9Un−5Un−1,n≥0, where the initial values are positive real numbers. Stability and periodicity are also investigated.


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## Body

## 1. Introduction

Solving the difference equation is one of the problems that is difficult to determine the solvability. The aim of this study is to solve difference equation of the tenth order and solve four specific cases of the following difference equation:(1)Un+1=Un−9Un−5Un−1Un−7Un−3±1±Un−9Un−5Un−1,n≥0,where the initial conditions U−9,U−8,U−7,U−6,U−5,U−4,U−3,U−2,U−1,U0 are the arbitrary positive real numbers. We also provide some properties of solutions such as periodicity in two cases and stability in the other two cases. Difference equations are used in a variety of probability problems such as hypergeometric, binomial, and poison distribution. Differential equations are related to difference equations in the same way that discrete mathematics and continuous mathematician are related. Difference equations are of importance to computer scientists for a variety of reasons. For example, when estimating the cost of an algorithm in big-O notation, converting a difficult differential problem to a nearly equivalent difference equation is the first step in solving. The study of asymptotic stability of nonlinear rational difference equations of high order is a difficult but rewarding task. It is particularly beneficial for analyzing the characteristics of mathematical models using different applications such as biological systems. The main topic in study is that the difference equations theory has been the asymptotic behavior of rational form of difference equation.In addition, various nonlinear trends in science and engineering can be modeled by this type of equation, and the solution of this type of equation provides a prototype for the development of theory [1]. In the literature, many applications theories’ differences equations have been investigated. El-Dessoky [2] investigated the behavior properties of the solutions of the rational difference equation:(2)Un+1=aUn+bUnUn−3cUn−4+dUn−3.Ghazela et al. [3] researched the analytic qualities of sixth-order difference equations:(3)Un+1=bUn−5cUn−2Un−5+d.Al-Matrafi and Al-Zubaidi [4] achieved global and local stability and forms of positive periodic solutions for two types of recursive equations:(4)Un+1=aUn−1±bUn−1Un−4cUn−4−dUn−6.Exploring some properties of the behavior of solutions appropriate to the class of recursive equation:(5)Un+1=aUn−1±bUn−1Un−3cUn−3−dUn−5,was the prime objective for Alayachi et al. in [5]. Sadiq and Kalim [6] studied solutions, equilibrium points, and periodicity of four types of difference equations:(6)Un+1=Un−20±1±Un−6Un−13Un−20.Elsayed in [7] was able to get the solutions to this difference questions:(7)Un+1=UnUn−2Un−4Un−1Un−3±1±UnUn−2Un−4.For more articles in this direction, we refer the reader to [8–14] and references cited therein.


## 2. The First Case:Un+1=Un−9Un−5Un−1/Un−7Un−31+Un−9Un−5Un−1

The aim of this section is studying the solutions form of the particular case:(8)Un+1=Un−9Un−5Un−1Un−7Un−31+Un−9Un−5Un−1,n=0,1,….Theorem 1.
Assume thatUnn=−9∞ are solutions of difference equations. Then, for n=0,1,2,…, we see that all solutions of equation (8) are given by the following formulas:(9)U12n−9=A∏k=0n−11+6kAEI∏k=0n−11+6k+2AEI,U12n−3=G∏k=0n−11+6k+3AEI∏k=0n−11+6k+5AEI,U12n−8=B∏k=0n−11+6kBFJ∏k=0n−11+6k+2BFJ,U12n−2=H∏k=0n−11+6k+3BFJ∏k=0n−11+6k+5BFJ,U12n−7=C∏k=0n−11+6k+1AEI∏k=0n−11+6k+3AEI,U12n−1=I∏k=0n−11+6k+4AEI∏k=0n−11+6k+6AEI,U12n−6=D∏k=0n−11+6k+1BFJ∏k=0n−11+6k+3BFJ,U12n=J∏k=0n−11+6k+4BFJ∏k=0n−11+6k+6BFJ,U12n−5=E∏k=0n−11+6k+2AEI∏k=0n−11+6k+4AEI,U12n+1=AEI∏k=0n−11+6k+5AEICG1+AEI∏k=0n−11+6k+7AEIU12n−4=F∏k=0n−11+6k+2BFJ∏k=0n−11+6k+4BFJ,U12n+2=BFJ∏k=0n−11+6k+5BFJDH1+BFJ∏k=0n−11+6k+7BFJ,where U−9=A, U−8=B, U−7=C, U−6=D, U−5=E, U−4=F, U−3=G, U−2=H, U−1=I, and U0=J.Proof.
Forn=1, the result holds. Now suppose that n>0 and that our assumption holds for n−1, that is,(10)U12n−21=A∏k=0n−21+6kAEI∏k=0n−21+6k+2AEI,U12n−15=G∏k=0n−21+6k+3AEI∏k=0n−21+6k+5AEI,U12n−20=B∏k=0n−21+6kBFJ∏k=0n−21+6k+2BFJ,U12n−14=H∏k=0n−21+6k+3BFJ∏k=0n−21+6k+5BFJ,U12n−19=C∏k=0n−21+6k+1AEI∏k=0n−21+6k+3AEI,U12n−13=I∏k=0n−21+6k+4AEI∏k=0n−21+6k+6AEI,U12n−18=D∏k=0n−21+6k+1BFJ∏k=0n−21+6k+3BFJ,U12n−12=J∏k=0n−21+6k+4BFJ∏k=0n−21+6k+6BFJ,U12n−17=E∏k=0n−21+6k+2AEI∏k=0n−21+6k+4AEI,U12n−11=AEI∏k=0n−21+6k+5AEICG1+AEI∏k=0n−21+6k+7AEI,U12n−16=F∏k=0n−21+6k+2BFJ∏k=0n−21+6k+4BFJ,U12n−10=BFJ∏k=0n−21+6k+5BFJDH1+BFJ∏k=0n−21+6k+7BFJ.
Now, we find from equation (8) that(11)U12n−9=U12n−19U12n−15U12n−11U12n−17U12n−131+U12n−19U12n−15U12n−11=C∏k=0n−21+6k+1AEI/∏k=0n−21+6k+3AEIG∏k=0n−21+6k+3AEI/C∏k=0n−21+6k+5AEIAEI∏k=0n−21+6k+5AEI/GC1+AEI∏k=0n−21+6k+7AEIE∏k=0n−21+6k+2AEI/∏k=0n−21+6k+4AEII∏k=0n−21+6k+4AEI/∏k=0n−21+6k+6AEI×11+C∏k=0n−21+6k+1AEI/∏k=0n−21+6k+3AEIG∏k=0n−21+6k+3AEI/∏k=0n−21+6k+5AEIAEI∏k=0n−21+6k+5AEI/CG1+AEI∏k=0n−21+6k+7AEI,U12n−9=AGCEI∏k=0n−21+6k+1AEI/GC1+AEI∏k=0n−21+6k+7AEIEI∏k=0n−21+6k+2AEI/∏k=0n−21+6k+6AEI1+AGCEI∏k=0n−21+6k+1AEI/GC1+AEI∏k=0n−21+6k+7AEI=A∏k=0n−21+6k+1AEI/1+AEI∏k=0n−21+6k+7AEI∏k=0n−21+6k+2AEI/∏k=0n−21+6k+6AEI1+AEI∏k=0n−21+6k+1AEI/1+AEI∏k=0n−21+6k+7AEI=A/1+6n−5AEI∏k=0n−21+6k+2AEI/∏k=0n−21+6k+6AEI1+AEI/1+6n−5AEI,U12n−9=A∏k=0n−21+6k+2AEI/∏k=0n−21+6k+6AEI1+6n−5AEI+AEI=A∏k=0n−21+6k+2AEI/∏k=0n−21+6k+6AEI1+6n−5AEI+AEI=A∏k=0n−11+6kAEI∏k=0n−11+6k+2AEI,U12n−8=U12n−18U12n−14U12n−10U12n−16U12n−121+U12n−18U12n−14U12n−10=D∏k=0n−21+6k+1BFJ/∏k=0n−21+6k+3BFJH∏k=0n−21+6k+3BFJ/∏k=0n−21+6k+5BFJBFJ∏k=0n−21+6k+5BFJ/DH1+BFJ∏k=0n−21+6k+7BFJF∏k=0n−21+6k+2BFJ/∏k=0n−21+6k+4BFJJ∏k=0n−21+6k+4BFJ/∏k=0n−21+6k+6BFJ×11+D∏k=0n−21+6k+1BFJ/∏k=0n−21+6k+3BFJH∏k=0n−21+6k+3BFJ/∏k=0n−21+6k+5BFJBFJ∏k=0n−21+6k+5BFJ/DH1+BFJ∏k=0n−21+6k+7BFJ=DHBFJ∏k=0n−21+6k+1BFJ/DH1+BFJ∏k=0n−21+6k+7BFJFJ∏k=0n−21+6k+2BFJ/∏k=0n−21+6k+6BFJ1+BFJ∏k=0n−21+6k+1BFJ/1+BFJ∏k=0n−21+6k+7BFJ=B∏k=0n−21+6k+1BFJ/1+BFJ∏k=0n−21+6k+7BFJ∏k=0n−21+6k+2BFJ/∏k=0n−21+6k+6BFJ1+BFJ∏k=0n−21+6k+1BFJ/1+BFJ∏k=0n−21+6k+7BFJ=B/1+6n−5BFJ∏k=0n−21+6k+2BFJ/∏k=0n−21+6k+6BFJ1+BFJ/1+6n−5BFJ,U12n−8=B∏k=0n−21+6k+2BFJ/∏k=0n−21+6k+6BFJ1+6n−5BFJ+BFJ=B∏k=0n−21+6k+2BFJ/∏k=0n−21+6k+6BFJ1+6n−5BFJ+BFJ=B∏k=0n−11+6kBFJ∏k=0n−11+6k+2BFJ,U12n−7=U12n−17U12n−13U12n−9U12n−15U12n−111+U12n−17U12n−13U12n−9=E∏k=0n−21+6k+2AEI/∏k=0n−21+6k+4AEII∏k=0n−21+6k+4AEI/∏k=0n−21+6k+6AEIA∏k=0n−11+6kAEI/∏k=0n−11+6k+2AEIG∏k=0n−21+6k+3AEI/∏k=0n−21+6k+5AEIAEI∏k=0n−21+6k+5AEI/CG1+AEI∏k=0n−21+6k+7AEI×11+E∏k=0n−21+6k+2AEI/∏k=0n−21+6k+4AEII∏k=0n−21+6k+4AEI/∏k=0n−21+6k+6AEIA∏k=0n−11+6kAEI/∏k=0n−11+6k+2AEI=∏k=0n−11+6kAEI/1+6n−4AEI∏k=0n−21+6k+6AEI∏k=0n−21+6k+3AEI/C1+AEI∏k=0n−21+6k+7AEI1+AEI∏k=0n−11+6kAEI/1+6n−4AEI∏k=0n−21+6k+6AEI=1/1+6n−4AEI∏k=0n−21+6k+3AEI/C1+AEI∏k=0n−21+6k+7AEI1+AEI/1+6n−4AEI=C1+AEI∏k=0n−21+6k+3AEI/∏k=0n−21+6k+7AEI1+6n−4AEI+AEI=C∏k=0n−11+6k+1AEI∏k=0n−11+6k+3AEI,U12n−6=U12n−16U12n−12U12n−8U12n−14U12n−101+U12n−16U12n−12U12n−8=F∏k=0n−21+6k+2BFJ/∏k=0n−21+6k+4BFJJ∏k=0n−21+6k+4BFJ/∏k=0n−21+6k+6BFJB∏k=0n−11+6kBFJ/∏k=0n−11+6k+2BFJH∏k=0n−21+6k+3BFJ/∏k=0n−21+6k+5BFJBFJ∏k=0n−21+6k+5BFJ/DH1+BFJ∏k=0n−21+6k+7BFJ×1/1+F∏k=0n−21+6k+2BFJ∏k=0n−21+6k+4BFJJ∏k=0n−21+6k+4BFJ∏k=0n−21+6k+6BFJB∏k=0n−11+6kBFJ/∏k=0n−11+6k+2BFJ=∏k=0n−11+6kBFJ/1+6n−4BFJ∏k=0n−21+6k+6BFJ∏k=0n−21+6k+3BFJ/D1+BFJ∏k=0n−21+6k+7BFJ1+BFJ∏k=0n−11+6kBFJ/1+6n−4BFJ∏k=0n−21+6k+6BFJ=1/1+6n−4BFJ∏k=0n−21+6k+3BFJ/D1+BFJ∏k=0n−21+6k+7BFJ1+BFJ/1+6n−4BFJ=C∏k=0n−11+6k+1BFJ∏k=0n−11+6k+3BFJ,
Also, we can prove the other relations. The proof is complete.Theorem 2.
Equation (8) has a unique equilibrium point U¯=0, which is nonhyperbolic.Proof.
To obtain equilibrium points of (8),(12)U¯=U¯3U¯21+U¯3.
Thus,(13)U¯31+U¯3=U¯3,U¯31+U¯3−1=0,U¯6=0.
Hence,U¯=0 is the equilibrium point of equation (8). Define a function h:0,∞5⟶0,∞, such that(14)hr,s,t,u,v=rstuv1+rst.
Then,(15)hrr,s,t,u,v=stuv1+rst2,hsr,s,t,u,v=rtuv1+rst2,htr,s,t,u,v=rsuv1+rst2,hur,s,t,u,v=−rstu2v1+rst,hvr,s,t,u,v=−rstuv21+rst.
Therefore,(16)hrU¯,U¯,U¯,U¯,U¯=1,hsr,s,t,u,v=1,htU¯,U¯,U¯,U¯,U¯=1,hur,s,t,u,v=−1,hvU¯,U¯,U¯,U¯,U¯=−1.
It follows the characteristic equation given by(17)λ10−λ9+λ8−λ6+λ4−λ2=0.
Hence,(18)λ8−λ7+λ6−λ4+λ2−1=0.
Clearly,λ=1 is one root of equation (17). Therefore, the equilibrium point is nonhyperbolic.
We provide numerical examples for equation (8) in order to confirm the results of this section.Figure 1
The solution of equation (8) when U−9=7, U−8=15, U−7=9, U−6=5, U−5=10, U−4=8, U−3=16, U−2=10, U−1=6, and U0=11.Figure 2
The local stability of equilibrium point of equation (8) when U−9=0.1, U−8=0.2, U−7=0.3, U−6=0.4, U−5=0.5, U−4=0.1, U−3=0.2, U−2=0.3, U−1=0.4, and U0=0.5.Example 1.
Assume the initial conditions areU−9=7, U−8=15, U−7=9, U−6=5, U−5=10, U−4=8, U−3=16, U−2=10, U−1=6, and U0=11 (Figure 1).Example 2.
Suppose thatU−9=0.1, U−8=0.2, U−7=0.3, U−6=0.4, U−5=0.5, U−4=0.1, U−3=0.2, U−2=0.3, U−1=0.4, and U0=0.5 (Figure 2).


## 3. Second Case:Un+1=Un−9Un−5Un−1/Un−7Un−31−Un−9Un−5Un−1

The solutions to difference equations(19)Un+1=Un−9Un−5Un−1Un−7Un−31−Un−9Un−5Un−1,n=0,1,…,are investigated in this section.Theorem 3.
Assume thatUnn=−9∞ are solutions of difference equations. Then, for n=0,1,2,…, we see that all solutions of equation (19) are given by the following formulas:(20)U12n−9=A∏k=0n−11−6kAEI∏k=0n−11−6k+2AEI,U12n−3=G∏k=0n−11−6k+3AEI∏k=0n−11−6k+5AEI,U12n−8=B∏k=0n−11−6kBFJ∏k=0n−11−6k+2BFJ,U12n−2=H∏k=0n−11−6k+3BFJ∏k=0n−11−6k+5BFJ,U12n−7=C∏k=0n−11−6k+1AEI∏k=0n−11−6k+3AEI,U12n−1=I∏k=0n−11−6k+4AEI∏k=0n−11−6k+6AEI,U12n−6=D∏k=0n−11−6k+3BFJ∏k=0n−11−6k+1BFJ,U12n=J∏k=0n−11−6k+4BFJ∏k=0n−11−6k+4BFJ,U12n−5=E∏k=0n−11−6k+2AEI∏k=0n−11−6k+4AEI,U12n+1=AEI∏k=0n−11−6k+5AEI1−AEI∏k=0n−11−6k+7AEI,U12n−4=F∏k=0n−11−6k+2BFJ∏k=0n−11−6k+4BFJ,U12n+2=BFJ∏k=0n−11−6k+5BFJ1−BFJ∏k=0n−11−6k+7BFJ,where U−9=A, U−8=B, U−7=C, U−6=D, U−5=E, U−4=F, U−3=G, U−2=H, U−1=I, and U0=J.Proof.
The proof is identical to the method to prove Theorem1.Theorem 4.
Equation (19) has a unique equilibrium point U¯=0, which is nonhyperbolic.Proof.
To obtain equilibrium points of equation (19),(21)U¯=U¯3U¯21−U¯3.
Thus,(22)U¯31−U¯3=U¯3,U¯31−U¯3−1=0,U¯6=0.
Hence,U¯=0 is the equilibrium point of equation (19). Define a function h:0,∞5⟶0,∞, such that(23)hr,s,t,u,v=rstuv1−rst.
Then,(24)hrr,s,t,u,v=stuv1−rst2,hsr,s,t,u,v=rtuv1−rst2,htr,s,t,u,v=rsuv1−rst2,hur,s,t,u,v=−rstu2v1−rst,hvr,s,t,u,v=−rstuv21−rst.
Therefore,(25)hrU¯,U¯,U¯,U¯,U¯=1,hsr,s,t,u,v=1,htU¯,U¯,U¯,U¯,U¯=1,hur,s,t,u,v=−1,hvU¯,U¯,U¯,U¯,U¯=−1.
It follows the characteristic equation given by(26)λ10−λ9+λ8−λ6+λ4−λ2=0.
Hence,(27)λ8−λ7+λ6−λ4+λ2−1=0.
Clearly,λ=1 is one root of equation (26). Therefore, the equilibrium point is nonhyperbolic.Figure 3
The solution of equation (19) when U−9=4, U−8=12, U−7=6, U−6=2, U−5=8, U−4=3, U−3=11, U−2=5, U−1=2, and U0=7.Figure 4
The local stability of equilibrium point of equation (19) with U−9=0.1, U−8=0.2, U−7=0.3, U−6=0.4, U−5=0.5, U−4=0.1, U−3=0.2, U−2=0.3, U−1=0.4, and U0=0.5.Example 3.
We consider the present numerical example for equation (19) for confirming the results of this section where the initial conditions are U−9=4, U−8=12, U−7=6, U−6=2, U−5=8, U−4=3, U−3=11, U−2=5, U−1=2, and U0=7 (Figure 3).Example 4.
We provide another numerical example for equation (19) with initial values U−9=0.1, U−8=0.2, U−7=0.3, U−6=0.4, U−5=0.5, U−4=0.1, U−3=0.2, U−2=0.3, U−1=0.4, and U0=0.5 (Figure 4).


## 4. Third Case:Un+1=Un−9Un−5Un−1/Un−7Un−3−1−Un−9Un−5Un−1

The goal of this section is to obtain the solutions form of the particular case:(28)Un+1=Un−9Un−5Un−1Un−7Un−3−1−Un−9Un−5Un−1,n=0,1,….Theorem 5.
Every solutionUnn=−9∞ of equation (28) is periodic with period twelve, and it is in the form(29)A,B,C,D,E,F,G,H,I,J,−AEICGAEI+1,−BFJDHBFJ+1,…or(30)U12n−9=A,U12n−8=B,U12n−7=C,U12n−6=D,U12n−5=E,U12n−4=F,U12n−3=G,U12n−2=H,U12n−1=I,U12n=J,U12n+1=−AEICGAEI+1,U12n+2=−BFJDHBFJ+1,where U−9=A, U−8=B, U−7=C, U−6=D, U−5=E, U−4=F, U−3=G, U−2=H, U−1=I, and U0=J.Proof.
Fromn=1,(31)U12n−21=A,U12n−20=B,U12n−19=C,U12n−18=D,U12n−17=E,U12n−16=F,U12n−15=G,U12n−14=H,U12n−13=I,U12n−12=J,U12n−11=−AEICGAEI+1,U12n−10=−BFJDHBFJ+1.
From equation (28), we see that(32)U12n−9=U12n−19U12n−15U12n−11U12n−17U12n−13−1−U12n−19U12n−15U12n−11=CG−AEI/CGAEI+1EI−1−CG−AEI/CGAEI+1=A,U12n−8=U12n−18U12n−14U12n−10U12n−16U12n−12−1−U12n−18U12n−14U12n−10=DH−BFJ/DHBFJ+1FJ−1−DH−BFJ/DHBFJ+1=B,U12n−7=U12n−17U12n−13U12n−9U12n−15U12n−11−1−U12n−17U12n−13U12n−9=AEIG−AEI/CGAEI+1−1−AEI=C,U12n−6=U12n−16U12n−12U12n−8U12n−14U12n−10−1−U12n−16U12n−12U12n−8=BFJH−BFJ/DHBFJ+1−1−BFJ=D,U12n−5=U12n−15U12n−11U12n−7U12n−13U12n−9−1−U12n−15U12n−11U12n−7=G−AEI/CGAEI+1CIA−1−G−AEI/CGAEI+1C=E,U12n−4=U12n−14U12n−10U12n−6U12n−12U12n−8−1−U12n−14U12n−10U12n−6=H−BFJ/DHBFJ+1DJB−1−H−BFJ/DHBFJ+1D=F,U12n−3=U12n−13U12n−9U12n−5U12n−11U12n−7−1−U12n−13U12n−9U12n−5=IAE−AEI/CGAEI+1C−1−IAE=G,U12n−2=U12n−12U12n−8U12n−4U12n−10U12n−6−1−U12n−12U12n−8U12n−4=JBF−BFJ/DHBFJ+1D−1−JBF=H.Figure 5
The solution of equation (28) with period twelve when U−9=8, U−8=16, U−7=10, U−6=6, U−5=11, U−4=9, U−3=17, U−2=11, U−1=7, and U0=12.Figure 6
The solution of equation (28) with period six when initial values satisfies the conditions in Theorem 6, that are U−9=1, U−8=2, U−7=3, U−6=4, U−5=−0.6667, U−4=−0.25, U−3=1, U−2=2, U−1=3, and U0=4.Theorem 6.
Every solutionUnn=−9∞ of equation (28) is periodic with period six, and it is of the form(33)A,B,C,D,E,F,A,…,iff(34)A=G,B=H,C=I,D=J,AEI=−2,BFJ=−2.Example 5.
We present numerical example for equation (28) for illustrating the results of this section where the initial conditions are U−9=8, U−8=16, U−7=10, U−6=6, U−5=11, U−4=9, U−3=17, U−2=11, U−1=7, and U0=12 (Figure 5).Example 6.
For confirming the results of this section, we consider numerical example for equation (28) where the initial conditions are U−9=1, U−8=2, U−7=3, U−6=4, U−5=−0.6667, U−4=−0.25, U−3=1, U−2=2, U−1=3, and U0=4 (Figure 6).


## 5. Fourth Case:Un+1=Un−9Un−5Un−1/Un−7Un−3−1+Un−9Un−5Un−1

The solutions to difference equations(35)Un+1=Un−9Un−5Un−1Un−7Un−3−1+Un−9Un−5Un−1,n=0,1,…,are studied in this section.Theorem 7.
Every solutionUnn=−9∞ of equation (35) is periodic with period twelve, and it is in the form(36)A,B,C,D,E,F,G,H,I,J,+AEICGAEI−1,+BFJDHBFJ−1,A,B,…,where U−9=A, U−8=B, U−7=C, U−6=D, U−5=E, U−4=F, U−3=G, U−2=H, U−1=I, and U0=J.Proof.
The proof is identical to the method to prove Theorem5.Theorem 8.
Every solutionUnn=−9∞ of equation (35) is periodic with period six, and it is in the form(37)A,B,C,D,E,F,A,….iff(38)A=G,B=H,C=I,D=J,AEI=2,BFJ=2.
We provide numerical examples of equation (35) for confirming our results.Figure 7
The solution of equation (35) with period twelve when U−9=8, U−8=16, U−7=10, U−6=6, U−5=11, U−4=9, U−3=17, U−2=11, U−1=7, and U0=12.Figure 8
The solution of equation (35) with period six when initial values satisfies the conditions in Theorem 8, that are U−9=1, U−8=2, U−7=3, U−6=4, U−5=0.6667, U−4=0.25, U−3=1, U−2=2, U−1=3, and U0=4.Example 7.
Assume that the starting conditions are as follows:U−9=8, U−8=16, U−7=10, U−6=6, U−5=11, U−4=9, U−3=17, U−2=11, U−1=7, and U0=12 (Figure 7).Example 8.
Let the initial conditions be given byU−9=1, U−8=2, U−7=3, U−6=4, U−5=0.6667, U−4=0.25, U−3=1, U−2=2, U−1=3, and U0=4 (Figure 8).


## 6. Conclusion

This research discussed the structure and behavior of solutions for four special cases of equation (1). In the second and third sections, we proved the stability of the equilibrium point. In the fourth and fifth sections, we obtained the periodic solutions to the equations with periodicity twelve. In addition, we studied the conditions of existence of the periodic solutions with period six.



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