# Solution Expressions of Discrete Systems of Difference Equations

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

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## Abstract

In this paper, we obtain the solution forms of fifth order systems of rational difference equationsPn+1=Pn4Sn2Qn/Sn−3Qn−11±Pn−4Sn−2Qn,Qn+1=Qn−4Pn−2Sn/Pn−3Sn−11±Qn−4Pn−2Sn, and Sn+1=Sn−4Qn−2Pn/Qn−3Pn−11±Sn−4Qn−2Pn. Where the initial values are nonzero real numbers. Numerical examples are also provided.


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## Body

## 1. Introduction

The applications of discrete dynamical systems and difference equations have grown rapidly, especially in economics and modern sciences. In [1], El-Metwally et al. investigated the asymptotic behavior of the population model:(1)Pn+1=bPn−1ePn+a.Elsayed et al. [2] studied the discrete population model(2)Pn+1=α0Pn−1+α1Pn−1Pn−4α2Pn−4+α3Pn−2.In economy, Askar [3] studied the discrete dynamical systems of duopoly games(3)Pn+1=Pn+kPna+Qna+q2−c1,Qn+1=Pn+ac2−Pn−a.Many of authors investigated the behavior of nonlinear systems of difference equations when the solutions are difficult to obtain. For example, Khaliq and Zubair in [4] studied properties of solutions such as persistence, boundedness, global stability, and locally asymptotically stable of equilibrium point(4)Pn+1=β1+α1e−Qnγ1+δ1Pn,Qn+1=β2+α2e−Snγ2+δ2Qn,Sn+1=β1+α3e−Pnγ3+δ3Sn.It is very interesting to get solutions of systems of nonlinear difference equations if possible. Other authors were able to obtain the solutions of difference equations. For instance, Alayachi et al. in [5] studied the formula of solutions of the following difference equations system of order three(5)Pn+1=Sn−1QnSn−1+Qn−2,Qn+1=Pn−1SnSn±Pn−2,Sn+1=Qn−1PnPn±Qn−2.Elsayed et al. [6] obtained the solution expressions and studied the behavior of three-dimensional systems of difference equations(6)Pn+1=a1Sn−1Qn−1Sn−1+Qn−1+Pn−1,Qn+1=a2Pn−1Sn−1Sn−1+Qn−1+Pn−1,Sn+1=a3Qn−1Pn−1Sn−1+Qn−1+Pn−1.See other related results on difference equations [7–17].Through such a paper, we will get the solution form for six cases of the fifth order systems given by following form:(7)Pn+1=Pn−4Sn−2QnSn−3Qn−11±Pn−4Sn−2Qn,Qn+1=Qn−4Pn−2SnPn−3Sn−11±Qn−4Pn−2Sn,Sn+1=Sn−4Qn−2PnQn−3Pn−11±Sn−4Qn−2Pn,where the initial conditions P−4,P−3,P−2,P−1,P0,Q−4,Q−3,Q−2,Q−1,Q0,S−4,S−3,S−2,S−1,S0 are any real number. Also, some numerical examples and figures will be presented to confirm the results of the obtained solutions.


## 2. The First Case

Through this section, we obtain the solution formula of system of the three difference equations(8)Pn+1=Pn−4Sn−2QnSn−3Qn−11+Pn−4Sn−2Qn,Qn+1=Qn−4Pn−2SnPn−3Sn−11+Qn−4Pn−2Sn,Sn+1=Sn−4Qn−2PnQn−3Pn−11+Sn−4Qn−2Pn.Theorem 1.
Assume thatPn,Qn,Sn are solutions of difference equations system. Then, for n=0,1,2,…, we see that all solutions of system (8) are given by the following formulas:(9)P6n+1=P−4n+1Q0n+1S−2n+1∏k=0n−11+6k+5P0Q−2S−4Q−1S−3P0nQ−2nS−4n∏k=0n1+6k+1P−4Q0S−2,P6n+2=P−2n+1Q−4n+1S0n+1∏k=0n−11+6k+6P−4Q0S−2P−4nQ0n+1S−2n+1∏k=0n1+6k+2P−2Q−4S0,P6n+3=P−3P0n+1Q−2n+1S−4n+1∏k=0n1+6k+1P−2Q−4S0P−2n+1Q−4n+1S0n+1∏k=0n1+6k+3P0Q−2S−4,P6n+4=P−2P−4n+1Q0n+1S−2n+1∏k=0n1+6k+2P0Q−2S−4P0n+1Q−2n+1S−4n+1∏k=0n1+6k+4P−4Q0S−2,P6n+5=P−1P−2n+1Q−4n+1S0n+1∏k=0n1+6k+3P−4Q0S−2P−4n+1Q0n+1S−2n+1∏k=0n1+6k+5P−2Q−4S−0,P6n+6=P0n+2Q−2n+1S−4n+1∏k=0n1+6k+4P−2Q−4S0P−2n+1Q−4n+1S0n+1∏k=0n1+6k+6P0Q−2S−4,Q6n+1=Q−4n+1S0n+1P−2n+1∏k=0n−11+6k+5Q0S−2P−4S−1P−3Q0nS−2nP−4n∏k=0n1+6k+1Q−4S0P−2,Q6n+2=Q−2n+1S−4n+1P0n+1∏k=0n−11+6k+6Q−4S0P−2Q−4nS0n+1P−2n+1∏k=0n1+6k+2Q−2S−4P0Q6n+3=Q−3Q0n+1S−2n+1P−4n+1∏k=0n1+6k+1Q−2S−4P0Q−2n+1S−4n+1P0n+1∏k=0n1+6k+3Q0S−2P−4,Q6n+4=Q−2Q−4n+1S0n+1P−2n+1∏k=0n1+6k+2Q0S−2P−4Q0n+1S−2n+1P−4n+1∏k=0n1+6k+4Q−4S0P−2,Q6n+5=Q−1Q−2n+1S−4n+1P0n+1∏k=0n1+6k+3Q−4S0P−2Q−4n+1S0n+1P−2n+1∏k=0n1+6k+5Q−2S−4P−0,Q6n+6=Q0n+2S−2n+1P−4n+1∏k=0n1+6k+4Q−2S−4P0Q−2n+1S−4n+1P0n+1∏k=0n1+6k+6Q0S−2P−4,S6n+1=S−4n+1P0n+1Q−2n+1∏k=0n−11+6k+5S0P−2Q−4P−1Q−3S0nP−2nQ−4n∏k=0n1+6k+1S−4P0Q−2,S6n+2=S−2n+1P−4n+1Q0n+1∏k=0n−11+6k+6S−4P0Q−2S−4nP0n+1Q−2n+1∏k=0n1+6k+2S−2P−4Q0,S6n+3=S−3S0n+1P−2n+1Q−4n+1∏k=0n1+6k+1S−2P−4Q0S−2n+1P−4n+1Q0n+1∏k=0n1+6k+3S0P−2Q−4,S6n+4=S−2S−4n+1P0n+1Q−2n+1∏k=0n1+6k+2S0P−2Q−4S0n+1P−2n+1Q−4n+1∏k=0n1+6k+4S−4P0Q−2,S6n+5=S−1S−2n+1P−4n+1Q0n+1∏k=0n1+6k+3S−4P0Q−2S−4n+1P0n+1Q−2n+1∏k=0n1+6k+5S−2P−4Q−0,S6n+6=S0n+2P−2n+1Q−4n+1∏k=0n1+6k+4S−2P−4Q0S−2n+1P−4n+1Q0n+1∏k=0n1+6k+6S0P−2Q−4.Proof.
Forn=1 the result holds. Now suppose that n>0 and that our assumption holds for n−1, that is,(10)P6n−5=P−4nQ0nS−2n∏k=0n−21+6k+5P0Q−2S−4Q−1S−3P0n−1Q−2n−1S−4n−1∏k=0n−11+6k+1P−4Q0S−2,P6n−4=P−2nQ−4nS0n∏k=0n−21+6k+6P−4Q0S−2P−4n−1Q0nS−2n∏k=0n−11+6k+2P−2Q−4S0,P6n−3=P−3P0nQ−2nS−4n∏k=0n−11+6k+1P−2Q−4S0P−2nQ−4nS0n∏k=0n−11+6k+3P0Q−2S−4,P6n−2=P−2P−4nQ0nS−2n∏k=0n−11+6k+2P0Q−2S−4P0nQ−2nS−4n∏k=0n−11+6k+4P−4Q0S−2,P6n−1=P−1P−2nQ−4nS0n∏k=0n−11+6k+3P−4Q0S−2P−4nQ0nS−2n∏k=0n−11+6k+5P−2Q−4S−0,P6n=P0n+1Q−2nS−4n∏k=0n−11+6k+4P−2Q−4S0P−2nQ−4nS0n∏k=0n−11+6k+6P0Q−2S−4,Q6n−5=Q−4nS0nP−2n∏k=0n−21+6k+5Q0S−2P−4S−1P−3Q0n−1S−2n−1P−4n−1∏k=0n1+6k+1Q−4S0P−2,Q6n−4=Q−2nS−4nP0n∏k=0n−21+6k+6Q−4S0P−2Q−4n−1S0nP−2n∏k=0n−11+6k+2Q−2S−4P0,Q6n−3=Q−3Q0nS−2nP−4n∏k=0n−11+6k+1Q−2S−4P0Q−2nS−4nP0n∏k=0n−11+6k+3Q0S−2P−4,Q6n−2=Q−2Q−4nS0nP−2n∏k=0n−11+6k+2Q0S−2P−4Q0nS−2nP−4n∏k=0n−11+6k+4Q−4S0P−2,Q6n−1=Q−1Q−2nS−4nP0n∏k=0n−11+6k+3Q−4S0P−2Q−4nS0nP−2n∏k=0n−11+6k+5Q−2S−4P−0,Q6n=Q0n+1S−2nP−4n∏k=0n−11+6k+4Q−2S−4P0Q−2nS−4nP0n∏k=0n−11+6k+6Q0S−2P−4,S6n−5=S−4nP0nQ−2n∏k=0n−21+6k+5S0P−2Q−4P−1Q−3S0n−1P−2n−1Q−4n−1∏k=0n−11+6k+1S−4P0Q−2,S6n−4=S−2nP−4nQ0n∏k=0n−21+6k+6S−4P0Q−2S−4n−1P0nQ−2n∏k=0n−11+6k+2S−2P−4Q0,S6n−3=S−3S0nP−2nQ−4n∏k=0n−11+6k+1S−2P−4Q0S−2nP−4nQ0n∏k=0n−11+6k+3S0P−2Q−4,S6n−2=S−2S−4nP0nQ−2n∏k=0n−11+6k+2S0P−2Q−4S0nP−2nQ−4n∏k=0n−11+6k+4S−4P0Q−2,Q6n−1=Q−1Q−2nS−4nP0n∏k=0n−11+6k+3Q−4S0P−2Q−4nS0nP−2n∏k=0n−11+6k+5Q−2S−4P−0,S6n=S0n+1P−2nQ−4n∏k=0n−11+6k+4S−2P−4Q0S−2nP−4nQ0n∏k=0n−11+6k+6S0P−2Q−4.
Now, we find from system (8) that(11)P6n+1=P6n−4S6n−2Q6nS6n−3Q6n−11+P6n−4S6n−2Q6n,=P−2nQ−4nS0n∏k=0n−21+6k+6P−4Q0S−2/P−4n−1Q0nS−2n∏k=0n−11+6k+2P−2Q−4S0S−2S−4nP0nQ−2n∏k=0n−11+6k+2S0P−2Q−4/S0nP−2nQ−4n∏k=0n−11+6k+4S−4P0Q−2,Q0n+1S−2nP−4n∏k=0n−11+6k+4Q−2S−4P0/Q−2nS−4nP0n∏k=0n−11+6k+6Q0S−2P−4S−3S0nP−2nQ−4n∏k=0n−11+6k+1S−2P−4Q0/S−2nP−4nQ0n∏k=0n−11+6k+3S0P−2Q−4Q−1Q−2nS−4nP0n∏k=0n−11+6k+3Q−4S0P−2/Q−4nS0nP−2n∏k=0n−11+6k+5Q−2S−4P−01+P−2nQ−4nS0n∏k=0n−21+6k+6P−4Q0S−2/P−4n−1Q0nS−2n∏k=0n−11+6k+2P−2Q−4S0S−2S−4nP0nQ−2n∏k=0n−11+6k+2S0P−2Q−4/S0nP−2nQ−4n∏k=0n−11+6k+4S−4P0Q−2Q0n+1S−2nP−4n∏k=0n−11+6k+4Q−2S−4P0/Q−2nS−4nP0n∏k=0n−11+6k+6Q0S−2P−4..=S−2Q0P−4/1+6nQ0S−2P−4S−3Q−1Q−2nS−4nP0n∏k=0n−11+6k+1S−2P−4Q01+6nQ0S−2P−4+Q0S−2P−4/S−2nP−4nQ0n∏k=0n−11+6k+5Q−2S−4P−01+6nQ0S−2P−4,=P−4n+1Q0n+1S−2n+1∏k=0n−11+6k+5P0Q−2S−4Q−1S−3P0nQ−2nS−4n∏k=0n1+6k+1P−4Q0S−2,Q6n+1=Q6n−4P6n−2S6nP6n−3S6n−11+Q6n−4P6n−2S6n,=Q−2nS−4nP0n∏k=0n−21+6k+6Q−4S0P−2/Q−4n−1S0nP−2n∏k=0n−11+6k+2Q−2S−4P0P−2P−4nQ0nS−2n∏k=0n−11+6k+2P0Q−2S−4/P0nQ−2nS−4n∏k=0n−11+6k+4P−4Q0S−2S0n+1P−2nQ−4n∏k=0n−11+6k+4S−2P−4Q0/S−2nP−4nQ0n∏k=0n−11+6k+6S0P−2Q−4P−3P0nQ−2nS−4n∏k=0n−11+6k+1P−2Q−4S0/P−2nQ−4nS0n∏k=0n−11+6k+3P0Q−2S−4S−1S−2nP−4nQ0n∏k=0n−11+6k+3S−4P0Q−2/S−4nP0nQ−2n∏k=0n−11+6k+5S−2P−4Q−01+Q−2nS−4nP0n∏k=0n−21+6k+6Q−4S0P−2/Q−4n−1S0nP−2n∏k=0n−11+6k+2Q−2S−4P0P−2P−4nQ0nS−2n∏k=0n−11+6k+2P0Q−2S−4/P0nQ−2nS−4n∏k=0n−11+6k+4P−4Q0S−2S0n+1P−2nQ−4n∏k=0n−11+6k+4S−2P−4Q0/S−2nP−4nQ0n∏k=0n−11+6k+6S0P−2Q−4,=P−2S0Q−4/1+6nS0P−2Q−4P−3S−1S−2nP−4nQ0n∏k=0n−11+6k+1P−2Q−4S01+6nS0P−2Q−4+S0P−2Q−4/P−2nQ−4nS0n∏k=0n−11+6k+5S−2P−4Q−01+6nS0P−2Q−4,=Q−4n+1S0n+1P−2n+1∏k=0n−11+6k+5Q0S−2P−4S−1P−3Q0nS−2nP−4n∏k=0n1+6k+1Q−4S0P−2,S6n+1=S6n−4Q6n−2P6nQ6n−3P6n−11+S6n−4Q6n−2P6n,=S−2nP−4nQ0n∏k=0n−21+6k+6S−4P0S]Q−2/S−4n−1P0nQ−2n∏k=0n−11+6k+2S−2P−4Q0S−2Q−4nS0nP−2n∏k=0n−11+6k+2Q0S−2P−4/Q0nS−2nP−4n∏k=0n−11+6k+4Q−4S0P−2P0n+1Q−2nS−4n∏k=0n−11+6k+4P−2Q−4S0/P−2nQ−4nS0n∏k=0n−11+6k+6P0Q−2S−4Q−3Q0nS−2nP−4n∏k=0n−11+6k+1Q−2S−4P0/Q−2nS−4nP0n∏k=0n−11+6k+3Q0S−2P−4P−1P−2nQ−4nS0n∏k=0n−11+6k+3P−4Q0S−2/P−4nQ0nS−2n∏k=0n−11+6k+5P−2Q−4S−01+S−2nP−4nQ0n∏k=0n−21+6k+6S−4P0S]Q−2/S−4n−1P0nQ−2n∏k=0n−11+6k+2S−2P−4Q0S−2Q−4nS0nP−2n∏k=0n−11+6k+2Q0S−2P−4/Q0nS−2nP−4n∏k=0n−11+6k+4Q−4S0P−2P0n+1Q−2nS−4n∏k=0n−11+6k+4P−2Q−4S0/P−2nQ−4nS0n∏k=0n−11+6k+6P0Q−2S−4,=Q−2P0S−4/1+6nP0S−2S−4Q−3P−1P−2nQ−4nS0n∏k=0n−11+6k+1S−2S−4P01+6nP0S−2S−4+P0S−2S−4/Q−2nS−4nP0n∏k=0n−11+6k+5P−2S−4S−01+6nP0S−2S−4,=S−4n+1P0n+1Q−2n+1∏k=0n−11+6k+5S0P−2Q−4P−1Q−3S0nP−2nQ−4n∏k=0n1+6k+1S−4P0Q−2.Example 1.
Figure1 shows numerical solution of system (8) with the initial conditions P−4=5,P−3=13,P−2=7,P−1=3,P0=19,Q−4=13,Q−3=5,Q−2=7,Q−1=3,Q0=10,S−4=10,S−3=12,S−2=10,S−1=6,S0=11.Figure 1
Sketch the numerical solution of system (8).Example 2.
This illustrates the numerical solution of system (8) with the initial conditions P−4=9,P−3=17,P−2=11,P−1=4,P0=16,Q−4=9,Q−3=1,Q−2=3,Q−1=1,Q0=6,S−4=6,S−3=7,S−2=8,S−1=10,S0=5 (See Figure 2).Figure 2
Sketch the numerical solution of system (8).


## 3. Second Case

Through this section, we obtain the solution form of the system of three difference equations given by the following:(12)Pn+1=Pn−4Sn−2QnSn−3Qn−11−Pn−4Sn−2Qn,Qn+1=Qn−4Pn−2SnPn−3Sn−11+Qn−4Pn−2Sn,Sn+1=Sn−4Qn−2PnQn−3Pn−11−Sn−4Qn−2Pn.Theorem 2.
Assume thatPn,Qn,Sn are solutions of difference equation system. Then for n=0,1,2,…, we see that all solutions of system (12) are given by the following formulas:(13)P6n+1=P−4n+1Q0n+1S−2n+1∏k=0n−11−6k+5P0Q−2S−4Q−1S−3P0nQ−2nS−4n∏k=0n1−6k+1P−4Q0S−2,P6n+2=P−2n+1Q−4n+1S0n+1∏k=0n−11−6k+6P−4Q0S−2P−4nQ0n+1S−2n+1∏k=0n1−6k+2P−2Q−4S0,P6n+3=P−3P0n+1Q−2n+1S−4n+1∏k=0n1−6k+1P−2Q−4S0P−2n+1Q−4n+1S0n+1∏k=0n1−6k+3P0Q−2S−4,P6n+4=P−2P−4n+1Q0n+1S−2n+1∏k=0n1−6k+2P0Q−2S−4P0n+1Q−2n+1S−4n+1∏k=0n1−6k+4P−4Q0S−2,P6n+5=P−1P−2n+1Q−4n+1S0n+1∏k=0n1−6k+3P−4Q0S−2P−4n+1Q0n+1S−2n+1∏k=0n1−6k+5P−2Q−4S−0,P6n+6=P0n+2Q−2n+1S−4n+1∏k=0n1−6k+4P−2Q−4S0P−2n+1Q−4n+1S0n+1∏k=0n1−6k+6P0Q−2S−4,Q6n+1=Q−4n+1S0n+1P−2n+1∏k=0n−11−6k+5Q0S−2P−4S−1P−3Q0nS−2nP−4n∏k=0n1−6k+1Q−4S0P−2,Q6n+2=Q−2n+1S−4n+1P0n+1∏k=0n−11−6k+6Q−4S0P−2Q−4nS0n+1P−2n+1∏k=0n1−6k+2Q−2S−4P0,Q6n+3=Q−3Q0n+1S−2n+1P−4n+1∏k=0n1−6k+1Q−2S−4P0Q−2n+1S−4n+1P0n+1∏k=0n1−6k+3Q0S−2P−4,Q6n+4=Q−2Q−4n+1S0n+1P−2n+1∏k=0n1−6k+2Q0S−2P−4Q0n+1S−2n+1P−4n+1∏k=0n1−6k+4Q−4S0P−2,Q6n+5=Q−1Q−2n+1S−4n+1P0n+1∏k=0n1−6k+3Q−4S0P−2Q−4n+1S0n+1P−2n+1∏k=0n1−6k+5Q−2S−4P−0,Q6n+6=Q0n+2S−2n+1P−4n+1∏k=0n1−6k+4Q−2S−4P0Q−2n+1S−4n+1P0n+1∏k=0n1−6k+6Q0S−2P−4,S6n+1=S−4n+1P0n+1Q−2n+1∏k=0n−11−6k+5S0P−2Q−4P−1Q−3S0nP−2nQ−4n∏k=0n1−6k+1S−4P0Q−2,S6n+2=S−2n+1P−4n+1Q0n+1∏k=0n−11−6k+6S−4P0Q−2S−4nP0n+1Q−2n+1∏k=0n1−6k+2S−2P−4Q0,S6n+3=S−3S0n+1P−2n+1Q−4n+1∏k=0n1−6k+1S−2P−4Q0S−2n+1P−4n+1Q0n+1∏k=0n1−6k+3S0P−2Q−4,S6n+4=S−2S−4n+1P0n+1Q−2n+1∏k=0n1−6k+2S0P−2Q−4S0n+1P−2n+1Q−4n+1∏k=0n1−6k+4S−4P0Q−2,S6n+5=S−1S−2n+1P−4n+1Q0n+1∏k=0n1−6k+3S−4P0Q−2S−4n+1P0n+1Q−2n+1∏k=0n1−6k+5S−2P−4Q−0,S6n+6=S0n+2P−2n+1Q−4n+1∏k=0n1−6k+4S−2P−4Q0S−2n+1P−4n+1Q0n+1∏k=0n1−6k+6S0P−2Q−4.Proof.
The proof is derived from the proof of the previous theorem.Example 3.
Figure3 shows numerical solution of system (12) with the initial conditions P−4=11,P−3=19,P−2=13,P−1=6,P0=18,Q−4=11,Q−3=3,Q−2=5,Q−1=3,Q0=8,S−4=8,S−3=9,S−2=10,S−1=12,S0=7.Figure 3
Drawing the numerical solution of system (12).Example 4.
This illustrates the numerical solution of system (12) with the initial conditions P−4=4,P−3=12,P−2=5,P−1=6,P0=14,Q−4=11,Q−3=6,Q−2=12,Q−1=2,Q0=9,S−4=9,S−3=3,S−2=9,S−1=11,S0=2 (See Figure 4).Figure 4
Drawing the numerical solution of system (12).


## 4. Third Case

In this section, we investigate the solutions of the difference equation system(14)Pn+1=Pn−4Sn−2QnSn−3Qn−1−1+Pn−4Sn−2Qn,Qn+1=Qn−4Pn−2SnPn−3Sn−1−1+Qn−4Pn−2Sn,Sn+1=Sn−4Qn−2PnQn−3Pn−1−1+Sn−4Qn−2Pn.Theorem 3.
Assume thatPn,Qn,Sn are solutions of difference equation system. Then for n=0,1,2,…, we see that all solutions of system (14) are given by the following formulas:(15)P6n+1=P−4n+1Q0n+1S−2n+1P0Q−2S−4−1nQ−1S−3P0nQ−2nS−4nP−4Q0S−2−1n+1,P6n+2=P−2n+1Q−4n+1S0n+1P−4nQ0n+1S−2n+1,P6n+3=P−3P0n+1Q−2n+1S−4n+1P−2Q−4S0−1n+1P−2n+1Q−4n+1S0n+1P0Q−2S−4−1n+1,P6n+4=P−2P−4n+1Q0n+1S−2n+1P0n+1Q−2n+1S−4n+1,P6n+5=P−1P−2n+1Q−4n+1S0n+1P−4Q0S−2−1n+1P−4n+1Q0n+1S−2n+1P−2Q−4S−0−1n+1,P6n+6=P0n+2Q−2n+1S−4n+1P−2n+1Q−4n+1S0n+1,Q6n+1=Q−4n+1S0n+1P−2n+1Q0S−2P−4−1nS−1P−3Q0nS−2nP−4nQ−4S0P−2−1n+1,Q6n+2=Q−2n+1S−4n+1P0n+1Q−4nS0n+1P−2n+1,Q6n+3=Q−3Q0n+1S−2n+1P−4n+1Q−2S−4P0−1n+1Q−2n+1S−4n+1P0n+1Q0S−2P−4−1n+1,Q6n+4=Q−2Q−4n+1S0n+1P−2n+1Q0n+1S−2n+1P−4n+1,Q6n+5=Q−1Q−2n+1S−4n+1P0n+1Q−4S0P−2−1n+1Q−4n+1S0n+1P−2n+1Q−2S−4P−0−1n+1,Q6n+6=Q0n+2S−2n+1P−4n+1Q−2n+1S−4n+1P0n+1,S6n+1=S−4n+1P0n+1Q−2n+1S0P−2Q−4−1nP−1Q−3S0nP−2nQ−4nS−4P0Q−2−1n+1,S6n+2=S−2n+1P−4n+1Q0n+1S−4nP0n+1Q−2n+1,S6n+3=S−3S0n+1P−2n+1Q−4n+1S−2P−4Q0−1n+1S−2n+1P−4n+1Q0n+1S0P−2Q−4−1n+1,S6n+4=S−2S−4n+1P0n+1Q−2n+1S0n+1P−2n+1Q−4n+1,S6n+5=S−1S−2n+1P−4n+1Q0n+1S−4P0Q−2−1n+1S−4n+1P0n+1Q−2n+1S−2P−4Q0−1n+1,S6n+6=S0n+2P−2n+1Q−4n+1S−2n+1P−4n+1Q0n+1.Proof.
This proof is obtained similar to the First Theory.Example 5.
Figure5 shows numerical solution of system (14) with the initial conditions P−4=5,P−3=13,P−2=8,P−1=6,P0=19,Q−4=13,Q−3=5,Q−2=7,Q−1=3,Q0=10,S−4=10,S−3=12,S−2=10,S−1=12,S0=6.Figure 5
Sketch the numerical solution of system (14).Example 6.
This illustrates the numerical solution of system (14) with the initial conditions P−4=10,P−3=16,P−2=10,P−1=3,P0=15,Q−4=8,Q−3=2,Q−2=4,Q−1=3,Q0=5,S−4=5,S−3=6,S−2=7,S−1=9,S0=3 (See Figure 6).Figure 6
Sketch the numerical solution of system (14).


## 5. Fourth Case

In this section, we investigate the solutions of the difference equation system:(16)Pn+1=Pn−4Sn−2QnSn−3Qn−1−1−Pn−4Sn−2Qn,Qn+1=Qn−4Pn−2SnPn−3Sn−1−1−Qn−4Pn−2Sn,Sn+1=Sn−4Qn−2PnQn−3Pn−1−1−Sn−4Qn−2Pn.Theorem 4.
Assume thatPn,Qn,Sn are solutions of difference equation system. Then for n=0,1,2,…, we see that all solutions of system (16) are given by the following formulas:(17)P6n+1=−P−4n+1Q0n+1S−2n+1P0Q−2S−4+1nQ−1S−3P0nQ−2nS−4nP−4Q0S−2+1n+1,P6n+2=P−2n+1Q−4n+1S0n+1P−4nQ0n+1S−2n+1,P6n+3=P−3P0n+1Q−2n+1S−4n+1P−2Q−4S0+1n+1P−2n+1Q−4n+1S0n+1P0Q−2S−4+1n+1,P6n+4=P−2P−4n+1Q0n+1S−2n+1P0n+1Q−2n+1S−4n+1,P6n+5=P−1P−2n+1Q−4n+1S0n+1P−4Q0S−2+1n+1P−4n+1Q0n+1S−2n+1P−2Q−4S−0+1n+1,P6n+6=P0n+2Q−2n+1S−4n+1P−2n+1Q−4n+1S0n+1,Q6n+1=−Q−4n+1S0n+1P−2n+1Q0S−2P−4+1nS−1P−3Q0nS−2nP−4nQ−4S0P−2+1n+1,Q6n+2=Q−2n+1S−4n+1P0n+1Q−4nS0n+1P−2n+1,Q6n+3=Q−3Q0n+1S−2n+1P−4n+1Q−2S−4P0+1n+1Q−2n+1S−4n+1P0n+1Q0S−2P−4+1n+1,Q6n+4=Q−2Q−4n+1S0n+1P−2n+1Q0n+1S−2n+1P−4n+1,Q6n+5=Q−1Q−2n+1S−4n+1P0n+1Q−4S0P−2+1n+1Q−4n+1S0n+1P−2n+1Q−2S−4P−0+1n+1,Q6n+6=Q0n+2S−2n+1P−4n+1Q−2n+1S−4n+1P0n+1,S6n+1=−S−4n+1P0n+1Q−2n+1S0P−2Q−4+1nP−1Q−3S0nP−2nQ−4nS−4P0Q−2+1n+1,S6n+2=S−2n+1P−4n+1Q0n+1S−4nP0n+1Q−2n+1,S6n+3=S−3S0n+1P−2n+1Q−4n+1S−2P−4Q0+1n+1S−2n+1P−4n+1Q0n+1S0P−2Q−4+1n+1,S6n+4=S−2S−4n+1P0n+1Q−2n+1S0n+1P−2n+1Q−4n+1,S6n+5=S−1S−2n+1P−4n+1Q0n+1S−4P0Q−2+1n+1S−4n+1P0n+1Q−2n+1S−2P−4Q0+1n+1,S6n+6=S0n+2P−2n+1Q−4n+1S−2n+1P−4n+1Q0n+1.Proof.
This proof is obtained similar to the First Theory.Example 7.
Figure7 shows numerical solution of system (16) with the initial conditions P−4=10,P−3=18,P−2=13,P−1=6,P0=16,Q−4=10,Q−3=3,Q−2=4,Q−1=2,Q0=6,S−4=7,S−3=8,S−2=9,S−1=11,S0=6.Figure 7
Drawing the numerical solution of system (16).Example 8.
For illustrates the numerical solution of system (16) with the initial conditions P−4=4,P−3=30,P−2=5,P−1=6,P0=14,Q−4=11,Q−3=16,Q−2=12,Q−1=11,Q0=19,S−4=9,S−3=3,S−2=−40,S−1=1,S0=30 (See Figure 8).Figure 8
Drawing the numerical solution of system (16).


## 6. Fifth Case

In this section, we investigate the solutions of the difference equation system:(18)Pn+1=Pn−4Sn−2QnSn−3Qn−1−1−Pn−4Sn−2Qn,Qn+1=Qn−4Pn−2SnPn−3Sn−1−1−Qn−4Pn−2Sn,Sn+1=Sn−4Qn−2PnQn−3Pn−1−1+Sn−4Qn−2Pn.Theorem 5.
Assume thatPn,Qn,Sn are solutions of difference equation system. Then for n=0,1,2,…, we see that all solutions of system (18) are given by the following formulas:(19)P6n+1=−P−4n+1Q0n+1S−2n+1P0Q−2S−4+1nQ−1S−3P0n+1Q−2nS−4nP−4Q0S−2+1n+1,P6n+2=P−2n+1Q−4n+1S0n+1P−4nQ0n+1S−2n+1,P6n+3=−1n+1P−3P0n+1Q−2n+1S−4n+1P−2Q−4S0+1n+1P−2n+1Q−4n+1S0n+1P0Q−2S−4+1n+1,P6n+4=−1n+1P−2P−4n+1Q0n+1S−2n+1P0Q−2S−2−1n+1P0n+1Q−2n+1S−4n+12P−4Q0S−2+1n+1,P6n+5=−1n+1P−1P−2n+1Q−4n+1S0n+13P−4Q0S−2+1n+1P−4n+1Q0n+1S−2n+1P−2Q−4S−0+1n+1,P6n+6=−1n+1P0n+2Q−2n+1S−4n+12P−2Q−4S0−1n+1P−2n+1Q−4n+1S0n+1,Q6n+1=−Q−4n+1S0n+1P−2n+1Q0S−2P−4+1nS−1P−3Q0nS−2nP−4nQ−4S0P−2+1n+1,Q6n+2=−1n+1Q−2n+1S−4n+1P0n+1Q−4nS0n+1P−2n+12P0Q−2S−4−1n+1,Q6n+3=−1n+1Q−3Q0n+1S−2n+1P−4n+1Q−2S−4P0+1n+1Q−2n+1S−4n+1P0n+13Q0S−2P−4+1n+1,Q6n+4=−1n+1Q−2Q−4n+1S0n+1P−2n+12P−4Q0S−2+1n+1Q0n+1S−2n+1P−4n+12P−2Q−4S0−1n+1,Q6n+5=−1n+1Q−1Q−2n+1S−4n+1P0n+1Q−4S0P−2−1n+1Q−4n+1S0n+1P−2n+1Q−2S−4P−0+1n+1,Q6n+6=Q0n+2S−2n+1P−4n+1Q−2n+1S−4n+1P0n+1,S6n+1=−1n+1S−4n+1P0n+1Q−2n+1S0P−2Q−4+1nP−1Q−3S0nP−2nQ−4nS−4P0Q−2+1n+1,S6n+2=S−2n+1P−4n+1Q0n+1S−4n+1P0n+1Q−2n+12S−2P−4Q0n+1,S6n+3=−1n+1S−3S0n+1P−2n+1Q−4n+12S−2P−4Q0+1n+1S−2n+1P−4n+1Q0n+1S0P−2Q−4+1n+1,S6n+4=S−2S−4n+1P0n+1Q−2n+1S0n+1P−2n+1Q−4n+1,S6n+5=−1n+1S−1S−2n+1P−4n+1Q0n+1S−4P0Q−2+1n+1S−4n+1P0n+1Q−2n+1S−2P−4Q0+1n+1,S6n+6=S0n+2P−2n+1Q−4n+12P−2Q0S−2n+1S−2n+1P−4n+1Q0n+1.Proof.
This proof is derived from the proof of First theorem.Example 9.
Figure9 shows the numerical solution of system (18) with the initial conditions P−4=−10,P−3=−18,P−2=−13,P−1=−6,P0=−16,Q−4=−10,Q−3=−3,Q−2=−4,Q−1=−2,Q0=−6,S−4=7,S−3=8,S−2=9,S−1=11,S0=6.Figure 9
Drawing the numerical solution of system (18).Example 10.
This illustrates the numerical solution of system (18) with the initial conditions P−4=−4,P−3=−30,P−2=−5,P−1=−6,P0=−14,Q−4=−11,Q−3=16,Q−2=−12,Q−1=−11,Q0=19,S−4=9,S−3=3,S−2=40,S−1=1,S0=30 (See Figure 10).Figure 10
Drawing the numerical solution of system (18).


## 7. Sixth Case

In this section, we investigate the solutions of the difference equation system(20)Pn+1=Pn−4Sn−2QnSn−3Qn−1−1+Pn−4Sn−2Qn,Qn+1=Qn−4Pn−2SnPn−3Sn−1−1+Qn−4Pn−2Sn,Sn+1=Sn−4Qn−2PnQn−3Pn−1−1−Sn−4Qn−2Pn.Theorem 6.
Assume thatPn,Qn,Sn are solutions of difference equation system. Then for n=0,1,2,…, we see that all solutions of system (20) are given by the following formulas:(21)P6n+1=P−4n+1Q0n+1S−2n+1P0Q−2S−4−1nQ−1S−3P0nQ−2nS−4nP−4Q0S−2−1n+1,P6n+2=P−2n+1Q−4n+1S0n+1P−4nQ0n+1S−2n+1,P6n+3=−1n+1P−3P0n+1Q−2n+1S−4n+1P−2Q−4S0−1n+1P−2n+1Q−4n+1S0n+1P0Q−2S−4+1n+1,P6n+4=−1n+1P−2P−4n+1Q0n+1S−2n+12P0Q−2S−4−1n+1P0n+1Q−2n+1S−4n+12P−4Q0S−2−1n+1,P6n+5=−1n+1P−1P−2n+1Q−4n+1S0n+13P−4Q0S−2−1n+1P−4n+1Q0n+1S−2n+1P−2Q−4S−0+1n+1,P6n+6=P0n+2Q−2n+1S−4n+12P−2Q−4S0+1n+1P−2n+1Q−4n+1S0n+1,Q6n+1=Q−4n+1S0n+1P−2n+1Q0S−2P−4−1nS−1P−3Q0nS−2nP−4nQ−4S0P−2−1n+1,Q6n+2=Q−2n+1S−4n+1P0n+1Q−4n+1S0n+1P−2n+12Q−2S−4P0n+1,Q6n+3=−1n+1Q−3Q0n+1S−2n+1P−4n+1Q−2S−4P0−1n+1Q−2n+1S−4n+1P0n+1Q0S−2P−4−1n+1,Q6n+4=−1n+1Q−2Q−4n+1S0n+1P−2n+1Q0S−2P−4−1n+1Q0n+1S−2n+1P−4n+1Q−4S0P−2+1n+1,Q6n+5=−1n+1Q−1Q−2n+1S−4n+1P0n+1Q−4S0P−2+1n+1Q−4n+1S0n+1P−2n+1Q−2S−4P−0−1n+1,Q6n+6=Q0n+2S−2n+1P−4n+1Q−2n+1S−4n+1P0n+1,S6n+1=−S−4n+1P0n+1Q−2n+1S0P−2Q−4−1nP−1Q−3S0nP−2nQ−4nS−4P0Q−2−1n+1,S6n+2=−1n+1S−2n+1P−4n+1Q0n+1S−4nP0n+1Q−2n+12P−4Q0S−2−1n+1,S6n+3=−1n+1S−3S0n+1P−2n+1Q−4n+1S−2P−4Q0−1n+1S−2n+1P−4n+1Q0n+1S0P−2Q−4−1n+1,S6n+4=S−2S−4n+1P0n+1Q−2n+1S0n+1P−2n+1Q−4n+1,S6n+5=−1n+1S−1S−2n+1P−4n+1Q0n+1S−4P0Q−2−1n+1S−4n+1P0n+1Q−2n+1S−2P−4Q0−1n+1,S6n+6=−1n+1S0n+2P−2n+1Q−4n+12P−4Q0S−2−1n+1S−2n+1P−4n+1Q0n+1.Proof.
This proof is obtained similar to the First theorem.Example 11.
Figure11 shows numerical solution of system (20) with the initial conditions P−4=−5,P−3=−13,P−2=−8,P−1=−6,P0=−19,Q−4=−13,Q−3=−5,Q−2=−7,Q−1=−3,Q0=−10,S−4=−10,S−3=−12,S−2=−10,S−1=−12,S0=−6.Figure 11
Sketch the numerical solution of system (20).Example 12.
This illustrates the numerical solution of system (20) with the initial conditions P−4=−8,P−3=16,P−2=−10,P−1=3,P0=−15,Q−4=−8,Q−3=2,Q−2=−4,Q−1=3,Q0=−5,S−4=−5,S−3=−6,S−2=−7,S−1=−9,S0=−3 (See Figure 12).Figure 12
Sketch the numerical solution of system (20).



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