# Displaying the Structure of the Solutions for Some Fifth-Order Systems of Recursive Equations

Authors: H. S. Alayachi; A. Q. Khan; M. S. M. Noorani; A. Khaliq
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

This paper presents the solutions to the following nonlinear systems of rational difference equations:xn+1=xn−3yn−4/yn1+xn−1yn−2xn−3yn−4,yn+1=yn−3xn−4/xn±1±yn−1xn−2yn−3xn−4 where initial conditions x−δ,y−δδ=4,3,…,0 are nonnegative real numbers. Finally some numerical simulations are presented to verify obtained theoretical results.


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## Body

## 1. Introduction

The purpose of present study is to solve and deal with the following difference equations systems:(1)xn+1=xn−3yn−4yn1+xn−1yn−2xn−3yn−4,yn+1=yn−3xn−4xn±1±yn−1xn−2yn−3xn−4,where x−δ,y−δδ=4,3,…,0 are arbitrary nonnegative real numbers.It is anticipated that discrete dynamical systems can be seen as discrete analogous of differential as well as delay differential equations. Moreover, these systems designate certain natural phenomena in economy, physics, biology, and many more. Many scholars and researchers have studied various dynamical properties of difference equations along their systems in recent years. For example, Asiri et al. [1] have investigated the periodicity nature of the following system:(2)xn+1=yn−2−yn−2xn−1yn+1,yn+1=xn−2±xn−2yn−1xn±1.Cinar et al. [2] have obtained the solution of the subsequent recursive system:(3)xn+1=xn−1ynxn−1−1,yn+1=yn−1xnyn−1−1,zn+1=xnynzn−1.Metwally and Elsayed [3] have explored the periodicity nature of the following system:(4)xn+1=yn−2−yn−2xn−1yn−1,yn+1=xn−2±xn−2yn−1xn±1.Touafek et al. [4] have explored the periodicity nature of the following system:(5)xn+1=xn−3±xn−3yn−1±1,yn+1=yn−3±yn−3xn−1±1.Elsayed [5] has obtained the solution of the following rational system:(6)xn+1=xn−1±xn−1yn±1,yn+1=yn−1yn−1xn∓1.For more interesting results regarding dynamical properties of difference and differential equations along their systems, we refer the readers to [2, 6–28] and the references cited therein.


## 2. On 1st System

(7)xn+1=xn−3yn−4yn1+xn−1yn−2xn−3yn−4,yn+1=yn−3xn−4xn1+yn−1xn−2yn−3xn−4.This section is about the investigation of the solution form of the following system:(8)xn+1=xn−3yn−4yn1+xn−1yn−2xn−3yn−4,yn+1=yn−3xn−4xn1+yn−1xn−2yn−3xn−4.The forms of solutions to (8) are given as Theorem 1.Theorem 1.
Letxn,yn be a solution to (8), and also let x−δ,y−δδ=4,3,…,0, respectively, be a,b,c,d,e,f,g,h,k,l. Then one has(9)x4n−4=en∏δ=0−1+n2δacgk+1an−1∏δ=0−1+n2δcegk+1,x4n−3=bfn∏δ=0−1+n2δbdhl+1ln∏δ=0−1+n2δ+1bdfh+1,x4n−2=cen∏δ=0−1+n1+2δacgk+1an∏δ=0−1+n1+2δcegk+1,x4n−1=dfn∏i=1−1+n1+2δbdhl+1ln∏δ=0−1+n2+2δbdfh+1,y4n−4=ln∏δ=0−1+n2δbdfh+1f−1+n∏δ=0−1+n2δbdhl+1,y4n−3=gan∏δ=0−1+n2δcegk+1en∏δ=0−1+n1+2δacgk+1,y4n−2=hln∏δ=0−1+n2δ+1bdfh+1fn∏δ=0−1+n1+2δbdhl+1,y4n−1=kan∏δ=0−1+n2δ+1cegk+1en∏δ=0−1+n2δ+2acgk+1.Proof.
Obviously results true ifn=0. Assuming that for n−1 and n−2 results hold, that is(10)x4n−9=df−2+n∏δ=0n−32δ+1bdhl+1ln−2∏δ=0n−32+2δbdfh+1,x4n−8=en−1∏δ=0−2+n2δacgki+1an−2∏δ=0−2+n2δcegk+1,x4n−7=bf−1+n∏δ=0−2+n2δbdhl+1ln−1∏δ=0−2+n1+2δbdfh+1,x4n−6=cen−1∏δ=0−2+n2δ+1acgk+1an−1∏δ=0−2+n2δ+1cegk+1,x4n−5=d×f−1+n∏δ=0−2+n2δ+1bdhl+1ln−1∏δ=0−2+n2+2δbdfh+1,y4n−9=kan−2∏δ=0n−32δ+1cegk+1en−2∏δ=0n−32+2δacgk+1,y4n−8=ln−1∏δ=0−2+n2δbdfh+1f−2+n∏δ=0−2+n2δbdhl+1,y4n−7=gan−1∏δ=0−2+n2δcegk+1en−1∏δ=0−2+n2δ+1acgk+1,y4n−6=hln−1∏δ=0−2+n2δ+1bdfh+1f−1+n∏δ=0−2+n1+2δbdhl+1,y4n−5=kan−1∏δ=0−2+n2δ+1cegk+1en−1∏δ=0−2+n2+2δacgk+1.
Now from (8), one has(11)x4n−4=x4n−8y4n−9y4n−51+x4n−6y4n−7x4n−8y4n−9=en−1∏δ=0−2+n2δacgk+1/an−2∏δ=0−2+n2δcegk+1kan−2∏δ=0n−31+2δcegk+1/en−2∏δ=0n−32δ+2acgk+1kan−1∏δ=0−2+n2δ+1cegk+1/en−1∏δ=0−2+n2δ+2acgk+11+cen−1∏δ=0−2+n2δ+1acgk+1/an−1∏δ=0−2+n2δ+1cegk+1gan−1∏δ=0n−22δcegk+1/en−1∏δ=0n−22δ+1acgk+1en−1∏δ=0−2+n2δacgk+1/an−2∏δ=0−2+n2δcegk+1kan−2∏δ=0n−32δ+1cegk+1/en−2∏δ=0n−32δ+2acgk+1=ek∏δ=0n−32δ+1cegk+1/∏δ=0−2+n2δcegk+1kan−1∏δ=0n−21+2δcegk+1/en−1∏δ=0−2+n2+2δacgk+11+cegk∏δ=0n−31+2δcegk+1/∏δ=0n−22δ+1cegk+1=en∏δ=0−2+n2+2δacgk+1an−12n−3cegk+1∏δ=0−2+n2δcegk+11+cegk/2n−3cegk+1=en∏δ=0−2+n2δ+2acgk+1an−1∏δ=0−2+n2δcegk+12n−3cegk+1+cegk=en∏δ=0−2+n2+2δacgk+1an−1∏δ=0−2+n2δcegk+1−2+2ncegk+1=en∏δ=0−1+n2δacgk+1an−1∏δ=0−1+n2δcegk+1,y4n−4=y4n−8x4n−9x4n−51+y4n−6x4n−7y4n−8x4n−9=ln−1∏δ=0−2+n2δbdfh+1/f−2+n∏δ=0−2+n2δbdhl+1df−2+n∏δ=0n−31+2δbdhl+1/ln−2∏δ=0n−32+2δbdfh+1df−1+n∏δ=0−2+n1+2δbdhl+1/ln−1∏δ=0−2+n2+2δbdfh+11+hln−1∏δ=0−2+n2δ+1bdfh+1/f−1+n∏δ=0−2+n1+2δbdhl+1bf−1+n∏δ=0n−22δbdhl+1/ln−1∏δ=0n−21+2δbdfh+1ln−1∏δ=0−2+n2δbdfh+1/f−2+n∏δ=0−2+n2δbdhl+1df−2+n∏δ=0n−31+2δbdhl+1/ln−2∏δ=0n−32+2δbdfh+1=ln−1∏δ=0−2+n2δbdfh+1/f−2+n∏δ=0−2+n2δbdhl+1df−2+n∏δ=0n−32δ+1bdhl+1/ln−2∏δ=0n−32δ+2bdfh+1df−1+n∏δ=0−2+n2δ+1bdhl+1/ln−1∏δ=0−2+n2δ+2bdfh+11+bdhl∏δ=0n−32δ+1bdhl+1/∏δ=0n−22δ+1bdhl+1=ln∏δ=0−2+n2δ+2bdfh+1f−1+n2n−3bdhl+1∏δ=0−2+n2δbdhl+11+bdhl/2n−3bdhl+1=ln∏δ=0−2+n2+2δbdfh+1f−1+n∏δ=0−2+n2δbdhl+12n−3bdhl+1+bdhl=ln∏δ=0−2+n2δ+2bdfh+1f−1+n∏δ=0−2+n2δbdhl+12n−2bdhl=ln∏δ=0−1+n2δbdfh+1f−1+n∏δ=0−1+n2δbdhl+1.
Again, from system (8), we obtain(12)x4n−3=x4n−7y4n−8y4n−41+x4n−5y4n−6x4n−7y4n−8=bf−1+n∏δ=0−2+n2δbdhl+1/ln−1∏δ=0−2+n2δ+1bdfh+1ln−1∏δ=0n−22δbdfh+1/f−2+n∏δ=0n−22δbdhl+1ln∏δ=0−1+n2δbdfh+1/f−1+n∏δ=0−1+n2δbdhl+11+df−1+n∏δ=0−2+n2δ+1bdhl+1/ln−1∏δ=0−2+n2δ+2bdfh+1hln−1∏δ=0n−22δ+1bdfh+1/f−1+n∏δ=0n−22δ+1bdhl+1bf−1+n∏δ=0−2+n2δbdhl+1/ln−1∏δ=0−2+n2δ+1bdfh+1ln−1∏δ=0n−22δbdfh+1/f−2+n∏δ=0n−22δbdhl+1=bfn∏δ=0−2+n2δbdfh+1∏δ=0−1+n2δbdhl+1ln∏δ=0−1+n2δbdfh+1∏δ=0−2+n2δ+1bdfh+11+bdfh∏δ=0n−22δbdfh+1/∏δ=0n−22δ+2bdfh+1=bfn∏δ=0−1+n2δbdhl+1ln2n−2bdfh+1∏δ=0−2+n2δ+1bdfh+11+bdfh/2n−2bdfh+1=bfn∏δ=0−1+n2δbdhl+1ln∏δ=0−2+n2δ+1bdfh+12n−2bdfh+1+bdfh=bfn∏δ=0−1+n2δbdhl+1ln∏δ=0−2+n2δ+1bdfh+1−1+2nbdfh+1=bfn∏δ=0−1+n2δbdhl+1ln∏δ=0−1+n2δ+1bdfh+1,y4n−3=y4n−7x4n−8x4n−41+y4n−5x4n−6y4n−7x4n−8=gan−1∏δ=0−2+n2δcegk+1/en−1∏δ=0−2+n2δ+1acgk+1en−1∏δ=0n−22δacgk+1/an−2∏δ=0n−22δcegk+1en∏δ=0−1+n2δacgk+1/an−1∏δ=0−1+n2δcegk+11+kan−1∏δ=0−2+n1+2δcegk+1/en−1∏δ=0−2+n2δ+2acgk+1cen−1∏δ=0n−21+2δacgk+1/an−1∏δ=0n−22δ+1cegk+1gan−1∏δ=0−2+n2δcegk+1/en−1∏δ=0−2+n2δ+1acgk+1en−1∏δ=0n−22δacgk+1/an−2∏δ=0n−22δcegk+1=gan∏δ=0−1+n2δcegk+1en∏δ=0−2+n2δ+1acgk+12n−2acgk+11+acgk/2n−2acgk+1=gan∏δ=0−1+n2δcegk+1en∏δ=0−2+n2δ+1acgk+12n−2acgk+1+acgk=gan∏δ=0−1+n2δcegk+1en∏δ=0−2+n2δ+1acgk+1−1+2nacgk+1=gan∏δ=0−1+n2δcegk+1en∏δ=0−1+n1+2δacgk+1.
Hence, the proof of other relations is obvious.


## 3. On 2nd System

(13)xn+1=xn−3yn−4yn1+xn−1yn−2xn−3yn−4,yn+1=yn−3xn−4xn1−yn−1xn−2yn−3xn−4.In this section, we explore the solutions of the recursive system:(14)xn+1=xn−3yn−4yn1+xn−1yn−2xn−3yn−4,yn+1=yn−3xn−4xn1−yn−1xn−2yn−3xn−4.Theorem 2.
Letxn,yn be a solution to (14), and also let x−δ,y−δδ=4,3,…,0, respectively, be a,b,c,d,e,f,g,h,k,l; then(15)x4n−4=en∏δ=0−1+n1−2δacgkan−1∏δ=0−1+n1+2δcegk,x4n−3=bfn∏δ=0−1+n1−2δbdhlln∏δ=0−1+n1+2δ+1bdfh,x4n−2=cen∏i=1n−11−2δ+1acgkan∏δ=0−1+n1+1+2δcegk,x4n−1=dfn∏δ=0−1+n1−2δ+1bdhlln∏δ=0−1+n1+2δ+2bdfh,y4n−4=ln∏δ=0−1+n1+2δbdfhf−1+n∏δ=0−1+n1−2δbdhl,y4n−3=gan∏δ=0−1+n1+2δcegken∏δ=0−1+n1−2δ+1acgk,y4n−2=hln∏δ=0−1+n1+2δ+1bdfhfn∏δ=0−1+n1−2δ+1bdhl,y4n−1=kan∏δ=0−1+n1+2δ+1cegken∏δ=0−1+n1−2δ+2acgk.Proof.
Obviously results true ifn=0. Assuming that for n−1 and n−2 results hold, that is(16)x4n−9=df−2+n∏δ=0n−31−2δ+1bdhlln−2∏δ=0n−31+2δ+2bdfh,x4n−8=en−1∏δ=0−2+n1−2δacgkan−2∏δ=0−2+n1+2δcegk,x4n−7=bf−1+n∏δ=0−2+n1−2δbdhlln−1∏δ=0−2+n1+2δ+1bdfh,x4n−6=cen−1∏i=1n−21−2δ+1acgkan−1∏δ=0−2+n1+1+2δcegk,x4n−5=df−1+n∏δ=0−2+n1−2δ+1bdhlln−1∏δ=0−2+n1+2+2δbdfh,y4n−9=kan−2∏δ=0n−31+2δ+1cegken−2∏δ=0n−31−2+2δacgk,y4n−8=ln−1∏δ=0−2+n1+2δbdfhf−2+n∏δ=0−2+n1−2δbdhl,y4n−7=gan−1∏δ=0−2+n1+2δcegken−1∏δ=0−2+n1−2δ+1acgk,y4n−6=hln−1∏δ=0−2+n1+2δ+1bdfhf−1+n∏δ=0−2+n1−2δ+1bdhl,y4n−5=kan−1∏δ=0−2+n1+2δ+1cegken−1∏δ=0−2+n1−2δ+2acgk.
Next, one can obtain from system (14) that(17)x4n−4=x4n−8y4n−9y4n−51+x4n−6y4n−7x4n−8y4n−9=en−1∏δ=0−2+n1−2δacgk/an−2∏δ=0−2+n1+2δcegkkan−2∏δ=0n−31+1+2δcegk/en−2∏δ=0n−31−2+2δacgkkan−1∏δ=0−2+n1+2δ+1cegk/en−1∏δ=0−2+n1−2δ+2acgk1+cen−1∏i=1n−21−2δ+1acgk/an−1∏δ=0−2+n1+2δ+1cegkgan−1∏δ=0n−21+2δcegk/en−1∏δ=0−2+n1−2δ+1acgken−1∏δ=0−2+n1−2δacgk/an−2∏δ=0−2+n1+2δcegkkan−2∏δ=0n−31+2δ+1cegk/en−2∏δ=0n−31−2δ+2acgk=en∏δ=0n−31+1+2δcegk∏δ=0−2+n1−2+2δacgkan−1∏δ=0−2+n1+2δ+1cegk∏δ=0−2+n1+2δcegk1+cegk/1+2n−3cegk=en∏δ=0−2+n1−2+2δacgkan−11+2n−3cegk∏δ=0−2+n1+2δcegk1+cegk/1+2n−3cegk=en∏δ=0−2+n1−2δ+2acgkan−1∏δ=0−2+n1+2δcegk1+2n−3cegk+cegk=en∏δ=0−2+n1−2+2δacgkan−1∏δ=0−2+n1+2δcegk1+2n−2cegk=en∏δ=0−1+n1−2δacgkan−1∏δ=0−1+n1+2δcegk.
Similarly, system (14) gives(18)y4n−4=y4n−8x4n−9x4n−51−y4n−6x4n−7y4n−8x4n−9=ln−1∏δ=0−2+n1+2δbdfh/f−2+n∏δ=0−2+n1−2δbdhldf−2+n∏δ=0n−31−2δ+1bdhl/ln−2∏δ=0n−31+2δ+2bdfhdf−1+n∏δ=0−2+n1−2δ+1bdhl/ln−1∏δ=0−2+n1+2δ+2bdfh1−hln−1∏δ=0−2+n1+2δ+1bdfh/f−1+n∏δ=0−2+n1−2δ+1bdhlbf−1+n∏δ=0n−21−2δbdhl/ln−1∏δ=0−2+n1+2δ+1bdfhln−1∏δ=0−2+n1+2δbdfh/f−2+n∏δ=0−2+n1−2δbdhldf−2+n∏δ=0n−31−2δ+1bdhl/ln−2∏δ=0n−31+2+2δbdfh=ln∏δ=0−2+n1+2δ+2bdfhf−1+n1−2n−3bdhl∏δ=0−2+n1−2δbdhl1−bdhl/1−2n−3bdhl=ln∏δ=0−2+n1+2δ+2bdfhf−1+n∏δ=0−2+n1−2δbdhl1−2n−3bdhl−bdhl=ln∏δ=0−2+n1+2+2δbdfhf−1+n∏δ=0−2+n1−2δbdhl1−2n−2bdhl=ln∏δ=0−1+n1+2δbdfhf−1+n∏δ=0−1+n1−2δbdhl.
Obviously, the subsequent unassertive relations follows, and this complete the proof.


## 4. On 3rd System

(19)xn+1=xn−3yn−4yn1+xn−1yn−2xn−3yn−4,yn+1=yn−3xn−4xn−1−yn−1xn−2yn−3xn−4.In this section, we will explore the solutions of following rational systems:(20)xn+1=xn−3yn−4yn1+xn−1yn−2xn−3yn−4,yn+1=yn−3xn−4xn−1−yn−1xn−2yn−3xn−4.Theorem 3.
Letxn,yn be a solution of (20), and also let x−δ,y−δδ=4,3,…,0, respectively, are a,b,c,d,e,f,g,h,k,l. Then one has(21)x4n−4=enan−1∏δ=0−1+n1+2δcegk,x4n−3=b×fnln∏δ=0n−11+1+2δbdfh,x4n−2=c−enacgk+1nan∏δ=0n−11+1+2δcegk,x4n−1=d−fn1+bdhlnln∏δ=0−1+n1+2+2δbdfh,y4n−4=ln∏δ=0−1+n1+2δbdfhf−1+n,y4n−3=g−an∏δ=0−1+n1+2δcegken1+acgkn,y4n−2=h−ln∏δ=0−1+n1+2δ+1bdfhfn1+bdhln,y4n−1=kan∏δ=0−1+n1+2δ+1cegken.Proof.
Obviously results true ifn=0. Assuming that for n−1 and n−2 results hold, that is(22)x4n−9=d−fn−21+bdhln−2ln−2∏Πn−3δ=01+2δ+2bdfh,x4n−8=en−1an−2∏δ=0−2+n1+2δcegk,x4n−7=b×f−1+nln−1∏δ=0n−21+2δ+1bdfh,x4n−6=c−en−11+acgkn−1an−1∏δ=0−2+n2δ+1cegk+1,x4n−5=d−fn−11+bdhln−1ln−1∏δ=0−2+n2+2δbdfh+1,y4n−9=kan−2∏δ=0n−31+1+2δcegken−2,y4n−8=ln−1∏δ=0−2+n1+2δbdfhf−2+n,y4n−7=g−an−1∏δ=0−2+n1+2δcegken−11+acgkn−1,y4n−6=h−ln−1∏δ=0−2+n1+2δ+1bdfhf−n+11+bdhln−1,y4n−5=kan−1∏δ=0−2+n1+2δ+1cegken−1.
Next, it can be seen from system (20) that(23)x4n−4=x4n−8y4n−9y4n−51+x4n−6y4n−7x4n−8y4n−9=/en−1an−2∏δ=0−2+n1+2δcegk/kan−2∏δ=0n−31+2δ+1cegken−2kan−1∏δ=0n−21+1+2δcegk/en−11+c−en−11+acgkn−1/an−1∏δ=0n−21+1+2δcegkg−an−1∏δ=0−2+n1+2δcegk/en−11+acgkn−1en−1/an−2∏δ=0−2+n1+2δcegkkan−2∏δ=0n−31+2δ+1cegk/en−2=en∏δ=0n−31+2δ+1cegkan−1∏δ=0−2+n1+2δcegk∏δ=0−2+n1+2δ+1cegk1+cegk∏δ=0n−31+1+2δcegk/∏δ=0n−21+1+2δcegk=enan−1∏δ=0−2+n1+2δcegk1+2n−3cegk1+cegk/1+2n−3cegk=enan−1∏δ=0−2+n1+2δcegk1+2n−3cegk+cegk=enan−1∏δ=0−2+n1+2δcegk1+2n−2cegk=enan−1∏δ=0−1+n2δ+1cegk,y4n−4=y4n−8x4n−9x4n−5−1−y4n−6x4n−7y4n−8x4n−9=ln−1∏δ=0−2+n1+2δbdfh/f−2+nd−fn−21+bdhln−2/ln−2∏δ=0n−31+2δ+2bdfhd−fn−11+bdhln−1/ln−1∏δ=0−2+n1+2δ+2bdfh−1−h−ln−1∏δ=0−2+n1+2δ+1bdfh/f−1+n1+bdhln−1bf−1+n/ln−1∏δ=0−2+n1+2δ+1bdfhln−1∏δ=0−2+n1+2δbdfh/f−2+nd−fn−21+bdhln−2/ln−2∏δ=0n−31+2δ+2bdfh=−1n−2ln∏δ=0−2+n1+2δ+2bdfh−fn−11+bdhl−1+bdhl/1+bdhl=−ln∏δ=0−2+n1+2δ+2bdfhf−1+n1+bdhl−1+bdhl/1+bdhl=−ln∏δ=0−1+n1+2δbdfhf−1+n−1−bdhl+bdhl=ln∏δ=0−1+n1+2δbdfhf−1+n.
Now, hereafter, we prove an extra result. It is noted that system (20) leads to(24)x4n−3=x4n−7y4n−8y4n−41+x4n−5y4n−6x4n−7y4n−8=bf−1+n/ln−1∏δ=0−2+n1+2δ+1bdfhln−1∏δ=0n−21+2δbdfh/f−2+nln∏δ=0−1+n1+2δbdfh/f−1+n1+d−fn−11+bdhln−1/ln−1∏δ=0n−21+2δ+2bdfhh−ln−1∏δ=0−2+n1+2δ+1bdfh/f−1+n1+bdhln−1bf−1+n/ln−1∏δ=0−2+n1+2δ+1bdfhln−1∏δ=0n−21+2δbdfh/f−2+n=bfn/∏δ=0−2+n1+2δ+1bdfhln∏δ=0n−11+2δbdfh/∏δ=0n−21+2δbdfh1+bdfh∏δ=0n−21+2δbdfh/∏δ=0−2+n1+2+2δbdfh=bfn/∏δ=0−2+n1+2δ+1bdfhln1+2n−2bdfh1+bdfh/1+2n−2bdfh=bfnln∏δ=0−2+n1+2δ+1bdfh1+2n−2bdfh+bdfh=bfnln∏δ=0−2+n1+2δ+1bdfh1+2n−1bdfh=bfnln∏δ=0−1+n1+2δ+1bdfh,y4n−3=y4n−7x4n−8x4n−4−1−y4n−5x4n−6y4n−7x4n−8=g−an−1∏δ=0−2+n1+2δcegk/en−11+acgkn−1en−1/an−2∏δ=0−2+n1+2δcegken/an−1∏δ=0−1+n1+2δcegk−1−kan−1∏δ=0−2+n1+2δ+1cegk/en−1c−en−11+acgkn−1/an−1∏δ=0−2+n1+2δ+1cegkg−an−1∏δ=0−2+n1+2δcegk/en−11+acgkn−1en−1/an−2∏δ=0−2+n1+2δcegk=ga−1n−1/1+acgkn−1/enan−1∏δ=0−1+n1+2δcegk−1−acgk=g−1n−1an∏δ=0−1+n1+2δcegken1+acgkn−1−1−acgk=g−an∏δ=0−1+n1+2δcegken1+acgkn.
In a similar way, one can establish other formulas.


## 5. On 4th System

(25)xn+1=xn−1yn−3yn−11+xn−1yn−3,yn+1=yn−1xn−3xn−1−1+yn−1xn−3.This section determines and studies the formulas for solutions of the following nonlinear system:(26)xn+1=xn−3yn−4yn1+xn−1yn−2xn−3yn−4,yn+1=yn−3xn−4xn−1+yn−1xn−2yn−3xn−4.Theorem 4.
Letxn,yn be a solution to (26), and also let x−δ,y−δδ=4,3,…,0, respectively, are a,b,c,d,e,f,g,h,k,l. Then one has(27)x4n−4=enan−1∏δ=0−1+n1+2δcegk,x4n−3=b×fnln∏δ=0n−11+2δ+1bdfh,x4n−2=cen−1+acgknan∏δ=0n−11+2δ+1cegk,x4n−1=dfn−1+bdhlnln∏δ=0−1+n1+2δ+2bdfh,y4n−4=ln∏δ=0−1+n1+2δbdfhf−1+n,y4n−3=gan∏δ=0n−11+2δcegken−1+acgkn,y4n−2=hln∏δ=0−1+n1+2δ+1bdfhfn−1+bdhln,y4n−1=kan∏δ=0−1+n1+2δ+1cegken.Proof.
Obviously results true ifn=0. Assuming that for n−1 and n−2 results hold, that is(28)x4n−9=df−2+n−1+bdhln−2ln−2∏δ=0n−31+2δ+2bdfh,x4n−8=en−1an−2∏δ=0−2+n1+2δcegk,x4n−7=bf−1+nln−1∏δ=0n−21+2δ+1bdfh,x4n−6=cen−1−1+acgkn−1an−1∏δ=0−2+n1+2δ+1cegk,x4n−5=df−1+n−1+bdhln−1ln−1∏δ=0−2+n1+2δ+2bdfh,y4n−9=kan−2∏δ=0n−31+2δ+1cegken−2,y4n−8=ln−1∏δ=0−2+n1+2δbdfhf−2+n,y4n−7=gan−1∏δ=0n−21+2δcegken−1−1+acgkn−1,y4n−6=hln−1∏δ=0−2+n1+2δ+1bdfhf−1+n−1+bdhln−1,y4n−5=kan−1∏δ=0−2+n1+2δ+1cegken−1.
Next, one can obtain from system (26) that(29)x4n−4=x4n−8y4n−9y4n−51+x4n−6y4n−7x4n−8y4n−9=en−1/an−2∏δ=0−2+n1+2δcegkkan−2∏δ=0n−31+2δ+1cegk/en−2kan−1∏δ=0n−21+2δ+1cegk/en−11+cen−1−1+acgkn−1/an−1∏δ=0n−21+2δ+1cegkgan−1∏δ=0−2+n1+2δcegk/en−1−1+acgkn−1en−1/an−2∏δ=0−2+n1+2δcegkkan−2∏δ=0n−31+2δ+1cegk/en−2=en∏δ=0n−31+2δ+1cegk/∏δ=0−2+n1+2δcegkan−1∏δ=0−2+n1+2δ+1cegk1+cegk∏δ=0n−31+2δ+1cegk/∏δ=0−2+n1+2δ+1cegk=enan−1∏δ=0−2+n1+2δcegk1+2n−3cegk1+cegk/1+2n−3cegk=enan−1∏δ=0−2+n1+2δcegk1+2n−3cegk+cegk=enan−1∏δ=0−2+n1+2δcegk1+2n−2cegk=enan−1∏δ=0−1+n1+2δcegk,y4n−4=y4n−8x4n−9x4n−5−1+y4n−6x4n−7y4n−8x4n−9=/ln−1∏δ=0−2+n1+2δbdfhf−2+n/df−2+n−1+bdhln−2ln−2∏δ=0n−31+2δ+2bdfhdf−1+n−1+bdhln−1/ln−1∏δ=0−2+n1+2δ+2bdfh−1+hln−1∏δ=0−2+n1+2δ+1bdfh/f−1+n−1+bdhln−1bf−1+n/ln−1∏δ=0−2+n1+2δ+1bdfhln−1∏δ=0−2+n1+2δbdfh/f−2+ndf−2+n−1+bdhln−2/ln−2∏δ=0n−31+2δ+2bdfh=ln∏δ=0−2+n1+2δ+2bdfhf−1+n−1+bdhl−1+bdhl/−1+bdhl=ln∏δ=0−2+n1+2δ+2bdfhf−1+n1−bdhl+bdhl=ln∏δ=0−1+n1+2δbdfhf−1+n.
Also from system (26) one gets:(30)x4n−3=x4n−7y4n−8y4n−41+x4n−5y4n−6x4n−7y4n−8=bf−1+n/ln−1∏δ=0−2+n1+2δ+1bdfhln−1∏δ=0n−21+2δbdfh/f−2+nln∏δ=0n−11+2δbdfh/f−1+n1+df−1+n−1+bdhln−1/ln−1∏δ=0n−21+2δ+2bdfhhln−1∏δ=0−2+n1+2δ+1bdfh/f−1+n−1+bdhln−1bf−1+n/ln−1∏δ=0−2+n1+2δ+1bdfhln−1∏δ=0n−21+2δbdfh/f−2+n=bfn/∏δ=0−2+n1+2δ+1bdfhln∏δ=0n−11+2δbdfh/∏δ=0−2+n1+2δbdfh1+bdfh∏δ=0n−21+2δbdfh/∏δ=0−2+n1+2δ+2bdfh=bfn/∏δ=0−2+n1+2δ+1bdfhln1+2n−2bdfh1+bdfh/1+2n−2bdfh=bfnln1+2n−2bdfh+bdfh∏δ=0−2+n1+2δ+1bdfh=bfnln1+2n−1bdfh∏δ=0n−21+2δ+1bdfh=bfnln∏δ=0−1+n1+2δ+1bdfh,y4n−3=y4n−7x4n−8x4n−4−1+y4n−5x4n−6y4n−7x4n−8=gan−1∏δ=0−2+n1+2δcegk/en−1−1+acgkn−1en−1/an−2∏δ=0−2+n1+2δcegken/an−1∏δ=0−1+n1+2δcegk−1+kan−1∏δ=0−2+n1+2δ+1cegk/en−1cen−1−1+acgkn−1/an−1∏δ=0−2+n1+2δ+1cegkgan−1∏δ=0−2+n1+2δcegk/en−1−1+acgkn−1en−1/an−2∏δ=0n−21+2δcegk=gan/−1+acgkn−1en/∏δ=0n−11+2δcegk−1+acgk=gan∏δ=0−1+n1+2δcegken−1+acgkn−1−1+acgk=gan∏δ=0−1+n1+2δcegken−1+acgkn.
Subsequent results follow from the above assertion.


## 6. Numerical Simulation

In order to verify our theoretical results, we consider four interesting numerical examples by fixing suitable initial values. These simulation shows the solutions of under consideration discrete-time systems, which are depicted in (8), (14), (20), and (26).Example 1.
This example presents that the solutions of discrete-time system (8) with x−δ,y−δδ=4,3,…,0, respectively, are 1,0.2,0.1,0.02,2,1.5,0.3,1,0.4,4 as shown in Figure 1.Figure 1
Plot of the solutions of1st system.Example 2.
This example shows the behavior of system (14). The relevant plot is given in Figure 2 under the initial conditions: x−δ,y−δδ=4,3,…,0, respectively, are 0.5,3.5,0.01,2,0.2,1.5,3,0.1,0.7,0.3.Figure 2
Plot of the solutions of2nd system.Example 3.
Figure3 depicts the dynamics of solutions to (20) when we randomly consider the values as follows: x−δ,y−δδ=4,3,…,0, respectively, are 0.1,0.9,0.01,0.3,0.2,0.4,0.3,0.03,0.2,0.8.Figure 3
Plot of the solutions of3rd system.Example 4.
The behavior of system (26) are plotted in Figure 4 with x−δ,y−δδ=4,3,…,0, respectively, are 0.1,0.05,0.06,0.2,0.2,0.4,0.2,0.04,0.08,0.81.Figure 4
Plot of the solutions of4th system.


## 7. Conclusion

In this paper, we deal with the form of the solutions of four cases of the nonlinear systems of difference equationsxn+1=xn−3yn−4/yn1+xn−1yn−2xn−3yn−4,yn+1=yn−3xn−4/xn±1±yn−1xn−2yn−3xn−4. Finally some numerical examples are giving by fixing suitable initial values to show the qualitative behavior of under consideration systems.



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