# Closed-Form Solution of a Rational Difference Equation

Authors: Tarek F. Ibrahim; Abdul Qadeer Khan; Burak Oğul; Dağistan Şimşek
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

In this paper, we study the solution of the difference equationΩm+1=Ωm−7q+6/1+∏t=05Ωm−q+1t−q, where the initials are positive real numbers.


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## Body

## 1. Introduction

Difference equations appear naturally as discrete analogues in many sciences such as biology, ecology, and physics. In recent years, many authors studied the solution form of difference equations. For instance, Cinar [1–3] studied(1)Ωm+1=Ωm−11+aΩmΩm−1,m=0,1,…,Ωm+1=Ωm−1−1+aΩmΩm−1,m=0,1,…,Ωm+1=aΩm−11+bΩmΩm−1,m=0,1,…,respectively.DeVault et al. [4] examined(2)Ωm+1=AΩm+1Ωm−2,m=0,1,….Elsayed [5] dealt with(3)Ωm+1=a+bΩm−l+cΩm−qdΩm−l+eΩm−q,m=0,1,….Simsek et al. [6–9] studied(4)Ωm+1=Ωm−31+Ωm−1,m=0,1,…,Ωm+1=Ωm−51+Ωm−2,m=0,1,…,Ωm+1=Ωm−51+Ωm−1Ωm−3,m=0,1,…,Ωm+1=Ωm−31+ΩmΩm−1Ωm−2,m=0,1,…,respectively. For some more results concerning difference equations, we refer the reader to [10–21].In this work, we deal with the following nonlinear difference equation:(5)Ωm+1=Ωm−7q+61+∏t=05Ωm−q+1t−q,m=0,1,…,where Ω−7q+6,Ω−7q+5,…,Ω−1,Ω0∈0,∞ is investigated.


## 2. Main Results

LetΩ¯ be the unique equilibrium of equation (5); then,(6)Ω¯=Ω¯1+Ω¯.Ω¯.Ω¯.Ω¯.Ω¯.Ω¯⇒Ω¯+Ω¯7=Ω¯⇒Ω¯7=0⇒Ω¯=0,so Ω¯=0 is obtained. For every q≥0 and m>q, notation i=q,m¯ means i=q,q+1,…,m.Theorem 1.
For (5), the following statements are true:(a)
The sequencesΩ7q+7n−7q+6, Ω7q+7n−7q+5, …, Ω7q+7n are decreased, and a1,a2,…,a7q+7≥0 exists such that(7)limn⟶∞Ω7q+7n−7q+6+t=a1+t,t=0,7q+6¯.(b)
(8)∏t=06limn⟶∞Ω7q+7n−q+1t−q+i=0,i=0,q¯,or(9)∏t=06aq+1t+i=0,i=1,q+1¯.(c)
If there existsn0∈ℕ such that(10)∏t=05Ωm−q+1t−q≥Ωm+1∏t=04Ωm−q+1t−qfor alln≥n0, then(11)limn⟶∞Ωm=0.(d)
We can generate the following formulas:(12)Ω7q+7n+rq+1+s+1=Ωr−7q+1+s+11−∏m=17Ω−mq+m−1+s/Ωr−7q+1+s+11+∏m=16Ω−mq+m−1+s×∑h=0n∏k=17h+r1∏m=16Ωq+1t−mq+m−1+s+1.(e)
IfΩ7q+7n+t−1q+t⟶at−1q+t≠0, then Ω7q+7n+6q+7⟶0 as n⟶∞.⋮
IfΩ7q+7n+tq+t⟶atq+t≠0, then Ω7q+7n+7q+7⟶0 as n⟶∞. t=1,6¯.Proof.
(a)
Firstly, for alln∈ℕ, from (5), one gets Ωm+1<Ωm−7q+6. So,(13)limn⟶∞Ω7q+7n−7q+6+t=a1+t,t=0,7q+6¯.(b)
In view of equation (5),(14)Ω7q+7n+1=Ω7q+7n−7q+61+∏t=05Ω7q+7n−q+1t−q.Then,(15)limn⟶∞Ω7q+7n+1=limn⟶∞Ω7q+7n−7q+61+∏t=05Ω7q+7n−q+1t−q,∏t=06limn⟶∞Ω7q+7n−q+1t−q+i=0,i=0,q¯,or(16)∏t=06aq+1t+i=0,i=1,q+1¯,is obtained.(c)
If there existsn0∈ℕ such that(17)∏t=05Ωm−q+1t−q≥Ωm+1∏t=04Ωm−q+1t−qfor alln≥n0,then ifi=0,q¯,(18)a1+i≤aq+2+i≤a2q+3+i≤a3q+4+i≤a4q+5+i≤a5q+6+i≤a6q+7+i≤a1+i,and we obtained the result.(d)
SubtractingΩm−7q+6 from both sides in (5), we have(19)Ωm+1−Ωm−7q+6=11+∏t=05Ωm−q+1t−qΩm−q−Ωm−8q+7,and the following formula, forn≥q+1,(20)Ωq+1n−q+12−1−Ωq+1n−q+22+5q+3=Ω1−Ω−7q+6∏k=1n−q+111+∏t=05Ωq+1k−q+1t−q,⋮Ωq+1n−q+12−q+1−Ωq+1n−q+22+4q+3,=Ωq+1−Ω−6q+6∏k=1n−q+111+∏t=05Ωq+1k−q+1t.From (20), we get(21)Ω7q+7n+1−Ω−7q+6=Ω1−Ω−7q+6∑h=0n∏k=17h1∏t=05Ωq+1k−q+1t−q+1,⋅⋅⋅Ω7q+7n+q+1−Ω−6q+6=Ωq+1−Ω−6q+6∑h=0n∏k=17h1∏t=05Ωq+1k−q+1t+1.So,(22)Ω7q+7n+q+2−Ω−6q+5=Ω1−Ω−7q+6∑h=0n∏k=17h+11∏t=05Ωq+1k−q+1t−q+1,⋅⋅⋅Ω7q+7n+2q+2−Ω−5q+5=Ωq+1−Ω−6q+6∑h=0n∏k=17h+11∏t=05Ωq+1k−q+1t+1.Also,(23)Ω7q+7n+2q+3−Ω−5q+4=Ω1−Ω−7q+6∑h=0n∏k=17h+21∏t=05Ωq+1k−q+1t−q+1,⋅⋅⋅Ω7q+7n+3q+3−Ω−4q+4=Ωq+1−Ω−6q+6∑h=0n∏k=17h+21∏t=05Ωq+1k−q+1t+1.Moreover,(24)Ω7q+7n+3q+4−Ω−4q+3=Ω1−Ω−7q+6∑h=0n∏k=17h+31∏t=05Ωq+1k−q+1t−q+1,⋅⋅⋅Ω7q+7n+4q+4−Ω−3q+3=Ωq+1−Ω−6q+6∑h=0n∏k=17h+31∏t=05Ωq+1k−q+1t+1.On the contrary,(25)Ω7q+7n+4q+5−Ω−3q+2=Ω1−Ω−7q+6∑h=0n∏k=17h+41∏t=05Ωq+1k−q+1t−q+1,⋅⋅⋅Ω7q+7n+5q+5−Ω−2q+2=Ωq+1−Ω−6q+6∑h=0n∏k=17h+41∏t=05Ωq+1k−q+1t+1.Also,(26)Ω7q+7n+5q+6−Ω−2q+1=Ω1−Ω−7q+6∑h=0n∏k=17h+51∏t=05Ωq+1k−q+1t−q+1,⋅⋅⋅Ω7q+7n+6q+6−Ω−q+1=Ωq+1−Ω−6q+6∑h=0n∏k=17h+51∏t=05Ωq+1k−q+1t+1.Moreover,(27)Ω7q+7n+6q+7−Ω−q+=Ω1−Ω−7q+6∑h=0n∏k=17h+61∏t=05Ωq+1k−q+1t−q+1,⋅⋅⋅Ω7q+7n+7q+7−Ω0=Ωq+1−Ω−6q+6∑h=0n∏k=17h+61∏t=05Ωq+1k−q+1t+1.Now, we get the above formulas:(28)Ω7q+7n+rq+1+s+1=Ωr−7q+1+s+11−∏m=17Ω−mq+m−1+s/Ωr−7q+1+s+11+∏m=16Ω−mq+m−1+s×∑h=0n∏k=17h+r1∏m=16Ωq+1t−mq+m−1+s+1,wherer=0,6¯ and s=0,q¯ hold.(e)
Suppose thata1=aq+2=a2q+3=a3q+4=a4q+5=a5q+6=a6q+7=0. By (d), we produce the following formulas:(29)limn⟶∞Ω7q+7n+1=limn⟶∞Ω−7q+61−∏t=05Ω−q+1t−q∏t=05Ω−q+1t−q+1×∑h=0n∏k=17h1∏t=05Ω−q+1k−q+1t−q+1,a1=Ω−7q+61−∏t=05Ω−q+1t−qΩ−11+q∏t=05Ω−q+1t−q+1∑h=0∞∏k=17h1∏t=05Ω−q+1k−q+1t−q+1,a1=0⇒∏t=05Ω−q+1t−q+1∏t=05Ω−q+1t−q=∑h=0∞∏k=17h1∏t=05Ωq+1k−q+1t−q+1.
Similarly,(30)limn⟶∞Ω7q+7n+q+2=limn⟶∞Ω−6q+51−∏t=06Ω−q+1t−q/Ω−6q+5∏t=05Ω−q+1t−q+1×∑h=0n∏k=17h+11∏t=05Ω−q+1k−q+1t−q+1,aq+2=Ω−6q+51−∏t=06Ω−q+1t−q/Ω−6q+5∏t=05Ω−q+1t−q+1×∑h=0∞∏k=17h+11∏t=05Ω−q+1k−q+1t−q+1,aq+2=0⇒∏t=05Ω−q+1t−q+1∏t=06Ω−q+1t−q/Ω−6q+5=∑h=0∞∏k=17h+11∏t=05Ωq+1k−q+1t−q+1.
Similarly,(31)limn⟶∞Ω7q+7n+2q+3=limn⟶∞Ω−5q+41−∏t=06Ω−q+1t−q/Ω−5q+4∏t=05Ω−q+1t−q+1×∑h=0n∏k=17h+21∏t=05Ω−q+1k−q+1t−q+1,a2q+3=Ω−5q+41−∏t=06Ω−q+1t−q/Ω−5q+4∏t=05Ω−q+1t−q+1×∑h=0n∏k=17h+21∏t=05Ω−q+1k−q+1t−q+1,a2q+3=0⇒∏t=05Ω−q+1t−q+1∏t=06Ω−q+1t−q/Ω−5q+4=∑h=0∞∏k=17h+21∏t=05Ωq+1k−q+1t−q+1.
Similarly,(32)limn⟶∞Ω7q+7n+3q+4=limn⟶∞Ω−4q+31−∏t=06Ω−q+1t−q/Ω−4q+3∏t=05Ω−q+1t−q+1×∑h=0n∏k=17h+31∏t=05Ω−q+1k−q+1t−q+1,a3q+4=Ω−4q+31−∏t=06Ω−q+1t−q/Ω−4q+3∏t=05Ω−q+1t−q+1×∑h=0n∏k=17h+31∏t=05Ω−q+1k−q+1t−q+1,a3q+4=0⇒∏t=05Ω−q+1t−q+1∏t=06Ω−q+1t−q/Ω−4q+3=∑h=0∞∏k=17h+31∏t=05Ωq+1k−q+1t−q+1.
Similarly,(33)limn⟶∞Ω7q+7n+4q+5=limn⟶∞Ω−3q+21−∏t=06Ω−q+1t−q/Ω−3q+2∏t=05Ω−q+1t−q+1×∑h=0n∏k=17h+41∏t=05Ω−q+1k−q+1t−q+1,a4q+5=Ω−3q+21−∏t=06Ω−q+1t−q/Ω−3q+2∏t=05Ω−q+1t−q+1×∑h=0n∏k=17h+41∏t=05Ω−q+1k−q+1t−q+1,a4q+5=0⇒∏t=05Ω−q+1t−q+1∏t=06Ω−q+1t−q/Ω−3q+2=∑h=0∞∏k=17h+41∏t=05Ωq+1k−q+1t−q+1.
Similarly,(34)limn⟶∞Ω7q+7n+5q+6=limn⟶∞Ω−2q+11−∏t=06Ω−q+1t−q/Ω−2q+1∏t=05Ω−q+1t−q+1×∑h=0n∏k=17h+51∏t=05Ω−q+1k−q+1t−q+1,a5q+6=Ω−2q+11−∏t=06Ω−q+1t−q/Ω−2q+1∏t=05Ω−q+1t−q+1×∑h=0n∏k=17h+51∏t=05Ω−q+1k−q+1t−q+1,a5q+6=0⇒∏t=05Ω−q+1t−q+1∏t=06Ω−q+1t−q/Ω−2q+1=∑h=0∞∏k=17h+51∏t=05Ωq+1k−q+1t−q+1.
Similarly,(35)limn⟶∞Ω7q+7n+6q+7=limn⟶∞Ω−q1−∏t=06Ω−q+1t−q/Ω−2q+1∏t=05Ω−q+1t−q+1××∑h=0n∏k=17h+61∏t=05Ω−q+1k−q+1t−q+1,a6q+7=Ω−q1−∏t=06Ω−q+1t−q/Ω−q∏t=05Ω−q+1t−q+1∑h=0n∏k=17h+61∏t=05Ω−q+1k−q+1t−q+1,a6q+7=0⇒∏t=05Ω−q+1t−q+1∏t=06Ω−q+1t−q/Ω−q=∑h=0∞∏k=17h+61∏t=05Ωq+1k−q+1t−q+1.
From equations (29) and (30),(36)∏t=05Ω−q+1t−q+1∏t=05Ω−q+1t−q=∑h=0∞∏k=17h1∏t=05Ωq+1k−q+1t−q+1>∏t=05Ω−q+1t−q+1∏t=06Ω−q+1t−q/Ω−6q+5=∑h=0∞∏k=17h+11∏t=05Ωq+1k−q+1t−q+1.
Thus,Ω−7q+6>Ω−6q+5.
From equations (30) and (31),(37)∏t=05Ω−q+1t−q+1∏t=06Ω−q+1t−q/Ω−6q+5=∑h=0∞∏k=17h+11∏t=05Ωq+1k−q+1t−q+1>∏t=05Ω−q+1t−q+1∏t=06Ω−q+1t−q/Ω−5q+4=∑h=0∞∏k=17h+21∏t=05Ωq+1k−q+1t−q+1.
Thus,Ω−6q+5>Ω−5q+4.
From equations (31) and (32),(38)∏t=05Ω−q+1t−q+1∏t=06Ω−q+1t−q/Ω−5q+4=∑h=0∞∏k=17h+21∏t=05Ωq+1k−q+1t−q+1>∏t=05Ω−q+1t−q+1+1∏t=06Ω−q+1t−q/Ω−4q+3=∑h=0∞∏k=17h+31∏t=05Ωq+1k−q+1t−q+1.
Thus,Ω−5q+4>Ω−4q+3.
From equations (32) and (33),(39)∏t=05Ω−q+1t−q+1∏t=06Ω−q+1t−q/Ω−4q+3=∑h=0∞∏k=17h+31∏t=05Ωq+1k−q+1t−q+1>∏t=05Ω−q+1t−q+1∏t=06Ω−q+1t−q/Ω−3q+2=∑h=0∞∏k=17h+41∏t=05Ωq+1k−q+1t−q+1.
Thus,Ω−4q+3>Ω−3q+2.
From equations (33) and (34),(40)∏t=05Ω−q+1t−q+1∏t=06Ω−q+1t−q/Ω−3q+2=∑h=0∞∏k=17h+41∏t=05Ωq+1k−q+1t−q+1>∏t=05Ω−q+1t−q+1∏t=06Ω−q+1t−q/Ω−2q+1=∑h=0∞∏k=17h+51∏t=05Ωq+1k−q+1t−q+1.
Thus,Ω−3q+2>Ω−2q+1.
From equations (34) and (35),(41)∏t=05Ω−q+1t−q+1∏t=06Ω−q+1t−q/Ω−2q+1=∑h=0∞∏k=17h+51∏t=05Ωq+1k−q+1t−q+1>∏t=05Ω−q+1t−q+1∏t=06Ω−q+1t−q/Ω−q=∑h=0∞∏k=17h+61∏t=05Ωq+1k−q+1t−q+1.
Thus,Ω−2q+1>Ω−q.(42)Ω−7q+6>Ω−6q+5>Ω−5q+4>Ω−4q+3>Ω−3q+2>Ω−2q+1>Ω−q.


## 3. Examples

In this section, we consider some numerical examples.Example 1.
Assume that, forq=1, we get Ωm+1=Ω−13/1+Ω−1Ω−3Ω−5Ω−7Ω−9Ω−11.(43)Ω−13=10,Ω−12=9,Ω−11=2,Ω−10=28,Ω−9=27,Ω−8=6.5,Ω−7=5.5,Ω−6=24,Ω−5=23,Ω−4=22,Ω−3=21,Ω−2=5,Ω−1=4,Ω0=3.
Then, we have the graph in Figure1.Figure 1
Dynamics of equation (5) with initial conditions in Example 1.Example 2.
If we select the initial conditions as follows,(44)Ω−13=29,Ω−12=28,Ω−11=27,Ω−10=26,Ω−9=25,Ω−8=24,Ω−7=23,Ω−6=22,Ω−5=21,Ω−4=20,Ω−3=19,Ω−2=18,Ω−1=17,Ω0=16,then we have the graph in Figure 2.Figure 2
Dynamics of equation (5) with initial conditions in Example 2.



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