# Forms of Solutions for Some Two-Dimensional Systems of Rational Partial Recursion Equations

Authors: Tarek F. Ibrahim; A. Q. Khan
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

In this paper, we offer the closed-form expressions of systems of second-order partial difference equations. We will utilize an alternative approach to verify the results by (odd-even) dual mathematical induction. We research and enforce the specific solutions of partial difference formulas and ordinary difference formulas as a straight effect.


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## Body

## 1. Introduction

Partial difference and differential equations are common in mathematically oriented scientific areas, such as physics and engineering. They are fundamental in the modern-day scientific understanding of mechanical engineering analysis, sound, heat, diffusion, electrodynamics, electrostatics, fluid dynamics, elasticity, general relativity, and also quantum mechanics (see for instance [1–3]). We discover both ordinary and partial difference formulas in probability, characteristics, and even other mathematical physics locations. Certainly, Laplace, as well as Lagrange, took into consideration the solution of partial difference formulas in their research of characteristics as well as probability. An instance of a partial difference formula is the following famous relationship:(1)Cst=Cs−1t−1+Cst−1,1≤s<t.Some authors research solutions for partial difference formulas.For instance, Heins [4] considered the solution of(2)Qt+1,s+Qt−1,s=2Qt,s+1,under some conditions.In [5], Carlitz researched a solution of(3)Qt,s−Qt,s−1−Qt−1,s−Qt,s−2+3Qt−1,s−1−Qt−2,s=0.For more outcomes on partial difference equations, we refer the reader to [3, 6–35].In our work, we have actually researched the solutions complying with systems of partial difference formulas:(4)αQt,s+βQt,sQt−2,s−2Rt−1,s−1−Qt−2,s−2=0,(5)γRt,s+δRt,sRt−2,s−2Qt−1,s−1−Rt−2,s−2=0,where t,s∈ℕ0, ℕ0=ℕ∪0, α,β,γ,δ∈1,−1, and initial Qt,0,Qt,−1,Q0,s,Q−1,s, Rt,0,Rt,−1,R0,s, and R−1,s are real numbers.As a straight effect, we can derive the solutions of a family of partial difference equations:(6)αQt,s+βQt,sQt−2,s−2Qt−1,s−1−Qt−2,s−2=0,where t,s∈ℕ0, ℕ0=ℕ∪0, α,β∈1,−1, and initials Qt,0,Qt,−1,Q0,s, and Q−1,s are real numbers.In addition, we can acquire the precise solution adhering to systems of ordinary difference formulas:(7)αQt+βQtQt−2Rt−1−Qt−2=0,γRt+δRtRt−2Qt−1−Rt−2=0,where t∈ℕ0, ℕ0=ℕ∪0, α,β,γ,δ∈1,−1, and initials Q0,Q−1,R0, and R−1 are real numbers.


## 2. Solution Forms

We give solutions of systems (4) and (5) for values of α,β,γ,δ∈1,−1. This system can be rewritten as(8)Qt,s=Qt−2,s−2α+βQt−2,s−2Rt−1,s−1,Rt,s=Rt−2,s−2γ+δRt−2,s−2Qt−1,s−1.


### 2.1. Whenα,β=−1,1andγ,δ=−1,−1

(9)Qt,s=Qt−2,s−2−1+Qt−2,s−2Rt−1,s−1,Rt,s=Rt−2,s−2−1−Rt−2,s−2Qt−1,s−1.Theorem 1.
LetQt,s,Rt,st,s=−k∞ be a solution of (9) with(10)Qt,0,Qt,−1,Q0,s,Q−1,s,Rt,0,Rt,−1,R0,s,R−1,s,where t,s∈ℕ0 and ℕ0=ℕ∪0. Suppose Q−1,s−2R0,s−1≠1, Qt−2,−1Rt−1,0≠1, R−1,s−2Q0,s−1≠−1, and Rt−2,−1Qt−1,0≠−1. Then, the solutions of (9), for t,s≥1 and t≥s, are(11)Qt,s=−1s/2Qt−s,0∏k=0s−2/21+2k+1Qt−s,0Rt−s−1,−11+2k+2Qt−s,0Rt−s−1,−1,seven,−1t+1/2Qt−s−1,−1∏k=0s−1/2−1+2kQt−s−1,−1Rt−s,0−1+2k+1Qt−s−1,−1Rt−s,0,sodd,(12)Rt,s=−1s/2Rt−s,0∏k=0s−2/2−1+2k+1Rt−s,0Qt−s−1,−1−1+2k+2Rt−s,0Qt−s−1,−1,seven,−1t+1/2Rt−s−1,−1∏k=0s−1/21+2kRt−s−1,−1Qt−s,01+2k+1Rt−s−1,−1Qt−s,0,sodd,(13)Qs,n=−1t+1/2Q−1,t−s−1∏k=0s−1/2−1+2kQ−1,t−s−1R0,t−s−1+2k+1Q−1,t−s−1R0,t−s,sodd,−1s/2Q0,t−s∏k=0s−2/21+2k+1Q0,t−sR−1,t−s−11+2k+2Q0,t−sR−1,t−s−1,seven,(14)Rs,n=−1t+1/2R−1,t−s−1∏k=0s−1/21+2kR−1,t−s−1Q0,t−s1+2k+1R−1,t−s−1Q0,t−s,sodd,−1s/2R0,t−s∏k=0s−2/2−1+2k+1R0,t−sQ−1,t−s−1−1+2k+2R0,t−sQ−1,t−s−1,seven.Proof.
From system (9), we see the following:(15)Q1,1=Q−1,−1−1+Q−1,−1R0,0=−Q−1,−1∏k=01−1/2−1+2kQ−1,−1R0,0−1+2k+1Q−1,−1R0,0,R1,1=R−1,−1−1−R−1,−1Q0,0=−R−1,−1∏k=01−1/21+2kR−1,−1Q0,01+2k+1R−1,−1Q0,0.
Now, we prove that (11)–(14) hold for t,s=2,2:(16)Q2,2=Q0,0−1+Q0,0R1,1=Q0,0−1+Q0,0R−1,−1/−1−R−1,−1Q0,0=−Q0,01+Q0,0R−1,−11+2Q0,0R−1,−1=−Q0,0∏k=02−2/21+2k+1Q0,0R−1,−11+2k+2Q0,0R−1,−1,R2,2=R0,0−1−R0,0Q1,1=R0,0−1−R0,0Q−1,−1/−1−Q−1,−1R0,0=−R0,01−R0,0Q−1,−11−2R0,0Q−1,−1=−R0,0∏k=02−2/2−1+2k+1R0,0Q−1,−1−1+2k+2R0,0Q−1,−1.
Moreover, we prove that (11)–(14) hold for t,s=1,2 and t,s=2,1:(17)Q1,2=Q−1,0−1+Q−1,0R0,1=−Q−1,0∏k=01−1/2−1+2kQ−1,0R0,1−1+2k+1Q−1,0R0,1,R1,2=R−1,0−1−R−1,0Q0,1=−R−1,0∏k=01−1/21+2kR−1,0Q0,11+2k+1R−1,0Q0,1,Q2,1=Q0,−1−1+Q0,−1R1,0=−Q0,−1∏k=01−1/2−1+2kQ0,−1R1,0−1+2k+1Q0,−1R1,0,R2,1=R0,−1−1−R0,−1Q1,0=−R0,−1∏k=01−1/21+2kR0,−1Q1,01+2k+1R0,−1Q1,0.
Now, suppose that (11)–(14) hold for s=1 and s=2 with t∈ℕ. So,(18)Qt,1=−Qt−2,−1∏k=00−1+2kQt−2,−1Rt−1,0−1+2k+1Qt−2,−1Rt−1,0=−Qt−2,−11−Qt−2,−1Rt−1,0,Rt,1=−Rt−2,−11+Rt−2,−1Qt−1,0,Qt,2=−Qt−2,01+Qt−2,0Rt−3,−11+2Qt−2,0Rt−3,−1,Rt,2=−Rt−2,01−Rt−2,0Qt−3,−11−2Rt−2,0Qt−3,−1.
Now, prove that (11)–(14) hold for s=1 with t+2:(19)Qt+2,1=Qt,−1−1+Qt,−1Rs+1,0=−Qt,−1∏k=00/21−2kQt,−1Rs+1,01−2k+1Qt,−1Rs+1,0,Rt+2,1=Rt,−1−1−Rt,−1Qs+1,0=−Rt,−1∏k=01−1/21+2kRt,−1Qs+1,01+2k+1Rt,−1Qs+1,0.
Now, we prove that (11)–(14) hold for s=2 with t+2:(20)Qt+2,2=Qt,0−1+Qt,0Rs+1,1=Qt,0−1+Qt,0Rt−1,−1/−1−Rt−1,−1Qt,0=−Qt,01+Rt−1,−1Qt,01+2Rt−1,−1Qt,0=−Qt,0∏k=02−2/21+2k+1Qt,0Rt−1,−11+2k+2Qt,0Rt−1,−1.Similarly,(21)Rt+2,2=Rt,0−1−Rt,0Qs+1,1=−Rt,0∏k=02−2/21−2k+1Rt,0Qt−1,−11−2k+2Rt,0Qt−1,−1.
From (9), we have(22)Qt,s+2=Qt−2,s−1+Qt−2,sRt−1,s+1.
There are four cases:(1)
Ift>s+2 and s is even,(23)Qt,s+2=Qt−2,s−1+Qt−2,sRt−1,s+1=−1s/2Qt−s−2,0∏k=0s−2/21+2k+1Φ/1+2k+2Φ−1+−1s/2Qt−s−2,0∏k=0s−2/21+2k+1Φ/1+2k+2Φ−1s+2/2Rt−s−3,−1∏∏k=0s/21+2kΦ/1+2k+1Φ=−1s+2/2Qt−s−2,0∏k=0s−2/21+2k+1Φ/1+2k+2Φ1+Φ/1+t+1Φ=−1s+2/2Qt−s−2,0∏k=0s/21+2k+1Φ1+2k+2Φ,whereΦ=Qt−s−2,0Rt−s−3,−1.(2)
Ift>s+2 and s is odd,(24)Qt,s+2=Qt−2,s−1+Qt−2,sRt−1,s+1=−1t+1/2Qt−s−3,−1∏k=0s−1/2−1+2kΨ/−1+2k+1Ψ−1+−1t+1/2Qt−s−3,−1∏k=0s−1/2−1+2kΨ/−1+2k+1Ψ−1t+1/2Rt−s−2,0∏k=0s−1/2−1+2k+1Ψ/−1+2k+2Ψ=−1t+1/2Qt−s−3,−1∏k=0s−1/2−1+2kΨ/−1+2k+1Ψ−1−Ψ/−1+t+1Ψ=−1s+3/2Qt−s−3,−1∏k=0t+1/2−1+2kΨ−1+2k+1Ψ,whereΨ=Qt−s−3,−1Rt−s−2,0.(3)
This is as in part (1)(4)
This is as in part (2)Proposition 1.
Properties for system (9) are as follows:(1)
Ifs is even and Qt−s,0=0, then Qt,s=0(2)
Ifs is odd and Qt−s,0=0, then Rt,s=−1t+1/2Rt−s−1,−1(3)
Ifs is even and Rt−s,0=0, then Rt,s=0(4)
Ifs is odd and Rt−s,0=0, then Qt,s=−1t+1/2Qt−s−1,−1(5)
Ifs is even and Qt−s−1,−1=0, then Rt,s=−1s/2Rt−s,0(6)
Ifs is odd and Qt−s−1,−1=0, then Qt,s=0(7)
Ifs is even and Rt−s−1,−1=0, then Qt,s=−1s/2Qt−s,0(8)
Ifs is odd and Rt−s−1,−1=0, then Rt,s=0Proposition 2.
We have the following properties for system (9):(1)
Ifs is even and Q0,t−s=0, then Qs,t=0(2)
Ifs is odd and Q0,t−s=0, then Rs,t=−1t+1/2R−1,t−s−1(3)
Ifs is even and R0,t−s=0, then Rs,t=0(4)
Ifs is odd and R0,t−s=0, then Qs,t=−1t+1/2Q−1,t−s−1(5)
Ifs is even and Q−1,t−s−1=0, then Rs,t=−1s/2R0,t−s(6)
Ifs is odd and Q−1,t−s−1=0, then Qs,t=0(7)
Ifs is even and R−1,t−s−1=0, then Qs,t=−1s/2Q0,t−s(8)
Ifs is odd and R−1,t−s−1=0, then Rs,t=0Remark 1.
As a special case of (9), we have(25)Qt=Qt−2−1+Qt−2Rt−1,Rt=Rt−2−1−Rt−2Qt−1.Corollary 1.
LetQt,Rtt=−k∞ be a solution of (25) with initials Q0,Q−1,R0,andR−1. Suppose Q−1R0≠1 and R−1Q0≠−1. Then,(26)Qt=−1t/2Q0∏k=0t−2/21+2k+1Q0R−11+2k+2Q0R−1,teven,−1t+1/2Q−1∏k=0t−1/2−1+2kQ−1R0−1+2k+1Q−1R0todd,Rt=−1t/2R0∏k=0t−2/2−1+2k+1R0Q−1−1+2k+2R0Q−1,teven,−1t+1/2R−1∏k=0t−1/21+2kR−1Q01+2k+1R−1Q0todd.


### 2.2. Whenα,β=γ,δ=−1,−1

(27)Qt,s=Qt−2,s−2−1−Qt−2,s−2Rt−1,s−1,Rt,s=Rt−2,s−2−1−Rt−2,s−2Qt−1,s−1.Theorem 2.
LetQt,s,Rt,st,s=−k∞ be a solution of (27) with Qt,0,Qt,−1,Q0,s,Q−1,s,Rt,0,Rt,−1,R0,s,andR−1,s where t,s∈ℕ0 and ℕ0=ℕ∪0. Suppose Q−1,s−2R0,s−1≠−1, Qt−2,−1Rt−1,0≠−1, R−1,s−2Q0,s−1≠−1, and Rt−2,−1Qt−1,0≠−1. Then,(28)Qt,s=−1t+1/2Qt−s−1,−11+Qt−s−1,−1Rt−s,0t+1/2,sodd;−1s/2Qt−s,01+Qt−s,0Rt−s−1,−1s/2,seven,Rt,s=−1t+1/2Rt−s−1,−11+Rt−s−1,−1Qt−s,0t+1/2,sodd;−1s/2Rt−s,01+Rt−s,0Qt−s−1,−1s/2,seven,Qs,t=−1t+1/2Q−1,t−s−11+Q−1,t−s−1R0,t−st+1/2,sodd;−1s/2Q0,t−s1+Q0,t−sR−1,t−s−1s/2,seven,Rs,t=−1t+1/2R−1,t−s−11+R−1,t−s−1Q0,t−st+1/2,sodd;−1s/2R0,t−s1+R0,t−sQ−1,t−s−1s/2,seven.Proof.
The proof is given as in Theorem (1).Remark 2.
If we consider(29)Qt,s=Qt−2,s−2−1−Qt−2,s−2Qt−1,s−1,then the solution of equation (29) is as follows.Corollary 2.
LetQt,st,s=−k∞ be a solution of (29) with Qt,0,Qt,−1,Q0,s,andQ−1,s where t,s∈ℕ0 and ℕ0=ℕ∪0. Suppose Q−1,s−2Q0,s−1≠−1 and Qt−2,−1Qt−1,0≠−1. Then,(30)Qt,s=−1t+1/2Qt−s−1,−11+Qt−s−1,−1Qt−s,0t+1/2,sodd,−1s/2Qt−s,01+Qt−s,0Qt−s−1,−1s/2,seven,Qs,t=−1t+1/2Q−1,t−s−11+Q−1,t−s−1Q0,t−st+1/2,sodd,−1s/2Q0,t−s1+Q0,t−sQ−1,t−s−1s/2,seven.Proposition 3.
We have the following:(1)
Ifs is even and Qt−s,0=0, then Qt,s=0(2)
Ifs is odd and Qt−s,0=0, then Qt,s=−1t+1/2Qt−s−1,−1(3)
Ifs is even and Qt−s−1,−1=0, then Qt,s=−1s/2Qt−s,0(4)
Ifs is odd and Qt−s−1,−1=0, then Qt,s=0(5)
Ifs is even and Q0,t−s=0, then Qs,t=0(6)
Ifs is odd and Q0,t−s=0, then Qs,t=−1t+1/2Q−1,t−s−1(7)
Ifs is even and Q−1,t−s−1=0, then Qs,t=−1s/2Q0,t−s(8)
Ifs is odd and Q−1,t−s−1=0, then Qs,t=0Remark 3.
As a special of system (27), we have(31)Qt=Qt−2−1−Qt−2Rt−1,Rt=Rt−2−1−Rt−2Qt−1.Corollary 3.
LetQt,Rtt=−k∞ be a solution of (31) with Q0,Q−1,R0,R−1, Q−1R0≠−1, and R−1Q0≠−1. Then,(32)Qt=−1t+1/2Q−11+Q−1R0t+1/2,todd,−1t/2Q01+Q0R−1t/2,teven,Rt=−1t+1/2R−11+R−1Q0t+1/2,todd,−1t/2R01+R0Q−1t/2,teven.Remark 4.
As a special case of system (31), we get(33)Qt=Qt−2−1−Qt−2Qt−1.
The closed-form solution of (33) is(34)Qt=−1t+1/2Q−11+Q−1Q0t+1/2,todd,−1t/2Q01+Q0Q−1t/2,teven,where t∈ℕ and Q−1Q0≠−1.
Ifn is even (or odd) and Q0=0, then Qt=0(Qt=−1t+1/2Q−1).
Also, ifn is even (or odd) and Q−1=0, then Qt=−1t/2Q0(Qt=0).


### 2.3. Whenα,β=1,1andγ,δ=−1,−1

(35)Qt,s=Qt−2,s−21+Qt−2,s−2Rt−1,s−1,Rt,s=Rt−2,s−2−1−Rt−2,s−2Qt−1,s−1.Theorem 3.
LetQt,s,Rt,st,s=−k∞ be a solution of (35) with Qt,0,Qt,−1,Q0,s,Q−1,s,Rt,0,Rt,−1,R0,s,andR−1,s where t,s∈ℕ0 and ℕ0=ℕ∪0. Suppose Q−1,s−2R0,s−1≠−1, Qt−2,−1Rt−1,0≠−1, R−1,s−2Q0,s−1≠−1, and Rt−2,−1Qt−1,0≠−1. Then,(36)Qt,s=Qt−s−1,−11+2Qt−s−1,−1Rt−s,0s−1/41+Qt−s−1,−1Rt−s,0t+1/2,s=4K+1,−1s−2/4Qt−s,0−1+Qt−s,0Rt−s−1,−1s−2/41+Qt−s,0Rt−s−1,−1s+2/4,s=4K+2,Qt−s−1,−11+2Qt−s−1,−1Rt−s,0t+1/41+Qt−s−1,−1Rt−s,0t+1/2,s=4K+3,−1s/4Qt−s,0−1+Qt−s,0Rt−s−1,−1s/41+Qt−s,0Rt−s−1,−1s/4,s=4K+4,Rt,s=−1s+3/4Rt−s−1,−1−1+Rt−s−1,−1Qt−s,0s−1/41+Rt−s−1,−1Qt−s,0s+3/4,s=4K+1,−1s+2/4Rt−s,01+Rt−s,0Qt−s−1,−1s/21+2Rt−s,0Qt−s−1,−1s+2/4,s=4K+2,−1t+1/4Rt−s−1,−1−1+Rt−s−1,−1Qt−s,0t+1/41+Rt−s−1,−1Qt−s,0t+1/4,s=4K+3,Rt−s,01+Rt−s,0Qt−s−1,−1s/21+2Rt−s,0Qt−s−1,−1s/4,s=4K+4,Qs,t=Q−1,t−s−11+2Q−1,t−s−1R0,t−ss−1/41+Q−1,t−s−1R0,t−st+1/2,s=4K+1,−1s−2/4Q0,t−s−1+Q0,t−sR−1,t−s−1s−2/41+Q0,t−sR−1,t−s−1s+2/4,s=4K+2,Q−1,t−s−11+2Q−1,t−s−1R0,t−st+1/41+Q−1,t−s−1R0,t−st+1/2,s=4K+3,−1s/4Q0,t−s−1+Q0,t−sR−1,t−s−1s/41+Q0,t−sR−1,t−s−1s/4,s=4K+4,Rs,n=−1s+3/4R−1,t−s−1−1+R−1,t−s−1Q0,t−ss−1/41+R−1,t−s−1Q0,t−ss+3/4,s=4K+1,−1s+2/4R0,t−s1+R0,t−sQ−1,t−s−1s/21+2R0,t−sQ−1,t−s−1s+2/4,s=4K+2,−1t+1/4R−1,t−s−1−1+R−1,t−s−1Q0,t−st+1/41+R−1,t−s−1Q0,t−st+1/4,s=4K+3,R0,t−s1+R0,t−sQ−1,t−s−1s/21+2R0,t−sQ−1,t−s−1s/4,s=4K+4,where k=0,1,2,3,….Proof.
The proof is given by piecewise double mathematical induction.Proposition 4.
We have the following:(1)
Ifs is even and Qt−s,0=0, then Qt,s=0(2)
Ifs is odd and Qt−s,0=0, then Rt,s=±Rt−s−1,−1(3)
Ifs is even and Rt−s,0=0, then Rt,s=0(4)
Ifs is odd and Rt−s,0=0, then Qt,s=Qt−s−1,−1(5)
Ifs is even and Qt−s−1,−1=0, then Rt,s=±Rt−s,0(6)
Ifs is odd and Qt−s−1,−1=0, then Qt,s=0(7)
Ifs is even and Rt−s−1,−1=0, then Qt,s=Qt−s,0(8)
Ifs is odd and Rt−s−1,−1=0, then Rt,s=0Proposition 5.
We have the following:(1)
Ifs is even and Q0,t−s=0, then Qs,t=0(2)
Ifs is odd and Q0,t−s=0, then Rs,t=±R−1,t−s−1(3)
Ifs is even and R0,t−s=0, then Rs,t=0(4)
Ifs is odd and R0,t−s=0, then Qs,t=Q−1,t−s−1(5)
Ifs is even and Q−1,t−s−1=0, then Rs,t=±R0,t−s(6)
Ifs is odd and Q−1,t−s−1=0, then Qs,t=0(7)
Ifs is even and R−1,t−s−1=0, then Qs,t=Q0,t−s(8)
Ifs is odd and R−1,t−s−1=0, then Rs,t=0Remark 5.
As a special case,(37)Qt=Qt−21+Qt−2Rt−1,Rt=Rt−2−1−Rt−2Qt−1.Corollary 4.
LetQt,Rtt=−k∞ be a solution of (37) with initials Q0,Q−1,R0,andR−1. Suppose Q−1R0≠−1 and R−1Q0≠−1. Then,(38)Qt=Q−11+2Q−1R0t−1/41+Q−1R0t+1/2,t=4K+1,−1t−2/4Q0−1+Q0R−1t−2/41+Q0R−1t+2/4,t=4K+2,Q−11+2Q−1R0t+1/41+Q−1R0t+1/2,t=4K+3,−1t/4Q0−1+Q0R−1t/41+Q0R−1t/4,t=4K+4,Rt=−1t+3/4R−1−1+R−1Q0t−1/41+R−1Q0t+3/4,t=4K+1,−1t+2/4R01+R0Q−1t/21+2R0Q−1t+2/4,t=4K+2,−1t+1/4R−1−1+R−1Q0t+1/41+R−1Q0t+1/4,t=4K+3,R01+R0Q−1t/21+2R0Q−1t/4,t=4K+4,where k=0,1,2,3….


## 2.1. Whenα,β=−1,1andγ,δ=−1,−1

(9)Qt,s=Qt−2,s−2−1+Qt−2,s−2Rt−1,s−1,Rt,s=Rt−2,s−2−1−Rt−2,s−2Qt−1,s−1.Theorem 1.
LetQt,s,Rt,st,s=−k∞ be a solution of (9) with(10)Qt,0,Qt,−1,Q0,s,Q−1,s,Rt,0,Rt,−1,R0,s,R−1,s,where t,s∈ℕ0 and ℕ0=ℕ∪0. Suppose Q−1,s−2R0,s−1≠1, Qt−2,−1Rt−1,0≠1, R−1,s−2Q0,s−1≠−1, and Rt−2,−1Qt−1,0≠−1. Then, the solutions of (9), for t,s≥1 and t≥s, are(11)Qt,s=−1s/2Qt−s,0∏k=0s−2/21+2k+1Qt−s,0Rt−s−1,−11+2k+2Qt−s,0Rt−s−1,−1,seven,−1t+1/2Qt−s−1,−1∏k=0s−1/2−1+2kQt−s−1,−1Rt−s,0−1+2k+1Qt−s−1,−1Rt−s,0,sodd,(12)Rt,s=−1s/2Rt−s,0∏k=0s−2/2−1+2k+1Rt−s,0Qt−s−1,−1−1+2k+2Rt−s,0Qt−s−1,−1,seven,−1t+1/2Rt−s−1,−1∏k=0s−1/21+2kRt−s−1,−1Qt−s,01+2k+1Rt−s−1,−1Qt−s,0,sodd,(13)Qs,n=−1t+1/2Q−1,t−s−1∏k=0s−1/2−1+2kQ−1,t−s−1R0,t−s−1+2k+1Q−1,t−s−1R0,t−s,sodd,−1s/2Q0,t−s∏k=0s−2/21+2k+1Q0,t−sR−1,t−s−11+2k+2Q0,t−sR−1,t−s−1,seven,(14)Rs,n=−1t+1/2R−1,t−s−1∏k=0s−1/21+2kR−1,t−s−1Q0,t−s1+2k+1R−1,t−s−1Q0,t−s,sodd,−1s/2R0,t−s∏k=0s−2/2−1+2k+1R0,t−sQ−1,t−s−1−1+2k+2R0,t−sQ−1,t−s−1,seven.Proof.
From system (9), we see the following:(15)Q1,1=Q−1,−1−1+Q−1,−1R0,0=−Q−1,−1∏k=01−1/2−1+2kQ−1,−1R0,0−1+2k+1Q−1,−1R0,0,R1,1=R−1,−1−1−R−1,−1Q0,0=−R−1,−1∏k=01−1/21+2kR−1,−1Q0,01+2k+1R−1,−1Q0,0.
Now, we prove that (11)–(14) hold for t,s=2,2:(16)Q2,2=Q0,0−1+Q0,0R1,1=Q0,0−1+Q0,0R−1,−1/−1−R−1,−1Q0,0=−Q0,01+Q0,0R−1,−11+2Q0,0R−1,−1=−Q0,0∏k=02−2/21+2k+1Q0,0R−1,−11+2k+2Q0,0R−1,−1,R2,2=R0,0−1−R0,0Q1,1=R0,0−1−R0,0Q−1,−1/−1−Q−1,−1R0,0=−R0,01−R0,0Q−1,−11−2R0,0Q−1,−1=−R0,0∏k=02−2/2−1+2k+1R0,0Q−1,−1−1+2k+2R0,0Q−1,−1.
Moreover, we prove that (11)–(14) hold for t,s=1,2 and t,s=2,1:(17)Q1,2=Q−1,0−1+Q−1,0R0,1=−Q−1,0∏k=01−1/2−1+2kQ−1,0R0,1−1+2k+1Q−1,0R0,1,R1,2=R−1,0−1−R−1,0Q0,1=−R−1,0∏k=01−1/21+2kR−1,0Q0,11+2k+1R−1,0Q0,1,Q2,1=Q0,−1−1+Q0,−1R1,0=−Q0,−1∏k=01−1/2−1+2kQ0,−1R1,0−1+2k+1Q0,−1R1,0,R2,1=R0,−1−1−R0,−1Q1,0=−R0,−1∏k=01−1/21+2kR0,−1Q1,01+2k+1R0,−1Q1,0.
Now, suppose that (11)–(14) hold for s=1 and s=2 with t∈ℕ. So,(18)Qt,1=−Qt−2,−1∏k=00−1+2kQt−2,−1Rt−1,0−1+2k+1Qt−2,−1Rt−1,0=−Qt−2,−11−Qt−2,−1Rt−1,0,Rt,1=−Rt−2,−11+Rt−2,−1Qt−1,0,Qt,2=−Qt−2,01+Qt−2,0Rt−3,−11+2Qt−2,0Rt−3,−1,Rt,2=−Rt−2,01−Rt−2,0Qt−3,−11−2Rt−2,0Qt−3,−1.
Now, prove that (11)–(14) hold for s=1 with t+2:(19)Qt+2,1=Qt,−1−1+Qt,−1Rs+1,0=−Qt,−1∏k=00/21−2kQt,−1Rs+1,01−2k+1Qt,−1Rs+1,0,Rt+2,1=Rt,−1−1−Rt,−1Qs+1,0=−Rt,−1∏k=01−1/21+2kRt,−1Qs+1,01+2k+1Rt,−1Qs+1,0.
Now, we prove that (11)–(14) hold for s=2 with t+2:(20)Qt+2,2=Qt,0−1+Qt,0Rs+1,1=Qt,0−1+Qt,0Rt−1,−1/−1−Rt−1,−1Qt,0=−Qt,01+Rt−1,−1Qt,01+2Rt−1,−1Qt,0=−Qt,0∏k=02−2/21+2k+1Qt,0Rt−1,−11+2k+2Qt,0Rt−1,−1.Similarly,(21)Rt+2,2=Rt,0−1−Rt,0Qs+1,1=−Rt,0∏k=02−2/21−2k+1Rt,0Qt−1,−11−2k+2Rt,0Qt−1,−1.
From (9), we have(22)Qt,s+2=Qt−2,s−1+Qt−2,sRt−1,s+1.
There are four cases:(1)
Ift>s+2 and s is even,(23)Qt,s+2=Qt−2,s−1+Qt−2,sRt−1,s+1=−1s/2Qt−s−2,0∏k=0s−2/21+2k+1Φ/1+2k+2Φ−1+−1s/2Qt−s−2,0∏k=0s−2/21+2k+1Φ/1+2k+2Φ−1s+2/2Rt−s−3,−1∏∏k=0s/21+2kΦ/1+2k+1Φ=−1s+2/2Qt−s−2,0∏k=0s−2/21+2k+1Φ/1+2k+2Φ1+Φ/1+t+1Φ=−1s+2/2Qt−s−2,0∏k=0s/21+2k+1Φ1+2k+2Φ,whereΦ=Qt−s−2,0Rt−s−3,−1.(2)
Ift>s+2 and s is odd,(24)Qt,s+2=Qt−2,s−1+Qt−2,sRt−1,s+1=−1t+1/2Qt−s−3,−1∏k=0s−1/2−1+2kΨ/−1+2k+1Ψ−1+−1t+1/2Qt−s−3,−1∏k=0s−1/2−1+2kΨ/−1+2k+1Ψ−1t+1/2Rt−s−2,0∏k=0s−1/2−1+2k+1Ψ/−1+2k+2Ψ=−1t+1/2Qt−s−3,−1∏k=0s−1/2−1+2kΨ/−1+2k+1Ψ−1−Ψ/−1+t+1Ψ=−1s+3/2Qt−s−3,−1∏k=0t+1/2−1+2kΨ−1+2k+1Ψ,whereΨ=Qt−s−3,−1Rt−s−2,0.(3)
This is as in part (1)(4)
This is as in part (2)Proposition 1.
Properties for system (9) are as follows:(1)
Ifs is even and Qt−s,0=0, then Qt,s=0(2)
Ifs is odd and Qt−s,0=0, then Rt,s=−1t+1/2Rt−s−1,−1(3)
Ifs is even and Rt−s,0=0, then Rt,s=0(4)
Ifs is odd and Rt−s,0=0, then Qt,s=−1t+1/2Qt−s−1,−1(5)
Ifs is even and Qt−s−1,−1=0, then Rt,s=−1s/2Rt−s,0(6)
Ifs is odd and Qt−s−1,−1=0, then Qt,s=0(7)
Ifs is even and Rt−s−1,−1=0, then Qt,s=−1s/2Qt−s,0(8)
Ifs is odd and Rt−s−1,−1=0, then Rt,s=0Proposition 2.
We have the following properties for system (9):(1)
Ifs is even and Q0,t−s=0, then Qs,t=0(2)
Ifs is odd and Q0,t−s=0, then Rs,t=−1t+1/2R−1,t−s−1(3)
Ifs is even and R0,t−s=0, then Rs,t=0(4)
Ifs is odd and R0,t−s=0, then Qs,t=−1t+1/2Q−1,t−s−1(5)
Ifs is even and Q−1,t−s−1=0, then Rs,t=−1s/2R0,t−s(6)
Ifs is odd and Q−1,t−s−1=0, then Qs,t=0(7)
Ifs is even and R−1,t−s−1=0, then Qs,t=−1s/2Q0,t−s(8)
Ifs is odd and R−1,t−s−1=0, then Rs,t=0Remark 1.
As a special case of (9), we have(25)Qt=Qt−2−1+Qt−2Rt−1,Rt=Rt−2−1−Rt−2Qt−1.Corollary 1.
LetQt,Rtt=−k∞ be a solution of (25) with initials Q0,Q−1,R0,andR−1. Suppose Q−1R0≠1 and R−1Q0≠−1. Then,(26)Qt=−1t/2Q0∏k=0t−2/21+2k+1Q0R−11+2k+2Q0R−1,teven,−1t+1/2Q−1∏k=0t−1/2−1+2kQ−1R0−1+2k+1Q−1R0todd,Rt=−1t/2R0∏k=0t−2/2−1+2k+1R0Q−1−1+2k+2R0Q−1,teven,−1t+1/2R−1∏k=0t−1/21+2kR−1Q01+2k+1R−1Q0todd.


## 2.2. Whenα,β=γ,δ=−1,−1

(27)Qt,s=Qt−2,s−2−1−Qt−2,s−2Rt−1,s−1,Rt,s=Rt−2,s−2−1−Rt−2,s−2Qt−1,s−1.Theorem 2.
LetQt,s,Rt,st,s=−k∞ be a solution of (27) with Qt,0,Qt,−1,Q0,s,Q−1,s,Rt,0,Rt,−1,R0,s,andR−1,s where t,s∈ℕ0 and ℕ0=ℕ∪0. Suppose Q−1,s−2R0,s−1≠−1, Qt−2,−1Rt−1,0≠−1, R−1,s−2Q0,s−1≠−1, and Rt−2,−1Qt−1,0≠−1. Then,(28)Qt,s=−1t+1/2Qt−s−1,−11+Qt−s−1,−1Rt−s,0t+1/2,sodd;−1s/2Qt−s,01+Qt−s,0Rt−s−1,−1s/2,seven,Rt,s=−1t+1/2Rt−s−1,−11+Rt−s−1,−1Qt−s,0t+1/2,sodd;−1s/2Rt−s,01+Rt−s,0Qt−s−1,−1s/2,seven,Qs,t=−1t+1/2Q−1,t−s−11+Q−1,t−s−1R0,t−st+1/2,sodd;−1s/2Q0,t−s1+Q0,t−sR−1,t−s−1s/2,seven,Rs,t=−1t+1/2R−1,t−s−11+R−1,t−s−1Q0,t−st+1/2,sodd;−1s/2R0,t−s1+R0,t−sQ−1,t−s−1s/2,seven.Proof.
The proof is given as in Theorem (1).Remark 2.
If we consider(29)Qt,s=Qt−2,s−2−1−Qt−2,s−2Qt−1,s−1,then the solution of equation (29) is as follows.Corollary 2.
LetQt,st,s=−k∞ be a solution of (29) with Qt,0,Qt,−1,Q0,s,andQ−1,s where t,s∈ℕ0 and ℕ0=ℕ∪0. Suppose Q−1,s−2Q0,s−1≠−1 and Qt−2,−1Qt−1,0≠−1. Then,(30)Qt,s=−1t+1/2Qt−s−1,−11+Qt−s−1,−1Qt−s,0t+1/2,sodd,−1s/2Qt−s,01+Qt−s,0Qt−s−1,−1s/2,seven,Qs,t=−1t+1/2Q−1,t−s−11+Q−1,t−s−1Q0,t−st+1/2,sodd,−1s/2Q0,t−s1+Q0,t−sQ−1,t−s−1s/2,seven.Proposition 3.
We have the following:(1)
Ifs is even and Qt−s,0=0, then Qt,s=0(2)
Ifs is odd and Qt−s,0=0, then Qt,s=−1t+1/2Qt−s−1,−1(3)
Ifs is even and Qt−s−1,−1=0, then Qt,s=−1s/2Qt−s,0(4)
Ifs is odd and Qt−s−1,−1=0, then Qt,s=0(5)
Ifs is even and Q0,t−s=0, then Qs,t=0(6)
Ifs is odd and Q0,t−s=0, then Qs,t=−1t+1/2Q−1,t−s−1(7)
Ifs is even and Q−1,t−s−1=0, then Qs,t=−1s/2Q0,t−s(8)
Ifs is odd and Q−1,t−s−1=0, then Qs,t=0Remark 3.
As a special of system (27), we have(31)Qt=Qt−2−1−Qt−2Rt−1,Rt=Rt−2−1−Rt−2Qt−1.Corollary 3.
LetQt,Rtt=−k∞ be a solution of (31) with Q0,Q−1,R0,R−1, Q−1R0≠−1, and R−1Q0≠−1. Then,(32)Qt=−1t+1/2Q−11+Q−1R0t+1/2,todd,−1t/2Q01+Q0R−1t/2,teven,Rt=−1t+1/2R−11+R−1Q0t+1/2,todd,−1t/2R01+R0Q−1t/2,teven.Remark 4.
As a special case of system (31), we get(33)Qt=Qt−2−1−Qt−2Qt−1.
The closed-form solution of (33) is(34)Qt=−1t+1/2Q−11+Q−1Q0t+1/2,todd,−1t/2Q01+Q0Q−1t/2,teven,where t∈ℕ and Q−1Q0≠−1.
Ifn is even (or odd) and Q0=0, then Qt=0(Qt=−1t+1/2Q−1).
Also, ifn is even (or odd) and Q−1=0, then Qt=−1t/2Q0(Qt=0).


## 2.3. Whenα,β=1,1andγ,δ=−1,−1

(35)Qt,s=Qt−2,s−21+Qt−2,s−2Rt−1,s−1,Rt,s=Rt−2,s−2−1−Rt−2,s−2Qt−1,s−1.Theorem 3.
LetQt,s,Rt,st,s=−k∞ be a solution of (35) with Qt,0,Qt,−1,Q0,s,Q−1,s,Rt,0,Rt,−1,R0,s,andR−1,s where t,s∈ℕ0 and ℕ0=ℕ∪0. Suppose Q−1,s−2R0,s−1≠−1, Qt−2,−1Rt−1,0≠−1, R−1,s−2Q0,s−1≠−1, and Rt−2,−1Qt−1,0≠−1. Then,(36)Qt,s=Qt−s−1,−11+2Qt−s−1,−1Rt−s,0s−1/41+Qt−s−1,−1Rt−s,0t+1/2,s=4K+1,−1s−2/4Qt−s,0−1+Qt−s,0Rt−s−1,−1s−2/41+Qt−s,0Rt−s−1,−1s+2/4,s=4K+2,Qt−s−1,−11+2Qt−s−1,−1Rt−s,0t+1/41+Qt−s−1,−1Rt−s,0t+1/2,s=4K+3,−1s/4Qt−s,0−1+Qt−s,0Rt−s−1,−1s/41+Qt−s,0Rt−s−1,−1s/4,s=4K+4,Rt,s=−1s+3/4Rt−s−1,−1−1+Rt−s−1,−1Qt−s,0s−1/41+Rt−s−1,−1Qt−s,0s+3/4,s=4K+1,−1s+2/4Rt−s,01+Rt−s,0Qt−s−1,−1s/21+2Rt−s,0Qt−s−1,−1s+2/4,s=4K+2,−1t+1/4Rt−s−1,−1−1+Rt−s−1,−1Qt−s,0t+1/41+Rt−s−1,−1Qt−s,0t+1/4,s=4K+3,Rt−s,01+Rt−s,0Qt−s−1,−1s/21+2Rt−s,0Qt−s−1,−1s/4,s=4K+4,Qs,t=Q−1,t−s−11+2Q−1,t−s−1R0,t−ss−1/41+Q−1,t−s−1R0,t−st+1/2,s=4K+1,−1s−2/4Q0,t−s−1+Q0,t−sR−1,t−s−1s−2/41+Q0,t−sR−1,t−s−1s+2/4,s=4K+2,Q−1,t−s−11+2Q−1,t−s−1R0,t−st+1/41+Q−1,t−s−1R0,t−st+1/2,s=4K+3,−1s/4Q0,t−s−1+Q0,t−sR−1,t−s−1s/41+Q0,t−sR−1,t−s−1s/4,s=4K+4,Rs,n=−1s+3/4R−1,t−s−1−1+R−1,t−s−1Q0,t−ss−1/41+R−1,t−s−1Q0,t−ss+3/4,s=4K+1,−1s+2/4R0,t−s1+R0,t−sQ−1,t−s−1s/21+2R0,t−sQ−1,t−s−1s+2/4,s=4K+2,−1t+1/4R−1,t−s−1−1+R−1,t−s−1Q0,t−st+1/41+R−1,t−s−1Q0,t−st+1/4,s=4K+3,R0,t−s1+R0,t−sQ−1,t−s−1s/21+2R0,t−sQ−1,t−s−1s/4,s=4K+4,where k=0,1,2,3,….Proof.
The proof is given by piecewise double mathematical induction.Proposition 4.
We have the following:(1)
Ifs is even and Qt−s,0=0, then Qt,s=0(2)
Ifs is odd and Qt−s,0=0, then Rt,s=±Rt−s−1,−1(3)
Ifs is even and Rt−s,0=0, then Rt,s=0(4)
Ifs is odd and Rt−s,0=0, then Qt,s=Qt−s−1,−1(5)
Ifs is even and Qt−s−1,−1=0, then Rt,s=±Rt−s,0(6)
Ifs is odd and Qt−s−1,−1=0, then Qt,s=0(7)
Ifs is even and Rt−s−1,−1=0, then Qt,s=Qt−s,0(8)
Ifs is odd and Rt−s−1,−1=0, then Rt,s=0Proposition 5.
We have the following:(1)
Ifs is even and Q0,t−s=0, then Qs,t=0(2)
Ifs is odd and Q0,t−s=0, then Rs,t=±R−1,t−s−1(3)
Ifs is even and R0,t−s=0, then Rs,t=0(4)
Ifs is odd and R0,t−s=0, then Qs,t=Q−1,t−s−1(5)
Ifs is even and Q−1,t−s−1=0, then Rs,t=±R0,t−s(6)
Ifs is odd and Q−1,t−s−1=0, then Qs,t=0(7)
Ifs is even and R−1,t−s−1=0, then Qs,t=Q0,t−s(8)
Ifs is odd and R−1,t−s−1=0, then Rs,t=0Remark 5.
As a special case,(37)Qt=Qt−21+Qt−2Rt−1,Rt=Rt−2−1−Rt−2Qt−1.Corollary 4.
LetQt,Rtt=−k∞ be a solution of (37) with initials Q0,Q−1,R0,andR−1. Suppose Q−1R0≠−1 and R−1Q0≠−1. Then,(38)Qt=Q−11+2Q−1R0t−1/41+Q−1R0t+1/2,t=4K+1,−1t−2/4Q0−1+Q0R−1t−2/41+Q0R−1t+2/4,t=4K+2,Q−11+2Q−1R0t+1/41+Q−1R0t+1/2,t=4K+3,−1t/4Q0−1+Q0R−1t/41+Q0R−1t/4,t=4K+4,Rt=−1t+3/4R−1−1+R−1Q0t−1/41+R−1Q0t+3/4,t=4K+1,−1t+2/4R01+R0Q−1t/21+2R0Q−1t+2/4,t=4K+2,−1t+1/4R−1−1+R−1Q0t+1/41+R−1Q0t+1/4,t=4K+3,R01+R0Q−1t/21+2R0Q−1t/4,t=4K+4,where k=0,1,2,3….



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*Source: 9966197-2021-04-28.xml*