# Behaviors of the Solutions via Their Closed-Form Formulas for Two Rational Third-Order Difference Equations

Authors: Raafat Abo-Zeid; Abdul Qadeer Khan
Journal: Discrete Dynamics in Nature and Society
 (2021)

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## Abstract

In this work, we derive the solution formulas and study their behaviors for the difference equationsxn+1=αxnxn−3/−βxn−3+γxn−2,n∈ℕ0 and xn+1=αxnxn−3/βxn−3−γxn−2,n∈ℕ0 with real initials and positive parameters. We show that there exist periodic solutions for the second equation under certain conditions when β2<4αγ. Finally, we give some illustrative examples.


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## Body

## 1. Introduction

In [1–5], the first author ([1] together with Kamal) solved and studied the solutions for the difference equations(1)xn+1=xnxn−1xn−xn−2,xn+1=xnxn−1−xn+xn−2,xn+1=axn−3b+cxn−1xn−3,xn+1=axnxn−1±bxn−1+cxn−2,xn+1=axnxn−kbxn−cxn−k−1,xn+1=xnxn−2−axn−1+bxn−2,where n∈ℕ0, with real initials and positive parameters.In [6], the authors explored the dynamics of the difference equation(2)xn+1=axn+bxnxn−2cxn−2+dxn−3,n∈ℕ0,where a,b,c,d and x−3,x−2,x−1,x0 are positive real numbers. They provided the solution of the mentioned equation when a=1, b=1, c=1, and d=1.By virtue of the wide applications in the last few decades, difference equations turned into one of the major areas of research. There are many books dealing with difference equations through the qualitative behavior of nonlinear equations (see [7–11]).Closed-form solutions for nonlinear difference are not available, except for some few equations (see, for example, [6, 7, 10, 12–26]). In [27], the authors studied the two recursive equations(3)xn+1=αxnxn−3−βxn−3+γxn−2,n∈ℕ0,(4)xn+1=αxnxn−3βxn−3−γxn−2,n∈ℕ0,with real initials and positive parameters. They provided the solutions for the two mentioned equations when α=β=γ.Motivated by [27], we shall solve, find the forbidden set, and study (∀ positive real values of α,β,γ) global behavior of the admissible solutions for equations (3) and (4) where α,β,γ are positive real numbers and x−3, x−2, x−1, x0 are nonzero real numbers.If we set(5)βα=a,γα=b,then equations (3) and (4) are reduced to(6)xn+1=xnxn−3−axn−3+bxn−2,n∈ℕ0,(7)xn+1=xnxn−3axn−3−bxn−2,n∈ℕ0.


## 2. The Difference Equationxn+1=xnxn−3/−axn−3+bxn−2

During this section, we suppose that(8)θn=λ+n−λ−na2+4b,where(9)λ−=−a2−a2+4b2,λ+=−a2+a2+4b2.


### 2.1. Solution of Equation (6)

According with the transformation,(10)yn=xn−1xn,withy−2=x−3x−2,y−1=x−2x−1,y0=x−1x0.Equation (6) becomes(11)yn+1=−a+byn−2,n=0,1,….By solving equation (11) and after some calculations, the solution of equation (6) can be obtained.Theorem 1.
Supposexnn=−3∞ is an admissible solution of equation (6). The solution of equation (6) is(12)xn=νμ0n−1/3μ−1n−1/3μ−2n+2/3,n=1,4,…,νμ0n−2/3μ−1n+1/3μ−2n+1/3,n=2,5,…,νμ0n/3μ−1n/3μ−2n/3,n=3,6,…,where ν=x0x−1x−2x−3 and(13)μ−3+in=bθnx−3+i+θn+1x−4+i,wherei=1,2,3andn=0,1,….Proof.
We can write the solution formula (12) as(14)x3m+1=νμ0mμ−1mμ−2m+1,x3m+2=νμ0mμ−1m+1μ−2m+1,x3m+3=νμ0m+1μ−1m+1μ−2m+1.
The proof is by using the mathematical induction onm.
Whenm=0,(15)x1=νμ00μ−10μ−21=x0x−1x−2x−3x−1x−2bx−2−ax−3=x0x−3−ax−3+bx−2,x2=νμ00μ−10μ−21=x0x−1x−2x−3x−1−ax−2+bx−1−ax−3+bx−2=x−2−ax−2+bx−1x0x−3−ax−3+bx−2=x1x−2−ax−2+bx−1,x3=νμ00μ−10μ−21=x−1x−2−ax−1+bx0−ax−2+bx−1x0x−3−ax−3+bx−2=x−1−ax−1+bx0x1x−2−ax−2+bx−1=x2x−1−ax−1+bx0.
Suppose that the equalities (14) are true for m0>0.
Then,(16)x3m0+1=x3m0x3m0−3−ax3m0−3+bx3m0−2=ν/μ0m0μ−1m0μ−2m0ν/μ0m0−1μ−1m0−1μ−2m0−1−aν/μ0m0−1μ−1m0−1μ−2m0−1+bν/μ0m0−1μ−1m0−1μ−2m0=νμ0m0μ−1m0−aμ−2m0+bμ−2m0−1.
But, we can see that(17)−aμ−2m0+bμ−2m0−1=μ−2m0+1.
Then,(18)x3m0+1=νμ0m0μ−1m0μ−2m0+1.
Therefore,(19)x3m+1=νμ0mμ−1mμ−2m+1,m=0,1,….
Similarly, we can obtainx3m+2 and x3m+3.
This completes the proof.For equation (6), it is clear that if x0=0 and x−1x−2x−3≠0, then x4 is not defined. Let x−1=0 and x0x−2x−3≠0, then x7 is not defined. Also, if x−2=0 and x0x−1x−3≠0, then x6 is not defined. Now, if x−3=0 and x0x−1x−2≠0, then x5 is undefined.This implies that the pointx0,x−1,x−2,x−3 with ∏j=03x−j=0 belongs to the forbidden point of equation (6).Theorem 2.
Equation (6) has the forbidden set(20)F=∪i=03u0,u−1,u−2,u−3∈ℝ4:u−i=0∪∪m=1∞u0,u−1,u−2,u−3∈ℝ4:u0=−1bθm+1θmu−1∪∪m=1∞u0,u−1,u−2,u−3∈ℝ4:u−1=−1bθm+1θmu−2∪∪m=1∞u0,u−1,u−2,u−3∈ℝ4:u−2=−1bθm+1θmu−3.


### 2.2. Global Dynamics of Equation (6)

Here, we illustrate the global behavior result and provide some examples.Theorem 3.
Assume thatxnn=−3∞ is an admissible solution of equation (6). Then, the following hold:(1)
Ifa+b>1, then the solution xnn=−3∞ converges to zero.(2)
Ifa+b=1, then the solution xnn=−3∞ converges to period-2 solution.(3)
Ifa+b<1, then the solution xnn=−3∞ is unbounded.Proof.
We can writeθm=λ−mλ+/λ−m−1/a2+4b.(1)
Ifa+b>1, then λ−>1. It follows that θm⟶∞ and μ−3+im⟶∞ as m⟶∞.Therefore,xnn=−3∞ converges to zero.(2)
Assume thata+b=1, then λ−=−1.This implies that(21)θ2m⟶−1a2+4b,θ2m+1⟶1a2+4b,asm⟶∞.Also,(22)μ−3+i2m⟶−bx−3+i+x−4+ia2+4b=L−3+i,asm⟶∞,i=1,2,3.Similarly,(23)μ−3+i2m+1⟶bx−3+i−x−4+ia2+4b=−L−3+i,asm⟶∞,i=1,2,3.Using formula (14), we can write(24)x6m+1=νμ02mμ−12mμ−22m+1⟶−νL0L−1L−2=−M,whereM=ν/L0L−1L−2.Similarly, we can getlimn⟶∞x6m+j for 2≤j≤6 in terms of M.The result is obtained by noting(25)M=νL0L−1L−2=a2+4b3/2ν∏t=13−bx−3+t+x−4+t,and asm⟶∞, we have that(26)x6m+j⟶−M,n=1,3,5,M,n=2,4,6.(3)
Ifa+b<1, then λ−>−1. That is, θm⟶0 as m⟶∞. Also, μ−2mm=0∞, μ−1mm=0∞, and μ0mm=0∞ are converging to zero.
This implies thatx3m+tm=−1∞ are unbounded, t=0,1,2.Example 1.
Consider the solutionxnn=−3∞ of equation (6) such that a=0.5, b=0.1 (a+b<1), with initial values x−3=2.1, x−2=−1, x−1=0.2, and x0=−1.8. Figure 1 shows that solution xnn=−3∞ is unbounded.Figure 1
xn+1=xnxn−3/−0.5xn−3+0.1xn−2.Example 2.
Considerxnn=−3∞ of equation (6) such that a=1, b=0.1 (a+b>1), with x−3=2.1, x−2=−1, x−1=0.2, and x0=−1.8. Figure 2 shows that xnn=−3∞ converges to zero.Figure 2
xn+1=xnxn−3/−xn−3+0.1xn−2.Example 3.
Considerxnn=−3∞ of equation (6) such that a=0.6, b=0.4 (a+b=1), with x−3=2.1, x−2=−1, x−1=0.2, and x0=−1.8. Figure 3 shows that xnn=−3∞ converges to the period-2 solution(27)…,M,−M,M,−M,…,where(28)M=a2+4b3/2ν∏t=13−bx−3+t+x−4+t=0.62+40.43/2−1.80.2−12.1−0.4−1.8+0.2−0.40.2−1−0.4−1+2.1≃−0.83513.Figure 3
xn+1=xnxn−3/−0.6xn−3+0.4xn−2.Example 4.
Considerxnn=−3∞ of equation (6) such that a=0.9, b=0.1 (a+b=1), with x−3=2.1, x−2=−1, x−1=0.2, and x0=−1.8. Figure 4 shows that the solution xnn=−3∞ converges to the period-2 solution(29)…,M,−M,M,−M,…,where(30)M=a2+4b3/2ν∏t=13−bx−3+t+x−4+t=0.92+40.13/2−1.80.2−12.1−0.1−1.8+0.2−0.10.2−1−0.1−1+2.1≃−1.18003.Figure 4
xn+1=xnxn−3/−0.9xn−3+0.1xn−2.


## 2.1. Solution of Equation (6)

According with the transformation,(10)yn=xn−1xn,withy−2=x−3x−2,y−1=x−2x−1,y0=x−1x0.Equation (6) becomes(11)yn+1=−a+byn−2,n=0,1,….By solving equation (11) and after some calculations, the solution of equation (6) can be obtained.Theorem 1.
Supposexnn=−3∞ is an admissible solution of equation (6). The solution of equation (6) is(12)xn=νμ0n−1/3μ−1n−1/3μ−2n+2/3,n=1,4,…,νμ0n−2/3μ−1n+1/3μ−2n+1/3,n=2,5,…,νμ0n/3μ−1n/3μ−2n/3,n=3,6,…,where ν=x0x−1x−2x−3 and(13)μ−3+in=bθnx−3+i+θn+1x−4+i,wherei=1,2,3andn=0,1,….Proof.
We can write the solution formula (12) as(14)x3m+1=νμ0mμ−1mμ−2m+1,x3m+2=νμ0mμ−1m+1μ−2m+1,x3m+3=νμ0m+1μ−1m+1μ−2m+1.
The proof is by using the mathematical induction onm.
Whenm=0,(15)x1=νμ00μ−10μ−21=x0x−1x−2x−3x−1x−2bx−2−ax−3=x0x−3−ax−3+bx−2,x2=νμ00μ−10μ−21=x0x−1x−2x−3x−1−ax−2+bx−1−ax−3+bx−2=x−2−ax−2+bx−1x0x−3−ax−3+bx−2=x1x−2−ax−2+bx−1,x3=νμ00μ−10μ−21=x−1x−2−ax−1+bx0−ax−2+bx−1x0x−3−ax−3+bx−2=x−1−ax−1+bx0x1x−2−ax−2+bx−1=x2x−1−ax−1+bx0.
Suppose that the equalities (14) are true for m0>0.
Then,(16)x3m0+1=x3m0x3m0−3−ax3m0−3+bx3m0−2=ν/μ0m0μ−1m0μ−2m0ν/μ0m0−1μ−1m0−1μ−2m0−1−aν/μ0m0−1μ−1m0−1μ−2m0−1+bν/μ0m0−1μ−1m0−1μ−2m0=νμ0m0μ−1m0−aμ−2m0+bμ−2m0−1.
But, we can see that(17)−aμ−2m0+bμ−2m0−1=μ−2m0+1.
Then,(18)x3m0+1=νμ0m0μ−1m0μ−2m0+1.
Therefore,(19)x3m+1=νμ0mμ−1mμ−2m+1,m=0,1,….
Similarly, we can obtainx3m+2 and x3m+3.
This completes the proof.For equation (6), it is clear that if x0=0 and x−1x−2x−3≠0, then x4 is not defined. Let x−1=0 and x0x−2x−3≠0, then x7 is not defined. Also, if x−2=0 and x0x−1x−3≠0, then x6 is not defined. Now, if x−3=0 and x0x−1x−2≠0, then x5 is undefined.This implies that the pointx0,x−1,x−2,x−3 with ∏j=03x−j=0 belongs to the forbidden point of equation (6).Theorem 2.
Equation (6) has the forbidden set(20)F=∪i=03u0,u−1,u−2,u−3∈ℝ4:u−i=0∪∪m=1∞u0,u−1,u−2,u−3∈ℝ4:u0=−1bθm+1θmu−1∪∪m=1∞u0,u−1,u−2,u−3∈ℝ4:u−1=−1bθm+1θmu−2∪∪m=1∞u0,u−1,u−2,u−3∈ℝ4:u−2=−1bθm+1θmu−3.


## 2.2. Global Dynamics of Equation (6)

Here, we illustrate the global behavior result and provide some examples.Theorem 3.
Assume thatxnn=−3∞ is an admissible solution of equation (6). Then, the following hold:(1)
Ifa+b>1, then the solution xnn=−3∞ converges to zero.(2)
Ifa+b=1, then the solution xnn=−3∞ converges to period-2 solution.(3)
Ifa+b<1, then the solution xnn=−3∞ is unbounded.Proof.
We can writeθm=λ−mλ+/λ−m−1/a2+4b.(1)
Ifa+b>1, then λ−>1. It follows that θm⟶∞ and μ−3+im⟶∞ as m⟶∞.Therefore,xnn=−3∞ converges to zero.(2)
Assume thata+b=1, then λ−=−1.This implies that(21)θ2m⟶−1a2+4b,θ2m+1⟶1a2+4b,asm⟶∞.Also,(22)μ−3+i2m⟶−bx−3+i+x−4+ia2+4b=L−3+i,asm⟶∞,i=1,2,3.Similarly,(23)μ−3+i2m+1⟶bx−3+i−x−4+ia2+4b=−L−3+i,asm⟶∞,i=1,2,3.Using formula (14), we can write(24)x6m+1=νμ02mμ−12mμ−22m+1⟶−νL0L−1L−2=−M,whereM=ν/L0L−1L−2.Similarly, we can getlimn⟶∞x6m+j for 2≤j≤6 in terms of M.The result is obtained by noting(25)M=νL0L−1L−2=a2+4b3/2ν∏t=13−bx−3+t+x−4+t,and asm⟶∞, we have that(26)x6m+j⟶−M,n=1,3,5,M,n=2,4,6.(3)
Ifa+b<1, then λ−>−1. That is, θm⟶0 as m⟶∞. Also, μ−2mm=0∞, μ−1mm=0∞, and μ0mm=0∞ are converging to zero.
This implies thatx3m+tm=−1∞ are unbounded, t=0,1,2.Example 1.
Consider the solutionxnn=−3∞ of equation (6) such that a=0.5, b=0.1 (a+b<1), with initial values x−3=2.1, x−2=−1, x−1=0.2, and x0=−1.8. Figure 1 shows that solution xnn=−3∞ is unbounded.Figure 1
xn+1=xnxn−3/−0.5xn−3+0.1xn−2.Example 2.
Considerxnn=−3∞ of equation (6) such that a=1, b=0.1 (a+b>1), with x−3=2.1, x−2=−1, x−1=0.2, and x0=−1.8. Figure 2 shows that xnn=−3∞ converges to zero.Figure 2
xn+1=xnxn−3/−xn−3+0.1xn−2.Example 3.
Considerxnn=−3∞ of equation (6) such that a=0.6, b=0.4 (a+b=1), with x−3=2.1, x−2=−1, x−1=0.2, and x0=−1.8. Figure 3 shows that xnn=−3∞ converges to the period-2 solution(27)…,M,−M,M,−M,…,where(28)M=a2+4b3/2ν∏t=13−bx−3+t+x−4+t=0.62+40.43/2−1.80.2−12.1−0.4−1.8+0.2−0.40.2−1−0.4−1+2.1≃−0.83513.Figure 3
xn+1=xnxn−3/−0.6xn−3+0.4xn−2.Example 4.
Considerxnn=−3∞ of equation (6) such that a=0.9, b=0.1 (a+b=1), with x−3=2.1, x−2=−1, x−1=0.2, and x0=−1.8. Figure 4 shows that the solution xnn=−3∞ converges to the period-2 solution(29)…,M,−M,M,−M,…,where(30)M=a2+4b3/2ν∏t=13−bx−3+t+x−4+t=0.92+40.13/2−1.80.2−12.1−0.1−1.8+0.2−0.10.2−1−0.1−1+2.1≃−1.18003.Figure 4
xn+1=xnxn−3/−0.9xn−3+0.1xn−2.


## 3. The Difference Equationxn+1=xnxn−3/axn−3−bxn−2

We discuss the behaviors of the solutions of equation (7). The transformation (10) reduces equation (7) into the recursive equation(31)yn+1=a−byn−2,n=0,1,….During this section, we suppose that(32)λ1=a2−a2−4b2,λ2=a2+a2−4b2.Clear thatλ1 and λ2 are the roots of the equation(33)λ2−aλ+b=0.


### 3.1. Casea2>4b

In this subsection, we assume thata2>4b. Clear that(34)0<λ1<a2<λ2<ba<a.During this subsection, we suppose that(35)θ^n=λ2n−λ1na2−4b,n=0,1,….Theorem 4.
Assume thatxnn=−3∞ is an admissible solution of equation (7), then(36)xn=νμ^0n−1/3μ^−1n−1/3μ^−2n+2/3,n=1,4,…,νμ^0n−2/3μ^−1n+1/3μ^−2n+1/3,n=2,5,…,νμ^0n/3μ^−1n/3μ^−2n/3,n=3,6,…,where ν=x0x−1x−2x−3 and(37)μ^−3+in=−bθ^nx−3+i+θ^n+1x−4+i,wherei=1,2,3andn=0,1,….Proof.
Its proof is same as the proof of Theorem1 and is omitted.Theorem 5.
Letxnn=−3∞ be an admissible solution of recursive equation (7). Then, the following hold:(1)
Ifa<b+1, then one has the following:(a)
Ifb<1, then the solution xnn=−3∞ is unbounded.(b)
Ifb>1, then the solution xnn=−3∞ converges to zero.(2)
Ifa=b+1, then one has the following:(a)
Ifb<1, then the solution xnn=−3∞ converges to a finite limit.(b)
Ifb>1, then the solution xnn=−3∞ converges to zero.(3)
Ifa>b+1, then the solution xnn=−3∞ converges to zero.Proof.
We can writeθm=λ2m1−λ1/λ2m/a2−4b.(1)
Ifa<b+1, then either λ1>1 or λ2<1.(a)
Ifb<1, then λ2<1. That is, θ^m⟶0 as m⟶∞. This implies that μ^−2mm=0∞, μ^−1mm=0∞, and μ^0mm=0∞ are converging to zero. Therefore, xnn=−3∞ is unbounded.(b)
Ifb>1, then λ1>1. That is, θ^m⟶∞ as m⟶∞. This implies that μ^−2m, μ^−1m, and μ^0m are unbounded. Therefore, xnn=−3∞ converges to zero.(2)
Suppose thata=b+1, then either λ1=1 or λ2=1.(a)
Ifb<1, then λ2=1. It follows that θ^m⟶2/a2−4b as m⟶∞. This implies that(38)μ^−3+im⟶−bx−3+i+x−4+ia2−4b=L^−3+i,asm⟶∞,i=1,2,3.Then,(39)x3m+1=νμ^0mμ^−1mμ^−2m+1⟶νL^0L^−1L^−2=M^,where(40)M^=νa2−4b3/2∏t=13−bx−3+t+x−4+t.Similarly, we can show thatx3m+2⟶M^ and x3m+3⟶M^ as m⟶∞.Therefore,xnn=−3∞ converges to M as m⟶∞.(b)
Ifb>1, then λ1=1. That is, θ^m⟶∞ as m⟶∞.This implies thatμ^−2mm=0∞, μ^−1mm=0∞, and μ^0mm=0∞ are unbounded.Therefore,xnn=−3∞ converges to zero.(3)
Ifa>b+1, then λ1<1<λ2. That is, θ^m⟶∞ as m⟶∞.
This implies thatμ^−2m, μ^−1m, and μ^0m are unbounded.
Therefore,xnn=−3∞ converges to zero.Example 5.
Considerxnn=−3∞ of recursive equation (7) where a=1, b=0.2 (a<b+1 and b<1), with x−3=1, x−2=−3, x−1=−1, and x0=1.8. Figure 5 shows the unbounded solution xnn=−3∞.Figure 5
xn+1=xnxn−3/xn−3−0.2xn−2.Example 6.
Considerxnn=−3∞ of recursive equation (7) such that a=1.2, b=0.1 (a>b+1), with x−3=1, x−2=−3, x−1=−1, and x0=1. Figure 6 shows that xnn=−3∞ converges to zero.Figure 6
xn+1=xnxn−3/1.2xn−3−0.1xn−2.Example 7.
Considerxnn=−3∞ of recursive equation (7) such that a=1.3, b=0.3 (a=b+1 and b<1), with x−3=−2, x−2=−2, x−1=1, and x0=2. Figure 7 shows that xnn=−3∞ converges to M^ where(41)M^=νa2−4b3/2∏t=13−bx−3+t+x−4+t=1.32−40.33/2−0.3−2−2−0.31−2−0.32+1≃2.1304.Figure 7
xn+1=xnxn−3/1.3xn−3−0.3xn−2.Example 8.
Considerxnn=−3∞ of recursive equation (7) such that a=2.01, b=1.01 (a=b+1 and b>1), with x−3=−2, x−2=−2, x−1=1, and x0=2. Figure 8 shows that the solution xnn=−3∞ converges to zero.Figure 8
xn+1=xnxn−3/2.01xn−3−1.01xn−2.


### 3.2. Casea2=4b

In this subsection, we assume thata2=4b. That is, λ1=λ2=a/2.Theorem 6.
Assume thatxnn=−3∞ is an admissible solution of recursive equation (7). Then,(42)xn=2an8νδ0n−1/3δ−1n−1/3δ−2n+2/3,n=1,4,…,2an8νμ^0n−2/3μ^−1n+1/3δ−2n+1/3,n=2,5,…,2an8νδ0n/3δ−1n/3δ−2n/3,n=3,6,…,where ν=x0x−1x−2x−3 and(43)δ−3+in=−anx−3+i+2n+1x−4+i,i=1,2,3andn=0,1,….Proof.
It is enough to see that(44)δ−3+in+1=2δ−3+in−δ−3+in−1,i=1,2,3andn=0,1,…,and its proof is same as of Theorem 1 and is omitted.Theorem 7.
Letxnn=−3∞ be an admissible solution of recursive equation (7). Then, the following hold:(1)
Ifa≥2, then solution xnn=−3∞ converges to zero.(2)
Ifa<2, then xnn=−3∞ is unbounded.Proof.
Clear thatδ−3+im⟶∞ as m⟶∞, i=1,2,3.(1)
Ifa>2, then 2/an⟶0 as n⟶∞. This implies that the solution xnn=−3∞ converges to zero.Now, suppose thata=2. Then,(45)x3m+1=8νδ0mδ−1mδ−2m+1⟶0,asm⟶∞.Similarly, forx3m+2 and x3m+3.Therefore,xnn=−3∞ converges to zero.(2)
Ifa<2, then 2/am⟶∞ as m⟶∞. Then,(46)x3m+1=2a3m+18νδ0mδ−1mδ−2m+1=2a3m+18νm3∏t=01−ax−t+21/m+1x−t−1−a1/m+1x−2+22/m+1x−3.
By applying L’Hospital’s rule, we get(47)x3m+1⟶∞,asm⟶∞.
Similarly, forx3m+2 and x3m+3.
Hence,xnn=−3∞ is unbounded.Example 9.
Considerxnn=−3∞ of recursive equation (7) such that a=1, b=0.25 (a2=4b and a<2), with x−3=0.2, x−2=−2, x−1=2, and x0=−1.2. Figure 9 shows that the unbounded solution xnn=−3∞.Figure 9
xn+1=xnxn−3/xn−3−0.25xn−2.Example 10.
Considerxnn=−3∞ of recursive equation (7) such that a=2, b=1 (a2=4b and a≥2), with x−3=0.2, x−2=−2, x−1=2, and x0=−1.2. Figure 10 shows that xnn=−3∞ converges to zero.Figure 10
xn+1=xnxn−3/2xn−3−xn−2.


### 3.3. Casea2<4b

Hereafter, we study the final case whena2<4b.During this subsection, we suppose that(48)λ1=a2−4b−a22i,λ2=a2+4b−a22i.That is,(49)λ1=λ2=b,sinϕ=4b−a22b,cosϕ=a2b.Theorem 8.
Letxnn=−2∞ be an admissible solution of recursive equation (3). Then,(50)xn=sin3αbn/2νξ0n−1/3ξ−1n−1/3ξ−2n+2/3,n=1,4,…,sin3αbn/2νξ0n−2/3ξ−1n+1/3ξ−2n+1/3,n=2,5,…,sin3αbn/2νξ0n/3ξ−1n/3ξ−2n/3,n=3,6,…,where ν=x0x−1x−2, ξ−3+in=−bx−3+isinnϕ+x−4+isinn+1ϕ, ϕ=tan−14b−a2/b∈0,π/2.Theorem 9.
Assume thatxnn=−3∞ is an admissible solution of equation (7). Then, we have the following:(1)
Ifb>1, then the solution xnn=−3∞ converges to zero.(2)
Ifb<1, then the solution xnn=−3∞ is unbounded.Proof.
Its proof is direct consequence and is omitted.Theorem 10.
Letxnn=−3∞ be an admissible solution of recursive equation (7) and let b=1. If ϕ=l/kπ (with 0<l<k/2), then xnn=−3∞ is periodic having prime period-6k.Proof.
Using formula (50), we can write(51)x3m+3k+1=sin3ϕνξ0m+kξ−1m+kξ−2m+1+k.
But, for eachi=1,2,3, we have(52)ξ−3+im+2k=−x−3+isinm+2kϕ+x−4+isinm+2k+1ϕ=−x−3+isinmϕ+2kϕ+x−4+isinm+1ϕ+2kϕ=−x−3+isinmϕ+2lπ+x−4+isinm+1ϕ+2lπ=−x−3+isinmϕ+x−4+isinm+1ϕ=ξ−3+im.
Then, form≥−1, we have(53)x3m+6k+1=sin3ϕνξ0m+2kξ−1m+2kξ−2m+1+2k=sin3ϕνξ0mξ−1mξ−2m+1=x3m+1.
Similarly, we can show that form≥−1, we have(54)x3m+6k+2=x3m+2,x3m+6k+3=x3m+3.
This completes the proof.Example 11.
Considerxnn=−3∞ of recursive equation (7) with x−3=−0.8, x−2=1.1, x−1=−0.9, and x0=−1.2. If a=2+2,b=1 (a2<4b and ϕ=π/8), then xnn=−3∞ is periodic with prime period 6k=48 (see Figure 11).Figure 11
xn+1=xnxn−3/2+2xn−3−xn−2.Example 12.
Considerxnn=−3∞ of recursive equation (7) with x−3=−0.5, x−2=2, x−1=0.5, and x0=−1.1. If a=b=1 (a2<4b and ϕ=π/3), then xnn=−3∞ is periodic having prime period-6k=18 (see Figure 12).Figure 12
xn+1=xnxn−3/xn−3−xn−2.


### 3.4. Forbidden Sets

In this subsection, we give the forbidden set of recursive equation (7) when a2>4b, a2=4b, and a2<4b.Clear that, ifx0=0 and x−1x−2x−3≠0, then x4 is undefined. If x−1=0 and x0x−2x−3≠0, then x7 is undefined. If x−2=0 and x0x−1x−3≠0, then x6 is undefined. If x−3=0 and x0x−1x−2≠0, then x5 is undefinedThe following result gives the forbidden sets of equation (7) for all values of a>0 and b>0.Theorem 11.
We have the following statements:(1)
Ifa2>4b, then equation (3) has the forbidden set(55)F1=∪i=03v0,v−1,v−2,v−3∈ℝ4:v−i=0∪∪m=1∞v0,v−1,v−2,v−3∈ℝ4:v0=1bθ^m+1θ^mv−1∪∪m=1∞v0,v−1,v−2,v−3∈ℝ4:v−1=1bθ^m+1θ^mv−2∪∪m=1∞v0,v−1,v−2,v−3∈ℝ4:v−2=1bθ^m+1θ^mv−3.(2)
Ifa2=4b, then equation (3) has the forbidden set(56)F2=∪i=03v0,v−1,v−2,v−3∈ℝ4:v−i=0∪∪m=1∞v0,v−1,v−2,v−3∈ℝ4:v0=2b1+mmv−1∪∪m=1∞v0,v−1,v−2,v−3∈ℝ4:v−1=2b1+mmv−2∪∪m=1∞v0,v−1,v−2,v−3∈ℝ4:v−2=2b1+mmv−3.(3)
Ifa2<4b, then equation (3) has the forbidden set(57)F3=∪i=03v0,v−1,v−2,v−3∈ℝ4:v−i=0∪∪m=1∞v0,v−1,v−2,v−3∈ℝ4:v0=1bsinm+1ϕsinmϕv−1∪∪m=1∞v0,v−1,v−2,v−3∈ℝ4:v−1=1bsinm+1ϕsinmϕv−2∪∪m=1∞v0,v−1,v−2,v−3∈ℝ4:v−2=1bsinm+1ϕsinmϕv−3.


## 3.1. Casea2>4b

In this subsection, we assume thata2>4b. Clear that(34)0<λ1<a2<λ2<ba<a.During this subsection, we suppose that(35)θ^n=λ2n−λ1na2−4b,n=0,1,….Theorem 4.
Assume thatxnn=−3∞ is an admissible solution of equation (7), then(36)xn=νμ^0n−1/3μ^−1n−1/3μ^−2n+2/3,n=1,4,…,νμ^0n−2/3μ^−1n+1/3μ^−2n+1/3,n=2,5,…,νμ^0n/3μ^−1n/3μ^−2n/3,n=3,6,…,where ν=x0x−1x−2x−3 and(37)μ^−3+in=−bθ^nx−3+i+θ^n+1x−4+i,wherei=1,2,3andn=0,1,….Proof.
Its proof is same as the proof of Theorem1 and is omitted.Theorem 5.
Letxnn=−3∞ be an admissible solution of recursive equation (7). Then, the following hold:(1)
Ifa<b+1, then one has the following:(a)
Ifb<1, then the solution xnn=−3∞ is unbounded.(b)
Ifb>1, then the solution xnn=−3∞ converges to zero.(2)
Ifa=b+1, then one has the following:(a)
Ifb<1, then the solution xnn=−3∞ converges to a finite limit.(b)
Ifb>1, then the solution xnn=−3∞ converges to zero.(3)
Ifa>b+1, then the solution xnn=−3∞ converges to zero.Proof.
We can writeθm=λ2m1−λ1/λ2m/a2−4b.(1)
Ifa<b+1, then either λ1>1 or λ2<1.(a)
Ifb<1, then λ2<1. That is, θ^m⟶0 as m⟶∞. This implies that μ^−2mm=0∞, μ^−1mm=0∞, and μ^0mm=0∞ are converging to zero. Therefore, xnn=−3∞ is unbounded.(b)
Ifb>1, then λ1>1. That is, θ^m⟶∞ as m⟶∞. This implies that μ^−2m, μ^−1m, and μ^0m are unbounded. Therefore, xnn=−3∞ converges to zero.(2)
Suppose thata=b+1, then either λ1=1 or λ2=1.(a)
Ifb<1, then λ2=1. It follows that θ^m⟶2/a2−4b as m⟶∞. This implies that(38)μ^−3+im⟶−bx−3+i+x−4+ia2−4b=L^−3+i,asm⟶∞,i=1,2,3.Then,(39)x3m+1=νμ^0mμ^−1mμ^−2m+1⟶νL^0L^−1L^−2=M^,where(40)M^=νa2−4b3/2∏t=13−bx−3+t+x−4+t.Similarly, we can show thatx3m+2⟶M^ and x3m+3⟶M^ as m⟶∞.Therefore,xnn=−3∞ converges to M as m⟶∞.(b)
Ifb>1, then λ1=1. That is, θ^m⟶∞ as m⟶∞.This implies thatμ^−2mm=0∞, μ^−1mm=0∞, and μ^0mm=0∞ are unbounded.Therefore,xnn=−3∞ converges to zero.(3)
Ifa>b+1, then λ1<1<λ2. That is, θ^m⟶∞ as m⟶∞.
This implies thatμ^−2m, μ^−1m, and μ^0m are unbounded.
Therefore,xnn=−3∞ converges to zero.Example 5.
Considerxnn=−3∞ of recursive equation (7) where a=1, b=0.2 (a<b+1 and b<1), with x−3=1, x−2=−3, x−1=−1, and x0=1.8. Figure 5 shows the unbounded solution xnn=−3∞.Figure 5
xn+1=xnxn−3/xn−3−0.2xn−2.Example 6.
Considerxnn=−3∞ of recursive equation (7) such that a=1.2, b=0.1 (a>b+1), with x−3=1, x−2=−3, x−1=−1, and x0=1. Figure 6 shows that xnn=−3∞ converges to zero.Figure 6
xn+1=xnxn−3/1.2xn−3−0.1xn−2.Example 7.
Considerxnn=−3∞ of recursive equation (7) such that a=1.3, b=0.3 (a=b+1 and b<1), with x−3=−2, x−2=−2, x−1=1, and x0=2. Figure 7 shows that xnn=−3∞ converges to M^ where(41)M^=νa2−4b3/2∏t=13−bx−3+t+x−4+t=1.32−40.33/2−0.3−2−2−0.31−2−0.32+1≃2.1304.Figure 7
xn+1=xnxn−3/1.3xn−3−0.3xn−2.Example 8.
Considerxnn=−3∞ of recursive equation (7) such that a=2.01, b=1.01 (a=b+1 and b>1), with x−3=−2, x−2=−2, x−1=1, and x0=2. Figure 8 shows that the solution xnn=−3∞ converges to zero.Figure 8
xn+1=xnxn−3/2.01xn−3−1.01xn−2.


## 3.2. Casea2=4b

In this subsection, we assume thata2=4b. That is, λ1=λ2=a/2.Theorem 6.
Assume thatxnn=−3∞ is an admissible solution of recursive equation (7). Then,(42)xn=2an8νδ0n−1/3δ−1n−1/3δ−2n+2/3,n=1,4,…,2an8νμ^0n−2/3μ^−1n+1/3δ−2n+1/3,n=2,5,…,2an8νδ0n/3δ−1n/3δ−2n/3,n=3,6,…,where ν=x0x−1x−2x−3 and(43)δ−3+in=−anx−3+i+2n+1x−4+i,i=1,2,3andn=0,1,….Proof.
It is enough to see that(44)δ−3+in+1=2δ−3+in−δ−3+in−1,i=1,2,3andn=0,1,…,and its proof is same as of Theorem 1 and is omitted.Theorem 7.
Letxnn=−3∞ be an admissible solution of recursive equation (7). Then, the following hold:(1)
Ifa≥2, then solution xnn=−3∞ converges to zero.(2)
Ifa<2, then xnn=−3∞ is unbounded.Proof.
Clear thatδ−3+im⟶∞ as m⟶∞, i=1,2,3.(1)
Ifa>2, then 2/an⟶0 as n⟶∞. This implies that the solution xnn=−3∞ converges to zero.Now, suppose thata=2. Then,(45)x3m+1=8νδ0mδ−1mδ−2m+1⟶0,asm⟶∞.Similarly, forx3m+2 and x3m+3.Therefore,xnn=−3∞ converges to zero.(2)
Ifa<2, then 2/am⟶∞ as m⟶∞. Then,(46)x3m+1=2a3m+18νδ0mδ−1mδ−2m+1=2a3m+18νm3∏t=01−ax−t+21/m+1x−t−1−a1/m+1x−2+22/m+1x−3.
By applying L’Hospital’s rule, we get(47)x3m+1⟶∞,asm⟶∞.
Similarly, forx3m+2 and x3m+3.
Hence,xnn=−3∞ is unbounded.Example 9.
Considerxnn=−3∞ of recursive equation (7) such that a=1, b=0.25 (a2=4b and a<2), with x−3=0.2, x−2=−2, x−1=2, and x0=−1.2. Figure 9 shows that the unbounded solution xnn=−3∞.Figure 9
xn+1=xnxn−3/xn−3−0.25xn−2.Example 10.
Considerxnn=−3∞ of recursive equation (7) such that a=2, b=1 (a2=4b and a≥2), with x−3=0.2, x−2=−2, x−1=2, and x0=−1.2. Figure 10 shows that xnn=−3∞ converges to zero.Figure 10
xn+1=xnxn−3/2xn−3−xn−2.


## 3.3. Casea2<4b

Hereafter, we study the final case whena2<4b.During this subsection, we suppose that(48)λ1=a2−4b−a22i,λ2=a2+4b−a22i.That is,(49)λ1=λ2=b,sinϕ=4b−a22b,cosϕ=a2b.Theorem 8.
Letxnn=−2∞ be an admissible solution of recursive equation (3). Then,(50)xn=sin3αbn/2νξ0n−1/3ξ−1n−1/3ξ−2n+2/3,n=1,4,…,sin3αbn/2νξ0n−2/3ξ−1n+1/3ξ−2n+1/3,n=2,5,…,sin3αbn/2νξ0n/3ξ−1n/3ξ−2n/3,n=3,6,…,where ν=x0x−1x−2, ξ−3+in=−bx−3+isinnϕ+x−4+isinn+1ϕ, ϕ=tan−14b−a2/b∈0,π/2.Theorem 9.
Assume thatxnn=−3∞ is an admissible solution of equation (7). Then, we have the following:(1)
Ifb>1, then the solution xnn=−3∞ converges to zero.(2)
Ifb<1, then the solution xnn=−3∞ is unbounded.Proof.
Its proof is direct consequence and is omitted.Theorem 10.
Letxnn=−3∞ be an admissible solution of recursive equation (7) and let b=1. If ϕ=l/kπ (with 0<l<k/2), then xnn=−3∞ is periodic having prime period-6k.Proof.
Using formula (50), we can write(51)x3m+3k+1=sin3ϕνξ0m+kξ−1m+kξ−2m+1+k.
But, for eachi=1,2,3, we have(52)ξ−3+im+2k=−x−3+isinm+2kϕ+x−4+isinm+2k+1ϕ=−x−3+isinmϕ+2kϕ+x−4+isinm+1ϕ+2kϕ=−x−3+isinmϕ+2lπ+x−4+isinm+1ϕ+2lπ=−x−3+isinmϕ+x−4+isinm+1ϕ=ξ−3+im.
Then, form≥−1, we have(53)x3m+6k+1=sin3ϕνξ0m+2kξ−1m+2kξ−2m+1+2k=sin3ϕνξ0mξ−1mξ−2m+1=x3m+1.
Similarly, we can show that form≥−1, we have(54)x3m+6k+2=x3m+2,x3m+6k+3=x3m+3.
This completes the proof.Example 11.
Considerxnn=−3∞ of recursive equation (7) with x−3=−0.8, x−2=1.1, x−1=−0.9, and x0=−1.2. If a=2+2,b=1 (a2<4b and ϕ=π/8), then xnn=−3∞ is periodic with prime period 6k=48 (see Figure 11).Figure 11
xn+1=xnxn−3/2+2xn−3−xn−2.Example 12.
Considerxnn=−3∞ of recursive equation (7) with x−3=−0.5, x−2=2, x−1=0.5, and x0=−1.1. If a=b=1 (a2<4b and ϕ=π/3), then xnn=−3∞ is periodic having prime period-6k=18 (see Figure 12).Figure 12
xn+1=xnxn−3/xn−3−xn−2.


## 3.4. Forbidden Sets

In this subsection, we give the forbidden set of recursive equation (7) when a2>4b, a2=4b, and a2<4b.Clear that, ifx0=0 and x−1x−2x−3≠0, then x4 is undefined. If x−1=0 and x0x−2x−3≠0, then x7 is undefined. If x−2=0 and x0x−1x−3≠0, then x6 is undefined. If x−3=0 and x0x−1x−2≠0, then x5 is undefinedThe following result gives the forbidden sets of equation (7) for all values of a>0 and b>0.Theorem 11.
We have the following statements:(1)
Ifa2>4b, then equation (3) has the forbidden set(55)F1=∪i=03v0,v−1,v−2,v−3∈ℝ4:v−i=0∪∪m=1∞v0,v−1,v−2,v−3∈ℝ4:v0=1bθ^m+1θ^mv−1∪∪m=1∞v0,v−1,v−2,v−3∈ℝ4:v−1=1bθ^m+1θ^mv−2∪∪m=1∞v0,v−1,v−2,v−3∈ℝ4:v−2=1bθ^m+1θ^mv−3.(2)
Ifa2=4b, then equation (3) has the forbidden set(56)F2=∪i=03v0,v−1,v−2,v−3∈ℝ4:v−i=0∪∪m=1∞v0,v−1,v−2,v−3∈ℝ4:v0=2b1+mmv−1∪∪m=1∞v0,v−1,v−2,v−3∈ℝ4:v−1=2b1+mmv−2∪∪m=1∞v0,v−1,v−2,v−3∈ℝ4:v−2=2b1+mmv−3.(3)
Ifa2<4b, then equation (3) has the forbidden set(57)F3=∪i=03v0,v−1,v−2,v−3∈ℝ4:v−i=0∪∪m=1∞v0,v−1,v−2,v−3∈ℝ4:v0=1bsinm+1ϕsinmϕv−1∪∪m=1∞v0,v−1,v−2,v−3∈ℝ4:v−1=1bsinm+1ϕsinmϕv−2∪∪m=1∞v0,v−1,v−2,v−3∈ℝ4:v−2=1bsinm+1ϕsinmϕv−3.



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