# On a Third-Order System of Difference Equations with Variable Coefficients

Authors: Stevo Stević; Josef Diblík; Bratislav Iričanin; Zdeněk Šmarda
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We show that the system of three difference equationsxn+1=an(1)xn-2/(bn(1)ynzn-1xn-2+cn(1)), yn+1=an(2)yn-2/(bn(2)znxn-1yn-2+cn(2)), and zn+1=an(3)zn-2/(bn(3)xnyn-1zn-2+cn(3)), n∈ℕ0, where all elements of the sequences an(i), bn(i), cn(i), n∈ℕ0, i∈{1,2,3}, and initial values x-j, y-j, z-j, j∈{0,1,2}, are real numbers, can be solved. Explicit formulae for solutions of the system are derived, and some consequences on asymptotic behavior of solutions for the case when coefficients are periodic with period three are deduced.


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## Body

## 1.  Introduction

Studying nonlinear difference equations and systems is an area of a great interest nowadays (see, e.g., [1–39] and the references therein).This paper studies the system of three difference equations(1.1)xn+1=an(1)xn-2bn(1)ynzn-1xn-2+cn(1),yn+1=an(2)yn-2bn(2)znxn-1yn-2+cn(2),zn+1=an(3)zn-2bn(3)xnyn-1zn-2+cn(3),n∈N0,
where all elements of the sequences an(i), bn(i), cn(i), n∈ℕ0, i∈{1,2,3}, and initial values x-j, y-j, z-j, j∈{0,1,2} are real numbers. The cases when both bn(i) and cn(i) are equal to zero for some fixed i∈{1,2,3} and an n∈ℕ0, are not interesting so they are excluded. In [37] we have shown that system (1.1) for the case when the sequences an(i), bn(i), cn(i), n∈ℕ0, i∈{1,2,3}, are constant can be explicitly solved (if solutions are well defined). Some recent results on solving difference equations can be found, for example, in [6, 7, 12, 24, 25, 34, 35, 38, 39]. For some old results see, for example, classic book [14].Note that the solutions of (1.1) such that all sequences an(i), bn(i), cn(i), n∈ℕ0, i∈{1,2,3}, and initial values in system (1.1) are positive, are also positive, that is,(1.2)xn>0,yn>0,zn>0foreveryn≥-2,
so that there are a lot of well defined solutions of the system. In fact, for “majority” initial values of system (1.1), solutions are well defined, but we will not discuss the problem here. Instead of that we assume, throughout the paper, that solutions of (1.1) are well defined. We also adopt the customary notation ∏i=k+1kgi=1 and ∑i=k+1kgi=0.We show that in the main case, system (1.1) is transformed to a third-order system of nonhomogeneous linear first-order difference equations, which can be explicitly solved.This idea appeared for the first time in [24] for the case of the scalar equation with constant coefficients corresponding to system (1.1) and was also used later in [1, 4]. Some related transformations are used also in [25, 30]. For a different approach in dealing with the scalar difference equation see [2, 3]. For some related scalar difference equations see, for example, [8, 13, 26] and the related references therein. Some related results on systems of difference equations can be found in [11, 15–22] (see also references therein).Here we give explicit formulae for solutions of system (1.1) and present some consequences on asymptotic behavior of the solutions for the case when coefficients are periodic with period three.


## 2.  Casean(i)=0 for Some i∈{1,2,3} and All n∈ℕ0

Ifan(1)=0,n∈ℕ0, then the first equation in (1.1) becomes(2.1)xn=0,n∈N,
so that from the second and the third equations and since the solution is well defined we get(2.2)yn+2=an+1(2)cn+1(2)yn-1,zn+1=an(3)cn(3)zn-2,n∈N,cn+1(2)≠0≠cn(3), n∈ℕ, from which it follows that(2.3)y3n+1=y1∏j=1na3j(2)c3j(2),y3n+2=y2∏j=1na3j+1(2)c3j+1(2),y3n=y0∏j=1na3j-1(2)c3j-1(2),n∈N,z3n+1=z1∏j=1na3j(3)c3j(3),z3n-1=z-1∏j=0n-1a3j+1(3)c3j+1(3),z3n=z0∏j=1na3j-1(3)c3j-1(3),n∈N.Ifan(2)=0, n∈ℕ0, then the second equation in (1.1) becomes(2.4)yn=0,n∈N,
so that from the first and the third equations we get(2.5)xn+1=an(1)cn(1)xn-2,zn+2=an+1(3)cn+1(3)zn-1,n∈N,cn(1)≠0≠cn+1(3), from which it follows that(2.6)x3n+1=x1∏j=1na3j(1)c3j(1),x3n-1=x-1∏j=0n-1a3j+1(1)c3j+1(1),x3n=x0∏j=1na3j-1(1)c3j-1(1),n∈N,z3n+1=z1∏j=1na3j(3)c3j(3),z3n+2=z2∏j=1na3j+1(3)c3j+1(3),z3n=z0∏j=1na3j-1(3)c3j-1(3),n∈N.Finally, ifan(3)=0, n∈ℕ0, then the third equation in (1.1) becomes(2.7)zn=0,n∈N,
so that from the first and the second equation we get(2.8)xn+2=an+1(1)cn+1(1)xn-1,yn+1=an(2)cn(2)yn-2,n∈N,cn+1(1)≠0≠cn(2), from which it follows that(2.9)x3n+1=x1∏j=1na3j(1)c3j(1),x3n+2=x2∏j=1na3j+1(1)c3j+1(1),x3n=x0∏j=1na3j-1(1)c3j-1(1),n∈N,y3n+1=y1∏j=1na3j(2)c3j(2),y3n-1=y-1∏j=0n-1a3j+1(2)c3j+1(2),y3n=y0∏j=1na3j-1(2)c3j-1(2),n∈N.


## 3. Explicit Formulae for the Casean(i)≠0 for All i∈{1,2,3} and n∈ℕ0

Here we consider system (1.1) in the case when an(i)≠0 for all i∈{1,2,3} and n∈ℕ0. Noticing that in this case, system (1.1) can be written in the form(3.1)xn+1=xn-2b̂n(1)ynzn-1xn-2+ĉn(1),yn+1=yn-2b̂n(2)znxn-1yn-2+ĉn(2),zn+1=zn-2b̂n(3)xnyn-1zn-2+ĉn(3),n∈N0,
where b̂n(i)=bn(i)/an(i), ĉn(i)=cn(i)/an(i), i∈{1,2,3}, we see that we may assume that an(i)=1, for every n∈ℕ0 and for each i∈{1,2,3}.Hence we consider, without loss of generality, the system(3.2)xn+1=xn-2bn(1)ynzn-1xn-2+cn(1),yn+1=yn-2bn(2)znxn-1yn-2+cn(2),zn+1=zn-2bn(3)xnyn-1zn-2+cn(3),n∈N0
using the same notation for coefficients as in (1.1) except for the coefficients an(i), assuming that an(i)=1 for all i∈{1,2,3} and n∈ℕ0.First we consider the case when some of initial values of solutions of system (3.2) is equal to zero.Ifxn0=0 for some n0∈ℕ then from (3.2) it follows that xn0-3k=0, for each k∈ℕ0 such that n0-3k≥-2. Hence, we have that x-2=0 or x-1=0 or x0=0.Ifx-2=0, then x3n-2=0, n∈ℕ0, which implies(3.3)y3n+3=1c3n+2(2)y3n,z3n+2=1c3n+1(3)z3n-1,n∈N0,
and consequently(3.4)y3n=y0∏j=0n-1c3j+2(2),z3n+2=z-1∏j=0nc3j+1(3),n∈N0.Ifx-1=0, then x3n-1=0, n∈ℕ0, which implies(3.5)y3n+1=1c3n(2)y3n-2,z3n+3=1c3n+2(3)z3n,n∈N0,
and consequently(3.6)y3n+1=y-2∏j=0nc3j(2),z3n=z0∏j=0n-1c3j+2(3),n∈N0.Ifx0=0, then x3n=0,n∈ℕ0, which implies(3.7)y3n+2=1c3n+1(2)y3n-1,z3n+1=1c3n(3)z3n-2,n∈N0,
and consequently(3.8)y3n+2=y-1∏j=0nc3j+1(2),z3n+1=z-2∏j=0nc3j(3),n∈N0.Ifyn1=0 or zn1=0 for some n1∈ℕ then similar results are obtained analogously.


### 3.1.  Main Case

Here we study well-defined solutions of system (1.1) when neither of the sequences (an(i))n∈ℕ0, i∈{1,2,3}, or initial conditions x-i, y-i, z-i, i∈{0,1,2}, is equal to zero. Recall that we may assume that an(i)=1, for every n∈ℕ0 and for each i∈{1,2,3}.Following the idea in [37], we use a transformation which reduces nonlinear systems (1.1) and (3.2) to third-order systems of nonhomogeneous linear difference equations.If we multiply the first equation in system (3.2) by ynzn-1, the second by znxn-1 and the third by xnyn-1, and then using in such obtained system the change of variables(3.9)un+1=1xn+1ynzn-1,vn+1=1yn+1znxn-1,wn+1=1zn+1xnyn-1,n≥-1,
the system is, for n∈ℕ0, transformed into(3.10)un+1=cn(1)vn+bn(1),vn+1=cn(2)wn+bn(2),wn+1=cn(3)un+bn(3).
System (3.10) implies that for n≥2(3.11)un+1=cn(1)cn-1(2)cn-2(3)un-2+cn(1)cn-1(2)bn-2(3)+cn(1)bn-1(2)+bn(1),(3.12)vn+1=cn-2(1)cn(2)cn-1(3)vn-2+cn(2)cn-1(3)bn-2(1)+cn(2)bn-1(3)+bn(2),(3.13)wn+1=cn-1(1)cn-2(2)cn(3)wn-2+cn-1(1)cn(3)bn-2(2)+cn(3)bn-1(1)+bn(3),
where values for u0, v0, w0 are computed by (3.9) with n=-1.Equation (3.11) implies that the sequences (u3n+i)n∈ℕ0, i∈{0,1,2}, are solutions of the first-order linear difference equation(3.14)u3n+i=c3n+i-1(1)c3n+i-2(2)c3n+i-3(3)u3(n-1)+i+c3n+i-1(1)c3n+i-2(2)b3n+i-3(3)+c3n+i-1(1)b3n+i-2(2)+b3n+i-1(1),n∈N.Applying the well-known formula for solutions of first-order difference equation we have that the general solution of equation (3.14) is(3.15)u3n+i=ui∏j=1n(c3j+i-1(1)c3j+i-2(2)c3j+i-3(3))+∑l=1n(c3l+i-1(1)c3l+i-2(2)b3l+i-3(3)+c3l+i-1(1)b3l+i-2(2)+b3l+i-1(1))∏j=l+1n(c3j+i-1(1)c3j+i-2(2)c3j+i-3(3))
for every n∈ℕ and each i∈{0,1,2}.From (3.12) we get that the sequences (v3n+i)n∈ℕ0, i∈{0,1,2}, are solutions of the first-order linear difference equation(3.16)v3n+i=c3n+i-3(1)c3n+i-1(2)c3n+i-2(3)v3(n-1)+i+c3n+i-1(2)c3n+i-2(3)b3n+i-3(1)+c3n+i-1(2)b3n+i-2(3)+b3n+i-1(2),n∈N,
from which it follows that(3.17)v3n+i=vi∏j=1n(c3j+i-3(1)c3j+i-1(2)c3j+i-2(3))+∑l=1n(c3l+i-1(2)c3l+i-2(3)b3l+i-3(1)+c3l+i-1(2)b3l+i-2(3)+b3l+i-1(2))∏j=l+1n(c3j+i-3(1)c3j+i-1(2)c3j+i-2(3)),
for every n∈ℕ and each i∈{0,1,2}.From (3.13) we get that the sequences (w3n+i)n∈ℕ0, i∈{0,1,2}, are solutions of the first-order linear difference equation(3.18)w3n+i=c3n+i-2(1)c3n+i-3(2)c3n+i-1(3)w3(n-1)+i+c3n+i-2(1)c3n+i-1(3)b3n+i-3(2)+c3n+i-1(3)b3n+i-2(1)+b3n+i-1(3),n∈N.Hence, for everyn∈ℕ and each i∈{0,1,2}, we have that(3.19)w3n+i=wi∏j=1n(c3j+i-2(1)c3j+i-3(2)c3j+i-1(3))+∑l=1n(c3l+i-2(1)c3l+i-1(3)b3l+i-3(2)+c3l+i-1(3)b3l+i-2(1)+b3l+i-1(3))∏j=l+1n(c3j+i-2(1)c3j+i-3(2)c3j+i-1(3)).Now note that from (3.9) we have(3.20)xn+1=1un+1ynzn-1=vnun+1xn-2,yn+1=1vn+1znxn-1=wnvn+1yn-2,zn+1=1wn+1xnyn-1=unwn+1zn-2,n∈N0.Hence(3.21)x3n+i+1=xi-2∏j=0nv3j+iu3j+i+1,n∈N0,i∈{0,1,2},(3.22)y3n+i+1=yi-2∏j=0nw3j+iv3j+i+1,n∈N0,i∈{0,1,2},(3.23)z3n+i+1=zi-2∏j=0nu3j+iw3j+i+1,n∈N0,i∈{0,1,2}.Applying (3.15), (3.17) and (3.19) in (3.21)–(3.23), we get explicit solutions of system (3.2) in terms of sequences bn(i), cn(i), n∈ℕ0, i∈{0,1,2}.The results in this section can be summed up in Table1.Table 1CaseFormulas for well-defined solutions of system (3.2)x-2=0x3n-2=0,y3n=y0∏j=0n-1c3j+2(2),z3n+2=z-1∏j=0nc3j+1(3),n∈ℕ0x-1=0x3n-1=0,y3n+1=y-2∏j=0nc3j(2),z3n=z0∏j=0n-1c3j+2(3),n∈ℕ0x0=0x3n=0,y3n+2=y-1∏j=0nc3j+1(2),z3n+1=z-2∏j=0nc3j(3),n∈ℕ0y-2=0y3n-2=0,z3n=z0∏j=0n-1c3j+2(3),x3n+2=x-1∏j=0nc3j+1(1),n∈ℕ0y-1=0y3n-1=0,z3n+1=z-2∏j=0nc3j(3),x3n=x0∏j=0n-1c3j+2(1),n∈ℕ0y0=0y3n=0,z3n+2=z-1∏j=0nc3j+1(3),x3n+1=x-2∏j=0nc3j(1),n∈ℕ0z-2=0z3n-2=0,x3n=x0∏j=0n-1c3j+2(1),y3n+2=y-1∏j=0nc3j+1(2),n∈ℕ0z-1=0z3n-1=0,x3n+1=x-2∏j=0nc3j(1),y3n=y0∏j=0n-1c3j+2(2),n∈ℕ0z0=0z3n=0,x3n+2=x-1∏j=0nc3j+1(1),y3n+1=y-2∏j=0nc3j(2),n∈ℕ0x3n+i+1=xi-2∏j=0nv3j+iu3j+i+1,y3n+i+1=yi-2∏j=0nw3j+iv3j+i+1,z3n+i+1=zi-2∏j=0nu3j+iw3j+i+1,n∈ℕ0,i∈{0,1,2}u3n+i=ui∏j=1n(c3j+i-1(1)c3j+i-2(2)c3j+i-3(3))x-i≠0≠y-i+∑l=1n(c3l+i-1(1)c3l+i-2(2)b3l+i-3(3)+c3l+i-1(1)b3l+i-2(2)+b3l+i-1(1))∏j=l+1n(c3j+i-1(1)c3j+i-2(2)c3j+i-3(3)),z-i≠0,i∈{0,1,2}v3n+i=vi∏j=1n(c3j+i-3(1)c3j+i-1(2)c3j+i-2(3))+∑l=1n(c3l+i-1(2)c3l+i-2(3)b3l+i-3(1)+c3l+i-1(2)b3l+i-2(3)+b3l+i-1(2))∏j=l+1n(c3j+i-3(1)c3j+i-1(2)c3j+i-2(3)),w3n+i=wi∏j=1n(c3j+i-2(1)c3j+i-3(2)c3j+i-1(3))+∑l=1n(c3l+i-2(1)c3l+i-1(3)b3l+i-3(2)+c3l+i-1(3)b3l+i-2(1)+b3l+i-1(3))∏j=l+1n(c3j+i-2(1)c3j+i-3(2)c3j+i-1(3))Remark 3.1.
Formulae for the solutions of system (3.2) when some of the numbers bn(i), cn(i), n∈ℕ0, i∈{0,1,2} are zero follow from the formulae given in Table 1.


## 3.1.  Main Case

Here we study well-defined solutions of system (1.1) when neither of the sequences (an(i))n∈ℕ0, i∈{1,2,3}, or initial conditions x-i, y-i, z-i, i∈{0,1,2}, is equal to zero. Recall that we may assume that an(i)=1, for every n∈ℕ0 and for each i∈{1,2,3}.Following the idea in [37], we use a transformation which reduces nonlinear systems (1.1) and (3.2) to third-order systems of nonhomogeneous linear difference equations.If we multiply the first equation in system (3.2) by ynzn-1, the second by znxn-1 and the third by xnyn-1, and then using in such obtained system the change of variables(3.9)un+1=1xn+1ynzn-1,vn+1=1yn+1znxn-1,wn+1=1zn+1xnyn-1,n≥-1,
the system is, for n∈ℕ0, transformed into(3.10)un+1=cn(1)vn+bn(1),vn+1=cn(2)wn+bn(2),wn+1=cn(3)un+bn(3).
System (3.10) implies that for n≥2(3.11)un+1=cn(1)cn-1(2)cn-2(3)un-2+cn(1)cn-1(2)bn-2(3)+cn(1)bn-1(2)+bn(1),(3.12)vn+1=cn-2(1)cn(2)cn-1(3)vn-2+cn(2)cn-1(3)bn-2(1)+cn(2)bn-1(3)+bn(2),(3.13)wn+1=cn-1(1)cn-2(2)cn(3)wn-2+cn-1(1)cn(3)bn-2(2)+cn(3)bn-1(1)+bn(3),
where values for u0, v0, w0 are computed by (3.9) with n=-1.Equation (3.11) implies that the sequences (u3n+i)n∈ℕ0, i∈{0,1,2}, are solutions of the first-order linear difference equation(3.14)u3n+i=c3n+i-1(1)c3n+i-2(2)c3n+i-3(3)u3(n-1)+i+c3n+i-1(1)c3n+i-2(2)b3n+i-3(3)+c3n+i-1(1)b3n+i-2(2)+b3n+i-1(1),n∈N.Applying the well-known formula for solutions of first-order difference equation we have that the general solution of equation (3.14) is(3.15)u3n+i=ui∏j=1n(c3j+i-1(1)c3j+i-2(2)c3j+i-3(3))+∑l=1n(c3l+i-1(1)c3l+i-2(2)b3l+i-3(3)+c3l+i-1(1)b3l+i-2(2)+b3l+i-1(1))∏j=l+1n(c3j+i-1(1)c3j+i-2(2)c3j+i-3(3))
for every n∈ℕ and each i∈{0,1,2}.From (3.12) we get that the sequences (v3n+i)n∈ℕ0, i∈{0,1,2}, are solutions of the first-order linear difference equation(3.16)v3n+i=c3n+i-3(1)c3n+i-1(2)c3n+i-2(3)v3(n-1)+i+c3n+i-1(2)c3n+i-2(3)b3n+i-3(1)+c3n+i-1(2)b3n+i-2(3)+b3n+i-1(2),n∈N,
from which it follows that(3.17)v3n+i=vi∏j=1n(c3j+i-3(1)c3j+i-1(2)c3j+i-2(3))+∑l=1n(c3l+i-1(2)c3l+i-2(3)b3l+i-3(1)+c3l+i-1(2)b3l+i-2(3)+b3l+i-1(2))∏j=l+1n(c3j+i-3(1)c3j+i-1(2)c3j+i-2(3)),
for every n∈ℕ and each i∈{0,1,2}.From (3.13) we get that the sequences (w3n+i)n∈ℕ0, i∈{0,1,2}, are solutions of the first-order linear difference equation(3.18)w3n+i=c3n+i-2(1)c3n+i-3(2)c3n+i-1(3)w3(n-1)+i+c3n+i-2(1)c3n+i-1(3)b3n+i-3(2)+c3n+i-1(3)b3n+i-2(1)+b3n+i-1(3),n∈N.Hence, for everyn∈ℕ and each i∈{0,1,2}, we have that(3.19)w3n+i=wi∏j=1n(c3j+i-2(1)c3j+i-3(2)c3j+i-1(3))+∑l=1n(c3l+i-2(1)c3l+i-1(3)b3l+i-3(2)+c3l+i-1(3)b3l+i-2(1)+b3l+i-1(3))∏j=l+1n(c3j+i-2(1)c3j+i-3(2)c3j+i-1(3)).Now note that from (3.9) we have(3.20)xn+1=1un+1ynzn-1=vnun+1xn-2,yn+1=1vn+1znxn-1=wnvn+1yn-2,zn+1=1wn+1xnyn-1=unwn+1zn-2,n∈N0.Hence(3.21)x3n+i+1=xi-2∏j=0nv3j+iu3j+i+1,n∈N0,i∈{0,1,2},(3.22)y3n+i+1=yi-2∏j=0nw3j+iv3j+i+1,n∈N0,i∈{0,1,2},(3.23)z3n+i+1=zi-2∏j=0nu3j+iw3j+i+1,n∈N0,i∈{0,1,2}.Applying (3.15), (3.17) and (3.19) in (3.21)–(3.23), we get explicit solutions of system (3.2) in terms of sequences bn(i), cn(i), n∈ℕ0, i∈{0,1,2}.The results in this section can be summed up in Table1.Table 1CaseFormulas for well-defined solutions of system (3.2)x-2=0x3n-2=0,y3n=y0∏j=0n-1c3j+2(2),z3n+2=z-1∏j=0nc3j+1(3),n∈ℕ0x-1=0x3n-1=0,y3n+1=y-2∏j=0nc3j(2),z3n=z0∏j=0n-1c3j+2(3),n∈ℕ0x0=0x3n=0,y3n+2=y-1∏j=0nc3j+1(2),z3n+1=z-2∏j=0nc3j(3),n∈ℕ0y-2=0y3n-2=0,z3n=z0∏j=0n-1c3j+2(3),x3n+2=x-1∏j=0nc3j+1(1),n∈ℕ0y-1=0y3n-1=0,z3n+1=z-2∏j=0nc3j(3),x3n=x0∏j=0n-1c3j+2(1),n∈ℕ0y0=0y3n=0,z3n+2=z-1∏j=0nc3j+1(3),x3n+1=x-2∏j=0nc3j(1),n∈ℕ0z-2=0z3n-2=0,x3n=x0∏j=0n-1c3j+2(1),y3n+2=y-1∏j=0nc3j+1(2),n∈ℕ0z-1=0z3n-1=0,x3n+1=x-2∏j=0nc3j(1),y3n=y0∏j=0n-1c3j+2(2),n∈ℕ0z0=0z3n=0,x3n+2=x-1∏j=0nc3j+1(1),y3n+1=y-2∏j=0nc3j(2),n∈ℕ0x3n+i+1=xi-2∏j=0nv3j+iu3j+i+1,y3n+i+1=yi-2∏j=0nw3j+iv3j+i+1,z3n+i+1=zi-2∏j=0nu3j+iw3j+i+1,n∈ℕ0,i∈{0,1,2}u3n+i=ui∏j=1n(c3j+i-1(1)c3j+i-2(2)c3j+i-3(3))x-i≠0≠y-i+∑l=1n(c3l+i-1(1)c3l+i-2(2)b3l+i-3(3)+c3l+i-1(1)b3l+i-2(2)+b3l+i-1(1))∏j=l+1n(c3j+i-1(1)c3j+i-2(2)c3j+i-3(3)),z-i≠0,i∈{0,1,2}v3n+i=vi∏j=1n(c3j+i-3(1)c3j+i-1(2)c3j+i-2(3))+∑l=1n(c3l+i-1(2)c3l+i-2(3)b3l+i-3(1)+c3l+i-1(2)b3l+i-2(3)+b3l+i-1(2))∏j=l+1n(c3j+i-3(1)c3j+i-1(2)c3j+i-2(3)),w3n+i=wi∏j=1n(c3j+i-2(1)c3j+i-3(2)c3j+i-1(3))+∑l=1n(c3l+i-2(1)c3l+i-1(3)b3l+i-3(2)+c3l+i-1(3)b3l+i-2(1)+b3l+i-1(3))∏j=l+1n(c3j+i-2(1)c3j+i-3(2)c3j+i-1(3))Remark 3.1.
Formulae for the solutions of system (3.2) when some of the numbers bn(i), cn(i), n∈ℕ0, i∈{0,1,2} are zero follow from the formulae given in Table 1.


## 4.  Some Consequences

### 4.1.  Casean(i)≠0 for i∈{1,2,3} and n∈ℕ0

First we use the formulae in Section3 to get solutions of system (1.1), when an(i)≠0 for i∈{1,2,3} and n∈ℕ0.For this we replace sequences(4.1)bn(i),cn(i),i∈{1,2,3},
in formulas of Section 3 with sequences(4.2)bn(i)an(i),cn(i)an(i),i∈{1,2,3}.
We arrange these formulae in Table 2.Table 2CaseFormulas for well-defined solutions of system (1.1)x-2=0x3n-2=0,y3n=y0∏j=0n-1(a3j+2(2)c3j+2(2)),z3n+2=z-1∏j=0n(a3j+1(3)c3j+1(3)),n∈ℕ0x-1=0x3n-1=0,y3n+1=y-2∏j=0n(a3j(2)c3j(2)),z3n=z0∏j=0n-1(a3j+2(3)c3j+2(3)),n∈ℕ0x0=0x3n=0,y3n+2=y-1∏j=0n(a3j+1(2)c3j+1(2)),z3n+1=z-2∏j=0n(a3j(3)c3j(3)),n∈ℕ0y-2=0y3n-2=0,z3n=z0∏j=0n-1(a3j+2(3)c3j+2(3)),x3n+2=x-1∏j=0n(a3j+1(1)c3j+1(1)),n∈ℕ0y-1=0y3n-1=0,z3n+1=z-2∏j=0n(a3j(3)c3j(3)),x3n=x0∏j=0n-1(a3j+2(1)c3j+2(1)),n∈ℕ0y0=0y3n=0,z3n+2=z-1∏j=0n(a3j+1(3)c3j+1(3)),x3n+1=x-2∏j=0n(a3j(1)c3j(1)),n∈ℕ0z-2=0z3n-2=0,x3n=x0∏j=0n-1(a3j+2(1)c3j+2(1)),y3n+2=y-1∏j=0n(a3j+1(2)c3j+1(2)),n∈ℕ0z-1=0z3n-1=0,x3n+1=x-2∏j=0n(a3j(1)c3j(1)),y3n=y0∏j=0n-1(a3j+2(2)c3j+2(2)),n∈ℕ0z0=0z3n=0,x3n+2=x-1∏j=0n(a3j+1(1)c3j+1(1)),y3n+1=y-2∏j=0n(a3j(2)c3j(2)),n∈ℕ0x3n+i+1=xi-2∏j=0nv3j+iu3j+i+1,y3n+i+1=yi-2∏j=0nw3j+iv3j+i+1,z3n+i+1=zi-2∏j=0nu3j+iw3j+i+1,n∈ℕ0,i∈{0,1,2}u3n+i=ui∏j=1nc3j+i-1(1)c3j+i-2(2)c3j+i-3(3)a3j+i-1(1)a3j+i-2(2)a3j+i-3(3)x-i≠0≠y-iz-i≠0,i∈{0,1,2}+∑l=1n(c3l+i-1(1)c3l+i-2(2)b3l+i-3(3)a3l+i-1(1)a3l+i-2(2)a3l+i-3(3)+c3l+i-1(1)b3l+i-2(2)a3l+i-1(1)a3l+i-2(2)+b3l+i-1(1)a3l+i-1(1))∏j=l+1nc3j+i-1(1)c3j+i-2(2)c3j+i-3(3)a3j+i-1(1)a3j+i-2(2)a3j+i-3(3),v3n+i=vi∏j=1nc3j+i-3(1)c3j+i-1(2)c3j+i-2(3)a3j+i-3(1)a3j+i-1(2)a3j+i-2(3)+∑l=1n(c3l+i-1(2)c3l+i-2(3)b3l+i-3(1)a3l+i-1(2)a3l+i-2(3)a3l+i-3(1)+c3l+i-1(2)b3l+i-2(3)a3l+i-1(2)a3l+i-2(3)+b3l+i-1(2)a3l+i-1(2))∏j=l+1nc3j+i-3(1)c3j+i-1(2)c3j+i-2(3)a3j+i-3(1)a3j+i-1(2)a3j+i-2(3),w3n+i=wi∏j=1nc3j+i-2(1)c3j+i-3(2)c3j+i-1(3)a3j+i-2(1)a3j+i-3(2)a3j+i-1(3)+∑l=1n(c3l+i-2(1)c3l+i-1(3)b3l+i-3(2)a3l+i-2(1)a3l+i-1(3)a3l+i-3(2)+c3l+i-1(3)b3l+i-2(1)a3l+i-1(3)a3l+i-2(1)+b3l+i-1(3)a3l+i-1(3))∏j=l+1nc3j+i-2(1)c3j+i-3(2)c3j+i-1(3)a3j+i-2(1)a3j+i-3(2)a3j+i-1(3).


### 4.2. Casebn(i), cn(i), n∈ℕ0, i∈{1,2,3} Are Period-Three Sequences

Now we get formulae for solutions of system (3.2) when the sequences bn(i), cn(i), n∈ℕ0, i∈{1,2,3} are periodic with period three.If this holds then from (3.15) we have that(4.3)u3n+i=ui∏j=1n(ci+2(1)ci+1(2)ci(3))+∑l=1n(ci+2(1)ci+1(2)bi(3)+ci+2(1)bi+1(2)+bi+2(1))∏j=l+1n(ci+2(1)ci+1(2)ci(3))=ui(ci+2(1)ci+1(2)ci(3))n+(ci+2(1)ci+1(2)bi(3)+ci+2(1)bi+1(2)+bi+2(1))∑l=1n(ci+2(1)ci+1(2)ci(3))n-l=ui(ci+2(1)ci+1(2)ci(3))n+(ci+2(1)ci+1(2)bi(3)+ci+2(1)bi+1(2)+bi+2(1))1-(ci+2(1)ci+1(2)ci(3))n1-ci+2(1)ci+1(2)ci(3)
for every n∈ℕ0, when ci+2(1)ci+1(2)ci(3)≠1, for some i∈{0,1,2}, while(4.4)u3n+i=ui+(ci+2(1)ci+1(2)bi(3)+ci+2(1)bi+1(2)+bi+2(1))n,
when ci+2(1)ci+1(2)ci(3)=1, for some i∈{0,1,2}. Here we regard that ck(j)=ci(j), for some j∈{1,2,3}, i∈{0,1,2} and k≥3, when k≡i(mod3).Similarly, we get(4.5)v3n+i=vi(ci(1)ci+2(2)ci+1(3))n+(ci+2(2)ci+1(3)bi(1)+ci+2(2)bi+1(3)+bi+2(2))1-(ci(1)ci+2(2)ci+1(3))n1-ci(1)ci+2(2)ci+1(3)
for every n∈ℕ0, if ci(1)ci+2(2)ci+1(3)≠1, for some i∈{0,1,2}, and(4.6)v3n+i=vi+(ci+2(2)ci+1(3)bi(1)+ci+2(2)bi+1(3)+bi+2(2))n
for every n∈ℕ0, when ci(1)ci+2(2)ci+1(3)=1, for some i∈{0,1,2}.Finally(4.7)w3n+i=wi(ci+1(1)ci(2)ci+2(3))n+(ci+1(1)ci+2(3)bi(2)+ci+2(3)bi+1(1)+bi+2(3))1-(ci+1(1)ci(2)ci+2(3))n1-ci+1(1)ci(2)ci+2(3),
for every n∈ℕ0, when ci+1(1)ci(2)ci+2(3)≠1, for some i∈{0,1,2}, and(4.8)w3n+i=wi+(ci+1(1)ci+2(3)bi(2)+ci+2(3)bi+1(1)+bi+2(3))n,
for every n∈ℕ0, when ci+1(1)ci(2)ci+2(3)=1, for some i∈{0,1,2}.If we assume thatan(i)≠0 for all i∈{1,2,3} and n∈ℕ0, that these three sequences are also periodic with period three, and replace the sequences in (4.1) with the corresponding in (4.2) we get formulae for solutions of system (1.1) when the sequences an(i), bn(i), and cn(i), n∈ℕ0, i∈{1,2,3} are periodic with period three.These formulae follows from above obtained ones and are summarized in Table3.Table 3CaseFormulas for well-defined solutions of system (1.1) with period-three coefficientsx-i≠0≠y-ix3n+i+1=xi-2∏j=0nv3j+iu3j+i+1,y3n+i+1=yi-2∏j=0nw3j+iv3j+i+1,z-i≠0,
i∈{0,1,2}z3n+i+1=zi-2∏j=0nu3j+iw3j+i+1,n∈ℕ0,i∈{0,1,2}q1=ci+2(1)ci+1(2)ci(3)ai+2(1)ai+1(2)ai(3)≠1u3n+i=uiq1n+(ci+2(1)ci+1(2)bi(3)ai+2(1)ai+1(2)ai(3)+ci+2(1)bi+1(2)ai+2(1)ai+1(2)+bi+2(1)ai+2(1))1-q1n1-q1,n∈ℕ0,i∈{0,1,2}q1=ci+2(1)ci+1(2)ci(3)ai+2(1)ai+1(2)ai(3)=1u3n+i=ui+(ci+2(1)ci+1(2)bi(3)ai+2(1)ai+1(2)ai(3)+ci+2(1)bi+1(2)ai+2(1)ai+1(2)+bi+2(1)ai+2(1))n,n∈ℕ0,i∈{0,1,2}q2=ci(1)ci+2(2)ci+1(3)ai(1)ai+2(2)ai+1(3)≠1v3n+i=viq2n+(ci+2(2)ci+1(3)bi(1)ai+2(2)ai+1(3)ai(1)+ci+2(2)bi+1(3)ai+2(2)ai+1(3)+bi+2(2)ai+2(2))1-q2n1-q2,n∈ℕ0,i∈{0,1,2}q2=ci(1)ci+2(2)ci+1(3)ai(1)ai+2(2)ai+1(3)=1v3n+i=vi+(ci+2(2)ci+1(3)bi(1)ai+2(2)ai+1(3)ai(1)+ci+2(2)bi+1(3)ai+2(2)ai+1(3)+bi+2(2)ai+2(2))n,n∈ℕ0,i∈{0,1,2}q3=ci+1(1)ci(2)ci+2(3)ai+1(1)ai(2)ai+2(3)≠1w3n+i=wiq3n+(ci+1(1)ci+2(3)bi(2)ai+1(1)ai+2(3)ai(2)+ci+2(3)bi+1(1)ai+2(3)ai+1(1)+bi+2(3)ai+2(3))1-q3n1-q3,n∈ℕ0,i∈{0,1,2}q3=ci+1(1)ci(2)ci+2(3)ai+1(1)ai(2)ai+2(3)=1w3n+i=wi+(ci+1(1)ci+2(3)bi(2)ai+1(1)ai+2(3)ai(2)+ci+2(3)bi+1(1)ai+2(3)ai+1(1)+bi+2(3)ai+2(3))n,n∈ℕ0,i∈{0,1,2}


## 4.1.  Casean(i)≠0 for i∈{1,2,3} and n∈ℕ0

First we use the formulae in Section3 to get solutions of system (1.1), when an(i)≠0 for i∈{1,2,3} and n∈ℕ0.For this we replace sequences(4.1)bn(i),cn(i),i∈{1,2,3},
in formulas of Section 3 with sequences(4.2)bn(i)an(i),cn(i)an(i),i∈{1,2,3}.
We arrange these formulae in Table 2.Table 2CaseFormulas for well-defined solutions of system (1.1)x-2=0x3n-2=0,y3n=y0∏j=0n-1(a3j+2(2)c3j+2(2)),z3n+2=z-1∏j=0n(a3j+1(3)c3j+1(3)),n∈ℕ0x-1=0x3n-1=0,y3n+1=y-2∏j=0n(a3j(2)c3j(2)),z3n=z0∏j=0n-1(a3j+2(3)c3j+2(3)),n∈ℕ0x0=0x3n=0,y3n+2=y-1∏j=0n(a3j+1(2)c3j+1(2)),z3n+1=z-2∏j=0n(a3j(3)c3j(3)),n∈ℕ0y-2=0y3n-2=0,z3n=z0∏j=0n-1(a3j+2(3)c3j+2(3)),x3n+2=x-1∏j=0n(a3j+1(1)c3j+1(1)),n∈ℕ0y-1=0y3n-1=0,z3n+1=z-2∏j=0n(a3j(3)c3j(3)),x3n=x0∏j=0n-1(a3j+2(1)c3j+2(1)),n∈ℕ0y0=0y3n=0,z3n+2=z-1∏j=0n(a3j+1(3)c3j+1(3)),x3n+1=x-2∏j=0n(a3j(1)c3j(1)),n∈ℕ0z-2=0z3n-2=0,x3n=x0∏j=0n-1(a3j+2(1)c3j+2(1)),y3n+2=y-1∏j=0n(a3j+1(2)c3j+1(2)),n∈ℕ0z-1=0z3n-1=0,x3n+1=x-2∏j=0n(a3j(1)c3j(1)),y3n=y0∏j=0n-1(a3j+2(2)c3j+2(2)),n∈ℕ0z0=0z3n=0,x3n+2=x-1∏j=0n(a3j+1(1)c3j+1(1)),y3n+1=y-2∏j=0n(a3j(2)c3j(2)),n∈ℕ0x3n+i+1=xi-2∏j=0nv3j+iu3j+i+1,y3n+i+1=yi-2∏j=0nw3j+iv3j+i+1,z3n+i+1=zi-2∏j=0nu3j+iw3j+i+1,n∈ℕ0,i∈{0,1,2}u3n+i=ui∏j=1nc3j+i-1(1)c3j+i-2(2)c3j+i-3(3)a3j+i-1(1)a3j+i-2(2)a3j+i-3(3)x-i≠0≠y-iz-i≠0,i∈{0,1,2}+∑l=1n(c3l+i-1(1)c3l+i-2(2)b3l+i-3(3)a3l+i-1(1)a3l+i-2(2)a3l+i-3(3)+c3l+i-1(1)b3l+i-2(2)a3l+i-1(1)a3l+i-2(2)+b3l+i-1(1)a3l+i-1(1))∏j=l+1nc3j+i-1(1)c3j+i-2(2)c3j+i-3(3)a3j+i-1(1)a3j+i-2(2)a3j+i-3(3),v3n+i=vi∏j=1nc3j+i-3(1)c3j+i-1(2)c3j+i-2(3)a3j+i-3(1)a3j+i-1(2)a3j+i-2(3)+∑l=1n(c3l+i-1(2)c3l+i-2(3)b3l+i-3(1)a3l+i-1(2)a3l+i-2(3)a3l+i-3(1)+c3l+i-1(2)b3l+i-2(3)a3l+i-1(2)a3l+i-2(3)+b3l+i-1(2)a3l+i-1(2))∏j=l+1nc3j+i-3(1)c3j+i-1(2)c3j+i-2(3)a3j+i-3(1)a3j+i-1(2)a3j+i-2(3),w3n+i=wi∏j=1nc3j+i-2(1)c3j+i-3(2)c3j+i-1(3)a3j+i-2(1)a3j+i-3(2)a3j+i-1(3)+∑l=1n(c3l+i-2(1)c3l+i-1(3)b3l+i-3(2)a3l+i-2(1)a3l+i-1(3)a3l+i-3(2)+c3l+i-1(3)b3l+i-2(1)a3l+i-1(3)a3l+i-2(1)+b3l+i-1(3)a3l+i-1(3))∏j=l+1nc3j+i-2(1)c3j+i-3(2)c3j+i-1(3)a3j+i-2(1)a3j+i-3(2)a3j+i-1(3).


## 4.2. Casebn(i), cn(i), n∈ℕ0, i∈{1,2,3} Are Period-Three Sequences

Now we get formulae for solutions of system (3.2) when the sequences bn(i), cn(i), n∈ℕ0, i∈{1,2,3} are periodic with period three.If this holds then from (3.15) we have that(4.3)u3n+i=ui∏j=1n(ci+2(1)ci+1(2)ci(3))+∑l=1n(ci+2(1)ci+1(2)bi(3)+ci+2(1)bi+1(2)+bi+2(1))∏j=l+1n(ci+2(1)ci+1(2)ci(3))=ui(ci+2(1)ci+1(2)ci(3))n+(ci+2(1)ci+1(2)bi(3)+ci+2(1)bi+1(2)+bi+2(1))∑l=1n(ci+2(1)ci+1(2)ci(3))n-l=ui(ci+2(1)ci+1(2)ci(3))n+(ci+2(1)ci+1(2)bi(3)+ci+2(1)bi+1(2)+bi+2(1))1-(ci+2(1)ci+1(2)ci(3))n1-ci+2(1)ci+1(2)ci(3)
for every n∈ℕ0, when ci+2(1)ci+1(2)ci(3)≠1, for some i∈{0,1,2}, while(4.4)u3n+i=ui+(ci+2(1)ci+1(2)bi(3)+ci+2(1)bi+1(2)+bi+2(1))n,
when ci+2(1)ci+1(2)ci(3)=1, for some i∈{0,1,2}. Here we regard that ck(j)=ci(j), for some j∈{1,2,3}, i∈{0,1,2} and k≥3, when k≡i(mod3).Similarly, we get(4.5)v3n+i=vi(ci(1)ci+2(2)ci+1(3))n+(ci+2(2)ci+1(3)bi(1)+ci+2(2)bi+1(3)+bi+2(2))1-(ci(1)ci+2(2)ci+1(3))n1-ci(1)ci+2(2)ci+1(3)
for every n∈ℕ0, if ci(1)ci+2(2)ci+1(3)≠1, for some i∈{0,1,2}, and(4.6)v3n+i=vi+(ci+2(2)ci+1(3)bi(1)+ci+2(2)bi+1(3)+bi+2(2))n
for every n∈ℕ0, when ci(1)ci+2(2)ci+1(3)=1, for some i∈{0,1,2}.Finally(4.7)w3n+i=wi(ci+1(1)ci(2)ci+2(3))n+(ci+1(1)ci+2(3)bi(2)+ci+2(3)bi+1(1)+bi+2(3))1-(ci+1(1)ci(2)ci+2(3))n1-ci+1(1)ci(2)ci+2(3),
for every n∈ℕ0, when ci+1(1)ci(2)ci+2(3)≠1, for some i∈{0,1,2}, and(4.8)w3n+i=wi+(ci+1(1)ci+2(3)bi(2)+ci+2(3)bi+1(1)+bi+2(3))n,
for every n∈ℕ0, when ci+1(1)ci(2)ci+2(3)=1, for some i∈{0,1,2}.If we assume thatan(i)≠0 for all i∈{1,2,3} and n∈ℕ0, that these three sequences are also periodic with period three, and replace the sequences in (4.1) with the corresponding in (4.2) we get formulae for solutions of system (1.1) when the sequences an(i), bn(i), and cn(i), n∈ℕ0, i∈{1,2,3} are periodic with period three.These formulae follows from above obtained ones and are summarized in Table3.Table 3CaseFormulas for well-defined solutions of system (1.1) with period-three coefficientsx-i≠0≠y-ix3n+i+1=xi-2∏j=0nv3j+iu3j+i+1,y3n+i+1=yi-2∏j=0nw3j+iv3j+i+1,z-i≠0,
i∈{0,1,2}z3n+i+1=zi-2∏j=0nu3j+iw3j+i+1,n∈ℕ0,i∈{0,1,2}q1=ci+2(1)ci+1(2)ci(3)ai+2(1)ai+1(2)ai(3)≠1u3n+i=uiq1n+(ci+2(1)ci+1(2)bi(3)ai+2(1)ai+1(2)ai(3)+ci+2(1)bi+1(2)ai+2(1)ai+1(2)+bi+2(1)ai+2(1))1-q1n1-q1,n∈ℕ0,i∈{0,1,2}q1=ci+2(1)ci+1(2)ci(3)ai+2(1)ai+1(2)ai(3)=1u3n+i=ui+(ci+2(1)ci+1(2)bi(3)ai+2(1)ai+1(2)ai(3)+ci+2(1)bi+1(2)ai+2(1)ai+1(2)+bi+2(1)ai+2(1))n,n∈ℕ0,i∈{0,1,2}q2=ci(1)ci+2(2)ci+1(3)ai(1)ai+2(2)ai+1(3)≠1v3n+i=viq2n+(ci+2(2)ci+1(3)bi(1)ai+2(2)ai+1(3)ai(1)+ci+2(2)bi+1(3)ai+2(2)ai+1(3)+bi+2(2)ai+2(2))1-q2n1-q2,n∈ℕ0,i∈{0,1,2}q2=ci(1)ci+2(2)ci+1(3)ai(1)ai+2(2)ai+1(3)=1v3n+i=vi+(ci+2(2)ci+1(3)bi(1)ai+2(2)ai+1(3)ai(1)+ci+2(2)bi+1(3)ai+2(2)ai+1(3)+bi+2(2)ai+2(2))n,n∈ℕ0,i∈{0,1,2}q3=ci+1(1)ci(2)ci+2(3)ai+1(1)ai(2)ai+2(3)≠1w3n+i=wiq3n+(ci+1(1)ci+2(3)bi(2)ai+1(1)ai+2(3)ai(2)+ci+2(3)bi+1(1)ai+2(3)ai+1(1)+bi+2(3)ai+2(3))1-q3n1-q3,n∈ℕ0,i∈{0,1,2}q3=ci+1(1)ci(2)ci+2(3)ai+1(1)ai(2)ai+2(3)=1w3n+i=wi+(ci+1(1)ci+2(3)bi(2)ai+1(1)ai+2(3)ai(2)+ci+2(3)bi+1(1)ai+2(3)ai+1(1)+bi+2(3)ai+2(3))n,n∈ℕ0,i∈{0,1,2}


## 5.  Some Applications

Using above listed formulae the behavior of solutions of system (1.1) or (3.2) can be described. We will present here some results which can be obtained from these formulae, to demonstrate how they can be used. Before we formulate the results note that if cn(j), n∈ℕ0, j∈{1,2,3} are periodic with period three, then(5.1)ci(1)ci+2(2)ci+1(3)=ci+3(1)ci+2(2)ci+1(3),ci+1(1)ci+3(2)ci+2(3)=ci+1(1)ci(2)ci+2(3),(5.2)ci+2(1)ci+1(2)ci(3)=ci+2(1)ci+1(2)ci+3(3).Theorem 5.1.
Consider system (3.2). Let the sequences bn(j), cn(j), n∈ℕ0, j∈{1,2,3} be periodic with period three,
(5.3)pi∶=ci+2(2)ci+1(3)bi(1)+ci+2(2)bi+1(3)+bi+2(2)ci+3(1)ci+2(2)bi+1(3)+ci+3(1)bi+2(2)+bi+3(1),qi∶=vi-ui+1ci+3(1)ci+2(2)bi+1(3)+ci+3(1)bi+2(2)+bi+3(1),ri∶=vi+ui+1ci+3(1)ci+2(2)bi+1(3)+ci+3(1)bi+2(2)+bi+3(1),
when ci+3(1)ci+2(2)bi+1(3)+ci+3(1)bi+2(2)+bi+3(1)≠0, and for some i∈{0,1,2} the following condition holds:
(5.4)ci(1)ci+2(2)ci+1(3)=1.
Then if x3n+i+1≠0 for every n∈ℕ0, the following statements hold: (a)
if|pi|<1, then x3n+i+1→0, as n→∞;(b)
if|pi|>1, then |x3n+i+1|→∞, as n→∞;(c)
ifpi=1 and qi<0, then x3n+i+1→0, as n→∞;(d)
ifpi=1 and qi>0, then |x3n+i+1|→∞, as n→∞;(e)
ifpi=1 and qi=0, then x3n+i+1 is convergent;(f)
ifpi=-1 and ri>0, then x3n+i+1→0, as n→∞;(g)
ifpi=-1 and ri<0, then |x3n+i+1|→∞, as n→∞, so that x6n+i+1→+∞ and x6n+i+4→-∞, or x6n+i+1→-∞ and x6n+i+4→+∞ as n→∞;(h)
Ifpi=-1 and ri=0, then x6n+i+1 and x6n+i+4 are convergent, as n→∞.
Ifci+3(1)ci+2(2)bi+1(3)+ci+3(1)bi+2(2)+bi+3(1)=0, then the following statements hold:(i)
ifci+2(2)ci+1(3)bi(1)+ci+2(2)bi+1(3)+bi+2(2)≠0, then |x3n+i+1|→∞, as n→∞;(j)
ifci+2(2)ci+1(3)bi(1)+ci+2(2)bi+1(3)+bi+2(2)=0, then
(5.5)x3n+i+1=xi-2(viui+1)n+1.Proof.
(a), (b) From (3.21), (5.4) and (5.1), we have that for each i∈{0,1,2}(5.6)x3n+i+1=xi-2∏j=0nv3j+iu3j+i+1=xi-2∏j=0nvi+(ci+2(2)ci+1(3)bi(1)+ci+2(2)bi+1(3)+bi+2(2))jui+1+(ci+3(1)ci+2(2)bi+1(3)+ci+3(1)bi+2(2)+bi+3(1))j,
for every n∈ℕ0.
Since(5.7)lim⁡n→∞vi+(ci+2(2)ci+1(3)bi(1)+ci+2(2)bi+1(3)+bi+2(2))nui+1+(ci+3(1)ci+2(2)bi+1(3)+ci+3(1)bi+2(2)+bi+3(1))n=pi,
the results in (a) and (b) follow from (5.6) easily.
(c), (d), (e) Sincepi=1, we have that
(5.8)v3n+iu3n+i+1=(1+vi(ci+3(1)ci+2(2)bi+1(3)+ci+3(1)bi+2(2)+bi+3(1))n)(1+ui+1(ci+3(1)ci+2(2)bi+1(3)+ci+3(1)bi+2(2)+bi+3(1))n)-1=1+qin+O(1n2)=exp⁡(qin+O(1n2)),n⟶∞.
From (5.8) and by using well-known asymptotic relation
(5.9)∑j=n0n1j~ln⁡n,
where we assume n0≥1 and n→∞; the results in (c), (d), and (e) easily follow.
(f), (g), (h) Sincepi=-1, we have that
(5.10)v3n+iu3n+i+1=-(1-vi(ci+3(1)ci+2(2)bi+1(3)+ci+3(1)bi+2(2)+bi+3(1))n)(1+ui+1(ci+3(1)ci+2(2)bi+1(3)+ci+3(1)bi+2(2)+bi+3(1))n)-1=-(1-rin+O(1n2))=-exp⁡(-rin+O(1n2)),n⟶∞.
From (5.6), (5.10) and asymptotic relation (5.9) the results in (f), (g), and (h) follow.
(i), (j) These two statements follow easily from (5.6).Remark 5.2.
Ifx3n0+i+1=0 for an n0∈ℕ0, then by (3.2) we get x3n+i+1=0, for n≥n0, which is the reason why we posed the condition x3n+i+1≠0, for every n∈ℕ0, in Theorem 5.1. Similar conditions will be posed in the theorems which follow.Theorem 5.3.
Consider system (3.2). Let the sequences bn(j), cn(j), n∈ℕ0, j∈{1,2,3} be periodic with period three,
(5.11)p̂i∶=ci+1(1)ci+2(3)bi(2)+ci+2(3)bi+1(1)+bi+2(3)ci+3(2)ci+2(3)bi+1(1)+ci+3(2)bi+2(3)+bi+3(2),q̂i∶=wi-vi+1ci+3(2)ci+2(3)bi+1(1)+ci+3(2)bi+2(3)+bi+3(2),r̂i∶=wi+vi+1ci+3(2)ci+2(3)bi+1(1)+ci+3(2)bi+2(3)+bi+3(2),
when ci+3(2)ci+2(3)bi+1(1)+ci+3(2)bi+2(3)+bi+3(2)≠0, and that for some i∈{0,1,2} the following condition holds
(5.12)ci+1(1)ci+3(2)ci+2(3)=1.
Then if y3n+i+1≠0 for every n∈ℕ0, the following statements hold: (a)
if|p̂i|<1, then y3n+i+1→0, as n→∞;(b)
if|p̂i|>1, then |y3n+i+1|→∞, as n→∞;(c)
ifp̂i=1 and q̂i<0, then y3n+i+1→0, as n→∞;(d)
ifp̂i=1 and q̂i>0, then |y3n+i+1|→∞, as n→∞;(e)
ifp̂i=1 and q̂i=0, then y3n+i+1 is convergent;(f)
ifp̂i=-1 and r̂i>0, then y3n+i+1→0, as n→∞;(g)
ifp̂i=-1 and r̂i<0, then |y3n+i+1|→∞, as n→∞, so that y6n+i+1→+∞ and y6n+i+4→-∞, or y6n+i+1→-∞ and y6n+i+4→+∞ as n→∞;(h)
ifp̂i=-1 and r̂i=0, then y6n+i+1 and y6n+i+4 are convergent, as n→∞.
Ifci+3(2)ci+2(3)bi+1(1)+ci+3(2)bi+2(3)+bi+3(2)=0, then the following statements hold:(i)
ifci+1(1)ci+2(3)bi(2)+ci+2(3)bi+1(1)+bi+2(3)≠0, then |y3n+i+1|→∞, as n→∞;(j)
ifci+1(1)ci+2(3)bi(2)+ci+2(3)bi+1(1)+bi+2(3)=0, then
(5.13)y3n+i+1=yi-2(wivi+1)n+1.Proof.
From (3.22), by using condition (5.12) and the second equality in (5.1), we have that
(5.14)y3n+i+1=yi-2∏j=0nwi+(ci+1(1)ci+2(3)bi(2)+ci+2(3)bi+1(1)+bi+2(3))jvi+1+(ci+3(2)ci+2(3)bi+1(1)+ci+3(2)bi+2(3)+bi+3(2))j,
for each i∈{0,1,2} and every n∈ℕ0, from which the results in this theorem follows similar to Theorem 5.1.Theorem 5.4.
Consider system (3.2). Let the sequences bn(j), cn(j), n∈ℕ0, j∈{1,2,3} be periodic with period three,
(5.15)pi*∶=ci+2(1)ci+1(2)bi(3)+ci+2(1)bi+1(2)+bi+2(1)ci+2(1)ci+3(3)bi+1(2)+ci+3(3)bi+2(1)+bi+3(3),qi*∶=ui-wi+1ci+2(1)ci+3(3)bi+1(2)+ci+3(3)bi+2(1)+bi+3(3),ri*∶=ui+wi+1ci+2(1)ci+3(3)bi+1(2)+ci+3(3)bi+2(1)+bi+3(3),
when ci+2(1)ci+3(3)bi+1(2)+ci+3(3)bi+2(1)+bi+3(3)≠0, and for some i∈{0,1,2} the following condition holds
(5.16)ci+2(1)ci+1(2)ci(3)=1.
Then if z3n+i+1≠0 for every n∈ℕ0, the following statements hold: (a)
if|pi*|<1, then z3n+i+1→0, as n→∞;(b)
if|pi*|>1, then |z3n+i+1|→∞, as n→∞;(c)
ifpi*=1 and qi*<0, then z3n+i+1→0, as n→∞;(d)
ifpi*=1 and qi*>0, then |z3n+i+1|→∞, as n→∞;(e)
ifpi*=1 and qi*=0, then z3n+i+1 is convergent;(f)
ifpi*=-1 and ri*>0, then z3n+i+1→0, as n→∞;(g)
ifpi*=-1 and ri*<0, then |z3n+i+1|→∞, as n→∞, so that z6n+i+1→+∞ and z6n+i+4→-∞, or z6n+i+1→-∞ and z6n+i+4→+∞ as n→∞;(h)
ifpi*=-1 and ri*=0, then z6n+i+1 and z6n+i+4 are convergent, as n→∞.
Ifci+2(1)ci+3(3)bi+1(2)+ci+3(3)bi+2(1)+bi+3(3)=0, then the following statements hold:(i)
ifci+2(1)ci+1(2)bi(3)+ci+2(1)bi+1(2)+bi+2(1)≠0, then |z3n+i+1|→∞, as n→∞;(j)
ifci+2(1)ci+1(2)bi(3)+ci+2(1)bi+1(2)+bi+2(1)=0, then
(5.17)z3n+i+1=zi-2(uiwi+1)n+1.Proof.
From (3.23), by using conditions (5.16) and (5.2), we have that
(5.18)z3n+i+1=zi-2∏j=0nui+(ci+2(1)ci+1(2)bi(3)+ci+2(1)bi+1(2)+bi+2(1))jwi+1+(ci+2(1)ci+3(3)bi+1(2)+ci+3(3)bi+2(1)+bi+3(3))j,
for each i∈{0,1,2} and every n∈ℕ0, from which the results in this theorem follows similar to Theorem 5.1.Theorem 5.5.
Consider system (3.2). Let the sequences bn(j), cn(j), n∈ℕ0, j∈{1,2,3} be periodic with period three and for some i∈{0,1,2} the following condition holds:
(5.19)ci(1)ci+2(2)ci+1(3)≠1.
Then if x3n+i+1≠0 for every n∈ℕ0, the following statements hold: (a)
if|ci(1)ci+2(2)ci+1(3)|>1,
(5.20)si∶=vi-ci+2(2)ci+1(3)bi(1)+ci+2(2)bi+1(3)+bi+2(2)1-ci(1)ci+2(2)ci+1(3)=0,ti+1∶=ui+1-ci+3(1)ci+2(2)bi+1(3)+ci+3(1)bi+2(2)+bi+3(1)1-ci+3(1)ci+2(2)ci+1(3)≠0
then x3n+i+1→0, as n→∞;(b)
if|ci(1)ci+2(2)ci+1(3)|>1, si≠0, and ti+1=0, then |x3n+i+1|→∞, as n→∞;(c)
ifci+3(1)ci+2(2)bi+1(3)+ci+3(1)bi+2(2)+bi+3(1)≠0, si=ti+1=0, then
(5.21)x3n+i+1=xi-2(ci+2(2)ci+1(3)bi(1)+ci+2(2)bi+1(3)+bi+2(2)ci+3(1)ci+2(2)bi+1(3)+ci+3(1)bi+2(2)+bi+3(1))n+1;(d)
ifci+3(1)ci+2(2)bi+1(3)+ci+3(1)bi+2(2)+bi+3(1)=ci+2(2)ci+1(3)bi(1)+ci+2(2)bi+1(3)+bi+2(2)=0 and (5.20) holds, then
(5.22)x3n+i+1=xi-2(siti+1)n+1;(e)
if|ci(1)ci+2(2)ci+1(3)|>1 and 0<|si|<|ti+1|, then x3n+i+1→0 as n→∞;(f)
if|ci(1)ci+2(2)ci+1(3)|>1 and |si|>|ti+1|>0, then |x3n+i+1|→+∞ as n→∞;(g)
if|ci(1)ci+2(2)ci+1(3)|>1 and si=ti+1≠0, then x3n+i+1 converges, as n→∞;(h)
if|ci(1)ci+2(2)ci+1(3)|>1 and si=-ti+1≠0, then x6n+i+1 and x6n+i+4 converge, as n→∞;(i)
if|ci(1)ci+2(2)ci+1(3)|<1 and 0<|ai|<|bi|, where
(5.23)ai∶=ci+2(2)ci+1(3)bi(1)+ci+2(2)bi+1(3)+bi+2(2)1-ci(1)ci+2(2)ci+1(3),bi=ci+3(1)ci+2(2)bi+1(3)+ci+3(1)bi+2(2)+bi+3(1)1-ci+3(1)ci+2(2)ci+1(3),
then x3n+i+1→0 as n→∞;(j)
if|ci(1)ci+2(2)ci+1(3)|<1 and |ai|>|bi|>0, then |x3n+i+1|→+∞ as n→∞;(k)
if|ci(1)ci+2(2)ci+1(3)|<1 and ai=bi≠0, then x3n+i+1 converges, as n→∞;(l)
if|ci(1)ci+2(2)ci+1(3)|<1 and ai=-bi≠0, then x6n+i+1 and x6n+i+4 converge, as n→∞.Proof.
From (3.21) and by using conditions (5.19) and (5.2), we have that
(5.24)x3n+i+1=xi-2∏j=0nsi(ci(1)ci+2(2)ci+1(3))j+((ci+2(2)ci+1(3)bi(1)+ci+2(2)bi+1(3)+bi+2(2))/(1-ci(1)ci+2(2)ci+1(3)))ti+1(ci+3(1)ci+2(2)ci+1(3))j+((ci+3(1)ci+2(2)bi+1(3)+ci+3(1)bi+2(2)+bi+3(1))/(1-ci+3(1)ci+2(2)ci+1(3))),
for every n∈ℕ0 and for each i∈{0,1,2}. Using (5.24) the statements in (a)–(d) easily follows.
(e)–(h) Letλ∶=ci(1)ci+2(2)ci+1(3). Then
(5.25)si(ci(1)ci+2(2)ci+1(3))j+((ci+2(2)ci+1(3)bi(1)+ci+2(2)bi+1(3)+bi+2(2))/(1-ci(1)ci+2(2)ci+1(3)))ti+1(ci+3(1)ci+2(2)ci+1(3))j+((ci+3(1)ci+2(2)bi+1(3)+ci+3(1)bi+2(2)+bi+3(1))/(1-ci+3(1)ci+2(2)ci+1(3)))=siλj+aiti+1λj+bi,
where ai, bi are defined by (5.23). We have
(5.26)siλj+aiti+1λj+bi=siti+1(1+aisiλj)(1+biti+1λj)-1=siti+1(1+(aisi-biti+1)1λj+O(1λ2j)).
From this and since |λ|>1, the results in these four cases easily follow.
(i)–(l) We have(5.27)siλj+aiti+1λj+bi=aibi(1+siaiλj)(1+ti+1biλj)-1=aibi(1+(siai-ti+1bi)λj+O(λ2j)).
From this and since |λ|<1, the results in these four cases easily follow.Theorem 5.6.
Consider system (3.2). Let the sequences bn(j), cn(j), n∈ℕ0, j∈{1,2,3} be periodic with period three and for some i∈{0,1,2} the following condition holds:
(5.28)ci+1(1)ci(2)ci+2(3)≠1.
Then if y3n+i+1≠0 for every n∈ℕ0, the following statements hold: (a)
if|ci+1(1)ci(2)ci+2(3)|>1,
(5.29)ŝi∶=wi-ci+1(1)ci+2(3)bi(2)+ci+2(3)bi+1(1)+bi+2(3)1-ci+1(1)ci(2)ci+2(3)=0,t̂i+1=vi+1-ci+3(2)ci+2(3)bi+1(1)+ci+3(2)bi+2(3)+bi+3(2)1-ci+1(1)ci+3(2)ci+2(3)≠0,
then y3n+i+1→0, as n→∞;(b)
if|ci+1(1)ci(2)ci+2(3)|>1, ŝi≠0, and t̂i+1=0, then |y3n+i+1|→∞, as n→∞;(c)
ifci+3(2)ci+2(3)bi+1(1)+ci+3(2)bi+2(3)+bi+3(2)≠0, ŝi=t̂i+1=0, then
(5.30)y3n+i+1=yi-2(ci+1(1)ci+2(3)bi(2)+ci+2(3)bi+1(1)+bi+2(3)ci+3(2)ci+2(3)bi+1(1)+ci+3(2)bi+2(3)+bi+3(2))n+1;(d)
ifci+1(1)ci+2(3)bi(2)+ci+2(3)bi+1(1)+bi+2(3)=ci+3(2)ci+2(3)bi+1(1)+ci+3(2)bi+2(3)+bi+3(2)=0 and (5.29) holds, then
(5.31)y3n+i+1=yi-2(ŝit̂i+1)n+1;(e)
if|ci+1(1)ci(2)ci+2(3)|>1 and 0<|ŝi|<|t̂i+1|, then y3n+i+1→0 as n→∞;(f)
if|ci+1(1)ci(2)ci+2(3)|>1 and |ŝi|>|t̂i+1|>0, then |y3n+i+1|→+∞ as n→∞;(g)
if|ci+1(1)ci(2)ci+2(3)|>1 and ŝi=t̂i+1≠0, then y3n+i+1 converges, as n→∞;(h)
if|ci+1(1)ci(2)ci+2(3)|>1 and ŝi=-t̂i+1≠0, then y6n+i+1 and y6n+i+4 converge, as n→∞;(i)
if|ci+1(1)ci(2)ci+2(3)|<1 and 0<|âi|<|b̂i|, where
(5.32)âi∶=ci+1(1)ci+2(3)bi(2)+ci+2(3)bi+1(1)+bi+2(3)1-ci+1(1)ci(2)ci+2(3),b̂i∶=ci+3(2)ci+2(3)bi+1(1)+ci+3(2)bi+2(3)+bi+3(2)1-ci+1(1)ci+3(2)ci+2(3),
then y3n+i+1→0 as n→∞;(j)
if|ci+1(1)ci(2)ci+2(3)|<1 and |âi|>|b̂i|>0, then |y3n+i+1|→+∞ as n→∞;(k)
if|ci+1(1)ci(2)ci+2(3)|<1 and âi=b̂i≠0, then y3n+i+1 converges, as n→∞;(l)
if|ci+1(1)ci(2)ci+2(3)|<1 and âi=-b̂i≠0, then y6n+i+1 and y6n+i+4 converge, as n→∞.Proof.
From (3.22) and by using condition (5.28), we have that
(5.33)y3n+i+1=yi-2∏j=0nŝi(ci+1(1)ci(2)ci+2(3))j+((ci+1(1)ci+2(3)bi(2)+ci+2(3)bi+1(1)+bi+2(3))/(1-ci+1(1)ci(2)ci+2(3)))t̂i+1(ci+1(1)ci+3(2)ci+2(3))j+((ci+3(2)ci+2(3)bi+1(1)+ci+3(2)bi+2(3)+bi+3(2))/(1-ci+1(1)ci+3(2)ci+2(3))),
for every n∈ℕ0 and for each i∈{0,1,2}, from which the statements in (a)–(l) follows similar as the corresponding ones in Theorem 5.5.Theorem 5.7.
Consider system (3.2). Let the sequences bn(j), cn(j), n∈ℕ0, j∈{1,2,3} be periodic with period three and for some i∈{0,1,2} the following condition holds
(5.34)ci+2(1)ci+1(2)ci(3)≠1.
Then if z3n+i+1≠0 for every n∈ℕ0, the following statements hold: (a)
if|ci+2(1)ci+1(2)ci(3)|>1,
(5.35)si*∶=ui-ci+2(1)ci+1(2)bi(3)+ci+2(1)bi+1(2)+bi+2(1)1-ci+2(1)ci+1(2)ci(3)=0,ti+1*∶=wi+1-ci+2(1)ci+3(3)bi+1(2)+ci+3(3)bi+2(1)+bi+3(3)1-ci+2(1)ci+1(2)ci+3(3)≠0
then z3n+i+1→0, as n→∞;(b)
if|ci+2(1)ci+1(2)ci(3)|>1, si*≠0,ti+1*=0, then |z3n+i+1|→∞, as n→∞;(c)
ifci+2(1)ci+3(3)bi+1(2)+ci+3(3)bi+2(1)+bi+3(3)≠0, si*=ti+1*=0, then
(5.36)z3n+i+1=zi-2(ci+2(1)ci+1(2)bi(3)+ci+2(1)bi+1(2)+bi+2(1)ci+2(1)ci+3(3)bi+1(2)+ci+3(3)bi+2(1)+bi+3(3))n+1;(d)
ifci+2(1)ci+1(2)bi(3)+ci+2(1)bi+1(2)+bi+2(1)=ci+2(1)ci+3(3)bi+1(2)+ci+3(3)bi+2(1)+bi+3(3)=0 and (5.35) holds, then
(5.37)z3n+i+1=zi-2(si*ti+1*)n+1;(e)
if|ci+2(1)ci+1(2)ci(3)|>1 and 0<|si*|<|ti+1*|, then z3n+i+1→0 as n→∞;(f)
if|ci+2(1)ci+1(2)ci(3)|>1 and |si*|>|ti+1*|>0, then |z3n+i+1|→+∞ as n→∞;(g)
if|ci+2(1)ci+1(2)ci(3)|>1 and si*=ti+1*≠0, then z3n+i+1 converges, as n→∞;(h)
if|ci+2(1)ci+1(2)ci(3)|>1 and si*=-ti+1*≠0, then z6n+i+1 and z6n+i+4 converge, as n→∞;(i)
if|ci+2(1)ci+1(2)ci(3)|<1 and 0<|ai*|<|bi*|, where
(5.38)ai*∶=ci+2(1)ci+1(2)bi(3)+ci+2(1)bi+1(2)+bi+2(1)1-ci+2(1)ci+1(2)ci(3),bi*∶=ci+2(1)ci+3(3)bi+1(2)+ci+3(3)bi+2(1)+bi+3(3)1-ci+2(1)ci+1(2)ci+3(3),
then z3n+i+1→0 as n→∞;(j)
if|ci+2(1)ci+1(2)ci(3)|<1 and |ai*|>|bi*|>0, then |z3n+i+1|→+∞ as n→∞;(k)
if|ci+2(1)ci+1(2)ci(3)|<1 and ai*=bi*≠0, then z3n+i+1 converges, as n→∞;(l)
if|ci+2(1)ci+1(2)ci(3)|<1 and ai*=-bi*≠0, then z6n+i+1 and z6n+i+4 converge, as n→∞.Proof.
From (3.23) and by using condition (5.34), we have that
(5.39)z3n+i+1=zi-2∏j=0nsi*(ci+2(1)ci+1(2)ci(3))j+((ci+2(1)ci+1(2)bi(3)+ci+2(1)bi+1(2)+bi+2(1))/(1-ci+2(1)ci+1(2)ci(3)))ti+1*(ci+2(1)ci+1(2)ci+3(3))j+((ci+2(1)ci+3(3)bi+1(2)+ci+3(3)bi+2(1)+bi+3(3))/(1-ci+2(1)ci+1(2)ci+3(3))),
for every n∈ℕ0 and for each i∈{0,1,2}, from which the statements in (a)–(l) follow similar as the corresponding ones in Theorem 5.5.



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*Source: 508523-2012-05-29.xml*