Analysis of positivity results for discrete fractional operators by means of exponential kernels

Pshtiwan Othman Mohammed; Donal O'Regan; Aram Bahroz Brzo; Khadijah M. Abualnaja; Dumitru Baleanu; Pshtiwan Othman Mohammed; Donal O'Regan; Aram Bahroz Brzo; Khadijah M. Abualnaja; Dumitru Baleanu

doi:10.3934/math.2022865

AIMS Mathematics

2022, Volume 7, Issue 9: 15812-15823. doi: 10.3934/math.2022865

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Research article Special Issues

Analysis of positivity results for discrete fractional operators by means of exponential kernels

1.
Department of Mathematics, College of Education, University of Sulaimani, Sulaimani 46001, Iraq
2.
School of Mathematical and Statistical Sciences, National University of Ireland, Galway, Ireland
3.
Department of Physics, College of Education, University of Sulaimani, Sulaimani 46001, Kurdistan Region, Iraq
4.
Department of Mathematics and Statistics, College of Science, Taif University, P.O. Box 11099, Taif 21944, Saudi Arabia
5.
Department of Mathematics, Cankaya University, 06530 Balgat, Ankara, Turkey
6.
Institute of Space Sciences, R76900 Magurele-Bucharest, Romania
7.
Department of Mathematics, King Abdul Aziz University, Jeddah 21577, Saudi Arabia

Received: 02 March 2022 Revised: 22 June 2022 Accepted: 23 June 2022 Published: 27 June 2022
MSC : 26A48, 26A51, 33B10, 39A12, 39B62

In this study, we consider positivity and other related concepts such as $\alpha-$ convexity and $\alpha-$ monotonicity for discrete fractional operators with exponential kernel. Namely, we consider discrete $\Delta$ fractional operators in the Caputo sense and we apply efficient initial conditions to obtain our conclusions. Note positivity results are an important factor for obtaining the composite of double discrete fractional operators having different orders.

Keywords:

Citation: Pshtiwan Othman Mohammed, Donal O'Regan, Aram Bahroz Brzo, Khadijah M. Abualnaja, Dumitru Baleanu. Analysis of positivity results for discrete fractional operators by means of exponential kernels[J]. AIMS Mathematics, 2022, 7(9): 15812-15823. doi: 10.3934/math.2022865

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Abstract

1. Introduction

This paper is devoted to study the expressions forms of the solutions and periodic nature of the following third-order rational systems of difference equations

$\begin{equation*} x_{n+1} = \dfrac{y_{n-1}z_{n}}{z_{n}\pm x_{n-2}}, \;y_{n+1} = \dfrac{z_{n-1}x_{n} }{x_{n}\pm y_{n-2}}, \ z_{n+1} = \dfrac{x_{n-1}y_{n}}{y_{n}\pm z_{n-2}}, \end{equation*}$

with initial conditions are non-zero real numbers.

In the recent years, there has been great concern in studying the systems of difference equations. One of the most important reasons for this is a exigency for some mechanization which can be used in discussing equations emerge in mathematical models characterizing real life situations in economic, genetics, probability theory, psychology, population biology and so on.

Difference equations display naturally as discrete peer and as numerical solutions of differential equations having more applications in ecology, biology, physics, economy, and so forth. For all that the difference equations are quite simple in expressions, it is frequently difficult to realize completely the dynamics of their solutions see ^{[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19]} and the related references therein.

There are some papers dealed with the difference equations systems, for example, The periodic nature of the solutions of the nonlinear difference equations system

$\begin{equation*} A_{n+1} = \dfrac{1}{C_{n}}, \;B_{n+1} = \dfrac{B_{n}}{A_{n-1}B_{n-1}}, \;C_{n+1} = \dfrac{1}{A_{n-1}}, \end{equation*}$

has been studied by Cinar in ^[7].

Almatrafi ^[3] determined the analytical solutions of the following systems of rational recursive equations

$\begin{equation*} x_{n+1} = \dfrac{x_{n-1}y_{n-3}}{y_{n-1}(\pm 1\pm x_{n-1}y_{n-3})}, \;y_{n+1} = \dfrac{y_{n-1}x_{n-3}}{x_{n-1}(\pm 1\pm y_{n-1}x_{n-3})}. \end{equation*}$

In ^[20], Khaliq and Shoaib studied the local and global asymptotic behavior of non-negative equilibrium points of a three-dimensional system of two order rational difference equations

$\begin{equation*} x_{n+1} = \dfrac{x_{n-1}}{\varepsilon +x_{n-1}y_{n-1}z_{n-1}}, \;y_{n+1} = \dfrac{ y_{n-1}}{\zeta +x_{n-1}y_{n-1}z_{n-1}}, \ z_{n+1} = \dfrac{z_{n-1}}{\eta +x_{n-1}y_{n-1}z_{n-1}}. \end{equation*}$

In ^[9], Elabbasy et al. obtained the form of the solutions of some cases of the following system of difference equations

$\begin{eqnarray*} x_{n+1} & = &\dfrac{a_{1}+a_{2}y_{n}}{a_{3}z_{n}+a_{4}x_{n-1}z_{n}}, \ y_{n+1} = \dfrac{b_{1}z_{n-1}+b_{2}z_{n}}{b_{3}x_{n}y_{n}+b_{4}x_{n}y_{n-1}}, \\ z_{n+1} & = &\dfrac{c_{1}z_{n-1}+c_{2}z_{n}}{ c_{3}x_{n-1}y_{n-1}+c_{4}x_{n-1}y_{n}+c_{5}x_{n}y_{n}}. \end{eqnarray*}$

In ^[12], Elsayed et al. have got the solutions of the systems of rational higher order difference equations

$\begin{equation*} A_{n+1} = \dfrac{1}{A_{n-p}B_{n-p}}, \;\;B_{n+1} = \dfrac{A_{n-p}B_{n-p}}{ A_{n-q}B_{n-q}}, \end{equation*}$

and

$\begin{equation*} A_{n+1} = \dfrac{1}{A_{n-p}B_{n-p}C_{n-p}}, \;\;B_{n+1} = \dfrac{ A_{n-p}B_{n-p}C_{n-p}}{A_{n-q}B_{n-q}C_{n-q}}, \;C_{n+1} = \dfrac{ A_{n-q}B_{n-q}C_{n-q}}{A_{n-r}B_{n-r}C_{n-r}}. \end{equation*}$

Kurbanli ^[25,26] investigated the behavior of the solutions of the following systems

$\begin{eqnarray*} A_{n+1} & = &\dfrac{A_{n-1}}{A_{n-1}B_{n}-1}, \;B_{n+1} = \dfrac{B_{n-1}}{ B_{n-1}A_{n}-1}, \ \ C_{n+1} = \dfrac{1}{C_{n}B_{n}}, \\ A_{n+1} & = &\dfrac{A_{n-1}}{A_{n-1}B_{n}-1}, \;B_{n+1} = \dfrac{B_{n-1}}{ B_{n-1}A_{n}-1}, \ \ C_{n+1} = \dfrac{C_{n-1}}{C_{n-1}B_{n}-1}. \end{eqnarray*}$

In ^[32], Yalçınkaya has obtained the conditions for the global asymptotically stable of the system

$\begin{equation*} A_{n+1} = \dfrac{B_{n}A_{n-1}+a}{B_{n}+A_{n-1}}, \;B_{n+1} = \dfrac{A_{n}B_{n-1}+a }{A_{n}+B_{n-1}}. \end{equation*}$

Zhang et al. ^[39] investigated the persistence, boundedness and the global asymptotically stable of the solutions of the following system

$\begin{equation*} R_{n} = A+\dfrac{1}{Q_{n-p}}, \ \ \ \;Q_{n} = A+\dfrac{Q_{n-1}}{R_{n-r}Q_{n-s}}. \end{equation*}$

Similar to difference equations and systems were studied see ^{[21,22,23,24,27,28,29,30,31,32,33,34,35,36,37,38]}.

2. The system: $x_{n+1}=\dfrac{y_{n-1}z_{n}}{z_{n}+x_{n-2}}, \; y_{n+1}= \dfrac{z_{n-1}x_{n}}{x_{n}+y_{n-2}},$ $z_{n+1}=\dfrac{x_{n-1}y_{n}}{ y_{n}+z_{n-2}}$

In this section, we obtain the expressions form of the solutions of the following three dimension system of difference equations

$\begin{equation} x_{n+1} = \dfrac{y_{n-1}z_{n}}{z_{n}+x_{n-2}}, \;y_{n+1} = \dfrac{z_{n-1}x_{n}}{ x_{n}+y_{n-2}}, \ z_{n+1} = \dfrac{x_{n-1}y_{n}}{y_{n}+z_{n-2}}, \end{equation}$

(1)

where $n\in \mathbb{N}_{0}$ and the initial conditions are non-zero real numbers.

Theorem 1. We assume that $\{x_{n}, y_{n}, z_{n}\}$ are solutions of system (1).Then

$\begin{eqnarray*} x_{6n-2} & = &\frac{ak^{3n}}{\prod \limits_{i = 0}^{n-1}(a+(6i)k)(a+(6i+2)k)(a+(6i+4)k)}, \\ x_{6n-1} & = &\frac{bf^{3n}}{\prod \limits_{i = 0}^{n-1}(g+(6i+1)f)(g+(6i+3)f)(g+(6i+5)f)}, \\ x_{6n} & = &\frac{c^{3n+1}}{\prod \limits_{i = 0}^{n-1}(d+(6i+2)c)(d+(6i+4)c)(d+(6i+6)c)}, \\ x_{6n+1} & = &\frac{ek^{3n+1}}{(a+k)\prod \limits_{i = 0}^{n-1}(a+(6i+3)k)(a+(6i+5)k)(a+(6i+7)k)}, \end{eqnarray*}$

$\begin{eqnarray*} x_{6n+2} & = &\frac{f^{3n+2}}{(g+2f)\prod \limits_{i = 0}^{n-1}(g+(6i+4)f)(g+(6i+6)f)(g+(6i+8)f)}, \\ x_{6n+3} & = &\frac{hc^{3n+2}}{(d+c)(d+3c)\prod \limits_{i = 0}^{n-1}(d+(6i+5)c)(d+(6i+7)c)(d+(6i+9)c)}, \end{eqnarray*}$

$\begin{eqnarray*} y_{6n-2} & = &\frac{dc^{3n}}{\prod \limits_{i = 0}^{n-1}(d+(6i)c)(d+(6i+2)c)(d+(6i+4)c)}, \\ y_{6n-1} & = &\frac{ek^{3n}}{\prod \limits_{i = 0}^{n-1}(a+(6i+1)k)(a+(6i+3)k)(a+(6i+5)k)}, \\ y_{6n} & = &\frac{f^{3n+1}}{\prod \limits_{i = 0}^{n-1}(g+(6i+2)f)(g+(6i+4)f)(g+(6i+6)f)}, \\ y_{6n+1} & = &\frac{hc^{3n+1}}{(d+c)\prod \limits_{i = 0}^{n-1}(d+(6i+3)c)(d+(6i+5)c)(d+(6i+7)c)}, \\ y_{6n+2} & = &\frac{k^{3n+2}}{(a+2k)\prod \limits_{i = 0}^{n-1}(a+(6i+4)k)(a+(6i+6)k)(a+(6i+8)k)}, \\ y_{6n+3} & = &\frac{bf^{3n+2}}{(g+f)(g+3f)\prod \limits_{i = 0}^{n-1}(g+(6i+5)f)(g+(6i+7)f)(g+(6i+9)f)}, \end{eqnarray*}$

and

$\begin{eqnarray*} z_{6n-2} & = &\frac{gf^{3n}}{\prod \limits_{i = 0}^{n-1}(g+(6i)f)(g+(6i+2)f)(g+(6i+4)f)}, \\ z_{6n-1} & = &\frac{hc^{3n}}{\prod \limits_{i = 0}^{n-1}(d+(6i+1)c)(d+(6i+3)c)(d+(6i+5)c)}, \\ z_{6n} & = &\frac{k^{3n+1}}{\prod \limits_{i = 0}^{n-1}(a+(6i+2)k)(a+(6i+4)k)(a+(6i+6)k)}, \\ z_{6n+1} & = &\frac{bf^{3n+1}}{(g+f)\prod \limits_{i = 0}^{n-1}(g+(6i+3)f)(g+(6i+5)f)(g+(6i+7)f)}, \end{eqnarray*}$

$\begin{eqnarray*} z_{6n+2} & = &\frac{c^{3n+2}}{(d+2c)\prod \limits_{i = 0}^{n-1}(d+(6i+4)c)(d+(6i+6)c)(d+(6i+8)c)}, \\ z_{6n+3} & = &\frac{ek^{3n+2}}{(a+k)(a+3k)\prod \limits_{i = 0}^{n-1}(a+(6i+5)k)(a+(6i+7)k)(a+(6i+9)k)}, \end{eqnarray*}$

where $x_{-2} = a, \ x_{-1} = b,$ $x_{0} = c,$ $y_{-2} = d,$ $y_{-1} = e,$ $y_{0} = f,$ $z_{-2} = g,$ $z_{-1} = h$ $and$ $z_{0} = k.$

Proof. For $n = 0$ the result holds. Now assume that $n > 1$ and that our assumption holds for $n-1$ , that is,

$\begin{eqnarray*} x_{6n-8} & = &\frac{ak^{3n-3}}{\prod \limits_{i = 0}^{n-2}(a+(6i)k)(a+(6i+2)k)(a+(6i+4)k)}, \\ x_{6n-7} & = &\frac{bf^{3n-3}}{\prod \limits_{i = 0}^{n-2}(g+(6i+1)f)(g+(6i+3)f)(g+(6i+5)f)}, \\ x_{6n-6} & = &\frac{c^{3n-2}}{\prod \limits_{i = 0}^{n-2}(d+(6i+2)c)(d+(6i+4)c)(d+(6i+6)c)}, \\ x_{6n-5} & = &\frac{ek^{3n-2}}{(a+k)\prod \limits_{i = 0}^{n-2}(a+(6i+3)k)(a+(6i+5)k)(a+(6i+7)k)}, \\ x_{6n-4} & = &\frac{f^{3n-1}}{(g+2f)\prod \limits_{i = 0}^{n-2}(g+(6i+4)f)(g+(6i+6)f)(g+(6i+8)f)}, \\ x_{6n-3} & = &\frac{hc^{3n-1}}{(d+c)(d+3c)\prod \limits_{i = 0}^{n-2}(d+(6i+5)c)(d+(6i+7)c)(d+(6i+9)c)}, \end{eqnarray*}$

$\begin{eqnarray*} y_{6n-8} & = &\frac{dc^{3n-3}}{\prod \limits_{i = 0}^{n-2}(d+(6i)c)(d+(6i+2)c)(d+(6i+4)c)}, \\ y_{6n-7} & = &\frac{ek^{3n-3}}{\prod \limits_{i = 0}^{n-2}(a+(6i+1)k)(a+(6i+3)k)(a+(6i+5)k)}, \\ y_{6n-6} & = &\frac{f^{3n-2}}{\prod \limits_{i = 0}^{n-2}(g+(6i+2)f)(g+(6i+4)f)(g+(6i+6)f)}, \end{eqnarray*}$

$\begin{eqnarray*} y_{6n-5} & = &\frac{hc^{3n-2}}{(d+c)\prod \limits_{i = 0}^{n-2}(d+(6i+3)c)(d+(6i+5)c)(d+(6i+7)c)}, \\ y_{6n-4} & = &\frac{k^{3n-1}}{(a+2k)\prod \limits_{i = 0}^{n-2}(a+(6i+4)k)(a+(6i+6)k)(a+(6i+8)k)}, \\ y_{6n-3} & = &\frac{bf^{3n-1}}{(g+f)(g+3f)\prod \limits_{i = 0}^{n-2}(g+(6i+5)f)(g+(6i+7)f)(g+(6i+9)f)}, \end{eqnarray*}$

and

$\begin{eqnarray*} z_{6n-8} & = &\frac{gf^{3n-3}}{\prod \limits_{i = 0}^{n-2}(g+(6i)f)(g+(6i+2)f)(g+(6i+4)f)}, \\ z_{6n-7} & = &\frac{hc^{3n-3}}{\prod \limits_{i = 0}^{n-2}(d+(6i+1)c)(d+(6i+3)c)(d+(6i+5)c)}, \\ z_{6n-6} & = &\frac{k^{3n-2}}{\prod \limits_{i = 0}^{n-2}(a+(6i+2)k)(a+(6i+4)k)(a+(6i+6)k)}, \\ z_{6n-5} & = &\frac{bf^{3n-2}}{(g+f)\prod \limits_{i = 0}^{n-2}(g+(6i+3)f)(g+(6i+5)f)(g+(6i+7)f)}, \\ z_{6n-4} & = &\frac{c^{3n-1}}{(d+2c)\prod \limits_{i = 0}^{n-2}(d+(6i+4)c)(d+(6i+6)c)(d+(6i+8)c)}, \\ z_{6n-3} & = &\frac{ek^{3n-1}}{(a+k)(a+3k)\prod \limits_{i = 0}^{n-2}(a+(6i+5)k)(a+(6i+7)k)(a+(6i+9)k)}. \end{eqnarray*}$

It follows from Eq (1) that

$\begin{eqnarray*} x_{6n-2} & = &\dfrac{y_{6n-4}z_{6n-3}}{z_{6n-3}+x_{6n-5}} \\ & = &\tfrac{\left( \frac{k^{3n-1}}{(a+2k)\prod \limits_{i = 0}^{n-2}(a+(6i+4)k)(a+(6i+6)k)(a+(6i+8)k)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \right) \left( \frac{ ek^{3n-1}}{(a+k)(a+3k)\prod \limits_{i = 0}^{n-2}(a+(6i+5)k)(a+(6i+7)k)(a+(6i+9)k)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \right) }{\left( \frac{ ek^{3n-1}}{(a+k)(a+3k)\prod \limits_{i = 0}^{n-2}(a+(6i+5)k)(a+(6i+7)k)(a+(6i+9)k)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \right) +\left( \frac{ ek^{3n-2}}{(a+k)\prod\limits_{i = 0}^{n-2}(a+(6i+3)k)(a+(6i+5)k)(a+(6i+7)k)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \right) } \\ & = &\dfrac{\left( \frac{k^{3n}}{(a+2k)\prod \limits_{i = 0}^{n-2}(a+(6i+4)k)(a+(6i+6)k)(a+(6i+8)k)}\right) }{ (a+3k)\prod\limits_{i = 0}^{n-2}(a+(6i+9)k)\left[ \left( \frac{k}{ (a+3k)\prod\limits_{i = 0}^{n-2}(a+(6i+9)k)}\right) +\left( \frac{1}{ \prod\limits_{i = 0}^{n-2}(a+(6i+3)k)}\right) \right] } \\ & = &\dfrac{\left( \frac{k^{3n}}{(a+2k)\prod \limits_{i = 0}^{n-2}(a+(6i+4)k)(a+(6i+6)k)(a+(6i+8)k)}\right) }{\left[ k+\left( \frac{(a+3k)\prod\limits_{i = 0}^{n-2}(a+(6i+9)k)}{ \prod\limits_{i = 0}^{n-2}(a+(6i+3)k)}\right) \right] } \\ & = &\dfrac{\left( \frac{k^{3n}}{(a+2k)\prod \limits_{i = 0}^{n-2}(a+(6i+4)k)(a+(6i+6)k)(a+(6i+8)k)}\right) }{\left[ k+\left( a+(6n-3)k\right) \right] } \\ & = &\frac{ak^{3n}}{a(a+2k)\left( a+(6n-2)k\right) \prod\limits_{i = 0}^{n-2}(a+(6i+4)k)(a+(6i+6)k)(a+(6i+8)k)}. \end{eqnarray*}$

Then we see that

$\begin{equation*} x_{6n-2} = \frac{k^{3n}}{\prod \limits_{i = 0}^{n-1}(a+(6i)k)(a+(6i+2)k)(a+(6i+4)k)}. \end{equation*}$

Also, we see from Eq (1) that

$\begin{eqnarray*} y_{6n-2} & = &\dfrac{z_{6n-4}x_{6n-3}}{x_{6n-3}+y_{6n-5}} \\ & = &\tfrac{\left( \frac{c^{3n-1}}{(d+2c)\prod \limits_{i = 0}^{n-2}(d+(6i+4)c)(d+(6i+6)c)(d+(6i+8)c)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \right) \left( \frac{ hc^{3n-1}}{(d+c)(d+3c)\prod \limits_{i = 0}^{n-2}(d+(6i+5)c)(d+(6i+7)c)(d+(6i+9)c)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \right) }{\left( \frac{ hc^{3n-1}}{(d+c)(d+3c)\prod \limits_{i = 0}^{n-2}(d+(6i+5)c)(d+(6i+7)c)(d+(6i+9)c)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \right) +\left( \frac{ hc^{3n-2}}{(d+c)\prod\limits_{i = 0}^{n-2}(d+(6i+3)c)(d+(6i+5)c)(d+(6i+7)c)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \right) } \\ & = &\dfrac{\left( \frac{c^{3n}}{(d+2c)\prod \limits_{i = 0}^{n-2}(d+(6i+4)c)(d+(6i+6)c)(d+(6i+8)c)}\right) }{ (d+3c)\prod\limits_{i = 0}^{n-2}(d+(6i+9)c)\left[ \left( \frac{c}{ (d+3c)\prod\limits_{i = 0}^{n-2}(d+(6i+9)c)}\right) +\left( \frac{1}{ \prod\limits_{i = 0}^{n-2}(d+(6i+3)c)}\right) \right] } \\ & = &\dfrac{\left( \frac{c^{3n}}{(d+2c)\prod \limits_{i = 0}^{n-2}(d+(6i+4)c)(d+(6i+6)c)(d+(6i+8)c)}\right) }{\left[ c+d+(6n-3)c\right] } \\ & = &\dfrac{c^{3n}}{\left[ d+(6n-2)c\right] (d+2c)\prod \limits_{i = 0}^{n-2}(d+(6i+4)c)(d+(6i+6)c)(d+(6i+8)c)}. \end{eqnarray*}$

Then

$\begin{equation*} y_{6n-2} = \dfrac{dc^{3n}}{\prod \limits_{i = 0}^{n-1}(d+(6i)c)(d+(6i+2)c)(d+(6i+4)c)}. \end{equation*}$

Finally from Eq (1), we see that

$\begin{eqnarray*} z_{6n-2} & = &\dfrac{x_{6n-4}y_{6n-3}}{y_{6n-3}+z_{6n-5}} \\ & = &\tfrac{\left( \frac{f^{3n-1}}{(g+2f)\prod \limits_{i = 0}^{n-2}(g+(6i+4)f)(g+(6i+6)f)(g+(6i+8)f)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \right) \left( \frac{ bf^{3n-1}}{(g+f)(g+3f)\prod \limits_{i = 0}^{n-2}(g+(6i+5)f)(g+(6i+7)f)(g+(6i+9)f)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \right) }{\left( \frac{ bf^{3n-1}}{(g+f)(g+3f)\prod \limits_{i = 0}^{n-2}(g+(6i+5)f)(g+(6i+7)f)(g+(6i+9)f)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \right) +\left( \frac{ bf^{3n-2}}{(g+f)\prod\limits_{i = 0}^{n-2}(g+(6i+3)f)(g+(6i+5)f)(g+(6i+7)f)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \right) } \\ & = &\dfrac{\left( \frac{f^{3n}}{(g+2f)\prod \limits_{i = 0}^{n-2}(g+(6i+4)f)(g+(6i+6)f)(g+(6i+8)f)}\right) }{ (g+3f)\prod\limits_{i = 0}^{n-2}(g+(6i+9)f)\left[ \left( \frac{f}{ (g+3f)\prod\limits_{i = 0}^{n-2}(g+(6i+9)f)}\right) +\left( \frac{1}{ \prod\limits_{i = 0}^{n-2}(g+(6i+3)f)}\right) \right] } \\ & = &\dfrac{\left( \frac{f^{3n}}{(g+2f)\prod \limits_{i = 0}^{n-2}(g+(6i+4)f)(g+(6i+6)f)(g+(6i+8)f)}\right) }{\left[ f+\left( \frac{(g+3f)\prod\limits_{i = 0}^{n-2}(g+(6i+9)f)}{ \prod\limits_{i = 0}^{n-2}(g+(6i+3)f)}\right) \right] } \\ & = &\dfrac{\left( \frac{f^{3n}}{(g+2f)\prod \limits_{i = 0}^{n-2}(g+(6i+4)f)(g+(6i+6)f)(g+(6i+8)f)}\right) }{\left[ f+(g+(6n-3)f)\right] } \\ & = &\dfrac{f^{3n}}{(g+(6n-2)f)(g+2f)\prod \limits_{i = 0}^{n-2}(g+(6i+4)f)(g+(6i+6)f)(g+(6i+8)f)}. \end{eqnarray*}$

Thus

$\begin{equation*} z_{3n-2} = \dfrac{gf^{3n}}{\prod \limits_{i = 0}^{n-1}(g+(6i)f)(g+(6i+2)f)(g+(6i+4)f)}. \end{equation*}$

By similar way, one can show the other relations. This completes the proof.

Lemma 1. Let $\{x_{n}, y_{n}, z_{n}\}$ be a positive solution of system (1), then all solution of (1) is bounded and approaching to zero.

Proof. It follows from Eq (1) that

$\begin{eqnarray*} x_{n+1} & = &\dfrac{y_{n-1}z_{n}}{z_{n}+x_{n-2}}\leq y_{n-1}, \ \ \ \ \ y_{n+1} = \dfrac{z_{n-1}x_{n}}{x_{n}+y_{n-2}}\leq z_{n-1}, \\ z_{n+1} & = &\dfrac{x_{n-1}y_{n}}{y_{n}+z_{n-2}}\leq x_{n-1}, \end{eqnarray*}$

we see that

$\begin{eqnarray*} x_{n+4} &\leq &y_{n+2}, \ \ \ \ \ y_{n+2}\leq z_{n}, \ \ z_{n}\leq x_{n-2}, \ \ \ \Rightarrow \ x_{n+4} < x_{n-2}, \\ y_{n+4} &\leq &z_{n+2}, \ \ \ z_{n+2}\leq x_{n}, \ \ \ x_{n}\leq y_{n-2}, \ \ \ \ \ \Rightarrow \ y_{n+4} < y_{n-2}, \\ z_{n+4} &\leq &x_{n+2}, \ \ \ x_{n+2}\leq y_{n}, \ \ \ y_{n}\leq z_{n-2}, \ \ \ \ \ \Rightarrow \ z_{n+4} < z_{n-2}, \end{eqnarray*}$

Then all subsequences of $\{x_{n}, y_{n}, z_{n}\}$ (i.e., for $\{x_{n}\}$ are $\{x_{6n-2}\}$ , $\{x_{6n-1}\}$ , $\{x_{6n}\},$ $\{x_{6n+1}\}$ , $\{x_{6n+2}\}$ , $\{x_{6n+3}\}\$ are decreasing and at that time are bounded from above by $K, L$ and $M$ since $K = \max \{x_{-2}, x_{-1}, x_{0}, x_{1}, x_{2}, x_{3}\},$ $L = \max \{y_{-2}, y_{-1}, y_{0}, y_{1}, y_{2}, y_{3}\}$ and $M = \max \{z_{-2}, z_{-1}, z_{0}, z_{1}, z_{2}, z_{3}\}$ .

Example 1. We assume an interesting numerical example for the system (1) with $x_{-2} = -.22, \; x_{-1} = -.4, \ x_{0} = .12, \; y_{-2} = -.62,$ $y_{-1} = 4, \ y_{0} = .3, \; z_{-2} = .4, \; z_{-1} = .53\ and\; z_{0} = -2$ . (See Figure 1).

Figure 1. This figure shows the behavior of the solutions of the system (1) with the initial conditions

$x_{-2} = -.22, \; x_{-1} = -.4, \ x_{0} = .12, \; y_{-2} = -.62,$

$y_{-1} = 4, \ y_{0} = .3, \; z_{-2} = .4, \; z_{-1} = .53\ and\; z_{0} = -2.$ (We see from this figure that all solutions converges to zero).

DownLoad: Full-Size Img PowerPoint

3. The system: $x_{n+1}=\dfrac{y_{n-1}z_{n}}{z_{n}+x_{n-2}}, \; y_{n+1}= \dfrac{z_{n-1}x_{n}}{x_{n}+y_{n-2}},$ $z_{n+1}=\dfrac{x_{n-1}y_{n}}{ y_{n}-z_{n-2}}$

In this section, we get the solution's form of the following system of difference equations

$\begin{equation} x_{n+1} = \dfrac{y_{n-1}z_{n}}{z_{n}+x_{n-2}}, \;y_{n+1} = \dfrac{z_{n-1}x_{n}}{ x_{n}+y_{n-2}}, \ z_{n+1} = \dfrac{x_{n-1}y_{n}}{y_{n}-z_{n-2}}, \end{equation}$

(2)

where $n\in \mathbb{N}_{0}$ and the initial values are non-zero real numbers with $x_{-2}\neq \pm z_{0}, \neq -2z_{0},$ $z_{-2}\neq y_{0}, \neq 2y_{0}, \neq 3y_{0}$ and $y_{-2}\neq 2x_{0}, \neq \pm x_{0}.$

Theorem 2. Assume that $\{x_{n}, y_{n}, z_{n}\}$ are solutions of (2). Then for $n = 0, 1, 2, ...,$

$\begin{eqnarray*} x_{6n-2} & = &\frac{(-1)^{n}k^{3n}}{a^{2n-1}(a+2k)^{n}}, \ x_{6n-1} = \frac{ (-1)^{n}bf^{3n}}{(f-g)^{2n}(3f-g)^{n}}, \ x_{6n} = \frac{(-1)^{n}c^{3n+1}}{ d^{2n}(2c-d)^{n}}, \\ x_{6n+1} & = &\frac{ek^{3n+1}}{(a-k)^{n}(a+k)^{2n+1}}, \ x_{6n+2} = \frac{ (-1)^{n}f^{3n+2}}{g^{n}(2f-g)^{2n+1}}, \ x_{6n+3} = \frac{(-1)^{n}hc^{3n+2}}{ (c-d)^{2n+1}(c+d)^{n+1}}, \end{eqnarray*}$

$\begin{eqnarray*} y_{6n-2} & = &\frac{(-1)^{n}c^{3n}}{d^{2n-1}(2c-d)^{n}}, \ y_{6n-1} = \frac{ ek^{3n}}{(a-k)^{n}(a+k)^{2n}}, \ y_{6n} = \frac{(-1)^{n}f^{3n+1}}{ g^{n}(2f-g)^{2n}}, \\ y_{6n+1} & = &\frac{(-1)^{n}hc^{3n+1}}{(c-d)^{2n}(c+d)^{n+1}}, \ y_{6n+2} = \frac{ (-1)^{n}k^{3n+2}}{a^{2n}(a+2k)^{n+1}}, \ y_{6n+3} = \frac{(-1)^{n}bf^{3n+2}}{ (f-g)^{2n+1}(3f-g)^{n+1}}, \end{eqnarray*}$

and

$\begin{eqnarray*} z_{6n-2} & = &\frac{(-1)^{n}f^{3n}}{g^{n-1}(2f-g)^{2n}}, \ z_{6n-1} = \frac{ (-1)^{n}hc^{3n}}{(c-d)^{2n}(c+d)^{n}}, \ z_{6n} = \frac{(-1)^{n}k^{3n+1}}{ a^{2n}(a+2k)^{n}}, \\ z_{6n+1} & = &\frac{(-1)^{n}bf^{3n+1}}{(f-g)^{2n+1}(3f-g)^{n}}, \ z_{6n+2} = \frac{(-1)^{n+1}c^{3n+2}}{d^{2n+1}(2c-d)^{n}}, \ z_{6n+3} = \frac{-ek^{3n+2}}{ (a-k)^{n}(a+k)^{2n+2}}, \end{eqnarray*}$

where $x_{-2} = a, \ x_{-1} = b,$ $x_{0} = c,$ $y_{-2} = d,$ $y_{-1} = e,$ $y_{0} = f,$ $z_{-2} = g,$ $z_{-1} = h$ $and$ $z_{0} = k.$

Proof. The result is true for $n = 0$ . Now suppose that $n > 0$ and that our claim verified for $n-1$ . That is,

$\begin{eqnarray*} x_{6n-8} & = &\frac{(-1)^{n-1}k^{3n-3}}{a^{2n-3}(a+2k)^{n-1}}, \ x_{6n-7} = \frac{ (-1)^{n-1}bf^{3n-3}}{(f-g)^{2n-2}(3f-g)^{n-1}}, \ x_{6n-6} = \frac{ (-1)^{n-1}c^{3n-2}}{d^{2n-2}(2c-d)^{n-1}}, \\ x_{6n-5} & = &\frac{ek^{3n-2}}{(a-k)^{n-1}(a+k)^{2n-1}}, \ x_{6n-4} = \frac{ (-1)^{n-1}f^{3n-1}}{g^{n-1}(2f-g)^{2n-1}}, \ x_{6n-3} = \frac{ (-1)^{n-1}hc^{3n-1}}{(c-d)^{2n-1}(c+d)^{n}}, \end{eqnarray*}$

$\begin{eqnarray*} y_{6n-8} & = &\frac{(-1)^{n-1}c^{3n-3}}{d^{2n-3}(2c-d)^{n-1}}, \ y_{6n-7} = \frac{ ek^{3n-3}}{(a-k)^{n-1}(a+k)^{2n-2}}, \ y_{6n-6} = \frac{(-1)^{n-1}f^{3n-2}}{ g^{n-1}(2f-g)^{2n-2}}, \\ y_{6n-5} & = &\frac{(-1)^{n-1}hc^{3n-2}}{(c-d)^{2n-2}(c+d)^{n}}, \ y_{6n-4} = \frac{(-1)^{n-1}k^{3n-1}}{a^{2n-2}(a+2k)^{n}}, \ y_{6n-3} = \frac{ (-1)^{n-1}bf^{3n-1}}{(f-g)^{2n-1}(3f-g)^{n}}, \end{eqnarray*}$

and

$\begin{eqnarray*} z_{6n-8} & = &\frac{(-1)^{n-1}f^{3n-3}}{g^{n-2}(2f-g)^{2n-2}}, \ z_{6n-7} = \frac{ (-1)^{n-1}hc^{3n-3}}{(c-d)^{2n-2}(c+d)^{n-1}}, \ z_{6n-6} = \frac{ (-1)^{n-1}k^{3n-2}}{a^{2n-2}(a+2k)^{n-1}}, \\ z_{6n-5} & = &\frac{(-1)^{n-1}bf^{3n-2}}{(f-g)^{2n-1}(3f-g)^{n-1}}, \ z_{6n-4} = \frac{(-1)^{n}c^{3n-1}}{d^{2n-1}(2c-d)^{n-1}}, \ z_{6n-3} = \frac{-ek^{3n-1}}{ (a-k)^{n-1}(a+k)^{2n}}. \end{eqnarray*}$

Now from Eq (2), it follows that

$\begin{eqnarray*} x_{6n-2} & = &\dfrac{y_{6n-4}z_{6n-3}}{z_{6n-3}+x_{6n-5}} \\ & = &\dfrac{\left( \dfrac{(-1)^{n-1}k^{3n-1}}{a^{2n-2}(a+2k)^{n}}\right) \left( \dfrac{-ek^{3n-1}}{(a-k)^{n-1}(a+k)^{2n}}\right) }{\left( \dfrac{ -ek^{3n-1}}{(a-k)^{n-1}(a+k)^{2n}}\right) +\left( \dfrac{ek^{3n-2}}{ (a-k)^{n-1}(a+k)^{2n-1}}\right) } \\ & = &\dfrac{\left( \dfrac{(-1)^{n}k^{3n}}{a^{2n-2}(a+2k)^{n}}\right) }{\left( -k+a+k\right) } = \dfrac{(-1)^{n}k^{3n}}{a^{2n-1}(a+2k)^{n}}, \\ y_{6n-2} & = &\dfrac{z_{6n-4}x_{6n-3}}{x_{6n-3}+y_{6n-5}} = \dfrac{\left( \frac{ (-1)^{n}c^{3n-1}}{d^{2n-1}(2c-d)^{n-1}}\right) \left( \frac{ (-1)^{n-1}hc^{3n-1}}{(c-d)^{2n-1}(c+d)^{n}}\right) }{\left( \frac{ (-1)^{n-1}hc^{3n-1}}{(c-d)^{2n-1}(c+d)^{n}}\right) +\left( \frac{ (-1)^{n-1}hc^{3n-2}}{(c-d)^{2n-2}(c+d)^{n}}\right) } \\ & = &\dfrac{\left( \frac{(-1)^{n}c^{3n}}{d^{2n-1}(2c-d)^{n-1}}\right) }{c+c-d} = \frac{(-1)^{n}c^{3n}}{d^{2n-1}(2c-d)^{n}}, \\ z_{6n-2} & = &\dfrac{x_{6n-4}y_{6n-3}}{y_{6n-3}-z_{6n-5}} = \dfrac{\left( \frac{ (-1)^{n-1}f^{3n-1}}{g^{n-1}(2f-g)^{2n-1}}\right) \left( \frac{ (-1)^{n-1}bf^{3n-1}}{(f-g)^{2n-1}(3f-g)^{n}}\right) }{\left( \frac{ (-1)^{n-1}bf^{3n-1}}{(f-g)^{2n-1}(3f-g)^{n}}\right) -\left( \frac{ (-1)^{n-1}bf^{3n-2}}{(f-g)^{2n-1}(3f-g)^{n-1}}\right) } \\ & = &\dfrac{\left( \frac{(-1)^{n-1}f^{3n}}{g^{n-1}(2f-g)^{2n-1}}\right) }{ \left( f-3f+g\right) } = \frac{(-1)^{n}f^{3n}}{g^{n-1}(2f-g)^{2n}}. \end{eqnarray*}$

Also, we see from Eq (2) that

$\begin{eqnarray*} x_{6n-1} & = &\dfrac{y_{6n-3}z_{6n-2}}{z_{6n-2}+x_{6n-4}} \\ & = &\dfrac{\left( \frac{(-1)^{n-1}bf^{3n-1}}{(f-g)^{2n-1}(3f-g)^{n}}\right) \left( \frac{(-1)^{n}f^{3n}}{g^{n-1}(2f-g)^{2n}}\right) }{\left( \frac{ (-1)^{n}f^{3n}}{g^{n-1}(2f-g)^{2n}}\right) +\left( \frac{(-1)^{n-1}f^{3n-1}}{ g^{n-1}(2f-g)^{2n-1}}\right) } \\ & = &\dfrac{\left( \frac{(-1)^{n}bf^{3n}}{(f-g)^{2n-1}(3f-g)^{n}}\right) }{ \left( -f+2f-g\right) } = \frac{(-1)^{n}bf^{3n}}{(f-g)^{2n}(3f-g)^{n}}, \\ y_{6n-1} & = &\dfrac{z_{6n-3}x_{6n-2}}{x_{6n-2}+y_{6n-4}} = \dfrac{\left( \frac{ -ek^{3n-1}}{(a-k)^{n-1}(a+k)^{2n}}\right) \left( \frac{(-1)^{n}k^{3n}}{ a^{2n-1}(a+2k)^{n}}\right) }{\left( \frac{(-1)^{n}k^{3n}}{a^{2n-1}(a+2k)^{n}} \right) +\left( \frac{(-1)^{n-1}k^{3n-1}}{a^{2n-2}(a+2k)^{n}}\right) } \\ & = &\dfrac{\left( \frac{ek^{3n}}{(a-k)^{n-1}(a+k)^{2n}}\right) }{-k+a} = \frac{ ek^{3n}}{(a-k)^{n}(a+k)^{2n}}, \\ z_{6n-1} & = &\dfrac{x_{6n-3}y_{6n-2}}{y_{6n-2}-z_{6n-4}} = \dfrac{\left( \frac{ (-1)^{n-1}hc^{3n-1}}{(c-d)^{2n-1}(c+d)^{n}}\right) \left( \frac{ (-1)^{n}c^{3n}}{d^{2n-1}(2c-d)^{n}}\right) }{\left( \frac{(-1)^{n}c^{3n}}{ d^{2n-1}(2c-d)^{n}}\right) -\left( \frac{(-1)^{n}c^{3n-1}}{ d^{2n-1}(2c-d)^{n-1}}\right) } \\ & = &\dfrac{\left( \frac{(-1)^{n-1}hc^{3n}}{(c-d)^{2n-1}(c+d)^{n}}\right) }{ c-\left( 2c-d\right) } = \frac{(-1)^{n}hc^{3n}}{(c-d)^{2n}(c+d)^{n}}. \end{eqnarray*}$

Also, we can prove the other relations.

Example 2. See below for system (2) with the initial conditions $x_{-2} = 11, \; x_{-1} = 5, \ x_{0} = 13, \; y_{-2} = 6,$ $y_{-1} = 7, \ y_{0} = 3, \; z_{-2} = 14,$ $z_{-1} = 9$ $and\; z_{0} = 2.$

Figure 2. This figure shows the behavior of solutions of the systems of rational recursive sequence

$x_{n+1} = \dfrac{y_{n-1}z_{n}}{z_{n}+x_{n-2}}, \; y_{n+1} = \dfrac{z_{n-1}x_{n}}{x_{n}+y_{n-2}}, \ z_{n+1} = \dfrac{x_{n-1}y_{n} }{y_{n}-z_{n-2}},$ when we take the initial conditions:

$x_{-2} = 11, \; x_{-1} = 5, \ x_{0} = 13, \; y_{-2} = 6,$

$y_{-1} = 7, \ y_{0} = 3, \; z_{-2} = 14,$

$z_{-1} = 9$

$and\; z_{0} = 2.$ (See the figure we can conclude that all the solutions unboundedness solutions).

DownLoad: Full-Size Img PowerPoint

4. The system: $x_{n+1}=\dfrac{y_{n-1}z_{n}}{z_{n}+x_{n-2}}, \; y_{n+1}= \dfrac{z_{n-1}x_{n}}{x_{n}-y_{n-2}},$ $z_{n+1}=\dfrac{x_{n-1}y_{n}}{ y_{n}+z_{n-2}}$

Here, we obtain the form of solutions of the system

$\begin{equation} x_{n+1} = \dfrac{y_{n-1}z_{n}}{z_{n}+x_{n-2}}, \;y_{n+1} = \dfrac{z_{n-1}x_{n}}{ x_{n}-y_{n-2}}, \ z_{n+1} = \dfrac{x_{n-1}y_{n}}{y_{n}+z_{n-2}}, \end{equation}$

(3)

where $n\in \mathbb{N}_{0}$ and the initial values are non-zero real numbers with $x_{-2}\neq \pm z_{0}, \neq 2z_{0},$ $z_{-2}\neq \pm y_{0}, \neq -2y_{0}$ and $y_{-2}\neq x_{0}, \neq 2x_{0}, \neq 3x_{0}.$

Theorem 3. If $\ \{x_{n}, y_{n}, z_{n}\}$ are solutions of system (3) where $x_{-2} = a, \ x_{-1} = b,$ $x_{0} = c,$ $y_{-2} = d,$ $y_{-1} = e,$ $y_{0} = f,$ $z_{-2} = g,$ $z_{-1} = h$ $and$ $z_{0} = k.$ Then for $n = 0, 1, 2, ...,$

$\begin{eqnarray*} x_{6n-2} & = &\frac{k^{3n}}{a^{2n-1}(a-2k)^{n}}, \ x_{6n-1} = \frac{ (-1)^{n}bf^{3n}}{(f-g)^{n}(f+g)^{2n}}, \ x_{6n} = \frac{(-1)^{n}c^{3n+1}}{ d^{n}(d-2c)^{2n}}, \\ x_{6n+1} & = &\frac{(-1)^{n}ek^{3n+1}}{(a-k)^{2n}(a+k)^{n+1}}, \ x_{6n+2} = \frac{ (-1)^{n}f^{3n+2}}{g^{2n}(2f+g)^{n+1}}, \ x_{6n+3} = \frac{(-1)^{n}hc^{3n+2}}{ (c-d)^{2n+1}(3c-d)^{n+1}}, \end{eqnarray*}$

$\begin{eqnarray*} y_{6n-2} & = &\frac{(-1)^{n}c^{3n}}{d^{n-1}(d-2c)^{2n}}, \ y_{6n-1} = \frac{ (-1)^{n}ek^{3n}}{(a-k)^{2n}(a+k)^{n}}, \ y_{6n} = \frac{(-1)^{n}f^{3n+1}}{ g^{2n}(2f+g)^{n}}, \\ y_{6n+1} & = &\frac{(-1)^{n}hc^{3n+1}}{(c-d)^{2n+1}(3c-d)^{n}}, \ y_{6n+2} = \frac{-k^{3n+2}}{a^{2n+1}(a-2k)^{n}}, \ y_{6n+3} = \frac{(-1)^{n}bf^{3n+2}}{ (f-g)^{n}(f+g)^{2n+2}}, \end{eqnarray*}$

and

$\begin{eqnarray*} z_{6n-2} & = &\frac{(-1)^{n}f^{3n}}{g^{2n-1}(2f+g)^{n}}, \ z_{6n-1} = \frac{ (-1)^{n}hc^{3n}}{(c-d)^{2n}(3c-d)^{n}}, \ z_{6n} = \frac{k^{3n+1}}{ a^{2n}(a-2k)^{n}}, \\ z_{6n+1} & = &\frac{(-1)^{n}bf^{3n+1}}{(f-g)^{n}(f+g)^{2n+1}}, \ z_{6n+2} = \frac{ (-1)^{n}c^{3n+2}}{d^{n}(2c-d)^{2n+1}}, \ z_{6n+3} = \frac{(-1)^{n+1}ek^{3n+2}}{ (a-k)^{2n+1}(a+k)^{n+1}}. \end{eqnarray*}$

Proof. As the proof of Theorem 2 and so will be left to the reader.

Example 3. We put the initials $x_{-2} = 8, \; x_{-1} = 15, \ x_{0} = 13, \; y_{-2} = 6, \; y_{-1} = 7,$ $y_{0} = 3, \; z_{-2} = 14, \; z_{-1} = 19$ $and\; z_{0} = 2,$ for the system (3), see Figure 3.

Figure 3. This figure shows the unstable of the solutions of the difference equations system (3) with the initial values

$x_{-2} = 8, \; x_{-1} = 15, \ x_{0} = 13, \; y_{-2} = 6, \; y_{-1} = 7,$

$y_{0} = 3, \; z_{-2} = 14, \; z_{-1} = 19$

$and\; z_{0} = 2$ .

DownLoad: Full-Size Img PowerPoint

The following systems can be treated similarly.

5. The system: $x_{n+1}=\dfrac{y_{n-1}z_{n}}{z_{n}-x_{n-2}}, \; y_{n+1}= \dfrac{z_{n-1}x_{n}}{x_{n}+y_{n-2}},$ $z_{n+1}=\dfrac{x_{n-1}y_{n}}{ y_{n}+z_{n-2}}$

In this section, we deal with the solutions of the following system

$\begin{equation} x_{n+1} = \dfrac{y_{n-1}z_{n}}{z_{n}-x_{n-2}}, \;y_{n+1} = \dfrac{z_{n-1}x_{n}}{ x_{n}+y_{n-2}}, \ z_{n+1} = \dfrac{x_{n-1}y_{n}}{y_{n}+z_{n-2}}, \end{equation}$

(4)

where $n\in \mathbb{N}_{0}$ and the initial values are non-zero real with $x_{-2}\neq z_{0}, \neq 2z_{0}, \neq 3z_{0},$ $z_{-2}\neq \pm y_{0}, \neq 2y_{0}$ and $y_{-2}\neq \pm x_{0}, \neq -2x_{0}.$

Theorem 4. The solutions of system (4) are given by

$\begin{eqnarray*} x_{6n-2} & = &\frac{(-1)^{n}k^{3n}}{a^{n-1}(a-2k)^{2n}}, \ x_{6n-1} = \frac{ (-1)^{n}bf^{3n}}{(f-g)^{2n}(f+g)^{n}}, \ x_{6n} = \frac{(-1)^{n}c^{3n+1}}{ d^{2n}(d+2c)^{n}}, \\ x_{6n+1} & = &\frac{-ek^{3n+1}}{(a-k)^{2n+1}(a-3k)^{n}}, \ x_{6n+2} = \frac{ (-1)^{n+1}f^{3n+2}}{g^{2n+1}(2f-g)^{n}}, \ x_{6n+3} = \frac{(-1)^{n+1}hc^{3n+2} }{(c-d)^{n}(c+d)^{2n+2}}, \end{eqnarray*}$

$\begin{eqnarray*} y_{6n-2} & = &\frac{(-1)^{n}c^{3n}}{d^{2n-1}(d+2c)^{n}}, \ y_{6n-1} = \frac{ ek^{3n}}{(a-k)^{2n}(a-3k)^{n}}, \ y_{6n} = \frac{(-1)^{n}f^{3n+1}}{ g^{2n}(2f-g)^{n}}, \\ y_{6n+1} & = &\frac{(-1)^{n}hc^{3n+1}}{(c+d)^{2n+1}(c-d)^{n}}, \ y_{6n+2} = \frac{ -k^{3n+2}}{a^{n}(a-2k)^{2n+1}}, \ y_{6n+3} = \frac{(-1)^{n}bf^{3n+2}}{ (f-g)^{2n+1}(f+g)^{n+1}}, \end{eqnarray*}$

and

$\begin{eqnarray*} z_{6n-2} & = &\frac{(-1)^{n}f^{3n}}{g^{2n-1}(2f-g)^{n}}, \ z_{6n-1} = \frac{ (-1)^{n}hc^{3n}}{(c+d)^{2n}(c-d)^{n}}, \ z_{6n} = \frac{(-1)^{n}k^{3n+1}}{ a^{n}(a-2k)^{2n}}, \\ z_{6n+1} & = &\frac{(-1)^{n}bf^{3n+1}}{(f-g)^{2n}(f+g)^{n+1}}, \ z_{6n+2} = \frac{ (-1)^{n}c^{3n+2}}{d^{2n}(2c+d)^{n+1}}, \ z_{6n+3} = \frac{ek^{3n+2}}{ (a-k)^{2n+1}(a-3k)^{n+1}}, \end{eqnarray*}$

where $x_{-2} = a, \ x_{-1} = b,$ $x_{0} = c,$ $y_{-2} = d,$ $y_{-1} = e,$ $y_{0} = f,$ $z_{-2} = g,$ $z_{-1} = h$ $and$ $z_{0} = k.$

Example 4. shows the behavior of the solution of system (4) with $x_{-2} = 18, \; x_{-1} = -15, \ x_{0} = 3, \; y_{-2} = 6,$ $y_{-1} = .7, \ y_{0} = -3,$ $z_{-2} = 4, \; z_{-1} = -9\ and\; z_{0} = 5$ .

Figure 4. This figure shows the behavior of the system

$x_{n+1} = \dfrac{ y_{n-1}z_{n}}{z_{n}-x_{n-2}}, \; y_{n+1} = \dfrac{z_{n-1}x_{n}}{x_{n}+y_{n-2}}, \ z_{n+1} = \dfrac{x_{n-1}y_{n}}{y_{n}+z_{n-2}}$ with the initial conditions:-

$x_{-2} = 18, \; x_{-1} = -15, \ x_{0} = 3,$

$y_{-2} = 6,$

$y_{-1} = .7,$

$y_{0} = -3, \ z_{-2} = 4, \; z_{-1} = -9\ and\; z_{0} = 5.-0.6, \ x_{-1} = 0.2, \ x_{0} = -5.$ (From the figure, we see that all solutions goes to zero).

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6. The system: $x_{n+1}=\dfrac{y_{n-1}z_{n}}{z_{n}-x_{n-2}}, \; y_{n+1}= \dfrac{z_{n-1}x_{n}}{x_{n}-y_{n-2}},$ $z_{n+1}=\dfrac{x_{n-1}y_{n}}{ y_{n}-z_{n-2}}$

In this section, we obtain the solutions of the difference system

$\begin{equation} x_{n+1} = \dfrac{y_{n-1}z_{n}}{z_{n}-x_{n-2}}, \;y_{n+1} = \dfrac{z_{n-1}x_{n}}{ x_{n}-y_{n-2}}, \ z_{n+1} = \dfrac{x_{n-1}y_{n}}{y_{n}-z_{n-2}}, \end{equation}$

(5)

where the initials are arbitrary non-zero real numbers with $x_{-2}\neq z_{0},$ $z_{-2}\neq y_{0}$ and $y_{-2}\neq x_{0}.$

Theorem 5. If $\ \{x_{n}, y_{n}, z_{n}\}$ are solutions of system (5) where $x_{-2} = a, \ x_{-1} = b,$ $x_{0} = c,$ $y_{-2} = d,$ $y_{-1} = e,$ $y_{0} = f,$ $z_{-2} = g,$ $z_{-1} = h$ $and$ $z_{0} = k$ . Then

$\begin{eqnarray*} x_{6n-2} & = &\frac{k^{3n}}{a^{3n-1}}, \ x_{6n-1} = \frac{bf^{3n}}{(f-g)^{3n}}, \ x_{6n} = \frac{c^{3n+1}}{d^{3n}}, \\ x_{6n+1} & = &\frac{ek^{3n+1}}{(k-a)^{3n+1}}, \ x_{6n+2} = \frac{f^{3n+2}}{ g^{3n+1}}, \ x_{6n+3} = \frac{hc^{3n+2}}{(c-d)^{3n+2}}, \end{eqnarray*}$

$\begin{eqnarray*} y_{6n-2} & = &\frac{c^{3n}}{d^{3n-1}}, \ y_{6n-1} = \frac{ek^{3n}}{(k-a)^{3n}}, \ y_{6n} = \frac{f^{3n+1}}{g^{3n}}, \\ y_{6n+1} & = &\frac{hc^{3n+1}}{(c-d)^{3n+1}}, \ y_{6n+2} = \frac{k^{3n+2}}{ a^{3n+1}}, \ y_{6n+3} = \frac{bf^{3n+2}}{(f-g)^{3n+2}}, \end{eqnarray*}$

and

$\begin{eqnarray*} z_{6n-2} & = &\frac{f^{3n}}{g^{3n-1}}, \ z_{6n-1} = \frac{hc^{3n}}{(c-d)^{3n}}, \ z_{6n} = \frac{k^{3n+1}}{a^{3n}}, \\ z_{6n+1} & = &\frac{bf^{3n+1}}{(f-g)^{3n+1}}, \ z_{6n+2} = \frac{c^{3n+2}}{ d^{3n+1}}, \ z_{6n+3} = \frac{ek^{3n+2}}{(k-a)^{3n+2}}. \end{eqnarray*}$

Example 5. shows the dynamics of the solution of system (5) with $x_{-2} = 18, \; x_{-1} = -15, \; x_{0} = 3, y_{-2} = 6, y_{-1} = .7,$ $y_{0} = -3, z_{-2} = 4, z_{-1} = -9\ and\; z_{0} = 5.$

Figure 5. This figure shows the behavior of the system of nonlinear difference equations (5) with the initial conditions considered as follows:-

$x_{-2} = 18, \; x_{-1} = -15, \ x_{0} = 3, \; y_{-2} = 6,$

$y_{-1} = .7, \ y_{0} = -3, \; z_{-2} = 4, \; z_{-1} = -9\ and\; z_{0} = 5$ .

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7. Conclusions

This paper discussed the expression's form and boundedness of some systems of rational third order difference equations. In Section 2, we studied the qualitative behavior of system $x_{n+1} = \dfrac{y_{n-1}z_{n}}{z_{n}+x_{n-2}}, \; y_{n+1} = \dfrac{z_{n-1}x_{n}}{x_{n}+y_{n-2}}, \ z_{n+1} = \dfrac{x_{n-1}y_{n} }{y_{n}+z_{n-2}},$ first we have got the form of the solutions of this system, studied the boundedness and gave numerical example and drew it by using Matlab. In Section 3, we have got the solution's of the system $x_{n+1} = \dfrac{y_{n-1}z_{n}}{z_{n}+x_{n-2}}, \; y_{n+1} = \dfrac{z_{n-1}x_{n}}{ x_{n}+y_{n-2}}, \ z_{n+1} = \dfrac{x_{n-1}y_{n}}{y_{n}-z_{n-2}},$ and take a numerical example. In Sections 4–6, we obtained the solution of the following systems respectively, $x_{n+1} = \dfrac{y_{n-1}z_{n}}{z_{n}+x_{n-2}}, \; y_{n+1} = \dfrac{z_{n-1}x_{n}}{ x_{n}-y_{n-2}}, \ z_{n+1} = \dfrac{x_{n-1}y_{n}}{y_{n}+z_{n-2}},$ $x_{n+1} = \dfrac{y_{n-1}z_{n}}{z_{n}-x_{n-2}}, \; y_{n+1} = \dfrac{z_{n-1}x_{n}}{ x_{n}+y_{n-2}}, \ z_{n+1} = \dfrac{x_{n-1}y_{n}}{y_{n}+z_{n-2}},$ and $x_{n+1} = \dfrac{y_{n-1}z_{n}}{z_{n}-x_{n-2}}, \; y_{n+1} = \dfrac{z_{n-1}x_{n}}{ x_{n}-y_{n-2}}, \ z_{n+1} = \dfrac{x_{n-1}y_{n}}{y_{n}-z_{n-2}}.$ Also, in each case we take a numerical example to illustrates the results.

Acknowledgements

This project was funded by the Deanship of Scientific Research (DSR) at King Abdulaziz University, Jeddah, under grant no. (G: 233–130–1441). The authors, therefore, acknowledge with thanks DSR for technical and financial support.

Conflict of interest

All authors declare no conflicts of interest in this paper.

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Analysis of positivity results for discrete fractional operators by means of exponential kernels

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Abstract

1. Introduction

2. The system: $x_{n+1}=\dfrac{y_{n-1}z_{n}}{z_{n}+x_{n-2}}, \; y_{n+1}= \dfrac{z_{n-1}x_{n}}{x_{n}+y_{n-2}},$ $z_{n+1}=\dfrac{x_{n-1}y_{n}}{ y_{n}+z_{n-2}}$

3. The system: $x_{n+1}=\dfrac{y_{n-1}z_{n}}{z_{n}+x_{n-2}}, \; y_{n+1}= \dfrac{z_{n-1}x_{n}}{x_{n}+y_{n-2}},$ $z_{n+1}=\dfrac{x_{n-1}y_{n}}{ y_{n}-z_{n-2}}$

4. The system: $x_{n+1}=\dfrac{y_{n-1}z_{n}}{z_{n}+x_{n-2}}, \; y_{n+1}= \dfrac{z_{n-1}x_{n}}{x_{n}-y_{n-2}},$ $z_{n+1}=\dfrac{x_{n-1}y_{n}}{ y_{n}+z_{n-2}}$

5. The system: $x_{n+1}=\dfrac{y_{n-1}z_{n}}{z_{n}-x_{n-2}}, \; y_{n+1}= \dfrac{z_{n-1}x_{n}}{x_{n}+y_{n-2}},$ $z_{n+1}=\dfrac{x_{n-1}y_{n}}{ y_{n}+z_{n-2}}$

6. The system: $x_{n+1}=\dfrac{y_{n-1}z_{n}}{z_{n}-x_{n-2}}, \; y_{n+1}= \dfrac{z_{n-1}x_{n}}{x_{n}-y_{n-2}},$ $z_{n+1}=\dfrac{x_{n-1}y_{n}}{ y_{n}-z_{n-2}}$

7. Conclusions

Acknowledgements

Conflict of interest

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AIMS Mathematics

Analysis of positivity results for discrete fractional operators by means of exponential kernels

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Abstract

1. Introduction

2. The system: xn+1=yn−1znzn+xn−2,yn+1=zn−1xnxn+yn−2, x_{n+1}=\dfrac{y_{n-1}z_{n}}{z_{n}+x_{n-2}}, \; y_{n+1}= \dfrac{z_{n-1}x_{n}}{x_{n}+y_{n-2}}, zn+1=xn−1ynyn+zn−2z_{n+1}=\dfrac{x_{n-1}y_{n}}{ y_{n}+z_{n-2}}

3. The system: xn+1=yn−1znzn+xn−2,yn+1=zn−1xnxn+yn−2, x_{n+1}=\dfrac{y_{n-1}z_{n}}{z_{n}+x_{n-2}}, \; y_{n+1}= \dfrac{z_{n-1}x_{n}}{x_{n}+y_{n-2}}, zn+1=xn−1ynyn−zn−2z_{n+1}=\dfrac{x_{n-1}y_{n}}{ y_{n}-z_{n-2}}

4. The system: xn+1=yn−1znzn+xn−2,yn+1=zn−1xnxn−yn−2, x_{n+1}=\dfrac{y_{n-1}z_{n}}{z_{n}+x_{n-2}}, \; y_{n+1}= \dfrac{z_{n-1}x_{n}}{x_{n}-y_{n-2}}, zn+1=xn−1ynyn+zn−2z_{n+1}=\dfrac{x_{n-1}y_{n}}{ y_{n}+z_{n-2}}

5. The system: xn+1=yn−1znzn−xn−2,yn+1=zn−1xnxn+yn−2, x_{n+1}=\dfrac{y_{n-1}z_{n}}{z_{n}-x_{n-2}}, \; y_{n+1}= \dfrac{z_{n-1}x_{n}}{x_{n}+y_{n-2}}, zn+1=xn−1ynyn+zn−2z_{n+1}=\dfrac{x_{n-1}y_{n}}{ y_{n}+z_{n-2}}

6. The system: xn+1=yn−1znzn−xn−2,yn+1=zn−1xnxn−yn−2, x_{n+1}=\dfrac{y_{n-1}z_{n}}{z_{n}-x_{n-2}}, \; y_{n+1}= \dfrac{z_{n-1}x_{n}}{x_{n}-y_{n-2}}, zn+1=xn−1ynyn−zn−2z_{n+1}=\dfrac{x_{n-1}y_{n}}{ y_{n}-z_{n-2}}

7. Conclusions

Acknowledgements

Conflict of interest

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2. The system: $x_{n+1}=\dfrac{y_{n-1}z_{n}}{z_{n}+x_{n-2}}, \; y_{n+1}= \dfrac{z_{n-1}x_{n}}{x_{n}+y_{n-2}},$ $z_{n+1}=\dfrac{x_{n-1}y_{n}}{ y_{n}+z_{n-2}}$

3. The system: $x_{n+1}=\dfrac{y_{n-1}z_{n}}{z_{n}+x_{n-2}}, \; y_{n+1}= \dfrac{z_{n-1}x_{n}}{x_{n}+y_{n-2}},$ $z_{n+1}=\dfrac{x_{n-1}y_{n}}{ y_{n}-z_{n-2}}$

4. The system: $x_{n+1}=\dfrac{y_{n-1}z_{n}}{z_{n}+x_{n-2}}, \; y_{n+1}= \dfrac{z_{n-1}x_{n}}{x_{n}-y_{n-2}},$ $z_{n+1}=\dfrac{x_{n-1}y_{n}}{ y_{n}+z_{n-2}}$

5. The system: $x_{n+1}=\dfrac{y_{n-1}z_{n}}{z_{n}-x_{n-2}}, \; y_{n+1}= \dfrac{z_{n-1}x_{n}}{x_{n}+y_{n-2}},$ $z_{n+1}=\dfrac{x_{n-1}y_{n}}{ y_{n}+z_{n-2}}$

6. The system: $x_{n+1}=\dfrac{y_{n-1}z_{n}}{z_{n}-x_{n-2}}, \; y_{n+1}= \dfrac{z_{n-1}x_{n}}{x_{n}-y_{n-2}},$ $z_{n+1}=\dfrac{x_{n-1}y_{n}}{ y_{n}-z_{n-2}}$