Mathematical analysis of tumor-free equilibrium in BCG treatment with effective IL-2 infusion for bladder cancer model

Irina Volinsky; Svetlana Bunimovich-Mendrazitsky; Irina Volinsky; Svetlana Bunimovich-Mendrazitsky

doi:10.3934/math.2022896

AIMS Mathematics

2022, Volume 7, Issue 9: 16388-16406. doi: 10.3934/math.2022896

Previous Article Next Article

Research article Special Issues

Mathematical analysis of tumor-free equilibrium in BCG treatment with effective IL-2 infusion for bladder cancer model

Department of Mathematics, Ariel University, Ariel 4076414, Israel

Received: 12 April 2022 Revised: 03 June 2022 Accepted: 19 June 2022 Published: 06 July 2022
MSC : 34K20, 34D20

We present a theoretical study of bladder cancer treatment with Bacillus Calmette-Guerin (BCG) and interleukin 2 (IL-2) using a system biology approach to translate the treatment process into a mathematical model. We investigated the influence of IL-2 on effector cell proliferation, presented as a distributed feedback control in integral form. The variables in the system of Ordinary Differential Equations (ODE) are the main participants in the immune response after BCG instillations: BCG, immune cells, tumor cells infected with BCG, and non-infected with BCG. IL-2 was involved in the tumor-immune response without adding a new equation. We use the idea of reducing the system of integro-differential equations (IDE) to a system of ODE and examine the local stability analysis of the tumor-free equilibrium state of the model. A significant result of the model analysis is the requirements for the IL-2 dose and duration, depending on the treatment regimen and tumor growth. We proved that the BCG+IL-2 treatment protocol is more effective in this model, using the spectral radius method. Moreover, we introduced a parameter for individual control of IL-2 in each injection using the Cauchy matrix for the IDE system, and we obtained conditions under which this system would be exponentially stable in a tumor-free equilibrium.

Keywords:

combined BCG + IL-2 treatment,
system of integro-differential equations,
exponential stability,
feedback control, Cauchy matrix

Citation: Irina Volinsky, Svetlana Bunimovich-Mendrazitsky. Mathematical analysis of tumor-free equilibrium in BCG treatment with effective IL-2 infusion for bladder cancer model[J]. AIMS Mathematics, 2022, 7(9): 16388-16406. doi: 10.3934/math.2022896

Related Papers:

[1]	Ahmed Ghezal, Mohamed Balegh, Imane Zemmouri . Solutions and local stability of the Jacobsthal system of difference equations. AIMS Mathematics, 2024, 9(2): 3576-3591. doi: 10.3934/math.2024175
[2]	Hashem Althagafi, Ahmed Ghezal . Solving a system of nonlinear difference equations with bilinear dynamics. AIMS Mathematics, 2024, 9(12): 34067-34089. doi: 10.3934/math.20241624
[3]	M. T. Alharthi . Correction: On the solutions of some systems of rational difference equations. AIMS Mathematics, 2025, 10(2): 2277-2278. doi: 10.3934/math.2025105
[4]	M. T. Alharthi . On the solutions of some systems of rational difference equations. AIMS Mathematics, 2024, 9(11): 30320-30347. doi: 10.3934/math.20241463
[5]	Eunjung Lee, Dojin Kim . Stability analysis of the implicit finite difference schemes for nonlinear Schrödinger equation. AIMS Mathematics, 2022, 7(9): 16349-16365. doi: 10.3934/math.2022893
[6]	Shulan Kong, Chengbin Wang, Yawen Sun . A recursive filter for a class of two-dimensional nonlinear stochastic systems. AIMS Mathematics, 2025, 10(1): 1741-1756. doi: 10.3934/math.2025079
[7]	Yeyang Jiang, Zhihua Liao, Di Qiu . The existence of entire solutions of some systems of the Fermat type differential-difference equations. AIMS Mathematics, 2022, 7(10): 17685-17698. doi: 10.3934/math.2022974
[8]	Abdulghani R. Alharbi . Traveling-wave and numerical solutions to nonlinear evolution equations via modern computational techniques. AIMS Mathematics, 2024, 9(1): 1323-1345. doi: 10.3934/math.2024065
[9]	Ibraheem M. Alsulami, E. M. Elsayed . On a class of nonlinear rational systems of difference equations. AIMS Mathematics, 2023, 8(7): 15466-15485. doi: 10.3934/math.2023789
[10]	Nan Li, Jiachuan Geng, Lianzhong Yang . Some results on transcendental entire solutions to certain nonlinear differential-difference equations. AIMS Mathematics, 2021, 6(8): 8107-8126. doi: 10.3934/math.2021470

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).

DownLoad: Full-Size Img PowerPoint

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:-

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

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

DownLoad: Full-Size Img PowerPoint

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.

References

[1]	V. A. Kuznetsov, I. A. Makalkin, M. A. Taylor, A. S. Perelson, Nonlinear dynamics of immunogenic tumours: parameter estimation and global analysis, B. Math. Biol., 56 (1994), 295–321. https://doi.org/10.1016/S0092-8240(05)80260-5 doi: 10.1016/S0092-8240(05)80260-5
[2]	L. G. De Pillis, A. E. Radunskaya, C. L. Wiseman, A validated mathematical model of cell-mediated immune response to tumor growth, Cancer Res., 65 (2005), 7950–7958. https://doi.org/10.1158/0008-5472.CAN-05-0564 doi: 10.1158/0008-5472.CAN-05-0564
[3]	L. G. De Pillis, W. Gu, A. E. Radunskaya, Mixed immnotherapy and chemotherapy of tumors: modeling, applications and biological interpretations, J. Theor. Biol., 238 (2006), 841–862. https://doi.org/10.1016/j.jtbi.2005.06.037 doi: 10.1016/j.jtbi.2005.06.037
[4]	L. G. De Pillis, A. Eladdadi, A. Radunskaya, Modeling cancer-immune responses to therapy, J. Pharmacokinet Phar., 41 (2014), 461–478. https://doi.org/10.1007/s10928-014-9386-9 doi: 10.1007/s10928-014-9386-9
[5]	D. Kirschner, J. C. Panetta, Modeling immunotherapy of the tumor–immune interaction, J. Math. Biol., 37 (1998), 235–252. https://doi.org/10.1007/s002850050127 doi: 10.1007/s002850050127
[6]	F. Castiglione, B. Piccoli, Cancer immunotherapy, mathematical modeling and optimal control, J. Theor. Biol., 247 (2007), 723–732. https://doi.org/10.1016/j.jtbi.2007.04.003 doi: 10.1016/j.jtbi.2007.04.003
[7]	H. M. Byrne, Dissecting cancer through mathematics: from the cell to the animal model, Nat. Rev. Cancer, 10 (2010), 221–230. https://doi.org/10.1038/nrc2808 doi: 10.1038/nrc2808
[8]	N. Kronik, Y. Kogan, V. Vainstein, Z. Agur, Improving alloreactive CTL immunotherapy for malignant gliomas using a simulation model of their interactive dynamics, Cancer Immunol. Immun., 57 (2008), 425–439. https://doi.org/10.1007/s00262-007-0387-z doi: 10.1007/s00262-007-0387-z
[9]	A. B. Weiner, A. S. Desai, J. J. Meeks, Tumor Location May Predict Adverse Pathology and Survival Following Definitive Treatment for Bladder Cancer: A National Cohort Study, Eur. Urol. Oncol., 2 (2019), 304–310. https://doi.org/10.1016/j.euo.2018.08.018 doi: 10.1016/j.euo.2018.08.018
[10]	F. Bray, J. Ferlay, I. Soerjomataram, R. L. Siegel, L. Torre, A. Jemal, Global cancer statistics 2018: Globocan estimates of incidence and mortality worldwide for 36 cancers in 185 countries, CA: a cancer journal for clinicians, 68 (2018), 394–424. https://doi.org/10.3322/caac.21492 doi: 10.3322/caac.21492
[11]	M. P. Zeegers, F. E. Tan, E. Dorant, P. A. van Den Brandt, The impact of characteristics of cigarette smoking on urinary tract cancer risk: a meta-analysis of epidemiological studies, Cancer, 89 (2000), 630–639. https://doi.org/10.1002/1097-0142(20000801)89:3<630::AID-CNCR19>3.0.CO;2-Q doi: 10.1002/1097-0142(20000801)89:3<630::AID-CNCR19>3.0.CO;2-Q
[12]	A. Morales, D. Eidinger, A. W. Bruce, Intracavity Bacillus Calmette-Guerin in the treatment of superficial bladder tumors, J. Urol., 116 (1976), 180–182. https://doi.org/10.1016/S0022-5347(17)58737-6 doi: 10.1016/S0022-5347(17)58737-6
[13]	C. Pettenati, M. A. Ingersoll, Mechanisms of BCG immunotherapy and its outlook for bladder cancer, Nat. Rev. Urol., 15 (2018), 615–625. https://doi.org/10.1038/s41585-018-0055-4 doi: 10.1038/s41585-018-0055-4
[14]	C. F. Lee, S. Y. Chang, D. S. Hsieh, D. S. Yu, Immunotherapy for bladder cancer using recombinant bacillus Calmette-Guerin DNA vaccines and interleukin-12 DNA vaccine, J. Urol., 171 (2004), 1343–1347. https://doi.org/10.1097/01.ju.0000103924.93206.93 doi: 10.1097/01.ju.0000103924.93206.93
[15]	R. L. Steinberg, L. J. Thomas, S. L. Mott, M. A. O'Donnell, Multi-perspective tolerance evaluation of bacillus Calmette-Guerin with interferon in the treatment of non-muscle invasive bladder cancer, Bladder Cancer, 5 (2019), 39–49. https://doi.org/10.3233/BLC-180203 doi: 10.3233/BLC-180203
[16]	A. Shapiro, O. Gofrit, D. Pode, The treatment of superficial bladder tumor with IL-2 and BCG, J. Urol., 177 (2007), 81–82. https://doi.org/10.1016/S0022-5347(18)30509-3 doi: 10.1016/S0022-5347(18)30509-3
[17]	S. Bunimovich-Mendrazitsky, E. Shochat, L. Stone, Mathematical Model of BCG Immunotherapy in Superficial Bladder Cancer, B. Math. Biol., 69 (2007), 1847–1870. https://doi.org/10.1007/s11538-007-9195-z doi: 10.1007/s11538-007-9195-z
[18]	O. Nave, S. Hareli, M. Elbaz, I. H. Iluz, S. Bunimovich-Mendrazitsky, BCG and IL-2 model for bladder cancer treatment with fast and slow dynamics based on SPVF method—stability analysis, Math. Biosci. Eng., 16 (2019), 5346–5379. https://doi.org/10.3934/mbe.2019267 doi: 10.3934/mbe.2019267
[19]	T. Lazebnik, N. Aaroni, S. Bunimovich-Mendrazitsky, PDE based geometry model for BCG immunotherapy of bladder cancer, Biosystems, 200 (2021), 104319. https://doi.org/10.1016/j.biosystems.2020.104319 doi: 10.1016/j.biosystems.2020.104319
[20]	E. Guzev, S. Halachmi, S. Bunimovich-Mendrazitsky, Additional extension of the mathematical model for BCG immunotherapy of bladder cancer and its validation by auxiliary tool, Int. J. Nonlin. Sci. Num., 20 (2019), 675–689. https://doi.org/10.1515/ijnsns-2018-0181 doi: 10.1515/ijnsns-2018-0181
[21]	S. Bunimovich-Mendrazitsky, I. Chaskalovic, J. C. Gluckman, A mathematical model of combined bacillus Calmette-Guerin (BCG) and interleukin (IL)-2 immunotherapy of superficial bladder cancer, J. Theor. Biol., 277 (2011), 27–40. https://doi.org/10.1016/j.jtbi.2011.02.008 doi: 10.1016/j.jtbi.2011.02.008
[22]	S. Bunimovich-Mendrazitsky, S. Halachmi, N. Kronik, Improving Bacillus Calmette Guérin (BCG) immunotherapy for bladder cancer by adding interleukin-2 (IL-2): a mathematical model, Math. Med. Biol., 33 (2016), 159–188. https://doi.org/10.1093/imammb/dqv007 doi: 10.1093/imammb/dqv007
[23]	L. Shaikhet, S. Bunimovich-Mendrazitsky, Stability analysis of delayed immune response BCG infection in bladder cancer treatment model by stochastic perturbations, Comput. Math. Method. M., 2018 (2018), 9653873. https://doi.org/10.1155/2018/9653873 doi: 10.1155/2018/9653873
[24]	E. Fridman, L. Shaikhet, Simple LMIs for stability of stochastic systems with delay term given by Stieltjes integral or with stabilizing delay, Syst. Control Lett., 124 (2019), 83–91. https://doi.org/10.1016/j.sysconle.2018.12.007 doi: 10.1016/j.sysconle.2018.12.007
[25]	I. Volinsky, S. D. Lombardo, P. Cheredman, Stability Analysis and Cauchy Matrix of a Mathematical Model of Hepatitis B Virus with Control on Immune System near Neighborhood of Equilibrium Free Point, Symmetry, 13 (2021), 166. https://doi.org/10.3390/sym13020166 doi: 10.3390/sym13020166
[26]	R. P. Agarwal, L. Berezansky, E. Braverman, A. Domoshnitsky, Nonoscillation Theory of Functional Differential Equations with Applications, Springer: New York, NY, USA, 2012.
[27]	I. Volinsky, Stability Analysis of a Mathematical Model of Hepatitis B Virus with Unbounded Memory Control on the Immune System in the Neighborhood of the Equilibrium Free Point, Symmetry, 13 (2021), 1437. https://doi.org/10.3390/sym13081437 doi: 10.3390/sym13081437
[28]	R. F. M. Bevers, K. H. Kurth, D. H. J. Schamhart, Role of urothelial cells in BCG immunotherapy for superficial bladder cancer, Brit. J. Cancer, 91 (2004), 607–612. https://doi.org/10.1038/sj.bjc.6602026 doi: 10.1038/sj.bjc.6602026
[29]	L. M. Wein, J. T. Wu, D. H. Kirn, Validation and analysis of a mathematical model of a replication-competent oncolytic virus for cancer treatment: Implications for virus design and delivery, Cancer Res., 63 (2003), 1317–1324.
[30]	J. Wigginton, D. Kirschner, A model to predict cell-mediated immune regulatory mechanisms during human infection with Mycobacterium tuberculosis, J. Immunol., 166 (2001), 1951–1967. https://doi.org/10.4049/jimmunol.166.3.1951 doi: 10.4049/jimmunol.166.3.1951
[31]	F. Biemar, M. Foti, Global progress against cancer-challenges and opportunities, Cancer Biol. Med., 10 (2013), 183–186.

This article has been cited by:

1.	Khalil S. Al-Basyouni, Elsayed M. Elsayed, On Some Solvable Systems of Some Rational Difference Equations of Third Order, 2023, 11, 2227-7390, 1047, 10.3390/math11041047
2.	Ibraheem M. Alsulami, E. M. Elsayed, On a class of nonlinear rational systems of difference equations, 2023, 8, 2473-6988, 15466, 10.3934/math.2023789
3.	E.M. Elsayed, B.S. Alofi, The periodic nature and expression on solutions of some rational systems of difference equations, 2023, 74, 11100168, 269, 10.1016/j.aej.2023.05.026
4.	Hashem Althagafi, Dynamics of difference systems: a mathematical study with applications to neural systems, 2025, 10, 2473-6988, 2869, 10.3934/math.2025134

Reader Comments

Your name:*

Email:*
© 2022 the Author(s), licensee AIMS Press. This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0)

通讯作者: 陈斌, bchen63@163.com

1.
沈阳化工大学材料科学与工程学院沈阳 110142

AIMS Mathematics

1.8 3.4

Metrics

Article views(1842) PDF downloads(93) Cited by(1)

Preview PDF

Download XML

Export Citation

Article outline

Show full outline

Figures and Tables

Figures(1) / Tables(1)

AIMS Mathematics

Mathematical analysis of tumor-free equilibrium in BCG treatment with effective IL-2 infusion for bladder cancer model

Related Papers:

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

References

This article has been cited by:

Reader Comments

通讯作者: 陈斌, bchen63@163.com

Metrics

Figures and Tables

Other Articles By Authors

Catalog

AIMS Mathematics

Mathematical analysis of tumor-free equilibrium in BCG treatment with effective IL-2 infusion for bladder cancer model

Related Papers:

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

References

This article has been cited by:

Reader Comments

通讯作者: 陈斌, bchen63@163.com

Metrics

Figures and Tables

Other Articles By Authors

Related pages

Tools

Export File

Citation

Format

Content

Catalog

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