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Dynamical behavior of a system of three-dimensional nonlinear difference equations

Advances in Difference Equations20182018:223

https://doi.org/10.1186/s13662-018-1667-y

  • Received: 23 February 2018
  • Accepted: 29 May 2018
  • Published:

Abstract

In this paper, we study the boundedness, persistence, and periodicity of the positive solutions and the global asymptotic stability of the positive equilibrium points of the system of difference equations
$$x_{n+1}=A+\frac{x_{n-1}}{z_{n}},\qquad y_{n+1}=A+ \frac{y_{n-1}}{z _{n}},\qquad z_{n+1}=A+\frac{z_{n-1}}{y_{n}},\quad n=0,1,\ldots , $$
where \(A\in ( 0,\infty ) \) and the initial conditions \(x_{i}\), \(y_{i}\), \(z_{i}\in ( 0,\infty ) \), \(i=-1,0\).

Keywords

  • System of difference equations
  • Solution
  • Boundedness
  • Equilibrium point
  • Stability
  • Global asymptotic stability

MSC

  • 39A10
  • 39A30

1 Introduction

Difference equation or discrete dynamical system is a diverse field which impacts almost every branch of pure and applied mathematics. Lately, there has been great interest in investigating the behavior of solutions of a system of nonlinear difference equations and discussing the asymptotic stability of their equilibrium points. One of the reasons for this is a necessity for some techniques which can be used in investigating equations arising in mathematical models that describe real life situations in population biology, economics, probability theory, genetics, psychology, and so forth, see [3, 5, 8, 9]. Also, similar works in two and three dimensions (limit behaviors) for more general cases, i.e., continuous and discrete cases, have been done by some authors, see [1, 1113, 16]. There are many papers in which systems of difference equations have been studied, as in the examples given below.

In [14], Papaschinopoulos and Schinas considered the system of difference equations
$$ x_{n+1}=A+\frac{y_{n}}{x_{n-p}},\qquad y_{n+1}=A+ \frac{x_{n}}{y _{n-q}},\quad n=0,1,\ldots, $$
(1)
where \(A\in ( 0,\infty ) \), p, q are positive integers and \(x_{-p},\ldots,x_{0}\), \(y_{-q},\ldots,y_{0}\) are positive numbers.
In [15], Papaschinopoulos and Schinas studied the system of difference equations
$$ x_{n+1}=A+\frac{x_{n-1}}{y_{n}},\qquad y_{n+1}=A+ \frac{y_{n-1}}{x _{n}},\quad n=0,1,\ldots, $$
(2)
where A is a positive constant and the initial conditions are positive numbers.
In [2], Bao investigated the local stability, oscillation, and boundedness character of positive solutions of the system of difference equations
$$ x_{n+1}=A+\frac{x_{n-1}^{p}}{y_{n}^{p}},\qquad y_{n+1}=A+ \frac{y _{n-1}^{p}}{x_{n}^{p}},\quad n=0,1,\ldots, $$
(3)
where \(A\in ( 0,\infty ) \), \(p\in [ 1,\infty ) \) and the initial conditions \(x_{i}\), \(y_{i}\in ( 0,\infty ) \), \(i=-1,0\).
In [7], Gümüş and Soykan considered the dynamical behavior of positive solutions for a system of rational difference equations of the following form:
$$ u_{n+1}=\frac{\alpha u_{n-1}}{\beta +\gamma v_{n-2}^{p}},\qquad v _{n+1}=\frac{\alpha_{1}v_{n-1}}{\beta_{1}+\gamma_{1}u_{n-2}^{p}}, \quad n=0,1,\ldots, $$
(4)
where the parameters α, β, γ, \(\alpha_{1}\), \(\beta_{1}\), \(\gamma_{1}\), p and the initial values \(u_{-i}\), \(v_{-i}\) for \(i=0,1,2\) are positive real numbers.
In [6], Göcen and Cebeci studied the general form of periodic solutions of some higher order systems of difference equations
$$ x_{n+1}=\frac{\pm x_{n-k}y_{n- ( 2k+1 ) }}{y_{n- ( 2k+1 ) }\mp y_{n-k}},\qquad y_{n+1}=\frac{\pm y_{n-k}x_{n- ( 2k+1 ) }}{x_{n- ( 2k+1 ) }\mp x_{n-k}}, \quad n,k\in \mathbb{N} _{0}, $$
(5)
where the initial values are arbitrary real numbers.

Also, for similar results in the area of difference equations and systems, see [4, 10, 1621].

In this paper, we investigate the stability, boundedness character, and periodicity of positive solutions of the system of difference equations
$$ \begin{aligned} &x_{n+1}=A+\frac{x_{n-1}}{z_{n}}, \qquad y_{n+1}=A+ \frac{y_{n-1}}{z _{n}}, \\ &z_{n+1}=A+\frac{z_{n-1}}{y_{n}},\quad n=0,1,\ldots , \end{aligned}$$
(6)
where A and the initial values \(x_{-1}\), \(x_{0}\), \(y_{-1}\), \(y_{0}\), \(z_{-1}\), \(z_{0}\) are positive real numbers.

2 Preliminaries

We recall some basic definitions that we afterwards need in the paper.

Let us introduce the discrete dynamical system:
$$\begin{aligned}& x_{n+1} =f_{1} ( x_{n},x_{n-1}, \ldots,x_{n-k},y_{n},y_{n-1},\ldots,y _{n-k},z_{n},z_{n-1},\ldots,z_{n-k} ), \\& y_{n+1} =f_{2} ( x_{n},x_{n-1}, \ldots,x_{n-k},y_{n},y_{n-1},\ldots,y _{n-k},z_{n},z_{n-1},\ldots,z_{n-k} ), \\& z_{n+1} =f_{3} ( x_{n},x_{n-1}, \ldots,x_{n-k},y_{n},y_{n-1},\ldots,y _{n-k},z_{n},z_{n-1},\ldots,z_{n-k} ), \end{aligned}$$
(7)
\(n\in \mathbb{N} \), where \(f_{1}:I_{1}^{k+1}\times I_{2}^{k+1}\times I_{3}^{k+1}\rightarrow I_{1}\), \(f_{2}:I_{1}^{k+1}\times I_{2}^{k+1} \times I_{3}^{k+1}\rightarrow I_{2}\), and \(f_{3}:I_{1}^{k+1}\times I _{2}^{k+1}\times I_{3}^{k+1}\rightarrow I_{3}\) are continuously differentiable functions and \(I_{1}\), \(I_{2}\), \(I_{3}\) are some intervals of real numbers. Also, a solution \(\{ ( x_{n},y_{n},z _{n} ) \}_{n=-k}^{\infty }\) of system (7) is uniquely determined by the initial values \(( x_{-i},y_{-i},z_{-i} ) \in I_{1}\times I_{2}\times I_{3}\) for \(i\in \{ 0,1,\ldots,k \} \).

Definition 1

An equilibrium point of system (7) is a point \(( \overline{x},\overline{y},\overline{z} ) \) that satisfies
$$\begin{aligned}& \overline{x} =f_{1} ( \overline{x},\overline{x},\ldots, \overline{x}, \overline{y},\overline{y},\ldots,\overline{y},\overline{z}, \overline{z},\ldots, \overline{z} ), \\& \overline{y} =f_{2} ( \overline{x},\overline{x},\ldots, \overline{x}, \overline{y},\overline{y},\ldots,\overline{y},\overline{z}, \overline{z},\ldots, \overline{z} ), \\& \overline{z} =f_{3} ( \overline{x},\overline{x},\ldots, \overline{x}, \overline{y},\overline{y},\ldots,\overline{y},\overline{z}, \overline{z},\ldots, \overline{z} ). \end{aligned}$$
Together with system (7), if we consider the associated vector map
$$F= ( f_{1},x_{n},x_{n-1},\ldots,x_{n-k},f_{2},y_{n},y_{n-1}, \ldots,y _{n-k},f_{3},z_{n-1},\ldots,z_{n-k} ), $$
then the point \(( \overline{x},\overline{y},\overline{z} ) \) is also called a fixed point of the vector map F.

Definition 2

([2, 3])

Let \((\overline{x}, \overline{y},\overline{z} ) \) be an equilibrium point of system (7).
  1. (a)
    An equilibrium point \(( \overline{x},\overline{y}, \overline{z} ) \) is called stable if, for every \(\varepsilon >0\), there exists \(\delta >0\) such that, for every initial value \(( x_{-i},y_{-i},z_{-i} ) \in I_{1}\times I_{2}\times I_{3}\), with
    $$\sum_{i=-k}^{0} \vert x_{i}- \overline{x} \vert < \delta,\qquad \sum_{i=-k}^{0} \vert y_{i}-\overline{y} \vert < \delta,\qquad \sum _{i=-k}^{0} \vert z_{i}-\overline{z} \vert < \delta $$
    implying \(\vert x_{n}-\overline{x} \vert <\varepsilon \), \(\vert y_{n}-\overline{y} \vert <\varepsilon \), \(\vert z _{n}-\overline{z} \vert <\varepsilon \) for \(n\in \mathbb{N} \).
     
  2. (b)

    An equilibrium point \(( \overline{x},\overline{y}, \overline{z} ) \) of system (7) is called unstable if it is not stable.

     
  3. (c)
    An equilibrium point \(( \overline{x},\overline{y}, \overline{z} ) \) of system (7) is called locally asymptotically stable if it is stable and if, in addition, there exists \(\gamma >0\) such that
    $$\sum_{i=-k}^{0} \vert x_{i}- \overline{x} \vert < \gamma,\qquad \sum_{i=-k}^{0} \vert y_{i}-\overline{y} \vert < \gamma,\qquad \sum _{i=-k}^{0} \vert z_{i}-\overline{z} \vert < \gamma $$
    and \(( x_{n},y_{n},z_{n} ) \rightarrow ( \overline{x}, \overline{y},\overline{z} ) \) as \(n\rightarrow \infty \).
     
  4. (d)

    An equilibrium point \(( \overline{x},\overline{y}, \overline{z} ) \) of system (7) is called a global attractor if \(( x_{n},y_{n},z_{n} ) \rightarrow ( \overline{x},\overline{y},\overline{z} ) \) as \(n\rightarrow \infty \).

     
  5. (e)

    An equilibrium point \(( \overline{x},\overline{y}, \overline{z} ) \) of system (7) is called globally asymptotically stable if it is stable and a global attractor.

     

Definition 3

Let \(( \overline{x},\overline{y},\overline{z} ) \) be an equilibrium point of the map F where \(f_{1}\), \(f_{2}\), and \(f_{3}\) are continuously differentiable functions at \(( \overline{x},\overline{y},\overline{z} ) \). The linearized system of system (7) about the equilibrium point \(( \overline{x},\overline{y},\overline{z} ) \) is
$$X_{n+1}=F ( X_{n} ) =BX_{n}, $$
where
$$X_{n}= \left ( \begin{matrix} x_{n} \\ \vdots \\ x_{n-k} \\ y_{n} \\ \vdots \\ y_{n-k} \\ z_{n} \\ \vdots \\ z_{n-k} \end{matrix} \right ) $$
and B is a Jacobian matrix of system (7) about the equilibrium point \(( \overline{x},\overline{y},\overline{z} ) \).

Definition 4

Assume that \(X_{n+1}=F ( X_{n} ) \), \(n=0,1,\ldots \) , is a system of difference equations such that is a fixed point of F. If no eigenvalues of the Jacobian matrix B about have absolute value equal to one, then is called hyperbolic. Otherwise, is said to be nonhyperbolic.

Theorem 5

(The linearized stability theorem [8], p. 11)

Assume that
$$X_{n+1}=F ( X_{n} ),\quad n=0,1,\ldots, $$
is a system of difference equations such that is a fixed point of F.
  1. (a)

    If all eigenvalues of the Jacobian matrix B about lie inside the open unit disk \(\vert \lambda \vert <1\), that is, if all of them have absolute value less than one, then is locally asymptotically stable.

     
  2. (b)

    If at least one of them has a modulus greater than one, then is unstable.

     
A positive solution \(\{ ( x_{n},y_{n},z_{n} ) \}_{n=-k}^{ \infty }\) of system (7) is bounded and persists if there exist positive constants M, N such that
$$M\leq x_{n},y_{n},z_{n}\leq N,\quad n=-m,-m+1, \ldots. $$
A positive solution \(\{ ( x_{n},y_{n},z_{n} ) \}_{n=-k}^{ \infty }\) of system (7) is periodic with period p if
$$x_{n+p}=x_{n},\qquad y_{n+p}=y_{n}, \qquad z_{n+p}=z_{n}\quad \text{for all }n\geq -1. $$

3 Main results

In this section, we prove our main results.

Theorem 6

The following statements are true:
  1. (i)
    If \(( \overline{x},\overline{y},\overline{z} ) \) is a positive equilibrium point of system (6), then
    $$( \overline{x},\overline{y},\overline{z} ) = \textstyle\begin{cases} ( A+1,A+1,A+1 ), & \textit{if }A\neq 1, \\ ( \mu,\mu,\frac{\mu }{\mu -1} ),\quad \mu \in ( 1, \infty ) & \textit{if }A=1. \end{cases} $$
     
  2. (ii)

    If \(A>1\), then the equilibrium point of system (6) is locally asymptotically stable.

     
  3. (iii)

    If \(0< A<1\), then the equilibrium point of system (6) is locally unstable.

     
  4. (iv)

    If \(A=1\), then for every \(\mu \in ( 1,\infty ) \) there exist positive solutions \(\{ ( x_{n},y_{n},z_{n} ) \}\) of system (6) which tend to the positive equilibrium point \(( \mu,\mu,\frac{\mu }{\mu -1} ) \).

     

Proof

(i) It is easily seen from the definition of equilibrium point that the equilibrium points of system (6) are the nonnegative solutions of the equations
$$\overline{x}=A+\frac{\overline{x}}{\overline{z}},\qquad \overline{y}=A+\frac{ \overline{y}}{\overline{z}}, \qquad \overline{z}=A+\frac{\overline{z}}{ \overline{y}}. $$
From this, we get
$$\begin{aligned}& \overline{x}.\overline{z} =A\overline{z}+\overline{x},\qquad \overline{y}. \overline{z}=A\overline{z}+\overline{y},\qquad \overline{z}.\overline{y}=A \overline{y}+\overline{z} \\& \quad \Rightarrow \quad \overline{x}.\overline{z}-\overline{x}=\overline{y}. \overline{z}-\overline{y},\qquad A\overline{z}+\overline{y}=A \overline{y}+ \overline{z} \\& \quad \Rightarrow \quad \overline{x} ( \overline{z}-1 ) =\overline{y} ( \overline{z}-1 ),\qquad \overline{z} ( A-1 ) =\overline{y} ( A-1 ). \end{aligned}$$
From which it follows that if \(A\neq 1\),
$$\overline{x}=\overline{y}=\overline{z}=A+1\quad \Rightarrow\quad ( \overline{x}, \overline{y},\overline{z} ) = ( A+1,A+1,A+1 ). $$
Also, we have
$$\begin{aligned}& \frac{\overline{x}.\overline{z}-\overline{x}}{\overline{z}} =A,\qquad \frac{\overline{y}.\overline{z}-\overline{y}}{\overline{z}}=A,\qquad \frac{\overline{z}.\overline{y}-\overline{z}}{\overline{y}}=A \\& \quad \Rightarrow \quad \frac{\overline{x}.\overline{z}-\overline{x}}{ \overline{z}}=\frac{\overline{y}.\overline{z}-\overline{y}}{ \overline{z}},\qquad \frac{\overline{y}.\overline{z}-\overline{y}}{ \overline{z}}=\frac{\overline{z}.\overline{y}-\overline{z}}{ \overline{y}} \\& \quad \Rightarrow \quad \overline{x}.\overline{z}-\overline{x}=\overline{y}. \overline{z}-\overline{y},\qquad \overline{y}^{2}\overline{z}- \overline{y}^{2}=\overline{z}^{2}\overline{y}- \overline{z}^{2} \\& \quad \Rightarrow \quad \overline{x} ( \overline{z}-1 ) =\overline{y} ( \overline{z}-1 ),\qquad \overline{y}.\overline{z} ( \overline{y}-\overline{z} ) = ( \overline{y}- \overline{z} ) ( \overline{y}+\overline{z} ). \end{aligned}$$
From which it follows that if \(A=1\),
$$\begin{aligned}& \overline{x}=\overline{y}\quad \text{and}\quad \overline{y}.\overline{z}= \overline{y}+\overline{z} \\& \quad \Rightarrow\quad ( \overline{x},\overline{y}, \overline{z} ) = \biggl( \mu,\mu,\frac{\mu }{\mu -1} \biggr),\quad \mu \in ( 1, \infty ). \end{aligned}$$
In that case, we have a continuum of positive equilibria which lie on the hyperboloid
$$ \overline{y}.\overline{z}=\overline{y}+\overline{z}. $$
(8)
(ii) We consider the following transformation to build the corresponding linearized form of system (6):
$$( x_{n},x_{n-1},y_{n},y_{n-1},z_{n},z_{n-1} ) \rightarrow ( f,f_{1},g,g_{1},h,h_{1} ), $$
where
$$\begin{aligned}& f =A+\frac{x_{n-1}}{z_{n}}, \\& f_{1} =x_{n}, \\& g =A+\frac{y_{n-1}}{z_{n}}, \\& g_{1} =y_{n}, \\& h =A+\frac{z_{n-1}}{y_{n}}, \\& h_{1} =z_{n}. \end{aligned}$$
The Jacobian matrix about the equilibrium point \(( \overline{x}, \overline{y},\overline{z} ) \) under the above transformation is given by
$$ B ( \overline{x},\overline{y},\overline{z} ) = \left ( \begin{matrix} 0 & \frac{1}{\overline{z}} & 0 & 0 & -\frac{\overline{x}}{ \overline{z}^{2}} & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{\overline{z}} & -\frac{\overline{y}}{ \overline{z}^{2}} & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & -\frac{\overline{z}}{\overline{y}^{2}} & 0 & 0 & \frac{1}{ \overline{y}} \\ 0 & 0 & 0 & 0 & 1 & 0 \end{matrix} \right ). $$
(9)
Hence, the linearized system of system (6) about the equilibrium point \(( \overline{x},\overline{y},\overline{z} ) =( A+1, A+1,A+1) \) is
$$X_{n+1}=B ( \overline{x},\overline{y},\overline{z} ) X_{n} , $$
where
$$X_{n}= \bigl( ( x_{n},x_{n-1},y_{n},y_{n-1},z_{n},z_{n-1} ) \bigr) ^{T} $$
and
$$B ( \overline{x},\overline{y},\overline{z} ) = \left ( \begin{matrix} 0 & \frac{1}{A+1} & 0 & 0 & -\frac{1}{A+1} & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{A+1} & -\frac{1}{A+1} & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & -\frac{1}{A+1} & 0 & 0 & \frac{1}{A+1} \\ 0 & 0 & 0 & 0 & 1 & 0 \end{matrix} \right ). $$
Then the characteristic equation of \(B ( \overline{x}, \overline{y},\overline{z} ) \) about \(( \overline{x}, \overline{y},\overline{z} ) = ( A+1,A+1,A+1 ) \) is
$$ \lambda^{6}-\frac{ ( 3A+4 ) }{ ( A+1 ) ^{2}}\lambda ^{4}+ \frac{ ( 3A+4 ) }{ ( A+1 ) ^{3}}\lambda^{2}-\frac{1}{ ( A+1 ) ^{3}}=0. $$
(10)
From this, the roots of characteristic equation (10) are
$$\begin{aligned}& \lambda_{1} =\frac{1}{\sqrt{A+1}}, \\& \lambda_{2} =-\frac{1}{\sqrt{A+1}}, \\& \lambda_{3} =\frac{1}{2}\frac{\sqrt{4A+5}-1}{A+1}, \\& \lambda_{4} =-\frac{1}{2}\frac{\sqrt{4A+5}+1}{A+1}, \\& \lambda_{5} =\frac{1}{2}\frac{\sqrt{4A+5}+1}{A+1}, \\& \lambda_{6} =-\frac{1}{2}\frac{\sqrt{4A+5}-1}{A+1}. \end{aligned}$$
From the linearized stability theorem, since \(A>1\), all roots of the characteristic equation lie inside the open unit disk \(\vert \lambda \vert <1\). Therefore, the positive equilibrium point of system (6) is locally asymptotically stable.

(iii) From the proof of (ii), it is true.

(iv) From (9), the linearized system of system (6) about the equilibrium point \(( \overline{x}, \overline{y},\overline{z} ) = ( \mu,\mu, \frac{\mu }{\mu -1} ) \) is
$$X_{n+1}=B ( \overline{x},\overline{y},\overline{z} ) X_{n} , $$
where
$$X_{n}= \bigl( ( x_{n},x_{n-1},y_{n},y_{n-1},z_{n},z_{n-1} ) \bigr) ^{T} $$
and
$$B ( \overline{x},\overline{y},\overline{z} ) = \left ( \begin{matrix} 0 & \frac{\mu -1}{\mu } & 0 & 0 & -\frac{ ( \mu -1 ) ^{2}}{ \mu } & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{\mu -1}{\mu } & -\frac{ ( \mu -1 ) ^{2}}{ \mu } & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & -\frac{1}{\mu ( \mu -1 ) } & 0 & 0 & \frac{1}{ \mu } \\ 0 & 0 & 0 & 0 & 1 & 0 \end{matrix} \right ). $$
Hence, the characteristic equation of the matrix B is
$$ \lambda^{6}- \biggl( \frac{2\mu^{2}-1}{\mu^{2}} \biggr) \lambda^{4}+ \biggl( \frac{\mu^{3}+\mu^{2}-3\mu +1}{\mu } \biggr) \lambda^{2}-\frac{ ( \mu -1 ) ^{2}}{\mu^{3}}=0. $$
(11)
Therefore, the roots of equation (11) are:
$$\begin{aligned}& \lambda_{1} =-1, \\& \lambda_{2} =1, \\& \lambda_{3} =\frac{\sqrt{\mu -1}}{\mu }, \\& \lambda_{4} =-\frac{\sqrt{\mu -1}}{\mu }, \\& \lambda_{5} =\sqrt{\frac{\mu -1}{\mu }}, \\& \lambda_{6} =-\sqrt{\frac{\mu -1}{\mu }}. \end{aligned}$$
Then the modulus of four of the roots of (11) is less than 1. So, there exist positive solutions of system (6) which tend to the positive equilibrium point \(( \mu,\mu,\frac{ \mu }{\mu -1} ) \) of system (6) (this follows from the following proposition). This completes the proof.

In the following proposition we find positive solutions of system (6) which tend to \(( \overline{x},\overline{y}, \overline{z} ) \) as \(n\rightarrow \infty \). □

Proposition 7

Let \(\{ ( x_{n},y_{n},z_{n} ) \}\) be a positive solution of system (6). Then, if there exists \(s\in \{ -1,0,\ldots \} \) such that, for \(n\geq s\), \(x_{n}\geq \overline{x}\), \(y_{n}\geq \overline{y}\), \(z_{n}\geq \overline{z}\) (resp., \(x_{n}< \overline{x}\), \(y_{n}<\overline{y}\), \(z_{n}<\overline{z}\)), the solution \(\{ ( x_{n},y_{n},z_{n} ) \}\) tends to the positive equilibrium \(( \overline{x},\overline{y},\overline{z} ) \) of system (6) as \(n\rightarrow \infty \).

Proof

Let \(\{ ( x_{n},y_{n},z_{n} ) \}\) be a positive solution of system (6) such that
$$ x_{n}\geq \overline{x},\qquad y_{n}\geq \overline{y}, \qquad z_{n} \geq \overline{z},\quad n\geq s, $$
(12)
where \(s\in \{ -1,0,\ldots \} \). Then from (6) and (12) we have
$$ x_{n+1}=A+\frac{x_{n-1}}{z_{n}}\leq A+\frac{x_{n-1}}{\overline{z}},\quad n\geq 1. $$
(13)
Set
$$ u_{n+1}=A+\frac{u_{n-1}}{\overline{z}},\quad n\geq 1 $$
(14)
such that
$$ u_{s}=x_{s},\qquad u_{s+1}=x_{s+1},\quad s\in \{ -1,0,1,\ldots \} , n\geq s. $$
(15)
Then the solution \(u_{n}\) of the difference equation (14) is as follows:
$$ u_{n}=c_{1} \biggl( \frac{1}{\sqrt{\overline{z}}} \biggr) ^{n}+c_{1} \biggl( -\frac{1}{\sqrt{\overline{z}}} \biggr) ^{n}+\frac{A \overline{z}}{A\overline{z}-1}=c_{1} \biggl( \frac{1}{\sqrt{ \overline{z}}} \biggr) ^{n}+c_{1} \biggl( -\frac{1}{ \sqrt{\overline{z}}} \biggr) ^{n}+\overline{x}, $$
(16)
where \(c_{1}\), \(c_{2}\) depend on \(x_{s}\), \(x_{s+1}\). In addition, relations (13) and (14) imply that
$$ x_{n+1}-u_{n+1}\leq \frac{x_{n-1}-u_{n-1}}{\overline{z}},\quad n>s. $$
(17)
Then, by using (15) and (17) and induction, we have
$$ x_{n}\leq u_{n},\quad n\geq s. $$
(18)
Therefore, from (12), (16), and (18), it is clear that
$$ \lim_{n\rightarrow \infty }x_{n}=\overline{x}. $$
(19)
Similarly, we can prove that
$$ \lim_{n\rightarrow \infty }y_{n}=\overline{y}\quad \text{and}\quad \lim_{n\rightarrow \infty }z_{n}=\overline{z}. $$
(20)
Thus, from (19) and (20), the solution \(\{ ( x_{n},y_{n},z_{n} ) \}\) tends to \(( \overline{x}, \overline{y},\overline{z} ) \) as \(n\rightarrow \infty \).

Arguing as above, we can show that if \(x_{n}<\overline{x}\), \(y_{n}< \overline{y}\), \(z_{n}<\overline{z}\) for \(n\geq s\), then \(\{ ( x _{n},y_{n},z_{n} ) \}\) tends to \(( \overline{x}, \overline{y},\overline{z} ) \) as \(n\rightarrow \infty \). The proof of the proposition is completed. □

Theorem 8

Assume that \(0< A<1\) and \(\{ ( x_{n},y_{n},z_{n} ) \}\) is an arbitrary positive solution of system (6). Then the following statements are true.
  1. (i)
    If
    $$ \begin{aligned} &x_{-1}< 1,\qquad y_{-1}< 1,\qquad z_{-1}< 1,\qquad x_{0}> \frac{1}{1-A}, \\ & y_{0}>\frac{1}{1-A},\qquad z_{0}>\frac{1}{1-A} , \end{aligned} $$
    (21)
    then
    $$\begin{aligned}& \lim_{n\rightarrow \infty }x_{2n+1} =A,\qquad \lim_{n\rightarrow \infty }y_{2n+1}=A, \qquad \lim_{n\rightarrow \infty }z_{2n+1}=A, \\& \lim_{n\rightarrow \infty }x_{2n} =\infty,\qquad \lim _{n\rightarrow \infty }y_{2n}=\infty,\qquad \lim_{n\rightarrow \infty }z_{2n}= \infty. \end{aligned}$$
     
  2. (ii)
    If
    $$ \begin{aligned}&x_{0}< 1,\qquad y_{0}< 1,\qquad z_{0}< 1,\qquad x_{-1}>\frac{1}{1-A}, \\ &y_{-1}>\frac{1}{1-A},\qquad z_{-1}> \frac{1}{1-A}, \end{aligned} $$
    (22)
    then
    $$\begin{aligned}& \lim_{n\rightarrow \infty }x_{2n+1} =\infty,\qquad \lim _{n\rightarrow \infty }y_{2n+1}=\infty,\qquad \lim_{n\rightarrow \infty }z_{2n+1}= \infty, \\& \lim_{n\rightarrow \infty }x_{2n} =A,\qquad \lim_{n\rightarrow \infty }y_{2n}=A, \qquad \lim_{n\rightarrow \infty }z_{2n}=A. \end{aligned}$$
     

Proof

(i) From (6) and (21), we get
$$\begin{aligned}& x_{1} =A+\frac{x_{-1}}{z_{0}}< A+\frac{1}{z_{0}}< A+ ( 1-A ) =1, \\& y_{1} =A+\frac{y_{-1}}{z_{0}}< A+\frac{1}{z_{0}}< A+ ( 1-A ) =1, \\& z_{1} =A+\frac{z_{-1}}{y_{0}}< A+\frac{1}{y_{0}}< A+ ( 1-A ) =1, \\& x_{2} =A+\frac{x_{0}}{z_{1}}>x_{0}>\frac{1}{1-A}, \\& y_{2} =A+\frac{y_{0}}{z_{1}}>y_{0}>\frac{1}{1-A}, \\& z_{2} =A+\frac{z_{0}}{y_{1}}>z_{0}>\frac{1}{1-A}. \end{aligned}$$
By induction, for \(n=0,1,2,\ldots\) , we obtain
$$\begin{aligned}& \begin{aligned}&x_{2n-1} < 1,\qquad y_{2n-1}< 1,\qquad z_{2n-1}< 1, \\ &x_{2n} >\frac{1}{1-A},\qquad y_{2n}> \frac{1}{1-A},\qquad z_{2n}> \frac{1}{1-A}. \end{aligned} \end{aligned}$$
(23)
Thus, relations (6) and (23) imply that
$$\begin{aligned}& x_{2n} =A+\frac{x_{2n-2}}{z_{2n-1}}>A+x_{2n-2}>2A+ \frac{x_{2n-4}}{z _{2n-3}}>2A+x_{2n-4}, \\& y_{2n} =A+\frac{y_{2n-2}}{z_{2n-1}}>A+y_{2n-2}>2A+ \frac{y_{2n-4}}{z _{2n-3}}>2A+y_{2n-4}, \\& z_{2n} =A+\frac{z_{2n-2}}{y_{2n-1}}>A+z_{2n-2}>2A+ \frac{z_{2n-4}}{y _{2n-3}}>2A+z_{2n-4}. \end{aligned}$$
From which we get
$$\lim_{n\rightarrow \infty }x_{2n}=\infty,\qquad \lim _{n\rightarrow \infty }y_{2n}=\infty,\qquad \lim_{n\rightarrow \infty }z_{2n}= \infty. $$
Noting that (23) and taking limits on both sides of three equations
$$x_{2n+1}=A+\frac{x_{2n-1}}{z_{2n}},\qquad y_{2n+1}=A+ \frac{y_{2n-1}}{z _{2n}},\qquad z_{2n+1}=A+\frac{z_{2n-1}}{y_{2n}}, $$
we have
$$\lim_{n\rightarrow \infty }x_{2n+1}=A,\qquad \lim_{n\rightarrow \infty }y_{2n+1}=A, \qquad \lim_{n\rightarrow \infty }z_{2n+1}=A. $$

(ii) The proof is similar to the proof of (i), so we leave it to readers. □

Theorem 9

Assume that \(A=1\). Then every positive solution of system (6) is bounded and persists.

Proof

Let \(\{ ( x_{n},y_{n},z_{n} ) \} \) be a positive solution of system (6).

Obviously, \(x_{n}>1\), \(y_{n}>1\), \(z_{n}>1\), for \(n\geq 1\). So, we have
$$x_{i},y_{i},z_{i}\in \biggl[ K, \frac{K}{K-1} \biggr] ,\quad i=1,2,\ldots,m+1, $$
where
$$K=\min \biggl\{ \alpha,\frac{\beta }{\beta -1} \biggr\} >1,\quad \alpha = \min _{1\leq i\leq m+1} \{ x_{i},y_{i},z_{i} \} , \beta = \max_{1\leq i\leq m+1} \{ x_{i},y_{i},z_{i} \} . $$
Then we obtain
$$\begin{aligned}& K =1+\frac{K}{K/ ( K-1 ) }\leq x_{m+2}=1+\frac{x_{1}}{z _{m+1}}\leq 1+ \frac{K/ ( K-1 ) }{K}=\frac{K}{K-1}, \\& K =1+\frac{K}{K/ ( K-1 ) }\leq y_{m+2}=1+\frac{y_{1}}{z _{m+1}}\leq 1+ \frac{K/ ( K-1 ) }{K}=\frac{K}{K-1}, \\& K =1+\frac{K}{K/ ( K-1 ) }\leq z_{m+2}=1+\frac{z_{1}}{y _{m+1}}\leq 1+ \frac{K/ ( K-1 ) }{K}=\frac{K}{K-1}. \end{aligned}$$
By induction, we get
$$x_{i},y_{i},z_{i}\in \biggl[ K, \frac{K}{K-1} \biggr] ,\quad i=1,2,\ldots . $$
The proof of the following theorem is seen easily and will be omitted. □

Theorem 10

Assume \(A=1\). Then every positive solution of system (6) is periodic of period 2.

Theorem 11

Assume \(A>1\). Then every positive solution of system (6) is bounded.

Proof

Let \(\{ ( x_{n},y_{n},z_{n} ) \}\) be a positive solution of system (6). Clearly,
$$ x_{n},y_{n},z_{n}>A>1\quad \text{for }n\geq 1. $$
(24)
From (24), we have
$$ x_{n+1}=A+\frac{x_{n-1}}{z_{n}}\leq A+\frac{x_{n-1}}{A},\quad n \geq 1. $$
(25)
Set
$$ u_{n+1}=A+\frac{u_{n-1}}{A},\quad n\geq 1, $$
(26)
such that
$$ u_{s}=x_{s},\qquad u_{s+1}=x_{s+1},\quad s\in \{ -1,0,1,\ldots \} ,n\geq s. $$
(27)
Then the solution \(u_{n}\) of the difference equation (26) is as follows:
$$ u_{n}=c_{1} \biggl( \frac{1}{\sqrt{A}} \biggr) ^{n}+c_{1} \biggl( -\frac{1}{ \sqrt{A}} \biggr) ^{n}+\frac{A^{2}}{A-1}. $$
(28)
Indeed, from (26) we get
$$\begin{aligned}& u_{n+1}-\frac{1}{A}u_{n-1} =0\quad \Rightarrow\quad \lambda^{2}-\frac{1}{A}=0 \\& \quad \Rightarrow \quad \lambda_{1,2}=\pm \frac{1}{\sqrt{A}}. \end{aligned}$$
The homogeneous solution of difference equation (26) is given by
$$u_{h}=c_{1} \biggl( \frac{1}{\sqrt{A}} \biggr) ^{n}+c_{2} \biggl( -\frac{1}{ \sqrt{A}} \biggr) ^{n}. $$
Also, from (26), the equilibrium solution of difference equation (26) is as follows:
$$\overline{x}-\frac{1}{A}\overline{x}=A\quad \Rightarrow\quad \overline{x}=\frac{A ^{2}}{A-1}. $$
In addition, relations (25) and (28) imply that
$$ x_{n+1}-u_{n+1}\leq \frac{x_{n-1}-u_{n-1}}{A},\quad n>s. $$
(29)
Then, by using (27) and (29) and induction, we have
$$ x_{n}\leq u_{n},\quad n\geq s. $$
(30)
Therefore, from (24), (28), and (30), we obtain
$$A< x_{n}\leq c_{1} \biggl( \frac{1}{\sqrt{A}} \biggr) ^{n}+c_{2} \biggl( -\frac{1}{ \sqrt{A}} \biggr) ^{n}+\frac{A^{2}}{A-1}, $$
where
$$\begin{aligned}& c_{1} =\frac{1}{2} \biggl( x_{0}+ \sqrt{A}x_{1}-\frac{A^{2}}{A-1} ( 1+\sqrt{A} ) \biggr), \\& c_{2} =\frac{1}{2} \biggl( x_{0}- \sqrt{A}x_{1}-\frac{A^{2}}{A-1} ( 1-\sqrt{A} ) \biggr). \end{aligned}$$
Similarly, we can prove that
$$\begin{aligned}& A < y_{n}\leq c_{3} \biggl( \frac{1}{\sqrt{A}} \biggr) ^{n}+c_{4} \biggl( -\frac{1}{\sqrt{A}} \biggr) ^{n}+\frac{A^{2}}{A-1}, \\& A < z_{n}\leq c_{5} \biggl( \frac{1}{\sqrt{A}} \biggr) ^{n}+c_{6} \biggl( -\frac{1}{\sqrt{A}} \biggr) ^{n}+\frac{A^{2}}{A-1}, \end{aligned}$$
where
$$\begin{aligned}& c_{3} =\frac{1}{2} \biggl( y_{0}+ \sqrt{A}y_{1}-\frac{A^{2}}{A-1} ( 1+\sqrt{A} ) \biggr), \\& c_{4} =\frac{1}{2} \biggl( y_{0}- \sqrt{A}y_{1}-\frac{A^{2}}{A-1} ( 1-\sqrt{A} ) \biggr), \\& c_{5} =\frac{1}{2} \biggl( z_{0}+ \sqrt{A}z_{1}-\frac{A^{2}}{A-1} ( 1+\sqrt{A} ) \biggr), \\& c_{6} =\frac{1}{2} \biggl( z_{0}- \sqrt{A}z_{1}-\frac{A^{2}}{A-1} ( 1-\sqrt{A} ) \biggr). \end{aligned}$$
 □

Theorem 12

Suppose that \(A>1\). Then the positive equilibrium point of system (6) is globally asymptotically stable.

Proof

By means of Theorem 11, we set
$$\begin{aligned}& \begin{aligned} &L_{1} =\lim_{n\rightarrow \infty }\sup x_{n},\qquad L_{2}= \lim_{n\rightarrow \infty }\sup y_{n},\qquad L_{3}= \lim_{n\rightarrow \infty }\sup z_{n}, \\ &m_{1} =\lim_{n\rightarrow \infty }\inf x_{n},\qquad m_{2}= \lim_{n\rightarrow \infty }\inf y_{n},\qquad m_{3}= \lim_{n\rightarrow \infty }\inf z_{n}. \end{aligned} \end{aligned}$$
(31)
Then, from (6) and (31), we have
$$\begin{aligned}& \begin{aligned}&L_{1} \leq A+\frac{L_{1}}{m_{3}},\qquad L_{2} \leq A+\frac{L_{2}}{m _{3}},\qquad L_{3}\leq A+\frac{L_{3}}{m_{2}}, \\ &m_{1} \geq A+\frac{m_{1}}{L_{3}},\qquad m_{2}\geq A+ \frac{m_{2}}{L _{3}},\qquad m_{3}\geq A+\frac{m_{3}}{L_{2}}. \end{aligned} \end{aligned}$$
(32)
Relations (32) imply that
$$AL_{2}+m_{3}\leq m_{3}L_{2}\leq Am_{3}+L_{2},\qquad AL_{3}+m_{2} \leq m_{2}L_{3}\leq Am_{2}+L_{3}, $$
from which we have
$$( A-1 ) ( L_{2}-m_{3} ) \leq 0,\qquad ( A-1 ) ( L_{3}-m_{2} ) \leq 0. $$
Since \(A>1\), we get
$$L_{2}\leq m_{3}\leq L_{3},\qquad L_{3}\leq m_{2}\leq L_{2}, $$
from this it is obvious that
$$ L_{2}=L_{3}=m_{2}=m_{3}. $$
(33)
Moreover, from (32) it follows that
$$L_{1}m_{3}\leq Am_{3}+L_{1},\qquad m_{1}L_{3}\leq AL_{3}+m_{1} , $$
from which
$$L_{1} ( m_{3}-1 ) \leq Am_{3},\qquad AL_{3}\leq m_{1} ( L_{3}-1 ). $$
Using (33), we have
$$L_{1} ( L_{3}-1 ) \leq m_{1} ( L_{3}-1 ), $$
then
$$L_{1}\leq m_{1}. $$
Since \(x_{n}\) is bounded, it implies that
$$L_{1}=m_{1}. $$
Hence, every positive solution \(\{ ( x_{n},y_{n},z_{n} ) \} \) of system (6) tends to the positive equilibrium system (6). So, the proof is completed. □

4 Future works

We will concentrate on the dynamical behavior of the following system of difference equations:
$$x_{n+1}=A+\frac{x_{n-m}}{z_{n}},\qquad y_{n+1}=A+ \frac{y_{n-m}}{z _{n}}, \qquad z_{n+1}=A+\frac{z_{n-m}}{y_{n}},\quad n=0,1, \ldots , $$
where \(A\in ( 0,\infty ) \) and \(x_{i},y_{i},z_{i}\in ( 0,\infty ) \), \(i=0,1,\ldots,m\), and the following cyclic system of difference equations:
$$x_{n+1}^{(i)}=A_{i}+\frac{x_{n-1}^{(i)}}{x_{n}^{(i+1)}},\quad i=1,2, \ldots,k. $$

Declarations

Acknowledgements

We are thankful to the referees for their careful reading of the manuscript and for valuable comments and suggestions that greatly improved the presentation of this work.

Funding

Not applicable.

Authors’ contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Competing interests

The authors declare that they have no competing interests.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Department of Mathematics, Art and Science Faculty, Bülent Ecevit University, Zonguldak, Turkey

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