1 Introduction

Recently, studying the qualitative behavior of difference equations and systems is a topic of great interest. Applications of discrete dynamical systems and difference equations have appeared recently in many areas such as ecology, population dynamics, queuing problems, statistical problems, stochastic time series, combinatorial analysis, number theory, geometry, electrical networks, neural networks, quanta in radiation, genetics in biology, economics, psychology, sociology, physics, engineering, economics, probability theory and resource management. Unfortunately, these are only considered as the discrete analogs of differential equations. It is a well-known fact that difference equations appeared much earlier than differential equations and were instrumental in paving the way for the development of the latter. It is only recently that difference equations have started receiving the attention they deserve. Perhaps this is largely due to the advent of computers where differential equations are solved by using their approximate difference equation formulations. The theory of discrete dynamical systems and difference equations developed greatly during the last twenty-five years of the twentieth century. The theory of difference equations occupies a central position in applicable analysis. There is no doubt that the theory of difference equations will continue to play an important role in mathematics as a whole. Nonlinear difference equations of order greater than one are of paramount importance in applications. It is very interesting to investigate the behavior of solutions of a system of higher-order rational difference equations and to discuss the local asymptotic stability of their equilibrium points. Systems of rational difference equations have been studied by several authors. Especially there has been a great interest in the study of the attractivity of the solutions of such systems. For more results on the qualitative behavior of nonlinear difference equations, we refer the interested reader to [119].

Zhang et al. [20] studied the dynamics of a system of rational third-order difference equations

x n + 1 = x n 2 B + y n y n 1 y n 2 , y n + 1 = y n 2 A + x n x n 1 x n 2 .

Din et al. [11] investigated the dynamics of a system of fourth-order rational difference equations

x n + 1 = α x n 3 β + γ y n y n 1 y n 2 y n 3 , y n + 1 = α 1 y n 3 β 1 + γ 1 x n x n 1 x n 2 x n 3 .

To be motivated by the above studies, our aim in this paper is to investigate the qualitative behavior of the following (k+1)th-order systems of rational difference equations:

x n + 1 = α x n k β + γ i = 0 k y n i , y n + 1 = α 1 y n k β 1 + γ 1 i = 0 k x n i ,n=0,1,,
(1)

where the parameters α, β, γ, α 1 , β 1 , γ 1 and initial conditions x 0 , x 1 ,, x k , y 0 , y 1 ,, y k are positive real numbers, and

x n + 1 = a y n k b + c i = 0 k x n i , y n + 1 = a 1 x n k b 1 + c 1 i = 0 k y n i ,n=0,1,,
(2)

where the parameters a, b, c, a 1 , b 1 , c 1 and initial conditions x 0 , x 1 ,, x k , y 0 , y 1 ,, y k are positive real numbers. This paper is a natural extension of [11, 20, 21].

Let us consider (2k+2)-dimensional discrete dynamical system of the form

x n + 1 = f ( x n , x n 1 , , x n k , y n , y n 1 , , y n k ) , y n + 1 = g ( x n , x n 1 , , x n k , y n , y n 1 , , y n k ) , n = 0 , 1 , ,
(3)

where f: I k + 1 × J k + 1 I and g: I k + 1 × J k + 1 J are continuously differentiable functions and I, J are some intervals of real numbers. Furthermore, a solution { ( x n , y n ) } n = k of system (3) is uniquely determined by initial conditions ( x i , y i )I×J for i{k,k+1,,1,0}. Along with system (3), we consider the corresponding vector map F=(f, x n , x n 1 ,, x n k ,g, y n , y n 1 ,, y n k ). An equilibrium point of (3) is a point ( x ¯ , y ¯ ) that satisfies

x ¯ = f ( x ¯ , x ¯ , , x ¯ , y ¯ , y ¯ , , y ¯ ) , y ¯ = g ( x ¯ , x ¯ , , x ¯ , y ¯ , y ¯ , , y ¯ ) .

The point ( x ¯ , y ¯ ) is also called a fixed point of the vector map F.

Definition 1 Let ( x ¯ , y ¯ ) be an equilibrium point of system (3).

  1. (i)

    An equilibrium point ( x ¯ , y ¯ ) is said to be stable if for every ε>0 there exists δ>0 such that for every initial condition ( x i , y i ), i{k,k+1,,1,0}, i = k 0 ( x i , y i )( x ¯ , y ¯ )<δ implies ( x n , y n )( x ¯ , y ¯ )<ε for all n>0, where is the usual Euclidian norm in R 2 .

  2. (ii)

    An equilibrium point ( x ¯ , y ¯ ) is said to be unstable if it is not stable.

  3. (iii)

    An equilibrium point ( x ¯ , y ¯ ) is said to be asymptotically stable if there exists η>0 such that i = k 0 ( x i , y i )( x ¯ , y ¯ )<η and ( x n , y n )( x ¯ , y ¯ ) as n.

  4. (iv)

    An equilibrium point ( x ¯ , y ¯ ) is called a global attractor if ( x n , y n )( x ¯ , y ¯ ) as n.

  5. (v)

    An equilibrium point ( x ¯ , y ¯ ) is called an asymptotic global attractor if it is a global attractor and stable.

Definition 2 Let ( x ¯ , y ¯ ) be an equilibrium point of the map

F=(f, x n , x n 1 ,, x n k ,g, y n , y n 1 ,, y n k ),

where f and g are continuously differentiable functions at ( x ¯ , y ¯ ). The linearized system of (3) about the equilibrium point ( x ¯ , y ¯ ) is

X n + 1 =F( X n )= F J X n ,

where

X n = ( x n x n 1 x n k y n y n 1 y n k )

and F J is the Jacobian matrix of system (3) about the equilibrium point ( x ¯ , y ¯ ).

Lemma 1 [22]

Assume that X n + 1 =F( X n ), n=0,1, , is a system of difference equations and X ¯ is the fixed point of F. If all eigenvalues of the Jacobian matrix J F about X ¯ lie inside an open unit disk |λ|<1, then X ¯ is locally asymptotically stable. If one of them has norm greater than one, then X ¯ is unstable.

Lemma 2 [23]

Assume that X n + 1 =F( X n ), n=0,1, , is a system of difference equations and X ¯ is the equilibrium point of this system. The characteristic polynomial of this system about the equilibrium point X ¯ is P(λ)= a 0 λ n + a 1 λ n 1 ++ a n 1 λ+ a n =0, with real coefficients and a 0 >0. Then all roots of the polynomial P(λ) lie inside the open unit disk |λ| if and only if Δ k >0 for k=0,1, , where Δ k is the principal minor of order k of the n×n matrix

Δ n = ( a 1 a 3 a 5 0 a 0 a 2 a 4 0 0 a 1 a 3 0 0 0 0 a n ) .
(4)

Let us consider a system of difference equations

X n + 1 = ( A + B ( n ) ) X n ,
(5)

where X n is an m-dimensional vector, A C m × m is a constant matrix, and B: Z + C m × m is a matrix function satisfying

B ( n ) 0
(6)

as n, where denotes any matrix norm which is associated with the vector norm

( x , y ) = x 2 + y 2 .

Proposition 1 (Perron’s theorem)[24]

Suppose that condition (6) holds. If X n is a solution of (5), then either X n =0 for all large n or

ρ= lim n ( X n ) 1 / n
(7)

exists and is equal to the modulus of one of the eigenvalues of matrix A.

Proposition 2 [24]

Suppose that condition (6) holds. If X n is a solution of (5), then either X n =0 for all large n or

ρ= lim n X n + 1 X n
(8)

exists and is equal to the modulus of one of the eigenvalues of matrix A.

2 On the system x n + 1 = α x n k β + γ i = 0 k y n i , y n + 1 = α 1 y n k β 1 + γ 1 i = 0 k x n i

In this section, we shall investigate the qualitative behavior of system (1). Let ( x ¯ , y ¯ ) be an equilibrium point of system (1), then for α>β and α 1 > β 1 , system (1) has two positive equilibrium points P 0 =(0,0), P 1 =(A,B), where A= ( α 1 β 1 γ 1 ) 1 k + 1 and B= ( α β γ ) 1 k + 1 .

To construct the corresponding linearized form of system (1), we consider the following transformation:

( x n , x n 1 , x n 2 ,, x n k , y n , y n 1 ,, y n k )(f, f 1 ,, f n k ,g, g 1 ,, g n k ),
(9)

where f= α x n k β + γ i = 0 k y n i , f 1 = x n , f 2 = x n 1 , … , f n k = x n ( k 1 ) and g= α 1 y n k β 1 + γ 1 i = 0 k x n i , g 1 = y n , g 2 = y n 1 , … , g n k = y n ( k 1 ) . The Jacobian matrix about the fixed point ( x ¯ , y ¯ ) under the transformation (9) is given by

F J ( x ¯ , y ¯ )= ( 0 0 0 A B B B B 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 C C C C 0 0 0 D 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 ) ,

where A= α β + γ y ¯ k + 1 , B= α γ x ¯ y ¯ k ( β + γ y ¯ k + 1 ) 2 , C= α 1 γ 1 y ¯ x ¯ k ( β 1 + γ 1 x ¯ k + 1 ) 2 and D= α 1 β 1 + γ 1 x ¯ k + 1 .

Theorem 1 Let α<β and α 1 < β 1 , then every solution {( x n , y n )} of system (1) is bounded.

Proof It is easy to verify that

0 x n ( α β ) m + 1 x k if  n = ( k + 1 ) m + 1 , 0 x n ( α β ) m + 1 x 1 k if  n = ( k + 1 ) m + 2 , 0 x n ( α β ) m + 1 x 1 if  n = ( k + 1 ) m + k , 0 x n ( α β ) m + 1 x 0 if  n = ( k + 1 ) m + ( k + 1 ) ,

and

0 y n ( α 1 β 1 ) m + 1 y k if  n = ( k + 1 ) m + 1 , 0 y n ( α 1 β 1 ) m + 1 y 1 k if  n = ( k + 1 ) m + 2 , 0 y n ( α 1 β 1 ) m + 1 y 1 if  n = ( k + 1 ) m + k , 0 y n ( α 1 β 1 ) m + 1 y 0 if  n = ( k + 1 ) m + k + 1 .

Take δ 1 =max{ x k ,, x 0 } and δ 2 =max{ y k ,, y 0 }. Then 0 x n < δ 1 and 0 y n < δ 2 for all n=0,1,2, . □

Theorem 2 The equilibrium point P 0 of system (1) is locally asymptotically stable.

Proof The linearized system of (1) about the equilibrium point (0,0) is given by

X n + 1 = F J (0,0) X n ,

where

X n = ( x n x n 1 x n k y n y n 1 y n k ) ,

and

E= F J ( P 0 )= ( e i j ) ( 2 k + 2 ) × ( 2 k + 2 ) = ( 0 0 0 α β 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 α 1 β 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 ) .

Let λ 1 , λ 2 ,, λ 2 k + 2 denote the 2k+2 eigenvalues of matrix E. Let D=diag( d 1 , d 2 ,, d 2 k + 2 ) be a diagonal matrix, where d 1 = d k + 2 =1, d 1 + m = d k + 2 + m =1mϵ, 1mk, and

ϵ=min { 1 k , 1 k ( 1 α β ) , 1 k ( 1 α 1 β 1 ) } .

Clearly, D is invertible. Computing DE D 1 , we obtain

DE D 1 = ( 0 0 0 α β d 1 d k + 1 1 0 0 0 0 d 2 d 1 1 0 0 0 0 0 0 0 0 0 d k + 1 d k 1 0 0 0 0 0 0 0 0 0 0 0 0 α 1 β 1 d k + 2 d 2 k + 2 1 0 0 0 0 d k + 3 d k + 2 1 0 0 0 0 0 0 0 0 0 d 2 k + 2 d 2 k + 1 1 0 ) .

We obtain the following two inequalities:

0< d k + 1 < d k << d 2 ,0< d 2 k + 2 < d 2 k + 1 << d k + 3 ,

which implies that

d 2 d 1 1 <1, d 3 d 2 1 <1,, d k + 1 d k 1 <1,

and

d k + 3 d k + 2 1 <1, d k + 4 d k + 3 1 <1,, d 2 k + 2 d 2 k + 1 1 <1.

Furthermore,

α β d 1 d k + 1 1 = α β d k + 1 1 = α β 1 1 k ϵ <1,

and

α 1 β 1 d k + 2 d 2 k + 2 1 = α 1 β 1 d 2 k + 2 1 = α 1 β 1 1 1 k ϵ <1.

It is a well-known fact that E has the same eigenvalues as DE D 1 . Hence, we obtain

max 1 m 2 k + 2 | λ m | = D E D 1 = max { d 2 d 1 1 , , d k + 1 d k 1 , d k + 3 d k + 2 1 , , d 2 k + 2 d 2 k + 1 1 , α β d 1 d k + 1 1 , α 1 β 1 d k + 2 d 2 k + 2 1 } < 1 .

Hence, the equilibrium point P 0 of system (1) is locally asymptotically stable. □

Theorem 3 The positive equilibrium point P 1 of system (1) is unstable.

Proof The linearized system of (1) about the equilibrium point P 1 is given by

X n + 1 = F J ( P 1 ) X n ,

where

X n = ( x n x n 1 x n k y n y n 1 y n k )

and

F J ( P 1 )= ( 0 0 0 1 L L L L 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 M M M M 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 ) ,

where L= ( γ γ 1 ) 1 k + 1 ( α 1 β 1 ) 1 k + 1 ( α β ) k k + 1 α and M= ( γ 1 γ ) 1 k + 1 ( α β ) 1 k + 1 ( α 1 β 1 ) k k + 1 α 1 . The characteristic polynomial of F J ( P 1 ) is given by

P ( λ ) = λ 2 k + 2 L M [ λ 2 k + 2 λ 2 k 1 + + k λ k + 1 + ( k + 1 ) λ k + k λ k 1 + ( k 1 ) λ k 2 + + 1 ] + 2 λ k + 1 + 1 .
(10)

From (10), we have

Δ ( 2 k + 2 ) × ( 2 k + 2 ) = ( L M 3 L M 5 L M 0 1 2 L M 4 L M 0 0 L M 3 L M 0 0 0 0 1 L M ) .
(11)

It is clear that not all of Δ ( 2 k + 2 ) × ( 2 k + 2 ) >0. Therefore, by Lemma 1, the unique positive equilibrium point ( x ¯ , y ¯ )=( ( α 1 β 1 γ 1 ) 1 k + 1 , ( α β γ ) 1 k + 1 ) is unstable. □

Theorem 4 Let α>β and α 1 > β 1 , and let {( x n , y n )} be a solution of system (1). Then, for i=0,1,,k, the following statements are true:

  1. (i)

    If ( x i , y i )(0, ( α 1 β 1 γ 1 ) 1 k + 1 )×( ( α β γ ) 1 k + 1 ,), then

    ( x n , y n ) ( 0 , ( α 1 β 1 γ 1 ) 1 k + 1 ) × ( ( α β γ ) 1 k + 1 , ) .
  2. (ii)

    If ( x i , y i )( ( α 1 β 1 γ 1 ) 1 k + 1 ,)×(0, ( α β γ ) 1 k + 1 ), then

    ( x n , y n ) ( ( α 1 β 1 γ 1 ) 1 k + 1 , ) × ( 0 , ( α β γ ) 1 k + 1 ) .

Proof It follows from induction. □

Theorem 5 Let α<β and α 1 < β 1 , then the equilibrium point P 0 of system (1) is globally asymptotically stable.

Proof For α<β and α 1 < β 1 , from Theorem 2, P 0 is locally asymptotically stable. From Theorem 1, every positive solution ( x n , y n ) is bounded, i.e., 0 x n μ and 0 y n ν for all n=0,1,2, , where μ=max{ x k , x k + 1 ,, x 1 , x 0 } and ν=max{ y k , y k + 1 ,, y 1 , y 0 }. So, it is sufficient to prove that {( x n , y n )} is decreasing. From system (1), one has

x n + 1 = α x n k β + γ i = 0 k y n i α x n k β < x n k .

This implies that x ( k + 1 ) n + 1 < x ( k + 1 ) n k and x ( k + 1 ) n + ( k + 2 ) < x ( k + 1 ) n + 1 . Hence, the subsequences

{ x ( k + 1 ) n + 1 },{ x ( k + 1 ) n + 2 },,{ x ( k + 1 ) n + k },{ x ( k + 1 ) n + ( k + 1 ) }

are decreasing, i.e., the sequence { x n } is decreasing. Also,

y n + 1 = α y n k β + γ i = 0 k x n i α y n k β < y n k .

This implies that y ( k + 1 ) n + 1 < y ( k + 1 ) n k and y ( k + 1 ) n + ( k + 2 ) < y ( k + 1 ) n + 1 . Hence, the subsequences

{ y ( k + 1 ) n + 1 },{ y ( k + 1 ) n + 2 },,{ y ( k + 1 ) n + k },{ y ( k + 1 ) n + ( k + 1 ) }

are decreasing, i.e., the sequence { y n } is decreasing. Hence, lim n x n = lim n y n =0. □

Theorem 6 Let α>β and α 1 > β 1 . Then, for a solution {( x n , y n )} of system (1), the following statements are true:

  1. (i)

    If x n 0, then y n .

  2. (ii)

    If y n 0, then x n .

2.1 Rate of convergence

We investigate the rate of convergence of a solution that converges to the equilibrium point P 0 of system (1).

Assume that lim n x n = x ¯ and lim n x n = y ¯ . First we will find a system of limiting equations for the map F. The error terms are given as

x n + 1 x ¯ = i = 0 k A i ( x n i x ¯ ) + i = 0 k B i ( y n i y ¯ ) , y n + 1 y ¯ = i = 0 k C i ( x n i x ¯ ) + i = 0 k D i ( y n i y ¯ ) .

Set e n 1 = x n x ¯ and e n 2 = y n y ¯ , one has

e n + 1 1 = i = 0 k A i e n i 1 + i = 0 k B i e n i 2 , e n + 1 2 = i = 0 k C i e n i 1 + i = 0 k D i e n i 2 ,

where A i =0 for i{0,1,,k1},

A k = α β + γ i = 0 k y n i , B 0 = α γ x ¯ y n 1 y n 2 y n k ( β + γ i = 0 k y n i ) ( β + γ y ¯ k + 1 ) , B 1 = α γ x ¯ y ¯ y n 2 y n 3 y n k ( β + γ i = 0 k y n i ) ( β + γ y ¯ k + 1 ) , B 2 = α γ x ¯ y ¯ 2 y n 3 y n k ( β + γ i = 0 k y n i ) ( β + γ y ¯ k + 1 ) , , B k 1 = α γ x ¯ y ¯ k 1 y n k ( β + γ i = 0 k y n i ) ( β + γ y ¯ k + 1 ) , B k = α γ x ¯ y ¯ k ( β + γ i = 0 k y n i ) ( β + γ y ¯ k + 1 ) , C 0 = α 1 γ 1 y ¯ x n 1 x n 2 x n k ( β 1 + γ 1 i = 0 k x n i ) ( β 1 + γ 1 x ¯ k + 1 ) , C 1 = α 1 γ 1 y ¯ x ¯ x n 2 x n 3 x n k ( β 1 + γ 1 i = 0 k x n i ) ( β 1 + γ 1 x ¯ k + 1 ) , C 2 = α 1 γ 1 y ¯ x ¯ 2 x n 3 x n k ( β 1 + γ 1 i = 0 k x n i ) ( β 1 + γ 1 x ¯ k + 1 ) , , C k 1 = α 1 γ 1 y ¯ x ¯ k 1 x n k ( β 1 + γ 1 i = 0 k x n i ) ( β 1 + γ 1 x ¯ k + 1 ) , C k = α 1 γ 1 y ¯ x ¯ k ( β 1 + γ 1 i = 0 k x n i ) ( β 1 + γ 1 x ¯ k + 1 ) , D i = 0 for  i { 0 , 1 , , k 1 } and D k = α 1 β 1 + γ 1 i = 0 k x n i .

Taking the limits, we obtain lim n A i =0 for i{0,1,,k1}, lim n A k = α β + γ y ¯ k + 1 , lim n B i = α γ x ¯ y ¯ k ( β + γ y ¯ k + 1 ) 2 for i{0,1,,k}, lim n C i = α 1 γ 1 y ¯ x ¯ k ( β 1 + γ 1 x ¯ k + 1 ) 2 for i{0,1,,k}, lim n D i =0 for i{0,1,,k1} and lim n D k = α 1 β 1 + γ 1 x ¯ k + 1 . Hence, the limiting system of error terms at ( x ¯ , y ¯ )=(0,0) can be written as

E n + 1 =K E n ,
(12)

where

E n = ( e n 1 e n 1 1 e n k 1 e n 2 e n 1 2 e n k 2 ) , K = ( 0 0 0 α β 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 α 1 β 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 ) ,

which is similar to the linearized system of (1) about the equilibrium point ( x ¯ , y ¯ )=(0,0). Using proposition (1), one has the following result.

Theorem 7 Assume that {( x n , y n )} is a positive solution of system (1) such that lim n x n = x ¯ , and lim n y n = y ¯ , where ( x ¯ , y ¯ )=(0,0). Then the error vector E n of every solution of (1) satisfies both of the following asymptotic relations:

lim n ( E n ) 1 n =|λ F J ( x ¯ , y ¯ )|, lim n E n + 1 E n =|λ F J ( x ¯ , y ¯ )|,

where λ F J ( x ¯ , y ¯ ) are the characteristic roots of the Jacobian matrix F J ( x ¯ , y ¯ ) about (0,0).

3 On the system x n + 1 = a y n k b + c i = 0 k x n i , y n + 1 = a 1 x n k b 1 + c 1 i = 0 k y n i

In this section, we shall investigate the qualitative behavior of system (2). Let ( x ¯ , y ¯ ) be an equilibrium point of system (2), then system (2) has a unique equilibrium point (0,0). To construct the corresponding linearized form of system (2), we consider the following transformation:

( x n , x n 1 , x n 2 ,, x n k , y n , y n 1 ,, y n k )(f, f 1 ,, f n k ,g, g 1 ,, g n k ),
(13)

f= a y n k b + c i = 0 k x n i , f 1 = x n , f 2 = x n 1 , …, f n k = x n ( k 1 ) and g= a 1 x n k b 1 + c 1 i = 0 k y n i , g 1 = y n , g 2 = y n 1 , … , g n k = y n ( k 1 ) . The Jacobian matrix about the fixed point ( x ¯ , y ¯ ) under the transformation (13) is given by

F J ( x ¯ , y ¯ )= ( A A A A 0 0 0 B 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 C D D D D 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 ) ,

where A= a c y ¯ x ¯ k ( b + c x ¯ k + 1 ) 2 , B= a b + c x ¯ k + 1 , C= a 1 b 1 + c 1 y ¯ k + 1 and D= a 1 c 1 x ¯ y ¯ k ( b 1 + c 1 y ¯ k + 1 ) 2 .

Theorem 8 Let {( x n , y n )} be a positive solution of system (2), then for every m1, the following results hold.

(i) 0 x n { ( a b ) m + 1 ( a 1 b 1 ) m y k if  n = ( 2 k + 2 ) m + 1 , ( a b ) m + 1 ( a 1 b 1 ) m y 1 k if  n = ( 2 k + 2 ) m + 2 , ( a b ) m + 1 ( a 1 b 1 ) m y 1 if  n = ( 2 k + 2 ) m + k , ( a b ) m + 1 ( a 1 b 1 ) m y 0 if  n = ( 2 k + 2 ) m + ( k + 1 ) , ( a a 1 b b 1 ) m + 1 x k if  n = ( 2 k + 2 ) m + ( k + 2 ) , ( a a 1 b b 1 ) m + 1 x 1 k if  n = ( 2 k + 2 ) m + ( k + 3 ) , ( a a 1 b b 1 ) m + 1 x 1 if  n = ( 2 k + 2 ) m + ( 2 k + 1 ) , ( a a 1 b b 1 ) m + 1 x 0 if  n = ( 2 k + 2 ) m + ( 2 k + 2 ) . (ii) 0 y n { ( a b ) m ( a 1 b 1 ) m + 1 x k if  n = ( 2 k + 2 ) m + 1 , ( a b ) m ( a 1 b 1 ) m + 1 x 1 k if  n = ( 2 k + 2 ) m + 2 , ( a b ) m ( a 1 b 1 ) m + 1 x 1 if  n = ( 2 k + 2 ) m + k , ( a b ) m ( a 1 b 1 ) m + 1 x 0 if  n = ( 2 k + 2 ) m + ( k + 1 ) , ( a a 1 b b 1 ) m + 1 y k if  n = ( 2 k + 2 ) m + ( k + 2 ) , ( a a 1 b b 1 ) m + 1 y 1 k if  n = ( 2 k + 2 ) m + ( k + 3 ) , ( a a 1 b b 1 ) m + 1 y 1 if  n = ( 2 k + 2 ) m + ( 2 k + 1 ) , ( a a 1 b b 1 ) m + 1 y 0 if  n = ( 2 k + 2 ) m + ( 2 k + 2 ) .

Lemma 3 Let a a 1 b b 1 <1, then every solution { x n , y n } n = k of system (2) is bounded.

Proof Assume that

λ 1 =max { b 1 a 1 y k , b 1 a 1 y k + 1 , , b 1 a 1 y 0 , x k , x k + 1 , , x 0 } ,

and

λ 2 =max { b a x k , b a x k + 1 , , b a x 0 , y k , y k + 1 , , y 0 } .

Then from Theorem 8 one can easily see that 0 x n < λ 1 and 0 y n < λ 2 for all n=0,1, . □

Theorem 9 The equilibrium point (0,0) of equation (2) is locally asymptotically stable.

Proof The linearized system of (2) about the equilibrium point (0,0) is given by

X n + 1 = F J (0,0) X n ,

where

X n = ( x n x n 1 x n k y n y n 1 y n k )

and

H= F J (0,0)= ( 0 0 0 0 0 0 0 a b 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 a 1 b 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 ) .

Let λ 1 , λ 2 ,, λ 2 k + 2 denote the 2k+2 eigenvalues of matrix E. Let D=diag( d 1 , d 2 ,, d 2 k + 2 ) be a diagonal matrix, where d 1 = d k + 2 =1, d 1 + m = d k + 2 + m =1mϵ, 1mk and

ϵ=min { 1 k , 1 k ( 1 a b ) , 1 k ( 1 a 1 b 1 ) } .

Clearly, D is invertible. Computing DH D 1 , we obtain

DH D 1 = ( 0 0 0 0 0 0 0 a b d 1 d 2 k + 2 1 d 2 d 1 1 0 0 0 0 0 0 0 0 0 d k + 1 d k 1 0 0 0 0 0 0 0 0 a 1 b 1 d k + 2 d k + 1 1 0 0 0 0 0 0 0 0 d k + 3 d k + 2 1 0 0 0 0 0 0 0 0 0 d 2 k + 2 d 2 k + 1 1 0 ) .

Next, we have the following two inequalities:

0< d k + 1 < d k << d 2 ,0< d 2 k + 2 < d 2 k + 1 << d k + 3 ,

which implies that

d 2 d 1 1 <1, d 3 d 2 1 <1,, d k + 1 d k 1 <1,

and

d k + 3 d k + 2 1 <1, d k + 4 d k + 3 1 <1,, d 2 k + 2 d 2 k + 1 1 <1.

Furthermore,

a b d 1 d 2 k + 2 1 = a b d 2 k + 2 1 = a b 1 1 k ϵ <1,

and

a 1 b 1 d k + 2 d k + 1 1 = a 1 b 1 d k + 1 1 = a 1 b 1 1 1 k ϵ <1.

Now H has the same eigenvalues as DH D 1 , we obtain that

max 1 m 2 k + 2 | λ m | = D H D 1 = max { d 2 d 1 1 , , d k + 1 d k 1 , d k + 3 d k + 2 1 , , d 2 k + 2 d 2 k + 1 1 , a b d 1 d 2 k + 2 1 , a 1 b 1 d k + 2 d k + 1 1 } < 1 .

Hence, the equilibrium point (0,0) of system (2) is locally asymptotically stable. □

Theorem 10 Let a<b and a 1 < b 1 , then the equilibrium point (0,0) of system (2) is globally asymptotically stable.

Proof Assume that a<b and a 1 < b 1 . Then from Theorem 9 the equilibrium point (0,0) of system (2) is locally asymptotically stable. Moreover, from Lemma 3 every positive solution ( x n , y n ) is bounded, i.e., 0 x n μ and 0 y n ν for all n=0,1,2, , where μ=max{ x k , x k + 1 ,, x 0 } and ν=max{ y k , y k + 1 ,, y 0 }. Now, it is sufficient to prove that ( x n , y n ) is decreasing. From system (2) one has

x n + 1 = a y n k b + c i = 0 k x n i a y n k b < y n k .

This implies that x ( 2 k + 2 ) n + 1 < y ( 2 k + 2 ) n k and x ( 2 k + 2 ) n + ( 2 k + 3 ) < y ( 2 k + 2 ) n + ( k + 2 ) .

y n + 1 = a 1 x n k b 1 + c 1 i = 0 k y n i a x n 4 b < x n k .

This implies that

y ( 2 k + 2 ) n + 1 < x ( 2 k + 2 ) n k and y ( 2 k + 2 ) n + ( 2 k + 3 ) < x ( 2 k + 2 ) n + ( k + 2 ) .

Hence, x ( 2 k + 2 ) n + ( 2 k + 3 ) < y ( 2 k + 2 ) n + ( k + 2 ) < x ( 2 k + 2 ) n + 1 and y ( 2 k + 2 ) n + ( 2 k + 3 ) < x ( 2 k + 2 ) n + ( k + 2 ) < y ( 2 k + 2 ) n + 1 . Hence, the subsequences

{ x ( 2 k + 2 ) n + 1 },{ x ( 2 k + 2 ) n + 2 },,{ x ( 2 k + 2 ) n + ( 2 k + 2 ) }

and

{ y ( 2 k + 2 ) n + 1 },{ y ( 2 k + 2 ) n + 2 },,{ y ( 2 k + 2 ) n + ( 2 k + 2 ) }

are decreasing. Therefore the sequences { x n } and { y n } are decreasing. Hence, lim n x n = lim n y n =0. □

3.1 Rate of convergence

Assume that lim n x n = x ¯ and lim n y n = y ¯ . First we will find a system of limiting equations for system (2). The error terms are given as

x n + 1 x ¯ = i = 0 k A i ( x n i x ¯ ) + i = 0 k B i ( y n i y ¯ ) , y n + 1 y ¯ = i = 0 k C i ( x n i x ¯ ) + i = 0 k D i ( y n i y ¯ ) .

Set e n 1 = x n x ¯ and e n 2 = y n y ¯ , then one has

e n + 1 1 = i = 0 k A i e n i 1 + i = 0 k B i e n i 2 , e n + 1 2 = i = 0 k C i e n i 1 + i = 0 k D i e n i 2 ,

where

A 0 = a c y ¯ x n 1 x n 2 x n k ( b + c i = 0 k x n i ) ( b + c x ¯ k + 1 ) , A 1 = a c x ¯ y ¯ x n 2 x n 3 x n k ( b + c i = 0 k x n i ) ( b + c x ¯ k + 1 ) , , A k 1 = a c x ¯ k 1 y ¯ x n k ( b + c i = 0 k x n i ) ( b + c x ¯ k + 1 ) , A k = a c x ¯ k y ¯ ( b + c i = 0 k x n i ) ( b + c x ¯ k + 1 ) ,

B i =0 for i{0,1,,k1}, B k = a b + c i = 0 k x n i , C i =0 for i{0,1,,k1},

C k = a 1 b 1 + c 1 i = 0 k y n i , D 0 = a 1 c 1 x ¯ y n 1 y n 2 y n k ( b 1 + c 1 i = 0 k y n i ) ( b 1 + c 1 y ¯ k + 1 ) , D 1 = a 1 c 1 x ¯ y ¯ y n 2 y n 3 y n k ( b 1 + c 1 i = 0 k y n i ) ( b 1 + c 1 y ¯ k + 1 ) , , D k 1 = a 1 c 1 x ¯ y ¯ k 1 y n k ( b 1 + c 1 i = 0 k y n i ) ( b 1 + c 1 y ¯ k + 1 ) ,

and D k = a 1 c 1 y ¯ y ¯ k ( b 1 + c 1 i = 0 k y n i ) ( b 1 + c 1 y ¯ k + 1 ) . Taking the limits, we obtain lim n A i = a c y ¯ x ¯ k ( b + c x ¯ k + 1 ) 2 for i{0,1,,k}, lim n B i =0 for i{0,1,,k1}, lim n B k = a b + c x ¯ k + 1 , lim n C i =0 for i{0,1,,k1}, lim n C k = a 1 b 1 + c 1 y ¯ k + 1 , lim n D i = a 1 c 1 x ¯ y ¯ k ( b 1 + c 1 y ¯ k + 1 ) 2 for i{0,1,,k}, So, the limiting system of error terms can be written as

E n + 1 =K E n ,

where

E n = ( e n 1 e n 1 1 e n k 1 e n 2 e n 1 2 e n k 2 )

and

K= ( 0 0 0 0 0 0 0 a b 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 a 1 b 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 ) ,

which is similar to the linearized system of (2) about the equilibrium point ( x ¯ , y ¯ )=(0,0). Using proposition (1), one has the following result.

Theorem 11 Assume that {( x n , y n )} is a positive solution of system (2) such that lim n x n = x ¯ , and lim n y n = y ¯ , where ( x ¯ , y ¯ )=(0,0). Then the error vector E n of every solution of (2) satisfies both of the following asymptotic relations:

lim n ( E n ) 1 n =|λ F J ( x ¯ , y ¯ )|, lim n E n + 1 E n =|λ F J ( x ¯ , y ¯ )|,

where λ F J ( x ¯ , y ¯ ) are the characteristic roots of the Jacobian matrix F J ( x ¯ , y ¯ ) about ( x ¯ , y ¯ )=(0,0).

4 Examples

In order to verify our theoretical results, we consider some interesting numerical examples in this section. These examples show that the equilibrium point (0,0) of both systems (1) and (2) is globally asymptotically stable.

Example 1 Consider system (1) with initial conditions x 8 =1.9, x 7 =1.7, x 6 =2.5, x 5 =0.9, x 4 =1.5, x 3 =10.4, x 2 =6.9, x 1 =0.6, x 0 =2.9, y 8 =2.8, y 7 =1.6, y 6 =1.8, y 5 =2.6, y 4 =2.8, y 3 =2.8, y 2 =3.5, y 1 =2.1, y 0 =1.6. Moreover, choose the parameters α=15.5, β=17, γ=27, α 1 =11.2, β 1 =12, γ 1 =23. Then system (1) can be written as

x n + 1 = 15.5 x n 8 17 + 27 i = 0 8 y n i , y n + 1 = 11.2 y n 8 12 + 23 i = 0 8 x n i ,n=0,1,,
(14)

with initial conditions x 8 =1.9, x 7 =1.7, x 6 =2.5, x 5 =0.9, x 4 =1.5, x 3 =10.4, x 2 =6.9, x 1 =0.6, x 0 =2.9, y 8 =2.8, y 7 =1.6, y 6 =1.8, y 5 =2.6, y 4 =2.8, y 3 =2.8, y 2 =3.85, y 1 =2.1, y 0 =1.6. Moreover, in Figure 1, the plot of x n is shown in Figure 1a, the plot of y n is shown in Figure 1b, and an attractor of system (14) is shown in Figure 1c.

Figure 1
figure 1

Plots for system ( 14 ).

Example 2 Consider system (1) with initial conditions x 15 =0.9, x 14 =1.9, x 13 =0.8, x 12 =0.3, x 11 =1.7, x 10 =2.9, x 9 =1.9, x 8 =2.9, x 7 =0.3, x 6 =3.5, x 5 =1.4, x 4 =1.5, x 3 =13.4, x 2 =17.9, x 1 =11.7, x 0 =9.8, y 15 =2.9, y 14 =3.8, y 13 =0.9, y 12 =0.8, y 11 =1.1, y 10 =3.9, y 9 =3.6, y 8 =1.9, y 7 =3.7, y 6 =2.9, y 5 =1.5, y 4 =0.9, y 3 =1.8, y 2 =3.9, y 1 =6.7, y 0 =2.9. Moreover, choose the parameters α=120, β=125, γ=11.5, α 1 =140, β 1 =145, γ 1 =14.5. Then system (1) can be written as

x n + 1 = 120 x n 15 125 + 11.5 i = 0 15 y n i , y n + 1 = 140 y n 15 145 + 14.5 i = 0 15 x n i ,n=0,1,,
(15)

with initial conditions x 15 =0.9, x 14 =1.9, x 13 =0.8, x 12 =0.3, x 11 =1.7, x 10 =2.9, x 9 =1.9, x 8 =2.9, x 7 =0.3, x 6 =3.5, x 5 =1.4, x 4 =1.5, x 3 =13.4, x 2 =17.9, x 1 =11.7, x 0 =9.8, y 15 =2.9, y 14 =3.8, y 13 =0.9, y 12 =0.8, y 11 =1.1, y 10 =3.9, y 9 =3.6, y 8 =1.9, y 7 =3.7, y 6 =2.9, y 5 =1.5, y 4 =0.9, y 3 =1.8, y 2 =3.9, y 1 =6.7, y 0 =2.9. Moreover, in Figure 2, the plot of x n is shown in Figure 2a, the plot of y n is shown in Figure 2b, and an attractor of system (15) is shown in Figure 2c.

Figure 2
figure 2

Plots for system ( 15 ).

Example 3 Consider system (2) with initial conditions x 6 =0.9, x 5 =2.9, x 4 =1.6, x 3 =9.4, x 2 =6.9, x 1 =2.8, x 0 =1.9, y 6 =1.9, y 5 =2.7, y 4 =2.9, y 3 =2.8, y 2 =3.9, y 1 =2.4, y 0 =0.8. Moreover, choose the parameters a=115, b=117, c=27, a 1 =111, b 1 =112, c 1 =23. Then system (2) can be written as

x n + 1 = 115 y n 6 117 + 27 i = 0 6 x n i , y n + 1 = 111 x n 6 112 + 23 i = 0 6 y n i ,n=0,1,,
(16)

with initial conditions x 6 =0.9, x 5 =2.9, x 4 =1.6, x 3 =9.4, x 2 =6.9, x 1 =2.8, x 0 =1.9, y 6 =1.9, y 5 =2.7, y 4 =2.9, y 3 =2.8, y 2 =3.9, y 1 =2.4, y 0 =0.8. Moreover, in Figure 3, the plot of x n is shown in Figure 3a, the plot of y n is shown in Figure 3b, and an attractor of system (16) is shown in Figure 3c.

Figure 3
figure 3

Plots for system ( 16 ).

Example 4 Consider system (2) with initial conditions x 12 =7.1, x 11 =2.8, x 10 =19.1, x 9 =20.9, x 8 =9.9, x 7 =13.4, x 6 =1.9, x 5 =10.9, x 4 =12.9, x 3 =0.9, x 2 =1.9, x 1 =15.8, x 0 =4.9, y 12 =10.9, y 11 =1.1, y 10 =3.7, y 9 =0.6, y 8 =1.7, y 7 =1.7, y 6 =1.9, y 5 =1.9, y 4 =1.9, y 3 =2.8, y 2 =17.9, y 1 =27.4, y 0 =0.04. Moreover, choose the parameters a=121, b=125, c=9, a 1 =129, b 1 =130, c 1 =7. Then system (2) can be written as

x n + 1 = 121 y n 12 125 + 9 i = 0 12 x n i , y n + 1 = 129 x n 12 130 + 7 i = 0 12 y n i ,n=0,1,,
(17)

with initial conditions x 12 =7.1, x 11 =2.8, x 10 =19.1, x 9 =20.9, x 8 =9.9, x 7 =13.4, x 6 =1.9, x 5 =10.9, x 4 =12.9, x 3 =0.9, x 2 =1.9, x 1 =15.8, x 0 =4.9, y 12 =10.9, y 11 =1.1, y 10 =3.7, y 9 =0.6, y 8 =1.7, y 7 =1.7, y 6 =1.9, y 5 =1.9, y 4 =1.9, y 3 =2.8, y 2 =17.9, y 1 =27.4, y 0 =0.04. Moreover, in Figure 4, the plot of x n is shown in Figure 4a, the plot of y n is shown in Figure 4b, and an attractor of system (17) is shown in Figure 4c.

Figure 4
figure 4

Plots for system ( 17 ).

Conclusion

This work is a natural extension of [11, 20, 21]. In the paper, we have investigated the qualitative behavior of (2k+2)-dimensional discrete dynamical systems. Each system has only one equilibrium point which is stable under some restriction to parameters. The linearization method is used to show that equilibrium point (0,0) is locally asymptotically stable. The main objective of dynamical systems theory is to predict the global behavior of a system based on the knowledge of its present state. An approach to this problem consists of determining the possible global behaviors of the system and determining which initial conditions lead to these long-term behaviors. In case of higher-order dynamical systems, it is crucial to discuss global behavior of the system. Some powerful tools such as semiconjugacy and weak contraction cannot be used to analyze global behavior of systems (1) and (2). In the paper, we prove the global asymptotic stability of equilibrium point (0,0) by using simple techniques. We have carried out a systematical local and global stability analysis of both systems. The most important finding here is that the unique equilibrium point (0,0) can be a global asymptotic attractor for systems (1) and (2). Moreover, we have determined the rate of convergence of a solution that converges to the equilibrium point (0,0) of systems (1) and (2). Some numerical examples are provided to support our theoretical results. These examples are experimental verifications of theoretical discussions.