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On the behavior of solutions of the system of rational difference equations x n + 1 = x n - 1 y n x n - 1 - 1 , y n + 1 = y n - 1 x n y n - 1 - 1 , z n + 1 = 1 y n z n

Abstract

In this article, we investigate the solutions of the system of difference equations x n + 1 = x n - 1 y n x n - 1 - 1 , y n + 1 = y n - 1 x n y n - 1 - 1 , z n + 1 = 1 y n z n where x 0, x -1, y 0, y -1, z 0, z -1 real numbers such that y 0 x -1 ≠ 1, x 0 y -1 ≠ 1 and y 0 z 0 ≠ 0.

1. Introduction

In [1], Kurbanli et al. studied the behavior of positive solutions of the system of rational difference equations

x n + 1 = x n - 1 y n x n - 1 + 1 , y n + 1 = y n - 1 x n y n - 1 + 1 .

In [2], Cinar studied the solutions of the systems of difference equations

x n + 1 = 1 y n , y n + 1 = y n x n - 1 y n - 1 .

In [3], Kurbanli, studied the behavior of solutions of the system of rational difference equations

x n + 1 = x n - 1 y n x n - 1 - 1 , y n + 1 = y n - 1 x n y n - 1 - 1 , z n + 1 = z n - 1 y n z n - 1 - 1 .

In [4], Papaschinnopoulos and Schinas proved the boundedness, persistence, the oscillatory behavior, and the asymptotic behavior of the positive solutions of the system of difference equations

x n + 1 = i = 0 k A i y n - i p i , y n + 1 = i = 0 k B i x n - i q i

In [5], Clark and Kulenović investigate the global stability properties and asymptotic behavior of solutions of the system of difference equations

x n + 1 = x n a + c y n , y n + 1 = y n b + d x n .

In [6], Camouzis and Papaschinnopoulos studied the global asymptotic behavior of positive solutions of the system of rational difference equations

x n + 1 =1+ x n y n - m , y n + 1 =1+ y n x n - m .

In [7], Kulenović and Nurkanović studied the global asymptotic behavior of solutions of the system of difference equations

x n + 1 = a + x n b + y n , y n + 1 = c + y n d + z n , z n + 1 = e + z n f + x n .

In [8], Özban studied the positive solutions of the system of rational difference equations

x n + 1 = 1 y n - k , y n + 1 = y n x n - m y n - m - k .

In [9], Zhang et al. investigated the behavior of the positive solutions of the system of the difference equations

x n = A + 1 y n - p , y n = A + y n - 1 x n - r y n - s .

In [10], Yalcinkaya studied the global asymptotic stability of the system of difference equations

z n + 1 = t n z n - 1 + a t n + z n - 1 , t n + 1 = z n t n - 1 + a z n + t n - 1

In [11], Irićanin and Stević studied the positive solutions of the system of difference equations

x n + 1 ( 1 ) = 1 + x n ( 2 ) x n - 1 ( 3 ) , x n + 1 ( 2 ) = 1 + x n ( 3 ) x n - 1 ( 4 ) , , x n + 1 ( k ) = 1 + x n ( 1 ) x n - 1 ( 2 ) , x n + 1 ( 1 ) = 1 + x n ( 2 ) + x n - 1 ( 3 ) x n - 2 ( 4 ) , x n + 1 ( 2 ) = 1 + x n ( 3 ) + x n - 1 ( 4 ) x n - 2 ( 5 ) , , x n + 1 ( k ) = 1 + x n ( 1 ) + x n - 1 ( 2 ) x n - 2 ( 3 )

Although difference equations are very simple in form, it is extremely difficult to understand throughly the global behavior of their solutions, for example, see Refs. [1234].

In this article, we investigate the behavior of the solutions of the difference equation system

x n + 1 = x n - 1 y n x n - 1 - 1 , y n + 1 = y n - 1 x n y n - 1 - 1 , z n + 1 = 1 y n z n
(1.1)

where x 0, x -1, y 0, y -1, z 0, z -1 real numbers such that y 0 x -1 ≠ 1, x 0 y -1 ≠ 1 and y 0 z 0 ≠ 0.

2. Main results

Theorem 1. Let y 0 = a, y -1 = b, x 0 = c, x -1 = d, z 0 = e, z -1 = f be real numbers such that y 0 x -1 ≠ 1, x 0 y -1 ≠ 1 and y 0 z 0 ≠ 0. Let {x n , y n , z n } be a solution of the system (1.1). Then all solutions of (1.1) are

x n = d ( a d - 1 ) n , n - - - o d d c ( c b - 1 ) n , n - - - e v e n
(1.2)
y n = b ( c b - 1 ) n , n - - - o d d a ( a d - 1 ) n , n - - - e v e n
(1.3)
z n = b n - 1 a n e a d - 1 c d - 1 i = 1 k i , n - - - o d d a n e ( a d - 1 ) i = 1 k ( i - 1 ) ( c b - 1 ) i = 1 k i b n , n - - - e v e n
(1.4)

Proof. For n = 0, 1, 2, 3, we have

x 1 = x - 1 y 0 x - 1 - 1 = d a d - 1 , y 1 = y - 1 x 0 y - 1 - 1 = b c b - 1 , z 1 = 1 y 0 z 0 = 1 a e , x 2 = x 0 y 1 x 0 - 1 = c b c b - 1 c - 1 = c ( c b - 1 ) , y 2 = y 0 x 1 y 0 - 1 = a d a d - 1 a - 1 = a ( a d - 1 ) z 2 = 1 y 1 z 1 = 1 b c b - 1 1 a e = ( c b - 1 ) a e b , x 3 = x 1 y 2 x 1 - 1 = d a d - 1 a a d - 1 d a d - 1 - 1 = d ( a d - 1 ) 2 , y 3 = y 1 x 2 y 1 - 1 = b c b - 1 c c b - 1 b c b - 1 - 1 = b ( c b - 1 ) 2 , z 3 = 1 y 2 z 2 = 1 a ( a d - 1 ) ( c b - 1 ) a e b = b a 2 e ( a d - 1 ) ( c b - 1 )

for n = k, assume that

x 2 k - 1 = x 2 k - 3 y 2 k - 2 x 2 k - 3 - 1 = d ( a d - 1 ) k , x 2 k = x 2 k - 2 y 2 k - 1 x 2 k - 2 - 1 = c ( c b - 1 ) k , y 2 k - 1 = y 2 k - 3 x 2 k - 2 y 2 k - 3 - 1 = b ( c b - 1 ) k , y 2 k = y 2 k - 2 x 2 k - 1 y 2 k - 2 - 1 = a ( a d - 1 ) k

and

z 2 k - 1 = b k - 1 a k e [ ( a d - 1 ) ( c b - 1 ) ] i = 1 k i , z 2 k = a k e ( a d - 1 ) i = 1 k ( i - 1 ) ( c b - 1 ) i = 1 k i b k

are true. Then, for n = k + 1 we will show that (1.2), (1.3), and (1.4) are true. From (1.1), we have

x 2 k + 1 = x 2 k - 1 y 2 k x 2 k - 1 - 1 = d ( a d - 1 ) k a a d - 1 k d ( a d - 1 ) k - 1 = d ( a d - 1 ) k + 1 , y 2 k + 1 = y 2 k - 1 x 2 k y 2 k - 1 - 1 = b ( c b - 1 ) k c c b - 1 k b ( c b - 1 ) k - 1 = b ( c b - 1 ) k + 1 .

Also, similarly from (1.1), we have

z 2 k + 1 = 1 y 2 k z 2 k = 1 a a d - 1 k a k e ( a d - 1 ) i = 1 k i - 1 ( c b - 1 ) i = 1 k i b k = b k a k + 1 e ( a d - 1 ) i = 1 k i ( c b - 1 ) i = 1 k i .

Also, we have

x 2 k + 2 = x 2 k y 2 k + 1 x 2 k - 1 = c c b - 1 k b ( c b - 1 ) k + 1 c ( c b - 1 ) k - 1 = c c b - 1 k b ( c b - 1 ) c - 1 = c ( c b - 1 ) k + 1 , y 2 k + 2 = y 2 k x 2 k + 1 y 2 k - 1 = a a d - 1 k d ( a d - 1 ) k + 1 a ( a d - 1 ) k - 1 = a a d - 1 k d ( a d - 1 ) a - 1 = a ( a d - 1 ) k + 1

and

z 2 k + 2 = 1 y 2 k + 1 z 2 k + 1 = 1 b ( c b - 1 ) k + 1 b k a k + 1 e ( a d - 1 ) i = 1 k i ( c b - 1 ) i = 1 k i = a k + 1 e ( a d - 1 ) i = 1 k i ( c b - 1 ) i = 1 k + 1 i b k + 1 = a k + 1 e ( a d - 1 ) i = 1 k + 1 ( i - 1 ) ( c b - 1 ) i = 1 k + 1 i b k + 1 .

Corollary 1. Let {x n , y n , z n } be a solution of the system (1.1). Let a, b, c, d, e, f be real numbers such that ad ≠ 1, cb ≠ 1, ae ≠ 0 and b ≠ 0. Also, if ad, cb (1, 2) and b > a then we have

lim n x 2 n - 1 = lim n y 2 n - 1 = lim n z 2 n - 1 =

and

lim n x 2 n = lim n y 2 n = lim n z 2 n =0.

Proof. From ad, cb (1, 2) and b > a we have 0 < ad -1 < 1 and 0 < cb - 1 < 1.

Hence, we obtain

lim n x 2 n - 1 = lim n d ( a d - 1 ) n = d lim n 1 ( a d - 1 ) n = d . = - , d < 0 + , d > 0 , lim n y 2 n - 1 = lim n b ( c b - 1 ) n = b lim n 1 ( c b - 1 ) n = b . = - , b < 0 + , b > 0

and

lim n z 2 n - 1 = lim n b n - 1 a n e [ ( a d - 1 ) ( c b - 1 ) ] i = 1 k i = 1 e . = - , e < 0 + , e > 0

Similarly, from ad, cb (1, 2) and b > a, we have 0 < ad - 1 < 1 and 0 < cb - 1 < 1.

Hence, we obtain

lim n x 2 n = lim n c ( c d - 1 ) n = c lim n ( c d - 1 ) n = c . 0 = 0 , lim n y 2 n = lim n a ( a f - 1 ) n = a lim n ( a f - 1 ) n = a . 0 = 0 .

and

lim n z 2 n = lim n a n e ( a d - 1 ) i = 1 k i - 1 ( c b - 1 ) i = 1 k i b n = 0 . e . 0 = 0 .

Corollary 2. Let {x n , y n , z n } be a solution of the system (1.1). Let a, b, c, d, e, f be real numbers such that ad ≠ 1, cb ≠ 1, ae ≠ 0 and b ≠ 0. If a = b and cb = ad = 2 then we have

lim n x 2 n - 1 = d , lim n y 2 n - 1 = b , lim n z 2 n - 1 = 1 a e

and

lim n x 2 n = c , lim n y 2 n = a , lim n z 2 n = e .

Proof. From a = b and cb = ad = 2 then we have, cb - 1 = ad - 1 = 1. Hence, we have

lim n ( c b - 1 ) n =1

and

lim n ( a d - 1 ) n =1.

Also, we have

lim n x 2 n - 1 = lim n d ( a d - 1 ) n = d lim n 1 ( a d - 1 ) n = d . 1 = d , lim n y 2 n - 1 = lim n b ( c b - 1 ) n = b lim n 1 ( c b - 1 ) n = b . 1 = b

and

lim n z 2 n - 1 = lim n b n - 1 a n e [ ( a d - 1 ) ( c b - 1 ) ] i = 1 K i = lim n 1 a e b n - 1 a n - 1 [ ( a d - 1 ) ( c b - 1 ) ] i = 1 k i = 1 a e .

Similarly, we have

lim n x 2 n = lim n c ( c b - 1 ) n = c lim n ( c b - 1 ) n = c . 1 = c , lim n y 2 n = lim n a ( a d - 1 ) n = a lim n ( a d - 1 ) n = a . 1 = a .

and

lim n z 2 n = lim n a n e ( a d - 1 ) i = 1 k ( i - 1 ) ( c b - 1 ) i = 1 k i b n =1.e=e.

Corollary 3. Let {x n , y n , z n } be a solution of the system (1.1). Let a, b, c, d, e, f be real numbers such that ad ≠ 1, cb ≠ 1, ae ≠ 0 and b ≠ 0. Also, if 0 < a, b, c, d, e, f < 1 then we have

lim n x 2 n = lim n y 2 n = lim n z 2 n =0

and

lim n x 2 n - 1 = lim n y 2 n - 1 = lim n z 2 n - 1 =.

Proof. From 0 < a, b, c, d, e, f < 1 we have -1 < ad - 1 < 0 and - 1 < cb - 1 < 0. Hence, we obtain

lim n x 2 n = lim n c ( b c - 1 ) n = c lim n ( b c - 1 ) n = c . 0 = 0 , lim n y 2 n = lim n a ( a d - 1 ) n = a lim n ( a d - 1 ) n = a . 0 = 0

and

lim n z 2 n = lim n a n e ( a d - 1 ) i = 1 k ( i - 1 ) ( c b - 1 ) i = 1 k i b n = e . 0 = 0 .

Similarly, we have

lim n x 2 n - 1 = lim n d ( a d - 1 ) n = d lim n 1 ( a d - 1 ) n = d lim n 1 ( a d - 1 ) n = d . = - , n - o d d + , n - e v e n , lim n y 2 n - 1 = lim n b ( b c - 1 ) n = b lim n 1 ( b c - 1 ) n = b . = - , n - o d d + , n - e v e n .

and

lim n z 2 n - 1 = lim n b n - 1 a n e [ ( a d - 1 ) ( c b - 1 ) ] i = 1 k i =+.

Corollary 4. Let {x n , y n , z n } be a solution of the system (1.1). Let a, b, c, d, e, f be real numbers such that ad ≠ 1, cb ≠ 1, ae ≠ 0, and b ≠ 0. Also, if 0 < a, b, c, d, e, f < 1 then we have

lim n x 2 n y 2 n - 1 = c b , lim n x 2 n - 1 y 2 n = a d

and

lim n z 2 n - 1 z 2 n =.

Proof. The proof is clear from Theorem 1. □

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Kurbanli, A.S. On the behavior of solutions of the system of rational difference equations x n + 1 = x n - 1 y n x n - 1 - 1 , y n + 1 = y n - 1 x n y n - 1 - 1 , z n + 1 = 1 y n z n . Adv Differ Equ 2011, 40 (2011). https://doi.org/10.1186/1687-1847-2011-40

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Keywords

  • Differential Equation
  • Real Number
  • Partial Differential Equation
  • Ordinary Differential Equation
  • Functional Analysis