On the global attractivity of positive solutions of a rational difference equation

Selçuk J. Appl. Math. Selçuk Journal of Vol. 9. No. 2. pp. 3 8, 2008 Applied Mathematics

On The Global Attractivity of Positive Solutions of A Rational Dif-ference Equation

·

Ibrahim Yalcinkaya

Selcuk University, Education Faculty, Mathematics Department, 42090, Konya, Turkey e-mail:iyalcinkaya1708@ yaho o.com

Received: October 07, 2007

Abstract. We study the positive solutions and attractivity of the di¤erence equation xn+1= xn (2k+1)=(1 + xn kxn (2k+1)) where x i(for i = 0; 1; :::; 2k +

1) are nonnegative real numbers.

Key words: Di¤erence Equation; Positive Solutions; Equilibrium Point; Global Attractivity.

2000 Mathematics Subject Classi…cation. 39A10 1. Introduction

Recently there has been a great interest in studying the attractivity, the solu-tions and the periodic nature of non-linear di¤erence equasolu-tions. For example [1-15].

In [3] Cinar studied the positive solutions of the di¤erence equation xn+1=

xn 1

1 + xnxn 1

where x 1and x0 are positive real numbers.

Moreover, in [14] Stevic gave some additional information about behavior of the solutions of this di¤erence equation with the initial values x 1and x0which are

real numbers.

In [11], [12], [13] we solved the following three problems

xn+1= xn 3 1 + xn 1 ; xn+1= xn 5 1 + xn 2 and xn+1= xn 5 1 + xn 1xn 3 for n = 0; 1; 2; :::: where the positive initial values.

Our aim in this paper is to investigate the global attractivity of the positive solutions of the di¤erence equation

(1.1) xn+1=

xn (2k+1)

1 + xn kxn (2k+1)

where

(1.2) x i (f or i = 0; 1; :::; 2k + 1) are nonnegative real numbers.

Similar to the references in this paper, we de…ne Eq.(1.1) with (1.2) and inves-tigate the solutions of this di¤erence equation. For k = 0; Eq.(1.1) turns into Eq.(1.1) in Cinar’s paper [3]. This paper is motivated by [3] and [14].

2. Main Results

Theorem 1. _{Suppose that (1:2) holds and let fx}ng1_{n= 2k 1} be a solution of

Eq.(1:1) with x i= ai(for i = 0; 1; :::; 2k + 1). Then for n = 0; 1; ::: all solutions

of Eq.(1:1) are the following:

x(2k+2)n+1= a2k+1 n Y i=1 (1+2ia2k+1ak) (1+a2k+1ak) n Y i=1 (1+(2i+1)a2k+1ak) ; x(2k+2)n+2= a2k n Y i=1 (1+2ia2kak 1) (1+a2kak 1) n Y i=1 (1+(2i+1)a2kak 1) ; :::; x(2k+2)n+k= ak+2 n Y i=1 (1+2iak+2a1) (1+ak+2a1) n Y i=1 (1+(2i+1)ak+2a1) ; x(2k+2)n+k+1= ak+1 n Y i=1 (1+2iak+1a0) (1+ak+1a0) n Y i=1 (1+(2i+1)ak+1a0) ; x(2k+2)n+k+2= ak(1+aka2k+1) n Y i=1 ((2i+1)aka2k+1+1) (2aka2k+1+1) n Y i=1 ((2i+2)aka2k+1+1) ; :::; x(2k+2)n+2k+2= a0(1+ak+1a0) n Y i=1 ((2i+1)ak+1a0+1) (2a0ak+1+1) n Y i=1 ((2i+2)a0ak+1+1) :

Proof. x1; x2; :::; x(2k+2)are clear from Eq.(1.1). From our assumption (1.2) all

solutions of Eq.(1.1) are positive. Also, for n = 1 the result holds. Now assume that n > 1 and our assumption holds for (n 1): We shall show that the result holds for n. From our assumption for (n 1); we have

x(2k+2)n (2k+1)= a2k+1 n 1 Y i=1 (1+2ia2k+1ak) (1+a2k+1ak) n_Y1 i=1 (1+(2i+1)a2k+1ak) ; :::; x(2k+2)n k= ak(1+aka2k+1) n 1 Y i=1 ((2i+1)aka2k+1+1) (2aka2k+1+1) n_Y1 i=1 ((2i+2)aka2k+1+1) ; :::; x(2k+2)n= a0(1+ak+1a0) n_Y1 i=1 ((2i+1)ak+1a0+1) (2a0ak+1+1) nY1 i=1 ((2i+2)a0ak+1+1) :

Then, from Eq.(1.1) and the above equality, we have x(2k+2)n+1= x(2k+2)n (2k+1) 1+x(2k+2)n kx(2k+2)n (2k+1) = a2k+1 n 1 Y i=1 (1+2ia2k+1ak) (1+a2k+1ak) n 1 Y i=1 (1+(2i+1)a2k+1ak) 1+ ak(1+aka2k+1) n 1 Y i=1 ((2i+1)aka2k+1+1) (2aka2k+1+1) n_Y1 i=1 ((2i+2)aka2k+1+1) a2k+1 n 1 Y i=1 (1+2ia2k+1ak) (1+a2k+1ak) n_Y1 i=1 (1+(2i+1)a2k+1ak) = a2k+1 n 1 Y i=1 (1+2ia2k+1ak) (1+a2k+1ak) n 1 Y i=1 (1+(2i+1)a2k+1ak) 2na2k+1ak+1 (2n+1)a2k+1ak+1 = a2k+1 n Y i=1 (1+2ia2k+1ak) (1+a2k+1ak) n Y i=1 (1+(2i+1)a2k+1ak) Hence, we have x(2k+2)n+1= a2k+1 n Y i=1 (1 + 2ia2k+1ak) (1 + a2k+1ak) n Y i=1 (1 + (2i + 1)a2k+1ak)

Similarly, one can easily obtain x(2k+2)n+i (for i = 2; 3; :::; k + 1): Therefore,

x(2k+2)n+k+2= x(2k+2)n k 1+x(2k+2)n+1x(2k+2)n k = ak(1+aka2k+1) n 1 Y i=1 ((2i+1)aka2k+1+1) (2aka2k+1+1) n 1 Y i=1 ((2i+2)aka2k+1+1) 1+ a2k+1 n Y i=1 (1+2ia2k+1ak) (1+a2k+1ak) n Y i=1 (1+(2i+1)a2k+1ak) ak(1+aka2k+1) n 1 Y i=1 ((2i+1)aka2k+1+1) (2aka2k+1+1) n 1 Y i=1 ((2i+2)aka2k+1+1) = ak(1+aka2k+1) n Y i=1 ((2i+1)aka2k+1+1) (2aka2k+1+1) n Y i=1 ((2i+2)aka2k+1+1) Hence, we have x(2k+2)n+k+2= ak(1 + aka2k+1) n Y i=1 ((2i + 1)aka2k+1+ 1) (2aka2k+1+ 1) n Y i=1 ((2i + 2)aka2k+1+ 1)

Similarly, one can easily obtain x(2k+2)n+i (for i = k + 3; k + 4; :::; 2k + 2):

Therefore, they will be omitted. Thus, the proof is completed by induction. Lemma 1. Eq.(1:1) has a unique equilibrium point which is the number zero. Corollary 1. If x i > 0 (for i = 0; 1; :::; 2k + 1); then 0 xn xn 2k for all

n > 0:

Theorem 2. Every positive solution of Eq.(1.1) converges to zero.

Proof. _{Let fx}ng1n= 2k 1 be an arbitrary solution of Eq.(1.1). It is enough to

prove that the subsequences x(2k+2)n+i(for i = 1; 2; :::; 2k + 2) converge to zero

as n ! 1: Since n X i=1 1

1+(2i+1)a2k+1ak ! +1 as n ! 1 and the series

1 X i=n0 (1 i2) is conver-gent, x(2k+2)n+1= a2k+1 1+a2k+1ak n Y i=1 (1+2ia2k+1ak) n Y i=1 (1+(2i+1)a2k+1ak) = a2k+1 1+a2k+1akexp n X i=1 ln( 1+2ia2k+1ak 1+(2i+1)a2k+1ak) !

= a2k+1 1+a2k+1akexp n X i=1 ln(1 a2k+1ak 1+(2i+1)a2k+1ak) ! = a2k+1 1+a2k+1akk(n0) exp a2k+1ak n X i=n0 (_1+(2i+1)a1 2k+1ak+ ( 1 i2)) ! ! 0; as n ! 1:

Here k(ni) is a positive constant depending on ni2 N (for i = 0; 1; :::; 2k + 1):

Similarly, we obtain x(2k+2)n+i! 0; as n ! 1 (for i = 2; 3; :::; k + 1). Also, x(2k+2)n+k+2 = ak(1+aka2k+1) (2a2k+1ak+1) n Y i=1 ((2i+1)a2k+1ak+1) n Y i=1 ((2i+2)a2k+1ak+1) = ak(1+aka2k+1) 2a2k+1ak+1 exp n X i=1 ln((2i+1)a2k+1ak+1 (2i+2)a2k+1ak+1) ! = ak(1+aka2k+1) 2a2k+1ak+1 exp n X i=1 ln(1 a2k+1ak 1+(2i+2)a2k+1ak) ! = a2k+1 1+a2k+1akk(nk+1) exp a2k+1ak n X i=nk+1 (_1+(2i+2)a1 2k+1ak + ( 1 i2)) ! ! 0; as n ! 1: Similarly, we obtain x(2k+2)n+k+i! 0; as n ! 1 (for i = 3; 4; :::; k + 2) as desired.

Remark 1. Note that the di¤erence equation xn+1=

xn (2k+1)

1 + xn kxn (2k+1)

for n = 0; 1; 2; :::

where and are positive real numbers, can be reduced to Eq.(1:1) by the change of variables xn=

q yn:

Acknowledgement

I am grateful to the anonymous referees for their valuable suggestions that improved the quality of this study.

References

1. Abu-Saris R., De Vault R. (2003): On the asymptotic behaviour of xn+1=a_xn_nxn₁;

Journal of Mathematical Analysis and Applications, 280, 148 -154.

2. Berenhaut K. S., Stevic S. (2007): The global attractivity of high order rational di¤erence equation, Journal of Mathematical Analysis and Applications, 326, 940 -944. 3. Cinar C. (2004): On the positive solutions of the di¤erence equation xn+1 =

xn 1

1+xnxn 1, Applied Mathematics and Computation, 150, 21-24.

4. Cinar C. (2004): On the periodic cycle ofxn+1=a_cn_n+b_xnnx₁n, Applied Mathematics

and Computation, 150, 1-4.

5. Cinar C. (2004): On the positive solutions of the di¤erence equation xn+1 = xn 1

1+xnxn 1, Applied Mathematics and Computation, 158, 813-816.

6. Cinar C. (2004): On the positive solutions of the di¤erence equation xn+1 = axn 1

1+bxnxn 1, Applied Mathematics and Computation, 156, 587-590.

7. Devault R., Kent C., Kosmala, W. (2003): On the recursive sequence xn+1 =

p +xn k

xn ;Journal of Di¤erence Equations and Applications, 9, 721 -730.

8. El-A…… M. M. (2004): On the recursive sequencexn+1= _Bx+ x_n+Cxn+ x_nn₁1;Applied

Mathematics and Computation, 147, 617-628.

9. Hamza E. H. (2006): On the recursive sequence xn+1 = + xnxn1; Journal of

Mathematical Analysis and Applications, 322, 668 -674.

10. Peny Y. (2006): Global asymptotic stability of nonlinear di¤erence equation, Applied Mathematics and Computation, 182, 67-72.

11. Simsek D., Cinar C., Yalcinkaya I. (2006): On the recursive sequencexn+1 = xn 3

1+xn 1;Int. Journal of Contemporary Math. Sciences, 10, 475-480.

12. Simsek D., Cinar C., Karatas R., Yalcinkaya I. (2006): On the recursive sequence xn+1=_1+xxn_n5₂;Int. Journal of Pure and Applied Math., 9-12, 481-487.

13. Simsek D., Cinar C., Karatas R., Yalcinkaya, I. (2006): On the recursive sequence xn+1=_1+xx_nn₁5_{xn 3};Int. Journal of Pure and Applied Math., 28, 117-124.

14. Stevic S. (2004): More on a rational recurrence relation, Applied Mathematics E-Notes, 4, 80-84.

15. Yang X., Evans D. J., Megson G. M. (2006): On two rational di¤erence equations, Applied Mathematics and Computation, 176, 422-430.