On the Solutions of the Recursive Sequence. i=0. Saniye Ergin and Ramazan Karataş 1

Transcription

1 Thai Journal of Mathematics Volume 14 (2016) Number 2 : ISSN On the Solutions of the Recursive Sequence ax n k a k x n i Saniye Ergin and Ramazan Karataş 1 Akdeniz University Education Faculty Mathematics Department Konyaaltı Antalya Turkiye saniye86@hotmailcomtr (S Ergin) rkaratas@akdenizedutr (R Karataş) Abstract : We obtain in this paper the solutions of the difference equation ax n k for n = 012 a k x n i where k is a positive number and initial conditions are non zero real numbers with k x i a Keywords : difference equation; solution; periodicity 2010 Mathematics Subject Classification : 39A11 1 Introduction Recently there has been a lot of interest in studying the solution of nonlinear difference equations For some results in this area see for example [1 14] Cinar [2] investigated the positive solutions of the rational difference equation 1 Corresponding author x 1+x n x Copyright c 2016 by the Mathematical Association of Thailand All rights reserved

2 392 Thai J Math 14 (2016)/ S Ergin and R Karataş Elsayed[3] investigated the qualitative behavior of the solution of the difference equation x n x (x n ±1) Aloqeili [1] studied the solutions stability character semi-cycle behavior of the difference equation and gave the following formulation x a x x n x n = n 2 x 0 n+1 2 x 1 a 2i 1 (1 a) (1 a 2i 1 )x 1x 0 a 2i (1 a) (1 a 2i )x 1x 0 n even a 2i 1 (1 a) (1 a 2i )x 1x 0 a 2i+1 (1 a) (1 a 2i+1 )x 1x 0 n odd Hamza et al [4] studied the global stability periodic nature oscillation and the boundedness of solutions of the difference equation A k i=l x n 2i 1 B +C k 1 i=l x n 2i Elabbasy et al [5] investigated some qualitative behavior of the solutions of the recursive sequence αx n k β +γ k x n i Karatas [6] studied the dynamics of the solution of the difference equation Ax n m B +C 2k+1 x n i Our goal in this paper is to obtain the solutions of the difference equation ax n k for n = 012 (11) a k x n i where k is a positive number and initial conditions are non zero real numbers with k x i a

3 On the Solutions of the Recursive Sequence 393 Let I be an interval of real numbers and let f : I k+1 I be a continuously differentiable function Then for every set of initial conditions x k x k+1 x 0 I the difference equation has a unique solution {x n } n= k [7] f (x n x x n k ) n = 01 (12) 2 Main Results Theorem 21 Let {x n } n= k be a solution of Eq(11) and assume that k x i = p and p a Then for n = 01 x (k+1)n+1 = x (k+1)n+2 = x (k+1)n+3 = x (k+1)n+4 = x (k+1)n+k = x (k+1)n+k+1 = n ax k [a (k +1)ip] n {a [(k +1)i+1]p} x (k 1) (a 2p) n {a [(k +1)i+1]p} n {a [(k +1)i+2]p} x (k 2) (a 2p) (a 3p) n {a [(k +1)i+2]p} n {a [(k +1)i+3]p} x (k 3) (a 3p) (a 4p) n {a [(k +1)i+3]p} n {a [(k +1)i+4]p} x 1[a (k 1)p] (a kp) x 0(a kp) [a (k +1)p] n {a [(k +1)i+k 1]p} n {a [(k +1)i+k]p} n {a [(k +1)i+k]p} n {a [(k +1)i+k+1]p}

4 394 Thai J Math 14 (2016)/ S Ergin and R Karataş Proof For n = 0 the result holds Now suppose that n > 0 and that our assumption holds for That is ax k [a (k +1)ip] x (k+1)n k = {a [(k +1)i+1]p} x (k+1)n (k 1) = x (k+1)n (k 2) = x (k+1)n (k 3) = x (k+1) = x (k 1) {a [(k +1)i+1]p} (a 2p) {a [(k +1)i+2]p} x (k 2) (a 2p) {a [(k +1)i+2]p} (a 3p) {a [(k +1)i+3]p} x (k 3) (a 3p) {a [(k +1)i+3]p} (a 4p) {a [(k +1)i+4]p} x 1[a (k 1)p] {a [(k +1)i+k 1]p} (a kp) {a [(k +1)i+k]p} x 0(a kp) {a [(k +1)i+k]p} x (k+1)n = [a (k +1)p] {a [(k +1)i+k+1]p} Now it follows from Eq(11) that ax (k+1)n k x (k+1)n+1 = a k x (k+1)n i

5 On the Solutions of the Recursive Sequence 395 Hence we have x (k+1)n+1 = = a a a ax k ap [a (k+1)p] ax k a ap {a [(k+1)i+1]p} {a [(k+1)i+1]p} i=2 i=2 {a [(k+1)i+k+1]p} ax k [a (k +1)ip] a (k +1)np = a [(k +1)n+1]p {a [(k +1)i+1]p} Hence we have x (k+1)n+1 = n ax k [a (k +1)ip] n {a [(k +1)i+1]p} Similarly we get from Eq(11) that x (k+1)n+2 = ax (k+1)n (k 1) k 1 a x (k+1)n i i= 1

6 396 Thai J Math 14 (2016)/ S Ergin and R Karataş Then x (k+1)n+2 = = a a Hence we have a ax (k 1) (a 2p) ap {a [(k+1)i+1]p} {a [(k+1)i+2]p} n {a [(k+1)n+1]p}[a (k+1)p] {a [(k+1)i+k+1]p} ax (k 1) {a [(k +1)i+1]p} (a 2p) {a [(k +1)i+2]p} a [(k +1)n+1]p a{a [(k +1)n+2]p} x (k+1)n+2 = n ax (k 1) {a [(k +1)i+1]p} (a 2p) n {a [(k +1)i+2]p} Similarly one can obtain the other cases Thus the proof is completed Theorem 22 Eq(11) has periodic solutions of period (k + 1) iff one of the initial condition is zero and will be take the form { x k x (k 1) x 1 x 0 x 1 x 2 x k+2 } Proof Firstly assume that there exists a prime period (k + 1) solution x k x (k 1) x 1 x 0 x 1 x 2 x k+2 of Eq(11) We have from the form of solution of Eq(11) that x k = ax k a p x (k 1) = x (k 1) x 0 = x 0(a kp) a 2p a (k +1)p Then p = 0 That is one of the initial condition is zero Now suppose that one of the initial condition is zero Then we have x (k+1)n+1 = x k x (k+1)n+2 = x (k 1) x (k+1)n+k+1 = x 0 x (k+1)n+k+2 = x k x (k+1)n+k+3 = x (k 1) x (k+1)n+2(k+1) = x 0 Thus we obtain a period (k +1) solution The proof is complete

7 On the Solutions of the Recursive Sequence 397 References [1] M Aloqeili Dynamics of a rational difference equation Appl Math Comput 176 (2) (2006) [2] C Cinar On the positive solutions of the difference equation x 1+x nx Appl Math Comput 158 (2004) [3] EM Elsayed On the solutions of the recursive sequence of order two Fasciculi Mathematici 40 (2008) 5 13 [4] AE Hamza R Khalaf-Allah On the recursive sequence A k i=l x n 2i 1 B+C k 1 Comput Math Appl 56 (2008) i=l x n 2i [5] EM Elabbasy H El-Metwally EM Elsayed On the difference equation αx n k β+γ k J Conc Appl Math 5 (2) (2007) x n i [6] R Karatas Global behavior of a higher order difference equation Comput Math Appl 60 (2010) [7] VL Kocic G Ladas Global Behavior of Nonlinear Difference Equations of High Order with Applications Kluwer Academic Publishers Dordrecht 1993 [8] EM Elabbasy H El-Metwally EM Elsayed Some properties and expressions of solutions for a class of nonlinear difference equation Utilitas Mathematica 87 (2012) [9] EM Elsayed Qualitative properties for a fourth order rational diffference equation Acta Appl Math 110 (2010) [10] EM Elsayed Qualitative behavior of difference equation of order two Math Comput Model 50 (2009) [11] AE Hamza R Khalaf-Allah Global behavior of a higher order difference equation J Math Stat 3 (1) (2007) [12] R Karatas On the solutions of the recursive sequence ax n (2k+1) a+x n k x n (2k+1) Fasciculi Mathematici 45 (2010) [13] C Karatas İ Yalçınkaya On the solutions of the difference equation xn+1 = Ax n (2k+1) Thai J Math 9 (1) (2011) k+1 x n i A+ [14] TI Saary Modern Nonlinear Equations McGraw Hill Newyork 1967 (Received 5 September 2012) (Accepted 3 September 2013) Thai J Math