9_700_elsayed.dvi EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 4, No. 3, 2011, 287-303 ISSN 1307-5543 – www.ejpam.com On the Solution of Some Difference Equations Elsayed M. Elsayed King AbdulAziz University, Faculty of Science, Mathematics Department, P. O. Box 80203, Jeddah 21589, Saudi Arabia Permanent address: Department of Mathematics, Faculty of Science, Mansoura University, Man- soura 35516, Egypt. Abstract. We obtain in this paper the solutions of the following difference equations xn+1 = xn−3 ±1± xn−1 xn−3 , n= 0,1, ..., where the initial conditions are arbitrary nonzero real numbers. 2000 Mathematics Subject Classifications: 39A10 Key Words and Phrases: difference equations, recursive sequences, periodic solution. 1. Introduction In this paper we obtain the solutions of the following difference equations xn+1 = xn−3 ±1± xn−1 xn−3 , n= 0,1, . . . , (1) where the initial conditions are arbitrary nonzero real numbers. The study of Difference Equations has been growing continuously for the last decade. This is largely due to the fact that difference equations manifest themselves as mathematical mod- els describing real life situations in probability theory, queuing theory, statistical problems, stochastic time series, combinatorial analysis, number theory, geometry, electrical network, quanta in radiation, genetics in biology, economics, psychology, sociology, etc. In fact, now it occupies a central position in applicable analysis and will no doubt continue to play an important role in mathematics as a whole. Recently there has been a lot of interest in studying the global attractivity, boundedness character, periodicity and the solution form of nonlinear difference equations. For some results Email addresses: emelsayed�mans.edu.eg, emmelsayed�yahoo. om http://www.ejpam.com 287 c© 2011 EJPAM All rights reserved. E. Elsayed / Eur. J. Pure Appl. Math, 4 (2011), 287-303 288 in this area, for example: Agarwal et al. [2] investigated the global stability, periodicity character and gave the solution of some special cases of the difference equation xn+1 = a+ d xn−l xn−k b− cxn−s . Aloqeili [4] has obtained the solutions of the difference equation xn+1 = xn−1 a− xn xn−1 . Cinar [6–8] obtained the solutions of the following difference equations xn+1 = xn−1 1+ xn xn−1 , xn+1 = xn−1 −1+ xn xn−1 , xn+1 = axn−1 1+ bxn xn−1 . Cinar et al. [9] studied the solutions and attractivity of the difference equation xn+1 = xn−3 −1+ xn xn−1 xn−2 xn−3 . Elabbasy et al. [11–12] investigated the global stability, periodicity character and gave the solution of some special cases of the following difference equations xn+1 = axn− bxn cxn − d xn−1 , xn+1 = αxn−k β + γ ∏k i=0 xn−i . In [19] Elsayed dealed with the dynamics and found the solution of the following rational recursive sequences xn+1 = xn−5 ±1± xn−1 xn−3 xn−5 . Karatas et al. [34] obtained the solution of the difference equation xn+1 = axn−(2k+2) −a+ ∏2k+2 i=0 xn−i . Simsek et al. [38]-[39] obtained the solutions of the following difference equations xn+1 = xn−3 1+ xn−1 , xn+1 = xn−5 1+ xn−1 xn−3 . In [40] Stevic solved the following problem xn+1 = xn−1 1+ xn . Yalçınkaya et al. [49] considered the dynamics of the difference equation xn+1 = α+ xn−m x k n . E. Elsayed / Eur. J. Pure Appl. Math, 4 (2011), 287-303 289 Zayed [52] considered the behavior of the following difference equation xn+1 = Axn+ Bxn−k + pxn + xn−k q+ xn−k . Other related results on rational difference equations can be found in refs. [2-51]. The study of these equations is quite challenging and rewarding and is still in its infancy. We believe that the nonlinear rational difference equations are of paramount importance in their own right, and furthermore we believe that these results about such equations over prototypes for the development of the basic theory of the global behavior of nonlinear rational difference equations. Let us introduce some basic definitions and some theorems that we need in the sequel. Let I be some interval of real numbers and let f : Ik+1→ I , be a continuously differentiable function. Then for every set of initial conditions x−k, x−k+1, . . . , x0 ∈ I , the difference equation xn+1 = f (xn, xn−1, . . . , xn−k), n= 0,1, . . . , (2) has a unique solution {xn}∞n=−k . Definition 1 (Equilibrium Point). A point x ∈ I is called an equilibrium point of Eq. (2) if x = f (x , x , . . . , x). That is, xn = x for n≥ 0, is a solution of Eq. (2), or equivalently, x is a fixed point of f . Definition 2 (Periodicity). A sequence {xn}∞n=−k is said to be periodic with period p if xn+p = xn for all n≥ −k. 2. On the Difference Equation xn+1 = xn−3 1+ xn−1 xn−3 In this section we give a specific form of the solutions of the difference equation xn+1 = xn−3 1+ xn−1 xn−3 , n= 0,1, . . . , (3) where the initial conditions are arbitrary nonzero positive real numbers. Theorem 1. Let {xn}∞n=−3 be a solution of Eq. (3). Then for n= 0,1, . . . x4n−3 = d n−1 ∏ i=0 (1+ 2i bd) n−1 ∏ i=0 (1+ (2i + 1)bd) , x4n−1 = b n−1 ∏ i=0 (1+ (2i+ 1)bd) n−1 ∏ i=0 (1+ (2i+ 2)bd) , E. Elsayed / Eur. J. Pure Appl. Math, 4 (2011), 287-303 290 x4n−2 = c n−1 ∏ i=0 (1+ 2iac) n−1 ∏ i=0 (1+ (2i + 1)ac) , x4n = a n−1 ∏ i=0 (1+ (2i+ 1)ac) n−1 ∏ i=0 (1+ (2i+ 2)ac) , where x−3 = d, x−2 = c, x−1 = b, x−0 = a, −1 ∏ i=0 Ai = 1. Proof. For n = 0 the result holds. Now suppose that n > 0 and that our assumption holds for n− 1. That is; x4n−7 = d n−2 ∏ i=0 (1+ 2i bd) n−2 ∏ i=0 (1+ (2i + 1)bd) , x4n−5 = b n−2 ∏ i=0 (1+ (2i+ 1)bd) n−2 ∏ i=0 (1+ (2i+ 2)bd) , x4n−6 = c n−2 ∏ i=0 (1+ 2iac) n−2 ∏ i=0 (1+ (2i + 1)ac) , x4n−4 = a n−2 ∏ i=0 (1+ (2i+ 1)ac) n−2 ∏ i=0 (1+ (2i+ 2)ac) . Now, it follows from Eq. (3) that x4n−3 = x4n−7 1+ x4n−5 x4n−7 = d n−2 ∏ i=0 (1+ 2i bd) n−2 ∏ i=0 (1+ (2i+ 1)bd) 1+ b n−2 ∏ i=0 (1+ (2i+ 1)bd) n−2 ∏ i=0 (1+ (2i + 2)bd) d n−2 ∏ i=0 (1+ 2i bd) n−2 ∏ i=0 (1+ (2i+ 1)bd) = d n−2 ∏ i=0 (1+ 2i bd) n−2 ∏ i=0 (1+ (2i+ 1)bd)        1+ bd n−2 ∏ i=0 (1+ 2i bd) n−2 ∏ i=0 (1+ (2i + 2)bd)        E. Elsayed / Eur. J. Pure Appl. Math, 4 (2011), 287-303 291 = d n−2 ∏ i=0 (1+ 2i bd) n−2 ∏ i=0 (1+ (2i+ 1)bd) � 1+ bd (1+ (2n− 2)bd) � = d n−2 ∏ i=0 (1+ 2i bd) n−2 ∏ i=0 (1+ (2i+ 1)bd) � 1+ bd (1+ (2n− 2)bd) � (1+ (2n− 2)bd) (1+ (2n− 2)bd) = d n−1 ∏ i=0 (1+ 2i bd) n−2 ∏ i=0 (1+ (2i+ 1)bd)((1+ (2n− 2)bd)+ bd) = d n−1 ∏ i=0 (1+ 2i bd) n−2 ∏ i=0 (1+ (2i+ 1)bd)(1+ (2n− 1)bd) . Hence, we have x4n−3 = d n−1 ∏ i=0 (1+ 2i bd) n−1 ∏ i=0 (1+ (2i+ 1)bd) . Similarly one can prove the other relations. The proof is complete. Theorem 2. Eq. (3) has a unique equilibrium point which is the number zero. Proof. For the equilibrium points of Eq. (3), we can write x = x 1+ x2 . Then x + x3 = x , or, x3 = 0. Thus the equilibrium point of Eq. (3) is x = 0. E. Elsayed / Eur. J. Pure Appl. Math, 4 (2011), 287-303 292 Theorem 3. Every positive solution of Eq. (3) is bounded and lim n→∞xn = 0. Proof. It follows from Eq. (3) that xn+1 = xn−3 1+ xn−1 xn−3 ≤ xn−3. Then the subsequences {x4n−3}∞n=0, {x4n−2}∞n=0, {x4n−1}∞n=0, {x4n}∞n=0 are decreasing and so are bounded from above by M =max{x−3, x−2, x−1, x0}. Lemma 1. Eq. (3) has no prime period two solution. Numerical Examples For confirming the results of this section, we consider numerical examples which represent different types of solutions to Eq. (3). Example 1. Consider x−3 = 4, x−2 = 9, x−1 = 6, x0 = 7. See Fig. 1. 0 5 10 15 20 25 30 35 40 45 50 0 1 2 3 4 5 6 7 8 9 n x( n) plot of x(n+1)= (x(n−3)/(1+x(n−1)*x(n−3)) Figure 1 Example 2. See Fig. 2, since x−3 = 1.4, x−2 = 0.9, x−1 = 0.6, x0 = 0.7. E. Elsayed / Eur. J. Pure Appl. Math, 4 (2011), 287-303 293 0 5 10 15 20 25 30 35 40 45 50 0 0.2 0.4 0.6 0.8 1 1.2 1.4 n x( n) plot of x(n+1)= (x(n−3)/(1+x(n−1)*x(n−3)) Figure 2 3. On the Difference Equation xn+1 = xn−3 1− xn−1 xn−3 In this section we give a specific form of the solutions of the difference equation xn+1 = xn−3 1− xn−1 xn−3 , n= 0,1, . . . , (4) where the initial conditions are arbitrary nonzero positive real numbers. Theorem 4. Let {xn}∞n=−3 be a solution of Eq. (4). Then for n= 0,1, . . . x4n−3 = d n−1 ∏ i=0 (1− 2i bd) n−1 ∏ i=0 (1− (2i + 1)bd) , x4n−1 = b n−1 ∏ i=0 (1− (2i+ 1)bd) n−1 ∏ i=0 (1− (2i+ 2)bd) , x4n−2 = c n−1 ∏ i=0 (1− 2iac) n−1 ∏ i=0 (1− (2i + 1)ac) , x4n = a n−1 ∏ i=0 (1− (2i+ 1)ac) n−1 ∏ i=0 (1− (2i+ 2)ac) , where x−3 = d, x−2 = c, x−1 = b, x−0 = a, −1 ∏ i=0 Ai = 1 and jbd 6= 1 jac 6= 1 for j = 1,2,3, . . .. Proof. As the proof of Theorem 1. E. Elsayed / Eur. J. Pure Appl. Math, 4 (2011), 287-303 294 Theorem 5. Eq. (4) has a unique equilibrium point which is the number zero. Proof. As the proof of Theorem 2. Numerical Examples Example 3. Consider x−3 = 0.7, x−2 = 0.5, x−1 = 3, x0 = 4. See Fig. 3. 0 10 20 30 40 50 60 70 80 −1 −0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 n x( n) plot of x(n+1)= (x(n−3)/(1−x(n−1)*x(n−3)) Figure 3 Example 4. See Fig. 4, since x−3 = 7, x−2 = 11, x−1 = 0.3, x0 = 4. 0 10 20 30 40 50 60 70 80 −8 −6 −4 −2 0 2 4 6 8 10 12 n x( n) plot of x(n+1)= (x(n−3)/(1−x(n−1)*x(n−3)) Figure 4 E. Elsayed / Eur. J. Pure Appl. Math, 4 (2011), 287-303 295 4. On the Difference Equation xn+1 = xn−3 −1+ xn−1xn−3 In this section we investigate the solutions of the following difference equation xn+1 = xn−3 −1+ xn−1 xn−3 , n= 0,1, . . . , (5) where the initial conditions are arbitrary non zero real numbers with x−3 x−1 6= 1, x−2 x0 6= 1. Theorem 6. Let {xn}∞n=−3 be a solution of Eq. (5). Then for n= 0,1, . . . x4n−3 = d (−1+ bd)n , x4n−1 = b (−1+ bd)n , x4n−2 = c (−1+ ac)n , x4n = a (−1+ ac)n , where x−3 = d, x−2 = c, x−1 = b, x−0 = a. Proof. For n = 0 the result holds. Now suppose that n > 0 and that our assumption holds for n− 1. That is; x4n−7 = d (−1+ bd)n−1 , x4n−5 = b (−1+ bd)n−1 , x4n−6 = c (−1+ ac)n−1 , x4n−4 = a (−1+ ac)n−1 . Now, it follows from Eq.(5) that x4n−3 = x4n−7 −1+ x4n−5 x4n−7 = d (−1+ bd)n−1 −1+ b (−1+ bd)n−1 d (−1+ bd)n−1 = d (−1+ bd)n−1 (−1+ bd) . Hence, we have x4n−3 = d (−1+ bd)n . Similarly x4n−2 = x4n−6 −1+ x4n−4 x4n−6 = c (−1+ ac)n−1 −1+ a (−1+ ac)n−1 c (−1+ ac)n−1 = c (−1+ ac)n−1 (−1+ ac) . E. Elsayed / Eur. J. Pure Appl. Math, 4 (2011), 287-303 296 Hence, we have x4n−3 = c (−1+ ac)n . Similarly, one can easily obtain the other relations. Thus, the proof is completed. Theorem 7. Eq. (5) has three equilibrium points which are 0, p 2, −p2. Proof. For the equilibrium points of Eq. (5), we can write x = x −1+ x2 . Thus we have −x + x3 = x , or, x(x2 − 2) = 0. Thus the equilibrium points of Eq. (5) are 0, p 2, −p2. Theorem 8. Eq. (5) has a periodic solutions of period four iff ac = bd = 2 and will be take the form {d , c, b, a, d , c, b, a, . . .}. Proof. First suppose that there exists a prime period four solution d , c, b, a, d , c, b, a, . . . , of Eq. (5), we see from Eq. (5) that d = d (−1+ bd)n , b = b (−1+ bd)n , c = c (−1+ ac)n , a = a(−1+ ac)n, or, (−1+ bd)n = 1, (−1+ ac)n = 1. Then bd = 2, ac = 2. Second suppose ac = 2, bd = 2. Then we see from Eq. (5) that x4n−3 = d , x4n−2 = c, x4n−1 = b, x4n = a. Thus we have a period four solution and the proof is complete. Lemma 2. Eq. (5) has no prime period two solution. Lemma 3. Assume that ac, bd 6= 1± 1. Then Eq. (5) has unbounded solutions. E. Elsayed / Eur. J. Pure Appl. Math, 4 (2011), 287-303 297 Numerical Examples Example 5. We consider x−3 = 0.4, x−2 = 0.9, x−1 = 0.16, x0 = 1.7. See Fig. 5. 0 10 20 30 40 50 60 70 −0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 x 10 4 n x( n) plot of x(n+1)= (x(n−3)/(−1+x(n−1)*x(n−3)) Figure 5 Example 6. See Fig. 6, since x−3 = 0.7, x−2 = 0.5, x−1 = 20/7, x0 = 4. 0 5 10 15 20 25 30 0.5 1 1.5 2 2.5 3 3.5 4 n x( n) plot of x(n+1)= (x(n−3)/(−1+x(n−1)*x(n−3)) Figure 6 Example 7. In Fig. 7, we assume x−3 = 0.7, x−2 = 0.5, x−1 = 3, x0 = 4. E. Elsayed / Eur. J. Pure Appl. Math, 4 (2011), 287-303 298 0 10 20 30 40 50 60 70 80 0 2 4 6 8 10 12 14 16 18 20 n x( n) plot of x(n+1)= (x(n−3)/(−1+x(n−1)*x(n−3)) Figure 7 5. On the Difference Equation xn+1 = xn−3 −1− xn−1xn−3 In this section we investigate the solutions of the following difference equation xn+1 = xn−3 −1− xn−1 xn−3 , n= 0,1, . . . , (6) where the initial conditions are arbitrary nonzero real numbers with x−3 x−1 6= −1, x−2 x0 6= −1. Theorem 9. Let {xn}∞n=−3 be a solution of Eq. (6). Then for n= 0,1, . . . x4n−3 = (−1)n d (1+ bd)n , x4n−1 = (−1)n b (1+ bd)n , x4n−2 = (−1)n c (1+ ac)n , x4n = (−1)n a (1+ ac)n , where x−3 = d, x−2 = c, x−1 = b, x−0 = a. Proof. As the proof of Theorem 6. Theorem 10. Eq. (6) has three equilibrium points which are 0, p 2, −p2. Proof. As the proof of Theorem 7. Theorem 11. Eq. (6) has a periodic solutions of period four iff ac = bd = −2 and will be take the form {d , c, b, a, d , c, b, a, . . .}. Proof. As the proof of Theorem 8. Lemma 4. Eq. (6) has no prime period two solution. Lemma 5. Assume that ac, bd 6= −1± 1. Then Eq. (6) has unbounded solutions. E. Elsayed / Eur. J. Pure Appl. Math, 4 (2011), 287-303 299 Numerical Examples Example 8. We consider x−3 = 0.7, x−2 = 0.6, x−1 = 0.3, x0 = 0.4. See Fig. 8. 0 20 40 60 80 100 120 140 160 180 −6000 −4000 −2000 0 2000 4000 6000 n x( n) plot of x(n+1)= (x(n−3)/(−1−x(n−1)*x(n−3)) Figure 8 Example 9. See Fig. 9, since x−3 = 0.7, x−2 = 6, x−1 = −3, x0 = −0.4. 0 10 20 30 40 50 60 70 80 −250 −200 −150 −100 −50 0 50 n x( n) plot of x(n+1)= (x(n−3)/(−1−x(n−1)*x(n−3)) Figure 9 Example 10. In Fig. 10, we assume x−3 = −2.5, x−2 = −6, x−1 = 0.8, x0 = 1/3. REFERENCES 300 0 5 10 15 20 25 30 −6 −5 −4 −3 −2 −1 0 1 n x( n) plot of x(n+1)= (x(n−3)/(−1−x(n−1)*x(n−3)) Figure 10 References [1] R P Agarwal. Difference Equations and Inequalities. 1st edition, Marcel Dekker, New York, 1992, 2nd edition, 2000. [2] R P Agarwal and E M Elsayed. Periodicity and stability of solutions of higher order rational difference equation. Advanced Studies in Contemporary Mathematics, 17(2):181- 201, 2008. [3] R P Agarwal and E M Elsayed. 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