DOI: 10.28919/ejma.2024.4.5 Eur. J. Math. Appl. (2024)4:5 URL: http://ejma.euap.org © 2024 European Journal of Mathematics and Applications ON A FAMILY OF FIFTEENTH-ORDER DIFFERENCE EQUATIONS N. MATHEBULA AND M. FOLLY-GBETOULA∗ Abstract. Difference equations are mathematical tools that are useful in modeling diverse dynamic systems because they represent how a variable changes across discrete time increments. Applying symmetries to complicated difference equations can be a valuable tool for simplifica- tion. Transformations based on symmetry allow one to lower the order of difference equations, making them more comprehensible and solvable. The primary purpose of this project is to gen- eralize and extend some results in [A. M. Ahmeda, S. Mohammadya, L. Aljoufia, Expressions and dynamical behavior of solutions of a class of rational difference equations of fifteenth-order, J. Math. Computer Sci. 25 (2022) 10–22] using symmetries. 1. Introduction The study of difference equations has attracted the attention of many researchers. They are employed to model phenomena in which the variable is discrete. Sophus Lie (1842-1899) pioneered the notion of continuous symmetry in the nineteenth century. He invented and applied symmetry analysis to differential equations between 1872 and 1899. This ground-breaking theory paved way for algorithms to be used to solve differential equations in a systematic manner. Maeda demonstrated in 1987 that an enhanced version of Lie’s approach could also be used to solve ordinary difference equations. In this study, we embark on a quest to unveil the profound connection between difference equations and Lie symmetry analysis. We will explore how the application of symmetry princi- ples can clarify the underlying structure of discrete dynamical systems and provide new tools for their analysis. Through a series of calculations and theoretical developments, we will demon- strate the power of symmetries in uncovering hidden patterns in the difference equation under- study, ultimately advancing our understanding of the behavior of the solution. This work is inspired by the work of Ahmeda et. al. [1], where the authors studied the difference equations (1) xn+1 = xn−14 ±1± xnxn−2xn−5xn−8xn−11xn−14 . They obtained the expressions and dynamical behavior of solutions of (1) using mostly proof by induction. We use symmetry methods to solve for the generalized difference equation below School of Mathematics, University of the Witwatersrand, 2050, Johannesburg, South Africa ∗Corresponding author E-mail addresses: 1846259@students.wits.ac.za, Mensah.Folly-Gbetoula@wits.ac.za. Key words and phrases. Difference equation; symmetry; reduction; solutions. Received 06/01/2024. 1 https://doi.org/10.28919/ejma.2024.4.5 http://ejma.euap.org Eur. J. Math. Appl. | https://doi.org/10.28919/ejma.2024.4.5 2 (2) xn+1 = xn−14 an + bnxnxn−2xn−5xn−8xn−11xn−14 , where An and Bn are real numbers. 1.1. Preliminaries on construction of symmetries for difference equations. To begin, consider the fifteenth-order ordinary difference equation (3) xn+15 = ω(n, xn, xn+3, xn+6, xn+9, xn+12), ∂ω ∂xn 6= 0, where ω represents a smooth function and n an independent variable. The general solution of (3) depends on arbitrary variables and may be expressed as (4) xn = f(n, c1, · · · , c15). Definition 1. The forward shift operator is given by (5) S : n 7→ n+ 1, Sixn = xn+i. The fifteenth-order ordinary difference equation (3) admits a symmetry generator X given by X = Q ∂ ∂xn + S3Q ∂ ∂xn+3 + S6Q ∂ ∂xn+6 + S9Q ∂ ∂xn+9 + S12Q ∂ ∂xn+12 (6) that meets the symmetry requirement (7) S(15)Q−Xω = 0. The function Q = Q(n, xn) is known as the characteristic of the group of transformations. For more details on this, please see [7]. 2. Main Results We will employ Lie point symmetry in this chapter to obtain generic solutions to the fifteen- order ordinary difference equation (2). Given the definitions and notation utilized in this work, we consider the equivalent fifteenth-order ordinary difference equation (8) xn+15 = xn An +Bnxnxn+3xn+6xn+9xn+12 of (2). Applying the symmetry condition (7) to (8), we get S15Q− ( Q ∂ω ∂xn + S3Q ∂w ∂xn+3 + S6Q ∂w ∂xn+6 + S9Q ∂w ∂xn+9 + S12Q ∂w ∂xn+12 ) = 0(9) where ω is the right hand side expression in (8) and noting that ω,y denotes the partial derivative of ω to y. Applying the operator L = ∂ ∂xn + An Bnx2nxn+6xn+9xn+12 ∂ ∂xn+3 (10) https://doi.org/10.28919/ejma.2024.4.5 Eur. J. Math. Appl. | https://doi.org/10.28919/ejma.2024.4.5 3 to (9), we obtain AnQ(n+ 12, xn+12) xn+12(An +Bnxnxn+3xn+6xn+9xn+12)2 + AnQ(n+ 9, xn+9) xn+9(An +Bnxnxn+3xn+6xn+9xn+12)2 + AnQ(n+ 6, xn+6) xn+6(An +Bnxnxn+3xn+6xn+9xn+12)2 + AnQ ′(n+ 3, xn+3) (An +Bnxnxn+3xn+6xn+9xn+12)2 + AnQ ′(n, xn) (An +Bnxnxn+3xn+6xn+9xn+12)2 + 2AnQ(n, xn) xn(An +Bnxnxn+3xn+6xn+9xn+12)2 = 0(11) or simply 2xn+6xn+9xn+12Q(n, xn)− xnxn+6xn+9xn+12(Q ′(n, xn)−Q′(n+ 3, xn+3))+ xn(xn+9xn+12Q(n+ 6, xn+6) + xn+6xn+12Q(n+ 9, xn+9) + xn+6xn+9Q(n+ 12, xn+12)) = 0.(12) To get around the difficulty of dealing with different arguments, we differentiate (12) twice in relation to xn. Thus, we get the following: xnxn+6xn+9xn+12Q ′′′(n, xn) = 0.(13) The generic solution is (14) Q(n, xn) = γnx 2 n + αnxn + βn where αn, βn and γn are some functions of n. By substituting Q(n, xn) in the symmetry condition (7), we get that γn = βn = 0 and αn must satisfy (15) αn+3+αn+6+αn+9+αn+12+αn+15 = 0 αn − αn+15 = 0. Hence, the symmetry generator is of the form (16) X = Q(n, xn) ∂ ∂xn = αnxn ∂ ∂xn , where , thanks to (15), (17) αn + αn+3 + αn+6 + αn+9 + αn+12 = 0. The solutions of (17) are (18) αn = e i(2knπ) 15 , k = 1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14. It follows form (16) and (18) that (19) Xk = ei 2knπ 15 xn ∂ ∂xn , k = 1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, are symmetries of (8). Introducing the canonical coordinate Sn = ∫ 1 Q(n, xn) dxn = ∫ 1 αnxn dxn,(20) we get (21) Snαn = ln |xn|. https://doi.org/10.28919/ejma.2024.4.5 Eur. J. Math. Appl. | https://doi.org/10.28919/ejma.2024.4.5 4 Let F̃n = Snαn + Sn+3αn+3 + Sn+6αn+6 + Sn+9αn+9 + Sn+12αn+12 and Fn = e−F̃n .(22) As a result, we obtain F̃n = ln (xnxn+3xn+6xn+9xn+12) and Fn = 1 xnxn+3xn+6xn+9xn+12 .(23) Now shifting equation (23) three times and substituting the expression of xn+15 given in (8), we get Fn+3 = AnFn +Bn.(24) By iterating (24), we get the following: F3n+j = Fj ( n−1∏ t=0 A3t+j ) + n−1∑ i=0 ( B3i+j n−1∏ k2=i+1 A3k2+j ) , j = 0, 1, 2.(25) Also, xn+15 = Fn Fn+3 xn(26) whose iteration yields x15n+k =xk ( n−1∏ s=0 F15s+k F15s+k+3 ) , k = 0, 1, 2, · · · , 14 =xk ( n−1∏ s=0 F15s+3b k 3 c+τ(k) F15s+3b k 3 c+τ(k)+3 ) =xk ( n−1∏ s=0 F3(5s+b k 3 c)+τ(k) F3(5s+1+b k 3 c)+τ(k) ) (27) since we can always write any integer in the form a = 3ba 3 c+ τ(i), where τ(i) is the remainder when a is divided by 3. Substituting (25) into equation (27), we obtain x15n+k =xk n−1∏ s=0 Fτ(k) ( 5s+b k 3 c−1∏ t=0 A3t+τ(k) ) + 5s+b k 3 c−1∑ i=0 ( B3i+τ(k) 5s+b k 3 c−1∏ k2=i+1 A3k2+τ(k) ) Fτ(k) ( 5s+b k 3 c∏ t=0 A3t+τ(k) ) + 5s+b k 3 c∑ i=0 ( B3i+τ(k) 5s+b k 3 c∏ k2=i+1 A3k2+τ(k) )(28) =xk n−1∏ s=0 ( 5s+b k 3 c−1∏ t=0 A3t+τ(k) ) + 5s+b k 3 c−1∑ i=0 ( B3i+τ(k) Fτ(k) 5s+b k 3 c−1∏ k2=i+1 A3k2+τ(k) ) Fτ(k) ( 5s+b k 3 c∏ t=0 A3t+τ(k) ) + 5s+b k 3 c∑ i=0 ( B3i+τ(k) Fτ(k) 5s+b k 3 c∏ k2=i+1 A3k2+τ(k) ) ,(29) where 1/Fτ(k) = xτ(k)xτ(k)+3xτ(k)+6xτ(k)+9xτ(k)+12. https://doi.org/10.28919/ejma.2024.4.5 Eur. J. Math. Appl. | https://doi.org/10.28919/ejma.2024.4.5 5 3. Case where An = A and Bn = B Substituting An = A and Bn = B into equation (28), we obtain x15n+k =xk  n−1∏ s=0 A5s+b k 3 c + B Fτ(k) 5s+b k 3 c−1∑ s=0 As A5s+1+b k 3 c + B Fτ(k) 5s+b k 3 c∑ s=0 As (30) =xk  n−1∏ s=0 A5s+b k 3 c +Bxτ(k)xτ(k)+3xτ(k)+6xτ(k)+9xτ(k)+12 5s+b k 3 c−1∑ s=0 As A5s+1+b k 3 c +Bxτ(k)xτ(k)+3xτ(k)+6xτ(k)+9xτ(k)+12 5s+b k 3 c∑ s=0 As (31) =  xk n−1∏ s=0 1+B(5s+b k 3 c)xτ(k)xτ(k)+3xτ(k)+6xτ(k)+9xτ(k)+12 1+B(5s+b k 3 c+1)xτ(k)xτ(k)+3xτ(k)+6xτ(k)+9xτ(k)+12 ,when A = 1; xk n−1∏ s=0 A5s+b k3 c+B(1−A5s+b k3 c) 1−A xτ(k)xτ(k)+3xτ(k)+6xτ(k)+9xτ(k)+12 A5s+1+b k3 c+B(1−A5s+1+b k3 c) 1−A xτ(k)xτ(k)+3xτ(k)+6xτ(k)+9xτ(k)+12 ,when A 6= 1 (32) for k = 0, 1, 2 · · · , 11. 3.1. Case where A = −1. For the special case A = −1, we have that x15n+k = xk n−1∏ s=0 (−1)s + B((−1)b k 3 c−(−1)s) 2 xτ(k)xτ(k)+3xτ(k)+6xτ(k)+9xτ(k)+12 −(−1)s + B((−1)b k 3 c+(−1)s) 2 xτ(k)xτ(k)+3xτ(k)+6xτ(k)+9xτ(k)+12  .(33) This means that x15n+k = xk if n even xk −1+Bxkxk+3xk+6xk+9xk+12 if n odd , k = 0, 1, 2;(34a) x15n+k = xk if n even xk(−1 +Bxk−3xkxk+3xk+6xk+9) if n odd , k = 3, 4, 5;(34b) x15n+k = xk if n even xk −1+Bxk−6xk−3xkxk+3xk+6 if n odd , k = 6, 7, 8;(34c) x15n+k = xk if n even xk(−1 +Bxk−9xk−6xk−3xkxk+3) if n odd , k = 9, 10, 11;(34d) x15n+k = xk if n even xk(−1 +Bxk−12xk−9xk−6xk−3xk) if n odd , k = 12, 13, 14.(34e) https://doi.org/10.28919/ejma.2024.4.5 Eur. J. Math. Appl. | https://doi.org/10.28919/ejma.2024.4.5 6 3.2. Special cases in literature. Remember that we shifted equation (2) forward 14 times to get (8), the solution of which is provided by (32). We now reverse the equations in (32) 14 times to find the solution of the difference equation (2), which is given by x15n+k−14 =xk−14 n−1∏ s=0 1 +B(5s+ bk 3 c)xτ(k)−14xτ(k)−11xτ(k)−8xτ(k)−5xτ(k)−2 1 +B(5s+ bk 3 c+ 1)xτ(k)−14xτ(k)−11xτ(k)−8xτ(k)−5xτ(k)−2 .(35) when A = 1; and x15n+k−14 = xk−14 n−1∏ s=0 A5s+b k 3 c + B(1−A5s+b k3 c) 1−A xτ(k)−14xτ(k)−11xτ(k)−8xτ(k)−5xτ(k)−2 A5s+1+b k 3 c + B(1−A5s+1+b k3 c) 1−A xτ(k)−14xτ(k)−11xτ(k)−8xτ(k)−5xτ(k)−2 (36) when A 6= 1. If we let j = 14− k,(37a) then bj 3 c = 4− bk 3 c(37b) and τ(j) = 2− τ(k)(37c) for k = 0, 1, . . . , 14. It follows from (35), (36) and (37) that x15n−j =x−j n−1∏ s=0 1 +B(5s+ 4− b j 3 c)x−τ(j)−12x−τ(j)−9x−τ(j)−6x−τ(j)−3x−τ(j) 1 +B(5s+ 5− b j 3 c)x−τ(j)−12x−τ(j)−9x−τ(j)−6x−τ(j)−3x−τ(j) ,(38) when A = 1; and x15n−j = x−j n−1∏ s=0 A5s+4−b j 3 c + B(1−A5s+4−b j3 c) 1−A x−τ(j)−12x−τ(j)−9x−τ(j)−6x−τ(j)−3x−τ(j) A5s+5−b j 3 c + B(1−A5s+5−b j3 c) 1−A x−τ(j)−12x−τ(j)−9x−τ(j)−6x−τ(j)−3x−τ(j) (39) when A 6= 1. 3.2.1. Case where A = 1 and B = 1. Let Mj = 5−b j 3 c, Pj = 4∏ k=0 amod(j,3)+3k and x−j = aj. We have (40) Pj = 4∏ k=0 amod(j,3)+3k = x−τ(j)−12x−τ(j)−9x−τ(j)−6x−τ(j)−3x−τ(j). Then equation 38 becomes (41) x15n−j = aj n−1∏ s=0 ( 1 + (5s+Mj − 1)Pj 1 + (5s+Mj)Pj ) which is the same as Theorem 2.1 in Lama’s paper given by the equation below: (42) x15n−k = ak n−1∏ i=0 ( 1 + (5i+Mk − 1)Pk 1 + (5i+Mk)Pk ) . https://doi.org/10.28919/ejma.2024.4.5 Eur. J. Math. Appl. | https://doi.org/10.28919/ejma.2024.4.5 7 3.2.2. Case where A = 1 and B = −1. Similarly, when A = 1 and B = −1, equation (38) is given by (43) x15n−j = aj n−1∏ s=0 ( −1 + (5s+Mj − 1)Pj −1 + (5s+Mj)Pj ) which is the same as Theorem 3.1 in Lama’s paper as given below: (44) x15n−k = ak n−1∏ i=0 ( −1 + (5i+Mk − 1)Pk −1 + (5i+Mk)Pk ) . 3.2.3. Case where A = −1 and B = 1. When we let A = −1 and B = 1, equation (39) becomes (45) x15n−j = aj n−1∏ s=0 (−1)5s+Mj−1 + (1−(−1)5s+Mj−1) 2 Pj (−1)5s+Mj + (1−(−1)5s+Mj ) 2 Pj which is the same as Theorem 4.1 in Lama’s paper given by (46) x15n−k = ak n−1∏ i=0 (−1)5i+Mk−1 + (1−(−1)5i+Mk−1) 2 Pk (−1)5i+Mk + (1−(−1)5i+Mk ) 2 Pk . Note that Mj = 5− b j 3 c, Pj = 4∏ k=0 amod(j,3)+3k and x−j = aj. 3.2.4. Case where A = −1 and B = −1. Similarly, when A = 1 and B = −1, equation (39) gives (47) x15n−j = aj n−1∏ s=0 (−1)5s+Mj−1 + (−1−(−1)5s+Mj−2) 2 Pj (−1)5s+Mj + (−1−(−1)5s+Mj−1) 2 Pj which is the same as Theorem 4.5 in Lama’s paper given by (48) x15n−k = ak n−1∏ i=0 (−1)5i+Mk−1 + (−1−(−1)5i+Mk−2) 2 Pk (−1)5i+Mk + (−1−(−1)5i+Mk−1) 2 Pk . https://doi.org/10.28919/ejma.2024.4.5 Eur. J. Math. Appl. | https://doi.org/10.28919/ejma.2024.4.5 8 4. Numerical Examples Figure 1. Graph of xn+15 = xn 4−3xnxn+3xn+6xn+9xn+12 . 4.1. Example 1. Figure 1 depicts the graph of (8) with the initial conditions x0 = 2;x1 = 1;x2 = −1;x3 = 1/2;x4 = 3/2;x5 = 1;x6 = 3;x7 = 3/4;x8 = 2;x9 = 1/3;x10 = 1/9;x11 = 7/2;x12 = 1;x13 = 8;x14 = −1/7 satisfying xτ(j)xτ(j)+3xτ(j)+6xτ(j)+9xτ(j)+12 = 1− A B (49) and xi 6= xi+3, xi 6= xi+5.(50) The answer is, as predicted, 15-periodic. https://doi.org/10.28919/ejma.2024.4.5 Eur. J. Math. Appl. | https://doi.org/10.28919/ejma.2024.4.5 9 Figure 2. Graph of xn+15 = xn 4−3xnxn+3xn+6xn+9xn+12 . 4.2. Example 2. Figure 2 depicts the graph of (8) with the initial conditions x0 = 2;x1 = 1;x2 = −1;x3 = 1/2;x4 = −1;x5 = 2;x6 = 1;x7 = −1;x8 = 1/2;x9 = −1;x10 = 2;x11 = 1;x12 = −1;x13 = 1/2;x14 = −1 satisfying xτ(j)xτ(j)+3xτ(j)+6xτ(j)+9xτ(j)+12 = 1− A B (51) and xi 6= xi+3, xi = xi+5.(52) The answer is, as predicted, 5-periodic. https://doi.org/10.28919/ejma.2024.4.5 Eur. J. Math. Appl. | https://doi.org/10.28919/ejma.2024.4.5 10 Figure 3. Graph of xn+15 = xn 4−3xnxn+3xn+6xn+9xn+12 . 4.3. Example 3. Figure 3 depicts the graph of (8) with the initial conditions x0 = 1;x1 = 1;x2 = 1; x3 = 1; x4 = 1; x5 = 1; x6 = 1; x7 = 1; x8 = 1; x9 = 1; x10 = 1; x11 = 1; x12 = 1; x13 = 1;x14 = 1 satisfying x5τ(j) = 1− A B (53) and xi = xi+3.(54) The answer is, as predicted, 1-periodic. https://doi.org/10.28919/ejma.2024.4.5 Eur. J. Math. Appl. | https://doi.org/10.28919/ejma.2024.4.5 11 Figure 4. Graph of xn+15 = xn −1+Bxnxn+3xn+6xn+9xn+12 . 4.4. Example 4. Figure 4 depicts the graph of (8) with the initial conditions x0 = 1;x1 = 1;x2 = 1; x3 = 1; x4 = 1; x5 = 1; x6 = 1; x7 = 1; x8 = 1; x9 = 1; x10 = 1; x11 = 1; x12 = 1; x13 = 1;x14 = 1 satisfying x5τ(j) 6= 2 B (55) and xi = xj.(56) The answer is, as predicted, 30-periodic. 5. Conclusion This investigation into the symmetry and exact Solutions of a fifteenth-Order difference equation has produced results. The major goal was to confirm and extend the findings of Lama et al. [1]. As a matter of fact, this goal has been achieved, it has been proved that the findings of this study are consistent with Lama’s work through analysis and mathematical inquiry. References [1] A. M. Ahmeda, S. Al Mohammadya, L. Sh. Aljoufia, Expressions and dynamical behavior of solutions of a class of rational difference equations of fifteenth-order, J. Math. Computer Sci. 25 (2022) 10–22. [2] M. Aloqeli, Dynamics of a rational difference equation, Appl. Math. Comput. 176 (2006) 768–774. [3] M. Folly-Gbetoula, Symmetry, Reductions and exact solutions of the difference equation un+2 = aun/(1 + bunun+1), J. Differ. Equ. Appl. 23 (2017) 1017–1024. [4] M. Folly-Gbetoula, K. Mkhwanazi, D. Nyirenda, On a study of a family of higher order recurrence relations, Math. Probl. Eng. 2022 (2022) 6770105. https://doi.org/10.28919/ejma.2024.4.5 Eur. J. Math. Appl. | https://doi.org/10.28919/ejma.2024.4.5 12 [5] M. Folly-Gbetoula, Dynamics and solutions of higher-order difference equations, Mathematics, 11 (2023) 3693. [6] M. Folly-Gbetoula, A. H. Kara, Symmetries, conservation laws, and integrability of difference equations, Adv. Differ. Equ. 2014 (2014) 224. [7] P. E. Hydon, Difference equations by differential equation methods, Cambridge University Press, Cam- bridge, 2014. [8] K. Mkhwanazi, M. Folly-Gbetoula, Symmetries and solvability of a class of higher order systems of ordinary difference equations, Symmetry, 14 (2022) 108. [9] G. R. W. Quispel, R. Sahadevan, Lie symmetries and the integration of difference equations, Phys. Lett. A 184 (1993) 64–70. https://doi.org/10.28919/ejma.2024.4.5 1. Introduction 1.1. Preliminaries on construction of symmetries for difference equations 2. Main Results 3. Case where An = A and Bn=B 3.1. Case where A = -1 3.2. Special cases in literature 4. Numerical Examples 4.1. Example 1 4.2. Example 2 4.3. Example 3 4.4. Example 4 5. Conclusion References