International Journal of Analysis and Applications Volume 18, Number 3 (2020), 493-512 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-18-2020-493 EXACT SOLUTIONS OF KUPERSHMIDT EQUATION, APPROXIMATE SOLUTIONS FOR TIME-FRACTIONAL KUPERSHMIDT EQUATION: A COMPARISON STUDY MEDJAHED DJILALI1,∗, ALI HAKEM2 AND ABDELKADER BENALI3 1Oran Higher School of Economics, Laboratory ACEDP, Djillali Liabes University, 22000 SIDI-BEL-ABBES, Algeria 2Laboratory ACEDP, Djillali Liabes University, 22000 SIDI-BEL-ABBES, Algeria 3Faculty of The Exact sciences And computer, Mathematics Department, University of Hassiba Benbouali, Chlef 02000, Algeria ∗Corresponding author: djilalimedjahed@yahoo.fr Abstract. In this article, a technique namely Tanh method is applied to obtain some traveling wave solutions for Kupershmidt equation, and by using LADM we obtain an approximate solution to time- fractional Kupershmidt equation. A comparison between the traveling wave solution (exact solution) and the approximate one of equation under study, indicate that Laplace Adomian Decomposition Method (LADM) is highly accurate and can be considered a very useful and valuable method. 1. Introduction The study of nonlinear evolution equations have attracted attention of many mathematicians and physicists. Many authors are interested to the research of the exact solutions [9, 21, 30], because the exact solutions to nonlinear evolution equations are the key tool to understand the various physical phenomena that govern the real world today. Hence, searching for exact traveling wave solutions to nonlinear evolution Received February 19th, 2020; accepted March 16th, 2020; published May 1st, 2020. 2010 Mathematics Subject Classification. 83C15, 35C07, 81Q05, 35L05, 47J35. Key words and phrases. Adomian polynomials; Caputo’s fractional derivative; Kupershmidt equation; Laplace transform Adomian decomposition method; Tanh method. ©2020 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 493 https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-18-2020-493 Int. J. Anal. Appl. 18 (3) (2020) 494 equations plays an important role in the study of nonlinear physical phenomena in many fields such as fluid dynamics, water wave mechanics, meteorology, electromagnetic theory, plasma physics and nonlinear optics [9,21]. In this paper, we will study an important nonlinear evolution equation called kueprshmidt equation (see [12,13] ) in the form ut −u5x − 5 2 uu3x − 25 4 uxu2x − 5 4 u2ux = 0. Many researchers have studied the general fifth order KdV equation in different contexts: ut + ωu5x + αuu3x + βuxu2x+ + γu 2ux = 0, where ω,α,β and γ real constants. This class includes the generalized Kaup-Kupershmidt equation [34] ut + 20a 2bu5x + 10abuu3x + 25abuxu2x + bu 2ux = 0. As the constants a 6= 0,b 6= 0 take different values, we retrieve different types of Kaup-Kupershmidt equation. For examples, in the case a = 1 20 ,b = 30 see [11,16,39], for a = 1 60 ,b = 180, see [38], Reyes [32] studied the case a = 1 10 ,b = −5, if we take a = − 1 30 ,b = 45 we will find the equation studied by Parker [19, 28], and when a = 1 30 ,b = 5 we get the equation treated in [7,17]. While, we obtain the Kupershmidt equation (the equation under study) by taking a = 1 5 ,b = −5 4 . To investigate the traveling wave solutions (soliton solutions) [10, 18], we propose in this work the Tanh method (or hyperbolic tangent method), because it is a powerful technique to search for traveling waves coming out from one-dimensional nonlinear wave and evolution equations. In particular, in those problems where dispersive effects, reaction, diffusion and/or convection play an important role. To show the strength of the method, an overview is given to find out which kind of problems are solved with this technique and how in some nontrivial cases this method, adapted to the problem at hand, still can be applied. Single as well as coupled equations, arising from wave phenomena which appears in different scientific domains such as physics, chemical kinetics, geochemistry and mathematical biology [15,24,25,45]. But some evolution problems do not admit the traveling wave solutions, due to that, we propose a semi- analytical method called Laplace Adomian Decomposition Method (LADM), it is a combination of the Adomian Decomposition Method (ADM) and Laplace transforms. This method was successfully used for solving different problems in [5,8,14,18,20,23,37,40]. The ADM was introduced by Adomian [1–4] and has been applied to a wide class of problems in physics, biology and chemical reactions. The method provides the solution in a rapid convergent series with computable terms. The underlying idea of the technique is to assume an infinite solution of the form u = ∑∞ n=0 un, then apply Laplace transformation to the differential equation. The nonlinear terms are then decomposed in terms of Adomian polynomials [6, 41, 42] and an Int. J. Anal. Appl. 18 (3) (2020) 495 iterative algorithm is constructed for the determination of the un in a recursive manner. Our goal is to obtained the approximate solutions of the time-fractional Kupershmidt equation, and compare this solution (in particular case) with the traveling wave solution of the equation to show that the proposed algorithm (LADM) is suitable for such problems and is very efficient. 2. Preliminaries Before the beginning of this research, we are trying in a hurry to get to know the supporting materials to accomplish this work. 2.1. The Tanh method. The non-linear wave and evolution equations (in principle, in one dimension) are commonly written as: ut = [u,ux,uxx, . . . ] or utt = [u,ux,uxx, . . . ] (2.1) We like to know whether traveling waves (or stationary waves) are solutions of (2.1). The first step is to unite the independent variables x and t into one particular variable through the definition ζ = c(x − µt). Here c(> 0) represents the wave number and µ is the (unknown) velocity of the traveling wave. Accordingly, the quantity u(x; t) is replaced by U(ζ), so that we deal with ODEs, rather than with PDEs. In this way, equations like (2.1) are transformed into −cµ dU dζ = [U,c dU dζ , d2U dζ2 , . . . ] or c2µ2 d2U dζ2 = [U,c dU dζ , d2U dζ2 , . . . ] (2.2) Our main goal is to derive exact or at least approximate solutions, if possible, for these ODEs. So we introduce a new variable Φ = tanh ζ in the ODE. The latter equation then solely depends on Φ, because all derivatives d dζ in (2.2) are now replaced by (1 − Φ2) d dζ The solution(s) we are looking for, will be written as a finite power series in Φ F(Φ) = N∑ n=0 anΦ n (2.3) To determine N (highest order of Φ), the following balancing procedure is used.At least two terms propor- tional to ΦN must appear after substitution of ansatz (2.3) into the equation under study. As a result of this analysis, we definitely require aN+1 = 0 and aN 6= 0 for a particular N. It turns out that N = 1 or 2 in most cases. This balance (and thus N) is obtained by comparing the behavior of ΦN in the highest derivative against its counterpart within the nonlinear term(s). As soon as N is determined in this way, we get after substitution of (2.3) into(2.2) (transformed to the Φ variable) algebraic equations for an (n = 0; 1; . . . ; N). Depending on the problem under study, the wave number c will remain fixed or undetermined. As already mentioned, the velocity µ of the traveling wave is always a function of c. If one is able to find nontrivial values for an (n = 0; 1; . . . ; N), in terms of known quantities, a solution is ultimately obtained (see [24]). Int. J. Anal. Appl. 18 (3) (2020) 496 2.2. Laplace transform. Given a suitable function F(t) the Laplace transform [35, 36], written f(s) is defined by L[F(t)] = f(s) = ∫ ∞ 0 F(t)e−stdt, (2.4) the inverse Laplace transform is defined by L−1[f(s)] = F(t). (2.5) The important properties of Laplace transform and it’s inverse that will be used in this paper are : • If F1(t) and F2(t) are two functions whose Laplace transform exists, then • L[aF1(t) + bF2(t)] = aL[F1(t)] + bL[F2(t)], • L(tα) = Γ(α + 1)s−α−1, α > 0, • L(tn) = n! sn+1 , n a positive integer. • The inverse Laplace transform is linear, i.e L−1[af1(s) + bf2(s)] = aL−1[f1(s)] + bL−1[f2(s)], • L−1( 1 sα ) = t α−1 Γ(α) , α > 0 2.3. Caputo derivative. There exists a vast literature on different definitions of fractional derivatives. The most popular ones are the Riemann-Liouville and the Caputo derivatives. The Caputo derivative of order α is defined by the formula [22,27,29]: Dα∗f(t) =   1 Γ(m−α) ∫ t 0 (t− τ)m−α−1f(m)(τ)dτ, if m− 1 < α < m dm dtm f(t), if α = m, (2.6) where m ∈ N∗ and Γ(.) denotes the Gamma function defined by Γ(x) = ∫∞ 0 tx−1e−tdt, x > 0. The important properties of the Caputo derivative that will be used in this paper are [23,26,31,33,43,44]: Dαtβ = Γ(1 + β) Γ(1 + β −α) tβ−α (2.7) Dαc = 0 (2.8) The Laplace transform of the Caputo derivative is: L[Dαt u(x,t)] = s αu(x,s) − n−1∑ i=0 u(i)(x, 0+)sα−1−i, n− 1 < α ≤ n (2.9) 2.4. The Adomian Decomposition Method Combined with Laplace Transform. The ADM is a method to solve ordinary and nonlinear differential equations. Using this method is possible to express analytic solutions in terms of a series. In a nutshell, the method identifies and separates the linear and nonlinear parts of a differential equation. Inverting and applying the highest order differential operator that is contained in the linear part of the equation, it is possible to express the solution in terms of the rest of the equation affected by the inverse operator. At this point, the solution is proposed by means of a series Int. J. Anal. Appl. 18 (3) (2020) 497 with terms that will be determined and that give rise to the Adomian polynomials. The nonlinear part can also be expressed in terms of these polynomials. The initial (or the border conditions) and the terms that contain the independent variables will be considered as the initial approximation. In this way and by means of a recurrence relations, it is possible to find the terms of the series that give the approximate solution of the differential equation (see [14]). Given a partial (or ordinary) differential equation Fu(x,t) = h(x,t) with initial condition u(x, 0) = f(x), (2.10) where F is a differential operator that could, in general, be nonlinear and therefore includes some linear and nonlinear terms. In general, Eq. (2.10) could be written as Ltu(x,t) = Ru(x,t) + Nu(x,t) + h(x,t) (2.11) where Lt = ∂ α ∂tα , 0 < α ≤ 1 (in this paper) , R is a linear operator that includes partial derivatives with respect to x , N is a nonlinear operator and h is a non-homogeneous term that is u -independent. The LADM consists of applying Laplace transform first on both sides of Eq. (2.11), obtaining L{Ltu(x,t)} = L{Ru(x,t) + Nu(x,t) + h(x,t)} . (2.12) An equivalent expression to (5 )is sαu(x,s) −u(x, 0)sα−1 = L{Ru(x,t) + Nu(x,t) + h(x,t)} . (2.13) In the homogeneous case, h(x,t) = 0, we have u(x,s) = f(x) s + 1 sα L{Ru(x,t) + Nu(x,t)} . (2.14) now, applying the inverse Laplace transform to E q. (2.14) u(x,t) = f(x) + L−1[ 1 sα L{Ru(x,t) + Nu(x,t)}]. (2.15) The ADM method proposes a series solution u (x , t ) given by, u(x,t) = ∞∑ n=0 un(x,t). (2.16) The nonlinear term Nu(x,t) is given by Nu(x,t) = ∞∑ n=0 Pn(u0,u1,u2, . . . ,un). (2.17) where {An} ∞ n=0 is the so-called Adomian polynomials sequence established in [42] , in general, give us term to term: Int. J. Anal. Appl. 18 (3) (2020) 498 P0 = N(u0) P1 = u1N ′(u0) P2 = u2N ′(u0) + 1 2 u21N ′′(u0) P3 = u3N ′(u0) + u1u2N ′′(u0) + 1 3! u31N (3)(u0) P4 = u4N ′(u0) + ( 1 2 u22 + u1u3)N ′′(u0) + 1 2! u21u2N (3)(u0) + 1 4! u41N (4)(u0) ... Other polynomials can be generated in a similar way. Some other approaches to obtain Adomian’s polyno- mials can be found in [42]. Using (2.16) and (2.17) into E q. (2.15), we obtain, ∞∑ n=0 un(x,t) = f(x) + L−1 [ 1 sα L { R ∞∑ n=0 un(x,t) + ∞∑ n=0 Pn(u0,u1,u2, . . . ,un) }] . (2.18) we deduce the following recurrence formulas  u0(x,t) = f(x) un+1(x,t) = L−1 [ 1 sα L{Run(x,t) + Pn(u0,u1,u2, . . . ,un)} ] , n = 0, 1, 2 . . . (2.19) Using (2.19) we can obtain an approximate solution of (2.10), using u(x,t) ≈ k∑ n=0 un(x,t), where lim t→∞ k∑ n=0 un(x,t) = u(x,t) (2.20) Remark 2.1. All results and plots bellow are obtained by using Mathematica software. 3. Main results 3.1. Kupershmidt equation solutions by using Tanh method. in this section, we will apply the Tanh method to find the axact solutions of Kupershmidt equation in the form, ∂u(x,t) ∂t − ∂5u(x,t) ∂x5 − 5 2 u(x,t) ∂3u(x,t) ∂x3 − 25 4 ∂u(x,t) ∂x ∂2u(x,t) ∂x2 − 5 4 u(x,t)2 ∂u(x,t) ∂x = 0. (3.1) We consider the traveling wave transformation defined by, U(ζ) = u(x,t), ζ = c(x−µt). (3.2) Using traveling wave Eqs. (3.2), then (3.1) transform into the following ordinary differential equations µU(1) + c4U(5) + 5 2 c2U(3)U + 25 4 c2U(2)U(1) + 5 4 U2U(1) = 0, (3.3) Int. J. Anal. Appl. 18 (3) (2020) 499 Now balancing the highest order derivative U(5) and nonlinear term U(2)U(1), we get 2N + 3 = N + 5 or equivalent to N = 2. Therefore, Eq. (2.3) reduces to U(ζ) = a0 + a1 tanh(ζ) + a2 tanh 2(ζ), (3.4) substituting Eq. (3.4) into Eq. (3.3) and using Mathematica software we get a polynomial of tanh(ζ)k, (k = 0, 1, 2, ...). Equating the coefficients of this polynomial of the same powers of tanh(ζ) to zero, we obtain a system of algebraic equations for a0,a1,a2,µ and c. − 16a1c5 + 5a0a1c3 − 25 2 a1a2c 3 −a1cµ− 5 4 a20a1c = 0 − 272a2c5 + 35 2 a21c 3 − 25a22c 3 + 40a0a2c 3 − 2a2cµ− 5 2 a0a 2 1c− 5 2 a20a2c = 0 136a1c 5 − 20a0a1c3 + 265 2 a1a2c 3 + a1cµ− 5a31c 4 + 5 4 a20a1c− 15 2 a0a1a2c = 0, 1232a2c 5 − 45a21c 3 + 165a22c 3 − 100a0a2c3 + 2a2cµ + 5 2 a0a 2 1c− 5a0a 2 2c + 5 2 a20a2c− 5a 2 1a2c = 0, − 240a1c5 + 15a0a1c3 − 515 2 a1a2c 3 + 5a31c 4 − 25 4 a1a 2 2c + 15 2 a0a1a2c = 0, − 1680a2c5 + + 55 2 a21c 3 − 275a22c 3 + 60a0a2c 3 − 5a32c 2 + 5a0a 2 2c + 5a 2 1a2c = 0, 120a1c 5 + 275 2 a1a2c 3 + 25 4 a1a 2 2c = 0, 720a2c 5 + 135a22c 3 + 5a32c 2 = 0, where a2 6= 0. Solving them by means of Mathematica gives:{ a0 → 4c2,a1 → 0,a2 →−6c2,µ →−c4 } ,{ a0 → 32c2,a1 → 0,a2 →−48c2,µ →−176c4 } , substituting into Eq. (3.4), it follows u1(x,t) = 4c 2 − 6c2 tanh2(cx + c5t), (3.5) u2(x,t) = 32c 2 − 48c2 tanh2(cx + 176c5t), (3.6) 3.2. The approximate solution of time-fractional Kupershmidt equation by LADM. Consider the time-fractional Kupershmidt equation ∂α ∂tα u(x,t) = ∂5u(x,t) ∂x5 + 5 2 u(x,t) ∂3u(x,t) ∂x3 + 25 4 ∂u(x,t) ∂x ∂2u(x,t) ∂x2 + 5 4 u(x,t)2 ∂u(x,t) ∂x (3.7) subject to the initial conditions u(x, 0) = f(x) = 4c2 − 6c2 tanh2(cx), (3.8) Int. J. Anal. Appl. 18 (3) (2020) 500 where 0 < α ≤ 1 and ∂ α ∂tα = Dαt the derivatives in the sens of Caputo. Comparing (3.7) with Eq. (2.11) we have that h(x,t) = 0, Lt and R becomes: Ltu = D α t u = ∂α ∂tα u, R(u) = ∂5u(x,t) ∂x5 = u5x(x,t), (3.9) while the nonlinear term are given by Nu = 5 2 u(x,t) ∂3u(x,t) ∂x3 + 25 4 ∂u(x,t) ∂x ∂2u(x,t) ∂x2 + 5 4 u(x,t)2 ∂u(x,t) ∂x := 5 2 u(x,t)u3x(x,t) + 25 4 ux(x,t)u2x(x,t) + 5 4 u(x,t)2ux(x,t), (3.10) By using now Eq. (2.19) through the LADM method we obtain recursively  u0(x,t) = f(x) un+1(x,t) = L−1 [ 1 sα L{R(un) + Pn(u0,u1,u2, . . . ,un)} ] , n = 0, 1, 2 . . . (3.11) from this, we will consider the decomposition of the nonlinear terms into Adomian polynomials as Nu = N1u + N2u + N3u = ∞∑ n=0 Pn(u0,u1,u2, . . . ,un). (3.12) Let N1u = 5 2 u(x,t)u3x(x,t) = 5 2 ∞∑ n=0 un ∞∑ n=0 un3x = ∞∑ n=0 An(u0,u1,u2, . . . ,un), (3.13) N2u = 25 4 ux(x,t)u2x(x,t) = 25 4 ∞∑ n=0 unx ∞∑ n=0 un2x = ∞∑ n=0 Bn(u0,u1,u2, . . . ,un), (3.14) N3u = 5 4 u(x,t)2ux(x,t) = 5 4 ( ∞∑ n=0 un )2 ∗ ∞∑ n=0 unx = ∞∑ n=0 Cn(u0,u1,u2, . . . ,un), (3.15) where Pn = An + Bn + Cn. Using ADM, Eq.(2.16) gives u(x,t) = ∞∑ n=0 un(x,t), (3.16) thus, the Adomian polynomials An are in the forms A0 = 5 2 u0u03x A1 = 5 2 u1u03x + 5 2 u0u13x A2 = 5 2 u2u03x + 5 2 u1u13x + 5 2 u0u23x A3 = 5 2 u3u03x + 5 2 u2u13x + 5 2 u1u23x + 5 2 u0u33x A4 = 5 2 u4u03x + 5 2 u3u13x + 5 2 u2u23x + 5 2 u1u33x + 5 2 u0u43x, ... (3.17) Int. J. Anal. Appl. 18 (3) (2020) 501 B0 = 25 4 u0xu02x B1 = 25 4 u1xu02x + 25 4 u0xu12x B2 = 25 4 u2xu02x + 25 4 u1xu12x + 25 4 u0xu22x B3 = 25 4 u3xu02x + 25 4 u2xu12x + 25 4 u1xu22x + 25 4 u0xu32x B4 = 25 4 u4xu02x + 25 4 u3xu12x + 25 4 u2xu22x + 25 4 u1xu32x + 25 4 u0xu42x, ... (3.18) and C0 = 5 4 u20u0x, C1 = 5 4 u20u1x + 5 2 u1u0u0x, C2 = 5 4 u20u2x + 5 2 u2u0u0x + 5 4 u21u0x C3 = 5 4 u20u3x + 5 2 u3u0u0x + 5 2 u2u0u1x + 5 2 u1u0u2x + 5 2 u1u2u0x + 5 4 u21u1x C4 = 5 4 u20u4x + 5 2 u4u0u0x + 5 2 u2u0u2x + 5 4 u22u0x + 5 2 u1u3u0x + 5 2 u1u2u1x + 5 4 u21u2x. ... (3.19) Through the LADM we obtain recursively u0(x,t) = f(x), u1(x,t) = L−1 [ 1 sα L{u05x + A0 + B0 + C0} ] , u2(x,t) = L−1 [ 1 sα L{u15x + A1 + B1 + C1} ] , u3(x,t) = L−1 [ 1 sα L{u25x + A2 + B2 + C2} ] , ... ... un+1(x,t) = L−1 [ 1 sα L{un5x + An + Bn + Cn} ] . (3.20) Int. J. Anal. Appl. 18 (3) (2020) 502 Besides A0 = − 1440c7 tanh3(cx)sech4(cx) + 960c7 tanh(cx)sech4(cx) + 720c7 tanh5(cx)sech2(cx) − 480c7 tanh3(cx)sech2(cx) B0 =900c 7 tanh(cx)sech6(cx) − 1800c7 tanh3(cx)sech4(cx) C0 = − 540c7 tanh5(cx)sech2(cx) + 720c7 tanh3(cx)sech2(cx) − 240c7 tanh(cx)sech2(cx). With the above, we have u0(x,t) =4c 2 − 6c2 tanh2(cx) u1(x,t) = − 732c7tα tanh(cx)sech6(cx) Γ(α + 1) − 744c7tα tanh3(cx)sech4(cx) Γ(α + 1) + 960c7tα tanh(cx)sech4(cx) Γ(α + 1) − 12c7tα tanh5(cx)sech2(cx) Γ(α + 1) + 240c7tα tanh3(cx)sech2(cx) Γ(α + 1) − 240c7tα tanh(cx)sech2(cx) Γ(α + 1) , (3.21) and proceeding in a similar way we get A1 = 101760c12tαsech10(cx) Γ(α + 1) − 328320c12tα tanh2(cx)sech10(cx) Γ(α + 1) + 1188720c12tα tanh4(cx)sech8(cx) Γ(α + 1) − 120000c12tαsech8(cx) Γ(α + 1) − 442560c12tα tanh2(cx)sech8(cx) Γ(α + 1) + 833760c12tα tanh6(cx)sech6(cx) Γ(α + 1) 19200c12tαsech6(cx) Γ(α + 1) − 2111040c12tα tanh4(cx)sech6(cx) Γ(α + 1) + 950400c12tα tanh2(cx)sech6(cx) Γ(α + 1) − 684720c12tα tanh8(cx)sech4(cx) Γ(α + 1) + 990240c12tα tanh6(cx)sech4(cx) Γ(α + 1) − 187200c12tα tanh4(cx)sech4(cx) Γ(α + 1) − 105600c12tα tanh2(cx)sech4(cx) Γ(α + 1) − 1440c12tα tanh10(cx)sech2(cx) Γ(α + 1) + 29760c12tα tanh8(cx)sech2(cx) Γ(α + 1) − 48000c12tα tanh6(cx)sech2(cx) Γ(α + 1) + 19200c12tα tanh4(cx)sech2(cx) Γ(α + 1) , B1 = 54900c12tαsech12(cx) Γ(α + 1) − 72000c12tαsech10(cx) Γ(α + 1) + 18000c12tαsech8(cx) Γ(α + 1) − 1035000c12tα tanh2(cx)sech10(cx) Γ(α + 1) + 202500c12tα tanh4(cx)sech8(cx) Γ(α + 1) + 1278000c12tα tanh2(cx)sech8(cx) Γ(α + 1) + 1299600c12tα tanh6(cx)sech6(cx) Γ(α + 1) − 1224000c12tα tanh4(cx)sech6(cx) Γ(α + 1) − 216000c12tα tanh2(cx)sech6(cx) Γ(α + 1) + 7200c12tα tanh8(cx)sech4(cx) Γ(α + 1) − 144000c12tα tanh6(cx)sech4(cx) Γ(α + 1) + 144000c12tα tanh4(cx)sech4(cx) Γ(α + 1) , Int. J. Anal. Appl. 18 (3) (2020) 503 C1 = − 14640c12tαsech8(cx) Γ(α + 1) + 19200c12tαsech6(cx) Γ(α + 1) − 4800c12tαsech4(cx) Γ(α + 1) − 164700c12tα tanh4(cx)sech8(cx) Γ(α + 1) + 131760c12tα tanh2(cx)sech8(cx) Γ(α + 1) − 36720c12tα tanh6(cx)sech6(cx) Γ(α + 1) + 175680c12tα tanh4(cx)sech6(cx) Γ(α + 1) − 129600c12tα tanh2(cx)sech6(cx) Γ(α + 1) + 129060c12tα tanh8(cx)sech4(cx) Γ(α + 1) − 270720c12tα tanh6(cx)sech4(cx) Γ(α + 1) + 162720c12tα tanh4(cx)sech4(cx) Γ(α + 1) − 19200c12tα tanh2(cx)sech4(cx) Γ(α + 1) + 1080c12tα tanh10(cx)sech2(cx) Γ(α + 1) − 23040c12tα tanh8(cx)sech2(cx) Γ(α + 1) + 50880c12tα tanh6(cx)sech2(cx) Γ(α + 1) − 38400c12tα tanh4(cx)sech2(cx) Γ(α + 1) + 9600c12tα tanh2(cx)sech2(cx) Γ(α + 1) , thus, u2 = 24c12t2αsech4(cx) Γ(2α + 1) + 12c12t2α cosh(2cx)sech4(cx) Γ(2α + 1) , (3.22) A2 = 115200c17sech4(cx) tanh11(cx)t2α Γ(α + 1)2 + 32371200c17sech6(cx) tanh9(cx)t2α Γ(α + 1)2 + 11520000c17sech10(cx) tanh7(cx)t2α Γ(α + 1)2 + 149932800c17sech8(cx) tanh7(cx)t2α Γ(α + 1)2 + 2304000c17sech4(cx) tanh7(cx)t2α Γ(α + 1)2 + 249523200c17sech12(cx) tanh5(cx)t2α Γ(α + 1)2 + 18316800c17sech8(cx) tanh5(cx)t2α Γ(α + 1)2 + 20736000c17sech6(cx) tanh5(cx)t2α Γ(α + 1)2 + 11520c17 cosh(2cx)sech4(cx) tanh5(cx)t2α Γ(2α + 1) + 137164320c17sech14(cx) tanh3(cx)t2α Γ(α + 1)2 + 308563200c17sech10(cx) tanh3(cx)t2α Γ(α + 1)2 + 6336000c17sech6(cx) tanh3(cx)t2α Γ(α + 1)2 + 23040c17sech6(cx) tanh3(cx)t2α Γ(2α + 1) + 15360c17sech4(cx) tanh3(cx)t2α Γ(2α + 1) + 960c17 cosh(2cx)sech4(cx) tanh3(cx)t2α Γ(2α + 1) + 4320c17sech6(cx) sinh(2cx) tanh2(cx)t2α Γ(2α + 1) + 10080c17sech4(cx) sinh(2cx) tanh2(cx)t2α Γ(2α + 1) + 960c17sech4(cx) sinh(2cx)t2α Γ(2α + 1) + 46382400c17sech14(cx) tanh(cx)t2α Γ(α + 1)2 + 11808000c17sech10(cx) tanh(cx)t2α Γ(α + 1)2 + 2880c17 cosh(2cx)sech8(cx) tanh(cx)t2α Γ(2α + 1) + 6720c17 cosh(2cx)sech6(cx) tanh(cx)t2α Γ(2α + 1) Int. J. Anal. Appl. 18 (3) (2020) 504 − 2880c17sech4(cx) tanh13(cx)t2α Γ(α + 1)2 − 1550880c17sech6(cx) tanh11(cx)t2α Γ(α + 1)2 − 83764800c17sech8(cx) tanh9(cx)t2α Γ(α + 1)2 − 1267200c17sech4(cx) tanh9(cx)t2α Γ(α + 1)2 − 57772800c17sech6(cx) tanh7(cx)t2α Γ(α + 1)2 − 294724800c17sech10(cx) tanh5(cx)t2α Γ(α + 1)2 − 1152000c17sech4(cx) tanh5(cx)t2α Γ(α + 1)2 − 366019200c17sech12(cx) tanh3(cx)t2α Γ(α + 1)2 − 86400000c17sech8(cx) tanh3(cx)t2α Γ(α + 1)2 − 18622080c17sech16(cx) tanh(cx)t2α Γ(α + 1)2 − 38419200c17sech12(cx) tanh(cx)t2α Γ(α + 1)2 − 1152000c17sech8(cx) tanh(cx)t2α Γ(α + 1)2 − 23040c17sech4(cx) tanh5(cx)t2α Γ(2α + 1) − 17280c17sech4(cx) sinh(2cx) tanh4(cx)t2α Γ(2α + 1) − 11520c17 cosh(2cx)sech6(cx) tanh3(cx)t2α Γ(2α + 1) − 2880c17sech6(cx) sinh(2cx)t2α Γ(2α + 1) − 5760c17sech8(cx) tanh(cx)t2α Γ(2α + 1) − 13440c17sech6(cx) tanh(cx)t2α Γ(2α + 1) − 5760c17 cosh(2cx)sech4(cx) tanh(cx)t2α Γ(2α + 1) , B2 = 288000c17sech4(cx) tanh11(cx)t2α Γ(α + 1)2 + 54576000c17sech6(cx) tanh9(cx)t2α Γ(α + 1)2 + 431856000c17sech8(cx) tanh7(cx)t2α Γ(α + 1)2 + 5760000c17sech4(cx) tanh7(cx)t2α Γ(α + 1)2 + 220665600c17sech12(cx) tanh5(cx)t2α Γ(α + 1)2 + 43200000c17sech6(cx) tanh5(cx)t2α Γ(α + 1)2 + 143305200c17sech14(cx) tanh3(cx)t2α Γ(α + 1)2 + 360144000c17sech10(cx) tanh3(cx)t2α Γ(α + 1)2 + 7200000c17sech6(cx) tanh3(cx)t2α Γ(α + 1)2 + 43200c17sech6(cx) tanh3(cx)t2α Γ(2α + 1) + 18000c17sech6(cx) sinh(2cx) tanh2(cx)t2α Γ(2α + 1) + 115956000c17sech14(cx) tanh(cx)t2α Γ(α + 1)2 + 29520000c17sech10(cx) tanh(cx)t2α Γ(α + 1)2 + 7200c17 cosh(2cx)sech8(cx) tanh(cx)t2α Γ(2α + 1) − 7200c17sech4(cx) tanh13(cx)t2α Γ(α + 1)2 − 2602800c17sech6(cx) tanh11(cx)t2α Γ(α + 1)2 − 210794400c17sech8(cx) tanh9(cx)t2α Γ(α + 1)2 − 3168000c17sech4(cx) tanh9(cx)t2α Γ(α + 1)2 − 177393600c17sech10(cx) tanh7(cx)t2α Γ(α + 1)2 − 102096000c17sech6(cx) tanh7(cx)t2α Γ(α + 1)2 Int. J. Anal. Appl. 18 (3) (2020) 505 − 154908000c17sech10(cx) tanh5(cx)t2α Γ(α + 1)2 − 128736000c17sech8(cx) tanh5(cx)t2α Γ(α + 1)2 − 2880000c17sech4(cx) tanh5(cx)t2α Γ(α + 1)2 − 416520000c17sech12(cx) tanh3(cx)t2α Γ(α + 1)2 − 95040000c17sech8(cx) tanh3(cx)t2α Γ(α + 1)2 − 46555200c17sech16(cx) tanh(cx)t2α Γ(α + 1)2 − 96048000c17sech12(cx) tanh(cx)t2α Γ(α + 1)2 − 2880000c17sech8(cx) tanh(cx)t2α Γ(α + 1)2 − 21600c17 cosh(2cx)sech6(cx) tanh3(cx)t2α Γ(2α + 1) − 1800c17sech8(cx) sinh(2cx)t2α Γ(2α + 1) − 14400c17sech8(cx) tanh(cx)t2α Γ(2α + 1) − 3600c17 cosh(2cx)sech6(cx) tanh(cx)t2α Γ(2α + 1) , C2 = 86400c17sech6(cx) tanh9(cx)t2α Γ(α + 1)2 + 5702400c17sech8(cx) tanh7(cx)t2α Γ(α + 1)2 + 26697600c17sech10(cx) tanh5(cx)t2α Γ(α + 1)2 + 1728000c17sech6(cx) tanh5(cx)t2α Γ(α + 1)2 + 4320c17sech4(cx) tanh5(cx)t2α Γ(2α + 1) + 1080c17sech4(cx) sinh(2cx) tanh4(cx)t2α Γ(2α + 1) + 21081600c17sech12(cx) tanh3(cx)t2α Γ(α + 1)2 + 6912000c17sech8(cx) tanh3(cx)t2α Γ(α + 1)2 + 4320c17sech6(cx) tanh3(cx)t2α Γ(2α + 1) + 2880c17 cosh(2cx)sech4(cx) tanh3(cx)t2α Γ(2α + 1) + 480c17sech4(cx) sinh(2cx)t2α Γ(2α + 1) + 1440c17 cosh(2cx)sech6(cx) tanh(cx)t2α Γ(2α + 1) + 1920c17sech4(cx) tanh(cx)t2α Γ(2α + 1) − 2160c17sech6(cx) tanh11(cx)t2α Γ(α + 1)2 − 267840c17sech8(cx) tanh9(cx)t2α Γ(α + 1)2 − 8566560c17sech10(cx) tanh7(cx)t2α Γ(α + 1)2 − 950400c17sech6(cx) tanh7(cx)t2α Γ(α + 1)2 − 16338240c17sech12(cx) tanh5(cx)t2α Γ(α + 1)2 − 12268800c17sech8(cx) tanh5(cx)t2α Γ(α + 1)2 − 8037360c17sech14(cx) tanh3(cx)t2α Γ(α + 1)2 − 19094400c17sech10(cx) tanh3(cx)t2α Γ(α + 1)2 − 864000c17sech6(cx) tanh3(cx)t2α Γ(α + 1)2 − 2160c17 cosh(2cx)sech4(cx) tanh5(cx)t2α Γ(2α + 1) − 2160c17 cosh(2cx)sech6(cx) tanh3(cx)t2α Γ(2α + 1) − 5760c17sech4(cx) tanh3(cx)t2α Γ(2α + 1) − 1440c17sech4(cx) sinh(2cx) tanh2(cx)t2α Γ(2α + 1) − 2880c17sech6(cx) tanh(cx)t2α Γ(2α + 1) − 960c17 cosh(2cx)sech4(cx) tanh(cx)t2α Γ(2α + 1) , Int. J. Anal. Appl. 18 (3) (2020) 506 u3 = − 1260c17Γ(2α + 1)t3α cosh(4cx) tanh(cx)sech8(cx) Γ(α + 1)2Γ(3α + 1) − 29700c17Γ(2α + 1)t3α tanh(cx)sech8(cx) Γ(α + 1)2Γ(3α + 1) + 2169c17t3α cosh(4cx) tanh(cx)sech8(cx) Γ(3α + 1) + 20880c17Γ(2α + 1)t3α cosh(2cx) tanh(cx)sech8(cx) Γ(α + 1)2Γ(3α + 1) − 73341c17t3α cosh(2cx) tanh(cx)sech8(cx) 2Γ(3α + 1) − 3c17t3α cosh(6cx) tanh(cx)sech8(cx) 2Γ(3α + 1) + 51879c17t3α tanh(cx)sech8(cx) Γ(3α + 1) . (3.23) Thus, the approximate solution of time-fractional Kupershmidt equation (3.7) with the first four terms is: u(x,t) = 20880c17Γ(2α + 1)t3α cosh(2cx) tanh(cx)sech8(cx) Γ(α + 1)2Γ(3α + 1) − 73341c17t3α cosh(2cx) tanh(cx)sech8(cx) 2Γ(3α + 1) − 29700c17Γ(2α + 1)t3α tanh(cx)sech8(cx) Γ(α + 1)2Γ(3α + 1) − 1260c17Γ(2α + 1)t3α cosh(4cx) tanh(cx)sech8(cx) Γ(α + 1)2Γ(3α + 1) + 2169c17t3α cosh(4cx) tanh(cx)sech8(cx) Γ(3α + 1) − 3c17t3α cosh(6cx) tanh(cx)sech8(cx) 2Γ(3α + 1) + 51879c17t3α tanh(cx)sech8(cx) Γ(3α + 1) + 12c12t2α cosh(2cx)sech4(cx) Γ(2α + 1) − 732c7tα tanh(cx)sech6(cx) Γ(α + 1) − 744c7tα tanh3(cx)sech4(cx) Γ(α + 1) + 960c7tα tanh(cx)sech4(cx) Γ(α + 1) − 24c12t2αsech4(cx) Γ(2α + 1) − 12c7tα tanh5(cx)sech2(cx) Γ(α + 1) + 240c7tα tanh3(cx)sech2(cx) Γ(α + 1) − 240c7tα tanh(cx)sech2(cx) Γ(α + 1) − 6c2 tanh2(cx) + 4c2. (3.24) Set u(x,t) = uα(x,t) and take in particular c = 12 , we have: u1(x,t) = − 2507t3 tanh ( x 2 ) sech8 ( x 2 ) 262144 + 3393t3 cosh(x) tanh ( x 2 ) sech8 ( x 2 ) 524288 − t3 cosh(3x) tanh ( x 2 ) sech8 ( x 2 ) 524288 − 117t3 cosh(2x) tanh ( x 2 ) sech8 ( x 2 ) 262144 + 3t2 cosh(x)sech4 ( x 2 ) 2048 − 183 32 t tanh (x 2 ) sech6 (x 2 ) − 93 16 t tanh3 (x 2 ) sech4 (x 2 ) + 15 2 t tanh (x 2 ) sech4 (x 2 ) − 3 32 t tanh5 (x 2 ) sech2 (x 2 ) − 3t2sech4 ( x 2 ) 1024 + 15 8 t tanh3 (x 2 ) sech2 (x 2 ) − 15 8 t tanh (x 2 ) sech2 (x 2 ) − 3 2 tanh2 (x 2 ) + 1, (3.25) Int. J. Anal. Appl. 18 (3) (2020) 507 u1 2 (x,t) = − 17293t3/2 tanh ( x 2 ) sech8 ( x 2 ) 32768 √ π − 2475t3/2 tanh ( x 2 ) sech8 ( x 2 ) 2048π3/2 + 435t3/2 cosh(x) tanh ( x 2 ) sech8 ( x 2 ) 512π3/2 + 723t3/2 cosh(2x) tanh ( x 2 ) sech8 ( x 2 ) 32768 √ π − 105t3/2 cosh(2x) tanh ( x 2 ) sech8 ( x 2 ) 2048π3/2 − 3 512 tsech4 (x 2 ) − 24447t3/2 cosh(x) tanh ( x 2 ) sech8 ( x 2 ) 65536 √ π − t3/2 cosh(3x) tanh ( x 2 ) sech8 ( x 2 ) 65536 √ π + 3t cosh(x)sech4 ( x 2 ) 1024 − 183 √ t tanh ( x 2 ) sech6 ( x 2 ) 16 √ π − 93 √ t tanh3 ( x 2 ) sech4 ( x 2 ) 8 √ π + 15 √ t tanh ( x 2 ) sech4 ( x 2 ) √ π − 3 √ t tanh5 ( x 2 ) sech2 ( x 2 ) 16 √ π + 15 √ t tanh3 ( x 2 ) sech2 ( x 2 ) 4 √ π − 15 √ t tanh ( x 2 ) sech2 ( x 2 ) 4 √ π − 3 2 tanh2 (x 2 ) + 1, (3.26) u3 4 (x,t) = t3/2 cosh(x)sech4 ( x 2 ) 256 √ π + 51879t9/4 tanh ( x 2 ) sech8 ( x 2 ) 131072Γ ( 13 4 ) − 22275√πt9/4 tanh (x2)sech8 (x2) 131072Γ ( 7 4 )2 Γ ( 13 4 ) − 183t3/4 tanh ( x 2 ) sech6 ( x 2 ) 32Γ ( 7 4 ) − 93t3/4 tanh3 (x2)sech4 (x2) 16Γ ( 7 4 ) + 15t3/4 tanh (x2)sech4 (x2) 2Γ ( 7 4 ) − 3t3/4 tanh5 ( x 2 ) sech2 ( x 2 ) 32Γ ( 7 4 ) + 15t3/4 tanh3 (x2)sech2 (x2) 8Γ ( 7 4 ) − 15t3/4 tanh (x2)sech2 (x2) 8Γ ( 7 4 ) + 2169t9/4 cosh(2x) tanh ( x 2 ) sech8 ( x 2 ) 131072Γ ( 13 4 ) + 3915√πt9/4 cosh(x) tanh (x2)sech8 (x2) 32768Γ ( 7 4 )2 Γ ( 13 4 ) − 73341t9/4 cosh(x) tanh ( x 2 ) sech8 ( x 2 ) 262144Γ ( 13 4 ) − 3t9/4 cosh(3x) tanh (x2)sech8 (x2) 262144Γ ( 13 4 ) − 945 √ πt9/4 cosh(2x) tanh ( x 2 ) sech8 ( x 2 ) 131072Γ ( 7 4 )2 Γ ( 13 4 ) − t3/2sech4 ( x 2 ) 128 √ π − 3 2 tanh2 (x 2 ) + 1. (3.27) Remark 3.1. Under the initial conditions (3.8) with c = 1 2 , if α = 1 then (3.7) becomes (3.1), and an exact solution is (3.5). Int. J. Anal. Appl. 18 (3) (2020) 508 t= 1 t = 3 t= 5 x U L A D M U E x a c t E r r o r U L A D M U E x a c t E r r o r U L A D M U E x a c t E r r o r -5 -0 .4 57 57 3 -0 .4 57 57 6 2 .2 3 6 7 4 ∗ 1 0 − 6 -0 .4 51 95 9 -0 .4 52 01 8 0. 00 00 59 24 62 -0 .4 45 47 7 -0 .4 45 74 6 0. 00 02 68 78 4 -4 -0 .3 87 43 8 -0 .3 87 45 0. 00 00 12 38 11 -0 .3 72 77 9 -0 .3 73 11 1 0. 00 03 32 43 7 -0 .3 55 51 -0 .3 57 04 0. 00 15 30 52 -3 -0 .2 13 17 9 -0 .2 13 21 5 0. 00 00 36 14 89 -0 .1 78 36 5 -0 .1 79 34 2 0. 00 09 77 48 6 -0 .1 37 53 3 -0 .1 42 06 8 0. 00 45 34 46 -2 0. 16 03 5 0. 16 03 97 0. 00 00 46 97 8 0. 22 26 02 0. 22 38 52 0. 00 12 49 45 0. 28 47 25 0. 29 04 17 0. 00 56 91 99 -1 0. 71 35 55 0. 71 32 99 0. 00 02 55 82 3 0. 78 42 49 0. 77 73 34 0. 00 69 14 62 0. 86 78 67 0. 83 58 36 0. 03 20 30 6 0 0. 99 85 35 0. 99 85 36 9 .5 3 1 4 7 ∗ 1 0 − 7 0. 98 68 16 0. 98 68 93 0. 00 00 76 86 46 0. 96 33 79 0. 96 39 67 0. 00 05 87 90 4 1 0. 64 49 6 0. 64 52 16 0. 00 02 55 43 4 0. 56 76 43 0. 57 45 26 0. 00 68 83 48 0. 47 07 78 0. 50 25 74 0. 03 17 95 8 2 0. 10 04 83 0. 10 04 36 0. 00 00 47 64 04 0. 04 55 15 8 0. 04 42 12 8 0. 00 13 03 07 -0 .0 02 03 72 9 -0 .0 08 14 25 2 0. 00 61 05 23 3 -0 .2 43 92 9 -0 .2 43 89 3 0. 00 00 36 11 07 -0 .2 72 56 9 -0 .2 71 59 4 0. 00 09 74 34 7 -0 .3 01 05 1 -0 .2 96 54 2 0. 00 45 09 47 4 -0 .4 00 24 -0 .4 00 22 7 0. 00 00 12 44 94 -0 .4 11 93 8 -0 .4 11 6 0. 00 03 37 96 8 -0 .4 23 28 5 -0 .4 21 71 2 0. 00 15 73 06 5 -0 .4 62 5 -0 .4 62 49 8 2 .2 7 7 6 2 ∗ 1 0 − 6 -0 .4 66 91 8 -0 .4 66 85 6 0. 00 00 62 55 81 -0 .4 71 00 6 -0 .4 70 71 2 0. 00 02 94 35 5 Table 1. A comparison between approximate solution and exact solution of (3.1) for t = 1, 3, 5. Int. J. Anal. Appl. 18 (3) (2020) 509 Exact Approx ...for α=1 Approx ...for α=0.5 Approx ...for α=0.75 -10 -5 5 10 x -0.5 0.5 1.0 U (a) UExact and ULADM Figure 1. Plot of the Exact solution (3.5) of Eq. (3.1) and Approximate solutions of Eq. (3.1), when c = 1 2 for t = 5 and x ∈ [−10, 10]. (a) u(x, t) (b) u1(x, t) Figure 2. Plot of the exact solution (3.5) given by Eq. (3.1) and approximate solution uLADM given by Eq. (3.7) when α = 1 and c = 12 , for (x,t) ∈ [−10, 10] × [0, 5] . 4. Conclusion In this paper, we discussed three stages related to the study of Kupershmidt equation. First we used the Tanh method to get the exact solution of the equation under study. In the second stage, thanks to the LADM method (ADM in combination with the Laplace transform), we obtain the approximate solutions to the Int. J. Anal. Appl. 18 (3) (2020) 510 (a) u 1 2 (x, t) (b) u 3 4 (x, t) Figure 3. Plot of the approximate solution uLADM given by Eq. (3.7) when α = 0.5,α = 0.75 and c = 1 2 , for (x,t) ∈ [−10, 10] × [0, 5]. time-fractional Kupershmidt equation. Finally, in order to show the accuracy and efficiency of our method, compare our results with the exact solution of the equation obtained by the Tanh method. Furthermore, we conclude that the LADM is a powerful tool that produces high quality approximate solutions for nonlinear partial differential equations using simple calculations and that attains converge with only few terms. Conflicts of Interest: The author(s) declare that there are no conflicts of interest regarding the publication of this paper. References [1] A. Abdelrazec, D. Pelinovsky , Convergence of the Adomian Decomposition Method for Initial-Value Problems, Numer. Methods Partial Differ. Equations, 27 (2011), 749–766. [2] G. Adomian , A Review of the Decomposition Method in Applied Mathematics , J. Math. Anal. Appl. 135 (1988), 501–544. [3] G. 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