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Article

A Novel Analytical View of Time-Fractional Korteweg-De Vries Equations via a New Integral Transform

1
Department of Mathematics, Government College University, Faisalabad 38000, Pakistan
2
Department of Mathematics, Government College Women University, Faisalabad 38000, Pakistan
3
Department of Mathematics, Imam Mohammad Ibn Saud Islamic University, Riyadh 12211, Saudi Arabia
4
Division of Applied Mathematics, Thu Dau Mot University, Thu Dau Mot City 75000, Vietnam
5
Department of Sciences, École Normale Superieure de Meknes, Moulay Ismail University, Meknes 50000, Morocco
6
Department of Mathematics, Abdul Wali Khan University Mardan (AWKUM), Mardan 23200, Pakistan
7
Department of Mathematics, Faculty of Science, Taif University, P.O. Box 11099, Taif 21944, Saudi Arabia
*
Author to whom correspondence should be addressed.
Submission received: 10 June 2021 / Revised: 5 July 2021 / Accepted: 6 July 2021 / Published: 13 July 2021
(This article belongs to the Special Issue Applied Mathematics and Fractional Calculus)

Abstract

:
We put into practice relatively new analytical techniques, the Shehu decomposition method and the Shehu iterative transform method, for solving the nonlinear fractional coupled Korteweg-de Vries (KdV) equation. The KdV equation has been developed to represent a broad spectrum of physics behaviors of the evolution and association of nonlinear waves. Approximate-analytical solutions are presented in the form of a series with simple and straightforward components, and some aspects show an appropriate dependence on the values of the fractional-order derivatives that are, in a certain sense, symmetric. The fractional derivative is proposed in the Caputo sense. The uniqueness and convergence analysis is carried out. To comprehend the analytical procedure of both methods, three test examples are provided for the analytical results of the time-fractional KdV equation. Additionally, the efficiency of the mentioned procedures and the reduction in calculations provide broader applicability. It is also illustrated that the findings of the current methodology are in close harmony with the exact solutions. It is worth mentioning that the proposed methods are powerful and are some of the best procedures to tackle nonlinear fractional PDEs.

1. Introduction

The formulation of exact and explicit PDE solutions is essential for a good perspective on the mechanisms of diverse physical processes. Hirota and Satsuma proposed a coupled KdV framework to address the effects of two long waves with independent dispersion correlations. It was developed as an evolution equation regulating the propagation of a one-dimensional, small-amplitude, long-surface gravity wave in a shallow water channel. The non-linear coupled system of partial differential equations (PDEs) has several applications in physical systems such as fluid mechanics, aquifers, chaos, thermodynamics, plasma physics and many more. By examining a spectral 4 × 4 equation with three possibilities, Wu et al. [1] established a unique hierarchy of nonlinear equations of evolution. Therefore, the action of the KdV solitons acknowledges the impact of the former’s existence. It is demonstrated that it determines the velocity [2,3] of the KdV subsystem. The fractional-order paired KdV equations are written as follows:
δ Φ t ¯ δ = σ 3 Φ x 3 6 σ Φ Φ x + 6 Ψ Ψ x , δ Ψ t ¯ δ = σ 3 Ψ x 3 3 ζ Φ Ψ x , t ¯ > 0 , 0 < δ 1 ,
where σ and ζ are constants and δ is the fractional order derivative of Φ ( x , t ¯ ) and Ψ ( x , t ¯ ) , respectively. The functions Φ ( x , t ¯ ) and Ψ ( x , t ¯ ) are regarded as essential functions of space and time, vanishing for t ¯ and x , respectively. The latter technique reduces to the conventional paired KdV equations since σ = ζ = 1 is utilized.
An exemplary equation in this scheme is the modified coupled KdV system (MCKdV). This equation is governed by the non-linear PDEs listed in [4]:
δ Φ t ¯ δ = 1 2 3 Φ t ¯ 3 3 Φ 2 Φ x + 3 2 Υ 2 Ψ x 2 + 3 Ψ x Υ x + 3 2 Ψ 2 Υ x 2 + 3 Ψ Υ Φ x + 3 Φ Υ Ψ x + 3 Φ Ψ Υ x , δ Ψ t ¯ δ = 3 Ψ x 3 3 Φ x Ψ x 3 Ψ 2 Φ x 2 3 Ψ 2 Υ x + 6 Φ Ψ Φ x + 3 Φ 2 Ψ x , δ Υ t ¯ δ = 3 Υ x 3 3 Φ x Υ x 3 Υ 2 Φ x 2 3 Υ 2 Ψ x + 6 Φ Υ Φ x + 3 Φ 2 Υ x , t ¯ > 0 , 0 < δ 1 .
The modified KdV equation in its usual form is simplified by the MCKdV Equation (2), with Ψ = Υ = 0 . KdV equations are a source of non evolution equations that have a variety of applications in technology and physical sciences. The KdV equations, for example, produce ion acoustic solutions in plasma physics [5,6]. Geophysical fluid dynamics in shallow waters and deep oceans are characterized by long waves [7,8].
Numerous researchers have proposed several schemes to solve the time-fractional KdV equation using different methods, such as the Adomian decomposition method (ADM) [9], differential transform method (DTM) [10], homotopy analysis method (HAM) [11], Natural decomposition method (NDM) [12], variational iteration method [13], Elzaki projected differential transform method (EPDTM) [14], modified tanh technique (MTT) [15], new iterative method (NIM) [16], Lie symmetry analysis (LSA) [17], spectral volume method (SVM) [18,19] and so on. Analogously, similar results for (2) have been proposed by Fan [20], Cavlak and Inc [21], Inc et al. [22], Lin et al. [23], Karczewska and Szczeciński [24] and Ghoreishi et al. [25].
In recent years, the modeling of dynamical processes has progressed by incorporating notions acquired from fractional-order differential equations (FDEs). Fractional calculus resulted in the emergence of the generalization of derivatives and integrals. However, fractional calculus has a long history. Recently, it has become popular in applied sciences such as viscoelastoplastic materials, random walks, optical fibers, solid state physics, plasma physics, chaos, bifurcation, condensed matter, electromagnetic flux, image processing, virology, and biological models; memory operators called fractional derivatives are used to describe damping impacts or deterioration. Several formulations and notions of fractional derivatives were introduced by Coimbra, Davison and Essex, Riesz, Riemann–Liouville, Hadamard, Weyl, Jumarie, Grünwald–Letnikov, and Liouville–Caputo [26,27,28,29], and the characteristics of these derivatives are investigated in [30,31,32,33]. Because of their prominent features and direct physical interpretation, the implementation of the Caputo fractional derivative is gaining popularity in physics, whereas the Liouville–Caputo has a singularity in its kernel.
Maitama and Zhao [34] recently identified the Shehu transformation as an important integral transformation. The Shehu transform (ST) is a modification of the Laplace transformation. Alternately, by inserting ω = 1 in ST, then we recapture the Laplace transform. Complex non-linear PDEs can be converted to simpler equations via this procedure.
Despite the tremendous boost that was provided by Gorge Adomian in 1980 is known as the Adomian decomposition method (ADM). It has been successfully applied to a variety of physical models of PDEs, such as Burger’s equation, a nonlinear second-order PDE with numerous applications in applied sciences. The ADM correlated with several integral transforms, such as Laplace, modified Laplace, Mohand, Aboodh, Elzaki and many more. Recently, modified Laplace ADM [35] for solving nonlinear Volterra integral and integro-differential equations based on the Newton–Raphson formula, Discrete ADM [36] used for solving time-fractional Navier–Stokes equation, Laplace ADM [37] for finding the numerical solution of a fractional order epidemic model of a vector-born disease and hence forth.
Daftardar-Gejji and Jafari [38] proposed a new recursive approach for solving functional equations with asymptotic solutions. The novel recursive process is framed on the basis of decaying the nonlinear terms is known as the iterative Laplace transform method [39]. This process is fast and precise, and it avoids the use of an unconditioned matrix, complicated integrals, and infinite series forms. This method does not necessitate any explicit settings for the problem. Several studies have considered NITM to solve PDEs, such as the KdV Equation [16], Fornberg–Whitham equation [40], and Klein–Gordon equations [41].
Despite the significant body of work on fractional PDEs models, estimating the approximate-analytical solutions of the corresponding governing PDE is not a trivial task. In this context, we aim to develop two efficient algorithms for estimating the approximate-analytical solutions of KdV and MCKdV equations that model the dynamics of the process under investigation. The ADM and NITM are modified with the ST, and the new method is known to be the Shehu decomposition method and Shehu iterative transform method. The novel methods are applied to examining the fractional-order of the system of KDV equations. The outcome of some test examples was examined in order to demonstrate the practicality of the proposed strategy. Innovative techniques are used to derive the outcomes of the fractional-order and integral-order models. The convergence and uniqueness analysis for SDM is also presented. Using synthetic trajectories derived from the KdV and MCKdV models, we demonstrate the validity and feasibility of the suggested algorithmic approaches to deriving the approximate-analytical solutions in a simulation study. The proposed method can be used to solve other fractional orders of linear and non-linear PDEs.

2. Preliminaries

Several definitions and axiom outcomes from the literature are prerequisites in our analysis.
Definition 1
([34]). Shehu transform (ST) for a function Φ ( t ¯ ) having exponential order over the set of functions is stated as
S = Φ ( t ¯ ) | K , k 1 , k 2 > 0 , | Φ ( t ¯ ) | < K exp ( | t ¯ | / k j ) , i f t ¯ ( 1 ) j × [ 0 , ) , j = 1 , 2 ; K , k 1 , k 2 > 0 ,
where Φ ( t ¯ ) is represented by S Φ ( t ¯ ) = S ( ξ , ω ) , is described as
S Φ ( t ¯ ) = 0 Φ ( t ¯ ) exp ξ ω t ¯ d t ¯ = S ( ξ , ω ) , t ¯ 0 , ω [ κ 1 , κ 2 ] .
A useful result of the ST is stated as:
S t ¯ δ = 0 exp ξ ω t ¯ t ¯ δ d t ¯ = ( δ + 1 ) ω ξ δ + 1 .
Definition 2
([34]). The inverse ST of a mapping Φ ( t ¯ ) is stated as
S 1 ω ξ m δ + 1 = t ¯ m δ Γ ( m δ + 1 ) , ( δ ) > 0 , a n d m > 0 .
Lemma 1.
(Linearity property of ST) Let ST of Φ 1 ( t ¯ ) and Φ 2 ( t ¯ ) are P ( ξ , ω ) and Q ( ξ , ω ) , respectively, [34],
S γ 1 Φ 1 ( t ¯ ) + γ 2 Φ 2 ( t ¯ ) = S γ 1 Φ 1 ( t ¯ ) + S γ 2 Φ 2 ( t ¯ ) = γ 1 P ( ξ , ω ) + γ 2 Q ( ξ , ω ) ,
where γ 1 and γ 2 are arbitrary constants.
Lemma 2
([34]). ST of Caputo fractional derivative of order δ is stated as
S D t ¯ δ Φ ( t ¯ ) = ξ ω δ S Φ ( x , t ¯ ) κ = 0 m 1 ξ ω δ κ 1 Φ ( κ ) ( x , 0 ) , m 1 δ m , m N .

3. Configuration of the SDM

Assume the nonlinear fractional PDE:
D t ¯ δ Φ ( x , t ¯ ) + L Φ ( x , t ¯ ) + N Φ ( x , t ¯ ) = F ( x , t ¯ ) , t ¯ > 0 , 0 < δ 1
subject to the condition
Φ ( x , 0 ) = G ( x ) ,
where D t ¯ δ = δ Φ ( x , t ¯ ) t ¯ δ denotes the fractional-order Caputo derivative operator with 0 < δ 1 while L and N are linear and nonlinear terms and F ( x , t ¯ ) indicates the source term.
Employing the Shehu transform to (9), and we acquire
S D t ¯ δ Φ ( x , t ¯ ) + L Φ ( x , t ¯ ) + N Φ ( x , t ¯ ) = S F ( x , t ¯ ) .
Taking differentiation property of Shehu transform, we find
ξ δ ω δ U ( ξ , ω ) = κ = 0 m 1 ξ ω δ κ 1 Φ ( κ ) ( 0 ) + S L Φ ( x , t ¯ ) + N Φ ( x , t ¯ ) + S F ( x , t ¯ ) .
Th inverse Shehu transform of (11) gives
Φ ( x , t ¯ ) = S 1 κ = 0 m 1 ξ ω δ κ 1 Φ ( κ ) ( 0 ) + ω δ ξ δ S F ( x , t ¯ ) S 1 ω δ ξ δ S L Φ ( x , t ¯ ) + N Φ ( x , t ¯ ) .
The Shehu decomposition method solution Φ ( x , t ¯ ) is represented by the following infinite series
Φ ( x , t ¯ ) = m = 0 Φ m ( x , t ¯ ) .
Thus, the nonlinear term N ( x , t ¯ ) can be evaluated by the Adomian decomposition method prescribed as
N Φ ( x , t ¯ ) = m = 0 A ˜ m ( Φ 0 , Φ 1 , . . . ) , m = 0 , 1 , . . . ,
where
A ˜ m ( Φ 0 , Φ 1 , . . . ) = 1 m ! d m d λ m N j = 0 λ j Φ j λ = 0 , m > 0 .
Substituting (13) and (14) into (12), we have
m = 0 Φ m ( x , t ¯ ) = G ( x ) + G ˜ ( x ) S 1 ω δ ξ δ S [ L Φ ( x , t ¯ ) + m = 0 A ˜ m ] .
Finally, the iterative procedure for (15) is obtained as follows:
Φ 0 ( x , t ¯ ) = G ( x ) + G ˜ ( x ) , m = 0 , Φ m + 1 ( x , t ¯ ) = S 1 ω δ ξ δ S L Φ m ( x , t ¯ ) + m = 0 A ˜ m , m 1 .

4. Basic Formulation of the SITM

Let us suppose the following general fractional PDE:
D t ¯ δ Φ ( x , t ¯ ) + L Φ ( x , t ¯ ) + N Φ ( x , t ¯ ) = F ( x , t ¯ ) , t ¯ > 0 , m 1 < δ m , m N
subject to the condition
Φ ( κ ) ( x , 0 ) = G κ ( x ) , κ = 0 , 1 , 2 , . . . , m 1 ,
where L and N are linear and nonlinear terms and F ( x , t ¯ ) indicates the source term.
Utilizing the Shehu transform to (17), we obtain
S D t ¯ δ Φ ( x , t ¯ ) + L Φ ( x , t ¯ ) + N Φ ( x , t ¯ ) = S F ( x , t ¯ ) .
Taking differentiation property of Shehu transform, we find
ξ δ ω δ U ( ξ , ω ) = κ = 0 m 1 ξ ω δ κ 1 Φ ( κ ) ( 0 ) + S L Φ ( x , t ¯ ) + N Φ ( x , t ¯ ) + S F ( x , t ¯ ) .
Th inverse Shehu transform of (19) gives
Φ ( x , t ¯ ) = S 1 κ = 0 m 1 ξ ω δ κ 1 Φ ( κ ) ( 0 ) + ω δ ξ δ S F ( x , t ¯ ) S 1 ω δ ξ δ S L Φ ( x , t ¯ ) + N Φ ( x , t ¯ ) .
From the recursive relation, we obtain
Φ ( x , t ¯ ) = m = 0 Φ m ( x , t ¯ ) .
Furthermore, the operator L is linear, therefore
L m = 0 Φ m ( x , t ¯ ) = m = 0 L Φ m ( x , t ¯ ) ,
and we decomposed the nonlinear operator N as in [38]
N m = 0 Φ m ( x , t ¯ ) = N Φ 0 ( x , t ¯ ) + m = 0 N κ = 0 Φ κ ( x , t ¯ ) N κ = 0 m 1 Φ κ ( x , t ¯ ) = N ( Φ 0 ) + κ = 1 D m ,
where D m = N κ = 0 m Φ κ N κ = 0 m 1 Φ κ .
By putting (21), (22) and (23) into (24), we obtain
m = 0 Φ m ( x , t ¯ ) = S 1 κ = 0 m 1 ξ ω δ κ 1 Φ ( κ ) ( 0 ) + ω δ ξ δ S F ( x , t ¯ ) S 1 ω δ ξ δ S L κ = 0 m Φ κ ( x , t ¯ ) + N ( Φ 0 ) + κ = 1 m D m .
Thus, we establish the subsequent iteration
Φ 0 ( x , t ¯ ) = S 1 κ = 0 m 1 ξ ω δ κ 1 Φ ( κ ) ( 0 ) + ω δ ξ δ S F ( x , t ¯ ) , Φ 1 ( x , t ¯ ) = S 1 ω δ ξ δ S L Φ 0 ( x , t ¯ ) + N Φ 0 ( x , t ¯ ) , Φ m + 1 ( x , t ¯ ) = S 1 ω δ ξ δ S L Φ m ( x , t ¯ ) + D m } , m 1 .
Finally, (17) and (18) yield the m-term solution in series form, described as
Φ ( x , t ¯ ) Φ 0 ( x , t ¯ ) + Φ 1 ( x , t ¯ ) + Φ 2 ( x , t ¯ ) + . . . + Φ m ( x , t ¯ ) , m N .

5. Existence and Uniqueness Results for Shehu Decomposition Method

In what follows, we will demonstrate that the sufficient conditions assure the existence of a unique solution. Our desired existence of solutions in the case of SDM follows by [42].
Theorem 1.
(Uniqueness theorem): Equation (16) has a unique solution whenever 0 < ϵ < 1 , where ϵ = ( L ˇ 1 + L ˇ 2 + L ˇ 3 ) t ¯ ( δ 1 ) δ ! .
Proof. 
Assume that M = C [ I ] , . represents all continuous mappings on the Banach space, defined on I = [ 0 , T ] having the norm . . For this we introduce a mapping W : M M , we have
Φ n + 1 ( x , t ¯ ) = Φ ( x , t ¯ ) + S 1 ω ξ δ S L Φ n ( x , t ¯ ) + R Φ n ( x , t ¯ ) + N Φ n ( x , t ¯ ) , n 0 ,
where L Φ ( x , t ¯ ) 3 Φ ( x , t ¯ ) x 2 and R Φ ( x , t ¯ ) Φ ( x , t ¯ ) x . Now assume that L Φ ( x , t ¯ ) and M Φ ( x , t ¯ ) are also Lipschitzian with | R Φ R Φ ˇ | < L ˇ 1 | Φ Φ ˇ | and | L Φ L Φ ˇ | < L ˇ 2 | Φ Φ ˇ | , where L ˇ 1 and L ˇ 2 are Lipschitz constant, respectively, and Φ , Φ ˇ are various values of the mapping.
W Φ W Φ ˇ = max t ¯ I S 1 ω ξ δ S L Φ ( x , t ¯ ) + R Φ ( x , t ¯ ) + N Φ ( x , t ¯ ) S 1 ω ξ δ S L Φ ˇ ( x , t ¯ ) + R Φ ˇ ( x , t ¯ ) + N Φ ˇ ( x , t ¯ )
max t ¯ I S 1 ω ξ δ S L Φ ( x , t ¯ ) L Φ ˇ ( x , t ¯ ) + S 1 ω ξ δ S R Φ ( x , t ¯ ) R Φ ˇ ( x , t ¯ ) + S 1 ω ξ δ S N Φ ( x , t ¯ ) N Φ ˇ ( x , t ¯ )
max t ¯ I L ˇ 1 S 1 ω ξ δ S | Φ ( x , t ¯ ) Φ ˇ ( x , t ¯ ) | + L ˇ 2 S 1 ω ξ δ S | Φ ( x , t ¯ ) Φ ˇ ( x , t ¯ ) | + L ˇ 3 S 1 ω ξ δ S | Φ ( x , t ¯ ) Φ ˇ ( x , t ¯ ) |
max t ¯ I L ˇ 1 + L ˇ 2 + L ˇ 3 S 1 ω ξ δ S | Φ ( x , t ¯ ) Φ ˇ ( x , t ¯ ) | L ˇ 1 + L ˇ 2 + L ˇ 3 S 1 ω ξ δ S Φ ( x , t ¯ ) Φ ˇ ( x , t ¯ ) = ( L ˇ 1 + L ˇ 2 + L ˇ 3 ) t ¯ ( δ 1 ) ( δ ) ! Φ ( x , t ¯ ) Φ ˇ ( x , t ¯ ) .
Under the assumption 0 < ϵ < 1 , the mapping is contraction. Thus, by Banach contraction fixed point theorem, there exists a unique solution to (9). Hence, this completes the proof. □
Theorem 2.
(Convergence Analysis) The general form solution of (9) will be convergent.
Proof. 
Suppose S ^ n be the n t h partial sum, that is S ^ n = m = 0 n Φ m ( x , t ¯ ) . Firstly, we show that { S ^ n } is a Cauchy sequence in Banach space in M . Taking into consideration a new representation of Adomian polynomials we obtain
R ¯ ( S ^ n ) = H ˇ n + p = 0 n 1 H ˇ p , N ¯ ( S ^ n ) = H ˇ n + c = 0 n 1 H ˇ c .
Now
S ^ n S ^ q = max t ¯ I | S ^ n S ^ q | = max t ¯ I | m = q + 1 n Φ ˇ ( x , t ¯ ) | , ( m = 1 , 2 , 3 , . . . )
max t ¯ I S 1 ω ξ δ S m = q + 1 n L Φ n 1 ( x , t ¯ ) + S 1 ω ξ δ S m = q + 1 n R Φ n 1 ( x , t ¯ ) + S 1 ω ξ δ S m = q + 1 n H ˇ n 1 ( x , t ¯ )
= max t ¯ I S 1 ω ξ δ S m = q n 1 L Φ n ( x , t ¯ ) + S 1 ω ξ δ S m = q n 1 R Φ n ( x , t ¯ ) + S 1 ω ξ δ S m = q n 1 H ˇ n ( x , t ¯ )
max t ¯ I S 1 ω ξ δ S m = q n 1 L ( S ^ n 1 ) L ( S ^ q 1 ) + S 1 ω ξ δ S m = q n 1 R ( S ^ n 1 ) R ( S ^ q 1 ) + S 1 ω ξ δ S m = q n 1 N ( S ^ n 1 ) N ( S ^ q 1 )
max t ¯ I S 1 ω ξ δ S L ( S ^ n 1 ) L ( S ^ q 1 ) + S 1 ω ξ δ S R ( S ^ n 1 ) R ( S ^ q 1 ) + S 1 ω ξ δ S N ( S ^ n 1 ) N ( S ^ q 1 )
L ˇ 1 max t ¯ I S 1 ω ξ δ S ( S ^ n 1 ) ( S ^ q 1 ) + L ˇ 2 max t ¯ I S 1 ω ξ δ S ( S ^ n 1 ) ( S ^ q 1 ) + L ˇ 3 max t ¯ I S 1 ω ξ δ S ( S ^ n 1 ) ( S ^ q 1 )
= ( L ˇ 1 + L ˇ 2 + L ˇ 3 ) t ¯ ( δ 1 ) δ ! S ^ n 1 S ^ q 1 .
Consider n = q + 1 ; then
S ^ q + 1 S ^ q ϵ S ^ q S ^ q 1 ϵ 2 S ^ q 1 S ^ q 2 . . . ϵ q S ^ 1 S ^ 0 ,
where ( L ˇ 1 + L ˇ 2 + L ˇ 3 ) t ¯ ( δ 1 ) δ ! . Analogously, from the triangular inequality we have
S ^ n S ^ q S ^ q + 1 S ^ q + S ^ q + 2 S ^ q + 1 + . . . + S ^ n S ^ n 1 ϵ q + ϵ q + 1 + . . . + ϵ n 1 S ^ 1 S ^ 0 ϵ q 1 ϵ n q ϵ Φ 1 ,
since 0 < ϵ < 1 , we have ( 1 ϵ n q ) < 1 , then
S ^ n S ^ q ϵ q 1 ϵ max t ¯ I Φ 1 .
However, | Φ 1 | < (since Φ ( x , t ¯ ) is bounded). Thus, as q , then S ^ n S ^ q 0 . Hence, { S ^ 1 } is a Cauchy sequence in K . As a result, the series n = 0 Φ n is convergent and this completes the proof. □
Theorem 3
([42]). (Error estimate) The maximum absolute truncation error of the series solution (9)–(16) is computed as
max t ¯ I | Φ ( x , t ¯ ) n = 1 q Φ n ( x , t ¯ ) | ϵ q 1 ϵ max t ¯ I Φ 1 .

6. Evaluation of the Fractional KdV Model

This section represents some test examples by employing two novel methods, SDM and SITM via the Caputo derivative operator. Furthermore, the convergence and stability of the method are elaborated on.
Problem 1
([16]). Assume the time-fractional coupled nonlinear KdV Equation (1) with σ = ζ = 1 , subject to the condition
Φ ( x , 0 ) = 2 sec h 2 β 2 + x 2 , Ψ ( x , 0 ) = σ 2 2 sec h 2 β 2 + x 2 .
Case I. First, we surmise the Shehu decomposition method for Problem 1.
Employing the Shehu transformation to (1), we find
ξ δ ω δ U ( ξ , ω ) κ = 0 m 1 ξ ω δ κ 1 Φ ( κ ) ( 0 ) = S σ 3 Φ x 3 6 σ Φ Φ x + 6 Ψ Ψ x , ξ δ ω δ V ( ξ , ω ) κ = 0 m 1 ξ ω δ κ 1 Ψ ( κ ) ( 0 ) = S σ 3 Ψ x 3 3 σ Φ Ψ x .
In view of (31) and simple computations yield
U ( ξ , ω ) = ω ξ Φ ( 0 ) ( x , 0 ) + ω δ ξ δ S σ 3 Φ x 3 6 σ Φ Φ x + 6 Ψ Ψ x , V ( ξ , ω ) = ω ξ Ψ ( 0 ) ( x , 0 ) + ω δ ξ δ S σ 3 Ψ x 3 3 σ Φ Ψ x .
Applying the inverse Shehu transform, we have
Φ ( x , t ¯ ) = S 1 ω ξ Φ ( x , 0 ) + S 1 ω δ ξ δ S σ 3 Φ x 3 6 σ Φ Φ x + 6 Ψ Ψ x , Ψ ( x , t ¯ ) = S 1 ω ξ Ψ ( x , 0 ) + S 1 ω δ ξ δ S σ 3 Ψ x 3 3 σ Φ Ψ x .
By virtue of the Shehu decomposition method, we have
Φ 0 ( x , t ¯ ) = S 1 ω ξ Φ ( x , 0 ) = S 1 ξ ω 2 sec h 2 β 2 + x 2 = 2 sec h 2 β 2 + x 2 , Ψ 0 ( x , t ¯ ) = S 1 ω ξ Ψ ( x , 0 ) = σ 2 2 sec h 2 β 2 + x 2 .
m = 0 Φ m + 1 ( x , t ¯ ) = S 1 ω δ ξ δ S σ m = 0 ( Φ x x x ) m 6 σ m = 0 A m + 6 m = 0 B m , m = 0 Ψ m + 1 ( x , t ¯ ) = S 1 ω δ ξ δ S σ m = 0 ( Ψ x x x ) m 3 σ m = 0 C m , m = 0 , 1 , 2 , . . . .
The first few Adomian polynomials are presented as follows:
A 0 ( Φ Φ x ) = Φ 0 Φ 0 x , A 1 ( Φ Φ x ) = Φ 0 Φ 1 x + Φ 1 Φ 0 x , A 2 ( Φ Φ x ) = Φ 1 Φ 2 x + Φ 1 Φ 1 x + Φ 2 Φ 0 x , B 0 ( Ψ Ψ x ) = Ψ 0 Ψ 0 x , B 1 ( Ψ Ψ x ) = Ψ 0 Ψ 1 x + Ψ 1 Ψ 0 x , B 2 ( Ψ Ψ x ) = Ψ 1 Ψ 2 x + Ψ 1 Ψ 1 x + Ψ 2 Ψ 0 x , C 0 ( Φ Ψ x ) = Φ 0 Ψ 0 x , C 1 ( Φ Ψ x ) = Φ 0 Ψ 1 x + Φ 1 Ψ 0 x , C 2 ( Φ Ψ x ) = Φ 1 Ψ 2 x + Φ 1 Ψ 1 x + Φ 2 Ψ 0 x .
For m = 0 , 1 , 2 , 3 , . . .
Φ 1 ( x , t ¯ ) = S 1 ω δ ξ δ S σ ( Φ x x x ) 0 6 σ A 0 + 6 B 0 = S 1 ω δ + 2 ξ δ + 2 5 σ tan h β 2 + x 2 sec h 2 β 2 + x 2 = 5 σ tan h β 2 + x 2 sec h 2 β 2 + x 2 t ¯ δ Γ ( δ + 1 ) , Ψ 1 ( x , t ¯ ) = S 1 ω δ ξ δ S σ ( Ψ x x x ) 0 3 σ C 0 = 5 σ 3 / 2 2 tan h β 2 + x 2 sec h 2 β 2 + x 2 t ¯ δ Γ ( δ + 1 ) . Φ 2 ( x , t ¯ ) = S 1 ω δ ξ δ S σ ( Φ x x x ) 1 6 σ A 1 + 6 B 1 = S 1 ω 2 δ + 2 ξ 2 δ + 2 8 σ 2 2 2 cosh 2 σ 2 + x 2 3 sec h 4 σ 2 + x 2 = 8 σ 2 2 2 cosh 2 σ 2 + x 2 3 sec h 4 σ 2 + x 2 t ¯ 2 δ Γ ( 2 δ + 1 ) , Ψ 2 ( x , t ¯ ) = S 1 ω δ ξ δ S σ ( Ψ x x x ) 1 3 σ C 1 = 5 σ 5 / 2 2 2 2 cosh 2 σ 2 + x 2 3 sec h 4 σ 2 + x 2 t ¯ 2 δ Γ ( 2 δ + 1 ) ,
Φ 3 ( x , t ¯ ) = S 1 ω δ ξ δ S σ ( Φ x x x ) 2 6 σ A 2 + 6 B 2 = σ 3 4 sin h σ 2 + x 2 2 Γ 2 ( δ + 1 ) Γ ( 3 δ + 1 ) cosh 7 β 2 + x 2 [ 2 Γ 2 ( δ + 1 ) cos h 4 σ 2 + x 2 18 Γ 2 ( δ + 1 ) cos h 2 σ 2 + x 2 + 6 Γ ( 2 δ + 1 ) cos h 2 σ 2 + x 2 + 18 Γ 2 ( δ + 1 ) 9 Γ ( 2 δ + 1 ) ] , Ψ 3 ( x , t ¯ ) = S 1 ω δ ξ δ S σ ( Ψ x x x ) 2 3 σ C 2 = σ 7 / 2 11 sin h σ 2 + x 2 2 2 Γ 2 ( δ + 1 ) Γ ( 3 δ + 1 ) cosh 7 β 2 + x 2 [ 2 Γ 2 ( δ + 1 ) cos h 4 σ 2 + x 2 18 Γ 2 ( δ + 1 ) cos h 2 σ 2 + x 2 + 6 Γ ( 2 δ + 1 ) cos h 2 σ 2 + x 2 + 18 Γ 2 ( δ + 1 ) 9 Γ ( 2 δ + 1 ) ] , .
The Shehu decomposition method solution for Problem 1 is presented as:
Φ ( x , t ¯ ) = Φ 0 ( x , t ¯ ) + Φ 1 ( x , t ¯ ) + Φ 2 ( x , t ¯ ) + Φ 3 ( x , t ¯ ) + . . . , = 2 sec h 2 β 2 + x 2 + 5 σ tan h β 2 + x 2 sec h 2 β 2 + x 2 t ¯ δ Γ ( δ + 1 ) + 8 σ 2 2 2 cosh 2 σ 2 + x 2 3 sec h 4 σ 2 + x 2 t ¯ 2 δ Γ ( 2 δ + 1 ) + σ 3 4 sin h σ 2 + x 2 2 Γ 2 ( δ + 1 ) Γ ( 3 δ + 1 ) cosh 7 β 2 + x 2 [ 2 Γ 2 ( δ + 1 ) cos h 4 σ 2 + x 2 18 Γ 2 ( δ + 1 ) cos h 2 σ 2 + x 2 + 6 Γ ( 2 δ + 1 ) cos h 2 σ 2 + x 2 + 18 Γ 2 ( δ + 1 ) 9 Γ ( 2 δ + 1 ) ] + . . . .
Analogously, we have
Ψ ( x , t ¯ ) = σ 2 2 sec h 2 β 2 + x 2 + 5 σ 3 / 2 2 tan h β 2 + x 2 sec h 2 β 2 + x 2 t ¯ δ Γ ( δ + 1 ) + 5 σ 5 / 2 2 2 2 cosh 2 σ 2 + x 2 3 sec h 4 σ 2 + x 2 t ¯ 2 δ Γ ( 2 δ + 1 ) + σ 7 / 2 11 sin h σ 2 + x 2 2 2 Γ 2 ( δ + 1 ) Γ ( 3 δ + 1 ) cosh 7 β 2 + x 2 [ 2 Γ 2 ( δ + 1 ) cos h 4 σ 2 + x 2 18 Γ 2 ( δ + 1 ) cos h 2 σ 2 + x 2 + 6 Γ ( 2 δ + 1 ) cos h 2 σ 2 + x 2 + 18 Γ 2 ( δ + 1 ) 9 Γ ( 2 δ + 1 ) ] + . . . .
By setting δ = 1 , we then obtain the exact solution of coupled KdV Equation (1)
Φ ( x , t ¯ ) = 2 sec h 2 β 2 + x 2 σ 3 t ¯ 2 , Ψ ( x , t ¯ ) = σ 2 2 sec h 2 β 2 + x 2 σ 3 t ¯ 2 .
Case II. Now, we surmise the Shehu iterative transform method on Problem 1.
Applying the proposed analytical approach to (33), yields
Φ 0 ( x , t ¯ ) = S 1 ω ξ Φ ( x , 0 ) = S 1 ξ ω 2 sec h 2 β 2 + x 2 = 2 sec h 2 β 2 + x 2 , Ψ 0 ( x , t ¯ ) = S 1 ω ξ Ψ ( x , 0 ) = σ 2 2 sec h 2 β 2 + x 2 . Φ 1 ( x , t ¯ ) = S 1 ω δ ξ δ S σ 3 Φ 0 x 3 6 σ Φ 0 Φ 0 x + 6 Ψ 0 Ψ 0 x = S 1 ω δ + 2 ξ δ + 2 5 σ tan h β 2 + x 2 sec h 2 β 2 + x 2 = 5 σ tan h β 2 + x 2 sec h 2 β 2 + x 2 t ¯ δ Γ ( δ + 1 ) , Ψ 1 ( x , t ¯ ) = S 1 ω δ ξ δ S σ 3 Ψ 0 x 3 3 σ Φ 0 Ψ 0 x = 5 σ 3 / 2 2 tan h β 2 + x 2 sec h 2 β 2 + x 2 t ¯ δ Γ ( δ + 1 ) . Φ 2 ( x , t ¯ ) = S 1 ω δ ξ δ S σ 3 Φ 1 x 3 6 σ Φ 1 Φ 1 x + 6 Ψ 1 Ψ 1 x = S 1 ω 2 δ + 2 ξ 2 δ + 2 8 σ 2 2 2 cosh 2 σ 2 + x 2 3 sec h 4 σ 2 + x 2 = 8 σ 2 2 2 cosh 2 σ 2 + x 2 3 sec h 4 σ 2 + x 2 t ¯ 2 δ Γ ( 2 δ + 1 ) , Ψ 2 ( x , t ¯ ) = S 1 ω δ ξ δ S σ 3 Ψ 1 x 3 3 σ Φ 1 Ψ 1 x = 5 σ 5 / 2 2 2 2 cosh 2 σ 2 + x 2 3 sec h 4 σ 2 + x 2 t ¯ 2 δ Γ ( 2 δ + 1 ) ,
Φ 3 ( x , t ¯ ) = S 1 ω δ ξ δ S σ 3 Φ 2 x 3 6 σ Φ 2 Φ 2 x + 6 Ψ 2 Ψ 2 x = σ 3 4 sin h σ 2 + x 2 2 Γ 2 ( δ + 1 ) Γ ( 3 δ + 1 ) cosh 7 β 2 + x 2 [ 2 Γ 2 ( δ + 1 ) cos h 4 σ 2 + x 2 18 Γ 2 ( δ + 1 ) cos h 2 σ 2 + x 2 + 6 Γ ( 2 δ + 1 ) cos h 2 σ 2 + x 2 + 18 Γ 2 ( δ + 1 ) 9 Γ ( 2 δ + 1 ) ] , Ψ 3 ( x , t ¯ ) = S 1 ω δ ξ δ S σ 3 Ψ 2 x 3 3 σ Φ 2 Ψ 2 x = σ 7 / 2 11 sin h σ 2 + x 2 2 2 Γ 2 ( δ + 1 ) Γ ( 3 δ + 1 ) cosh 7 β 2 + x 2 [ 2 Γ 2 ( δ + 1 ) cos h 4 σ 2 + x 2 18 Γ 2 ( δ + 1 ) cos h 2 σ 2 + x 2 + 6 Γ ( 2 δ + 1 ) cos h 2 σ 2 + x 2 + 18 Γ 2 ( δ + 1 ) 9 Γ ( 2 δ + 1 ) ] ,
Φ n ( x , t ¯ ) = S 1 ω δ ξ δ S σ 3 Φ m 1 x 3 6 σ Φ m 1 Φ m 1 x + 6 Ψ m 1 Ψ m 1 x , Ψ m ( x , t ¯ ) = S 1 ω δ ξ δ S σ 3 Ψ m 1 x 3 3 σ Φ m 1 Ψ m 1 x .
The series of solutions for Problem 1 is presented as:
Φ ( x , t ¯ ) = Φ 0 ( x , t ¯ ) + Φ 1 ( x , t ¯ ) + Φ 2 ( x , t ¯ ) + Φ 3 ( x , t ¯ ) + . . . Φ m ( x , t ¯ ) , Ψ ( x , t ¯ ) = Ψ 0 ( x , t ¯ ) + Ψ 1 ( x , t ¯ ) + Ψ 2 ( x , t ¯ ) + Ψ 3 ( x , t ¯ ) + . . . Ψ m ( x , t ¯ ) .
Consequently, we have
Φ ( x , t ¯ ) = 2 sec h 2 β 2 + x 2 + 5 σ tan h β 2 + x 2 sec h 2 β 2 + x 2 t ¯ δ Γ ( δ + 1 ) + 8 σ 2 2 2 cosh 2 σ 2 + x 2 3 sec h 4 σ 2 + x 2 t ¯ 2 δ Γ ( 2 δ + 1 ) + σ 3 4 sin h σ 2 + x 2 2 Γ 2 ( δ + 1 ) Γ ( 3 δ + 1 ) cosh 7 β 2 + x 2 [ 2 Γ 2 ( δ + 1 ) cos h 4 σ 2 + x 2 18 Γ 2 ( δ + 1 ) cos h 2 σ 2 + x 2 + 6 Γ ( 2 δ + 1 ) cos h 2 σ 2 + x 2 + 18 Γ 2 ( δ + 1 ) 9 Γ ( 2 δ + 1 ) ] + . . . , Ψ ( x , t ¯ ) = σ 2 2 sec h 2 β 2 + x 2 + 5 σ 3 / 2 2 tan h β 2 + x 2 sec h 2 β 2 + x 2 t ¯ δ Γ ( δ + 1 ) + 5 σ 5 / 2 2 2 2 cosh 2 σ 2 + x 2 3 sec h 4 σ 2 + x 2 t ¯ 2 δ Γ ( 2 δ + 1 ) + σ 7 / 2 11 sin h σ 2 + x 2 2 2 Γ 2 ( δ + 1 ) Γ ( 3 δ + 1 ) cosh 7 β 2 + x 2 [ 2 Γ 2 ( δ + 1 ) cos h 4 σ 2 + x 2 18 Γ 2 ( δ + 1 ) cos h 2 σ 2 + x 2 + 6 Γ ( 2 δ + 1 ) cos h 2 σ 2 + x 2 + 18 Γ 2 ( δ + 1 ) 9 Γ ( 2 δ + 1 ) ] + . . . .
By setting δ = 1 , we then obtain the exact solution of coupled KdV Equation (1)
Φ ( x , t ¯ ) = 2 sec h 2 β 2 + x 2 σ 3 t ¯ 2 , Ψ ( x , t ¯ ) = σ 2 2 sec h 2 β 2 + x 2 σ 3 t ¯ 2 .
In Figure 1 and Figure 2, the exact and approximate results of Φ ( x , t ¯ ) and Ψ ( x , t ¯ ) are demonstrated at = 1 , σ = 0.5 and β = 2 . In Figure 3 and Figure 4, the surface and 2D graph for Φ ( x , t ¯ ) and Ψ ( x , t ¯ ) for various fractional order are presented which shows that the SDM/SITM approximated results derived are in a strong agreement with the exact and the numerical ones. This comparison represents a strong correlation between the SDM and exact findings. Therefore, the SDM/SITM are reliable novel approaches which require less computation time and is quite straightforward and more flexible than the homotopy perturbation method and homotopy analysis method.
Problem 2
([16]). Assume the time-fractional coupled nonlinear KdV equation is presented as:
δ Φ t ¯ δ = Ψ x 1 2 Φ 2 x δ Ψ t ¯ δ = Φ x 3 Φ x 3 Φ Ψ x , t ¯ > 0 , 0 < δ 1
subject to the condition
Φ ( x , 0 ) = σ tan h 2 + σ x 2 + 1 , Ψ ( x , 0 ) = σ 2 2 sec h 2 2 + σ x 2 1 .
Case I. First, we surmise the Shehu decomposition method for Problem 2.
Employing the Shehu transformation to (34), we find
ξ δ ω δ U ( ξ , ω ) κ = 0 m 1 ξ ω δ κ 1 Φ ( κ ) ( 0 ) = S Ψ x 1 2 Φ 2 x , ξ δ ω δ V ( ξ , ω ) κ = 0 m 1 ξ ω δ κ 1 Ψ ( κ ) ( 0 ) = S Φ x 3 Φ x 3 Φ Ψ x .
In view of (35) and simple computations yield
U ( ξ , ω ) = ω ξ Φ ( 0 ) ( x , 0 ) + ω δ ξ δ S Ψ x 1 2 Φ 2 x , V ( ξ , ω ) = ω ξ Ψ ( 0 ) ( x , 0 ) + ω δ ξ δ S Φ x 3 Φ x 3 Φ Ψ x .
Applying the inverse Shehu transform, we have
Φ ( x , t ¯ ) = S 1 ω ξ Φ ( x , 0 ) + S 1 ω δ ξ δ S Ψ x 1 2 Φ 2 x , Ψ ( x , t ¯ ) = S 1 ω ξ Ψ ( x , 0 ) + S 1 ω δ ξ δ S Φ x 3 Φ x 3 Φ Ψ x .
By virtue of the Shehu decomposition method, we have
Φ 0 ( x , t ¯ ) = S 1 ω ξ Φ ( x , 0 ) = S 1 ω ξ σ tan h 2 + σ x 2 + 1 = σ tan h 2 + σ x 2 + 1 , Ψ 0 ( x , t ¯ ) = S 1 ω ξ Ψ ( x , 0 ) = σ 2 2 sec h 2 2 + σ x 2 1 .
It follows that
m = 0 Φ m + 1 ( x , t ¯ ) = S 1 ω δ ξ δ S σ m = 0 ( Ψ x ) m 1 2 m = 0 D m , m = 0 Ψ m + 1 ( x , t ¯ ) = S 1 ω δ ξ δ S m = 0 ( Φ x ) m m = 0 ( Ψ x x x ) m m = 0 ( Φ Ψ ) x m , m = 0 , 1 , 2 , . . . .
The first few Adomian polynomials are presented as follows:
D 0 ( Φ 2 ) = Φ 0 2 , D 1 ( Φ 2 ) = 2 Φ 0 Φ 1 , D 2 ( Φ 2 ) = 2 Φ 0 Φ 2 + Φ 1 2 .
For m = 0 , 1 , 2 , . . .
Φ 1 ( x , t ¯ ) = S 1 ω δ ξ δ S σ ( Ψ x ) 0 1 2 D 0 = σ 2 2 S 1 ω δ + 2 ξ δ + 2 sec h 2 2 + σ x 2 = σ 2 2 sec h 2 2 + σ x 2 t ¯ δ Γ ( δ + 1 ) , Ψ 1 ( x , t ¯ ) = S 1 ω δ ξ δ S ( Φ x ) 0 ( Ψ x x x ) 0 ( Φ Ψ ) x 0 = σ 3 2 sin h 2 + σ x 2 sec h 3 2 + σ x 2 t ¯ δ Γ ( δ + 1 ) .
Φ 2 ( x , t ¯ ) = S 1 ω δ ξ δ S σ ( Ψ x ) 1 1 2 D 1 = S 1 σ 5 4 ω 2 δ + 2 ξ 2 δ + 2 sec h 2 2 + σ x 2 + 3 σ 5 4 ω 2 δ + 2 ξ 2 δ + 2 sin h 2 2 + σ x 2 sec h 4 2 + σ x 2 + σ 7 4 S 1 Γ ( 2 δ + 1 ) Γ 2 ( δ + 1 ) ω 3 δ + 2 ξ 3 δ + 2 sin h 2 + σ x 2 sec h 5 2 + σ x 2 = σ 5 4 sec h 2 2 + σ x 2 + 3 σ 5 4 sin h 2 2 + σ x 2 sec h 4 2 + σ x 2 t ¯ 2 δ Γ ( 2 δ + 1 ) + σ 7 4 sin h 2 + σ x 2 sec h 5 2 + σ x 2 Γ ( 2 δ + 1 ) t ¯ 3 δ Γ 2 ( δ + 1 ) Γ ( 3 δ + 1 ) ,
Ψ 2 ( x , t ¯ ) = S 1 ω δ ξ δ ( Φ x ) 1 ( Ψ x x x ) 1 ( Φ Ψ ) x 1 = σ 6 4 2 cosh 2 2 + σ x 2 3 sec h 4 2 + σ x 2 t ¯ 2 δ Γ ( 2 δ + 1 ) , .
The Shehu decomposition method solution for Problem 2 is presented as:
Φ ( x , t ¯ ) = Φ 0 ( x , t ¯ ) + Φ 1 ( x , t ¯ ) + Φ 2 ( x , t ¯ ) + . . . , = σ tan h 2 + σ x 2 + 1 σ 2 2 sec h 2 2 + σ x 2 t ¯ δ Γ ( δ + 1 ) + σ 5 4 sec h 2 2 + σ x 2 + 3 σ 5 4 sin h 2 2 + σ x 2 sec h 4 2 + σ x 2 t ¯ 2 δ Γ ( 2 δ + 1 ) + σ 7 4 sin h 2 + σ x 2 sec h 5 2 + σ x 2 Γ ( 2 δ + 1 ) t ¯ 3 δ Γ 2 ( δ + 1 ) Γ ( 3 δ + 1 ) + . . . .
Analogously, we have
Ψ ( x , t ¯ ) = 1 + σ 2 2 sec h 2 2 + σ x 2 + σ 3 2 sin h 2 + σ x 2 sec h 3 2 + σ x 2 t ¯ δ Γ ( δ + 1 ) + σ 6 4 2 cosh 2 2 + σ x 2 3 sec h 4 2 + σ x 2 t ¯ 2 δ Γ ( 2 δ + 1 ) + . . . .
By setting δ = 1 , we obtain the exact solution of the coupled KdV Equation (34)
Φ ( x , t ¯ ) = σ tanh 2 + σ x 2 σ 2 t ¯ 2 + 1 , Ψ ( x , t ¯ ) = σ 2 2 sec h 2 2 + σ x 2 σ 2 t ¯ 2 1 .
Case II. Now, we surmise the Shehu iterative transform method on Problem 2.
Applying the proposed analytical approach to (38) yields
Φ 0 ( x , t ¯ ) = S 1 ω ξ Φ ( x , 0 ) = S 1 ω ξ σ tan h 2 + σ x 2 + 1 = σ tan h 2 + σ x 2 + 1 , Ψ 0 ( x , t ¯ ) = S 1 ω ξ Ψ ( x , 0 ) = σ 2 2 sec h 2 2 + σ x 2 1 Φ 1 ( x , t ¯ ) = S 1 ω δ ξ δ S Ψ 0 x 1 2 Φ 0 2 x = σ 2 2 S 1 ω δ + 2 ξ δ + 2 sec h 2 2 + σ x 2 = σ 2 2 sec h 2 2 + σ x 2 t ¯ δ Γ ( δ + 1 ) ,
Ψ 1 ( x , t ¯ ) = S 1 ω δ ξ δ S Φ 0 x 3 Φ 0 x 3 Φ 0 Ψ 0 x = σ 3 2 sin h 2 + σ x 2 sec h 3 2 + σ x 2 t ¯ δ Γ ( δ + 1 ) ,
Φ 2 ( x , t ¯ ) = S 1 ω δ ξ δ S Ψ 1 x 1 2 Φ 1 2 x = S 1 σ 5 4 ω 2 δ + 2 ξ 2 δ + 2 sec h 2 2 + σ x 2 + 3 σ 5 4 ω 2 δ + 2 ξ 2 δ + 2 sin h 2 2 + σ x 2 sec h 4 2 + σ x 2 + σ 7 4 S 1 Γ ( 2 δ + 1 ) Γ 2 ( δ + 1 ) ω 3 δ + 2 ξ 3 δ + 2 sin h 2 + σ x 2 sec h 5 2 + σ x 2 = σ 5 4 sec h 2 2 + σ x 2 + 3 σ 5 4 sin h 2 2 + σ x 2 sec h 4 2 + σ x 2 t ¯ 2 δ Γ ( 2 δ + 1 ) + σ 7 4 sin h 2 + σ x 2 sec h 5 2 + σ x 2 Γ ( 2 δ + 1 ) t ¯ 3 δ Γ 2 ( δ + 1 ) Γ ( 3 δ + 1 ) ,
Ψ 2 ( x , t ¯ ) = S 1 ω δ ξ δ Φ 1 x 3 Φ 1 x 3 Φ 1 Ψ 1 x = σ 6 4 2 cosh 2 2 + σ x 2 3 sec h 4 2 + σ x 2 t ¯ 2 δ Γ ( 2 δ + 1 ) , Φ m ( x , t ¯ ) = S 1 ω δ ξ δ S Ψ m 1 x 1 2 Φ m 1 2 x , Ψ m ( x , t ¯ ) = S 1 ω δ ξ δ Φ m 1 x 3 Φ m 1 x 3 Φ m 1 Ψ m 1 x .
The series of solution for Problem 2 is presented as:
Φ ( x , t ¯ ) = Φ 0 ( x , t ¯ ) + Φ 1 ( x , t ¯ ) + Φ 2 ( x , t ¯ ) + . . . Φ m ( x , t ¯ ) , Ψ ( x , t ¯ ) = Ψ 0 ( x , t ¯ ) + Ψ 1 ( x , t ¯ ) + Ψ 2 ( x , t ¯ ) + . . . Ψ m ( x , t ¯ ) .
Consequently, we have
Φ ( x , t ¯ ) = σ tan h 2 + σ x 2 + 1 σ 2 2 sec h 2 2 + σ x 2 t ¯ δ Γ ( δ + 1 ) + σ 5 4 sec h 2 2 + σ x 2 + 3 σ 5 4 sin h 2 2 + σ x 2 sec h 4 2 + σ x 2 t ¯ 2 δ Γ ( 2 δ + 1 ) + σ 7 4 sin h 2 + σ x 2 sec h 5 2 + σ x 2 Γ ( 2 δ + 1 ) t ¯ 3 δ Γ 2 ( δ + 1 ) Γ ( 3 δ + 1 ) + . . . , Ψ ( x , t ¯ ) = 1 + σ 2 2 sec h 2 2 + σ x 2 + σ 3 2 sin h 2 + σ x 2 sec h 3 2 + σ x 2 t ¯ δ Γ ( δ + 1 ) + σ 6 4 2 cosh 2 2 + σ x 2 3 sec h 4 2 + σ x 2 t ¯ 2 δ Γ ( 2 δ + 1 ) + . . . .
By setting δ = 1 , we then obtain the exact solution of coupled KdV Equation (34)
Φ ( x , t ¯ ) = σ tanh 2 + σ x 2 σ 2 t ¯ 2 + 1 , Ψ ( x , t ¯ ) = σ 2 2 sec h 2 2 + σ x 2 σ 2 t ¯ 2 1 .
In Figure 5 and Figure 6, the exact and approximate results of Φ ( x , t ¯ ) and Ψ ( x , t ¯ ) are demonstrated at = 1 , σ = 0.5 and β = 2 . In Figure 7 and Figure 8, the surface and 2D graph for Φ ( x , t ¯ ) and Ψ ( x , t ¯ ) for various fractional order are presented which shows that the SDM/SITM approximated results derived are in a strong agreement with the exact and the numerical ones. This comparison represents a strong correlation between the SDM and exact findings. Therefore, the SDM/SITM are reliable novel approaches which require less computation time and are quite straightforward and more flexible than the homotopy perturbation method and the homotopy analysis method.
Problem 3
([16]). Assume the time-fractional coupled nonlinear MCKdV equations is presented as (2) subject to the condition
Φ ( x , 0 ) = 2 + tanh x 2 , Ψ ( x , 0 ) = 2 tanh x 4 , Υ ( x , 0 ) = 2 tanh x .
Case I. First, we surmise the Shehu decomposition method for Problem 3.
Employing the Shehu transformation to (2), we find
ξ δ ω δ U ( ξ , ω ) κ = 0 m 1 ξ ω δ κ 1 Φ ( κ ) ( 0 ) = S [ 1 2 3 Φ t ¯ 3 3 Φ 2 Φ x + 3 2 Υ 2 Ψ x 2 + 3 Ψ x Υ x + 3 2 Ψ 2 Υ x 2 + 3 Ψ Υ Φ x + 3 Φ Υ Ψ x + 3 Φ Ψ Υ x ] , ξ δ ω δ V ( ξ , ω ) κ = 0 m 1 ξ ω δ κ 1 Ψ ( κ ) ( 0 ) = S 3 Ψ x 3 3 Φ x Ψ x 3 Ψ 2 Φ x 2 3 Ψ 2 Υ x + 6 Φ Ψ Φ x + 3 Φ 2 Ψ x , ξ δ ω δ W ( ξ , ω ) κ = 0 m 1 ξ ω δ κ 1 Υ ( κ ) ( 0 ) = S 3 Υ x 3 3 Φ x Υ x 3 Υ 2 Φ x 2 3 Υ 2 Ψ x + 6 Φ Υ Φ x + 3 Φ 2 Υ x .
In view of (39) and simple computations yield
U ( ξ , ω ) = ω ξ Φ ( 0 ) ( x , 0 ) + ω δ ξ δ S [ 1 2 3 Φ t ¯ 3 3 Φ 2 Φ x + 3 2 Υ 2 Ψ x 2 + 3 Ψ x Υ x + 3 2 Ψ 2 Υ x 2 + 3 Ψ Υ Φ x + 3 Φ Υ Ψ x + 3 Φ Ψ Υ x ] , V ( ξ , ω ) = ω ξ Ψ ( 0 ) ( x , 0 ) + ω δ ξ δ S 3 Ψ x 3 3 Φ x Ψ x 3 Ψ 2 Φ x 2 3 Ψ 2 Υ x + 6 Φ Ψ Φ x + 3 Φ 2 Ψ x , W ( ξ , ω ) = ω ξ Υ ( 0 ) ( x , 0 ) + ω δ ξ δ S 3 Υ x 3 3 Φ x Υ x 3 Υ 2 Φ x 2 3 Υ 2 Ψ x + 6 Φ Υ Φ x + 3 Φ 2 Υ x .
Applying the inverse Shehu transform, we have
Φ ( x , t ¯ ) = S 1 ω ξ Φ ( x , 0 ) + S 1 [ ω δ ξ δ S [ 1 2 3 Φ t ¯ 3 3 Φ 2 Φ x + 3 2 Υ 2 Ψ x 2 + 3 Ψ x Υ x + 3 2 Ψ 2 Υ x 2 + 3 Ψ Υ Φ x + 3 Φ Υ Ψ x + 3 Φ Ψ Υ x ] ] , Ψ ( x , t ¯ ) = S 1 ω ξ Ψ ( x , 0 ) + S 1 ω δ ξ δ S 3 Ψ x 3 3 Φ x Ψ x 3 Ψ 2 Φ x 2 3 Ψ 2 Υ x + 6 Φ Ψ Φ x + 3 Φ 2 Ψ x , Υ ( x , t ¯ ) = S 1 ω ξ Υ ( x , 0 ) + S 1 ω δ ξ δ S 3 Υ x 3 3 Φ x Υ x 3 Υ 2 Φ x 2 3 Υ 2 Ψ x + 6 Φ Υ Φ x + 3 Φ 2 Υ x .
By virtue of the Shehu decomposition method, we have
Φ 0 ( x , t ¯ ) = S 1 ω ξ Φ ( x , 0 ) = 1 2 S 1 ω ξ 2 + tanh x = 1 2 2 + tanh x , Ψ 0 ( x , t ¯ ) = S 1 ω ξ Ψ ( x , 0 ) = 1 4 2 tanh x , Υ 0 ( x , t ¯ ) = S 1 ω ξ Υ ( x , 0 ) = 2 tanh x .
It follows that
m = 0 Φ m + 1 ( x , t ¯ ) = S 1 [ ω δ ξ δ S [ 1 2 m = 0 ( Φ x x x ) m 3 m = 0 E m + 3 2 m = 0 F m + 3 m = 0 G m + 3 2 m = 0 H m + 3 m = 0 I m + 3 m = 0 J m + 3 m = 0 K m ] ] , m = 0 Ψ m + 1 ( x , t ¯ ) = S 1 ω δ ξ δ S m = 0 ( Ψ x x x ) m 3 m = 0 M m 3 m = 0 N m 3 m = 0 O m + 6 m = 0 P m + 3 m = 0 Q m , m = 0 Υ m + 1 ( x , t ¯ ) = S 1 ω δ ξ δ S m = 0 ( Υ x x x ) m 3 m = 0 R m 3 m = 0 S m 3 m = 0 T m + 6 m = 0 X m + 3 m = 0 Y m , m = 0 , 1 , 2 , . . . .
The first few Adomian polynomials are presented as follows:
E j ( Φ 2 Φ x ) = Φ 0 2 Φ 0 x , f o r j = 0 ( 2 Φ 0 Φ 1 ) Φ 0 x + Φ 0 2 Φ 1 x , f o r j = 1 ( 2 Φ 0 Φ 2 + Φ 1 2 ) Φ 0 x + ( 2 Φ 0 Φ 1 ) Φ 1 x + Φ 0 2 Φ 2 x , f o r j = 2 F j ( Υ Ψ x x ) = Υ 0 Ψ 0 x x , f o r j = 0 Υ 1 Ψ 0 x x + Υ 0 Ψ 1 x x , f o r j = 1 Υ 2 Ψ 0 x x + Υ 1 Ψ 1 x x + Υ 0 Ψ 2 x x , f o r j = 2 G j ( Ψ x Υ x ) = Ψ 0 x Υ 0 x , f o r j = 0 Ψ 0 x Υ 1 x + Ψ 1 x Υ 0 x f o r j = 1 Ψ 2 x Υ 0 x + Ψ 1 x Υ 1 x + Ψ 0 x Υ 2 x f o r j = 2 H j ( Ψ x Υ x x ) = Ψ 0 x Υ 0 x x , f o r j = 0 Ψ 0 x Υ 1 x x + Ψ 1 x Υ 0 x x , f o r j = 1 Ψ 2 x Υ 0 x x + Ψ 1 x Υ 1 x x + Ψ 0 x Υ 2 x x , f o r j = 2 I j ( Ψ z Φ x ) = ( Ψ Υ ) 0 Φ 0 x , f o r j = 0 ( Ψ Υ ) 0 Φ 1 x + ( Ψ Υ ) 1 Φ 0 x , f o r j = 1 ( Ψ Υ ) 0 Φ 2 x + ( Ψ Υ ) 1 Φ 1 x + ( Ψ Υ ) 2 Φ 0 x , f o r j = 2 J j ( Φ z Ψ x ) = ( Φ Υ ) 0 Ψ 0 x , f o r j = 0 ( Φ Υ ) 0 Ψ 1 x + ( Φ Υ ) 1 Ψ 0 x , f o r j = 1 ( Φ Υ ) 0 Ψ 2 x + ( Φ Υ ) 1 Ψ 1 x + ( Φ Υ ) 2 Ψ 0 x , f o r j = 2 K j ( Φ Ψ Υ x ) = ( Φ Ψ ) 0 Υ 0 x , f o r j = 0 ( Φ Ψ ) 0 Υ 1 x + ( Φ Ψ ) 1 Υ 0 x , f o r j = 1 ( Φ Ψ ) 0 Υ 2 x + ( Φ Ψ ) 1 Υ 1 x + ( Φ Ψ ) 2 Υ 0 x , f o r j = 2 M j ( Φ x Ψ x ) = Φ 0 x Ψ 0 x , f o r j = 0 Φ 0 x Ψ 1 x + Φ x Ψ 0 x , f o r j = 1 Φ 2 x Ψ 0 x + Φ 1 x Ψ 1 x + Φ 0 x Ψ 1 x , f o r j = 2 N j ( Ψ Φ x x ) = Ψ 0 Φ 0 x x , f o r j = 0 Ψ 0 Φ 1 x x + Ψ 1 Φ 0 x x , f o r j = 1 Ψ 2 Φ 0 x x + Ψ 1 Φ 1 x x + Ψ 0 Φ 2 x x , f o r j = 2 O j ( Ψ 2 Υ x ) = Ψ 0 2 Υ 0 x , f o r j = 0 ( 2 Ψ 0 Ψ 1 ) Υ 0 x + Ψ 0 2 Υ 1 x , f o r j = 1 ( 2 Ψ 0 Ψ 2 + Ψ 1 2 ) Υ 0 x + ( 2 Ψ 0 Ψ 1 ) Υ 1 x + Ψ 0 2 Υ 2 x , f o r j = 2
P j ( Φ Ψ Φ x ) = ( Φ Ψ ) 0 Φ 0 x , f o r j = 0 ( Φ Ψ ) 0 Φ 1 x + ( Φ Ψ ) 0 Φ 1 x , f o r j = 1 ( Φ Ψ ) 0 Φ 2 x + ( Φ Ψ ) 1 Φ 1 x + ( Φ Ψ ) 2 Φ 0 x , f o r j = 2 Q j ( Φ 2 Ψ x ) = Φ 0 2 Ψ 0 x , f o r j = 0 ( 2 Φ 0 Φ 1 ) Ψ 0 x + Φ 0 2 Ψ 1 x , f o r j = 1 ( 2 Φ 0 Φ 2 + Φ 1 2 ) Ψ 0 x + ( 2 Φ 0 Φ 1 ) Ψ 1 x + Φ 0 2 Ψ 2 x , f o r j = 2
R 0 ( Φ x Ψ x ) = Φ 0 x Ψ 0 x , f o r j = 0 Φ 0 x Ψ 1 x + Φ 1 x Ψ 0 x , f o r j = 1 Φ 2 x Ψ 0 x + Φ 1 x Ψ 1 x + Φ 0 x Ψ 2 x , f o r j = 2 S j ( Υ Φ x x ) = Υ 0 Φ 0 x x , f o r j = 0 Υ 0 Φ 1 x x + Υ 1 Φ 0 x x , f o r j = 1 Υ 2 Φ 0 x x + Υ 1 Φ 1 x x + Υ 0 Φ 2 x x , f o r j = 2 T j ( Υ 2 Ψ x ) = Υ 0 2 Ψ 0 x , f o r j = 0 ( 2 Υ 0 Υ 1 ) Ψ 0 x + Υ 0 2 Ψ 1 x , f o r j = 1 ( 2 Υ 0 Υ 2 + Υ 1 2 ) Ψ 0 x + ( 2 Υ 0 Υ 1 ) Ψ 1 x + Υ 0 2 Ψ 2 x , f o r j = 2 X j ( Φ Υ 1 Φ x ) = ( Φ Υ ) 0 Φ 0 x , f o r j = 0 ( Φ Υ ) 0 Φ 1 x + ( Φ Υ ) 1 Φ 0 x , f o r j = 1 ( Φ Υ ) 2 Φ 0 x + ( Φ Υ ) 1 Φ 1 x + ( Φ Υ ) 2 Φ 0 x , f o r j = 2 Y j ( Φ 2 Υ x ) = Φ 0 2 Υ 0 x , f o r j = 0 ( 2 Φ 0 Φ 1 ) Υ 0 x + Φ 0 2 Υ 1 x , f o r j = 1 ( 2 Φ 0 Φ 2 + Φ 1 2 ) Υ 0 x + ( 2 Φ 0 Φ 1 ) Υ 1 x + Φ 0 2 Υ 2 x , f o r j = 2 .
For m = 0 , 1 , 2 , 3 , . . .
Φ 1 ( x , t ¯ ) = S 1 ω δ ξ δ S 1 2 ( Φ x x x ) 0 3 E 0 + 3 2 F 0 + 3 G 0 + 3 2 H 0 + 3 I 0 + 3 J 0 + 3 K 0 = 11 2 sec h 2 ( x ) S 1 ω δ + 2 ξ δ + 2 = 11 2 sec h 2 ( x ) t ¯ δ Γ ( δ + 1 ) , Ψ 1 ( x , t ¯ ) = S 1 ω δ ξ δ S ( Ψ x x x ) 0 3 M 0 3 N 0 3 O 0 + 6 P 0 + 3 Q 0 = 11 8 sec h 2 ( x ) t ¯ δ Γ ( δ + 1 ) , Υ 1 ( x , t ¯ ) = S 1 ω δ ξ δ S ( Υ x x x ) 0 3 R 0 3 S ^ 0 3 T 0 + 6 X 0 + 3 Y 0 = 11 2 sec h 2 ( x ) t ¯ δ Γ ( δ + 1 ) .
Φ 2 ( x , t ¯ ) = S 1 ω δ ξ δ S 1 2 ( Φ x x x ) 1 3 E 1 + 3 2 F 1 + 3 G 1 + 3 2 H 1 + 3 I 1 + 3 J 1 + 3 K 1 = 121 8 tan h ( x ) sec h 2 ( x ) t ¯ 2 δ Γ ( 2 δ + 1 ) , Ψ 2 ( x , t ¯ ) = S 1 ω δ ξ δ S ( Ψ x x x ) 1 3 M 1 3 N 1 3 O 1 + 6 P 1 + 3 Q 1 = 121 8 tan h ( x ) sec h 2 ( x ) t ¯ 2 δ Γ ( 2 δ + 1 ) , Υ 2 ( x , t ¯ ) = S 1 ω δ ξ δ S ( Υ x x x ) 1 3 R 1 3 S ^ 1 3 T 1 + 6 X 1 + 3 Y 1 = 242 8 tan h ( x ) sec h 2 ( x ) t ¯ 2 δ Γ ( 2 δ + 1 ) ,
Φ 3 ( x , t ¯ ) = S 1 ω δ ξ δ S 1 2 ( Φ x x x ) 2 3 E 2 + 3 2 F 2 + 3 G 2 + 3 2 H 2 + 3 I 2 + 3 J 2 + 3 K 2 = 1331 48 sec h 4 ( x ) cosh ( 2 x ) 2 t ¯ 3 δ Γ ( 3 δ + 1 ) , Ψ 3 ( x , t ¯ ) = S 1 ω δ ξ δ S ( Ψ x x x ) 2 3 M 2 3 N 2 3 O 2 + 6 P 2 + 3 Q 2 = 2662 96 sec h 4 ( x ) cosh ( 2 x ) 2 t ¯ 3 δ Γ ( 3 δ + 1 ) , Υ 3 ( x , t ¯ ) = S 1 ω δ ξ δ S ( Υ x x x ) 2 3 R 2 3 S ^ 2 3 T 2 + 6 X 2 + 3 Y 2 = 2662 48 sec h 4 ( x ) cosh ( 2 x ) 2 t ¯ 3 δ Γ ( 3 δ + 1 ) , .
The Shehu decomposition method solution for Problem 3 is presented as:
Φ ( x , t ¯ ) = Φ 0 ( x , t ¯ ) + Φ 1 ( x , t ¯ ) + Φ 2 ( x , t ¯ ) + Φ 3 ( x , t ¯ ) . . . , = 1 2 2 + tanh x + 11 2 sec h 2 ( x ) t ¯ δ Γ ( δ + 1 ) 121 8 tan h ( x ) sec h 2 ( x ) t ¯ 2 δ Γ ( 2 δ + 1 ) + 1331 48 sec h 4 ( x ) cosh ( 2 x ) 2 t ¯ 3 δ Γ ( 3 δ + 1 ) + . . . .
Analogously, we have
Ψ ( x , t ¯ ) = 1 4 2 tanh x 11 8 sec h 2 ( x ) t ¯ δ Γ ( δ + 1 ) + 121 8 tan h ( x ) sec h 2 ( x ) t ¯ 2 δ Γ ( 2 δ + 1 ) 1331 48 sec h 4 ( x ) cosh ( 2 x ) 2 t ¯ 3 δ Γ ( 3 δ + 1 ) + . . . , Υ ( x , t ¯ ) = 2 tanh x 11 2 sec h 2 ( x ) t ¯ δ Γ ( δ + 1 ) + 121 4 tan h ( x ) sec h 2 ( x ) t ¯ 2 δ Γ ( 2 δ + 1 ) 2662 48 sec h 4 ( x ) cosh ( 2 x ) 2 t ¯ 3 δ Γ ( 3 δ + 1 ) + . . . .
By Setting δ = 1 , we then obtain the exact solution of coupled KdV Equation (2)
Φ ( x , t ¯ ) = 1 2 2 + tanh x 11 t ¯ 2 , Ψ ( x , t ¯ ) = 1 4 2 tanh x 11 t ¯ 2 , Υ ( x , t ¯ ) = 2 tanh x 11 t ¯ 2 .
Case II. Now, we surmise the new iterative transform method for Problem 3.
Applying the proposed analytical approach to (42) yields
Φ 0 ( x , t ¯ ) = 1 2 2 + tanh x , Ψ 0 ( x , t ¯ ) = 1 4 2 tanh x , Υ 0 ( x , t ¯ ) = 2 tanh x , Φ 1 ( x , t ¯ ) = S 1 [ ω δ ξ δ S [ 1 2 3 Φ 0 t ¯ 3 3 Φ 0 2 Φ 0 x + 3 2 Υ 0 2 Ψ 0 x 2 + 3 Ψ 0 x Υ 0 x + 3 2 Ψ 0 2 Υ 0 x 2 + 3 Ψ 0 Υ 0 Φ 0 x + 3 Φ 0 Υ 0 Ψ 0 x + 3 Φ 0 Ψ 0 Υ 0 x ] ] = 11 2 sec h 2 ( x ) S 1 ω δ + 2 ξ δ + 2 = 11 2 sec h 2 ( x ) t ¯ δ Γ ( δ + 1 ) ,
Ψ 1 ( x , t ¯ ) = S 1 ω δ ξ δ S 3 Ψ 0 x 3 3 Φ 0 x Ψ 0 x 3 Ψ 0 2 Φ 0 x 2 3 Ψ 0 2 Υ 0 x + 6 Φ 0 Ψ 0 Φ 0 x + 3 Φ 0 2 Ψ 0 x = 11 8 sec h 2 ( x ) t ¯ δ Γ ( δ + 1 ) , Υ 1 ( x , t ¯ ) = S 1 ω δ ξ δ S 3 Υ 0 x 3 3 Φ 0 x Υ 0 x 3 Υ 0 2 Φ 0 x 2 3 Υ 0 2 Ψ 0 x + 6 Φ 0 Υ 0 Φ 0 x + 3 Φ 0 2 Υ 0 x = 11 2 sec h 2 ( x ) t ¯ δ Γ ( δ + 1 ) ,
Φ 2 ( x , t ¯ ) = S 1 [ ω δ ξ δ S [ 1 2 3 Φ 1 t ¯ 3 3 Φ 1 2 Φ 1 x + 3 2 Υ 1 2 Ψ 1 x 2 + 3 Ψ 1 x Υ 1 x + 3 2 Ψ 1 2 Υ 1 x 2 + 3 Ψ 1 Υ 1 Φ 1 x + 3 Φ 1 Υ 1 Ψ 1 x + 3 Φ 1 Ψ 1 Υ 1 x ] ] = 121 8 tan h ( x ) sec h 2 ( x ) t ¯ 2 δ Γ ( 2 δ + 1 ) ,
Ψ 2 ( x , t ¯ ) = S 1 ω δ ξ δ S 3 Ψ 1 x 3 3 Φ 1 x Ψ 1 x 3 Ψ 1 2 Φ 1 x 2 3 Ψ 1 2 Υ 1 x + 6 Φ 1 Ψ 1 Φ 1 x + 3 Φ 1 2 Ψ 1 x = 121 8 tan h ( x ) sec h 2 ( x ) t ¯ 2 δ Γ ( 2 δ + 1 ) , Υ 2 ( x , t ¯ ) = S 1 ω δ ξ δ S 3 Υ 1 x 3 3 Φ 1 x Υ 1 x 3 Υ 1 2 Φ 1 x 2 3 Υ 1 2 Ψ 1 x + 6 Φ 1 Υ 1 Φ 1 x + 3 Φ 1 2 Υ 1 x = 242 8 tan h ( x ) sec h 2 ( x ) t ¯ 2 δ Γ ( 2 δ + 1 ) ,
Φ 3 ( x , t ¯ ) = S 1 [ ω δ ξ δ S [ 1 2 3 Φ 2 t ¯ 3 3 Φ 2 2 Φ 2 x + 3 2 Υ 2 2 Ψ 2 x 2 + 3 Ψ 2 x Υ 2 x + 3 2 Ψ 2 2 Υ 2 x 2 + 3 Ψ 2 Υ 2 Φ 2 x + 3 Φ 2 Υ 2 Ψ 2 x + 3 Φ 2 Ψ 2 Υ 2 x ] ] = 1331 48 sec h 4 ( x ) cosh ( 2 x ) 2 t ¯ 3 δ Γ ( 3 δ + 1 ) , Ψ 3 ( x , t ¯ ) = S 1 ω δ ξ δ S 3 Ψ 2 x 3 3 Φ 2 x Ψ 2 x 3 Ψ 2 2 Φ 2 x 2 3 Ψ 2 2 Υ 2 x + 6 Φ 2 Ψ 2 Φ 2 x + 3 Φ 2 2 Ψ 2 x = 2662 96 sec h 4 ( x ) cosh ( 2 x ) 2 t ¯ 3 δ Γ ( 3 δ + 1 ) , Υ 3 ( x , t ¯ ) = S 1 ω δ ξ δ S 3 Υ 2 x 3 3 Φ 2 x Υ 2 x 3 Υ 2 2 Φ 2 x 2 3 Υ 2 2 Ψ 2 x + 6 Φ 2 Υ 2 Φ 2 x + 3 Φ 2 2 Υ 2 x = 2662 48 sec h 4 ( x ) cosh ( 2 x ) 2 t ¯ 3 δ Γ ( 3 δ + 1 ) , Φ m ( x , t ¯ ) = S 1 [ ω δ ξ δ S [ 1 2 3 Φ m 1 t ¯ 3 3 Φ m 1 2 Φ m 1 x + 3 2 Υ m 1 2 Ψ m 1 x 2 + 3 Ψ m 1 x Υ m 1 x + 3 2 Ψ m 1 2 Υ m 1 x 2 + 3 Ψ m 1 Υ m 1 Φ m 1 x + 3 Φ m 1 Υ m 1 Ψ m 1 x + 3 Φ m 1 Ψ m 1 Υ m 1 x ] ] , Ψ m ( x , t ¯ ) = S 1 [ ω δ ξ δ S [ 3 Ψ m 1 x 3 3 Φ m 1 x Ψ m 1 x 3 Ψ m 1 2 Φ m 1 x 2 3 Ψ m 1 2 Υ m 1 x + 6 Φ m 1 Ψ m 1 Φ m 1 x + 3 Φ m 1 2 Ψ m 1 x ] ] , Υ m ( x , t ¯ ) = S 1 [ ω δ ξ δ S [ 3 Υ m 1 x 3 3 Φ m 1 x Υ m 1 x 3 Υ m 1 2 Φ m 1 x 2 3 Υ m 1 2 Ψ m 1 x + 6 Φ m 1 Υ m 1 Φ m 1 x + 3 Φ m 1 2 Υ m 1 x ] ] .
The series solution for Problem 3 is presented as:
Φ ( x , t ¯ ) = Φ 0 ( x , t ¯ ) + Φ 1 ( x , t ¯ ) + Φ 2 ( x , t ¯ ) + Φ 3 ( x , t ¯ ) + . . . Φ m ( x , t ¯ ) , = 1 2 2 + tanh x + 11 2 sec h 2 ( x ) t ¯ δ Γ ( δ + 1 ) 121 8 tan h ( x ) sec h 2 ( x ) t ¯ 2 δ Γ ( 2 δ + 1 ) + 1331 48 sec h 4 ( x ) cosh ( 2 x ) 2 t ¯ 3 δ Γ ( 3 δ + 1 ) + . . . .
Analogously, we have
Ψ ( x , t ¯ ) = 1 4 2 tanh x 11 8 sec h 2 ( x ) t ¯ δ Γ ( δ + 1 ) + 121 8 tan h ( x ) sec h 2 ( x ) t ¯ 2 δ Γ ( 2 δ + 1 ) 1331 48 sec h 4 ( x ) cosh ( 2 x ) 2 t ¯ 3 δ Γ ( 3 δ + 1 ) + . . . , Υ ( x , t ¯ ) = 2 tanh x 11 2 sec h 2 ( x ) t ¯ δ Γ ( δ + 1 ) + 121 4 tan h ( x ) sec h 2 ( x ) t ¯ 2 δ Γ ( 2 δ + 1 ) 2662 48 sec h 4 ( x ) cosh ( 2 x ) 2 t ¯ 3 δ Γ ( 3 δ + 1 ) + . . . .
By setting δ = 1 , we then obtain the exact solution of MCKdV Equation (2)
Φ ( x , t ¯ ) = 1 2 2 + tanh x 11 t ¯ 2 , Ψ ( x , t ¯ ) = 1 4 2 tanh x 11 t ¯ 2 , Υ ( x , t ¯ ) = 2 tanh x 11 t ¯ 2 .
In Figure 9, Figure 10 and Figure 11 the exact and approximate results of Φ ( x , t ¯ ) , Ψ ( x , t ¯ ) and Υ ( x , t ¯ ) are demonstrated at = 1 , σ = 0.5 and β = 2 , respectively. In Figure 12, Figure 13 and Figure 14, the surface and 2D graph for Φ ( x , t ¯ ) , Ψ ( x , t ¯ ) and Υ ( x , t ¯ ) for various fractional orders are presented which show that the SDM/SITM approximated results derived are in a strong agreement with the exact and the numerical ones. This comparison represents a strong correlation between the SDM and exact findings. Therefore, the SDM/SITM are reliable novel approaches which require less computation time and is quite straightforward and more flexible than the homotopy perturbation method or homotopy analysis method, because the ST permits one of several scenarios to reduce the deficiency mainly occurs because of unsatisfied initial conditions that appear in other semi-analytical methods such as the SDM/SITM.

7. Conclusions

Understanding complex nonlinear PDEs remains a difficult challenge when their generative model is unknown. This challenge becomes more complex when it comes to evaluating time fractional nonlinear PDEs, surmising the model that governs their evolution. To cope with this difficulty, numerous numerical methods have been employed for dealing with nonlinear physical phenomena. Toward addressing this goal, in this paper, we have considered a time-fractional KdV equation and have developed effective, rigorous and robust algorithmic strategies (Shehu decompsition method and Shehu iterative transform method) to estimate approximate-analytical solutions and so identify the main numerical solutions appearing in the literature. In this approach, we do not need the Lagrange multiplier, correction functional, stationary conditions, or to calculate heavy integrals because the results established are noise free, which overcomes the shortcomings of existing methods. It is remarkable that the projected approaches are well-organized analytical methods for finding approximate-analytical solutions to complex nonlinear PDEs. Finally, we conclude that this scheme will be taken into account in order to cope with other complex non-linear fractional order systems of equations.

Author Contributions

All authors contributed equally. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Acknowledgments

This research was supported by Taif University Researchers Supporting Project Number (TURSP-2020/96), Taif University, Taif, Saudi Arabia.

Conflicts of Interest

The authors declare no conflict of interest.

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Figure 1. The exact and approximate (SDM/SITM) solution graph at Φ ( x , t ¯ ) of Problem 1 for = 1 , σ = 0.5 and β = 2 .
Figure 1. The exact and approximate (SDM/SITM) solution graph at Φ ( x , t ¯ ) of Problem 1 for = 1 , σ = 0.5 and β = 2 .
Symmetry 13 01254 g001
Figure 2. The exact and approximate (SDM/SITM) solution graph at Ψ ( x , t ¯ ) of Problem 1 for = 1 , σ = 0.5 and β = 2 .
Figure 2. The exact and approximate (SDM/SITM) solution graph at Ψ ( x , t ¯ ) of Problem 1 for = 1 , σ = 0.5 and β = 2 .
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Figure 3. Numerical evaluation of graph of Ψ ( x , t ¯ ) for Problem 1 for various fractional order δ = 0.4, 0.6, 0.8, 1, = 1 , σ = 0.5 and β = 2 .
Figure 3. Numerical evaluation of graph of Ψ ( x , t ¯ ) for Problem 1 for various fractional order δ = 0.4, 0.6, 0.8, 1, = 1 , σ = 0.5 and β = 2 .
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Figure 4. Numerical evaluation of graph Φ ( x , t ¯ ) for Problem 1 for various fractional order δ = 0.4 , 0.6 , 0.8 , 1 , = 1 , σ = 0.5 and β = 2 .
Figure 4. Numerical evaluation of graph Φ ( x , t ¯ ) for Problem 1 for various fractional order δ = 0.4 , 0.6 , 0.8 , 1 , = 1 , σ = 0.5 and β = 2 .
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Figure 5. The exact and approximate (SDM/SITM) solution graph at Φ ( x , t ¯ ) of Problem 2 for = 1 , σ = 0.5 and β = 2 .
Figure 5. The exact and approximate (SDM/SITM) solution graph at Φ ( x , t ¯ ) of Problem 2 for = 1 , σ = 0.5 and β = 2 .
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Figure 6. The exact and approximate (SDM/SITM) solution graph at Ψ ( x , t ¯ ) of Problem 2 for = 1 , σ = 0.5 and β = 2 .
Figure 6. The exact and approximate (SDM/SITM) solution graph at Ψ ( x , t ¯ ) of Problem 2 for = 1 , σ = 0.5 and β = 2 .
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Figure 7. Numerical evaluation of graph of Ψ ( x , t ¯ ) for Problem 2 for various fractional order δ = 0.4, 0.6, 0.8, 1, = 1 , σ = 0.5 and β = 2 .
Figure 7. Numerical evaluation of graph of Ψ ( x , t ¯ ) for Problem 2 for various fractional order δ = 0.4, 0.6, 0.8, 1, = 1 , σ = 0.5 and β = 2 .
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Figure 8. Numerical evaluation of graph of Ψ ( x , t ¯ ) for Problem 2 for various fractional order δ = 0.4 , 0.6 , 0.8 , 1 , = 1 , σ = 0.5 and β = 2 .
Figure 8. Numerical evaluation of graph of Ψ ( x , t ¯ ) for Problem 2 for various fractional order δ = 0.4 , 0.6 , 0.8 , 1 , = 1 , σ = 0.5 and β = 2 .
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Figure 9. The exact and analytical solution graph at Φ ( x , t ¯ ) of Problem 3 for = 1 , σ = 0.5 and β = 2 .
Figure 9. The exact and analytical solution graph at Φ ( x , t ¯ ) of Problem 3 for = 1 , σ = 0.5 and β = 2 .
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Figure 10. The exact and analytical solution graph at Ψ ( x , t ¯ ) of Problem 3 for = 1 , σ = 0.5 and β = 2 .
Figure 10. The exact and analytical solution graph at Ψ ( x , t ¯ ) of Problem 3 for = 1 , σ = 0.5 and β = 2 .
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Figure 11. The exact and analytical solution graph at Υ ( x , t ¯ ) of Problem 3 for = 1 , σ = 0.5 and β = 2 .
Figure 11. The exact and analytical solution graph at Υ ( x , t ¯ ) of Problem 3 for = 1 , σ = 0.5 and β = 2 .
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Figure 12. Numerical evaluation of graph at Ψ ( x , t ¯ ) Problem 3 for various fractional order δ = 0.4 , 0.6 , 0.8 , 1 , = 1 , σ = 0.5 and β = 2 .
Figure 12. Numerical evaluation of graph at Ψ ( x , t ¯ ) Problem 3 for various fractional order δ = 0.4 , 0.6 , 0.8 , 1 , = 1 , σ = 0.5 and β = 2 .
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Figure 13. Numerical evaluation of graph at Φ ( x , t ¯ ) Problem 3 for various fractional order δ = 0.4 , 0.6 , 0.8 , 1 , = 1 , σ = 0.5 and β = 2 .
Figure 13. Numerical evaluation of graph at Φ ( x , t ¯ ) Problem 3 for various fractional order δ = 0.4 , 0.6 , 0.8 , 1 , = 1 , σ = 0.5 and β = 2 .
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Figure 14. Numerical evaluation of graph at Υ ( x , t ¯ ) Problem 3 for various fractional order δ = 0.4 , 0.6 , 0.8 , 1 , = 1 , σ = 0.5 and β = 2 .
Figure 14. Numerical evaluation of graph at Υ ( x , t ¯ ) Problem 3 for various fractional order δ = 0.4 , 0.6 , 0.8 , 1 , = 1 , σ = 0.5 and β = 2 .
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Rashid, S.; Khalid, A.; Sultana, S.; Hammouch, Z.; Shah, R.; Alsharif, A.M. A Novel Analytical View of Time-Fractional Korteweg-De Vries Equations via a New Integral Transform. Symmetry 2021, 13, 1254. https://0-doi-org.brum.beds.ac.uk/10.3390/sym13071254

AMA Style

Rashid S, Khalid A, Sultana S, Hammouch Z, Shah R, Alsharif AM. A Novel Analytical View of Time-Fractional Korteweg-De Vries Equations via a New Integral Transform. Symmetry. 2021; 13(7):1254. https://0-doi-org.brum.beds.ac.uk/10.3390/sym13071254

Chicago/Turabian Style

Rashid, Saima, Aasma Khalid, Sobia Sultana, Zakia Hammouch, Rasool Shah, and Abdullah M. Alsharif. 2021. "A Novel Analytical View of Time-Fractional Korteweg-De Vries Equations via a New Integral Transform" Symmetry 13, no. 7: 1254. https://0-doi-org.brum.beds.ac.uk/10.3390/sym13071254

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