Computational Analysis of Fractional-Order KdV Systems in the Sense of the Caputo Operator via a Novel Transform

Abstract

The main features of scientific efforts in physics and engineering are the development of models for various physical issues and the development of solutions. In order to solve the time-fractional coupled Korteweg–De Vries (KdV) equation, we combine the novel Yang transform, the homotopy perturbation approach, and the Adomian decomposition method in the present investigation. KdV models are crucial because they can accurately represent a variety of physical problems, including thin-film flows and waves on shallow water surfaces. The fractional derivative is regarded in the Caputo meaning. These approaches apply straightforward steps through symbolic computation to provide a convergent series solution. Different nonlinear time-fractional KdV systems are used to test the effectiveness of the suggested techniques. The symmetry pattern is a fundamental feature of the KdV equations and the symmetrical aspect of the solution can be seen from the graphical representations. The numerical outcomes demonstrate that only a small number of terms are required to arrive at an approximation that is exact, efficient, and trustworthy. Additionally, the system’s approximative solution is illustrated graphically. The results show that these techniques are extremely effective, practically applicable for usage in such issues, and adaptable to other nonlinear issues.

1. Introduction

Fractional calculus (FC) theory offers a foundation for the extension of differentiation to non-integer orders. The fractional derivative operators [1] provide a concise explanation of modelling issues involving the concepts of non-locality and memory effects that are not adequately explained by integer-order operators. The discussion of a variety of issues, including viscoelastic systems and electrode–electrolyte polarisation, which are modelled by fractional equations, is made possible by the use of fractional calculus [2]. FC more precisely and aggressively defines physical phenomena than classical calculus. Fractional-order nonlinear models are widely used in many areas and are also important in nonlinear wave phenomena [3,4]. Fractional derivatives have been defined in several ways, including the Riemann–Liouville, Caputo, and Grunwald–Letnikov operators, the most well known of which are the Caputo and Riemann–Liouville fractional derivatives, which have often been employed in recent research. Many scientific domains have successfully used the idea of fractional derivatives to represent a variety of real-world occurrences in recent years [5]. The evolution of a physical phenomenon from its original state to its current state is included in fractional-order operators. Thus, model systems that explain the impact of memory effects are often evaluated through the application of fractional-order operators [6].

Differential equations that involve unknown multivariable functions and their fractional or fractional-integer partial derivatives with respect to those variables are known as fractional partial differential equations (PDEs). These kinds of equations are used to formulate, and afterwards aid in the solution of, problems involving functions of numerous variables, such as sound or heat propagation, electrostatics, electrodynamics, fluid flow, elasticity, and many more [7,8,9,10,11,12]. These two physically different occurrences can be formulated similarly using fractional PDEs. Fractional-order partial differential equations (FPDEs) have been suggested and studied during the past few decades in a variety of scientific areas, including biology, plasma physics, finance, chemistry, fluid mechanics, and mechanics of materials. In order to better express physical and control systems, systems of fractional partial differential equations have become more and more popular [13,14,15,16,17]. Approximative or numerical techniques are typically used because some fractional-order partial differential equations do not have accurate analytic solutions. In the literature, there are numerous analytical and numerical techniques for handling FPDEs, such as the multiple exp-function algorithm for solving the nonlinear fractional Sharma–Tasso–Olver equation [18] and the fractional-order Hirota–Satsuma coupled KdV [19]; the homotopy perturbation transform method for solving time-fractional Belousov–Zhabotinsky reactions [20] and diffusion equations of fractional-order in a plasma and fluids [21]; the fractional complex transformation for fractional nonlinear partial differential equations in mathematical physics [22] and the time-fractional heat conduction equation [23]; the variational iteration transform method for the investigation of the Newell–Whitehead–Segel equations having fractional order [24]; the homotopy analysis method for solving fractional Lorenz systems [25] and time-fractional Schrödinger equations [26]; the finite element method for parabolic equations of fractional order [27] and the time-fractional Fokker–Planck equation [28]; the fractional sub-equation method for generalised fractional KdV–Zakharov–Kuznetsov equations [29]; and so on.

A dimensionless form of the equation known as the KdV equation was derived by “Korteweg” and “de Vries” for the study of dispersive wave events in plasma physics and quantum mechanics. Korteweg and de Vries developed the traditional KdV equation in 1895 as a nonlinear partial differential equation to simulate waves on shallow water surfaces. Numerous research papers have been written regarding this specific solvable model. Numerous researchers have recently proposed novel uses for the classical KdV equation, including its usage to represent long internal waves in a density-stratified ocean, ion-acoustic waves in a plasma, and acoustic waves on a crystal lattice. In this study, we determine approximations to the nonlinear coupled time-fractional KdV equations that are given by

$$\begin{array}{cc}\hfill & D_t^{\varsigma }\mathbf{J}(x,t)=\mathbf{a}{\mathbf{J}}_{xxx}(x,t)+\mathbf{b}\mathbf{J}(x,t){\mathbf{J}}_x(x,t)+\mathbf{c}\mathbf{K}(x,t){\mathbf{K}}_x(x,t)\hfill \\ \hfill & D_t^{\beta }\mathbf{K}(x,t)=\mathbf{d}{\mathbf{J}}_{xxx}(x,t)-\mathbf{e}\mathbf{K}(x,t){\mathbf{K}}_x(x,t),0<\varsigma ,\beta \le 1,\hfill \end{array}$$

(1)

subjected to the initial sources

$$\mathbf{J}(x,0)=h_1\left(x\right),\mathbf{K}(x,0)=h_2\left(x\right),$$

(2)

where $\mathbf{a},\mathbf{d}<0$; $\mathbf{b},\mathbf{c}$, and $\mathbf{e}$ are constant parameters. The KdV equation has been the subject of a considerable number of studies because it has applications in the analysis of shallow-water waves and many other physical phenomena. It is possible that the precise solution to the KdV equation does not exist. For its numerical solution, many analytical techniques have been employed [30]. The numerical solutions of the third- and fifth-order terms, which serve as the primary dispersive terms in the KdV equation, were examined in [31]. However, plasma physics is described by the fifth-order KdV equation [32]. Nonlinear KdV reductive perturbation theories were investigated in [33]. The higher-order nonlinear KdV equation was investigated in [34] using a variational iteration method. A compact-type interpolation of constrained profiles was used in [35] to obtain numerical solutions for the KdV–Burgers equation. In [30], the approach of converting homotopy perturbation is used to analyse the numerical results of fifth-order KdV equations. Two computational techniques were applied to the third- and fifth-order KdV equations in [31].

The main objective of this study is to apply the homotopy perturbation transform method (HPTM) and the Yang transform decomposition method (YTDM) to the time-fractional coupled KdV equations in the frame of the Caputo derivative. To create the YTDM, the decomposition method and the Yang transform are merged. The Yang transform was introduced by Xiao-Jun Yang and can be used to solve a variety of differential equations with constant coefficients [36]. George Adomian introduced the Adomian decomposition method [37,38]. An extensive class of linear and nonlinear equations that are helpful for numerous research projects have been subjected to the Adomian decomposition approach. This method’s ability to correctly and accurately solve these equations is its key characteristic. The time-fractional Fisher’s equation [39], time-fractional Lax’s Korteweg–De Vries equation [40], time-fractional Phi-four equation [41], and many other differential equations have been solved using the YTDM in the literature. A key role is played by the homotopy perturbation approach, which was first presented by J.H. He in 1998. This is because it solves the issue at hand without the need for any kind of transformation, linearisation, or discrimination. Analytical findings for the suggested strategy show how computationally advantageous and simple to use the method is.

The introduction and purpose of the article are presented in Section 1 of the paper. The fundamental principles of fractional calculus are covered in Section 2. We outline the notion of the suggested methods in Section 3. In Section 4, a few numerical examples are used to illustrate how effective the suggested solutions are. The manuscript’s conclusion is presented in Section 5.

2. Basic Concept

Some fractional calculus definitions and notation needed in the course of this work are discussed in this section.

Definition 1.

The fractional Caputo derivative is stated as [42]

$$\begin{array}{cc}\hfill & D_t^{\varsigma }\mathbf{J}(x,t)=\frac{1}{\mathrm{\Gamma }(k-\varsigma )}{\int }_0^t{(t-\tau )}^{k-\varsigma -1}{\mathbf{J}}^{\left(k\right)}(x,\tau )d\tau ,k-1<\varsigma \le k,k\in N.\hfill \end{array}$$

(3)

Definition 2.

The Yang transform (YT) of the function is [43]

$$Y\left\{\mathbf{J}\left(t\right)\right\}=M\left(u\right)={\int }_0^{\infty }{e}^{\frac{-t}{u}}\mathbf{J}\left(t\right)dt,t>0,$$

(4)

with u illustrating the transform variable.

Some basic properties of the YT are the following:

$$\begin{array}{cc}\hfill Y\left[1\right]=& u,\hfill \\ \hfill Y\left[t\right]=& u^2,\hfill \\ \hfill Y\left[t^q\right]=& \mathrm{\Gamma }(q+1)u^{q+1}.\hfill \end{array}$$

(5)

and the inverse YT is

$$Y^{-1}\left\{M\left(u\right)\right\}=\mathbf{J}\left(t\right).$$

(6)

Definition 3.

The YT of a function of nth order is [43]

$$Y\left\{{\mathbf{J}}^n\left(t\right)\right\}=\frac{M\left(u\right)}{u^n}-\sum _{k=0}^{n-1}\frac{{\mathbf{J}}^k\left(0\right)}{u^{n-k-1}},\forall n=1,2,3,\cdots$$

(7)

Definition 4.

The YT of a function of fractional order is [43]

$$Y\left\{{\mathbf{J}}^{\varsigma }\left(t\right)\right\}=\frac{M\left(u\right)}{u^{\varsigma }}-\sum _{k=0}^{n-1}\frac{{\mathbf{J}}^k\left(0\right)}{u^{\varsigma -(k+1)}},n-1<\varsigma \le n.$$

(8)

3. Algorithm of the HPTM

The algorithm of the HPTM on a general nonlinear fractional model is illustrated in the following section.

$$D_t^{\varsigma }\mathbf{J}(x,t)=\mathcal{L}\left[x\right]\mathbf{J}(x,t)+\mathcal{M}\left[x\right]\mathbf{J}(x,t),0<\varsigma \le 1,$$

(9)

with the initial condition

$$\mathbf{J}(x,0)=\vartheta \left(x\right).$$

Here, $D_t^{\varsigma }=\frac{{\partial }^{\varsigma }}{\partial t^{\varsigma }}$ denotes the fractional Caputo operator, $\mathcal{L}\left[x\right]$ is the linear, and $\mathcal{M}\left[x\right]$ is the nonlinear operator.

Computing the YT, we obtain

$$Y\left[D_t^{\varsigma }\mathbf{J}(x,t)\right]=Y[\mathcal{L}\left[x\right]\mathbf{J}(x,t)+\mathcal{M}\left[x\right]\mathbf{J}(x,t)],$$

(10)

$$\frac{1}{u^{\varsigma }}\{M\left(u\right)-u\mathbf{J}\left(0\right)\}=Y[\mathcal{L}\left[x\right]\mathbf{J}(x,t)+\mathcal{M}\left[x\right]\mathbf{J}(x,t)].$$

(11)

After that, we have

$$M\left(\mathbf{J}\right)=u\mathbf{J}\left(0\right)+u^{\varsigma }Y[\mathcal{L}\left[x\right]\mathbf{J}(x,t)+\mathcal{M}\left[x\right]\mathbf{J}(x,t)].$$

(12)

Operating the inverse YT, we have

$$\mathbf{J}(x,t)=\mathbf{J}\left(0\right)+Y^{-1}\left[u^{\varsigma }Y[\mathcal{L}\left[x\right]\mathbf{J}(x,t)+\mathcal{M}\left[x\right]\mathbf{J}(x,t)]\right].$$

(13)

By means of homotopy perturbation method [44], we have

$$\mathbf{J}(x,t)=\mathbf{J}\left(0\right)+ϵ\left[Y^{-1}\left[u^{\varsigma }Y[\mathcal{L}\left[x\right]\mathbf{J}(x,t)+\mathcal{M}\left[x\right]\mathbf{J}(x,t)]\right]\right],$$

(14)

with parameter $ϵ\in [0,1]$.

The solution is expanded in series form as

$$\mathbf{J}(x,t)=\sum _{k=0}^{\infty }{ϵ}^k{\mathbf{J}}_k(x,t),$$

(15)

with

$$\mathcal{M}\left[x\right]\mathbf{J}(x,t)=\sum _{k=0}^{\infty }{ϵ}^k{H}_n\left(\mathbf{J}\right).$$

(16)

The following strategy can be operated to acquire He’s polynomials [45] as

$$H_n({\mathbf{J}}_0,{\mathbf{J}}_1,...,{\mathbf{J}}_n)=\frac{1}{\mathrm{\Gamma }(n+1)}{D}_{ϵ}^k\left(\mathcal{M}\left(\sum _{k=0}^{\infty }{ϵ}^i{\mathbf{J}}_i\right)\right){|}_{ϵ=0},$$

(17)

where $D_{ϵ}^k=\frac{{\partial }^k}{\partial {ϵ}^k}.$

Now, we substitute (15) and (16) into (14) to obtain

$$\sum _{k=0}^{\infty }{ϵ}^k{\mathbf{J}}_k(x,t)=\mathbf{J}\left(0\right)+ϵ\times \left(Y^{-1}\left[u^{\varsigma }Y\{\mathcal{L}\sum _{k=0}^{\infty }{ϵ}^k{\mathbf{J}}_k(x,t)+\sum _{k=0}^{\infty }{ϵ}^k{H}_k\left(\mathbf{J}\right)\}\right]\right).$$

(18)

Equating the similar components of $ϵ$, we obtain

$$\begin{array}{cc}\hfill & {ϵ}^0:{\mathbf{J}}_0(x,t)=\mathbf{J}\left(0\right),\hfill \\ \hfill & {ϵ}^1:{\mathbf{J}}_1(x,t)=Y^{-1}\left[u^{\varsigma }Y(\mathcal{L}\left[x\right]{\mathbf{J}}_0(x,t)+H_0\left(\mathbf{J}\right))\right],\hfill \\ \hfill & {ϵ}^2:{\mathbf{J}}_2(x,t)=Y^{-1}\left[u^{\varsigma }Y(\mathcal{L}\left[x\right]{\mathbf{J}}_1(x,t)+H_1\left(\mathbf{J}\right))\right],\hfill \\ \hfill & \cdots \hfill \\ \hfill & {ϵ}^k:{\mathbf{J}}_k(x,t)=Y^{-1}\left[u^{\varsigma }Y(\mathcal{L}\left[x\right]{\mathbf{J}}_{k-1}(x,t)+H_{k-1}\left(\mathbf{J}\right))\right],\hfill \\ \hfill k>0,k\in N.\end{array}$$

(19)

Lastly, the analytical solution is stated as

$$\mathbf{J}(x,t)=\underset{ϵ\to 1}{lim}\sum _{k=0}^{\infty }{ϵ}^k{\mathbf{J}}_k(x,t).$$

(20)

4. Algorithm of the YTDM

The algorithm of the YTDM on a general nonlinear fractional model is illustrated in the following section.

$$D_t^{\varsigma }\mathbf{J}(x,t)=\mathcal{L}(x,t)+\mathcal{M}(x,t),0<\varsigma \le 1,$$

(21)

with the initial condition

$$\mathbf{J}(x,0)=\vartheta \left(x\right).$$

Here, $D_t^{\varsigma }=\frac{{\partial }^{\varsigma }}{\partial t^{\varsigma }}$ denotes the fractional Caputo operator, $\mathcal{L}$ is the linear, and $\mathcal{M}$ is the nonlinear operator.

Computing the YT, we obtain

$$\begin{array}{cc}\hfill & Y\left[D_t^{\varsigma }\mathbf{J}(x,t)\right]=Y[\mathcal{L}(x,t)+\mathcal{M}(x,t)],\hfill \\ \hfill & \frac{1}{u^{\varsigma }}\{M\left(u\right)-u\mathbf{J}\left(0\right)\}=Y[\mathcal{L}(x,t)+\mathcal{M}(x,t)].\hfill \end{array}$$

(22)

After that, we have

$$M\left(\mathbf{J}\right)=u\mathbf{J}\left(0\right)+u^{\varsigma }Y[\mathcal{L}(x,t)+\mathcal{M}(x,t)],$$

(23)

Operating the inverse YT, we have

$$\mathbf{J}(x,t)=\mathbf{J}\left(0\right)+Y^{-1}[u^{\varsigma }Y[\mathcal{L}(x,t)+\mathcal{M}(x,t)].$$

(24)

The unknown function $\mathbf{J}(x,t)$ in terms of infinite series [37] is stated as

$$\mathbf{J}(x,t)=\sum _{m=0}^{\infty }{\mathbf{J}}_m(x,t).$$

(25)

Here, we illustrate the nonlinear term as

$$\mathcal{M}(x,t)=\sum _{m=0}^{\infty }{\mathcal{A}}_m,$$

(26)

with

$${\mathcal{A}}_m=\frac{1}{m!}{\left[\frac{{\partial }^m}{\partial {\ell }^m}\left\{\mathcal{M}\left(\sum _{k=0}^{\infty }{\ell }^k{x}_k,\sum _{k=0}^{\infty }{\ell }^k{t}_k\right)\right\}\right]}_{\ell =0}.$$

(27)

Now, we substitute (25) and (26) into (24) to obtain

$$\sum _{m=0}^{\infty }{\mathbf{J}}_m(x,t)=\mathbf{J}\left(0\right)+Y^{-1}{u}^{\varsigma }\left[Y\left\{\mathcal{L}(\sum _{m=0}^{\infty }{x}_m,\sum _{m=0}^{\infty }{t}_m)+\sum _{m=0}^{\infty }{\mathcal{A}}_m\right\}\right].$$

(28)

So, we obtain

$${\mathbf{J}}_0(x,t)=\mathbf{J}\left(0\right),$$

(29)

$${\mathbf{J}}_1(x,t)=Y^{-1}\left[u^{\varsigma }Y\{\mathcal{L}(x_0,t_0)+{\mathcal{A}}_0\}\right].$$

Finally, in general for $m\ge 1$, we have

$${\mathbf{J}}_{m+1}(x,t)=Y^{-1}\left[u^{\varsigma }Y\{\mathcal{L}(x_m,t_m)+{\mathcal{A}}_m\}\right].$$

5. Application

Example 1.

Let us assume a fractional coupled KdV Equation (1) with $\mathit{a}=-\kappa ,\mathit{b}=-6\kappa$, $\mathit{c}=2\nu ,\mathit{d}=-\mu ,$ and $\mathit{e}=3\mu ,$ having initial condition

$$\mathbf{J}(x,0)=\frac{\zeta }{\kappa }{\left(sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)}^2,\mathbf{K}(x,0)=\frac{\zeta }{\sqrt{2\kappa }}{\left(sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)}^2.$$

Case I: Solution by HPTM

Computing the YT, we obtain

$$\begin{array}{cc}\hfill & Y\left[\frac{{\partial }^{\varsigma }\mathbf{J}}{\partial t^{\varsigma }}\right]=Y\left[-\kappa {\mathbf{J}}_{xxx}(x,t)-6\kappa \mathbf{J}(x,t){\mathbf{J}}_x(x,t)+2\nu \mathbf{K}(x,t){\mathbf{K}}_x(x,t)\right],\hfill \\ \hfill & Y\left[\frac{{\partial }^{\beta }\mathbf{K}}{\partial t^{\beta }}\right]=Y\left[-\mu {\mathbf{J}}_{xxx}(x,t)-3\mu \mathbf{J}(x,t){\mathbf{K}}_x(x,t)\right].\hfill \end{array}$$

(30)

After that, we have

$$\begin{array}{cc}\hfill & \frac{1}{u^{\varsigma }}\{M\left(u\right)-u\mathbf{J}\left(0\right)\}=Y\left[-\kappa {\mathbf{J}}_{xxx}(x,t)-6\kappa \mathbf{J}(x,t){\mathbf{J}}_x(x,t)+2\nu \mathbf{K}(x,t){\mathbf{K}}_x(x,t)\right],\hfill \\ \hfill & \frac{1}{u^{\beta }}\{M\left(u\right)-u\mathbf{K}\left(0\right)\}=Y\left[-\mu {\mathbf{J}}_{xxx}(x,t)-3\mu \mathbf{J}(x,t){\mathbf{K}}_x(x,t)\right],\hfill \end{array}$$

(31)

$$\begin{array}{cc}\hfill & M\left(u\right)=u\mathbf{J}\left(0\right)+u^{\varsigma }\left[-\kappa {\mathbf{J}}_{xxx}(x,t)-6\kappa \mathbf{J}(x,t){\mathbf{J}}_x(x,t)+2\nu \mathbf{K}(x,t){\mathbf{K}}_x(x,t)\right],\hfill \\ \hfill & M\left(u\right)=u\mathbf{K}\left(0\right)+u^{\beta }\left[-\mu {\mathbf{J}}_{xxx}(x,t)-3\mu \mathbf{J}(x,t){\mathbf{K}}_x(x,t)\right].\hfill \end{array}$$

(32)

Operating the inverse YT, we have

$$\begin{array}{cc}\hfill & \mathbf{J}(x,t)=\mathbf{J}\left(0\right)+Y^{-1}\left[u^{\varsigma }\left\{Y\left(-\kappa {\mathbf{J}}_{xxx}(x,t)-6\kappa \mathbf{J}(x,t){\mathbf{J}}_x(x,t)+2\nu \mathbf{K}(x,t){\mathbf{K}}_x(x,t)\right)\right\}\right],\hfill \\ \hfill & \mathbf{K}(x,t)=\mathbf{K}\left(0\right)+Y^{-1}\left[u^{\beta }\left\{Y\left(-\mu {\mathbf{J}}_{xxx}(x,t)-3\mu \mathbf{J}(x,t){\mathbf{K}}_x(x,t)\right)\right\}\right],\hfill \\ \hfill & \mathbf{J}(x,t)=\frac{\zeta }{\kappa }{\left(sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)}^2+Y^{-1}\left[u^{\varsigma }\left\{Y\left(-\kappa {\mathbf{J}}_{xxx}(x,t)-6\kappa \mathbf{J}(x,t){\mathbf{J}}_x(x,t)+2\nu \mathbf{K}(x,t){\mathbf{K}}_x(x,t)\right)\right\}\right],\hfill \\ \hfill & \mathbf{K}(x,t)=\frac{\zeta }{\sqrt{2\kappa }}{\left(sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)}^2+Y^{-1}\left[u^{\beta }\left\{Y\left(-\mu {\mathbf{J}}_{xxx}(x,t)-3\mu \mathbf{J}(x,t){\mathbf{K}}_x(x,t)\right)\right\}\right].\hfill \end{array}$$

(33)

By means of the homotopy perturbation method, we have

$$\begin{array}{cc}& \sum _{k=0}^{\infty }{ϵ}^k{\mathbf{J}}_k(x,t)=\frac{\zeta }{\kappa }{\left(sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)}^2+ϵ\left(Y^{-1}\right[u^{\varsigma }Y[-\kappa {\left(\sum _{k=0}^{\infty }{ϵ}^k{\mathbf{J}}_k(x,t)\right)}_{xxx}-6\kappa \left(\sum _{k=0}^{\infty }{ϵ}^k{H}_k(x,t)\right)+\hfill \\ & 2\nu \left(\sum _{k=0}^{\infty }{ϵ}^k{H}_k(x,t)\right)\left]\right]),\hfill \\ & \sum _{k=0}^{\infty }{ϵ}^k{\mathbf{K}}_k(x,t)=\frac{\zeta }{\sqrt{2\kappa }}{\left(sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)}^2+ϵ\left(Y^{-1}\left[u^{\beta }Y\left[-\mu {\left(\sum _{k=0}^{\infty }{ϵ}^k{\mathbf{J}}_k(x,t)\right)}_{xxx}-3\mu \left(\sum _{k=0}^{\infty }{ϵ}^k{H}_k(x,t)\right)\right]\right]\right).\hfill \end{array}$$

(34)

Equating the similar components of ϵ, we obtain

$$\begin{array}{cc}\hfill & {ϵ}^0:{\mathbf{J}}_0(x,t)=\frac{\zeta }{\kappa }{\left(sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)}^2,\hfill \\ \hfill & {ϵ}^0:{\mathbf{K}}_0(x,t)=\frac{\zeta }{\sqrt{2\kappa }}{\left(sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)}^2,\hfill \\ \hfill & {ϵ}^1:{\mathbf{J}}_1(x,t)=\left(\frac{1}{2}\zeta {\left(\frac{\zeta }{\kappa }\right)}^{\frac{3}{2}}\left(7-2\nu +cosh\left(\sqrt{\frac{\zeta }{\kappa }}x\right)\right){sech}^4\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)tanh\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)\frac{t^{\varsigma }}{\mathrm{\Gamma }(\varsigma +1)},\hfill \\ \hfill & {ϵ}^1:{\mathbf{K}}_1(x,t)=\left(4\sqrt{2}\sqrt{\kappa }\mu {\left(\frac{\zeta }{\kappa }\right)}^{\frac{5}{2}}{csch}^3\left(\sqrt{\frac{\zeta }{\kappa }}x\right){sinh}^4\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)\frac{t^{\beta }}{\mathrm{\Gamma }(\beta +1)},\hfill \\ \hfill & {ϵ}^2:{\mathbf{J}}_2(x,t)=\frac{t^{\varsigma }{\zeta }^4}{32{\kappa }^3\mathrm{\Gamma }(2\varsigma +1)\mathrm{\Gamma }(\varsigma +\beta +1)}(8\nu t^{\beta }\mu \left(2+cosh\left(\sqrt{\frac{\zeta }{\kappa }}x\right)-cosh\left(2\sqrt{\frac{\zeta }{\kappa }}x\right)\right)\mathrm{\Gamma }(2\varsigma +1)\hfill \\ \hfill & \kappa t^{\varsigma }\left(40-32\nu +(-345+104\nu )cosh\left(\sqrt{\frac{\zeta }{\kappa }}x\right)-8(-15+4\nu )cosh\left(2\sqrt{\frac{\zeta }{\kappa }}x\right)+cosh\left(3\sqrt{\frac{\zeta }{\kappa }}x\right)\right)\hfill \\ \hfill & \mathrm{\Gamma }(\varsigma +\beta +1)){sech}^8\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right),\hfill \\ \hfill & {ϵ}^2:{\mathbf{K}}_2(x,t)=\frac{-1}{8\sqrt{\kappa }{\kappa }^{\frac{7}{2}}\mathrm{\Gamma }(2\beta +1)\mathrm{\Gamma }(\varsigma +\beta +1)}(\mu {\zeta }^4{t}^{\beta }sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)(-\mu t^{\beta }(9-14cosh\left(\sqrt{\frac{\zeta }{\kappa }}x\right)+\hfill \\ \hfill & cosh\left(2\sqrt{\frac{\zeta }{\kappa }}x\right))\mathrm{\Gamma }(\varsigma +\beta +1)-12\kappa t^{\varsigma }\left(7-2\nu +cosh\left(\sqrt{\frac{\zeta }{\kappa }}x\right)\right)\mathrm{\Gamma }(2\beta +1){tanh}^2\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right))),\hfill \\ \hfill & \cdots \hfill \end{array}$$

The solution we obtained is taken in series form as

$$\begin{array}{cc}\hfill & \mathbf{J}(x,t)={\mathbf{J}}_0(x,t)+{\mathbf{J}}_1(x,t)+{\mathbf{J}}_2(x,t)+\cdots \hfill \\ \hfill & \mathbf{J}(x,t)=\frac{\zeta }{\kappa }{\left(sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)}^2+(\frac{1}{2}\zeta {\left(\frac{\zeta }{\kappa }\right)}^{\frac{3}{2}}\left(7-2\nu +cosh\left(\sqrt{\frac{\zeta }{\kappa }}x\right)\right){sech}^4\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\hfill \\ \hfill & tanh\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right))\frac{t^{\varsigma }}{\mathrm{\Gamma }(\varsigma +1)}+\frac{t^{\varsigma }{\zeta }^4}{32{\kappa }^3\mathrm{\Gamma }(2\varsigma +1)\mathrm{\Gamma }(\varsigma +\beta +1)}(8\nu t^{\beta }\mu \left(2+cosh\left(\sqrt{\frac{\zeta }{\kappa }}x\right)-cosh\left(2\sqrt{\frac{\zeta }{\kappa }}x\right)\right)\hfill \\ \hfill & \mathrm{\Gamma }(2\varsigma +1)\kappa t^{\varsigma }\left(40-32\nu +(-345+104\nu )cosh\left(\sqrt{\frac{\zeta }{\kappa }}x\right)-8(-15+4\nu )cosh\left(2\sqrt{\frac{\zeta }{\kappa }}x\right)+cosh\left(3\sqrt{\frac{\zeta }{\kappa }}x\right)\right)\hfill \\ \hfill & \mathrm{\Gamma }(\varsigma +\beta +1)){sech}^8\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)+\cdots \hfill \\ \hfill & \mathbf{K}(x,t)={\mathbf{K}}_0(x,t)+{\mathbf{K}}_1(x,t)+{\mathbf{K}}_2(x,t)+\cdots \hfill \\ \hfill & \mathbf{K}(x,t)=\frac{\zeta }{\sqrt{2\kappa }}{\left(sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)}^2+\left(4\sqrt{2}\sqrt{\kappa }\mu {\left(\frac{\zeta }{\kappa }\right)}^{\frac{5}{2}}{csch}^3\left(\sqrt{\frac{\zeta }{\kappa }}x\right){sinh}^4\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)\frac{t^{\beta }}{\mathrm{\Gamma }(\beta +1)}+\hfill \\ \hfill & \frac{-1}{8\sqrt{\kappa }{\kappa }^{\frac{7}{2}}\mathrm{\Gamma }(2\beta +1)\mathrm{\Gamma }(\varsigma +\beta +1)}(\mu {\zeta }^4{t}^{\beta }sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)(-\mu t^{\beta }\left(9-14cosh\left(\sqrt{\frac{\zeta }{\kappa }}x\right)+cosh\left(2\sqrt{\frac{\zeta }{\kappa }}x\right)\right)\hfill \\ \hfill & \mathrm{\Gamma }(\varsigma +\beta +1)-12\kappa t^{\varsigma }\left(7-2\nu +cosh\left(\sqrt{\frac{\zeta }{\kappa }}x\right)\right)\mathrm{\Gamma }(2\beta +1){tanh}^2\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\left)\right)+\cdots \hfill \end{array}$$

Case II: Solution by YTDM

Computing the YT, we obtain

$$\begin{array}{cc}\hfill & Y\left[\frac{{\partial }^{\varsigma }\mathbf{J}}{\partial t^{\varsigma }}\right]=Y\left[-\kappa {\mathbf{J}}_{xxx}(x,t)-6\kappa \mathbf{J}(x,t){\mathbf{J}}_x(x,t)+2\nu \mathbf{K}(x,t){\mathbf{K}}_x(x,t)\right],\hfill \\ \hfill & Y\left[\frac{{\partial }^{\beta }\mathbf{K}}{\partial t^{\beta }}\right]=Y\left[-\mu {\mathbf{J}}_{xxx}(x,t)-3\mu \mathbf{J}(x,t){\mathbf{K}}_x(x,t)\right].\hfill \end{array}$$

(35)

After that, we have

$$\begin{array}{cc}\hfill & \frac{1}{u^{\varsigma }}\{M\left(u\right)-u\mathbf{J}\left(0\right)\}=Y\left[-\kappa {\mathbf{J}}_{xxx}(x,t)-6\kappa \mathbf{J}(x,t){\mathbf{J}}_x(x,t)+2\nu \mathbf{K}(x,t){\mathbf{K}}_x(x,t)\right],\hfill \\ \hfill & \frac{1}{u^{\beta }}\{M\left(u\right)-u\mathbf{K}\left(0\right)\}=Y\left[-\mu {\mathbf{J}}_{xxx}(x,t)-3\mu \mathbf{J}(x,t){\mathbf{K}}_x(x,t)\right],\hfill \end{array}$$

(36)

$$\begin{array}{cc}\hfill & M\left(u\right)=u\mathbf{J}\left(0\right)+u^{\varsigma }\left[-\kappa {\mathbf{J}}_{xxx}(x,t)-6\kappa \mathbf{J}(x,t){\mathbf{J}}_x(x,t)+2\nu \mathbf{K}(x,t){\mathbf{K}}_x(x,t)\right],\hfill \\ \hfill & M\left(u\right)=u\mathbf{K}\left(0\right)+u^{\beta }\left[-\mu {\mathbf{J}}_{xxx}(x,t)-3\mu \mathbf{J}(x,t){\mathbf{K}}_x(x,t)\right].\hfill \end{array}$$

(37)

Operating the inverse YT, we have

$$\begin{array}{cc}\hfill & \mathbf{J}(x,t)=\mathbf{J}\left(0\right)+Y^{-1}\left[u^{\varsigma }\left\{Y\left(-\kappa {\mathbf{J}}_{xxx}(x,t)-6\kappa \mathbf{J}(x,t){\mathbf{J}}_x(x,t)+2\nu \mathbf{K}(x,t){\mathbf{K}}_x(x,t)\right)\right\}\right],\hfill \\ \hfill & \mathbf{K}(x,t)=\mathbf{K}\left(0\right)+Y^{-1}\left[u^{\beta }\left\{Y\left(-\mu {\mathbf{J}}_{xxx}(x,t)-3\mu \mathbf{J}(x,t){\mathbf{K}}_x(x,t)\right)\right\}\right],\hfill \\ \hfill & \mathbf{J}(x,t)=\frac{\zeta }{\kappa }{\left(sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)}^2+Y^{-1}\left[u^{\varsigma }\left\{Y\left(-\kappa {\mathbf{J}}_{xxx}(x,t)-6\kappa \mathbf{J}(x,t){\mathbf{J}}_x(x,t)+2\nu \mathbf{K}(x,t){\mathbf{K}}_x(x,t)\right)\right\}\right],\hfill \\ \hfill & \mathbf{K}(x,t)=\frac{\zeta }{\sqrt{2\kappa }}{\left(sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)}^2+Y^{-1}\left[u^{\beta }\left\{Y\left(-\mu {\mathbf{J}}_{xxx}(x,t)-3\mu \mathbf{J}(x,t){\mathbf{K}}_x(x,t)\right)\right\}\right].\hfill \end{array}$$

(38)

Thus, the required solution in terms of infinite series form is

$$\mathbf{J}(x,t)=\sum _{m=0}^{\infty }{\mathbf{J}}_m(x,t),\mathbf{K}(x,t)=\sum _{m=0}^{\infty }{\mathbf{K}}_m(x,t).$$

(39)

Assume the nonlinear terms by the Adomian polynomial as $\mathbf{J}(x,t){\mathbf{J}}_x(x,t)={\sum }_{m=0}^{\infty }{\mathcal{A}}_m,\mathbf{K}(x,t){\mathbf{K}}_x(x,t)={\sum }_{m=0}^{\infty }{\mathcal{B}}_m,\mathbf{J}(x,t){\mathbf{K}}_x(x,t)={\sum }_{m=0}^{\infty }{\mathcal{C}}_m.$ So, we obtain

$$\begin{array}{cc}\hfill & \sum _{m=0}^{\infty }{\mathbf{J}}_m(x,t)=\mathbf{J}(x,0)+Y^{-1}\left[u^{\varsigma }Y\left[-\kappa {\mathbf{J}}_{xxx}(x,t)-6\kappa \sum _{m=0}^{\infty }{\mathcal{A}}_m+2\nu \sum _{m=0}^{\infty }{\mathcal{B}}_m\right]\right],\hfill \\ \hfill & \sum _{m=0}^{\infty }{\mathbf{K}}_m(x,t)=\mathbf{K}(x,0)+Y^{-1}\left[u^{\varsigma }Y\left[-\mu {\mathbf{J}}_{xxx}(x,t)-3\mu \sum _{m=0}^{\infty }{\mathcal{C}}_m\right]\right],\hfill \\ \hfill & \sum _{m=0}^{\infty }{\mathbf{J}}_m(x,t)=\frac{\zeta }{\kappa }{\left(sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)}^2+Y^{-1}\left[u^{\varsigma }Y\left[-\kappa {\mathbf{J}}_{xxx}(x,t)-6\kappa \sum _{m=0}^{\infty }{\mathcal{A}}_m+2\nu \sum _{m=0}^{\infty }{\mathcal{B}}_m\right]\right],\hfill \\ \hfill & \sum _{m=0}^{\infty }{\mathbf{K}}_m(x,t)=\frac{\zeta }{\sqrt{2\kappa }}{\left(sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)}^2+Y^{-1}\left[u^{\varsigma }Y\left[-\mu {\mathbf{J}}_{xxx}(x,t)-3\mu \sum _{m=0}^{\infty }{\mathcal{C}}_m\right]\right].\hfill \end{array}$$

(40)

By equating both sides, we obtain

$${\mathbf{J}}_0(x,t)=\frac{\zeta }{\kappa }{\left(sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)}^2,$$

$${\mathbf{K}}_0(x,t)=\frac{\zeta }{\sqrt{2\kappa }}{\left(sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)}^2.$$

On $m=0$

$${\mathbf{J}}_1(x,t)=\left(\frac{1}{2}\zeta {\left(\frac{\zeta }{\kappa }\right)}^{\frac{3}{2}}\left(7-2\nu +cosh\left(\sqrt{\frac{\zeta }{\kappa }}x\right)\right){sech}^4\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)tanh\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)\frac{t^{\varsigma }}{\mathrm{\Gamma }(\varsigma +1)},$$

$${\mathbf{K}}_1(x,t)=\left(4\sqrt{2}\sqrt{\kappa }\mu {\left(\frac{\zeta }{\kappa }\right)}^{\frac{5}{2}}{csch}^3\left(\sqrt{\frac{\zeta }{\kappa }}x\right){sinh}^4\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)\frac{t^{\beta }}{\mathrm{\Gamma }(\beta +1)}.$$

On $m=1$

$$\begin{array}{cc}\hfill & {\mathbf{J}}_2(x,t)=\frac{t^{\varsigma }{\zeta }^4}{32{\kappa }^3\mathrm{\Gamma }(2\varsigma +1)\mathrm{\Gamma }(\varsigma +\beta +1)}(8\nu t^{\beta }\mu \left(2+cosh\left(\sqrt{\frac{\zeta }{\kappa }}x\right)-cosh\left(2\sqrt{\frac{\zeta }{\kappa }}x\right)\right)\mathrm{\Gamma }(2\varsigma +1)\hfill \\ \hfill & \kappa t^{\varsigma }\left(40-32\nu +(-345+104\nu )cosh\left(\sqrt{\frac{\zeta }{\kappa }}x\right)-8(-15+4\nu )cosh\left(2\sqrt{\frac{\zeta }{\kappa }}x\right)+cosh\left(3\sqrt{\frac{\zeta }{\kappa }}x\right)\right)\hfill \\ \hfill & \mathrm{\Gamma }(\varsigma +\beta +1)){sech}^8\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right),\hfill \end{array}$$

$$\begin{array}{cc}\hfill & {\mathbf{K}}_2(x,t)=\frac{-1}{8\sqrt{\kappa }{\kappa }^{\frac{7}{2}}\mathrm{\Gamma }(2\beta +1)\mathrm{\Gamma }(\varsigma +\beta +1)}(\mu {\zeta }^4{t}^{\beta }sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)(-\mu t^{\beta }(9-14cosh\left(\sqrt{\frac{\zeta }{\kappa }}x\right)+\hfill \\ \hfill & cosh\left(2\sqrt{\frac{\zeta }{\kappa }}x\right))\mathrm{\Gamma }(\varsigma +\beta +1)-12\kappa t^{\varsigma }\left(7-2\nu +cosh\left(\sqrt{\frac{\zeta }{\kappa }}x\right)\right)\mathrm{\Gamma }(2\beta +1){tanh}^2\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right))).\hfill \end{array}$$

In this manner, we can easily obtain the other terms for $(m\ge 3)$, so

$$\begin{array}{cc}\hfill & \mathbf{J}(x,t)={\mathbf{J}}_0(x,t)+{\mathbf{J}}_1(x,t)+{\mathbf{J}}_2(x,t)+\cdots \hfill \\ \hfill & \mathbf{J}(x,t)=\frac{\zeta }{\kappa }{\left(sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)}^2+(\frac{1}{2}\zeta {\left(\frac{\zeta }{\kappa }\right)}^{\frac{3}{2}}\left(7-2\nu +cosh\left(\sqrt{\frac{\zeta }{\kappa }}x\right)\right){sech}^4\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\hfill \\ \hfill & tanh\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right))\frac{t^{\varsigma }}{\mathrm{\Gamma }(\varsigma +1)}+\frac{t^{\varsigma }{\zeta }^4}{32{\kappa }^3\mathrm{\Gamma }(2\varsigma +1)\mathrm{\Gamma }(\varsigma +\beta +1)}(8\nu t^{\beta }\mu \left(2+cosh\left(\sqrt{\frac{\zeta }{\kappa }}x\right)-cosh\left(2\sqrt{\frac{\zeta }{\kappa }}x\right)\right)\hfill \\ \hfill & \mathrm{\Gamma }(2\varsigma +1)\kappa t^{\varsigma }\left(40-32\nu +(-345+104\nu )cosh\left(\sqrt{\frac{\zeta }{\kappa }}x\right)-8(-15+4\nu )cosh\left(2\sqrt{\frac{\zeta }{\kappa }}x\right)+cosh\left(3\sqrt{\frac{\zeta }{\kappa }}x\right)\right)\hfill \\ \hfill & \mathrm{\Gamma }(\varsigma +\beta +1)){sech}^8\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)+\cdots \hfill \\ \hfill & \mathbf{K}(x,t)={\mathbf{K}}_0(x,t)+{\mathbf{K}}_1(x,t)+{\mathbf{K}}_2(x,t)+\cdots \hfill \\ \hfill & \mathbf{K}(x,t)=\frac{\zeta }{\sqrt{2\kappa }}{\left(sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)}^2+\left(4\sqrt{2}\sqrt{\kappa }\mu {\left(\frac{\zeta }{\kappa }\right)}^{\frac{5}{2}}{csch}^3\left(\sqrt{\frac{\zeta }{\kappa }}x\right){sinh}^4\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\right)\frac{t^{\beta }}{\mathrm{\Gamma }(\beta +1)}+\hfill \\ \hfill & \frac{-1}{8\sqrt{\kappa }{\kappa }^{\frac{7}{2}}\mathrm{\Gamma }(2\beta +1)\mathrm{\Gamma }(\varsigma +\beta +1)}(\mu {\zeta }^4{t}^{\beta }sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)(-\mu t^{\beta }\left(9-14cosh\left(\sqrt{\frac{\zeta }{\kappa }}x\right)+cosh\left(2\sqrt{\frac{\zeta }{\kappa }}x\right)\right)\hfill \\ \hfill & \mathrm{\Gamma }(\varsigma +\beta +1)-12\kappa t^{\varsigma }\left(7-2\nu +cosh\left(\sqrt{\frac{\zeta }{\kappa }}x\right)\right)\mathrm{\Gamma }(2\beta +1){tanh}^2\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}x\right)\left)\right)+\cdots \hfill \end{array}$$

Inserting $\varsigma =\beta =1$, we have

$$\mathbf{J}(x,t)=\frac{\zeta }{\kappa }{\left(sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}(x-\zeta t)\right)\right)}^2,\mathbf{K}(x,t)=\frac{\zeta }{\sqrt{2\kappa }}{\left(sech\left(\frac{1}{2}\sqrt{\frac{\zeta }{\kappa }}(x-\zeta t)\right)\right)}^2.$$

(41)

Example 2.

Let us assume fractional coupled KdV Equation (1) with $\mathit{a}=-1,\mathit{b}=-6,\mathit{c}=3$, $\mathit{d}=-1,$ and $\mathit{e}=3,$ having initial condition

$$\mathbf{J}(x,0)=\frac{4{\varrho }^2{e}^{\varrho x}}{{(1+e^{\varrho x})}^2},\mathbf{K}(x,0)=\frac{4{\varrho }^2{e}^{\varrho x}}{{(1+e^{\varrho x})}^2}.$$

Case I: Solution by HPTM

Computing the YT, we obtain

$$\begin{array}{cc}\hfill & Y\left[\frac{{\partial }^{\varsigma }\mathbf{J}}{\partial t^{\varsigma }}\right]=Y\left[-{\mathbf{J}}_{xxx}(x,t)-6\mathbf{J}(x,t){\mathbf{J}}_x(x,t)+3\mathbf{K}(x,t){\mathbf{K}}_x(x,t)\right],\hfill \\ \hfill & Y\left[\frac{{\partial }^{\beta }\mathbf{K}}{\partial t^{\beta }}\right]=Y\left[-{\mathbf{J}}_{xxx}(x,t)-3\mathbf{J}(x,t){\mathbf{K}}_x(x,t)\right].\hfill \end{array}$$

(42)

After that, we have

$$\begin{array}{cc}\hfill & \frac{1}{u^{\varsigma }}\{M\left(u\right)-u\mathbf{J}\left(0\right)\}=Y\left[-{\mathbf{J}}_{xxx}(x,t)-6\mathbf{J}(x,t){\mathbf{J}}_x(x,t)+3\mathbf{K}(x,t){\mathbf{K}}_x(x,t)\right],\hfill \\ \hfill & \frac{1}{u^{\beta }}\{M\left(u\right)-u\mathbf{K}\left(0\right)\}=Y\left[-{\mathbf{J}}_{xxx}(x,t)-3\mathbf{J}(x,t){\mathbf{K}}_x(x,t)\right],\hfill \end{array}$$

(43)

$$\begin{array}{cc}\hfill & M\left(u\right)=u\mathbf{J}\left(0\right)+u^{\varsigma }\left[-{\mathbf{J}}_{xxx}(x,t)-6\mathbf{J}(x,t){\mathbf{J}}_x(x,t)+3\mathbf{K}(x,t){\mathbf{K}}_x(x,t)\right],\hfill \\ \hfill & M\left(u\right)=u\mathbf{K}\left(0\right)+u^{\beta }\left[-{\mathbf{J}}_{xxx}(x,t)-3\mathbf{J}(x,t){\mathbf{K}}_x(x,t)\right].\hfill \end{array}$$

(44)

Operating the inverse YT, we have

$$\begin{array}{cc}\hfill & \mathbf{J}(x,t)=\mathbf{J}\left(0\right)+Y^{-1}\left[u^{\varsigma }\left\{Y\left(-{\mathbf{J}}_{xxx}(x,t)-6\mathbf{J}(x,t){\mathbf{J}}_x(x,t)+3\mathbf{K}(x,t){\mathbf{K}}_x(x,t)\right)\right\}\right],\hfill \\ \hfill & \mathbf{K}(x,t)=\mathbf{K}\left(0\right)+Y^{-1}\left[u^{\beta }\left\{Y\left(-{\mathbf{J}}_{xxx}(x,t)-3\mathbf{J}(x,t){\mathbf{K}}_x(x,t)\right)\right\}\right],\hfill \\ \hfill & \mathbf{J}(x,t)=\frac{4{\varrho }^2{e}^{\varrho x}}{{(1+e^{\varrho x})}^2}+Y^{-1}\left[u^{\varsigma }\left\{Y\left(-{\mathbf{J}}_{xxx}(x,t)-6\mathbf{J}(x,t){\mathbf{J}}_x(x,t)+3\mathbf{K}(x,t){\mathbf{K}}_x(x,t)\right)\right\}\right],\hfill \\ \hfill & \mathbf{K}(x,t)=\frac{4{\varrho }^2{e}^{\varrho x}}{{(1+e^{\varrho x})}^2}+Y^{-1}\left[u^{\beta }\left\{Y\left(-{\mathbf{J}}_{xxx}(x,t)-3\mathbf{J}(x,t){\mathbf{K}}_x(x,t)\right)\right\}\right].\hfill \end{array}$$

(45)

By means of the homotopy perturbation method, we have

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }{ϵ}^k{\mathbf{J}}_k(x,t)=\frac{4{\varrho }^2{e}^{\varrho x}}{{(1+e^{\varrho x})}^2}+ϵ\left(Y^{-1}\right[u^{\varsigma }Y[-{\left(\sum _{k=0}^{\infty }{ϵ}^k{\mathbf{J}}_k(x,t)\right)}_{xxx}-6\left(\sum _{k=0}^{\infty }{ϵ}^k{H}_k(x,t)\right)+\hfill \\ \hfill & 3\left(\sum _{k=0}^{\infty }{ϵ}^k{H}_k(x,t)\right)\left]\right]),\hfill \\ \hfill & \sum _{k=0}^{\infty }{ϵ}^k{\mathbf{K}}_k(x,t)=\frac{4{\varrho }^2{e}^{\varrho x}}{{(1+e^{\varrho x})}^2}+ϵ\left(Y^{-1}\left[u^{\beta }Y\left[-{\left(\sum _{k=0}^{\infty }{ϵ}^k{\mathbf{J}}_k(x,t)\right)}_{xxx}-3\left(\sum _{k=0}^{\infty }{ϵ}^k{H}_k(x,t)\right)\right]\right]\right).\hfill \end{array}$$

(46)

Equating the similar components of ϵ, we obtain

$$\begin{array}{cc}\hfill & {ϵ}^0:{\mathbf{J}}_0(x,t)=\frac{4{\varrho }^2{e}^{\varrho x}}{{(1+e^{\varrho x})}^2},\hfill \\ \hfill & {ϵ}^0:{\mathbf{K}}_0(x,t)=\frac{4{\varrho }^2{e}^{\varrho x}}{{(1+e^{\varrho x})}^2},\hfill \\ \hfill & {ϵ}^1:{\mathbf{J}}_1(x,t)=\left(\frac{4{\varrho }^5{e}^{\varrho x}(-1+e^{\varrho x})}{{(1+e^{\varrho x})}^3}\right)\frac{t^{\varsigma }}{\mathrm{\Gamma }(\varsigma +1)},\hfill \\ \hfill & {ϵ}^1:{\mathbf{K}}_1(x,t)=\left(\frac{4{\varrho }^5{e}^{\varrho x}(-1+e^{\varrho x})}{{(1+e^{\varrho x})}^3}\right)\frac{t^{\beta }}{\mathrm{\Gamma }(\beta +1)},\hfill \\ \hfill & {ϵ}^2:{\mathbf{J}}_2(x,t)=\frac{1}{{(1+e^{\varrho x})}^6\mathrm{\Gamma }(2\varsigma +1)\mathrm{\Gamma }(\varsigma +\beta +1)}\left(4{\varrho }^8{e}^{\varrho x}{t}^{\varsigma }\right(-24e^{\varrho x}(1-3e^{\varrho x}+e^{2\varrho x})t^{\beta }\mathrm{\Gamma }(2\varsigma +1)+\hfill \\ \hfill & (1+22e^{\varrho x}-78e^{2\varrho x}+22e^{3\varrho x}+e^{4\varrho x})t^{\varsigma }\mathrm{\Gamma }(\varsigma +\beta +1)\left)\right),\hfill \\ \hfill & {ϵ}^2:{\mathbf{K}}_2(x,t)=\frac{1}{{(1+e^{\varrho x})}^6\mathrm{\Gamma }(2\beta +1)\mathrm{\Gamma }(\varsigma +\beta +1)}\left(4{\varrho }^8{e}^{\varrho x}{t}^{\beta }\right((1-14e^{\varrho x}+18e^{2\varrho x}-14e^{3\varrho x}+e^{4\varrho x})t^{\beta }\hfill \\ \hfill & \mathrm{\Gamma }(\varsigma +\beta +1)+12e^{\varrho x}{(1+e^{\varrho x})}^2{t}^{\varsigma }\mathrm{\Gamma }(2\beta +1)\left)\right),\hfill \\ \hfill & \cdots \hfill \end{array}$$

The solution we obtained is taken in series form as

$$\begin{array}{cc}\hfill & \mathbf{J}(x,t)={\mathbf{J}}_0(x,t)+{\mathbf{J}}_1(x,t)+{\mathbf{J}}_2(x,t)+\cdots \hfill \\ \hfill & \mathbf{J}(x,t)=\frac{4{\varrho }^2{e}^{\varrho x}}{{(1+e^{\varrho x})}^2}+\left(\frac{4{\varrho }^5{e}^{\varrho x}(-1+e^{\varrho x})}{{(1+e^{\varrho x})}^3}\right)\frac{t^{\varsigma }}{\mathrm{\Gamma }(\varsigma +1)}+\frac{1}{{(1+e^{\varrho x})}^6\mathrm{\Gamma }(2\varsigma +1)\mathrm{\Gamma }(\varsigma +\beta +1)}(4{\varrho }^8{e}^{\varrho x}{t}^{\varsigma }\hfill \\ \hfill & \left(-24e^{\varrho x}(1-3e^{\varrho x}+e^{2\varrho x})t^{\beta }\mathrm{\Gamma }(2\varsigma +1)+(1+22e^{\varrho x}-78e^{2\varrho x}+22e^{3\varrho x}+e^{4\varrho x})t^{\varsigma }\mathrm{\Gamma }(\varsigma +\beta +1)\right))+\cdots \hfill \\ \hfill & \mathbf{K}(x,t)={\mathbf{K}}_0(x,t)+{\mathbf{K}}_1(x,t)+{\mathbf{K}}_2(x,t)+\cdots \hfill \\ \hfill & \mathbf{K}(x,t)=\frac{4{\varrho }^2{e}^{\varrho x}}{{(1+e^{\varrho x})}^2}+\left(\frac{4{\varrho }^5{e}^{\varrho x}(-1+e^{\varrho x})}{{(1+e^{\varrho x})}^3}\right)\frac{t^{\beta }}{\mathrm{\Gamma }(\beta +1)}+\frac{1}{{(1+e^{\varrho x})}^6\mathrm{\Gamma }(2\beta +1)\mathrm{\Gamma }(\varsigma +\beta +1)}(4{\varrho }^8{e}^{\varrho x}{t}^{\beta }\hfill \\ \hfill & \left((1-14e^{\varrho x}+18e^{2\varrho x}-14e^{3\varrho x}+e^{4\varrho x})t^{\beta }\mathrm{\Gamma }(\varsigma +\beta +1)+12e^{\varrho x}{(1+e^{\varrho x})}^2{t}^{\varsigma }\mathrm{\Gamma }(2\beta +1)\right))+\cdots \hfill \end{array}$$

Case II: Solution by YTDM

Computing the YT, we obtain

$$\begin{array}{cc}\hfill & Y\left[\frac{{\partial }^{\varsigma }\mathbf{J}}{\partial t^{\varsigma }}\right]=Y\left[-{\mathbf{J}}_{xxx}(x,t)-6\mathbf{J}(x,t){\mathbf{J}}_x(x,t)+3\mathbf{K}(x,t){\mathbf{K}}_x(x,t)\right],\hfill \\ \hfill & Y\left[\frac{{\partial }^{\beta }\mathbf{K}}{\partial t^{\beta }}\right]=Y\left[-{\mathbf{J}}_{xxx}(x,t)-3\mathbf{J}(x,t){\mathbf{K}}_x(x,t)\right].\hfill \end{array}$$

(47)

After that, we have

$$\begin{array}{cc}\hfill & \frac{1}{u^{\varsigma }}\{M\left(u\right)-u\mathbf{J}\left(0\right)\}=Y\left[-{\mathbf{J}}_{xxx}(x,t)-6\mathbf{J}(x,t){\mathbf{J}}_x(x,t)+3\mathbf{K}(x,t){\mathbf{K}}_x(x,t)\right],\hfill \\ \hfill & \frac{1}{u^{\beta }}\{M\left(u\right)-u\mathbf{K}\left(0\right)\}=Y\left[-{\mathbf{J}}_{xxx}(x,t)-3\mathbf{J}(x,t){\mathbf{K}}_x(x,t)\right],\hfill \end{array}$$

(48)

$$\begin{array}{cc}\hfill & M\left(u\right)=u\mathbf{J}\left(0\right)+u^{\varsigma }\left[-{\mathbf{J}}_{xxx}(x,t)-6\mathbf{J}(x,t){\mathbf{J}}_x(x,t)+3\mathbf{K}(x,t){\mathbf{K}}_x(x,t)\right],\hfill \\ \hfill & M\left(u\right)=u\mathbf{K}\left(0\right)+u^{\beta }\left[-{\mathbf{J}}_{xxx}(x,t)-3\mathbf{J}(x,t){\mathbf{K}}_x(x,t)\right].\hfill \end{array}$$

(49)

Operating the inverse YT, we have

$$\begin{array}{cc}\hfill & \mathbf{J}(x,t)=\mathbf{J}\left(0\right)+Y^{-1}\left[u^{\varsigma }\left\{Y\left(-{\mathbf{J}}_{xxx}(x,t)-6\mathbf{J}(x,t){\mathbf{J}}_x(x,t)+3\mathbf{K}(x,t){\mathbf{K}}_x(x,t)\right)\right\}\right],\hfill \\ \hfill & \mathbf{K}(x,t)=\mathbf{K}\left(0\right)+Y^{-1}\left[u^{\beta }\left\{Y\left(-{\mathbf{J}}_{xxx}(x,t)-3\mathbf{J}(x,t){\mathbf{K}}_x(x,t)\right)\right\}\right],\hfill \\ \hfill & \mathbf{J}(x,t)=\frac{4{\varrho }^2{e}^{\varrho x}}{{(1+e^{\varrho x})}^2}+Y^{-1}\left[u^{\varsigma }\left\{Y\left(-{\mathbf{J}}_{xxx}(x,t)-6\mathbf{J}(x,t){\mathbf{J}}_x(x,t)+3\mathbf{K}(x,t){\mathbf{K}}_x(x,t)\right)\right\}\right],\hfill \\ \hfill & \mathbf{K}(x,t)=\frac{4{\varrho }^2{e}^{\varrho x}}{{(1+e^{\varrho x})}^2}+Y^{-1}\left[u^{\beta }\left\{Y\left(-{\mathbf{J}}_{xxx}(x,t)-3\mathbf{J}(x,t){\mathbf{K}}_x(x,t)\right)\right\}\right].\hfill \end{array}$$

(50)

Thus, the required solution in terms of infinite series form is

$$\mathbf{J}(x,t)=\sum _{m=0}^{\infty }{\mathbf{J}}_m(x,t),\mathbf{K}(x,t)=\sum _{m=0}^{\infty }{\mathbf{K}}_m(x,t).$$

(51)

Assume the nonlinear terms by the Adomian polynomial as $\mathbf{J}(x,t){\mathbf{J}}_x(x,t)={\sum }_{m=0}^{\infty }{\mathcal{A}}_m,\mathbf{K}(x,t){\mathbf{K}}_x(x,t)={\sum }_{m=0}^{\infty }{\mathcal{B}}_m,\mathbf{J}(x,t){\mathbf{K}}_x(x,t)={\sum }_{m=0}^{\infty }{\mathcal{C}}_m.$ So, we get

$$\begin{array}{cc}\hfill & \sum _{m=0}^{\infty }{\mathbf{J}}_m(x,t)=\mathbf{J}(x,0)+Y^{-1}\left[u^{\varsigma }Y\left[-{\mathbf{J}}_{xxx}(x,t)-6\sum _{m=0}^{\infty }{\mathcal{A}}_m+3\sum _{m=0}^{\infty }{\mathcal{B}}_m\right]\right],\hfill \\ \hfill & \sum _{m=0}^{\infty }{\mathbf{K}}_m(x,t)=\mathbf{K}(x,0)+Y^{-1}\left[u^{\varsigma }Y\left[-{\mathbf{J}}_{xxx}(x,t)-3\sum _{m=0}^{\infty }{\mathcal{C}}_m\right]\right],\hfill \\ \hfill & \sum _{m=0}^{\infty }{\mathbf{J}}_m(x,t)=\frac{4{\varrho }^2{e}^{\varrho x}}{{(1+e^{\varrho x})}^2}+Y^{-1}\left[u^{\varsigma }Y\left[-{\mathbf{J}}_{xxx}(x,t)-6\sum _{m=0}^{\infty }{\mathcal{A}}_m+3\sum _{m=0}^{\infty }{\mathcal{B}}_m\right]\right],\hfill \\ \hfill & \sum _{m=0}^{\infty }{\mathbf{K}}_m(x,t)=\frac{4{\varrho }^2{e}^{\varrho x}}{{(1+e^{\varrho x})}^2}+Y^{-1}\left[u^{\varsigma }Y\left[-{\mathbf{J}}_{xxx}(x,t)-3\sum _{m=0}^{\infty }{\mathcal{C}}_m\right]\right].\hfill \end{array}$$

(52)

By equating both sides, we obtain

$${\mathbf{J}}_0(x,t)=\frac{4{\varrho }^2{e}^{\varrho x}}{{(1+e^{\varrho x})}^2},$$

$${\mathbf{K}}_0(x,t)=\frac{4{\varrho }^2{e}^{\varrho x}}{{(1+e^{\varrho x})}^2}.$$

On $m=0$

$${\mathbf{J}}_1(x,t)=\left(\frac{4{\varrho }^5{e}^{\varrho x}(-1+e^{\varrho x})}{{(1+e^{\varrho x})}^3}\right)\frac{t^{\varsigma }}{\mathrm{\Gamma }(\varsigma +1)},$$

$${\mathbf{K}}_1(x,t)=\left(\frac{4{\varrho }^5{e}^{\varrho x}(-1+e^{\varrho x})}{{(1+e^{\varrho x})}^3}\right)\frac{t^{\beta }}{\mathrm{\Gamma }(\beta +1)}.$$

On $m=1$

$$\begin{array}{cc}\hfill & {\mathbf{J}}_2(x,t)=\frac{1}{{(1+e^{\varrho x})}^6\mathrm{\Gamma }(2\varsigma +1)\mathrm{\Gamma }(\varsigma +\beta +1)}\left(4{\varrho }^8{e}^{\varrho x}{t}^{\varsigma }\right(-24e^{\varrho x}(1-3e^{\varrho x}+e^{2\varrho x})t^{\beta }\mathrm{\Gamma }(2\varsigma +1)+\hfill \\ \hfill & (1+22e^{\varrho x}-78e^{2\varrho x}+22e^{3\varrho x}+e^{4\varrho x})t^{\varsigma }\mathrm{\Gamma }(\varsigma +\beta +1)\left)\right),\hfill \end{array}$$

$$\begin{array}{cc}\hfill & {\mathbf{K}}_2(x,t)=\frac{1}{{(1+e^{\varrho x})}^6\mathrm{\Gamma }(2\beta +1)\mathrm{\Gamma }(\varsigma +\beta +1)}\left(4{\varrho }^8{e}^{\varrho x}{t}^{\beta }\right((1-14e^{\varrho x}+18e^{2\varrho x}-14e^{3\varrho x}+e^{4\varrho x})t^{\beta }\hfill \\ \hfill & \mathrm{\Gamma }(\varsigma +\beta +1)+12e^{\varrho x}{(1+e^{\varrho x})}^2{t}^{\varsigma }\mathrm{\Gamma }(2\beta +1)\left)\right).\hfill \end{array}$$

In this manner, we can easily obtain the other terms for $(m\ge 3)$, so

$$\begin{array}{cc}\hfill & \mathbf{J}(x,t)={\mathbf{J}}_0(x,t)+{\mathbf{J}}_1(x,t)+{\mathbf{J}}_2(x,t)+\cdots \hfill \\ \hfill & \mathbf{J}(x,t)=\frac{4{\varrho }^2{e}^{\varrho x}}{{(1+e^{\varrho x})}^2}+\left(\frac{4{\varrho }^5{e}^{\varrho x}(-1+e^{\varrho x})}{{(1+e^{\varrho x})}^3}\right)\frac{t^{\varsigma }}{\mathrm{\Gamma }(\varsigma +1)}+\frac{1}{{(1+e^{\varrho x})}^6\mathrm{\Gamma }(2\varsigma +1)\mathrm{\Gamma }(\varsigma +\beta +1)}(4{\varrho }^8{e}^{\varrho x}{t}^{\varsigma }\hfill \\ \hfill & \left(-24e^{\varrho x}(1-3e^{\varrho x}+e^{2\varrho x})t^{\beta }\mathrm{\Gamma }(2\varsigma +1)+(1+22e^{\varrho x}-78e^{2\varrho x}+22e^{3\varrho x}+e^{4\varrho x})t^{\varsigma }\mathrm{\Gamma }(\varsigma +\beta +1)\right))+\cdots \hfill \\ \hfill & \mathbf{K}(x,t)={\mathbf{K}}_0(x,t)+{\mathbf{K}}_1(x,t)+{\mathbf{K}}_2(x,t)+\cdots \hfill \\ \hfill & \mathbf{K}(x,t)=\frac{4{\varrho }^2{e}^{\varrho x}}{{(1+e^{\varrho x})}^2}+\left(\frac{4{\varrho }^5{e}^{\varrho x}(-1+e^{\varrho x})}{{(1+e^{\varrho x})}^3}\right)\frac{t^{\beta }}{\mathrm{\Gamma }(\beta +1)}+\frac{1}{{(1+e^{\varrho x})}^6\mathrm{\Gamma }(2\beta +1)\mathrm{\Gamma }(\varsigma +\beta +1)}(4{\varrho }^8{e}^{\varrho x}{t}^{\beta }\hfill \\ \hfill & \left((1-14e^{\varrho x}+18e^{2\varrho x}-14e^{3\varrho x}+e^{4\varrho x})t^{\beta }\mathrm{\Gamma }(\varsigma +\beta +1)+12e^{\varrho x}{(1+e^{\varrho x})}^2{t}^{\varsigma }\mathrm{\Gamma }(2\beta +1)\right))+\cdots \hfill \end{array}$$

Inserting $\varsigma =\beta =1$ we have

$$\mathbf{J}(x,t)=\mathbf{K}(x,t)=\frac{4{\varrho }^2{e}^{\varrho (x-{\varrho }^{2}t)}}{{(1+e^{\varrho (x-{\varrho }^{2}t)})}^2}.$$

(53)

6. Results Discussion

The graphical and numerical analysis presented in this section offers valuable insights into the behavior and accuracy of our proposed solution method for the coupled Korteweg-de Vries (KdV) equations. Figure 1a depict the behavior of the exact solution and Figure 1b depict the behavior of our approaches solution for $\mathbf{J}(x,t)$ at $\varsigma =1$. Figure 2a,b show the outcomes of suggested techniques at different orders of $\varsigma =0.80,0.60$. The graphs in Figure 3a,b illustrate the nature of the precise and proposed approaches solution for $\mathbf{K}(x,t)$ at $\varsigma =1$ at $\varsigma =1$. The approximate solutions using the suggested methods are depicted in Figure 4 in both 3D and 2D graphs, with distinct values of fractional order $\varsigma =0.40$, 0.60, 0.80, 1. Table 1 displays the accurate and proposed methods estimated solution at various values of $\varsigma$ for $\mathbf{J}(x,t)$. Table 2 displays the accurate and proposed methods estimated solution at various values of $\varsigma$ for $\mathbf{K}(x,t)$. Figure 5a depict the behavior of the exact solution and Figure 5b depict the behavior of our approaches solution for $\mathbf{J}(x,t)$ and $\mathbf{K}(x,t)$ at $\varsigma =1$. Figure 6a,b explore the behavior of our approaches solutions at different orders of $\varsigma =0.80,0.60$ for $\mathbf{J}(x,t)$ and $\mathbf{K}(x,t)$. Similarly, Figure 7a,b show the outcomes of suggested techniques through 3D and 2D graphs, with distinct values of fractional order $\varsigma =0.40,0.60,0.80,1$. Table 3 displays the accurate and proposed methods estimated solution at various values of $\varsigma$ for $\mathbf{J}(x,t)$. In summary, our graphical and numerical analysis demonstrates the effectiveness of our proposed methods in approximating the solutions of coupled KdV equations. It has been proven that the suggested approaches are the most effective means of resolving FPDEs.

Figure 1. Graph (a) demonstrating the accurate solution, (b) demonstrating our technique’s solution for $\mathbf{J}(x,t)$.

Figure 2. Graph (a) demonstrating our solution at $\varsigma =0.8$, (b) demonstrating our solution at $\varsigma =0.6$ for $\mathbf{J}(x,t)$.

Figure 3. Graph (a) demonstrating the accurate solution, (b) demonstrating our technique’s solution for $\mathbf{K}(x,t)$.

Figure 4. Graph demonstrating our technique’s solution for various order of $\varsigma$ for $\mathbf{K}(x,t)$.

Table 1. Analysis of the exact and our technique’s solution at numerous orders of $\varsigma$ for $\mathbf{J}(x,t)$.

$\mathit{\varsigma }$	$\mathit{\varsigma }=0.85$	$\mathit{\varsigma }=0.90$	$\mathit{\varsigma }=0.95$	$\mathit{\varsigma }=1\left(\mathit{Appro}\right)$	$\mathit{\varsigma }=1\left(\mathit{Exact}\right)$
0.0	1.00000000	1.00000000	1.00000000	1.00000000	0.99999975
0.1	0.99765268	0.99760754	0.99757599	0.99755399	0.99755374
0.2	0.99036039	0.99027100	0.99020853	0.99016496	0.99016472
0.3	0.97826716	0.97813527	0.97804311	0.97797883	0.97797860
0.4	0.96160834	0.96143650	0.96131642	0.96123266	0.96123245
0.5	0.94070103	0.94049247	0.94034673	0.94024507	0.94024488
0.6	0.91593153	0.91569002	0.91552126	0.91540355	0.91540338
0.7	0.88774061	0.88747037	0.88728152	0.88714980	0.88714966
0.8	0.85660774	0.85631323	0.85610742	0.85596388	0.85596376
0.9	0.82303486	0.82272069	0.82250115	0.82234803	0.82234793
1.0	0.78753093	0.78720170	0.78697163	0.78681116	0.78681109

Table 2. Analysis of the exact and our technique’s solution at numerous orders of $\varsigma$ for $\mathbf{K}(x,t)$.

$\mathit{\varsigma }$	$\mathit{\varsigma }=0.85$	$\mathit{\varsigma }=0.90$	$\mathit{\varsigma }=0.95$	$\mathit{\varsigma }=1\left(\mathit{Appro}\right)$	$\mathit{\varsigma }=1\left(\mathit{Exact}\right)$
0.1	0.70565703	0.70556126	0.70549434	0.70544766	0.70537701
0.2	0.70070648	0.70051685	0.70038434	0.70029191	0.70015219
0.3	0.69235298	0.69207321	0.69187771	0.69174135	0.69153530
0.4	0.68075931	0.68039479	0.68014006	0.67996239	0.67969398
0.5	0.66614650	0.66570406	0.66539490	0.66517925	0.66485353
0.6	0.64878509	0.64827277	0.64791477	0.64766507	0.64728793
0.7	0.62898482	0.62841154	0.62801094	0.62773152	0.62730954
0.8	0.60708344	0.60645870	0.60602213	0.60571762	0.60525778
0.9	0.58343532	0.58276886	0.58230315	0.58197832	0.58148780
1.0	0.55840034	0.55770193	0.55721388	0.55687347	0.55635945

Figure 5. Graph (a) demonstrating the accurate solution, (b) demonstrating our technique’s solution for $\mathbf{J}(x,t)$ and $\mathbf{K}(x,t)$.

Figure 6. Graph (a) demonstrating our solution at $\varsigma =0.8$, (b) demonstrating our solution at $\varsigma =0.6$ for $\mathbf{J}(x,t)$ and $\mathbf{K}(x,t)$.

Figure 7. Graph demonstrating our technique’s solution various order of $\varsigma$ for $\mathbf{J}(x,t)$ and $\mathbf{K}(x,t)$.

Table 3. Analysis of the exact and our technique’s solution at numerous orders of $\varsigma$ for $\mathbf{J}(x,t)$ and $\mathbf{K}(x,t)$.

$\mathit{\varsigma }$	$\mathit{\varsigma }=0.85$	$\mathit{\varsigma }=0.90$	$\mathit{\varsigma }=0.95$	$\mathit{\varsigma }=1\left(\mathit{Appro}\right)$	$\mathit{\varsigma }=1\left(\mathit{Exact}\right)$
0.0	1.00000000	1.00000000	1.00000000	1.00000000	0.99999974
0.1	0.99765268	0.99760754	0.99757599	0.99755399	0.99755374
0.2	0.99036039	0.99027100	0.99020853	0.99016496	0.99016472
0.3	0.97826716	0.97813527	0.97804311	0.97797883	0.97797860
0.4	0.96160834	0.96143650	0.96131642	0.96123266	0.96123245
0.5	0.94070103	0.94049247	0.94034673	0.94024507	0.94024488
0.6	0.91593153	0.91569002	0.91552126	0.91540355	0.91540338
0.7	0.88774061	0.88747037	0.88728152	0.88714980	0.88714966
0.8	0.85660773	0.85631323	0.85610742	0.85596388	0.85596376
0.9	0.82303486	0.82272069	0.82250115	0.82234803	0.82234793
1.0	0.78753093	0.78720170	0.78697163	0.78681116	0.78681109

7. Conclusions

The YTDM and HPTM have been successfully used in this study to approximate the solution of the time-fractional coupled KdV equation. This study highlights the significance of fractional derivatives and the methods for handling the recurrence relation. The results obtained provide the highest degree of agreement regarding the exact solutions to the test issues. By using the suggested approaches to test issues, it was confirmed that the mathematical model of arbitrary order could interpret any experimental data more accurately than the model of integer-order. The series solutions are consequently executed by employing both techniques. With the aid of the MAPLE 2015 software, all iterations were computed. Tables and graphs are also used to present the numerical solutions. We have seen that the results of the arbitrary order converge to the integer-order solution for the test issues as the arbitrary order approaches integer order. These methods prove useful for a wide range of nonlinear fractional differential equations in science and engineering, as shown by the solution graphs and tables. This strategy could be expanded in future studies to address diverse nonlinear obstacle issues.