Certain Analytic Solutions for Time-Fractional Models Arising in Plasma Physics via a New Approach Using the Natural Transform and the Residual Power Series Methods

Abstract

This paper studies three time-fractional models that arise in plasma physics: the modified Korteweg–deVries–Zakharov–Kuznetsov equation, the stochastic potential Korteweg–deVries equation, and the forced Korteweg–deVries equation. These equations are significant in plasma physics for modeling nonlinear ion acoustic waves and thus helping us to understand wave dynamics in plasmas. We introduce a new approach that relies on a new fractional expansion in the natural transform space and residual power series method to construct analytical solutions to the governing models. We investigate the theoretical analysis of the proposed method for these equations to expose this approach’s applicability, efficiency, and effectiveness in constructing analytical solutions to the governing equations. Moreover, we present a comparative discussion between the solutions derived during the work and those given in the literature to confirm that the proposed approach generates analytical solutions that rapidly converge to exact solutions, which proves the effectiveness of the proposed method.

1. Introduction

Nonlinear wave phenomena are of fundamental importance in the context of physics in general and plasma dynamics in particular [1]. A comprehensive understanding of ion acoustic waves and their behavior is essential for many applications in the context of plasma physics, such as energy transfer, plasma containment, and wave–particle interactions [2,3]. In light of this, many mathematical models have been presented to describe such phenomena, including the Korteweg–de Vries (KdV) equation [4], the nonlinear Schrödinger equation [5], Burgers’ equation [6], the Zakharov system [7], the sine-Gordon equation [8], the Benjamin–Bona–Mahony equation [9], the Kadomtsev–Petviashvili equation [10], and others. Since the emergence of the concept of fractional differential and integral calculus and the work carried out by researchers in studying its impact on mathematical systems in various fields of science, researchers have turned their attention to researching the importance of fractional differential and integral calculus in plasma physics [11,12]. It has been shown that fractional calculations have great importance for the effects of memory and generic properties of systems, which play a prominent role in providing an accurate description of complex plasma dynamics [13,14]. Considering the long-term interactions in plasma systems, the fractional derivative provides a more general approach than classical derivatives, which in turn helps in modeling the wave propagation, diffusion processes, and anomalous transport phenomena that may not be available in standard models to capture the complex behavior of plasma [15]. This enables scientists to gain a deeper understanding of plasma behavior in physical environments and laboratories by developing fractional mathematical models for ion acoustic waves, plasma turbulence, and other nonlinear phenomena [16]. The main interest of this work is to study three important equations in the context of plasma physics: the fractional modified Korteweg–deVries–Zakharov–Kuznetsov (mKdV-ZK) equation, the fractional stochastic potential Korteweg–deVries (spKdV) equation, and the fractional forced Korteweg–deVries (FKdV) equation. We consider the following time-fractional KdV-ZK equation [17]:

$${\mathcal{D}}_t^{\alpha }\psi +{\mu }^2\psi {\displaystyle \frac{\partial \psi }{\partial x}}+{\displaystyle \frac{{\partial }^3\psi }{\partial x^3}}+{\displaystyle \frac{{\partial }^3\psi }{\partial x\partial y^2}}+{\displaystyle \frac{{\partial }^3\psi }{\partial x\partial z^2}}=0, t\ge 0, x,y,z\in \mathbb{R},$$

(1.1)

subject to the initial condition:

$$\psi (x,y,z,0)=f\left(x,y,z\right),x,y,z\in \mathbb{R},$$

(1.2)

where ${\mathcal{D}}_t^{\alpha }\psi$ represents the Caputo derivative of the electric field potential $\psi \left(x,y,z,t\right)$ with respect to the time $t$ of the fractional derivative order $\alpha \in (0,1]$. The variables $x,y$ and $z$ are the scaled space coordinates, while $\mu$ is the dispersion coefficient that stands for positive correlation with negative correlation, Boltzmann distribution, and fluid species. The Korteweg–DeVries–Zakharov–Kuznetsov equation was introduced by the authors [18] with the aim of using the reductive perturbation approach for mixing both warm adiabatic fluids and cold stationary isothermal solid background species in a magnetized plasma where they discussed acoustic electron, acoustic ion, and acoustic dust waves. In Ref. [19], the mKdV-ZK equation is presented using the reductive perturbation approach to control the diagonal diffusion in nonlinear electrostatic modes. In Ref. [20], multiple-soliton solutions for the mKdV-ZK equation were obtained. Solitary solutions were explored using the modified extended direct algebraic method for the mKdV-ZK equation in Ref. [21]. Younas et al. constructed traveling wave solutions by utilizing two different approaches to the mKdV-ZK equation [22]. The second equation considered in this work is the time-fractional spKdV equation [23]:

$${\mathcal{D}}_t^{\alpha }\psi +\omega {\displaystyle \frac{\partial \psi }{\partial x}}+\beta {\left({\displaystyle \frac{\partial \psi }{\partial x}}\right)}^2+\gamma {\displaystyle \frac{{\partial }^3\psi }{\partial x^3}}=0, t\ge 0, x\in \mathbb{R},$$

(1.3)

subject to the initial condition:

$$\psi \left(x,0\right)=g\left(x\right),x\in \mathbb{R},$$

(1.4)

where $\omega$ is the stochastic parameter, $\beta$ is the nonlinearity coefficient, and $\gamma$ refers to the dispersion coefficient. Equation (1.3) is a nonlinear model that appears in applications of multicomponent plasmas and electrical circuits and is used to predict the weak dispersive effects on the propagation of nonlinear optical waves and photons. The spKdV equation is presented in Ref. [23], where the authors obtained lumps, breathers, and multi-soliton solutions using the simplified Hirota method and the Cole–Hopf transformation. In Ref. [24], the authors suggested the fractional spKdV equation and they constructed closed-form solutions for the proposed model. In Ref. [25], the spKdV equation was investigated and the Lie symmetry approach was utilized to carry out the symmetry generators to obtain traveling wave solutions and conservation laws for the proposed model. The final equation that will be investigated in our paper is the following fractional FKdV equation [26]:

$${\displaystyle \frac{1}{c}}{\mathcal{D}}_t^{\alpha }\psi +\left(\left({\mathcal{F}}_r-1\right)-{\displaystyle \frac{3}{2\chi }}\psi \right){\displaystyle \frac{\partial \psi }{\partial x}}-{\displaystyle \frac{{\chi }^2}{6}}{\displaystyle \frac{{\partial }^3\psi }{\partial x^3}}={\displaystyle \frac{1}{2}}{\displaystyle \frac{d\varphi \left(x\right)}{dx}}, t\ge 0, x\in \mathbb{R},$$

(1.5)

subject to the initial condition:

$$\psi \left(x,0\right)=h\left(x\right),x\in \mathbb{R}.$$

(1.6)

The term $\left(\left({\mathcal{F}}_r-1\right)-{\displaystyle \frac{3}{2\chi }}\psi \right){\displaystyle \frac{\partial \psi }{\partial x}}$ describes nonlinear interactions in a wave that are measured using the parameter $\chi$ which arises from the amplitude of the plasma wave where higher amplitudes affect the wave velocity. The parameter ${\mathcal{F}}_r$ represents some force or property of the plasma. The term ${\displaystyle \frac{1}{2}}{\displaystyle \frac{d\varphi \left(x\right)}{dx}}$ represents an external influence on the plasma wave such as an electric or magnetic field or any other disturbance. The importance of Equation (1.5) lies in allowing the integration of external forces, memory effects, nonlinearity, and dispersion to model the propagation of nonlinear ion-acoustic waves, as these components are necessary to accurately describe the wave behavior in plasma environments. This model is of great value in many contexts such as astrophysical plasmas affected by large-scale forces, fusion plasmas affected by electric and magnetic fields, and the study of stability and turbulence in plasma systems. In Ref. [26], the authors investigated the fractional FKdV equation to construct traveling wave solutions using the fractional natural decomposition method. The distributed-order time-fractional FKdV equation is presented in Ref. [27], where the authors used the shifted Legendre operational matrix to infer approximate solutions for the proposed model. In Ref. [28], the authors studied the FKdV equation and obtained approximate solutions for the model using the q-homotopy analysis transform technique. The fractional FKdV equation was discussed with the help of the Yang homotopy perturbation method and the Yang transform decomposition method to construct approximate solutions for the model in Ref. [29].

With the wide development in scientific research regarding mathematical systems that describe and model physical phenomena and the emergence and discovery of complex systems in various fields of science, the need to develop methods to derive accurate and approximate solutions for these presented systems arose. Researchers worked to develop such techniques, including the reproducing kernel Hilbert space method [30], the multistep reduced differential transformation method [31], the Atangana–Baleanu fractional framework for the reproducing kernel technique [32], the multistep generalized differential transform method [33], the homotopy analysis method [34], the shifted Jacobi polynomials method [35], the Chebyshev polynomials method [36], auxiliary equation method [37], the B-spline schemes method [38], the fractional variational iteration method [39,40,41], the Kudryashov method [42], and others. Abu Arqub et al. [43] presented a new analytical method, the residual power series (RPS) method, for obtaining solitary wave solutions of time-fractional dispersive partial differential equations based on the generalized Taylor series formula and residual error function. Researchers have employed this method to derive analytical solutions for a wide range of fractional equations as it is an effective method [44,45]. In Ref. [46], El-Ajou proposed a new technique that combines the Laplace transform and RPS method, called the Laplace residual power series (LRPS) method, where the proposed method was utilized to explore exact solitary solutions to the nonlinear time-fractional dispersive partial differential equations (PDEs). The main goal of this work is to suggest a new attractive approach, the natural residual power series (NRPS) method, that depends on the natural transformation and RPS method. Moreover, we seek to utilize the NRPS method to infer analytical solutions for the governing Equations (1.1)–(1.6). The NRPS method is based on using natural transformation to transform the time-fractional differential equation into the natural space, then using a suitable expansion to deal with the newly obtained equation together with the RPS method, which in turn helps in deducing the expansion coefficients based on the concept of limit, which reduces the calculations required to reach accurate approximate solutions to the governing equation.

We organized the paper with the introduction as the first section. Section 2 presents some preliminaries about the Caputo fractional derivative. Section 3 is devoted to introducing the definition of natural transformation and investigating its essential properties, then presenting the natural fractional expansion. In Section 4, we introduce the steps of the proposed method. Applications of the NRPS method to construct approximate solutions for the governing models are introduced in Section 5. Some conclusions are presented in Section 6.

2. Preliminaries

Since the advent of fractional calculus, many fractional operators have appeared, such as the Riemann–Liouville operator [47], the Caputo operator [48], Atangana–Baleanu operator [49], and others. Here, we present the definition of the Caputo fractional derivative and some of its essential properties that are useful in our work.

Definition 1

([49]). The Caputo derivative of fractional order $\alpha >0$ of an integrable function $\psi \left(t\right)$ is given by:

$${\mathcal{D}}_t^{\alpha }\psi \left(t,x\right)={\displaystyle \frac{1}{\mathsf{\Gamma }\left(k-\alpha \right)}}\underset{0}{\overset{t}{\int }}{\displaystyle \frac{\frac{{\partial }^k\psi (\tau ,x)}{\partial {\tau }^k }}{{\left(t-\tau \right)}^{\alpha -k+1}}}d\tau ,$$

(2.1)

where $t\ge 0, k=\left[\alpha \right]+1$. The Riemann–Liouville fractional integral of order $\alpha , Re\left(\alpha \right)>0$ is given by:

$${\mathcal{J}}_t^{\alpha }\psi \left(t\right)={\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}\underset{0}{\overset{t}{\int }}{\left(t-\tau \right)}^{\alpha -1}\psi \left(\tau ,x\right)d\tau .$$

(2.2)

Lemma 1

([49]). For $\alpha >0, \eta >-1, p\in \mathbb{R}$ and $t\ge 0$, the Caputo fractional derivative has the property:

$$\begin{array}{}\hfill \text{(2.3)}\hfill & \hfill \text{(a)}\hfill & {\mathcal{D}}_t^{\alpha }p=0.\hfill \hfill \text{(2.4)}\hfill & \hfill \text{(b)}\hfill & {\mathcal{D}}_t^{\alpha }{t}^{\eta }={\displaystyle \frac{\mathsf{\Gamma }\left(\eta +1\right)}{\mathsf{\Gamma }\left(\eta +1-\alpha \right)}}{t}^{\eta -\alpha }.\hfill \hfill \text{(2.5)}\hfill & \hfill \text{(c)}\hfill & {\mathcal{D}}_t^{\alpha }{\mathcal{J}}_t^{\alpha }\psi \left(t,x\right)=\psi \left(t,x\right).\hfill \hfill \text{(2.6)}\hfill & \hfill \text{(d)}\hfill & {\mathcal{J}}_t^{\alpha }{\mathcal{D}}_t^{\alpha }\psi \left(t,x\right)=\psi \left(t,x\right)-{\displaystyle \sum _{i=0}^{k-1}}{\displaystyle \frac{{\partial }^i\psi \left(0^{+},x\right)}{\partial t^i}} {\displaystyle \frac{t^i}{i!}}.\hfill \end{array}$$

3. Natural Transformation and Fractional Expansion

The natural transformation combines the features of the Laplace and the Sumudu transforms into a powerful tool for dealing with a variety of differential and integral equations. Natural transformation provides a simple and flexible approach to transforming functions from the time domain to the complex domain, making it powerful for dealing with problems involving fractional time models. Furthermore, the natural transform can handle a wide range of functions with singularities and simplifies the process of finding analytical solutions to complex problems. Therefore, it is considered a valuable tool widely used in applied mathematics, engineering, and physics [50,51,52,53,54]. The general integral transform of the function $\psi \left(t,x\right), t,x\in \mathbb{R}$ is defined as:

$$\mathcal{T}\left\{\psi \left(t,x\right)\right\}=\underset{-\infty }{\overset{\infty }{\int }}\mathcal{K}\left(s,t\right)\psi \left(t,x\right)dt,$$

(3.1)

where $\mathcal{K}\left(s,t\right)$ is the kernel of the transform and $s$ represents the real (complex) number that is independent of $t$. The Laplace transform, Mellin transform, and Hankel transform can be obtained by letting the kernel of the transform in (3.1) be: $e^{-st}, t^{s-1}\left(st\right)$, and $t{J}_n\left(st\right)$, respectively. To define the natural transformation, we define the following set:

$$\mathcal{C}=\left\{\psi \left(t,x\right):\exists B, {\sigma }_1,{\sigma }_2, \left|\psi \left(t,x\right)\right|<B{e}^{\left({\displaystyle \frac{\left|t\right|}{{\sigma }_i}}\right)},t\in {\left(-1\right)}^i\times \left[0,\infty \right), x\in A\subseteq \mathbb{R},i\in {\mathbb{Z}}^{+}\right\}$$

(3.2)

Definition 2

([51]). Let $\psi \left(t,x\right)\in \mathcal{C}$. The natural transformation of the function $\psi \left(t,x\right)$ is defined as:

$$\mathsf{\Psi }\left(s,w,x\right)=\mathcal{N}\left\{\psi \left(t,x\right)\right\}=\underset{0}{\overset{\infty }{\int }}{e}^{-st}\psi \left(wt,x\right)dt, s,w\in (0,\infty ),$$

(3.3)

where $\mathcal{N}\left\{\psi \left(t,x\right)\right\}$ is the natural transformation of $\psi \left(t,x\right)$ and the variables $s$ and $w$ are the natural transform variables. The inverse natural transformation of the function $\Psi \left(s,w,x\right)$ is defined as follows:

$$\psi \left(t,x\right)={\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x\right)\right\}={\displaystyle \frac{1}{2\pi i}}\underset{u-i\infty }{\overset{u+i\infty }{\int }}{e}^{-{\displaystyle \frac{st}{w}}}\mathsf{\Psi }\left(s,w,x\right)ds, t\in \left[0,\infty \right),$$

(3.4)

where the integral is applied for a complex number $s=a+ib$ along $s=u$.

The following theorem shows the existence of natural transformations.

Theorem 1

([51]). Let $\psi \left(t,x\right)\in \mathcal{C}$ be piecewise continuous on $\left[0,\infty \right)\times A$ and of an exponential order ${\displaystyle \frac{\theta }{w}}$ as $t\to \infty$. Then, the natural transformation of $\psi \left(t,x\right)$ exists for all $s\in \left(\theta ,\infty \right),w\in \left(0,\infty \right)$.

Remark 1.

The Laplace and Sumudu transforms can be obtained by letting $w=1$ and $s=1$, respectively, in the natural transform (3.3).

Remark 2.

Let $\mathsf{\Psi }\left(s,w,x\right)=\mathcal{N}\left\{\psi \left(t,x\right)\right\}$ be the natural transform of the function and let

$${\mathsf{\Psi }}_L\left(s,x\right)=\mathcal{L}\left\{\psi \left(t,x\right)\right\}=\underset{0}{\overset{\infty }{\int }}{e}^{-st}\psi \left(t,x\right)dt,$$

$${\mathsf{\Psi }}_S\left(w,x\right)=\mathcal{S}\left\{\psi \left(t,x\right)\right\}=\underset{0}{\overset{\infty }{\int }}{e}^{-t}\psi \left(wt,x\right)dt,$$

be the Laplace and Sumudu transforms of the function $\psi \left(t,x\right)$, respectively. Then, the duality relationship between the Natural-Laplace and Natural-Sumudu transforms is given by:

$$\mathcal{N}\left\{\psi \left(t,x\right)\right\}=\mathsf{\Psi }\left(s,w,x\right)={\displaystyle \frac{1}{w}}{\mathsf{\Psi }}_L\left({\displaystyle \frac{s}{w}},x\right),$$

(3.5)

$$\mathcal{N}\left\{\psi \left(t,x\right)\right\}=\mathsf{\Psi }\left(s,w,x\right)={\displaystyle \frac{1}{s}}{\mathsf{\Psi }}_S\left({\displaystyle \frac{w}{s}},x\right).$$

(3.6)

Lemma 2

([52]). Let ${\psi }_1\left(t,x\right),{\psi }_2\left(t,x\right)\in \mathcal{C}$ be piecewise continuous on $\left[0,\infty \right)\times A$ and of an exponential order ${\displaystyle \frac{{\theta }_1}{w}}$ and ${\displaystyle \frac{{\theta }_2}{w}}$, respectively, with ${\theta }_1<{\theta }_2$. Then, the natural transform satisfies the following properties:

$$\begin{array}{}\hfill \text{(3.7)}\hfill & \hfill \text{(i)}\hfill & \mathcal{N}\left(1\right)={\displaystyle \frac{1}{s}}.\hfill \hfill \text{(3.8)}\hfill & \hfill \text{(ii)}\hfill & \mathcal{N}\left(t^{k\eta }\right)={\displaystyle \frac{\mathsf{\Gamma }\left(k\eta +1\right)w^{k\eta }}{s^{k\eta +1}}}, \eta >-1.\hfill \hfill \text{(3.9)}\hfill & \hfill \text{(iii)}\hfill & \mathcal{N}\left(p_1{\psi }_1\left(t,x\right)+p_2{\psi }_2\left(t,x\right)\right)=p_1{\mathsf{\Psi }}_1\left(s,w,x\right)+p_2{\mathsf{\Psi }}_2\left(s,w,x\right), p_1,p_2\in \mathbb{R}.\hfill \hfill \text{(3.10)}\hfill & \hfill \text{(iv)}\hfill & {\mathcal{N}}^{-1}\left(p_1{\mathsf{\Psi }}_1\left(s,w,x\right)+p_2{\mathsf{\Psi }}_2\left(s,w,x\right)\right)=p_1{\psi }_1\left(t,x\right)+p_2{\psi }_1\left(t,x\right), p_1,p_2\in \mathbb{R}.\hfill \end{array}$$

Lemma 3.

Let $\psi \left(t,x\right)\in \mathcal{C}$ be piecewise continuous on $\left[0,\infty \right)\times A$ and of exponential order ${\displaystyle \frac{\theta }{w}}$. Then, the natural transform satisfies the following properties:

$$\begin{array}{}\hfill \text{(3.11)}\hfill & \hfill \text{(i)}\hfill & \mathcal{N}\left\{{\displaystyle \frac{{\partial }^k}{\partial t^k}}\psi \left(t,x\right)\right\}={\displaystyle \frac{s^k}{w^k}}\mathsf{\Psi }\left(s,w,x\right)-{\displaystyle \sum _{i=0}^{k-1}}{\displaystyle \frac{s^{k-i-1}}{w^{k-i}}}{\displaystyle \frac{{\partial }^i}{\partial t^i}}\psi \left(0,x\right),k\in \mathbb{Z}.\hfill \hfill \text{(3.12)}\hfill & \hfill \text{(ii)}\hfill & \underset{s\to \infty }{\lim }s\mathsf{\Psi }\left(s,w,x\right)=\psi \left(0,x\right).\hfill \hfill \text{(3.13)}\hfill & \hfill \text{(iii)}\hfill & \mathcal{N}\left\{{\mathcal{D}}_t^{\alpha }\psi \left(t,x\right)\right\}={\displaystyle \frac{s^{\alpha }}{w^{\alpha }}}\mathsf{\Psi }\left(s,w,x\right)-{\displaystyle \frac{s^{\alpha -1}}{w^{\alpha }}}\psi \left(0,x\right),0<\alpha <1. \hfill \hfill \text{(3.14)}\hfill & \hfill \text{(iv)}\hfill & \mathcal{N}\left\{{\mathcal{D}}_t^{k\alpha }\psi \left(t,x\right)\right\}={\displaystyle \frac{s^{k\alpha }}{w^{k\alpha }}}\mathsf{\Psi }\left(s,w,x\right)-{\displaystyle \sum _{i=0}^{k-1}}{\displaystyle \frac{s^{\left(k-i\right)\alpha -1}}{w^{\left(k-i\right)\alpha }}}{\mathcal{D}}_t^{i\alpha }\psi \left(0,x\right),0<\alpha <1.\hfill \end{array}$$

where ${\mathcal{D}}_t^{k\alpha }={\mathcal{D}}_t^{\alpha }{\mathcal{D}}_t^{\alpha }\dots {\mathcal{D}}_t^{\alpha }$ ( $k$-times).

Proof. 

The properties (i) can be found in Ref. [48]. To prove (ii), we use the property (i) at $k=1$, we obtain:

$$\mathcal{N}\left\{{\displaystyle \frac{\partial }{\partial t}}\psi \left(t,x\right)\right\}={\displaystyle \frac{s}{w}}\mathsf{\Psi }\left(s,w,x\right)-{\displaystyle \frac{1}{w}}\psi \left(0,x\right).$$

(3.15)

Which gives:

$$s\mathsf{\Psi }\left(s,w,x\right)=w\underset{-\infty }{\overset{\infty }{\int }}{e}^{-st}{\displaystyle \frac{\partial }{\partial t}}\psi \left(wt,x\right)dt+\psi \left(0,x\right).$$

(3.16)

Taking the limit for both sides of (3.16) as $s\to \infty$ to obtain the property (ii). To prove (iii), we use the following fact for the Sumudu transform for the Caputo fractional derivative:

$${\mathsf{\Psi }}_{\mathrm{S}}^{*}\left(w,x\right)=\mathcal{S}\left\{{\mathcal{D}}_t^{\alpha }\psi \left(t,x\right)\right\}={\displaystyle \frac{1}{w^{\alpha }}}\left({\mathsf{\Psi }}_S\left(w,x\right)-\psi \left(0,x\right)\right).$$

(3.17)

Substitute $w={\displaystyle \frac{w}{s}}$ in (3.17) to obtain:

$${\mathsf{\Psi }}_{\mathrm{S}}^{*}\left({\displaystyle \frac{w}{s}},x\right)={\displaystyle \frac{s^{\alpha }}{w^{\alpha }}}\left({\mathsf{\Psi }}_S\left({\displaystyle \frac{w}{s}},x\right)-\psi \left(0,x\right)\right).$$

(3.18)

Multiply (3.18) by ${\displaystyle \frac{1}{s}}$, we have:

$${\displaystyle \frac{1}{s}}{\mathsf{\Psi }}_{\mathrm{S}}^{*}\left({\displaystyle \frac{w}{s}},x\right)={\displaystyle \frac{s^{\alpha }}{w^{\alpha }}}{\displaystyle \frac{1}{s}} {\mathsf{\Psi }}_S\left({\displaystyle \frac{w}{s}},x\right)-{\displaystyle \frac{s^{\alpha -1}}{w^{\alpha }}}\psi \left(0,x\right).$$

(3.19)

Therefore, using the relationship (3.6) to obtain the property (iii):

$$\mathcal{N}\left\{{\mathcal{D}}_t^{\alpha }\psi \left(t,x\right)\right\}={\displaystyle \frac{s^{\alpha }}{w^{\alpha }}}\mathsf{\Psi }\left(s,w,x\right)-{\displaystyle \frac{s^{\alpha -1}}{w^{\alpha }}}\psi \left(0,x\right).$$

(3.20)

We use mathematical induction to verify the property (iv). For $k=1$, the property is proved in (iii). We assume the formula is satisfied at $k=n$, then we have:

$$\mathcal{N}\left\{{\mathcal{D}}_t^{n\alpha }\psi \left(t,x\right)\right\}={\displaystyle \frac{s^{n\alpha }}{w^{n\alpha }}}\mathsf{\Psi }\left(s,w,x\right)-\sum _{i=0}^{n-1}{\displaystyle \frac{s^{\left(n-i\right)\alpha -1}}{w^{\left(n-i\right)\alpha }}}{\mathcal{D}}_t^{i\alpha }\psi \left(0,x\right).$$

(3.21)

We seek to verify it at $k=n+1$. Now,

$$\mathcal{N}\left\{{\mathcal{D}}_t^{\left(n+1\right)\alpha }\psi \left(t,x\right)\right\}=\mathcal{N}\left\{{\mathcal{D}}_t^{\alpha }\left({\mathcal{D}}_t^{n\alpha }\psi \left(t,x\right)\right)\right\}=\mathcal{N}\left\{{\mathcal{D}}_t^{\alpha }\varphi \left(t,x\right)\right\},$$

(3.22)

where $\varphi \left(t,x\right)={\mathcal{D}}_t^{n\alpha }\psi \left(t,x\right)$. Using (iii), we obtain:

$$\mathcal{N}\left\{{\mathcal{D}}_t^{\alpha }\varphi \left(t,x\right)\right\}={\displaystyle \frac{s^{\alpha }}{w^{\alpha }}}\mathcal{N}\left\{{\mathcal{D}}_t^{n\alpha }\psi \left(t,x\right)\right\}-{\displaystyle \frac{s^{\alpha -1}}{w^{\alpha }}}{\mathcal{D}}_t^{n\alpha }\psi \left(0,x\right).$$

(3.23)

Using (3.21) into (3.23) to infer:

$$\begin{array}{cc}\hfill \mathcal{N}\left\{{\mathcal{D}}_t^{\alpha }\varphi \left(t,x\right)\right\}=& {\displaystyle \frac{s^{\alpha }}{w^{\alpha }}}\left({\displaystyle \frac{s^{n\alpha }}{w^{n\alpha }}}\mathsf{\Psi }\left(s,w,x\right)-{\displaystyle \sum _{i=0}^{n-1}}{\displaystyle \frac{s^{\left(n-i\right)\alpha -1}}{w^{\left(n-i\right)\alpha }}}{\mathcal{D}}_t^{i\alpha }\psi \left(0,x\right)\right)-{\displaystyle \frac{s^{\alpha -1}}{w^{\alpha }}}{\mathcal{D}}_t^{n\alpha }\psi \left(0,x\right)\hfill \\ & ={\displaystyle \frac{s^{\left(n+1\right)\alpha }}{w^{\left(n+1\right)\alpha }}}\mathsf{\Psi }\left(s,w,x\right)-{\displaystyle \sum _{i=0}^{n-1}}{\displaystyle \frac{s^{\left(n+1-i\right)\alpha -1}}{w^{\left(n+1-i\right)\alpha }}}{\mathcal{D}}_t^{i\alpha }\psi \left(0,x\right)-{\displaystyle \frac{s^{\alpha -1}}{w^{\alpha }}}{\mathcal{D}}_t^{n\alpha }\psi \left(0,x\right)\hfill \\ & ={\displaystyle \frac{s^{\left(n+1\right)\alpha }}{w^{\left(n+1\right)\alpha }}}\mathsf{\Psi }\left(s,w,x\right)-{\displaystyle \sum _{i=0}^n}{\displaystyle \frac{s^{\left(n+1-i\right)\alpha -1}}{w^{\left(n+1-i\right)\alpha }}}{\mathcal{D}}_t^{i\alpha }\psi \left(0,x\right).\hfill \end{array}$$

(3.24)

From (3.22) and (3.24), we obtain:

$$\mathcal{N}\left\{{\mathcal{D}}_t^{\left(n+1\right)\alpha }\psi \left(t,x\right)\right\}={\displaystyle \frac{s^{\left(n+1\right)\alpha }}{w^{\left(n+1\right)\alpha }}}\mathsf{\Psi }\left(s,w,x\right)-\sum _{i=0}^n{\displaystyle \frac{s^{\left(n+1-i\right)\alpha -1}}{w^{\left(n+1-i\right)\alpha }}}{\mathcal{D}}_t^{i\alpha }\psi \left(0,x\right).$$

(3.25)

Therefore, the poof is complete. □

The next theorem presents a new fractional expansion which is essential for our proposed method, the NRPS method.

Theorem 2.

Let $\psi \left(t,x\right)\in \mathcal{C}$ be piecewise continuous on $\left[0,\infty \right)\times A$ and of an exponential order ${\displaystyle \frac{\theta }{w}}$, and let $\mathsf{\Psi }\left(s,w,x\right)=\mathcal{N}\left\{\psi \left(t,x\right)\right\}$ have the following fractional expansion:

$$\mathsf{\Psi }\left(s,w,x\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x\right)w^{k\alpha }}{s^{1+k\alpha }}}, \alpha \in \left(0,1\right],x\in A, s\in \left(\theta ,\infty \right),w\in \left(0,\infty \right).$$

(3.26)

Then, the coefficients ${\Pi }_k\left(x\right)={\mathcal{D}}_t^{k\alpha }\psi \left(0,x\right),k=0,1,2,\dots$. Consequently, $\psi (t,x)$ can be written in the following fractional Taylor’s formula:

$$\psi \left(t,x\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{{\mathcal{D}}_t^{k\alpha }\psi \left(0,x\right)t^{k\alpha }}{\mathsf{\Gamma }\left(1+k\alpha \right)}},\alpha \in \left(0,1\right],x\in \mathbb{R},t\in \left(0,\infty \right).$$

(3.27)

Proof. 

Assume that $\mathsf{\Psi }\left(s,w,x\right)$ has the fractional expansion in (3.26). Multiply both sides of (3.26) by $s$, then take the limit as $s\to \infty$ to obtain:

$$\underset{s\to \infty }{\lim }s\mathsf{\Psi }\left(s,w,x\right)=\underset{s\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}s\sum _{k=0}^{\infty }{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x\right)w^{k\alpha }}{s^{1+k\alpha }}}=\underset{s\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\left({\mathsf{\Pi }}_0\left(x\right)+\sum _{k=1}^{\infty }{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x\right)w^{k\alpha }}{s^{k\alpha }}}\right)={\mathsf{\Pi }}_0\left(x\right).$$

Using (3.12), it is possible to obtain ${\mathsf{\Pi }}_0\left(x\right)=\psi (0,x)$. Hence, we can write the fractional expansion (3.26) in the form:

$$\mathsf{\Psi }\left(s,w,x\right)={\displaystyle \frac{\psi \left(0,x\right)}{s}}+\sum _{k=1}^{\infty }{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x\right)w^{k\alpha }}{s^{1+k\alpha }}}, \alpha \in \left(0,1\right],x\in \mathbb{R},s\in \left(\theta ,\infty \right),w\in \left(0,\infty \right).$$

(3.28)

Now, multiply both sides of (3.28) by $s^{\alpha +1}$ to obtain:

$$s^{\alpha +1}\mathsf{\Psi }\left(s,w,x\right)=s^{\alpha }\psi \left(0,x\right)+{\mathsf{\Pi }}_1\left(x\right)w^{\alpha }+\sum _{k=2}^{\infty }{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x\right)w^{k\alpha }}{s^{\left(k-1\right)\alpha }}}.$$

(3.29)

Consequently, with the aid of (3.13), we have:

$$\begin{array}{l}{\mathsf{\Pi }}_1\left(x\right)={\displaystyle \frac{s^{\alpha +1}}{w^{\alpha }}}\mathsf{\Psi }\left(s,w,x\right)-{\displaystyle \frac{s^{\alpha }}{w^{\alpha }}}\psi \left(0,x\right)-{\displaystyle \sum _{k=2}^{\infty }}{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x\right)w^{\left(k-1\right)\alpha }}{s^{\left(k-1\right)\alpha }}}\\ \begin{array}{cc}\hspace{1em}\hspace{1em}& =s\left({\displaystyle \frac{s^{\alpha }}{w^{\alpha }}}\mathsf{\Psi }\left(s,w,x\right)-{\displaystyle \frac{s^{\alpha -1}}{w^{\alpha }}}\psi \left(0,x\right)\right)-{\displaystyle \sum _{k=2}^{\infty }}{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x\right)w^{(k-1)\alpha }}{s^{\left(k-1\right)\alpha }}}\hfill \\ \hspace{1em}\hspace{1em}& =s\mathcal{N}\left\{{\mathcal{D}}_t^{\alpha }\psi \left(t,x\right)\right\}-{\displaystyle \sum _{k=2}^{\infty }}{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x\right)w^{(k-1)\alpha }}{s^{\left(k-1\right)\alpha }}}.\hfill \end{array}\end{array}$$

(3.30)

Take the limit as $s\to \infty$ for both sides of (3.30) with the aid of (3.12) to infer:

$${\mathsf{\Pi }}_1\left(x\right)=\underset{s\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}s\mathcal{N}\left\{{\mathcal{D}}_t^{\alpha }\psi \left(t,x\right)\right\}={\mathcal{D}}_t^{\alpha }\psi \left(0,x\right).$$

(3.31)

To find ${\mathsf{\Pi }}_2\left(x\right)$, multiply both sides of (3.28) by $s^{2\alpha +1}$ to obtain:

$$s^{2\alpha +1}\mathsf{\Psi }\left(s,w,x\right)=s^{2\alpha }\psi \left(0,x\right)+s^{\alpha }{\mathsf{\Pi }}_1\left(x\right)w^{\alpha }+{\mathsf{\Pi }}_2\left(x\right)w^{2\alpha }+\sum _{k=3}^{\infty }{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x\right)w^{k\alpha }}{s^{\left(k-2\right)\alpha }}}.$$

(3.32)

Thus, using (3.31) and (3.14), we infer:

$$\begin{array}{l}{\mathsf{\Pi }}_2\left(x\right)={\displaystyle \frac{s^{2\alpha +1}}{w^{2\alpha }}}\mathsf{\Psi }\left(s,w,x\right)-{\displaystyle \frac{s^{2\alpha }}{w^{2\alpha }}}\psi \left(0,x\right)-{\displaystyle \frac{s^{\alpha }{\mathsf{\Pi }}_1\left(x\right)}{w^{\alpha }}}-{\displaystyle \sum _{k=3}^{\infty }}{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x\right)w^{\left(k-2\right)\alpha }}{s^{\left(k-2\right)\alpha }}}\hfill \\ \begin{array}{cc}\hspace{1em}\hspace{1em}& =s\left({\displaystyle \frac{s^{2\alpha }}{w^{2\alpha }}}\mathsf{\Psi }\left(s,w,x\right)-{\displaystyle \frac{s^{2\alpha -1}}{w^{2\alpha }}}\psi \left(0,x\right)-{\displaystyle \frac{s^{\alpha -1}{\mathcal{D}}_t^{\alpha }\psi \left(0,x\right)}{w^{\alpha }}}\right)-{\displaystyle \sum _{k=3}^{\infty }}{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x\right)w^{\left(k-2\right)\alpha }}{s^{\left(k-2\right)\alpha }}}\hfill \\ \hspace{1em}\hspace{1em}& =s\mathcal{N}\left\{{\mathcal{D}}_t^{2\alpha }\psi \left(t,x\right)\right\}-{\displaystyle \sum _{k=3}^{\infty }}{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x\right)w^{\left(k-2\right)\alpha }}{s^{\left(k-2\right)\alpha }}}.\hfill \end{array}\end{array}$$

(3.33)

Take the limit as $s\to \infty$ for both sides of (3.33) with the aid of (3.12) to obtain:

$${\mathsf{\Pi }}_2\left(x\right)=\underset{s\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}s\mathcal{N}\left\{{\mathcal{D}}_t^{2\alpha }\psi \left(t,x\right)\right\}={\mathcal{D}}_t^{2\alpha }\psi \left(0,x\right).$$

(3.34)

Now, we use the induction to complete the proof. Then, assume that ${\mathsf{\Pi }}_k\left(x\right)={\mathcal{D}}_t^{k\alpha }\psi \left(0,x\right), k=0,1,2,\dots ,n$. Then, the fractional expansion (3.28) can be written as:

$$\mathsf{\Psi }\left(s,w,x\right)={\displaystyle \frac{\psi \left(0,x\right)}{s}}+\sum _{k=1}^n{\displaystyle \frac{{\mathcal{D}}_t^{k\alpha }\psi \left(0,x\right)w^{k\alpha }}{s^{1+k\alpha }}}+{\displaystyle \frac{{\mathsf{\Pi }}_{n+1}\left(x\right)w^{\left(n+1\right)\alpha }}{s^{1+\left(n+1\right)\alpha }}}+\sum _{k=n+2}^{\infty }{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x\right)w^{k\alpha }}{s^{1+k\alpha }}}.$$

(3.35)

To find the coefficient ${\mathsf{\Pi }}_{n+1}\left(x\right)$, we multiply both sides of (3.35) by $s^{\left(n+1\right)\alpha +1}$, one has:

$$\begin{array}{cc}\hfill s^{\left(n+1\right)\alpha +1}\mathsf{\Psi }\left(s,w,x\right)& \\ & =s^{\left(n+1\right)\alpha }\psi \left(0,x\right)+{\displaystyle \sum _{k=1}^n}{\displaystyle \frac{{\mathcal{D}}_t^{k\alpha }\psi \left(0,x\right)w^{k\alpha }}{s^{\left(k-n-1\right)\alpha }}}+{\mathsf{\Pi }}_{n+1}\left(x\right)w^{\left(n+1\right)\alpha }\\ & +{\displaystyle \sum _{k=n+2}^{\infty }}{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x\right)w^{k\alpha }}{s^{\left(k-n-1\right)\alpha }}}.\end{array}$$

(3.36)

Consequently, with the aid of (3.14), we obtain:

$$\begin{array}{c}{\mathsf{\Pi }}_{n+1}\left(x\right)={\displaystyle \frac{s^{\left(n+1\right)\alpha +1}}{w^{\left(n+1\right)\alpha }}}\mathsf{\Psi }\left(s,w,x\right)-{\displaystyle \frac{s^{\left(n+1\right)\alpha }}{w^{\left(n+1\right)\alpha }}}\psi \left(0,x\right)-{\displaystyle \sum _{k=1}^n}{\displaystyle \frac{{\mathcal{D}}_t^{k\alpha }\psi \left(0,x\right)s^{\left(n+1-k\right)\alpha }}{w^{\left(n+1-k\right)\alpha }}}-{\displaystyle \sum _{k=n+2}^{\infty }}{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x\right)w^{\left(k-n-1\right)\alpha }}{s^{\left(k-n-1\right)\alpha }}}\hfill \\ \begin{array}{cc} & =s\left({\displaystyle \frac{s^{\left(n+1\right)\alpha }}{w^{\left(n+1\right)\alpha }}}\mathsf{\Psi }\left(s,w,x\right)-{\displaystyle \frac{s^{\left(n+1\right)\alpha -1}}{w^{\left(n+1\right)\alpha }}}\psi \left(0,x\right)-{\displaystyle \sum _{k=1}^n}{\displaystyle \frac{{\mathcal{D}}_t^{k\alpha }\psi \left(0,x\right)s^{\left(n+1-k\right)\alpha -1}}{w^{\left(n+1-k\right)\alpha }}}\right)\\ & -{\displaystyle \sum _{k=n+2}^{\infty }}{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x\right)w^{\left(k-n-1\right)\alpha }}{s^{\left(k-n-1\right)\alpha }}}=s\mathcal{N}\left\{{\mathcal{D}}_t^{\left(n+1\right)\alpha }\psi \left(t,x\right)\right\}-{\displaystyle \sum _{k=n+2}^{\infty }}{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x\right)w^{\left(k-n-1\right)\alpha }}{s^{\left(k-n-1\right)\alpha }}}.\end{array}\end{array}$$

(3.37)

Take the limit as $s\to \infty$ for both sides of (3.37) with the aid of (3.12) to obtain:

$${\mathsf{\Pi }}_{n+1}\left(x\right)=\underset{s\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}s\mathcal{N}\left\{{\mathcal{D}}_t^{\left(n+1\right)\alpha }\psi \left(t,x\right)\right\}={\mathcal{D}}_t^{\left(n+1\right)\alpha }\psi \left(0,x\right).$$

(3.38)

Which prove that ${\mathsf{\Pi }}_k\left(x\right)={\mathcal{D}}_t^{k\alpha }\psi \left(0,x\right),k=0, 1, 2,\dots .$. Hence, the function $\mathsf{\Psi }\left(s,w,x\right)$ can be written as:

$$\mathsf{\Psi }\left(s,w,x\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{{\mathcal{D}}_t^{k\alpha }\psi \left(0,x\right)w^{k\alpha }}{s^{1+k\alpha }}}, \alpha \in \left(0,1\right],x\in A, s\in \left(\theta ,\infty \right),w\in \left(0,\infty \right).$$

(3.39)

Take the inverse natural transform of (3.39) with the aid of (3.8) to obtain:

$$\psi \left(t,x\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{{\mathcal{D}}_t^{k\alpha }\psi \left(0,x\right)t^{k\alpha }}{\mathsf{\Gamma }\left(1+k\alpha \right)}},\alpha \in \left(0,1\right],x\in \mathbb{R},t\in \left(0,\infty \right).$$

(3.40)

Which completes the proof. □

We are now going to prove that the fractional expansion for $\mathsf{\Psi }\left(s,w,x\right)$ in (3.26) converges under certain conditions.

Theorem 3.

Let $\psi \left(t,x\right)\in \mathcal{C}$ be piecewise continuous on $\left[0,\infty \right)\times A$ and of an exponential order ${\displaystyle \frac{\theta }{w}}$, and let $\mathsf{\Psi }\left(s,w,x\right)=\mathcal{N}\left\{\psi \left(t,x\right)\right\}$ have the fractional expansion (3.26). Then, the reminder ${\mathfrak{R}}_{\mathcal{K}}\left(s,w,x\right)$ of (3.26) is satisfied:

$$\left|{\mathfrak{R}}_{\mathcal{K}}\left(s,w,x\right)\right|\le {\displaystyle \frac{w^{\left(\mathcal{K}+1\right)\alpha }}{s^{\left(\mathcal{K}+1\right)\alpha +1}}}\mathfrak{M}\left(x\right), \alpha \in \left(0,1\right],x\in A, s\in \left(\theta ,{\theta }^{*}\right],w\in \left(0,\infty \right),$$

(3.41)

provided that:

$$\left|s\mathcal{N}\left\{{\mathcal{D}}_t^{\left(\mathcal{K}+1\right)\alpha }\psi \left(t,x\right)\right\}\right|\le \mathfrak{M}\left(x\right), \alpha \in \left(0,1\right],x\in A, s\in \left(\theta ,{\theta }^{*}\right],w\in \left(0,\infty \right).$$

(3.42)

Proof. 

The reminder ${\mathfrak{R}}_k\left(s,w,x\right)$ of the fractional expansion (3.26) can be written as:

$${\mathfrak{R}}_{\mathcal{K}}\left(s,w,x\right)=\mathsf{\Psi }\left(s,w,x\right)-\sum _{k=0}^{\mathcal{K}}{\displaystyle \frac{{\mathcal{D}}_t^{k\alpha }\psi \left(0,x\right)w^{k\alpha }}{s^{1+k\alpha }}}.$$

(3.43)

Multiply both sides of (3.43) by ${\displaystyle \frac{s^{\left(\mathcal{K}+1\right)\alpha +1}}{w^{\left(\mathcal{K}+1\right)\alpha }}}$; with the aid of (3.14) we obtain:

$$\begin{array}{l}{\displaystyle \frac{s^{\left(\mathcal{K}+1\right)\alpha +1}}{w^{\left(\mathcal{K}+1\right)\alpha }}}{\mathfrak{R}}_{\mathcal{K}}\left(s,w,x\right)={\displaystyle \frac{s^{\left(\mathcal{K}+1\right)\alpha +1}}{w^{\left(\mathcal{K}+1\right)\alpha }}}\mathsf{\Psi }\left(s,w,x\right)-{\displaystyle \sum _{k=0}^{\mathcal{K}}}{\displaystyle \frac{{\mathcal{D}}_t^{k\alpha }\psi \left(0,x\right)s^{\left(\mathcal{K}+1-k\right)\alpha }}{w^{\left(\mathcal{K}+1-k\right)\alpha }}}\\ \begin{array}{cc}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}& =s\left({\displaystyle \frac{s^{\left(\mathcal{K}+1\right)\alpha }}{w^{\left(\mathcal{K}+1\right)\alpha }}}\mathsf{\Psi }\left(s,w,x\right)-{\displaystyle \sum _{k=0}^{\mathcal{K}}}{\displaystyle \frac{{\mathcal{D}}_t^{k\alpha }\psi \left(0,x\right)s^{\left(\mathcal{K}+1-k\right)\alpha -1}}{w^{\left(\mathcal{K}+1-k\right)\alpha }}}\right)\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}& =s\mathcal{N}\left\{{\mathcal{D}}_t^{\left(\mathcal{K}+1\right)\alpha }\psi \left(t,x\right)\right\}.\hfill \end{array}\end{array}$$

(3.44)

Using (3.42) and (3.44), we obtain the result in (3.41). □

4. A Description of the Proposed NRPS Method

This section is devoted to presenting the NRPS method to establish approximate solutions to the nonlinear fractional PDE:

$${\mathcal{D}}_t^{\alpha }\psi \left(t,x\right)={\mathcal{G}}_x\left[\psi \left(t,x\right)\right], \alpha \in \left(0,1\right],x\in A,t\ge 0,$$

(4.1)

with the initial condition:

$$\psi \left(0,x\right)=g\left(x\right),x\in A.$$

(4.2)

The symbol ${\mathcal{G}}_x$ denotes a nonlinear operator relative to the $x$ of degree $r,x\in A,t\ge 0$, ${\mathcal{D}}_t^{\alpha }$ refers to the $\alpha$-th Caputo fractional derivative for $\alpha \in (0,1]$, and $\psi \left(t,x\right)$ is an unknown function to be determined. To obtain the desired analytical solution for (4.1)–(4.2) using the NRPS method, we can perform the following steps:

Step 1. Apply the natural transformation for both sides of (4.1) by taking advantage of the facts in Lemmas 2 and 3, we obtain:

$$\begin{gathered} \mathsf{\Psi }\left(s,w,x\right)-{\displaystyle \frac{1}{s}}g\left(x\right)-{\displaystyle \frac{w^{\alpha }}{s^{\alpha }}}\mathcal{N}\left\{{\mathcal{G}}_x\left[{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x\right)\right\}\right]\right\}=0, \alpha \in \left(0,1\right],x\in A, s \\ \in \left(\theta ,\infty \right),w\in \left(0,\infty \right), \end{gathered}$$

(4.3)

where $\mathsf{\Psi }\left(s,w,x\right)=\mathcal{N}\{\psi \left(t,x\right),x\in A,s\in \left(\theta ,\mathrm{\infty }\right),w\in \left(0,\mathrm{\infty }\right)$.

Step 2. Suppose that the solution of (4.3) can be written in the following expansion:

$$\mathsf{\Psi }\left(s,w,x\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x\right)w^{k\alpha }}{s^{1+k\alpha }}}, \alpha \in \left(0,1\right],x\in A, s\in \left(\theta ,\infty \right),w\in \left(0,\infty \right).$$

(4.4)

Define the $\mathcal{K}$th truncated series of $\mathsf{\Psi }\left(s,w,x\right)$ as:

$${\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x\right)=\sum _{k=0}^{\mathcal{K}}{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x\right)w^{k\alpha }}{s^{1+k\alpha }}}, \alpha \in \left(0,1\right],x\in A, s\in \left(\theta ,\infty \right),w\in \left(0,\infty \right).$$

(4.5)

Using the results in Theorem 2, we deduce that the initial guess ${\mathsf{\Pi }}_0\left(x\right)=g\left(x\right)$. Upon this fact, we write the $\mathcal{K}$th truncated series of $\mathsf{\Psi }\left(s,w,x\right)$ as:

$${\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x\right)={\displaystyle \frac{g\left(x\right)}{s}}+\sum _{k=1}^{\mathcal{K}}{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x\right)w^{k\alpha }}{s^{1+k\alpha }}}, \alpha \in \left(0,1\right],x\in A, s\in \left(\theta ,\infty \right),w\in \left(0,\infty \right).$$

(4.6)

Step 3. We define the so-called natural residual function (NRF) as:

$$\begin{gathered} \mathcal{N}Res\left(s,w,x\right)=\mathsf{\Psi }\left(s,w,x\right)-{\displaystyle \frac{1}{s}}g\left(x\right)-{\displaystyle \frac{w^{\alpha }}{s^{\alpha }}}\mathcal{N}\left\{{\mathcal{G}}_x\left[{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x\right)\right\}\right]\right\}, \alpha \\ \in \left(0,1\right],x\in A, s\in \left(\theta ,\infty \right),w\in \left(0,\infty \right), \end{gathered}$$

(4.7)

and the $\mathcal{K}$th truncated NRF as:

$$\begin{gathered} \mathcal{N}Re{s}_{\mathcal{K}}\left(s,w,x\right)={\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x\right)-{\displaystyle \frac{1}{s}}g\left(x\right)-{\displaystyle \frac{w^{\alpha }}{s^{\alpha }}}\mathcal{N}\left\{{\mathcal{G}}_x\left[{\mathcal{N}}^{-1}\left\{{\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x\right)\right\}\right]\right\}, \alpha \\ \in \left(0,1\right],x\in A, s\in \left(\theta ,\infty \right),w\in \left(0,\infty \right), \end{gathered}$$

(4.8)

Now, we introduce some essential facts to obtain the desired coefficients ${\mathsf{\Pi }}_k\left(x\right)$ in the expansion (4.6).

Remark 3.

The NRF and the $\mathcal{K}$th truncated NRF in (4.7) and (4.8), respectively, satisfy the following properties:

$$\begin{array}{}\hfill \text{(4.9)}\hfill & \hfill \text{(i)}\hfill & \mathcal{N}Res\left(s,w,x\right)=0,x\in A,s\in \left(\theta ,\mathrm{\infty }\right),w\in \left(0,\mathrm{\infty }\right).\hfill \hfill \text{(4.10)}\hfill & \hfill \text{(ii)}\hfill & \underset{k\to \infty }{\lim }\mathcal{N}Re{s}_{\mathcal{K}}\left(s,w,x\right)=\mathcal{N}Res\left(s,w,x\right),x\in A, s\in \left(\theta ,\infty \right),w\in \left(0,\infty \right).\hfill \hfill \text{(4.11)}\hfill & \hfill \text{(iii)}\hfill & \underset{s\to \infty }{\lim }{s}^{1+\mathcal{K}\alpha }\mathcal{N}Re{s}_{\mathcal{K}}\left(s,w,x\right)=0, \mathcal{K}=1,2,\dots ,x\in A, s\in \left(\theta ,\infty \right),w\in \left(0,\infty \right).\hfill \end{array}$$

Proof. 

The properties (i) and (ii) are clear. To prove (iii), we represent the NRF (4.7) in fractional expansion as:

$$\begin{gathered} \mathcal{N}Res\left(s,w,x\right)=\sum _{k=1}^{\infty }{\displaystyle \frac{\left({\mathsf{\Pi }}_k\left(x\right)-{\mathcal{G}}_x^{\left(k\right)}\left[{\mathsf{\Pi }}_i\left(x\right)\right]\right)w^{k\alpha }}{s^{1+k\alpha }}}, \alpha \in \left(0,1\right],x\in A, s\in \left(\theta ,\infty \right),w \\ \in \left(0,\infty \right), i\in \left\{0,1,2,\dots ,k-1\right\}, \end{gathered}$$

(4.12)

where ${\mathcal{G}}_x^{\left(k\right)},k=1,2,3,\dots$ are differential operators related to the $x$ of degree $r$. Using (4.9), we obtain ${\mathsf{\Pi }}_k\left(x\right)-{\mathcal{G}}_x^{\left(k\right)}\left[{\mathsf{\Pi }}_i\left(x\right)\right]=0$ for $k=1,2,3,\dots$ and $i\in \left\{0,1,2,\dots ,k-1\right\}$. Substitute the $\mathcal{K}$th truncated series ${\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x\right)$ into (4.12) that can be written in the following expression:

$$\mathcal{N}Re{s}_{\mathcal{K}}\left(s,w,x\right)=\sum _{k=1}^{\mathcal{K}}{\displaystyle \frac{\left({\mathsf{\Pi }}_k\left(x\right)-{\mathcal{G}}_x^{\left(k\right)}\left[{\mathsf{\Pi }}_i\left(x\right)\right]\right)w^{k\alpha }}{s^{1+k\alpha }}}+\sum _{k=\mathcal{K}+1}^{\mathcal{K}r+1}{\displaystyle \frac{\left({\mathsf{\Pi }}_k\left(x\right)-{{\mathcal{G}}^{*}}_x^{\left(\mathcal{K}k\right)}\left[{\mathsf{\Pi }}_j\left(x\right)\right]\right)w^{k\alpha }}{s^{1+k\alpha }}},$$

4.13

where $i\in \left\{0,1,2,\dots ,k-1\right\},j\in \left\{0,1,2,\dots ,\mathcal{K}\right\},$ and ${{\mathcal{G}}^{\mathrm{*}}}_x^{\left(\mathcal{K}k\right)},k=\mathcal{K}+1,\mathcal{K}+2,\dots ,\mathcal{K}r+1$ are differential operators related to $x$ and ${\mathsf{\Pi }}_k\left(x\right)-{{\mathcal{G}}^{\mathrm{*}}}_x^{\left(\mathcal{K}k\right)}\left[{\mathsf{\Pi }}_j\left(x\right)\right]\ne 0$. By multiplying both sides of (4.13) by $s^{1+k\alpha }$ and taking the limit as $s\to \mathrm{\infty }$, we obtain:

$$\begin{gathered} \underset{s\to \infty }{\lim }{s}^{1+k\alpha }\mathcal{N}Re{s}_k\left(s,w,x\right)=\left({\mathsf{\Pi }}_k\left(x\right)-{\mathcal{G}}_x^{\left(k\right)}\left[{\mathsf{\Pi }}_i\left(x\right)\right]\right)w^{k\alpha }=0, k=1,2,\dots ,i \\ \in \left\{0,1,2,\dots ,k-1\right\}. \end{gathered}$$

4.14

Step 4. Substitute the $\mathcal{K}$th truncated series ${\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x\right)$ in (4.6) into the $\mathcal{K}$th truncated NRF (4.8).

Step 5. To determine the coefficients ${\mathsf{\Pi }}_k\left(x\right), k=1,2,\dots ,\mathcal{K}$, we solve the system that was obtained using $\underset{s\to \infty }{\lim }{s}^{1+k\alpha }\mathcal{N}Re{s}_k\left(s,w,x\right)=0,k=1,2,\dots ,\mathcal{K}$ and collect the results into the $\mathcal{K}$th truncated series ${\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x\right)$ in (4.6).

Step 6. Apply the inverse natural operator for both sides of the inferred $\mathcal{K}$th truncated series ${\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x\right)$ to obtain the $\mathcal{K}$th approximate solution for the nonlinear fractional PDE (4.1). □

5. Applications

In this section, we investigate the time-fractional KdV-ZK Equations (1.1) and (1.2), the time-fractional spKdV Equations (1.3) and (1.4), and the fractional FKdV Equations (1.5) and (1.6) using the proposed NRPS method. We used the mathematical software package Mathematica 13 to implement all symbolic and numerical computations.

Application 1.

Consider the time-fractional KdV-ZK equation [17]:

$${\mathcal{D}}_t^{\alpha }\psi +36\psi {\displaystyle \frac{\partial \psi }{\partial x}}+{\displaystyle \frac{{\partial }^3\psi }{\partial x^3}}+{\displaystyle \frac{{\partial }^3\psi }{\partial x\partial y^2}}+{\displaystyle \frac{{\partial }^3\psi }{\partial x\partial z^2}}=0, t\ge 0, x,y,z\in \mathbb{R},$$

(5.1)

subject to the initial condition:

$$\psi \left(0,x,y,z\right)={\displaystyle \frac{1}{\sqrt{2}}}\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}\left(x+y+z\right),x,y,z\in \mathbb{R}.$$

(5.2)

Apply the natural transformation for both sides of (5.1) with the aid of (3.13) to obtain:

$$\begin{array}{c}{\displaystyle \frac{s^{\alpha }}{w^{\alpha }}}\mathsf{\Psi }\left(s,w,x,y,z\right)-{\displaystyle \frac{s^{\alpha -1}}{w^{\alpha }}}\psi \left(0,x,y,z\right)+36\mathcal{N}\left\{{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x,y,z\right)\right\}{\displaystyle \frac{\partial }{\partial x}}{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x,y,z\right)\right\}\right\}\\ \begin{array}{cc} & +\mathcal{N}\left\{{\displaystyle \frac{{\partial }^3}{\partial x^3}}{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x,y,z\right)\right\}\right\}+\mathcal{N}\left\{{\displaystyle \frac{{\partial }^3}{\partial x\partial y^2}}{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x,y,z\right)\right\}\right\}\hfill \\ & +\mathcal{N}\left\{{\displaystyle \frac{{\partial }^3}{\partial x\partial z^2}}{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x,y,z\right)\right\}\right\}=0,\hfill \end{array}\end{array}$$

(5.3)

where $\Psi \left(s,w,x,y,z\right)=\mathcal{N}\left\{\psi \left(t,x,y,z\right)\right\}$. Using (5.2) in (5.3), we obtain:

$$\begin{array}{cc}\hfill \mathsf{\Psi }\left(s,w,x,y,z\right)-& {\displaystyle \frac{1}{s\sqrt{2}}}\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}\left(x+y+z\right)\hfill \\ & +{\displaystyle \frac{w^{\alpha }}{s^{\alpha }}}(36\mathcal{N}\left\{{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x,y,z\right)\right\}{\displaystyle \frac{\partial }{\partial x}}{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x,y,z\right)\right\}\right\}\hfill \\ & +\mathcal{N}\left\{{\displaystyle \frac{{\partial }^3}{\partial x^3}}{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x,y,z\right)\right\}\right\}\hfill \\ & +\mathcal{N}\left\{{\displaystyle \frac{{\partial }^3}{\partial x\partial y^2}}{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x,y,z\right)\right\}\right\}\hfill \\ & +\mathcal{N}\left\{{\displaystyle \frac{{\partial }^3}{\partial x\partial z^2}}{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x,y,z\right)\right\}\right\})=0,\hfill \end{array}$$

(5.4)

Suppose that the solution of (5.4) can be written in the following expansion:

$$\mathsf{\Psi }\left(s,w,x,y,z\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x,y,z\right)w^{k\alpha }}{s^{1+k\alpha }}}, \alpha \in \left(0,1\right],x,y,z\in A, s\in \left(\theta ,\infty \right),w\in \left(0,\infty \right).$$

(5.5)

Define the $\mathcal{K}$th truncated series of $\Psi \left(s,w,x,y,z\right)$ as:

$${\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x,y,z\right)=\sum _{k=0}^{\mathcal{K}}{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x,y,z\right)w^{k\alpha }}{s^{1+k\alpha }}}, \alpha \in \left(0,1\right],x,y,z\in A, s\in \left(\theta ,\infty \right),w\in \left(0,\infty \right).$$

(5.6)

Using the results in Theorem 2, we deduce that the initial guess ${\Pi }_0\left(x,y,z\right)=\sqrt{{\displaystyle \frac{1}{2}}}sech\left(\left(x+y+z\right)\right)$. Upon this fact, we write the $\mathcal{K}$th truncated series of $\Psi \left(s,w,x,y,z\right)$ as:

$${\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x,y,z\right)={\displaystyle \frac{1}{s\sqrt{2}}}\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}\left(x+y+z\right)+\sum _{k=1}^{\mathcal{K}}{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x,y,z\right)w^{k\alpha }}{s^{1+k\alpha }}}.$$

(5.7)

We define the NRF as:

$$\begin{array}{cc}\hfill \mathcal{N}Res\left(s,w,x,y,z\right)& \\ & =\mathsf{\Psi }\left(s,w,x,y,z\right)-{\displaystyle \frac{1}{s\sqrt{2}}}\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}\left(x+y+z\right)\hfill \\ & +{\displaystyle \frac{w^{\alpha }}{s^{\alpha }}}(36\mathcal{N}\left\{{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x,y,z\right)\right\}{\displaystyle \frac{\partial }{\partial x}}{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x,y,z\right)\right\}\right\}\hfill \\ & +\mathcal{N}\left\{{\displaystyle \frac{{\partial }^3}{\partial x^3}}{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x,y,z\right)\right\}\right\}\hfill \\ & +\mathcal{N}\left\{{\displaystyle \frac{{\partial }^3}{\partial x\partial y^2}}{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x,y,z\right)\right\}\right\}\hfill \\ & +\mathcal{N}\left\{{\displaystyle \frac{{\partial }^3}{\partial x\partial z^2}}{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x,y,z\right)\right\}\right\}),\hfill \end{array}$$

(5.8)

and the $\mathcal{K}$th truncated NRF as:

$$\begin{array}{cc}\hfill \mathcal{N}Re{s}_{\mathcal{K}}\left(s,w,x,y,z\right)& \\ & ={\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x,y,z\right)-{\displaystyle \frac{1}{s\sqrt{2}}}\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}\left(x+y+z\right)\hfill \\ & +{\displaystyle \frac{w^{\alpha }}{s^{\alpha }}}(36\mathcal{N}\left\{{\mathcal{N}}^{-1}\left\{{\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x,y,z\right)\right\}{\displaystyle \frac{\partial }{\partial x}}{\mathcal{N}}^{-1}\left\{{\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x,y,z\right)\right\}\right\}\hfill \\ & +\mathcal{N}\left\{{\displaystyle \frac{{\partial }^3}{\partial x^3}}{\mathcal{N}}^{-1}\left\{{\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x,y,z\right)\right\}\right\}\hfill \\ & +\mathcal{N}\left\{{\displaystyle \frac{{\partial }^3}{\partial x\partial y^2}}{\mathcal{N}}^{-1}\left\{{\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x,y,z\right)\right\}\right\}\hfill \\ & +\mathcal{N}\left\{{\displaystyle \frac{{\partial }^3}{\partial x\partial z^2}}{\mathcal{N}}^{-1}\left\{{\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x,y,z\right)\right\}\right\}).\hfill \end{array}$$

(5.9)

For $\mathcal{K}=1$, we obtain the 1st NRF with using (5.7) as follows:

$$\begin{array}{cc}\hfill \mathcal{N}Re{s}_1\left(s,w,x,y,z\right)& \\ & ={\displaystyle \frac{{\mathsf{\Pi }}_1\left(x,y,z\right)w^{\alpha }}{s^{1+\alpha }}}\hfill \\ & +{\displaystyle \frac{w^{\alpha }}{s^{\alpha }}}\left(36\mathcal{N}\right\{{\mathcal{N}}^{-1}\left\{{\displaystyle \frac{1}{s\sqrt{2}}}\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}\left(x+y+z\right)+{\displaystyle \frac{{\mathsf{\Pi }}_1\left(x,y,z\right)w^{\alpha }}{s^{1+\alpha }}}\right\}{\displaystyle \frac{\partial }{\partial x}}{\mathcal{N}}^{-1}\{{\displaystyle \frac{1}{s\sqrt{2}}}\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}\left(x+y+z\right)\hfill \\ & +{\displaystyle \frac{{\mathsf{\Pi }}_1\left(x,y,z\right)w^{\alpha }}{s^{1+\alpha }}}\left\}\right\}+\mathcal{N}\left\{{\displaystyle \frac{{\partial }^3}{\partial x^3}}{\mathcal{N}}^{-1}\left\{{\displaystyle \frac{1}{s\sqrt{2}}}\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}\left(x+y+z\right)+{\displaystyle \frac{{\mathsf{\Pi }}_1\left(x,y,z\right)w^{\alpha }}{s^{1+\alpha }}}\right\}\right\}\hfill \\ & +\mathcal{N}\left\{{\displaystyle \frac{{\partial }^3}{\partial x\partial y^2}}{\mathcal{N}}^{-1}\left\{{\displaystyle \frac{1}{s\sqrt{2}}}\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}\left(x+y+z\right)+{\displaystyle \frac{{\mathsf{\Pi }}_1\left(x,y,z\right)w^{\alpha }}{s^{1+\alpha }}}\right\}\right\}\hfill \\ & +\mathcal{N}\left\{{\displaystyle \frac{{\partial }^3}{\partial x\partial z^2}}{\mathcal{N}}^{-1}\right\{{\displaystyle \frac{1}{s\sqrt{2}}}\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}\left(x+y+z\right)+{\displaystyle \frac{{\mathsf{\Pi }}_1\left(x,y,z\right)w^{\alpha }}{s^{1+\alpha }}}\left\}\right\}).\hfill \end{array}$$

(5.10)

By multiplying both sides of (5.10) by $s^{1+\alpha }$ and using $\underset{s\to \infty }{\mathit{lim}}{s}^{1+\alpha }\mathcal{N}Re{s}_1\left(s,w,x\right)=0$, we obtain:

$${\mathsf{\Pi }}_1\left(x,y,z\right)=3\sqrt{3}\mathrm{sech}\left(x+y+z\right)\tanh \left(x+y+z\right).$$

(5.11)

Similarly, we can obtain the following:

$${\mathsf{\Pi }}_2\left(x,y,z\right)={\displaystyle \frac{9}{2}}\sqrt{3}\left(-3+\cosh \left(2\left(x+y+z\right)\right)\right){\mathrm{sech}\left(x+y+z\right)}^3.$$

(5.12)

$${\mathsf{\Pi }}_3\left(x,y,z\right)={\displaystyle \frac{27}{4}}\sqrt{3}{\mathrm{sech}\left(x+y+z\right)}^4\left(-23\sinh \left(x+y+z\right)+\sinh \left(3\left(x+y+z\right)\right)\right).$$

(5.13)

$${\mathsf{\Pi }}_4\left(x,y,z\right)={\displaystyle \frac{81\sqrt{3}}{8}}\left(115-76\cosh \left(2\left(x+y+z\right)\right)+\cosh \left(4\left(x+y+z\right)\right)\right){\mathrm{sech}\left(x+y+z\right)}^5.$$

(5.14)

The CPU time that is needed to obtain these results (5.11)–(5.14) was $35.390625$ s. Substitute (5.11)–(5.14) into (5.7) to obtain the $\mathcal{K}$th truncated series of $\Psi \left(s,w,x\right)$ as:

$$\begin{gathered} {\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x,y,z\right)={\displaystyle \frac{1}{s\sqrt{2}}}\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}\left(x+y+z\right)+{\displaystyle \frac{{\mathsf{\Pi }}_1\left(x,y,z\right)w^{\alpha }}{s^{1+\alpha }}}+{\displaystyle \frac{{\mathsf{\Pi }}_2\left(x,y,z\right)w^{2\alpha }}{s^{1+2\alpha }}} \\ +{\displaystyle \frac{{\mathsf{\Pi }}_3\left(x,y,z\right)w^{3\alpha }}{s^{1+3\alpha }}}+{\displaystyle \frac{{\mathsf{\Pi }}_4\left(x,y,z\right)w^{4\alpha }}{s^{1+4\alpha }}}+\dots +{\displaystyle \frac{{\mathsf{\Pi }}_{\mathcal{K}}\left(x,y,z\right)w^{\mathcal{K}\alpha }}{s^{1+\mathcal{K}\alpha }}}. \end{gathered}$$

(5.15)

Apply the inverse natural operator of (5.15) to obtain the $\mathcal{K}$th NRPS solution for the time-fractional KdV-ZK equation as:

$$\begin{gathered} {\psi }_{\mathcal{K}}\left(t,x,y,z\right)={\displaystyle \frac{1}{\sqrt{2}}}\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}\left(x+y+z\right)+{\displaystyle \frac{{\mathsf{\Pi }}_1\left(x,y,z\right)t^{\alpha }}{\mathsf{\Gamma }\left(1+\alpha \right)}}+{\displaystyle \frac{{\mathsf{\Pi }}_2\left(x,y,z\right)t^{2\alpha }}{\mathsf{\Gamma }\left(1+2\alpha \right)}} \\ +{\displaystyle \frac{{\mathsf{\Pi }}_3\left(x,y,z\right)t^{3\alpha }}{\mathsf{\Gamma }\left(1+3\alpha \right)}}+{\displaystyle \frac{{\mathsf{\Pi }}_4\left(x,y,z\right)t^{4\alpha }}{\mathsf{\Gamma }\left(1+4\alpha \right)}}+\dots +{\displaystyle \frac{{\mathsf{\Pi }}_{\mathcal{K}}\left(x,y,z\right)t^{\mathcal{K}\alpha }}{\mathsf{\Gamma }\left(1+\mathcal{K}\alpha \right)}}. \end{gathered}$$

(5.16)

We present a numerical simulation of the results obtained. We assume that $y=z=0$.

Table 1. A comparison between the exact solution and the NRPS solution for the time-fractional KdV-ZK equation.

$$\mathit{x}$$	$$\mathit{t}$$	Exact Solution	NRPS Solution	Absolute Error	Relative Error	CPU Time
$-1$	$\begin{array}{c}0.0\\ 0.1\\ 0.2\\ 0.3\end{array}$	$\begin{array}{c}1.122462928047994\\ 0.878805774963547\\ 0.671997937091220\\ 0.506783751732631\end{array}$	$\begin{array}{c}1.122462928047995\\ 0.878736016940593\\ 0.669464247695219\\ 0.486544390701797\end{array}$	$\begin{array}{c}2.220446\times {10}^{-16}\\ 6.975802\times {10}^{-5}\\ 2.533689\times {10}^{-3}\\ 2.023936\times {10}^{-2}\end{array}$	$\begin{array}{c}1.978190\times {10}^{-16}\\ 7.937820\times {10}^{-5}\\ 3.770382\times {10}^{-3}\\ 3.993687\times {10}^{-2}\end{array}$	$0.015625$
$0$	$\begin{array}{c}0.0\\ 0.1\\ 0.2\\ 0.3\end{array}$	$\begin{array}{c}1.732050807568877\\ 1.656928147349753\\ 1.461072649720631\\ 1.208615771635065\end{array}$	$\begin{array}{c}1.732050807568877\\ 1.657031356966050\\ 1.467047034010839\\ 1.267319925263052\end{array}$	$\begin{array}{c}2.220446\times {10}^{-16}\\ 1.032096\times {10}^{-5}\\ 5.974384\times {10}^{-3}\\ 5.870415\times {10}^{-2}\end{array}$	$\begin{array}{c}1.281975\times {10}^{-16}\\ 6.228973\times {10}^{-5}\\ 4.089039\times {10}^{-3}\\ 4.857139\times {10}^{-2}\end{array}$	$0.015625$
$1$	$\begin{array}{c}0.0\\ 0.1\\ 0.2\\ 0.3\end{array}$	$\begin{array}{c}1.122462928047994\\ 1.379934335375192\\ 1.602159904080592\\ 1.723426491772224\end{array}$	$\begin{array}{c}1.122462928047995\\ 1.379959423861089\\ 1.601751160457130\\ 1.709575007146963\end{array}$	$\begin{array}{c}2.220446\times {10}^{-16}\\ 2.508848\times {10}^{-5}\\ 4.087436\times {10}^{-4}\\ 1.385148\times {10}^{-2}\end{array}$	$\begin{array}{c}1.978190\times {10}^{-16}\\ 1.818092\times {10}^{-5}\\ 2.551203\times {10}^{-4}\\ 8.037177\times {10}^{-3}\end{array}$	$0.015625$

compares the exact solution for the time-fractional KdV-ZK Equation (5.1) and the inferred fourth NRPS solution (5.16) where the absolute and relative errors are presented. Figure 1 presents the absolute error between the exact solution and the explored approximate fourth NRPS solution (5.16) for the time-fractional KdV-ZK Equation (5.1). To our knowledge, no works in the literature have been presented to deduce approximate solutions to Equation (5.1). Still, all the works presented are concerned with deriving exact solutions to the equation. From Table 1 and Figure 1, we can see the convergence of the numerical solutions derived through the proposed method with the exact solutions, which provides insight into the high accuracy of the proposed method. Figure 2 shows the surfaces of the exact and the fourth NRPS solutions (5.16) at $\alpha =1$. Figure 3 presents the fourth NRPS solution (5.16) surfaces at different fractional orders $\alpha =0.95$ and $\alpha =0.9$ to demonstrate the effect of the fractional derivative on the established surfaces. For more explanation, we depicted, in Figure 4, Figure 5 and Figure 6, the NRPS solutions explored in 2D plots at different fractional orders $\alpha$ where the time $t$ is considered at various values.

Applications 2.

Consider the following time-fractional spKdV equation:

$${\mathcal{D}}_t^{\alpha }\psi -8{\displaystyle \frac{\partial \psi }{\partial x}}+{\left({\displaystyle \frac{\partial \psi }{\partial x}}\right)}^2+2{\displaystyle \frac{{\partial }^3\psi }{\partial x^3}}=0, t\ge 0, x\in \mathbb{R},$$

(5.17)

subject to the initial condition:

$$\psi \left(x,0\right)=g\left(x\right)=1+12\tanh \left(x\right),x\in \mathbb{R}.$$

(5.18)

Apply the natural transformation for both sides of (5.17) with the aid of (3.13) to obtain:

$$\begin{gathered} {\displaystyle \frac{s^{\alpha }}{w^{\alpha }}}\mathsf{\Psi }\left(s,w,x\right)-{\displaystyle \frac{s^{\alpha -1}}{w^{\alpha }}}\psi \left(0,x\right)-8\mathcal{N}\left\{{\displaystyle \frac{\partial }{\partial x}}{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x\right)\right\}\right\}+\mathcal{N}\left\{{\left({\displaystyle \frac{\partial }{\partial x}}{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x\right)\right\}\right)}^2\right\} \\ +2\mathcal{N}\left\{{\displaystyle \frac{{\partial }^3}{\partial x^3}}{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x\right)\right\}\right\}=0, \end{gathered}$$

(5.19)

where $\Psi \left(s,w,x\right)=\mathcal{N}\left\{\psi \left(t,x\right)\right\}$. Using (5.18) in (5.19), we obtain:

$$\begin{array}{cc}\hfill \mathsf{\Psi }\left(s,w,x\right)-{\displaystyle \frac{1}{s}}(1& +12\tanh \left(x\right))\hfill \\ & +{\displaystyle \frac{w^{\alpha }}{s^{\alpha }}}(-8\mathcal{N}\left\{{\displaystyle \frac{\partial }{\partial x}}{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x\right)\right\}\right\}+\mathcal{N}\left\{{\left({\displaystyle \frac{\partial }{\partial x}}{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x\right)\right\}\right)}^2\right\}\hfill \\ & +2\mathcal{N}\left\{{\displaystyle \frac{{\partial }^3}{\partial x^3}}{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x\right)\right\}\right\})=0,\hfill \end{array}$$

(5.20)

Suppose that the solution of (5.20) can be written in the following expansion:

$$\mathsf{\Psi }\left(s,w,x\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x\right)w^{k\alpha }}{s^{1+k\alpha }}}, \alpha \in \left(0,1\right],x\in A, s\in \left(\theta ,\infty \right),w\in \left(0,\infty \right).$$

(5.21)

Define the $\mathcal{K}$th truncated series of $\Psi \left(s,w,x\right)$ as:

$${\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x\right)=\sum _{k=0}^{\mathcal{K}}{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x\right)w^{k\alpha }}{s^{1+k\alpha }}}, \alpha \in \left(0,1\right],x\in A, s\in \left(\theta ,\infty \right),w\in \left(0,\infty \right).$$

(5.22)

Using the results in Theorem 2, we deduce that the initial guess ${\Pi }_0\left(x\right)=1+12\mathit{tanh}\left(x\right)$. Based on this fact, we write the $\mathcal{K}$th truncated series of $\Psi \left(s,w,x\right)$ as:

$${\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x\right)={\displaystyle \frac{1}{s}}\left(1+12\tanh \left(x\right)\right)+\sum _{k=1}^{\mathcal{K}}{\displaystyle \frac{{\mathsf{\Pi }}_k\left(x\right)w^{k\alpha }}{s^{1+k\alpha }}}.$$

(5.23)

We define the NRF as:

$$\begin{array}{cc}\hfill \mathcal{N}Res\left(s,w,x\right)=& \mathsf{\Psi }\left(s,w,x\right)-{\displaystyle \frac{1}{s}}\left(1+12\tanh \left(x\right)\right)\hfill \\ & +{\displaystyle \frac{w^{\alpha }}{s^{\alpha }}}(-8\mathcal{N}\left\{{\displaystyle \frac{\partial }{\partial x}}{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x\right)\right\}\right\}+\mathcal{N}\left\{{\left({\displaystyle \frac{\partial }{\partial x}}{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x\right)\right\}\right)}^2\right\}\hfill \\ & +2\mathcal{N}\left\{{\displaystyle \frac{{\partial }^3}{\partial x^3}}{\mathcal{N}}^{-1}\left\{\mathsf{\Psi }\left(s,w,x\right)\right\}\right\},\hfill \end{array}$$

(5.24)

and the $\mathcal{K}$th truncated NRF as:

$$\begin{array}{cc}\hfill \mathcal{N}Re{s}_{\mathcal{K}}\left(s,w,x\right)& ={\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x\right)-{\displaystyle \frac{1}{s}}\left(1+12\tanh \left(x\right)\right)\hfill \\ & +{\displaystyle \frac{w^{\alpha }}{s^{\alpha }}}(-8\mathcal{N}\left\{{\displaystyle \frac{\partial }{\partial x}}{\mathcal{N}}^{-1}\left\{{\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x\right)\right\}\right\}+\mathcal{N}\left\{{\left({\displaystyle \frac{\partial }{\partial x}}{\mathcal{N}}^{-1}\left\{{\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x\right)\right\}\right)}^2\right\}\hfill \\ & +2\mathcal{N}\left\{{\displaystyle \frac{{\partial }^3}{\partial x^3}}{\mathcal{N}}^{-1}\left\{{\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x\right)\right\}\right\},\hfill \end{array}$$

(5.25)

Using the fact that $\underset{s\to \infty }{\mathit{lim}}{s}^{1+k\alpha }\mathcal{N}Re{s}_k\left(s,w,x\right)=0, k=1,2,\dots$, we obtain:

$${\mathsf{\Pi }}_1\left(x\right)=8g^{\prime }\left(x\right)-{\left(g^{\prime }\left(x\right)\right)}^2-2g^{\prime \prime \prime }\left(x\right).$$

(5.26)

$${\mathsf{\Pi }}_2\left(x\right)=8{\mathsf{\Pi }}_1^{\prime }\left(x\right)-2g^{\prime }\left(x\right){\mathsf{\Pi }}_1^{\prime }\left(x\right)-2{\mathsf{\Pi }}_1^{\prime \prime \prime }\left(x\right).$$

(5.27)

$${\mathsf{\Pi }}_3\left(x\right)={\displaystyle \frac{-\mathsf{\Gamma }\left(1+2\alpha \right){\left({\mathsf{\Pi }}_1^{\prime }\left(x\right)\right)}^2-{\left(\mathsf{\Gamma }\left(1+\alpha \right)\right)}^2\left(-8{\mathsf{\Pi }}_2^{\prime }\left(x\right)+2g^{\prime }\left(x\right){\mathsf{\Pi }}_2^{\prime }\left(x\right)+2{\mathsf{\Pi }}_2^{\prime \prime \prime }\left(x\right)\right)}{{\left(\mathsf{\Gamma }\left(1+\alpha \right)\right)}^2}}.$$

(5.28)

$${\mathsf{\Pi }}_4\left(x\right)={\displaystyle \frac{-2\mathsf{\Gamma }\left(1+3\alpha \right){\mathsf{\Pi }}_1^{\prime }\left(x\right){\mathsf{\Pi }}_2^{\prime }\left(x\right)+8\mathsf{\Gamma }\left(1+\alpha \right)\mathsf{\Gamma }\left(1+2\alpha \right)\left(-8{\mathsf{\Pi }}_3^{\prime }\left(x\right)+2g^{\prime }\left(x\right){\mathsf{\Pi }}_3^{\prime }\left(x\right)+2{\mathsf{\Pi }}_3^{\prime \prime \prime }\left(x\right)\right)}{\mathsf{\Gamma }\left(1+\alpha \right)\mathsf{\Gamma }\left(1+2\alpha \right)}}.$$

(5.29)

The CPU time needed to obtain these results (5.26)–(5.29) was $8.53125$ s. Substitute (5.26)–(5.29) into (5.23) to obtain the $\mathcal{K}$th truncated series of $\Psi \left(s,w,x\right)$ as:

$${\mathsf{\Psi }}_{\mathcal{K}}\left(s,w,x\right)={\displaystyle \frac{1}{s}}\left(1+12\tanh \left(x\right)\right)+{\displaystyle \frac{{\mathsf{\Pi }}_1\left(x\right)w^{\alpha }}{s^{1+\alpha }}}+{\displaystyle \frac{{\mathsf{\Pi }}_2\left(x\right)w^{2\alpha }}{s^{1+2\alpha }}}+{\displaystyle \frac{{\mathsf{\Pi }}_3\left(x\right)w^{3\alpha }}{s^{1+3\alpha }}}+{\displaystyle \frac{{\mathsf{\Pi }}_4\left(x\right)w^{4\alpha }}{s^{1+4\alpha }}}+\dots +{\displaystyle \frac{{\mathsf{\Pi }}_{\mathcal{K}}\left(x\right)w^{\mathcal{K}\alpha }}{s^{1+\mathcal{K}\alpha }}}.$$

(5.30)

Apply the inverse natural operator of (5.30) to obtain the $\mathcal{K}$th NRPS solution for the time-fractional spKdV equation as:

$${\psi }_{\mathcal{K}}\left(t,x,y,z\right)=1+12\tanh \left(x\right)+{\displaystyle \frac{{\mathsf{\Pi }}_1\left(x\right)t^{\alpha }}{\mathsf{\Gamma }\left(1+\alpha \right)}}+{\displaystyle \frac{{\mathsf{\Pi }}_2\left(x\right)t^{2\alpha }}{\mathsf{\Gamma }\left(1+2\alpha \right)}}+{\displaystyle \frac{{\mathsf{\Pi }}_3\left(x\right)t^{3\alpha }}{\mathsf{\Gamma }\left(1+3\alpha \right)}}+{\displaystyle \frac{{\mathsf{\Pi }}_4\left(x\right)t^{4\alpha }}{\mathsf{\Gamma }\left(1+4\alpha \right)}}+\dots +{\displaystyle \frac{{\mathsf{\Pi }}_{\mathcal{K}}\left(x\right)t^{\mathcal{K}\alpha }}{\mathsf{\Gamma }\left(1+\mathcal{K}\alpha \right)}}.$$

(5.31)

Table 2. Comparison between the exact solution and NRPS solution for the time-fractional spKdV equation.

$\mathit{x}$	$\mathit{t}$	Exact Solution	NRPS Solution	Absolute Error	Relative Error	CPU Time
$-5$	$\begin{array}{c}0.0\\ 0.1\\ 0.2\\ 0.3\\ 0.4\\ 0.5\end{array}$	$\begin{array}{c}-10.99891045115114\\ -10.99866923540665\\ -10.99837462040912\\ -10.99801478665156\\ -10.99757529934621\\ -10.99703853017633\end{array}$	$\begin{array}{c}-10.998910451151142\\ -10.998910451151142\\ -10.998910451151142\\ -10.998910451151142\\ -10.998910451151142\\ -10.998910451151142\end{array}$	$\begin{array}{c}0.0\\ 2.412157\times {10}^{-4}\\ 5.358307\times {10}^{-4}\\ 8.956644\times {10}^{-4}\\ 1.335151\times {10}^{-3}\\ 1.871920\times {10}^{-3}\end{array}$	$\begin{array}{c}0.0\\ 2.193135\times {10}^{-5}\\ 4.871908\times {10}^{-5}\\ 8.143874\times {10}^{-5}\\ 1.214041\times {10}^{-4}\\ 1.702204\times {10}^{-4}\end{array}$	$2.328125$
$5$	$\begin{array}{c}0.0\\ 0.1\\ 0.2\\ 0.3\\ 0.4\\ 0.5\end{array}$	$\begin{array}{c}12.99891045115114\\ 12.99910794550953\\ 12.99926964263438\\ 12.99940203066545\\ 12.99951042190528\\ 12.99959916587564\end{array}$	$\begin{array}{c}12.998910451151142\\ 12.998910451151142\\ 12.998910451151142\\ 12.998910451151142\\ 12.998910451151142\\ 12.998910451151142\end{array}$	$\begin{array}{c}0.0\\ 1.974943\times {10}^{-4}\\ 3.591914\times {10}^{-4}\\ 4.915795\times {10}^{-4}\\ 5.999707\times {10}^{-4}\\ 6.887147\times {10}^{-4}\end{array}$	$\begin{array}{c}0.0\\ 1.519291\times {10}^{-5}\\ 2.763166\times {10}^{-5}\\ 3.781554\times {10}^{-5}\\ 4.615333\times {10}^{-5}\\ 5.297968\times {10}^{-5}\end{array}$	$1.703125$

shows the absolute and relative errors for the constructed numerical results to demonstrate the accuracy of the NRPS method. Figure 7 presents the absolute error between the exact solution and the explored approximate 4th NRPS solution (5.31) for the time-fractional spKdV Equation (5.17). When comparing the solution explored using the proposed method and the exact solution and calculating the absolute error and relative error, it becomes clear that the proposed method achieves an accuracy within ${10}^{-4}$ in dealing with Equation (5.17), which is a suitable level for numerical modeling applications that require accurate and stable solutions. It is worth noting here that based on the literature review, it was found that there are no works that focused on extracting approximate solutions to Equation (5.17), which prompted us to compare the approximate solutions derived here with the exact solutions to the equation. This may be a strong motivation to work on presenting other works that focus on deducing approximate solutions to the equation using new techniques and comparing them with the numerical solutions presented in this work. Figure 8 depicts the exact and 4th NRPS solution surfaces in (5.31) for the time-fractional spKdV Equation (5.17) where we can note the harmony between them, which ensures the efficiency of the proposed method. In Figure 9, we present the 2D plots for the exact and 4th NRPS solutions for the time-fractional spKdV Equation (5.17) at $t=0.1$ and $t=1$. Note that the harmony between the depicted solutions reduces with time. This harmony can be increased for a longer time by increasing the terms of the NRPS solution (5.17). To illustrate the effect of the fractional order $\alpha$ on the behavior of the constructed approximate solution, we present Figure 10 which depicts the 4th NRPS solution (5.17) at different fractional orders $\alpha =0.95,0.75,0.5,0.25$.

Application 3.

Consider the time-fractional FKdV equation:

$${\displaystyle \frac{1}{\sqrt{9.8}}}{\mathcal{D}}_t^{\alpha }\psi +\left(-2-{\displaystyle \frac{3}{2}}\psi \right){\displaystyle \frac{\partial \psi }{\partial x}}-{\displaystyle \frac{1}{6}}{\displaystyle \frac{{\partial }^3\psi }{\partial x^3}}={\displaystyle \frac{-x}{40}}{e}^{{\displaystyle \frac{-x^2}{4}}}, t\ge 0, x\in \mathbb{R},$$

(5.32)

where $\varphi \left(x\right)=1+0.1e^{{\displaystyle \frac{-x^2}{4}}}$. We assume the initial condition as follows:

$$\psi \left(x,0\right)=h\left(x\right)={\displaystyle \frac{-2e^x}{{\left(1+e^x\right)}^2}},x\in \mathbb{R}.$$

(5.33)

By applying the same argument above for the NRPS method, we obtain the following results:

$${\mathsf{\Pi }}_1\left(x\right)={\displaystyle \frac{\sqrt{9.8}}{6}}\left(3{\varphi }^{\prime }\left(x\right)+6h^{\prime }\left(x\right)+6h^{\prime }\left(x\right)+9h\left(x\right)h^{\prime }\left(x\right)+h^{\prime \prime \prime }\left(x\right)\right).$$

(5.34)

$${\mathsf{\Pi }}_2\left(x\right)={\displaystyle \frac{\sqrt{9.8}}{6}}\left(6{\mathsf{\Pi }}_1^{\prime }\left(x\right)+6{\mathsf{\Pi }}_1^{\prime }\left(x\right)+9h\left(x\right){\mathsf{\Pi }}_1^{\prime }\left(x\right)+9{\mathsf{\Pi }}_1\left(x\right)h\prime \left(\mathrm{x}\right)+{\mathsf{\Pi }}_1^{\prime \prime \prime }\left(x\right)\right).$$

(5.35)

$$\begin{gathered} {\mathsf{\Pi }}_3\left(x\right)={\displaystyle \frac{\sqrt{9.8}}{{6\left(\mathsf{\Gamma }\left(1+\alpha \right)\right)}^2}}\left(9\mathsf{\Gamma }\left(1+2\alpha \right){\mathsf{\Pi }}_1\left(x\right){\mathsf{\Pi }}_1^{\prime }\left(x\right) \\ +{\left(\mathsf{\Gamma }\left(1+\alpha \right)\right)}^2\left(12{\mathsf{\Pi }}_2^{\prime }\left(x\right)+9h\left(x\right){\mathsf{\Pi }}_2^{\prime }\left(x\right)+9{\mathsf{\Pi }}_2\left(x\right)h^{\prime }\left(x\right)+{\mathsf{\Pi }}_2^{\prime \prime \prime }\left(x\right)\right)\right). \end{gathered}$$

(5.36)

$$\begin{gathered} {\mathsf{\Pi }}_4\left(x\right)={\displaystyle \frac{\sqrt{9.8}}{6\mathsf{\Gamma }\left(1+\alpha \right)\mathsf{\Gamma }\left(1+2\alpha \right)}}\left(9\mathsf{\Gamma }\left(1+3\alpha \right)\left({\mathsf{\Pi }}_2\left(x\right){\mathsf{\Pi }}_1^{\prime }\left(x\right)+{\mathsf{\Pi }}_1\left(x\right){\mathsf{\Pi }}_2^{\prime }\left(x\right)\right) \\ +\mathsf{\Gamma }\left(1+\alpha \right)\mathsf{\Gamma }\left(1+2\alpha \right)\left(12{\mathsf{\Pi }}_3^{\prime }\left(x\right)+9h\left(x\right){\mathsf{\Pi }}_3^{\prime }\left(x\right)+9h^{\prime }\left(x\right){\mathsf{\Pi }}_3\left(x\right)+{\mathsf{\Pi }}_3^{\prime \prime \prime }\left(x\right)\right)\right). \end{gathered}$$

(5.37)

The CPU time needed to obtain these results (5.34)–(5.37) was $16.59375$ s. Using these results, we obtain the $\mathcal{K}$th NRPS solution for the time-fractional FKdV equation as:

$${\psi }_{\mathcal{K}}\left(t,x,y,z\right)={\displaystyle \frac{-2e^x}{{\left(1+e^x\right)}^2}}+{\displaystyle \frac{{\mathsf{\Pi }}_1\left(x\right)t^{\alpha }}{\mathsf{\Gamma }\left(1+\alpha \right)}}+{\displaystyle \frac{{\mathsf{\Pi }}_2\left(x\right)t^{2\alpha }}{\mathsf{\Gamma }\left(1+2\alpha \right)}}+{\displaystyle \frac{{\mathsf{\Pi }}_3\left(x\right)t^{3\alpha }}{\mathsf{\Gamma }\left(1+3\alpha \right)}}+{\displaystyle \frac{{\mathsf{\Pi }}_4\left(x\right)t^{4\alpha }}{\mathsf{\Gamma }\left(1+4\alpha \right)}}+\dots +{\displaystyle \frac{{\mathsf{\Pi }}_{\mathcal{K}}\left(x\right)t^{\mathcal{K}\alpha }}{\mathsf{\Gamma }\left(1+\mathcal{K}\alpha \right)}}.$$

(5.38)

Table 3. A comparison between the exact solution and NRPS solution for the time-fractional FKdV equation.

$\mathit{x}$	$\mathit{t}$	Exact Solution	NRPS Solution	Absolute Error	Relative Error	CPU Time
$-0.05$	$\begin{array}{c}0.0\\ 0.01\\ 0.02\\ 0.03\\ 0.04\\ 0.05\end{array}$	$\begin{array}{c}-0.49968763016223\\ -0.49968763016223\\ -0.49968763016223\\ -0.49968763016223\\ -0.49968763016223\\ -0.49968763016223\end{array}$	$\begin{array}{c}-0.49968763016223\\ -0.49987856920723\\ -0.49981332801340\\ -0.49949449146807\\ -0.49892579802621\\ -0.49811213971041\end{array}$	$\begin{array}{c}0.0\\ 1.909390\times {10}^{-4}\\ 1.256978\times {10}^{-4}\\ 1.931386\times {10}^{-4}\\ 7.618321\times {10}^{-4}\\ 1.575490\times {10}^{-3}\end{array}$	$\begin{array}{c}0.0\\ 3.821168\times {10}^{-4}\\ 2.515528\times {10}^{-4}\\ 3.865188\times {10}^{-4}\\ 1.524616\times {10}^{-3}\\ 3.152950\times {10}^{-3}\end{array}$	$41.265625$
$0$	$\begin{array}{c}0.0\\ 0.01\\ 0.02\\ 0.03\\ 0.04\\ 0.05\end{array}$	$\begin{array}{c}-0.5\\ -0.5\\ -0.5\\ -0.5\\ -0.5\\ -0.5\end{array}$	$\begin{array}{c}-0.5\\ -0.49987108520902\\ -0.49948494667779\\ -0.49884340193134\\ -0.49794948017807\\ -0.49680742230976\end{array}$	$\begin{array}{c}0.0\\ 1.289147\times {10}^{-4}\\ 5.150533\times {10}^{-4}\\ 1.156598\times {10}^{-3}\\ 2.050519\times {10}^{-3}\\ 3.192577\times {10}^{-3}\end{array}$	$\begin{array}{c}0.0\\ 2.578295\times {10}^{-4}\\ 1.030106\times {10}^{-3}\\ 2.313196\times {10}^{-3}\\ 4.101039\times {10}^{-3}\\ 6.385155\times {10}^{-3}\end{array}$	$58.0625$
$0.05$	$\begin{array}{c}0.0\\ 0.01\\ 0.02\\ 0.03\\ 0.04\\ 0.05\end{array}$	$\begin{array}{c}-0.49968763016223\\ -0.49968763016223\\ -0.49968763016223\\ -0.49968763016223\\ -0.49968763016223\\ -0.49968763016223\end{array}$	$\begin{array}{c}-0.49968763016223\\ -0.49923907955869\\ -0.49853263964456\\ -0.49756918623542\\ -0.49635074871447\\ -0.49488051003255\end{array}$	$\begin{array}{c}0.0\\ 4.485506\times {10}^{-4}\\ 1.154990\times {10}^{-3}\\ 2.118443\times {10}^{-3}\\ 3.336881\times {10}^{-3}\\ 4.807120\times {10}^{-3}\end{array}$	$\begin{array}{c}0.0\\ 8.976620\times {10}^{-4}\\ 2.311425\times {10}^{-3}\\ 4.239536\times {10}^{-3}\\ 6.677934\times {10}^{-3}\\ 9.620250\times {10}^{-3}\end{array}$	$39.9375$

shows the numerical results at $x\in \left\{-0.05,0,0.05\right\}$ and $t\in \left\{0,0.01,0.02,0.03,0.04,0.05\right\}$ and the absolute and relative errors correspond to these numerical results which show convergences between the approximate and exact solutions for the time-fractional FKdV Equation (5.38). Figure 11 presents the absolute error between the exact solution and the explored approximate fourth NRPS solution (5.31) for the time-fractional FKdV Equation (5.32). It is clear from Table 3 that the absolute and relative errors were calculated at different values of the variable x and the variable t, and the accuracy of the proposed method was within ${10}^{-3}$, which is a suitable level to confirm that the proposed method in this work is effective and suitable for dealing with such issues in the context of numerical modeling. Moreover, the approximate solutions presented here are in agreement with those presented in Ref. [28]. It is worth noting here that the work in Ref. [28] did not present the absolute or relative error of the derived solutions and did not provide a reading for comparing the solutions with other approximate solutions of Equation (5.32). Figure 12 and Figure 13 present the surface of the fourth NRPS approximate solution and the exact solution to be able to observe the harmony between them. In Figure 9, we consider $\alpha =1$, while we consider different fractional orders $\alpha \in \left\{0.9,0.8\right\}$ in Figure 13. Figure 14 compares the exact solution and the fourth NRPS solution (5.38) at $t=0.05$ and $t=0.15$. The harmony between the depicted solutions is a little larger at $t=0.05$ than at $t=0.15$. To further illustrate the effect of the fractional derivative on the behavior of the solution derived using the proposed method for the time-fractional FKdV Equation (5.32), we present Figure 15, Figure 16 and Figure 17. Figure 15 was performed by considering $t\in \{0,0.1,0.2\}$ at $\alpha =0.9$ and $\alpha =0.75$, while we performed Figure 16 by considering $t\in \{0,0.15,0.3,0.5\}$ where $\alpha =0.95$ and $\alpha =0.75$. Figure 17 depicts the fourth NRPS solution at $t\in \{0.05,0.15,0.25,0.5\}$ where $\alpha \in \{0.95,0.8,0.7,0.6\}$.

From the introduced numerical and graphical results, we can deduce that the proposed NRPS method is efficient in dealing with the governing models by noticing the harmony and convergences between the approximate and exact solutions. We utilized four terms of the NRPS solutions to show their efficiency and we can achieve higher efficiency by obtaining further terms of the approximate solutions.

6. Conclusions

This work investigated three time-fractional models that arise in plasma physics: the mKdV-ZK equation, the spKdV equation, and the FKdV equation. The governing equations in this work are of considerable importance in plasma physics by modeling nonlinear ion acoustic waves which play a fundamental role in understanding wave dynamics in plasma. We considered the fractional derivative in the Caputo sense. In this work, we investigate a new work based on the new fractional expansion in the natural transformation space and the RPS method to construct analytical solutions for the governing models. The RPS method is one of the important methods in deriving approximate solutions for fractional models, but it may require complex calculations to deal with more complex fractional models. To address this point, the Laplace transform was integrated with the RPS method in the literature. In this work, we seek to compose the natural transformation with the RPS method, as the natural transformation is more general than the Laplace transform, which enables us to deal with a wider range of complex fractional systems. To achieve our goal, we investigated the theoretical analysis of the proposed method to reveal the applicability, efficiency, and effectiveness of this approach to dealing with the governing models. Clear steps were set for the proposed method to obtain analytical solutions that are consistent and close to the exact solutions. These steps were implemented for the three governing equations, and an analytical study of the derived solutions for each equation was presented to confirm that the proposed approach generates analytical solutions that converge quickly to exact solutions, which proves the effectiveness of the proposed method.