Investigation of the Time-Fractional Generalized Burgers–Fisher Equation via Novel Techniques

Abstract

Numerous applied mathematics and physical applications, such as the simulation of financial mathematics, gas dynamics, nonlinear phenomena in plasma physics, fluid mechanics, and ocean engineering, utilize the time-fractional generalized Burgers–Fisher equation (TF-GBFE). This equation describes the concept of dissipation and illustrates how reaction systems can be coordinated with advection. To examine and analyze the present evolution equation (TF-GBFE), the modified forms of the Adomian decomposition method (ADM) and homotopy perturbation method (HPM) with Yang transform are utilized. When the results are achieved, they are connected to exact solutions of the $\sigma =1$ order and even for different values of $\sigma$ to verify the technique’s validity. The results are represented as two- and three-dimensional graphs. Additionally, the study of the precise and suggested technique solutions shows that the suggested techniques are very accurate.

1. Introduction

Fractional order calculus is a very dated topic in mathematics. In 1695, fractional derivatives made their appearance, notably in the argument concerning a meaning extension. Compared to classical calculus, derivatives and integrals of any order provide more accurate representations of the phenomena that occur in the real world [1]. A significant amount of research on fractional calculus was published during the 20th century by numerous pioneers, including Miller and Ross [2], Caputo [3], Liao [4], Podlubny [5], and others.

Numerous related fields of science and engineering and many natural physical phenomena, such as nanotechnology [6], electrodynamics [7], and finance [8], are associated with problems involving fractional calculus applications. The nonlinear fractional partial differential equations (NFPDEs) analytical and numerical solutions are crucial for understanding the characteristics of nonlinear issues that occur in daily life [9,10,11]. The exact solution of the NFPDEs has been determined using a variety of mathematical techniques that have been developed and examined [12,13,14,15]. Examples include the Elzaki homotopy perturbation method for fractional-order regularized long-wave models [16], Yang transform decomposition method for Noyes–Field model for time-fractional Belousov–Zhabotinsky reaction [17] and time-fractional Fisher’s equation [18], natural transform decomposition method for fractional-order Kaup–Kupershmidt equation [19] and fractional modified Boussinesq and approximate long-wave equations [20], q-homotopy analysis transform method for fractional coupled Navier–Stokes equations [21], fractional reduced differential transformation method fractional Klein–Gordon equations [22] and multiterm time-fractional diffusion equations [23], variational iteration transform method for fractional-order nonlinear systems of third-order Korteweg–de Vries equation (KdV) and Burgers equations [24] and fractional-order Newell–Whitehead–Segel equations [25], the first-integral method to study the Burgers–KdV (KdVB) equation [26] and Sub-equation method for the conformable fractional generalized Kuramoto–Sivashinsky equation [27], and many more [28,29,30,31,32,33]. The fractional differential equation is a helpful tool for representing nonlinear events in scientific and engineering models. In applied mathematics and engineering, partial differential equations, particularly nonlinear ones, have been utilized to simulate a wide range of scientific phenomena. The NFPDEs allowed researchers to recognize and model a wide range of significant and real-world physical issues in parallel with their work in the physical sciences [34,35,36]. It has always been claimed how important it is to obtain approximations for them using either numerical or analytical methods. As a result, symmetry analysis is an excellent tool for studying partial differential equations, particularly when looking at equations derived from mathematical ideas related to accounting. Despite the idea that symmetry is the basis of nature, symmetry is absent from the majority of observations in the natural world. Including unexpected symmetry-breaking events is a complete way to mask symmetry. The two types of symmetry are finite and infinitesimal. Discrete and continuous finite symmetries come in two different varieties. Natural symmetries such as parity and temporal inversion are discrete transformations of space, whereas the other is continuous. Patterns have always captivated mathematicians. In the seventeenth century, the classification of spatial and planar patterns developed rapidly. Regrettably, achieving an accurate solution to fractional nonlinear differential equations has become more and more challenging.

Numerous integer-order differential equation phenomena are difficult to describe adequately. Therefore, fractional nonlinear differential equations have research value. Recently, the NFPDEs have attracted the attention of academics. As a result of its use in numerous fields of research and engineering, NFPDEs have become incredibly popular in recent years. To better understand the mechanism controlled by the NFPDEs, we conducted this study on the time-fractional generalized Burgers–Fisher equation (TF-GBFE). The reaction, advection, and dissipation mechanisms are all included in the nonlinear equation known as the TF-GBFE. Burgers and Fisher diffusion transfer features and reaction form features are used in this nonlinear equation.

A number of academics have considered investigating the TF-GBFE, a crucial fluid dynamic model, to help with the mathematical analysis of physical flows and the investigation of various numerical techniques. Due to the inclusion of reaction, convection, and diffusion mechanisms, the TF-GBFE is extremely nonlinear. Burgers–Fisher is the name of a model that combines the diffusion and convective qualities of Burgers’ equation with the diffusion and reaction aspects of Fisher’s equation.

The TF-GBFE, which simulates the ocean’s far field of wave propagation, is a key nonlinear diffusion equation in ocean engineering. Strong turbulent diffusion is the cause of the traveling waves as a result of the interactions between nonlinear radiation, turbulent diffusion, and convection on the irregularity of the sea surface temperature (SST). According to [37], wind-induced currents affect the speed and direction of traveling wavefronts. The convection-diffusion model is driven by the TF-GBFE, which may describe underwater landslides and, as a result, simulate the most dangerous tsunamis in the coastal regions [38].

The world has witnessed an influx of crucial information, for example, tsunami wave fields brought on by the earthquakes in Japan in 2011. These incredible capabilities of ocean waves can be utilized to provide a variety of energy sources for a variety of uses. We must look at the mathematical assumptions of such natural problems to reduce the immense force of such massive disasters or use them as useful energy sources. If we approach these problems in a different way, we might select the best method for analyzing such possible crises and implement the appropriate safety measures. It is possible to broaden the current study to incorporate oceanic modeling, nonlinear mechanical models, and thermal systems [39,40].

The fractional order $\sigma$ and any real constants $a,b$, and $\delta$ are used to express the TF-GBFE as follows:

$$D_{\varsigma }^{\sigma }\mathbb{U}(\mu ,\varsigma )-{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+a{\mathbb{U}}^{\delta }(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )+b\mathbb{U}(\mu ,\varsigma )({\mathbb{U}}^{\delta }(\mu ,\varsigma )-1)=0,$$

(1)

where $0<\sigma \le 1,0\le \mu \le 1,\varsigma \ge 0$ and $a,b,\delta$ are non-zero parameters, concerning initial source as:

$$\mathbb{U}(\mu ,0)={\left[\frac{1}{2}\left(1-tanh\left(\frac{a\delta }{2(1+\delta )}\mu \right)\right)\right]}^{\frac{1}{\delta }}.$$

(2)

having an exact solution:

$$\mathbb{U}(\mu ,\varsigma )={\left[\frac{1}{2}\left(1-tanh\left(\frac{a\delta }{2(1+\delta )}\left(\mu -\left(\frac{a^2+b{(1+\delta )}^2}{a(1+\delta )}\right)\varsigma \right)\right)\right)\right]}^{\frac{1}{\delta }}.$$

(3)

This study attempts to solve the TF-GBFE mentioned above using the Yang transform decomposition technique (YTDM) and homotopy perturbation transform method (HPTM). It uses 2D–3D plots to represent the results of the precise solution. The solutions are obtained according to the orders $\sigma =0.40,\sigma =0.60,\sigma =0.80$, and $\sigma =1$ at various $\mu$ and time values $\varsigma$. The errors between the exact and suggested techniques solutions for $\sigma =1$ are also compared to assess the effectiveness of the proposed methods, serving as a standard for the solution’s precision and stability.

This paper is organized as follows: In Section 2, this study aims to provide detailed definitions and properties of fractional calculus. The suggested techniques are presented in Section 3 and Section 4, and Section 5 explains how to apply these approaches to resolve various cases. We finish the paper with the main conclusion in Section 6.

2. Preliminaries

Some of the basic ideas and characteristics of the fractional calculus theory are explained in this section.

Definition 1.

In Caputo terms, the fractional derivative is as follows [41,42]:

$$D_{\varsigma }^{\sigma }\mathbb{U}(\mu ,\varsigma )=\frac{1}{\Gamma (k-\sigma )}{\int }_0^{\varsigma }{(\varsigma -\sigma )}^{k-\sigma -1}{\mathbb{U}}^{\left(k\right)}(\mu ,\sigma )d\sigma ,$$

(4)

where $k-1<\sigma \le k$ for $k\in \mathbb{N}.$

Definition 2.

The Yang transform is represented as [42]

$$Y\left\{\mathbb{U}\left(\varsigma \right)\right\}=M\left(u\right)={\int }_0^{\infty }{e}^{\frac{-\varsigma }{u}}\mathbb{U}\left(\varsigma \right)d\varsigma ,\varsigma >0,u\in (-{\varsigma }_1,{\varsigma }_2)$$

(5)

and the Yang inverse transform is given as

$$Y^{-1}\left\{M\left(u\right)\right\}=\mathbb{U}\left(\varsigma \right).$$

(6)

Definition 3.

The Yang transform of the n-th derivative is represented as [42]

$$Y\left\{{\mathbb{U}}^n\left(\varsigma \right)\right\}=\frac{M\left(u\right)}{u^n}-\sum _{k=0}^{n-1}\frac{{\mathbb{U}}^k\left(0\right)}{u^{n-k-1}},n=1,2,3,\dots .$$

(7)

Definition 4.

The Yang transform in terms of the fractional-order derivative is represented as [42]

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

(8)

3. Fundamental Idea of HPTM

To present the fundamental idea of HPTM, we consider a general nonlinear fractional partial differential equation of the form:

$$D_{\varsigma }^{\sigma }\mathbb{U}(\mu ,\varsigma )={\mathcal{P}}_1\left[\mu \right]\mathbb{U}(\mu ,\varsigma )+{\mathcal{Q}}_1\left[\mu \right]\mathbb{U}(\mu ,\varsigma ),0<\sigma \le 1$$

(9)

concerning initial conditions

$$\mathbb{U}(\mu ,0)=\xi \left(\mu \right).$$

where $D_{\varsigma }^{\sigma }=\frac{{\partial }^{\sigma }}{\partial {\varsigma }^{\sigma }}$ is the Caputo operator and ${\mathcal{P}}_1\left[\mu \right],{\mathcal{Q}}_1\left[\mu \right]$ are linear and nonlinear terms.

On taking the Yang transform, we obtain

$$Y\left\{D_{\varsigma }^{\sigma }\mathbb{U}(\mu ,\varsigma )\right\}=Y\left\{{\mathcal{P}}_1\left[\mu \right]\mathbb{U}(\mu ,\varsigma )+{\mathcal{Q}}_1\left[\mu \right]\mathbb{U}(\mu ,\varsigma )\right\},$$

(10)

that is

$$\frac{1}{u^{\sigma }}\{M\left(u\right)-u\mathbb{U}\left(0\right)\}=Y\left\{{\mathcal{P}}_1\left[\mu \right]\mathbb{U}(\mu ,\varsigma )+{\mathcal{Q}}_1\left[\mu \right]\mathbb{U}(\mu ,\varsigma )\right\}.$$

(11)

On simplification, we obtain

$$M\left(u\right)=u\mathbb{U}\left(0\right)+u^{\sigma }Y\left\{{\mathcal{P}}_1\left[\mu \right]\mathbb{U}(\mu ,\varsigma )+{\mathcal{Q}}_1\left[\mu \right]\mathbb{U}(\mu ,\varsigma )\right\}.$$

(12)

By employing the inverse Yang transform, we obtain

$$\mathbb{U}(\mu ,\varsigma )=\mathbb{U}\left(0\right)+Y^{-1}\left\{u^{\sigma }Y[{\mathcal{P}}_1\left[\mu \right]\mathbb{U}(\mu ,\varsigma )+{\mathcal{Q}}_1\left[\mu \right]\mathbb{U}(\mu ,\varsigma )]\right\}.$$

(13)

Thus, by HPM,

$$\mathbb{U}(\mu ,\varsigma )=\sum _{k=0}^{\infty }{ϵ}^k{\mathbb{U}}_k(\mu ,\varsigma )$$

(14)

with parameter $ϵ\in [0,1]$.

The nonlinear term is determined as

$${\mathcal{Q}}_1\left[\mu \right]\mathbb{U}(\mu ,\varsigma )=\sum _{k=0}^{\infty }{ϵ}^k{H}_n\left(\mathbb{U}\right).$$

(15)

In addition, He’s polynomials $H_k\left(\mathbb{U}\right)$ are as

$$H_n({\mathbb{U}}_0,{\mathbb{U}}_1,\dots ,{\mathbb{U}}_n)=\frac{1}{\Gamma (n+1)}{D}_{ϵ}^k{\left[{\mathcal{Q}}_1\left(\sum _{k=0}^{\infty }{ϵ}^i{\mathbb{U}}_i\right)\right]}_{ϵ=0}$$

(16)

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

On substituting (14) and (15) in (12), we obtain

$$\sum _{k=0}^{\infty }{ϵ}^k{\mathbb{U}}_k(\mu ,\varsigma )=\mathbb{U}\left(0\right)+ϵY^{-1}\left\{u^{\sigma }Y\left\{{\mathcal{P}}_1\sum _{k=0}^{\infty }{ϵ}^k{\mathbb{U}}_k(\mu ,\varsigma )+\sum _{k=0}^{\infty }{ϵ}^k{H}_k\left(\mathbb{U}\right)\right\}\right\}.$$

(17)

On comparing of $ϵ$ coefficients, we obtain

$$\begin{array}{cc}\hfill {ϵ}^0:{\mathbb{U}}_0(\mu ,\varsigma )& =\mathbb{U}\left(0\right),\hfill \\ \hfill {ϵ}^1:{\mathbb{U}}_1(\mu ,\varsigma )& =Y^{-1}\left\{u^{\sigma }Y\left\{{\mathcal{P}}_1\left[\mu \right]{\mathbb{U}}_0(\mu ,\varsigma )+H_0\left(\mathbb{U}\right)\right\}\right\},\hfill \\ \hfill {ϵ}^2:{\mathbb{U}}_2(\mu ,\varsigma )& =Y^{-1}\left\{u^{\sigma }Y\left\{{\mathcal{P}}_1\left[\mu \right]{\mathbb{U}}_1(\mu ,\varsigma )+H_1\left(\mathbb{U}\right)\right\}\right\},\hfill \\ \hfill & ⋮\hfill \\ \hfill {ϵ}^k:{\mathbb{U}}_k(\mu ,\varsigma )& =Y^{-1}\left\{u^{\sigma }Y\left\{{\mathcal{P}}_1\left[\mu \right]{\mathbb{U}}_{k-1}(\mu ,\varsigma )+H_{k-1}\left(\mathbb{U}\right)\right\}\right\}\hfill \end{array}$$

(18)

for $k\in \mathbb{N}.$ Thus, the ${\mathbb{U}}_k(\mu ,\varsigma )$ solution is determined as

$$\mathbb{U}(\mu ,\varsigma )=\underset{M\to \infty }{lim}\sum _{k=1}^M{\mathbb{U}}_k(\mu ,\varsigma ).$$

(19)

4. Fundamental Idea of YTDM

To present the fundamental idea of YTDM, we consider a general nonlinear fractional partial differential equation of the form:

$$D_{\varsigma }^{\sigma }\mathbb{U}(\mu ,\varsigma )={\mathcal{P}}_1(\mu ,\varsigma )+{\mathcal{Q}}_1(\mu ,\varsigma ),0<\sigma \le 1$$

(20)

concerning initial conditions

$$\mathbb{U}(\mu ,0)=\xi \left(\mu \right),$$

where $D_{\varsigma }^{\sigma }=\frac{{\partial }^{\sigma }}{\partial {\varsigma }^{\sigma }}$ is the Caputo operator and ${\mathcal{P}}_1,{\mathcal{Q}}_1$ are linear and nonlinear terms.

On taking the Yang transform, we obtain

$$Y\left\{D_{\varsigma }^{\sigma }\mathbb{U}(\mu ,\varsigma )\right\}=Y\left\{{\mathcal{P}}_1(\mu ,\varsigma )+{\mathcal{Q}}_1(\mu ,\varsigma )\right\},$$

(21)

that is

$$\frac{1}{u^{\sigma }}\{M\left(u\right)-u\mathbb{U}\left(0\right)\}=Y\left\{{\mathcal{P}}_1(\mu ,\varsigma )+{\mathcal{Q}}_1(\mu ,\varsigma )\right\}.$$

(22)

For simplification, we obtain

$$M\left(u\right)=u\mathbb{U}\left(0\right)+u^{\sigma }Y\left\{{\mathcal{P}}_1(\mu ,\varsigma )+{\mathcal{Q}}_1(\mu ,\varsigma )\right\}.$$

(23)

By employing the inverse Yang transform, we have

$$\mathbb{U}(\mu ,\varsigma )=\mathbb{U}\left(0\right)+Y^{-1}\left\{u^{\sigma }Y\left\{{\mathcal{P}}_1(\mu ,\varsigma )+{\mathcal{Q}}_1(\mu ,\varsigma )\right\}\right\}.$$

(24)

Thus, by YTDM,

$$\mathbb{U}(\mu ,\varsigma )=\sum _{m=0}^{\infty }{\mathbb{U}}_m(\mu ,\varsigma ).$$

(25)

The nonlinear term is determined as

$${\mathcal{Q}}_1(\mu ,\varsigma )=\sum _{m=0}^{\infty }{\mathcal{A}}_m$$

(26)

having

$${\mathcal{A}}_m=\frac{1}{m!}{\left[\frac{{\partial }^m}{\partial {\ell }^m}\left\{{\mathcal{Q}}_1\left(\sum _{k=0}^{\infty }{\ell }^k{\mu }_k,\sum _{k=0}^{\infty }{\ell }^k{\varsigma }_k\right)\right\}\right]}_{\ell =0}.$$

(27)

On substituting (25) and (27) into (24), we obtain

$$\sum _{m=0}^{\infty }{\mathbb{U}}_m(\mu ,\varsigma )=\mathbb{U}\left(0\right)+Y^{-1}\left\{u^{\sigma }Y\left\{{\mathcal{P}}_1\left(\sum _{m=0}^{\infty }{\mu }_m,\sum _{m=0}^{\infty }{\varsigma }_m\right)+\sum _{m=0}^{\infty }{\mathcal{A}}_m\right\}\right\}.$$

(28)

Hence, we have

$$\begin{array}{cc}\hfill {\mathbb{U}}_0(\mu ,\varsigma )& =\mathbb{U}\left(0\right),\hfill \\ \hfill {\mathbb{U}}_1(\mu ,\varsigma )& =Y^{-1}\left\{u^{\sigma }{Y}^{+}\{{\mathcal{P}}_1({\mu }_0,{\varsigma }_0)+{\mathcal{A}}_0\}\right\}.\hfill \end{array}$$

(29)

Thus, in general, for $m\ge 1$, we can obtain

$${\mathbb{U}}_{m+1}(\mu ,\varsigma )=Y^{-1}\left\{u^{\sigma }{Y}^{+}\{{\mathcal{P}}_1({\mu }_m,{\varsigma }_m)+{\mathcal{A}}_m\}\right\}.$$

5. Numerical Examples

Example 1.

Let us assume $a=-1$ and $b=\delta =1$ in (1), we have

$$D_{\varsigma }^{\sigma }\mathbb{U}(\mu ,\varsigma )-{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )-\mathbb{U}(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )+\mathbb{U}(\mu ,\varsigma )(\mathbb{U}(\mu ,\varsigma )-1)=0,0<\sigma \le 1$$

(30)

concerning initial condition

$$\mathbb{U}(\mu ,0)=\frac{1}{2}\left(1+tanh\left(\frac{\mu }{4}\right)\right).$$

On taking the Yang transform, we have

$$Y\left\{D_{\varsigma }^{\sigma }\mathbb{U}(\mu ,\varsigma )\right\}=Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+\mathbb{U}(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-\mathbb{U}(\mu ,\varsigma )(\mathbb{U}(\mu ,\varsigma )-1)\right\}.$$

(31)

For simplification, we obtain

$$\frac{1}{u^{\sigma }}\{M\left(u\right)-u\mathbb{U}\left(0\right)\}=Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+\mathbb{U}(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-\mathbb{U}(\mu ,\varsigma )(\mathbb{U}(\mu ,\varsigma )-1)\right\},$$

(32)

that is

$$M\left(u\right)=u\mathbb{U}\left(0\right)+u^{\sigma }Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+\mathbb{U}(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-\mathbb{U}(\mu ,\varsigma )(\mathbb{U}(\mu ,\varsigma )-1)\right\}.$$

(33)

By employing the inverse Yang transform, we obtain

$$\mathbb{U}(\mu ,\varsigma )=\mathbb{U}\left(0\right)+Y^{-1}\left\{u^{\sigma }Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+\mathbb{U}(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-\mathbb{U}(\mu ,\varsigma )(\mathbb{U}(\mu ,\varsigma )-1)\right\}\right\},$$

that is

$$\begin{array}{cc}\hfill \mathbb{U}(\mu ,\varsigma )& =\frac{1}{2}\left(1+tanh\left(\frac{\mu }{4}\right)\right)\hfill \\ \hfill & +Y^{-1}\left\{u^{\sigma }Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+\mathbb{U}(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-\mathbb{U}(\mu ,\varsigma )(\mathbb{U}(\mu ,\varsigma )-1)\right\}\right\}.\hfill \end{array}$$

(34)

Thus, by HPM, we have

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{ϵ}^k{\mathbb{U}}_k(\mu ,\varsigma )& =\frac{1}{2}\left(1+tanh\left(\frac{\mu }{4}\right)\right)\hfill \\ \hfill & +ϵY^{-1}\left\{u^{\sigma }Y\left\{{\left(\sum _{k=0}^{\infty }{ϵ}^k{\mathbb{U}}_k(\mu ,\varsigma )\right)}_{\mu \mu }+\left(\sum _{k=0}^{\infty }{ϵ}^k{H}_k\left(\mathbb{U}\right)\right)+\left(\sum _{k=0}^{\infty }{ϵ}^k{\mathbb{U}}_k(\mu ,\varsigma )\right)\right\}\right\}.\hfill \end{array}$$

(35)

Using the proposed algorithm, He’s polynomial $H_k\left(\mathbb{U}\right)$ can be expressed in nonlinear terms as

$$\sum _{k=0}^{\infty }{ϵ}^k{H}_k\left(\mathbb{U}\right)=\mathbb{U}(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-{\mathbb{U}}^2(\mu ,\varsigma ).$$

(36)

Some of He’s polynomials are determined as:

$$\begin{array}{cc}\hfill H_0\left(\mathbb{U}\right)& ={\mathbb{U}}_0{\left({\mathbb{U}}_0\right)}_{\mu }-{\mathbb{U}}_0^2,\hfill \\ \hfill H_1\left(\mathbb{U}\right)& ={\mathbb{U}}_0{\left({\mathbb{U}}_1\right)}_{\mu }+{\mathbb{U}}_1{\left({\mathbb{U}}_0\right)}_{\mu }-2{\mathbb{U}}_0{\mathbb{U}}_1,\hfill \\ \hfill H_2\left(\mathbb{U}\right)& ={\mathbb{U}}_0{\left({\mathbb{U}}_2\right)}_{\mu }+{\mathbb{U}}_1{\left({\mathbb{U}}_1\right)}_{\mu }+{\mathbb{U}}_2{\left({\mathbb{U}}_0\right)}_{\mu }-2{\mathbb{U}}_0{\mathbb{U}}_2+{\left({\mathbb{U}}_1\right)}^2.\hfill \end{array}$$

On comparing ϵ coefficients, we obtain

$$\begin{array}{cc}\hfill {ϵ}^0:{\mathbb{U}}_0(\mu ,\varsigma )& =\frac{1}{2}\left(1+tanh\left(\frac{\mu }{4}\right)\right),\hfill \\ \hfill {ϵ}^1:{\mathbb{U}}_1(\mu ,\varsigma )& =Y^{-1}\left\{u^{\sigma }Y\left\{{\left({\mathbb{U}}_0\right)}_{\mu \mu }-H_0\left(\mathbb{U}\right)+{\mathbb{U}}_0\right\}\right\}=\frac{5{\varsigma }^{\sigma }}{16cosh{\left(\frac{1}{4}\mu \right)}^2},\hfill \\ \hfill {ϵ}^2:{\mathbb{U}}_2(\mu ,\varsigma )& =Y^{-1}\left\{u^{\sigma }Y\left\{{\left({\mathbb{U}}_1\right)}_{\mu \mu }-H_1\left(\mathbb{U}\right)+{\mathbb{U}}_1\right\}\right\}=-\frac{25sinh\left(\frac{\mu }{4}\right){\varsigma }^{2\sigma }}{128cosh{\left(\frac{\mu }{4}\right)}^3},\hfill \\ \hfill {ϵ}^3:{\mathbb{U}}_3(\mu ,\varsigma )& =Y^{-1}\left\{u^{\sigma }Y\left\{{\left({\mathbb{U}}_2\right)}_{\mu \mu }-H_2\left(\mathbb{U}\right)+{\mathbb{U}}_2\right\}\right\}=\frac{125\left(2cosh{\left(\frac{\mu }{4}\right)}^2-3\right){\varsigma }^{3\sigma }}{3072cosh{\left(\frac{\mu }{4}\right)}^4},\hfill \\ \hfill & ⋮.\hfill \end{array}$$

Thus, the solution is determined as

$$\begin{array}{cc}\hfill \mathbb{U}(\mu ,\varsigma )& ={\mathbb{U}}_0(\mu ,\varsigma )+{\mathbb{U}}_1(\mu ,\varsigma )+{\mathbb{U}}_2(\mu ,\varsigma )+{\mathbb{U}}_3(\mu ,\varsigma )+\cdots \hfill \\ \hfill & =\frac{1}{2}\left(1+tanh\left(\frac{\mu }{4}\right)\right)\hfill \\ \hfill & +\frac{5{\varsigma }^{\sigma }}{16cosh{\left(\frac{1}{4}\mu \right)}^2}-\frac{25sinh\left(\frac{\mu }{4}\right){\varsigma }^{2\sigma }}{128cosh{\left(\frac{\mu }{4}\right)}^3}+\frac{125\left(2cosh{\left(\frac{\mu }{4}\right)}^2-3\right){\varsigma }^{3\sigma }}{3072cosh{\left(\frac{\mu }{4}\right)}^4}+\cdots .\hfill \end{array}$$

By implementing

 YTDM

On taking

the Yang transform, we obtain

$$Y\left\{\frac{{\partial }^{\sigma }\mathbb{U}}{\partial {\varsigma }^{\sigma }}\right\}=Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+\mathbb{U}(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-\mathbb{U}(\mu ,\varsigma )(\mathbb{U}(\mu ,\varsigma )-1)\right\}.$$

(37)

For simplification, we obtain

$$\frac{1}{u^{\sigma }}\{M\left(u\right)-u\mathbb{U}\left(0\right)\}=Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+\mathbb{U}(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-\mathbb{U}(\mu ,\varsigma )(\mathbb{U}(\mu ,\varsigma )-1)\right\},$$

(38)

that is

$$M\left(u\right)=u\mathbb{U}\left(0\right)+u^{\sigma }Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+\mathbb{U}(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-\mathbb{U}(\mu ,\varsigma )(\mathbb{U}(\mu ,\varsigma )-1)\right\}.$$

(39)

By employing the inverse Yang transform, we obtain

$$\begin{array}{cc}\hfill \mathbb{U}(\mu ,\varsigma )& =\mathbb{U}\left(0\right)+Y^{-1}\left\{u^{\sigma }Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+\mathbb{U}(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-\mathbb{U}(\mu ,\varsigma )(\mathbb{U}(\mu ,\varsigma )-1)\right\}\right\}\hfill \\ \hfill & =\frac{1}{2}\left(1+tanh\left(\frac{\mu }{4}\right)\right)\hfill \\ \hfill & +Y^{-1}\left\{u^{\sigma }Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+\mathbb{U}(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-\mathbb{U}(\mu ,\varsigma )(\mathbb{U}(\mu ,\varsigma )-1)\right\}\right\}.\hfill \end{array}$$

(40)

The solution in terms of series is stated as:

$$\mathbb{U}(\mu ,\varsigma )=\sum _{m=0}^{\infty }{\mathbb{U}}_m(\mu ,\varsigma ).$$

(41)

The nonlinear terms are determined as:

$$\mathbb{U}(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-{\mathbb{U}}^2(\mu ,\varsigma )=\sum _{m=0}^{\infty }{\mathcal{A}}_m.$$

Thus, we have

$$\begin{array}{cc}\hfill \sum _{m=0}^{\infty }{\mathbb{U}}_m(\mu ,\varsigma )& =\mathbb{U}(\mu ,0)+Y^{-1}\left\{u^{\sigma }Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+\sum _{m=0}^{\infty }{\mathcal{A}}_m+\mathbb{U}(\mu ,\varsigma )\right\}\right\}\hfill \\ \hfill & =\frac{1}{2}\left(1+tanh\left(\frac{\mu }{4}\right)\right)+Y^{-1}\left\{u^{\sigma }Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+\sum _{m=0}^{\infty }{\mathcal{A}}_m+\mathbb{U}(\mu ,\varsigma )\right\}\right\}.\hfill \end{array}$$

(42)

Few nonlinear terms are calculated as

$$\begin{array}{cc}\hfill {\mathcal{A}}_0& ={\mathbb{U}}_0{\left({\mathbb{U}}_0\right)}_{\mu }-{\mathbb{U}}_0^2,\hfill \\ \hfill {\mathcal{A}}_1& ={\mathbb{U}}_0{\left({\mathbb{U}}_1\right)}_{\mu }+{\mathbb{U}}_1{\left({\mathbb{U}}_0\right)}_{\mu }-2{\mathbb{U}}_0{\mathbb{U}}_1,\hfill \\ \hfill {\mathcal{A}}_2& ={\mathbb{U}}_0{\left({\mathbb{U}}_2\right)}_{\mu }+{\mathbb{U}}_1{\left({\mathbb{U}}_1\right)}_{\mu }+{\mathbb{U}}_2{\left({\mathbb{U}}_0\right)}_{\mu }-2{\mathbb{U}}_0{\mathbb{U}}_2+{\left({\mathbb{U}}_1\right)}^2.\hfill \end{array}$$

Now, by comparing both sides, we obtain

$${\mathbb{U}}_0(\mu ,\varsigma )=\frac{1}{2}\left(1+tanh\left(\frac{\mu }{4}\right)\right).$$

For $m=0$, we have

$${\mathbb{U}}_1(\mu ,\varsigma )=\frac{5{\varsigma }^{\sigma }}{16cosh{\left(\frac{1}{4}\mu \right)}^2}.$$

For $m=1$, we have

$${\mathbb{U}}_2(\mu ,\varsigma )=-\frac{25sinh\left(\frac{\mu }{4}\right){\varsigma }^{2\sigma }}{128cosh{\left(\frac{\mu }{4}\right)}^3}.$$

For $m=2$, we have

$${\mathbb{U}}_3(\mu ,\varsigma )=\frac{125\left(2cosh{\left(\frac{\mu }{4}\right)}^2-3\right){\varsigma }^{3\sigma }}{3072cosh{\left(\frac{\mu }{4}\right)}^4}.$$

Therefore, the other terms for $m\ge 3$ are easy to obtain

$$\begin{array}{cc}\hfill \mathbb{U}(\mu ,\varsigma )& =\sum _{m=0}^{\infty }{\mathbb{U}}_m(\mu ,\varsigma )\hfill \\ \hfill & ={\mathbb{U}}_0(\mu ,\varsigma )+{\mathbb{U}}_1(\mu ,\varsigma )+{\mathbb{U}}_2(\mu ,\varsigma )+{\mathbb{U}}_3(\mu ,\varsigma )+\cdots \hfill \\ \hfill & =\frac{1}{2}\left(1+tanh\left(\frac{\mu }{4}\right)\right)\hfill \\ \hfill & +\frac{5{\varsigma }^{\sigma }}{16cosh{\left(\frac{1}{4}\mu \right)}^2}-\frac{25sinh\left(\frac{\mu }{4}\right){\varsigma }^{2\sigma }}{128cosh{\left(\frac{\mu }{4}\right)}^3}+\frac{125\left(2cosh{\left(\frac{\mu }{4}\right)}^2-3\right){\varsigma }^{3\sigma }}{3072cosh{\left(\frac{\mu }{4}\right)}^4}+\cdots .\hfill \end{array}$$

On substituting $\sigma =1$, we obtain

$$\mathbb{U}(\mu ,\varsigma )=\frac{1}{2}\left(1+tanh\left(\frac{\mu }{4}+\frac{5\varsigma }{8}\right)\right).$$

(43)

In Figure 1, show that the exact and analytical solutions and Figure 2, represent the approximate solution of fractional order at $\sigma =0.8$ and $0.6$ of Example 1. In Figure 3, show that the different fractional order of two and three dimensional figures. In

Table 1. Comparison of the exact and suggested solutions numerically at various orders of $\sigma$ for Example 1.

$\mathit{\varsigma }$	$\mathit{\mu }$	$\mathit{\sigma }=0.97$	$\mathit{\sigma }=0.98$	$\mathit{\sigma }=0.99$	$\mathit{\sigma }=1$ (approx)	$\mathit{\sigma }=1$ (exact)
	0.2	0.525326	0.525308	0.525298	0.525290	0.525290
	0.4	0.550192	0.550167	0.550155	0.550143	0.550143
0.01	0.6	0.574789	0.574771	0.574760	0.574748	0.574748
	0.8	0.5989008	0.598999	0.598993	0.598987	0.598987
	1	0.622794	0.622778	0.622765	0.622753	0.622753
	0.2	0.525639	0.525623	0.525609	0.525602	0.525602
	0.4	0.550492	0.550471	0.550459	0.550452	0.550452
0.02	0.6	0.575088	0.575067	0.575058	0.575053	0.575053
	0.8	0.599321	0.599304	0.599294	0.599288	0.599288
	1	0.623089	0.623072	0.623058	0.623046	0.623046
	0.2	0.525952	0.525935	0.525924	0.525914	0.525914
	0.4	0.550798	0.550784	0.550773	0.550762	0.550762
0.03	0.6	0.575389	0.575375	0.575364	0.575358	0.575358
	0.8	0.599629	0.599608	0.599595	0.599588	0.599588
	1	0.623378	0.623358	0.623349	0.623340	0.623340
	0.2	0.526263	0.526242	0.526231	0.526225	0.526225
	0.4	0.551109	0.551087	0.551079	0.551071	0.551071
0.04	0.6	0.575708	0.575688	0.575672	0.575664	0.575664
	0.8	0.599907	0.599899	0.599893	0.599888	0.599888
	1	0.623689	0.623657	0.623642	0.623633	0.623633
	0.2	0.526583	0.526559	0.526545	0.526537	0.526537
	0.4	0.551406	0.551395	0.551386	0.551380	0.551380
0.05	0.6	0.575997	0.575986	0.575974	0.575969	0.575969
	0.8	0.600206	0.600199	0.600193	0.600188	0.600188
	1	0.623962	0.623941	0.623932	0.623926	0.623926

, comparison of the exact and suggested solutions numerically at various orders of σ for Example 1.

Example 2.

Let us assume $a=-1,b=1$ and $\delta =2$ in (1), we have

$$D_{\varsigma }^{\sigma }\mathbb{U}(\mu ,\varsigma )-{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )-{\mathbb{U}}^2(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )+\mathbb{U}(\mu ,\varsigma )({\mathbb{U}}^2(\mu ,\varsigma )-1)=0,0<\sigma \le 1$$

(44)

concerning initial condition

$$\mathbb{U}(\mu ,0)=\sqrt{\frac{1}{2}\left(1+tanh\left(\frac{\mu }{3}\right)\right)}.$$

On taking the Yang transform, we obtain

$$Y\left\{D_{\varsigma }^{\sigma }\mathbb{U}(\mu ,\varsigma )\right\}=Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+{\mathbb{U}}^2(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-\mathbb{U}(\mu ,\varsigma )({\mathbb{U}}^2(\mu ,\varsigma )-1)\right\}.$$

(45)

For simplification, we obtain

$$\frac{1}{u^{\sigma }}\{M\left(u\right)-u\mathbb{U}\left(0\right)\}=Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+{\mathbb{U}}^2(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-\mathbb{U}(\mu ,\varsigma )\left({\mathbb{U}}^2(\mu ,\varsigma )-1\right)\right\},$$

(46)

that is

$$M\left(u\right)=u\mathbb{U}\left(0\right)+u^{\sigma }Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+{\mathbb{U}}^2(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-\mathbb{U}(\mu ,\varsigma )\left({\mathbb{U}}^2(\mu ,\varsigma )-1\right)\right\}.$$

(47)

By employing the inverse Yang transform, we obtain

$$\begin{array}{cc}\hfill \mathbb{U}(\mu ,\varsigma )& =\mathbb{U}\left(0\right)+Y^{-1}\left\{u^{\sigma }Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+{\mathbb{U}}^2(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-\mathbb{U}(\mu ,\varsigma )\left({\mathbb{U}}^2(\mu ,\varsigma )-1\right)\right\}\right\}\hfill \\ \hfill & =\sqrt{\frac{1}{2}\left(1+tanh\left(\frac{\mu }{3}\right)\right)}\hfill \\ \hfill & +Y^{-1}\left\{u^{\sigma }Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+{\mathbb{U}}^2(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-\mathbb{U}(\mu ,\varsigma )\left({\mathbb{U}}^2(\mu ,\varsigma )-1\right)\right\}\right\}.\hfill \end{array}$$

(48)

Thus, by HPM, we have

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{ϵ}^k{\mathbb{U}}_k(\mu ,\varsigma )& =\sqrt{\frac{1}{2}\left(1+tanh\left(\frac{\mu }{3}\right)\right)}\hfill \\ \hfill & +ϵY^{-1}\left\{u^{\sigma }Y\left\{{\left(\sum _{k=0}^{\infty }{ϵ}^k{\mathbb{U}}_k(\mu ,\varsigma )\right)}_{\mu \mu }+\left(\sum _{k=0}^{\infty }{ϵ}^k{H}_k\left(\mathbb{U}\right)\right)+\left(\sum _{k=0}^{\infty }{ϵ}^k{\mathbb{U}}_k(\mu ,\varsigma )\right)\right\}\right\}.\hfill \end{array}$$

(49)

Using the proposed algorithm, He’s polynomial $H_k\left(\mathbb{U}\right)$ can be expressed in nonlinear terms as:

$$\sum _{k=0}^{\infty }{ϵ}^k{H}_k\left(\mathbb{U}\right)={\mathbb{U}}^2(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-{\mathbb{U}}^3(\mu ,\varsigma ).$$

(50)

Some of He’s polynomials are determined as:

$$\begin{array}{cc}\hfill H_0\left(\mathbb{U}\right)& ={\mathbb{U}}_0^2{\left({\mathbb{U}}_0\right)}_{\mu }-{\mathbb{U}}_0^3,\hfill \\ \hfill H_1\left(\mathbb{U}\right)& ={\mathbb{U}}_0^2{\left({\mathbb{U}}_1\right)}_{\mu }+2{\mathbb{U}}_0{\mathbb{U}}_1{\left({\mathbb{U}}_0\right)}_{\mu }-3{\mathbb{U}}_0^2{\mathbb{U}}_1,\hfill \\ \hfill H_2\left(\mathbb{U}\right)& ={\mathbb{U}}_0^2{\left({\mathbb{U}}_2\right)}_{\mu }+{\mathbb{U}}_2{\mathbb{U}}_0{\left({\mathbb{U}}_0\right)}_{\mu }+{\mathbb{U}}_1^2{\left({\mathbb{U}}_0\right)}_{\mu }-3{\mathbb{U}}_0^2{\mathbb{U}}_2-3{\mathbb{U}}_0{\mathbb{U}}_1^2.\hfill \end{array}$$

On comparing ϵ coefficients, we obtain

$$\begin{array}{cc}\hfill {ϵ}^0:{\mathbb{U}}_0(\mu ,\varsigma )& =\sqrt{\frac{1}{2}\left(1+tanh\left(\frac{\mu }{3}\right)\right)},\hfill \\ \hfill {ϵ}^1:{\mathbb{U}}_1(\mu ,\varsigma )& =Y^{-1}\left\{u^{\sigma }Y\left\{{\left({\mathbb{U}}_0\right)}_{\mu \mu }+H_0\left(\mathbb{U}\right)+{\mathbb{U}}_0\right\}\right\}\hfill \\ \hfill & =\frac{5}{18}\frac{\sqrt{2}}{cosh{\left(\frac{1}{3}\mu \right)}^2\sqrt{\frac{cosh\left(\frac{1}{3}\mu \right)+sinh\left(\frac{1}{3}\mu \right)}{cosh\left(\frac{1}{3}\mu \right)}}},\hfill \\ \hfill {ϵ}^2:{\mathbb{U}}_2(\mu ,\varsigma )& =Y^{-1}\left\{u^{\sigma }Y\left\{{\left({\mathbb{U}}_1\right)}_{\mu \mu }+H_1\left(\mathbb{U}\right)+{\mathbb{U}}_1\right\}\right\}\hfill \\ \hfill & =-\frac{25}{324}\frac{\sqrt{2}\left(8sinh\left(\frac{1}{3}\mu \right)cosh{\left(\frac{1}{3}\mu \right)}^2+8cosh{\left(\frac{1}{3}\mu \right)}^3-3sinh\left(\frac{1}{3}\mu \right)-7cosh\left(\frac{1}{3}\mu \right)\right)}{cosh{\left(\frac{1}{3}\mu \right)}^3{\left(cosh\left(\frac{1}{3}\mu \right)+sinh\left(\frac{1}{3}\mu \right)\right)}^2\sqrt{\frac{cosh\left(\frac{1}{3}\mu \right)+sinh\left(\frac{1}{3}\mu \right)}{cosh\left(\frac{1}{3}\mu \right)}}}\hfill \\ \hfill {ϵ}^3:{\mathbb{U}}_3(\mu ,\varsigma )& =Y^{-1}\left\{u^{\sigma }Y\left\{{\left({\mathbb{U}}_2\right)}_{\mu \mu }+H_2\left(\mathbb{U}\right)+{\mathbb{U}}_2\right\}\right)\}\hfill \\ \hfill & =\frac{125}{8748}\frac{\sqrt{2}\left(32sinh\left(\frac{1}{3}\mu \right)cosh{\left(\frac{1}{3}\mu \right)}^3+32cosh{\left(\frac{1}{3}\mu \right)}^4-36sinh\left(\frac{1}{3}\mu \right)cosh\left(\frac{1}{3}\mu \right)-52cosh{\left(\frac{1}{3}\mu \right)}^2+15\right)}{{\left(cosh\left(\frac{1}{3}\mu \right)+sinh\left(\frac{1}{3}\mu \right)\right)}^{2}cosh{\left(\frac{1}{3}\mu \right)}^4\sqrt{\frac{cosh\left(\frac{1}{3}\mu \right)+sinh\left(\frac{1}{3}\mu \right)}{cosh\left(\frac{1}{3}\mu \right)}}}\hfill \\ \hfill & ⋮.\hfill \end{array}$$

Thus, the solution is determined as:

$$\begin{array}{cc}\hfill \mathbb{U}(\mu ,\varsigma )& ={\mathbb{U}}_0(\mu ,\varsigma )+{\mathbb{U}}_1(\mu ,\varsigma )+{\mathbb{U}}_2(\mu ,\varsigma )+{\mathbb{U}}_3(\mu ,\varsigma )+\cdots \hfill \\ \hfill & =\sqrt{\frac{1}{2}\left(1+tanh\left(\frac{\mu }{3}\right)\right)}+\frac{5}{18}\frac{\sqrt{2}}{cosh{\left(\frac{1}{3}\mu \right)}^2\sqrt{\frac{cosh\left(\frac{1}{3}\mu \right)+sinh\left(\frac{1}{3}\mu \right)}{cosh\left(\frac{1}{3}\mu \right)}}}\hfill \\ \hfill & -\frac{25}{324}\frac{\sqrt{2}\left(8sinh\left(\frac{1}{3}\mu \right)cosh{\left(\frac{1}{3}\mu \right)}^2+8cosh{\left(\frac{1}{3}\mu \right)}^3-3sinh\left(\frac{1}{3}\mu \right)-7cosh\left(\frac{1}{3}\mu \right)\right)}{cosh{\left(\frac{1}{3}\mu \right)}^3{\left(cosh\left(\frac{1}{3}\mu \right)+sinh\left(\frac{1}{3}\mu \right)\right)}^2\sqrt{\frac{cosh\left(\frac{1}{3}\mu \right)+sinh\left(\frac{1}{3}\mu \right)}{cosh\left(\frac{1}{3}\mu \right)}}}\hfill \\ \hfill & +\frac{125}{8748}\frac{\sqrt{2}\left(32sinh\left(\frac{1}{3}\mu \right)cosh{\left(\frac{1}{3}\mu \right)}^3+32cosh{\left(\frac{1}{3}\mu \right)}^4-36sinh\left(\frac{1}{3}\mu \right)cosh\left(\frac{1}{3}\mu \right)-52cosh{\left(\frac{1}{3}\mu \right)}^2+15\right)}{{\left(cosh\left(\frac{1}{3}\mu \right)+sinh\left(\frac{1}{3}\mu \right)\right)}^{2}cosh{\left(\frac{1}{3}\mu \right)}^4\sqrt{\frac{cosh\left(\frac{1}{3}\mu \right)+sinh\left(\frac{1}{3}\mu \right)}{cosh\left(\frac{1}{3}\mu \right)}}}+\cdots .\hfill \end{array}$$

By implementing

 YTDM

On taking

the Yang transform, we obtain

$$Y\left\{\frac{{\partial }^{\sigma }\mathbb{U}}{\partial {\varsigma }^{\sigma }}\right\}=Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+{\mathbb{U}}^2(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-\mathbb{U}(\mu ,\varsigma )\left({\mathbb{U}}^2(\mu ,\varsigma )-1\right)\right\}.$$

(51)

For simplification, we obtain

$$\frac{1}{u^{\sigma }}\{M\left(u\right)-u\mathbb{U}\left(0\right)\}=Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+{\mathbb{U}}^2(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-\mathbb{U}(\mu ,\varsigma )\left({\mathbb{U}}^2(\mu ,\varsigma )-1\right)\right\},$$

(52)

that is

$$M\left(u\right)=u\mathbb{U}\left(0\right)+u^{\sigma }Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+{\mathbb{U}}^2(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-\mathbb{U}(\mu ,\varsigma )\left({\mathbb{U}}^2(\mu ,\varsigma )-1\right)\right\}.$$

(53)

By employing the inverse Yang transform, we obtain

$$\begin{array}{cc}\hfill \mathbb{U}(\mu ,\varsigma )& =\mathbb{U}\left(0\right)+Y^{-1}\left\{u^{\sigma }Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+{\mathbb{U}}^2(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-\mathbb{U}(\mu ,\varsigma )\left({\mathbb{U}}^2(\mu ,\varsigma )-1\right)\right\}\right\}\hfill \\ \hfill & =\sqrt{\frac{1}{2}\left(1+tanh\left(\frac{\mu }{3}\right)\right)}\hfill \\ \hfill & +Y^{-1}\left\{u^{\sigma }Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+{\mathbb{U}}^2(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-\mathbb{U}(\mu ,\varsigma )\left({\mathbb{U}}^2(\mu ,\varsigma )-1\right)\right\}\right\}.\hfill \end{array}$$

(54)

The solution in terms of series is stated as:

$$\mathbb{U}(\mu ,\varsigma )=\sum _{m=0}^{\infty }{\mathbb{U}}_m(\mu ,\varsigma ).$$

(55)

The nonlinear terms are determined as

$${\mathbb{U}}^2(\mu ,\varsigma ){\mathbb{U}}_{\mu }(\mu ,\varsigma )-{\mathbb{U}}^3(\mu ,\varsigma )=\sum _{m=0}^{\infty }{\mathcal{A}}_m.$$

Thus, we have

$$\begin{array}{cc}\hfill \sum _{m=0}^{\infty }{\mathbb{U}}_m(\mu ,\varsigma )& =\mathbb{U}(\mu ,0)+Y^{-1}\left\{u^{\sigma }Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+\sum _{m=0}^{\infty }{\mathcal{A}}_m+\mathbb{U}(\mu ,\varsigma )\right\}\right\}\hfill \\ \hfill & =\sqrt{\frac{1}{2}\left(1+tanh\left(\frac{\mu }{3}\right)\right)}+Y^{-1}\left\{u^{\sigma }Y\left\{{\mathbb{U}}_{\mu \mu }(\mu ,\varsigma )+\sum _{m=0}^{\infty }{\mathcal{A}}_m+\mathbb{U}(\mu ,\varsigma )\right\}\right\}.\hfill \end{array}$$

(56)

Few nonlinear terms are calculated as

$$\begin{array}{cc}\hfill {\mathcal{A}}_0& ={\mathbb{U}}_0^2{\left({\mathbb{U}}_0\right)}_{\mu }-{\mathbb{U}}_0^3,\hfill \\ \hfill {\mathcal{A}}_1& ={\mathbb{U}}_0^2{\left({\mathbb{U}}_1\right)}_{\mu }+2{\mathbb{U}}_0{\mathbb{U}}_1{\left({\mathbb{U}}_0\right)}_{\mu }-3{\mathbb{U}}_0^2{\mathbb{U}}_1,\hfill \\ \hfill {\mathcal{A}}_2& ={\mathbb{U}}_0^2{\left({\mathbb{U}}_2\right)}_{\mu }+{\mathbb{U}}_2{\mathbb{U}}_0{\left({\mathbb{U}}_0\right)}_{\mu }+{\mathbb{U}}_1^2{\left({\mathbb{U}}_0\right)}_{\mu }-3{\mathbb{U}}_0^2{\mathbb{U}}_2-3{\mathbb{U}}_0{\mathbb{U}}_1^2.\hfill \end{array}$$

Now, by comparing both sides, we obtain

$${\mathbb{U}}_0(\mu ,\varsigma )=\sqrt{\frac{1}{2}\left(1+tanh\left(\frac{\mu }{3}\right)\right)}.$$

For $m=0$, we have

$${\mathbb{U}}_1(\mu ,\varsigma )=\frac{5}{18}\frac{\sqrt{2}}{cosh{\left(\frac{1}{3}\mu \right)}^2\sqrt{\frac{cosh\left(\frac{1}{3}\mu \right)+sinh\left(\frac{1}{3}\mu \right)}{cosh\left(\frac{1}{3}\mu \right)}}}.$$

For $m=1$, we have

$${\mathbb{U}}_2(\mu ,\varsigma )=-\frac{25}{324}\frac{\sqrt{2}\left(8sinh\left(\frac{1}{3}\mu \right)cosh{\left(\frac{1}{3}\mu \right)}^2+8cosh{\left(\frac{1}{3}\mu \right)}^3-3sinh\left(\frac{1}{3}\mu \right)-7cosh\left(\frac{1}{3}\mu \right)\right)}{cosh{\left(\frac{1}{3}\mu \right)}^3{\left(cosh\left(\frac{1}{3}\mu \right)+sinh\left(\frac{1}{3}\mu \right)\right)}^2\sqrt{\frac{cosh\left(\frac{1}{3}\mu \right)+sinh\left(\frac{1}{3}\mu \right)}{cosh\left(\frac{1}{3}\mu \right)}}}.$$

For $m=2$, we have

$${\mathbb{U}}_3(\mu ,\varsigma )=\frac{125}{8748}\frac{\sqrt{2}\left(32sinh\left(\frac{1}{3}\mu \right)cosh{\left(\frac{1}{3}\mu \right)}^3+32cosh{\left(\frac{1}{3}\mu \right)}^4-36sinh\left(\frac{1}{3}\mu \right)cosh\left(\frac{1}{3}\mu \right)-52cosh{\left(\frac{1}{3}\mu \right)}^2+15\right)}{{\left(cosh\left(\frac{1}{3}\mu \right)+sinh\left(\frac{1}{3}\mu \right)\right)}^{2}cosh{\left(\frac{1}{3}\mu \right)}^4\sqrt{\frac{cosh\left(\frac{1}{3}\mu \right)+sinh\left(\frac{1}{3}\mu \right)}{cosh\left(\frac{1}{3}\mu \right)}}}.$$

Therefore, the other terms for $m\ge 3$ are easy to obtain

$$\begin{array}{cc}\hfill \mathbb{U}(\mu ,\varsigma )& =\sum _{m=0}^{\infty }{\mathbb{U}}_m(\mu ,\varsigma )={\mathbb{U}}_0(\mu ,\varsigma )+{\mathbb{U}}_1(\mu ,\varsigma )+{\mathbb{U}}_2(\mu ,\varsigma )+{\mathbb{U}}_3(\mu ,\varsigma )+\cdots \hfill \\ \hfill & ={\mathbb{U}}_0(\mu ,\varsigma )+{\mathbb{U}}_1(\mu ,\varsigma )+{\mathbb{U}}_2(\mu ,\varsigma )+{\mathbb{U}}_3(\mu ,\varsigma )+\cdots \hfill \\ \hfill & =\sqrt{\frac{1}{2}\left(1+tanh\left(\frac{\mu }{3}\right)\right)}+\frac{5}{18}\frac{\sqrt{2}}{cosh{\left(\frac{1}{3}\mu \right)}^2\sqrt{\frac{cosh\left(\frac{1}{3}\mu \right)+sinh\left(\frac{1}{3}\mu \right)}{cosh\left(\frac{1}{3}\mu \right)}}}\hfill \\ \hfill & -\frac{25}{324}\frac{\sqrt{2}\left(8sinh\left(\frac{1}{3}\mu \right)cosh{\left(\frac{1}{3}\mu \right)}^2+8cosh{\left(\frac{1}{3}\mu \right)}^3-3sinh\left(\frac{1}{3}\mu \right)-7cosh\left(\frac{1}{3}\mu \right)\right)}{cosh{\left(\frac{1}{3}\mu \right)}^3{\left(cosh\left(\frac{1}{3}\mu \right)+sinh\left(\frac{1}{3}\mu \right)\right)}^2\sqrt{\frac{cosh\left(\frac{1}{3}\mu \right)+sinh\left(\frac{1}{3}\mu \right)}{cosh\left(\frac{1}{3}\mu \right)}}}\hfill \\ \hfill & +\frac{125}{8748}\frac{\sqrt{2}\left(32sinh\left(\frac{1}{3}\mu \right)cosh{\left(\frac{1}{3}\mu \right)}^3+32cosh{\left(\frac{1}{3}\mu \right)}^4-36sinh\left(\frac{1}{3}\mu \right)cosh\left(\frac{1}{3}\mu \right)-52cosh{\left(\frac{1}{3}\mu \right)}^2+15\right)}{{\left(cosh\left(\frac{1}{3}\mu \right)+sinh\left(\frac{1}{3}\mu \right)\right)}^{2}cosh{\left(\frac{1}{3}\mu \right)}^4\sqrt{\frac{cosh\left(\frac{1}{3}\mu \right)+sinh\left(\frac{1}{3}\mu \right)}{cosh\left(\frac{1}{3}\mu \right)}}}+\cdots .\hfill \end{array}$$

On substituting $\sigma =1$, we obtain

$$\mathbb{U}(\mu ,\varsigma )=\sqrt{\frac{1}{2}\left(1+tanh\left(\frac{\mu }{3}+\frac{10\varsigma }{9}\right)\right)}.$$

(57)

In Figure 4, show that the exact and analytical solutions and Figure 5, represent the approximate solution of fractional order at $\sigma =0.8$ and $0.6$ of Example 2. In Figure 6, show that the different fractional order of two and three dimensional figures. In

Table 2. Comparison of the exact and our solutions numerically at various orders of $\sigma$ for Example 2.

$\mathit{\varsigma }$	$\mathit{\mu }$	$\mathit{\sigma }=0.4$	$\mathit{\sigma }=0.6$	$\mathit{\sigma }=0.8$	$\mathit{\sigma }=1$ (approx)	$\mathit{\sigma }=1$ (exact)
	0.2	0.730689	0.730663	0.730649	0.730641	0.730641
	0.4	0.752908	0.752898	0.752887	0.752874	0.752874
0.01	0.6	0.774134	0.774115	0.774103	0.774093	0.774093
	0.8	0.794267	0.794239	0.794226	0.794215	0.794215
	1	0.813214	0.813198	0.813183	0.813175	0.813175
	0.2	0.731079	0.731054	0.731033	0.731019	0.731019
	0.4	0.753278	0.753258	0.753245	0.753236	0.753236
0.02	0.6	0.774489	0.774462	0.774449	0.774438	0.774438
	0.8	0.794589	0.794567	0.794552	0.794540	0.794540
	1	0.813532	0.813505	0.813494	0.813481	0.813481
	0.2	0.731439	0.731418	0.731408	0.731397	0.731397
	0.4	0.753645	0.753624	0.753611	0.753598	0.753598
0.03	0.6	0.774829	0.774808	0.774792	0.774782	0.774782
	0.8	0.794914	0.794892	0.794879	0.794866	0.794866
	1	0.813834	0.813811	0.813796	0.813786	0.813786
	0.2	0.731827	0.731803	0.731789	0.731775	0.731775
	0.4	0.754015	0.753996	0.753978	0.753960	0.753960
0.04	0.6	0.775194	0.775155	0.775139	0.775126	0.775126
	0.8	0.795246	0.795215	0.795204	0.795191	0.795191
	1	0.814147	0.814123	0.814104	0.814092	0.814092
	0.2	0.732217	0.732189	0.732167	0.732153	0.732153
	0.4	0.754393	0.754357	0.754337	0.754321	0.754321
0.05	0.6	0.775519	0.775499	0.775483	0.775470	0.775470
	0.8	0.795587	0.795552	0.795534	0.795515	0.795515
	1	0.814449	0.814424	0.814407	0.814397	0.814397

, comparison of the exact and suggested solutions numerically at various orders of σ for Example 2.

6. Conclusions

In this study, the time fractional generalized Burgers–Fisher equation (TF-GBFE) was effectively solved using two established analytical techniques. The Yang transform decomposition method (YTDM) and the homotopy perturbation transform method (HPTM), both of which have a clear understanding, can be used to solve time-dependent equations approximatively in series. Adomian polynomials and He’s polynomials are used to express the nonlinear terms in the problems that are being studied. The suggested hybrid approach makes obtaining the solution to fractional problems simpler and more direct. Its graphical form is used to plot the results that were obtained. A powerful association between the actual and suggested approaches’ solutions is demonstrated using graphs and tables. The behavior of various dynamics of the specified physical phenomena is depicted using fractional solutions. With the help of Maple, the above problems can be seen in tabular and graphical forms. It is also important to note that the results from the YTDM and HPTM correspond quite well to the precise solutions of the representative examples considered for comparison. Moreover, all the proposed methods are not limited to the analysis of TF-GBFE but extend to include many fractional differential equations related to some strong nonlinear phenomena that arise in different plasma physics systems.