A Fractional Semi-Analytical Iterative Method for the Approximate Treatment of Fisher’s Equations_prueba_latex


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==A Fractional Semi-Analytical Iterative Method for the Approximate Treatment of Fisher's Equations ==

'''Areej Almuneef <math>{ }^{1}</math> and Ahmed Hagag <math>{ }^{2}</math>,

<math>{ }^{1}</math> Department of Mathematical Sciences, College of Science, Princess Nourah bint Abdulrahman University, P.O. Box 84428, Riyadh, 11671, Saudi Arabia

<math>{ }^{2}</math> Department of Basic Science, Faculty of Engineering, Sinai University, Kantara Branch, Ismailia, 41636, Egypt'''
-->

==Abstract==

This study presents a novel fractional semi-analytical iterative approach for solving nonlinear fractional Fisher's equations using the Caputo fractional operator. The primary objective is to provide a method that yields exact solutions to nonlinear fractional equations without requiring assumptions about nonlinear terms. By applying the Temimi-Ansari Method (TAM) with fractional calculus, this approach offers a robust solution to the time-fractional nonlinear Fisher's equation, a model relevant in fields such as population dynamics, tumor growth, and gene propagation. In this work, tables and graphical illustrations show that the proposed method minimizes computational complexity and delivers significant accuracy across multiple cases of Fisher's equations. The findings indicate that TAM with fractional order derivatives provides accurate, efficient approximations with reduced computational workload, showcasing the technique's potential for addressing a wide range of nonlinear fractional differential equations.

Revista Internacional de Métodos Numéricos para Cálculo y Diseño en Ingeniería

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<span style="text-align: center; font-size: 75%;">([[#fnc-1|<sup>1</sup>]]) </span>

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<span style="text-align: center; font-size: 75%;">([[#fnc-2|<sup>2</sup>]]) #1</span>

==OPEN ACCESS==

Received: 19/07/2024

Accepted: 01/11/2024

==DOI==

10.23967/j.rimni.2024.10.56315

==Keywords:==

Fractional calculus fisher's equation temimi-ansari method (TAM) semi-analytical iterative method numerical results

==1 Introduction==

Recently, it has been shown that fractional partial differential equations (FPDEs), which date back to 1695 , are superior in simulating the memory and hereditary features of complex materials. As a result, FPDEs are increasingly being included in the investigation of practical applications such as system identification, seepage in fractal media, anomalous diffusion, polymers and proteins, viscoelastic mechanics, and more [1-5].

With the rapid growth of nonlinear science, engineers and scientists have become increasingly concerned with approximate and asymptotic methods for solving nonlinear problems, especially in fields like solid-state physics, plasma physics, and fluid mechanics. In many branches of science and engineering, the exact or numerical solutions of FPDEs are crucial. However, finding these solutions remains challenging, necessitating the development of novel approaches. Most new nonlinear equations lack exact analytic solutions, leading to the widespread use of numerical and analytical

approaches, such as the expansion method [6], Adomian decomposition method [7], perturbation techniques [8,9], Lyapunov's artificial small parameter method [10], homotopy perturbation Sumudu transform method [11], weighted finite difference method [12], homotopy analysis transform method [13-15], and He's semi-inverse method [16], fractional sub-equation method [17], Laplace residual power series method [18], homotopy transform perturbation method [19], fractional reduced differential transform method [20] and so on. These techniques often have inherent fiaws, such as the calculation of Lagrange multipliers, divergent results, Adomian's polynomials, and extensive computational work. To address these issues, Temimi and Ansari developed the Temimi-Ansari method (TAM) [21-23], which overcomes many of the limitations of existing analytical methods. This work aims to extend the use of TAM, employing Caputo's operator to derive approximate solutions to the time-fractional nonlinear Fisher's equation.

Over the past few decades, numerous researchers have worked on numerically investigating Fisher's reaction-diffusion equation. For instance, Kudreyko and Cattani examined solutions to Fisher's equation using the wavelet-Galerkin approach [24], while Ablowitz and Zepetella explored traveling wave solutions [25]. Other researchers used the allowable stress design (ASD) method, Sinc collocation methods, and least-squares finite element methods [26-28]. Larson studied the transient behavior and time-dependent asymptotic convergence of solutions [29], while Mickens created a novel category of finite difference techniques to address the problem [30]. For the nonlinear Fisher's reactiondiffusion problem, the modified cubic B-spline differential quadrature technique has been expanded to include a classic differential quadrature technique [31].

Many researchers have offered diverse approaches for solving the Fisher's equations analytically. For example, Yadav et al. used the fractional-order homotopy analysis method (HAM) [32], while Mirzazadeh used the fractional differential transform method to solve the nonlinear Fisher's equation [33]. The homotopy perturbation method (HPM) method was utilized to find analytical solutions and calculate the absolute error of the solution [34]. Fisher-type equations have also been studied analytically using the Laplace homotopy perturbation approach [35]. Additionally, the optimum homotopy asymptotic approach has been utilized to provide approximate solutions to the Fisher equation [36]. The Shehu transformation and homotopy perturbation approach have been employed to describe the fractional analysis of Fisher's equations [37], a nonlinear model that combines linear diffusion and nonlinear growth, taking the following form:

<math display="inline">\varpi _{\tau }=\varpi _{\zeta \zeta }+\alpha \left(1-\varpi ^{\beta }\right)(\varpi{-\xi}), 0<\xi{<1}</math>.

Fisher presented the equation as a gene selection model where the positive constant <math display="inline">\alpha </math> represents the reaction factor, <math display="inline">\beta </math> represents the growth rates or reaction terms, <math display="inline">\xi </math> indicates a spatial or nondimensional variable or parameter and <math display="inline">\omega </math> indicates population density. This equation describes populations, which can refer to tumor cells, humans, or fish. Fisher's equation explains the wave of beneficial genes advancing. The equation has garnered interest from scholars due to its numerous applications across various fields of engineering and science. It appears in contexts such as neurophysiology and nuclear reactor theory.

The arrangement of the study is as follows: The introduction is given in the first part. We present some necessary fractional calculus principles in the second part, which will be applied in this study. The foundational idea of the fractional order TAM, which is utilized to forecast fractional differential equation solutions, is covered in the third part. The suggested method is used to solve the nonlinear Fisher's problem in the fourth part. The results are finally presented in Section 5.

==2 Preliminaries==

Distinct thoughts of fractional calculus have been developed during the last hundreds of years, among them the AB fractional derivative operator, <math display="inline">\mathrm{C}-\mathrm{F}</math> fractional integral operator, Caputo fractional derivative operator, Riemann-Liouville (R-L) fractional derivative operator and others [38,39].

==Definition 1==

The fractional integral operator of <math display="inline">\mathrm{R}-\mathrm{L}</math> of a function <math display="inline">\bar{\sigma }(\tau ) \in C_{r}, v \geq{-1}</math> is acquainted as

<math display="inline">\mathcal{J}_{0}^{\theta } \varpi (\tau )= \begin{cases}\frac{1}{\Gamma (\theta{+1)}} \int _{0}^{\tau } \varpi (\tau )(d \tau )^{\theta }=\frac{1}{\Gamma (\theta )} \int _{0}^{\tau }(\tau{-\varepsilon})^{\theta{-1}} \varpi (\varepsilon ) d \varepsilon , & \tau , \theta{>0}, \\ \varpi (t), & \theta=0 .\end{cases}</math>

==Definition 2==

The Caputo fractional differential operator of order <math display="inline">\theta{>0}</math> is defined as

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math> \mathfrak{J}^{k-\theta } \mathfrak{D}^{k} \varpi (\tau )=\mathfrak{D}_{*}^{\theta } \varpi (\tau )= \begin{cases}\frac{d^{k}}{d \tau ^{k}} \varpi (\tau ), & \theta =k \in \mathbb{N}, \\ \frac{1}{\Gamma (k-\theta )} \int _{\varepsilon }^{\tau }(\tau{-\varepsilon})^{k-\theta{-1}} \varpi ^{(k)}(\varepsilon ) d \varepsilon , & k-1<\theta \leq k \in \mathbb{N} .\end{cases} </math>
|}
|}

==3 Construction of Fractional TAM==

To characterize the essential concept of the suggested technique, we consider the generic nonhomogeneous FPDE as [21-23]

<math display="inline">\Psi (\varpi (\zeta , \tau ))+\Theta (\varpi (\zeta , \tau ))=p(\zeta , \tau ), k-1<\theta \leq k</math>

with the boundary conditions

<math display="inline">\mathfrak{B}\left(\varpi , \frac{\partial \varpi }{\partial \zeta }\right)=0</math>,

where the Caputo fractional operator of <math display="inline">\varpi (\zeta , \tau )</math> is indicated by <math display="inline">\Psi =\mathfrak{D}_{\tau }^{\theta }=\frac{\partial ^{\theta }}{\partial \tau ^{\ominus }}</math>, with <math display="inline">\Theta </math> indicating the general differential operators, <math display="inline">\varpi (\zeta , \tau )</math> represents the nameless function, the independent variable is denoted by <math display="inline">\zeta </math>, the dependent variable is denoted by <math display="inline">\tau </math>. The recognized continuous functions are represented by <math display="inline">\mathrm{H}(\zeta , \tau )</math>, and the boundary operator is indicated by <math display="inline">\mathfrak{B}</math>. <math display="inline">\Psi </math> is the main requirement here, and it is the general fractional differential operator. Along with the nonlinear expressions, we may arrange various linear expressions as appropriate. The suggested methodology begins by obtaining the initial condition through the elimination of the nonlinear part, as follows:

<math display="inline">\mathfrak{D}_{\tau }^{\theta } \omega _{0}(\zeta , \tau )=p(\zeta , \tau ), \mathfrak{B}\left(\omega _{0}, \frac{\partial \omega _{0}}{\partial \zeta }\right)=0</math>.

To get the next iteration of the solution, we solve the following equation:

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math> \mathfrak{D}_{\tau }^{\theta } \varpi _{1}(\zeta , \tau )+\Theta \left(\varpi _{0}(\zeta , \tau )\right)=p(\zeta , \tau ), \mathfrak{B}\left(\varpi _{1}, \frac{\partial \varpi _{1}}{\partial \zeta }\right)=0 </math>
|}
|}

As a result, we have a simple iterative stride <math display="inline">\omega _{k+1}(\zeta , \tau )</math> which is the adequate approach to a linear and nonlinear set of problems

<math display="inline">\mathcal{D}_{1}^{\theta } \varpi _{k+1}(\zeta , \tau )+\Theta \left(\varpi _{k}(\zeta , \tau )\right)=p(\zeta , \tau ), \mathfrak{R}\left(\varpi _{k+1}, \frac{\partial \varpi _{k+1}}{\partial \zeta }\right)=0</math>

In this approach, it is important to note that <math display="inline">\varpi _{k+1}(\zeta , \tau )</math> solves the problem (4) separately at each step. The iterative approach is straightforward to apply, and each iteration yields a solution closer to the exact one. By continuing this method, an ideal approximate solution corresponding to the exact solution can be obtained. Therefore, the solution to Eq. (4) can be expressed as follows:

<math display="inline">\varpi (\zeta , \tau )=\lim _{k \rightarrow \infty } \varpi _{k}(\zeta , \tau )</math>

==4 Applications by TAM==

Numerous genuine physical implementations involving FPDEs are difficult to solve exactly. Because of this, an approximate solution usually suffices to solve the problem. To obtain such approximate solutions, the method developed here (FTAM) may be utilized. Here, we analyze four examples to show that our approach to solving the nonlinear Fisher's equation is more efficient and effective than other widely used approaches.

==Case 1==

We can look at the nonlinear fractional Fisher equation shown below [40]:

<math display="inline">\frac{\partial ^{\theta } \varpi (\zeta , \tau )}{\partial \tau ^{\theta }}=\frac{\partial ^{2} \varpi (\zeta , \tau )}{\partial \zeta ^{2}}+\varpi (\zeta , \tau )(1-\varpi (\zeta , \tau )), 0<\theta \leq 1, \zeta \in \mathbb{I}, \tau{>0}</math>,

subject to a constant initial condition

<math display="inline">\omega (\zeta , 0)=\xi </math>.

By first rewriting the problem as a semi-analytical iterative method (FTAM),

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>  \Omega (\varpi (\zeta , \tau ))=D_{\tau }^{\theta } \varpi (\zeta , \tau )=\frac{\partial ^{\theta } \varpi (\zeta , \tau )}{\partial \tau ^{\theta }}, \Theta (\varpi (\zeta , \tau ))=\frac{\partial ^{2} \varpi (\zeta , \tau )}{\partial \zeta ^{2}}+\varpi (\zeta , \tau )(1-\varpi (\zeta , \tau )), </math>
|-
| style="text-align: center;" | <math> H(\zeta , \tau )=0 .  </math>
|}
|}

The initial issue that must be addressed is

<math display="inline">\Omega \left(\varpi _{0}(\zeta , \tau )\right)=0, \varpi _{0}(\zeta , 0)=\xi </math>.

Eq. (13) can solved by doing the following basic manipulation:

<math display="inline">I^{\theta }\left(D_{\tau }^{\theta } \varpi _{0}(\zeta , \tau )\right)=0, \varpi _{0}(\zeta , 0)=\xi </math>.

The essential features of definition (2) are used to generate the main iteration

<math display="inline">\varpi _{0}(\zeta , \tau )=\xi </math>.

The second iteration maybe calculated as follows:

<math display="inline">\Omega \left(\varpi _{1}(\zeta , \tau )\right)+\Theta \left(\varpi _{0}(\zeta , \tau )\right)+w(\zeta , \tau )=0, \varpi _{1}(\zeta , 0)=\xi </math>.

With the definition (2) applied and both sides of the previous equation integrated, we obtain

<math display="inline">I^{\theta }\left(D_{\tau }^{\theta } \varpi _{1}(\zeta , \tau )\right)=I^{\theta }\left(\frac{\partial ^{2} \varpi _{0}(x, \tau )}{\partial \zeta ^{2}}+\varpi _{0}(\zeta , \tau )\left(1-\varpi _{0}(\zeta , \tau )\right)\right), \varpi _{1}(\zeta , 0)=\xi </math>.

The following iteration is then obtained as

<math display="inline">\varpi _{1}(\zeta , \tau )=\xi -\frac{\xi (\xi{-1)} \tau ^{\theta }}{\Gamma (\theta{+1)}}</math>.

Calculating the third iteration may be done as

<math display="inline">\Omega \left(\varpi _{2}(\zeta , \tau )\right)+\Theta \left(\varpi _{1}(\zeta , \tau )\right)+w(\zeta , \tau )=0, \varpi _{2}(\zeta , 0)=\xi </math>.

With the definition (2) applied and both sides of the previous equation integrated, we obtain

<math display="inline">I^{\theta }\left(D_{\tau }^{\theta } \varpi _{2}(\zeta , \tau )\right)=I^{\theta }\left(\frac{\partial ^{2} \varpi _{1}(\zeta , \tau )}{\partial \zeta ^{2}}+\varpi _{1}(\zeta , \tau )\left(1-\varpi _{1}(\zeta , \tau )\right)\right), \varpi _{2}(\zeta , 0)=\xi </math>.

Then we acquire the next iteration as

<math display="inline">\varpi _{2}(\zeta , \tau )=\xi -\frac{\xi (\xi{-1)} \tau ^{\theta }}{\Gamma (\theta{+1)}}+\frac{(\xi{-1)} \xi (2 \xi{-1)} \tau ^{2 \theta }}{\Gamma (2 \theta{+1)}}-\frac{(\xi{-1)}^{2} \xi ^{2} \Gamma (2 \theta{+1)} \tau ^{3 \theta }}{\Gamma (\theta{+1)}^{2} \Gamma (3 \theta{+1)}}</math>.

Every <math display="inline">\varpi _{k}(\zeta , \tau )</math> repetition yields a rough solution to Eq. (10) based on Eq. (9). As the iteration count increases, the analytical solution approaches the exact solution more closely. The following analytical solution in series form can be produced by repeating this method:

<math display="inline">\varpi (\zeta , \tau )=\lim _{k \rightarrow \infty } \varpi _{k}(\zeta , \tau ) \simeq \varpi _{2}(\zeta , \tau )</math>,

which contains the exact solution as [40]

<math display="inline">\varpi (\zeta , \tau )=\frac{\xi e^{\tau }}{1-\xi{+\xi}e^{\tau }}</math>.

==Case 2==

We can look at the nonlinear fractional Fisher equation shown below [40]:

<math display="inline">\frac{\partial ^{\theta } \varpi (\zeta , \tau )}{\partial \tau ^{\theta }}=\frac{\partial ^{2} \varpi (\zeta , \tau )}{\partial \zeta ^{2}}+6 \varpi (\zeta , \tau )(1-\varpi (\zeta , \tau )), 0<\theta \leq 1, \zeta \in \mathbb{I}, \tau{>0}</math>,

subject to the initial condition

<math display="inline">\varpi (\zeta , 0)=\frac{1}{\left(1+e^{\zeta }\right)^{2}}</math>.

By first rewriting the problem as semi-analytical iterative method (FTAM)

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math> \begin{align} \Omega (\varpi (\zeta , \tau ))= & D_{\tau }^{\theta } \varpi (\zeta , \tau )=\frac{\partial ^{\theta } \varpi (\zeta , \tau )}{\partial \tau ^{\theta }}, \Theta (\varpi (\zeta , \tau ))=\frac{\partial ^{2} \varpi (\zeta , \tau )}{\partial \zeta ^{2}}+6 \varpi (\zeta , \tau )(1-\varpi (\zeta , \tau )) \\ & H(\zeta , \tau )=0 \end{align} </math>
|}
|}

The initial issue that must be addressed is

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math> \Omega \left(\varpi _{0}(\zeta , \tau )\right)=0, \varpi _{0}(\zeta , 0)=\frac{1}{\left(1+e^{\zeta }\right)^{2}} . </math>
|}
|}

Eq. (27) can solve by doing the following basic manipulation:

<math display="inline">I^{\theta }\left(D_{\tau }^{\theta } \varpi _{0}(\zeta , \tau )\right)=0, \varpi _{0}(\zeta , 0)=\frac{1}{\left(1+e^{\zeta }\right)^{2}}</math>.

The essential features of definition (2) are used to generate the main iteration

<math display="inline">\varpi _{0}(\zeta , \tau )=\frac{1}{\left(1+e^{\zeta }\right)^{2}}</math>.

The second iteration may be calculated as follows:

<math display="inline">\Omega \left(\varpi _{1}(\zeta , \tau )\right)+\Theta \left(\varpi _{0}(\zeta , \tau )\right)+w(\zeta , \tau )=0, \varpi _{1}(\zeta , 0)=\frac{1}{\left(1+e^{\zeta }\right)^{2}}</math>.

With the definition (2) applied and both sides of the previous equation integrated, we obtain

<math display="inline">I^{\theta }\left(D_{\tau }^{\theta } \varpi _{1}(\zeta , \tau )\right)=I^{\theta }\left(\frac{\partial ^{2} \varpi _{0}(\zeta , \tau )}{\partial \zeta ^{2}}+6 \varpi _{0}(\zeta , \tau )\left(1-\varpi _{0}(\zeta , \tau )\right)\right), \varpi _{1}(\zeta , 0)=\frac{1}{\left(1+e^{\zeta }\right)^{2}}</math>.

Then, we acquire the next iteration as

<math display="inline">\varpi _{1}(\zeta , \tau )=\frac{1}{\left(e^{\zeta }+1\right)^{2}}+\frac{10 e^{\zeta } \tau ^{\theta }}{\left(e^{\zeta }+1\right)^{3} \Gamma (\theta{+1)}}</math>

The third iteration can be calculated as

<math display="inline">\Omega \left(\varpi _{2}(\zeta , \tau )\right)+\Theta \left(\varpi _{1}(\zeta , \tau )\right)+w(\zeta , \tau )=0, \varpi _{2}(\zeta , 0)=\frac{1}{\left(1+e^{\zeta }\right)^{2}}</math>.

With the definition (2) applied and both sides of the previous equation integrated, we obtain

<math display="inline">I^{\theta }\left(D_{\tau }^{\theta } \varpi _{2}(\zeta , \tau )\right)=I^{\theta }\left(\frac{\partial ^{2} \varpi _{1}(\zeta , \tau )}{\partial \zeta ^{2}}+6 \varpi _{1}(\zeta , \tau )\left(1-\varpi _{1}(\zeta , \tau )\right)\right), \varpi _{2}(\zeta , 0)=\frac{1}{\left(1+e^{\zeta }\right)^{2}}</math>.

Then we acquire the next iteration as

<math display="inline">\varpi _{2}(\zeta , \tau )=\frac{1}{\left(e^{\zeta }+1\right)^{2}}+\frac{10 e^{\zeta } \tau ^{\theta }}{\left(e^{\zeta }+1\right)^{3} \Gamma (\theta{+1)}}+\frac{50 e^{\zeta }\left(2 e^{\zeta }-1\right)\tau ^{2 \theta }}{\left(e^{\zeta }+1\right)^{4} \Gamma (2 \theta{+1)}}-\frac{600 e^{2 \zeta } \Gamma (2 \alpha{+1)} \tau ^{3 \alpha }}{\left(e^{\zeta }+1\right)^{6} \Gamma (\alpha{+1)}^{2} \Gamma (3 \alpha{+1)}}</math>

Every <math display="inline">\varpi _{k}(\zeta , \tau )</math> repetition yields a rough solution to Eq. (24) based on Eq. (9). As the iteration count increases, the analytical solution approaches the exact solution more closely. The following analytical solution in series form can be produced by repeating this method:

<math display="inline">\varpi (\zeta , \tau )=\lim _{k \rightarrow \infty } \varpi _{k}(\zeta , \tau ) \simeq \varpi _{2}(\zeta , \tau )</math>,

which contains the exact solution as [40]

<math display="inline">\varpi (\zeta , \tau )=\frac{1}{\left(1+e^{\zeta{-5} \tau }\right)^{2}}</math>.

==Case 3==

In this case, we look at the fractional Fisher-type nonlinear diffusion equation [40]

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math> \frac{\partial ^{\theta } \varpi (\zeta , \tau )}{\partial \tau ^{\theta }}=\frac{\partial ^{2} \varpi (\zeta , \tau )}{\partial \zeta ^{2}}+\varpi (\zeta , \tau )(1-\varpi (\zeta , \tau ))(\varpi (\zeta , \tau )-\xi ), 0<\theta , \xi \leq 1, </math>
|}
|}

subject to the initial condition

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math> \varpi (\zeta , 0)=\frac{1}{1+e^{\frac{-\zeta }{\sqrt{2}}}} . </math>
|}
|}

By first rewriting the problem as semi-analytical iterative method (FTAM)

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math> \begin{align} \Omega (\varpi (\zeta , \tau ))= & D_{t}^{\theta } \varpi (\zeta , \tau )=\frac{\partial ^{\theta } \varpi (\zeta , t)}{\partial \tau ^{\theta }}, \Theta (\varpi (\zeta , \tau ))=\frac{\partial ^{2} \varpi (\zeta , \tau )}{\partial \zeta ^{2}} \\ & +\varpi (\zeta , \tau )(1-\varpi (\zeta , \tau ))(\varpi (\zeta , \tau )-\xi ), H(\zeta , \tau )=0 . \end{align} </math>
|}
|}

The initial issue that must be addressed is

<math display="inline">\Omega \left(\varpi _{0}(\zeta , \tau )\right)=0, \omega _{0}(\zeta , 0)=\frac{1}{1+e^{\frac{-\zeta }{\sqrt{2}}}}</math>.

Eq. (42) can solve by doing the following basic manipulation

<math display="inline">I^{\theta }\left(D_{\tau }^{\theta } \varpi _{0}(\zeta , \tau )\right)=0, \varpi _{0}(\zeta , 0)=\frac{1}{1+e^{\frac{-\zeta }{\sqrt{2}}}}</math>.

The essential features of definition (2) are used to generate the main iteration

<math display="inline">\omega _{0}(\zeta , \tau )=\frac{1}{1+e^{\frac{-\zeta }{\sqrt{2}}}}</math>.

The second iteration may be calculated as follows:

<math display="inline">\Omega \left(\varpi _{1}(\zeta , \tau )\right)+\Theta \left(\varpi _{0}(\zeta , \tau )\right)+w(\zeta , \tau )=0, \varpi _{1}(\zeta , 0)=\frac{1}{1+e^{\frac{-\zeta }{\sqrt{2}}}}</math>.

With the definition (2) applied and both sides of the previous equation integrated, we obtain

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math> I^{\theta }\left(D_{\tau }^{\theta } \varpi _{1}(\zeta , \tau )\right)=I^{\theta }\left(\frac{\partial ^{2} \varpi _{0}(\zeta , \tau )}{\partial \zeta ^{2}}+\varpi _{0}(\zeta , \tau )\left(1-\varpi _{0}(\zeta , \tau )\right)\left(\varpi _{0}(\zeta , \tau )-\xi \right)\right), \varpi _{1}(\zeta , 0)=\frac{1}{1+e^{\frac{-\zeta }{\sqrt{2}}}} . </math>
|}
|}

Then, we acquire the next iteration as

<math display="inline">\varpi _{1}(\zeta , \tau )=\frac{1}{1+e^{\frac{-\zeta }{\sqrt{2}}}}+\frac{(1-2 \xi ) \tau ^{\theta }}{4\left(\cosh \left(\frac{\zeta }{\sqrt{2}}\right)+1\right)\Gamma (\theta{+1)}}</math>.

The third iteration can be calculated as

<math display="inline">\Omega \left(\varpi _{2}(\zeta , \tau )\right)+\Theta \left(\varpi _{1}(\zeta , \tau )\right)+w(\zeta , \tau )=0, \varpi _{2}(\zeta , 0)=\frac{1}{1+e^{\frac{-\zeta }{\sqrt{2}}}}</math>.

With the definition (2) applied and both sides of the previous equation integrated, we obtain

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math> I^{\theta }\left(D_{\tau }^{\theta } \varpi _{2}(\zeta , \tau )\right)=I^{\theta }\left(\frac{\partial ^{2} \varpi _{1}(\zeta , \tau )}{\partial \zeta ^{2}}+\varpi _{1}(\zeta , \tau )\left(1-\varpi _{1}(\zeta , \tau )\right)\left(\varpi _{1}(\zeta , \tau )-\xi \right)\right), \varpi _{2}(\zeta , 0)=\frac{1}{1+e^{\frac{-\zeta }{\sqrt{2}}}} . </math>
|}
|}

Then we acquire the next iteration as

<math display="inline">\omega _{2}(\zeta , \tau )=\omega _{1}+\frac{(1-2 \xi )^{2} A \tau ^{2 \theta }\left(\frac{2(\xi{-2)} A+\xi{+1)} \Gamma (2 \theta{+1)}(A+B) \tau ^{\theta }}{\Gamma (\theta{+1)}^{2} \Gamma (3 \theta{+1)}}+\frac{(2 \xi{-1)} B \Gamma 3 \theta{+1))}^{2 \theta }}{\Gamma (\theta{+1)}^{3} \Gamma (4 \theta{+1)}}-\frac{2(A-1)(A+1)^{3}}{\Gamma (2 \theta{+1)}}\right)}{8(A+1)^{6}}</math>,

where <math display="inline">A=e^{\frac{\zeta }{\sqrt{2}}}</math> and <math display="inline">B=e^{\sqrt{25}}</math>. Every <math display="inline">\varpi _{k}(\zeta , \tau )</math> repetition yields a rough solution to Eq. (52) based on Eq. (9). As the iteration count increases, the analytical solution approaches the exact solution more closely. The following analytical solution in series form can be produced by repeating this method:

<math display="inline">\varpi (\zeta , \tau )=\lim _{k \rightarrow \infty } \varpi _{k}(\zeta , \tau ) \simeq \varpi _{2}(\zeta , \tau )</math>,

which contains the exact solution as [40]

<math display="inline">\varpi (\zeta , \tau )=\frac{1}{1+e^{\frac{-(\zeta{+}c \tau )}{\sqrt{2}}}}</math>.

where <math display="inline">c=\frac{\sqrt{2}}{2-\xi }</math>.

==Case 4==

We can look at the nonlinear fractional Fisher equation shown below [41]:

<math display="inline">\frac{\partial ^{\theta } \varpi (\zeta , \tau )}{\partial \tau ^{\theta }}=\frac{\partial ^{2} \varpi (\zeta , \tau )}{\partial \zeta ^{2}}+\left(1-\varpi ^{2}(\zeta , \tau )\right)(2 \varpi (\zeta , \tau )+\xi ), 0<\theta \leq 1,|\xi |<1</math>.

subject to the initial condition

<math display="inline">\varpi (\zeta , 0)=\tanh (\zeta ) \&  \varpi (\zeta , 0)=\operatorname{coth}(\zeta )</math>.

We obtain the following analytical solutions by applying the same fundamental idea of the TAM of fractional order

<math display="inline">\varpi _{0}(\zeta , \tau )=\tanh (\zeta ), \varpi _{1}(\zeta , \tau )=\tanh (\zeta )+\frac{\xi \tau ^{\theta } \operatorname{sech}^{2}(\zeta )}{\Gamma (\theta{+1)}}</math>

<math display="inline">\varpi _{2}(\zeta , \tau )=\varpi _{1}(\zeta , \tau )-\xi ^{2} \tau ^{2 \theta } \operatorname{sech}^{2}(\zeta )\left(\frac{2 \tanh (\zeta )}{\Gamma (2 \theta{+1)}}+\frac{\Gamma (2 \theta{+1))}^{\theta } \operatorname{sech}^{2}(\zeta )(\xi{+6} \tanh (\zeta ))}{\Gamma (\theta{+1)}^{2} \Gamma (3 \theta{+1)}}+\frac{2 \varepsilon \Gamma (3 \theta{+1))}^{2} \theta ^{\operatorname{sech}}{ }^{4}(\zeta )}{\Gamma (\theta{+1)}^{3} \Gamma (\theta \theta{+1)}}\right), \cdots </math>

Every <math display="inline">\varpi _{k}(\zeta , \tau )</math> repetition yields a rough solution to Eq. (53) based on Eq. (9). As the iteration count increases, the analytical solution approaches the exact solution more closely. The following

analytical solution in series form can be produced by repeating this method:

<math display="inline">\varpi (\zeta , \tau )=\lim _{m \rightarrow \infty } \varpi _{m}(\zeta , \tau ) \simeq \varpi _{2}(\zeta , \tau )</math>

which contains the exact solution as [41]

<math display="inline">\omega (\zeta , \tau )=\tanh (\zeta{+\xi}\tau ) \&  \varpi (\zeta , \tau )=\operatorname{coth}(\zeta{+\xi}\tau )</math>.

The analytical results for the particular state <math display="inline">\theta=1</math> demonstrate that the general style of the approximate solution is the same as that of the exact solution. To geometrically demonstrate the behavior of the analytical solution TAM, the exact solution was compared with the fifth iteration of the approximate solution in two and three dimensions, as shown in Figs. 1-5. Tables 1 and 2 present the numerical outcomes and errors of the suggested method for four cases with varying time and spatial variables.

[[Image:Draft_Sanchez Pinedo_470453401-2025_04_01_641976f6b73e20cd7663g-09.png|500px|center|2025_04_01_641976f6b73e20cd7663g-09]]

Figure 1: The behavior of a collection of approximate solutions obtained by TAM for Case 1

==SCIPEDIA==

[[Image:Draft_Sanchez Pinedo_470453401-2025_04_01_641976f6b73e20cd7663g-10(4).png|500px|center|2025_04_01_641976f6b73e20cd7663g-10(4)]]

(a) Exact solution

[[Image:Draft_Sanchez Pinedo_470453401-2025_04_01_641976f6b73e20cd7663g-10(5).png|500px|center|2025_04_01_641976f6b73e20cd7663g-10(5)]]

(c) Exemplification of the TAM with the exact solution.

[[Image:Draft_Sanchez Pinedo_470453401-2025_04_01_641976f6b73e20cd7663g-10(2).png|500px|center|2025_04_01_641976f6b73e20cd7663g-10(2)]]

(e) The solution of the power series

[[Image:Draft_Sanchez Pinedo_470453401-2025_04_01_641976f6b73e20cd7663g-10.png|500px|center|2025_04_01_641976f6b73e20cd7663g-10]]

(b) FTAM solution

[[Image:Draft_Sanchez Pinedo_470453401-2025_04_01_641976f6b73e20cd7663g-10(3).png|500px|center|2025_04_01_641976f6b73e20cd7663g-10(3)]]

(d) Comparison between various amounts of <math display="inline">\theta </math>

[[Image:Draft_Sanchez Pinedo_470453401-2025_04_01_641976f6b73e20cd7663g-10(1).png|500px|center|2025_04_01_641976f6b73e20cd7663g-10(1)]]

(f) Comparison at <math display="inline">\zeta=1</math>

Figure 2: The behavior of a collection of approximate solutions obtained by TAM for Case 2

==SCIPEDIA==

[[Image:Draft_Sanchez Pinedo_470453401-2025_04_01_641976f6b73e20cd7663g-11.png|500px|2025_04_01_641976f6b73e20cd7663g-11]]
{| class="floating_imageSCP" style="text-align: center; border: 1px solid #BBB; margin: 1em auto; width: 100%;max-width: 100%;"
|-
|[[Image:Draft_Sanchez Pinedo_470453401-2025_04_01_641976f6b73e20cd7663g-11.png|600px|]]
|}

Figure 3: The behavior of a collection of approximate solutions obtained by TAM for Case 3

==SCIPEDIA==

[[Image:Draft_Sanchez Pinedo_470453401-2025_04_01_641976f6b73e20cd7663g-12.png|500px|2025_04_01_641976f6b73e20cd7663g-12]]
{| class="floating_imageSCP" style="text-align: center; border: 1px solid #BBB; margin: 1em auto; width: 100%;max-width: 100%;"
|-
|[[Image:Draft_Sanchez Pinedo_470453401-2025_04_01_641976f6b73e20cd7663g-12.png|600px|]]
|}

Figure 4: The behavior of a collection of approximate solutions obtained by TAM for Case 4

[[Image:Draft_Sanchez Pinedo_470453401-2025_04_01_641976f6b73e20cd7663g-13.png|500px|center|2025_04_01_641976f6b73e20cd7663g-13]]

Figure 5: The absolute error between exact and TAM solutions for three cases

Table 1: Comparison of approximate solutions acquired by TAM with exact solution for Case 1 at <math display="inline">\xi=0.5</math>


{|  class="floating_tableSCP wikitable" style="text-align: left; margin: 1em auto;min-width:50%;"
|- style="border-top: 2px solid;"
|  t 
| 0.0 
| 0.1 
| 0.2 
| 0.3 
| 0.4 
| 0.5 
| 0.6 
| 0.7 
| 0.8 
| 0.9 
| 1.0 
|- style="border-top: 2px solid;"
| <math display="inline">\varpi _{E x}</math> 
| 0.5 
| 0.5249 
| 0.5498 
| 0.5744 
| 0.5986 
| 0.6224 
| 0.6456 
| 0.6681 
| 0.6899 
| 0.7109 
| 0.7310 
|-
| <math display="inline">\varpi _{T A M}</math> 
| 0.5 
| 0.5249 
| 0.5498 
| 0.5744 
| 0.5986 
| 0.6224 
| 0.6456 
| 0.6681 
| 0.6900 
| 0.7110 
| 0.7311 
|- style="border-bottom: 2px solid;"
| <math display="inline">E_{\sigma }</math> 
| 0.0 
| <math display="inline">1.4 \mathrm{E}-11</math> 
| <math display="inline">1.8 \mathrm{E}-09</math> 
| <math display="inline">3.2 \mathrm{E}-08</math> 
| <math display="inline">2.3 \mathrm{E}-07</math> 
| <math display="inline">1.1 \mathrm{E}-06</math> 
| <math display="inline">3.9 \mathrm{E}-06</math> 
| <math display="inline">1.1 \mathrm{E}-05</math> 
| <math display="inline">2.8 \mathrm{E}-05</math> 
| <math display="inline">6.3 \mathrm{E}-05</math> 
| <math display="inline">1.2 \mathrm{E}-04</math> 

|}

Table 2: Comparison of approximate solutions acquired by TAM with exact solution


{|  class="floating_tableSCP wikitable" style="text-align: center; margin: 1em auto;min-width:50%;"
|- style="border-top: 2px solid;"
| rowspan='2' style="border-left: 2px solid;border-right: 2px solid;" | X
| colspan='3' style="text-align: left;border-left: 2px solid;border-right: 2px solid;border-left: 2px solid;border-right: 2px solid;" | Case <math>2(\xi=0.1)</math>
| colspan='3' style="text-align: left;border-left: 2px solid;border-right: 2px solid;border-left: 2px solid;border-right: 2px solid;" | Case <math>3(\xi=0.001)</math>
| colspan='3' style="text-align: left;border-left: 2px solid;border-right: 2px solid;border-left: 2px solid;border-right: 2px solid;" | Case <math>4(\xi=0.1)</math>
|- style="border-top: 2px solid;"
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">\bar{\omega }_{E x}</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">\varpi _{{TAM }}</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">E_{{б }}</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">\bar{\omega }_{E x}</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">\varpi _{{TAM }}</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">E_{{б }}</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">\bar{\omega }_{E x}</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">\varpi _{{TAM }}</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">E_{{б }}</math> 
|- style="border-top: 2px solid;"
| style="border-left: 2px solid;border-right: 2px solid;" |  0.0 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.387456 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.387313 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">1.42 \mathrm{E}-4</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.512504 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.512472 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">3.12 \mathrm{E}-5</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.00999 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.009999 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">5.00 \mathrm{E}-8</math> 
|- style="border-top: 2px solid;"
| style="border-left: 2px solid;border-right: 2px solid;" |  0.1 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.358427 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.358299 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">1.27 \mathrm{E}-4</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.530147 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.530116 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">3.14 \mathrm{E}-5</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.109558 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.109556 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">2.00 \mathrm{E}-6</math> 
|- style="border-top: 2px solid;border-bottom: 2px solid;"
| style="border-left: 2px solid;border-right: 2px solid;" |  0.2 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.329984 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.329873 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">1.10 \mathrm{E}-4</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.547716 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.547684 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">3.15 \mathrm{E}-5</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.206966 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.206963 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">3.71 \mathrm{E}-6</math> 

|}

(Continued)

Table 2 (continued)


{|  class="floating_tableSCP wikitable" style="text-align: center; margin: 1em auto;min-width:50%;"
|- style="border-top: 2px solid;"
| rowspan='2' style="border-left: 2px solid;border-right: 2px solid;" | x
| colspan='3' style="text-align: left;border-left: 2px solid;border-right: 2px solid;border-left: 2px solid;border-right: 2px solid;" | Case <math>2(\xi=0.1)</math>
| colspan='3' style="text-align: left;border-left: 2px solid;border-right: 2px solid;border-left: 2px solid;border-right: 2px solid;" | Case <math>3(\xi=0.001)</math>
| colspan='3' style="text-align: left;border-left: 2px solid;border-right: 2px solid;border-left: 2px solid;border-right: 2px solid;" | Case <math>4(\xi=0.1)</math>
|- style="border-top: 2px solid;"
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">\omega _{E x}</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">\varpi _{{TAM }}</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">E_{{бб }}</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">\varpi _{E x}</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">\varpi _{{TAM }}</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">E_{{ш }}</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">\varpi _{E x}</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">\varpi _{{TAM }}</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">E_{{б }}</math> 
|- style="border-top: 2px solid;"
| style="border-left: 2px solid;border-right: 2px solid;" |  0.3 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.302317 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.302223 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">9.48 \mathrm{E}-5</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.5 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.565135 
| style="border-left: 2px solid;border-right: 2px solid;" | 3.1 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.300437 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.300432 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">4.97 \mathrm{E}-6</math> 
|- style="border-top: 2px solid;"
| style="border-left: 2px solid;border-right: 2px solid;" |  0.4 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.275603 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.275521 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">8.19 \mathrm{E}-5</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.582457 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.582425 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">3.16 \mathrm{E}-</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.388473 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.38846 
| style="border-left: 2px solid;border-right: 2px solid;" | 5.67 
|- style="border-top: 2px solid;"
| style="border-left: 2px solid;border-right: 2px solid;" |  0.5 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.250000 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.249926 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">7.43 \mathrm{E}-5</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.599547 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.599516 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">3.15 \mathrm{E}-5</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.469945 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.469939 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">5.85 \mathrm{E}-6</math> 
|- style="border-top: 2px solid;"
| style="border-left: 2px solid;border-right: 2px solid;" |  0.6 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.225645 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.225572 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">7.30 \mathrm{E}-5</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.616398 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.616367 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">3.13 \mathrm{E}-5</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.544127 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.544122 
| style="border-left: 2px solid;border-right: 2px solid;" | 5.59 E 
|- style="border-top: 2px solid;"
| style="border-left: 2px solid;border-right: 2px solid;" |  0.7 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.202649 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.202571 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">7.84 \mathrm{E}-5</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | . 632975 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.632944 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">3.13 \mathrm{E}-5</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.610677 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.610672 
| style="border-left: 2px solid;border-right: 2px solid;" | 5.03 
|- style="border-top: 2px solid;"
| style="border-left: 2px solid;border-right: 2px solid;" |  0.8 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.181099 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.181009 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">8.99 \mathrm{E}-5</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.649242 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.649212 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">3.07 \mathrm{E}-5</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.669590 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.669586 
| style="border-left: 2px solid;border-right: 2px solid;" | 4.32 E 
|- style="border-top: 2px solid;"
| style="border-left: 2px solid;border-right: 2px solid;" |  0.9 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.161052 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.160945 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">1.06 \mathrm{E}-4</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.665170 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.665140 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">3.03 \mathrm{E}-5</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.721132 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.721129 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">3.56 \mathrm{E}-6</math> 
|- style="border-top: 2px solid;border-bottom: 2px solid;"
| style="border-left: 2px solid;border-right: 2px solid;" |  1.0 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.142537 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.142411 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">1.26 \mathrm{E}-4</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.680731 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.680701 
| style="border-left: 2px solid;border-right: 2px solid;" | <math display="inline">2.98 \mathrm{E}-5</math> 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.765762 
| style="border-left: 2px solid;border-right: 2px solid;" | 0.765759 
| style="border-left: 2px solid;border-right: 2px solid;" | 2.85 E 

|}

==5 Conclusion==

The primary contribution of this labor is the successful demonstration of TAM with fractional time derivatives to obtain analytical solutions to distinct nonlinear fractional differential equations. The outcomes indicate that a small number of approximate expressions yield highly reliable outcomes, and the error in the approximate solution decreases rapidly as the number of these expressions increases. Furthermore, compared to other approaches, this one uses less computing power and a central processing unit (CPU). The effectiveness of this strategy has been shown to confirm TAM's accuracy and dependability. The results show that this approach works very well for solving a class of fractional operator nonlinear FPD problems. The suggested approach solves a variety of fractional equations and systems in an efficient and effective manner.

Acknowledgement: Not applicable.

Funding Statement: This research project was funded by the Deanship of Scientific Research and Libraries, Princess Nourah bint Abdulrahman University, through the Program of Research Project Funding After Publication, grant No. (RPFAP-15-1445).

Author Contributions: The authors confirm contribution to the paper as follows: study conception and design: Ahmed Hagag; data collection: Areej Almuneef; analysis and interpretation of results: Ahmed Hagag; draft manuscript preparation: Areej Almuneef. All authors reviewed the results and approved the final version of the manuscript.

Availability of Data and Materials: Not applicable.

Ethics Approval: Not applicable.

Conflicts of Interest: The authors declare no conflicts of interest to report regarding the present study.

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