Fractional-View Analysis of Fokker-Planck Equations by ZZ Transform with Mittag-Leffler Kernel

Abstract

This work combines a ZZ transformation with the Adomian decomposition method to solve the fractional-order Fokker-Planck equations. The fractional derivative is represented in the Atangana-Baleanu derivative. It is looked at with graphs that show that the accurate and estimated results are close to each other, indicating that the method works. Fractional-order solutions are the most in line with the dynamics of the targeted problems, and they provide an endless number of options for an optimal mathematical model solution for a particular physical phenomenon. This analytical approach produces a series type result that quickly converges to actual answers. The acquired outcomes suggest that the novel analytical solution method is simple to use and very successful at assessing complicated equations that occur in related research and engineering fields.

1. Introduction

Fractional calculus (FC) depicts the facts of existence in a more straightforward manner. Making this issue accessible to the world of engineering and science as a common topic offers another layer for better understanding or expressing fundamental natures. Because FC is the language that nature knows, connecting with it in this manner is effective. This issue of FC has been with mathematicians over the previous three centuries, and it has just recently been pulled into other practical sectors of engineering, science, and economics [1,2,3,4,5,6]. FC deals with integrals and derivatives in any real or complex order. Atangana-Baleanu, Caputo, Caputo Fabrizio, Hilfer, Katugampola, and other fractional operators have recently been suggested and implemented to cope with real-world applications. In this paper, we looked at a non-singular and non-local fractional operator using the Mittag-Leffler function as its kernel, which may be used to mimic a variety of real-world issues. Due to its non-singularity and non-locality, it has been termed the Atangana-Baleanu fractional operator, and there are a number of illustrative uses in the literature [7,8,9,10,11]. Several major strategies have been utilised, such as the fractional operational matrix technique [12], Shehu decomposition technique [13,14], q-homotopy transformation method [15], homotopy perturbation method [16,17], and variational iteration transform technique [18]. The foundation of nature is symmetry, however the great majority of natural occurrences lack symmetry. In order to conceal symmetry, it is effective to break unexpected symmetry. There are two kinds of symmetry: finite and infinitesimal. There can be discrete or continuous symmetries in finite symmetries. Natural symmetries include symmetry and time reversal are discontinuous, whereas space is a constant transformation. Mathematicians have been fascinated by patterns for millennia. The eighteenth century witnessed the emergence of systematic categorization of planar and spatial patterns.

The link between the Aboodh transformation (AT), the ZZ transformation (ZZT), and the Laplace transformation (LT) is established in this study, with numerous applications described in [19,20,21]. Later, we used this ZZ transform to solve various test examples described in the Atangana-Baleanu sense. The current author’s contributions to this work are first establishing the link among the Laplace, Aboodh, and ZZ transformations, and second, using the ZZ transform to solve fractional differential equations expressed in the Atangana-Baleanu derivative. The ZZ transform generalizes a few well-known transforms, which can be related to other well-known ones. The Natural transform is obtained by dividing the ZZ transform by the modified variable. This work also includes relationships with other integral transformations in terms of theorems. This transformation has the virtue of converging to the Sumudu transformation, which is useful when solving fractional differential equations with variable coefficients [22,23,24]. Adomian (1980) established the Adomian decomposition technique (ADM), an efficient method for finding explicit and numerical solutions to a larger and more general class of differential systems representing real-world issues [25,26,27]. This strategy effectively addresses initial and boundary value problems, linear and nonlinear, ordinary and partial differential equations, and stochastic systems. Furthermore, this approach does not require any linearization or perturbation. ADM has been used extensively in the last two decades, since it yields approximate analytical solutions for nonlinear problems, and there has been much interest in utilizing it to solve fractional differential equations [28,29,30].

The Fokker-Planck equation was mainly implemented by Fokker and Planck to recognize the Brownian suggestion of particles. It is notable that the fractional order Fokker-Planck (FP) equations are a mathematical concept used in both the biological and physical sciences. The spread of its Markovian and continuous character leads to the approximation of some randomizations, in acceptable processes and procedures [31]. The Fokker-Planck equations are used in various applied sciences, such as electron relaxation, polymer dynamics, quantum optics, solid-state system and probability flux [32].

Consider the fractional-order FP equation in the general form, as:

$${}^{ABC}{\nu }_{\rho }^{\gamma }(\phi ,\rho )={\nu }_{\phi }(\phi ,\rho )+{\nu }_{\phi \phi }(\phi ,\rho ),\phi \in R\rho >0,\gamma \in (0,1],$$

(1)

with the initial condition:

$$\nu (\phi ,0)=\zeta \left(\phi \right).$$

The fractional-order Fokker-Planck equations have been successfully implemented in different areas, such as chemical physics, biological molecules, engineering, and energy consumption. In fact, fractional diffusion, a specific type of fractional-order Fokker-Planck equation, has also been applied in a wide range of situations, including diffusion processes, viscoelasticity, and frequency-dependent material damping behavior [31]. However, it is difficult to analysis a precise solution for fractional problems. Several analytical and numerical techniques applied for fractional-order Fokker-Planck equations include the Laplace transformation technique [32], the multi-step reduced differential transformation technique [33], the predictor corrector method [34], the Adomain decomposition technique [35], and the variational iteration technique [36].

Here, papers [37,38,39,40,41] address the application of ADM to various transport models, also including fractional and nonlinear cases; works [42,43,44,45,46] provide reviews or/and developments of various computational approaches to linear [43,44,45,46] and nonlinear [42,43,45] transport/advection-diffusion problems; [47] introduces a homotopy perturbation method for nonlinear transport equations, while [48] proposes perturbational approach to construct analytical approximations based on the double-parameter transformation perturbation expansion method. Finally, the review paper [49] contains an exhaustive review of various modern fractional calculus applications, and paper [50] discusses some non-standard definitions of Caputo fractional derivatives, while paper [51] provides a comprehensive overview of computational practices used in fractional calculus. In this study, we utilized the Modified decomposition method (MDM) to solve fractional-order Fokker-Planck equations. The Adomian decomposition technique [52,53] is a very good technique for solving linear and non-linear partial differential equations, as well as integral differential and fractional differential equations, in a series type solution. The obtained solutions to the fractional problems applying the suggested technique are also utilized the analysis of the equations. It has been proven that the suggested method can be used to analyze a variety of fractional partial differential equations and systems.

2. Preliminaries

Definition 1.

An Aboodh transformation (AT) is given for terms of exponential order. We consider term in the set B, expressed by:

$$B=\left\{h:\left|h\left(\rho \right)\right|<M{e}^{n_j\left|\rho \right|},\mathrm{if}\rho \in {(-1)}^j\times [0,\infty ),j=1,2;\left(M,n_1,n_2>0\right)\right\}.$$

M must be a finite number for a given function in set B; $n_1,n_2$ may be finite or infinite.

Then, the AT denote by the operator $A(.)$ is expressed as the integral equation [19,20]:

$$A\left\{h\left(\rho \right)\right\}=\frac{1}{\varsigma }{\int }_0^{\infty }h\left(\rho \right)e^{-\varsigma \rho }d\rho ,\rho \ge 0,n_1\le \varsigma \le n_2.$$

Theorem 1.

Now, we consider G and F as the Laplace and Aboodh transformations of $h\left(\rho \right)\in B$, then [21,22]:

$$G\left(\varsigma \right)=\frac{F\left(\varsigma \right)}{\varsigma }.$$

Zain U. Zafar [23] was the first to develop the ZZ transform. It is a mixture of the Laplace transform and AT. The ZZ transformation is expressed.

Definition 2.

(ZZ transform) Suppose that $h\left(\rho \right)\forall \rho \ge 0$ is a function, then the ZZ transformation $\mathcal{Z}(\varrho ,\varsigma )$ of $h\left(\rho \right)$ is defined as [23]:

$$\mathcal{Z}\left(h\left(\rho \right)\right)=\mathcal{Z}(\varrho ,\varsigma )=\varsigma {\int }_0^{\infty }h\left(\varrho \rho \right)e^{-\varsigma \rho }d\rho .$$

The ZZ transformation is linear, as are the Aboodh and Laplace transforms. The Mittag-Leffler function is given as

$$E_{\delta }\left(z\right)=\sum _{ȷ=0}^{\infty }\frac{z^{ȷ}}{\Gamma (1+ȷ\delta )},Re\left(\delta \right)>0.$$

Definition 3.

The Atangana-Baleanu derivative (ABD) of a term $\nu (\phi ,\rho )\in H^1(a,b)$, then for $\delta \in$$(0,1)$, is defined as [24]:

$${}_a^{ABC}{D}_{\rho }^{\delta }\nu (\phi ,\rho )=\frac{\psi \left(\delta \right)}{1-\delta }{\int }_a^{\rho }{\nu }^{\prime }(\phi ,\rho )E_{\delta }\left(\frac{-\delta {(\rho -\eta )}^{\delta }}{1-\delta }\right)d\eta .$$

Definition 4.

Let $\nu (\phi ,\rho )\in H^1(a,b)$, then for $\delta \in (0,1)$, the Riemann-Liouville ABD is defined as [24]:

$${}_a^{ABR}{D}_{\rho }^{\delta }\nu (\phi ,\eta )=\frac{\psi \left(\delta \right)}{1-\delta }\frac{d}{d\rho }{\int }_a^{\rho }\nu (\phi ,\eta )E_{\delta }\left(\frac{-\delta {(\rho -\eta )}^{\delta }}{1-\delta }\right)d\eta ,$$

with the condition $\psi \left(0\right)=\psi \left(1\right)=1$, $\psi \left(\delta \right)$ is a normalization function, and $b>a$.

Theorem 2.

The Laplace transform of the Riemann-Liouville Atangana-Baleanu derivative is expressed as [24]:

$$\mathrm{L}\left\{{}_a^{ABC}{D}_{\rho }^{\delta }\nu (\phi ,\rho )\right\}\left(\varsigma \right)=\frac{\psi \left(\delta \right)}{1-\delta }\frac{{\varsigma }^{\delta }L\left\{\nu (\phi ,\rho )\right\}-{\varsigma }^{\delta -1}\nu (\phi ,0)}{{\varsigma }^{\delta }+\frac{\delta }{1-\delta }}$$

and

$$L\left\{\begin{array}{c}ABR\hfill \\ a\hfill \end{array}{D}_{\rho }^{\delta }\nu (\phi ,\rho )\right\}\left(\varsigma \right)=\frac{\psi \left(\delta \right)}{1-\delta }\frac{{\varsigma }^{\delta }L\left\{\nu (\phi ,\rho )\right\}}{{\varsigma }^{\delta }+\frac{\delta }{1-\delta }}$$

The theorems that follow are based on the idea that $h\left(\rho \right)\in H^1(a,b),b>a$ and $\delta \in (0,1)$.

3. Methodology of MDM

The general implementation of the given method

$${}^{ABC}{D}_{\rho }^{\delta }\nu (\phi ,\rho )=\overline{\mathcal{H}}(\phi ,\rho )+\mathcal{N}(\phi ,\rho ),0<\delta \le 1,$$

(2)

with the initial condition:

$$\nu (\phi ,0)=\xi \left(\phi \right),$$

where ${}^{ABC}{D}_{\rho }^{\delta }$ is the Atangana-Baleanu fractional derivative of order $\delta$; $\overline{\mathcal{H}}$ is a linear and $\mathcal{N}$ a nonlinear term.

On both sides, using the ZZ transformation of Equation (2), to achieve:

$$\mathcal{Z}\left[\nu (\phi ,\rho )\right]=\mathcal{Z}[\overline{\mathcal{H}}(\phi ,\rho )+\mathcal{N}(\phi ,\rho )].$$

(3)

By the differentiation property of the ZZ transformation, we get:

$$\frac{\psi \left(\delta \right)}{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}\mathcal{Z}\{\nu (\phi ,\rho )-\frac{\varrho }{\varsigma }\nu (\phi ,0)\}=\mathcal{Z}[\overline{\mathcal{H}}(\phi ,\rho )+\mathcal{N}(\phi ,\rho )].$$

(4)

Equation (4) implies that:

$$\mathcal{Z}\left(\nu (\phi ,\rho )\right)=\frac{\varrho }{\varsigma }\nu (\phi ,0)+\frac{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}{\psi \left(\delta \right)}\mathcal{Z}[\overline{\mathcal{H}}(\phi ,\rho )+\mathcal{N}(\phi ,\rho )],$$

(5)

Applying the ZZ inverse transformation of Equation (5), we get:

$$\nu (\phi ,\rho )=\nu (\phi ,0)+{\mathcal{Z}}^{-1}[\frac{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}{\psi \left(\delta \right)}\mathcal{Z}\left\{\overline{\mathcal{H}}(\phi ,\rho )+\mathcal{N}(\phi ,\rho )\right\}].$$

(6)

MDM determines the infinite sequence’s result of $\nu (\phi ,\rho )$:

$$\nu (\phi ,\rho )=\sum _{ȷ=0}^{\infty }{\nu }_{ȷ}(\phi ,\rho ).$$

(7)

The non-linear functions $\mathcal{N}$ is expressed as:

$$\mathcal{N}(\phi ,\rho )=\sum _{ȷ=0}^{\infty }{\mathcal{A}}_{ȷ}.$$

(8)

The Adomian polynomials can show all types of non-linear functions, as:

$${\mathcal{A}}_{ȷ}=\frac{1}{ȷ!}{\left[\frac{{\partial }^{ȷ}}{\partial {\ell }^{ȷ}}\left\{\mathcal{N}\left(\sum _{ȷ=0}^{\infty }{\ell }^{ȷ}{\nu }_{ȷ}\right)\right\}\right]}_{\ell =0},$$

(9)

putting Equations (7) and (9) into (6), gives:

$$\begin{array}{c}\hfill \sum _{ȷ=0}^{\infty }{\nu }_{ȷ}(\phi ,\rho )={\nu }_0(\phi ,0)+{\mathcal{Z}}^{-1}\left[\frac{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}{\psi \left(\delta \right)}\mathcal{Z}\left\{\sum _{ȷ=0}^{\infty }{\overline{\mathcal{H}}}_{ȷ}+\sum _{ȷ=0}^{\infty }{\mathcal{A}}_{ȷ}\right\}\right].\end{array}$$

(10)

The following terms are described.

$${\nu }_0(\phi ,\rho )=\xi \left(\phi \right),$$

(11)

$${\nu }_1(\phi ,\rho )={\mathcal{Z}}^{-1}\left[\frac{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}{\psi \left(\delta \right)}\mathcal{Z}\{{\overline{\mathcal{H}}}_0+{\mathcal{A}}_0\}\right],$$

where the general for $ȷ\ge 1$, is determined as:

$${\nu }_{ȷ+1}(\phi ,\rho )={\mathcal{Z}}^{-1}\left[\frac{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}{\psi \left(\delta \right)}\mathcal{Z}\{{\overline{\mathcal{H}}}_{ȷ}+{\mathcal{A}}_{ȷ}\}\right].$$

4. Numerical Results

Problem 1.

Consider the fractional-order FP equation:

$$\frac{{\partial }^{\delta }}{\partial {\rho }^{\delta }}\nu (\phi ,\rho )+\frac{\partial }{\partial \phi }\left(\frac{\phi }{6}\nu (\phi ,\rho )\right)-\frac{{\partial }^2}{\partial {\phi }^2}\left(\frac{{\phi }^2}{12}\nu (\phi ,\rho )\right)=0,\phi ,\rho >0,\delta \in (0,1],$$

(12)

with the initial condition:

$$\nu (\phi ,0)={\phi }^2$$

On employing the ZZ transform, we get:

$$\mathcal{Z}\left\{\frac{{\partial }^{\delta }\nu }{\partial {\rho }^{\delta }}\right\}=\mathcal{Z}\left[-\frac{\partial }{\partial \phi }\left(\frac{\phi }{6}\nu (\phi ,\rho )\right)+\frac{{\partial }^2}{\partial {\phi }^2}\left(\frac{{\phi }^2}{12}\nu (\phi ,\rho )\right)\right].$$

(13)

By using the ZZ differentiation property, we have:

$$\frac{\psi \left(\delta \right)}{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}\mathcal{Z}\left\{\nu (\phi ,\rho )-\frac{\varrho }{\varsigma }\nu (\phi ,0)\right\}=\mathcal{Z}\left[-\frac{\partial }{\partial \phi }\left(\frac{\phi }{6}\nu (\phi ,\rho )\right)+\frac{{\partial }^2}{\partial {\phi }^2}\left(\frac{{\phi }^2}{12}\nu (\phi ,\rho )\right)\right].$$

(14)

On applying the inverse ZZ transform, we have:

$$\begin{array}{cc}\hfill & \nu (\phi ,\rho )=\nu (\phi ,0)+{\mathcal{Z}}^{-1}\left[\frac{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}{\psi \left(\delta \right)}\left\{\mathcal{Z}\left(-\frac{\partial }{\partial \phi }\left(\frac{\phi }{6}\nu (\phi ,\rho )\right)+\frac{{\partial }^2}{\partial {\phi }^2}\left(\frac{{\phi }^2}{12}\nu (\phi ,\rho )\right)\right)\right\}\right],\hfill \\ \hfill & \nu (\phi ,\rho )={\phi }^2+{\mathcal{Z}}^{-1}\left[\frac{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}{\psi \left(\delta \right)}\left\{\mathcal{Z}\left(-\frac{\partial }{\partial \phi }\left(\frac{\phi }{6}\nu (\phi ,\rho )\right)+\frac{{\partial }^2}{\partial {\phi }^2}\left(\frac{{\phi }^2}{12}\nu (\phi ,\rho )\right)\right)\right\}\right].\hfill \end{array}$$

(15)

The solution in terms of infinite sequence $\nu (\phi ,\rho )$ by means of MDM as:

$$\begin{array}{cc}\hfill & \nu (\phi ,\rho )=\sum _{ȷ=0}^{\infty }{\nu }_{ȷ}(\phi ,\rho ).\hfill \\ \hfill & \sum _{ȷ=0}^{\infty }{\nu }_{ȷ}(\phi ,\rho )={\phi }^2+{\mathcal{Z}}^{-1}\left[\frac{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}{\psi \left(\delta \right)}\mathcal{Z}\left[-\frac{\partial }{\partial \phi }\left(\frac{\phi }{6}\nu (\phi ,\rho )\right)+\frac{{\partial }^2}{\partial {\phi }^2}\left(\frac{{\phi }^2}{12}\nu (\phi ,\rho )\right)\right]\right].\hfill \end{array}$$

(16)

On both sides, comparing Equation (16), we get:

$${\nu }_0(\phi ,\rho )={\phi }^2,$$

For$ȷ=0$

$${\nu }_1(\phi ,\rho )={\phi }^2\frac{1}{2B\left(\delta \right)}\left(1-\delta +\frac{\delta {\rho }^{\delta }}{\Gamma (\delta +1)}\right),$$

For$ȷ=1$

$${\nu }_2(\phi ,\rho )={\phi }^2\frac{1}{8{\left(\mathrm{B}\left(\delta \right)\right)}^2}\left({(1-\delta )}^2+\frac{2\delta (1-\delta ){\rho }^{\delta }}{\Gamma (\delta +1)}+\frac{{\delta }^2{\rho }^{2\delta }}{\Gamma (2\delta +1)}\right),$$

For$ȷ=2$

$$\begin{array}{cc}\hfill & {\nu }_3(\phi ,\rho )={\phi }^2\frac{1}{24{\left(\mathrm{B}\left(\delta \right)\right)}^3}\left[{(1-\delta )}^3+\frac{3\delta {(1-\delta )}^2{\rho }^{\delta }}{\Gamma (\delta +1)}+\frac{{\delta }^2(1-\delta ){\rho }^{2\delta +1}}{\Gamma (2\delta +2)}+\frac{2{\delta }^2(1-\delta ){\rho }^{2\delta }}{\Gamma (2\delta +1)}+\frac{{\delta }^3{\rho }^{2\delta +1}}{\Gamma (2\delta +2)}\right].\hfill \end{array}$$

The remaining components of the MDM solution, ${\nu }_{ȷ}$ for $(ȷ\ge 3)$ are determined simply.

$$\nu (\phi ,\rho )=\sum _{ȷ=0}^{\infty }{\nu }_{ȷ}(\phi ,\rho )={\nu }_0(\phi ,\rho )+{\nu }_1(\phi ,\rho )+{\nu }_2(\phi ,\rho )+{\nu }_3(\phi ,\rho )+\cdots .$$

$$\begin{array}{cc}\hfill & \nu (\phi ,\rho )={\phi }^2+{\phi }^2\frac{1}{2B\left(\delta \right)}\left(1-\delta +\frac{\delta {\rho }^{\delta }}{\Gamma (\delta +1)}\right)+{\phi }^2\frac{1}{8{\left(\mathrm{B}\left(\delta \right)\right)}^2}\left({(1-\delta )}^2+\frac{2\delta (1-\delta ){\rho }^{\delta }}{\Gamma (\delta +1)}+\frac{{\delta }^2{\rho }^{2\delta }}{\Gamma (2\delta +1)}\right)+\hfill \\ \hfill & {\phi }^2\frac{1}{24{\left(\mathrm{B}\left(\delta \right)\right)}^3}\left[{(1-\delta )}^3+\frac{3\delta {(1-\delta )}^2{\rho }^{\delta }}{\Gamma (\delta +1)}+\frac{{\delta }^2(1-\delta ){\rho }^{2\delta +1}}{\Gamma (2\delta +2)}+\frac{2{\delta }^2(1-\delta ){\rho }^{2\delta }}{\Gamma (2\delta +1)}+\frac{{\delta }^3{\rho }^{2\delta +1}}{\Gamma (2\delta +2)}\right]+\cdots .\hfill \end{array}$$

The MDM solution at $\delta =1$ is:

$$\nu (\phi ,\rho )={\phi }^2{e}^{\frac{\rho }{2}}.$$

(17)

In Figure 1, represents the actual solution and the 2nd depicts the analytic result, which are in close contact. Figure 2 depicts fractional-order graphs at $\delta =0.8$ and $0.6$ for Problem 1. Similarly, in Figure 3, it is demonstrated that the Problem 1 fractional-order is $\delta =0.4$. The data demonstrate that the proposed method corresponds to the real solution for the given problem.

Figure 1. The exact and analytical solution at $\delta =1$ of Problem 1.

Figure 2. Fractional solution at $\delta =0.8$ and $0.6$ of Problem 1.

Figure 3. Fractional solution at $\delta =0.4$ of Problem 1.

Problem 2.

Consider the fractional-order FP equation:

$$\frac{{\partial }^{\delta }}{\partial {\rho }^{\delta }}\nu (\phi ,\rho )+\frac{\partial }{\partial \phi }\left(\phi \nu (\phi ,\rho )\right)-\frac{{\partial }^2}{\partial {\phi }^2}\left(\frac{{\phi }^2}{2}\nu (\phi ,\rho )\right)=0,\phi ,\rho >0,\delta \in (0,1],$$

(18)

with the initial condition:

$$\nu (\phi ,0)=\phi .$$

On employing the ZZ transform, we get:

$$\mathcal{Z}\left\{\frac{{\partial }^{\delta }\nu }{\partial {\rho }^{\delta }}\right\}=\mathcal{Z}\left[-\frac{\partial }{\partial \phi }\left(\phi \nu (\phi ,\rho )\right)+\frac{{\partial }^2}{\partial {\phi }^2}\left(\frac{{\phi }^2}{2}\nu (\phi ,\rho )\right)\right],$$

(19)

By using the ZZ differentiation property, we have:

$$\frac{B\left(\delta \right)}{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}\mathcal{Z}\left\{\nu (\phi ,\rho )-\frac{\varrho }{\varsigma }\nu (\phi ,0)\right\}=\mathcal{Z}\left[-\frac{\partial }{\partial \phi }\left(\phi \nu (\phi ,\rho )\right)+\frac{{\partial }^2}{\partial {\phi }^2}\left(\frac{{\phi }^2}{2}\nu (\phi ,\rho )\right)\right].$$

(20)

On applying the inverse ZZ transform, we have:

$$\begin{array}{cc}\hfill & \nu (\phi ,\rho )=\nu (\phi ,0)+{\mathcal{Z}}^{-1}\left[\frac{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}{B\left(\delta \right)}\left\{\mathcal{Z}\left(-\frac{\partial }{\partial \phi }\left(\phi \nu (\phi ,\rho )\right)+\frac{{\partial }^2}{\partial {\phi }^2}\left(\frac{{\phi }^2}{2}\nu (\phi ,\rho )\right)\right)\right\}\right],\hfill \\ \hfill & \nu (\phi ,\rho )=\phi +{\mathcal{Z}}^{-1}\left[\frac{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}{B\left(\delta \right)}\left\{\mathcal{Z}\left(-\frac{\partial }{\partial \phi }\left(\phi \nu (\phi ,\rho )\right)+\frac{{\partial }^2}{\partial {\phi }^2}\left(\frac{{\phi }^2}{2}\nu (\phi ,\rho )\right)\right)\right\}\right].\hfill \end{array}$$

(21)

Using MDM, the solution in term of the infinite series $\nu (\phi ,\rho )$ can be expressed as:

$$\nu (\phi ,\rho )=\sum _{ȷ=0}^{\infty }{\nu }_{ȷ}(\phi ,\rho ).$$

(22)

$$\begin{array}{cc}\hfill & \sum _{ȷ=0}^{\infty }{\nu }_{ȷ}(\phi ,\rho )=\phi +{\mathcal{Z}}^{-1}\left[\frac{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}{B\left(\delta \right)}\mathcal{Z}\left[-\frac{\partial }{\partial \phi }\left(\phi \nu (\phi ,\rho )\right)+\frac{{\partial }^2}{\partial {\phi }^2}\left(\frac{{\phi }^2}{2}\nu (\phi ,\rho )\right)\right]\right].\hfill \end{array}$$

(23)

By comparing Equation (23) on both sides, we get:

$${\nu }_0(\phi ,\rho )=\phi ,$$

For $ȷ=0$

$${\nu }_1(\phi ,\rho )=\phi \frac{1}{B\left(\delta \right)}\left(1-\delta +\frac{\delta {\rho }^{\delta }}{\Gamma (\delta +1)}\right),$$

For $ȷ=1$

$${\nu }_2(\phi ,\rho )=\phi \frac{1}{{\left(\mathrm{B}\left(\delta \right)\right)}^2}\left({(1-\delta )}^2+\frac{2\delta (1-\delta ){\rho }^{\delta }}{\Gamma (\delta +1)}+\frac{{\delta }^2{\rho }^{2\delta }}{\Gamma (2\delta +1)}\right),$$

For $ȷ=2$

$$\begin{array}{cc}\hfill & {\nu }_3(\phi ,\rho )=\phi \frac{1}{{\left(\mathrm{B}\left(\delta \right)\right)}^3}\left({(1-\delta )}^3+\frac{3\delta {(1-\delta )}^2{\rho }^{\delta }}{\Gamma (\delta +1)}+\frac{{\delta }^2(1-\delta ){\rho }^{2\delta +1}}{\Gamma (2\delta +2)}+\frac{2{\delta }^2(1-\delta ){\rho }^{2\delta }}{\Gamma (2\delta +1)}+\frac{{\delta }^3{\rho }^{2\delta +1}}{\Gamma (2\delta +2)}\right).\hfill \end{array}$$

The MDM solution remaining components ${\nu }_{ȷ}$ with $(ȷ\ge 3)$ are simply determined.

$$\nu (\phi ,\rho )=\sum _{ȷ=0}^{\infty }{\nu }_{ȷ}(\phi ,\rho )={\nu }_0(\phi ,\rho )+{\nu }_1(\phi ,\rho )+{\nu }_2(\phi ,\rho )+{\nu }_3(\phi ,\rho )+\cdots .$$

$$\begin{array}{cc}\hfill & \nu (\phi ,\rho )=\phi +\phi \frac{1}{B\left(\delta \right)}\left(1-\delta +\frac{\delta {\rho }^{\delta }}{\Gamma (\delta +1)}\right)+\phi \frac{1}{{\left(\mathrm{B}\left(\delta \right)\right)}^2}\left({(1-\delta )}^2+\frac{2\delta (1-\delta ){\rho }^{\delta }}{\Gamma (\delta +1)}+\frac{{\delta }^2{\rho }^{2\delta }}{\Gamma (2\delta +1)}\right)\hfill \\ \hfill & +\phi \frac{1}{{\left(\mathrm{B}\left(\delta \right)\right)}^3}\left({(1-\delta )}^3+\frac{3\delta {(1-\delta )}^2{\rho }^{\delta }}{\Gamma (\delta +1)}+\frac{{\delta }^2(1-\delta ){\rho }^{2\delta +1}}{\Gamma (2\delta +2)}+\frac{2{\delta }^2(1-\delta ){\rho }^{2\delta }}{\Gamma (2\delta +1)}+\frac{{\delta }^3{\rho }^{2\delta +1}}{\Gamma (2\delta +2)}\right)+\cdots .\hfill \end{array}$$

The MDM solution at $\delta =1$ is:

$$\nu (\phi ,\rho )=\phi e^{\rho }.$$

(24)

In Figure 4, represents the actual solution and the 2nd depicts the analytic result, which are in close contact. Figure 5 depicts fractional-order graphs at $\delta =0.8$ and $0.6$ for Problem 2. Similarly, in Figure 6, it is demonstrated that the Problem 2 fractional-order is $\delta =0.4$. The data demonstrate that the proposed method corresponds to the real solution for the given problem.

Figure 4. The exact and analytical solution at $\delta =1$ of Problem 2.

Figure 5. Fractional solution at $\delta =0.8$ and $0.6$ of Problem 2.

Figure 6. Fractional solution at $\delta =0.4$ of Problem 2.

Problem 3.

Consider the fractional-order FP equation:

$$\frac{{\partial }^{\delta }}{\partial {\rho }^{\delta }}\nu (\phi ,\rho )+\frac{\partial }{\partial \phi }\left(\frac{4}{\phi }{\nu }^2(\phi ,\rho )\right)-\frac{\partial }{\partial \phi }\left(\frac{\phi }{3}\nu (\phi ,\rho )\right)-\frac{{\partial }^2}{\partial {\phi }^2}\left({\nu }^2(\phi ,\rho )\right)=0,\phi ,\rho >0,\delta \in (0,1],$$

(25)

with the initial condition:

$$\nu (\phi ,0)={\phi }^2.$$

On employing the ZZ transform, we get:

$$\mathcal{Z}\left\{\frac{{\partial }^{\delta }\nu }{\partial {\rho }^{\delta }}\right\}=\mathcal{Z}\left[\frac{\partial }{\partial \phi }\left(\frac{\phi }{3}\nu (\phi ,\rho )\right)+\frac{{\partial }^2}{\partial {\phi }^2}\left({\nu }^2(\phi ,\rho )\right)-\frac{\partial }{\partial \phi }\left(\frac{4}{\phi }{\nu }^2(\phi ,\rho )\right)\right].$$

(26)

By using the ZZ differentiation property, we have:

$$\frac{B\left(\delta \right)}{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}\mathcal{Z}\left\{\nu (\phi ,\rho )-\frac{\varrho }{\varsigma }\nu (\phi ,0)\right\}=\mathcal{Z}\left[\frac{\partial }{\partial \phi }\left(\frac{\phi }{3}\nu (\phi ,\rho )\right)+\frac{{\partial }^2}{\partial {\phi }^2}\left({\nu }^2(\phi ,\rho )\right)-\frac{\partial }{\partial \phi }\left(\frac{4}{\phi }{\nu }^2(\phi ,\rho )\right)\right].$$

(27)

On applying the inverse ZZ transform, we have:

$$\begin{array}{cc}\hfill & \nu (\phi ,\rho )=\nu (\phi ,0)+{\mathcal{Z}}^{-1}\left[\frac{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}{B\left(\delta \right)}\left\{\mathcal{Z}\left(\frac{\partial }{\partial \phi }\left(\frac{\phi }{3}\nu (\phi ,\rho )\right)+\frac{{\partial }^2}{\partial {\phi }^2}\left({\nu }^2(\phi ,\rho )\right)-\frac{\partial }{\partial \phi }\left(\frac{4}{\phi }{\nu }^2(\phi ,\rho )\right)\right)\right\}\right],\hfill \\ \hfill & \nu (\phi ,\rho )={\phi }^2+{\mathcal{Z}}^{-1}\left[\frac{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}{B\left(\delta \right)}\left\{\mathcal{Z}\left(\frac{\partial }{\partial \phi }\left(\frac{\phi }{3}\nu (\phi ,\rho )\right)+\frac{{\partial }^2}{\partial {\phi }^2}\left({\nu }^2(\phi ,\rho )\right)-\frac{\partial }{\partial \phi }\left(\frac{4}{\phi }{\nu }^2(\phi ,\rho )\right)\right)\right\}\right].\hfill \end{array}$$

(28)

Now, we assume that the infinite series from the function $\nu (\phi ,\rho )$, which is unknown, has the solution as:

$$\nu (\phi ,\rho )=\sum _{ȷ=0}^{\infty }{\nu }_{ȷ}(\phi ,\rho ).$$

(29)

The nonlinear terms are therefore described by the Adomian polynomial ${\nu }^2={\sum }_{ȷ=0}^{\infty }{\mathcal{A}}_{ȷ}$. Applying particular concepts, Equation (28) can be expressed as:

$$\begin{array}{cc}\hfill & \sum _{ȷ=0}^{\infty }{\nu }_{ȷ}(\phi ,\rho )=\nu (\phi ,0)+{\mathcal{Z}}^{-1}\left[\frac{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}{B\left(\delta \right)}\mathcal{Z}\left[\frac{\partial }{\partial \phi }\left(\frac{\phi }{3}\nu (\phi ,\rho )\right)+\sum _{ȷ=0}^{\infty }{\mathcal{A}}_{ȷ}\right]\right],\hfill \\ \hfill & \sum _{ȷ=0}^{\infty }{\nu }_{ȷ}(\phi ,\rho )={\phi }^2+{\mathcal{Z}}^{-1}\left[\frac{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}{B\left(\delta \right)}\mathcal{Z}\left[\frac{\partial }{\partial \phi }\left(\frac{\phi }{3}\nu (\phi ,\rho )\right)+\sum _{ȷ=0}^{\infty }{\mathcal{A}}_{ȷ}\right]\right].\hfill \end{array}$$

(30)

Now, by Adomian polynomials $\mathcal{A}$, the non-linear term is decomposed as,

$${\mathcal{A}}_0={\nu }_0^2,{\mathcal{A}}_1=2{\nu }_0{\nu }_1,{\mathcal{A}}_2=2{\nu }_0{\nu }_2+{\left({\nu }_2\right)}^2.$$

By comparing Equation (30) both sides, we get:

$${\nu }_0(\phi ,\rho )={\phi }^2,$$

On $ȷ=0$

$${\nu }_1(\phi ,\rho )={\phi }^2\frac{1}{B\left(\delta \right)}\left(1-\delta +\frac{\delta {\rho }^{\delta }}{\Gamma (\delta +1)}\right),$$

On $ȷ=1$

$${\nu }_2(\phi ,\rho )={\phi }^2\frac{1}{{\left(\mathrm{B}\left(\delta \right)\right)}^2}\left({(1-\delta )}^2+\frac{2\delta (1-\delta ){\rho }^{\delta }}{\Gamma (\delta +1)}+\frac{{\delta }^2{\rho }^{2\delta }}{\Gamma (2\delta +1)}\right),$$

On $ȷ=2$

$${\nu }_3(\phi ,\rho )={\phi }^2\frac{1}{{\left(\mathrm{B}\left(\delta \right)\right)}^3}\left({(1-\delta )}^3+\frac{3\delta {(1-\delta )}^2{\rho }^{\delta }}{\Gamma (\delta +1)}+\frac{{\delta }^2(1-\delta ){\rho }^{2\delta +1}}{\Gamma (2\delta +2)}+\frac{2{\delta }^2(1-\delta ){\rho }^{2\delta }}{\Gamma (2\delta +1)}+\frac{{\delta }^3{\rho }^{2\delta +1}}{\Gamma (2\delta +2)}\right).$$

The MDM solution remaining components ${\nu }_{ȷ}$ with $(ȷ\ge 3)$, are simply calculated.

$$\nu (\phi ,\rho )=\sum _{ȷ=0}^{\infty }{\nu }_{ȷ}(\phi ,\rho )={\nu }_0(\phi ,\rho )+{\nu }_1(\phi ,\rho )+{\nu }_2(\phi ,\rho )+{\nu }_3(\phi ,\rho )+\cdots .$$

$$\begin{array}{c}\hfill \nu (\phi ,\rho )={\phi }^2+{\phi }^2\frac{1}{B\left(\delta \right)}\left(1-\delta +\frac{\delta {\rho }^{\delta }}{\Gamma (\delta +1)}\right)+{\phi }^2\frac{1}{{\left(\mathrm{B}\left(\delta \right)\right)}^2}\left({(1-\delta )}^2+\frac{2\delta (1-\delta ){\rho }^{\delta }}{\Gamma (\delta +1)}+\frac{{\delta }^2{\rho }^{2\delta }}{\Gamma (2\delta +1)}\right)+\\ \hfill {\phi }^2\frac{1}{{\left(\mathrm{B}\left(\delta \right)\right)}^3}\left({(1-\delta )}^3+\frac{3\delta {(1-\delta )}^2{\rho }^{\delta }}{\Gamma (\delta +1)}+\frac{{\delta }^2(1-\delta ){\rho }^{2\delta +1}}{\Gamma (2\delta +2)}+\frac{2{\delta }^2(1-\delta ){\rho }^{2\delta }}{\Gamma (2\delta +1)}+\frac{{\delta }^3{\rho }^{2\delta +1}}{\Gamma (2\delta +2)}\right)+\cdots .\end{array}$$

The MDM solution at $\delta =1$ is:

$$\nu (\phi ,\rho )={\phi }^2{e}^{\rho }.$$

(31)

In Figure 7, represents the actual solution and the 2nd depicts the analytic result, which are in close contact. Figure 8 depicts fractional-order graphs at $\delta =0.8$ and $0.6$ for Problem 3. Similarly, in Figure 9, it is demonstrated that the Problem 3 fractional-order is $\delta =0.4$. The data demonstrate that the proposed method corresponds to the real solution for the given problem.

Figure 7. The exact and analytical solution at $\delta =1$ of Problem 3.

Figure 8. The first graph of the fractional order solution at $\delta =0.8$, and the second at $0.6$ of Problem 3.

Figure 9. The fractional order solution at $\delta =0.4$ of Problem 3.

Problem 4.

Consider the fractional-order FP equation:

$$\frac{{\partial }^{\delta }}{\partial {\rho }^{\delta }}\nu (\phi ,\rho )-\frac{\partial }{\partial \phi }\nu (\phi ,\rho )-\frac{{\partial }^2}{\partial {\phi }^2}\nu (\phi ,\rho )=0,\rho >0,\delta \in (0,1],$$

(32)

with the initial condition:

$$\nu (\phi ,0)=\phi .$$

On employing the ZZ transform, we get:

$$\mathcal{Z}\left\{\frac{{\partial }^{\delta }\nu }{\partial {\rho }^{\delta }}\right\}=\mathcal{Z}\left[\frac{\partial }{\partial \phi }\nu (\phi ,\rho )+\frac{{\partial }^2}{\partial {\phi }^2}\nu (\phi ,\rho )\right],$$

(33)

By using the ZZ differentiation property, we have:

$$\frac{B\left(\delta \right)}{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}\mathcal{Z}\left\{\nu (\phi ,\rho )-\frac{\varrho }{\varsigma }\nu (\phi ,0)\right\}=\mathcal{Z}\left[\frac{\partial }{\partial \phi }\nu (\phi ,\rho )+\frac{{\partial }^2}{\partial {\phi }^2}\nu (\phi ,\rho )\right].$$

(34)

On applying the inverse ZZ transform, we have:

$$\begin{array}{cc}\hfill & \nu (\phi ,\rho )=\nu \left(0\right)+{\mathcal{Z}}^{-1}\left[\frac{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}{B\left(\delta \right)}\left\{\mathcal{Z}\left(\frac{\partial }{\partial \phi }\nu (\phi ,\rho )+\frac{{\partial }^2}{\partial {\phi }^2}\nu (\phi ,\rho )\right)\right\}\right],\hfill \\ \hfill & \nu (\phi ,\rho )=\phi +{\mathcal{Z}}^{-1}\left[\frac{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}{B\left(\delta \right)}\left\{\mathcal{Z}\left(\frac{\partial }{\partial \phi }\nu (\phi ,\rho )+\frac{{\partial }^2}{\partial {\phi }^2}\nu (\phi ,\rho )\right)\right\}\right].\hfill \end{array}$$

(35)

Using MDM, the solution in terms of the infinite series $\nu (\phi ,\rho )$ can be expressed as:

$$\nu (\phi ,\rho )=\sum _{ȷ=0}^{\infty }{\nu }_{ȷ}(\phi ,\rho ).$$

(36)

$$\begin{array}{cc}\hfill & \sum _{ȷ=0}^{\infty }{\nu }_{ȷ}(\phi ,\rho )=\phi +{\mathcal{Z}}^{-1}\left[\frac{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}{B\left(\delta \right)}\mathcal{Z}\left[\frac{\partial }{\partial \phi }\nu (\phi ,\rho )+\frac{{\partial }^2}{\partial {\phi }^2}\nu (\phi ,\rho )\right]\right].\hfill \end{array}$$

(37)

By comparing Equation (37) on both sides, we get:

$${\nu }_0(\phi ,\rho )=\phi ,$$

For $ȷ=0$

$${\nu }_1(\phi ,\rho )=\frac{1}{B\left(\delta \right)}\left(1-\delta +\frac{\delta {\rho }^{\delta }}{\Gamma (\delta +1)}\right),$$

For $ȷ=1$

$${\nu }_2(\phi ,\rho )=0,$$

For $ȷ=2$

$$\begin{array}{cc}\hfill & {\nu }_3(\phi ,\rho )=0.\hfill \end{array}$$

The MDM solution remaining components ${\nu }_{ȷ}$ with $(ȷ\ge 3)$ are simply calculated.

$$\nu (\phi ,\rho )=\sum _{ȷ=0}^{\infty }{\nu }_{ȷ}(\phi ,\rho )={\nu }_0(\phi ,\rho )+{\nu }_1(\phi ,\rho )+{\nu }_2(\phi ,\rho )+{\nu }_3(\phi ,\rho )+\cdots .$$

$$\begin{array}{cc}\hfill & \nu (\phi ,\rho )=\phi +\frac{1}{B\left(\delta \right)}\left(1-\delta +\frac{\delta {\rho }^{\delta }}{\Gamma (\delta +1)}\right)+0+0+\cdots .\hfill \end{array}$$

The MDM solution at $\delta =1$ is:

$$\nu (\phi ,\rho )=\phi +\rho .$$

(38)

In Figure 10, represents the actual solution and the 2nd depicts the analytic result, which are in close contact. Figure 11 depicts fractional-order graphs at $\delta =0.8$ and $0.6$ for Problem 4. Similarly, in Figure 12, it is demonstrated that the Problem 4 fractional-order is $\delta =0.4$. The data demonstrate that the proposed method corresponds to the real solution for the given problem.

Figure 10. The exact and analytical solution at $\delta =1$ of Problem 4.

Figure 11. Fractional solution at $\delta =0.8$ and $0.6$ of Problem 4.

Figure 12. Fractional solution at $\delta =0.4$ of Example 4.

Problem 5.

Consider the fractional-order FP equation:

$$\frac{{\partial }^{\delta }}{\partial {\rho }^{\delta }}\nu (\phi ,\rho )-(1-\phi )\frac{\partial }{\partial \phi }\nu (\phi ,\rho )-\left(e^{\rho }{\phi }^2\right)\frac{{\partial }^2}{\partial {\phi }^2}\nu (\phi ,\rho )=0,\rho >0,\delta \in (0,1],$$

(39)

with the initial condition:

$$\nu (\phi ,0)=1+\phi .$$

On employing the ZZ transform, we get:

$$\mathcal{Z}\left\{\frac{{\partial }^{\delta }\nu }{\partial {\rho }^{\delta }}\right\}=\mathcal{Z}\left[(1-\phi )\frac{\partial }{\partial \phi }\nu (\phi ,\rho )+\left(e^{\rho }{\phi }^2\right)\frac{{\partial }^2}{\partial {\phi }^2}\nu (\phi ,\rho )\right].$$

(40)

By using the ZZ differentiation property, we have:

$$\frac{B\left(\delta \right)}{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}\mathcal{Z}\left\{\nu (\phi ,\rho )-\frac{\varrho }{\varsigma }\nu (\phi ,0)\right\}=\mathcal{Z}\left[(1-\phi )\frac{\partial }{\partial \phi }\nu (\phi ,\rho )+\left(e^{\rho }{\phi }^2\right)\frac{{\partial }^2}{\partial {\phi }^2}\nu (\phi ,\rho )\right].$$

(41)

On applying the inverse ZZ transform, we have:

$$\begin{array}{cc}\hfill & \nu (\phi ,\rho )=\nu \left(0\right)+{\mathcal{Z}}^{-1}\left[\frac{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}{B\left(\delta \right)}\left\{\mathcal{Z}\left((1-\phi )\frac{\partial }{\partial \phi }\nu (\phi ,\rho )+\left(e^{\rho }{\phi }^2\right)\frac{{\partial }^2}{\partial {\phi }^2}\nu (\phi ,\rho )\right)\right\}\right],\hfill \\ \hfill & \nu (\phi ,\rho )=(1+\phi )+{\mathcal{Z}}^{-1}\left[\frac{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}{B\left(\delta \right)}\left\{\mathcal{Z}\left((1-\phi )\frac{\partial }{\partial \phi }\nu (\phi ,\rho )+\left(e^{\rho }{\phi }^2\right)\frac{{\partial }^2}{\partial {\phi }^2}\nu (\phi ,\rho )\right)\right\}\right].\hfill \end{array}$$

(42)

The solution in terms of infinite sequence $\nu (\phi ,\rho )$ by means of MDM is:

$$\nu (\phi ,\rho )=\sum _{ȷ=0}^{\infty }{\nu }_{ȷ}(\phi ,\rho ).$$

(43)

$$\begin{array}{cc}\hfill & \sum _{ȷ=0}^{\infty }{\nu }_{ȷ}(\phi ,\rho )=(1+\phi )+{\mathcal{Z}}^{-1}\left[\frac{\left(1-\delta +\delta {\left(\frac{\varrho }{\varsigma }\right)}^{\delta }\right)}{B\left(\delta \right)}\mathcal{Z}\left[(1-\phi )\frac{\partial }{\partial \phi }\nu (\phi ,\rho )+\left(e^{\rho }{\phi }^2\right)\frac{{\partial }^2}{\partial {\phi }^2}\nu (\phi ,\rho )\right]\right].\hfill \end{array}$$

(44)

By comparing Equation (44) on both sides, we get:

$${\nu }_0(\phi ,\rho )=1+\phi ,$$

For $ȷ=0$

$${\nu }_1(\phi ,\rho )=(1+\phi )\frac{1}{B\left(\delta \right)}\left(1-\delta +\frac{\delta {\rho }^{\delta }}{\Gamma (\delta +1)}\right).$$

For $ȷ=1$

$${\nu }_2(\phi ,\rho )=(1+\phi )\frac{1}{{\left(\mathrm{B}\left(\delta \right)\right)}^2}\left({(1-\delta )}^2+\frac{2\delta (1-\delta ){\rho }^{\delta }}{\Gamma (\delta +1)}+\frac{{\delta }^2{\rho }^{2\delta }}{\Gamma (2\delta +1)}\right).$$

For $ȷ=2$

$$\begin{array}{cc}\hfill & {\nu }_3(\phi ,\rho )=(1+\phi )\frac{1}{{\left(\mathrm{B}\left(\delta \right)\right)}^3}\left({(1-\delta )}^3+\frac{3\delta {(1-\delta )}^2{\rho }^{\delta }}{\Gamma (\delta +1)}+\frac{{\delta }^2(1-\delta ){\rho }^{2\delta +1}}{\Gamma (2\delta +2)}+\frac{2{\delta }^2(1-\delta ){\rho }^{2\delta }}{\Gamma (2\delta +1)}+\frac{{\delta }^3{\rho }^{2\delta +1}}{\Gamma (2\delta +2)}\right).\hfill \end{array}$$

The MDM solution remaining components ${\nu }_{ȷ}$ with $(ȷ\ge 3)$ are easily calculated.

$$\nu (\phi ,\rho )=\sum _{ȷ=0}^{\infty }{\nu }_{ȷ}(\phi ,\rho )={\nu }_0(\phi ,\rho )+{\nu }_1(\phi ,\rho )+{\nu }_2(\phi ,\rho )+{\nu }_3(\phi ,\rho )+\cdots .$$

$$\begin{array}{cc}\hfill & \nu (\phi ,\rho )=(1+\phi )+(1+\phi )\frac{1}{B\left(\delta \right)}\left(1-\delta +\frac{\delta {\rho }^{\delta }}{\Gamma (\delta +1)}\right)+(1+\phi )\frac{1}{{\left(\mathrm{B}\left(\delta \right)\right)}^2}\left({(1-\delta )}^2+\frac{2\delta (1-\delta ){\rho }^{\delta }}{\Gamma (\delta +1)}+\frac{{\delta }^2{\rho }^{2\delta }}{\Gamma (2\delta +1)}\right)\hfill \\ \hfill & +(1+\phi )\frac{1}{{\left(\mathrm{B}\left(\delta \right)\right)}^3}\left({(1-\delta )}^3+\frac{3\delta {(1-\delta )}^2{\rho }^{\delta }}{\Gamma (\delta +1)}+\frac{{\delta }^2(1-\delta ){\rho }^{2\delta +1}}{\Gamma (2\delta +2)}+\frac{2{\delta }^2(1-\delta ){\rho }^{2\delta }}{\Gamma (2\delta +1)}+\frac{{\delta }^3{\rho }^{2\delta +1}}{\Gamma (2\delta +2)}\right)+\cdots .\hfill \end{array}$$

The MDM solution at $\delta =1$ is:

$$\nu (\phi ,\rho )=e^{\rho }(1+\phi ).$$

(45)

In Figure 13, represents the actual solution and the 2nd depicts the analytic result, which are in close contact. Figure 14 depicts fractional-order graphs at $\delta =0.8$ and $0.6$ for Problem 5. Similarly, in Figure 15, it is demonstrated that the Problem 5 fractional-order is $\delta =0.4$. The data demonstrate that the proposed method corresponds to the real solution for the given problem.

Figure 13. The exact and analytical solution at $\delta =1$ of Problem 5.

Figure 14. Fractional solution at $\delta =0.8$ and $0.6$ of Problem 5.

Figure 15. Fractional solution at $\delta =0.4$ of Problem 5.

5. Conclusions

In this study, we successfully apply the modified decomposition method to solve the fractional-order Fokker-Planck equations analytically. Five different issues involving fractional-order Fokker-Planck equations were explored to demonstrate the viability of the recommended method. The numerical simulations demonstrate that the outcomes of our method are in close accordance with the exact answer. For getting numerical solutions to fractional-order Fokker-Planck equations, the current method is relatively simple, efficient, and appropriate. The primary advantage of the proposed approach is the series form solution, which rapidly converges to the exact solution.