# Solution of Fractional Partial Differential Equations Using Fractional Power Series Method

Authors: Asif Iqbal Ali; Muhammad Kalim; Adnan Khan
Journal: International Journal of Differential Equations
 (2021)

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## Abstract

In this paper, we are presenting our work where the noninteger order partial differential equation is studied analytically and numerically using the noninteger power series technique, proposed to solve a noninteger differential equation. We are familiar with a coupled system of the nonlinear partial differential equation (NLPDE). Noninteger derivatives are considered in the Caputo operator. The fractional-order power series technique for finding the nonlinear fractional-order partial differential equation is found to be relatively simple in implementation with an application of the direct power series method. We obtained the solution of nonlinear dispersive equations which are used in electromagnetic and optics signal transformation. The proposed approach of using the noninteger power series technique appears to have a good chance of lowering the computational cost of solving such problems significantly. How to paradigm an initial representation plays an important role in the subsequent process, and a few examples are provided to clarify the initial solution collection.


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## Body

## 1. Introduction

Mathematical equations containing two or more independent variables are called partial differential equations (PDEs). The highest derivative in the PDEs is called the order of the equation. The specific solution of the PDEs is a function that solves the equation, which can be verified by substituting the solution into the equation and obtaining an identity. The solution containing all the particular solutions of the equation in question is called the general solution. The term used to express a specific solution for second-order and higher-order nonlinear partial differential equations is called an exact solution. Problems involving the function of many variables, such as heat or sound propagation, static electricity, electrodynamics, and fluid flow, are solved by PDE [1, 2].Generally, the solution with unknown coefficients is assumed to be a power series, and then the solution is substituted into the differential equation to find the recurrence relationship of the coefficients. New applications in fluid dynamics, viscoelasticity, mathematical biology, electrochemistry, and physics have led to the latest development in fractional DE. Using data from experiments, fractional DEs have recently proved to be useful tools for simulating many physical phenomena and fractional PDEs have been suggested for filtering flows in porous media. The examples that have been studied and solved include the space-time fractional diffusion wave equation, fractional advection scattering equation, fractional telegraph equation, fractional KdV equation, and linear nonhomogeneous fractional partial differential equations [2,3].Nowadays, the uses of nonlinear partial (or ordinary) differential equations in the practice of multidisciplinary research are considered to be an important technique for solving real-life problems that have social and public-health implications. Researchers in different scientific areas like biology, physiology, cell biology, chemistry, chemical physics, engineering, and systematic natural sciences are including applied mathematical methods in their practices. These applications are not only useful for technological advances in computer science to study cancers, hereditary diseases, tumor formation, kinematic processes, and DNA communication processes but also in mathematical controls to supply a successful cancer therapy. Indeed, there are many biomedical engineering applications for studying biological problems involving highly nonlinear and sophisticated collaborative processes, where the most promising theory available to conduct such studies is complex dynamical systems theory. The description of this theory almost entirely depends on nonlinear partial differential equations (NLPDE). Power series solution (PSS) technique has been restricted to unravel the linear differential equations, i.e., ordinary differential equations (ODEs) [1, 2] and partial differential equations (PDEs) [4, 5].Linear partial differential equations have traditionally been overcome using the variable separation method because it creates an ODE system that is easier to decipher with PSSM. Examples of them are the spherical harmonics used and the Legendre polynomials in the Bessel equation in cylindrical coordinates or the Laplace equation in spherical coordinates [4, 5]. It is understood that in NLPDE, as we remember, variable separation technology cannot be used. The way to unlock NLPDE is to use approximate analysis techniques to get the answer, that is, semianalytical or nondigital, direct or indirect. Directly, there are methods such as loose operator formalism [6] or inverse dispersion transformation [7]. In a direct way, for instance, the PSS can be used in an asymptotic calculation of the Hirota technique linking a bilinear operator method [8] and the homotopy analysis method (HAM) [9, 10] and the Adomian decomposition technique [11]. The latest approach requires a sequence of expansions with a nonsmall perturbation parameter estimate to change the convergence. This approach is distinct from the standard perturbation theory. Methods, also simpler, to estimate the representation in NLPDE are the Taylor polynomial approximation (TPA) technique [12, 13] and therefore for the power series solution method (PSSM). In each technique, a semianalytical representation is attained to implement the PSS technique. However, no use has been made of the PSSM to decipher the nonlinear ODE [14–17] or NLPDE [18–20]. Because NLPDEs do not have exact solutions in general, it is hard to define the most efficient method to be used; indeed, the only way to establish the efficiencies of the relevant methods involves employing experimental methods.Further information on how to solve NLPDEs by using a point contact, Bäcklund, hodograph, Legendre, or Euler transformations or by applying Lie algebras and groups can be found in reference [21]. There is an immense work on noninteger calculus, and this has been grown exponentially in recent years, with noninteger ODEs becoming commonly used as mathematical prototypes [22, 23]. Recent studies consist of the solution of fractal media [24], viscoelastic materials [23], economics and finance [22, 25], compartment models [26], in porous media [27], epidemiology [28], nerve cell signaling [29], and anomalous diffusion system [30]. From a modeling perspective, the interpretation of the representation as an algebraic countenance involving functions of structure strictures is commonly pursued. Important progress has been made concerning linear noninteger ODEs by expanding proven techniques of elucidation for linear nonfractional ODEs. The Laplace transform technique can be extended to noninteger linear ODEs, with constant coefficients (see [31–33]).The disadvantage of this method depends on the complexity of the Laplace transform of such functions, regardless of whether the equations are homogeneous or not, or by inverting the Laplace transform based on a calibrated computable function. In the noninteger linear ODE, with variable coefficients, the Laplace transform method has an additional limitation, that is, the Laplace transform of the product of the function is only available in special cases. In nonfractional regular linear ODEs, with different coefficients, the series expansion method is considered to help clarify the problem. Examples of power series can be seen in [34–48]. The description of this theory is also related to the use of the residual power series method [49] to solve space-time fractional PDEs. The comparison of Aboodh transformation and differential transformation method (DTM) numerically in result solution obtained is compared the solution by DTM is rapid convergent [50]. Several papers have been dedicated to the application of series expansion approaches to linear noninteger ODEs which basically is the subject of this article.


## 2. Fundamental Concepts

Definition 1.
The Caputo noninteger derivative operator of orderμ with respect to “t” is defined in the subsequent arrangement.(1)Dtμhu=Dμhu=Jn−μDnhu,Dμhu=Γn−μ−1∫0uu−t−μ+n−1hntdt,where μ>0,u>0,n≥μ>n−1,n∈ℕ.
In a comparable manner of nonfractional-order differentiation, the Caputo noninteger derivative operator is a linear operation.(2)Dμαfu+βgu=αDμfu+βDμgu,where α and β are constants. We have Dμk=0 for the Caputo’s derivative, if k is constant.(3)Dμum=0,form∈ℕ0andm<μ,Γm+1Γm−μ+1um−μ,form∈ℕ0andm≥μ.
We practice the maximum functionμ to signify the least integer larger than or equivalent to μ, and ℕ0=0,1,2,…. Evoke that for μ∈ℕ, the Caputo differential operator corresponds with the normal differential operator of nonfractional order.Definition 2.
For the variableu and coefficients ann=0,1,…,∞, if u>u0, the fractional power series (FPS) about the point u0 is defined as(4)∑n=0∞anu−u0nμ=a0+u−u0µa1+u−u02µa2+u−u03µa3+⋯,where μ>0,m≥μ>m−1,m∈ℕ+.Theorem 1.
Let the radius of convergence for the function with fractional power series (FPS) representationhu=∑n=0∞anunμ, 0≤u<R, be greater than zero (i.e., R>0).Then, form∈ℕ+ and m−1<μ≤m, the following expression holds true:(5)Dμhu=∑m=1∞amΓmμ+1um−1μΓm−1μ+1.Proof.
From the linearity concept of Caputo derivative and the idea of power series derivative, we have(6)Dμhu=Dμ∑m=0∞amumμ=Dμa0+a1uμ+a2u2μ+⋯=Dμa0+a1Dμuμ+a2Dμu2μ+⋯=a1Dμuμ+a2Dμu2μ+⋯Dμhu=∑m=1∞amDμumμ.
From the power rule of the Caputo derivative, we have(7)Dμumμ=0,formμ∈ℕ0andmμ<μ,Γmμ+1Γm−1μ+1um−1μ,formμ∈ℕ0andmμ≥μ,ℕ0=0,1,2,…,μ∈Z,μ≥μ.
From equations (3) and (5), we arrive at(8)Dμhu=∑m=1∞amDμumμ=∑m=1∞amΓmμ+1Γm−1μ+1um−1μ,where hu=∑m=0∞amumμ.


### 2.1. Application

In order to validate the high degree of efficiency and precision of the projected FPS approach for unraveling fractional-order systems, numerical forms and instances are pragmatic. The reader can discover a sketch and applications for this technique in [42]. Computations were accomplished by using MATLAB. Examples are taken from [32].Example 1.
Consider the succeeding noninteger order structure:(9)D0μux,t=1+vdudx+u,D0μvx,t=1−udvdx−v,subject to the initial conditions(10)u0,x=e−x,v0,x=ex.
The closed-form result of this arrangement whenμ=1 is(11)ut,x=et−x,vt,x=ex−t.
Solution:(12)Letut,x=∑k=0∞akxtμk=a0x+tμa1x+t2μa2x+⋯+tnμanx+⋯,(13)vt,x=∑k=0∞bkxtμk=b0x+tμb1x+t2μb2x+⋯+tnμbnx+⋯.
So using Theorem1, equations (12) and (13) become(14)Dμux,t=Dμ∑k=0∞akxtμk,OrDμux,t=∑k=1∞akxΓkμ+1Γk−1μ+1tμk−1,(15)Dμvx,t=Dμ∑k=0∞bkxtμk,OrDμvx,t=∑k=1∞Γkμ+1Γk−1μ+1bkxtμkk−1.
Similarly,(16)S∂ux,t∂x=∑k=0∞∂akx∂xtμk,(17)∂vx,t∂x=∑k=0∞∂bkx∂xtμk.
Apply (12)–(17) on the first and second equation of system (9), respectively, system (9) becomes(18)∑k=1∞akΓkμ+1Γk−1μ+1tμk−1=1+∑k=0∞tμkak+∑k=0∞tμk∂ak∂x∑k=0∞bktμk,∑k=1∞Γkμ+1Γk−1μ+1bktμk−1=1+∑k=0∞tμkbk+∑k=0∞tμk∂bk∂x∑k=0∞aktμk.
Comparing the different powers oft to both sides of system (18), for t0,(19)a1x=1Γμ+11+b0∂a0∂x+a0,b1x=1Γμ+11−a0∂b0∂x−b0.
Buta0=u0,x=e−x, v0,x=b0=ex, ∂a0/∂x=−e−x, and ∂b0/∂x=−ex, so (19) becomes(20)a1x=1Γμ+11+ex−e−x+e−x=e−xΓμ+1,b1x=1Γμ+11−e−xex−ex=−exΓμ+1.
Fortμ,(21)a2x=Γμ+1Γ2μ+1a1+b1∂a0∂x+b0∂a1∂x,b2x=Γμ+1Γ2μ+1−b1−a1∂b0∂x−a0∂b1∂x.
But∂a0/∂x=−e−x, ∂b0/∂x=ex, ∂a1/∂x=e−x/Γμ+1, ∂b1/∂x=−ex/Γμ+1, so (21) becomes(22)a2x=Γμ+1Γ2μ+1e−xΓμ+1+ex−e−xΓμ+1−exΓμ+1−e−x=e−xΓ2μ+1,b2x=Γμ+1Γ2μ+1exΓμ+1−e−x−exΓμ+1−e−xΓμ+1ex=exΓ2μ+1.
Fort2μ,(23)a3x=Γ2μ+1Γ3μ+1a2+b2∂a0∂x+b0∂a2∂x+b1∂a1∂x,b3x=Γ2μ+1Γ3μ+1−b2−a2∂b0∂x−a0∂b2∂x−a1∂b1∂x.
But∂a0/∂x=−e−x, ∂b0/∂x=ex, ∂a1/∂x=e−x/Γμ+1, ∂b1/∂x=−ex/Γμ+1,∂a2/∂x=−e−x/Γ2μ+1, ∂b2/∂x=ex/Γ2μ+1, so (26) becomes(24)a3x=Γ2μ+1Γ3μ+1e−xΓ2μ+1+−exΓμ+1−e−xΓμ+1+exΓ2μ+1−e−x+ex−e−xΓ2μ+1,b3x=Γ2μ+1Γ3μ+1−exΓ2μ+1−e−xΓ2μ+1ex−e−xexΓ2μ+1−e−xΓμ+1−exΓμ+1,or(25)a3x=1Γ3μ+1e−x−2+Γ2μ+1Γμ+12,b3x=1Γ3μ+1−ex−2+Γ2μ+1Γμ+12.
Fort3μ,(26)a4x=Γ3μ+1Γ4μ+1a3+∂a3∂xb0+∂a2∂xb1+∂a1∂xb2+∂a0∂xb3,b4x=Γ3μ+1Γ4μ+1−b3−∂b3∂xa0−∂b2∂xa1−∂b1∂xa2−∂b0∂xa3,a4x=−0.5Γ4μ+1Γ2μ+1/2Γμ+12π−6πe−xΓμμΓ2μ+12+4Γ2μ+12Γμμπ+e−xΓμ+14Γμ+34Γμ+12216μ−Γμ+14Γμ+34Γμ+12216μΓμμ,b4x=−0.5Γ4μ+1Γ2μ+1/2Γμ+12π−6πe−xΓμμΓ2μ+12+4Γ2μ+12Γμμπ+e−xΓμ+14Γμ+34Γμ+12216μ+Γμ+14Γμ+34Γμ+12216μΓμμ.
In a similar way, we can compare the values fort4μ, t5μ, …
So,(27)ut,x=e−x+e−xΓμ+1tμ+e−xΓ2μ+1t2μ+1Γ3μ+1e−x−2+Γ2μ+1Γμ+12t3μ+−0.5Γ4μ+1Γ2μ+1/2Γμ+12π−6πe−xΓμμΓ2μ+12+4Γ2μ+12Γμμπ+e−xΓμ+14Γμ+34Γμ+12216μ−Γμ+14Γμ+34Γμ+12216μΓμμt4μ+⋯,(28)vt,x=e−x+e−xΓμ+1tμ+e−xΓ2μ+1t2μ+1Γ3μ+1e−x−2+Γ2μ+1Γμ+12t3μ+−0.5Γ4μ+1Γ2μ+1/2Γμ+12π−6πe−xΓμμΓ2μ+12+4Γ2μ+12Γμμπ+e−xΓμ+14Γμ+34Γμ+12216μ+Γμ+14Γμ+34Γμ+12216μΓμμt4μ+⋯,
Ifμ=1, then equations (27) and (28) give(29)ut,x=e−x1+tΓ2+t2Γ3+t3Γ4+t4Γ5+…=et−x,vt,x=ex1−tΓ2+t2Γ3−t3Γ4+t4Γ5+…=e−t+x.Example 2.
Consider the following fractional-order system:(30)D0μut,x=2−2vdudx+u,D0μvt,x=3+3udvdx−v,subject to the initial conditions(31)u0,x=ex,v0,x=e−x.
The closed-form result of this system whenμ=1 is(32)ut,x=ex+t,vt,x=e−x−t.
Solution:
Let(33)ut,x=∑k=0∞akxtμk=a0+tμa1+t2μa2+⋯+tnμan+⋯,(34)vt,x=∑k=0∞bkxtμk=b0+tμb1+t2μb2+⋯+tnμbn+⋯.
Apply (12)–(17) on the first and second equations of the system (30), respectively. So, system (30) becomes(35)∑k=1∞akΓkμ+1Γk−1μ+1tμk−1=2−2∑k=0∞bktμk∑k=0∞∂ak∂xtμk+∑k=0∞aktμk,∑k=1∞bkΓkμ+1Γk−1μ+1tμk−1=3+3∑k=0∞aktμk∑k=0∞∂bk∂xtμk+∑k=0∞bktμk.
Comparing the different powers oft to both sides of the system (35), for t0,(36)a1x=1Γμ+12+a0−2b0∂a0∂x,b1x=1Γμ+13−b0+3a0∂b0∂x.
After substituting the valuesa0b0, ∂a0/∂x, and ∂b0/∂x, we obtain(37)a1x=exΓμ+1,b1x=−e−xΓμ+1.
Fortμ,(38)a2x=Γμ+1Γ2μ+1a1−2b0∂a1∂x−2b1∂a0∂x,b2x=Γμ+1Γ2μ+1−b1+3a1∂b0∂x+3a0∂b1∂x.
After substituting the valuesa0a1b0, b1, ∂a0/∂x, ∂a1/∂x, ∂b0/∂x, and ∂b1/∂x, we obtain(39)a2x=exΓ2μ+1,b2x=e−xΓ2μ+1.
Fort2μ,(40)a3x=Γ2μ+1Γ3μ+1a2−2∂a2∂xb0−2∂a1∂xb1−2∂a0∂xb2,b3x=Γ2μ+1Γ3μ+1−b2+3∂b2∂xa0+3∂b1∂xa1+3∂b0∂xa2.
After substituting the valuesa0a1, a2b0, b1b2, ∂a0/∂x, ∂a1/∂x, ∂a2/∂x, ∂b0/∂x∂b1/∂x, and ∂b2/∂x, we obtain(41)a3x=1Γ3μ+1ex−4+2Γ2μ+1Γμ+12,b3x=1Γ3μ+1−e−x−6+3Γ2μ+1Γμ+12.
Fort3μ,(42)a4x=Γ3μ+1Γ4μ+1a3−2∂a3∂xb0−2∂a2∂xb1−2∂a1∂xb2−2∂a0∂xb3,b4x=Γ3μ+1Γ4μ+1−b3+3∂b3∂xa0+3∂b2∂xa1+3∂b1∂xa2+3∂b0∂xa3.
After inserting the valuesa0a1, a2a3b0, b1b2b3, ∂a0/∂x, ∂a1/∂x, ∂a2/∂x∂a3/∂x, ∂b0/∂x∂b1/∂x, ∂b2/∂x, and ∂b3/∂x, we obtain(43)a4x=−1Γ4μ+1Γ2μ+1/2Γμ+12π−13πexΓμμΓ2μ+12+4Γ2μ+12Γμμπ+3exΓμ+14Γμ+34Γμ+12216μ−Γμ+14Γμ+34Γμ+12216μΓμμ,b4x=−0.5Γ4μ+1Γ2μ+1/2Γμ+12π−26πe−xΓμμΓ2μ+12−12Γ2μ+12Γμμπ+6e−xΓμ+14Γμ+34Γμ+12216μ+3Γμ+14Γμ+34Γμ+12216μΓμμ.
In a similar way, we can compare the values fort4μ, t5μ, …
So, by inserting the values ofa0, a1, a2, a3, a4 , b0, b1, b2, b3, and b4 in (33 and 34), we have(44)ut,x=ex+exΓμ+1tμ+exΓ2μ+1t2μ+1Γ3μ+1ex−4+2Γ2μ+1Γμ+12t3μ+−1Γ4μ+1Γ2μ+1/2Γμ+12π−13πexΓμμΓ2μ+12+4Γ2μ+12Γμμπ+3exΓμ+14Γμ+34Γμ+12216μ−Γμ+14Γμ+34Γμ+12216μΓμμt4μ+⋯,(45)vt,x=e−x−e−xΓμ+1tμ+e−xΓ2μ+1t2μ+1Γ3μ+1−e−x−6+3Γ2μ+1Γμ+12t3μ+−0.5Γ4μ+1Γ2μ+1/2Γμ+12π−26πe−xΓμμΓ2μ+12−12Γ2μ+12Γμμπ+6e−xΓμ+14Γμ+34Γμ+12216μ+3Γμ+14Γμ+34Γμ+12216μΓμμt4μ+⋯.
Ifμ=1, then equations (44) and (45) give(46)ut,x=1+tΓ2+t2Γ3+t3Γ4+t4Γ5+⋯ex=ex+t,vt,x=1−tΓ2+t2Γ3−t3Γ4+t4Γ5+⋯e−x=e−x−t.Example 3.
Consider the following fractional-order system:(47)D0μut,x=2+3u−vdudx,D0μvt,x=2−3v+udvdx,subject to the initial conditions(48)u0,x=e2x,v0,x=e−2x.
The exact solution of this system whenμ=1 is(49)ut,x=e2x+3t,vt,x=e−2x−3t.
Solution:
Let(50)ux,t=∑k=0∞akxtμk=a0x+tμa1x+t2μa2x+⋯+tnμanx+⋯,(51)vx,t=∑k=0∞bkxtμk=b0x+tμb1x+t2μb2x+⋯+tnμbnx+⋯.
Apply (12)–(17) on the first and second equations of the system (47), respectively, system (47) becomes(52)∑k=1∞akΓkμ+1Γk−1μ+1tμk−1=2−∑k=0∞bktμk∑k=0∞∂ak∂xtμk+3∑k=0∞aktμk,∑k=1∞bkΓkμ+1Γk−1μ+1tμk−1=2+∑k=0∞aktμk∑k=0∞∂bk∂xtμk+3∑k=0∞bktμk.
Comparing the different powers oft to both sides of the system (52), for t0,(53)a1x=1Γμ+12−b0∂a0∂x+3a0,b1x=1Γμ+12+a0∂b0∂x−3b0.
After substituting the valuesa0xb0x, ∂a0x/∂x, and ∂b0x/∂x, we obtain(54)a1x=3e2xΓμ+1,b1x=−3e−2xΓμ+1.
Fortμ,(55)a2x=Γμ+1Γ2μ+13a1−∂a0∂xb1−∂a1∂xb0,b2x=Γμ+1Γ2μ+1−3b1+∂b0∂xa1+∂b1∂xa0.
After substituting the valuesa0a1b0, b1, ∂a0/∂x, ∂a1/∂x, ∂b0/∂x, and ∂b1/∂x, we obtain(56)a2x=9e2xΓ2μ+1,b2x=9e−2xΓ2μ+1.
Fort2μ,(57)a3x=Γ2μ+1Γ3μ+13a2−∂a2∂xb0−∂a1∂xb1−∂a0∂xb2,b3x=Γ2μ+1Γ3μ+1−3b2+∂b2∂xa0+∂b1∂xa1+∂b0∂xa2.
After substituting the valuesa0a1, a2b0, b1b2, ∂a0/∂x, ∂a1/∂x, ∂a2/∂x, ∂b0/∂x∂b1/∂x, and ∂b2/∂x, we obtain(58)a3x=9Γ3μ+13e2x−4+2Γ2μ+1Γμ+12,b3x=9Γ3μ+1−3e−2x−4+2Γ2μ+1Γμ+12.
Fort3μ,(59)a4x=Γ3μ+1Γ4μ+13a3−∂a3∂xb0−∂a2∂xb1−∂a1∂xb2−∂a0∂xb3,b4x=Γ3μ+1Γ4μ+1−3b3+∂b3∂xa0+∂b2∂xa1+∂b1∂xa2+∂b0∂xa3.
After inserting the valuesa0a1, a2a3b0, b1b2b3, ∂a0/∂x, ∂a1/∂x, ∂a2/∂x∂a3/∂x, ∂b0/∂x∂b1/∂x, ∂b2/∂x, and ∂b3/∂x, we obtain(60)a4x=1Γ4μ+1Γ2μ+1/2Γμ+12π917πe2xΓμμΓ2μ+12−12Γ2μ+12Γμμπ−2e2xΓμ+14Γμ+34Γμ+12216μ+3Γμ+14Γμ+34Γμ+12216μΓμμ,b4x=1Γ4μ+1Γ2μ+1/2Γμ+12π9−17πe2xΓμμΓ2μ+12−12Γ2μ+12Γμμπ+2e2xΓμ+14Γμ+34Γμ+12216μ+3Γμ+14Γμ+34Γμ+12216μΓμμ.
In a similar way, we can compare the values fort4μ, t5μ, …...
So, by inserting the values ofa0, a1, a2, a3, a4 , b0, b1, b2, b3, and b4 in (51) and (52), we have(61)ux,t=e2x+3e2xΓμ+1tμ+9e2xΓ2μ+1t2μ+9Γ3μ+13e2x−4+2Γ2μ+1Γμ+12t3μ+1Γ4μ+1Γ2μ+1/2Γμ+12π917πe2xΓμμΓ2μ+12−12Γ2μ+12Γμμπ−2e2xΓμ+14Γμ+34Γμ+12216μ+3Γμ+14Γμ+34Γμ+12216μΓμμt4μ+⋯,(62)vx,t=e−2x−3e−2xΓμ+1tμ+9e−2xΓ2μ+1t2μ+9Γ3μ+1−3e−2x−4+2Γ2μ+1Γμ+12t3μ+1Γ4μ+1Γ2μ+1/2Γμ+12π9−17πe2xΓμμΓ2μ+12−12Γ2μ+12Γμμπ+2e2xΓμ+14Γμ+34Γμ+12216μ+3Γμ+14Γμ+34Γμ+12216μΓμμt4μ+⋯.
Ifμ=1, then equations (61) and (62) give(63)ux,t=e2x1+3tΓ2+9t2Γ3+27t3Γ4+81t4Γ5+⋯=e2x+3t,vx,t=e−2x1−3tΓ2+9t2Γ3−27t3Γ4+81t4Γ5−⋯=e−2x−3t.


## 2.1. Application

In order to validate the high degree of efficiency and precision of the projected FPS approach for unraveling fractional-order systems, numerical forms and instances are pragmatic. The reader can discover a sketch and applications for this technique in [42]. Computations were accomplished by using MATLAB. Examples are taken from [32].Example 1.
Consider the succeeding noninteger order structure:(9)D0μux,t=1+vdudx+u,D0μvx,t=1−udvdx−v,subject to the initial conditions(10)u0,x=e−x,v0,x=ex.
The closed-form result of this arrangement whenμ=1 is(11)ut,x=et−x,vt,x=ex−t.
Solution:(12)Letut,x=∑k=0∞akxtμk=a0x+tμa1x+t2μa2x+⋯+tnμanx+⋯,(13)vt,x=∑k=0∞bkxtμk=b0x+tμb1x+t2μb2x+⋯+tnμbnx+⋯.
So using Theorem1, equations (12) and (13) become(14)Dμux,t=Dμ∑k=0∞akxtμk,OrDμux,t=∑k=1∞akxΓkμ+1Γk−1μ+1tμk−1,(15)Dμvx,t=Dμ∑k=0∞bkxtμk,OrDμvx,t=∑k=1∞Γkμ+1Γk−1μ+1bkxtμkk−1.
Similarly,(16)S∂ux,t∂x=∑k=0∞∂akx∂xtμk,(17)∂vx,t∂x=∑k=0∞∂bkx∂xtμk.
Apply (12)–(17) on the first and second equation of system (9), respectively, system (9) becomes(18)∑k=1∞akΓkμ+1Γk−1μ+1tμk−1=1+∑k=0∞tμkak+∑k=0∞tμk∂ak∂x∑k=0∞bktμk,∑k=1∞Γkμ+1Γk−1μ+1bktμk−1=1+∑k=0∞tμkbk+∑k=0∞tμk∂bk∂x∑k=0∞aktμk.
Comparing the different powers oft to both sides of system (18), for t0,(19)a1x=1Γμ+11+b0∂a0∂x+a0,b1x=1Γμ+11−a0∂b0∂x−b0.
Buta0=u0,x=e−x, v0,x=b0=ex, ∂a0/∂x=−e−x, and ∂b0/∂x=−ex, so (19) becomes(20)a1x=1Γμ+11+ex−e−x+e−x=e−xΓμ+1,b1x=1Γμ+11−e−xex−ex=−exΓμ+1.
Fortμ,(21)a2x=Γμ+1Γ2μ+1a1+b1∂a0∂x+b0∂a1∂x,b2x=Γμ+1Γ2μ+1−b1−a1∂b0∂x−a0∂b1∂x.
But∂a0/∂x=−e−x, ∂b0/∂x=ex, ∂a1/∂x=e−x/Γμ+1, ∂b1/∂x=−ex/Γμ+1, so (21) becomes(22)a2x=Γμ+1Γ2μ+1e−xΓμ+1+ex−e−xΓμ+1−exΓμ+1−e−x=e−xΓ2μ+1,b2x=Γμ+1Γ2μ+1exΓμ+1−e−x−exΓμ+1−e−xΓμ+1ex=exΓ2μ+1.
Fort2μ,(23)a3x=Γ2μ+1Γ3μ+1a2+b2∂a0∂x+b0∂a2∂x+b1∂a1∂x,b3x=Γ2μ+1Γ3μ+1−b2−a2∂b0∂x−a0∂b2∂x−a1∂b1∂x.
But∂a0/∂x=−e−x, ∂b0/∂x=ex, ∂a1/∂x=e−x/Γμ+1, ∂b1/∂x=−ex/Γμ+1,∂a2/∂x=−e−x/Γ2μ+1, ∂b2/∂x=ex/Γ2μ+1, so (26) becomes(24)a3x=Γ2μ+1Γ3μ+1e−xΓ2μ+1+−exΓμ+1−e−xΓμ+1+exΓ2μ+1−e−x+ex−e−xΓ2μ+1,b3x=Γ2μ+1Γ3μ+1−exΓ2μ+1−e−xΓ2μ+1ex−e−xexΓ2μ+1−e−xΓμ+1−exΓμ+1,or(25)a3x=1Γ3μ+1e−x−2+Γ2μ+1Γμ+12,b3x=1Γ3μ+1−ex−2+Γ2μ+1Γμ+12.
Fort3μ,(26)a4x=Γ3μ+1Γ4μ+1a3+∂a3∂xb0+∂a2∂xb1+∂a1∂xb2+∂a0∂xb3,b4x=Γ3μ+1Γ4μ+1−b3−∂b3∂xa0−∂b2∂xa1−∂b1∂xa2−∂b0∂xa3,a4x=−0.5Γ4μ+1Γ2μ+1/2Γμ+12π−6πe−xΓμμΓ2μ+12+4Γ2μ+12Γμμπ+e−xΓμ+14Γμ+34Γμ+12216μ−Γμ+14Γμ+34Γμ+12216μΓμμ,b4x=−0.5Γ4μ+1Γ2μ+1/2Γμ+12π−6πe−xΓμμΓ2μ+12+4Γ2μ+12Γμμπ+e−xΓμ+14Γμ+34Γμ+12216μ+Γμ+14Γμ+34Γμ+12216μΓμμ.
In a similar way, we can compare the values fort4μ, t5μ, …
So,(27)ut,x=e−x+e−xΓμ+1tμ+e−xΓ2μ+1t2μ+1Γ3μ+1e−x−2+Γ2μ+1Γμ+12t3μ+−0.5Γ4μ+1Γ2μ+1/2Γμ+12π−6πe−xΓμμΓ2μ+12+4Γ2μ+12Γμμπ+e−xΓμ+14Γμ+34Γμ+12216μ−Γμ+14Γμ+34Γμ+12216μΓμμt4μ+⋯,(28)vt,x=e−x+e−xΓμ+1tμ+e−xΓ2μ+1t2μ+1Γ3μ+1e−x−2+Γ2μ+1Γμ+12t3μ+−0.5Γ4μ+1Γ2μ+1/2Γμ+12π−6πe−xΓμμΓ2μ+12+4Γ2μ+12Γμμπ+e−xΓμ+14Γμ+34Γμ+12216μ+Γμ+14Γμ+34Γμ+12216μΓμμt4μ+⋯,
Ifμ=1, then equations (27) and (28) give(29)ut,x=e−x1+tΓ2+t2Γ3+t3Γ4+t4Γ5+…=et−x,vt,x=ex1−tΓ2+t2Γ3−t3Γ4+t4Γ5+…=e−t+x.Example 2.
Consider the following fractional-order system:(30)D0μut,x=2−2vdudx+u,D0μvt,x=3+3udvdx−v,subject to the initial conditions(31)u0,x=ex,v0,x=e−x.
The closed-form result of this system whenμ=1 is(32)ut,x=ex+t,vt,x=e−x−t.
Solution:
Let(33)ut,x=∑k=0∞akxtμk=a0+tμa1+t2μa2+⋯+tnμan+⋯,(34)vt,x=∑k=0∞bkxtμk=b0+tμb1+t2μb2+⋯+tnμbn+⋯.
Apply (12)–(17) on the first and second equations of the system (30), respectively. So, system (30) becomes(35)∑k=1∞akΓkμ+1Γk−1μ+1tμk−1=2−2∑k=0∞bktμk∑k=0∞∂ak∂xtμk+∑k=0∞aktμk,∑k=1∞bkΓkμ+1Γk−1μ+1tμk−1=3+3∑k=0∞aktμk∑k=0∞∂bk∂xtμk+∑k=0∞bktμk.
Comparing the different powers oft to both sides of the system (35), for t0,(36)a1x=1Γμ+12+a0−2b0∂a0∂x,b1x=1Γμ+13−b0+3a0∂b0∂x.
After substituting the valuesa0b0, ∂a0/∂x, and ∂b0/∂x, we obtain(37)a1x=exΓμ+1,b1x=−e−xΓμ+1.
Fortμ,(38)a2x=Γμ+1Γ2μ+1a1−2b0∂a1∂x−2b1∂a0∂x,b2x=Γμ+1Γ2μ+1−b1+3a1∂b0∂x+3a0∂b1∂x.
After substituting the valuesa0a1b0, b1, ∂a0/∂x, ∂a1/∂x, ∂b0/∂x, and ∂b1/∂x, we obtain(39)a2x=exΓ2μ+1,b2x=e−xΓ2μ+1.
Fort2μ,(40)a3x=Γ2μ+1Γ3μ+1a2−2∂a2∂xb0−2∂a1∂xb1−2∂a0∂xb2,b3x=Γ2μ+1Γ3μ+1−b2+3∂b2∂xa0+3∂b1∂xa1+3∂b0∂xa2.
After substituting the valuesa0a1, a2b0, b1b2, ∂a0/∂x, ∂a1/∂x, ∂a2/∂x, ∂b0/∂x∂b1/∂x, and ∂b2/∂x, we obtain(41)a3x=1Γ3μ+1ex−4+2Γ2μ+1Γμ+12,b3x=1Γ3μ+1−e−x−6+3Γ2μ+1Γμ+12.
Fort3μ,(42)a4x=Γ3μ+1Γ4μ+1a3−2∂a3∂xb0−2∂a2∂xb1−2∂a1∂xb2−2∂a0∂xb3,b4x=Γ3μ+1Γ4μ+1−b3+3∂b3∂xa0+3∂b2∂xa1+3∂b1∂xa2+3∂b0∂xa3.
After inserting the valuesa0a1, a2a3b0, b1b2b3, ∂a0/∂x, ∂a1/∂x, ∂a2/∂x∂a3/∂x, ∂b0/∂x∂b1/∂x, ∂b2/∂x, and ∂b3/∂x, we obtain(43)a4x=−1Γ4μ+1Γ2μ+1/2Γμ+12π−13πexΓμμΓ2μ+12+4Γ2μ+12Γμμπ+3exΓμ+14Γμ+34Γμ+12216μ−Γμ+14Γμ+34Γμ+12216μΓμμ,b4x=−0.5Γ4μ+1Γ2μ+1/2Γμ+12π−26πe−xΓμμΓ2μ+12−12Γ2μ+12Γμμπ+6e−xΓμ+14Γμ+34Γμ+12216μ+3Γμ+14Γμ+34Γμ+12216μΓμμ.
In a similar way, we can compare the values fort4μ, t5μ, …
So, by inserting the values ofa0, a1, a2, a3, a4 , b0, b1, b2, b3, and b4 in (33 and 34), we have(44)ut,x=ex+exΓμ+1tμ+exΓ2μ+1t2μ+1Γ3μ+1ex−4+2Γ2μ+1Γμ+12t3μ+−1Γ4μ+1Γ2μ+1/2Γμ+12π−13πexΓμμΓ2μ+12+4Γ2μ+12Γμμπ+3exΓμ+14Γμ+34Γμ+12216μ−Γμ+14Γμ+34Γμ+12216μΓμμt4μ+⋯,(45)vt,x=e−x−e−xΓμ+1tμ+e−xΓ2μ+1t2μ+1Γ3μ+1−e−x−6+3Γ2μ+1Γμ+12t3μ+−0.5Γ4μ+1Γ2μ+1/2Γμ+12π−26πe−xΓμμΓ2μ+12−12Γ2μ+12Γμμπ+6e−xΓμ+14Γμ+34Γμ+12216μ+3Γμ+14Γμ+34Γμ+12216μΓμμt4μ+⋯.
Ifμ=1, then equations (44) and (45) give(46)ut,x=1+tΓ2+t2Γ3+t3Γ4+t4Γ5+⋯ex=ex+t,vt,x=1−tΓ2+t2Γ3−t3Γ4+t4Γ5+⋯e−x=e−x−t.Example 3.
Consider the following fractional-order system:(47)D0μut,x=2+3u−vdudx,D0μvt,x=2−3v+udvdx,subject to the initial conditions(48)u0,x=e2x,v0,x=e−2x.
The exact solution of this system whenμ=1 is(49)ut,x=e2x+3t,vt,x=e−2x−3t.
Solution:
Let(50)ux,t=∑k=0∞akxtμk=a0x+tμa1x+t2μa2x+⋯+tnμanx+⋯,(51)vx,t=∑k=0∞bkxtμk=b0x+tμb1x+t2μb2x+⋯+tnμbnx+⋯.
Apply (12)–(17) on the first and second equations of the system (47), respectively, system (47) becomes(52)∑k=1∞akΓkμ+1Γk−1μ+1tμk−1=2−∑k=0∞bktμk∑k=0∞∂ak∂xtμk+3∑k=0∞aktμk,∑k=1∞bkΓkμ+1Γk−1μ+1tμk−1=2+∑k=0∞aktμk∑k=0∞∂bk∂xtμk+3∑k=0∞bktμk.
Comparing the different powers oft to both sides of the system (52), for t0,(53)a1x=1Γμ+12−b0∂a0∂x+3a0,b1x=1Γμ+12+a0∂b0∂x−3b0.
After substituting the valuesa0xb0x, ∂a0x/∂x, and ∂b0x/∂x, we obtain(54)a1x=3e2xΓμ+1,b1x=−3e−2xΓμ+1.
Fortμ,(55)a2x=Γμ+1Γ2μ+13a1−∂a0∂xb1−∂a1∂xb0,b2x=Γμ+1Γ2μ+1−3b1+∂b0∂xa1+∂b1∂xa0.
After substituting the valuesa0a1b0, b1, ∂a0/∂x, ∂a1/∂x, ∂b0/∂x, and ∂b1/∂x, we obtain(56)a2x=9e2xΓ2μ+1,b2x=9e−2xΓ2μ+1.
Fort2μ,(57)a3x=Γ2μ+1Γ3μ+13a2−∂a2∂xb0−∂a1∂xb1−∂a0∂xb2,b3x=Γ2μ+1Γ3μ+1−3b2+∂b2∂xa0+∂b1∂xa1+∂b0∂xa2.
After substituting the valuesa0a1, a2b0, b1b2, ∂a0/∂x, ∂a1/∂x, ∂a2/∂x, ∂b0/∂x∂b1/∂x, and ∂b2/∂x, we obtain(58)a3x=9Γ3μ+13e2x−4+2Γ2μ+1Γμ+12,b3x=9Γ3μ+1−3e−2x−4+2Γ2μ+1Γμ+12.
Fort3μ,(59)a4x=Γ3μ+1Γ4μ+13a3−∂a3∂xb0−∂a2∂xb1−∂a1∂xb2−∂a0∂xb3,b4x=Γ3μ+1Γ4μ+1−3b3+∂b3∂xa0+∂b2∂xa1+∂b1∂xa2+∂b0∂xa3.
After inserting the valuesa0a1, a2a3b0, b1b2b3, ∂a0/∂x, ∂a1/∂x, ∂a2/∂x∂a3/∂x, ∂b0/∂x∂b1/∂x, ∂b2/∂x, and ∂b3/∂x, we obtain(60)a4x=1Γ4μ+1Γ2μ+1/2Γμ+12π917πe2xΓμμΓ2μ+12−12Γ2μ+12Γμμπ−2e2xΓμ+14Γμ+34Γμ+12216μ+3Γμ+14Γμ+34Γμ+12216μΓμμ,b4x=1Γ4μ+1Γ2μ+1/2Γμ+12π9−17πe2xΓμμΓ2μ+12−12Γ2μ+12Γμμπ+2e2xΓμ+14Γμ+34Γμ+12216μ+3Γμ+14Γμ+34Γμ+12216μΓμμ.
In a similar way, we can compare the values fort4μ, t5μ, …...
So, by inserting the values ofa0, a1, a2, a3, a4 , b0, b1, b2, b3, and b4 in (51) and (52), we have(61)ux,t=e2x+3e2xΓμ+1tμ+9e2xΓ2μ+1t2μ+9Γ3μ+13e2x−4+2Γ2μ+1Γμ+12t3μ+1Γ4μ+1Γ2μ+1/2Γμ+12π917πe2xΓμμΓ2μ+12−12Γ2μ+12Γμμπ−2e2xΓμ+14Γμ+34Γμ+12216μ+3Γμ+14Γμ+34Γμ+12216μΓμμt4μ+⋯,(62)vx,t=e−2x−3e−2xΓμ+1tμ+9e−2xΓ2μ+1t2μ+9Γ3μ+1−3e−2x−4+2Γ2μ+1Γμ+12t3μ+1Γ4μ+1Γ2μ+1/2Γμ+12π9−17πe2xΓμμΓ2μ+12−12Γ2μ+12Γμμπ+2e2xΓμ+14Γμ+34Γμ+12216μ+3Γμ+14Γμ+34Γμ+12216μΓμμt4μ+⋯.
Ifμ=1, then equations (61) and (62) give(63)ux,t=e2x1+3tΓ2+9t2Γ3+27t3Γ4+81t4Γ5+⋯=e2x+3t,vx,t=e−2x1−3tΓ2+9t2Γ3−27t3Γ4+81t4Γ5−⋯=e−2x−3t.


## 3. Results and Discussion

Tables1and 2 show the values of ut,x and vt,x for Example 1 with different values of x, t, and μ. Afterward, the graphical simulations are drawn with the help of MATLAB. The graphs are shown in 2D and 3D, and these can be seen in Figures 1 and 2. Tables 3 and 4 elaborate the values of ut,x and vt,x for Example 2 and graphical imitations can be seen in Figures 3 and 4 for different values of t, x, and μ. Besides these, Tables 5 and 6 represent the values of t, x, and μ and simulations can be viewed in Figures 5 and 6.Table 1
Numerical values foru (t, x) of equation (27) for different values of t and x.
xtμ=1μ=0.8μ=0.6μ=0.4μ=0.20.20.251.0512641451.1778004831.3766470531.6900579852.1908960070.501.3496264761.5527829621.8256547732.1784056872.5981777880.751.7314076621.9957458622.3053608902.6367536072.91355918512.2173957892.5243611622.8336317703.0913143443.1836703710.50.250.7787956340.8712983531.0100066241.1995020451.4164390460.500.9998278841.1432525151.3147439421.4813052241.5920010810.751.2826583431.4584592581.6237290601.7238273481.71767061611.6426872031.8277320631.9494693411.9498422161.8199936330.80.250.5769459960.6442359910.7383927140.8360892640.8427072200.500.7406907140.8398649020.9362518910.9648804990.8466070440.750.9502166721.0604275541.1187637831.0475149360.83173457611.2169326111.3116565341.2944657041.1042188620.809757060Table 2
Numerical values forv (t, x) of equation (28) for different values of t and x.
xtμ=1μ=0.8μ=0.6μ=0.4μ=0.20.20.250.9512389640.8658897870.7771389340.7439449131.1253970150.500.7411115690.6781522610.6517996980.8266381981.5562862900.750.5790928120.5570612970.6386770231.0665820271.98700080610.4580260340.4948694680.7362967721.4299000922.4116262650.50.251.2840382951.1701237051.0570854001.0316119601.5632069980.501.0003959790.9215082020.9032056901.1628053822.1411166460.750.7816935320.7663399700.9027300121.4972373062.71160950310.6182704770.6934340871.0505155551.9896433413.2704395800.80.251.7332704001.5807965381.4349736031.4199218582.1541886530.501.3503933241.2500043621.2425682831.6165836122.9305550540.751.0551759001.0488366291.2591642652.0785611233.68972893410.8345778480.9614682881.4746665482.7452176894.429716292Figure 1
2D graphical representation forut,x and vt,x of equation (9).Figure 2
3D graphical representation forut,x and vt,x of equation (9).Table 3
Numerical values foru (t, x) of equation (44) for different values of t and x.
xtμ=1μ=0.8μ=0.6μ=0.4μ=0.20.20.251.5683018161.7579265622.0800123792.7980196754.9128189440.502.0134061092.3327587282.8901797524.1411456756.7919366460.752.5829567113.0484102103.9109449815.6858154468.41800121913.3079658023.9599632555.2036328027.4341366919.9065070370.50.252.1169860212.3762941072.8342844633.9187306747.1894184900.502.7178139703.1680018764.0032067595.94766753110.066557720.753.4866268664.1689869445.5063353618.29470822212.5530058414.4652867755.4597531577.42841639210.9536402114.826257650.80.252.8576322243.2110029843.8524452755.43153228210.262506430.503.6686651244.2954621925.5056360618.38621695914.486833820.754.7064539845.6816073117.65988711611.8163451118.1346782216.0275066807.48425776010.4315601015.7044730021.46722633Table 4
Numerical values forv (t, x) of equation (45) for different values of t and x.
xtμ=1μ=0.8μ=0.6μ=0.4μ=0.20.20.250.6376345470.5740012900.4978348820.5708264971.8346709720.500.4967819410.4341840610.4425926151.0130333163.1664816570.750.3881775200.3515991730.5845124131.8082778984.44436035310.3070240320.3481393600.9677993362.9015543695.6807553960.50.250.4723712900.4223530920.3508975780.3620025981.2611920500.500.3680251140.3080980470.2759460700.6461058942.2561114940.750.2875689800.2284991250.3427719941.2118092683.22703636310.2274489970.2013905640.5911305012.0173325084.1748822130.80.250.3499412590.3100093440.2420437470.2073020480.8363484190.500.2726397100.2146912300.1524912730.3742793731.5816927010.750.2130363400.1373043670.1636862840.7669934442.32522057610.1684983620.0926763810.3120873661.3622848403.059303926Figure 3
2D graphical representation forut,x and vt,x of equation (30).Figure 4
3D graphical representation forut,x and vt,x of equation (30).Table 5
Numerical values foru (x, t) of equation (61) for different values of x and t.
xtμ=1μ=0.8μ=0.6μ=0.4μ=0.20.20.253.1548304534.6919592858.86613577622.7216678567.890037710.506.56169769611.2007559922.7568301151.52487616108.69541010.7513.0498239623.1439854344.9751097787.11237439144.2819487124.4286294442.6396770176.52840205128.5195969176.93054140.50.255.7484758778.63587763816.9378191946.24073749145.43304550.5011.9561927220.9593371044.78616917107.0281709234.34938280.7523.7783295743.8493566889.83870690182.5069498311.9830971144.5118649481.45614191154.1476188270.5167864383.24550960.80.2510.4744059715.8221654231.6453853189.0952764728672561760.5021.7856035638.7406312284.92624204208.1617677463.30584860.7543.3269413481.57700295171.5855108356.3271994617.5545123181.10590594152.1843524295.5790527529.2525354759.1758921Table 6
The numerical value forv (x, t) of equation (62) for different values of x and t.
xtμ=1μ=0.8μ=0.6μ=0.4μ=0.20.20.250.3178128740.2479437260.5258721814.85561292731.241567990.500.1832906380.4169430673.08609020816.9696634957.094625350.750.3021022481.7091403359.53234339034.8594005281.0132343910.9216900635.08028166020.9035502057.73320775103.71814250.50.250.1744194030.1147466440.2228085563.03894277621.573527760.500.1005920350.1674629401.79117421511.2846976139.888340290.750.1657972280.897605834598226159023.6538007056.9049433310.5058342312.93483950013.5354444239.5851403473.096003690.80.250.0957233980.0416465360.0564837132.04193306916.267594780.500.0552060790.0305453461.0805092568.16472220030.445330860.750.0909914480.4522262584.03393541017.5040371743.6740327910.2776077111.7573958809.49174225029.6252697356.29021759Figure 5
2D graphical representation forux,t and vx,tof equation (47).Figure 6
3D graphical representation forux,t and vx,t of equation (47).Fractional physical equations can be solved successfully by using analytical and approximate solutions method called fractional power series method (FPSM). The Caputo operator is presented as fractional derivatives. As compared to other methods to solve nonlinear equations, the power series method is employed efficiently to obtain the solution. Results thus obtained show that FPSM can be applied to solve the system of nonlinear partial differential equations (PDEs) with accuracy and effectiveness.


## 4. Conclusion

Numerical simulation results prove the validity and reliability of the power series method. Therefore, it is more convenient to describe that PSM is very powerful in constructing fractional power series solutions for fractional PDEs in scientific problems of any order and time of space. As per the work done, we had offered the fractional-order power series method for finding the nonlinear fractional-order partial differential equation with comparable easy implementation. In the tables, reasonable values are presented for fractional and integer cases. 2D and 3D graphical presentations are given for fractional-orderμ=0.1 for t=0.25, 0.5, 0.75, and 1. The proposed method is quite promising in reducing the computational cost of solving such problems to a great extent.



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*Source: 6385799-2021-11-05.xml*