# A Novel Numerical Technique for Fractional Ordinary Differential Equations with Proportional Delay

Authors: Muhammad Imran Liaqat; Adnan Khan; Ali Akgül; Md. Shajib Ali
Journal: Journal of Function Spaces
 (2022)

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## Abstract

Some researchers have combined two powerful techniques to establish a new method for solving fractional-order differential equations. In this study, we used a new combined technique, known as the Elzaki residual power series method (ERPSM), to offer approximate and exact solutions for fractional multipantograph systems (FMPS) and pantograph differential equations (PDEs). In Caputo logic, the fractional-order derivative operator is measured. The Elzaki transform method and the residual power series method (RPSM) are combined in this novel technique. The suggested technique is based on a new version of Taylor’s series that generates a convergent series as a solution. Establishing the coefficients for a series, like the RPSM, necessitates computing the fractional derivatives each time. As ERPSM just requires the concept of a zero limit, we simply need a few computations to get the coefficients. The novel technique solves nonlinear problems without the need for He’s and Adomian polynomials, which is an advantage over the other combined methods based on homotopy perturbation and Adomian decomposition methods. The relative, recurrence, and absolute errors of the problems are analyzed to evaluate the efficiency and consistency of the presented method. Graphical significances are also identified for various values of fractional-order derivatives. As a result, the procedure is quick, precise, and easy to implement, and it yields outstanding results.


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## Body

## 1. Introduction

Many differential equations (DEs) that arise in applications are sufficiently complicated that closed-form solutions are not always feasible. Numerical methods offered a powerful substitute means for solving the DEs under the given initial conditions. Numerous methods have been developed in recent years to solve fractional-order differential equations (FODEs), including the homotopy perturbation method [1], the differential transform method [2], the operational matrix method [3], the conformable Shehu transform decomposition method [4], the variational iteration method [5], the Jacobi collocation method [6], the conformable Shehu transform iterative method [7], the spectral tau method [8], the Legendre wavelet method [9], the fractional natural decomposition method [10], the power series method with the conformable operator [11], and the Chebyshev polynomial method [12].Integral equations (IEs), DEs, and delay differential equations (DDEs) are all solved by employing integral transforms [13–17], which are among the most valuable techniques in mathematics. The conversion of DEs and IEs into terms of a simple algebraic equation is enabled by the appropriate selection of integral transform. The origins of integral transforms can be traced back to Laplace’s work in the 1780s and Fourier’s work in 1822 [18]. In the beginning, ordinary and partial DEs were solved using the Laplace transform and the Fourier transform, which are two well-known transforms. These transforms were then applied to FODEs [19–24]. In recent years, researchers have proposed lots of new different transformations to solve a variety of mathematical problems. FODEs are solved using the Aboodh transform [25], fractional complex transform [26], travelling wave transform [27], Sumudu transform [28], and ZZ transform [29]. These transformations are paired with additional analytical, numerical, or homotopy-based techniques to handle FODEs [30–35]. Numerous mathematicians have recently become interested in a transformation known as the Elzaki transform (ET) [36–41]. The ET was introduced by Elzaki to facilitate the process of solving ordinary and partial DEs in the time domain [42]. The ET is derived from the classical Fourier integral transform.We examine the functions in setH, which are described as
(1)H=Θτ∃M,ϒ1,ϒ2>0,Θτ<Meτ/ϒjifτ∈−1j×0,∞.The formula for E-T is as follows:(2)EΘτ=℘υ=υ∫0∞Θτe−τ/υdτ,ϒ1≤υ≤ϒ2.The following are the key advantages of the ET [36–45]:
(i)
The ET can easily be applied to the initial value problems with less computational work(ii)
The ET has unit-preserving properties and may be used to solve problems without resorting to the frequency domain(iii)
Numerous nonlinear DEs with variable coefficients, namely, the time-fractional wavelike equations, can be solved with it(iv)
It may handle a variety of difficult problems in engineering, physics, fluid mechanics, chemistry, and dynamics, such as Maxwell’s equations and fluid flow problemsThe Jordanian mathematician, Arqub, created the RPSM in 2013 [46]. The RPSM is a semianalytical method; it is a combination of Taylor’s series and the residual error function. It provides series solutions of linear and nonlinear DEs in the form of convergence series. In 2013, RPSM was implemented for the first time to find solutions to fuzzy DEs. Furthermore, this method has been successfully used to solve a wide range of FODEs, including time-fractional KdV-Burgers equations [47], time-fractional Schrödinger equations [48], the SIR epidemic model of fractional order [49], conformable-type Coudrey–Dodd–Gibbon–Sawada–Kotera [50], time-fractional Swift–Hohenberg problems [51], time-fractional Phi-4 equation [52], and the Zakharov–Kuznetsov equation [53].Researchers combined two powerful methods to develop a new method for solving FODEs. Some of these groups are described as a combination of the Adomian decomposition method and the Sumudu transform [54], as well as the homotopy analysis method and the natural transform [55] and the Laplace transformation with homotopy perturbation approach [56]. In this study, we applied the novel combined technique, known as the ERPSM, to provide approximate and exact solutions for FMPS and PDEs. To assess the efficiency and consistency of the proposed method, the relative, recurrence, and absolute errors of the problems are examined. Graphical significance is also found for various values of fractional-order derivatives. As a result, the technique is rapid, precise, and simple to use, and it produces excellent results. The set of rules for this new technique depends on transforming the given equation into the ET space, in the second step; establishing a series solution by using the new form of the Taylor series; and then acquiring the solution in the real space of the equation by applying the inverse ET.This novel technique can be used to construct power series expansion solutions for linear and nonlinear FODEs without perturbation, linearization, or discretization. Unlike the classical power series method, this method does not need to match the coefficients of the corresponding terms, and a recursion relation is not needed. The new method handles nonlinear problems without the need for He’s and Adomian polynomials, which is an advantage over existing combination methods based on homotopy perturbation and Adomian decomposition methods. This technique finds the coefficients of the series, relying on the limit concept but not the fractional derivatives as in the RPSM. Thus, only a few calculations are required to determine the coefficients related to RPSM. The closed-form and approximate solutions can be obtained by the proposed method through a quick convergence series.A PDE is a special kind of delay differential equation (DDE) with a proportional delay. In 1851, the first-time device named “pantograph” was used in the construction of an electric locomotive, which is where this name originated from that time. British Railways decided to make a new kind of electric locomotive in 1960. The target was to construct a new kind of electric locomotive that moves trains faster. The pantograph was a prominent part of the new fast-speed electric locomotive. Pantographs take current from an overhead wire, which is necessary for the locomotive to move. Therefore, Ockendon and Taylor observed the mechanism of the pantograph. As a result, they built a special kind of DDE form:(3)dΘdτ=ΩΘτ+σΘƛτ,τ≥0,where Ω,σ are real constants and 0<k<1, ƛ∈R. This article was first published in 1971 [57]. Then, such a type of DDE was named PDE. Various studies on PDEs have recently been published in the scientific literature [58–60]. PDEs are widely used in probability theory, nonlinear dynamical systems, astrophysics, quantum mechanics, electrodynamics, and cell growth [61–64]. In this study, we consider the following FMPS:
(4)DτωΘ1τ=Ω1Θ1τ+Λ1τ,Θ1τ,Θ2τ,Θ1ϖ1τ,Θ2ϖ2τ,(5)DτωΘ2τ=Ω2Θ2τ+Λ2τ,Θ1τ,Θ2τ,Θ1ϖ1τ,Θ2ϖ2τ,subject to the initial conditions
(6)Θ10=I1,Θ20=I2,where 0<ϖ1,ϖ2≤1,I1,I2 are finite constants; Λ1,Λ2 are analytical functions; and Dτω is the Caputo fractional derivative (CFD). The FMPS is a type of DDE that arises in a number of physics and engineering applications, including quantum mechanics, dynamical systems, electronic systems, and population dynamics [65, 66].The following linear PDE is given as(7)DτωΘτ=lΘτ+∑j=1smjΘujτ+∑j=1tpjDτωΘqjτ+gτ,where 0<ω≤1, τ≥0, l,mj,pj∈R, 0<ω≤1, uj>0, and qj<1. Dτω is the CFD of order ω, with the following initial condition:
(8)Θ0=ϕ.The nonlinear PDE is as follows:(9)DτωΘτ=ξτω,Θτ,Θƛτ,DτωΘƛτ,with the initial condition
(10)Θ0=I.Definition 1 (see [67]) (a novel fractional Taylor series formula in E-T).
Assume thatΘτ is a piecewise continuous and exponential order and that the E-T of Θτ,EΦτ=℘υ is provided by a fractional Taylor series.
(11)℘υ=∑ν=0∞ℏνυνω+2,where ℏν is the νth coefficient of the novel fractional Taylor series formula in E-T.Lemma 2.
Assume thatΘ1τ and Θ2τ are piecewise continuous and of exponential-order functions, and EΘ1τ=℘1υ, EΘ2τ=℘2υ, and λ1,λ2 are constants. Then, the following axioms hold [67]:
(i)
limυ⟶01/υ2℘υ=ℏ0(ii)
EDτωΘτ=℘υ/υω−∑r=0ν−1υr−ω+2Θr0,ν−1<ω≤ν(iii)
EDτνωΘτ=℘υ/υνω−∑r=0ν−1υr−νω+2Θr0,0<ω≤1The framework of this study is as follows. The fundamental recommendation beyond the ERPSM with convergence and absolute error analysis for FMPS is demonstrated in Section2. We also demonstrated numerical examples of FMPS to exemplify the competency, potential, and straightforwardness of the new combined technique. The linear and nonlinear PDEs are discussed in Sections 3 and 4, respectively. Finally, our findings are summarized in Section 5.


## 2. The ERPSM to Demonstrate the FMPS

In this section, we use ERPSM to construct the solutions of the FMPS as in Equations (4) and (5). The main algorithm of this method for solving the FMPS can be summarized by the following steps: applies the E-T to Equation (4). As a result, we get an algebraic form in E-T space. In the second step, using the novel fractional Taylor’s series formula in E-T, we represent the solution in the E-T space of the algebraic equation obtained in the first step. The coefficients of this expansion are determined with the help of residual function and limit concept. As a result, we have found a solution to the problem in its original space by taking the inverse E-T.In the next subsection, we derive the main algorithms of the ERPSM for the FMPS.


### 2.1. The Algorithm of ERPSM for Solving FMPS

We use the following algorithm to create the solution with the help of ERPSM for the FMPS as shown in Equations (4) and (5):Step 1.
Rewrite Equations (4) and (5). We have
(12)DτωΘ1τ−Ω1Θ1τ−Λ1τ,Θ1τ,Θ2τ,Θ1ϖ1τ,Θ2ϖ2τ=0,(13)DτωΘ2τ−Ω2Θ2τ−Λ2τ,Θ1τ,Θ2τ,Θ1ϖ1τ,Θ2ϖ2τ=0.Step 2.
We get the following result by implementing the E-T on both sides of Equations (12) and (13).
(14)EDτωΘ1τ−EΩ1Θ1τ−EΛ1τ,Θ1τ,Θ2τ,Θ1ϖ1τ,Θ2ϖ2τ=0,(15)EDτωΘ2τ−EΩ2Θ2τ−EΛ2τ,Θ1τ,Θ2τ,Θ1ϖ1τ,Θ2ϖ2τ=0.By utilizing the second part of Lemma2, Equations (14) and (15) become as follows:
(16)℘1υ=υ2Θ0+υωΩ1℘1υ+υωΛ1/υ,(17)℘2υ=υ2Θ0+υωΩ2℘2υ+υωΛ2/υ,where
(18)EΛ1τ,Θ1τ,Θ2τ,Θ1ϖ1τ,Θ2ϖ2τ=Λ1/υ,EΛ2τ,Θ1τ,Θ2τ,Θ1ϖ1τ,Θ2ϖ2τ=Λ2/υ,EΘ1τ=℘1υ,EΘ2τ=℘2υ.Step 3.
Assume that algebraic equations (16) and (17) have the solution in the expansion form as
(19)℘1υ=∑ν=0∞ℏ1νυνω+2,℘2υ=∑ν=0∞ℏ2νυνω+2.Theκth-truncated series of ℘1υ and ℘2υ are as
(20)℘1κυ=∑ν=0κℏ1νυνω+2,℘2κυ=∑ν=0κℏ2νυνω+2.Step 4.
By utilizing the following lemma,(21)ℏ1,0=limυ⟶01υ2Θ1τ=Θ10,ℏ2,0=limυ⟶01υ2Θ2τ=Θ20.Theκth-truncated series becomes
(22)℘κ,1υ=Θ10υ2+∑ν=1κℏ1νυνω+2,(23)℘κ,2υ=Θ20υ2+∑ν=1κℏ2,νυνω+2.Step 5.
Consider the Elzaki residual function (ERF) of Equations (22) and (23) separately, as well as the κth-truncated Elzaki residual functions (ERFs), so that
(24)ERes1υ=℘1υ−υ2Θ0−υωΩ1℘1υ−υωΛ1/υ,(25)ERes2υ=℘2υ−υ2Θ0−υωΩ2℘2υ−υωΛ2/υ,(26)ERes1,κυ=℘1,κυ−υ2Θ0−υωΩ1℘1,κυ−υωΛ1/υ,(27)ERes2,κυ=℘2,κυ−υ2Θ0−υωΩ2℘2,κυ−υωΛ2/υ.Step 6.
Replace the succession arrangement of℘1,κυ and ℘2,κυ in Equations (26) and (27).Step 7.
To highlight important facts, we extend some features that arise in the RPSM [67–69].That is obvious.(28)EResυ=0.Therefore,(29)limκ⟶∞EReskυ=EResυ.Since,limυ⟶01/υ2EResυ=0. It is clear that limυ⟶01/υ2EReskυ=0. As a result, we get the following:
(30)limυ⟶01υkω+2ERes1υ=limυ⟶01υkω+2ERes1,κυ=0,k=1,2,3,⋯,limυ⟶01υkω+2ERes2υ=limυ⟶01υkω+2ERes2,κυ=0,k=1,2,3,⋯.Step 8.
Replace theκth attained values of ℏ1,κ and ℏ2,κ into the κth-truncated series of ℘1,κυ and ℘2,κυ to become the κth-approximate explanation of Equations (16) and (17).Step 9.
Use the inverse E-T on℘1,κυ and ℘2,κυ to obtain the κth-approximate solution Θ1,κτ and Θ2,κτ in the original space.The next theorem clarifies and establishes the conditions for the series solutions to converge.Theorem 3.
LetZ be a Banach space denoted with a suitable norm . over which the sequence of partial sums ∑κ=0∞℘κυ is defined. Assume that the initial guess ℘0 remains inside the ball Br℘ of the solution ℘υ. Then, the series solution ∑κ=0∞℘κυ converges if ∃θ>0 such that ℘ν+1υ≤θ℘νυ.Proof.
A sequence of partial sums is defined as(31)Y0=℘0υ,Y1=℘0υ+℘1υ,Y2=℘0υ+℘1υ+℘2υ,Y3=℘0υ+℘1υ+℘2υ+℘3υ,⋮⋮⋮Yν=℘0υ+℘1υ+℘2υ+℘3υ+⋯+℘νυ.
Next is that we would have to show thatYνν=0∞ is a Cauchy sequence in Z. To demonstrate this, consider the following relationship:
(32)Yν+1−Yν=℘ν+1υ≤θ℘νυ≤θ2℘ν−1υ≤θ3℘ν−2υ≤θ4℘ν−3υ⋯≤θν+1℘0υ,where ν=0,1,2,⋯.
For everyν,κ∈Ν,ν≥κ, we have
(33)Yν−Yκ=Yν−Yν−1+Yν−1−Yν−2+Yν−2−Yν−3+Yν−3−Yν−4⋯Yκ+1−Yκ.
From triangle inequality, we have(34)Yν−Yν−1+Yν−1−Yν−2+Yν−2−Yν−3+Yν−3−Yν−4+⋯Yκ+1−Yκ≤Yν−Yν−1+Yν−1−Yν−2+Yν−2−Yν−3+Yν−3−Yν−4+⋯+Yκ+1−Yκ,and
(35)Yν−Yν−1+Yν−1−Yν−2+Yν−2−Yν−3+Yν−3−Yν−4+⋯+Yκ+1−Yκ≤θν℘0υ+θν−1℘0υ+θν−2℘0υ+⋯+θκ+1℘0υ=1−θν−κ1−θθκ+1℘0υ.
Therefore,(36)Yν−Yκ=1−θν−κ1−θθκ+1℘0υ.
Showing that the sequence is bounded, we can obtain for0<θ<1 that
(37)limν,κ⟶∞Yν−Yκ=0.This proves that the sequence of partial sums generated by ERPSM is Cauchy and hence convergent.In the next theorem, we determine the maximum truncation error.Theorem 4.
Let℘υ be the approximate solution of the truncated finite series ∑ν=0∞℘νυ. Assume it is attainable to acquire a real number θ∈0,1, in order that ℘ν+1υ≤θ℘νυ,∀ν∈Ν; furthermore, the utmost absolute error is
(38)℘υ−∑ν=0κ℘νυ≤θκ+11−κ℘0υ.Proof.
Let the series∑ν=0κ℘νυ be finite, then
(39)℘υ−∑ν=0κ℘νυ=∑κ+1∞℘νυ,≤∑κ+1∞℘νυ,≤∑κ+1∞θκ℘0υ,≤θκ+11+θ+θ2+θ3+⋯℘0υ,≤θκ+11−θ℘0υ.
This proof is complete.In the next subsection, two problems of FMPS are established to illustrate the performance and appropriateness of the proposed method.


### 2.2. Numerical Examples

To demonstrate the execution and capability of ERPSM, we investigated two interesting and important problems for FMPS:Problem 5.
Consider the following FMPS:(40)DτωΘ1τ−Θ1τ+Θ2τ−Θ1τ2+eτω/2−e−τω=0,(41)DτωΘ2τ+Θ1τ+Θ2τ+Θ2τ2−e−τω/2−eτω=0,with the initial conditions Θ10=1 and Θ20=1.Applying the E-T to Equations (40) and (41), we get
(42)EDτωΘ1τ−Θ1τ+Θ2τ−Θ1τ2+eτω/2−e−τω=0,(43)EDτωΘ2τ+Θ1τ+Θ2τ+Θ2τ2−e−τω/2−eτω=0.We have the following results from Equations (42) and (43) using the procedure mentioned in Subsection 2.1:
(44)℘1υ=υ2Θ10+υω℘1υ−υω℘2υ+4υω℘1υ2−2υω+22−υ+υω+21+υ,(45)℘2υ=υ2Θ20−υω℘1υ−4υω℘2υ2+2υω+22+υ+υω+21−υ.Assume that Equations (44) and (45) have a series solution in the following form:
(46)℘1υ=∑ν=0∞ℏ1,νυνω+2,℘2υ=∑ν=0∞ℏ2,νυνω+2.Theκth-truncated expansion is as follows:
(47)℘1,κυ=∑ν=0κℏ1,νυνω+2,℘2,κυ=∑ν=0κℏ2,νυνω+2.By using the first part of Lemma2, the κth-truncated series becomes
(48)℘1,κυ=υ2+∑ν=1κℏ1,νυνω+2,℘2,κυ=υ2+∑ν=1κℏ2,νυνω+2.The ERFs are formulated as(49)ERes1υ=℘1υ−υ2Θ10−υω℘1υ+υω℘2υ−4υω℘1υ2+2υω+22−υ−υω+21+υ,ERes2υ=℘2υ−υ2Θ20+υω℘1υ+4υω℘2υ2−2υω+22+υ−υω+21−υ.Theκth-truncated ERF takes the following form:
(50)ERes1,κυ=℘1,κυ−υ2Θ10−υω℘1,κυ+υω℘2,κυ−4υω℘1,κυ2+2υω+22−υ−υω+21+υ,ERes2,κυ=℘2,κυ−υ2Θ20+υω℘1,κυ+4υω℘2,κυ2−2υω+22+υ−υω+21−υ.Substituteκ=1,2,3,4,5 into Equations (65) and (66) and solve the following expression to find the unknown coefficients.
(51)limυ⟶01υkω+2ERes1,κυ=0,limυ⟶01υkω+2ERes2,κυ=0.Thus, we have(52)ℏ1,1=1,ℏ2,1=−1,ℏ1,2=2+12ω−3Γω+12,ℏ2,2=12ω+Γω+12,ℏ1,3=181+ω8+2ω+4+23ω+4−2ω+23+4ω+1Γω+1+38ωΓ2ω+1,ℏ2,3=−2−21−ω−2−3ω+1−2−2ω−1Γω+1+58Γ2ω+1,ℏ1,4=−143ω+2−16−2ω+5−24ω+5−23ω+6−25ω+5−26ω+6+2ω+33+23ω+1+325ω+1+22ω+2×Γω+1+23ω+123ω+1−3Γ2ω+1+326ωΓ3ω+1,ℏ2,4=143ω+216+22ω+5+23ω+5−24ω+5+25ω+5+2ω+31−22ω+1+23ω+2+25ω+1Γω+1−23ω+18ω+1+5Γ2ω+1+73Γ3ω+126ω,ℏ1,5=4+2−9ω+1+2−7ω+2+2−6ω+1+2−5ω+2+2−4ω+1+2−3ω+1+4−ω+1+2−10ω−13Γ3ω+1−3×2−4ω+1Γ3ω+1+5×27Γ4ω+1−4Γω+1−13Γ3ω+1−3×2−4ω+1Γ3ω+1+5×27Γ4ω+1−4Γω+1−32−9ω−1Γω+1−2−7ω+1Γω+1−2−5ω+1Γω+1−2−6ωΓω+1−32−4ωΓω+1−34−ωΓω+1−8−ωΓω+1+34Γ2ω+1+−32−4ωΓω+1−34−ωΓω+1−8−ωΓω+1,ℏ2,5=−4−2−8ω+1−2−7ω+1−2−5ω+2−2−4ω+1−3×2−3ω+1−2−ω+2−2−10ω+4ω+1Γω+127ω−7Γ3ω+13×24ω+4+2Γω+1−Γω+129ω+1−Γω+126ω−1−2−4ωΓω+1−4−ωΓω+1+38−ωΓω+1+54Γ2ω+1+5×Γ2ω+127ω+3+Γ2ω+124ω+Γ3ω+124+17Γ4ω+1384.We have the following 5th-order approximate solution in the original space for Equations (40) and (41) when ω=1 uses the procedure mentioned in Subsection 2.1:
(53)Θ1,5τ=1+τ+τ22!+τ33!+τ44!+τ55!,(54)Θ2,5τ=1−τ+τ22!−τ33!+τ44!−τ55!.Equations (53) and (54) represent the first five terms of eτ and e−τ, respectively, and therefore the exact solutions of Equations (40) and (41).The following 2-D graphs show the absolute and relative error for Example 1.Figures1 and 2 demonstrate the 2-D graphs of absolute and relative errors in the intervals τ∈0,1 over the ten-step approximate and exact solutions of Equation (40) at ω=1, respectively. According to the figures, the approximate solution is extremely close to the precise solution. Figures 3 and 4 are graphs of absolute and relative errors in the intervals τ∈0,1 over the ten-step approximate and exact solutions of Equation (41) at ω=1, respectively. One can perceive the equivalent verdicts depicted for Equation (40).Figure 1Figure 2Figure 3Figure 4Error functions are presented to observe the exactness and capability of the numerical method. To prove the exactness and capability of ERPSM, we selected three kinds of error functions, such as absolute, residual, and error functions.Table1 shows the absolute and relative errors at reasonable nominated grid points in the interval τ∈0,1 among the five-step approximate and exact solutions of Equations (40) and (41) at ω=1 attained using ERPSM. Table 1 shows that the approximate and exact solutions are quite close to each other, confirming the effectiveness of the recommended strategy.Problem 6.
Consider the following FMPS:(55)DτωΘ1τ+Θ1τ+e−τωcosτω2Θ2τ2+2e−3/4τωcosτω2sinτω4Θ1τ4=0,(56)DτωΘ2τ−eτωΘ12τ2+Θ22τ2=0,with the initial conditions Θ10=1,Θ20=0.Table 1
The absolute and relative error of Example 1.
τΘ1Abs.errorΘ1Rel.errorΘ2Abs.errorΘ2Rel.error0.29.149350321813188×10−87.490854479152386×10−88.641131510334077×10−81.055430186034679×10−70.46.030974603721262×10−64.042683174006208×10−65.379368972713294×10−68.025075491218842×10−60.67.080039050877396×10−53.885607815121622×10−55.963609402637182×10−51.086640481073082×10−40.84.102618258010615×10−41.843425212040309×10−43.262974505549021×10−47.261883310726820×10−41.01.615161333333059×10−35.941846488086295×10−41.212774566900720×10−33.296663066666799×10−3Applying the E-T to Equations (55) and (56), we get
(57)EDτωΘ1τ+Θ1τ+e−τωcosτω2Θ2τ2+2e−3/4τωcosτω2sinτω4Θ1τ4=0,(58)EDτωΘ2τ−eτωΘ12τ2+Θ22τ2=0.Using the approach mentioned in Subsection2.1, we obtain the following results from Equations (57) and (58) as
(59)℘1υ=υ2Θ10−υω℘1υ−υωEE−1υ21+υE−14υ24+υ2E−14℘2υ2−2υωEE−14υ24+3υE−14υ24+υ2E−14υ316+υ2E−116℘1υ4,(60)℘2υ=υ2Θ20+υωEE−1υ21−υE−14℘1υ22−E−14℘2υ22.Assume that Equations (59) and (60) have the series solution in the following form:
(61)℘1υ=∑ν=0∞ℏ1,νυνω+2,℘2υ=∑ν=0∞ℏ2,νυνω+2.Theκth-truncated expansions are as
(62)℘1,κυ=∑ν=0κℏ1,νυνω+2,℘2,κυ=∑ν=0κℏ2,νυνω+2.By using the first part of Lemma2, the κth-truncated expansions become
(63)℘1,κυ=υ2+∑ν=1κℏ1,νυνω+2,℘2,κυ=υ2+∑ν=1κℏ2,νυνω+2.The ERFs are formulated as(64)ERes1υ=℘1υ−υ2+υω℘1υ+υωEE−1υ21+υE−14υ24+υ2E−14℘2υ2+2υωEE−14υ24+3υE−14υ24+υ2E−14υ316+υ2E−116℘1υ4,ERes2υ=℘2υ−υωEE−1υ21−υE−14℘1υ22+E−14℘2υ22.Theκth-truncated ERF takes the following form:
(65)ERes1,κυ=℘1,κυ−υ2+υω℘1,κυ+υωEE−1υ21+υE−14υ24+υ2E−14℘2,κυ2+2υωEE−14υ24+3υE−14υ24+υ2E−14υ316+υ2E−116℘1,κυ4,(66)ERes2,κυ=℘2,κυ−υωEE−1υ21−υE−14℘1,κυ22+E−14℘2,κυ22.Substituteκ=1,2,3,4,5 into Equations (65) and (66) and solve the following expression to find the unknown coefficients.
(67)limυ⟶01υkω+2ERes1,κυ=0,limυ⟶01υkω+2ERes2,κυ=0.Thus, we have(68)ℏ1,1=−1,ℏ2,1=1,ℏ1,2=1−12ω−Γω+12,ℏ2,2=−12ω−1+Γω+1,ℏ1,3=38Γ2ω+1+2−12ω+1+1Γ1/2+ωπ+Γω+112−122ω−1+21−3ω+2−ω,ℏ2,3=18ω−2+2ω+1−24ω+1Γ1/2+ωπ−2ωΓω+1+12Γ2ω+1,ℏ1,4=1+2−6ω+1−2−5ω+1−2−3ω+1−2−ω−12Γω+1−12×8ωΓ2ω+1−2−2ω−14ω+23ω+1−4Γ1/2+ωπ−38Γ2ω+1−796Γ3ω+1+πΓ3ω+143ω+1Γ1/2+ω−3Γ3ω+12ω+122ω+3Γω+1−Γ3ω+14ω+1+2ω−125ω+1Γ2ω+1+Γω+18ω+1Γ2ω+1+8ωΓ3ω+125ωΓ2ω+1,ℏ2,4=−Γ3ω+12ω−324ω−1Γ2ω+1Γω+1−Γ3ω+1−2ω+1+3+2ωΓω+18ωΓ2ω+1+Γ3ω+14ωΓω+1Γω+1+Γ3ω+16+3Γ2ω+123ω+2+Γω+18ω−Γ3ω+12ωΓω+1+2ω+1+1Γ1/2+ω8ωπ−Γω+125ω−1+2+4ω−8ω26ω−1.We have the following 5th-order approximate solution in the original space for Equations (55) and (56) when ω=1 using the procedure mentioned in Subsection 2.1:
(69)Θ1,5τ=1−τ+τ33−τ46+τ530,(70)Θ2,5τ=τ−τ36+τ5120.Equations (69) and (70) represent the first five terms of e−τcosτ and sinτ, respectively, and therefore the exact solutions of Equations (55) and (56).In the next section, we will use our new ERPSM to find approximate and exact solutions to linear PDE.


## 2.1. The Algorithm of ERPSM for Solving FMPS

We use the following algorithm to create the solution with the help of ERPSM for the FMPS as shown in Equations (4) and (5):Step 1.
Rewrite Equations (4) and (5). We have
(12)DτωΘ1τ−Ω1Θ1τ−Λ1τ,Θ1τ,Θ2τ,Θ1ϖ1τ,Θ2ϖ2τ=0,(13)DτωΘ2τ−Ω2Θ2τ−Λ2τ,Θ1τ,Θ2τ,Θ1ϖ1τ,Θ2ϖ2τ=0.Step 2.
We get the following result by implementing the E-T on both sides of Equations (12) and (13).
(14)EDτωΘ1τ−EΩ1Θ1τ−EΛ1τ,Θ1τ,Θ2τ,Θ1ϖ1τ,Θ2ϖ2τ=0,(15)EDτωΘ2τ−EΩ2Θ2τ−EΛ2τ,Θ1τ,Θ2τ,Θ1ϖ1τ,Θ2ϖ2τ=0.By utilizing the second part of Lemma2, Equations (14) and (15) become as follows:
(16)℘1υ=υ2Θ0+υωΩ1℘1υ+υωΛ1/υ,(17)℘2υ=υ2Θ0+υωΩ2℘2υ+υωΛ2/υ,where
(18)EΛ1τ,Θ1τ,Θ2τ,Θ1ϖ1τ,Θ2ϖ2τ=Λ1/υ,EΛ2τ,Θ1τ,Θ2τ,Θ1ϖ1τ,Θ2ϖ2τ=Λ2/υ,EΘ1τ=℘1υ,EΘ2τ=℘2υ.Step 3.
Assume that algebraic equations (16) and (17) have the solution in the expansion form as
(19)℘1υ=∑ν=0∞ℏ1νυνω+2,℘2υ=∑ν=0∞ℏ2νυνω+2.Theκth-truncated series of ℘1υ and ℘2υ are as
(20)℘1κυ=∑ν=0κℏ1νυνω+2,℘2κυ=∑ν=0κℏ2νυνω+2.Step 4.
By utilizing the following lemma,(21)ℏ1,0=limυ⟶01υ2Θ1τ=Θ10,ℏ2,0=limυ⟶01υ2Θ2τ=Θ20.Theκth-truncated series becomes
(22)℘κ,1υ=Θ10υ2+∑ν=1κℏ1νυνω+2,(23)℘κ,2υ=Θ20υ2+∑ν=1κℏ2,νυνω+2.Step 5.
Consider the Elzaki residual function (ERF) of Equations (22) and (23) separately, as well as the κth-truncated Elzaki residual functions (ERFs), so that
(24)ERes1υ=℘1υ−υ2Θ0−υωΩ1℘1υ−υωΛ1/υ,(25)ERes2υ=℘2υ−υ2Θ0−υωΩ2℘2υ−υωΛ2/υ,(26)ERes1,κυ=℘1,κυ−υ2Θ0−υωΩ1℘1,κυ−υωΛ1/υ,(27)ERes2,κυ=℘2,κυ−υ2Θ0−υωΩ2℘2,κυ−υωΛ2/υ.Step 6.
Replace the succession arrangement of℘1,κυ and ℘2,κυ in Equations (26) and (27).Step 7.
To highlight important facts, we extend some features that arise in the RPSM [67–69].That is obvious.(28)EResυ=0.Therefore,(29)limκ⟶∞EReskυ=EResυ.Since,limυ⟶01/υ2EResυ=0. It is clear that limυ⟶01/υ2EReskυ=0. As a result, we get the following:
(30)limυ⟶01υkω+2ERes1υ=limυ⟶01υkω+2ERes1,κυ=0,k=1,2,3,⋯,limυ⟶01υkω+2ERes2υ=limυ⟶01υkω+2ERes2,κυ=0,k=1,2,3,⋯.Step 8.
Replace theκth attained values of ℏ1,κ and ℏ2,κ into the κth-truncated series of ℘1,κυ and ℘2,κυ to become the κth-approximate explanation of Equations (16) and (17).Step 9.
Use the inverse E-T on℘1,κυ and ℘2,κυ to obtain the κth-approximate solution Θ1,κτ and Θ2,κτ in the original space.The next theorem clarifies and establishes the conditions for the series solutions to converge.Theorem 3.
LetZ be a Banach space denoted with a suitable norm . over which the sequence of partial sums ∑κ=0∞℘κυ is defined. Assume that the initial guess ℘0 remains inside the ball Br℘ of the solution ℘υ. Then, the series solution ∑κ=0∞℘κυ converges if ∃θ>0 such that ℘ν+1υ≤θ℘νυ.Proof.
A sequence of partial sums is defined as(31)Y0=℘0υ,Y1=℘0υ+℘1υ,Y2=℘0υ+℘1υ+℘2υ,Y3=℘0υ+℘1υ+℘2υ+℘3υ,⋮⋮⋮Yν=℘0υ+℘1υ+℘2υ+℘3υ+⋯+℘νυ.
Next is that we would have to show thatYνν=0∞ is a Cauchy sequence in Z. To demonstrate this, consider the following relationship:
(32)Yν+1−Yν=℘ν+1υ≤θ℘νυ≤θ2℘ν−1υ≤θ3℘ν−2υ≤θ4℘ν−3υ⋯≤θν+1℘0υ,where ν=0,1,2,⋯.
For everyν,κ∈Ν,ν≥κ, we have
(33)Yν−Yκ=Yν−Yν−1+Yν−1−Yν−2+Yν−2−Yν−3+Yν−3−Yν−4⋯Yκ+1−Yκ.
From triangle inequality, we have(34)Yν−Yν−1+Yν−1−Yν−2+Yν−2−Yν−3+Yν−3−Yν−4+⋯Yκ+1−Yκ≤Yν−Yν−1+Yν−1−Yν−2+Yν−2−Yν−3+Yν−3−Yν−4+⋯+Yκ+1−Yκ,and
(35)Yν−Yν−1+Yν−1−Yν−2+Yν−2−Yν−3+Yν−3−Yν−4+⋯+Yκ+1−Yκ≤θν℘0υ+θν−1℘0υ+θν−2℘0υ+⋯+θκ+1℘0υ=1−θν−κ1−θθκ+1℘0υ.
Therefore,(36)Yν−Yκ=1−θν−κ1−θθκ+1℘0υ.
Showing that the sequence is bounded, we can obtain for0<θ<1 that
(37)limν,κ⟶∞Yν−Yκ=0.This proves that the sequence of partial sums generated by ERPSM is Cauchy and hence convergent.In the next theorem, we determine the maximum truncation error.Theorem 4.
Let℘υ be the approximate solution of the truncated finite series ∑ν=0∞℘νυ. Assume it is attainable to acquire a real number θ∈0,1, in order that ℘ν+1υ≤θ℘νυ,∀ν∈Ν; furthermore, the utmost absolute error is
(38)℘υ−∑ν=0κ℘νυ≤θκ+11−κ℘0υ.Proof.
Let the series∑ν=0κ℘νυ be finite, then
(39)℘υ−∑ν=0κ℘νυ=∑κ+1∞℘νυ,≤∑κ+1∞℘νυ,≤∑κ+1∞θκ℘0υ,≤θκ+11+θ+θ2+θ3+⋯℘0υ,≤θκ+11−θ℘0υ.
This proof is complete.In the next subsection, two problems of FMPS are established to illustrate the performance and appropriateness of the proposed method.


## 2.2. Numerical Examples

To demonstrate the execution and capability of ERPSM, we investigated two interesting and important problems for FMPS:Problem 5.
Consider the following FMPS:(40)DτωΘ1τ−Θ1τ+Θ2τ−Θ1τ2+eτω/2−e−τω=0,(41)DτωΘ2τ+Θ1τ+Θ2τ+Θ2τ2−e−τω/2−eτω=0,with the initial conditions Θ10=1 and Θ20=1.Applying the E-T to Equations (40) and (41), we get
(42)EDτωΘ1τ−Θ1τ+Θ2τ−Θ1τ2+eτω/2−e−τω=0,(43)EDτωΘ2τ+Θ1τ+Θ2τ+Θ2τ2−e−τω/2−eτω=0.We have the following results from Equations (42) and (43) using the procedure mentioned in Subsection 2.1:
(44)℘1υ=υ2Θ10+υω℘1υ−υω℘2υ+4υω℘1υ2−2υω+22−υ+υω+21+υ,(45)℘2υ=υ2Θ20−υω℘1υ−4υω℘2υ2+2υω+22+υ+υω+21−υ.Assume that Equations (44) and (45) have a series solution in the following form:
(46)℘1υ=∑ν=0∞ℏ1,νυνω+2,℘2υ=∑ν=0∞ℏ2,νυνω+2.Theκth-truncated expansion is as follows:
(47)℘1,κυ=∑ν=0κℏ1,νυνω+2,℘2,κυ=∑ν=0κℏ2,νυνω+2.By using the first part of Lemma2, the κth-truncated series becomes
(48)℘1,κυ=υ2+∑ν=1κℏ1,νυνω+2,℘2,κυ=υ2+∑ν=1κℏ2,νυνω+2.The ERFs are formulated as(49)ERes1υ=℘1υ−υ2Θ10−υω℘1υ+υω℘2υ−4υω℘1υ2+2υω+22−υ−υω+21+υ,ERes2υ=℘2υ−υ2Θ20+υω℘1υ+4υω℘2υ2−2υω+22+υ−υω+21−υ.Theκth-truncated ERF takes the following form:
(50)ERes1,κυ=℘1,κυ−υ2Θ10−υω℘1,κυ+υω℘2,κυ−4υω℘1,κυ2+2υω+22−υ−υω+21+υ,ERes2,κυ=℘2,κυ−υ2Θ20+υω℘1,κυ+4υω℘2,κυ2−2υω+22+υ−υω+21−υ.Substituteκ=1,2,3,4,5 into Equations (65) and (66) and solve the following expression to find the unknown coefficients.
(51)limυ⟶01υkω+2ERes1,κυ=0,limυ⟶01υkω+2ERes2,κυ=0.Thus, we have(52)ℏ1,1=1,ℏ2,1=−1,ℏ1,2=2+12ω−3Γω+12,ℏ2,2=12ω+Γω+12,ℏ1,3=181+ω8+2ω+4+23ω+4−2ω+23+4ω+1Γω+1+38ωΓ2ω+1,ℏ2,3=−2−21−ω−2−3ω+1−2−2ω−1Γω+1+58Γ2ω+1,ℏ1,4=−143ω+2−16−2ω+5−24ω+5−23ω+6−25ω+5−26ω+6+2ω+33+23ω+1+325ω+1+22ω+2×Γω+1+23ω+123ω+1−3Γ2ω+1+326ωΓ3ω+1,ℏ2,4=143ω+216+22ω+5+23ω+5−24ω+5+25ω+5+2ω+31−22ω+1+23ω+2+25ω+1Γω+1−23ω+18ω+1+5Γ2ω+1+73Γ3ω+126ω,ℏ1,5=4+2−9ω+1+2−7ω+2+2−6ω+1+2−5ω+2+2−4ω+1+2−3ω+1+4−ω+1+2−10ω−13Γ3ω+1−3×2−4ω+1Γ3ω+1+5×27Γ4ω+1−4Γω+1−13Γ3ω+1−3×2−4ω+1Γ3ω+1+5×27Γ4ω+1−4Γω+1−32−9ω−1Γω+1−2−7ω+1Γω+1−2−5ω+1Γω+1−2−6ωΓω+1−32−4ωΓω+1−34−ωΓω+1−8−ωΓω+1+34Γ2ω+1+−32−4ωΓω+1−34−ωΓω+1−8−ωΓω+1,ℏ2,5=−4−2−8ω+1−2−7ω+1−2−5ω+2−2−4ω+1−3×2−3ω+1−2−ω+2−2−10ω+4ω+1Γω+127ω−7Γ3ω+13×24ω+4+2Γω+1−Γω+129ω+1−Γω+126ω−1−2−4ωΓω+1−4−ωΓω+1+38−ωΓω+1+54Γ2ω+1+5×Γ2ω+127ω+3+Γ2ω+124ω+Γ3ω+124+17Γ4ω+1384.We have the following 5th-order approximate solution in the original space for Equations (40) and (41) when ω=1 uses the procedure mentioned in Subsection 2.1:
(53)Θ1,5τ=1+τ+τ22!+τ33!+τ44!+τ55!,(54)Θ2,5τ=1−τ+τ22!−τ33!+τ44!−τ55!.Equations (53) and (54) represent the first five terms of eτ and e−τ, respectively, and therefore the exact solutions of Equations (40) and (41).The following 2-D graphs show the absolute and relative error for Example 1.Figures1 and 2 demonstrate the 2-D graphs of absolute and relative errors in the intervals τ∈0,1 over the ten-step approximate and exact solutions of Equation (40) at ω=1, respectively. According to the figures, the approximate solution is extremely close to the precise solution. Figures 3 and 4 are graphs of absolute and relative errors in the intervals τ∈0,1 over the ten-step approximate and exact solutions of Equation (41) at ω=1, respectively. One can perceive the equivalent verdicts depicted for Equation (40).Figure 1Figure 2Figure 3Figure 4Error functions are presented to observe the exactness and capability of the numerical method. To prove the exactness and capability of ERPSM, we selected three kinds of error functions, such as absolute, residual, and error functions.Table1 shows the absolute and relative errors at reasonable nominated grid points in the interval τ∈0,1 among the five-step approximate and exact solutions of Equations (40) and (41) at ω=1 attained using ERPSM. Table 1 shows that the approximate and exact solutions are quite close to each other, confirming the effectiveness of the recommended strategy.Problem 6.
Consider the following FMPS:(55)DτωΘ1τ+Θ1τ+e−τωcosτω2Θ2τ2+2e−3/4τωcosτω2sinτω4Θ1τ4=0,(56)DτωΘ2τ−eτωΘ12τ2+Θ22τ2=0,with the initial conditions Θ10=1,Θ20=0.Table 1
The absolute and relative error of Example 1.
τΘ1Abs.errorΘ1Rel.errorΘ2Abs.errorΘ2Rel.error0.29.149350321813188×10−87.490854479152386×10−88.641131510334077×10−81.055430186034679×10−70.46.030974603721262×10−64.042683174006208×10−65.379368972713294×10−68.025075491218842×10−60.67.080039050877396×10−53.885607815121622×10−55.963609402637182×10−51.086640481073082×10−40.84.102618258010615×10−41.843425212040309×10−43.262974505549021×10−47.261883310726820×10−41.01.615161333333059×10−35.941846488086295×10−41.212774566900720×10−33.296663066666799×10−3Applying the E-T to Equations (55) and (56), we get
(57)EDτωΘ1τ+Θ1τ+e−τωcosτω2Θ2τ2+2e−3/4τωcosτω2sinτω4Θ1τ4=0,(58)EDτωΘ2τ−eτωΘ12τ2+Θ22τ2=0.Using the approach mentioned in Subsection2.1, we obtain the following results from Equations (57) and (58) as
(59)℘1υ=υ2Θ10−υω℘1υ−υωEE−1υ21+υE−14υ24+υ2E−14℘2υ2−2υωEE−14υ24+3υE−14υ24+υ2E−14υ316+υ2E−116℘1υ4,(60)℘2υ=υ2Θ20+υωEE−1υ21−υE−14℘1υ22−E−14℘2υ22.Assume that Equations (59) and (60) have the series solution in the following form:
(61)℘1υ=∑ν=0∞ℏ1,νυνω+2,℘2υ=∑ν=0∞ℏ2,νυνω+2.Theκth-truncated expansions are as
(62)℘1,κυ=∑ν=0κℏ1,νυνω+2,℘2,κυ=∑ν=0κℏ2,νυνω+2.By using the first part of Lemma2, the κth-truncated expansions become
(63)℘1,κυ=υ2+∑ν=1κℏ1,νυνω+2,℘2,κυ=υ2+∑ν=1κℏ2,νυνω+2.The ERFs are formulated as(64)ERes1υ=℘1υ−υ2+υω℘1υ+υωEE−1υ21+υE−14υ24+υ2E−14℘2υ2+2υωEE−14υ24+3υE−14υ24+υ2E−14υ316+υ2E−116℘1υ4,ERes2υ=℘2υ−υωEE−1υ21−υE−14℘1υ22+E−14℘2υ22.Theκth-truncated ERF takes the following form:
(65)ERes1,κυ=℘1,κυ−υ2+υω℘1,κυ+υωEE−1υ21+υE−14υ24+υ2E−14℘2,κυ2+2υωEE−14υ24+3υE−14υ24+υ2E−14υ316+υ2E−116℘1,κυ4,(66)ERes2,κυ=℘2,κυ−υωEE−1υ21−υE−14℘1,κυ22+E−14℘2,κυ22.Substituteκ=1,2,3,4,5 into Equations (65) and (66) and solve the following expression to find the unknown coefficients.
(67)limυ⟶01υkω+2ERes1,κυ=0,limυ⟶01υkω+2ERes2,κυ=0.Thus, we have(68)ℏ1,1=−1,ℏ2,1=1,ℏ1,2=1−12ω−Γω+12,ℏ2,2=−12ω−1+Γω+1,ℏ1,3=38Γ2ω+1+2−12ω+1+1Γ1/2+ωπ+Γω+112−122ω−1+21−3ω+2−ω,ℏ2,3=18ω−2+2ω+1−24ω+1Γ1/2+ωπ−2ωΓω+1+12Γ2ω+1,ℏ1,4=1+2−6ω+1−2−5ω+1−2−3ω+1−2−ω−12Γω+1−12×8ωΓ2ω+1−2−2ω−14ω+23ω+1−4Γ1/2+ωπ−38Γ2ω+1−796Γ3ω+1+πΓ3ω+143ω+1Γ1/2+ω−3Γ3ω+12ω+122ω+3Γω+1−Γ3ω+14ω+1+2ω−125ω+1Γ2ω+1+Γω+18ω+1Γ2ω+1+8ωΓ3ω+125ωΓ2ω+1,ℏ2,4=−Γ3ω+12ω−324ω−1Γ2ω+1Γω+1−Γ3ω+1−2ω+1+3+2ωΓω+18ωΓ2ω+1+Γ3ω+14ωΓω+1Γω+1+Γ3ω+16+3Γ2ω+123ω+2+Γω+18ω−Γ3ω+12ωΓω+1+2ω+1+1Γ1/2+ω8ωπ−Γω+125ω−1+2+4ω−8ω26ω−1.We have the following 5th-order approximate solution in the original space for Equations (55) and (56) when ω=1 using the procedure mentioned in Subsection 2.1:
(69)Θ1,5τ=1−τ+τ33−τ46+τ530,(70)Θ2,5τ=τ−τ36+τ5120.Equations (69) and (70) represent the first five terms of e−τcosτ and sinτ, respectively, and therefore the exact solutions of Equations (55) and (56).In the next section, we will use our new ERPSM to find approximate and exact solutions to linear PDE.


## 3. The ERPSM to Demonstrate the Linear PDE

In this section, we use our new ERPSM to construct the solutions of the linear PDE as in Equation (7). In the next subsection, we derive the main algorithms of the ERPSM for the linear PDE.


### 3.1. The Algorithm of ERPSM for Solving Linear PDE

In this subsection, we exploit ERPSM to create the solution to the linear PDE. We start by taking E-T on Equation (7). We get the following:
(71)EDτωΘτ=lEΘτ+∑j=1smjEΘujτ+∑j=1tpjEDτωΘqjτ+Egτ.By using the second part of Lemma2, Equation (71) becomes
(72)℘υ=υ2ϕ+υωl℘υ+υω∑j=1smi℘ujυui2+∑j=1tpj℘qjυqj2−υ2ϕ+υωGυ.Assuming that algebraic equation (72) has the solution in the expansion form as
(73)℘υ=∑ν=0∞ℏνυνω+2.Theκth-truncated series of ℘υ is as follows:
(74)℘κυ=∑ν=0κℏνυνω+2.By using the first part of Lemma2, we have the following:
(75)ℏ0=ϕ.Theκth-truncated series becomes
(76)℘kυ=ϕυ2+∑ν=1kℏνυνω+2.Now, define the ERF in the following form:(77)EResυ=℘υ−υ2ϕ−υωl℘υ−υω∑j=1smj℘ujυuj2−∑j=1tpj℘qjυqj2−υ2ϕ−υϖGυ.Theκth-ERF is as follows:
(78)EReskυ=℘kυ−υ2ϕ−υϖl℘kυ−υϖ∑j=1smi℘kujυuj2−∑j=1tpj℘kqjυqj2−υ2ϕ−υϖGυ.To highlight basic points, we generalize certain features that arise in the RPSM [67–69].It is understandable that(79)EResυ=0.Therefore,(80)limκ⟶∞EReskυ=EResυ.Since,limυ⟶01/υ2EResυ=0. It is obvious that limυ⟶01/υ2EReskυ=0. As a result, we get the following:
(81)limυ⟶01υkω+2EResυ=limυ⟶01υkω+2EReskυ=0,k=1,2,3,⋯.To determine the first undefined coefficientℏ1, substitute k=1, in Equations (76) and (78). As a result, we obtain as following:
(82)ERes1υ=℘1υ−υ2ϕ−υωl℘1υ−υω∑j=1smj℘1ujυuj2−∑j=1spj℘1qjυqj2−υ2ϕ−υωGυ,(83)℘1υ=ϕυ2+ℏ1υ2+ω,(84)℘1ujυ=ϕuj2υ2+ℏ1ujω+2υω+2,(85)℘1qjυ=ϕqj2υ2+ℏ1qjω+2υω+2.Using Equations (83), (84), and (85) in (82), we get the following:
(86)ERes1υ=ϕυ2+ℏ1υω+2−υ2ϕ−υωlϕυ2+ℏ1υω+2−υω∑j=1smiϕuj2υ2+ℏ1ujω+2υω+2uj2−∑j=1tpjϕqj2υ2+ℏ1qjω+2υω+2qj2−υ2ϕ−υωGυ,(87)ERes1υ=ℏ1υω+2−υωlϕυ2+ℏ1υω+2−υω∑j=1smjϕυ2+ℏ1ujωυω+2−∑j=1tpjℏ1qjωυω+2−υωGυ.Dividingυ2+ω by Equation (87), we get
(88)1υ2+ωERes1υ=ℏ11υ2+ωυω+2−1υ2+ωυωlϕυ2+ℏ1υω+2−1υ2+ωυω∑j=1smjϕυ2+ℏ1ujωυω+2−1υ2+ω∑j=1tpjℏ1qjωυω+2−1υ2+ωυωGυ.By usinglimυ⟶0 to the Equation (88), we get
(89)limυ⟶01υ2+ωERes1υ=ℏ1−lϕ+ℏ1limυ⟶0υ2−∑j=1smjϕ+ℏ1ujωlimυ⟶0υ2−∑j=1tpjℏ1qjω−limυ⟶01υ2Gυ.When we putlimυ⟶01/υ2+ωERes1υ=0 into Equation (89), we get
(90)ℏ1=lϕ+∑j=1smjϕ+g01−∑j=1tpjqjω.In the same manner, to determine the second undefined coefficient, putk=2 in Equations (76) and (78); we get the following:
(91)℘kυ=ϕυ2+ℏ1υω+2+ℏ2υ2ω+2,(92)℘kujυ=ϕuj2υ2+ℏ1ujω+2υω+2+ℏ2uj2ω+2υ2ω+2,(93)℘kqjυ=ϕqj2υ2+ℏ1qjω+2υω+2+ℏ2qj2ω+2υ2ω+2,(94)ERes2υ=℘2υ−υ2ϕ−υϖl℘2υ−υϖ∑j=1smi℘2ujυuj2−∑j=1tpj℘2qjυqj2−υ2ϕ−υϖGυ.Using Equations (91), (92), and (93) in (94), we get the following:
(95)ERes2υ=ℏ1υω+2+ℏ2υ2ω+2−lϕυ2+ω+lℏ1υ2ω+2+lℏ2υ3ω+2−υω∑j=1smjϕuj2υ2+ℏ1ujϖ+2υϖ+2+ℏ2uj2ω+2υ2ω+2ρi2−∑j=1tpjℏ1qjω+2υω+2+ℏ2qj2ω+2υ2ω+2qj2−υωGυ.Dividingυ2+2ω on Eq. (95), we have the following results:
(96)1υ2ω+2ERes2υ=ℏ2−lℏ1−lℏ2υω+∑j=1smjω1ujω+∑j=1smjℏ2uj2ωυω+∑j=1tpjℏ2qj2ω−1υ21υωGυ−g0.Applylimυ⟶0 to Equation (96), we get
(97)limυ⟶01υ2ω+2ERes2υ=ℏ2−lℏ1−lℏ2υω+∑j=1smjℏ1ujω+∑j=1smjϕℏ2uj2ωlimυ⟶0υω+∑j=1tpjℏ2qj2ω−limυ⟶01υ2EDτωgτ.By using the second part of Lemma2, we have
(98)1υωGυ−g0=EDτωgτ.Putlimυ⟶01/υ2ω+2ERes2υ=0 in Equation (97), and we have the second coefficient in the following form:
(99)ℏ2=lℏ1+∑j=1smjℏ1ujω+Dτωgτ01−∑j=1tpjqj2ω.Now, to define the third coefficient, putk=3 in Equations (76) and (78), and repeat the same steps. We get the following:
(100)ERes3υ=ℏ1υω+21−∑j=1tpjqjω+ℏ2υ2ω+21−∑j=1trjqj2ω+ℏ3υ3ω+2−lϕυ2+ω−lℏ1υ2ω+2−lℏ2υ3ω+2−lℏ3υ4ω+2−ϕυω+2∑j=1smj−ℏ1υ2ω+2∑j=1smjujω−ℏ2υ3ω+2∑j=1smjuj2ω−ℏ3υ4ω+2∑j=1smjuj3ω−ℏ3υ3ω+2∑j=1tpjqj3ω−υωGυ.Divideυ2+3ω by Equation (100).
(101)1υ3ω+2ERes3υ=ℏ3−lℏ2−lℏ3υω−ℏ2∑j=1smjuj2ω−ℏ3υω∑j=1smjuj3ω−ℏ3∑j=1tpjqj3ω−1υ21υ2ωGυ−1υ2ω−2g0−1υω−2Dτωgτ0,By usinglimυ⟶0to Equation (101), we get the third part of Lemma 2 in the following form:
(102)Dτ2ωgτ0=limυ⟶01υ21υ2ωGυ−1υ2ω−2g0−1υω−2Dτωgτ0.Putlimυ⟶01/υ3ω+2ERes3υ=0, in Equation (101); as a result, we get
(103)ℏ3=lℏν−1+∑j=1sℏν−1mjujν−1ω+Dτν−1ωgτ01−∑j=1tpjqjνω.As a consequence of the generalization, we acquire the following:(104)ℏν=ℏ0,ν=0,lℏν−1+∑i=1lℏν−1mjujν−1ω+Dϒν−1ωgτ01−∑j=1tpjqjνω,ν=1,2,3,⋯.Theκth-approximate solution of Equation (72) is as follows:
(105)℘kυ=ℏ0υ2+lϕ+∑j=1sβmjϕ+g01−∑j=1tpjqjωυ2+ϖ+lℏ1+∑j=1smjℏ1ujω+Dτωgτ01−∑j=1tpjqj2ωυ2+2ω+lℏ2+∑j=1sℏ2mjuj2ω+Dτ2ωgτ01−∑j=1tpjqj3ωυ2+3ω+⋯+lℏκ−1+∑i=1lℏκ−1mjujκ−1ω+Dτκ−1ωgτ01−∑j=1tpjqjκωυ2+κω.By taking inverse E-T on Equation (105), we get the κth-step approximate solution in the original space of Equation (7).
(106)Θκτ=ℏ0+lϕ+∑j=1smjϕ+g01−∑j=1tpjqjωτωΓω+1+lℏ1+∑j=1smjℏ1ujω+Dτωgτ01−∑j=1tpjqj2ωτ2ωΓ2ω+1+αℏ2+∑j=1sℏ2mjuj2ω+Dτ2ωgτ01−∑j=1tpjqj3ωτ3ωΓ3ω+1+⋯+lℏκ−1+∑i=1lℏκ−1mjujκ−1ω+Dτκ−1ωgτ01−∑j=1tpjqjκωτκωΓκω+1.In the next subsection, one PDE problem is established to illustrate the performance and appropriateness of the suggested method.


### 3.2. Numerical Example

In this subsection, we provided a problem to validate the performance and effectiveness of ERPSM for linear PDE.Example 1.
Consider the following linear PDE:(107)DτωΘτ=−Θτ+110Θ45τ+1254ωDτωΘ4τ5+8τω25−12e−4τω/5+e−τω,0<ω≤1,subject to the initial condition
(108)Θ0=0.
By comparing Equation (107) with (7), we get
(109)ℏ0=0,l=−1,gτ=8τω25−12e−4τω/5+e−τω,t=s=1,m1=0.1,u1=q1=0.8,p1=1254w.The recurrence relation that defines the values of the series coefficients as a result of Equation (104) is
(110)ℏν=0,ν=0,1/104/5ν−1ω−1ℏν−1+Dτν−1ωg01−1/25/4ω4/5νω,ν=1,2,3,⋯.First of all, we will compute a few of the terms ofDτν−1ωg0.Expand the following form in a few terms to accomplish this:(111)gτ=8τω25−12e−4τω/5+e−τω,gτ=12−7τω25+21τ2ω250−27τ3ω1250+437τ4ω75000−113τ5ω75000,Dτωgτ=−7Γω+125+21τωΓ2ω+1250Γω+1−27τ2ωΓ3ω+11250Γ2ω+1+437τ3ωΓ4ω+175000Γ3ω+1−113τ4ωΓ5ω+175000Γ4ω+1,Dτ2ωgτ=21Γ2ω+1250−27τωΓ3ω+11250Γω+1+437τ2ωΓ4ω+175000Γ2ω+1−113τ3ωΓ5ω+175000Γ3ω+1,Dτ3ωgτ=−27Γ3ω+11250+437τωΓ4ω+175000Γω+1−113τ2ωΓ5ω+175000Γ2ω+1,Dτ4ωgτ=437Γ4ω+175000−113τωΓ5ω+175000Γω+1,Dτ5ωgτ=−113Γ5ω+175000.Returning to Equation (110), we can easily find the values of ℏν, for ν=1,2,3,4,5, respectively, as
(112)ℏ0=0,ℏ1=1,ℏ2=25ω25+7Γω+1−54ω254ω−25ω,ℏ3=16−251+2ω25ω54ω−25ω25+7Γω+11254ω−25ω16ω−225ω+21Γ2ω+1250−53−2ω16ω,ℏ4=27Γ3ω+162564ω125−ω−1250−64ω−10125ω564ω−10125ω×64ϖ−10100ϖ−54ϖ5−ϖ+50+14Γϖ+11252−4ϖ5−ϖ64ϖ−2100ϖ+21Γ2ϖ+1250−53−2ϖ16ϖ,ℏ5=−437Γ4ω+137500256ω625−ω−75000−256ω−10625ω5256ω−10625ω×27Γ3ω+162564ω125−ω−1250−64ω−10125ω564ω−10125ω×64ω−10100ω−54ω5−ω+50+14Γω+11252−4ω5−ω64ω−2100ω+21Γ2ω+1250−53−2ω16ω.Forω=1, we have the following:
(113)ℏ0=0,ℏ1=1,ℏ2=−2,ℏ3=3,ℏ4=−4,ℏ5=5.The 5th-term approximate solution is as follows:(114)℘5υ=υ2+1−2υ2+2+3υ2+3−4υ2+4+5υ2+5.By taking inverse E-T on both sides of Equation (114),
(115)Θ5τ=τ−τ2+τ32−τ46+τ524.Equation (115) represents first five terms of τe−τ; therefore, the exact solution of Equation (107) is τe−τ at ω=1.The following 2-D graphs show the absolute and relative errors for Example1.Figures5 and 6 demonstrate the 2-D graphs of absolute and relative errors in the intervals τ∈0,1 over the ten-step approximate and exact solutions of Equation (107) at ω=1, respectively. According to the figures, the approximate solution is extremely close to the exact solution.Figure 5Figure 6Table2 shows the absolute and relative errors at reasonable nominated grid points in the interval τ∈0,1 among the five-step approximate and exact solutions of Equation (107) at ω=1 attained using ERPSM. From Table 2, it can be perceived that the approximate solution is in eminent contract with the exact solution, and this sanctions the efficiency of the recommended method. The convergence of the approximate solution to the exact solution for Equation (107) has been shown numerically as in Table 3. From the obtained results, it is evident that the present technique is an effective and convenient algorithm to solve certain classes of fractional-order DEs with fewer calculations and iteration steps.Table 2
The absolute and relative errors of Example1.
τAbs.errorRel.error0.25.1605107029284450×10−73.1515310030359390×10−60.43.1981585744245145×10−51.1927729870749216×10−40.63.5301834358408835×10−41.0720689345454725×10−30.81.9234953728893833×10−35.3510220976640050×10−31.07.1205587664326390×10−31.9355685668873270×10−2Table 3
The recurrence errorsΘ5τ−Θ4τ of the five-step approximate solution with different values of ω for Example 1.
τω=0.7ω=0.8ω=0.9ω=1.00.25.5251129381940380×10−73.557469322866979×10−82.210719076023441×10−91.337989787908603×10−100.46.2509517206696540×10−65.691950916587166×10−75.002286239854830×10−84.281567321307531×10−90.62.5838396081572233×10−52.881550151522252×10−63.101552986356802×10−73.251315184617906×10−80.87.0721445608883680×10−59.107121466539466×10−61.131888166924002×10−71.370101542818409×10−71.01.5443160141456830×10−42.223418326791861×10−53.089573833214964×10−64.181218087214387×10−7In the next section, we will use our new ERPSM to find approximate and exact solutions to nonlinear PDE.


## 3.1. The Algorithm of ERPSM for Solving Linear PDE

In this subsection, we exploit ERPSM to create the solution to the linear PDE. We start by taking E-T on Equation (7). We get the following:
(71)EDτωΘτ=lEΘτ+∑j=1smjEΘujτ+∑j=1tpjEDτωΘqjτ+Egτ.By using the second part of Lemma2, Equation (71) becomes
(72)℘υ=υ2ϕ+υωl℘υ+υω∑j=1smi℘ujυui2+∑j=1tpj℘qjυqj2−υ2ϕ+υωGυ.Assuming that algebraic equation (72) has the solution in the expansion form as
(73)℘υ=∑ν=0∞ℏνυνω+2.Theκth-truncated series of ℘υ is as follows:
(74)℘κυ=∑ν=0κℏνυνω+2.By using the first part of Lemma2, we have the following:
(75)ℏ0=ϕ.Theκth-truncated series becomes
(76)℘kυ=ϕυ2+∑ν=1kℏνυνω+2.Now, define the ERF in the following form:(77)EResυ=℘υ−υ2ϕ−υωl℘υ−υω∑j=1smj℘ujυuj2−∑j=1tpj℘qjυqj2−υ2ϕ−υϖGυ.Theκth-ERF is as follows:
(78)EReskυ=℘kυ−υ2ϕ−υϖl℘kυ−υϖ∑j=1smi℘kujυuj2−∑j=1tpj℘kqjυqj2−υ2ϕ−υϖGυ.To highlight basic points, we generalize certain features that arise in the RPSM [67–69].It is understandable that(79)EResυ=0.Therefore,(80)limκ⟶∞EReskυ=EResυ.Since,limυ⟶01/υ2EResυ=0. It is obvious that limυ⟶01/υ2EReskυ=0. As a result, we get the following:
(81)limυ⟶01υkω+2EResυ=limυ⟶01υkω+2EReskυ=0,k=1,2,3,⋯.To determine the first undefined coefficientℏ1, substitute k=1, in Equations (76) and (78). As a result, we obtain as following:
(82)ERes1υ=℘1υ−υ2ϕ−υωl℘1υ−υω∑j=1smj℘1ujυuj2−∑j=1spj℘1qjυqj2−υ2ϕ−υωGυ,(83)℘1υ=ϕυ2+ℏ1υ2+ω,(84)℘1ujυ=ϕuj2υ2+ℏ1ujω+2υω+2,(85)℘1qjυ=ϕqj2υ2+ℏ1qjω+2υω+2.Using Equations (83), (84), and (85) in (82), we get the following:
(86)ERes1υ=ϕυ2+ℏ1υω+2−υ2ϕ−υωlϕυ2+ℏ1υω+2−υω∑j=1smiϕuj2υ2+ℏ1ujω+2υω+2uj2−∑j=1tpjϕqj2υ2+ℏ1qjω+2υω+2qj2−υ2ϕ−υωGυ,(87)ERes1υ=ℏ1υω+2−υωlϕυ2+ℏ1υω+2−υω∑j=1smjϕυ2+ℏ1ujωυω+2−∑j=1tpjℏ1qjωυω+2−υωGυ.Dividingυ2+ω by Equation (87), we get
(88)1υ2+ωERes1υ=ℏ11υ2+ωυω+2−1υ2+ωυωlϕυ2+ℏ1υω+2−1υ2+ωυω∑j=1smjϕυ2+ℏ1ujωυω+2−1υ2+ω∑j=1tpjℏ1qjωυω+2−1υ2+ωυωGυ.By usinglimυ⟶0 to the Equation (88), we get
(89)limυ⟶01υ2+ωERes1υ=ℏ1−lϕ+ℏ1limυ⟶0υ2−∑j=1smjϕ+ℏ1ujωlimυ⟶0υ2−∑j=1tpjℏ1qjω−limυ⟶01υ2Gυ.When we putlimυ⟶01/υ2+ωERes1υ=0 into Equation (89), we get
(90)ℏ1=lϕ+∑j=1smjϕ+g01−∑j=1tpjqjω.In the same manner, to determine the second undefined coefficient, putk=2 in Equations (76) and (78); we get the following:
(91)℘kυ=ϕυ2+ℏ1υω+2+ℏ2υ2ω+2,(92)℘kujυ=ϕuj2υ2+ℏ1ujω+2υω+2+ℏ2uj2ω+2υ2ω+2,(93)℘kqjυ=ϕqj2υ2+ℏ1qjω+2υω+2+ℏ2qj2ω+2υ2ω+2,(94)ERes2υ=℘2υ−υ2ϕ−υϖl℘2υ−υϖ∑j=1smi℘2ujυuj2−∑j=1tpj℘2qjυqj2−υ2ϕ−υϖGυ.Using Equations (91), (92), and (93) in (94), we get the following:
(95)ERes2υ=ℏ1υω+2+ℏ2υ2ω+2−lϕυ2+ω+lℏ1υ2ω+2+lℏ2υ3ω+2−υω∑j=1smjϕuj2υ2+ℏ1ujϖ+2υϖ+2+ℏ2uj2ω+2υ2ω+2ρi2−∑j=1tpjℏ1qjω+2υω+2+ℏ2qj2ω+2υ2ω+2qj2−υωGυ.Dividingυ2+2ω on Eq. (95), we have the following results:
(96)1υ2ω+2ERes2υ=ℏ2−lℏ1−lℏ2υω+∑j=1smjω1ujω+∑j=1smjℏ2uj2ωυω+∑j=1tpjℏ2qj2ω−1υ21υωGυ−g0.Applylimυ⟶0 to Equation (96), we get
(97)limυ⟶01υ2ω+2ERes2υ=ℏ2−lℏ1−lℏ2υω+∑j=1smjℏ1ujω+∑j=1smjϕℏ2uj2ωlimυ⟶0υω+∑j=1tpjℏ2qj2ω−limυ⟶01υ2EDτωgτ.By using the second part of Lemma2, we have
(98)1υωGυ−g0=EDτωgτ.Putlimυ⟶01/υ2ω+2ERes2υ=0 in Equation (97), and we have the second coefficient in the following form:
(99)ℏ2=lℏ1+∑j=1smjℏ1ujω+Dτωgτ01−∑j=1tpjqj2ω.Now, to define the third coefficient, putk=3 in Equations (76) and (78), and repeat the same steps. We get the following:
(100)ERes3υ=ℏ1υω+21−∑j=1tpjqjω+ℏ2υ2ω+21−∑j=1trjqj2ω+ℏ3υ3ω+2−lϕυ2+ω−lℏ1υ2ω+2−lℏ2υ3ω+2−lℏ3υ4ω+2−ϕυω+2∑j=1smj−ℏ1υ2ω+2∑j=1smjujω−ℏ2υ3ω+2∑j=1smjuj2ω−ℏ3υ4ω+2∑j=1smjuj3ω−ℏ3υ3ω+2∑j=1tpjqj3ω−υωGυ.Divideυ2+3ω by Equation (100).
(101)1υ3ω+2ERes3υ=ℏ3−lℏ2−lℏ3υω−ℏ2∑j=1smjuj2ω−ℏ3υω∑j=1smjuj3ω−ℏ3∑j=1tpjqj3ω−1υ21υ2ωGυ−1υ2ω−2g0−1υω−2Dτωgτ0,By usinglimυ⟶0to Equation (101), we get the third part of Lemma 2 in the following form:
(102)Dτ2ωgτ0=limυ⟶01υ21υ2ωGυ−1υ2ω−2g0−1υω−2Dτωgτ0.Putlimυ⟶01/υ3ω+2ERes3υ=0, in Equation (101); as a result, we get
(103)ℏ3=lℏν−1+∑j=1sℏν−1mjujν−1ω+Dτν−1ωgτ01−∑j=1tpjqjνω.As a consequence of the generalization, we acquire the following:(104)ℏν=ℏ0,ν=0,lℏν−1+∑i=1lℏν−1mjujν−1ω+Dϒν−1ωgτ01−∑j=1tpjqjνω,ν=1,2,3,⋯.Theκth-approximate solution of Equation (72) is as follows:
(105)℘kυ=ℏ0υ2+lϕ+∑j=1sβmjϕ+g01−∑j=1tpjqjωυ2+ϖ+lℏ1+∑j=1smjℏ1ujω+Dτωgτ01−∑j=1tpjqj2ωυ2+2ω+lℏ2+∑j=1sℏ2mjuj2ω+Dτ2ωgτ01−∑j=1tpjqj3ωυ2+3ω+⋯+lℏκ−1+∑i=1lℏκ−1mjujκ−1ω+Dτκ−1ωgτ01−∑j=1tpjqjκωυ2+κω.By taking inverse E-T on Equation (105), we get the κth-step approximate solution in the original space of Equation (7).
(106)Θκτ=ℏ0+lϕ+∑j=1smjϕ+g01−∑j=1tpjqjωτωΓω+1+lℏ1+∑j=1smjℏ1ujω+Dτωgτ01−∑j=1tpjqj2ωτ2ωΓ2ω+1+αℏ2+∑j=1sℏ2mjuj2ω+Dτ2ωgτ01−∑j=1tpjqj3ωτ3ωΓ3ω+1+⋯+lℏκ−1+∑i=1lℏκ−1mjujκ−1ω+Dτκ−1ωgτ01−∑j=1tpjqjκωτκωΓκω+1.In the next subsection, one PDE problem is established to illustrate the performance and appropriateness of the suggested method.


## 3.2. Numerical Example

In this subsection, we provided a problem to validate the performance and effectiveness of ERPSM for linear PDE.Example 1.
Consider the following linear PDE:(107)DτωΘτ=−Θτ+110Θ45τ+1254ωDτωΘ4τ5+8τω25−12e−4τω/5+e−τω,0<ω≤1,subject to the initial condition
(108)Θ0=0.
By comparing Equation (107) with (7), we get
(109)ℏ0=0,l=−1,gτ=8τω25−12e−4τω/5+e−τω,t=s=1,m1=0.1,u1=q1=0.8,p1=1254w.The recurrence relation that defines the values of the series coefficients as a result of Equation (104) is
(110)ℏν=0,ν=0,1/104/5ν−1ω−1ℏν−1+Dτν−1ωg01−1/25/4ω4/5νω,ν=1,2,3,⋯.First of all, we will compute a few of the terms ofDτν−1ωg0.Expand the following form in a few terms to accomplish this:(111)gτ=8τω25−12e−4τω/5+e−τω,gτ=12−7τω25+21τ2ω250−27τ3ω1250+437τ4ω75000−113τ5ω75000,Dτωgτ=−7Γω+125+21τωΓ2ω+1250Γω+1−27τ2ωΓ3ω+11250Γ2ω+1+437τ3ωΓ4ω+175000Γ3ω+1−113τ4ωΓ5ω+175000Γ4ω+1,Dτ2ωgτ=21Γ2ω+1250−27τωΓ3ω+11250Γω+1+437τ2ωΓ4ω+175000Γ2ω+1−113τ3ωΓ5ω+175000Γ3ω+1,Dτ3ωgτ=−27Γ3ω+11250+437τωΓ4ω+175000Γω+1−113τ2ωΓ5ω+175000Γ2ω+1,Dτ4ωgτ=437Γ4ω+175000−113τωΓ5ω+175000Γω+1,Dτ5ωgτ=−113Γ5ω+175000.Returning to Equation (110), we can easily find the values of ℏν, for ν=1,2,3,4,5, respectively, as
(112)ℏ0=0,ℏ1=1,ℏ2=25ω25+7Γω+1−54ω254ω−25ω,ℏ3=16−251+2ω25ω54ω−25ω25+7Γω+11254ω−25ω16ω−225ω+21Γ2ω+1250−53−2ω16ω,ℏ4=27Γ3ω+162564ω125−ω−1250−64ω−10125ω564ω−10125ω×64ϖ−10100ϖ−54ϖ5−ϖ+50+14Γϖ+11252−4ϖ5−ϖ64ϖ−2100ϖ+21Γ2ϖ+1250−53−2ϖ16ϖ,ℏ5=−437Γ4ω+137500256ω625−ω−75000−256ω−10625ω5256ω−10625ω×27Γ3ω+162564ω125−ω−1250−64ω−10125ω564ω−10125ω×64ω−10100ω−54ω5−ω+50+14Γω+11252−4ω5−ω64ω−2100ω+21Γ2ω+1250−53−2ω16ω.Forω=1, we have the following:
(113)ℏ0=0,ℏ1=1,ℏ2=−2,ℏ3=3,ℏ4=−4,ℏ5=5.The 5th-term approximate solution is as follows:(114)℘5υ=υ2+1−2υ2+2+3υ2+3−4υ2+4+5υ2+5.By taking inverse E-T on both sides of Equation (114),
(115)Θ5τ=τ−τ2+τ32−τ46+τ524.Equation (115) represents first five terms of τe−τ; therefore, the exact solution of Equation (107) is τe−τ at ω=1.The following 2-D graphs show the absolute and relative errors for Example1.Figures5 and 6 demonstrate the 2-D graphs of absolute and relative errors in the intervals τ∈0,1 over the ten-step approximate and exact solutions of Equation (107) at ω=1, respectively. According to the figures, the approximate solution is extremely close to the exact solution.Figure 5Figure 6Table2 shows the absolute and relative errors at reasonable nominated grid points in the interval τ∈0,1 among the five-step approximate and exact solutions of Equation (107) at ω=1 attained using ERPSM. From Table 2, it can be perceived that the approximate solution is in eminent contract with the exact solution, and this sanctions the efficiency of the recommended method. The convergence of the approximate solution to the exact solution for Equation (107) has been shown numerically as in Table 3. From the obtained results, it is evident that the present technique is an effective and convenient algorithm to solve certain classes of fractional-order DEs with fewer calculations and iteration steps.Table 2
The absolute and relative errors of Example1.
τAbs.errorRel.error0.25.1605107029284450×10−73.1515310030359390×10−60.43.1981585744245145×10−51.1927729870749216×10−40.63.5301834358408835×10−41.0720689345454725×10−30.81.9234953728893833×10−35.3510220976640050×10−31.07.1205587664326390×10−31.9355685668873270×10−2Table 3
The recurrence errorsΘ5τ−Θ4τ of the five-step approximate solution with different values of ω for Example 1.
τω=0.7ω=0.8ω=0.9ω=1.00.25.5251129381940380×10−73.557469322866979×10−82.210719076023441×10−91.337989787908603×10−100.46.2509517206696540×10−65.691950916587166×10−75.002286239854830×10−84.281567321307531×10−90.62.5838396081572233×10−52.881550151522252×10−63.101552986356802×10−73.251315184617906×10−80.87.0721445608883680×10−59.107121466539466×10−61.131888166924002×10−71.370101542818409×10−71.01.5443160141456830×10−42.223418326791861×10−53.089573833214964×10−64.181218087214387×10−7In the next section, we will use our new ERPSM to find approximate and exact solutions to nonlinear PDE.


## 4. The ERPSM to Demonstrate the Nonlinear PDE

In this section, we use our new ERPSM to construct the solutions of the nonlinear PDE as in Equation (9). In the next subsection, we derive the main algorithms of the ERPSM for the nonlinear PDE.


### 4.1. The Algorithm of ERPSM for Solving Nonlinear PDE

We summarized the method of finding the ERPS solution for the nonlinear PDE with the following algorithm:Step 1.
Rewriting Equation (9), we have
(116)DτωΘτ−ξτω,Θτ,Θƛτ,DτωΘƛτ=0.Step 2.
E-T is used on both sides of Equation (116).
(117)EDτωΘτ−Eξτω,Θτ,Θƛτ,DτωΘƛτ=0.By using the second part of Lemma2, we get the following:
(118)℘υ−υ2ℏ−υω℘υ=0,where EΘτ=℘υ and Eξτω,Θτ,Θƛτ,DτωΘƛτ=℘υ.Step 3.
Assume that the solution to Equation (118) has the following extension:
(119)℘υ=∑ν=0∞ℏνυνω+2.
Introduce theκth-truncated series in the following form:
(120)℘kυ=∑ν=0kℏνυνω+2.Step 4.
By using the first part of Lemma2, we have
(121)ℏ0=Θ0=I.Step 5.
Theκth-truncated series of ℘υ becomes as follows:
(122)℘kυ=Iυ2+∑ν=1kℏνυνω+2.Step 6.
Describe the ERF and theκth-ERF of Equation (118), respectively, as follows:
(123)EResυ=℘υ−υ2I−υω℘υ,(124)EResκυ=℘κυ−υ2I−υω℘κυ.Step 7.
Replace the succession arrangement of℘kυ as in Equation (120) with (124).Step 8.
Solve the succeeding expression forν=1,2,3,⋯,κ step by step to obtain the unknown coefficients
(125)limυ⟶01υκω+2EResκυ=0,κ=1,2,3,⋯.Step 9.
Replace the attained values ofℏν into the κth-truncated series of ℘κυ to obtain the κth-approximate solution of Equation (118).Step 10.
Use the inverse E-T on℘kυ to obtain the κth-approximate solution Θκτ in the original space.In the next subsection, we determine the appropriateness of the recommended method for nonlinear PDFs.


### 4.2. Numerical Examples

In this subsection, we will provide two problems with nonlinear PDEs to validate the performance and effectiveness of ERPSM.Example 2.
Examine the following nonlinear fractional pantograph differential equation:(126)DτωΘτ=1−2Θ2τ2,1<ω≤2,0≤τ≤1,subject to the initial condition
(127)Θ0=1.By employing E-T on Equation (126),
(128)℘υ−υ2−υ2+ω+2υωEE−14℘2υ2=0.Assume that algebraic equation (128) has the solution in the expansion form as
(129)℘υ=∑ν=0∞ℏνυνω+2.Theκth-truncated series of Equation (128) is as follows:
(130)℘kυ=∑ν=0kℏνυνω+2.By using the first part of Lemma2, we get
(131)ℏ0=Θ0=1.Theκth-truncated series becomes as follows:
(132)℘kυ=υ2+∑ν=1kℏνυνω+2.Now, define the ERF in the following form:(133)EResυ=℘υ−υ2−υ2+ω+2υωEE−14℘2υ2.Theκth-ERF is as follows:
(134)EResκυ=℘κυ−υ2−υ2+ω+2υωEE−14℘κ2υ2.Therefore,(135)limυ⟶01υκω+2EResκυ=0.To determine the undefined coefficientsℏν, substitute κ=1,2,3,4,5, in Equations (132) and (134) and then utilize (135); thus, we have
(136)ℏ1=−1,ℏ2=42−ω,ℏ3=−2Γ1/2+ωπΓ1+ω−168ω,ℏ4=8327ωΓ2/3+ωΓ1/3+ω64ωπΓ1+ωΓ1/2+ω+6464ω+8Γ1/2+ω8ωπΓ1+ω,ℏ5=−1283Γ1/2+ωΓ1/2+2ω354ωΓ1/3+ωΓ2/3+ωΓ1+ω−32Γ1/2+2ω16ωΓ1/2+ωΓ1+ω−1634ωΓ1/2+ω2Γ1/2+2ω3π27ωΓ1/3+ωΓ2/3+ωΓ1+ω2−2561024ω−32327ωΓ1/3+ωΓ2/3+ωπ1024ωΓ1/2+ωΓ1+ω−32Γ1/2+ω128ωπΓ1+ω.The five-step approximate solution of Equation (128) is as follows:
(137)℘5υ=υ2−υ2+ω+42−ωυ2+2ω+ℏ3υ2+3ω+ℏ4υ2+4ω+ℏ5υ2+5ω.By taking inverse E-T at both sides of Equation (137),
(138)Θ5τ=1−τωΓ1+ω+42−ωϒ2ωΓ1+2ω+ℏ3τ3ωΓ1+3ω+ℏ4τ4ωΓ1+4ω+ℏ5τ5ωΓ1+5ω.Forω=2, Equation (138) becomes
(139)Θ5τ=1−τ22!+τ44!−τ66!+τ88!−τ1010!.Equation (139) represents the first six terms of cosτ; therefore, the exact solution of Equation (126) is cosτ at ω=2.The following 2-D graphs show the behavior of approximate and exact solutions to Example2.Figure7 shows the performance of the five-step approximate solution of Equation (126) for various values of ω and exact solution at ω=2. Clearly, the findings for various fractional ω values converge to the result in the case of ω=2. Furthermore, the approximate solution does overlap with the exact solution, which indicates the accuracy and effectiveness of the suggested method.Figure 7The following 2-D graphs show the absolute and relative errors for Example2.Figures8 and 9 demonstrate the 2-D graphs of absolute and relative errors in the intervals τ∈0,1 over the five-step approximate and exact solutions of Equation (126) at ω=1, respectively. The approximate solution is extremely close to the exact solution, as seen in the figures.Figure 8Figure 9Table4 shows the absolute and relative errors at reasonable nominated grid points in the interval τ∈0,1 among the five-step approximate and exact solutions of Equation (126) at ω=1 attained using ERPSM. From Table 4, it can be perceived that the approximate and exact solutions are in very good agreement, and this sanctions the efficiency of the recommended method. The convergence of the approximate solution to the exact solution for Equation (126) has been shown numerically as in Table 5. From the obtained results, it is evident that the present technique is an effective and convenient algorithm to solve certain classes of fractional-order DEs with fewer calculations and iteration steps.Example 3.
Consider the following nonlinear PDE:(140)DτωΘτ=12Θτ+121−ωΘτ2DτωΘτ2,τ≥0,0<ω≤1,subject to the initial condition
(141)Θ0=1.Table 4
The absolute and relative errors of Example2.
τAbs.errorRel.error0.20.00.00.43.508304757815494×10−143.808982011678268×10−140.64.535483100198689×10−125.495319744192834×10−120.81.429608653680247×10−102.051951896683681×10−101.07.120558766432639×10−92.051765518966810×10−9Table 5
The recurrence errorsΘ5τ−Θ4τ of the five-step approximate solution with different values of ω for Example 2.
τω=0.7ω=0.8ω=0.9ω=1.00.26.296560270772086×10−201.969389500244771×10−215.554963006701661×10−222.821869488536155×10−240.42.279598318393725×10−171.008327424125322×10−184.022222859549857×10−192.889594356261023×10−210.67.155401866274551×10−163.876349353331781×10−171.893795546581747×10−171.666285714285714×10−190.88.253027478106390×10−155.162636411521651×10−162.912400444857598×10−162.958944620811287×10−181.05.499819058543307×10−143.846463867665569×10−152.426030253023737×10−152.755731922398591×10−17By employing the E-T on Equation (140),
(142)℘υ=υ2+υω2℘υ+υω21−ωEE−14℘υ2E−14℘υ/2υω−υ2−ω.Assume that algebraic equation (142) has the solution in the expansion form as
(143)℘υ=∑ν=0∞ℏνυνω+2.Theκth-truncated series of Equation (142) is as follows:
(144)℘kυ=∑ν=0kℏνυνω+2.By employing the first part of Lemma2, we get
(145)ℏ0=Θ0=1.Theκth-truncated series of Equation (142) becomes as follows:
(146)℘kυ=υ2+∑ν=1kℏνυνω+2.Now, define the ERF in the following form:(147)EResυ=℘υ−υ2−υω2℘υ−υω21−ωEE−14℘υ2E−14℘υ/2υω−υ2−ω.Theκth-ERF is as follows:
(148)EResκυ=℘κυ−υ2−υω2℘κυ−υω21−ωEE−14℘κυ2E−14℘κυ/2υω−υ2−ω.To determine the undefined coefficientsℏν, substitute κ=1,2,3,4 in Equations (146) and (148) and then solve the equation limυ⟶01/υκω+2EResκυ=0. By utilizing this algorithm, we get the following first four coefficients of the series Equation (146):
(149)ℏ1=1,ℏ2=1+2ω−1+2ω+1,ℏ3=−8ω−16ω−232ωΓ1/2+ω+π2ω−4ω−8ω−16ω−232ωΓ1+ωπΓ1+ω832ω+22ω+2−28ω−816ω−2ω,ℏ4=π2−2−ω−24ω+422ω+52ωΓ1+3ω22ω−1228ω−124ω−1Γ1/2+ωΓ1+ω2+4ω16ω+8ω+2ω+1Γ1/2+ωπ22ω−128ω−124ω−1Γ1+ω+1+28ω+4ω+32ω+64ω+16ω+2ω22ω−128ω−124ω−1+3+32ωΓ3ω22ω−128ω−124ω−1Γ1+ω3.The four-step approximate solution of Equation (142) is as follows:
(150)℘4υ=υ2+υ2+ω+1+2ω1−2ω+1υ2+2ω+ℏ3υ2+3ω+ℏ4υ2+4ω.By employing inverse E-T on both sides of Equation (150),
(151)Θ4τ=1+τωΓ1+ω+1+2ω1−2ω+1τ2ωΓ1+2ω+ℏ3τ3ωΓ1+3ω+ℏ4τ4ωΓ1+4ω.Forω=1in Equation (151), it becomes
(152)Θ4τ=1+τ+τ22!+τ33!+τ44!.Equation (152) represents the first five terms of eτ; therefore, the exact solution of Equation (140) is eτ at ω=1.Table6 shows the absolute and relative errors of the five-step approximate solution of the nonlinear PDE of Example 3 at ω=1 on various chosen grid points in the interval τ∈0,1. The numerical findings in Table 6 prove that the approximate solution converges to the exact solution quickly. This indicates the accuracy and effectiveness of the suggested scheme.Table 6
The absolute and relative errors of Example3.
τAbs.errorRel.error0.29.149350321813188×10−87.490854479152386×10−80.46.030974603721262×10−64.042683174006208×10−60.67.080039050877396×10−53.885607815121622×10−50.84.102618258010615×10−41.843425212040309×10−41.01.615161333333059×10−35.941846488086295×10−4The following 2-D graphs show the behavior of approximate and exact solutions to Example3.Figure10 exemplifies the performance of the five-step approximate solutions of Equation (140) for some values of ω and exact solution at ω=1. Apparently, results in cases of fractional values of ω converging result in the case of ω=1. Furthermore, the approximate result overlaps with the precise result at ω=1, and this once more agrees with the efficiency and precision of the recommended scheme.Figure 10The following 2-D graphs show the absolute and relative errors for Example3.Figures11 and 12 demonstrate the 2-D graphs of absolute and relative errors in the intervals τ∈0,1 over the nine-step approximate and exact solutions of Equation (140) at ω=1, respectively. The approximate solution is extremely close to the exact solution, as seen in the figures.Figure 11Figure 12


## 4.1. The Algorithm of ERPSM for Solving Nonlinear PDE

We summarized the method of finding the ERPS solution for the nonlinear PDE with the following algorithm:Step 1.
Rewriting Equation (9), we have
(116)DτωΘτ−ξτω,Θτ,Θƛτ,DτωΘƛτ=0.Step 2.
E-T is used on both sides of Equation (116).
(117)EDτωΘτ−Eξτω,Θτ,Θƛτ,DτωΘƛτ=0.By using the second part of Lemma2, we get the following:
(118)℘υ−υ2ℏ−υω℘υ=0,where EΘτ=℘υ and Eξτω,Θτ,Θƛτ,DτωΘƛτ=℘υ.Step 3.
Assume that the solution to Equation (118) has the following extension:
(119)℘υ=∑ν=0∞ℏνυνω+2.
Introduce theκth-truncated series in the following form:
(120)℘kυ=∑ν=0kℏνυνω+2.Step 4.
By using the first part of Lemma2, we have
(121)ℏ0=Θ0=I.Step 5.
Theκth-truncated series of ℘υ becomes as follows:
(122)℘kυ=Iυ2+∑ν=1kℏνυνω+2.Step 6.
Describe the ERF and theκth-ERF of Equation (118), respectively, as follows:
(123)EResυ=℘υ−υ2I−υω℘υ,(124)EResκυ=℘κυ−υ2I−υω℘κυ.Step 7.
Replace the succession arrangement of℘kυ as in Equation (120) with (124).Step 8.
Solve the succeeding expression forν=1,2,3,⋯,κ step by step to obtain the unknown coefficients
(125)limυ⟶01υκω+2EResκυ=0,κ=1,2,3,⋯.Step 9.
Replace the attained values ofℏν into the κth-truncated series of ℘κυ to obtain the κth-approximate solution of Equation (118).Step 10.
Use the inverse E-T on℘kυ to obtain the κth-approximate solution Θκτ in the original space.In the next subsection, we determine the appropriateness of the recommended method for nonlinear PDFs.


## 4.2. Numerical Examples

In this subsection, we will provide two problems with nonlinear PDEs to validate the performance and effectiveness of ERPSM.Example 2.
Examine the following nonlinear fractional pantograph differential equation:(126)DτωΘτ=1−2Θ2τ2,1<ω≤2,0≤τ≤1,subject to the initial condition
(127)Θ0=1.By employing E-T on Equation (126),
(128)℘υ−υ2−υ2+ω+2υωEE−14℘2υ2=0.Assume that algebraic equation (128) has the solution in the expansion form as
(129)℘υ=∑ν=0∞ℏνυνω+2.Theκth-truncated series of Equation (128) is as follows:
(130)℘kυ=∑ν=0kℏνυνω+2.By using the first part of Lemma2, we get
(131)ℏ0=Θ0=1.Theκth-truncated series becomes as follows:
(132)℘kυ=υ2+∑ν=1kℏνυνω+2.Now, define the ERF in the following form:(133)EResυ=℘υ−υ2−υ2+ω+2υωEE−14℘2υ2.Theκth-ERF is as follows:
(134)EResκυ=℘κυ−υ2−υ2+ω+2υωEE−14℘κ2υ2.Therefore,(135)limυ⟶01υκω+2EResκυ=0.To determine the undefined coefficientsℏν, substitute κ=1,2,3,4,5, in Equations (132) and (134) and then utilize (135); thus, we have
(136)ℏ1=−1,ℏ2=42−ω,ℏ3=−2Γ1/2+ωπΓ1+ω−168ω,ℏ4=8327ωΓ2/3+ωΓ1/3+ω64ωπΓ1+ωΓ1/2+ω+6464ω+8Γ1/2+ω8ωπΓ1+ω,ℏ5=−1283Γ1/2+ωΓ1/2+2ω354ωΓ1/3+ωΓ2/3+ωΓ1+ω−32Γ1/2+2ω16ωΓ1/2+ωΓ1+ω−1634ωΓ1/2+ω2Γ1/2+2ω3π27ωΓ1/3+ωΓ2/3+ωΓ1+ω2−2561024ω−32327ωΓ1/3+ωΓ2/3+ωπ1024ωΓ1/2+ωΓ1+ω−32Γ1/2+ω128ωπΓ1+ω.The five-step approximate solution of Equation (128) is as follows:
(137)℘5υ=υ2−υ2+ω+42−ωυ2+2ω+ℏ3υ2+3ω+ℏ4υ2+4ω+ℏ5υ2+5ω.By taking inverse E-T at both sides of Equation (137),
(138)Θ5τ=1−τωΓ1+ω+42−ωϒ2ωΓ1+2ω+ℏ3τ3ωΓ1+3ω+ℏ4τ4ωΓ1+4ω+ℏ5τ5ωΓ1+5ω.Forω=2, Equation (138) becomes
(139)Θ5τ=1−τ22!+τ44!−τ66!+τ88!−τ1010!.Equation (139) represents the first six terms of cosτ; therefore, the exact solution of Equation (126) is cosτ at ω=2.The following 2-D graphs show the behavior of approximate and exact solutions to Example2.Figure7 shows the performance of the five-step approximate solution of Equation (126) for various values of ω and exact solution at ω=2. Clearly, the findings for various fractional ω values converge to the result in the case of ω=2. Furthermore, the approximate solution does overlap with the exact solution, which indicates the accuracy and effectiveness of the suggested method.Figure 7The following 2-D graphs show the absolute and relative errors for Example2.Figures8 and 9 demonstrate the 2-D graphs of absolute and relative errors in the intervals τ∈0,1 over the five-step approximate and exact solutions of Equation (126) at ω=1, respectively. The approximate solution is extremely close to the exact solution, as seen in the figures.Figure 8Figure 9Table4 shows the absolute and relative errors at reasonable nominated grid points in the interval τ∈0,1 among the five-step approximate and exact solutions of Equation (126) at ω=1 attained using ERPSM. From Table 4, it can be perceived that the approximate and exact solutions are in very good agreement, and this sanctions the efficiency of the recommended method. The convergence of the approximate solution to the exact solution for Equation (126) has been shown numerically as in Table 5. From the obtained results, it is evident that the present technique is an effective and convenient algorithm to solve certain classes of fractional-order DEs with fewer calculations and iteration steps.Example 3.
Consider the following nonlinear PDE:(140)DτωΘτ=12Θτ+121−ωΘτ2DτωΘτ2,τ≥0,0<ω≤1,subject to the initial condition
(141)Θ0=1.Table 4
The absolute and relative errors of Example2.
τAbs.errorRel.error0.20.00.00.43.508304757815494×10−143.808982011678268×10−140.64.535483100198689×10−125.495319744192834×10−120.81.429608653680247×10−102.051951896683681×10−101.07.120558766432639×10−92.051765518966810×10−9Table 5
The recurrence errorsΘ5τ−Θ4τ of the five-step approximate solution with different values of ω for Example 2.
τω=0.7ω=0.8ω=0.9ω=1.00.26.296560270772086×10−201.969389500244771×10−215.554963006701661×10−222.821869488536155×10−240.42.279598318393725×10−171.008327424125322×10−184.022222859549857×10−192.889594356261023×10−210.67.155401866274551×10−163.876349353331781×10−171.893795546581747×10−171.666285714285714×10−190.88.253027478106390×10−155.162636411521651×10−162.912400444857598×10−162.958944620811287×10−181.05.499819058543307×10−143.846463867665569×10−152.426030253023737×10−152.755731922398591×10−17By employing the E-T on Equation (140),
(142)℘υ=υ2+υω2℘υ+υω21−ωEE−14℘υ2E−14℘υ/2υω−υ2−ω.Assume that algebraic equation (142) has the solution in the expansion form as
(143)℘υ=∑ν=0∞ℏνυνω+2.Theκth-truncated series of Equation (142) is as follows:
(144)℘kυ=∑ν=0kℏνυνω+2.By employing the first part of Lemma2, we get
(145)ℏ0=Θ0=1.Theκth-truncated series of Equation (142) becomes as follows:
(146)℘kυ=υ2+∑ν=1kℏνυνω+2.Now, define the ERF in the following form:(147)EResυ=℘υ−υ2−υω2℘υ−υω21−ωEE−14℘υ2E−14℘υ/2υω−υ2−ω.Theκth-ERF is as follows:
(148)EResκυ=℘κυ−υ2−υω2℘κυ−υω21−ωEE−14℘κυ2E−14℘κυ/2υω−υ2−ω.To determine the undefined coefficientsℏν, substitute κ=1,2,3,4 in Equations (146) and (148) and then solve the equation limυ⟶01/υκω+2EResκυ=0. By utilizing this algorithm, we get the following first four coefficients of the series Equation (146):
(149)ℏ1=1,ℏ2=1+2ω−1+2ω+1,ℏ3=−8ω−16ω−232ωΓ1/2+ω+π2ω−4ω−8ω−16ω−232ωΓ1+ωπΓ1+ω832ω+22ω+2−28ω−816ω−2ω,ℏ4=π2−2−ω−24ω+422ω+52ωΓ1+3ω22ω−1228ω−124ω−1Γ1/2+ωΓ1+ω2+4ω16ω+8ω+2ω+1Γ1/2+ωπ22ω−128ω−124ω−1Γ1+ω+1+28ω+4ω+32ω+64ω+16ω+2ω22ω−128ω−124ω−1+3+32ωΓ3ω22ω−128ω−124ω−1Γ1+ω3.The four-step approximate solution of Equation (142) is as follows:
(150)℘4υ=υ2+υ2+ω+1+2ω1−2ω+1υ2+2ω+ℏ3υ2+3ω+ℏ4υ2+4ω.By employing inverse E-T on both sides of Equation (150),
(151)Θ4τ=1+τωΓ1+ω+1+2ω1−2ω+1τ2ωΓ1+2ω+ℏ3τ3ωΓ1+3ω+ℏ4τ4ωΓ1+4ω.Forω=1in Equation (151), it becomes
(152)Θ4τ=1+τ+τ22!+τ33!+τ44!.Equation (152) represents the first five terms of eτ; therefore, the exact solution of Equation (140) is eτ at ω=1.Table6 shows the absolute and relative errors of the five-step approximate solution of the nonlinear PDE of Example 3 at ω=1 on various chosen grid points in the interval τ∈0,1. The numerical findings in Table 6 prove that the approximate solution converges to the exact solution quickly. This indicates the accuracy and effectiveness of the suggested scheme.Table 6
The absolute and relative errors of Example3.
τAbs.errorRel.error0.29.149350321813188×10−87.490854479152386×10−80.46.030974603721262×10−64.042683174006208×10−60.67.080039050877396×10−53.885607815121622×10−50.84.102618258010615×10−41.843425212040309×10−41.01.615161333333059×10−35.941846488086295×10−4The following 2-D graphs show the behavior of approximate and exact solutions to Example3.Figure10 exemplifies the performance of the five-step approximate solutions of Equation (140) for some values of ω and exact solution at ω=1. Apparently, results in cases of fractional values of ω converging result in the case of ω=1. Furthermore, the approximate result overlaps with the precise result at ω=1, and this once more agrees with the efficiency and precision of the recommended scheme.Figure 10The following 2-D graphs show the absolute and relative errors for Example3.Figures11 and 12 demonstrate the 2-D graphs of absolute and relative errors in the intervals τ∈0,1 over the nine-step approximate and exact solutions of Equation (140) at ω=1, respectively. The approximate solution is extremely close to the exact solution, as seen in the figures.Figure 11Figure 12


## 5. Conclusion

In this paper, we describe a novel method for solving FMPS and PDEs that combines the RPSM with the E-T. To assess the effectiveness and reliability of ERPSM for FMPS and PDEs, the absolute, relative, and recurrence errors of linear and nonlinear problems are studied graphically and numerically. The interpretation of the 2-D plots and tables for different values of fractional order also validates that the approximate solution is rapidly convergent to the exact solution. The numerical and graphical consequences confirm that the ERPSM is extremely effective and precise.The ERPSM stands apart from other numerical methods in four major ways. This method has the advantage of requiring no minor or major physical parameter assumptions in the problem. As a result, it applies to both weakly and strongly nonlinear problems, overcoming some of the inherent limits of traditional perturbation approaches. Second, while addressing nonlinear problems, the ERPSM does not require He’s polynomials or Adomian polynomials. To solve nonlinear DEs, only a very small number of calculations are needed. As a consequence, it outperforms homotopy analysis and Adomian decomposition methods significantly. Third, the ERPSM provided a simple and rapid way to observe the coefficients of the recommended series as a solution to the problem. In contrast to the traditional RPSM, establishing the coefficients for a series requires computing the fractional derivative every time, while the ERPSM only requires the concept of the limit at zero in establishing the coefficients for the series. Finally, unlike conventional analytic approximation techniques, the ERPSM can create expansion solutions for linear and nonlinear FODEs without the need for perturbation, linearization, or discretization.Therefore, we conclude that our novel technique is simple to apply, accurate, adaptive, and efficient according to the results. It is significant to consider that implementing the ERPSM to solve other kinds of ordinary and partial DEs of noninteger order is actively attainable. For example, fractional KdV equations, fractional phi-4 equations, and fractional Schrodinger equations.



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