# New Numerical Algorithm to Solve Variable-Order Fractional Integrodifferential Equations in the Sense of Hilfer-Prabhakar Derivative

Authors: MohammadHossein Derakhshan
Journal: Abstract and Applied Analysis
 (2021)

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## Abstract

In this article, a numerical technique based on the Chebyshev cardinal functions (CCFs) and the Lagrange multiplier technique for the numerical approximation of the variable-order fractional integrodifferential equations are shown. The variable-order fractional derivative is considered in the sense of regularized Hilfer-Prabhakar and Hilfer-Prabhakar fractional derivatives. To solve the problem, first, we obtain the operational matrix of the regularized Hilfer-Prabhakar and Hilfer-Prabhakar fractional derivatives of CCFs. Then, this matrix and collocation method are used to reduce the solution of the nonlinear coupled variable-order fractional integrodifferential equations to a system of algebraic equations which is technically simpler for handling. Convergence and error analysis are examined. Finally, some examples are given to test the proposed numerical method to illustrate its accuracy and efficiency.


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## Body

## 1. Introduction

Fractional calculus as a branch of mathematical analysis due to important applications in computational science, engineering [1–4], chemistry, and many physical equations such as Burger’s equation [5] and Korteweg-de Vries equation [6], and viscous fluid [7] has been broadly significant. Recently, obtaining solution of the linear and nonlinear differential equations has important roles in fractional calculus [8, 9], because the behavior of the equations is determined by their analytical solutions. In this paper, a differential equation including a variable-order fractional operator is considered in which the equation is an extension of the fractional differential equations of fractional order. Getting exact solutions of the equations in mathematical sciences and physics is not easy, so approximate solutions to these equations using numerical methods are obtained. Different numerical methods for obtaining solutions of the linear and nonlinear equations in the papers are presented. For instance, in [10], a solution of the nonlinear variable-order fractional Riccati differential equations was obtained by using the spectral technique. A finite difference method for getting approximate solution of the variable-order fractional Volterra integrodifferential equations (VO-FVIDEs) has been considered in [11]. The numerical solution of the variable-order fractional integrodifferential equation consisting of cubic spline interpolation can be obtained in [12]. In paper [13], a collocation technique to obtain solution of the variable-order nonlinear cable equation was found. Obtaining the numerical solution of a class of nonlinear variable-order fractional differential equations (FDEs) by using the Legendre wavelets in [14] is studied. Also, Bhrawy and Zaky [15] used the shifted Jacobi polynomials to obtain the solution for variable-order fractional Schrödinger equations, Bhrawy and Zaky [16] studied the Jacobi-Gauss-Lobatto collocation method to obtain the solution for variable-order fractional Schrödinger equations, Mahmoud et al. [17] proposed the Jacobi wavelet collocation method to obtain the solution for variable-order fractional equations, Zaky et al. [18] proposed the shifted Chebyshev polynomials to obtain the solution for variable-order fractional equations, and in [19], a proper discrete form of fractional Gronwall-type inequality is introduced and other methods such as the shifted Jacobi collocation method [13] and shifted Jacobi method [20]. Other methods are proposed to obtain numerical solutions of the variable-order fractional integrodifferential equations in the literatures [21–27]. In this paper, we obtain the solution for the following fractional differential equation of order 0<μt,ν≤1 by using of the Chebyshev cardinal functions (CCFs) and the Lagrange multiplier method in which fx,t is known and ux,t is unknown:
(1)CDρ,ω,0+γ,μt,νux,t+Dρ,ω,0+γ,μt,νux,t=fx,t,ux,t.The initial condition for this equation is defined by(2)ux,0=d.Here,CDρ,ω,0+γ,μt,ν and Dρ,ω,0+γ,μt,ν are regularized Hilfer-Prabhakar fractional derivative [28] and Hilfer-Prabhakar fractional derivative [28] of order 0<μt≤1,0<ν≤1, respectively. According to the definition of the Hilfer fractional derivative, this Hilfer-Prabhakar fractional derivative is defined but instead of Riemann integral operator in the Hilfer fractional derivative, Prabhakar integral operator [28] of variable-order fractional is replaced, in which this Prabhakar integral operator for 0<μt≤1 is given by
(3)Eρ,μt,ω,0+γux,t=∫0tt−τμt−1Eρ,μtγωt−τρux,τdτ,where Eρ,μtγωtρ is the Prabhakar function and it is a generalization of the two-parametric Mittag-Leffler function Eα,βt,α>0,β∈ℂ, for γ=1 and the classical Mittag-Leffler function Eαt,α>0, for γ=μt=1 is considered which is defined by [28]
(4)Eρ,μtγt=∑n=0∞γnn!Γρn+μttn,Rρ,Rμt>0,γ>0,t∈ℂ,where γn=Γγ+n/Γγ and Γ. are the gamma function; moreover, this Prabhakar function for Rμt>0 is an entire function [29]. Our interest in this type of fractional derivative is related to its application in physical phenomena, for example, linear viscoelastic constitutive equations [30], anomalous dielectric relaxation of Havriliak-Negami function [31, 32], fractional viscoelasticity [33], spherical stellar systems, stochastic processes [34], telegraph equations [34], and anomalous relaxation in dielectrics [35]. Since this mathematical system which is given in Equation (1), due to the variable-order fractional operators and nonlinearity, is very complex, we need to reduce it by using a highly accurate and efficient expansion scheme. Thus, for this aim, we first extract an operational matrix of variable-order fractional derivatives for the CCFs; then, we use them for extending the unknown solution. In other words, the problem shown in Equation (1) is converted into an algebraic system of equations by exploiting the operational matrix of variable-order derivative. In this paper, we consider a class of variable-order fractional integrodifferential equations. For solving the given equations, operational matrices based on the CCFs are applied. First, we approximate the unknown function and its derivatives in terms of the CCFs. Then, by substituting these approximations into the equation and applying the properties of the CCFs together with the collocation points, the main problem is reduced to a set of nonlinear algebraic equations. By solving this system, the approximate solution is calculated. This article is divided into four sections as follows. In Section 2, some important and application definitions, lemma, and theorem which are applied in the next section are expressed. In Section 3, properties of the Chebyshev cardinal functions (CCFs) and the Lagrange multipliers are introduced. In Section 4, we use the properties described in Section 3 to solve the variable-order fractional differential equations. In Section 5, to illustrate the efficiency and accuracy of the proposed method in this article, four examples are displayed.


## 2. Definition and Fundamental Properties of Fractional Calculus

This section refers to the definitions and lemmas which are used for the next sections, and these definitions and lemmas are taken from articles [36, 37].
(5)Dρ,ω,0+γ,μt,νux,t=Eρ,1−μtν,ω,0+−νγddtEρ,1−μt1−ν,ω,0+−1−νγux,t,where ux,t∈Cxmin,xmax×0,tmax and d/dtux,t is an absolutely continuous function. Also, the Hilfer-Prabhakar derivative of a regularized type with similar properties above is defined as [37]
(6)CDρ,ω,0+γ,μt,νux,t=Eρ,1−μt,ω,0+−γddtux,t.Definition 1.
For0<μt≤1,0<ν≤1,ρ>0 and ω,γ∈ℝ, the Hilfer-Prabhakar derivative can be defined as [28, 37]Lemma 2.
For0<μt≤1,0<ν≤1,ρ>0 and ω,γ∈ℝ, the following formula holds [36]:
(7)Eρ,μt,ω,0+γtq−1x=Γqxμt+q−1Eρ,μt+qγωxρ,q>1.
Also, the following relation is obtained for fractional operatorDρ,ω,0+γ,μt,ν on the function tq−1[37]:
(8)Dρ,ω,0+γ,μt,νtq−1x=xq−μt−1Eρ,q−μt−γωxρ,q>1.Lemma 3.
For0<μt≤1,0<ν≤1,ρ>0 and ω,γ∈ℝ, the Hilfer-Prabhakar derivative of a regularized model is calculated as follows:
(9)CDρ,ω,0+γ,μt,νtq−1x=Γq−2xq−μt−1Eρ,q−μt−γωxρ,q>2.Proof. Applying the normal derivative definition and formulas (6) and (7) on the function t with the power of q, the result is obtained.


## 3. Approximation Function and Properties of the Lagrange and the CCF Polynomials

Any functionux,t∈Cxmin,xmax×0,tmax can be approximated in terms of Chebyshev cardinal functions of order m and Lagrange interpolating polynomials of order n as follows:
(10)ux,t≃∑i=1m+1∑j=1n+1uxi,tjφixψjt,where φix is the Chebyshev cardinal functions of order m on the interval x∈xmin,xmax which is defined by [38]
(11)φix=∏k=1,k≠im+1x−xkxi−xk,i=1,2,⋯,m+1,xk=xmincos2k−1π2m+2=xmax+xmin−xmax−xmin2cos2k−1π2m+2,and ψjt is the Lagrange interpolating polynomials of order n on the interval 0,tmax which is given by [38]
(12)ψjt=∏k=1,k≠jn+1t−tktj−tk,j=1,2,⋯,n+1,where
(13)ψjti=δj,i=1,i=j,0,i≠j.Also, in relation (10), tj is considered
(14)tj=tmax21−cos2j−1π2n+2.Functionux,t in Equation (10) can be rewritten as follows:
(15)ux,t≃∑i=1m+1∑j=1n+1uxi,tjφixψjt≜YmTx×Δ×Λnt,where Ymx,Δ,Λnt is defined by
(16)Ymx=φ1x,φ2x,⋯,φm+1xT,Δ=uijm+1×n+1,Λnt=ψ1t,ψ2t,⋯,ψn+1tT.The purpose of this paper is to express the functionψjt,j=0,1,2,⋯,n+1, as an operational matrix as follows [39]:
(17)ψjt=1γj×bj,0tn+bj,1tn−1+⋯+bj,n=1γj×bj,0,bj,1,⋯,bj,n1×n+1×tntn−1⋮1n+1×1,where γj,bj,k for k=0,1,2,⋯,n are given by
(18)γj=∏r=1,r≠jn+1tj−tr,bj,0=1,bj,k=−∑r=1k∑l=1n+1aj,rbj,k−rtlrk,1≤k≤n,r≠j.Theorem 4.
Supposeψjt is relation (17). Then, the Hilfer-Prabhakar derivative of a regularized type and the Hilfer-Prabhakar derivative on the function ψjt are obtained, respectively, as follows:
(19)CDρ,ω,0+γ,μt,νψjt=1γj×bj,0,bj,1,⋯,bj,n1×n+1×Γn−1tn−μtEρ,n−μt+1−γωtρΓn−2tn−μt−1Eρ,n−μt−γωtρ⋮0n+1×1=1γj∑k=0n−1bj,kn−k−1!tn−μt−kEρ,n−μt−k+1−γωtρ,(20)Dρ,ω,0+γ,μt,νψjt=1γj×bj,0,bj,1,⋯,bj,n1×n+1×tn−μtEρ,n−μt+1−γωtρtn−μt−1Eρ,n−μt−γωtρ⋮t−μtEρ,1−μt−γωtρn+1×1=1γj∑k=0nbj,ktn−μt−kEρ,n−μt−k+1−γωtρ.Proof. By taking the Hilfer-Prabhakar derivative of a regularized type on both sides of Equation (17), we get
(21)CDρ,ω,0+γ,μt,vψjt=1γj×bj,0CDρ,ω,0+γ,μt,vtn+bj,1CDρ,ω,0+γ,μt,vtn−1+⋯+bj,nCDρ,ω,0+γ,μt,v1=1γj×bj,0,bj,1,⋯,bj,n1×n+1×CDρ,ω,0+γ,μt,vtnCDρ,ω,0+γ,μt,vtn−1⋮CDρ,ω,0+γ,μt,v1n+1×1,applying Equation (9) on Equation (21) and the result is achieved. Prove relation (20) like the same process is considered for the Hilfer-Prabhakar derivative of a regularized type.Now, using Theorem4, the function Hilfer-Prabhakar derivative of a regularized-type ψjt can be approximated as a finite series in the form of the Chebyshev cardinal functions as follows:
(22)CDρ,ω,0+γ,μt,νψjt≃∑r=1n+11γj∑k=0n−1bj,kn−k−1!trn−μt−kEρ,n−μtr−k+1−γωtrρψrt=∑r=1n+1qj,rμtψrt,j=1,2,⋯,n+1.Also, using Theorem4, the function Hilfer-Prabhakar derivative ψjt can be approximated as a finite series in the form of the Chebyshev cardinal functions as follows:
(23)Dρ,ω,0+γ,μt,νψjt≃∑r=1n+11γj∑k=0nbj,ktrn−μt−kEρ,n−μtr−k+1−γωtrρψrt=∑r=1n+1pj,rμtψrt,j=1,2,⋯,n+1.So according to relations (22) and (23), we can get the following results:
(24)CDρ,ω,0+γ,μt,νΛnt=CDρ,ω,0+γ,μt,νψ1t,CDρ,ω,0+γ,μt,νψ2t,⋯,CDρ,ω,0+γ,μt,νψn+1tT=EμtΛnt,Eμt=qj,rμtn+1×n+1,j=1,2,⋯,n+1,r=1,2,⋯,n+1,(25)Dρ,ω,0+γ,μt,νΛnt=Dρ,ω,0+γ,μt,νψ1t,Dρ,ω,0+γ,μt,νψ2t,⋯,Dρ,ω,0+γ,μt,νψn+1tT=EμtΛnt,Eμt=pj,rμtn+1×n+1,j=1,2,⋯,n+1,r=1,2,⋯,n+1.Theorem 5.
Supposeux,t is an exact solution and umx,t is a numerical solution of Equation (1) in which ux,t,umx,t∈C0,xmax×0,tmax and M=maxx,t∈0,xmax×0,tmax∣∂m+1uρ,ς/∂tr∂xm+1−r∣ exist. Then, the following relation holds:
(26)∥ux,t−umx,t∥2≤Mxmax+tmaxm+1tmaxxmaxm+1!.Proof. To prove, Taylor’s expansion functionux,t around a point 0,0 can be written as
(27)ux,t=u0,0+x∂u0,0∂x+t∂u0,0∂t+12!x2∂2u0,0∂x2+t2∂2u0,0∂t2+xt∂2u0,0∂t∂x+⋯+1m!x∂u0,0∂x+t∂u0,0∂tm+1m+1!x∂uρ,ς∂x+t∂uρ,ς∂tm+1.We consider the following finite expansion as an approximation of the functionux,t:
(28)ux,t≃umx,t=u0,0+x∂u0,0∂x+t∂u0,0∂t+12!x2∂2u0,0∂x2+t2∂2u0,0∂t2+xt∂2u0,0∂t∂x+⋯+1m!x∂u0,0∂x+t∂u0,0∂tm.Then, we have(29)∥ux,t−umx,t∥2=∥1m+1!x∂uρ,ς∂x+t∂uρ,ς∂tm+1∥2∫0tmax∫0xmax1m+1!x∂uρ,ς∂x+t∂uρ,ς∂tm+12dxdt1/2,considering
(30)x∂uρ,ς∂x+t∂uρ,ς∂tm+1=∑r=0m+1mr∂m+1uρ,ς∂tr∂xm+1−rtrxm+1−r,and we obtain
(31)∥ux,t−umx,t∥2=∫0tmax∫0xmax1m+1!∑r=0m+1mr∂m+1uρ,ς∂tr∂xm+1−rtrxm+1−r2dxdt1/2≤maxx,t∈0,xmax×0,tmax∣∂m+1uρ,ς∂tr∂xm+1−r∣∫0tmax∫0xmax1m+1!∑r=0m+1mrtrxm+1−r2dxdt1/2=Mm+1!∫0tmax∫0xmaxx+t2m+2dxdt12≤Mxmax+tmaxm+1tmaxxmaxm+1!.For convergence of the function, sequenceumx,t→ux,t sufficient to show that the sequence umx,t is a Cauchy sequence in space C0,xmax×0,tmax. So, we have
(32)∥unx,t−umx,t∥2≤∥ux,t−umx,t∥2+∥ux,t−unx,t∥2≤M1xmax+tmaxm+1tmaxxmaxm+1!+M2xmax+tmaxn+1tmaxxmaxn+1!,M1,M2>0.According to the convergence growth ratexmax+tmaxm+1≪m+1!, as m,n→+∞, we have ∥unx,t−umx,t∥→0; therefore, the sequence umx,t is a Cauchy sequence in space C0,xmax×0,tmax. Since C0,xmax×0,tmax is complete, then umx,t converges. To prove the uniqueness of the solutions, we assume that u,u¯ are two different solutions of limm→∞umx,t; then, we have
(33)∥ux,t−u¯x,t∥2≤∥ux,t−umx,t∥2+∥u¯−umx,t∥2.Due to∥ux,t−umx,t∥→0 and ∥u¯x,t−umx,t∥→0, we have ∥ux,t−u¯x,t∥2→0, then u=u¯.


## 4. A Description of the Proposed Method for Solving the System

In this section, using relations described in Section3, we solve the proposed system stated in Section 1 which is introduced with (1). Using relation (15) and applying relations (24) and (25) on Equation (1), in this case, we obtain
(34)CDρ,ω,0+γ,μt,νux,t+Dρ,ω,0+γ,μt,νux,t=fx,t,ux,t,YmTxΔEμtΛnt+YmTxΔEμtΛnt=fx,t,ux,t.Using relation (15) for the function fx,t, in this case, the following relation is obtained:
(35)YmTxΔEμtΛnt+YmTxΔEμtΛnt=YmTxMΛnt,YmTxΔEμt+EμtΛnt=YmTxMΛnt,YmTxΔEμt+Eμt−MΛnt≜YmTxWΛnt≃0,where M=fxi,tj,uxi,tjm+1×n+1 and W=ΔEμt+Eμt−M and the initial condition for this equation is defined by
(36)ux,0=d≃YmTx×Δ×Λn0=YmTxd,YmTxΔΛn0−d≜YmTxΞ≃0,Here,dx=∑i=1m+1dxiφix≜dTYmx,d=dx1,dx2,⋯,dxm+1T, and Ξ=ΔΛn0−d are considered. By solving the following equations, the unknown function ux,t is determined:
(37)Wi,j=0,i=2,3,⋯,m,j=2,3,⋯,n,Ξi=0,i=1,2,3,⋯,m+1.


## 5. Show Numerical Examples Using the Proposed Method

In this section, we show the numerical results of the suggested method on four examples to illustrate the efficiency and accuracy of our scheme, where each of these examples is taken from various articles and all these examples using the method described in this article have been resolved for different values of a continuous functionμt and m,n. Here, obtained numerical results of some test problems are showed to illustrate the computational efficiency and applicability of the reported method. All numerical examples are done by applying MATLAB (R2020b) software. Also, the convergence order (C-order) of the depicted method is calculated by the following relation:
(38)C‐order=∣logE1/E2∣logm2+1m2+1/m1+1m1+1.Here,E1 and E2 show the first and the second values of the maximum absolute error (MAE) provided by the proposed method, respectively. Moreover, mi+1mi+1 for i=1,2 are the number of the basis functions used in the first and the second implementations, respectively. Here, the absolute error which is the difference between a numerical and exact solution is defined by
(39)E=uexact−unumerical.Example 6.
Letμt=6/10−3/10sinπt. Then, we consider the following linear fractional system of orders μt:
(40)CDρ,ω,0+γ,μt,νux,t+12Dρ,ω,0+γ,μt,νux,t=fx,t,ux,t,ux,0=e−x,where fx,t,ux,t is given by
(41)fx,t=e−x+te−x+t−sint−1+∑k=0∞−1ktk+1−μtEρ,k+2−μt−γωtρe−x+12∑k=0∞−1ktk+1−μtEρ,k+1−μt−γωtρe−x.The exact solution isux,t=e−x+t. The upper part in Figure 1 illustrates the numerical solution of Example 6 with m=4 and n=10, and the lower part shows the absolute error of this example. Table 1 shows the absolute error values for different values of m,n. Also, The values of the C-order of the presented method are listed in Table 2. From the obtained numerical results in Table 1, we see that the proposed method is an effective and good tool for solving this problem.Example 7.
Letμt=sin10πt/6+2/3. Then, we state the following nonlinear fractional system:
(42)CDρ,ω,0+γ,μt,νux,t+12Dρ,ω,0+γ,μt,νux,t=fx,t,ux,t,ux,0=0,where fx,t,ux,t is given by
(43)fx,t,ux,t=sinux,t−∑k=0∞−1kt2k+2−μtEρ,2k+3−μt−γωtρx3−12∑k=0∞−1kt2k+2−μtEρ,2k+3−μt−γωtρx3−sinx3sint+6xt3sint+x3cost1+e−xt.Figure 1
Portrait of the numerical solution and the absolute error for Example6 with m=4,n=10.Table 1
Absolute error for different values ofm,n for Example 6.
x
,
t
∣
u
x
,
t
−
u
∗
x
,
t
∣
∣
u
x
,
t
−
u
∗
x
,
t
∣
m
=
4
,
n
=
10
m
=
10
,
n
=
20
m
=
15
,
n
=
30
m
=
20
,
n
=
40
(0.68, 0.09)
4.9947
e
−
6
5.0090
e
−
7
1.8612
e
−
7
4.5778
e
−
8
(0.86, 0.18)
4.1719
e
−
6
4.1838
e
−
7
1.5546
e
−
7
3.8237
e
−
9
(1.04, 0.27)
3.4846
e
−
6
3.4946
e
−
7
1.2985
e
−
7
3.1938
e
−
9
(1.31, 0.09)
2.6601
e
−
6
2.6677
e
−
7
9.9127
e
−
8
2.4381
e
−
9
(2.21, 0.36)
1.0815
e
−
6
1.0846
e
−
7
4.8250
e
−
8
9.9127
e
−
10
(3.11, 0.45)
4.3971
e
−
7
4.4097
e
−
8
1.96172
e
−
8
4.0302
e
−
10Table 2
C-order of the presented method for Example 6.
C-order
C-order
m
=
4
,
n
=
10
m
=
10
,
n
=
20
m
=
15
,
n
=
30
m
=
20
,
n
=
40
8.56
9.23
11.54
13.65The analytical solution for this question isux,t=x3sint. The upper part in Figure 2 illustrates the numerical solution of Example 7 with m=4 and n=10, and the lower part shows the absolute error of this example. Table 3 shows the absolute error values for different values of m,n.Example 8.
We consider a fractional system of ordersμt=t2−t+0.8 as follows:
(44)CDρ,ω,0+γ,μt,νux,t−Dρ,ω,0+γ,μt,νux,t+ux,t=fx,t,ux,0=0,where fx,t is given as
(45)fx,t=2x2−xt2−tlnt+t22+2x−x2,where for this problem, analytical solution is ux,t=t22x−x2.The upper part in Figure 3 illustrates the numerical solution of Example 8 with m=4,n=10, and the lower part shows the absolute error of this example. Table 4 shows the absolute error values for different values of m,n.Example 9.
We consider a fractional system as follows:(46)CDρ,ω,0+γ,μt,νux,t+Dρ,ω,0+γ,μt,νux,t−∂2∂x2+∂2∂t2ux,t=fx,t,ux,0=0,μt=510−3e−t10,where fx,t is given as
(47)fx,t=2ltl−1t−1Γ1+llnt+tsinx+1,l>0,where for this problem, analytical solution is ux,t=2ltlsinx+1. The upper part in Figure 4 illustrates the numerical solution of Example 9 with m=4 and n=10, and the lower part shows the absolute error of this example. Table 5 shows the absolute error values for different values of m,n. Also, The values of the C-order of the presented method are listed in Table 6.Figure 2
Portrait of the numerical solution and the absolute error for Example7 with m=4,n=10.Table 3
Absolute error for different values ofm,n for Example 7.
x
,
t
∣
u
x
,
t
−
u
∗
x
,
t
∣
∣
u
x
,
t
−
u
∗
x
,
t
∣
m
=
4
,
n
=
10
m
=
10
,
n
=
20
m
=
15
,
n
=
30
m
=
20
,
n
=
40
(0.68, 0.09)
2.7916
e
−
8
2.7082
e
−
8
5.3945
e
−
9
2.5191
e
−
9
(0.86, 0.18)
1.0649
e
−
8
5.4784
e
−
9
1.09125
e
−
9
5.0958
e
−
10
(1.04, 0.27)
9.5832
e
−
9
9.6885
e
−
10
1.9298
e
−
10
9.0120
e
−
11
(1.31, 0.09)
1.9152
e
−
9
1.9363
e
−
10
3.8569
e
−
11
1.8011
e
−
11
(2.21, 0.36)
9.1957
e
−
10
9.2968
e
−
11
1.8518
e
−
11
8.6477
e
−
12
(3.11, 0.45)
2.5626
e
−
10
1.9729
e
−
11
5.1607
e
−
12
2.4099
e
−
12Figure 3
Portrait of the numerical solution and the absolute error for Example8 with m=4,n=10.Table 4
Absolute error for different values ofm,n for Example 8.
x
,
t
∣
u
x
,
t
−
u
∗
x
,
t
∣
∣
u
x
,
t
−
u
∗
x
,
t
∣
m
=
4
,
n
=
10
m
=
10
,
n
=
20
m
=
15
,
n
=
30
m
=
20
,
n
=
40
(0.68, 0.09)
6.9725
e
−
7
7.0321
e
−
8
2.8128
e
−
8
8.5088
e
−
9
(0.86, 0.18)
7.6156
e
−
8
7.6807
e
−
9
3.0723
e
−
9
9.2937
e
−
10
(1.04, 0.27)
7.7555
e
−
9
7.8218
e
−
10
3.1287
e
−
10
9.4643
e
−
11
(1.31, 0.09)
7.0214
e
−
9
7.0814
e
−
10
2.8325
e
−
10
8.5685
e
−
11
(2.21, 0.36)
3.6050
e
−
9
3.6359
e
−
10
1.4543
e
−
10
4.3994
e
−
11
(3.11, 0.45)
2.6815
e
−
9
2.7044
e
−
10
1.0817
e
−
10
3.2724
e
−
11Figure 4
Portrait of the numerical solution and the absolute error for Example9 with m=4,n=10.Table 5
Absolute error for different values ofm,n for Example 9.
x
,
t
∣
u
x
,
t
−
u
∗
x
,
t
∣
∣
u
x
,
t
−
u
∗
x
,
t
∣
m
=
4
,
n
=
10
m
=
10
,
n
=
20
m
=
15
,
n
=
30
m
=
20
,
n
=
40
(0.68, 0.09)
1.5851
e
−
7
9.5658
e
−
8
8.6331
e
−
8
2.3914
e
−
8
(0.86, 0.18)
1.2029
e
−
7
7.2593
e
−
8
6.5515
e
−
8
1.8148
e
−
8
(1.04, 0.27)
8.0869
e
−
8
4.8802
e
−
8
4.4043
e
−
8
1.2200
e
−
8
(1.31, 0.09)
2.0344
e
−
8
1.2277
e
−
8
1.1080
e
−
8
3.0692
e
−
9
(2.21, 0.36)
1.7705
e
−
8
1.0684
e
−
8
9.6429
e
−
9
2.6711
e
−
9
(3.11, 0.45)
3.3110
e
−
9
1.9981
e
−
9
1.8032
e
−
9
4.9952
e
−
10Table 6
C-order of the presented method for Example 9.
C-order
C-order
m
=
4
,
n
=
10
m
=
10
,
n
=
20
m
=
15
,
n
=
30
m
=
20
,
n
=
40
18.13
19.11
20.34
21.54



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*Source: 8817794-2021-02-24.xml*