# Numerical Approach Based on Two-Dimensional Fractional-Order Legendre Functions for Solving Fractional Differential Equations

Authors: Qingxue Huang; Fuqiang Zhao; Jiaquan Xie; Lifeng Ma; Jianmei Wang; Yugui Li
Journal: Discrete Dynamics in Nature and Society
 (2017)

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## Abstract

In this paper, a robust, effective, and accurate numerical approach is proposed to obtain the numerical solution of fractional differential equations. The principal characteristic of the approach is the new orthogonal functions based on shifted Legendre polynomials to the fractional calculus. Also the fractional differential operational matrix is driven. Then the matrix with the Tau method is utilized to transform this problem into a system of linear algebraic equations. By solving the linear algebraic equations, the numerical solution is obtained. The approach is tested via some examples. It is shown that the FLF yields better results. Finally, error analysis shows that the algorithm is convergent.


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## Body

## 1. Introduction

Fractional-order calculus has a long history. Compared with integer order, fractional differential equations show many advantages over the simulation of physical phenomena [1]. In recent decades, fractional partial differential equations (FPDEs) [2] are widely used to describe engineering processes and dynamical systems; more and more researchers have devoted to study a variety of methods for solving fractional differential equations. The most commonly used methods are generalized differential transform method [3], Adomian decomposition method [4], finite difference method [5], and so on. Recently, the authors have proposed wavelet methods for solving a class of fractional convection-diffusion equations. In [6], different families of wavelets [7–11] have been widely applied for solving problems of partial differential equations. In [12], the authors have obtained the approximate solutions of fractional differential equations using the generalized block pulse operational matrix.The methods based on the orthogonal functions [13–15] are powerful and wonderful for solving FPDEs and have achieved great success in this field. Now the operational matrices of fractional derivative and integral [14–16] for Bernstein polynomials and Jacobi polynomials [15] have been derived. However, these polynomials use integer power series to approximate fractional ones; it cannot accurately represent properties of fractional integral and differential. Recently, in [17], Rida and Yousef have proposed a fractional extension of classical Legendre polynomials by replacing the integer order derivative in Rodrigues formula [18] with fractional-order derivatives. These functions are complex, so they are not unsuitable for solving FPDEs. Subsequently, Kazem et al. put forward the orthogonal fractional-order Legendre functions based on shifted Legendre polynomials. In [19], they used these functions to obtain the numerical solutions of fractional partial differential equations. Numerical results show that their methods are effective, accurate, and easy to implement. In this paper, FLF is suitable to characterize the properties of fractional partial differential equations; thus we attempt to use fractional-order Legendre functions to acquire numerical solution of the FPDEs.The article is organized as follows. In Section2, we introduce some basic definitions of fractional derivatives and integrals. In Section 3, we construct the FLFs and introduce some properties. In Section 4, the fractional operational matrices of FLFs are obtained. In Section 5, we use the proposed method to solve FPDEs. In Section 6, the proposed method is tested through several numerical examples. In Section 7, the error analysis is given. Finally Section 8 concludes the paper.


## 2. Preliminaries and Notations

In this section, we recall the essentials of the fractional calculus theory that will be used in this article.Definition 1.
The Riemann-Liouville fractional integral operator of orderα>0 is defined as(1)Iαut=1Γα∫0tt-sα-1usds,α>0,I0ut=ut.Definition 2.
The Riemann-Liouville definition of fractional differential operator is given as follows:(2)D∗αut=1Γm-αdmdtm∫0tust-sα-m+1ds,α>0,m-1≤α<m,dmutdtm,α=m.Definition 3.
The Caputo definition of fractional differential operator is defined as(3)Dαut=1Γm-α∫0tumτt-τα-m+1dτ,m-1≤α<m,dmutdtm,α=m.The relation between the Riemann-Liouville operator and Caputo operator is given by the following expressions:(4)DαIαut=ut,IαDαut=ut-∑k=0m-1uk0+tkk!.Forα≥0,v≥-1, and constant C, Caputo fractional derivative has some basic properties which are needed here as follows:(5)iDαC=0,iiDαtv=0,forv∈N0,v<α,Γv+1Γv+1-αtv-α,forv∈N0,v≥αorv∉N0,v>α.iiiDα∑i=0mciuit=∑i=0mciDαuit,wherecii=0mareconstants.Definition 4 (generalized Taylor’s formula).
Suppose thatDiαut∈C0,1 for i=0,1,…,m-1, then one has(6)ut=∑i=0m-1tiαΓiα+1Diαu0++tmαΓmα+1Dmαuξ,where 0<ξ≤t,∀t∈0,1. Also, one has(7)ut-∑i=0m-1tiαΓiα+1Diαu0+≤MαtmαΓmα+1,where Mα≥Dmαuξ.In caseα=1, the generalized Taylor’s formula (6) is the classical Taylors formula.


## 3. Definitions of Fractional-Order Legendre Definition and Function Approximation

### 3.1. Shifted Legendre Polynomials

The well-known Legendre polynomials are orthogonal with the weight functionωx=1, which is defined on the interval -1,1; namely,(8)∫-11PmxPnxωxdx=22n+1,m=n,0,m≠n.In order to use these polynomials on the interval0,1, let t=2x-1, and then we derived the shifted Legendre polynomials Pn2x-1 which are denoted by Lnt. The shifted Legendre polynomials are orthogonal with the weight function ωlt=1 over 0,1, with the orthogonally property(9)∫01LitLjtωltdt=0,i≠j,12i+1,i=j.ThenLit can be obtained as follows:(10)Li+1t=2i+1i+1Lit-ii+1Li-1t,i=1,2,…,L0t=1,L1t=2t-1.The analytic form of the shifted Legendre polynomial Lnt of degree n is given by(11)Lnt=∑i=0nbniti,Ln0=-1n,Ln1=1,where bni=-1n+in+i!/n-i!(i!)2.


### 3.2. Fractional-Order Legendre Definition

We define the fractional-order Legendre functions (FLFs) by introducing the change of variablet=xα and α>0 on shifted Legendre polynomials. The fractional-order Legendre function Lixα is denoted by Fliαx. The fractional-order Legendre functions are a particular solution of the normalized eigenfunctions of the singular Sturm-Liouville problem. (12)x-x1+αFli′αx′+α2ii+1xα-1Flix=0,x∈0,1.ThenFliαx can be obtained as follows:(13)Fli+1αx=2i+12xα-1i+1Fliαx-ii+1Fli-1αx,i=1,2,…,Fl0αx=1,Fl1αx=2xα-1.We can derive the analytic form ofFliαx of degree iα as follows:(14)Fliαx=∑s=0ibs,ixsα,where bs,i=-1i+si+s!/i-s!s!2 and Fliα0=-1i,Fliα1=1.Theorem 5.
The FLFs are orthogonal with the weight functionωlαx=xα-1 on the interval 0,1; then the orthogonally condition is(15)∫01FlnαxFlmαxωlαxdx=12n+1αδnm.Proof.
With∫01LntLmtωltdt=1/2i+1δnm, where δnm is the Kronecker function, let t=xα, and then we have(16)∫01LntLmtdt=∫01LnxαLmxααxα-1dx=12n+1δnm,∫01LnxαLmxααxα-1dx=∫01FlnαxFlmαxαxα-1dx=12n+1δnm,∫01FlnαxFlmαxxα-1dx=12n+1αδnm.


### 3.3. Function Approximation

Supposeux∈L20,1; it can be expanded in terms of the fractional-order Legendre functions as follows:(17)ux=∑i=0∞ciFliαx,where the coefficients ci are obtained by(18)ci=α2i+1∫01Fliαxuxωlαxdx,i=0,1,2,….If we consider truncated series in (17), we obtain(19)ux≈umx=∑i=0m-1ciFliαx=CTΦx,where C and Φx are m×1 vector given by(20)C=c0,c1,…,cm-1T,Φx=Fl0αx,Fl1αx,…,Flm-1αxT.Theorem 6.
SupposeDiαux∈C0,1 for i=0,1,2,…,m-1. 2m+1α≥1 and Pmα=spanFl0αx,Fl1αx,…,Flm-1αx. If umx=CTΦx is the best approximation to ux from Pmα, then the error bound is presented as follows:(21)ux-umxω≤MαΓmα+112m+1α,where Mα≥Dmαux,x∈0,1.Proof.
Consider the generalized Taylor’s formula(22)ux=∑i=0m-1xiαΓiα+1Diαu0++xmαΓmα+1Dmαuξ,where 0<ξ<x,x∈0,1.
With Definition4(23)ux-∑i=0m-1xiαΓiα+1Diαf0+≤MαxmαΓmα+1.Since umx=CTΦx is the best approximation to u from Pmα and ∑i=0m-1xiα/Γiα+1Diαf0+∈Pmα, hence(24)ux-umxω2≤ux-∑i=0m-1xiαΓiα+1Diαu0+ω2≤Mα2Γmα+12∫01x2mαxα-1dx,ux-umxω2≤Mα2Γmα+12∫01x2m+1α-1dx=Mα2Γmα+122m+1α.
Now by taking the square roots, then the theorem is proved.For arbitrary functionux,t∈L20,1×0,1, they can be expanded as the following formula:(25)ux,t=∑i=0∞∑j=0∞uijFliaxFljβt,where(26)uij=2i+12j+1αβ∫01∫01ux,tFliαxFljβtωlαxωlβtdxdt,i,j=0,1,….Theorem 7.
If the series∑i=0∞∑j=0∞uijFliaxFljβt converges uniformly to ux,t on the square 0,1×0,1, then one has (27)uij=2i+12j+1αβ∫01∫01ux,tFliαxFljβtωlαxωlβtdxdt.Proof.
By multiplyingωlαxωlβtFlmαxFlnβt on both sides of (25), where i and j are fixed and integrating term wise with regard to x and t on 0,1×0,1, then(28)∫01∫01ux,tFlmαxFlnβtωlαxωlβtdxdt=∑i=0∞∑j=0∞uij∫01∫01FliαxFljβtFlmαxFlnβtωlαxωlβtdxdt=∑i=0∞∑j=0∞uij∫01ωlαxFliαxFlmαxdx∫01ωlβtFliβtFlmαtdt=uij∫01ωlαxFliαx2dx∫01ωlβtFliβt2dt=uij12i+1α12j+1β.Theorem 8.
If the functionux,t is a continuous function on 0,1×0,1 and the series ∑i=0∞∑j=0∞uijFliaxFljβt converges uniformly to ux,t, then ∑i=0∞∑j=0∞uijFliaxFljβt is the FLFs expansion of ux,t.Proof.
Using contradiction, let(29)ux,t=∑i=0∞∑j=0∞uijFliaxFljβt,ux,t≈∑i=0∞∑j=0∞gijFliaxFljβt.Then there is at least one coefficient such that unm≠gnm; however(30)unm=2m+12n+1αβ∫01∫01ux,tFlmaxFlnβtωlαxωlβtdxdt=gnm.Theorem 9.
If two continuous functions defined on0,1×0,1 have the identical FLFs expansions, then these two functions are identical.Proof.
Suppose thatux,t and fx,t can be expended by FLFs as follows:(31)ux,t≈∑i=0∞∑j=0∞uijFliaxFljβt,fx,t≈∑i=0∞∑j=0∞uijFliaxFljβt.
By subtracting the above two equations with each other, we have(32)ux,t-fx,t≈∑i=0∞∑j=0∞uij-uijFliaxFljβt=0=∑i=0∞∑j=0∞0FliaxFljβt.Theorem 10.
If the sum of the absolute values of the FLFs coefficients of a continuous functionux,t forms a convergent series, then the FLFs expansion is absolutely uniformly convergent and converges to the function ux,t.Proof.
Consider(33)∑i=0∞∑j=0∞uijFliaxFljβt≤∑i=0∞∑j=0∞uijFliaxFljβt≤∑i=0∞∑j=0∞uij.
If we consider truncated series in (25), we obtain(34)ux,t≈∑i=0m-1∑j=0n-1uijFliaxFljβt=ΦTxUΦt,where(35)Φx=Fl0αx,Fl1αx,…,Flm-1αxT;Φt=Fl0βt,Fl1βt,…,Fln-1βtT,U=uiji,j=0m-1,n-1.
In this paper, we use the Tau method [20] to compute the coefficients uij.


## 3.1. Shifted Legendre Polynomials

The well-known Legendre polynomials are orthogonal with the weight functionωx=1, which is defined on the interval -1,1; namely,(8)∫-11PmxPnxωxdx=22n+1,m=n,0,m≠n.In order to use these polynomials on the interval0,1, let t=2x-1, and then we derived the shifted Legendre polynomials Pn2x-1 which are denoted by Lnt. The shifted Legendre polynomials are orthogonal with the weight function ωlt=1 over 0,1, with the orthogonally property(9)∫01LitLjtωltdt=0,i≠j,12i+1,i=j.ThenLit can be obtained as follows:(10)Li+1t=2i+1i+1Lit-ii+1Li-1t,i=1,2,…,L0t=1,L1t=2t-1.The analytic form of the shifted Legendre polynomial Lnt of degree n is given by(11)Lnt=∑i=0nbniti,Ln0=-1n,Ln1=1,where bni=-1n+in+i!/n-i!(i!)2.


## 3.2. Fractional-Order Legendre Definition

We define the fractional-order Legendre functions (FLFs) by introducing the change of variablet=xα and α>0 on shifted Legendre polynomials. The fractional-order Legendre function Lixα is denoted by Fliαx. The fractional-order Legendre functions are a particular solution of the normalized eigenfunctions of the singular Sturm-Liouville problem. (12)x-x1+αFli′αx′+α2ii+1xα-1Flix=0,x∈0,1.ThenFliαx can be obtained as follows:(13)Fli+1αx=2i+12xα-1i+1Fliαx-ii+1Fli-1αx,i=1,2,…,Fl0αx=1,Fl1αx=2xα-1.We can derive the analytic form ofFliαx of degree iα as follows:(14)Fliαx=∑s=0ibs,ixsα,where bs,i=-1i+si+s!/i-s!s!2 and Fliα0=-1i,Fliα1=1.Theorem 5.
The FLFs are orthogonal with the weight functionωlαx=xα-1 on the interval 0,1; then the orthogonally condition is(15)∫01FlnαxFlmαxωlαxdx=12n+1αδnm.Proof.
With∫01LntLmtωltdt=1/2i+1δnm, where δnm is the Kronecker function, let t=xα, and then we have(16)∫01LntLmtdt=∫01LnxαLmxααxα-1dx=12n+1δnm,∫01LnxαLmxααxα-1dx=∫01FlnαxFlmαxαxα-1dx=12n+1δnm,∫01FlnαxFlmαxxα-1dx=12n+1αδnm.


## 3.3. Function Approximation

Supposeux∈L20,1; it can be expanded in terms of the fractional-order Legendre functions as follows:(17)ux=∑i=0∞ciFliαx,where the coefficients ci are obtained by(18)ci=α2i+1∫01Fliαxuxωlαxdx,i=0,1,2,….If we consider truncated series in (17), we obtain(19)ux≈umx=∑i=0m-1ciFliαx=CTΦx,where C and Φx are m×1 vector given by(20)C=c0,c1,…,cm-1T,Φx=Fl0αx,Fl1αx,…,Flm-1αxT.Theorem 6.
SupposeDiαux∈C0,1 for i=0,1,2,…,m-1. 2m+1α≥1 and Pmα=spanFl0αx,Fl1αx,…,Flm-1αx. If umx=CTΦx is the best approximation to ux from Pmα, then the error bound is presented as follows:(21)ux-umxω≤MαΓmα+112m+1α,where Mα≥Dmαux,x∈0,1.Proof.
Consider the generalized Taylor’s formula(22)ux=∑i=0m-1xiαΓiα+1Diαu0++xmαΓmα+1Dmαuξ,where 0<ξ<x,x∈0,1.
With Definition4(23)ux-∑i=0m-1xiαΓiα+1Diαf0+≤MαxmαΓmα+1.Since umx=CTΦx is the best approximation to u from Pmα and ∑i=0m-1xiα/Γiα+1Diαf0+∈Pmα, hence(24)ux-umxω2≤ux-∑i=0m-1xiαΓiα+1Diαu0+ω2≤Mα2Γmα+12∫01x2mαxα-1dx,ux-umxω2≤Mα2Γmα+12∫01x2m+1α-1dx=Mα2Γmα+122m+1α.
Now by taking the square roots, then the theorem is proved.For arbitrary functionux,t∈L20,1×0,1, they can be expanded as the following formula:(25)ux,t=∑i=0∞∑j=0∞uijFliaxFljβt,where(26)uij=2i+12j+1αβ∫01∫01ux,tFliαxFljβtωlαxωlβtdxdt,i,j=0,1,….Theorem 7.
If the series∑i=0∞∑j=0∞uijFliaxFljβt converges uniformly to ux,t on the square 0,1×0,1, then one has (27)uij=2i+12j+1αβ∫01∫01ux,tFliαxFljβtωlαxωlβtdxdt.Proof.
By multiplyingωlαxωlβtFlmαxFlnβt on both sides of (25), where i and j are fixed and integrating term wise with regard to x and t on 0,1×0,1, then(28)∫01∫01ux,tFlmαxFlnβtωlαxωlβtdxdt=∑i=0∞∑j=0∞uij∫01∫01FliαxFljβtFlmαxFlnβtωlαxωlβtdxdt=∑i=0∞∑j=0∞uij∫01ωlαxFliαxFlmαxdx∫01ωlβtFliβtFlmαtdt=uij∫01ωlαxFliαx2dx∫01ωlβtFliβt2dt=uij12i+1α12j+1β.Theorem 8.
If the functionux,t is a continuous function on 0,1×0,1 and the series ∑i=0∞∑j=0∞uijFliaxFljβt converges uniformly to ux,t, then ∑i=0∞∑j=0∞uijFliaxFljβt is the FLFs expansion of ux,t.Proof.
Using contradiction, let(29)ux,t=∑i=0∞∑j=0∞uijFliaxFljβt,ux,t≈∑i=0∞∑j=0∞gijFliaxFljβt.Then there is at least one coefficient such that unm≠gnm; however(30)unm=2m+12n+1αβ∫01∫01ux,tFlmaxFlnβtωlαxωlβtdxdt=gnm.Theorem 9.
If two continuous functions defined on0,1×0,1 have the identical FLFs expansions, then these two functions are identical.Proof.
Suppose thatux,t and fx,t can be expended by FLFs as follows:(31)ux,t≈∑i=0∞∑j=0∞uijFliaxFljβt,fx,t≈∑i=0∞∑j=0∞uijFliaxFljβt.
By subtracting the above two equations with each other, we have(32)ux,t-fx,t≈∑i=0∞∑j=0∞uij-uijFliaxFljβt=0=∑i=0∞∑j=0∞0FliaxFljβt.Theorem 10.
If the sum of the absolute values of the FLFs coefficients of a continuous functionux,t forms a convergent series, then the FLFs expansion is absolutely uniformly convergent and converges to the function ux,t.Proof.
Consider(33)∑i=0∞∑j=0∞uijFliaxFljβt≤∑i=0∞∑j=0∞uijFliaxFljβt≤∑i=0∞∑j=0∞uij.
If we consider truncated series in (25), we obtain(34)ux,t≈∑i=0m-1∑j=0n-1uijFliaxFljβt=ΦTxUΦt,where(35)Φx=Fl0αx,Fl1αx,…,Flm-1αxT;Φt=Fl0βt,Fl1βt,…,Fln-1βtT,U=uiji,j=0m-1,n-1.
In this paper, we use the Tau method [20] to compute the coefficients uij.


## 4. Fractional Differential Operational Matrix of FLFs

The derivative of theΦx can be approximated(36)DγΦx≈DγΦx,where Dγ is called the FLFs operational matrix of derivative.Theorem 11.
SupposeDγ is the m×m operational matrix of Caputo fractional derivatives of order γ>0,α>γ/2, when α∉N; then the elements of Dγare obtained as(37)diji,jm-1;m-1=2j+1α∑s=0i∑r=0jbr,jbs,i′Γsα+1Γsα-γ+11s+r+1α-γ,where(38)bs,i′=0,sα∈N0,sα<γ,bs,i′=bs,isα∉N0,sα≥γorsα∈N0,sα≥γ.Proof.
With the properties of the derivative (ii) and the orthogonally of FLFs, we have(39)DγFliαx=∑s=0ibs,i′Γsα+1Γsα-γ+1xsα-γ.Let(40)xsα-γ=∑j=0m-1djFljαx.Multiplying ωlαxFljαx on both sides of (40), we have (41)dj=2j+1α∑r=0jbr,j1s+r+1α-γ.Substituting (40) and (41) into (39), we have (42)DγFliαx=2j+1α∑j=0m-1∑s=0i∑r=0jbr,jbs,i′Γsα+1Γsα-γ+11s+r+1α-γFljαx.Hence, (43)dij=2j+1α∑s=0i∑r=0jbr,jbs,i′Γsα+1Γsα-γ+11s+r+1α-γ,i,j=0,1,…,m-1.


## 5. Description of the Proposed Method

We consider a class of fractional partial differential equations as follows:(44)∂vux,t∂xv+∂γux,t∂tγ=gx,t,subject to the initial conditions(45)u0,t=ux,0=0,where 0≤x,t≤1,n-1<v,γ≤n,∂vux,t/∂xv and ∂γux,t/∂tγ are fractional derivative in Caputo sense, gx,t is the known continuous function, and ux,t is the unknown function.If we approximate functionux,t by a terms of the FLFs, it can be written as (34). Then we have(46)∂vux,t∂xv≈∂vΦTxUΦt∂xv=∂vΦTx∂xvUΦt=∂vΦx∂xvTUΦt≈ΦTxDvTUΦt,∂γux,t∂tγ≈∂γΦTxUΦt∂tγ=ΦTxU∂γΦt∂tγ≈ΦTxUDγΦt,gx,t≈ΦTxGΦt,where(47)G=giji,j=0m-1;n-1,gij=2i+12j+1αβ∫01∫01gx,tFliαxFljβtωlαxωlβtdxdt.Substituting (46) into (44), we have(48)ΦTxDvTUΦt+ΦTxUDγΦt≈ΦTxGΦt.Similar to the typical Tau method [20], then we generate mn-n-m+1 algebraic equations by multiplying ωlαxωlβtFliαxFljβt (for i=0,1,2,…,m-2;j=0,1,2,…,n-2), integrating from 0 to1 and using the orthogonal property, we have(49)H=DvTU+UDγ-G=0.For condition. (45), we have(50)u0,t≈ΦT0UΦt≈CTΦt,ux,0≈ΦTxUΦ0≈ΦTxF,with(51)CT=cii=0n-1,ci=2i+1β∫01u0,tFliβtωlβtdt,F=fjj=0m-1,fj=2j+1α∫01ux,0Fljαxωlαxdx.So we obtain(52)ΦT0U≈CT;UΦ0≈F.Equations (49) and (52) generate mn+1 linear algebraic equations. By solving the linear algebraic equations, the unknown coefficients uij can be obtained.


## 6. Error Analysis

We considerex,t=ux,t-umnx,t as the error function of the approximate solution umnx,t for ux,t, where ux,t is the exact solution of (44) [21].Therefore,umnx,t satisfies the following problem:(53)∂vumnx,t∂xv+∂γumnx,t∂tγ+Rmnx,t=gx,t,where Rmnx,t is the residual function,(54)Rmnx,t=∂vumnx,t∂xv+∂γumnx,t∂tγ-gx,t.We find an approximationemn′x,t to the error function emnx,t in the same way as we did before for the solution of the problem, the error function satisfies the problem(55)∂vemnx,t∂xv+∂γemnx,t∂tγ=Rmnx,t.We should note that, in order to construct the approximateemn′x,t to emnx,t, only (55) needs to be recalculated in the same way as we did before for the solution of (44).


## 7. Numerical Examples

Example 1.
We consider the following FPDE:(56)∂1/4ux,t∂x1/4+∂1/2ux,t∂t1/2=gx,t,u0,t=0,ux,0=0,where gx,t=8x2t3/2/3π+32t2x7/4/21Γ3/4; using the method presented in Section 5, the equation is transformed as follows:(57)D1/4TU+UD1/2≈G,ΦT0U≈CT;UΦ0≈F.
By solving (57), we can obtain the numerical solution of the problem.
Whenn=6,m=6, the matrix U is get as follows:(58)U=0.01980.03960.02820.00980.001400.04860.09730.06940.02420.003400.0540.1080.0770.0270.003800.04120.08240.05880.02060.002900.02210.04430.03160.01100.001500.00620.01230.00870.003000.
Whenm=8,n=6, the matrix U is get as follows:(59)U=0.01980.03960.02820.00990.001300.04860.09710.06930.02410.003400.05400.10790.07700.02690.003800.04070.08140.05800.02030.002900.02250.04490.03200.01120.001600.00920.01830.01310.0046000.00270.00540.00380.0013000.00050.000900.000200.
Whenn=6, Figures 1, 2, and 3 show the numerical solution of (56) for α=1/4,β=1/2 with m=6,m=8,m=10. The exact solution is x2t2 which is shown in Figure 4. We can see clearly that the numerical solutions are very good agreement with the exact solution. Particularly, when m=10 the numerical solution matches the exact solution very well.Figure 1
Numerical solution withm=6 for Example 1.Figure 2
Numerical solution withm=8 for Example 1.Figure 3
Numerical solution withm=10 for Example 1.Figure 4
Exact solution for Example1.Example 2.
We consider the following equation:(60)∂vux,t∂xv+∂γux,t∂tγ=gx,t,u0,t=0.
The exact solution of this problem isux,t=x1/2t, where v=γ=1/2,gx,t=πt/2+2tx/π. The equation is transformed as follows:(61)D1/2TU+UD1/2≈G;ΦT0U≈CT,when m=3,n=3, and m=5,n=5. The results obtained using our proposed method and ADM at t=0.3 are shown in Figures 5 and 6. Comparing the two figures, it can be concluded that the FLFM has a better advantage than ADM for solving fractional partial differential equations and the numerical results have achieved a good convergence precision when small m and n are selected.Figure 5
The comparison between FLFM and ADM withm=3,n=3 for Example 2.Figure 6
The comparison between FLFM and ADM withm=5,n=5 for Example 2.Example 3.
Consider the following fractional Poisson equation:(62)∂vux,t∂xv+∂γux,t∂tγ=gx,t,u0,t=0,ux,0=0.
(1) When v=γ=5/4, (63)gx,t=5Γ1/416x5/2+5Γ1/416t15/4+30πx5/4t5/416Γ9/4+2t5/2x5/4Γ15/4π.
The exact solution isx5/2t5/4+x5/4t15/4. When n=4, the absolute errors for m=5,m=6,m=7 are shown in Figures 7, 8, and 9.
(2) When v=γ=1/2,gx,t=sin⁡x+t. The exact solution is unknown.
Therefore, in order to show the efficiency of the present method, we define the error function as follows:(64)ex,t=ΦTxDvTUΦt+ΦTxUDγΦt-ΦTxGΦt.
Letα=β=1/2 with m=n=8,m=n=9, and m=n=10; the absolute error ex,t is shown in Figures 10, 11, and 12.
Whenm=n=7, we can have the coefficients matrix U. (65)U=0.15700.17080.019800.0060000.17080.1482-0.01520-0.0051-0.00010.00020.0198-0.0153-0.0381-0.00070.0019-0.00040.0001-0.00030.0025-0.0007-0.00310.00060.0001-0.0001-0.0061-0.00510.00180.0006-0.00030.00010.00010-0.0002-0.00030.00010.0001-0.0001-3.029-0.00020.00020.000100.0001-3.0290.Then we have the numerical solution ux,t=ΦTxUΦt.Figure 7
The absolute error for Example3 with m=5.Figure 8
The absolute error for Example3 with m=6.Figure 9
The absolute error for Example3 with m=7.Figure 10
The absolute error for Example3 with m=n=8.Figure 11
The absolute error for Example3 with m=n=9.Figure 12
The absolute error for Example3 with m=n=10.


## 8. Conclusions

In this paper, we define a basis of FLFs and have derived its fractional differential operational matrix. The fractional derivatives are described in the Caputo sense. The matrix together with Tau method is used to effectively approximate the exact solution of fractional partial differential equations. Compared with other numerical approach, the FLFs-Tau method can accurately represent properties of fractional calculus. Moreover, only a small number of FLFs are needed to obtain a satisfactory result. Numerical results which are given in the previous section demonstrate good accuracy of the algorithm. Finally, error analysis shows that the algorithm is convergent.



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