# Left- and Right-Shifted Fractional Legendre Functions with an Application for Fractional Differential Equations

Authors: Haidong Qu; Xiaopeng Yang; Zihang She
Journal: Advances in Mathematical Physics
 (2020)

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## Abstract

Two new orthogonal functions named the left- and the right-shifted fractional-order Legendre polynomials (SFLPs) are proposed. Several useful formulas for the SFLPs are directly generalized from the classic Legendre polynomials. The left and right fractional differential expressions in Caputo sense of the SFLPs are derived. As an application, it is effective for solving the fractional-order differential equations with the initial value problem by using the SFLP tau method.


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## Body

## 1. Introduction

Legendre polynomials are a family of complete and orthogonal functions discovered by Adrien-Marie Legendre in 1782. As a very important application, Legendre spectral methods are successfully used to obtain numerical solutions of the various differential equations. Through Google Scholar search, there are almost 54,000 articles from 1980 to 2019 on the use of the Legendre spectral methods in the study of various problems, such as numerical solving for integrodifferential equations ([1–6]) and ordinary differential equations with fractional order ([7–9]) and integer order ([10]). Recently, the Legendre spectral method was proved to be an effective method to solve fractional differential equations, which has been studied by many scholars ([7, 8, 11–13]). More recently, many authors ([14–19]) applied Müntz orthogonal polynomials to solve the fractional-order differential equations (FDEs). Motivated by this literature, we define the left SFLPs by introducing the change of variable zL=2x−a/b−aα−1. In particular, when a=−1, b=1, and α=1, the left SFLPs degenerate the classic Legendre polynomials; while a=0, b=1, and 0<α<1, the left SFLPs are transformed into the fractional-order Legendre polynomials proposed in [7]. Similarly, the right SFLPs can be also obtained by introducing the change of variable zR=−2b−x/b−aα+1. Furthermore, to solve some FDEs, the SFLP tau method is better than the method based on the other orthogonal polynomials.


## 2. Shifted Fractional-Order Legendre Polynomials

In this section, we introduce some definitions, notations, and useful formulas about the shifted fractional-order Legendre polynomials. For the properties of classical Legendre polynomials, please refer to the literature [7, 11]. Now, we begin with the definition of Caputo fractional derivative.Definition 1. (see [20]).
Form−1<α≤m, m∈Z+, a,b∈R, the left side and the right side Caputo fractional derivative is defined by
(1)DaCxαux=1Γm−α∫axx−ξm−1−αumξdξ,DxCbαux=−1mΓm−α∫xbx−ξm−1−αumξdξ.Then, forα,β>0 and constant C, we have the following properties:
(2)DaCxαC=DxCbαC=0,(3)DaCxαx−aβ=0,β∈N+,α>β,Γβ+1Γβ−α+1x−aβ−α,(4)DxCbαb−xβ=0,β∈N+,α>β,Γβ+1Γβ−α+1b−xβ−α.Next, let us recall the definition of the classic Legendre polynomials. The classic Legendre polynomials, denoted byLnz,n=0,1,…, are orthogonal on the interval −1,1 with the orthogonality property
(5)∫−11LnzLmzdz=22n+1δnm,where δnm is the Kronecker function. Now, in order to apply Legendre polynomials on the finite interval a,b, we define the left and right SFLPs by introducing the change of variable z=zL and z=zR, respectively. Then, these two functions, denoted by LLnαx and RLnαx, n=0,1,…, are orthogonal polynomials with the weight function wLx=x−aα−1 and wRx=b−xα−1, respectively, those are
(6)∫abLLnαxLLmαxwLxdx=b−aαα2n+1δnm,∫abRLnαxRLmαxwRxdx=b−aαα2n+1δnm.Leta=−1, b=1, we plot the first six terms of the left and right SFLPs for α=0.5 in Figure 1, and in Figure 2(b), the sixth term of the left SFLPs is shown for different values of the order α. For the classic Legendre polynomials, see Figure 2(a). The following are some useful formulas about the left and right SFLPs on the interval a,b, which are directly generalized from the classic Legendre polynomials.
(1)
The analytic forms of the left SFLPs(7)LLnαx=∑j=0nLcj,nx−ajα,Lcj,n=−1n+jn+j!b−ajαn−j!j!2,and the right SFLPs
(8)RLnαx=∑j=0nRcj,nb−xjα,Rcj,n=−1jn+j!b−ajαn−j!j!2
(2)
Three-term recurrence relations for the left SFLPs(9)n+1LLn+1αx=2n+12x−ab−aα−1LLnαx−nLLn−1αx,with LL0αx=1 and LL1αx=2x−a/b−aα−1, and the right SFLPs
(10)n+1RLn+1αx=2n+1−2b−xb−aα+1RLnαx−nRLn−1αx,with RL0αx=1 and RL1αx=−2b−x/b−aα+1
(3)
Derivative recurrence relations for the left SFLPs(11)4n+2LLnαx=b−ax−ab−a1−αLLn+1αx−LLn−1αx′,and the right SFLPs
(12)4n+2RLnαx=b−ab−xb−a1−αRLn+1αx−LLn−1αx′
(4)
The boundary values of the left and right SFLPs(13)LLnαa=RLnαa=−1n,LLnαb=RLnαb=1,n=0,1,2,⋯
(5)
The left and right Legendre’s differential equations of fractional order(14)x−ab−a−x−ab−aα+1LLnαx′′+nn+1αb−a2x−ab−aα−1LLnαx=0,b−xb−a−b−xb−aα+1RLnαx′′+nn+1αb−a2b−xb−aα−1RLnαx=0Figure 1
(a) The first six terms of the left SFLPs forα=0.5. (b) The first six terms of the right SFLPs for α=0.5.
(a)
(b)Figure 2
(a) The first six terms of the left SFLPs forα=1. (b) The sixth term of the left SFLPs for α=0.5, 0.6, 0.7, 0.8, 0.9, and 1.
(a)
(b)In the following lemmas, we derive the fractional differential expressions of the left and right SFLPs in Caputo sense.Lemma 2.
Letα>0 and
(15)Ldi,jα=α2j+1b−aα∫abDaCxαLLiαxLLjαxwLxdx,i,j=0,1,2,⋯.
Then, we have(16)Ld0,j=0,Ldi,jα=∑n=1i∑m=0jLcn,iLcm,jα2j+1b−aαΓnα+1b−aαn+mαΓn−1α+1n+m,i≥1,where Lcn,i and Lcm,j are given by (7).Proof.
By (2), we have Ld0,j=0. Then, for i≥1, formulas (7), (2), and (3) lead to
(17)Ldi,jα=∫ab∑n=0iLcn,iDaCxαx−anα∑m=0jLcm,jx−amαwLxdx=∫ab∑n=1iLcn,iΓnα+1Γn−1α+1x−an−1α∑m=0jLcm,jx−amαwLxdx=∑n=1i∑m=0jLcn,iLcm,jα2j+1b−aαΓnα+1b−aαn+mαΓn−1α+1n+m.
From Lemma2, it is elementary to get
(18)DaCxαLLiαx=∑j=0i−1Ldi,jαLLjαx,with Ldi,jα given by (17). The following lemma on fractional differential expressions for the right SFLPs can be obtained similarly.Lemma 3.
Letα>0 and
(19)Rdi,jα=α2j+1b−aα∫abDxCbαRLiαxRLjαxwRxdx,i,j=0,1,2,⋯.
Then, we have(20)Rd0,j=0,Rdi,jα=∑n=1i∑m=0jRcn,iRcm,jα2j+1b−aαΓnα+1b−aαn+mαΓn−1α+1n+m,i≥1,where Rcn,i and Rcm,j are given by (8).


## 3. Application

In this section, we give two examples to illustrate that our methods are effective. First, we apply the left SFLP tau method to solve the fractional-order differential equation of the following form:(21)D−1Cxαux+ux=fx,−1<x<b,u−1=0.Supposefx=Γ2α+1/Γα+1⋅x+1α+x+12α. Then, the exact solution of (21) is ux=x+12α. Now, we use the left SFLP tau method to obtain it. Let
(22)ux≃∑i=0n−1ciLLiαx=CTϕx,with CT=c0,c1,⋯,cn−1 and ϕxT=LL0αx,LL1αx,⋯,LLn−1αx. From Lemma 2, we have
(23)D−1Cxαux≃∑i=0n−1ciD−1CxαLLiαx=CTDnαϕx,with the matrix Dnα=Ldi,jn×n, where Ldi,j is given by (17). Assume
(24)fx≃∑i=0n−1fiLLiαx=FTϕx,with FT=f0,f1,⋯,fn−1, where
(25)fi=α2i+1b+1α∫01fxLLiαxwxdx,i=0,1,⋯,n−1.Setn=3. Then
(26)LL0αx=1,LL1αx=2x+1b+1α−1,LL2αx=6x+1b+12α−6x+1b+1α+1,f0=Γ2αΓαb+1α+13b+12α,f1=Γ2αΓαb+1α+12b+12α,f2=16b+12α.By (22), (23), and (24), we have
(27)CTD3α+CT=FT,with three-order matrix
(28)D3α=000d1,0d1,1d1,2d2,0d2,1d2,2,which, in accordance with u−1=∑i=02ciLLiα−1=c0−c1+c2=0, yields
(29)CT=b+12α13,12,16.Using (22), we obtain the exact solution of (21). Obviously, we cannot get this exact solution by the classic Legendre tau method.Next, we apply the right SFLP tau method to solve the fractional-order differential equation of the following form:(30)DxC12αux+DxC1αux+ux=fx,a<x<1,DxC1αu1=u1=0,where DxC12αux≔DxC1αDxC1αux and fx=Γ3α+1/Γα+1⋅1−xα+Γ3α+1/Γ2α+1⋅1−x2α+1−x3α. Then, the exact solution of (30) is ux=1−x3α. Now, we use the right SFLP tau method to obtain it. Let
(31)ux≃∑i=0n−1ciRLiαx=CTφx,with CT=c0,c1,⋯,cn−1 and φxT=RL0αx,RL1αx,⋯,RLn−1αx. From Lemma 3, we have
(32)D−1Cxαux≃∑i=0n−1ciD−1CxαRLiαx=CTDnαφx,with the matrix Dnα=Rdi,jn×n, where Rdi,j is given by Lemma 3. Similar to the calculation of the matrix Dnα above, we get the fractional derivative matrix Dn2α, that is
(33)D−1Cx2αux≃∑i=0n−1ciD−1Cx2αRLiαx=CTDn2αφx.Assume(34)fx≃∑i=0n−1fiRLiαx=FTφx,with FT=f0,f1,⋯,fn−1, where
(35)fi=α2i+11−aα∫01fxRLiαxwxdx,i=0,1,⋯,n−1.Setn=4. Then
(36)RL0αx=1,RL1αx=−21−x1−aα−1,RL2αx=61−x1−a2α−61−x1−aα+1,RL3αx=−201−x1−a3α+301−x1−a2α−121−x1−aα+1,f0=5201−a3α+Γ3α+13Γ2α+11−a2α+Γ3α+12Γα+11−aα,f1=−9201−a3α−Γ3α+12Γ2α+11−a2α−Γ3α+12Γα+11−aα,f2=5201−a3α+Γ3α+16Γ2α+11−a2α,f3=−1201−a3α.By (31), (32), (33), and (34), we have
(37)CTD42α+CTD4α+CT=FT,with the fourth-order matrix
(38)D4α=0000Rd1,0Rd1,1Rd1,2Rd1,3Rd2,0Rd2,1Rd2,2Rd2,3Rd3,0Rd3,1Rd3,2Rd3,3,D42α=00000000R~d2,0R~d2,1R~d2,2R~d2,3R~d3,0R~d3,1R~d3,2R~~d3,3,which, in accordance with boundary value conditions
(39)u1=∑i=03ciRLiα1=c0+c1+c2+c3=0,DxC1αu1=∑i=03ciDxC1αRLiα1=c1+3c2+6c3=0,yields
(40)CT=1−a3α520,−920,520,−120.Finally, using (31), we obtain the exact solution of (30).


## 4. Conclusions

In this paper, the left- and right-shifted fractional-order Legendre polynomials are proposed by substituting the variables of the classic Legendre polynomials. Correspondingly, the differential expressions of these new polynomials for the left and right fractional derivatives in Caputo sense are derived, based on which the tau method can be used to solve the FDEs on the arbitrary finite intervala,b. Moreover, the results in this article are easy to generalize to the case of the other orthogonal functions, e.g., Chebyshev polynomials, which will be studied later.



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