# Using Reproducing Kernel for Solving a Class of Fractional Order Integral Differential Equations

Authors: Zhiyuan Li; Meichun Wang; Yulan Wang; Jing Pang
Journal: Advances in Mathematical Physics
 (2020)

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## Abstract

This paper is devoted to the numerical scheme for a class of fractional order integrodifferential equations by reproducing kernel interpolation collocation method with reproducing kernel function in the form of Jacobi polynomials. Reproducing kernel function in the form of Jacobi polynomials is established for the first time. It is implemented as a reproducing kernel method. The numerical solutions obtained by taking the different values of parameter are compared; Schmidt orthogonalization process is avoided. It is proved that this method is feasible and accurate through some numerical examples.


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## Body

## 1. Introduction

In this paper, the reproducing kernel interpolation collocation method with reproducing kernel function in the form of Jacobi polynomials is applied to solve the following linear fractional integrodifferential equations (FIDEs):(1)Dμu1x+∫0tk11x,tu1t+k12x,tu2tdt=f1x,Dμu2x+∫0tk21x,tu1t+k22x,tu2tdt=f2x,0<x,t≤1,u10=0,u20=0,where 0<μ≤1, fnx,n=1,2, and kijx,t, i,j=1,2, are given functions. Dμunx indicates that μ is the Caputo fractional derivative defined by unx, n=1,2.Fractional order integrodifferential equation appears in the formulation process of applied science, such as physics and finance. However, it is very difficult to obtain the analytic solution of linear integrodifferential equations of fractional order, so many researchers try their best to study numerical solution of linear FIDEs and system of linear FIDEs in recent years [1–5]. Since the reproducing kernel method can not only obtain the exact solution in the form of series but also obtain the approximate solution with higher accuracy, the method has been widely used in linear and nonlinear problems, integral and differential equations, fractional partial differential equation, and so on [6–15]. But there are no scholars that use the reproducing kernel interpolation collocation method to solve the linear integrodifferential equations of fractional order. In this paper, linear integrodifferential equations of fractional order are solved by the reproducing kernel interpolation collocation method with reproducing kernel function in the form of Jacobi polynomials for the first time. The fractional derivative is described in the Caputo sense.


## 2. Preliminaries

Definition 1.
The Caputo fractional derivative operator of order0<μ≤1 is defined as
(2)Dμut=1Γ1−μ∫0tt−τ−μ∂uτ∂τdτ,0<μ<1,∂ut∂t,α=1.Definition 2.
LetH be the real Hilbert spaces of functions f:Ω⟶R. A function K:Ω×Ω⟶R is called reproducing kernel for H if
(i)
Kx,·∈H for all x∈Ω
(ii)
fx=f,K·,xH for all f∈H and all x∈Ω


### 2.1. The Shifted Jacobi Polynomials

The shifted Jacobi polynomialsP1,iα,βx of degree i is given [16] by
(3)P1,iα,βx=∑k=0i−1i−kΓi+β+1Γi+k+1+α+βΓk+1+βΓi+α+β+1i−k!k!xk,where
(4)P1,iα,β0=−1iΓi+β+1Γ1+βi!,P1,iα,β1=Γi+α+1Γ1+αi!.The shifted Jacobi polynomials on the intervalx∈0,1 are orthogonal with the orthogonality condition which is
(5)∫01P1,nα,βxP1,mα,βxωα,βxdx=hk,where ωx=xβ1−xα is a weight function, and
(6)hk=Γk+α+1Γk+β+12k+α+β+1k!Γk+α+β+1,n=m,0,n≠m.


### 2.2. Reproducing Kernel Space

Definition 3.
Let(7)Hn0,1=P1,iα,βx∫01ωxP1,nα,βx2dx<∞,i=1,⋯,nbe the weighted inner product space of the shifted Jacobi polynomials on 0,1. The inner product and norm are defined as
(8)P1,mα,βx,P1,nα,βx=∫01ωxP1,mα,βxP1,nα,βxdx,P1,nα,βxHn=P1,nα,βx,P1,nα,βxHn,∀P1,nα,βx,P1,mα,βx∈Hn0,1.LetL20,1=fx∫01ωxfx2dx<∞. From [17–20], we can prove that Hn0,1 is a reproducing kernel Hilbert space. Its reproducing kernel is
(9)Rx,y=Rxy=∑i=0neixeiy,where eix=2k+α+β+1k!Γk+α+β+1/Γk+α+1Γk+β+1P1,iα,βx. Reproducing kernel is shown in Figures 1–4.Definition 4.
Let(10)H¯n0,1=u∣u∈Hn0,1,u0=0.Figure 1
Reproducing kernelRxy with n=3,α=β=0.Figure 2
The set of reproducing kernelRxy with n=1,2⋯,7,α=β=0.Figure 3
Reproducing kernelRxy with n=3,α=β=−0.5.Figure 4
Reproducing kernelRxy with n=3,α=β=0.5.Its norm is the same as the norm ofHn0,1. It can easily be shown that H¯n0,1 is a reproducing kernel Hilbert space. According to [18–22], the reproducing kernel of H¯n0,1 is
(11)Kx,y=Kxy=Rx,y−R0,xRy,0R0,02.Definition 5.
The inner product space is defined as(12)H¯n0,1⊕H¯n0,1=Ux=u1x,u2xTu1x,u2x∈H¯n0,1.Its inner product and norm are defined by
(13)Ux,Vx=∑i=12uix,vixH¯n0,1⊕H¯n0,1,Ux2=∑i=12uixH¯n0,1⊕H¯n0,1.It is easy to verify thatH¯n0,1⊕H¯n0,1 is a Hilbert space with the definition of inner product (13). Similarly, L20,1⊕L20,1 is also a Hilbert space.


## 2.1. The Shifted Jacobi Polynomials

The shifted Jacobi polynomialsP1,iα,βx of degree i is given [16] by
(3)P1,iα,βx=∑k=0i−1i−kΓi+β+1Γi+k+1+α+βΓk+1+βΓi+α+β+1i−k!k!xk,where
(4)P1,iα,β0=−1iΓi+β+1Γ1+βi!,P1,iα,β1=Γi+α+1Γ1+αi!.The shifted Jacobi polynomials on the intervalx∈0,1 are orthogonal with the orthogonality condition which is
(5)∫01P1,nα,βxP1,mα,βxωα,βxdx=hk,where ωx=xβ1−xα is a weight function, and
(6)hk=Γk+α+1Γk+β+12k+α+β+1k!Γk+α+β+1,n=m,0,n≠m.


## 2.2. Reproducing Kernel Space

Definition 3.
Let(7)Hn0,1=P1,iα,βx∫01ωxP1,nα,βx2dx<∞,i=1,⋯,nbe the weighted inner product space of the shifted Jacobi polynomials on 0,1. The inner product and norm are defined as
(8)P1,mα,βx,P1,nα,βx=∫01ωxP1,mα,βxP1,nα,βxdx,P1,nα,βxHn=P1,nα,βx,P1,nα,βxHn,∀P1,nα,βx,P1,mα,βx∈Hn0,1.LetL20,1=fx∫01ωxfx2dx<∞. From [17–20], we can prove that Hn0,1 is a reproducing kernel Hilbert space. Its reproducing kernel is
(9)Rx,y=Rxy=∑i=0neixeiy,where eix=2k+α+β+1k!Γk+α+β+1/Γk+α+1Γk+β+1P1,iα,βx. Reproducing kernel is shown in Figures 1–4.Definition 4.
Let(10)H¯n0,1=u∣u∈Hn0,1,u0=0.Figure 1
Reproducing kernelRxy with n=3,α=β=0.Figure 2
The set of reproducing kernelRxy with n=1,2⋯,7,α=β=0.Figure 3
Reproducing kernelRxy with n=3,α=β=−0.5.Figure 4
Reproducing kernelRxy with n=3,α=β=0.5.Its norm is the same as the norm ofHn0,1. It can easily be shown that H¯n0,1 is a reproducing kernel Hilbert space. According to [18–22], the reproducing kernel of H¯n0,1 is
(11)Kx,y=Kxy=Rx,y−R0,xRy,0R0,02.Definition 5.
The inner product space is defined as(12)H¯n0,1⊕H¯n0,1=Ux=u1x,u2xTu1x,u2x∈H¯n0,1.Its inner product and norm are defined by
(13)Ux,Vx=∑i=12uix,vixH¯n0,1⊕H¯n0,1,Ux2=∑i=12uixH¯n0,1⊕H¯n0,1.It is easy to verify thatH¯n0,1⊕H¯n0,1 is a Hilbert space with the definition of inner product (13). Similarly, L20,1⊕L20,1 is also a Hilbert space.


## 3. The Reproducing Kernel Interpolation Collocation Method

To solve equation (1), let
(14)l11u1=Dμu1x+∫0tk11x,tu1tdt,l12u2=∫0tk12x,tu2x,tdt,l21u1=∫0tk21x,tu1tdt,l22u2=Dμu2x+∫0tk22x,tu2tdt.So, equation (1) can be turned into
(15)LUx=Fx,where
(16)L=l11l12l21l22.The operatorL:H¯n0,1⊕H¯n0,1⟶L20,1⊕L20,1 is a bounded linear operator.Assuming thatxii=1∞ is dense on the interval 0,1, put ϕijk=lij∗kxkx, where lij∗ is the adjoint operator of lij. From [23–25], we have
(17)ϕijkx=lijKxxk,i,j=1,2,k=1,2,⋯.Putting(18)Ψi1x=ϕ11ix,ϕ12ixT,Ψi2x=ϕ21ix,ϕ22ixT,i,=1,2,⋯.Theorem 6.
For each fixedn, Ψij1,1n,2 is linearly independent in H¯n0,1⊕H¯n0,1.Proof.
Letting(19)0=∑i=1∞ci1Ψi1x+ci2Ψi2x,Ukx=uk,1x,uk,2xT,where uk,1x∈L20,1, when x=x1,x2,⋯,xk−1,xk+1,⋯xn,uk,1x=0. But uk,1x≠0, when x take other value, 0≐uk,2x∈L20,1.
WhenUk∈H¯n0,1⊕H¯n0,1, we have Luk=Fk. So,
(20)0=Uk,∑i=1nci1Ψi1x+ci2Ψi2x=∑i=1nci1Uk,Ψi1+ci2Uk,Ψi2=∑i=1nci1uk,1,ϕ11i+uk,2,ϕ12i+ci2uk,1,ϕ21i+uk,2,ϕ22i=∑i=1nci1l11uk,1xi+l12uk,2xi+ci2l21uk,1xi+l22uk,2xi=∑i=1nci1uk,1xi+ci2l21uk,1xi=ck1uk,1xk.
So,ck1=0,k=1,2,⋯,n. Similarly, we have ck2=0.Theorem 7.
Ψij1,1∞,2 is complete in space in H¯n0,1⊕H¯n0,1.Proof.
For each(21)Ux=u1x,u2xT∈H¯n0,1⊕H¯n0,1,it follows that Ux,Ψijx=0, for every i=1,2,⋯,j=1,2.
(22)0=Ux,Ψi1xH¯n0,1⊕H¯n0,1=u1x,l11∗KxixH¯n0,1+u2x,l12∗KxixH¯n0,1=l11u1xi+l12u2xi,0=Ux,Ψi2xH¯n,n0,1=u1x,l21∗KxixH¯n0,1+u2x,l22∗KxixH¯n0,1=l21u1xi+l22u2xi.
Since equation (1) has a unique solution, it follows that UX=0.The exact solution of equation (1) can be expressed as
(23)UX=∑i=1∞∑j=12cijΨijx,and truncating the infinite series of the analytic solution, we obtain the approximate solution of equation (1).
(24)UmX=∑i=1m∑j=12cijΨijx.Theorem 8.
LetU∈H¯n0,1⊕H¯n0,1 be the exact solution of equation (1), Um be the approximate solution of U, then Um converges uniformly to U.Proof.
(25)u1x−u1,mx=u1−u1,m,Kx≤u1−u1,mH¯n0,1KxH¯n0,1≤Mu1−u1,mH¯n0,1.
Similarly,(26)u2x−u2,mx≤Mu2−u2,mH¯n0,1.If we can obtain the coefficients of eachΨijx, the approximate solution Umx can be obtained as well. Using Ψijx to do the inner products with both sides of equation (24), we have
(27)∑i=1mci1Ψi1,Ψn1+∑j=1mcj2Ψj2,Ψn1=f1xn,n=1,2,⋯,m,∑i=1mci1Ψi1,Ψn2+∑j=1mcj2Ψj2,Ψn2=f2xn,n=1,2,⋯,m.Letting(28)L2m=Ψi1,Ψm1⋯Ψj2,Ψm1⋯⋯⋯Ψi1,Ψm2⋯Ψj2,Ψm2i,j,n=1,2,⋯,m,F=f1x1,⋯,f1xm,f2x1,⋯,f1xmT.It is obvious that the inverse ofA2m exists by Theorem 6. So, we have
(29)c11,c12,⋯,c1m,c21,c22,⋯,c2mT=L2m−1F.


## 4. Numerical Experiment

Example 1.
We consider the following linear integrodifferential equations of fractional order [5]:
(30)Dμu1x−∫01x+tu1t+x+tu2tdt=−120−x12+4x1/415−23x215Γ1/4,Dμu2x−∫01xt2u1t−xt2u2tdt=5x36+9x4/32Γ1/3,u10=0,u20=0,where the exact solution Ux=x−x3,x2−xT. The numerical results are given in Tables 1 and 2, and the absolute errors of Example 1 for m=10,n=3,μ=3/4,α=β=1/2 are plotted in Figures 5 and 6. Comparisons are made between the approximate and the exact solution for m=10,n=3,μ=3/4,α=β=1/2 in Figures 7 and 8. Errors of u1 and u2 for m=10,n=3,μ=3/4,α=β=1/2 are plotted in Figures 9 and 10.Example 2.
We consider the following linear integrodifferential equations of fractional order [5]:
(31)Dμu1x−∫012xtu1t+2xtu2x,tdt=83x80+25x6/511+15x33Γ1/5,Dμu2x−∫01x+tu1t−x+tu2x,tdt=5x36+9x4/32Γ1/3,u10=0,u20=0,where the exact solution is Ux=x3−x2,15/8x2T. We obtain the numerical results which are given in Tables 3 and 4, and the absolute errors of Example 2 for m=10,n=3,μ=4/5,α=β=1/2 are plotted in Figures 11 and 12. Comparisons are made between the approximate and exact solutions for m=10,n=3,μ=4/5,α=β=1/2 in Figures 13 and 14. Absolute errors of u1 for m=10,n=3,μ=3/4 are plotted in Figure 15. Absolute errors of u2 for m=10,n=3,μ=3/4 are showed in Figure 16.Example 3.
We consider the following linear integrodifferential equations of fractional order [4].
(32)Dμu1x−∫0tx−tu1t+x−tu2x,tdt=f1t,Dμu2x−∫0tx−tu1t−x−tu2x,tdt=f2t,u10=0,u20=0,
(i)
wheref1t=1+x−t3/3,f2t=1−t−t4/12, the exact solution is Ux=x+x2/2,x−x2/2T. By the proposed algorithm, we obtain the numerical results which are given in Tables 5 and 6, and the absolute errors of this example for m=10,n=2,μ=0.5,α=β=0 are plotted in Figures 17 and 18. Comparisons are made between the approximate and exact solutions in Figures 19 and 20. When taking different values of α,β, errors of u1 and u1 for m=10,n=2,μ=0.5 are plotted in Figures 21–24. Figures 25 and 26 illustrate the approximate solutions of u1 and u2 using the present method for different values of μ which are in agreement with the exact solution. Figures 27 and 28 illustrate the approximate solutions for different values of μ compared with the exact solution in Ref. [4]
(ii)
wheref1t=0.506344−1.86957x1.5+1.81521x1.7/x0.5, f2t=1.81521−0.239522−0.42377x+x1.2, the exact solution is Ux=x0.3+x2,x2T. The absolute errors of this example for m=200,n=2,μ=0.2,α=β=0.5 are given in Figures 29 and 30Table 1
The numerical solutions ofu1x for Example 1.
x
Exact solution
Approximate solution
u1x−u1,10x
u1x−u1,10xu1x
0.1
0.099
0.099
6.85563E−15
7.37345E−14
0.2
0.192
0.192
1.05471E−15
3.15141E−14
0.3
0.273
0.273
4.77396E−15
1.22003E−15
0.4
0.336
0.336
3.05311E−15
3.48597E−14
0.5
0.375
0.375
1.24345E−14
7.10543E−15
0.6
0.384
0.384
1.09912E−14
8.67362E−16
0.7
0.357
0.357
5.43454E−14
1.14910E−13
0.8
0.288
0.288
1.09912E−14
7.17019E−14
0.9
0.171
0.171
1.87628E−14
2.66194E−14Table 2
The numerical solutions ofu2x for Example 1.
x
Exact solution
Approximate solution
u2x−u2,10x
u2x−u2,10xu2x
0.1
−0.09
−0.09
3.55271E−15
9.15934E−14
0.2
−0.16
−0.16
3.85803E−15
7.11237E−15
0.3
−0.21
−0.21
6.82787E−15
3.46284E−14
0.4
−0.24
−0.24
5.55112E−16
2.33609E−14
0.5
−0.25
−0.25
8.21565E−15
1.17684E−14
0.6
−0.24
−0.24
8.82627E−15
9.25186E−16
0.7
−0.21
−0.21
1.88183E−14
4.83740E−14
0.8
−0.16
−0.16
3.66374E−15
2.35922E−14
0.9
−0.09
−0.09
9.88098E−15
2.71388E−14Figure 5
Absolute errors ofu1 obtained by the present method for Example 1.Figure 6
Absolute errors ofu2 obtained by the present method for Example 1.Figure 7
Comparisons between numerical and exact solutions ofu1 obtained by the present method for Example 1.Figure 8
Comparisons between numerical and exact solutions ofu2 obtained by the present method for Example 1.Figure 9
Absolute errors ofu1 and u2 obtained by the present method for Example 1.Figure 10
Relative errors ofu1 and u2 obtained by the present method for Example 1.Table 3
The numerical solutions ofu1x for Example 2.
x
Exact solution
Approximate solution
∣u1x−u1,10x∣
0.1
−0.009
−0.009
3.54577E−15
0.2
−0.032
−0.032
2.24126E−15
0.3
−0.063
−0.063
4.76008E−15
0.4
−0.096
−0.096
2.56739E−15
0.5
−0.125
−0.125
1.11022E−16
0.6
−0.144
−0.144
2.19269E−15
0.7
−0.147
−0.147
2.08167E−14
0.8
−0.128
−0.128
1.69864E−14
0.9
−0.081
−0.081
1.88738E−15Table 4
The numerical solutions ofu2x for Example 2.
x
Exact solution
Approximate solution
u2x−u2,10x
0.1
0.01875
0.01875
2.26208E−15
0.2
0.07500
0.07500
3.88578E−16
0.3
0.16875
0.16875
4.85723E−15
0.4
0.30000
0.30000
5.21805E−15
0.5
0.46875
0.46875
2.22045E−16
0.6
0.67500
0.67500
1.44329E−15
0.7
0.91875
0.91875
3.03091E−14
0.8
1.20000
1.20000
2.75335E−15
0.9
1.51875
1.51875
2.88658E−15
1.0
1.87500
1.87500
2.48690E−14Figure 11
Absolute errors ofu1 obtained by the present method with α=β=1/2 for Example 2.Figure 12
Absolute errors ofu2 obtained by the present method with α=β=1/2 for Example 2.Figure 13
Comparisons between numerical and exact solutions ofu1 obtained by the present method for Example 2.Figure 14
Comparisons between numerical and exact solutions ofu2 obtained by the present method for Example 2.Figure 15
Absolute errors ofu1 obtained by the present method with α=β=1/2 (red), α=β=0 (purple) for Example 2.Figure 16
Absolute errors ofu2 obtained by the present method with α=β=1/2 (red), α=β=0 (purple) for Example 2.Table 5
Comparison of the numerical result ofu1x in Example 3.
x
Exact solution
Present method approximate solution
Present methodu1x−u1,10x
Ref. [4] approximate solution
0.1
0.105
0.105
4.16334E−17
0.332
0.2
0.220
0.220
2.22045E−16
0.492
0.3
0.345
0.345
0.0000
0.630
0.4
0.480
0.480
1.66533E−16
0.759
0.5
0.625
0.625
1.11022E−16
0.884
0.6
0.780
0.780
6.66134E−16
1.007
0.7
0.945
0.945
6.66134E−16
1.129
0.8
1.120
1.120
1.77636E−15
1.252
0.9
1.305
1.305
2.22045E−15
1.376
1.0
1.500
1.500
2.22045E−15
1.500Table 6
Comparison of the numerical result ofu2x in Example 3.
x
Exact solution
Present method approximate solution
Present methodu2x−u2,10x
Ref. [4] approximate solution
0.1
0.095
0.095
9.71445E−17
0.300
0.2
0.180
0.180
2.49800E−16
0.402
0.3
0.255
0.255
3.33067E−16
0.466
0.4
0.320
0.320
3.88578E−16
0.506
0.5
0.375
0.375
1.11022E−16
0.530
0.6
0.420
0.420
2.22045E−16
0.542
0.7
0.455
0.455
3.88578E−16
0.544
0.8
0.480
0.480
6.10623E−16
0.537
0.9
0.495
0.495
7.77156E−16
0.522
1.0
0.500
0.500
1.11022E−15
0.500Figure 17
Absolute errors ofu1 obtained by the present method for Example 3.Figure 18
Absolute errors ofu2 obtained by the present method for Example 3.Figure 19
Comparisons between numerical and exact solutions ofu1 obtained by the present method for Example 3.Figure 20
Comparisons between numerical and exact solutions ofu2 obtained by the present method for Example 3.Figure 21
Absolute errors ofu1 obtained by the present method for Example 3.Figure 22
Absolute errors ofu2 obtained by the present method for Example 3.Figure 23
Relative errors ofu1 obtained by the present method with α=β=0 (red), α=β=0.5 (purple), α=0.5,β=−0.5 (green), and α=2/3,β=−1/3 (black) for Example 3.Figure 24
Relative errors ofu2 obtained by the present method with α=β=0 (red), α=β=0.5 (purple), α=0.5,β=−0.5 (green), and α=2/3,β=−1/3 (black) for Example 3.Figure 25
The numerical solution ofu1 obtained by the present method with μ=0.75,0.5,0.25,1 and exact solutions for Example 3.Figure 26
The numerical solution ofu2 obtained by the present method with μ=0.75,0.5,0.25,1 and exact solutions for Example 3.Figure 27
The numerical solution ofu1 with μ=0.75,0.5,0.25,1 for Example 3 in Ref. [4].Figure 28
The numerical solution ofu2 with μ=0.75,0.5,0.25,1 for Example 3 in Ref. [4].Figure 29
Absolute errors ofu1 obtained by the present method for Example 3.Figure 30
Absolute errors ofu2 obtained by the present method for Example 3.


## 5. Conclusions and Remarks

In this paper, linear integrodifferential equations of fractional order have been solved by the reproducing kernel interpolation collocation method with reproducing kernel function in the form of Jacobi polynomials for the first time. Comparisons are made between the approximate and exact solutions. We verify the feasibility of this method by selecting different parametersμ,α,β. From all tables and figures, we obtain that the algorithm is remarkably accurate and effective.All computations are performed by the Mathematica 7.0 software package.



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