# Numerical Solution of a Class of Time-Fractional Order Diffusion Equations in a New Reproducing Kernel Space

Authors: Xiaoli Zhang; Haolu Zhang; Lina Jia; Yulan Wang; Wei Zhang
Journal: Journal of Function Spaces
 (2020)

---

## Abstract

In this paper, we structure some new reproducing kernel spaces based on Jacobi polynomial and give a numerical solution of a class of time fractional order diffusion equations using piecewise reproducing kernel method (RKM). Compared with other methods, numerical results show the reliability of the present method.


---

## Body

## 1. Introduction

In this paper, we consider the following time-fractional order diffusion equation:(1)Dtγux,t+β1x,t∂ux,t∂x+β2x,t∂2ux,t∂x2=fx,t,x,t∈Ω=0,1×0,1,ux,0=ϕx,0≤x≤1,u0,t=φ1t,u1,t=φ2t,0≤t≤1,where β1x,t,β2x,t, fx,t are known functions, and Dtγux,t is the variable order Caputo fractional derivative of order γ:(2)Dtγux,t=1Γ1−γ∫0tt−τ−γ∂ux,τ∂τdτ.The time-fractional order equation [1–4] has wide applications in viscoelastic materials, signal processing, fluid mechanics, and dynamic of viscoelastic materials. The analytic solution to this equation is almost impossible to obtain. In recent years, several numerical methods [5–11] have been proposed. In previous work, the author used Taylor’s formula or Delta function to construct reproducing kernel space [12–17]. In this paper, we structure some new reproductive kernel spaces based on Jacobi polynomials and give a numerical solution of a class of time-space fractional order diffusion equation using piecewise reproducing kernel method.Definition 1.
LetH be a 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. Structing Reproductive Kernel Space Based on the Shifted Jacobi Polynomials

The shifted Jacobi polynomialsP1,iα,βx of degree i is given [18] 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(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.Definition 2.
Let(7)H¯n0,1=P1,iα,βx∫01ωxP1,nα,βx2dx<∞,i=1,…,n,is 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,where(9)P1,nα,βxH¯n=P1,nα,βx,P1,nα,βxH¯n,∀P1,nα,βx,P1,mα,βx∈H¯n0,1.
From [17, 19], we can prove that H¯n0,1 is a reproducing kernel Hilbert space. Its reproducing kernel is(10)Rnx,y=∑i=0neixeiy,where eix=2k+α+β+1k!Γk+α+β+1/Γk+α+1Γk+β+1P1,iα,βx. Using Ref. [5] and the reproducing kernel of H¯n0,1, we can get following reproducing kernel spaces.(1)
SpaceH2=uxux∈H¯n,u0=0, Hn with the same inner product as H¯n and Hn is a reproducing kernel space and its reproducing kernel is(11)K2x,y=Rnx,y−Rn0,xRny,0Rn0,02.(2)
SpaceH3=uxux∈H¯n+1,u0=0,u1=0, Hn+1 with the same inner product as H¯n+1 and Hn+1 is a reproducing kernel space and its reproducing kernel is(12)K3x,y=kn+1x,y−kn+11,xkn+1y,1k51,12,whereKn+1x,y=Rn+1x,y−Rn+10,xRn+1y,0/Rn+10,02.(3)
SpaceHΩ=H2⊗H3=ux,tux,t∈H2⊗H3,u0,t=u1,t=ux,0=0, and its reproducing kernel is(13)Kx,t,y,s=Knx,y×Kn+1t,s,x,y,t,s∈Ω,where K2x,y,K3x,y from (11) and (12). Reproducing kernels with different α,β are shown in Figures 1–8.Figure 1
Reproducing kernelR3x,y of space H¯30,1 with α=β=1.Figure 2
The set of reproducing kernelRnx,y with n=1,2,…,7,α=β=1.Figure 3
Reproducing kernelR3x,y of space H¯30,1 with α=2,β=3.Figure 4
The set of reproducing kernelRnx,y with n=1,2,…,7,α=2,β=3.Figure 5
Reproducing kernelR3x,y of space H¯30,1 with α=−0.5,β=1.Figure 6
The set of reproducing kernelRnx,y with n=1,2,…,7,α=0.5,β=1.Figure 7
Reproducing kernelR3x,y of space H¯30,1 with α=β=0.5.Figure 8
Reproducing kernelsRxy with n=1,2,…,7,α=β=−0.5.


## 3. Piecewise Reproducing Kernel Method

After homogenization, equation (1) is converted to the following form:(14)Ax,tvx,t=gx,t,x,t∈Ω=0,1×0,1,vx,0=0,0≤x≤1,v0,t=0,v1,t=0,0≤t≤1,where A is a operator, and(15)Ax,tux,t≜Dtγux,t+β1x,t∂ux,t∂x+β2x,t∂2ux,t∂x2.v
x
,
t
=
u
x
,
t
−
U
x
,
t
−
u
0
x
+
U
0
x, Ux,t=φ1t1−x+φ2tx,U0x=Ux,0,u0x=ux,0, gx,t=fx,t−Ax,tUx,t+u0x−U0x.Letxi,tii=1∞ be nodes in interval 0,1×0,1, ψix,t=Ay,sKy,sx,ty,s=xi,ti,i=1,2,…,∞.(16)ψ¯ix,t=∑k=1iβikψkx,t,βii>0,i=1,2,…,∞,where the βik are the coefficients resulting from Gram–Schmidt orthonormalization.Theorem 1.
(see [11, 19–23]). If A−1 is existing and xi,tii=1∞ is denumerable dense points in Ω, then(17)vx,t=∑i=1∞∑k=1iβikgxk,tkψ¯ix,t,is an analytical solution of (14).
Deriving from the form of (17), we get the approximate solution of (14) as(18)vNx,t=∑i=1N∑k=1iβikgxk,tkψ¯ix,t.
However, the direct application of (17) could not have a good numerical simulation effect possibly for (1). The focus of this paper is to fill this gap, so we use the piecewise reproducing kernel method. The main technique of the piecewise reproducing kernel method see Ref. [16, 21, 24, 25]. More about convergence theorem and error estimation, those detailed proof can be seen in [23–26].


## 4. Numerical Experiments

In this section, some numerical experiments are studied to demonstrate the accuracy of the present method.Experiment 1.
We consider the following time fractional reaction-diffusion equation:(19)Dtγux,t=x2Γ3−γ∂2ux,t∂x2−xΓ3−γ∂ux,t∂x+fx,t,0<x<1,0≤t≤1,ux,0=x21−x,u0,t=0,u1,t=0,where fx,t=8x21−xt2−γ+3x34t2+1/Γ3−γ, the exact solution uTx,t=x21−x4t2+1. Numerical solution of Experiment 1 is shown in Figures 9–11 and Table 1. From Table 1, we can see that the absolute error is getting smaller and smaller when h is smaller. Figure 10 shows the relationship between absolute error and reproducing kernel. Figure 11 shows the relationship between absolute error and γ.Figure 9
Numerical results of Experiment1 for α=β=0.5,γ=0.8,n=4.Figure 10
Absolute errors of Experiment1 with different reproducing kernels for t=0.1,h=0.0001, γ=0.5,n=2.Figure 11
Absolute errors of Experiment1 with different fractional derivatives for h=0.01,t=0.1,α=β=1,n=2.Table 1
Comparison of absolute errors obtained for Experiment1 at α=β=0.5,γ=0.8, t=0.01,n=4.
x
Exact solution
Traditional RKMN=2
Present methodh=0.001 N=2
Present methodh=0.000001 N=2
0.1
0.0090
2.272e − 03
4.565e − 06
6.536e − 11
0.2
0.0320
4.032e − 03
9.332e − 06
1.164e − 09
0.3
0.0630
5.284e − 03
1.384e − 05
3.207e − 09
0.4
0.0960
6.029e − 03
1.765e − 05
5.585e − 09
0.5
0.1251
6.270e − 03
2.028e − 05
7.818e − 09
0.6
0.1441
6.010e − 03
2.129e − 05
9.425e − 09
0.7
0.1471
5.250e − 03
2.023e − 05
9.927e − 09
0.8
0.1281
3.994e − 03
1.663e − 05
8.843e − 09
0.9
0.0810
2.243e − 03
1.004e − 05
5.694e − 09Experiment 2.
We consider the following time fractional reaction-diffusion equation [7](20)Dtγux,t+x∂ux,t∂x−∂2ux,t∂x2=fx,t,0<x<1,0≤t≤1,ux,0=x2−x3,u0,t=u1,t=0,where fx,t=2t2−γ/Γ3−γx2−x3+t2+12x2−3x3+6x−2, the exact solution uTx,t=t2+1x2−x3. Numerical solution of Experiment 2 is shown in Table 2. From Table 2, we can see that the absolute error obtained by the present method is smaller than the absolute error obtained by Ref. [7].Table 2
Comparison of absolute errors obtained for Experiment2 with α=β=0.5,n=4.
x
,
t
Present methodγ=0.5 h=1
Present methodγ=0.5 h=0.000001
Ref. [7] γ=0.5
Present methodγ=0.7 h=1
Present methodγ=0.7 h=0.000001
Ref. [7] γ=0.7 0.1,0.12.085e − 031.767e − 096.317e − 063.315e − 036.246e − 101.048e − 050.2,0.27.020e − 031.183e − 109.014e − 061.074e − 024.179e − 099.436e − 060.3,0.31.179e − 021.283e − 081.091e − 051.787e − 022.083e − 081.086e − 050.4,0.41.294e − 023.875e − 081.222e − 052.015e − 025.094e − 081.086e − 050.5,0.57.814e − 037.546e − 081.212e − 051.464e − 029.133e − 089.833e − 060.6,0.64.288e − 031.157e − 071.136e − 056.253e − 041.340e − 078.678e − 060.7,0.72.088e − 021.475e − 079.658e − 061.888e − 021.662e − 077.024e − 060.8,0.83.512e − 021.541e − 077.072e − 063.600e − 021.703e − 074.934e − 060.9,0.93.456e – 021.137e − 073.757e − 063.678e − 021.240e − 072.535e − 06Experiment 3.
We consider the following time-space fractional advection-reaction-diffusion equation(21)Dtγux,t=xγt2−γ∂γux,t∂xγ+5Γ2−γΓ3−γ∂γ+1ux,t∂xγ+1+fx,t,0<x<1,0≤t≤1,ux,0=x2,u0,t=0,u1,t=4t2+1,where fx,t=−32x2t4−γ/Γ3−γ, the exact solution uTx,t=x24t2+1. Numerical solution of Experiment 3 is shown in Figure 12 and Table 3.Figure 12
(a) The absolute error of Experiment3 for γ=0.3,n=4. (b) The absolute error of Experiment 3 for γ=0.6,n=4.
(a)
(b)Table 3
Comparison of absolute errors obtained for Experiment3 in t=0.02,n=4.
x
γ
=
0.3 h=1
γ
=
0.3 h=0.001
γ
=
0.3 h=0.000001
γ
=
0.6 h=1
γ
=
0.6 h=0.001
γ
=
0.6 h=0.000001
0.1
1.977e – 03
9.912e – 06
6.282e – 09
1.865e – 03
2.126e – 05
9.670e – 09
0.2
3.514e – 03
1.762e – 05
1.117e – 08
3.283e – 03
3.779e – 05
1.719e – 08
0.3
4.612e – 03
2.313e – 05
1.466e – 08
4.308e – 03
4.960e – 05
2.256e – 08
0.4
5.271e – 03
2.643e – 05
1.675e – 08
4.924e – 03
5.668e – 05
2.579e – 08
0.5
5.490e – 03
2.753e – 05
1.745e – 08
5.129e – 03
5.905e – 05
2.686e – 08
0.6
5.271e – 03
2.643e – 05
1.675e – 08
4.923e – 03
5.668e – 05
2.579e – 08
0.7
4.612e – 03
2.313e – 05
1.466e – 08
4.308e – 03
4.960e – 05
2.256e – 08
0.8
3.514e – 03
1.765e – 05
1.117e – 08
3.283e – 03
3.779e – 05
1.719e – 08
0.9
1.977e – 03
9.912e – 06
6.282e – 09
1.846e – 03
2.126e – 05
9.670e – 09Experiment 4.
We consider the following time-space fractional advection-reaction-diffusion equation:(22)Dtγux,t=∂2ux,t∂x2+fx,t,0<x<1,0≤t≤1,ux,0=0,u0,t=u1,t=0,where fx,t=3π/4Γ2.5−αt1.5−αx4x−1, the exact solution uTx,t=x41−xt1.5. By mathematical 8.0, the numerical comparison of absolute errors by present method are given in Figures 13–20 at n=2,N=5. The reproducing kernel of Experiment 4 with different α,β is shown in Table 4 by present method at n=2,N=5. Figures 13 and 14 show the relationship between absolute error and α,β. From Figures 15 and 16, we can see that the absolute error is small. Figures 17–20 show the relationship between absolute error and γ.Figure 13
Absolute errors of Experiment4 with different reproducing kernels for t=0.1,h=0.01, γ=0.5.Figure 14
Absolute errors of Experiment4 with different reproducing kernels for t=0.1,h=0.0001, γ=0.5.Figure 15
Absolute errors of Experiment4 by the present method for t=0.1,h=0.00001, α=β=1.Figure 16
Absolute errors of Experiment1 by the present method for t=0.1, α=β=1,γ=0.5.Figure 17
Absolute errors of Experiment4 with different fractional derivatives at h=0.001, t=0.1,α=1/2,β=1.Figure 18
Absolute errors of Experiment4 with different fractional derivatives at h=0.001, t=0.1,α=β=0.Figure 19
Absolute errors of Experiment4 with different fractional derivatives at h=0.01,t=0.1, α=β=1.Figure 20
Absolute errors of Experiment4 with different fractional derivatives at h=0.001, t=0.1,α=β=1.Table 4
The reproducing kernel of Experiments1 and 4 with different α,β.
α
β
K
2
x
,
y
K
3
x
,
y α=0β=04xy12−15y+5x−3+4y60xy−1+x−1+y4−7y+7x−1+2yα=1/2β=1/2128xy/7π21−28y+4x−7+10y1024y/5π−1+xx−1+y9−16y+16x−1+2yα=1β=130xy10−14y+7x−2+3y280−1+xx−1+yy5−9y+9x−1+2yα=1/2β=11155/512xy80−104y+13x−8+11y6435/512−1+xx−1+yy80−136y+17x−8+15yα=2β=3504xy7−10y+5x−2+3y2772−1+xx−1+yy7−12y+2x−6+11y


## 5. Conclusions

In this paper, some new reproductive kernels are given. The numerical results of some models show that the present method has high precision compared with traditional RKM, and has a better convergence for this kind of model. Besides, the method can also be used to study other time variable fractional order advection-dispersion model.



---

*Source: 1794975-2020-04-10.xml*