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

1. Introduction

In this paper, we consider the following time-fractional order diffusion equation:

[mathematical expression not reproducible], (1)

where [[beta].sub.1](x, t), [[beta].sub.2](x, t), f(x, t) are known functions, and [D.sup.[gamma].sub.t]u (x, t) is the variable order Caputo fractional derivative of order [gamma]:

[mathematical expression not reproducible]. (2)

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. Let H be a real Hilbert spaces of functions f: [OMEGA] [right arrow] R. A function K: [OMEGA] x [OMEGA] [right arrow] R is called reproducing kernel for H if

(i) K (x, *) [member of] H for all x [member of] [OMEGA],

(ii) f(x) = [(f, K(*, x)).sub.H] for all f [member of] H and all x [member of] [OMEGA].

2. Structing Reproductive Kernel Space Based on the Shifted Jacobi Polynomials

The shifted Jacobi polynomials [p.sup.[alpha],[beta].sub.1.i] (x) of degree i is given  by

[p.sup.[alpha],[beta].sub.1.i](x) = [i.summation over (k=0)] [(-1).sup.(i-k)] [GAMMA](i + [beta] + 1)[GAMMA](i + k + 1 + [alpha] + [beta])/[GAMMA](k + 1 + [beta])[GAMMA](I + [alpha] + [beta] + 1)(i-k)!k! [x.sup.k], (3)

where

[mathematical expression not reproducible]. (4)

The shifted Jacobi polynomials on the interval x e [ 0, 1] are orthogonal with the orthogonality condition

[mathematical expression not reproducible], (5)

where w(x) = [x.sup.[beta]] [(1 - x).sup.[alpha]] is a weight function and

[mathematical expression not reproducible]. (6)

Definition 2. Let

[mathematical expression not reproducible], (7)

is the weighted inner product space of the shifted Jacobi polynomials on [0,1]. The inner product and norm are defined as

[mathematical expression not reproducible], (8)

where

[mathematical expression not reproducible], (9)

From [17, 19], we can prove that [[bar.H].sub.n][0,1] is a reproducing kernel Hilbert space. Its reproducing kernel is

[R.sub.n](x, y) = [n.summation over (i=0)] [e.sub.i](x) [e.sub.i](y), (10)

where [e.sub.i](x) = ([square root of (((2k + [alpha] + [beta] + 1))] k![GAMMA](k + [alpha] + [beta] + 1))/ [GAMMA](k + [alpha] + 1)[GAMMA](k + [beta] + 1)) [p.sup.[alpha],[beta].sub.1.i](x). Using Ref.  and the reproducing kernel of [[bar.H].sub.n][0,1], we can get following reproducing kernel spaces.

(1) Space [H.sub.2] = {u (x)| u(x) [member of] [[bar.H].sub.n], u (0) = 0}, [H.sub.n] with the same inner product as [[bar.H].sub.n] and [H.sub.n] is a reproducing kernel space and its reproducing kernel is

[K.sub.2](x, y)= [R.sub.n](x, y) - [R.sub.n](0, x)([R.sub.n] (y, 0)/[parallel][R.sub.n](0, 0)[[parallel].sup.2]. (11)

(2) Space [H.sub.3] = {u(x)| u(x) [member of] [[bar.H].sub.n+1], u(0) = 0, u(1) = 0}, [H.sub.n+1] with the same inner product as [[bar.H].sub.n+1] and [H.sub.n+1] is a reproducing kernel space and its reproducing kernel is

[K.sub.3](x, y) = [k.sub.n+1](x, y) - [k.sub.n+1](1, x) [k.sub.n+1](y, 1)/[parallel][k.sub.5](1,1)[[parallel].sup.2], (12)

where [k.sub.n+1] (x, y) = [R.sub.n+1] (x, y)-([R.sub.n+1] (0, x)[R.sub.n+1] (y, 0)/ [parallel][R.sub.n+1](0,0)[[parallel].sup.2]).

(3) Space H([OMEGA]) = [H.sub.2] [cross product] [H.sub.3] = {u(x, t)|u(x, t) [member of] [H.sub.2] [cross product] [H.sub.3], u(0, t) = u(1, t) = u(x, 0) = 0}, and its reproducing kernel is

K (x, t, y, s) = [K.sub.n](x, y) x [K.sub.n+1] (t, s), (x, y), (t, s) [member of] Q, (13)

where [K.sub.2](x, y), [K.sub.3](x, y) from (11) and (12). Reproducing kernels with different [alpha], [beta] are shown in Figures 1-8.

3. Piecewise Reproducing Kernel Method

After homogenization, equation (1) is converted to the following form:

[mathematical expression not reproducible], (14)

where A is a operator, and

[mathematical expression not reproducible]. (15)

v(x, t) = u(x, t) - U(x, t) - [u.sub.0] (x) + [U.sub.0] (x), U(x, t) = [[phi].sub.1] (t)(1 - x) + [phi] [sub.2] (t) x, [U.sub.0](x) = U (x, 0), [u.sub.0](x) = u(x, 0), g (x, t) = f (x, t) - [A.sub.(x, t)] (U (x, t) + [u.sub.0] (x) - [U.sub.0](x)).

Let [{[x.sub.i], [t.sub.i]}.sup.[infinity].sub.i=1] be nodes in interval [0,1] x [0,1], [[psi].sub.i](x, t) = [A.sub.(y, s)][K.sub.(y, s)](x,t)[|.sub.(y, s)=([x.sub.i], [t.sub.i])], i = 1 2,.., [infinity].

[[bar.[psi]].sub.i](x, t) = [i.summation over (k=1)][[beta].sub.ik][[psi].sub.k](x, t), [[beta].sub.ii] > 0, i = 1, 2, ..., [infinity], (16)

where the [[beta].sub.ik] are the coefficients resulting from Gram-Schmidt orthonormalization.

Theorem 1 (see [11, 19-23]). If [A.sup.-1] is existing and [{[x.sub.i], [t.sub.i]}.sup.[infinity].sub.i=1] is denumerable dense points in Q, then

v(x, t) = [[infinity].summation over (i=1)] [i.summation over (k=1)] [[beta].sub.ik]g([x.sub.k], [t.sub.k])[[bar.[psi]].sub.i](x, t), (17)

is an analytical solution of (14).

Deriving from the form of (17), we get the approximate solution of (14) as

[v.sub.N](x, t) = [N.summation over (i=1)] [i.summation over (k=1)] [[beta].sub.ik]g([x.sub.k], [t.sub.k])[[bar.[psi]].sub.i](x, t). (18)

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:

[mathematical expression not reproducible], (19)

where f(x, t) = ((8[x.sup.2](1 - x)[t.sup.2-[gamma]] + 3[x.sup.3] (4[t.sup.2] + 1))/[GAMMA](3 - [gamma])), the exact solution [u.sub.T](x, t) = [x.sub.2] (1 - x)(4[t.sup.2] + 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 [gamma].

Experiment 2. We consider the following time fractional reaction-diffusion equation 

[mathematical expression not reproducible], (20)

where f (x, t) = ((2[t.sup.2-[gamma]])/[GAMMA](3 - [gamma])) ([x.sup.2] - [x.sup.3]) + ([t.sup.2] + 1) (2[x.sup.2] - 3[x.sup.3] + 6x - 2), the exact solution [u.sub.T] (x, t) = ([t.sup.2] + 1)([x.sup.2] - [x.sup.3]). 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. .

Experiment 3. We consider the following time-space fractional advection-reaction-diffusion equation

[mathematical expression not reproducible], (21)

where f(x, t) = -(32[x.sup.2][t.sup.4-[gamma]]/[GAMMA](3 - [gamma])), the exact solution [u.sub.T] (x, t) = [x.sup.2] (4t2 + 1). Numerical solution of Experiment 3 is shown in Figure 12 and Table 3.

Experiment 4. We consider the following time-space fractional advection-reaction-diffusion equation:

[mathematical expression not reproducible], (22)

where f(x, t) = (3 [square root of ([pi])]/4[GAMMA](2.5 - [alpha]))[t.sup.1.5-[alpha]] [x.sup.4] (x - 1), the exact solution [u.sub.T](x, t) = [x.sup.4](1 - x)[t.sup.1.5]. 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 a, fi is shown in Table 4 by present method at n = 2, N = 5. Figures 13 and 14 show the relationship between absolute error and [alpha], [beta]. From Figures 15 and 16, we can see that the absolute error is small. Figures 17-20 show the relationship between absolute error and [gamma].

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.

https://doi.org/10.1155/2020/1794975

Data Availability

The data used to support the findings of this study are available from the corresponding author upon request.

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this article.

Acknowledgments

The authors would like to express our deep thanks to the anonymous reviewers whose valuable comments and suggestions helped us improve this article greatly. This paper was supported by the Natural Science Foundation of Inner Mongolia (2017MS0103).

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Xiaoli Zhang, (1,2) Haolu Zhang, (3) Lina Jia, (2) Yulan Wang [ID], (2) and Wei Zhang [ID] (1)

(1) Institute of Economics and Management, Jining Normal University, Jining 012000, Inner Mongolia, China

(2) Department of Mathematics, Inner Mongolia University of Technology, Hohhot 010051, China

(3) School of Civil Engineering, Inner Mongolia University of Technology, Hohhot 010051, China

Correspondence should be addressed to Yulan Wang; wylnei@163.com

Received 17 December 2019; Accepted 10 March 2020; Published 10 April 2020

Caption: FIGURE 1: Reproducing kernel [R.sub.3] (x,y) of space [[bar.H].sub.3] [0,1] witl [alpha] = [beta] = 1.

Caption: FIGURE 2: The set of reproducing kernel [R.sub.n](x, y) with n = 1, 2, ...,7, [alpha] = [beta] = 1.

Caption: FIGURE 3: Reproducing kernel R3 (x,y) of space [[bar.H].sub.3] [0,1] with [alpha] = 2, [beta] = 3.

Caption: FIGURE 4: The set of reproducing kernel [R.sub.n](x, y) with n = 1, 2, ...,7, [alpha] = 2, [beta] = 3.

Caption: FIGURE 5: Reproducing kernel [R.sub.3](x, y) of space [[bar.H].sub.3] [0,1] with [alpha] = - 0.5, [beta] = 1.

Caption: FIGURE 6: The set of reproducing kernel [R.sub.n](x, y) with n = 1, 2, ...,7, [alpha] = 0.5, [beta] = 1.

Caption: FIGURE 7: Reproducing kernel [R.sub.3] (x, y) of space [[bar.H].sub.3] [0,1] with [alpha] = [beta] = 0.5.

Caption: FIGURE 8: Reproducing kernels [R.sub.x](y) with n = 1, 2, ...,7, [alpha] = [beta] = -0.5.

Caption: FIGURE 9: Numerical results of Experiment 1 for [alpha] = [beta] = 0.5, [gamma] = 0.8, n = 4.

Caption: FIGURE 10: Absolute errors of Experiment 1 with different reproducing kernels for t = 0.1, h = 0.0001, [gamma] = 0.5, n = 2.

Caption: FIGURE 11: Absolute errors of Experiment 1 with different fractional derivatives for h = 0.01, t = 0.1, [alpha] = [beta] = 1, n = 2.

Caption: FIGURE 12: (a) The absolute error of Experiment 3 for [gamma] = 0.3, n = 4. (b) The absolute error of Experiment 3 for [gamma] = 0.6, n = 4.

Caption: FIGURE 13: Absolute errors of Experiment 4 with different reproducing kernels for t = 0.1, h = 0.01, [gamma] = 0.5.

Caption: FIGURE 14: Absolute errors of Experiment 4 with different reproducing kernels for t = 0.1, h = 0.0001, [gamma] = 0.5.

Caption: FIGURE 15: Absolute errors of Experiment 4 by the present method for t = 0.1, h = 0.00001, [alpha] = [beta] = 1.

Caption: FIGURE 16: Absolute errors of Experiment 1 by the present method for t = 0.1, [alpha] = [beta] = 1, [gamma] = 0.5.

Caption: FIGURE 17: Absolute errors of Experiment 4 with different fractional derivatives at h = 0.001, t = 0.1, [alpha] = (1/2), [beta] = 1.

Caption: FIGURE 18: Absolute errors of Experiment 4 with different fractional derivatives at h = 0.001, t = 0.1, [alpha] = [beta] = 0.

Caption: FIGURE 19: Absolute errors of Experiment 4 with different fractional derivatives at h = 0.01, t = 0.1, [alpha] = [beta] = 1.

Caption: FIGURE 20: Absolute errors of Experiment 4 with different fractional derivatives at h = 0.001, t = 0.1, [alpha] = [beta] = 1.
```TABLE 1: Comparison of absolute errors obtained for Experiment 1
at [alpha] = [beta] = 0.5, [gamma] = 0.8, t = 0.01, n = 4.

x      Exact     Traditional   Present method   Present method
solution    RKM N = 2     (h = 0.001)     (h = 0.000001)
N = 2            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 - 09

TABLE 2: Comparison of absolute errors obtained for
Experiment 2 with [alpha] = [beta] = 0.5, n = 4.

(x, t)       Present method   Present method     Ref. 
[gamma] = 0.5    [gamma] = 0.5    [gamma] = 0.5
h = 1         h = 0.000001

(0.1, 0.1)    2.085e - 03      1.767e - 09      6.317e - 06
(0.2, 0.2)    7.020e - 03      1.183e - 10      9.014e - 06
(0.3, 0.3)    1.179e - 02      1.283e - 08      1.091e - 05
(0.4, 0.4)    1.294e - 02      3.875e - 08      1.222e - 05
(0.5, 0.5)    7.814e - 03      7.546e - 08      1.212e - 05
(0.6, 0.6)    4.288e - 03      1.157e - 07      1.136e - 05
(0.7, 0.7)    2.088e - 02      1.475e - 07      9.658e - 06
(0.8, 0.8)    3.512e - 02      1.541e - 07      7.072e - 06
(0.9, 0.9)    3.456e - 02      1.137e - 07      3.757e - 06

(x, t)       Present method   Present method    Ref. 
[gamma] = 0.7    [gamma] = 0.7       II O
h = 1         h = 0.000001

(0.1, 0.1)    3.315e - 03      6.246e - 10     1.048e - 05
(0.2, 0.2)    1.074e - 02      4.179e - 09     9.436e - 06
(0.3, 0.3)    1.787e - 02      2.083e - 08     1.086e - 05
(0.4, 0.4)    2.015e - 02      5.094e - 08     1.086e - 05
(0.5, 0.5)    1.464e - 02      9.133e - 08     9.833e - 06
(0.6, 0.6)    6.253e - 04      1.340e - 07     8.678e - 06
(0.7, 0.7)    1.888e - 02      1.662e - 07     7.024e - 06
(0.8, 0.8)    3.600e - 02      1.703e - 07     4.934e - 06
(0.9, 0.9)    3.678e - 02      1.240e - 07     2.535e - 06

TABLE 3: Comparison of absolute errors obtained for
Experiment 3 in t = 0.02, n = 4.

x     [gamma] = 0.3    [gamma] = 0.3    [gamma] = 0.3
h = 1          h = 0.001       h = 0.000001

0.1    1.977e - 03      9.912e - 06      6.282e - 09
0.2    3.514e - 03      1.762e - 05      1.117e - 08
0.3    4.612e - 03      2.313e - 05      1.466e - 08
0.4    5.271e - 03      2.643e - 05      1.675e - 08
0.5    5.490e - 03      2.753e - 05      1.745e - 08
0.6    5.271e - 03      2.643e - 05      1.675e - 08
0.7    4.612e - 03      2.313e - 05      1.466e - 08
0.8    3.514e - 03      1.765e - 05      1.117e - 08
0.9    1.977e - 03      9.912e - 06      6.282e - 09

x     [gamma] = 0.6    [gamma] = 0.6    [gamma] = 0.6
h = 1          h = 0.001       h = 0.000001

0.1    1.865e - 03      2.126e - 05      9.670e - 09
0.2    3.283e - 03      3.779e - 05      1.719e - 08
0.3    4.308e - 03      4.960e - 05      2.256e - 08
0.4    4.924e - 03      5.668e - 05      2.579e - 08
0.5    5.129e - 03      5.905e - 05      2.686e - 08
0.6    4.923e - 03      5.668e - 05      2.579e - 08
0.7    4.308e - 03      4.960e - 05      2.256e - 08
0.8    3.283e - 03      3.779e - 05      1.719e - 08
0.9    1.846e - 03      2.126e - 05      9.670e - 09

TABLE 4: The reproducing kernel of Experiments 1 and 4 with different
[alpha], [beta].

[alpha]            [beta]                   [K.sub.2] (x, y)

[alpha] = 0        [beta] = 0         4xy (12 - 15y + 5x(-3 + 4y))
[alpha] = (1/2)    [beta] = (1/2)            (128xy/7[pi])
(21 - 28y + 4x(-7 + 10y))
[alpha] = 1        [beta] = 1         30xy(10 - 14y + 7x(-2 + 3y))
[alpha] = (1/2)    [beta] = (1/2)             (1155/512)xy
(80 - 104y + 13x(-8 + 11y))
[alpha] = 2        [beta] = 2        504xy (7 - 10y + 5x(-2 + 3y))

[alpha]            [beta]                   [K.sub.3] (x, y)

[alpha] = 0        [beta] = 0             60xy(-1 + x)(-1 + y)
(4 - 7y + 7x(-1 + 2y))
[alpha] = (1/2)    [beta] = (1/2)    (1024y/5[pi])(-1 + x)x(-1 + y)
(9 - 16y + 16x(-1 + 2y))
[alpha] = 1        [beta] = 1           280 (-1 + x) * (-1 + y)y
(5 - 9y + 9x(-1 + 2y))
[alpha] = (1/2)    [beta] = (1/2)     (6435/512)(-1 + x)x(-1 + y)y
(80 - 136y + 17x(-8 + 15y))
[alpha] = 2        [beta] = 2           2772 (-1 + x) * (-1 + y)y
(7 - 12y + 2x(-6 + 11y))
```