Difference Approximation for 2D Time-Fractional Integro-Differential Equation with Given Initial and Boundary Conditions

Abstract

In this investigation, a new algorithm based on the compact difference method is proposed. The purpose of this investigation is to solve the 2D time-fractional integro-differential equation. The Riemann–Liouville derivative was utilized to define the time-fractional derivative. Meanwhile, the weighted and shifted Grünwald difference operator and product trapezoidal formula were utilized to construct a high-order numerical scheme. Also, we analyzed the stability and convergence. The convergence order was $O({\tau }^2+h_x^4+h_y^4)$, where $\tau$ is the time step size, $h_x$ and $h_y$ are the spatial step sizes. Furthermore, several examples were provided to verify the correctness of our theoretical reasoning.

1. Introduction

Fractional differential equations (FDEs) have attracted much attention for modeling various phenomena in many scientific fields, such as physics [1,2,3], medicine [4,5,6], and electrochemistry [7]. However, analytical solutions in the majority of practical problems are difficult to find. Researchers have used numerical approximations instead of analytical ones to solve FDEs. For this, numerical methods were presented, for example, the finite difference method (FDM) [8,9], finite element method (FEM) [10,11,12], and other efficient methods [13,14,15,16].

In the realm of FDEs, the Caputo and Riemann–Liouville (RL) derivatives are mostly used. Furthermore, RL and Caputo derivatives have different properties. Comparing the definition of the Caputo derivative with the RL one, functions which are derivable in the Caputo sense are much “fewer” than those which are derivable in the RL sense. Physical and geometric interpretations for the RL derivative can be found in [17]. Here, we have no intention of mentioning which derivative is more widely utilized, but we must stress that every derivative has its own serviceable range. Additionally, there are some other special differential definitions, such as Caputo–Fabrizio and Atangana–Baleanu derivatives, without singular kernels [18,19].

Numerical investigations for the time-fractional integro-differential equation (TFIDE) have been studied by some scholars. For the nonlinear time-fractional partial integro-differential equation, Rawani et al. [20] proposed a hybrid method. Cao et al. [21] studied an accurate localized meshless collocation approach. For the linear time-fractional partial integro-differential equation, in [22], Fakhar-Izadi proposed a space–time Spectral–Galerkin method. Kamran et al. [23] proposed a localized transform method to approximate the solution of the equation. Generally speaking, the approximation of the integral term is our focus of research for solving the integro-differential equations. To approximate the integral term, in [24], the second-order fractional quadrature rule was proposed by Lubich. Dehghan and Abbaszadeh [25] studied a new numerical algorithm that combined the finite difference and finite element, with the time convergence order being estimated to be $O\left({\tau }^2\right)$. Qiao et al. [26] proposed an integral term processing method based on the trapezoidal rule and provided a high-precision ADI difference scheme. Based on the mixed finite element method, Wang et al. [27] considered the second-order scheme in the time direction. Huang et al. [28] studied the orthogonal spline collocation ADI scheme based on Lubich’s second-order quadrature convolution rule. Wang et al. [29,30] considered the trapezoidal PI rule and integration-by-parts method to improve the temporal accuracy, and proposed compact ADI schemes. In [31], Gu and Wu proposed a parallel-in-time iterative algorithm for Volterra partial integral-differential problems with a weakly singular kernel. Since there are much more studies on properties of the Caputo derivative, we focused on the studying the RL derivative in this article, which is helpful in understanding and modeling fractional equations with it.

This study investigates the compact finite difference technique for 2D fractional integro-differential equations. Consider the equation for

$$\left\{\begin{array}{cc}\hfill & {}_{0}D_t^{\alpha }u(x,y,t)=\gamma \Delta u(x,y,t)+{}_{0}I{}_t{}^{\beta }\Delta u(x,y,t)+f(x,y,t),(x,y)\in \Omega ,t\in (0,T],\hfill \\ \hfill & u(x,y,t)=\varphi (x,y,t),(x,y)\in \partial \Omega ,t\in (0,T],\hfill \\ \hfill & u(x,y,0)=\phi (x,y),(x,y)\in \Omega \cup \partial \Omega ,\hfill \end{array}\right.$$

(1)

where $\alpha ,\beta \in (0,1)$, $\Delta u(x,y,t)$ is the laplace operator, $\gamma$ is a positive constant, $\Omega =\{(x,y)\mid 0<x<L_1,0<y<L_2\}$ with $\partial \Omega$ being the boundary, and $f(x,y,t)$, $\varphi (x,y,t)$ and $\phi (x,y)$ are given functions.

In this article, the RL derivative ${}_{0}D_t^{\alpha }u(x,y,t)$ with $0<\alpha <1$ is expressed as (one can refer to [32]) follows:

$${}_{0}D_t^{\alpha }u(x,y,t)={\displaystyle \frac{1}{\Gamma (1-\alpha )}}{\displaystyle \frac{d}{dt}}{\int }_0^t{(t-s)}^{-\alpha }u(x,y,s)ds,$$

and the Rimann–Liouville fractional integral ${}_{0}I_t^{\beta }\Delta u(x,y,t)$ with $0<\beta <1$ is expressed as

$${}_{0}I_t^{\beta }\Delta u(x,y,t)={\displaystyle \frac{1}{\Gamma \left(\beta \right)}}{\int }_0^t{(t-s)}^{\beta -1}\Delta u(x,y,s)ds,$$

where $\Gamma (·)$ is the Gamma function.

In this article, firstly, we propose a new approach to approximate the RL derivative based on the weighted and shifted Grünwald difference operator. Secondly, a numerical scheme based on compact difference method and product trapezoidal formula is constructed for a 2D TFIDE. Finally, we used theoretical analysis and numerical examples to verify the effectiveness of the proposed method. We need to point out that the proposed method is discussed in view of the sufficient smoothness assumptions on the solution; thus, it is difficult to deal with the typical weak initial singularity of the solution. One can consider the variable-step or the graded mesh method for the typical weak initial singularity of the solution. See [33] for more about the variable-step L1 ${ }^{+}$ method. The following remark ensures the existence and uniqueness of the solution for Equation (1), which provides a guarantee for the subsequent work in this article.

Remark 1

([29,34]). For the sufficiently smooth $f(x,t)$, Equation (1) uses a unique solution and satisfies the regularities as $u(x,y,t)\in C^2([0,T];H_2\left(\overline{\Omega }\right)\cap H_0^1\left(\overline{\Omega }\right))$, for $(x,y,t)\in \overline{\Omega }\times [0,T],\overline{\Omega }=\Omega \cup \partial \Omega$.

This article is divided into different sections. The compact finite difference scheme studied in this article is proposed in the next section. Section 3 presents the stability analysis and error estimates of the proposed technique. In Section 4, some numerical tests are presented for comparison with the exact solution in order to verify the effectiveness of the proposed discrete scheme. A brief conclusion is given in the Section 5 of this article.

2. The Numerical Scheme

The compact difference scheme of Equation (1) will be considered in this section. Let $t_n=n\tau (0\le n\le N)$, where $\tau =T/N$ is the time step size and N is a positive integer. Let $x_i=i{h}_x(0\le i\le M_1)$ and $y_j=j{h}_y(0\le j\le M_2)$, where $h_x=L_1/M_1$ and $h_y=L_2/M_2$ are the spatial step sizes, and $M_1$ and $M_2$ are positive integers. The discrete grid ${\Omega }_h=\{(x_i,y_j)|1\le i\le M_1-1,1\le j\le M_2-1\}$, $\partial {\Omega }_h=\{(x_i,y_j)|i=0ori=M_1,j=0orj=M_2\}$ is the discrete boundary of ${\Omega }_h$, and ${\overline{\Omega }}_h={\Omega }_h\cup \partial {\Omega }_h$. Then, for the grid function $\mathcal{W}=\{{\mathcal{W}}_{ij}|1\le i\le M_1-1,1\le j\le M_2-1\}$ on ${\Omega }_h$, if $v\in \mathcal{W}$, we have

$${\delta }_x{v}_{i-{\displaystyle \frac{1}{2}},j}={\displaystyle \frac{1}{h_x}}(v_{ij}-v_{i-1,j}),{\delta }_x^2{v}_{i,j}={\displaystyle \frac{1}{h_x}}({\delta }_x{v}_{i-{\displaystyle \frac{1}{2}},j}-{\delta }_x{v}_{i+{\displaystyle \frac{1}{2}},j}),$$

(2)

$${\delta }_y{\delta }_x{v}_{i-{\displaystyle \frac{1}{2}},j-{\displaystyle \frac{1}{2}}}={\displaystyle \frac{1}{h_y}}({\delta }_x{v}_{i-{\displaystyle \frac{1}{2}},j}-{\delta }_x{v}_{i-{\displaystyle \frac{1}{2}},j-1}),{\delta }_y{\delta }_x^2{v}_{i,j-{\displaystyle \frac{1}{2}}}={\displaystyle \frac{1}{h_y}}({\delta }_x^2{v}_{i,j}-{\delta }_x^2{v}_{i,j-1}).$$

Particularly, for C in this article, C at different positions can represent different constants. Thus, $u_{ij}^n$ and $U_{ij}^n$ define the exact and approximate solutions of function u at the point $(x_i,y_j,t_n)$, respectively. In order to achieve fourth-order accuracy in the spatial direction, the compact difference operator is required.

$$H_x{U}_j=\left\{\begin{array}{cc}\hfill & (1+{\displaystyle \frac{h_x^2}{12}}{\delta }_x^2)U_{ij},1\le i\le M_1-1,0\le j\le M_2,\hfill \\ \hfill & U_{ij},i=0or{M}_1,0\le j\le M_2.\hfill \end{array}\right.$$

(3)

In Equation (3), $H_x$ is the compact difference operator in the x direction, and the definition of $H_y$ is similar to $H_x$. Further, for any grid function $u_{ij}\in \mathcal{W}$, define that $H{u}_{ij}=H_x{H}_y{u}_{ij}$.

The weighted and shifted Grünwald difference operator and the product trapezoidal formula are used in the following section. Next, some lemmas are introduced.

As in [35], we assume that $u\in$$L^{\alpha +1}\left(R\right)$ and let

$${}_{-\infty }D_t^{\alpha }u\left(t\right)={\displaystyle \frac{1}{\Gamma (1-\alpha )}}{\displaystyle \frac{d}{dt}}{\int }_{-\infty }^t{(t-s)}^{-\alpha }u\left(s\right)ds.$$

The following is the shifted Grünwald difference operator:

$$A_{\tau ,p}^{\left(\alpha \right)}u\left(t\right)={\displaystyle \frac{1}{{\tau }^{\alpha }}}\sum _{i=0}^{\infty }{g}_i^{\left(\alpha \right)}u(t-(i-p)\tau ),$$

(4)

and it approximates the RL derivative uniformly with first-order accuracy as $\tau \to 0$, i.e.,

$$A_{\tau ,p}^{\left(\alpha \right)}u\left(t\right)={}_{-\infty }D{}_t{}^{\alpha }u\left(t\right)+O\left(\tau \right),$$

where p is an integer, and the coefficients $g_i^{\left(\alpha \right)}(0<\alpha \le 1)$ are given by

$$g_0^{\left(\alpha \right)}=1,g_i^{\left(\alpha \right)}=(1-{\displaystyle \frac{\alpha +1}{i}})g_{i-1}^{\alpha },i=1,2,\cdots .$$

Moreover, we stipulate $g_0^{\left(\alpha \right)}=1$ and $g_i^{\left(\alpha \right)}$=0 $(i\ge 1)$ when $\alpha =0$.

Lemma 1

([36]). Let $u\in$$L^1\left(R\right)$, and define the weighted and shifted Grünwald difference approximation operator by

$${}_{L}D_{\tau ,p,q}^{\alpha }u\left(t\right)=\frac{\alpha -2q}{2(p-q)}{A}_{\tau ,p}^{\left(\alpha \right)}u\left(t\right)+\frac{2p-\alpha }{2(p-q)}{A}_{\tau ,q}^{\left(\alpha \right)}u\left(t\right).$$

(5)

Then, we have

$${}_{L}D_{\tau ,p,q}^{\alpha }u\left(t\right)={}_{-\infty }D{}_t{}^{\alpha }u\left(t\right)+O\left({\tau }^2\right)$$

uniformly for $t\in R$, where p, q are integers and $p\ne q$.

Moreover, p and q are symmetric, i.e., ${}_{L}D_{\tau ,p,q}^{\alpha }$= ${}_{L}D_{\tau ,q,p}^{\alpha }$. In [37], the RL derivative combined with Equation (5) was studied. Thus, we have

$$(1+{\displaystyle \frac{\alpha }{2}})A_{\tau ,0}^{\left(\alpha \right)}u\left(t\right)-{\displaystyle \frac{\alpha }{2}}{A}_{\tau ,-1}^{\left(\alpha \right)}u\left(t\right)={\displaystyle \frac{1}{{\tau }^{\alpha }}}\sum _{k=0}^{\infty }{w}_k^{\left(\alpha \right)}u(t-k\tau )={}_{-\infty }D{}_t{}^{\alpha }u\left(t\right)+O\left({\tau }^2\right),$$

where

$$\begin{array}{cc}\hfill & w_0^{\left(\alpha \right)}=(1+{\displaystyle \frac{\alpha }{2}})g_0^{\left(\alpha \right)}=1+{\displaystyle \frac{\alpha }{2}},\hfill \\ \hfill & w_k^{\left(\alpha \right)}=(1+{\displaystyle \frac{\alpha }{2}})g_k^{\left(\alpha \right)}-{\displaystyle \frac{\alpha }{2}}{g}_{k-1}^{\left(\alpha \right)},k\ge 1,\hfill \end{array}$$

(6)

with these conditions, we have

$${}_{0}D_t^{\alpha }u(x_i,y_j,t_n)={\tau }^{-\alpha }\sum _{k=0}^n{w}_k^{\left(\alpha \right)}u(x_i,y_j,t_{n-k})+O\left({\tau }^2\right).$$

Lemma 2

([38,39]). Suppose $u\left(x\right)\in C^6[x_{i-1},x_{i+1}]$ and $\varsigma \left(s\right)=5{(1-s)}^3-3{(1-s)}^5$; thus, we have

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{12}}(u_{xx}\left(x_{i+1}\right)+10u_{xx}\left(x_i\right)+u_{xx}\left(x_{i-1}\right))\hfill \\ \hfill & ={\displaystyle \frac{1}{h_x^2}}(u\left(x_{i+1}\right)-2u\left(x_i\right)+u\left(x_{i-1}\right))+{\displaystyle \frac{h_x^4}{360}}\underset{0}{\overset{1}{\int }}[f^6(x_i-s{h}_x)+f^6(x_i+s{h}_x)]\varsigma \left(s\right)ds.\hfill \end{array}$$

Lemma 2 gives that

$$H(u_{xx}(x_i,y_j)+u_{yy}(x_i,y_j))=H_y{\delta }_x^{2}u(x_i,y_j)+H_x{\delta }_y^{2}u(x_i,y_j)+R_{ij},$$

(7)

where $|R_{ij}|\le C(h_x^4+h_y^4).$

Lemma 3

([40,41]). Suppose $u\left(t\right)\in C^2[0,T]$; thus, there is a positive constant C related only to β when $1\le n\le N$. Therefore,

$$|{\int }_0^{t_n}{(t_n-s)}^{\beta -1}u\left(s\right)ds-\sum _{k=0}^n{a}_{n-k,n}^{n}u\left(t_{n-k}\right)|\le C\underset{0\le t\le T}{max}|u^{\prime \prime }\left(t\right)|t_n^{\beta }{\tau }^2,$$

where

$$a_{n-k,n}^n={\displaystyle \frac{{\tau }^{\beta }}{\beta (\beta +1)}}\times \left\{\begin{array}{cc}\hfill & 1,k=0,\hfill \\ \hfill & {(k+1)}^{\beta +1}-2k^{\beta +1}+{(k-1)}^{\beta +1},1\le k\le n-1,\hfill \\ \hfill & {(n-1)}^{\beta +1}-(n-1-\beta )n^{\beta },k=n.\hfill \end{array}\right.$$

(8)

Lemma 4.

Let $a_{n-k,n}^n$ under the definition of Equation (8); the sequence $\left\{a_{n-k,n}^n\right\}$ is monotonically decreasing concerning k.

Proof. 

From Equation (8), we have

$$a_{n-k,n}^n={\displaystyle \frac{{\tau }^{\beta }({(k+1)}^{\beta +1}-2k^{\beta +1}+{(k-1)}^{\beta +1})}{\beta (\beta +1)}},\mathrm{with}1\le k\le n-1\mathrm{and}2\le n\le N.$$

Let $h\left(ϵ\right)={\displaystyle \frac{{\tau }^{\beta }}{\beta (\beta +1)}}\times ({(ϵ+1)}^{\beta +1}-2{ϵ}^{\beta +1}+{(ϵ-1)}^{\beta +1})$. Then,

$$h^{\prime }\left(ϵ\right)={\displaystyle \frac{{\tau }^{\beta }}{\beta }}\times ({(ϵ+1)}^{\beta }-2{ϵ}^{\beta }+{(ϵ-1)}^{\beta }).$$

According to the mean value theorem, we have

$$h^{\prime }\left(ϵ\right)={\tau }^{\beta }({\xi }_1^{\beta -1}-{\xi }_2^{\beta -1}),$$

where ${\xi }_1\in (ϵ,ϵ+1)$, ${\xi }_2\in (ϵ-1,ϵ)$; thus, $h^{\prime }\left(ϵ\right)\le 0$. Therefore, the sequence $\left\{a_{n-k,n}^n\right\}$ on $[1,\infty )$ is monotonically decreasing. □

Lemma 5

([41]). Let β $(0<\beta <1)$ and $a_{k,n}^n$$(1\le n\le N)$ under the definition of Equation (8); thus,

$$\begin{array}{c}\hfill 0<a_{0,n}^n<{\tau }^{\beta }{(\beta +1)}^{-1},\sum _{k=1}^{n-1}|a_{k,n}^n|\le T^{\beta }{\beta }^{-1}.\end{array}$$

Assume that $u(x,y,t)\in C_{x,y,t}^{6,6,2}([0,L_1]\times [0,L_2]\times [0,T])$ and consider Equation (1) on grid point $(x_i,y_j,t_n)$. Thus, with Lemma 1, the left term in Equation (1) has the following form:

$${}_{0}D_t^{\alpha }u(x_i,y_j,t_n)={\displaystyle \frac{1}{{\tau }^{\alpha }}}\sum _{k=0}^n{w}_k^{\left(\alpha \right)}u(x_i,y_j,t_{n-k})+O\left({\tau }^2\right).$$

(9)

From Lemma 3, the second term to the right in Equation (1) can be expressed as follows:

$${\displaystyle \frac{1}{\Gamma \left(\beta \right)}}{\int }_0^{t_n}{(t_n-s)}^{\beta -1}\Delta u(x_i,y_j,s)ds={\displaystyle \frac{1}{\Gamma \left(\beta \right)}}\sum _{k=0}^n{a}_{n-k,n}^n\Delta u(x_i,y_j,t_{n-k})+O\left({\tau }^2\right).$$

(10)

Then, with Equations (9) and (10), Equation (1) can be rewritten in the following form:

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{{\tau }^{\alpha }}}\sum _{k=0}^n{w}_k^{\left(\alpha \right)}u(x_i,y_j,t_{n-k})\hfill \\ \hfill & =\gamma \Delta u(x_i,y_j,t_n)+{\displaystyle \frac{1}{\Gamma \left(\beta \right)}}\sum _{k=0}^n{a}_{n-k,n}^n\Delta u(x_i,y_j,t_{n-k})+f(x_i,y_j,t_n)+O\left({\tau }^2\right).\hfill \end{array}$$

(11)

Further, applying the compact difference operator H to Equation (11) and by using Equation (7), we can obtain the following difference scheme:

$$\begin{array}{cc}\hfill & H{\tau }^{-\alpha }\sum _{k=0}^n{w}_k^{\left(\alpha \right)}{u}_{ij}^{n-k}\hfill \\ \hfill & =\gamma (H_y{\delta }_x^2{u}_{ij}^n+H_x{\delta }_y^2{u}_{ij}^n)+\sum _{k=0}^n{\displaystyle \frac{a_{n-k,n}^n}{\Gamma \left(\beta \right)}}(H_y{\delta }_x^2{u}_{ij}^{n-k}+H_x{\delta }_y^2{u}_{ij}^{n-k})+H{f}_{ij}^n+R_{ij}^n,\hfill \end{array}$$

(12)

where $|R_{ij}^n|\le C({\tau }^2+h_x^4+h_y^4)$.

With Equation (12), by omitting the truncation error $R_{ij}^n$, the compact difference scheme for Equation (1) is as follows:

$$\left\{\begin{array}{cc}\hfill & H\sum _{k=0}^n{\displaystyle \frac{w_k^{\left(\alpha \right)}}{{\tau }^{\alpha }}}{U}_{ij}^{n-k}=\gamma (H_y{\delta }_x^2+H_x{\delta }_y^2)U_{ij}^n+\sum _{k=0}^n{\displaystyle \frac{a_{n-k,n}^n}{\Gamma \left(\beta \right)}}(H_y{\delta }_x^2+H_x{\delta }_y^2)U_{ij}^{n-k}+H{f}_{ij}^n,\hfill \\ \hfill & (x_i,y_j)\in {\Omega }_h,n\in [1,N],\hfill \\ \hfill & U_{ij}^0=\varphi (x_i,y_j),(x_i,y_j)\in {\overline{\Omega }}_h,\hfill \\ \hfill & U_{ij}^n=\phi (x_i,y_j,t_n),(x_i,y_j)\in \partial \Omega ,n\in [1,N].\hfill \end{array}\right.$$

(13)

It is clear to see that the coefficient matrices are strictly diagonally dominant for the compact difference scheme Equation (13). Thus, we have the following result as in [42].

Theorem 1.

The compact difference scheme Equation (13) is uniquely solvable.

3. Stability Analysis and Error Estimates

Now, stability and convergence of the given algorithm are considered. Below are some notations and lemmas to be used. For any grid function $U,V\in \mathcal{W}$, we have

$$\begin{array}{c}\hfill (U,V)=h_x{h}_y\sum _{i=1}^{M_1-1}\sum _{j=1}^{M_2-1}{U}_{ij}{V}_{ij}{,\parallel U\parallel }^2={(U,U),\parallel U\parallel }_{H_x}^2=(H_{x}U,U){,\parallel u\parallel }_{\infty }=\underset{0\le j\le M}{max}|u_j|.\end{array}$$

Lemma 6

([30]). Assume that $V\in \mathcal{W}$; thus,

$${\displaystyle \frac{2}{3}}{\parallel V\parallel }^2\le {\parallel V\parallel }_{H_x}^2\le {\parallel V\parallel }^{2}and\left({\displaystyle \frac{6}{L_1^2}}+{\displaystyle \frac{6}{L_2^2}}\right){\parallel V\parallel }^2\le \parallel {\delta }_x{V\parallel }^2+{\parallel {\delta }_{y}V\parallel }^2.$$

Lemma 7

([37,43]). If ${\left\{w_k^{\left(\alpha \right)}\right\}}_{n=0}^{\infty }$ under the definition of Equation (6), then for any positive integer m and real vector ${(u^0,u^1,\cdots ,u^m)}^T\in R^{m+1}$, it holds that

$$\sum _{n=0}^m\left(\sum _{k=0}^n{w}_k^{\left(\alpha \right)}{u}^{n-k}\right)u^n\ge 0.$$

Lemma 8

([38]). Suppose $W,V\in \mathcal{W}$, then

$$\begin{array}{c}\hfill ({\delta }_x^{2}W,V)=-({\delta }_{x}W,{\delta }_{x}V).\end{array}$$

Lemma 9

([38,44]). Let $U\in \mathcal{W}$; thus,

$$\parallel {\delta }_x{U\parallel }^2\le {\displaystyle \frac{4}{h_x^2}}{\parallel U\parallel }^{2}and\parallel H_{y}U\parallel \le \parallel U\parallel .$$

Lemma 10

([41,45]). Assume that $\left\{{\mu }_n\right\}$ is a non-negative sequence, which satisfies

$${\mu }_n\le \sum _{k=0}^{n-1}{\eta }_k{\mu }_k+b_{n}withn\ge 0,$$

where $b_n$ is non-decreasing sequence of non-negative numbers, and ${\eta }_k\ge 0$. Thus,

$${\mu }_n\le b_{n}exp\left(\sum _{k=0}^{n-1}{\eta }_k\right)withn\ge 0.$$

Lemma 11

([46]). If $\rho =({\rho }_1,{\rho }_2,\dots ,{\rho }_n)$ is a non-increasing sequence and $\varrho =({\varrho }_1,{\varrho }_2,\dots ,{\varrho }_n)$ is a non-decreasing sequence, then the following $\check{C}eby\check{s}ev$ inequality holds

$$\sum _{j=1}^m{\rho }_j{\varrho }_j\le {\displaystyle \frac{1}{m}}\sum _{j=1}^m{\rho }_j\sum _{j=1}^m{\varrho }_j.$$

Lemma 12.

Let $a_{n-m,n}^n$ under the definition of Equation (8). Thus,

$$\sum _{m=1}^{n-1}|a_{n-m,n}^n|\parallel U^{n-m}\parallel \le {\displaystyle \frac{M^{*}{T}^{\beta }}{(n-1)\beta }}\sum _{m=1}^{n-1}{\displaystyle \frac{\parallel U^m\parallel }{M_m}}withn=2,3,\dots ,N,$$

where $M^{*}$= $\underset{1\le m\le n-1}{max}\parallel U^{n-m}\parallel$ and $M_m=\parallel U^m\parallel$.

Proof. 

Let $M^{*}$= $\underset{1\le m\le n-1}{max}\parallel U^{n-m}\parallel$ and $M_m=\parallel U^m\parallel$. With Lemmas 4 and 11, we have

$$\sum _{m=1}^{n-1}|a_{n-m,n}^n|\parallel U^{n-m}\parallel \le \sum _{m=1}^{n-1}|a_{n-m,n}^n|M^{*}\le {\displaystyle \frac{1}{n-1}}\sum _{m=1}^{n-1}|a_{n-m,n}^n|\sum _{m=1}^{n-1}{M}^{*}.$$

From Lemma 5, we have

$$\sum _{m=1}^{n-1}|a_{n-m,n}^n|\parallel U^{n-m}\parallel \le {\displaystyle \frac{T^{\beta }}{(n-1)\beta }}\sum _{m=1}^{n-1}{M}^{*}.$$

Since the function u continues over $\Omega \times [0,T]$ ( $\Omega =\{(x,y)\mid 0\le x\le L_1,0\le y\le L_2\}$), the function u is bounded on $\Omega \times [0,T]$. If $M=\underset{1\le m\le n-1}{max}{M}^{*}/M_m$, then we have

$$\begin{array}{cc}\hfill \sum _{m=1}^{n-1}|a_{n-m,n}^n|\parallel U^{n-m}\parallel & \le {\displaystyle \frac{T^{\beta }{\beta }^{-1}}{(n-1)}}\left(M^{*}{\displaystyle \frac{\parallel U^1\parallel }{M_1}}+\dots +M^{*}{\displaystyle \frac{\parallel U^{n-1}\parallel }{M_{n-1}}}\right)\hfill \\ \hfill & ={\displaystyle \frac{T^{\beta }{\beta }^{-1}}{(n-1)}}\sum _{m=1}^{n-1}{\displaystyle \frac{M^{*}\parallel U^m\parallel }{M_m}}\hfill \\ \hfill & \le {\displaystyle \frac{M{T}^{\beta }}{(n-1)\beta }}\sum _{m=1}^{n-1}\parallel U^{n-m}\parallel .\hfill \end{array}$$

This concludes the proof. □

Theorem 2.

Let $U^n$ be the solutions of Equation (13); if ${\tau }^{\beta /2}/h^2\le C_1<\infty$ and ${\tau }^{\beta /2}{N}^{\beta }\le C_2<\infty$, then for all $\tau >0$, the following estimate holds:

$$\tau \sum _{n=0}^m\parallel U^n\parallel \le C\left(T\parallel U^0\parallel +\tau \sum _{n=0}^m\underset{0\le n\le m}{max}\parallel f^n\parallel \right),$$

where C, $C_1$ and $C_2$ are constants, and C and $C_2$ are dependent on T.

Proof. 

From Equation (13), we have

$$\begin{array}{c}\hfill H{\displaystyle \frac{1}{{\tau }^{\alpha }}}\sum _{k=0}^n{w}_k^{\left(\alpha \right)}{U}^{n-k}=\gamma (H_y{\delta }_x^2+H_x{\delta }_y^2)U^n+\sum _{k=0}^n{\displaystyle \frac{a_{n-k,n}^n}{\Gamma \left(\beta \right)}}(H_y{\delta }_x^2+H_x{\delta }_y^2)U^{n-k}+H{f}^n.\end{array}$$

(14)

Then, if we take the inner product of Equation (14) with $\tau U^n$, we can obtain the following:

$$\begin{array}{cc}\hfill & {\tau }^{1-\alpha }\sum _{k=0}^n{w}_k^{\left(\alpha \right)}(H{U}^{n-k},U^n)\hfill \\ \hfill & =\gamma \tau (H_y{\delta }_x^2{U}^n,U^n)+\gamma \tau (H_x{\delta }_y^2{U}^n,U^n)\hfill \\ \hfill & +{\displaystyle \frac{\tau }{\Gamma \left(\beta \right)}}\sum _{k=0}^n{a}_{n-k,n}^n\left((H_y{\delta }_x^2{U}^{n-k},U^n)+(H_x{\delta }_y^2{U}^{n-k},U^n)\right)+\tau (H{f}^n,U^n).\hfill \end{array}$$

(15)

Since the compact difference operators $H_x$ and $H_y$ are positive definite and self-joint, there are two positive operators $L_x$ and $L_y$, so $(Hu,w)=(L_x{L}_{y}u,L_x{L}_{y}w)=(Qu,Qw)$. Then, for the left term of Equation (15), we have

$${\tau }^{1-\alpha }\sum _{k=0}^n{w}_k^{\left(\alpha \right)}(H{U}^{n-k},U^n)={\tau }^{1-\alpha }\sum _{k=0}^n{w}_k^{\left(\alpha \right)}(Q{U}^{n-k},Q{U}^n).$$

(16)

For the first term on the right in Equation (15), with Lemmas 6 and 8, we have

$$\begin{array}{cc}\hfill & \gamma \tau (H_y{\delta }_x^2{U}^n,U^n)+\gamma \tau (H_x{\delta }_y^2{U}^n,U^n)\hfill \\ \hfill & =-\gamma \tau (H_y{\delta }_x{U}^n,{\delta }_x{U}^n)-\gamma \tau (H_x{\delta }_y{U}^n,{\delta }_y{U}^n)\hfill \\ \hfill & =-\gamma \tau \left(\parallel {\delta }_x{U}^n{\parallel }_{H_y}^2+{\parallel {\delta }_y{U}^n\parallel }_{H_x}^2\right)\hfill \\ \hfill & \le -{\displaystyle \frac{2\gamma \tau }{3}}\left(\parallel {\delta }_x{U}^n{\parallel }^2+{\parallel {\delta }_y{U}^n\parallel }^2\right)\hfill \\ \hfill & \le -{\displaystyle \frac{4\gamma \tau (L_2^2+L_1^2)}{\left(L_1^2{L}_2^2\right)}}{\parallel U^n\parallel }^2.\hfill \end{array}$$

(17)

For the second term on the right in Equation (15), with Lemmas 6 and 8, we have

$$\begin{array}{cc}\hfill & {\displaystyle \frac{\tau }{\Gamma \left(\beta \right)}}\sum _{k=0}^n{a}_{n-k,n}^n\left((H_y{\delta }_x^2{U}^{n-k},U^n)+(H_x{\delta }_y^2{U}^{n-k},U^n)\right)\hfill \\ \hfill & ={\displaystyle \frac{\tau }{\Gamma \left(\beta \right)}}\sum _{k=1}^n{a}_{n-k,n}^n\left((H_y{\delta }_x^2{U}^{n-k},U^n)+(H_x{\delta }_y^2{U}^{n-k},U^n)\right)\hfill \\ \hfill & +{\displaystyle \frac{\tau a_{n,n}^n}{\Gamma \left(\beta \right)}}\left((H_y{\delta }_x^2{U}^n,U^n)+(H_x{\delta }_y^2{U}^n,U^n)\right)\hfill \\ \hfill & ={\displaystyle \frac{\tau }{\Gamma \left(\beta \right)}}\sum _{k=1}^n{a}_{n-k,n}^n\left((H_y{\delta }_x^2{U}^{n-k},U^n)+(H_x{\delta }_y^2{U}^{n-k},U^n)\right)\hfill \\ \hfill & -{\displaystyle \frac{\tau a_{n,n}^n}{\Gamma \left(\beta \right)}}\left(\parallel {\delta }_x{U}^n{\parallel }_{H_y}^2+{\parallel {\delta }_y{U}^n\parallel }_{H_x}^2\right)\hfill \\ \hfill & \le {\displaystyle \frac{\tau }{\Gamma \left(\beta \right)}}\sum _{k=1}^n{a}_{n-k,n}^n\left((H_y{\delta }_x^2{U}^{n-k},U^n)+(H_x{\delta }_y^2{U}^{n-k},U^n)\right).\hfill \end{array}$$

With Lemma 8, Lemma 9, and the Cauchy–Schwarz inequality, we have

$$\begin{array}{cc}\hfill & {\displaystyle \frac{\tau }{\Gamma \left(\beta \right)}}\sum _{k=1}^n{a}_{n-k,n}^n\left((H_y{\delta }_x^2{U}^{n-k},U^n)+(H_x{\delta }_y^2{U}^{n-k},U^n)\right)\hfill \\ \hfill & =-{\displaystyle \frac{\tau }{\Gamma \left(\beta \right)}}\sum _{k=1}^n{a}_{n-k,n}^n\left((H_y{\delta }_x{U}^{n-k},{\delta }_x{U}^n)+(H_x{\delta }_y{U}^{n-k},{\delta }_y{U}^n)\right)\hfill \\ \hfill & \le {\displaystyle \frac{\tau }{\Gamma \left(\beta \right)}}\sum _{k=1}^n|a_{n-k,n}^n|\left(\parallel H_y{\delta }_x{U}^{n-k}\parallel \parallel {\delta }_x{U}^n\parallel +\parallel H_x{\delta }_y{U}^{n-k}\parallel \parallel {\delta }_y{U}^n\parallel \right)\hfill \\ \hfill & \le {\displaystyle \frac{\tau }{\Gamma \left(\beta \right)}}\sum _{k=1}^n|a_{n-k,n}^n|\left(\parallel {\delta }_x{U}^{n-k}\parallel \parallel {\delta }_x{U}^n\parallel +\parallel {\delta }_y{U}^{n-k}\parallel \parallel {\delta }_y{U}^n\parallel \right)\hfill \\ \hfill & \le {\displaystyle \frac{4\tau }{\Gamma \left(\beta \right)}}\sum _{k=1}^n|a_{n-k,n}^n|\left({\displaystyle \frac{\parallel U^{n-k}\parallel \parallel U^n\parallel }{h_x^2}}+{\displaystyle \frac{\parallel U^{n-k}\parallel \parallel U^n\parallel }{h_y^2}}\right).\hfill \end{array}$$

Let $h=min\{h_x,h_y\}$; thus, we have

$$\begin{array}{cc}\hfill & {\displaystyle \frac{4\tau }{\Gamma \left(\beta \right)}}\sum _{k=1}^n|a_{n-k,n}^n|\left({\displaystyle \frac{\parallel U^{n-k}\parallel \parallel U^n\parallel }{h_x^2}}+{\displaystyle \frac{\parallel U^{n-k}\parallel \parallel U^n\parallel }{h_y^2}}\right)\hfill \\ \hfill & \le {\displaystyle \frac{8\tau }{\Gamma \left(\beta \right)h^2}}\sum _{k=1}^n|a_{n-k,n}^n|\parallel U^{n-k}\parallel \parallel U^n\parallel \hfill \\ \hfill & \le {\displaystyle \frac{8C_1{\tau }^{1-\frac{\beta }{2}}}{\Gamma \left(\beta \right)}}\sum _{k=1}^n|a_{n-k,n}^n|\parallel U^{n-k}\parallel \parallel U^n\parallel .\hfill \end{array}$$

(18)

By inequality (16), inequality (17), and inequality (18), we get

$$\begin{array}{cc}\hfill & {\tau }^{1-\alpha }\sum _{k=0}^n{w}_k^{\left(\alpha \right)}(Q{U}^{n-k},Q{U}^n)+{\displaystyle \frac{4\gamma \tau (L_2^2+L_1^2)}{\left(L_1^2{L}_2^2\right)}}{\parallel U^n\parallel }^2\hfill \\ \hfill & \le {\displaystyle \frac{8C_1{\tau }^{1-\frac{\beta }{2}}}{\Gamma \left(\beta \right)}}\sum _{k=1}^n|a_{n-k,n}^n|\parallel U^{n-k}\parallel \parallel U^n\parallel +\tau \parallel f^n\parallel \parallel U^n\parallel .\hfill \end{array}$$

After deleting factor $\parallel U^n\parallel$, the following inequality holds:

$$\begin{array}{cc}\hfill & {\displaystyle \frac{{\tau }^{1-\alpha }}{\parallel U^n\parallel }}\sum _{k=0}^n{w}_k^{\left(\alpha \right)}(Q{U}^{n-k},Q{U}^n)+{\displaystyle \frac{4\gamma \tau (L_2^2+L_1^2)}{\left(L_1^2{L}_2^2\right)}}\parallel U^n\parallel \hfill \\ \hfill & \le {\displaystyle \frac{8C_1{\tau }^{1-\frac{\beta }{2}}}{\Gamma \left(\beta \right)}}\sum _{k=1}^n|a_{n-k,n}^n|\parallel U^{n-k}\parallel +\tau \parallel f^n\parallel .\hfill \end{array}$$

(19)

Summing up inequality (19) for n from 0 to m, the above inequality becomes the following form:

$$\begin{array}{cc}\hfill & {\tau }^{1-\alpha }\sum _{n=0}^m\sum _{k=0}^n{\displaystyle \frac{w_k^{\left(\alpha \right)}}{\parallel U^n\parallel }}(Q{U}^{n-k},Q{U}^n)+{\displaystyle \frac{4\gamma \tau (L_2^2+L_1^2)}{\left(L_1^2{L}_2^2\right)}}\sum _{n=0}^m\parallel U^n\parallel \hfill \\ \hfill & \le {\displaystyle \frac{8C_1{\tau }^{1-\frac{\beta }{2}}}{\Gamma \left(\beta \right)}}\sum _{n=0}^m\sum _{k=1}^n|a_{n-k,n}^n|\parallel U^{n-k}\parallel +\tau \sum _{n=0}^m\parallel f^n\parallel .\hfill \end{array}$$

(20)

From Lemma 7 and inequality (20), we have

$$\begin{array}{c}\hfill \tau \sum _{n=0}^m\parallel U^n\parallel \le {\displaystyle \frac{2C_1{\tau }^{1-\frac{\beta }{2}}{L}_1^2{L}_2^2}{\gamma \Gamma \left(\beta \right)(L_1^2+L_2^2)}}\sum _{n=0}^m\sum _{k=1}^n|a_{n-k,n}^n|\parallel U^{n-k}\parallel +{\displaystyle \frac{\tau L_1^2{L}_2^2}{4\gamma (L_1^2+L_2^2)}}\sum _{n=0}^m\parallel f^n\parallel .\end{array}$$

(21)

With Lemma 5, Lemma 12, and inequality (21), we have

$$\begin{array}{cc}\hfill & \tau \sum _{n=0}^m\parallel U^n\parallel \hfill \\ \hfill & ={\displaystyle \frac{2C_1{\tau }^{1-\frac{\beta }{2}}{L}_1^2{L}_2^2}{\gamma \Gamma \left(\beta \right)(L_1^2+L_2^2)}}\sum _{n=0}^m\left(\sum _{k=1}^{n-1}|a_{n-k,n}^n|\parallel U^{n-k}\parallel +|a_{0,n}^n|\parallel U^0\parallel \right)+{\displaystyle \frac{\tau {\gamma }^{-1}{L}_1^2{L}_2^2}{4(L_1^2+L_2^2)}}\sum _{n=0}^m\parallel f^n\parallel \hfill \\ \hfill & \le {\displaystyle \frac{2C_1{\tau }^{1-\frac{\beta }{2}}{L}_1^2{L}_2^2}{\gamma \Gamma \left(\beta \right)(L_1^2+L_2^2)}}\sum _{n=0}^m\left({\displaystyle \frac{M{T}^{\beta }}{(n-1)\beta }}\sum _{k=1}^{n-1}\parallel U^{n-k}\parallel +{\displaystyle \frac{{\tau }^{\beta }\parallel U^0\parallel }{\beta +1}}\right)+{\displaystyle \frac{\tau L_1^2{L}_2^2}{4\gamma (L_1^2+L_2^2)}}\sum _{n=0}^m\parallel f^n\parallel \hfill \\ \hfill & \le \sum _{k=1}^{n-1}{\displaystyle \frac{2\tau C_1{C}_{2}M{L}_1^2{L}_2^2}{\gamma \Gamma (\beta +1)(n-1)(L_1^2+L_2^2)}}\sum _{n=0}^m\parallel U^{n-k}\parallel \hfill \\ \hfill & +\sum _{n=0}^m\left({\displaystyle \frac{2{\tau }^{\frac{\beta }{2}+1}{C}_1{L}_1^2{L}_2^2}{\gamma \Gamma \left(\beta \right)(\beta +1)(L_1^2+L_2^2)}}\parallel U^0\parallel +{\displaystyle \frac{\tau L_1^2{L}_2^2}{4\gamma (L_1^2+L_2^2)}}\underset{0\le n\le m}{max}\parallel f^n\parallel \right).\hfill \end{array}$$

(22)

Let $b_n=\left({\displaystyle \frac{2{\tau }^{\beta /2+1}{C}_1{L}_1^2{L}_2^2}{\gamma \Gamma \left(\beta \right)(\beta +1)(L_2^2+L_2^2)}}\right){\sum }_{n=0}^m\parallel U^0\parallel +{\displaystyle \frac{\tau L_1^2{L}_2^2}{4\gamma (L_1^2+L_2^2)}}{\sum }_{n=0}^m\underset{0\le n\le m}{max}\parallel f^n\parallel$; thus, we can check if $b_n$ is a positive non-decreasing sequence concerning n. From Lemma 10 and inequality (22), we have

$$\begin{array}{cc}\hfill \tau \sum _{n=0}^m\parallel U^n\parallel & \le ({\displaystyle \frac{2{\tau }^{\frac{\beta }{2}+1}{C}_1{L}_1^2{L}_2^2}{\gamma \Gamma \left(\beta \right)(\beta +1)(L_2^2+L_2^2)}}\sum _{n=0}^m\parallel U^0\parallel \hfill \\ \hfill & +{\displaystyle \frac{\tau L_1^2{L}_2^2}{4\gamma (L_1^2+L_2^2)}}\sum _{n=0}^m\underset{0\le n\le m}{max}\parallel f^n\parallel )exp\left({\displaystyle \frac{2C_1{C}_{2}M{L}_1^2{L}_2^2}{\gamma \Gamma (\beta +1)(n-1)(L_2^2+L_2^2)}}\right).\hfill \end{array}$$

Let $C=max\{{\displaystyle \frac{2{\tau }^{\beta /2}{C}_1{L}_1^2{L}_2^2}{\gamma \Gamma \left(\beta \right)(\beta +1)(L_2^2+L_2^2)}},{\displaystyle \frac{L_1^2{L}_2^2}{4\gamma (L_1^2+L_2^2)}}\}exp\left({\displaystyle \frac{2C_1{C}_{2}M{L}_1^2{L}_2^2}{\gamma \Gamma (\beta +1)(n-1)(L_2^2+L_2^2)}}\right)$; thus, the following inequality holds:

$$\begin{array}{c}\hfill \tau \sum _{n=0}^m\parallel U^n\parallel \le C\left(T\parallel U^0\parallel +\tau \sum _{n=0}^m\underset{0\le n\le m}{max}\parallel f^n\parallel \right).\end{array}$$

This concludes the proof. □

Let $e_{ij}^n=u_{ij}^n-U_{ij}^n\{e_{ij}^n|0\le i\le M_1,0\le j\le M_2,\mathrm{and}0\le n\le N\}$. Now, we consider the convergence theorem of the proposed algorithm.

Theorem 3.

Assume that $u^n\in C_{x,y,t}^{6,6,2}([0,L_1]\times [0,L_2]\times [0,T])$ and $U^n\in {\overline{\Omega }}_h$ be the analytical solution and approximation solutions of Equation (1). If ${\tau }^{\beta /2}/h^2\le C_1<\infty$ and ${\tau }^{\beta /2}{N}^{\beta }\le C_2<\infty$, then the following holds:

$$\tau \sum _{n=0}^m\parallel e^n\parallel \le CT({\tau }^2+h_x^4+h_y^4),$$

where C, $C_1$, and $C_2$ are constants, and C and $C_2$ are dependent on T.

Proof. 

With the definition of $e^n$, we can obtain

$$\begin{array}{c}\hfill H{\displaystyle \frac{1}{{\tau }^{\alpha }}}\sum _{k=0}^n{w}_k^{\left(\alpha \right)}{e}^{n-k}=\gamma (H_y{\delta }_x^2+H_x{\delta }_y^2)e^n+\sum _{k=0}^n{\displaystyle \frac{a_{n-k,n}^n}{\Gamma \left(\beta \right)}}(H_y{\delta }_x^2+H_x{\delta }_y^2)e^{n-k}+R^n.\end{array}$$

(23)

From Theorem 2 and Equation (23), the following inequality holds:

$$\begin{array}{cc}\hfill \tau \sum _{n=0}^m\parallel e^n\parallel & \le C\left(T\parallel e^0\parallel +\tau \sum _{n=0}^m\parallel R^n\parallel \right)\hfill \\ \hfill & \le C\left(\tau \sum _{n=0}^m\parallel R^n\parallel \right)\hfill \\ \hfill & \le CT({\tau }^2+h_x^4+h_y^4).\hfill \end{array}$$

This concludes proving. □

4. Numerical Examples

To support our theoretical analysis, two tests are used in this section. All tests were carried out in MATLAB 2020 using the AMD Ryzen 4700U with 2.00 GHz and 16 GB of RAM as the hardware configuration.

The $L^{\infty }$ norm error (one can refer to [47]) between the exact and the numerical solutions is shown in the following tables:

$${\mathrm{Error}}_{L^{\infty }}=\underset{1\le n\le N}{max}|u(x_i,y_j,t_n)-U_{ij}^n|,$$

where $u(x_i,y_j,t_n)$ and $U_{ij}^n$ are the exact solution and numerical solution, respectively. In addition, the time convergence rate is

$${\mathrm{Rate}}^t=log(E_1^t/E_2^t)/log({\tau }_1/{\tau }_2),$$

where $E_1^t$ and $E_2^t$ are errors that correspond to the grids with mesh sizes ${\tau }_1$ and ${\tau }_2$ in time, respectively. Moreover, the space convergence rate is

$${\mathrm{Rate}}^x=log(E_1^x/E_2^x)/log(h_1/h_2),$$

where $E_1^x$ and $E_2^x$ are errors corresponding to the grids with mesh sizes $h_1$ and $h_2$ in space, respectively.

Example 1.

For the first model, we use $h=h_x=h_y$, $T=1$, $\Omega =[0,1]\times [0,1]$ and $\gamma =1$ with the exact solution as

$$u(x,y,t)=t^4{x}^4{(1-x)}^4{y}^4{(1-y)}^4,$$

then, we consider the following force term:

$$\begin{array}{cc}\hfill f(x,y,t)& ={\displaystyle \frac{24t^{4-\alpha }}{\Gamma (5-\alpha )}}\left(x^4{(1-x)}^4{y}^4{(1-y)}^4\right)\hfill \\ \hfill & -\left(t^4+{\displaystyle \frac{24t^{\beta +4}}{\Gamma (5+\beta )}}\right)(y^4{(1-y)}^4(12x^2(1-x^4)-32x^3{(1-x)}^3+12x^4{(1-x)}^2)\hfill \\ \hfill & +\left(x^4{(1-x)}^4\right)(12y^2(1-y^4)-32y^3{(1-y)}^3+12y^4{(1-y)}^2)),\hfill \end{array}$$

The above condition $\Omega =[0,1]\times [0,1]$ means that $M_1=M_2$. Let $M=M_1=M_2$;

Table 1. Example 1: Errors—computational orders with $M=32$ and $\beta =0.2$.

N	$\mathit{\alpha }$ = 0.2		$\mathit{\alpha }$ = 0.5		$\mathit{\alpha }$ = 0.8
	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{t}}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{t}}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{t}}$
70	9.1433 × 10−10	–	8.1616 × 10−10	–	6.6074 × 10 ${ }^{-10}$	–
80	7.0624 × 10−10	1.93	6.2979 × 10−10	1.94	5.0900 × 10 ${ }^{-10}$	1.95
90	5.6170 × 10 ${ }^{-10}$	1.94	5.0017 × 10 ${ }^{-10}$	1.96	4.0318 × 10 ${ }^{-10}$	1.98
100	4.5715 × 10 ${ }^{-10}$	1.96	4.0631 × 10 ${ }^{-10}$	1.97	3.2638 × 10 ${ }^{-10}$	2.01

,

Table 2. Example 1: Errors—computational orders with $M=32$ and $\beta =0.5$.

N	$\mathit{\alpha }$ = 0.2		$\mathit{\alpha }$ = 0.5		$\mathit{\alpha }$ = 0.8
	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{t}}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{t}}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{t}}$
50	2.0804 × 10 ${ }^{-9}$	–	1.8721 × 10 ${ }^{-9}$	–	1.5490 × 10 ${ }^{-9}$	–
60	1.4484 × 10 ${ }^{-9}$	1.99	1.3016 × 10 ${ }^{-9}$	1.99	1.0744 × 10 ${ }^{-9}$	2.01
70	1.0651 × 10 ${ }^{-9}$	1.99	9.5551 × 10 ${ }^{-10}$	2.01	7.8626 × 10 ${ }^{-10}$	2.03
80	8.1516 × 10 ${ }^{-10}$	2.00	7.2984 × 10 ${ }^{-10}$	2.02	5.9828 × 10 ${ }^{-10}$	2.05

,

Table 3. Example 1: Errors—computational orders with $M=32$ and $\beta =0.8$.

N	$\mathit{\alpha }$ = 0.2		$\mathit{\alpha }$ = 0.5		$\mathit{\alpha }$ = 0.8
	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{t}}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{t}}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{t}}$
10	4.3329 × 10 ${ }^{-8}$	–	3.8059 × 10 ${ }^{-8}$	–	2.9837 × 10 ${ }^{-8}$	–
20	1.0931 × 10 ${ }^{-8}$	1.99	9.5612 × 10 ${ }^{-9}$	1.99	7.4416 × 10 ${ }^{-9}$	2.00
30	4.8663 × 10 ${ }^{-9}$	2.00	4.2472 × 10 ${ }^{-9}$	2.00	3.2920 × 10 ${ }^{-9}$	2.01
40	2.7359 × 10 ${ }^{-9}$	2.00	2.3829 × 10 ${ }^{-9}$	2.01	1.8394 × 10 ${ }^{-9}$	2.02

,

Table 4. Example 1: Errors—computational orders with $N=400$ and $\beta =0.2$.

M	$\mathit{\alpha }$ = 0.2		$\mathit{\alpha }$ = 0.5		$\mathit{\alpha }$ = 0.8
	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{x}}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{x}}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{x}}$
12	1.6977 × 10 ${ }^{-8}$	–	1.6792 × 10 ${ }^{-8}$	–	1.6534 × 10 ${ }^{-8}$	–
16	5.4909 × 10 ${ }^{-9}$	3.92	5.4417 × 10 ${ }^{-9}$	3.92	5.3726 × 10 ${ }^{-9}$	3.91
20	2.2737 × 10 ${ }^{-9}$	3.95	2.2499 × 10 ${ }^{-9}$	3.96	2.2164 × 10 ${ }^{-9}$	3.97
24	1.0913 × 10 ${ }^{-9}$	4.03	1.0808 × 10 ${ }^{-9}$	4.02	1.0660 × 10 ${ }^{-9}$	4.02

,

Table 5. Example 1: Errors—computational orders with $N=400$ and $\beta =0.5$.

M	$\mathit{\alpha }$ = 0.2		$\mathit{\alpha }$ = 0.5		$\mathit{\alpha }$ = 0.8
	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{x}}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{x}}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{x}}$
12	1.6913 × 10 ${ }^{-8}$	–	1.6700 × 10 ${ }^{-8}$	–	1.6405 × 10 ${ }^{-8}$	–
16	5.4740 × 10 ${ }^{-9}$	3.92	5.4172 × 10 ${ }^{-9}$	3.91	5.3382 × 10 ${ }^{-9}$	3.90
20	2.2657 × 10 ${ }^{-9}$	3.96	2.2384 × 10 ${ }^{-9}$	3.96	2.2002 × 10 ${ }^{-9}$	3.97
24	1.0879 × 10 ${ }^{-9}$	4.02	1.0759 × 10 ${ }^{-9}$	4.02	1.0589 × 10 ${ }^{-9}$	4.01

and

Table 6. Example 1: Errors—computational orders with $N=400$ and $\beta =0.8$.

M	$\mathit{\alpha }$ = 0.2		$\mathit{\alpha }$ = 0.5		$\mathit{\alpha }$ = 0.8
	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{x}}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{x}}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{x}}$
12	1.6856 × 10 ${ }^{-8}$	–	1.6617 × 10 ${ }^{-8}$	–	1.6287 × 10 ${ }^{-8}$	–
16	5.4588 × 10 ${ }^{-9}$	3.92	5.3952 × 10 ${ }^{-9}$	3.91	5.3069 × 10 ${ }^{-9}$	3.90
20	2.2583 × 10 ${ }^{-9}$	3.96	2.2276 × 10 ${ }^{-9}$	3.96	2.1849 × 10 ${ }^{-9}$	3.98
24	1.0847 × 10 ${ }^{-9}$	4.02	1.0712 × 10 ${ }^{-9}$	4.02	1.0523 × 10 ${ }^{-9}$	4.01

present errors and computational orders for Example 1. Table 1, Table 2 and Table 3 confirm the second-order accuracy in the temporal direction. In Table 1, by setting $M=32$, $\beta =0.2$, we obtain a different $\alpha$ and N. Similar to $\beta =0.2$, the results of $\beta =0.5$ and $\beta =0.8$ under the same conditions are shown in Table 2 and Table 3, respectively. It can be seen that the numerical results agree with the theoretical ones. Table 4, Table 5 and Table 6 confirm the fourth-order accuracy in the spatial direction. For $N=400$, in Table 4, by setting $\beta =0.2$, we obtain a different $\alpha$ and M. Similar to $\beta =0.2$, the results of $\beta =0.5$ and $\beta =0.8$ under the same conditions are shown in Table 5 and Table 6, respectively. The numerical results obtained are consistent with those obtained theoretically.

In Figure 1a, by setting $\beta =0.5$, we obtain a different $\alpha$ and demonstrate the results with the $L^{\infty }$-norm, which attains the accuracy in the temporal direction with $O\left({\tau }^2\right)$. In Figure 1b, by setting $\beta =0.2$, we obtain a different $\alpha$. This demonstrates the results in the $L^{\infty }$-norm, which attains the accuracy in the spatial direction with $O\left(h^4\right)$. In Figure 2, by setting $M_1=M_2=100$ and $N=100$, this demonstrates a graph comprising a solution and error with $\alpha =\beta =0.99$.

For the compact finite difference scheme, the desired accuracy in the temporal direction and the spatial direction can achieve $O\left({\tau }^2\right)$ and $O\left(h^4\right)$ for any parameter $\alpha$ and $\beta$, which implies that the numerical solution has favorable stability.

Example 2.

For the second example, we consider $h=h_x=h_y$, $T=1$, $\Omega =[0,1]\times [0,1]$ and $\gamma =1$. This example is used to test the case that the exact solution is unknown. In this example, the initial term is $\phi (x,y)=sin\left(\pi x\right)sin\left(\pi y\right)$ and the force term is $f(x,y,t)=0$.

Table 7. Example 2: Errors—computational orders with $M=32$ and $\alpha =0.1$.

N	$\mathit{\beta }$ = 0.1		$\mathit{\beta }$ = 0.5		$\mathit{\beta }$ = 0.9
	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{t}}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{t}}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{t}}$
130	7.9699 × 10 ${ }^{-5}$	–	3.2922 × 10 ${ }^{-4}$	–	7.1783 × 10 ${ }^{-4}$	–
140	6.9555 × 10 ${ }^{-5}$	1.84	2.8754 × 10 ${ }^{-4}$	1.83	6.2756 × 10 ${ }^{-4}$	1.81
150	6.0783 × 10 ${ }^{-6}$	2.00	2.5147 × 10 ${ }^{-4}$	1.95	5.4930 × 10 ${ }^{-4}$	1.93
160	5.3123 × 10 ${ }^{-6}$	2.01	2.1993 × 10 ${ }^{-4}$	2.01	4.8079 × 10 ${ }^{-4}$	2.06

,

Table 8. Example 2: Errors—computational orders with $M=32$ and $\alpha =0.5$.

N	$\mathit{\beta }$ = 0.1		$\mathit{\beta }$ = 0.5		$\mathit{\beta }$ = 0.9
	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{t}}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{t}}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{t}}$
140	8.7400 × 10 ${ }^{-5}$	–	4.8492 × 10 ${ }^{-4}$	–	1.0980 × 10 ${ }^{-3}$	–
150	7.7033 × 10 ${ }^{-5}$	1.83	4.2813 × 10 ${ }^{-4}$	1.81	9.7004 × 10 ${ }^{-4}$	1.80
160	6.7865 × 10 ${ }^{-5}$	1.96	3.7777 × 10 ${ }^{-4}$	1.94	8.5644 × 10 ${ }^{-4}$	1.93
170	5.9694 × 10 ${ }^{-5}$	2.11	3.3277 × 10 ${ }^{-4}$	2.09	7.5481 × 10 ${ }^{-4}$	2.08

and

Table 9. Example 2: Errors—computational orders with $M=32$ and $\alpha =0.9$.

N	$\mathit{\beta }$ = 0.1		$\mathit{\beta }$ = 0.5		$\mathit{\beta }$ = 0.9
	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{t}}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{t}}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{t}}$
150	1.3776 × 10 ${ }^{-4}$	–	6.9117 × 10 ${ }^{-4}$	–	1.5866 × 10 ${ }^{-3}$	–
160	1.2261 × 10 ${ }^{-4}$	1.81	6.1698 × 10 ${ }^{-4}$	1.80	1.4272 × 10 ${ }^{-3}$	2.02
170	1.0889 × 10 ${ }^{-4}$	1.96	5.4946 × 10 ${ }^{-4}$	1.91	1.2628 × 10 ${ }^{-3}$	2.02
180	9.6385 × 10 ${ }^{-5}$	2.13	4.8762 × 10 ${ }^{-4}$	2.08	1.1212 × 10 ${ }^{-3}$	2.03

confirm the computational orders in the temporal direction with $O\left({\tau }^2\right)$. In order to verify the accuracy and convergence rate in the temporal direction, we consider the solution on the fine grid $(N=300$ and $M=32$) as the exact solution. The solutions on coarse grids $(M=32$ and different N) are used as numerical solutions. According to the data in Table 7, Table 8 and Table 9, it can be clearly seen that the second-order accuracy in the temporal direction is verified.

Table 10. Example 2: Errors—computational orders with $N=500$ and $\alpha =0.1$.

M	$\mathit{\beta }$ = 0.1		$\mathit{\beta }$ = 0.5		$\mathit{\beta }$ = 0.9
	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{x}}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{x}}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}^{\mathit{x}}$
4	3.4174 × 10 ${ }^{-9}$	–	1.7859 × 10 ${ }^{-8}$	–	4.5051 × 10 ${ }^{-8}$	–
8	2.0959 × 10 ${ }^{-10}$	4.03	1.0955 × 10 ${ }^{-9}$	4.03	2.7635 × 10 ${ }^{-9}$	4.03
16	1.2991 × 10 ${ }^{-11}$	4.01	6.7905 × 10 ${ }^{-11}$	4.01	1.7130 × 10 ${ }^{-10}$	4.01
32	7.6333 × 10 ${ }^{-13}$	4.08	3.9099 × 10 ${ }^{-12}$	4.08	1.0065 × 10 ${ }^{-11}$	4.08

,

Table 11. Example 2: Errors—computational orders with $N=500$ and $\alpha =0.5$.

M	$\mathit{\beta }$ = 0.1		$\mathit{\beta }$ = 0.5		$\mathit{\beta }$ = 0.9
	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}_{{\mathit{L}}^{\infty }}$
4	6.2105 × 10 ${ }^{-8}$	–	5.9446 × 10 ${ }^{-7}$	–	1.4047 × 10 ${ }^{-6}$	–
8	3.8108 × 10 ${ }^{-9}$	4.02	3.6463 × 10 ${ }^{-8}$	4.03	8.6159 × 10 ${ }^{-8}$	4.03
16	2.3622 × 10 ${ }^{-10}$	4.01	2.2602 × 10 ${ }^{-9}$	4.01	5.3406 × 10 ${ }^{-9}$	4.01
32	1.3880 × 10 ${ }^{-11}$	4.08	1.3280 × 10 ${ }^{-10}$	4.08	3.1380 × 10 ${ }^{-10}$	4.08

and

Table 12. Example 2: Errors—computational orders with $N=500$ and $\alpha =0.9$.

M	$\mathit{\beta }$ = 0.1		$\mathit{\beta }$ = 0.5		$\mathit{\beta }$ = 0.9
	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Error}}_{{\mathit{L}}^{\infty }}$	${\mathbf{Rate}}_{{\mathit{L}}^{\infty }}$
4	1.5936 × 10 ${ }^{-6}$	–	9.8327 × 10 ${ }^{-6}$	–	2.3640 × 10 ${ }^{-5}$	–
8	9.7745 × 10 ${ }^{-8}$	4.03	6.0304 × 10 ${ }^{-7}$	4.03	1.4498 × 10 ${ }^{-6}$	4.03
16	6.0587 × 10 ${ }^{-9}$	4.01	3.7379 × 10 ${ }^{-8}$	4.01	8.9865 × 10 ${ }^{-8}$	4.01
32	3.5599 × 10 ${ }^{-10}$	4.08	2.1963 × 10 ${ }^{-9}$	4.08	5.2802 × 10 ${ }^{-9}$	4.08

show that the computational order in the spatial direction is $O\left(h^4\right)$. In order to verify the accuracy and convergence rate in the spatial direction, we relate to the solution on the fine grid $(N=500$ and $M=64$) as the exact solution. The solutions on coarse grids $(N=500$ and different $M=4,8,16,32$) are used as numerical solutions. According to the data in Table 10, Table 11 and Table 12, it can be clearly seen that the fourth-order accuracy in the spatial direction is verified.

The results of the second example show that the proposed method is still effective as the exact solution is unknown. As the exact solution of this example is unknown, the orders of convergence in the computed solutions can be estimated by using the two-mesh principle; for more detailed information about the two-mesh principle, one can refer to [48,49]. This approach can enhance the reliability of the results, and we will use the two-mesh principle in our future work.

5. Conclusions

A new algorithm for 2D time-fractional integro-differential equations is studied in this manuscript, which is based on a compact difference operator. The time-fractional derivative was approximated by the weighted and shifted Grünwald difference operator; meanwhile, the integral term was approximated with a product of the trapezoidal formula. This paper also discussed the stability and error bounds of the presented method. Numerical investigations showed that the compact finite difference scheme converged with order $O({\tau }^2+h_x^4+h_y^4)$, which is in excellent agreement with our theoretical analysis. Note that the numerical accuracy of our proposed method does not depend on parameters $\alpha$ and $\beta$. The scientific content of this article is based on uniform meshes. Non-uniform meshes are very meaningful and interesting, and we will study then deeply in the future. In our future work, we would like to investigate other fractional derivatives in temporal and spatial directions, as well as algorithms comprising nonlinear and multi-kernel fractional integro-differential equations.