A new fully discrete finite difference/element approximation for fractional cable equation

Abstract

A novel fully discrete Crank–Nicolson finite element method, which is obtained by finite difference in time and finite element in space, is presented to approximate the fractional Cable equation. Compared to the L1-formula for discretizing fractional derivatives at time $t_{n+1}$, the proposed approximate scheme is directly obtained at time $t_{n+\frac{1}{2}}$, in which some new coefficients $(k+\frac{1}{2})^{1-\alpha }-(k-\frac{1}{2})^{1-\alpha }$ instead of $(k+1)^{1-\alpha }-k^{1-\alpha }$ are derived. Based on the new approximate formula, the stability and error estimate are analyzed in detail and the optimal convergence rate $O(\tau ^{\min \{1+\alpha _{1},1+\alpha _{2}\}}+h^{r+1})$ is obtained. Numerical examples in one-dimensional and two-dimensional spaces are shown to illustrate the effectiveness and feasibility of the studied algorithm.

1 Introduction

Fractional partial differential equations (PDEs), which mainly include time, space and space-time fractional PDEs, have a lot of applications (such as in chaos, mechanical systems, control, continuous-time random walks and so forth). Recently, some numerical solutions for Fractional PDEs have been obtained based on some different numerical methods. These numerical methods mainly cover finite element methods [6, 8, 15, 16, 18, 22, 27–29, 38, 42–44, 46, 47, 49], finite difference methods [1, 3, 7, 10, 11, 13, 17, 23, 26, 30–35, 37, 39, 41, 45, 48], spectral approximations [2, 24, 25], DG methods [9, 40] and so forth.

Here, we mainly give some introductions on finite element approximations of fractional PDEs in detail. Deng [8] discussed a finite element method for the space-time fractional Fokker–Planck equation. Zhang et al. [42] presented a Galerkin finite element scheme for symmetric space-fractional PDEs. In 2011, Li et al. [22] gave some detailed numerical analysis of finite element methods for fractional subdiffusion and superdiffusion equations. In 2012, Zhao and Li [46] gave some numerical analysis of time-space fractional telegraph equation based on finite element approximations. Wang et al. [38] studied space-fractional diffusion equations’ inhomogeneous Dirichlet boundary-value problems and analyzed their finite element approximations. In 2011, Jiang and Ma [15] considered a high-order finite element method for one-dimensional time-fractional PDE. Jiang and Ma [16] presented the moving finite element method for a time fractional PDE. In [44], Zhang et al. gave some analysis and numerical results on finite element method for Grwünwald–Letnikov time-fractional PDE. In [28, 29], some mixed finite element methods are studied for second and fourth order fractional PDEs. In 2010, Zheng et al. [47] gave a note on the finite element method for the advection diffusion equation with space-fractional derivative. Jin et al. [18] studied lumped mass Galerkin finite element method for the homogeneous diffusion equation with time-fractional derivative.

In this paper, our main purpose is to introduce finite element (FE) method to solve the following time-fractional Cable equation

$$\begin{aligned} \frac{\partial u(x,t)}{\partial t}=-\gamma {_0D_t^{1-\alpha _{1}}}u+{_0D_t^{1-\alpha _{2}}}\varDelta u, \quad x\in \varOmega , \, t\in [0,T], \end{aligned}$$

(1)

subject to the initial condition

$$\begin{aligned} u(x,0)=u_0(x), \quad x\in \varOmega , \end{aligned}$$

(2)

and the boundary condition

$$\begin{aligned} u(x,t)=0, \quad x\in \partial \varOmega , \, t\in [0,T]. \end{aligned}$$

(3)

Here $\varOmega \subset \mathbb {R}^d(d=1,2)$ is a bounded space domain with the boundary $\partial \varOmega$, $\varDelta =\frac{\partial ^2}{\partial x^2}(d=1)$ and $\varDelta =\frac{\partial ^2}{\partial x_1^2}+\frac{\partial ^2}{\partial x_2^2}(d=2)$ represent the Laplacian operators, and ${_0D_t^{\alpha }}$ means to take the left Riemann–Liouville fractional derivative with respect to the time variable t by order $\alpha$ $(0<\alpha <1)$, which is defined by

$$\begin{aligned} {_0D_t^{\alpha }}f(x,t)=\frac{1}{\varGamma (1-\alpha )}\frac{\partial }{\partial t}\int _{0}^{t}\frac{f(x,s)}{(t-s)^{\alpha }}\hbox {d}s, \end{aligned}$$

(4)

where $\varGamma (\cdot )$ denotes the Gamma function.

Based on the anomalous electrodiffusion in nerve cells, an important model–the fractional Cable equation has been derived and developed by many authors [5, 12, 19–21]. Recently, some numerical methods for numerically solving the fractional Cable equation have been developed. Lin and Xu [24] discussed the spectral discrete method for fractional Cable equation, some detailed error analysis are made and some calculated data are provided to verify their theoretical results. Bhrawy and Zaky [4] considered an accurate spectral collocation algorithm for solving nonlinear fractional Cable equations in one- and two-dimensional cases. Zhuang et al. [49] considered a Galerkin finite element approximation for one-dimensional time fractional Cable equation and obtained a good approximation accuracy in time. Hu and Zhang [14] solved numerically the fractional Cable equation by using implicit compact difference schemes. In [32], Quintana-Murillo and Yuste gave the explicit numerical method for fractional Cable equation including two temporal Riemann–Liouville derivatives.

In this discussion, we mainly present a novel discrete scheme of fractional derivative differing from the approximate method in [24, 35], and formulate then analyze a new fully discrete Crank–Nicolson type Galerkin finite element scheme, which is different from the one in [49]. In [24, 35], the time fractional derivative was discretized at time $t=t_{n+1}$. Compared to the approximate formula proposed in [24, 35], our approximate method is made directly at time $t_{n+\frac{1}{2}}$ and a new discrete scheme is formulated based on some different coefficients $(k+\frac{1}{2})^{1-\alpha }-(k-\frac{1}{2})^{1-\alpha }$ from the ones $(k+1)^{1-\alpha }-k^{1-\alpha }$ in [24, 35]. In the full text, we give some detailed analysis on stability and error estimates for two-dimensional problem of the fractional Cable equation. Moreover, two numerical tests are made to confirm our theoretical analysis.

In this paper, the functional spaces we adopted are the standard Sobolev spaces as follow

$$\begin{aligned} H_0^1(\varOmega ):= & {} \{v\in H^1(\varOmega ),\, v|_{\partial \varOmega }=0\}, \\ H^m(\varOmega ):= & {} \{v\in L^2(\varOmega ),\, D^{\beta }v\in L^2(\varOmega ),\text {for all} \,\, |\beta | \le m\}, \end{aligned}$$

where $L^2(\varOmega )$ is the space of measurable functions whose squares are Lebesgue integrable in $\varOmega$. The inner products of $L^2(\varOmega )$ and $H^1(\varOmega )$ are defined by

$$\begin{aligned} (u,v)=\int _{\varOmega }uv\hbox {d}x, \quad (u,v)_1=(u,v)+(\nabla u,\nabla v), \end{aligned}$$

and the corresponding norms are defined as

$$\begin{aligned} \Vert u\Vert _0=(u,u)^{\frac{1}{2}} , \quad \Vert u\Vert _1=(u,u)_1^{\frac{1}{2}}, \end{aligned}$$

respectively. Furthermore, $H^m(\varOmega )$ is equipped with the norm

$$\begin{aligned} \Vert u\Vert _m=\Big (\sum _{0 \le |\beta | \le m}\Vert D^{\beta }u\Vert _0^2\Big )^{\frac{1}{2}}. \end{aligned}$$

2 The fully discrete scheme

Set $\tau =\frac{T}{N}$ and let $0=t_0<t_1<\cdots <t_N=T$ be the uniform partition of the time interval [0, T], where $t_n=n\tau$. In the following, we denote $u^n=u(\cdot ,t_n)$ and the terms of time derivatives will be approximated at intermediate nodes $t_{n+\frac{1}{2}}=\frac{t_n+t_{n+1}}{2}$.

The first order time derivative $\frac{\partial u}{\partial t}$ at $t=t_{n+\frac{1}{2}}$ can be approximated as

$$\begin{aligned} \frac{\partial u(x,t_{n+\frac{1}{2}})}{\partial t}=\frac{u^{n+1}-u^n}{\tau }+O(\tau ^2), \quad n=0,1,\ldots , N-1. \end{aligned}$$

(5)

Recalling the relationship between the Caputo fractional derivative and Riemann–Liouville fractional derivative

$$\begin{aligned} {_0D_t^{\alpha }}f(x,t) = {}^C_0D_t^{\alpha }f(x,t) + \frac{f(x,0)}{\varGamma (1-\alpha )t^{\alpha }}, \quad 0<\alpha <1, \end{aligned}$$

(6)

as well as the definition of Caputo fractional derivative of order $\alpha$

$$\begin{aligned} {^C_0D_t^{\alpha }}f(x,t)=\frac{1}{\varGamma (1-\alpha )}\int _0^t\frac{\partial f(x,s)}{\partial s}\frac{\hbox {d}s}{(t-s)^{\alpha }}, \quad 0<\alpha <1, \end{aligned}$$

we can approximate the Caputo fractional derivative as the following

$$\begin{aligned}&{^{C}_0D_t^{\alpha }}f(x,t_{n+\frac{1}{2}}) \nonumber \\&\quad = \frac{1}{\varGamma (1-\alpha )} \left[ \sum \limits _{j=0}^{n-1}\int _{t_j}^{t_{j+1}} \frac{\partial f}{\partial s}\frac{\hbox {d}s}{\left( t_{n+\frac{1}{2}}-s\right) ^{\alpha }} +\int _{t_n}^{t_{n+\frac{1}{2}}}\frac{\partial f}{\partial s}\frac{\hbox {d}s}{\left( t_{n+\frac{1}{2}}-s\right) ^{\alpha }}\right] \nonumber \\&\quad = \frac{1}{\varGamma (1-\alpha )} \left[ \sum \limits _{j=0}^{n-1}\frac{f^{j+1}-f^j}{\tau } \int _{t_j}^{t_{j+1}}\frac{\hbox {d}s}{\left( t_{n+\frac{1}{2}}-s\right) ^{\alpha }} \right. \nonumber \\&\qquad \left. +\,\frac{f^{n+1}-f^n}{\tau }\int _{t_n}^{t_{n+\frac{1}{2}}} \frac{\hbox {d}s}{\left( t_{n+\frac{1}{2}}-s\right) ^{\alpha }}\right] +r_t \nonumber \\&\quad = \frac{\tau ^{-\alpha }}{\varGamma (2-\alpha )} \left\{ \sum \limits _{j=0}^{n-1}(f^{j+1}-f^j) \left[ \left( n+\frac{1}{2}-j\right) ^{1-\alpha }- \left( n-\frac{1}{2}-j\right) ^{1-\alpha } \right] \right. \nonumber \\&\qquad \left. +\,(f^{n+1}-f^n)\frac{1}{2^{1-\alpha }} \right\} +r_t \nonumber \\&\quad = \frac{\tau ^{-\alpha }}{\varGamma (2-\alpha )}\sum \limits _{k=0}^{n}c_k(f^{n+1-k}-f^{n-k})+r_t, \quad \quad n=0,1,2,\ldots ,N-1, \end{aligned}$$

(7)

where $c_k=(k+\frac{1}{2})^{1-\alpha }-(k-\frac{1}{2})^{1-\alpha }, \,( k=1,2,\ldots ),\,\, c_0=\frac{1}{2^{1-\alpha }}$ and $r_t=O(\tau ^{2-\alpha })$, which can be proved by the similar method in [24, 35].

Combining (6) and (7), we have that for $n=0,1,2,\ldots ,N-1$

$$\begin{aligned} {_0D_t^{1-\alpha _{1}}}u\left( t_{n+\frac{1}{2}}\right)= & {} \frac{\tau ^{\alpha _{1}-1}}{\varGamma (1+\alpha _{1})} \sum \limits _{k=0}^{n}c^1_k(u^{n+1-k}-u^{n-k}) \nonumber \\&+\frac{u(x,0)}{\varGamma (\alpha _{1})\left[ (n+\frac{1}{2})\tau \right] ^{1-\alpha _{1}}} + O(\tau ^{1+\alpha _1}), \nonumber \\ {_0D_t^{1-\alpha _{2}}}\varDelta u(t_{n+\frac{1}{2}})= & {} \frac{\tau ^{\alpha _{2}-1}}{\varGamma (1+\alpha _{2})}\sum \limits _{k=0}^{n}c^2_k \left( \varDelta u^{n+1-k}-\varDelta u^{n-k}\right) \nonumber \\&+\frac{\varDelta u(x,0)}{\varGamma (\alpha _{2}) \left[ (n+\frac{1}{2})\tau \right] ^{1-\alpha _{2}}}+O(\tau ^{1+\alpha _2}), \end{aligned}$$

(8)

where

$$\begin{aligned} c^1_k= \left( k+\frac{1}{2}\right) ^{\alpha _1}- \left( k-\frac{1}{2}\right) ^{\alpha _1}, \quad c^1_0=\frac{1}{2^{\alpha _1}}; \nonumber \\ c^2_k = \left( k+\frac{1}{2}\right) ^{\alpha _2}- \left( k-\frac{1}{2}\right) ^{\alpha _2}, \quad c^2_0=\frac{1}{2^{\alpha _2}}. \end{aligned}$$

(9)

Denote ${{\hat{\alpha }}_{1}}=\frac{\tau ^{\alpha _{1}}}{\varGamma (1+\alpha _{1})}$, ${{\hat{\alpha }}_{2}}=\frac{\tau ^{\alpha _{2}}}{\varGamma (1+\alpha _{2})}$, ${{\hat{\beta }}_{n}^{1}}=\frac{\tau }{\varGamma (\alpha _{1})[(n+\frac{1}{2})\tau ]^{1-\alpha _{1}}}$, ${{\hat{\beta }}_{n}^{2}}=\frac{\tau }{\varGamma (\alpha _{2})[(n+\frac{1}{2})\tau ]^{1-\alpha _{2}}}$, and $R_t=O(\tau ^2+\tau ^{1+\alpha _1}+\tau ^{1+\alpha _2})$, then the variational problem can be stated as: Find $u^{n+1}\in H^1_0(\varOmega )$ such that for all $v\in H^1_0(\varOmega )$,

$$\begin{aligned} (u^{n+1},v)= & {} (u^n,v)-\gamma {{\hat{\alpha }}_{1}}\sum \limits _{k=0}^{n}c^1_k(u^{n+1-k}-u^{n-k},v) \nonumber \\&-{{\hat{\alpha }}_{2}}\sum \limits _{k=0}^{n}c^2_k(\nabla u^{n+1-k}-\nabla u^{n-k},\nabla v) -\gamma {{\hat{\beta }}_{n}^{1}} (u^0,v) \nonumber \\&-{{\hat{\beta }}_{n}^{2}}(\nabla u^0,\nabla v)+\tau (R_t,v), \quad n=0,1,2,\ldots ,N-1. \end{aligned}$$

(10)

The following properties can be proved directly.

Lemma 1

The coefficients $c^1_k$ and $c^2_k$ defined by (9) satisfy that

$$\begin{aligned} \begin{aligned}&c^1_k>0, \quad c^2_k>0,\quad j=0,1,2,\ldots \\&c^1_0>c^1_1>c^1_2>\cdots >c^1_k>0, \,\, c^1_k\rightarrow 0\,(k\rightarrow \infty ); \\&c^2_0>c^2_1>c^2_2>\cdots >c^2_k>0, \,\, c^2_k\rightarrow 0\,(k\rightarrow \infty ). \end{aligned} \end{aligned}$$

(11)

Lemma 2

${{\hat{\alpha }}_{1}}=\frac{\tau ^{\alpha _{1}}}{\varGamma (1+\alpha _{1})}$, ${{\hat{\alpha }}_{2}}=\frac{\tau ^{\alpha _{2}}}{\varGamma (1+\alpha _{2})}$, ${{\hat{\beta }}_{n}^{1}}=\frac{\tau }{\varGamma (\alpha _{1})[(n+\frac{1}{2})\tau ]^{1-\alpha _{1}}}$, and ${{\hat{\beta }}_{n}^{2}}=\frac{\tau }{\varGamma (\alpha _{2})[(n+\frac{1}{2})\tau ]^{1-\alpha _{2}}}$ satisfy that

$$\begin{aligned} {{\hat{\beta }}_{n}^{1}} \le {{\hat{\alpha }}_{1}} c^1_n, \quad {{\hat{\beta }}_{n}^{2}} \le {{\hat{\alpha }}_{2}} c^2_n, \quad n\ge 1. \end{aligned}$$

(12)

Proof

Firstly we prove that ${{\hat{\beta }}_{n}^{1}} \le {{\hat{\alpha }}_{1}} c^1_n$.

From the definition of $c^1_n$, we have

$$\begin{aligned}&{{\hat{\alpha }}_{1}} c^1_n-{{\hat{\beta }}_{n}^{1}} \nonumber \\&\quad = \frac{\tau ^{\alpha _{1}}}{\varGamma (1+\alpha _{1})}\left[ \left( n + \frac{1}{2}\right) ^{\alpha _{1}} - \left( n-\frac{1}{2}\right) ^{\alpha _{1}}\right] - \frac{\tau }{\varGamma (\alpha _{1}) \left[ \left( n+\frac{1}{2}\right) \tau \right] ^{1-\alpha _{1}}} \nonumber \\&\quad = \frac{\tau ^{\alpha _{1}}\left[ \left( n+\frac{1}{2}\right) \tau \right] ^{1-\alpha _{1}} \left[ \left( n+\frac{1}{2}\right) ^{\alpha _{1}}-\left( n-\frac{1}{2}\right) ^{\alpha _{1}} \right] - \alpha _{1}\tau }{\varGamma (1+\alpha _{1}) \left[ \left( n+\frac{1}{2}\right) \tau \right] ^{1-\alpha _{1}}} \nonumber \\&\quad = \tau \frac{\left( n+\frac{1}{2}\right) ^{1-\alpha _{1}} \left[ \left( n+\frac{1}{2}\right) ^{\alpha _{1}} - \left( n-\frac{1}{2}\right) ^{\alpha _{1}}\right] - \alpha _{1}}{\varGamma (1+\alpha _{1})\left[ \left( n+\frac{1}{2}\right) \tau \right] ^{1-\alpha _{1}}}. \end{aligned}$$

(13)

Considering the differential mean-value theorem, we know

$$\begin{aligned}&\left( n+\frac{1}{2}\right) ^{1-\alpha _{1}} \left[ \left( n+\frac{1}{2}\right) ^{\alpha _{1}}- \left( n-\frac{1}{2}\right) ^{\alpha _{1}}\right] - \alpha _{1}\nonumber \\&\quad = \left( n+\frac{1}{2}\right) ^{1-\alpha _{1}}\alpha _{1}\xi ^{\alpha _{1}-1}-\alpha _{1}, \quad \xi \in \left( n-\frac{1}{2},n+\frac{1}{2}\right) \nonumber \\&\quad \ge \alpha _{1}-\alpha _{1}=0. \end{aligned}$$

(14)

After substituting the result of (14) to (13), we reach that ${{\hat{\alpha }}_{1}} c^1_n-{{\hat{\beta }}_{n}^{1}} \ge 0$, which is ${{\hat{\beta }}_{n}^{1}}\le {{\hat{\alpha }}_{1}} c^1_n$.

Similarly, we can prove that ${{\hat{\beta }}_{n}^{2}} \le {{\hat{\alpha }}_{2}} c^2_n$. $\square$

Now we consider approximating the variational problem in space. Let $\mathfrak {I}=\{\mathfrak {I}_h\}$ be a quasi-uniform partition of $\varOmega$. $\mathfrak {I}_h$ is a triangle of $\varOmega$, $h_e$ is the diameter of $\mathfrak {I}_h$, and $h=\max \{h_e\}$. Let $S^h\subset H_0^1(\varOmega )$ be a family of finite element space with the accuracy of order $r\ge 1$, i.e.,

$$\begin{aligned} S^h=\{\chi \in H_0^1(\varOmega )\cap C(\varOmega ): \chi |_{\mathfrak {I}_h}\in P_{r},\,\, \forall \mathfrak {I}_h\in \mathfrak {I}\}, \end{aligned}$$

where $P_r$ denotes the set of polynomials of degree up to r. The Crank–Nicolson type fully discrete finite element scheme can be stated as following: for $n\ge 1$, to find $u_h^{n+1}\in S^h$ such that $\forall v_h\in S^h$ satisfies

$$\begin{aligned}&\left( 1+\gamma {{\hat{\alpha }}_{1}} c^1_0\right) \left( u_h^{n+1}, v_h\right) + {{\hat{\alpha }}_{2}} c^2_0 \left( \nabla u_h^{n+1},\nabla v_h\right) \nonumber \\&\quad = \left( u_h^n,v_h\right) +\gamma {{\hat{\alpha }}_{1}}\sum \limits _{k=0}^{n-1}\left( c^1_k-c^1_{k+1}\right) \left( u_h^{n-k}, v_h\right) +\gamma \left( {{\hat{\alpha }}_{1}} c^1_n-{{\hat{\beta }}_{n}^{1}}\right) \left( u_h^0,v_h\right) \nonumber \\&\qquad + \, {{\hat{\alpha }}_{2}}\sum \limits _{k=0}^{n-1}\left( c^2_k-c^2_{k+1}\right) \left( \nabla u_h^{n-k}, \nabla v_h\right) + \left( {{\hat{\alpha }}_{2}} c^2_n-{{\hat{\beta }}_{n}^{2}}\right) \left( \nabla u_h^0,\nabla v_h\right) . \end{aligned}$$

(15)

When $n=0$, it equals that the two terms of sum in the above formulation don’t appear, which means: to find $u_h^1\in S^h$ such that for $\forall v_h\in S^h$ it holds

$$\begin{aligned}&(1+\gamma {{\hat{\alpha }}_{1}} c^1_0)(u_h^1,v_h)+{{\hat{\alpha }}_{2}} c^2_0(\nabla u_h^1,\nabla v_h) \nonumber \\&\quad = (u_h^0,v_h) +\gamma ({{\hat{\alpha }}_{1}} c^1_0-\hat{\beta ^1_0})(u_h^0,v_h)+({{\hat{\alpha }}_{2}} c^2_0-\hat{\beta ^2_0})(\nabla u_h^0,\nabla v_h), \end{aligned}$$

(16)

here $u_h^0$ is the approximation to the original value $u_0(x)$.

3 Stability and error estimation for the fully discrete scheme

In this section, we will give the stability analysis and error estimation for the fully discrete scheme (15) and (16). Throughout the following C will denote positive constants, not necessarily the same at different occurrences, which is independent of the finite element mesh size h and time step size $\tau$.

Theorem 1

The fully discrete formulations (15) and (16) are unconditionally stable, and it holds that

$$\begin{aligned} \omega ^{n} \le \Vert u_h^0\Vert ^2 +\frac{\gamma \tau ^{\alpha _{1}}}{\varGamma (1+\alpha _{1})}\Vert u_h^0\Vert ^2+ \frac{\tau ^{\alpha _{2}}}{\varGamma (1+\alpha _{2})}\Vert \nabla u_h^0\Vert ^2,\quad n\ge 1, \end{aligned}$$

(17)

where $\omega ^n=\Vert u_h^n\Vert ^2+\gamma {{\hat{\alpha }}_{1}}\sum \limits _{k=0}^{n-1}c^1_k\Vert u_h^{n-k}\Vert ^2+{{\hat{\alpha }}_{2}}\sum \limits _{k=0}^{n-1}c^2_k\Vert \nabla u_h^{n-k}\Vert ^2$.

Proof

When $n\ge 1$, taking $v_h=u_h^{n+1}$ in (15), we get

$$\begin{aligned}&\left( 1+\gamma {{\hat{\alpha }}_{1}} c^1_0\right) \Vert u_h^{n+1}\Vert ^2+{{\hat{\alpha }}_{2}} c^2_0\Vert \nabla u_h^{n+1}\Vert ^2 \nonumber \\&\quad = \left( u_h^{n},u_h^{n+1}\right) + \gamma {{\hat{\alpha }}_{1}}\sum \limits _{k=0}^{n-1}\left( c^1_k-c^1_{k+1}\right) \left( u_h^{n-k},u_h^{n+1}\right) +\gamma \left( {{\hat{\alpha }}_{1}} c^1_n-{{\hat{\beta }}_{n}^{1}}\right) \left( u_h^0,u_h^{n+1}\right) \nonumber \\&\qquad \;\;+\,{{\hat{\alpha }}_{2}}\sum \limits _{k=0}^{n-1}\left( c^2_k-c^2_{k+1}\right) \left( \nabla u_h^{n-k},\nabla u_h^{n+1}\right) + \left( {{\hat{\alpha }}_{2}} c^2_n-{{\hat{\beta }}_{n}^{2}}\right) \left( \nabla u_h^0,\nabla u_h^{n+1}\right) . \end{aligned}$$

(18)

Noting (11) and (12), we can get the following by Hölder inequality and Young inequality

$$\begin{aligned}&\left( 1+\gamma {{\hat{\alpha }}_{1}} c^1_0\right) \Vert u_h^{n+1}\Vert ^2+{{\hat{\alpha }}_{2}} c^2_0\Vert \nabla u_h^{n+1}\Vert ^2 \nonumber \\&\quad \le \frac{\Vert u_h^n\Vert ^2}{2} + \frac{\Vert u_h^{n+1}\Vert ^2}{2} + \gamma {{\hat{\alpha }}_{1}} \left[ \sum \limits _{k=0}^{n-1}\left( c^1_k-c^1_{k+1}\right) \frac{\Vert u_h^{n-k}\Vert ^2}{2} + \left( c^1_0-c^1_n\right) \frac{\Vert u_h^{n+1}\Vert ^2}{2}\right] \nonumber \\&\qquad +\,{{\hat{\alpha }}_{2}} \left[ \sum \limits _{k=0}^{n-1}\left( c^2_k-c^2_{k+1}\right) \frac{\Vert \nabla u_h^{n-k}\Vert ^2}{2}+ \left( c^2_0-c^2_n\right) \frac{\Vert \nabla u_h^{n+1}\Vert ^2}{2}\right] \nonumber \\&\qquad +\,\gamma \left( {{\hat{\alpha }}_{1}} c^1_n-{{\hat{\beta }}_{n}^{1}}\right) \left( \frac{\Vert u_h^0\Vert ^2}{2} +\frac{\Vert u_h^{n+1}\Vert ^2}{2}\right) \nonumber \\&\qquad +\, \left( {{\hat{\alpha }}_{2}} c^2_n-{{\hat{\beta }}_{n}^{2}}\right) \left( \frac{\Vert \nabla u_h^0\Vert ^2}{2}+\frac{\Vert \nabla u_h^{n+1}\Vert ^2}{2}\right) . \end{aligned}$$

(19)

Rearranging the terms, we obtain

$$\begin{aligned}&\Vert u_h^{n+1}\Vert ^2+\gamma {{\hat{\alpha }}_{1}}\sum _{n=0}^{n} c^1_k\Vert u_h^{n+1-k}\Vert ^2 \nonumber \\&\qquad + \, {{\hat{\alpha }}_{2}}\sum _{n=0}^{n} c^2_k\Vert \nabla u_h^{n+1-k}\Vert ^2 +\gamma {{\hat{\beta }}_{n}^{1}}\Vert u_h^{n+1}\Vert ^2+{{\hat{\beta }}_{n}^{2}}\Vert \nabla u_h^{n+1}\Vert ^2 \nonumber \\&\quad \le \Vert u_h^n\Vert ^2+\gamma {{\hat{\alpha }}_{1}}\sum \limits _{k=0}^{n-1}c^1_k\Vert u_h^{n-k}\Vert ^2+{{\hat{\alpha }}_{2}}\sum \limits _{k=0}^{n-1}c^2_k\Vert \nabla u_h^{n-k}\Vert ^2. \end{aligned}$$

(20)

Denoting $\omega ^n=\Vert u_h^n\Vert ^2+\gamma {{\hat{\alpha }}_{1}}\sum \limits _{k=0}^{n-1}c^1_k\Vert u_h^{n-k}\Vert ^2+{{\hat{\alpha }}_{2}}\sum \limits _{k=0}^{n-1}c^2_k\Vert \nabla u_h^{n-k}\Vert ^2$, the above formula becomes

$$\begin{aligned} \omega ^{n+1} \le \omega ^n \le \omega ^{n-1} \le \cdots \le \omega ^1. \end{aligned}$$

(21)

When $n=0$, taking $v_h=u_h^1$ in (15), we have

$$\begin{aligned}&\left( 1+\gamma {{\hat{\alpha }}_{1}} c^1_0\right) \Vert u_h^1\Vert ^2+{{\hat{\alpha }}_{2}} c^2_0\Vert \nabla u_h^1\Vert ^2 \nonumber \\&\quad = \left( u_h^0,u_h^1\right) + \gamma \left( {{\hat{\alpha }}_{1}} c^1_0-\hat{\beta ^1_0}\right) \left( u_h^0,u_h^1\right) + \left( {{\hat{\alpha }}_{2}} c^2_0-\hat{\beta ^2_0}\right) \left( \nabla u_h^0,\nabla u_h^1\right) \nonumber \\&\quad \le \frac{\Vert u_h^0\Vert ^2}{2}+\frac{\Vert u_h^1\Vert ^2}{2} +\gamma \frac{\left( {{\hat{\alpha }}_{1}} c^1_0-\hat{\beta ^1_0}\right) ^2}{{{\hat{\alpha }}_{1}} c^1_0}\frac{\Vert u_h^0\Vert ^2}{2}+\gamma {{\hat{\alpha }}_{1}} c^1_0\frac{\Vert u_h^1\Vert ^2}{2} \nonumber \\&\qquad +\frac{\left( {{\hat{\alpha }}_{2}} c^2_0-\hat{\beta ^2_0}\right) ^2}{{{\hat{\alpha }}_{2}} c^2_0}\frac{\Vert \nabla u_h^0\Vert ^2}{2}+{{\hat{\alpha }}_{2}} c^2_0\frac{\Vert \nabla u_h^1\Vert ^2}{2}, \end{aligned}$$

(22)

where

$$\begin{aligned} \frac{\left( {{\hat{\alpha }}_{1}} c^1_0-\hat{\beta ^1_0}\right) ^2}{{{\hat{\alpha }}_{1}} c^1_0}= & {} \Big (\frac{\tau ^{\alpha _{1}}}{\varGamma (1+\alpha _{1})2^{\alpha _{1}}} - \frac{\tau }{\varGamma (\alpha _{1}) (\frac{\tau }{2})^{1-\alpha _{1}}}\Big )^2\cdot \frac{\varGamma (1+\alpha _{1})2^{\alpha _{1}}}{\tau ^{\alpha _{1}}}\\= & {} \frac{\tau ^{2\alpha _{1}}(1-2\alpha _{1})^2}{(\varGamma (1+\alpha _{1})2^{\alpha _{1}})^2} \cdot \frac{\varGamma (1+\alpha _{1})2^{\alpha _{1}}}{\tau ^{\alpha _{1}}} \\= & {} \frac{(1-2\alpha _{1})^2}{\varGamma (1+\alpha _{1})}\left( \frac{\tau }{2}\right) ^{\alpha _{1}} \le \frac{\tau ^{\alpha _{1}}}{\varGamma (1+\alpha _{1})}, \end{aligned}$$

and

$$\begin{aligned} \frac{({{\hat{\alpha }}_{2}} c^2_0-\hat{\beta ^2_0})^2}{{{\hat{\alpha }}_{2}} c^2_0} \le \frac{\tau ^{\alpha _{2}}}{\varGamma (1+\alpha _{2})}. \end{aligned}$$

Thus we obtain from (22) that

$$\begin{aligned} \omega ^n\le & {} \omega ^1 =(1+\gamma {{\hat{\alpha }}_{1}} c^1_0)\Vert u_h^1\Vert ^2+{{\hat{\alpha }}_{2}} c^2_0\Vert \nabla u_h^1\Vert ^2 \nonumber \\\le & {} \Vert u_h^0\Vert ^2+\frac{\gamma \tau ^{\alpha _{1}}}{\varGamma (1+\alpha _{1})} \Vert u_h^0\Vert ^2+ \frac{\tau ^{\alpha _{2}}}{\varGamma (1+\alpha _{2})}\Vert \nabla u_h^0\Vert ^2. \end{aligned}$$

(23)

This complete the proof of (17). $\square$

Next, we define the Ritz projection operator $P^h: H_0^1(\varOmega )\rightarrow S^h$ as

$$\begin{aligned} (\nabla P^hu,\nabla v)=(\nabla u,\nabla v), \quad \forall v\in S^h. \end{aligned}$$

(24)

This projection operator $P^h$ satisfies the following estimation (see [36])

$$\begin{aligned} \Vert P^hv-v\Vert +h\Vert \nabla (P^hv-v)\Vert \le Ch^s\Vert v\Vert _s, \quad \forall v\in H^s\cap H_0^1, \,\, s\ge 1. \end{aligned}$$

(25)

Lemma 3

Suppose

$$\begin{aligned} R_s^{n}=\frac{u(t_{n+1})-u(t_n)}{\tau }-\frac{P^hu(t_{n+1}) - P^hu(t_{n})}{\tau }, n\ge 0, \end{aligned}$$

then it satisfies the following estimate

$$\begin{aligned} |R_s^{n}| \le C \left( \tau ^2+h^{r+1}\right) , \quad n \ge 0. \end{aligned}$$

(26)

Proof

For $n\ge 0$, by (5) and (25), we get

$$\begin{aligned} |R^{n}_{s}|= & {} \left| \frac{u(t_{n+1})-u(t_n)}{\tau }-\frac{P^hu(t_{n+1}) - P^hu(t_n)}{\tau }\right| \nonumber \\\le & {} \left| \frac{u(t_{n+1})-u(t_n)}{\tau }-\frac{\partial u\left( t_{n+\frac{1}{2}}\right) }{\partial t}\right| +\left| \frac{\partial u\left( t_{n+\frac{1}{2}}\right) }{\partial t} -P^h\frac{\partial u\left( t_{n+\frac{1}{2}}\right) }{\partial t}\right| \nonumber \\&+\left| P^h\frac{\partial u\left( t_{n+\frac{1}{2}}\right) }{\partial t}-\frac{P^hu(t_{n+1})-P^hu(t_n)}{\tau }\right| \nonumber \\\le & {} C(\tau ^2+h^{r+1}). \end{aligned}$$

(27)

$\square$

Lemma 4

Suppose

$$\begin{aligned} Q_s^{n}=\frac{{{\hat{\alpha }}_{1}}}{\tau }\sum \limits _{k=0}^{n}c_k \left( u^{n+1-k}-u^{n-k}\right) - \frac{{{\hat{\alpha }}_{1}}}{\tau }\sum \limits _{k=0}^{n}c_k \left( P^hu^{n+1-k}-P^hu^{n-k}\right) , \,\, n\ge 0, \end{aligned}$$

then it satisfies the following estimate

$$\begin{aligned} |Q_s^{n}| \le C \left( \tau ^{1+\alpha _{1}}+h^{r+1}\right) , \,\, n \ge 0. \end{aligned}$$

(28)

Proof

For $n\ge 0$, by (7) and (25), we obtain

$$\begin{aligned} |Q^{n}_{s}|\le & {} \left| \frac{{{\hat{\alpha }}_{1}}}{\tau }\sum \limits _{k=0}^{n}c_k \left( u^{n+1-k}-u^{n-k}\right) - {_0D_t}^{1-\alpha _{1}}u \left( t_{n+\frac{1}{2}}\right) \right| \nonumber \\&+\left| {_0D_t}^{1-\alpha _{1}}u \left( t_{n+\frac{1}{2}}\right) - P^h{_0D_t}^{1-\alpha _{1}} u \left( t_{n+\frac{1}{2}}\right) \right| \nonumber \\&+ \left| P^h{_0D_t}^{1-\alpha _{1}}u \left( t_{n+\frac{1}{2}}\right) - \frac{{{\hat{\alpha }}_{1}}}{\tau } \sum \limits _{k=0}^{n}c_k \left( P^hu^{n+1-k}-P^hu^{n-k}\right) \right| \nonumber \\\le & {} C \left( \tau ^{1+\alpha _{1}}+h^{r+1}\right) . \end{aligned}$$

(29)

$\square$

Theorem 2

Assume $u^{n}=u(x,y,t_{n})$ is the exact solution of (1), (2) and (3), $u_h^{n}$ is the FE approximate solution of the fully discrete scheme (15) and (16). Then $u_h^{n}$ satisfies the following error estimation

$$\begin{aligned} \max _{1\le n \le N} \Vert u^{n}-u_h^{n}\Vert \le C\sqrt{\frac{\varGamma (\alpha _{1})}{\gamma }} T^{1-\frac{\alpha _{1}}{2}} (\tau ^{\min \{1+\alpha _{1},1+\alpha _{2}\}}+h^{r+1}). \end{aligned}$$

(30)

Proof

Splitting $e^n=u(t_n)-u_h^n = (u(t_n) - P^hu(t_n)) + (P^hu(t_n)-u_h^n) = \rho ^n+\theta ^n$, we only need to estimate $\Vert \theta ^n\Vert$ by virtue of (25).

Noting that $u^n$ satisfies (10), it holds that

$$\begin{aligned}&(1+\gamma {{\hat{\alpha }}_{1}} c^1_0)(u^1,v)+{{\hat{\alpha }}_{2}} c^2_0(\nabla u^1,\nabla v) \nonumber \\&\quad = (u^0,v)+\gamma ({{\hat{\alpha }}_{1}} c^1_0-\hat{\beta ^1_0})(u^0,v)+({{\hat{\alpha }}_{2}} c^2_0-\hat{\beta ^2_0})(\nabla u^0,\nabla v)+\tau (R_t,v). \end{aligned}$$

(31)

Therefore, subtracting (16) from (31), then setting $v_h=\theta ^1$, we get

$$\begin{aligned}&(1+\gamma {{\hat{\alpha }}_{1}} c^1_0)\Vert \theta ^{1}\Vert ^2+{{\hat{\alpha }}_{2}} c^2_0\Vert \nabla \theta ^1\Vert ^2 \nonumber \\&\quad = (\theta ^0,\theta ^1)+\gamma ({{\hat{\alpha }}_{1}} c^1_0-\hat{\beta ^1_0})(\theta ^0,\theta ^1) +({{\hat{\alpha }}_{2}} c^2_0-\hat{\beta ^2_0})(\nabla \theta ^0,\nabla \theta ^1) \nonumber \\&\qquad +\tau (R_t+R^n_s+Q^n_s,\theta ^1), \nonumber \\&\quad \le \frac{\Vert \theta ^1\Vert ^2}{2}+\frac{\Vert \theta ^0\Vert ^2}{2}+\gamma ({{\hat{\alpha }}_{1}} c^1_0-\hat{\beta ^1_0}) \left( \frac{\Vert \theta ^0\Vert ^2}{2} + \frac{\Vert \theta ^1\Vert ^2}{2}\right) \nonumber \\&\qquad +\,({{\hat{\alpha }}_{2}} c^2_0-\hat{\beta ^2_0}) \left( \frac{\Vert \nabla \theta ^0\Vert ^2}{2} + \frac{\Vert \nabla \theta ^1\Vert ^2}{2}\right) \nonumber \\&\qquad +\frac{\tau ^2}{\gamma \hat{\beta ^1_0}} \frac{\Vert R_t+R^n_s+Q^n_s\Vert ^2}{2} + \gamma \hat{\beta ^1_0} \frac{\Vert \theta ^1\Vert ^2}{2}. \end{aligned}$$

(32)

Just take $u_h^0=P^hu_0$ for simplicity, then $\theta ^0=0$ and consequently,

$$\begin{aligned} (1+\gamma {{\hat{\alpha }}_{1}} c^1_0)\Vert \theta ^{1}\Vert ^2 +{{\hat{\alpha }}_{2}} c^2_0\Vert \nabla \theta ^1\Vert ^2\le & {} \frac{\tau ^2}{\gamma \hat{\beta ^1_0}}\Vert R_t+R^n_s+Q^n_s\Vert ^2 \nonumber \\\le & {} C\frac{\varGamma (\alpha _{1})}{\gamma 2^{1-\alpha _{1}}} \tau ^{2-\alpha _{1}} \left( \tau ^{\min \{1+\alpha _{1},1+\alpha _{2}\}} + h^{r+1}\right) ^2.\nonumber \\ \end{aligned}$$

(33)

When $n\ge 1$, subtracting (10) by (15) and taking $v_h=\theta ^{n+1}$, we get

$$\begin{aligned}&(1+\gamma {{\hat{\alpha }}_{1}} c^1_0)\Vert \theta ^{n+1}\Vert ^2+{{\hat{\alpha }}_{2}} c^2_0\Vert \nabla \theta ^{n+1}\Vert ^2 \nonumber \\&\quad = (\theta ^{n},\theta ^{n+1}) + \gamma {{\hat{\alpha }}_{1}}\sum \limits _{k=0}^{n-1}(c^1_k-c^1_{k+1}) (\theta ^{n-k},\theta ^{n+1}) \nonumber \\&\qquad +\, \gamma ({{\hat{\alpha }}_{1}} c^1_n-{{\hat{\beta }}_{n}^{1}}) (\theta ^0,\theta ^{n+1}) + {{\hat{\alpha }}_{2}}\sum \limits _{k=0}^{n-1}(c^2_k-c^2_{k+1})(\nabla \theta ^{n-k},\nabla \theta ^{n+1}) \nonumber \\&\qquad +\, ({{\hat{\alpha }}_{2}} c^2_n-{{\hat{\beta }}_{n}^{2}}) (\nabla \theta ^0,\nabla \theta ^{n+1}) + \tau (R_t+R^n_s+Q^n_s,\theta ^{n+1}). \end{aligned}$$

(34)

In the above deduction, (24) is used. Using Hölder inequality, Young inequality, (12) and (13), then we have

$$\begin{aligned}&(1+\gamma {{\hat{\alpha }}_{1}} c^1_0)\Vert \theta ^{n+1}\Vert ^2+{{\hat{\alpha }}_{2}} c^2_0\Vert \nabla \theta ^{n+1}\Vert ^2 \nonumber \\&\quad \le \Vert \theta ^n\Vert ^2+\gamma {{\hat{\alpha }}_{1}}\sum \limits _{k=0}^{n-1}(c^1_k-c^1_{k+1})\Vert \theta ^{n-k}\Vert ^2 +\gamma ({{\hat{\alpha }}_{1}} c^1_n-{{\hat{\beta }}_{n}^{1}})\Vert \theta ^0\Vert ^2 \nonumber \\&\qquad -\gamma {{\hat{\beta }}_{n}^{1}}\Vert \theta ^{n+1}\Vert ^2 -{{\hat{\beta }}_{n}^{2}}\Vert \nabla \theta ^{n+1}\Vert ^2 + {{\hat{\alpha }}_{2}}\sum \limits _{k=0}^{n-1}(c^2_k-c^2_{k+1})\Vert \nabla \theta ^{n-k}\Vert ^2 \nonumber \\&\qquad +\, ({{\hat{\alpha }}_{2}} c^2_n-{{\hat{\beta }}_{n}^{2}})\Vert \nabla \theta ^0\Vert ^2 + 2\tau |(R_t+R^n_s+Q^n_s,\theta ^{n+1})|. \end{aligned}$$

(35)

Since we have taken $u_h^0=P^hu_0$, $\theta ^0=0$ and $\nabla \theta ^0=0$ are used in the above derivation process. After transposition, it holds

$$\begin{aligned}&\Vert \theta ^{n+1}\Vert ^2+\gamma {{\hat{\alpha }}_{1}}\sum _{n=0}^{n+1} c^1_k\Vert \theta ^{n+1-k}\Vert ^2 +{{\hat{\alpha }}_{2}}\sum _{n=0}^{n+1} c^2_k\Vert \nabla \theta ^{n+1-k}\Vert ^2 \nonumber \\&\quad \le \Vert \theta ^n\Vert ^2+\gamma {{\hat{\alpha }}_{1}}\sum \limits _{k=0}^{n}c^1_k\Vert \theta ^{n-k}\Vert ^2+{{\hat{\alpha }}_{2}}\sum \limits _{k=0}^{n}c^2_k\Vert \nabla \theta ^{n-k}\Vert ^2 \nonumber \\&\qquad -\gamma {{\hat{\beta }}_{n}^{1}}\Vert \theta ^{n+1}\Vert ^2+2\tau |(R_t+R^n_s+Q^n_s, \theta ^{n+1})| \nonumber \\&\quad \le \Vert \theta ^n\Vert ^2+\gamma {{\hat{\alpha }}_{1}}\sum \limits _{k=0}^{n}c^1_k\Vert \theta ^{n-k}\Vert ^2+{{\hat{\alpha }}_{2}}\sum \limits _{k=0}^{n}c^2_k\Vert \nabla \theta ^{n-k}\Vert ^2 \nonumber \\&\qquad +\frac{1}{\gamma {{\hat{\beta }}_{n}^{1}}}\tau ^2\Vert R_t+R^n_s+Q^n_s\Vert ^2. \end{aligned}$$

(36)

Denotes $\eta ^n=\Vert \theta ^n\Vert ^2+\gamma {{\hat{\alpha }}_{1}}\sum \limits _{k=0}^{n}c_k\Vert \theta ^{n-k}\Vert ^2+{{\hat{\alpha }}_{2}}\sum \limits _{k=0}^{n}d_k\Vert \nabla \theta ^{n-k}\Vert ^2$, and the above formula becomes

$$\begin{aligned} \eta ^{n+1}\le & {} \eta ^n+\frac{C}{\gamma {{\hat{\beta }}_{n}^{1}}}\tau ^2 \left( \tau ^{\min \{1+\alpha _{1},1+\alpha _{2}\}}+h^{r+1}\right) ^2 \le \cdots \nonumber \\\le & {} \eta ^1+\frac{C}{\gamma } \left( \frac{1}{{{\hat{\beta }}_{n}^{1}}} + \frac{1}{\hat{\beta ^1_{n-1}}} +\cdots +\frac{1}{\hat{\beta ^1_1}} \right) \tau ^2 \left( \tau ^{\min \{1+\alpha _{1},1+\alpha _{2}\}} +h^{r+1}\right) ^2 \nonumber \\\le & {} \eta ^1 + \frac{C\varGamma (\alpha _{1})}{\gamma } \left\{ \left( n+\frac{1}{2}\right) ^{1-\alpha _{1}} + \left( n-\frac{1}{2}\right) ^{1-\alpha _{1}} +\cdots \right. \nonumber \\&\qquad \left. + \left( \frac{3}{2}\right) ^{1-\alpha _{1}}\right\} \tau ^{2-\alpha _{1}} \left( \tau ^{\min \{1+\alpha _{1},1+\alpha _{2}\}} + h^{r+1}\right) ^2 \nonumber \\\le & {} \eta ^1 + \frac{C\varGamma (\alpha _{1})}{\gamma }n \left( n+\frac{1}{2}\right) ^{1-\alpha _{1}}\tau ^{2-\alpha _{1}} \left( \tau ^{\min \{1+\alpha _{1}, 1+\alpha _{2}\}} + h^{r+1}\right) ^2 \nonumber \\\le & {} \eta ^1 + \frac{C\varGamma (\alpha _{1})}{\gamma }T^{2-\alpha _{1}} \left( \tau ^{\min \{1+\alpha _{1},1+\alpha _{2}\}}+h^{r+1}\right) ^2. \end{aligned}$$

(37)

Substitute (33) to the above inequality, we obtain

$$\begin{aligned} \eta ^{n+1}\le & {} C\frac{\varGamma (\alpha _{1})}{\gamma 2^{1-\alpha _{1}}}\tau ^{2-\alpha _{1}} \left( \tau ^{\min \{1+\alpha _{1},1+\alpha _{2}\}}+h^{r+1}\right) ^2 \nonumber \\&+\frac{C\varGamma (\alpha _{1})}{\gamma }T^{2-\alpha _{1}} \left( \tau ^{\min \{1+\alpha _{1}, 1+\alpha _{2}\}}+h^{r+1}\right) ^2 \nonumber \\\le & {} C\frac{\varGamma (\alpha _{1})T^{2-\alpha _{1}}}{\gamma } \left( \tau ^{\min \{1+\alpha _{1},1+\alpha _{2}\}}+h^{r+1}\right) ^2, \end{aligned}$$

(38)

so

$$\begin{aligned} \Vert e^{n+1}\Vert= & {} \Vert \theta ^{n+1}+\rho ^{n+1}\Vert \le \Vert \theta ^{n+1}\Vert +\Vert \rho ^n\Vert \nonumber \\\le & {} C\sqrt{\frac{\varGamma (\alpha _{1})}{\gamma }} T^{1-\frac{\alpha _{1}}{2}} \left( \tau ^{\min \{1+\alpha _{1},1+\alpha _{2}\}}+h^{r+1}\right) . \end{aligned}$$

(39)

Thus (30) is proved. $\square$

4 Numerical examples

In this section, we show two numerical examples to verify and illustrate the theoretical results of finite element approximation for the fractional Cable equation. Now we compute some numerical data based on one-dimensional case and two-dimensional case, respectively.

4.1 One-dimensional example

Based on the fully discrete scheme (15) and (16), we first check the convergence rate by the example of one-dimensional space. Therefore, we consider the Cable equation with a forcing term f:

$$\begin{aligned} u_t=-\gamma {_0D_t^{1-\alpha _{1}}}u+{_0D_t^{1-\alpha _{2}}}u_{xx}+f(x,t), \quad x\in [0,1], \, t\in [0,T]; \end{aligned}$$

(40)

Considering the following exact solution

$$\begin{aligned} u(x,t)=t^2\sin (2\pi x), \end{aligned}$$

the associated forcing term is

$$\begin{aligned} f(x,t) = 2 \left( t+\frac{\gamma }{\varGamma (2+\alpha _{1})}t^{1+\alpha _{1}} + \frac{4\pi ^2}{\varGamma (2+\alpha _{2})}t^{1+\alpha _{2}}\right) \sin (2\pi x). \end{aligned}$$

Let $S^h$ be the space of continuous piecewise linear functions on $\mathfrak {I}$. We first make the step of space $h=1/1000$ to be small enough to test the convergence order in time. The value of $\gamma$ is taken as $\gamma =1$ and $\gamma =1000$, respectively. From the results listed in Tables 1 and 2, we can see that the value of $\gamma$ affects the convergence order in time, which also verifies that the error results is related to the constant $\gamma$. Taking $\gamma =1000, \tau =1/1000$ for the purpose of testing the convergence order in space, we find from Table 3 that the convergence rates, which bring into correspondence with the theoretical convergence results (TCRs), almost reach to the second order for different values of $\alpha _{1}$ and $\alpha _{2}$.

Table 1 The convergence results in time when $\gamma =1, \,h=1/1000$ at $T=1$

Table 2 The convergence results in time when $\gamma =1000, \,h=1/1000$ at $T=1$

Table 3 The convergence results in space when $\gamma =1000, \,\tau =1/1000$ at $T=1$

4.2 Two-dimensional example

For validating the correctness of convergence results for the case in two-dimensional space, we need to consider the following two-dimensional problem with an exact solution

$$\begin{aligned} \left\{ \begin{array}{l l} \frac{\partial u}{\partial t}=-{_0D_t^{1-\alpha _{1}}}u+{_0D_t^{1-\alpha _{2}}}\varDelta u+f(x,y,t), &{} (x,y)\in \varOmega , \, t\in [0,T]; \\ u(x,y,t)=0, &{} (x,y)\in \partial \varOmega , \, t\in [0,T]; \\ u(x,y,0)=0, &{} (x,y)\in \varOmega , \end{array} \right. \end{aligned}$$

(41)

where the source term is chosen as

$$\begin{aligned} f(x,y,t) = 2 \left( t+\frac{1}{\varGamma (2+\alpha _{1})}t^{1+\alpha _{1}} + \frac{2\pi ^2}{\varGamma (2+\alpha _{2})}t^{1+\alpha _{2}}\right) \sin (\pi x)\sin (\pi y), \end{aligned}$$

which results in an exact solution of (41)

$$\begin{aligned} u(x,y,t)=t^2\sin (\pi x)\sin (\pi y). \end{aligned}$$

Now we take the isosceles right triangle mesh for the spatial domain $\varOmega =[0,1]\times [0,1]$ and choose the continuous piecewise linear space to get the convergence results. In Table 4, we give some numerical results of errors and orders of convergence with different space-time mesh $\tau =h=1/8,1/16/1/32,1/64$ and changed values of parameters $\alpha _1,\alpha _2$. From the results in Table 4, we easily find that on every case the rate of convergence, which are inclined to the spatial convergence order, is higher than the theoretical convergence results (TCRs) $\min \{1+\alpha _{1},1+\alpha _{2}\}$. This is due to the dominant position of space in our numerical example of two-dimensional space.

Table 4 The convergence results at $T=1$

In view of the discussion on the numerical results for two examples in one and two dimensional spaces, we claim that the theoretical results derived in this paper are correct.