A Mixed Finite Volume Element Method for Nonlinear Time Fractional Fourth-Order Reaction–Diffusion Models

Abstract

In this paper, a linearized mixed finite volume element (MFVE) scheme is proposed to solve the nonlinear time fractional fourth-order reaction–diffusion models with the Riemann–Liouville time fractional derivative. By introducing an auxiliary variable $\sigma =\Delta u$, the original fourth-order model is reformulated into a lower-order coupled system. The first-order time derivative and the time fractional derivative are discretized by using the BDF2 formula and the weighted and shifted Grünwald difference (WSGD) formula, respectively. Then, a fully discrete MFVE scheme is constructed by using the primal and dual grids. The existence and uniqueness of a solution for the MFVE scheme are proven based on the matrix theories. The scheme’s unconditional stability is rigorously derived by using the Gronwall inequality in detail. Moreover, the optimal error estimates for u in the discrete $L^{\infty }\left(L^2(\Omega )\right)$ and $L^2\left(H^1(\Omega )\right)$ norms and for $\sigma$ in the discrete $L^2\left(L^2(\Omega )\right)$ norm are obtained. Finally, three numerical examples are given to confirm its feasibility and effectiveness.

1. Introduction

Reaction–diffusion models can be used to describe many practical phenomena in scientific and engineering fields [1,2,3]. For example, in population biology applications, the reaction term captures growth dynamics, and the diffusion term characterizes migration processes [4]. Specifically, fourth-order reaction–diffusion models have an important role in some specialized fields, such as wave propagation in beams, brain warping, the generation of grooves on a flat surface, strain gradient elasticity, fluids in lungs, pattern formation in bistable systems, and so on. One can refer to references [5,6,7,8,9,10] for the corresponding applications. However, some anomalous diffusion phenomena no longer follow Gauss’s statistical law or Fick’s law [11,12], and cannot be adequately simulated by classical diffusion or reaction–diffusion models, which require the adoption of fractional models. Based on the above background and practical applications, in the current work, our goal is to numerically solve the following nonlinear time fractional fourth-order reaction–diffusion (FORD) models:

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}\frac{\partial u(\mathit{x},t)}{\partial t}-\frac{{\partial }^{\alpha }\Delta u(\mathit{x},t)}{\partial t^{\alpha }}-\Delta u(\mathit{x},t)+{\Delta }^{2}u(\mathit{x},t)+g\left(u(\mathit{x},t)\right)=f(\mathit{x},t),\hfill & (\mathit{x},t)\in \Omega \times J,\hfill \\ u(\mathit{x},t)=\Delta u(\mathit{x},t)=0,\hfill & (\mathit{x},t)\in \partial \Omega \times J,\hfill \\ u(\mathit{x},0)=u_0\left(\mathit{x}\right),\hfill & \mathit{x}\in \Omega ,\hfill \end{array}\right.\end{array}$$

(1)

where $\Delta$ is the Laplacian operator, $J=(0,T]$ is the time interval with $0<T<\infty$, and $\Omega \subset {\mathbb{R}}^2$ is a bounded convex polygonal domain with a boundary $\partial \Omega$. The source function $f(\mathit{x},t)$ and initial data $u_0\left(\mathit{x}\right)$ are known functions. In (1), the Riemann–Liouville time fractional derivative term is presented as follows:

$$\begin{array}{c}\hfill \frac{{\partial }^{\alpha }\Delta u(\mathit{x},t)}{\partial t^{\alpha }}=\frac{1}{\Gamma (1-\alpha )}\frac{\partial }{\partial t}{\int }_0^t\frac{\Delta u(\mathit{x},s)}{{(t-s)}^{\alpha }}\mathrm{d}s,\end{array}$$

(2)

where $0<\alpha <1$. Moreover, we assume that the nonlinear term $g\left(u\right)$ satisfies the Lipschitz condition, that is, $|g\left(u_1\right)-g\left(u_2\right)|\le L|u_1-u_2|$, where L is a positive constant.

In recent years, numerous numerical methods have been employed to solve fractional-order fourth-order PDEs. Vong and Wang [13] established a high-order compact finite difference (FD) scheme to solve the fourth-order fractional subdiffusion model. Liu et al. [14] designed an MFE scheme to solve a time fractional FORD equation. Zhang and Pu [15] constructed a second-order compact FD scheme to solve the fourth-order fractional subdiffusion equation by using the $L2-1_{\sigma }$ formula. Liu et al. [16] proposed an FE method to solve a time fractional fourth-order diffusion model with a nonlinear term. Nikan et al. [17] proposed a local radial basis function method to solve the time fractional FORD equation. Wang et al. [18] constructed a fully discrete MFE scheme based on the modified $L1$ Crank–Nicolson method to solve a nonlinear fourth-order fractional diffusion-wave equation. Guo et al. [19] proposed a double exponential Sinc-Galerkin method to solve the nonlinear fourth-order time fractional equation. Zhang and Feng [20] proposed a mixed virtual element algorithm to solve the time fractional fourth-order subdiffusion model. Haghi et al. [21] designed a fourth-order compact FD scheme to solve the nonlinear time fractional FORD equation. Zhang et al. [22] constructed a block-centered FD scheme to solve time fractional fourth-order parabolic equations. An and Huang [23] constructed a nonuniform Alikhanov scheme to solve the time fractional fourth-order diffusion-wave model. Here, we will develop a time second-order MFVE method to solve the nonlinear time fractional FORD models with the aid of the WSGD formula.

The WSGD formula was proposed by Tian et al. [24] to approximate the Riemann–Liouville fractional derivative. In contrast to the $L1$ formula [25,26], the WSGD formula can achieve the second-order convergence accuracy that does not depend on fractional parameters. Since then, this approximation formula has been rapidly developed. Wang and Vong [27] constructed compact FD schemes based on the WSGD formula to solve two fractional models, including the subdiffusion model and the diffusion-wave model. Wang et al. [28] proposed a $H^1$-Galerkin MFE method to solve the nonlinear time fractional convection–diffusion equation. Liu et al. [29] proposed some second-order $\theta$ schemes to solve the nonlinear fractional cable equation by combining it with the WSGD formula. Fang et al. [30] constructed a fast TT-M FVE algorithm to solve the nonlinear time fractional coupled diffusion equation. Zhang et al. [31] proposed a compact FD scheme with the WSGD formula to solve the time fractional integro-differential model. Fang et al. [32] designed fast two-grid FVE algorithms to solve the nonlinear time fractional mobile/immobile transport model by combining the WSGD formula with Crank–Nicolson scheme. Ali et al. [33] considered augmented FV methods with the WSGD formula to solve nonlinear time fractional degenerate parabolic equation. Wang et al. [34] constructed a two-grid MFE algorithm based on the WSGD formula and BDF2-$\theta$ scheme to solve the nonlinear fractional pseudo-hyperbolic wave model.

In this paper, our main purpose is to present a time second-order MFVE method for solving the nonlinear time fractional FORD models with the Riemann–Liouville time fractional derivative by combining the WSGD formula and the FVE method [35,36,37,38,39]. We introduce an auxiliary variable $\sigma (\mathit{x},t)$ to express (1) as the lower-order couple system, employ the WSGD formula to approximate the Riemann–Liouville time fractional derivative, design primal and dual grids for a space region $\Omega$, and establish the MFVE scheme through an interpolation operator $I_h^{\ast }$. In our theoretical analysis, we prove the existence and uniqueness of the solution for the MFVE scheme by means of the matrix theories, give the unconditional stability results, and obtain the optimal error estimates for u in the discrete $L^{\infty }\left(L^2(\Omega )\right)$ and $L^2\left(H^1(\Omega )\right)$ norms and for $\sigma$ in the discrete $L^2\left(L^2(\Omega )\right)$ norm. Finally, we give three numerical examples to validate the effectiveness and convergence accuracy of the proposed MFVE method.

The rest of this paper is organized as follows. In Section 2, we construct a second-order fully discrete MFVE scheme for the nonlinear time fractional FORD equations. In Section 3, we give the proof of the existence and uniqueness of discrete solution based on the matrix theories in detail. The unconditional stability and optimal error estimates are derived in Section 4 and Section 5, respectively. Finally, some numerical results are presented to demonstrate the feasibility and effectiveness in Section 6. Moreover, the notation C is denoted as a generic positive constant, which is independent of the grid parameters and has different values at different occurrences.

2. Linearized MFVE Scheme

We first define a new auxiliary variable $\sigma (\mathit{x},t)=\Delta u(\mathit{x},t)$ and subsequently reformulate (1) as follows:

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}\frac{\partial u(\mathit{x},t)}{\partial t}-\frac{{\partial }^{\alpha }\Delta u(\mathit{x},t)}{\partial t^{\alpha }}-\Delta u(\mathit{x},t)+\Delta \sigma (\mathit{x},t)+g\left(u(\mathit{x},t)\right)=f(\mathit{x},t),\hfill & (\mathit{x},t)\in \Omega \times J,\hfill \\ \sigma (\mathit{x},t)=\Delta u(\mathit{x},t),\hfill & (\mathit{x},t)\in \Omega \times J,\hfill \\ u(\mathit{x},t)=\sigma (\mathit{x},t)=0,\hfill & (\mathit{x},t)\in \partial \Omega \times J,\hfill \\ u(\mathit{x},0)=u_0\left(\mathit{x}\right),\hfill & \mathit{x}\in \Omega .\hfill \end{array}\right.\end{array}$$

(3)

Next, we introduce an equidistant partition of $\overline{J}=[0,T]$ given by $0=t_0<t_1<\cdots <t_N=T$; here, $t_n=n\tau$ and $\tau =T/N$ with a positive integer N, $n=0,1,\cdots ,N$. Given a function $\phi$ defined on $[0,T]$, we set ${\phi }^n=\phi \left(t_n\right)$ and

$$\begin{array}{c}\hfill {\partial }_t^2{\varphi }^{n+1}=\left\{\begin{array}{cc}\frac{{\varphi }^1-{\varphi }^0}{\tau },\hfill & \mathrm{if} n=0,\hfill \\ \frac{3{\varphi }^{n+1}-4{\varphi }^n+{\varphi }^{n-1}}{2\tau },\hfill & \mathrm{if} n\ge 1.\hfill \end{array}\right.\end{array}$$

(4)

Then, we apply ${\partial }_t^2{u}^{n+1}$ to approximate $\frac{\partial u(\mathit{x},t_{n+1})}{\partial t}$, that is

$$\begin{array}{c}\hfill \frac{\partial u(\mathit{x},t_{n+1})}{\partial t}={\partial }_t^2{u}^{n+1}+E_{u,t}^{n+1},\end{array}$$

(5)

where the truncation error $E_{u,t}^{n+1}$ satisfies the following estimate:

$$\begin{array}{c}\hfill E_{u,t}^{n+1}=\frac{\partial u(\mathit{x},t_{n+1})}{\partial t}-{\partial }_t^2{u}^{n+1}=\left\{\begin{array}{cc}O\left(\tau \right),\hfill & \mathrm{if} n=0,\hfill \\ O\left({\tau }^2\right),\hfill & \mathrm{if} n\ge 1.\hfill \end{array}\right.\end{array}$$

(6)

For approximating the fractional derivative $\frac{{\partial }^{\alpha }\Delta u(\mathit{x},t)}{\partial t^{\alpha }}$ at time $t=t_{n+1}$, we select the WSGD formula [24] as follows:

$$\begin{array}{c}\hfill \frac{{\partial }^{\alpha }\Delta u(\mathit{x},t_{n+1})}{\partial t^{\alpha }}={\tau }^{-\alpha }{\displaystyle \sum _{k=0}^{n+1}}{q}_{\alpha }\left(k\right)\Delta u^{n-k+1}+E_{u,\alpha }^{n+1},\end{array}$$

(7)

with the truncation error $E_{u,\alpha }^{n+1}=O\left({\tau }^2\right)$, and

$$\begin{array}{c}\hfill q_{\alpha }\left(k\right)=\left\{\begin{array}{cc}\frac{\alpha +2}{2}{g}_0^{\alpha },\hfill & \mathrm{if} k=0,\hfill \\ \frac{\alpha +2}{2}{g}_k^{\alpha }+\frac{-\alpha }{2}{g}_{k-1}^{\alpha },\hfill & \mathrm{if} k>0,\hfill \end{array}\right.\end{array}$$

(8)

$$\begin{array}{c}\hfill g_0^{\alpha }=1, g_k^{\alpha }=\frac{\Gamma (k-\alpha )}{\Gamma (-\alpha )\Gamma (k+1)}, g_k^{\alpha }=\left(1-\frac{\alpha +1}{k}\right)g_{k-1}^{\alpha }, k\ge 1.\end{array}$$

(9)

From [28], we use the linearized formulation denoted by $G\left[u^{n+1}\right]$ to approximate the nonlinear item $g\left(u\right)$ at $t=t_{n+1}$, that is,

$$\begin{array}{c}\hfill G\left[u^{n+1}\right]=\left\{\begin{array}{cc}g\left(u^0\right),\hfill & \mathrm{if} n=0,\hfill \\ 2g\left(u^n\right)-g\left(u^{n-1}\right),\hfill & \mathrm{if} n\ge 1,\hfill \end{array}\right.\end{array}$$

(10)

where the truncation error denoted by $E_g^{n+1}$ satisfies

$$\begin{array}{c}\hfill E_g^{n+1}=g\left(u^{n+1}\right)-G\left[u^{n+1}\right]=\left\{\begin{array}{cc}O\left(\tau \right),\hfill & \mathrm{if} n=0,\hfill \\ O\left({\tau }^2\right),\hfill & \mathrm{if} n\ge 1.\hfill \end{array}\right.\end{array}$$

(11)

Thus, we can rewrite the system (3) at $t=t_{n+1}$ as follows:

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}\left(a\right)\hfill & {\partial }_t^2{u}^{n+1}-{\tau }^{-\alpha }{\displaystyle {\sum }_{k=0}^{n+1}}{q}_{\alpha }\left(k\right)\Delta u^{n-k+1}-\Delta u^{n+1}+\Delta {\sigma }^{n+1}+G\left[u^{n+1}\right]\hfill \\ & =f^{n+1}-E_{u,t}^{n+1}-E_{u,\alpha }^{n+1}-E_g^{n+1},\hfill \\ \left(a\right)\hfill & {\sigma }^{n+1}=\Delta u^{n+1}.\hfill \end{array}\right.\end{array}$$

(12)

Let ${\mathcal{T}}_h=\left\{K\right\}$ be a quasi-uniform triangular grid with the maximum diameter $h=\max \left\{h_K\right\}$ for the domain $\Omega$, and denote $Z_h$ as the set of all vertices in ${\mathcal{T}}_h$. We also need to construct the corresponding dual grid ${\mathcal{T}}_h^{\ast }$. Observing Figure 1 (see [30]), for a node $z_0\in Z_h^0$ with its immediate nodes $z_i$$(i=1,2,\dots ,k)$, denote $M_i$ as the midpoint of the edge $\overline{z_0{z}_i}$ and adopt a special point $Q_i$ (such as barycenter in Figure 1 (left) or circumcenter in Figure 1 (right)) in a triangle $▵z_0{z}_i{z}_{i+1}$ with $z_{k+1}=z_1$. A control volume $K_{z_0}^{\ast }$ can then be constructed by connecting $M_1,Q_1,\dots ,M_k,Q_k,M_1$. In addition, if $z_0$ is on the boundary, we can similarly construct the corresponding control volume.

Base on the above partitions, we select the following linear element space $V_h$ as the trial function space:

$$\begin{array}{c}\hfill V_h=\{v\in H_0^1(\Omega ):v{|}_K\in P_1\left(K\right), \forall K\in {\mathcal{T}}_h\},\end{array}$$

(13)

and select $V_h^{\ast }$ as the test function space, that is,

$$\begin{array}{c}\hfill V_h^{\ast }=\{v\in L^2{(\Omega ):v|}_{K_z^{\ast }}\in P_0\left(K_z^{\ast }\right), \forall K_z^{\ast }\in {\mathcal{T}}_h^{\ast },\mathrm{and} v{|}_{\partial \Omega }=0\}.\end{array}$$

(14)

Let ${\phi }_z$ be the standard nodal linear basis function with respect to the node z, and ${\psi }_z^{\ast }$ be the characteristic function of the control volume $K_z^{\ast }$; then, $V_h=\mathrm{span}\{{\phi }_z\left(\mathit{x}\right):z\in Z_h^0\}$ and $V_h^{\ast }=\mathrm{span}\{{\psi }_z^{\ast }\left(\mathit{x}\right):z\in Z_h^0\}$. Consider two interpolation operators $I_h:C(\overline{\Omega })\to V_h$ and $I_h^{\ast }:C(\overline{\Omega })\to V_h^{\ast }$, defined as follows:

$$\begin{array}{cc}\hfill & I_{h}v\left(\mathit{x}\right)=\sum _{z\in Z_h^0}v\left(z\right){\phi }_z\left(\mathit{x}\right),\hfill \end{array}$$

(15)

$$\begin{array}{cc}\hfill & I_h^{\ast }v\left(\mathit{x}\right)=\sum _{z\in Z_h^0}v\left(z\right){\psi }_z^{\ast }\left(\mathit{x}\right).\hfill \end{array}$$

(16)

From [35], we know that

$$\begin{array}{c}\hfill {\parallel v-I_{h}v\parallel }_k\le C{h}^{2-k}{\parallel v\parallel }_2, k=0,1, \forall v\in H^2(\Omega ),\end{array}$$

(17)

$$\begin{array}{c}\hfill \parallel v-I_h^{\ast }v\parallel \le Ch{\parallel v\parallel }_1, \forall v\in H^1(\Omega ).\end{array}$$

(18)

Now, integrating (3) over every control volume $K_z^{\ast }$, and applying the Green formula and operator $I_h^{\ast }$, we have

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}\left(a\right)\hfill & ({\partial }_t^2{u}^{n+1},I_h^{\ast }{w}_h)+{\tau }^{-\alpha }{\sum }_{k=0}^{n+1}{q}_{\alpha }\left(k\right)a(u^{n-k+1},I_h^{\ast }{w}_h)\hfill \\ & +a(u^{n+1},I_h^{\ast }{w}_h)-a({\sigma }^{n+1},I_h^{\ast }{w}_h)+(G\left[u^{n+1}\right],I_h^{\ast }{w}_h)\hfill \\ & =(f^{n+1},I_h^{\ast }{w}_h)-(E_{u,t}^{n+1},I_h^{\ast }{w}_h)-(E_{u,\alpha }^{n+1},I_h^{\ast }{w}_h)-(E_g^{n+1},I_h^{\ast }{w}_h),\forall w_h\in V_h,\hfill \\ \left(b\right)\hfill & ({\sigma }^{n+1},I_h^{\ast }{v}_h)+a(u^{n+1},I_h^{\ast }{v}_h)=0,\forall v_h\in V_h,\hfill \end{array}\right.\end{array}$$

(19)

where

$$\begin{array}{c}\hfill a(\overline{u},\overline{v})=\left\{\begin{array}{cc}-\sum _{z\in Z_h}\overline{v}\left(z\right){\int }_{\partial K_z^{\ast }}\nabla \overline{u}·\mathit{n}\mathrm{d}s,\hfill & \forall \overline{u}\in V_h,\overline{v}\in V_h^{\ast },\hfill \\ {\int }_{\Omega }\nabla \overline{u}·\nabla \overline{v}\mathrm{d}\mathit{x},\hfill & \forall \overline{u},\overline{v}\in H_0^1(\Omega ),\hfill \end{array}\right.\end{array}$$

(20)

and $\mathit{n}$ stands for the outer-normal direction on $\partial K_z^{\ast }$.

Let $u_h^n$ and ${\sigma }_h^n$ be the discrete solutions of u and $\sigma$ at $t=t_n$, respectively. Thus, we can obtain a linearized MFVE scheme: find $(u_h^{n+1},{\sigma }_h^{n+1})\in V_h\times V_h,(n=0,1,\dots ,N-1)$, such that

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}\left(a\right)\hfill & ({\partial }_t^2{u}_h^{n+1},I_h^{\ast }{w}_h)+{\tau }^{-\alpha }{\sum }_{k=0}^{n+1}{q}_{\alpha }\left(k\right)a(u_h^{n-k+1},I_h^{\ast }{w}_h)+a(u_h^{n+1},I_h^{\ast }{w}_h)\hfill \\ & -a({\sigma }_h^{n+1},I_h^{\ast }{w}_h)=-(G\left[u_h^{n+1}\right],I_h^{\ast }{w}_h)+(f^{n+1},I_h^{\ast }{w}_h), \forall w_h\in V_h,\hfill \\ \left(b\right)\hfill & ({\sigma }_h^{n+1},I_h^{\ast }{v}_h)+a(u_h^{n+1},I_h^{\ast }{v}_h)=0, \forall v_h\in V_h,\hfill \end{array}\right.\end{array}$$

(21)

where $(u_h^0,{\sigma }_h^0)\in V_h\times V_h$ satisfies

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}\left(a\right)a({\sigma }_h^0,I_h^{\ast }{w}_h)=a({\sigma }_0,I_h^{\ast }{w}_h),\hfill & \forall w_h\in V_h,\hfill \\ \left(b\right)a(u_h^0,I_h^{\ast }{v}_h)+({\sigma }_h^0,I_h^{\ast }{v}_h)=0,\hfill & \forall v_h\in V_h,\hfill \end{array}\right.\end{array}$$

(22)

and ${\sigma }_0(\mathit{x},t)=\Delta u_0(\mathit{x},t)$.

Remark 1.

In the above MFVE scheme, we specifically select $(u_h^0,{\sigma }_h^0)$, which satisfies (22), that is, $u_h^0=R_h{u}_0$ and ${\sigma }_h^0=R_h{\sigma }_0$, where $R_h$ is the MFVE projection defined in (54). This choice is extremely necessary in error estimates in Section 5.

3. Existence and Uniqueness

We first give some lemmas to prove the uniqueness of existence and subsequent theoretical analysis.

Lemma 1 

([35]). For the bilinear form $(·,I_h^{\ast }·)$, it can be obtained that

$$\begin{array}{c}\hfill (w_h,I_h^{\ast }{z}_h)=(w_h,I_h^{\ast }{z}_h), \forall w_h,z_h\in V_h.\end{array}$$

(23)

Moreover, there exist two constants ${\mu }_1>0$ and ${\mu }_2>0$ independent of h such that

$$\begin{array}{cc}& (z_h,I_h^{\ast }{z}_h)\ge {\mu }_1{\parallel z_h\parallel }^2, \forall z_h\in V_h,\hfill \end{array}$$

(24)

$$\begin{array}{cc}& (z_h,I_h^{\ast }{w}_h)\le {\mu }_2\parallel z_h\parallel \parallel w_h\parallel , \forall z_h,w_h\in V_h.\hfill \end{array}$$

(25)

Lemma 2 

([35,36]). For the bilinear form $a(·,I_h^{\ast }·)$, the following property holds:

$$\begin{array}{c}\hfill a(w_h,I_h^{\ast }{z}_h)=a(z_h,I_h^{\ast }{w}_h), \forall w_h,z_h\in V_h.\end{array}$$

(26)

Moreover, there exist three positive constants $h_0,{\mu }_3$ and ${\mu }_4$ such that, for $0<h\le h_0$,

$$\begin{array}{cc}& a(z_h,I_h^{\ast }{z}_h)\ge {\mu }_3{\parallel z_h\parallel }_1^2, \forall z_h\in V_h,\hfill \end{array}$$

(27)

$$\begin{array}{cc}& a(z_h,I_h^{\ast }{w}_h)\le {\mu }_4{\parallel z_h\parallel }_1{\parallel w_h\parallel }_1, \forall z_h,w_h\in V_h.\hfill \end{array}$$

(28)

Lemma 3 

([24,27,30]). For a sequence ${\left\{q_{\alpha }\left(k\right)\right\}}_{k=1}^{\infty }$ defined by (11), and an arbitrary positive integer P, if a real vector $\forall ({\varphi }^0,{\varphi }^1,\dots ,{\varphi }^P)\in {\mathbb{R}}^{P+1}$, then

$$\begin{array}{c}\hfill {\displaystyle \sum _{n=0}^P}{\displaystyle \sum _{k=0}^n}{q}_{\alpha }\left(k\right)({\varphi }^{n-k},I_h^{\ast }{\varphi }^n)\ge 0.\end{array}$$

(29)

Lemma 4 

([30]). For a function sequence ${\left\{{\psi }^n\right\}}_{n=0}^{\infty }$ in $L^2(\Omega )$, it can be obtained that

$$\begin{array}{c}\hfill ({\partial }_t^2{\psi }^{n+1},I_h^{\ast }{\psi }^{n+1})\ge \left\{\begin{array}{cc}\frac{1}{2\tau }[({\psi }^1,I_h^{\ast }{\psi }^1)-({\psi }^0,I_h^{\ast }{\psi }^0)], \hfill & \mathit{if} n=0,\hfill \\ \\ \frac{1}{4\tau }(\Lambda [{\psi }^{n+1},{\psi }^n]-\Lambda [{\psi }^n,{\psi }^{n-1}]), \hfill & \mathit{if} n\ge 1,\hfill \end{array}\right.\end{array}$$

(30)

where

$$\begin{array}{c}\hfill \Lambda [{\psi }^n,{\psi }^{n-1}]\triangleq ({\psi }^n,I_h^{\ast }{\psi }^n)+(2{\psi }^n-{\psi }^{n-1},I_h^{\ast }(2{\psi }^n-{\psi }^{n-1})).\end{array}$$

(31)

Theorem 1.

The MFVE scheme (21) has a unique solution.

Proof. 

Let $\{{\phi }_i:i=1,2\cdots ,M_Z^0\}$ be the basis $V_h$, where $M_Z^0$ is the number of all interior vertices. Then, $\{u_h^n,{\sigma }_h^n\}$∈$V_h\times V_h$ can be expanded as follows:

$$\begin{array}{c}\hfill u_h^n\left(\mathit{x}\right)={\displaystyle \sum _{i=1}^{M_Z^0}}{\tilde{u}}_i^n{\phi }_i\left(\mathit{x}\right), {\sigma }_h^n\left(\mathit{x}\right)={\displaystyle \sum _{i=1}^{M_Z^0}}{\tilde{\sigma }}_i^n{\phi }_i\left(\mathit{x}\right).\end{array}$$

(32)

Substituting (32) into (21) and selecting $w_h={\phi }_j,v_h={\phi }_j$$(j=1,2,\cdots ,M_Z^0)$, we can obtain the following matrix form:

$$\left[\begin{array}{cc}{A}_1+{\tau }^{1-\alpha }{q}_{\alpha }\left(0\right)A_2+\tau A_2& -\tau A_2\\ A_2& A_1\end{array}\right]\left[\begin{array}{c}{\tilde{\mathit{u}}}^1\\ {\tilde{\sigma }}^1\end{array}\right]=\left[\begin{array}{c}C\\ \mathbf{0}\end{array}\right],$$

(33)

and

$$\left[\begin{array}{cc}\frac{3}{2}{A}_1+{\tau }^{1-\alpha }{q}_{\alpha }\left(0\right)A_2+\tau A_2& -\tau A_2\\ A_2& A_1\end{array}\right]\left[\begin{array}{c}{\tilde{\mathit{u}}}^{n+1}\\ {\tilde{\sigma }}^{n+1}\end{array}\right]=\left[\begin{array}{c}D\\ \mathbf{0}\end{array}\right], n\ge 1,$$

(34)

where

$$\begin{array}{cc}{\tilde{\mathit{u}}}^n={({\tilde{u}}_1^n,{\tilde{u}}_2^n,\cdots ,{\tilde{u}}_{M_Z^0}^n)}^T, \hfill & {\tilde{\sigma }}^n={({\tilde{\sigma }}_1^n,{\tilde{\sigma }}_2^n,\cdots ,{\tilde{\sigma }}_{M_Z^0}^n)}^T,\hfill \\ A_1={\left(({\phi }_i,I_h^{\ast }{\phi }_j)\right)}_{i,j=1,\cdots ,M_Z^0},\hfill & A_2={\left(a({\phi }_i,I_h^{\ast }{\phi }_j)\right)}_{i,j=1,\cdots ,M_Z^0},\hfill \\ \tilde{G}\left({\tilde{\mathit{u}}}^n\right)={\left((g\left(u_h^n\right),I_h^{\ast }{\phi }_j)\right)}_{j=1,\cdots ,M_Z^0}^T,\hfill & F^n={\left((f\left(t_n\right),I_h^{\ast }{\phi }_j)\right)}_{j=1,\cdots ,M_Z^0}^T,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill C=& A_1{\tilde{\mathit{u}}}^0-{\tau }^{1-\alpha }{q}_{\alpha }\left(1\right)A_2{\tilde{\mathit{u}}}^0-\tau \tilde{G}\left({\tilde{\mathit{u}}}^0\right)+\tau F^1,\hfill \\ \hfill D=& 2A_1{\tilde{\mathit{u}}}^n-\frac{1}{2}{A}_1{\tilde{\mathit{u}}}^{n-1}-{\tau }^{1-\alpha }{\displaystyle \sum _{k=1}^{n+1}}{q}_{\alpha }\left(k\right)A_2{\tilde{\mathit{u}}}^{n-k+1}\hfill \\ & -2\tau \tilde{G}\left({\tilde{\mathit{u}}}^n\right)+\tau \tilde{G}\left({\tilde{\mathit{u}}}^{n-1}\right)+\tau F^{n+1}.\hfill \end{array}$$

Noting that matrices $A_1$ and $A_2$ are symmetric positive, we have

$$\left[\begin{array}{cc}E& \tau A_2{A}_1^{-1}\\ 0& E\end{array}\right]\left[\begin{array}{cc}{A}_1+{\tau }^{1-\alpha }{q}_{\alpha }\left(0\right)A_2+\tau A_2& -\tau A_2\\ A_2& A_1\end{array}\right]=\left[\begin{array}{cc}{B}_1& 0\\ A_2& A_1\end{array}\right],$$

(35)

and

$$\left[\begin{array}{cc}E& \tau A_2{A}_1^{-1}\\ 0& E\end{array}\right]\left[\begin{array}{cc}\frac{3}{2}{A}_1+{\tau }^{1-\alpha }{q}_{\alpha }\left(0\right)A_2+\tau A_2& -\tau A_2\\ A_2& A_1\end{array}\right]=\left[\begin{array}{cc}{B}_2& 0\\ A_2& A_1\end{array}\right],$$

(36)

where $B_1=A_1+{\tau }^{1-\alpha }{q}_{\alpha }\left(0\right)A_2+\tau A_2+\tau A_2{A}_1^{-1}{A}_2$ and $B_2=\frac{3}{2}{A}_1+{\tau }^{1-\alpha }{q}_{\alpha }\left(0\right)A_2+\tau A_2+\tau A_2{A}_1^{-1}{A}_2$. It is easy to see that $B_1$ and $B_2$ are invertible, so the coefficient matrices of (33) and (34) are invertible. This conclusion implies that the MFVE scheme (21) has a unique solution. □

4. Stability Analysis

In this section, we intend to present the unconditional stability results for the MFVE scheme (21) and (22).

Theorem 2.

Let $(u_h^{n+1},{\sigma }_h^{n+1})\in V_h\times V_h$ be the solutions of the MFVE system (21), then it follows that

$$\begin{array}{cc}& \parallel u_h^{n+1}\parallel +{\left(\tau {\displaystyle \sum _{k=0}^n}{\parallel u_h^{k+1}\parallel }_1^2\right)}^{\frac{1}{2}}+{\left(\tau {\displaystyle \sum _{k=0}^n}{\parallel {\sigma }_h^{k+1}\parallel }^2\right)}^{\frac{1}{2}}\hfill \\ \hfill \le & C\left(\parallel u_h^0\parallel +{\tau }^{\frac{1-\alpha }{2}}{\parallel u_h^0\parallel }_1+\parallel g\left(0\right)\parallel +\underset{t\in [0,T]}{\sup }\parallel f\left(t\right)\parallel \right).\hfill \end{array}$$

Proof. 

Taking $w_h=u_h^{n+1}$ and $v_h={\sigma }_h^{n+1}$ in (21), and applying Lemma 2, we have

$$\begin{array}{cc}& ({\partial }_t^2{u}_h^{n+1},I_h^{\ast }{u}_h^{n+1})+{\tau }^{-\alpha }{\displaystyle \sum _{k=0}^{n+1}}{q}_{\alpha }\left(k\right)a(u_h^{n-k+1},I_h^{\ast }{u}_h^{n+1})\hfill \\ & +a(u_h^{n+1},I_h^{\ast }{u}_h^{n+1})+({\sigma }_h^{n+1},I_h^{\ast }{\sigma }_h^{n+1})\hfill \\ \hfill =& -(G\left[u_h^{n+1}\right],I_h^{\ast }{u}_h^{n+1})+(f^{n+1},I_h^{\ast }{u}_h^{n+1}).\hfill \end{array}$$

(37)

Utilizing Lemmas 1 and 2 and the Young inequality, we obtain

$$\begin{array}{cc}& ({\partial }_t^2{u}_h^{n+1},I_h^{\ast }{u}_h^{n+1})+{\tau }^{-\alpha }{\displaystyle \sum _{k=0}^{n+1}}{q}_{\alpha }\left(k\right)a(u_h^{n-k+1},I_h^{\ast }{u}_h^{n+1})+{\mu }_3{\parallel u_h^{n+1}\parallel }_1^2+({\sigma }_h^{n+1},I_h^{\ast }{\sigma }_h^{n+1})\hfill \\ \hfill \le & C({\parallel G\left[u_h^{n+1}\right]\parallel }^2+{\parallel f^{n+1}\parallel }^2)+\frac{{\mu }_3}{2}{\parallel u_h^{n+1}\parallel }^2.\hfill \end{array}$$

(38)

Noting that $g\left(\tilde{u}\right)$ satisfies the Lipschitz condition, we have $\parallel g\left(\tilde{u}\right)\parallel \le L\parallel \tilde{u}\parallel +\parallel g\left(0\right)\parallel$. From the expression of $G\left[u_h^{n+1}\right]$ and the triangle inequality, when $n\ge 1$, we have

$$\begin{array}{c}\hfill {\parallel G\left[u_h^{n+1}\right]\parallel }^2\le {\parallel 2g\left(u_h^n\right)-g\left(u_h^{n-1}\right)\parallel }^2\le C({\parallel u_h^n\parallel }^2+{\parallel u_h^{n-1}\parallel }^2+{\parallel g\left(0\right)\parallel }^2),\end{array}$$

(39)

and when $n=0$, we have

$$\begin{array}{c}\hfill {\parallel G\left[u_h^1\right]\parallel }^2={\parallel g\left(u_h^0\right)\parallel }^2\le C({\parallel u_h^0\parallel }^2+{\parallel g\left(0\right)\parallel }^2).\end{array}$$

(40)

Next, when $n\ge 1$ in (37), utilizing Lemma 4, we have

$$\begin{array}{cc}& \frac{1}{4\tau }(\Lambda [u_h^{n+1},u_h^n]-\Lambda [u_h^n,u_h^{n-1}])+{\tau }^{-\alpha }{\displaystyle \sum _{k=0}^{n+1}}{q}_{\alpha }\left(k\right)a(u_h^{n-k+1},I_h^{\ast }{u}_h^{n+1})\hfill \\ & +\frac{{\mu }_3}{2}{\parallel u_h^{n+1}\parallel }_1^2+({\sigma }_h^{n+1},I_h^{\ast }{\sigma }_h^{n+1})\hfill \\ \hfill \le & C({\parallel u_h^n\parallel }^2+{\parallel u_h^{n-1}\parallel }^2+{\parallel g\left(0\right)\parallel }^2+{\parallel f^{n+1}\parallel }^2).\hfill \end{array}$$

(41)

Multiplying (41) by $4\tau$ and summing over n from 1 to M, we have

$$\begin{array}{cc}& \Lambda [u_h^{M+1},u_h^M]+4{\tau }^{1-\alpha }{\displaystyle \sum _{n=1}^M}{\displaystyle \sum _{k=0}^{n+1}}{q}_{\alpha }\left(k\right)a(u_h^{n-k+1},I_h^{\ast }{u}_h^{n+1})\hfill \\ & +2{\mu }_3\tau {\displaystyle \sum _{n=1}^M}{\parallel u_h^{n+1}\parallel }_1^2+4\tau {\displaystyle \sum _{n=1}^M}({\sigma }_h^{n+1},I_h^{\ast }{\sigma }_h^{n+1})\hfill \\ \hfill \le & \Lambda [u_h^1,u_h^0]+C\tau {\displaystyle \sum _{n=1}^M}({\parallel u_h^n\parallel }^2+{\parallel u_h^{n-1}\parallel }^2)+C\tau {\displaystyle \sum _{n=1}^M}({\parallel g\left(0\right)\parallel }^2+{\parallel f^{n+1}\parallel }^2).\hfill \end{array}$$

(42)

When $n=0$ in (37), noting that $({\partial }_t^2{u}_h^1,I_h^{\ast }{u}_h^1)\ge \frac{1}{2\tau }[(u_h^1,I_h^{\ast }{u}_h^1)-(u_h^0,I_h^{\ast }{u}_h^0)]$, multiplying (37) by $2\tau$, and applying Lemmas 1 and 2, we obtain

$$\begin{array}{cc}& (u_h^1,I_h^{\ast }{u}_h^1)+2{\tau }^{1-\alpha }{\displaystyle \sum _{k=0}^1}{q}_{\alpha }\left(k\right)a(u_h^{1-k},I_h^{\ast }{u}_h^1)+2{\mu }_3\tau {\parallel u_h^1\parallel }_1^2+2\tau ({\sigma }_h^1,I_h^{\ast }{\sigma }_h^1)\hfill \\ \hfill \le & (u_h^0,I_h^{\ast }{u}_h^0)+C\tau ({\parallel G\left[u_h^1\right]\parallel }^2+{\parallel f^1\parallel }^2)+{\mu }_3\tau {\parallel u_h^1\parallel }^2.\hfill \end{array}$$

(43)

Substituting (40) into (43) yields

$$\begin{array}{cc}& (u_h^1,I_h^{\ast }{u}_h^1)+2{\tau }^{1-\alpha }{\displaystyle \sum _{k=0}^1}{q}_{\alpha }\left(k\right)a(u_h^{1-k},I_h^{\ast }{u}_h^1)+{\mu }_3\tau {\parallel u_h^1\parallel }_1^2+2\tau ({\sigma }_h^1,I_h^{\ast }{\sigma }_h^1)\hfill \\ \hfill \le & (u_h^0,I_h^{\ast }{u}_h^0)+C\tau ({\parallel u_h^0\parallel }^2+{\parallel g\left(0\right)\parallel }^2+{\parallel f^1\parallel }^2).\hfill \end{array}$$

(44)

Then, (44) is rewritten as the following result:

$$\begin{array}{cc}& (u_h^1,I_h^{\ast }{u}_h^1)+2{\tau }^{1-\alpha }{\displaystyle \sum _{n=0}^1}{\displaystyle \sum _{k=0}^n}{q}_{\alpha }\left(k\right)a(u_h^{1-k},I_h^{\ast }{u}_h^1)+{\mu }_3\tau {\parallel u_h^1\parallel }_1^2+2\tau ({\sigma }_h^1,I_h^{\ast }{\sigma }_h^1)\hfill \\ \hfill \le & (u_h^0,I_h^{\ast }{u}_h^0)+2{\tau }^{1-\alpha }{q}_{\alpha }\left(0\right)a(u_h^0,I_h^{\ast }{u}_h^0)+C\tau ({\parallel u_h^0\parallel }^2+{\parallel g\left(0\right)\parallel }^2+{\parallel f^1\parallel }^2).\hfill \end{array}$$

(45)

Applying Lemma 3 in (45), we have

$$\begin{array}{c}\hfill (u_h^1,I_h^{\ast }{u}_h^1)\le (u_h^0,I_h^{\ast }{u}_h^0)+2{\tau }^{1-\alpha }{q}_{\alpha }\left(0\right)a(u_h^0,I_h^{\ast }{u}_h^0)+C\tau ({\parallel u_h^0\parallel }^2+{\parallel g\left(0\right)\parallel }^2+{\parallel f^1\parallel }^2).\end{array}$$

(46)

Now, from (42) and (44), we obtain

$$\begin{array}{cc}& \Lambda [u_h^{M+1},u_h^M]+2(u_h^1,I_h^{\ast }{u}_h^1)+4{\tau }^{1-\alpha }{\displaystyle \sum _{n=0}^M}{\displaystyle \sum _{k=0}^{n+1}}{q}_{\alpha }\left(k\right)a(u_h^{n-k+1},I_h^{\ast }{u}_h^{n+1})\hfill \\ & +2{\mu }_3\tau {\displaystyle \sum _{n=0}^M}{\parallel u_h^{n+1}\parallel }_1^2+4\tau {\displaystyle \sum _{n=0}^M}({\sigma }_h^{n+1},I_h^{\ast }{\sigma }_h^{n+1})\hfill \\ \hfill \le & \Lambda [u_h^1,u_h^0]+2(u_h^0,I_h^{\ast }{u}_h^0)+C\tau {\displaystyle \sum _{n=0}^M}{\parallel u_h^n\parallel }^2+C\tau {\displaystyle \sum _{n=0}^M}({\parallel g\left(0\right)\parallel }^2+{\parallel f^{n+1}\parallel }^2).\hfill \end{array}$$

(47)

Applying Lemma 4, we have

$$\begin{array}{cc}\hfill \Lambda [u_h^1,u_h^0]& =(u_h^1,I_h^{\ast }{u}_h^1)+(2u_h^1-u_h^0,I_h^{\ast }(2u_h^1-u_h^0))\hfill \\ & \le 9(u_h^1,I_h^{\ast }{u}_h^1)+2(u_h^0,I_h^{\ast }{u}_h^0).\hfill \end{array}$$

(48)

Substituting (48) into (47), and making use of (46), we rewrite (47) as the following result:

$$\begin{array}{cc}& \Lambda [u_h^{M+1},u_h^M]+2(u_h^1,I_h^{\ast }{u}_h^1)+4{\tau }^{1-\alpha }{\displaystyle \sum _{n=0}^{M+1}}{\displaystyle \sum _{k=0}^n}{q}_{\alpha }\left(k\right)a(u_h^{n-k},I_h^{\ast }{u}_h^n)\hfill \\ & +2{\mu }_3\tau {\displaystyle \sum _{n=0}^M}{\parallel u_h^{n+1}\parallel }_1^2+4\tau {\displaystyle \sum _{n=0}^M}({\sigma }_h^{n+1},I_h^{\ast }{\sigma }_h^{n+1})\hfill \\ \hfill \le & 13(u_h^0,I_h^{\ast }{u}_h^0)+18{\tau }^{1-\alpha }{q}_{\alpha }\left(0\right)a(u_h^0,I_h^{\ast }{u}_h^0)\hfill \\ & +C\tau {\displaystyle \sum _{n=0}^M}{\parallel u_h^n\parallel }^2+C\tau {\displaystyle \sum _{n=0}^M}{(\parallel g\left(0\right)\parallel }^2+{\parallel f^{n+1}\parallel }^2).\hfill \end{array}$$

(49)

Making use of the discrete Gronwall lemma and Lemma 3, we have

$$\begin{array}{cc}& \Lambda [u_h^{M+1},u_h^M]+2(u_h^1,I_h^{\ast }{u}_h^1)+2{\mu }_3\tau {\displaystyle \sum _{n=0}^M}{\parallel u_h^{n+1}\parallel }_1^2+4\tau {\displaystyle \sum _{n=0}^M}({\sigma }_h^{n+1},I_h^{\ast }{\sigma }_h^{n+1})\hfill \\ \hfill \le & C\left({\parallel u_h^0\parallel }^2+{\tau }^{1-\alpha }{\parallel u_h^0\parallel }_1^2+{\parallel g\left(0\right)\parallel }^2+\underset{t\in [0,T]}{\sup }{\parallel f\left(t\right)\parallel }^2\right).\hfill \end{array}$$

(50)

Then, utilizing Lemma 1 finalizes the proof. □

5. Convergence Analysis

In order to obtain the error estimates for the MFVE scheme (21)–(22), we first introduce a projection operator $P_h:H_0^1(\Omega )\cap H^2(\Omega )\to V_h$ as follows:

$$\begin{array}{c}\hfill a(v-P_{h}v,I_h^{\ast }{w}_h)=0, \forall w_h\in V_h.\end{array}$$

(51)

Lemma 5 

([35]). For the projection operator $P_h$, the following estimates can be obtained:

$$\begin{array}{cc}& {\parallel v-P_{h}v\parallel }_1\le Ch{\left|v\right|}_2, \forall v\in H_0^1(\Omega )\cap H^2(\Omega ),\hfill \end{array}$$

(52)

$$\begin{array}{cc}& \parallel v-P_{h}v\parallel \le C{h}^2{\parallel v\parallel }_{3,p}, \forall v\in H_0^1(\Omega )\cap W^{3,p}(\Omega ), p>1.\hfill \end{array}$$

(53)

For $(u,\sigma ):[0,T]\to H_0^1(\Omega )\times H_0^1(\Omega )$, we also need to introduce an MFVE projection $(R_{h}u,R_h\sigma ):[0,T]\to V_h\times V_h$ such that

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}\left(a\right)a(u-R_{h}u,I_h^{\ast }{v}_h)+(\sigma -R_h\sigma ,I_h^{\ast }{v}_h)=0,\hfill & \forall v_h\in V_h,\hfill \\ \left(b\right)a(\sigma -R_h\sigma ,I_h^{\ast }{w}_h)=0,\hfill & \forall w_h\in V_h.\hfill \end{array}\right.\end{array}$$

(54)

Then, the MFVE projection $R_h$ satisfies the following estimates.

Lemma 6.

For the MFVE projection $R_h$, we have

$$\begin{array}{cc}& {\parallel \sigma -R_h\sigma \parallel }_1\le Ch{\left|\sigma \right|}_2, \forall \sigma \in H_0^1(\Omega )\cap H^2(\Omega ),\hfill \end{array}$$

(55)

$$\begin{array}{cc}& \parallel \sigma -R_h\sigma \parallel \le C{h}^2{\parallel \sigma \parallel }_{3,p}, \forall \sigma \in H_0^1(\Omega )\cap W^{3,p}(\Omega ),\hfill \end{array}$$

(56)

$$\begin{array}{cc}& {\parallel u-R_{h}u\parallel }_1\le {Ch\left|u\right|}_2+C{h}^2{\parallel \sigma \parallel }_{3,p}, \forall u\in H_0^1(\Omega )\cap H^2(\Omega ), \forall \sigma \in H_0^1(\Omega )\cap W^{3,p}(\Omega ),\hfill \end{array}$$

(57)

$$\begin{array}{cc}& \parallel u-R_{h}u\parallel \le C{h}^2{(\parallel u\parallel }_{3,p}+{\parallel \sigma \parallel }_{3,p}), \forall u,\sigma \in H_0^1(\Omega )\cap W^{3,p}(\Omega ),\hfill \end{array}$$

(58)

where $p>1$ is a positive integer.

Proof. 

From (51) and (54)$\left(b\right)$, we can easily see that $R_h\sigma =P_h\sigma$. Then, we obtain the estimates (55) and (56). Next, making use of (54)$\left(a\right)$, we have

$$\begin{array}{c}\hfill a(u-P_{h}u+P_{h}u-R_{h}u,I_h^{\ast }{v}_h)+(\sigma -R_h\sigma ,I_h^{\ast }{v}_h)=0, \forall v_h\in V_h.\end{array}$$

(59)

From the definition of $P_h$, we have

$$\begin{array}{c}\hfill a(P_{h}u-R_{h}u,I_h^{\ast }{v}_h)+(\sigma -R_h\sigma ,I_h^{\ast }{v}_h)=0, \forall v_h\in V_h.\end{array}$$

(60)

Take $v_h=P_{h}u-R_{h}u$ in (60) to obtain

$$\begin{array}{c}\hfill a(P_{h}u-R_{h}u,I_h^{\ast }(P_{h}u-R_{h}u))\le C\parallel \sigma -R_h\sigma \parallel \parallel P_{h}u-R_{h}u\parallel .\end{array}$$

(61)

Noting that $a(P_{h}u-R_{h}u,I_h^{\ast }(P_{h}u-R_{h}u))\ge {\mu }_3{\parallel P_{h}u-R_{h}u\parallel }_1^2$ in (61), and making use of (56), we have

$$\begin{array}{c}\hfill {\parallel P_{h}u-R_{h}u\parallel }_1\le C\parallel \sigma -R_h\sigma \parallel \le C{h}^2{\parallel \sigma \parallel }_{3,p},p>1.\end{array}$$

(62)

Finally, apply the triangle inequality and Lemma 5 to complete the proof. □

Now, we give the error analysis, and separate the errors as follows:

$$\begin{array}{cc}\hfill & u\left(t_n\right)-u_h^n=u\left(t_n\right)-R_{h}u\left(t_n\right)+R_{h}u\left(t_n\right)-u_h^n={\rho }^n+{\theta }^n,\hfill \\ \hfill & \sigma \left(t_n\right)-{\sigma }_h^n=\sigma \left(t_n\right)-R_h\sigma \left(t_n\right)+R_h\sigma \left(t_n\right)-{\sigma }_h^n={\xi }^n+{\eta }^n.\hfill \end{array}$$

Making use of the MFVE projection (54), we give the following error equations:

$$\begin{array}{cc}\hfill & ({\partial }_t^2{\theta }^{n+1},I_h^{\ast }{w}_h)+{\tau }^{-\alpha }{\displaystyle \sum _{k=0}^{n+1}}{q}_{\alpha }\left(k\right)a({\theta }^{n-k+1},I_h^{\ast }{w}_h)\hfill \\ \hfill & +a({\theta }^{n+1},I_h^{\ast }{w}_h)-a({\eta }^{n+1},I_h^{\ast }{w}_h)\hfill \\ \hfill =& -({\partial }_t^2{\rho }^{n+1},I_h^{\ast }{w}_h)+{\tau }^{-\alpha }{\displaystyle \sum _{k=0}^{n+1}}{q}_{\alpha }\left(k\right)({\xi }^{n-k+1},I_h^{\ast }{w}_h)\hfill \\ \hfill & -(G\left[u^{n+1}\right]-G\left[u_h^{n+1}\right],I_h^{\ast }{w}_h)+({\xi }^{n+1},I_h^{\ast }{w}_h)\hfill \\ \hfill & -(E_{u,t}^{n+1},I_h^{\ast }{w}_h)-(E_{u,\alpha }^{n+1},I_h^{\ast }{w}_h)-(E_g^{n+1},I_h^{\ast }{w}_h), \forall w_h\in V_h,\hfill \end{array}$$

(63)

and

$$\begin{array}{c}\hfill ({\eta }^{n+1},I_h^{\ast }{v}_h)+a({\theta }^{n+1},I_h^{\ast }{v}_h)=0, \forall v_h\in V_h,\end{array}$$

(64)

where ${\eta }^0=0$ and ${\theta }^0=0$.

Theorem 3.

Let $(u,\sigma )$ be the solution of the lower-order model (3) and $(u_h^n,{\sigma }_h^n)$ be the solution of the MFVE scheme (21) and (22). Then, it follows that

$$\begin{array}{cc}& \underset{1\le n\le N}{\max }\parallel u\left(t_n\right)-u_h^n\parallel \le C(h^2+{\tau }^2),\hfill \\ & {\left(\tau {\displaystyle \sum _{n=1}^N}{\parallel u\left(t_n\right)-u_h^n\parallel }_1^2\right)}^{\frac{1}{2}}\le C(h+{\tau }^2),\hfill \\ & {\left(\tau {\displaystyle \sum _{n=1}^N}{\parallel \sigma \left(t_n\right)-{\sigma }_h^n\parallel }^2\right)}^{\frac{1}{2}}\le C(h^2+{\tau }^2).\hfill \end{array}$$

Proof. 

Choosing $w_h={\theta }^{n+1}$ and $v_h={\eta }^{n+1}$ in (63) and (64), respectively, we obtain

$$\begin{array}{cc}& ({\partial }_t^2{\theta }^{n+1},I_h^{\ast }{\theta }^{n+1})+{\tau }^{-\alpha }{\displaystyle \sum _{k=0}^{n+1}}{q}_{\alpha }\left(k\right)a({\theta }^{n-k+1},I_h^{\ast }{\theta }^{n+1})\hfill \\ & +({\eta }^{n+1},I_h^{\ast }{\eta }^{n+1})+a({\theta }^{n+1},I_h^{\ast }{\theta }^{n+1})\hfill \\ \hfill =& -({\partial }_t^2{\rho }^{n+1},I_h^{\ast }{\theta }^{n+1})+{\tau }^{-\alpha }{\displaystyle \sum _{k=0}^{n+1}}{q}_{\alpha }\left(k\right)({\xi }^{n-k+1},I_h^{\ast }{\theta }^{n+1})+({\xi }^{n+1},I_h^{\ast }{\theta }^{n+1})\hfill \\ & -(G\left[u^{n+1}\right]-G\left[u_h^{n+1}\right],I_h^{\ast }{\theta }^{n+1})-(E_{u,t}^{n+1},I_h^{\ast }{\theta }^{n+1})\hfill \\ & -(E_{u,\alpha }^{n+1},I_h^{\ast }{\theta }^{n+1})-(E_g^{n+1},I_h^{\ast }{\theta }^{n+1}).\hfill \end{array}$$

(65)

Noting that ${\tau }^{-\alpha }{\sum }_{k=0}^{n+1}{q}_{\alpha }\left(k\right){\xi }^{n-k+1}=\frac{{\partial }^{\alpha }{\xi }^{n+1}}{\partial t^{\alpha }}-E_{\xi ,\alpha }^{n+1}$ and applying Lemmas 1 and 2 and the Young inequality in (65), we have

$$\begin{array}{cc}& ({\partial }_t^2{\theta }^{n+1},I_h^{\ast }{\theta }^{n+1})+{\tau }^{-\alpha }{\displaystyle \sum _{k=0}^{n+1}}{q}_{\alpha }\left(k\right)a({\theta }^{n-k+1},I_h^{\ast }{\theta }^{n+1})+{\mu }_1{\parallel {\eta }^{n+1}\parallel }^2+{\mu }_3{\parallel {\theta }^{n+1}\parallel }_1^2\hfill \\ \hfill \le & C[{\parallel {\partial }_t^2{\rho }^{n+1}\parallel }^2+{\parallel \frac{{\partial }^{\alpha }{\xi }^{n+1}}{\partial t^{\alpha }}\parallel }^2+{\parallel E_{\xi ,\alpha }^{n+1}\parallel }^2+{\parallel {\xi }^{n+1}\parallel }^2\hfill \\ & +{\parallel G\left[u^{n+1}\right]-G\left[u_h^{n+1}\right]\parallel }^2+{\parallel E_{u,t}^{n+1}\parallel }^2+{\parallel E_{u,\alpha }^{n+1}\parallel }^2+{\parallel E_g^{n+1}\parallel }^2]+\frac{{\mu }_3}{2}{\parallel {\theta }^{n+1}\parallel }^2.\hfill \end{array}$$

(66)

For the item ${\parallel {\partial }_t^2{\rho }^{n+1}\parallel }^2$, when $n\ge 1$, applying the triangle inequality and Lemma 6, we have

$$\begin{array}{cc}\hfill {\parallel {\partial }_t^2{\rho }^{n+1}\parallel }^2& ={\parallel \frac{3{\rho }^{n+1}-3{\rho }^n}{2\tau }-\frac{{\rho }^n-{\rho }^{n-1}}{2\tau }\parallel }^2\le \frac{C}{\tau }{\int }_{t_{n-1}}^{t_{n+1}}{\parallel {\rho }_t\parallel }^2\mathrm{d}t\hfill \\ & \le C({\parallel u_t\parallel }_{L^{\infty }\left(W^{3,p}\right)}+{{\parallel {\sigma }_t\parallel }_{L^{\infty }\left(W^{3,p}\right)})}^2{h}^4, p\ge 1.\hfill \end{array}$$

(67)

When $n=0$, we have

$$\begin{array}{cc}\hfill {\parallel {\partial }_t^2{\rho }^1\parallel }^2& ={\parallel \frac{{\rho }^1-{\rho }^0}{\tau }\parallel }^2\le \frac{C}{\tau }{\int }_{t_0}^{t_1}{\parallel {\rho }_t\parallel }^2\mathrm{d}t\hfill \\ & \le C({\parallel u_t\parallel }_{L^{\infty }\left(W^{3,p}\right)}+{{\parallel {\sigma }_t\parallel }_{L^{\infty }\left(W^{3,p}\right)})}^2{h}^4,p\ge 1.\hfill \end{array}$$

(68)

Moreover, when $n\ge 1$, for the item ${\parallel G\left[u^{n+1}\right]-G\left[U^{n+1}\right]\parallel }^2$, we obtain

$$\begin{array}{cc}\hfill {\parallel G\left[u^{n+1}\right]-G\left[u_h^{n+1}\right]\parallel }^2& \le {(2\parallel g\left(u^n\right)-g\left(u_h^n\right)\parallel +\parallel g\left(u^{n-1}\right)-g\left(u_h^{n-1}\right)\parallel )}^2\hfill \\ & \le C{(\parallel u^n-u_h^n\parallel +\parallel u^{n-1}-u_h^{n-1}\parallel )}^2\hfill \\ & \le C({\parallel {\rho }^n\parallel }^2+{\parallel {\rho }^{n-1}\parallel }^2+{\parallel {\theta }^n\parallel }^2+{\parallel {\theta }^{n-1}\parallel }^2),\hfill \end{array}$$

(69)

and when $n=0$, we obtain

$$\begin{array}{cc}\hfill {\parallel G\left[u^1\right]-G\left[u_h^1\right]\parallel }^2& ={\parallel g\left(u^0\right)-g\left(u_h^0\right)\parallel }^2\hfill \\ & {\le C\parallel u^0-u_h^0\parallel }^2\hfill \\ & {\le C\parallel {\rho }^0\parallel }^2.\hfill \end{array}$$

(70)

Now, applying Lemma 4, when $n\ge 1$ in (66), we obtain

$$\begin{array}{cc}& \frac{1}{4\tau }\left(\Lambda [{\theta }^{n+1},{\theta }^n]-\Lambda [{\theta }^n,{\theta }^{n-1}]\right)+\frac{{\mu }_3}{2}{\parallel {\theta }^{n+1}\parallel }_1^2\hfill \\ & +{\tau }^{-\alpha }{\displaystyle \sum _{k=0}^{n+1}}{q}_{\alpha }\left(k\right)a({\theta }^{n-k+1},I_h^{\ast }{\theta }^{n+1})+{\mu }_1{\parallel {\eta }^{n+1}\parallel }^2\hfill \\ \hfill \le & C(h^4+{\tau }^4+{\parallel {\theta }^n\parallel }^2+{\parallel {\theta }^{n-1}\parallel }^2).\hfill \end{array}$$

(71)

Multiplying (71) by $4\tau$, and summing over n from 1 to M, we obtain

$$\begin{array}{cc}& \Lambda [{\theta }^{M+1},{\theta }^M]+4{\tau }^{1-\alpha }{\displaystyle \sum _{n=1}^M}{\displaystyle \sum _{k=0}^{n+1}}{q}_{\alpha }\left(k\right)a({\theta }^{n-k+1},I_h^{\ast }{\theta }^{n+1})\hfill \\ & +2{\mu }_3\tau {\displaystyle \sum _{n=1}^M}{\parallel {\theta }^{n+1}\parallel }_1^2+4{\mu }_1\tau {\displaystyle \sum _{n=1}^M}{\parallel {\eta }^{n+1}\parallel }^2\hfill \\ \hfill \le & \Lambda [{\theta }^1,{\theta }^0]+C\tau {\displaystyle \sum _{n=1}^M}\left(h^4+{\tau }^4+{\parallel {\theta }^n\parallel }^2+{\parallel {\theta }^{n-1}\parallel }^2\right).\hfill \end{array}$$

(72)

When $n=0$ in (66), applying Lemma 4, we have

$$\begin{array}{cc}& \frac{1}{2\tau }[({\theta }^1,I_h^{\ast }{\theta }^1)-({\theta }^0,I_h^{\ast }{\theta }^0)]+a({\theta }^1,I_h^{\ast }{\theta }^1)\hfill \\ & +{\tau }^{-\alpha }{\displaystyle \sum _{k=0}^1}{q}_{\alpha }\left(k\right)a({\theta }^{1-k},I_h^{\ast }{\theta }^1)+({\eta }^1,I_h^{\ast }{\eta }^1)\hfill \\ \hfill \le & -({\partial }_t^2{\rho }^1,I_h^{\ast }{\theta }^1)+(\frac{{\partial }^{\alpha }{\xi }^1}{\partial t^{\alpha }}-E_{\xi ,\alpha }^1,I_h^{\ast }{\theta }^1)-(G\left[u^1\right]-G\left[u_h^1\right],I_h^{\ast }{\theta }^1)\hfill \\ & +({\xi }^1,I_h^{\ast }{\theta }^1)-(E_{u,t}^1,I_h^{\ast }{\theta }^1)-(E_{u,\alpha }^1,I_h^{\ast }{\theta }^1)-(E_g^1,I_h^{\ast }{\theta }^1).\hfill \end{array}$$

(73)

Multiplying (73) by $2\tau$, applying Lemmas 1 and 2 and the Young inequality, we have

$$\begin{array}{cc}& ({\theta }^1,I_h^{\ast }{\theta }^1)+2{\tau }^{1-\alpha }{\displaystyle \sum _{k=0}^1}{q}_{\alpha }\left(k\right)a({\theta }^{1-k},I_h^{\ast }{\theta }^1)+2{\mu }_3\tau {\parallel {\theta }^1\parallel }_1^2+2{\mu }_1\tau {\parallel {\eta }^1\parallel }^2\hfill \\ \hfill \le & ({\theta }^0,I_h^{\ast }{\theta }^0)+C{\tau }^2({\parallel E_{u,t}^1\parallel }^2+{\parallel E_{u,\alpha }^1\parallel }^2+{\parallel E_g^1\parallel }^2)+{\mu }_3\tau {\parallel {\theta }^1\parallel }^2+\frac{1}{2}({\theta }^1,I_h^{\ast }{\theta }^1)\hfill \\ & +C\tau \left({\parallel {\partial }_t^2{\rho }^1\parallel }^2+{\parallel \frac{{\partial }^{\alpha }{\xi }^1}{\partial t^{\alpha }}\parallel }^2+{\parallel E_{\xi ,\alpha }^1\parallel }^2+{\parallel {\xi }^1\parallel }^2+{\parallel G\left[u^1\right]-G\left[u_h^1\right]\parallel }^2\right).\hfill \end{array}$$

(74)

Applying Lemma 6, we have

$$\begin{array}{cc}& ({\theta }^1,I_h^{\ast }{\theta }^1)+4{\tau }^{1-\alpha }{\displaystyle \sum _{k=0}^1}{q}_{\alpha }\left(k\right)a({\theta }^{1-k},I_h^{\ast }{\theta }^1)+2{\mu }_3\tau {\parallel {\theta }^1\parallel }_1^2+4{\mu }_1\tau {\parallel {\eta }^1\parallel }^2\hfill \\ \hfill \le & 2({\theta }^0,I_h^{\ast }{\theta }^0)+C\tau (h^4+{\tau }^4)+C{\tau }^4.\hfill \end{array}$$

(75)

Then, we rewrite (75) as the following result:

$$\begin{array}{cc}& ({\theta }^1,I_h^{\ast }{\theta }^1)+4{\tau }^{1-\alpha }{\displaystyle \sum _{n=0}^1}{\displaystyle \sum _{k=0}^n}{q}_{\alpha }\left(k\right)a({\theta }^{1-k},I_h^{\ast }{\theta }^1)+2{\mu }_3\tau {\parallel {\theta }^1\parallel }_1^2+4{\mu }_1\tau {\parallel {\eta }^1\parallel }^2\hfill \\ \hfill \le & 2({\theta }^0,I_h^{\ast }{\theta }^0)+C\tau (h^4+{\tau }^4)+C{\tau }^4+4{\tau }^{1-\alpha }{q}_{\alpha }\left(0\right)a({\theta }^0,I_h^{\ast }{\theta }^0).\hfill \end{array}$$

(76)

Noting that ${\theta }^0=0$, from Lemma 3, we have

$$\begin{array}{c}\hfill ({\theta }^1,I_h^{\ast }{\theta }^1)\le C\tau (h^4+{\tau }^4)+C{\tau }^4.\end{array}$$

(77)

Now, adding (72) and (76), we have

$$\begin{array}{cc}& \Lambda [{\theta }^{M+1},{\theta }^M]+({\theta }^1,I_h^{\ast }{\theta }^1)+4{\tau }^{1-\alpha }{\displaystyle \sum _{n=0}^M}{\displaystyle \sum _{k=0}^{n+1}}{q}_{\alpha }\left(k\right)a({\theta }^{n-k+1},I_h^{\ast }{\theta }^{n+1})\hfill \\ & +2{\mu }_3\tau {\displaystyle \sum _{n=0}^M}{\parallel {\theta }^{n+1}\parallel }_1^2+4{\mu }_1\tau {\displaystyle \sum _{n=0}^M}{\parallel {\eta }^{n+1}\parallel }^2\hfill \\ \hfill \le & \Lambda [{\theta }^1,{\theta }^0]+C(h^4+{\tau }^4)+C\tau {\displaystyle \sum _{n=0}^M}{\parallel {\theta }^n\parallel }^2,\hfill \end{array}$$

(78)

Noting that $\Lambda [{\theta }^1,{\theta }^0]=5({\theta }^1,I_h^{\ast }{\theta }^1)\le C\tau (h^4+{\tau }^4)+C{\tau }^4$, we have

$$\begin{array}{cc}& \Lambda [{\theta }^{M+1},{\theta }^M]+({\theta }^1,I_h^{\ast }{\theta }^1)+4{\tau }^{1-\alpha }{\displaystyle \sum _{n=0}^{M+1}}{\displaystyle \sum _{k=0}^n}{q}_{\alpha }\left(k\right)a({\theta }^{n-k},I_h^{\ast }{\theta }^n)\hfill \\ & +2{\mu }_3\tau {\displaystyle \sum _{n=0}^M}{\parallel {\theta }^{n+1}\parallel }_1^2+4{\mu }_1\tau {\displaystyle \sum _{n=0}^M}{\parallel {\eta }^{n+1}\parallel }^2\hfill \\ \hfill \le & C(h^4+{\tau }^4)+C\tau {\displaystyle \sum _{n=0}^M}{\parallel {\theta }^n\parallel }^2+4{\tau }^{1-\alpha }{q}_{\alpha }\left(0\right)({\theta }^0,I_h^{\ast }{\theta }^0).\hfill \end{array}$$

(79)

Noting that ${\theta }^0=0$, utilizing Lemma 3 and the discrete Gronwall lemma, we have

$$\begin{array}{c}\hfill \Lambda [{\theta }^{M+1},{\theta }^M]+({\theta }^1,I_h^{\ast }{\theta }^1)+2{\mu }_3\tau {\displaystyle \sum _{n=0}^M}{\parallel {\theta }^{n+1}\parallel }_1^2+4{\mu }_1\tau {\displaystyle \sum _{n=0}^M}{\parallel {\eta }^{n+1}\parallel }^2\le C(h^4+{\tau }^4).\end{array}$$

(80)

Finally, Lemmas 4 and 6 are applied to finish the proof. □

6. Numerical Examples

In this section, we provide three tests conducted to examine the effectiveness and convergence accuracy of the MFVE scheme (21) and (22).

Example 1.

In (1), select $\Omega ={(0,1)}^2$, $J=(0,1]$, $g\left(u\right)=\sin \left(u\right)$, $u_0\left(\mathit{x}\right)=0$, and $f(\mathit{x},t)$ as follows:

$$\begin{array}{cc}\hfill f(\mathit{x},t)=& \left(2t+\frac{16{\pi }^2{t}^{2-\alpha }}{\Gamma (3-\alpha )}+8{\pi }^2{t}^2+64{\pi }^4{t}^2\right)\sin \left(2\pi x_1\right)\sin \left(2\pi x_2\right)\hfill \\ & +\sin \left(t^2\sin \left(2\pi x_1\right)\sin \left(2\pi x_2\right)\right), \forall \mathit{x}=(x_1,x_2)\in \Omega .\hfill \end{array}$$

The corresponding exact solutions of u and $\sigma$ are obtained as follows:

$$\begin{array}{cc}\hfill & u(\mathit{x},t)=t^2\sin \left(2\pi x_1\right)\sin \left(2\pi x_2\right),\hfill \\ \hfill & \sigma (\mathit{x},t)=-8{\pi }^2{t}^2\sin \left(2\pi x_1\right)\sin \left(2\pi x_2\right).\hfill \end{array}$$

In the actual numerical calculation, for different fractional parameters, we conducted many numerical experiments, and provide herein the error behaviors in the following discrete norms:

$$\begin{array}{cc}\hfill & {\parallel u-u_h\parallel }_{{\tilde{L}}^{\infty }\left(L^2(\Omega )\right)}=\underset{1\le n\le N}{\max }\parallel u\left(t_n\right)-u_h^n\parallel ,\hfill \\ \hfill & {\parallel u-u_h\parallel }_{{\tilde{L}}^2\left(H^1(\Omega )\right)}={\left(\tau {\displaystyle \sum _{n=1}^N}{\parallel u\left(t_n\right)-u_h^n\parallel }_1^2\right)}^{\frac{1}{2}},\hfill \\ \hfill & {\parallel \sigma -{\sigma }_h\parallel }_{{\tilde{L}}^2\left(L^2(\Omega )\right)}={\left(\tau {\displaystyle \sum _{n=1}^N}{\parallel \sigma \left(t_n\right)-{\sigma }_h^n\parallel }^2\right)}^{\frac{1}{2}}.\hfill \end{array}$$

We first take the fractional parameter $\alpha =0.2, 0.4, 0.6, 0.8$; select the time and space grid parameters $(h,\tau )=(\frac{\sqrt{2}}{10},\frac{1}{10})$, $(\frac{\sqrt{2}}{20},\frac{1}{20})$, $(\frac{\sqrt{2}}{40},\frac{1}{40})$, and $(\frac{\sqrt{2}}{80},\frac{1}{80})$ which satisfy $h=\sqrt{2}\tau$; and provide the numerical results and error behaviors in

Table 1. Error behaviors for u in discrete $L^{\infty }\left(L^2\right)$ norm with $h=\sqrt{2}\tau$ in Example 1.

$\mathit{\alpha }$	$(\mathit{h},\mathit{\tau })$	$(\frac{\sqrt{2}}{10},\frac{1}{10})$	$(\frac{\sqrt{2}}{20},\frac{1}{20})$	$(\frac{\sqrt{2}}{40},\frac{1}{40})$	$(\frac{\sqrt{2}}{80},\frac{1}{80})$
0.2	${\parallel u-u_h\parallel }_{{\tilde{L}}^{\infty }\left(L^2\right)}$	$5.766763\times {10}^{-2}$	$1.523696\times {10}^{-2}$	$3.862442\times {10}^{-3}$	$9.689649\times {10}^{-4}$
	Order		$1.920187$	$1.979989$	$1.994997$
0.4	${\parallel u-u_h\parallel }_{{\tilde{L}}^{\infty }\left(L^2\right)}$	$5.761139\times {10}^{-2}$	$1.522022\times {10}^{-2}$	$3.858078\times {10}^{-3}$	$9.678625\times {10}^{-4}$
	Order		$1.920365$	$1.980035$	$1.995008$
0.6	${\parallel u-u_h\parallel }_{{\tilde{L}}^{\infty }\left(L^2\right)}$	$5.755558\times {10}^{-2}$	$1.520359\times {10}^{-2}$	$3.853742\times {10}^{-3}$	$9.667672\times {10}^{-4}$
	Order		$1.920544$	$1.980080$	$1.995019$
0.8	${\parallel u-u_h\parallel }_{{\tilde{L}}^{\infty }\left(L^2\right)}$	$5.750367\times {10}^{-2}$	$1.518807\times {10}^{-2}$	$3.849693\times {10}^{-3}$	$9.657440\times {10}^{-4}$
	Order		$1.920715$	$1.980124$	$1.995030$

,

Table 2. Error behaviors for u in discrete $L^2\left(H^1\right)$ norm with $h=\sqrt{2}\tau$ in Example 1.

$\mathit{\alpha }$	$(\mathit{h},\mathit{\tau })$	$(\frac{\sqrt{2}}{10},\frac{1}{10})$	$(\frac{\sqrt{2}}{20},\frac{1}{20})$	$(\frac{\sqrt{2}}{40},\frac{1}{40})$	$(\frac{\sqrt{2}}{80},\frac{1}{80})$
0.2	${\parallel u-u_h\parallel }_{{\tilde{L}}^2\left(H^1\right)}$	$6.521600\times {10}^{-1}$	$3.115402\times {10}^{-1}$	$1.515934\times {10}^{-1}$	$7.469699\times {10}^{-2}$
	Order		1.065808	1.039212	1.021085
0.4	${\parallel u-u_h\parallel }_{{\tilde{L}}^2\left(H^1\right)}$	$6.521290\times {10}^{-1}$	$3.115354\times {10}^{-1}$	$1.515928\times {10}^{-1}$	$7.469691\times {10}^{-2}$
	Order		1.065762	1.039195	1.021080
0.6	${\parallel u-u_h\parallel }_{{\tilde{L}}^2\left(H^1\right)}$	$6.520967\times {10}^{-1}$	$3.115303\times {10}^{-1}$	$1.515921\times {10}^{-1}$	$7.469683\times {10}^{-2}$
	Order		1.065714	1.039178	1.021076
0.8	${\parallel u-u_h\parallel }_{{\tilde{L}}^2\left(H^1\right)}$	$6.520650\times {10}^{-1}$	$3.115253\times {10}^{-1}$	$1.515914\times {10}^{-1}$	$7.469674\times {10}^{-2}$
	Order		1.065667	1.039161	1.021071

and

Table 3. Error behaviors for $\sigma$ in discrete $L^2\left(L^2\right)$ norm with $h=\sqrt{2}\tau$ in Example 1.

$\mathit{\alpha }$	$(\mathit{h},\mathit{\tau })$	$(\frac{\sqrt{2}}{10},\frac{1}{10})$	$(\frac{\sqrt{2}}{20},\frac{1}{20})$	$(\frac{\sqrt{2}}{40},\frac{1}{40})$	$(\frac{\sqrt{2}}{80},\frac{1}{80})$
0.2	${\parallel \sigma -{\sigma }_h\parallel }_{{\tilde{L}}^2\left(L^2\right)}$	$1.455052\times {10}^{+0}$	$3.429433\times {10}^{-1}$	$8.318115\times {10}^{-2}$	$2.047806\times {10}^{-2}$
	Order		2.085028	2.043642	2.022177
0.4	${\parallel \sigma -{\sigma }_h\parallel }_{{\tilde{L}}^2\left(L^2\right)}$	$1.452570\times {10}^{+0}$	$3.422371\times {10}^{-1}$	$8.300050\times {10}^{-2}$	$2.043282\times {10}^{-2}$
	Order		2.085540	2.043804	2.022232
0.6	${\parallel \sigma -{\sigma }_h\parallel }_{{\tilde{L}}^2\left(L^2\right)}$	$1.449962\times {10}^{+0}$	$3.414908\times {10}^{-1}$	$8.280916\times {10}^{-2}$	$2.038484\times {10}^{-2}$
	Order		2.086096	2.043984	2.022293
0.8	${\parallel \sigma -{\sigma }_h\parallel }_{{\tilde{L}}^2\left(L^2\right)}$	$1.447383\times {10}^{+0}$	$3.407477\times {10}^{-1}$	$8.261807\times {10}^{-2}$	$2.033688\times {10}^{-2}$
	Order		2.086671	2.044175	2.022359

. It is easy to observe that the convergence orders for u in discrete $L^{\infty }\left(L^2(\Omega )\right)$ norm are close to 2, and that for u in discrete $L^2\left(H^1(\Omega )\right)$ norm and for $\sigma$ in discrete $L^2\left(L^2(\Omega )\right)$, the convergence orders are close to 1. These error behaviors are in agreement with the numerical theoretical results in Theorem 3. Furthermore, to test whether the fractional parameter $\alpha$ has an impact on the algorithm when the parameters are sufficiently small and large, we specifically selected this parameter with values of $0.001$ and $0.999$ to carry out the above experiment. We provide the numerical results and error behaviors in

Table 4. Error behaviors with $h=\sqrt{2}\tau$ in Example 1.

$\mathit{\alpha }$	$(\mathit{h},\mathit{\tau })$	$(\frac{\sqrt{2}}{10},\frac{1}{10})$	$(\frac{\sqrt{2}}{20},\frac{1}{20})$	$(\frac{\sqrt{2}}{40},\frac{1}{40})$	$(\frac{\sqrt{2}}{80},\frac{1}{80})$
0.001	${\parallel u-u_h\parallel }_{{\tilde{L}}^{\infty }\left(L^2\right)}$	$5.772113\times {10}^{-2}$	$1.525288\times {10}^{-2}$	$3.866594\times {10}^{-3}$	$9.700137\times {10}^{-4}$
	Order		1.920018	1.979946	1.994986
	${\parallel u-u_h\parallel }_{{\tilde{L}}^2\left(H^1\right)}$	$6.521881\times {10}^{-1}$	$3.115446\times {10}^{-1}$	$1.515940\times {10}^{-1}$	$7.469706\times {10}^{-2}$
	Order		1.065849	1.039226	1.021089
	${\parallel \sigma -{\sigma }_h\parallel }_{{\tilde{L}}^2\left(L^2\right)}$	$1.457286\times {10}^{+0}$	$3.435762\times {10}^{-1}$	$8.334267\times {10}^{-2}$	$2.051847\times {10}^{-2}$
	Order		2.084582	2.043503	2.022132
0.999	${\parallel u-u_h\parallel }_{{\tilde{L}}^{\infty }\left(L^2\right)}$	$5.745960\times {10}^{-2}$	$1.517481\times {10}^{-2}$	$3.846222\times {10}^{-3}$	$9.648663\times {10}^{-4}$
	Order		1.920870	1.980164	1.995041
	${\parallel u-u_h\parallel }_{{\tilde{L}}^2\left(H^1\right)}$	$6.520366\times {10}^{-1}$	$3.115207\times {10}^{-1}$	$1.515908\times {10}^{-1}$	$7.469667\times {10}^{-2}$
	Order		1.065625	1.039146	1.021067
	${\parallel \sigma -{\sigma }_h\parallel }_{{\tilde{L}}^2\left(L^2\right)}$	$1.445043\times {10}^{+0}$	$3.400646\times {10}^{-1}$	$8.244164\times {10}^{-2}$	$2.029258\times {10}^{-2}$
	Order		2.087232	2.044364	2.022421

, and obtained the same conclusions with respect to the convergence orders as that discussed on the other selections of fractional-order parameters.

Example 2.

In this example, we take the same Ω, J, and $u_0\left(\mathit{x}\right)$ as in Example 1, and select $g\left(u\right)=u^3-u$ and $f(\mathit{x},t)$ as follows:

$$\begin{array}{cc}\hfill f(\mathit{x},t)=& \left((1+\alpha )t^{\alpha }+\frac{2{\pi }^2\Gamma (2+\alpha )t}{\Gamma \left(2\right)}+2{\pi }^2{t}^{1+\alpha }+4{\pi }^4{t}^{1+\alpha }\right)\sin \left(\pi x_1\right)\sin \left(\pi x_2\right)\hfill \\ & +{\left(t^{1+\alpha }\sin \left(\pi x_1\right)\sin \left(\pi x_2\right)\right)}^3-t^{1+\alpha }\sin \left(\pi x_1\right)\sin \left(\pi x_2\right), \forall \mathit{x}=(x_1,x_2)\in \Omega .\hfill \end{array}$$

The corresponding exact solutions for u and $\sigma$ are as follows:

$$\begin{array}{cc}\hfill & u(\mathit{x},t)=t^{1+\alpha }\sin \left(\pi x_1\right)\sin \left(\pi x_2\right),\hfill \\ \hfill & \sigma (\mathit{x},t)=-2{\pi }^2{t}^{1+\alpha }\sin \left(\pi x_1\right)\sin \left(\pi x_2\right).\hfill \end{array}$$

We first point out that the exact solutions here are dependent on the fractional parameter $\alpha$ and have lower regularity than in Example 1. Now, for different fractional parameters $\alpha =0.2, 0.4, 0.6, 0.8$, we carried out the numerical tests with time and space grid parameters $h=\sqrt{2}\tau$ as in Example 1. We also observed that the convergence orders for u in discrete $L^{\infty }\left(L^2(\Omega )\right)$ norm are close to 2 (

Table 5. Error behaviors for u in discrete $L^{\infty }\left(L^2\right)$ norm with $h=\sqrt{2}\tau$ in Example 2.

$\mathit{\alpha }$	$(\mathit{h},\mathit{\tau })$	$(\frac{\sqrt{2}}{10},\frac{1}{10})$	$(\frac{\sqrt{2}}{20},\frac{1}{20})$	$(\frac{\sqrt{2}}{40},\frac{1}{40})$	$(\frac{\sqrt{2}}{80},\frac{1}{80})$
0.2	${\parallel u-u_h\parallel }_{{\tilde{L}}^{\infty }\left(L^2\right)}$	$1.471101\times {10}^{-2}$	$3.723406\times {10}^{-3}$	$9.336329\times {10}^{-4}$	$2.335700\times {10}^{-4}$
	Order		1.982202	1.995696	1.999001
0.4	${\parallel u-u_h\parallel }_{{\tilde{L}}^{\infty }\left(L^2\right)}$	$1.465757\times {10}^{-2}$	$3.708416\times {10}^{-3}$	$9.297074\times {10}^{-4}$	$2.325687\times {10}^{-4}$
	Order		1.982771	1.995955	1.999120
0.6	${\parallel u-u_h\parallel }_{{\tilde{L}}^{\infty }\left(L^2\right)}$	$1.459159\times {10}^{-2}$	$3.689962\times {10}^{-3}$	$9.248730\times {10}^{-4}$	$2.313357\times {10}^{-4}$
	Order		1.983459	1.996279	1.999267
0.8	${\parallel u-u_h\parallel }_{{\tilde{L}}^{\infty }\left(L^2\right)}$	$1.451090\times {10}^{-2}$	$3.667521\times {10}^{-3}$	$9.189935\times {10}^{-4}$	$2.298343\times {10}^{-4}$
	Order		1.984260	1.996679	1.999461

), and that for u in discrete $L^2\left(H^1(\Omega )\right)$ norm and for $\sigma$ in discrete $L^2\left(L^2(\Omega )\right)$, the convergence orders are close to 1 (

Table 6. Error behaviors for u in discrete $L^2\left(H^1\right)$ norm with $h=\sqrt{2}\tau$ in Example 2.

$\mathit{\alpha }$	$(\mathit{h},\mathit{\tau })$	$(\frac{\sqrt{2}}{10},\frac{1}{10})$	$(\frac{\sqrt{2}}{20},\frac{1}{20})$	$(\frac{\sqrt{2}}{40},\frac{1}{40})$	$(\frac{\sqrt{2}}{80},\frac{1}{80})$
0.2	${\parallel u-u_h\parallel }_{{\tilde{L}}^2\left(H^1\right)}$	$1.929246\times {10}^{-1}$	$9.292205\times {10}^{-2}$	$4.554283\times {10}^{-2}$	$2.253807\times {10}^{-2}$
	Order		1.053944	1.028797	1.014860
0.4	${\parallel u-u_h\parallel }_{{\tilde{L}}^2\left(H^1\right)}$	$1.841831\times {10}^{-1}$	$8.831872\times {10}^{-2}$	$4.318488\times {10}^{-2}$	$2.134527\times {10}^{-2}$
	Order		1.060349	1.032193	1.016610
0.6	${\parallel u-u_h\parallel }_{{\tilde{L}}^2\left(H^1\right)}$	$1.768079\times {10}^{-1}$	$8.441087\times {10}^{-2}$	$4.117758\times {10}^{-2}$	$2.032851\times {10}^{-2}$
	Order		1.066682	1.035570	1.018355
0.8	${\parallel u-u_h\parallel }_{{\tilde{L}}^2\left(H^1\right)}$	$1.704919\times {10}^{-1}$	$8.104293\times {10}^{-2}$	$3.944269\times {10}^{-2}$	$1.944856\times {10}^{-2}$
	Order		1.072945	1.038928	1.020094

and

Table 7. Error behaviors for $\sigma$ in discrete $L^2\left(L^2\right)$ norm with $h=\sqrt{2}\tau$ in Example 2.

$\mathit{\alpha }$	$(\mathit{h},\mathit{\tau })$	$(\frac{\sqrt{2}}{10},\frac{1}{10})$	$(\frac{\sqrt{2}}{20},\frac{1}{20})$	$(\frac{\sqrt{2}}{40},\frac{1}{40})$	$(\frac{\sqrt{2}}{80},\frac{1}{80})$
0.2	${\parallel \sigma -{\sigma }_h\parallel }_{{\tilde{L}}^2\left(L^2\right)}$	$1.011536\times {10}^{-1}$	$2.423441\times {10}^{-2}$	$5.935657\times {10}^{-3}$	$1.475471\times {10}^{-3}$
	Order		2.061419	2.029577	2.008232
0.4	${\parallel \sigma -{\sigma }_h\parallel }_{{\tilde{L}}^2\left(L^2\right)}$	$9.600324\times {10}^{-2}$	$2.288820\times {10}^{-2}$	$5.591883\times {10}^{-3}$	$1.386020\times {10}^{-3}$
	Order		2.068479	2.033198	2.012386
0.6	${\parallel \sigma -{\sigma }_h\parallel }_{{\tilde{L}}^2\left(L^2\right)}$	$9.148462\times {10}^{-2}$	$2.170218\times {10}^{-2}$	$5.286904\times {10}^{-3}$	$1.306137\times {10}^{-3}$
	Order		2.075689	2.037345	2.017116
0.8	${\parallel \sigma -{\sigma }_h\parallel }_{{\tilde{L}}^2\left(L^2\right)}$	$8.743685\times {10}^{-2}$	$2.063873\times {10}^{-2}$	$5.014040\times {10}^{-3}$	$1.236173\times {10}^{-3}$
	Order		2.082887	2.041309	2.020093

). Otherwise, for the sufficiently small and large parameters $\alpha =0.001$ and $0.999$, we also carried out experiments and obtained the same conclusions as in Example 1 and observed that the convergence orders are independent of the fractional parameter $\alpha$ (

Table 8. Error behaviors with $h=\sqrt{2}\tau$ in Example 2.

$\mathit{\alpha }$	$(\mathit{h},\mathit{\tau })$	$(\frac{\sqrt{2}}{10},\frac{1}{10})$	$(\frac{\sqrt{2}}{20},\frac{1}{20})$	$(\frac{\sqrt{2}}{40},\frac{1}{40})$	$(\frac{\sqrt{2}}{80},\frac{1}{80})$
0.001	${\parallel u-u_h\parallel }_{{\tilde{L}}^{\infty }\left(L^2\right)}$	$1.475184\times {10}^{-2}$	$3.734884\times {10}^{-3}$	$9.366412\times {10}^{-4}$	$2.343395\times {10}^{-4}$
	Order		1.981760	1.995495	1.998896
	${\parallel u-u_h\parallel }_{{\tilde{L}}^2\left(H^1\right)}$	$2.034201\times {10}^{-1}$	$9.841665\times {10}^{-2}$	$4.834960\times {10}^{-2}$	$2.395604\times {10}^{-2}$
	Order		1.047488	1.025398	1.013114
	${\parallel \sigma -{\sigma }_h\parallel }_{{\tilde{L}}^2\left(L^2\right)}$	$1.071138\times {10}^{-1}$	$2.578425\times {10}^{-2}$	$6.325435\times {10}^{-3}$	$1.566549\times {10}^{-3}$
	Order		2.054582	2.027253	2.013575
0.999	${\parallel u-u_h\parallel }_{{\tilde{L}}^{\infty }\left(L^2\right)}$	$1.441263\times {10}^{-2}$	$3.640531\times {10}^{-3}$	$9.119422\times {10}^{-4}$	$2.280341\times {10}^{-4}$
	Order		1.985113	1.997135	1.999693
	${\parallel u-u_h\parallel }_{{\tilde{L}}^2\left(H^1\right)}$	$1.650419\times {10}^{-1}$	$7.811777\times {10}^{-2}$	$3.793154\times {10}^{-2}$	$1.868106\times {10}^{-2}$
	Order		1.079110	1.042253	1.021821
	${\parallel \sigma -{\sigma }_h\parallel }_{{\tilde{L}}^2\left(L^2\right)}$	$8.377340\times {10}^{-2}$	$1.967926\times {10}^{-2}$	$4.768831\times {10}^{-3}$	$1.173929\times {10}^{-3}$
	Order		2.089816	2.044969	2.022290

), which is consistent with the truncation error of the WSGD formula. Finally, from the above numerical experiments, we can see that the linearized MFVE method for the time fractional FORD models with the nonlinear term is feasible and effective.

Example 3.

Consider the following nonlinear time fractional FORD model:

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}\frac{\partial u(\mathit{x},t)}{\partial t}-\frac{{\partial }^{\alpha }\Delta u(\mathit{x},t)}{\partial t^{\alpha }}-\Delta u(\mathit{x},t)+{\Delta }^{2}u(\mathit{x},t)+\sin \left(u(\mathit{x},t)\right)=0,\hfill & (\mathit{x},t)\in \Omega \times J,\hfill \\ u(\mathit{x},t)=\Delta u(\mathit{x},t)=0,\hfill & (\mathit{x},t)\in \partial \Omega \times J,\hfill \\ u(\mathit{x},0)=u_0\left(\mathit{x}\right),\hfill & \mathit{x}\in \Omega ,\hfill \end{array}\right.\end{array}$$

where $\Omega ={(-1,1)}^2$, $J=(0,1]$, and the initial data for  $u_0\left(\mathit{x}\right)$ are as follows:

$$\begin{array}{c}\hfill u_0\left(\mathit{x}\right)=\left\{\begin{array}{cc}k\exp (-1/(r^2-x_1^2-x_2^2)),\hfill & \mathrm{if} x_1^2+x_2^2<r^2,\hfill \\ 0,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.\end{array}$$

Then, we can obtain the corresponding ${\sigma }_0\left(\mathit{x}\right)=\Delta u_0\left(\mathit{x}\right)$.

In this example, it should be pointed out that the source function $f(\mathit{x},t)=0$, and it is difficult to obtain the exact solutions. Therefore, we mainly observe the change in numerical solutions at the initial and final moments and the influence of fractional parameters $\alpha$. In order to conduct this experiment, we considered $k=100$ and $r=0.35$ in the initial data, and the spatial and temporal step sizes $h=\sqrt{2}/50$ and $\tau =1/100$. In Figure 2, we show the projections of the initial data $u_0\left(\mathit{x}\right)$ and ${\sigma }_0\left(\mathit{x}\right)$ to depict the initial state, respectively. Now, we take the fractional parameters $\alpha =0.001$ and $0.999$, and show the projections of the numerical solutions u and $\sigma$ at $t=1$ in Figure 3 and Figure 4, respectively. It is easy to see that fractional parameters have a significant impact on the reaction–diffusion process.

7. Conclusions

In this paper, we design a time second-order linearized MFVE scheme to solve nonlinear time fractional FORD equations by using the BDF2 and WSGD formulas. This scheme harnesses the advantages of the FVE method to simultaneously compute two physical quantities with good accuracy. We give the proof of the existence and uniqueness of the discrete solution by using the matrix theories, derive the unconditional stability result in detail, and obtain the optimal error estimates for u and $\sigma$, where the temporal discretization achieves second-order accuracy unrelated to the fractional parameter $\alpha$. Furthermore, we present some numerical results to validate the theoretical analysis, thereby demonstrating the advantages of the WSGD formula. In future work, we will consider combining the FVE method with some fast computational techniques to solve more fractional PDEs.