A Mixed Element Algorithm Based on the Modified L1 Crank–Nicolson Scheme for a Nonlinear Fourth-Order Fractional Diffusion-Wave Model

Abstract

In this article, a new mixed finite element (MFE) algorithm is presented and developed to find the numerical solution of a two-dimensional nonlinear fourth-order Riemann–Liouville fractional diffusion-wave equation. By introducing two auxiliary variables and using a particular technique, a new coupled system with three equations is constructed. Compared to the previous space–time high-order model, the derived system is a lower coupled equation with lower time derivatives and second-order space derivatives, which can be approximated by using many time discrete schemes. Here, the second-order Crank–Nicolson scheme with the modified $L1$-formula is used to approximate the time direction, while the space direction is approximated by the new MFE method. Analyses of the stability and optimal $L^2$ error estimates are performed and the feasibility is validated by the calculated data.

1. Introduction

Fourth-order fractional partial differential equations (PDEs) including fourth-order fractional subdiffusion models [1,2,3] and fourth-order fractional diffusion-wave models [2,4,5] can be founded in many fields of science and engineering. Thus far, there have been many efficient numerical algorithms for solving linear or nonlinear fourth-order fractional subdiffusion and diffusion-wave models. Liu et al. [6], Liu et al. [7], and Liu et al. [8] considered different mixed element methods to solve fourth-order nonlinear fractional subdiffusion models with the first-order time derivative and developed numerical theories including stability and convergence. Liu et al. [3] introduced a mixed element algorithm with a new approximation of the fractional derivative. Ji et al. [9], Ran et al. [10], Nandal and Pandey [11], Sun et al. [12], and Huang et al. [13] considered some difference schemes for linear or nonlinear fourth-order fractional diffusion or diffusion-wave models. Abbaszadeh and Dehghan [14] studied the direct meshless local Petrov–Galerkin method for solving fourth-order reaction-diffusion problems with a time-fractional derivative. Yang et al. [15] and Zhang et al. [16] found the numerical solutions for a fourth-order fractional model by using the orthogonal spline collocation method. Tariq and Akram [17] considered a quintic spline technique to solve a fourth-order time-fractional subdiffusion model. Guo et al. [18] and Du et al. [19] studied the LDG methods for solving some time-fractional subdiffusion models with fourth-order spatial derivative terms, respectively. In [1], Nikan et al. developed a local radial basis function generated by the finite difference scheme for a time-fractional fourth-order reaction-diffusion model. In [5], Jafari et al. solved a fourth-order fractional diffusion-wave equation by the decomposition method. Hu and Zhang [2] implemented numerical calculations via finite difference methods for fourth-order time fractional subdiffusion and diffusion-wave models. Li and Wong [20] developed an efficient numerical algorithm for a fourth-order time-fractional diffusion-wave model.

Here, we propose a new mixed element algorithm to solve the following nonlinear fourth-order time-fractional diffusion-wave model:

$$\left\{\begin{array}{c}\frac{{\partial }^{2}u}{\partial t^2}+\frac{{\partial }_{RL}^{\beta }u}{\partial t^{\beta }}+\frac{\partial u}{\partial t}+{▵}^{2}u-▵f\left(u\right)=g(\mathbf{x},t),(\mathbf{x},t)\in \mathsf{\Omega }\times J,\hfill \\ u(\mathbf{x},t)=\Delta u(\mathbf{x},t)=0,t\in \overline{J},\hfill \\ u(\mathbf{x},0)=0,\frac{\partial u}{\partial t}(\mathbf{x},0)=u_1\left(\mathbf{x}\right),\mathbf{x}\in \overline{\mathsf{\Omega }},\hfill \end{array}\right.$$

(1)

where $\mathsf{\Omega }\subset R^d(d\le 2)$ and $J=(0,T]$ with $0<T<\infty$ are the spatial domain and time interval, respectively. $u_1\left(\mathbf{x}\right)$ is an initial value function, $g(\mathbf{x},t)$ is a given source term, $f\left(u\right)$ is a polynomial function or bounded function on u satisfying $f\in C^2\left(R\right)$, and the Riemann–Liouville fractional derivative is defined by

$$\begin{array}{c}\frac{{\partial }_{RL}^{\beta }u}{\partial t^{\beta }}=\frac{1}{\mathsf{\Gamma }(2-\beta )}\frac{{\partial }^2}{\partial t^2}{\int }_0^t\frac{u\left(s\right)ds}{{(t-s)}^{\beta -1}},1<\beta <2,\hfill \end{array}$$

(2)

where the nonlinear fourth-order fractional diffusion-wave model (1) can be generated by the classical fourth-order hyperbolic wave equation. When $\beta \to 1$ or 2 and $f\left(u\right)=u^3-u$, the model (1) can be reduced to an important Cahn–Hilliard equation model [21].

Recently, Zeng and Li [22] developed a new Crank–Nicolson scheme based on a modified $L1$-formula, whose coefficients are different from the famous $L1$-formula (see [23,24] for the fractional parameter $\alpha \in (0,1)$). One should note that this modified $L1$-formula can only approximate the Caputo or Riemann–Liouville fractional derivative with parameter $\alpha \in (0,1)$, and it cannot approximate the case $\beta \in (1,2)$. Here, we will develop the modified $L1$-formula for the case of $\beta \in (1,2)$ by using some techniques.

In this article, by introducing two auxiliary functions and using some techniques, we propose a new mixed element algorithm. Here, our major contributions are as follows: (1) by the introduction of two auxiliary functions, we reduce the nonlinear fourth-order time-fractional diffusion-wave model to a low-order coupled system; (2) we turn order $\beta \in (1,2)$ into order $\alpha \in (0,1)$ for the Riemann–Liouville fractional derivative; (3) we approximate the derived coupled system with a fractional derivative with order $\alpha \in (0,1)$ by the modified $L1$ Crank–Nicolson scheme with the developed new mixed element method; (4) we derive the stability of the new mixed element scheme and optimal error estimates in the $L^2$-norm for three functions.

The structure of this article is as follows: in Section 2, we provide some numerical approximation formulas, propose a new mixed element scheme, and prove the stability of the derived scheme; in Section 3, we derive optimal error estimates for three variables; in Section 4, some numerical data are computed and discussed; Finally, in Section 5, we give some concluding remarks.

2. Numerical Approximation and Stability

Based on the relation between the Riemann–Liouville fractional derivative and Caputo fractional derivative, we take $\alpha =\beta -1$ and $v=\frac{\partial u}{\partial t}$ to obtain

$$\begin{array}{cc}\frac{{\partial }_{RL}^{\beta }u}{\partial t^{\beta }}\hfill & =\frac{1}{\mathsf{\Gamma }(2-\beta )}{\int }_0^t\frac{\frac{{\partial }^{2}u\left(s\right)}{\partial s^2}ds}{{(t-s)}^{\beta -1}}+\frac{u\left(0\right)}{\mathsf{\Gamma }(1-\beta )}{t}^{-\beta }+\frac{\frac{\partial u\left(0\right)}{\partial t}}{\mathsf{\Gamma }(2-\beta )}{t}^{1-\beta }\hfill \\ & =\frac{1}{\mathsf{\Gamma }(1-\alpha )}{\int }_0^t\frac{\frac{\partial v}{\partial s}ds}{{(t-s)}^{\alpha }}+\frac{v\left(0\right)}{\mathsf{\Gamma }(1-\alpha )}{t}^{-\alpha }\hfill \\ & =\frac{{\partial }_{RL}^{\alpha }v}{\partial t^{\alpha }},0<\alpha <1.\hfill \end{array}$$

(3)

Let $\sigma =▵u-f\left(u\right)$; (1) can be rewritten as the following coupled system:

$$\left\{\begin{array}{c}v=\frac{\partial u}{\partial t},(\mathbf{x},t)\in \mathsf{\Omega }\times J,\hfill \\ \frac{\partial \sigma }{\partial t}=▵v-f_u\left(u\right)v,(\mathbf{x},t)\in \mathsf{\Omega }\times J,\hfill \\ \frac{\partial v}{\partial t}+\frac{{\partial }_{RL}^{\alpha }v}{\partial t^{\alpha }}+v+▵\sigma =g(\mathbf{x},t),(\mathbf{x},t)\in \mathsf{\Omega }\times J.\hfill \end{array}\right.$$

(4)

For formulating the fully discrete scheme, we insert the nodes $t_n=n\Delta t(n=0,1,2,\cdots ,N)$ in time interval $[0,T]$, where $t_n$ satisfy $0=t_0<t_1<t_2<\cdots <t_N=T$ and the time step length size $\Delta t=T/N$, for some positive integer N. For a smooth function $\varphi$ defined on the time interval $[0,T]$, we denote ${\varphi }^n=\varphi \left(t_n\right)$.

Now, we need to introduce some lemmas on integer and fractional derivatives.

Lemma 1.

At $t_{k+\frac{1}{2}}$, the following relation holds:

$$\begin{array}{cc}\frac{\partial \varphi }{\partial t}(t_{k+\frac{1}{2}})\hfill & =\frac{{\varphi }^{k+1}-{\varphi }^k}{\Delta t}+O(\Delta t^2)\hfill \\ & \triangleq P_{\Delta t}{\varphi }^{k+\frac{1}{2}}+O(\Delta t^2).\hfill \end{array}$$

(5)

Lemma 2.

At $t_{k+\frac{1}{2}}$, we have

$$\begin{array}{cc}\varphi (t_{k+\frac{1}{2}})\hfill & =\frac{{\varphi }^{k+1}+{\varphi }^k}{2}+O(\Delta t^2)\hfill \\ & \triangleq {\varphi }^{k+\frac{1}{2}}+O(\Delta t^2).\hfill \end{array}$$

(6)

Lemma 3.

([22]). At $t_{k+\frac{1}{2}}$, the Caputo fractional derivative has the following form:

$$\begin{array}{cc}\frac{{\partial }_C^{\alpha }\varphi }{\partial t^{\alpha }}(t_{k+\frac{1}{2}})\hfill & =\frac{1}{\mathsf{\Gamma }(1-\alpha )}{\int }_0^{t_{k+\frac{1}{2}}}\frac{\frac{\partial \varphi }{\partial s}ds}{{(t_{k+\frac{1}{2}}-s)}^{\alpha }}\hfill \\ & =\frac{\Delta t^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}[a_0\varphi (t_{k+\frac{1}{2}})-{\displaystyle \sum _{j=1}^k}(a_{k-j}-a_{k-j+1})\varphi (t_{j-\frac{1}{2}})\hfill \\ & -(a_k-b_k)\varphi (t_{\frac{1}{2}})-b_k\varphi \left(t_0\right)]+O(\Delta t^{2-\alpha }),\hfill \end{array}$$

(7)

for $k\ge 0$, we have

$$\begin{array}{c}{b}_k=2[{(k+\frac{1}{2})}^{1-\alpha }-k^{1-\alpha }],a_{k-j}=[{(k-j+1)}^{1-\alpha }-{(k-j)}^{1-\alpha }].\hfill \end{array}$$

(8)

By Lemma 3, we have

Lemma 4.

([22]). At $t_{k+\frac{1}{2}}$, the Riemann–Liouville fractional derivative has the following approximation:

$$\begin{array}{cc}\frac{{\partial }_{RL}^{\alpha }\varphi }{\partial t^{\alpha }}(t_{k+\frac{1}{2}})=\hfill & \frac{\Delta t^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}[a_0\varphi (t_{k+\frac{1}{2}})-{\displaystyle {\displaystyle \sum _{j=1}^k}}(a_{k-j}-a_{k-j+1})\varphi (t_{j-\frac{1}{2}})\hfill \\ & -(a_k-b_k)\varphi (t_{\frac{1}{2}})-{\widehat{b}}_k\varphi \left(t_0\right)]+O(\Delta t^{2-\alpha }),\hfill \end{array}$$

(9)

where ${\widehat{b}}_k=b_k-(1-\alpha ){(k+\frac{1}{2})}^{-\alpha }$.

Remark 1.

In [22], the authors provided the $L1$-formula above, which is different from the usual $L1$-formula and called the modified $L1$-formula.

By the approximation scheme above, we arrive at

$$\left\{\begin{array}{cc}\left(a\right)\hfill & P_{\Delta t}{u}^{n+\frac{1}{2}}=v^{n+\frac{1}{2}}+R_1^{n+\frac{1}{2}},\hfill \\ \left(b\right)\hfill & P_{\Delta t}{\sigma }^{n+\frac{1}{2}}=▵v^{n+\frac{1}{2}}-\frac{f_u\left(u^{n+1}\right)v^{n+1}+f_u\left(u^n\right)v^n}{2}+R_2^{n+\frac{1}{2}},\hfill \\ \left(c\right)\hfill & P_{\Delta t}{v}^{n+\frac{1}{2}}+\frac{\Delta t^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}[a_0{v}^{n+\frac{1}{2}}-{\displaystyle {\displaystyle \sum _{j=1}^n}}(a_{n-j}-a_{n-j+1})v^{j-\frac{1}{2}}\hfill \\ & -(a_n-b_n)v^{\frac{1}{2}}-{\widehat{b}}_n{v}^0]+v^{n+\frac{1}{2}}+▵{\sigma }^{n+\frac{1}{2}}=g(\mathbf{x},t_{n+\frac{1}{2}})+R_3^{n+\frac{1}{2}},\hfill \end{array}\right.$$

(10)

where

$$\begin{array}{cc}\hfill & R_1^{n+\frac{1}{2}}=P_{\Delta t}{u}^{n+\frac{1}{2}}-\frac{\partial u}{\partial t}(t_{n+\frac{1}{2}})+(v(t_{n+\frac{1}{2}})-v^{n+\frac{1}{2}})=O(\Delta t^2),\hfill \\ & R_2^{n+\frac{1}{2}}=P_{\Delta t}{\sigma }^{n+\frac{1}{2}}-\frac{\partial \sigma }{\partial t}(t_{n+\frac{1}{2}})+(▵v(t_{n+\frac{1}{2}})-▵v^{n+\frac{1}{2}})\hfill \\ & +f_u(u(t_{n+\frac{1}{2}}))v(t_{n+\frac{1}{2}})-\frac{f_u\left(u^{n+1}\right)v^{n+1}+f_u\left(u^n\right)v^n}{2}=O(\Delta t^2),\hfill \\ & R_3^{n+\frac{1}{2}}=P_{\Delta t}{v}^{n+\frac{1}{2}}-\frac{\partial v}{\partial t}(t_{n+\frac{1}{2}})+O(\Delta t^{2-\alpha })+(v^{n+\frac{1}{2}}-v(t_{n+\frac{1}{2}}))\hfill \\ & +(▵{\sigma }^{n+\frac{1}{2}}-▵\sigma (t_{n+\frac{1}{2}}))=O(\Delta t^{2-\alpha }).\hfill \end{array}$$

For $(\phi ,\psi ,\chi )\in L^2\times H_0^1\times H_0^1$, we have the following mixed weak formulation:

$$\left\{\begin{array}{cc}\left(a\right)\hfill & (P_{\Delta t}{u}^{n+\frac{1}{2}},\phi )=(v^{n+\frac{1}{2}},\phi )+(R_1^{n+\frac{1}{2}},\phi ),\hfill \\ \left(b\right)\hfill & (P_{\Delta t}{\sigma }^{n+\frac{1}{2}},\psi )+(\nabla v^{n+\frac{1}{2}},\nabla \psi )+(\frac{f_u\left(u^{n+1}\right)v^{n+1}+f_u\left(u^n\right)v^n}{2},\psi )=(R_2^{n+\frac{1}{2}},\psi ),\hfill \\ \left(c\right)\hfill & (P_{\Delta t}{v}^{n+\frac{1}{2}},\chi )+\left(\frac{\Delta t^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}\right[a_0{v}^{n+\frac{1}{2}}-{\displaystyle {\displaystyle \sum _{j=1}^n}}(a_{n-j}-a_{n-j+1})v^{j-\frac{1}{2}}\hfill \\ \hfill -& (a_n-b_n)v^{\frac{1}{2}}-{\widehat{b}}_n{v}^0],\chi )+(v^{n+\frac{1}{2}},\chi )-(\nabla {\sigma }^{n+\frac{1}{2}},\nabla \chi )=(g(\mathbf{x},t_{n+\frac{1}{2}}),\chi )+(R_3^{n+\frac{1}{2}},\chi ).\hfill \end{array}\right.$$

(11)

For $({\phi }_h,{\psi }_h,{\chi }_h)\in L_h\times V_h\times V_h\subset L^2\times H_0^1\times H_0^1$, based on the mixed weak formulation above, we formulate the following new mixed element system:

$$\left\{\begin{array}{cc}\left(a\right)\hfill & (P_{\Delta t}{u}_h^{n+\frac{1}{2}},{\phi }_h)=(v_h^{n+\frac{1}{2}},{\phi }_h),\hfill \\ \left(b\right)\hfill & (P_{\Delta t}{\sigma }_h^{n+\frac{1}{2}},{\psi }_h)+(\nabla v_h^{n+\frac{1}{2}},\nabla {\psi }_h)+(\frac{f_u\left(u_h^{n+1}\right)v_h^{n+1}+f_u\left(u_h^n\right)v_h^n}{2},{\psi }_h)=0,\hfill \\ \left(c\right)\hfill & (P_{\Delta t}{v}_h^{n+\frac{1}{2}},{\chi }_h)+\left(\frac{\Delta t^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}\right[a_0{v}_h^{n+\frac{1}{2}}-{\displaystyle {\displaystyle \sum _{j=1}^n}}(a_{n-j}-a_{n-j+1})v_h^{j-\frac{1}{2}}\hfill \\ & -(a_n-b_n)v_h^{\frac{1}{2}}-{\widehat{b}}_n{v}_h^0],{\chi }_h)+(v_h^{n+\frac{1}{2}},{\chi }_h)-(\nabla {\sigma }_h^{n+\frac{1}{2}},\nabla {\chi }_h)=(g(\mathbf{x},t_{n+\frac{1}{2}}),{\chi }_h).\hfill \end{array}\right.$$

(12)

Lemma 5.

(See [22]). For ${\widehat{b}}_{n-j}$, the following important inequality holds:

$$\frac{\Delta t^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )}{\displaystyle \sum _{j=0}^{n-1}}{\widehat{b}}_{n-j}\le \frac{C{T}^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )},$$

(13)

where C is a positive constant that is independent of space–time step length sizes h and $\Delta t$.

Proof. 

Applying the Taylor formula, we have

$$\begin{array}{cc}\hfill & \frac{\Delta t^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )}{\displaystyle \sum _{j=0}^{n-1}}{\widehat{b}}_{n-j}\hfill \\ =\hfill & \frac{\Delta t^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )}{\displaystyle \sum _{j=0}^{n-1}}[2{(n-j+\frac{1}{2})}^{1-\alpha }-2{(n-j)}^{1-\alpha }-(1-\alpha ){(n-j+\frac{1}{2})}^{-\alpha }]\hfill \\ =\hfill & \frac{2\Delta t^{1-\alpha }}{\Gamma (2-\alpha )}{\displaystyle \sum _{j=0}^{n-1}}{(n-j)}^{1-\alpha }[{(1+\frac{1}{2(n-j)})}^{1-\alpha }-1-\frac{1-\alpha }{2(n-j+\frac{1}{2})}{(1+\frac{1}{2(n-j)})}^{1-\alpha }]\hfill \\ =\hfill & \frac{\Delta t^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )}{\displaystyle \sum _{j=0}^{n-1}}{(n-j)}^{1-\alpha }[\frac{1-\alpha }{2(n-j)}+\frac{(1-\alpha )\alpha }{2!}(1+\kappa \frac{1}{2(n-j)}{)}^{1-\alpha }\frac{1}{4{(n-j)}^2}\hfill \\ & -\frac{1-\alpha }{2(n-j+\frac{1}{2})}(1+\frac{1-\alpha }{2(n-j)}+\frac{(1-\alpha )\alpha }{2!}(1+\kappa \frac{1}{2(n-j)}{)}^{1-\alpha }\frac{1}{4{(n-j)}^2})]\hfill \\ =\hfill & \frac{\Delta t^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )}{\displaystyle \sum _{j=0}^{n-1}}{(n-j)}^{1-\alpha }[\frac{1-\alpha }{2(n-j)}-\frac{1-\alpha }{2(n-j+\frac{1}{2})}+O(\frac{1}{{(n-j)}^2})]\hfill \\ =\hfill & \frac{\Delta t^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )}{\displaystyle \sum _{j=0}^{n-1}}{(n-j)}^{1-\alpha }[\frac{1-\alpha }{2(n-j+\frac{1}{2})(n-j)}+O(\frac{1}{{(n-j)}^2})].\hfill \end{array}$$

(14)

Noting that $\Delta t=\frac{T}{N}$ and $n-j\le N$, we have

$$\begin{array}{cc}\hfill & \frac{\Delta t^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )}{\displaystyle \sum _{j=0}^{n-1}}{(n-j)}^{1-\alpha }[\frac{1-\alpha }{2(n-j+\frac{1}{2})(n-j)}+O(\frac{1}{{(n-j)}^2})]\hfill \\ =\hfill & \frac{1}{\mathsf{\Gamma }(2-\alpha )}{\displaystyle \sum _{j=0}^{n-1}}{T}^{1-\alpha }{(\frac{n-j}{N})}^{1-\alpha }[\frac{1-\alpha }{2(n-j+\frac{1}{2})(n-j)}+O(\frac{1}{{(n-j)}^2})]\hfill \\ \le \hfill & \frac{T^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )}{\displaystyle \sum _{j=0}^{n-1}}[\frac{1}{{(n-j)}^2}+O(\frac{1}{{(n-j)}^2})]\le \frac{C{T}^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )}{\displaystyle \sum _{n=1}^{+\infty }}\frac{1}{n^2}\le \frac{C{T}^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )}.\hfill \end{array}$$

(15)

Substitute (15) into (14) to obtain the conclusion. □

Next, we will prove the stability.

Theorem 1.

For $n\ge 0$, the stability for the fully discrete system (12) holds:

$$\begin{array}{cc}\hfill & \left(a\right).\parallel v_h^{n+1}\parallel +\parallel {\sigma }_h^{n+1}\parallel +(\frac{\Delta t^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )}{\displaystyle \sum _{j=1}^{n+1}}{a}_{n-j+1}{\parallel v_h^{j-\frac{1}{2}}\parallel }^2{)}^{\frac{1}{2}}\hfill \\ \le \hfill & C(\parallel v_h^0\parallel +\parallel {\sigma }_h^0\parallel +{\displaystyle \underset{0\le j\le n}{max}}\{\parallel g(\mathbf{x},t_{j+\frac{1}{2}})\parallel \}),\hfill \\ \hfill & \left(b\right).\parallel u_h^{n+1}\parallel \le C(\parallel u_h^0\parallel +\parallel v_h^0\parallel +\parallel {\sigma }_h^0\parallel +{\displaystyle \underset{0\le j\le n}{max}}\{\parallel g(\mathbf{x},t_{j+\frac{1}{2}})\parallel \}).\hfill \end{array}$$

(16)

Proof. 

In (12) $\left(a\right)$, we take ${\phi }_h=u_h^{n+\frac{1}{2}}$, and use Cauchy–Schwarz inequality as well as Young inequality to obtain

$$\begin{array}{cc}\frac{1}{2}(\parallel v_h^{n+\frac{1}{2}}{\parallel }^2+\parallel u_h^{n+\frac{1}{2}}{\parallel }^2)\ge (v_h^{n+\frac{1}{2}},u_h^{n+\frac{1}{2}})\hfill & =(P_{\Delta t}{u}_h^{n+\frac{1}{2}},u_h^{n+\frac{1}{2}})\hfill \\ & \ge \frac{\parallel u_h^{n+1}{\parallel }^2-{\parallel u_h^n\parallel }^2}{2\Delta t}.\hfill \end{array}$$

(17)

In (12) $\left(b\right)$, set ${\psi }_h={\sigma }_h^{n+\frac{1}{2}}$ and make use of Cauchy–Schwarz inequality to arrive at

$$\begin{array}{cc}\hfill & (\nabla v_h^{n+\frac{1}{2}},\nabla {\sigma }_h^{n+\frac{1}{2}})\hfill \\ =\hfill & -(P_{\Delta t}{\sigma }_h^{n+\frac{1}{2}},{\sigma }_h^{n+\frac{1}{2}})-(\frac{f_u\left(u_h^{n+1}\right)v_h^{n+1}+f_u\left(u_h^n\right)v_h^n}{2},{\sigma }_h^{n+\frac{1}{2}})\hfill \\ \le \hfill & -\frac{\parallel {\sigma }_h^{n+1}{\parallel }^2-{\parallel {\sigma }_h^n\parallel }^2}{2\Delta t}+\frac{1}{2}(\parallel f_u\left(u_h^{n+1}\right){\parallel }_{\infty }\parallel v_h^{n+1}\parallel +\parallel f_u\left(u_h^n\right){\parallel }_{\infty }\parallel v_h^n\parallel )\parallel {\sigma }_h^{n+\frac{1}{2}}\parallel \hfill \\ \le \hfill & -\frac{\parallel {\sigma }_h^{n+1}{\parallel }^2-{\parallel {\sigma }_h^{n+1}\parallel }^2}{2\Delta t}+C(\parallel v_h^{n+1}{\parallel }^2+\parallel v_h^n{\parallel }^2+\parallel {\sigma }_h^{n+1}{\parallel }^2+\parallel {\sigma }_h^n{\parallel }^2).\hfill \end{array}$$

(18)

In (12) $\left(c\right)$, set ${\chi }_h=v_h^{n+\frac{1}{2}}$ and use Cauchy–Schwarz inequality to obtain

$$\begin{array}{cc}\hfill & -(\nabla {\sigma }_h^{n+\frac{1}{2}},\nabla v_h^{n+\frac{1}{2}})\hfill \\ =\hfill & -(P_{\Delta t}{v}_h^{n+\frac{1}{2}},v_h^{n+\frac{1}{2}})-\left(\frac{\Delta t^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}\right[a_0{v}_h^{n+\frac{1}{2}}-{\displaystyle \sum _{j=1}^n}(a_{n-j}-a_{n-j+1})v_h^{j-\frac{1}{2}}\hfill \\ & -(a_n-b_n)v_h^{\frac{1}{2}}-b_n{v}_h^0],v_h^{n+\frac{1}{2}})-{\parallel v_h^{n+\frac{1}{2}}\parallel }^2+(g(\mathbf{x},t_{n+\frac{1}{2}}),v_h^{n+\frac{1}{2}})\hfill \\ \le \hfill & -\frac{\parallel v_h^{n+1}{\parallel }^2-{\parallel v_h^n\parallel }^2}{2\Delta t}-\left(\frac{\Delta t^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}\right[a_0{v}_h^{n+\frac{1}{2}}-{\displaystyle \sum _{j=1}^n}(a_{n-j}-a_{n-j+1})v_h^{j-\frac{1}{2}}\hfill \\ & -(a_n-b_n)v_h^{\frac{1}{2}}-{\widehat{b}}_n{v}_h^0],v_h^{n+\frac{1}{2}})-\frac{1}{2}\parallel v_h^{n+\frac{1}{2}}{\parallel }^2+\frac{1}{2}{\parallel g(\mathbf{x},t_{n+\frac{1}{2}})\parallel }^2.\hfill \end{array}$$

(19)

Add (18) and (19) to obtain

$$\begin{array}{cc}\hfill & \frac{\parallel v_h^{n+1}{\parallel }^2-{\parallel v_h^n\parallel }^2}{2\Delta t}+\frac{\parallel {\sigma }_h^{n+1}{\parallel }^2-{\parallel {\sigma }_h^n\parallel }^2}{2\Delta t}+\frac{1}{2}{\parallel v_h^{n+\frac{1}{2}}\parallel }^2\hfill \\ \le \hfill & -\left(\frac{\Delta t^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}\right[a_0{v}_h^{n+\frac{1}{2}}-{\displaystyle \sum _{j=1}^n}(a_{n-j}-a_{n-j+1})v_h^{j-\frac{1}{2}}-(a_n-b_n)v_h^{\frac{1}{2}}-{\widehat{b}}_n{v}_h^0],v_h^{n+\frac{1}{2}})\hfill \\ & +C(\parallel v_h^{n+1}{\parallel }^2+\parallel v_h^n{\parallel }^2+\parallel {\sigma }_h^{n+1}{\parallel }^2+\parallel {\sigma }_h^n{\parallel }^2+\parallel g(\mathbf{x},t_{n+\frac{1}{2}}){\parallel }^2).\hfill \end{array}$$

(20)

Refer to Lemma 4.2 in [22] to easily obtain

$$\begin{array}{cc}\hfill & -\left(\frac{\Delta t^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}\right[a_0{v}_h^{n+\frac{1}{2}}-{\displaystyle \sum _{j=1}^n}(a_{n-j}-a_{n-j+1})v_h^{j-\frac{1}{2}}-(a_n-b_n)v_h^{\frac{1}{2}}-b_n{v}_h^0],v_h^{n+\frac{1}{2}})\hfill \\ \le \hfill & \frac{\Delta t^{-\alpha }}{2\mathsf{\Gamma }(2-\alpha )}({\displaystyle \sum _{j=1}^n}{a}_{n-j}\parallel v_h^{j-\frac{1}{2}}{\parallel }^2-{\displaystyle \sum _{j=1}^{n+1}}{a}_{n-j+1}\parallel v_h^{j-\frac{1}{2}}{\parallel }^2+{\widehat{b}}_n{\parallel v_h^0\parallel }^2).\hfill \end{array}$$

(21)

Combine (20) with (21) to obtain

$$\begin{array}{cc}\hfill & \parallel v_h^{n+1}{\parallel }^2+\parallel {\sigma }_h^{n+1}{\parallel }^2+\Delta t\parallel v_h^{n+\frac{1}{2}}{\parallel }^2+\frac{\Delta t^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )}{\displaystyle \sum _{j=1}^{n+1}}{a}_{n-j+1}{\parallel v_h^{j-\frac{1}{2}}\parallel }^2\hfill \\ \le \hfill & \parallel v_h^n{\parallel }^2+\parallel {\sigma }_h^n{\parallel }^2+\frac{\Delta t^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )}{\displaystyle \sum _{j=1}^n}{a}_{n-j}\parallel v_h^{j-\frac{1}{2}}{\parallel }^2+\frac{\Delta t^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )}{\widehat{b}}_n{\parallel v_h^0\parallel }^2\hfill \\ & +C\Delta t(\parallel v_h^{n+1}{\parallel }^2+\parallel v_h^n{\parallel }^2+\parallel {\sigma }_h^{n+1}{\parallel }^2+\parallel {\sigma }_h^n{\parallel }^2+\parallel g(\mathbf{x},t_{n+\frac{1}{2}}){\parallel }^2).\hfill \end{array}$$

(22)

We denote

$$\begin{array}{cc}\hfill & \mathsf{\Xi }(v_h^{n+1},{\sigma }_h^{n+1})=\parallel v_h^{n+1}{\parallel }^2+\parallel {\sigma }_h^{n+1}{\parallel }^2+\frac{\Delta t^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )}{\displaystyle \sum _{j=1}^{n+1}}{a}_{n-j+1}{\parallel v_h^{j-\frac{1}{2}}\parallel }^2.\hfill \end{array}$$

(23)

Remove the non-negative term to obtain

$$\begin{array}{cc}\mathsf{\Xi }(v_h^{n+1},{\sigma }_h^{n+1})\hfill & \le \left(\frac{1+\Delta t}{1-\Delta t}\right)\mathsf{\Xi }(v_h^n,{\sigma }_h^n)+\frac{1}{1-\Delta t}\frac{\Delta t^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )}{\widehat{b}}_n\parallel v_h^0{\parallel }^2+\frac{C\Delta t}{1-\Delta t}{\parallel g(\mathbf{x},t_{n+\frac{1}{2}})\parallel }^2\hfill \\ & \le (\frac{1+\Delta t}{1-\Delta t}{)}^2\mathsf{\Xi }(v_h^{n-1},{\sigma }_h^{n-1})+\frac{1}{1-\Delta t}\frac{\Delta t^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )}{\parallel v_h^0\parallel }^2{\displaystyle \sum _{j=0}^1}{\widehat{b}}_{n-j}(\frac{1+\Delta t}{1-\Delta t}{)}^j\hfill \\ & +\frac{C\Delta t}{1-\Delta t}{\displaystyle \sum _{j=0}^1}(\frac{1+\Delta t}{1-\Delta t}{)}^j{\parallel g(\mathbf{x},t_{n-j+\frac{1}{2}})\parallel }^2\hfill \\ & \le \cdots \cdots \hfill \\ & \le (\frac{1+\Delta t}{1-\Delta t}{)}^n\mathsf{\Xi }(v_h^1,{\sigma }_h^1)+\frac{1}{1-\Delta t}\frac{\Delta t^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )}{\parallel v_h^0\parallel }^2{\displaystyle \sum _{j=0}^{n-1}}{\widehat{b}}_{n-j}(\frac{1+\Delta t}{1-\Delta t}{)}^j\hfill \\ & +\frac{C\Delta t}{1-\Delta t}{\displaystyle \sum _{j=0}^{n-1}}(\frac{1+\Delta t}{1-\Delta t}{)}^j{\parallel g(\mathbf{x},t_{n-j+\frac{1}{2}})\parallel }^2.\hfill \end{array}$$

(24)

Noting that $\left(\frac{1+\Delta t}{1-\Delta t}\right)>1$, $\Delta t=T/N\le T/n$, we have

$$\begin{array}{cc}(\frac{1+\Delta t}{1-\Delta t}{)}^n\hfill & \le (\frac{1+\Delta t}{1-\Delta t}{)}^{n+1}\le \cdots \le (1+\frac{2\Delta t}{1-\Delta t}{)}^{\frac{T}{\Delta t}}\hfill \\ & \le \underset{\Delta t\to 0}{lim}(1+\frac{2\Delta t}{1-\Delta t}{)}^{\frac{T(1-\Delta t)}{2\Delta t}\frac{2}{1-\Delta t}}=e^2.\hfill \end{array}$$

(25)

Further, noting that ${\widehat{b}}_{n-j}>0$ and using Lemma 5, we have

$$\begin{array}{c} \frac{1}{1-\Delta t}\frac{\Delta t^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )}\parallel v_h^0{\parallel }^2{\displaystyle \sum _{j=0}^{n-1}}{b}_{n-j}(\frac{1+\Delta t}{1-\Delta t}{)}^j+\frac{C\Delta t}{1-\Delta t}{\displaystyle \sum _{j=0}^{n-1}}(\frac{1+\Delta t}{1-\Delta t}{)}^j{\parallel g(\mathbf{x},t_{n-j+\frac{1}{2}})\parallel }^2\hfill \\ \le \frac{e^2}{1-\Delta t}\frac{\Delta t^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )}\parallel v_h^0{\parallel }^2{\displaystyle \sum _{j=0}^{n-1}}{\widehat{b}}_{n-j}+\frac{C\Delta t}{1-\Delta t}{\displaystyle \sum _{j=0}^{n-1}}{\parallel g(\mathbf{x},t_{n-j+\frac{1}{2}})\parallel }^2\hfill \\ \le C(\frac{T^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )}\parallel v_h^0{\parallel }^2+\underset{1\le j\le n}{max}\{\parallel g(\mathbf{x},t_{j+\frac{1}{2}}){\parallel }^2\}).\hfill \end{array}$$

(26)

Substitute (25) and (26) into (24) to arrive at

$$\begin{array}{cc}\mathsf{\Xi }(v_h^{n+1},{\sigma }_h^{n+1})\le \hfill & C(\mathsf{\Xi }(v_h^1,{\sigma }_h^1)+\frac{T^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )}\parallel v_h^0{\parallel }^2+{\displaystyle \underset{1\le j\le n}{max}}\{\parallel g(\mathbf{x},t_{j+\frac{1}{2}}){\parallel }^2\}).\hfill \end{array}$$

(27)

Now, we estimate $\mathsf{\Xi }(v_h^1,{\sigma }_h^1)$. Using (12) $\left(c\right)$, taking ${\chi }_h=v_h^{\frac{1}{2}}$, and using Cauchy–Schwarz inequality, we have

$$\begin{array}{cc}\hfill & -(\nabla {\sigma }_h^{\frac{1}{2}},\nabla v_h^{\frac{1}{2}})\hfill \\ =\hfill & -(P_{\Delta t}{v}_h^{\frac{1}{2}},v_h^{\frac{1}{2}})-\left(\frac{\Delta t^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}\right[{(\frac{1}{2})}^{-\alpha }{v}_h^{\frac{1}{2}}-\alpha {(\frac{1}{2})}^{-\alpha }{v}_h^0],v_h^{\frac{1}{2}})-{\parallel v_h^{\frac{1}{2}}\parallel }^2+(g(\mathbf{x},t_{\frac{1}{2}}),v_h^{\frac{1}{2}})\hfill \\ \le \hfill & -\frac{\parallel v_h^1{\parallel }^2-{\parallel v_h^0\parallel }^2}{2\Delta t}-{(\frac{1}{2})}^{-\alpha }\frac{\Delta t^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}{\parallel v_h^{\frac{1}{2}}\parallel }^2\hfill \\ & +\frac{\Delta t^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}\alpha {(\frac{1}{2})}^{-\alpha }(\parallel v_h^0{\parallel }^2+\parallel v_h^{\frac{1}{2}}{\parallel }^2)-\frac{1}{2}\parallel v_h^{\frac{1}{2}}{\parallel }^2+\frac{1}{2}{\parallel g(\mathbf{x},t_{\frac{1}{2}})\parallel }^2.\hfill \end{array}$$

(28)

For $n=0$, we sum for (18) and (28) to obtain

$$\begin{array}{cc}\hfill & \frac{\parallel v_h^1{\parallel }^2-{\parallel v_h^0\parallel }^2}{2\Delta t}+\frac{\parallel {\sigma }_h^1{\parallel }^2-{\parallel {\sigma }_h^0\parallel }^2}{2\Delta t}+(\frac{1}{2}+\frac{2^{\alpha }\Delta t^{-\alpha }}{\mathsf{\Gamma }(1-\alpha )}){\parallel v_h^{\frac{1}{2}}\parallel }^2\hfill \\ \le \hfill & \frac{\alpha 2^{\alpha }\Delta t^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}\parallel v_h^0{\parallel }^2+C(\parallel v_h^1{\parallel }^2+\parallel v_h^0{\parallel }^2+\parallel {\sigma }_h^1{\parallel }^2+\parallel {\sigma }_h^0{\parallel }^2+\parallel g(\mathbf{x},t_{\frac{1}{2}}){\parallel }^2).\hfill \end{array}$$

(29)

Noting that $1-\alpha \le 2^{\alpha }$, ($0<\alpha <1$) and (23), we have, for sufficiently small $\Delta t$,

$$\begin{array}{cc}\mathsf{\Xi }(v_h^1,{\sigma }_h^1)\hfill & =\parallel v_h^1{\parallel }^2+\parallel {\sigma }_h^1{\parallel }^2+\frac{\Delta t^{1-\alpha }}{\mathsf{\Gamma }(2-\alpha )}{a}_0{\parallel v_h^{\frac{1}{2}}\parallel }^2\hfill \\ & \le \parallel v_h^1{\parallel }^2+\parallel {\sigma }_h^1{\parallel }^2+(1+\frac{2\Delta t^{1-\alpha }}{\mathsf{\Gamma }(1-\alpha )})\parallel v_h^{\frac{1}{2}}{\parallel }^2\le C(\parallel v_h^0{\parallel }^2+\parallel {\sigma }_h^0{\parallel }^2+\parallel g(\mathbf{x},t_{\frac{1}{2}}){\parallel }^2).\hfill \end{array}$$

(30)

Substitute (30) into (27) to obtain

$$\begin{array}{cc}\mathsf{\Xi }(v_h^{n+1},{\sigma }_h^{n+1})\hfill & \le C(\parallel v_h^0{\parallel }^2+\parallel {\sigma }_h^0{\parallel }^2+{\displaystyle \underset{0\le j\le n}{max}}\{\parallel g(\mathbf{x},t_{j+\frac{1}{2}}){\parallel }^2\}),n\ge 0.\hfill \end{array}$$

(31)

Combine (31) with (17) and use the Gronwall lemma to obtain

$$\begin{array}{cc}\parallel u_h^{n+1}{\parallel }^2\hfill & \le C(\parallel u_h^0{\parallel }^2+\parallel v_h^0{\parallel }^2+\parallel {\sigma }_h^0{\parallel }^2+{\displaystyle \underset{0\le j\le n}{max}}\{\parallel g(\mathbf{x},t_{j+\frac{1}{2}}){\parallel }^2\}),n\ge 0.\hfill \end{array}$$

(32)

Using (31) and (32), we obtain the conclusion. □

3. A Priori Error Estimate

Now, we provide two projection operators [25] to derive a priori error estimates of our mixed finite element method.

Lemma 6.

Define the $L^2$ projection ${\mathcal{P}}_h:L^2\left(\mathsf{\Omega }\right)\to L_h$ as

$$(u-{\mathcal{P}}_{h}u,{\phi }_h)=0,\forall {\phi }_h\in L_h,$$

(33)

with the estimate inequality

$$\parallel u-{\mathcal{P}}_{h}u\parallel +\parallel u_t-{\mathcal{P}}_h{u}_t\parallel \le C{h}^{m+1}{\parallel u\parallel }_{m+1},\forall u\in L^2\left(\mathsf{\Omega }\right).$$

(34)

Lemma 7.

Define the elliptic projection ${\mathcal{Q}}_h:H_0^1\left(\mathsf{\Omega }\right)\to V_h$ as

$$(\nabla (v-{\mathcal{Q}}_{h}v),\nabla {\varphi }_h)=0,\forall {\varphi }_h\in V_h,$$

(35)

with the following inequality:

$$\begin{array}{cc}\hfill & \parallel v-{\mathcal{Q}}_{h}v\parallel +\parallel v_t-{\mathcal{Q}}_h{v}_t\parallel +h\parallel v-{\mathcal{Q}}_h{v\parallel }_1\le C{h}^{k+1}{(\parallel v\parallel }_{k+1}+\parallel v_t{\parallel }_{k+1}),\hfill \\ & \forall v\in H_0^1\left(\mathsf{\Omega }\right)\cap H^{k+1}\left(\mathsf{\Omega }\right).\hfill \end{array}$$

(36)

In what follows, we derive the proof of error estimates in $L^2$-norm in detail.

Theorem 2.

For ${\mathcal{P}}_{h}u\left(0\right)=u_h^0$, ${\mathcal{Q}}_{h}v\left(0\right)=v_h^0$ and ${\mathcal{Q}}_h\sigma \left(0\right)={\sigma }_h^0$, there exists a positive constant C that is independent of space–time step length sizes $(h,\Delta t)$ and we have for $n\ge 0$

$$\begin{array}{cc}\hfill & \parallel u\left(t_{n+1}\right)-u_h^{n+1}\parallel +\parallel v\left(t_{n+1}\right)-v_h^{n+1}\parallel +\parallel \sigma \left(t_{n+1}\right)-{\sigma }_h^{n+1}\parallel \hfill \\ & \le C[(1+\mu t_{n+\frac{1}{2}}^{1-\beta })h^{k+1}+\Delta t^{3-\beta }+h^{m+1}],\hfill \end{array}$$

(37)

where, for the Caputo fractional derivative, we take μ as 0; for the Riemann–Liouville fractional derivative, we take μ as 1.

Proof. 

For convenience, we write

$$\begin{array}{cc}\hfill & u\left(t_n\right)-u_h^n=(u\left(t_n\right)-{\mathcal{P}}_h{u}^n)+({\mathcal{P}}_h{u}^n-u_h^n)={\mathcal{E}}^n+{\mathfrak{E}}^n,\hfill \\ & v\left(t_n\right)-v_h^n=(v\left(t_n\right)-{\mathcal{Q}}_h{v}^n)+({\mathcal{Q}}_h{v}^n-v_h^n)={\mathcal{F}}^n+{\mathfrak{F}}^n,\hfill \\ & \sigma \left(t_n\right)-{\sigma }_h^n=(\sigma \left(t_n\right)-{\mathcal{Q}}_h{\sigma }^n)+({\mathcal{Q}}_h{\sigma }^n-{\sigma }_h^n)={\mathcal{H}}^n+{\mathfrak{H}}^n.\hfill \end{array}$$

Applying triangle inequality, we have

$$\begin{array}{cc}\hfill & \parallel u\left(t_n\right)-u_h^n\parallel \le \parallel {\mathcal{E}}^n\parallel +\parallel {\mathfrak{E}}^n\parallel ,\hfill \\ & \parallel v\left(t_n\right)-v_h^n\parallel \le \parallel {\mathcal{F}}^n\parallel +\parallel {\mathfrak{F}}^n\parallel ,\hfill \\ & \parallel \sigma \left(t_n\right)-{\sigma }_h^n\parallel \le \parallel {\mathcal{H}}^n\parallel +\parallel {\mathfrak{H}}^n\parallel .\hfill \end{array}$$

(38)

Using Lemmas 6 and 7, we arrive at the estimates of $\parallel {\mathcal{E}}^n\parallel$, $\parallel {\mathcal{F}}^n\parallel$, and $\parallel {\mathcal{H}}^n\parallel$. Consequently, in the discussion below, we only need to derive the estimates of $\parallel {\mathfrak{E}}^n\parallel$, $\parallel {\mathfrak{F}}^n\parallel$, and $\parallel {\mathfrak{H}}^n\parallel$. Using projections (33) and (35), we have error equations as follows:

$$\left\{\begin{array}{cc}\left(a\right)\hfill & (P_{\Delta t}{\mathfrak{E}}^{n+\frac{1}{2}},{\phi }_h)=-(P_{\Delta t}{\mathcal{E}}^{n+\frac{1}{2}},{\phi }_h)+({\mathcal{F}}^{n+\frac{1}{2}}+{\mathfrak{F}}^{n+\frac{1}{2}},{\phi }_h)+(R_1^{n+\frac{1}{2}},{\phi }_h),\hfill \\ \left(b\right)\hfill & (P_{\Delta t}{\mathfrak{H}}^{n+\frac{1}{2}},{\psi }_h)+(\nabla {\mathfrak{F}}^{n+\frac{1}{2}},\nabla {\psi }_h)\hfill \\ \hfill =& -(\frac{f_u\left(u^{n+1}\right)v^{n+1}+f_u\left(u^n\right)v^n}{2}-\frac{f_u\left(u_h^{n+1}\right)v_h^{n+1}+f_u\left(u_h^n\right)v_h^n}{2},{\psi }_h)\hfill \\ & -(P_{\Delta t}{\mathcal{H}}^{n+\frac{1}{2}},{\psi }_h)+(R_2^{n+\frac{1}{2}},{\psi }_h),\hfill \\ \left(c\right)\hfill & (P_{\Delta t}{\mathfrak{F}}^{n+\frac{1}{2}},{\chi }_h)+\left(\frac{\Delta t^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}\right[a_0{\mathfrak{F}}^{n+\frac{1}{2}}-{\displaystyle \sum _{j=1}^n}(a_{n-j}-a_{n-j+1}){\mathfrak{F}}^{j-\frac{1}{2}}\hfill \\ & -(a_n-b_n){\mathfrak{F}}^{\frac{1}{2}}-{\widehat{b}}_n{\mathfrak{F}}^0],{\chi }_h)+({\mathcal{F}}^{n+\frac{1}{2}}+{\mathfrak{F}}^{n+\frac{1}{2}},{\psi }_h)-(\nabla {\mathfrak{H}}^{n+\frac{1}{2}},\nabla {\chi }_h)\hfill \\ \hfill =& -(P_{\Delta t}{\mathcal{F}}^{n+\frac{1}{2}},{\chi }_h)-\left(\frac{\Delta t^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}\right[a_0{\mathcal{F}}^{n+\frac{1}{2}}-{\displaystyle \sum _{j=1}^n}(a_{n-j}-a_{n-j+1}){\mathcal{F}}^{j-\frac{1}{2}}\hfill \\ & -(a_n-b_n){\mathcal{F}}^{\frac{1}{2}}-{\widehat{b}}_n{\mathcal{F}}^0],{\chi }_h)+(R_3^{n+\frac{1}{2}},{\chi }_h).\hfill \end{array}\right.$$

(39)

In (39), we set ${\phi }_h={\mathfrak{E}}^{n+\frac{1}{2}}$, ${\chi }_h={\mathfrak{F}}^{n+\frac{1}{2}}$, and ${\psi }_h={\mathfrak{H}}^{n+\frac{1}{2}}$, and add the resulting equations to obtain

$$\begin{array}{cc}\hfill & (P_{\Delta t}{\mathfrak{E}}^{n+\frac{1}{2}},{\mathfrak{E}}^{n+\frac{1}{2}})+(P_{\Delta t}{\mathfrak{F}}^{n+\frac{1}{2}},{\mathfrak{F}}^{n+\frac{1}{2}})+(P_{\Delta t}{\mathfrak{H}}^{n+\frac{1}{2}},{\mathfrak{H}}^{n+\frac{1}{2}})\hfill \\ & +\left(\frac{\Delta t^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}\right[a_0{\mathfrak{F}}^{n+\frac{1}{2}}-{\displaystyle \sum _{j=1}^n}(a_{n-j}-a_{n-j+1}){\mathfrak{F}}^{j-\frac{1}{2}}-(a_n-b_n){\mathfrak{F}}^{\frac{1}{2}}-{\widehat{b}}_n{\mathfrak{F}}^0],{\mathfrak{F}}^{n+\frac{1}{2}})\hfill \\ =\hfill & -(P_{\Delta t}{\mathcal{E}}^{n+\frac{1}{2}},{\mathfrak{E}}^{n+\frac{1}{2}})-(P_{\Delta t}{\mathcal{F}}^{n+\frac{1}{2}},{\mathfrak{F}}^{n+\frac{1}{2}})-(P_{\Delta t}{\mathcal{H}}^{n+\frac{1}{2}},{\mathfrak{H}}^{n+\frac{1}{2}})\hfill \\ & +({\mathcal{F}}^{n+\frac{1}{2}}+{\mathfrak{F}}^{n+\frac{1}{2}},{\mathfrak{E}}^{n+\frac{1}{2}})+({\mathcal{F}}^{n+\frac{1}{2}}+{\mathfrak{F}}^{n+\frac{1}{2}},{\mathfrak{F}}^{n+\frac{1}{2}})\hfill \\ & -(\frac{f_u\left(u^{n+1}\right)v^{n+1}+f_u\left(u^n\right)v^n}{2}-\frac{f_u\left(u_h^{n+1}\right)v_h^{n+1}+f_u\left(u_h^n\right)v_h^n}{2},{\mathfrak{H}}^{n+\frac{1}{2}})\hfill \\ & -\left(\frac{\Delta t^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}\right[a_0{\mathcal{F}}^{n+\frac{1}{2}}-{\displaystyle \sum _{j=1}^n}(a_{n-j}-a_{n-j+1}){\mathcal{F}}^{j-\frac{1}{2}}-(a_n-b_n){\mathcal{F}}^{\frac{1}{2}}-{\widehat{b}}_n{\mathcal{F}}^0],{\mathfrak{F}}^{n+\frac{1}{2}})\hfill \\ & +(R_1^{n+\frac{1}{2}},{\mathfrak{E}}^{n+\frac{1}{2}})+(R_2^{n+\frac{1}{2}},{\mathfrak{H}}^{n+\frac{1}{2}})+(R_3^{n+\frac{1}{2}},{\mathfrak{F}}^{n+\frac{1}{2}}).\hfill \end{array}$$

(40)

Now, we need to estimate all terms on the right-hand side of (40). Using Cauchy–Schwarz inequality, we have

$$\begin{array}{cc}\hfill & -(P_{\Delta t}{\mathcal{E}}^{n+\frac{1}{2}},{\mathfrak{E}}^{n+\frac{1}{2}})-(P_{\Delta t}{\mathcal{F}}^{n+\frac{1}{2}},{\mathfrak{F}}^{n+\frac{1}{2}})-(P_{\Delta t}{\mathcal{H}}^{n+\frac{1}{2}},{\mathfrak{H}}^{n+\frac{1}{2}})\hfill \\ & +({\mathcal{F}}^{n+\frac{1}{2}}+{\mathfrak{F}}^{n+\frac{1}{2}},{\mathfrak{E}}^{n+\frac{1}{2}})+({\mathcal{F}}^{n+\frac{1}{2}}+{\mathfrak{F}}^{n+\frac{1}{2}},{\mathfrak{F}}^{n+\frac{1}{2}})\hfill \\ \le \hfill & C(\parallel P_{\Delta t}{\mathcal{E}}^{n+\frac{1}{2}}{\parallel }^2+\parallel P_{\Delta t}{\mathcal{F}}^{n+\frac{1}{2}}{\parallel }^2+\parallel P_{\Delta t}{\mathcal{H}}^{n+\frac{1}{2}}{\parallel }^2+\parallel {\mathcal{F}}^{n+\frac{1}{2}}{\parallel }^2)\hfill \\ & +C(\parallel {\mathfrak{E}}^{n+\frac{1}{2}}{\parallel }^2+\parallel {\mathfrak{F}}^{n+\frac{1}{2}}{\parallel }^2+\parallel {\mathfrak{H}}^{n+\frac{1}{2}}{\parallel }^2).\hfill \end{array}$$

(41)

Applying the mean value theorem and Cauchy–Schwarz inequality, we have

$$\begin{array}{cc}\hfill & -(\frac{f_u\left(u^{n+1}\right)v^{n+1}+f_u\left(u^n\right)v^n}{2}-\frac{f_u\left(u_h^{n+1}\right)v_h^{n+1}+f_u\left(u_h^n\right)v_h^n}{2},{\mathfrak{H}}^{n+\frac{1}{2}})\hfill \\ =\hfill & -\frac{1}{2}(f_u\left(u^{n+1}\right)(v^{n+1}-v_h^{n+1})+(f_u\left(u^{n+1}\right)-f_u\left(u_h^{n+1}\right))v_h^{n+1}\hfill \\ & +f_u\left(u^n\right)(v^n-v_h^n)+(f_u\left(u^n\right)-f_u\left(u_h^n\right))v_h^n,{\mathfrak{H}}^{n+\frac{1}{2}})\hfill \\ \le \hfill & \frac{1}{2}(\parallel f_u\left(u^{n+1}\right){\parallel }_{\infty }\parallel v^{n+1}-v_h^{n+1}\parallel +\parallel f_{uu}\left({\overline{\theta }}^{n+1}\right){\parallel }_{\infty }\parallel u^{n+1}-u_h^{n+1}\parallel \parallel v_h^{n+1}{\parallel }_{\infty }\hfill \\ & +\parallel f_u\left(u^n\right){\parallel }_{\infty }\parallel v^n-v_h^n\parallel +\parallel f_{uu}\left({\overline{\theta }}^n\right){\parallel }_{\infty }\parallel u^n-u_h^n\parallel \parallel v_h^n{\parallel }_{\infty })\parallel {\mathfrak{H}}^{n+\frac{1}{2}}\parallel .\hfill \\ \le \hfill & C(\parallel {\mathcal{E}}^{n+1}{\parallel }^2+\parallel {\mathcal{F}}^{n+1}{\parallel }^2+\parallel {\mathcal{E}}^n{\parallel }^2+\parallel {\mathcal{F}}^n{\parallel }^2+\parallel {\mathfrak{E}}^{n+1}{\parallel }^2\hfill \\ & +\parallel {\mathfrak{F}}^{n+1}{\parallel }^2+\parallel {\mathfrak{H}}^{n+1}{\parallel }^2+\parallel {\mathfrak{E}}^n{\parallel }^2+\parallel {\mathfrak{F}}^n{\parallel }^2+\parallel {\mathfrak{H}}^n{\parallel }^2),\hfill \end{array}$$

(42)

where we use the boundedness of $\parallel f_u\left(u^n\right){\parallel }_{\infty }$ and the following bounded inequality:

$$\begin{array}{c}\parallel f_{uu}\left({\overline{\theta }}^n\right){\parallel }_{\infty }+{\parallel v_h^n\parallel }_{\infty }\le C,\hfill \end{array}$$

(43)

where one can apply inverse inequality [25], and use a similar method as the one in [7,26].

Making use of (9), (3), Cauchy–Schwarz inequality, as well as Young inequality, we have

$$\begin{array}{cc}\hfill & -\left(\frac{\Delta t^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}\right[a_0{\mathcal{F}}^{n+\frac{1}{2}}-{\displaystyle \sum _{j=1}^n}(a_{n-j}-a_{n-j+1}){\mathcal{F}}^{j-\frac{1}{2}}-(a_n-b_n){\mathcal{F}}^{\frac{1}{2}}-{\widehat{b}}_n{\mathcal{F}}^0],{\mathfrak{F}}^{n+\frac{1}{2}})\hfill \\ & +(R_1^{n+\frac{1}{2}},{\mathfrak{E}}^{n+\frac{1}{2}})+(R_2^{n+\frac{1}{2}},{\mathfrak{H}}^{n+\frac{1}{2}})+(R_3^{n+\frac{1}{2}},{\mathfrak{F}}^{n+\frac{1}{2}})\hfill \\ =\hfill & -(\frac{1}{\mathsf{\Gamma }(1-\alpha )}{\int }_0^{t_{n+\frac{1}{2}}}\frac{\frac{\partial \mathcal{F}}{\partial s}ds}{{(t_{n+\frac{1}{2}}-s)}^{\alpha }}+\frac{\mu {\mathcal{F}}^0}{\mathsf{\Gamma }(1-\alpha )}{t}_{n+\frac{1}{2}}^{-\alpha }+O(\Delta t^{2-\alpha }),{\mathfrak{F}}^{n+\frac{1}{2}})\hfill \\ & +(R_1^{n+\frac{1}{2}},{\mathfrak{E}}^{n+\frac{1}{2}})+(R_2^{n+\frac{1}{2}},{\mathfrak{H}}^{n+\frac{1}{2}})+(R_3^{n+\frac{1}{2}},{\mathfrak{F}}^{n+\frac{1}{2}})\hfill \\ \le \hfill & C[(1+\mu t_{n+\frac{1}{2}}^{-\alpha })h^{2k+2}+\Delta t^{4-2\alpha }+\parallel {\mathfrak{E}}^{n+\frac{1}{2}}{\parallel }^2+\parallel {\mathfrak{F}}^{n+\frac{1}{2}}{\parallel }^2+{\parallel {\mathfrak{H}}^{n+\frac{1}{2}}\parallel }^2].\hfill \end{array}$$

(44)

Making a combination for (41)–(44) and using (18), we have

$$\begin{array}{cc}\hfill & \frac{(\parallel {\mathfrak{E}}^{n+1}{\parallel }^2+\parallel {\mathfrak{F}}^{n+1}{\parallel }^2+\parallel {\mathfrak{H}}^{n+1}{\parallel }^2)-(\parallel {\mathfrak{E}}^n{\parallel }^2+\parallel {\mathfrak{F}}^n{\parallel }^2+\parallel {\mathfrak{H}}^n{\parallel }^2)}{2\Delta t}\hfill \\ & +\frac{\Delta t^{-\alpha }}{2\mathsf{\Gamma }(2-\alpha )}{\displaystyle \sum _{j=1}^{n+1}}{a}_{n-j+1}{\parallel {\mathfrak{F}}^{j-\frac{1}{2}}\parallel }^2\hfill \\ =\hfill & \frac{\Delta t^{-\alpha }}{2\mathsf{\Gamma }(2-\alpha )}{\displaystyle \sum _{j=1}^n}{a}_{n-j}\parallel {\mathfrak{F}}^{j-\frac{1}{2}}{\parallel }^2+\frac{\Delta t^{-\alpha }}{2\mathsf{\Gamma }(2-\alpha )}{\widehat{b}}_n{\parallel {\mathfrak{F}}^0\parallel }^2+C[(1+\mu t_{n+\frac{1}{2}}^{-\alpha })h^{2k+2}+\Delta t^{4-2\alpha }\hfill \\ & +\parallel P_{\Delta t}{\mathcal{E}}^{n+\frac{1}{2}}{\parallel }^2+\parallel P_{\Delta t}{\mathcal{F}}^{n+\frac{1}{2}}{\parallel }^2+\parallel P_{\Delta t}{\mathcal{H}}^{n+\frac{1}{2}}{\parallel }^2+\parallel {\mathcal{E}}^{n+1}{\parallel }^2+{\parallel {\mathcal{F}}^{n+1}\parallel }^2\hfill \\ & +\parallel {\mathcal{E}}^n{\parallel }^2+\parallel {\mathcal{F}}^n{\parallel }^2+\parallel {\mathfrak{E}}^{n+1}{\parallel }^2+\parallel {\mathfrak{F}}^{n+1}{\parallel }^2+\parallel {\mathfrak{H}}^{n+1}{\parallel }^2+\parallel {\mathfrak{E}}^n{\parallel }^2+\parallel {\mathfrak{F}}^n{\parallel }^2+{\parallel {\mathfrak{H}}^n\parallel }^2].\hfill \end{array}$$

(45)

With given conditions ${\mathfrak{E}}^0=0$, ${\mathfrak{F}}^0=0$, ${\mathfrak{H}}^0=0$, we use (23) to arrive at

$$\begin{array}{cc}\hfill & \mathsf{\Xi }({\mathfrak{F}}^{n+1},{\mathfrak{H}}^{n+1})+{\parallel {\mathfrak{E}}^{n+1}\parallel }^2\hfill \\ \le \hfill & \mathsf{\Xi }({\mathfrak{F}}^n,{\mathfrak{H}}^n)+{\parallel {\mathfrak{E}}^n\parallel }^2+C\Delta t[(1+\mu t_{n+\frac{1}{2}}^{-\alpha })h^{2k+2}+\Delta t^{4-2\alpha }\hfill \\ & +h^{2m+2}+\parallel {\mathfrak{E}}^{n+1}{\parallel }^2+\parallel {\mathfrak{F}}^{n+1}{\parallel }^2+\parallel {\mathfrak{H}}^{n+1}{\parallel }^2+\parallel {\mathfrak{E}}^n{\parallel }^2+\parallel {\mathfrak{F}}^n{\parallel }^2+{\parallel {\mathfrak{H}}^n\parallel }^2].\hfill \end{array}$$

(46)

Sum for (46) with respect to n to arrive at

$$\begin{array}{cc}\hfill & \mathsf{\Xi }({\mathfrak{F}}^{n+1},{\mathfrak{H}}^{n+1})+{\parallel {\mathfrak{E}}^{n+1}\parallel }^2\hfill \\ \le \hfill & \mathsf{\Xi }({\mathfrak{F}}^1,{\mathfrak{H}}^1)+{\parallel {\mathfrak{E}}^1\parallel }^2+C\Delta t{\displaystyle \sum _{j=1}^n}[(1+\mu t_{j+\frac{1}{2}}^{-\alpha })h^{2k+2}+\Delta t^{4-2\alpha }+h^{2m+2}]\hfill \\ & +C\Delta t{\displaystyle \sum _{j=1}^{n+1}}[\parallel {\mathfrak{E}}^j{\parallel }^2+\parallel {\mathfrak{F}}^j{\parallel }^2+{\parallel {\mathfrak{H}}^j\parallel }^2].\hfill \end{array}$$

(47)

For $n=0$, we use a similar derivation to the one of $n\ge 1$ and apply triangle inequality to arrive at

$$\begin{array}{cc}\hfill & \mathsf{\Xi }({\mathfrak{F}}^1,{\mathfrak{H}}^1)+{\parallel {\mathfrak{E}}^1\parallel }^2\le C[(1+\mu t_{\frac{1}{2}}^{-\alpha })h^{2k+2}+\Delta t^{4-2\alpha }+h^{2m+2}].\hfill \end{array}$$

(48)

Substitute (48) into (47) and use the Gronwall lemma to obtain

$$\begin{array}{cc}\mathsf{\Xi }({\mathfrak{F}}^{n+1},{\mathfrak{H}}^{n+1})+{\parallel {\mathfrak{E}}^{n+1}\parallel }^2\le \hfill & C[(1+\mu t_{n+\frac{1}{2}}^{-\alpha })h^{2k+2}+\Delta t^{4-2\alpha }+h^{2m+2}],\forall n\ge 0.\hfill \end{array}$$

(49)

Combining (49), (34), and (36) with (38) and noting that $\alpha =\beta -1$, we complete the proof of the theorem. □

Remark 2.

Compared with the classical mixed element method for fourth-order partial differential equations, our method can approximate simultaneously three variables with optimal error estimates in $L^2$-norm. More importantly, we can obtain directly optimal error estimates in $L^2$-norm for auxiliary variables in solving fourth-order PDEs, which are difficult to achieve by using classical mixed element methods [6,7,8].

4. Numerical Tests

Here, we will verify the theoretical results by numerical computing. In (1), we take space domain $\overline{\mathsf{\Omega }}={[0,1]}^2$, time interval $\overline{J}=[0,1]$, nonlinear term $f\left(u\right)=u^3-u$, initial conditions with $u(x,y,0)=0$, $u_1(x,y)=0$, and exact solution $u=t^{3}sin\left(2\pi x\right)sin\left(2\pi y\right)$; we can obtain the source term $g(x,y,t)$ and two auxiliary variables $v=3t^{2}sin\left(2\pi x\right)sin\left(2\pi y\right)$ and $\sigma =-t^{3}sin\left(2\pi x\right)sin\left(2\pi y\right)(8{\pi }^2+t^{6}sin{\left(2\pi x\right)}^{2}sin{\left(2\pi y\right)}^2-1)$. In the following numerical calculations, the order of convergence in space is calculated by the following formula with a sufficiently small time step size $\Delta t$

$$\mathrm{Order}={log}_{\frac{h_1}{h_2}}\frac{\parallel \varphi -{\varphi }_{h_1}\parallel }{\parallel \varphi -{\varphi }_{h_2}\parallel },$$

where $h_k$$(k=1,2)$ represents different space mesh step lengths.

For implementing the new mixed element algorithm, we approximate the spatial direction by the finite element method with the basis function $P(x,y)=a+bx+cy+dxy$ and discretize the time direction by using the modified $L1$ Crank–Nicolson scheme. In

Table 1. The convergence results for u with $\Delta t=1/200$.

$\mathit{\beta }$	h	$\parallel \mathit{u}-{\mathit{u}}_{\mathit{h}}\parallel$	Order	CPU-Time (s)
1.1	$\sqrt{2}/9$	4.1181 $\times {10}^{-2}$		1.13
	$\sqrt{2}/16$	1.3341 $\times {10}^{-2}$	1.9590	4.04
	$\sqrt{2}/25$	5.5001 $\times {10}^{-3}$	1.9855	19.03
1.5	$\sqrt{2}/9$	4.1175 $\times {10}^{-2}$		1.10
	$\sqrt{2}/16$	1.3339 $\times {10}^{-2}$	1.9590	4.02
	$\sqrt{2}/25$	5.4991 $\times {10}^{-3}$	1.9855	19.40
1.9	$\sqrt{2}/9$	4.1169 $\times {10}^{-2}$		1.14
	$\sqrt{2}/16$	1.3336 $\times {10}^{-2}$	1.9591	4.07
	$\sqrt{2}/25$	5.4977 $\times {10}^{-3}$	1.9857	19.33

, by taking the fixed time mesh parameter $\Delta t=1/200$, changed spatial step length sizes $h=\sqrt{2}/9$, $\sqrt{2}/16$ and $\sqrt{2}/25$, and different parameters $\beta =1.1,1.5,1.9$, we show the $L^2$-norm error estimates and second-order convergence data in space. In

Table 2. The convergence results for v with $\Delta t=1/200$.

$\mathit{\beta }$	h	$\parallel \mathit{v}-{\mathit{v}}_{\mathit{h}}\parallel$	Order	CPU-Time (s)
1.1	$\sqrt{2}/9$	1.2293 $\times {10}^{-1}$		1.13
	$\sqrt{2}/16$	3.9815 $\times {10}^{-2}$	1.9594	4.04
	$\sqrt{2}/25$	1.6417 $\times {10}^{-2}$	1.9851	19.03
1.5	$\sqrt{2}/9$	1.2292 $\times {10}^{-1}$		1.10
	$\sqrt{2}/16$	3.9816 $\times {10}^{-2}$	1.9593	4.02
	$\sqrt{2}/25$	1.6417 $\times {10}^{-2}$	1.9852	19.40
1.9	$\sqrt{2}/9$	1.2293 $\times {10}^{-1}$		1.14
	$\sqrt{2}/16$	3.9819 $\times {10}^{-2}$	1.9592	4.07
	$\sqrt{2}/25$	1.6418 $\times {10}^{-2}$	1.9852	19.33

and

Table 3. The convergence results for $\sigma$ with $\Delta t=1/200$.

$\mathit{\beta }$	h	$\parallel \mathit{\sigma }-{\mathit{\sigma }}_{\mathit{h}}\parallel$	Order	CPU-Time (s)
1.1	$\sqrt{2}/9$	1.8858 $\times {10}^{+0}$		1.13
	$\sqrt{2}/16$	5.9584 $\times {10}^{-1}$	2.0024	4.04
	$\sqrt{2}/25$	2.4398 $\times {10}^{-1}$	2.0007	19.03
1.5	$\sqrt{2}/9$	1.8853 $\times {10}^{+0}$		1.10
	$\sqrt{2}/16$	5.9568 $\times {10}^{-1}$	2.0025	4.02
	$\sqrt{2}/25$	2.4391 $\times {10}^{-1}$	2.0007	19.40
1.9	$\sqrt{2}/9$	1.8849 $\times {10}^{+0}$		1.14
	$\sqrt{2}/16$	5.9550 $\times {10}^{-1}$	2.0026	4.07
	$\sqrt{2}/25$	2.4381 $\times {10}^{-1}$	2.0010	19.33

, we compute the convergence results v and $\sigma$, respectively. From Table 1, Table 2 and Table 3, one can see that the numerical method is effective for solving nonlinear fourth-order fractional diffusion-wave equation models with a smooth solution.

5. Concluding Remarks

From the calculated results in Table 1, Table 2 and Table 3, one can see that our method for solving fourth-order fractional diffusion-wave equations in this article can obtain optimal error estimates in $L^2$-norm for three variables, which is in agreement with the derived theoretical results. These results for auxiliary variables are difficult to achieve directly by using classical mixed element methods [6,7,8].

In the future, we will improve our mixed element method by combining other techniques [7,27,28] with high-order time approximate schemes and develop their optimal numerical theories.