Mixed element algorithm based on a second-order time approximation scheme for a two-dimensional nonlinear time fractional coupled sub-diffusion model

Abstract

In this article, a numerical algorithm is presented to solve a two-dimensional nonlinear time fractional coupled sub-diffusion problem, where the second-order Crank–Nicolson scheme with a second-order WSGD formula is used in the time direction, and a mixed element method is applied in the space direction. The existence and uniqueness of the mixed element solution and the stability for fully discrete scheme are proven. In addition, the optimal a priori error estimates for unknown scalar function u and v in $L^2$ and $H^1$-norms and a priori error estimates for their fluxes $\sigma$ and $\lambda$ in $(L^2)^2$-norm are obtained. Finally, some numerical calculations are presented to illustrate the validity for the proposed numerical algorithm.

1 Introduction

Fractional calculus and models [1,2,3,4,5,6,7,8,9,10,11,12] have been studied and developed by a large number of authors. Especially, fractional diffusion models including time, space and space-time fractional cases have attracted a lot of attention. Time fractional diffusion models have many different forms of expression based on different application areas, such as the fractional Cable model [13,14,15,16,17], fractional mobile/immobile transport equation [18], fractional reaction-diffusion model [19,20,21,22,23,24,25,26,27,28,29], fractional fourth-order diffusion system [30, 31] and multi-term fractional subdiffusion model [32]. Space fractional diffusion problems can be expressed as fractional Allen–Cahn models [33,34,35,36], space-fractional reaction-diffusion models [37,38,39,40,41], space-time fractional diffusion problems [42,43,44,45,46]. In these fractional diffusion models, the coupled time-fractional diffusion systems are a kind of important mathematical models, which can reflect the interaction between different substances. Recently, some authors considered this kind of fractional coupled diffusion system. Hou et al. [47] developed a mixed finite element method for a coupled time fractional diffusion equation with a nonlinear term. Authors studied the L1-approximation for the Caputo time fractional derivative and obtained the temporal error result with $O(\tau ^{2-\alpha }+\tau ^{2-\beta })$. In [48], Kumar et al. considered the Galerkin finite element schemes based on the Crank–Nicolson algorithm for a coupled time-fractional nonlinear diffusion model.

In this paper, we consider a mixed element method to solve the following time fractional coupled sub-diffusion system

$$\begin{aligned} \left\{ \begin{aligned}&\frac{\partial u}{\partial t}+k_1\frac{\partial ^\alpha u}{\partial t^\alpha }=a \Delta u+f(u,v)+\bar{f}({\mathbf{x}} ,t),\quad ({\mathbf{x}} ,t)\in \Omega \times J,\\&\frac{\partial v}{\partial t}+k_2\frac{\partial ^\beta v}{\partial t^\beta }=c\Delta v+g(u,v)+\bar{g}({\mathbf{x}} ,t),\quad ({\mathbf{x}} ,t)\in \Omega \times J,\\&u({\mathbf{x}} ,t)=v({\mathbf{x}} ,t)=0,\quad ({\mathbf{x}} ,t)\in \partial \Omega \times \bar{J},\\&u({\mathbf{x}} ,0)=u_0({\mathbf{x}} ),v({\mathbf{x}} ,0)=v_0({\mathbf{x}} ),\quad {\mathbf{x}} \in \bar{\Omega }, \end{aligned}\right. \end{aligned}$$

(1)

where $\Omega \subset R^d (d\leqslant 2)$ is a bounded convex polygonal region satisfying Lipschitz continuous boundary $\partial \Omega$, and $J=(0,T]$ is the time interval with $0<T<\infty$. $u_0({\mathbf{x}} ), v_0({\mathbf{x}} ), \bar{f}({\mathbf{x}} ,t)$ and $\bar{g}({\mathbf{x}} ,t)$ are known functions. Parameters $k_1, k_2, a$ and c are positive constants. The nonlinear terms f(u, v) and g(u, v) are two different quadratic polynomials without constant terms about u and v. The notation $\frac{\partial ^\gamma u}{\partial t^\gamma }$ is the Caputo fractional derivative defined by

$$\begin{aligned} \frac{\partial ^\gamma u({\mathbf{x}} ,t)}{\partial t^\gamma }=\frac{1}{\Gamma (1-\gamma )}\int _0^t\frac{\partial u({\mathbf{x}} ,s)}{\partial s}\frac{{\text {d}}s}{(t-s)^\gamma },\quad (0<\gamma <1), \end{aligned}$$

(2)

where $\gamma =\alpha$ or $\beta$.

In this article, our main purpose is to present a second-order Crank–Nicolson mixed element algorithm for solving the two-dimensional nonlinear time fractional coupled sub-diffusion system, where the time fractional derivative is approximated by a weighted and shifted Grünwald difference (WSGD) formula, which was proposed by Tian et al. [49] and were also developed by Wang and Vong [50], Liu et al. [13], Liu et al. [15], Du et al. [51]. There exist some researches on the mixed element methods, see [47, 52, 53, 56]. In [47], Hou et al. developed a mixed element method for a one-dimensional time fractional coupled system, and obtained a lower time convergence rate. In [52], authors considered the mixed element method with a first-order time approximation scheme for a linear fractional sub-diffusion problem. In [53], Zhao et al. used the L1 approximation with the mixed element method for a linear fractional diffusion model. In [54], Li et al. solved a fractional diffusion wave model by using a mixed element method. In [56], Abbaszadeh and Dehghan solved a linear fractional reaction-diffusion model by using a mixed element method, derived the error estimate, and implemented the numerical tests. Here, we will show the detailed numerical analyses and calculations of the mixed element method with a time second-order convergence order for solving the two-dimensional nonlinear coupled time fractional diffusion model covering initial-boundary conditions. Based on these considerations, our main results are as the following:

• Compared with the standard finite element method, the approximation solutions for unknown scalar functions and their fluxes can be obtained by the mixed element method.

• The existence and uniqueness of the mixed element solution, the stability of the considered mixed element system and the a priori error estimates for unknown scalar functions and their fluxes are proven or derived.

• A two-dimensional example for verifying the spatial convergence rate and a one-dimensional example for checking the second-order time convergence order are provided, respectively.

Throughout the full text, we will use $C>0$ as a positive constant, which is independent of space mesh h, time step $\tau$ and fractional parameters $\alpha ,\beta$. The structure of the article is as follows: In Sect. 2, based on the definition of the Caputo fractional derivative, some approximation formulas for the time fractional derivatives are provided and the mixed finite element scheme is proposed. In Sect. 3, the existence and uniqueness of the mixed element solution are proved and the stability is derived. In Sect. 4, some a priori error estimates are obtained. In Sect. 5, the numerical examples are provided to verify the theory results. In Sect. 6, the remarking conclusions are summarized.

2 Mixed element approximation

By introducing two auxiliary variables $\sigma =\nabla u$ and $\lambda =\nabla v$, the coupled system (1) can be rewritten as

$$\begin{aligned} \left\{ \begin{aligned}&\frac{\partial u}{\partial t}+k_1\frac{\partial ^\alpha u}{\partial t^\alpha }=a\nabla \cdot \sigma +f(u,v)+\bar{f}({\mathbf{x}} ,t),\\&\frac{\partial v}{\partial t}+k_2\frac{\partial ^\beta v}{\partial t^\beta }=c\nabla \cdot \lambda +g(u,v)+\bar{g}({\mathbf{x}} ,t),\\&\sigma -\nabla u=0,\\&\lambda -\nabla v=0. \end{aligned}\right. \end{aligned}$$

(3)

In the following content, we will give the fully discrete scheme.

Let $0=t_0<t_1<t_2<\cdots <t_M=T$ be a grid partition for the temporal interval [0, T], where $t_n=n\tau$, $\tau =\frac{T}{M}$ is the time step and M is a positive integer. With $u^{n}=u({\mathbf{x}} ,t_{n})$, we introduce the following notations:

$$\begin{aligned} \begin{aligned}&D_tl^{n+\frac{1}{2}}=\frac{l^{n+1}-l^{n}}{\tau },\quad l^{n+\frac{1}{2}}=\frac{l^{n+1}+l^n}{2},\\&\sum _{i=0}^{n}p_\gamma (i)l^{n+\frac{1}{2}-i} =\frac{1}{2}\left( \sum _{i=0}^{n+1}p_\gamma (i)l^{n+1-i}+\sum _{i=0}^np_\gamma (i)l^{n-i}\right) ,\\&l(u^{n+\frac{1}{2}},v^{n+\frac{1}{2}}) ={\left\{ \begin{array}{ll}\frac{3}{2}l(u^n,v^n)-\frac{1}{2}l(u^{n-1},v^{n-1}),\quad &{}n\geqslant 1,\\ l(u^0,v^0),\quad &{}n=0.\end{array}\right. }\\ \end{aligned} \end{aligned}$$

(4)

Lemma 1

Following Refs.[15, 50], we can get the discrete formulation of the fractional derivative (2) as follows

$$\begin{aligned} \begin{aligned} \frac{\partial ^\gamma l({\mathbf{x}} ,t_{n+\frac{1}{2}})}{\partial t^\gamma }&=\sum _{i=0}^{n}\frac{p_\gamma (i)}{\tau ^\gamma } l^{n+\frac{1}{2}-i}+O(\tau ^2), \end{aligned} \end{aligned}$$

(5)

where

$$\begin{aligned}\begin{aligned} p_\gamma (i)&=\left\{ \begin{array}{ll} \frac{\gamma +2}{2}g_0^\gamma ,\quad &{}i=0,\\ \frac{\gamma +2}{2}g_i^\gamma +\frac{-\gamma }{2}g_{i-1}^\gamma ,\quad &{}i\geqslant 1, \end{array} \right. \\&g_0^\gamma =1,g_i^\gamma =\frac{\Gamma (i-\gamma )}{\Gamma (-\gamma )\Gamma (i+1)},g_i^\gamma =\left( 1-\frac{\gamma +1}{i}\right) g_{i-1}^\gamma ,\quad i\geqslant 1. \end{aligned}\end{aligned}$$

Based on Lemma 1 and Refs. [30, 50], we can get the following time discrete formulation of the coupled system (3) at time $t_{n+\frac{1}{2}}$

Case 1: $n\geqslant 1$

$$\begin{aligned} \left\{ \begin{aligned} (a)&D_tu^{n+\frac{1}{2}}+k_1\tau ^{-\alpha }\sum _{i=0}^{n}p_\alpha (i)u^{n+\frac{1}{2}-i}-a\nabla \cdot \sigma ^{n+\frac{1}{2}} =f(u^{n+\frac{1}{2}},v^{n+\frac{1}{2}})+\bar{f}^{n+\frac{1}{2}}+R_1,\\ (b)&\sigma ^{n+\frac{1}{2}}-\nabla u^{n+\frac{1}{2}}=R_{2},\\ (c)&D_tv^{n+\frac{1}{2}},+k_2\tau ^{-\beta }\sum _{i=0}^{n}p_\beta (i)v^{n+\frac{1}{2}-i}-c\nabla \cdot \lambda ^{n+\frac{1}{2}} =g(u^{n+\frac{1}{2}},v^{n+\frac{1}{2}})+\bar{g}^{n+\frac{1}{2}}+R_3,\\ (d)&\lambda ^{n+\frac{1}{2}}-\nabla v^{n+\frac{1}{2}}=R_{4}, \end{aligned}\right. \end{aligned}$$

(6)

Case 2: $n=0$

$$\begin{aligned} \left\{ \begin{aligned} (a)&D_tu^{\frac{1}{2}}+k_1\tau ^{-\alpha }\sum _{i=0}^{0}p_\alpha (i)u^{\frac{1}{2}-i}-a\nabla \cdot \sigma ^{\frac{1}{2}} =f(u^{\frac{1}{2}},v^{\frac{1}{2}})+\bar{f}^{\frac{1}{2}}+R_5,\\ (b)&\sigma ^{\frac{1}{2}}-\nabla u^{\frac{1}{2}}=R_{6},\\ (c)&D_tv^{\frac{1}{2}},+k_2\tau ^{-\beta }\sum _{i=0}^{0}p_\beta (i)v^{\frac{1}{2}-i}-c\nabla \cdot \lambda ^{\frac{1}{2}} =g(u^{\frac{1}{2}},v^{\frac{1}{2}})+\bar{g}^{\frac{1}{2}}+R_7,\\ (d)&\lambda ^{\frac{1}{2}}-\nabla v^{\frac{1}{2}}=R_{8}, \end{aligned}\right. \end{aligned}$$

(7)

where

$$\begin{aligned}&R_1=E_1^{n+\frac{1}{2}}+E_3^{n+\frac{1}{2}}+E_5^{n+\frac{1}{2}} +E_{11}^{n+\frac{1}{2}}+E_{13}^{n+\frac{1}{2}}=O(\tau ^2), R_2=E_7^{n+\frac{1}{2}}+E_9^{n+\frac{1}{2}}=O(\tau ^2),\\&R_3=E_2^{n+\frac{1}{2}}+E_4^{n+\frac{1}{2}}+E_6^{n+\frac{1}{2}} +E_{12}^{n+\frac{1}{2}}+E_{14}^{n+\frac{1}{2}}=O(\tau ^2), R_4=E_8^{n+\frac{1}{2}}+E_{10}^{n+\frac{1}{2}}=O(\tau ^2),\\&R_5=E_1^{\frac{1}{2}}+E_3^{\frac{1}{2}}+E_5^{\frac{1}{2}} +E_{11}^{\frac{1}{2}}+E_{13}^{\frac{1}{2}}=O(\tau ), R_6=E_7^{\frac{1}{2}}+E_9^{\frac{1}{2}}=O(\tau ^2),\\&R_7=E_2^{\frac{1}{2}}+E_4^{\frac{1}{2}}+E_6^{\frac{1}{2}} +E_{12}^{\frac{1}{2}}+E_{14}^{\frac{1}{2}}=O(\tau ), R_8=E_8^{\frac{1}{2}}+E_{10}^{\frac{1}{2}}=O(\tau ^2),\\&E_1^{n+\frac{1}{2}}=u({t_{n+\frac{1}{2}}})-D_tu^{n+\frac{1}{2}} ={\left\{ \begin{array}{ll}O(\tau ^2),n\geqslant 1,\\ O(\tau ),n=0,\end{array}\right. }\\&E_2^{n+\frac{1}{2}}=v({t_{n+\frac{1}{2}}})-D_tv^{n+\frac{1}{2}} ={\left\{ \begin{array}{ll}O(\tau ^2),n\geqslant 1,\\ O(\tau ),n=0,\end{array}\right. }\\&E_3^{n+\frac{1}{2}}=k_1\frac{\partial ^\alpha u(t_{n+\frac{1}{2}})}{\partial t^\alpha }-k_1\sum _{i=0}^{n}\frac{p_\alpha (i)}{\tau ^\alpha }u^{n+\frac{1}{2}-i}=O(\tau ^2),\\&E_4^{n+\frac{1}{2}}=k_2\frac{\partial ^\beta v(t_{n+\frac{1}{2}})}{\partial t^\beta }-k_2\sum _{i=0}^{n}\frac{p_\beta (i)}{\tau ^\beta }v^{n+\frac{1}{2}-i}=O(\tau ^2),\\&E_5^{n+\frac{1}{2}}=a\nabla \cdot \sigma (t_{n+\frac{1}{2}})-a\nabla \cdot \sigma ^{n+\frac{1}{2}}=O(\tau ^2),\\&E_6^{n+\frac{1}{2}}=c\nabla \cdot \lambda (t_{n+\frac{1}{2}})-c\nabla \cdot \lambda ^{n+\frac{1}{2}}=O(\tau ^2),\\&E_7^{n+\frac{1}{2}}=\sigma (t_{n+\frac{1}{2}})-\sigma ^{n+\frac{1}{2}}=O(\tau ^2), E_8^{n+\frac{1}{2}}=\lambda (t_{n+\frac{1}{2}})-\lambda ^{n+\frac{1}{2}}=O(\tau ^2),\\&E_9^{n+\frac{1}{2}}=\nabla u(t_{n+\frac{1}{2}})-\nabla u^{n+\frac{1}{2}}=O(\tau ^2), E_{10}^{n+\frac{1}{2}}=\nabla v(t_{n+\frac{1}{2}})-\nabla v^{n+\frac{1}{2}}=O(\tau ^2),\\&E_{11}^{n+\frac{1}{2}}=f(u(t_{n+\frac{1}{2}}),v(t_{n+\frac{1}{2}}))-f(u^{n+\frac{1}{2}},v^{n+\frac{1}{2}}) ={\left\{ \begin{array}{ll}O(\tau ^2),n\geqslant 1,\\ O(\tau ),n=0,\end{array}\right. }\\&E_{12}^{n+\frac{1}{2}}=g(u(t_{n+\frac{1}{2}}),v(t_{n+\frac{1}{2}}))-g(u^{n+\frac{1}{2}},v^{n+\frac{1}{2}}) ={\left\{ \begin{array}{ll}O(\tau ^2),n\geqslant 1,\\ O(\tau ),n=0,\end{array}\right. }\\&E_{13}^{n+\frac{1}{2}}=\bar{f}(t_{n+\frac{1}{2}})-\bar{f}^{n+\frac{1}{2}}=O(\tau ^2), E_{14}^{n+\frac{1}{2}}=\bar{g}(t_{n+\frac{1}{2}})-\bar{g}^{n+\frac{1}{2}}=O(\tau ^2). \end{aligned}$$

Remark 1

Based on the fourth formula in (4), we easily see that the coupled system (6)–(7) is a linearized system. In actual calculation, we can also approximate the nonlinear term $l(u(t_{n+\frac{1}{2}}),v(t_{n+\frac{1}{2}}))$ by the linearized term $l(\frac{3}{2}u^n-\frac{1}{2}u^{n-1},\frac{3}{2}v^n-\frac{1}{2}v^{n-1})$, where l is taken as f or g and $n\geqslant 1$ is a positive integer.

Based on the coupled system (6)–(7), we find $\{u,v;\sigma ,\lambda \}:[0,T]\mapsto H_0^1\times (L^2(\Omega ))^2$ satisfying the following mixed weak formulation.

Case 1: $n\geqslant 1$

$$\begin{aligned} \left\{ \begin{aligned} (a)&(D_tu^{n+\frac{1}{2}},w)+k_1\tau ^{-\alpha }\left( \sum _{i=0}^{n}p_\alpha (i) u^{n+\frac{1}{2}-i},w\right) +(a\sigma ^{n+\frac{1}{2}},\nabla w)\\&=(f(u^{n+\frac{1}{2}},v^{n+\frac{1}{2}}),w)+(\bar{f}^{n+\frac{1}{2}},w)+(R_1,w), \quad \forall w\in H_0^1,\\ (b)&(\sigma ^{n+\frac{1}{2}},\varvec{z})-(\nabla u^{n+\frac{1}{2}},\varvec{z})=(R_{2},\varvec{z}),\quad \forall \varvec{z}\in (L^2(\Omega ))^2,\\ (c)&(D_tv^{n+\frac{1}{2}},w)+k_2\tau ^{-\beta }\left( \sum _{i=0}^{n}p_\beta (i) v^{n+\frac{1}{2}-i},w\right) +(c\lambda ^{n+\frac{1}{2}},\nabla w)\\&=(g(u^{n+\frac{1}{2}},v^{n+\frac{1}{2}}),w)+(\bar{g}^{n+\frac{1}{2}},w)+(R_3,w), \quad \forall w\in H_0^1,\\ (d)&(\lambda ^{n+\frac{1}{2}},\varvec{z})-(\nabla v^{n+\frac{1}{2}},\varvec{z})=(R_{4},\varvec{z}), \quad \forall \varvec{z}\in (L^2(\Omega ))^2. \end{aligned}\right. \end{aligned}$$

(8)

Case 2: $n=0$

$$\begin{aligned} \left\{ \begin{aligned} (a)&(D_tu^{\frac{1}{2}},w)+k_1\tau ^{-\alpha }\left( \sum _{i=0}^{0}p_\alpha (i) u^{\frac{1}{2}-i},w\right) +(a\sigma ^{\frac{1}{2}},\nabla w)\\&=(f(u^{\frac{1}{2}},v^{\frac{1}{2}}),w)+(\bar{f}^{\frac{1}{2}},w)+(R_5,w),\quad \forall w\in H_0^1,\\ (b)&(\sigma ^{\frac{1}{2}},\varvec{z})-(\nabla u^{\frac{1}{2}},\varvec{z})=(R_6,\varvec{z}), \quad \forall \varvec{z}\in (L^2(\Omega ))^2,\\ (c)&(D_tv^{\frac{1}{2}},w)+k_2\tau ^{-\beta } \left( \sum _{i=0}^{0}p_\beta (i) v^{\frac{1}{2}-i},w\right) +(c\lambda ^{\frac{1}{2}},\nabla w) \\&=(g(u^{\frac{1}{2}},v^{\frac{1}{2}}),w)+(\bar{g}^{\frac{1}{2}},w)+(R_7,w),\quad \forall w\in H_0^1,\\ (d)&(\lambda ^{\frac{1}{2}},\varvec{z})-(\nabla v^{\frac{1}{2}},\varvec{z})=(R_8,\varvec{z}), \quad \forall \varvec{z}\in (L^2(\Omega ))^2. \end{aligned}\right. \end{aligned}$$

(9)

Then we introduce the mixed element space $(W_h,\varvec{Z}_h)$ as follows

$$\begin{aligned} \begin{aligned}&W_h=\{w_h\in C^0(\Omega )\cap H_0^1|w_h\in P_1(K),\forall K\in {\mathcal {K}}_h\},\\&\varvec{Z}_h=\{\varvec{z}_h=(z_{1h},z_{2h})\in (L^2(\Omega ))^2|z_{1h},z_{2h}\in P_0(K),\forall K\in {\mathcal {K}}_h\}. \end{aligned}\end{aligned}$$

(10)

Considering the above mixed element space, the fully discrete scheme of (8)–(9) is to find $\{u_h,v_h;\sigma _h,\lambda _h\}:[0,T]\mapsto W_h\times \varvec{Z}_h$ such that

Case 1: $n\geqslant 1$

$$\begin{aligned} \left\{ \begin{aligned} (a)&(D_tu_h^{n+\frac{1}{2}},w_h)+k_1\tau ^{-\alpha }\left( \sum _{i=0}^{n}p_\alpha (i) u_h^{n+\frac{1}{2}-i},w_h\right) +(a\sigma _h^{n+\frac{1}{2}},\nabla w_h)\\&=(f(u_h^{n+\frac{1}{2}},v_h^{n+\frac{1}{2}}),w_h)+(\bar{f}^{n+\frac{1}{2}},w_h), \quad \forall w_h\in W_h,\\ (b)&(\sigma _h^{n+\frac{1}{2}},\varvec{z}_h)-(\nabla u_h^{n+\frac{1}{2}}, \varvec{z}_h)=0, \quad \forall \varvec{z}_h\in \varvec{Z}_h,\\ (c)&(D_tv_h^{n+\frac{1}{2}},w_h)+k_2\tau ^{-\beta } \left( \sum _{i=0}^{n}p_\beta (i) v_h^{n+\frac{1}{2}-i},w_h\right) +(c\lambda _h^{n+\frac{1}{2}},\nabla w_h)\\&=(g(u_h^{n+\frac{1}{2}},v_h^{n+\frac{1}{2}}),w_h)+(\bar{g}^{n+\frac{1}{2}}, w_h),\quad \forall w_h\in W_h,\\ (d)&(\lambda _h^{n+\frac{1}{2}},\varvec{z}_h)-(\nabla v_h^{n+\frac{1}{2}}, \varvec{z}_h)=0, \quad \forall \varvec{z}_h\in \varvec{Z}_h. \end{aligned}\right. \end{aligned}$$

(11)

Case 2: $n=0$

$$\begin{aligned} \left\{ \begin{aligned} (a)&(D_tu_h^{\frac{1}{2}},w_h)+k_1\tau ^{-\alpha }\left( \sum _{i=0}^{0}p_{\alpha }(i) u_h^{\frac{1}{2}-i},w_h\right) +(a\sigma _h^{\frac{1}{2}},\nabla w_h)\\&=(f(u^{\frac{1}{2}}_h,v^{\frac{1}{2}}_h),w_h)+(\bar{f}^{\frac{1}{2}},w_h),\quad \forall w_h\in W_h,\\ (b)&(\sigma ^{\frac{1}{2}}_h,\varvec{z}_h)-(\nabla u_h^{\frac{1}{2}},\varvec{z}_h)=0, \quad \forall \varvec{z}_h\in \varvec{Z}_h,\\ (c)&(D_tv_h^{\frac{1}{2}},w_h)+k_2\tau ^{-\beta }\left( \sum _{i=0}^{0}p_{\beta }(i) v_h^{\frac{1}{2}-i},w_h\right) +(c\lambda _h^{\frac{1}{2}},\nabla w_h)\\&=(g(u^{\frac{1}{2}}_h,v^{\frac{1}{2}}_h),w_h)+(\bar{g}^{\frac{1}{2}},w_h),\quad \forall w_h\in W_h,\\ (d)&(\lambda _h^{\frac{1}{2}},\varvec{z}_h)-(\nabla v_h^{\frac{1}{2}},\varvec{z}_h)=0, \quad \forall \varvec{z}_h\in \varvec{Z}_h. \end{aligned}\right. \end{aligned}$$

(12)

3 Existence, uniqueness and stability

3.1 Existence and uniqueness of the mixed element solution

Here, we will discuss the existence and uniqueness of the mixed element solution.

Theorem 1

There exists a unique solution for the coupled mixed element system (11)–(12).

Proof

Now take FE space $W_h=span\{{\phi _i}\}_{i=1}^{M_h}$, $\varvec{Z}_h=span\{\varvec{\psi }_{j}\}_{j=1}^{N_h}$. For any $u_h,v_h\in W_h$, and any $\varvec{\sigma }_h,\varvec{\lambda }_h\in \varvec{Z}_h$, we have

$$\begin{aligned} u_h^n=\sum _{i=1}^{M_h}u_i^n\phi _i, v_h^n=\sum _{i=1}^{M_h}v_i^n\phi _i, \varvec{\sigma }_h^n=\sum _{j=1}^{N_h}\sigma _j^n\varvec{\psi }_j, \varvec{\lambda }_h^n=\sum _{j=1}^{N_h}\lambda _j^n\varvec{\psi }_j. \end{aligned}$$

Choosing $w_h=\phi _m$, $\varvec{z}_h=\varvec{\psi }_l$ $(1\leqslant m\leqslant M_h, 1\leqslant l\leqslant N_h)$, and substituting them into the mixed element system (11), we have for $n\geqslant 1$

$$\begin{aligned} \left\{ \begin{aligned} (a)&\,\sum _{i=1}^{M_h}D_tu_i^{n+\frac{1}{2}}(\phi _i,\phi _m)+k_1\tau ^{-\alpha }\sum _{k=0}^{n}p_\alpha (k) \left( \sum _{i=1}^{M_h}u_i^{n+\frac{1}{2}-k}(\phi _i,\phi _m)\right) +a\sum _{j=1}^{N_h}\sigma _j^{n+\frac{1}{2}}(\varvec{\psi }_j,\nabla \phi _m)\\&\,=\left( \frac{3}{2}f(\sum _{i=1}^{M_h}u_i^n\phi _i,\sum _{i=1}^{M_h}v_i^n\phi _i) -\frac{1}{2}f(\sum _{i=1}^{M_h}u_i^{n-1}\phi _i,\sum _{i=1}^{M_h}v_i^{n-1}\phi _i),\phi _m\right) +(f^{n+\frac{1}{2}},\phi _m),\\ (b)&\,\sum _{j=1}^{N_h}\sigma _j^{n+\frac{1}{2}}(\varvec{\psi }_j,\varvec{\psi }_l) -\sum _{i=1}^{M_h}u_i^{n+\frac{1}{2}}(\nabla \phi _i,\varvec{\psi }_l)=0,\\ (c)&\,\sum _{i=1}^{M_h}D_tv_i^{n+\frac{1}{2}}(\phi _i,\phi _m)+k_2\tau ^{-\beta }\sum _{k=0}^{n}p_{\beta }(k) \left( \sum _{i=1}^{M_h}v_i^{n+\frac{1}{2}-k}(\phi _i,\phi _m)\right) +c\sum _{j=1}^{N_h}\lambda _j^{n+\frac{1}{2}}(\varvec{\psi }_j,\nabla \phi _m)\\&\,=\left( \frac{3}{2}g(\sum _{i=1}^{M_h}u_i^n\phi _i,\sum _{i=1}^{M_h}v_i^n\phi _i) -\frac{1}{2}g(\sum _{i=1}^{M_h}u_i^{n-1}\phi _i,\sum _{i=1}^{M_h}v_i^{n-1}\phi _i), \phi _m\right) +(g^{n+\frac{1}{2}},\phi _m),\\ (d)&\,\sum _{j=1}^{N_h}\lambda _j^{n+\frac{1}{2}}(\varvec{\psi }_j,\varvec{\psi }_l) -\sum _{i=1}^{M_h}v_i^{n+\frac{1}{2}}(\nabla \phi _i,\varvec{\psi }_l)=0.\\ \end{aligned}\right. \end{aligned}$$

(13)

According to the above formula, we get the following matrix equations for $n\geqslant 1$

$$\begin{aligned} \left\{ \begin{aligned} (a)&\,\left( \tau ^{-1}+\frac{k_1}{2}p_{\alpha }(0)\tau ^{-\alpha }\right) \varvec{A}\varvec{u}^{n+1} +\frac{a}{2}\varvec{B}\varvec{\sigma }^{n+1}\\&\,=\tau ^{-1}\varvec{A}\varvec{u}^{n}-\frac{k_1}{2}\tau ^{-\alpha }\sum _{k=0}^{n} (p_\alpha (k+1)+p_\alpha (k)) \varvec{A}\varvec{u}^{n-k}-\frac{a}{2}\varvec{B}\varvec{\sigma }^{n} +\frac{3}{2}\varvec{F}^{n}-\frac{1}{2}\varvec{F}^{n-1} +\bar{\varvec{F}}^{n+\frac{1}{2}}\\&\,\triangleq \varvec{H}(\varvec{u}^{n},\varvec{u}^{n-1},\cdots ,\varvec{u}^{0}; \varvec{\sigma }^{n};\varvec{F}^{n},\varvec{F}^{n-1};\bar{\varvec{F}}^{n+\frac{1}{2}}) \triangleq \varvec{H}^{n},\\ (b)&\,\varvec{C}\varvec{\sigma }^{n+1}-\varvec{D}\varvec{u}^{n+1}=-\varvec{C}\varvec{\sigma }^{n} +\varvec{D}\varvec{u}^{n}\triangleq \varvec{I}(\varvec{u}^{n},\varvec{\sigma }^{n})\triangleq \varvec{I}^{n},\\ (c)&\,\left( \tau ^{-1}+\frac{k_2}{2}p_{\beta }(0)\tau ^{-\beta }\right) \varvec{A}\varvec{v}^{n+1} +\frac{c}{2}\varvec{B}\varvec{\lambda }^{n+1}\\&\,=\tau ^{-1}\varvec{A}\varvec{v}^{n}-\frac{k_2}{2}\tau ^{-\beta }\sum _{k=0}^{n}(p_\beta (k+1)+p_\beta (k)) \varvec{A}\varvec{v}^{n-k}-\frac{c}{2}\varvec{B}\varvec{\lambda }^{n} +\frac{3}{2}\varvec{G}^{n}-\frac{1}{2}\varvec{G}^{n-1}+\bar{\varvec{G}}^{n+\frac{1}{2}}\\&\,\triangleq \varvec{J}(\varvec{v}^{n},\varvec{v}^{n-1},\cdots ,\varvec{v}^{0}; \varvec{\lambda }^{n};\varvec{G}^{n},\varvec{G}^{n-1};\bar{\varvec{G}}^{n+\frac{1}{2}}) \triangleq \varvec{J}^{n},\\ (d)&\,\varvec{C}\varvec{\lambda }^{n+1}-\varvec{D}\varvec{v}^{n+1}=-\varvec{C}\varvec{\lambda }^{n} +\varvec{D}\varvec{v}^{n}\triangleq \varvec{K}(\varvec{v}^{n},\varvec{\lambda }^{n})\triangleq \varvec{K}^{n},\\ \end{aligned}\right. \end{aligned}$$

(14)

where

$$\begin{aligned}&\,\varvec{A}=\Big [(\phi _i,\phi _m)\Big ]^T_{1\leqslant i,m \leqslant M_h}, \varvec{B}=\Big [(\varvec{\psi }_j,\nabla \phi _m)\Big ]^T_{1\leqslant j \leqslant N_h,1\leqslant m \leqslant M_h,},\\&\,\varvec{C}=\Big [(\varvec{\psi }_j,\varvec{\psi }_l)\Big ]^T_{1\leqslant j,l \leqslant N_h,}, \varvec{D}=\Big [(\nabla \phi _i,\varvec{\psi }_l)\Big ]^T_{1\leqslant i \leqslant M_h,1\leqslant l \leqslant N_h,},\\&\,\varvec{F}^n=\Big [\left( f(\sum _{i=0}^{M_h}u_i^n\phi _i,\sum _{i=0}^{M_h}v_i^n\phi _i),\phi _1\right) ,\cdots , \left( f(\sum _{i=0}^{M_h}u_i^n\phi _i,\sum _{i=0}^{M_h}v_i^n\phi _i),\phi _{M_h}\right) \Big ]^{T},\\&\,\varvec{G}^n=\Big [\left( g(\sum _{i=0}^{M_h}u_i^n\phi _i,\sum _{i=0}^{M_h}v_i^n\phi _i),\phi _1\right) ,\cdots , \left( g(\sum _{i=0}^{M_h}u_i^n\phi _i,\sum _{i=0}^{M_h}v_i^n\phi _i),\phi _{M_h}\right) \Big ]^{T},\\&\,\bar{\varvec{F}}=\Big [(\bar{f}^n,\phi _1),\cdots ,(\bar{f}^n,\phi _{M_h})\Big ]^{T}, \bar{\varvec{G}}=\Big [(\bar{g}^n,\phi _1),\cdots ,(\bar{g}^n,\phi _{M_h})\Big ]^{T},\\&\,\varvec{u}^n=[u_1^n,u_2^n,\cdots ,u_{M_h}^n]^T, \varvec{v}^n=[v_1^n,v_2^n,\cdots ,v_{M_h}^n]^T,\\&\,\varvec{\sigma }^n=[\sigma _1^n,\sigma _2^n,\cdots ,\sigma _{N_h}^n]^T, \varvec{\lambda }^n=[\lambda _1^n,\lambda _2^n,\cdots ,\lambda _{N_h}^n]^T. \end{aligned}$$

We rewrite (14) as the following matrix formulation for $n\geqslant 1$

$$\begin{aligned} \varvec{L}\varvec{X}^{n+1}=\varvec{Y}^n, \end{aligned}$$

(15)

where

$$\begin{aligned}\begin{aligned}&\varvec{L}=\left[ \begin{array}{cccc} (\tau ^{-1}+\frac{k_1}{2}p_{\alpha }(0)\tau ^{-\alpha })\varvec{A}&{}\frac{a}{2}\varvec{B}&{}&{}\\ -\varvec{D}&{}\varvec{C}&{}&{}\\ &{}&{}(\tau ^{-1}+\frac{k_2}{2}p_{\beta }(0)\tau ^{-\beta })\varvec{A}&{}\frac{c}{2}\varvec{B}\\ &{}&{}-\varvec{D}&{}\varvec{C} \end{array} \right] ,\\&\varvec{X}^{n+1}=\left[ \begin{array}{c} \varvec{u}^{n+1}\\ \varvec{\sigma }^{n+1}\\ \varvec{v}^{n+1}\\ \varvec{\lambda }^{n+1} \end{array} \right] ,\qquad \varvec{Y}^n= \left[ \begin{array}{c} \varvec{H}^{n}\\ \varvec{I}^{n}\\ \varvec{J}^{n}\\ \varvec{K}^{n} \end{array} \right] . \end{aligned} \end{aligned}$$

Next, we need to consider the boundary condition $w|_{\partial \Omega }=0,\forall w\in W_h$. Without loss of generality, we assume that all the boundary points are at the front, i.e.

$$\begin{aligned} \begin{aligned}&\varvec{u}^n=[0,\cdots ,0,u_{N+1}^n,u_{N+2}^n,\cdots ,u_{M_h}^n]^T =[\varvec{0},\tilde{\varvec{u}}^n]^T,\\&\varvec{v}^n=[0,\cdots ,0,v_{N+1}^n,v_{N+2}^n,\cdots ,v_{M_h}^n]^T =[\varvec{0},\tilde{\varvec{v}}^n]^T. \end{aligned} \end{aligned}$$

(16)

Based on (16), the $p_i$ th row and the $q_i$ th column of $\varvec{L}$ should be crossed off, where $p_i$ and $q_i$ satisfy $1\leqslant p_1,q_1 \leqslant N$ and $M_h+N_h+1\leqslant p_2,q_2 \leqslant M_h+N_h+N$. At the same time, the $p_i$ th row of $\varvec{Y}^{n}$ and $\varvec{Y}^{n}$ should be removed. By modifying the dimensions of total stiffness matrix $\varvec{L}$ and total load vector $\varvec{Y}^{n}$, we can get the following matrix equations for $n \geqslant 1$

$$\begin{aligned} \begin{aligned} \left[ \begin{array}{cccc} (\tau ^{-1}+\frac{k_1}{2}p_{\alpha }(0)\tau ^{-\alpha })\tilde{\varvec{A}}&{}\frac{a}{2}\tilde{\varvec{B}}&{}&{}\\ -\tilde{\varvec{D}}&{}\varvec{C}&{}&{}\\ &{}&{}(\tau ^{-1}+\frac{k_2}{2}p_{\beta }(0)\tau ^{-\beta })\tilde{\varvec{A}}&{}\frac{c}{2}\tilde{\varvec{B}}\\ &{}&{}-\tilde{\varvec{D}}&{}\varvec{C} \end{array} \right] \left[ \begin{array}{c} \tilde{\varvec{u}}^{n+1}\\ {\varvec{\sigma }}^{n+1}\\ \tilde{\varvec{v}}^{n+1}\\ {\varvec{\lambda }}^{n+1} \end{array} \right] = \left[ \begin{array}{c} \tilde{\varvec{H}}^{n}\\ \varvec{I}^{n}\\ \tilde{\varvec{J}}^{n}\\ \varvec{K}^{n} \end{array} \right] . \end{aligned} \end{aligned}$$

(17)

The form of matrix equations for $n=0$ is similar to the above. Let $\varepsilon _{\alpha }=\frac{1}{\tau ^{-1}+\frac{k_1}{2}p_{\alpha }(0)\tau ^{-\alpha }}$ and $\varepsilon _{\beta }=\frac{1}{\tau ^{-1}+\frac{k_2}{2}p_{\beta }(0)\tau ^{-\beta }}$ and (17) is equivalent to

$$\begin{aligned} \left\{ \begin{aligned}&(a)~\tilde{\varvec{A}}\tilde{\varvec{u}}^{n+1}+\frac{a\varepsilon _{\alpha }}{2}\tilde{\varvec{B}}\varvec{\sigma }^{n+1} =\varepsilon _{\alpha }\tilde{\varvec{H}}^n,\\&(b)~-\tilde{\varvec{D}}\tilde{\varvec{u}}^{n+1}+\varvec{C}\varvec{\sigma }^{n+1}=\varvec{I}^{n},\\&(c)~\tilde{\varvec{A}}\tilde{\varvec{v}}^{n+1}+\frac{c\varepsilon _{\beta }}{2}\tilde{\varvec{B}} \varvec{\lambda }^{n+1}=\varepsilon _{\beta }\tilde{\varvec{J}}^n,\\&(d)~-\tilde{\varvec{D}}\tilde{\varvec{v}}^{n+1}+\varvec{C}\varvec{\lambda }^{n+1}=\varvec{K}^{n}. \end{aligned}\right. \end{aligned}$$

(18)

We know that $\tilde{\varvec{A}}$ and $\varvec{C}$ in (18) are invertible matrixes (Gram matrixes) which are formulated by the inner products of basis functions $\{\phi _i\}_{i=N+1}^{M_h}$ and $\{\varvec{\psi }_j\}_{j=1}^{N_h}$, respectively. So, we can rewrite (18) as

$$\begin{aligned} \left\{ \begin{aligned}&(a)~\left( \varvec{E}+\frac{a\varepsilon _{\alpha }}{2}\tilde{\varvec{A}}^{-1}\tilde{\varvec{B}}\varvec{C}^{-1} \tilde{\varvec{D}}\right) \tilde{\varvec{u}}^{n+1} =\varepsilon _{\alpha }\tilde{\varvec{A}}^{-1}\tilde{\varvec{H}}^{n}-\frac{a\varepsilon _{\alpha }}{2}\tilde{\varvec{A}}^{-1} \tilde{\varvec{B}}\varvec{C}^{-1}\varvec{I}^{n},\\&(b)~\varvec{\sigma }^{n+1}=\varvec{C}^{-1}(\varvec{I}^{n}+\tilde{\varvec{D}}\tilde{\varvec{u}}^{n+1}),\\&(c)~\left( \varvec{E}+\frac{c\varepsilon _{\beta }}{2}\tilde{\varvec{A}}^{-1}\tilde{\varvec{B}}\varvec{C}^{-1} \tilde{\varvec{D}}\right) \tilde{\varvec{v}}^{n+1} =\varepsilon _{\beta }\tilde{\varvec{A}}^{-1}\tilde{\varvec{J}}^{n}-\frac{c\varepsilon _{\beta }}{2}\tilde{\varvec{A}}^{-1} \tilde{\varvec{B}}\varvec{C}^{-1}\varvec{K}^{n}, \\&(d)~\varvec{\lambda }^{n+1}=\varvec{C}^{-1}(\varvec{K}^{n}+\tilde{\varvec{D}}\tilde{\varvec{v}}^{n+1}), \end{aligned}\right. \end{aligned}$$

(19)

where $\varvec{E}$ is an identity matrix. By taking two sufficiently small parameters $\varepsilon _{\alpha }$ and $\varepsilon _{\beta }$, we easily know that $\varvec{E}+\frac{a\varepsilon_{\alpha} }{2}\tilde{\varvec{A}}^{-1}\tilde{\varvec{B}}\varvec{C}^{-1}\tilde{\varvec{D}}$ is an invertible matrix. So, we can obtain a unique iterative solution vector $[\tilde{\varvec{u}}^{n+1},\varvec{\sigma }^{n+1},\tilde{\varvec{v}}^{n+1},\varvec{\lambda }^{n+1}]^{T}$ in (19) based on the computed solution vector $[\tilde{\varvec{u}}^{k},\varvec{\sigma }^{k},\tilde{\varvec{v}}^{k},\varvec{\lambda }^{k}]^{T}$, $k=0,1,2,\cdots ,n$, where the solution vector with $k=1$ can be solved uniquely by adopting a similar process to the case above. Based on the above discussions, one can know that mixed element system has a unique solution.

3.2 Stability analysis

Next, we will give the stability analysis.

Lemma 2

(See [13, 50]) Let $\left\{ p_\gamma (i)\right\}$ be defined as in Lemma 1. Then for any positive integer L and real vector $(w^0,w^1,\cdots ,w^L)\in R^{L+1}$, it holds that

$$\begin{aligned} \begin{aligned} \sum _{n=0}^L \left( \sum _{i=0}^np_\gamma (i)l^{n-i}\right) l^n\geqslant 0. \end{aligned} \end{aligned}$$

(20)

Theorem 2

Let $\{u_h^{n+1}\}_{n=0}^{L}$, $\{v_h^{n+1}\}_{n=0}^{L}$, $\{\sigma _h^{n+1}\}_{n=0}^{L}$, and $\{\lambda _h^{n+1}\}_{n=0}^{L}$ be the solution of the fully discrete scheme (11)–(12). There holds true for $\forall n=0,1,\ldots ,L$ $(0\leqslant L\leqslant M-1)$

$$\begin{aligned} \begin{aligned}&\Vert u_h^{L+1}\Vert ^2+\Vert v_h^{L+1}\Vert ^2+ \tau \sum _{n=0}^L(\Vert \sigma _h^{n+\frac{1}{2}}\Vert ^2+\Vert \lambda _h^{n+\frac{1}{2}}\Vert ^2)\\&\quad \leqslant C\left( \Vert u_h^0\Vert ^2+\Vert v_h^0\Vert ^2+ \max _{0\leqslant n \leqslant M-1}\{\Vert \bar{f}^{n+\frac{1}{2}}\Vert ^2+\Vert \bar{g}^{n+\frac{1}{2}}\Vert ^2\}\right) . \end{aligned} \end{aligned}$$

(21)

Proof

Taking $w_h=2\tau u^{n+\frac{1}{2}}_h$ in (11)(a) and $z_h=2a\tau \sigma ^{n+\frac{1}{2}}_h$ in (11)(b), we can get

$$\begin{aligned} \left\{ \begin{aligned} (a)&\left( \frac{u_h^{n+1}-u_h^n}{\tau },2\tau u_h^{n+\frac{1}{2}}\right) +k_1\tau ^{-\alpha }\left( \sum _{i=0}^{n}p_\alpha (i)u_h^{n+\frac{1}{2}-i},2\tau u_h^{n+\frac{1}{2}}\right) \\&+(a\sigma _h^{n+\frac{1}{2}}, 2\tau \nabla u_h^{n+\frac{1}{2}}) =(f(u_h^{n+\frac{1}{2}},v_h^{n+\frac{1}{2}}),2\tau u^{n+\frac{1}{2}}_h)+(\bar{f}^{n+\frac{1}{2}},2\tau u_h^{n+\frac{1}{2}}),\\ (b)&(\sigma _h^{n+\frac{1}{2}},2a\tau \sigma _h^{n+\frac{1}{2}})-(\nabla u_h^{n+\frac{1}{2}},2a\tau \sigma _h^{n+\frac{1}{2}})=0. \end{aligned}\right. \end{aligned}$$

(22)

Adding (22)(a) and (22)(b), the above formula can be rewritten as

$$\begin{aligned} \begin{aligned}&\Vert u_h^{n+1}\Vert ^2-\Vert u_h^{n}\Vert ^2+2k_1\tau ^{1-\alpha } \left( \sum _{i=0}^{n}p_\alpha (i)u_h^{n+\frac{1}{2}-i},u_h^{n+\frac{1}{2}}\right) +2a\tau \Vert \sigma _h^{n+\frac{1}{2}}\Vert ^2\\&\quad =2\tau (f(u_h^{n+\frac{1}{2}},v_h^{n+\frac{1}{2}}),u_h^{n+\frac{1}{2}})+2\tau (\bar{f}^{n+\frac{1}{2}},u_h^{n+\frac{1}{2}}). \end{aligned} \end{aligned}$$

(23)

Sum (23) for n from 1 to L to get

$$\begin{aligned} \begin{aligned}&\sum _{n=1}^{L}(\Vert u_h^{n+1}\Vert ^2-\Vert u_h^{n}\Vert ^2)+2k_1\tau ^{1-\alpha }\sum _{n=1}^{L}\left( \sum _{i=0}^{n} p_\alpha (i)u_h^{n+\frac{1}{2}-i},u_h^{n+\frac{1}{2}}\right) \\&\quad +2a\tau \sum _{n=1}^{L}\Vert \sigma _h^{n+\frac{1}{2}}\Vert ^2=2\tau \sum _{n=1}^{L} (f(u_h^{n+\frac{1}{2}},v_h^{n+\frac{1}{2}}),u_h^{n+\frac{1}{2}}) +2\tau \sum _{n=1}^{L}(\bar{f}^{n+\frac{1}{2}},u_h^{n+\frac{1}{2}}). \end{aligned} \end{aligned}$$

(24)

Using Hölder inequality inequality and Young inequality and noting that the fourth formula in (4), we have

$$\begin{aligned} \begin{aligned}&\Vert u_h^{L+1}\Vert ^2+2a\tau \sum _{n=1}^{L}\Vert \sigma _h^{n+\frac{1}{2}}\Vert ^2 +2k_1\tau ^{1-\alpha }\sum _{n=1}^{L}\left( \sum _{i=0}^{n}p_\alpha (i)u_h^{n+\frac{1}{2}-i},u_h^{n+\frac{1}{2}}\right) \\&\quad \leqslant \Vert u_h^1\Vert ^2 +\frac{3}{2}\tau \sum _{n=1}^{L}(\Vert u_h^{n}\Vert _{\infty }+\Vert v_h^{n}\Vert _{\infty }) (\Vert u_h^{n}\Vert +\Vert v_h^{n}\Vert )\Vert u_h^{n+\frac{1}{2}}\Vert \\&\qquad +\frac{1}{2}\tau \sum _{n=1}^{L}(\Vert u_h^{n-1}\Vert _{\infty }+\Vert v_h^{n-1}\Vert _{\infty }) (\Vert u_h^{n-1}\Vert +\Vert v_h^{n-1}\Vert )\Vert u_h^{n+\frac{1}{2}}\Vert \\&\qquad +2\tau \sum _{n=1}^{L}\Vert u_h^{n+\frac{1}{2}}\Vert ^2+2\tau \sum _{n=1}^L\Vert \bar{f}^{n+\frac{1}{2}}\Vert ^2 \\&\quad \leqslant \Vert u_h^1\Vert ^2+C\tau \sum _{n=1}^{L}\Vert u_h^{n+\frac{1}{2}}\Vert ^2+C\tau \sum _{n=0}^{L}(\Vert u_h^{n}\Vert ^2+\Vert v_h^{n}\Vert ^2)+2\tau \sum _{n=1}^L\Vert \bar{f}^{n+\frac{1}{2}}\Vert ^2. \end{aligned} \end{aligned}$$

(25)

Similarly, we take $w_h=2\tau v^{n+\frac{1}{2}}_h$ in (11)(c) and $z_h=2\tau c \lambda ^{n+\frac{1}{2}}_h$ in (11)(d), and use Hölder inequality inequality and Young inequality to obtain

$$\begin{aligned} \begin{aligned}&\Vert v_h^{L+1}\Vert ^2+2c\tau \sum _{n=1}^{L}\Vert \lambda _h^{n+\frac{1}{2}}\Vert ^2 +2k_2\tau ^{1-\beta }\sum _{n=1}^{L}\left( \sum _{i=0}^{n}p_\beta (i)v_h^{n+\frac{1}{2}-i},v_h^{n+\frac{1}{2}}\right) \\&\quad \leqslant \Vert v_h^1\Vert ^2+C\tau \sum _{n=1}^{L}\Vert v_h^{n+\frac{1}{2}}\Vert ^2+C\tau \sum _{n=0}^{L}(\Vert u_h^{n}\Vert ^2+\Vert v_h^{n}\Vert ^2) +2\tau \sum _{n=1}^{L}\Vert \bar{g}^{n+\frac{1}{2}}\Vert ^2. \end{aligned} \end{aligned}$$

(26)

Adding (25) and (26), we can get

$$\begin{aligned} \begin{aligned}&\Vert u_h^{L+1}\Vert ^2+\Vert v_h^{L+1}\Vert ^2 +2\tau \sum _{n=1}^L(a\Vert \sigma _h^{n+\frac{1}{2}}\Vert ^2+c\Vert \lambda _h^{n+\frac{1}{2}}\Vert ^2)\\&\qquad +2k_1\tau ^{1-\alpha }\sum _{n=1}^{L}\left( \sum _{i=0}^{n}p_\alpha (i)u_h^{n+\frac{1}{2}-i},u_h^{n+\frac{1}{2}}\right) +2k_2\tau ^{1-\beta }\sum _{n=1}^{L}\left( \sum _{i=0}^{n}p_\beta (i)v_h^{n+\frac{1}{2}-i},v_h^{n+\frac{1}{2}}\right) \\&\quad \leqslant \Vert u_h^1\Vert ^2+\Vert v_h^1\Vert ^2+C\tau \sum _{n=1}^L(\Vert u_h^{n+\frac{1}{2}}\Vert ^2+\Vert v_h^{n+\frac{1}{2}}\Vert ^2)+C\tau \sum _{n=0}^{L}(\Vert u_h^{n}\Vert ^2+\Vert v_h^{n}\Vert ^2)\\&\qquad +2\tau \sum _{n=1}^L(\Vert \bar{f}^{n+\frac{1}{2}}\Vert ^2+\Vert \bar{g}^{n+\frac{1}{2}}\Vert ^2). \end{aligned} \end{aligned}$$

(27)

By the similar way to (27), when $n=0$, we take $w_h=2\tau u_h^{\frac{1}{2}}$ in (12)(a), $z_h=2a\tau \sigma _h^{\frac{1}{2}}$ in (12)(b), $w_h=2\tau v_h^{\frac{1}{2}}$ in (12)(c) and $z_h=2c\tau \lambda _h^{n+\frac{1}{2}}$ in (12)(d) to easily get

$$\begin{aligned} \begin{aligned}&\Vert u_h^{1}\Vert ^2+\Vert v_h^{1}\Vert ^2 +2\tau (a\Vert \sigma _h^{\frac{1}{2}}\Vert ^2+c\Vert \lambda _h^{\frac{1}{2}}\Vert ^2)\\&\qquad +2k_1\tau ^{1-\alpha }\left( \sum _{i=0}^{0}p_\alpha (i)u_h^{\frac{1}{2}-i},u_h^{\frac{1}{2}}\right) +2k_2\tau ^{1-\beta }\left( \sum _{i=0}^{0}p_\beta (i)v_h^{\frac{1}{2}-i},v_h^{\frac{1}{2}}\right) \\&\quad \leqslant C(\Vert u_h^0\Vert ^2+\Vert v_h^0\Vert ^2)+C\tau (\Vert u_h^{\frac{1}{2}}\Vert ^2+\Vert v_h^{\frac{1}{2}}\Vert ^2) +2\tau (\Vert \bar{f}^{\frac{1}{2}}\Vert ^2+\Vert \bar{g}^{\frac{1}{2}}\Vert ^2). \end{aligned} \end{aligned}$$

(28)

By substituting (28) into (27), dropping two nonnegative terms, and using Lemma 2 and triangle inequality, we can get

$$\begin{aligned} \begin{aligned}&(1-C\tau )(\Vert u_h^{L+1}\Vert ^2+\Vert v_h^{L+1}\Vert ^2) +2\tau \sum _{n=0}^L(a\Vert \sigma _h^{n+\frac{1}{2}}\Vert ^2+c\Vert \lambda _h^{n+\frac{1}{2}}\Vert ^2)\\&\quad \leqslant C(\Vert u_h^0\Vert ^2+\Vert v_h^0\Vert ^2)+C\tau \sum _{n=0}^L(\Vert u_h^{n+\frac{1}{2}}\Vert ^2+\Vert v_h^{n+\frac{1}{2}}\Vert ^2)+\\&\qquad +C\tau \sum _{n=0}^{L}(\Vert u_h^{n}\Vert ^2+\Vert v_h^{n}\Vert ^2) +2\tau \sum _{n=0}^L(\Vert \bar{f}^{n+\frac{1}{2}}\Vert ^2+\Vert \bar{g}^{n+\frac{1}{2}}\Vert ^2) \\&\quad \leqslant C(\Vert u_h^0\Vert ^2+\Vert v_h^0\Vert ^2)+C\tau \sum _{n=0}^L(\Vert u_h^{n}\Vert ^2+\Vert v_h^{n}\Vert ^2) +2\tau \sum _{n=0}^L(\Vert \bar{f}^{n+\frac{1}{2}}\Vert ^2+\Vert \bar{g}^{n+\frac{1}{2}}\Vert ^2). \end{aligned} \end{aligned}$$

(29)

For the small enough $\tau$, we can use Gronwall lemma to yield the result.

Remark 2

Here, based on the boundness assumptions for both $\Vert u^n\Vert _{\infty }$ and $\Vert v^n\Vert _{\infty }$, we can prove that $\Vert u_h^n\Vert _{\infty }$ and $\Vert v_h^n\Vert _{\infty }$ are bounded. For the related discussion, one can see Remark 3.3 in Ref. [31].

4 Error estimates of the mixed element scheme

In what follows, the error estimate theorem will be given. Firstly, we need to introduce two lemmas.

Lemma 3

(See [57]) Let $(P_hu,P_h\sigma ):[0,T]\mapsto W_h\times \varvec{Z}_h$ be given by the following mixed projections

$$\begin{aligned} \left\{ \begin{aligned}&(a)(\sigma -P_h\sigma ,\nabla w_h)=0,\quad \forall w_h\in W_h,\\&(b)(\sigma -P_h\sigma ,\varvec{z}_h)-(\nabla (u-P_hu),\varvec{z}_h)=0,\quad \forall \varvec{z}_h\in \varvec{Z}_h,\\ \end{aligned}\right. \end{aligned}$$

(30)

then, there exists a constant C independent of h such that

$$\begin{aligned}&\Vert u-P_hu\Vert +h(\Vert \sigma -P_h\sigma \Vert +\Vert u-P_hu\Vert _1)\leqslant Ch^2(\Vert u\Vert _{H^2}+\Vert \sigma \Vert _{(H^1)^2}), \end{aligned}$$

(31)

$$\begin{aligned}&\Vert u_t-P_hu_t\Vert \leqslant Ch^2(\Vert u_t\Vert _{H^2}+\Vert \sigma _t\Vert _{(H^1)^2}). \end{aligned}$$

(32)

Lemma 4

Referring to Ref.[15], we can easily get the following error inequality

$$\begin{aligned} \begin{aligned} \left( \sum _{i=0}^{n}\frac{p_{\alpha }(i)}{\tau ^{\alpha }}(u^{n+\frac{1}{2}} -P_hu^{n+\frac{1}{2}}),w_h\right) \leqslant&C(h^{4}+\tau ^4)+\Vert w_h\Vert ^2,\quad \forall w_h\in W_h.\\ \end{aligned} \end{aligned}$$

(33)

Theorem 3

If $\{u^{n+1}\}_{n=0}^{L}$, $\{v^{n+1}\}_{n=0}^{L}$, $\{\sigma ^{n+1}\}_{n=0}^{L}$, and $\{\lambda ^{n+1}\}_{n=0}^{L}$ are the exact solutions of (8)–(9), $\{u_h^{n+1}\}_{n=0}^{L}$, $\{v_h^{n+1}\}_{n=0}^{L}$, $\{\sigma _h^{n+1}\}_{n=0}^{L}$, and $\{\lambda _h^{n+1}\}_{n=0}^{L}$ are the numerical solutions of (11)–(12), then for $\forall n=0,1,\cdots ,L(0\leqslant L\leqslant M-1)$ there exists a positive constant C such that

$$\begin{aligned} \begin{aligned}&\Vert u^{L+1}-u_h^{L+1}\Vert +\Vert v^{L+1}-v_h^{L+1}\Vert \\&\quad \leqslant C(h^{2}+\tau ^2), \left( \tau \sum _{n=1}^L\Vert \sigma ^{n+\frac{1}{2}}-\sigma _h^{n+\frac{1}{2}}\Vert ^2\right) ^{\frac{1}{2}} +\left( \tau \sum _{n=1}^L\Vert \lambda ^{n+\frac{1}{2}} -\lambda _h^{n+\frac{1}{2}}\Vert ^2\right) ^{\frac{1}{2}}\\&\quad \leqslant C(h+\tau ^2), \left( \tau \sum _{n=1}^L\Vert u^{n+\frac{1}{2}}-u_h^{n+\frac{1}{2}}\Vert _1^2\right) ^{\frac{1}{2}}\\&\quad +\left( \tau \sum _{n=1}^L\Vert v^{n+\frac{1}{2}}-v_h^{n+\frac{1}{2}}\Vert _1^2\right) ^{\frac{1}{2}} \leqslant C(h+\tau ^2). \end{aligned} \end{aligned}$$

(34)

Proof

For simplicity, we introduce the following notations

$$\begin{aligned} \begin{aligned}&u(t_n)-u_h^n=u(t_n)-P_hu^n+P_hu^n-u_h^n=\phi ^n+\theta ^n,\\&\sigma (t_n)-\sigma _h^n=\sigma (t_n)-P_h\sigma ^n+P_h\sigma ^n-\sigma _h^n=\psi ^n+\omega ^n,\\&v(t_n)-v_h^n=v(t_n)-P_hv^n+P_hv^n-v_h^n=\eta ^n+\zeta ^n,\\&\lambda (t_n)-\lambda _h^n=\lambda (t_n)-P_h\lambda ^n+P_h\lambda ^n-\lambda _h^n=\rho ^n+\xi ^n. \end{aligned}\end{aligned}$$

(35)

Combining (8), (11), (35) with Lemma 3, we can get the error equations for $n\geqslant 1$

$$\begin{aligned} \left\{ \begin{aligned} (a)&(D_t\theta ^{n+\frac{1}{2}},w_h)+k_1\tau ^{-\alpha }\left( \sum _{i=0}^{n}p_\alpha (i)\theta ^{n+\frac{1}{2}-i},w_h\right) +(a\omega ^{n+\frac{1}{2}},\nabla w_h)\\ =&-(D_t\phi ^{n+\frac{1}{2}},w_h)-k_1\tau ^{-\alpha }\left( \sum _{i=0}^{n}p_\alpha (i)\phi ^{n+\frac{1}{2}-i},w_h\right) \\&+(f(u^{n+\frac{1}{2}},v^{n+\frac{1}{2}})-f(u_h^{n+\frac{1}{2}},v_h^{n+\frac{1}{2}}),w_h)+(R_1,w_h),\\ (b)&(\omega ^{n+\frac{1}{2}},z_h)-(\nabla \theta ^{n+\frac{1}{2}},z_h)=(R_2,\varvec{z}_h),\\ (c)&(D_t\zeta ^{n+\frac{1}{2}},w_h)+k_2\tau ^{-\beta }\left( \sum _{i=0}^{n}p_\beta (i)\zeta ^{n+\frac{1}{2}-i},w_h\right) +(c\xi ^{n+\frac{1}{2}},\nabla w_h)\\ =&-(D_t\eta ^{n+\frac{1}{2}},w_h)-k_2\tau ^{-\beta }\left( \sum _{i=0}^{n}p_\beta (i)\eta ^{n+\frac{1}{2}-i},w_h\right) \\&+(g(u^{n+\frac{1}{2}},v^{n+\frac{1}{2}})-g(u_h^{n+\frac{1}{2}},v_h^{n+\frac{1}{2}}),w_h)+(R_3,w_h),\\ (d)&(\xi ^{n+\frac{1}{2}},z_h)-(\nabla \zeta ^{n+\frac{1}{2}},z_h)=(R_4,\varvec{z}_h).\\ \end{aligned}\right. \end{aligned}$$

(36)

Setting $w_h=2\tau \theta ^{n+\frac{1}{2}}$ in (36)(a), $z_h=2a\tau \omega ^{n+\frac{1}{2}}$ in (36)(b), we add the resulting equations to get

$$\begin{aligned} \begin{aligned}&\Vert \theta ^{n+1}\Vert ^2-\Vert \theta ^n\Vert ^2+2k_1\tau ^{1-\alpha }\left( \sum _{i=0}^{n}p_\alpha (i)\theta ^{n+\frac{1}{2}-i}, \theta ^{n+\frac{1}{2}}\right) +2a\tau \Vert \omega ^{n+\frac{1}{2}}\Vert ^2\\&\quad =-2\tau \left( \frac{\phi ^{n+1}-\phi ^n}{\tau },\theta ^{n+\frac{1}{2}}\right) -2k_1\tau ^{1-\alpha }\left( \sum _{i=0}^{n}p_{\alpha }(i)\phi ^{n+\frac{1}{2}-i},\theta ^{n+\frac{1}{2}}\right) \\&\qquad +2\tau (f(u^{n+\frac{1}{2}},v^{n+\frac{1}{2}})-f(u_h^{n+\frac{1}{2}},v_h^{n+\frac{1}{2}}),\theta ^{n+\frac{1}{2}}) +(R_1+R_2,2\tau \theta ^{n+\frac{1}{2}}). \end{aligned} \end{aligned}$$

(37)

Summing (37) for n from 1 to L, we arrive at

$$\begin{aligned} \begin{aligned}&\Vert \theta ^{L+1}\Vert ^2-\Vert \theta ^1\Vert ^2+2k_1\tau ^{1-\alpha }\sum _{n=1}^L \left( \sum _{i=0}^{n}p_\alpha (i)\theta ^{n+\frac{1}{2}-i}, \theta ^{n+\frac{1}{2}}\right) +2a\tau \sum _{n=1}^{L}\Vert \omega ^{n+\frac{1}{2}}\Vert ^2\\&\quad =-2\tau \sum _{n=1}^L\left( \frac{\phi ^{n+1}-\phi ^n}{\tau },\theta ^{n+\frac{1}{2}}\right) -2k_1\tau ^{1-\alpha }\sum _{n=1}^L\left( \sum _{i=0}^{n}p_{\alpha }(i)\phi ^{n+\frac{1}{2}-i},\theta ^{n+\frac{1}{2}}\right) \\&\qquad +2\tau \sum _{n=1}^L(f(u^{n+\frac{1}{2}},v^{n+\frac{1}{2}})-f(u_h^{n+\frac{1}{2}},v_h^{n+\frac{1}{2}}),\theta ^{n+\frac{1}{2}}) +2\tau \sum _{n=1}^L(R_1+R_2,\theta ^{n+\frac{1}{2}}). \end{aligned} \end{aligned}$$

(38)

Noting that the nonlinear term f(u, v) is a quadratic polynomial without constant terms about u and v, we use Hölder inequality, triangle inequality, (31) and Remark 2 to get

$$\begin{aligned} \begin{aligned}&(f(u^{n+\frac{1}{2}},v^{n+\frac{1}{2}})-f(u_h^{n+\frac{1}{2}}, v_h^{n+\frac{1}{2}}),\theta ^{n+\frac{1}{2}})\\&\quad \leqslant \frac{3}{2}\Big [\sum _{k=n-1}^{n}(\Vert u^k\Vert _{\infty }+\Vert u^k_h\Vert _{\infty }+\Vert v^k_h\Vert _{\infty })\Vert u^k-u_h^k\Vert \Big ]\Vert \theta ^{n+\frac{1}{2}}\Vert \\&\qquad + \frac{3}{2}\Big [\sum _{k=n-1}^{n}(\Vert v^k\Vert _{\infty }+\Vert v^k_h\Vert _{\infty }+\Vert u^k\Vert _{\infty })\Vert v^k-v_h^k\Vert \Big ]\Vert \theta ^{n+\frac{1}{2}}\Vert \\&\quad \leqslant C\Big [\sum _{k=n-1}^{n}(\Vert \phi ^k+\theta ^k\Vert +\Vert \eta ^k+\zeta ^k\Vert )\Big ]\Vert \theta ^{n+\frac{1}{2}}\Vert \\&\quad \leqslant C\Big [\sum _{k=n-1}^{n}(\Vert \theta ^k\Vert ^2+\Vert \zeta ^k\Vert ^2+\Vert \phi ^k\Vert ^2+\Vert \eta ^k\Vert ^2)+\Vert \theta ^{n+\frac{1}{2}}\Vert ^2\Big ]). \end{aligned} \end{aligned}$$

(39)

For (38), we use Cauchy–Schwarz inequality, Young inequality, Lemma 4 and inequality (39), we have

$$\begin{aligned} \begin{aligned}&\Vert \theta ^{L+1}\Vert ^2-\Vert \theta ^1\Vert ^2+2k_1\tau ^{1-\alpha }\sum _{n=1}^L \left( \sum _{i=0}^{n}p_\alpha (i)\theta ^{n+\frac{1}{2}-i}, \theta ^{n+\frac{1}{2}}\right) +2a\tau \sum _{n=1}^{L}\Vert \omega ^{n+\frac{1}{2}}\Vert ^2\\&\quad \leqslant C\int _{t_1}^{t_{L+1}}\Vert \phi _{t}\Vert ^2{\text {d}}s+C\tau \sum _{n=1}^{L}(h^{4}+\tau ^4)+ C\tau \sum _{n=1}^L(\Vert R_1\Vert ^2+\Vert R_2\Vert ^2)\\&\qquad +C\tau \sum _{n=1}^L(\Vert \phi ^{n+\frac{1}{2}}\Vert ^2 +\Vert \theta ^{n+\frac{1}{2}}\Vert ^2+\Vert \eta ^{n+\frac{1}{2}}\Vert ^2+\Vert \zeta ^{n+\frac{1}{2}}\Vert ^2)\\&\qquad +C\tau \sum _{n=0}^L(\Vert \phi ^{n}\Vert ^2 +\Vert \theta ^{n}\Vert ^2+\Vert \eta ^{n}\Vert ^2+\Vert \zeta ^{n}\Vert ^2). \end{aligned} \end{aligned}$$

(40)

We set $w_h=2\tau \zeta ^{n+\frac{1}{2}}$ in (36)(c) and $z_h=2c\tau \xi ^{n+\frac{1}{2}}$ in (36)(d), and use the similar process to the derivation of (37)–(40) to obtain

$$\begin{aligned} \begin{aligned}&\Vert \zeta ^{L+1}\Vert ^2-\Vert \zeta ^1\Vert ^2+2k_2\tau ^{1-\beta }\sum _{n=1}^L \left( \sum _{i=0}^{n}p_\beta (i)\zeta ^{n+\frac{1}{2}-i}, \zeta ^{n+\frac{1}{2}}\right) +2c\tau \sum _{n=1}^{L}\Vert \xi ^{k+\frac{1}{2}}\Vert ^2\\&\quad \leqslant C\int _{t_1}^{t_{L+1}}\Vert \eta _{t}\Vert ^2{\text {d}}s+C\tau \sum _{n=1}^L(h^{4}+\tau ^4)+ C\tau \sum _{n=1}^L(\Vert R_3\Vert ^2+\Vert R_4\Vert ^2)\\&\qquad +C\tau \sum _{n=1}^L(\Vert \phi ^{n+\frac{1}{2}}\Vert ^2 +\Vert \theta ^{n+\frac{1}{2}}\Vert ^2+\Vert \eta ^{n+\frac{1}{2}}\Vert ^2+\Vert \zeta ^{n+\frac{1}{2}}\Vert ^2)\\&\qquad +C\tau \sum _{n=0}^L(\Vert \phi ^{n}\Vert ^2 +\Vert \theta ^{n}\Vert ^2+\Vert \eta ^{n}\Vert ^2+\Vert \zeta ^{n}\Vert ^2). \end{aligned} \end{aligned}$$

(41)

Summing (40) and (41) and using triangle inequality, we have the following inequality

$$\begin{aligned} \begin{aligned}&\Vert \theta ^{L+1}\Vert ^2+\Vert \zeta ^{L+1}\Vert ^2+2\tau \left( a\sum _{n=1}^{L}\Vert \omega ^{n+\frac{1}{2}}\Vert ^2+c\sum _{n=1}^{L}\Vert \xi ^{n+\frac{1}{2}}\Vert ^2\right) \\&\qquad +2k_1\tau ^{1-\alpha }\sum _{n=1}^L \left( \sum _{i=0}^{n}p_\alpha (i)\theta ^{n+\frac{1}{2}-i}, \theta ^{n+\frac{1}{2}}\right) +2k_2\tau ^{1-\beta }\sum _{n=1}^L \left( \sum _{i=0}^{n}p_\beta (i)\zeta ^{n+\frac{1}{2}-i}, \zeta ^{n+\frac{1}{2}}\right) \\&\quad \leqslant \Vert \theta ^1\Vert ^2+\Vert \zeta ^1\Vert ^2+C\int _{t_1}^{t_{L+1}}(\Vert \phi _{t}\Vert ^2+\Vert \eta _{t}\Vert ^2){\text {d}}s+C\tau \sum _{n=1}^L(h^{4}+\tau ^4)\\&\qquad +C\tau \sum _{n=1}^{L}(\Vert R_{1}\Vert ^2+\Vert R_{2}\Vert ^2+\Vert R_{3}\Vert ^2+\Vert R_{4}\Vert ^2)+C\tau \sum _{n=0}^{L+1}(\Vert \phi ^{n}\Vert ^2 +\Vert \theta ^{n}\Vert ^2+\Vert \eta ^{n}\Vert ^2+\Vert \zeta ^{n}\Vert ^2). \end{aligned} \end{aligned}$$

(42)

Subtracting (12) from (9), noticing (35) and Lemma 3, we obtain

$$\begin{aligned} \left\{ \begin{aligned} (a)&(D_t\theta ^{\frac{1}{2}},w_h)+k_1\tau ^{-\alpha }\left( \sum _{i=0}^{0}p_\alpha (i)\theta ^{\frac{1}{2}-i},w_h\right) +(a\omega ^{\frac{1}{2}},\nabla w_h)\\ =&-(D_t\phi ^{\frac{1}{2}},w_h)-k_1\tau ^{-\alpha }\left( \sum _{i=0}^{0}p_\alpha (i)\phi ^{\frac{1}{2}-i},w_h\right) +(f(u^{\frac{1}{2}},v^{\frac{1}{2}})-f(u_h^{\frac{1}{2}},v_h^{\frac{1}{2}}),w_h)+(R_5,w_h),\\ (b)&(\omega ^{\frac{1}{2}},z_h)-(\nabla \theta ^{\frac{1}{2}},z_h)=(R_6,\varvec{z}_h),\\ (c)&(D_t\zeta ^{\frac{1}{2}},w_h)+k_2\tau ^{-\beta }\left( \sum _{i=0}^{0}p_\beta (i)\zeta ^{\frac{1}{2}-i},w_h\right) +(c\xi ^{\frac{1}{2}},\nabla w_h)\\ =&-(D_t\eta ^{\frac{1}{2}},w_h)-k_2\tau ^{-\beta }\left( \sum _{i=0}^{0}p_\beta (i)\eta ^{\frac{1}{2}-i},w_h\right) +(g(u^{\frac{1}{2}},v^{\frac{1}{2}})-g(u_h^{\frac{1}{2}},v_h^{\frac{1}{2}}),w_h)+(R_7,w_h),\\ (d)&(\xi ^{\frac{1}{2}},z_h)-(\nabla \zeta ^{\frac{1}{2}},z_h)=(R_8,\varvec{z}_h). \end{aligned}\right. \end{aligned}$$

(43)

By the similar means to (42), we set $w_h=2\tau \theta ^{\frac{1}{2}}$ in (43)(a), $z_h=2a\tau \omega ^{\frac{1}{2}}$ in (43)(b), $w_h=2\tau \zeta ^{\frac{1}{2}}$ in (43)(c), $z_h=2c\tau \xi ^{\frac{1}{2}}$ in (43)(d), and consider $\Vert \theta ^{0}\Vert ^2=0,\Vert \zeta ^{0}\Vert ^2=0$ to have

$$\begin{aligned} \begin{aligned}&\Vert \theta ^{1}\Vert ^2+\Vert \zeta ^{1}\Vert ^2+ 2\tau (a\Vert \omega ^{\frac{1}{2}}\Vert ^2+c\Vert \xi ^{\frac{1}{2}}\Vert ^2)\\&\qquad +2k_1\tau ^{1-\alpha }\left( \sum _{i=0}^{0}p_\alpha (i)\theta ^{\frac{1}{2}-i},\theta ^{\frac{1}{2}}\right) +2k_2\tau ^{1-\beta }\left( \sum _{i=0}^{0}p_\beta (i)\zeta ^{\frac{1}{2}-i},\zeta ^{\frac{1}{2}}\right) \\&\quad \leqslant C\int _{t_0}^{t_{1}}(\Vert \phi _{t}\Vert ^2+\Vert \eta _{t}\Vert ^2){\text {d}}s+C\tau (h^{4}+\tau ^4)+C(\tau ^2\Vert R_5\Vert ^2+\Vert R_6\Vert ^2+ \tau ^2\Vert R_7\Vert ^2+\Vert R_8\Vert ^2)\\&\qquad +C\tau (\Vert \phi ^{\frac{1}{2}}\Vert ^2+\Vert \theta ^{\frac{1}{2}}\Vert ^2+\Vert \eta ^{\frac{1}{2}}\Vert ^2+\Vert \zeta ^{\frac{1}{2}}\Vert ^2). \end{aligned} \end{aligned}$$

(44)

By Lemma 2, substitute (44) into (42) to get

$$\begin{aligned} \begin{aligned}&(1-C\tau )(\Vert \theta ^{L+1}\Vert ^2+\Vert \zeta ^{L+1}\Vert ^2)+2\tau \left( a\sum _{n=0}^{L}\Vert \omega ^{n+\frac{1}{2}}\Vert ^2+c\sum _{n=0}^{L}\Vert \xi ^{n+\frac{1}{2}}\Vert ^2\right) \\ \leqslant&C\int _{t_0}^{t_{L+1}}(\Vert \phi _{t}\Vert ^2+\Vert \eta _{t}\Vert ^2){\text {d}}s+C\tau \sum _{n=0}^L(h^{4}+\tau ^4)\\&\qquad +C\tau \sum _{n=1}^{L}(\Vert R_{1}\Vert ^2+\Vert R_{2}\Vert ^2+ \Vert R_{3}\Vert ^2+\Vert R_{4}\Vert ^2)\\&\qquad +C(\tau ^2\Vert R_5\Vert ^2+\Vert R_6\Vert ^2+ \tau ^2\Vert R_7\Vert ^2+\Vert R_8\Vert ^2)\\&+C\tau \sum _{n=0}^{L}(\Vert \phi ^{n}\Vert ^2 +\Vert \theta ^{n}\Vert ^2+\Vert \eta ^{n}\Vert ^2+\Vert \zeta ^{n}\Vert ^2). \end{aligned} \end{aligned}$$

(45)

Take $z_h=2a\tau \nabla \theta ^{n+\frac{1}{2}}$ in (36)(b), $z_h=2c\tau \nabla \zeta ^{n+\frac{1}{2}}$ in (36)(d), $z_h=2a\tau \nabla \theta ^{\frac{1}{2}}$ in (43)(b) and $z_h=2c\tau \nabla \zeta ^{\frac{1}{2}}$ in (43)(d) and use (45) to get

$$\begin{aligned} \begin{aligned}&2\tau \left( a\sum _{n=0}^{L}\Vert \nabla \theta ^{n+\frac{1}{2}}\Vert ^2+c\sum _{n=0}^{L}\Vert \nabla \zeta ^{n+\frac{1}{2}}\Vert ^2\right) \\ \leqslant&C\tau \left( a\sum _{n=0}^{L}\Vert \omega ^{n+\frac{1}{2}}\Vert ^2+c\sum _{n=0}^{L}\Vert \xi ^{n+\frac{1}{2}}\Vert ^2\right) +C\sum _{n=0}^{L+1}(\Vert \theta ^{n}\Vert ^2+\Vert \zeta ^{n}\Vert ^2)+C\tau ^4. \end{aligned} \end{aligned}$$

(46)

Finally, for a small enough $\tau$, combining (45)–(46), Gronwall lemma, Lemma 3 with triangle inequality, we complete the proof of Theorem 3.

5 Numerical tests

In this section, we will provide two numerical examples to verify the correctness of the theory results.

5.1 One-dimensional case

For calculating the time convergence order, we choose the following one-dimensional example

$$\begin{aligned} \left\{ \begin{aligned}&\frac{\partial u}{\partial t}+\frac{\partial ^\alpha u}{\partial t^\alpha }= u_{xx}+(u^2-v)+\bar{f}(x,t),(x,t)\in (0,1)\times (0,1],\\&\frac{\partial v}{\partial t}+\frac{\partial ^\beta v}{\partial t^\beta }= v_{xx}+(v-u^2)+\bar{g}(x,t),(x,t)\in (0,1)\times (0,1],\\&u(0,t)=u(1,t)=v(0,t)=v(1,t)=0,t\in [0,1],\\&u(x,0)=0,v(x,0)=0,x\in [0,1], \end{aligned}\right. \end{aligned}$$

(47)

where

$$\begin{aligned} \bar{f}(x,t)= & {} \left( 2t+\frac{2 t^{2-\alpha }}{\Gamma (3-\alpha )}+4\pi ^2t^2\right) \sin {(2\pi x)}-t^4\sin ^2{(2\pi x)}+t^2\sin (\pi x),\\ \bar{g}(x,t)= & {} \left( 2t+\frac{2 t^{2-\beta }}{\Gamma (3-\beta )}+\pi ^2t^2\right) \sin {(\pi x)}+t^4\sin ^2{(2\pi x)}-t^2\sin (\pi x). \end{aligned}$$

So we can easily obtain that the exact solution is

$$\begin{aligned} \left\{ \begin{aligned}&u=t^2\sin (2\pi x),\\&v=t^2\sin (\pi x),\\&\sigma =2\pi t^2\cos (2\pi x),\\&\lambda =\pi t^2\cos (\pi x). \end{aligned}\right. \end{aligned}$$

(48)

By choosing a fixed space step $h=1/400$, different time discrete parameters $\tau =1/10$, 1/20, 1/30, 1/40, and changed fractional parameters $\alpha =\beta =0.1,0.5,0.9$, we get some computing data in Tables 1, 2 and 3. From these computing results, one can easily see the convergence order in time for unknown functions and their derivatives are close to 2.

Table 1 Time convergence results with $h=1/400, \alpha =0.1$, $\beta =0.1$

Table 2 Time convergence results with $h=1/400, \alpha =0.5$, $\beta =0.5$

Table 3 Time convergence results with $h=1/400, \alpha =0.9$, $\beta =0.9$

5.2 Two-dimensional example

We choose a two-dimensional nonlinear time fractional coupled sub-diffusion system

$$\begin{aligned} \left\{ \begin{aligned}&\frac{\partial u}{\partial t}+\frac{\partial ^\alpha u}{\partial t^\alpha }= \Delta u+(u^2-v)+\bar{f}({\mathbf{x}} ,t),({\mathbf{x}} ,t)\in \Omega \times J,\\&\frac{\partial v}{\partial t}+\frac{\partial ^\beta v}{\partial t^\beta }= \Delta v+(v-u^2)+\bar{g}({\mathbf{x}} ,t),({\mathbf{x}} ,t)\in \Omega \times J,\\&u({\mathbf{x}} ,t)=v({\mathbf{x}} ,t)=0,({\mathbf{x}} ,t)\in \partial \Omega \times \bar{J},\\&u({\mathbf{x}} ,0)=0,v({\mathbf{x}} ,0)=0,{\mathbf{x}} \in \bar{\Omega }, \end{aligned}\right. \end{aligned}$$

(49)

where $\Omega =(0,1)\times (0,1),J=(0,1], k_1=k_2=a=c=1$, and $\bar{f}({\mathbf{x}} ,t)$ and $\bar{g}({\mathbf{x}} ,t)$ are chosen as

$$\begin{aligned} \begin{aligned} \bar{f}({\mathbf{x}} ,t)=&\left( 2t+\frac{2 t^{2-\alpha }}{\Gamma (3-\alpha )}+8\pi ^2t^2\right) \sin {(2\pi x)}\sin {(2\pi y)}\\&-t^4\sin ^2{(2\pi x)}\sin ^2{(2\pi y)}+t^2\sin (4\pi x)\sin (4\pi y), \bar{g}({\mathbf{x}} ,t)=&\left( 2t+\frac{2 t^{2-\beta }}{\Gamma (3-\beta )}+32\pi ^2t^2\right) \sin {(4\pi x)}\sin {(4\pi y)}\\&\quad +t^4\sin ^2{(2\pi x)}\sin ^2{(2\pi y)}-t^2\sin (4\pi x)\sin (4\pi y). \end{aligned} \end{aligned}$$

Considering the given function $\bar{f}({\mathbf{x}} ,t)$ and $\bar{g}({\mathbf{x}} ,t)$, we can easily test and verify the exact solution as follows

$$\begin{aligned} \left\{ \begin{aligned}&u=t^2\sin (2\pi x)\sin (2\pi y), \\&v=t^2\sin (4\pi x)\sin (4\pi y),\\&\sigma = (2\pi t^2\cos (2\pi x)\sin (2\pi y),2\pi t^2\sin (2\pi x)\cos (2\pi y)),\\&\lambda =(4\pi t^2\cos (4\pi x)\sin (4\pi y),4\pi t^2\sin (4\pi x)\cos (4\pi y)). \end{aligned}\right. \end{aligned}$$

(50)

For solving the equations above, we use triangular elements. Selecting any unit $e=\Delta P_iP_jP_k$, we choose

$$\begin{aligned} N(x,y)=\{[N_i(x,y),N_j(x,y),N_k(x,y)]^T|N_i=S_i/S,N_j=S_j/S,N_k=S_k/S\} \end{aligned}$$

as the linear interpolation basis function of finite element space $V_h$ and choose the constant basis function of finite element space $Z_h$. So we can get $u_h=N_iu_i+N_ju_j+N_ku_k$, $v_h=N_iv_i+N_jv_j+N_kv_k$, $\sigma _h=[\sigma _{1h},\sigma _{2h}]$ and $\lambda _h=[\lambda _{1h},\lambda _{2h}]$. After unit analysis, we can synthesize stiffness matrices

$$\begin{aligned} \begin{aligned}&\Delta _e=\frac{1}{2}\left| \begin{array}{ccc} x_i&{}y_i&{}1\\ x_j&{}y_j&{}1\\ x_k&{}y_k&{}1 \end{array} \right| , A^{(e)}=\Delta _e\left[ \begin{array}{ccc} \frac{1}{6}&{}\frac{1}{12}&{}\frac{1}{12}\ \\ \frac{1}{12}&{}\frac{1}{6}&{}\frac{1}{12}\\ \frac{1}{12}&{}\frac{1}{12}&{}\frac{1}{6} \end{array} \right] ,\\&C^{(e)}=\Delta _e\left[ \begin{array}{cc} 1&{}0\\ 0&{}1 \end{array} \right] , B^{(e)}=D^{T(e)}=\frac{1}{2}\left[ \begin{array}{ccc} a_i&{}a_j&{}a_k\\ b_i&{}b_j&{}b_k \end{array} \right] ,\\&a_i=\left| \begin{array}{cc} y_j&{}1\\ y_k&{}1 \end{array} \right| , a_j=\left| \begin{array}{cc} y_k&{}1\\ y_i&{}1 \end{array} \right| , a_k=\left| \begin{array}{cc} y_i&{}1\\ y_j&{}1 \end{array} \right| ,\\&b_i=-\left| \begin{array}{cc} x_j&{}1\\ x_k&{}1 \end{array} \right| , b_j=-\left| \begin{array}{cc} x_k&{}1\\ x_i&{}1 \end{array} \right| , b_k=-\left| \begin{array}{cc} x_i&{}1\\ x_j&{}1 \end{array} \right| .\\ \end{aligned} \end{aligned}$$

(51)

For testing the space convergence order, we take changed step sizes $h=\sqrt{2}\tau =\frac{\sqrt{2}}{8}$, $\frac{\sqrt{2}}{12}$, $\frac{\sqrt{2}}{16}$, $\frac{\sqrt{2}}{20}$, different fractional parameters $\alpha =\beta =0.01,0.5,0.99$ to get the computing data in Tables 4, 5 and 6. We also show the computing data with $\alpha =0.25,\beta =0.75$ in Table 7. From Tables 4, 5, 6 and 7, one can find the space convergence orders of u, v in $L^2$-norm are close to 2. At the same time, the space convergence orders of u, v in $H^1$-norm and $\sigma ,\lambda$ in $(L^2)^2$-norm tend to 1. By the computed data, it is easy to check that the numerical convergence orders are in agreement with the theory results. Besides, by choosing $h=\frac{\sqrt{2}}{35}$, $\tau =\frac{1}{35}$ and $\alpha =\beta =0.5$, the comparison surfaces between the numerical solution and the exact solution for unknown scalar functions and their fluxes are provided in Figs. 1, 2, 3, 4, 5, 6, 7 and 8, from which one can clearly see that the behaviors of the numerical solution under our mixed element spaces are almost consistent with the ones of exact solution which illustrates that our algorithm is effective for numerically solving nonlinear fractional coupled diffusion system.

Table 4 Space convergence results with $\alpha =0.01$, $\beta =0.01$

Table 5 Space convergence results with $\alpha =0.5$, $\beta =0.5$

Table 6 Space convergence results with $\alpha =0.99$, $\beta =0.99$

Table 7 Space convergence results with $\alpha =0.25$, $\beta =0.75$

Fig. 1

Surface for exact solution u

Fig. 2

Surface for numerical solution $u_h$

Fig. 3

Surface for exact solution v

Fig. 4

Surface for numerical solution $v_h$

Fig. 5

Surface for exact solution $\sigma =(\sigma _1,\sigma _2)$

Fig. 6

Surface for numerical solution $\sigma _h=(\sigma _{1h},\sigma _{2h})$

Fig. 7

Surface for exact solution $\lambda =(\lambda _1,\lambda _2)$

Fig. 8

Surface for numerical solution $\lambda _h=(\lambda _{1h},\lambda _{2h})$

Remark 3

In this article, we implement numerical computing by choosing triangular elements. One can also consider rectangular elements [55].

6 Conclusions

In this paper, we solve a time fractional coupled sub-diffusion system by a mixed element method with the second-order time approximation. By making use of this numerical algorithm, we can arrive at the approximation solutions of four functions. We provide the detailed proof of existence and uniqueness of the mixed element solution and stability analysis, and derive optimal a priori error estimates in $L^2$ and $H^1$-norm for the unknown u, v and a priori error estimates in $(L^2)^2$-norm for the fluxes $\sigma$ and $\lambda$. Finally, via some numerical results, we illustrate the validity for the proposed numerical method.