Error Analysis of a PFEM Based on the Euler Semi-Implicit Scheme for the Unsteady MHD Equations

Abstract

In this article, we mainly consider a first order penalty finite element method (PFEM) for the 2D/3D unsteady incompressible magnetohydrodynamic (MHD) equations. The penalty method applies a penalty term to relax the constraint “$\nabla ·\mathit{u}=0$”, which allows us to transform the saddle point problem into two smaller problems to solve. The Euler semi-implicit scheme is based on a first order backward difference formula for time discretization and semi-implicit treatments for nonlinear terms. It is worth mentioning that the error estimates of the fully discrete PFEM are rigorously derived, which depend on the penalty parameter $ϵ$, the time-step size $\tau$, and the mesh size h. Finally, two numerical tests show that our scheme is effective.

1. Introduction

The magnetohydrodynamic (MHD) describes the dynamic behavior of a conducting fluid under external electromagnetic field, which is the coupling of the Navier–Stokes (NS) system and Maxwell’s system. It has wide practical applications in geophysics, astrophysics, and confinement for controlled thermonuclear fusion (cf. [1,2,3]). Concerning the corresponding extensive theoretical modeling/numerical analysis of the MHD system, we refer to [1,2,3,4,5,6,7,8,9,10,11,12] and the references therein.

In this paper, we mainly consider the 2D/3D unsteady incompressible MHD equations. This model is a coupled strongly nonlinear system, and it is a saddle point problem due to the incompressible constraint. Therefore, it is necessary to construct unconditionally stable and decoupled algorithms for our model. For the time discretization, it is well-known that simple discretizations, like fully explicit or implicit type schemes, can lead to considerable instabilities or suffer from costly time expense. Recently, Euler semi-implicit schemes for some evolution differential equations have been given in [4,13,14]. This method is unconditionally stable. For the saddle point problem, there are many methods to release the incompressibility constraint for incompressible flow such as the projection method, the pressure stabilization method, the artificial compressibility method, and the penalty method (see also [15,16,17,18,19,20]).

It is worth mentioning that the penalty method is the simplest and the most basic of these methods mentioned above. For the penalty method, it can be traced back to [21]. Then, the optimal error estimate of the unsteady NS system based on the penalty finite element method (PFEM) was given in [22]. A PFEM of a Euler implicit/explicit scheme for the unsteady NS system was proposed in [23]. An error estimate of the unsteady NS system based on the $P_1$ nonconforming PFEM was given in [24]. The authors study the PFEM of the steady MHD equations in [25]. A decoupling PFEM for the steady incompressible MHD equations was given in [18].

The aim of this paper is to develop a first order linear and decoupled scheme. We adopt an implicit scheme for the linear terms and semi-implicit treatments for nonlinear terms. Meanwhile, the penalty method is used for fluid equations. This method decouples the MHD equations into two small equations; one is the equations of the velocity and magnetic field $(\mathit{u},\mathit{B})$, and the other is the equation of pressure p. Then, the error estimates for the developed scheme are present, which depend on the penalty parameter $ϵ$, the time-step size $\tau$, and the mesh size h. Finally, we give two numerical tests to verify the theoretical results of our method.

The structure of the paper is as follows: in Section 2, we present the model and estimates for the solutions of penalty MHD equations. In Section 3, we give the Euler semi-implicit scheme in the case of time discretization and the convergence rate of the semi-discrete solutions. In Section 4, we give the PFEM based on the Euler semi-implicit scheme. In Section 5, we give the error estimates of the fully discrete solutions. In Section 6, the error estimates are obtained for our scheme. In Section 7, two numerical tests show that our scheme is effective. Finally, we give some conclusions.

2. Functional Setting of the Unsteady MHD Equation

In this paper, we consider the unsteady incompressible MHD equations as follows:

$$\left\{\begin{array}{c}{\mathit{u}}_t-\nu \Delta \mathit{u}+(\mathit{u}·\nabla )\mathit{u}+\nabla p+S\mathit{B}\times \nabla \times \mathit{B}=\mathit{f},\hspace{1em}\mathrm{in} {\mathcal{D}}_{\mathcal{T}},\hfill \\ \nabla ·\mathit{u}=0,\hspace{1em}\mathrm{in} {\mathcal{D}}_{\mathcal{T}},\hfill \\ {\mathit{B}}_t+\mu \nabla \times \nabla \times \mathit{B}-\nabla \times (\mathit{u}\times \mathit{B})=\mathit{g},\hspace{1em}\mathrm{in} {\mathcal{D}}_{\mathcal{T}},\hfill \\ \nabla ·\mathit{B}=0,\hspace{1em}\mathrm{in} {\mathcal{D}}_{\mathcal{T}},\hfill \\ \mathit{u}\left(0\right)={\mathit{u}}^0,\hspace{1em}\mathit{B}\left(0\right)={\mathit{B}}^0,\hspace{1em}\mathrm{in} \mathcal{D},\hfill \\ \mathit{u}=\mathbf{0},\hspace{1em}\mathit{B}\times \mathit{n}=\mathbf{0},\hspace{1em}\mathrm{on} \gamma ,\hfill \end{array}\right.$$ (1)

where ${\mathcal{D}}_{\mathcal{T}}=\mathcal{D}\times [0,\mathcal{T}]$, $\Gamma =\partial \mathcal{D}\times [0,\mathcal{T}]$, $\mathcal{D}\subset {\mathbb{R}}^d(d=2\mathrm{or}3)$ stands for a bounded, convex, and open domain with the boundary $\partial \mathcal{D}$, $\mathcal{T}$ is the final time. Here, $\mathit{u},p,\mathit{B}$ are the velocity, the pressure, and the magnetic field, $\mathit{f}$ is the external force term, $\mathit{g}$ is the known applied current with $\nabla ·\mathit{g}=0$, and $\mathit{n}$ denotes the outward normal on $\partial \mathcal{D}$. For the physical parameters, ${\nu }^{-1}=R_e$ (fluid Reynolds number), ${\mu }^{-1}=R_m$ (magnetic Reynolds number) and S is the coupling coefficient.

Next, we give a penalty method for the unsteady MHD equations. Instead of solving (1), we solve $({\mathit{u}}_{ϵ},p_{ϵ},{\mathit{B}}_{ϵ})$ from the penalty MHD equations:

$$\left\{\begin{array}{c}{\mathit{u}}_{ϵt}-\nu \Delta {\mathit{u}}_{ϵ}+\tilde{b}({\mathit{u}}_{ϵ},{\mathit{u}}_{ϵ})+\nabla p_{ϵ}+S{\mathit{B}}_{ϵ}\times \nabla \times {\mathit{B}}_{ϵ}=\mathit{f},\hspace{1em}\mathrm{in} {\mathcal{D}}_{\mathcal{T}},\hfill \\ \nabla ·{\mathit{u}}_{ϵ}+\frac{ϵ}{\nu }{p}_{ϵ}=0,\hspace{1em}\mathrm{in} {\mathcal{D}}_{\mathcal{T}},\hfill \\ {\mathit{B}}_{ϵt}+\mu \nabla \times \nabla \times {\mathit{B}}_{ϵ}-\nabla \times ({\mathit{u}}_{ϵ}\times {\mathit{B}}_{ϵ})=\mathit{g},\hspace{1em}\mathrm{in} {\mathcal{D}}_{\mathcal{T}},\hfill \\ \nabla ·{\mathit{B}}_{ϵ}=0,\hspace{1em}\mathrm{in} {\mathcal{D}}_{\mathcal{T}},\hfill \\ {\mathit{u}}_{ϵ}\left(0\right)={\mathit{u}}^0,\hspace{1em}{\mathit{B}}_{ϵ}\left(0\right)={\mathit{B}}^0,\hspace{1em}\mathrm{in} \mathcal{D},\hfill \\ {\mathit{u}}_{ϵ}=\mathbf{0},\hspace{1em}{\mathit{B}}_{ϵ}\times \mathit{n}=\mathbf{0},\hspace{1em}\mathrm{on} \gamma ,\hfill \end{array}\right.$$ (2)

where $0<ϵ<1$ is the penalty parameter; $\tilde{b}(\mathit{u},\mathit{v})=(\mathit{u}·\nabla )\mathit{v}+\frac{1}{2}(\nabla ·\mathit{u})\mathit{v}$ is the modified nonlinear term.

Then, we give some notations and estimates for MHD equations. For $1\le r\le \infty$, $L^r\left(\mathcal{D}\right)$ denotes the usual Lebesgue space on $\mathcal{D}$ with the norm ${\parallel ·\parallel }_{L^r}$. The inner product of the space $L^2\left(\mathcal{D}\right)$ is denoted by $(·,·)$ that is $(u,v)={\int }_{\mathcal{D}}uvd\mathbf{x}$, and the norm of the space $L^2\left(\mathcal{D}\right)$ is denoted by $\parallel ·\parallel$. For all non-negative integers k and r, $W^{k,r}\left(\mathcal{D}\right)$ stands for the standard Sobolev space equipped with the standard Sobolev norm ${\parallel ·\parallel }_{k,r}$. The norm of the space $W^{k,2}\left(\mathcal{D}\right)$ is represented by ${\parallel ·\parallel }_k$. The functions and spaces of vectors are represented in boldface.

Next, we give several function spaces

$$\begin{array}{cc}\hfill \hspace{1em}& \tilde{\mathbf{H}}:=\{\mathit{v}\in {\mathbf{L}}^2\left(\mathcal{D}\right):\nabla ·\mathit{v}=0,\mathit{v}·\mathit{n}{\mid }_{\partial \mathcal{D}}=0\},\hfill \\ \hfill \hspace{1em}& \mathbf{X}:={\mathbf{H}}_0^1\left(\mathcal{D}\right)=\{\mathit{v}\in H^1{\left(\mathcal{D}\right)}^d:\mathit{v}{\mid }_{\partial \mathcal{D}}=\mathbf{0}\},\hfill \\ \hfill \hspace{1em}& \mathrm{M}:=L_0^2\left(\mathcal{D}\right)=\{q\in L^2\left(\mathcal{D}\right):{\int }_{\mathcal{D}}qd\mathit{x}=0\},\hfill \\ \hfill \hspace{1em}& \mathbf{W}:={\mathbf{H}}_n^1\left(\mathcal{D}\right)=\{\mathit{w}\in H^1{\left(\mathcal{D}\right)}^d:\mathit{w}·\mathit{n}{\mid }_{\partial \mathcal{D}}=0\},\hfill \\ \hfill \hspace{1em}& {\mathbf{X}}_0:=\{\mathit{v}\in \mathbf{X}:\nabla ·\mathit{v}=0\},\hfill \\ \hfill \hspace{1em}& {\mathbf{W}}_0:=\{\mathit{w}\in \mathbf{W}:\nabla ·\mathit{w}=0\}.\hfill \end{array}$$

We define $A_1\mathit{u}=-\Delta \mathit{u}$ and $A_{1ϵ}\mathit{u}=-\Delta \mathit{u}-\frac{1}{ϵ}\nabla \nabla ·\mathit{u}$, which are the operators associated with NS equations and the penalty NS equations. They are the positive self-adjoint operators from $D\left(A_1\right)={\mathbf{H}}^2\left(\mathcal{D}\right)\cap \mathbf{X}$ onto ${\mathbf{L}}^2\left(\mathcal{D}\right)$ and the powers $A_1^{\alpha }$ and $A_{1ϵ}^{\alpha }$$(\alpha \in \mathbb{R})$ are well defined. Similarly, we define the operator $A_2=P_H(\nabla \times \nabla \times -\nabla \nabla ·):D\left(A_2\right)\to \tilde{\mathbf{H}}$, where $D\left(A_2\right)={\mathbf{H}}^2\left(\mathcal{D}\right)\cap \mathbf{W}$ and $P_H$ is the $L^2$-orthogonal projector (cf. [7,26,27]). Thus, we have

$$\begin{array}{c}(A_1\mathit{u},\mathit{v})=(A_1^{\frac{1}{2}}\mathit{u},A_1^{\frac{1}{2}}\mathit{v})=(\nabla \mathit{u},\nabla \mathit{v}),\hspace{1em}\forall \mathit{u},\mathit{v}\in \mathbf{X},\hfill \\ (A_{1ϵ}\mathit{u},\mathit{v})=(A_{1ϵ}^{\frac{1}{2}}\mathit{u},A_{1ϵ}^{\frac{1}{2}}\mathit{v})=(\nabla \mathit{u},\nabla \mathit{v})+\frac{1}{ϵ}(\nabla ·\mathit{u},\nabla ·\mathit{v}),\hspace{1em}\forall \mathit{u},\mathit{v}\in \mathbf{X},\hfill \\ (A_2\mathit{B},\mathit{C})=(A_2^{\frac{1}{2}}\mathit{B},A_2^{\frac{1}{2}}\mathit{C})=(\nabla \times \mathit{B},\nabla \times \mathit{C})+(\nabla ·\mathit{B},\nabla ·\mathit{C}),\hspace{1em}\forall \mathit{B},\mathit{C}\in \mathbf{W}.\hfill \end{array}$$

Define

$$\begin{array}{cc}\hfill b(\mathit{u},\mathit{v},\mathbf{w})& =(u·\nabla \mathit{v},\mathbf{w})+\frac{1}{2}((\nabla ·\mathit{u})\mathit{v},\mathbf{w})=\frac{1}{2}b(\mathit{u},\mathit{v},\mathbf{w})-\frac{1}{2}b(\mathit{u},\mathbf{w},\mathit{v}),\hspace{1em}\forall \mathit{u},\mathit{v},\mathbf{w}\in \mathbf{X}.\hfill \end{array}$$

Therefore, the trilinear form $b(·,·,·)$ satisfies

$$\begin{array}{c}\hfill b(\mathit{u},\mathit{v},\mathit{v})=0,\hspace{1em}\forall \mathit{u},\mathit{v}\in \mathbf{X}.\end{array}$$ (3)

A weak formulation for (1) is as follows: find $(\mathit{u},p,\mathit{B})\in L^2(0,\mathcal{T};\mathbf{X})\times L^2(0,\mathcal{T};\mathrm{M})\times L^2(0,\mathcal{T};\mathbf{W})$ such that, for all $(\mathit{v},q,\mathit{C})\in \mathbf{X}\times \mathrm{M}\times \mathbf{W}$ (cf. [1,4]),

$$\left\{\begin{array}{c}({\mathit{u}}_t,\mathit{v})+\nu (\nabla \mathit{u},\nabla \mathit{v})+b(\mathit{u},\mathit{u},\mathit{v})-(\nabla ·\mathit{v},p)+(\nabla ·\mathit{u},q)+S(\mathit{B}\times \nabla \times \mathit{B},\mathit{v})=(\mathit{f},\mathit{v}),\hfill \\ ({\mathit{B}}_t,\mathit{C})+\mu (\nabla \times \mathit{B},\nabla \times \mathit{C})-(\mathit{u}\times \mathit{B},\nabla \times \mathit{C})=(\mathit{g},\mathit{C}),\hfill \\ \mathit{u}\left(0\right)={\mathit{u}}^0,\hspace{1em}\mathit{B}\left(0\right)={\mathit{B}}^0,\hfill \end{array}\right.$$ (4)

where ${\mathit{u}}_t\in L^{4/d}(0,\mathcal{T};{\mathbf{X}}^{\prime })$, ${\mathit{B}}_t\in L^{4/d}(0,\mathcal{T};{\mathbf{W}}^{\prime })$ (${\mathbf{X}}^{\prime }$ and ${\mathbf{W}}^{\prime }$ are the dual spaces of $\mathbf{X}$ and $\mathbf{W}$, respectively), $\nabla ·{\mathit{u}}^0=\nabla ·{\mathit{B}}^0=0$ and the weak formulation for (2) is as follows: find $({\mathit{u}}_{ϵ},p_{ϵ},{\mathit{B}}_{ϵ})\in L^2(0,\mathcal{T};\mathbf{X})\times L^2(0,\mathcal{T};\mathrm{M})\times L^2(0,\mathcal{T};\mathbf{W})$ such that, for all $(\mathit{v},q,\mathit{C})\in \mathbf{X}\times \mathrm{M}\times \mathbf{W}$

$$\left\{\begin{array}{c}({\mathit{u}}_{ϵt},\mathit{v})-\nu (\nabla {\mathit{u}}_{ϵ},\nabla \mathit{v})+b({\mathit{u}}_{ϵ},{\mathit{u}}_{ϵ},\mathit{v})-(\nabla ·\mathit{v},p_{ϵ})+(\nabla ·{\mathit{u}}_{ϵ},q)+S({\mathit{B}}_{ϵ}\times \nabla \times {\mathit{B}}_{ϵ},\mathit{v})\hfill \\ \hspace{1em}+\frac{ϵ}{\nu }(p_{ϵ},q)=(\mathit{f},\mathit{v}),\hfill \\ ({\mathit{B}}_{ϵt},\mathit{C})+\mu (\nabla \times {\mathit{B}}_{ϵ},\nabla \times \mathit{C})-({\mathit{u}}_{ϵ}\times {\mathit{B}}_{ϵ},\nabla \times \mathit{C})=(\mathit{g},\mathit{C}),\hfill \\ {\mathit{u}}_{ϵ}\left(0\right)={\mathit{u}}^0,\hspace{1em}{\mathit{B}}_{ϵ}\left(0\right)={\mathit{B}}^0,\hfill \end{array}\right.$$ (5)

where ${\mathit{u}}_{ϵt}\in L^{4/d}(0,\mathcal{T};{\mathbf{X}}^{\prime })$, ${\mathit{B}}_{ϵt}\in L^{4/d}(0,\mathcal{T};{\mathbf{W}}^{\prime })$.

Remark 1.

Taking the $L^2$ inner product of the first equation in (1) with $\mathit{v}$, the second equation in (1) with q and summing up the two relations, we obtain the first equation of (4). Taking the $L^2$ inner product of the third equation in (1) with $\mathit{C}$, we obtain the second equation of (4). Equation (5) can be obtained similarly.

Remark 2.

For $\varphi \in {\mathbf{H}}_0^2\left(\mathcal{D}\right)$, taking $\mathit{C}=\nabla \varphi$ in the second equation of (4), we easily obtain ${(\nabla ·\mathit{B})}_t=0$, which implies $\nabla ·\mathit{B}\left(t\right)=\nabla ·{\mathit{B}}^0=0$. In addition, the second equation of (4) has an equivalent form (cf. [1,7])

$$\begin{array}{c}\hfill ({\mathit{B}}_t,\mathit{C})+\mu (\nabla \times \mathit{B},\nabla \times \mathit{C})+\mu (\nabla ·\mathit{B},\nabla ·\mathit{C})-(\mathit{u}\times \mathit{B},\nabla \times \mathit{C})=(\mathit{g},\mathit{C}),\end{array}$$

in which $\nabla ·\mathit{B}$ is used as a penalty term. For $\forall t\in (0,\mathcal{T})$, we choose $\mathit{C}=\nabla \varphi \left(t\right)\in \mathbf{W}$ in the second equation of (4); here, $\varphi \left(t\right)$ is generated by the boundary value problem

$$\left\{\begin{array}{c}\Delta \varphi \left(t\right)=\nabla ·\mathit{B}\left(t\right),\hfill \\ \frac{\partial \varphi }{\partial \mathit{n}}{|}_{\partial \mathcal{D}}=0.\hfill \end{array}\right.$$

We can obtain $\frac{1}{2}{\parallel \nabla \varphi \left(t\right)\parallel }^2+\mu {\int }_0^{\mathcal{T}}{\parallel \nabla ·\mathit{B}\left(t\right)\parallel }^{2}dt=0$, which also implies $\nabla ·\mathit{B}\left(t\right)=0$.

Using the operators $A_{1ϵ},A_2$, we can rewrite the penalized system (5) as

$$\left\{\begin{array}{c}{\mathit{u}}_{ϵt}+\nu A_{1ϵ}{\mathit{u}}_{ϵ}+\tilde{b}({\mathit{u}}_{ϵ},{\mathit{u}}_{ϵ})+S({\mathit{B}}_{ϵ}\times \nabla \times {\mathit{B}}_{ϵ})=\mathit{f},\hfill \\ {\mathit{B}}_{ϵt}+\mu A_2{\mathit{B}}_{ϵ}-\nabla \times ({\mathit{u}}_{ϵ}\times {\mathit{B}}_{ϵ})=\mathit{g}.\hfill \end{array}\right.$$ (6)

Referring to [2,6,13,28], the following estimates hold:

$$\begin{array}{ccc}& & {\parallel \mathit{v}\parallel }_{{\mathbf{L}}^p}\le c{\parallel \mathit{v}\parallel }_1,\hspace{1em}2\le p\le 6,\hspace{1em}\mathit{v}\in {\mathbf{H}}^1\left(\mathcal{D}\right),\hfill \end{array}$$ (7)

$$\begin{array}{ccc}& & {\parallel \mathit{v}\parallel }_{{\mathbf{L}}^{\infty }}+{\parallel \nabla \mathit{v}\parallel }_{{\mathbf{L}}^3}\le {c\parallel \mathit{v}\parallel }_1^{\frac{1}{2}}{\parallel \mathit{v}\parallel }_2^{\frac{1}{2}},\hspace{1em}\mathit{v}\in {\mathbf{H}}^2\left(\mathcal{D}\right),\hfill \end{array}$$ (8)

$$\begin{array}{ccc}& & {\parallel \mathit{C}\parallel }_1\le c\parallel \nabla \times \mathit{C}\parallel +c\parallel \nabla ·\mathit{C}\parallel ,\hspace{1em}\forall \mathit{C}\in \mathbf{W},\hfill \end{array}$$ (9)

$$\begin{array}{ccc}& & (\mathit{B}\times \nabla \times \mathit{C},\mathit{v})=(\mathit{v}\times \mathit{B},\nabla \times \mathit{C}),\hspace{1em}\forall \mathit{B},\mathit{C}\in \mathbf{W},\hspace{1em}\mathit{v}\in \mathbf{X},\hfill \end{array}$$ (10)

$$\begin{array}{ccc}& & \parallel \nabla \times \mathit{C}\parallel \le \sqrt{2}\parallel \nabla \mathit{C}\parallel ,\hspace{1em}\parallel \nabla ·\mathit{C}\parallel \le \sqrt{d}\parallel \nabla \mathit{C}\parallel ,\hspace{1em}\forall \mathit{C}\in \mathbf{W},\hfill \end{array}$$ (11)

$$\begin{array}{ccc}& & \nabla \times (\mathit{v}\times \mathit{C})=\mathit{v}(\nabla ·\mathit{C})-\mathit{C}(\nabla ·\mathit{v})+(\mathit{C}·\nabla )\mathit{v}-(\mathit{v}·\nabla )\mathit{C},\hspace{1em}\forall \mathit{v},\mathit{C}\in {\mathbf{H}}^1\left(\mathcal{D}\right),\hfill \end{array}$$ (12)

where $c\left(\mathcal{D}\right)>0$ is a constant, which have different values in different cases.

In this paper, $C>0$ denotes a constant depending on $(\nu ,\mu ,S,\mathcal{D},\mathcal{T},{\mathit{u}}_0,{\mathit{B}}_0,\mathit{f},\mathit{g})$, which may have different values in different cases. We make the following assumptions for (1), which specify the regularity of the data and the smoothness of the domain $\mathcal{D}$ (cf. [1,4]).

Assumption 1.

The initial data ${\mathit{u}}_0\in {\mathit{X}}_0\cap {\mathit{H}}^2(\Omega )$ and ${\mathit{B}}_0\in {\mathit{W}}_0\cap {\mathit{H}}^2(\Omega ),$ the external force $\mathit{f}$, and the applied current $\mathit{g}$ satisfy the following bound:

$$\parallel {\mathit{u}}_0{\parallel }_2+{\parallel {\mathit{B}}_0\parallel }_2+\underset{0\le t\le \mathcal{T}}{sup}\left\{\parallel \mathit{f}\left(t\right)\parallel +\parallel \mathit{g}\left(t\right)\parallel +\parallel {\mathit{f}}_t\left(t\right)\parallel +\parallel {\mathit{g}}_t\left(t\right)\parallel \right\}\le C.$$

Assumption 1 ensures that there is a unique strong solution for (1) over some time interval $[0,\mathcal{T})$; we have (cf. [5])

$$\begin{array}{c}\hfill \begin{array}{c}\mathit{u}\in C(0,\mathcal{T};\mathbf{X})\cap L^2(0,\mathcal{T};{\mathbf{H}}^2\left(\mathcal{D}\right)),\hspace{1em}p\in L^2(0,\mathcal{T};H^1\left(\mathcal{D}\right)\cap M),\\ \mathit{B}\in C(0,\mathcal{T};\mathbf{W})\cap L^2(0,\mathcal{T};{\mathbf{H}}^2\left(\mathcal{D}\right)),\end{array}\end{array}$$

such that ${\mathit{u}}_t,{\mathit{B}}_t\in L^2(0,\mathcal{T};L^2\left(\mathcal{D}\right))$, and Equation (4) holds for almost all $t\in [0,\mathcal{T})$. If the data ${\mathit{u}}_0,{\mathit{B}}_0,\mathit{f}$ and $\mathit{g}$ are sufficiently small, then the solution exists for any $\mathcal{T}>0$ and satisfies

$$\begin{array}{c}\hfill \underset{0\le t\le \mathcal{T}}{sup}{(\parallel \mathit{u}\left(t\right)\parallel }_1+{\parallel \mathit{B}\left(t\right)\parallel }_1)<\infty .\end{array}$$ (13)

Assumption 2.

The problem (4) has a weak solution $\left(\mathit{u}\right(t),p(t),\mathit{B}(t\left)\right)$ satisfying $\mathit{u}\in L^2(0,\mathcal{T};{\mathit{X}}_0)$, $p\in L^2(0,\mathcal{T};M)$ and $\mathit{B}\in L^2(0,\mathcal{T};{\mathit{W}}_0)$ such that

$$\begin{array}{c}\hfill {\int }_0^{\mathcal{T}}{(\parallel \nabla \mathit{u}\left(t\right)\parallel }^4{+\parallel \nabla \times \mathit{B}\left(t\right)\parallel }^4)dt\le C.\end{array}$$

Remark 3.

Instead of assuming the data are small or strong condition (13) holds, we give the Assumption 2 to guarantee the uniqueness of weak solution to the 3D MHD problem on interval $[0,\mathcal{T}]$ (see [4,5,28]).

Assumption 3.

Assume that the boundary of $\mathcal{D}$ is smooth so that the unique solution $(\mathit{v},q)\in \mathit{X}\times M$ of the steady Stokes problem

$$\begin{array}{c}\hfill {-\Delta \mathit{v}+\nabla q=\mathit{f},\hspace{1em}\nabla ·\mathit{v}=0,\hspace{1em}\mathrm{in}\mathcal{D},\hspace{1em}\mathit{v}|}_{\partial \mathcal{D}}=\mathbf{0},\end{array}$$

for prescribed $\mathit{f}\in {\mathit{L}}^2\left(\mathcal{D}\right)$ satisfies

$$\begin{array}{c}\hfill {\parallel \mathit{v}\parallel }_2+{\parallel q\parallel }_1\le c\parallel \mathit{f}\parallel ;\end{array}$$

and Maxwell’s equations

$$\begin{array}{c}\hfill {\nabla \times \nabla \times \mathit{C}=\mathit{g},\hspace{1em}\nabla ·\mathit{C}=0,\hspace{1em}\mathrm{in}\mathcal{D},\hspace{1em}\mathit{C}\times \mathit{n}|}_{\partial \mathcal{D}}=\mathbf{0},\end{array}$$

for the prescribed $\mathit{g}\in {\mathit{L}}^2\left(\mathcal{D}\right)$ admit a unique solution $\mathit{C}\in {\mathit{W}}_0$, which satisfies

$$\begin{array}{c}\hfill {\parallel \mathit{C}\parallel }_2\le c\parallel \mathit{g}\parallel .\end{array}$$

Remark 4.

The validity of Assumption 3 is known if $\partial \mathcal{D}$ is of $C^2$, or if $\mathcal{D}$ is a convex polyhedron (see [4]).

Next, we need the following lemma given in [26].

Lemma 1.

There exists a constant $c_1>0$ depending only on $\mathcal{D}$ and such that, for sufficiently small ϵ, we have

$$\begin{array}{ccc}& & \parallel \Delta \mathit{u}\parallel \le c_1\parallel A_{1ϵ}\mathit{u}\parallel ,\hspace{1em}\forall \mathit{u}\in {\mathbf{H}}^2\left(\mathcal{D}\right)\cap \mathbf{X},\hfill \\ & & \parallel \nabla \mathit{u}\parallel \le c_1\parallel A_{1ϵ}^{\frac{1}{2}}\mathit{u}\parallel ,\hspace{1em}\forall \mathit{u}\in \mathbf{X},\hfill \\ & & \parallel A_{1ϵ}^{-1}\mathit{u}\parallel \le c_1{\parallel \mathit{u}\parallel }_{-2},\hspace{1em}\forall \mathit{u}\in {\mathbf{H}}^{-2}\left(\mathcal{D}\right),\hfill \end{array}$$

where ${\mathbf{H}}^{-2}\left(\mathcal{D}\right)$ is the dual space of ${\mathbf{H}}^2\left(\mathcal{D}\right)\cap \mathbf{X}$, and ${\parallel ·\parallel }_{-2}$ is the corresponding norm.

Theorem 1.

Under Assumptions 1–3, the solution $\left(\mathit{u}\right(t),p(t),\mathit{B}(t\left)\right)$ of the problem (4) satisfies the estimates

$$\begin{array}{ccc}& & \underset{0\le t\le \mathcal{T}}{sup}\{\parallel {\mathit{u}}_t{\left(t\right)\parallel }^2+\parallel {\mathit{B}}_t{\left(t\right)\parallel }^2+{\parallel \mathit{u}\left(t\right)\parallel }_2^2+{\parallel p\left(t\right)\parallel }_1^2+{\parallel \mathit{B}\left(t\right)\parallel }_2^2\}\hfill \\ & & \hspace{1em}+{\int }_0^{\mathcal{T}}(\parallel {\mathit{u}}_{tt}{\parallel }_{{\mathit{X}}_0^{\prime }}^2+\parallel {\mathit{B}}_{tt}{\parallel }_{{\mathit{W}}^{\prime }}^2+\parallel \nabla {\mathit{u}}_t{\parallel }^2+\parallel \nabla {\mathit{B}}_t{\parallel }^2)dt\le C,\hfill \\ & & \underset{0\le t\le \mathcal{T}}{sup}\{\sigma \left(t\right)\parallel {\mathit{u}}_t{\left(t\right)\parallel }_1^2+\sigma \left(t\right)\parallel {\mathit{B}}_t{\left(t\right)\parallel }_1^2\}+{\int }_0^{\mathcal{T}}\sigma \left(t\right)(\parallel {\mathit{u}}_{tt}{\parallel }^2+\parallel {\mathit{B}}_{tt}{\parallel }^2)dt\hfill \\ & & \hspace{1em}+{\int }_0^{\mathcal{T}}\sigma \left(t\right)(\parallel {\mathit{u}}_t{\parallel }_2^2+\parallel {\mathit{B}}_t{\parallel }_2^2+\parallel p_t{\parallel }_1^2)dt\le C,\hfill \end{array}$$

where

$$\begin{array}{c}\hfill \parallel {\mathit{u}}_{tt}{\parallel }_{{\mathit{X}}_0^{\prime }}=\underset{\mathit{v}\in {\mathit{X}}_0}{sup}\frac{({\mathit{u}}_{tt},\mathit{v})}{\parallel \nabla \mathit{v}\parallel },\hspace{1em}{\parallel {\mathit{B}}_{tt}\parallel }_{{\mathit{W}}^{\prime }}=\underset{\mathit{C}\in \mathit{W}}{sup}\frac{({\mathit{B}}_{tt},\mathit{C})}{\parallel \nabla \mathit{C}\parallel }.\end{array}$$

For the proof of these results, we can refer to [1,4].

Theorem 2.

Under Assumptions 1–3, the solution $({\mathit{u}}_{ϵ}\left(t\right),p_{ϵ}\left(t\right),{\mathit{B}}_{ϵ}\left(t\right))$ of the problem (3) satisfies the estimates

$$\begin{array}{ccc}& & \underset{0\le t\le \mathcal{T}}{sup}\{\parallel {\mathit{u}}_{ϵt}{\left(t\right)\parallel }^2+\parallel {\mathit{B}}_{ϵt}{\left(t\right)\parallel }^2+\parallel {\mathit{u}}_{ϵ}{\left(t\right)\parallel }_2^2+\parallel p_{ϵ}{\left(t\right)\parallel }_1^2+\parallel {\mathit{B}}_{ϵ}\left(t\right){\parallel }_2^2\}\hfill \\ & & \hspace{1em}+{\int }_0^{\mathcal{T}}(\parallel A_{1ϵ}^{-\frac{1}{2}}{\mathit{u}}_{ϵtt}{\parallel }^2+\parallel A_2^{-\frac{1}{2}}{\mathit{B}}_{ϵtt}{\parallel }^2+\parallel \nabla {\mathit{u}}_{ϵt}{\parallel }^2+\parallel \nabla {\mathit{B}}_{ϵt}{\parallel }^2)dt\le C,\hfill \\ & & \underset{0\le t\le \mathcal{T}}{sup}\{\sigma \left(t\right)\parallel {\mathit{u}}_{ϵt}{\left(t\right)\parallel }_1^2+\sigma \left(t\right)\parallel {\mathit{B}}_{ϵt}{\left(t\right)\parallel }_1^2\}+{\int }_0^{\mathcal{T}}\sigma \left(t\right)(\parallel {\mathit{u}}_{ϵtt}{\parallel }^2+\parallel {\mathit{B}}_{ϵtt}{\parallel }^2+\parallel p_{ϵt}{\parallel }^2)dt\le C.\hfill \end{array}$$

We can finish the proof by a similar technique used in the proof of Theorem 1.

Theorem 3.

Under Assumptions 1–3, we have the following estimate (cf. [26,29])

$$\begin{array}{ccc}& & \underset{0\le t\le \mathcal{T}}{sup}{\{\sigma \left(t\right)}^{\frac{1}{2}}(\parallel \mathit{u}\left(t\right)-{\mathit{u}}_{ϵ}\left(t\right)\parallel +\parallel \mathit{B}\left(t\right)-{\mathit{B}}_{ϵ}\left(t\right)\parallel )+\sigma \left(t\right)(\parallel \mathit{u}\left(t\right)-{\mathit{u}}_{ϵ}{\left(t\right)\parallel }_1+\parallel \mathit{B}\left(t\right)-{\mathit{B}}_{ϵ}{\left(t\right)\parallel }_1\left)\right\}\hfill \\ & & \hspace{1em}+{\left({\int }_0^{\mathcal{T}}{\sigma }^2\left(t\right){\parallel p\left(t\right)-p_{ϵ}\left(t\right)\parallel }^{2}dt\right)}^{\frac{1}{2}}\le Cϵ.\hfill \end{array}$$

Proof. 

We first consider the linear form of MHD equations. Then, we subtract the penalized linear MHD equations from the linear MHD equations to obtain their error equations. The error estimate in linear form is obtained through its dual problem

$$\begin{array}{c}\hfill {\int }_0^{\mathcal{T}}(\parallel \mathit{u}\left(t\right)-{\mathit{u}}_{ϵ}\left(t\right)\parallel +\parallel \mathit{B}\left(t\right)-{\mathit{B}}_{ϵ}\left(t\right)\parallel )dt\le Cϵ.\end{array}$$

Next, we obtain the following error estimate by choosing an appropriate $L^2$ inner product for the error equations

$$\begin{array}{cc}\hfill \hspace{1em}& \underset{0\le t\le \mathcal{T}}{sup}{\{\sigma \left(t\right)}^{\frac{1}{2}}(\parallel \mathit{u}\left(t\right)-{\mathit{u}}_{ϵ}\left(t\right)\parallel +\parallel \mathit{B}\left(t\right)-{\mathit{B}}_{ϵ}\left(t\right)\parallel )+\sigma \left(t\right)(\parallel \mathit{u}\left(t\right)-{\mathit{u}}_{ϵ}{\left(t\right)\parallel }_1+\parallel \mathit{B}\left(t\right)-{\mathit{B}}_{ϵ}{\left(t\right)\parallel }_1\left)\right\}\hfill \\ \hfill \hspace{1em}& \hspace{1em}+{\left({\int }_0^{\mathcal{T}}{\sigma }^2\left(t\right){\parallel p\left(t\right)-p_{ϵ}\left(t\right)\parallel }^{2}dt\right)}^{\frac{1}{2}}\le Cϵ.\hfill \end{array}$$

Finally, we transform the nonlinear MHD equations into an intermediate linear equations, and then obtain Theorem 3 by applying a suitable $L^2$ inner product to this system and using the previous result. □

In addition, we need the following discrete Gronwall lemma (cf. [4,30,31]).

Lemma 2.

Let $a_n,b_n,d_n$ and $C_0$ be nonnegative numbers for integer $n\ge 0$, such that

$$\begin{array}{c}\hfill a_m+\tau \sum _{n=1}^m{b}_n\le \tau \sum _{n=0}^{m-1}{d}_n{a}_n+C_0,\hspace{1em}\forall m\ge 1.\end{array}$$ (14)

Then

$$\begin{array}{c}\hfill a_m+\tau \sum _{n=1}^m{b}_n\le C_{0}exp(\tau \sum _{n=0}^{m-1}{d}_n),\hspace{1em}\forall m\ge 1.\end{array}$$ (15)

3. The Euler Semi-Implicit Scheme and Its Error Estimates: Time Discretization

In this section, we consider a time discretization for the penalty MHD system (6). Let $\tau =\frac{\mathcal{T}}{N}$ be the time-step size and $N>0$ is an integer. Then, $t_n=n\tau$, $n=1,2,\cdots ,N$ denote the discrete time levels. The time-discrete approximations to $({\mathit{u}}_{ϵ}\left(t_n\right),p_{ϵ}\left(t_n\right),{\mathit{B}}_{ϵ}\left(t_n\right))$ will be denoted by $({\mathit{u}}_{ϵ}^n,p_{ϵ}^n,{\mathit{B}}_{ϵ}^n)$ for all $1\le n\le N$. Consider the Euler semi-implicit time-stepping algorithm: Given $({\mathit{u}}_{ϵ}^{n-1},{\mathit{B}}_{ϵ}^{n-1})\in \mathbf{X}\times \mathbf{W}$, find $({\mathit{u}}_{ϵ}^n,{\mathit{B}}_{ϵ}^n)\in \mathbf{X}\times \mathbf{W}$, $p_{ϵ}^n\in \mathrm{M}$ such that

$$\left\{\begin{array}{c}(d_t{\mathit{u}}_{ϵ}^n,\mathit{v})-\nu (\nabla {\mathit{u}}_{ϵ}^n,\nabla \mathit{v})+b({\mathit{u}}_{ϵ}^{n-1},{\mathit{u}}_{ϵ}^n,\mathit{v})-(\nabla ·\mathit{v},p_{ϵ}^n)+(\nabla ·{\mathit{u}}_{ϵ}^n,q)\hfill \\ +S({\mathit{B}}_{ϵ}^{n-1}\times \nabla \times {\mathit{B}}_{ϵ}^n,\mathit{v})+\frac{ϵ}{\nu }(p_{ϵ}^n,q)=(\mathit{f}\left(t_n\right),\mathit{v}),\hfill \\ (d_t{\mathit{B}}_{ϵ}^n,\mathit{C})+\mu (\nabla \times {\mathit{B}}_{ϵ}^n,\nabla \times \mathit{C})-({\mathit{u}}_{ϵ}^n\times {\mathit{B}}_{ϵ}^{n-1},\nabla \times \mathit{C})=(\mathit{g}\left(t_n\right),\mathit{C}),\hfill \end{array}\right.$$ (16)

where $({\mathit{u}}_{ϵ}^0,{\mathit{B}}_{ϵ}^0)=({\mathit{u}}^0,{\mathit{B}}^0)$, $d_t{\mathit{u}}^n=\frac{1}{\tau }({\mathit{u}}^n-{\mathit{u}}^{n-1})$. We can rewrite (16) as

$$\left\{\begin{array}{c}{d}_t{\mathit{u}}_{ϵ}^n+\nu A_{1ϵ}{\mathit{u}}_{ϵ}^n+\tilde{b}({\mathit{u}}_{ϵ}^{n-1},{\mathit{u}}_{ϵ}^n)+S({\mathit{B}}_{ϵ}^{n-1}\times \nabla \times {\mathit{B}}_{ϵ}^n)={\mathit{f}}^n,\hfill \\ d_t{\mathit{B}}_{ϵ}^n+\mu A_2{\mathit{B}}_{ϵ}^n-\nabla \times ({\mathit{u}}_{ϵ}^n\times {\mathit{B}}_{ϵ}^{n-1})={\mathit{g}}^n.\hfill \end{array}\right.$$ (17)

Remark 5.

Since $\nabla ·{\mathit{B}}^0=0$, we take $\mathit{C}=\nabla \varphi$ with $\varphi \in {\mathit{H}}_0^2\left(\mathcal{D}\right)$ and deduce from the second equation in (16) and the identity $\nabla \times \nabla \varphi =\mathbf{0}$ that $\nabla ·{\mathit{B}}^n=0$ for all $0\le n\le N$.

Next, we give a priori bound of the scheme (17).

Theorem 4.

Under Assumptions 1–3, we have a priori bound

$$\begin{array}{c}\parallel {\mathit{u}}_{ϵ}^m{\parallel }^2+S\parallel {\mathit{B}}_{ϵ}^m{\parallel }^2+\tau \sum _{n=1}^m(\nu \parallel A_{1ϵ}^{\frac{1}{2}}{\mathit{u}}_{ϵ}^n{\parallel }^2+S\mu \parallel A_2^{\frac{1}{2}}{\mathit{B}}_{ϵ}^n{\parallel }^2)+\tau \sum _{n=1}^m(\parallel d_t{\mathit{u}}_{ϵ}^n{\parallel }^2+S\parallel d_t{\mathit{B}}_{ϵ}^n{\parallel }^2)\tau \le C,\end{array}$$

for all $1\le m\le N$.

Proof. 

Taking the $L^2$ inner product of the first equation in (17) with $2{\mathit{u}}_{ϵ}^n\tau$, and the second equation in (17) with $2S{\mathit{B}}_{ϵ}^n\tau$, we obtain

$$\begin{array}{c}\parallel {\mathit{u}}_{ϵ}^n{\parallel }^2+S\parallel {\mathit{B}}_{ϵ}^n{\parallel }^2-(\parallel {\mathit{u}}_{ϵ}^{n-1}{\parallel }^2+S\parallel {\mathit{B}}_{ϵ}^{n-1}{\parallel }^2)+(\parallel d_t{\mathit{u}}_{ϵ}^n{\parallel }^2+S\parallel d_t{\mathit{B}}_{ϵ}^n{\parallel }^2){\tau }^2\\ \hspace{1em}+2\nu \parallel A_{1ϵ}^{\frac{1}{2}}{\mathit{u}}_{ϵ}^n{\parallel }^2\tau +2S\mu \parallel A_2^{\frac{1}{2}}{\mathit{B}}_{ϵ}^n{\parallel }^2\tau \le 2({\mathit{f}}^n,{\mathit{u}}_{ϵ}^n)\tau +2S({\mathit{g}}^n,{\mathit{B}}_{ϵ}^n)\tau .\end{array}$$ (18)

By using (7), (9), Lemma 1 and the Young inequality, we have

$$\begin{array}{c}2|({\mathit{f}}^n,{\mathit{u}}_{ϵ}^n)|\le 2\parallel {\mathit{f}}^n\parallel \parallel {\mathit{u}}_{ϵ}^n\parallel \le \nu \parallel A_{1ϵ}^{\frac{1}{2}}{\mathit{u}}_{ϵ}^n{\parallel }^2+\frac{c}{\tau }{\int }_{t_{n-1}}^{t_n}{\parallel \mathit{f}\left(t\right)\parallel }^{2}dt,\\ 2S|({\mathit{g}}^n,{\mathit{B}}_{ϵ}^n)|\le 2S\parallel {\mathit{g}}^n\parallel \parallel {\mathit{B}}_{ϵ}^n\parallel \le S\mu \parallel A_2^{\frac{1}{2}}{\mathit{B}}_{ϵ}^n{\parallel }^2+\frac{c}{\tau }{\int }_{t_{n-1}}^{t_n}{\parallel \mathit{g}\left(t\right)\parallel }^{2}dt.\end{array}$$

Combining the above inequalities with (18), we obtain

$$\begin{array}{c}\parallel {\mathit{u}}_{ϵ}^n{\parallel }^2+S\parallel {\mathit{B}}_{ϵ}^n{\parallel }^2-(\parallel {\mathit{u}}_{ϵ}^{n-1}{\parallel }^2+S\parallel {\mathit{B}}_{ϵ}^{n-1}{\parallel }^2)+(\parallel d_t{\mathit{u}}_{ϵ}^n{\parallel }^2+S\parallel d_t{\mathit{B}}_{ϵ}^n{\parallel }^2){\tau }^2\\ \hspace{1em}+\nu \parallel A_{1ϵ}^{\frac{1}{2}}{\mathit{u}}_{ϵ}^n{\parallel }^2\tau +S\mu \parallel A_2^{\frac{1}{2}}{\mathit{B}}_{ϵ}^n{\parallel }^2\tau \le c{\int }_{t_{n-1}}^{t_n}{(\parallel \mathit{f}\left(t\right)\parallel }^2+{\parallel \mathit{g}\left(t\right)\parallel }^2)dt.\end{array}$$

Summing the above inequality from 1 to m, we derive

$$\begin{array}{ccc}\hfill & & \parallel {\mathit{u}}_{ϵ}^m{\parallel }^2+S\parallel {\mathit{B}}_{ϵ}^m{\parallel }^2+\tau \sum _{n=1}^m(\parallel d_t{\mathit{u}}_{ϵ}^n{\parallel }^2+S\parallel d_t{\mathit{B}}_{ϵ}^n{\parallel }^2)\tau +\tau \sum _{n=1}^m(\nu \parallel A_{1ϵ}^{\frac{1}{2}}{\mathit{u}}_{ϵ}^n{\parallel }^2\tau +S\mu \parallel A_2^{\frac{1}{2}}{\mathit{B}}_{ϵ}^n{\parallel }^2)\hfill \\ & & \hspace{1em}\le \parallel {\mathit{u}}_{ϵ}^0{\parallel }^2+S\parallel {\mathit{B}}_{ϵ}^0{\parallel }^2+c{\int }_0^{\mathcal{T}}{(\parallel \mathit{f}\left(t\right)\parallel }^2+{\parallel \mathit{g}\left(t\right)\parallel }^2)dt,\hfill \end{array}$$ (19)

for all $1\le m\le N$. Using Assumption 1 and (19), we obtain Theorem 4. □

Next, we establish the error estimates in time for the Euler semi-implicit scheme (17). To do this, subtracting (17) from (6) and setting $e_u^n={\mathit{u}}_{ϵ}\left(t_n\right)-{\mathit{u}}_{ϵ}^n$, $e_B^n={\mathit{B}}_{ϵ}\left(t_n\right)-{\mathit{B}}_{ϵ}^n$, we have

$$\begin{array}{ccc}\hfill & & d_t{e}_u^n+\nu A_{1ϵ}{e}_u^n+\tilde{b}(e_u^{n-1},{\mathit{u}}_{ϵ}\left(t_n\right))+\tilde{b}({\mathit{u}}_{ϵ}^{n-1},e_u^n)+S{e}_B^{n-1}\times \nabla \times {\mathit{B}}_{ϵ}\left(t_n\right)\hfill \\ & & \hspace{1em}+S{\mathit{B}}_{ϵ}^{n-1}\times \nabla \times e_B^n=R_1^n,\hfill \end{array}$$ (20)

$$\begin{array}{cc}\hfill \hspace{1em}& d_t{e}_B^n+\mu A_2{e}_B^n-\nabla \times (e_u^n\times {\mathit{B}}_{ϵ}\left(t_{n-1}\right))-\nabla \times ({\mathit{u}}_{ϵ}^n\times e_B^{n-1})=R_2^n,\hfill \end{array}$$ (21)

where

$$\begin{array}{cc}\hfill R_1^n=& -\frac{1}{\tau }{\int }_{t_{n-1}}^{t_n}(t-t_{n-1}){\mathit{u}}_{ϵtt}\left(t\right)dt+\tilde{b}({\mathit{u}}_{ϵ}\left(t_{n-1}\right)-{\mathit{u}}_{ϵ}\left(t_n\right),{\mathit{u}}_{ϵ}\left(t_n\right))\hfill \\ \hfill \hspace{1em}& +S(({\mathit{B}}_{ϵ}\left(t_{n-1}\right)-{\mathit{B}}_{ϵ}\left(t_n\right))\times \nabla \times {\mathit{B}}_{ϵ}\left(t_n\right)),\hfill \end{array}$$ (22)

$$\begin{array}{cc}\hfill R_2^n=& -\frac{1}{\tau }{\int }_{t_{n-1}}^{t_n}(t-t_{n-1}){\mathit{B}}_{ϵtt}\left(t\right)dt-\nabla \times ({\mathit{u}}_{ϵ}\left(t_n\right)\times ({\mathit{B}}_{ϵ}\left(t_{n-1}\right)-{\mathit{B}}_{ϵ}\left(t_n\right))).\hfill \end{array}$$ (23)

We are now in a position to state and prove two error estimates for the Euler semi-implicit scheme (17).

Theorem 5.

Under Assumptions 1–3, we obtain

$$\begin{array}{c}\hfill \underset{1\le n\le N}{sup}(\parallel {\mathit{u}}_{ϵ}\left(t_n\right)-{\mathit{u}}_{ϵ}^n\parallel +\sqrt{S}\parallel {\mathit{B}}_{ϵ}\left(t_n\right)-{\mathit{B}}_{ϵ}^n\parallel )+{\tau }^{\frac{1}{2}}\sum _{n=1}^N(\sqrt{\nu }\parallel {\mathit{u}}_{ϵ}\left(t_n\right)-{\mathit{u}}_{ϵ}^n{\parallel }_1+\sqrt{S\mu }\parallel {\mathit{B}}_{ϵ}\left(t_n\right)-{\mathit{B}}_{ϵ}^n{\parallel }_1)\le C\tau .\end{array}$$

Proof. 

Taking the inner $L^2$ product of (20) with $2e_u^n\tau$, (21) with $2e_B^n\tau$, thanks to (4) and (10), we deduce that

$$\begin{array}{cc}\hfill \hspace{1em}& \parallel e_u^n{\parallel }^2+S\parallel e_B^n{\parallel }^2-(\parallel e_u^{n-1}{\parallel }^2+S\parallel e_B^{n-1}{\parallel }^2)+2\nu \parallel A_{1ϵ}^{\frac{1}{2}}{e}_u^n{\parallel }^2\tau +2S\mu \parallel A_2^{\frac{1}{2}}{e}_B^n{\parallel }^2\tau +2b(e_u^{n-1},{\mathit{u}}_{ϵ}\left(t_n\right),e_u^n)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}+2S(e_B^{n-1}\times \nabla \times {\mathit{B}}_{ϵ}\left(t_n\right),e_u^n)\tau -2S({\mathit{u}}_{ϵ}\left(t_n\right)\times e_B^{n-1},\nabla \times e_B^n)\tau \le 2(R_1^n,e_u^n)\tau +2(R_2^n,e_B^n)\tau .\hfill \end{array}$$ (24)

By using (7)–(11) and Lemma 1, we obtain

$$\begin{array}{cc}\hfill 2\left|b\right(e_u^{n-1},{\mathit{u}}_{ϵ}\left(t_n\right),e_u^n\left)\right|& \le c\parallel e_u^{n-1}\parallel \parallel e_u^n{\parallel }_1\parallel {\mathit{u}}_{ϵ}\left(t_n\right){\parallel }_2\le \frac{\nu }{6}\parallel A_{1ϵ}^{\frac{1}{2}}{e}_u^n{\parallel }^2+c\parallel {\mathit{u}}_{ϵ}\left(t_n\right){\parallel }_2^2{\parallel e_u^{n-1}\parallel }^2,\hfill \\ \hfill 2S\left|\right(e_B^{n-1}\times \nabla \times {\mathit{B}}_{ϵ}\left(t_n\right),e_u^n\left)\right|& \le c\parallel e_B^{n-1}\parallel \parallel e_u^n{\parallel }_1\parallel {\mathit{B}}_{ϵ}\left(t_n\right){\parallel }_2\le \frac{\nu }{6}\parallel A_{1ϵ}^{\frac{1}{2}}{e}_u^n{\parallel }^2+c\parallel {\mathit{B}}_{ϵ}\left(t_n\right){\parallel }_2^2{\parallel e_B^{n-1}\parallel }^2,\hfill \\ \hfill 2S\left|\right({\mathit{u}}_{ϵ}\left(t_n\right)\times e_B^{n-1},\nabla \times e_B^n\left)\right|& \le c\parallel {\mathit{u}}_{ϵ}\left(t_n\right){\parallel }_2\parallel e_B^n{\parallel }_1\parallel e_B^{n-1}\parallel \le \frac{S\mu }{4}\parallel A_2^{\frac{1}{2}}{e}_B^n{\parallel }^2+c\parallel {\mathit{u}}_{ϵ}\left(t_n\right){\parallel }_2^2{\parallel e_B^{n-1}\parallel }^2.\hfill \end{array}$$

Similarly, we can derive

$$\begin{array}{cc}\hfill 2\left|\right(R_1^n,e_u^n\left)\right|\le & \frac{\nu }{6}\parallel A_{1ϵ}^{\frac{1}{2}}{e}_u^n{\parallel }^2+c\tau {\int }_{t_{n-1}}^{t_n}\parallel A_{1ϵ}^{-\frac{1}{2}}{\mathit{u}}_{ϵtt}{\parallel }^{2}dt+c\tau \parallel {\mathit{u}}_{ϵ}\left(t_n\right){\parallel }_1^2{\int }_{t_{n-1}}^{t_n}{\parallel {\mathit{u}}_{ϵt}\left(t\right)\parallel }_1^{2}dt\hfill \\ \hfill \hspace{1em}& +c\tau \parallel {\mathit{B}}_{ϵ}\left(t_n\right){\parallel }_1^2{\int }_{t_{n-1}}^{t_n}{\parallel {\mathit{B}}_{ϵt}\left(t\right)\parallel }_1^{2}dt,\hfill \\ \hfill 2\left|\right(R_2^n,e_B^n\left)\right|\le & \frac{\nu }{6}\parallel A_2^{\frac{1}{2}}{e}_B^n{\parallel }^2+c\tau {\int }_{t_{n-1}}^{t_n}\parallel A_2^{-\frac{1}{2}}{\mathit{B}}_{ϵtt}{\parallel }^{2}dt+c\tau \parallel {\mathit{u}}_{ϵ}\left(t_n\right){\parallel }_1^2{\int }_{t_{n-1}}^{t_n}{\parallel {\mathit{B}}_{ϵt}\left(t\right)\parallel }_1^{2}dt.\hfill \end{array}$$

Combining the above inequalities with (24), we obtain

$$\begin{array}{cc}\hfill \hspace{1em}& \parallel e_u^n{\parallel }^2+S\parallel e_B^n{\parallel }^2-(\parallel e_u^{n-1}{\parallel }^2+S\parallel e_B^{n-1}{\parallel }^2)+\nu \parallel A_{1ϵ}^{\frac{1}{2}}{e}_u^n{\parallel }^2\tau +S\mu \parallel A_2^{\frac{1}{2}}{e}_B^n{\parallel }^2\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}\le c{d}_{n-1}(\parallel e_u^{n-1}{\parallel }^2+\parallel e_B^{n-1}{\parallel }^2)\tau +c{\tau }^2{\int }_{t_{n-1}}^{t_n}(\parallel A_{1ϵ}^{-\frac{1}{2}}{\mathit{u}}_{ϵtt}{\parallel }^2+\parallel A_2^{-\frac{1}{2}}{\mathit{B}}_{ϵtt}{\parallel }^2)dt\hfill \\ & \hspace{1em}+c{\tau }^2(\parallel {\mathit{u}}_{ϵ}\left(t_n\right){\parallel }_1^2+\parallel {\mathit{B}}_{ϵ}\left(t_n\right){\parallel }_1^2){\int }_{t_{n-1}}^{t_n}(\parallel {\mathit{u}}_{ϵt}\left(t\right){\parallel }_1^2+\parallel {\mathit{B}}_{ϵt}\left(t\right){\parallel }_1^2)dt,\hfill \end{array}$$ (25)

where $d_{n-1}=\parallel {\mathit{u}}_{ϵ}\left(t_n\right){\parallel }_2^2+{\parallel {\mathit{B}}_{ϵ}\left(t_n\right)\parallel }_2^2$. Summing (25) from $n=1$ to m, due to Theorem 2, we obtain

$$\begin{array}{cc}\hfill \hspace{1em}& \parallel e_u^m{\parallel }^2+S\parallel e_B^m{\parallel }^2+\tau \sum _{n=1}^m(\nu \parallel A_{1ϵ}^{\frac{1}{2}}{e}_u^n{\parallel }^2+S\mu \parallel A_2^{\frac{1}{2}}{e}_B^n{\parallel }^2)\le c\tau \sum _{n=0}^{m-1}{d}_n(\parallel e_u^n{\parallel }^2+\parallel e_B^n{\parallel }^2)+C{\tau }^2.\hfill \end{array}$$ (26)

Then, by applying Lemma 2 to (26) and Theorem 2, we have

$$\begin{array}{c}\hfill \parallel e_u^m{\parallel }^2+S\parallel e_B^m{\parallel }^2+\tau \sum _{n=1}^m(\nu \parallel A_{1ϵ}^{\frac{1}{2}}{e}_u^n{\parallel }^2+S\mu \parallel A_2^{\frac{1}{2}}{e}_B^n{\parallel }^2)\le C{\tau }^{2}exp\left(\tau \sum _{n=0}^{m-1}{d}_n\right)\le C{\tau }^2,\end{array}$$ (27)

for any $1\le m\le N$. Using (27), (9) and Lemma 1, we obtain Theorem 5. □

Theorem 6.

Under Assumptions 1–3, we have

$$\begin{array}{c}\hfill \underset{1\le n\le N}{sup}{\sigma }^{\frac{1}{2}}\left(t_n\right)(\parallel {\mathit{u}}_{ϵ}\left(t_n\right)-{\mathit{u}}_{ϵ}^n{\parallel }_1+\sqrt{S\mu }\parallel {\mathit{B}}_{ϵ}\left(t_n\right)-{\mathit{B}}_{ϵ}^n{\parallel }_1)+{\left(\tau \sum _{n=1}^N\sigma \left(t_n\right){\parallel p_{ϵ}\left(t_n\right)-p_{ϵ}^n\parallel }^2\right)}^{\frac{1}{2}}\le C\tau .\end{array}$$

Proof. 

Taking the inner product of (20) with $2d_t{e}_u^n\tau$, (21) with $2d_t{e}_B^n\tau$, we deduce that

$$\begin{array}{cc}\hfill \hspace{1em}& 2\parallel d_t{e}_u^n{\parallel }^2\tau +2\parallel d_t{e}_B^n{\parallel }^2\tau +\nu (\parallel A_{1ϵ}^{\frac{1}{2}}{e}_u^n{\parallel }^2-\parallel A_{1ϵ}^{\frac{1}{2}}{e}_u^{n-1}{\parallel }^2+\parallel A_{1ϵ}^{\frac{1}{2}}(e_u^n-e_u^{n-1}){\parallel }^2)\hfill \\ \hfill \hspace{1em}& \hspace{1em}+\mu (\parallel A_2^{\frac{1}{2}}{e}_B^n{\parallel }^2-\parallel A_2^{\frac{1}{2}}{e}_B^{n-1}{\parallel }^2+\parallel A_2^{\frac{1}{2}}(e_B^n-e_B^{n-1}){\parallel }^2)+2b(e_u^{n-1},{\mathit{u}}_{ϵ}\left(t_n\right),d_t{e}_u^n)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}+2b({\mathit{u}}_{ϵ}\left(t_{n-1}\right),e_u^n,d_t{e}_u^n)\tau +2b(e_u^{n-1},e_u^n,d_t{e}_u^n)\tau +2S(e_B^{n-1}\times \nabla \times {\mathit{B}}_{ϵ}\left(t_n\right),d_t{e}_u^n)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}+2S({\mathit{B}}_{ϵ}\left(t_{n-1}\right)\times \nabla \times e_B^n,d_t{e}_u^n)\tau +2S(e_B^{n-1}\times \nabla \times e_B^n,d_t{e}_u^n)\tau +2(\nabla \times ({\mathit{u}}_{ϵ}\left(t_n\right)\times e_B^{n-1}),d_t{e}_B^n)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}+2(\nabla \times (e_u^n\times {\mathit{B}}_{ϵ}\left(t_{n-1}\right)),d_t{e}_B^n)\tau +2(\nabla \times (e_u^n\times e_B^{n-1}),d_t{e}_B^n)\tau =2(R_1^n,d_t{e}_u^n)\tau +2(R_2^n,d_t{e}_B^n)\tau .\hfill \end{array}$$ (28)

Using (7)–(12) and Lemma 1, we obtain

$$\begin{array}{cc}\hfill 2\left|b\right(e_u^{n-1},{\mathit{u}}_{ϵ}\left(t_n\right),d_t{e}_u^n\left)\right|& \le \frac{1}{6}\parallel d_t{e}_u^n{\parallel }^2+c\parallel {\mathit{u}}_{ϵ}\left(t_n\right){\parallel }_2^2\parallel A_{1ϵ}^{\frac{1}{2}}{e}_u^{n-1}{\parallel }^2,\hfill \\ \hfill 2\left|b\right({\mathit{u}}_{ϵ}\left(t_{n-1}\right),e_u^n,d_t{e}_u^n\left)\right|& \le \frac{1}{6}\parallel d_t{e}_u^n{\parallel }^2+c\parallel {\mathit{u}}_{ϵ}\left(t_{n-1}\right){\parallel }_2^2\parallel A_{1ϵ}^{\frac{1}{2}}{e}_u^{n-1}{\parallel }^2,\hfill \\ \hfill 2\left|b\right(e_u^{n-1},e_u^n,d_t{e}_u^n\left)\right|& \le \frac{\nu }{6\tau }\parallel A_{1ϵ}^{\frac{1}{2}}(e_u^n-e_u^{n-1}){\parallel }^2+c\frac{1}{\tau }\parallel e_u^n{\parallel }_1^2\parallel A_{1ϵ}^{\frac{1}{2}}{e}_u^{n-1}{\parallel }^2,\hfill \\ \hfill 2S\left|\right(e_B^{n-1}\times \nabla \times {\mathit{B}}_{ϵ}\left(t_n\right),d_t{e}_u^n\left)\right|& \le \frac{1}{6}\parallel d_t{e}_u^n{\parallel }^2+c\parallel {\mathit{B}}_{ϵ}\left(t_n\right){\parallel }_2^2\parallel A_2^{\frac{1}{2}}{e}_B^{n-1}{\parallel }^2,\hfill \\ \hfill 2S\left|\right({\mathit{B}}_{ϵ}\left(t_{n-1}\right)\times \nabla \times e_B^n,d_t{e}_u^n\left)\right|& \le \frac{1}{6}\parallel d_t{e}_u^n{\parallel }^2+c\parallel {\mathit{B}}_{ϵ}\left(t_{n-1}\right){\parallel }_2^2\parallel A_2^{\frac{1}{2}}{e}_B^n{\parallel }^2,\hfill \\ \hfill 2S\left|\right(e_B^{n-1}\times \nabla \times e_B^n,d_t{e}_u^n\left)\right|& \le \frac{1}{6\tau }\parallel A_{1ϵ}^{\frac{1}{2}}(e_u^n-e_u^{n-1}){\parallel }^2+c\frac{1}{\tau }\parallel e_B^n{\parallel }_1^2\parallel A_{1ϵ}^{\frac{1}{2}}{e}_B^{n-1}{\parallel }^2,\hfill \\ \hfill 2\left|\right(\nabla \times ({\mathit{u}}_{ϵ}\left(t_n\right)\times e_B^{n-1}),d_t{e}_B^n\left)\right|& \le \frac{1}{6}\parallel d_t{e}_B^n{\parallel }^2+c\parallel {\mathit{u}}_{ϵ}\left(t_n\right){\parallel }_2^2\parallel A_2^{\frac{1}{2}}{e}_B^{n-1}{\parallel }^2,\hfill \\ \hfill 2\left|\right(\nabla \times (e_u^n\times {\mathit{B}}_{ϵ}\left(t_{n-1}\right)),d_t{e}_B^n\left)\right|& \le \frac{1}{6}\parallel d_t{e}_B^n{\parallel }^2+c\parallel {\mathit{B}}_{ϵ}\left(t_{n-1}\right){\parallel }_2^2\parallel A_{1ϵ}^{\frac{1}{2}}{e}_u^n{\parallel }^2,\hfill \\ \hfill 2\left|\right(\nabla \times (e_u^n\times e_B^{n-1}),d_t{e}_B^n\left)\right|& \le \frac{1}{6\tau }\parallel A_2^{\frac{1}{2}}(e_B^n-e_B^{n-1}){\parallel }^2+c\frac{1}{\tau }\parallel e_u^n{\parallel }_1^2\parallel A_{1ϵ}^{\frac{1}{2}}{e}_B^{n-1}{\parallel }^2.\hfill \end{array}$$

Similarly, we have

$$\begin{array}{cc}\hfill 2\left|\right(R_1^n,d_t{e}_u^n\left)\right|\le & \frac{1}{6}\parallel d_t{e}_u^n{\parallel }^2+c\tau {\int }_{t_{n-1}}^{t_n}\parallel {\mathit{u}}_{ϵtt}{\parallel }^{2}dt+c\tau \parallel {\mathit{u}}_{ϵ}\left(t_n\right){\parallel }_2^2{\int }_{t_{n-1}}^{t_n}{\parallel {\mathit{u}}_{ϵt}\left(t\right)\parallel }_1^{2}dt\hfill \\ \hfill \hspace{1em}& +c\tau \parallel {\mathit{B}}_{ϵ}\left(t_n\right){\parallel }_2^2{\int }_{t_{n-1}}^{t_n}{\parallel {\mathit{B}}_{ϵt}\left(t\right)\parallel }_1^{2}dt,\hfill \\ \hfill 2\left|\right(R_2^n,d_t{e}_B^n\left)\right|\le & \frac{1}{6}\parallel d_t{e}_B^n{\parallel }^2+c\tau {\int }_{t_{n-1}}^{t_n}\parallel {\mathit{B}}_{ϵtt}{\parallel }^{2}dt+c\tau \parallel {\mathit{u}}_{ϵ}\left(t_n\right){\parallel }_2^2{\int }_{t_{n-1}}^{t_n}{\parallel {\mathit{B}}_{ϵt}\left(t\right)\parallel }_1^{2}dt.\hfill \end{array}$$

Combining the above inequalities with (28), and using Theorem 2, we can derive

$$\begin{array}{cc}\hfill \hspace{1em}& \parallel d_t{e}_u^n{\parallel }^2\tau +\parallel d_t{e}_B^n{\parallel }^2\tau +\nu (\parallel A_{1ϵ}^{\frac{1}{2}}{e}_u^n{\parallel }^2-\parallel A_{1ϵ}^{\frac{1}{2}}{e}_u^{n-1}{\parallel }^2)+\mu (\parallel A_2^{\frac{1}{2}}{e}_B^n{\parallel }^2-\parallel A_2^{\frac{1}{2}}{e}_B^{n-1}{\parallel }^2)\hfill \\ \hfill \hspace{1em}& \hspace{1em}\le c(\parallel A_{1ϵ}^{\frac{1}{2}}{e}_u^{n-1}{\parallel }^2+\parallel A_2^{\frac{1}{2}}{e}_B^{n-1}{\parallel }^2+\parallel A_{1ϵ}^{\frac{1}{2}}{e}_u^n{\parallel }^2+\parallel A_2^{\frac{1}{2}}{e}_B^n{\parallel }^2)\tau +c{d}_{n-1}(\parallel A_{1ϵ}^{\frac{1}{2}}{e}_u^{n-1}{\parallel }^2+\parallel A_2^{\frac{1}{2}}{e}_B^{n-1}{\parallel }^2)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}+c{\tau }^2{\int }_{t_{n-1}}^{t_n}(\parallel {\mathit{u}}_{ϵtt}{\parallel }^2+\parallel {\mathit{B}}_{ϵtt}{\parallel }^2)dt+c{\tau }^2(\parallel {\mathit{u}}_{ϵ}\left(t_n\right){\parallel }_2^2+\parallel {\mathit{B}}_{ϵ}\left(t_n\right){\parallel }_2^2){\int }_{t_{n-1}}^{t_n}(\parallel {\mathit{u}}_{ϵt}\left(t\right){\parallel }_1^2+\parallel {\mathit{B}}_{ϵt}\left(t\right){\parallel }_1^2)dt,\hfill \end{array}$$

where $d_{n-1}={\tau }^{-1}(\parallel e_u^n{\parallel }_1^2+\parallel e_B^n{\parallel }_1^2)$. Multiplying this inequality by $\sigma \left(t_n\right)$ and taking the sum with respect to n from 1 to m, thanks to Theorems 2 and 5, we obtain

$$\begin{array}{cc}\hfill \hspace{1em}& \sigma \left(t_m\right)\left(\nu \right(\parallel A_{1ϵ}^{\frac{1}{2}}{e}_u^m{\parallel }^2+\mu (\parallel A_2^{\frac{1}{2}}{e}_B^m{\parallel }^2)+\tau \sum _{n=1}^m\sigma \left(t_n\right)(\parallel d_t{e}_u^n{\parallel }^2+\parallel d_t{e}_B^n{\parallel }^2)\hfill \\ \hfill \hspace{1em}& \hspace{1em}\le c\tau \sum _{n=0}^{m-1}{d}_n(\parallel A_{1ϵ}^{\frac{1}{2}}{e}_u^n{\parallel }^2+\parallel A_2^{\frac{1}{2}}{e}_B^n{\parallel }^2)+C{\tau }^2.\hfill \end{array}$$

Then, by applying Lemma 2 to this inequality and using Theorem 5, we have

$$\begin{array}{cc}\hfill \hspace{1em}& \sigma \left(t_m\right)\left(\nu \right(\parallel A_{1ϵ}^{\frac{1}{2}}{e}_u^m{\parallel }^2+\mu (\parallel A_2^{\frac{1}{2}}{e}_B^m{\parallel }^2)+\tau \sum _{n=1}^m\sigma \left(t_n\right)(\parallel d_t{e}_u^n{\parallel }^2+\parallel d_t{e}_B^n{\parallel }^2)\le C{\tau }^2,\hfill \end{array}$$ (29)

for all $1\le m\le N$.

Finally, using (7)–(12) and the LBB condition (cf. [4,13])

$$\begin{array}{c}\hfill \beta \parallel q\parallel \le \underset{\mathit{v}\in \mathbf{X}}{sup}\frac{(\nabla ·\mathit{v},q)}{\parallel \nabla \mathit{v}\parallel },\hspace{1em}\forall q\in M,\end{array}$$ (30)

we derive

$$\begin{array}{cc}\hfill \beta \parallel p_{ϵ}\left(t_n\right)-p_{ϵ}^n\parallel & \le c\parallel d_t{e}_u^n\parallel +\nu \parallel e_u^n{\parallel }_1+c\parallel e_u^{n-1}{\parallel }_1\parallel {\mathit{u}}_{ϵ}\left(t_n\right){\parallel }_1+c\parallel {\mathit{u}}_{ϵ}^{n-1}{\parallel }_1{\parallel e_u^n\parallel }_1\hfill \\ \hfill \hspace{1em}& +c\parallel e_B^{n-1}{\parallel }_1\parallel \nabla \times {\mathit{B}}_{ϵ}\left(t_n\right)\parallel +c\parallel {\mathit{B}}_{ϵ}^{n-1}{\parallel }_1\parallel \nabla \times e_B^n\parallel +c\parallel R_1^n\parallel \hfill \\ \hfill \hspace{1em}& \le c\parallel d_t{e}_u^n\parallel +\nu \parallel e_u^n{\parallel }_1+c\parallel e_u^{n-1}{\parallel }_1\parallel {\mathit{u}}_{ϵ}\left(t_n\right){\parallel }_1+c\parallel {\mathit{u}}_{ϵ}^{n-1}{\parallel }_1{\parallel e_u^n\parallel }_1\hfill \\ \hfill \hspace{1em}& +c\parallel e_B^{n-1}{\parallel }_1\parallel {\mathit{B}}_{ϵ}\left(t_n\right){\parallel }_1+c\parallel {\mathit{B}}_{ϵ}^{n-1}{\parallel }_1\parallel e_B^n{\parallel }_1+c\tau {\int }_{t_{n-1}}^{t_n}{\parallel {\mathit{u}}_{ϵtt}\parallel }^{2}dt\hfill \\ \hfill \hspace{1em}& +c\tau \parallel {\mathit{u}}_{ϵ}\left(t_n\right){\parallel }_2^2{\int }_{t_{n-1}}^{t_n}\parallel {\mathit{u}}_{ϵt}\left(t\right){\parallel }_1^{2}dt+c\tau \parallel {\mathit{B}}_{ϵ}\left(t_n\right){\parallel }_2^2{\int }_{t_{n-1}}^{t_n}{\parallel {\mathit{B}}_{ϵt}\left(t\right)\parallel }_1^{2}dt.\hfill \end{array}$$

Multiplying this inequality by $\sigma \left(t_n\right)$ and taking the sum with respect to n from 1 to m, due to Theorems 2 and 5 and (29), we obtain

$$\begin{array}{cc}\hfill \tau \sum _{n=1}^m\sigma \left(t_n\right){\parallel p_{ϵ}\left(t_n\right)-p_{ϵ}^n\parallel }^2\hspace{1em}& \le C{\tau }^2,\hfill \end{array}$$ (31)

for all $1\le m\le N$. Using (29), (31), (9), and Lemma 1, we obtain Theorem 6. □

4. PFEM for the MHD Equations

We further consider a spatial discretization for the penalty MHD system of time discretization in this section (cf. [4]). ${\mathcal{J}}_h$ is a family of quasi-uniformly regular partitions of $\mathcal{D}$ into triangles or tetrahedron elements K with the diameter $h_K$. Let the mesh size $h=\underset{K\in {\mathcal{J}}_h}{max}{h}_K$. We give three finite element spaces ${\mathbf{X}}_h$, ${\mathrm{M}}_h$, ${\mathbf{W}}_h$ with ${\mathbf{X}}_h\subset \mathbf{X}$, ${\mathrm{M}}_h\subset \mathrm{M}$ and ${\mathbf{W}}_h\subset \mathbf{W}$.

Let ${\rho }_h:\mathrm{M}\to {\mathrm{M}}_h$ denote the $L^2$-orthogonal projector which is defined by

$$\begin{array}{c}\hfill ({\rho }_{h}q,q_h)=(q,q_h),\hspace{1em}\forall q\in \mathrm{M},q_h\in {\mathrm{M}}_h.\end{array}$$ (32)

Assumption 4.

The finite element space $({\mathbf{X}}_h,M_h)$ satisfies the discrete LBB condition (cf. [4,22])

$$\begin{array}{c}\hfill \underset{{\mathit{v}}_h\in {\mathbf{X}}_h}{sup}\frac{d({\mathit{v}}_h,q_h)}{\parallel \nabla {\mathit{v}}_h\parallel }\ge {\beta }_1\parallel q_h\parallel ,\hspace{1em}{q}_h\in M_h,\end{array}$$ (33)

where ${\beta }_1$ is a positive constant depending on $\mathcal{D}$. For each $\mathit{v}\in {\mathit{H}}^2\left(\mathcal{D}\right)\cap \mathit{X}$, $q\in H^1\left(\mathcal{D}\right)\cap M,\mathit{C}\in {\mathit{H}}^2\left(\mathcal{D}\right)\cap \mathit{W}$, there exist ${\pi }_h\mathit{v}\in {\mathit{X}}_h$, ${\rho }_{h}q\in M_h$ and $J_h\mathit{C}\in {\mathit{W}}_h$ such that

$$\begin{array}{cc}\hfill \hspace{1em}& (\nabla ·(\mathit{v}-{\pi }_h\mathit{v}),q_h)=0,\hspace{1em}{q}_h\in M_h,\hfill \end{array}$$ (34)

$$\begin{array}{cc}\hfill \hspace{1em}& \parallel \nabla (\mathit{v}-{\pi }_h\mathit{v}){\parallel \le ch\parallel \mathit{v}\parallel }_2,\hspace{1em}\parallel q-{\rho }_h{q\parallel \le ch\parallel q\parallel }_1,\hspace{1em}\parallel \nabla (\mathit{C}-J_h\mathit{C}){\parallel \le ch\parallel \mathit{C}\parallel }_2,\hfill \end{array}$$ (35)

together with the inverse inequalities

$$\begin{array}{c}\hfill \parallel \nabla {\mathit{v}}_h\parallel \le c{h}^{-1}\parallel {\mathit{v}}_h\parallel ,\hspace{1em}{\mathit{v}}_h\in {\mathit{X}}_h,\parallel \nabla {\mathit{C}}_h\parallel \le c{h}^{-1}\parallel {\mathit{C}}_h\parallel ,\hspace{1em}{\mathit{C}}_h\in {\mathit{W}}_h.\end{array}$$ (36)

Next, to obtain an approximation of $(\mathit{u},p,\mathit{B})$, we consider the following finite element pairs:

$$\begin{array}{c}{\mathbf{X}}_h={(P_{1,h}^b)}^d\cap \mathbf{X},\hfill \\ {\mathrm{M}}_h=\{q_h\in C^0\left(\mathcal{D}\right)\cap \mathrm{M}:q_h{|}_K\in P_1\left(K\right),\forall K\in {\mathcal{J}}_h\},\hfill \\ {\mathbf{W}}_h=\{{\mathit{C}}_h\in C^0\left(\mathcal{D}\right)\cap \mathbf{W}:{\mathit{C}}_h{|}_K\in P_1{\left(K\right)}^d,\forall K\in {\mathcal{J}}_h\},\hfill \end{array}$$

where

$$\begin{array}{c}\hfill P_{1,h}^b=\{v_h\in C^0\left(\mathcal{D}\right):v_h{|}_K\in P_1\left(K\right)\oplus span\left\{b\right\},\forall K\in {\mathcal{J}}_h\},\end{array}$$

$\left\{b\right\}$ is a bubble function. Let $b\in {\mathbf{H}}_0^1\left(K\right)$ take the value 1 at the barycentre of K and satisfy $0\le b\left(\mathbf{x}\right)\le 1$, which is called a “bubble function” (cf. [22]). Furthermore, we denote the discrete subspace ${\mathbf{X}}_{0h}$ of ${\mathbf{X}}_0$ as

$$\begin{array}{c}\hfill {\mathbf{X}}_{0h}=\{{\mathit{v}}_h\in {\mathbf{X}}_h:d({\mathit{v}}_h,q_h)=0,q_h\in M_h\}.\end{array}$$

The finite element approximation for (16) based on ${\mathbf{X}}_h\times {\mathrm{M}}_h\times {\mathbf{W}}_h$ is given as follows: find $({\mathit{u}}_{ϵh}^n,p_{ϵh}^n,{\mathit{B}}_{ϵh}^n)$ such that for all $1\le n\le N$ and $({\mathit{v}}_h,q_h,{\mathit{C}}_h)\in {\mathbf{X}}_h\times {\mathrm{M}}_h\times {\mathbf{W}}_h$,

$$\begin{array}{cc}\hfill \hspace{1em}& (d_t{\mathit{u}}_{ϵh}^n,\mathit{v})+\nu (\nabla {\mathit{u}}_{ϵh}^n,\nabla \mathit{v})+b({\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵh}^n,\mathit{v})-(\nabla ·\mathit{v},p_{ϵh}^n)+(\nabla ·{\mathit{u}}_{ϵh}^n,q)+S({\mathit{B}}_{ϵh}^{n-1}\times \nabla \times {\mathit{B}}_{ϵh}^n,\mathit{v})\hfill \\ \hfill \hspace{1em}& \hspace{1em}+\frac{ϵ}{\nu }(p_{ϵh}^n,q)=(\mathit{f}\left(t_n\right),\mathit{v}),\hfill \end{array}$$ (37)

$$\begin{array}{cc}\hfill \hspace{1em}& (d_t{\mathit{B}}_{ϵh}^n,\mathit{C})+\mu (\nabla \times {\mathit{B}}_{ϵh}^n,\nabla \times \mathit{C})+\mu (\nabla ·{\mathit{B}}_{ϵh}^n,\nabla ·\mathit{C})-({\mathit{u}}_{ϵh}^n\times {\mathit{B}}_{ϵh}^{n-1},\nabla \times \mathit{C})=(\mathit{g}\left(t_n\right),\mathit{C}),\hfill \end{array}$$ (38)

$$\begin{array}{cc}\hfill \hspace{1em}& {\mathit{u}}_{ϵh}^0=r_{1h}{\mathit{u}}^0,\hspace{1em}{\mathbf{B}}_{ϵh}^0=r_{2h}{\mathbf{B}}^0,\hfill \end{array}$$ (39)

where $r_{1h}:{\mathbf{L}}^2\left(\mathcal{D}\right)\to {\mathbf{X}}_{0h}$ and $r_{2h}:{\mathbf{L}}^2\left(\mathcal{D}\right)\to {\mathbf{W}}_h$ are $L^2$-orthogonal projectors. According to (34)–(36), these operators satisfy the following properties (cf. [3,4,22]):

$$\begin{array}{cc}\hfill \hspace{1em}& \parallel \mathit{v}-r_{1h}\mathit{v}\parallel +h\parallel \nabla (\mathit{v}-r_{1h}\mathit{v})\parallel \le c{h}^2\parallel A_1\mathit{v}\parallel ,\hspace{1em}\forall \mathit{v}\in {\mathbf{H}}^2\left(\mathcal{D}\right)\cap \mathbf{X},\hfill \end{array}$$ (40)

$$\begin{array}{cc}\hfill \hspace{1em}& \parallel \mathit{w}-r_{2h}\mathit{w}\parallel +h\parallel \nabla (\mathit{w}-r_{2h}\mathit{w})\parallel \le c{h}^i{\parallel \mathit{w}\parallel }_i,\hspace{1em}\forall \mathit{w}\in {\mathbf{H}}^2\left(\mathcal{D}\right)\cap {\mathbf{W}}_0,\hspace{1em}i=1,2.\hfill \end{array}$$ (41)

Here, we define the discrete Stokes operator $A_{1h}=-r_{1h}{\Delta }_h$, which is defined by (see [4])

$$\begin{array}{c}\hfill (-{\Delta }_h{\mathit{u}}_h,v_h)=(\nabla {\mathit{u}}_h,\nabla {\mathit{v}}_h),\hspace{1em}{\mathit{u}}_h,{\mathit{v}}_h\in {\mathbf{X}}_h,\end{array}$$

its discrete norm $\parallel {\mathit{v}}_h{\parallel }_{k,2}=\parallel A_{1h}^{\frac{k}{2}}{\mathit{v}}_h\parallel$ of the $k\in \mathbb{R}$ order can be defined, where

$$\begin{array}{c}\hfill \parallel {\mathit{v}}_h{\parallel }_{1,2}=\parallel \nabla {\mathit{v}}_h\parallel ,\hspace{1em}\parallel {\mathit{v}}_h{\parallel }_{2,2}=\parallel A_{1h}{\mathit{v}}_h\parallel ,\hspace{1em}\parallel {\mathit{v}}_h{\parallel }_{-1,2}=\parallel A_{1h}^{-\frac{1}{2}}{\mathit{v}}_h\parallel =\underset{{\mathit{w}}_h\in {\mathbf{X}}_h}{sup}\frac{({\mathit{v}}_h,{\mathit{w}}_h)}{\parallel A_{1h}^{\frac{1}{2}}{\mathit{w}}_h\parallel },\hspace{1em}\forall {\mathit{v}}_h\in {\mathbf{X}}_{0h}.\end{array}$$

Meanwhile, we define the discrete operator $A_{2h}{\mathit{B}}_h=r_{2h}({\nabla }_h\times \nabla \times {\mathit{B}}_h+{\nabla }_h\nabla ·{\mathit{B}}_h)\in {\mathbf{W}}_h$ as follows:

$$\begin{array}{c}\hfill (A_{2h}{\mathit{B}}_h,C_h)=(A_{2h}^{\frac{1}{2}}{\mathit{B}}_h,A_{2h}^{\frac{1}{2}}{C}_h)=(\nabla \times {\mathit{B}}_h,\nabla \times {\mathit{C}}_h)+(\nabla ·{\mathit{B}}_h,\nabla ·{\mathit{C}}_h),\end{array}$$

and its discrete norm $\parallel {\mathit{B}}_h{\parallel }_k=\parallel A_{2h}^{\frac{k}{2}}{\mathit{B}}_h\parallel$ of the $k\in \mathbb{R}$ order can be defined, where

$$\begin{array}{c}\hfill \parallel {\mathit{B}}_h{\parallel }_{1,2}=\parallel A_{2h}^{\frac{1}{2}}{\mathit{B}}_h\parallel ,\hspace{1em}\parallel {\mathit{B}}_h{\parallel }_{2,2}=\parallel A_{2h}{\mathit{B}}_h\parallel ,\hspace{1em}\parallel {\mathit{B}}_h{\parallel }_{-1,2}=\parallel A_{2h}^{-\frac{1}{2}}{\mathit{B}}_h\parallel =\underset{{\mathit{C}}_h\in {\mathbf{W}}_h}{sup}\frac{({\mathit{B}}_h,{\mathit{C}}_h)}{\parallel A_{2h}^{\frac{1}{2}}{\mathit{C}}_h\parallel },\hspace{1em}\forall {\mathit{B}}_h\in {\mathbf{W}}_h.\end{array}$$

To obtain the error analysis of the scheme in the following section, we need the following discrete estimates which are obtained from [4,7]).

Lemma 3.

The estimates of ${\mathit{v}}_h$ and $C_h$ are as follows:

$$\begin{array}{c}\parallel {\mathit{v}}_h{\parallel }_{L^6}\le \parallel {\mathit{v}}_h{\parallel }_{1,2},\hspace{1em}\parallel {\mathit{v}}_h{\parallel }_{L^3}\le c\parallel {\mathit{v}}_h{\parallel }^{\frac{1}{2}}{\parallel {\mathit{v}}_h\parallel }_{1,2}^{\frac{1}{2}},\hfill \\ \parallel \nabla {\mathit{v}}_h{\parallel }_{L^6}\le \parallel {\mathit{v}}_h{\parallel }_{2,2},\hspace{1em}\parallel \nabla {\mathit{v}}_h{\parallel }_{L^3}+\parallel {\mathit{v}}_h{\parallel }_{L^{\infty }}\le c\parallel {\mathit{v}}_h{\parallel }_{1,2}^{\frac{1}{2}}{\parallel {\mathit{v}}_h\parallel }_{2,2}^{\frac{1}{2}},\hspace{1em}\forall {\mathit{v}}_h\in {\mathit{X}}_{0h},\hfill \\ \parallel {\mathit{C}}_h{\parallel }_{L^6}\le \parallel {\mathit{C}}_h{\parallel }_{1,2},\hspace{1em}\parallel {\mathit{C}}_h{\parallel }_{L^3}\le c\parallel {\mathit{C}}_h{\parallel }^{\frac{1}{2}}{\parallel {\mathit{C}}_h\parallel }_{1,2}^{\frac{1}{2}},\hfill \\ \parallel \nabla {\mathit{C}}_h{\parallel }_{L^6}\le \parallel C_h{\parallel }_{2,2},\hspace{1em}\parallel \nabla {\mathit{C}}_h{\parallel }_{L^3}+\parallel {\mathit{C}}_h{\parallel }_{L^{\infty }}\le c\parallel {\mathit{C}}_h{\parallel }_{1,2}^{\frac{1}{2}}{\parallel {\mathit{C}}_h\parallel }_{2,2}^{\frac{1}{2}},\hspace{1em}\forall {\mathit{C}}_h\in {\mathit{W}}_h.\hfill \end{array}$$

For the finite element space ${\mathbf{X}}_h\times {\mathrm{M}}_h\times {\mathbf{W}}_h$ given above, problems (37) and (38) allow us to calculate velocity and pressure separately, i.e., (37) and (38) can be reduced as follows: find $({\mathit{u}}_{ϵh}^n,p_{ϵh}^n,{\mathit{B}}_{ϵh}^n)\in {\mathbf{X}}_h\times {\mathrm{M}}_h\times {\mathbf{W}}_h$ such that

$$\left\{\begin{array}{c}(d_t{\mathit{u}}_{ϵh}^n,\mathit{v})+\nu (A_{1h}^{\frac{1}{2}}{\mathit{u}}_{ϵh}^n,A_{1h}^{\frac{1}{2}}\mathit{v})+b({\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵh}^n,\mathit{v})+S({\mathit{B}}_{ϵh}^{n-1}\times \nabla \times {\mathit{B}}_{ϵh}^n,\mathit{v})\hfill \\ \hspace{1em}+\frac{\nu }{ϵ}(\nabla ·\mathit{v},{\rho }_h\nabla ·{\mathit{u}}_{ϵh}^n)=(\mathit{f}\left(t_n\right),\mathit{v}),\hfill \\ (d_t{\mathit{B}}_{ϵh}^n,\mathit{C})+\mu (A_{2h}^{\frac{1}{2}}{\mathit{B}}_{ϵh}^n,A_{2h}^{\frac{1}{2}}\mathit{C})-({\mathit{u}}_{ϵh}^n\times {\mathit{B}}_{ϵh}^{n-1},\nabla \times \mathit{C})=(\mathit{g}\left(t_n\right),\mathit{C}),\hfill \\ p_{ϵh}^n=-\frac{\nu }{ϵ}{\rho }_h\nabla ·{\mathit{u}}_{ϵh}^n.\hfill \end{array}\right.$$ (42)

In the following, we give the algebraic matrix form for the 2D case, and the algebraic matrix form for the 3D case is similar. In order to analyze the detailed form of the coefficient matrix for this scheme, we write the fluid velocity and magnetic field vectors

$${\mathit{u}}_{ϵh}^n=(u_{1ϵh}^n,u_{2ϵh}^n),\hspace{1em}{\mathit{B}}_{ϵh}^n=(B_{1ϵh}^n,B_{2ϵh}^n),$$

and the corresponding test function vectors

$$\begin{array}{c}\hfill {\mathit{v}}_h^n=(v_{1h}^n,v_{2h}^n),\hspace{1em}{\mathit{C}}_h^n=(C_{1h}^n,C_{2h}^n),\end{array}$$

then, we expand (42) and obtain that

Step 1. Find $({\mathit{u}}_{ϵh}^n,{\mathit{B}}_{ϵh}^n)$ from

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\tau }^{-1}(u_{1ϵh}^n,v_{1h})+\nu ({\partial }_x{u}_{1ϵh}^n,{\partial }_x{v}_{1h})+\nu ({\partial }_y{u}_{1ϵh}^n,{\partial }_y{v}_{1h})+(u_{1ϵh}^{n-1}{\partial }_x{u}_{1ϵh}^n,v_{1h})\hfill \\ \hspace{1em}+(u_{2ϵh}^{n-1}{\partial }_y{u}_{1ϵh}^n,v_{1h})+S(B_{2ϵh}^{n-1}{\partial }_x{B}_{2ϵh}^n,v_{1h})-S(B_{2ϵh}^{n-1}{\partial }_y{B}_{1ϵh}^n,v_{1h})\hfill \\ \hspace{1em}+\frac{\nu }{ϵ}({\rho }_h({\partial }_x{u}_{1ϵh}^n+{\partial }_y{u}_{2ϵh}^n),{\partial }_x{v}_{1h})=(f_1\left(t_n\right),v_{1h})+{\tau }^{-1}(u_{1ϵh}^{n-1},v_{1h}),\hfill \\ {\tau }^{-1}(u_{2ϵh}^n,v_{2h})+\nu ({\partial }_x{u}_{2ϵh}^n,{\partial }_x{v}_{2h})+\nu ({\partial }_y{u}_{2ϵh}^n,{\partial }_y{v}_{2h})+(u_{1ϵh}^{n-1}{\partial }_x{u}_{2ϵh}^n,v_{2h})\hfill \\ \hspace{1em}+(u_{2ϵh}^{n-1}{\partial }_y{u}_{2ϵh}^n,v_{2h})-S(B_{1ϵh}^{n-1}{\partial }_x{B}_{2ϵh}^n,v_{2h})+S(B_{1ϵh}^{n-1}{\partial }_y{B}_{1ϵh}^n,v_{2h})\hfill \\ \hspace{1em}+\frac{\nu }{ϵ}({\rho }_h({\partial }_y{u}_{2ϵh}^n+{\partial }_y{u}_{2ϵh}^n),{\partial }_x{v}_{1h})=(f_2\left(t_n\right),v_{2h})+{\tau }^{-1}(u_{2ϵh}^{n-1},v_{2h}),\hfill \\ {\tau }^{-1}(B_{1ϵh}^n,C_{1h})-\eta ({\partial }_x{B}_{2ϵh}^n-{\partial }_y{B}_{1ϵh}^n,{\partial }_y{C}_{1h})+\eta ({\partial }_x{B}_{1ϵh}^n+{\partial }_y{B}_{2ϵh}^n,{\partial }_x{C}_{1h})\hfill \\ \hspace{1em}+(u_{1ϵh}^n{B}_{2ϵh}^{n-1}-u_{2ϵh}^n{B}_{1ϵh}^{n-1},{\partial }_y{C}_{1h})=(g_1\left(t_n\right),C_{1h})+{\tau }^{-1}(B_{1ϵh}^{n-1},C_{1h}),\hfill \\ {\tau }^{-1}(B_{2ϵh}^n,C_{2h})+\eta ({\partial }_x{B}_{2ϵh}^n-{\partial }_y{B}_{1ϵh}^n,{\partial }_x{C}_{2h})+\eta ({\partial }_x{B}_{1ϵh}^n+{\partial }_y{B}_{2ϵh}^n,{\partial }_y{C}_{2h})\hfill \\ \hspace{1em}-(u_{1ϵh}^n{B}_{2ϵh}^{n-1}-u_{2ϵh}^n{B}_{1ϵh}^{n-1},{\partial }_x{C}_{2h})=(g_2\left(t_n\right),C_{2h})+{\tau }^{-1}(B_{2ϵh}^{n-1},C_{2h}).\hfill \end{array}\right.\end{array}$$

Step 2. Find $p_{ϵh}^n$ from

$$\begin{array}{c}\hfill p_{ϵh}^n=-\frac{\nu }{ϵ}{\rho }_h({\partial }_x{u}_{1ϵh}^n+{\partial }_y{u}_{2ϵh}^n).\end{array}$$

Here, ${\partial }_x{v}_{1h}=\partial v_{1h}/\partial x$ and ${\partial }_y{v}_{2h}=\partial v_{2h}/\partial y$. Next, we assume the spaces ${\mathbf{X}}_h$ and ${\mathbf{W}}_h$ are combined with the basis functions

$$\begin{array}{c}\hfill {\mathbf{X}}_h=span\{{\phi }_i:i=1,\cdots ,\mathcal{N}\},\hspace{1em}{\mathbf{W}}_h=span\{{\psi }_i:i=1,\cdots ,\mathcal{M}\},\end{array}$$

where $\mathcal{N}$ and $\mathcal{M}$ denote the number of the basis functions in each of spaces. Then,

$$\begin{array}{cc}\hfill \hspace{1em}& u_{lϵh}^n=\sum _{i=1}^{\mathcal{N}}{u}_{lϵh,i}^n{\phi }_i,\hspace{1em}{v}_{lh}={\phi }_j,\hspace{1em}l=1,2,\hspace{1em}j=1,\cdots ,\mathcal{N},\hfill \\ \hfill \hspace{1em}& B_{lϵh}^n=\sum _{i=1}^{\mathcal{M}}{B}_{lϵh,i}^n{\psi }_i,\hspace{1em}{C}_{lh}={\psi }_j,\hspace{1em}l=1,2,\hspace{1em}j=1,\cdots ,\mathcal{M}.\hfill \end{array}$$

Next, we show the relationship between the $({\mathit{u}}_{ϵh}^{n-1},{\mathit{B}}_{ϵh}^{n-1})$ and $({\mathit{u}}_{ϵh}^n,{\mathit{B}}_{ϵh}^n)$. Apparently, step 1 of the fully discrete PFEM generates an algebraic system as follows:

$$\left[\begin{array}{cc}\mathbb{A}& {\mathbb{B}}_1\\ {\mathbb{B}}_2& \mathbb{C}\end{array}\right]\left[\begin{array}{c}\mathit{u}\\ \mathit{B}\end{array}\right]=\left[\begin{array}{c}\mathit{F}\\ \mathit{G}\end{array}\right],$$

where

$$\begin{array}{c}\mathit{u}={(u_{1ϵh,i=1,\mathcal{N}}^n,u_{2ϵh,i=1,\mathcal{N}}^n)}^T,\hspace{1em}\mathit{B}={(B_{1ϵh,i=1,\mathcal{M}}^n,B_{2ϵh,i=1,\mathcal{M}}^n)}^T,\hfill \\ \mathit{F}={((f_1(t_n),{\phi }_j)+{\tau }^{-1}(u_{1ϵh,i}^{n-1},{\phi }_j),(f_2(t_n),{\phi }_j)+{\tau }^{-1}(u_{2ϵh,i}^{n-1},{\phi }_j))}^T,\hfill \\ \mathit{G}={((g_1(t_n),{\psi }_j)+{\tau }^{-1}(B_{1ϵh,i}^{n-1},{\psi }_j),(g_2(t_n),{\psi }_j)+{\tau }^{-1}(B_{2ϵh,i}^{n-1},{\psi }_j))}^T.\hfill \end{array}$$

Specifically, detailed calculation for PFEM gives

$$\mathbb{A}=\left[\begin{array}{cc}{a}_{ij}+b_{ij}& c_{ij}\\ d_{ij}& a_{ij}+e_{ij}\end{array}\right],{\mathbb{B}}_1=\left[\begin{array}{cc}{g}_{ij}& f_{ij}\\ k_{ij}& h_{ij}\end{array}\right],\mathbb{C}=\left[\begin{array}{cc}{l}_{ij}& m_{ij}\\ m_{ij}^T& l_{ij}\end{array}\right],{\mathbb{B}}_2=-S^{-1}{\mathbb{B}}_1^T,$$

where

$$\begin{array}{cc}\hfill \hspace{1em}& a_{ij}={\tau }^{-1}({\phi }_i,{\phi }_j)+\nu (({\partial }_x{\phi }_i,{\partial }_x{\phi }_j)+({\partial }_y{\phi }_i,{\partial }_y{\phi }_j))+(u_{1ϵh}^{n-1}{\partial }_x{\phi }_i+u_{2ϵh}^{n-1}{\partial }_y{\phi }_i,{\phi }_j),\hfill \\ \hfill \hspace{1em}& b_{ij}=\frac{\nu }{ϵ}({\rho }_h{\partial }_x{\phi }_i,{\partial }_x{\phi }_j),\hspace{1em}{c}_{ij}=\frac{\nu }{ϵ}({\rho }_h{\partial }_y{\phi }_i,{\partial }_x{\phi }_j),\hspace{1em}{d}_{ij}=\frac{\nu }{ϵ}({\rho }_h{\partial }_x{\phi }_i,{\partial }_y{\phi }_j),\hfill \\ \hfill \hspace{1em}& e_{ij}=\frac{\nu }{ϵ}({\rho }_h{\partial }_y{\phi }_i,{\partial }_y{\phi }_j),\hspace{1em}{f}_{ij}=S(B_{2ϵh}^{n-1}{\partial }_x{\psi }_i,{\phi }_j),\hspace{1em}{g}_{ij}=-S(B_{2ϵh}^{n-1}{\partial }_y{\psi }_i,{\phi }_j),\hfill \\ \hfill \hspace{1em}& l_{ij}={\tau }^{-1}({\psi }_i,{\psi }_j)+\eta ({\partial }_y{\psi }_i,{\partial }_y{\psi }_j)+\eta ({\partial }_x{\psi }_i,{\partial }_x{\psi }_j),\hspace{1em}{h}_{ij}=-S(B_{1ϵh}^{n-1}{\partial }_x{\psi }_i,{\phi }_j),\hfill \\ \hfill \hspace{1em}& k_{ij}=S(B_{1ϵh}^{n-1}{\partial }_y{\psi }_i,{\phi }_j),\hspace{1em}{m}_{ij}=\eta ({\partial }_y{\psi }_i,{\partial }_x{\psi }_j)-\eta ({\partial }_x{\psi }_i,{\partial }_y{\psi }_j).\hfill \end{array}$$

Then, we obtain $p_{ϵ}^n$ by step 2.

Arguing in exactly the same way as in the proof of Theorem 4, using (9) and Lemma 1, we obtain a priori bound of schemes (37)–(39).

Theorem 7.

Under Assumptions 1–3, we have

$$\begin{array}{c}\hfill \parallel {\mathit{u}}_{ϵh}^m{\parallel }^2+S\parallel {\mathit{B}}_{ϵh}^m{\parallel }^2+\tau \sum _{n=1}^m(\nu \parallel A_{1h}^{\frac{1}{2}}{\mathit{u}}_{ϵh}^n{\parallel }^2+S\mu \parallel A_{2h}^{\frac{1}{2}}{\mathit{B}}_{ϵh}^n{\parallel }^2)+\tau \sum _{n=1}^m(\parallel d_t{\mathit{u}}_{ϵh}^n{\parallel }^2+S\parallel d_t{\mathit{B}}_{ϵh}^n{\parallel }^2)\le C,\end{array}$$

for all $1\le m\le N$.

5. Error Analysis for the Fully Discrete Euler Semi-Implicit Scheme

In this section, we establish the error estimates for $({\mathit{u}}_{ϵh}^n,p_{ϵh}^n,{\mathit{B}}_{ϵh}^n)$ of the fully discrete Euler semi-implicit scheme (37)–(39). To this end, subtracting (37)–(38) from (16), we have

$$\begin{array}{cc}\hfill \hspace{1em}& (d_t({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n),\mathit{v})+\nu (\nabla ({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n),\nabla \mathit{v})+b({\mathit{u}}_{ϵ}^{n-1}-{\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵh}^n,\mathit{v})+b({\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n,\mathit{v})\hfill \\ \hfill \hspace{1em}& \hspace{1em}-(\nabla ·\mathit{v},p_{ϵ}^n-p_{ϵh}^n)+(\nabla ·({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n),q)+S(({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1})\times \nabla \times {\mathit{B}}_{ϵh}^n,\mathit{v})\hfill \\ \hfill \hspace{1em}& \hspace{1em}+S({\mathit{B}}_{ϵh}^{n-1}\times \nabla \times ({\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n),\mathit{v})+\frac{ϵ}{\nu }(p_{ϵ}^n-p_{ϵh}^n,q)=(\mathit{f}(t_n),\mathit{v}),\hfill \end{array}$$ (43)

$$\begin{array}{cc}\hfill \hspace{1em}& (d_t({\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n),\mathit{C})+\mu (\nabla \times ({\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n),\nabla \times \mathit{C})+\mu (\nabla ·({\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n),\nabla ·\mathit{C})\hfill \\ \hfill \hspace{1em}& \hspace{1em}-(({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n)\times {\mathit{B}}_{ϵh}^{n-1},\nabla \times \mathit{C})-({\mathit{u}}_{ϵh}^n\times ({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1}),\nabla \times \mathit{C})=(\mathit{g}(t_n),\mathit{C}).\hfill \end{array}$$ (44)

In order to derive estimates of the error, we need the following regularity results.

Lemma 4.

Under Assumptions 1–3, we have the following estimates:

$$\begin{array}{cc}\hfill \hspace{1em}& \parallel A_{1ϵ}^{\frac{1}{2}}{\mathit{u}}_{ϵ}^m{\parallel }^2+\parallel A_2^{\frac{1}{2}}{\mathit{B}}_{ϵ}^m{\parallel }^2+\tau \sum _{n=1}^m(\parallel A_{1ϵ}{\mathit{u}}_{ϵ}^n{\parallel }^2+\parallel A_2{\mathit{B}}_{ϵ}^n{\parallel }^2+\parallel p_{ϵ}^n{\parallel }_1^2)\le C,\hfill \\ \hfill \hspace{1em}& \parallel A_{1ϵ}{\mathit{u}}_{ϵ}^m{\parallel }^2+\parallel A_2{\mathit{B}}_{ϵ}^m{\parallel }^2+\parallel p_{ϵ}^m{\parallel }_1^2+\parallel d_t{\mathit{u}}_{ϵ}^m{\parallel }^2+\parallel d_t{\mathit{B}}_{ϵ}^m{\parallel }^2+\tau \sum _{n=1}^m(\parallel A_{1ϵ}^{\frac{1}{2}}{d}_t{\mathit{u}}_{ϵ}^n{\parallel }^2+\parallel A_2^{\frac{1}{2}}{d}_t{\mathit{B}}_{ϵ}^n{\parallel }^2)\le C,\hfill \\ \hfill \hspace{1em}& \sigma (t_m)(\parallel A_{1ϵ}^{\frac{1}{2}}{d}_t{\mathit{u}}_{ϵ}^m{\parallel }^2+\parallel A_2^{\frac{1}{2}}{d}_t{\mathit{B}}_{ϵ}^m{\parallel }^2)+\tau \sum _{n=1}^m\sigma (t_n)(\parallel A_{1ϵ}{d}_t{\mathit{u}}_{ϵ}^n{\parallel }^2+\parallel A_2{d}_t{\mathit{B}}_{ϵ}^n{\parallel }^2+\parallel d_t{p}_{ϵ}^n{\parallel }_1^2)\le C,\hfill \end{array}$$

for all $1\le n\le N$.

We refer to [4,32] for the proof of these results.

Theorem 8.

Under Assumptions 1–3, we have

$$\begin{array}{c}\hfill \parallel {\mathit{u}}_{ϵ}^m-{\mathit{u}}_{ϵh}^m\parallel +\sqrt{S}\parallel {\mathit{B}}_{ϵ}^m-{\mathit{B}}_{ϵh}^m\parallel +{\tau }^{\frac{1}{2}}\sum _{n=1}^m(\sqrt{\nu }\parallel \nabla ({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n)\parallel +\sqrt{S\mu }\parallel \nabla ({\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n)\parallel )\le Ch.\end{array}$$

Proof. 

Setting ${\eta }_u^n=r_{1h}{\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n$, ${\eta }_p^n=\rho p_{ϵ}^n-p_{ϵh}^n$ and ${\eta }_B^n=r_{2h}{\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n$ and taking $({\mathit{v}}_h,q_h)=2({\eta }_u^n,{\eta }_p^n)\tau$ in (43), ${\mathit{C}}_h=2S{\eta }_B^n\tau$ in (44), thanks to (4) and (10), we deduce that

$$\begin{array}{cc}\hfill \hspace{1em}& \parallel {\eta }_u^n{\parallel }^2-\parallel {\eta }_u^{n-1}{\parallel }^2+S(\parallel {\eta }_B^n{\parallel }^2-\parallel {\eta }_B^{n-1}{\parallel }^2)+\parallel d_t{\eta }_u^n{\parallel }^2{\tau }^2+S{\parallel d_t{\eta }_B^n\parallel }^2{\tau }^2\hfill \\ \hfill \hspace{1em}& \hspace{1em}+\nu \parallel A_{1h}^{\frac{1}{2}}{\eta }_u^n{\parallel }^2\tau +\nu \parallel A_{1h}^{\frac{1}{2}}({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n){\parallel }^2\tau +S\mu \parallel A_{2h}^{\frac{1}{2}}{\eta }_B^n{\parallel }^2+S\mu \parallel A_{2h}^{\frac{1}{2}}({\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n){\parallel }^2\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}+\frac{ϵ}{\nu }(\parallel {\eta }_p^n{\parallel }^2+\parallel p_{ϵ}-p_{ϵ}^n{\parallel }^2)\tau +2b({\mathit{u}}_{ϵ}^{n-1}-{\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵ}^n,{\eta }_u^n)\tau +2b({\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵ}^n-r_{1h}{\mathit{u}}_{ϵ}^n,{\eta }_u^n)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}+2S(({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1})\times \nabla \times {\mathit{B}}_{ϵ}^n,{\eta }_u^n)\tau -2S({\mathit{u}}_{ϵ}^n\times ({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1}),\nabla \times {\eta }_B^n)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}+2S({\mathit{B}}_{ϵh}^{n-1}\times \nabla \times ({\mathit{B}}_{ϵ}^n-r_{2h}{\mathit{B}}_{ϵ}^n),{\eta }_u^n)\tau -2S(({\mathit{u}}_{ϵ}^n-r_{1h}{\mathit{u}}_{ϵ}^n)\times {\mathit{B}}_{ϵ}^{n-1},\nabla \times {\eta }_B^n)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}+2S(({\mathit{u}}_{ϵ}^n-r_{1h}{\mathit{u}}_{ϵ}^n)\times ({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1}),\nabla \times {\eta }_B^n)\tau =\nu \parallel A_1^{\frac{1}{2}}({\mathit{u}}_{ϵ}^n-r_{1h}{\mathit{u}}_{ϵ}^n){\parallel }^2\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}+S\mu \parallel A_2^{\frac{1}{2}}({\mathit{B}}_{ϵ}^n-r_{2h}{\mathit{B}}_{ϵ}^n){\parallel }^2\tau -2(d_t({\mathit{u}}_{ϵ}^n-r_{1h}{\mathit{u}}_{ϵ}^n),{\eta }_u^n)\tau -2(d_t({\mathit{B}}_{ϵ}^n-r_{2h}{\mathit{B}}_{ϵ}^n),{\eta }_B^n)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}+2d({\eta }_u^n,p_{ϵ}^n-\rho p_{ϵ}^n)\tau +\frac{ϵ}{\nu }{\parallel p_{ϵ}^n-\rho p_{ϵ}^n\parallel }^2\tau .\hfill \end{array}$$ (45)

Combining (7)–(12) and Lemma 3, we obtain

$$\begin{array}{cc}\hfill \hspace{1em}& 2|b({\mathit{u}}_{ϵ}^{n-1}-{\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵ}^n,{\eta }_u^n)|\le \frac{\nu }{8}\parallel A_{1h}^{\frac{1}{2}}{\eta }_u^n{\parallel }^2+c\parallel {\mathit{u}}_{ϵ}^n{\parallel }_2^2{\parallel {\mathit{u}}_{ϵ}^{n-1}-{\mathit{u}}_{ϵh}^{n-1}\parallel }^2,\hfill \\ \hfill \hspace{1em}& 2|b({\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵ}^n-r_{1h}{\mathit{u}}_{ϵ}^n,{\eta }_u^n)|\le \frac{\nu }{8}\parallel A_{1h}^{\frac{1}{2}}{\eta }_u^n{\parallel }^2+c\parallel {\mathit{u}}_{ϵh}^{n-1}{\parallel }_{1,2}^2{\parallel \nabla ({\mathit{u}}_{ϵ}^n-r_{1h}{\mathit{u}}_{ϵ}^n)\parallel }^2,\hfill \\ \hfill \hspace{1em}& 2S|(({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1})\times \nabla \times {\mathit{B}}_{ϵ}^n,{\eta }_u^n)|\le \frac{\nu }{8}\parallel A_{1h}^{\frac{1}{2}}{\eta }_u^n{\parallel }^2+c\parallel {\mathit{B}}_{ϵ}^n{\parallel }_2^2{\parallel {\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1}\parallel }^2,\hfill \\ \hfill \hspace{1em}& 2S|({\mathit{B}}_{ϵh}^{n-1}\times \nabla \times ({\mathit{B}}_{ϵ}^n-r_{2h}{\mathit{B}}_{ϵ}^n),{\eta }_u^n)|\le \frac{\nu }{8}\parallel A_{1h}^{\frac{1}{2}}{\eta }_u^n{\parallel }^2+c\parallel {\mathit{B}}_{ϵh}^{n-1}{\parallel }_{1,2}^2{\parallel \nabla ({\mathit{B}}_{ϵ}^n-r_{2h}{\mathit{B}}_{ϵ}^n)\parallel }^2,\hfill \\ \hfill \hspace{1em}& 2S|({\mathit{u}}_{ϵ}^n\times ({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1}),\nabla \times {\eta }_B^n)|\le \frac{s\mu }{8}\parallel A_{2h}^{\frac{1}{2}}{\eta }_B^n{\parallel }^2+c\parallel {\mathit{u}}_{ϵ}^n{\parallel }_2^2{\parallel {\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1}\parallel }^2,\hfill \\ \hfill \hspace{1em}& 2S|(({\mathit{u}}_{ϵ}^n-r_{1h}{\mathit{u}}_{ϵ}^n)\times {\mathit{B}}_{ϵ}^{n-1},\nabla \times {\eta }_B^n)|\le \frac{s\mu }{8}\parallel A_{2h}^{\frac{1}{2}}{\eta }_B^n{\parallel }^2+c\parallel \nabla {\mathit{B}}_{ϵ}^{n-1}{\parallel }^2{\parallel \nabla ({\mathit{u}}_{ϵ}^n-r_{1h}{\mathit{u}}_{ϵ}^n)\parallel }^2,\hfill \\ \hfill \hspace{1em}& 2S|(({\mathit{u}}_{ϵ}^n-r_{1h}{\mathit{u}}_{ϵ}^n)\times ({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1}),\nabla \times {\eta }_B^n)|\hfill \\ \hfill \hspace{1em}& \hspace{1em}\le \frac{s\mu }{8}\parallel A_{2h}^{\frac{1}{2}}{\eta }_B^n{\parallel }^2+c(\parallel {\mathit{B}}_{ϵ}^{n-1}{\parallel }_1^2+\parallel {\mathit{B}}_{ϵh}^{n-1}{\parallel }_{1,2}^2)\parallel \nabla ({\mathit{u}}_{ϵ}^n-r_{1h}{\mathit{u}}_{ϵ}^n){\parallel }^2,\hfill \\ \hfill \hspace{1em}& 2|(d_t({\mathit{u}}_{ϵ}^n-r_{1h}{\mathit{u}}_{ϵ}^n),{\eta }_u^n)\le \frac{\nu }{8}\parallel A_{1h}^{\frac{1}{2}}{\eta }_u^n{\parallel }^2+c{\parallel d_t({\mathit{u}}_{ϵ}^n-r_{1h}{\mathit{u}}_{ϵ}^n)\parallel }^2,\hfill \\ \hfill \hspace{1em}& 2|(d_t({\mathit{B}}_{ϵ}^n-r_{2h}{\mathit{B}}_{ϵ}^n),{\eta }_B^n)|\le \frac{s\mu }{8}\parallel A_{2h}^{\frac{1}{2}}{\eta }_B^n{\parallel }^2+c{\parallel d_t({\mathit{B}}_{ϵ}^n-r_{2h}{\mathit{B}}_{ϵ}^n)\parallel }^2,\hfill \\ \hfill \hspace{1em}& 2|d({\eta }_u^n,p_{ϵ}^n-\rho p_{ϵ}^n)|\le \frac{\nu }{8}\parallel A_{1h}^{\frac{1}{2}}{\eta }_u^n{\parallel }^2+c{\parallel p_{ϵ}^n-\rho p_{ϵ}^n\parallel }^2.\hfill \end{array}$$

Combining the above inequalities with (45), and using (40) and (41), we can derive

$$\begin{array}{cc}\hfill \hspace{1em}& \parallel {\eta }_u^n{\parallel }^2-\parallel {\eta }_u^{n-1}{\parallel }^2+S(\parallel {\eta }_B^n{\parallel }^2-\parallel {\eta }_B^{n-1}{\parallel }^2)+\parallel d_t{\eta }_u^n{\parallel }^2{\tau }^2+S\parallel d_t{\eta }_B^n{\parallel }^2{\tau }^2+\nu \parallel A_{1h}^{\frac{1}{2}}({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n){\parallel }^2\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}+S\mu \parallel A_{2h}^{\frac{1}{2}}({\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n){\parallel }^2\tau \le c{h}^2(\parallel {\mathit{u}}_{ϵh}^{n-1}{\parallel }_{1,2}^2+\parallel {\mathit{B}}_{ϵh}^{n-1}{\parallel }_{1,2}^2+\parallel \nabla {\mathit{B}}_{ϵ}^{n-1}{\parallel }^2)(\parallel {\mathit{u}}_{ϵ}^n{\parallel }_2^2+\parallel {\mathit{B}}_{ϵ}^n{\parallel }_2^2)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}+d_{n-1}(\parallel {\mathit{u}}_{ϵ}^{n-1}-{\mathit{u}}_{ϵh}^{n-1}{\parallel }^2+\parallel {\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1}{\parallel }^2)\tau +c{h}^2(\parallel {\mathit{u}}_{ϵ}^n{\parallel }_2^2+\parallel {\mathit{B}}_{ϵ}^n{\parallel }_2^2+\parallel d_t{\mathit{u}}_{ϵ}^n{\parallel }_1^2+\parallel d_t{\mathit{B}}_{ϵ}^n{\parallel }_1^2+\parallel p_{ϵ}^n{\parallel }_1^2)\tau ,\hfill \end{array}$$

where $d_{n-1}=\parallel {\mathit{u}}_{ϵ}^n{\parallel }^2+{\parallel {\mathit{B}}_{ϵ}^n\parallel }^2$. Summing this inequality from $n=1$ to m, and using (40) and (41), Theorem 7, and Lemmas 4 and 2, we have

$$\begin{array}{cc}\hfill \hspace{1em}& \parallel {\mathit{u}}_{ϵ}^m-{\mathit{u}}_{ϵh}^m{\parallel }^2-S\parallel {\mathit{B}}_{ϵ}^m-{\mathit{B}}_{ϵh}^m{\parallel }^2+{\tau }^2\sum _{n=1}^m(\parallel d_t{\eta }_u^n{\parallel }^2+S\parallel d_t{\eta }_B^n{\parallel }^2)\hfill \\ \hfill \hspace{1em}& \hspace{1em}+\tau \sum _{n=1}^m(\nu \parallel A_{1h}^{\frac{1}{2}}({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n){\parallel }^2+S\mu \parallel A_{2h}^{\frac{1}{2}}({\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n){\parallel }^2)\hfill \\ \hfill \hspace{1em}& \hspace{1em}\le \tau \sum _{n=0}^{m-1}{d}_n(\parallel {\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n{\parallel }^2+\parallel {\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n{\parallel }^2)+C{h}^2\le C{h}^2,\hfill \end{array}$$

for all $1\le n\le N$. The proof is thus complete. □

Theorem 9.

Under Assumptions 1–4, we have

$$\begin{array}{ccc}\hfill & & {\sigma }^{\frac{1}{2}}(t_m)(\sqrt{\nu }\parallel {\mathit{u}}_{ϵ}^m-{\mathit{u}}_{ϵh}^m{\parallel }_1+\sqrt{\mu }\parallel {\mathit{B}}_{ϵ}^m-{\mathit{B}}_{ϵh}^m{\parallel }_1)\hfill \\ & & \hspace{1em}+{\tau }^{\frac{1}{2}}\sum _{n=1}^m{\sigma }^{\frac{1}{2}}(t_n)(\parallel d_t({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n)\parallel +\parallel d_t({\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n)\parallel )\le Ch,\hfill \end{array}$$ (46)

$${\tau }^{\frac{1}{2}}\sum _{n=1}^m{\sigma }^{\frac{1}{2}}(t_n)\parallel p_{ϵ}^n-p_{ϵh}^n\parallel \le Ch.$$ (47)

Proof. 

To obtain the error estimates of the PFEM, we give the Galerkin projector $R_{1h}:(\mathbf{X},\mathrm{M})\to {\mathbf{X}}_h$, $Q_h:(\mathbf{X},\mathrm{M})\to {\mathrm{M}}_h$, which satisfies (cf. [22])

$$\begin{array}{cc}\hfill (\nabla (R_{1h}(\mathit{u},p)-\mathit{u}),\nabla {\mathit{v}}_h)& -(\nabla ·{\mathit{v}}_h,Q_h(\mathit{u},p)-p)+(\nabla ·(R_{1h}(\mathit{u},p)-\mathit{u}),q_h)\hfill \\ \hfill \hspace{1em}& +\frac{ϵ}{\nu }(Q_h(\mathit{u},p)-p,q_h)=0,\hspace{1em}\forall ({\mathit{v}}_h,q_h)\in ({\mathbf{X}}_h,{\mathrm{M}}_h),\hfill \end{array}$$

for all $(\mathit{u},p)\in (\mathbf{X},\mathrm{M})$ with $\nabla ·\mathit{u}+\frac{ϵ}{\nu }p=0$. In addition, $R_{2h}:\mathbf{W}\to {\mathbf{W}}_h$ is the $H^1$-orthogonal projector defined by (cf. [4])

$$\begin{array}{c}\hfill (\nabla \times (R_{2h}\mathit{B}-\mathit{B}),\nabla \times {\mathit{C}}_h)+(\nabla ·(R_{2h}\mathit{B}-\mathit{B}),\nabla ·{\mathit{C}}_h)=0,\hspace{1em}\forall {\mathit{C}}_h\in {\mathbf{W}}_h,\end{array}$$

for all $\mathit{B}\in \mathbf{W}$. Due to the properties of $R_{1h}(\mathit{u},p)$, $Q_h(\mathit{u},p)$ and $R_{2h}$, we have

$$\parallel \mathit{u}-R_{1h}(\mathit{u},p)\parallel +h\parallel \nabla (\mathit{u}-R_{1h}(\mathit{u},p))\parallel +h\parallel p-Q_h(\mathit{u},p)\parallel \le C{h}^i{(\parallel \mathit{u}\parallel }_i+{\parallel p\parallel }_{i-1}),\hspace{1em}i=1,2,$$ (48)

and

$$\parallel R_{2h}\mathit{B}-\mathit{B}\parallel +h\parallel \nabla (R_{2h}\mathit{B}-\mathit{B})\parallel \le C{h}^2{\parallel \mathit{B}\parallel }_2.$$ (49)

Letting ${\xi }_u^n=R_{1h}({\mathit{u}}_{ϵ}^n,p_{ϵ}^n)-{\mathit{u}}_{ϵh}^n$, ${\xi }_p^n=Q_h({\mathit{u}}_{ϵ}^n,p_{ϵ}^n)-p_{ϵh}^n$, and ${\xi }_B^n=R_{2h}{\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n$, we derive from (43) and (44) that

$$\begin{array}{cc}\hfill \hspace{1em}& (d_t({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n),{\mathit{v}}_h)+\nu (A_{1h}^{\frac{1}{2}}{\xi }_u^n,A_{1h}^{\frac{1}{2}}{\mathit{v}}_h)+b({\xi }_u^{n-1},{\mathit{u}}_{ϵ}^n,{\mathit{v}}_h)+b({\mathit{u}}_{ϵh}^{n-1},{\xi }_u^n,{\mathit{v}}_h)-(\nabla ·{\mathit{v}}_h,{\xi }_p^n)\hfill \\ \hfill \hspace{1em}& \hspace{1em}+(\nabla ·d_t{\xi }_u^n,q_h)+\frac{ϵ}{\nu }(d_t{\xi }_p^n,q_h)+S({\xi }_B^n\times \nabla \times {\mathit{B}}_{ϵ}^n,{\mathit{v}}_h)+S({\mathit{B}}_{ϵh}^n\times \nabla \times {\xi }_B^n,{\mathit{v}}_h)\hfill \\ \hfill \hspace{1em}& \hspace{1em}=b(R_{1h}({\mathit{u}}_{ϵ}^{n-1},p_{ϵ}^{n-1})-{\mathit{u}}_{ϵ}^{n-1},{\mathit{u}}_{ϵ}^n,{\mathit{v}}_h)+b({\mathit{u}}_{ϵh}^{n-1},R_{1h}({\mathit{u}}_{ϵ}^n,p_{ϵ}^n)-{\mathit{u}}_{ϵ}^n,{\mathit{v}}_h)\hfill \\ \hfill \hspace{1em}& \hspace{1em}+S(R_{2h}{\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵ}^{n-1}\times \nabla \times {\mathit{B}}_{ϵ}^n,{\mathit{v}}_h)+S({\mathit{B}}_{ϵh}^{n-1}\times \nabla \times (R_{2h}{\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵ}^n),v_h),\hfill \end{array}$$ (50)

$$\begin{array}{cc}\hfill \hspace{1em}& (d_t({\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n),{\mathit{C}}_h)+\mu (A_{2h}^{\frac{1}{2}}{\xi }_B^n,A_{2h}^{\frac{1}{2}}{\mathit{C}}_h)-({\xi }_u^n\times {\mathit{B}}_{ϵ}^{n-1},\nabla \times {\mathit{C}}_h)-({\mathit{u}}_{ϵh}^n\times {\xi }_B^{n-1},\nabla \times {\mathit{C}}_h)\hfill \\ \hfill \hspace{1em}& \hspace{1em}=-((R_{1h}({\mathit{u}}_{ϵ}^n,p_{ϵ}^n)-{\mathit{u}}_{ϵ}^n)\times {\mathit{B}}_{ϵh}^{n-1},\nabla \times {\mathit{C}}_h)-({\mathit{u}}_{ϵh}^n\times (R_{2h}{\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵ}^{n-1}),\nabla \times {\mathit{C}}_h).\hfill \end{array}$$ (51)

Taking ${\mathit{v}}_h=2d_t{\xi }_u^n\tau$ and $q_h=2{\xi }_p^n\tau$ in (50) and ${\mathit{C}}_h=2d_t{\xi }_B^n\tau$ in (51), we obtain

$$\begin{array}{cc}\hfill \hspace{1em}& 2\parallel d_t{\xi }_u^n{\parallel }^2\tau +2\parallel d_t{\xi }_B^n{\parallel }^2\tau +\nu \parallel A_{1h}^{\frac{1}{2}}{\xi }_u^n{\parallel }^2+\mu \parallel A_{2h}^{\frac{1}{2}}{\xi }_B^n{\parallel }^2-\nu \parallel A_{1h}^{\frac{1}{2}}{\xi }_u^{n-1}{\parallel }^2-\mu \parallel A_{2h}^{\frac{1}{2}}{\xi }_B^{n-1}{\parallel }^2\hfill \\ \hfill \hspace{1em}& \hspace{1em}+\frac{ϵ}{\nu }(\parallel {\xi }_p^n{\parallel }^2-\parallel {\xi }_p^{n-1}{\parallel }^2)+2b({\mathit{u}}_{ϵ}^{n-1}-{\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵ}^n,d_t{\xi }_u^n)\tau +2b({\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n,d_t{\xi }_u^n)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}+2S(({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1})\times \nabla \times {\mathit{B}}_{ϵ}^{n-1},d_t{\xi }_u^n)\tau +2S({\mathit{B}}_{ϵh}^{n-1}\times \nabla \times ({\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n),d_t{\xi }_u^n)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}-2(({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n)\times {\mathit{B}}_{ϵ}^{n-1},\nabla \times d_t{\xi }_B^n)\tau -2({\mathit{u}}_{ϵh}^n\times ({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1}),\nabla \times d_t{\xi }_B^n)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}\le 2(d_t(R_{1h}({\mathit{u}}_{ϵ}^n,p_{ϵ}^n)-{\mathit{u}}_{ϵ}^n),d_t{\xi }_u^n)\tau +2(d_t(R_{2h}{\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵ}^n),d_t{\xi }_B^n)\tau .\hfill \end{array}$$ (52)

By using (7)–(12), (48) and (49), we obtain

$$\begin{array}{cc}\hfill \hspace{1em}& 2|b({\mathit{u}}_{ϵ}^{n-1}-{\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵ}^n,d_t{\xi }_u^n)|+2|b({\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n,d_t{\xi }_u^n)|\hfill \\ \hfill \hspace{1em}& \hspace{1em}\le 2|b({\mathit{u}}_{ϵ}^{n-1}-{\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵ}^n,d_t{\xi }_u^n)|+2|b({\mathit{u}}_{ϵ}^{n-1},{\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n,d_t{\xi }_u^n)|+2|b({\mathit{u}}_{ϵ}^{n-1}-{\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n,d_t{\xi }_u^n)|\hfill \\ \hfill \hspace{1em}& \hspace{1em}\le \frac{1}{8}\parallel d_t{\xi }_u^n{\parallel }^2+c\parallel {\mathit{u}}_{ϵ}^n{\parallel }_2^2\parallel \nabla ({\mathit{u}}_{ϵ}^{n-1}-{\mathit{u}}_{ϵh}^{n-1}){\parallel }^2+c\parallel {\mathit{u}}_{ϵ}^{n-1}{\parallel }_2^2{\parallel \nabla ({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n)\parallel }^2\hfill \\ \hfill \hspace{1em}& \hspace{1em}+c{h}^{-2}\parallel \nabla ({\mathit{u}}_{ϵ}^{n-1}-{\mathit{u}}_{ϵh}^{n-1}){\parallel }^2{\parallel \nabla ({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n)\parallel }^2,\hfill \\ \hfill \hspace{1em}& 2S|(({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1})\times \nabla \times {\mathit{B}}_{ϵ}^{n-1},d_t{\xi }_u^n)|+2S|({\mathit{B}}_{ϵh}^{n-1}\times \nabla \times ({\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n),d_t{\xi }_u^n)|\hfill \\ \hfill \hspace{1em}& \hspace{1em}\le \frac{1}{8}\parallel d_t{\xi }_u{\parallel }^2+c\parallel {\mathit{B}}_{ϵ}^n{\parallel }_2^2\parallel \nabla ({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1}){\parallel }^2+c\parallel {\mathit{B}}_{ϵ}^{n-1}{\parallel }_2^2{\parallel \nabla ({\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n)\parallel }^2\hfill \\ \hfill \hspace{1em}& \hspace{1em}+c{h}^{-2}\parallel \nabla ({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1}){\parallel }^2{\parallel \nabla ({\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n)\parallel }^2,\hfill \\ \hfill \hspace{1em}& 2|(({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n)\times {\mathit{B}}_{ϵ}^{n-1},\nabla \times d_t{\xi }_B^n)|+2|({\mathit{u}}_{ϵh}^n\times ({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1}),\nabla \times d_t{\xi }_B^n)|\hfill \\ \hfill \hspace{1em}& \hspace{1em}\le \frac{1}{8}\parallel d_t{\xi }_B^n{\parallel }^2+c\parallel {\mathit{B}}_{ϵ}^{n-1}{\parallel }_2^2\parallel \nabla ({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n){\parallel }^2+c\parallel {\mathit{u}}_{ϵ}^n{\parallel }_2^2{\parallel \nabla ({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1})\parallel }^2\hfill \\ \hfill \hspace{1em}& \hspace{1em}+c{h}^{-2}\parallel \nabla ({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n){\parallel }^2{\parallel \nabla ({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1})\parallel }^2,\hfill \\ \hfill \hspace{1em}& 2|(d_t(R_{1h}({\mathit{u}}_{ϵ}^n,p_{ϵ}^n)-{\mathit{u}}_{ϵ}^n),d_t{\xi }_u^n)|+2|(d_t(R_{2h}{\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵ}^n),d_t{\xi }_B^n)|\hfill \\ \hfill \hspace{1em}& \hspace{1em}\le \frac{1}{8}\parallel d_t{\xi }_u^n{\parallel }^2+\frac{1}{8}\parallel d_t{\xi }_B^n{\parallel }^2+c{h}^4(\parallel d_t{\mathit{u}}_{ϵ}^n{\parallel }_2^2+\parallel d_t{\mathit{B}}_{ϵ}^n{\parallel }_2^2+\parallel d_t{p}_{ϵ}^n{\parallel }_1^2).\hfill \end{array}$$

Combining the above inequalities with (52), and using Theorems 4 and 8, we can derive

$$\begin{array}{cc}\hfill \hspace{1em}& \parallel d_t{\xi }_u^n{\parallel }^2\tau +\parallel d_t{\xi }_B^n{\parallel }^2\tau +\nu (\parallel A_{1h}^{\frac{1}{2}}{\xi }_u^n{\parallel }^2-\parallel A_{1h}^{\frac{1}{2}}{\xi }_u^{n-1}{\parallel }^2)+\mu (\parallel A_{2h}^{\frac{1}{2}}{\xi }_B^n{\parallel }^2-\parallel A_{2h}^{\frac{1}{2}}{\xi }_B^{n-1}{\parallel }^2)+\frac{ϵ}{\nu }(\parallel {\xi }_p^n{\parallel }^2-\parallel {\xi }_p^{n-1}{\parallel }^2)\hfill \\ \hfill \hspace{1em}& \hspace{1em}\le c(\parallel \nabla ({\mathit{u}}_{ϵ}^{n-1}-{\mathit{u}}_{ϵh}^{n-1}){\parallel }^2+\parallel \nabla ({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n){\parallel }^2+\parallel \nabla ({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1}){\parallel }^2+\parallel \nabla ({\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n){\parallel }^2)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}+c{h}^4(\parallel d_t{\mathit{u}}_{ϵ}^n{\parallel }_2^2+\parallel d_t{\mathit{B}}_{ϵ}^n{\parallel }_2^2+\parallel d_t{p}_{ϵ}^n{\parallel }_1^2)\tau +c{h}^{-2}{d}_{n-1}(\parallel \nabla {\xi }_u^{n-1}{\parallel }^2+\parallel \nabla {\xi }_B^{n-1}{\parallel }^2)\tau ,\hfill \end{array}$$

where $d_{n-1}=c{h}^{-2}(\parallel \nabla ({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n){\parallel }^2+\parallel \nabla ({\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n){\parallel }^2)$. Multiplying this inequality by $\sigma (t_n)$ and taking the sum with respect to n from 1 to m, thanks to Theorems 4 and 8 and Lemma 2, we obtain

$$\begin{array}{cc}\hfill \hspace{1em}& \sigma (t_m)(\nu \parallel A_{1h}^{\frac{1}{2}}{\xi }_u^n{\parallel }^2+\mu (\parallel A_{2h}^{\frac{1}{2}}{\xi }_B^n{\parallel }^2)+\tau \sum _{n=1}^m\sigma (t_n)(\parallel d_t{\xi }_u^n{\parallel }^2+\parallel d_t{\xi }_B^n{\parallel }^2)\hfill \\ \hfill \hspace{1em}& \hspace{1em}\le \tau \sum _{n=0}^{m-1}{d}_n(\parallel \nabla {\xi }_u^n{\parallel }^2+\parallel \nabla {\xi }_B^n{\parallel }^2)+C{h}^2\le C{h}^2.\hfill \end{array}$$ (53)

Moreover, by using (48) and (49) and Theorem 4, we obtain

$$\begin{array}{c}\hfill \parallel \nabla (R_{1h}({\mathit{u}}_{ϵ}^n,p_{ϵ}^n)-{\mathit{u}}_{ϵ}^n){\parallel }^2+\parallel \nabla (R_{2h}{\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵ}^n){\parallel }^2\le c{h}^2(\parallel {\mathit{u}}_{ϵ}^n{\parallel }_2^2+\parallel p_{ϵ}^n{\parallel }_1^2+\parallel {\mathit{B}}_{ϵ}^n{\parallel }_2^2)\le C{h}^2.\end{array}$$ (54)

Hence, by combining (53) with (54), we obtain (46).

Next, using (7)–(12), (33) and (43), we derive

$$\begin{array}{ccc}& & \beta \parallel {\xi }_p^n\parallel \le c\parallel d_t({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n)\parallel +\nu \parallel \nabla ({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n)\parallel +c\parallel \nabla ({\mathit{u}}_{ϵ}^{n-1}-{\mathit{u}}_{ϵh}^{n-1})\parallel \parallel \nabla {\mathit{u}}_{ϵ}^n\parallel \hfill \\ & & \hspace{1em}+c(\parallel \nabla {\mathit{u}}_{ϵ}^{n-1}\parallel +\parallel \nabla ({\mathit{u}}_{ϵ}^{n-1}-{\mathit{u}}_{ϵh}^{n-1})\parallel )\parallel \nabla ({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n)\parallel +c\parallel p_{ϵ}^n-Q_h({\mathit{u}}_{ϵ}^n,p_{ϵ}^n)\parallel \hfill \\ & & \hspace{1em}+c\parallel \nabla ({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1})\parallel \parallel \nabla {\mathit{B}}_{ϵ}^n\parallel +c(\parallel \nabla {\mathit{B}}_{ϵ}^{n-1}\parallel +\parallel \nabla ({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1})\parallel )\parallel \nabla ({\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n)\parallel .\hfill \end{array}$$

Multiplying this inequality by $\sigma (t_n)$ and taking the sum with respect to n from 1 to m, due to Theorem 8 and (46), we have

$$\begin{array}{ccc}\hfill & & \tau \sum _{n=1}^m\sigma (t_n){\parallel {\xi }_p^n\parallel }^2\le C{h}^2.\hfill \end{array}$$ (55)

Using (55) and (48) and (43), we obtain (47). Thus, this proof is thus complete. □

Next, we make the $L^2$-error estimates. Taking ${\mathit{v}}_h=2A_{1h}^{-1}{\eta }_u^n\tau$ and $q_h=0$ in (43) and ${\mathit{C}}_h=2A_{2h}^{-1}{\eta }_B^n\tau$ in (44), we obtain

$$\begin{array}{cc}\hfill \hspace{1em}& \parallel {\eta }_u^n{\parallel }_{-1,2}^2+\parallel {\eta }_B^n{\parallel }_{-1,2}^2-(\parallel {\eta }_u^{n-1}{\parallel }_{-1,2}^2+\parallel {\eta }_B^{n-1}{\parallel }_{-1,2}^2)+\parallel d_t{\eta }_u^n{\parallel }_{-1,2}^2{\tau }^2+\parallel d_t{\eta }_B^n{\parallel }_{-1,2}^2{\tau }^2+\mu \parallel {\eta }_B^n{\parallel }^2\tau +\nu {\parallel {\eta }_u^n\parallel }^2\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}+\nu \parallel {\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n{\parallel }^2\tau +\mu {\parallel {\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n\parallel }^2\tau +2b({\mathit{u}}_{ϵ}^{n-1}-{\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵ}^n,A_{1h}^{-1}{\eta }_u^n)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}+2b({\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n,A_{1h}^{-1}{\eta }_u^n)\tau +2S(({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1})\times \nabla \times {\mathit{B}}_{ϵ}^{n-1},A_{1h}^{-1}{\eta }_u^n)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}+2S({\mathit{B}}_{ϵh}^{n-1}\times \nabla \times ({\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n),A_{1h}^{-1}{\eta }_u^n)\tau -2(({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n)\times {\mathit{B}}_{ϵ}^{n-1},\nabla \times A_{2h}^{-1}{\eta }_B^n)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}-2({\mathit{u}}_{ϵh}^n\times ({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1}),\nabla \times A_{2h}^{-1}{\eta }_B^n)\tau =-2(d_t({\mathit{u}}_{ϵ}^n-r_{1h}{\mathit{u}}_{ϵ}^n),A_{1h}^{-1}{\eta }_u^n)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}-2(d_t({\mathit{B}}_{ϵ}^n-r_{2h}{\mathit{B}}_{ϵ}^n),A_{2h}^{-1}{\eta }_B^n)\tau +\nu \parallel r_{1h}{\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵ}^n{\parallel }^2\tau +\mu {\parallel r_{2h}{\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵ}^n\parallel }^2\tau .\hfill \end{array}$$ (56)

By using (11), (12), and Lemma 3, we have

$$\begin{array}{cc}\hfill \hspace{1em}& 2|b({\mathit{u}}_{ϵ}^{n-1}-{\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵ}^n,A_{1h}^{-1}{\eta }_u^n)|\hfill \\ \hfill \hspace{1em}& \hspace{1em}\le 2|b({\mathit{u}}_{ϵ}^{n-1}-r_{1h}{\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵ}^n,A_{1h}^{-1}{\eta }_u^n)|+2|b({\eta }_u^{n-1},{\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n,A_{1h}^{-1}{\eta }_u^n)|+2|b({\eta }_u^{n-1},{\mathit{u}}_{ϵh}^n,A_{1h}^{-1}{\eta }_u^n)|\hfill \\ \hfill \hspace{1em}& \hspace{1em}\le \frac{\nu }{8}\parallel {\eta }_u^n{\parallel }^2+c\parallel {\mathit{u}}_{ϵ}^n{\parallel }_1^2\parallel {\mathit{u}}_{ϵ}^{n-1}-r_{1h}{\mathit{u}}_{ϵ}^{n-1}{\parallel }^2+c\parallel {\mathit{u}}_{ϵh}^n{\parallel }_{2,2}^2\parallel {\eta }_u^{n-1}{\parallel }_{-1,2}^2+c\parallel {\eta }_u^{n-1}{\parallel }^2{\parallel {\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n\parallel }_1^2,\hfill \\ \hfill \hspace{1em}& 2|b({\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n,A_{1h}^{-1}{\eta }_u^n)|\hfill \\ \hfill \hspace{1em}& \hspace{1em}\le 2|b({\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n,A_{1h}^{-1}({\eta }_u^n-{\eta }_u^{n-1}))|+2|b({\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵ}^n-r_{1h}{\mathit{u}}_{ϵ}^n,A_{1h}^{-1}{\eta }_u^{n-1})|+2|b({\mathit{u}}_{ϵh}^{n-1},{\eta }_u^n,A_{1h}^{-1}{\eta }_u^{n-1})|\hfill \\ \hfill \hspace{1em}& \hspace{1em}\le \frac{1}{8\tau }\parallel {\eta }_u^n-{\eta }_u^{n-1}{\parallel }_{-1,2}^2+\frac{\nu }{8}(\parallel {\eta }_u^n{\parallel }^2+\parallel {\mathit{u}}_{ϵ}^n-r_{1h}{\mathit{u}}_{ϵ}^n{\parallel }^2)+c\tau \parallel {\mathit{u}}_{ϵh}^{n-1}{\parallel }_{1,2}^2\parallel {\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n{\parallel }_1^2+c\parallel {\mathit{u}}_{ϵh}^{n-1}{\parallel }_{2,2}^2{\parallel {\eta }_u^{n-1}\parallel }_{-1,2}^2,\hfill \\ \hfill \hspace{1em}& 2S|(({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1})\times \nabla \times {\mathit{B}}_{ϵ}^{n-1},A_{1h}^{-1}{\eta }_u^n)|\hfill \\ \hfill \hspace{1em}& \hspace{1em}\le \frac{\nu }{8}\parallel {\eta }_u^n{\parallel }^2+c\parallel {\mathit{B}}_{ϵ}^{n-1}-r_{2h}{\mathit{B}}_{ϵ}^{n-1}{\parallel }^2\parallel {\mathit{B}}_{ϵ}^n{\parallel }_1^2+c\parallel {\eta }_B^{n-1}{\parallel }^2\parallel {\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n{\parallel }_1^2+c\parallel {\mathit{B}}_{ϵh}^n{\parallel }_{2,2}^2{\parallel {\eta }_B^{n-1}\parallel }_{-1,2}^2,\hfill \\ \hfill \hspace{1em}& 2S|({\mathit{B}}_{ϵh}^{n-1}\times \nabla \times ({\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n),A_{1h}^{-1}{\eta }_u^n)|\hfill \\ \hfill \hspace{1em}& \hspace{1em}\le \frac{1}{8\tau }\parallel {\eta }_u^n-{\eta }_u^{n-1}{\parallel }_{-1,2}^2+\frac{\mu }{8}(\parallel {\eta }_B^n{\parallel }^2+\parallel {\mathit{B}}_{ϵ}^n-r_{2h}{\mathit{B}}_{ϵ}^n{\parallel }^2)+c\tau \parallel {\mathit{B}}_{ϵh}^{n-1}{\parallel }_{1,2}^2\parallel {\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n{\parallel }_1^2+c\parallel {\mathit{B}}_{ϵh}^{n-1}{\parallel }_{2,2}^2{\parallel {\eta }_u^{n-1}\parallel }_{-1,2}^2,\hfill \\ \hfill \hspace{1em}& 2|(({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n)\times {\mathit{B}}_{ϵ}^{n-1},\nabla \times A_{2h}^{-1}{\eta }_B^n)|\hfill \\ \hfill \hspace{1em}& \hspace{1em}\le \frac{1}{8\tau }\parallel {\eta }_B^n-{\eta }_B^{n-1}{\parallel }_{-1,2}^2+c\tau \parallel {\mathit{B}}_{ϵ}^{n-1}{\parallel }_1^2\parallel {\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n{\parallel }_1^2+\frac{\mu }{8}(\parallel {\eta }_u^n{\parallel }^2+\parallel {\mathit{u}}_{ϵ}^n-r_{1h}{\mathit{u}}_{ϵ}^n{\parallel }^2)+c\parallel {\mathit{B}}_{ϵ}^{n-1}{\parallel }_2^2{\parallel {\eta }_B^{n-1}\parallel }_{-1,2}^2,\hfill \\ \hfill \hspace{1em}& 2|({\mathit{u}}_{ϵh}^n\times ({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1}),\nabla \times A_{2h}^{-1}{\eta }_B^n)|\le \frac{\mu }{8}\parallel {\eta }_B^n{\parallel }^2+c\parallel {\mathit{B}}_{ϵ}^{n-1}-r_{2h}{\mathit{B}}_{ϵ}^{n-1}{\parallel }^2\parallel {\mathit{B}}_{ϵh}^n{\parallel }_{1,2}^2+c\parallel {\mathit{u}}_{ϵh}^n{\parallel }_{2,2}^2{\parallel {\eta }_B^{n-1}\parallel }_{-1,2}^2,\hfill \\ \hfill \hspace{1em}& 2|(d_t({\mathit{u}}_{ϵ}^n-r_{1h}{\mathit{u}}_{ϵ}^n),A_{1h}^{-1}{\eta }_u^n)|+2|(d_t({\mathit{B}}_{ϵ}^n-r_{2h}{\mathit{B}}_{ϵ}^n),A_{2h}^{-1}{\eta }_B^n)|\hfill \\ \hfill \hspace{1em}& \hspace{1em}\le \frac{1}{8}(\nu \parallel {\eta }_u^n{\parallel }^2+\mu \parallel {\eta }_B^n{\parallel }^2)+c\parallel d_t({\mathit{u}}_{ϵ}^n-r_{1h}{\mathit{u}}_{ϵ}^n){\parallel }^2+c{\parallel d_t({\mathit{B}}_{ϵ}^n-r_{2h}{\mathit{B}}_{ϵ}^n)\parallel }^2.\hfill \end{array}$$

Combining these inequalities with (56), using (40) and (41), we obtain

$$\begin{array}{cc}\hfill \hspace{1em}& \parallel {\eta }_u^n{\parallel }_{-1,2}^2+\parallel {\eta }_B^n{\parallel }_{-1,2}^2-(\parallel {\eta }_u^{n-1}{\parallel }_{-1,2}^2+\parallel {\eta }_B^{n-1}{\parallel }_{-1,2}^2)+\mu \parallel {\eta }_B^n{\parallel }^2\tau +\nu {\parallel {\eta }_u^n\parallel }^2\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}\le d_{n-1}(\parallel {\eta }_u^{n-1}{\parallel }_{-1,2}^2+\parallel {\eta }_B^{n-1}{\parallel }_{-1,2}^2)\tau +c{h}^4(\parallel {\mathit{u}}_{ϵ}^n{\parallel }_2^2+\parallel {\mathit{B}}_{ϵ}^n{\parallel }_2^2+\parallel p_{ϵ}^n{\parallel }_1^2)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}+c{h}^4(\parallel {\mathit{u}}_{ϵh}^n{\parallel }_{1,2}^2+\parallel {\mathit{u}}_{ϵ}^n{\parallel }_1^2+\parallel {\mathit{B}}_{ϵ}^n{\parallel }_1^2)(\parallel {\mathit{u}}_{ϵ}^n{\parallel }_2^2+\parallel {\mathit{B}}_{ϵ}^n{\parallel }_2^2+\parallel p_{ϵ}^n{\parallel }_1^2)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}+c\parallel {\eta }_u^{n-1}{\parallel }^2\parallel {\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n{\parallel }^2\tau +c\parallel {\eta }_B^{n-1}{\parallel }^2{\parallel {\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n\parallel }_1^2\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}+c(\parallel {\mathit{u}}_{ϵh}^n{\parallel }^2+\parallel {\mathit{B}}_{ϵh}^{n-1}{\parallel }^2+\parallel {\mathit{B}}_{ϵ}^{n-1}{\parallel }^2)(\parallel {\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n{\parallel }_1^2+\parallel {\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n{\parallel }_1^2){\tau }^2,\hfill \end{array}$$ (57)

where $d_{n-1}=\parallel {\mathit{u}}_{ϵh}^n{\parallel }_{2,2}^2+\parallel {\mathit{B}}_{ϵh}^n{\parallel }_{2,2}^2+\parallel {\mathit{u}}_{ϵh}^{n-1}{\parallel }_{2,2}^2+\parallel {\mathit{B}}_{ϵh}^{n-1}{\parallel }_{2,2}^2+{\parallel {\mathit{B}}_{ϵ}^{n-1}\parallel }_2^2$. In addition, we have

$$\begin{array}{c}\hfill \parallel {\mathit{u}}_{ϵh}^n{\parallel }_{2,2}\le c{h}^{-1}\parallel {\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n{\parallel }_1+\parallel {\mathit{u}}_{ϵ}^n{\parallel }_2,\hspace{1em}\parallel {\mathit{B}}_{ϵh}^n{\parallel }_{2,2}\le c{h}^{-1}\parallel {\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n{\parallel }_1+{\parallel {\mathit{B}}_{ϵ}^n\parallel }_2.\end{array}$$

Summing (57) from $n=1$ to m, and using Lemma 4 and Theorems 8 and 9, we have

$$\begin{array}{cc}\hfill \hspace{1em}& \parallel {\eta }_u^n{\parallel }_{-1,2}^2+\parallel {\eta }_B^n{\parallel }_{-1,2}^2+\tau \sum _{n=1}^m(\mu \parallel {\eta }_B^n{\parallel }^2+\nu \parallel {\eta }_u^n{\parallel }^2)\le \tau \sum _{n=0}^{m-1}{d}_n(\parallel {\eta }_u^n{\parallel }_{-1,2}^2+\parallel {\eta }_B^n{\parallel }_{-1,2}^2)+C{h}^4+C{h}^2\tau .\hfill \end{array}$$ (58)

Then, applying Lemma 2 to (58) and using (40) and (41), we obtain

$$\begin{array}{cc}\hfill \hspace{1em}& \parallel {\eta }_u^n{\parallel }_{-1,2}^2+\parallel {\eta }_B^n{\parallel }_{-1,2}^2+\tau \sum _{n=1}^m(\mu \parallel {\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n{\parallel }^2+\nu \parallel {\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n{\parallel }^2)\le C{h}^4+C{\tau }^2.\hfill \end{array}$$ (59)

Theorem 10.

Under the assumption of Lemma 4, we have

$$\begin{array}{c}\hfill {\sigma }^{\frac{1}{2}}(t_m)(\parallel {\mathit{u}}_{ϵ}^m-{\mathit{u}}_{ϵh}^m\parallel +\parallel {\mathit{B}}_{ϵ}^m-{\mathit{B}}_{ϵh}^m\parallel )\le C(h^2+\tau ).\end{array}$$ (60)

Proof. 

Taking ${\mathit{v}}_h=2{\xi }_u^n\tau$ and $q_h=2{\xi }_p^n\tau$ in (43) and ${\mathit{C}}_h=2{\xi }_B^n\tau$ in (44), and adding these equations together, we have

$$\begin{array}{cc}\hfill \hspace{1em}& \parallel {\xi }_u^n{\parallel }^2+\parallel {\xi }_B^n{\parallel }^2-(\parallel {\xi }_u^{n-1}{\parallel }^2+\parallel {\xi }_B^{n-1}{\parallel }^2)+2(\nu \parallel {\xi }_u^n{\parallel }_1^2+\mu \parallel {\xi }_B^n{\parallel }_1^2+\frac{\nu }{ϵ}\parallel {\xi }_p^n{\parallel }^2)\tau +2b({\mathit{u}}_{ϵ}^{n-1}-{\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵ}^n,{\xi }_u^n)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}+2b({\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n,{\xi }_u^n)\tau +2S(({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1})\times \nabla \times {\mathit{B}}_{ϵ}^n,{\xi }_u^n)\tau +2S({\mathit{B}}_{ϵh}^{n-1}\times \nabla \times ({\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n),{\xi }_u^n)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}-2(({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n)\times {\mathit{B}}_{ϵ}^{n-1},\nabla \times {\xi }_B^n)\tau -2({\mathit{u}}_{ϵh}^n\times ({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1}),\nabla \times {\xi }_B^n)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}\le 2(d_t(R_{1h}({\mathit{u}}_{ϵ}^n,p_{ϵ}^n)-{\mathit{u}}_{ϵ}^n),{\xi }_u^n)\tau +2(d_t(R_{2h}{\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵ}^n),{\xi }_B^n)\tau .\hfill \end{array}$$ (61)

By using (11), (12), and Lemma 3, we obtain

$$\begin{array}{cc}\hfill \hspace{1em}& 2|b({\mathit{u}}_{ϵ}^{n-1}-{\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵ}^n,{\xi }_u^n)|+2|b({\mathit{u}}_{ϵh}^{n-1},{\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n,{\xi }_u^n)|\hfill \\ \hfill \hspace{1em}& \hspace{1em}\le \frac{\nu }{8}\parallel {\xi }_u^n{\parallel }_1^2+c\parallel {\mathit{u}}_{ϵ}^n{\parallel }_2^2\parallel {\mathit{u}}_{ϵ}^{n-1}-{\mathit{u}}_{ϵh}^{n-1}{\parallel }^2+c\parallel {\mathit{u}}_{ϵ}^{n-1}{\parallel }_2^2\parallel {\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n{\parallel }^2+c\parallel {\mathit{u}}_{ϵ}^{n-1}-{\mathit{u}}_{ϵh}^{n-1}{\parallel }_1^2{\parallel {\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n\parallel }_1^2,\hfill \\ \hfill \hspace{1em}& 2S|(({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1})\times \nabla \times {\mathit{B}}_{ϵ}^n,{\xi }_u^n)|+2S|({\mathit{B}}_{ϵh}^{n-1}\times \nabla \times ({\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n),{\xi }_u^n)|\hfill \\ \hfill \hspace{1em}& \hspace{1em}\le \frac{\nu }{8}\parallel {\xi }_u^n{\parallel }_1^2+c\parallel {\mathit{B}}_{ϵ}^n{\parallel }_2^2\parallel {\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1}{\parallel }^2+c\parallel {\mathit{B}}_{ϵ}^{n-1}{\parallel }_2^2\parallel {\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n{\parallel }^2+c\parallel {\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1}{\parallel }_1^2{\parallel {\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n\parallel }_1^2,\hfill \\ \hfill \hspace{1em}& 2|(({\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n)\times {\mathit{B}}_{ϵ}^{n-1},\nabla \times {\xi }_B^n)|+2|({\mathit{u}}_{ϵh}^n\times ({\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1}),\nabla \times {\xi }_B^n)|\hfill \\ \hfill \hspace{1em}& \hspace{1em}\le \frac{\mu }{8}\parallel {\xi }_B^n{\parallel }_1^2+c\parallel {\mathit{B}}_{ϵ}^{n-1}{\parallel }_2^2\parallel {\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n{\parallel }^2+c\parallel {\mathit{u}}_{ϵ}^n{\parallel }_2^2\parallel {\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1}{\parallel }^2+c\parallel {\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1}{\parallel }_1^2{\parallel {\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n\parallel }_1^2,\hfill \\ \hfill \hspace{1em}& 2|(d_t(R_{1h}({\mathit{u}}_{ϵ}^n,p_{ϵ}^n)-{\mathit{u}}_{ϵ}^n),{\xi }_u^n)|+2|(d_t(R_{2h}{\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵ}^n),{\xi }_B^n)|\hfill \\ \hfill \hspace{1em}& \hspace{1em}\le \frac{\nu }{8}\parallel {\xi }_u^n{\parallel }_1^2+\frac{\mu }{8}\parallel {\xi }_B^n{\parallel }_1^2+c\parallel d_t(R_{1h}({\mathit{u}}_{ϵ}^n,p_{ϵ}^n)-{\mathit{u}}_{ϵ}^n){\parallel }^2+c\parallel d_t{(R_{2h}{\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵ}^n\parallel }^2.\hfill \end{array}$$

Combining these inequalities with (61) and using (40) and (41), we have

$$\begin{array}{cc}\hfill \hspace{1em}& \parallel {\xi }_u^n{\parallel }^2+\parallel {\xi }_B^n{\parallel }^2-(\parallel {\xi }_u^{n-1}{\parallel }^2+\parallel {\xi }_B^{n-1}{\parallel }^2)+(\nu \parallel {\xi }_u^n{\parallel }_1^2+\mu \parallel {\xi }_B^n{\parallel }_1^2)\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}\le c(\parallel {\mathit{u}}_{ϵ}^n{\parallel }_2^2+\parallel {\mathit{B}}_{ϵ}^n{\parallel }_2^2)(\parallel {\mathit{u}}_{ϵ}^{n-1}-{\mathit{u}}_{ϵh}^{n-1}\parallel +\parallel {\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1}\parallel )\tau +c{h}^4(\parallel d_t{\mathit{u}}_{ϵ}^n{\parallel }_2^2+\parallel d_t{\mathit{B}}_{ϵ}^n{\parallel }_2^2+\parallel d_t{p}_{ϵ}^n{\parallel }_1^2)\hfill \\ \hfill \hspace{1em}& \hspace{1em}+c(\parallel {\mathit{u}}_{ϵ}^{n-1}{\parallel }_2^2+\parallel {\mathit{B}}_{ϵ}^{n-1}{\parallel }_2^2)(\parallel {\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n\parallel +\parallel {\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n\parallel )\tau \hfill \\ \hfill \hspace{1em}& \hspace{1em}+c(\parallel {\mathit{u}}_{ϵ}^{n-1}-{\mathit{u}}_{ϵh}^{n-1}{\parallel }_1^2+\parallel {\mathit{B}}_{ϵ}^{n-1}-{\mathit{B}}_{ϵh}^{n-1}{\parallel }_1^2)(\parallel {\mathit{u}}_{ϵ}^n-{\mathit{u}}_{ϵh}^n{\parallel }_1^2+\parallel {\mathit{B}}_{ϵ}^n-{\mathit{B}}_{ϵh}^n{\parallel }_1^2)\tau .\hfill \end{array}$$

Multiplying this inequality by $\sigma (t_n)$ and taking the sum with respect to n from 1 to m, thanks to Theorems 4 and 8 and Lemma 2, we obtain

$$\begin{array}{c}\hfill \sigma (t_m)(\parallel {\xi }_u^m{\parallel }^2+\parallel {\xi }_B^m{\parallel }^2)+\tau \sum _{n=1}^m\sigma (t_n)(\nu \parallel {\xi }_u^n{\parallel }_1^2+\mu \parallel {\xi }_B^n{\parallel }_1^2)\tau \le C{h}^4+C{\tau }^2.\end{array}$$ (62)

Then, by applying (40), (41), Theorem 4, and (62), we obtain Theorem 10. □

6. Error Estimates

Combining Theorem 3 and the results in Section 3, Section 4 and Section 5, we obtain the following results on convergence of the fully discrete Euler semi-implicit scheme.

Theorem 11.

Under Assumptions 1–4, we have the following error estimates

$$\begin{array}{cc}\hfill \hspace{1em}& {\sigma }^{\frac{1}{2}}(t_m)(\sqrt{\nu }\parallel \mathit{u}(t_m)-{\mathit{u}}_{ϵh}^m{\parallel }_1+\sqrt{\mu }\parallel \mathit{B}(t_m)-{\mathit{B}}_{ϵh}^m{\parallel }_1)+{\left(\tau \sum _{n=1}^m\sigma (t_n){\parallel p(t_n)-p_{ϵh}^n\parallel }^2\right)}^{\frac{1}{2}}\le C(h+\tau +ϵ),\hfill \\ \hfill \hspace{1em}& {\sigma }^{\frac{1}{2}}(t_m)(\parallel \mathit{u}(t_m)-{\mathit{u}}_{ϵh}^m\parallel +\parallel \mathit{B}(t_m)-{\mathit{B}}_{ϵh}^m\parallel )\le C(h^2+\tau +ϵ).\hfill \end{array}$$

7. Numerical Example

In this part, we present two numerical tests to validate the accuracy and performance of our scheme. We use the $P_1^b/P_1$ element that satisfies the LBB condition for velocity and pressure $(\mathit{u},p)$, and the $P_1$ element for magnetic field $\mathit{B}$. The penalty parameter is selected as $ϵ=\tau$ in all the numerical tests.

7.1. Convergence Tests

We verify the convergence rates of the PFEM based on the Euler implicit scheme in this example. We use the computational domain ${[0,1]}^d,d=2,3$ and set parameters $\nu =\mu =S=1$. The source terms are given by the following exact solutions:

$$\left\{\begin{array}{c}\mathit{u}=(y(y-1)(2y-1)x^2{(x-1)}^{2}cos(t),-x(x-1)(2x-1)y^2{(y-1)}^{2}cos(t)),\hfill \\ p=(2y-1)(2x-1)cos(t),\hfill \\ \mathit{B}=(cos(\pi y)sin(\pi x)cos(t),-cos(\pi x)sin(\pi y)cos(t)),\hfill \end{array}\right.$$

and

$$\left\{\begin{array}{c}\mathit{u}=((y^4+z^2)exp(-t),(z^4+x^2)exp(-t),(x^4+y^2)exp(-t)),\hfill \\ p=(2x-1)(2y-1)(2z-1)exp(-t),\hfill \\ \mathit{B}=((sin(y)+z)exp(-t),(sin(z)+x)exp(-t),(sin(x)+y)exp(-t)).\hfill \end{array}\right.$$

We choose $\tau =h^2$ and $h=1/n$ ($n=8,16,32,64,128$ in ${\mathbb{R}}^2$ or $n=4,8,12,16,20$${\mathbb{R}}^3$). The numerical errors and the space convergence rates of the PFEM based on the semi-implicit scheme at $t_n=1$ s are presented in Table 1 and Table 2. We observe the first order accuracy for $H^1$ errors of $\mathit{u},\mathit{B}$, and the second order accuracy asymptotically for $L^2$ errors of $\mathit{u},\mathit{B}$, which are consistent with our theoretical results. These convergence rates are consistent with the expected orders. Notice that $L^2$ errors of p has a faster convergence rate than the theoretical results.

Table 1.

The convergence rates of our scheme at $t_n=1$ s (2D).

h	$\parallel \mathit{u}-{\mathit{u}}_{\mathit{ϵ}\mathit{h}}^{\mathit{n}}\parallel$	Ratio	$\parallel \mathit{u}-{\mathit{u}}_{\mathit{ϵ}\mathit{h}}^{\mathit{n}}{\parallel }_1$	Ratio	$\parallel \mathit{p}-{\mathit{p}}_{\mathit{ϵ}\mathit{h}}^{\mathit{n}}\parallel$	Ratio	$\parallel \mathit{B}-{\mathit{B}}_{\mathit{ϵ}\mathit{h}}^{\mathit{n}}\parallel$	Ratio	$\parallel \mathit{B}-{\mathit{B}}_{\mathit{ϵ}\mathit{h}}^{\mathit{n}}{\parallel }_1$	Ratio
1/8	2.46 × 10${}^{-4}$		5.34 × 10${}^{-3}$		5.01 × 10${}^{-3}$		1.39 × 10${}^{-2}$		3.30 × 10${}^{-1}$
1/16	6.18 × 10${}^{-5}$	2.00	2.59 × 10${}^{-3}$	1.04	1.43 × 10${}^{-3}$	1.81	3.55 × 10${}^{-3}$	1.97	1.66 × 10${}^{-1}$	0.99
1/32	1.53 × 10${}^{-5}$	2.01	1.28 × 10${}^{-3}$	1.02	4.29 × 10${}^{-4}$	1.74	8.92 × 10${}^{-4}$	1.99	8.33 × 10${}^{-2}$	1.00
1/64	3.80 × 10${}^{-6}$	2.01	6.34 × 10${}^{-4}$	1.01	1.37 × 10${}^{-4}$	1.65	2.23 × 10${}^{-4}$	2.00	4.17 × 10${}^{-2}$	1.00
1/128	9.45 × 10${}^{-6}$	2.01	3.16 × 10${}^{-4}$	1.00	4.56 × 10${}^{-5}$	1.58	5.59 × 10${}^{-5}$	2.00	2.08 × 10${}^{-2}$	1.00

Table 2.

The convergence rates of our scheme at $t_n=1$ s (3D).

h	$\parallel \mathit{u}-{\mathit{u}}_{\mathit{ϵ}\mathit{h}}^{\mathit{n}}\parallel$	Ratio	$\parallel \mathit{u}-{\mathit{u}}_{\mathit{ϵ}\mathit{h}}^{\mathit{n}}{\parallel }_1$	Ratio	$\parallel \mathit{p}-{\mathit{p}}_{\mathit{ϵ}\mathit{h}}^{\mathit{n}}\parallel$	Ratio	$\parallel \mathit{B}-{\mathit{B}}_{\mathit{ϵ}\mathit{h}}^{\mathit{n}}\parallel$	Ratio	$\parallel \mathit{B}-{\mathit{B}}_{\mathit{ϵ}\mathit{h}}^{\mathit{n}}{\parallel }_1$	Ratio
1/4	2.41 × 10${}^{-2}$		2.49 × 10${}^{-1}$		9.05 × 10${}^{-2}$		1.53 × 10${}^{-3}$		2.44 × 10${}^{-2}$
1/8	6.11 × 10${}^{-3}$	1.98	1.26 × 10${}^{-1}$	0.99	2.95 × 10${}^{-2}$	1.61	3.60 × 10${}^{-4}$	2.08	1.21 × 10${}^{-2}$	1.01
1/12	2.72 × 10${}^{-3}$	2.00	8.38 × 10${}^{-2}$	1.00	1.46 × 10${}^{-2}$	1.73	1.58 × 10${}^{-4}$	2.03	8.02 × 10${}^{-3}$	1.01
1/16	1.53 × 10${}^{-3}$	2.00	6.29 × 10${}^{-2}$	1.00	8.84 × 10${}^{-3}$	1.75	8.83 × 10${}^{-5}$	2.02	6.01 × 10${}^{-3}$	1.00
1/20	9.80 × 10${}^{-4}$	2.00	5.03 × 10${}^{-2}$	1.00	6.00 × 10${}^{-3}$	1.74	5.64 × 10${}^{-5}$	2.01	4.81 × 10${}^{-3}$	1.00

7.2. Two-Sided Lid-Driven Square Cavity Flow

In this example, we test the 2D/3D two-sided driven cavity flow problem (cf. [33]). The problem we study is the incompressible viscous flow in a square cavity whose top and bottom walls move in the same (parallel) or opposite (antiparallel) direction. We take the initial values ${\mathit{u}}^0={\mathit{B}}^0=\mathbf{0}$ and the source terms $\mathit{f}=\mathit{g}=\mathbf{0}$. In the 2D case, we set a computational domain as $\mathcal{D}={[0,1]}^2$. The two boundary conditions are shown below:

$$\left\{\begin{array}{c}\mathit{u}=\mathbf{0},\hspace{1em}\mathrm{on}x=0,1,\hfill \\ \mathit{u}=(1,0),\hspace{1em}\mathrm{on}y=0,1,\hfill \\ \mathit{n}\times \mathit{B}=(1,0)\times \mathit{B}\hspace{1em}\mathrm{on}\Gamma ,\hfill \end{array}\right.\hspace{1em}\left\{\begin{array}{c}\mathit{u}=\mathbf{0},\hspace{1em}\mathrm{on}x=0,1,\hfill \\ \mathit{u}=(1,0),\hspace{1em}\mathrm{on}y=1,\hfill \\ \mathit{u}=(-1,0),\hspace{1em}\mathrm{on}y=0,\hfill \\ \mathit{n}\times \mathit{B}=(1,0)\times \mathit{B}\hspace{1em}\mathrm{on}\Gamma .\hfill \end{array}\right.$$

We set $h=\frac{1}{60}$, $\tau =\frac{1}{3600}$. First, we consider the upper and lower walls moving in the same direction at the same speed along the x-axis. Figure 1 shows the velocity streamlines for the fluid Reynolds number $R_e=1,100,1000$, the magnetic Reynolds number $R_m=1$, and the coupling coefficient $S=10$. It can be observed that the velocity streamlines are symmetric lines parallel to these walls and pass through the center of the cavity. With the increase of the fluid Reynolds number $R_e$, the centers of the two symmetric vortices move to the right, and the two symmetric vortices become four symmetric vortices. Figure 2 shows the velocity streamlines for the fluid Reynolds number $R_e=100$, the magnetic Reynolds number $R_m=1$ and the coupling coefficient $S=1,100,1000$. With the increase of coupling coefficient S, the two symmetric large vortices can split into more and more small vortices.

Figure 1.

The velocity streamlines of the upper and lower walls moving in the same direction for $R_e=1,100,1000,R_m=1,S=10$.

Figure 2.

The velocity streamlines of the upper and lower walls moving in the same direction for $R_e=100,R_m=1,S=1,100,1000$.

Then, we consider the upper and lower walls moving in the opposite direction at the same speed along the x-axis. Figure 3 gives the velocity streamlines for the fluid Reynolds number $R_e=1,100,1000$, the magnetic Reynolds number $R_m=1$ and the coupling coefficient $S=1$. We find that, with the increase of the fluid Reynolds number $R_e$, the centers of the two symmetric vortices shift to the upper right corner and the lower left corner, respectively. Figure 4 presents the velocity streamlines for the fluid Reynolds number $R_e=1$, the magnetic Reynolds number $R_m=1$ and the coupling coefficient $S=100,1000,10000$. It can be observed from the figure that, with the increase of coupling coefficient S, the two symmetric large vortices become four small vortices.

Figure 3.

The velocity streamlines of the upper and lower walls moving in the opposite direction for $R_e=1,100,1000,R_m=1,S=1$.

Figure 4.

The velocity streamlines of the upper and lower walls moving in the opposite direction for $R_e=1,R_m=1,S=100,1000,\mathrm{10,000}$.

In the 3D case, we set a calculational domain is $\mathcal{D}={[0,1]}^3$. The two boundary conditions are shown below:

$$\left\{\begin{array}{c}\mathit{u}=\mathbf{0},\hspace{1em}\mathrm{on}x=0,1,\hfill \\ \mathit{u}=(1,0,0),\hspace{1em}\mathrm{on}y=0,1,\hfill \\ \mathit{n}\times \mathit{B}=(1,0,0)\times \mathit{B}\hspace{1em}\mathrm{on}\Gamma ,\hfill \end{array}\right.\hspace{1em}\left\{\begin{array}{c}\mathit{u}=\mathbf{0},\hspace{1em}\mathrm{on}x=0,1,\hfill \\ \mathit{u}=(1,0,0),\hspace{1em}\mathrm{on}y=1,\hfill \\ \mathit{u}=(-1,0,0),\hspace{1em}\mathrm{on}y=0,\hfill \\ \mathit{n}\times \mathit{B}=(1,0,0)\times \mathit{B}\hspace{1em}\mathrm{on}\Gamma .\hfill \end{array}\right.$$

We set $h=\frac{1}{12}$, $\tau =\frac{1}{600}$. First, we consider the top and bottom walls moving in the same direction at the same speed. Figure 5 shows the velocity streamlines at plane $y=0.5$ for the fluid Reynolds number $R_e=1,100,500$, the magnetic Reynolds number $R_m=1$ and the coupling coefficient $S=1$. We find that, with the increase of the fluid Reynolds number $R_e$, the centers of the two symmetric vortices move to the right. Figure 6 shows the velocity streamlines at plane $y=0.5$ for the fluid Reynolds number $R_e=10$, the magnetic Reynolds number $R_m=1$ and the the coupling coefficient $S=1,100,500$. We find that the two symmetric large vortices can split into more and more small vortices.

Figure 5.

The velocity streamlines of the upper and lower walls moving in the same direction for $R_e=1,100,500,R_m=1,S=1$.

Figure 6.

The velocity streamlines of the upper and lower walls moving in the same direction for $R_e=10,R_m=1,S=1,100,500$.

Then, we consider the top and bottom walls moving in the opposite direction at the same speed. Figure 7 gives the velocity streamlines at plane $y=0.5$ for the fluid Reynolds number $R_e=1,100,500$, the magnetic Reynolds number $R_m=1$, and the coupling coefficient $S=1$. We can see the centers of the two symmetric vortices shift to the upper right corner and the lower left corner, respectively. Figure 8 presents the velocity streamlines at plane $y=0.5$ for the fluid Reynolds number $R_e=1$, the magnetic Reynolds number $R_m=1$, and the coupling coefficient $S=1,100,500$. With the increase of coupling coefficient S, the centers of the two symmetric vortices are slightly offset.

Figure 7.

The velocity streamlines of the upper and lower walls moving in the opposite direction for $R_e=1,100,500,R_m=1,S=1$.

Figure 8.

The velocity streamlines of the upper and lower walls moving in the opposite direction for $R_e=1,R_m=1,S=1,100,500$.

8. Conclusions

We present the fully discrete PFEM for the 2D/3D unsteady MHD equations in this paper. We introduce a penalty term to decouple the MHD equations into two small equations: one is the equations of velocity and magnetic field $(\mathit{u},\mathit{B})$, and the other is the equation of pressure p. Furthermore, we derive the error estimates for our scheme. Finally, two 2D/3D numerical experiments are given to verify the theoretical results.

Author Contributions

Conceptualization, H.S.; Data curation, K.S.; Investigation, H.S.; Methodology, K.S.; Software, K.S.; Validation, H.S.; Writing—original draft, K.S.; Writing—review & editing, X.F. All authors have read and agreed to the published version of the manuscript.

Conflicts of Interest

The authors declare no conflict of interest.

Funding Statement

This work is partly supported by the NSF of China (Nos. 12061076, 12126361, 11701493), the Scientific Research Plan of Universities in the Autonomous Region (No. XJEDU2020I 001), and the Key Laboratory Open Project of Xinjiang Province (No. 2020D04002).