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AIMS Neuroscience

2014, Volume 1, Issue 1: 1-3. doi: 10.3934/Neuroscience.2014.1.1
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Editorial

Debating New Theory in Neuroscience

  • 1.
    Department of Psychology at Bon Secours St. Francis Hospital, Greenville, SC.USA;
  • 2.
    Rutgers State University, Center for Computational & Integrative Biology, 315 Penn St, Camden, NJ 08102 USA

  • Citation: Robert A Moss, Joseph V Martin. Debating New Theory in Neuroscience[J]. AIMS Neuroscience, 2014, 1(1): 1-3. doi: 10.3934/Neuroscience.2014.1.1

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  • 1.   Introduction

    Magnetohydrodynamics(MHD) is a subject that studies the motion of conductive fluids in magnetic fields. It is mainly used in astrophysics, controlled thermonuclear reactions and industry. The system of MHD equations is a coupled equation system of Navier-Stokes equations of fluid mechanics and Maxwell's equations of electrodynamics (see[1,2]). In this paper, we consider the unsteady incompressible MHD equations as follows:

    $ {ut+(u⋅∇)u−ν△u+μB×curlB+∇p=f,x∈Ω,∇⋅u=0,x∈Ω,Bt+Rm−1curlcurlB−curl(u×B)=g,x∈Ω,∇⋅B=0,x∈Ω, $ (1.1)

    where $ u $ is velocity, $ \nu $ kinematic viscosity, $ \mu $ magnetic permeability, $ p $ hydrodynamic pressure, $ B $ magnetic field, $ f $ body force, and $ g $ applied current. The coefficients are the magnetic Reynolds number $ Rm $. As in [3], we assume that $ \Omega $ is a convex domain in $ R^d $, $ d = 2, 3 $ and $ t\in [0, T] $. Here, $ u_t = \frac{\partial u}{\partial t} $, $ B_t = \frac{\partial B}{\partial t} $. The term $ u\times B $ explains the effect of hydrodynamic flow on current. The magnetic Reynolds number $ Rm $ is moderate or low (from $ 10^{-2} $ to 1) (see[4]). For the purposes of simplicity, we choose $ Rm = 1 $ in our numerical analysis.

    System (1.1) is considered in conjunction with the following initial boundary conditions

    $ u(x,0)=u0(x),B(x,0)=B0(x),∀x∈Ω,u=0,B⋅n=0,n×curlB=0,∀x∈∂Ω,t>0, $

    where $ {\bf{n}} $ denotes the outer unit normal of $ \partial \Omega $.

    Over that last few decades, there has been a lot of works devoted on the numerical solution of MHD flows. In [5], some mathematical equations related to the MHD equations have been studied. In [6], a weighted regularization approach was applied to incompressible MHD problems. Reference [7,8] decoupled the linear MHD problem involving conducting and insulating regions. A mixed finite element method for stationary incompressible MHD equations was given in [9]. In order to avoid in-sup condition, Gerbeau [10] presented a stabilized finite element method and Salah et al. [11] proposed a Galerkin least-square method. In [12], a two-level finite element method with a stabilizing sub-grid for the incompressible MHD equations has been shown.

    Recently, a convergence analysis of three finite element iterative methods for 2D/3D stationary incompressible MHD equations has been given by Dong and He (see[13]). In order to continue the in-depth explore of three-dimensional incompressible MHD system, an unconditional convergence of the Euler semi-implicit scheme for the three-dimensional incompressible MHD equations was studied by He [14]. In [15], a two-level Newton iteration method for 2D/3D steady incompressible MHD equation was proposed by Dong and He. In addition, Yuksel and Ingram have discussed the full discretization of Grank-Nicolson scheme at small magnetic Reynolds numbers (see[16]). The defect correction finite element method for the MHD equations was shown by Si et. al. [17,18,19]. In [20], a decoupling penalty finite element method for the stationary incompressible MHD equation was presented. To further study this method, we can refer to [21,22,23].

    In this paper, a modified characteristics projection finite element method for the unsteady incompressible magnetohydrodynamics(MHD) equations will be given. In this method, modified characteristics finite element method and the projection method will be combined for solving the incompressible MHD equations. Both the stability and the optimal error estimates both in $ L^2 $ and $ H^1 $ norms for the modified characteristics projection finite element method will be derived. In order to demonstrate the effectiveness of our method, we will present some numerical results at the end of the manuscript. The contents of this paper are divided into sections as follows. To obtain the unconditional stability and optimal error estimates, in section 2, we introduce some notations and present a modified characteristics projection algorithm for the MHD equations. Then, in section 3, we give stability and error analysis and show that this method has optimal convergence order by mathematical induction. In order to demonstrate the effectiveness of our method, some numerical results are presented in the last section.

    2.   Notation and preliminaries

    In this section, we introduce some notations. We employ the standard Sobolev spaces $ H^{k}(\Omega) = W^{k, 2} $, for nonnegative $ k $ with norm $ \|v\|_{k} = (\sum_{|\beta| = 0}^{k}\|D^{\beta}v\|_{0}^{2})^{1/2} $. For vector-value function, we use the Sobolev spaces $ \mbox{H}^{k}(\Omega) = (H^{k}(\Omega))^{d} $ with norm $ \|\mbox{v}\|_{k} = (\sum_{i = 1}^{d}\|v_{i}\|_{k}^{2})^{1/2} $(see [24,25]). For simplicity, we set:

    $ X=H10(Ω)2,X0={v∈X:∇⋅v=0},M=L20(Ω)={q∈L2(Ω)d:∫Ωq(x)dx=0},H={v∈L2(Ω)d:∇⋅v=0},W={ψ∈(H1(Ω))d:ψ⋅n|∂Ω=0},W0={ψ∈W:∇⋅ψ=0}, $

    which are equipped with the norms

    $ ‖v‖X=‖∇v‖0,‖q‖M=‖q‖0,‖ψ‖W=‖ψ‖1. $

    In this paper, we will use the following equality

    $ curl(ψ×B)=B⋅∇ψ−ψ⋅∇B,∀ψ,B∈W0. $

    Then, we recall the following Poincaré-Friedrichs inequality in $ W $: there holds

    $ ‖c‖0≤C‖curlc‖0,∀c∈W, $

    with a constant $ C > 0 $ solely depending on the domain $ \Omega $; (see[26]). Here, we define the following trilinear form $ a_1(\cdot, \cdot, \cdot) $ as follows, for all $ u\in V $, $ w, v\in X $,

    $ a1(u,v,w)=(u⋅∇w+12(∇⋅u)w,v)=12(u⋅∇w,v)−12(u⋅∇v,w) $

    and the properties of the trilinear form $ a_1(\cdot, \cdot, \cdot) $ [27] are helpful in our analysis

    $ a1(u,v,w)=−a1(u,w,v),  u∈X0,v,w∈X,|a1(u,v,w)|≤N2‖u‖0(‖∇v‖0‖w‖L∞+‖v‖L6‖∇w‖L3),  u∈L2(Ω)2, v∈X,w∈L∞(Ω)2∩X,|a1(u,v,w)|≤N2(‖u‖L∞‖∇v‖0+‖∇u‖L3‖v‖L6)‖w‖0,  u∈L∞(Ω)2∩X, v∈X,w∈L2(Ω)2,‖v‖0≤γ0‖∇v‖0,‖v‖L6≤C‖∇v‖0,‖w‖L4≤C0‖∇w‖0,  ∀w∈X, $

    where $ N > 0 $ is a constant, $ \gamma_0 $ (only dependent on $ \Omega $) is a positive constant, and $ C_0 $ (only dependent on $ \Omega $) is an embedding constant of $ H^{1}(\Omega)\hookrightarrow L^{4}(\Omega) $ ($ \hookrightarrow $ denotes the continuous embedding).

    Furthermore, we define the Stokes operator $ \mathcal{A}:D(\mathcal{A})\longrightarrow \tilde{H} $ such that $ \mathcal{A} = -P\triangle $, where $ D(\mathcal{A}) = H^{2}(\Omega)^{d}\cap X_{0} $ and $ P $ is an $ L^2 $-projection from $ L^2(\Omega)^d $ to $ \tilde{H} = \{v\in L^2(\Omega)^d; \nabla\cdot v = 0, v\cdot n|_{\partial\Omega} = 0\} $.

    Next, we recall a discrete version of Gronwall's inequality established in [14].

    Lemma 2.1 Let $ C_{0}, a_n, b_n $ and $ d_n $, for integers $ n\geq0 $, be the non-negative numbers such that

    $ am+τm∑n=1bn≤τm−1∑n=1dnan+C0   m≥1. $ (2.1)

    Then

    $ am+τm∑n=1bn≤C0exp(τm−1∑n=1dn)   m≥1. $ (2.2)

    Throughout this paper, we make the following assumptions on the prescribed date for the problem (1.1), which satisfies the regularity of the data needed for our main results.

    Assumption (A1)(see[27,28]): Assume that the boundary of $ \Omega $ is smooth so that the unique solution $ (v, q) $ of the steady Stokes problem

    $ −Δv+∇q=f,∇⋅v=0inΩ,v|∂Ω=0, $

    for prescribed $ f\in L^{2}(\Omega)^{3} $ satisfies

    $ ‖v‖2+‖q‖1≤c‖f‖0, $

    and Maxwell's equations

    $ curlcurlB=g,∇⋅B=0inΩ,n×curlB=0,B⋅∇nonΩ, $

    for the prescribed $ g\in L^{2}(\Omega)^{3} $ admits a unique solution $ B\in W_{0} $ which satisfies

    $ ‖B‖2≤c‖g‖0. $

    Assumption (A2): The initial data $ u_0, B_0, f $ satisfy

    $ ‖Au0‖0+‖B0‖2+sup0≤t≤T‖f‖0+‖ft‖0≤C. $ (2.3)

    Assumption (A3): We assume that the MHD problem admits a unique local strong solution $ (u, p, B) $ on $ (0, T) $ such that

    $ sup0≤t≤T‖Aut‖0+‖Bt‖2+‖p‖1+‖ut‖0+‖Bt‖0+∫T0(‖ut‖21+‖Bt‖21+‖pt‖21+‖utt‖20+‖Btt‖20)dt≤C. $ (2.4)

    Let $ 0 = t_{0} < t_{1} < \cdots < t_{N} = T $ be a uniform partition of the time interval $ [0, T] $ with $ t_{n} = n\tau $, and $ N $ being a positive integer. Set $ u^{n} = (\cdot, t_{n}) $, For any sequence of functions $ \{f^{n}\}_{n = 0}^{N} $, we define

    $ Dτfn=fn−fn−1τ. $

    At the initial time step, we start with $ U^{0} = \tilde{U}^{0} = u_{0}, B^{0} = \tilde{B}^{0} = b_{0} $ and an arbitrary $ P^{0}\in M\cap H^{1}(\Omega) $. Then, we can give

    $ ∇P0=f0−u0⋅∇u0+νΔu0−μB0×curlB0. $

    Then, we give the algorithm as follows

    Algorithm 2.1 (Time-discrete modified characteristics projection finite element method)

    Step 1. Solve $ B^{n+1} $, for instance

    $ Bn+1−ˆBnτ+curlcurlBn+1−(Bn⋅∇)Un=gn+1, $ (2.5)
    $ Bn+1⋅n=0,n×curlBn+1=0, on ∂Ω, $ (2.6)

    where

    $ ˆln={ln(ˆx),   ˆx=x−Unτ∈Ω,0,   othrewise. $

    Step 2. Find $ \tilde{U}^{n+1} $, such that

    $ ˜Un+1−ˆUnτ−νΔ˜Un+1+∇Pn+μBn×curlBn+1=fn+1, $ (2.7)
    $ ˜Un+1=0, on ∂Ω. $ (2.8)

    Step 3. Calculate $ (U^{n+1}, P^{n+1}) $, for example

    $ Un+1−˜Un+1τ+∇(Pn+1−Pn)=0, $ (2.9)
    $ ∇⋅Un+1=0, $ (2.10)
    $ Un+1=0 on ∂Ω,. $ (2.11)

    The corresponding weak form is

    Step 1. Solve $ B^{n+1} $, for instance

    $ (Bn+1−ˆBnτ,ψ)+(curlBn+1,curlψ)−((Bn⋅∇)Un,ψ)=(gn+1,ψ),   ∀ψ∈W. $ (2.12)

    Step 2. Find $ \tilde{U}^{n+1} $, such that

    $ (˜Un+1−ˆUnτ,v)+ν(∇˜Un+1,∇v)+(∇Pn,v)+μ(Bn⋅∇v,Bn+1)−μ(v⋅∇Bn,Bn)=(fn+1,v),   ∀v∈X, $ (2.13)

    Step 3. Calculate $ (U^{n+1}, P^{n+1}) $, for example

    $ (Un+1−˜Un+1τ,v)+(∇(Pn+1−Pn),v)=0,   ∀v∈X, $ (2.14)
    $ (∇⋅Un+1,ϕ)=0,   ∀ϕ∈M. $ (2.15)

    Remark 2.1 As we all know $ (B^{n}\times \mbox{curl}B^{n+1}, v) = (B^{n}\cdot\nabla v, B^{n+1})-(v\cdot\nabla B^{n}, B^{n+1}) $. In order to improve our algorithm, we change it as $ (B^{n}\cdot\nabla v, B^{n+1})-(v\cdot\nabla B^{n}, B^{n}) $.

    We denote $ T_{h} $ be a regular and quasi-uniform partition of the domain $ \Omega $ into the triangles for $ d = 2 $ or tetrahedra for $ d = 3 $ with diameters by a real positive parameter $ h(h\rightarrow 0) $. The finite element pair $ (X_{h}, M_{h}, W_{h}) $ is constructed based on $ T_{h} $.

    Let $ W_{0n}^h = X_h\times W_h $. There exits a constant $ \beta_0 > 0 $ (only dependent on $ \Omega $) such that

    $ sup0≠(v,ψ)∈Wh0nd(v,q)‖∇v‖0≥β0‖q‖0,  ∀q∈Mh. $ (2.16)

    There exits a mapping $ r_h\in \mathcal{L}(H^2(\Omega)^2\cap V, X_h) $ satisfying

    $ (∇⋅(v−rhv),q)=0,‖∇(v−rhv)‖0≤Ch‖v‖2,  ∀v∈H2(Ω)2∩V,  q∈Mh, $

    and a $ L^2 $-projection operator $ \rho_h:M\rightarrow M_h $ satisfying

    $ ‖q−ρhq‖0≤Ch‖q‖1,  ∀q∈H1(Ω)∩M, $

    and a mapping $ R_h\in \mathcal{L}(H^2(\Omega)^2\cap V_n, W_h) $ satisfying

    $ (∇×RhΦ,∇×Ψ)+(∇⋅RhΦ,∇⋅Ψ)=(∇×Φ,∇×Ψ)+(∇⋅Φ,∇Ψ)=(∇×Φ,∇×Ψ),  ∀Ψ∈Wh,‖Φ−RhΦ‖0+h‖Φ−RhΦ‖1≤Ch2‖Φ‖2,  ∀H2(Ω)∩Vh. $

    Here, we define the discrete Stokes operator $ A_{h} = -P_h\triangle_h $ and $ \triangle_h $ (see [15] and the references therein) defined by

    $ −(△huh,vh)=(∇uh,∇vh),∀uh,vh∈Vh. $

    Meanwhile, we define the discrete operator $ A_{1h} B_h = R_{1h}(\nabla_h\times \nabla\times B_h+\nabla_h\nabla\cdot B_h)\in W_h $ (see [15]) as follows

    $ (A1hBh,Ψ)=(A121hBh,A121hBh)=(∇×Bh,∇×Ψ)+(∇⋅Bh,∇⋅Ψ),∀Bh,Ψ∈Wh. $

    We define the Stokes projection $ (R_{h}(u, p), Q_{h}(u, P)):(X, M)\rightarrow (X_{h}, M_{h}) $ by

    $ ν(Rh(u,p)−u,∇vh)−(Qh(u,p)−p,∇⋅vh)=0, $ (2.17)
    $ (∇⋅(Rh(u,p)−u),ϕh)=0. $ (2.18)

    By classical FEM theory ([25,29,30]), we have the following results.

    Lemma 2.2 Assume that $ u \in H_{0}^{1}(\Omega)^{d}\cap H^{r+1}(\Omega)^{d} $ and $ p \in L_{0}^{2}(\Omega)\cap H^{r}(\Omega) $. Then,

    $ ‖Rh(u,p)−u‖0+h(‖∇(Rh(u,p)−u)‖0+‖Qh(u,p)−p‖0)≤Chr+1(‖u‖r+1+‖p‖r), $ (2.19)

    and

    $ ‖Rh(u,p)‖L∞≤C(‖u‖2+‖p‖1). $ (2.20)

    Lemma 2.3 If $ (u, p)\in W^{2, k}(\Omega)^{d}\times W^{1, k}(\Omega) $ for $ k > d $,

    $ ‖∇Rh(u,p)‖L∞≤C(‖u‖W1,∞+‖p‖L∞). $ (2.21)

    Algorithm 2.2 (The fully discrete modified characteristics projection finite element method)

    Step 1. Solve $ B_{h}^{n+1} $, for instance

    $ (Bn+1h−ˆBnhτ,ψh)+(curlBn+1h,curlψh)−((Bnh⋅∇)unh,ψh)=(gn+1,ψh),   ∀ψh∈Wh. $ (2.22)

    where

    $ ˆlnh={lnh(ˆx),   ˆx=x−unhτ∈Ω,0,   othrewise. $

    Step 2. Find $ \tilde{U}^{n+1} $, such that

    $ (˜un+1h−ˆunhτ,vh)+ν(∇˜un+1h,∇vh)+(∇pnh,vh)+μ(Bnh⋅∇vh,Bn+1h)−μ(vh⋅∇Bnh,Bnh)=(fn+1,vh),   ∀vh∈Xh. $ (2.23)

    Step 3. Calculate $ (u_{h}^{n+1}, p_{h}^{n+1}) $, for example

    $ (un+1h−˜un+1hτ,vh)+(∇(pn+1h−pnh),vh)=0,   ∀vh∈Xh, $ (2.24)
    $ (∇⋅un+1h,ϕh)=0,   ∀ϕh∈Mh. $ (2.25)

    3.   Unconditional stability and convergence

    3.1.   Error estimation for the time-discrete method

    Here, we make use of the following notation:

    $ en=un−Un,˜e=un−˜Un,n=0,1,...,N,εn=bn−Bn,ˆεn=bn−ˆBn,n=0,1,...,N,ξn=pn−Pn,n=0,1,...,N. $

    Lemma 3.1 (see[31]). Assume that $ g_{1}, g_{2}, \rho $ are three functions defined in $ \Omega $ and $ \rho|_{\partial\Omega} = 0 $. If

    $ τ(‖g1‖W1,∞+‖g2‖W1,∞)≤12, $

    then

    $ ‖ρ(x−g1(x)τ)−ρ(x−g2(x)τ)‖Lq≤Cτ‖ρ‖W1,q2‖g1−g2‖Lq1,‖ρ(x−g1(x)τ)−ρ(x−g2(x)τ)‖−1≤Cτ‖ρ‖0‖g1−g2‖W1,∞, $

    where $ 1/q_{1}+1/q_{2} = 1/q, 1 < q < \infty $.

    The following theorem gives insight on the error estimate between the time-discrete solution and the solution of the time-dependent MHD system (1.1).

    Theorem 3.1 Suppose that assumption A2–A3 are satisfied, there exists a positive $ C > 0 $ such that

    $ max0≤n≤m(‖em+1‖20+μ‖εm+1‖20)+m∑n=0(‖en+1−en‖20+μ‖εn+1−εn‖20),+τm∑n=0(ν‖∇˜en+1‖20+μ‖curlεn+1‖20)≤Cτ2,max0≤n≤m(ν‖em+1‖21+μ‖curlεm+1‖20)+τm∑n=1(ν‖en+1−en‖21+μ‖curlεn+1−curlεn‖20)+τm∑n=1(‖Dτen+1‖20+μ‖Dτεn+1‖20)≤Cτ2,τm∑n=0(‖en+1‖22+‖˜en+1‖22+‖εn+1‖22)≤Cτ2,τm∑n=0(‖DτUn+1‖22+‖Dτ˜Un+1‖22+‖DτPn+1‖21+‖DτBn+1‖22+‖Un+1‖2W2,d∗+‖Bn+1‖2W2,d∗)+max0≤n≤m(‖Un+1‖2+‖˜Un+1‖2+‖Pn+1‖1+‖Bn+1‖2)≤C. $

    Proof. Firstly, we prove the following inequality

    $ ‖em‖2+τ3/4‖Um‖W2,d∗≤1, $ (3.1)
    $ ‖εm‖2+τ3/4‖Bm‖W2,d∗≤1, $ (3.2)

    by mathematical induction for $ m = 0, 1, ..., N $.

    Since $ U^{0} = u^{0}, B^{0} = b^{0} $ the above inequality hold for $ m = 0 $.

    We assume that (3.1) and (3.2) hold for $ m\leq n $ for some integer $ n\geq0 $. By Sobolev embedding theory, we get

    $ ‖Um‖L∞≤‖um‖L∞+C‖em‖2≤C, $ (3.3)
    $ ‖Bm‖L∞≤‖bm‖L∞+C‖εm‖2≤C, $ (3.4)

    and

    $ τ‖Um‖W1,∞≤1/4, $ (3.5)
    $ τ‖Bm‖W1,∞≤1/4, $ (3.6)

    when $ \tau\leq\tau_{1} $ for some positive constant $ \tau_{1} $.

    To prove (3.1) and (3.2) for $ m = n+1 $, we rewrite (1.1) by

    $ Dτun+1−νΔun+1+∇pn+1=fn+1−un−ˉunτ+Rn+1tr1−μbn+1×curlbn+1, $ (3.7)
    $ ∇⋅un+1=0, $ (3.8)
    $ Dτbn+1+curlcurlbn+1=gn+1−bn−ˉbnτ+Rn+1tr2+bn+1⋅∇un+1. $ (3.9)

    where $ u^{n+1} = U(t_{n+1}), b^{n+1} = B(t_{n+1}) $, and

    $ ˉln={ln(ˉx),ˉx=x−unτ∈Ω,0,otherwise. $

    and

    $ Rn+1tr1=un+1−ˉunτ−un+1t−(un+1⋅∇)un+1,Rn+1tr2=bn+1−ˉbnτ−bn+1t−(un+1⋅∇)bn+1, $

    define the truncation error. We can refer to ([31]), there holds that

    $ τN∑m=1‖Rmtr1‖20≤Cτ2,τN∑m=1‖Rmtr2‖20≤Cτ2. $ (3.10)

    The corresponding weak form is

    $ (Dτun+1,v)+ν(∇un+1,∇v)−(pn+1,∇⋅v)=(fn+1,v)−(un−ˉunτ,v)+(Rn+1tr1,v)−μ(bn+1×curlbn+1,v), $ (3.11)
    $ (∇⋅un+1,ϕ)=0, $ (3.12)
    $ (Dτbn+1,ψ)+(curlbn+1,curlψ)=(gn+1,ψ)−(bn−ˉbnτ,ψ)+(Rn+1tr2,ψ)+(bn+1⋅∇(un+1−un),ψ)+(bn+1⋅∇un,ψ). $ (3.13)

    Combining (2.13) and (2.14), we obtain

    $ (Un+1−ˆUnτ,v)+ν(∇˜Un+1,∇v)−(Pn+1,∇⋅v)+μ(Bn⋅∇v,Bn+1)−μ(v⋅∇Bn,Bn)=(fn+1,v). $ (3.14)

    Then, we rewrite (3.14) and (2.12) as

    $ (DτUn+1,v)+ν(∇˜Un+1,∇v)−(Pn+1,∇⋅v)=(fn+1,v)−(Un−ˆUnτ,v)−μ(Bn⋅∇v,Bn+1)+μ(v⋅∇Bn,Bn), $ (3.15)

    and

    $ (DτBn+1,ψ)+(curlBn+1,curlψ)=(gn+1,ψ)+(Bn⋅∇Un,ψ)−(Bn−ˆBnτ,ψ). $ (3.16)

    Subtracting (3.15) and (3.16) from (3.11) and (3.13), respectively, leads to

    $ (Dτen+1,v)+ν(∇˜en+1,∇v)−(ξn+1,∇⋅v)=−(un−ˉun−(Un−ˆUn)τ,v)+(Rn+1tr1,v)−μ((bn+1−bn)⋅∇v−v⋅∇(bn+1−bn),bn+1)+μ(v⋅∇bn,bn+1−bn)+μ(v⋅∇εn,bn)+μ(v⋅∇Bn,εn)−μ(εn⋅∇v,bn+1)−μ(Bn⋅∇v,εn+1), $ (3.17)

    and

    $ (Dτεn+1,ψ)+(curlεn+1,curlψ)=−(bn−ˉbn−(Bn−ˆBn)τ,ψ)+(Rn+1tr2,ψ)+(bn+1⋅∇(un+1−un)+(bn+1−bn)⋅∇un+εn⋅∇un+Bn⋅∇en,ψ). $ (3.18)

    Testing (2.9) by $ \tau e^{n+1} $, since $ \nabla\cdot e^{n+1} = 0 $, we arrive at

    $ (en+1,˜en+1)=(en+1,en+1). $ (3.19)

    Testing (2.9) by $ \tau e^{n} $, we get

    $ (en,˜en+1)=(en,en+1). $ (3.20)

    Subtracting (3.20) from (3.19), we can obtain

    $ (en+1−en,˜en+1)=(en+1−en,en+1). $ (3.21)

    Let $ v = 2\tau \tilde{e}^{n+1} $ in (3.17), we obtain

    $ ‖en+1‖20−‖en‖20+‖en+1−en‖20+2ντ‖∇˜en+1‖20−2τ(ξn+1,∇⋅˜en+1)=−2((en−(ˉun−ˆUn),˜en+1)+2τ(Rn+1tr1,˜en+1)−2μτ((bn+1−bn)⋅∇˜en+1−˜en+1⋅∇(bn+1−bn),bn)+2μτ(˜en+1⋅∇bn,bn+1−bn)+2μτ(˜en+1⋅∇εn,bn)+2μτ(˜en+1⋅∇Bn,εn)−2μτ(εn⋅∇˜en+1,bn+1)−2μτ(Bn⋅∇˜en+1,εn+1). $ (3.22)

    Let $ \psi = 2\mu\tau \varepsilon^{n+1} $ in (3.18), we get

    $ μ‖εn+1‖20−μ‖εn‖20+μ‖εn+1−εn‖20+2μτ‖curlεn+1‖20=−2μ(εn−(ˉbn−ˆBn),εn+1)+2μτ(Rn+1tr2,εn+1)+2μτ(bn+1⋅∇(un+1−un)+(bn+1−bn)⋅∇un+εn⋅∇un+Bn⋅∇en,εn+1). $ (3.23)

    Taking sum of (3.22) and (3.23) yields

    $ ‖en+1‖20−‖en‖20+μ‖εn+1‖20−μ‖εn‖20+‖en+1−en‖20+μ‖εn+1−εn‖20+2ντ‖∇˜en+1‖20+2μτ‖curlεn+1‖20−2τ(ξn+1,∇⋅˜en+1)=−2(en−ˆen,˜en+1)−2(ˆun−ˉun,˜en+1)+2τ(Rn+1tr1,˜en+1)−2μτ((bn+1−bn)⋅∇˜en+1,bn)+2μτ(˜en+1⋅∇(bn+1−bn),bn)+2μτ(˜en+1⋅∇bn,bn+1−bn)+2μτ(˜en+1⋅∇εn,bn)+2μτ(˜en+1⋅∇Bn,εn)−2μτ(εn⋅∇˜en+1,bn+1)−2μτ(Bn⋅∇˜en+1,εn+1)−2μ(εn−ˆεn,εn+1)−2μ(ˆbn−ˉbn,εn+1)+2μτ(Rn+1tr2,εn+1)+2μτ(bn+1⋅∇(un+1−un),εn+1)+2μτ((bn+1−bn)⋅∇un,εn+1)+2μτ(εn⋅∇unεn+1)+2μτ(Bn⋅∇en,εn+1). $ (3.24)

    From (2.9), we have

    $ ˜en+1=en+1+τ∇(ξn+1−ξn)−τ∇(pn+1−pn). $

    Then, we can obtain

    $ −2τ(ξn+1,∇⋅˜en+1)=−2τ(ξn+1,∇⋅en+1)+2τ2(∇ξn+1,∇(ξn+1−ξn))−4τ2(∇ξn+1,∇(pn+1−pn))=τ2(‖∇ξn+1‖20−‖∇ξn‖20+‖∇ξn+1−∇ξn‖20)−2τ2(∇ξn+1,∇(pn+1−pn))≥τ2(‖∇ξn+1‖20−‖∇ξn‖20+‖∇(ξn+1−ξn)‖20)−Cτ2‖∇ξn+1‖0‖∇(pn+1−pn)‖0=τ2(‖∇ξn+1‖20−‖∇ξn‖20+‖∇(ξn+1−ξn)‖20)−τ(C‖τ∇ξn+1‖0‖∇(pn+1−pn)‖0)≥τ2(‖∇ξn+1‖20−‖∇ξn‖20+‖∇(ξn+1−ξn)‖20)−τ(Cτ2‖∇ξn+1‖20+14‖∇(pn+1−pn)‖20)=τ2(‖∇ξn+1‖20−‖∇ξn‖20+‖∇(ξn+1−ξn)‖20)−Cτ3‖∇ξn+1‖20+14‖∇(pn+1−pn)‖20). $

    On the other hand, using Lemma 3.1, we deduce that

    $ 2|(en−ˆen,˜en+1)|≤2‖en−ˆen‖−1‖∇˜en+1‖0≤Cτ‖en‖0‖Un‖W1,∞‖∇˜en+1‖0≤ντ12‖∇˜en+1‖20+Cτ‖en‖20,2|(ˆun−ˉun,˜en+1)|≤C‖ˆun−ˉun‖L6/5‖˜en+1‖L6≤Cτ‖∇un‖L3‖en‖0‖∇˜en+1‖0≤ντ12‖∇˜en+1‖20+Cτ‖en‖20,2τ|(Rn+1tr1,˜en+1)|≤‖Rn+1tr1‖0‖˜en+1‖0≤ντ12‖∇˜en+1‖20+Cτ‖Rn+1tr1‖20,2μτ|((bn+1−bn)⋅∇˜en+1,bn+1)|≤Cτ‖bn+1−bn‖0‖∇˜en+1‖0‖bn+1‖2≤ντ12‖∇˜en+1‖20+Cτ3‖bn+1t‖20‖bn+1‖22,2μτ|(˜en+1⋅∇(bn+1−bn),bn+1)|≤Cτ‖˜en+1‖L6‖∇bn+1‖L3‖bn+1−bn‖0≤ντ12‖∇˜en+1‖20+Cτ3‖bn+1t‖20‖bn+1‖22,2μτ|(˜en+1⋅∇bn,bn+1−bn)|≤Cτ‖˜en+1‖L6‖∇bn+1‖L3‖bn+1−bn‖0≤ντ12‖∇˜en+1‖20+Cτ3‖bn+1t‖20‖bn+1‖22,2μτ|(˜en+1⋅∇εn,bn)|≤Cτ‖∇˜en+1‖0‖bn‖2‖εn‖0≤ντ12‖∇˜en+1‖20+Cτ‖εn‖20‖bn‖22,2μτ|(˜en+1⋅∇Bn,εn)|≤Cτ‖∇˜en+1‖0‖Bn‖L∞‖εn‖0≤ντ12‖∇˜en+1‖20+Cτ‖εn‖20‖Bn‖2L∞,2μτ|(εn⋅∇˜en+1,bn+1)|≤Cτ‖εn‖0‖∇˜en+1‖0‖bn+1‖L∞≤ντ12‖∇˜en+1‖20+Cτ‖εn‖20‖bn+1‖22,2μτ|(Bn⋅∇˜en+1,εn+1)|≤Cτ‖Bn‖L∞‖∇˜en+1‖0‖εn+1‖0≤ντ12‖∇˜en+1‖20+Cτ‖εn‖20,2μτ|(εn−ˆεn,εn+1)|≤Cτ‖εn‖0‖Bn‖W1,∞‖∇εn+1‖0≤μτ8‖curlεn+1‖20+Cτ‖εn‖20,2μτ|(ˆbn−ˉbn,εn+1)|≤Cτ‖ˆbn−ˉbn‖L6/5‖εn+1‖L6≤Cτ‖∇bn‖L3‖εn‖0‖∇εn+1‖0≤μτ8‖curlεn+1‖20+Cτ‖εn‖20,2μτ|(Rn+1tr2,εn+1)|≤Cτ‖Rn+1tr2‖0‖εn+1‖0≤μτ8‖curlεn+1‖20+Cτ‖Rn+1tr2‖20,2μτ|(bn+1⋅∇(un+1−un),εn+1)|≤Cτ‖bn+1‖2‖un+1−un‖0‖curlεn+1‖0≤μτ8‖curlεn+1‖20+Cτ‖un+1−un‖20,2μτ|((bn+1−bn)⋅∇un,εn+1)|≤Cτ‖bn+1−bn‖0‖∇un‖L∞‖εn+1‖0≤μτ8‖curlεn+1‖20+Cτ3‖bn+1t‖20‖∇un‖2L∞,2μτ|(εn⋅∇un,εn+1)|≤Cτ‖εn‖0‖∇un‖L3‖curlεn+1‖0≤μτ8‖curlεn+1‖20+Cτ‖εn‖20‖∇un‖2L3,2μτ|(Bn⋅∇en,εn+1)|≤Cτ‖Bn‖L∞‖en‖0‖curlεn+1‖0≤μτ8‖curlεn+1‖20+Cτ‖en‖20. $

    Substituting these above inequality into (3.24), we obtain

    $ ‖en+1‖20−‖en‖20+μ‖εn+1‖20−μ‖εn‖20+τ2‖∇ξn+1‖20−τ2‖∇ξn‖20+‖en+1−en‖20+μ‖εn+1−εn‖20+τ2‖∇(ξn+1−ξn)‖20+ντ‖∇˜en+1‖20+μτ‖curlεn+1‖20≤Cτ(‖en‖20+‖εn‖20+‖εn+1‖20)+Cτ(‖Rn+1tr1‖20+‖Rn+1tr2‖20)+Cτ3(‖∇ξn+1‖20+‖∇ξn‖20)+Cτ3. $ (3.25)

    Taking sum of the (3.25) for $ n $ from $ 0 $ to $ m\leq N $ and using discrete Gronwall's inequality Lemma 2.1, we get

    $ max0≤n≤m(‖em+1‖20+μ‖εm+1‖20+τ2‖∇ξm+1‖20)+m∑n=0(‖en+1−en‖20+μ‖εn+1−εn‖20)+τm∑n=0(ν‖∇˜en+1‖20+μ‖curlεn+1‖20)≤Cτ2. $ (3.26)

    Owing to (3.21), we have

    $ (∇(en+1−en),∇˜en+1)=(∇(en+1−en),∇en+1). $ (3.27)

    To acquire the $ H^{1} $ estimates, we take $ v = 2\tau D_{\tau}e^{n+1} $ in (3.17) to get

    $ ν(‖en+1‖21−‖en‖21+‖en+1−en‖21)+2τ‖Dτen+1‖20=−2((en−(ˉun−ˆUn),Dτen+1)+2τ(Rn+1tr1,Dτen+1)−2μτ((bn+1−bn)⋅∇Dτen+1−Dτen+1⋅∇(bn+1−bn),bn+1)+2μτ(Dτen+1⋅∇bn,bn+1−bn)+2μτ(Dτen+1⋅∇εn,bn)+2μτ(Dτen+1⋅∇Bn,εn)−2μτ(εn⋅∇Dτen+1,bn+1)−2μτ(Bn⋅∇Dτen+1,εn+1). $ (3.28)

    Then, we take $ \psi = 2\mu\tau D_{\tau}\varepsilon^{n+1} $ in (3.18) to get

    $ μ(‖curlεn+1‖20−‖curlεn‖20+‖curlεn+1−curlεn‖20)+2μτ‖Dτεn+1‖20=−2μ(εn−(ˉbn−ˆBn),Dτεn+1)+2μτ(Rn+1tr2,Dτεn+1)+2μτ(bn+1⋅∇(un+1−un)+(bn+1−bn)⋅∇un+εn⋅∇un+Bn⋅∇en,Dτεn+1). $ (3.29)

    Combining (3.28) and (3.29), we obtain

    $ ν(‖en+1‖21−‖en‖21+‖en+1−en‖21)+μ(‖curlεn+1‖20−‖curlεn‖20+‖curlεn+1−curlεn‖20)+2τ‖Dτen+1‖20+2ντ‖Dτεn+1‖20=−2(en−ˆen,Dτen+1)−2(ˆun−ˉun,Dτen+1)+2τ(Rn+1tr1,Dτen+1)−2μτ((bn+1−bn)⋅∇Dτen+1,bn+1)+2μτ(Dτen+1⋅∇(bn+1−bn),bn+1)+2μτ(Dτen+1⋅∇bn,bn+1−bn)+2μτ(Dτen+1⋅∇εn,bn)+2μτ(Dτen+1⋅∇Bn,εn)−2μτ(εn⋅∇Dτen+1,bn+1)−2μτ(Bn⋅∇Dτen+1,εn+1)−2μ(εn−ˆεn,εn+1)−2μ(ˆbn−ˉbn,Dτεn+1)+2μτ(Rn+1tr2,Dτεn+1)+2μτ(bn+1⋅∇(un+1−un),Dτεn+1)+2μτ((bn+1−bn)⋅∇un,Dτεn+1)+2μτ(εn⋅∇un,Dτεn+1)+2μτ(Bn⋅∇en,Dτεn+1). $ (3.30)

    Similarly, by Lemma 3.1, we have

    $ 2|(en−ˆen,Dτen+1)|≤2‖en‖W1,2‖Un‖L∞‖Dτen+1‖0≤τ16‖Dτen+1‖20+Cτ‖en‖21,2|(ˆun−ˉun,Dτen+1)|≤C‖un‖W1,4‖en‖L4‖Dτen+1‖0≤C‖un‖W1,4‖en‖1‖Dτen+1‖0≤τ16‖Dτen+1‖20+Cτ‖en‖21,2τ|(Rn+1tr1,Dτen+1)|≤‖Rn+1tr1‖0‖Dτen+1‖0≤τ16‖Dτen+1‖20+Cτ‖Rn+1tr1‖20,2μτ|((bn+1−bn)⋅∇Dτen+1,bn+1)|≤Cτ‖∇(bn+1−bn)‖0‖Dτen+1‖0‖bn+1‖2≤τ16‖Dτen+1‖20+Cτ3‖∇bn+1t‖20‖bn+1‖22,2μτ|(Dτen+1⋅∇(bn+1−bn),bn+1)|≤Cτ‖Dτen+1‖0‖∇(bn+1−bn)‖0‖bn+1‖2≤τ16‖Dτen+1‖20+Cτ3‖bn+1t‖20‖bn+1‖22,2μτ|(Dτen+1⋅∇bn,bn+1−bn)|≤Cτ‖Dτen+1‖0‖∇bn‖L∞‖bn+1−bn‖0≤τ16‖Dτen+1‖20+Cτ3‖∇bn+1t‖20‖∇bn‖2L∞,2μτ|(Dτen+1⋅∇εn,bn)|≤Cτ‖Dτen+1‖0‖∇εn‖0‖bn‖L∞≤τ16‖Dτen+1‖20+Cτ‖curlεn‖20‖bn‖2L∞,2μτ|(Dτen+1⋅∇Bn,εn)|≤Cτ‖Dτen+1‖0‖∇Bn‖L3‖εn‖L6≤τ16‖Dτen+1‖20+Cτ‖∇εn‖20‖∇Bn‖2L3,≤τ16‖Dτen+1‖20+Cτ‖curlεn‖20‖∇Bn‖2L3,2μτ|(εn⋅∇Dτen+1,bn+1)|≤Cτ‖∇εn‖0‖Dτen+1‖0‖bn+1‖2≤τ16‖Dτen+1‖20+Cτ‖curlεn‖20‖bn+1‖22,2μτ|(Bn⋅∇Dτen+1,εn+1)|≤Cτ‖Bn‖L∞‖∇εn+1‖0‖Dτen+1‖0≤τ16‖Dτen+1‖20+Cτ‖curlεn+1‖20‖Bn‖2L∞,2μτ|(εn−ˆεn,Dτεn+1)|≤Cτ‖εn‖W1,2‖Bn‖L∞‖Dτεn+1‖0≤μτ8‖Dτεn+1‖20+Cτ‖curlεn‖20,2μτ|(ˆbn−ˉbn,Dτεn+1)|≤Cτ‖∇bn‖L4‖εn‖L4‖Dτεn+1‖0≤μτ8‖Dτεn+1‖20+Cτ‖curlεn‖20,2μτ|(Rn+1tr2,Dτεn+1)|≤Cτ‖Rn+1tr2‖0‖Dτεn+1‖0≤μτ8‖Dτεn+1‖20+Cτ‖Rn+1tr2‖20,2μτ|(bn+1⋅∇(un+1−un),Dτεn+1)|≤Cτ‖bn+1‖2‖∇(un+1−un)‖0‖Dτεn+1‖0≤μτ8‖Dτεn+1‖20+Cτ‖∇(un+1−un)‖20,2μτ|((bn+1−bn)⋅∇un,Dτεn+1)|≤Cτ‖bn+1−bn‖0‖∇un‖L∞‖Dτεn+1‖0≤μτ8‖Dτεn+1‖20+Cτ3‖bn+1t‖20‖∇un‖2L∞,2μτ|(εn⋅∇un,Dτεn+1)|≤Cτ‖εn‖0‖∇un‖L∞‖Dτεn+1‖0≤μτ8‖Dτεn+1‖20+Cτ‖curlεn‖20‖∇un‖2L∞,2μτ|(Bn⋅∇en,Dτεn+1)|≤Cτ‖Bn‖L∞‖∇en‖0‖Dτεn+1‖0≤μτ8‖Dτεn+1‖20+Cτ‖∇en‖20‖Bn‖2L∞. $

    Substituting these above inequality into (3.30), we obtain

    $ ν(‖en+1‖21−‖en‖21+‖en+1−en‖21)+μ(‖curlεn+1‖20−‖curlεn‖20+‖curlεn+1−curlεn‖20)+τ‖Dτen+1‖20+μτ‖Dτεn+1‖20≤Cτ(‖en‖21+‖curlεn+1‖20+‖curlεn‖20)+Cτ(‖Rn+1tr1‖20+‖Rn+1tr2‖20)+Cτ3. $ (3.31)

    Taking sum of the (3.31) for $ n $ from $ 0 $ to $ m\leq N $ and using discrete Gronwall's inequality Lemma 2.1, we arrive at

    $ max0≤n≤m(ν‖em+1‖21+μ‖curlεm+1‖20)+τm∑n=1(ν‖en+1−en‖21+μ‖curlεn+1−curlεn‖20)+τm∑n=1(‖Dτen+1‖20+μ‖Dτεn+1‖20)≤Cτ2. $ (3.32)

    Furthermore, the standard result for (3.17) and (3.18) with $ p = 2 $, respectively, leads to

    $ ‖˜en+1‖2≤C‖Dτen+1‖0+Cτ‖en−ˆen‖0+Cτ‖ˆUn−ˉUn‖0+C‖Rn+1tr1‖0+C‖bn+1−bn‖0‖∇bn+1‖L∞+C‖bn+1−bn‖0‖∇bn+1‖L∞+C‖∇bn‖L∞‖bn+1−bn‖0+C‖∇εn‖0‖bn‖L∞+C‖Bn‖W1,∞‖εn‖0+C‖εn‖0‖∇bn+1‖L∞+C‖Bn‖L∞‖curlεn+1‖0+‖ξn+1‖1≤C‖Dτen+1‖0+C‖Rn+1tr1‖0+‖ξn+1‖1+Cτ, $ (3.33)
    $ ‖εn+1‖2≤C‖Dτεn+1‖0+Cτ‖εn−ˆεn‖0+Cτ‖ˆbn−ˉbn‖0+C‖Rn+1tr2‖0+C‖bn+1‖L∞‖un+1−un‖0+C‖bn+1−bn‖0‖∇un‖L∞+C‖εn‖0‖∇un‖L∞+C‖Bn‖L∞‖∇en‖0≤C‖Dτen+1‖0+C‖Rn+1tr‖0+Cτ. $ (3.34)

    Thanks to (3.19), we can obtain

    $ (Δen+1,Δen+1)=(Δ˜en+1,Δen+1),‖en+1‖2≤‖˜en+1‖2. $

    Then, we have

    $ τm∑n=0(‖en+1‖22+‖˜en+1‖22+‖εn+1‖22)≤Cτ2. $ (3.35)

    From (3.35), we can arrive at

    $ max0≤n≤m‖Un+1‖2≤max0≤n≤m(‖un+1‖2+‖en+1‖2)≤C, $ (3.36)
    $ max0≤n≤m‖˜Un+1‖2≤max0≤n≤m(‖un+1‖2+‖˜en+1‖2)≤C, $ (3.37)
    $ max0≤n≤m‖Pn+1‖1≤max0≤n≤m(‖pn+1‖1+‖ξn+1‖1)≤C, $ (3.38)
    $ max0≤n≤m‖Bn+1‖2≤max0≤n≤m(‖bn+1‖2+‖εn+1‖2)≤C, $ (3.39)
    $ τm∑n=0‖DτUn+1‖22≤2τm∑n=0(‖Dτun+1‖22+‖Dτen+1‖22)≤C, $ (3.40)
    $ τm∑n=0‖Dτ˜Un+1‖22≤2τm∑n=0(‖Dτun+1‖22+‖Dτ˜en+1‖22)≤C, $ (3.41)
    $ τm∑n=0‖DτPn+1‖21≤2τm∑n=0(‖Dτpn+1‖21+‖Dτξn+1‖21)≤C, $ (3.42)
    $ τm∑n=0‖DτBn+1‖22≤2τm∑n=0(‖Dτbn+1‖22+‖Dτεn+1‖22)≤C. $ (3.43)

    From (2.7) and (2.5), we have

    $ DτUn+1−νΔ˜Un+1+∇Pn+1=fn+1−Un−ˆUnτ−μBn×curlBn+1, $ (3.44)

    and

    $ DτBn+1+curlcurlBn+1=gn+1+(Bn⋅∇)Un−Bn+1−ˆBnτ. $ (3.45)

    By (2.9), we have

    $ −ΔUn+1=∇×∇טUn+1. $ (3.46)

    Considering the identity $ \nabla^2\tilde{U}^{n+1} = \nabla\nabla\cdot\tilde{U}^{n+1}-\nabla\times\nabla\times\tilde{U}^{n+1} $ and (3.46), we can obtain

    $ Δ˜Un+1=∇∇⋅˜Un+1−ΔUn+1. $ (3.47)

    Now, the standard result for the Stokes system (3.17) and (3.18) with $ p = d^{*} > 2 $, respectively, leads to

    $ ‖Un+1‖W2,d∗+‖Pn+1−∇⋅˜Un+1‖W1,d∗≤C‖DτUn+1‖Ld∗+Cτ‖ˆUn−Un‖Ld∗+C‖fn+1‖Ld∗+C‖Bn×curlBn+1‖Ld∗≤C(‖DτUn+1‖Ld∗+‖∇Un‖Ld∗‖Un‖L∞+‖fn+1‖Ld∗+‖Bn‖Ld∗‖curlBn+1‖Ld∗), $ (3.48)
    $ ‖Bn+1‖W2,d∗≤C‖DτBn+1‖Ld∗+C‖gn+1‖Ld∗+Cτ‖ˆBn−Bn‖Ld∗+C‖Bn⋅∇Un‖Ld∗≤C(‖DτBn+1‖Ld∗+‖gn+1‖Ld∗+‖∇Bn‖Ld∗‖Bn‖L∞+‖Bn‖Ld∗‖∇Un‖Ld∗). $ (3.49)

    By (3.48) and (3.49), we get

    $ τm∑n=0(‖Un+1‖2W2,d∗+‖Bn+1‖2W2,d∗)≤C. $ (3.50)

    Thus, we can obtain

    $ ‖en+1‖2+τ3/4‖Un+1‖W2,d∗≤1, $ (3.51)
    $ ‖εn+1‖2+τ3/4‖Bn+1‖W2,d∗≤1. $ (3.52)

    The proof is complete.

    3.2.   Error estimation for the fully discrete method

    For the sake of simplicity, we denote $ R_{h}^{n} = R_{h}(U^{n}, P^{n}), Q_{h}^{n} = Q_{h}(U^{n}, P^{n}), R_{0h} = R_{0h}B^{n}, n = 1, 2, ..., N $, and $ e_{h}^{n} = u_{h}^{n}-R_{h}^{n}, \tilde{e}_{h}^{n} = \tilde{u}_{h}^{n}-R_{h}^{n}, \varepsilon_{h}^{n} = B_{h}^{n}-R_{0h}^{n}, n = 1, 2, ..., N $, where $ R_{0h}:L^{2}(\Omega)^{3} \rightarrow W_{h} $ is the $ L^{2} $-orthogonal projection, where

    $ ‖R0hw−w‖0+h‖∇(R0hw−w)‖0≤Chr+1‖w‖r+1, $ (3.53)
    $ ‖R0hw‖L∞≤C‖w‖2, $ (3.54)
    $ ‖∇R0hw‖L∞≤C‖w‖W1,∞. $ (3.55)

    Theorem 3.2 Suppose that assumption A2–A3 are satisfied, there exists a positive $ C > 0 $ such that

    $ ‖um+1h‖20+μ‖Bm+1h‖20+m∑n=0(‖un+1h−unh‖20+μ‖Bn+1h−Bnh‖20)+τm∑n=0(ν‖∇˜un+1h‖20+μ‖curlBn+1h‖20)≤C,‖um+1h‖21+μ‖curlBm+1h‖20+m∑n=0(‖un+1h−unh‖21+μ‖curlBn+1h−curlBnh‖20)+τm∑n=0(ν‖Aun+1h‖20+μ‖curlBn+1h‖20)≤C,‖em+1h‖20+μ‖εm+1h‖20+m∑n=0(‖en+1h−enh‖20+μ‖εn+1h−εnh‖20)+τm∑n=0(ν‖∇˜en+1h‖20+μ‖curlεn+1h‖20)≤Ch4,‖un+1h‖2L∞+μ‖Bn+1h‖2L∞+τm∑n=0(‖un+1h‖2W1,∞+μ‖Bn+1h‖2W1,∞)≤C. $

    Proof. We prove following estimates

    $ ‖emh‖20+μ‖εmh‖20+τn∑m=0(ν‖∇emh‖20+μ‖curlεmh‖20)≤Ch4, $ (3.56)

    by mathematical induction for $ m = 0, 1, ..., N $.

    When $ m = 0, u_{h}^{0} = u_{0}, B_{h}^{0} = B_{0} $. It is obvious (3.56) holds at the initial time step. We assume that (3.56) holds for $ 0\leq m\leq n $ for some integer $ n\geq0 $. By (2.20), (2.21), (3.54), (3.55), Theorem 3.1 and inverse inequality, we infer

    $ τ‖umh‖W1,∞≤(‖emh‖W1,∞+‖Rmh‖W1,∞)≤C(h−d2‖∇emh‖0+‖Um‖W1,∞+‖Pm‖L∞)≤14, $ (3.57)
    $ τ‖Bmh‖W1,∞≤(‖εmh‖W1,∞+‖Rm0h‖W1,∞)≤C(h−d2‖∇εmh‖0+‖Bm‖W1,∞)≤14, $ (3.58)

    and

    $ τ‖umh‖L∞≤(‖emh‖L∞+‖Rmh‖L∞)≤C(h−d2‖emh‖0+‖Um‖2+‖Pm‖1)≤C, $ (3.59)
    $ τ‖Bmh‖L∞≤(‖εmh‖L∞+‖Rm0h‖L∞)≤C(h−d2‖εmh‖0+‖Bm‖2)≤C. $ (3.60)

    Combining (2.23) and (2.24), we obtain

    $ (un+1h−ˆunhτ,vh)+ν(∇˜un+1h,∇vh)−(pn+1h,∇⋅vh)+μ(Bnh⋅∇vh,Bn+1h)−μ(vh⋅∇Bnh,Bnh)=(fn+1,vh). $ (3.61)

    When $ m = n+1 $, letting $ v_{h} = 2\tau \tilde{u}_{h}^{n+1} $ in (3.61), we can obtain

    $ 2(un+1h−unh+unh−ˆunh,˜un+1h)+2ντ(∇˜un+1h,∇˜un+1h)−2τ(pn+1h,∇⋅˜un+1h)+2μτ(Bnh⋅∇˜un+1h,Bn+1h)−2μτ(˜un+1h⋅∇Bnh,Bnh)=2τ(fn+1,˜un+1h). $ (3.62)

    Letting $ v_{h} = \tau u_{h}^{n+1} $ in (2.24), since $ \nabla\cdot u_{h}^{n+1} = 0 $, it follows that

    $ (un+1h,˜un+1h)=(un+1h,un+1h). $ (3.63)

    Taking $ v_{h} = \tau u_{h}^{n} $ in (2.24), we get

    $ (unh,˜un+1h)=(unh,un+1h). $ (3.64)

    Subtracting (3.64) from (3.63), we can obtain

    $ (un+1h−unh,˜un+1h)=(un+1h−unh,un+1h). $ (3.65)

    Using $ 2(a-b, a) = \|a\|_{0}^{2}-\|b\|_{0}^{2}+\|a-b\|_{0}^{2} $, leads to

    $ ‖un+1h‖20−‖unh‖20+‖un+1h−unh‖20+2ντ‖∇˜un+1h‖20−2τ(pn+1h,∇⋅˜un+1h)=2τ(fn+1,˜un+1h)+2(ˆunh−unh,˜un+1h)−2μτ(Bnh⋅∇˜un+1h,Bn+1h). $ (3.66)

    Taking $ v_{h} = \nabla p_{h}^{n+1} $ in (2.24), we deduce

    $ −2τ(pn+1h,∇⋅˜un+1)=2τ2‖∇pn+1h‖20−2τ2‖∇pnh‖20+2τ2‖∇pn+1h−∇pnh‖20. $ (3.67)

    When $ m = n+1 $, letting $ \psi_{h} = 2\mu\tau B_{h}^{n+1} $ in (2.22), we can obtain

    $ 2μ(Bn+1h−Bnh+Bnh−ˆBnh,Bn+1h)+2μτ(curlBn+1h,curlBn+1h)−2μτ(Bnh⋅∇unh,Bn+1h)=2μτ(gn+1,Bn+1h). $ (3.68)

    Using $ 2(a-b, a) = \|a\|_{0}^{2}-\|b\|_{0}^{2}+\|a-b\|_{0}^{2} $, we have

    $ μ‖Bn+1h‖20−μ‖Bnh‖20+μ‖Bn+1h−Bnh‖20+2μτ‖curlBn+1h‖20=2μτ(gn+1,Bn+1h)+2μτ(Bnh⋅∇unh,Bn+1h)+2μ(ˆBnh−Bnh,Bn+1h). $ (3.69)

    Taking sum of (3.66) and (3.69) yields

    $ ‖un+1h‖20−‖unh‖20+μ‖Bn+1h‖20−μ‖Bnh‖20+2τ2‖∇pn+1h‖20−2τ2‖∇pnh‖20+‖un+1h−unh‖20+μ‖Bn+1h−Bnh‖20+2τ2‖∇pn+1h−∇pnh‖20+2ντ‖∇˜un+1h‖20+2μτ‖curlBn+1h‖20=2τ(fn+1,˜un+1h)+2μτ(gn+1,Bn+1h)+2(ˆunh−unh,˜un+1h)−2μτ(Bn⋅∇˜un+1h,Bn+1h)+2μ(ˆBnh−Bnh,Bn+1h)+2μτ(Bnh⋅∇unh,Bn+1h). $ (3.70)

    Then, we can obtain

    $ 2τ|(fn+1,˜un+1h)|≤Cτ‖fn+1‖0‖˜un+1h‖0≤ντ4‖∇˜un+1h‖20+Cτ‖fn+1‖20,2μτ|(gn+1,Bn+1h)|≤Cτ‖gn+1‖0‖Bn+1h‖0≤μτ4‖curlBn+1h‖20+Cτ‖gn+1‖20,2|(ˆunh−unh,˜un+1h)|≤Cτ‖ˆunh−unh‖6/5‖˜un+1h‖L6≤Cτ‖∇unh‖L3‖unh‖0‖∇˜un+1h‖0≤ντ4‖∇˜un+1h‖20+Cτ‖unh‖20,2μτ|(Bn⋅∇˜un+1h,Bn+1h)|≤Cτ‖Bn‖L∞‖∇˜un+1h‖0‖Bn+1h‖0≤ντ4‖∇˜un+1h‖20+Cτ‖Bn+1h‖20,2μ|(ˆBnh−Bnh,Bn+1h)|≤Cτ‖ˆBnh−Bnh‖−1‖Bn+1h‖1≤Cτ‖Bnh‖0‖Bnh‖W1,∞‖curlBn+1h‖0≤μτ4‖curlBn+1h‖20+Cτ‖Bnh‖20‖Bnh‖2W1,∞,2μτ|(Bn⋅∇unh,Bn+1h)|≤Cτ‖Bn‖L∞‖unh‖0‖curlBn+1h‖0≤μτ4‖curlBn+1h‖20+Cτ‖unh‖20. $

    Substituting these above estimates into (3.70), we obtain

    $ ‖un+1h‖20−‖unh‖20+μ‖Bn+1h‖20−μ‖Bnh‖20+τ2‖∇pn+1h‖20−τ2‖∇pnh‖20+‖un+1h−unh‖20+μ‖Bn+1h−Bnh‖20+2τ2‖∇pn+1h−∇pnh‖20+ντ‖∇˜un+1h‖20+μτ‖curlBn+1h‖20≤Cτ‖fn+1‖20+Cτ‖gn+1‖20+Cτ‖unh‖20+Cτ‖Bnh‖20. $ (3.71)

    Taking sum of the (3.71) for $ n $ from $ 0 $ to $ m\leq N $ and using discrete Gronwall's inequality Lemma 2.1, we get

    $ ‖um+1h‖20+μ‖Bm+1h‖20+τ2‖∇pm+1h‖20+m∑n=0(‖un+1h−unh‖20+μ‖Bn+1h−Bnh‖20)+τm∑n=0(ν‖∇˜un+1h‖20+μ‖curlBn+1h‖20)≤C. $

    Letting $ v_{h} = 2\tau\mathcal{A}u_{h}^{n+1} $ in (3.61), we obtain

    $ 2(un+1h−unh+unh−ˆunh,Aun+1h)+2ντ(A˜un+1h,Aun+1h)+2μτ(Bnh⋅∇Aun+1h,Bn+1h)−2μτ(Aun+1h⋅∇Bnh,Bnh)=2τ(fn+1,Aun+1h). $ (3.72)

    From (3.63), we arrive at

    $ (A˜un+1h,Aun+1h)=(Aun+1h,Aun+1h),(A˜un+1h,Aun+1h)=‖Aun+1h‖20. $

    Using $ 2(a-b, a) = \|a\|_{0}^{2}-\|b\|_{0}^{2}+\|a-b\|_{0}^{2} $, leads to

    $ ‖un+1h‖21−‖unh‖21+‖un+1h−unh‖21+2ντ‖Aun+1h‖20=2τ(fn+1,Aun+1h)+2(ˆunh−unh,Aun+1h)−2μτ(Bnh⋅∇Aun+1h,Bn+1h). $ (3.73)

    Letting $ \psi_{h} = 2\mu\tau \mbox{curlcurl}B_{h}^{n+1} $ in (2.22), we can obtain

    $ μ‖curlBn+1h‖20−μ‖curlBnh‖20+μ‖curlBn+1h−curlBnh‖20+2μτ‖curlcurlBn+1h‖20=2μτ(gn+1,curlcurlBn+1h)+2μτ(Bnh⋅∇unh,curlcurlBn+1h)+2μ(ˆBnh−Bnh,curlcurlBn+1h). $ (3.74)

    Taking sum of (3.73) and (3.74) yields

    $ ‖un+1h‖21−‖unh‖21+μ‖curlBn+1h‖20−μ‖curlBnh‖20+‖un+1h−unh‖21+μ‖curlBn+1h−curlBnh‖20+2ντ‖Aun+1h‖20+2μτ‖curlcurlBn+1h‖20=2τ(fn+1,Aun+1h)+2(ˆunh−unh,Aun+1h)+2τ(gn+1,curlcurlBn+1h)−2μτ(Bnh⋅∇Aun+1h,Bn+1h)+2μτ(Bnh⋅∇unh,curlcurlBn+1h)+2μ(ˆBnh−Bnh,curlcurlBn+1h). $ (3.75)

    Then, we get

    $ 2τ|(fn+1,Aun+1h)|≤Cτ‖fn+1‖0‖Aun+1h‖0≤ντ6‖Aun+1h‖20+Cτ‖fn+1‖20,2μτ|(gn+1,curlcurlBn+1h)|≤Cτ‖gn+1‖0‖curlcurlBn+1h‖0≤μτ4‖curlcurlBn+1h‖20+Cτ‖gn+1‖20,2|(ˆunh−unh,Aun+1h)|≤Cτ‖unh‖W1,4‖unh‖L4‖Aun+1h‖0≤ντ6‖Aun+1h‖20+Cτ‖unh‖21,2μτ|(Bnh⋅∇Aun+1h,Bn+1h)|≤Cτ‖Bnh‖W1,∞‖Aun+1h‖0‖Bn+1h‖0≤ντ6‖Aun+1h‖20+Cτ‖curlBn+1h‖20,2μτ|(Bnh⋅∇unh,curlcurlBn+1h)|≤Cτ‖Bnh‖L∞‖∇unh‖0‖curlcurlBn+1h‖0≤μτ4‖curlcurlBn+1h‖20+Cτ‖∇unh‖20,2μ|(ˆBnh−Bnh,curlcurlBn+1h)|≤Cτ‖Bnh‖W1,4‖Bnh‖L4‖curlcurlBn+1h‖0≤μτ4‖curlcurlBn+1h‖20+Cτ‖Bnh‖2W1,4‖Bnh‖2L4≤μτ4‖curlcurlBn+1h‖20+Cτ‖curlBnh‖20. $

    Substituting these above estimates into (3.75), we obtain

    $ ‖un+1h‖21−‖unh‖21+μ‖curlBn+1h‖20−μ‖curlBnh‖20+‖un+1h−unh‖21+μ‖curlBn+1h−curlBnh‖20+ντ‖Aun+1h‖20+μτ‖curlcurlBn+1h‖20≤Cτ‖fn+1‖20+Cτ‖gn+1‖20+Cτ‖unh‖21+Cτ(‖curlBn+1h‖20+‖curlBnh‖20). $ (3.76)

    Taking sum of the (3.76) for $ n $ from $ 0 $ to $ m\leq N $ and using discrete Gronwall's inequality Lemma 2.1, we get

    $ ‖um+1h‖21+μ‖curlBm+1h‖20+m∑n=0(‖un+1h−unh‖21+μ‖curlBn+1h−curlBnh‖20)+τm∑n=0(ν‖Aun+1h‖20+μ‖curlcurlBn+1h‖20)≤C. $

    Then, we rewrite (3.61) and (2.22) as

    $ (Dτun+1h,vh)+ν(∇˜un+1h,∇vh)−(Pn+1h,∇⋅vh)=(fn+1,vh)−(unh−ˆunhτ,vh)−μ(Bnh⋅∇vh,Bn+1h)+μ(vh⋅∇Bnh,Bnh), $ (3.77)

    and

    $ (DτBn+1h,ψh)+(curlBn+1h,curlψh)=(gn+1,ψh)+(Bnh⋅∇unh,ψh)−(Bnh−ˆBnhτ,ψh). $ (3.78)

    Subtracting (3.15) and (3.16) from (3.77) and (3.78), respectively, leads to

    $ (Dτen+1h,vh)+ν(∇˜en+1h,∇vh)−(pn+1h−Qn+1h,∇⋅vh)=(Dτ(Un+1−Rn+1h),vh)+(ˆunh−unhτ,vh)+(Un−ˆUnτ,vh)−μ(εnh⋅∇vh,Bn+1h)−μ((Rn0h−Bn)⋅∇vh,Bn+1h)−μ(Bn⋅∇vh,εn+1h)−μ(Bn⋅∇vh,Rn+10h−Bn+1)+μ(vh⋅∇εnh,Bnh)+μ(vh⋅∇Bn,εnh)−μ(vh⋅∇Bn,Rn0h−Bn)+μ(vh⋅∇(Rn0h−Bn),Bnh), $ (3.79)

    and

    $ (Dτεn+1h,ψh)+(curlεn+1h,curlψnh)=(Dτ(Bn+1−Rn+10h),ψh)+(Bnh−ˆBnhτ,ψh)+(ˆBn−Bnτ,ψh)+(εnh⋅∇unh,ψh)+((Rn0h−Bn)⋅∇unh,ψh)+(Bn⋅∇(Rnh−Un),ψh)+(Bn⋅∇enh,ψh). $ (3.80)

    Taking $ v_{h} = 2\tau e_{h}^{n+1} $ in (2.24), we can obtain

    $ (en+1h,˜en+1h)=(en+1h,en+1h),(Dτen+1h,˜en+1h)=(Dτen+1h,en+1h),(∇en+1h,∇˜en+1h)=(∇en+1h,∇en+1h)=‖∇en+1h‖20. $

    Let $ \theta^{n} = U^{n}-R_{h}^{n}, \eta^{n} = B^{n}-R_{0h}^{n} $. By taking $ v_{h} = 2\tau e_{h}^{n+1} $ in (3.79), we deduce

    $ ‖en+1h‖20−‖enh‖20+‖en+1h−enh‖20+2ντ‖∇en+1h‖20=2τ(DτUn+1−Rh(DτUn+1,DτPn+1),en+1h)+2(ˆenh−enh,en+1h)+2(θn−ˆθn,en+1h)−2μτ(εnh⋅∇en+1h,Bn+1h)−2μτ(ηn⋅∇en+1h,Bn+1h)−2μτ(Bn⋅∇en+1h,εn+1h)−2μτ(Bn⋅∇en+1h,ηn+1)+2μτ(en+1h⋅∇εnh,Bnh)+2μτ(en+1h⋅∇Bn,εnh)+2μτ(en+1h⋅∇Bn,ηn)+2μτ(en+1h⋅∇ηn,Bnh). $ (3.81)

    By taking $ \psi_{h} = 2\mu\tau \varepsilon_{h}^{n+1} $ in (3.80), we have

    $ μ‖εn+1h‖20−μ‖εnh‖20+μ‖εn+1h−εnh‖20+2μτ‖curlεn+1h‖20=2τ(DτBn+1−R0hDτBn+1),εn+1h)+2μ(εnh−ˆεnh,εn+1h)+2μ(ηn−ˆηn,εn+1h)+2μτ(εnh⋅∇unh,εn+1h)−2μτ(ηn⋅∇unh,εn+1h)+2μτ(Bn⋅∇(Rnh−Un),εn+1h)+2μτ(Bn⋅∇enh,εn+1h). $ (3.82)

    Combining (3.81) and (3.82), we obtain

    $ ‖en+1h‖20−‖enh‖20+μ‖εn+1h‖20−μ‖εnh‖20+‖en+1h−enh‖20+S‖εn+1h−εnh‖20+2ντ‖∇en+1h‖20+2μτ‖curlεn+1h‖20=2τ(DτUn+1−Rh(DτUn+1,DτPn+1),en+1h)+2(ˆenh−enh,en+1h)+2(θn−ˆθn,en+1h)−2μτ(εnh⋅∇en+1h,Bn+1h)−2μτ(ηn⋅∇en+1h,Bn+1h)−2μτ(Bn⋅∇en+1h,Bn+1h)−2μτ(Bn⋅∇en+1h,ηn+1)+2μτ(en+1h⋅∇εnh,Bnh)+2μτ(en+1h⋅∇Bn,εnh)+2μτ(en+1h⋅∇Bn,ηn)+2μτ(en+1h⋅∇ηn,Bnh)+2τ(DτBn+1−R0hDτBn+1),εn+1h)+2μ(εnh−ˆεnh,εn+1h)+2μ(ηn−ˆηn,εn+1h)+2μτ(εnh⋅∇unh,εn+1h)−2μτ(ηn⋅∇unh,εn+1h)+2μτ(Bn⋅∇(Rnh−Un),εn+1h)−2μτ(Bn⋅∇enh,Bn+1h). $ (3.83)

    Due to (2.19) and Theorem 3.1, we can get $ \|\theta^{n}\|_{0}\leq Ch^{2}(\|U^{n}\|_{2}+\|P^{n}\|_{1})\leq Ch^{2} $, $ \|\eta^{n}\|_{0}\leq Ch^{2}\|B^{n}\|_{2} $. By Lemma 3.1, there holds that

    $ 2τ|(DτUn+1−Rh(DτUn+1,DτPn+1),en+1h)|≤Cτh2‖en+1h‖0(‖DτUn+1‖2+DτPn+1‖1)≤ντ20‖∇en+1h‖20+Cτh4(‖DτUn+1‖22+DτPn+1‖21),2|(ˆenh−enh,en+1h)|≤C‖enh−ˆenh‖−1‖∇en+1h‖0≤Cτ‖∇enh‖0‖unh‖W1,∞‖∇en+1h‖0≤ντ20‖∇en+1h‖20+Cτ‖enh‖20,2|(θn−ˆθn,en+1h)|≤C‖θn−ˆθn‖−1‖∇en+1h‖0≤Cτ‖θn‖0‖unh‖W1,∞‖∇en+1h‖0≤ντ20‖∇en+1h‖20+Cτ‖θn‖20,2μτ|(εnh⋅∇en+1h,Bn+1h)|≤Cτ‖εnh‖0‖∇en+1h‖0‖Bn+1h‖L∞≤ντ20‖∇en+1h‖20+Cτ‖εnh‖20,2μτ|(ηn⋅∇en+1h,Bn+1h)|≤Cτ‖ηn‖0‖∇en+1h‖0‖Bn+1h‖L∞≤ντ20‖∇en+1h‖20+Cτ‖ηn‖20,2μτ|(Bn⋅∇en+1h,εn+1h)|≤Cτ‖Bn‖L∞‖∇en+1h‖0‖εn+1h‖0≤ντ20‖∇en+1h‖20+Cτ‖εn+1h‖20,2μτ|(Bn⋅∇en+1h,ηn+1)|≤Cτ‖Bn‖2‖∇en+1h‖0‖ηn+1‖0≤ντ20‖∇en+1h‖20+Cτ‖ηn+1‖20,2μτ|(en+1h⋅∇εnh,Bnh)|≤Cτ‖∇en+1h‖0‖εnh‖0‖Bnh‖L∞≤ντ20‖∇en+1h‖20+Cτ‖εnh‖20,2μτ|(en+1h⋅∇Bn,εn+1h)|≤Cτ‖∇en+1h‖0‖Bn‖2‖εnh‖0≤ντ20‖∇en+1h‖20+Cτ‖εnh‖20,2μτ|(en+1h⋅∇Bn,ηn)|≤Cτ‖∇en+1h‖0‖Bn‖2‖ηn‖0≤ντ20‖∇en+1h‖20+Cτ‖ηn‖20,2μτ|(en+1h⋅∇ηn,Bnh)|≤Cτ‖∇en+1h‖0‖ηn‖0‖Bnh‖L∞≤ντ20‖∇en+1h‖20+Cτ‖ηn‖20,2τ|(DτBn+1−R0hDτBn+1,εn+1h)|≤Cτh2‖εn+1h‖0‖DτBn+1‖2‖ηn‖0≤μτ16‖curlεn+1h‖20+Cτh4‖DτBn+1‖22,2μ|(εnh−ˆεnh,εn+1h)|≤C‖εnh−ˆεnh‖0‖εn+1h‖0≤Cτ‖curlεnh‖0‖Bnh‖L∞‖εn+1h‖0≤μτ16‖curlεn+1h‖20+Cτ‖εn+1h‖20,2μ|(ηn−ˆηn,εn+1h)|≤C‖ηn−ˆηn‖−1‖∇εn+1h‖0≤Cτ‖ηn‖0‖Bnh‖W1,∞‖curlεn+1h‖0≤μτ16‖curlεn+1h‖20+Cτ‖ηn‖20,2μτ|(εnh⋅∇unh,εn+1h)|≤Cτ‖εnh‖0‖∇unh‖L3‖εn+1h‖L6≤μτ16‖curlεn+1h‖20+Cτ‖εn+1h‖20,2μτ|(ηn⋅∇unh,εn+1h)|≤Cτ‖ηn‖0‖∇unh‖L3‖εn+1h‖L6≤μτ16‖curlεn+1h‖20+Cτ‖ηn‖20,2μτ|(Bn⋅∇(Rnh−Un),εn+1h)|≤Cτ‖Bn‖2‖θn‖0‖∇εn+1h‖0≤μτ16‖curlεn+1h‖20+Cτ‖θn‖20,2μτ|(Bn⋅∇enh,εn+1h)|≤Cτ‖Bn‖L∞‖enh‖0‖curlεn+1h‖0≤μτ16‖curlεn+1h‖20+Cτ‖enh‖20. $

    Substituting these above estimates into (3.83), we obtain

    $ \begin{align} &\|e_{h}^{n+1}\|_{0}^{2}-\|e_{h}^{n}\|_{0}^{2}+\mu\|\varepsilon_{h}^{n+1}\|_{0}^{2}-\mu\|\varepsilon_{h}^{n}\|_{0}^{2} +\|e_{h}^{n+1}-e_{h}^{n}\|_{0}^{2} +\mu\|\varepsilon_{h}^{n+1}-\varepsilon_{h}^{n}\|_{0}^{2} \\ &+\nu\tau\| \nabla e_{h}^{n+1}\|_{0}^{2}+\mu\tau\|\mbox{curl}\varepsilon_{h}^{n+1}\|_{0}^{2} \\ \leq & C\tau(\|e_{h}^{n}\|_{0}^{2}+\|\varepsilon_{h}^{n}\|_{0}^{2}+\|\varepsilon_{h}^{n+1}\|_{0}^{2}) + C\tau h^{4}. \end{align} $ (3.84)

    Taking the sum of the (3.76) for $ n $ from $ 0 $ to $ m\leq N $ and using the discrete Gronwall's inequality Lemma 2.1, we get

    $ \begin{align*} &\|e_{h}^{m+1}\|_{0}^{2}+\mu\|\varepsilon_{h}^{m+1}\|_{0}^{2} +\sum\limits_{n = 0}^{m}(\|e_{h}^{n+1}-e_{h}^{n}\|_{0}^{2} +\mu\|\varepsilon_{h}^{n+1}-\varepsilon_{h}^{n}\|_{0}^{2}) \nonumber\\ &+\tau \sum\limits_{n = 0}^m\left(\nu\|\nabla e_{h}^{n+1}\|_{0}^{2} +\mu\|\mbox{curl}\varepsilon_{h}^{n+1}\|_{0}^{2}\right)\leq C h^{4}. \end{align*} $

    Then, we can deduce

    $ \begin{align*} \max\limits_{0\leq n\leq m}\|u_{h}^{n+1}\|_{L^{\infty}} &\leq\max\limits_{0\leq n\leq m}(\|R_{h}^{n+1}\|_{L^{\infty}}+\|e_{h}^{n+1}\|_{L^{\infty}}) \nonumber\\ &\leq C\max\limits_{0\leq n\leq m}\|U^{n+1}\|_{2}+C\max\limits_{0\leq n\leq m}\|P^{n+1}\|_{1}+Ch^{-\frac{1}{2}}\max\limits_{0\leq n\leq m}\|e_{h}^{n+1}\|_{0} \leq C, \nonumber\\ \max\limits_{0\leq n\leq m}\|B_{h}^{n+1}\|_{L^{\infty}} &\leq\max\limits_{0\leq n\leq m}(\|R_{0h}^{n+1}\|_{L^{\infty}}+\|\varepsilon_{h}^{n+1}\|_{L^{\infty}}) \nonumber\\ &\leq C\max\limits_{0\leq n\leq m}\|B^{n+1}\|_{2}+Ch^{-\frac{1}{2}}\max\limits_{0\leq n\leq m}\|\varepsilon_{h}^{n+1}\|_{0} \leq C, \nonumber\\ \tau\sum\limits_{n = 0}^m\|u_{h}^{n+1}\|_{W^{1,\infty}}^{2} &\leq 2\tau\sum\limits_{n = 0}^m(\|R_{h}^{n+1}\|_{W^{1,\infty}}^{2}+\|e_{h}^{n+1}\|_{W^{1,\infty}}^{2}) \nonumber\\ &\leq C\tau\sum\limits_{n = 0}^m(\|U^{n+1}\|_{W^{1,\infty}}^{2}+\|P^{n+1}\|_{L^{\infty}}^{2}) +C\tau h^{-2}\sum\limits_{n = 0}^m\|\nabla e_{h}^{n+1}\|_{0}^{2}\leq C, \nonumber\\ \tau\sum\limits_{n = 0}^m\|B_{h}^{n+1}\|_{W^{1,\infty}}^{2} &\leq 2\tau\sum\limits_{n = 0}^m(\|R_{0h}^{n+1}\|_{W^{1,\infty}}^{2}+\|\varepsilon_{h}^{n+1}\|_{W^{1,\infty}}^{2}) \nonumber\\ &\leq C\tau\sum\limits_{n = 0}^m\|B^{n+1}\|_{W^{1,\infty}}^{2} +C\tau h^{-2}\sum\limits_{n = 0}^m\|\nabla \varepsilon_{h}^{n+1}\|_{0}^{2} \leq C. \end{align*} $

    Thus, all the results in Theorem 3.2 have been proved.

    4.   Numerical results

    In order to show the effect of our method, we present some numerical results. Here, we consider the

    $ \begin{align*} u_1& = (y+y^4)*cos(t),\\ u_2& = (x+x^2)*cos(t),\\ p& = (2.0*x-1.0)*(2.0*y-1.0)*cos(t),\\ B_1& = (sin(y)+y)*cos(t),\\ B_2& = (sin(x)+x^2)*cos(t). \end{align*} $

    The boundary conditions and the forcing terms are given by the exact solution. The finite element spaces are chosen as P1b-P1-P1b finite element spaces. Here, we choose $ \tau = h^2 $ and $ h = 1/n $, $ n = 8, 16, 24, 32, 40, 48 $. Different Reynolds number $ Re $ and magnetic Reynolds number $ Rm $ are chosen to show the effect of our method, the numerical results were shown in Tables 1–6. Here, we use the software package FreeFEM++ [32] for our program.

    Table 1.  The numerical results for $ Re = 1.0 $ and $ Rm = 1.0 $ for different $ h $ at $ T = 1.0 $.
    $ 1/h$ $ \|{\bf{u}}_h^1-{\bf{u}}\|_0$ $ \|\nabla({\bf{u}}_{h}^1-{\bf{u}}_1)\|_0$ $ \|p_h^1-p\|_0$ $ \|{\bf{B}}_h^1-{\bf{B}}\|_0$ $ \|\nabla({\bf{B}}_h^1-{\bf{B}})\|_0$ $ \|E_h^1-E\|_0$ $ |\int_{\Omega} \nabla\cdot B dx|$ $ |\int_{\Omega} \nabla\cdot {\bf{u}} dx |$
    8 0.00395939 0.104518 0.0633311 0.00146071 0.0443105 0.0402426 4.95209e-17 1.99024e-14
    16 0.00109704 0.0527177 0.0212408 0.000384435 0.0221793 0.0208612 1.41995e-16 4.15508e-15
    24 0.000473611 0.0347495 0.0118957 0.000175071 0.0148256 0.0141051 5.03032e-17 5.95751e-15
    32 0.000256132 0.0258999 0.010131 9.82405e-05 0.0111161 0.010656 3.35182e-16 7.38832e-15
    40 0.000160786 0.0206654 0.00831346 6.2778e-05 0.00889124 0.00856254 7.97451e-16 5.55468e-15
    48 0.000110796 0.0171977 0.00648101 4.35568e-05 0.00740848 0.00715659 1.00015e-16 4.15077e-15
     | Show Table
    DownLoad: CSV
    Table 2.  Convergence order for $ Re = 1.0 $ and $ Rm = 1.0 $ for different $ h $ at $ T = 1.0 $.
    $ 1/h$ $ {\bf{u}}-L^2$ $ {\bf{u}}-H_1$ $ P-L^2$ $ {\bf{B}}-L^2$ $ {\bf{B}}-H_1$
    16 1.85166 0.987389 1.57608 1.92586 0.998435
    24 2.07166 1.02792 1.42982 1.93995 0.99343
    32 2.13671 1.0217 0.5582 2.00837 1.00097
    40 2.08665 1.01181 0.88607 2.00685 1.00084
    48 2.04246 1.00748 1.36571 2.00491 1.00066
     | Show Table
    DownLoad: CSV
    Table 3.  The numerical results for $ Re = 100.0 $ and $ Rm = 100.0 $ for different $ h $ at $ T = 1.0 $.
    $ 1/h$ $ \|{\bf{u}}_h^1-{\bf{u}}\|_0$ $ \|\nabla({\bf{u}}_{h}^1-{\bf{u}}_1)\|_0$ $ \|p_h^1-p\|_0$ $ \|{\bf{B}}_h^1-{\bf{B}}\|_0$ $ \|\nabla({\bf{B}}_h^1-{\bf{B}})\|_0$ $ \|E_h^1-E\|_0$ $ |\int_{\Omega} \nabla\cdot B dx|$ $ |\int_{\Omega} \nabla\cdot {\bf{u}} dx |$
    8 0.0172667 0.279161 0.0354359 0.0127351 0.19731 0.00779509 3.41253e-17 3.17954e-14
    16 0.00233046 0.0697115 0.01409 0.00240034 0.0479792 0.00159673 9.63754e-17 1.3337e-14
    24 0.000831767 0.0398145 0.00892872 0.000977778 0.0269431 0.000711847 8.03834e-18 8.42217e-15
    32 0.000422379 0.0280313 0.00649219 0.000527044 0.0184095 0.000385869 2.36579e-16 5.31439e-15
    40 0.000256947 0.0217553 0.00511173 0.000330314 0.013926 0.000239563 7.44033e-16 2.85026e-15
    48 0.000173343 0.0178297 0.00421705 0.000226765 0.0112143 0.000162916 7.14376e-17 2.41418e-15
     | Show Table
    DownLoad: CSV
    Table 4.  Convergence order for $ Re = 100.0 $ and $ Rm = 100.0 $ for different $ h $ at $ T = 1.0 $.
    $ 1/h$ $ {\bf{u}}-L^2$ $ {\bf{u}}-H_1$ $ P-L^2$ $ {\bf{B}}-L^2$ $ {\bf{B}}-H_1$
    16 2.88931 2.00163 1.33054 2.4075 2.03998
    24 2.54095 1.38147 1.12512 2.21495 1.42315
    32 2.35555 1.21978 1.10773 2.1482 1.32391
    40 2.22742 1.13589 1.07134 2.09391 1.2508
    48 2.1588 1.09144 1.05529 2.063 1.1878
     | Show Table
    DownLoad: CSV
    Table 5.  The numerical results for $ Re = 1000.0 $ and $ Rm = 1000.0 $ for different $ h $ at $ T = 1.0 $.
    $ 1/h$ $ \|{\bf{u}}_h^1-{\bf{u}}\|_0$ $ \|\nabla({\bf{u}}_{h}^1-{\bf{u}}_1)\|_0$ $ \|p_h^1-p\|_0$ $ \|{\bf{B}}_h^1-{\bf{B}}\|_0$ $ \|\nabla({\bf{B}}_h^1-{\bf{B}})\|_0$ $ \|E_h^1-E\|_0$ $ |\int_{\Omega} \nabla\cdot B dx|$ $ |\int_{\Omega} \nabla\cdot {\bf{u}} dx |$
    8 0.0397337 1.68458 0.0330041 0.029124 1.17999 0.0213862 1.41814e-16 1.99053e-14
    16 0.00470743 0.368592 0.0137702 0.00403407 0.219747 0.00228304 7.44847e-17 1.26213e-14
    24 0.00158913 0.177215 0.00901776 0.00143967 0.106852 0.000843484 8.59434e-17 8.328e-15
    32 0.000750714 0.102493 0.00644056 0.000756826 0.0713168 0.000456685 2.75943e-16 5.37049e-15
    40 0.000430867 0.0675245 0.00509083 0.000460115 0.0510193 0.000286991 8.76007e-16 2.73806e-15
    48 0.000276193 0.0481754 0.00421193 0.000306103 0.038122 0.000195792 1.25266e-16 2.4751e-15
     | Show Table
    DownLoad: CSV
    Table 6.  Convergence order for $ Re = 1000.0 $ and $ Rm = 1000.0 $ for different $ h $ at $ T = 1.0 $.
    $ 1/h$ $ {\bf{u}}-L^2$ $ {\bf{u}}-H_1$ $ P-L^2$ $ {\bf{B}}-L^2$ $ {\bf{B}}-H_1$
    16 3.07735 2.19229 1.26109 2.8519 2.42487
    24 2.6783 1.80614 1.04402 2.54119 1.77829
    32 2.60675 1.90339 1.16997 2.23523 1.40541
    40 2.48819 1.8701 1.05392 2.23021 1.50095
    48 2.43909 1.85191 1.03949 2.23536 1.59834
     | Show Table
    DownLoad: CSV

    Tables 1 and 2 give the numerical results for $ Re = 1.0 $ and $ Rm = 1.0 $. In order to show the effect of our method for high Reynolds number, we give the numerical results for $ Re = 100.0, \ 1000.0 $ and $ Rm = 100.0, \ 1000.0 $. Tables 3 and 4 give the numerical results for $ Re = 100.0 $ and $ Rm = 100.0 $. Tables 5 and 6 give the numerical results for $ Re = 1000.0 $ and $ Rm = 1000.0 $. It shows that our method is effect for high Reynolds numbers. It shows that the errors are goes small as the space step goes small and the convergence orders are optimal. We can see that $ |\int_{\Omega} \nabla\cdot B dx| $ and $ |\int_{\Omega} \nabla\cdot {{\bf u}} dx| $ are small, which means that our method can conserve the Gauss's law very well.

    5.   Conclusions

    In this paper, we have given a modified characteristics projection finite element method for the unsteady incompressible magnetohydrodynamics(MHD) equations. In this method, the modified characteristics finite element method and the projection method were combined for solving the incompressible MHD equations. Both the stability and the optimal error estimates in $ L^2 $ and $ H^1 $ norms for the modified characteristics projection finite element method have been given. In order to demonstrate the effectiveness of our method, some numerical results were given at the end of the manuscript.

    Acknowledgments

    The authors would like to thank the anonymous referees for their valuable suggestions and comments, which helped to improve the quality of the paper. This work is supported in part by National Natural Science Foundation of China (No. 11971152), China Postdoctoral Science Foundation (No. 2018M630907) and the Key scientific research projects Henan Colleges and Universities (No. 19B110007).

    Conflict of interest

    The authors declare there is no conflict of interest in this paper.

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Article outline
   Introduction
   Notation and preliminaries
   Unconditional stability and convergence
   Numerical results
   Conclusions
Acknowledgments
Conflict of interest
References

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