Optimal Error Estimates of Linearized Crank–Nicolson Galerkin Method for Landau–Lifshitz Equation

Abstract

This paper focuses on the optimal error estimates of a linearized Crank–Nicolson scheme for the Landau–Lifshitz (LL) equation describing the evolution of spin fields in continuum ferromagnets. We present a rigorous analysis for the regularity of the local strong solution to LL equation with Neumann boundary conditions. The proof of the optimal error estimates are based upon an error splitting technique proposed by Li and Sun. Numerical results are provided to confirm our theoretical analysis.

1 Introduction

The Landau–Lifshitz (LL) equation derived by Landau–Lifshitz [26] describes the evolution of magnetization in continuum ferromagnets and plays a fundamental role in the understanding of non-equilibrium magnetism. Let $\Omega \subset {\mathbb {R}}^d, d=2,3$ be a bounded and convex domain with smooth boundary $\partial \Omega$. The unknown magnetization field $\mathbf{m }: {\mathbb {R}}^+\times \Omega \longrightarrow {\mathbb {S}}^2$, where ${\mathbb {S}}^2$ is the unit sphere in ${\mathbb {R}}^3$, satisfies the following LL equation with an exchange fields:

$$\begin{aligned} \mathbf{m }_t=\gamma \mathbf{m }\times \Delta \mathbf{m }-\lambda \gamma \mathbf{m }\times (\mathbf{m }\times \Delta \mathbf{m }), \end{aligned}$$

(1.1)

where $\lambda >0$ represents the Glibert damping constant. $\gamma$ denotes the gyromagnetic factor. After time rescaling where time is measured in units of $\gamma ^{-1}$, the gyromagnetic factor $\gamma$ in the right-hand side of LL equation can be neglected. The initial and boundary conditions are taken to be

$$\begin{aligned} \left\{ \begin{array}{ll} {\nabla \mathbf{m }}\cdot { \mathbf{n }}=0,\quad &{}\hbox {on} \ {\mathbb {R}}^+\times \partial \Omega ,\\ \mathbf{m }(0,\mathbf{x })=\mathbf{m }_0,\quad &{}\hbox {in} \ \Omega , \end{array}\right. \end{aligned}$$

(1.2)

where $\mathbf{n }$ denotes the unit outward normal vector on $\partial \Omega$. We suppose that $\mathbf{m }_0$ satisfies the following compatibility conditions:

$$\begin{aligned} |\mathbf{m }_0|=1\quad \hbox {in} \ \Omega ,\quad \hbox {and}\quad \nabla \mathbf{m }_0\cdot \mathbf{n }=0 \ \hbox {on} \ \partial \Omega . \end{aligned}$$

(1.3)

By using $|\mathbf{m }|=1$ and the vector formulation below:

$$\begin{aligned} \mathbf{a }\times (\mathbf{b }\times \mathbf{c })=(\mathbf{a }\cdot \mathbf{c })\mathbf{b } -(\mathbf{a }\cdot \mathbf{b })\mathbf{c }, \quad \mathbf{a },\mathbf{b },\mathbf{c }\in {\mathbb {R}}^d, \end{aligned}$$

the LL equation (1.1) is equivalent to

$$\begin{aligned} \mathbf{m }_t-\lambda \Delta \mathbf{m }- \mathbf{m }\times \Delta \mathbf{m }=\lambda |\nabla \mathbf{m }|^2\mathbf{m }. \end{aligned}$$

(1.4)

From (1.1) and (1.4), other equivalent formulations [5] of the LL equation are

$$\begin{aligned}&\mathbf{m }_t+\lambda \mathbf{m }\times \mathbf{m }_t=(1+\lambda ^2)\mathbf{m }\times \Delta \mathbf{m }, \end{aligned}$$

(1.5)

$$\begin{aligned}&\lambda \mathbf{m }_t-\mathbf{m }\times \mathbf{m }_t=(1+\lambda ^2)(\Delta \mathbf{m }+|\nabla \mathbf{m }|^2\mathbf{m }). \end{aligned}$$

(1.6)

The first theoretical analysis for the LL equation was investigated by Visintin [32], where the existence result based on the energy estimate and some complementary properties were proved for the system of the LL equation coupled with Maxwell’s equations. The existence of the global weak solution for the LL equation was studied by Alouges-Soyeur for 3D model [4] and Guo-Hong for 2D model [21]. Meanwhile, the existence and uniqueness of the local strong solution were shown in [21]. Carbou-Fabrie in [9] proved the existence of the local strong solution in 3D bounded domain, and the existence of the global strong solution if $||\nabla \mathbf{m }_0||_{H^1}$ is small enough in 2D bounded domain. Some other theoretical results for the LL equation can be founded in [10–12, 15, 20, 33, 34] and references cited therein.

The numerical methods for the LL equation have been investigated by many authors. For a review of various numerical methods for LL equation, we refer to Cimrák [13], Kruzík-Prohl [25] and Prohl [31]. A key issue in constructing the fully discrete schemes is that we have to take into account the nonlinear constraint $|\mathbf{m }|=1$. A natural approach is to approximate the constraint $|\mathbf{m }|=1$ by a penalty function. In [31], based upon some penalty strategies, some different schemes for (1.1) were proposed. The convergence of their numerical solutions to the local strong solutions was proved. For the equivalent problem (1.4), Pistella-Valente [30] used Ginzburg-Landau penalty function $\frac{1}{\varepsilon }(|\mathbf{m }|^2-1)\mathbf{m }$ to relax the nonlinear constraint and gave an explicit finite difference scheme to solve the penalty problem numerically. The authors proved the convergence of the numerical solution under the time step restriction $\tau \le C(\varepsilon )h^2$. As pointed in [17], the penalty methods result in the use of a very small time step as h and $\varepsilon$ tend to zero, and extremely time-consuming in the practical computations. An alternative numerical tool compared to penalty method is the projection method proposed by E-Wang [16], where the numerical solution was projected on the sphere ${\mathbb {S}}^2$ at each time step. Motivated by E-Wang [16], an explicit scheme for (1.5) was proposed by Alouges-Jaisson in [3]. Subsequently, Bartels-Ko-Prohl in [5] improved the convergence results in [3] under the time step restriction $\tau \le o(h^{1+d/2})$. Alouges extended the explicit scheme in [3] to a new $\theta -$scheme and proved that the new scheme was unconditionally stable for $1/2\le \theta \le 1$ in [2]. However, at each time step, one has to build a new FE space which is orthogonal point-wisely to the FE solution at the previous time step. Besides the above works, Bartels-Lubich-Prohl in [6] introduced a finite element method which exactly preserves $|\mathbf{m }|=1$ on the nodal points for solving the equations of harmonic map heat flow which can be viewed as a limiting case of the Landau–Lifshitz equation, and proved the convergence of the finite element solution under minimal regularity assumptions. Bartels-Prohl in [7] studied a fully implicit and nonlinear scheme. Although this fully implicit scheme was unconditionally stable, at each time step, one has to solve a nonlinear problem which is solved by using a fixed point strategy. Bartels-Prohl proved the convergence of the inner iteration under the time step restriction $\tau \le O(h^2)$. On the other hand, for (1.4) in 2D bounded domain, Prohl in [31] studied a nonlinear semi-discrete scheme

$$\begin{aligned} D_\tau \mathbf{m }^{j+1}-\lambda \Delta \mathbf{m }^{j+1}- \mathbf{m }^{j+1}\times \Delta \mathbf{m }^{j+1}=\lambda |\nabla \mathbf{m }^j|^2\mathbf{m }^{j+1} \end{aligned}$$

(1.7)

for $0\le j\le J$, where $D_\tau$ is the backward Euler difference operator. The optimal error estimate $O(\tau )$ in time for the scheme (1.7) was proved. Motivated by the projection method in [16], a finite element discretization scheme corresponding to (1.7) was proposed by using the linear finite element approximation in the spatial discretization in [31]. Meanwhile, a suboptimal error estimate $O(\tau +h)$ in $\mathbf{L }^2$-norm was derived under the condition $\frac{1}{\tau }=o(\frac{1}{h^2})$. (see Theorem 4.10 in [31]). Instead of solving a nonlinear problem (1.7), Cimrák in [14] modified (1.7) to a linear semi-discrete scheme:

$$\begin{aligned} D_\tau \mathbf{m }^{j+1}-\lambda \Delta \mathbf{m }^{j+1}- \mathbf{m }^{j}\times \Delta \mathbf{m }^{j+1}=\lambda |\nabla \mathbf{m }^j|^2\mathbf{m }^{j+1}, \end{aligned}$$

(1.8)

for which the optimal error $O(\tau )$ was proved only in temporal direction.

Based upon the recent works by Li-Sun [27, 28] (also see [19, 24, 29]), where the finite element error are split into the temporal error, the spatial error and the projection error, Gao in [18] proposed a linearized semi-implicit scheme for (1.4) and proved the unconditionally stable without any time step restriction. In this paper, we will give and establish the unconditionally optimal error estimates of a linearized Crank–Nicolson scheme for the approximation of the solution to the LL equation (1.2–1.4). Using $\mathbf{P }_r$ finite element to approximate $\mathbf{m }$ with $r=1,2$, the optimal error estimates $O(\tau ^2+h^{r+1})$ in $\mathbf{L }^2$-norm and $O(\tau ^2+h^r)$ in $\mathbf{H }^1$-norm are derived by using the error splitting technique under the assumption $\mathbf{m }\in \mathbf{L }^\infty (0,T;\mathbf{H }^3(\Omega ))$. Meanwhile, the convergence rate between the nonlinear constraint $|\mathbf{m }|=1$ and the modulus of the approximation solution is obtained in the sense of $\mathbf{L }^2$-norm. Since these optimal error estimates are shown to hold without any time step restriction, therefore, the Crank–Nicolson scheme is unconditionally stable.

The rest of the paper is organized as follows. In Sect. 2, we recall some known results and present some regularities of $\mathbf{m }$ which are used in the proof of the optimal error estimates. The linearized Crank–Nicolson finite element scheme and the main result in this paper are presented in Sect. 3. Meanwhile, the discrete parabolic system corresponding to (1.4) is introduced. To prove the optimal error estimates by the error splitting technique, the temporal errors and the spatial errors are shown in Sects. 4 and 5, respectively. Numerical results are provided to confirm the theoretical analysis in Sect. 7.

2 Preliminaries and Regularity Results

For $k\in {\mathbb {N}}^+, 1\le p\le +\infty$, let $W^{k,p}(\Omega )$ denote the standard Sobolev space. For $p=2$, we use $H^k(\Omega )$ to denote $W^{k,2}(\Omega )$. The boldface Sobolev spaces $\mathbf{H }^k(\Omega ), \mathbf{W }^{k,p}(\Omega )$ and $\mathbf{L }^p(\Omega )$ are used to denote the vector Sobolev spaces $H^k(\Omega )^d, W^{k,p}(\Omega )^d$ and $L^p(\Omega )^d$, respectively. In particular, $(\cdot ,\cdot )$ denotes the $\mathbf{L }^2(\Omega )$ inner product. The symbols $C, C_0, C_1, \ldots$ are used to denote a generic positive constant which may depend on $\Omega , \mathbf{m }, \mathbf{m }_0$, $\lambda$ and are independent of the mesh size h and the time step $\tau$.

Throughout this paper, the following known inequalities will be frequently used (cf. [1]):

$$\begin{aligned} ||\mathbf{v }||_{L^r}\le C ||\mathbf{v }||_{H^1}(2\le r\le 6),\quad&\forall \ \mathbf{v }\in \mathbf{H }^1(\Omega ), \end{aligned}$$

(2.1)

$$\begin{aligned} ||\mathbf{v }||_{L^4}\le C ||\mathbf{v }||_{L^2}^{\frac{4-d}{4}}||\mathbf{v }||_{H^1}^{\frac{d}{4}},\quad&\forall \ \mathbf{v }\in \mathbf{H }^1(\Omega ), \end{aligned}$$

(2.2)

$$\begin{aligned} ||\mathbf{v }||_{L^3}\le C ||\mathbf{v }||_{L^2}^{\frac{6-d}{6}}||\mathbf{v }||_{H^1}^{\frac{d}{6}},\quad&\forall \ \mathbf{v }\in \mathbf{H }^1(\Omega ), \end{aligned}$$

(2.3)

$$\begin{aligned} ||\nabla \mathbf{v }||_{L^3}\le C ||\nabla \mathbf{v }||_{L^2}^{\frac{6-d}{6}}||\Delta \mathbf{v }||_{L^2}^{\frac{d}{6}}+||\nabla \mathbf{v }||_{L^2},\quad&\forall \ \mathbf{v }\in \mathbf{H }^2(\Omega ), \end{aligned}$$

(2.4)

$$\begin{aligned} ||\mathbf{v }||_{L^\infty }\le C ||\mathbf{v }||_{L^2}^{1/2}||\mathbf{v }||_{H^2}^{1/2} ,\quad&\forall \ \mathbf{v }\in \mathbf{H }^2(\Omega ), \end{aligned}$$

(2.5)

$$\begin{aligned} ||\nabla ^2\mathbf{v }||_{L^2}\le C ||\Delta \mathbf{v }||_{L^2},\quad&\forall \ \mathbf{v }\in \mathbf{H }^2(\Omega ) \ \hbox {with} \ \nabla \mathbf{v }\cdot \mathbf{n }|_{\partial \Omega }=0, \end{aligned}$$

(2.6)

$$\begin{aligned} || \mathbf{v }||_{L^\infty }\le C ||\mathbf{v }||_{L^2}^{\frac{4-d}{4}}(||\mathbf{v }||_{L^2}^2+||\Delta \mathbf{v }||_{L^2}^2)^{\frac{d}{8}}, \quad&\forall \ \mathbf{v }\in \mathbf{H }^2(\Omega ) \ \hbox {with} \ \nabla \mathbf{v }\cdot \mathbf{n }|_{\partial \Omega }=0, \end{aligned}$$

(2.7)

$$\begin{aligned} || \nabla ^3\mathbf{v }||_{L^2}\le C ||\nabla \Delta \mathbf{v }||_{L^2} +||\Delta \mathbf{v }||_{L^2} , \quad&\forall \ \mathbf{v }\in \mathbf{H }^3(\Omega ) \ \hbox {with} \ \nabla \mathbf{v }\cdot \mathbf{n }|_{\partial \Omega }=0. \end{aligned}$$

(2.8)

Carbou-Fabrie in [9] proved the following existence result of the local strong solution to the LL equation (1.4) and (1.2).

Theorem 2.1

Suppose $\mathbf{m }_0\in \mathbf{H }^2(\Omega )$. Then there exists a positive number T such that the solution $\mathbf{m }$ to the LL equation (1.4) and (1.2) satisfies the following estimate:

$$\begin{aligned} \max \limits _{t\in [0,T]}||\mathbf{m }(t)||_{H^2}\le C. \end{aligned}$$

(2.9)

Subsequently, Cimrák in [14] derived the estimates for $\mathbf{m }_t$.

Theorem 2.2

For the solution $\mathbf{m }$ to the LL equation (1.4) and (1.2) and taking T from Theorem 2.1, the following estimates are valid:

$$\begin{aligned} \max \limits _{t\in [0,T]}||\mathbf{m }_t(t)||_{L^2}+\displaystyle \int _0^T||\nabla \mathbf{m }_t||_{L^2}^2dt&\le C, \end{aligned}$$

(2.10)

$$\begin{aligned} \max \limits _{t\in [0,T]}\sqrt{\kappa (t)}||\nabla \mathbf{m }_t(t)||_{L^2} +\displaystyle \int _0^T\kappa (t)\left\{ ||\mathbf{m }_{tt}||_{L^2}^2+||\Delta \mathbf{m }_t||_{L^2}^2\right\} dt&\le C, \end{aligned}$$

(2.11)

where $\kappa (t)$ is a time weight function defined by $\kappa (t)=\min \{1, t\}$.

We can improve the regularity results in Theorem 2.2 under the strong assumption about $\mathbf{m }_0$. The improved regularity will be used to prove the optimal error estimates of the Crank–Nicolson scheme proposed in next section.

Theorem 2.3

Suppose that $\mathbf{m }_0\in \mathbf{H }^5(\Omega )$ satisfies the compatibility conditions:

$$\begin{aligned} \nabla \mathbf{m }_0\cdot \mathbf{n }=0 \quad \hbox {and} \quad \nabla (\lambda \Delta \mathbf{m }_0 +\mathbf{m }_0\times \Delta \mathbf{m }_0+\lambda |\nabla \mathbf{m }_0|^2\mathbf{m }_0)\cdot \mathbf{n }=0 \quad \hbox {on} \ \partial \Omega . \end{aligned}$$

Taking T from Theorem 2.1, we have

$$\begin{aligned} \max \limits _{t\in [0,T]}\left( ||\mathbf{m }||_{H^3}+ || \mathbf{m }_t||_{H^2}+|| \mathbf{m }_{tt}||_{H^1}\right) +\displaystyle \int _0^T \left( || \mathbf{m }_t||_{H^3}^2+|| \mathbf{m }_{tt}||_{H^2}^2\right) dt\le C. \end{aligned}$$

(2.12)

Proof

The proof of this theorem is too long and is postponed to the Appendix. $\square$

3 Main Results

Let $T_h=\{K\}$ be a quasi-uniform triangular or tetrahedral partition of $\Omega$ into triangles or tetrahedrons of diameters bounded by $h, 0<h<1$. For every $K\in \mathcal {T}_h$ and a nonnegative integer r, $P_{r}(K)$ denotes the space of the polynomials on K of degree at most r. Noticing the regularities of the solution mentioned in the above section, in this paper, we introduce the following finite element space

$$\begin{aligned} \mathbf{V }_h=\{\mathbf{v }_h\in \mathcal {C}(\overline{\Omega }), \ \mathbf{v }_h\in \mathbf{P }_r(K), \ r=1, 2, \quad \forall \ K\in T_h\}. \end{aligned}$$

Let $0=t_0<t_1<\cdots <t_N=T$ be a uniform partition of the time interval [0, T] with time step $\tau =T/N$ and $t_n=n\tau , 0\le n\le N$, where [0, T] is a time interval such that a strong solution exists. For $0\le n\le N$, let us denote $\mathbf{m }^n=\mathbf{m }(t_n,\mathbf{x })$. For any sequence of functions $\{f^n\}_{n=0}^N$, we define

$$\begin{aligned} D_\tau f^{n+1}=\displaystyle \frac{f^{n+1}-f^{n}}{\tau },\quad \overline{f}^{n+1/2}=\displaystyle \frac{f^{n+1}+f^{n}}{2},\quad \hat{f}^{n+1/2}=\displaystyle \frac{1}{2}(3f^n-f^{n-1}) \end{aligned}$$

for $1\le n\le N-1$. Under the above notations, we propose a linearized Crank–Nicolson Galerkin finite element scheme for the LL equation (1.4) and (1.2), which is to find $\mathbf{M }^{n+1}_h\in \mathbf{V }_h$ such that

$$\begin{aligned}&\left( D_\tau \mathbf{M }_h^{n+1},\mathbf{v }_h\right) +\lambda \left( \nabla \overline{\mathbf{M }}_h^{n+1/2},\nabla \mathbf{v }_h\right) + \left( \hat{\mathbf{M }}_h^{n+1/2}\times \nabla \overline{\mathbf{M }}_h^{n+1/2},\nabla \mathbf{v }_h\right) \nonumber \\&\quad =\lambda \left( |\nabla \hat{\mathbf{M }}_h^{n+1/2}|^2\hat{\mathbf{M }}_h^{n+1/2},\mathbf{v }_h\right) ,\quad \forall \ \mathbf{v }_h\in \mathbf{V }_h, \end{aligned}$$

(3.1)

for $n=0,1,\ldots ,N-1$, where $\hat{\mathbf{M }}_h^{1/2}$ is defined by

$$\begin{aligned}&\left( \displaystyle \frac{\hat{\mathbf{M }}_h^{1/2} -\mathbf{M }_h^0}{\tau /2},\mathbf{v }_h\right) +\lambda \left( \nabla \hat{\mathbf{M }}_h^{1/2},\nabla \mathbf{v }_h\right) + \left( \mathbf{M }_h^0\times \nabla \hat{\mathbf{M }}_h^{1/2},\nabla \mathbf{v }_h\right) \nonumber \\&\quad =\lambda \left( |\nabla \mathbf{M }_h^0|^2 \mathbf{M }_h^0,\mathbf{v }_h\right) ,\quad \forall \ \mathbf{v }_h\in \mathbf{V }_h \end{aligned}$$

(3.2)

with $\mathbf{M }_h^0=\Pi _h\mathbf{m }_0$. Here $\Pi _h$ is the Ritz projection from $\mathbf{H }^1(\Omega )$ onto $\mathbf{V }_h$ defined by

$$\begin{aligned} (\nabla \Pi _h\mathbf{u },\nabla \mathbf{v }_h)+(\Pi _h\mathbf{u },\mathbf{v }_h)=(\nabla \mathbf{u },\nabla \mathbf{v }_h)+( \mathbf{u },\mathbf{v }_h),\quad \forall \ \mathbf{u }\in \mathbf{H }^1(\Omega ), \mathbf{v }_h\in \mathbf{V }_h. \end{aligned}$$

Moreover, the following approximation is valid (see [8]):

$$\begin{aligned} ||\mathbf{E }^0||_{L^p}+h||\mathbf{E }^0||_{W^{1,p}}\le C h^{r+1} ,\quad \hbox {for} \ 2\le p\le +\infty , \end{aligned}$$

(3.3)

where $\mathbf{E }^0=\mathbf{m }_0-\mathbf{M }_h^0$. Taking $\mathbf{v }_h=\mathbf{M }_h^{n+1}$ in (3.1) and $\mathbf{v }_h=\hat{\mathbf{M }}_h^{1/2}$ in (3.2), the existence and uniqueness of $\mathbf{M }_h^{n+1}$ and $\hat{\mathbf{M }}_h^{1/2}$ are from the Lax–Milgram theorem due to the identity $(\mathbf{u }\times \nabla \mathbf{v },\nabla \mathbf{v })=0$.

The emphasis of this paper is to show the optimal error estimates for Crank–Nicolson Galerkin scheme (3.1–3.2). Then we assume the LL equation (1.4) and (1.2) admit a unique solution $\mathbf{m }$ satisfying (2.12) mentioned in Theorem 2.3. The main result derived in this paper is presented in the following theorem.

Theorem 3.1

Suppose that the solution $\mathbf{m }$ to (1.4) and (1.2) satisfies (2.12). Then the discrete system (3.1–3.2) admits a unique solution $\mathbf{M }_h^{n}$ for $1\le n\le N$. Moreover, there exists two positive $h_0$ and $\tau _0$ such that when $h<h_0$ and $\tau <\tau _0$, the following optimal error estimates hold:

$$\begin{aligned} \max \limits _{1\le n\le N}\left( ||\mathbf{m }^n-\mathbf{M }^n_h||_{L^2}+||1-|\mathbf{M }^n_h|^2||_{L^2}\right)&\le C_0(\tau ^2+h^{r+1}), \end{aligned}$$

(3.4)

$$\begin{aligned} \max \limits _{1\le n\le N}||\mathbf{m }^n-\mathbf{M }^n_h||_{H^1}&\le C_0(\tau ^2+h^r), \end{aligned}$$

(3.5)

where $C_0>0$ is some positive constant independent of h and $\tau$.

Theorem 3.1 will be proved by using the temporal-spatial error splitting technique proposed in [27, 28] in next three sections. Before the beginning of the proof, we recall other known inequalities frequently used in our proof. The following inverse inequality holds for $\mathbf{v }_h\in \mathbf{V }_h$ [8]:

$$\begin{aligned} ||\mathbf{v }_h||_{W^{l,q_1}}\le Ch^{m-l+d\min \left\{ \frac{1}{q_1}-\frac{1}{q_2},0\right\} }||\mathbf{v }_h||_{W^{m,q_2}},\quad \forall \ 1\le q_1,q_2\le \infty , \ 0\le m\le l. \end{aligned}$$

(3.6)

By the triangular and Cauchy inequalities, we have the following inequality [19]:

Lemma 3.1

Let $\{v^n\}_{n=0}^N$ be a sequence of function on $\Omega$. Then for any norm $||\cdot ||$, there holds

$$\begin{aligned} \tau ||v^n||\le 2\tau \displaystyle \sum \limits _{m=0}^{n-1}||\overline{v}^{m+1/2}||+\tau ||v^0|| \le 2\sqrt{T}\sqrt{\tau \displaystyle \sum \limits _{m=0}^{n-1}||\overline{v}^{m+1/2}||^2}+\tau ||v^0||. \end{aligned}$$

(3.7)

Finally, we recall a discrete version of Gronwall’s inequality established in [23].

Lemma 3.2

Let $a_k, b_k, c_k$ and $\gamma _k$, for integers $k\ge 0$, be the nonnegative numbers such that

$$\begin{aligned} a_n+\tau \displaystyle \sum \limits _{k=0}^n b_k\le \tau \displaystyle \sum \limits _{k=0}^n \gamma _ka_k+ \tau \displaystyle \sum \limits _{k=0}^n c_k+B, \quad \hbox {for} \ n\ge 0. \end{aligned}$$

(3.8)

Suppose that $\tau \gamma _k<1$, for all k, and set $\sigma _k=(1-\tau \gamma _k)^{-1}$. Then

$$\begin{aligned} a_n+\tau \displaystyle \sum \limits _{k=0}^n b_k\le \exp \left( \tau \displaystyle \sum \limits _{k=0}^n \gamma _k \sigma _k\right) \left( \tau \displaystyle \sum \limits _{k=0}^n c_k+B \right) , \quad \hbox {for} \ n\ge 0. \end{aligned}$$

(3.9)

Remark 3.1

If the first sum on the right in (3.8) extends only up to $n-1$, then estimate (3.9) holds for all $k>0$ with $\sigma _k=1$.

4 Temporal Error Analysis

In this section, we begin to estimate the temporal errors. For $\mathbf{M }^0=\mathbf{m }_0$, we define $\mathbf{M }^{n+1}, 0\le n\le N-1$, to be the solutions of the following discrete parabolic (or elliptic) system corresponding to the LL equation (1.4) and (1.2):

$$\begin{aligned}&D_\tau \mathbf{M }^{n+1}-\lambda \Delta \overline{\mathbf{M }}^{n+1/2} - \hat{\mathbf{M }}^{n+1/2}\times \Delta \overline{\mathbf{M }}^{n+1/2}\nonumber \\&\qquad - \nabla \hat{\mathbf{M }}^{n+1/2}\times \nabla \overline{\mathbf{M }}^{n+1/2} =\lambda |\nabla \hat{\mathbf{M }}^{n+1/2}|^2\hat{\mathbf{M }}^{n+1/2} \end{aligned}$$

(4.1)

with homogeneous Neumann boundary condition $\nabla \mathbf{M }^{n+1}\cdot \mathbf{n }|_{\partial \Omega }=0$, where $\hat{\mathbf{M }}^{1/2}$ is the solution to

$$\begin{aligned} \displaystyle \frac{\hat{\mathbf{M }}^{1/2}-\mathbf{m }_0}{\tau /2}-\lambda \Delta \hat{\mathbf{M }}^{1/2}- \mathbf{m }_0\times \Delta \hat{\mathbf{M }}^{1/2}- \nabla \mathbf{m }_0\times \nabla \hat{\mathbf{M }}^{1/2}=\lambda |\nabla \mathbf{m }_0|^2 \mathbf{m }_0 \end{aligned}$$

(4.2)

with homogeneous Neumann boundary condition $\nabla \hat{\mathbf{M }}^{1/2}\cdot \mathbf{n }|_{\partial \Omega }=0$. For all $\mathbf{v }\in \mathbf{H }^1(\Omega )$, the weak formulations of (4.1) and (4.2) are defined as follows: find $\mathbf{M }^{n+1}\in \mathbf{H }^1(\Omega )$ such that

$$\begin{aligned} (D_\tau \mathbf{M }^{n+1},\mathbf{v })\!+\!\lambda \left( \nabla \overline{\mathbf{M }}^{n+1/2},\nabla \mathbf{v }\right) \!+\! \left( \hat{\mathbf{M }}^{n\!+\!1/2}\times \nabla \overline{\mathbf{M }}^{n+1/2},\nabla \mathbf{v }\right) \!=\!\lambda \left( |\nabla \hat{\mathbf{M }}^{n+1/2}|^2\hat{\mathbf{M }}^{n+1/2},\mathbf{v }\right) , \end{aligned}$$

(4.3)

and $\hat{\mathbf{M }}^{1/2}\in \mathbf{H }^1(\Omega )$ such that

$$\begin{aligned} \left( \displaystyle \frac{\hat{\mathbf{M }}^{1/2}-\mathbf{m }_0}{\tau /2},\mathbf{v }\right) +\lambda \left( \nabla \hat{\mathbf{M }}^{1/2},\nabla \mathbf{v }\right) + \left( \mathbf{m }_0\times \nabla \hat{\mathbf{M }}^{1/2},\nabla \mathbf{v }\right) = \lambda \left( |\nabla \mathbf{m }_0|^2 \mathbf{m }_0,\mathbf{v }\right) . \end{aligned}$$

(4.4)

The existence and uniqueness of the solution to the problems (4.3) and (4.4) can be easily proved by using Lax–Milgram theorem. For $n=0,1,\ldots ,N-1$, the solution $\mathbf{m }^{n+1}$ satisfies the following discrete parabolic system:

$$\begin{aligned}&D_\tau \mathbf{m }^{n+1}-\lambda \Delta \overline{\mathbf{m }}^{n+1/2} - \hat{\mathbf{m }}^{n+1/2}\times \Delta \overline{\mathbf{m }}^{n+1/2}\nonumber \\&\qquad - \nabla \hat{\mathbf{m }}^{n+1/2}\times \nabla \overline{\mathbf{m }}^{n+1/2} =\mathbf{R }^n+\lambda |\nabla \hat{\mathbf{m }}^{n+1/2}|^2 \hat{\mathbf{m }}^{n+1/2} \end{aligned}$$

(4.5)

with Neumann boundary condition $\nabla \mathbf{m }^{n+1}\cdot \mathbf{n }=0$ on $\partial \Omega$, where

$$\begin{aligned} \mathbf{R }^n&=D_\tau \mathbf{m }^{n+1}-\mathbf{m }_t^{n+1/2}-\lambda \Delta (\overline{\mathbf{m }}^{n+1/2}-\mathbf{m }^{n+1/2})- \hat{\mathbf{m }}^{n+1/2}\times \Delta (\overline{\mathbf{m }}^{n+1/2}-\mathbf{m }^{n+1/2})\\&\quad - ( \hat{\mathbf{m }}^{n+1/2}-\mathbf{m }^{n+1/2})\times \Delta \mathbf{m }^{n+1/2}- \nabla \hat{\mathbf{m }}^{n+1/2}\times \nabla (\overline{\mathbf{m }}^{n+1/2}-\mathbf{m }^{n+1/2})\\&\quad - \nabla (\hat{\mathbf{m }}^{n+1/2}-\mathbf{m }^{n+1/2})\times \nabla \mathbf{m }^{n+1/2}+\lambda |\nabla \mathbf{m }^{n+1/2}|^2 \mathbf{m }^{n+1/2}-\lambda |\nabla \hat{\mathbf{m }}^{n+1/2}|^2 \hat{\mathbf{m }}^{n+1/2}. \end{aligned}$$

Moreover, $\hat{\mathbf{m }}^{1/2}$ satisfies

$$\begin{aligned} \displaystyle \frac{\hat{\mathbf{m }}^{1/2}-\mathbf{m }_0}{\tau /2}-\lambda \Delta \hat{\mathbf{m }}^{1/2}- \mathbf{m }_0\times \Delta \hat{\mathbf{m }}^{1/2}- \nabla \mathbf{m }_0\times \nabla \hat{\mathbf{m }}^{1/2} =\hat{\mathbf{R }}^0, \end{aligned}$$

(4.6)

with Neumann boundary condition $\nabla \hat{\mathbf{m }}^{1/2}\cdot \mathbf{n }=0$ on $\partial \Omega$, where

$$\begin{aligned} \hat{\mathbf{R }}^0=&\displaystyle \frac{ \mathbf{m }^{1/2}-\mathbf{m }_0}{\tau /2}-\mathbf{m }_t^{1/2} - ( \mathbf{m }_0-\mathbf{m }^{1/2})\times \Delta \mathbf{m }^{1/2}\\&- \nabla (\mathbf{m }_0-\mathbf{m }^{1/2})\times \nabla \mathbf{m }^{1/2}+\lambda |\nabla \mathbf{m }^{1/2}|^2 \mathbf{m }^{1/2}. \end{aligned}$$

By using the regularity (2.12) for $\mathbf{m }$ and Taylor formulation, we derive

$$\begin{aligned} \left( \tau \displaystyle \sum \limits _{n=0}^{N-1}||\mathbf{R }^n||_{H^1}^2\right) ^{1/2} +\tau ||\hat{\mathbf{R }}^0||_{H^1}\le C\tau ^2. \end{aligned}$$

(4.7)

For $0\le n\le N-1$, let us denote

$$\begin{aligned} \mathbf{e }^{n+1}=&\mathbf{m }^{n+1}-\mathbf{M }^{n+1}, \ \overline{\mathbf{e }}^{n+1/2}=\overline{\mathbf{m }}^{n+1/2}-\overline{\mathbf{M }}^{n+1/2}, \\ \hat{\mathbf{e }}^{n+1/2}=&\hat{\mathbf{m }}^{n+1/2}-\hat{\mathbf{M }}^{n+1/2}, \hat{\mathbf{e }}^{1/2}=\hat{\mathbf{m }}^{1/2}-\hat{\mathbf{M }}^{1/2}. \end{aligned}$$

Subtracting (4.2) from (4.6) leads to

$$\begin{aligned}&2\hat{\mathbf{e }}^{1/2}- {\lambda \tau }\Delta \hat{\mathbf{e }}^{1/2}- { \tau }\mathbf{m }_0\times \Delta \hat{\mathbf{e }}^{1/2}- { \tau } \nabla \mathbf{m }_0\times \nabla \hat{\mathbf{e }}^{1/2}\nonumber \\&\quad =\tau \hat{\mathbf{R }}^0+\tau \lambda \left( |\nabla \mathbf{m }^{1/2}|^2 \mathbf{m }^{1/2}-|\nabla \mathbf{m }_0|^2 \mathbf{m }_0\right) . \end{aligned}$$

(4.8)

Then we can prove

Lemma 4.1

Suppose that the solution $\mathbf{m }$ to (1.4) and (1.2) satisfies the regularity (2.12). Then there exist some $C>0$ independent of $\tau$ such that $\hat{\mathbf{e }}^{1/2}$ satisfies

$$\begin{aligned} ||\hat{\mathbf{e }}^{1/2}||_{L^2}+\tau ^{1/2}|| \hat{\mathbf{e }}^{1/2}||_{H^2}+\tau || \hat{\mathbf{e }}^{1/2}||_{H^3}\le C\tau ^2. \end{aligned}$$

(4.9)

Proof

Multiplying (4.8) by $\hat{\mathbf{e }}^{1/2}$ and integrating over $\Omega$, we get

$$\begin{aligned} ||\hat{\mathbf{e }}^{1/2}||_{L^2}^2+2\lambda \tau ||\nabla \hat{\mathbf{e }}^{1/2}||_{L^2}^2\le \tau ^2||\hat{\mathbf{R }}^0||_{L^2}^2 +\lambda \tau ^2|||\nabla \mathbf{m }^{1/2}|^2 \mathbf{m }^{1/2}-|\nabla \mathbf{m }_0|^2 \mathbf{m }_0||_{L^2}^2, \end{aligned}$$

where we use Young’s inequality and $(\mathbf{m }_0\times \Delta \hat{\mathbf{e }}^{1/2},\hat{\mathbf{e }}^{1/2})+(\nabla \mathbf{m }_0\times \nabla \hat{\mathbf{e }}^{1/2},\hat{\mathbf{e }}^{1/2})=0$. By the regularities of $\mathbf{m }$ and $\mathbf{m }_0$, using Taylor formulation, we obtain

$$\begin{aligned}&\ ||\nabla \mathbf{m }^{1/2}|^2 \mathbf{m }^{1/2}-|\nabla \mathbf{m }_0|^2 \mathbf{m }_0||_{L^2}^2\\&\quad \le ||(\nabla \mathbf{m }^{1/2}+\nabla \mathbf{m }_0)(\nabla \mathbf{m }^{1/2}-\nabla \mathbf{m }_0) \mathbf{m }^{1/2}||_{L^2}^2 +|||\nabla \mathbf{m }_0|^2( \mathbf{m }^{1/2}- \mathbf{m }_0)||_{L^2}^2\le C\tau ^2. \end{aligned}$$

Thus, there holds

$$\begin{aligned} ||\hat{\mathbf{e }}^{1/2}||_{L^2}\le C\tau ^2\quad \hbox {and} \quad ||\nabla \hat{\mathbf{e }}^{1/2}||_{L^2}^2\le C\tau ^3. \end{aligned}$$

(4.10)

Testing (4.8) by $-\Delta \hat{\mathbf{e }}^{1/2}$ and observing $(\mathbf{m }_0\times \Delta \hat{\mathbf{e }}^{1/2},\Delta \hat{\mathbf{e }}^{1/2})=0$ yield

$$\begin{aligned}&\ 2||\nabla \hat{\mathbf{e }}^{1/2}||_{L^2}^2+\lambda \tau ||\Delta \hat{\mathbf{e }}^{1/2}||_{L^2}^2\\&\quad \!=\!-\!{\gamma \tau } \left( \nabla \mathbf{m }_0\times \nabla \hat{\mathbf{e }}^{1/2},\Delta \hat{\mathbf{e }}^{1/2}\right) \!-\!\tau \left( \hat{\mathbf{R }}^0,\Delta \hat{\mathbf{e }}^{1/2}\right) \!-\!\lambda \tau \left( |\nabla \mathbf{m }^{1/2}|^2 \mathbf{m }^{1/2} \!-\!|\nabla \mathbf{m }_0|^2 \mathbf{m }_0,\Delta \hat{\mathbf{e }}^{1/2}\right) \\&\quad \le C\tau \left( ||\nabla \mathbf{m }_0||_{L^\infty }||\nabla \hat{\mathbf{e }}^{1/2}||_{L^2}+||\hat{\mathbf{R }}^0||_{L^2}+ ||\nabla \mathbf{m }^{1/2}|^2 \mathbf{m }^{1/2}-|\nabla \mathbf{m }_0|^2 \mathbf{m }_0||_{L^2} \right) ||\Delta \hat{\mathbf{e }}^{1/2}||_{L^2}\\&\quad \le \displaystyle \frac{\lambda \tau }{2}||\Delta \hat{\mathbf{e }}^{1/2}||_{L^2}^2 + C\tau \left( ||\nabla \hat{\mathbf{e }}^{1/2}||_{L^2}^2+||\hat{\mathbf{R }}^0||_{L^2}^2+ ||\nabla \mathbf{m }^{1/2}|^2 \mathbf{m }^{1/2}-|\nabla \mathbf{m }_0|^2 \mathbf{m }_0||_{L^2}^2\right) , \end{aligned}$$

which together with (4.10) implies that $||\Delta \hat{\mathbf{e }}^{1/2}||_{L^2}\le C\tau$. From (2.6) and (4.10) again, one has $||\hat{\mathbf{e }}^{1/2}||_{H^2}\le C\tau .$ Differentiating (4.8) with respect to $\mathbf{x }$ and testing the resulting equation by $-\nabla \Delta \hat{\mathbf{e }}^{1/2}$ lead to

$$\begin{aligned}&||\Delta \hat{\mathbf{e }}^{1/2}||_{L^2}^2 +\lambda \tau ||\nabla \Delta \hat{\mathbf{e }}^{1/2}||_{L^2}^2\nonumber \\&\quad \le C\tau ||\nabla \hat{\mathbf{R }}_0||_{L^2}^2+ C \tau ||\nabla \mathbf{m }_0\times \Delta \hat{\mathbf{e }}^{1/2}||_{L^2}^2+ C\tau \left| |\nabla ^2\mathbf{m }_0\times \nabla \hat{\mathbf{e }}^{1/2}\right| |_{L^2}^2\nonumber \\&\quad \quad + C\tau ||\nabla \mathbf{m }_0\times \nabla ^2 \hat{\mathbf{e }}^{1/2}||_{L^2}^2+ C\tau ||(\nabla ^2\mathbf{m }^{1/2}\cdot \nabla \mathbf{m }^{1/2})\mathbf{m }^{1/2} -(\nabla ^2\mathbf{m }_0\cdot \nabla \mathbf{m }_0)\mathbf{m }_0||_{L^2}^2\nonumber \\&\qquad + C\tau |||\nabla \mathbf{m }^{1/2}|^2\nabla \mathbf{m }^{1/2} -|\nabla \mathbf{m }_0|^2\nabla \mathbf{m }_0||_{L^2}^2\nonumber \\&\quad \le C\tau ^3+ C\tau ||(\nabla ^2\mathbf{m }^{1/2}\cdot \nabla \mathbf{m }^{1/2})\mathbf{m }^{1/2} -(\nabla ^2\mathbf{m }_0\cdot \nabla \mathbf{m }_0)\mathbf{m }_0||_{L^2}^2\nonumber \\&\qquad + C\tau |||\nabla \mathbf{m }^{1/2}|^2\nabla \mathbf{m }^{1/2} -|\nabla \mathbf{m }_0|^2\nabla \mathbf{m }_0||_{L^2}^2. \end{aligned}$$

(4.11)

The last two terms in the right-hand side of (4.11) are bounded, respectively, by

$$\begin{aligned}&\tau ||(\nabla ^2\mathbf{m }^{1/2}\cdot \nabla \mathbf{m }^{1/2})\mathbf{m }^{1/2} -(\nabla ^2\mathbf{m }_0\cdot \nabla \mathbf{m }_0)\mathbf{m }_0||_{L^2}^2\\&\quad \le C\tau ||(\nabla ^2\mathbf{m }^{1/2}\cdot (\nabla \mathbf{m }^{1/2}-\nabla \mathbf{m }_0))\mathbf{m }^{1/2}||_{L^2}^2 + C\tau ||(\nabla ^2\mathbf{m }^{1/2}\cdot \nabla \mathbf{m }_0)(\mathbf{m }^{1/2}-\mathbf{m }_0)||_{L^2}^2\\&\quad \quad + C\tau ||((\nabla ^2\mathbf{m }^{1/2}-\nabla ^2\mathbf{m }_0)\cdot \mathbf{m }_0)\mathbf{m }_0||_{L^2}^2\\&\quad \le C\tau ^3\left( ||\mathbf{m }^{1/2}||_{H^3}^4+||\mathbf{m }^{1/2}||_{H^2}^2||\mathbf{m }_0||_{H^3}^2+||\mathbf{m }_0||_{H^3}^4\right) ||\mathbf{m }_t||_{L^\infty (0,T;H^2(\Omega ))}^2, \end{aligned}$$

and

$$\begin{aligned}&\tau |||\nabla \mathbf{m }^{1/2}|^2\nabla \mathbf{m }^{1/2}-|\nabla \mathbf{m }_0|^2\nabla \mathbf{m }_0||_{L^2}^2\\&\quad \le C\tau ||(\nabla \mathbf{m }^{1/2}+\nabla \mathbf{m }_0)(\nabla \mathbf{m }^{1/2}-\nabla \mathbf{m }_0) \nabla \mathbf{m }^{1/2}||_{L^2}^2 + C\tau |||\nabla \mathbf{m }_0|^2\nabla ( \mathbf{m }^{1/2}- \mathbf{m }_0)||_{L^2}^2\\&\quad \le C\tau ^3\left( ||\mathbf{m }^{1/2}+\mathbf{m }_0||_{H^3}^2||\mathbf{m }^{1/2}||_{H^3}^2 +||\mathbf{m }_0||_{H^3}^4\right) ||\mathbf{m }_t||_{L^\infty (0,T;H^1(\Omega ))}^2. \end{aligned}$$

Combining these estimates into (4.11), we conclude that

$$\begin{aligned} ||\Delta \hat{\mathbf{e }}^{1/2}||_{L^2}^2\le C\tau ^3,\quad \hbox {and}\quad ||\nabla \Delta \hat{\mathbf{e }}^{1/2}||_{L^2}\le C\tau , \end{aligned}$$

which together with (4.10) completes the proof of (4.9). $\square$

As a consequence of Lemma 4.1 and (2.12), we get

$$\begin{aligned} ||\hat{\mathbf{M }}^{1/2}||_{H^3}\le ||\hat{\mathbf{e }}^{1/2}||_{H^3}+||\hat{\mathbf{m }}^{1/2}||_{H^3}&\le C. \end{aligned}$$

(4.12)

For $n=0,\ldots , N-1$, subtracting (4.1) from (4.5) leads to

$$\begin{aligned} D_\tau \mathbf{e }^{n+1}-\lambda \Delta \overline{\mathbf{e }}^{n+1/2}&=\mathbf{R }^n+ \left( \hat{\mathbf{m }}^{n+1/2}\times \Delta \overline{\mathbf{m }}^{n+1/2} -\hat{\mathbf{M }}^{n+1/2}\times \Delta \overline{\mathbf{M }}^{n+1/2}\right) \nonumber \\&\quad + \left( \nabla \hat{\mathbf{m }}^{n+1/2}\times \nabla \overline{\mathbf{m }}^{n+1/2}- \nabla \hat{\mathbf{M }}^{n+1/2}\times \nabla \overline{\mathbf{M }}^{n+1/2}\right) \nonumber \\&\quad +\lambda |\nabla \hat{\mathbf{m }}^{n+1/2}|^2 \hat{\mathbf{m }}^{n+1/2}-\lambda |\nabla \hat{\mathbf{M }}^{n+1/2}|^2\hat{\mathbf{M }}^{n+1/2}. \end{aligned}$$

(4.13)

Based upon (4.9) and (4.12), we have

Theorem 4.1

Suppose that the solution $\mathbf{m }$ to (1.4) and (1.2) satisfies the regularities (2.12). Then there exist some constant $C>0$ independent of $\tau$ such that $\mathbf{e }^{1}$ satisfies

$$\begin{aligned} || \mathbf{e }^{1}||_{L^2}+\tau ^{1/2}||\mathbf{e }^{1}||_{H^1} +\tau ||\mathbf{e }^{1}||_{H^2}+\tau ^{3/2}||\mathbf{e }^{1}||_{H^3}\le C\tau ^{5/2}. \end{aligned}$$

(4.14)

Proof

Taking $n=0$ in (4.13) and observing $\mathbf{e }^0=0$, we have

$$\begin{aligned}&{\mathbf{e }^1} -\lambda \tau \Delta \mathbf{e }^1 =\tau \mathbf{R }^0+\tau \left( \hat{\mathbf{m }}^{1/2}\times \Delta \overline{\mathbf{m }}^{1/2} -\hat{\mathbf{M }}^{1/2}\times \Delta \overline{\mathbf{M }}^{1/2}\right) \nonumber \\&\qquad +\tau \left( \nabla \hat{\mathbf{m }}^{1/2}\times \nabla \overline{\mathbf{m }}^{1/2}- \nabla \hat{\mathbf{M }}^{1/2}\times \nabla \overline{\mathbf{M }}^{1/2}\right) +\lambda \tau \left( |\nabla \hat{\mathbf{m }}^{1/2}|^2 \hat{\mathbf{m }}^{1/2}- |\nabla \hat{\mathbf{M }}^{1/2}|^2\hat{\mathbf{M }}^{1/2}\right) . \end{aligned}$$

(4.15)

Testing the above equation by $\mathbf{e }^1$, we obtain

$$\begin{aligned}&||\mathbf{e }^1||_{L^2}^2+\lambda \tau ||\nabla \mathbf{e }^1||_{L^2}^2\nonumber \\&\quad =\tau (\mathbf{R }^0,\mathbf{e }^1)+ \tau (\hat{\mathbf{m }}^{1/2}\times \Delta \overline{\mathbf{m }}^{1/2} -\hat{\mathbf{M }}^{1/2}\times \Delta \overline{\mathbf{M }}^{1/2},\mathbf{e }^1)\nonumber \\&\quad + \tau \left( \nabla \hat{\mathbf{m }}^{1/2}\times \nabla \overline{\mathbf{m }}^{1/2}- \nabla \hat{\mathbf{M }}^{1/2}\times \nabla \overline{\mathbf{M }}^{1/2},\mathbf{e }^1\right) \nonumber \\&\quad +\lambda \tau \left( |\nabla \hat{\mathbf{m }}^{{1/2}}|^2 \hat{\mathbf{m }}^{1/2} - |\nabla \hat{\mathbf{M }}^{1/2}|^2\hat{\mathbf{M }}^{1/2},\mathbf{e }^1\right) \nonumber \\&\quad =J_1+J_2+J_3+J_4. \end{aligned}$$

(4.16)

From (4.7), it is easy to obtain

$$\begin{aligned} J_1=\tau (\mathbf{R }^0,\mathbf{e }^1)\le \displaystyle \frac{1}{4}||\mathbf{e }^1||_{\mathbf{L }^2}^2+\tau ^2||\mathbf{R }^0||_{L^2}^2\le \displaystyle \frac{1}{4}||\mathbf{e }^1||_{L^2}^2+ C\tau ^5. \end{aligned}$$

By integrating by parts, there holds

$$\begin{aligned} J_2+J_3&=- \tau \left( \hat{\mathbf{e }}^{1/2}\times \nabla \overline{\mathbf{m }}^{1/2},\nabla \mathbf{e }^1\right) \\&\le \tau ||\nabla \overline{\mathbf{m }}^{1/2}||_{L^\infty }||\hat{\mathbf{e }}^{1/2}||_{L^2}||\nabla \mathbf{e }^1||_{L^2} \le \displaystyle \frac{\lambda \tau }{2}||\nabla \mathbf{e }^1||_{L^2}^2+ C\tau ^5, \end{aligned}$$

where we use $\overline{\mathbf{m }}^{1/2}-\overline{\mathbf{M }}^{1/2}=\mathbf{e }^1$. In order to bound $I_4$, according to

$$\begin{aligned}&|\nabla \hat{\mathbf{m }}^{{1/2}}|^2 \hat{\mathbf{m }}^{{1/2}}- |\nabla \hat{\mathbf{M }}^{1/2}|^2\hat{\mathbf{M }}^{1/2} \end{aligned}$$

(4.17)

$$\begin{aligned}&\quad =\left( \nabla \hat{\mathbf{e }}^{1/2}\cdot (\nabla \hat{\mathbf{m }}^{{1/2}}+ \nabla \hat{\mathbf{M }}^{{1/2}})\right) \hat{\mathbf{m }}^{{1/2}}+|\nabla \hat{\mathbf{M }}^{{1/2}}|^2\hat{\mathbf{e }}^{1/2}, \end{aligned}$$

(4.18)

we have

$$\begin{aligned} I_4&\le \tau \left( ||\nabla \hat{\mathbf{m }}^{1/2}+\nabla \hat{\mathbf{M }}^{1/2}||_{L^\infty }||\nabla \hat{\mathbf{e }}^{1/2}||_{L^2}|| \hat{\mathbf{m }}^{1/2}|_{L^\infty }+||\nabla \hat{\mathbf{M }}^{1/2}||_{L^\infty }^2|| \hat{\mathbf{e }}^{1/2}||_{L^2} \right) ||\mathbf{e }^1||_{L^2}\\&\le \displaystyle \frac{1}{4}||\mathbf{e }^1||_{L^2}^2+ C\tau ^5. \end{aligned}$$

Combining these estimates into (4.16), we conclude that

$$\begin{aligned} || \mathbf{e }^1||_{L^2}+\tau ^{1/2}|| \mathbf{e }^1||_{H^1} \le C\tau ^{5/2}. \end{aligned}$$

(4.19)

We test (4.15) by $-\Delta \mathbf{e }^1$ to get

$$\begin{aligned}&\ ||\nabla \mathbf{e }^1||_{\mathbf{L }^2}^2+\lambda \tau ||\Delta \mathbf{e }^1||_{\mathbf{L }^2}^2\\&\quad =-\tau (\mathbf{R }^0,\Delta \mathbf{e }^1)- \tau \left( \hat{\mathbf{m }}^{1/2}\times \Delta \overline{\mathbf{m }}^{1/2}-\hat{\mathbf{M }}^{1/2}\times \Delta \overline{\mathbf{M }}^{1/2}, \Delta \mathbf{e }^1\right) \\&\qquad - \tau \left( \nabla \hat{\mathbf{m }}^{1/2}\times \nabla \overline{\mathbf{m }}^{1/2}- \nabla \hat{\mathbf{M }}^{1/2}\times \nabla \overline{\mathbf{M }}^{1/2},\Delta \mathbf{e }^1\right) \\&\qquad -\lambda \tau \left( |\nabla \hat{\mathbf{m }}^{{1/2}}|^2 \hat{\mathbf{m }}^{{1/2}} - |\nabla \hat{\mathbf{M }}^{1/2}|^2\hat{\mathbf{M }}^{1/2},\Delta \mathbf{e }^1\right) \\&\quad =J_5+J_6+J_7+J_8. \end{aligned}$$

Following the arguments for $J_1$ and $J_4$, $J_5$ and $J_8$ are estimated by

$$\begin{aligned} J_5+J_8 \le \displaystyle \frac{\lambda \tau }{4}||\Delta \mathbf{e }^1||_{L^2}^2+ C\tau ^4. \end{aligned}$$

$J_6$ and $J_7$ can be bounded, respectively, by

$$\begin{aligned} J_6&=- \tau \left( \hat{\mathbf{e }}^{1/2}\times \Delta \overline{\mathbf{m }}^{1/2},\Delta \mathbf{e }^1\right) \\&\le \displaystyle \frac{\lambda \tau }{4}||\Delta \mathbf{e }^1||_{L^2}^2 +\tau ||\Delta \overline{\mathbf{m }}^{1/2}||_{L^3}^2||\nabla \hat{\mathbf{e }}^{1/2}||^2_{L^2} \le \displaystyle \frac{\lambda \tau }{4}||\Delta \mathbf{e }^1||_{L^2}^2+ C\tau ^4, \end{aligned}$$

and

$$\begin{aligned} J_7&=- \tau \left( \nabla \hat{\mathbf{e }}^{1/2}\times \nabla \overline{\mathbf{m }}^{1/2}-\displaystyle \frac{1}{2}\nabla \hat{\mathbf{e }}^{1/2}\times \nabla \mathbf{e }^1+\displaystyle \frac{1}{2} \nabla \hat{\mathbf{m }}^{1/2}\times \nabla \mathbf{e }^1,\Delta \mathbf{e }^1\right) \\&\le \tau \left( ||\nabla \hat{\mathbf{e }}^{1/2}||_{L^2}||\nabla \overline{\mathbf{m }}^{1/2}||_{L^\infty }\!+\! ||\nabla \hat{\mathbf{e }}^{1/2}||_{\mathbf{L }^6}||\nabla \mathbf{e }^1||_{L^3}\!+\! ||\nabla \hat{\mathbf{m }}^{1/2}||_{L^\infty } ||\nabla \mathbf{e }^{1}||_{L^2}\right) ||\Delta \mathbf{e }^1||_{L^2}\\&\le \displaystyle \frac{\lambda \tau }{4}||\Delta \mathbf{e }^1||_{L^2}^2+ C\tau ^4. \end{aligned}$$

Thus, we obtain

$$\begin{aligned} || \mathbf{e }^1||_{H^1}+\tau ^{1/2}||\mathbf{e }^1||_{H^2} \le C\tau ^2. \end{aligned}$$

(4.20)

Differentiating (4.15) with respect to $\mathbf{x }$ and testing the resulting equation by $-2\nabla \Delta \mathbf{e }^1$, and using Young inequality, we get

$$\begin{aligned}&||\Delta \mathbf{e }^1||_{\mathbf{L }^2}^2+ \tau ||\nabla \Delta \mathbf{e }^1||_{\mathbf{L }^2}^2\nonumber \\&\quad \le \tau ||\nabla \mathbf{R }^0||_{L^2}^2+ \tau \left| \left( \nabla \left( \hat{\mathbf{m }}^{1/2}\times \Delta \overline{\mathbf{m }}^{1/2} -\hat{\mathbf{M }}^{1/2}\times \Delta \overline{\mathbf{M }}^{1/2}\right) ,\nabla \Delta \mathbf{e }^1\right) \right| \nonumber \\&\qquad + \tau ||\nabla \left( \nabla \hat{\mathbf{m }}^{1/2}\times \nabla \overline{\mathbf{m }}^{1/2}- \nabla \hat{\mathbf{M }}^{1/2}\times \nabla \overline{\mathbf{M }}^{1/2}\right) ||_{L^2}^2\nonumber \\&\qquad +\tau ||\nabla \left( |\nabla \hat{\mathbf{m }}^{1/2}|^2 \hat{\mathbf{m }}^{1/2}- |\nabla \hat{\mathbf{M }}^{1/2}|^2\hat{\mathbf{M }}^{1/2}\right) ||_{L^2}^2. \end{aligned}$$

(4.21)

From (4.7), one has $\tau ||\nabla \mathbf{R }^0||_{L^2}^2\le C \tau ^4$. According to the following formulation

$$\begin{aligned}&\nabla \left( \hat{\mathbf{m }}^{1/2}\times \Delta \overline{\mathbf{m }}^{1/2} -\hat{\mathbf{M }}^{1/2}\times \Delta \overline{\mathbf{M }}^{1/2}\right) \\&\quad =\nabla \hat{\mathbf{e }}^{1/2}\times \Delta \overline{\mathbf{m }}^{1/2}+\nabla \hat{\mathbf{M }}^{1/2}\times \Delta \overline{\mathbf{e }}^1 + \hat{\mathbf{e }}^{1/2}\times \nabla \Delta \overline{\mathbf{m }}^{1/2}+ \hat{\mathbf{M }}^{1/2}\times \nabla \Delta \overline{\mathbf{e }}^1, \end{aligned}$$

we have

$$\begin{aligned}&\tau \left| \left( \nabla \left( \hat{\mathbf{m }}^{1/2}\times \Delta \overline{\mathbf{m }}^{1/2} -\hat{\mathbf{M }}^{1/2}\times \Delta \overline{\mathbf{M }}^{1/2}\right) ,\nabla \Delta \mathbf{e }^1\right) \right| \\&\quad \le \tau \left( || \hat{\mathbf{e }}^{1/2}||_{H^3}||\overline{\mathbf{m }}^{1/2}||_{H^2}+ ||\hat{\mathbf{M }}^{1/2}||_{H^3}|| \mathbf{e }^1||_{H^2} + || \hat{\mathbf{e }}^{1/2}||_{H^2}||\overline{\mathbf{m }}^{1/2}||_{H^3}\right) ||\nabla \Delta \mathbf{e }^1||_{L^2}\\&\quad \le \displaystyle \frac{\tau }{2}||\nabla \Delta \mathbf{e }^1||_{\mathbf{L }^2}^2+ C\tau ^3. \end{aligned}$$

The other two terms in the right-hand side of (4.21) are bounded, respectively, by

$$\begin{aligned}&\tau ||\nabla \left( \nabla \hat{\mathbf{m }}^{1/2}\times \nabla \overline{\mathbf{m }}^{1/2}- \nabla \hat{\mathbf{M }}^{1/2}\times \nabla \overline{\mathbf{M }}^{1/2}\right) ||_{L^2}^2\\&\quad \le \tau \left( || \hat{\mathbf{e }}^{1/2}||_{H^2}||\overline{\mathbf{m }}^{1/2}||_{H^3}+ ||\hat{\mathbf{M }}^{1/2}||_{H^3}|| \mathbf{e }^1||_{H^2} +|| \hat{\mathbf{e }}^{1/2}||_{H^3}||\overline{\mathbf{m }}^{1/2}||_{H^2} \right) ^2\le C\tau ^3, \end{aligned}$$

and, from (4.17),

$$\begin{aligned}&\tau ||\nabla \left( |\nabla \hat{\mathbf{m }}^{1/2}|^2 \hat{\mathbf{m }}^{1/2}- |\nabla \hat{\mathbf{M }}^{1/2}|^2\hat{\mathbf{M }}^{1/2}\right) ||_{L^2}^2\\&\quad \le \tau \left( || \hat{\mathbf{m }}^{1/2}+ \hat{\mathbf{M }}^{1/2}||_{H^3}|| \hat{\mathbf{m }}^{1/2}||_{H^2}|| \hat{\mathbf{e }}^{1/2}||_{H^2} +|| \hat{\mathbf{m }}^{1/2}+ \hat{\mathbf{M }}^{1/2}||_{H^2}|| \hat{\mathbf{m }}^{1/2}||_{H^2}|| \hat{\mathbf{e }}^{1/2}||_{H^3}\right. \\&\qquad +|| \hat{\mathbf{m }}^{1/2}+ \hat{\mathbf{M }}^{1/2}||_{H^3}|| \hat{\mathbf{m }}^{1/2}||_{H^3}|| \hat{\mathbf{e }}^{1/2}||_{H^1}+ +|| \hat{\mathbf{M }}^{1/2}||^2_{H^3}|| \hat{\mathbf{e }}^{1/2}||_{H^1}\\&\qquad +\left. ||\nabla \hat{\mathbf{M }}^{1/2}||_{H^3}^2|| \hat{\mathbf{e }}^{1/2}||_{H^2} \right) ^2\le C\tau ^3. \end{aligned}$$

Combining these estimates into (4.21), we obtain $||\nabla \Delta \mathbf{e }^1||_{L^2}\le C\tau$, which together with (2.8) implies that $||\mathbf{e }^1||_{H^3}\le C\tau$. $\square$

From (4.14), we have

$$\begin{aligned} ||\mathbf{M }^1||_{H^3}&\le ||\mathbf{m }^1||_{H^3}+||\mathbf{e }^1||_{H^3}\le C, \end{aligned}$$

(4.22)

$$\begin{aligned} ||D_\tau \mathbf{M }^1||_{H^3}&\le ||D_\tau \mathbf{m }^1||_{H^3}+\tau ^{-1}||\mathbf{e }^1||_{H^3}\le C. \end{aligned}$$

(4.23)

The main result in this section is the following temporal error estimates and the regularities of the solution $\mathbf{M }^n$ to the problem (4.1).

Theorem 4.2

Suppose that the solution $\mathbf{m }$ to (1.4) and (1.2) satisfies the regularities (2.12). Then there exist $C_1>0, C_2>0$ and $\tau _1>0$ such that when $\tau <\tau _1$, there holds

$$\begin{aligned} \max \limits _{1\le m\le N} \left( ||\mathbf{e }^{m}||_{H^1}^2+\tau \displaystyle \sum \limits _{k=1}^m||\mathbf{e }^k||_{H^2}^2+\tau ^2||\mathbf{e }^m||_{H^3}^2 \right)&\le C_1\tau ^4, \end{aligned}$$

(4.24)

$$\begin{aligned} \max \limits _{1\le m\le N} \left( ||\mathbf{M }^m||_{H^3}+||D_\tau \mathbf{M }^m||_{H^3}\right)&\le C_2. \end{aligned}$$

(4.25)

Proof

In view of (4.14) and (4.22–4.23), the inequalities (4.24) and (4.25) obviously hold for $m=1$. Now, we suppose that (4.24) and (4.25) hold for $m\le n$ with $n\le N-1$. Then we need to show that these inequalities also hold for $m= n+1$. From the assumption, there exists some small $\tau _{11}>0$ satisfying $C_1\tau _{11}^2<1$ such that when $\tau <\tau _{11}$, one has $||\mathbf{e }^m||_{H^3}\le 1$ and $||\mathbf{M }^m||_{H^3}\le ||\mathbf{e }^m||_{H^3}+||\mathbf{m }^m||_{H^3}\le C_3$ and $||\hat{\mathbf{M }}^{m+1/2}||_{H^3}\le C_3$ for $1\le m\le n$, where $C_3>0$ is independent of $C_1$. By a simple calculation, we have

$$\begin{aligned} \hat{\mathbf{e }}^{m+1/2} =\hat{\mathbf{m }}^{m+1/2}-\hat{\mathbf{M }}^{m+1/2}=3\overline{\mathbf{e }}^{m-1/2}-2\mathbf{e }^{m-1},\quad 1\le m\le N-1. \end{aligned}$$

From (4.24) for $m\le n$, when $\tau <\tau _{11}$, we get

$$\begin{aligned} \max \limits _{1\le m\le n} \left( ||\hat{\mathbf{e }}^{m+1/2}||_{H^1}^2 +\tau \displaystyle \sum \limits _{k=1}^m||\hat{\mathbf{e }}^{k+1/2}||_{H^2}^2 +\tau ^2||\hat{\mathbf{e }}^{m+1/2}||_{H^3}^2 \right) \le C_4C_1\tau ^4\le C_4\tau ^2, \end{aligned}$$

(4.26)

where $C_4>0$ is independent of $C_1$. Instead of n by m in (4.13), multiplying it by $\overline{\mathbf{e }}^{m+1/2}$ and integrating over $\Omega$, we get

$$\begin{aligned}&\displaystyle \frac{1}{2}D_\tau \left( ||\mathbf{e }^{m+1}||_{L^2}^2\right) +\lambda ||\nabla \overline{\mathbf{e }}^{m+1/2}||_{L^2}^2 =(\mathbf{R }^m,\overline{\mathbf{e }}^{m+1/2}) -\left( \hat{\mathbf{e }}^{m+1/2}\times \nabla \overline{\mathbf{m }}^{m+1/2}, \nabla \overline{\mathbf{e }}^{m+1/2}\right) \nonumber \\&\qquad + \lambda \left( |\nabla \hat{\mathbf{m }}^{m+1/2}|^2 \hat{\mathbf{m }}^{m+1/2}- |\nabla \hat{\mathbf{M }}^{m+1/2}|^2\hat{\mathbf{M }}^{m+1/2},\overline{\mathbf{e }}^{m+1/2}\right) , \end{aligned}$$

(4.27)

where we use

$$\begin{aligned}&\left( \hat{\mathbf{m }}^{m+1/2}\times \Delta \overline{\mathbf{m }}^{m+1/2} -\hat{\mathbf{M }}^{m+1/2}\times \Delta \overline{\mathbf{M }}^{m+1/2},\overline{\mathbf{e }}^{m+1/2}\right) \\&\qquad +\left( \nabla \hat{\mathbf{m }}^{m+1/2}\times \nabla \overline{\mathbf{m }}^{m+1/2}- \nabla \hat{\mathbf{M }}^{m+1/2}\times \nabla \overline{\mathbf{M }}^{m+1/2}, \overline{\mathbf{e }}^{m+1/2}\right) \\&\quad =- \left( \hat{\mathbf{e }}^{m+1/2}\times \nabla \overline{\mathbf{m }}^{m+1/2}, \nabla \overline{\mathbf{e }}^{m+1/2}\right) . \end{aligned}$$

An alternative to $|\nabla \hat{\mathbf{m }}^{m+1/2}|^2 \hat{\mathbf{m }}^{m+1/2}- |\nabla \hat{\mathbf{M }}^{m+1/2}|^2\hat{\mathbf{M }}^{m+1/2}$ is

$$\begin{aligned}&|\nabla \hat{\mathbf{m }}^{m+1/2}|^2 \hat{\mathbf{m }}^{m+1/2}- |\nabla \hat{\mathbf{M }}^{m+1/2}|^2\hat{\mathbf{M }}^{m+1/2}\nonumber \\&\quad =\left( \nabla \hat{\mathbf{e }}^{m+1/2}\cdot \left( \nabla \hat{\mathbf{m }}^{m+1/2}+\nabla \hat{\mathbf{M }}^{m+1/2}\right) \right) \nabla \hat{\mathbf{m }}^{m+1/2}+ |\nabla \hat{\mathbf{M }}^{m+1/2}|^2\hat{\mathbf{e }}^{m+1/2}, \end{aligned}$$

(4.28)

then the right-hand side of (4.27) can be bounded by

$$\begin{aligned}&||\mathbf{R }^m||_{L^2}||\overline{\mathbf{e }}^{m+1/2}||_{L^2}+||\hat{\mathbf{e }}^{m+1/2}||_{L^2}||\nabla \overline{\mathbf{m }}^{m+1/2}||_{L^\infty } ||\nabla \overline{\mathbf{e }}^{m+1/2}||_{L^2}\\&\qquad +\lambda \left( ||\nabla \hat{\mathbf{m }}^{m+1/2}+\nabla \hat{\mathbf{M }}^{m+1/2}||_{L^\infty }||\nabla \hat{\mathbf{m }}^{m+1/2}||_{L^\infty } ||\nabla \hat{\mathbf{e }}^{m+1/2}||_{L^2}\right. \\&\left. \qquad +||\nabla \hat{\mathbf{M }}^{m+1/2}||_{L^\infty }^2 || \hat{\mathbf{e }}^{m+1/2}||_{L^2}\right) || \overline{\mathbf{e }}^{m+1/2}||_{L^2}\\&\quad \le C_{5}\left( ||\overline{\mathbf{e }}^{m+1/2}||_{L^2}^2 +||\mathbf{R }^m||_{L^2}^2+|| \hat{\mathbf{e }}^{m+1/2}||_{L^2}^2\right) +\displaystyle \frac{\lambda }{2}||\nabla \overline{\mathbf{e }}^{m+1/2}||_{L^2}^2 +\displaystyle \frac{\lambda }{4}||\nabla \hat{\mathbf{e }}^{m+1/2}||_{L^2}^2, \end{aligned}$$

where $C_{5}>0$ is independent of $C_1$. Substituting the above inequality into (4.27) and using (4.7) and the discrete Gronwall inequality, we conclude that there exists some $\tau _{12}>0$ satisfying $2C_{5}\tau _{12}<1$, such that when $\tau <\tau _{12}$, there holds

$$\begin{aligned} \max \limits _{0\le m\le n}||\mathbf{e }^{m+1}||_{L^2}^2+\tau \displaystyle \sum \limits _{m=0}^{n} ||\nabla \overline{\mathbf{e }}^{m+1/2}||_{L^2}^2\le C_{6}\tau ^4. \end{aligned}$$

(4.29)

Instead of n by m in (4.13), multiplying it by $-\Delta \overline{\mathbf{e }}^{m+1/2}$ and integrating over $\Omega$ give

$$\begin{aligned}&\displaystyle \frac{1}{2}D_\tau \left( ||\nabla \mathbf{e }^{m+1}||_{L^2}^2\right) \!+\!\lambda ||\Delta \overline{\mathbf{e }}^{m+1/2}||_{L^2}^2\!=\!-\!\left( \mathbf{R }^m,\Delta \overline{\mathbf{e }}^{m+1/2}\right) \!-\! \left( \hat{\mathbf{e }}^{m+1/2}\times \Delta \overline{\mathbf{m }}^{m+1/2}, \Delta \overline{\mathbf{e }}^{m+1/2}\right) \nonumber \\&\qquad - \left( \nabla \hat{\mathbf{m }}^{m+1/2}\times \nabla \overline{\mathbf{m }}^{m+1/2}- \nabla \hat{\mathbf{M }}^{m+1/2}\times \nabla \overline{\mathbf{M }}^{m+1/2},\Delta \overline{\mathbf{e }}^{m+1/2}\right) \nonumber \\&\qquad -\lambda \left( |\nabla \hat{\mathbf{m }}^{m+1/2}|^2 \hat{\mathbf{m }}^{m+1/2}- |\nabla \hat{\mathbf{M }}^{m+1/2}|^2\hat{\mathbf{M }}^{m+1/2},\Delta \overline{\mathbf{e }}^{m+1/2}\right) . \end{aligned}$$

(4.30)

By Young’s inequality, we have

$$\begin{aligned}&(\mathbf{R }^m,\Delta \overline{\mathbf{e }}^{m+1/2}) + \left( \hat{\mathbf{e }}^{m+1/2}\times \Delta \overline{\mathbf{m }}^{m+1/2} ,\Delta \overline{\mathbf{e }}^{m+1/2}\right) \\&\quad \le \displaystyle \frac{\lambda }{6}||\Delta \overline{\mathbf{e }}^{m+1/2}||_{L^2}^2+C_{7}\left( ||\mathbf{R }^m||_{L^2}^2 +||\nabla \hat{\mathbf{e }}^{m+1/2}||_{L^2}^2\right) , \end{aligned}$$

where $C_{7}>0$ is independent of $C_1$. On the other hand, one has

$$\begin{aligned}&\left( \nabla \hat{\mathbf{m }}^{m+1/2}\times \nabla \overline{\mathbf{m }}^{m+1/2}- \nabla \hat{\mathbf{M }}^{m+1/2}\times \nabla \overline{\mathbf{M }}^{m+1/2},\Delta \overline{\mathbf{e }}^{m+1/2}\right) \\&\quad = \left( \nabla \hat{\mathbf{e }}^{m+1/2}\times \nabla \overline{\mathbf{m }}^{m+1/2}- \nabla \hat{\mathbf{e }}^{m+1/2}\times \nabla \overline{\mathbf{e }}^{m+1/2} +\nabla \hat{\mathbf{m }}^{m+1/2}\times \nabla \overline{\mathbf{e }}^{m+1/2},\Delta \overline{\mathbf{e }}^{m+1/2}\right) \\&\quad \le \left( ||\nabla \hat{\mathbf{e }}^{m+1/2}||_{L^2}\!+\!||\nabla \overline{\mathbf{e }}^{m+1/2}||_{L^2}\right) ||\Delta \overline{\mathbf{e }}^{m+1/2}||_{L^2}\!+\! ||\nabla \hat{\mathbf{e }}^{m+1/2}||_{L^\infty }||\nabla \overline{\mathbf{e }}^{m+1/2}||_{L^2} ||\Delta \overline{\mathbf{e }}^{m+1/2}||_{L^2} \\&\quad \le \displaystyle \frac{\lambda }{6}||\Delta \overline{\mathbf{e }}^{m+1/2}||_{L^2}^2+ C_{8} \left( ||\nabla \overline{\mathbf{e }}^{m+1/2}||_{L^2}^2+||\nabla \hat{\mathbf{e }}^{m+1/2}||_{L^2}^2+ C_4\tau ||\nabla \overline{\mathbf{e }}^{m+1/2}||_{L^2}^2\right) , \end{aligned}$$

where $C_{8}>0$ is independent of $C_1$. From (4.28), the last term in the right-hand side of (4.30) can be bounded by

$$\begin{aligned}&\lambda \left( |\nabla \hat{\mathbf{m }}^{m+1/2}|^2 \hat{\mathbf{m }}^{m+1/2}- |\nabla \hat{\mathbf{M }}^{m+1/2}|^2\hat{\mathbf{M }}^{m+1/2},\Delta \overline{\mathbf{e }}^{m+1/2}\right) \\&\quad \le \lambda \left( ||\nabla \hat{\mathbf{m }}^{m+1/2}+\nabla \hat{\mathbf{M }}^{m+1/2}||_{L^\infty }||\nabla \hat{\mathbf{m }}^{m+1/2}||_{L^\infty } ||\nabla \hat{\mathbf{e }}^{m+1/2}||_{L^2}\right. \\&\quad \quad +\left. ||\nabla \hat{\mathbf{M }}^{m+1/2}||_{L^\infty }^2 || \hat{\mathbf{e }}^{m+1/2}||_{L^2}\right) || \Delta \overline{\mathbf{e }}^{m+1/2}||_{L^2}\\&\quad \le \displaystyle \frac{\lambda }{6}||\Delta \overline{\mathbf{e }}^{m+1/2}||_{L^2}^2+C_{9}\left( || \hat{\mathbf{e }}^{m+1/2}||_{L^2}^2+ ||\nabla \hat{\mathbf{e }}^{m+1/2}||_{L^2}^2\right) , \end{aligned}$$

where $C_{9}>0$ is independent of $C_1$. Combining these estimates into (4.30) and using the discrete Gronwall inequality, we conclude that there exists some small $\tau _{13}>0$ satisfying

$$\begin{aligned} 2(C_{7}+2C_{8}+2C_{9}+C_{8}C_4\tau _{13})\tau _{13}<1, \end{aligned}$$

such that when $\tau <\tau _{13}$, there holds

$$\begin{aligned} \max \limits _{0\le m\le n}||\nabla \mathbf{e }^{m+1}||_{L^2}^2+\tau \displaystyle \sum \limits _{m=0}^{n}|| \Delta \overline{\mathbf{e }}^{m+1/2}||_{L^2}^2\le C_{10}\tau ^4, \end{aligned}$$

which together with (2.5) and (4.29) implies that

$$\begin{aligned} \max \limits _{0\le m\le n}|| \mathbf{e }^{m+1}||_{H^1}^2+\tau \displaystyle \sum \limits _{m=0}^{n}|| \overline{\mathbf{e }}^{m+1/2}||_{H^2}^2\le C_{11} \tau ^4. \end{aligned}$$

(4.31)

Differentiating (4.13) with respect to $\mathbf{x }$ and multiplying the resulting equation by $-\nabla \Delta \overline{\mathbf{e }}^{m+1/2}$ and integrating over $\Omega$ give

$$\begin{aligned}&\displaystyle \frac{1}{2}D_\tau (||\Delta \mathbf{e }^{m+1}||_{L^2}^2)+\lambda ||\nabla \Delta \overline{\mathbf{e }}^{m+1/2}||_{L^2}^2\nonumber \\&\quad =-(\nabla \mathbf{R }^n,\nabla \Delta \overline{\mathbf{e }}^{m+1/2}) - \left( \nabla \left( \hat{\mathbf{m }}^{m+1/2}\times \Delta \overline{\mathbf{m }}^{m+1/2}- \hat{\mathbf{M }}^{m+1/2}\times \Delta \overline{\mathbf{M }}^{m+1/2}\right) ,\nabla \Delta \overline{\mathbf{e }}^{m+1/2}\right) \nonumber \\&\qquad - \left( \nabla \left( \nabla \hat{\mathbf{m }}^{m+1/2}\times \nabla \overline{\mathbf{m }}^{m+1/2}- \nabla \hat{\mathbf{M }}^{m+1/2}\times \nabla \overline{\mathbf{M }}^{m+1/2}\right) ,\nabla \Delta \overline{\mathbf{e }}^{m+1/2}\right) \nonumber \\&\qquad -\lambda \left( \nabla \left( |\nabla \hat{\mathbf{m }}^{m+1/2}|^2 \hat{\mathbf{m }}^{m+1/2}- |\nabla \hat{\mathbf{M }}^{m+1/2}|^2\hat{\mathbf{M }}^{m+1/2}\right) ,\nabla \Delta \overline{\mathbf{e }}^{m+1/2}\right) . \end{aligned}$$

(4.32)

It is easy to show

$$\begin{aligned} (\nabla \mathbf{R }^m,\nabla \Delta \overline{\mathbf{e }}^{m+1/2})\le \displaystyle \frac{\lambda }{8}||\nabla \Delta \overline{\mathbf{e }}^{m+1/2}||_{L^2}^2 +C_{12}||\nabla \mathbf{R }^m||_{L^2}^2, \end{aligned}$$

where $C_{12}>0$ is independent of $C_1$. The second term in the right-hand side of (4.32) is bounded by

$$\begin{aligned}&\left( \nabla \left( \hat{\mathbf{m }}^{m+1/2}\times \Delta \overline{\mathbf{m }}^{m+1/2}- \hat{\mathbf{M }}^{m+1/2}\times \Delta \overline{\mathbf{M }}^{m+1/2}\right) ,\nabla \Delta \overline{\mathbf{e }}^{m+1/2}\right) \\&\quad \le \left( ||\hat{\mathbf{e }}^{m+1/2}||_{H^2}||\overline{\mathbf{m }}^{m+1/2}||_{H^3} +||\overline{\mathbf{e }}^{m+1/2}||_{H^2}||\hat{\mathbf{M }}^{m+1/2}||_{H^3}\right) ||\nabla \Delta \overline{\mathbf{e }}^{m+1/2}||_{L^2}\\&\quad \le \displaystyle \frac{\lambda }{8}||\nabla \Delta \overline{\mathbf{e }}^{m+1/2}||_{L^2}^2+C_{13}\left( ||\overline{\mathbf{e }}^{m+1/2}||_{H^2}^2 + ||\hat{\mathbf{e }}^{m+1/2}||_{H^2}^2\right) , \end{aligned}$$

where $C_{13}>0$ is independent of $C_1$. By a similar way, we have

$$\begin{aligned}&\left( \nabla \left( \nabla \hat{\mathbf{m }}^{m+1/2}\times \nabla \overline{\mathbf{m }}^{m+1/2}- \nabla \hat{\mathbf{M }}^{m+1/2}\times \nabla \overline{\mathbf{M }}^{m+1/2}\right) ,\nabla \Delta \overline{\mathbf{e }}^{m+1/2}\right) \\&\quad \le \left( ||\hat{\mathbf{e }}^{m+1/2}||_{H^2}||\overline{\mathbf{m }}^{m+1/2}||_{H^3}+||\overline{\mathbf{e }}^{m+1/2}||_{H^2}||\hat{\mathbf{M }}^{m+1/2}||_{H^3}\right) ||\nabla \Delta \overline{\mathbf{e }}^{m+1/2}||_{L^2}\\&\quad \le \displaystyle \frac{\lambda }{8}||\nabla \Delta \overline{\mathbf{e }}^{m+1/2}||_{L^2}^2+C_{13}\left( ||\overline{\mathbf{e }}^{m+1/2}||_{H^2}^2 + ||\hat{\mathbf{e }}^{m+1/2}||_{H^2}^2\right) . \end{aligned}$$

Finally, from (4.28), we have

$$\begin{aligned}&\lambda \left( \nabla \left( |\nabla \hat{\mathbf{m }}^{m+1/2}|^2 \hat{\mathbf{m }}^{m+1/2}- |\nabla \hat{\mathbf{M }}^{m+1/2}|^2 \hat{\mathbf{M }}^{m+1/2}\right) ,\nabla \Delta \overline{\mathbf{e }}^{m+1/2}\right) \\&\quad =\lambda \left( \left( \nabla ^2\hat{\mathbf{e }}^{m+1/2}\cdot \left( \nabla \hat{\mathbf{m }}^{m+1/2}+\nabla \hat{\mathbf{M }}^{m+1/2}\right) \right) \nabla \hat{\mathbf{m }}^{m+1/2},\nabla \Delta \overline{\mathbf{e }}^{m+1/2})\right) \\&\qquad +\lambda \left( \left( \nabla \hat{\mathbf{e }}^{m+1/2}\cdot \left( \nabla ^2 \hat{\mathbf{m }}^{m+1/2}+\nabla ^2 \hat{\mathbf{M }}^{m+1/2}\right) \right) \nabla \hat{\mathbf{m }}^{m+1/2},\nabla \Delta \overline{\mathbf{e }}^{m+1/2})\right) \\&\qquad +\lambda \left( \left( \nabla \hat{\mathbf{e }}^{m+1/2}\cdot \left( \nabla \hat{\mathbf{m }}^{m+1/2}+\nabla \hat{\mathbf{M }}^{m+1/2}\right) \right) \nabla ^2 \hat{\mathbf{m }}^{m+1/2},\nabla \Delta \overline{\mathbf{e }}^{m+1/2})\right) \\&\qquad +\lambda \left( |\nabla \hat{\mathbf{M }}^{m+1/2}|^2\nabla \hat{\mathbf{e }}^{m+1/2},\nabla \Delta \overline{\mathbf{e }}^{m+1/2}\right) \\&\qquad +2\lambda \left( \left( \nabla ^2\hat{\mathbf{M }}^{m+1/2}\cdot \nabla \hat{\mathbf{M }}^{m+1/2}\right) \hat{\mathbf{e }}^{m+1/2},\nabla \Delta \overline{\mathbf{e }}^{m+1/2}\right) \\&\quad \le C_{14} ||\hat{\mathbf{e }}^{m+1/2}||_{H^2}||\hat{\mathbf{m }}^{m+1/2}||_{H^3}||\hat{\mathbf{m }}^{m+1/2}+\hat{\mathbf{M }}^{m+1/2}||_{H^3}||\nabla \Delta \overline{\mathbf{e }}^{m+1/2}||_{L^2}\\&\qquad +C_{14}||\hat{\mathbf{e }}^{m+1/2}||_{H^2}||\hat{\mathbf{M }}^{m+1/2}||_{H^3}^2||\nabla \Delta \overline{\mathbf{e }}^{m+1/2}||_{L^2}\\&\quad \le \displaystyle \frac{\lambda }{8}||\nabla \Delta \overline{\mathbf{e }}^{m+1/2}||_{L^2}^2+C_{15} ||\hat{\mathbf{e }}^{m+1/2}||_{H^2}^2. \end{aligned}$$

where $C_{14}>0$ and $C_{15}>0$ are independent of $C_1$. Thus, using (2.6), (4.26), (4.31), (4.32) and the discrete Gronwall inequality, we conclude that there exists some $\tau _{14}>0$ satisfying $2(4C_{13}+C_{15})\tau _{14}<1$, such that when $\tau <\tau _{14}$, there holds

$$\begin{aligned} \max \limits _{0\le m\le n}|| \mathbf{e }^{m+1}||_{H^2}^2+\tau \displaystyle \sum \limits _{m=0}^{n}|| \nabla \Delta \overline{\mathbf{e }}^{m+1/2}||_{L^2}^2\le C_{16}\tau ^4. \end{aligned}$$

(4.33)

By using (3.7), we have

$$\begin{aligned} \tau ||\nabla \Delta \mathbf{e }^{n+1}||_{L^2}\le 2\sqrt{T}\sqrt{\tau \displaystyle \sum \limits _{m=0}^{n}||\nabla \Delta \overline{\mathbf{e }}^{m+1/2}||_{L^2}^2}\le C_{17}\tau ^2, \end{aligned}$$

for any $0<n\le N-1$, where we use $\mathbf{e }^0=0$. From (2.8) and (4.33), we obtain

$$\begin{aligned} \max \limits _{0\le n\le N-1} ||\mathbf{e }^{n+1}||_{H^3}\le C_{18}\tau . \end{aligned}$$

This completes the proof of (4.24) with $\tau =\{ \tau _{11}, \tau _{12}, \tau _{13}, \tau _{14}\}$ for sufficiently large $C_1$. The estimate (4.25) is a direct result of (4.24) if we observe the definition of $\mathbf{e }^n$ and the regularities (2.12). $\square$

5 Spatial Error Analysis

In this section, we begin to estimate spatial errors. To do it, we define the projection operator $P_h^0$ and $P_h^{n+1}, 0\le n\le N-1$, from $\mathbf{H }^1(\Omega )$ onto $\mathbf{V }_h$, respectively, by

$$\begin{aligned} \lambda (\nabla (P_h^0\mathbf{u }-\mathbf{u }),\nabla \mathbf{v }_h)+\lambda (P_h^0\mathbf{u }-\mathbf{u },\mathbf{v }_h)+ (\mathbf{m }_0\times \nabla (P_h^0\mathbf{u }-\mathbf{u }),\nabla \mathbf{v }_h) =0 , \end{aligned}$$

and

$$\begin{aligned} \lambda (\nabla (P_h^{n+1}\mathbf{u }-\mathbf{u }),\nabla \mathbf{v }_h)+\lambda (P_h^{n+1}\mathbf{u }-\mathbf{u },\mathbf{v }_h)+ \left( \hat{\mathbf{M }}^{n+1/2}\times \nabla (P_h^{n+1}\mathbf{u }-\mathbf{u }),\nabla \mathbf{v }_h\right) =0 \end{aligned}$$

for all $\mathbf{v }_h\in \mathbf{V }_h$. Then by the classical finite element theory [8], for $0\le n\le N$, we have

$$\begin{aligned} ||\mathbf{u }-P_h^{n}\mathbf{u }||_{L^2}+h||\mathbf{u }-P_h^{n}\mathbf{u }||_{H^1}&\le C h^{r+1}||\mathbf{u }||_{H^3}, \end{aligned}$$

(5.1)

$$\begin{aligned} ||P_h^{n}\mathbf{u }||_{W^{m,p}}&\le C||\mathbf{u }||_{W^{m,p}},\quad m=0,1,2, \ 2\le p\le +\infty , \end{aligned}$$

(5.2)

$$\begin{aligned} ||\mathbf{u }-P_h^{n}\mathbf{u }||_{W^{1,p}}&\le Ch||\mathbf{u }||_{W^{2,p}},\quad 2\le p\le 6. \end{aligned}$$

(5.3)

For $0\le n\le N-1$, let us denote

$$\begin{aligned} \mathbf{e }^{n+1}_h&=P_h^{n+1}\mathbf{M }^{n+1}-\mathbf{M }^{n+1}_h, \ \mathbf{E }^{n+1}=P_h^{n+1}\mathbf{M }^{n+1}-\mathbf{M }^{n+1},\\ \overline{\mathbf{e }}_h^{n+1/2}&=P_h^{n+1}\overline{\mathbf{M }}^{n+1/2}-\overline{\mathbf{M }}^{n+1/2}_h, \ \overline{\mathbf{E }}^{n+1/2}=P_h^{n+1}\overline{\mathbf{M }}^{n+1/2} -\overline{\mathbf{M }}^{n+1/2},\\ \hat{\mathbf{e }}_h^{n+1/2}&=P_h^{n}\hat{\mathbf{M }}^{n+1/2}-\hat{\mathbf{M }}^{n+1/2}_h, \ \ \hat{\mathbf{E }}^{n+1/2}=P_h^{n}\hat{\mathbf{M }}^{n+1/2}-\hat{\mathbf{M }}^{n+1/2},\\ \hat{\mathbf{e }}_h^{1/2}&=P_h^0\hat{\mathbf{M }}^{1/2}-\hat{\mathbf{M }}_h^{1/2}, \ \hat{\mathbf{E }}^{1/2}=P_h^0\hat{\mathbf{M }}^{1/2}-\hat{\mathbf{M }}^{1/2}. \end{aligned}$$

From the regularities for $\hat{\mathbf{M }}^{1/2}$ and $\mathbf{M }^{n+1}$ derived in Sect. 4, we have

$$\begin{aligned} ||\mathbf{E }^{n+1}||_{L^2}+||\overline{\mathbf{E }}^{n+1/2}||_{L^2}+||\hat{\mathbf{E }}^{n+1/2}||_{L^2}+||\hat{\mathbf{E }}^{1/2}||_{L^2}&\le C h^{r+1}, \end{aligned}$$

(5.4)

$$\begin{aligned} ||\mathbf{E }^{n+1}||_{H^1}+||\overline{\mathbf{E }}^{n+1/2}||_{H^1}+||\hat{\mathbf{E }}^{n+1/2}||_{H^1}+||\hat{\mathbf{E }}^{1/2}||_{H^1}&\le Ch^{r}, \end{aligned}$$

(5.5)

$$\begin{aligned} ||D_\tau \mathbf{E }^{n+1}||_{L^2}+||D_\tau \overline{\mathbf{E }}^{n+1/2}||_{L^2}+||D_\tau \hat{\mathbf{E }}^{n+1/2}||_{L^2}&\le C h^{r+1} \end{aligned}$$

(5.6)

for $0\le n\le N-1$ and $r=1,2$. First, we prove the following lemma.

Lemma 5.1

Suppose that the solution $\mathbf{m }$ to (1.4) and (1.2) satisfies the regularities (2.12). Then there exists some $C>0$ independent of $\tau$ and h such that $\hat{\mathbf{e }}_h^{1/2}$ satisfies

$$\begin{aligned} ||\hat{\mathbf{e }}_h^{1/2}||_{L^2}+\tau ||\nabla \hat{\mathbf{e }}_h^{1/2}||_{L^2}+h||\hat{\mathbf{e }}_h^{1/2}||_{H^1}\le C h^{r+1}. \end{aligned}$$

(5.7)

Proof

Subtracting (4.4) from (3.2), we get

$$\begin{aligned}&2( {\hat{\mathbf{e }}_h^{1/2}},\mathbf{v }_h)+\lambda \tau (\nabla \hat{\mathbf{e }}_h^{1/2},\nabla \mathbf{v }_h) +\tau (\mathbf{m }_0\times \nabla \hat{\mathbf{e }}_h^{1/2},\nabla \mathbf{v }_h) = 2( {\hat{\mathbf{E }}^{1/2}} ,\mathbf{v }_h)\\&\qquad + \lambda \tau (|\nabla \mathbf{m }_0|^2 \mathbf{m }_0-|\nabla \mathbf{M }_h^0|^2 \mathbf{M }_h^0,\mathbf{v }_h)-\lambda \tau (\hat{\mathbf{E }}^{1/2},\mathbf{v }_h) -\tau (\mathbf{E }^0\times \nabla \hat{\mathbf{M }}_h^{1/2},\nabla \mathbf{v }_h). \end{aligned}$$

Setting $\mathbf{v }_h= \hat{\mathbf{e }}_h^{1/2}$ in the above equation leads to

$$\begin{aligned}&2||\hat{\mathbf{e }}_h^{1/2}||_{L^2}^2+\lambda \tau ||\nabla \hat{\mathbf{e }}_h^{1/2}||_{L^2}^2\nonumber \\&\quad =(2-\lambda \tau )(\hat{\mathbf{E }}^{1/2},\hat{\mathbf{e }}_h^{1/2}) + \lambda \tau \left( |\nabla \mathbf{m }_0|^2 \mathbf{m }_0-|\nabla \mathbf{M }_h^0|^2 \mathbf{M }_h^0,\hat{\mathbf{e }}_h^{1/2}\right) \nonumber \\&\qquad - \tau \left( \mathbf{E }^0\times \nabla \hat{\mathbf{M }}_h^{1/2},\nabla \hat{\mathbf{e }}_h^{1/2}\right) =L_1+L_2+L_3. \end{aligned}$$

(5.8)

It follows from (5.1), (5.2) and (5.4) that

$$\begin{aligned} L_1\le \displaystyle \frac{1}{2}||\hat{\mathbf{e }}_h^{1/2}||_{L^2}^2+ C ||\hat{\mathbf{E }}^{1/2}||_{L^2}^2 \le \displaystyle \frac{1}{2}||\hat{\mathbf{e }}_h^{1/2}||_{L^2}^2+ C h^{2(r+1)}, \end{aligned}$$

and

$$\begin{aligned} L_3&=-\tau \left( \mathbf{E }^0\times \nabla P_h^0 \hat{\mathbf{M }}^{1/2},\nabla \hat{\mathbf{e }}_h^{1/2}\right) \\&\le \tau ||\mathbf{E }^0||_{L^2}||\nabla P_h^0 \hat{\mathbf{M }}^{1/2}||_{L^\infty }||\nabla \hat{\mathbf{e }}_h^{1/2}||_{L^2} \le \displaystyle \frac{\lambda \tau }{4}||\nabla \hat{\mathbf{e }}_h^{1/2}||_{L^2}^2+ C \tau h^{2(r+1)}, \end{aligned}$$

where we use

$$\begin{aligned} ||\nabla P_h^0 \hat{\mathbf{M }}^{1/2}||_{L^\infty }\le C ||\nabla \hat{\mathbf{M }}^{1/2}||_{L^\infty }\le C || \hat{\mathbf{M }}^{1/2}||_{H^3}\le C . \end{aligned}$$

We rewrite $|\nabla \mathbf{m }_0|^2 \mathbf{m }_0-|\nabla \mathbf{M }_h^0|^2 \mathbf{M }_h^0$ as

$$\begin{aligned}&|\nabla \mathbf{m }_0|^2 \mathbf{m }_0-|\nabla \mathbf{M }_h^0|^2 \mathbf{M }_h^0\\&\quad =|\nabla \mathbf{m }_0|^2\mathbf{E }^0- 2\nabla \mathbf{E }^0\cdot \nabla \mathbf{m }_0\mathbf{E }^0 +2\nabla \mathbf{E }^0\cdot \nabla \mathbf{m }_0\mathbf{m }_0 +|\nabla \mathbf{E }^0|^2\mathbf{E }^0-|\nabla \mathbf{E }^0|^2\mathbf{m }_0. \end{aligned}$$

Then $L_2$ is bounded by

$$\begin{aligned} L_2&\le \lambda \tau ||\nabla \mathbf{m }_0||_{L^\infty }^2||\mathbf{E }^0||_{L^2}||\hat{\mathbf{e }}_h^{1/2}||_{L^2} +\lambda \tau ||\nabla \mathbf{m }_0||_{L^\infty } ||\nabla \mathbf{E }^0||_{L^\infty } ||\mathbf{E }^0||_{L^2}||\hat{\mathbf{e }}_h^{1/2}||_{L^2}\\&\quad +\lambda \tau ||\nabla \mathbf{E }^0||_{L^\infty }^2||\mathbf{E }^0||_{\mathbf{L }^2}||\hat{\mathbf{e }}_h^{1/2}||_{L^2} +\lambda \tau ||\mathbf{m }_0||_{L^\infty }||\nabla \mathbf{E }^0||_{L^4}^2 ||\hat{\mathbf{e }}_h^{1/2}||_{L^2}\\&\quad + \lambda \tau | (\nabla \mathbf{E }^0\cdot \nabla \mathbf{m }_0\mathbf{m }_0,\hat{\mathbf{e }}_h^{1/2})|. \end{aligned}$$

Using integrating by parts, the last term in the above inequality is estimated by

$$\begin{aligned}&\ \lambda \tau | (\nabla \mathbf{E }^0\cdot \nabla \mathbf{m }_0\mathbf{m }_0,\hat{\mathbf{e }}_h^{1/2})|\\&\quad \le \lambda \tau \left( ||\mathbf{m }_0||_{L^\infty }||\Delta \mathbf{m }_0||_{L^3}||\mathbf{E }^0||_{L^2}||\hat{\mathbf{e }}_h^{1/2}||_{L^6}+ ||\nabla \mathbf{m }_0||_{L^\infty }^2||\mathbf{E }^0||_{L^2}||\hat{\mathbf{e }}_h^{1/2}||_{L^2}\right. \\&\qquad \left. +||\mathbf{m }_0||_{L^\infty }||\nabla \mathbf{m }_0||_{L^\infty }||\mathbf{E }^0||_{L^2}||\nabla \hat{\mathbf{e }}_h^{1/2}||_{L^2}\right) \end{aligned}$$

Thus, from Yong’s inequality, we get

$$\begin{aligned} L_2&\le C\tau ||\mathbf{E }^0||_{L^2}||\hat{\mathbf{e }}_h^{1/2}||_{L^2}+ C \tau ||\nabla \mathbf{E }^0||_{L^4}^2 ||\hat{\mathbf{e }}_h^{1/2}||_{L^2} + C \tau ||\mathbf{E }^0||_{L^2}||\nabla \hat{\mathbf{e }}_h^{1/2}||_{L^2}\\&\le \displaystyle \frac{1}{2}||\hat{\mathbf{e }}_h^{1/2}||_{L^2}^2+\displaystyle \frac{\lambda \tau }{4}||\nabla \hat{\mathbf{e }}_h^{1/2}||_{L^2}^2 + C\tau ||\mathbf{E }^0||_{L^2}^2+ C\tau ||\nabla \mathbf{E }^0||_{L^4}^4\\&\le \displaystyle \frac{1}{2}||\hat{\mathbf{e }}_h^{1/2}||_{L^2}^2+\displaystyle \frac{\lambda \tau }{4}||\nabla \hat{\mathbf{e }}_h^{1/2}||_{L^2}^2+ C\tau h^{2(r+1)}. \end{aligned}$$

Combining these estimates into (5.8), we get

$$\begin{aligned} ||\hat{\mathbf{e }}_h^{1/2}||_{L^2}+\tau ||\nabla \hat{\mathbf{e }}_h^{1/2}||_{L^2} \le C h^{r+1}, \end{aligned}$$

which together with inverse inequality yields $||\hat{\mathbf{e }}_h^{1/2}||_{H^1} \le Ch^{r}$. This completes the proof of this lemma. $\square$

To use the mathematical induction, we estimate $\mathbf{L }^2$-norm and $\mathbf{H }^1$-norm errors of $\mathbf{e }_h^1$ in Theorem 5.1.

Theorem 5.1

Suppose that the solution $\mathbf{m }$ to (1.4) and (1.2) satisfies the regularities (2.12). Then there exists some $C >0$ such that

$$\begin{aligned} || \mathbf{e }_h^{1}||_{L^2}+\tau ||\nabla \overline{\mathbf{e }}_h^{1/2}||_{L^2}^2+h|| \mathbf{e }_h^{1}||_{H^1}\le C h^{r+1}. \end{aligned}$$

(5.9)

Proof

Let $n=0$ in (3.1) and (4.3). Subtracting the resulting equations and setting $\mathbf{v }_h=\overline{\mathbf{e }}_h^{1/2}$ lead to

$$\begin{aligned}&\displaystyle \frac{1}{2}D_\tau (||\mathbf{e }_h^1||_{L^2}^2) +\lambda ||\nabla \overline{\mathbf{e }}_h^{1/2}||_{L^2}^2\nonumber \\&\quad = (D_\tau \mathbf{E }^1,\overline{\mathbf{e }}_h^{1/2})-\lambda ( \overline{\mathbf{E }}^{1/2}, \overline{\mathbf{e }}_h^{1/2}) - \left( (\hat{\mathbf{e }}^{1/2}_h-\hat{\mathbf{E }}^{1/2}) \times \nabla \overline{\mathbf{M }}_h^{1/2},\nabla \overline{\mathbf{e }}_h^{1/2}\right) \nonumber \\&\qquad +\lambda \left( |\nabla \hat{\mathbf{M }}^{1/2}|^2\hat{\mathbf{M }}^{1/2}-|\nabla \hat{\mathbf{M }}_h^{1/2}|^2\hat{\mathbf{M }}_h^{1/2},\overline{\mathbf{e }}_h^{1/2}\right) . \end{aligned}$$

(5.10)

Observing $\overline{\mathbf{e }}_h^{1/2}=\displaystyle \frac{\mathbf{e }_h^1+\mathbf{E }^0}{2}$, we have

$$\begin{aligned}&(D_\tau \mathbf{E }^1,\overline{\mathbf{e }}_h^{1/2})-\lambda ( \overline{\mathbf{E }}^{1/2}, \overline{\mathbf{e }}_h^{1/2})\\&\quad =\displaystyle \frac{1}{2} (D_\tau \mathbf{E }^1,\mathbf{e }_h^1+\mathbf{E }^0)+\displaystyle \frac{\lambda }{2}( \overline{\mathbf{E }}^{1/2},\mathbf{e }_h^1+\mathbf{E }^0)\le \displaystyle \frac{1}{4} ||\mathbf{e }_h^1||_{L^2}^2+ C h^{2(r+1)}. \end{aligned}$$

From the definition of $\overline{\mathbf{e }}_h^{1/2}$ and (5.2) and (4.25), we have

$$\begin{aligned}&((\hat{\mathbf{e }}^{1/2}_h-\hat{\mathbf{E }}^{1/2})\times \nabla \overline{\mathbf{M }}_h^{1/2},\nabla \overline{\mathbf{e }}_h^{1/2})\\&\quad = ((\hat{\mathbf{e }}^{1/2}_h-\hat{\mathbf{E }}^{1/2})\times \nabla P_h^1\overline{\mathbf{M }}^{1/2},\nabla \overline{\mathbf{e }}_h^{1/2}) \le \displaystyle \frac{\lambda }{4}||\nabla \overline{\mathbf{e }}_h^{1/2}||_{\mathbf{L }^2}^2+C h^{2(r+1)}. \end{aligned}$$

Rewrite $\hat{\mathbf{M }}^{n+1/2}-\hat{\mathbf{M }}_h^{n+1/2}=\hat{\mathbf{e }}_h^{n+1/2}-\hat{\mathbf{E }}^{n+1/2}$ as

$$\begin{aligned}&|\nabla \hat{\mathbf{M }}^{n+1/2}|^2\hat{\mathbf{M }}^{n+1/2}-|\nabla \hat{\mathbf{M }}_h^{n+1/2}|^2\hat{\mathbf{M }}_h^{n+1/2}\nonumber \\&\quad =|\nabla \hat{\mathbf{M }}^{n+1/2}|^2(\hat{\mathbf{e }}_h^{n+1/2}-\hat{\mathbf{E }}^{n+1/2})+ |\nabla (\hat{\mathbf{e }}_h^{n+1/2}-\hat{\mathbf{E }}^{n+1/2})|^2\hat{\mathbf{e }}_h^{n+1/2}\nonumber \\&\qquad -|\nabla (\hat{\mathbf{e }}_h^{n+1/2}-\hat{\mathbf{E }}^{n+1/2})|^2P_h^n\hat{\mathbf{M }}^{n+1/2} +2(\nabla \hat{\mathbf{M }}^{n+1/2}\cdot \nabla (\hat{\mathbf{e }}_h^{n+1/2}-\hat{\mathbf{E }}^{n+1/2}))\hat{\mathbf{e }}_h^{n+1/2}\nonumber \\&\qquad -2(\nabla \hat{\mathbf{M }}^{n+1/2}\cdot \nabla (\hat{\mathbf{e }}_h^{n+1/2}-\hat{\mathbf{E }}^{n+1/2}))P_h^n\hat{\mathbf{M }}^{n+1/2}. \end{aligned}$$

(5.11)

Taking $n=0$ in (5.11), we have

$$\begin{aligned}&\lambda \left( |\nabla \hat{\mathbf{M }}^{1/2}|^2\hat{\mathbf{M }}^{1/2}-|\nabla \hat{\mathbf{M }}_h^{1/2}|^2\hat{\mathbf{M }}_h^{1/2},\overline{\mathbf{e }}_h^{1/2}\right) \\&\quad \le ||\nabla \hat{\mathbf{M }}^{1/2}||_{L^6}^2||\hat{\mathbf{e }}_h^{1/2}-\hat{\mathbf{E }}^{1/2}||_{L^2}|| \overline{\mathbf{e }}_h^{1/2}||_{L^6}\\&\qquad + ||\hat{\mathbf{e }}_h^{1/2}||_{L^\infty } ||\nabla (\hat{\mathbf{e }}_h^{1/2}-\hat{\mathbf{E }}^{1/2})||_{L^2}||\nabla (\hat{\mathbf{e }}_h^{1/2}-\hat{\mathbf{E }}^{1/2})||_{L^3} || \overline{\mathbf{e }}_h^{1/2}||_{L^6}\\&\qquad + ||P_h^0\hat{\mathbf{M }}^{1/2}||_{L^\infty } ||\nabla (\hat{\mathbf{e }}_h^{1/2}-\hat{\mathbf{E }}^{1/2})||_{L^2}||\nabla (\hat{\mathbf{e }}_h^{1/2}-\hat{\mathbf{E }}^{1/2})||_{L^3} || \overline{\mathbf{e }}_h^{1/2}||_{L^6}\\&\qquad + ||\nabla \hat{\mathbf{M }}^{1/2}||_{L^\infty } ||\nabla (\hat{\mathbf{e }}_h^{1/2}-\hat{\mathbf{E }}^{1/2})||_{L^2}|| \hat{\mathbf{e }}_h^{1/2}||_{L^3} || \overline{\mathbf{e }}_h^{1/2}||_{L^6}\\&\qquad +|((\nabla \hat{\mathbf{M }}^{1/2}\cdot \nabla (\hat{\mathbf{e }}_h^{1/2}-\hat{\mathbf{E }}^{1/2}))P_h^0\hat{\mathbf{M }}^{1/2},\overline{\mathbf{e }}_h^{1/2})|\\&=L_4+\cdots +L_8. \end{aligned}$$

By (5.4–5.7) and the inverse inequality (3.6), $L_4,\ldots ,L_8$ can be estimated, respectively, by

$$\begin{aligned} L_4&\le \displaystyle \frac{\lambda }{20}||\nabla \overline{\mathbf{e }}_h^{1/2}||_{L^2}^2+C ||\hat{\mathbf{e }}_h^{1/2}-\hat{\mathbf{E }}^{1/2}||_{L^2}^2 \le \displaystyle \frac{\lambda }{20}||\nabla \overline{\mathbf{e }}_h^{1/2}||_{L^2}^2+C h^{2(r+1)}, \end{aligned}$$

and

$$\begin{aligned} L_5&\le C h^{-d/2}||\hat{\mathbf{e }}_h^{1/2}||_{L^2}||\nabla (\hat{\mathbf{e }}_h^{1/2}-\hat{\mathbf{E }}^{1/2})||_{L^2}(h^{-d/6}||\nabla \hat{\mathbf{e }}_h^{1/2}||_{L^2}+||\hat{\mathbf{E }}^{1/2}||_{L^3})||\nabla \overline{\mathbf{e }}_h^{1/2}||_{L^2}\\&\le C h^{r+1} h^{\min \{2r-2d/3, r-d/2\}} ||\nabla \overline{\mathbf{e }}_h^{1/2}||_{L^2} \le \displaystyle \frac{\lambda }{20}||\nabla \overline{\mathbf{e }}_h^{1/2}||_{L^2}^2+C h^{2(r+1)}, \end{aligned}$$

and

$$\begin{aligned} L_6&\le C (||\nabla \hat{\mathbf{e }}_h^{1/2}||_{L^2}+h^r)(h^{-d/6}||\nabla \hat{\mathbf{e }}_h^{1/2}||_{L^2}+h)||\nabla \overline{\mathbf{e }}_h^{1/2}||_{L^2}\\&\le C h^{r+1}||\nabla \overline{\mathbf{e }}_h^{1/2}||_{L^2}+C (h^{r-d/6}+h)||\nabla \hat{\mathbf{e }}_h^{1/2}||_{L^2}||\nabla \overline{\mathbf{e }}_h^{1/2}||_{L^2}\\&\le \displaystyle \frac{\lambda }{20}||\nabla \overline{\mathbf{e }}_h^{1/2}||_{L^2}^2+C (||\nabla \hat{\mathbf{e }}_h^{1/2}||_{L^2}^2+h^{2(r+1)}), \end{aligned}$$

and

$$\begin{aligned} L_7&\le C h^{-d/6}||\nabla (\hat{\mathbf{e }}_h^{1/2}-\hat{\mathbf{E }}^{1/2})||_{L^2}|| \hat{\mathbf{e }}_h^{1/2}||_{L^2} ||\nabla \overline{\mathbf{e }}_h^{1/2}||_{L^2}\le \displaystyle \frac{\lambda }{20} ||\nabla \overline{\mathbf{e }}_h^{1/2}||_{L^2}^2+C h^{2(r+1)}, \end{aligned}$$

and

$$\begin{aligned} L_8 \le \,&|((\nabla ^2\hat{\mathbf{M }}^{1/2}\cdot (\hat{\mathbf{e }}_h^{1/2}-\hat{\mathbf{E }}^{1/2}))P_h^0\hat{\mathbf{M }}^{1/2},\overline{\mathbf{e }}_h^{1/2})|\\&+|((\nabla \hat{\mathbf{M }}^{1/2}\cdot (\hat{\mathbf{e }}_h^{1/2}-\hat{\mathbf{E }}^{1/2}))\nabla P_h^0\hat{\mathbf{M }}^{1/2},\overline{\mathbf{e }}_h^{1/2})|\\&+|((\nabla \hat{\mathbf{M }}^{1/2}\cdot (\hat{\mathbf{e }}_h^{1/2}-\hat{\mathbf{E }}^{1/2}))P_h^0\hat{\mathbf{M }}^{1/2},\nabla \overline{\mathbf{e }}_h^{1/2})|\\ \le \,&||\nabla ^2\hat{\mathbf{M }}^{1/2}||_{L^3}||P_h^0\hat{\mathbf{M }}^{1/2}||_{L^\infty } ||\hat{\mathbf{e }}_h^{1/2}-\hat{\mathbf{E }}^{1/2}||_{L^2}||\overline{\mathbf{e }}_h^{1/2}||_{L^6}\\&+||\nabla \hat{\mathbf{M }}^{1/2}||_{L^3}||\nabla P_h^0\hat{\mathbf{M }}^{1/2}||_{L^\infty } ||\hat{\mathbf{e }}_h^{1/2}-\hat{\mathbf{E }}^{1/2}||_{L^2}||\overline{\mathbf{e }}_h^{1/2}||_{L^6}\\&+||\nabla \hat{\mathbf{M }}^{1/2}||_{L^\infty }||P_h^0\hat{\mathbf{M }}^{1/2}||_{L^\infty } ||\hat{\mathbf{e }}_h^{1/2}-\hat{\mathbf{E }}^{1/2}||_{L^2}||\overline{\mathbf{e }}_h^{1/2}||_{L^6}\\ \le \,&C ||\hat{\mathbf{e }}_h^{1/2}-\hat{\mathbf{E }}^{1/2}||_{L^2}||\nabla \overline{\mathbf{e }}_h^{1/2}||_{L^2}\le \displaystyle \frac{\lambda }{20} ||\nabla \overline{\mathbf{e }}_h^{1/2}||_{L^2}^2+C h^{2(r+1)}. \end{aligned}$$

Substituting these estimates into (5.10), we get $||\mathbf{e }_h^1||_{L^2}+\tau ||\nabla \overline{\mathbf{e }}_h^{1/2}||_{L^2}^2\le C h^{r+1}$, which together with the inverse inequality yields $||\mathbf{e }_h^1||_{H^1}\le C h^{r}$. $\square$

The main result in this section is the following theorem.

Theorem 5.2

Suppose that the solution $\mathbf{m }$ to (1.4) and (1.2) satisfies the regularities (2.12). Then there exist some $C_{19}>0$, $h_2>0$ and $\tau _2>0$ such that when $h<h_2$ and $\tau <\tau _2$, there holds

$$\begin{aligned} \max \limits _{1\le m\le N}\left( ||\mathbf{e }_h^m||_{L^2}^2+h^2||\mathbf{e }^m_h||_{H^1}^2 +\tau \displaystyle \sum \limits _{k=0}^{m-1}||\nabla \overline{\mathbf{e }}_h^{k+1/2}||_{L^2}^2\right) \le C_{19}h^{2(r+1)}. \end{aligned}$$

(5.12)

Proof

In terms of (5.9) derived in Theorem 5.1, the error estimate (5.12) holds for $m=1$. Now, we suppose that (5.12) holds for $m\le n, n\le N-1$. Then we need to show it also hold for $m= n+1$. Under this assumption, one has

$$\begin{aligned} \max \limits _{1\le m\le n}( ||\hat{\mathbf{e }}_h^{m+1/2}||_{L^2}^2+h^2 ||\hat{\mathbf{e }}_h^{m+1/2}||_{H^1}^2)\le C_{20}C_{19}h^{2(r+1)}, \end{aligned}$$

(5.13)

where $C_{20}>0$ is independent of $C_{19}$. Instead of n by m in (3.1) from (4.3), subtracting (3.1) from (4.3) and setting $\mathbf{v }_h=\overline{\mathbf{e }}_h^{m+1/2}$ lead to

$$\begin{aligned}&\displaystyle \frac{1}{2}D_\tau (||\mathbf{e }_h^{m+1}||_{L^2}^2) +\lambda ||\nabla \overline{\mathbf{e }}_h^{m+1/2}||_{L^2}^2\nonumber \\&\quad = (D_\tau \mathbf{E }^{m+1},\overline{\mathbf{e }}_h^{m+1/2})-\lambda ( \overline{\mathbf{E }}^{m+1/2}, \overline{\mathbf{e }}_h^{m+1/2}) - \left( (\hat{\mathbf{e }}^{m+1/2}_h-\hat{\mathbf{E }}^{m+1/2})\times \nabla \overline{\mathbf{M }}_h^{m+1/2},\nabla \overline{\mathbf{e }}_h^{m+1/2}\right) \nonumber \\&\qquad +\lambda \left( |\nabla \hat{\mathbf{M }}^{m+1/2}|^2\hat{\mathbf{M }}^{m+1/2}-|\nabla \hat{\mathbf{M }}_h^{m+1/2}|^2\hat{\mathbf{M }}_h^{m+1/2},\overline{\mathbf{e }}_h^{m+1/2}\right) . \end{aligned}$$

(5.14)

All terms in the right-hand side of (5.14) can be bounded by a similar way for Theorem 5.1. From (5.4), (5.6) and the definition of $\overline{\mathbf{e }}_h^{m+1/2}$, we have

$$\begin{aligned}&\quad \ (D_\tau \mathbf{E }^{m+1},\overline{\mathbf{e }}_h^{m+1/2})-\lambda ( \overline{\mathbf{E }}^{m+1/2}, \overline{\mathbf{e }}_h^{m+1/2}) \le \displaystyle \frac{\lambda }{4}||\nabla \overline{\mathbf{e }}_h^{m+1/2}||_{\mathbf{L }^2}^2+ C_{21} h^{2(r+1)}, \end{aligned}$$

and

$$\begin{aligned} \left( (\hat{\mathbf{e }}^{m+1/2}_h-\hat{\mathbf{E }}^{m+1/2})\times \nabla \overline{\mathbf{M }}_h^{m+1/2},\nabla \overline{\mathbf{e }}_h^{m+1/2}\right) \le \displaystyle \frac{\lambda }{4}||\nabla \overline{\mathbf{e }}_h^{m+1/2}||_{\mathbf{L }^2}^2+C_{21} h^{2(r+1)}, \end{aligned}$$

where $C_{21}>0$ is independent of $C_{19}$. From (5.11), the last term is estimated by

$$\begin{aligned}&\displaystyle \frac{\lambda }{4}||\nabla \overline{\mathbf{e }}_h^{m+1/2}||_{L^2}^2+C_{22}||\hat{\mathbf{e }}^{m+1/2}_h||_{L^2}^2+C_{22}h^{2(r+1)}\\&\qquad +C_{22}(1+C_{20}^{1/2}C_{19}^{1/2})^4h^{2\min \{2r-2d/3, r-d/2\}}h^{2(r+1)}\\&\qquad + C_{22}(h+(1+C_{20}^{1/2}C_{19}^{1/2})h^{r-d/6})^2||\nabla \hat{\mathbf{e }}_h^{n+1/2}||_{L^2}^2, \end{aligned}$$

where $C_{22}>0$ is independent of $C_{19}$. We take sufficiently small h to satisfy

$$\begin{aligned} (1+C_{20}^{1/2}C_{19}^{1/2})^4h^{2\min \{2r-2d/3, r-d/2\}}<1\quad \hbox {and} \quad C_{22}(h+(1+C_{20}^{1/2}C_{19}^{1/2})h^{r-d/6})^2<\displaystyle \frac{\lambda }{4}. \end{aligned}$$

Then from discrete Gronwall inequality, there exists some small $\tau _{2}>0$ satisfying $2C_{22} \tau _2<1$ such that when $\tau <\tau _2$, there holds

$$\begin{aligned} \max \limits _{0\le m\le n}||\mathbf{e }_h^{m+1}||_{L^2}^2 +\tau \displaystyle \sum \limits _{m=0}^{n}||\nabla \overline{\mathbf{e }}_h^{m+1/2}||_{L^2}^2\le \exp (TC_{23})h^{2(r+1)}. \end{aligned}$$

We take sufficiently large $C_{19}$ to satisfy $C_{19}>\exp (TC_{23})$ and complete the proof of Theorem 5.2 by using the inverse inequality (3.6). $\square$

6 Proof of Theorem 3.1

The existence and uniqueness of the solution to (3.1) can be proved by a similar way in [18]. Here we omit it and only prove the optimal error estimates (3.4) and (3.5), which is based upon the following error splitting:

$$\begin{aligned} ||\mathbf{m }^n-\mathbf{M }_h^n||_{H^m}&\le ||\mathbf{e }^n||_{H^m}+||\mathbf{E }^n||_{H^m}+||\mathbf{e }_{h}^n||_{H^m}, \quad m=0,1, \end{aligned}$$

(6.1)

where $||\mathbf{e }^n||_{H^m}, ||\mathbf{e }_{h}^n||_{H^m}$ and $||\mathbf{E }^n||_{H^m}$ are the temporal error, the spatial error and the projection error, respectively. In terms of (6.1) and (5.4–5.5), the optimal $\mathbf{L }^2$-norm and $\mathbf{H }^1$-norm error estimates (3.4) and (3.5) are the direct consequences of (4.24) (5.4), (5.5) and (5.12) with $\tau _0=\min \{\tau _1, \tau _2\}$ and $h_0=h_2$. Although the FEM scheme (3.1) does not satisfy the nonlinear constraint $|\mathbf{M }_h^n|=1$, the convergence rate between 1 and $|\mathbf{M }_h^n|^2$ in $L^2$-norm can be obtained. Indeed, by using

$$\begin{aligned} ||\mathbf{M }_h^n||_{L^\infty }\le ||P_h^{n}\mathbf{M }^n||_{L^\infty }+||\mathbf{e }_h^n||_{L^\infty } \le C||\mathbf{M }^n||_{L^\infty }+Ch^{r+1-d/2}\le C, \end{aligned}$$

then the convergence rate between 1 and $|\mathbf{M }_h^n|^2$ in the sense of $\mathbf{L }^2$-norm is from

$$\begin{aligned} ||1-|\mathbf{M }^n_h|^2||_{L^2}&=|||\mathbf{m }^n|^2-|\mathbf{M }^n_h|^2||_{L^2}\\&=||\mathbf{m }^n-\mathbf{M }^n_h||_{L^2}||\mathbf{m }^n+\mathbf{M }^n_h||_{L^\infty }\\&\le C||\mathbf{m }^n-\mathbf{M }^n_h||_{L^2}\le C_0(\tau ^2+h^{r+1}). \end{aligned}$$

This completes the proof of Theorem 3.1.

7 Numerical Results

In this section, we present the numerical results to verify the optimal error estimates derived in Theorem 3.1 by using linear FEM. All programs are implemented by the free finite element software FreeFem++[22]. We consider the LL equation in the unit circle $\Omega =\{(x,y): \ x^2+y^2<1\}$. The initial value $\mathbf{m }_0$ is taken as

$$\begin{aligned} \mathbf{m }_0=(\sin (x)\cos (y), \cos (x)\cos (y), \sin (y)). \end{aligned}$$

Gilbert damping constant is set as $\lambda =1$. We take a uniform triangular partition with M nodes on $\partial \Omega$. Then a class of uniform meshes of the unit circle is made by a mesh generator in FreeFem++; see Fig. 1 for illustration.

Fig. 1

The FEM meshes of the unit circle with $M=50$

We solve the LL equation (1.4) and (1.2) by using the linearized Crank–Nicolson scheme (3.1–3.2) with linear FEM. Based upon our theoretical analysis, in this case, we have

$$\begin{aligned} \max \limits _{1\le n\le N}\left( ||\mathbf{m }^n-\mathbf{M }^n_h||_{L^2}+||1-|\mathbf{M }^n_h|^2||_{L^2}\right)&\le C_0(\tau ^2+h^2),\\ \max \limits _{1\le n\le N}||\mathbf{m }^n-\mathbf{M }^n_h||_{H^1}&\le C_0(\tau ^2+h). \end{aligned}$$

To confirm the optimal convergence rates for the errors $||\mathbf{m }^n-\mathbf{M }_h^n||_{L^2}$, $||\mathbf{m }^n-\mathbf{M }_h^n||_{H^1}$ and $||1-|\mathbf{M }_h^n|^2||_{L^2}$, the time step is taken as $\tau =0.1/M$. On the other hand, since no exact solution exists, the reference solution is taken as the numerical solution corresponding to $M=300$ and $\tau =0.1/M$. The numerical errors are displayed in Table 1, from which we can see $\mathbf{L }^2$ convergence rates are in good agreement with our theoretical analysis. $\mathbf{H }^1$ convergence rate is better than that in the theoretical analysis.

Table 1 Numerical $\mathbf{L }^2$ and $H^1$ errors and convergence rates for linear FEM

Table 2 Numerical $\mathbf{H }^1$ and $\mathbf{L }^2$ errors and convergence rates for different $\tau$

To confirm the convergence rate $O(\tau ^2)$ of the temporal error, the reference solution is taken as the numerical solution corresponding to fixed $M=120$ and $\tau =2.5\times 10^{-4}$. For different time step $\tau _{i+1}=0.5\tau _i$ for $i=1,2,3$ with $\tau _1=10^{-2}$, the numerical $\mathbf{H }^1$ and $\mathbf{L }^2$ errors are displayed in Table 2. It can be observed that the Crank–Nicolson scheme (3.1–3.2) is of the $O(\tau ^2)$ convergence rates of the temporal error which coincides with the ones predicted in Theorem 3.1.

These numerical results show the efficiency of the Crank–Nicolson scheme (3.1–3.2) and confirm our theoretical analysis.

Appendix: Proof of Theorem 2.3

Taking $t=0$ at (1.4), we get

$$\begin{aligned} \mathbf{m }_t(0)=\lambda \Delta \mathbf{m }_0 + \mathbf{m }_0\times \Delta \mathbf{m }_0+\lambda |\nabla \mathbf{m }_0|^2\mathbf{m }_0. \end{aligned}$$

By a classical argument, we can prove

$$\begin{aligned} ||\mathbf{m }_t(0)||_{H^3}\le C. \end{aligned}$$

(8.1)

Differentiating (1.4) with respect to t leads to

$$\begin{aligned} \mathbf{m }_{tt}-\lambda \Delta \mathbf{m }_t= \mathbf{m }_t\times \Delta \mathbf{m }+\mathbf{m }\times \Delta \mathbf{m }_t+\lambda |\nabla \mathbf{m }|^2\mathbf{m }_t+ 2\lambda (\nabla \mathbf{m }\cdot \nabla \mathbf{m }_t)\mathbf{m }. \end{aligned}$$

(8.2)

The following theorem improves the regularity (2.11) in Theorem 2.2.

Theorem 8.1

Suppose $\mathbf{m }_0\in \mathbf{H }^5(\Omega )$. Then we have

$$\begin{aligned} \max \limits _{t\in [0,T]} || \mathbf{m }_t(t)||_{H^1} +\displaystyle \int _0^T ||\mathbf{m }_{tt}||_{L^2}^2+|| \mathbf{m }_t||_{H^2}^2 dt\le C. \end{aligned}$$

(8.3)

Proof

Multiplying (8.2) by $-\Delta \mathbf{m }_t$ and integrating over $\Omega$ yield

$$\begin{aligned}&\ \displaystyle \frac{1}{2}\displaystyle \frac{d}{dt}||\nabla \mathbf{m }_t||_{L^2}^2+\lambda ||\Delta \mathbf{m }_t||_{L^2}^2\nonumber \\&\quad =-(\mathbf{m }_t\times \Delta \mathbf{m },\Delta \mathbf{m }_t)-\lambda (|\nabla \mathbf{m }|^2\mathbf{m }_t,\Delta \mathbf{m }_t)- 2\lambda ((\nabla \mathbf{m }\cdot \nabla \mathbf{m }_t)\mathbf{m },\Delta \mathbf{m }_t)\nonumber \\&\quad =I_1+I_2+I_3, \end{aligned}$$

(8.4)

where we use $(\mathbf{m }\times \Delta \mathbf{m }_t,\Delta \mathbf{m }_t)=0$. By Hölder inequality, Young inequality, (2.7) and (2.9), $I_1$ is bounded as

$$\begin{aligned} |I_1|&\le ||\mathbf{m }_t||_{L^\infty }||\Delta \mathbf{m }||_{L^2}||\Delta \mathbf{m }_t||_{L^2}\\&\le C||\mathbf{m }_t||_{L^2}^{\frac{4-d}{4}}(||\mathbf{m }_t||_{L^2}^2+||\Delta \mathbf{m }_t||_{L^2}^2)^{\frac{d}{8}}||\Delta \mathbf{m }_t||_{L^2}\\&\le \varepsilon _1||\Delta \mathbf{m }_t||_{L^2}^2+C ||\mathbf{m }_t||_{L^2}^{\frac{4-d}{2}} (||\mathbf{m }_t||_{L^2}^2+||\Delta \mathbf{m }_t||_{L^2}^2)^{\frac{d}{4}}\\&\le 2\varepsilon _1||\Delta \mathbf{m }_t||_{L^2}^2+C ||\mathbf{m }_t||_{L^2}^2, \end{aligned}$$

where $\varepsilon _1>0$ is a small constant determined later. It follows from (2.1) and (2.9) that

$$\begin{aligned} |I_2|\le ||\mathbf{m }_t||_{L^6}||\nabla \mathbf{m }||_{L^6}^2||\Delta \mathbf{m }_t||_{L^2}\le \varepsilon _1||\Delta \mathbf{m }_t||_{L^2}^2+C||\nabla \mathbf{m }_t||_{L^2}^2. \end{aligned}$$

By a similar way, we have

$$\begin{aligned} |I_3|&\le C ||\nabla \mathbf{m }_t||_{L^2}^{\frac{6-d}{6}}||\Delta \mathbf{m }_t||_{L^2}^{1+\frac{d}{6}}+C||\nabla \mathbf{m }_t||_{L^2}||\Delta \mathbf{m }_t||_{L^2}\\&\le \varepsilon _1||\Delta \mathbf{m }_t||_{L^2}^2+C||\nabla \mathbf{m }_t||_{L^2}^2. \end{aligned}$$

Let $\varepsilon _1<\lambda /4$. Then combining these estimates into (8.4), and using Gronwall inequality, we conclude that

$$\begin{aligned} \max \limits _{t\in [0,T]} || \mathbf{m }_t(t)||_{H^1} +\displaystyle \int _0^T || \mathbf{m }_t||_{H^2}^2 dt\le C, \end{aligned}$$

(8.5)

where we use (2.10) and (8.1). From (8.2), one has

$$\begin{aligned} ||\mathbf{m }_{tt}||_{L^2}\le&||\Delta \mathbf{m }_t||_{L^2}+||\mathbf{m }_t||_{L^\infty }||\Delta \mathbf{m }||_{L^2}+||\mathbf{m }||_{L^\infty }||\Delta \mathbf{m }_t||_{L^2}\\&+||\nabla \mathbf{m }||_{L^6}^2||\mathbf{m }_t||_{L^6}+||\nabla \mathbf{m }||_{L^6}||\nabla \mathbf{m }_t||_{L^3}\\ \le&C ||\Delta \mathbf{m }_t||_{L^2}+C||\mathbf{m }_t||_{L^2}^{\frac{4-d}{4}}(||\mathbf{m }_t||_{L^2}^2+||\Delta \mathbf{m }_t||_{L^2}^2)^{\frac{d}{8}}\\&+C|| \mathbf{m }_t||_{H^1}+C||\nabla \mathbf{m }_t||_{L^2}^{\frac{6-d}{6}}||\Delta \mathbf{m }_t||_{L^2}^{\frac{d}{6}}+C||\nabla \mathbf{m }_t||_{L^2}\\ \le&C||\mathbf{m }_t||_{H^2}+C(||\mathbf{m }_t||_{L^2}^2+||\Delta \mathbf{m }_t||_{L^2}^2)^{1/2}+C||\Delta \mathbf{m }_t||_{L^2}, \end{aligned}$$

which implies that $\mathbf{m }_{tt}\in \mathbf{L }^2(0,T;\mathbf{L }^2(\Omega ))$ after integrating above inequality from 0 to T and using (8.5). $\square$

Taking $t=0$ in (8.2) and using (8.1), we have

$$\begin{aligned} ||\mathbf{m }_{tt}(0)||_{H^1}\le C. \end{aligned}$$

(8.6)

Differentiating (8.2) with respect to t, again, we get

$$\begin{aligned} \mathbf{m }_{ttt}-\lambda \Delta \mathbf{m }_{tt}=&\mathbf{m }_{tt}\times \Delta \mathbf{m }+ 2\mathbf{m }_t\times \Delta \mathbf{m }_t+\mathbf{m }\times \Delta \mathbf{m }_{tt}+2\lambda |\nabla \mathbf{m }_t|^2\mathbf{m }\nonumber \\&+\lambda |\nabla \mathbf{m }|^2\mathbf{m }_{tt}+ 2\lambda (\nabla \mathbf{m }\cdot \nabla \mathbf{m }_{tt})\mathbf{m }+4\lambda (\nabla \mathbf{m }\cdot \nabla \mathbf{m }_{t})\mathbf{m }_t. \end{aligned}$$

(8.7)

Then we can show the following estimate for $\mathbf{m }_{tt}$.

Theorem 8.2

Suppose $\mathbf{m }_0\in \mathbf{H }^5(\Omega )$. Then we have

$$\begin{aligned} \max \limits _{t\in [0,T]} || \mathbf{m }_{tt}(t)||_{L^2} +\displaystyle \int _0^T || \mathbf{m }_{tt}||_{H^1}^2 dt\le C. \end{aligned}$$

(8.8)

Proof

Testing (8.7) by $\mathbf{m }_{tt}$ and using $(\mathbf{m }_{tt}\times \Delta \mathbf{m },\mathbf{m }_{tt})=0$ yield

$$\begin{aligned}&\displaystyle \frac{1}{2}\displaystyle \frac{d}{dt}|| \mathbf{m }_{tt}||_{L^2}^2+\lambda ||\nabla \mathbf{m }_{tt}||_{L^2}^2\nonumber \\&\quad =(\mathbf{m }\times \Delta \mathbf{m }_{tt},\mathbf{m }_{tt})+2(\mathbf{m }_{t}\times \Delta \mathbf{m }_t, \mathbf{m }_{tt})+\lambda (|\nabla \mathbf{m }|^2\mathbf{m }_{tt}, \mathbf{m }_{tt})\nonumber \\&\qquad +2\lambda (|\nabla \mathbf{m }_t|^2\mathbf{m }, \mathbf{m }_{tt}) +2\lambda ((\nabla \mathbf{m }\cdot \nabla \mathbf{m }_{tt})\mathbf{m },\mathbf{m }_{tt})+4\lambda ((\nabla \mathbf{m }\cdot \nabla \mathbf{m }_{t})\mathbf{m }_t,\mathbf{m }_{tt})\nonumber \\&\quad =I_4+\cdots +I_9. \end{aligned}$$

(8.9)

By integrating by parts, we estimate $I_4$ by

$$\begin{aligned} |I_4|\le ||\nabla \mathbf{m }||_{L^6}||\nabla \mathbf{m }_{tt}||_{L^2}||\mathbf{m }_{tt}||_{L^3} \le \varepsilon _2 ||\nabla \mathbf{m }_{tt}||_{L^2}^2+C||\mathbf{m }_{tt}||_{L^2}^2 \end{aligned}$$

for some small $\varepsilon _2>0$ determined later, where we use Hölder inequality, Young inequality, (2.1), (2.3), (2.9). From (2.1–2.9) and (8.5), other terms in the right-hand side of (8.9) are bounded, respectively, by

$$\begin{aligned} |I_5|&\le ||\mathbf{m }_t||_{L^3}||\Delta \mathbf{m }_{t}||_{L^2}||\mathbf{m }_{tt}||_{L^6}\le \varepsilon _2 ||\nabla \mathbf{m }_{tt}||_{L^2}^2+ C ||\Delta \mathbf{m }_{t}||_{L^2}^2,\\ |I_6|&\le ||\nabla \mathbf{m }||_{L^6}^2||\mathbf{m }_{tt}||_{L^2}||\mathbf{m }_{tt}||_{L^6}\le \varepsilon _2 ||\nabla \mathbf{m }_{tt}||_{L^2}^2+ C || \mathbf{m }_{tt}||_{L^2}^2,\\ |I_7|&\le ||\mathbf{m }||_{L^\infty }||\nabla \mathbf{m }_t||_{L^3}||\nabla \mathbf{m }_t||_{L^2}||\mathbf{m }_{tt}||_{L^6}\le C||\nabla \mathbf{m }_t||_{L^2}^{\frac{6-d}{6}}|| \mathbf{m }_t||_{H^2}^{\frac{d}{6}}||\nabla \mathbf{m }_{tt}||_{L^2}\\&\le \varepsilon _2 ||\nabla \mathbf{m }_{tt}||_{L^2}^2+ C || \mathbf{m }_t||_{H^2}^2+ C ,\\ |I_8|&\le ||\nabla \mathbf{m }||_{L^6}||\mathbf{m }||_{L^\infty }||\nabla \mathbf{m }_{tt}||_{L^2}||\mathbf{m }_{tt}||_{L^3}\\&\le C ||\nabla \mathbf{m }_{tt}||_{L^2}||\mathbf{m }_{tt}||_{L^2}+ C ||\mathbf{m }_{tt}||_{L^2}^{\frac{6-d}{6}}||\nabla \mathbf{m }_{tt}||_{L^2}^{1+\frac{d}{6}}\\&\le \varepsilon _2 ||\nabla \mathbf{m }_{tt}||_{L^2}^2+ C || \mathbf{m }_{tt}||_{L^2}^2,\\ |I_9|&\le ||\nabla \mathbf{m }||_{L^2}||\nabla \mathbf{m }_t||_{L^6}||\mathbf{m }_t||_{L^6}||\mathbf{m }_{tt}||_{L^6}\\&\le C ||\mathbf{m }_t||_{H^2}||\nabla \mathbf{m }_{tt}||_{L^2}\le \varepsilon _2 ||\nabla \mathbf{m }_{tt}||_{L^2}^2+ C || \mathbf{m }_{t}||_{H^2}^2. \end{aligned}$$

Combining these estimates into (8.9) and taking $\varepsilon _2<\lambda /6$, the estimate (8.8) follows from Gronwall inequality. $\square$

To improve the regularity of the solution $\mathbf{m }$ to the LL equation (1.4) and (1.2), the following estimates (8.10) and (8.12) for $\mathbf{m }\times \Delta \mathbf{m }$ are needed.

Lemma 8.1

Suppose $\mathbf{m }_0\in \mathbf{H }^5(\Omega )$. Then we have

$$\begin{aligned} \max \limits _{t\in [0,T]} || (\mathbf{m }\times \Delta \mathbf{m })(t)||_{H^1} +\displaystyle \int _0^T ||\mathbf{m }\times \Delta \mathbf{m }||_{H^2}^2 dt\le C . \end{aligned}$$

(8.10)

Proof

It follows from (1.5) and (8.3) that

$$\begin{aligned} ||\mathbf{m }\times \Delta \mathbf{m }||_{H^1}&\le ||\mathbf{m }_t||_{H^1}+||\mathbf{m }\times \mathbf{m }_t||_{H^1}\nonumber \\&\le C||\mathbf{m }_t||_{H^1}+||\nabla \mathbf{m }\times \mathbf{m }_t||_{L^2}+C||\mathbf{m }\times \nabla \mathbf{m }_t||_{L^2}\nonumber \\&\le C||\mathbf{m }_t||_{H^1}+C||\mathbf{m }||_{H^2}||\mathbf{m }_t||_{H^1}\le C. \end{aligned}$$

(8.11)

Differentiating (1.5) with respect to $\mathbf{x }$ twice, we obtain

$$\begin{aligned} (1+\lambda ^2)\nabla ^2(\mathbf{m }\times \Delta \mathbf{m })=\nabla ^2\mathbf{m }_t+\lambda (\nabla ^2\mathbf{m }\times \mathbf{m }_t+\nabla \mathbf{m }\times \nabla \mathbf{m }_t+\mathbf{m }\times \nabla ^2 \mathbf{m }_t). \end{aligned}$$

Then one has

$$\begin{aligned}&\displaystyle \int _0^T||\nabla ^2(\mathbf{m }\times \Delta \mathbf{m })||_{L^2}^2dt\\&\quad \le \displaystyle \int _0^T\left( ||\nabla ^2\mathbf{m }_t||_{L^2}^2+||\nabla ^2\mathbf{m }\times \mathbf{m }_t||_{L^2}^2 +||\nabla \mathbf{m }\times \nabla \mathbf{m }_t||_{L^2}^2+||\mathbf{m }\times \nabla ^2 \mathbf{m }_t||_{L^2}^2\right) dt\\&\quad \le C \displaystyle \int _0^T\left( ||\mathbf{m }_t||_{H^2}^2+||\mathbf{m }||_{H^2}^2||\mathbf{m }_t||_{L^\infty }^2 +||\mathbf{m }||_{H^2}^2||\mathbf{m }_t||_{H^2}^2+||\mathbf{m }||_{L^\infty }^2||\mathbf{m }_t||_{H^2}^2\right) dt\le C, \end{aligned}$$

which together with (8.11) completes the proof of (8.10). $\square$

Lemma 8.2

Suppose $\mathbf{m }_0\in \mathbf{H }^5(\Omega )$. Then we have

$$\begin{aligned} \max \limits _{t\in [0,T]} || (\mathbf{m }\times \Delta \mathbf{m })_t||_{L^2} +\displaystyle \int _0^T || (\mathbf{m }\times \Delta \mathbf{m })_t||_{H^1}^2 dt\le C. \end{aligned}$$

(8.12)

Proof

Differentiating (1.5) with respect to t yields

$$\begin{aligned} (1+\lambda ^2)(\mathbf{m }\times \Delta \mathbf{m })_t=\mathbf{m }_{tt}+\lambda \mathbf{m }\times \mathbf{m }_{tt}. \end{aligned}$$

From (2.9) and (8.8), it is easy to show

$$\begin{aligned} \max \limits _{t\in [0,T]}|| (\mathbf{m }\times \Delta \mathbf{m })_t||_{L^2}\le C. \end{aligned}$$

Moreover, from (2.9) and (8.8) one has

$$\begin{aligned}&\displaystyle \int _0^T ||\nabla (\mathbf{m }\times \Delta \mathbf{m })_t||_{L^2}^2 dt\\&\quad \le \displaystyle \int _0^T ||\nabla \mathbf{m }_{tt}||_{L^2}^2+||\nabla \mathbf{m }\times \mathbf{m }_{tt}||_{L^2}^2+||\mathbf{m }\times \nabla \mathbf{m }_{tt}||_{L^2}^2 dt\\&\quad \le \displaystyle \int _0^T C||\nabla \mathbf{m }_{tt}||_{L^2}^2+||\mathbf{m }||_{H^2}^2||\nabla \mathbf{m }_{tt}||_{L^2}^2+||\mathbf{m }||_{L^\infty }^2|| \nabla \mathbf{m }_{tt}||_{L^2}^2 dt \le C, \end{aligned}$$

which completes the proof of (8.12). $\square$

Based upon the above results, The regularities of the solution $\mathbf{m }$ can be improved.

Theorem 8.3

Suppose $\mathbf{m }_0\in \mathbf{H }^5(\Omega )$. Then we have

$$\begin{aligned} \max \limits _{t\in [0,T]} || \mathbf{m }(t)||_{H^3} \le C. \end{aligned}$$

(8.13)

Proof

Differentiating (1.4) with respect to $\mathbf{x }$ yields:

$$\begin{aligned} -\lambda \nabla \Delta \mathbf{m }=\nabla (\mathbf{m }\times \Delta \mathbf{m })-\nabla \mathbf{m }_t+2\lambda (\nabla ^2\mathbf{m }\cdot \nabla \mathbf{m })\mathbf{m }+\lambda |\nabla \mathbf{m }|^2\nabla \mathbf{m }. \end{aligned}$$

From (2.8), we have

$$\begin{aligned} ||\nabla ^3\mathbf{m }||_{L^2}&\le ||\nabla \mathbf{m }_t||_{L^2}+||\nabla (\mathbf{m }\times \Delta \mathbf{m })||_{L^2}+||\Delta \mathbf{m }||_{L^2} +||(\nabla ^2\mathbf{m }\cdot \nabla \mathbf{m })\mathbf{m }||_{L^2}+|||\nabla \mathbf{m }|^2\nabla \mathbf{m }||_{L^2}\\&\le C\left( ||\nabla \mathbf{m }_t||_{L^2}+||\nabla (\mathbf{m }\times \Delta \mathbf{m })||_{L^2}+||\Delta \mathbf{m }||_{L^2} +||\nabla ^2\mathbf{m }||_{L^3}||\nabla \mathbf{m }||_{L^6}+||\nabla \mathbf{m }||_{L^6}^2 \right) \\&\le C ||\nabla ^2\mathbf{m }||_{L^2}^{\frac{6-d}{6}}||\nabla ^3\mathbf{m }||_{L^2}^{\frac{d}{6}}+ C \le \displaystyle \frac{1}{2}||\nabla ^3\mathbf{m }||_{L^2}+ C||\nabla ^2\mathbf{m }||_{L^2}+ C, \end{aligned}$$

which leads to $||\nabla ^3\mathbf{m }||_{L^2}\le C$ and $\mathbf{m }\in \mathbf{L }^\infty (0,T;\mathbf{H }^3(\Omega ))$. $\square$

An alternative to (8.2 ) is

$$\begin{aligned} \mathbf{m }_{tt}-\lambda \Delta \mathbf{m }_t= (\mathbf{m }\times \Delta \mathbf{m })_t+\lambda |\nabla \mathbf{m }|^2\mathbf{m }_t+ 2\lambda (\nabla \mathbf{m }\cdot \nabla \mathbf{m }_t)\mathbf{m }. \end{aligned}$$

(8.14)

We can derive the following estimates for $\mathbf{m }_t$ and $\mathbf{m }_{tt}$.

Theorem 8.4

Suppose $\mathbf{m }_0\in \mathbf{H }^5(\Omega )$. Then we have

$$\begin{aligned} \max \limits _{t\in [0,T]} || \mathbf{m }_t(t)||_{H^2} +\displaystyle \int _0^T || \mathbf{m }_t||_{H^3}^2 dt\le C. \end{aligned}$$

(8.15)

Proof

In view of (2.5), (8.8) and (8.12), it is easy to show

$$\begin{aligned} ||\nabla ^2\mathbf{m }_t||_{L^2}&\le C\left( ||\mathbf{m }_{tt}||_{L^2}+||(\mathbf{m }\times \Delta \mathbf{m })_t||_{L^2}+|||\nabla \mathbf{m }|^2\mathbf{m }_t||_{L^2}+||(\nabla \mathbf{m }\cdot \nabla \mathbf{m }_t)\mathbf{m }||_{L^2}\right) \\&\le C \left( ||\mathbf{m }_{tt}||_{L^2}+||(\mathbf{m }\times \Delta \mathbf{m })_t||_{L^2}+||\mathbf{m }||_{H^3}^2||\mathbf{m }_t||_{L^2}+ ||\mathbf{m }||_{H^3}||\nabla \mathbf{m }_t||_{L^2}\right) \le C, \end{aligned}$$

which leads to $\mathbf{m }_t\in \mathbf{L }^\infty (0,T;\mathbf{H }^2(\Omega ))$. Differentiating (8.14) with respect to $\mathbf{x }$ yields:

$$\begin{aligned} \nabla \mathbf{m }_{tt}-\lambda \nabla \Delta \mathbf{m }_t&= \nabla (\mathbf{m }\times \Delta \mathbf{m })_t+ 2\lambda (\nabla ^2\mathbf{m }\cdot \nabla \mathbf{m })\mathbf{m }_t+\lambda |\nabla \mathbf{m }|^2\nabla \mathbf{m }_t\\&\quad +2\lambda (\nabla ^2\mathbf{m }\cdot \nabla \mathbf{m }_t)\mathbf{m }+2\lambda (\nabla \mathbf{m }\cdot \nabla ^2\mathbf{m }_t)\mathbf{m }+2\lambda (\nabla \mathbf{m }\cdot \nabla \mathbf{m }_t)\nabla \mathbf{m }. \end{aligned}$$

By using (2.8), we get

$$\begin{aligned} ||\nabla ^3\mathbf{m }_t||_{L^2}&\le C\left( ||\nabla \mathbf{m }_{tt}||_{L^2}+|| \nabla (\mathbf{m }\times \Delta \mathbf{m })_t||_{L^2}+||\mathbf{m }||_{H^3}^2||\nabla \mathbf{m }_t||_{L^2}\right. \\&\quad \left. +||\Delta \mathbf{m }_t||_{L^2}+||\mathbf{m }||_{H^3}|| \mathbf{m }_t||_{H^2}\right) , \end{aligned}$$

which together with (8.8) and (8.12) yields $\mathbf{m }_t\in \mathbf{L }^2(0,T;\mathbf{H }^3(\Omega ))$. $\square$

Theorem 8.5

Suppose $\mathbf{m }_0\in \mathbf{H }^5(\Omega )$. Then we have

$$\begin{aligned} \max \limits _{t\in [0,T]} || \mathbf{m }_{tt}(t)||_{H^1} +\displaystyle \int _0^T || \mathbf{m }_{tt}||_{H^2}^2 dt\le C. \end{aligned}$$

(8.16)

Proof

Testing (8.7) by $-\Delta \mathbf{m }_{tt}$ and using Young inequality yield

$$\begin{aligned}&\displaystyle \frac{1}{2}\displaystyle \frac{d}{dt}||\nabla \mathbf{m }_{tt}||_{L^2}^2+\lambda ||\Delta \mathbf{m }_{tt}||_{L^2}^2\\&\quad \le \left( ||\mathbf{m }_{tt}||_{L^6}||\mathbf{m }||_{H^3}+||\mathbf{m }_t||_{L^\infty }||\Delta \mathbf{m }_t||_{L^2}+||\nabla \mathbf{m }||_{L^\infty }^2||\mathbf{m }_{tt}||_{L^2}\right. \\&\qquad \left. +||\mathbf{m }_t||_{H^2}^2+||\mathbf{m }||_{H^3}^2||\nabla \mathbf{m }_{tt}||_{L^2}+||\mathbf{m }||_{L^\infty }||\mathbf{m }_t||_{L^\infty }||\nabla \mathbf{m }_t||_{L^2}\right) ||\Delta \mathbf{m }_{tt}||_{L^2}\\&\quad \le \displaystyle \frac{\lambda }{2}||\Delta \mathbf{m }_{tt}||_{L^2}^2+C||\nabla \mathbf{m }_{tt}||_{L^2}^2+ C. \end{aligned}$$

Integrating the above inequality from 0 to T and using (8.6), we complete the proof of (8.16). $\square$