Singular limits of the equations of compressible ideal magneto-hydrodynamics in a domain with boundaries

Abstract

Singular limits of the equations of compressible ideal magneto-hydrodynamics with physical boundary conditions are considered. The uniform existence of classical solutions with respect to the Mach number and Alfvén number is established by energy methods. Under appropriate conditions on the initial data, convergence of the solutions of the original system is proved as the two small parameters tend to zero with the limiting solutions satisfying the two-dimensional incompressible flow.

Keywords

Compressible ideal magneto-hydrodynamics Mach number Alfvén number singular limits

1. Introduction

We consider the equations of compressible ideal magneto-hydrodynamics (ideal MHD) in three space dimensions $\begin{matrix} (1.1) & \{\begin{matrix} a (\partial_{t} p + u \cdot \nabla p) + \nabla \cdot u = 0, \\ ρ (\partial_{t} u + u \cdot \nabla u) + \frac{1}{ε_{M}^{2}} \nabla p + \frac{1}{ε_{A}^{2}} B \times (\nabla \times B) = 0, \\ \partial_{t} B - \nabla \times (u \times B) = 0, \nabla \cdot B = 0 . \end{matrix} \end{matrix}$ Here the unknown functions $p = p (t, x), u = u (t, x) \in R^{3}$ and $B = B (t, x) \in R^{3}$ stand for pressure, fluid velocity, and magnetic field, respectively. Density ρ is determined by the equation of state $ρ = f (p)$ , where $f (\cdot)$ is a given function so that $f > 0$ and $\frac{\partial f}{\partial p} > 0$ . The function a is given by $a = \frac{1}{ρ} \frac{\partial f}{\partial p} > 0$ . The small parameters $ε_{M}$ and $ε_{A}$ denote the Mach number and the Alfvén number, respectively. Here, we assume that $ε_{M} = ε_{A} : = ε$ .

Introducing the scalings for pressure and magnetic field $\begin{matrix} p = 1 + ε r, B = e_{1} + ε b, \end{matrix}$ with $e_{1} = {(1, 0, 0)}^{⊤}$ , we can reformulate (1.1) into $\begin{matrix} (1.2) & \{\begin{matrix} a (\partial_{t} r + u \cdot \nabla r) + \frac{1}{ε} \nabla \cdot u = 0, \\ ρ (\partial_{t} u + u \cdot \nabla u) + \frac{1}{ε} \nabla r + b \times (\nabla \times b) = \frac{1}{ε} (\partial_{1} b - \nabla b_{1}), \\ \partial_{t} b + u \cdot \nabla b - b \cdot \nabla u + b \nabla \cdot u = \frac{1}{ε} (\partial_{1} u - e_{1} \nabla \cdot u), \nabla \cdot b = 0 . \end{matrix} \end{matrix}$

The underlying physical space we consider here can be three-dimensional half space $Ω_{1} : = R_{+}^{3} = {(x_{1}, x_{2}, x_{3}) \in R^{3} ∣ x_{1} > 0}$ or an infinite slab bounded above and below by two parallel planes $Ω_{2} : = {(x_{1}, x_{2}, x_{3}) \in R^{3} ∣ h > x_{1} > 0}$ . And we can also assume the horizontal variable $(x_{2}, x_{3})$ belongs to the two-dimensional torus $T^{2}$ , that is $Ω_{3} : = (0, \infty) \times T^{2}$ or $Ω_{4} : = (0, h) \times T^{2}$ . Throughout this paper, we will use Ω to denote any one of these four domains. The boundaries of the domains will be denoted by Γ.

We supply the system (1.2) with the following initial data $\begin{matrix} (1.3) & (r, u, b) |_{t = 0} = (r_{0}, u_{0}, b_{0}) (x), \end{matrix}$ and boundary conditions [14,15] $\begin{matrix} (1.4) & u \cdot n |_{Γ} = 0, b \times n |_{Γ} = 0, \end{matrix}$ or $\begin{matrix} (1.5) & u |_{Γ} = 0, \end{matrix}$ where $n = (n_{1}, n_{2}, n_{3})$ is the unit outward normal of Ω and $b_{0}$ satisfies the divergence-free condition $\nabla \cdot b_{0} = 0$ .

In the present paper, our goal is to study the well-posedness and singular limit $ε \to 0$ of the initial boundary value problem (1.2), (1.3) and (1.4) (or (1.2), (1.3) and (1.5)). First, we discuss the local well-posedness for fixed ε. Set $U = {(p, u, b)}^{t}$ and rewrite (1.2) into the symmetric form $\begin{matrix} A_{0} (U) \partial_{t} U + \sum_{i = 1}^{3} A_{i} (U) \partial_{i} U = 0, \end{matrix}$ with $A_{0} (U) = diag {a, ρ, ρ, ρ, 1, 1, 1}$ . After applying the boundary condition $u_{1} = 0$ that holds for both (1.4) and (1.5), the boundary matrix of the equation (1.2) is $\begin{array}{l} A_{n} & = \sum_{i = 1}^{3} A_{i} (U) n_{i} |_{u_{1} = 0} = \mp A_{1} (U) |_{u_{1} = 0} \\ = \mp (\begin{matrix} 0 & \frac{1}{ε} & 0 & 0 & 0 & 0 & 0 \\ \frac{1}{ε} & 0 & 0 & 0 & 0 & b_{2} & b_{3} \\ 0 & 0 & 0 & 0 & 0 & - \frac{1}{ε} - b_{1} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & - \frac{1}{ε} - b_{1} \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & b_{2} & - \frac{1}{ε} - b_{1} & 0 & 0 & 0 & 0 \\ 0 & b_{3} & 0 & - \frac{1}{ε} - b_{1} & 0 & 0 & 0 \end{matrix}) . \end{array}$ The vector $(0, 0, 0, 0, 1, 0, 0)$ is an eigenvector of the symmetric matrix $A_{n}$ with eigenvalue zero. Restricting to the subspace orthogonal to that vector by deleting the fifth row and fifth column leaves a matrix whose determinant is $- \frac{{(1 + ε b_{1})}^{4}}{ε^{6}}$ . Hence for any bounded set of initial data the boundary matrix will have constant rank 6 on the boundary as long as the solution remains in a neighborhood of the initial data and satisfies the boundary condition, provided that ε is restricted to be sufficiently small. In particular, for any sufficiently small but fixed positive ε, the local well-posedness of classical solutions to the initial boundary problem (1.2), (1.3) and (1.4) (or (1.2), (1.3) and (1.5)) can be established in a similar fashion as Yanagisawa and Matsumura [14,15], who proved the existence of solutions of the original system (1.1) with boundary conditions analogous to those used here for the system (1.2). Our main efforts here are to establish the uniform existence with respect to ε of classical solutions to the initial boundary value problem (1.2), (1.3) and (1.4) (or (1.2), (1.3) and (1.5)) and prove the convergence to the solution of a limit equation as $ε \to 0$ .

In a domain without boundary (the whole space or high-dimensional torus), there have been some mathematical studies of the singular limits of ideal MHD. When $ε_{M}$ tends to zero but $ε_{A}$ remains positive, Klainerman and Majda [7,8] proved the convergence of the solutions to isentropic ideal MHD for the well-prepared initial data. And recently, Jiang, Ju and Li [6] investigated the convergence of solutions to nonisentropic ideal MHD with general initial data. When $ε_{A}$ tends to zero but $ε_{M}$ remains positive, the related results are quite limited. Browning and Kreiss [2] studied the limit of compressible ideal MHD with more assumptions on the initial data. Goto [4] discussed the singular limit of the incompressible ideal magneto-fluid motion with respect to the Alfvén number in the three dimensional torus $T^{3}$ . See also [10] where both viscous and inviscid MHD system are considered in torus. When $ε_{M}$ and $ε_{A}$ tend to zero simultaneously, we didn’t find any previous works even in the periodic domain. Recently, Cheng, Ju and Schochet developed the general theory of singular limits of a class of evolutionary partial differential equations having two small parameters in [3]. This theory can be applied directly to the singular limits of MHD when Alfvén number is much smaller than Mach number, still without boundary.

For the singular limits of ideal MHD in presence of boundaries, there have been very few rigorous studies, at least to the best of our knowledge. The mathematical analysis is more complicated and difficult than initial value problem because of the interaction between large operators and the boundary. Schochet [11,12] investigated the singular limits in bounded domain for quasilinear symmetric hyperbolic systems having a vorticity equations. Low Mach number limit was proved for compressible Euler equations in an exterior domain in [1] and [5]. Unfortunately, singular limits of the initial boundary value problem (1.2), (1.3) and (1.4) (or (1.2), (1.3) and (1.5)) is beyond the works [1,5,11,12] since we are considering the characteristic boundary conditions and there are no vorticity equations in the system.

Throughout this paper, C is a genetic constant which may change from line to line and is independent of ε. $L^{p} (Ω)$ ( $1 ⩽ p < \infty$ ) denotes the space of measurable functions whose p-powers are integrable on Ω with the norm $‖ \cdot ‖_{L^{p}}$ , and $L^{\infty} (R^{3})$ is the space of bounded measurable functions on Ω, with the norm $‖ \cdot ‖_{L^{\infty}}$ and also simply denote $‖ \cdot ‖_{L^{2}}$ by $‖ \cdot ‖$ . $W^{s, p}$ denotes the usual Sobolev space. In particular, we use the notation $H^{s} = W^{s, 2}$ with the norm $‖ \cdot ‖_{s}$ . We also use the sub-script h to indicate the “horizontal” variable. For example, we set $u_{h} = {(u_{2}, u_{3})}^{⊤}$ , $\nabla_{h} = {(\partial_{2}, \partial_{3})}^{⊤}$ , $\nabla_{h} \cdot = \partial_{2} + \partial_{3}$ and $Δ_{h} = \partial_{22} + \partial_{33}$ .

The rest parts of this paper are arranged as follows. The main results are stated in Section 2. In Section 3, we shall establish the uniform existence of the solution to the initial boundary value problem (1.2), (1.3) and (1.4) (or (1.2), (1.3) and (1.5)). Based on the results in Section 3, convergence of the solutions is proved as the small parameter ε tends to zero, and the limit system is identified in Section 4. Some lemmas that are used in Section 3 are given in the Appendix.

2. Main results

Define $\begin{matrix} X_{m} (T) = ⋂_{j = 0}^{m} C^{j} (0, T; H^{m - j} (Ω)), \end{matrix}$ with norm $\begin{matrix} ‖ u ‖_{X_{m} (T)} = sup_{t \in [0, T]} \sum_{j = 0}^{m} {‖ \partial_{t}^{j} u ‖}_{H^{m - j} (Ω)} . \end{matrix}$ For small but fixed ε, we have the following existence theorem.

Theorem 2.1 (Slight variant of [14,15]).

Let $m ⩾ 3$ being an integer. Assume that the initial data $(r_{0}, u_{0}, b_{0}) \in H^{m} (Ω)$ satisfies the compatibility conditions of order $m - 1$ . Then there exists a positive constant $T_{0}$ such that the initial boundary value problem ( 1.2 ), ( 1.3 ) and ( 1.4 ) (or ( 1.2 ), ( 1.3 ) and ( 1.5 )) has a unique solution $(r, u, b)$ which belongs to $X_{m} (T_{0})$ .

Before stating our main results on the uniform existence, let us introduce the following weighted norms $\begin{array}{l} (2.1) & | | | w | | |_{3, ε, T} \equiv sup_{0 ⩽ t ⩽ T} | | | w | | |_{3, ε}, \\ (2.2) & | | | w | | |_{3, ε} \equiv \sum_{| α | ⩽ 3} ε^{α_{0}} ‖ \partial_{t}^{α_{0}} \partial_{1}^{α_{1}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} w ‖, \\ (2.3) & | | | w | | |_{3, ε, T}^{*} \equiv sup_{0 ⩽ t ⩽ T} | | | w | | |_{3, ε}^{*}, \\ (2.4) & | | | w | | |_{3, ε}^{*} \equiv \sum_{| α | ⩽ 3} ε^{{[α_{0} - 1]}_{+}} ‖ \partial_{t}^{α_{0}} \partial_{1}^{α_{1}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} w ‖ . \end{array}$ Here, $α = (α_{0}, α_{1}, α_{2}, α_{3})$ is a multi-index with the length $| α | = α_{0} + α_{1} + α_{2} + α_{3}$ . And for an integer m, $\begin{matrix} {[m]}_{+} = \{\begin{matrix} 0, & m ⩽ 0, \\ m, & m > 0 . \end{matrix} \end{matrix}$ Now, we have the following theorem.

Theorem 2.2.
Assume that the hypothesis of Theorem 2.1 hold and that the initial data satisfy $\begin{matrix} (2.5) & {‖ (r_{0}, u_{0}, b_{0}) ‖}_{3} ⩽ Δ_{1}, \end{matrix}$ for some constant $Δ_{1} > 0$ . Then there exist constants $0 < T_{} < T_{0}$ and* $Δ_{2} > 0$ independent of ε such that the solution of the initial boundary value problem ( 1.2 ), ( 1.3 ) and ( 1.4 ) (or ( 1.2 ), ( 1.3 ) and ( 1.5 )) exists on $[0, T_{}]$ and satisfies* $\begin{matrix} (2.6) & | | | (r, u, b) | {| |}_{3, ε, T_{}} ⩽ Δ_{2} . \end{matrix}$

In the above theorem, in view of the assumptions (2.5) on initial data, we can deduce that $| | | (r, u, b) | | |_{3, ε} |_{t = 0} ⩽ M$ for some constant $M > 0$ independent of ε through the equations (1.2). In particular, we do not need to prepare the initial data here. On the other hand, we have got the uniform estimates (2.6) of solution in the weighted norms (2.1) with the weight of the term $\partial_{t}^{k} w$ and its spatial derivatives equal to $ε^{k}$ . This implies that we haven’t got any uniform bounds for some norm of the first order time derivative of solution which is important for the convergence theory. To prove the convergence of solution, we need to consider the well-prepared initial data.
Theorem 2.3.
Under the assumptions of Theorem* 2.2 , assume further that there exists some constant $Δ_{3} > 0$ such that $\begin{matrix} (2.7) & {‖ \nabla (r_{0} + b_{01}) ‖}_{2} + ‖ \nabla \cdot u_{0} ‖_{2} + ‖ \partial_{1} b_{0} ‖_{2} + {‖ (\partial_{1} u_{0} - e_{1} \nabla \cdot u_{0}) ‖}_{2} ⩽ ε Δ_{3}, \end{matrix}$ where $b_{01}$ denotes the first component of $b_{0}$ . Then there exist constants $T^{} > 0$ and* $Δ_{4} > 0$ independent of ε such that the solution (r,u,b) to the initial boundary value problem ( 1.2 ), ( 1.3 ) and ( 1.4 ) (or ( 1.2 ), ( 1.3 ) and ( 1.5 )) exists on $[0, T^{}]$ and satisfies* $\begin{matrix} (2.8) & | | | (r, u, b) | | |_{3, ε, T^{}}^{} ⩽ Δ_{4} . \end{matrix}$

From the equation (1.2), we deduce from (2.7) that $| | | (r, u, b) | | |_{3, ε}^{} |_{t = 0} ⩽ M$ for some $M > 0$ . That implies that we need to prepare the first order time derivative of initial data to be bounded. Once we have the uniform estimates (2.8), we can prove the following convergence results.
Theorem 2.4.
Under the assumptions of Theorem* 2.3 and restrict our underlying physical space Ω to be $Ω_{4}$ . Assume further that there exist ${\bar{r}}_{0}, {\bar{u}}_{0}$ and ${\bar{b}}_{0}$ such that $\begin{matrix} (2.9) & (r_{0}, b_{0}, u_{0}) \to ({\bar{r}}_{0}, {\bar{u}}_{0}, {\bar{b}}_{0}), \end{matrix}$ in $H^{3} (Ω)$ . Then there exists $(\bar{r}, \bar{u}, \bar{b}) \in C ([0, T^{}]; H^{3 - δ} (Ω)) \cap Lip ([0, T^{}]; H^{2} (Ω))$ for any $δ > 0$ such that the unique solution $(r, u, b)$ of the initial boundary value problem ( 1.2 ), ( 1.3 ) and ( 1.4 ) satisfies $\begin{matrix} (2.10) & (r, u, b) \to (\bar{r}, \bar{u}, \bar{b}), \end{matrix}$ ${weak}^{}$ in $L^{\infty} ([0, T^{}]; H^{3} (Ω))$ and strongly in $C ([0, T^{}]; H^{3 - δ} (Ω))$ as $ε \to 0$ . Moreover,* $(\bar{r}, \bar{u}, \bar{b})$ satisfies $\begin{matrix} (2.11) & \begin{matrix} {\bar{u}}_{1} = {\bar{b}}_{2} = {\bar{b}}_{3} = 0, \\ \{\begin{matrix} \bar{ρ} (\partial_{t} {\bar{u}}_{h} + ({\bar{u}}_{h} \cdot \nabla_{h}) {\bar{u}}_{h}) + \nabla_{h} \bar{π} = 0, \\ \nabla_{h} \cdot {\bar{u}}_{h} = 0, \end{matrix} \end{matrix} \end{matrix}$ and $\begin{matrix} (2.12) & \{\begin{matrix} \partial_{t} \bar{r} + ({\bar{u}}_{h} \cdot \nabla_{h}) \bar{r} = 0, \\ {\bar{b}}_{1} = - \bar{r} + constant . \end{matrix} \end{matrix}$ Here, $\overline{ρ} = ρ (1)$ is a constant, and $\bar{π}$ is determined by $\begin{matrix} (2.13) & - Δ_{h} \bar{π} = \bar{ρ} \nabla_{h} \cdot (({\bar{u}}_{h} \cdot \nabla_{h}) {\bar{u}}_{h}) on T^{2} . \end{matrix}$
Remark 2.5.
Note that the flow described by (2.11) is the two dimensional incompressible Euler flow in $T^{2}$ , and both $\bar{r}$ and ${\bar{b}}_{1}$ satisfy a linear transport equation since one can easily find that $\begin{matrix} \partial_{t} {\bar{b}}_{1} + ({\bar{u}}_{h} \cdot \nabla_{h}) {\bar{b}}_{1} = 0 . \end{matrix}$
Remark 2.6.
To obtain the convergence of solutions and identify the limit equations in the above theorem, we have restricted our physical domain to be $(0, h) \times T^{2}$ in order to be able to take average value of the solution. For instance, to determine the limit of the singular term $\frac{1}{ε} \nabla (r + b_{1})$ , we need to use the Poincaré-type inequality to get the uniform estimates of $\frac{1}{ε} (r + b_{1}) - \frac{1}{ε} av (r + b_{1})$ on the finite domain, where $av f$ denotes the spatial average of f. When the domain is infinite, there are still some problems to identify the limit equations.
Remark 2.7.
As can be seen from Theorem 2.4, the boundary condition we used here is given by (1.4). The case when the alternative boundary condition (1.5) holds is left for future study.

We now briefly describe the strategy of the proof of our main results. The uniform estimates in Theorems 2.2 and 2.3 are obtained by making use of energy estimate and the special structure of the equation. Roughly speaking, the proof of the uniform estimates (2.6) or (2.8) can be divided into two parts, namely, the tangential derivative estimates and normal derivative estimates. By direct energy estimate, we can get the tangential derivative estimates of the solution because all the singular terms and boundary terms vanish. However, for the normal derivative estimates, we can not use similar method because differentiation with respect to the normal variable destroys the boundary condition. To establish the normal derivative estimates of the solution, one has to represent the normal derivatives of the solution by some combinations of the tangential derivatives of the solution. And this is why we need to adopt weighted norms like (2.1)–(2.4), which is very different from previous works [4,10] where the underlying physical space is torus. However, since we are dealing with the characteristic boundary conditions with constant multiplicity (rank of the boundary matrix is the constant value 6), it is impossible to do this for all the components of the solution. Fortunately, if we separate the solution into “noncharacteristic” part and the “characteristic” part, the characteristic part has only one element $b_{1}$ , while the normal derivative of $b_{1}$ can be obtained by using the divergence-free condition of the magnetic field b. Thus, all the normal derivative estimates of the solution can be achieved through tangential derivative estimates.

The proof of the convergence part (Theorem 2.4) is based on the compactness argument. Since the uniform estimates established in Theorem 2.3 are not enough to obtain the convergence of the singular terms in equation (1.2), we need to prove Proposition 4.2 which concerns the uniform estimates of the singular terms. Moreover, the boundary conditions must be determined for the singular terms, and this has been done in Proposition 4.3. Based on the uniform estimates in Theorem 2.3 and Proposition 4.2, we can take limit in system (1.2) to obtain equation (4.11) below, where $\bar{K}, \partial_{1} \bar{L}$ and $\partial_{1} \bar{N}$ are from the singular terms $\frac{1}{ε} \nabla \cdot u, \frac{1}{ε} \partial_{1} b$ and $\frac{1}{ε} \partial_{1} u$ , respectively. We complete the proof of Theorem 2.4 by showing that $\partial_{1} \bar{L} = \partial_{1} \bar{N} = \bar{K} = 0$ through the following strategies. After showing a proper form of (4.11) holds in the classical sense, we prove that the limit function $(\bar{r}, \bar{u}, \bar{b})$ is actually independent of $x_{1}$ , and this is achieved by giving a full description of the kernel of the large operator $L$ defined by (4.12). Thus, taking $\partial_{1}$ of (4.30), we obtain that $\partial_{11} \bar{L} = \partial_{11} \bar{N} = 0$ . Then using the boundary conditions that we have determined in Proposition 4.3, we conclude that $\partial_{1} \bar{L} = \partial_{1} \bar{N} = 0$ . Finally, we obtain $\bar{K} = 0$ by showing that $\bar{r} + {\bar{b}}_{1}$ is always a constant.
3. Uniform existence theorems

We first give the proof of Theorem 2.2. The key part is to obtain the uniform estimates in ε of the solution of the initial boundary value problem (1.2), (1.3) and (1.4) (or (1.2), (1.3) and (1.5)).

In order to show that (2.6) holds, following the strategy of [11], it suffices to find norms $| | | \cdot | | |_{E_{1}}$ and $| | | \cdot | | |_{E_{2}}$ satisfying $\begin{matrix} (3.1) & k_{1} (| | | \cdot | | |_{E_{1}} + | | | \cdot | | |_{E_{2}}) ⩽ | | | \cdot | | |_{3, ε} ⩽ k_{2} (| | | \cdot | | |_{E_{1}} + | | | \cdot | | |_{E_{2}}), \end{matrix}$ for which we can show that $\begin{matrix} (3.2) & \frac{d}{d t} | | | \cdot | | |_{E_{1}}^{2} ⩽ F_{1} (| | | \cdot | | |_{3, ε}) \end{matrix}$ and $\begin{matrix} (3.3) & | | | \cdot | | |_{E_{2}} ⩽ F_{2} (| | | \cdot | | |_{E_{1}}) + ε F_{3} (| | | \cdot | | |_{3, ε}), \end{matrix}$ where $k_{1}$ and $k_{2}$ are positive constants, and $F_{i}$ are continuous nondecreasing functions on $(0, \infty)$ throughout this paper. For ε sufficiently small, (3.3) can be solved for $| | | \cdot | | |_{E_{2}}$ in terms of $| | | \cdot | | |_{E_{1}}$ . Thus (3.2) yields a differential inequality for $| | | \cdot | | |_{E_{1}}$ that shows it to be bounded for some time T independent of ε, and then (3.1) and (3.3) show that $| | | \cdot | | |_{3, ε}$ is bounded independent of ε, since $| | | \cdot | | |_{3, ε}$ is bounded initially.

Here, we set $V = (r, u, b)$ and define $\begin{array}{l} (3.4) & | | | V | | |_{E_{1}} = \sum_{α_{0} + α_{2} + α_{3} ⩽ 3} ε^{α_{0}} ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} V ‖, \\ (3.5) & | | | V | | |_{E_{2}} = \sum_{| α | ⩽ 3, α_{1} ⩾ 1} ε^{α_{0}} ‖ \partial_{t}^{α_{0}} \partial_{1}^{α_{1}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} V ‖ . \end{array}$ Obviously, (3.1) holds true for the above norms. Hence, the rest part of this section is devoted to deriving estimates (3.2) and (3.3).

Lemma 3.1.
Under the assumptions of Theorem 2.2 , we have $\begin{matrix} (3.6) & \frac{d}{d t} \sum_{α_{2} + α_{3} ⩽ 3} {‖ \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} V ‖}^{2} ⩽ F_{4} (| | | V | | |_{3, ε}) . \end{matrix}$
Proof.
For $α_{2} + α_{3} ⩽ 3$ , taking $\partial_{2}^{α_{2}} \partial_{3}^{α_{3}}$ of (1.2) gives $\begin{matrix} (3.7) & \{\begin{matrix} a (\partial_{t} + u \cdot \nabla) \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} r + \frac{1}{ε} \nabla \cdot \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} u = C_{r}, \\ ρ (\partial_{t} + u \cdot \nabla) \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} u + \frac{1}{ε} \nabla \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} r + b \times (\nabla \times \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} b) \\ = \frac{1}{ε} (\partial_{1} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} b - \nabla \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} b_{1}) + C_{u}, \\ (\partial_{t} + u \cdot \nabla) \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} b - b \cdot \nabla \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} u + b \nabla \cdot \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} u \\ = \frac{1}{ε} (\partial_{1} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} u - e_{1} \nabla \cdot \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} u) + C_{b}, \\ \nabla \cdot \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} b = 0 . \end{matrix} \end{matrix}$ Here, the commutators $(C_{r}, C_{u}, C_{b})$ are given by $\begin{array}{l} C_{r} = - [\partial_{2}^{α_{2}} \partial_{3}^{α_{3}}, a (\partial_{t} + u \cdot \nabla)] r, \\ C_{u} = - [\partial_{2}^{α_{2}} \partial_{3}^{α_{3}}, ρ (\partial_{t} + u \cdot \nabla)] u - [\partial_{2}^{α_{2}} \partial_{3}^{α_{3}}, b \times] (\nabla \times b), \\ C_{b} = - [\partial_{2}^{α_{2}} \partial_{3}^{α_{3}}, u \cdot \nabla] b + [\partial_{2}^{α_{2}} \partial_{3}^{α_{3}}, b \cdot \nabla] u - [\partial_{2}^{α_{2}} \partial_{3}^{α_{3}}, b \nabla \cdot] u . \end{array}$ Multiplying (3.7) by $\partial_{2}^{α_{2}} \partial_{3}^{α_{3}} V$ , integrating over Ω, and integrating by parts give $\begin{array}{l} \frac{1}{2} \frac{d}{d t} \int_{Ω} (a {| \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} r |}^{2} + ρ {| \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} u |}^{2} + {| \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} b |}^{2}) - \frac{1}{2} \int_{Ω} (\partial_{t} a {| \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} r |}^{2} + \partial_{t} ρ {| \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} u |}^{2}) \\ - \frac{1}{2} \int_{Ω} (\nabla \cdot (a u) {| \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} r |}^{2} + \nabla \cdot (ρ u) {| \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} u |}^{2}) - \frac{1}{2} \int_{Ω} \nabla \cdot u {| \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} b |}^{2} \\ + \int_{Ω} b \times (\nabla \times \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} b) \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} u - \int_{Ω} b \cdot \nabla \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} u \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} b + \int_{Ω} b \nabla \cdot \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} u \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} b \\ (3.8) & ⩽ \int_{Ω} (C_{r} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} r + C_{u} \cdot \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} u + C_{b} \cdot \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} b), \end{array}$ because the singular terms and boundary terms vanish by the boundary condition (1.4) or (1.5). By using Lemma A.1 in the Appendix, we get from (3.8) that $\begin{array}{l} \frac{d}{d t} \int_{Ω} (a {| \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} r |}^{2} + ρ {| \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} u |}^{2} + {| \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} b |}^{2}) \\ ⩽ C {‖ (\partial_{t} a, \partial_{t} ρ, \nabla \cdot (a u), \nabla \cdot (ρ u), \nabla \cdot u, \nabla b) ‖}_{L^{\infty}} {‖ \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} V ‖}^{2} \\ (3.9) & + \int_{Ω} (C_{r} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} r + C_{u} \cdot \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} u + C_{b} \cdot \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} b) . \end{array}$ Using the Moser-type calculus inequalities of Lemma A.2 in the Appendix, we get the estimates of the commutators $\begin{array}{l} ‖ C_{r} ‖ & ⩽ ‖ [\partial_{2}^{α_{2}} \partial_{3}^{α_{3}}, a \partial_{t}] r ‖ + ‖ [\partial_{2}^{α_{2}} \partial_{3}^{α_{3}}, a u \cdot \nabla] r ‖ \\ (3.10) & ⩽ ‖ \nabla a ‖_{L^{\infty}} ‖ \partial_{t} r ‖_{2} + ‖ \partial_{t} r ‖_{L^{\infty}} ‖ \nabla a ‖_{2} + ‖ a u ‖_{L^{\infty}} ‖ r ‖_{3} + ‖ \nabla r ‖_{L^{\infty}} ‖ a u ‖_{3} . \end{array}$ Since a is a smooth function of $ε r$ , we get from the Sobolev inequality of Lemma A.3 and Lemma A.4 in the Appendix that $\begin{array}{l} (3.11) & \begin{matrix} ‖ \nabla a ‖_{L^{\infty}} ⩽ ε f_{1} (‖ r ‖_{W^{1, \infty}}) ⩽ ε f_{1} (‖ r ‖_{3}), \\ ‖ \nabla a ‖_{2} ⩽ ε f_{2} (‖ r ‖_{3}), \\ ‖ a u ‖_{L^{\infty}} ⩽ ‖ a ‖_{L^{\infty}} ‖ u ‖_{L^{\infty}} ⩽ f_{3} ({‖ (r, u) ‖}_{3}), \\ ‖ a u ‖_{3} ⩽ ‖ a ‖_{L^{\infty}} ‖ u ‖_{3} + ‖ a ‖_{3} ‖ u ‖_{L^{\infty}} ⩽ f_{4} ({‖ (r, u) ‖}_{3}) . \end{matrix} \end{array}$ Here and in the rest part, $f_{i} (\cdot)$ are also continuous nondecreasing functions on $(0, \infty)$ . Thus, combining the estimates (3.10) and (3.11), one has $\begin{array}{l} ‖ C_{r} ‖ & ⩽ f_{1} (‖ r ‖_{3}) ‖ ε \partial_{t} r ‖_{2} + ‖ ε \partial_{t} r ‖_{2} f_{2} (‖ r ‖_{3}) + f_{3} ({‖ (r, u) ‖}_{3}) ‖ r ‖_{3} + f_{4} ({‖ (r, u) ‖}_{3}) ‖ \nabla r ‖_{2} \\ (3.12) & ⩽ f_{5} (‖ V ‖_{3, ε}) . \end{array}$ The estimates of $C_{u}$ and $C_{b}$ can be given in a similar fashion, so we have $\begin{matrix} (3.13) & ‖ C_{u} ‖ + ‖ C_{b} ‖ ⩽ f_{6} (| | | V | | |_{3, ε}) . \end{matrix}$ Since a and ρ are smooth functions of $ε r$ , using the Sobolev inequality, $\begin{array}{l} {‖ (\partial_{t} a, \partial_{t} ρ) ‖}_{L^{\infty}} ⩽ ε f_{7} ({‖ (r, \partial_{t} r) ‖}_{L^{\infty}}) ⩽ f_{7} (| | | V | | |_{3, ε}), \\ {‖ (\nabla \cdot (a u), \nabla \cdot (ρ u), \nabla \cdot u, \nabla b) ‖}_{L^{\infty}} ⩽ f_{8} (| | | V | | |_{3, ε}) . \end{array}$ As a result, the first term on the right-hand side of (3.9) can be bounded by $\begin{matrix} (3.14) & {‖ (\partial_{t} a, \partial_{t} ρ, \nabla \cdot (a u), \nabla \cdot (ρ u), \nabla \cdot u, \nabla b) ‖}_{L^{\infty}} {‖ \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} V ‖}^{2} ⩽ f_{9} (| | | V | | |_{3, ε}) . \end{matrix}$ Substituting (3.12), (3.13) and (3.14) into (3.9) gives $\begin{matrix} \frac{d}{d t} {‖ \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} V ‖}^{2} ⩽ f_{10} (| | | V | | |_{3, ε}) . \end{matrix}$ Thus, taking summation with respect to $α_{2}, α_{3}$ yields the estimates of Lemma 3.1. The proof is completed. □

We proceed to derive the estimates of the tangential derivatives of the solution V of (1.2).
Lemma 3.2.
Under the assumptions of Theorem 2.2 , we have $\begin{matrix} (3.15) & \frac{d}{d t} \sum_{α_{0} + α_{2} + α_{3} ⩽ 3, α_{0} ⩾ 1} ε^{2 α_{0}} {‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} V ‖}^{2} ⩽ F_{5} (| | | V | | |_{3, ε}) . \end{matrix}$
Proof.
For $α_{0} ⩾ 1$ , $α_{0} + α_{2} + α_{3} ⩽ 3$ , taking $\partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}}$ of (1.2), multiplying by $ε^{2 α_{0}} \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} V$ , integrating over Ω, and integrating by parts give $\begin{array}{l} ε^{2 α_{0}} \frac{d}{d t} \int_{Ω} (a {| \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} r |}^{2} + ρ {| \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} u |}^{2} + {| \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} b |}^{2}) \\ ⩽ {‖ (\partial_{t} a, \partial_{t} ρ, \nabla \cdot (a u), \nabla \cdot (ρ u), \nabla \cdot u, \nabla b) ‖}_{L^{\infty}} {‖ ε^{α_{0}} \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} V ‖}^{2} \\ + ε^{2 α_{0}} \int_{Ω} ({\tilde{C}}_{r} \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} r + {\tilde{C}}_{u} \cdot \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} u + {\tilde{C}}_{b} \cdot \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} b) \\ (3.16) & : = R_{1} + R_{2}, \end{array}$ where, still, the large terms and boundary terms vanish. The commutators here are given by $\begin{array}{l} {\tilde{C}}_{r} = - [\partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}}, a] \partial_{t} r - [\partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}}, a u \cdot \nabla] r, \\ {\tilde{C}}_{u} = - [\partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}}, ρ] \partial_{t} u - [\partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}}, ρ u \cdot \nabla] u - [\partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}}, b \times] (\nabla \times b), \\ {\tilde{C}}_{b} = - [\partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}}, u \cdot \nabla] b + [\partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}}, b \cdot \nabla] u - [\partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}}, b \nabla \cdot] u . \end{array}$ Using an estimate similar to (3.14), $R_{1}$ can be easily bounded by $f_{11} (| | | V | | |_{3, ε})$ . Now, we need to give the estimate of $R_{2}$ which is the most involved because we can not use the commutator estimate in Lemma A.2 directly. It is obvious that $\begin{matrix} (3.17) & R_{2} ⩽ ‖ ε^{α_{0}} {\tilde{C}}_{r} ‖ ‖ ε^{α_{0}} \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} r ‖ + ‖ ε^{α_{0}} {\tilde{C}}_{u} ‖ ‖ ε^{α_{0}} \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} u ‖ + ‖ ε^{α_{0}} {\tilde{C}}_{b} ‖ ‖ ε^{α_{0}} \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} b ‖ . \end{matrix}$ Let us consider the estimate of the first term on the right-hand side of (3.17). From the expression of ${\tilde{C}}_{r}$ and the Leibniz formula, for $α = (α_{0}, α_{2}, α_{3})$ and $β = (β_{0}, β_{2}, β_{3})$ , we have $\begin{array}{l} ‖ ε^{α_{0}} {\tilde{C}}_{r} ‖ & ⩽ C \sum_{β ⩽ α, | β | ⩾ 1} ‖ ε^{β_{0}} \partial_{t}^{β_{0}} \partial_{2}^{β_{2}} \partial_{3}^{β_{3}} a \cdot ε^{α_{0} - β_{0}} \partial_{t}^{α_{0} - β_{0} + 1} \partial_{2}^{α_{2} - β_{2}} \partial_{3}^{α_{3} - β_{3}} r ‖ \\ + C \sum_{β ⩽ α, | β | ⩾ 1} ‖ ε^{β_{0}} \partial_{t}^{β_{0}} \partial_{2}^{β_{2}} \partial_{3}^{β_{3}} (a u) \cdot ε^{α_{0} - β_{0}} \partial_{t}^{α_{0} - β_{0}} \partial_{2}^{α_{2} - β_{2}} \partial_{3}^{α_{3} - β_{3}} \nabla r ‖ \\ (3.18) & : = J_{1} + J_{2} . \end{array}$ We give the estimate of $J_{1}$ in detail now. Recall that $| α | = α_{0} + α_{2} + α_{3}$ and $α_{0} ⩾ 1$ , we need to consider the following three cases.

Case 1: $| α | = | α_{0} | = 1$ . In this case, we have $α_{2} = α_{3} = β_{2} = β_{3} = 0$ and $β_{0} = 1$ . As a result, we get that $\begin{matrix} | J_{1} | ⩽ C ‖ ε \partial_{t} a \cdot \partial_{t} r ‖ ⩽ C ‖ \partial_{t} a ‖_{L^{\infty}} ‖ ε \partial_{t} r ‖ ⩽ f_{12} (‖ r ‖_{3, ε}), \end{matrix}$ where we have used the fact the a is a smooth function of $ε r$ .

Case 2: $| α | = 2$ . In this case, $α_{0}$ can be one or two. If $α_{0} = 1$ , then we get $α_{2} = 1$ or $α_{3} = 1$ . We assume $α_{2} = 1$ without loss of generality. From (3.18), we have $\begin{array}{l} J_{1} & ⩽ C \sum_{β_{0}, β_{2} ⩽ 1, | β | ⩾ 1} ‖ ε^{β_{0}} \partial_{t}^{β_{0}} \partial_{2}^{β_{2}} a \cdot ε^{1 - β_{0}} \partial_{t}^{2 - β_{0}} \partial_{2}^{1 - β_{2}} r ‖ \\ = C (‖ \partial_{2} a \cdot ε \partial_{t}^{2} r ‖ + ‖ ε \partial_{t} a \cdot \partial_{t} \partial_{2} r ‖ + ‖ ε \partial_{t} \partial_{2} a \cdot \partial_{t} r ‖) \\ ⩽ C ({‖ \partial a (\cdot) ‖}_{L^{\infty}} ‖ ε^{2} \partial_{t}^{2} r ‖ + ‖ \partial_{t} a ‖_{L^{\infty}} ‖ ε \partial_{t} \partial_{2} r ‖ + ‖ ε \partial_{t} r ‖_{L^{\infty}} ‖ \partial_{t} \partial_{2} a ‖) ⩽ f_{13} (‖ r ‖_{3, ε}) . \end{array}$ Next, if $α_{0} = 2$ , then $α_{2} = α_{3} = β_{2} = β_{3} = 0$ . We have that $\begin{array}{l} J_{1} & ⩽ C \sum_{1 ⩽ β_{0} ⩽ 2} ‖ ε^{β_{0}} \partial_{t}^{β_{0}} a \cdot ε^{2 - β_{0}} \partial_{t}^{3 - β_{0}} r ‖ = C (‖ \partial_{t} a \cdot ε^{2} \partial_{t}^{2} r ‖ + ‖ ε^{2} \partial_{t}^{2} a \cdot \partial_{t} r ‖) \\ ⩽ C (‖ \partial_{t} a ‖_{L^{\infty}} ‖ ε^{2} \partial_{t}^{2} r ‖ + ‖ ε \partial_{t}^{2} a ‖ ‖ ε \partial_{t} r ‖_{L^{\infty}}) ⩽ f_{14} (‖ r ‖_{3, ε}) . \end{array}$

Case 3: $| α | = 3$ . There are three subcases here. First, when $α_{0} = 1$ , we have $α_{2} + α_{3} = 2$ . Without loss of generality, we assume that $α_{2} = 2, α_{3} = 0$ . From (3.18), we have $\begin{array}{l} J_{1} & ⩽ C \sum_{β_{0} ⩽ 1, β_{2} ⩽ 2, | β | ⩾ 1} ‖ ε^{β_{0}} \partial_{t}^{β_{0}} \partial_{2}^{β_{2}} a \cdot ε^{1 - β_{0}} \partial_{t}^{2 - β_{0}} \partial_{2}^{2 - β_{2}} r ‖ \\ (3.19) & = C \sum_{1 ⩽ β_{2} ⩽ 2} ‖ \partial_{2}^{β_{2}} a \cdot ε \partial_{t}^{2} \partial_{2}^{2 - β_{2}} r ‖ + C \sum_{β_{2} ⩽ 2} ‖ \partial_{t} \partial_{2}^{β_{2}} a \cdot ε \partial_{t} \partial_{2}^{2 - β_{2}} r ‖ . \end{array}$ The difficulty lies in the estimate of the first term when $β_{2} = 2$ on the right-hand side of (3.19) because we can not bound $‖ \partial_{2}^{2} r ‖_{L^{\infty}}$ or $‖ \partial_{t}^{2} r ‖_{L^{\infty}}$ . The idea is to use the original equation (1.2) to express $\partial_{t}^{2} r$ in terms of expression having at most one time derivative. For $β_{2} = 2$ in the first term, we get from the first equation of (1.2) that $\begin{array}{l} ‖ \partial_{2}^{2} a \cdot ε \partial_{t}^{2} r ‖ & = ‖ \partial_{2}^{2} a \cdot ε \partial_{t} (u \cdot \nabla r + \frac{1}{ε a} \nabla \cdot u) ‖ \\ ⩽ ‖ \partial_{2}^{2} a \cdot ε \partial_{t} (u \cdot \nabla r) ‖ + ‖ \partial_{2}^{2} a \cdot \partial_{t} (\frac{1}{a} \nabla \cdot u) ‖ \\ ⩽ ‖ \partial_{2}^{2} a \cdot ε \partial_{t} u \cdot \nabla r ‖ + ‖ \partial_{2}^{2} a \cdot u \cdot ε \partial_{t} \nabla r ‖ \\ + ‖ \frac{1}{a^{2}} \partial_{2}^{2} a \cdot (\partial_{t} a \nabla \cdot u) ‖ + ‖ \frac{1}{a} \partial_{2}^{2} a \cdot \partial_{t} \nabla \cdot u ‖ \\ ⩽ ‖ \partial_{2}^{2} a ‖ ‖ ε \partial_{t} u ‖_{L^{\infty}} ‖ \nabla r ‖_{L^{\infty}} + {‖ \partial_{2}^{2} a ‖}_{L^{4}} ‖ u ‖_{L^{\infty}} ‖ ε \partial_{t} \nabla r ‖_{L^{4}} \\ (3.20) & + {‖ \frac{1}{a^{2}} (\partial_{t} a \nabla \cdot u) ‖}_{L^{\infty}} ‖ \partial_{2}^{2} a ‖ + {‖ \frac{1}{a} ‖}_{L^{\infty}} {‖ \partial_{2}^{2} a ‖}_{L^{4}} ‖ \partial_{t} \nabla \cdot u ‖_{L^{4}} . \end{array}$ Using the Nirenberg inequality in Lemma A.5, we can control the $L^{4}$ norms of (3.20) as $\begin{array}{l} {‖ \partial_{2}^{2} a ‖}_{L^{4}} ⩽ C ‖ \partial_{2} a ‖_{L^{\infty}}^{\frac{1}{2}} {‖ \partial_{2}^{3} a ‖}_{L^{2}}^{\frac{1}{2}}, \\ ‖ \partial_{t} \nabla r ‖_{L^{4}} ⩽ C ‖ \partial_{t} a ‖_{L^{\infty}}^{\frac{1}{2}} {‖ \partial_{t} \nabla^{2} a ‖}_{L^{2}}^{\frac{1}{2}}, \\ ‖ \partial_{t} \nabla \cdot u ‖_{L^{4}} ⩽ C ‖ \partial_{t} u ‖_{L^{\infty}}^{\frac{1}{2}} {‖ \partial_{t} \nabla^{2} u ‖}_{L^{2}}^{\frac{1}{2}} . \end{array}$ Combining the estimates (3.20) and the above $L^{4}$ estimates gives $\begin{matrix} ‖ \partial_{2}^{2} a \cdot ε \partial_{t}^{2} r ‖ ⩽ f_{15} (‖ V ‖_{3, ε}) . \end{matrix}$ If $β_{2} = 1$ , the first term on the right-hand side of (3.19) can be easily bounded by $\begin{matrix} ‖ \partial_{2} a \cdot ε \partial_{t}^{2} \partial_{2} r ‖ ⩽ {‖ \partial a (\cdot) ‖}_{L^{\infty}} ‖ ε^{2} \partial_{t}^{2} \partial_{2} r ‖ . \end{matrix}$ Next, let us consider the second term on the right-hand side of (3.19). In fact, this term can be bounded directly from its expression since it does not includes twice time derivatives like the first term. One has $\begin{array}{l} \sum_{β_{2} ⩽ 2} ‖ \partial_{t} \partial_{2}^{β_{2}} a \cdot ε \partial_{t} \partial_{2}^{2 - β_{2}} r ‖ & = ‖ \partial_{t} a \cdot ε \partial_{t} \partial_{2}^{2} r ‖ + ‖ \partial_{t} \partial_{2} a \cdot ε \partial_{t} \partial_{2} r ‖ + ‖ \partial_{t} \partial_{2}^{2} a \cdot ε \partial_{t} r ‖ \\ ⩽ ‖ \partial_{t} a ‖_{L^{\infty}} ‖ ε \partial_{t} \partial_{2}^{2} r ‖ + ‖ \partial_{t} \partial_{2} a ‖_{L^{4}} ‖ ε \partial_{t} \partial_{2} r ‖_{L^{4}} + ‖ \partial_{t} \partial_{2}^{2} a ‖ ‖ ε \partial_{t} r ‖_{L^{\infty}} \\ ⩽ f_{16} (‖ V ‖_{3, ε}) . \end{array}$

Second, when $α_{0} = 2$ , we have $α_{2} + α_{3} = 1$ . Similarly, we can assume that $α_{2} = 1$ , $α_{3} = 0$ without loss of generality. From (3.18), we have $\begin{array}{l} J_{1} & ⩽ C \sum_{β_{0} ⩽ 2, β_{2} ⩽ 1, | β | ⩾ 1} ‖ ε^{β_{0}} \partial_{t}^{β_{0}} \partial_{2}^{β_{2}} a \cdot ε^{2 - β_{0}} \partial_{t}^{2 - β_{0} + 1} \partial_{2}^{1 - β_{2}} r ‖ \\ (3.21) & = C ‖ \partial_{2} a \cdot ε^{2} \partial_{t}^{3} r ‖ + C \sum_{β_{2} ⩽ 1} ‖ ε \partial_{t} \partial_{2}^{β_{2}} a \cdot ε \partial_{t}^{2} \partial_{2}^{1 - β_{2}} r ‖ + C \sum_{β_{2} ⩽ 1} ‖ ε^{2} \partial_{t}^{2} \partial_{2}^{β_{2}} a \cdot \partial_{t} \partial_{2}^{1 - β_{2}} r ‖ . \end{array}$ Obviously, the first term on the right-hand side of (3.21) can be bounded by $\begin{matrix} ‖ \partial_{2} a \cdot ε^{2} \partial_{t}^{3} r ‖ ⩽ ‖ \partial_{2} a ‖_{L^{\infty}} ‖ ε^{2} \partial_{t}^{3} r ‖ ⩽ {‖ \partial a (\cdot) ‖}_{L^{\infty}} ‖ ε^{3} \partial_{t}^{3} r ‖ . \end{matrix}$ Let us turn to the second term. When $β_{2} = 0$ , we have $\begin{matrix} ‖ ε \partial_{t} a \cdot ε \partial_{t}^{2} \partial_{2} r ‖ ⩽ ‖ \partial_{t} a ‖_{L^{\infty}} ‖ ε^{2} \partial_{t}^{2} \partial_{2} r ‖ . \end{matrix}$ While when $β_{2} = 1$ , we get from equation (1.2) that $\begin{array}{l} ‖ ε \partial_{t} \partial_{2} a \cdot ε \partial_{t}^{2} r ‖ & = ‖ ε \partial_{t} \partial_{2} a \cdot ε \partial_{t} (u \cdot \nabla r) ‖ + ‖ ε \partial_{t} \partial_{2} a \cdot \partial_{t} (\frac{1}{a} \nabla \cdot u) ‖ \\ ⩽ ‖ ε \partial_{t} \partial_{2} a \cdot ε \partial_{t} u \cdot \nabla r ‖ + ‖ ε \partial_{t} \partial_{2} a \cdot u \cdot ε \partial_{t} \nabla r ‖ \\ (3.22) & + ‖ ε \partial_{t} \partial_{2} a \cdot \partial_{t} (\frac{1}{a}) \nabla \cdot u ‖ + ‖ \partial_{t} \partial_{2} a \cdot (\frac{1}{a}) ε \partial_{t} \nabla \cdot u ‖ . \end{array}$ By the method we have used in (3.20), we can control all the right-hand side terms of (3.22). The estimate for the third term on the right-hand side of (3.21) is very similar to the second one, so we omit it for simplicity.

Finally, we shall discuss the case when $α_{0} = 3$ . Notice that now we have $α_{2} = α_{3} = 0$ . Thus, we get from (3.18) that $\begin{matrix} J_{1} ⩽ C \sum_{1 ⩽ β_{0} ⩽ 3} ‖ ε^{β_{0}} \partial_{t}^{β_{0}} a \cdot ε^{3 - β_{0}} \partial_{t}^{4 - β_{0}} r ‖ . \end{matrix}$ When $β_{0} = 1$ or $β_{0} = 3$ , one can easily get $\begin{matrix} ‖ \partial_{t} a \cdot ε^{3} \partial_{t}^{3} r ‖ + ‖ ε^{2} \partial_{t}^{3} a \cdot ε \partial_{t} r ‖ ⩽ ‖ \partial_{t} a ‖_{L^{\infty}} ‖ ε^{3} \partial_{t}^{3} r ‖ + ‖ ε^{2} \partial_{t}^{3} a ‖ ‖ ε \partial_{t} r ‖_{L^{\infty}} . \end{matrix}$ When $β_{0} = 2$ , we need to turn to the original equation (1.2) again to reformulate $\partial_{t}^{2} a \cdot \partial_{t}^{2} r$ into $\begin{array}{l} \partial_{t}^{2} a \cdot \partial_{t}^{2} r & = ε \partial_{t} (\partial a \cdot \partial_{t} r) \cdot \partial_{t} \partial_{t} r \\ (3.23) & = ε \partial_{t} (\partial a \cdot (u \cdot \nabla r + \frac{1}{ε a} \nabla \cdot u)) \cdot \partial_{t} (u \cdot \nabla r + \frac{1}{ε a} \nabla \cdot u) . \end{array}$ Notice that all the right-hand side terms of (3.23) possess at most one time derivative, so we can use similar method as (3.20) and (3.22) to control $ε^{3} ‖ \partial_{t}^{2} a \cdot \partial_{t}^{2} r ‖$ because we have necessary powers of ε.

Collecting all the discussions on the above three cases, we can see that $J_{1}$ can be bound by $f_{17} (‖ V ‖_{3, ε})$ . The second term $J_{2}$ on the right-hand side of (3.18) can be estimate in a similar fashion as the first term we just discussed. Namely, following the strategy above, we have $\begin{matrix} J_{2} ⩽ f_{18} (‖ V ‖_{3, ε}) . \end{matrix}$ Moreover, the estimates of the other two terms ${\tilde{C}}_{u}$ and ${\tilde{C}}_{b}$ in $R_{2}$ are similar to that of ${\tilde{C}}_{r}$ , we omit them for the sake of simplicity. Putting the commutator estimates into (3.17) yields the estimate of $R_{2}$ , $\begin{matrix} | R_{2} | ⩽ f_{19} (| | | V | | |_{3, ε}) . \end{matrix}$ Combining (3.16) and the estimate for $R_{1}$ , $R_{2}$ gives the estimate (3.15). Thus, we complete the proof of Lemma 3.2. □

Next, we are going to derive the estimate of the normal derivatives of the solution. Note that this can not be achieved by using energy estimate like the tangential derivative estimates in Lemma 3.1 and Lemma 3.2, since differentiation with respect to $x_{1}$ destroys the boundary condition, boundary terms that cannot be controlled appear from integration by parts. The main idea here is to represent the normal derivatives of the solution in terms of the tangential derivatives by making full use of the structure of the equation. More specifically, since the rank of the boundary matrix is the constant value 6, we can represent the normal derivatives of $(r, u_{1}, u_{2}, u_{3}, b_{2}, b_{3})$ (non-characteristic part) by the combinations of the tangential derivatives of the solution. Thus, it remains to handle the normal derivative of $b_{1}$ (characteristic part), but this is not a problem because we have the divergence-free condition on the magnetic field b.
Lemma 3.3.
Under the assumptions of Theorem 2.2 , we have $\begin{matrix} (3.24) & \sum_{| α | ⩽ 3, α_{1} ⩾ 1} ε^{α_{0}} ‖ \partial_{t}^{α_{0}} \partial_{1}^{α_{1}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} V ‖ ⩽ F_{6} (| | | V | | |_{E_{1}}) + ε F_{7} (| | | V | | |_{3, ε}) . \end{matrix}$
Proof.
First, for $α_{1} = 1$ , $α_{0} + α_{2} + α_{3} ⩽ 2$ , we get from the divergence-free condition of b that $\begin{matrix} ε^{α_{0}} \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} \partial_{1} b_{1} = - ε^{α_{0}} \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} (\partial_{2} b_{2} + \partial_{3} b_{3}) . \end{matrix}$ Thus, we have $\begin{array}{l} ε^{α_{0}} ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} \partial_{1} b_{1} ‖ & ⩽ ε^{α_{0}} ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} (\partial_{2} b_{2} + \partial_{3} b_{3}) ‖ \\ (3.25) & ⩽ ε^{α_{0}} (‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2} + 1} \partial_{3}^{α_{3}} b_{2} ‖ + ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3} + 1} b_{3} ‖) ⩽ C | | | V | | |_{E_{1}} . \end{array}$ Next, we get from the system (1.2) that $\begin{matrix} \partial_{1} r = - ε ρ (\partial_{t} + u \cdot \nabla) u_{1} + ε (b \cdot \nabla b_{1} - \frac{1}{2} \partial_{1} | b |^{2}) . \end{matrix}$ As a result, one has $\begin{array}{l} ε^{α_{0}} ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} \partial_{1} r ‖ & ⩽ ε ε^{α_{0}} ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} (ρ (\partial_{t} + u \cdot \nabla) u_{1}) ‖ \\ (3.26) & + ε ε^{α_{0}} ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} (b_{2} \partial_{2} b_{1} + b_{3} \partial_{3} b_{1} - b_{2} \partial_{1} b_{2} - b_{3} \partial_{1} b_{3}) ‖ . \end{array}$ Using product estimate in Lemma A.2, we can see that $\begin{matrix} (3.27) & \begin{matrix} ε^{α_{0} + 1} ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} (ρ \partial_{t} u_{1}) ‖ ⩽ f_{20} (| | | V | | |_{E_{1}}), \\ ε^{α_{0} + 1} ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} (ρ u \cdot \nabla) u_{1}) ‖ ⩽ ε f_{21} (| | | V | | |_{3, ε}), \\ ε^{α_{0} + 1} ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} (b_{2} \partial_{2} b_{1} + b_{3} \partial_{3} b_{1} - b_{2} \partial_{1} b_{2} - b_{3} \partial_{1} b_{3}) ‖ ⩽ ε f_{22} (| | | V | | |_{3, ε}) . \end{matrix} \end{matrix}$ Combining estimates (3.26) and (3.27) yields $\begin{matrix} (3.28) & ε^{α_{0}} ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} \partial_{1} r ‖ ⩽ f_{23} (| | | V | | |_{3, ε}) + ε f_{24} (| | | V | | |_{3, ε}) . \end{matrix}$ From the first equation of (1.2), we have $\begin{matrix} \partial_{1} u_{1} = - ε a (\partial_{t} r + u \cdot \nabla r) - (\partial_{2} u_{2} + \partial_{3} u_{3}) . \end{matrix}$ By a similar calculation as (3.27), we obtain $\begin{array}{l} ε^{α_{0}} ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} \partial_{1} u_{1} ‖ & ⩽ ε ε^{α_{0}} ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} (a (\partial_{t} + u \cdot \nabla) r) ‖ + ε^{α_{0}} ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} (\partial_{2} u_{2} + \partial_{3} u_{3}) ‖ \\ (3.29) & ⩽ f_{25} (| | | V | | |_{E_{1}}) + ε f_{26} (| | | V | | |_{3, ε}) . \end{array}$ From the equation satisfied by $b_{2}$ in (1.2), we get $\begin{matrix} \partial_{1} u_{2} = ε (\partial_{t} + u \cdot \nabla) b_{2} + ε (b_{2} \partial_{1} u_{1} + b_{2} \partial_{3} u_{3} - b_{1} \partial_{1} u_{2} - b_{3} \partial_{3} u_{2}) . \end{matrix}$ Thus, a direct calculation gives $\begin{matrix} (3.30) & ε^{α_{0}} ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} \partial_{1} u_{2} ‖ ⩽ f_{27} (| | | V | | |_{E_{1}}) + ε f_{28} (| | | V | | |_{3, ε}) . \end{matrix}$ The estimate of $\partial_{1} u_{3}$ is very similar to $\partial_{1} u_{2}$ , we omit it for the sake of simplicity. So it remains to estimate $\partial_{1} b_{2}$ and $\partial_{1} b_{3}$ . Since $\partial_{1} b_{2}$ satisfies $\begin{matrix} \partial_{1} b_{2} = ε ρ (\partial_{t} + u \cdot \nabla) u_{2} + (\partial_{2} r + \partial_{2} b_{1}) + ε (b_{2} \partial_{2} b_{1} + b_{3} \partial_{2} b_{3} - b_{1} \partial_{1} b_{2} - b_{3} \partial_{3} b_{2}), \end{matrix}$ we get $\begin{matrix} (3.31) & ε^{α_{0}} ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} \partial_{1} b_{2} ‖ ⩽ f_{29} (| | | V | | |_{E_{1}}) + ε f_{30} (| | | V | | |_{3, ε}) . \end{matrix}$ The estimate of $\partial_{1} b_{3}$ can be obtained in a similar fashion. Thus, we have proved that (3.24) holds when $α_{1} = 1$ , $α_{0} + α_{2} + α_{3} ⩽ 2$ .

Clearly, to complete the proof of (3.24), we need to handle the cases when $α_{1} = 2$ , $α_{0} + α_{2} + α_{3} ⩽ 1$ and $α_{1} = 3$ , $α_{0} + α_{2} + α_{3} = 0$ . This can be done by induction on $α_{1}$ . Now, assume that the estimate (3.24) holds when $α_{1} ⩽ k$ ( $k ⩾ 1$ ), we consider the case $α_{1} = k + 1$ .

Still, we start with the estimate of $\partial_{1}^{k + 1} b_{1}$ . Using the divergence-free condition of b, one has $\begin{matrix} ε^{α_{0}} \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} \partial_{1}^{k + 1} b_{1} = - ε^{α_{0}} \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} \partial_{1}^{k} (\partial_{2} b_{2} + \partial_{3} b_{3}) . \end{matrix}$ Taking $L^{2}$ norm yields $\begin{array}{l} ε^{α_{0}} ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} \partial_{1}^{k + 1} b_{1} ‖ & = ε^{α_{0}} ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} \partial_{1}^{k} (\partial_{2} b_{2} + \partial_{3} b_{3}) ‖ \\ ⩽ ε^{α_{0}} ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2} + 1} \partial_{3}^{α_{3}} \partial_{1}^{k} b_{2} ‖ + ε^{α_{0}} ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3} + 1} \partial_{1}^{k} b_{3} ‖ \\ (3.32) & ⩽ f_{27} (| | | V | | |_{E_{1}}) + ε f_{28} (| | | V | | |_{3, ε}), \end{array}$ where we have used the assumptions that (3.24) holds for $α_{1} = k$ . For the estimate of $\partial^{k + 1} r$ , we have $\begin{array}{l} ε^{α_{0}} ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} \partial_{1}^{k + 1} r ‖ & ⩽ ε^{α_{0}} ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} \partial_{1}^{k} (ρ (\partial_{t} + u \cdot \nabla) u_{1}) ‖ \\ + ε^{α_{0}} ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} \partial_{1}^{k} (b_{2} \partial_{2} b_{1} + b_{3} \partial_{3} b_{1} - b_{2} \partial_{1} b_{2} - b_{3} \partial_{1} b_{3}) ‖ \\ (3.33) & ⩽ f_{29} (| | | V | | |_{E_{1}}) + ε f_{30} (| | | V | | |_{3, ε}) . \end{array}$ Following a similar line as (3.26)–(3.33), we can get the estimates of other terms by using equation (1.2) to represent the normal derivatives of the solution by tangential derivatives. Namely, we can obtain the following estimates $\begin{matrix} ε^{α_{0}} ‖ \partial_{t}^{α_{0}} \partial_{2}^{α_{2}} \partial_{3}^{α_{3}} \partial_{1}^{k + 1} (u_{1}, u_{2}, u_{3}, b_{2}, b_{3}) ‖ ⩽ f_{31} (| | | V | | |_{E_{1}}) + ε f_{32} (| | | V | | |_{3, ε}) . \end{matrix}$ Finally, adding the above estimates for $α_{1} = 1, 2, 3$ gives the results of Lemma 3.3. The proof is completed. □

Combining the estimates of Lemma 3.1–Lemma 3.3, we can show that (2.6) holds for some $T_{*}$ and $Δ_{2}$ independent of ε. So Theorem 2.2 is proved. Theorem 2.3 can be proved in a similar fashion as Theorem 2.2 because it is concerned with the well-prepared initial data case. So we omit it for simplicity.
4. Convergence as $ε \to 0$

In this section, we prove the convergence of the solution obtained in Theorem 2.3 as $ε \to 0$ . First, we need the following Aubin–Lions compactness lemma.

Lemma 4.1 ([13]).

Let ${u^{ε} (t, x)}$ be the sequences of functions satisfying $\begin{matrix} u^{ε} \in C ([0, T]; H^{s} (Ω)) \cap C^{1} ([0, T]; H^{s - 1} (Ω)), \end{matrix}$ and there exists a constant M, independent of ε, such that $\begin{matrix} {‖ u^{ε} ‖}_{s} + {‖ \partial_{t} u^{ε} ‖}_{s - 1} ⩽ M . \end{matrix}$ Then, by passing to a subsequence, there exists a function $u^{0} \in C ([0, T] \times Ω)$ such that, as ε goes to zero, $u^{ε}$ converges to $u^{0}$ ${weak}^{*}$ in $L^{\infty} ([0, T]; H^{s} (Ω))$ and strongly in $C ([0, T]; H^{s - δ} (Ω))$ for any $δ > 0$ , and moreover, $\begin{matrix} u^{0} \in C_{w} ([0, T]; H^{s} (Ω)) \cap Lip ([0, T]; H^{s - 1} (Ω)), \end{matrix}$ $\partial_{t} u^{ε}$ converges to $\partial_{t} u^{0}$ ${weak}^{*}$ in $L^{\infty} ([0, T]; H^{s - 1} (Ω))$ .

In view of the uniform estimates obtained in Theorem 2.3 and the Aubin–Lions Lemma 4.1, we can find functions $\begin{matrix} (\bar{r}, \bar{u}, \bar{b}) \in C ([0, T^{*}]; H^{3} (Ω)) \cap Lip ([0, T^{*}]; H^{2} (Ω)) \end{matrix}$ such that, by passing to a subsequence, $\begin{matrix} (r, u, b) \to (\bar{r}, \bar{u}, \bar{b}), \end{matrix}$ ${weak}^{*}$ in $L^{\infty} ([0, T^{*}]; H^{3} (Ω))$ as $ε \to 0$ . Moreover, for any $δ > 0$ , the strong convergence results hold in $C ([0, T^{*}]; H^{3 - δ} (Ω))$ . Thus, we can show that $\begin{array}{l} (a \partial_{t} r, ρ \partial_{t} u, \partial_{t} b) \to (\bar{a} \partial_{t} \bar{r}, \bar{ρ} \partial_{t} \bar{u}, \partial_{t} \bar{b}), \\ (a u \cdot \nabla r, ρ u \cdot \nabla u, u \cdot \nabla b) \to (\bar{a} \bar{u} \cdot \nabla \bar{r}, \bar{ρ} \bar{u} \cdot \nabla \bar{u}, \bar{u} \cdot \nabla \bar{b}), \\ b \times (\nabla \times b) \to \bar{b} \times (\nabla \times \bar{b}), \\ (b \cdot \nabla u, b \nabla \cdot u) \to (\bar{b} \cdot \nabla \bar{u}, \bar{b} \nabla \cdot \bar{u}), \end{array}$ ${weak}^{*}$ in $L^{\infty} ([0, T^{*}]; H^{2} (Ω))$ . Here, $\overline{a} = a (1)$ , $\overline{ρ} = ρ (1)$ are constants.

Next, we study the convergence of the other terms in equation (1.2). From now on, we restrict our attention to the domain $Ω = Ω_{4}$ and boundary condition (1.4). Precisely, on the boundary $\begin{matrix} Γ = {x_{1} = 0, (x_{2}, x_{3}) \in T^{2}} \cup {x_{1} = h, (x_{2}, x_{3}) \in T^{2}}, \end{matrix}$ it holds that $\begin{matrix} (4.1) & u_{1} = 0, b_{2} = 0, b_{3} = 0 . \end{matrix}$ First, we have the following proposition.

Proposition 4.2.

Under the assumptions of Theorem 2.3 , we have the following estimates $\begin{array}{l} (4.2) & sup_{t \in [0, T^{*}]} \frac{1}{ε} ‖ \nabla \cdot u ‖_{2} ⩽ C, \\ (4.3) & sup_{t \in [0, T^{*}]} \frac{1}{ε} {‖ \nabla (r + b_{1}) ‖}_{2} ⩽ C, \end{array}$ and $\begin{matrix} (4.4) & sup_{t \in [0, T^{*}]} \frac{1}{ε} {‖ \partial_{1} (u, b) ‖}_{2} ⩽ C . \end{matrix}$

Proof.

Clearly, (4.2) is a direct consequence of (2.8) and the first equation of (1.2). Next, set $\bar{H} = {(L^{2} (Ω))}^{3}$ , then we have [9] $\begin{matrix} \bar{H} = H_{σ} \oplus H_{σ}^{⊥}, \end{matrix}$ where $\begin{array}{l} H_{σ} = {w \in {(L^{2} (Ω))}^{3} | \int_{Ω} w \cdot \nabla ϕ d x = 0, \forall ϕ \in H^{1}}, \\ H_{σ}^{⊥} = {f ∣ f = \nabla g, g \in H^{1}} . \end{array}$ Now, we can introduce the Leray operator $P$ as orthogonal projection from $\bar{H}$ to $H_{σ}$ , $\begin{matrix} P : \bar{H} \to H_{σ} . \end{matrix}$ Applying the Leray projection $P$ to the second equation of (1.2), we get $\begin{matrix} (4.5) & P [ρ (\partial_{t} u + u \cdot \nabla u) + b \times (\nabla \times b)] = \frac{1}{ε} P \partial_{1} b . \end{matrix}$ We claim that $\partial_{1} b \in H_{σ}$ , that is $P \partial_{1} b = \partial_{1} b$ . Actually, for any $ϕ \in H^{1}$ , we have $\begin{matrix} \int_{Ω} \partial_{1} b \cdot \nabla ϕ = \int_{Γ} \partial_{1} b \cdot n ϕ - \int_{Ω} \partial_{1} \nabla \cdot b ϕ = \int_{Γ} \partial_{1} b \cdot n ϕ . \end{matrix}$ Since $n = \pm (1, 0, 0)$ here, it holds on the boundary Γ that $\begin{matrix} \partial_{1} b \cdot n = \pm \partial_{1} b_{1} = \mp (\partial_{2} b_{2} + \partial_{3} b_{3}) = 0, \end{matrix}$ where the divergence-free condition of b and the boundary condition (4.1) are used. This implies $\begin{matrix} \int_{Ω} \partial_{1} b \cdot \nabla ϕ = 0, \end{matrix}$ which yields that $\partial_{1} b \in H_{σ}$ . Thus, from (4.5), we get $\begin{matrix} \frac{1}{ε} \partial_{1} b = P [ρ (\partial_{t} u + u \cdot \nabla u) + b \times (\nabla \times b)] . \end{matrix}$ Using the boundedness of the operator $P$ on $H^{s}$ , we get that $\begin{matrix} \frac{1}{ε} ‖ \partial_{1} b ‖_{2} ⩽ {‖ ρ (\partial_{t} u + u \cdot \nabla u) + b \times (\nabla \times b) ‖}_{2} ⩽ C . \end{matrix}$ As a result, we have $\begin{matrix} \frac{1}{ε} {‖ \nabla (r + b_{1}) ‖}_{2} ⩽ \frac{1}{ε} ‖ \partial_{1} b ‖_{2} + {‖ ρ (\partial_{t} u + u \cdot \nabla u) + b \times (\nabla \times b) ‖}_{2} ⩽ C . \end{matrix}$ It remains to estimate $\frac{1}{ε} ‖ \partial_{1} u ‖_{2}$ . From the last equation in (1.2), one has $\begin{matrix} (4.6) & \frac{1}{ε} {‖ (\partial_{1} u - e_{1} \nabla \cdot u) ‖}_{2} = ‖ \partial_{t} b + u \cdot \nabla b - b \cdot \nabla u + b \nabla \cdot u ‖_{2} ⩽ C . \end{matrix}$ Combining (4.2) and (4.6) gives the desired estimate of $\frac{1}{ε} ‖ \partial_{1} u ‖_{2}$ . This completes the proof of Proposition 4.2. □

From the boundary condition (4.1) and the equation (1.2), we have the following proposition.

Proposition 4.3.

On the boundary Γ, it holds that $\begin{matrix} \partial_{1} r = \partial_{1} b_{1} = 0, \partial_{1} u_{2} = \partial_{1} u_{3} = 0 . \end{matrix}$

Proof.

From the divergence-free condition of b and the boundary condition (4.1), it is easy to find that $\begin{matrix} \partial_{1} b_{1} = - (\partial_{2} b_{2} + \partial_{3} b_{3}) = 0, \end{matrix}$ holds on the boundary Γ. Using the equation satisfied by $u_{1}$ in (1.2), we have $\begin{matrix} ρ (\partial_{t} u_{1} + u \cdot \nabla u_{1}) + \frac{1}{ε} \partial_{1} r + b_{2} (\partial_{1} b_{2} - \partial_{2} b_{1}) - b_{3} (\partial_{3} b_{1} - \partial_{1} b_{3}) = 0 . \end{matrix}$ Setting $x_{1} = 0$ (or $x_{1} = h$ ) in the above equation yields $\begin{matrix} \partial_{1} r = 0 . \end{matrix}$ Similarly, using the equation satisfied by $b_{h}$ in (1.2), we get $\begin{matrix} \partial_{t} b_{h} + u \cdot \nabla b_{h} - b \cdot \nabla u_{h} + b_{h} \nabla \cdot u = \frac{1}{ε} \partial_{1} u_{h} . \end{matrix}$ Letting $x_{1} = 0$ (or $x_{1} = h$ ) in the above equation, one has $\begin{matrix} - b_{1} \partial_{1} u_{h} = \frac{1}{ε} \partial_{1} u_{h}, \end{matrix}$ which further gives $\partial_{1} u_{h} = 0$ because ε is small enough. We complete the proof of Proposition 4.3. □

Now based on the above two propositions, we are going to establish the convergence of the singular terms in the system (1.2).

Lemma 4.4.

By passing to a subsequence, there exists a scalar function $\begin{matrix} \bar{q} (t, x) \in L^{\infty} ([0, T^{*}]; L^{2} (Ω)), \end{matrix}$ such that as $ε \to 0$ , $\begin{array}{l} \frac{1}{ε} (r + b_{1}) - \frac{1}{ε} av (r + b_{1}) \to \bar{q} {weak}^{*} in L^{\infty} ([0, T^{*}]; L^{2} (Ω)), \\ \frac{1}{ε} \nabla (r + b_{1}) \to \nabla \bar{q} {weak}^{*} in L^{\infty} ([0, T^{*}]; H^{2} (Ω)), \end{array}$ where $\begin{matrix} av (r + b_{1}) = | Ω |^{- 1} \int_{Ω} (r + b_{1}) (t, x) d x, \end{matrix}$ and it holds that $\partial_{1} \bar{q} |_{Γ} = 0$ .

Proof.

Using the Poincaré inequality on Ω, we get $\begin{matrix} {‖ \frac{1}{ε} (r + b_{1}) - \frac{1}{ε} av (r + b_{1}) ‖}_{L^{2} (Ω)} ⩽ C (\frac{1}{ε} {‖ \nabla (r + b_{1}) ‖}_{L^{2} (Ω)}^{2}) ⩽ C . \end{matrix}$ So, by passing to a subsequence, there exists a function $\begin{matrix} \bar{q} (t, x) \in L^{\infty} ([0, T^{*}]; L^{2} (Ω)) \end{matrix}$ such that $\begin{matrix} (4.7) & \frac{1}{ε} (r + b_{1}) - \frac{1}{ε} av (r + b_{1}) \to \bar{q} (t, x), \end{matrix}$ ${weak}^{*}$ in $L^{\infty} ([0, T^{*}]; L^{2} (Ω))$ . By the estimate (4.3) in Proposition 4.2 and (4.7), we have $\begin{matrix} \frac{1}{ε} \nabla (r + b_{1}) = \frac{1}{ε} \nabla (r + b_{1}) - \frac{1}{ε} \nabla av (r + b_{1}) \to \nabla \bar{q}, \end{matrix}$ ${weak}^{*}$ in $L^{\infty} ([0, T^{*}]; H^{2} (Ω))$ . The boundary condition of the limit function $\bar{q}$ is a direct consequence of Proposition 4.3 because $\begin{matrix} \partial_{1} (r + b_{1}) = \partial_{1} r + \partial_{1} b_{1} = 0, \end{matrix}$ holds on the boundary. This completes the proof of Lemma 4.4. □

Lemma 4.5.

By passing to a subsequence, there exists a vector field $\begin{matrix} \bar{N} (t, x) \in L^{\infty} ([0, T^{*}]; L^{2} (Ω)) \end{matrix}$ such that as $ε \to 0$ , $\begin{array}{l} \frac{1}{ε} \tilde{u} \to \bar{N} {weak}^{*} in L^{\infty} ([0, T^{*}]; L^{2} (Ω)), \\ \frac{1}{ε} \partial_{1} u \to \partial_{1} \bar{N} {weak}^{*} in L^{\infty} ([0, T^{*}]; H^{2} (Ω)), \end{array}$ where $\begin{matrix} (4.8) & \tilde{u} (t, x_{1}, x_{2}, x_{3}) = u (t, x_{1}, x_{2}, x_{3}) - u (t, 0, x_{2}, x_{3}), \end{matrix}$ and it holds that ${\bar{N}}_{1} |_{Γ} = \partial_{1} {\bar{N}}_{h} |_{Γ} = 0$ .

Proof.

From (4.8), we find that $\tilde{u} (t, 0, x_{2}, x_{3}) = 0$ . Thus, we get $\begin{matrix} (4.9) & \frac{1}{ε} | \tilde{u} (t, x_{1}, x_{2}, x_{3}) | = \frac{1}{ε} | \int_{0}^{x_{1}} \partial_{1} \tilde{u} (t, ξ, x_{2}, x_{3}) d ξ | . \end{matrix}$ Integrating (4.9) over Ω gives $\begin{array}{l} \frac{1}{ε} \int_{Ω} {| \tilde{u} (t, x_{1}, x_{2}, x_{3}) |}^{2} d x & = \frac{1}{ε} \int_{Ω} {| \int_{0}^{x_{1}} \partial_{1} \tilde{u} (t, ξ, x_{2}, x_{3}) d ξ |}^{2} d x \\ ⩽ C \frac{1}{ε} \int_{Ω} \int_{0}^{h} {| \partial_{1} \tilde{u} (t, ξ, x_{2}, x_{3}) |}^{2} d ξ d x \\ ⩽ C \frac{1}{ε} ‖ \partial_{1} u ‖^{2} ⩽ C, \end{array}$ where we have used (4.4) in Proposition 4.2. Thus, by passing to a subsequence, there exists a vector field $\bar{N} (t, x)$ such that $\begin{matrix} \frac{1}{ε} \tilde{u} \to \bar{N}, \end{matrix}$ ${weak}^{*}$ in $L^{\infty} ([0, T^{*}]; L^{2} (Ω))$ . Using the uniform estimates (4.4) again, we get $\begin{matrix} \frac{1}{ε} \partial_{1} u = \frac{1}{ε} \partial_{1} \tilde{u} \to \partial_{1} \bar{N}, \end{matrix}$ ${weak}^{*}$ in $L^{\infty} ([0, T^{*}]; H^{2} (Ω))$ . Moreover, from Proposition 4.3 and the boundary condition (4.1), we obtain that the limit function $\bar{N}$ satisfies $\begin{matrix} {\bar{N}}_{1} |_{Γ} = \partial_{1} {\bar{N}}_{h} |_{Γ} = 0 . \end{matrix}$ We complete the proof of Lemma 4.5. □

Lemma 4.6.

By passing to a subsequence, there exists a vector field $\begin{matrix} \bar{L} (t, x) \in L^{\infty} ([0, T^{*}]; L^{2} (Ω)) \end{matrix}$ such that as $ε \to 0$ , $\begin{array}{l} \frac{1}{ε} \tilde{b} \to \bar{L} {weak}^{*} in L^{\infty} ([0, T^{*}]; L^{2} (Ω)), \\ \frac{1}{ε} \partial_{1} b \to \partial_{1} \bar{L} {weak}^{*} in L^{\infty} ([0, T^{*}]; H^{2} (Ω)), \end{array}$ and it holds that ${\bar{L}}_{h} |_{Γ} = \partial_{1} {\bar{L}}_{1} |_{Γ} = 0$ , $\nabla \cdot \partial_{1} \bar{L} = 0$ . Here, $\begin{matrix} \tilde{b} = b (t, x_{1}, x_{2}, x_{3}) - b (t, 0, x_{2}, x_{3}) . \end{matrix}$

Proof.

This lemma can be proved in a similar fashion as Lemma 4.4 and 4.5 except that we need to show that $\partial_{1} \bar{L}$ is divergence-free. However, this is apparent because $\partial_{1} b$ is divergence-free. So we omit the details of the proof of this lemma for the sake of simplicity. □

Now, we are ready to prove our main Theorem 2.4. First, since (4.2) holds, we can find a scalar function $\bar{K}$ such that $\begin{matrix} (4.10) & \frac{1}{ε} \nabla \cdot u \to \bar{K}, \end{matrix}$ ${weak}^{*}$ in $L^{\infty} ([0, T^{*}]; H^{2} (Ω))$ . Taking $ε \to 0$ in the equation (1.2), and using Lemma 4.4–Lemma 4.6, we get $\begin{matrix} (4.11) & \{\begin{matrix} \bar{a} (\partial_{t} \bar{r} + \bar{u} \cdot \nabla \bar{r}) + \bar{K} = 0, \\ \bar{ρ} (\partial_{t} \bar{u} + \bar{u} \cdot \nabla \bar{u}) + \nabla \bar{q} + \bar{b} \times (\nabla \times \bar{b}) = \partial_{1} \bar{L}, \\ \partial_{t} \bar{b} + \bar{u} \cdot \nabla \bar{b} - \bar{b} \cdot \nabla \bar{u} + \bar{b} \nabla \cdot \bar{u} = \partial_{1} \bar{N} - e_{1} \bar{K}, \\ \nabla \cdot \bar{b} = 0 . \end{matrix} \end{matrix}$ Before going further, we need to give a description of the kernel of the operator $L$ which is defined as $\begin{array}{l} (4.12) & L (\begin{matrix} r \\ u \\ b \end{matrix}) : = (\begin{matrix} \nabla \cdot u \\ \nabla r + \nabla b_{1} - \partial_{1} b \\ e_{1} \nabla \cdot u - \partial_{1} u \end{matrix}) . \end{array}$ We also define $\begin{matrix} H_{mhd} : = \{W = (\begin{matrix} r \\ u \\ b \end{matrix}) \in {(L^{2} (Ω))}^{7} | \nabla \cdot b = 0\} . \end{matrix}$ Then we have the following proposition.

Proposition 4.7.

The kernel of $L$ can be characterized as follows $\begin{matrix} ker L = \{(\begin{matrix} r \\ u \\ b \end{matrix}) \in H_{mhd} | \partial_{1} (r, u, b) = 0, \partial_{2} u_{2} + \partial_{3} u_{3} = 0, \nabla (r + b_{1}) = 0\} . \end{matrix}$

Proof.

If we have $\begin{matrix} L (\begin{matrix} r \\ u \\ b \end{matrix}) = 0, \end{matrix}$ then $\begin{matrix} (4.13) & \nabla \cdot u = 0, \partial_{1} u = 0, \nabla r + \nabla b_{1} = \partial_{1} b . \end{matrix}$ As a result, we get $\begin{matrix} (4.14) & \partial_{2} u_{2} + \partial_{3} u_{3} = 0, \partial_{1} r = 0, \nabla_{h} (r + b_{1}) = \partial_{1} b_{h} . \end{matrix}$ From the last equality, we have $\begin{matrix} (4.15) & \{\begin{matrix} \partial_{2} r + \partial_{2} b_{1} = \partial_{1} b_{2}, \\ \partial_{3} r + \partial_{3} b_{1} = \partial_{1} b_{3} . \end{matrix} \end{matrix}$ Taking $\partial_{12} {(4.15)}_{2} + \partial_{13} {(4.15)}_{3}$ , and using $\partial_{1} r = 0$ gives $\begin{matrix} (4.16) & \{\begin{matrix} \partial_{122} b_{1} = \partial_{112} b_{2}, \\ \partial_{133} b_{1} = \partial_{113} b_{3} . \end{matrix} \end{matrix}$ From the divergence free condition of b, we have $\begin{matrix} (4.17) & \partial_{11} (\partial_{1} b_{1} + \partial_{2} b_{2} + \partial_{3} b_{3}) = 0 . \end{matrix}$ Combining equality (4.16) and (4.17) yields $\begin{matrix} (4.18) & △ (\partial_{1} b_{1}) = 0 . \end{matrix}$ Using the boundary condition (4.1) and the divergence free condition of b, we get that on the boundary $\begin{matrix} (4.19) & \partial_{1} b_{1} |_{Γ} = - (\partial_{2} b_{2} + \partial_{3} b_{3}) |_{Γ} = 0 . \end{matrix}$ Multiplying (4.18) by $\partial_{1} b_{1}$ and integrating over Ω gives $\begin{matrix} \int_{Ω} {| \nabla (\partial_{1} b_{1}) |}^{2} = 0, \end{matrix}$ because the boundary term vanishes. Thus, we have $\partial_{1} b_{1} = 0$ . Applying $\partial_{1}$ to (4.15) and using $\partial_{1} r = \partial_{1} b_{1} = 0$ , one has $\begin{matrix} \partial_{11} b_{2} = \partial_{11} b_{3} = 0 . \end{matrix}$ Thus, by using the boundary conditions on $b_{2}$ and $b_{3}$ , we have $\begin{matrix} \partial_{1} b_{2} = \partial_{1} b_{3} = 0 . \end{matrix}$ Finally, as we have known that r and $b_{1}$ is independent of $x_{1}$ , we get from the last equality in (4.14) that $\begin{matrix} \nabla (r + b_{1}) = 0 . \end{matrix}$ This completes the proof of the Proposition 4.7. □

We continue our proof of the main Theorem 2.4. Let us reformulate (1.2) into $\begin{matrix} \{\begin{matrix} ε [a (\partial_{t} r + u \cdot \nabla r)] + \nabla \cdot u = 0, \\ ε [ρ (\partial_{t} u + u \cdot \nabla u) + b \times (\nabla \times b)] + \nabla r = \partial_{1} b - \nabla b_{1}, \\ ε [\partial_{t} b + u \cdot \nabla b - b \cdot \nabla u + b \nabla \cdot u] = \partial_{1} u - e_{1} \nabla \cdot u, \nabla \cdot b = 0 . \end{matrix} \end{matrix}$ Using the uniform estimates in Theorem 2.3, letting $ε \to 0$ in the above system, we find that $\begin{matrix} (4.20) & \{\begin{matrix} \nabla \cdot \bar{u} = 0, \\ \nabla \bar{r} = \partial_{1} \bar{b} - \nabla {\bar{b}}_{1}, \\ \partial_{1} \bar{u} - e_{1} \nabla \cdot \bar{u} = 0 . \end{matrix} \end{matrix}$ This implies that the limit function $(\bar{r}, \bar{u}, \bar{b})$ belongs to $ker L$ . In particular, we have $\begin{matrix} \partial_{1} ({\bar{u}}_{1}, {\bar{b}}_{2}, {\bar{b}}_{3}) = 0 . \end{matrix}$ On the other hand, recall that on the boundary Γ, $u_{1} = b_{2} = b_{3} = 0$ , so we get $\begin{matrix} (4.21) & {\bar{u}}_{1} = {\bar{b}}_{2} = {\bar{b}}_{3} \equiv 0 . \end{matrix}$

Back to equation (4.11), let us discuss the regularity of the solution of (4.11). First, since (4.21) holds, we find that $\begin{matrix} (4.22) & \partial_{1} {\bar{N}}_{h} = 0 . \end{matrix}$ Second, we want to show that the second equation in (4.11) holds in classical sense. To this end, we need to define another projection operator $\bar{P}$ . Set $\begin{matrix} M : = {\partial_{1} w ∣ w \in {(H^{1} (Ω))}^{3}, w_{2} = w_{3} = 0 on the boundary}, \end{matrix}$ and denote the orthogonal complement of M in ${(L^{2} (Ω))}^{3}$ by $M^{⊥}$ . Then we can define $\bar{P}$ as the orthogonal projection from ${(L^{2} (Ω))}^{3}$ to $M^{⊥}$ .

Applying $\bar{P} P$ to the second equation of (4.11) gives $\begin{matrix} (4.23) & \bar{ρ} \bar{P} P \partial_{t} \bar{u} + \bar{ρ} \bar{P} P (\bar{u} \cdot \nabla \bar{u}) + \bar{P} P (\bar{b} \times (\nabla \times \bar{b})) = 0 . \end{matrix}$ We claim that $\bar{P} P \partial_{t} \bar{u} = \partial_{t} \bar{u}$ , i.e. $\partial_{t} \bar{u} \in H_{σ} \cap M^{⊥}$ . First, it is obvious that $\partial_{t} \bar{u} \in H_{σ}$ because $\nabla \cdot \bar{u} = 0$ and ${\bar{u}}_{1} = 0$ . Next, for any $\partial_{1} w \in M$ , we have $\begin{matrix} \int_{Ω} \partial_{t} \bar{u} \cdot \partial_{1} w = \int_{Γ} \partial_{t} \bar{u} \cdot w - \int_{Ω} \partial_{t} \partial_{1} \bar{u} \cdot w = 0, \end{matrix}$ where we have used $\partial_{1} \bar{u} = 0$ and $({\bar{u}}_{1}, w_{2}, w_{3})$ vanishes on the boundary. So we get $\partial_{t} \bar{u} \in H_{σ} \cap M^{⊥}$ . Thus, from (4.23), we have $\begin{matrix} (4.24) & \bar{ρ} \partial_{t} \bar{u} + \bar{ρ} \bar{P} P (\bar{u} \cdot \nabla \bar{u}) + \bar{P} P (\bar{b} \times (\nabla \times \bar{b})) = 0 . \end{matrix}$ Since $(\bar{u}, \bar{b}) \in C ([0, T]; H^{3 - δ} (Ω))$ for any $δ > 0$ , we can choose sufficiently small δ such that $3 - δ - 1 > \frac{3}{2}$ and $\begin{matrix} (\bar{u} \cdot \nabla \bar{u}, \bar{b} \times (\nabla \times \bar{b})) \in C ([0, T]; H^{2 - δ} (Ω)) . \end{matrix}$ Thus, using the boundedness of the projection operators $P$ and $\bar{P}$ on $H^{s}$ with $s ⩾ 0$ , we get from (4.24) that $\begin{matrix} (4.25) & \partial_{t} \bar{u} \in C ([0, T]; H^{2 - δ} (Ω)) . \end{matrix}$ Applying $P$ to the second equation of (4.11), we have $\begin{matrix} (4.26) & \bar{ρ} \partial_{t} \bar{u} + \bar{ρ} P (\bar{u} \cdot \nabla \bar{u}) + P (\bar{b} \times (\nabla \times \bar{b})) = \partial_{1} \bar{L} . \end{matrix}$ Combining (4.25) and (4.26) yields $\begin{matrix} (4.27) & \partial_{1} \bar{L} \in C ([0, T]; H^{2 - δ} (Ω)) . \end{matrix}$ Furthermore, it follows easily that $\begin{matrix} (4.28) & \nabla \bar{q} \in C ([0, T]; H^{2 - δ} (Ω)) . \end{matrix}$ Now, from (4.25), (4.27) and (4.28), we can see that the second equation of (4.11) actually holds in the classical sense.

However, we can not prove that the first equation and the equation satisfied by ${\bar{b}}_{1}$ in (4.11) hold in the classical sense. The reason is that we can not find proper projection to eliminant $\bar{K}$ . To circumvent this difficulty, we consider the difference of these two equations $\begin{matrix} (4.29) & \bar{a} (\partial_{t} \bar{r} + \bar{u} \cdot \nabla \bar{r}) - (\partial_{t} {\bar{b}}_{1} + \bar{u} \cdot \nabla {\bar{b}}_{1}) = \partial_{1} {\bar{N}}_{1} . \end{matrix}$ We now prove the orthogonality of $\bar{a} \partial_{t} \bar{r} - \partial_{t} {\bar{b}}_{1}$ and $\partial_{1} {\bar{N}}_{1}$ in $L^{2} (Ω)$ . In fact, since $\bar{r}$ and ${\bar{b}}_{1}$ are independent of $x_{1}$ , we have $\begin{matrix} \int_{Ω} (\bar{a} \partial_{t} \bar{r} - \partial_{t} {\bar{b}}_{1}) \partial_{1} {\bar{N}}_{1} = \int_{T^{2}} (\bar{a} \partial_{t} \bar{r} - \partial_{t} {\bar{b}}_{1}) \int_{0}^{h} \partial_{1} {\bar{N}}_{1} = 0, \end{matrix}$ where the boundary condition of ${\bar{N}}_{1}$ in Lemma 4.5 is used. Therefore we can follow a similar procedure as (4.23)–(4.28) to prove that $\begin{matrix} \bar{a} \partial_{t} \bar{r} - \partial_{t} {\bar{b}}_{1} \in C ([0, T]; H^{2 - δ} (Ω)), \partial_{1} \bar{L} \in C ([0, T]; H^{2 - δ} (Ω)) . \end{matrix}$ In other words, equation (4.29) also holds in classical sense. Collecting the discussions above, we have that $\begin{matrix} (4.30) & \{\begin{matrix} \bar{ρ} (\partial_{t} \bar{u} + \bar{u} \cdot \nabla \bar{u}) + \nabla \bar{q} + \bar{b} \times (\nabla \times \bar{b}) = \partial_{1} \bar{L}, \\ \bar{a} (\partial_{t} \bar{r} + \bar{u} \cdot \nabla \bar{r}) - (\partial_{t} {\bar{b}}_{1} + \bar{u} \cdot \nabla {\bar{b}}_{1}) = \partial_{1} {\bar{N}}_{1} . \\ \nabla \cdot \bar{u} = 0 . \end{matrix} \end{matrix}$ holds in classical sense. Applying $\partial_{1}$ to the system (4.30) and using the fact that $\partial_{1} (\bar{r}, \bar{u}, {\bar{b}}_{1}) = 0$ , we get $\begin{matrix} (4.31) & \{\begin{matrix} \nabla \partial_{1} \bar{q} = \partial_{11} \bar{L}, \\ \partial_{11} {\bar{N}}_{1} = 0 . \end{matrix} \end{matrix}$ Since $\nabla \cdot \partial_{1} \bar{L} = 0$ , we get from the first equation of (4.31) that $\begin{matrix} \nabla \cdot \nabla \partial_{1} \bar{q} = △ \partial_{1} \bar{q} = 0 . \end{matrix}$ Recall that we have $\partial_{1} \bar{q} = 0$ on the boundary Γ. Then $\partial_{1} \bar{q} \equiv 0$ . As a result, we have $\begin{matrix} (4.32) & \partial_{11} \bar{L} = 0 . \end{matrix}$ From Lemma 4.6, we know that on the boundary it holds that $\begin{matrix} {\bar{L}}_{h} |_{Γ} = \partial_{1} {\bar{L}}_{1} |_{Γ} = 0 . \end{matrix}$ Multiplying (4.32) by $\bar{L}$ , integrating over Ω, one has $\begin{matrix} 0 = \int_{Ω} \bar{L} \partial_{11} \bar{L} = - \int_{Ω} | \partial_{1} \bar{L} |^{2}, \end{matrix}$ since the boundary term vanishes. So we have $\partial_{1} \bar{L} = 0$ . By the second equality in (4.31) and the boundary condition ${\bar{N}}_{1} |_{Γ} = 0$ in Lemma 4.5, we get $\begin{matrix} 0 = \int_{Ω} {\bar{N}}_{1} \partial_{11} {\bar{N}}_{1} = - \int_{Ω} | \partial_{1} {\bar{N}}_{1} |^{2} . \end{matrix}$ This implies $\partial_{1} {\bar{N}}_{1} = 0$ . Recall $\partial_{1} {\bar{N}}_{h} = 0$ in (4.22), thus we have that $\partial_{1} \bar{N} = 0$ .

Now, substituting (4.21), $\partial_{1} \bar{L} = \partial_{1} \bar{N} = 0$ and $\nabla \cdot \bar{u} = 0$ into (4.11) gives $\begin{matrix} (4.33) & \{\begin{matrix} \bar{a} (\partial_{t} \bar{r} + {\bar{u}}_{h} \cdot \nabla_{h} \bar{r}) + \bar{K} = 0, \\ \bar{ρ} (\partial_{t} {\bar{u}}_{h} + {\bar{u}}_{h} \cdot \nabla_{h} {\bar{u}}_{h}) + \nabla_{h} \bar{q} + \frac{1}{2} \nabla_{h} {({\bar{b}}_{1})}^{2} = 0, \\ \partial_{t} {\bar{b}}_{1} + {\bar{u}}_{h} \cdot \nabla_{h} {\bar{b}}_{1} + \bar{K} = 0, \\ \nabla_{h} \cdot {\bar{u}}_{h} = 0 . \end{matrix} \end{matrix}$ Next, to complete the proof of our main Theorem 2.4, we need to show that $\bar{K} = 0$ . Integrating the first equation in (1.2) over Ω and using the boundary condition (4.1) we get $\begin{array}{l} 0 & = \int_{Ω} a (\partial_{t} r + u \cdot \nabla r) = \int_{Ω} (a - \bar{a}) (\partial_{t} r + u \cdot \nabla r) + \int_{Ω} \bar{a} (\partial_{t} r + u \cdot \nabla r) \\ (4.34) & = \bar{a} \int_{Ω} \partial_{t} r - \bar{a} \int_{Ω} r \nabla \cdot u + \int_{Ω} (a - \bar{a}) (\partial_{t} r + u \cdot \nabla r) . \end{array}$ From the uniform estimates in Theorem 2.3 and Proposition 4.2, one has $\begin{matrix} a - \bar{a} = O (ε), \nabla \cdot u = O (ε) . \end{matrix}$ Letting $ε \to 0$ in (4.34), we have $\begin{matrix} (4.35) & \partial_{t} \int_{Ω} \bar{r} = 0 . \end{matrix}$ Subtracting the third equation of (4.33) from the first equation there yields $\begin{matrix} (4.36) & (\partial_{t} + {\bar{u}}_{h} \cdot \nabla_{h}) (\bar{a} \bar{r} - {\bar{b}}_{1}) = 0 . \end{matrix}$ On the other hand, Proposition 4.7 implies that $\begin{matrix} (4.37) & {\bar{b}}_{1} = - \bar{r} + α (t), \end{matrix}$ for some function $α (t)$ . Substituting (4.37) into (4.36) yields $\begin{matrix} (4.38) & (\bar{a} + 1) (\partial_{t} + {\bar{u}}_{h} \cdot \nabla_{h}) \bar{r} = \frac{d}{d t} α (t) . \end{matrix}$ Integrating (4.38) over $T^{2}$ and using (4.35), we have $\begin{matrix} | T^{2} | \frac{d}{d t} α (t) = (\bar{a} + 1) \int_{T^{2}} (\partial_{t} + {\bar{u}}_{h} \cdot \nabla_{h}) \bar{r} = - (\bar{a} + 1) \int_{T^{2}} (\nabla_{h} \cdot {\bar{u}}_{h}) \bar{r} = 0 . \end{matrix}$ So it implies that $α (t)$ is a constant. As a result, (4.37) can be written into $\begin{array}{l} (4.39) & {\bar{b}}_{1} = - \bar{r} + constant, \end{array}$ and (4.38) is reduced into $\begin{matrix} (\partial_{t} + {\bar{u}}_{h} \cdot \nabla_{h}) \bar{r} = 0, \end{matrix}$ which further gives $\bar{K} = 0$ due to (4.33).

Finally, it is straightforward to reduce (4.33) into the desired form (2.11) and (2.12) with $\bar{π} : = \bar{q} + \frac{1}{2} {({\bar{b}}_{1})}^{2}$ .

As usual, the uniqueness of the limit functions implies the convergence holds as $ε \to 0$ without restricting to a subsequence.

Footnotes

Acknowledgements

This research was supported by the ISF-NSFC joint research program (NSFC Grant No. 11761141008 and ISF Grant No. 2519/17). Ju is supported in part by NSFC (Grants Nos. 11571046, 11471028, and 11671225) and BJNSF (Grant No. 1182004). We thank the referee for several suggestions that helped to improve the presentation of the paper.

Appendix

In this section, we will give some useful lemmas, the first of which is obtained by integration by parts while taking the boundary conditions into account.

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