Global well-posedness of the MHD equations via the comparison principle

TL;DR

This paper establishes the global well-posedness of the incompressible MHD equations near equilibrium in certain domains using the comparison principle and a comparison function.

Contribution

It introduces a novel approach employing the comparison principle to prove global well-posedness for MHD equations in mixed domain settings.

Findings

Proves global existence and uniqueness of solutions near equilibrium.

Develops a new method based on the comparison principle.

Applicable to domains of the form R^k x T^{d-k}.

Abstract

In this paper, we prove the global well-posedness of the incompressible MHD equations near a homogeneous equilibrium in the domain $R^k\times T^{d-k}, d\geq2,k\geq1$ by using the comparison principle and constructing the comparison function.

Equations

$$\displaystyle\left\{\begin{array}[]{l}\partial_{t}v-\nu\Delta v+v\cdot\nabla v+\nabla p=b\cdot\nabla b,\\ \partial_{t}b-\mu\Delta b+v\cdot\nabla b=b\cdot\nabla v,\\ \text{div }v=\text{div }b=0,\end{array}\right.$$

$$Z_{+}=v+b,Z_{-}=v-b,$$

$$\displaystyle\left\{\begin{array}[]{l}\partial_{t}Z_{+}+Z_{-}\cdot\nabla Z_{+}=\mu\Delta Z_{+}-\nabla p,\\ \partial_{t}Z_{-}+Z_{+}\cdot\nabla Z_{-}=\mu\Delta Z_{-}-\nabla p,\\ \text{div}Z_{+}=\text{div}Z_{-}=0.\end{array}\right.$$

$$z_{+}=Z_{+}-B_0,z_{-}=Z_{-}+B_0.$$

$$\displaystyle\left\{\begin{array}[]{l}\partial_{t}z_{+}+Z_{-}\cdot\nabla z_{+}=\mu\Delta z_{+}-\nabla p,\\ \partial_{t}z_{-}+Z_{+}\cdot\nabla z_{-}=\mu\Delta z_{-}-\nabla p,\\ \text{div}z_{+}=\text{div}z_{-}=0.\end{array}\right.$$

$${\rho }_{\pm }(t,X{)}^2=\sum _{k=0}^N{\int }_{{\mathbb{R}}^d}|{\nabla }^k{z}_{\pm }{|}^2(t,Y)\theta (|X-Y|)dY,$$

$$\theta(r)=\left\{\begin{array}[]{l}1\quad\text{for}\ |r|\leq 1,\\ 0\quad\text{for}\ |r|\geq 2,\end{array}\right.\ \ |\theta^{\prime}(r)|^{2}\leq C\theta(r).$$

$$J_{\pm }(0)={\int }_{\mathbb{R}}\underset{y\in {\mathbb{R}}^{d-1}}{\sup }{\rho }_{\pm }(0,x,y)dx.$$

$$\Vert z_{\pm }(t){\Vert }_{H^N({\mathbb{R}}^k\times {\mathbb{T}}^{d-k})}\le C\Vert z_{\pm }(0){\Vert }_{H^N({\mathbb{R}}^k\times {\mathbb{T}}^{d-k})}\text{for any}t\in [0,+\infty ).$$

$${\partial }_t{\rho }_{\pm }\mp B_0\cdot \nabla {\rho }_{\pm }-\mu △{\rho }_{\pm }\le CF,$$

$$\displaystyle F(t,X)=\sum_{k+j\leq N+1,0\leq k,j\leq N}\big{\|}|\nabla^{k}z_{+}||\nabla^{j}z_{-}|(t)\big{\|}_{L^{2}(B(X,2))}+\sum_{k=0}^{N}\big{\|}\nabla^{k+1}p(t)\big{\|}_{L^{2}(B(X,2))}.$$

$$\displaystyle\left\{\begin{array}[]{l}\partial_{t}\nabla^{a}z_{+}-\mu\Delta\nabla^{a}z_{+}-B_{0}\cdot\nabla\nabla^{a}z_{+}+\dfrac{}{}\sum\limits_{b+c=a}C_{a}^{b}(\nabla^{b}z_{-}\cdot\nabla\nabla^{c}z_{+})+\nabla\nabla^{a}p=0,\\ \partial_{t}\nabla^{a}z_{-}-\mu\Delta\nabla^{a}z_{-}+B_{0}\cdot\nabla\nabla^{a}z_{-}+\dfrac{}{}\sum\limits_{b+c=a}C_{a}^{b}(\nabla^{b}z_{+}\cdot\nabla\nabla^{c}z_{-})+\nabla\nabla^{a}p=0,\\ \text{div}\nabla^{a}z_{+}=\text{div}\nabla^{a}z_{-}=0.\end{array}\right.$$

$${\partial }_t|{\nabla }^a{z}_{+}{|}^2$$

$$+2\sum _{b+c=a}{C}_a^b({\nabla }^b{z}_{-}\cdot \nabla {\nabla }^c{z}_{+}{\nabla }^a{z}_{+})+2{\nabla }^a{z}_{+}\cdot \nabla {\nabla }^{a}p=0,$$

$$({\partial }_t{\rho }_{+}^2-\mu △{\rho }_{+}^2-B_0\cdot \nabla {\rho }_{+}^2)(t,X)+2\mu {\int }_{{\mathbb{R}}^d}\sum _{|a|\le N}|\nabla {\nabla }^a{z}_{+}{|}^2(t,Y)\theta (|X-Y|)dY$$

$$+\frac{}{}2{\int }_{{\mathbb{R}}^d}\sum _{|a|\le N}\sum _{b+c=a}{C}_a^b({\nabla }^b{z}_{-}\cdot \nabla {\nabla }^c{z}_{+}{\nabla }^a{z}_{+})(t,Y)\theta (|X-Y|)dY$$

$$+2{\int }_{{\mathbb{R}}^d}\sum _{|a|\le N}{\nabla }^a{z}_{+}\cdot \nabla {\nabla }^{a}p(t,Y)\theta (|X-Y|)dY=0.$$

$$2({\rho }_{+}^{(\epsilon )}\nabla {\rho }_{+}^{(\epsilon )})(t,X)=(\nabla {\rho }_{+}^2)(t,X)=2{\int }_{{\mathbb{R}}^d}\sum _{|a|\le N}(\nabla {\nabla }^a{z}_{+}\cdot {\nabla }^a{z}_{+}{)}^2(t,Y)\theta (|X-Y|)dY$$

$$\le 2{\left({\int }_{{\mathbb{R}}^d}\sum _{|a|\le N}|\nabla {\nabla }^a{z}_{+}{|}^2(t,Y)\theta (|X-Y|)dY{\int }_{{\mathbb{R}}^d}\sum _{|a|\le N}|{\nabla }^a{z}_{+}{|}^2(t,Y)\theta (|X-Y|)dY\right)}^{\frac{1}{2}}$$

$$=2{\left({\int }_{{\mathbb{R}}^d}\sum _{|a|\le N}|\nabla {\nabla }^a{z}_{+}{|}^2(t,Y)\theta (|X-Y|)dY\right)}^{\frac{1}{2}}{\rho }_{+}(t,X),$$

$${\int }_{{\mathbb{R}}^d}\sum _{|a|\le N}|\nabla {\nabla }^a{z}_{+}{|}^2(t,Y)\theta (|X-Y|)dY\ge |\nabla {\rho }_{+}^{(\epsilon )}(t,X){|}^2.$$

$$-{\int }_{{\mathbb{R}}^d}({\nabla }^b{z}_{-}\cdot \nabla {\nabla }^c{z}_{+}{\nabla }^a{z}_{+})(t,Y)\theta (|X-Y|)dY$$

$$\le {\left({\int }_{B(X,2)}|{\nabla }^{|b|}{z}_{-}{|}^2|{\nabla }^{|c|+1}{z}_{+}{|}^2(t,Y)dY{\int }_{{\mathbb{R}}^d}|{\nabla }^a{z}_{+}{|}^2(t,Y)\theta (|X-Y|)dY\right)}^{\frac{1}{2}}$$

$$\le F(t,X){\rho }_{+}(t,X).$$

$$-2{\int }_{{\mathbb{R}}^d}({\nabla }^b{z}_{-}\cdot \nabla {\nabla }^c{z}_{+}{\nabla }^a{z}_{+})(t,Y)\theta (|X-Y|)dY$$

$$=-{\int }_{{\mathbb{R}}^d}(z_{-}\cdot \nabla |{\nabla }^a{z}_{+}{|}^2)(t,Y)\theta (|X-Y|)dY$$

$$=-{\int }_{{\mathbb{R}}^d}{z}_{-}\cdot \nabla \theta (|X-Y|)|{\nabla }^a{z}_{+}{|}^2(t,Y)dY$$

$$\le C{\left({\int }_{B(X,2)}|z_{-}{|}^2|{\nabla }^{|a|}{z}_{+}{|}^2(t,Y)dY{\int }_{{\mathbb{R}}^d}|{\nabla }^a{z}_{+}{|}^2(t,Y)\theta (|X-Y|)dY\right)}^{\frac{1}{2}}$$

$$\le CF(t,X){\rho }_{+}(t,X).$$

$$-{\int }_{{\mathbb{R}}^d}{\nabla }^a{z}_{+}\cdot \nabla {\nabla }^{a}p(t,Y)\theta (|X-Y|)dY$$

$$\le {\left({\int }_{B(X,2)}|{\nabla }^{|a|+1}p{|}^2(t,Y)dY{\int }_{{\mathbb{R}}^d}|{\nabla }^a{z}_{+}{|}^2(t,Y)\theta (|X-Y|)dY\right)}^{\frac{1}{2}}$$

$$\le F(t,X){\rho }_{+}(t,X).$$

$$({\partial }_t{\rho }_{+}^2-\mu △{\rho }_{+}^2-B_0\cdot \nabla {\rho }_{+}^2)(t,X)+2\mu |\nabla {\rho }_{+}^{(\epsilon )}(t,X){|}^2\le CF(t,X){\rho }_{+}(t,X),$$

$$({\partial }_t{\rho }_{+}^{(\epsilon )}-\mu △{\rho }_{+}^{(\epsilon )}-B_0\cdot \nabla {\rho }_{+}^{(\epsilon )})(t,X)\le CF(t,X).$$

$$F(t,X)\le C{\int }_{{\mathbb{R}}^d}\frac{{\rho }_{+}(t,Y){\rho }_{-}(t,Y)}{1+|X-Y{|}^{d+1}}dY.$$

$$\Vert f{\Vert }_{L^2(B(X,2))}\le C{\int }_{B(X,3)}\Vert f{\Vert }_{L^2(B(Y,\frac{1}{2}))}dY.$$

$${\int }_{B(X,\frac{1}{2})}\Vert f{\Vert }_{L^2(B(X,2)\cap B(Y,3))}^{2}dY$$

Full text

1. Introduction

In this paper, we consider the incompressible magneto-hydrodynamics (MHD) equations in $[0,T)\times\Omega,\Omega\subseteq\mathbb{R}^{d},d\geq 2$:

[TABLE]

where $v$ denotes the velocity field and $b$ denotes the magnetic field, and $\nu\geq 0$ is the viscosity coefficient, $\mu\geq 0$ is the resistivity coefficient. If $\nu=\mu=0$, (1.4) is the so called ideal MHD equations; If $\nu>0$ and $b=0$, (1.4) is reduced to the Navier-Stokes equations.

It is well-known that the 2-D MHD equations with full viscosities(i.e., $\nu>0$ and $\mu>0$) have global smooth solution. In the absence of resistivity(i.e., $\mu=0$), the global existence of weak solution and strong solution of the MHD equations is still an open question. Recently, Cao and Wu [5] proved the global regularity of the 2-D MHD equations with partial dissipation and magnetic diffusion. Motivated by numerical observation [3]: the energy of the MHD equations is dissipated at a rate independent of the ohmic resistivity, there are a lot of works [1, 7, 8, 9, 11] devoted to the global well-posedness of the MHD equations without resistivity in a homogeneous magnetic field $B_{0}$.

In high temperature plasmas, both $\nu$ and $\mu$ are usually very small. Thus, it is very interesting to investigate the long-time dynamics of the MHD equations in such case. In this paper, we consider the case of $\mu=\nu$. In terms of the Els$\ddot{a}$sser variables

[TABLE]

the MHD equations (1.4) can be written as

[TABLE]

We introduce the fluctuations

[TABLE]

Then the system (1.8) can be reformulated as

[TABLE]

Bardos-Sulem-Sulem [2] proved the global well-posedness of (1.12) for $\Omega=\mathbb{R}^{d}$ and $\mu=0$ when the initial data is small in a weighed Hölder space. Recently, He-Xu-Yu [6] proved the global well-posedness of (1.12) for any $\mu\geq 0$ and $\Omega=\mathbb{R}^{3}$ by using some ideas from nonlinear stability of Minkowski space-time in general relativity. Cai and Lei [4] proved simliar result for $\Omega=\mathbb{R}^{d},d=2,3$ by using Alinhac’s ghost weight method. Wei and Zhang [10] dealt with more physical case, which allows $\nu\neq\mu$ and $\Omega$ to be a strip. For all these results, the key mechanism leading to the global well-posedness is that the nonlinear terms $z_{-}\cdot\nabla z_{+}$ and $z_{+}\cdot\nabla z_{-}$ are essentially neglected after a long time, because $z_{\pm}$ are transported along the opposite direction.

The goal of this paper is twofold: (1) include the domain $\mathbb{R}^{k}\times\mathbb{T}^{d-k},k\geq 1$. Previous results required $k\geq 2$ at least; (2) develop an elementary and much simpler method via the comparison principle.

Without loss of generality, we take the background magnetic field $B_{0}=(1,0,\cdots,0)$. We will identify a function in $\mathbb{R}^{k}\times\mathbb{T}^{d-k}$ with a function in $\mathbb{R}^{d}$ periodic in $d-k$ directions $e_{1},\cdots,e_{d-k}$, where $e_{1},\cdots,e_{d-k},B_{0}$ are orthogonal.

We introduce

[TABLE]

where $\theta(r)$ is a smooth cut-off function so that

[TABLE]

Let

[TABLE]

Our result is stated as follows.

Theorem 1.1**.**

Let $d\geq 2,1\leq k\leq d$ and $z_{\pm}(0)\in H^{N}(\mathbb{R}^{k}\times\mathbb{T}^{d-k})$ for some integer $N>\frac{d}{2}+1$. There exists $\epsilon_{1}>0$ so that if $J_{\pm}(0)\leq\epsilon_{1}$, then there exists a global unique solution (z_{+},z_{-})\in C\big{(}[0,+\infty),H^{N}(\mathbb{R}^{k}\times\mathbb{T}^{d-k})\big{)} to the MHD equations (1.12) satisfying

[TABLE]

Let us give some remarks on our result.

We only require the initial data to decay at infinity in $B_{0}$ direction, which is the key for the global well-posedness in $\mathbb{R}\times\mathbb{T}^{d-1}$.

• 2.

It is easy to check that $J_{\pm}(0)\leq C\epsilon$ for the initial data considered in [4] and [6]. Indeed, for the data in [4], we have $\rho_{\pm}(0,X)\leq C\epsilon(1+|X|)^{-\delta}$. For the data in [6], we have $\rho_{\pm}(0,X)\leq C\epsilon(R^{2}+|X|^{2})^{-\frac{1}{2}}(\ln(R^{2}+|X|^{2}))^{-2}.$ Here $\delta>1,\ R\geq 100$ and $\epsilon>0$ is small.

• 3.

At a first glance, there seems no difference between $\mathbb{R}^{d}$ and $\mathbb{R}^{k}\times\mathbb{T}^{d-k}$. From the proofs in [4] and [6], it seems that the case of $d=3$ is easier than the case of $d=2$. However, for a solution (z_{+},z_{-})\in C\big{(}[0,+\infty),H^{N}(\mathbb{R}^{2})\big{)}, \big{(}z_{+}(t,x),0,z_{-}(t,x),0\big{)} is also a solution in C\big{(}[0,+\infty),H^{N}(\mathbb{R}^{2}\times\mathbb{T})\big{)}. Thus, the case of $\mathbb{R}^{2}\times\mathbb{T}$ is not easier than the case of $\mathbb{R}^{2}$. In this sense, the case of $\mathbb{R}\times\mathbb{T}^{d-1}$ may be harder.

Throughout this paper, we denote by $C$ a constant independent of $t,\mu$, which may be different from line to line.

2. Local energy inequality

In this section, we derive the following local energy inequalities

[TABLE]

where $F(t,X)$ is given by

[TABLE]

Here $B(X,r)$ denotes a ball in $\mathbb{R}^{d}$ with the center at $X$ and radius $r$.

We only prove (2.1) for $\rho_{+}$, the case of $\rho_{-}$ is similar. For any multi-index $a\in\mathbb{N}^{d}$, using the same notations as in [4], one can easily deduce from (1.12) that

[TABLE]

Taking inner product of the first equation with $\nabla^{a}z_{+}$, we obtain

[TABLE]

from which, we deduce that

[TABLE]

Let $\rho_{+}^{(\varepsilon)}=(\rho_{+}^{2}+\varepsilon)^{\frac{1}{2}}$ for $\varepsilon>0.$ We have

[TABLE]

which implies that

[TABLE]

If $|a|\leq N,\ b+c=a,\ c<a$, then $|b|,|c|+1\leq N$ and

[TABLE]

If $|a|\leq N,\ b+c=a,\ c=a$, then $b=(0,\cdots,0)$ and

[TABLE]

If $|a|\leq N,$ then

[TABLE]

Summing up, we obtain

[TABLE]

which gives

[TABLE]

Now, the local energy inequality (2.1) follows by letting $\varepsilon\to 0$.

Let us conclude this section by the following estimate for $F(t,X)$:

[TABLE]

We need the following fact.

Lemma 2.1**.**

It holds that

[TABLE]

Proof.

By Fubini theorem, we have

[TABLE]

As $B(X,2)\cap B(Y,3)=B(X,2)$ for $Y\in B(X,\frac{1}{2})$, we infer that

[TABLE]

where $\omega_{d}$ is the volume of the unit ball in $\mathbb{R}^{d}$, and this gives the result. ∎

Now let us prove (2.4). By Sobolev embedding, we have

[TABLE]

where $\dfrac{1}{p_{k}}=\dfrac{k-1}{2(N-1)}$ for $1\leq k\leq N$ and $\dfrac{1}{p_{k}}=0$ for $k=0.$ Thus, for $k+j\leq N+1,0\leq k,j\leq N$, we have $\dfrac{1}{p_{k}}+\dfrac{1}{p_{j}}\leq\frac{1}{2}$ and

[TABLE]

Due to $\text{div}z_{\pm}=0,$ we infer from the first equation of (1.12) that

[TABLE]

Then by the interior elliptic estimates, we get

[TABLE]

from which and Lemma 2.1, we infer that

[TABLE]

It remains to estimate $\|\nabla p(t)\|_{L^{\infty}(B(X,4))}$. For this, we use the following representation formula of the pressure $p(t,X)$:

[TABLE]

where $N(X)$ is the Newton potential. Thanks to $|\nabla^{k}N(X)|\leq C|X|^{2-d-k}$ and Sobolev embedding, we obtain

[TABLE]

Notice that $\|\triangle p(t)\|_{H^{N-1}(B(Y,\frac{1}{2}))}\leq C\rho_{+}(t,Y)\rho_{-}(t,Y)$, which along with Lemma 2.1 gives

[TABLE]

This shows that

[TABLE]

Thanks to $\dfrac{1}{1+|X^{\prime}-Y|^{d+1}}\leq\dfrac{C}{1+|X-Y|^{d+1}}$ for $X^{\prime}\in B(X,4)$, we have

[TABLE]

Inserting this into (2.5), we arrive at (2.4).

3. Comparison principle

It follows from (2.1) and (2.4) that

[TABLE]

where $N_{1}(X)=(1+|X|^{d+1})^{-1}$.

To control $\rho_{\pm}$, we establish the following comparison principle.

Lemma 3.1**.**

Let $0\leq\rho_{\pm}^{1}\in L^{\infty}\cap C^{0}([0,T)\times\mathbb{R}^{d})$ satisfy

[TABLE]

If $\rho_{\pm}(0)\leq\rho_{\pm}^{1}(0)$ in $\mathbb{R}^{d}$, then $\rho_{\pm}\leq\rho_{\pm}^{1}$ in $[0,T)\times\mathbb{R}^{d}$.

Proof.

It follows from (3.1) and (3.2) that

[TABLE]

Let $f^{+}=\max(f,0)$. By maximum principle, we deduce that for $t\in[0,T)$,

[TABLE]

where $M=\|\rho_{+}+\rho_{-}^{1}\|_{L^{\infty}([0,T)\times\mathbb{R}^{d})}<+\infty$. This implies that $\|(\rho_{\pm}-\rho_{\pm}^{1})^{+}(t)\|_{L^{\infty}(\mathbb{R}^{d})}=0,$ hence, $\rho_{\pm}(t)\leq\rho_{\pm}^{1}(t)$ for $t\in[0,T)$. ∎

To construct comparison functions $\rho_{\pm}^{1}$, we need the following observation.

Lemma 3.2**.**

There exists an absolute constant $C_{0}>1$ so that

[TABLE]

Here $\|N_{1}\|_{L^{1}(\mathbb{R}^{d})}\leq C_{0}$.

Proof.

As $N_{1}(X)\leq CN_{1}(Y)$ for $|X-Y|\leq 1$, we have

[TABLE]

Using the fact that

[TABLE]

we infer that

[TABLE]

which gives the first inequality.

Using $\|z_{\pm}(0)\|_{H^{N}(B(Y,\frac{1}{2}))}\leq C\rho_{\pm}(0,Y)$ and Lemma 2.1, we deduce that

[TABLE]

which gives the second inequality. ∎

Now let us construct the comparison function.

Lemma 3.3**.**

Let $\rho_{\pm}^{0}\in L^{1}\cap C(\mathbb{R}),\ \rho_{\pm}^{00}\in L^{2}\cap L^{\infty}\cap C(\mathbb{R}^{k}\times\mathbb{T}^{d-k})$. Assume that

[TABLE]

Then there exists $\epsilon_{0}>0$ depending only on $C_{0},C_{1}$ such that if $\|\rho_{\pm}^{0}\|_{L^{1}(\mathbb{R})}<\epsilon_{0}$, then there exists 0\leq\rho_{\pm}^{1}\in L^{\infty}\cap C\big{(}[0,+\infty)\times\mathbb{R}^{d}\big{)} satisfying (3.2) and $\rho_{\pm}^{00}\leq\rho_{\pm}^{1}(0)$. Moreover,

[TABLE]

Proof.

Step 1. Construction of the data

For $\mu=0$, we set $g_{\pm}^{00}=\rho_{\pm}^{00},\ g_{\pm}^{0}=\rho_{\pm}^{0}$. For $\mu>0$, we set

[TABLE]

It is easy to check that

[TABLE]

and $0\leq g_{\pm}^{00}(x,y)\leq g_{\pm}^{0}(x)$ for $x\in\mathbb{R},y\in\mathbb{R}^{d-1}.$

Let

[TABLE]

Then we have

[TABLE]

Step 2. Construction of comparison function

Let $(\rho_{\pm}^{01},g_{\pm}^{01})(t,X)$ be the solution to

[TABLE]

and $(\rho_{\pm}^{11},g_{\pm}^{1},h_{\pm}^{1})(t,x)$ be the solution to

[TABLE]

Thanks to the construction of the data, we find that

[TABLE]

Now we take $\rho_{\pm}^{1}(t)=C_{0}\rho_{\pm}^{10}(t)*N_{1},$ where

[TABLE]

Step 3. Verification of the conditions

By our construction, it is easy to check that

[TABLE]

which implies that

[TABLE]

As $0\leq h_{\pm}^{1}<1$, we have

[TABLE]

Since $\rho_{\pm}^{01}(t)$ and $\rho_{\pm}^{01}(t)*N_{1}$ satisfy $\partial_{t}f\mp B_{0}\cdot\nabla f-\mu\triangle f=0$ and $\rho_{\pm}^{01}(0)*N_{1}=\rho_{\pm}^{00}*N_{1}\leq C_{0}\rho_{\pm}^{00}=C_{0}\rho_{\pm}^{01}(0)$, we conclude that $\rho_{\pm}^{01}(t)*N_{1}\leq C_{0}\rho_{\pm}^{01}(t).$ Thanks to $g_{\pm}^{00}*N_{1}\leq C_{0}g_{\pm}^{00}$, we similarly have $g_{\pm}^{01}(t)*N_{1}\leq C_{0}g_{\pm}^{01}(t)$. Thus,

[TABLE]

which along with (3.3) gives

[TABLE]

Taking $\epsilon_{0}=1/(2C_{0}^{3}C_{1})$, we find that 0\leq\rho_{\pm}^{1}\in L^{\infty}\cap C\big{(}[0,+\infty)\times\mathbb{R}^{d}\big{)} satisfies (3.2). As $\rho_{\pm}^{10}(0)\geq\rho_{\pm}^{01}(0)=\rho_{\pm}^{00},$ we have $\rho_{\pm}^{1}(0)=C_{0}\rho_{\pm}^{10}(0)*N_{1}\geq C_{0}\rho_{\pm}^{00}*N_{1}\geq\rho_{\pm}^{00}.$

By standard energy estimate, we can deduce that

[TABLE]

For $\mu=0$, $\|g_{\pm}^{00}\|_{L^{2}(\mathbb{R}^{k}\times\mathbb{T}^{d-k})}=\|\rho_{\pm}^{00}\|_{L^{2}(\mathbb{R}^{k}\times\mathbb{T}^{d-k})}$, and for $\mu>0$,

[TABLE]

Thus, we obtain

[TABLE]

This completes the proof. ∎

4. Proof of Theorem 1.1

By the local well-posedness result, there exists a unique solution z_{\pm}\in C\big{(}[0,T^{*}),H^{N}(\mathbb{R}^{k}\times\mathbb{T}^{d-k})\big{)} to the MHD equations (1.12), where $T^{*}$ is the maximal existence time of the solution.

Let $\rho_{\pm}^{00}=C_{0}\rho_{\pm}(0)*N_{1}$. Then Lemma 3.2 ensures that

[TABLE]

Let

[TABLE]

where $\rho_{\pm}^{*}(x)=\sup\limits_{y\in\mathbb{R}^{d-1}}\rho_{\pm}(0,x,y)$, thus $J_{\pm}(0)=\|\rho_{\pm}^{*}\|_{L^{1}(\mathbb{R})}.$ Thanks to

[TABLE]

we find that $\rho_{\pm}^{00}(x,y)\leq\rho_{\pm}^{0}(x)$ and

[TABLE]

if $J_{\pm}(0)\leq\epsilon_{1}=\epsilon_{0}/2C_{0}^{2}$.

Now, Lemma 3.3 ensures that there exists 0\leq\rho_{\pm}^{1}\in L^{\infty}\cap C\big{(}[0,+\infty)\times\mathbb{R}^{d}\big{)}, which satisfies (3.2), and $\rho_{\pm}(0)\leq\rho_{\pm}^{00}\leq\rho_{\pm}^{1}(0)$ in $\mathbb{R}^{d}$, and

[TABLE]

Then we infer from Lemma 3.1 that $0\leq\rho_{\pm}\leq\rho_{\pm}^{1}$ in $[0,T)\times\mathbb{R}^{d}$ for $0<T<T^{*}$. Hence,

[TABLE]

for any $t\in[0,T)$, which implies $T^{*}=+\infty.$

5. Decay estimates

In this section, we provide some decay estimates of the solution in time when $\mu>0$. Here we use the notations in section 3 and section 4.

Using the fact that $(\rho_{\pm}^{01},g_{\pm}^{01})(t,X)=e^{\mu t\triangle}(\rho_{\pm}^{00},g_{\pm}^{00})(X-B_{0}t)$, we have

[TABLE]

which gives

[TABLE]

Thus, we deduce that for $2\leq p\leq\infty$,

[TABLE]

which along with the fact that $e^{\mu t\triangle}\rho_{\pm}^{00}=C_{0}e^{\mu t\triangle}\rho_{\pm}(0)*N_{1}$, we infer that

[TABLE]

for $2\leq p\leq\infty.$ Thus, we obtain

[TABLE]

If $\rho_{\pm}(0)\in L^{1}(\mathbb{R}^{d})$, we similarly have

[TABLE]

Acknowledgements

This work was supported by NSF of China under Grant 11425103.