Boundary layer problems for the two-dimensional inhomogeneous incompressible magnetohydrodynamics equations

Abstract

In this paper, we study the well-posedness of boundary layer problems for the inhomogeneous incompressible magnetohydrodynamics (MHD) equations, which are derived from the two-dimensional density-dependent incompressible MHD equations. Under the assumption that initial tangential magnetic field is not zero and density is a small perturbation of the outer constant flow in supernorm, the local-in-time existence and uniqueness of inhomogeneous incompressible MHD boundary layer equations are established in weighted conormal Sobolev space by energy method.

Appendices

Appendix A: Calculus inequalities

In this appendix, we will introduce some basic inequalities that be used frequently in this paper. First of all, we introduce the following Hardy type inequality, which can refer to [36].

Lemma A.1

Let the proper function $f:{\mathbb {T}}\times {\mathbb {R}}^+ \rightarrow {\mathbb {R}}$, and satisfies $f(x, y)|_{y=0}=0$ and $\underset{{y \rightarrow +\infty }}{\lim } f(x,y) = 0$. If $\lambda > - \frac{1}{2}$, then it holds true

$$\begin{aligned} \Vert f\Vert _{L^2_\lambda ({\mathbb {T}}\times {\mathbb {R}}^+)} \le \frac{2}{2\lambda +1} \Vert \partial _y f\Vert _{L^2_{\lambda +1} ({\mathbb {T}}\times {\mathbb {R}}^+)}. \end{aligned}$$

(A.1)

Next, we will state the following Sobolev-type inequality.

Lemma A.2

Let the proper function $f:{\mathbb {T}}\times {\mathbb {R}}^+ \rightarrow {\mathbb {R}}$, and satisfies $\underset{{y \rightarrow +\infty }}{\lim } f(x,y) = 0$. Then there exists a universal constant $C>0$ such that

$$\begin{aligned} \Vert f\Vert _{L^\infty _0({\mathbb {T}}\times {\mathbb {R}}^+)}\le & {} C(\Vert \partial _y f\Vert _{L^2_0({\mathbb {T}}\times {\mathbb {R}}^+)}+\Vert \partial _{xy}^2 f\Vert _{L^2_0({\mathbb {T}}\times {\mathbb {R}}^+)})^{\frac{1}{2}} (\Vert f\Vert _{L^2_0({\mathbb {T}}\times {\mathbb {R}}^+)}\nonumber \\&\quad +\Vert \partial _x f\Vert _{L^2_0({\mathbb {T}}\times {\mathbb {R}}^+)})^{\frac{1}{2}}, \end{aligned}$$

(A.2)

or equivalently

$$\begin{aligned} \Vert f\Vert _{L^\infty _0({\mathbb {T}}\times {\mathbb {R}}^+)}\le & {} C(\Vert f\Vert _{L^2_0({\mathbb {T}}\times {\mathbb {R}}^+)}+\Vert \partial _x f\Vert _{L^2_0({\mathbb {T}}\times {\mathbb {R}}^+)} +\Vert \partial _y f\Vert _{L^2_0({\mathbb {T}}\times {\mathbb {R}}^+)}\nonumber \\&\quad +\Vert \partial _{xy}^2 f\Vert _{L^2_0({\mathbb {T}}\times {\mathbb {R}}^+)}). \end{aligned}$$

(A.3)

Proof

Indeed, the estimate (A.3) follows directly from estimate (A.2) and the Cauchy–Schwartz inequality. Hence, we only give the proof for the estimate (A.2). On one hand, thanks to the one-dimensional Sobolev inequality for the $y-$variable, we get

$$\begin{aligned} |f(x, y)|^2 \le C\left( \int _0^\infty |\partial _\xi f(x, \xi )|^2d\xi \right) ^{\frac{1}{2}} \left( \int _0^\infty |f(x, \xi )|^2d\xi \right) ^{\frac{1}{2}}. \end{aligned}$$

(A.4)

On the other hand, we apply the following one-dimensional Sobolev inequality for $x-$variable to get

$$\begin{aligned} |f(x, y)|^2\le & {} C(\Vert f(y)\Vert _{L^2({\mathbb {T}})}^2+\Vert \partial _x f(y)\Vert _{L^2({\mathbb {T}})}^2), \quad |\partial _y f(x, y)|^2\nonumber \\\le & {} C(\Vert \partial _y f(y)\Vert _{L^2({\mathbb {T}})}^2+\Vert \partial _{xy} f(y)\Vert _{L^2({\mathbb {T}})}^2). \end{aligned}$$

(A.5)

Therefore, substituting the estimate (A.5) into (A.4), we complete the proof of estimate (A.2). $\square$

Now we will state the Moser type inequality as follow:

Lemma A.3

Denote $\Omega :={\mathbb {T}}\times {\mathbb {R}}^+$, let the proper functions $f(t, x, y): {\mathbb {R}}^+\times \Omega \rightarrow {\mathbb {R}}$ and $g(t, x, y): {\mathbb {R}}^+ \times \Omega \rightarrow {\mathbb {R}}$. Then, there exists a constant $C_m>0$ such that

$$\begin{aligned}&\int _0^t \Vert ({\mathcal {Z}}^{\beta } f {\mathcal {Z}}^{\gamma }g)(\tau )\Vert _{L^2_l(\Omega )}^2d\tau \le C_m(\Vert \langle y \rangle ^{l_1} f\Vert _{L^\infty _{x,y,t}}^2\nonumber \\&\quad \int _0^t \Vert g\Vert _{{\mathcal {H}}^{m}_{l_2}}^2 d\tau +\Vert \langle y \rangle ^{l_1} g\Vert _{L^\infty _{x,y,t}}^2\int _0^t \Vert f\Vert _{{\mathcal {H}}^{m}_{l_2}}^2 d\tau ), \end{aligned}$$

(A.6)

where $|\beta +\gamma |=m$ and $l_1+l_2=l$.

Proof

For any $p\ge 2$, due to the relation $|Z_2 f|^p=Z_2(f Z_2 f |Z_2 f|^{p-2})-(p-1)f Z_2^2 f|Z_2 f|^{p-2}$, we find

$$\begin{aligned}&\int _{{\mathbb {R}}^+} \langle y \rangle ^{\theta l p}|Z_2 f|^p dy =\int _{{\mathbb {R}}^+}\langle y \rangle ^{\theta l p} Z_2(f Z_2 f |Z_2 f|^{p-2}) dy -(p-1) \\&\quad \int _{{\mathbb {R}}^+}\langle y \rangle ^{\theta l p}f Z_2^2 f|Z_2 f|^{p-2}dy. \end{aligned}$$

Integrating by part and applying the Hölder inequality, we find for $0 \le \theta \le 1$ and $0\le \theta _1 \le \frac{\theta }{2}$ that

$$\begin{aligned} \begin{aligned}&\Vert \langle y \rangle ^{\theta l} Z_2 f\Vert _{L_0^p({\mathbb {R}}^+)}^p\\&\quad \le C_p \int _{{\mathbb {R}}^+}\langle y \rangle ^{\theta l p-1} |f|(|Z_2 f|+|Z_2^2 f|)|Z_2 f|^{p-2} dy\\&\quad \le C_p\Vert \langle y \rangle ^{\theta l} Z_2 f\Vert _{L_0^p({\mathbb {R}}^+)}^{p-2} \Vert \langle y \rangle ^{\theta _1 l} f\Vert _{L_0^q({\mathbb {R}}^+)} \Vert \langle y \rangle ^{(2\theta -\theta _1) l}(|Z_2 f|,|Z_2^2 f|)\Vert _{L_0^r({\mathbb {R}}^+)}, \end{aligned} \end{aligned}$$

and hence, it follows

$$\begin{aligned} \Vert \langle y \rangle ^{\theta l} Z_2 f\Vert _{L_0^p({\mathbb {R}}^+)}^2 \le C_p \Vert \langle y \rangle ^{\theta _1 l} f\Vert _{L_0^q({\mathbb {R}}^+)} \sum _{1\le k \le 2}\Vert \langle y \rangle ^{(2\theta -\theta _1) l}Z_2^k f\Vert _{L_0^r({\mathbb {R}}^+)}. \end{aligned}$$

Here $\frac{1}{q}+\frac{1}{r}=\frac{2}{p}$. Integrating with respect to t and x variables, and applying Hölder inequality, we get

$$\begin{aligned} \Vert \langle y \rangle ^{\theta l} Z_2 f\Vert _{L^p(Q_T)}^2 \le C_p \Vert \langle y \rangle ^{\theta _1 l} f\Vert _{L^q(Q_T)} \sum _{1\le k \le 2}\Vert \langle y \rangle ^{(2\theta -\theta _1) l}Z_2^k f\Vert _{L^r(Q_T)}. \end{aligned}$$

Here $Q_T=\Omega \times [0, T]$. Similarly, it is easy to justify for $i=0,1$,

$$\begin{aligned} \Vert \langle y \rangle ^{\theta l} Z_i f\Vert _{L^p_0(Q_T)}^2 \le C_p \Vert \langle y \rangle ^{\theta _1 l} f\Vert _{L^q_0(Q_T)} \sum _{1\le k \le 2}\Vert \langle y \rangle ^{(2\theta -\theta _1) l}Z_i^k f\Vert _{L^r_0(Q_T)}. \end{aligned}$$

Here $\frac{1}{q}+\frac{1}{r}=\frac{2}{p}$. By multiple application of the above inequality, we get(proof by induction)

$$\begin{aligned}&\Vert \langle y \rangle ^{(|\alpha |\theta -(|\alpha |-1)\theta _1)l} {\mathcal {Z}}^\alpha f\Vert _{L^{p_1}_0(Q_T)} \le C_p \Vert \langle y \rangle ^{\theta _1 l} f\Vert _{L^{q_1}_0(Q_T)}^{1-\frac{|\alpha |}{m}} \\&\quad \sum _{1\le |\beta | \le m}\Vert \langle y \rangle ^{(m\theta -(m-1)\theta _1) l} {\mathcal {Z}}^{\beta } f\Vert _{L^{r_1}_0(Q_T)}^{{\frac{|\alpha |}{m}}}, \end{aligned}$$

where $\frac{1}{p_1}=\frac{1}{q_1}(1-\frac{|\alpha |}{m})+\frac{|\alpha |}{r_1 m}$ and $1\le |\alpha | \le m-1$. Then, we get for $|\beta |+|\gamma |=m$ that

$$\begin{aligned} \begin{aligned}&\Vert \langle y \rangle ^{l}{\mathcal {Z}}^{\beta }f {\mathcal {Z}}^{\gamma } g\Vert _{L^2_0(Q_T)}^2 \\&\quad \le C \Vert \langle y \rangle ^{\frac{|\beta |}{m}l+(1-\frac{2|\beta |}{m})l_1} {\mathcal {Z}}^{\beta }f \Vert _{L^{\frac{2m}{|\beta |}}_0(Q_T)}^2 \Vert \langle y \rangle ^{\frac{|\gamma |}{m}l+(\frac{2|\beta |}{m}-1)l_1}{\mathcal {Z}}^{\gamma } f\Vert _{L^{\frac{2m}{|\gamma |}}_0(Q_T)}^2\\&\quad \le C_m \Vert \langle y \rangle ^{l_1}f\Vert _{L^\infty _0(Q_T)}^{2(1-\frac{|\beta |}{m})} \sum _{1\le |\beta | \le m}\Vert \langle y \rangle ^{l-l_1}{\mathcal {Z}}^{\beta }f\Vert _{L^2_0(Q_T)}^{\frac{2|\beta |}{m}}\\&\qquad \times \Vert \langle y \rangle ^{l_1} g\Vert _{L^\infty _0(Q_T)}^{2(1- \frac{|\gamma |}{m})} \sum _{1\le |\gamma | \le m}\Vert \langle y \rangle ^{l-l_1}{\mathcal {Z}}^{\gamma } g\Vert _{L^2_0(Q_T)}^{\frac{2|\gamma |}{m}}\\&\quad \le C_m \Vert f\Vert _{L^\infty _{l_1}(Q_T)}^2\sum _{1\le |\beta | \le m}\Vert {\mathcal {Z}}^\beta g\Vert _{L^2_{l-l_1}(Q_T)}^2 +C_m \Vert g\Vert _{L^\infty _{l_1}(Q_T)}^2\sum _{1\le |\beta | \le m}\Vert {\mathcal {Z}}^\beta f\Vert _{L^2_{l-l_1}(Q_T)}^2. \end{aligned} \end{aligned}$$

Therefore, we complete the proof of this lemma. $\square$

Finally, we establish the following $L^\infty -$estimate with weight for the heat equation.

Lemma A.4

For the heat equation $\partial _t F(t, x)-\epsilon \partial _x^2 F(t, x)=G(t, x), \ (t, x) \in {\mathbb {R}}^+\times {\mathbb {R}}^+$; with the boundary condition $F(t, x)|_{x=0}=0$ and initial data $F(t, x)|_{t=0}=F_0$. Then, it holds true

$$\begin{aligned} \Vert x\partial _x F\Vert _{L^\infty _0({\mathbb {R}}^+)}\le & {} C(\Vert F_0\Vert _{L^\infty _0({\mathbb {R}}^+)}+\Vert x\partial _x F_0\Vert _{L^\infty _0({\mathbb {R}}^+)}) \nonumber \\&\quad +\,C\int _0^t (\Vert G\Vert _{L^\infty _0({\mathbb {R}}^+)}+\Vert x\partial _x G\Vert _{L^\infty _0({\mathbb {R}}^+)}) d\tau , \end{aligned}$$

(A.7)

where C is a constant independent of the parameter $\epsilon$.

Proof

First of all, let us consider the heat equation

$$\begin{aligned} \partial _t H(t, x)-\epsilon \partial _x^2 H(t, x)=0,\quad (t, x) \in {\mathbb {R}}^+\times {\mathbb {R}}^+; \end{aligned}$$

(A.8)

with the initial data and boundary condition

$$\begin{aligned} H(t, x)|_{t=0}=H_0(x),\quad x \in {\mathbb {R}}^+; \quad H(t, x)|_{x=0}=0,\quad t \in {\mathbb {R}}^+. \end{aligned}$$

In order to transform the problem (A.8) into a problem in the whole space, let us define ${\widetilde{H}}(t, x)$ by

$$\begin{aligned} {\widetilde{H}}(t, x)= H(t, x), \ x>0;\quad {\widetilde{H}}(t, x)= -H(t, -x), \ x<0, \end{aligned}$$

and define the initial data ${\widetilde{H}}_0(x)$ by

$$\begin{aligned} {\widetilde{H}}_0(x)= H_0(x), \ x>0;\quad {\widetilde{H}}_0(x)= -H_0(-x), \ x<0. \end{aligned}$$

It is easy to justify that the function ${\widetilde{H}}(t, x)$ solves the following evolution equation

$$\begin{aligned} \partial _t {\widetilde{H}}(t, x)-\epsilon \partial _x^2 {\widetilde{H}}(t, x)=0,\quad (t, x) \in {\mathbb {R}}^+\times {\mathbb {R}}; \quad {\widetilde{H}}(t, x)|_{t=0}={\widetilde{H}}_0(x), \quad x \in {\mathbb {R}}.\nonumber \\ \end{aligned}$$

(A.9)

Define $S(t, x)=\frac{1}{\sqrt{4\pi \epsilon t}}e^{-\frac{|x|^2}{\sqrt{4\epsilon t}}}$, then the solution of evolution (A.9) can be expressed as

$$\begin{aligned} {\widetilde{H}}(t, x)=\int _{{\mathbb {R}}} {\widetilde{H}}_0(\xi )S(t, x-\xi )d\xi , \end{aligned}$$

(A.10)

which implies directly

$$\begin{aligned} x \partial _x {\widetilde{H}}(t, x)=\int _{{\mathbb {R}}} {\widetilde{H}}_0(\xi )x \partial _x S(t, x-\xi )d\xi . \end{aligned}$$

In view of the relation $x \partial _x S(t, x-\xi )=(x-\xi )\partial _x S(t, x-\xi )+\xi \partial _x S(t, x-\xi )$, we get

$$\begin{aligned} x \partial _x {\widetilde{H}}(t, x)=\int _{{\mathbb {R}}} {\widetilde{H}}_0(\xi )(x-\xi ) \partial _x S(t, x-\xi )d\xi +\int _{{\mathbb {R}}} {\widetilde{H}}_0(\xi )\xi \partial _x S(t, x-\xi )d\xi . \end{aligned}$$

Due to $\int _{{\mathbb {R}}}|(x-\xi ) \partial _x S(t, x-\xi )|d\xi \le C$, it follows

$$\begin{aligned} \Bigg |\int _{{\mathbb {R}}} {\widetilde{H}}_0(\xi )(x-\xi ) \partial _x S(t, x-\xi )d\xi \Bigg | \le C\Vert {\widetilde{H}}_0\Vert _{L^\infty _0({\mathbb {R}})}. \end{aligned}$$

Using the equality $\partial _x S(t, x-\xi )=-\partial _{\xi } S(t, x-\xi )$, the integration by part yields directly

$$\begin{aligned} \Bigg |\int _{{\mathbb {R}}}{\widetilde{H}}_0(\xi )\xi \partial _x S(t, x-\xi )d\xi \Bigg | \le C(\Vert {\widetilde{H}}_0\Vert _{L^\infty _0({\mathbb {R}})} +\Vert x \partial _x {\widetilde{H}}_0\Vert _{L^\infty _0({\mathbb {R}})}), \end{aligned}$$

and hence, we get

$$\begin{aligned} \begin{aligned} \Vert x \partial _x H (t, x)\Vert _{L^\infty _0({\mathbb {R}}^+)}&\le \Vert x \partial _x {\widetilde{H}}(t, x)\Vert _{L^\infty _0({\mathbb {R}})} \le C \Vert ({\widetilde{H}}_0, x \partial _x {\widetilde{H}}_0)\Vert _{L^\infty _0({\mathbb {R}})})\\&\le C (\Vert {H}_0\Vert _{L^\infty _0({\mathbb {R}}^+)} +\Vert x \partial _x {H}_0\Vert _{L^\infty _0({\mathbb {R}}^+)}). \end{aligned} \end{aligned}$$

This, along with representation (A.10) and the well-known Duhamel formula, we complete the proof of this lemma. $\square$

Appendix B: Almost equivalence of weighted norms

In this subsection we will use the quantity $h^\epsilon _m$ in weighted norm, $h^\epsilon$ and its derivatives in $L^\infty$ norm to control the quantities $Z_\tau ^{\alpha _1}h^\epsilon$ and $Z_\tau ^{\alpha _1}\psi ^\epsilon$ in weighted norm. To derive these estimates, we shall apply the Lemma A.1, which has been introduced previously in Appendix A.

Lemma B.1

Let the stream function $\psi ^\epsilon (t, x, y)$ satisfies $\partial _y \psi ^\epsilon =h^\epsilon , \partial _x \psi ^\epsilon = -g^\epsilon , \psi ^\epsilon |_{y=0}=0$. There exists a constant $\delta \in (0, 1)$, such that $h^\epsilon (t, x, y)+1\ge \delta , \forall (t, x, y)\in [0, T]\times \Omega$. Then, for $l \ge 1$ and $|\alpha _1|=m$, we have the following estimates:

$$\begin{aligned}&\Vert \frac{Z^{\alpha _1}_\tau \psi ^\epsilon }{h^\epsilon +1}\Vert _{L^2_{l-1}(\Omega )}\le \frac{2 \delta ^{-1}}{2l-1} \Vert h^\epsilon _m \Vert _{L^2_l(\Omega )}, \end{aligned}$$

(B.1)

$$\begin{aligned}&\Vert Z^{\alpha _1}_\tau h^\epsilon \Vert _{L^2_l}(\Omega ) \le \Vert h^\epsilon _m\Vert _{L^2_l(\Omega )} +\frac{2 \delta ^{-1}}{2l-1} \Vert \partial _y h^\epsilon \Vert _{L^\infty _1(\Omega )} \Vert h^\epsilon _m \Vert _{L^2_l(\Omega )}, \end{aligned}$$

(B.2)

$$\begin{aligned}&\Vert \frac{\partial _x Z^{\alpha _1}_\tau \psi ^\epsilon }{h^\epsilon +1}\Vert _{L^2_{l-1}(\Omega )} \le \frac{2\delta ^{-1}}{2l-1} \Vert \partial _x h^\epsilon _m\Vert _{L^2_{l}(\Omega )} +\frac{4\delta ^{-2}}{2l-1} \Vert \partial _x h^\epsilon \Vert _{L^\infty _0(\Omega )}\Vert h^\epsilon _m\Vert _{L^2_{l}(\Omega )}, \end{aligned}$$

(B.3)

$$\begin{aligned}&\Vert \partial _y Z^{\alpha _1}_\tau h^\epsilon \Vert _{L^2_l(\Omega )} \le \Vert \partial _y h^\epsilon _m \Vert _{L^2_l(\Omega )}+C_l \delta ^{-1}(\Vert \partial _y h^\epsilon \Vert _{L^\infty _0(\Omega )} \nonumber \\&\quad \quad +\Vert Z_2 \partial _y h^\epsilon \Vert _{L^\infty _1(\Omega )})\Vert h^\epsilon _m\Vert _{L^2_l(\Omega )}, \end{aligned}$$

(B.4)

where the constant $C_l$ depends only on l.

Proof

(i) By virtue of the definition $h^\epsilon _m=Z_\tau ^{\alpha _1}h^\epsilon -\frac{\partial _y h^\epsilon }{h^\epsilon +1} Z_\tau ^{\alpha _1} \psi ^\epsilon$, it is easy to obtain $h^\epsilon _m=(h^\epsilon +1)\partial _y(\frac{Z_\tau ^{\alpha _1} \psi ^\epsilon }{h^\epsilon +1})$. Integrating over [0, y] and applying the boundary condition $\psi ^\epsilon |_{y=0}=0$, we have

$$\begin{aligned} \frac{Z_\tau ^{\alpha _1} \psi ^\epsilon }{h^\epsilon +1}= \int _0^y \frac{h^\epsilon _m}{h^\epsilon +1}d\xi , \end{aligned}$$

(B.5)

and along with the Hardy inequality (A.1), yields directly

$$\begin{aligned} \Vert \frac{Z_\tau ^{\alpha _1} \psi ^\epsilon }{h^\epsilon +1} \Vert _{L^2_{l-1}(\Omega )}\le & {} \Bigg \Vert \int _0^y \frac{h^\epsilon _m}{h^\epsilon +1}d\xi \Bigg \Vert _{L^2_{l-1}(\Omega )} \le \frac{2}{2l-1}\Vert \frac{h^\epsilon _m}{h^\epsilon +1} \Vert _{L^2_l(\Omega )} \nonumber \\\le & {} \frac{2 \delta ^{-1}}{2l-1} \Vert h^\epsilon _m \Vert _{L^2_l(\Omega )}, \end{aligned}$$

(B.6)

where we have used the fact $h^\epsilon +1\ge \delta$ in the last inequality.

(ii) In view of the relation $Z_\tau ^{\alpha _1}h^\epsilon =h^\epsilon _m+\frac{\partial _y h^\epsilon }{h^\epsilon +1} Z_\tau ^{\alpha _1} \psi ^\epsilon$, we get

$$\begin{aligned} \Vert Z^{\alpha _1}_\tau h^\epsilon \Vert _{L^2_l(\Omega )} \le \Vert h^\epsilon _m\Vert _{L^2_l(\Omega )}+\Vert \partial _y h^\epsilon \Vert _{L^\infty _1(\Omega )} \Bigg \Vert \frac{ Z^{\alpha _1}_\tau \psi ^\epsilon }{h^\epsilon +1}\Bigg \Vert _{L^2_{l-1}(\Omega )}, \end{aligned}$$

which, together with estimate (B.6), yields directly

$$\begin{aligned} \Vert Z^{\alpha _1}_\tau h^\epsilon \Vert _{L^2_l(\Omega )} \le \Vert h^\epsilon _m\Vert _{L^2_l(\Omega )}+\frac{2 \delta ^{-1}}{2l-1} \Vert \partial _y h^\epsilon \Vert _{L^\infty _1(\Omega )} \Vert h^\epsilon _m \Vert _{L^2_l(\Omega )}. \end{aligned}$$

(B.7)

(iii) Differentiating the equality ${Z_\tau ^{\alpha _1} \psi ^\epsilon }=(h^\epsilon +1)\int _0^y \frac{h^\epsilon _m}{h^\epsilon +1}d\xi$ with respect to x variable, we find

$$\begin{aligned} \partial _x Z_\tau ^{\alpha _1} \psi ^\epsilon =\partial _x h^\epsilon \int _0^y \frac{h^\epsilon _m}{h^\epsilon +1} d\xi +(h^\epsilon +1)\int _0^y \frac{\partial _x h^\epsilon _m}{h^\epsilon +1}d\xi -(h^\epsilon +1)\int _0^y \frac{h^\epsilon _m \partial _x h^\epsilon }{(h^\epsilon +1)^2}d\xi , \end{aligned}$$

which, implies that

$$\begin{aligned} \frac{\partial _x Z_\tau ^{\alpha _1} \psi ^\epsilon }{h^\epsilon +1} =\frac{\partial _x h^\epsilon }{h^\epsilon +1}\int _0^y \frac{h^\epsilon _m}{h^\epsilon +1} d\xi +\int _0^y \frac{\partial _x h^\epsilon _m}{h^\epsilon +1}d\xi -\int _0^y \frac{h^\epsilon _m \partial _x h^\epsilon }{(h^\epsilon +1)^2}d\xi , \end{aligned}$$

and hence, we apply the Hardy inequality (A.1) and $h^\epsilon +1 \ge \delta$ to get

$$\begin{aligned} \begin{aligned} \Vert \frac{\partial _x Z^{\alpha _1}_\tau \psi ^\epsilon }{h^\epsilon +1}\Vert _{L^2_{l-1}(\Omega )}&\le \frac{4}{2l-1}\Vert \frac{\partial _x h^\epsilon }{h^\epsilon +1}\Vert _{L^\infty _0(\Omega )}\Vert \frac{h^\epsilon _m}{h^\epsilon +1}\Vert _{L^2_{l}(\Omega )} +\frac{2}{2l-1} \Vert \frac{\partial _x h^\epsilon _m}{h^\epsilon +1} \Vert _{L^2_{l}(\Omega )}\\&\le \frac{2\delta ^{-1}}{2l-1} \Vert \partial _x h^\epsilon _m\Vert _{L^2_{l}(\Omega )} +\frac{4\delta ^{-2}}{2l-1} \Vert \partial _x h^\epsilon \Vert _{L^\infty _0(\Omega )}\Vert h^\epsilon _m\Vert _{L^2_{l}(\Omega )}. \end{aligned}\nonumber \\ \end{aligned}$$

(B.8)

(iv)Differentiating the equality $Z_\tau ^{\alpha _1}h^\epsilon =h^\epsilon _m+\frac{\partial _y h^\epsilon }{h^\epsilon +1} Z_\tau ^{\alpha _1} \psi ^\epsilon$ with the y variable, it follows

$$\begin{aligned} \partial _y Z^{\alpha _1}_\tau h^\epsilon =\partial _y h^\epsilon _m +\partial _y^2 h^\epsilon \frac{Z^{\alpha _1}_\tau \psi ^\epsilon }{h^\epsilon +1}+\partial _y h^\epsilon \partial _y \left( \frac{Z^{\alpha _1}_\tau \psi ^\epsilon }{h^\epsilon +1}\right) , \end{aligned}$$

which, together with the relation (B.5), yields

$$\begin{aligned} \begin{aligned} \partial _y Z^{\alpha _1}_\tau h^\epsilon&=\partial _y h^\epsilon _m +\partial _y^2 h^\epsilon \int _0^y \frac{h^\epsilon _m}{h^\epsilon +1}d\xi +\eta _h h^\epsilon _m\\&=\partial _y h^\epsilon _m +Z_2 \partial _y h^\epsilon \frac{1}{\varphi (y)}\int _0^y \frac{h^\epsilon _m}{h^\epsilon +1}d\xi +\eta _h h^\epsilon _m, \end{aligned} \end{aligned}$$

where $\varphi (y)=\frac{y}{1+y}$ and $\eta _h=\frac{\partial _y h^\epsilon }{h^\epsilon +1}$. The application of Hardy inequality (A.1) and $h^\epsilon +1\ge \delta$ yields

$$\begin{aligned} \begin{aligned} \Vert \partial _y Z^{\alpha _1}_\tau h^\epsilon \Vert _{L^2_l(\Omega )}&\le \Vert \partial _y h^\epsilon _m \Vert _{L^2_l(\Omega )}+\frac{2\delta ^{-1}}{2l-1}\Vert Z_2 \partial _y h^\epsilon \Vert _{L^\infty _1(\Omega )} \Vert h^\epsilon _m \Vert _{L^2_{l}(\Omega )}\\&\quad +\delta ^{-1}\Vert \partial _y h^\epsilon \Vert _{L^\infty _0(\Omega )}\Vert h^\epsilon _m\Vert _{L^2_l(\Omega )}\\&\le \Vert \partial _y h^\epsilon _m \Vert _{L^2_l(\Omega )} +\frac{(2l+1) \delta ^{-1}}{2l-1}(\Vert \partial _y h^\epsilon \Vert _{L^\infty _0(\Omega )} \\&\qquad +\Vert Z_2 \partial _y h^\epsilon \Vert _{L^\infty _1(\Omega )})\Vert h^\epsilon _m\Vert _{L^2_l(\Omega )}. \end{aligned} \end{aligned}$$

(B.9)

Therefore, the estimates (B.6)–(B.9) imply the estimates (B.1)–(B.4). $\square$

Let us define

$$\begin{aligned} Y_{m,l}(t):=1+{{\Vert (\varrho ^\epsilon , u^\epsilon , h^\epsilon )(t)\Vert _{{\mathcal {H}}^m_l}^2}}+\Vert \partial _y(\varrho ^\epsilon , u^\epsilon , h^\epsilon )(t)\Vert _{{\mathcal {H}}^{m-1}_l}^2 +\Vert \partial _y \varrho ^\epsilon (t)\Vert _{{\mathcal {H}}^{1,\infty }_1}^2, \end{aligned}$$

(B.10)

and hence we will establish the following almost equivalent relation.

Lemma B.2

Let $(\varrho ^\epsilon , u^\epsilon , v^\epsilon , h^\epsilon , g^\epsilon )$ be sufficiently smooth solution, defined on $[0, T^\epsilon ]$, to the regularized MHD boundary layer equations (3.2)–(3.3). There exists a constant $\delta \in (0, 1)$, such that $h^\epsilon (t, x, y)+1\ge \delta , \forall (t, x, y)\in [0, T]\times \Omega$. Then, for $m \ge 4$ and $l \ge 1$, it holds true

$$\begin{aligned} \Theta _{m,l}(t) \le C\Vert (\rho _0, u_{10}, h_{10})\Vert _{\overline{{\mathcal {B}}}^m_l}^4 +C t \Vert (\rho _0, u_{10}, h_{10})\Vert _{\widehat{{\mathcal {B}}}^m_l}^4 +C_l \delta ^{-8}{\mathcal {N}}_{m,l}^{12}(t), \end{aligned}$$

(B.11)

and

$$\begin{aligned} {\mathcal {N}}_{m,l}(t) \le C\Vert (\rho _0, u_{10}, h_{10})\Vert _{\overline{{\mathcal {B}}}^m_l}^4 +C t \Vert (\rho _0, u_{10}, h_{10})\Vert _{\widehat{{\mathcal {B}}}^m_l}^4 +C\delta ^{-8} \Theta _{m,l}^{12}(t), \end{aligned}$$

(B.12)

where $\Theta _{m,l}(t)$ and ${\mathcal {N}}_{m,l}(t)$ are defined in (3.5) and (3.114) respectively.

Proof

By virtue of the definition $\varrho ^\epsilon _m=Z_\tau ^{\alpha _1} \varrho ^\epsilon -\frac{\partial _y \varrho ^\epsilon }{h^\epsilon +1}Z_\tau ^{\alpha _1}\psi ^\epsilon$ and the estimate (B.1), we find

$$\begin{aligned} \begin{aligned} \Vert Z_\tau ^{\alpha _1} \varrho ^\epsilon \Vert _{L^2_l(\Omega )}^2&\le \Vert \varrho ^\epsilon _m\Vert _{L^2_l(\Omega )}^2+\Vert \partial _y \varrho ^\epsilon \Vert _{L^\infty _1(\Omega )}^2\Vert \frac{Z_\tau ^{\alpha _1 }\psi ^\epsilon }{h^\epsilon +1}\Vert _{L^2_{l-1}(\Omega )}^2\\&\le \Vert \varrho ^\epsilon _m\Vert _{L^2_l(\Omega )}^2+C_l \delta ^{-2}\Vert \partial _y \varrho ^\epsilon \Vert _{L^\infty _1(\Omega )}^2\Vert h^\epsilon _m\Vert _{L^2_{l}(\Omega )}^2. \end{aligned} \end{aligned}$$

Similarly, we can obtain for $|\alpha _1|=m$ that

$$\begin{aligned} \Vert Z_\tau ^{\alpha _1}(u^\epsilon , h^\epsilon )\Vert _{L^2_l(\Omega )}^2 \le \Vert (u^\epsilon _m, h^\epsilon _m)\Vert _{L^2_l(\Omega )}^2+C_l \delta ^{-2}(1+\Vert \partial _y (u^\epsilon , h^\epsilon )\Vert _{L^\infty _1(\Omega )}^2)\Vert h^\epsilon _m\Vert _{L^2_{l}(\Omega )}^2. \end{aligned}$$

The combination of the above two estimates yields directly

$$\begin{aligned} \Vert Z_\tau ^{\alpha _1}(\varrho ^\epsilon , u^\epsilon , h^\epsilon )\Vert _{L^2_l(\Omega )}^2 \le C_l \delta ^{-2}(1+\Vert \partial _y (\varrho ^\epsilon , u^\epsilon , h^\epsilon )\Vert _{L^\infty _1(\Omega )}^2)\Vert (\varrho ^\epsilon _m, u^\epsilon _m, h^\epsilon _m)\Vert _{L^2_l(\Omega )}^2,\nonumber \\ \end{aligned}$$

(B.13)

and hence, we have for $m \ge 4, l \ge 1$

$$\begin{aligned} \begin{aligned}&\Vert Z_\tau ^{\alpha _1}(\varrho ^\epsilon , u^\epsilon , h^\epsilon )\Vert _{L^2_l(\Omega )}^2 \\&\quad \le C\Vert \partial _y(u^\epsilon , h^\epsilon )\Vert _{L^\infty _1(\Omega )}^4 +C\delta ^{-4}(1+\Vert (\varrho ^\epsilon _m, u^\epsilon _m, h^\epsilon _m)\Vert _{L^2_l(\Omega )}^4+\Vert \partial _y \varrho ^\epsilon \Vert _{L^\infty _1(\Omega )}^4)\\&\quad \le C\Vert (\rho _0, u_{10}, h_{10})\Vert _{\overline{{\mathcal {B}}}^m_l}^2 +C t \Vert (\rho _0, u_{10}, h_{10})\Vert _{\widehat{{\mathcal {B}}}^m_l}^2+C_l \delta ^{-4}(1+X_{m,l}^6(t)). \end{aligned}\qquad \end{aligned}$$

(B.14)

Due to the definition of $X_{m,l}(t)$ and $Y_{m,l}(t)$ in (3.56) and (B.10) respectively, we get from (B.14) that

$$\begin{aligned} Y_{m,l}(t)\le C\Vert (\rho _0, u_{10}, h_{10})\Vert _{\overline{{\mathcal {B}}}^m_l}^2 +C t \Vert (\rho _0, u_{10}, h_{10})\Vert _{\widehat{{\mathcal {B}}}^m_l}^2+C_l \delta ^{-4}(1+X_{m,l}^6(t)).\qquad \end{aligned}$$

(B.15)

On the other hand, by virtue of the definition of $\varrho ^\epsilon _m(t)$ and the estimate (B.1), we find

$$\begin{aligned} \begin{aligned} \Vert \varrho ^\epsilon _m(t)\Vert _{L^2_l(\Omega )}^2&\le \Vert Z_\tau ^{\alpha _1} \varrho ^\epsilon (t)\Vert _{L^2_l(\Omega )}^2 +\Vert \partial _y \varrho ^\epsilon (t)\Vert _{L^\infty _1(\Omega )}^2\Vert \frac{Z_\tau ^{\alpha _1 }\psi ^\epsilon }{h^\epsilon +1}(t)\Vert _{L^2_{l-1}(\Omega )}^2\\&\le \Vert Z_\tau ^{\alpha _1} \varrho ^\epsilon (t)\Vert _{L^2_l(\Omega )}^2+C_l\delta ^{-2}\Vert \partial _y \varrho ^\epsilon (t)\Vert _{L^\infty _1(\Omega )}^2\Vert Z_\tau ^{\alpha _1}h^\epsilon (t)\Vert _{L^2_{l}(\Omega )}^2, \end{aligned} \end{aligned}$$

and hence, we also have

$$\begin{aligned} \begin{aligned} \Vert (u^\epsilon _m, h^\epsilon _m)(t)\Vert _{L^2_l(\Omega )}^2&\le \Vert Z_\tau ^{\alpha _1}(u^\epsilon , h^\epsilon )(t)\Vert _{L^2_l(\Omega )}^2 \\&\quad +C\delta ^{-2}(1+\Vert \partial _y (u^\epsilon , h^\epsilon )(t)\Vert _{L^\infty _1(\Omega )}^2)\Vert Z_\tau ^{\alpha _1}h^\epsilon (t)\Vert _{L^2_{l}(\Omega )}^2. \end{aligned} \end{aligned}$$

Then, the combination of the above estimates yields directly

$$\begin{aligned}&\Vert (\varrho ^\epsilon _m, u^\epsilon _m, h^\epsilon _m)(t)\Vert _{L^2_l}^2 \le C\Vert (\rho _0, u_{10}, h_{10})\Vert _{\overline{{\mathcal {B}}}^m_l}^2 +C t \Vert (\rho _0, u_{10}, h_{10})\Vert _{\widehat{{\mathcal {B}}}^m_l}^2 \nonumber \\&\quad +C_l\delta ^{-4}(1+Y_{m,l}^6(t)), \end{aligned}$$

(B.16)

where $m \ge 4, l \ge 1$. According to the definition of $X_{m,l}(t)$ and $Y_{m,l}(t)$, we get from (B.16) that

$$\begin{aligned} X_{m,l}(t)\le C\Vert (\rho _0, u_{10}, h_{10})\Vert _{\overline{{\mathcal {B}}}^m_l}^2 +C t \Vert (\rho _0, u_{10}, h_{10})\Vert _{\widehat{{\mathcal {B}}}^m_l}^2+C_l \delta ^{-4}(1+Y_{m,l}^6(t)).\quad \end{aligned}$$

(B.17)

Next, by virtue of the definition $\varrho ^\epsilon _m(t)$ and estimate (3.89), we find

$$\begin{aligned} \begin{aligned} \Vert \partial _x Z_\tau ^{\alpha _1} \varrho ^\epsilon \Vert _{L^2_l(\Omega )}^2&\le \Vert \partial _x \varrho ^\epsilon _m\Vert _{L^2_l(\Omega )}^2+C_l \delta ^{-2}\Vert \partial _y \varrho ^\epsilon \Vert _{L^\infty _1(\Omega )}^2\Vert \partial _x h^\epsilon _m\Vert _{L^2_l(\Omega )}^2\\&\quad +C_l \delta ^{-4}(\Vert \partial _{xy} \varrho ^\epsilon \Vert _{L^\infty _1(\Omega )}^2 +\Vert \partial _{y} \varrho ^\epsilon \Vert _{L^\infty _1(\Omega )}^2\Vert \partial _{x} h^\epsilon \Vert _{L^\infty _1(\Omega )}^2)\Vert h^\epsilon _m\Vert _{L^2_l(\Omega )}^2\\&\le \Vert \partial _x \varrho ^\epsilon _m\Vert _{L^2_l(\Omega )}^2+C_l \delta ^{-2} X_{m,l}(t) \Vert \partial _x h^\epsilon _m\Vert _{L^2_l(\Omega )}^2 +C_l \delta ^{-4}(1+ X_{m,l}^3(t)). \end{aligned} \end{aligned}$$

Similarly, by routine checking, we may conclude that

$$\begin{aligned} \begin{aligned}&\Vert \partial _x Z_\tau ^{\alpha _1} (u^\epsilon , h^\epsilon )\Vert _{L^2_l(\Omega )}^2 \\&\quad \le C(\Vert (\rho _0, u_{10}, h_{10})\Vert _{\overline{{\mathcal {B}}}^m_l}^2 +t \Vert (\rho _0, u_{10}, h_{10})\Vert _{\widehat{{\mathcal {B}}}^m_l}^2) +C_l\delta ^{-8}(1+X_{m,l}^6(t))\\&\qquad \!+\!C_l \delta ^{-2}(1+\Vert (\rho _0, u_{10}, h_{10})\Vert _{\overline{{\mathcal {B}}}^m_l} +t \Vert (\rho _0, u_{10}, h_{10})\Vert _{\widehat{{\mathcal {B}}}^m_l}\\&\qquad +X_{m,l}^3(t) )\Vert \partial _x h^\epsilon _m\Vert _{L^2_l(\Omega )}^2\\&\qquad +\Vert \partial _x (u^\epsilon _m, h^\epsilon _m)\Vert _{L^2_l(\Omega )}^2. \end{aligned}\qquad \end{aligned}$$

for $m\ge 5, l \ge 1$, and hence it follows

$$\begin{aligned} \begin{aligned}&\epsilon \Vert \partial _x(\varrho ^\epsilon , u^\epsilon , h^\epsilon )\Vert _{{\mathcal {H}}^m_l}^2 \\&\quad \le C_l \delta ^{-2}(1+\Vert (\rho _0, u_{10}, h_{10})\Vert _{\overline{{\mathcal {B}}}^m_l} +t \Vert (\rho _0, u_{10}, h_{10})\Vert _{\widehat{{\mathcal {B}}}^m_l}+X_{m,l}^3(t) )D_x^{m,l}(t)\\&\qquad +C(\Vert (\rho _0, u_{10}, h_{10})\Vert _{\overline{{\mathcal {B}}}^m_l}^2 +t \Vert (\rho _0, u_{10}, h_{10})\Vert _{\widehat{{\mathcal {B}}}^m_l}^2)+C_l \delta ^{-8}(1+X_{m,l}^6(t)), \end{aligned} \end{aligned}$$

(B.18)

where $D_x^{m,l}(t)$ is defined in (3.115). By virtue of the definition $\varrho ^\epsilon _m(t)$ and estimate (3.86), we get

$$\begin{aligned} \begin{aligned} \Vert \partial _x \varrho ^\epsilon _m\Vert _{L^2_l(\Omega )}^2&\le \Vert \partial _x Z_\tau ^{\alpha _1} \varrho ^\epsilon \Vert _{L^2_l(\Omega )}^2 +C\delta ^{-2}\Vert \partial _y \varrho ^\epsilon \Vert _{L^\infty _1(\Omega )}^2\Vert \partial _x Z_\tau ^{\alpha _1}h^\epsilon \Vert _{L^2_{l}(\Omega )}^2\\&\quad +C\delta ^{-4}(\Vert \partial _{xy}\varrho ^\epsilon \Vert _{L^\infty _1(\Omega )}^2 +\Vert \partial _y \varrho ^\epsilon \Vert _{L^\infty _1(\Omega )}^2\Vert \partial _x h^\epsilon \Vert _{L^\infty _0(\Omega )}^2)\Vert Z_\tau ^{\alpha _1}h^\epsilon \Vert _{L^2_l(\Omega )}\\&\le \Vert \partial _x Z_\tau ^{\alpha _1} \varrho ^\epsilon \Vert _{L^2_l(\Omega )}^2 +C\delta ^{-2} Y_{m,l}(t) \Vert \partial _x Z_\tau ^{\alpha _1}h^\epsilon \Vert _{L^2_{l}}^2+C\delta ^{-4}(1+ Y_{m,l}^3(t)). \end{aligned} \end{aligned}$$

Similarly, by routine checking, we may conclude that

$$\begin{aligned} \begin{aligned}&\Vert \partial _x (u^\epsilon _m, h^\epsilon _m)\Vert _{L^2_l(\Omega )}^2 \\&\quad \le C\delta ^{-2}(1+\Vert (\rho _0, u_{10}, h_{10})\Vert _{\overline{{\mathcal {B}}}^m_l} +t \Vert (\rho _0, u_{10}, h_{10})\Vert _{\widehat{{\mathcal {B}}}^m_l}\\&\qquad +Y_{m,l}^3(t) ) \Vert \partial _x Z_\tau ^{\alpha _1}h^\epsilon \Vert _{L^2_{l}(\Omega )}^2\\&\quad +C(\Vert (\rho _0, u_{10}, h_{10})\Vert _{\overline{{\mathcal {B}}}^m_l}^2 +t \Vert (\rho _0, u_{10}, h_{10})\Vert _{\widehat{{\mathcal {B}}}^m_l}^2)+C\delta ^{-8}(1+Y_{m,l}^6(t))\\&\quad +\Vert \partial _x Z_\tau ^{\alpha _1} (u^\epsilon , h^\epsilon )\Vert _{L^2_l(\Omega )}^2, \end{aligned} \end{aligned}$$

and hence, it follows

$$\begin{aligned} \begin{aligned} D_x^{m,l}(t)&\le C\delta ^{-2}(1+\Vert (\rho _0, u_{10}, h_{10})\Vert _{\overline{{\mathcal {B}}}^m_l} +t \Vert (\rho _0, u_{10}, h_{10})\Vert _{\widehat{{\mathcal {B}}}^m_l}\\&\quad + Y_{m,l}^3(t) )\epsilon \Vert \partial _x(\varrho ^\epsilon , u^\epsilon , h^\epsilon )\Vert _{{\mathcal {H}}^m_l}^2\\&\quad +C(\Vert (\rho _0, u_{10}, h_{10})\Vert _{\overline{{\mathcal {B}}}^m_l}^2 +t \Vert (\rho _0, u_{10}, h_{10})\Vert _{\widehat{{\mathcal {B}}}^m_l}^2)+C\delta ^{-8}(1+Y_{m,l}^6(t)). \end{aligned} \end{aligned}$$

(B.19)

where $D_x^{m,l}(t)$ is defined in (3.115). Similarly, we can justify the estimates

$$\begin{aligned} \begin{aligned} \Vert \partial _y(\sqrt{\epsilon } \varrho ^\epsilon , \sqrt{\mu } u^\epsilon , \sqrt{\kappa } h^\epsilon )\Vert _{{\mathcal {H}}^m_l}^2&\le C(\Vert (\rho _0, u_{10}, h_{10})\Vert _{\overline{{\mathcal {B}}}^m_l}^4 +t \Vert (\rho _0, u_{10}, h_{10})\Vert _{\widehat{{\mathcal {B}}}^m_l}^4)\\&\quad +D_y^{m,l}(t)+C_l \delta ^{-8}(1+X_{m,l}^{12}(t)), \end{aligned} \end{aligned}$$

(B.20)

and

$$\begin{aligned} \begin{aligned} D_y^{m,l}(t)&\le C(\Vert (\rho _0, u_{10}, h_{10})\Vert _{\overline{{\mathcal {B}}}^m_l}^4 +t \Vert (\rho _0, u_{10}, h_{10})\Vert _{\widehat{{\mathcal {B}}}^m_l}^4)\\&\quad +\Vert \partial _y(\sqrt{\epsilon } \varrho ^\epsilon , \sqrt{\mu } u^\epsilon , \sqrt{\kappa } h^\epsilon )\Vert _{{\mathcal {H}}^m_l}^2+C\delta ^{-8}(1+Y_{m,l}^{12}(t)). \end{aligned} \end{aligned}$$

(B.21)

Therefore, the combination of estimates (B.15), (B.17), (B.18), (B.19), (B.20) and (B.21) can establish the estimates (B.11) and (B.12). $\square$