In this section, we start to derive the lower-order energy inequality in the a priori estimates for the transformed MHD problem. To this end, let \((\eta,u)\) be a solution of the transformed MHD problem with perturbed pressure q, such that
$$ \sqrt{\mathcal{G}_{1}(T)+ \sup_{0\leq\tau\leq T} \mathcal{E}^{H}(\tau)}\leq\delta\in(0,1) \quad\mbox{for some }T>0, $$
(3.1)
where δ is sufficiently small. It should be noted that the smallness depends on the known physical parameters in (1.4), and will be repeatedly used in what follows. Moreover, we assume that the solution \((\eta,{u},q)\) possesses proper regularity, so that the formal calculation makes sense. We remind the reader that in the calculations, we shall repeatedly use Cauchy-Schwarz’s inequality, Hölder’s inequality, the embedding inequalities (see [16], 4.12 Theorem)
$$ \|f\|_{L^{p}}\lesssim\| f\|_{1} \quad\mbox{for }2 \leq p\leq6\quad\mbox{and}\quad\| f\|_{L^{\infty}}\lesssim\| f\|_{2} , $$
(3.2)
and the interpolation inequalities (see [16], 5.2 Theorem)
$$ \|f\|_{j}\lesssim\|f\|_{0}^{1-\frac{j}{i}}\|f \|_{i}^{\frac{j}{i}}\leq C_{\epsilon}\|f\|_{0} + \epsilon\|f\|_{H^{i}} $$
(3.3)
for any \(0\leq j< i\) and any constant \(\epsilon>0\), where the constant \(C_{\epsilon}\) depends on the domain Ω and ϵ. In addition, we shall also frequently use the following two estimates:
$$ \|fg\|_{j}\lesssim\|f\|_{j}\|g \|_{\kappa(j)} \quad\mbox{for }j\geq0 $$
(3.4)
and
$$ \|f\|_{0}\lesssim\|\partial_{3} f \|_{0} \quad\mbox{for }f\in H^{1}_{0}, $$
(3.5)
where \(\kappa(j)=j\) for \(j\geq2\) and \(\kappa(j)=2\) for \(j\leq1\). We also introduce the following inequality, see (3.4) in [17]:
$$ \|f\|_{0}\leq h\|\bar{M}\cdot\nabla f \|_{0}/{ \pi}. $$
(3.6)
Before deriving the lower-order energy inequality defined on \((0,T]\), we first give some preliminary estimates, temporal derivative estimates, horizontal spatial estimates and Stokes estimates in sequence.
Preliminary estimates
In this subsection, we derive some preliminary estimates for \(\mathcal {A}\). To begin with, we give an expression of \(\mathcal{A}\). Using (2.15)1, we have
$$ \partial_{t} \det(I+\nabla\eta)=\sum _{1\leq i,j\leq3}\partial_{t}\partial_{j} \eta_{i} A_{ij}^{*}=\sum_{1\leq i,j\leq3} A_{ij}^{*}\partial_{j}u_{i} , $$
(3.7)
where \(A^{*}_{ij}\) is the algebraic complement minor of the \((i,j)\)th entry in the matrix \(I+\nabla\eta\). Recalling the definition of \(\mathcal{A}\), we see that
$$\mathcal{A}=\bigl(A^{*}_{ij}\bigr)_{3\times3}/\det(I+\nabla\eta). $$
Inserting this relation into (3.7), we get
$$\partial_{t} \det(I+\nabla\eta)= {\det(I+\nabla\eta)}\sum _{1\leq i,j,\leq3} \mathcal{A}_{ij}\partial_{j}u_{i}=0, $$
which, together with initial condition \(\det(I+\nabla\eta_{0})=1\), implies
$$\det(I+\nabla\eta)=1. $$
Thus we obtain
$$ \mathcal{A}=\bigl(A^{*}_{ij}\bigr)_{3\times3}. $$
(3.8)
Now, exploiting (2.15)1, (3.1), (3.4) and (3.8), we easily see that
$$\begin{aligned}& \|\mathcal{A}\|_{j} \lesssim1 +\|\eta \|_{j+1}\bigl(1+\| \eta\|_{j+1}\bigr)\lesssim1 \quad\mbox{for }0\leq j \leq6, \\& \|\nabla\mathcal{A} \|_{j}\leq\| \eta\|_{{j+2}}\bigl(1+\| \eta\| _{j+1}\bigr) \lesssim\| \eta\|_{{j+2}} \quad\mbox{for }0\leq j \leq5. \end{aligned}$$
(3.9)
Similarly, we further deduce that
$$\big\| \partial_{t}^{i}\mathcal{A}\big\| _{j} \lesssim\sum _{k=0}^{i-1} \big\| \partial_{t}^{k} \nabla u\big\| _{j} \quad\mbox{for any } 1\leq i\leq4\mbox{ and } 0\leq j \leq8-2i. $$
Letting \(\tilde{\mathcal{A}}:=\mathcal{A}-I\), we next bound \(\tilde {\mathcal{A}}\). To this end, we assume that δ is so small that the following expansion holds:
$$\mathcal{A}^{T}=I-\nabla\eta+(\nabla\eta)^{2}\sum _{i=0}^{\infty}(-\nabla\eta)^{i}=I-\nabla \eta+(\nabla\eta)^{2}\mathcal{A}^{T}, $$
whence
$$\tilde{\mathcal{A}}^{T} =(\nabla\eta)^{2} \mathcal{A}^{T}-\nabla\eta. $$
Using (3.1), (3.4) and (3.9), we find that
$$\|\tilde{\mathcal{A}}\|_{j} \lesssim\|\nabla\eta\|_{j} \quad\mbox{for }0\leq j\leq6. $$
Temporal derivative estimates
In this subsection, we try to control temporal derivatives. For this purpose, we apply \(\partial_{t}^{j}\) to (2.15) to get
$$ \left\{\textstyle\begin{array}{l} \partial_{t}^{j+1}\eta=\partial_{t}^{j} u, \\ {\rho}\partial_{t}^{j+1} u-\mu\Delta_{\mathcal{A}} \partial _{t}^{j}u+\nabla_{\mathcal{A}} \partial_{t}^{j} q \\ \quad=\lambda_{0} \partial_{t}^{j} \bar{M}\cdot\nabla(\bar{M}\cdot\nabla \zeta)+ 2\rho\omega\partial_{t}^{j} (u_{2}e_{1}-u_{1}e_{2}) +\mu N^{t,j}_{u}+N^{t,j}_{q},\\ \operatorname {div}_{\mathcal{A}} \partial_{t}^{j} u =\operatorname{div} D_{u}^{t,j} , \end{array}\displaystyle \right. $$
(3.10)
where
$$\begin{aligned}& N^{t,j}_{u}:= \sum_{0\leq m< j, 0\leq n\leq j} \partial_{t}^{j-m-n }\mathcal{A}_{il} \partial_{l}\bigl(\partial_{t}^{n}\mathcal {A}_{ik}\partial_{t}^{m} \partial_{k}u \bigr), \\& N^{t,j}_{q}:=-\sum_{0\leq l< j}\bigl( \partial^{j-l}_{t}\mathcal{A}_{ik} \partial^{l}_{t} \partial_{k} q \bigr)_{3\times1}, \\& D^{t,j}_{u}:= \biggl(-\sum _{0\leq l< j}C_{j}^{j-l}\partial_{t}^{j-l} \mathcal{A}_{ki}\partial_{t}^{l} u_{k} \biggr)_{3\times1}, \\& C_{j}^{j-l}\mbox{ denotes the number of }(j-l) \mbox{-combinations from a given set }S\mbox{ of }j \mbox{ elements}, \end{aligned}$$
(3.11)
and we have used relation (2.7) in (3.11). Then from (3.10) we show the following estimates:
Lemma 3.1
It holds that for
\(j=0\)
and 1,
$$\begin{aligned}& \frac{\mathrm{d}}{\mathrm{d}t} \bigl(\big\| \sqrt{{\rho}} \partial_{t}^{j} u\big\| _{0}^{2}+\lambda_{0} \big\| \bar{M}\cdot\nabla \partial_{t}^{j} \eta\big\| ^{2}_{0} \bigr) +\mu\big\| \nabla_{\mathcal{A}}\partial_{t}^{j}u \big\| ^{2}_{0}\lesssim\sqrt{\mathcal{E}^{H}} \mathcal{D}^{L}, \end{aligned}$$
(3.12)
$$\begin{aligned}& \frac{\mathrm{d}}{\mathrm{d}t} \| \nabla _{\mathcal{A}} u_{t} \|_{0}^{2} +\frac {1}{\mu}\| \sqrt{{\rho}} u_{tt}\|^{2}_{0}\lesssim\| u \|_{2}^{2}+\| u_{t}\|_{0}^{2}+ \sqrt{\mathcal{E}^{H}} \mathcal{D}^{L}. \end{aligned}$$
(3.13)
Proof
(1) We only prove (3.12) for \(j=1\), since the derivation of the case \(j=0\) is similar. Multiplying (3.10)2 with \(j=1\) by \(u_{t}\), integrating (by parts) the resulting equality over Ω, and using (3.10)1, we get
$$ \begin{aligned}[b] &\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \bigl(\|\sqrt{{\rho}} u_{t}\| _{0}^{2}+ \lambda_{0} \|\bar{M} \cdot\eta_{t}\|^{2}_{0} \bigr) +\mu\| \nabla_{\mathcal{A}}u_{t}\|^{2}_{0} \\ &\quad = 2\rho\omega \int\partial_{t} (u_{2}e_{1}-u_{1}e_{2}) \cdot u_{t} \,\mathrm{d}y + \int q_{t} \operatorname{div}_{\mathcal{A}}u_{t}\,\mathrm{d}y \\ &\qquad{} + \mu \int N^{t,1}_{u}\cdot u_{t}\,\mathrm{d}y+ \int N^{t,1}_{q} \cdot u_{t}\,\mathrm{d}y\\ &\quad:=\sum _{k=1}^{4}I^{L}_{k}. \end{aligned} $$
(3.14)
The last three integrals \(I_{2}^{L},\ldots,I^{L}_{4}\) can be estimated as follows:
$$\begin{aligned}& \begin{aligned}[b] I^{L}_{2}&:= - \int\nabla q_{t} \cdot D^{t,1}_{u}\,\mathrm{d}y \lesssim\|\nabla q_{t}\|_{0}\|\mathcal{A}_{t} \|_{L^{4}}\|u\|_{L^{4}}\lesssim\| \nabla q_{t} \|_{0}\|\mathcal{A}_{t}\|_{1}\|u\|_{1}\\ & \lesssim\sqrt{\mathcal{E}^{H}} \mathcal{D}^{L}, \end{aligned} \end{aligned}$$
(3.15)
$$\begin{aligned}& I^{L}_{3}\lesssim \| \mathcal{A} \|_{2} \|\mathcal{A}_{t}\|_{2}\| u \|_{2}\|u_{t}\|_{0}\lesssim\sqrt{\mathcal {E}^{H}}\mathcal{D}^{L}, \end{aligned}$$
(3.16)
$$\begin{aligned}& I^{L}_{4}\lesssim\|\mathcal{A}_{t} \|_{0}\|\nabla q\| _{L^{\infty}}\|u_{t} \|_{0}\lesssim\|\mathcal{A}_{t}\|_{0}\|\nabla q \| _{2}\|u_{t}\|_{0}\lesssim\sqrt{ \mathcal{E}^{H}} \mathcal{D}^{L}, \end{aligned}$$
(3.17)
where we have used (3.10)3 in (3.15). Consequently, the desired estimate (3.12) follows from (3.15)-(3.17).
(2) Now we turn to the proof of (3.13). Multiplying (3.10)2 with \(j=1\) by \(u_{tt}\), integrating (by parts) the resulting equality over Ω, and using (3.10)1, we conclude
$$ \begin{aligned}[b] &\frac{\mu}{2}\frac{\mathrm{d}}{\mathrm{d}t}\| \nabla_{\mathcal{A}} u_{t}\| ^{2}_{0} +\| \sqrt{{\rho}} u_{tt} \|_{0}^{2}\\&\quad = \int \bigl(2\rho\omega \partial_{t} (u_{2}e_{1}-u_{1}e_{2}) \cdot u_{tt} + \lambda_{0} \bar{M}\cdot\nabla(\bar{M}\cdot \nabla u)\cdot u_{tt}\bigr)\,\mathrm{d}y \\ &\qquad{} + \int q_{t} \operatorname{div}_{\mathcal{A}} u_{tt}\,\mathrm{d}y+ \mu \int N^{t,1}_{u}\cdot u_{tt}\,\mathrm{d}y+ \int N^{t,1}_{q} \cdot u_{tt} \,\mathrm{d}y \\ &\qquad{}+{\mu} \int\nabla_{\mathcal{A}}u_{t}: \nabla_{\mathcal{A}_{t}} u_{t}\,\mathrm{d}y\\ &\quad:=\sum_{k=1}^{5}J^{H}_{k}. \end{aligned} $$
(3.18)
On the other hand, the five integrals \(J^{H}_{1},\ldots,J^{H}_{5}\) can be bounded as follows:
$$\begin{aligned}& J^{H}_{1}\lesssim\bigl(\|u_{t} \|_{0}+\| u\|_{2}\bigr)\| u_{tt} \|_{0}, \end{aligned}$$
(3.19)
$$\begin{aligned}& \begin{aligned}[b] J^{H}_{2}&= - \int\nabla q_{t} \cdot D_{u}^{t,2} \,\mathrm{d}x \lesssim\| \nabla q_{t} \|_{0}\bigl( \|\mathcal{A}_{tt} \|_{0}\|u\|_{2}+ \|\mathcal{A}_{t}\|_{2} \|u_{t}\|_{0}\bigr) \\&\lesssim\sqrt{ {\mathcal{E}^{H}}} \mathcal{D}^{L} , \end{aligned} \end{aligned}$$
(3.20)
$$\begin{aligned}& J^{H}_{3}\lesssim\|\mathcal{A}\|_{2} \| \mathcal{A}_{t}\|_{2}\|u\|_{2} \| u_{tt} \|_{0}\lesssim\sqrt{\mathcal{E}^{H}} \mathcal{D}^{L}, \end{aligned}$$
(3.21)
$$\begin{aligned}& J^{H}_{4}\lesssim\|\mathcal{A}_{t}\|_{0} \|\nabla q\|_{2} \| u_{tt} \|_{0} \lesssim\sqrt{ \mathcal{E}^{H}}\mathcal{D}^{L}, \end{aligned}$$
(3.22)
$$\begin{aligned}& J^{H}_{5}\lesssim\|\mathcal{A}\|_{2} \|\mathcal{A}_{t}\|_{2}\| u_{t}\|_{1}^{2} \lesssim\sqrt{\mathcal{E}^{H}} \mathcal{D}^{L}. \end{aligned}$$
(3.23)
Thus, substituting (3.19)-(3.23) into (3.18) and using Cauchy-Schwarz’s inequality, we immediately get (3.13). □
Horizontal spatial estimates
In this subsection, we establish the estimates of horizontal spatial derivatives. For this purpose, we rewrite (2.15) as the following non-homogeneous linear form:
$$ \left\{\textstyle\begin{array}{l} \eta_{t}= u, \\ {\rho} u_{t}-\mu\Delta u+\nabla q-\lambda_{0} \bar{M}\cdot\nabla (\bar{M}\cdot\nabla\eta)=2\rho\omega (u_{2}e_{1}-u_{1}e_{2}) + \mu N^{h}_{u}+N^{h}_{q},\\ \operatorname {div}u =D^{h}_{u}, \end{array}\displaystyle \right. $$
(3.24)
where
$$\begin{gathered}N^{h}_{u}:= \partial_{l}\bigl[(\tilde{\mathcal{A}}_{jl} \tilde{ \mathcal{A}}_{jk}+ \tilde{\mathcal{A}}_{lk}+\tilde{ \mathcal{A}}_{kl})\partial_{k}u_{i} \bigr]_{3\times1}, \\ N^{h}_{q}:=-(\tilde{\mathcal{A}}_{ik} \partial_{k} q)_{3\times1} \quad \mbox{and}\quad D^{h}_{u}:= \tilde{\mathcal{A}}_{lk}\partial_{k} u_{l}. \end{gathered} $$
Then we have the following estimate on horizontal spatial derivatives of η.
Lemma 3.2
It holds that
$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{d}t} \biggl( \int{\rho}\partial_{h}^{j} \eta\cdot \partial_{h}^{j} u\,\mathrm{d}y +\frac{\mu}{2}\big\| \nabla \partial_{h}^{j}\eta\big\| _{0}^{2} \biggr) +\lambda_{0} \big\| \bar{M}\cdot\nabla\partial_{h}^{j} \eta\big\| _{0}^{2} \\ \quad \lesssim\sqrt{\mathcal{E}^{H}} \mathcal{D}^{L} +\|u \| _{j,0}^{2},\quad0\leq j\leq3. \end{gathered} $$
Proof
We only show the case \(j=3\); the remaining three cases can be verified similarly. Applying \(\partial_{h}^{3}\) to (3.24)2, multiplying the resulting equation by \(\partial^{3}_{h}\eta\), and then using (3.24)1, we get
$$\begin{gathered} {\rho} \partial_{t}\bigl( \partial_{h}^{3}\eta\cdot\partial_{h}^{3} u \bigr)-\bigl(\mu\Delta\partial_{h}^{3} \eta_{t}+\lambda_{0} \bar{M}\cdot\nabla\bigl(\bar{M}\cdot \nabla\partial_{h}^{3}\eta\bigr)\bigr)\cdot \partial_{h}^{3}\eta \\ \quad= \bigl( 2\rho\omega\partial_{h}^{3} (u_{2}e_{1}-u_{1}e_{2}) +\mu \partial_{h}^{3}N^{h}_{u}+\partial _{h}^{3}N^{h}_{q}-\nabla \partial_{h}^{3} q \bigr)\cdot\partial_{h}^{3} \eta+{\rho}\big|\partial_{h}^{3} u\big|^{2}. \end{gathered} $$
If we integrate (by parts) the above identity over Ω, we obtain
$$ \begin{aligned}[b] &\frac{\mathrm{d}}{\mathrm{d}t} \biggl( \int{\rho} \partial_{h}^{3} \eta\cdot \partial_{h}^{3} u \,\mathrm{d}y+\frac{\mu}{2}\big\| \nabla \partial_{h}^{3} \eta\big\| _{0}^{2} \biggr) +\lambda_{0} \big\| \bar{M}\cdot\nabla\partial_{h}^{3} \eta\big\| _{0}^{2} \\ &\quad= 2\rho\omega \int\partial_{h}^{3} (u_{2}e_{1}-u_{1}e_{2}) \cdot\partial_{h}^{3}\eta\,\mathrm{d}y+ \int\bigl(\mu\partial_{h}^{3}N^{h}_{u}+ \partial_{h}^{3}N^{h}_{q}\bigr)\cdot \partial_{h}^{3}\eta\,\mathrm{d}y \\ &\quad\quad{}+ \int\partial_{h}^{3} q\operatorname{div}\partial^{3}_{h} \eta\,\mathrm{d}y +\big\| \sqrt{{\rho}}\partial_{h}^{3} u \big\| ^{2}_{0}\\&\quad \lesssim\sum_{k=1}^{3}K^{L}_{k}+ \| u\|^{2}_{3,0}, \end{aligned} $$
(3.25)
where the first three integrals on the right-hand side of the first equality in (3.25) are denoted by \(K^{L}_{1}\), \(K^{L}_{2}\) and \(K^{L}_{3}\), respectively.
We have the boundedness
$$ K^{L}_{1} \leq \big\| \partial_{h}^{3} u\big\| _{0}\big\| \partial_{h}^{3} \eta\big\| _{0}. $$
(3.26)
Noting that
$$\begin{aligned} \big\| \mu\partial_{h}^{3}N^{h}_{u}+ \partial_{h}^{3}N^{h}_{q} \big\| _{0} & \lesssim\|\tilde {\mathcal {A}}\| _{4}\|u\|_{3}+\|\tilde {\mathcal {A}}\|_{2}\|u\|_{5} +\|\tilde {\mathcal {A}}\|_{2}\|\nabla q \|_{3}+\|\tilde {\mathcal {A}}\|_{4}\| \nabla q\|_{1} \\ &\lesssim\sqrt{\mathcal{E}^{H}\mathcal{D}^{L}}, \end{aligned} $$
we find that
$$ K^{L}_{2} \lesssim\big\| \mu\partial_{h}^{3}N^{h}_{u}+ \partial_{h}^{3}N^{h}_{q} \big\| _{0}\big\| \partial_{h}^{3}\eta\big\| _{0} \lesssim\sqrt{\mathcal{E}^{H}} \mathcal{D}^{L}. $$
(3.27)
Next we estimate the third integral \(K_{3}^{L}\). To start with, we analyze the property of divη. Since
$$\det(I+\nabla\eta)=1, $$
we have by Sarrus’ rule
$$\begin{aligned} \operatorname{div}\eta={}&\partial_{1} \eta_{2}\partial_{2} \eta_{1}+ \partial_{2} \eta_{3}\partial_{3}\eta_{2}+\partial_{3} \eta_{1}\partial_{1}\eta_{3} -\partial_{1} \eta_{1}\partial_{2} \eta_{2}- \partial_{1} \eta_{1}\partial_{3}\eta_{3}-\partial_{2} \eta_{2}\partial_{3}\eta_{3} \\ &+ \partial_{1}\eta_{1}(\partial_{2} \eta_{3}\partial_{3}\eta_{2} -\partial _{2}\eta_{2}\partial_{3}\eta_{3})+ \partial_{2}\eta_{1}(\partial_{1}\eta_{2} \partial_{3}\eta_{3} -\partial_{1} \eta_{3}\partial_{3}\eta_{2}) \\&+\partial_{3} \eta_{1}(\partial_{1}\eta_{3}\partial_{2} \eta_{2} -\partial_{1}\eta_{2}\partial_{2} \eta_{3}). \end{aligned} $$
Multiplying the above identity by a smooth test function ϕ, and then integrating (by parts) the resulting identity over Ω, we derive that
$$\int\phi\operatorname{div}\eta\,\mathrm{d}y=- \int\nabla\phi\cdot\psi\,\mathrm{d}y, $$
where
$$\psi:=\left ( \textstyle\begin{array}{c} \eta_{1}(\partial_{2}\eta_{2}+\partial_{3}\eta_{3} )+ \eta_{1}(\partial_{2}\eta_{3} \partial_{3}\eta_{2} - \partial_{2}\eta_{2}\partial_{3}\eta_{3}) \\ \eta_{2}\partial_{3}\eta_{3}-\eta_{1}\partial_{1}\eta_{2}+ \eta_{1}( \partial_{1}\eta_{2} \partial_{3}\eta_{3} -\partial_{1}\eta_{3}\partial_{3}\eta_{2})\\ -\eta_{1}\partial_{1}\eta_{3} -\eta_{2}\partial_{2}\eta_{3}+\eta_{1}(\partial_{1}\eta_{3} \partial_{2}\eta_{2} -\partial_{1}\eta_{2}\partial_{2}\eta_{3}) \end{array}\displaystyle \right ). $$
This means that
$$\operatorname{div}\eta=\operatorname{div}\psi. $$
Thus, it follows immediately that
$$ \begin{aligned}[b] K^{L}_{3} &= - \int\partial_{h}^{3} \nabla q\cdot \partial_{h}^{3}\psi\,\mathrm{d}y=- \int\partial_{h} \nabla q\cdot\partial_{h}^{5} \psi\,\mathrm{d}y \\ &\lesssim\|\nabla q\|_{1}\big\| \partial_{h}^{5}\psi \big\| _{0} \lesssim\| \eta\| _{6}\|\nabla q\|_{1}\| \eta\|_{3} \lesssim\sqrt{\mathcal{E}^{H}} \mathcal{D}^{L}. \end{aligned} $$
(3.28)
Now, substituting (3.26), (3.27) and (3.28) into (3.25), we immediately obtain the desired estimate for the case \(j=3\). □
Similarly, we also establish the following estimates of horizontal spatial derivatives of u:
Lemma 3.3
We have
$$\frac{\mathrm{d}}{\mathrm{d}t} \bigl(\big\| \sqrt{ {\rho} } \partial_{h}^{j} u\big\| ^{2}_{0}+ \lambda_{0} \big\| \bar{M}\cdot\nabla\partial_{h}^{j} \eta\big\| _{0}^{2} \bigr) + \mu\big\| \nabla\partial_{h}^{j} u\big\| _{0}^{2} \lesssim\sqrt{\mathcal{E}^{H}} \mathcal{D}^{L}, \quad j=1,2,3. $$
Proof
We only prove the case \(j=3\); the remaining two cases can be shown similarly. Applying \(\partial_{h}^{3}\) to (3.24)2, and multiplying the resulting equality by \(\partial^{3}_{h} u\), we make use of (3.24)1 to get
$$\begin{gathered} {{\rho}} \partial_{h}^{3}u_{t} \cdot\partial_{h}^{3}u- \mu\Delta\partial _{h}^{3} u \cdot\partial_{h}^{3} u- \lambda_{0} \bar{M}\cdot\nabla\bigl(\bar{M}\cdot\nabla \partial_{h}^{3}\eta\bigr)\cdot\partial_{h}^{3} \eta_{t} \\ \quad=\bigl( 2\rho\omega\partial_{h}^{3}(u_{2}e_{1}-u_{1}e_{2}) +\mu\partial_{h}^{3}N^{h}_{u}+ \partial_{h}^{3}N^{h}_{q} -\nabla \partial_{h}^{3} q\bigr)\cdot\partial_{h}^{3}u. \end{gathered} $$
Integrating (by parts) the above identity over Ω, we have
$$ \begin{aligned}[b] &\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \biggl( \int{\rho} \big|\partial_{h}^{3} u\big|^{2} \,\mathrm{d}y+\lambda_{0} \big\| \bar{M}\cdot\nabla\partial_{h}^{3} \eta\big\| _{0}^{2} \biggr)+\frac{\mu}{2}\big\| \nabla \partial_{h}^{3} u\big\| _{0}^{2} \\ &= \int\bigl(\mu\partial_{h}^{3}N^{h}_{u}+ \partial_{h}^{3}N^{h}_{q}\bigr)\cdot \partial_{h}^{3} u\,\mathrm{d}y + \int\partial_{h}^{3} q\operatorname{div}\partial^{3}_{h} u\,\mathrm{d}y =: M^{L}_{1}+M^{L}_{2}. \end{aligned} $$
(3.29)
On the other hand, similarly to (3.27) and (3.28), the two integrals \(M^{L}_{1}\) and \(M^{L}_{2}\) can be estimated as follows:
$$\begin{aligned}& M^{L}_{1} \lesssim\big\| \mu\partial_{h}^{3}N^{h}_{u}+ \partial_{h}^{3}N^{h}_{q} \big\| _{0}\big\| \partial_{h}^{3}u\big\| _{0}\lesssim \sqrt{\mathcal{E}^{H}} \mathcal{D}^{L}, \end{aligned}$$
(3.30)
$$\begin{aligned}& M^{L}_{2} =- \int\partial_{h}^{4} q \partial^{2}_{h}D^{h}_{u} \,\mathrm{d}y\lesssim\|\nabla q\|_{3}\|\tilde {\mathcal {A}}\|_{2}\|u \|_{3}\lesssim\sqrt{\mathcal{E}^{H}} \mathcal{D}^{L}, \end{aligned}$$
(3.31)
where we have used (3.24)3 in (3.31). Consequently, putting the above two estimates into (3.29), we obtain Lemma 3.3 for the case \(j=3\). □
Stokes problem and stability condition
In this subsection, we use the regularity theory of the Stokes problem to derive more estimates of \((\eta,u)\). To this end, we rewrite (3.24)2 and (3.24)3 as the following Stokes problem:
$$ \left\{\textstyle\begin{array}{l} -\Delta w+\nabla q\\ \quad = \lambda_{0} \bar{M}\cdot\nabla(\bar{M}\cdot \nabla\eta)-\lambda_{0} m_{3}^{2} \Delta\eta+ 2\rho\omega (u_{2}e_{1}-u_{1}e_{2})-{\rho} u_{t}+ \mu N^{h}_{u}+ N^{h}_{q},\\ \operatorname {div}w =\mu D^{h}_{u}+ \lambda_{0} m^{2}_{3}\operatorname{div}\eta, \end{array}\displaystyle \right. $$
(3.32)
coupled with boundary condition
$$ \omega|_{\partial\Omega}=0, $$
(3.33)
where \(\omega=\lambda_{0} m^{2}_{3} \eta+\mu u\).
Now, applying \(\partial^{k}_{h}\) to (3.32) and (3.33), we get
$$ \left\{\textstyle\begin{array}{l} -\Delta\partial^{k}_{h} w+\nabla\partial^{k}_{h}q\\ \quad =\partial ^{k}_{h}(\lambda_{0} \bar{M}\cdot\nabla(\bar{M}\cdot\nabla\eta )-\lambda_{0} m_{3}^{2} \Delta\eta + 2\rho\omega (u_{2}e_{1}-u_{1}e_{2}))-{\rho} \partial^{k}_{h} u_{t}+ \mu\partial ^{k}_{h}N^{h}_{u}+N^{h}_{q},\\ \operatorname {div}\partial^{k}_{h}w =\mu\partial^{k}_{h}D^{h}_{u}+\lambda_{0} m^{2}_{3}\operatorname{div}\partial^{k}_{h} \eta,\\ \partial^{k}_{h}\omega|_{\partial\Omega}=0. \end{array}\displaystyle \right. $$
Then we apply the classical regularity theory to the Stokes problem as in [18], Proposition 2.3, to deduce that
$$ \|\omega\|_{k,i-k+2}^{2}+\|\nabla q \|_{k,i-k}^{2} \lesssim\| \nabla\eta\|_{k+1,i-k}^{2} +\big\| ( u, u_{t})\big\| _{k,i-k}^{2} +S_{k ,i}^{\omega}, $$
(3.34)
where
$$S_{k,i}^{\omega}:=\big\| \bigl(N^{h}_{u}, N^{h}_{q}\bigr)\big\| _{k,i-k}^{2} +\big\| \bigl(D^{h}_{u} ,\operatorname{div}\eta\bigr) \big\| _{k,i-k+1}^{2}. $$
In addition, applying \(\partial_{t}^{k}\) to (3.24)2-(3.24)3, we see that
$$\left\{\textstyle\begin{array}{l} -\mu\Delta\partial_{t}^{k} u+\nabla\partial_{t}^{k} q\\ \quad= \lambda_{0} \bar{M}\cdot \partial_{t}^{k} \nabla(\bar{M}\cdot\nabla\eta) + 2\rho\omega\partial_{t}^{k} (u_{2}e_{1}-u_{1}e_{2})-{\rho} \partial_{t}^{k+1}u +\mu\partial_{t}^{k}N^{h}_{u}+ \partial_{t}^{k}N^{h}_{q}, \\ \operatorname {div}\partial_{t}^{k}u = \partial_{t}^{k}D^{h}_{u}, \\ \partial_{t}^{k}u|_{\partial\Omega}=0 . \end{array}\displaystyle \right. $$
Hence, we apply again the classical regularity theory to the Stokes problem to get
$$ \big\| \partial_{t}^{k} u\big\| _{i-2k+2}^{2}+ \big\| \nabla\partial_{t}^{k} q\big\| _{i-2k}^{2} \lesssim\big\| \partial^{k}_{t}\bigl( \nabla^{2} \eta,u,u_{t}\bigr)\big\| _{i-2k}^{2} +S_{k,i}^{u}, $$
(3.35)
where \(S_{k,i}^{u}: =\|\partial_{t}^{k} ( N^{h}_{u}, N^{h}_{q})\|_{i-2k}^{2}+\| \partial_{t}^{k} D^{h}_{u}\|_{i-2k+1}^{2}\). As a result of (3.34) and (3.35), one has the following estimates.
Lemma 3.4
We have
$$\begin{aligned}& \frac{\mathrm{d}}{\mathrm{d}t} \|\eta\| _{3,*}^{2}+c\bigl(\big\| (\eta,u)\big\| _{3}^{2} +\| \nabla q\|_{1}^{2} \bigr) \lesssim\big\| (u,u_{t}) \big\| _{1}^{2}+\| \eta \|_{2,1}^{2} +{\mathcal{E}^{H} } { { \mathcal{D}}^{L}}, \end{aligned}$$
(3.36)
$$\begin{aligned}& \| u\|_{3}^{2}+\|\nabla q \|_{1}^{2} \lesssim\| \eta\|_{3}^{2} +\big\| (u,u_{t})\big\| _{1}^{2}+ {\mathcal{E}^{H}}E^{L}, \end{aligned}$$
(3.37)
$$\begin{aligned}& \|u_{t}\|_{2}^{2}+\|\nabla q_{t}\|_{0}^{2} \lesssim\| u \|_{2}^{2}+\big\| (u_{t},u_{tt}) \big\| _{0}^{2}+ {\mathcal{E}^{H}}\mathcal{D}^{L}, \end{aligned}$$
(3.38)
where
\(E^{L}:=\mathcal{E}^{L}-\|\nabla\eta\|_{3,0}^{2}\)
and
\(\|\eta\| _{3,*}\)
is equivalent to
\(\|\eta\|_{3}\).
Proof
Noting that, by virtue of (2.15)1,
$$\|\omega\|_{ k,i- k+2}^{2} =\big\| \bigl( \lambda_{0} m^{2}_{3} \eta,\mu u\bigr) \big\| _{{ k,i- k+2} }^{2} +\frac{\lambda_{0} m^{2}_{3}\mu}{2}\frac{\mathrm{d}}{\mathrm {d}t} \|\eta \|_{{ k,i- k+2}}^{2}, $$
we deduce from (3.34) that
$$ \begin{aligned}[b] & \frac{\mathrm{d}}{\mathrm{d}t} \|\eta \|_{{ k,i- k+2}}^{2} +c\bigl(\big\| (\eta, u)\big\| _{ k,i- k+2 }^{2}+ \|\nabla q\|_{ k,i- k}^{2}\bigr) \\ &\quad\lesssim\|u\|_{k,i-k}^{2}+ \| \eta\|_{ k+1, i- k+1}^{2}+ \| u_{t}\| _{i}^{2}+S_{ k,i}^{\omega}. \end{aligned} $$
(3.39)
In particular, we take \((i,k)=(1,0)\) and \((i,k)=(1,1)\) to get
$$ \frac{\mathrm{d}}{\mathrm{d}t}\|\eta\|_{3}^{2} +c\bigl(\big\| (\eta, u)\big\| _{3}^{2}+\| \nabla q\|_{1}^{2} \bigr)\lesssim\big\| (u ,u_{t})\big\| _{1}^{2}+\|\eta \|_{1,2}^{2}+ S_{0,1}^{\omega}$$
(3.40)
and
$$ \frac{\mathrm{d}}{\mathrm{d}t}\|\eta\|_{1,2}^{2} +c\bigl(\big\| (\eta, u)\big\| _{1,2}^{2}+\| \nabla q\|_{1,0}^{2} \bigr)\lesssim\big\| (u ,u_{t})\big\| _{1}^{2}+\|\eta \|_{2,1}^{2}+ S_{1,1}^{\omega}. $$
(3.41)
On the other hand, it is easy to show that
$$ \begin{aligned}[b] S_{0,1}^{\omega}+ S_{1,1}^{\omega}&\leq\big\| \bigl( N^{h}_{u},N^{h}_{q} \bigr)\big\| _{1}^{2}+\big\| \bigl(D^{h}_{u}, \operatorname{div} \eta\bigr)\big\| _{2}^{2} \lesssim\|\tilde {\mathcal {A}}\|_{2}^{2}\bigl(\|u\|_{3}^{2}+\| \nabla q \|_{1}^{2}\bigr)+\|\eta\|_{3}^{4} \\ &\lesssim {\mathcal{E}^{H}{{E}}^{L}}\lesssim{ \mathcal{E}^{H} {\mathcal{D}}^{L}}. \end{aligned} $$
(3.42)
Thus we immediately obtain (3.36) from (3.40)-(3.42).
Now we turn to the derivation of (3.37). In view of (3.35) with \((i,k)=(1,0)\), we have
$$\| u\|_{3}^{2}+\|\nabla q\|_{1}^{2} \lesssim\| \eta\|_{3}^{2} +\big\| (u,u_{t}) \big\| _{1}^{2} +S_{0,1}^{u}. $$
On the other hand, we can use (3.42) to infer that
$$S_{0,1}^{u}=\big\| \bigl(N^{h}_{u} ,N^{h}_{q}\bigr)\big\| _{1}^{2}+\big\| D^{h}_{u} \big\| _{2}^{2} \lesssim{ \mathcal{E}^{H} { {E}}^{L} }. $$
Hence, (3.37) follows from the above two estimates.
Finally, to show (3.38), we take \((i,k)=(2,1)\) in (3.35) to deduce that
$$\|u_{t}\|_{2}^{2}+\|\nabla q_{t} \|_{0}^{2} \lesssim\big\| \bigl(\nabla^{2} u,u_{t},u_{tt} \bigr)\big\| _{0}^{2}+ S_{1,2}^{u} \lesssim\| u\| _{2}^{2}+\big\| ( u_{t},u_{tt})\big\| _{0}^{2} + S_{1,2}^{u}. $$
Keeping in mind that
$$\begin{aligned} S_{1,2}^{u}&\lesssim\big\| \partial_{t}\bigl( N^{h}_{u},N^{h}_{q} \bigr)\big\| _{0}^{2}+\big\| \partial_{t} D^{h}_{u} \big\| _{1}^{2} \\ &\lesssim \|\tilde {\mathcal {A}}_{t}\|_{2}^{2}\bigl(\|u \|_{2}^{2}+\|\nabla q\|_{0}^{2}\bigr)+\| \tilde {\mathcal {A}}\|_{2}^{2}\bigl(\| u_{t}\|_{2}^{2}+ \|\nabla q_{t}\|_{0}^{2}\bigr) \lesssim{ \mathcal{E}^{H}{\mathcal{D}}^{L} }, \end{aligned} $$
we get (3.38) from the above two estimates. □
Lower-order energy inequality
Now, we are able to build the lower-order energy inequality. In what follows, the letters \(c_{i}^{L}\) and \(i=1,\ldots,7\) will denote generic positive constants which may depend on the domain Ω and some physical parameters in the transformed MHD equations (2.15).
Proposition 3.1
Under the assumption (3.1), if
δ
is sufficiently small, then there is an energy functional
\(\tilde{\mathcal{E}}^{L}\)
which is equivalent to
\(\mathcal{E}^{L}\), such that
$$ \frac{\mathrm{d}}{\mathrm{d}t} \tilde{\mathcal {E}}^{L}+\mathcal {D}^{L}\leq0\quad\textit{on }(0,T]. $$
(3.43)
Proof
We choose δ so small that
$$ \big\| \nabla\partial_{t}^{j} u \big\| ^{2}_{0}\lesssim\big\| \nabla_{\mathcal{A}} \partial_{t}^{j} u\big\| ^{2}_{0},\quad1 \leq j\leq3. $$
(3.44)
Then, thanks to (3.5), (3.6) and (3.44), we deduce from (3.12) and Lemmas 3.2-3.3 that there are constants \(c_{1}^{L}\), \(c_{2}^{L}\) and \(\sigma_{0}\geq1\), such that
$$ \frac{\mathrm{d}}{\mathrm{d}t}\tilde{\mathcal{E}}^{L}_{1} +\tilde{\mathcal{D}}^{L}_{1} \leq c_{1}^{L} (1+ \sigma) \sqrt{\mathcal{E}^{H}} \mathcal{D}^{L} \quad \mbox{for any }\sigma\geq\sigma_{0} , $$
(3.45)
where \(\sigma_{0}\) depends on the domain and the known physical parameters, and
$$\begin{aligned}& \begin{aligned} \tilde{\mathcal{E}}^{L}_{1}:={}& \|\sqrt{{\rho}} u_{t}\|_{0}^{2}+ \lambda_{0} \| \bar{M}\cdot\nabla u\|^{2}_{0} + \sigma\sum_{k=0}^{3}\bigl( \|\sqrt{ {\rho} } u\|^{2}_{k,0} + \lambda_{0} \|\bar{M}\cdot\nabla \eta\|_{k,0}^{2} \bigr) \\ &+\sum_{k=0}^{3} \biggl(\sum _{\alpha_{1}+\alpha_{2}=k} \int{\rho} \partial_{1}^{\alpha_{1}}\partial_{2}^{\alpha_{2}} \eta\cdot\partial_{1}^{\alpha_{1}}\partial_{2}^{\alpha_{2}} u\,\mathrm{d}y +\frac{ \mu}{2}\|\nabla\eta\|_{k,0}^{2} \biggr), \end{aligned}\\& \tilde{{\mathcal{D}}}^{L}_{1}:= c_{2}^{L} \Biggl(\| u_{t}\|^{2}_{1}+ \sum _{k=0}^{3} \bigl(\big\| (\eta,\bar{M}\cdot\nabla\eta) \big\| _{k,0}^{2}+\|u\| _{k,1}^{2}\bigr) \Biggr). \end{aligned}$$
Utilizing (3.13), (3.36), the interpolation inequality, and the estimate
$$ \|\eta\|_{k,1}^{2}\lesssim\|\eta \|_{k+1,0}^{2}+\|\bar{M}\cdot\nabla\eta\|_{k,0}^{2} \quad\mbox{for any }0\leq k\leq6, $$
(3.46)
we find that
$$ \begin{aligned}[b] &\frac{\mathrm{d}}{\mathrm{d}t} \bigl( \|\eta \|_{3,*}^{2}+c_{3}^{L} \| \nabla_{\mathcal {A}} u_{t}\|_{0}^{2} \bigr)+ c_{4}^{L}\bigl( \big\| (\eta, u)\big\| _{3}^{2}+ \|\nabla q\|_{1}^{2} +\| u_{tt} \|^{2}_{0}\bigr) \\ & \quad\leq c_{5}^{L}\bigl(\big\| (u,u_{t}) \big\| _{1}^{2} +\|\eta\|_{3,0}^{2}+\|\bar{M} \cdot\nabla\eta\|_{2,0}^{2}+\sqrt{\mathcal{E}^{H}} \mathcal{D}^{L}\bigr). \end{aligned} $$
(3.47)
Now, multiplying (3.47) by \(c_{2}^{L}/(2c_{5}^{L})\) and adding the resulting inequality to (3.45), we obtain
$$ \begin{aligned} &\frac{\mathrm{d}}{\mathrm{d}t} \tilde{{ \mathcal{E}}}^{L}_{2} +\tilde{\mathcal{D}}^{L} \leq c_{6}^{L}(1+\sigma)\sqrt{\mathcal{E}^{H}} \mathcal{D}^{L}, \end{aligned} $$
(3.48)
where \(\tilde{{\mathcal{E}}}^{L}_{2}\) and \(\tilde{\mathcal{D}}^{L} \) are defined by
$$\begin{gathered} \tilde{ {\mathcal{E}}}^{L}_{2}:=\tilde{\mathcal {E}}^{L}_{1}+ c_{2}^{L}\bigl(\|\eta \|_{3,*}^{2} + c_{3}^{L}\| \nabla_{\mathcal{A}} u_{t}\| _{0}^{2} \bigr)/ \bigl(2c_{5}^{L}\bigr), \\ \tilde{\mathcal{D}}^{L}:= \tilde{{\mathcal{D}}}^{L}_{1}/2+ c_{2}^{L} c_{4}^{L} \bigl(\big\| (\eta,u) \big\| _{3}^{2}+\|\nabla q\|_{1}^{2} + \|u_{tt}\|^{2}_{0}\bigr)/\bigl(2c_{5}^{L} \bigr). \end{gathered}$$
On the other hand, from (3.38) we get \(\mathcal{D}^{L}\lesssim\tilde{\mathcal{D}}^{L}+\sqrt{\mathcal {E}^{H}}\mathcal {D}^{L}\), which implies \(\mathcal{D}^{L}\lesssim\tilde{\mathcal{D}}^{L}\) for sufficiently small δ. Therefore, (3.48) can be rewritten as follows:
$$ \begin{aligned} &\frac{\mathrm{d}}{\mathrm{d}t} \tilde{{ \mathcal{E}}}^{L}_{2} +c_{7}^{L}{ \mathcal{D}}^{L} \leq c_{6}^{L}(1+\sigma)\sqrt{ \mathcal{E}^{H}} \mathcal{D}^{L}. \end{aligned} $$
(3.49)
Next we show that \(\mathcal{E}^{L}\) can be controlled by \(\tilde {{\mathcal{E}}}^{L}_{2}\). By virtue of Cauchy-Schwarz’s inequality, (3.5) and (3.44), there is an appropriately large constant σ, depending on ρ, μ and Ω, such that
$$\|\nabla\eta\|_{3,0}^{2}+ \|\eta\|_{3}^{2}+ \|u_{t}\|_{1}^{2} \lesssim\tilde{{ \mathcal{E}}}^{L}_{2}. $$
In view of (3.37), we further have
$${\mathcal{E}}^{L} \lesssim\tilde{{\mathcal{E}}}^{L}_{2} +\sqrt{\mathcal{E}^{H}} {\mathcal{E}}^{L}, $$
which implies \({\mathcal{E}}^{L}\lesssim\tilde{\mathcal{E}}^{L}_{2}\) for sufficiently small δ. In addition, obviously \(\tilde{{\mathcal{E}}}^{L}_{2}\lesssim\mathcal{E}^{L}\). Hence, the energy functional \(\tilde{{\mathcal{E}}}^{L}_{2}\) is equivalent to \(\mathcal{E}^{L}\). Finally, letting \(\delta\leq c_{7}^{L} /2 c_{6}^{L}(1+\sigma)\) and noting \(\tilde {\mathcal{E}}^{L} =2\tilde{{\mathcal{E}}}^{L}_{2}/c_{7}^{L}\), we see that (3.49) immediately implies (3.43). Obviously, the energy functional \(\tilde{\mathcal{E}}^{L} \) is still equivalent to \(\mathcal{E}^{L}\). This completes the proof. □