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Time-decay solutions of the initial-boundary value problem of rotating magnetohydrodynamic fluids
- Weiwei Wang1Email authorView ORCID ID profile and
- Youyi Zhao1
Received: 5 May 2017
Accepted: 21 July 2017
Published: 11 August 2017
Abstract
We have investigated an initial-boundary problem for the perturbation equations of rotating, incompressible, and viscous magnetohydrodynamic (MHD) fluids with zero resistivity in a horizontally periodic domain. The velocity of the fluid in the domain is non-slip on both upper and lower flat boundaries. We switch the analysis of the initial-boundary problem from Euler coordinates to Lagrangian coordinates under proper initial data, and get a so-called transformed MHD problem. Then, we exploit the two-tiers energy method. We deduce the time-decay estimates for the transformed MHD problem which, together with a local well-posedness result, implies that there exists a unique time-decay solution to the transformed MHD problem. By an inverse transformation of coordinates, we also obtain the existence of a unique time-decay solution to the original initial-boundary problem with proper initial data.
Keywords
magnetohydrodynamic fluid equilibrium state magnetic field decay estimates rotation1 Introduction
The three-dimensional (3D) rotating, incompressible and viscous magnetohydrodynamic (MHD) equations with zero resistivity in a domain \(\Omega\subset\mathbb{R}^{3}\) read as follows: Here the unknowns \({v}= {v}(x,t)\), \(M:= {M}(x,t)\) and \(p=p(x,t)\) denote the velocity, the magnetic field, and the pressure of the incompressible MHD fluid respectively; \(\mu>0\), ρ and \(\lambda_{0}\) stand for the coefficients of the shear viscosity, the density constant, and the permeability of vacuum, respectively. \(2\rho(\vec{\omega}\times v)\) represents the Coriolis force, and \(\vec{\omega}=(0,0,\omega)\) denotes the constant angular velocity in the vertical direction. In system (1.1), equation (1.1)1 describes the balance law of momentum, while (1.1)2 is called the induction equation. As for the constraint \(\operatorname{div} M=0\), it can be seen just as a restriction on the initial value of M since \((\operatorname{div} M)_{t}=0\) due to (1.1)2.
$$ \left \{ \textstyle\begin{array}{l} \rho v_{t}+\rho v\cdot\nabla v +\nabla( p+\lambda_{0} |M|^{2}/2)+2\rho (\vec{\omega}\times v)=\mu\Delta v + \lambda_{0}M\cdot\nabla M, \\ M_{t}=M\cdot\nabla v-v\cdot\nabla M, \\ \operatorname{div} {v}=\operatorname{div} M=0 . \end{array}\displaystyle \right . $$
(1.1)
Let \(\bar{M}:=(m_{1},m_{2},m_{3})\) be a constant vector with \(m_{3}\neq0\), and \((0,\bar{M},\bar{p})\) be a rest state of the system (1.1). We denote the perturbation to an equilibrium state \((0,\bar{M})\) by Then, \(({v},N,q)\) satisfies the perturbation equations where we have used the relation \(\vec{\omega}\times v=\omega( v_{1}e_{2}-v_{2} e_{1})\). For system (1.2), we impose the initial and the boundary conditions:
where \(v_{0}\) and \(N_{0}\) should satisfy the compatibility conditions \(\operatorname {div}v_{0}=\operatorname{div}N_{0}=0\). We call the initial-boundary value problem (1.2)-(1.4) the MHD problem (with rotation) for simplicity. In this article, we always assume that the domain is horizontally periodic with finite height, i.e., where \(\mathcal{T} :=(2\pi L_{1}\mathbb{T})\times(2\pi L_{2}\mathbb{T}) \), \(\mathbb{T}=\mathbb{R}/\mathbb{Z}\), and \(2\pi L_{1}, 2\pi L_{2}>0\) are the periodicity lengths.
$${v}= {v}- {0},\qquad N= M -\bar{M},\qquad\tilde{q}=p-\bar{p}. $$
$$ \left \{ \textstyle\begin{array}{l} \rho{ v}_{t}+\rho{ v}\cdot\nabla { v}+\nabla(\tilde{q}+\lambda_{0} | N+\bar{M} |^{2}/2) \\ \quad=\mu\Delta v + \lambda_{0}(N+\bar{M})\cdot\nabla(N+\bar {M})+ 2 \rho\omega(v_{2} e_{1}- v_{1}e_{2}),\\ N_{t}=(N+\bar{M})\cdot\nabla v- v\cdot\nabla(N+\bar{M}),\\ \operatorname{div}v= \operatorname{div} {N}=0, \end{array}\displaystyle \right . $$
(1.2)
$$\begin{aligned}& (v,N)|_{t=0}=(v_{0},N_{0})\quad \mbox{in } \Omega, \end{aligned}$$
(1.3)
$$\begin{aligned}& v(\cdot,t)|_{\partial\Omega}={ 0}\quad\mbox{for any }t>0, \end{aligned}$$
(1.4)
$$\Omega:=\bigl\{ x:=\bigl(x',x_{3}\bigr)\in \mathbb{R}^{3}\mid x'\in\mathcal{T},0< x_{3}< h \bigr\} \quad\mbox{with }h>0, $$
The effects of magnetic fields and rotation on the motion of pure fluids were widely investigated; see [1–10] and the references cited therein. In particular, Tan and Wang [11] showed that the well-posedness problem of the initial-boundary problem (1.2)-(1.4) for \(\omega=0\) (i.e., without the effect of rotation). In this article, we further consider \(\omega\neq0\), and show that there also exists a unique time-decay solution to the initial-boundary problem (1.2)-(1.4) in Lagrangian coordinates (see Theorem 2.1), which, together with the inverse transformation of coordinates, implies the existence of a time-decay solution to the original initial-boundary problem (1.2)-(1.4) with proper initial data in \(H^{7}(\Omega)\). Our result also holds for the case \(\omega=0\), thus improves Tan and Wang’s result in [11], in which the sufficiently small initial data at least belong to \(H^{16}(\Omega)\).
In the next section we introduce the form of the initial-boundary problem (1.2)-(1.4) in Lagrangian coordinates, and the details of our result.
2 Main results
2.1 Reformulation
In general, it is difficult to directly show the existence of a unique global-in-time solution to (1.2)-(1.4). Instead, we switch our analysis to Lagrangian coordinates as in [12, 13]. To this end, we assume that there is an invertible mapping \(\zeta_{0}:=\zeta_{0}(y):\Omega\to\Omega\), such that where \(\zeta_{3}^{0}\) denotes the third component of \(\zeta_{0}\). We define the flow map ζ as the solution to We denote the Eulerian coordinates by \((x,t)\) with \(x=\zeta(y,t)\), where \((y,t)\in\Omega\times\mathbb{R}^{+}\) stand for the Lagrangian coordinates. In order to switch back and forth from Lagrangian to Eulerian coordinates, we assume that \(\zeta(\cdot,t)\) is invertible and \(\Omega=\zeta(\Omega, t)\). In other words, the Eulerian domain of the fluid is the image of Ω under mapping ζ. In view of the non-slip boundary condition \(v|_{\partial\Omega}=0\), we have \(\partial\Omega=\zeta(\partial\Omega, t)\). In addition, since \(\det\nabla\zeta_{0}=1\), we have due to \(\operatorname{div}v=0\); see [14], Proposition 1.4.
$$ \partial\Omega=\zeta_{0}(\partial\Omega) \quad\mbox{and}\quad \det\nabla\zeta_{0}\equiv1, $$
(2.1)
$$ \left \{ \textstyle\begin{array}{l} \zeta_{t} (y,t)=v(\zeta(y,t),t) , \\ \zeta(y,0)= {\zeta_{0}}. \end{array}\displaystyle \right . $$
(2.2)
$$ \det(\nabla\zeta)=1 $$
(2.3)
Now, we further define the Lagrangian unknowns by Thus in Lagrangian coordinates the evolution equations for u, p̃ and B read as with initial and boundary conditions Moreover, \(\operatorname {div}_{\mathcal{A}}B=0\) if the initial data \(\zeta_{0}\) and \(B_{0}\) satisfy Here \(\mathcal{A}_{0}\) denotes the initial value of \(\mathcal{A}\), the matrix \(\mathcal{A}:=(\mathcal{A}_{ij})_{3\times3}\) via \(\mathcal{A}^{T}=(\nabla\zeta)^{-1}:=(\partial_{j} \zeta _{i})^{-1}_{3\times 3}\), and the differential operators \(\nabla_{\mathcal{A}}\), \(\operatorname {div}\mathcal{A}\) and \(\Delta_{\mathcal{A}}\) are defined by \(\nabla_{\mathcal {A}}f:= (\mathcal {A}_{1k}\partial_{k}f, \mathcal{A}_{2k}\partial_{k}f,\mathcal{A}_{3k}\partial_{k}f)^{T}\), \(\operatorname{div}_{\mathcal {A}}(X_{1},X_{2},X_{3})^{T}:=\mathcal{A}_{lk}\partial_{k} X_{l}\), and \(\Delta _{\mathcal{A}}f:=\operatorname{div}_{\mathcal{A}}\nabla_{\mathcal{A}}f\) for appropriate f and X. It should be noted that we have used the Einstein convention of summation over repeated indices, and \(\partial _{k}=\partial_{y_{k}}\). In addition, in view of the definition of \(\mathcal{A}\) and (2.3), we can see that \(\mathcal{A}=(A^{*}_{ij})_{3\times3}\), where \(A^{*}_{ij}\) is the algebraic complement minor of the \((i,j)\)th entry \(\partial_{j} \zeta_{i}\). Since \(\partial_{k}A^{*}_{ik}=0\), we can get an important relation which will be used in the derivation of temporal derivative estimates.
$$ (u,\tilde{p},B) (y,t)=\bigl( v,p+\lambda_{0} |M|^{2}/2,M\bigr) \bigl(\zeta(y,t),t\bigr) \quad\mbox{for } (y,t)\in \Omega\times\mathbb{R}^{+} . $$
(2.4)
$$ \left \{ \textstyle\begin{array}{ll} \zeta_{t}=u, \\ {\rho} u_{t}-\mu\Delta_{\mathcal{A}}u+\nabla_{\mathcal{A}}\tilde{p}= \lambda_{0} B\cdot\nabla_{\mathcal{A}} B+2\rho\omega (u_{2}e_{1}-u_{1}e_{2}), \\ B_{t}-B\cdot\nabla_{\mathcal{A}}u=0, \\ \operatorname {div}_{\mathcal{A}}u=0, \end{array}\displaystyle \right . $$
(2.5)
$$(u,\zeta-y)|_{\partial\Omega}=0 \quad\mbox{and}\quad (\zeta, u, B)|_{t=0}=(\zeta_{0}, u_{0}, B_{0}). $$
$$ \operatorname {div}_{\mathcal{A}_{0}}B_{0}=0. $$
(2.6)
$$ \operatorname{div}_{\mathcal{A}}u=\partial_{l} ( \mathcal{A}_{kl}u_{k})=0, $$
(2.7)
Our next goal is to eliminate B by expressing it in terms of ζ. This can be achieved in the same manner as in [12, 13]. For the reader’s convenience, we give the derivation here. In view of the definition of \(\mathcal{A}\), one has where \(\delta_{ij}=0\) for \(i\neq j\), and \(\delta_{ij}=1\) for \(i=j\). Thus, applying \(\mathcal{A}_{jl}\) to (2.5)4, we obtain which implies that \(\partial_{t}(\mathcal{A}_{jl}B_{j})=0\) (i.e., \((\mathcal{A}^{T} B)_{t}=0\)). Hence, which yields \(B_{i}=\partial_{l} \zeta_{i}\mathcal{A}_{jl}^{0}B_{j}^{0}\), i.e., Here and in what follows, the notation \(f^{0}\) also denotes the initial data of the function f. To obtain the asymptotic stability in time, we naturally expect Thus (2.9) formally implies Putting the above expression of \(B_{0}\) into (2.9), we get Moreover, in view of (2.8), (2.11) and (2.12), the Lorentz force term can be represented by
$$\partial_{i}\zeta_{k}\mathcal{A}_{kj}= \mathcal{A}_{ik}\partial_{k}\zeta_{j}= \delta_{ij}, $$
$$ \mathcal{A}_{jl}\partial_{t} B_{j}=B_{i} \mathcal{A}_{ik}\partial_{k} u_{j} \mathcal{A}_{jl}=B_{i}\mathcal{A}_{ik}\partial _{t}(\partial_{k} \zeta_{j})\mathcal{A}_{jl}=-B_{i} \mathcal{A}_{ik}\partial_{k} \zeta_{j} \partial_{t} \mathcal{A}_{jl}=-B_{j} \partial_{t}\mathcal{A}_{jl}, $$
$$ \mathcal{A}_{jl}B_{j}=\mathcal{A}_{jl}^{0}B_{j}^{0}, $$
(2.8)
$$ B=\nabla\zeta\mathcal{A}^{T}_{0} B_{0}. $$
(2.9)
$$ (\zeta,B)\mbox{ converges to }(y,\bar{M}) \quad\mbox{as }t\to\infty. $$
(2.10)
$$ \mathcal{A}^{T}_{0} B_{0}=\bar{M}, \quad\textit{i.e., } B_{0}= \bar{M}\cdot\nabla\zeta_{0}. $$
(2.11)
$$ B=\bar{M}\cdot\nabla\zeta. $$
(2.12)
$$ \begin{aligned}B\cdot\nabla_{\mathcal{A}} B=B_{l} \mathcal{A}_{lk}\partial_{k} B = \mathcal{A}_{lk}^{0} B_{l}^{0} \partial_{k}(\bar{M}\cdot\nabla\zeta)=\bar{M} \cdot\nabla(\bar {M}\cdot \nabla\zeta). \end{aligned} $$
(2.13)
Summing up the above analyses, we can see that, under the initial conditions (2.1) and (2.11), one can use the relation (2.13) to change (2.5) into the following Navier-Stokes system: and B is defined by (2.12). Now, we introduce the shift functions Then the evolution equations for the shift functions \((\eta,q)\) and u read as where \(\mathcal{A}=(I+\nabla\eta)^{-1}\), \(I=(\delta_{ij})_{3\times3}\). The associated initial and boundary conditions read as follows: It should be noted that the shift function q is the sum of the perturbed fluid and the magnetic pressures in Lagrangian coordinates. Hence we still call q the perturbation pressure for the sake of simplicity. In this article, we call the initial-boundary value problem (2.15)-(2.16) the transformed MHD problem.
$$\left\{ \textstyle\begin{array}{l} \zeta_{t}= u, \\ {\rho}u_{t}-\mu\Delta_{\mathcal{A}} u+\nabla_{\mathcal{A}} \tilde{ p}=\lambda_{0} \bar{M}\cdot\nabla(\bar{M}\cdot\nabla\zeta)+2\rho \omega(u_{2}e_{1}-u_{1}e_{2}),\\ \operatorname {div}_{\mathcal{A}} u =0, \end{array}\displaystyle \right. $$
$$ \eta=\zeta-y \quad\mbox{and}\quad q=\tilde{p}-\bar{p}. $$
(2.14)
$$ \left\{\textstyle\begin{array}{l} \eta_{t}= u, \\ {\rho} u_{t}-\mu\Delta_{\mathcal{A}} u+\nabla_{\mathcal{A}} q-\lambda_{0} \bar{M}\cdot\nabla(\bar{M}\cdot\nabla\zeta)= 2\rho\omega(u_{2}e_{1}-u_{1}e_{2}) ,\\ \operatorname {div}_{\mathcal{A}} u =0, \end{array}\displaystyle \right. $$
(2.15)
$$ (\eta,u)|_{t=0}=(\eta_{0},u_{0}), \qquad(\eta,u)|_{\partial\Omega}=0. $$
(2.16)
2.2 Main results
Before stating our first main result on the transformed MHD problem in detail, we introduce some simplified notations that shall be used throughout this paper: where, and in what follows, the letter c denotes a generic constant which may depend on the domain Ω and some physical parameters, such as \(\lambda_{0}\), M̄, g, μ and ρ in the MHD equations (2.15). It should be noted that a product space \((X)^{n}\) of vector functions is still denoted by X, for example, a vector function \({u}\in(H^{2})^{3}\) is denoted by \({u}\in H^{2}\) with norm \(\| {u}\|_{H^{2}}:=(\sum_{k=1}^{3}\|u_{k}\| _{H^{2}}^{2})^{1/2}\). Finally, we define some functionals: Next, we introduce our main result.
$$ \begin{gathered} \mathbb{R}^{+}_{0}:=[0,\infty),\qquad \int:= \int_{\Omega},\qquad L^{p}:=L^{p} ( \Omega):=W^{0,p}(\Omega)\quad\mbox{for }1< p\leq\infty, \\ {H}^{1}_{0}:=W^{1,2}_{0}(\Omega),\qquad {H}^{k}:=W^{k,2}(\Omega), \qquad\|\cdot\|_{k} :=\|\cdot \|_{H^{k}(\Omega)}\quad\mbox{for }k\geq1, \\ \partial_{h}^{k}\mbox{ denotes }\partial_{ 1}^{\alpha_{1}} \partial_{ 2}^{\alpha_{2}}\quad\mbox{for any }\alpha_{1}+ \alpha_{2}=k, \qquad\|\cdot\|_{m,k}:=\sum _{\alpha_{1}+\alpha_{2}=m} \|\partial_{1}^{\alpha_{1}} \partial_{2}^{\alpha_{2}}\cdot\|_{k}^{2}, \\ a\lesssim b \mbox{ means that } a\leq cb, \end{gathered} $$
$$ \begin{gathered} \mathcal{E}^{L}:=\|\nabla\eta \|_{3,0}^{2}+\big\| (\eta,u)\big\| _{3}^{2}+\big\| (u_{t},\nabla q)\big\| _{1}^{2}, \\ \mathcal{D}^{L}:=\big\| (\eta,\bar{M}\cdot\nabla\eta, \nabla u) \big\| ^{2}_{3,0}+ \big\| (\eta,u)\big\| _{3}^{2}+\sum _{k=1}^{2}\big\| \partial_{t}^{k}u \big\| _{4-2k}^{2}+\|\nabla q\|_{1}^{2}+\| \nabla q_{t}\|_{0}^{2}, \\ \mathcal{E}^{H}:=\|\nabla\eta\|^{2}_{6,0}+\|\eta \|_{6}^{2} +\sum_{k=0}^{3} \big\| \partial_{t}^{k} u\big\| _{6-2k}^{2}+\sum _{k=0}^{2} \big\| \nabla\partial_{t}^{k} q\big\| _{4-2k}^{2}, \\ \mathcal{D}^{H}:= \big\| (\eta,\bar{M}\cdot \nabla\eta,\nabla u)\big\| ^{2}_{6,0}+\big\| (\eta, u) \big\| _{6}^{2} +\sum_{k=1}^{3} \big\| \partial_{t}^{k} u\big\| _{7-2k}^{2}+\sum _{k=1}^{2}\big\| \nabla\partial_{t}^{k}q \big\| _{5-2k}^{2}+\|\nabla q\|_{4}^{2}, \\ \mathcal{G}_{1}(t)=\sup_{0\leq\tau< t}\big\| \eta(\tau) \big\| _{7}^{2},\qquad\mathcal{G}_{2}(t)= \int_{0}^{t}\frac{\|(\eta,u)(\tau)\|_{7}^{2}}{(1+\tau)^{3/2}} \,\mathrm{d}\tau, \\ \mathcal{G}_{3}(t)=\sup_{0\leq\tau< t} \mathcal{E}^{H}(\tau)+ \int_{0}^{t} \mathcal{D}^{H}(\tau)\,\mathrm{d} \tau,\qquad\mathcal{G}_{4}(t)=\sup_{0\leq \tau< t}(1+ \tau)^{3}\mathcal{E}^{L}(\tau). \end{gathered} $$
Theorem 2.1
Let Ω be a horizontally periodic domain with finite height, ω
be an arbitrary real number, and
\(m_{3}\neq0\). Then there is a sufficiently small constant
\(\delta>0\), such that for any
\((\eta_{0}, u_{0})\in H^{7}\times H^{6}\)
satisfying the following conditions:
Here the positive constants
δ
and
c
depend on the domain Ω and other known physical parameters
\(\lambda_{0}\), M̄, μ
and
ρ.
- (1)
\(\| \eta_{0} \|_{7}^{2}+\|u_{0}\|_{6}^{2} \leq\delta\);
- (2)
\(\zeta_{0}:=y+\eta_{0}\) satisfies (2.1);
- (3)
\((\eta_{0}, u_{0})\) satisfies necessary compatibility conditions (i.e., \(\partial_{t}^{j} u(x,0)|_{\partial\Omega }=0\) for \(j=1\) and 2),
$$ \mathcal{G}(\infty):=\sum_{k=1}^{4} \mathcal{G}_{k}(\infty) \leq c \bigl(\| \eta_{0} \|_{7}^{2}+\|u_{0}\|_{6}^{2} \bigr). $$
(2.17)
Remark 2.1
Exploiting the inverse transformation of ζ, we can easily deduce from Theorem 2.1 the global well-posedness of the original MHD problem (1.2)-(1.4). More precisely, there is a sufficiently small constant \(\delta_{0}> 0\), such that, for any \(( v_{0},N_{0})\in H^{6}\) satisfying the following conditions:
- (1)
there exists an invertible mapping \(\zeta_{0}:=\zeta _{0}(x):\Omega\to\Omega\), such that (2.1) holds, where \(\mathcal{A}^{T}_{0}=(\nabla\zeta_{0})^{-1}\);
- (2)
\((\bar{M}+N_{0})(\zeta_{0})=\bar{M}\cdot\nabla\zeta_{0} \);
- (3)
\(\| \zeta_{0} -x\|_{7}^{2}+\|v_{0}\|_{6}^{2} \leq\delta_{0}\);
- (4)
the initial data \(\varrho_{0}\), \(v_{0}\), \(N_{0}\) satisfy necessary compatibility conditions (i.e., \(\partial _{t}^{j}v(x,0)|_{\partial\Omega}=0\) for \(j=1\) and 2),
$$ \begin{aligned}[b] &\sup_{0\leq t< \infty} \Biggl( \|N\|_{6}^{2}+\sum_{k=0}^{3} \big\| \partial_{t}^{k} {v}(t)\big\| _{6-2k}^{2}+ \sum_{k=0}^{2} \big\| \nabla\partial_{t}^{k} \tilde{q}(t)\big\| _{4-2k}^{2} \Biggr) \\ &\quad+\sup_{0\leq t< \infty}(1+t)^{3} \bigl( \big\| ( v,N) \big\| _{3}^{2}+\big\| (v_{t},\nabla\tilde{q}) \big\| _{1}^{2} \bigr) \leq c \bigl( \|\zeta_{0}-x \|_{7}^{2}+\| v_{0}\|_{6}^{2} \bigr). \end{aligned} $$
(2.18)
Now we briefly describe the basic idea in the proof of Theorem 2.1. By the standard energy method, there are two functionals \(\tilde {\mathcal{E}}^{L}\) and Q of \((\eta,u)\) satisfying the lower-order energy inequality (see Proposition 3.1) where the functional \(\tilde{\mathcal{E}}^{L}\) is equivalent to \(\mathcal{E}^{L}\). Unfortunately, we can not close the energy estimates only based on (2.19), since \(\mathcal{Q}\) can not be controlled by \(\tilde{\mathcal{E}}^{L}\). However, we observe that the structure of the energy inequality above is very similar to that of the surface wave problem [15], for which Guo and Tice developed a two-tier energy method to overcome this difficulty. In the spirit of the two-tier energy method, we look after a higher-order energy inequality to match the lower-order energy inequality (2.19). Since \(\tilde{\mathcal{E}}^{L}\) contains \(\|\eta\|_{3}\), we find that the higher-order energy at least includes \(\|\eta\|_{6}\). Thus, similar to (2.19), we establish the higher-order energy inequality (see Proposition 4.1) where the functional \(\tilde{\mathcal{E}}^{H}\) is equivalent to \(\mathcal{E}^{H}\). Moreover, the highest-order norm \(\|(\eta,u)\|_{7}^{2}\) enjoys the highest-order energy inequality where the norm \(\|\eta\|_{7,*} \) is equivalent to \(\|\eta\|_{7} \). In the derivation of the a priori estimates, we have \(\mathcal {Q}\lesssim\mathcal{E}^{H}\), and thus (2.19) implies (see Proposition 3.1) Consequently, by the two-tier energy method, we can deduce the global-in-time stability estimate (2.17) based on (2.20)-(2.22).
$$ \frac{\mathrm{d}}{\mathrm{d}t}\tilde{\mathcal{E}}^{L}+ \mathcal{D}^{L}\leq\mathcal{Q}\mathcal{D}^{L}, $$
(2.19)
$$ \frac{\mathrm{d}}{\mathrm{d}t}\tilde{\mathcal {E}}^{H}+ {\mathcal {D}}^{H}\leq\sqrt{\mathcal{E}^{L}}\big\| (\eta,u) \big\| _{7}^{2}, $$
(2.20)
$$ \frac{\mathrm{d}}{\mathrm{d}t} \|\eta\|_{7,*}^{2}+ \big\| ( \eta,u)\big\| _{7}^{2}\lesssim\mathcal{E}^{H}+ \mathcal{D}^{H}, $$
(2.21)
$$ \frac{\mathrm{d}}{\mathrm{d}t}\tilde {\mathcal{E}}^{L}+ \mathcal{D}^{L}\leq0. $$
(2.22)
The rest of the sections are mainly devoted to the proof of Theorem 2.1. In Section 3, we first derive the lower-order energy inequality (2.22) for the transformed MHD problem. Then in Section 4 we derive the higher-order energy inequality (2.20) and the highest-order energy inequality (2.21). Based on these three energy inequalities, we prove Theorem 2.1 by adapting the two-tier energy method in Section 5.
3 Lower-order energy inequality
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 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) and the interpolation inequalities (see [16], 5.2 Theorem) 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: and 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]:
$$ \sqrt{\mathcal{G}_{1}(T)+ \sup_{0\leq\tau\leq T} \mathcal{E}^{H}(\tau)}\leq\delta\in(0,1) \quad\mbox{for some }T>0, $$
(3.1)
$$ \|f\|_{L^{p}}\lesssim\| f\|_{1} \quad\mbox{for }2 \leq p\leq6\quad\mbox{and}\quad\| f\|_{L^{\infty}}\lesssim\| f\|_{2} , $$
(3.2)
$$ \|f\|_{j}\lesssim\|f\|_{0}^{1-\frac{j}{i}}\|f \|_{i}^{\frac{j}{i}}\leq C_{\epsilon}\|f\|_{0} + \epsilon\|f\|_{H^{i}} $$
(3.3)
$$ \|fg\|_{j}\lesssim\|f\|_{j}\|g \|_{\kappa(j)} \quad\mbox{for }j\geq0 $$
(3.4)
$$ \|f\|_{0}\lesssim\|\partial_{3} f \|_{0} \quad\mbox{for }f\in H^{1}_{0}, $$
(3.5)
$$ \|f\|_{0}\leq h\|\bar{M}\cdot\nabla f \|_{0}/{ \pi}. $$
(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.
3.1 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 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 Inserting this relation into (3.7), we get which, together with initial condition \(\det(I+\nabla\eta_{0})=1\), implies Thus we obtain
$$ \partial_{t} \det(I+\nabla\eta)=\sum _{1\leq i,j\leq3}\partial_{t}\partial_{j} \eta_{i} A_{ij}^{*}=\sum_{1\leq i,j\leq3} A_{ij}^{*}\partial_{j}u_{i} , $$
(3.7)
$$\mathcal{A}=\bigl(A^{*}_{ij}\bigr)_{3\times3}/\det(I+\nabla\eta). $$
$$\partial_{t} \det(I+\nabla\eta)= {\det(I+\nabla\eta)}\sum _{1\leq i,j,\leq3} \mathcal{A}_{ij}\partial_{j}u_{i}=0, $$
$$\det(I+\nabla\eta)=1. $$
$$ \mathcal{A}=\bigl(A^{*}_{ij}\bigr)_{3\times3}. $$
(3.8)
Now, exploiting (2.15)1, (3.1), (3.4) and (3.8), we easily see that Similarly, we further deduce 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)
$$\big\| \partial_{t}^{i}\mathcal{A}\big\| _{j} \lesssim\sum _{k=0}^{i-1} \big\| \partial_{t}^{k} \nabla u\big\| _{j} \quad\mbox{for any } 1\leq i\leq4\mbox{ and } 0\leq j \leq8-2i. $$
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: whence Using (3.1), (3.4) and (3.9), we find that
$$\mathcal{A}^{T}=I-\nabla\eta+(\nabla\eta)^{2}\sum _{i=0}^{\infty}(-\nabla\eta)^{i}=I-\nabla \eta+(\nabla\eta)^{2}\mathcal{A}^{T}, $$
$$\tilde{\mathcal{A}}^{T} =(\nabla\eta)^{2} \mathcal{A}^{T}-\nabla\eta. $$
$$\|\tilde{\mathcal{A}}\|_{j} \lesssim\|\nabla\eta\|_{j} \quad\mbox{for }0\leq j\leq6. $$
3.2 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 where and we have used relation (2.7) in (3.11). Then from (3.10) we show the following estimates:
$$ \left\{\textstyle\begin{array}{l} \partial_{t}^{j+1}\eta=\partial_{t}^{j} u, \\ {\rho}\partial_{t}^{j+1} u-\mu\Delta_{\mathcal{A}} \partial _{t}^{j}u+\nabla_{\mathcal{A}} \partial_{t}^{j} q \\ \quad=\lambda_{0} \partial_{t}^{j} \bar{M}\cdot\nabla(\bar{M}\cdot\nabla \zeta)+ 2\rho\omega\partial_{t}^{j} (u_{2}e_{1}-u_{1}e_{2}) +\mu N^{t,j}_{u}+N^{t,j}_{q},\\ \operatorname {div}_{\mathcal{A}} \partial_{t}^{j} u =\operatorname{div} D_{u}^{t,j} , \end{array}\displaystyle \right. $$
(3.10)
$$\begin{aligned}& N^{t,j}_{u}:= \sum_{0\leq m< j, 0\leq n\leq j} \partial_{t}^{j-m-n }\mathcal{A}_{il} \partial_{l}\bigl(\partial_{t}^{n}\mathcal {A}_{ik}\partial_{t}^{m} \partial_{k}u \bigr), \\& N^{t,j}_{q}:=-\sum_{0\leq l< j}\bigl( \partial^{j-l}_{t}\mathcal{A}_{ik} \partial^{l}_{t} \partial_{k} q \bigr)_{3\times1}, \\& D^{t,j}_{u}:= \biggl(-\sum _{0\leq l< j}C_{j}^{j-l}\partial_{t}^{j-l} \mathcal{A}_{ki}\partial_{t}^{l} u_{k} \biggr)_{3\times1}, \\& C_{j}^{j-l}\mbox{ denotes the number of }(j-l) \mbox{-combinations from a given set }S\mbox{ of }j \mbox{ elements}, \end{aligned}$$
(3.11)
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 The last three integrals \(I_{2}^{L},\ldots,I^{L}_{4}\) can be estimated as follows:
where we have used (3.10)3 in (3.15). Consequently, the desired estimate (3.12) follows from (3.15)-(3.17).
$$ \begin{aligned}[b] &\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \bigl(\|\sqrt{{\rho}} u_{t}\| _{0}^{2}+ \lambda_{0} \|\bar{M} \cdot\eta_{t}\|^{2}_{0} \bigr) +\mu\| \nabla_{\mathcal{A}}u_{t}\|^{2}_{0} \\ &\quad = 2\rho\omega \int\partial_{t} (u_{2}e_{1}-u_{1}e_{2}) \cdot u_{t} \,\mathrm{d}y + \int q_{t} \operatorname{div}_{\mathcal{A}}u_{t}\,\mathrm{d}y \\ &\qquad{} + \mu \int N^{t,1}_{u}\cdot u_{t}\,\mathrm{d}y+ \int N^{t,1}_{q} \cdot u_{t}\,\mathrm{d}y\\ &\quad:=\sum _{k=1}^{4}I^{L}_{k}. \end{aligned} $$
(3.14)
$$\begin{aligned}& \begin{aligned}[b] I^{L}_{2}&:= - \int\nabla q_{t} \cdot D^{t,1}_{u}\,\mathrm{d}y \lesssim\|\nabla q_{t}\|_{0}\|\mathcal{A}_{t} \|_{L^{4}}\|u\|_{L^{4}}\lesssim\| \nabla q_{t} \|_{0}\|\mathcal{A}_{t}\|_{1}\|u\|_{1}\\ & \lesssim\sqrt{\mathcal{E}^{H}} \mathcal{D}^{L}, \end{aligned} \end{aligned}$$
(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)
(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 On the other hand, the five integrals \(J^{H}_{1},\ldots,J^{H}_{5}\) can be bounded as follows:
Thus, substituting (3.19)-(3.23) into (3.18) and using Cauchy-Schwarz’s inequality, we immediately get (3.13). □
$$ \begin{aligned}[b] &\frac{\mu}{2}\frac{\mathrm{d}}{\mathrm{d}t}\| \nabla_{\mathcal{A}} u_{t}\| ^{2}_{0} +\| \sqrt{{\rho}} u_{tt} \|_{0}^{2}\\&\quad = \int \bigl(2\rho\omega \partial_{t} (u_{2}e_{1}-u_{1}e_{2}) \cdot u_{tt} + \lambda_{0} \bar{M}\cdot\nabla(\bar{M}\cdot \nabla u)\cdot u_{tt}\bigr)\,\mathrm{d}y \\ &\qquad{} + \int q_{t} \operatorname{div}_{\mathcal{A}} u_{tt}\,\mathrm{d}y+ \mu \int N^{t,1}_{u}\cdot u_{tt}\,\mathrm{d}y+ \int N^{t,1}_{q} \cdot u_{tt} \,\mathrm{d}y \\ &\qquad{}+{\mu} \int\nabla_{\mathcal{A}}u_{t}: \nabla_{\mathcal{A}_{t}} u_{t}\,\mathrm{d}y\\ &\quad:=\sum_{k=1}^{5}J^{H}_{k}. \end{aligned} $$
(3.18)
$$\begin{aligned}& J^{H}_{1}\lesssim\bigl(\|u_{t} \|_{0}+\| u\|_{2}\bigr)\| u_{tt} \|_{0}, \end{aligned}$$
(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)
3.3 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: where Then we have the following estimate on horizontal spatial derivatives of η.
$$ \left\{\textstyle\begin{array}{l} \eta_{t}= u, \\ {\rho} u_{t}-\mu\Delta u+\nabla q-\lambda_{0} \bar{M}\cdot\nabla (\bar{M}\cdot\nabla\eta)=2\rho\omega (u_{2}e_{1}-u_{1}e_{2}) + \mu N^{h}_{u}+N^{h}_{q},\\ \operatorname {div}u =D^{h}_{u}, \end{array}\displaystyle \right. $$
(3.24)
$$\begin{gathered}N^{h}_{u}:= \partial_{l}\bigl[(\tilde{\mathcal{A}}_{jl} \tilde{ \mathcal{A}}_{jk}+ \tilde{\mathcal{A}}_{lk}+\tilde{ \mathcal{A}}_{kl})\partial_{k}u_{i} \bigr]_{3\times1}, \\ N^{h}_{q}:=-(\tilde{\mathcal{A}}_{ik} \partial_{k} q)_{3\times1} \quad \mbox{and}\quad D^{h}_{u}:= \tilde{\mathcal{A}}_{lk}\partial_{k} u_{l}. \end{gathered} $$
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 If we integrate (by parts) the above identity over Ω, we obtain 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.
$$\begin{gathered} {\rho} \partial_{t}\bigl( \partial_{h}^{3}\eta\cdot\partial_{h}^{3} u \bigr)-\bigl(\mu\Delta\partial_{h}^{3} \eta_{t}+\lambda_{0} \bar{M}\cdot\nabla\bigl(\bar{M}\cdot \nabla\partial_{h}^{3}\eta\bigr)\bigr)\cdot \partial_{h}^{3}\eta \\ \quad= \bigl( 2\rho\omega\partial_{h}^{3} (u_{2}e_{1}-u_{1}e_{2}) +\mu \partial_{h}^{3}N^{h}_{u}+\partial _{h}^{3}N^{h}_{q}-\nabla \partial_{h}^{3} q \bigr)\cdot\partial_{h}^{3} \eta+{\rho}\big|\partial_{h}^{3} u\big|^{2}. \end{gathered} $$
$$ \begin{aligned}[b] &\frac{\mathrm{d}}{\mathrm{d}t} \biggl( \int{\rho} \partial_{h}^{3} \eta\cdot \partial_{h}^{3} u \,\mathrm{d}y+\frac{\mu}{2}\big\| \nabla \partial_{h}^{3} \eta\big\| _{0}^{2} \biggr) +\lambda_{0} \big\| \bar{M}\cdot\nabla\partial_{h}^{3} \eta\big\| _{0}^{2} \\ &\quad= 2\rho\omega \int\partial_{h}^{3} (u_{2}e_{1}-u_{1}e_{2}) \cdot\partial_{h}^{3}\eta\,\mathrm{d}y+ \int\bigl(\mu\partial_{h}^{3}N^{h}_{u}+ \partial_{h}^{3}N^{h}_{q}\bigr)\cdot \partial_{h}^{3}\eta\,\mathrm{d}y \\ &\quad\quad{}+ \int\partial_{h}^{3} q\operatorname{div}\partial^{3}_{h} \eta\,\mathrm{d}y +\big\| \sqrt{{\rho}}\partial_{h}^{3} u \big\| ^{2}_{0}\\&\quad \lesssim\sum_{k=1}^{3}K^{L}_{k}+ \| u\|^{2}_{3,0}, \end{aligned} $$
(3.25)
We have the boundedness Noting that we find that
$$ K^{L}_{1} \leq \big\| \partial_{h}^{3} u\big\| _{0}\big\| \partial_{h}^{3} \eta\big\| _{0}. $$
(3.26)
$$\begin{aligned} \big\| \mu\partial_{h}^{3}N^{h}_{u}+ \partial_{h}^{3}N^{h}_{q} \big\| _{0} & \lesssim\|\tilde {\mathcal {A}}\| _{4}\|u\|_{3}+\|\tilde {\mathcal {A}}\|_{2}\|u\|_{5} +\|\tilde {\mathcal {A}}\|_{2}\|\nabla q \|_{3}+\|\tilde {\mathcal {A}}\|_{4}\| \nabla q\|_{1} \\ &\lesssim\sqrt{\mathcal{E}^{H}\mathcal{D}^{L}}, \end{aligned} $$
$$ K^{L}_{2} \lesssim\big\| \mu\partial_{h}^{3}N^{h}_{u}+ \partial_{h}^{3}N^{h}_{q} \big\| _{0}\big\| \partial_{h}^{3}\eta\big\| _{0} \lesssim\sqrt{\mathcal{E}^{H}} \mathcal{D}^{L}. $$
(3.27)
Next we estimate the third integral \(K_{3}^{L}\). To start with, we analyze the property of divη. Since we have by Sarrus’ rule Multiplying the above identity by a smooth test function ϕ, and then integrating (by parts) the resulting identity over Ω, we derive that where This means that Thus, it follows immediately that
$$\det(I+\nabla\eta)=1, $$
$$\begin{aligned} \operatorname{div}\eta={}&\partial_{1} \eta_{2}\partial_{2} \eta_{1}+ \partial_{2} \eta_{3}\partial_{3}\eta_{2}+\partial_{3} \eta_{1}\partial_{1}\eta_{3} -\partial_{1} \eta_{1}\partial_{2} \eta_{2}- \partial_{1} \eta_{1}\partial_{3}\eta_{3}-\partial_{2} \eta_{2}\partial_{3}\eta_{3} \\ &+ \partial_{1}\eta_{1}(\partial_{2} \eta_{3}\partial_{3}\eta_{2} -\partial _{2}\eta_{2}\partial_{3}\eta_{3})+ \partial_{2}\eta_{1}(\partial_{1}\eta_{2} \partial_{3}\eta_{3} -\partial_{1} \eta_{3}\partial_{3}\eta_{2}) \\&+\partial_{3} \eta_{1}(\partial_{1}\eta_{3}\partial_{2} \eta_{2} -\partial_{1}\eta_{2}\partial_{2} \eta_{3}). \end{aligned} $$
$$\int\phi\operatorname{div}\eta\,\mathrm{d}y=- \int\nabla\phi\cdot\psi\,\mathrm{d}y, $$
$$\psi:=\left ( \textstyle\begin{array}{c} \eta_{1}(\partial_{2}\eta_{2}+\partial_{3}\eta_{3} )+ \eta_{1}(\partial_{2}\eta_{3} \partial_{3}\eta_{2} - \partial_{2}\eta_{2}\partial_{3}\eta_{3}) \\ \eta_{2}\partial_{3}\eta_{3}-\eta_{1}\partial_{1}\eta_{2}+ \eta_{1}( \partial_{1}\eta_{2} \partial_{3}\eta_{3} -\partial_{1}\eta_{3}\partial_{3}\eta_{2})\\ -\eta_{1}\partial_{1}\eta_{3} -\eta_{2}\partial_{2}\eta_{3}+\eta_{1}(\partial_{1}\eta_{3} \partial_{2}\eta_{2} -\partial_{1}\eta_{2}\partial_{2}\eta_{3}) \end{array}\displaystyle \right ). $$
$$\operatorname{div}\eta=\operatorname{div}\psi. $$
$$ \begin{aligned}[b] K^{L}_{3} &= - \int\partial_{h}^{3} \nabla q\cdot \partial_{h}^{3}\psi\,\mathrm{d}y=- \int\partial_{h} \nabla q\cdot\partial_{h}^{5} \psi\,\mathrm{d}y \\ &\lesssim\|\nabla q\|_{1}\big\| \partial_{h}^{5}\psi \big\| _{0} \lesssim\| \eta\| _{6}\|\nabla q\|_{1}\| \eta\|_{3} \lesssim\sqrt{\mathcal{E}^{H}} \mathcal{D}^{L}. \end{aligned} $$
(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 Integrating (by parts) the above identity over Ω, we have
$$\begin{gathered} {{\rho}} \partial_{h}^{3}u_{t} \cdot\partial_{h}^{3}u- \mu\Delta\partial _{h}^{3} u \cdot\partial_{h}^{3} u- \lambda_{0} \bar{M}\cdot\nabla\bigl(\bar{M}\cdot\nabla \partial_{h}^{3}\eta\bigr)\cdot\partial_{h}^{3} \eta_{t} \\ \quad=\bigl( 2\rho\omega\partial_{h}^{3}(u_{2}e_{1}-u_{1}e_{2}) +\mu\partial_{h}^{3}N^{h}_{u}+ \partial_{h}^{3}N^{h}_{q} -\nabla \partial_{h}^{3} q\bigr)\cdot\partial_{h}^{3}u. \end{gathered} $$
$$ \begin{aligned}[b] &\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \biggl( \int{\rho} \big|\partial_{h}^{3} u\big|^{2} \,\mathrm{d}y+\lambda_{0} \big\| \bar{M}\cdot\nabla\partial_{h}^{3} \eta\big\| _{0}^{2} \biggr)+\frac{\mu}{2}\big\| \nabla \partial_{h}^{3} u\big\| _{0}^{2} \\ &= \int\bigl(\mu\partial_{h}^{3}N^{h}_{u}+ \partial_{h}^{3}N^{h}_{q}\bigr)\cdot \partial_{h}^{3} u\,\mathrm{d}y + \int\partial_{h}^{3} q\operatorname{div}\partial^{3}_{h} u\,\mathrm{d}y =: M^{L}_{1}+M^{L}_{2}. \end{aligned} $$
(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:
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 \(j=3\). □
$$\begin{aligned}& M^{L}_{1} \lesssim\big\| \mu\partial_{h}^{3}N^{h}_{u}+ \partial_{h}^{3}N^{h}_{q} \big\| _{0}\big\| \partial_{h}^{3}u\big\| _{0}\lesssim \sqrt{\mathcal{E}^{H}} \mathcal{D}^{L}, \end{aligned}$$
(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)
3.4 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: coupled with boundary condition where \(\omega=\lambda_{0} m^{2}_{3} \eta+\mu u\).
$$ \left\{\textstyle\begin{array}{l} -\Delta w+\nabla q\\ \quad = \lambda_{0} \bar{M}\cdot\nabla(\bar{M}\cdot \nabla\eta)-\lambda_{0} m_{3}^{2} \Delta\eta+ 2\rho\omega (u_{2}e_{1}-u_{1}e_{2})-{\rho} u_{t}+ \mu N^{h}_{u}+ N^{h}_{q},\\ \operatorname {div}w =\mu D^{h}_{u}+ \lambda_{0} m^{2}_{3}\operatorname{div}\eta, \end{array}\displaystyle \right. $$
(3.32)
$$ \omega|_{\partial\Omega}=0, $$
(3.33)
Now, applying \(\partial^{k}_{h}\) to (3.32) and (3.33), we get Then we apply the classical regularity theory to the Stokes problem as in [18], Proposition 2.3, to deduce that where
$$ \left\{\textstyle\begin{array}{l} -\Delta\partial^{k}_{h} w+\nabla\partial^{k}_{h}q\\ \quad =\partial ^{k}_{h}(\lambda_{0} \bar{M}\cdot\nabla(\bar{M}\cdot\nabla\eta )-\lambda_{0} m_{3}^{2} \Delta\eta + 2\rho\omega (u_{2}e_{1}-u_{1}e_{2}))-{\rho} \partial^{k}_{h} u_{t}+ \mu\partial ^{k}_{h}N^{h}_{u}+N^{h}_{q},\\ \operatorname {div}\partial^{k}_{h}w =\mu\partial^{k}_{h}D^{h}_{u}+\lambda_{0} m^{2}_{3}\operatorname{div}\partial^{k}_{h} \eta,\\ \partial^{k}_{h}\omega|_{\partial\Omega}=0. \end{array}\displaystyle \right. $$
$$ \|\omega\|_{k,i-k+2}^{2}+\|\nabla q \|_{k,i-k}^{2} \lesssim\| \nabla\eta\|_{k+1,i-k}^{2} +\big\| ( u, u_{t})\big\| _{k,i-k}^{2} +S_{k ,i}^{\omega}, $$
(3.34)
$$S_{k,i}^{\omega}:=\big\| \bigl(N^{h}_{u}, N^{h}_{q}\bigr)\big\| _{k,i-k}^{2} +\big\| \bigl(D^{h}_{u} ,\operatorname{div}\eta\bigr) \big\| _{k,i-k+1}^{2}. $$
In addition, applying \(\partial_{t}^{k}\) to (3.24)2-(3.24)3, we see that Hence, we apply again the classical regularity theory to the Stokes problem to get 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.
$$\left\{\textstyle\begin{array}{l} -\mu\Delta\partial_{t}^{k} u+\nabla\partial_{t}^{k} q\\ \quad= \lambda_{0} \bar{M}\cdot \partial_{t}^{k} \nabla(\bar{M}\cdot\nabla\eta) + 2\rho\omega\partial_{t}^{k} (u_{2}e_{1}-u_{1}e_{2})-{\rho} \partial_{t}^{k+1}u +\mu\partial_{t}^{k}N^{h}_{u}+ \partial_{t}^{k}N^{h}_{q}, \\ \operatorname {div}\partial_{t}^{k}u = \partial_{t}^{k}D^{h}_{u}, \\ \partial_{t}^{k}u|_{\partial\Omega}=0 . \end{array}\displaystyle \right. $$
$$ \big\| \partial_{t}^{k} u\big\| _{i-2k+2}^{2}+ \big\| \nabla\partial_{t}^{k} q\big\| _{i-2k}^{2} \lesssim\big\| \partial^{k}_{t}\bigl( \nabla^{2} \eta,u,u_{t}\bigr)\big\| _{i-2k}^{2} +S_{k,i}^{u}, $$
(3.35)
Lemma 3.4
We have
where
\(E^{L}:=\mathcal{E}^{L}-\|\nabla\eta\|_{3,0}^{2}\)
and
\(\|\eta\| _{3,*}\)
is equivalent to
\(\|\eta\|_{3}\).
$$\begin{aligned}& \frac{\mathrm{d}}{\mathrm{d}t} \|\eta\| _{3,*}^{2}+c\bigl(\big\| (\eta,u)\big\| _{3}^{2} +\| \nabla q\|_{1}^{2} \bigr) \lesssim\big\| (u,u_{t}) \big\| _{1}^{2}+\| \eta \|_{2,1}^{2} +{\mathcal{E}^{H} } { { \mathcal{D}}^{L}}, \end{aligned}$$
(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)
Proof
Noting that, by virtue of (2.15)1, we deduce from (3.34) that In particular, we take \((i,k)=(1,0)\) and \((i,k)=(1,1)\) to get and On the other hand, it is easy to show that Thus we immediately obtain (3.36) from (3.40)-(3.42).
$$\|\omega\|_{ k,i- k+2}^{2} =\big\| \bigl( \lambda_{0} m^{2}_{3} \eta,\mu u\bigr) \big\| _{{ k,i- k+2} }^{2} +\frac{\lambda_{0} m^{2}_{3}\mu}{2}\frac{\mathrm{d}}{\mathrm {d}t} \|\eta \|_{{ k,i- k+2}}^{2}, $$
$$ \begin{aligned}[b] & \frac{\mathrm{d}}{\mathrm{d}t} \|\eta \|_{{ k,i- k+2}}^{2} +c\bigl(\big\| (\eta, u)\big\| _{ k,i- k+2 }^{2}+ \|\nabla q\|_{ k,i- k}^{2}\bigr) \\ &\quad\lesssim\|u\|_{k,i-k}^{2}+ \| \eta\|_{ k+1, i- k+1}^{2}+ \| u_{t}\| _{i}^{2}+S_{ k,i}^{\omega}. \end{aligned} $$
(3.39)
$$ \frac{\mathrm{d}}{\mathrm{d}t}\|\eta\|_{3}^{2} +c\bigl(\big\| (\eta, u)\big\| _{3}^{2}+\| \nabla q\|_{1}^{2} \bigr)\lesssim\big\| (u ,u_{t})\big\| _{1}^{2}+\|\eta \|_{1,2}^{2}+ S_{0,1}^{\omega}$$
(3.40)
$$ \frac{\mathrm{d}}{\mathrm{d}t}\|\eta\|_{1,2}^{2} +c\bigl(\big\| (\eta, u)\big\| _{1,2}^{2}+\| \nabla q\|_{1,0}^{2} \bigr)\lesssim\big\| (u ,u_{t})\big\| _{1}^{2}+\|\eta \|_{2,1}^{2}+ S_{1,1}^{\omega}. $$
(3.41)
$$ \begin{aligned}[b] S_{0,1}^{\omega}+ S_{1,1}^{\omega}&\leq\big\| \bigl( N^{h}_{u},N^{h}_{q} \bigr)\big\| _{1}^{2}+\big\| \bigl(D^{h}_{u}, \operatorname{div} \eta\bigr)\big\| _{2}^{2} \lesssim\|\tilde {\mathcal {A}}\|_{2}^{2}\bigl(\|u\|_{3}^{2}+\| \nabla q \|_{1}^{2}\bigr)+\|\eta\|_{3}^{4} \\ &\lesssim {\mathcal{E}^{H}{{E}}^{L}}\lesssim{ \mathcal{E}^{H} {\mathcal{D}}^{L}}. \end{aligned} $$
(3.42)
Now we turn to the derivation of (3.37). In view of (3.35) with \((i,k)=(1,0)\), we have On the other hand, we can use (3.42) to infer that Hence, (3.37) follows from the above two estimates.
$$\| u\|_{3}^{2}+\|\nabla q\|_{1}^{2} \lesssim\| \eta\|_{3}^{2} +\big\| (u,u_{t}) \big\| _{1}^{2} +S_{0,1}^{u}. $$
$$S_{0,1}^{u}=\big\| \bigl(N^{h}_{u} ,N^{h}_{q}\bigr)\big\| _{1}^{2}+\big\| D^{h}_{u} \big\| _{2}^{2} \lesssim{ \mathcal{E}^{H} { {E}}^{L} }. $$
Finally, to show (3.38), we take \((i,k)=(2,1)\) in (3.35) to deduce that Keeping in mind that we get (3.38) from the above two estimates. □
$$\|u_{t}\|_{2}^{2}+\|\nabla q_{t} \|_{0}^{2} \lesssim\big\| \bigl(\nabla^{2} u,u_{t},u_{tt} \bigr)\big\| _{0}^{2}+ S_{1,2}^{u} \lesssim\| u\| _{2}^{2}+\big\| ( u_{t},u_{tt})\big\| _{0}^{2} + S_{1,2}^{u}. $$
$$\begin{aligned} S_{1,2}^{u}&\lesssim\big\| \partial_{t}\bigl( N^{h}_{u},N^{h}_{q} \bigr)\big\| _{0}^{2}+\big\| \partial_{t} D^{h}_{u} \big\| _{1}^{2} \\ &\lesssim \|\tilde {\mathcal {A}}_{t}\|_{2}^{2}\bigl(\|u \|_{2}^{2}+\|\nabla q\|_{0}^{2}\bigr)+\| \tilde {\mathcal {A}}\|_{2}^{2}\bigl(\| u_{t}\|_{2}^{2}+ \|\nabla q_{t}\|_{0}^{2}\bigr) \lesssim{ \mathcal{E}^{H}{\mathcal{D}}^{L} }, \end{aligned} $$
3.5 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 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 where \(\sigma_{0}\) depends on the domain and the known physical parameters, and Utilizing (3.13), (3.36), the interpolation inequality, and the estimate we find that Now, multiplying (3.47) by \(c_{2}^{L}/(2c_{5}^{L})\) and adding the resulting inequality to (3.45), we obtain where \(\tilde{{\mathcal{E}}}^{L}_{2}\) and \(\tilde{\mathcal{D}}^{L} \) are defined by 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:
$$ \big\| \nabla\partial_{t}^{j} u \big\| ^{2}_{0}\lesssim\big\| \nabla_{\mathcal{A}} \partial_{t}^{j} u\big\| ^{2}_{0},\quad1 \leq j\leq3. $$
(3.44)
$$ \frac{\mathrm{d}}{\mathrm{d}t}\tilde{\mathcal{E}}^{L}_{1} +\tilde{\mathcal{D}}^{L}_{1} \leq c_{1}^{L} (1+ \sigma) \sqrt{\mathcal{E}^{H}} \mathcal{D}^{L} \quad \mbox{for any }\sigma\geq\sigma_{0} , $$
(3.45)
$$\begin{aligned}& \begin{aligned} \tilde{\mathcal{E}}^{L}_{1}:={}& \|\sqrt{{\rho}} u_{t}\|_{0}^{2}+ \lambda_{0} \| \bar{M}\cdot\nabla u\|^{2}_{0} + \sigma\sum_{k=0}^{3}\bigl( \|\sqrt{ {\rho} } u\|^{2}_{k,0} + \lambda_{0} \|\bar{M}\cdot\nabla \eta\|_{k,0}^{2} \bigr) \\ &+\sum_{k=0}^{3} \biggl(\sum _{\alpha_{1}+\alpha_{2}=k} \int{\rho} \partial_{1}^{\alpha_{1}}\partial_{2}^{\alpha_{2}} \eta\cdot\partial_{1}^{\alpha_{1}}\partial_{2}^{\alpha_{2}} u\,\mathrm{d}y +\frac{ \mu}{2}\|\nabla\eta\|_{k,0}^{2} \biggr), \end{aligned}\\& \tilde{{\mathcal{D}}}^{L}_{1}:= c_{2}^{L} \Biggl(\| u_{t}\|^{2}_{1}+ \sum _{k=0}^{3} \bigl(\big\| (\eta,\bar{M}\cdot\nabla\eta) \big\| _{k,0}^{2}+\|u\| _{k,1}^{2}\bigr) \Biggr). \end{aligned}$$
$$ \|\eta\|_{k,1}^{2}\lesssim\|\eta \|_{k+1,0}^{2}+\|\bar{M}\cdot\nabla\eta\|_{k,0}^{2} \quad\mbox{for any }0\leq k\leq6, $$
(3.46)
$$ \begin{aligned}[b] &\frac{\mathrm{d}}{\mathrm{d}t} \bigl( \|\eta \|_{3,*}^{2}+c_{3}^{L} \| \nabla_{\mathcal {A}} u_{t}\|_{0}^{2} \bigr)+ c_{4}^{L}\bigl( \big\| (\eta, u)\big\| _{3}^{2}+ \|\nabla q\|_{1}^{2} +\| u_{tt} \|^{2}_{0}\bigr) \\ & \quad\leq c_{5}^{L}\bigl(\big\| (u,u_{t}) \big\| _{1}^{2} +\|\eta\|_{3,0}^{2}+\|\bar{M} \cdot\nabla\eta\|_{2,0}^{2}+\sqrt{\mathcal{E}^{H}} \mathcal{D}^{L}\bigr). \end{aligned} $$
(3.47)
$$ \begin{aligned} &\frac{\mathrm{d}}{\mathrm{d}t} \tilde{{ \mathcal{E}}}^{L}_{2} +\tilde{\mathcal{D}}^{L} \leq c_{6}^{L}(1+\sigma)\sqrt{\mathcal{E}^{H}} \mathcal{D}^{L}, \end{aligned} $$
(3.48)
$$\begin{gathered} \tilde{ {\mathcal{E}}}^{L}_{2}:=\tilde{\mathcal {E}}^{L}_{1}+ c_{2}^{L}\bigl(\|\eta \|_{3,*}^{2} + c_{3}^{L}\| \nabla_{\mathcal{A}} u_{t}\| _{0}^{2} \bigr)/ \bigl(2c_{5}^{L}\bigr), \\ \tilde{\mathcal{D}}^{L}:= \tilde{{\mathcal{D}}}^{L}_{1}/2+ c_{2}^{L} c_{4}^{L} \bigl(\big\| (\eta,u) \big\| _{3}^{2}+\|\nabla q\|_{1}^{2} + \|u_{tt}\|^{2}_{0}\bigr)/\bigl(2c_{5}^{L} \bigr). \end{gathered}$$
$$ \begin{aligned} &\frac{\mathrm{d}}{\mathrm{d}t} \tilde{{ \mathcal{E}}}^{L}_{2} +c_{7}^{L}{ \mathcal{D}}^{L} \leq c_{6}^{L}(1+\sigma)\sqrt{ \mathcal{E}^{H}} \mathcal{D}^{L}. \end{aligned} $$
(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 In view of (3.37), we further have 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. □
$$\|\nabla\eta\|_{3,0}^{2}+ \|\eta\|_{3}^{2}+ \|u_{t}\|_{1}^{2} \lesssim\tilde{{ \mathcal{E}}}^{L}_{2}. $$
$${\mathcal{E}}^{L} \lesssim\tilde{{\mathcal{E}}}^{L}_{2} +\sqrt{\mathcal{E}^{H}} {\mathcal{E}}^{L}, $$
4 Higher-order and highest-order energy inequalities
In this section, we derive the estimates (2.20) and (2.21) for the transformed magnetic RT problem. First we shall establish the higher-order version of Lemmas 3.1-3.4 in the following:
Lemma 4.1
We have
$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{d}t} \biggl(\big\| \sqrt {{\rho}} \partial_{t}^{j } u \big\| _{0}^{2}+ \lambda_{0} \big\| \bar{M}\cdot\nabla\partial_{t}^{j-1} u \big\| ^{2}_{0} + \int\nabla\partial_{t}^{j-1}q\cdot D^{t,j}_{u}\,\mathrm{d}y \biggr)+\mu\big\| \nabla_{\mathcal{A}}\partial_{t}^{j} u \big\| ^{2}_{0} \\ \quad\lesssim\sqrt{\mathcal{E}^{L}} \mathcal{D}^{H},\quad j=2,3. \end{gathered} $$
Proof
We only prove the case \(j=3\), and the case \(j=2\) can be shown similarly. Multiplying (3.10)2 with \(j=3\) by \(\partial_{t}^{3}u\), integrating (by parts) the resulting equation over Ω, and using (3.10)1 and (3.10)3, we obtain On the other hand, the integrals \(I^{H}_{1}\), \(I_{2}\) and \(I^{H}_{3}\) can be estimated as follows: Inserting the above three inequalities into (4.1), we get the desired conclusion immediately. □
$$ \begin{aligned}[b] &\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \bigl(\big\| \sqrt{{\rho}}\partial_{t}^{3 } u \big\| _{0}^{2}+\lambda_{0} \|\bar{M}\cdot\nabla u_{tt}\| ^{2}_{0} \bigr) +\mu\big\| \nabla_{\mathcal{A}} \partial_{t}^{3 }u\big\| ^{2}_{0} \\ &\quad = \int\partial_{t}^{3 }q\operatorname{div} D_{u}^{t,3} \,\mathrm{d}y + \mu \int N^{t,3}_{u}\cdot\partial_{t}^{3 }u \,\mathrm{d}y+ \int N^{t,3}_{q} \cdot\partial_{t}^{3 }u \,\mathrm{d}y:=\sum_{k=1}^{3}I^{H}_{k}. \end{aligned} $$
(4.1)
$$\begin{aligned}& \begin{aligned}I^{H}_{1} ={} & {-} \int\nabla\partial_{t}^{3}q\cdot D^{t,3}_{u}\,\mathrm{d}y=-\frac{\mathrm{d}}{\mathrm{d}t} \int\nabla\partial_{t}^{2}q\cdot D^{t,3}_{u}\,\mathrm{d}y + \int\nabla\partial_{t}^{2}q\cdot \partial_{t}D^{t,3}_{u}\,\mathrm{d}y \\ \leq{}& {-}\frac{\mathrm{d}}{\mathrm{d}t} \int\nabla\partial_{t}^{2}q\cdot D^{t,3}_{u}\,\mathrm{d}y \\ & +c\big\| \nabla\partial_{t}^{2}q \big\| _{0}\bigl(\big\| \partial_{t}^{4 }\mathcal{A} \big\| _{0}\|u\|_{2}+\big\| \partial_{t}^{3}\mathcal{A}\big\| _{1} \|u_{t}\|_{1}+\|\mathcal{A}_{tt}\|_{0}\| u_{tt}\|_{2} +\|\mathcal{A}_{t}\|_{1}\big\| \partial_{t}^{3}u\big\| _{1}\bigr) \\ \leq{}& {-}\frac{\mathrm{d}}{\mathrm{d}t} \int\nabla\partial_{t}^{2}q\cdot D^{t,3}_{u}\,\mathrm{d}y+c\sqrt{\mathcal{E}^{L}} \mathcal{D}^{H}, \end{aligned} \\& \begin{aligned} I^{H}_{2} \lesssim{}& \bigl[\big\| \partial_{t}^{3}\mathcal{A}\big\| _{0}\|\mathcal{A}\| _{2}\|u\|_{3} +\|\mathcal{A}_{tt} \|_{0}\bigl( \|\mathcal{A}_{t}\|_{2}\|u\| _{3} +\|\mathcal{A}\|_{2}\|u_{t}\|_{3}\bigr) \\ & + \|\mathcal{A}_{t}\|_{2}\bigl( \|\mathcal{A}_{tt} \|_{2} \|u\|_{2} + \|\mathcal{A}_{t}\| _{2} \|u_{t}\|_{2}+\|\mathcal{A}\|_{2}\|u_{tt} \|_{2}\bigr) \\ & + \|\mathcal{A}\|_{2}\bigl(\big\| \partial_{t}^{3}\mathcal{A} \big\| _{1}\| u\|_{2}+\|\mathcal{A}_{tt}\|_{2} \|u_{t}\|_{1}+\|\mathcal{A}_{tt}\|_{0} \|u_{t}\|_{3} + \|\mathcal{A}_{t}\| _{1} \|u_{tt}\|_{2}\bigl)\bigr]\big\| \partial_{t}^{3}u \big\| _{1} \\ \lesssim{}& \sqrt{\mathcal{E}^{L}}\mathcal{D}^{H}, \end{aligned} \\& I^{H}_{3} \lesssim\bigl(\big\| \partial_{t}^{3}\mathcal{A}\big\| _{0}\|\nabla q \|_{1}+\|\mathcal{A}_{tt}\|_{0}\|\nabla q_{t}\|_{1} +\|\mathcal{A}_{t}\|_{1}\| \nabla q_{tt}\|_{0}\bigr)\big\| \partial_{t}^{3}u \big\| _{1}\lesssim\sqrt{\mathcal{E}^{L}} \mathcal{D}^{H}. \end{aligned}$$
Lemma 4.2
We have
$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{d}t} \biggl( \int{\rho} \partial_{h}^{k}\eta\cdot \partial_{h}^{k} u \,\mathrm{d}y+\frac{\mu}{2}\big\| \nabla \partial_{h}^{k} \eta\big\| _{0}^{2} \biggr) +\lambda_{0} \big\| \bar{M}\cdot\nabla\partial_{h}^{k} \eta\big\| _{0}^{2} \\ \quad\lesssim\sqrt{\mathcal{E}^{L}}\bigl(\big\| (\eta,u)\big\| _{7}^{2}+ \mathcal{D}^{H}\bigr) +\big\| \partial_{h}^{k}u \big\| _{0}^{2}\quad\textit{for any }k=5\textit{ and }6. \end{gathered} $$
Proof
We only show the case \(k=6\), since the derivation of the case \(k=5\) is similar. Since the derivation involves the norm \(\|\nabla q\|_{5}\), we first estimate \(\|\nabla q\|_{5}\). It follows from (3.35) that The last two terms on the right-hand side of the above inequality can be bounded as follows: Hence, if δ is sufficiently small, then one gets from the above two estimates that
$$ \| u\|_{7}+\|\nabla q\|_{5} \lesssim\|\eta\|_{7} +\big\| (u,u_{t})\big\| _{5} + \big\| \bigl(N^{h}_{u}, N^{h}_{q}\bigr) \big\| _{5}+\big\| D^{h}_{u}\big\| _{6}. $$
$$\big\| \bigl(N^{h}_{u}, N^{h}_{q}\bigr)\big\| _{5} + \big\| D^{h}_{u} \big\| _{6} \lesssim \|\tilde {\mathcal {A}}\|_{6}\|u\| _{7} +\|\tilde {\mathcal {A}}\|_{5}\|\nabla q\|_{5}. $$
$$ \|u\|_{7}+\|\nabla q\|_{5} \lesssim\|\eta \|_{7} +\sqrt{\mathcal{D}^{H}}. $$
(4.2)
Now, applying \(\partial_{h}^{6}\) to (3.24)2, multiplying the resulting equality by \(\partial^{6}_{h}\eta\), we utilize (3.24)1 to have Integrating (by parts) the above identity over Ω, we obtain where the first four integrals on the right-hand side are denoted by \(J^{H}_{1},\ldots,J^{H}_{4}\), respectively.
$$\begin{gathered} {\rho} \partial_{t}\bigl( \partial_{h}^{6}\eta\cdot\partial_{h}^{6} u \bigr)-\bigl(\mu\Delta\partial_{h}^{6} \eta_{t}+\lambda_{0} \bar{M}\cdot\nabla\bigl(\bar{M}\cdot \nabla\partial_{h}^{6}\eta\bigr)\bigr)\cdot \partial_{h}^{6}\eta \\ \quad =\partial_{h}^{6}\bigl( 2\rho\omega( u_{2}e_{1}-u_{1}e_{2}) +\mu N^{h}_{u}+ N^{h}_{q} -\nabla q \bigr) \cdot\partial_{h}^{6}\eta+{\rho}\big|\partial_{h}^{6} u\big|^{2}. \end{gathered} $$
$$ \begin{aligned}[b] &\frac{\mathrm{d}}{\mathrm{d}t} \biggl( \int{\rho} \partial_{h}^{6} \eta\cdot \partial_{h}^{6} u \,\mathrm{d}y+\frac{\mu}{2}\big\| \nabla \partial_{h}^{6} \eta\big\| _{0}^{2} \biggr) +\lambda_{0} \big\| \bar{M}\cdot\nabla\partial_{h}^{6} \eta\big\| _{0}^{2} \\ &\quad=2\rho\omega \int\partial_{h}^{6} ( u_{2}e_{1}-u_{1}e_{2}) \cdot\partial_{h}^{6}\eta\,\mathrm{d}y+ \mu \int\partial_{h}^{6}N^{h}_{u} \cdot\partial_{h}^{6}\eta\,\mathrm{d}y- \int\partial_{h}^{5}N^{h}_{q} \cdot\partial_{h}^{7}\eta\,\mathrm{d}y \\ &\quad\quad{}- \int\partial_{h}^{6} q\operatorname{div}\partial^{6}_{h} \eta\,\mathrm{d}y +\big\| \sqrt{{\rho}}\partial_{h}^{6} u \big\| ^{2}_{0}\\ &\quad \lesssim\sum_{k=1}^{4}J^{H}_{k} +\big\| \partial_{h}^{6}u\big\| ^{2}_{0}, \end{aligned} $$
(4.3)
On the other hand, the four integrals \(J^{H}_{1},\ldots,J^{H}_{4}\) can be bounded as follows:
where the interpolation inequality (3.3) has been employed in (4.6) and (4.7). Consequently, putting the above four estimates into (4.3), we obtain Lemma 4.2. □
$$\begin{aligned}& J^{H}_{1}\lesssim\big\| \partial_{h}^{6} \eta\big\| _{0}\big\| \partial_{h}^{6}u\big\| _{0} , \end{aligned}$$
(4.4)
$$\begin{aligned}& \begin{aligned}[b] J^{H}_{2}={}&{-}\mu \int\partial_{h}^{6} \bigl((\tilde{ \mathcal{A}}_{jl} \tilde{\mathcal{A}}_{jk} + \tilde{ \mathcal{A}}_{lk}+\tilde{\mathcal{A}}_{kl})\partial_{k}u_{i} \bigr) \partial_{l}\partial_{h}^{6} \eta_{i}\,\mathrm{d}y \\ \lesssim{}& \bigl(\|\tilde {\mathcal {A}}\|_{6}\|u\|_{3}+\|\tilde {\mathcal {A}}\|_{4}\|u \|_{5}+\|\tilde {\mathcal {A}}\|_{2}\|u\|_{7}\bigr)\| \eta \|_{6,1} \\ \lesssim{}&\bigl(\|\tilde {\mathcal {A}}\|_{6}\|u\|_{3} + \|\tilde {\mathcal {A}}\|_{2}^{\frac{1}{2}}\|\tilde {\mathcal {A}}\| _{6}^{\frac{1}{2}}\|u \|_{3}^{\frac{1}{2}} \|u\|_{7}^{\frac{1}{2}}+\|\tilde {\mathcal {A}}\|_{2}\|u\|_{7}\bigr)\|\eta\|_{6,1} \\ \lesssim{}&\sqrt{\mathcal{E}^{L}}\big\| (\eta,u)\big\| _{7}^{2}, \end{aligned} \end{aligned}$$
(4.5)
$$\begin{aligned}& \begin{aligned}[b] J^{H}_{3} \lesssim{}& \bigl(\|\tilde {\mathcal {A}}\|_{6}\|\nabla q\|_{1}+ \|\tilde {\mathcal {A}}\|_{3}\|\nabla q\|_{4}+\| \tilde {\mathcal {A}}\|_{2}\|\nabla q \|_{5}\bigr)\|\eta\|_{7} \\ \lesssim{}&\bigl(\|\tilde {\mathcal {A}}\|_{6}\|\nabla q\|_{1}+ \|\tilde {\mathcal {A}}\|_{2}^{\frac{3}{4}} \|\tilde {\mathcal {A}}\|_{6}^{\frac{1}{4}}\|\nabla q\| _{1}^{\frac{1}{4}}\|\nabla q\|_{5}^{\frac{3}{4}}+\|\tilde {\mathcal {A}}\|_{2}\|\nabla q\| _{5}\bigr)\|\eta\|_{7} \\ \lesssim{}&\sqrt{\mathcal{E}^{L}}\bigl(\|\eta\|_{7}^{2}+ \mathcal{D}^{H}\bigr), \end{aligned} \end{aligned}$$
(4.6)
$$\begin{aligned}& \begin{aligned}[b] J^{H}_{4}={}& \int\partial_{h}^{6} q\cdot\partial_{h}^{6} \operatorname{div}\eta\,\mathrm{d}y \lesssim\|\nabla q\|_{5}\bigl(\| \eta \|_{3}\|\eta\|_{7}+\|\eta\|_{5}^{2}\bigr) \\ \lesssim{}&\|\eta\|_{3}\|\nabla q\|_{5}\bigl( \|\eta \|_{5}+\|\eta\|_{7}\bigr) \lesssim\sqrt{\mathcal{E}^{L}} \bigl(\|\eta\|_{7}^{2}+\mathcal{D}^{H}\bigr), \end{aligned} \end{aligned}$$
(4.7)
Lemma 4.3
We have
$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{d}t} \bigl(\big\| \sqrt{{\rho} } \partial _{h}^{k} u \big\| ^{2}_{0}+ \lambda_{0} \big\| \bar{M}\cdot\nabla \partial_{h}^{k}\eta\big\| _{0}^{2} \bigr) + \mu\big\| \nabla\partial_{h}^{k} u\big\| _{0}^{2}\\ \quad\lesssim\sqrt{\mathcal{E}^{L}}\bigl(\big\| (\eta,u)\big\| _{7}^{2}+ \mathcal{D}^{H}\bigr)\quad\textit{for }k=5\textit{ and }6. \end{gathered}$$
Proof
We only show the case \(k=6\), since the derivation of the case \(k=5\) is similar. Applying \(\partial_{h}^{6}\) to (3.24)2, multiplying the resulting equality by \(\partial^{6}_{h} u\), we make use of (3.24)1 to have Integrating (by parts) the above identity over Ω, we get
$$\begin{gathered} {{\rho}} \partial_{h}^{6}u_{t} \cdot\partial_{h}^{6}u - \mu\Delta\partial _{h}^{6} u\cdot\partial_{h}^{6} u- \lambda_{0} m^{2}\partial_{3}^{2}\partial _{h}^{6}\eta\cdot\partial_{h}^{6} \eta_{t} \\ \quad =\bigl(2\rho\omega\partial_{h}^{6} ( u_{2}e_{1}-u_{1}e_{2}) +\mu\partial _{h}^{6}N^{h}_{u}+ \partial_{h}^{6}N^{h}_{q}-\nabla \partial_{h}^{6} q \bigr)\cdot\partial_{h}^{6}u. \end{gathered} $$
$$ \begin{aligned}[b] &\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \biggl( \int{\rho} \big|\partial_{h}^{6} u\big|^{2} \,\mathrm{d}y+\lambda_{0} \big\| \bar{M}\cdot\nabla\partial_{h}^{6} \eta\big\| _{0}^{2} \biggr)+\frac{\mu}{2}\big\| \nabla \partial_{h}^{6} u\big\| _{0}^{2} \\ &\quad= \mu \int\partial_{h}^{6}N^{h}_{u} \cdot\partial_{h}^{6} u\,\mathrm{d}y - \int\partial_{h}^{5}N^{h}_{q} \cdot\partial_{h}^{7} u\,\mathrm{d}y+ \int\partial_{h}^{6} q\operatorname{div} \partial^{6}_{h} u\,\mathrm{d}y =:\sum _{k=1}^{3}M^{H}_{k}. \end{aligned} $$
(4.8)
Analogously to (4.4)-(4.7), the three integrals \(M^{H}_{1}\), \(M^{H}_{2}\) and \(M^{H}_{3}\) can be bounded as follows:
$$\begin{gathered}M^{H}_{1} =- \int\partial_{h}^{6} \bigl[(\tilde{ \mathcal{A}}_{jl} \tilde{\mathcal{A}}_{jk}+ \tilde{ \mathcal{A}}_{lk}+\tilde{\mathcal{A}}_{kl})\partial_{k}u_{i} \bigr]\cdot\partial_{l}\partial_{h}^{6} u\,\mathrm{d}y \lesssim\sqrt{\mathcal{E}^{L}}\big\| (\eta,u)\big\| _{7}^{2}, \\ M^{H}_{2} \lesssim\sqrt{ \mathcal{E}^{L}}\bigl(\big\| (\eta,u)\big\| _{7}^{2}+ \mathcal{D}^{H}\bigr), \\ \begin{aligned} M^{H}_{3} = {}& \int\partial_{h}^{6} q\cdot\partial_{h}^{6}D^{h}_{u} \,\mathrm{d}y \lesssim\|\nabla q\|_{5}\bigl(\|\tilde {\mathcal {A}}\|_{6}\|u \|_{3}+\|\tilde {\mathcal {A}}\|_{4}\|u\|_{5}+\|\tilde {\mathcal {A}}\| _{2}\|u\|_{7}\bigr) \\ \lesssim{}&\|\nabla q\|_{5}\bigl(\|\eta\|_{7}\|u \|_{3}+\|\tilde {\mathcal {A}}\|_{2}^{\frac{1}{2}}\| \tilde {\mathcal {A}}\|_{6}^{\frac{1}{2}} \|u\|_{3}^{\frac{1}{2}}\|u\|_{7}^{\frac{1}{2}}+\|\tilde {\mathcal {A}}\|_{2}\|u\|_{7}\bigr) \lesssim\sqrt{\mathcal{E}^{L}} \bigl(\|\eta\|_{7}^{2}+\mathcal{D}^{H}\bigr). \end{aligned} \end{gathered} $$
Substituting the above three inequalities into (4.8), we obtain Lemma 4.3. □
Lemma 4.4
The following estimates hold:
where the norm
\(\|\eta\|_{6,*}^{2}\)
is equivalent to
\(\|\eta\|_{6}^{2}\)
and
\({{E}^{H}}:=\mathcal{E}^{H}-\|\nabla\eta\|_{6,0}^{2}\).
$$\begin{aligned}& \sum_{k=0}^{2}\bigl(\big\| \partial_{t}^{k} u\big\| _{6-2k}^{2}+ \big\| \nabla \partial_{t}^{k} q\big\| _{4-2k}^{2}\bigr) \lesssim\| \eta\|_{6}^{2}+ \big\| \bigl( u,u_{t},u_{tt}, \partial_{t}^{3} u\bigr)\big\| _{0}^{2}+ { \mathcal{E}^{L}} {E}^{H}, \end{aligned}$$
(4.9)
$$\begin{aligned}& \begin{aligned}[b] &\frac{\mathrm{d}}{\mathrm{d}t} \|\eta \|_{6,*}^{2}+\| \eta\|_{6}^{2} + \sum _{k=0}^{2}\bigl(\big\| \partial_{t}^{k} u\big\| _{6-2k}^{2}+ \big\| \nabla\partial_{t}^{k} q\big\| _{4-2k}^{2}\bigr) \\ & \quad\lesssim\| \eta\|_{5,1}^{2}+\big\| \bigl(\eta, u, u_{t},u_{tt},\partial_{t}^{3} u\bigr) \big\| _{0}^{2} + {\mathcal{E}^{L}} \mathcal{D}^{H}, \end{aligned} \end{aligned}$$
(4.10)
$$\begin{aligned}& \sum_{k=1}^{2}\bigl( \big\| \partial_{t}^{k} u\big\| _{7-2k}^{2}+\big\| \nabla \partial_{t}^{k} q\big\| _{5-2k}^{2}\bigr) \lesssim\|u\|_{5}^{2}+\big\| \bigl(u_{t},u_{tt}, \partial_{t}^{3} u\bigr)\big\| _{1}^{2}+ \mathcal{E}^{H}\mathcal{D}^{H}, \end{aligned}$$
(4.11)
Proof
(1) We begin with the derivation of (4.9). Taking \((i,k)=(4,0)\) in the Stokes estimate (3.35), we have Noting that we get
$$ \begin{aligned} & \| u\|_{6}^{2}+ \|\nabla q\|_{4}^{2}\lesssim\| \eta\|_{6}^{2}+ \big\| (u,u_{t})\big\| _{4}^{2} +S_{0,4}^{u}. \end{aligned} $$
$$\begin{aligned} S_{0,4}^{u}={} &\big\| \bigl(N^{h}_{u}, N^{h}_{q}\bigr) \big\| _{4}^{2}+\big\| D^{h}_{u} \big\| _{5}^{2} \\ \lesssim{}&\|\tilde {\mathcal {A}}\|_{5}^{2}\|u\|_{3}^{2}+ \|\tilde {\mathcal {A}}\|_{2}^{2}\|u\|_{6}^{2}+ \|\tilde {\mathcal {A}}\|_{5}^{2}\| \nabla q\|_{1}^{2}+ \|\tilde {\mathcal {A}}\|_{2}^{2}\|\nabla q\|_{4}^{2}\lesssim \mathcal{E}^{L} { {E}}^{H} , \end{aligned} $$
$$ \| u\|_{6}^{2}+ \|\nabla q\|_{4}^{2}\lesssim\| \eta\|_{6}^{2}+ \big\| (u,u_{t})\big\| _{4}^{2} +\mathcal{E}^{L} { {E}}^{H}. $$
(4.12)
Using the recursion formula (3.35) for \(i=4\) from \(k=1\) to 2, we obtain On the other hand, \(S_{k,4}^{u}\) can be bounded from above by Therefore, which, together with (4.12), yields In view of the interpolation inequality (3.3), we have Thus, inserting (4.15) into (4.14), one gets (4.9).
$$\sum_{k=1}^{2}\bigl(\big\| \partial_{t}^{k} u\big\| _{6-2k}^{2}+ \big\| \nabla \partial_{t}^{k} q\big\| _{4-2k}^{2}\bigr) \lesssim\big\| \partial_{t}^{2}(u,u_{t}) \big\| _{0}^{2}+ \sum_{k=1}^{2} \bigl( \big\| \partial_{t}^{k-1} u\big\| _{6-2k}^{2}+S_{k,4}^{u} \bigr). $$
$$\begin{aligned} \sum_{k=1}^{2} S_{k,4}^{u}={}& \sum_{k=1}^{2} \bigl( \big\| \partial_{t}^{k} N^{h}_{u} \big\| _{4-2k}^{2}+\big\| \partial_{t}^{k} D^{h}_{u}\big\| _{5-2k}^{2}+\big\| \partial_{t}^{k} N^{h}_{q} \big\| _{4-2k}^{2}\bigr) \\ \lesssim{}& \|\tilde {\mathcal {A}}\|_{2}^{2}\bigl(\|u_{t} \|_{4}^{2}+\|\nabla q_{t}\|_{2}^{2} \bigr) + \|\tilde {\mathcal {A}}\| _{2}^{2}\bigl(\|u_{tt} \|_{2}^{2}+\|\nabla q_{tt}\|_{0}^{2} \bigr) \\ & +\|\tilde {\mathcal {A}}\|_{5}^{2}\|u_{t}\|_{1}^{2} +\|\tilde {\mathcal {A}}_{t}\|_{2}^{2}\bigl(\|u \|_{4}^{2}+ \|\nabla q\| _{2}^{2}\bigr)+ \|\tilde {\mathcal {A}}_{t}\|_{2}^{2}\bigl(\|u_{t} \|_{2}^{2}+\|\nabla q_{t}\|_{0}^{2} \bigr) \\ & +\|\tilde {\mathcal {A}}_{t}\|_{3}^{2} \|u\|_{3}^{2}+ \|\tilde {\mathcal {A}}_{tt}\|_{1}^{2}\bigl(\|u\|_{3}^{2} + \|\nabla q\|_{1}^{2}\bigr)\\ \lesssim{}&\mathcal{E}^{L} E^{H}. \end{aligned} $$
$$ \sum_{k=1}^{2} \bigl(\big\| \partial_{t}^{k} u\big\| _{6-2k}^{2}+ \big\| \nabla\partial_{t}^{k} q\big\| _{4-2k}^{2} \bigr) \lesssim\big\| \partial_{t}^{2}(u,u_{t}) \big\| _{0}^{2} +\sum_{k=1}^{2} \big\| \partial_{t}^{k-1} u\big\| _{6-2k}^{2} + \mathcal{E}^{L} E^{H}, $$
(4.13)
$$ \sum_{k=0}^{2}\bigl(\big\| \partial_{t}^{k} u\big\| _{6-2k}^{2}+ \big\| \nabla \partial_{t}^{k} q\big\| _{4-2k}^{2}\bigr) \lesssim\|\eta\|_{6}^{2}+ \big\| \partial_{t}^{2}(u,u_{t}) \big\| _{0}^{2}+\sum_{k=1}^{2} \big\| \partial_{t}^{k-1} u\big\| _{6-2k}^{2} + \mathcal{E}^{L} E^{H}. $$
(4.14)
$$ \|u\|_{4}\leq C_{\epsilon}\|u\|_{0}+ \epsilon\|u\|_{6} \quad\mbox{and}\quad \| u_{t}\|_{2} \lesssim C_{\epsilon}\|u_{t}\|_{0}+\epsilon \|u_{t}\|_{4} . $$
(4.15)
(2) We proceed to prove the estimate (4.10). Exploiting the recursion formula (3.39) for \(i=4\) from \(k=0\) to 4, we see that there are positive constants \(c_{k}\), \(k=0,\ldots,4\), such that which combined with (4.13) gives where we have used the fact that \(S_{k ,4}^{\omega}\leq S_{0,4}^{\omega}\) for \(0\leq k\leq4\). Since and \(E^{H}\leq\mathcal{D}^{H}\), we further infer that Using the interpolation inequality (3.3), we get (4.10) from (4.16), where \(\|\eta\|_{6,*}^{2}\) equals to \(\sum_{k=0}^{4}c_{k}\|\eta\|_{{k,6-k}}^{2}\) multiplied by some positive constant.
$$\frac{\mathrm{d}}{\mathrm{d}t} \sum_{k=0}^{4} c_{k}\|\eta\|_{{k,6-k}}^{2} + \big\| (\eta, u) \big\| _{6}^{2}+\|\nabla q\|_{4}^{2} \lesssim\| u_{t} \|_{4}^{2}+ \| \eta \|_{5,1}^{2}+\sum_{k=0}^{4} S_{k ,4}^{\omega}, $$
$$ \begin{gathered} \frac{\mathrm{d}}{\mathrm{d}t} \sum _{k=0}^{4}c_{k}\|\eta\|_{{k,6-k}}^{2} +c\Biggl(\| \eta\|_{6}^{2} + \sum _{k=0}^{2}\bigl(\big\| \partial_{t}^{k} u\big\| _{6-2k}^{2}+ \big\| \nabla\partial_{t}^{k} q\big\| _{4-2k}^{2}\bigr)\Biggr) \\ \quad \lesssim\|\eta\|_{5,1}^{2} + \big\| \partial_{t}^{2}(u,u_{t}) \big\| _{0}^{2}+\sum_{k=1}^{2} \big\| \partial_{t}^{k-1} u\big\| _{6-2k}^{2} + \mathcal{E}^{L} E^{H} + S_{0,4}^{\omega}, \end{gathered} $$
$$\begin{aligned} S_{0,4}^{\omega}= {}& \big\| \bigl( N^{h}_{u}, N^{h}_{q}\bigr) \big\| _{4}^{2}+\big\| \bigl(D^{h}_{u} ,\operatorname {div}\eta \bigr)\big\| _{5}^{2} \\ \lesssim{}&\bigl(\|\tilde {\mathcal {A}}\|_{5}^{2}\|u\|_{3}^{2}+ \|\tilde {\mathcal {A}}\|_{2}^{2}\|u\|_{6}^{2}+\|\eta \|_{3}^{2}\| \eta\|_{6}^{2} + \|\tilde {\mathcal {A}}\|_{5}^{2}\|\nabla q\|_{1}^{2}+ \|\tilde {\mathcal {A}}\|_{2}^{2}\|\nabla q\|_{4}^{2}\bigr) \\ \lesssim{}& \mathcal{E}^{L} { {E}}^{H} \end{aligned} $$
$$ \begin{aligned}[b] & \frac{\mathrm{d}}{\mathrm{d}t} \sum _{k=0}^{4}c_{k}\|\eta\|_{{k,6-k}}^{2}+c \Biggl(\| \eta\|_{6}^{2} + \sum _{k=0}^{2}\bigl(\big\| \partial_{t}^{k} u\big\| _{6-2k}^{2}+ \big\| \nabla\partial_{t}^{k} q\big\| _{4-2k}^{2}\bigr) \Biggr) \\ &\quad\lesssim\| \eta\|_{5,1}^{2} + \big\| \partial_{t}^{2}( u,u_{t})\big\| _{0}^{2}+\sum _{k=1}^{2} \big\| \partial_{t}^{k-1}u \big\| _{6-2k}^{2} + {\mathcal{E}^{L}} \mathcal{D}^{H}. \end{aligned} $$
(4.16)
(3) Finally, we derive the estimate (4.11) for higher-order dissipation estimates. We use the recursion formula (3.35) for \(i=5\) from \(k=1\) to 2 to deduce that Noting that we obtain (4.11) from (4.17). □
$$ \sum_{k=1}^{2} \bigl( \big\| \partial_{t}^{k} u\big\| _{7-2k}^{2}+ \big\| \nabla\partial_{t}^{k} q\big\| _{5-2k}^{2} \bigr) \lesssim\big\| \partial_{t}^{2}( u, u_{t}) \big\| _{1}^{2} +\sum_{k=1}^{2} \bigl(\big\| \partial_{t}^{k-1}u\big\| _{7-2k}^{2} +S_{k,5}^{u}\bigr). $$
(4.17)
$$ \begin{aligned}\sum_{k=1}^{2} S_{k,5}^{u}={}& \sum_{k=1}^{2} \bigl( \big\| \partial_{t}^{k} \bigl(N^{h}_{u} , N^{h}_{q}\bigr)\big\| _{5-2k}^{2}+\big\| \partial_{t}^{k} D^{h}_{u} \big\| _{6-2k}^{2}\bigr) \\ \lesssim{}&\|\tilde {\mathcal {A}}_{t}\|_{4}^{2}\|u \|_{5}^{2}+\|\tilde {\mathcal {A}}\|_{4}^{2} \|u_{t}\|_{5}^{2}+ \|\tilde {\mathcal {A}}_{t} \|_{3}^{2}\|\nabla q\|_{3}^{2}+\|\tilde {\mathcal {A}}\|_{3}^{2}\|\nabla q_{t}\|_{3}^{2} \\ &+ \|\tilde {\mathcal {A}}_{tt}\|_{2}^{2}\|u\|_{3}^{2}+ \|\tilde {\mathcal {A}}_{t}\|_{2}^{2}\|u_{t} \|_{3}^{2}+ \|\tilde {\mathcal {A}}\| _{2}^{2} \|u_{tt}\|_{3}^{2} \\ &+ \|\tilde {\mathcal {A}}_{tt}\|_{2}^{2}\|\nabla q \|_{1}^{2}+ \|\tilde {\mathcal {A}}_{t}\|_{2}^{2} \|\nabla q_{t}\| _{1}^{2}+ \|\tilde {\mathcal {A}}\|_{2}^{2}\|\nabla q_{tt}\|_{1}^{2} \\ \lesssim{}&\mathcal{E}^{H}\mathcal{D}^{H}, \end{aligned} $$
Now we are in a position to build the higher-order and highest-order energy inequalities. In what follows, the letters \(c_{i}^{H}\) and \(i=1,\ldots,6\) will denote generic constants which may depend on the domain Ω and some physical parameters in the transformed MHD equations (2.15).
Proposition 4.1
Under the assumption (3.1), if
δ
is sufficiently small, then there are two norms
\(\tilde{\mathcal{E}}^{H}\)
and
\(\|\eta\|_{7,*}^{2}\)
which are equivalent to
\(\mathcal{E}^{H}\)
and
\(\|\eta\|_{7}^{2}\)
respectively, such that
$$\begin{aligned}& \frac{\mathrm{d}}{\mathrm{d}t}\tilde {\mathcal{E}}^{H}+\mathcal {D}^{H}\lesssim\sqrt{\mathcal{E}^{L}}\big\| (\eta,u) \big\| _{7}^{2}, \end{aligned}$$
(4.18)
$$\begin{aligned}& \frac{\mathrm{d}}{\mathrm{d}t} \|\eta\| _{7,*}^{2}+ \big\| ( \eta,u)\big\| _{7}^{2}\lesssim\mathcal{E}^{H}+ \mathcal{D}^{H} \quad\textit{on } (0,T]. \end{aligned}$$
(4.19)
Proof
(1) We first prove (4.18). Similarly to (3.48), we make use of (3.5), (3.6), (3.44) and (3.46) to deduce from Lemmas 4.1-4.3 and (4.10) that there are constants \(c_{1}^{H}\), \(c_{2}^{H}\), \(c_{3}^{H}\) and \(\alpha_{0}\geq1\), such that where \(\alpha_{0}\) depends on the domain and the known physical parameters, and Moreover, by (4.20) and Proposition 3.1 we find that where \(\tilde{\mathcal{E}}^{H}_{2}:=\tilde{\mathcal{E}}^{H}_{1}+ c_{4}^{H}\tilde{\mathcal{E}}^{L}\) and \(\tilde{\mathcal{D}}^{H} := \tilde{\mathcal{D}}^{H}_{1} + c_{5}^{H} {\mathcal{D}}^{L}\). By (4.11), we have \(\mathcal{D}^{H}\lesssim\tilde {\mathcal{D}}^{H}+{\mathcal{E}^{H}}\mathcal{D}^{H}\). So, (4.21) can be rewritten as
$$ \frac{\mathrm{d}}{\mathrm{d}t} \tilde{\mathcal{E}}^{H}_{1} +\tilde{\mathcal{D}}^{H}_{1}\leq c_{1}^{H} \bigl[ (1+\alpha) \sqrt{\mathcal{E}^{L}} \bigl(\big\| (\eta,u) \big\| _{7}^{2}+\mathcal{D}^{H}\bigr)+\big\| (\eta, u, u_{t})\big\| _{0}^{2} \bigr] \quad\mbox{for any } \alpha\geq\alpha_{0}, $$
(4.20)
$$\begin{aligned}& \begin{aligned} \tilde{\mathcal{E}}^{H}_{1}:={}& \sum_{k=2}^{3} \biggl(\big\| \sqrt{{\rho}} \partial_{t}^{k} u \big\| _{0}^{2}+ \lambda_{0} \big\| \bar{M}\cdot\nabla\partial_{t}^{k-1} u\big\| ^{2}_{0} + \int\nabla\partial_{t}^{k-1}q\cdot D^{t,k}_{u}\,\mathrm{d} y \biggr) \\ & +\sum_{5\leq\alpha_{1}+\alpha_{2}\leq6} \int{\rho} \partial_{1}^{\alpha_{1}}\partial_{2}^{\alpha_{2}} \eta\cdot\partial_{1}^{\alpha _{1}}\partial_{2}^{\alpha_{2}} u\,\mathrm{d}y +\sum_{k=5}^{6}\frac{ \mu}{2} \| \nabla\eta\|_{k,0}^{2} \\ &+\alpha\bigl( \|\sqrt{{\rho}} u\|_{6,0}^{2}+ \lambda_{0} \|\bar{M}\cdot\nabla\eta\|_{6,0}^{2} \bigr) +c_{2}^{H}\|\eta\|^{2}_{6,*}, \end{aligned}\\& \begin{aligned} \tilde{\mathcal{D}}^{H}_{1}:={}& c_{3}^{H} \Biggl(\sum_{k=2}^{3}\big\| \partial_{t}^{k} u\big\| ^{2}_{1}+ \sum _{k=5}^{6}\big\| (\eta,\bar{M}\cdot\nabla\eta, \nabla u)\big\| _{k,0}^{2} +\|\eta\|_{6}^{2}\\ &+ \sum_{k=0}^{2}\bigl(\big\| \partial_{t}^{k} u\big\| _{6-2k}^{2}+\big\| \nabla\partial_{t}^{k} q\big\| _{4-2k}^{2}\bigr) \Biggr). \end{aligned} \end{aligned}$$
$$ \frac{\mathrm{d}}{\mathrm{d}t} \tilde{ \mathcal{E}}^{H}_{2} +\tilde{\mathcal{D}}^{H} \leq c_{1}^{H} (1+ \sigma) \sqrt{\mathcal{E}^{L}} \bigl(\big\| (\eta,u)\big\| _{7}^{2}+\mathcal{D}^{H}\bigr) , $$
(4.21)
$$ \frac{\mathrm{d}}{\mathrm{d}t}\tilde {\mathcal{E}}^{H}_{2} +c_{6}^{H}{\mathcal{D}}^{H} \leq c_{1}^{H}(1+\sigma)\sqrt{\mathcal{E}^{L}}\big\| ( \eta,u)\big\| _{7}^{2}\quad\mbox{for sufficiently small } \delta. $$
(4.22)
Next, we show that \(\mathcal{E}^{H}\) can be controlled by \(\tilde {\mathcal{E}}^{H}_{2}\). Keeping in mind that we use Cauchy-Schwarz’s inequality, (3.5) and (4.23) to infer that there is an appropriately large constant α, depending on ρ, μ and Ω, such that Recalling (4.9) and the fact that \(E^{H}\leq\mathcal {E}^{H}\), we have which implies \({\mathcal{E}}^{H} \lesssim\tilde{\mathcal{E}}^{H}_{2}\) for sufficiently small δ.
$$ \begin{aligned}[b] \sum_{k=2}^{3} \int\nabla\partial_{t}^{k-1}q\cdot D^{t,k}_{u}\,\mathrm{d}y\lesssim{}&\sum_{k=2}^{3} \big\| \nabla\partial_{t}^{k-1}q\big\| _{0} \big\| D^{t,k}_{u}\big\| _{0} \\ \lesssim{}& \sqrt{\mathcal{E}^{H}} \Biggl(\big\| (u,u_{t}) \big\| _{2}\sum_{k=2}^{3}\big\| \partial_{t}^{k}\mathcal{A}\big\| _{0}+\| \mathcal{A}_{t}\|_{2}\|u_{tt}\|_{0} \Biggr)\\ \lesssim{}&\bigl(\mathcal{E}^{H}\bigr)^{\frac{3}{2}}, \end{aligned} $$
(4.23)
$$\|\nabla\eta\|^{2}_{6,0}+\|\eta\|_{6}^{2} +\sum_{k=0}^{3} \big\| \partial_{t}^{k} u\big\| _{0}^{2} \lesssim\tilde{\mathcal{E}}^{H}_{2}+ \bigl(\mathcal{E}^{H}\bigr)^{\frac{3}{2}}. $$
$${\mathcal{E}}^{H} \lesssim\tilde{\mathcal{E}}^{H}_{2} + {\mathcal{E}^{L}} { \mathcal{E}}^{H}+\bigl( \mathcal{E}^{H}\bigr)^{2}\leq\tilde{\mathcal {E}}^{H}_{2}+2\bigl(\mathcal{E}^{H} \bigr)^{\frac{3}{2}}, $$
On the other hand, \(\tilde{\mathcal{E}}^{H}_{2}\lesssim\mathcal{E}^{H}\) obviously. Hence, the energy functional \(\tilde{\mathcal{E}}^{H}_{2}\) is equivalent to \(\mathcal{E}^{H}\). Finally, letting \(\delta\leq c_{6}^{H} /2 c_{3}^{H}(1+\sigma)\) and denoting \(\tilde{\mathcal{E}}^{H}=2\tilde{\mathcal{E}}_{2}^{H}/c_{6}^{H}\), we see that (4.22) implies (4.18).
(2) We proceed to derive the highest-order estimate (4.19). We start with employing the recursion formula (3.39) with \(i=5\) from \(k=0\) to 4 to find that there are positive constants \(d_{k}\) (\(k=0,1,2\)), such that On the other hand, by virtue of (4.2), Thus, we conclude Hence, (4.19) holds by defining \(\|\eta\|_{7,*}^{2}: =2\sum _{k=0}^{5}d_{k}\|\eta\|_{k,7-k}^{2}\), provided δ is sufficiently small. □
$$ \frac{\mathrm{d}}{\mathrm{d}t} \sum_{k=0}^{5} d_{k}\|\eta\|_{{k,7-k}}^{2} +\big\| (\eta, u) \big\| _{7}^{2} \lesssim\|u\|_{5}^{2} +\|\eta \|_{6,1}^{2} +\|u_{t}\|_{5}^{2} +S_{0,5}^{\omega}. $$
$$ \begin{aligned} S_{0,5}^{\omega}={} & \big\| \bigl(N^{h}_{u},N^{h}_{q}\bigr) \big\| _{5}^{2} +\big\| \bigl( D^{h}_{u} ,\operatorname {div}\eta\bigr)\big\| _{6}^{2} \\ \lesssim{}&\|\tilde {\mathcal {A}}\|_{6}^{2}\|u\|_{3}^{2} +\|\tilde {\mathcal {A}}\|_{3}^{2}\|u\|_{7}^{2} + \|\tilde {\mathcal {A}}\| _{5}^{2}\|\nabla q\|_{5}^{2}+ \|\eta \|_{4}^{2}\|\eta\|_{7}^{2} \lesssim \mathcal{D}^{H} + \mathcal{E}^{H} \| \eta \|_{7}^{2}. \end{aligned} $$
$$\frac{\mathrm{d}}{\mathrm{d}t} \sum_{k=0}^{5} d_{k}\|\eta\|_{{k,7-k}}^{2} + \big\| (\eta, u) \big\| _{7 }^{2} \lesssim\mathcal{E}^{H}+ \mathcal{D}^{H}+\mathcal{E}^{H}\|\eta\|_{7}^{2}. $$
The following lemma will be needed in the next section.
Lemma 4.5
Under the assumption (3.1), if
δ
is sufficiently small, then
$$ \mathcal{E}^{H}\lesssim\mathcal{E}:=\|\eta \|_{7}^{2} +\|u\|_{6}^{2}. $$
(4.24)
Proof
In view of (4.9), we have which results in for sufficiently small δ. Below we show that the \(L^{2}\)-norm of \(u_{t}\) and \(\partial_{t}^{3}u\) can be controlled by \(\sqrt{\mathcal{E}}\).
$$\mathcal{E}^{H}\lesssim\| \eta\|_{7}^{2}+ \big\| \bigl(u,u_{t},u_{tt},\partial_{t}^{3}u \bigr)\big\| _{0}^{2}+\mathcal{E}^{L} \mathcal{E}^{H}, $$
$$ \mathcal{E}^{H}\lesssim\| \eta\|_{7}^{2}+ \big\| \bigl(u,u_{t},u_{tt},\partial_{t}^{3}u \bigr)\big\| _{0}^{2} $$
(4.25)
(1) First, to bound \(u_{t}\), we multiply (3.10)2 with \(j=0\) by \(u_{t}\) and integrate the resulting equation over Ω to obtain where the last term on the right-hand side can be bounded from above by \(\mathcal{E}\). Hence,
$$ \begin{aligned} \|\sqrt{ {\rho} }u_{t}\|_{0}^{2} ={} & \int\bigl(\mu\Delta_{\mathcal{A}} u +\lambda_{0} \bar{M}\cdot \nabla( \bar{M}\cdot\nabla u)+2\rho\omega\partial_{t} (u_{2}e_{1}-u_{1}e_{2})\bigr)\cdot u_{t}\,\mathrm{d}y\\ &- \int\nabla q\cdot D_{u}^{t,1} \,\mathrm{d}y \\ \lesssim{}& \mathcal{E}+\|\nabla q\|_{0}\big\| D^{t,1}_{u} \big\| _{0}, \end{aligned}$$
$$ \| u_{t}\|_{0}^{2}\lesssim \mathcal{E}. $$
(4.26)
(2) Then we control the term \(u_{tt}\). Multiplying (3.10)2 with \(j=1\) by \(u_{tt}\) and integrating the resulting equation over Ω, we infer that which yields Noting that and we see that
$$\begin{aligned} \|\sqrt{{\rho}} u_{tt}\|_{0}^{2} ={}& \int\bigl(\mu\Delta_{\mathcal{A}} \partial_{t}u+ \lambda_{0} \bar{M}\cdot\nabla(\bar{M}\cdot\nabla u)+2\rho\omega (u_{2}e_{1}-u_{1}e_{2})+ \mu N^{t,1}_{u} +N^{t,1}_{q}\bigr)\\ &\cdot \partial_{t}^{2} u \,\mathrm{d}y \\ &- \int\nabla q_{t}\cdot D^{t,2}_{u} \,\mathrm{d}y, \end{aligned}$$
$$\| u_{tt}\|_{0}^{2} \lesssim\big\| (u,u_{t}) \big\| _{2}^{2}+\big\| \bigl(N_{u}^{t,1},N_{q}^{t,1} \bigr)\big\| _{0}^{2}+\|\nabla q_{t}\|_{0} \big\| D_{u}^{t,2}\big\| _{0}. $$
$$\big\| \bigl(N_{u}^{t,1},N_{q}^{t,1}\bigr) \big\| _{0}^{2}\lesssim\mathcal{E} $$
$$\|\nabla q_{t}\|_{0}\big\| D_{u}^{t,2} \big\| _{0}\lesssim\|\mathcal{A}_{t}\|_{1} \|u_{t}\| _{1}+\|\mathcal{A}_{tt}\|_{0} \|u\|_{2}\lesssim\|u_{t}\|_{1}^{2}+ \mathcal{E}, $$
$$ \| u_{tt}\|_{0}^{2} \lesssim \|u_{t}\|_{2}^{2} +\mathcal{E}. $$
(4.27)
(3) Finally, we estimate the term \(\partial_{t}^{3}u\). If we multiply (3.10)2 with \(j=2\) by \(\partial_{t}^{3}u\) in \(L^{2}(\Omega)\), we obtain whence
$$\begin{aligned} \big\| \sqrt{{\rho}}\partial_{t}^{3} u\big\| _{0}^{2} ={}& \int\bigl(\mu\Delta_{\mathcal{A}} \partial_{t}^{2}u+ \lambda_{0} \bar{M}\cdot\nabla(\bar{M}\cdot\nabla u_{t})\\ &+2 \rho\omega\partial_{t}^{2} (u_{2}e_{1}-u_{1}e_{2}) +\mu N^{t,2}_{u} +N^{t,2}_{q}\bigr)\cdot \partial_{t}^{3} u \,\mathrm{d}y \\ &- \int\nabla q_{tt}\cdot D^{t,3}_{u} \,\mathrm{d}y, \end{aligned} $$
$$\big\| \partial_{t}^{3} u\big\| _{0}^{2}\lesssim \|u\|_{1}^{2}+ \big\| (u_{t},u_{tt}) \big\| _{2}^{2}+\big\| \bigl(N_{u}^{t,2},N_{q}^{t,2} \bigr)\big\| ^{2}_{0}+\|\nabla q_{tt}\|_{0} \big\| D_{u}^{t,3}\big\| _{0}. $$
On the other hand, it is easy to show that and Therefore,
$$\big\| \bigl(N_{u}^{t,2},N_{q}^{t,2}\bigr) \big\| ^{2}_{0}\lesssim\|u_{t}\|_{1}^{2}+ \mathcal{E} $$
$$\|\nabla q_{tt}\|_{0}\big\| D_{u}^{t,3} \big\| _{0}\lesssim\|u_{t}\|_{2}^{2} + \|u_{tt}\| _{1}^{2}+\mathcal{E}. $$
$$ \big\| \partial_{t}^{3} u\big\| _{0}^{2} \lesssim\|u_{t}\|_{2}^{2}+ \| u_{tt} \|_{2}^{2} +\mathcal{E}. $$
(4.28)
Now, we control the term \(\|u_{tt}\|_{2}\) in (4.28). By (3.10)2 with \(j=1\), we deduce that where the last term on the right-hand side can be bounded by \(\mathcal {E}\). Consequently, To estimate \(\nabla q_{t}\), we rewrite (3.24)2 as follows: with boundary condition where In view of the standard elliptic regularity estimates [19], we find that Putting the above inequality into (4.29), we obtain which, together with the interpolation inequality (3.3), yields Consequently, from (4.28) and (4.30) it follows that
$$\| u_{tt}\|_{2}^{2}\lesssim\| u_{t} \|_{4}^{2}+\| \nabla q_{t}\|_{2}^{2}+ \big\| \bigl( \nabla^{2} u , N^{t,1}_{u}, N^{t,1}_{q}\bigr)\big\| _{2}^{2}, $$
$$ \| u_{tt}\|_{2}^{2}\lesssim\| u_{t}\|_{4}^{2}+\|\nabla q_{t} \|_{2}^{2} +\mathcal{E}. $$
(4.29)
$$\Delta q_{t} = f_{1}, $$
$$\nabla{ q_{t}}\cdot\nu=f_{2}\cdot\nu\quad\mbox{on }\partial \Omega, $$
$$\begin{gathered} f_{1}:=\operatorname{div}\bigl(\mu \partial_{t} N^{h}_{u}+\partial_{t} N^{h}_{q}-{\rho} u_{tt}+\mu\Delta u_{t}+\lambda_{0} \bar{M}\cdot\nabla(\bar{M}\cdot\nabla u)+2\rho\omega\partial_{t} (u_{2}e_{1}-u_{1}e_{2}) \bigr), \\ f_{2}:= \mu\partial_{t} N^{h}_{u}+ \partial_{t} N^{h}_{q} +\mu\Delta u_{t}+\lambda_{0} \bar{M}\cdot\nabla(\bar{M}\cdot\nabla u)+2 \rho\omega\partial_{t} (u_{2}e_{1}-u_{1}e_{2}). \end{gathered} $$
$$\|\nabla q_{t}\|_{2}^{2}\lesssim\|f_{1} \|_{1}^{2}+ \|f_{2}\|_{2}^{2} \leq\|u_{t}\|_{4}^{2}+\| u_{tt} \|_{1}^{2}+\mathcal{E}. $$
$$\| u_{tt}\|_{2}^{2}\lesssim\| u_{t} \|_{4}^{2}+\|u_{tt}\|_{1}^{2}+ \mathcal{E}, $$
$$ \| u_{tt}\|_{2}^{2}\lesssim\| u_{t}\|_{4}^{2}+\|u_{tt} \|_{0}^{2}+\mathcal{E}. $$
(4.30)
$$ \big\| \partial_{t}^{3} u\big\| _{0}^{2} \lesssim\| u_{tt}\|_{0}^{2}+\|u_{t} \|_{4}^{2}+ \mathcal{E} . $$
(4.31)
Next, we control \(\|u_{t}\|_{4}\) in (4.31). Using (2.15)2, we derive that where, by virtue of (3.24)2, ∇q satisfies with boundary condition and Thus, we can use the elliptic regularity estimate to get Hence, from (4.32) we get which, together with the interpolation inequality (3.3), implies that Inserting (4.33) into (4.31), we conclude
$$ \| u_{t}\|_{4}^{2} \lesssim\big\| \bigl(\Delta_{\mathcal{A}} u, \nabla^{2}\eta,u,\nabla_{\mathcal {A}} q \bigr)\big\| _{4}^{2}\lesssim\|\nabla q\|_{4}^{2}+ \mathcal{E}, $$
(4.32)
$$\Delta q = f_{3} $$
$$\nabla{ q}\cdot\nu=f_{4}\cdot\nu \quad\mbox{on }\partial\Omega, $$
$$\begin{gathered} f_{3}:=\operatorname{div}\bigl( \mu N^{h}_{u}+N^{h}_{q}- { \rho}u_{t}+\mu\Delta u+\lambda_{0} \bar{M}\cdot\nabla(\bar{M} \cdot\nabla\eta)+2\rho\omega(u_{2}e_{1}-u_{1}e_{2}) \bigr), \\ f_{4}:=\bigl( \mu N^{h}_{u}+ N^{h}_{q}+\mu\Delta u+\lambda_{0}\bar{M}\cdot \nabla(\bar{M}\cdot\nabla\eta)\bigr). \end{gathered} $$
$$\|\nabla q\|_{4}^{2}\lesssim\| f_{3} \|_{3}^{2}+ \|f_{4}\|_{4}^{2} \lesssim\|u_{t}\| _{3}^{2}+\mathcal{E}. $$
$$\| u_{t}\|_{4}^{2} \lesssim\|u_{t} \|_{3}^{2}+ \mathcal{E}, $$
$$ \| u_{t}\|_{4}^{2} \lesssim \|u_{t}\|_{0}^{2}+ \mathcal{E} . $$
(4.33)
$$ \big\| \partial_{t}^{3} u\big\| _{0}^{2} \lesssim\big\| (u_{t}, u_{tt})\big\| _{0}^{2} + \mathcal{E}. $$
(4.34)
5 Proof of Theorem 2.1
This section is devoted to the proof of Theorem 2.1. Roughly speaking, Theorem 2.1 is shown by combining the a priori stability estimate (2.17) and the local well-posedness of the transformed MHD problem. Before we derive the a priori stability estimate (2.17), we begin with estimating the terms \(\mathcal{G}_{1},\ldots,\mathcal{G}_{4}\).
Using (4.19), and recalling the equivalence of \(\|\eta\| _{7,*}^{2}\) and \(\|\eta\|_{7}^{2}\), we deduce that which yields
$$\begin{aligned} \|\eta\|_{7}^{2}\lesssim{}& \|\eta_{0} \|_{7}^{2}e^{-t}+ \int_{0}^{t}e^{-(t-\tau)}\bigl( \mathcal{E}^{H}(\tau)+\mathcal{D}^{H}(\tau)\bigr)\,\mathrm{d}\tau \\ \lesssim{}& \|\eta_{0}\|_{7}^{2}e^{-t}+ \sup_{0\leq\tau\leq t}\mathcal{E}^{H}(\tau) \int_{0}^{t}e^{-(t-\tau)}\,\mathrm{d}\tau+ \int_{0}^{t}\mathcal{D}^{H}(\tau)\,\mathrm{d} \tau \\ \lesssim{}&\|\eta_{0}\|_{7}^{2}e^{-t}+ \mathcal{G}_{3}(t), \end{aligned}$$
$$ \mathcal{G}_{1}(t)\lesssim\|\eta_{0} \|_{7}^{2} +\mathcal{G}_{3}(t). $$
(5.1)
Multiplying (4.19) by \((1+t)^{-3/2}\), we get which implies that
$$ \frac{\mathrm{d}}{\mathrm{d}t}\frac{ \|\eta\|_{7,*}^{2}}{(1+t)^{3/2}}+ \frac{3}{2}\frac{\|\eta\| _{7,*}^{2}}{(1+t)^{5/2}}+ \frac{\|(\eta,u)\| _{7}^{2}}{(1+t )^{3/2}}\lesssim\frac{\mathcal{E}^{H}}{(1+t)^{3/2}} +\frac{\mathcal {D}^{H}}{(1+t)^{3/2}}, $$
$$ \mathcal{G}_{2}(t)\lesssim\|\eta_{0} \|_{7}^{2}+\mathcal{G}_{3}(t). $$
(5.2)
An integration of (4.18) with respect to t gives Let From now on, we further assume \(\sqrt{\mathcal{G}_{5}(T)}\leq\delta\), which is a stronger requirement than (3.1). Thus, we make use of (5.2) to find that which implies
$$\mathcal{G}_{3}(t)\lesssim\mathcal{E}^{H}(0)+ \int_{0}^{t}\sqrt{\mathcal{E}^{L}(\tau)} \big\| (\eta,u) (\tau)\big\| _{7}^{2}\,\mathrm{d}\tau. $$
$$\mathcal{G}_{5}(t):=\mathcal{G}_{1}(t)+ \sup _{0\leq\tau\leq t}\mathcal{E}^{H}(\tau)+\mathcal{G}_{4}(t). $$
$$\begin{aligned}\mathcal{G}_{3}(t) \lesssim{}& \mathcal{E}^{H}(0)+ \int_{0}^{t} {\delta}(1+\tau)^{-3/2}\big\| ( \eta,u) (\tau)\big\| _{7}^{2}\,\mathrm{d}\tau \\ \lesssim{}&\mathcal{E}^{H}(0)+ {\delta} \bigl( \|\eta_{0} \|_{7}^{2}+\mathcal{G}_{3}(t) \bigr), \end{aligned} $$
$$ \mathcal{G}_{3}(t) \lesssim\|\eta_{0} \|_{7}^{2}+\mathcal{E}^{H}(0). $$
(5.3)
Finally, we show the time decay behavior of \(\mathcal{G}_{4}(t)\), noting that \(\mathcal{E}^{L}\) can be controlled by \(\mathcal{D}^{L}\), except the term \(\|\nabla\eta\|_{3,0}\) in \(\mathcal{E}^{L}\). To deal with \(\|\nabla\eta\|_{3,0}\), we use (3.3) to get On the other hand, we combine (5.1) with (5.3) to get Thus, Putting the above estimate into the lower-order energy inequality (3.43), we obtain which yields with \(\mathcal{I}_{0}:=c( \mathcal{E}^{H}(0)+ \|\eta_{0}\|_{7}^{2})\) for some positive constant c. Therefore,
$$\|\nabla\eta\|_{3,0}\lesssim\|\eta\|_{3,0}^{\frac{3}{4}}\| \eta\| _{3 ,4}^{\frac{1}{4}}. $$
$${\mathcal{E}}^{L}+\|\eta\|_{7 }^{2}\lesssim \tilde{\mathcal{E}}^{L}+\| \eta\|_{7 }^{2}\lesssim \|\eta_{0}\|_{7}^{2}+\mathcal{E}^{H}(0) . $$
$$\tilde{\mathcal{E}}^{L}\lesssim{\mathcal{E}}^{L}\lesssim \bigl(\mathcal{D}^{L}\bigr)^{\frac{3}{4}}\bigl({\mathcal{E}}^{L}+ \|\eta\|_{7 }^{2}\bigr)^{\frac{1}{4}} \lesssim\bigl( \mathcal{D}^{L}\bigr)^{\frac{3}{4}} \bigl(\|\eta_{0} \|_{7}^{2}+\mathcal{E}^{H}(0)\bigr)^{\frac{1}{4}}. $$
$$\frac{\mathrm{d}}{\mathrm{d}t} \tilde{\mathcal{E}}^{L}+ \frac{ (\tilde{\mathcal{E}}^{L})^{\frac{4}{3}}}{\mathcal{I}_{0} ^{1/{3}}} \lesssim0, $$
$${\mathcal{E}}^{L}\lesssim\tilde{\mathcal{E}}^{L}\lesssim \frac {\mathcal{I}_{0}}{ ((\mathcal{I}_{0}/\mathcal{E}^{L}(0))^{1/3}+ t/3 )^{3}} \lesssim\frac{ \|\eta_{0}\|_{7}^{2}+ \mathcal{E}^{H}(0)}{ 1 +t^{3}} $$
$$ \mathcal{G}_{4}(t)\lesssim\|\eta_{0} \|_{7}^{2} + \mathcal{E}^{H}(0). $$
(5.4)
Now we sum up the estimates (5.1)-(5.4) to conclude that where (4.24) has been also used. Consequently, we have proved the following a priori stability estimate.
$$ \mathcal{G}(t): =\sum_{k=1}^{4} \mathcal{G}_{k}(t)\lesssim\|\eta_{0}\| _{7}^{2}+ \mathcal{E}^{H}(0)\lesssim\|\eta_{0}\|_{7}^{2}+ \|u_{0}\|_{6}^{2}, $$
Proposition 5.1
Let
\((\eta,u)\)
be a solution of the transformed MHD problem with an associated perturbation pressure
q. Then there is a sufficiently small
\(\delta_{1}\), such that
\((\eta,u,q)\)
enjoys the following a priori stability estimate:
provided that
\(\sqrt{\mathcal{G}_{5}(T_{1})}\leq\delta_{1}\)
for some
\(T_{1}>0\). Here
\(C_{1}\geq1\)
denotes a constant depending on the domain Ω and other physical parameters in the transformed MHD equations.
$$ \mathcal{G}(T_{1})\leq C_{1}\bigl( \| \eta_{0}\|_{7}^{2}+\|u_{0} \|_{6}^{2} \bigr), $$
(5.5)
In view of the a priori stability estimate in Proposition 5.1 and the following result of local existence of a small solution to the transformed MHD problem, we immediately obtain Theorem 2.1.
Proposition 5.2
There is a sufficiently small
\(\delta_{2}\), such that for any given initial data
\((\eta_{0},u_{0})\in H^{7}\times H^{6}\)
satisfying
and the compatibility conditions (i.e., \(\partial_{t}^{j} u(x,0)|_{\partial\Omega}=0\), \(j=1,2\)), there exist a
\(T_{2}:=T_{2}(\delta_{2})>0\)
which depends on
\(\delta_{2}\), the domain Ω and other known physical parameters, and a unique classical solution
\((\eta, u)\in C^{0}([0,T_{2}],H^{7}\times H^{6} )\)
to the transformed MHD problem (2.15), (2.16) with an associated perturbation pressure
q. Moreover, \(\partial_{t}^{i}u\in C^{0}([0,T_{2}],H^{6-2i})\)
for
\(1\leq i\leq3\), \(q\in C^{0}([0,T_{2}],H^{5})\), \(\mathcal{E}^{H}(0)\lesssim\|\eta_{0}\|_{7}^{2}+\|u_{0}\| _{6}^{2}\), and
and
\(\mathcal{G}(t)\)
is continuous on
\([0,T_{2}]\).
$$ \sqrt{\|\eta_{0}\|_{7}^{2}+ \|{u}_{0}\|_{6}^{2}}\leq\delta_{2} $$
$$\sup_{0\leq\tau\leq T_{2}}\bigl(\big\| \eta(\tau)\big\| _{7}^{2}+ \mathcal{E}^{H}(\tau)\bigr)+ \int_{0}^{T_{2}} \bigl( \mathcal{D}^{H}( \tau)+\big\| u(\tau)\big\| _{7}^{2}+\big\| q(\tau)\big\| _{6}^{2} \bigr) \,\mathrm{d}\tau< \infty, $$
Proof
The transformed MHD problem is very similar to the surface wave problem (1.4) in [20]. Moreover, the current problem is indeed simpler than the surface wave problem due to the non-slip boundary condition \(u|_{\partial\Omega}=0\). Using the standard method in [20], one can easily establish Proposition 5.2, hence we omit its proof here. In addition, the continuity, such as \((\eta, u,q)\in C^{0}([0,T],H^{7}\times H^{6}\times H^{6})\), \(\mathcal{G}(t)\) and so on, can be verified by using the regularity of \((\eta,u,q)\), the transformed MHD equations and a standard regularized method. □
6 Conclusion
We have proved the existence of a unique time-decay solution to the initial-boundary problem (1.2)-(1.4) of rotating MHD fluids in Lagrangian coordinates, which, together with the inverse transformation of coordinates, implies the existence of a time-decay solution to the original initial-boundary problem (1.2)-(1.4) with proper initial data in \(H^{7}(\Omega)\). Our result also holds for the case \(\omega=0\) (i.e., the absence of rotation), thus it improves Tan and Wang’s result in [11], in which the sufficiently small initial data at least belongs to \(H^{16}(\Omega)\). Hence our result reveals that rotation does not affect the existence of solutions of rotating MHD fluids. We mention that the phenomenon of rotating MHD fluids widely exists in nature, so our result has potential applications. In addition, based on Theorem 2.1, we will further study the Rayleigh-Taylor problem of rotating MHD fluids in the future; please refer to [21–28] for relevant results on the Rayleigh-Taylor problem.
Declarations
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Authors’ Affiliations
(1)
College of Mathematics and Computer Science, Fuzhou University
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