First of all, let us give the local existence of the solution for problem (1.1).
Lemma 2.1
Assume that the initial data \((\rho _{0}, \mathbf{u}_{0},\mathbf{B}_{0})\) satisfy \((\rho _{0}-1,\mathbf{u}_{0},\mathbf{B}_{0})\in H^{4} \times H^{3}\times H^{3}\) with \(0< \underline{\rho }\leq \rho _{0}\). Then there exists a time \(T^{*}>0\) such that in \(0\leq t\leq T^{*}\) Cauchy problem (1.1)–(1.2) possesses a unique solution \((\rho ,\mathbf{u},\mathbf{B})\) satisfying
$$\begin{aligned} \begin{aligned} & \rho -1 \in L^{\infty }\bigl(0,T^{*};H^{4} \bigr) \cap L^{2} \bigl(0,T^{*};H^{5} \bigr), \\ &\mathbf{u}, \mathbf{B}\in L^{\infty }\bigl(0,T^{*};H^{3} \bigr), \qquad \nabla \mathbf{u}, \nabla \mathbf{B}\in L^{2} \bigl(0,T^{*};H^{3}\bigr). \end{aligned} \end{aligned}$$
Proof
Following the fixed point argument (for instance, Theorem 2.1 in [5] and Theorem 2.1 in [23]), using the technique in [9] to deal with the higher-order derivatives of density and the energy method to estimate the magnetic field B (see Lemma 2.3 in [6]), the local existence and uniqueness are quite standard. For completeness, we outline the proof here. Let us begin with the following auxiliary system:
$$ \textstyle\begin{cases} \rho _{t}+\operatorname{div}(\rho \tilde{u})=0, \\ (\rho \mathbf{u})_{t}+\operatorname{div}(\rho \tilde{u}\otimes \mathbf{u})-\mu \Delta \mathbf{u}-(\mu +\lambda ) \nabla \operatorname{div}\mathbf{u} +\nabla P(\rho )- \frac{\hbar ^{2}}{2}\rho \nabla ( \frac{\Delta \sqrt{\rho }}{\sqrt{\rho }} ) \\ \quad =(\operatorname{curl} \mathbf{B})\times \mathbf{B}, \\ \mathbf{B}_{t}-v\Delta \mathbf{B}=\operatorname{curl}( \tilde{u}\times \mathbf{B}), \qquad \operatorname{div} \mathbf{B} = 0, \end{cases} $$
(2.1)
with initial condition (1.2). Here, \(\tilde{u} \in R_{T^{*}}\) is a known function with \(\tilde{u}(x,0)=\mathbf{u}_{0}(x)\) and
$$ R_{T^{*}}= \biggl\{ z \Bigm| \sup_{0\leq t\leq T^{*}} \Vert z \Vert ^{2}_{H^{3}} + \int ^{T^{*}}_{0} \Vert \nabla z \Vert ^{2}_{H^{3}}\,dt \leq R \biggr\} , $$
where \(R>1\) and \(T^{*}>0\) will be chosen later.
It is obvious that (2.1)1 is a linear transport equation with regular ũ. The existence and uniqueness are well known. As did Lemma 2.2 in [6], let
$$ \frac{d x(X,t)}{dt}=\tilde{u}\bigl(x(X,t),t\bigr) \quad \text{and}\quad x(X,0)=X, $$
then we have
$$ \frac{d \rho (x(X,t), t)}{dt}=-\rho \operatorname{div}\tilde{u}, $$
that is,
$$ \rho (x,t)=\rho _{0}\exp \biggl(- \int ^{t}_{0} \operatorname{div}\tilde{u}\,ds \biggr). $$
Therefore,
$$ \rho (x,t)\le \rho _{0}\exp \biggl( \int ^{t}_{0} \Vert \operatorname{div} \tilde{u} \Vert _{L^{\infty }}\,ds\biggr) \le \rho _{0}\exp \biggl( \int ^{t}_{0} C \Vert \tilde{u} \Vert _{H^{3}}\,ds\biggr). $$
Thus, if \(T^{*}>0\) is suitably small, we can get the bound of ρ. On the other hand, (2.1)3 is a linear parabolic equation with respect to the function B. Following the standard energy argument for linear parabolic equation, we have the existence of a unique solution B and the desired estimates. With the estimates for \((\rho , \mathbf{B})\) at hand and using the classical theory of linear parabolic equation again, we can get the existence and uniqueness of u by (2.1)2, which implies that we obtain the existence and uniqueness for the linearized system (2.1).
Next, we can define a map Λ on \(R_{T^{*}}\) to the solution of (2.1)2 as
$$ \Lambda (\tilde{u}) = \mathbf{u}. $$
Taking advantage of the energy argument, one can get some desired estimates for \((\rho , \mathbf{u}, \mathbf{B})\), which yields that u belongs to \(R_{T^{*}}\) for some large positive constant R and an appropriately small \(T^{*}\). Finally, Schauder’s fixed point theorem gives us the local well-posedness of Cauchy problem (1.1)–(1.2). □
Now, we are ready to show a priori estimates on the solutions. Firstly, we need to rewrite system (1.1)–(1.2) as follows:
$$ \textstyle\begin{cases} \rho _{t}+\operatorname{div}(\rho \mathbf{u})=0, \\ \rho \mathbf{u}_{t}-{\mu \Delta \mathbf{u}-(\mu + \lambda )\nabla \operatorname{div}\mathbf{u}}+{\nabla P}- \frac{\hbar ^{2}}{4}{\Delta \nabla \rho }=-\rho \mathbf{u} \cdot \nabla \mathbf{u}+F_{1}, \\ \mathbf{B}_{t}-v\Delta \mathbf{B}=\operatorname{curl}( \mathbf{u}\times \mathbf{B}), \qquad \operatorname{div} \mathbf{B} = 0, \end{cases} $$
(2.2)
where
$$ F_{1}=\frac{\hbar ^{2}}{4} \biggl( \frac{ \vert \nabla \rho \vert ^{2}\nabla \rho }{\rho ^{2}}+ \frac{\nabla \rho \Delta \rho }{\rho }- \frac{\nabla \rho \nabla ^{2}\rho }{\rho } \biggr) +{(\operatorname{curl} \mathbf{B})\times \mathbf{B}}, $$
with the prescribed initial condition
$$ (\rho ,\mathbf{u},\mathbf{B}) (x,t)|_{t=0} = \bigl( \rho _{0}(x),\mathbf{u}_{0}(x),\mathbf{B}_{0}(x) \bigr)\rightarrow (1,0,0) \quad \text{as } \vert x \vert \to \infty . $$
(2.3)
Define
$$ A(T)=\sup_{0\le t\le T} \int \bigl( \bigl\vert \nabla ^{2}\rho \bigr\vert ^{2}+ \vert \nabla \mathbf{u} \vert ^{2}+ \vert \mathbf{B} \vert ^{4} \bigr)\,dx. $$
In what follows, the generic constants \(C>0\) and \(C_{i}>0\) (\(i=1,2,3\)) and a suitably small constant \(\delta _{1}>0\) are dependent on some known constants \(\underline{\rho }\), ρ̄, μ, λ, γ, ħ, and v, but independent of t, respectively. Particularly, we write \(C(M)\) to emphasize that C may depend on M, where \(M=(1+C_{3})M_{1}\).
Next, let us establish necessary lower-order estimates for the global solution \((\rho ,\mathbf{u},\mathbf{B})\) independent of time. We assume that \((\rho ,\mathbf{u},\mathbf{B})\) is a solution of system (2.2)–(2.3) on \(\mathbb{R}^{3} \times (0,T)\) for positive time \(T>0\). Here, we set \(E_{0}\le 1\) without loss of generality.
Proposition 2.1
Under the assumptions of Theorem 1.1, assume that the solution \((\rho ,\mathbf{u},\mathbf{B})\) satisfies
$$ A(T)\le 2M, \qquad \frac{1}{4} \underline{\rho }\le \rho \le 2 \bar{\rho }, $$
(2.4)
for \((x,t)\in \mathbb{R}^{3} \times (0,T)\), then it holds that
$$ A(T)\le \frac{3}{2}M,\qquad \frac{1}{2}\underline{\rho }\le \rho \le \frac{3}{2}\bar{\rho }, $$
(2.5)
provided \(E_{0}\le \delta \), where δ is a positive constant depending on \(\underline{\rho }\), ρ̄, μ, λ, γ, v, and M but independent of t.
In order to prove Proposition 2.1, we firstly investigate the following lemmas. Then, we finish the proof of Proposition 2.1 after Lemma 2.5.
Lemma 2.2
Let \((\rho ,\mathbf{u},\mathbf{B})(x,t)\) be the solution of problem (2.2)–(2.3). Then it holds that
$$ \begin{aligned} &\sup_{0\le t\le T} \int \biggl(\frac{1}{2}\rho \vert \mathbf{u} \vert ^{2}+G(\rho )+\frac{\hbar ^{2}}{8\rho } \vert \nabla \rho \vert ^{2}+v \vert \mathbf{B} \vert ^{2} \biggr)\,dx \\ &\quad {}+ \int _{0}^{T} \int \bigl(\mu \vert \nabla \mathbf{u} \vert ^{2}+( \lambda +\mu ) \vert \operatorname{div}\mathbf{u} \vert ^{2}+v \vert \nabla \mathbf{B} \vert ^{2} \bigr)\,dx\,dt \le E_{0}. \end{aligned} $$
(2.6)
Proof
Multiplying (2.2)1–(2.2)3 by \(G^{\prime }(\rho )\), ρu and B respectively, summing the resulting equations up and integrating them by parts, we obtain (2.6). □
Lemma 2.3
Under the assumptions of Proposition 2.1, it holds that
$$ \frac{1}{2}\underline{\rho }\le \rho \le \frac{3}{2}\bar{\rho } $$
(2.7)
for \((x,t)\in \mathbb{R}^{3} \times (0,T)\), provided \(E_{0}\le \min \{ (\frac{1-\frac{1}{2}\underline{\rho }}{C(M)} )^{4}, (\frac{\frac{3}{2}\bar{\rho }-1}{C(M)} )^{4} \}\).
Proof
By (2.4), (2.6), and the Sobolev inequality, we obtain
$$ \Vert \rho -1 \Vert _{L^{\infty }}\le C \Vert \rho -1 \Vert _{L^{2}}^{\frac{1}{4}} \bigl\Vert \nabla ^{2} \rho \bigr\Vert _{L^{2}}^{\frac{3}{4}}\le C(M)E_{0}^{\frac{1}{4}}, $$
which implies
$$ 1-C(M)E_{0}^{\frac{1}{4}}< \rho < 1+C(M)E_{0}^{\frac{1}{4}}. $$
When \(E_{0}\le \min \{ (\frac{1-\frac{1}{2}\underline{\rho }}{C(M)} )^{4}, (\frac{\frac{3}{2}\bar{\rho }-1}{C(M)} )^{4} \}\), we get (2.7). □
Lemma 2.4
Under the assumptions of Proposition 2.1, it holds that
$$ \int _{0}^{T} \Vert \nabla \rho \Vert _{L^{2}}^{2}\,dt+ \int _{0}^{T} \bigl\Vert \nabla ^{2} \rho \bigr\Vert _{L^{2}}^{2}\le C(M)E_{0}, $$
(2.8)
provided \(E_{0}\le (\frac{\hbar ^{2}}{32C(M)} )^{4}\).
Proof
Multiplying (2.2)2 by \(\frac{1}{\rho }\nabla \rho \) and using (2.2)1, then integrating the resulting equation over \(\mathbb{R}^{3}\), one has
$$ \begin{aligned} & \int \gamma \rho ^{\gamma -2} \vert \nabla \rho \vert ^{2}\,dx+ \int \frac{\hbar ^{2}}{4}\rho ^{-1} \bigl\vert \nabla ^{2}\rho \bigr\vert ^{2}\,dx \\ &\quad =-\frac{d}{dt} \int \mathbf{u}\cdot \nabla \rho \,dx+ \int \operatorname{div}\mathbf{u}\cdot \operatorname{div}(\rho \mathbf{u})\,dx \\ &\qquad {}+ \int \frac{1}{\rho }\nabla \rho \cdot \bigl( \mu \Delta \mathbf{u}+(\mu +\lambda )\nabla \operatorname{div} \mathbf{u} \bigr)\,dx- \int \nabla \rho \cdot \mathbf{u}\cdot \nabla \mathbf{u}\,dx \\ &\qquad {} + \int \frac{1}{\rho }\nabla \rho \cdot F_{1} \,dx+ \frac{\hbar ^{2}}{4} \int \biggl(\frac{1}{\rho ^{2}} \biggr)\nabla \rho \cdot \nabla \rho \cdot \nabla ^{2}\rho \,dx \\ &\quad =-\frac{d}{dt} \int \mathbf{u}\cdot \nabla \rho \,dx+\sum_{i=1}^{5}J_{i}. \end{aligned} $$
(2.9)
In view of (2.4), (2.6), and (2.7), and using Hölder, Young, and Sobolev inequalities, we obtain
$$\begin{aligned}& \begin{aligned} J_{1}+J_{2}+J_{3}\le {}& C \Vert \nabla \mathbf{u} \Vert _{L^{2}} \Vert \nabla \rho \Vert _{L^{6}} \Vert \mathbf{u} \Vert _{L^{3}} +C \Vert \nabla \mathbf{u} \Vert _{L^{2}}^{2}+C \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}} \Vert \nabla \mathbf{u} \Vert _{L^{2}} \\ \le{} &\frac{\hbar ^{2}}{16} \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}}^{2}+ C(M) \Vert \nabla \mathbf{u} \Vert _{L^{2}}^{2}, \end{aligned} \\& \begin{aligned} J_{4}+J_{5}\le{} & C \bigl( \Vert \nabla \rho \Vert _{L^{4}} \Vert \nabla \rho \Vert _{L^{4}} \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}}+ \Vert \nabla \rho \Vert _{L^{4}}^{4} + \Vert \nabla \rho \Vert _{L^{6}} \Vert \mathbf{B} \Vert _{L^{3}} \Vert \nabla \mathbf{B} \Vert _{L^{2}} \bigr) \\ \le{} & C(M)E_{0}^{\frac{1}{4}} \bigl( \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}}^{2}+ \Vert \nabla \mathbf{B} \Vert _{L^{2}}^{2} \bigr). \end{aligned} \end{aligned}$$
Substituting the estimates of \(J_{i}\) (\(i=1,\ldots ,5\)) into (2.9), for \(E_{0}\le (\frac{\hbar ^{2}}{32C(M)} )^{4}\), it leads to
$$ \Vert \nabla \rho \Vert _{L^{2}}^{2}+ \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}}^{2} \le -C \frac{d}{dt} \int \mathbf{u}\cdot \nabla \rho \,dx+ C(M) \bigl( \Vert \nabla \mathbf{u} \Vert _{L^{2}}^{2}+ \Vert \nabla \mathbf{B} \Vert _{L^{2}}^{2}\bigr). $$
(2.10)
Integrating (2.10) from 0 to T, combining with (2.6) and (2.7), we obtain the desired result (2.8). The proof of this lemma is complete. □
Lemma 2.5
Under the assumptions of Proposition 2.1, it holds that
$$ \int _{0}^{T} \bigl\Vert \nabla ^{3}\rho \bigr\Vert _{L^{2}}^{2}\,dt\le C(M) \biggl(1+ \int _{0}^{T} \bigl\Vert \nabla ^{2}\mathbf{u} \bigr\Vert _{L^{2}}^{2}\,dt \biggr)+C \int _{0}^{T} \Vert \nabla \mathbf{B}\cdot \mathbf{B} \Vert _{L^{2}}^{2}\,dt, $$
(2.11)
provided \(E_{0}\le ( \frac{\hbar ^{2}\underline{\rho }^{\frac{9}{2}}}{16(\underline{\rho }^{2} +\hbar ^{\frac{3}{2}}(\underline{\rho }\bar{\rho })^{\frac{1}{4}})C(M)} )^{4}\).
Proof
Multiplying (2.2)2 by \(-\frac{1}{\rho }\Delta \nabla \rho \), integrating by parts over \(\mathbb{R}^{3}\), we have
$$\begin{aligned} \begin{aligned} &\frac{\hbar ^{2}}{4} \int \frac{1}{\rho } \vert \Delta \nabla \rho \vert ^{2}\,dx \\ &\quad =\frac{d}{dt} \int \Delta \nabla \rho \cdot \mathbf{u}\,dx+ \int \Delta \nabla \operatorname{div}(\rho \mathbf{u})\cdot \mathbf{u}\,dx+ \int \Delta \nabla \rho \cdot \bigl(\gamma \rho ^{ \gamma -2}\nabla \rho \bigr)\,dx \\ &\qquad {}+ \int \Delta \nabla \rho \cdot \biggl(\mathbf{u}\cdot \nabla \mathbf{u} - \frac{1}{\rho } \bigl(\mu \Delta \mathbf{u}+(\mu +\lambda )\nabla \operatorname{div} \mathbf{u} \bigr) \biggr)\,dx - \int \frac{1}{\rho }\Delta \nabla \rho \cdot F_{1}\,dx \\ &\quad =\frac{d}{dt} \int \Delta \nabla \rho \cdot \mathbf{u}\,dx+ \sum _{i=1}^{4}L_{i}. \end{aligned} \end{aligned}$$
(2.12)
Utilizing (2.4), (2.6), (2.7), Hölder, Sobolev, and Young inequalities, then integrating by parts, we have
$$\begin{aligned}& \begin{aligned} L_{1}={}& \int \Delta \nabla \operatorname{div}(\rho \mathbf{u}) \cdot \mathbf{u} \,dx = \int \nabla \operatorname{div}(\rho \mathbf{u})\cdot \Delta \mathbf{u} \,dx \\ ={}& \int \nabla ^{2}\rho \cdot \mathbf{u}\Delta \mathbf{u} \,dx +2 \int \nabla \rho \cdot \nabla \mathbf{u}\Delta \mathbf{u}\,dx + \int \rho \bigl\vert \nabla ^{2} \mathbf{u} \bigr\vert ^{2}\,dx \\ \le {}& C \Vert \rho \Vert _{L^{\infty }} \bigl\Vert \nabla ^{2}\mathbf{u} \bigr\Vert ^{2}_{L^{2}} +C \Vert \nabla \rho \Vert _{L^{3}} \Vert \nabla \mathbf{u} \Vert _{L^{6}} \bigl\Vert \nabla ^{2}\mathbf{u} \bigr\Vert _{L^{2}} +C \Vert \mathbf{u} \Vert _{L^{\infty }} \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}} \bigl\Vert \nabla ^{2}\mathbf{u} \bigr\Vert _{L^{2}} \\ \le {}&C \bigl\Vert \nabla ^{2}\mathbf{u} \bigr\Vert ^{2}_{L^{2}}+C \Vert \nabla \rho \Vert ^{\frac{1}{2}}_{L^{2}} \bigl\Vert \nabla ^{2}\rho \bigr\Vert ^{\frac{1}{2}}_{L^{2}} \bigl\Vert \nabla ^{2} \mathbf{u} \bigr\Vert ^{2}_{L^{2}} +C \Vert \nabla \mathbf{u} \Vert ^{ \frac{1}{2}}_{L^{2}} \bigl\Vert \nabla ^{2}\mathbf{u} \bigr\Vert ^{\frac{3}{2}}_{L^{2}} \bigl\Vert \nabla ^{2}\rho \bigr\Vert ^{\frac{1}{2}}_{L^{2}} \bigl\Vert \nabla ^{2}\rho \bigr\Vert ^{\frac{1}{2}}_{L^{2}} \\ \le {}& C(M) \bigl( \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla ^{2} \mathbf{u} \bigr\Vert _{L^{2}}^{2}\bigr), \end{aligned} \\& \begin{aligned} L_{2}={}& \int \Delta \nabla \rho \cdot \bigl(\gamma \rho ^{\gamma -2}\nabla \rho \bigr)\,dx \\ \le {}&\gamma \bigl(\bar{\rho }^{\gamma -2}+\underline{\rho }^{\gamma -2}\bigr) \Vert \Delta \nabla \rho \Vert _{L^{2}} \Vert \nabla \rho \Vert _{L^{2}} \\ \le {}&\frac{\varepsilon }{2} \Vert \Delta \nabla \rho \Vert _{L^{2}}^{2} + \gamma ^{2}\bigl(\bar{\rho }^{\gamma -2}+\underline{\rho }^{\gamma -2}\bigr)^{2} \frac{1}{2\varepsilon } \Vert \nabla \rho \Vert _{L^{2}}^{2}, \end{aligned} \\& \begin{aligned} L_{3}={}& \int \Delta \nabla \rho \cdot \biggl(\mathbf{u}\cdot \nabla \mathbf{u} - \frac{1}{\rho }\bigl(\mu \Delta \mathbf{u}+(\mu +\lambda )\nabla { \operatorname{div}}\mathbf{u}\bigr) \biggr)\,dx \\ \le {}& \Vert \Delta \nabla \rho \Vert _{L^{2}} \Vert \mathbf{u} \Vert _{L^{3}} \Vert \nabla \mathbf{u} \Vert _{L^{6}} + \frac{(2\mu +\lambda )}{\bar{\rho }} \Vert \Delta \nabla \rho \Vert _{L^{2}}\bigl( \Vert \Delta \mathbf{u} \Vert _{L^{2}} + \Vert \nabla { \operatorname{div}} \mathbf{u} \Vert _{L^{2}}\bigr) \\ \le {}&C \Vert \mathbf{u} \Vert _{L^{2}}^{\frac{1}{2}} \Vert \nabla \mathbf{u} \Vert _{L^{2}}^{\frac{1}{2}} \bigl\Vert \nabla ^{2} \mathbf{u} \bigr\Vert _{L^{2}} \Vert \Delta \nabla \rho \Vert _{L^{2}} + \frac{\varepsilon }{2} \Vert \Delta \nabla \rho \Vert _{L^{2}}^{2} + \frac{(2\mu +\lambda )^{2}}{\bar{\rho }^{2}2\varepsilon } \bigl\Vert \nabla ^{2} \mathbf{u} \bigr\Vert _{L^{2}}^{2} \\ \le {}&\frac{C(M)}{\underline{\rho }^{\frac{1}{4}}}E_{0}^{\frac{1}{4}}\bigl( \Vert \Delta \nabla \rho \Vert _{L^{2}}^{2} + \bigl\Vert \nabla ^{2} \mathbf{u} \bigr\Vert _{L^{2}}^{2}\bigr) +\frac{\varepsilon }{2} \Vert \Delta \nabla \rho \Vert _{L^{2}}^{2} + \frac{(2\mu +\lambda )^{2}}{\bar{\rho }^{2}2\varepsilon } \bigl\Vert \nabla ^{2} \mathbf{u} \bigr\Vert _{L^{2}}^{2}, \end{aligned} \end{aligned}$$
and
$$\begin{aligned} L_{4} =& - \int \frac{1}{\rho }\Delta \nabla \rho \cdot F_{1}\,dx \\ =& - \int \frac{1}{\rho }\Delta \nabla \rho \cdot \biggl( \frac{\hbar ^{2}}{4} \biggl( \frac{ \vert \nabla \rho \vert ^{2}\nabla \rho }{\rho ^{2}}+ \frac{\nabla \rho \Delta \rho }{\rho } - \frac{\nabla \rho \nabla ^{2}\rho }{\rho } \biggr)+({\operatorname{curl}} \mathbf{B})\times \mathbf{B} \biggr)\,dx \\ \le& C\frac{\hbar ^{2}}{\underline{\rho }^{3}} \Vert \Delta \nabla \rho \Vert _{L^{2}} \Vert \nabla \rho \Vert _{L^{6}}^{3} +C \frac{\hbar ^{2}}{\underline{\rho }^{2}} \Vert \Delta \nabla \rho \Vert _{L^{2}}^{2} \Vert \nabla \rho \Vert _{L^{3}} +\frac{1}{\underline{\rho }} \Vert \Delta \nabla \rho \Vert _{L^{2}} \Vert \nabla \mathbf{B}\cdot \mathbf{B} \Vert _{L^{2}} \\ \le& C\frac{\hbar ^{2}}{\underline{\rho }^{3}} \Vert \Delta \nabla \rho \Vert _{L^{2}} \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}}^{3} +C \frac{\hbar ^{2}}{\underline{\rho }^{2}} \Vert \Delta \nabla \rho \Vert ^{2}_{L^{2}} \Vert \nabla \rho \Vert _{L^{2}}^{\frac{1}{2}} \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}}^{\frac{1}{2}} \\ &{}+ \frac{1}{\underline{\rho }} \Vert \Delta \nabla \rho \Vert _{L^{2}} \Vert \nabla \mathbf{B}\cdot \mathbf{B} \Vert _{L^{2}} \\ \le& C(M)\frac{\hbar ^{2}}{\underline{\rho }^{3}} \Vert \Delta \nabla \rho \Vert _{L^{2}} \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}} +C(M) \frac{\hbar ^{\frac{3}{2}}\bar{\rho }^{\frac{1}{4}}}{\underline{\rho }^{2}}E_{0}^{\frac{1}{4}} \Vert \Delta \nabla \rho \Vert ^{2}_{L^{2}} + \frac{1}{\underline{\rho }} \Vert \Delta \nabla \rho \Vert _{L^{2}} \Vert \nabla \mathbf{B}\cdot \mathbf{B} \Vert _{L^{2}} \\ \le& \frac{\varepsilon }{2} \Vert \Delta \nabla \rho \Vert _{L^{2}}^{2}+C(M) \frac{\hbar ^{4}}{\underline{\rho }^{6}} \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}}^{2} +C(M) \frac{\hbar ^{\frac{3}{2}}\bar{\rho }^{\frac{1}{4}}}{\underline{\rho }^{2}}E_{0}^{\frac{1}{4}} \Vert \Delta \nabla \rho \Vert ^{2}_{L^{2}} +\frac{\varepsilon }{2} \Vert \Delta \nabla \rho \Vert _{L^{2}}^{2} \\ &{} +\frac{1}{\underline{\rho }^{2}2\varepsilon } \Vert \nabla \mathbf{B}\cdot \mathbf{B} \Vert _{L^{2}}^{2}. \end{aligned}$$
Therefore, it can be obtained that
$$\begin{aligned} &L_{2}+L_{3}+L_{4} \\ &\quad \le \biggl(2\varepsilon +\biggl(\frac{1}{\underline{\rho }^{\frac{1}{4}}} + \frac{\hbar ^{\frac{3}{2}}\bar{\rho }^{\frac{1}{4}}}{\underline{\rho }^{2}} \biggr)C(M)E_{0}^{\frac{1}{4}} \biggr) \Vert \Delta \nabla \rho \Vert _{L^{2}}^{2} +\gamma ^{2}\bigl( \bar{\rho }^{\gamma -2}+\underline{\rho }^{\gamma -2}\bigr)^{2} \frac{1}{2\varepsilon } \Vert \nabla \rho \Vert _{L^{2}}^{2} \\ &\qquad {}+C(M)\frac{\hbar ^{4}}{\underline{\rho }^{6}} \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}}^{2} + \biggl(\frac{C(M)}{\underline{\rho }^{\frac{1}{4}}}E_{0}^{\frac{1}{4}} + \frac{(2\mu +\lambda )^{2}}{\bar{\rho }^{2}2\varepsilon } \biggr) \bigl\Vert \nabla ^{2}\mathbf{u} \bigr\Vert _{L^{2}}^{2} + \frac{1}{\underline{\rho }^{2}2\varepsilon } \Vert \nabla \mathbf{B}\cdot \mathbf{B} \Vert _{L^{2}}^{2} \\ &\quad \le \frac{\hbar ^{2}}{8} \Vert \Delta \nabla \rho \Vert _{L^{2}}^{2} +C(M) \bigl( \Vert \nabla \rho \Vert _{H^{1}}^{2}+ \bigl\Vert \nabla ^{2}\mathbf{u} \bigr\Vert _{L^{2}}^{2} \bigr)+C \Vert \nabla \mathbf{B} \cdot \mathbf{B} \Vert _{L^{2}}^{2} \end{aligned}$$
provided
$$ \varepsilon =\frac{\hbar ^{2}}{32}\quad \text{and} \quad E_{0}\le \biggl( \frac{\hbar ^{2}\underline{\rho }^{\frac{9}{2}}}{16(\underline{\rho }^{2}+\hbar ^{\frac{3}{2}}(\underline{\rho }\bar{\rho })^{\frac{1}{4}})C(M)} \biggr)^{4}. $$
Substituting the estimates of \(L_{i}\) (\(i=1,\ldots ,4\)) into (2.12), we obtain
$$ \begin{aligned} &\frac{\hbar ^{2}}{8} \int \vert \Delta \nabla \rho \vert ^{2}\,dx \\ &\quad \le \frac{d}{dt} \int \Delta \nabla \rho \cdot \mathbf{u}\,dx+C(M) \bigl( \Vert \nabla \rho \Vert _{H^{1}}^{2}+ \bigl\Vert \nabla ^{2}\mathbf{u} \bigr\Vert _{L^{2}}^{2} \bigr) +C \Vert \nabla \mathbf{B}\cdot \mathbf{B} \Vert _{L^{2}}^{2}. \end{aligned} $$
(2.13)
Integrating the above inequality over \((0,T)\), by (2.4), (2.6), and (2.8), we get (2.11). □
Lemma 2.6
Under the assumptions of Proposition 2.1, it holds that
$$ \frac{d}{dt} \Vert \mathbf{B} \Vert _{L^{4}}^{4}+v \Vert \mathbf{B}\cdot \nabla \mathbf{B} \Vert _{L^{2}}^{2}\le 0, $$
(2.14)
provided \(E_{0}\le (\frac{2v}{C(M)} )^{2}\).
Proof
Multiplying (2.2)3 by \(|\mathbf{B}|^{2}\mathbf{B}\), integrating the resulting equation over \(\mathbb{R}^{3}\), combining with (2.4), we have
$$ \frac{1}{4}\frac{d}{dt} \Vert \mathbf{B} \Vert _{L^{4}}^{4}+3v \Vert \mathbf{B}\cdot \nabla \mathbf{B} \Vert _{L^{2}}^{2} \le C \Vert \mathbf{B}\cdot \nabla \mathbf{B} \Vert _{L^{2}} \bigl\Vert \vert \mathbf{B} \vert ^{2} \bigr\Vert _{L^{6}} \Vert \mathbf{u} \Vert _{L^{3}} \le C(M)E_{0}^{\frac{1}{2}} \Vert \mathbf{B} \cdot \nabla \mathbf{B} \Vert _{L^{2}}^{2}. $$
Therefore, letting \(E_{0}\le (\frac{2v}{C(M)} )^{2}\), we obtain (2.14). □
Proof of Proposition 2.1
Multiplying (2.2)2 by \(-\frac{1}{\rho }\Delta (\rho \mathbf{u} )\), using (2.2)1 and integrating it over \(\mathbb{R}^{3}\), we have
$$ \begin{aligned} &\frac{1}{2} \frac{d}{dt}\biggl( \Vert \nabla \mathbf{u} \Vert _{L^{2}}^{2}+ \frac{\hbar ^{2}}{4} \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}}^{2}\biggr) + \int \frac{1}{\rho }\Delta \mathbf{u}\cdot \bigl(\mu \Delta \mathbf{u}+(\mu +\lambda )\nabla \operatorname{div} \mathbf{u} \bigr)\,dx \\ &\quad = \int \mathbf{u}_{t}\cdot \bigl( \Delta (\rho \mathbf{u})- \Delta \mathbf{u} \bigr)\,dx \\ &\qquad {}- \int \frac{1}{\rho } \bigl(\mu \Delta \mathbf{u}+(\mu +\lambda ) \nabla \operatorname{div}\mathbf{u} \bigr)\cdot (\Delta \rho \mathbf{u} +2\nabla \rho \cdot \nabla \mathbf{u} )\,dx \\ &\qquad {}+\frac{\hbar ^{2}}{4} \int \biggl(1-\frac{1}{\rho }\biggr)\nabla \Delta \rho \cdot \Delta (\rho \mathbf{u} )\,dx + \int ( \mathbf{u}\cdot \nabla \mathbf{u} )\cdot \Delta (\rho \mathbf{u} )\,dx \\ &\qquad {} + \int \frac{1}{\rho }\nabla P\Delta (\rho \mathbf{u} )\,dx - \int \frac{1}{\rho } F_{1}\cdot \Delta (\rho \mathbf{u} )\,dx \\ &\quad = \sum_{i=1}^{6}K_{i}. \end{aligned} $$
(2.15)
Taking advantage of (2.4) and (2.6), we give the estimates about \(\|\mathbf{u}_{t}\|_{L^{2}}\) and \(\|\Delta (\rho \mathbf{u} )\|_{L^{2}}\) as follows:
$$\begin{aligned}& \Vert \mathbf{u}_{t} \Vert _{L^{2}} \le C\bigl( \bigl\Vert \nabla ^{2} \mathbf{u} \bigr\Vert _{L^{2}}+ \bigl\Vert \nabla ^{3}\rho \bigr\Vert _{L^{2}} + \Vert \nabla \rho \Vert _{L^{2}}+ \Vert \mathbf{B}\cdot \nabla \mathbf{B} \Vert _{L^{2}}\bigr), \end{aligned}$$
(2.16)
$$\begin{aligned}& \begin{aligned} \bigl\Vert \Delta (\rho \mathbf{u} ) \bigr\Vert _{L^{2}}&\le C\bigl( \bigl\Vert \nabla ^{2} \mathbf{u} \bigr\Vert _{L^{2}}+ \Vert \nabla \rho \Vert _{L^{3}} \Vert \nabla \mathbf{u} \Vert _{L^{6}}+ \Vert \mathbf{u} \Vert _{L^{\infty }} \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}}\bigr) \\ &\le C \bigl\Vert \nabla ^{2}\mathbf{u} \bigr\Vert _{L^{2}}+C(M) \bigl\Vert \nabla ^{2} \rho \bigr\Vert _{L^{2}}. \end{aligned} \end{aligned}$$
(2.17)
Recalling (2.4), (2.7), (2.15), and (2.16) and utilizing Hölder, Young, and Sobolev inequalities, we obtain
$$\begin{aligned}& \begin{aligned} K_{1}\le {}& C \Vert \mathbf{u}_{t} \Vert _{L^{2}} \bigl( \Vert \rho -1 \Vert _{L^{\infty }} \bigl\Vert \nabla ^{2}\mathbf{u} \bigr\Vert _{L^{2}}+ \Vert \nabla \rho \Vert _{L^{3}} \Vert \nabla \mathbf{u} \Vert _{L^{6}} + \Vert \mathbf{u} \Vert _{L^{3}} \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{6}} \bigr) \\ \le {}& C(M)E_{0}^{\frac{1}{2}} \Vert \mathbf{u}_{t} \Vert _{L^{2}} \bigl( \bigl\Vert \nabla ^{2}\mathbf{u} \bigr\Vert _{L^{2}}+ \bigl\Vert \nabla ^{3}\rho \bigr\Vert _{L^{2}} \bigr) \\ \le {}& C(M)E_{0}^{\frac{1}{4}} \bigl( \bigl\Vert \nabla ^{2}\mathbf{u} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla ^{3}\rho \bigr\Vert _{L^{2}}^{2}+ \Vert \nabla \rho \Vert _{L^{2}}^{2} + \Vert \mathbf{B}\cdot \nabla \mathbf{B} \Vert _{L^{2}}^{2} \bigr), \end{aligned} \\& \begin{aligned} K_{2}\le {}& C \bigl\Vert \nabla ^{2}\mathbf{u} \bigr\Vert _{L^{2}} \bigl( \Vert \nabla \mathbf{u} \Vert _{L^{6}} \Vert \nabla \rho \Vert _{L^{3}}+ \Vert \mathbf{u} \Vert _{L^{\infty }} \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}} \bigr) \\ \le {}& \biggl(\frac{\mu }{8}+C(M)E_{0}^{\frac{1}{4}} \biggr) \bigl\Vert \nabla ^{2} \mathbf{u} \bigr\Vert _{L^{2}}^{2}+C(M) \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}}^{2} \end{aligned} \end{aligned}$$
and
$$\begin{aligned}& K_{3}+K_{4}+K_{5}+K_{6} \\& \quad \le C \bigl( \Vert \rho -1 \Vert _{L^{\infty }} \bigl\Vert \nabla ^{3}\rho \bigr\Vert _{L^{2}}+ \Vert \mathbf{u} \Vert _{L^{3}} \Vert \nabla \mathbf{u} \Vert _{L^{6}} \bigr) \bigl( \bigl\Vert \nabla ^{2}\mathbf{u} \bigr\Vert _{L^{2}}+C(M) \bigl\Vert \nabla ^{2} \rho \bigr\Vert _{L^{2}} \bigr) \\& \qquad {} +C \bigl( \Vert \mathbf{B}\cdot \nabla \mathbf{B} \Vert _{L^{2}}+ \Vert \nabla \rho \Vert _{L^{6}}^{3}+ \Vert \nabla \rho \Vert _{L^{3}} \bigl\Vert \nabla ^{2} \rho \bigr\Vert _{L^{6}} \bigr) \bigl( \bigl\Vert \nabla ^{2}\mathbf{u} \bigr\Vert _{L^{2}}+C(M) \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}} \bigr) \\& \quad \le C \bigl( \Vert \nabla \rho \Vert _{L^{2}}^{\frac{1}{2}} \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}}^{\frac{1}{2}} \bigl\Vert \nabla ^{3}\rho \bigr\Vert _{L^{2}} + \Vert \mathbf{u} \Vert _{L^{2}}^{\frac{1}{2}} \Vert \nabla \mathbf{u} \Vert _{L^{2}}^{\frac{1}{2}} \bigl\Vert \nabla ^{2} \mathbf{u} \bigr\Vert _{L^{2}} \bigr) \\& \qquad {}\times \bigl( \bigl\Vert \nabla ^{2} \mathbf{u} \bigr\Vert _{L^{2}}+C(M) \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}} \bigr) \\& \qquad {} + C \bigl( \Vert \mathbf{B}\cdot \nabla \mathbf{B} \Vert _{L^{2}} + \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}}^{3} \bigr) \bigl( \bigl\Vert \nabla ^{2} \mathbf{u} \bigr\Vert _{L^{2}}+C(M) \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}} \bigr) \\& \quad \le \biggl(\frac{\mu }{8}+C(M)E_{0}^{\frac{1}{2}} \biggr) \bigl\Vert \nabla ^{2} \mathbf{u} \bigr\Vert _{L^{2}}^{2} +C \Vert \mathbf{B}\cdot \nabla \mathbf{B} \Vert _{L^{2}}^{2}+C(M)E_{0}^{-\frac{1}{2}} \Vert \nabla \rho \Vert _{H^{1}}^{2}. \end{aligned}$$
Substituting the estimates of \(K_{i}\) (\(i=1,\ldots ,6\)) into (2.15), by (2.6) and (2.8), one has
$$ \begin{aligned} &\frac{d}{dt} \bigl( \Vert \nabla \mathbf{u} \Vert _{L^{2}}^{2}+ \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}}^{2} \bigr) +2C_{1} \bigl\Vert \nabla ^{2} \mathbf{u} \bigr\Vert _{L^{2}}^{2} \\ &\quad \le C_{2} \Vert \mathbf{B}\cdot \nabla \mathbf{B} \Vert _{L^{2}}^{2}+C(M)E_{0}^{\frac{1}{2}} \bigl\Vert \nabla ^{3}\rho \bigr\Vert _{L^{2}}^{2}+C(M)E_{0}^{-\frac{1}{2}} \Vert \nabla \rho \Vert _{H^{1}}^{2}. \end{aligned} $$
(2.18)
Multiplying (2.14) by \(\frac{2C_{2}}{v}\), then substituting the resulting equation and (2.18) into (2.11), combining with (2.8), we obtain
$$ \begin{aligned} & \Vert \mathbf{B} \Vert _{L^{4}}^{4}+ \Vert \nabla \mathbf{u} \Vert _{L^{2}}^{2}+ \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}}^{2}+C_{1} \int _{0}^{T} \bigl\Vert \nabla ^{2} \mathbf{u} \bigr\Vert _{L^{2}}^{2}\,dt \\ &\quad \le C(M)E_{0}^{\frac{1}{2}}+C_{3} \Vert \mathbf{B}_{0} \Vert _{L^{4}}^{4}+ \Vert \nabla \mathbf{u}_{0} \Vert _{L^{2}}^{2}+ \bigl\Vert \nabla ^{2}\rho _{0} \bigr\Vert _{L^{2}}^{2} \\ &\quad \le \frac{3}{2}(C_{3}+1)M_{1}. \end{aligned} $$
(2.19)
Thus we complete the proof of Proposition 2.1. □
Lemma 2.7
Under the assumptions of Theorem 1.1, it holds that
$$\begin{aligned} & \Vert \nabla \mathbf{B} \Vert _{L^{2}}^{2}+ \int _{0}^{T} \bigl( \Vert \mathbf{B}_{t} \Vert _{L^{2}}^{2}+ \bigl\Vert \nabla ^{2} \mathbf{B} \bigr\Vert _{L^{2}}^{2} \bigr)\,dt\le C(M)E_{0}+C \Vert \nabla \mathbf{B}_{0} \Vert _{L^{2}}^{2}, \end{aligned}$$
(2.20)
$$\begin{aligned} & \bigl\Vert \nabla ^{2}\mathbf{B} \bigr\Vert _{L^{2}}^{2}+ \int _{0}^{T} \bigl( \Vert \nabla \mathbf{B}_{t} \Vert _{L^{2}}^{2}+ \bigl\Vert \nabla ^{3} \mathbf{B} \bigr\Vert _{L^{2}}^{2} \bigr)\,dt \le C(M) \bigl(1+ \Vert \nabla \mathbf{B}_{0} \Vert _{H^{1}}^{2} \bigr). \end{aligned}$$
(2.21)
Proof
Squaring both sides of (2.2)3, then integrating the resulting equation over \(\mathbb{R}^{3}\), by (2.4) and (2.6), we have
$$ \begin{aligned} & v\frac{d}{dt} \Vert \nabla \mathbf{B} \Vert _{L^{2}}^{2}+ \Vert \mathbf{B}_{t} \Vert _{L^{2}}^{2}+v^{2} \bigl\Vert \nabla ^{2} \mathbf{B} \bigr\Vert _{L^{2}}^{2} \\ &\quad \le C \bigl( \Vert \nabla \mathbf{u} \Vert _{L^{2}}^{2} \Vert \mathbf{B} \Vert _{L^{\infty }}^{2}+ \Vert \nabla \mathbf{B} \Vert _{L^{6}}^{2} \Vert \mathbf{u} \Vert _{L^{6}} \Vert \mathbf{u} \Vert _{L^{2}} \bigr) \\ &\quad \le C \bigl( \Vert \nabla \mathbf{u} \Vert _{L^{2}}^{2} \Vert \nabla \mathbf{B} \Vert _{L^{2}} \bigl\Vert \nabla ^{2}\mathbf{B} \bigr\Vert _{L^{2}} + \bigl\Vert \nabla ^{2}\mathbf{B} \bigr\Vert _{L^{2}}^{2} \Vert \nabla \mathbf{u} \Vert _{L^{2}} \Vert \mathbf{u} \Vert _{L^{2}} \bigr) \\ &\quad \le \biggl(C(M)E_{0}^{\frac{1}{2}}+\frac{v^{2}}{4} \biggr) \bigl\Vert \nabla ^{2} \mathbf{B} \bigr\Vert _{L^{2}}^{2}+C(M) \Vert \nabla \mathbf{B} \Vert _{L^{2}}^{2}. \end{aligned} $$
(2.22)
Integrating inequality (2.22) from 0 to T, we obtain (2.20). Similarly, combining with (2.4), (2.6), and (2.20), we obtain (2.21). Thus, we complete the proof of this lemma. □
At last, we prove the high-order estimates depending on time t.
Lemma 2.8
Under the assumptions of Theorem 1.1, it holds that
$$\begin{aligned} & { \bigl\Vert \nabla ^{3}\rho \bigr\Vert ^{2}_{H^{1}}+ \bigl\Vert \nabla ^{2} \mathbf{u} \bigr\Vert ^{2}_{H^{1}} + \int _{0}^{T}\bigl( \bigl\Vert \nabla ^{4}\rho \bigr\Vert ^{2}_{H^{1}}+ \bigl\Vert \nabla ^{3}\mathbf{u} \bigr\Vert ^{2}_{H^{1}} \bigr)\,dt\le C_{T}, } \end{aligned}$$
(2.23)
$$\begin{aligned} & \bigl\Vert \nabla ^{3}\mathbf{B} \bigr\Vert _{L^{2}}^{2}+ \int _{0}^{T} \bigl\Vert \nabla ^{4} \mathbf{B} \bigr\Vert _{L^{2}}^{2}\,dt\le C_{T}. \end{aligned}$$
(2.24)
Proof
Similar to the proofs of Lemma 3.10, Lemma 3.11, and Lemma 4.3 in [10], we can obtain (2.23). In fact, following Lemma 3.10 in [10], multiplying (2.2)2 by \(-\frac{\Delta ^{2}(\rho u)}{\rho }\), integrating by part over \(\mathbb{R}^{3}\times (0,T)\), we have
$$ \begin{aligned} &\sup_{0\le t\le T} \bigl( \bigl\Vert \nabla ^{3}\rho \bigr\Vert ^{2}_{L^{2}}+ \bigl\Vert \nabla ^{2} u \bigr\Vert ^{2}_{L^{2}} \bigr) + \int _{0}^{T} \bigl\Vert \nabla ^{3} u \bigr\Vert ^{2}_{L^{2}}\,dt \\ &\quad \le C\delta \int _{0}^{T} \bigl\Vert \nabla ^{4}\rho \bigr\Vert _{L^{2}}^{2}\,dt+C(M) - \int _{0}^{T} \int {(\operatorname{curl}\mathbf{B})\times \mathbf{B}}\cdot \frac{\Delta ^{2}(\rho u)}{\rho }\,dx\,dt \\ &\quad = C\delta \int _{0}^{T} \bigl\Vert \nabla ^{4}\rho \bigr\Vert _{L^{2}}^{2}\,dt+C(M)+I_{1}, \end{aligned} $$
(2.25)
where δ is small enough and determined later. Similar to Lemma 3.11 and Lemma 4.3 in [10], multiplying \(-\frac{\nabla \Delta ^{2}\rho }{\rho }\), integrating by part over \(\mathbb{R}^{3}\times (0,T)\), we get
$$ \begin{aligned} & \int _{0}^{T} \bigl\Vert \nabla ^{4}\rho \bigr\Vert _{L^{2}}^{2}\,dt \\ &\quad \le C\sup_{0\le t\le T}\bigl( \bigl\Vert \nabla ^{3} \rho \bigr\Vert ^{2}_{L^{2}}+ \bigl\Vert \nabla ^{2} u \bigr\Vert ^{2}_{L^{2}}\bigr) +C \int _{0}^{T} \bigl\Vert \nabla ^{3} u \bigr\Vert ^{2}_{L^{2}}\,dt+C \\ &\qquad {}- \int _{0}^{T} \int {(\operatorname{curl}\mathbf{B})\times \mathbf{B}}\cdot \frac{\nabla \Delta ^{2}\rho }{\rho }\,dx\,dt \\ &\quad = C\sup_{0\le t\le T}\bigl( \bigl\Vert \nabla ^{3} \rho \bigr\Vert ^{2}_{L^{2}}+ \bigl\Vert \nabla ^{2} u \bigr\Vert ^{2}_{L^{2}}\bigr) +C \int _{0}^{T} \bigl\Vert \nabla ^{3} u \bigr\Vert ^{2}_{L^{2}}\,dt+C+I_{2}. \end{aligned} $$
(2.26)
Substituting (2.26) into (2.25) and choosing \(\delta >0\) sufficiently small gives
$$ \sup_{0\le t\le T}\bigl( \bigl\Vert \nabla ^{3}\rho \bigr\Vert ^{2}_{L^{2}}+ \bigl\Vert \nabla ^{2} u \bigr\Vert ^{2}_{L^{2}} \bigr) + \int _{0}^{T} \bigl\Vert \nabla ^{3} u \bigr\Vert ^{2}_{L^{2}}\,dt \le C(M)+I_{1}+C+CI_{2}. $$
(2.27)
According to integrating by parts and the Cauchy inequality, we obtain
$$\begin{aligned} I_{1} =&- \int _{0}^{T} \int {(\operatorname{curl}\mathbf{B})\times \mathbf{B}}\cdot \frac{\Delta ^{2}(\rho u)}{\rho }\,dx\,dt \\ =&- \int _{0}^{T} \int \Delta \biggl( \frac{(\operatorname{curl}\mathbf{B}) \times \mathbf{B}}{\rho } \biggr) \cdot \Delta (\rho u)\,dx\,dt \\ \le &C \int _{0}^{T} \int \biggl\vert \nabla ^{2} \biggl( \frac{(\operatorname{curl}\mathbf{B})\times \mathbf{B}}{\rho } \biggr) \biggr\vert ^{2}\,dx\,dt + \int _{0}^{T} \int \bigl\vert \nabla ^{2}(\rho u) \bigr\vert ^{2}\,dx\,dt. \end{aligned}$$
Utilizing (2.19), (2.20), (2.21), Hölder, Young, and Sobolev inequalities, we get
$$\begin{aligned} &\int \biggl\vert \nabla ^{2} \biggl( \frac{(\operatorname{curl}\mathbf{B})\times \mathbf{B}}{\rho } \biggr) \biggr\vert ^{2}\,dx \\ &\quad \le C \Vert \nabla \rho \Vert _{L^{6}}^{4} \Vert \nabla \mathbf{B} \Vert _{L^{6}}^{2} \Vert \mathbf{B} \Vert _{L^{\infty }}^{2} +C \bigl\Vert \nabla ^{2} \rho \bigr\Vert _{L^{6}}^{2} \Vert \nabla \mathbf{B} \Vert _{L^{6}}^{2} \Vert \mathbf{B} \Vert _{L^{6}}^{2} +C \Vert \nabla \rho \Vert _{L^{6}}^{2} \Vert \nabla \mathbf{B} \Vert _{L^{6}}^{4} \\ &\qquad {}+C \bigl\Vert \nabla ^{3}\mathbf{B} \bigr\Vert _{L^{2}}^{2} \Vert \mathbf{B} \Vert _{L^{\infty }}^{2} +C \bigl\Vert \nabla ^{2} \mathbf{B} \bigr\Vert _{L^{6}}^{2} \Vert \nabla \mathbf{B} \Vert _{L^{6}} \Vert \nabla \mathbf{B} \Vert _{L^{2}} \\ &\quad \le C \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}}^{4} \bigl\Vert \nabla ^{2} \mathbf{B} \bigr\Vert _{L^{2}}^{2} \Vert \nabla \mathbf{B} \Vert _{H^{1}}^{2} +C \bigl\Vert \nabla ^{3}\rho \bigr\Vert _{L^{2}}^{2} \bigl\Vert \nabla ^{2}\mathbf{B} \bigr\Vert _{L^{2}}^{2} \Vert \nabla \mathbf{B} \Vert _{L^{2}}^{2} +C \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}}^{2} \bigl\Vert \nabla ^{2}\mathbf{B} \bigr\Vert _{L^{2}}^{4} \\ &\qquad {}+C \bigl\Vert \nabla ^{3}\mathbf{B} \bigr\Vert _{L^{2}}^{2} \Vert \nabla \mathbf{B} \Vert _{H^{1}}^{2} +C \bigl\Vert \nabla ^{3} \mathbf{B} \bigr\Vert _{L^{2}}^{2} \bigl\Vert \nabla ^{2}\mathbf{B} \bigr\Vert _{L^{2}} \Vert \nabla \mathbf{B} \Vert _{L^{2}} \\ &\quad \le C(M) \bigl( \bigl\Vert \nabla ^{2}\mathbf{B} \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla ^{3} \rho \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla ^{3}\mathbf{B} \bigr\Vert _{L^{2}}^{2}\bigr). \end{aligned}$$
Thus, by (2.11), (2.14), (2.19), and (2.20), it is obtained that
$$ \begin{aligned} &\int _{0}^{T} \int \biggl\vert \nabla ^{2} \biggl( \frac{(\operatorname{curl}\mathbf{B})\times \mathbf{B}}{\rho } \biggr) \biggr\vert ^{2}\,dx\,dt \\ &\quad \le C(M) \biggl( \int _{0}^{T} \bigl\Vert \nabla ^{2}\mathbf{B} \bigr\Vert _{L^{2}}^{2}\,dt+ \int _{0}^{T} \bigl\Vert \nabla ^{3}\rho \bigr\Vert _{L^{2}}^{2}\,dt + \int _{0}^{T} \bigl\Vert \nabla ^{3}\mathbf{B} \bigr\Vert _{L^{2}}^{2}\,dt \biggr) \\ &\quad \le C(M) \biggl(C(M)+C \Vert \nabla \mathbf{B}_{0} \Vert _{L^{2}}^{2} + \int _{0}^{T} \bigl\Vert \nabla ^{2}u \bigr\Vert _{L^{2}}^{2}\,dt \\ &\qquad {}+ \int _{0}^{T} \Vert \nabla \mathbf{B}\cdot \mathbf{B} \Vert _{L^{2}}^{2}\,dt + \Vert \nabla \mathbf{B}_{0} \Vert _{H^{1}}^{2}\biggr) \\ &\quad \le C(M) \bigl(C(M)+ \Vert \mathbf{B}_{0} \Vert _{L^{4}}^{4}+ \Vert \nabla \mathbf{B}_{0} \Vert _{H^{1}}^{2}\bigr) \\ &\quad \le C(M). \end{aligned} $$
(2.28)
In addition, using (2.4) and (2.19), we have
$$\begin{aligned}& \int \bigl\vert \nabla ^{2}(\rho u) \bigr\vert ^{2}\,dx \\& \quad \le C \int \bigl\vert \nabla ^{2}\rho \bigr\vert ^{2}u^{2}\,dx+C \int \vert \nabla \rho \vert ^{2} \vert \nabla u \vert ^{2}\,dx+C \int \rho ^{2} \bigl\vert \nabla ^{2} u \bigr\vert \,dx \\& \quad \le C \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}} \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{6}} \Vert u \Vert _{L^{6}}^{2} +C \Vert \nabla u \Vert _{L^{2}} \Vert \nabla u \Vert _{L^{6}} \Vert \nabla \rho \Vert _{L^{6}}^{2}+C \bigl\Vert \nabla ^{2} u \bigr\Vert _{L^{2}}^{2} \\& \quad \le C \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}} \bigl\Vert \nabla ^{3}\rho \bigr\Vert _{L^{2}} \Vert \nabla u \Vert _{L^{2}}^{2} +C \Vert \nabla u \Vert _{L^{2}} \bigl\Vert \nabla ^{2} u \bigr\Vert _{L^{2}} \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}} \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}}+C \bigl\Vert \nabla ^{2} u \bigr\Vert _{L^{2}}^{2} \\& \quad \le C(M) \bigl( \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla ^{3}\rho \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla ^{2} u \bigr\Vert _{L^{2}}^{2}\bigr). \end{aligned}$$
Hence, using (2.8), (2.20), (2.14), and (2.19), it leads to
$$ \begin{aligned} \int _{0}^{T} \int \bigl\vert \nabla ^{2}(\rho u) \bigr\vert ^{2}\,dx\,dt &\le C(M) \int _{0}^{T}\bigl( \bigl\Vert \nabla ^{2}\rho \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla ^{3}\rho \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla ^{2} u \bigr\Vert _{L^{2}}^{2} \bigr)\,dt \\ &\le C(M) \bigl(C(M)+E_{0}+ \Vert \mathbf{B}_{0} \Vert _{L^{4}}^{4}\bigr) \\ &\le C(M). \end{aligned} $$
(2.29)
Therefore,
$$ I_{1}\le C(M). $$
(2.30)
On the other hand, for \(I_{2}\), integrating by parts, the Cauchy inequality, (2.14), (2.20), (2.26), and (2.28), one has
$$ \begin{aligned} I_{2}&=- \int _{0}^{T} \int {(\operatorname{curl}\mathbf{B})\times \mathbf{B}}\cdot \frac{\nabla \Delta ^{2}\rho }{\rho }\,dx\,dt \\ &=- \int _{0}^{T} \int \Delta \biggl( \frac{(\operatorname{curl}\mathbf{B})\times \mathbf{B}}{\rho } \biggr) \cdot \nabla \Delta \rho \,dx\,dt \\ &\le C \int _{0}^{T} \int \biggl\vert \nabla ^{2} \biggl( \frac{(\operatorname{curl}\mathbf{B})\times \mathbf{B}}{\rho } \biggr) \biggr\vert ^{2}\,dx\,dt +C \int _{0}^{T} \int \bigl\vert \nabla ^{3}\rho \bigr\vert ^{2}\,dx\,dt \\ &\le C(M). \end{aligned} $$
(2.31)
Substituting (2.30) and (2.31) into (2.27), we obtain
$$ \sup_{0\le t\le T}\bigl( \bigl\Vert \nabla ^{3}\rho \bigr\Vert ^{2}_{L^{2}}+ \bigl\Vert \nabla ^{2} u \bigr\Vert ^{2}_{L^{2}}\bigr) + \int _{0}^{T} \bigl\Vert \nabla ^{3} u \bigr\Vert ^{2}_{L^{2}}\,dt \le C(M). $$
Together with (2.26), it yields
$$ \int _{0}^{T} \bigl\Vert \nabla ^{4}\rho \bigr\Vert _{L^{2}}^{2}\,dt\le C(M). $$
Therefore, we get the desired estimate (2.23).
Using (2.4), (2.6), (2.20), (2.21), and (2.23), similar to the proof of Lemma 2.7, one can obtain (2.24). □