Firstly, using similar methods to [27] and the references therein, we can obtain the local existence and the uniqueness of the solutions to (1.1)-(1.3) with the regularities as in Theorem 1.1. We omit it here for brevity. In the following of this section, we derive some a priori estimates for a local classical solution of problem (1.1)-(1.3) on \(\Omega\times[0,T]\) for some \(T>0\), with the initial data \((m_{0},n_{0},u_{0})\) satisfying (1.10)-(1.14).
Next, we give the well-known Gagliardo-Nirenberg inequality, which will be used frequently later.
Lemma 2.1
(Gagliardo-Nirenberg inequality) [14]
For any
\(p,b',r'\in[1,\infty]\)
and any integer
l
and
j, there exist some generic constant
\(\alpha\in[0,1]\)
and
\(C>0\), for every function
\(u\in C_{0}^{\infty}\), such that
$$ \bigl\| \nabla^{j} u\bigr\| _{L^{p}(\mathbb{R}^{N})}\leq C\bigl\| \nabla^{l} u\bigr\| _{L^{b'}(\mathbb {R}^{N})}^{\alpha}\|u\|_{L^{r'}(\mathbb{R}^{N})}^{1-\alpha}, $$
(2.1)
where
$$ \frac{1}{p}=\frac{j}{N}+\alpha \biggl( \frac{1}{b'}-\frac{l}{N} \biggr)+(1-\alpha)\frac{1}{r'},\quad \frac{j}{l}\leq\alpha\leq1. $$
(2.2)
\(\alpha\neq1\), when
\(l-\frac{N}{b'}=j\)
and
\(1< p<\infty\).
Next, the following Zlotnik inequality will be used to get the uniform (in time) upper bound of the m and n.
Lemma 2.2
([28])
Let the function
y
satisfy
$$ y'(t)= g(y)+b'(t), \quad\textit{on } [0,T] ,\qquad y(0)=y^{0}, $$
with
\(g\in C(\mathbb{R})\)
and
\(y,b\in W^{1,1}(0,T)\). If
\(g(\infty)=-\infty\)
and
$$ b(t_{2}) -b(t_{1}) \le N_{0} +N_{1}(t_{2}-t_{1}) $$
(2.3)
for all
\(0\le t_{1}< t_{2}\le T\)
with some
\(N_{0}\ge0\)
and
\(N_{1}\ge0\), then
$$ y(t)\le\max \bigl\{ y^{0},\overline{\zeta} \bigr\} +N_{0}< \infty, \quad\textit{on } [0,T], $$
where
\(\overline{\zeta} \)
is a constant such that
$$ g(\zeta)\le-N_{1} \quad\textit{for } \zeta\ge\overline{\zeta}. $$
(2.4)
Lemma 2.3
([29])
Assume
\(X\subset E\subset Y\)
are Banach spaces and
\(X\hookrightarrow\hookrightarrow E\). Then the following embeddings are compact:
$$\begin{aligned}& (\mathrm{i})\quad \biggl\{ \varphi:\varphi\in L^{q}(0,T; X), \frac{\partial\varphi}{\partial t}\in L^{1}(0,T;Y) \biggr\} \hookrightarrow\hookrightarrow L^{q}(0,T; E), \quad\textit{if } 1\leq q\leq\infty; \\& (\mathrm{ii}) \quad \biggl\{ \varphi:\varphi\in L^{\infty}(0,T; X), \frac{\partial\varphi}{\partial t} \in L^{r}(0,T;Y) \biggr\} \hookrightarrow\hookrightarrow C\bigl([0,T]; E \bigr), \quad \textit{if } 1< r\leq\infty. \end{aligned}$$
Firstly, we denote
$$\begin{aligned}& A_{1}(T)=\sup_{0< t\leq T}\sigma\int|\nabla u|^{2}\,dx+\int_{0}^{T}\int\sigma m| \dot{u}|^{2}\,dx\,ds, \\& A_{2}(T)=\sup_{0< t\leq T}\sigma^{3}\int m| \dot{u}|^{2}\,dx+\int_{0}^{T}\int \sigma ^{3}|\nabla\dot{u}|^{2}\,dx\,ds, \end{aligned}$$
here \(\sigma(t)=\min\{1,t\}\). For any \((x,t)\in \Omega\times[0,T]\), we make the following a priori assumptions:
$$ 0\leq m(x,t)\leq2\bar{m} $$
(2.5)
and
$$ A_{1}(T)+A_{2}(T)\leq2C_{0}^{\frac{1}{2}}. $$
(2.6)
Under the assumption (2.5), we make the following remark.
Remark 2.1
Under the conditions of Theorem 1.1, for any \(0\leq t\leq T\), we have
$$ 0\leq\underline{s}_{0}m\leq n\leq\frac{\tilde{n}}{\tilde{m}}m, \quad x \in\Omega. $$
(2.7)
Proof
In fact, define the particle trajectories \(x=X(t,y)\) given by
$$ \left \{ \textstyle\begin{array}{@{}l} \frac{d}{dt}X(t,y)=u(X(t,y),t),\\ X(0,y)=y. \end{array}\displaystyle \right . $$
(2.8)
From (1.1)1 and (1.1)2, we have
$$ \biggl(\frac{n}{m} \biggr)_{t}+u\cdot\nabla \biggl(\frac{n}{m} \biggr)=0, $$
(2.9)
which implies
$$ \frac{n(x,t)}{m(x,t)}=\frac{n_{0}}{m_{0}} \bigl(X^{-1}(t,x) \bigr):=s_{0}=s_{0}(x,t), $$
(2.10)
where \(X^{-1}\) denotes the inverse of X. We obtain (2.7) together with (1.13). □
In the following, C denotes a generic constant depending only on \(\bar{m}\), \(\tilde{m}\), \(\tilde{n}\), \(a_{0}\), \(C^{0}\), \(\underline{s}_{0}\), μ, λ, and the initial data may vary in different estimates; we write \(C(\alpha)\) to emphasize that C depends on α.
Remark 2.2
Under the conditions of Theorem 1.1, we have
$$\begin{aligned}& 0\leq P_{m}\leq C\bigl(C^{0}\bigr), \quad\mbox{in }\Omega \times[0,T], \end{aligned}$$
(2.11)
$$\begin{aligned}& 2a_{0}C^{0}\leq P_{n}\leq C \bigl(C^{0}, a_{0}, \underline{s}_{0}\bigr), \quad\mbox{in }\Omega \times[0,T]. \end{aligned}$$
(2.12)
The above remark plays a key role in the proof of the upper bound of the density \((m,n)\). Now, the standard energy estimate is given as follows.
Proposition 2.1
If
\((m,n,u)(x,t)\)
is a classical solution of (1.1)-(1.3) in
\([0,T]\), satisfying the assumptions (2.5)-(2.6), then we have
$$\begin{aligned} & \sup_{0\leq t\leq T}\int \biggl(\frac{1}{2}m|u|^{2}+G(m,n) \biggr)\,dx+\int_{0}^{T} \int\bigl(\mu|\nabla u|^{2}+(\lambda+\mu)|\operatorname{div}u|^{2}\bigr)\,dx\,dt \\ &\quad{} +\beta^{-1}\int_{0}^{T}\int _{\partial\Omega}|u|^{2}\,dS_{x}\,dt \leq{C_{0}}. \end{aligned}$$
(2.13)
Proof
In [14, 20], one obtained this standard energy estimate in \(\mathbb{R}^{N}\) (\(N=2,3\)). Using a similar argument to [14, 20], we can easily obtain this energy estimate and omit the details. □
Notice that the function \(G(m,n)\) is equivalent to \((m-\tilde {m})^{2}+(n-\tilde{n})^{2}\), which implies
$$\begin{aligned} &\sup_{0\leq t\leq T}\int \bigl( m|u|^{2}+(m- \tilde{m})^{2}+(n-\tilde{n})^{2} \bigr)\,dx \\ &\quad{}+\int _{0}^{T}\int|\nabla u|^{2}\,dx\,dt+ \beta^{-1}\int_{0}^{T}\int _{\partial\Omega}|{u}|^{2}\,dS_{x}\,dt\leq CC_{0}. \end{aligned}$$
(2.14)
Now, we use ω and F, denoting the vorticity matrix and the effective viscous flux, defined in the following form:
$$ F\triangleq(2\mu+\lambda)\operatorname{div}u-P(m,n)+P(\tilde{m},\tilde {n}), \qquad \omega \triangleq\nabla\times{u}, $$
then we can rewrite (1.1)3 in the form
$$ m\dot{u}^{j}=\partial_{j}F+\mu \partial_{k}\omega^{j,k}, $$
(2.15)
which implies
$$ \Delta F=\operatorname{div}(m\dot{u}),\qquad \mu\Delta \omega^{j,k}= \partial_{k}\bigl(m\dot{u}^{j}\bigr)-\partial_{j} \bigl(m\dot{u}^{k}\bigr) $$
(2.16)
and
$$ (\mu+\lambda)\Delta u^{j}=\partial_{j}F+ ( \mu+\lambda)\partial _{k}w^{j,k}+\partial_{j} \bigl(P(m, n)-P(\tilde{m}, \tilde{n})\bigr). $$
(2.17)
Together with the energy estimate, we can get the following lemma.
Lemma 2.4
If
\((m,n,u)\)
is a classical solution of (1.1)-(1.3), we have for any
\(p\in[2, 6]\)
$$\begin{aligned}& \|u\|_{L^{p}}\leq C(\bar{m})C_{0}^{\frac{6-p}{4p}} \|\nabla u\|_{L^{2}}^{\frac{3p-6}{2p}}+C_{0}^{\frac{6-p}{6p}}\|\nabla u\|_{L^{2}}, \end{aligned}$$
(2.18)
$$\begin{aligned}& \|\nabla F\|_{L^{p}}+\|\nabla\omega\|_{L^{p}}\leq C\bigl(\|m\dot{u}\|_{L^{p}}+\| \nabla{u}\|_{L^{p}}\bigr), \end{aligned}$$
(2.19)
$$\begin{aligned}& \begin{aligned}[b] \|F\|_{L^{p}}+\|\omega\|_{L^{p}}\leq{}& C\|m\dot{u} \|_{L^{2}}^{\frac {3p-6}{2p}}\bigl(\|\nabla u\|_{L^{2}}+\bigl\| P(m,n)-P( \tilde{m},\tilde{n})\bigr\| _{L^{2}}\bigr)^{\frac{6-p}{2p}}\\ &{}+C\bigl(\|\nabla{u}\|_{L^{2}}+\bigl\| P(m,n)-P(\tilde{m},\tilde{n})\bigr\| _{L^{2}}^{\frac{6-p}{2p}}\|\nabla{u}\|_{L^{2}}^{\frac{3p-6}{2p}}\bigr), \end{aligned} \end{aligned}$$
(2.20)
$$\begin{aligned}& \|\nabla u\|_{L^{p}}\leq C\bigl(\|F\|_{L^{p}}+\|\omega \|_{L^{p}}\bigr)+C\bigl\| P(m,n)-P(\tilde{m},\tilde{n})\bigr\| _{L^{p}}, \end{aligned}$$
(2.21)
$$\begin{aligned}& \begin{aligned}[b] \|\nabla u\|_{L^{p}}\leq{}&C\bigl(\|\nabla u \|_{L^{2}}^{\frac{6-p}{2p}}\bigl(\|m\dot {u}\|_{L^{2}}+\|\nabla{u} \|_{L^{2}}\bigr)\bigr)\\ &{}+C\bigl(\bigl\| P(m,n)-P(\tilde{m},\tilde{n})\bigr\| _{L^{2}}+\bigl\| P(m,n)-P(\tilde {m},\tilde {n})\bigr\| _{L^{6}}\bigr)^{\frac{3p-6}{2p}}. \end{aligned} \end{aligned}$$
(2.22)
Also, for
\(0\leq t_{1}\leq t_{2}\leq T\), we have for any
\(p\geq2\)
and
\(r\geq0\)
$$\begin{aligned} \int_{t_{1}}^{t_{2}}\int \sigma^{r}\bigl|P(m,n)-P(\tilde{m},\tilde {n})\bigr|^{p}\,dx\,ds\leq C \biggl(\int_{t_{1}}^{t_{2}}\int\sigma^{r}|F|^{p} \,dx\,ds+C_{0} \biggr). \end{aligned}$$
(2.23)
Proof
Using a similar argument to that in [22] (Lemma 4.3) and [30] (Lemma 3.3), we can obtain this lemma and omit the details. □
The proofs of the following estimates are similar to [14, 24, 25, 30, 31].
Lemma 2.5
If
\((m,n,u)\)
is the classical solution of (1.1)-(1.3) in
\([0,T]\)
as in Proposition
2.1, then we have
$$ \begin{aligned}[b] A_{1}(T)&=\sup_{0< t\leq T}\sigma\int| \nabla u|^{2}\,dx+\int_{0}^{T} \int\sigma m|\dot{u}|^{2}\,dx\,dt\\ &\leq CC_{0}+C\int_{0}^{T} \int\sigma| \nabla u|^{3}\,dx\,dt+C\int_{0}^{T} \int \sigma\bigl(|u|^{2}|\nabla{u}|+|u||\nabla{u}|^{2}\bigr)\,dx \,dt \end{aligned} $$
(2.24)
and
$$\begin{aligned} A_{2}(T) =&\sup_{0< t\leq T} \sigma^{3} \int m|\dot{u}|^{2}\,dx+\int_{0}^{T} \int\sigma^{3} |\nabla u|^{2}\,dx\,dt \\ \leq&CC_{0}+CA_{1}(T) \\ &{}+C\int_{0}^{T} \int\sigma^{3}\bigl[|u|^{4}+|\nabla u|^{4}+|\dot {u}||\nabla{u}||u|+|\dot{u}||\nabla{u}|^{2}\bigr]\,dx\,dt. \end{aligned}$$
(2.25)
Proof
Multiplying (1.1)3 by \(\sigma\dot{u}\), and integrating over Ω, we obtain
$$\begin{aligned} \int\sigma m|\dot{u}|^{2}\,dx =&\int\bigl(-\sigma\dot{u} \cdot \nabla P+\mu\sigma\Delta u\cdot\dot{u}+(\lambda+\mu)\sigma\nabla \operatorname{div}u\cdot\dot{u}\bigr)\,dx \\ =&\sum_{i=1}^{3}J_{i}. \end{aligned}$$
(2.26)
Integrating by parts and using (1.1)1, (1.1)2, and the Cauchy inequality, we have
$$\begin{aligned} J_{1} =&\int-\sigma\dot{u}\cdot\nabla P\,dx \\ =& \biggl(\int\sigma\operatorname{div}u\bigl(P(m,n)-P(\tilde{m},\tilde {n})\bigr)\,dx \biggr)_{t} \\ &{}-\int \bigl\{ \sigma'\operatorname{div}u\bigl(P(m,n)-P(\tilde{m}, \tilde {n})\bigr)+\sigma P(m,n)_{t}\operatorname{div}u+\sigma(u\cdot\nabla u) \cdot \nabla P \bigr\} \,dx \\ =& \biggl(\int\sigma\operatorname{div}u\bigl(P(m,n)-P(\tilde{m},\tilde {n})\bigr)\,dx \biggr)_{t} -\int\sigma'\operatorname{div}u\bigl(P(m,n)-P( \tilde{m},\tilde {n})\bigr)\,dx \\ &{}+\int\sigma \bigl\{ (mP_{m}+nP_{n}) ( \operatorname{div}u)^{2}-P(m,n) (\operatorname{div}u)^{2}+P(m,n) \partial_{i} u^{j}\partial_{j}u^{i} \bigr\} \,dx \\ \leq& \biggl(\int\sigma\operatorname{div}u\bigl(P(m,n)-P(\tilde{m},\tilde {n})\bigr) \,dx \biggr)_{t}+C\|\nabla{u}\|_{L^{2}}^{2}+C \sigma'C_{0}, \end{aligned}$$
(2.27)
and noting the boundary condition (2.13) and the outer normal vector \(N=(N^{1},N^{2},N^{3})=(0,0,-1)\), we get
$$\begin{aligned} J_{2} =&\mu\sigma\int\Delta{u}\cdot\dot{u}\,dx \\ =&\mu\sigma \int\Delta {u}\cdot (u_{t}+u\cdot\nabla{u})\,dx \\ =&-\frac{\mu}{2} \biggl(\int\sigma|\nabla u|^{2}\,dx \biggr)_{t}+\frac{\mu }{2}\int \sigma'|\nabla u|^{2}\,dx-\mu\sigma\int\partial_{i}u^{j}\partial _{i}u^{k}\partial _{k}u^{j}\,dx \\ &{}+ \frac{\mu}{2}\int\sigma\partial_{k}u^{k}\bigl(\partial _{i}u^{j}\bigr)^{2}\,dx +\mu\sigma\int_{\partial\Omega}\partial_{i}u^{j} \dot {u}^{j}N^{i}\,dS_{x} \\ =&-\frac{\mu}{2} \biggl(\int\sigma|\nabla{u}|^{2}\,dx \biggr)_{t}+\frac {\mu }{2}\int\sigma'| \nabla{u}|^{2}\,dx -\mu\sigma\int\partial_{i}u^{j} \partial_{i}u^{k}\partial_{k}u^{j}\,dx \\ &{}+\frac {\mu }{2}\int\sigma\partial_{k}u^{k}\bigl( \partial_{i}u^{j}\bigr)^{2}\,dx -\frac{\mu}{2} \biggl(\sigma\int_{\partial\Omega}\beta ^{-1}|u|^{2}\,dS_{x} \biggr)_{t} \\ &{}-\mu \sigma\int_{\partial\Omega}\beta^{-1}u^{j}u^{i}u_{i}^{j}\,dS_{x} +\frac{\mu}{2}\sigma'\int_{\partial\Omega}\beta ^{-1}|u|^{2}\,dS_{x} \\ \triangleq&-\frac{\mu}{2} \biggl(\int\sigma|\nabla{u}|^{2}\,dx \biggr)_{t}+\frac {\mu}{2}\int\sigma'| \nabla{u}|^{2}\,dx -\mu\sigma\int\partial_{i}u^{j} \partial_{i}u^{k}\partial_{k}u^{j}\,dx \\ &{}+ \frac {\mu }{2}\int\sigma\partial_{k}u^{k}\bigl( \partial_{i}u^{j}\bigr)^{2}\,dx -\frac{\mu}{2} \biggl(\sigma\int_{\partial\Omega}\beta ^{-1}|u|^{2}\,dS_{x} \biggr)_{t}+J_{2}^{1}+J_{2}^{2}. \end{aligned}$$
(2.28)
Now we need to estimate the boundary term \(J_{2}^{1}\) and \(J_{2}^{2}\). We apply the fact that for \(h\in(C^{1}\cap W^{1,1})(\overline{\Omega})\),
$$\begin{aligned} \int_{\partial\Omega}h(x)\,dS =\int_{\Omega\bigcap\{0\leq x_{3}\leq1\} } \bigl[h(x)+(x_{3}-1)h_{x_{3}}(x)\bigr]\,dx. \end{aligned}$$
(2.29)
Since \(j,k\in\{1,2\}\), using the fact (2.29), and integrating by parts in the \(x_{1}\) and \(x_{2}\) directions imply
$$\begin{aligned} J_{2}^{1} =&-\mu\sigma\int _{\partial\Omega}\beta ^{-1}u^{j}u^{i}u_{i}^{j}\,dS_{x} \\ =&-\mu\sigma\int_{\Omega\bigcap\{0\leq x_{3}\leq1\}}\beta^{-1} \bigl[u^{j}u^{i}u_{i}^{j} +(x_{3}-1) \bigl(u^{j}u^{i}u_{i}^{j} \bigr)_{x_{3}}\bigr]\,dx \\ =&-\mu\sigma\int_{\Omega\bigcap\{0\leq x_{3}\leq1\}}\beta^{-1}\bigl\{ u^{j}u^{i}u_{i}^{j}+(x_{3}-1) \bigl(u_{x_{3}}^{j}u^{i}u_{i}^{j}+u^{j}u_{x_{3}}^{i}u_{i}^{j} \bigr)\bigr\} \\ &{}+\mu\sigma\int_{\Omega\bigcap\{0\leq x_{3}\leq1\}}\beta ^{-1}(x_{3}-1) \bigl(u_{i}^{j}u^{i}u_{x_{3}}^{j}+u^{j}u_{i}^{i}u_{x_{3}}^{j} \bigr)\,dx \\ \leq&C\sigma\int_{\Omega}\bigl(|u|^{2}| \nabla{u}|+|u||\nabla{u}|^{2}\bigr)\,dx, \end{aligned}$$
(2.30)
$$\begin{aligned} J_{2}^{2} =&\frac{\mu}{2} \sigma'\int_{\partial\Omega}\beta^{-1}|u|^{2} \,dS_{x} \\ =&\frac{\mu}{2}\sigma'\int _{\Omega\bigcap\{0\leq x_{3}\leq1\}}\beta ^{-1}\bigl[|u|^{2}+2(x_{3}-1)u \cdot{u_{x_{3}}}\bigr]\,dx \\ \leq&C\sigma'\int|u|^{2}\,dx+C\|\nabla{u} \|_{L^{2}}^{2} \leq C\sigma'\biggl(C_{0}+C_{0}^{\frac{2}{3}} \int|\nabla{u}|^{2}\,dx\biggr)+C\|\nabla {u}\| _{L^{2}}^{2} \\ \leq&CC_{0}\sigma'+C\|\nabla{u}\|_{L^{2}}^{2}, \end{aligned}$$
(2.31)
here we have used (2.18), then we can estimate \(J_{2}\) as
$$\begin{aligned} J_{2} \leq&-\frac{\mu}{2} \biggl(\int\sigma| \nabla{u}|^{2}\,dx+\sigma \int_{\partial\Omega} \beta^{-1}|u|^{2}\,dS_{x} \biggr)_{t} +C \|\nabla{u}\|_{L^{2}}^{2} \\ &{}+C\sigma\int|\nabla{u}|^{3}\,dx+C\sigma\int_{\Omega} \bigl(|u|^{2}|\nabla {u}|+|u||\nabla{u}|^{2}\bigr) \,dx+CC_{0}\sigma'. \end{aligned}$$
(2.32)
Similarly, we have
$$\begin{aligned} J_{3} =&(\lambda+\mu)\int\sigma\nabla(\operatorname{div}u) \cdot\dot {u}\,dx \\ \leq&-\frac{(\lambda+\mu)}{2} \biggl(\int\sigma|\operatorname{div}u|^{2}\,dx \biggr)_{t}+C\|\nabla{u}\|_{L^{2}}^{2} +\int\sigma| \nabla{u}|^{3}\,dx \\ \leq&-\frac{(\lambda+\mu)}{2} \biggl(\int\sigma|\operatorname{div}u|^{2}\,dx \biggr)_{t}+\int\sigma|\nabla{u}|^{3}\,dx+C\|\nabla{u} \|_{L^{2}}^{2}. \end{aligned}$$
(2.33)
Combing (2.28)-(2.33), then integrating the resulting inequality over \([0,T]\) and using Young’s inequality, we can obtain
$$\begin{aligned} &\sup_{0\leq{t}\leq{T}}\sigma\|\nabla{u} \|_{L^{2}}^{2}+\int_{0}^{T}\int \sigma {m}|\dot{u}|^{2}\,dx\,dt +\int_{0}^{T} \int\sigma|\operatorname{div}u|^{2}\,dx\,dt \\ &\qquad{}+\int_{0}^{T} \int_{\partial \Omega }\sigma\beta^{-1}|u|^{2} \,dS_{x}\,dt \\ &\quad\leq CC_{0}+C\int_{0}^{T}\int\sigma| \nabla{u}|^{3}\,dx\,dt+C\int_{0}^{T}\int \sigma \bigl(|u|^{2}|\nabla{u}|+|\nabla{u}|^{2}|u|\bigr)\,dx \,dt. \end{aligned}$$
(2.34)
Multiplying \(\sigma^{3}\dot{u}^{j}(\frac{\partial}{\partial t} +\operatorname{div}(u\cdot))\) to \(\mbox{(1.1)}_{3}^{j}\), summing with respect to j, and integrating the resulting equation over Ω, we have
$$\begin{aligned} \biggl(\frac{\sigma^{3}}{2}\int m|\dot{u}|^{2}\,dx \biggr)_{t} =&\frac {3}{2}\int \sigma^{2} \sigma_{t}m|\dot{u}|^{2}\,dx+\mu\sigma^{3}\int \dot{u}^{j}\bigl[\Delta {u_{t}^{j}}+\operatorname{div} \bigl(u\Delta u^{j}\bigr)\bigr]\,dx \\ &{}-\sigma^{3}\int \dot{u}^{j}\bigl[\partial_{j}P_{t}+\operatorname{div}( \partial _{j}Pu)\bigr]\,dx \\ &{}+(\lambda+\mu)\sigma^{3}\int\dot{u}^{j}\bigl[ \partial_{t}\partial _{j}\operatorname{div}u+\operatorname{div}(u \partial_{j}\operatorname{div}u)\bigr]\,dx =\sum _{i=1}^{4}H_{i}. \end{aligned}$$
(2.35)
Integrating by parts and using Young’s inequality again, we have
$$\begin{aligned} H_{2} =&\mu\int\sigma^{3}\dot{u}^{j} \bigl[\Delta u_{t}^{j}+\operatorname{div}\bigl(u\Delta u^{j}\bigr)\bigr]\,dx \\ =&-\mu\int\sigma^{3}\bigl[|\nabla\dot{u}|^{2}+ \partial_{i}\dot {u}^{j}\partial _{k}u^{k} \partial_{i}u^{j} -\partial_{i} \dot{u}^{j}\partial_{i}u^{k}\partial_{k}u^{j} -\partial_{i}u^{j}\partial_{i}u^{k} \partial_{k}\dot{u}^{j}\bigr]\,dx \\ &{}-\mu\int_{\partial\Omega}\sigma^{3}\beta^{-1}| \dot{u}|^{2}\,dS_{x} +\mu\int_{\partial\Omega} \sigma^{3}\dot{u}^{j}u_{k}^{j}u^{k}\,dS_{x} -\mu\int_{\partial\Omega}\sigma^{3}\partial_{k} \dot{u}^{j}u^{k}\partial _{3}u^{j}\,dS_{x} \\ \leq&-\frac{\mu}{2}\int\sigma^{3}|\nabla\dot{u}|^{2} \,dx+C\int\sigma ^{3}|\nabla u|^{4}\,dx -\mu\int _{\partial\Omega}\sigma^{3}\beta^{-1}|\dot {u}|^{2}\,dS_{x} \\ &{}+C\mu\int\sigma^{3}\bigl[|u||\nabla{u}||\dot{u}|+|u||\nabla {u}|| \nabla\dot {u}|+|\nabla{u}|^{2}|\dot{u}|\bigr]\,dx \\ \leq&-\frac{\mu}{4}\int\sigma^{3}|\nabla\dot{u}|^{2} \,dx+C\int\sigma ^{3}|\nabla u|^{4}\,dx -\mu\int _{\partial\Omega}\sigma^{3}\beta^{-1}|\dot {u}|^{2}\,dS_{x} \\ &{}+C\int\sigma^{3}\bigl(|\nabla{u}|^{4}+|u|^{4} \bigr)\,dx+C\mu\int\sigma ^{3}\bigl[|u||\nabla {u}||\dot{u}|+| \nabla{u}|^{2}|\dot{u}|\bigr]\,dx, \end{aligned}$$
(2.36)
where we have used
$$\begin{aligned} \int_{\partial\Omega}\sigma^{3} \beta^{-1}\dot{u}^{j}u_{k}^{j}u^{k} \partial_{3}u^{j} \,dS_{x}\leq C\int _{\Omega}\sigma^{3}|\nabla{u}||u||\nabla\dot{u}|\,dx \end{aligned}$$
(2.37)
and
$$\begin{aligned} \int_{\partial\Omega}\sigma^{3} \partial_{k}\dot{u}^{j}u^{k}\,dS_{x}\leq C \int_{\Omega}\sigma^{3}\bigl[|u||\nabla{u}|| \dot{u}|+|u||\nabla{u}||\nabla \dot {u}|+|\nabla{u}|^{2}|\dot{u}|\bigr] \,dx. \end{aligned}$$
(2.38)
From (1.1)1 and (1.1)2, we have
$$\begin{aligned} H_{3} =&-\int\sigma^{3}\dot{u}^{j} \bigl[\partial_{j}P_{t}+\operatorname{div}(\partial _{j}Pu)\bigr]\,dx \\ =&-\int\sigma^{3}\bigl[P_{m}(m\operatorname{div}u+u\cdot\nabla m) \partial_{j}\dot {u}^{j}+P_{n}(n\operatorname{div}u+u \cdot\nabla n)\partial_{j}\dot {u}^{j}\bigr]\,dx \\ &{}-\int\sigma^{3}P(m,n)\partial_{j}\bigl( \partial_{k}\dot {u}^{j}u^{k}\bigr)\,dx \\ =&\int\sigma^{3}\bigl[-P_{m}m\operatorname{div}u \partial_{j}\dot{u}^{j}-P_{n}n\operatorname{div}u \partial_{j}\dot{u}^{j}+\partial_{k}\bigl( \partial_{j}\dot{u}^{j}u^{k}\bigr)P -P\bigl( \partial_{j}\bigl(\partial_{k}\dot{u}^{j}u^{k} \bigr)\bigr)\bigr]\,dx \\ =&\int\sigma^{3}\bigl[-P_{m}m\operatorname{div}u \partial_{j}\dot{u}^{j}-P_{n}n\operatorname{div}u \partial_{j}\dot{u}^{j}+\partial_{j} \dot{u}^{j}\operatorname{div}uP -\partial_{k}\dot{u}^{j} \partial_{j}u^{k}P\bigr]\,dx \\ \leq&C\int\sigma^{3}|\nabla u||\nabla\dot{u}|\,dx\leq \frac{\mu }{8}\int \sigma^{3}|\nabla\dot{u}|^{2}\,dx+C\int \sigma^{3}|\nabla u|^{2}\,dx. \end{aligned}$$
(2.39)
Similarly,
$$\begin{aligned} H_{4}\leq-\frac{\lambda+\mu}{2}\int\sigma|\operatorname{div}\dot {u}|^{2}\,dx+C\int \sigma|\nabla{u}|^{4}\,dx. \end{aligned}$$
(2.40)
Combining (2.35)-(2.40) and integrating the result inequality over \((0,T)\), noting that \(\int_{0}^{T}\int\sigma'm|\dot{u}|^{2}\,dx\,dt \leq\int{m}|\dot{u}|^{2}\,dx\leq A_{1}(T)\), we can obtain (2.25), thus we complete the proof of Lemma 2.5. □
The following lemma will be applied to bound the higher-order terms occurring on the right hand side of (2.24) and (2.25).
Lemma 2.6
If
\((m,n,u)\)
is the classical solution of (1.1)-(1.3) in
\([0,T]\)
as in Proposition
2.1, then there is a positive constant
\(T_{1}\)
such that
$$ \sup_{t\in[0,T_{1}\wedge{T}]}\int|\nabla u|^{2}\,dx+\int _{0}^{T_{1}\wedge T}\int m|\dot{u}^{2}|\,dx\,dt\leq C(1+M). $$
(2.41)
Proof
In fact, multiplying (1.1)3 by \(\dot{u}\), then integrating the resulting equality over \(\Omega\times[0,t]\), we have
$$ \int_{0}^{t}\int m|\dot{u}|^{2}\,dx\,ds= \int_{0}^{t}\int \bigl\{ -\dot{u}\cdot\nabla P(m,n)+ \mu\Delta u\cdot u+(\mu+\lambda) \nabla(\operatorname{div}u)\cdot\dot{u} \bigr\} \,dx\,ds. $$
Using a similar argument to that in the proof of (2.24), we have
$$\begin{aligned} &\int|\nabla u|^{2}\,dx+\int_{0}^{t} \int m|\dot{u}|^{2}\,dx\,ds \\ &\quad\leq C(C_{0}+M)+C\int_{0}^{t}\int| \nabla{u}|^{3}\,dx\,ds+\int_{0}^{t}\int |u|^{2}|\nabla {u}|+|u||\nabla{u}|^{2}\,dx\,ds. \end{aligned}$$
From (2.5), (2.6), (2.14), (2.19)-(2.23), using Young’s inequality and the Hölder inequality, we obtain
$$\begin{aligned} \int_{0}^{t}\int| \nabla{u}|^{3}\,dx\,ds \leq&C\int_{0}^{t} \int \bigl(|F|^{3}+|\omega |^{3} \bigr)\,dx\,ds +C\int _{0}^{t}\int\bigl[P(m,n)-P(\tilde{m},\tilde{n}) \bigr]^{3}\,dx\,ds \\ \leq&C\int_{0}^{t}\bigl(\|F\|_{L^{2}}^{\frac{3}{2}} \|\nabla{F}\|_{L^{2}}^{\frac{3}{2}} +\|\omega\|_{L^{2}}^{\frac{3}{2}} \|\nabla{\omega}\|_{L^{2}}^{\frac {3}{2}}\bigr)\,ds+CC_{0} \\ \leq&C\int_{0}^{t} \bigl(\|\nabla{u} \|_{L^{2}}+\bigl\| P(m,n)-P(\tilde {m},\tilde {n})\bigr\| _{L^{2}} \bigr)^{\frac{3}{2}} \\ &{}\times\bigl(\|m\dot{u}\|_{L^{2}}+\|\nabla{u} \|_{L^{2}}\bigr)^{\frac {3}{2}}\,ds+CC_{0} \\ \leq&\frac{1}{4}\int_{0}^{t}\int m| \dot{u}|^{2}\,dx\,ds+C\int_{0}^{t}\| \nabla{u}\|_{L^{2}}^{6}\,ds+CC_{0} \end{aligned}$$
(2.42)
and
$$\begin{aligned} &\int_{0}^{t} \int|u|^{2}|\nabla{u}|+|u||\nabla{u}|^{2}\,dx\,ds \\ &\quad\leq C\int_{0}^{t}\int\bigl(| \nabla{u}|^{2}+|u|^{4}\bigr)\,dx\,ds+C\int _{0}^{t}\|\nabla {u}\| _{L^{3}}\|\nabla{u} \|_{L^{2}}^{2}\,ds \\ &\quad\leq C\int_{0}^{t}\bigl[C_{0}^{\frac{1}{2}} \|\nabla{u}\|_{L^{2}}^{3}+C_{0}^{\frac {1}{3}}\| \nabla{u}\|_{L^{2}}^{4}\bigr]\,ds +C\int_{0}^{t} \bigl(\|\nabla{u}\|_{L^{3}}^{3}+\|\nabla{u}\|_{L^{2}}^{3} \bigr)\,ds \\ &\quad\leq\frac{1}{4}\int_{0}^{t}\int{m} \dot{u}\,dx\,ds+C\int_{0}^{t}\bigl(\|\nabla {u}\| _{L^{2}}^{3}+\|\nabla{u}\|_{L^{2}}^{4}+\| \nabla{u}\|_{L^{2}}^{6}\bigr)\,ds +CC_{0}, \end{aligned}$$
(2.43)
which implies
$$\begin{aligned} &\int|\nabla u|^{2}\,dx+\int_{0}^{t} \int m|\dot{u}|^{2}\,dx\,ds \\ &\quad\leq C(1+M)+Ct\Bigl(1+\sup_{s\in[0,t]}\bigl\| \nabla{u}(\cdot,t)\bigr\| _{L^{2}}^{6}\Bigr) \\ &\quad\leq C(1+M)+Ct\sup_{s\in[0,t]}\bigl\| \nabla{u}(\cdot,t) \bigr\| _{L^{2}}^{6}\leq C(1+M), \end{aligned}$$
we can easily obtain (2.41) when we choose \(T_{1}=\min \{1,\frac{1}{8C^{3}(1+M)^{2}} \}\). □
Proposition 2.2
If
\((m,n,u)\)
is the classical solution of (1.1)-(1.3) in
\([0,T]\)
as in Proposition
2.1, then there is a positive constant
C
such that
$$ A_{1}(T)+A_{2}(T)\leq C_{0}^{\frac{1}{2}}, $$
(2.44)
provided
\(C_{0}\leq\varepsilon_{1}\).
Proof
From Lemma 2.5, we get
$$\begin{aligned} A_{1}(T)+A_{2}(T) \leq& CC_{0}+C\int _{0}^{T}\int\sigma^{3}|\nabla u|^{4}\,dx\,dt+C\int_{0}^{T}\int\sigma| \nabla u|^{3}\,dx\,dt \\ &{}+C\int_{0}^{T}\sigma^{3}\|u \|_{L^{4}}^{4}\,dt +C\int_{0}^{T} \int\sigma^{3}\bigl[|u||\nabla{u}||\dot{u}|+|\nabla{u}|^{2}| \dot {u}|\bigr]\,dx\,dt \\ &{}+C\int_{0}^{T}\int\sigma\bigl(| \nabla{u}||u|^{2}+|\nabla{u}||u|\bigr)\,dx\,dt. \end{aligned}$$
(2.45)
We now estimate the term of the right hand sides of (2.45). Using (2.21), we have
$$ \int_{0}^{T}\int \sigma^{3}|\nabla u|^{4}\,dx\,dt\leq C\int _{0}^{T}\int\sigma ^{3} \bigl[|F|^{4}+|\omega|^{4}+\bigl|P(m,n)-P(\tilde{m}, \tilde{n})\bigr|^{4}\bigr]\,dx\,dt. $$
(2.46)
From (2.5)-(2.6), (2.14), and (2.20)-(2.22), we have
$$\begin{aligned} &\int_{0}^{T}\int \sigma^{3}\bigl(|F|^{4}+|\omega|^{4}\bigr)\,dx\,dt \\ &\quad\leq C\int_{0}^{T}\sigma^{3} \bigl(\| \nabla u\|_{L^{2}}+\bigl\| P(m,n)-P(\tilde {m},\tilde{n})\bigr\| _{L^{2}} \bigr) \|m\dot{u}\|_{L^{2}}^{3}\,dt \\ &\qquad{}+C\int_{0}^{T}\sigma^{3}\|\nabla{u} \|_{L^{2}}^{4}\,dt +C\int_{0}^{T} \sigma^{3}\bigl\| P(m,n)-P(\tilde{m},\tilde{n})\bigr\| _{L^{2}}\|\nabla {u}\| _{L^{2}}^{3}\,dt \\ &\quad\leq C\sup_{0< t\leq T}\bigl[\sigma^{\frac{3}{2}}\|\sqrt{m}\dot{u} \| _{L^{2}}\bigl(\sigma^{\frac{1}{2}}\|\nabla u\|_{L^{2}}+C_{0}^{\frac {1}{2}} \bigr)\bigr]\int_{0}^{T}\int\sigma m| \dot{u}|^{2}\,dx\,dt \\ &\qquad{}+CC_{0}^{\frac{3}{2}}\sup_{0< t\leq T}\bigl( \sigma^{\frac{1}{2}}\|\nabla {u}\| _{L^{2}}\bigr)+CC_{0}\sup _{0< t\leq T}\bigl(\sigma\|\nabla{u}\|_{L^{2}}^{2} \bigr) \\ &\quad\leq C\bigl(A_{1}^{\frac{1}{2}}(T)+C_{0}^{\frac{1}{2}} \bigr)A_{2}^{\frac {1}{2}}(T)A_{1}(T)+CC_{0}^{\frac{3}{2}}A_{1}(T)^{\frac {1}{2}}+CC_{0}A_{1}(T) \\ &\quad\leq CC_{0}, \end{aligned}$$
(2.47)
using (2.23) and (2.47), we get
$$\begin{aligned} \int_{0}^{T}\int \sigma^{3}\bigl|P(m,n)-P(\tilde{m},\tilde{n})\bigr|^{4}\,ds\leq C \biggl(\int_{0}^{T}\int\sigma^{3}|F|^{4} \,dx\,ds+C_{0} \biggr) \leq CC_{0}. \end{aligned}$$
(2.48)
Notice that
$$\begin{aligned} \int_{0}^{T}\int\sigma| \nabla{u}|^{3}\,dx\,ds=\int_{0}^{T_{1}\wedge T}\int \sigma |\nabla{u}|^{3}\,dx\,ds+\int_{T_{1}\wedge T}^{T} \int\sigma|\nabla{u}|^{3}\,dx\,ds. \end{aligned}$$
(2.49)
By Young’s inequality, (2.14) and (2.46)-(2.47), we obtain
$$\begin{aligned} \int_{T_{1}\wedge T}^{T}\int\sigma|\nabla u|^{3}\,dx\,ds \leq&\int_{T_{1}\wedge T}^{T}\int \sigma\bigl(|\nabla u|^{4}+|\nabla u|^{2}\bigr)\,dx\,ds \\ \leq&C\int_{T_{1}\wedge T}^{T}\int\sigma^{3}| \nabla u|^{4}\,dx\,ds \\ &{}+\int_{T_{1}\wedge T}^{T}\int| \nabla u|^{2}\,dx\,ds\leq CC_{0}, \end{aligned}$$
(2.50)
and by (2.14), (2.22), and Lemma 2.6, we get
$$\begin{aligned} \int_{0}^{T_{1}\wedge T}\sigma\|\nabla{u} \|_{L^{3}}^{3}\,dt \leq&C\int_{0}^{T_{1}\wedge T} \sigma\|\nabla u\|_{L^{2}}^{\frac{3}{2}} \bigl(\|m\dot{u} \|_{L^{2}}^{\frac{3}{2}}+\|\nabla{u}\|_{L^{2}}^{\frac {3}{2}}+C_{0}^{\frac{1}{4}}+C_{0}^{\frac{3}{4}} \bigr)\,dt \\ \leq&C\int_{0}^{T_{1}\wedge T}\bigl(\sigma^{\frac{1}{4}}\| \nabla{u}\| _{L^{2}}^{\frac{3}{2}}\bigr) \biggl(\sigma\int m| \dot{u}|^{2}\,dx \biggr)^{\frac {3}{4}}\,dt \\ &{}+C\int_{0}^{T_{1}\wedge T} \sigma\|\nabla u\| _{L^{2}}^{3}\,ds+CC_{0} \\ \leq&C\sup_{t\in(0,T_{1}\wedge{T}]}\bigl(\bigl(\sigma\|\nabla{u}\| _{L^{2}}^{2}\bigr)^{\frac {1}{4}}\|\nabla{u} \|_{L^{2}}^{\frac{1}{2}}\bigr) \\ &{}\times\int_{0}^{T_{1}\wedge T} \|\nabla{u}\|_{L^{2}}^{\frac{1}{2}} \biggl(\sigma\int m|\dot{u}|^{2}\,dx \biggr)^{\frac{3}{4}}\,dt \\ &{}+\sup_{t\in(0,T_{1}\wedge{T}]}\bigl(\sigma\|\nabla{u}\| _{L^{2}}^{2} \bigr)^{\frac {1}{2}}\int_{0}^{T_{1}\wedge{T}} \sigma^{\frac{1}{2}}\|\nabla{u}\| _{L^{2}}^{2} \,dt+CC_{0} \\ \leq&CA_{1}(T)C_{0}^{\frac{1}{4}}+CC_{0}\leq CC_{0}^{\frac{3}{4}}. \end{aligned}$$
(2.51)
Using (2.14), (2.18)-(2.22) we get
$$\begin{aligned}& \int_{0}^{T}\int \sigma^{3}|u|^{4}\,dx\,dt\leq C\int_{0}^{T} \sigma^{3}\bigl(C_{0}^{\frac {1}{2}}\| \nabla{u} \|_{L^{2}}^{3}+C_{0}^{\frac{1}{3}}\|\nabla{u} \|_{L^{2}}^{4}\bigr)\,dt\leq CC_{0}, \end{aligned}$$
(2.52)
$$\begin{aligned}& \int_{0}^{T}\int\sigma\bigl(| \nabla{u}||u|^{2}+|\nabla {u}|^{2}|u|\bigr)\,dx \\& \quad\leq C\int_{0}^{T}\int|\nabla{u}|^{2} \,dx\,dt+\int_{0}^{T}\int\sigma^{2}|u|^{4} \,dx\,dt +\int_{0}^{T}\sigma\|\nabla{u} \|_{L^{3}}\|\nabla{u}\|_{L^{2}}^{2}\,dt \\& \quad\leq C\int_{0}^{T}\int|\nabla{u}|^{2} \,dx\,dt+\int_{0}^{T}\int\sigma ^{2}|u|^{4}\,dx\,dt+C\int \sigma\bigl(\|\nabla{u} \|_{L^{3}}^{3} +\|\nabla{u}\|_{L^{2}}^{3}\bigr) \,dt \\& \quad\leq CC_{0}^{\frac{3}{4}}, \end{aligned}$$
(2.53)
and
$$\begin{aligned} &\int_{0}^{T}\int\sigma\bigl(|u|| \nabla{u}|\dot{u}|+|\nabla{u}|^{2}|\dot {u}|\bigr)\,dx\,dt \\ &\quad\leq\int_{0}^{T}\sigma^{3}\|u \|_{L^{3}}\|\nabla{u}\|_{L^{2}}\|\nabla\dot {u}\|_{L^{2}} \,dt +C\int_{0}^{T}\sigma^{3}\|\nabla{u} \|_{L^{3}}\|\nabla{u}\|_{L^{2}}\|\nabla \dot {u}\|_{L^{2}} \,dt \\ &\quad\leq C\int_{0}^{T}\sigma^{3} \bigl[C_{0}^{\frac{1}{4}}\|\nabla{u}\| _{L^{2}}^{\frac {1}{2}}+C_{0}^{\frac{1}{6}} \|\nabla{u}\|_{L^{2}}\bigr] \|\nabla{u}\|_{L^{2}}\|\nabla\dot{u} \|_{L^{2}}\,dt \\ &\qquad{}+C\int_{0}^{T}\sigma^{3}\|\nabla{u} \|_{L^{3}}\|\nabla{u}\|_{L^{2}}\|\nabla \dot {u}\|_{L^{2}} \,dt \\ &\quad\leq CC_{0}^{\frac{1}{4}}\int_{0}^{T} \sigma^{3}\|\nabla{u}\|_{L^{2}}^{\frac {3}{2}}\|\nabla\dot{u} \|_{L^{2}}\,dt +CC_{0}^{\frac{1}{6}}\int_{0}^{T} \sigma^{3}\|\nabla{u}\|_{L^{2}}^{2}\|\nabla \dot {u} \|_{L^{2}}\,dt \\ &\qquad{}+C\int_{0}^{T}\sigma^{3}\|\nabla{u} \|_{L^{3}}\|\nabla{u}\|_{L^{2}}\|\nabla \dot {u}\|_{L^{2}} \,dt \\ &\quad\leq CC_{0}^{\frac{3}{4}}, \end{aligned}$$
(2.54)
here we have used the following fact:
$$ \begin{aligned}[b] &\int_{0}^{T}\sigma^{3} \|\nabla{u}\|_{L^{3}}\|\nabla{u}\|_{L^{2}}\|\nabla \dot {u} \|_{L^{2}}\,dt\\ &\quad\leq C\int_{0}^{T}\sigma^{3}\| \nabla{u}\|_{L^{3}}^{3}\,ds+C\int_{0}^{T} \sigma^{3}\| \nabla {u}\|_{L^{2}}\|\nabla{u}\|_{L^{2}}^{\frac{1}{2}} \|\nabla\dot{u}\|_{L^{2}}^{\frac{3}{2}}\,ds\\ &\quad\leq CC_{0}^{\frac{3}{4}}+CA_{1}(T)^{\frac{1}{2}} \biggl( \int_{0}^{T}\sigma \| \nabla{u} \|_{L^{2}}^{2}\,ds \biggr)^{\frac{1}{4}} \biggl(\int _{0}^{T}\sigma^{3}\|\nabla\dot{u} \|_{L^{2}}^{2}\,ds \biggr)^{\frac {3}{4}}\leq CC_{0}^{\frac{3}{4}}, \end{aligned} $$
(2.55)
which together with (2.45)-(2.54) give
$$\begin{aligned} A_{1}(T)+A_{2}(T) \leq&CC_{0}^{\frac{3}{4}} \leq C_{0}^{\frac{1}{2}}, \end{aligned}$$
(2.56)
when we choose \(\varepsilon_{1}\leq C(\bar{m},M)^{-4}\). This completes the proof of Proposition 2.2. □
From Proposition 2.2, we can obtain the following corollary.
Corollary 2.1
If
\((m,n,u)\)
is the classical solution of (1.1)-(1.3) in
\([0,T]\)
as in Proposition
2.1, then there is a positive constant
C
such that
$$\begin{aligned}& \sup_{0< t\leq T}\|\nabla u\|_{L^{2}}^{2}+ \int_{0}^{T}\int m|\dot {u}|^{2}\,dx\,dt \leq C(\bar{m},M), \end{aligned}$$
(2.57)
$$\begin{aligned}& \sup_{0< t\leq T}\int\sigma m|\dot{u}|^{2} \,dx+\int_{0}^{T}\int\sigma |\nabla\dot {u}|^{2}\,dx\,dt\leq C(\bar{m},M), \end{aligned}$$
(2.58)
provided
\(C_{0}<\varepsilon_{1}\).
Proof
Obviously, we can get (2.57) by Lemma 2.6 and Proposition 2.2.
Next, we prove (2.58). Applying the operator \(\sigma\dot {u}(\frac{\partial}{\partial t}+\operatorname{div}(u\cdot))\) to (1.1)3 and integrating the resulting equality over \(\Omega\times[0,T]\) and by using integration by parts, (1.1)1, (1.1)2, Young’s inequality, (2.14), (2.22), and (2.57), we have
$$\begin{aligned} &\sup_{0< t\leq T}\int\sigma m| \dot{u}|^{2}\,dx+\int_{0}^{T}\int\sigma | \nabla \dot{u}|^{2}\,dx\,dt \\ &\quad\leq\int_{0}^{T}\int\sigma_{t}m| \dot{u}|^{2}\,dx\,dt+C\int_{0}^{T}\int \sigma |\nabla u|^{4}\,dx\,dt+C(\bar{m})C_{0} \\ &\qquad{}+C\int_{0}^{T}\int\sigma|u|^{4} \,dx+C\int_{0}^{T}\int\sigma\bigl[|u||\nabla {u}|| \dot {u}|+|\nabla{u}|^{2}|\dot{u}|\bigr]\,dx\,dt \\ &\quad\leq\int_{0}^{\sigma(T)}\int m|\dot{u}|^{2} \,dx\,dt+C\int_{T_{1}\wedge {T}}^{T}\int\sigma|\nabla u|^{4}\,dx\,dt+C\int_{0}^{T_{1}\wedge{T}}\int\sigma | \nabla u|^{4}\,dx\,dt \\ &\qquad{}+C\int_{0}^{T}\int\sigma^{3}|u|^{4} \,dx\,dt+C\int_{0}^{T}\int\sigma \bigl[|u||\nabla {u}||\dot{u}|+|\nabla{u}|^{2}|\dot{u}|\bigr]\,dx\,dt+C( \bar{m})C_{0} \\ &\quad\leq C(\bar{m},M)+C\int_{T_{1}\wedge{T}}^{T}\int \sigma^{3}|\nabla u|^{4}\,dx\,dt+C\int_{0}^{T_{1}\wedge{T}} \int\sigma|\nabla u|^{4}\,dx\,dt \\ &\qquad{}+C\int_{0}^{T}\int\sigma\bigl[|u||\nabla{u}|| \dot{u}|+|\nabla {u}|^{2}|\dot {u}|\bigr]\,dx\,dt \\ &\quad\leq C(\bar{m},M)+C\int_{0}^{T_{1}\wedge{T}}\sigma\|\nabla{u} \| _{L^{2}} \bigl(\|m\dot{u}\|_{L^{2}}^{3}+\bigl\| P(m,n)-P( \tilde{m},\tilde{n})\bigr\| _{L^{6}}^{3} \\ &\qquad{}+\| \nabla {u} \|_{L^{2}}^{3} +\bigl\| P(m,n)-P(\tilde{m},\tilde{n}) \bigr\| _{L^{2}}^{3} \bigr)\,dt \\ &\qquad{}+C\int_{0}^{T}\int\sigma\bigl[|u||\nabla{u}|| \dot{u}|+|\nabla {u}|^{2}|\dot {u}|\bigr]\,dx\,dt \\ &\quad\leq C(\bar{m},M)+C(\bar{m})\sup_{t\in(0,T_{1}\wedge{T}]} \bigl[\bigl(\sigma ^{\frac{1}{2}}\|\nabla{u}\|_{L^{2}}\bigr) \bigl(\sigma^{\frac{1}{2}}\|m \dot{u}\|_{L^{2}}\bigr) \bigr] \\ &\qquad{}\times\int_{0}^{T_{1}\wedge{T}} \|m\dot{u}\|_{L^{2}}^{2}\,dt+\sup_{t\in (0,T)}\bigl( \sigma\|\nabla{u}\|_{L^{2}}^{2}\bigr)\int_{0}^{T_{1}\wedge{T}} \|\nabla {u}\| _{L^{2}}^{2}\,dt +\frac{1}{2}\int_{0}^{T}\sigma\|\nabla \dot{u}\|_{L^{2}}^{2}\,dt \\ &\quad\leq C(\bar{m},M)+C(\bar{m},M)\sup_{t\in(0,T]} \sigma^{\frac {1}{2}}\| m\dot{u}\|_{L^{2}}+\frac{1}{2}\int _{0}^{T}\sigma\|\nabla\dot{u}\|_{L^{2}}^{2} \,dt. \end{aligned}$$
(2.59)
This completes the proof of Corollary 2.1. □
Next, we give the time-independent upper bound of m and n by using similar arguments to [22, 24]. It is noted that (2.11) plays a key role here.
Proposition 2.3
If
\((m,n,u)\)
is the classical solution of (1.1)-(1.3) in
\([0,T]\)
as in Proposition
2.1, then there is a positive constant
C
such that
$$ \sup_{0\leq t\leq T}\bigl\| m(t)\bigr\| _{L^{\infty}}\leq \frac{7\bar {m}}{4}, \qquad\sup_{0\leq t\leq T}\bigl\| n(t)\bigr\| _{L^{\infty}}\leq \frac {7\bar{m}}{4}\frac{\tilde{n}}{\tilde{m}},\quad (x,t)\in\Omega \times[0,T], $$
(2.60)
provided that
\(C_{0}\leq\varepsilon\).
Proof
Rewrite the equation of the mass conservation (1.1)1 as
$$ D_{t} m=g(m)+b'(t), $$
where
$$\begin{aligned}& D_{t}m\triangleq m_{t}+u \cdot\nabla m ,\qquad g(m)\triangleq- \frac{m}{2\mu+\lambda}\bigl(P(m, n)-P(\tilde{m}, \tilde{n})\bigr) ,\\& b(t)\triangleq- \frac{1}{2\mu+\lambda} \int_{0}^{t}m F\,dt. \end{aligned}$$
For \(t\in[0,\sigma(T)]\), by Lemma 2.1, (2.14), (2.19), and (2.20), we have, for all \(0\leq t_{1}< t_{2}\leq\sigma{(T)}\),
$$\begin{aligned} &\bigl|b(t_{2})-b(t_{1})\bigr|\\ &\quad\leq C(\bar{m})\int_{t_{1}}^{t_{2}} \bigl\| F(\cdot,t)\bigr\| _{L^{\infty}}\,dt\\ &\quad\leq C(\bar{m})\int_{0}^{\sigma(T)}\bigl\| F(\cdot,t) \bigr\| ^{1/4}_{L^{2}}\bigl\| \nabla {F}(\cdot,t)\bigr\| ^{3/4}_{L^{6}} \,dt\\ &\quad\leq C(\bar{m})\int_{0}^{\sigma(T)}\bigl(\|\nabla{u} \|_{L^{2}}^{\frac{1}{4}} +\bigl\| P(m,n)-P(\tilde{m},\tilde{n}) \bigr\| _{L^{2}}^{\frac{1}{4}}\bigr)\\ &\qquad{}\times \bigl(\|\nabla\dot{u}\|_{L^{2}}^{\frac{3}{4}}+\|m\dot{u}\| _{L^{2}}^{\frac {3}{4}}+\|\nabla{u}\|_{L^{2}}^{\frac{3}{4}} + \bigl\| P(m,n)-P(\tilde{m},\tilde{n})\bigr\| _{L^{2}}^{\frac{3}{4}}\\ &\qquad{}+\bigl\| P(m,n)-P(\tilde {m},\tilde{n})\bigr\| _{L^{6}}^{\frac{3}{4}}\bigr)\,dt\\ &\quad\leq C(\bar{m})\int_{0}^{\sigma(T)}\bigl( \sigma^{-\frac{1}{2}}\bigl(\sigma ^{\frac {1}{2}}\|\nabla{u}\|_{L^{2}} \bigr)^{\frac{1}{4}} +C_{0}^{\frac{1}{8}}\sigma^{-\frac{3}{8}}\bigr) \bigl[\bigl(\sigma\|\nabla\dot{u}\| _{L^{2}}^{2} \bigr)^{\frac{3}{8}} +\bigl(\bigl(\sigma\|m\dot{u}\|_{L^{2}}^{2} \bigr)^{\frac{3}{8}}\bigr)\bigr]\,dt\\ &\qquad{}+C(\bar{m})\int_{0}^{\sigma(T)}\bigl( \sigma^{\frac{1}{2}}\|\nabla {u}\| _{L^{2}}\bigr)^{\frac{1}{4}} \sigma^{-\frac{1}{8}}\bigl(\bigl\| P(m,n)-P(\tilde{m},\tilde{n})\bigr\| _{L^{2}}^{2} \bigr)^{\frac{3}{8}}\,dt +C(\bar{m})C_{0}^{\frac{1}{2}}\\ &\quad\leq C(\bar{m})C_{0}^{\frac{1}{16}}\int_{0}^{\sigma(T)} \bigl(\sigma ^{-\frac{1}{2}}+1\bigr) \bigl[\bigl(\sigma\|\nabla\dot{u} \|_{L^{2}}^{2}\bigr)^{\frac{3}{8}}+\bigl(\sigma\|m\dot {u}\| _{L^{2}}^{2}\bigr)^{\frac{3}{8}}\bigr]\,dt\\ &\qquad{}+C(\bar{m})C_{0}^{\frac{1}{16}} \int_{0}^{\sigma(T)} \sigma^{-\frac{1}{8}}\bigl(\bigl\| P(m,n)-P(\tilde {m},\tilde {n})\bigr\| _{L^{2}}^{2} \bigr)^{\frac{3}{8}}\,dt +C(\tilde{m})C_{0}^{\frac{1}{2}}\\ &\quad\leq C(\bar{m})C_{0}^{\frac{1}{16}} \biggl(1+\int _{0}^{1}\sigma^{-\frac{4}{5}}\,dt \biggr)^{\frac{5}{8}} \biggl[ \biggl(\int_{0}^{\sigma(T)} \sigma\|\nabla \dot{u}\|_{L^{2}}^{2}\,dt \biggr)^{\frac{3}{8}} + \biggl(\int_{0}^{\sigma (T)}\sigma\| m\dot{u} \|_{L^{2}}^{2}\,dt \biggr)^{\frac{3}{8}} \biggr]\\ &\qquad{}+C(\bar{m})C_{0}^{\frac{1}{16}} \biggl(\int_{0}^{\sigma(T)} \sigma ^{-\frac {1}{5}}\,dt \biggr)^{\frac{5}{8}} \biggl(\int_{0}^{\sigma(T)} \bigl\| P(m,n)-P(\tilde{m},\tilde{n})\bigr\| _{L^{2}}^{2}\,dt \biggr)^{\frac{3}{8}} +C(\tilde{m})C_{0}^{\frac{1}{2}}\\ &\quad\leq C(\bar{m},M)C_{0}^{\frac{1}{16}}, \end{aligned}$$
provided that \(C_{0}\leq\varepsilon_{1}\). Therefore, for \(t\in[0, \sigma(T)]\), choose \(N_{1}=0\) and \(N_{0}=C(\bar{m},M)C_{0}^{\frac{1}{16}}\) and \(\bar{\xi}=2\tilde{m}\). Then
$$\begin{aligned} g(\xi) =&-\frac{\xi}{2\mu+\lambda} \bigl(P(\xi, \xi s_{0})-P( \tilde{m}, \tilde{n}) \bigr) \\ =& -\frac{\xi}{2\mu+\lambda} \bigl(P(\xi, s_{0} \xi)-P(\xi, \tilde{n})+P( \xi, \tilde{n})-P(\tilde{m}, \tilde{n}) \bigr) \\ =& -\frac{\xi}{2\mu+\lambda} \bigl(P_{m}\bigl(\tilde{m}+ \theta_{1}(\xi -\tilde{m}), \tilde{n}\bigr) (\xi-\tilde{m})+P_{n} \bigl(\xi,\tilde{n}+(s_{0}\xi-\tilde {n})\theta _{2}\bigr) (s_{0}\xi-\tilde{n}) \bigr) \\ \triangleq& -\frac{1}{2\mu+\lambda}z(\xi), \end{aligned}$$
(2.61)
where \(\theta_{1}, \theta_{2}\in(0, 1)\) are constants. From Remark 2.2 and (1.14), we obtain
$$ z(\xi)\geq2a_{0}C^{0} \xi(\underline{s}_{0} \xi- \tilde{n})\geq4a_{0}C^{0} \tilde{m}(2\underline{s}_{0}\tilde{m}-\tilde{n})\geq0, \quad\mbox{for all }\xi\geq\bar{\xi}= 2\tilde{m}, $$
and Lemma 2.2 yields
$$\begin{aligned} \sup_{t\in[0,\sigma(T)]}\|m\|_{L^{\infty}} \leq& \max\{ \bar{m}, 2\tilde{m}\}+ C(\bar{m},M)C_{0}^{\frac{1}{16}} \leq \bar{m}+C(\bar{m},M)C_{0}^{\frac{1}{16}}\leq\frac{3}{2} \bar{m}, \end{aligned}$$
(2.62)
provided that
$$ C_{0}\leq\min\{\varepsilon_{1},\varepsilon_{2}\},\quad \mbox{where }\varepsilon_{2}= \biggl(\frac{\bar{m}}{2C(\bar{m},M)} \biggr)^{16}. $$
When \(t\in[\sigma(T), T]\), by using Lemma 2.1, Proposition 2.2, (2.14), (2.19), and (2.58), we have
$$\begin{aligned} \bigl|b(t_{2})-b(t_{1})\bigr| \leq& C(\bar{m})\int_{t_{1}}^{t_{2}} \bigl\| F(\cdot,t)\bigr\| _{L^{\infty}}\,dt \\ \leq& \frac{a_{0}C^{0}}{2\mu+\lambda}(t_{2}-t_{1})+C(\bar{m})\int _{t_{2}}^{t_{1}}\bigl\| F(\cdot,t)\bigr\| _{L^{\infty}}^{\frac{8}{3}} \,dt \\ \leq&\frac{a_{0}C^{0}}{2\mu+\lambda}(t_{2}-t_{1}) +C(\bar{m})\int _{t_{2}}^{t_{1}}\bigl\| F(\cdot,t)\bigr\| ^{2/3}_{L^{2}} \bigl\| \nabla F(\cdot,t)\bigr\| ^{2}_{L^{6}}\,dt \\ \leq&\frac{a_{0}C^{0}}{2\mu+\lambda}(t_{2}-t_{1}) +C(\bar{m})\int _{t_{2}}^{t_{1}}\bigl(\|\nabla\dot{u}\|_{L^{2}}^{\frac {2}{3}}+ \bigl\| P(m,n)-P(\tilde{m},\tilde{n})\bigr\| _{L^{2}}^{\frac{2}{3}}\bigr) \\ &{}\times\bigl(\|m\dot{u}\|_{L^{6}}^{2}+\|\nabla{u} \|_{L^{2}}^{2}+\bigl\| P(m,n)-P(\tilde {m},\tilde{n}) \bigr\| _{L^{6}}^{2}\bigr)\,dt \\ \leq&\frac{a_{0}C^{0}}{2\mu+\lambda}(t_{2}-t_{1}) +C( \bar{m})C_{0}^{\frac{1}{6}}\int_{\sigma(T)}^{T} \|\nabla\dot{u}\|_{L^{2}}^{2}\,dt\\ &{} +C(\bar{m})C_{0}^{\frac{1}{6}}+C( \bar{m})C_{0}^{\frac {1}{2}}(t_{2}-t_{1})+C_{0}^{\frac{1}{6}} \int_{\sigma(T)}^{T}\|m\dot{u}\| _{L^{2}}^{2}\, dt \\ \leq&\biggl(\frac{a_{0}C^{0}}{2\mu+\lambda}+C(\bar{m})C_{0}^{\frac {1}{2}}\biggr) (t_{2}-t_{1})+C(\bar{m})C_{0}^{\frac{2}{3}} \\ \leq&\frac{2a_{0}C^{0}}{2\mu+\lambda}(t_{2}-t_{1})+C( \bar{m})C_{0}^{\frac{2}{3}}, \end{aligned}$$
provided that \(C_{0}\leq\{\varepsilon_{1},\varepsilon_{2},\varepsilon_{3}\}\), where \(\varepsilon_{3}= (\frac{a_{0}C^{0}}{C(\bar{m})(2\mu+\lambda )} )^{2}\). Therefore, for \(t\in[\sigma(T), T]\), choose \(N_{1}= \frac{2a_{0}C^{0}}{2\mu+\lambda}\) and \(N_{0}=C(\bar{m})C_{0}^{\frac{2}{3}}\) and \(\bar{\xi}=(2+\delta_{1})\tilde{m}\). From Remark 2.2, (1.9), and (1.14), we obtain
$$\begin{aligned} &z(\xi)\geq2a_{0}C^{0} (2+\delta_{1})\tilde{m} \bigl((2+\delta_{1})\underline{s}_{0} \tilde{m}-\tilde{n}\bigr) \geq2a_{0}C^{0}(2+\delta_{1}) \tilde{m}\tilde{n} \frac{\delta_{1}}{2}\geq2a_{0} C^{0}, \\ &\quad\mbox{for all }\xi\geq\bar{\xi}= (2+\delta_{1})\tilde{m}, \end{aligned}$$
where \(\delta_{1} \in(0, 1]\) is a small constant. Then Lemma 2.3 yields
$$\begin{aligned} \sup_{t\in[\sigma(T),T]}\|m\|_{L^{\infty}} \leq& \max \biggl\{ \frac{3}{2}\bar{m}, (2+\delta_{1})\tilde{m}\biggr\} + C \bigl(\bar{m}, \tilde{m},\tilde{n}, C^{0}, a_{0}, \underline{s}_{0}, \mu, \lambda\bigr) C_{0}^{\frac{2}{3}} \\ \leq&\frac{3}{2}\bar{m}+C(\bar{m})C_{0}^{\frac{2}{3}}\leq \frac{7}{4}\bar{m}, \end{aligned}$$
(2.63)
provided that
$$ C_{0}\leq\varepsilon \triangleq\min\{\varepsilon_{1}, \varepsilon_{2},\varepsilon _{3},\varepsilon _{4}\}, \quad\mbox{where }\varepsilon_{4}= \biggl(\frac{\bar{m}}{4C(\bar{m})} \biggr)^{\frac {3}{2}}. $$
Then (2.62) and (2.63) complete the proof of Proposition 2.3. □
From now on, we will assume that the initial energy \(C_{0}<\varepsilon\) and the constant C may depend on T, \(\|\sqrt{m}g\|_{L^{2}}\), \(\|\nabla g\|_{L^{2}}\), \(\|(m_{0}-\tilde{m},n_{0}-\tilde{n})\|_{H^{3}}\), \(\|u_{0}\|_{D^{1}\cap D^{3}}\), besides μ, λ, \(\tilde{m}\), \(\tilde{n}\), \(\bar{m}\), \(a_{0}\), \(C^{0}\), \(\underline{s}_{0}\), and M, where g is the same as in (1.12).
Finally, we give the proof of the high-order regularity estimates of \((m,n,u)\), which is due to [22–24, 30] for the single-fluid Navier-Stokes equations.
Proposition 2.4
([20, 22])
If
\((m,n,u)\)
is a classical solution of (1.1)-(1.3) in
\([0,T]\), we have the following estimates:
$$\begin{aligned}& \sup_{0< t\leq T}\int m|\dot{u}|^{2}\,dx+ \int_{0}^{T}\int|\nabla\dot {u}|^{2}\,dx \,dt\leq C, \end{aligned}$$
(2.64)
$$\begin{aligned}& \sup_{0< t\leq T}\bigl(\|\nabla m\|_{L^{2}\cap L^{6}}+\| \nabla n\|_{L^{2}\cap L^{6}}+\|\nabla u\|_{H^{1}}\bigr)+\int_{0}^{T} \|\nabla u\|_{L^{\infty}}\,dt\leq C. \end{aligned}$$
(2.65)
Proof
Applying the operator \(\dot{u}(\frac{\partial}{\partial t}+\operatorname{div}(u\cdot))\) to (1.1)3 and integrating the resulting equality over \([0,T]\), we have
$$\begin{aligned} \biggl(\frac{1}{2}\int m|\dot{u}|^{2}\,dx \biggr)_{t} =&- \int\dot{u}^{j}\bigl[\partial_{j}P_{t}+ \operatorname{div}(\partial _{j}Pu)\bigr]\,dx+\int\mu\dot{u}^{j} \bigl(\Delta u_{t}^{j}+\operatorname{div}\bigl(u\Delta u^{j}\bigr) \bigr)\,dx \\ &{}+\int(\lambda+\mu)\dot{u}^{j} \bigl(\partial_{j} \partial_{t}(\operatorname{div}u)+\operatorname{div}\bigl(u\partial_{j}( \operatorname{div}u)\bigr) \bigr)\,dx. \end{aligned}$$
Using integration by parts, (1.1)1, (1.1)2, and Young’s inequality, we have
$$\begin{aligned} & \biggl(\int m|\dot{u}|^{2}\,dx \biggr)_{t}+\int|\nabla \dot{u}|^{2}\,dx \\ &\quad\leq C\bigl(\|\nabla{u}\|_{L^{4}}^{4}+\|\nabla{u} \|_{L^{2}}^{2}+\|u\| _{L^{4}}^{4}\bigr)+C\int \bigl(|u||\nabla{u}||\dot{u}|+|\nabla{u}|^{2}|\dot {u}|\bigr)\,dx \\ &\quad\leq C\|\nabla u\|_{L^{2}}\|\nabla u\|_{L^{6}}^{3}+C\| \nabla{u}\| _{L^{2}}^{3}+C\int\bigl(|u||\nabla{u}||\dot{u}|+| \nabla{u}|^{2}|\dot {u}|\bigr)\,dx+C \\ &\quad\leq C \bigl(\|F\|_{L^{6}}^{3}+\|\omega\|_{L^{6}}^{3}+ \bigl\| P(m,n)-P(\tilde {m},\tilde{n})\bigr\| _{L^{6}}^{3} \bigr)+\delta\| \nabla\dot{u}\|_{L^{2}}^{2} +C\|\nabla{u}\|_{L^{3}}^{3}+C \\ &\quad\leq C \bigl(\|\nabla{F}\|_{L^{2}}^{3}+\|\nabla\omega \|_{L^{2}}^{3} \bigr)+\frac {1}{2}\|\nabla\dot{u} \|_{L^{2}}^{2} +C\|\nabla{u}\|_{L^{3}}^{3}+C \\ &\quad\leq C\|\sqrt{m}\dot{u}\|_{L^{2}}^{4}+C\|\nabla{u} \|_{L^{3}}^{3}+\frac{1}{2}\| \nabla\dot{u} \|_{L^{2}}^{2}+C, \end{aligned}$$
using (2.22) and
$$\begin{aligned} \int_{0}^{T}\|\nabla{u}\|_{L^{3}}^{3} \,ds \leq&\int_{0}^{T\wedge T_{1}}\|\nabla {u}\| _{L^{3}}^{3}\,ds+\int_{T\wedge T_{1}}^{T} \sigma^{3}\|\nabla{u}\| _{L^{3}}^{3}\,ds\\ \leq&C+C\int_{0}^{T\wedge T_{1}}\|\nabla{u} \|_{L^{2}}^{\frac{3}{2}}\bigl(\| m\dot {u}\|_{L^{2}}^{\frac{3}{2}}+ \bigl\| P(m,n)-P(\tilde{m},\tilde{n})\bigr\| _{L^{2}}^{\frac{3}{2}}\\ &{}\times \bigl\| P(m,n)-P( \tilde{m},\tilde{n})\bigr\| _{L^{6}}^{\frac{3}{2}}+\|\nabla{u}\| _{L^{2}}^{\frac{3}{2}}\bigr)\,ds\\ \leq&C \end{aligned}$$
and the compatibility condition, we can define
$$ \sqrt{m}\dot{u} |_{t=0}=-\sqrt{m_{0}}g. $$
Then Gronwall’s inequality gives (2.64).
Next, we prove (2.65). For \(p\in[2,6]\), differentiating (1.1)1 with respect to \(x_{i}\), and then multiplying both sides of the result equation by \(p|\partial_{i}m|^{p-2}\partial_{i}m\), we get
$$\begin{aligned} &\bigl(|\nabla m|^{p}\bigr)_{t}+ \operatorname{div}\bigl(|\nabla m|^{p}u\bigr)+(p-1)|\nabla m|^{p} \operatorname{div}u \\ &\quad{}+p|\nabla m|^{p-2}(\nabla m)^{T}\nabla u(\nabla m)+pm| \nabla m|^{p-2} \nabla m\cdot\nabla\operatorname{div}u=0. \end{aligned}$$
(2.66)
Similarly, we can obtain
$$ \begin{aligned}[b] &\bigl(|\nabla n|^{p}\bigr)_{t}+ \operatorname{div}\bigl(|\nabla n|^{p}u\bigr)+(p-1)|\nabla n|^{p} \operatorname{div}u\\ &\quad{}+p|\nabla n|^{p-2}(\nabla n)^{T}\nabla u(\nabla n)+pn| \nabla n|^{p-2} \nabla n\cdot\nabla\operatorname{div}u=0. \end{aligned} $$
(2.67)
By the standard \(L^{p}\)-estimate for an elliptic system,
$$\begin{aligned}& -\mu\Delta u-(\lambda+\mu)\nabla\operatorname{div}u=m\dot{u}+\nabla P, \quad \mbox{in }\Omega, \\& (u_{1},u_{2},u_{3})=\beta \bigl(u_{x_{3}}^{1},u_{x_{3}}^{2},0\bigr) \quad \mbox{in }\partial \Omega, \end{aligned}$$
(2.68)
we obtain
$$ \bigl\| \nabla^{2}u\bigr\| _{L^{p}}\leq C\bigl(\|m\dot{u} \|_{L^{p}}+\|\nabla P\|_{L^{p}}\bigr). $$
(2.69)
Now, we give the estimate for \(\|\nabla{u}\|_{L^{\infty}}\), which is crucial to obtain the estimate of \(\|(\nabla{m},\nabla{n})\|_{L^{p}}\). As in [22], we set \(w=u-v\); w and v satisfy
$$ \left \{ \textstyle\begin{array}{@{}l} -\mu\Delta{v}-(\mu+\lambda)\nabla\operatorname{div}v=-\nabla (P(m,n)-P(\tilde {m},\tilde{n})), \quad \mbox{in } \Omega,\\ (v^{1}(x),v^{2}(x),v^{3}(x))=\beta(v_{x_{3}}^{1}(x),v_{x_{3}}^{2}(x),0)\quad \mbox{on }\partial\Omega, t>0, \end{array}\displaystyle \right . $$
then, by the standard regularity estimate for elliptic systems, we have
$$ \begin{aligned} &\|\nabla{v}\|_{L^{q}}\leq C\bigl(\bigl\| P(m,n)-P(\tilde{m}, \tilde{n})\bigr\| _{L^{q}}\bigr), \\ & \bigl\| \nabla^{2}v\bigr\| _{L^{q}} \leq C\bigl\| \nabla\bigl(P(m,n)-P(\tilde{m},\tilde {n})\bigr)\bigr\| _{L^{q}}, \quad \mbox{for }q\in[2,\infty) \end{aligned} $$
(2.70)
and w satisfies
$$ \left \{ \textstyle\begin{array}{@{}l} -\mu\Delta{w}-(\mu+\lambda)\nabla\operatorname{div}w=m\dot{u}, \quad \mbox{in }\Omega,\\ (w^{1}(x),w^{2}(x),w^{3}(x))=\beta(w_{x_{3}}^{1}(x),w_{x_{3}}^{2}(x),0) \quad\mbox{on }\partial\Omega, t>0. \end{array}\displaystyle \right . $$
Similarly,
$$ \bigl\| \nabla^{2}w\bigr\| _{L^{q}}\leq C\|m\dot{u} \|_{L^{q}},\qquad \|\nabla{w}\| _{L^{\infty}}\leq\bigl(\|m{\cdot{u}} \|_{L^{2}}+\|m{\dot{u}}\|_{L^{6}}\bigr) \quad\mbox{for }q\in(1,\infty). $$
(2.71)
In order to obtain \(\|\nabla{v}\|_{L^{\infty}}\), we give the following fact.
Remark 2.3
Let \(\Omega=\{x\in\mathbb{R}^{3}\mid x_{3}>0\}\) and \(\nabla{v}\in W^{1,q}(\Omega )\) with \(q\in(3,\infty)\). There exists a constant C depending only on q such that
$$ \|\nabla{v}\|_{L^{\infty}}\leq C\bigl(1+\ln\bigl(e+\bigl\| \nabla^{2}v\bigr\| _{L^{q}}\bigr)\bigr)\| \nabla {v} \|_{BMO},\quad \mbox{with }q\in(3,\infty); $$
(2.72)
here
$$\begin{aligned}& \|\nabla{v}\|_{BMO}=\|\nabla{v}\|_{L^{2}}+[ \nabla{v}]_{BMO}, \qquad [\nabla{v}]_{BMO}=\sup _{r>0,x\in\Omega}\frac{1}{\Omega_{r}(x)}\int_{\Omega _{r}(x)}\bigl| \nabla{v}(y)-\nabla{v_{r}}(x)\bigr|\,dy,\\& \nabla{v_{r}}(x)=\frac{1}{|\Omega_{r}(x)|}\int_{\Omega_{r}(x)} \nabla{v}(y)\,dy. \end{aligned}$$
Then, by using the classical theory for elliptic systems, we have
$$ \|\nabla{v}\|_{BMO}\leq C\bigl(\bigl\| P(m,n)-P(\tilde{m},\tilde{n}) \bigr\| _{L^{2}}+\bigl\| P(m,n)-P(\tilde{m},\tilde{n})\bigr\| _{L^{\infty}}\bigr)\leq C( \bar{m}), $$
which together with (2.72) implies
$$ \|\nabla{v}\|_{L^{\infty}}\leq C\bigl(1+\ln\bigl(e+\bigl\| \nabla^{2}v\bigr\| _{L^{q}}\bigr)\bigr). $$
(2.73)
From (2.66) and (2.67) we have
$$\begin{aligned} \partial_{t}\bigl(\|\nabla m\|_{L^{p}}+\|\nabla n \|_{L^{p}}\bigr) \leq&C\|\nabla u\| _{L^{\infty}} \bigl(\|\nabla m \|_{L^{p}}+\|\nabla n\|_{L^{p}}\bigr)+C\bigl\| \nabla^{2}u \bigr\| _{L^{p}} \\ \leq&C\bigl(1+\|\nabla u\|_{L^{\infty}}\bigr) \bigl(\|\nabla m\|_{L^{p}}+\|\nabla n \|_{L^{p}}\bigr)+C\|m\dot{u}\|_{L^{p}}. \end{aligned}$$
(2.74)
Then, using Remark 2.3 and (2.70), we obtain
$$\begin{aligned} \|\nabla u\|_{L^{\infty}} \leq&C\bigl(\|\nabla{w} \|_{L^{\infty}}+\|\nabla {v}\| _{L^{\infty}}\bigr) \\ \leq&C\|\nabla{w}\|_{L^{\infty}}+\bigl(1+\ln\bigl(e+\bigl\| \nabla^{2}v \bigr\| _{L^{q}}\bigr)\bigr) \\ \leq&C\bigl(1+\|m\dot{u}\|_{L^{6}}+\ln\bigl(e+\|\nabla{m}\|_{q}+ \|\nabla{n}\|_{L^{q}}\bigr)\bigr). \end{aligned}$$
(2.75)
We set
$$ f(t)=e+\|\nabla m\|_{L^{6}}+\|\nabla n\|_{L^{6}}, \qquad g(t)=1+\|m \dot{u}\|_{L^{6}}. $$
Substituting (2.74) into (2.75) with \(p=6\), we have
$$ f'(t)\leq Cg(t)f(t)+Cf(t)\ln f(t)+Cg(t), $$
which yields
$$ \bigl(\ln f(t)\bigr)'\leq Cg(t)+C\ln f(t). $$
(2.76)
Then Lemma 2.4, (2.60), and (2.64) imply
$$\begin{aligned} \int_{0}^{T}g(t)\,dt\leq C\int _{0}^{T}\bigl(1+\|m\dot{u}\|_{L^{6}}\bigr)\,dt\leq C \int_{0}^{T}\bigl(1+\| \nabla\dot{u}\|_{L^{2}}\bigr) \,dt\leq C, \end{aligned}$$
(2.77)
and we get by using Gronwall’s inequality and (2.76)
$$ \sup_{0\leq t\leq T}f(t)\leq C, $$
i.e.,
$$ \sup_{0\leq t\leq T}\bigl(\|\nabla m\|_{L^{6}}+\| \nabla n\|_{L^{6}}\bigr)\leq C. $$
(2.78)
From (2.75), (2.77), and (2.78), we obtain
$$ \int_{0}^{T}\|\nabla u \|_{L^{\infty}}\,dt\leq C. $$
(2.79)
Setting \(p=2\) in (2.74), we have
$$ \sup_{0\leq t\leq T}\bigl(\|\nabla m\|_{L^{2}}+\| \nabla n\|_{L^{2}}\bigr)\leq C. $$
(2.80)
This completes the proof of Proposition 2.4. □
Corollary 2.2
If
\((m,n,u)\)
is a classical solution of (1.1)-(1.3) in
\([0,T]\), we have the following estimates:
$$ \sup_{0\leq t\leq T}\int m|u_{t}|^{2} \,dx+\int_{0}^{T}\int|\nabla u_{t}|^{2} \,dx\,ds\leq C. $$
(2.81)
In order to get the higher-order regularity estimates of the solution, we need to bound \(P_{mm}\), \(P_{mn}\), \(P_{nn}\), \(P_{mmm}\), \(P_{mmn}\), \(P_{mnn}\), and \(P_{nnn}\).
Lemma 2.7
Under the conditions of Theorem
1.1, we have
$$\begin{aligned}& |P_{mm}|\leq C,\qquad |P_{mn}|\leq C,\qquad |P_{nn}|\leq C, \end{aligned}$$
(2.82)
$$\begin{aligned}& \begin{aligned} &|P_{mmm}|\leq C,\qquad |P_{mmn}|\leq C,\qquad |P_{nnm}|\leq C, \\ & |P_{nnn}|\leq C, \quad\textit{in }\mathbb{R}^{3} \times[0,T]. \end{aligned} \end{aligned}$$
(2.83)
Proof
Since
$$\begin{aligned}& P_{mm}=C^{0}\frac{c}{(b^{2}+c)^{\frac{3}{2}}},\qquad P_{nn}=-C^{0} \frac{4k_{0}a_{0}^{2} m}{(b^{2}+c)^{\frac{3}{2}}},\qquad P_{mn}=C^{0}2k_{0}a_{0} \frac{(k_{0}-m)+a_{0}n}{(b^{2}+c)^{\frac{3}{2}}}, \\& P_{mmn}=\frac{4k_{0}a_{0}C^{0}}{(b^{2}+c)^{\frac{3}{2}}}-\frac {12C^{0}k_{0}a_{0}n(k_{0}a_{0}+a_{0}m+a_{0}^{2}n)}{(b^{2}+c)^{\frac{5}{2}}},\qquad P_{mmm}= \frac{12C^{0}k_{0}a_{0}nb}{(b^{2}+c)^{\frac{5}{2}}}, \\& P_{nnn}=\frac{12C^{0}k_{0}a_{0}^{2}m(k_{0}a_{0}+a_{0}m+a_{0}^{2}n)}{(b^{2}+c)^{\frac {5}{2}}},\\& P_{mnn}=\frac{2k_{0}a_{0}^{2}C^{0}}{(b^{2}+c)^{\frac{3}{2}}}- \frac {6C^{0}k_{0}a_{0}^{2}(k_{0}^{2}-m^{2}+2a_{0}k_{0}n+a_{0}^{2}n)}{(b^{2}+c)^{\frac{5}{2}}}, \end{aligned}$$
by (2.7) and (2.60), we can obtain (2.82) and (2.83) easily. □
Proposition 2.5
If
\((m,n,u)\)
is a classical solution of (1.1)-(1.3) in
\([0,T]\), we have the following estimates:
$$ \sup_{0\leq t\leq T}\bigl(\|m-\tilde{m} \|_{H^{2}}+\|n-\tilde{n}\|_{H^{2}}+\bigl\| P(m, n)-P(\tilde{m}, \tilde{n})\bigr\| _{H^{2}}\bigr)\leq C. $$
(2.84)
Proof
At first, we give the elliptic estimate as follows:
$$ \|\nabla u\|_{H^{2}}\leq C\bigl(\|F\|_{H^{2}}+\| \omega\| _{H^{2}}+\bigl\| P(m, n)-P(\tilde{m},\tilde{n})\bigr\| _{H^{2}} \bigr). $$
(2.85)
Then from (1.1)1, (1.1)2, and the above estimate, we have
$$ \begin{aligned}[b] &\frac{d}{dt} \bigl(\bigl\| \nabla^{2} m\bigr\| _{L^{2}}^{2}+ \bigl\| \nabla^{2} n\bigr\| _{L^{2}}^{2} \bigr)\\ &\quad\leq C\bigl(1+\|\nabla u\|_{L^{\infty}}\bigr) \bigl(\bigl\| \nabla^{2} m \bigr\| _{L^{2}}^{2}+\bigl\| \nabla^{2} n\bigr\| _{L^{2}}^{2} \bigr)+C\|\nabla{u}\|_{H^{2}}^{2}+C. \end{aligned} $$
(2.86)
Using Proposition 2.4 and the same idea as the proof in [22] (Lemma 4.3), we have
$$\begin{aligned} &\|F\|_{H^{2}}+\|\omega\|_{H^{2}}+\bigl\| P(m, n)-P(\tilde{m}, \tilde{n})\bigr\| _{H^{2}}\\ &\quad\leq C\bigl(\|F\|_{H^{1}}+\|\omega\| _{H^{1}}+\| m\dot{u}\|_{H^{1}}+\bigl\| P(m, n)-P(\tilde{m},\tilde{n}) \bigr\| _{H^{1}}\bigr)\\ &\qquad{}+C\bigl(\bigl\| \nabla^{2}m\bigr\| _{L^{2}}\bigl\| \nabla^{2}n \bigr\| _{L^{2}}\bigr)\\ &\quad\leq C\bigl(1+\|m\dot{u}\|_{L^{2}}+\bigl\| \nabla(m\dot{u})\bigr\| _{L^{2}} \bigr)+C\bigl(\bigl\| \nabla ^{2}m\bigr\| _{L^{2}}+\bigl\| \nabla^{2}n \bigr\| _{L^{2}}\bigr) \\ &\quad\leq C\bigl(1+\|\nabla\dot{u}\|_{L^{2}}+\|\nabla{m}\|_{L^{3}}\| \dot{u}\| _{L^{6}}\bigr)+C\bigl(\bigl\| \nabla^{2}m\bigr\| _{L^{2}}+\bigl\| \nabla^{2}n\bigr\| _{L^{2}}\bigr), \end{aligned}$$
which together with Proposition 2.4 and Gronwall’s inequality yields
$$ \sup_{0\leq t\leq T}\int \bigl(\bigl\| \nabla^{2} m \bigr\| _{L^{2}}+\bigl\| \nabla^{2} n\bigr\| _{L^{2}} \bigr)\,dx\leq C. $$
From the lemma, we have \(\sup_{0\leq{t}\leq{T}}\|P(m,n)-P(\tilde {m},\tilde{n})\|_{H^{2}}\leq C\). Thus we finish the proof of Proposition 2.5. □
Proposition 2.6
If
\((m,n,u)\)
is a classical solution of (1.1)-(1.3) in
\([0,T]\), we have the following estimates:
$$ \sup_{0\leq t\leq T}\bigl(\|m_{t} \|_{H^{1}}+\|n_{t}\|_{H^{1}}+\|P_{t} \|_{H^{1}}\bigr)+\int_{0}^{T}\bigl( \|m_{tt}\| _{L^{2}}^{2}+\|n_{tt} \|_{L^{2}}^{2}+\|P_{tt}\|_{L^{2}}^{2} \bigr)\,dt\leq C. $$
(2.87)
Proof
From (1.1)1, (2.1), (2.60) and (2.65), we get
$$ \|m_{t}\|_{L^{2}}\leq C\|u\|_{L^{\infty}}\| \nabla{m}\|_{L^{2}}+C\|\nabla {u}\| _{L^{2}}\leq C. $$
(2.88)
Similarly, we have
$$ \|n_{t}\|_{L^{2}}\leq C\|u\|_{L^{\infty}}\| \nabla{n}\|_{L^{2}}+C\|\nabla {u}\| _{L^{2}}\leq C. $$
(2.89)
Applying the ∇ operator to (1.1)1 yields
$$ \partial_{j}m_{t}+\partial_{j}u^{i} \partial_{i}m+u^{i}\partial_{i}\partial _{j}m+\partial_{j}m\operatorname{div}u+m\partial_{j} \operatorname{div}u=0. $$
(2.90)
By (2.1), (2.60), (2.65), and (2.84), we can obtain
$$\begin{aligned} \|\nabla{m_{t}}\|_{L^{2}} \leq&C\|\nabla{u} \|_{L^{3}}\|\nabla{m}\| _{L^{6}}+C\| u\|_{L^{\infty}}\bigl\| \nabla^{2}m\bigr\| _{L^{2}}+\bigl\| \nabla^{2}u\bigr\| _{L^{2}} \\ \leq&C\|\nabla{u}\|_{L^{2}}^{\frac{1}{2}}\bigl\| \nabla^{2}{u} \bigr\| _{L^{2}}^{\frac {1}{2}}\bigl\| \nabla^{2}m\bigr\| _{L^{2}}+C \|u\|_{L^{2}}^{\frac{1}{4}}\bigl\| \nabla^{2}u\bigr\| _{L^{2}}^{\frac{3}{4}} +C\bigl\| \nabla^{2}u\bigr\| _{L^{2}} \leq C. \end{aligned}$$
(2.91)
Similarly, we have
$$ \|\nabla{n_{t}}\|_{L^{2}}\leq C. $$
(2.92)
Next, differentiating (1.1)1 with respect to t yields
$$ m_{tt}+u_{t}\cdot\nabla{m}+u\cdot \nabla{m_{t}}+m_{t}\operatorname{div}u+m\operatorname{div}u_{t}=0. $$
(2.93)
Thus, we get from Lemma (2.1), (2.65), (2.81), (2.84), (2.92), and (2.93)
$$\begin{aligned} \int_{0}^{T}\|m_{tt} \|_{L^{2}}^{2}\,dt \leq&C\int_{0}^{T} \bigl(\|u_{t}\|_{L^{6}}^{2}\| \nabla {m} \|_{L^{3}}^{2}+\|\nabla{m_{t}}\|_{L^{2}}^{2} +\|m_{t}\|_{L^{6}}^{2}\|\nabla{u} \|_{L^{3}}^{2}+ \|\nabla{u_{t}}\|_{L^{2}}^{2} \bigr)\,dt \\ \leq&C. \end{aligned}$$
(2.94)
Similarly, we have
$$ \int_{0}^{T}\| \nabla{n_{tt}}\|_{L^{2}}^{2}\,dt\leq C. $$
(2.95)
The estimates for \(P_{t}\) and \(P_{tt}\) can be dealt with in a similar way. □
Corollary 2.3
[20, 22]
If
\((m,n,u)\)
is a classical solution of (1.1)-(1.3) in
\([0,T]\), we have the following estimates:
$$ \sup_{0\leq t\leq T}\int|\nabla{u_{t}}|^{2} \,dx+\int_{0}^{T}\int {m}u_{tt}^{2} \,dx\,dt\leq C. $$
(2.96)
Proposition 2.7
If
\((m,n,u)\)
is a classical solution of (1.1)-(1.3) in
\([0,T]\), we have the following estimates:
$$\begin{aligned}& \sup_{0\leq t\leq T}\bigl(\|m-\tilde{m}\|_{H^{3}}+ \|n-\tilde{n}\|_{H^{3}}+\bigl\| P(m,n)-P(\tilde{m},\tilde{n})\bigr\| _{H^{3}}\bigr) \leq C, \end{aligned}$$
(2.97)
$$\begin{aligned}& \sup_{0\leq t\leq T}\|\nabla u\|_{H^{2}}+\int _{0}^{T}\bigl(\|\nabla u\|_{H^{3}}^{2} +\|\nabla u_{t}\|_{H^{1}}^{2}\bigr)\,dt\leq C. \end{aligned}$$
(2.98)
Proof
By using the Hölder inequality, Young’s inequality, Lemma 2.1, (2.60), (2.65), (2.84), and (2.87), we have
$$\begin{aligned} \bigl\| \nabla(m\dot{u})\bigr\| _{L^{2}} =&\bigl\| \nabla(mu_{t}+mu \cdot\nabla{u})\bigr\| _{L^{2}}\leq C \bigl\| |\nabla{m}||{u_{t}}| \bigr\| _{L^{2}}+C\|m\nabla{u_{t}}\| _{L^{2}} \\ &{}+C \bigl\| |\nabla{m}||u||\nabla{u}| \bigr\| _{L^{2}}+C \bigl\| m|\nabla {u}|^{2} \bigr\| _{L^{2}}+C \bigl\| m|u||\nabla^{2}{u}| \bigr\| _{L^{2}} \\ \leq&C\|\nabla{m}\|_{L^{3}}\|u_{t}\|_{L^{6}}+C\| \nabla{u_{t}}\|_{L^{2}}+C\|u\| _{L^{\infty}}\|\nabla{m} \|_{L^{3}}\|\nabla{u}\|_{L^{6}} \\ &{}+C\|\nabla{u}\|_{L^{3}}\|\nabla{u}\|_{L^{6}}+C\|u \|_{L^{\infty}}\bigl\| \nabla ^{2}u\bigr\| _{L^{2}} \\ \leq&C, \end{aligned}$$
(2.99)
which together with (2.64) imply
$$ \sup_{0\leq t\leq T}\|m\dot{u}\|_{H^{1}}\leq C. $$
(2.100)
The standard \(H^{1}\)-estimate for the elliptic equations gives
$$\begin{aligned} \bigl\| \nabla^{2}{u}\bigr\| _{H^{1}} \leq&C\bigl\| \mu \Delta{u}+(\mu+\lambda)\nabla \operatorname{div}u\bigr\| _{H^{1}} \\ \leq&C\|m\dot{u}+\nabla{P}\|_{H^{1}} \\ \leq&C\|m\dot{u}\|_{H^{1}}+C\|\nabla{P}\|_{H^{1}}\leq C, \end{aligned}$$
(2.101)
where we have used (1.1)3, Lemma 2.1, (2.60), (2.65), (2.84), and (2.87) give
$$ \sup_{0< t\leq T}\|\nabla{u}\|_{H^{2}}\leq C. $$
(2.102)
Next, by using the standard \(L^{2}\)-estimate for the elliptic system, we can obtain
$$\begin{aligned} \bigl\| \nabla^{2}u_{t}\bigr\| _{L^{2}} \leq&C\bigl\| \mu\Delta{u_{t}}+(\mu+\lambda)\nabla \operatorname{div}u_{t} \bigr\| _{L^{2}} \\ =&C\|mu_{tt}+m_{t}u_{t}+m_{t}u\cdot \nabla{u}+mu_{t}\cdot\nabla{u}+mu\cdot \nabla {u_{t}}+ \nabla{P_{t}}\|_{L^{2}} \\ \leq&C\|mu_{tt}\|_{L^{2}}+C\|m_{t} \|_{L^{3}}\|u_{t}\|_{L^{6}}+C\|u\|_{L^{\infty }} \|m_{t}\|_{L^{3}}\|\nabla{u}\|_{L^{6}} \\ &{}+\|u_{t}\|_{L^{6}}\|\nabla{u}\|_{L^{3}}+C\|u \|_{L^{\infty}}\|\nabla{u_{t}}\| _{L^{2}}+C\| \nabla{P_{t}}\|_{L^{2}} \\ \leq&C\|mu_{tt}\|_{L^{2}}+C, \end{aligned}$$
(2.103)
which together with (2.96) imply
$$ \int_{0}^{T}\| \nabla{u_{t}}\|_{H^{1}}^{2}\,dt\leq C. $$
(2.104)
In order to estimate \(\|\nabla^{2}u\|_{H^{2}}\), we use the standard \(H^{2}\)-estimate of the elliptic equations again to get
$$\begin{aligned} \bigl\| \nabla^{2}u\bigr\| _{H^{2}} \leq&C\bigl\| \mu\Delta{u}+( \mu+\lambda)\nabla \operatorname{div}u\bigr\| _{H^{2}} \\ \leq&C\|m\dot{u}\|_{H^{2}}+C\|\nabla{P}\|_{H^{2}} \\ \leq&C+C\|\nabla{u_{t}}\|_{H^{1}}+C\bigl(\bigl\| \nabla^{3}m\bigr\| _{L^{2}}+\bigl\| \nabla^{3}n\bigr\| _{L^{2}} \bigr), \end{aligned}$$
(2.105)
here we also have used (1.1)3, Lemma 2.1, Lemma 2.4, (2.101), and the following simple facts:
$$\begin{aligned} \bigl\| \nabla^{2}(mu_{t})\bigr\| _{L^{2}} \leq&C \bigl\| \bigl|\nabla^{2}m\bigr||u_{t}| \bigr\| _{L^{2}} +C \bigl\| | \nabla{m}||\nabla{u_{t}}| \bigr\| _{L^{2}}+C\bigl\| \nabla^{2}u_{t} \bigr\| _{L^{2}} \\ \leq&C\bigl\| \nabla^{2}m\bigr\| _{L^{2}}\|\nabla{u_{t}} \|_{L^{2}}+C\|\nabla{m}\| _{L^{3}}\| \nabla{u_{t}} \|_{L^{6}}+C\bigl\| \nabla^{2}u\bigr\| _{L^{2}} \\ \leq&C+C\|\nabla{u_{t}}\|_{L^{2}} \end{aligned}$$
(2.106)
and
$$\begin{aligned} \bigl\| \nabla^{2}(mu\cdot\nabla{u})\bigr\| _{L^{2}} \leq&C \bigl\| \bigl|\nabla ^{2}(mu)\bigr||\nabla {u}| \bigr\| _{L^{2}} +C \bigl\| \bigl| \nabla(mu)\bigr|\bigl|\nabla^{2}{u}\bigr| \bigr\| _{L^{2}}+C\bigl\| \nabla^{3}u \bigr\| _{L^{2}} \\ \leq&C+C\bigl\| \nabla^{2}(mu)\bigr\| _{L^{2}}\|\nabla{u} \|_{H^{2}}+C\bigl\| \nabla(mu)\bigr\| _{L^{3}}\bigl\| \nabla^{2}{u} \bigr\| _{L^{6}} \\ \leq&C+C\bigl\| \nabla^{2}{m}\bigr\| _{L^{2}}\|u\|_{L^{\infty}}+C\| \nabla{m}\| _{L^{6}}\| \nabla{u}\|_{L^{3}}+C\bigl\| \nabla^{2}u\bigr\| _{L^{2}} \\ \leq&C. \end{aligned}$$
(2.107)
Applying the \(\nabla^{3}\) operator to (1.1)1, multiplying with \(\nabla^{3}m\), and integrating the resulting equation over \(\mathbb{R}^{3}\times[0,T]\), we obtain
$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\int\bigl|\nabla^{3}m\bigr|^{2} \,dx =&-\int_{0}^{T}\int\nabla ^{3}\bigl( \operatorname{div}(mu)\bigr)\cdot\nabla^{3}m\,dx\,dt \\ \leq&C \bigl( \bigl\| \bigl|\nabla^{3}u\bigr||\nabla{m}| \bigr\| _{L^{2}}+ \bigl\| \bigl|\nabla ^{2}u\bigr|\bigl|\nabla^{2}m\bigr| \bigr\| _{L^{2}} + \bigl\| | \nabla{u}|\bigl|\nabla^{3}m\bigr| \bigr\| _{L^{2}} \\ &{}+\bigl\| \nabla^{4}u\bigr\| _{L^{2}} \bigr)\bigl\| \nabla^{3}m \bigr\| _{L^{2}} \\ \leq&C \bigl(\bigl\| \nabla^{3}u\bigr\| _{L^{2}}\|\nabla{m} \|_{H^{2}}+\bigl\| \nabla^{2}u\bigr\| _{L^{3}}\bigl\| \nabla^{2}{m}\bigr\| _{L^{6}} \\ &{}+\bigl\| \nabla^{3}m\bigr\| _{L^{2}} \|\nabla{u}\|_{L^{\infty}} \bigr)\bigl\| \nabla^{3}m\bigr\| _{L^{2}} \\ &{}+C \bigl(1+\bigl\| \nabla^{2}u_{t}\bigr\| _{L^{2}}+\bigl\| \nabla^{3}m\bigr\| _{L^{2}} \bigr)\bigl\| \nabla ^{3}m\bigr\| _{L^{2}} \\ \leq&C+C\bigl\| \nabla^{2}u_{t}\bigr\| _{L^{2}}^{2}+C \bigl\| \nabla^{3}m\bigr\| _{L^{2}}^{2}. \end{aligned}$$
(2.108)
Similarly, we have
$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\int\bigl|\nabla^{3}n\bigr|^{2} \,dx \leq C+C\bigl\| \nabla^{2}u_{t}\bigr\| _{L^{2}}^{2}+C \bigl\| \nabla^{3}n\bigr\| _{L^{2}}^{2}, \end{aligned}$$
(2.109)
where we have used (2.105). From (2.108)-(2.109), we have
$$ \frac{d}{dt}\bigl(\bigl\| \nabla^{3}m \bigr\| _{L^{2}}^{2}+\bigl\| \nabla^{3}n\bigr\| _{L^{2}}^{2} \bigr) \leq C+C\bigl\| \nabla^{2}u_{t}\bigr\| _{H^{2}}^{2}+C \bigl(\bigl\| \nabla^{3}m\bigr\| _{L^{2}}^{2}+\bigl\| \nabla^{3}n\bigr\| _{L^{2}}^{2}\bigr). $$
(2.110)
Then by using Gronwall’s inequality and (2.104), we can get
$$ \sup_{0\leq t\leq T}\bigl(\bigl\| \nabla^{3}m \bigr\| _{L^{2}}+\bigl\| \nabla^{3}n\bigr\| _{L^{2}}\bigr) \leq C. $$
(2.111)
Collecting the estimates (2.105)-(2.101), we obtain
$$ \int_{0}^{T}\|\nabla{u} \|_{H^{3}}^{2}\,dt\leq C. $$
(2.112)
This completes the proof of Proposition 2.7. □
Proposition 2.8
([22])
If
\((m,n,u)\)
is a classical solution of (1.1)-(1.3) in
\([0,T]\), we have the following estimates:
$$ \sup_{0\leq t\leq T}\sigma\bigl(\bigl\| \nabla^{2} u_{t}\bigr\| _{L^{2}}+\bigl\| \nabla ^{4} u\bigr\| _{L^{2}} \bigr)+\int_{0}^{T} \sigma^{2}\|\nabla u_{tt}\|_{L^{2}}^{2}\,dt\leq C. $$
(2.113)