Blow-up criteria for 3D nematic liquid crystal models in a bounded domain

Abstract

In this paper we prove some blow-up criteria for two 3D density-dependent nematic liquid crystal models in a bounded domain.

MSC:35Q30, 76D03, 76D09.

1 Introduction

Let $\mathrm{\Omega }\subseteq {\mathbb{R}}^3$ be a bounded domain with smooth boundary ∂ Ω, and let ν be the unit outward normal vector on ∂ Ω. We consider the regularity criterion to the density-dependent incompressible nematic liquid crystal model as follows [1–4]:

$$divu=0,$$

(1.1)

$${\partial }_t\rho +div(\rho u)=0,$$

(1.2)

$${\partial }_t(\rho u)+div(\rho u\otimes u)+\mathrm{\nabla }\pi -\mathrm{\Delta }u=-\mathrm{\nabla }\cdot (\mathrm{\nabla }d\odot \mathrm{\nabla }d),$$

(1.3)

$${\partial }_{t}d+u\cdot \mathrm{\nabla }d+({|d|}^2-1)d-\mathrm{\Delta }d=0,$$

(1.4)

in $(0,\mathrm{\infty })\times \mathrm{\Omega }$ with initial and boundary conditions

$$(\rho ,u,d)(\cdot ,0)=({\rho }_0,u_0,d_0)\text{in }\mathrm{\Omega },$$

(1.5)

$$u=0,{\partial }_{\nu }d=0\text{on }\partial \mathrm{\Omega },$$

(1.6)

where ρ denotes the density, u the velocity, π the pressure, and d represents the macroscopic molecular orientations, respectively. The symbol $\mathrm{\nabla }d\odot \mathrm{\nabla }d$ denotes a matrix whose $(i,j)$th entry is ${\partial }_{i}d{\partial }_{j}d$, and it is easy to find that $\mathrm{\nabla }d\odot \mathrm{\nabla }d=\mathrm{\nabla }{d}^T\mathrm{\nabla }d$.

When d is a given constant vector, then (1.1)-(1.3) represent the well-known density-dependent Navier-Stokes system, which has received many studies; see [5–7] and references therein.

When $\rho =1$, Guillén-González et al. [8] proved the blow-up criterion

$${\int }_0^T({\parallel u(t)\parallel }_{L^q}^{\frac{2q}{q-3}}+{\parallel \mathrm{\nabla }d(t)\parallel }_{L^q}^{\frac{2q}{q-3}})dt<\mathrm{\infty }\text{with }3<q\le \mathrm{\infty }$$

(1.7)

and $0<T<\mathrm{\infty }$.

It is easy to prove that the problem (1.1)-(1.6) has a unique local-in-time strong solution [6, 9], and thus we omit the details here. The aim of this paper is to consider the regularity criterion; we will prove the following theorem.

Theorem 1.1 Let ${\rho }_0\in W^{1,q}(\mathrm{\Omega })$, $u_0\in H_0^1(\mathrm{\Omega })\cap H^2(\mathrm{\Omega })$, $d_0\in H^3(\mathrm{\Omega })$ with $3<q\le 6$ and ${\rho }_0\ge 0$, $div{u}_0=0$ in Ω and ${\partial }_{\nu }{d}_0=0$ on ∂ Ω. We also assume that the following compatibility condition holds true: $\mathrm{\exists }(\mathrm{\nabla }{\pi }_0,g)\in L^2(\mathrm{\Omega })$ such that

$$\mathrm{\nabla }{\pi }_0-\mathrm{\Delta }{u}_0+\mathrm{\nabla }\cdot (\mathrm{\nabla }{d}_0\odot \mathrm{\nabla }{d}_0)=\sqrt{{\rho }_0}g\mathit{\text{in}}\mathrm{\Omega }.$$

Let $(\rho ,u,d)$ be a local strong solution to the problem (1.1)-(1.6). If u satisfies

$${\int }_0^T{\parallel u(t)\parallel }_{L^q}^{\frac{2q}{q-3}}dt<\mathrm{\infty }\mathit{\text{with}}3<q\le \mathrm{\infty }$$

(1.8)

and $0<T<\mathrm{\infty }$, then the solution $(\rho ,u,d)$ can be extended beyond $T>0$.

Remark 1.1 When $\rho \equiv 1$, our result improves (1.7) to (1.8).

Remark 1.2 By similar calculations as those in [6], we can replace $L^q$-norm in (1.8) by $L_w^q$-norm, and thus we omit the details here.

Remark 1.3 When the space dimension $n=2$, we can prove that the problem (1.1)-(1.6) has a unique global-in-time strong solution by the same method as that in [10], and thus we omit the details here.

Next we consider another liquid model: (1.1), (1.2), (1.3), (1.5), (1.6) and

$${\partial }_{t}d+u\cdot \mathrm{\nabla }d-\mathrm{\Delta }d={|\mathrm{\nabla }d|}^{2}d,$$

(1.9)

with $|d|\equiv 1$ in $(0,\mathrm{\infty })\times \mathrm{\Omega }$. Li and Wang [9] proved that the problem has a unique local strong solution. When $\mathrm{\Omega }:={\mathbb{R}}^3$, Fan et al. [11] proved a regularity criterion. The aim of this paper is to study the regularity criterion of the problem in a bounded domain. We will prove the following theorem.

Theorem 1.2 Let the initial data satisfy the same conditions in Theorem  1.1 and $|d_0|\equiv 1$ in Ω. Let $(\rho ,u,d)$ be a local strong solution to the problem (1.1)-(1.3), (1.5), (1.6) and (1.9). If u and ∇d satisfy

$${\int }_0^T({\parallel u(t)\parallel }_{L^q}^{\frac{2q}{q-3}}+{\parallel \mathrm{\nabla }d\parallel }_{L^q}^{\frac{2q}{q-3}})dt<\mathrm{\infty }\mathit{\text{with}}3<q\le \mathrm{\infty }$$

(1.10)

and $0<T<\mathrm{\infty }$, then the solution $(\rho ,u,d)$ can be extended beyond $T>0$.

2 Proof of Theorem 1.1

We only need to establish a priori estimates.

Below we shall use the notation

$$\int ={\int }_{\mathrm{\Omega }}.$$

First, thanks to the maximum principle, it follows from (1.1) and (1.2) that

$$0\le \rho \le {\parallel {\rho }_0\parallel }_{L^{\mathrm{\infty }}}<\mathrm{\infty }.$$

(2.1)

Testing (1.3) by u and using (1.1) and (1.2), we see that

$$\frac{1}{2}\frac{d}{dt}\int \rho u^{2}dx+\int {|\mathrm{\nabla }u|}^{2}dx=-\int (u\cdot \mathrm{\nabla })d\cdot \mathrm{\Delta }ddx.$$

(2.2)

Testing (1.4) by $-\mathrm{\Delta }d+({|d|}^2-1)d$ and using (1.1), we find that

$$\begin{array}{r}\frac{d}{dt}\int (\frac{1}{2}{|\mathrm{\nabla }d|}^2+\frac{1}{4}{({|d|}^2-1)}^2)dx+\int {(-\mathrm{\Delta }d+({|d|}^2-1)d)}^{2}dx\\ =\int (u\cdot \mathrm{\nabla })d\cdot \mathrm{\Delta }ddx.\end{array}$$

(2.3)

Summing up (2.2) and (2.3), we have the well-known energy inequality

$$\begin{array}{l}\frac{1}{2}\frac{d}{dt}\int (\rho u^2+{|\mathrm{\nabla }d|}^2+\frac{1}{2}{({|d|}^2-1)}^2)dx+\int ({|\mathrm{\nabla }u|}^2+{(-\mathrm{\Delta }d+({|d|}^2-1)d)}^2)dx\\ \le 0.\end{array}$$

(2.4)

Next, we prove the following estimate:

$${\parallel d\parallel }_{L^{\mathrm{\infty }}(0,T;L^{\mathrm{\infty }})}\le max(1,{\parallel d_0\parallel }_{L^{\mathrm{\infty }}}).$$

(2.5)

Without loss of generality, we assume that $1\le {\parallel d_0\parallel }_{L^{\mathrm{\infty }}}$. Multiplying (1.4) by 2d, we get

$${\partial }_t\varphi +u\cdot \mathrm{\nabla }\varphi -\mathrm{\Delta }\varphi +2{|d|}^2\varphi =-2{|d|}^2({\parallel d_0\parallel }_{L^{\mathrm{\infty }}}^2-1)-2{|\mathrm{\nabla }d|}^2\le 0$$

(2.6)

with $\varphi :={|d|}^2-{\parallel d_0\parallel }_{L^{\mathrm{\infty }}}^2$ and $\varphi (\cdot ,0)={|d_0|}^2-{\parallel d_0\parallel }_{L^{\mathrm{\infty }}}^2\le 0$ and ${\partial }_{\nu }\varphi =0$ on $\partial \mathrm{\Omega }\times (0,\mathrm{\infty })$. Then (2.5) follows from (2.6) by the maximum principle.

In the following calculations, we use the following Gauss-Green formula [12]:

$$\begin{array}{rcl}-{\int }_{\mathrm{\Omega }}\mathrm{\Delta }f\cdot f{|f|}^{p-2}dx& =& \frac{1}{2}{\int }_{\mathrm{\Omega }}{|f|}^{p-2}{|\mathrm{\nabla }f|}^{2}dx+4\frac{p-2}{p^2}{\int }_{\mathrm{\Omega }}{\left|\mathrm{\nabla }{|f|}^{p/2}\right|}^{2}dx\\ -{\int }_{\partial \mathrm{\Omega }}{|f|}^{p-2}(f\cdot \mathrm{\nabla })f\cdot \nu dS-{\int }_{\partial \mathrm{\Omega }}{|f|}^{p-2}(curlf\times \nu )\cdot fdS\end{array}$$

(2.7)

and the following estimate [13, 14]:

$${\parallel f\parallel }_{L^p(\partial \mathrm{\Omega })}\le C{\parallel f\parallel }_{L^p(\mathrm{\Omega })}^{1-\frac{1}{p}}{\parallel f\parallel }_{W^{1,p}(\mathrm{\Omega })}^{\frac{1}{p}}$$

(2.8)

with $1<p<\mathrm{\infty }$.

Taking ∇ to (1.4) _i , we deduce that

$${\partial }_t\mathrm{\nabla }{d}_i+(u\cdot \mathrm{\nabla })\mathrm{\nabla }{d}_i+\mathrm{\nabla }\left(({|d|}^2-1)d_i\right)-\mathrm{\Delta }\mathrm{\nabla }{d}_i=\sum _j\mathrm{\nabla }{u}_j{\partial }_j{d}_i.$$

Testing the above equation by ${|\mathrm{\nabla }{d}_i|}^{p-2}\mathrm{\nabla }{d}_i$ ($2\le p\le 6$), using (1.1), (2.7), (2.8), (2.5) and summing over i, we derive

$$\begin{array}{c}\frac{1}{p}\frac{d}{dt}{\int }_{\mathrm{\Omega }}{|\mathrm{\nabla }d|}^{p}dx+\frac{1}{2}{\int }_{\mathrm{\Omega }}{|\mathrm{\nabla }d|}^{p-2}{\left|{\mathrm{\nabla }}^{2}d\right|}^{2}dx+4\frac{p-2}{p^2}{\int }_{\mathrm{\Omega }}{\left|\mathrm{\nabla }{|\mathrm{\nabla }d|}^{p/2}\right|}^{2}dx\hfill \\ =-\sum _i{\int }_{\partial \mathrm{\Omega }}{|\mathrm{\nabla }{d}_i|}^{p-2}(\mathrm{\nabla }{d}_i\cdot \mathrm{\nabla })\nu \cdot \mathrm{\nabla }{d}_{i}dS+\sum _{i,j}{\int }_{\mathrm{\Omega }}\left[\mathrm{\nabla }{u}_j{\partial }_j{d}_i{|\mathrm{\nabla }{d}_i|}^{p-2}\right]\cdot \mathrm{\nabla }{d}_{i}dx\hfill \\ -\sum _i{\int }_{\mathrm{\Omega }}\mathrm{\nabla }\left(({|d|}^2-1)d_i\right)\cdot {|\mathrm{\nabla }{d}_i|}^{p-2}\mathrm{\nabla }{d}_{i}dx\hfill \\ \le C{\int }_{\partial \mathrm{\Omega }}{|\mathrm{\nabla }d|}^{p}dS-\sum _{i,j}{\int }_{\mathrm{\Omega }}{u}_j\mathrm{\nabla }\cdot \left({\partial }_j{d}_i{|\mathrm{\nabla }{d}_i|}^{p-2}\mathrm{\nabla }{d}_i\right)dx+C{\int }_{\mathrm{\Omega }}{|\mathrm{\nabla }d|}^{p}dx\hfill \\ \le C{\int }_{\partial \mathrm{\Omega }}{|\mathrm{\nabla }d|}^{p}dS+C{\int }_{\mathrm{\Omega }}|u|{|\mathrm{\nabla }d|}^{p/2}\cdot \left|\mathrm{\nabla }{|\mathrm{\nabla }d|}^{p/2}\right|dx+C{\int }_{\mathrm{\Omega }}{|\mathrm{\nabla }d|}^{p}dx\hfill \\ +{\int }_{\mathrm{\Omega }}|u|{|\mathrm{\nabla }d|}^{\frac{p}{2}}\cdot {|\mathrm{\nabla }d|}^{\frac{p}{2}-1}\left|{\mathrm{\nabla }}^{2}d\right|dx\hfill \\ \le C{\int }_{\partial \mathrm{\Omega }}{w}^{2}dS+C{\int }_{\mathrm{\Omega }}|u|w|\mathrm{\nabla }w|dx+C{\int }_{\mathrm{\Omega }}{w}^{2}dx+{\int }_{\mathrm{\Omega }}|u|w{|\mathrm{\nabla }d|}^{\frac{p}{2}-1}\left|{\mathrm{\nabla }}^{2}d\right|dx\hfill \\ (w:={|\mathrm{\nabla }d|}^{p/2})\hfill \\ \le C{\parallel w\parallel }_{L^2}{\parallel w\parallel }_{H^1}+C{\parallel u\parallel }_{L^q}{\parallel w\parallel }_{L^{\frac{2q}{q-2}}}{\parallel \mathrm{\nabla }w\parallel }_{L^2}+C{\parallel w\parallel }_{L^2}^2\hfill \\ +C{\parallel u\parallel }_{L^q}{\parallel \mathrm{\nabla }w\parallel }_{L^{\frac{2q}{q-2}}}{\parallel {|\mathrm{\nabla }d|}^{\frac{p}{2}-1}\left|{\mathrm{\nabla }}^{2}d\right|\parallel }_{L^2}\hfill \\ \le 2\frac{p-2}{p^2}{|\mathrm{\nabla }w|}_{L^2}^2+\frac{1}{4}{\parallel {|\mathrm{\nabla }d|}^{\frac{p}{2}-1}\left|{\mathrm{\nabla }}^{2}d\right|\parallel }_{L^2}^2+C{\parallel w\parallel }_{L^2}^2+C{\parallel u\parallel }_{L^q}^{\frac{2q}{q-3}}{\parallel w\parallel }_{L^2}^2,\hfill \end{array}$$

which gives

$$\begin{array}{c}\frac{d}{dt}{\int }_{\mathrm{\Omega }}{w}^{2}dx+C{\int }_{\mathrm{\Omega }}{|\mathrm{\nabla }w|}^{2}dx+C{\int }_{\mathrm{\Omega }}{|\mathrm{\nabla }d|}^{p-2}{\left|{\mathrm{\nabla }}^{2}d\right|}^{2}dx\hfill \\ \le C{\parallel w\parallel }_{L^2}^2+C{\parallel u\parallel }_{L^q}^{\frac{2q}{q-3}}{\parallel w\parallel }_{L^2}^2.\hfill \end{array}$$

Therefore,

$${\parallel \mathrm{\nabla }d\parallel }_{L^{\mathrm{\infty }}(0,T;L^p)}\le C\text{with }2\le p\le 6.$$

(2.9)

Testing (1.3) by $u_t$, using (1.1), (1.2), (2.1) and (2.9), we have

$$\begin{array}{l}\frac{1}{2}\frac{d}{dt}\int {|\mathrm{\nabla }u|}^{2}dx+\int \rho {|u_t|}^{2}dx\\ =-\int \rho u\cdot \mathrm{\nabla }u\cdot u_{t}dx+\frac{d}{dt}\int \mathrm{\nabla }d\odot \mathrm{\nabla }d:\mathrm{\nabla }udx-2\int \mathrm{\nabla }{d}_t\odot \mathrm{\nabla }d:\mathrm{\nabla }udx\\ \le {\parallel \sqrt{\rho }\parallel }_{L^{\mathrm{\infty }}}{\parallel u\parallel }_{L^q}{\parallel \mathrm{\nabla }u\parallel }_{L^{\frac{2q}{q-2}}}{\parallel \sqrt{\rho }{u}_t\parallel }_{L^2}\\ +\frac{d}{dt}\int \mathrm{\nabla }d\odot \mathrm{\nabla }d:\mathrm{\nabla }udx+2{\parallel \mathrm{\nabla }{d}_t\parallel }_{L^2}{\parallel \mathrm{\nabla }d\parallel }_{L^6}{\parallel \mathrm{\nabla }u\parallel }_{L^3}\\ \le C{\parallel u\parallel }_{L^q}{\parallel \mathrm{\nabla }u\parallel }_{L^2}^{1-\frac{3}{q}}{\parallel u\parallel }_{H^2}^{\frac{3}{q}}{\parallel \sqrt{\rho }{u}_t\parallel }_{L^2}\\ +\frac{d}{dt}\int \mathrm{\nabla }d\odot \mathrm{\nabla }d:\mathrm{\nabla }udx+C{\parallel \mathrm{\nabla }{d}_t\parallel }_{L^2}{\parallel \mathrm{\nabla }u\parallel }_{L^2}^{1/2}{\parallel u\parallel }_{H^2}^{1/2}.\end{array}$$

(2.10)

By the $H^2$-regularity theory of the Stokes system, it follows from (1.3) that

$$\begin{array}{rcl}{\parallel u\parallel }_{H^2}& \le & C{\parallel \mathrm{\nabla }{d}^T\cdot \mathrm{\Delta }d\parallel }_{L^2}+C{\parallel \rho u_t+\rho u\cdot \mathrm{\nabla }u\parallel }_{L^2}\\ \le & C{\parallel \mathrm{\nabla }d\parallel }_{L^6}{\parallel \mathrm{\Delta }d\parallel }_{L^3}+C{\parallel \sqrt{\rho }{u}_t\parallel }_{L^2}+C{\parallel u\parallel }_{L^q}{\parallel \mathrm{\nabla }u\parallel }_{L^{\frac{2q}{q-2}}}\\ \le & C{\parallel \mathrm{\Delta }d\parallel }_{L^3}+C{\parallel \sqrt{\rho }{u}_t\parallel }_{L^2}+C{\parallel u\parallel }_{L^q}{\parallel \mathrm{\nabla }u\parallel }_{L^2}^{1-\frac{3}{q}}{\parallel u\parallel }_{H^2}^{\frac{3}{q}}\\ \le & \frac{1}{2}{\parallel u\parallel }_{H^2}+C{\parallel \mathrm{\Delta }d\parallel }_{L^3}+C{\parallel \sqrt{\rho }{u}_t\parallel }_{L^2}+C{\parallel u\parallel }_{L^q}^{\frac{q}{q-3}}{\parallel \mathrm{\nabla }u\parallel }_{L^2},\end{array}$$

which yields

$${\parallel u\parallel }_{H^2}\le C{\parallel \sqrt{\rho }{u}_t\parallel }_{L^2}+C{\parallel u\parallel }_{L^q}^{\frac{q}{q-3}}{\parallel \mathrm{\nabla }u\parallel }_{L^2}+C{\parallel \mathrm{\Delta }d\parallel }_{L^3}.$$

(2.11)

Testing (1.4) by $-\mathrm{\Delta }{d}_t$, using (2.5) and (2.9), we obtain

$$\begin{array}{l}\frac{1}{2}\frac{d}{dt}\int {|\mathrm{\Delta }d|}^{2}dx+\int {|\mathrm{\nabla }{d}_t|}^{2}dx\\ =\int (({|d|}^2-1)d+u\cdot \mathrm{\nabla }d)\mathrm{\Delta }{d}_{t}dx\\ \le \int \left|[\mathrm{\nabla }({|d|}^{2}d-d)+\mathrm{\nabla }(u\cdot \mathrm{\nabla }d)]\mathrm{\nabla }{d}_t\right|dx\\ \le C(1+{\parallel u\parallel }_{L^q}{\parallel \mathrm{\Delta }d\parallel }_{L^{\frac{2q}{q-2}}}+{\parallel \mathrm{\nabla }u\parallel }_{L^3}{\parallel \mathrm{\nabla }d\parallel }_{L^6}){\parallel \mathrm{\nabla }{d}_t\parallel }_{L^2}\\ \le C(1+{\parallel u\parallel }_{L^q}{\parallel \mathrm{\Delta }d\parallel }_{L^2}^{1-\frac{3}{q}}{\parallel d\parallel }_{H^3}^{\frac{3}{q}}+{\parallel \mathrm{\nabla }u\parallel }_{L^2}^{1/2}{\parallel u\parallel }_{H^2}^{1/2}){\parallel \mathrm{\nabla }{d}_t\parallel }_{L^2}.\end{array}$$

(2.12)

On the other hand, by the $H^3$-regularity theory of the elliptic equation, from (1.4), (2.5) and (2.9) we infer that

$$\begin{array}{rcl}{\parallel d\parallel }_{H^3}& \le & C({\parallel d\parallel }_{L^2}+{\parallel \mathrm{\nabla }\mathrm{\Delta }d\parallel }_{L^2})\\ \le & C(1+{\parallel \mathrm{\nabla }({\partial }_{t}d+u\cdot \mathrm{\nabla }d+{|d|}^{2}d-d)\parallel }_{L^2})\\ \le & C(1+{\parallel \mathrm{\nabla }{d}_t\parallel }_{L^2}+{\parallel \mathrm{\nabla }u\parallel }_{L^3}{\parallel \mathrm{\nabla }d\parallel }_{L^6}+{\parallel u\parallel }_{L^q}{\parallel \mathrm{\Delta }d\parallel }_{L^{\frac{2q}{q-2}}})\\ \le & C(1+{\parallel \mathrm{\nabla }{d}_t\parallel }_{L^2}+{\parallel \mathrm{\nabla }u\parallel }_{L^3}+{\parallel u\parallel }_{L^q}{\parallel \mathrm{\Delta }d\parallel }_{L^2}^{1-\frac{3}{q}}{\parallel d\parallel }_{H^3}^{\frac{3}{q}}),\end{array}$$

which gives

$${\parallel d\parallel }_{H^3}\le C(1+{\parallel \mathrm{\nabla }{d}_t\parallel }_{L^2}+{\parallel \mathrm{\nabla }u\parallel }_{L^3}+{\parallel u\parallel }_{L^q}^{\frac{q}{q-3}}{\parallel \mathrm{\Delta }d\parallel }_{L^2}).$$

(2.13)

Combining (2.11) and (2.13), we have

$$\begin{array}{rl}{\parallel u\parallel }_{H^2}+{\parallel d\parallel }_{H^3}\le & C+C{\parallel \sqrt{\rho }{u}_t\parallel }_{L^2}+C{\parallel \mathrm{\nabla }{d}_t\parallel }_{L^2}+C{\parallel \mathrm{\nabla }u\parallel }_{L^2}\\ +C{\parallel u\parallel }_{L^q}^{\frac{q}{q-3}}{\parallel \mathrm{\nabla }u\parallel }_{L^2}+C{\parallel \mathrm{\Delta }d\parallel }_{L^2}+C{\parallel u\parallel }_{L^q}^{\frac{q}{q-3}}{\parallel \mathrm{\Delta }d\parallel }_{L^2}.\end{array}$$

(2.14)

Putting (2.14) into (2.10) and (2.12) and summing up, we arrive at

$$\begin{array}{c}\frac{1}{2}\frac{d}{dt}\int ({|\mathrm{\nabla }u|}^2+{|\mathrm{\Delta }d|}^2)dx+\int (\rho {|u_t|}^2+{|\mathrm{\nabla }{d}_t|}^2)dx-\frac{d}{dt}\int \mathrm{\nabla }d\odot \mathrm{\nabla }d:\mathrm{\nabla }udx\hfill \\ \le \frac{1}{4}\int \rho {|u_t|}^{2}dx+\frac{1}{4}\int {|\mathrm{\nabla }{d}_t|}^{2}dx+C+C{\parallel \mathrm{\nabla }u\parallel }_{L^2}^2\hfill \\ +C{\parallel u\parallel }_{L^q}^{\frac{2q}{q-3}}{\parallel \mathrm{\nabla }u\parallel }_{L^2}^2+C{\parallel \mathrm{\Delta }d\parallel }_{L^2}^2+C{\parallel u\parallel }_{L^q}^{\frac{2q}{q-3}}{\parallel \mathrm{\Delta }d\parallel }_{L^2}^2,\hfill \end{array}$$

which leads to

$${\parallel u\parallel }_{L^{\mathrm{\infty }}(0,T;H^1)}\le C,{\parallel \sqrt{\rho }{u}_t\parallel }_{L^2(0,T;L^2)}\le C,$$

(2.15)

$${\parallel d\parallel }_{L^{\mathrm{\infty }}(0,T;H^2)}+{\parallel d_t\parallel }_{L^2(0,T;H^1)}\le C.$$

(2.16)

It follows from (2.14), (2.15) and (2.16) that

$${\parallel u\parallel }_{L^2(0,T;H^2)}+{\parallel d\parallel }_{L^2(0,T;H^3)}\le C.$$

(2.17)

Taking ${\partial }_t$ to (1.3), testing by $u_t$, using (1.1), (1.2) and (2.15), we have

$$\begin{array}{l}\frac{1}{2}\frac{d}{dt}\int \rho {|u_t|}^{2}dx+\int {|\mathrm{\nabla }{u}_t|}^{2}dx\\ \le |\int \rho u\cdot \mathrm{\nabla }(u_t^2+u\cdot \mathrm{\nabla }u\cdot u_t)dx|+|\int \rho u_t\cdot \mathrm{\nabla }u\cdot u_{t}dx|\\ +2|\int \mathrm{\nabla }{d}_t\odot \mathrm{\nabla }d:\mathrm{\nabla }{u}_{t}dx|\\ \le C{\parallel \sqrt{\rho }{u}_t\parallel }_{L^2}{\parallel u\parallel }_{L^{\mathrm{\infty }}}{\parallel \mathrm{\nabla }{u}_t\parallel }_{L^2}+C{\parallel u\parallel }_{L^6}^2{\parallel \mathrm{\nabla }u\parallel }_{L^6}{\parallel \mathrm{\nabla }{u}_t\parallel }_{L^2}\\ +C{\parallel u\parallel }_{L^6}^2{\parallel \mathrm{\Delta }u\parallel }_{L^2}{\parallel u_t\parallel }_{L^6}+C{\parallel \sqrt{\rho }{u}_t\parallel }_{L^2}{\parallel u\parallel }_{L^6}{\parallel \mathrm{\nabla }u\parallel }_{L^6}^2\\ +C{\parallel \sqrt{\rho }{u}_t\parallel }_{L^2}{\parallel u_t\parallel }_{L^6}{\parallel \mathrm{\nabla }u\parallel }_{L^3}+C{\parallel \mathrm{\nabla }{d}_t\parallel }_{L^2}{\parallel \mathrm{\nabla }d\parallel }_{L^{\mathrm{\infty }}}{\parallel \mathrm{\nabla }{u}_t\parallel }_{L^2}\\ \le \frac{1}{4}{\parallel \mathrm{\nabla }{u}_t\parallel }_{L^2}^2+C{\parallel u\parallel }_{L^{\mathrm{\infty }}}{\parallel \sqrt{\rho }{u}_t\parallel }_{L^2}^2+C{\parallel u\parallel }_{H^2}^2+C{\parallel u\parallel }_{H^2}^2{\parallel \sqrt{\rho }{u}_t\parallel }_{L^2}^2\\ +C{\parallel \mathrm{\nabla }d\parallel }_{L^{\mathrm{\infty }}}^2{\parallel \mathrm{\nabla }{d}_t\parallel }_{L^2}^2.\end{array}$$

(2.18)

Taking ${\partial }_t$ to (1.4), testing by $-\mathrm{\Delta }{d}_t$, using (2.5), (2.15), (2.16) and (2.17), we arrive at

$$\begin{array}{l}\frac{1}{2}\frac{d}{dt}\int {|\mathrm{\nabla }{d}_t|}^{2}dx+\int {|\mathrm{\Delta }{d}_t|}^{2}dx\\ =\int {(u\cdot \mathrm{\nabla }d+{|d|}^{2}d-d)}_t\cdot \mathrm{\Delta }{d}_{t}dx\\ \le \int [u_t\cdot \mathrm{\nabla }d+u\cdot \mathrm{\nabla }{d}_t+{({|d|}^{2}d-d)}_t]\mathrm{\Delta }{d}_{t}dx\\ =-\int \mathrm{\nabla }(u_t\cdot \mathrm{\nabla }d)\cdot \mathrm{\nabla }{d}_{t}dx+\int [u\cdot \mathrm{\nabla }{d}_t+{({|d|}^{2}d-d)}_t]\mathrm{\Delta }{d}_{t}dx\\ \le ({\parallel \mathrm{\nabla }{u}_t\parallel }_{L^2}{\parallel \mathrm{\nabla }d\parallel }_{L^{\mathrm{\infty }}}+{\parallel u_t\parallel }_{L^6}{\parallel \mathrm{\Delta }d\parallel }_{L^3}){\parallel \mathrm{\nabla }{d}_t\parallel }_{L^2}\\ +{\parallel u\parallel }_{L^6}{\parallel \mathrm{\nabla }{d}_t\parallel }_{L^3}{\parallel \mathrm{\Delta }{d}_t\parallel }_{L^2}+C{\parallel \mathrm{\nabla }{d}_t\parallel }_{L^2}^2\\ \le \frac{1}{4}{\parallel \mathrm{\nabla }{u}_t\parallel }_{L^2}^2+\frac{1}{4}{\parallel \mathrm{\Delta }{d}_t\parallel }_{L^2}^2+C{\parallel \mathrm{\nabla }d\parallel }_{L^{\mathrm{\infty }}}^2{\parallel \mathrm{\nabla }{d}_t\parallel }_{L^2}^2+C{\parallel \mathrm{\Delta }d\parallel }_{L^3}^2{\parallel \mathrm{\nabla }{d}_t\parallel }_{L^2}^2\\ +C{\parallel \mathrm{\nabla }{d}_t\parallel }_{L^2}^2.\end{array}$$

(2.19)

Combining (2.18) and (2.19), we have

$${\parallel \sqrt{\rho }{u}_t\parallel }_{L^{\mathrm{\infty }}(0,T;L^2)}+{\parallel u_t\parallel }_{L^2(0,T;H^1)}\le C,$$

(2.20)

$${\parallel d_t\parallel }_{L^{\mathrm{\infty }}(0,T;H^1)}+{\parallel d_t\parallel }_{L^2(0,T;H^2)}\le C.$$

(2.21)

It follows from (1.4), (2.21) and (2.16) that

$${\parallel d\parallel }_{L^{\mathrm{\infty }}(0,T;H^2)}\le C.$$

(2.22)

It follows from (2.14), (2.15), (2.20) and (2.21) that

$${\parallel u\parallel }_{L^{\mathrm{\infty }}(0,T;H^2)}+{\parallel d\parallel }_{L^{\mathrm{\infty }}(0,T;H^3)}\le C.$$

(2.23)

It follows from (1.3), (2.20) and (2.23) that

$${\parallel u\parallel }_{L^2(0,T;W^{2,6})}\le C,$$

(2.24)

from which it follows that

$${\parallel \mathrm{\nabla }u\parallel }_{L^2(0,T;L^{\mathrm{\infty }})}\le C.$$

(2.25)

Applying ∇ to (1.2), testing by ${|\mathrm{\nabla }\rho |}^{q-2}\mathrm{\nabla }\rho$ ($2\le q\le 6$) and using (2.25), we have

$$\frac{d}{dt}{\parallel \mathrm{\nabla }\rho \parallel }_{L^q}^q\le C{\parallel \mathrm{\nabla }u\parallel }_{L^{\mathrm{\infty }}}{\parallel \mathrm{\nabla }\rho \parallel }_{L^q}^q,$$

which implies

$${\parallel \mathrm{\nabla }\rho \parallel }_{L^{\mathrm{\infty }}(0,T;L^q)}\le C,$$

(2.26)

and therefore

$$\begin{array}{rcl}{\parallel {\rho }_t\parallel }_{L^{\mathrm{\infty }}(0,T;L^q)}& =& {\parallel u\mathrm{\nabla }\rho \parallel }_{L^{\mathrm{\infty }}(0,T;L^q)}\\ \le & {\parallel u\parallel }_{L^{\mathrm{\infty }}(0,T;L^{\mathrm{\infty }})}{\parallel \mathrm{\nabla }\rho \parallel }_{L^{\mathrm{\infty }}(0,T;L^q)}\\ \le & C.\end{array}$$

(2.27)

This completes the proof.

3 Proof of Theorem 1.2

This section is devoted to the proof of Theorem 1.2. We only need to establish a priori estimates.

First, we still have (2.1) and (2.2).

Next, we easily infer that

$$|d|\equiv 1\text{in }(0,\mathrm{\infty })\times \mathrm{\Omega }.$$

(3.1)

Testing (1.9) by $-\mathrm{\Delta }d-{|\mathrm{\nabla }d|}^{2}d$ and using (1.1) and (3.1), we find that

$$\frac{1}{2}\frac{d}{dt}\int {|\mathrm{\nabla }d|}^{2}dx+\int {|\mathrm{\Delta }d+{|\mathrm{\nabla }d|}^{2}d|}^{2}dx=\int (u\cdot \mathrm{\nabla })d\cdot \mathrm{\Delta }ddx.$$

(3.2)

Summing up (2.2) and (3.2), we have the well-known energy inequality

$$\frac{1}{2}\frac{d}{dt}\int (\rho u^2+{|\mathrm{\nabla }d|}^2)dx+\int ({|\mathrm{\nabla }u|}^2+{|\mathrm{\Delta }d+{|\mathrm{\nabla }d|}^{2}d|}^2)dx\le 0.$$

(3.3)

Taking ∇ to (1.9) _i , testing by ${|\mathrm{\nabla }{d}_i|}^{p-2}\mathrm{\nabla }{d}_i$ ($2\le p\le 6$), using (1.1), (2.7), (2.8) and (3.1), similarly to (2.9), we deduce that

$$\begin{array}{c}\frac{1}{p}\frac{d}{dt}{\int }_{\mathrm{\Omega }}{|\mathrm{\nabla }d|}^{p}dx+\frac{1}{2}{\int }_{\mathrm{\Omega }}{|\mathrm{\nabla }d|}^{p-2}{\left|{\mathrm{\nabla }}^{2}d\right|}^{2}dx+4\frac{p-2}{p^2}{\int }_{\mathrm{\Omega }}{\left|\mathrm{\nabla }{|\mathrm{\nabla }d|}^{p/2}\right|}^{2}dx\hfill \\ =-\sum _i{\int }_{\partial \mathrm{\Omega }}{|\mathrm{\nabla }{d}_i|}^{p-2}(\mathrm{\nabla }{d}_i\cdot \mathrm{\nabla })\nu \cdot \mathrm{\nabla }{d}_{i}dS+\sum _{i,j}{\int }_{\mathrm{\Omega }}\mathrm{\nabla }{u}_j{\partial }_j{d}_i{|\mathrm{\nabla }{d}_i|}^{p-2}\mathrm{\nabla }{d}_{i}dx\hfill \\ +{\int }_{\mathrm{\Omega }}\mathrm{\nabla }\left({|\mathrm{\nabla }d|}^{2}d\right)\cdot {|\mathrm{\nabla }d|}^{p-2}\mathrm{\nabla }ddx\hfill \\ \le \frac{p-2}{p^2}{\int }_{\mathrm{\Omega }}{|\mathrm{\nabla }w|}^{2}dx+C{\parallel w\parallel }_{L^2}^2+C{\parallel u\parallel }_{L^q}^{\frac{2q}{q-3}}{\parallel w\parallel }_{L^2}^2\hfill \\ +{\int }_{\mathrm{\Omega }}{|\mathrm{\nabla }d|}^2{w}^{2}dx+C{\int }_{\mathrm{\Omega }}|\mathrm{\nabla }d|\left|{\mathrm{\nabla }}^{2}d\right|{|\mathrm{\nabla }d|}^{\frac{p}{2}-1}{|\mathrm{\nabla }d|}^{\frac{p}{2}}dx(w:={|\mathrm{\nabla }d|}^{p/2})\hfill \\ \le \frac{p-2}{p^2}{\int }_{\mathrm{\Omega }}{|\mathrm{\nabla }w|}^{2}dx+C{\parallel w\parallel }_{L^2}^2+C{\parallel u\parallel }_{L^q}^{\frac{2q}{q-3}}{\parallel w\parallel }_{L^2}^2\hfill \\ +{\parallel \mathrm{\nabla }d\parallel }_{L^q}^2{\parallel w\parallel }_{L^{\frac{2q}{q-2}}}^2+\frac{1}{4}{\int }_{\mathrm{\Omega }}{|\mathrm{\nabla }d|}^{p-2}{\left|{\mathrm{\nabla }}^{2}d\right|}^{2}dx\hfill \\ \le 2\frac{p-2}{p^2}{\int }_{\mathrm{\Omega }}{|\mathrm{\nabla }w|}^{2}dx+C{\parallel w\parallel }_{L^2}^2+C{\parallel u\parallel }_{L^q}^{\frac{2q}{q-3}}{\parallel w\parallel }_{L^2}^2\hfill \\ +C{\parallel \mathrm{\nabla }d\parallel }_{L^q}^{\frac{2q}{q-3}}{\parallel w\parallel }_{L^2}^2+\frac{1}{4}{\int }_{\mathrm{\Omega }}{|\mathrm{\nabla }d|}^{p-2}{\left|{\mathrm{\nabla }}^{2}d\right|}^{2}dx,\hfill \end{array}$$

which yields

$${\parallel \mathrm{\nabla }d\parallel }_{L^{\mathrm{\infty }}(0,T;L^p)}+{\int }_0^T\int {|\mathrm{\nabla }d|}^2{\left|{\mathrm{\nabla }}^{2}d\right|}^{2}dxdt\le C.$$

(3.4)

We still have (2.10) and (2.11).

Similarly to (2.12), testing (1.9) by $-\mathrm{\Delta }{d}_t$, using (3.1) and (3.4), we get

$$\begin{array}{l}\frac{1}{2}\frac{d}{dt}\int {|\mathrm{\Delta }d|}^{2}dx+\int {|\mathrm{\nabla }{d}_t|}^{2}dx\\ =\int (u\cdot \mathrm{\nabla }d-{|\mathrm{\nabla }d|}^{2}d)\mathrm{\Delta }{d}_{t}dx\\ ={\int }_{\mathrm{\Omega }}\mathrm{\nabla }({|\mathrm{\nabla }d|}^{2}d-u\cdot \mathrm{\nabla }d)\cdot \mathrm{\nabla }{d}_{t}dx\\ \le C(1+{\parallel u\parallel }_{L^q}{\parallel \mathrm{\Delta }d\parallel }_{L^{\frac{2q}{q-2}}}+{\parallel \mathrm{\nabla }u\parallel }_{L^3}{\parallel \mathrm{\nabla }d\parallel }_{L^6}){\parallel \mathrm{\nabla }{d}_t\parallel }_{L^2}\\ +C({\parallel \mathrm{\nabla }d\parallel }_{L^6}^3+{\parallel \mathrm{\nabla }{|\mathrm{\nabla }d|}^2\parallel }_{L^2}){\parallel \mathrm{\nabla }{d}_t\parallel }_{L^2}\\ \le C(1+{\parallel u\parallel }_{L^q}{\parallel \mathrm{\Delta }d\parallel }_{L^2}^{1-\frac{3}{q}}{\parallel d\parallel }_{H^3}^{\frac{3}{q}}+{\parallel \mathrm{\nabla }u\parallel }_{L^2}^{1/2}{\parallel u\parallel }_{H^2}^{1/2}+{\parallel \mathrm{\nabla }{|\mathrm{\nabla }d|}^2\parallel }_{L^2}){\parallel \mathrm{\nabla }{d}_t\parallel }_{L^2}.\end{array}$$

(3.5)

Similarly to (2.13), we have

$$\begin{array}{rcl}{\parallel d\parallel }_{H^3}& \le & C(1+{\parallel \mathrm{\nabla }{d}_t\parallel }_{L^2}+{\parallel \mathrm{\nabla }u\parallel }_{L^3}+{\parallel u\parallel }_{L^q}^{\frac{q}{q-3}}{\parallel \mathrm{\Delta }d\parallel }_{L^2}+{\parallel \mathrm{\nabla }\left({|\mathrm{\nabla }d|}^{2}d\right)\parallel }_{L^2})\\ \le & C(1+{\parallel \mathrm{\nabla }{d}_t\parallel }_{L^2}+{\parallel \mathrm{\nabla }u\parallel }_{L^3}+{\parallel u\parallel }_{L^q}^{\frac{q}{q-3}}{\parallel \mathrm{\Delta }d\parallel }_{L^2}+{\parallel \mathrm{\nabla }{|\mathrm{\nabla }d|}^2\parallel }_{L^2}).\end{array}$$

(3.6)

Combining (2.11) and (3.6), we have

$${\parallel u\parallel }_{H^2}+{\parallel d\parallel }_{H^3}\le \text{ the right hand side of (2.14)}+C{\parallel \mathrm{\nabla }{|\mathrm{\nabla }d|}^2\parallel }_{L^2}.$$

(3.7)

Putting (3.7) into (3.5) and (2.10) and using the Gronwall inequality, we still have (2.15), (2.16), (2.17) and (2.18).

Similarly to (2.19), applying ${\partial }_t$ to (1.9), testing by $-\mathrm{\Delta }{d}_t$ and using (3.4), we have

$$\begin{array}{l}\frac{1}{2}\frac{d}{dt}\int {|\mathrm{\nabla }{d}_t|}^{2}dx+\int {|\mathrm{\Delta }{d}_t|}^{2}dx\\ =\int {(u\cdot \mathrm{\nabla }d-{|\mathrm{\nabla }d|}^{2}d)}_t\cdot \mathrm{\Delta }{d}_{t}dx\\ =\int (u_t\cdot \mathrm{\nabla }d+u\cdot \mathrm{\nabla }{d}_t-{|\mathrm{\nabla }d|}^2{d}_t-d{\partial }_t{|\mathrm{\nabla }d|}^2)\mathrm{\Delta }{d}_{t}dx\\ =-\int \mathrm{\nabla }(u_t\cdot \mathrm{\nabla }d)\cdot \mathrm{\nabla }{d}_{t}dx+\int (u\cdot \mathrm{\nabla }{d}_t-{|\mathrm{\nabla }d|}^2{d}_t-d{\partial }_t{|\mathrm{\nabla }d|}^2)\mathrm{\Delta }{d}_{t}dx\\ \le \frac{1}{4}{\parallel \mathrm{\nabla }{u}_t\parallel }_{L^2}^2+\frac{1}{4}{\parallel \mathrm{\Delta }{d}_t\parallel }_{L^2}^2+C{\parallel \mathrm{\nabla }d\parallel }_{L^{\mathrm{\infty }}}^2{\parallel \mathrm{\nabla }{d}_t\parallel }_{L^2}^2\\ +C{\parallel \mathrm{\Delta }d\parallel }_{L^3}^2{\parallel \mathrm{\nabla }{d}_t\parallel }_{L^2}^2+C{\parallel \mathrm{\nabla }{d}_t\parallel }_{L^2}^2.\end{array}$$

(3.8)

Combining (2.18) and (3.8) and using the Gronwall inequality, we still obtain (2.20) and (2.21).

By similar calculations as those in (2.22)-(2.27), we still arrive at (2.22)-(2.27).

This completes the proof.