Open Access

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

Boundary Value Problems20132013:176

DOI: 10.1186/1687-2770-2013-176

Received: 9 April 2013

Accepted: 11 July 2013

Published: 26 July 2013

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.

Keywords

liquid crystal blow-up criterion bounded domain

1 Introduction

Let Ω R 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq1_HTML.gif 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 [14]:
div u = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ1_HTML.gif
(1.1)
t ρ + div ( ρ u ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ2_HTML.gif
(1.2)
t ( ρ u ) + div ( ρ u u ) + π Δ u = ( d d ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ3_HTML.gif
(1.3)
t d + u d + ( | d | 2 1 ) d Δ d = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ4_HTML.gif
(1.4)
in ( 0 , ) × Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq2_HTML.gif with initial and boundary conditions
( ρ , u , d ) ( , 0 ) = ( ρ 0 , u 0 , d 0 ) in  Ω , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ5_HTML.gif
(1.5)
u = 0 , ν d = 0 on  Ω , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ6_HTML.gif
(1.6)

where ρ denotes the density, u the velocity, π the pressure, and d represents the macroscopic molecular orientations, respectively. The symbol d d https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq3_HTML.gif denotes a matrix whose ( i , j ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq4_HTML.gifth entry is i d j d https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq5_HTML.gif, and it is easy to find that d d = d T d https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq6_HTML.gif.

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 [57] and references therein.

When ρ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq7_HTML.gif, Guillén-González et al. [8] proved the blow-up criterion
0 T ( u ( t ) L q 2 q q 3 + d ( t ) L q 2 q q 3 ) d t < with  3 < q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ7_HTML.gif
(1.7)

and 0 < T < https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq8_HTML.gif.

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 ρ 0 W 1 , q ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq9_HTML.gif, u 0 H 0 1 ( Ω ) H 2 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq10_HTML.gif, d 0 H 3 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq11_HTML.gif with 3 < q 6 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq12_HTML.gif and ρ 0 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq13_HTML.gif, div u 0 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq14_HTML.gif in Ω and ν d 0 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq15_HTML.gif on Ω. We also assume that the following compatibility condition holds true: ( π 0 , g ) L 2 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq16_HTML.gif such that
π 0 Δ u 0 + ( d 0 d 0 ) = ρ 0 g in Ω . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equa_HTML.gif
Let ( ρ , u , d ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq17_HTML.gif be a local strong solution to the problem (1.1)-(1.6). If u satisfies
0 T u ( t ) L q 2 q q 3 d t < with 3 < q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ8_HTML.gif
(1.8)

and 0 < T < https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq8_HTML.gif, then the solution ( ρ , u , d ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq17_HTML.gif can be extended beyond T > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq18_HTML.gif.

Remark 1.1 When ρ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq19_HTML.gif, our result improves (1.7) to (1.8).

Remark 1.2 By similar calculations as those in [6], we can replace L q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq20_HTML.gif-norm in (1.8) by L w q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq21_HTML.gif-norm, and thus we omit the details here.

Remark 1.3 When the space dimension n = 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq22_HTML.gif, 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
t d + u d Δ d = | d | 2 d , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ9_HTML.gif
(1.9)

with | d | 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq23_HTML.gif in ( 0 , ) × Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq2_HTML.gif. Li and Wang [9] proved that the problem has a unique local strong solution. When Ω : = R 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq24_HTML.gif, 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 | 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq25_HTML.gif in Ω. Let ( ρ , u , d ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq26_HTML.gif be a local strong solution to the problem (1.1)-(1.3), (1.5), (1.6) and (1.9). If u and d satisfy
0 T ( u ( t ) L q 2 q q 3 + d L q 2 q q 3 ) d t < with 3 < q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ10_HTML.gif
(1.10)

and 0 < T < https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq8_HTML.gif, then the solution ( ρ , u , d ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq17_HTML.gif can be extended beyond T > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq18_HTML.gif.

2 Proof of Theorem 1.1

We only need to establish a priori estimates.

Below we shall use the notation
= Ω . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equb_HTML.gif
First, thanks to the maximum principle, it follows from (1.1) and (1.2) that
0 ρ ρ 0 L < . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ11_HTML.gif
(2.1)
Testing (1.3) by u and using (1.1) and (1.2), we see that
1 2 d d t ρ u 2 d x + | u | 2 d x = ( u ) d Δ d d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ12_HTML.gif
(2.2)
Testing (1.4) by Δ d + ( | d | 2 1 ) d https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq27_HTML.gif and using (1.1), we find that
d d t ( 1 2 | d | 2 + 1 4 ( | d | 2 1 ) 2 ) d x + ( Δ d + ( | d | 2 1 ) d ) 2 d x = ( u ) d Δ d d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ13_HTML.gif
(2.3)
Summing up (2.2) and (2.3), we have the well-known energy inequality
1 2 d d t ( ρ u 2 + | d | 2 + 1 2 ( | d | 2 1 ) 2 ) d x + ( | u | 2 + ( Δ d + ( | d | 2 1 ) d ) 2 ) d x 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ14_HTML.gif
(2.4)
Next, we prove the following estimate:
d L ( 0 , T ; L ) max ( 1 , d 0 L ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ15_HTML.gif
(2.5)
Without loss of generality, we assume that 1 d 0 L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq28_HTML.gif. Multiplying (1.4) by 2d, we get
t ϕ + u ϕ Δ ϕ + 2 | d | 2 ϕ = 2 | d | 2 ( d 0 L 2 1 ) 2 | d | 2 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ16_HTML.gif
(2.6)

with ϕ : = | d | 2 d 0 L 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq29_HTML.gif and ϕ ( , 0 ) = | d 0 | 2 d 0 L 2 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq30_HTML.gif and ν ϕ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq31_HTML.gif on Ω × ( 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq32_HTML.gif. Then (2.5) follows from (2.6) by the maximum principle.

In the following calculations, we use the following Gauss-Green formula [12]:
Ω Δ f f | f | p 2 d x = 1 2 Ω | f | p 2 | f | 2 d x + 4 p 2 p 2 Ω | | f | p / 2 | 2 d x Ω | f | p 2 ( f ) f ν d S Ω | f | p 2 ( curl f × ν ) f d S https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ17_HTML.gif
(2.7)
and the following estimate [13, 14]:
f L p ( Ω ) C f L p ( Ω ) 1 1 p f W 1 , p ( Ω ) 1 p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ18_HTML.gif
(2.8)

with 1 < p < https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq33_HTML.gif.

Taking to (1.4) i , we deduce that
t d i + ( u ) d i + ( ( | d | 2 1 ) d i ) Δ d i = j u j j d i . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equc_HTML.gif
Testing the above equation by | d i | p 2 d i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq34_HTML.gif ( 2 p 6 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq35_HTML.gif), using (1.1), (2.7), (2.8), (2.5) and summing over i, we derive
1 p d d t Ω | d | p d x + 1 2 Ω | d | p 2 | 2 d | 2 d x + 4 p 2 p 2 Ω | | d | p / 2 | 2 d x = i Ω | d i | p 2 ( d i ) ν d i d S + i , j Ω [ u j j d i | d i | p 2 ] d i d x i Ω ( ( | d | 2 1 ) d i ) | d i | p 2 d i d x C Ω | d | p d S i , j Ω u j ( j d i | d i | p 2 d i ) d x + C Ω | d | p d x C Ω | d | p d S + C Ω | u | | d | p / 2 | | d | p / 2 | d x + C Ω | d | p d x + Ω | u | | d | p 2 | d | p 2 1 | 2 d | d x C Ω w 2 d S + C Ω | u | w | w | d x + C Ω w 2 d x + Ω | u | w | d | p 2 1 | 2 d | d x ( w : = | d | p / 2 ) C w L 2 w H 1 + C u L q w L 2 q q 2 w L 2 + C w L 2 2 + C u L q w L 2 q q 2 | d | p 2 1 | 2 d | L 2 2 p 2 p 2 | w | L 2 2 + 1 4 | d | p 2 1 | 2 d | L 2 2 + C w L 2 2 + C u L q 2 q q 3 w L 2 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equd_HTML.gif
which gives
d d t Ω w 2 d x + C Ω | w | 2 d x + C Ω | d | p 2 | 2 d | 2 d x C w L 2 2 + C u L q 2 q q 3 w L 2 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Eque_HTML.gif
Therefore,
d L ( 0 , T ; L p ) C with  2 p 6 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ19_HTML.gif
(2.9)
Testing (1.3) by u t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq36_HTML.gif, using (1.1), (1.2), (2.1) and (2.9), we have
1 2 d d t | u | 2 d x + ρ | u t | 2 d x = ρ u u u t d x + d d t d d : u d x 2 d t d : u d x ρ L u L q u L 2 q q 2 ρ u t L 2 + d d t d d : u d x + 2 d t L 2 d L 6 u L 3 C u L q u L 2 1 3 q u H 2 3 q ρ u t L 2 + d d t d d : u d x + C d t L 2 u L 2 1 / 2 u H 2 1 / 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ20_HTML.gif
(2.10)
By the H 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq37_HTML.gif-regularity theory of the Stokes system, it follows from (1.3) that
u H 2 C d T Δ d L 2 + C ρ u t + ρ u u L 2 C d L 6 Δ d L 3 + C ρ u t L 2 + C u L q u L 2 q q 2 C Δ d L 3 + C ρ u t L 2 + C u L q u L 2 1 3 q u H 2 3 q 1 2 u H 2 + C Δ d L 3 + C ρ u t L 2 + C u L q q q 3 u L 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equf_HTML.gif
which yields
u H 2 C ρ u t L 2 + C u L q q q 3 u L 2 + C Δ d L 3 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ21_HTML.gif
(2.11)
Testing (1.4) by Δ d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq38_HTML.gif, using (2.5) and (2.9), we obtain
1 2 d d t | Δ d | 2 d x + | d t | 2 d x = ( ( | d | 2 1 ) d + u d ) Δ d t d x | [ ( | d | 2 d d ) + ( u d ) ] d t | d x C ( 1 + u L q Δ d L 2 q q 2 + u L 3 d L 6 ) d t L 2 C ( 1 + u L q Δ d L 2 1 3 q d H 3 3 q + u L 2 1 / 2 u H 2 1 / 2 ) d t L 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ22_HTML.gif
(2.12)
On the other hand, by the H 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq39_HTML.gif-regularity theory of the elliptic equation, from (1.4), (2.5) and (2.9) we infer that
d H 3 C ( d L 2 + Δ d L 2 ) C ( 1 + ( t d + u d + | d | 2 d d ) L 2 ) C ( 1 + d t L 2 + u L 3 d L 6 + u L q Δ d L 2 q q 2 ) C ( 1 + d t L 2 + u L 3 + u L q Δ d L 2 1 3 q d H 3 3 q ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equg_HTML.gif
which gives
d H 3 C ( 1 + d t L 2 + u L 3 + u L q q q 3 Δ d L 2 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ23_HTML.gif
(2.13)
Combining (2.11) and (2.13), we have
u H 2 + d H 3 C + C ρ u t L 2 + C d t L 2 + C u L 2 + C u L q q q 3 u L 2 + C Δ d L 2 + C u L q q q 3 Δ d L 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ24_HTML.gif
(2.14)
Putting (2.14) into (2.10) and (2.12) and summing up, we arrive at
1 2 d d t ( | u | 2 + | Δ d | 2 ) d x + ( ρ | u t | 2 + | d t | 2 ) d x d d t d d : u d x 1 4 ρ | u t | 2 d x + 1 4 | d t | 2 d x + C + C u L 2 2 + C u L q 2 q q 3 u L 2 2 + C Δ d L 2 2 + C u L q 2 q q 3 Δ d L 2 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equh_HTML.gif
which leads to
u L ( 0 , T ; H 1 ) C , ρ u t L 2 ( 0 , T ; L 2 ) C , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ25_HTML.gif
(2.15)
d L ( 0 , T ; H 2 ) + d t L 2 ( 0 , T ; H 1 ) C . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ26_HTML.gif
(2.16)
It follows from (2.14), (2.15) and (2.16) that
u L 2 ( 0 , T ; H 2 ) + d L 2 ( 0 , T ; H 3 ) C . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ27_HTML.gif
(2.17)
Taking t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq40_HTML.gif to (1.3), testing by u t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq36_HTML.gif, using (1.1), (1.2) and (2.15), we have
1 2 d d t ρ | u t | 2 d x + | u t | 2 d x | ρ u ( u t 2 + u u u t ) d x | + | ρ u t u u t d x | + 2 | d t d : u t d x | C ρ u t L 2 u L u t L 2 + C u L 6 2 u L 6 u t L 2 + C u L 6 2 Δ u L 2 u t L 6 + C ρ u t L 2 u L 6 u L 6 2 + C ρ u t L 2 u t L 6 u L 3 + C d t L 2 d L u t L 2 1 4 u t L 2 2 + C u L ρ u t L 2 2 + C u H 2 2 + C u H 2 2 ρ u t L 2 2 + C d L 2 d t L 2 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ28_HTML.gif
(2.18)
Taking t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq40_HTML.gif to (1.4), testing by Δ d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq38_HTML.gif, using (2.5), (2.15), (2.16) and (2.17), we arrive at
1 2 d d t | d t | 2 d x + | Δ d t | 2 d x = ( u d + | d | 2 d d ) t Δ d t d x [ u t d + u d t + ( | d | 2 d d ) t ] Δ d t d x = ( u t d ) d t d x + [ u d t + ( | d | 2 d d ) t ] Δ d t d x ( u t L 2 d L + u t L 6 Δ d L 3 ) d t L 2 + u L 6 d t L 3 Δ d t L 2 + C d t L 2 2 1 4 u t L 2 2 + 1 4 Δ d t L 2 2 + C d L 2 d t L 2 2 + C Δ d L 3 2 d t L 2 2 + C d t L 2 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ29_HTML.gif
(2.19)
Combining (2.18) and (2.19), we have
ρ u t L ( 0 , T ; L 2 ) + u t L 2 ( 0 , T ; H 1 ) C , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ30_HTML.gif
(2.20)
d t L ( 0 , T ; H 1 ) + d t L 2 ( 0 , T ; H 2 ) C . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ31_HTML.gif
(2.21)
It follows from (1.4), (2.21) and (2.16) that
d L ( 0 , T ; H 2 ) C . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ32_HTML.gif
(2.22)
It follows from (2.14), (2.15), (2.20) and (2.21) that
u L ( 0 , T ; H 2 ) + d L ( 0 , T ; H 3 ) C . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ33_HTML.gif
(2.23)
It follows from (1.3), (2.20) and (2.23) that
u L 2 ( 0 , T ; W 2 , 6 ) C , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ34_HTML.gif
(2.24)
from which it follows that
u L 2 ( 0 , T ; L ) C . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ35_HTML.gif
(2.25)
Applying to (1.2), testing by | ρ | q 2 ρ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq41_HTML.gif ( 2 q 6 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq42_HTML.gif) and using (2.25), we have
d d t ρ L q q C u L ρ L q q , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equi_HTML.gif
which implies
ρ L ( 0 , T ; L q ) C , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ36_HTML.gif
(2.26)
and therefore
ρ t L ( 0 , T ; L q ) = u ρ L ( 0 , T ; L q ) u L ( 0 , T ; L ) ρ L ( 0 , T ; L q ) C . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ37_HTML.gif
(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 | 1 in  ( 0 , ) × Ω . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ38_HTML.gif
(3.1)
Testing (1.9) by Δ d | d | 2 d https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq43_HTML.gif and using (1.1) and (3.1), we find that
1 2 d d t | d | 2 d x + | Δ d + | d | 2 d | 2 d x = ( u ) d Δ d d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ39_HTML.gif
(3.2)
Summing up (2.2) and (3.2), we have the well-known energy inequality
1 2 d d t ( ρ u 2 + | d | 2 ) d x + ( | u | 2 + | Δ d + | d | 2 d | 2 ) d x 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ40_HTML.gif
(3.3)
Taking to (1.9) i , testing by | d i | p 2 d i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq34_HTML.gif ( 2 p 6 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq35_HTML.gif), using (1.1), (2.7), (2.8) and (3.1), similarly to (2.9), we deduce that
1 p d d t Ω | d | p d x + 1 2 Ω | d | p 2 | 2 d | 2 d x + 4 p 2 p 2 Ω | | d | p / 2 | 2 d x = i Ω | d i | p 2 ( d i ) ν d i d S + i , j Ω u j j d i | d i | p 2 d i d x + Ω ( | d | 2 d ) | d | p 2 d d x p 2 p 2 Ω | w | 2 d x + C w L 2 2 + C u L q 2 q q 3 w L 2 2 + Ω | d | 2 w 2 d x + C Ω | d | | 2 d | | d | p 2 1 | d | p 2 d x ( w : = | d | p / 2 ) p 2 p 2 Ω | w | 2 d x + C w L 2 2 + C u L q 2 q q 3 w L 2 2 + d L q 2 w L 2 q q 2 2 + 1 4 Ω | d | p 2 | 2 d | 2 d x 2 p 2 p 2 Ω | w | 2 d x + C w L 2 2 + C u L q 2 q q 3 w L 2 2 + C d L q 2 q q 3 w L 2 2 + 1 4 Ω | d | p 2 | 2 d | 2 d x , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equj_HTML.gif
which yields
d L ( 0 , T ; L p ) + 0 T | d | 2 | 2 d | 2 d x d t C . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ41_HTML.gif
(3.4)

We still have (2.10) and (2.11).

Similarly to (2.12), testing (1.9) by Δ d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq38_HTML.gif, using (3.1) and (3.4), we get
1 2 d d t | Δ d | 2 d x + | d t | 2 d x = ( u d | d | 2 d ) Δ d t d x = Ω ( | d | 2 d u d ) d t d x C ( 1 + u L q Δ d L 2 q q 2 + u L 3 d L 6 ) d t L 2 + C ( d L 6 3 + | d | 2 L 2 ) d t L 2 C ( 1 + u L q Δ d L 2 1 3 q d H 3 3 q + u L 2 1 / 2 u H 2 1 / 2 + | d | 2 L 2 ) d t L 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ42_HTML.gif
(3.5)
Similarly to (2.13), we have
d H 3 C ( 1 + d t L 2 + u L 3 + u L q q q 3 Δ d L 2 + ( | d | 2 d ) L 2 ) C ( 1 + d t L 2 + u L 3 + u L q q q 3 Δ d L 2 + | d | 2 L 2 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ43_HTML.gif
(3.6)
Combining (2.11) and (3.6), we have
u H 2 + d H 3  the right hand side of (2.14) + C | d | 2 L 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ44_HTML.gif
(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 t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq40_HTML.gif to (1.9), testing by Δ d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_IEq38_HTML.gif and using (3.4), we have
1 2 d d t | d t | 2 d x + | Δ d t | 2 d x = ( u d | d | 2 d ) t Δ d t d x = ( u t d + u d t | d | 2 d t d t | d | 2 ) Δ d t d x = ( u t d ) d t d x + ( u d t | d | 2 d t d t | d | 2 ) Δ d t d x 1 4 u t L 2 2 + 1 4 Δ d t L 2 2 + C d L 2 d t L 2 2 + C Δ d L 3 2 d t L 2 2 + C d t L 2 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-176/MediaObjects/13661_2013_Article_427_Equ45_HTML.gif
(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.

Declarations

Acknowledgements

This work is partially supported by the Zhejiang Innovation Project (Grant No. T200905), the ZJNSF (Grant No. R6090109) and the NSFC (Grant No. 11171154). The authors are indebted to the referee for some helpful suggestions.

Authors’ Affiliations

(1)
Department of Applied Mathematics, Nanjing Forestry University
(2)
Department of Mathematics, Inha University
(3)
Department of Mathematics, Zhejiang Normal University

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