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

Jishan Fan1, Gen Nakamura2 and Yong Zhou3*

Author Affiliations

1 Department of Applied Mathematics, Nanjing Forestry University, Nanjing, P.R. China

2 Department of Mathematics, Inha University, Incheon, 402-751, Republic of Korea

3 Department of Mathematics, Zhejiang Normal University, Jinhua, P.R. China

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Boundary Value Problems 2013, 2013:176  doi:10.1186/1687-2770-2013-176

 Received: 9 April 2013 Accepted: 11 July 2013 Published: 26 July 2013

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

### 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 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]:

div u = 0 , (1.1)

t ρ + div ( ρ u ) = 0 , (1.2)

t ( ρ u ) + div ( ρ u u ) + π Δ u = ( d d ) , (1.3)

t d + u d + ( | d | 2 1 ) d Δ d = 0 , (1.4)

in ( 0 , ) × Ω with initial and boundary conditions

( ρ , u , d ) ( , 0 ) = ( ρ 0 , u 0 , d 0 ) in  Ω , (1.5)

u = 0 , ν d = 0 on  Ω , (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 denotes a matrix whose ( i , j ) th entry is i d j d , and it is easy to find that d d = d T 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 ρ = 1 , 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 (1.7)

and 0 < T < .

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 1.1Let ρ 0 W 1 , q ( Ω ) , u 0 H 0 1 ( Ω ) H 2 ( Ω ) , d 0 H 3 ( Ω ) with 3 < q 6 and ρ 0 0 , div u 0 = 0 in Ω and ν d 0 = 0 onΩ. We also assume that the following compatibility condition holds true: ( π 0 , g ) L 2 ( Ω ) such that

π 0 Δ u 0 + ( d 0 d 0 ) = ρ 0 g in Ω .

Let ( ρ , u , d ) be a local strong solution to the problem (1.1)-(1.6). Ifusatisfies

0 T u ( t ) L q 2 q q 3 d t < with 3 < q (1.8)

and 0 < T < , then the solution ( ρ , u , d ) can be extended beyond T > 0 .

Remark 1.1 When ρ 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

t d + u d Δ d = | d | 2 d , (1.9)

with | d | 1 in ( 0 , ) × Ω . Li and Wang [9] proved that the problem has a unique local strong solution. When Ω : = 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 1.2Let the initial data satisfy the same conditions in Theorem 1.1 and | d 0 | 1 in Ω. Let ( ρ , u , d ) be a local strong solution to the problem (1.1)-(1.3), (1.5), (1.6) and (1.9). Ifuanddsatisfy

0 T ( u ( t ) L q 2 q q 3 + d L q 2 q q 3 ) d t < with 3 < q (1.10)

and 0 < T < , then the solution ( ρ , 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

= Ω .

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

0 ρ ρ 0 L < . (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 . (2.2)

Testing (1.4) by Δ d + ( | d | 2 1 ) d 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 . (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 . (2.4)

Next, we prove the following estimate:

d L ( 0 , T ; L ) max ( 1 , d 0 L ) . (2.5)

Without loss of generality, we assume that 1 d 0 L . 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 (2.6)

with ϕ : = | d | 2 d 0 L 2 and ϕ ( , 0 ) = | d 0 | 2 d 0 L 2 0 and ν ϕ = 0 on Ω × ( 0 , ) . 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 (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 (2.8)

with 1 < p < .

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 .

Testing the above equation by | d i | p 2 d i ( 2 p 6 ), 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 ,

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 .

Therefore,

d L ( 0 , T ; L p ) C with  2 p 6 . (2.9)

Testing (1.3) by u t , 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 . (2.10)

By the H 2 -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 ,

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 . (2.11)

Testing (1.4) by Δ d t , 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 . (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

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 ) ,

which gives

d H 3 C ( 1 + d t L 2 + u L 3 + u L q q q 3 Δ d L 2 ) . (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 . (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 ,

u L ( 0 , T ; H 1 ) C , ρ u t L 2 ( 0 , T ; L 2 ) C , (2.15)

d L ( 0 , T ; H 2 ) + d t L 2 ( 0 , T ; H 1 ) C . (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 . (2.17)

Taking t to (1.3), testing by u t , 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 . (2.18)

Taking t to (1.4), testing by Δ d t , 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 . (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 , (2.20)

d t L ( 0 , T ; H 1 ) + d t L 2 ( 0 , T ; H 2 ) C . (2.21)

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

d L ( 0 , T ; H 2 ) C . (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 . (2.23)

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

u L 2 ( 0 , T ; W 2 , 6 ) C , (2.24)

from which it follows that

u L 2 ( 0 , T ; L ) C . (2.25)

Applying ∇ to (1.2), testing by | ρ | q 2 ρ ( 2 q 6 ) and using (2.25), we have

d d t ρ L q q C u L ρ L q q ,

which implies

ρ L ( 0 , T ; L q ) C , (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 . (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 . (3.1)

Testing (1.9) by Δ d | d | 2 d 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 . (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 . (3.3)

Taking ∇ to (1.9)i, testing by | d i | p 2 d i ( 2 p 6 ), 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 ,

which yields

d L ( 0 , T ; L p ) + 0 T | d | 2 | 2 d | 2 d x d t C . (3.4)

We still have (2.10) and (2.11).

Similarly to (2.12), testing (1.9) by Δ d t , 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 . (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 ) . (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 . (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 to (1.9), testing by Δ d t 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 . (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.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors read and approved the final manuscript.

### 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.

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