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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 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 [14]:

divu=0,
(1.1)
t ρ+div(ρu)=0,
(1.2)
t (ρu)+div(ρuu)+πΔu=(dd),
(1.3)
t d+ud+ ( | d | 2 1 ) dΔd=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=0on Ω,
(1.6)

where ρ denotes the density, u the velocity, π the pressure, and d represents the macroscopic molecular orientations, respectively. The symbol dd denotes a matrix whose (i,j)th entry is i d j d, and it is easy to find that dd= 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 [57] 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 ) dt<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 1.1 Let ρ 0 W 1 , q (Ω), u 0 H 0 1 (Ω) H 2 (Ω), d 0 H 3 (Ω) with 3<q6 and ρ 0 0, div u 0 =0 in Ω and ν d 0 =0 on Ω. 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). If u satisfies

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

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

Remark 1.1 When ρ1, 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+udΔd= | 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 1.2 Let 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). If u and d satisfy

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

0ρ 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 dx+ | u | 2 dx=(u)dΔddx.
(2.2)

Testing (1.4) by Δd+( | 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+uϕΔϕ+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 (2p6), 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 ) Cwith 2p6.
(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 ,

which leads to

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 (2q6) 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|1in (0,)×Ω.
(3.1)

Testing (1.9) by Δd | d | 2 d and using (1.1) and (3.1), we find that

1 2 d d t | d | 2 dx+ | Δ d + | d | 2 d | 2 dx=(u)dΔddx.
(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 ) dx+ ( | u | 2 + | Δ d + | d | 2 d | 2 ) dx0.
(3.3)

Taking to (1.9) i , testing by | d i | p 2 d i (2p6), 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 dxdtC.
(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.

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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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Correspondence to Yong Zhou.

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Keywords

  • liquid crystal
  • blow-up criterion
  • bounded domain