Open Access
Issue
Wuhan Univ. J. Nat. Sci.
Volume 30, Number 3, June 2025
Page(s) 235 - 240
DOI https://doi.org/10.1051/wujns/2025303235
Published online 16 July 2025

© Wuhan University 2025

Licence Creative CommonsThis is an Open Access article distributed under the terms of the Creative Commons Attribution License (https://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

0 Introduction

In this paper, we study the following 3D stationary Q-tensor system of liquid crystal:

{ u u - μ   Δ u + P = - ( Q Q ) - λ ( | Q | H ) + ( Q Δ Q - Δ Q Q ) , u = 0 , u Q + Q   Ω - Ω Q - λ | Q | D = Γ H , (1)

with

H = Δ Q - a Q + b ( Q 2 - t r ( Q 2 ) 3 I 3 × 3 ) - c Q t r ( Q 2 ) .

Here uR3,PR and

Q S 0 3 { A = ( a i j ) 3 × 3 | a i j = a j i , t r ( A ) = 0 }

stand for the flow velocity, the scalar pressure and the nematic tensor order parameter, respectively. The parameters μ>0, Γ-1>0 and λR represent the viscosity coefficient, the rotational viscosity and the nematic alignment, respectively. The coefficients a,b,cR with c>0 are constants. (QQ)i,j=xiQ:xjQ is the symmetric additional stress tensor. D12(u+uT), and Ω12(u-uT) are the symmetric and skew symmetric, respectively, where the notation T represents the transposition of a matrix.

When Q=0, system (1) reduces to the 3D stationary Navier-Stokes system

{ u u - Δ u + P = 0 , u = 0 . (2)

For system (2), a well-known result on the Liouville theorem is given by Galdi[1], which concludes if uL92(R 3), then u=0. Chae-Wölf[2] gave the following logarithmic improvement

R 3 | u ( x ) | 9 2 { l o g ( 2 + 1 | u ( x ) | ) } - 1 d x < .

Kozono-Terasawa-Wakasugi[3] extended Galdi's result[1] to the Lorentz spaces L92,(R3). Luo and Yin[4] showed that if the solution to system (2) satisfies

u i L x 1 p i L x 2 q i L x 3 r i ( R   3 ) ,   p i , q i , r i [ 1 , ) ,   1 p i + 1 q i + 1 r i = 2 3 ( i = 1,2 , 3 ) , (3)

then u=0, where

    f L x 1 p L x 2 q L x 3 r   ( R ( R ( R | f   | p d x 1 ) q p d x 2 ) r q d x 3 ) 1 r .

For more Liouville theorem results of system (2), one could refer to Refs. [5-9] and references therein.

In recent years, the Q-tensor system of liquid crystal (1) has received much attention. However, there are few results on its Liouville theorem. Gong et al[10] proved that if

u L 9 2 ( R   3 ) H ˙ 1 ( R   3 ) , Q H 2 ( R   3 ) , b 2 - 24 a c 0 ,

then u=0,Q=0. Later, Lai and Wu[11] generalized the conditions to

u L 9 2 , ( R   3 ) H ˙ 1 ( R   3 ) , Q H 2 ( R   3 ) , b 2 - 24 a c 0 . (4)

On Liouville theorem for many other models, there are also numerous results (see for example Refs. [12-15]).

Inspired by the works mentioned above in this paper, we will prove that there is only the trivial solution for the steady Q-tensor system of liquid crystal model in mixed Lorentz spaces. First of all, we recall the definition of mixed Lorentz spaces.

Definition 1   Assume the indexes p=(p1,p2,p3) and q=(q1,q2,q3) satisfying 1pi<, 1qi, or pi=qi= (i=1,2,3). A mixed Lorentz space Lp1,q1(Rx1;Lp2,q2(Rx2;Lp3,q3(Rx3))) is the set of functions for which the following norm is finite:

    f L x 1 p 1 , q 1 L x 2 p 2 , q 2 L x 3 p 3 , q 3 { ( 0 ( 0 ( 0 | t 1 1 p 1 t 2 1 p 2 t 3 1 p 3 f ( t 1 , t 2 , t 3 ) | q 1 d t 1 t 1 ) q 2 q 1 d t 2 t 2 ) p 3 p 2 d t 3 t 3 ) 1 p 3 ,              1 p i < , s u p t > 0 | { x 3 : s u p γ γ > 0 | { x 2 : s u p λ λ > 0 | { x 1 : | f ( x 1 , x 2 , x 3 ) | λ } | 1 p 1 γ } | 1 p 2 t } | 1 p 3 ,   p i = .

The main result of the paper is stated in the following theorem.

Theorem 1   Let (u,Q) be a smooth solution to system (1) in R3. If

u L x 1 p , L x 2 q , L x 3 r , ( R 3 ) H ˙ 1 ( R 3 ) , Q H 2 ( R 3 ) , p , q , r ( 3 , ] , 1 p + 1 q + 1 r 2 3 ,   a n d   b 2 - 24 a c 0 ,

then u=0,Q=0.

Remark 1   In the case of p=q=r=92 in Theorem 1, the sufficient condition coincides with (4) obtained in previous work[11]. In addition, due to the following embedding (see for example Ref. [16])

L 9 2 ( R   3 ) L 9 2 , q ( R   3 )   ( 9 2 q ) , (5)

we deduce that our result extends Gong et al 's result in Ref. [10].

When Q=0, we obtain the following Liouville theorem for the Navier-Stokes equations (2), which is a supplement to the result of Ref. [4] (see (2)).

Corollary 1   Let uLx1p,Lx2q,Lx3r,(R3)H˙1(R3) be a smooth solution to the Navier-Stokes equations (2) in R3. If  p,q,r(3,],1p+1q+1r23, then u=0.

1 Proof of Theorem 1

In this section, we give the proof of Theorem 1. To streamline the presentation, set

B 1 { x R   3 |   x ( R , 2 R ) } ,   B 2 { x i R |   R < | x i | < 2 R ,   i = 1,2 , 3 } ,

    L x 1 p , L x 2 q , L x 3 r , ( B 2 )     L x 1 p , ( R | x 1 | 2 R ) L x 2 q , ( R | x 2 | 2 R ) L x 3 r , ( R | x 3 | 2 R ) .

We firstly recall the following lemma.

Lemma 1[17] Let β(Q)=1-6[tr(Q3)]2|Q|6,QS03, then 0β(Q)1.

Proof   We consider a smooth cut-off function ϕCc(R) such that

ϕ ( y ) = { 1 ,    | y | < 1 , 0 ,    | y | 2 .

For each R>0, defining

ϕ R ( x ) = ϕ ( x 1 R )   ϕ ( x 2 R )   ϕ ( x 3 R ) ,   x = ( x 1 , x 2 , x 3 ) R 3 ,

then, one has

ϕ R ( x ) = { 1 ,    | x | < R , 0 ,    | x | 2 R .

Moreover, there is a constant C independent of R such that |kϕR|CRk for k{0,1,2,3}.

Multiplying the first equation and the second equation of (1) by u(x)ϕR(x) and -H(x)ϕR(x), respectively, we know

μ   3 | u | 2 ϕ R ( x ) d x + Γ   3 | H | 2 ϕ R ( x ) d x =   3 ( u Q ) : Δ Q ϕ R ( x ) d x -   3 ( Ω Q - Q Ω ) : Δ Q ϕ R ( x ) d x -   3 ( u Q ) : [ a Q + b ( Q 2 - t r ( Q 2 ) 3 I 3 × 3 ) + c Q | Q | 2 ] ϕ R ( x ) d x   +   3 ( Ω Q - Q Ω ) : [ a Q + b ( Q 2 - t r ( Q 2 ) 3 I 3 × 3 ) + c Q | Q | 2 ] ϕ R ( x ) d x - λ   3 | Q | D : H ϕ R ( x ) d x -   3 ( Q Q ) u ϕ R ( x ) d x   + λ   3 | x | H : ( u   ϕ R ( x ) ) d x -   3 ( Q Δ Q - Δ Q Q ) : ( u   ϕ R ( x ) ) d x + R   3 1 2 | u | 2 u ϕ R ( x ) d x +   3 P   u ϕ R ( x ) + μ 2 | u | 2 Δ ϕ R ( x ) d x i = 1 10 I i . (6)

Firstly, since Q is symmetric and Ω is skew symmetric, we deduce I4=0. For I1+I6,I2+I8 and I5+I7, using Hölder inequality, we obtain

I 1 + I 6 = R   3 u Q : Δ Q ϕ R ( x ) d x -   3 ( Q Q ) u ϕ R ( x ) d x =   3 ( u Q ) : Q ϕ R ( x ) d x -   3 ( u Q ) Δ Q ϕ R ( x ) d x - R   3 u 1 2 | Q | 2 ϕ R ( x ) d x = R   3 1 2 | Q | 2 u ϕ R ( x ) d x

C R | Q | 2 L 2 , ( B 1 )       u   L x 1 p , L x 2 q , L x 3 r , ( B 2 )       ϕ ( R )   L x 1 2 p p - 2 , 1 L x 2 2 q q - 2 , 1 L x 3 2 r r - 2 , 1 ( B 2 ) C R 1 2 - ( 1 p + 1 q + 1 r )   | Q | 2 L 2 ( B 1 )       u   L x 1 p , L x 2 q , L x 3 r , ( B 2 )       ϕ   L x 1 2 p p - 2 , 1 L x 2 2 q q - 2 , 1 L x 3 2 r r - 2 , 1 ( B 2 ) , (7)

I 2 + I 8 = -   3 ( Ω Q - Q Ω ) : Δ Q ϕ R ( x ) d x -   3 ( Q Q ) u ϕ R ( x ) d x = -   3 ( Q Δ Q - Δ Q Q ) : u ϕ R ( x ) d x C R 2 Q L 2 , ( B 1 ) Q L ( B 1 )       u   L x 1 p , L x 2 q , L x 3 r , ( B 2 )       ϕ ( R )   L x 1 2 p p - 2 , 1 L x 2 2 q q - 2 , 1 L x 3 2 r r - 2 , 1 ( B 2 ) C R 1 2 - ( 1 p + 1 q + 1 r ) 2 Q L 2 ( B 1 )       u   L x 1 p , L x 2 q , L x 3 r , ( B 2 )       ϕ   L x 1 2 p p - 2 , 1 L x 2 2 q q - 2 , 1 L x 3 2 r r - 2 , 1 ( B 2 ) , (8)

I 5 + I 7 = - λ   3 | Q | D : H ϕ R ( x ) d x + λ   3 | Q | H : ( u ϕ R ( x ) ) d x =   3 | Q | H : u ϕ R ( x ) d x C R H L 2 ( B 1 ) Q L ( B 1 )       u   L x 1 p , L x 2 q , L x 3 r , ( B 2 )       ϕ ( R )   L x 1 2 p p - 2 , 1 L x 2 2 q q - 2 , 1 L x 3 2 r r - 2 , 1 ( B 2 ) C R 1 2 - ( 1 p + 1 q + 1 r ) H L 2 ( B 1 )       u   L x 1 p , L x 2 q , L x 3 r , ( B 2 )       ϕ   L x 1 2 p p - 2 , 1 L x 2 2 q q - 2 , 1 L x 3 2 r r - 2 , 1 ( B 2 ) . (9)

For I3 and I9, along the same line, we have

I 3 = - 3 ( u Q ) : [ a Q + b ( Q 2 - t r ( Q 2 ) 3 I 3 × 3 ) + c Q | Q | 2 ] ϕ R ( x ) d x = - 3 ( Q Δ Q - Δ Q Q ) : u ϕ R ( x )   d x C R ( 1 + Q L ( B 1 ) + Q L ( B 1 ) 2 )     u L x 1 p , L x 2 q , L x 3 r , ( B 2 ) Q L 4 , ( B 1 ) 2       ϕ ( R )   L x 1 2 p p - 2 , 1 L x 2 2 q q - 2 , 1 L x 3 2 r r - 2 , 1 ( B 2 ) C R 1 2 - ( 1 p + 1 q + 1 r )       u   L x 1 p , L x 2 q , L x 3 r , ( B 2 ) Q L 4 ( B 1 ) 2     ϕ L x 1 2 p p - 2 , 1 L x 2 2 q q - 2 , 1 L x 3 2 r r - 2 , 1 ( B 2 ) , (10)

I 9 = R   3 1 2 | u | 2 u ϕ R ( x )   d x C R     u L x 1 p , L x 2 q , L x 3 r , ( B 2 ) 3     ϕ ( R ) L x 1 p p - 3 , 1 L x 2 q q - 3 , 1 L x 3 r r - 3 , 1 ( B 2 ) C R 2 - ( 3 p + 3 q + 3 r )     u L x 1 p , L x 2 q , L x 3 r , ( B 2 ) 3     ϕ L x 1 p p - 3 , 1 L x 2 q q - 3 , 1 L x 3 r r - 3 , 1 ( B 2 ) . (11)

For I10, note that

Δ P = - d i v d i v ( u u + Q Q + λ | Q | H + Q Δ Q - Δ Q Q ) .

Let P=P1+P2 such that ΔP1=-divdiv(f1) and ΔP2=-divdiv(f2), where

f 1 u u ,   f 2 Q Q + λ | Q | H + Q Δ Q - Δ Q Q .

In view of the conditions uLx1p,s1Lx2q,s2Lx3r,s3(R3) and QH2(R3), we obtain that

f 1 L x 1 p 2 , L x 2 q 2 , L x 3 r 2 , ( R   3 ) ,   f 2 L 2 ( R   3 ) .

By Calderron-Zygmund theorem, it follows that

P 1 L x 1 p 2 , L x 2 q 2 , L x 3 r 2 , ( R   3 ) ,   P 2 L 2 ( R   3 ) .

Then,

I 10 = R   3 P   u ϕ R ( x ) + μ 2 | u | 2 Δ ϕ R ( x ) d x = R   3 P 1 u ϕ R ( x ) d x + R   3 P 2 u ϕ R ( x ) d x + R   3 μ 2 | u | 2 Δ ϕ R ( x ) d x C R     u L x 1 p , L x 2 q , L x 3 r , ( B 2 ) 3     ϕ ( R ) L x 1 p p - 3 , 1 L x 2 q q - 3 , 1 L x 3 r r - 3 , 1 ( B 2 ) + C R     u L x 1 p , L x 2 q , L x 3 r , ( B 2 )     ϕ ( R ) L x 1 2 p p - 2 , 1 L x 2 2 q q - 2 , 1 L x 3 2 r r - 2 , 1 ( B 2 ) P 2 L 2 , ( B 1 ) + C R 2     u L x 1 p , L x 2 q , L x 3 r , ( B 2 ) 2     Δ ϕ ( R ) L x 1 p p - 2 , 1 L x 2 q q - 2 , 1 L x 3 r r - 2 , 1 ( B 2 ) C R 2 - ( 3 p + 3 q + 3 r )     u L x 1 p , L x 2 q , L x 3 r , ( B 2 ) 3     ϕ L x 1 p p - 3 , 1 L x 2 q q - 3 , 1 L x 3 r r - 3 , 1 ( B 2 ) + C R 1 2 - ( 1 p + 1 q + 1 r )     u L x 1 p , L x 2 q , L x 3 r , ( B 2 ) P 2 L 2 ( B 1 )     ϕ ( R ) L x 1 2 p p - 2 , 1 L x 2 2 q q - 2 , 1 L x 3 2 r r - 2 , 1 ( B 2 ) + C R 1 - ( 2 p + 2 q + 2 r )     u L x 1 p , L x 2 q , L x 3 r , ( B 2 ) 2     ϕ ( R ) L x 1 p p - 2 , 1 L x 2 q q - 2 , 1 L x 3 r r - 2 , 1 ( B 2 ) . (12)

Therefore, when R, we can obtain Ii0 (i=1,,10). Thanks to the Sobolev embedding, we have

u L 6 ( R   3 ) C u L 2 ( R   3 ) ,

then u=0 and H=0. By the definition of H, we observe

- Δ Q = - a Q + b [ Q 2 - t r ( Q 2 ) 3 I 3 × 3 ] - c Q   t r ( Q 2 ) . (13)

Taking the inner product of the equation (13) with QϕR(x), and by Lemma 1, one yields

R   3 | Q | 2 ϕ R ( x )   d x =   3 | Q | 2 Δ ϕ R ( x )   d x -   3 [ a | Q | 2 - b   t r ( | Q | 3 ) + c | Q | 4 ] ϕ R ( x )   d x C R 2   3 | Q | 2 d x + R   3 ( b 2 - 24 a c 24 c ) | Q | 2 ϕ R ( x )   d x C R 2   3 | Q | 2 d x 0   ( R ) ,

which combined with QL2<, concludes Q=0. The proof of Theorem 1 is completed.

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