# Regularity Criterion for the Nematic Liquid Crystal Flows in Terms of Velocity

Authors: Ruiying Wei; Zheng-an Yao; Yin Li
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We study the regularity criterion for the 3D nematic liquid crystal flows in the framework of anisotropic Lebesgue space. More precisely, we proved some sufficient conditions in terms of velocity or the fractional derivative of velocity in one direction.


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## Body

## 1. Introduction

This paper is devoted to the regularity criterion for the three-dimensional nematic liquid crystal flows:(1)ut+(u·∇)u+∇p-νΔu=-λ∇·(∇d⊙∇d),hhhhhhhhhhhhhhhhhhhhhhhiix∈R3,t>0,dt+(u·∇)d=γ(Δd-f(d)),x∈R3,t>0,∇·u=0,x∈R3,t>0,
with initial data
(2)(u,d)∣t=0=(u0,d0),x∈R3,
where u(x,t) is the velocity field, d(x,t) represents the macroscopic average of the nematic liquid crystal orientation field, and p(x,t) is the scalar pressure. The symbol ∇d⊙∇d denotes a matrix whose (i,j)th entry is given by ∂id·∂jd for 1≤i,j≤3; here f(d)=(1/ε2)(|d|2-1)d. Since the sizes of the viscosity constants ν,λ,γ,ɛ do not play important roles in our proof, for simplicity, we assume all these positive constants to be one.The hydrodynamic theory of liquid crystals was established by Ericksen and Leslie [1–4]; the model (1) is a simplified system of Ericksen-Leslie model which was first introduced by Lin in [5], and one of the most significant works is given by Lin and Liu [6]; more precisely, they established global existence for weak solutions and classical solutions. Recently, Liu et al. in [7] established the regularity criterion for (1) as follows:
(3)∫0T‍∥∂3u(τ)∥Lαβdτ<∞,with2β+3α≤1,α>3.
One may refer to some interesting and important regularity criteria of nematic liquid crystal flows studied by many authors (see, e.g., [8–13] and the references therein). When d is constant, the system (1) becomes the well-known Navier-Stokes equations. The regularity of solutions to the 3D NS equations has been widely investigated during the past fifty years; see, for example, [14–22] and so on. The aim of this paper is to establish a new regularity criterion by providing sufficient condition in terms of velocity or the fractional derivative of velocity in one direction in the framework of anisotropic Lebesgue space.Throughout the paper, the norm of the Lebesgue spacesLp(R3) is denoted by ∥·∥Lp and denoted the directional derivatives of a function ϕ by ∂iϕ=(∂ϕ/∂xi)(i=1,2,3), the symbol  ∫f(x)dx=∫R3‍f(x)dx‍, Λi=-∂i2,Λ=-Δ, ∥∥ϕ∥Lip∥Lj,kq=(∫R2‍(∫R‍|ϕ(x)|pdxi)q/pdxjdxk)1/q, and (i,j,k) belongs to the permutation group S=span⁡{1,2,3}. Denote
(4)E1={(γ,α)∈(2,∞]2,1γ+2α<1},E2={α∈(2,∞],3-αrα<1,1-αrα-2<1}.Theorem 1.
Let(u0,d0)∈H1(R3)×H2(R3) with the initial data div⁡u0=0, and let the pair (u,b) be the weak solution to the liquid crystal flows (1)-(2) on [0,T) for some 0<T<∞. If u satisfies
(5)∫0T‍∥∥u(τ)∥Liγ∥Lj,kαβdτ<∞,with2β+2α+1γ≤1,hhhhhhhhhhhhhhhhhhhhhhhhhhhii(γ,α)∈E1,
then (u,d) can be extended beyond T.Theorem 2.
Let(u0,d0)∈H1(R3)×H2(R3) with the initial data div⁡u0=0, and let the pair (u,b) be the weak solution to the liquid crystal flows (1)-(2) on [0,T) for some 0<T<∞. If u satisfies
(6)∫0T‍∥Λiru(τ)∥Lαβdτ<∞,withhhhhh2β+3α≤{1+1α,α∈(2,∞),ifr∈(1α,1],or1+r,α∈E2,ifr∈[0,1α),
then (u,d) can be extended beyond T.Corollary 3.
Under the assumption of Theorem2, if we fix r=1, then the sufficient condition is that
(7)∫0T‍∥∂iu(τ)∥Lαβdτ<∞,with2β+3α≤1+1α,hhhhhhhhhhhhhhhhhhhhhhhhiiiiiα∈(2,∞).Remark 4.
Comparing with the corresponding results in [7], it is obvious that the conclusion of Corollary 3 is an improvement version of Theorem 1.1 in [7] in some sense.


## 2. The Proof of Theorems1 and 2

In this section, we will prove Theorems1 and 2. For convenience, we first recall the following three-dimensional Sobolev and Ladyzhenskaya inequalities in the whole space (see, e.g., [23–25]).Lemma 5.
Let2≤q≤6, 2≤p<∞, and 2<r, s≤∞. There hold that
(8)∥ϕ∥Lq≤C∥ϕ∥L2(6-q)/2q∥∂1ϕ∥L2(q-2)/2q∥∂2ϕ∥L2(q-2)/2q∥∂3ϕ∥L2(q-2)/2q,∥∥ϕ∥Li,jp∥Lk2≤C∥ϕ∥L22/p∥∂iϕ∥L2(p-2)/2p∥∂jϕ∥L2(p-2)/2p,∥∥ϕ∥Li2r/(r-2)∥Lj,k2s/(s-2)≤C∥ϕ∥L21-(1/r)-(2/s)∥∂iϕ∥L21/r∥∂jϕ∥L21/s∥∂kϕ∥L21/s.Proof of Theorem1.
Suppose that[0,T*) is the maximal interval of the existence of the local smooth solution. If T*≥T, then there is nothing to prove; on the other side, for T*<T, our strategy is to show that
(9)limsup⁡t→T*(∥∇u(·,t)∥L22+∥Δd(·,t)∥L22)≤C,
under the assumption (5). As a result, the interval [0,T*) cannot be a maximal interval of existence, which leads to a contradiction.
We multiply (1)1 by u and integrate over R3 and, similarly, multiply (1)2 by -Δd+f(d) and integrate over R3 and then by adding two results above and using the fact that ∇·(∇d⊙∇d)=∇(|∇d|2/2)+Δd∇d, we obtain
(10)12ddt∫‍(|u|2+|∇d|2+12(|d|2-1)2)dxhhhhh+∫‍(|∇u|2+|Δd-f(d)|2)dx=0.
Here we used the facts that div u=0 and (u·∇u,u)=(u,∇p)=(u·∇d,f(d))=(u,∇(|d|2/2))=0; here (·,·) denotes the usual inner product of L2(R3), which implies
(11)∥u∥L∞(0,T;L2)+∥u∥L2(0,T;H1)≤C.
Besides, we multiply (1)2 by |d|4d and integrate over R3 and get
(12)16ddt∫‍|u|6dx+∫‍(5|d|4|∇d|2+|d|8)dx=∫‍|d|6dx,
which implies
(13)∥d(·,t)∥L∞(0,T;L6)≤C∥d0∥L6≤C∥d0∥H1.
Multiplying the first equation of (1) by -Δu and integrating over R3. Similarly, by taking Δ on both sides of the second equation of (1), by multiplying the resulting equation by Δd, by integrating over R3, and then by adding two results above and taking the divergence-free condition divu=0 into account, we obtain
(14)12ddt(∥∇u∥L22+∥Δd∥L22)+∥Δu∥L22+∥∇Δd∥L22=∫‍(u·∇)u·Δudx-2∑i=13‍∫‍∇ui∂i∇dΔddx-∫‍Δf(d)Δddx=I1+I2+I3.
In the following, we establish the bounds ofI1–I3, for the first term I1; thanks to Lemma 5 and using Young’s inequality, we have
(15)I1≤∫‍|u||∇u||Δu|dx≤∥∥u∥L3γ∥L1,2α∥∥∇u∥L32γ/(γ-2)∥L1,22α/(α-2)∥Δu∥L2≤C∥∥u∥L3γ∥L1,2α∥∇u∥L21-(1/γ)-(2/α)×∥∂1∇u∥L21/α∥∂2∇u∥L21/α∥∂3∇u∥L21/γ∥Δu∥L2≤C∥∥u∥L3γ∥L1,2α∥∇u∥L21-(1/γ)-(2/α)∥Δu∥L2(1/γ)+(2/α)+1≤12∥Δu∥L22+C∥∥u∥L3γ∥L1,2α2αγ/(αγ-2γ-α)∥∇u∥L22.
For the second term I2, similar to estimate of I1, we have
(16)I2=-2∑i=13‍∫‍∇ui∂i∇dΔddx=2∑i,j=13‍∫‍(ui∂j∂i∂jdΔddx+ui∂i∂jd∂jΔd)dx≤C∫‍|u||∇2d||∇Δd|dx≤C∥∥u∥L3γ∥L1,2α∥Δd∥L21-(1/γ)-(2/α)×∥∂1Δd∥L21/α∥∂2Δd∥L21/α∥∂3Δd∥L21/γ∥∇Δd∥L2≤14∥∇Δd∥L22+C∥∥u∥L3γ∥L1,2α2αγ/(αγ-2γ-α)∥Δd∥L22.
For the term I3, using Hölder’s inequality, Young’s inequality, and (13), one has
(17)I3=-∫‍Δf(d)Δddx=∫‍∇(|d|2d)·∇Δddx+∥Δd∥L22=3∫‍|d|2∇d·∇Δddx+∥Δd∥L22≤C∥d∥L62∥∇d∥L6∥∇Δd∥L2+∥Δd∥L22≤14∥∇Δd∥L22+C∥Δd∥L22.
Substituting the above estimates (15)–(17) into (14), we obtain
(18)ddt(∥∇u∥L22+∥Δd∥L22)+∥Δu∥L22+∥∇Δd∥L22≤C(1+∥∥u∥L3γ∥L1,2α2αγ/(αγ-2γ-α))∥∇u∥L22+C∥Δd∥L22.
Integrating (18) from 0 to t, using Hölder’s inequality and Young’s inequality, one has
(19)∥∇u∥L22+∥Δd∥L22+∫0t‍(∥Δu∥L22+∥∇Δd∥L22)dτ≤C∫0t‍(1+∥∥u∥L3γ∥L1,2α2αγ/(αγ-2γ-α))×(∥∇u∥L22+∥Δd∥L22)dτ+∥∇u0∥L22+∥Δd0∥L22.
Finally, applying Gronwall’s inequality and using condition (5), then (u,d) can be extended beyond T. This completes the proof of Theorem 1.Proof of Theorem2.
Whenr∈[0,1/α), combining Theorem 1 and using the following imbedding theorem, one can get the conclusion that
(20)∥∥u∥Liα/(1-αr)∥Lj,kα≤C∥Λiru∥Lα.
When r∈(1/α,1), our strategy is to show that
(21)∫0t‍∥Λiru(τ)∥Lαβdτ<∞,2β+3α≤1+1α,α∈(2,∞)
is a sufficient condition. We can verify that integral term ∫0t‍∥∥u(τ)∥Li∞∥Lj,kδ2δ/(δ-2)dτ satisfies the conditions of Theorem 1 with δ∈(2,∞). Applying Lemma 5, Hölder’s inequality, and the interpolation theorem, one can conclude that, for δ∈[α(2r+1)/rα,α],
(22)∥∥u∥Li∞∥Lj,kδ≤C∥∥u∥Li2θ∥Λiru∥Liα1-θ∥Lj,kδ≤C∥∥u∥Li2∥Lj,kpθ∥∥Λiru∥Liα∥Lj,kα1-θ≤C∥∥u∥Lj,kp∥Li2θ∥Λiru∥Lα1-θ≤C∥u∥L22θ/p∥∂ju∥L2(p-2)θ/2p×∥∂ku∥L2(p-2)θ/2p∥Λiru∥Lα1-θ≤C∥u∥L22θ/p∥∇u∥L2(p-2)θ/p∥Λiru∥Lα1-θ,
where 1/δ=(θ/p)+((1-θ)/α) with θ=2(rα-1)/(2(rα-1)+α) and we have used the fact that δ≥((2r+1)α-2)/αr implies p≥2. Using Hölder’s inequality, one has
(23)∫0t‍∥∥u(τ)∥Li∞∥Lj,kδ2δ/(δ-2)dτ≤C∫0t‍∥u∥L24δθ/p(δ-2)∥∇u∥L22δθ(p-2)/p(δ-2)×∥Λiru∥Lα2(1-θ)δ/(δ-2)dτ≤C∥u∥Lt∞L24δθ/p(δ-2)∥∇u∥Lt2L22δθ(p-2)/p(δ-2)×(∫0t‍∥Λiru∥Lα(2(1-θ)δ/(δ-2))ηdτ)1/η,
where η=p(δ-2)/(p(δ-2)-δθ(p-2)).
According to the fact that1/δ=(θ/p)+((1-θ)/α) and θ=2(rα-1)/(2(rα-1)+α), we have
(24)2(1-θ)δδ-2η=2pδ(1-θ)p(δ-2)-δθ(p-2)=2δ(1-θ)δ-2-δθ(1-(2/p))=2δ(1-θ)δ-2-δθ+2δ(1/δ)-2δ((1-θ)/α)=2δ(1-θ)δ-δθ-(2δ(1-θ)/α)=2(1-θ)1-θ-(2(1-θ)/α)=2α(1-θ)α-αθ-2(1-θ)=2αα-2.
This together with Theorem 1 gives the desired result of Theorem 2.



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