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On well-posedness of Ericksen-Leslie’sparabolic-hyperbolic liquid crystal model
Ning JiangSchool of Mathematics and Statistics
Wuhan University(Joint work with Yilong Luo)
February 20th, 2017, Basque Center for Applied Mathematics
Ericksen-Leslie’s system for liquid crystals
Hydrodynamic theory of liquid crystals – Ericksen-Leslie’ssystem:
∂tρ + div(ρu) = 0 ,ρu = ρF + divσ ,
ρ1ω = ρ1G + g + divπ .(0.1)
A = 12 (∇u + ∇>u), B = 1
2 (∇u − ∇>u), ω = d = ∂td + (u · ∇)d,N = ω − Bd .
The constitutive relations for σ, π and g are given by:
σij= −pδij + σij − ρ∂W∂dk ,i
dk ,j , πij = βidj + ρ ∂W∂dj,i
,
gij= γdi − βjdi,j − ρ∂W∂di
+ gi .(0.2)
p: the pressure, vector β = (β1 , · · · , βn)> and the scalarfunction γ: Lagrangian multipliers for the constraint |d| = 1.
Ericksen-Leslie’s system for liquid crystals
W : the Oseen-Frank energy functional
2W =k1(divd)2 + k2|d · (∇ × d)|2 + k3|d × (∇ × d)|2
+(k2 + k4)[tr(∇d)2 − (divd)2
].
g: the kinematic transport gi = λ1Ni + λ2djAij , whichrepresents the effect of the macroscopic flow field on themicroscopic structure.σ: the stress tensor σij =µ1dk Akpdpdidj + µ2Nidj + µ3diNj + µ4Aij + µ5Aik dk dj + µ6diAjk dk .
coefficients µi(1 ≤ i ≤ 6) are called Leslie coefficients. Sodivσ = 1
2µ4∆u + divσ .λ1 = µ2 − µ3 , λ2 = µ5 − µ6 , µ2 + µ3 = µ6 − µ5 .
For simplicity, we assume F = 0, G = 0, the density isconstant, i.e. ρ = 1, which yields incompressibility divu = 0 .We also take k1 = k2 = k3 = 1, k4 = 0.
Ericksen-Leslie’s system for liquid crystals
Ericksen-Leslie’s parabolic-hyperbolic liquid crystal modelreduces to the following form:
∂tu + u · ∇u − 12µ4∆u + ∇p = −div(∇d ∇d) + divσ ,
divu = 0 ,ρ1d = ∆d + γd + λ1(d − Bd) + λ2Ad ,
(0.3)on Rn × R+ with the constraint |d| = 1, where the Lagrangianmultiplier γ is given by
γ ≡ γ(u, d, d) = −ρ1|d|2 + |∇d|2 − λ2d>Ad . (0.4)
Initial data:
u|t=0 = uin(x), d|t=0 = din(x), d|t=0 = din(x) , (0.5)
where din and din satisfy the constraint and compatibilitycondition:
|din | = 1 , din · din = 0 . (0.6)
Special case: Navier-Stokes coupled with wave map
A particularly important special case of the parabolic-hyperbolicsystem of Ericksen-Leslie’s model is that the term divσ vanishes.Namely, the coefficients µi
′s, (1 ≤ i ≤ 6, i , 4) of divσ are chosenas 0, which immediately implies λ1 = λ2 = 0. Consequently, thesystem (0.3) reduces to a model which is Navier-Stokes equationscoupled with a wave map from Rn to S2:
∂tu + u · ∇u + ∇p = 12µ4∆u − div(∇d ∇d) ,
divu = 0 ,ρ1d = ∆d + (−ρ1|d|2 + |∇d|2)d .
(0.7)
if u = 0, then d is a wave map from Rn to S2;
if d = 0, then u satisfies the standard Navier-Stokesequations.
ρ1 = 0, λ1 = −1, parabolic model
If ρ1 = 0 and λ1 = −1 in the 3rd equation of (0.3), the systemreduces to the parabolic Ericksen-Leslie’s system.
The static analogue of the parabolic Ericksen-Leslie’s systemis Oseen-Frank model (Hardt-Kinderlehrer-Lin, 80’s).
Ginzburg-Landau approximation : partial regularity andregularity (Lin-Liu, 90’s and early 00’s).
Global weak solutions with at most a finite number of singulartimes (2D, Lin-Lin-Wang, 2010, 3D, Lin-Wang, 2015).
∂tu + u · ∇u + ∇p = ∆u − div(∇d ∇d) ,divu = 0 ,
∂td + u · ∇d = ∆d + |∇d|2d , |d| = 1 ,(0.8)
More general parabolic Ericksen-Leslie’s system: manypeople · · ·
ρ1 > 0, parabolic-hyperbolic model (inertialEricksen-Leslie model)
Very few works for general model, except for very special case:u = 0 (physically, inertia effects dominate viscosity) and 1-D, eventhis simplest case is very subtle.
Singularities of variational wave equation. (Saxton 1989,Glassey-Hunter-Zheng 1996, Chen-Huang-Liu 2015 · · · )
Orientational waves: splay and twist waves. (Ali-Hunter 2007,2009)
Dissipative and energy conservative solution to variationalwave equations. (Zheng-Zhang, Bressan, · · · )
utt − c(u)(c(u)ux)x = 0 . (0.9)
Inertial model in Q-tensor framework
It is the system coupled a forced incompressibleNavier-Stokes equations, modeling the flow, with a hyperbolicconvection-diffusion system for matrix-valued functions thatmodel the evolution of the orientations of the nematicmolecules. The inertial term is responsible for the hyperboliccharacter.
Qian-Sheng model:
u + ∇p−β42 ∆u = div (−L∇Q ∇Q + β1QtrQA + β5AQ + β6QA)
+div(µ22 (Q − [Ω,Q] + µ1[Q , Q − [Ω,Q]])
),
divu = 0 ,
JQ + µ1Q= L∆Q − aQ + b(Q2 − 1
d |Q |2Id
)− cQ |Q |2 + µ2
2 A
+µ1[Ω,Q] .(0.10)
Mathematical works for inertial Qian-Sheng model
Global existence and uniqueness for small initial data (DeAnna-Zarnescu 2016).
key assumption 1: Newtonian viscosity β4 is large enough.key assumption 2: the coefficient a > 0 which gives “damping”.(it is physical but not the most physical regime)technical trick: higher-order commutator estimate.
A class of global “twist waves” (De Anna-Zarnescu 2016).
A global existence of the dissipative solution which is inspiredfrom that of incompressible Euler equation defined by P-L.Lions (Feireisl-Rocca-Schimperna-Zarnescu 2016).
First results involving second order material derivative, manyopen questions left in this direction.
Geometric constraint |d| = 1
The constraint |d| = 1 brings analytic difficulties (particularly inhigher order nonlinearities) on the Ericksen-Leslie’s system(0.3), even in the parabolic case ρ1 = 0.
It is impossible to construct approximate solutions (uε , dε) withthis unit-length constraint.
A priori estimate without this unit-length constraint.
The unit-length constraint should be put on the initial data, i.e.initially |din | = 1, at later time, |d(·, t)| should be forced to be 1.
Lagrangian multiplier γ and constraint |d| = 1
The following lemma is on the relation between the Lagrangianmultiplier γ and the geometric constraint |d| = 1.
Lemma
Assume (u, d) is a classical solution to the Ericksen-Leslie’sparabolic-hyperbolic system (0.3)-(0.5) satisfyingu ∈ L∞(0,T ; Hs) ∩ L2(0,T ; Hs+1), ∇d ∈ L∞(0,T ; Hs) andd ∈ L∞(0,T ; Hs) for some T ∈ (0,∞), where s > n
2 + 2.If the constraint |d| = 1 is required, then the Lagrangian multiplier γis
γ = −ρ1|d|2 + |∇d|2 − λ2d>Ad . (0.11)
Conversely, if we give the form of γ as (0.11) and d satisfies theinitial data conditions din · din = 0, |din | = 1 and |d|L∞([0,T ]×Rn) < ∞,then |d| = 1 .
Key of the proof: |d|2 − 1 satisfies a wave equation with 0 initialdatum, then Gronwall inequality to show it is constantly 0.
Proof of the Lemma
Let h = |d|2 − 1. Then h solves the following Cauchy problem for agiven smooth u:
ρ1h − λ1h −∆h = 2γh ,h|t=0 = 0 ,h|t=0 = 0 .
(0.12)
For a given smooth divergence-free u we consider the followingflow
∂tX(t , x) = u(t ,X(t , x)) ,X(0, x) = x .
By letting ψ(t , x) = h(t ,X(t , x)),ρ1∂
2t ψ − λ1∂tψ −∆ψ = 2γ(t ,X(t , x))ψ ,
∂tψ|t=0 = 0 ,ψ|t=0 = 0 .
(0.13)
Our goal is to verify ψ(t , x) = 0 for all times t ≥ 0.
Main Theorem (I)
s > n2 + 2, uin, din ∈ Hs , ∇din ∈ Hs , |din | = 1, din · din = 0. The
initial energy E in ≡ |uin |2Hs + ρ1|din |2Hs + |∇din |2Hs . Then, ifβ ≡ µ4 − 4µ6 > 0, and E in is small enough, namely,
E in < ε0 ≡ min1, ρ1β
2
[96C(√ρ1(µ1+µ6)+|λ1 |−λ2)]2
,ρ2
1β4
[96C(√ρ1(µ1+µ6)+|λ1 |−λ2)]4
,
where C = C(n, s) > 0 is a constant which will be determined inLemma 3, then, there exists a unique solutionu ∈ L∞(0,T ; Hs) ∩ L2(0,T ; Hs+1), ∇d ∈ L∞(0,T ; Hs) andd ∈ L∞(0,T ; Hs) to the system (0.3)-(0.5), where0 < T ≤ 1
48C1ln 1
E in . Moreover,(u, d) satisfies the energy bound(|d − din |2L2 + |u|2Hs + ρ1|d|2Hs + |∇d|2Hs
)(t) + 1
4β
∫ t
0|∇u|2Hs (τ)dτ
≤ E in + 12C1T√
E in , for all 0 ≤ t ≤ T ,
where C1 = C[1 +
1+2(|λ1 |−λ2)√ρ1
+ 1(√ρ1)3 +
(√ρ1−λ2)
2+(|λ1 |−λ2)2
ρ1β
]> 0.
Main Theorem (II)
If µ1 = µ2 = µ3 = µ5 = µ6 = 0 and the initial energy E in < ∞, thenthere is a unique solution u ∈ L∞(0,T ; Hs) ∩ L2(0,T ; Hs+1),∇d ∈ L∞(0,T ; Hs) and d ∈ L∞(0,T ; Hs) to the system (0.7)-(0.5),
where 0 < T < 14C2
ln (E in+1)2
E in(E in+2) . Moreover, (u, d) satisfies
(|d − din |2L2 + |u|2Hs + ρ1|d|2Hs + |∇d|2Hs
)(t) + 1
2µ4
∫ t
0|∇u|2Hs (τ)dτ
≤ C2(E in,T) for all 0 ≤ t ≤ T ,
where C2 = C(1 + 1µ4
+ |∇din |Hs ) > 0, and
C2(E in,T) = E in + 2C2T2∏
i=0
[(1 − Y(E in)e4C2T
)− 12 + i − 1
]> 0,
Y(E in) =E in(E in+2)(E in+1)2 ∈ (0, 1) and C = C(n, s) > 0 is determined in
Lemma 3.
Main Theorem (III)
Assume that α ≡ µ4 − 4µ6 −(|λ1 |−7λ2)
2
η −2(7|λ1 |−2λ2)
2
|λ1 |> 0 and
µ2 < µ3 (i.e. λ1 < 0), where η = 12 min
1, 1
ρ1, |λ1 |
ρ1
∈ (0, 1
2 ]. If the
initial data satisfy E in ≤ ε1 ≡1
|λ1 |+2 min
12ε0,
θ2
(8C3)2
, where
C3 = 4C ′(1 + 1√
ρ1
)(1 + µ1 + |λ1| − λ2 + µ6 − ρ1λ2 + ρ1 + 1√
ρ1
)> 0,
θ = minα, η, 1
2 |λ1|> 0 and the constant C ′ = C ′(n, s) > 0, then
there exists a unique global solutionu ∈ L∞(0,∞; Hs) ∩ L2(0,∞; Hs+1), ∇d ∈ L∞(0,∞; Hs) andd ∈ L∞(0,∞; Hs) to the parabolic-hyperbolic system (0.3)-(0.5).Moreover, the solution (u, d) satisfies
supt≥0
(|u|2Hs + ρ1|d|2Hs + |∇d|2Hs
)+
∫ ∞
0|∇u|2Hs dt ≤ 2(|λ1|+ 2)E in .
The approximate system of (0.3).
We first construct the approximate system of (0.3): ddt (uε , dε , dε)> = Fε(uε , dε , dε) ,(uε , dε , dε)>
∣∣∣t=0 = (Jεuin,Jε din,Jεdin)> .
(0.14)
where
Fε(uε , dε , dε) = (Fε(uε , dε , dε), 1ρ1
Gε(uε , dε , dε),Hε(uε , dε , dε))>
Fε(uε , dε , dε) = 12µ4Jε∆uε − PJε [Jεuε · ∇Jεuε ]
−PJεdiv(∇Jεdε ∇Jεdε) + PJεdivσ(Jεuε ,Jεdε ,Jε dε) ,
Gε(uε , dε , dε) = Jε∆dε − ρ1Jε [Jεuε · ∇Jε dε ]
+Jε(γ(Jεuε ,Jεdε ,Jε dε)Jεdε)
+λ1(Jε dε − Jε(JεBεJεdε)) + λ2Jε(JεAεJεdε) ,
Hε(uε , dε , dε) = dε − Jε [Jεuε · ∇Jεdε ] .
Here the mollifier operator Jε is defined as Jε f = F −1(1|ξ|≤ 1εF (f))
and F is the standard Fourier transform.
Local-wellposedness of the approximate system
Lemma
Let s > n2 + 2. Then there exists a unique solution
(uε , dε , dε)> ∈ C([0,Tε); Hs × Hs × Hs) to the system (0.14) for themaximal Tε > 0.
As J2ε = Jε , (Jεuε ,Jεdε) is also a solution to the system (0.14).
Then (uε , dε) = (Jεuε ,Jεdε) and dε = Jε dε . So (uε , dε) solves
∂tuε = −JεP(uε · ∇uε) + 12µ4∆uε − JεPdiv(∇dε ∇dε) +JεPdivσ ,
divuε = 0 ,ρ1∂t dε =−ρ1Jε(uε ·∇dε)+∆dε+Jε(γ
εdε)+λ1(dε−Jε(Bεdε))+λ2Jε(Aεdε) ,
(uε , dε , dε)>∣∣∣t=0 = (Jεuin,Jε din,Jεdin)> ,
(0.15)dε = ∂tdε +Jε(uε · ∇dε), γε = −ρ1|dε |2 + |∇dε |2 − λ2(dε)>Aε(dε) ,
(σ)ji = µ1dεk dεpAεkpdεi dεj + µ2dεj (dεi + Bε
kidεk ) + µ5dεj dεk Aε
ki + µ6dεi dεk Aεkj .
Uniform energy estimate
Let Eε(t) ≡ |dε − Jεdin |2L2 + |uε |2Hs + ρ1|dε |2Hs + |∇dε |2Hs , and
Fε(t) ≡ |∇uε |2Hs be the energy and energy dissipation.
Lemma
Assume β = µ4 − 4µ6 > 0 and (uε , dε) is a smooth solution to(0.15). Then there exists C = C(n, s) > 0 s.t. for all t ∈ [0,Tε)
12
ddt Eε(t) + 1
4βFε(t) ≤ P(Eε(t)) + Q(Eε(t))Fε(t) , (0.16)
Q(Eε) = C(µ1 + µ6 + |λ1 |−λ2√ρ1
)(1 + |∇din |2Hs )(|∇din |Hs +5∑
i=1
Ei2ε ) ,
P(Eε) = C0(Eε + 1)(Eε + 2) ,
C0 = C[
1+|λ1 |−λ2+(|λ1 |−λ2)|∇din |2Hs
√ρ1
+ |∇din |Hs
+ 1(√ρ1)3 +
(√ρ1−λ2)
2+(|λ1 |−λ2)2
ρ1β
].
Uniform energy estimate—continued
If µ1 = µ2 = µ3 = µ5 = µ6 = 0, we have Q(Eε) = 0 and β = µ4.Thus
12
ddt Eε(t) + 1
4µ4Fε(t) ≤ P(Eε(t)) , (0.17)
where
P(Eε) = C( 1√ρ1
+ 1µ4
+ |∇din |Hs )Eε(Eε + 1)(Eε + 2) .
Proof of Theorem (I)
Note Eε(0) ≤ E in . If E in < min1, ρ1β
2
[96C(√ρ1(µ1+µ6)+|λ1 |−λ2)]2
, we
have Q(Eε(0)) ≤ 18β. Define
T∗ε = supT > 0; Eε(t) ≤ 2 and Q(Eε(t)) ≤ 1
4β hold for all t ∈ [0,T ].
Lemma 2 implies that Eε(t) is continuous, thus T∗ε > 0. So for anyfixed ε > 0 and for all t ∈ [0,T∗ε ]
12
ddt
Eε(t) +[
14β − Q(Eε(t))
]Fε(t) ≤ 12C1Eε(t) ,
which implies Eε(t) ≤ E ine24C1t holds for all ε > 0 and t ∈ [0,T∗ε ].Thus let 0 < T ≤ 1
48C1ln 1
E in such that for all t ∈ [0,minT ,T∗ε ]
Eε(t) ≤ E ine24C1T ≤√
E in < 1 ,
which consequently implies that for all t ∈ [0,minT ,T∗ε ]
Q(Eε(t)) ≤ 12C(µ1 + µ6 + |λ1 |−λ2√
ρ1
)(E in)
14 .
Proof of Theorem (I)– continued
Let ε0 = min1, ρ1β
2
[96C(√ρ1(µ1+µ6)+|λ1 |−λ2)]2
,ρ2
1β4
[96C(√ρ1(µ1+µ6)+|λ1 |−λ2)]4
,
s.t. if E in < ε0, then for all t ∈ [0,minT ,T∗ε ], Eε(t) ≤√
E in < 1, andQ(Eε(t)) ≤ 1
8β . Thus, by the continuity of Eε(t), T∗ε ≥ T . As aconsequence, for all t ∈ [0,T ],
Eε(t) + 14β
∫ t
0|∇uε |2Hs (τ)dτ ≤ C1(E in,T) , (0.18)
where C1(E in,T) = E in + 12C1T√
E in > 0. Thus the energyestimate is closed. The rest convergence proof is tedious butstandard.
Proof of Theorem (III)–Global solution
Need a new a priori estimate: For 1 ≤ k ≤ s, take ∇k in 3rdequation of (0.3), multiply by ∇k d. (we did it for d in localexistence.)
12
ddt
(ρ1|d + d|2
Hs + (|λ1| − ρ1)|d|2Hs − ρ1|d|2Hs
)− ρ1|d|2Hs + |∇d|2
Hs
≤(|λ1| − 7λ2)|∇u|Hs |∇d|Hs + Cρ1|d|Hs |∇u|Hs |∇d|Hs
+C(1 + ρ1)(|d|2Hs + |∇d|2Hs )(|∇d|Hs + |∇d|2Hs ) (0.19)
+C(|λ1| − λ2)|∇u|Hs |∇d|Hs (|∇d|Hs + |∇d|2Hs + |∇d|3Hs ) .
Taking a positive constant η = 12 min
1, 1
ρ1, |λ1 |
ρ1
∈ (0, 1
2 ], wemultiply by η in the inequality (0.19) and then add it to the energyestimate obtained in the local existence.
Proof of Theorem (III)–Global solution
12
ddt
(|u|2Hs + ρ1(1 − η)|d|2Hs + (1 − ηρ1)|∇d|2Hs
+ ηρ1|d + d|2H2 + ηρ1|∇
s+1d|2L2 + ηρ1|λ1||d|2Hs
)+ 1
2α|∇u|2Hs + (|λ1| − ηρ1)|d|2Hs + η|∇d|2Hs
≤C ′(1 + µ1 + |λ1| − λ2 + µ6 − ρ1λ2 + ρ1 + 1√ρ1
)(|u|Hs + |d|Hs +
4∑i=1
|∇d|iHs
)×
(|∇u|Hs + |d|Hs + |∇d|Hs
)|∇u|Hs ,
where α = µ4 − 4µ6 −(|λ1 |−7λ2)
2
η −2(7|λ1 |−2λ2)
2
|λ1 |> 0.
Proof of Theorem (III)–Global solution
We denote by
E(t) ≡ |u|2Hs + ρ1(1 − η)|d|2Hs + (1 − ηρ1)|∇d|2Hs + ηρ1|d + d|2Hs
+ ηρ1|∇s+1d|2L2 + ηρ1|λ1||d|2Hs
(0.20)and
D(t) ≡ |∇u|2Hs + |d|2Hs + |∇d|2Hs ,
We have a new energy estimate
ddtE(t) + θD(t) ≤ C3
4∑q=1
[E(t)]q2D(t) , (0.21)
where θ = minα, η, 1
2 |λ1|> 0. The rest is continuity argument
similar as before.
Some open problems
For the system (0.7) which is the Navier-Stokes equationscoupled with a wave map from Rn to S2, can we prove globalin time classical solution with small initial data? So far we justshowed the local existence with large data.
Physically, the inertial constant ρ1 is very small. Do thesolutions constructed here (uρ1 , dρ1) converge to that ofparabolic Ericksen-Leslie system as ρ1 → 0?
Global classical solutions without the constraint µ2 < µ3?
Weak solutions? For these solutions, the constraint |d| = 1 willbring serious difficulties.
