Some results on the nematic liquid crystals theory

Transcription

1 Some results on the nematic liquid crystals theory Marius Paicu University of Bordeaux joint work with Arghir Zarnescu Mathflows 2015, Porquerolles September 17, 2015

2 Complex fluids: Basic laws Incompressibility: u = 0 (1) where u is a vector valued function expressing the velocity of the fluid at a point in space.

3 Complex fluids: Basic laws Incompressibility: u = 0 (1) where u is a vector valued function expressing the velocity of the fluid at a point in space. The balance of momentum is ( ) u ρ + (u )u = T p (2) t where ρ is the density, T is the stress tensor and p is an isotropic pressure. The stress tensor T represents the forces which the material develops in response to being deformed.

4 Complex fluids: Basic laws Incompressibility: u = 0 (1) where u is a vector valued function expressing the velocity of the fluid at a point in space. The balance of momentum is ( ) u ρ + (u )u = T p (2) t where ρ is the density, T is the stress tensor and p is an isotropic pressure. The stress tensor T represents the forces which the material develops in response to being deformed. We need a constitutive relation relating T to the motion of the fluid. The constitutive law for the classical Newtonian fluid is T = ν ( u + ( u) tr ) where ν is the viscosity. In this case the system (1), (2) becomes the celebrated Navier-Stokes system of equations.

5 The additional stress tensor and energy dissipation In the case of non-newtonian fluids containing suspensions of liquid crystal molecules the stress has an additional component representing forces due to the liquid crystal molecules. On the other hand one should have an equation for the liquid crystals, showing how the flow affects the orientation and distribution of the molecules.

6 The additional stress tensor and energy dissipation In the case of non-newtonian fluids containing suspensions of liquid crystal molecules the stress has an additional component representing forces due to the liquid crystal molecules. On the other hand one should have an equation for the liquid crystals, showing how the flow affects the orientation and distribution of the molecules. The additional stress tensor encodes the coupling between the flow and the molecules. The form of the additional stress tensor is directly related to energy dissipation. More precisely the content of the stress tensor should be such that the total energy of the fluid E(t) }{{} total energy decreases in time. = 1 u(x, t) 2 dx + F(t) 2 R }{{}} d {{} free energy of the molecules kinetic energy of the flow

7 Some types of complex fluids Oldroyd-B: { tu + u u ν u + p = µ τ tτ + u τ + aτ + τω Ωτ b(dτ + τd) = µ 2 D The formal energy estimate is: 1 d 2 dt (µ 2 u(t) 2 L 2 + µ 1 τ(t) 2 L 2 ) + νµ 2 u(t) 2 L 2 + aµ 1 τ(t) 2 L 2 b D(t) L τ(t) 2 L 2 -Lions-Masmoudi, Chemin-Masmoudi, Guillopé-Saut Smoluchowski Navier-Stokes systems v + v v ν v + p t = τ in Ω (0, T ) f + v f + g (Wf ) g f t = 0 in Ω (0, T ) v = 0 in Ω (0, T ), where τ ij = M γ(1) ij (m)f (t, x, m)dm + M M γ(2) ij (m 1, m 2 )f (t, x, m 1 )f (t, x, m 2 )dm and W = cα ij j v i. The energy E(t) = 1 2 R d u(t, x) 2 dx + R d M f log f dx dm decreases in time. Constantin-Fefferman-Titi-Zarnescu, Constantin-Masmoudi, Lin-Zhang-Zhang, Masmoudi-Zhang

8 Simplified model of liquid crystals Ericksen-Leslie: d(t, x) the director of the crystals molecules with d(t, x) = 1. v t + v v ν v + p = ( d d), in Ω (0, T ) t d + v d d = d 2 d, in Ω (0, T ) v = 0 d = 1 in Ω (0, T ), - The energy E(t) = 1 2 ( u(t, x) 2 + d(t, x) 2 ) dx decreases in time. R d E(t) + d d 2 d 2 dx = E(0) R d - F. Lin and C. Liu studied the global weak solutions and global strong solutions for small data.

9 Q-tensor system and Navier-Stokes. The equations The flow equations: { tu + u u = ν u + p + τ + σ u = 0 where we have the symmetric part of the additional stress tensor: τ = ξ (Q + 13 ) Id H ξh (Q + 13 ) Id +2ξ(Q + 1 ( ) 3 Id)QH L Q Q + tr(q2 ) Id 3 and an antisymmetric part σ = QH HQ where H = L Q aq + b[q 2 tr(q2 ) Id] cqtr(q 2 ) 3 The equation for the liquid crystal molecules, represented by functions with values in the space of Q-tensors (i.e. symmetric and traceless d d matrices): with ( t + u )Q S( u, Q) = ΓH S( u, Q) def = (ξd + Ω)(Q Id) + (Q Id)(ξD Ω) 2ξ(Q Id)tr(Q u)

10 Energy dissipation and weak solutions-apriori bounds I The total energy E(t) def L = R d 2 Q 2 + a 2 tr(q2 ) b 3 tr(q3 ) + c 4 tr2 (Q 2 ) dx }{{} is decreasing d E(t) 0. dt free energy of the liquid crystal molecules + 1 u 2 (t, x) dx 2 R } d {{} kinetic energy of the flow Simple proof: multiply the first equation in the system to the right by H, take the trace, integrate over R d and by parts and sum with the second equation multiplied by u and integrated over R d and by parts. In the process maximal derivatives are cancelled and you observe suprizing non-trivial cancellations

11 Energy dissipation and weak solutions-apriori bounds II ( Γ tr R d d E(t) = ν u 2 dx dt R d 2 L Q aq + b[q 2 tr(q2 ) Id] cqtr(q )) 2 dx 0 d Note that this does not readily provide L p norm estimates. Proposition For d = 2, 3 there exists a weak solution (Q, u) of the coupled system, with restriction c > 0, subject to initial conditions Q(0, x) = Q(x) H 1 (R d ), u(0, x) = ū(x) L 2 (R d ), ū = 0 in D (R d ) (3) The solution (Q, u) is such that Q L loc(r +; H 1 ) L 2 loc(r +; H 2 ) and u L loc(r +; L 2 ) L 2 loc(r +; H 1 ). Remark: -similar results by Guillen-Gonalez and Rodriguez-Bellido in bounded domains - weak solutions in the periodic case by M. Wilkinson

12 Regularity difficulties: the maximal derivatives and the co-rotational parameter Recall the system: ( t + u )Q ( ξd(u) + Ω(u) ) (Q Id) + (Q Id)( ξd(u) Ω(u) ) 2ξ(Q + 1 Id)tr(Q u) = ΓH 3 ( ) tu + u u = ν u + p + QH HQ (ξ ( Q Id) H + ξh ( ) Q + 1 ) 3 Id) +2ξ ((Q )QH L ( Q Q tr(q2 ) ) u = 0 with H = L Q aq + b[q 2 tr(q2 ) 3 Id] cqtr(q 2 ). Worse than Navier-Stokes Where s the difficulty?

13 Regularity difficulties: the maximal derivatives and the co-rotational parameter Recall the system: ( t + u )Q ( ξd(u) + Ω(u) ) (Q Id) + (Q Id)( ξd(u) Ω(u) ) 2ξ(Q + 1 Id)tr(Q u) = ΓH 3 ( ) tu + u u = ν u + p + QH HQ (ξ ( Q Id) H + ξh ( ) Q + 1 ) 3 Id) +2ξ ((Q )QH L ( Q Q tr(q2 ) ) u = 0 with H = L Q aq + b[q 2 tr(q2 ) 3 Id] cqtr(q 2 ). Worse than Navier-Stokes Where s the difficulty? If ξ = 0 maximal derivatives only

14 Regularity difficulties: the maximal derivatives and the co-rotational parameter Recall the system: ( t + u )Q ( ξd(u) + Ω(u) ) (Q Id) + (Q Id)( ξd(u) Ω(u) ) 2ξ(Q + 1 Id)tr(Q u) = ΓH 3 ( ) tu + u u = ν u + p + QH HQ (ξ ( Q Id) H + ξh ( ) Q + 1 ) 3 Id) +2ξ ((Q )QH L ( Q Q tr(q2 ) ) u = 0 with H = L Q aq + b[q 2 tr(q2 ) 3 Id] cqtr(q 2 ). Worse than Navier-Stokes Where s the difficulty? If ξ = 0 maximal derivatives only If ξ 0 maximal derivatives+high power of Q

15 Regularity difficulties: the maximal derivatives and the co-rotational parameter Recall the system: ( t + u )Q ( ξd(u) + Ω(u) ) (Q Id) + (Q Id)( ξd(u) Ω(u) ) 2ξ(Q + 1 Id)tr(Q u) = ΓH 3 ( ) tu + u u = ν u + p + QH HQ (ξ ( Q Id) H + ξh ( ) Q + 1 ) 3 Id) +2ξ ((Q )QH L ( Q Q tr(q2 ) ) u = 0 with H = L Q aq + b[q 2 tr(q2 ) 3 Id] cqtr(q 2 ). Worse than Navier-Stokes Where s the difficulty? If ξ = 0 maximal derivatives only If ξ 0 maximal derivatives+high power of Q

16 Energy dissipation revisited I Cancellations that appear in the energy dissipation destroy some high derivatives...

17 Energy dissipation revisited I Cancellations that appear in the energy dissipation destroy some high derivatives... The cancellation lemma: Lemma For any symmetric matrices Q, Q R d d and Ω αβ = 1 2 (u α,β u β,α ) R d d (decaying fast enough at infinity so that we can integrate by parts, in the formula below, without boundary terms) we have: tr ( (ΩQ Q Ω) Q ) dx β (Q R d R d αγ Q γβ Q αγq γβ )uα dx = 0

18 Energy dissipation revisited I Cancellations that appear in the energy dissipation destroy some high derivatives... The cancellation lemma: Lemma For any symmetric matrices Q, Q R d d and Ω αβ = 1 2 (u α,β u β,α ) R d d (decaying fast enough at infinity so that we can integrate by parts, in the formula below, without boundary terms) we have: tr ( (ΩQ Q Ω) Q ) dx β (Q R d R d αγ Q γβ Q αγq γβ )uα dx = 0 Idea: differentiate the equations and to the highest derivatives the equations will keep the same structure (as the initial system) plus lower-order derivatives perturbation; thus we can use the high-derivatives cancellations available for the initial system to avoid the maximal derivatives Unlike in previous approaches (in complex fluids) we do not estimate the two equations separately but always together.

19 Energy dissipation revisited II Littlewood-Paley language: take take χ D(B(0, 1)) such that χ 1 on B(0, 1/2) and let ϕ(ξ) = χ(ξ/2) χ(ξ). Define q u = F 1 (ϕ(2 q ξ)fu), S q u = F 1 (χ(2 q ξ)fu) Then we have in the sense of distributions u = S 0 u + q>0 qu. Bony s paraproduct decomposition: q (ab) = S q 1 a q b + Σ q q 5[ q, S q 1a] q b +Σ q q 5(S q 1a S q 1 a) q q + Σ q >q 5 q (S q +2b q a) In our case, the equation in Q becomes: t q u ν q u = q p + L (S q 1 Q q Q q QS q 1 Q) ( q (u u) L q Q Q 1 ) 3 tr( Q Q) + perturbative (lower derivatives) terms

20 The rate of increase of the high norms- the Brezis-Gallouet trick and beyond Let y(t) = u(t) 2 H s + Q(t) 2 Hs, s > 1. An estimate of the form y (t) f (t)y(t) log(1 + y(t)) and f (t) Ce t would give y(t) Ce eet.

21 The rate of increase of the high norms- the Brezis-Gallouet trick and beyond Let y(t) = u(t) 2 H s + Q(t) 2 Hs, s > 1. An estimate of the form y (t) f (t)y(t) log(1 + y(t)) and f (t) Ce t would give y(t) Ce eet. Where does the logarithm come from? Brezis-Gallouet trick-logarithmic embedding! ( ) u L (R 2 ) u H ln(e + u H 1+ε) f (t) ln(1 + y) This works in the co-rotational case ξ = 0 after peeling out the maximal derivatives. If ξ 0 turns out that we can obtain an estimate of the form: ( ) y (t) f (t)y(t) 1 + Cξ log(1 + y(t))(1 + ln(1 + ln y))

22 The rate of increase of the high norms- the Brezis-Gallouet trick and beyond Let y(t) = u(t) 2 H s + Q(t) 2 Hs, s > 1. An estimate of the form y (t) f (t)y(t) log(1 + y(t)) and f (t) Ce t would give y(t) Ce eet. Where does the logarithm come from? Brezis-Gallouet trick-logarithmic embedding! ( ) u L (R 2 ) u H ln(e + u H 1+ε) f (t) ln(1 + y) This works in the co-rotational case ξ = 0 after peeling out the maximal derivatives. If ξ 0 turns out that we can obtain an estimate of the form: ( ) y (t) f (t)y(t) 1 + Cξ log(1 + y(t))(1 + ln(1 + ln y)) One logarithm is produced through through the logarithmic embedding. But the double logarithm?

23 The rate of increase of the high norms- the Brezis-Gallouet trick and beyond Let y(t) = u(t) 2 H s + Q(t) 2 Hs, s > 1. An estimate of the form y (t) f (t)y(t) log(1 + y(t)) and f (t) Ce t would give y(t) Ce eet. Where does the logarithm come from? Brezis-Gallouet trick-logarithmic embedding! ( ) u L (R 2 ) u H ln(e + u H 1+ε) f (t) ln(1 + y) This works in the co-rotational case ξ = 0 after peeling out the maximal derivatives. If ξ 0 turns out that we can obtain an estimate of the form: ( ) y (t) f (t)y(t) 1 + Cξ log(1 + y(t))(1 + ln(1 + ln y)) One logarithm is produced through through the logarithmic embedding. But the double logarithm?

24 The regularity result, in 2D Theorem Let s > 1 and ( Q, ū) H s+1 (R 2 ) H s (R 2 ). There exists a global a solution (Q(t, x), u(t, x)) of the coupled system, with restriction c > 0, subject to initial conditions Q(0, x) = Q(x), u(0, x) = ū(x) and Q L 2 loc (R +; H s+2 (R 2 )) L loc (R +; H s+1 (R 2 )), u L 2 loc (R +; H s+1 (R 2 ) L loc (R +; H s ). Moreover, we have: ) e e ect L Q(t, ) 2 H s (R 2 ) + u(t, ) 2 H s (R 2 ) (e C + Q H s+1 (R 2 ) + ū H s (R 2 ) (4) where the constant C depends only on Q, ū, a, b, c, Γ and L. If ξ = 0 the increase in time of the norms above can be made to be only doubly exponential. Remark: results locally in time are obtained by H. Abels, G. Dolzmann, Y. Liu in bounded domains

25 The difference between ξ = 0 (co-rotational) and ξ 0 Recall the system: ( t + u )Q ( ξd(u) + Ω(u) ) (Q Id) + (Q Id)( ξd(u) Ω(u) ) 2ξ(Q + 1 Id)tr(Q u) = ΓH 3 ( ) tu + u u = ν u α + p + QH HQ (ξ ( Q Id) H + ξh ( ) Q + 1 ) 3 Id) +2ξ ((Q )QH L ( Q Q tr(q2 ) ) u = 0 with H = L Q aq + b[q 2 tr(q2 ) 3 Id] cqtr(q 2 ).

26 A technical trick-how the double logarithm appears I We want to obtain y + ν u 2 2 H s + ΓL ( 2 Q 2 H s y ln(e + y) 1 + ln ( e + ln(e + y) )) with y = u 2 H s + Q 2 H s.

27 A technical trick-how the double logarithm appears I We want to obtain y + ν u 2 2 H s + ΓL ( 2 Q 2 H s y ln(e + y) 1 + ln ( e + ln(e + y) )) with y = u 2 H s + Q 2 H s. Attempt to estimate I def = S q Q 1 q u L q 2 Q L q Q L 2 q 5 (I q ( worst term) so we want to estimate 2 2qs I y) We have I S q Q 1 q 2 Q L q u 2 q Q L 2 (5) q 5 L ε L 1 ε Using the interpolation inequality with p = 1 ɛ 1 [1, 2], we obtain: f L 2p C p p p f L 2 f L 2 I C Sq Q q 2 q 5 L ε Q L q u 1 ε L 2 q u ε L 2 q Q L 2, where C > 0 is constant independent of ε (0, 1 2 ). Using Young s inequality we obtain: ( ( ) 2 I C Sq Q q 2 Q L 1 ε 2 q u L q 5 L ε 2 + ν q u L 2 + ΓL2 q Q 2 ) 100 L 2 2 2qs ( ( C Sq Q 2 L ε ) 2 Q L 1 ε 2 u H s + ν u 2 H 100 s + ΓL2 100 ) Q H s

28 A technical trick -how the double logarithm appears II We have obtained: y ( Q 2 L ε On the other hand using again the interpolation inequality Q L ) 2 1 ε y(t) (6) g L 2p C p p p g L 2 g L 2 we get: 2 ( 1 ε 1 ) 1 2ε Q 1 ε Q 2 L ε ε 1 ε Q 2 L 2 ( 1 ε ) 1 1 ε 2 (1 + Q L 2 ) Q 2 L 2 } {{ } def = f (t) where for the last inequality we assumed 0 < ε < 1. Then (6) becomes: 2 y (t) C(1 + f (t)) Q 2 L 2 [ (1 + f (t)) ln(e + y(t)) ] 1 1 ε ( 1 ε ) 1 1 ε y(t) Observing that the constants in the interpolation inequality do not depend on the space L p that we work with and denoting N def = ln(e + y) we choose ε def = (1 + ln N) 1 and observing that [N(1 + ln N)] 1+ 1 ln N CN(1 + ln N) for some constant C independent of N, the last inequality becomes: ϕ (t) C(1 + f (t)) 3 Q 2 ) (e ( L 2 ϕ(t) ln + ϕ(t) 1 + ln(e + ln(ϕ(t) + e) ))

29 Uniqueness of weak solutions - recently obtained by F. de Anna and A. Zarnescu Let (u 1, Q 1 ) and (u 2, Q 2 ) two weak solutions. We denote δu := u 1 u 2 and δq := Q 1 Q 2 while δs(q, u) = S(Q 1, u 1 ) S(Q 2, u 2 ). Similarly, we define δh(q), δf (Q), δτ and δσ. Thus (δu, δq) is a weak solution of t δq L δq = δs(q, u) + ΓδH(Q) δu Q 1 u 2 δq t δu δu + δπ = Ldiv{δτ + δσ} δu u 1 u 2 δu (δu, δq) t=0 = (0, 0). By L 2 energy estimates d dt ( δu 2 L 2 + δq 2 L 2) C (u, Q) L ( δu 2 L 2 + δq 2 L 2). Difficulty: we can not control u L by u L 2. Idea: We have to control a lower regularity energy E s (t) = ( δu(t) 2 H s + δq(t) 2 H s ), s (0, 1). We take s = 1 2.

30 Good terms: I = (δu Q 2, δq) H 1 C δu 2 Q 2 H cν δu 2 +cl δq 2. H 1 2 H 1 2 H 2 1 One logarithmic estimates: Let J = δq tr(q 2 u 2 ), δq. By using dyadic decomposition and we obtain S N (Q 2 ) L C N Q 2 H 1 J C NE 1 2 (t) + 2 N Q 2 H 1. We take N = ln(e 1 2 (t)) to obtain J f (t)( ln(e 1 2 (t))).

31 ( ((δq ) Two logarithmic estimates: K = 2ξ + 1 Id)tr(δQδH)), δu. 2 H 2 1 using the decomposition in low frequencies part and high frequencies part and the interpolation inequality f L 2 ɛ C f ɛ ɛ L2 f 1 ɛ L S 2 N f L C N f H 1 we obtain: K ( ) 1 CN 1 ε f (t) δu N u 2 ε H 2 1 L + ν δu 2. H 1 2 We chose N = ln δϕ H 2 1 with a double logarithm. Finally and ε = (1 + ln N) 1, we have a inequality E 1 2 (t) Cf (t)( ln E 1 2 (t)) ln( ln(e 1 2 (t))). The uniqueness follows by using Osgood theorem.

32 Nematic Phase - Ericksen Leslie Model Coupled system between the inhomonegenous and incompressible Navier-Stokes equation and the heat flow of harmonic maps into sphere. t ρ + div (ρu) = 0 t (ρu) + div(ρu u) ν u + Π = div ( d d ) (EL) t d + u d d = d 2 d divu = 0, d = 1 (u, d) t=0 = (u 0, d 0 ). ρ = ρ(t, x) R + is the density, u = u(t, x) R N the velocity field, Π = Π(t, x) R the pressure and d = d(t, x) S N 1 is the molecular orientation of the liquid crystal. - the inhomogeneous case by M. Hieber, M. Nesensohn, J. Prüss, and K. Schade by using a quasilinear approach

33 L p L q and Besov spaces Theorem (F. de Anna) Let 1 < p < d, ρ 0 L (R N N ) and (u 0, d 0) Ḃ 1+ p p,r (R N ) such that µ 1 ρ 1 0 L + (u 0, d 0) Ḃ 1+ N p p,r (EL) has a global weak solution. If (u 0, d 0) Ḃ 1+ N p +ɛ p,r c 0µ, Let ρ 1 = 1 + a. We recall that we want to solve the equation where tu ν u + Π = f (a, u) f (a, u) = a(ν u + Π + div( d d)) u u, for a density with a 0 L and u 0 B 1+ d p p,r. We note that a verify a transport equation so a L a 0 L. the solution is unique. Difficulty: define products as a u without any regularity on a and with a regularity as u in some Besov spaces? Idea: u 0 B 2σ p,r can be expressed as t σ S(t)u 0 L p L r (R +; dt t ), S(t) = et. In particular, (u 0, d 0) B 2 r p,r if (S(t)u 0, S(t)d 0) L r (R +; L p (R N )). (u L, d L ) L p 1 t (L q 1 x ), (u L, d L ) Lp 2 t (L q 2 ), (u L, d L ) L p 3 t (L q 3 Finally,we use maximal regularity t 0 S(t S)F (a, u)ds L p L q to have u L p L q. x ) then f (a, u) Lp 3 t (L q 3 x ) small.