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Some results on the nematic liquid crystals theory Marius Paicu University of Bordeaux joint work with Arghir Zarnescu Mathflows 2015, Porquerolles September 17, 2015 Complex fluids: Basic laws Incompressibility: u = 0 (1) where u


  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 ∂ t + ( u · ∇ ) u = ∇ · T − ∇ p (2) 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 ∂ t + ( u · ∇ ) u = ∇ · T − ∇ p (2) 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 � ∇ u + ( ∇ u ) tr � T = ν 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 � 1 R d | u ( x , t ) | 2 dx E ( t ) = + F ( t ) 2 ���� ���� � �� � total energy free energy of the molecules kinetic energy of the flow decreases in time.

  7. Some types of complex fluids ◮ Oldroyd-B: � ∂ t u + u ∇ u − ν ∆ u + ∇ p = µ ∇ · τ ∂ t τ + u ∇ τ + a τ + τ Ω − Ω τ − b ( D τ + τ D ) = µ 2 D ◮ The formal energy estimate is: 1 d dt ( µ 2 � u ( t ) � 2 L 2 + µ 1 � τ ( t ) � 2 L 2 ) + νµ 2 �∇ u ( t ) � 2 L 2 + a µ 1 � τ ( t ) � 2 L 2 2 ≤ | b |� D ( t ) � L ∞ � τ ( t ) � 2 L 2 -Lions-Masmoudi, Chemin-Masmoudi, Guillop´ e-Saut ◮ Smoluchowski Navier-Stokes systems ∂ v  ∂ t + v · ∇ v − ν ∆ v + ∇ p = ∇ · τ in Ω × (0 , T )      ∂ f ∂ t + v ∇ f + ∇ g · ( Wf ) − ∆ g f = 0 in Ω × (0 , T )     ∇ · v = 0 in Ω × (0 , T ) ,  where M γ (1) M γ (2) � � � τ ij = ij ( m ) f ( t , x , m ) dm + ij ( m 1 , m 2 ) f ( t , x , m 1 ) f ( t , x , m 2 ) dm and M W = c ij α ∂ j v i . R d | u ( t , x ) | 2 dx + ◮ The energy E ( t ) = 1 � � � M f log f dx dm decreases in time. R d 2 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 ) , � R d ( | u ( t , x ) | 2 + |∇ d ( t , x ) | 2 ) dx decreases in time. - The energy E ( t ) = 1 2 � R d | ∆ d − |∇ d | 2 d | 2 dx = E (0) E ( t ) + - 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: ◮ � ∂ t u + u ∇ u = ν ∆ u + ∇ p + ∇ · τ + ∇ · σ ∇ · u = 0 where we have the symmetric part of the additional stress tensor: � Q + 1 � � Q + 1 � τ = − ξ 3 Id H − ξ H 3 Id ∇ Q ⊙ ∇ Q + tr ( Q 2 ) +2 ξ ( Q + 1 � � 3 Id ) QH − L Id 3 and an antisymmetric part σ = QH − HQ where H = L ∆ Q − aQ + b [ Q 2 − tr ( Q 2 ) Id ] − cQ tr ( 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): ( ∂ t + u · ∇ ) Q − S ( ∇ u , Q ) = Γ H with = ( ξ D + Ω)( Q + 1 3 Id ) + ( Q + 1 3 Id )( ξ D − Ω) − 2 ξ ( Q + 1 S ( ∇ u , Q ) def 3 Id ) tr ( Q ∇ u )

  10. Energy dissipation and weak solutions-apriori bounds I ◮ The total energy � 2 |∇ Q | 2 + a L 2 tr ( Q 2 ) − b 3 tr ( Q 3 ) + c def 4 tr 2 ( Q 2 ) dx E ( t ) = R d � �� � free energy of the liquid crystal molecules � 1 R d | u | 2 ( t , x ) dx + 2 � �� � kinetic energy of the flow d is decreasing dt E ( t ) ≤ 0. ◮ 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 ◮ � d R d |∇ u | 2 dx dt E ( t ) = − ν � � 2 � L ∆ Q − aQ + b [ Q 2 − tr ( Q 2 ) Id ] − cQ tr ( Q 2 ) − Γ dx ≤ 0 R d tr 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 ) = ¯ u ( x ) ∈ L 2 ( R d ) , ∇ · ¯ u = 0 in D ′ ( R d ) (3) loc ( R + ; H 1 ) ∩ L 2 loc ( R + ; H 2 ) and The solution ( Q , u ) is such that Q ∈ L ∞ loc ( R + ; L 2 ) ∩ L 2 loc ( R + ; H 1 ) . u ∈ L ∞ 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: ( Q + 1 3 Id ) + ( Q + 1  � � � � ( ∂ t + u · ∇ ) Q − ξ D ( u ) + Ω( u ) 3 Id ) ξ D ( u ) − Ω( u )   − 2 ξ ( Q + 1 3 Id ) tr ( Q ∇ u ) = Γ H         � �  ∂ t u + u ∇ u = ν ∆ u + ∇ p + ∇ · QH − HQ  � � � Q + 1 Q + 1 � � �  −∇ · ξ 3 Id H + ξ H 3 Id     � �  ( Q + 1 ∇ Q ⊙ ∇ Q + 1 � 3 tr ( Q 2 ) �  +2 ξ ∇ · 3 ) QH − L ∇ ·      ∇ · u = 0 with H = L ∆ Q − aQ + b [ Q 2 − tr ( Q 2 ) Id ] − cQ tr ( Q 2 ). 3 ◮ Worse than Navier-Stokes ◮ Where’s the difficulty?

  13. Regularity difficulties: the maximal derivatives and the “co-rotational parameter” ◮ Recall the system: ( Q + 1 3 Id ) + ( Q + 1  � � � � ( ∂ t + u · ∇ ) Q − ξ D ( u ) + Ω( u ) 3 Id ) ξ D ( u ) − Ω( u )   − 2 ξ ( Q + 1 3 Id ) tr ( Q ∇ u ) = Γ H         � �  ∂ t u + u ∇ u = ν ∆ u + ∇ p + ∇ · QH − HQ  � � � Q + 1 Q + 1 � � �  −∇ · ξ 3 Id H + ξ H 3 Id     � �  ( Q + 1 ∇ Q ⊙ ∇ Q + 1 � 3 tr ( Q 2 ) �  +2 ξ ∇ · 3 ) QH − L ∇ ·      ∇ · u = 0 with H = L ∆ Q − aQ + b [ Q 2 − tr ( Q 2 ) Id ] − cQ tr ( Q 2 ). 3 ◮ 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: ( Q + 1 3 Id ) + ( Q + 1  � � � � ( ∂ t + u · ∇ ) Q − ξ D ( u ) + Ω( u ) 3 Id ) ξ D ( u ) − Ω( u )   − 2 ξ ( Q + 1 3 Id ) tr ( Q ∇ u ) = Γ H         � �  ∂ t u + u ∇ u = ν ∆ u + ∇ p + ∇ · QH − HQ  � � � Q + 1 Q + 1 � � �  −∇ · ξ 3 Id H + ξ H 3 Id     � �  ( Q + 1 ∇ Q ⊙ ∇ Q + 1 � 3 tr ( Q 2 ) �  +2 ξ ∇ · 3 ) QH − L ∇ ·      ∇ · u = 0 with H = L ∆ Q − aQ + b [ Q 2 − tr ( Q 2 ) Id ] − cQ tr ( Q 2 ). 3 ◮ Worse than Navier-Stokes ◮ Where’s the difficulty? ◮ If ξ = 0 maximal derivatives only ◮ If ξ � = 0 maximal derivatives+high power of Q

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