Conference in Honour of Jerry Ericksen s 90th Birthday

Transcription

1 On the K 13 problem in the Oseen-Frank theory of nematic liquid crystals Arghir Dani Zarnescu University of Sussex and Simion Stoilow Institute of the Romanian Academy joint work with Stuart Day (U of Sussex) 24 October 2015 Conference in Honour of Jerry Ericksen s 90th Birthday

2 Nematic liquid crystals modelling 1 1 Simulation by C. Zannoni group

3 Nematic liquid crystals modelling 1 Continuum level modelling through functions n : Ω R 3 S 2 1 Simulation by C. Zannoni group

4 Nematic liquid crystals modelling 1 Continuum level modelling through functions n : Ω R 3 S 2 Many deficiencies, particularly related to discontinuities and defect patterns, but still the most widely-accepted theory. 1 Simulation by C. Zannoni group

5 Nematic liquid crystals modelling 1 Continuum level modelling through functions n : Ω R 3 S 2 Many deficiencies, particularly related to discontinuities and defect patterns, but still the most widely-accepted theory. Good experimental relevance. 1 Simulation by C. Zannoni group

6 Energy functionals Equilibrium configurations obtained as minimizers of the energy functional: F [n] = g(n, n) dx with g satisfying the physical requirements: Frame indiference: g(qn(p), Q n(p)q t ) = g(n(p), n(p)), for all Q SO(3) and at any point P Ω Material symmetry: g( n, n) = g(n, n) Ω

7 Energy functionals Equilibrium configurations obtained as minimizers of the energy functional: F [n] = g(n, n) dx with g satisfying the physical requirements: Frame indiference: g(qn(p), Q n(p)q t ) = g(n(p), n(p)), for all Q SO(3) and at any point P Ω Material symmetry: g( n, n) = g(n, n) One also needs the energy to be bounded from below in order to have a minimizer... Ω

8 Energy functionals Equilibrium configurations obtained as minimizers of the energy functional: F [n] = g(n, n) dx with g satisfying the physical requirements: Frame indiference: g(qn(p), Q n(p)q t ) = g(n(p), n(p)), for all Q SO(3) and at any point P Ω Material symmetry: g( n, n) = g(n, n) One also needs the energy to be bounded from below in order to have a minimizer... Energy density g: Oseen (1933), Frank (1958)-phenomenological derivation Nehring & Saupe (1971)-molecular expansion derivation Ω

9 The molecular expansions: the K αβ, I αβ terms (I) The average energy between the two volume elements at points P and P is given by f dτ dτ, where, in cylindrical coordinates f depends on the four quantities defining the position and direction of two vectors relative to each others, namely (r, x 3, n r, n α ).

10 The molecular expansions: the K αβ, I αβ terms (I) The average energy between the two volume elements at points P and P is given by f dτ dτ, where, in cylindrical coordinates f depends on the four quantities defining the position and direction of two vectors relative to each others, namely (r, x 3, n r, n α ). Switching back to cartesian coordinates a Taylor expansion gives: f = f f ν a νi n i,p x p p=1 ν=r,α i=1,2 3 p,q=1 f µν a νi a µj n i,p n j,q + f ν a νi n i,pq x p x q +... ν=r,α i=1,2 ν=r,α i=1,2

11 The molecular expansions: the K αβ, I αβ terms (I) The average energy between the two volume elements at points P and P is given by f dτ dτ, where, in cylindrical coordinates f depends on the four quantities defining the position and direction of two vectors relative to each others, namely (r, x 3, n r, n α ). Switching back to cartesian coordinates a Taylor expansion gives: f = f f ν a νi n i,p x p p=1 ν=r,α i=1,2 3 p,q=1 f µν a νi a µj n i,p n j,q + f ν a νi n i,pq x p x q +... ν=r,α i=1,2 ν=r,α i=1,2 Note that linear terms of the second derivatives of n make contributions of same order as quadratic terms of first derivatives.

12 The molecular expansions: the K αβ, I αβ terms (I) The average energy between the two volume elements at points P and P is given by f dτ dτ, where, in cylindrical coordinates f depends on the four quantities defining the position and direction of two vectors relative to each others, namely (r, x 3, n r, n α ). Switching back to cartesian coordinates a Taylor expansion gives: f = f f ν a νi n i,p x p p=1 ν=r,α i=1,2 3 p,q=1 f µν a νi a µj n i,p n j,q + f ν a νi n i,pq x p x q +... ν=r,α i=1,2 ν=r,α i=1,2 Note that linear terms of the second derivatives of n make contributions of same order as quadratic terms of first derivatives. The energy density at P is obtained by integrating on a sphere around P: g(p) = 3 p=1 i=1 2 c ip n i,p ( c ijpq n i,p n j,q + p,q=1 i,j=1 2 c ipq n i,pq ) +... i=1

13 The molecular expansions: the K αβ, I αβ terms (II) Coefficients and their symmetry: c 11 = c 22 = K 1 c 12 = c 21 = K 2 c 1111 = c 2222 = 1 2 K 11 c 1122 = c 2211 = 1 2 K 22 c 1133 = c 2233 = 1 2 K 33 c 1212 = c = 1 2 K 24 = 1 4 (K 11 K 22 ) c 113 = c 223 = 1 2 K 13 c 123 = c 213 = 1 2 K 23

14 The molecular expansions: the K αβ, I αβ terms (II) Coefficients and their symmetry: c 11 = c 22 = K 1 c 12 = c 21 = K 2 c 1111 = c 2222 = 1 2 K 11 c 1122 = c 2211 = 1 2 K 22 c 1133 = c 2233 = 1 2 K 33 c 1212 = c = 1 2 K 24 = 1 4 (K 11 K 22 ) c 113 = c 223 = 1 2 K 13 c 123 = c 213 = 1 2 K 23 Then the energy density is: g =K 1 div n K 2 (n curl n) K 11 (div n) 2 } {{ } :=I K 22 (n curl n) 2 } {{ } :=I K 33 (n n)2 } {{ } :=I 33 K 12 (n curl n)(div n) (K 22 + K 24 ) 2 K 23 n (n curl n) ((div n) 2 tr( n) 2 ) } {{ } :=I 24 +K 13 n div n } {{ } :=I 13

15 A bit of history Oseen(1933) introduced the K 13 term phenomenologically Frank (1958) ignored the K 13 term in his phenomenological theory Nehring and Saupe (1971) reintroduced the K 13 term on a molecular basis

16 A bit of history Oseen(1933) introduced the K 13 term phenomenologically Frank (1958) ignored the K 13 term in his phenomenological theory Nehring and Saupe (1971) reintroduced the K 13 term on a molecular basis Oldano and Barbero (1985) found an example involving the K 13 term and making the energy unbounded from below!

17 Types of distortions:twist,bend and splay K 11 div n 2 +K 22 n curl n 2 + K 33 n curl n 2 + (K 22 + K 24 ) ( tr( n) 2 (div n) 2) +K 13 div [(div n)n] } {{ }} {{ } saddle-splay(?) splay-bend(?) Twist Bend Splay

18 The equal elastic constants reduction The identity valid for n C 1 (Ω; S 2 ) gives: tr( n) 2 + n curl n 2 + n curl n 2 = n 2 3 K ii I i + K 13 I 13 + K 24 I 24 =(K 11 K 22 K 24 )(div n) 2 + K 22 n 2 + K 24 tr( n) 2 i=1 + (K 33 K 22 ) n curl n 2 + K 13 div [(div n)n]

19 The equal elastic constants reduction The identity valid for n C 1 (Ω; S 2 ) gives: tr( n) 2 + n curl n 2 + n curl n 2 = n 2 3 K ii I i + K 13 I 13 + K 24 I 24 =(K 11 K 22 K 24 )(div n) 2 + K 22 n 2 + K 24 tr( n) 2 i=1 + (K 33 K 22 ) n curl n 2 + K 13 div [(div n)n] Taking K 11 = K 22 = K 33 = K gives the energy that we will focus on: K F [n] = 2 n 2 + K 24 (tr( n) 2 (div n) 2 ) + K 13 div [(div n)n] dx Ω

20 Null Lagrangians and Nilpotent energies: Ericksen s work In 1962 J.L. Ericksen showed that an energy density that depends just on n or n (on Ω!) does not contribute to the Euler-Lagrange equations (is a nilpotent energy ) if and only if it can be expressed in the form: A + B i n j,i + 2 B pqi n i,p n j,q + Bdet( n) n j n j with A a constant and Bs functions of n only, such that B pqi = B qpi.

21 Null Lagrangians and Nilpotent energies: Ericksen s work In 1962 J.L. Ericksen showed that an energy density that depends just on n or n (on Ω!) does not contribute to the Euler-Lagrange equations (is a nilpotent energy ) if and only if it can be expressed in the form: A + B i n j,i + 2 B pqi n i,p n j,q + Bdet( n) n j n j with A a constant and Bs functions of n only, such that B pqi = B qpi. Furthermore, assuming the physical invariances under orthogonal transformations he reduced it to: H [(n k,k n i,i n i,k n k,i ] + 2H [n i,j n p,q n p n q n q,p n j,q n j n p ]

22 Null Lagrangians and Nilpotent energies: Ericksen s work In 1962 J.L. Ericksen showed that an energy density that depends just on n or n (on Ω!) does not contribute to the Euler-Lagrange equations (is a nilpotent energy ) if and only if it can be expressed in the form: A + B i n j,i + 2 B pqi n i,p n j,q + Bdet( n) n j n j with A a constant and Bs functions of n only, such that B pqi = B qpi. Furthermore, assuming the physical invariances under orthogonal transformations he reduced it to: H [(n k,k n i,i n i,k n k,i ] + 2H [n i,j n p,q n p n q n q,p n j,q n j n p ] But no second order derivatives...

23 The tame null-lagrangian I 24 Let K 24 (tr( n) 2 (div n) 2 ) = K 24 div (( n)n (div n)n) be the surface gradient of n and s n = n(id ν ν) div s n = tr( s n). These depend only on the boundary values of n!!! Then one can check that (( n)n (div n)n) ν = (( s n)n (div s n)n) ν

24 For K 13 0 the energy is unbounded from below (Oldano and Barbero 85): Take parallel plates at height 0 and d and consider configurations depending just on one variable, say z: n(z) = (cos(θ(z)), sin(θ(z))), z [0, d] where θ is the angle between n and the z axis (with z axis parallel on the plates). Consider equal elastic constants, K 24 = 0 but K Then the energy reduces to: d 0 K 2 ( ) 2 dθ + K 13 [sin(2θ(0))θ (0) sin(2θ(d))θ (d)] dz so we can get arbitrarily negative energy...

25 For K 13 0 the energy is unbounded from below (Oldano and Barbero 85): Take parallel plates at height 0 and d and consider configurations depending just on one variable, say z: n(z) = (cos(θ(z)), sin(θ(z))), z [0, d] where θ is the angle between n and the z axis (with z axis parallel on the plates). Consider equal elastic constants, K 24 = 0 but K Then the energy reduces to: d 0 K 2 ( ) 2 dθ + K 13 [sin(2θ(0))θ (0) sin(2θ(d))θ (d)] dz so we can get arbitrarily negative energy.....as long as θ(0), θ(d) {0, π 2 }.

26 For K 13 0 the energy is unbounded from below (Oldano and Barbero 85): Take parallel plates at height 0 and d and consider configurations depending just on one variable, say z: n(z) = (cos(θ(z)), sin(θ(z))), z [0, d] where θ is the angle between n and the z axis (with z axis parallel on the plates). Consider equal elastic constants, K 24 = 0 but K Then the energy reduces to: d 0 K 2 ( ) 2 dθ + K 13 [sin(2θ(0))θ (0) sin(2θ(d))θ (d)] dz so we can get arbitrarily negative energy.....as long as θ(0), θ(d) {0, π 2 }. But for some special boundary conditions, hometropic and tangential (θ(0), θ(d) {0, π } ) things are OK in this example. 2

27 A formal reduction: the K24, Ĩ13 K 24 I 24 + K 13 I 13 = K 24 [tr( n) 2 (div n) 2 ] + K 13 div [(div n)n] = K 24 [tr( n) 2 (div n) 2 ] K 13 [tr( n) 2 (div n) 2 ] + K 13 [ (div n) 2 + (n )(div n) ]

28 A formal reduction: the K24, Ĩ13 K 24 I 24 + K 13 I 13 = K 24 [tr( n) 2 (div n) 2 ] + K 13 div [(div n)n] = K 24 [tr( n) 2 (div n) 2 ] K 13 [tr( n) 2 (div n) 2 ] + K 13 [ (div n) 2 + (n )(div n) ] = (K 24 K 13 ) [tr( n) 2 (div n) 2 ] +K } {{ }} {{ } 13[tr( n) 2 + (n )(div n)] def = I K24 24

29 A formal reduction: the K24, Ĩ13 K 24 I 24 + K 13 I 13 = K 24 [tr( n) 2 (div n) 2 ] + K 13 div [(div n)n] = K 24 [tr( n) 2 (div n) 2 ] K 13 [tr( n) 2 (div n) 2 ] + K 13 [ (div n) 2 + (n )(div n) ] = (K 24 K 13 ) [tr( n) 2 (div n) 2 ] +K } {{ }} {{ } 13[tr( n) 2 + (n )(div n)] def = I K24 24 = K24 I 24 + K 13 (n i,j n j,i + n i n j,ji ) = K24 I 24 + K 13 (n i n j,i ),j = K24 I 24 + K 13 div [(n )n] } {{ } = K24 I 24 + K 13 Ĩ 13 def =Ĩ13

30 Natural Question: given that the energy is unbounded from below is there anything to do so to salvage the theory...?

31 Natural Question: given that the energy is unbounded from below is there anything to do so to salvage the theory...? Natural answer: how about local minimizers?

32 Natural Question: given that the energy is unbounded from below is there anything to do so to salvage the theory...? Natural answer: how about local minimizers? Second Natural Question: Local minimizers in what sense?

33 Functional spaces and critical points Question: What is the natural function space in which to study the Oseen-Frank functional? F [n] = Ω K 2 n 2 + K 24 (tr( n) 2 (div n) 2 ) + K 13 [(div n) 2 + (n )div n] dx

34 Functional spaces and critical points Question: What is the natural function space in which to study the Oseen-Frank functional? F [n] = Ω K 2 n 2 + K 24 (tr( n) 2 (div n) 2 ) + K 13 [(div n) 2 + (n )div n] dx The function spaces should be chosen so that the energy makes sense for functions in it...

35 Functional spaces and critical points Question: What is the natural function space in which to study the Oseen-Frank functional? F [n] = Ω K 2 n 2 + K 24 (tr( n) 2 (div n) 2 ) + K 13 [(div n) 2 + (n )div n] dx The function spaces should be chosen so that the energy makes sense for functions in it... One possible choice: A = W 1,2 (Ω; S 2 ) W 2,1 (Ω; S 2 )

36 Functional spaces and critical points Question: What is the natural function space in which to study the Oseen-Frank functional? F [n] = Ω K 2 n 2 + K 24 (tr( n) 2 (div n) 2 ) + K 13 [(div n) 2 + (n )div n] dx The function spaces should be chosen so that the energy makes sense for functions in it... One possible choice: A = W 1,2 (Ω; S 2 ) W 2,1 (Ω; S 2 ) Natural question revisited: Is this a good space in which to look for local minimizers/critical points?

37 Critical points and the relevance of the W 2,1 (Ω; S 2 ) space Take ϕ C c 1 (Ω; R 3 ). Then we have: d dt t=0e[ n + tϕ n + tϕ ] = K 2 Ω n : ϕ n 2 n ϕ dx + K }{{} =0 +K 13 }{{}... = 0 =0

38 Critical points and the relevance of the W 2,1 (Ω; S 2 ) space Take ϕ C c 1 (Ω; R 3 ). Then we have: d dt t=0e[ n + tϕ n + tϕ ] = K 2 Ω n : ϕ n 2 n ϕ dx + K }{{} =0 +K 13 }{{}... = 0 =0 with the K 24 term and the K 13 terms being zero because both n+tϕ n+tϕ and ( n+tϕ do not change on the boundary as t varies! n+tϕ )

39 Critical points and the relevance of the W 2,1 (Ω; S 2 ) space Take ϕ C c 1 (Ω; R 3 ). Then we have: d dt t=0e[ n + tϕ n + tϕ ] = K 2 Ω n : ϕ n 2 n ϕ dx + K }{{} =0 +K 13 }{{}... = 0 =0 with the K 24 term and the K 13 terms being zero because both n+tϕ n+tϕ and ( n+tϕ do not change on the boundary as t varies! n+tϕ ) Using that n has a bit more regularity than the standard harmonic maps, namely n W 2,1 we can integrate by parts, to get: K ( n + n n 2 ) ϕ dx = 0 2 Ω and thus, since ϕ was arbitrary: n + n n 2 = 0, a.e. in Ω.

40 The boundary reduction We take now test functions ϕ C 1 (Ω; R 3 ) with ϕ(x) = 0, x Ω (thus this time the gradient of ϕ might be non-zero on the boundary). We have: d dt t=0e[ n + tϕ n + tϕ ] = K 2 and the K and K 24 terms are zero. n : ϕ n 2 n ϕ dx +K 24 }{{}... Ω } {{ } =0 =0 3 + K 13 ϕ β,α n α (ν β n β n, ν ) ds = 0 Ω β,α=1

41 The boundary reduction We take now test functions ϕ C 1 (Ω; R 3 ) with ϕ(x) = 0, x Ω (thus this time the gradient of ϕ might be non-zero on the boundary). We have: d dt t=0e[ n + tϕ n + tϕ ] = K 2 n : ϕ n 2 n ϕ dx +K 24 }{{}... Ω } {{ } =0 =0 3 + K 13 ϕ β,α n α (ν β n β n, ν ) ds = 0 Ω β,α=1 and the K and K 24 terms are zero. We are left with ϕ β ν n, ν (ν β n β n, ν ) ds = 0 for β = 1, 2, 3. Ω

42 The boundary reduction We take now test functions ϕ C 1 (Ω; R 3 ) with ϕ(x) = 0, x Ω (thus this time the gradient of ϕ might be non-zero on the boundary). We have: d dt t=0e[ n + tϕ n + tϕ ] = K 2 n : ϕ n 2 n ϕ dx +K 24 }{{}... Ω } {{ } =0 =0 3 + K 13 ϕ β,α n α (ν β n β n, ν ) ds = 0 Ω β,α=1 and the K and K 24 terms are zero. We are left with ϕ β ν n, ν (ν β n β n, ν ) ds = 0 Ω for β = 1, 2, 3. Choosing ϕ suitably we get: n, ν {0, ±1}

43 Homeotropic boundary conditions and null Lagrangians We take n = ν on Ω. Then we have: Ω Ĩ 13 = [n i n j,i ],j dx Ω = n i n j,i ν j dσ = n i n j,i n j dσ Ω Ω 1 = 2 n i(n j n j ),i dσ = 0 Ω

44 Homeotropic boundary conditions and null Lagrangians We take n = ν on Ω. Then we have: Ω Ĩ 13 = [n i n j,i ],j dx Ω = n i n j,i ν j dσ = n i n j,i n j dσ Ω Ω 1 = 2 n i(n j n j ),i dσ = 0 Ω In terms of the original variables, we have: K 24 I 24 + K 13 I 13 dx = (K 24 K 13 ) [tr( n) 2 (div n) 2 ] Ω Ω } {{ } I 24

45 Tangential boundary conditions and null Lagrangians We take n ν = 0 on Ω. Then we have directly in terms of the original variables: K 24 I 24 + K 13 I 13 dx = K 24 (( n)n) ν K 24 (div n) }{{} n ν +K 13 ( n) }{{} n ν dσ Ω Ω =0 =0 = K 24 n i,j n j ν i dσ Ω = K 24 ν i,j n i n j dσ Ω

46 Tangential boundary conditions, I 24 and the second fundamental form Some differential geometry: for a smooth surface, S = Ω the normal ν : S S 2 is the Gauss map. dν P : T P S T ν(p) S 2 T P S The second fundamental form Π p is related to dν P through: II p (n, n) = dν P (n) n and for n = 1 this gives the curvature of the normal section. The last expression is ν i,j n i n j, i.e. the I 24 term for tangential boundary conditions 2 2 Diagram from Wikipedia-on curvature

47 The homeotropic versus normal boundary conditions, distinguishing K 13 and K 24 Thus we have: For homeotropic boundary conditions the K 13 contribution is the same as the K 24 contribution but with a negative sign. For tangential boundary conditions the K 13 contribution vanishes and the left K 24 contribution has a geometrical meaning. Thus one needs a suitable combination of the two types of boundary conditions to detect a non-trivial K 13 contribution!

48 Conclusions A natural question: Are there local minimizers for the Oseen-Frank energy with the K 13 0?

49 Conclusions A natural question: Are there local minimizers for the Oseen-Frank energy with the K 13 0? The answer depends on the functional space... A natural choice, W 1,2 (Ω; S 2 ) W 2,1 (Ω; S 2 ): forces the critical points to have homeotropic or tangential boundary conditions. For these boundary conditions the K 13 term either vanishes (tangential) or becomes part of K 24 (homeotropic). Possible relevance to nematic droplets... Open question: could a different functional spaces provide a different answer? Note though that it is a matter also of geometry not only of the functional space, i.e. some geometries and associated boundary conditions might be incompatible with the existence of local minimizers.

50 THANK YOU! and Happy 90th Birthday to Jerry!