GGlobal minimisers of cholesteric liquid crystal systems
Stephen BedfordOctober 16, 2018
Abstract
In this paper we examine the modelling and minimisation of cholesteric liquid crystals systemswithin the Oseen-Frank theory. We focus on a cuboid domain with the frustrated boundary condi-tions n (0 , x , y ) = (1 , ,
0) and n (1 , x , y ) = (0 , , ffi ciently long. Finally we show that our analysiseasily translates over the study the global stability of the constant state n ( x , y , z ) = (0 , ,
1) withunfrustrated boundary conditions.
The mathematical study of liquid crystals through the calculus of variations has received increasingattention in recent decades. In this area, the main supposition is that the free energy of the systems canbe modelled by an integral functional. Unfortunately there is no broad consensus as to the correct formof the free energy density or even what the order parameter should be. The Oseen-Frank theory uses theanisotropic axis of the molecules, the director, to model the energy [17, 9]. Ericksen’s theory uses thedirector and a scalar order parameter which measures the variance from the director [8]. The Landau-de Gennes and Q-Tensor theories use matrix order parameters, which are symmetric and traceless, toproperly respect the head-to-tail symmetry of the constituent molecules [6, 2]. In general there hasbeen a great deal of attention on how to best model liquid crystal systems. However there are far fewermathematical studies of these theories for a given setup. The works of Majumdar, Robbins, Zyskin andNewton are notable exceptions [14, 15]. In this paper we use the Oseen-Frank free energy functionaland study two specific problems. We will focus on cholesteric liquid crystals because they are moregeneral, and less well understood, than nematic liquid crystals. We consider a liquid crystal sample heldin a general cuboid domain
Ω = ( − L , L ) × ( − L , L ) × (0 , L ) . Then we shall investigate global minimisers with two di ff erent sets of boundary conditions on the topand bottom plates. The first has homeotropic boundary conditions on both faces while the second usesplanar non-degenerate boundary conditions on the bottom face instead, so that n ( x , y , = e . On theother four faces we will simply impose periodic boundary conditions. To the best of our knowledge,the frustrated boundary condition case has not been studied rigorously with the vectorial calculus ofvariations before. The other case involving homeotropic anchoring was considered by Gartland et al.[10] in a purely one-dimensional formulation. In contrast, the results that we will prove in Section 5apply in three spatial dimensions but they still qualitatively agree with the conclusions from [10]. Animportant aspect of the Oseen-Frank model that we shall consider is the assumptions on the elasticconstants, K i . The general free energy density is given by w ( n , ∇ n ) : = K ( ∇ · n ) + K ( n · ∇ × n + t ) + K | n × ∇ × n | + ( K + K ) (cid:16) tr (cid:16) ∇ n (cid:17) − ( ∇ · n ) (cid:17) . a r X i v : . [ m a t h . A P ] N ov INTRODUCTION
Naturally the values of the elastic constants are an important factor in determining the minimisers. Theusual assumption made in the literature, see for example [21, 22], is that it should be the case that w ( n , ∇ n ) (cid:62)
0, and equals zero if and only if n is any fixed rotation of cos( tz )sin( tz )0 . This logically leads to the following set of inequalities, called Ericksen’s inequalities [7, 22], K , K , K (cid:62) , K + K = t (cid:44) K (cid:62) K + K , K (cid:62) | K | , K (cid:62) t = . (1)This logic has possible flaws for two reasons. The first is that it is slightly illogical from a mathematicalpoint of view to derive an integral functional assuming a given minimum of the Lagrangian; it is prefer-able for the minimum state to emerge naturally from the form of the free energy density. Secondly thetwo sets of inequalities in (1) do not exhibit a continuous dependence on the cholesteric twist parameter t . Nematic liquid crystals are, in some sense, a limit of cholesteric liquid crystals as t →
0, hence onewould expect a set of inequalities which would respect this limit. The restriction that K + K =
0, sothat we have no contribution from the saddle-splay term, is particularly limiting for cholesterics. Inter-estingly, Nehring & Saupe [16] showed that using a di ff erent approach to deriving the cholesteric freeenergy density, one deduces that K + K (cid:44)
0. In this paper we will keep the contributions from thesaddle-splay term and when we apply a one constant approximation, we will suppose that K = K = K = K , K = n = e ), so long as the height of the cell is su ffi ciently small. It is generally thought thatthe fingers arise from the competition of two local energy minimisers; the constant state and a morehelical state (an example of which we will find explicitly later in the paper). Near the isotropic transitiontemperature, some cholesteric liquid crystals form cubic lattices of cylinders called blue phases [12, 23]which possess interesting optical properties. They are called blue phases because the periodicity ofthe lattice is the same length as the wavelength of visible light, so it reflects a certain visible colour.The first such studied liquid crystals reflected blue light, hence their name. Even more recently, localstructures called torons, which comprise a small volume of tightly twisted molecules, have been inducedby a high power laser and shown to be stable [19]. These torons appear to be formed when the buildingblock of the blue phase, the double twist cylinder, is twisted upon itself to create a torus. This structurenecessarily results in two defects.Modern technological advances have made it possible to unambiguously reconstruct the director fieldwithin any of the above cholesteric structures [20]. A experimental technique called fluoresence confocalpolarizing microscopy (FCPM) is used to perform this reconstruction. As a result, the molecular makeupof all of these structures have been found either experimentally or through simulations. However therehas been little analytical progress using them to find local or global energy minimisers of an entiresystem. The work which follows in this paper is aimed at being able to predict physically viable statesof cholesteric liquid crystal problems. 2 NOTATION AND PRELIMINARIES
Although we will not be able to prove the stability of multi-dimensional minimisers in this paper directly,we will establish a great deal of groundwork by finding one-dimensional minimisers and investigatingtheir stability. In Section 3 we study the Oseen-Frank free energy with the frustrated boundary conditions n | z = = e and n | z = L = e . We find the unique global minimisers of the one dimensional problem forgeneral elastic constants. From then on we consider the one constant Oseen-Frank energy I ( n ) = (cid:90) Ω |∇ n | + t n · ∇ × n + t dx . (2)In Section 4 we prove the first of our two main stability results, Theorem 11: these one-dimensionalstates are in fact the global minimisers of (2) so long as t is su ffi ciently small. This analytical predictionis consistent with the notion that cholesterics liquid crystals with suitably long pitch can be considered asa perturbation of nematic liquid crystals. We then investigate the relationship between the saddle-splayterm, the function space and periodicity in the lateral directions with Proposition 12. This result givesa subtle indication that there may be a function space modelling issue for cholesterics, this idea will befurther explored in an upcoming paper [4].In the final section we examine the same problem but with homeotropic boundary conditions on boththe top and bottom faces of the cuboid. This problem is a little di ff erent in that for any t ∈ R , n = e isalways a solution of the Euler-Lagrange equation. Just as in the frustrated case we prove that n = e isa global minimiser of the problem provided that t is su ffi ciently small. This is an interesting supplementto the local stability result [3, Section 7] which demonstrated that n = e is a strong local minimiser if t < π and not a local minimiser when t > π . All of these stability results give credence to the notionthat cholesteric liquid crystals with long pitch can be understood as a perturbation of the more easilyunderstood nematic liquid crystal systems. Throughout this paper, unless stated otherwise, Ω ⊂ R is the cuboid domain Ω : = ( − L , L ) × ( − L , L ) × (0 , . We note that we can assume, without loss of generality, that the domain has height 1 because scaling thedomain simply corresponds to a linear scaling of the variable t . Then the problem that we are consideringis the minimisation of I ( n ) : = (cid:90) Ω w ( n , ∇ n ) dx (3)over A : = (cid:26) n ∈ W , (cid:16) Ω , S (cid:17) (cid:12)(cid:12)(cid:12)(cid:12) n | z = = e , n | z = = e , n | x = − L = n | x = L , n | y = − L = n | y = L (cid:27) , (4)where w ( n , ∇ n ) : = K ( ∇ · n ) + K ( n · ∇ × n + t ) + K | n × ∇ × n | + ( K + K ) (cid:16) tr( ∇ n ) − ( ∇ · n ) (cid:17) . The Sobolev space W , (cid:16) Ω , S (cid:17) is simply the set of functions n ∈ W , (cid:16) Ω , R (cid:17) such that | n | = Definition 1.
A fucntion n ∈ A is a strong local minimiser of the functional I if there exists an (cid:15) > such that if m ∈ A and || n − m || ∞ < (cid:15) then I ( m ) (cid:62) I ( n ) . Definition 2.
A fucntion n ∈ A is a weak local minimiser of the functional I if there exists an (cid:15) > suchthat if m ∈ A and || n − m || , ∞ < (cid:15) then I ( m ) (cid:62) I ( n ) . ONE VARIABLE MINIMISERS
We define the set of test functions of our problem to be Var A whereVar A : = (cid:26) v ∈ C ∞ (cid:16) Ω , R (cid:17) ∩ W , (cid:16) Ω , R (cid:17) (cid:12)(cid:12)(cid:12)(cid:12) v | z = = v | z = = , v | x = − L = v | x = L , v | y = − L = v | y = L (cid:27) . Theorem 3. [3] Suppose that n ∈ A ∩ W , ∞ (cid:16) Ω , R (cid:17) is a weak local minimiser of I. Then for every v ∈ Var A , dd (cid:15) I (cid:32) n + (cid:15) v | n + (cid:15) v | (cid:33)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:15) = = and d d (cid:15) I (cid:32) n + (cid:15) v | n + (cid:15) v | (cid:33)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:15) = (cid:62) . Theorem 4. [3] Suppose that n ∈ A ∩ C (cid:16) Ω , R (cid:17) . If there exists some δ > such thatdd (cid:15) I (cid:32) n + (cid:15) v | n + (cid:15) v | (cid:33)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:15) = = and d d (cid:15) I (cid:32) n + (cid:15) v | n + (cid:15) v | (cid:33)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:15) = (cid:62) δ || v − ( v · n ) n || , for every v ∈ Var A , then n is a strong local minimiser of I. From here on we will be assuming that t (cid:62) t simplydetermines the direction of the helix that the molecules prefer to form. Thus if t < x (cid:48) = − x to derive a situation where t (cid:62)
0. In this section we will assume that n is a function of z only, which will simplify the free energy significantly, and to avoid degeneracy wealso assume that K , K , K >
0. For maximum simplification we rewrite the director as a function of itsEuler angles θ ( z ) , φ ( z ). By substituting n = cos φ cos θ sin φ cos θ sin θ , into (3), we lose any contribution from the saddle-splay term since it vanishes for functions whichdepend on only one variable. Then we simplify to find I ( n ) = L L (cid:90) (cid:16) K cos θ + K sin θ (cid:17) θ (cid:48) + cos θ (cid:104)(cid:16) K cos θ + K sin θ (cid:17) φ (cid:48) − tK φ (cid:48) (cid:105) + t K dz . (5)Converting the problem into one involving the Euler angles is convenient from a computational stand-point as it deals with the unit vector constraint but there are regularity issues involved, especially at thepoles where φ is undefined. Intuitively it seems sensible that θ and φ should have a comparable regularityto the director field n . However by simple calculations we can see that (cid:90) Ω |∇ n | dx = (cid:90) Ω |∇ θ | + cos θ |∇ φ | dx < ∞ ∀ n ∈ W , ( Ω , S ) . This means that formally we might only expect ∇ θ, cos θ ∇ φ ∈ L ( Ω , R ) so that θ ∈ W , but we neednot have that ∇ φ ∈ L . The following proposition proves this regularity for θ and fortunately the preciseform of the Lagrangian in (5) will then allow us to deal with the terms involving φ relatively simply. Proposition 5.
Let d ∈ N and suppose that Ω ⊂ R d is an open set. If n ∈ W , (cid:16) Ω , S (cid:17) then θ ∈ W , (cid:18) Ω , (cid:20) − π , π (cid:21)(cid:19) where θ is the Euler angle out of the ( x , y ) -plane associated with n . ONE VARIABLE MINIMISERS
Proof
The first element of the proof is to show the following pointwise inequality for any n , n ∈ S : | n − n | (cid:62) C | θ − θ | . (6)We take n , n ∈ S then in terms of their Euler angles n = cos θ cos φ cos θ sin φ sin θ n = cos θ cos φ cos θ sin φ sin θ , (7)where θ , θ ∈ (cid:104) − π , π (cid:105) and φ , φ ∈ [0 , π ). We substitute (7) into the left hand side of (6) to find | n − n | = − n · n = − θ sin θ − θ cos θ cos( φ − φ ) (cid:62) − θ sin θ − θ cos θ = − cos( θ − θ )) . (8)A simple exercise in L’Hˆoptial’s rule gives us thatlim x → − cos( x ) x = . Therefore inf x ∈ [ − π,π ] − cos( x ) x : = C > . (9)Then by combining (9) and (8) we see that | n − n | (cid:62) C | θ − θ | . We can now conclude the assertionby using the di ff erence quotient characterisation of W , [13, Thm 10.55]. We take our n ∈ W , (cid:16) Ω , S (cid:17) and let x ∈ Ω be a point where n is well defined, then (cid:90) Ω | n ( x ) − n ( x ) | dx (cid:62) C (cid:90) Ω | θ ( x ) − θ ( x ) | dx < ∞ , so that θ ∈ L ( Ω , R ). The form of the inequality (6) allows us to go further and say that || n ( · + h e j ) − n ( · ) || , Ω (cid:48) (cid:62) C || θ ( · + h e j ) − θ ( · ) || , Ω (cid:48) for every Ω (cid:48) ⊂⊂ Ω , h ∈ R and j ∈ { , . . . , n } . Therefore as n satisfies the di ff erence quotient characteri-sation, so must θ , which gives us the result. (cid:3) Now that we know the precise regularity of θ we can return to (5). Clearly with regards to minimisation,the final constant term in (5) is unimportant as its energy contribution is the same for all states. Weapply a very direct approach to minimising the contributions from φ . If we look simply at the centralterm involving φ we can minimise it formally. To do this we note that f ( x ) : = ax − bx (cid:62) f (cid:32) b a (cid:33) = − b a . Thus if we simply set φ (cid:48) = K t (cid:16) K cos θ + K sin θ (cid:17) φ (0) = , (10)then we have minimised the contribution with regards to φ (cid:48) and have the following inequality I ( n ) (cid:62) L L (cid:90) (cid:16) K cos θ + K sin θ (cid:17) θ (cid:48) − K t cos θ (cid:16) K cos θ + K sin θ (cid:17) + K t dz = : 4 L L (cid:90) f ( θ ) θ (cid:48) − g ( θ ) + K t dz . ONE VARIABLE MINIMISERS
This reduces the problem to a one-dimensional functional of one variable which is easier to investigate.In order to find its minimum we first need to prove a preliminary lemma whereby we will be able showthat it is not energetically favourable for θ to achieve either of its bounds, ± π , except at z =
1. For thelemma we introduce the following notation: F α ( v ) : = (cid:90) α f ( v ) v (cid:48) − g ( v ) + K t dz , A α : = (cid:26) v ∈ W , (0 , α ) (cid:12)(cid:12)(cid:12) v (0) = , v ( α ) = π (cid:27) . Lemma 6. If < α < β < ∞ then inf v ∈A α F α ( v ) > inf w ∈A β F β ( w ) . Proof
We know that inf v ∈ R f ( v ) = min { K , K } > α , theproblem of minimising F α over A α has a smooth minimiser [5, Ch.4.2] which satisfies the followingEuler-Lagrange equation (2 f ( v ) v (cid:48) ) (cid:48) = − g (cid:48) ( v ) . This equation has the first integral f ( v ) v (cid:48) + g ( v ) = C α , where C α is a constant. For the consistency of this equation C α (cid:62) K t because f ( v ( z )) > , g ( v ( z )) (cid:54) K t , and g ( v (0)) = K t . Importantly we need to eliminate the possibility that C α = K t so that we can deduce v (cid:48) has a fixedsign. If C α = K t , then we would be looking to solve v (cid:48) = K t − g ( v ) f ( v ) v (0) = . (11)We know that the minimiser we are searching for is smooth, hence a simple argument looking at intervalswhere v (cid:48) has fixed sign shows that the unique solution to (11) is v = C α > K t and v (cid:48) has a fixed sign, so taking the square-root of (11) gives v (cid:48) = (cid:18) C α − K t cos v ( K cos v + K sin v ) (cid:19) (cid:16) K cos v + K sin v (cid:17) : = k ( z , v ) v (0) = . We can use the Picard-Lindel¨of Theorem here to deduce existence and uniqueness of this ODE [11, p.8].It is applicable since the function k ( z , v ) is smooth in its arguments, hence Lipschitz on any bounded set.This means that we have a unique solution of this ODE, v α , given implicitly by (cid:90) v α ( z )0 (cid:16) K cos u + K sin u (cid:17) (cid:18) C α − K t cos u ( K cos u + K sin u ) (cid:19) du = z . We finalise the definition of v α by applying the top boundary condition of v α ( α ) = π . This is possiblebecause the functional η ( C ) : = (cid:90) π (cid:16) K cos u + K sin u (cid:17) (cid:18) C − K t cos u ( K cos u + K sin u ) (cid:19) du , ONE VARIABLE MINIMISERS is a strictly monotone decreasing function satisfying η ( K t ) = ∞ and η ( C ) → C → ∞ . This means that there is a unique solution to the equation η ( C α ) = α . Therefore we have established thatfor all α > v α satisfying F α ( v α ) = inf v ∈A α F α ( v ) . Now we note that for β > α > v ( z ) : = z ∈ (0 , β − α ) v α ( z − ( β − α )) if z ∈ ( β − α, β ) ∈ A β , (12)with the property that F β ( v ) = F α ( v α ) = inf w ∈A α F α ( w ) . (13)An immediate consequence of (13) is that F β ( v β ) = inf u ∈A β F β ( u ) (cid:54) F β ( v ) = inf w ∈A α F α ( w ) . (14)To conclude that this inequality is strict we need to show that v (cid:44) v β . However we know this to be truebecause C β > K t tells us that v (cid:48) β ( z ) (cid:44) z ∈ [0 ,
1] whereas v (cid:48) ( z ) = z ∈ (0 , β − α ) in (12).Hence the inequality in (14) is strict. (cid:3) We can now prove the main result of this section by finding the one variable global minimisers.
Theorem 7.
The function θ ∗ = θ ∗ ( z ) given implicitly by the equation (cid:90) θ ∗ (cid:16) K cos u + K sin u (cid:17) (cid:18) C − K t cos u ( K cos u + K sin u ) (cid:19) du = z θ ∗ (1) = π , (15) with φ ∗ as defined by (10) , determines a director field n ∗ which is the global minimum of I over admis-sible functions which depend only on z. Moreover the constant C in (15) is determined by the uniquesolution to (cid:90) π (cid:16) K cos u + K sin u (cid:17) (cid:18) C − K t cos u ( K cos u + K sin u ) (cid:19) du = . (16) Proof
We take an arbitrary n ∈ A which depends only on z . We know that I ( n ) (cid:62) L L (cid:82) (cid:16) K cos θ + K sin θ (cid:17) θ (cid:48) − K t cos θ ( K cos θ + K sin θ ) + K t dz = L L (cid:82) f ( θ ) θ (cid:48) − g ( θ ) + K t dz . (17)We also know that θ ∈ W , (0 , ⊂⊂ C [0 , θ is a continuous function. Lemma 6 tells us that if θ ( α ) = ± π for some 0 < α < (cid:90) f ( θ ) θ (cid:48) − g ( θ ) + K t dz (cid:62) (cid:90) f ( θ ∗ ) θ ∗(cid:48) − g ( θ ∗ ) + K t dz . (18)7 LONG CHOLESTERIC PITCH where θ ∗ = v in the notation of Lemma 6. Additionally the uniqueness of the minimiser from Lemma6 for each positive α means that we have equality in (18) if and only if θ = θ ∗ . For completeness weshould note that given a smooth θ ∗ , the equation involving φ , (10), defines a unique smooth function φ ∗ ( z ) by Picard-Lindel¨of [11, p.8]. So now we can say that I ( n ) (cid:62) L L (cid:90) f ( θ ∗ ) θ ∗(cid:48) − g ( θ ∗ ) + K t dz = I ( n ∗ ) , with equality if and only if θ = θ ∗ and also φ (cid:48) = K t (cid:16) K cos θ + K sin θ (cid:17) and φ (0) = . Hence I ( n ) (cid:62) I ( n ∗ ) with equality if and only if θ = θ ∗ and φ = φ ∗ , or equivalently n = n ∗ . (cid:3) Remark 8.
Theorem 7 concludes the discussion on the minimisers of functions of z alone. Howevernot having a closed form for these minimisers is a significant problem since we want to find explicitminimisers in order to compare with experimental observations. In order to achieve this we need toassign values to the elastic constants, we will use the one constant approximation because it makes theanalysis most tractable.
When we apply the one constant approximation to the general Oseen-Frank free energy, the cholestericfree energy reduces to I ( n ) = K (cid:90) Ω |∇ n | + t n · ∇ × n + t dx , (19)with the same set of admissible functions A : = (cid:110) n ∈ W , ( Ω , S ) (cid:12)(cid:12)(cid:12) n | z = = e , n | z = = e , n | x = − L = n | x = L , n | y = − L = n | y = L (cid:111) . (20)For ease we will drop the now unimportant elastic constant K from our functional. We already know theform of the minimiser amongst functions of one variable of this functional from Theorem 7. Hence webegin this section with a simple corollary which is just a specific case of Theorem 7. Corollary 9.
The function, θ ( z ) , given implicitly by the equation (cid:90) θ ( z )0 (cid:0) D − t cos u (cid:1) du = z , (21) where the constant D is defined by (cid:90) π (cid:0) D − t cos u (cid:1) du = , (22) and φ = tz , (23) determines a director field n ∗ which is the unique global minimum of (19) over n ∈ A which dependonly on z. Proof LONG CHOLESTERIC PITCH
By substituting K = K = K = K into (15) and (16), and by setting D = CK , we immediately obtain the result (cid:3) Before we proceed further, to better understand our one variable minimiser it is constructive to computethe explicit solutions of (21) and (22) for various values of t . For all of the numerical calculationsthe standard ODE solver in Matlab was used, a fourth order Runge-Kutta method. The following fourfigures: Figures 1a, 1b, 1c, and 1d, each show a plot of θ against the height of the cell, each for a di ff erentvalue of the cholesteric twist strength, t . (a) t = . t = t =
10 (d) t = Figure 1: θ plots for various values of t The trend is clear to see. Small values of t correspond to a slight bending of the case t = ⇒ θ = π z ).Larger values of t have much more curvature towards the top of the cell. The reason for this can be seenby looking at (19) and substituting φ (cid:48) = t . When we do so, we find that the energy, purely in terms of θ ,is given by (cid:90) θ (cid:48) + t sin θ dz . Hence it is no surprise that as t increases the second term becomes more dominant, meaning that θ remains small so as to minimise sin θ .We return now to the problem set out in (19) and (20). We have found the global minimiser of this prob-lem over functions of z alone. The very natural question from that point is to ask whether these functions,obtained numerically in Figure 1, are the global minimisers of the entire problem. The question has been9 LONG CHOLESTERIC PITCH partly answered using Theorem 11 below.Firstly and most importantly, in order to check whether our candidate minimiser is a global minimiserwe need to check that it satisfies the Euler-Lagrange equation. Hence we need to derive the equation,which we do in the following lemma.
Lemma 10.
If we have n ∈ A ∩ W , ∞ (cid:16) Ω , S (cid:17) which is a weak local minimiser of (19) . Then it satisfiesthe following Euler-Lagrange equation (in the sense of distributions) ∆ n − t ∇ × n + (cid:16) |∇ n | + t n · ∇ × n (cid:17) n = , (24) together with the natural boundary conditions ( ∇ n ) ν | x = − L = ( ∇ n ) ν | x = L , ( ∇ n ) ν | y = − L = ( ∇ n ) ν | y = L . (25) Proof
We take some φ ∈ Var A and note that n + (cid:15) φ | n + (cid:15) φ | → n in W , ∞ as (cid:15) → . Therefore, because n is a weak local minimiser, it must be the case that the first variation vanishes0 = dd (cid:15) I (cid:32) n + (cid:15) φ | n + (cid:15) φ | (cid:33)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:15) = = (cid:90) Ω ∇ φ : ∇ n − ( n · φ ) (cid:16) |∇ n | + t n · ∇ × n (cid:17) + t ( φ · ∇ × n + n · ∇ × φ ) dx . By using integration by parts, together with the boundary conditions for both φ and n , we can deducethat (cid:82) Ω φ · ∇ × n dx = (cid:82) Ω n · ∇ × φ dx . Hence0 = (cid:90) Ω − φ · (cid:16) ∆ n − t ∇ × n + n (cid:16) |∇ n | + t n · ∇ × n (cid:17)(cid:17) dx + (cid:90) ∂ Ω φ · [( ∇ n ) ν ] dS , for any φ ∈ Var A . This proves equations (24) and (25) and completes the proof. (cid:3) Theorem 11.
There exists some t ∗ > such that whenever t (cid:54) t ∗ , the function defined by (21) , (22) , and (23) is the unique global minimiser ofI ( n ) = (cid:90) Ω |∇ n | + t n · ∇ × n + t dx , over all admissible functions n ∈ A . Proof
We have our candidate minimiser of this functional n ∗ which is given by the equation n ∗ = cos θ cos( tz )cos θ sin( tz )sin θ , (26)where θ is a solution of θ (cid:48)(cid:48) = t cos θ sin θ, θ (0) = , θ (1) = π . (27)However we can find the first integral of (27) by multiplying by θ (cid:48) and integrating to deduce that (for δ t > θ (cid:48) + t cos θ = δ t + t , θ (0) = , θ (1) = π . (28)The proof that this is indeed the global minimum has two main calculations.10 LONG CHOLESTERIC PITCH • Firstly we will show the following relation for an arbitrary n ∈ A H ( n − n ∗ ) = I ( n ) − I ( n ∗ ) , where the functional H is given by H ( v ) : = (cid:90) Ω |∇ v | + t v · ∇ × v − (cid:16) |∇ n ∗ | + t n ∗ · ∇ × n ∗ (cid:17) | v | dx . H takes functions from B which is the set given by B : = (cid:26) v ∈ W , (cid:16) Ω , R (cid:17) (cid:12)(cid:12)(cid:12)(cid:12) v | z = = , v | z = = , v | x = − L = v | x = L , v | y = − L = v | y = L (cid:27) . • Secondly we will show that if t is small enough then for all v ∈ B H ( v ) (cid:62) γ t (cid:90) Ω |∇ v | dx , for some γ t > I ( n ) − I ( n ∗ ) Calculation
Take an arbitrary n ∈ A then (cid:90) Ω |∇ n − ∇ n ∗ | dx = (cid:90) Ω |∇ n | + |∇ n ∗ | − ∇ n : ∇ n ∗ dx . (29)Now we can use the Euler-Lagrange equation to rewrite the final term in the relation above. The strongform of the Euler-Lagrange equation which is satisfied by our function n ∗ is ∆ n ∗ − t ∇ × n ∗ + (cid:16) |∇ n ∗ | + t n ∗ · ∇ × n ∗ (cid:17) n ∗ = . We multiply this equation by n and then we can use integration by parts to rewrite the equation as follows0 = (cid:90) Ω n · ∆ n ∗ − t n · ∇ × n ∗ + λ ( n · n ∗ ) dx = (cid:90) Ω n i n ∗ i , − t n · ∇ × n ∗ + λ ( n · n ∗ ) dx = (cid:90) Ω − n i , n ∗ i , − t n · ∇ × n ∗ + λ ( n · n ∗ ) dx = (cid:90) Ω −∇ n : ∇ n ∗ − t n · ∇ × n ∗ + λ ( n · n ∗ ) dx , where λ : = |∇ n ∗ | + t n ∗ · ∇ × n ∗ . We know that the integration by parts holds in the above calculationdue to the definitions of n and n ∗ . n i n ∗ i , (0) = n ∗ , (0) = − t sin ( tz ) cos θ − θ (cid:48) cos ( tz ) sin θ (cid:12)(cid:12)(cid:12) z = = , and similarly n i n ∗ i , (1) = n ∗ , (1) = θ (cid:48) cos θ (cid:12)(cid:12)(cid:12) z = = . Thus (29) can be written as (cid:90) Ω |∇ n − ∇ n ∗ | dx = (cid:90) Ω |∇ n | + |∇ n ∗ | + t n · ∇ × n ∗ − λ ( n · n ∗ ) dx . LONG CHOLESTERIC PITCH
Using the facts that 2(1 − n · n ∗ ) = | n − n ∗ | and that I ( n ∗ ) = (cid:82) Ω λ + t dx , we deduce the followingcalculations: (cid:90) Ω |∇ n − ∇ n ∗ | dx = (cid:90) Ω |∇ n | + |∇ n ∗ | + t n · ∇ × n ∗ − λ n · n ∗ dx = (cid:90) Ω |∇ n | + |∇ n ∗ | + t n · ∇ × n ∗ + λ (1 − n · n ∗ ) − λ dx = (cid:90) Ω |∇ n | + |∇ n ∗ | + t n · ∇ × n ∗ + λ | n − n ∗ | + t dx − I ( n ∗ ) = I ( n ) + I ( n ∗ ) + t (cid:90) Ω n · ∇ × n ∗ ) − ( n · ∇ × n ) − ( n ∗ · ∇ × n ∗ ) + λ | n − n ∗ | dx − I ( n ∗ ) = I ( n ) − I ( n ∗ ) + t (cid:90) Ω n · ∇ × n ∗ ) − ( n · ∇ × n ) − ( n ∗ · ∇ × n ∗ ) dx + (cid:90) Ω λ | n − n ∗ | dx . (30)Looking at the central integral in the equation above, we can rearrange the expression to get (cid:90) Ω n · ∇ × n ∗ ) − ( n · ∇ × n ) − ( n ∗ · ∇ × n ∗ ) dx = (cid:90) Ω (cid:2) ( n − n ∗ ) · ∇ × ( n ∗ − n ) (cid:3) + ( n · ∇ × n ∗ ) − ( n ∗ · ∇ × n ) dx . (31)Then using integration by parts we find that (cid:90) Ω n · ∇ × n ∗ dx = (cid:90) Ω n i (cid:15) i jk n ∗ k , j = (cid:90) ∂ Ω n i n ∗ k (cid:15) i jk ν j dS − (cid:90) Ω n ∗ k (cid:15) i jk n i , j dx = (cid:90) Ω n ∗ · ∇ × n dx . (32)The surface integral is easily seen to be zero by applying the boundary conditions from the set A . Weare now in a position to conclude the calculation by combining the final line of (30), (31) and (32): (cid:90) Ω |∇ n − ∇ n ∗ | + t (cid:2) ( n − n ∗ ) · ∇ × ( n − n ∗ ) (cid:3) − λ | n − n ∗ | dx = H ( n − n ∗ ) = I ( n ) − I ( n ∗ ) . Integral estimate
Now we look to find a lower bound for the integral functional H and we begin by noting the followingrelation obtained just by simple algebra (using (26) and (28)) |∇ n ∗ | + t n ∗ · ∇ × n ∗ = δ t − t + t sin θ. (33)Thus we can rewrite H in the following, more concrete, form H ( v ) = (cid:90) Ω |∇ v | + t v · ∇ × v − ( δ t − t + t sin θ ) | v | dx . Firstly we infer that 0 < δ t (cid:54) π . This is clearly true because looking at (28) simply shows that θ (cid:48) > δ t .Hence if δ t > π then necessarily θ (1) > π . It is also an elementary fact that |∇ × v | (cid:54) √ |∇ v | , so H ( v ) (cid:62) (cid:90) Ω |∇ v | + t v · ∇ × v − (cid:32) π + t (cid:33) | v | dx (cid:62) (cid:90) Ω |∇ v | − t | v ||∇ × v | − (cid:32) π + t (cid:33) | v | dx (cid:62) (cid:90) Ω |∇ v | − t √ | v ||∇ v | − (cid:32) π + t (cid:33) | v | dx . LONG CHOLESTERIC PITCH
From here we can use Cauchy’s inequality, ab (cid:54) ( (cid:15) a + b (cid:15) ), with (cid:15) = t to obtain H ( v ) (cid:62) (cid:90) Ω |∇ v | (cid:32) − √ (cid:33) − (cid:32) π + t + √ t (cid:33) | v | dx . (34)As can be seen by the argument below, (cid:15) = t is not the optimal value to use in Cauchy’s inequality forproving the global minimum property for the widest range of t values. A simple calculation shows that (cid:15) = (cid:18) √ + √ (cid:19) t ≈ . t is in fact the optimal constant but we use 4 t for ease here. Having established(34) we apply the Poincar´e inequality with the sharp constant π , to deduce that H ( v ) (cid:62) (cid:34) π (cid:32) − √ (cid:33) − (cid:32) π + t + √ t (cid:33)(cid:35) (cid:90) Ω | v | dx = : γ t (cid:90) Ω | v | dx ≈ (cid:16) . − . t (cid:17) (cid:90) Ω | v | dx . Conclusion
We are now in a position to conclude the proof by drawing the two strands together. By taking anarbitrary n ∈ A we know that I ( n ) − I ( n ∗ ) = H ( n − n ∗ ) (cid:62) γ t (cid:90) Ω | n − n ∗ | dx . The definition of γ t shows that if t (cid:54) .
766 then γ t >
0. This means that there exists some t ∗ > t < t ∗ , n ∗ is indeed the unique global minimiser of the problem. (cid:3) Theorem 11 is a very interesting, if slightly unsurprising, result. For nematic liquid crystals, usingthe one constant approximation results in the very simple Lagrangian |∇ n | . Clearly if we studied thenematic problem in isolation with the same boundary conditions we would be able to conclude that thefunction cos (cid:16) π z (cid:17) (cid:16) π z (cid:17) , would be the unique global minimum. Theorem 11 proves that this behaviour propagates into cholestericliquid crystals for long pitch lengths. This is the first time that cholesteric liquid crystals with long pitchlengths have been analytically shown to have similar qualitative behaviour to nematic liquid crystals. Inthe final result of this section we prove an interesting result which shows that the combination of theLagrangian our particular set of admissible functions A , (4), removes any contribution of the saddlesplay term. Proposition 12.
Let n ∈ A . Then (cid:90) Ω tr (cid:16) ∇ n (cid:17) − ( ∇ · n ) dx = . Proof
To avoid technical issues over density of smooth functions in Sobolev spaces between manifolds we willshow the slightly stronger statement that the following integral is zero J ( n ) : = (cid:90) Ω tr (cid:16) ∇ n (cid:17) − ( ∇ · n ) dx , (35)13 LONG CHOLESTERIC PITCH for all unconstrained functions n ∈ A (cid:48) : = (cid:26) v ∈ W , (cid:16) Ω , R (cid:17) (cid:12)(cid:12)(cid:12)(cid:12) v | z = = e , v | z = = e , v | x = − L = v | x = L , v | y = − L = v | y = L (cid:27) . First we look to show that expression (35) is zero for a dense subset B , of A (cid:48) . B : = (cid:26) v ∈ C ∞ (cid:16) Ω , R (cid:17) (cid:12)(cid:12)(cid:12)(cid:12) v | z = = e , v | z = = e , v | x = − L = v | x = L , v | y = − L = v | y = L (cid:27) . So we take an arbitrary element v ∈ B and we observe that since we are in the special case of a rect-angular domain, the periodic boundary conditions imply that for arbitrary y ∈ ( − L , L ), z ∈ (0 , ff erentiability up to the boundary is ensured by [1, p.705].) ∂ v ∂ y ( − L , y , z ) = lim h → v ( − L , y + h , z ) − v ( − L , y , z ) h = lim h → v ( L , y + h , z ) − v ( L , y , z ) h = ∂ v ∂ y ( L , y , z ) . By very similar arguments we conclude the following set of equalities ∂ v ∂ y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x = − L = ∂ v ∂ y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x = L , ∂ v ∂ z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x = − L = ∂ v ∂ z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x = L ,∂ v ∂ x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) y = − L = ∂ v ∂ x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) y = L , ∂ v ∂ z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) y = − L = ∂ v ∂ z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) y = L . (36)This puts us in a position where we can calculate the integral; we rearrange the saddle-splay term into adivergence and use the divergence theorem to find J ( v ) = (cid:90) Ω tr (cid:16) ∇ v (cid:17) − ( ∇ · v ) dx = (cid:90) Ω ∇ · (( v · ∇ ) v − ( ∇ · v ) v ) dx = (cid:90) ∂ Ω (( v · ∇ ) v − ( ∇ · v ) v ) · N dS = (cid:90) x = − L (( v · ∇ ) v − ( ∇ · v ) v ) · ( − e ) + (cid:90) x = L (( v · ∇ ) v − ( ∇ · v ) v ) · e + (cid:90) y = − L (( v · ∇ ) v − ( ∇ · v ) v ) · ( − e ) + (cid:90) y = L (( v · ∇ ) v − ( ∇ · v ) v ) · e + (cid:90) z = (( v · ∇ ) v − ( ∇ · v ) v ) · ( − e ) + (cid:90) z = (( v · ∇ ) v − ( ∇ · v ) v ) · e . (37)Where, as usual, e i represent the usual R basis vectors. In order to deduce that each of the above linesin the previous equation are equal to zero we note that(( v · ∇ ) v − ( ∇ · v ) v ) · e = v v , + v v , − v v , − v v , (( v · ∇ ) v − ( ∇ · v ) v ) · e = v v , + v v , − v v , − v v , (( v · ∇ ) v − ( ∇ · v ) v ) · e = v v , + v v , − v v , − v v , . (38)The first two equalities in (38) show that the first two lines of (37) are indeed zero by utilising (36).Finally we note that since v is prescribed and constant in the z = , ∂ v ∂ x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z = = ∂ v ∂ y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z = = ∂ v ∂ x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z = = ∂ v ∂ y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z = = , meaning that the final line in (37) is zero as both of the integrands are identically zero. So we have that J ( v ) = ∀ v ∈ B . ALTERNATIVE PROBLEM
So now if we take an arbitrary element n ∈ A . , the density of B implies that ∃ ( v j ) ⊂ B such that || v j − n || , −→ as j → ∞ . With this information it is clear when writing (35) in component form and using the fact that f j → f in L and g j → g in L ⇒ f j g j → f g in L , we deduce the convergence of the integral values so that0 = J ( v j ) −→ J ( n ) . Therefore J ( n ) = n ∈ A (cid:48) and A ⊂ A (cid:48) , so the statement is proved. (cid:3)
This proposition indicates that there may be a subtle flaw somewhere in our modelling of the cholestericliquid crystal problem. We know that there are periodic structures that can form in cholesteric liquidcrystals which make good energetic use of the saddle-splay term (blue phases for example). Howeverthe function space choice of W , (cid:16) Ω , S (cid:17) would not allow us to be able to predict such a structure. Thisissue will be explored in depth a subsequent paper [4] where we show that considering director fields inSBV( Ω , S ) more easily allows for the prediction of multi-dimensional cholesteric structures. Up to this point we have exclusively focused our attentions on the problem with perpendicular boundaryconditions applied to the top and bottom faces of the cell. However a more commonly studied problemis that of homeotropic boundary conditions on both faces. In this final section we show that much of ouranalysis from Theorem 11 can be easily applied to such a situation. Therefore we consider the problemof minimising I ( n ) = (cid:90) Ω |∇ n | + t n · ∇ × n + t dx (39)over A : = (cid:26) n ∈ W , (cid:16) Ω , S (cid:17) (cid:12)(cid:12)(cid:12)(cid:12) n | z = = n | z = = e , n | x = − L = n | x = L , n | y = − L = n | y = L (cid:27) . (40)This problem is much more approachable in that for every value of t we have a simple solution of theEuler-Lagrange equation. The constant function n = e satisfies ∆ n − t ∇ × n + n (cid:16) |∇ n | + t n · ∇ × n (cid:17) = , for any t . Accordingly this simplifies much of the analysis to deduce whether this a global or localminimiser. This constant state is often called the unwound cholesteric state because of its deviation fromthe helical configuration n ( z ) : = cos( tz )sin( tz )0 . The stability of the unwound state has been approached both experimentally [10], and with simulations.However with the local minimiser results Theorems 3 and 4 we can precisely quantify its stability, bothlocally and globally, with the two results below.
Theorem 13.
There exists some t ∗ > such that whenever t < t ∗ , the function n = e is the uniqueglobal minimiser of (39) over (40) 15 ALTERNATIVE PROBLEM
Proof
The main steps of this proof closely resemble those of the proof of Theorem 11, however this case ismuch simpler because our candidate minimiser is constant. Let n ∗ : = e , and n ∈ A . Then I ( n ) − I ( n ∗ ) = (cid:90) Ω |∇ n | + t n · ∇ × n dx = (cid:90) Ω |∇ ( n − n ∗ ) | + t ( n − n ∗ ) · ∇ × ( n − n ∗ ) + t n ∗ · ∇ × n dx = (cid:90) Ω |∇ ( n − n ∗ ) | + t ( n − n ∗ ) · ∇ × ( n − n ∗ ) dx = : H ( n − n ∗ ) (41)The final equality holds in (41) because of the periodicity in the x and y directions for n . From here wecan follow the same logic as can be found from (33) to (34) to deduce that I ( n ) − I ( n ∗ ) = H ( n − n ∗ ) (cid:62) (cid:90) Ω |∇ ( n − n ∗ ) | (cid:32) − √ (cid:33) − t √ | n − n ∗ | dx (cid:62) (cid:32) π − π √ − t √ (cid:33) (cid:90) Ω | n − n ∗ | dx . Therefore when t < t ∗ : = π √ √ − ≈ .
06, we have that I ( n ) − I ( n ∗ ) (cid:62) γ t (cid:90) Ω | n − n ∗ | dx for some γ t >
0. This proves the global stability of n ∗ . (cid:3) Theorem 14. [3] Consider the variational problem as described in (39) and (40) . If t < π then n = e is a strict strong local minimiser of I. If t > π then n = e is not a weak local minimiser of I. Remark.
As was mentioned in [3] it is unknown whether the constant state e is a local minimiser ornot at the critical value of t = π . The problem smoothly depends on the cholesteric twist t, howeverunless we can show that e is a global minimiser up to t = π we know of no way to deduce its stabilityat the bifurcation point. These results tie in with Theorem 11 in that they too demonstrate how cholesteric liquid crystals arerelated to nematic liquid crystals for long pitch lengths. It is immediately obvious that if we studiednematic liquid crystals with the set of admissible functions as given in (40), the unique global minimiseris n = e . According to Theorem 13, this property propagates into the cholesteric setting for smallvalues of t . Theorem 14 extends this result and shows that e is a strong local minimiser for the problemup to t = π , but not a weak local minimum thereafter. It is interesting to note that n = e is a globalminimum at least up to t = .
06 and a strong local minimiser up to t = π . This obviously allows thepossibility of there being a range of t values where the unwound state is a local minimiser but not aglobal minimiser. As far as we are aware, the theorems presented in this paper are the first to show thisqualitative correspondence between nematic liquid crystals and low chirality cholesteric liquid crystalsTo the best of our knowledge it is still an open problem to analytically prove the existence of multi-dimensional minimisers for a specific cholesteric liquid crystal problem when the director field n lies inthe space W , (cid:16) Ω , S (cid:17) . It may indeed be possible to prove this fact, whether analytically or numerically,using Sobolev regularity director fields. However in a forthcoming paper [4], we will show that wecan predict the existence of these structures much more easily when we relax the problem so that n ∈ S BV (cid:16) Ω , S (cid:17) . The technique has potential because it allows the Oseen-Frank theory to better respectthe head-to-tail symmetry of the molecules. 16 EFERENCES
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