The Rayleigh problem for nematic liquid crystals. Short time solution

The Rayleigh problem for nematic liquid crystals.Similarity solutions

M. D�õez a,1, C. Atkinson b,*

a Escuela Superior de Ingenieros Industriales, Universidad de Navarra, P. Manuel Lardiz�abal 13, 20018 San Sebasti�an,

Spainb Department of Mathematics, Imperial College of Sciences, Technology and Medicine, 180 Queen's Gate, London SW7

2BZ, UK

Received 2 February 2000; accepted 2 February 2000

Abstract

The exact solution of the full Navier±Stokes equations for the problem of an in®nite plate on z � 0 in aNewtonian ¯uid given an impulsive motion in its plane (the Rayleigh problem) is extended here to thesituation where the ¯uid is that of nematic liquid crystal.

If the initial director direction is parallel to the plate in the x direction with the surface prepared in orderto maintain that initial alignment and the plate is given an impulsive motion in the y direction, then theclassical solution goes through with a constant viscosity.

However in the more complicated situation in which the plate has been prepared so that the directors arealigned or perpendicular to it and the directors for z > 0 have an initial alignment given by h0 (the tilt angleof the directors out of the plate) with 06 h0 < p=2, assuming the same uniform twist initial alignment/�z; 0� � /0 as the preferred alignment for / on the surface plate, then the velocity ®eld does not satisfy adi�usion equation. Here the situation when the plate is given an impulsive motion of the form vx � u0=t1=2

and vy � v0=t1=2 for t > 0 is considered. When the approximation of neglecting ¯uid inertia is made, theequation for the variation of the tilt angle of the directors in time and position becomes a non lineardi�usion equation with the di�usion coe�cient depending on h. The twist angle, /, of the directors alsosatis®es a di�usion equation with a di�usion coe�cient varying in space and time, this being determined bythe behaviour of the tilt angle. However, the inertia is needed to satisfy the initial static condition of thenematic and it is included by means of a singular perturbation method.

An approximate solution of the problem when the initial alignment for / is di�erent from the preferreddirection for / on the surface, that is, when /�z; 0� � /0 is di�erent from /�0; t� � /1 is also given.

The e�ect of the director inertia is neglected throughout this paper, in common with approximationsusually made in the nematic literature. Ó 2001 Elsevier Science Ltd. All rights reserved.

www.elsevier.com/locate/ijengsci

International Journal of Engineering Science 39 (2001) 245±284

*Corresponding author. Tel.: +1-44-0171-589-5111; fax: +1-44-0171-589-8517.

1 Present address: GKN-GTE R&D Department, Apartado 37, 20750 Zumaia, Spain.

0020-7225/01/$ - see front matter Ó 2001 Elsevier Science Ltd. All rights reserved.

PII: S0020-7225(00)00042-2

1. Introduction

Suppose that the initial state of the director ®eld of a nematic liquid crystal occupying thehalfspace z > 0 is such that the directors are aligned parallel to an in®nite plate which lies onz � 0, i.e. the director ®eld is in the x direction everywhere. Let the plate be given a velocityvy�0; t� � u0=t1=2 for all time (see Fig. 1). Then one can show that the velocity ®eld in the in®nitebody of ¯uid can be deduced from the full balance and constitutive equations of the ¯uid, whichare set out in Section 2, to satisfy a parabolic di�erential equation for vy�z; t�, with a constantdi�usion coe�cient related to the viscosity of the ¯uid (cf. Section 3). Thus in this simple case theexact transient solution of the full equations is exactly analogous to the classical Rayleighproblem for a Newtonian ¯uid.

On the other hand, if the initial director ®eld is such that the directors are all initially alignedwith either the z direction or the y direction (the direction of imposed motion of the plate) and noprescribed precondition is given to the directors of the plate surface then no such simple reductionseems possible. The angular momentum balance equations preclude such a simple reduction.

We are assuming throughout this paper that we can neglect the e�ects of the director inertia, incommon with approximations usually made in the nematic literature.

The continuum theory of nematic liquid crystals leads to four coupled non-linear di�erentialequations.

For the initial and boundary conditions considered here (see Section 4), the directors may berepresented in terms of a tilt angle h�z; t� out of the plane z � 0 and /�z; t�, the twist from the x-axis about the normal (see Fig. 2):

Fig. 1. Halfspace of a nematic liquid crystal resting on an in®nite plate on z � 0 which is pulled with a velocity

vy�0; t� � u0=t1=2 in the y direction.

Fig. 2. Tilt angle of the director h�z; t� out of the plane and twist about the normal /�z; t�.

246 M. D�õez, C. Atkinson / International Journal of Engineering Science 39 (2001) 245±284

nx � cos�h�z; t�� cos�/�z; t��;ny � cos�h�z; t�� sin�/�z; t��; �1�nz � sin�h�z; t��:

The two non-zero velocity components may be written vx�z; t� and vy�z; t� while vz is zero be-cause of the incompressibility assumption (see Eqs. (20)±(24)).

We consider the problem of a halfspace of a nematic liquid crystal resting on an in®nite plate onz � 0 and suppose that at time t � 0 the director ®eld is uniform throughout all z > 0 due to anapplied electromagnetic ®eld which is switched o� at the time t � 0�. This initial alignment may beprescribed as h�z; 0� � h0, with 06 h0 < p=2, and /�z; 0� � /0 (see Fig. 3 for example). The surfaceis prepared so that the directors are parallel ÿh�0; t� � 0 and /�0; t� � /0 ± or perpendicular to itÿh�0; t� � p=2 and /�0; t� � /0. More generally the surface may be prepared so that there arepreferred alignment directions given by h�0; t� � h1 and /�0; t� � /0, with h1 di�erent from h0 (thechange of variable (46) cannot be applied when h1 � h0). Then, the spacetime variations of thedirector and velocity ®elds produced by these conditions are analysed in Section 4 for the casewhere the plate is pulled with a velocity vx � u0=t1=2 and vy � v0=t1=2 for all time t > 0. The morecomplicated situation when the initial alignment for / is di�erent from the preferred direction for/ on the surface, that is, when /�z; 0� � /0 is di�erent from /�0; t� � /1 is treated in Appendix A.

If we set the density q to zero in the di�erential equations, then, after a little algebra, we candeduce a single non-linear partial di�erential equation for the tilt angle of the directors as afunction of z and t (as happened in the problem treated in [1]). This is always possible when theboundary and initial conditions for the angular coordinate / are of the form/�z; 0� � /�0; t� � /0, so that the solution for /, when the density is neglected, is /�z; t� � /0. Theresulting non-linear parabolic di�erential equation suggests that the information about the an-gular coordinate h on the surface di�uses throughout the liquid crystal as a function of space andtime exactly analogous to that of di�usion of mass and temperature.

The boundary and initial conditions of the problem treated in Section 4 (and see Section 5.1 fora particular example) allow similarity substitutions (see Eqs. (26)±(30)) which simplify theproblem, i.e. the governing partial di�erential equations of the problem reduce to a system of

Fig. 3. Halfspace of a nematic liquid crystal resting on an in®nite plate on z � 0. The plate is given a prescribed velocity

vx�0; t� � u0=t1=2 and vy�0; t� � v0=t1=2. The directors are initially oriented along the directions h�z; 0� � h0

and /�z; 0� � /0 and the surface is prepared so that the directors are aligned with the plate.

M. D�õez, C. Atkinson / International Journal of Engineering Science 39 (2001) 245±284 247

ordinary di�erential equations. In particular, when the density q is set to zero, the equation for thetilt angle of the directors h reduces to a non-linear ordinary di�erential equation (see Eq. (47)) forwhich dual and complementary principles would be useful (see, e.g. [2]). Here, we only use amaximum principle for the partial di�erential equation (see, e.g. [3]) to note that the result of this`di�usion' must be between the di�usion obtained by bounding the generalised di�usion coe�cientabove and below. In addition, this also motivates a selection of an average di�usion coe�cient.The generalised di�usion coe�cient turns out to be a complicated function of the materialproperties of the liquid crystal and depends on the angular coordinate h (analogously to a con-centration dependent di�usion coe�cient in matter transport, see Eq. (48)) and we ®nd that theuse of an average coe�cient is quite accurate.

In the more complicated situations when the initial alignment for / is di�erent from the pre-ferred direction for / on the surface (see Appendix A), if we set the density q to zero in thegoverning di�erential equations, then we can also deduce a non-linear partial di�erential equationfor the twist angle of the directors. This non-linear parabolic di�erential equation is similar to theequation of heat conduction with `thermal di�usivity' and heat capacity both dependent on h, andtherefore on z and t.

The boundary and initial conditions of the problem considered in Appendix A (see Section 5.2for a particular example) allow similarity substitutions which reduce the partial di�erentialequation for / to an ordinary di�erential equation. We give an approximate solution of thisequation (see Eq. (128) or Eq. (A.12)) which indicates how information about the angle / to bein¯uenced by the variation in h and how this information varies through space and time.

The complete problem of the variations in the angle h and / is complicated due to the non-linear interactions of the full equations. However, the approximate solutions (A.12) or (128) for /and (A.20) for h, when inertia is neglected, show the in¯uence of Q�ave on the space-time variationof / and Qave and on the space-time variation of h, each of these being related to the di�usion-likebehaviour (Q being related to a di�usion coe�cient).

We use singular perturbation theory, both, when there is and is not a discontinuity in theboundary and initial conditions for /, in order to calculate the velocity ®eld accurately. This isbecause the direct use of our ®rst approximation to the tilt angle and twist angle solutions leads toan expression for the velocity ®eld that cannot satisfy all boundary conditions (see Eqs. (52) and(53) for the ®rst case and Eqs. (A.21) and (A.22) for the second one). With this method, we in-troduce the in¯uence of the ¯uid inertia terms to obtain the velocity ®eld and also to determine thecorrection to the angular coordinates.

Although the matching procedure carries through quite well for the cases in this paper, the oneexample in which the method fails is where the initial director direction is perpendicular to theplate and the surface plate has been prepared so that the directors lie ¯at on it (see Eq. (75), forexample, when h0 � p=2). We conjecture that this `instability' would be removed if we includedthe director inertia, which has been neglected throughout this paper as it is an approximationusually made in the nematic literature.

The detailed plan of the paper is as follows: Section 2 describes the equations of motion of anematic liquid crystal in Cartesian coordinates. Section 3 extends the classical solution of theRayleigh problem in a Newtonian ¯uid to the situation where the ¯uid is that of nematic liquidcrystal. Section 4 considers a ®rst problem of a halfspace nematic liquid crystal resting on anin®nite plate on z � 0 where the directors have an initial uniform alignment given by h�z; 0� � h0


and /�z; 0� � /0 and the surface is prepared so that the directors have preferred directions givenby h�0; t� � h1 and /�0; t� � /0. This section studies the subsequent motion when the plate ispulled with a velocity vx � u0=t1=2 and vy � v0=t1=2 for all time t > 0. Section 5 gives some resultsobtained when the perturbation theory described in Section 4 is applied to a particular set ofconstants and to some particular boundary and initial conditions. An approximate solution, wheninertia due to q is neglected, is also given in this section for a particular problem with disconti-nuity in the boundary and initial conditions in the angular coordinate /. Appendix A outlineshow to proceed in these cases, that is, when /�z; 0� � /0 is di�erent from /�0; t� � /1. Section 6gives some concluding remarks. Finally, Appendix B studies the nature of the partial di�erentialequations for the angular coordinates h and / as functions of z and t.

2. Fundamental equations

The starting point for our analysis is the phenomenological equations of Ericksen [4] and Leslie[5] for ¯ow problems in nematic liquid crystals. For completeness these equations are repeatedhere. The ¯ow of a nematic liquid crystal is described by the velocity vector v together with adirector n which is a unit vector giving the orientation of the anisotropic axis in these trans-versely isotropic liquids. With the assumption of incompressibility, the equations are the con-straints

vi;i � 0; nini � 1 �2�

together with the balance laws:

q_vi � Fi � tij;j;

r�ni � Gi � gi � Sij;j�3�

and the constitutive relations:

tij � ÿpdij ÿ oWonk;j

nk;i � ~tij;

Sij � nibj �oWoni;j

; �4�

gi � cni ÿ �nibj�;j ÿoWoni� ~gi;

where

2W � k1�ni;i�2 � k2�eijknink;j�2 � k3ninjnk;ink;j � �k2 � k4��ni;jnj;i ÿ �ni;i�2� �5�

and


~tij � l1nknpAkpninj � l2Ninj � l3Njni � l4Aij � l5Aiknknj � l6Ajknkni;

~gi � k1Ni � k2Aiknk;

2Aij � vi;j � vj;i;

2Ni � 2 _ni � �vj;i ÿ vi;j�nj

�6�

and

k1 � l2 ÿ l3; k2 � l5 ÿ l6: �7�

Cartesian tensor notation is used together with the usual summation convention and a commadenotes partial di�erentiation and the superposed dot a material time derivative. The ®rst part ofEq. (3) represents conservation of linear momentum, q is the density, F any external body forceand t the stress tensor (asymmetric). The second part of (3) represents conservation of angularmomentum, r is a constant inertial coe�cient, G a generalised body force arising from a bodycouple present due to magnetic or electric ®elds, g is a generalised intrinsic body force whichdepends upon the director ®eld n through third part of Eq. (4), and S a generalised stress tensor.The inertial term in this equation r�ni is usually assumed negligible and hence omitted in mosttreatments. In the constitutive relations (4) the scalar p is the arbitrary pressure following from theassumption of incompressibility while c and b represent similar indeterminacies due to the con-straint on the additional kinematic variable n. W is the energy function and the coe�cients inexpressions (5) and (6) are constants if thermal e�ects are ignored. Various authors have discussedthe possibility of restrictions on these coe�cients for physically acceptable behaviour. In partic-ular, the coe�cients in (6) are to be consistent with the thermodynamic inequality

~tijvi;j ÿ ~gi _ni > 0: �8�

Other restrictions have been suggested, e.g. the Parodi [6] relationship between the viscous co-e�cients

l2 � l3 � l6 ÿ l5: �9�

3. Rayleigh problem for nematic liquid crystals

We consider the particular problem of a halfspace of positive nematic liquid crystal resting onan in®nite plate on z � 0 (see Fig. 1) which, initially at rest, is pulled with a velocity u0=t1=2 in the ydirection for all time. The z-axis is normal to the plate and the x-axis parallel to the ®eld freealignment.

We suppose an initial alignment h � 0 in the x direction for all values of y and z at the timet � 0.

Then we seek a solution of the balance laws subject to the boundary and initial conditions:

vy�0; t� � u0=t1=2; t > 0; vy�z; 0� � 0; z > 0: �10�


All the conditions of the problem are satis®ed if only the non-zero velocity component is vy de-pending on �z; t� and if nx � 1 and ny � nz � 0 for all time. Then the balance laws (3) reduce to thefollowing parabolic di�erential equation for vy with a constant coe�cient related to the viscositiesof the ¯uid:

qovy

ot� ga

o2vy

oz2; �11�

where

ga �l4

2: �12�

Introducing the change of variable

g � qga

� �1=2 z2t1=2

�13�

and looking for solutions in the form

vy � u0

t1=2f1�g�: �14�

Eq. (11) is reduced to

f 001 �g� � 2d

dg�gf1�g�� 0: �15�

This di�erential equation can be integrated with the assumptions f 01�1� � 0 and f1�1� � 0 whichare consistent with the condition of zero initial velocity, to give

f 01�g� � 2gf1�g� � 0: �16�

The initial and boundary conditions of the problem reduce to:

f1 � 1 when g � 0;f1 � 0 when g � 1: �17�

The solution obtained for vy�z; t� turns out to be

vy�z; t� � u0

t1=2exp

�ÿ q

ga

� �z2

4t

�: �18�

4. Rayleigh problem for nematic liquid crystals: Surface prepared

We now consider a halfspace of positive nematic liquid crystal resting on an in®nite plate onz � 0 which, initially at rest, is pulled with a prescribed velocity vx � u0=t1=2 and vy � v0=t1=2 for all


time t > 0. We suppose the initial director directions are constant throughout the ¯uid due to anapplied electromagnetic ®eld which is switched o� at the time t � 0, we also assume that thesurface has been prepared so that there are preferred alignments for the directors in contact withthe boundary plate. Special attention is however given when the directors on the surface areperpendicular or parallel to it (see Fig. 2 for example).

We have two non-zero velocity components, vx�z; t� and vy�z; t�. The third component of thevelocity vz is zero because of the incompressibility assumption.

We need two angles to de®ne the problem: the tilt of the director out of the plane h�z; t� and thetwist about the normal /�z; t�, measured as indicated in the ®gure. Thus we assume the followingform of solution:

nx � cos�h�z; t�� cos�/�z; t�� vx � vx�z; t�;ny � cos�h�z; t�� sin�/�z; t�� vy � vy�z; t�;nz � sin�h�z; t�� vz � 0:

�19�

Note that these equations for nx; ny and nz satisfy the condition n2x � n2

y � n2z � 1 given by (2).

Assuming no body forces and neglecting director inertia, the linear momentum equations turnout to be

qovx

ot

� �� o

ozovx

oz�a�h� cos2 /

�� b�h�� ovy

oza�h� cos / sin /ÿ oh

otm�h� cos /

ÿ o/ot

c�h� sin /

��20�

and

qovy

ot

� �� o

ozovx

oza�h� cos / sin /

�� ovy

oz�a�h� sin2 /� b�h�� ÿ oh

otm�h� sin /

� o/ot

c�h� cos /

��21�

and the angular momentum equations

ÿ k1 cos ho/ot� �k2 ÿ k1�

2sin h

ovx

ozsin /

�ÿ ovy

ozcos /

�� 2s�h� sin hÿ s0�h� cos h� o/

oz

� �ohoz

� �ÿ s�h� cos h

o2/oz2

� �� 0 �22�

and

2k1

ohot� h�h� ovx

ozcos /

�� ovy

ozsin /

�� f 0�h� oh

oz

� �2

� 2f �h� o2hoz2

� �� 2s�h� sin hÿ s0�h� cos h� cos h

o/oz

� �2

� 0; �23�


where

a�h� � l1 sin2 h

�� l3 � l6�

2

�cos2 h;

b�h� � �l5 ÿ l2�2

sin2 h� l4

2;

m�h� � l2 sin2 hÿ l3 cos2 h;

c�h� � l2 sin h cos h; �24�

s�h� � k2 cos2 h� k3 sin2 h;

f �h� � k1 cos2 h� k3 sin2 h;

h�h� � k1 � k2 cos�2h�:

We follow the procedure outlined in [1]. We consider situations in which the boundary and initialconditions are of the form

h�0; t� � h1; /�0; t� � /0; vy�0; t� � v0

t1=2; vx�0; t� � u0

t1=2; t > 0;

h�z; 0� � h0; /�z; 0� � /0; vy�z; 0� � 0; vx�z; 0� � 0; z > 0;�25�

where 06 h0 < p=2, as explained below (Eq. (75)) and h1 di�erent from h0 as explained below (Eq.(46)). As mentioned above, special attention is given when h1 � 0 or h1 � p=2. Appendix Aoutlines how to proceed in cases when /�z; 0� � /0 is di�erent from /�0; t� � /1.

The boundary and initial conditions given by (25) allow the change of variable

f � z

�kt�1=2; k � ÿ k3

k1

> 0 �26�

(note that we use k for f to be a dimensionless variable).We then look for solutions of the form

vx � kt

� �1=2

G�f�; �27�

vy � kt

� �1=2

H�f�; �28�

h � F �f�; �29�


/ � U�f�: �30�

These functional forms are consistent with the above boundary and initial conditions (25). Then,the problem to be solved is

ÿ qkf2

G�f� � G0�f��a�h� cos2 /� b�h�� H 0�f�a�h� cos / sin /� f2

F 0�f�m�h� cos /

� f2

U 0�f�c�h� sin / �31�

and

ÿ qkf2

H�f� � G0�f�a�h� cos / sin /� H 0�f��a�h� sin2 /� b�h�� f2

F 0�f�m�h� sin /

ÿ f2

U 0�f�c�h� cos /: �32�

In obtaining (31) and (32) we have integrated the di�erential equations with the assumptionsG�1� � 0; G0�1� � 0, H�1� � 0; H 0�1� � 0; F 0�1� � 0 and U 0�1� � 0 which are consistentwith the conditions of zero initial velocity and constant initial values of h and /.

The equations corresponding to the conservation of the angular momentum become

k1k2

cos hfU 0�f� � �k2 ÿ k1�2

k sin h�G0�f� sin /ÿ H 0�f� cos /�� 2s�h� sin hÿ s0�h� cos h�F 0�f�U 0�f� ÿ s�h� cos hU 00�f�� 0 �33�

and

k1kfF 0�f� ÿ h�h�k�cos /G0�f� � sin /H 0�f�� ÿ f 0�h��F 0�f��2 ÿ 2f �h�F 00�f�ÿ �2s�h� sin hÿ s0�h� cos h� cos h�U 0�f��2� 0 �34�

subject to the remaining boundary and initial conditions

F � h1; U � /0; G � u0=k1=2; H � v0=k1=2 when f � 0;

F � h0; U � /0; G � 0; H � 0 when f � 0: �35�

We assume an asymptotic solution of the form

F �f� � F0�f� � e1=2F1�f� � eF2�f� �O�e3=2�;U�f� � /0 � e1=2U1�f� � eU2�f� �O�e3=2�; �36�G�f� � G0�f� � e1=2G1�f� � eG2�f� �O�e3=2�;H�f� � H0�f� � e1=2H1�f� � eH2�f� �O�e3=2�;


where

e � qkgc

with gc ��l5 ÿ l2 � l4�

2: �37�

Note that in the earlier paper [1] the parameter e could have also been used.We introduce these forms of solution into the di�erential equations of the problem, say (31)±

(34), obtaining the following problem to order e0:

�a�F0� cos2 /0 � b�F0��G00�f� � a�F0� cos /0 sin /0H 00�f� �f2

m�F0� cos /0F 00�f� � 0; �38�

a�F0� cos /0 sin /0G00�f� � �a�F0� sin2 /0 � b�F0��H 00�f� �f2

m�F0� sin /0F 00�f� � 0; �39�

2f �F0�F 000 �f� � f 0�F0��F 00�f��2 ÿ k1F 00�f�fk � h�F0�k�cos /0G00�f� � sin /0H 00�f�� 0 �40�

and

�k2 ÿ k1�2

k sin F0�sin /0G00�f� ÿ cos /0H 00�f�� 0: �41�

Eqs. (38) and (39) can be combined to give

cos /0G00�f� � sin /0H 00�f� � ÿf2

F 00�f�m�F0�g�F0� : �42�

This expression is introduced into (40). Taking into account that f �h� > 0, as k1 and k3 are bothpositive, (40) can be written as

d

df�f 1=2�F0�F 00�f��

jk1jk2

a�F0�f 1=2�F0� fF 00�f� � 0; �43�

where

g�h� � a�h� � b�h� � 1

2�2l1 sin2 h cos2 h� �l3 � l6� cos2 h� �l5 ÿ l2� sin2 h� l4� �44�

and a�h� is de®ned as

a�h� � 1� m�h�h�h�2k1g�h� � 1� �m�h��

2

k1g�h� 6 1 �45�

since k1 < 0. Further use of the entropy production inequality restricts a�h� to be positive, as g�h�is positive.


The change of variable

u0 �R h0

h f 1=2�h�dhR h0

h1f 1=2�h�dh

�46�

lets us write (43) in the following suitable form for applying the variational procedure outlined in[1,3] or [7]. Note that this change of variable cannot be applied when h0 � h1

d

du0

du0

df

� �� kQ�u0�

2f � 0; �47�

where

Q�u� � jk1j a�u�f �u� : �48�

With this new variable, the boundary conditions take the form

u0 � 1 when f � 0;

u0 � 0 when f!1: �49�

If we assume Q�u0� as constant, say

Q�u0� � Qave where Qave �Z 1

0

uQ�u�du �50�

we can get an exact solution of (47)

u0 � erfc

��kQave

p2

f

� �: �51�

Once we know F0�f�, we can work out G0�f� and H0�f� from (38) and (39), respectively, to obtain

G0�f� � u0

k1=2ÿZ f

0

m�F0�f1��g�F0�f1��

f1

2F 00�f1� cos /0 df1 �52�

and

H0�f� � v0

k1=2ÿZ f

0

m�F0�f1��g�F0�f1��

f1

2F 00�f1� sin /0 df1: �53�

Both solutions G0�f� and H0�f� can only satisfy the conditions at f � 0; G0�0� � u0=k1=2 andH0�0� � v0=k1=2. The conditions at in®nity are not satis®ed. Then (52) and (53) are not uniformlyvalid through the full range of f. We will have to give a di�erent treatment for f small and f


large. In order to investigate the behaviour of the solution near h � h0 a new coordinate system �fis used

eaf � �f; �54�

where a > 0 is expected so for e� 1; f� 1 brings �f of order 1. Eqs. (31)±(34) become

ÿ e1ÿagc�f

2�G��f� � ea �G0��f��a��h� cos2 �/� �b��h�� ea �H 0��f��a��h� cos �/ sin �/

��f2

�F 0��f� �m��h� cos �/��f2

�U 0��f��c��h� sin �/; �55�

ÿ e1ÿagc�f

2�H��f� � ea �H 0��f��a��h� sin2 �/� �b��h�� ea �G0��f��a��h� cos �/ sin �/

��f2

�F 0��f� �m��h� sin �/ÿ�f2

�U 0�f��c��h� cos �/; �56�

k1k2

cos �h �f �U 0��f� � �k2 ÿ k1�2

k sin �h� �G0��f� sin �/ÿ �H 0��f� cos �/�ea

� �2�s��h� sin �hÿ �s0��h� cos �h� �F 0��f� �U 0��f�e2a ÿ �s��h� cos �h �U 00��f�e2a � 0 �57�

and

k1k�f �F 0��f� ÿ �h��h�k�cos �/ �G0��f� � sin �/ �H 0��f��ea ÿ �f 0��h�� F 0��f��2e2a ÿ 2 �f ��h� �F 00��f�e2a

ÿ �2�s��h� sin �hÿ �s0��h� cos �h� cos �h� �U 0��f��2e2a � 0; �58�

where

a�h�f�� a�h��f=qa�� a��h�;b�h�f�� b�h��f=qa�� b��h� � � �

�59�

For the solution valid at in®nity we look for solutions of the form

�F ��f� � h0 � ea �F1��f� � e2a �F2��f� �O�e3a�;�U��f� � /0 � ea �U1��f� � e2a �U2��f� �O�e3a�;�G��f� � �G0��f� � ea �G1��f� � e2a �G2��f� �O�e3a�;�H��f� � �H0��f� � ea �H1��f� � e2a �H2��f� �O�e3a�:

�60�


In order to satisfy the necessary conditions at in®nity, we require:

�F1�1� � 0; �F2�1� � 0;

�U1�1� � 0; �U2�1� � 0;

�G0�1� � 0; �G1�1� � 0;

�H0�1� � 0; �H1�1� � 0:

�61�

It is anticipated that the conditions at f � 0 may not be able to be satis®ed due to the singularperturbation character of the problem.

Introducing (60) into (55) and (58) and expanding the coe�cients depending on �h and �/ byTaylor's series about h0 and /0, respectively by using the expression �h��f� � �F ��f� and �/��f� � �U��f�from (60), we get problems of di�erent orders. Selecting a � 1=2, we have the following outerproblem (valid for f large) to order e1=2:

ÿ gc�f

2�G0��f� � ��a�h0� cos2 /0 � �b�h0�� G00��f� � �a�h0� cos /0 sin /0

�H 00��f�

��f2

�m�h0� cos /0�F 01�f� �

�f2

�c�h0� sin /0�U 01�f�; �62�

ÿ gc�f

2�H0��f� � �a�h0� cos /0 sin /0

�G00��f� � ��a�h0� sin2 /0 � �b�h0�� H 00��f�

��f2

�m�h0� sin /0�F 01�f� ÿ

�f2

�c�h0� cos /0�U 01�f�; �63�

k1 cos h0�f �U 01��f� � �k2 ÿ k1� sin h0� �G00��f� sin /0 ÿ �H 00��f� cos /0� � 0 �64�

and

k1�f �F 01��f� ÿ �h�h0�� G00��f� cos /0 � �H 00��f� sin /0� � 0: �65�

Solving (64) and (65) for �F1��f� and �U1��f� and substituting in (62) and (63) gives:

ÿ gc�f

2�Q10��f� � �g�h0��a�h0� �Q010 �66�

and

ÿ gc�f

2�Q20��f� � �b�h0��p�h0� �Q020; �67�


where

�Q1i��f� � �Gi��f� cos /0 � �Hi��f� sin /0 with i � 0; 1; 2; . . . ;

�Q2i��f� � �Gi��f� sin /0 ÿ �Hi��f� cos /0 with i � 0; 1; 2; . . .�68�

and

p�h� � 1� �k1 ÿ k2�2k1

l2 sin2 hb�h� � 1� l2

2 sin2 hk1b�h� �69�

using the Parodi relation for the last expression. Note 0 < p�h�6 1 as b�h� > 0 and k1 < 0. Theuse of the entropy production inequality restricts b�h� to be positive.

The di�erential equations (66) and (67) can be integrated satisfying the boundary conditions ofthe velocity ®eld at in®nity to give

�Q10��f� � B10 exp

ÿ gc

4�g�h0��a�h0��f2

!�70�

and

�Q20��f� � B20 exp

ÿ gc

4�p�h0��b�h0��f2

!: �71�

The leading order outer solution, valid for f large, for the velocity ®eld turns out to be

�G0��f� � B10 cos /0 exp

ÿ gc

4�g�h0��a�h0��f2

!� B20 sin /0 exp

ÿ gc

4�b�h0��p�h0��f2

!�72�

and

�H0��f� � B10 sin /0 exp

ÿ gc

4�g�h0��a�h0��f2

!ÿ B20 cos /0 exp

ÿ gc

4�b�h0��p�h0��f2

!; �73�

where the unknown constants B10 and B20 are to be determined by the matching procedure.The outer solutions (valid for f large) for the tilt angle of the directors �F ��f� and for the twist

about the normal �U��f�, to order e1=2 turn out to be

�F1��f� � B10

�h�h0�2k1

��gcp

�g�h0��a�h0�r

erfc

��gc

�g�h0��a�h0�r �f

2

0@ 1A �74�


and

�U1��f� � B20

�k1 ÿ k2�2k1

tan h0

��gcp

�b�h0��p�h0�r

erfc

��gc

�b�h0��p�h0�r �f

2

0@ 1A: �75�

Note both conditions at f � 0; �U1�0� � 0 and �H1�0� � 0, are not satis®ed due to the perturbationcharacter of the problem. Note also that this solution for �U1��f� converges, as we restrict h0 to be06 h0 < p=2.

If we rewrite these expressions in terms of the coordinate f and expand for e� 1, we get

�G��f� � B10 cos /0 � B20 sin /0 �O�e1=2�;

�H��f� � B10 sin /0 ÿ B20 cos /0 �O�e1=2�;

�F ��f� � h0 � e1=2B10

�h�h0�2k1

��gcp

�g�h0��a�h0�r

�O�e�; �76�

�U��f� � /0 � e1=2B20

�k1 ÿ k2�2k1

tan h0

��gcp

�b�h0��p�h0�r

�O�e�:

Writing G�f� and H�f� given in (52) and (53) (solution valid for f small) in terms of the coordinate�f and expanding for e small at ®xed �f gives

G�f� � u0

k1=2ÿZ 1

0

m�F0�f1��g�F0�f1��

f1

2F 00�f1� cos /0 df1 �O�exp�ÿ�f2=e�� 77�

and

H�f� � V0

k1=2ÿZ 1

0

m�F0�f1��g�F0�f1��

f1

2F 00�f1� sin /0 df1 �O�exp�ÿ�f2=e��: �78�

Matching the expressions for G and �G and H and �H , respectively to zero order in e gives

B10 � �u0 cos /0 � v0 sin /0�k1=2

ÿZ 1

0

m�F0�f��g�F0�f��

f2

F 00�f�df �79�

and

B20 � �u0 sin /0 ÿ v0 cos /0�k1=2

: �80�


Before investigating corrections to the inner solution assumed valid for f small, we consider theouter solution to order e. �G1�f�; �H1�f�; �F2�f� and �U2�f�, taking into account (68), have to satisfythe following equations:

ÿ gc�f

2�Q11��f� � �g�h0� �Q011��f� � �g0�h0� �Q010��f� �F1��f� ÿ �a�h0� �Q020��f� �U1��f� �

�f2

�m0�h0� �F 01��f� �F1��f�

��f2

�c�h0� �U 01��f� �U1��f� ��f2

�m�h0� �F 02��f�; �81�

ÿ gc�f

2�Q21��f� � �b�h0� �Q021��f� ÿ �a�h0� �Q010��f� �U1��f� � �b0�h0� �Q020��f� �F1��f� ÿ

�f2

�m�h0� �F 01��f� �U1��f�

��f2

�c0�h0� �U 01��f� �F1��f� ��f2

�c�h0� �U 02��f�; �82�

ÿk1 sin h0�f �F1��f� �U 01��f� � k1 cos h0

�f �U 02��f� � �k2 ÿ k1� sin h0�U1��f� �Q010��f�

� �k2 ÿ k1� sin h0�Q021��f� � �k2 ÿ k1� cos h0

�F1��f� �Q020��f�� 0 �83�

and

k1�f �F 02��f� ÿ �h0�h0�F1��f� �Q010��f� ÿ �h�h0� �Q011��f� � �h�h0� �Q020��f�U1��f� � 0: �84�

�F 02��f� and �U 02��f� can be found from (82) and (83), respectively and introducing their expressionsinto (80) and (81), gives

ÿ gc�f

2�Q11��f� � ÿ��g�h0��a�h0� ÿ �b�h0��p�h0�� Q020��f� �U1��f� � �g�h0��a�h0� �Q011��f�

� d

dh��g�h��a�h��

� �h0

�Q010��f� �F1��f� �85�

and

ÿ gc�f

2�Q21��f� � ÿ��g�h0��a�h0� ÿ �b�h0��p�h0�� Q010��f� �U1��f� � �b�h0��p�h0� �Q021��f�

� d

dh��b�h��p�h��

� �h0

�Q020��f� �F1��f� �86�

since �U1 and �F1 are given from (74) and (75), these equations can be solved to give


�Q11��f� � D10 exp

ÿ gc

4�g�h0��a�h0��f2

!� exp

ÿ gc

4�g�h0��a�h0��f2

!

�8<:ÿ B2

10

�h�h0�4k1

��g�h��a�h��0h0

��g�h0��a�h0��2��

g3cp

�g�h0��a�h0�

s Z �f

0

�f1 erfc

��gc

�g�h0��a�h0�r �f1

2

0@ 1Ad�f1

ÿ �k1 ÿ k2�4k1

B220 tan h0gc

1�b�h0��p�h0�

"ÿ 1

�g�h0��a�h0�

# ��gcp

�b�h0��p�h0�r

�Z �f

0

exp

ÿ gc

�f21

4

1�b�h0��p�h0�

ÿ 1

�g�h0��a�h0�

!!�f1 erfc

��gc

�b�h0��p�h0�r �f1

2

0@ 1Ad�f1

9=;�87�

and

�Q21��f� � D20 exp

ÿ gc

4�b�h0��p�h0��f2

!� exp

ÿ gc

4�b�h0��p�h0��f2

!8<:ÿ B10B20

�h�h0�4k1

��b�h��p�h��0h0

��b�h0��p�h0��2��

g3cp

�g�h0��a�h0�

s Z 1

�f

�f1 erfc

��gc

�g�h0��a�h0�r �f1

2

0@ 1Ad�f1

� B10B20

�k1 ÿ k2�4k1

gc tan h0

1�b�h0��p�h0�

"ÿ 1

�g�h0��a�h0�

# ��gcp

�b�h0��p�h0�r

�Z 1

�fexp

ÿ gc

�f21

4

1

�g�h0��a�h0�

ÿ 1

�b�h0��p�h0�

!!�f1 erfc

��gc

�b�h0��p�h0�r �f1

2

0@ 1Ad�f1

9=;:�88�

�G1��f� and �H1��f� will be given by

�G1��f� � �Q11 cos /0 � �Q21 sin /0 �89�

and

�H1��f� � �Q11 sin /0 ÿ �Q21 cos /0: �90�

The conditions at in®nity are satis®ed. Note, however, that we cannot satisfy the conditions atf � 0. The unknown constants D10 and D20 are to be determined by matching with the innersolution. If we write the outer solution (valid for f large) in inner coordinate system and weexpand them for e� 1, we get


�G��f� � B10 cos /0 � B20 sin /0 � e1=2�D10 cos /0 � D20 sin /0� �O�e�;�H��f� � B10 sin /0 ÿ B20 cos /0 � e1=2�D10 sin /0 ÿ D20 cos /0� �O�e�;

�F ��f� � h0 � e1=2B10

�h�h0�2k1

��gcp

�g�h0��a�h0�r

�O�e�; �91�

�U��f� � /0 � e1=2B20

�k1 ÿ k2�2k1

tan h0

��gcp

�b�h0��p�h0�r

�O�e�:

This suggests that the next approximation in the expansion for f small is of order e1=2 as we hadassumed. In order to obtain the inner solution to order e1=2 we work in a slightly di�erent way. Ifwe neglect the inertia in (31) and (32) we get:

H 0�f� � ÿ f2

F 0�f�m�h�g�h� sin /� f

2U 0�f� c�h�

b�h� cos /; �92�

G0�f� � ÿ f2

F 0�f�m�h�g�h� cos /ÿ f

2U 0�f� c�h�

b�h� sin / �93�

both expressions being valid up to order q, that is, valid up to order e. From (91) and (92) we canwrite

cos /G0�f� � sin /H 0�f� � ÿ f2F 0�f�m�h�

g�h� ;

sin /G0�f� ÿ cos /H 0�f� � ÿ f2U 0�f� c�h�

b�h� :�94�

Introducing these expressions into (34) and considering a form of solution given by (36), weobtain the equation

2f �h�F 00�f� � f 0�h��F 0�f��2 ÿ k1ka�h�fF 0�f� � 0 �95�

valid up to order e. This di�erential equation can be written in the form

d

dfdudf

� �� kQ�u�

2f

dudf� 0: �96�

If we now expand

u�f� � u0�f� � e1=2u1�f� � � � � �97�

we get

d

dfdu0

df

� �� kQ�u0�

2f

du0

df� 0; �98�


which is the variational problem we studied for the zero order case and to order e1=2

d

dfdu1

df

� �� kf

2

d

df�Q�u0�u1� � 0: �99�

Just as happened in the problem studied in [1], when we introduce the e�ect of inertia, we have acorrection to the condition at in®nity to order e1=2, that is, the condition u � 0 when f!1 isreplaced by u � 0� Aq1=2 when f!1, where A is determined by matching. If we approximateQ�u0� � Qave, we get

u1 � Aerf

��kQave

p2

f

� �: �100�

The connection between u1�f� and F1�f� is given by

F1�f� � ÿA

R h0

h1f 1=2�h�dh

f 1=2�F0� erf

��kQave

p2

f

� �: �101�

In order to work out U1�f�, we introduce (94) into (33) obtaining the expression

k1k2

p�h� cos hfU 0�f� ÿ s�h� cos hU 00�f� � �2s�h� sin hÿ s0�h� cos h�F 0�f�U 0�f� � 0 �102�

valid up to order e. If we consider solutions of the form (36), the problem corresponding to ordere1=2 is

k1k2

p�F0� cos �F0�fU 01�f� ÿ s�F0� cos �F0�U 001 �f�� 2s�F0� sin�F0� ÿ s0�F0� cos �F0��F 00�f�U 01�f� � 0: �103�

Multiplying this equation by cos�F0�, we get

d

df�s�F0� cos2�F0�U 01�f��

kQ��F0�2

fs�F0� cos2�F0�U 01�f� � 0; �104�

where

Q��h� � jk1jp�h�s�h� : �105�

If we approximate Q��F0� � Q�ave, the solution obtained for U1�f� with the condition U1�0� � 0is

U1�f� � CZ f

0

exp ÿ �kQ�ave=4�f21

ÿ �s�F0� cos2�F0� df1: �106�


The condition at in®nity is not satis®ed due to the perturbation character of the problem.Once we know the solutions to order e1=2 of h�f� and /�f�, we can work out the solutions for the

velocity components G1�f� and H1�f�. Taking into account that (92) and (93) are valid up to ordere and that we look for a solution in the form (36), the problem we get to order e1=2 is

H 01�f� � ÿf2

m�F0�g�F0� F 00�f�U1�f� cos /0 �

f2

c�F0�b�F0�U

01�f� cos /0 ÿ

f2

d

dfm�F0�g�F0� F1�f�

� �sin /0

�107�

and

G01�f� � ÿf2

d


� �cos /0 �

f2

m�F0�g�F0� F 00�f�U1�f� sin /0 ÿ

f2

c�F0�b�F0�U

01�f� sin /0:

�108�

Then, using the boundary conditions at f � 0, we get

H1�f� � ÿZ f

0

f1

2

m�F0�g�F0� F 00�f1�U1�f1� cos /0 df1 �

Z f

0

f1

2

c�F0�b�F0�U

01�f1� cos /0 df1

ÿZ f

0

f1

2

d

dfm�F0�g�F0� F1�f1�

� �sin /0 df1 �109�

and

G1�f� � ÿZ f

0

f1

2

d

df1

m�F0�g�F0� F1�f1�

� �cos /0 df1

Z f

0

f1

2

m�F0�g�F0� F 00�f1�U1�f1� sin /0 df1

ÿZ f

0

f1

2

c�F0�b�F0�U

01�f1� sin /0 df1: �110�

Again the conditions at in®nity are not satis®ed.Now we can write the inner solution, valid for f small, in the outer coordinate system �f and

expand for e� 1 in order to calculate the unknown constants. The expressions obtained for thevelocity ®eld turn out to be

G�f� � u0

k1=2ÿZ 1

0

f1

2

m�F0�g�F0� F 00�f1� cos /0 df1 � e1=2

Z 1

0

f1

2

m�F0�g�F0� F 00�f1�U1�f1� sin /0 df

�ÿZ 1

0

f1

2

d

df1

m�F0�g�F0� F1�f1�

� �cos /0 df1 ÿ

Z f

0

f1

2

c�F0�b�F0�U

01�f1� sin /0 df

��O�e� �111�

and


H�f� � v0

k1=2ÿZ 1

0

f1

2

m�F0�g�F0� F 00�f1� sin /0 df1 � e1=2

�ÿZ 1

0

f1

2

m�F0�g�F0� F 00�f1�U1�f1� cos /0 df

ÿZ 1

0

f1

2

d

df1

m�F0�g�F0� F1�f1�

� �sin /0 df1 �

Z f

0

f1

2

c�F0�b�F0�U

01�f1� cos /0 df

��O�e�

�112�

and for the angular coordinates

F �f� � h0 ÿ e1=2A

R h0

h1f 1=2�h�dh

f 1=2�h0� �O�e�;

U�f� � /0 � e1=2CZ 1

0


ÿ �s�F0� cos2�F0� df1 �O�e�:

�113�

Matching these expressions with (91) we obtain

C � B20

�k1 ÿ k2�2k1

tan h0

��gcp

�b�h0��p�h0�r

1

Z 1

0


ÿ �s�F0� cos2�F0� df1

, ! ; �114�

D10 � ÿZ 1

0

f2

d


� �; �115�

D20 �Z 1

0

f2

F 00�f1�m�F0�g�F0� U1�f1�

�ÿ f

2

c�F0�b�F0�U

01�f1�

�df1 �116�

and

A � ÿB10

h�h0�2k1

f 1=2�h0�R h0

h1f 1=2�h�dh

��gc p

�g�h0��a�h0�r

: �117�

5. Results

In this section we outline how the previous analysis can be used for two particular problems,one of them with discontinuity in the boundary and initial conditions for /.

5.1. Continuity in the boundary and initial conditions for /

Let us consider the following boundary and initial conditions


h�0; t� � 0; /�0; t� � p6; vy�0; t� � 10ÿ3

t1=2; vx�0; t� � 2� 10ÿ3

t1=2; t > 0;

h�z; 0� � p4; /�z; 0� � p

6; vy�z; 0� � 0; vx�z; 0� � 0; z > 0:

�118�

The calculations have been performed using the data for MBBA near 25�C, namely q �1 g cmÿ3; k1 � k3 � 6 � 10ÿ7 dyn; l1 � 0:065 P; l2 � ÿ0:775 P; l3 � ÿ0:01 P; l4 � 0:83 P;l5 � 0:46 P and l6 � ÿ0:35 P [8]. The general method will of course apply to any boundary andinitial conditions of the form (25). For the Frank elastic constants, the one constant approxi-mation, namely k1 � k2 � k3 is seen to be reasonable.

If we ®rst neglect the inertia term (i.e. set q � 0 in Eqs. (20)±(24)), an approximate solution isobtained when the function Q�u0� in Eq. (47) is bounded by constants, i.e.

Qmin6Q�u0�6Qmax �119�

or

Q u0� � � Qave where Qave �Z 1

0

Q�u0�du0: �120�

Note that this choice of Q�u0� will lie somewhere within the bounds of (119) (see Fig. 4).Fig. 5 shows the variation of the tilt angle h of the directors with respect to the dimensionless

variable f when inertial terms are neglected and Q�u0� is assumed constant (Q�u0� � Qave;Qmax

and Qmin).Once we have F0�f�, we proceed to obtain the velocity ®elds. Recall Eqs. (27) and (28) that G0�f�

and H0�f�, expressions valid for f small when the inertia q is neglected, are shown in Figs. 6 and 7,respectively. To obtain the expressions valid for f large when the inertia q is neglected, i.e. �G0�f�and �H0�f�, it has been necessary to calculate the constants B10 and B20 involved in their expressions(see Eqs. (72) and (73)). �G0��f� and �H0��f� are shown in Figs. 8 and 9, respectively. One can easilyappreciate how the inner solution, valid for f small, does not satisfy the velocity condition at

Fig. 4. k � Q�u�h�� versus h for MBBA near 25�C when h�z; 0� � p=4 and h�0; t� � 0.


in®nity, as well as how the outer solution, valid for f large, for the velocity does not satisfy thecondition in the neighbourhood of f � 0.

Now we proceed to demonstrate that the higher order approximations for the angular coor-dinates, i.e., F1�f� and U1�f�, as well as for the velocity ®elds G1�f� and H1�f�, all produce verysmall corrections to the ®rst order approximations. Remember that we have assumed the fol-lowing asymptotic expansion of the inner solution, valid for f small

F �f� � F0�f� � e1=2F1�f� �O�e�;U�f� � U0�f� � e1=2U1�f� �O�e�;G�f� � G0�f� � e1=2G1�f� �O�e�;H�f� � H0�f� � e1=2H1�f� �O�e�;

�121�

Fig. 6. Variation of G0�f� ÿ vx � �2� 10ÿ3=t1=2�G0�f�, valid for f small, when the plate is pulled with a velocity

vx � 2� 10ÿ3=t1=2; vy � 10ÿ3=t1=2 and Q�u� is replaced by Qave.

Fig. 5. Variation of the tilt angle of the director h0�f� when inertia q is neglected and Q�u� is assumed constant

(Q�u� � Qmax; Q�u� � Qave and Q�u� � Qmin).


Fig. 8. (a) Variation of G0�f� valid for f large when the plate is pulled with a velocity vx � 2� 10ÿ3=t1=2; vy � 10ÿ3=t1=2

using the exact solution for the tilt angle of the directors when q is neglected F0�f� obtained when Q�u� is replaced by

constants (Q�u� � Qmax; Q�u� � Qave and Q�u� � Qmin). (b) Detailed plot of G0�f�.

Fig. 7. Variation of H0�f� ÿ vy � �1� 10ÿ3=t1=2�H0�f�, valid for f small, when the plate is pulled with a velocity

vx � 2� 10ÿ3=t1=2; vy � 10ÿ3=t1=2 and Q�u� is replaced by Qave.


where e1=2 � 8:7� 10ÿ4 for this particular problem.Assuming Q�u0� � Qave, the next order approximation for the tilt of the director angle valid for

f small is given by Eq. (101). The constant A is calculated using the matching procedure and theexpression obtained is given by (117). The resulting plot for F1�f� is shown in Fig. 10 whenQ�u0� � Qave; Q�u0� � Qmax and Q�u0� � Qmin. One can easily see the small correction F1�f� in-troduces in F �f�, taking into account the small value of e1=2.

In order to calculate the next approximation for the twist angle of the directors about thenormal, U1�f�, we have assumed that the variable coe�cient Q��h� given by Eq. (105) is constantQ��h� � Q�ave. Fig. 11 shows k � Q��h� versus h, as well as the average value taken for this variablecoe�cient. The constant C involved in the expression of U1�f�, recall that U1�f� is given by Eq.(106) has been calculated by the asymptotic matching procedure. Its value is given by Eq. (114).The resulting values of U1�f� have been plotted in Fig. 12. One can also appreciate here the small

Fig. 9. (a) Variation of H 0�f� versus �f when the plate is pulled with a velocity vx � 2� 10ÿ3=t1=2; vy � 10ÿ3=t1=2 using

the exact solution for the tilt angle of the director F0�f� when Q�u� is assumed constant. (b) Detailed plot of H 0��f�.


Fig. 11. Plot of k � Q��h� versus h.

Fig. 10. Solution for the tilt angle of the director F1�f� to order e1=2 when Q�f� is assuming constant. The inner solution

for the tilt angle (valid for f small) is given by F �f� � F0�f� � e1=2F1�f� �O�e� and e1=2 � 8:7� 10ÿ4.

Fig. 12. Solution for the twist angle of the directors about the normal U1�f� to order e1=2 when Q�f� � Qave. The inner

solution for the polar angle (valid for f small) is given by U�f� � p=6� e1=2U1�f� �O�e� and e1=2 � 8:7� 10ÿ4.


correction that U1�f� introduces to p=6 (the zero order solution for the twist angle of the direc-tors).

Figs. 13(a) and 14(a) show the next order approximation for the velocity ®elds G1�f� and H1�f�;respectively. Detailed plots of them are also shown in Figs. 13(b) and 14(b). Here, also smallcorrections are introduced to the zero order solution.

One can easily appreciate how the inner solutions to order e1=2 (for the angular coordinatesF1�f� and U1�f� and the velocity ®elds G1�f� and H1�f�) do not satisfy the condition at in®nity.

Figs. 15 and 16 show the outer solutions for the angular coordinates to order e1=2, that is, F1�f�and U1�f�, respectively. This introduces small corrections to the solution to zero order in e (that is,neglecting inertia) taking into account that we have assumed an asymptotic expansion of the outersolution, valid for f large, of the form

Fig. 13. (a) Variation of G1�f� when Q�f� is approximated by Q�f� � Qave. The component of the velocity in x direction

up to order e is given by vx � �k=t�1=2G�f� where G�f� � G0�f� � e1=2G1�f� �O�e� with e1=2 � 8:7� 10ÿ4 for the par-

ticular problem analysed. (b) A detailed plot of G1�f�.


�F ��f� � h0 � e1=2 �F1��f� �O�e�;�U��f� � /0 � e1=2 �U1��f� �O�e�;�G��f� � �G0��f� � e1=2 �G1��f� �O�e�;�H��f� � �H0��f� � e1=2 �H1��f� �O�e�

�122�

and that e1=2 � 8:7� 10ÿ4.These outer angular coordinates to order e1=2 do not also satisfy the conditions in the neigh-

bourhood of f � 0 due to the singular perturbation character of the problem.

Fig. 14. (a) Variation of H1�f� when Q�f� is approximated by Q�f� � Qave. The component of the velocity in y direction

up to order e is given by vy � �k=t�1=2H�f� where H�f� � H0�f� � e1=2H1�f� �O�e� with e1=2 � 8:7� 10ÿ4 for the par-

ticular problem analysed. (b) A detailed plot of H1�f�.


5.2. Discontinuity in the boundary and initial conditions for /

Let us consider the following boundary and initial conditions:

h�0; t� � 0; /�0; t� � p3; vy�0; t� � 10ÿ3

t1=2; vx�0; t� � 2� 10ÿ3

t1=2; t > 0;

h�z; 0� � p4; /�z; 0� � p

6; vy�z; 0� � 0; vx�z; 0� � 0; z > 0;

�123�

Fig. 15. Outer solution (valid for f large) for the tilt angle to order e1=2. The outer solution �F ��f� has been expanded as

follows: �F ��f� � p=4� e1=2 �F1��f� �O�e�, with e1=2 � 8:7� 10ÿ4.

Fig. 16. Outer solution (valid for f large) for the twist angle to order e1=2. The outer solution �U��f� has been expanded as

follows: �U��f� � p=6� e1=2 �U1��f� �O�e�, with e1=2 � 8:7� 10ÿ4.


An approximate solution of this problem can be obtained by following the matching procedureoutlined in Appendix A. Here, we only show the zero order inner and outer solutions of theangular coordinates h0�f� and U0�f� and those of the velocity ®eld G0�f� and H0�f�.

An approximate zero order inner solution of the tilt angle of the director F0�f� is given by(A.20), say

F0�f� � h1 � A1

Z f

0

1

k1=23

exp

�ÿ k

4Qavef

22

�df2 ÿ

Z f

0

1

k1=23

� exp

�ÿ k

4Qavef

22

�Z 1

f2

H�f1�A20 exp

�ÿ k

2Q�ave

�ÿ Qave

2

�f2

1

�df1 df2

when the common one constant approximation for the Frank constants has been assumed, as wellas Q�F0� � Qave and Q��F0� � Q�ave. In order to calculate the approximate solution of F0�f�, webound H�F0�f1�� as follows:

Hmin < H�F0�f1�� < Hmax: �124�

We have also considered

H�f� � Have with Have � 1

2k3=23

1

�h0 ÿ h1�1

cos2 h1

�ÿ 1

cos2 h0

�; �125�

where

h0 � h�z; 0� and h1 � h�0; t�: �126�

Fig. 17 shows the variation of the tilt angle of the directors, when inertia q is neglected, withrespect to the dimensionless variable f when all of these approximations have been assumed. The

Fig. 17. Approximate solution for the tilt angle of the director F0�f�, when inertia q is neglected, for the problem

studied in Section 5.2.


corresponding curves when H�F0�f1�� Hmin; H�F0�f1�� Hmax and H�F0�f1�� Have is assumedare really very close.

The approximations Q��F0� � Q�ave, as well as k1 � k2 � k3, have also been used for the curve ofthe variation of the twist angle of the directors about the normal U0, when inertia q is neglected, asshown in Fig. 18.

One can easily appreciate how both inner solutions, F0 and U0 match with the zero outer so-lution h0 and /0, respectively.

Figs. 19±22 show the inner and outer solutions for the components of the velocity ®eld, sayG0�f�; H0�f�; �G0��f� and �H0��f�, respectively. These curves use all the mentioned approxima-tions, as they involve in their expressions the approximate solutions calculated for F0�f� andU0�f�.

Fig. 18. Approximate solution for the twist angle of the director U0�f�, when inertia q is neglected, for the problem

studied in Section 5.2.

Fig. 19. Approximate solution for the velocity component G0�f�, when inertia q is neglected, for the problem studied in

Section 5.2.


Fig. 20. Approximate solution for the velocity component H0�f�, when inertia q is neglected, for the problem studied in

Section 5.2.

Fig. 22. Approximate solution for the outer solution of the velocity component in the y direction, �H0��f�, when inertia qis neglected, for the problem studied in Section 5.2.

Fig. 21. Approximate solution for the outer solution of the velocity component in the x direction, �G0��f�, when inertia qis neglected, for the problem studied in Section 5.2.


6. Concluding remarks

When there is not discontinuity in the boundary and initial conditions for / and the inertia q isneglected, the equation for the variation in the tilt angle of the director becomes a non-lineardi�usion equation (see Eq. (B.8)). This non-linear di�usion equation suggests that the informationof the angular coordinate h on the surface di�uses throughout the liquid crystal as a function of zand t, exactly analogous to that of di�usion of mass and temperature.

The introduction of the e�ect of the inertia is included by means of a singular perturbationmethod. The in¯uence of the ¯uid inertia on the higher order approximation to the angular co-ordinates (the tilt angle of the directors h and the twist about the normal /) is very small as can beappreciated in Figs. 10 and 12. Also small is the in¯uence of the inertia on the higher order ve-locity ®eld (see Figs. 13(a) and 14(a)) although, in this case, it is necessary to include the inertia inorder to satisfy the condition at in®nity (see Eqs. (52) and (53)), e.g. the static ¯uid condition attime t � 0.

In the more complicated situation treated in Appendix A where the initial alignment for / isdi�erent from the preferred direction on the surface for /, if we set the density q to zero, we canalso deduce a non-linear partial di�erential equation for the twist angle of the directors about thenormal. This non-linear parabolic di�erential equation is similar to the equation of heat con-duction with ``thermal di�usivity'' and ``heat capacity'', both dependent on h�z; t� and therefore,spatially and time dependent.

For our problem with similarity variables f � z=�kt�1=2, Eq. (A.10) can be integrated to give

s�F0� cos2�F0�U 00�f� � C exp

�ÿZ f

0

kQ��F0�f1��2

f1 df1

�; �127�

where C is a constant. Approximating Q��h� by some constant average, we obtain the approxi-mate solution

U0�f� � /1 � CZ f

0

exp ÿ �k=4�Q�avef21

ÿ �s�F0�f1�� cos2�F0�f1��

df1; �128�

which gives an indication of how information about the angle / is in¯uenced by the variation in hand how this information varies through space and time.

The complete problem of the variations in the angles h and / is complicated due to the non-linear interactions of the full equations. However, the results (125) for U0�f� and (A.20) for F0�f�show the in¯uence of Q�ave on the time-space variation of U0�f� and Qave on the time-space vari-ation of F0�f�, each of these relating to the di�usion-like behaviour.

Acknowledgements

One of the authors (M.D.) wishes to thank the Asociaci�on de Amigos de la Universidad deNavarra for the grant received to carry out her Ph.D. work.


Appendix A

The aim of this appendix is to outline how to proceed when the initial / alignment of thedirectors, /�z; 0� � /0 is di�erent from the / alignment on the surface /�0; t� � /1.

In this case, the inner solution, valid for f small, is expanded as follows.

F �f� � F0�f� � e1=2F1�f� � eF2�f� �O�e3=2�;U�f� � U0�f� � e1=2U1�f� � eU2�f� �O�e3=2�;G�f� � G0�f� � e1=2G1�f� � eG2�f� �O�e3=2�;H�f� � H0�f� � e1=2H1�f� � eH2�f� �O�e3=2�:

�A:1�

These asymptotic expansions are valid when we neglect ¯uid inertia.The inner problem to order e0 turns out to be

�a�F0� cos2�U0� � b�F0��G00�f� � a�F0� cos�U0� sin�U0�H 00�f�f2

m�F0� cos�U0�F 00�f�

� f2

c�F0� sin�U0�U 00�f� � 0; �A:2�

a�F0� cos�U0� sin�U0�G00�f� � �a�F0� sin2�U0� � b�F0��H 00�f� �f2

m�F0� sin�U0�F 00�f�

ÿ f2

c�F0� cos�U0�U 00�f� � 0; �A:3�

2f �F0�F 000 �f� � f 0�F0��F 00�f��2 ÿ k1F 00�f�fk � h�F0�k�cos�U0�G00�f� � sin�U0�H 00�f�� 2s�F0� sin F0 ÿ s0�F0� cos F0� cos F0�U 00�2 � 0 �A:4�

and

k1k2

cos F0 fU 00�f� ��k2 ÿ k1�

2k sin F0�sin /0 G00�f� ÿ cos /0 H 00�f��

� �2s�F0� sin F0 ÿ s0�F0� cos F0�F 00�f�U 00�f� ÿ s�F0� cos F0 U 000 �f� � 0: �A:5�

Combining (A.1) and (A.2) we get

cos�U0�G00�f� � sin�U0�H 00�f� � ÿf2

F 00�f�m�F0�g�F0� �A:6�

and

sin�U0�G00�f� ÿ cos�U0�H 00�f� � ÿf2

U 00�f�c�F0�b�F0� : �A:7�


Introducing these expressions into (A.4) and (A.5), we get

k1k2

p�F0� cos F0fU 00�f� � �2s�F0� sin F0 ÿ s0�F0� cos F0�F 00�f�U 00�f� ÿ s�F0� cos F0 U 000 �f� � 0

�A:8�and

2f �F0�F 000 �f� � f 0�F0��F 00�f��2 ÿ k1ka�F0�fF 00�f�� 2s�F0� sin F0 ÿ s0�F0� cos F0� cos F0�U 00�f��2 � 0: �A:9�

Multiplying (A.8) by cos F0, we get

d

df�s�F0� cos2�F0�U 00�f��

kQ��F0�2

fs�F0� cos2�F0�U 00�f� � 0; �A:10�

where

Q��h� � jk1jp�h�s�h� : �A:11�

The di�erential equation (A.10) can be integrated to satisfy the boundary conditions at f � 0 andat f � 1, that is, U�0� � /1 and U�1� � /0, assuming F0�f� known, to obtain

U0�f� � /1 � A0

Z f

0

exp ÿ �k=2� R f1

0Q��F0�f2��f2 df2

� �s�F0�f1�� cos2�F0�f1��

df1; �A:12�

where

A0 � �/0 ÿ /1�R10

exp ÿ k2

R f1

0Q��F0�f2��f2 df2

� �s�F0�f1�� cos2�F0�f1��= df1

: �A:13�

On the other hand, if we divide (A.9) by 2f 1=2�h�, it can be written as

kQ�F0�2

f 1=2�F0�fF 00�f� �d

df�f 1=2�F0�F 00�f�� J�F0�f��; �A:14�

where

J�F0�f�� H�F0�f��A20 exp

�ÿ k

2

Z f

0

Q��F0�f1��f1 df1

��A:15�

and

H�F0�f�� 1

2f 1=2�F0�d

dF0

�ÿ 1

s�F0� cos2 F0

�: �A:16�


Integrating (A.14) gives

f 1=2�F0�F 00�f� � ÿZ 1

fJ�F0�f1�� exp

k2

Z f1

ff2Q�F0�f2��df2

� �df1

� A1 exp

�ÿ k

2

Z f

0

f2Q�F0�f2��df2

�; �A:17�

with A1 being an unknown constant to be determined by matching. We can calculated an ap-proximate solution of (A.17) by assuming J�f1� a known function of f1, e.g. we can boundH�F0�f1�� as follows:

Hmin < H�F0�f1�� < Hmax; �A:18�

where Hmin and Hmax could be functions of f. In the same way, we can assumeQ�F0�f�� Qave and Q�F0�f�� Q�ave. Then

f 1=2�F0�F 00�f� � A1 exp

�ÿ k

4Qavef

2

�ÿZ 1

fH�f1�A2

0 exp

�ÿ k

2Q�ave

�ÿ Qave

2

�f2

1

�df1

� exp

�ÿ k

4Qavef

2

�: �A:19�

For the convergence of this integral, we require that Q�ave ÿ �Qave=2� > 0. Finally integrating(A.19) gives

F0�f� � h1 � A1

Z f

0

1

k1=23

exp

�ÿ k

4Qavef

22

�df2 ÿ

Z f

0

1

k1=23

exp

�ÿ k

4Qavef

22

��Z 1

f2

H�f1�A20 exp

�ÿ k

2Q�ave

�ÿ Qave

2

�f2

1

�df1 df2; �A:20�

where the common one constant approximation for the Frank constants has been assumed.Once F0�f� and U0�f� are known, we can integrate (A.6) and (A.7) to obtain G0�f� and H0�f�:

G0�f� � u0

k1=2ÿZ f

0

f1

2F 00�f1�

m�F0�g�F0� cos�U0�df1 ÿ

Z f

0

f1

2U 00�f1�

c�F0�b�F0� sin�U0�df1 �A:21�

and

H0�f� � v0

k1=2ÿZ f

0

f1

2F 00�f1�

m�F0�g�F0� sin�U0�df1 �

Z f

0

f1

2U 00�f1�

c�F0�b�F0� cos�U0�df1: �A:22�

Note that these solutions cannot satisfy the boundary conditions at in®nity.Now write the inner solution, valid for f small, in the outer coordinate system �f and expand for

e� 1:


G�f� � u0

k1=2ÿZ 1

0

f1

2

m�F0�g�F0� F 00�f1� cos U0 df1 ÿ

Z 1

0

f1

2

c�F0�b�F0�U

00�f1� sin U0 df1 �O�e1=2�;

H�f� � v0

k1=2ÿZ 1

0

f1

2

m�F0�g�F0� F 00�f1� sin U0 df1 �

Z 1

0

f1

2

c�F0�b�F0�U

00�f1� cos U0 df1 �O�e1=2�;

F �f� � h1 � A1

Z 1

0

1

k1=23

exp

�ÿ k

4Qavef

22

�df2 �A:23�

ÿZ 1

0

exp ÿ k4Qavef

22

ÿ �k1=2

3

Z 1

f2

H�f1�A20 exp

�ÿ k

2Q�ave

�ÿ Qave

2

�f2

1

�df1 df2 �O�e1=2�;

U�f� � /0 �O�e1=2�:

There is no modi®cation in the outer solution with respect to the problem treated in the maintext. Therefore, writing the outer solution in the inner coordinate system and expanding for e� 1gives:

�G��f� � B10 cos /0 � B20 sin /0 �O�e1=2�;�H��f� � B10 sin /0 ÿ B20 cos /0 �O�e1=2�;�F ��f� � h0 � e1=2B10

�h�h0�2k1

��gcp

�g�h0��a�h0�r

�O�e�;

�U��f� � /0 � e1=2B20

�k1 ÿ k2�2k1

tan h0

��gcp

�b�h0��p�h0�r

�O�e�:

�A:24�

Matching (A.23) with (A.24) gives

B10 � �u0 cos /0 � v0 sin /0�k1=2

ÿZ 1

0

f1

2

c�F0�b�F0�U

00�f1� sin�U0 ÿ /0�df1

ÿZ 1

0

f1

2

m�F0�g�F0� F 00�f1� cos�U0 ÿ /0�df1; �A:25�

B20 � �u0 sin /0 ÿ v0 cos /0�k1=2

ÿZ 1

0

f1

2

c�F0�b�F0�U

00�f1� cos�U0 ÿ /0�df1

�Z 1

0

f1

2

m�F0�g�F0� F 00�f1� sin�U0 ÿ /0�df1 �A:26�

and

A1 �h0 ÿ h1 �

R10

exp ÿ k4Qavef

22

ÿ �=k1=2

3

� � R1f2

H�f1�A20 exp ÿ k

2�Q�ave ÿ Qave

2�f2

1

ÿ �df1 df2R1

01

k1=2

3

exp ÿ k4Qavef

22

ÿ �df2

:

�A:27�


Appendix B

The aim of this appendix is to study the nature of the partial di�erential equations for h and /as functions of z and t.

A common approximation in nematic liquid crystal literature is to neglect inertia in (20) and(21). Thus, setting the left-hand side of these equations to zero and integrating over z gives

A�t� � ovx

oz�a�h� cos2 /� b�h�� ovy

oza�h� cos / sin /ÿ oh

otm�h� cos /ÿ o/

otc�h� sin / �B:1�

and

B�t� � ovx

oza�h� cos / sin /� ovy

oz�a�h� sin2 /� b�h�� ÿ oh

otm�h� sin /� o/

otc�h� cos /; �B:2�

with A�t� and B�t� being arbitrary functions of t. Clearly, for conditions in whichovx=oz and ovy=oz both tend to zero at in®nity and h and / both tend to a constant there, we canset A�t� � 0 and B�t� � 0. Then (B.1) and (B.2) can be combined to give

cos /ovx

oz� sin /

ovy

oz� oh

otm�h�g�h� �B:3�

and

sin /ovx

ozÿ cos /

ovy

oz� o/

otc�h�b�h� : �B:4�

Introducing (B.3) and (B.4) into the di�erential equations corresponding to the conservation ofthe angular momentum, that is, Eqs. (22) and (23), we get

2k1a�h� ohot� f 0�h� oh

oz

� �2

� 2f �h� o2hoz2

� �� 2s�h� sin hÿ s0�h� cos h� cos h

o/oz

� �2

� 0

�B:5�and

k1p�h� cos ho/otÿ �2s�h� sin hÿ s0�h� cos h� o/

oz

� �ohoz

� �� s�h� cos h

o2/oz2

� �� 0: �B:6�

Eq. (B.6) can be written as

ooz

s�h� cos2 ho/oz

� �� Q��h�s�h� cos2 h

o/ot: �B:7�

This equation is reminiscent of heat conduction with a thermal di�usivity and ``heat capacity'',both dependent on h, and therefore, spatially and time dependent. Eq. (B.7) gives an indication of


how information about the angle / is in¯uenced by the variations in h and how this informationvaries through space and time. Recall that s�h� and Q��h� are de®ned in Eqs. (24) and (105),respectively.

On the other hand, when the boundary and initial conditions for the angular coordinate / areof the form /�z; 0� � /�0; t� � /0 so that the solution for /, when the density q is neglected, is/�z; t� � /0 for t > 0 and z > 0, (B.5) can be written as

ooz

f 1=2�h� ohoz

� �� Q�h�f 1=2�h� oh

ot: �B:8�

This non-linear di�usion equation suggests that the information of the angular coordinate h onthe surface di�uses throughout the liquid crystal as a function of space and time exactly analogousto that of di�usion of temperature with variable conductivity and heat capacity. Recall thatf �h� and Q�h� are de®ned in Eqs. (24) and (48), respectively.

References

[1] C. Atkinson, M. D�õez, Int. J. Eng. Sci. 37 (1999) 575±607.

[2] C. Atkinson, J.P. Bourne, Int. J. Eng. Sci. 34 (1996) 21±46.

[3] C. Atkinson, Int. J. Eng. Sci. 26 (1988) 1071±1085.

[4] J.L. Ericksen, Trans. Soc. Rheol. 5 (1961) 23±34.

[5] F.M. Leslie, Arch. Rational Mech. Anal. 28 (1968) 265±283.

[6] O. Parodi, J. Phys. (Paris) 31 (1970) 581±584.

[7] A.M. Arthurs, Complementary Variational Principles, Clarendon Press, Oxford, 1980.

[8] M.J. Stephen, J.P. Straley, Rev. Mod. Phys. 46 (1974) 617±704.


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The Rayleigh problem for nematic liquid crystals. Short time solution

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