Liquid Layers, Capillary Interfaces,

8/12/2019 Liquid Layers, Capillary Interfaces,

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Liquid Layers, Capillary Interfaces,and Floating Bodies

Lecture Notes

Erich MiersemannMathematisches Institut

Universit at Leipzig

Version March, 2013


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Contents

1 Introduction 9

1.1 Mean curvature, Gauss curvature . . . . . . . . . . . . . . . . 91.2 Liquid layers . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.3 Capillary interfaces . . . . . . . . . . . . . . . . . . . . . . . . 121.4 Floating bodies . . . . . . . . . . . . . . . . . . . . . . . . . . 141.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2 Liquid layers 192.1 Governing formulas . . . . . . . . . . . . . . . . . . . . . . . . 212.2 Explicit solutions . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.2.1 Adsorption in slit cavities . . . . . . . . . . . . . . . . 222.2.2 Adsorption in cylinder cavities . . . . . . . . . . . . . 25

2.2.3 Adsorption on a cylinder . . . . . . . . . . . . . . . . 272.2.4 Adsorption in spherical cavities . . . . . . . . . . . . . 282.2.5 Adsorption on a sphere . . . . . . . . . . . . . . . . . 29

2.3 Pores of general geometry . . . . . . . . . . . . . . . . . . . . 302.4 Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.4.1 Bounded surfaces . . . . . . . . . . . . . . . . . . . . . 312.4.2 Unbounded surfaces . . . . . . . . . . . . . . . . . . . 34

2.5 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3 Capillary interfaces 43

3.1 Governing energy . . . . . . . . . . . . . . . . . . . . . . . . . 433.2 Equilibrium conditions . . . . . . . . . . . . . . . . . . . . . . 44

3.2.1 The capillary tube . . . . . . . . . . . . . . . . . . . . 473.3 Explicit solutions . . . . . . . . . . . . . . . . . . . . . . . . . 49

3.3.1 Ascent of a liquid at a vertical wall . . . . . . . . . . . 493.3.2 Ascent of a liquid between two parallel plates . . . . . 52

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4 CONTENTS

3.3.3 Zero gravity solutions . . . . . . . . . . . . . . . . . . 54

3.4 Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603.4.1 Strong minimizers . . . . . . . . . . . . . . . . . . . . 653.4.2 On the existence of an embedding foliation . . . . . . 69


4 Floating bodies 934.1 Governing energy . . . . . . . . . . . . . . . . . . . . . . . . . 934.2 Equilibrium conditions . . . . . . . . . . . . . . . . . . . . . . 95

4.2.1 Restricted movements . . . . . . . . . . . . . . . . . . 1034.2.2 Explicit solutions . . . . . . . . . . . . . . . . . . . . . 106

4.3 Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1094.3.1 The oating ball . . . . . . . . . . . . . . . . . . . . . 122


5 Wetting barriers 1355.1 Equilibrium conditions . . . . . . . . . . . . . . . . . . . . . . 1365.2 Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1405.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

6 Asymptotic formulas 145

6.1 Comparison principles . . . . . . . . . . . . . . . . . . . . . . 1456.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

6.2.1 An interior estimate . . . . . . . . . . . . . . . . . . . 1556.2.2 Narrow circular tube, interior ascent . . . . . . . . . . 1556.2.3 Wide circular tube . . . . . . . . . . . . . . . . . . . . 1616.2.4 Ascent at a needle . . . . . . . . . . . . . . . . . . . . 1646.2.5 Narrow tube of general bounded cross section, interior

ascent . . . . . . . . . . . . . . . . . . . . . . . . . . . 1666.2.6 Ascent in a wedge . . . . . . . . . . . . . . . . . . . . 1696.2.7 A numerical method . . . . . . . . . . . . . . . . . . . 173

6.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

7 Surface tension 1877.1 Wilhelmys plate method . . . . . . . . . . . . . . . . . . . . 1897.2 A rod method . . . . . . . . . . . . . . . . . . . . . . . . . . . 1897.3 Paddays method . . . . . . . . . . . . . . . . . . . . . . . . . 1907.4 A new method . . . . . . . . . . . . . . . . . . . . . . . . . . 190


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Preface

In these notes we study liquid layers, capillary interfaces and oating bodies.The leading term in the associated equilibrium equation for the interface is

the mean curvature. In the case of liquid layers no volume constraint orcontact angle occur. Thus the admissible variations which yield necessaryor sufficient conditions are much easier than the variations which we will usefor the oating body problem where we have to take into account contactangles between the capillary surface and the solid materials as well as rigidmotions of the body.

The chapter on liquid layers is based on a cooperation with Peter Schillerand Hans-Jorg Mogel from the Institute of Physical Chemistry at the TUBergakademie Freiberg. Concerning capillary interfaces we recommend theinspiring book Equilibrium Capillary Surfaces of Robert Finn [21]. Therecent interest concerning oating bodies was probably initiated through the

paper Convex particles at interfaces of E. Raphael, J.-M. di Meglio, M.Berger and E. Calabi [57] which was suggested by P.-G. de Gennes.

For the convenience of the reader we tried to keep the Chapter 2 (liquidlayers), Chapter 3 (capillary interfaces) and Chapter 4 (oating bodies)independently from each other with the consequence that some formulasused in Chapter 2 and Chapter 3 follow from more general formulas derivedin Chapter 4. Each of the Chapters 24 contains an appendix with thelemmas, including proofs, which are used in that chapter.

Some of the exercises in the sections Problems are just exercises andothers are open problems.

The content of these notes is not encyclopedic at all.First of all I would like to thank Robert Finn for many inspiring con-

versations over many years, and I am thankful one of the authors of thearticle [57] for sending me a reprint years ago. Part I of the book MinimalSurfaces I of U. Dierkes, S. Hildebrandt, A. K uster and O. Wohlrab [18](second edition of U. Dierkes, S. Hildebrandt, F. Sauvigny [19]) was an es-sential help for me to calculate some variational formulas. Moreover I thank

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8 CONTENTS

Stefan Ackermann for his kindly and permanent help to produce source les

for these and other lecture notes.


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Chapter 1

Introduction

1.1 Mean curvature, Gauss curvature

Assume x : u U R 3, u = ( u1, u2) or u = ( , ), denes a regularsurface S , where U R 2 is a domain in R 2. The normal N = N (u) on S isdened through

N = xu1 xu2|xu1 xu2 |

.

The coefficients of the rst fundamental form are given by

E = x , x , F = x , x , G = x , x ,

and the coefficients of the second fundamental form are

l = N , x , m = N , x = N , x , n = N , x .The mean curvature H = H (u) is dened by

H = lG + nE 2mF

2(EG F 2) ,

and the Gauss curvature by

K = ln

m2

EG F 2 .Remark. In terms of the principle curvatures R1, R2, see [7], pp. 57, onehas

H = 12

1R1

+ 1R2

, K = 1R1R2

.

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10 CHAPTER 1. INTRODUCTION

An important formula which we will use frequently is

x = 2HN, (1.1)

where is the Laplace-Beltrami operator on S , see [18], [19] p.71, p.72,resp., for a proof of this formula.

The following formula will be used concerning the question whether ornot a given equilibrium surface denes a strong minimizer of the associatedenergy functional. Consider a family {F c} of regular surfaces which can bedescribed as its level surfaces

F c = {xR 3 : s(x) = c}.Suppose that

s(x) = 0, then

Q(x) := s(x)|s(x)|

is orthogonal to all surfaces F c, see Figure 1.1, anddiv Q(x) = 2H (x), (1.2)

where H (x) is the mean curvature at x of the surface of the family whichcontains x, see [18], p.77, for a proof of this formula, where x = x(u) is aregular C 2-parametrization of F c such that N (u) = Q(x(u)).

Q(x)

F

F

x

c

c1

2

Figure 1.1: Level curves

1.2 Liquid layers

Porous materials have a large amount of cavities different in size and geom-etry. Such materials swell and shrink in dependence on air humidity. Herewe consider an isolated cavity, see [64] for some cavities of special geometry.


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1.2. LIQUID LAYERS 11

Let s

R 3 be a domain occupied by homogeneous solid material. The

question is whether or not liquid layers l on s are stable, where v isthe domain lled with vapour and S is the capillary surface which is theinterface between liquid and vapour, see Figure 1.2.

N

s

v

lliquid

vapour

solid

S

Figure 1.2: Liquid layer in a pore

Let

E (S , ) = |S|+ w(S ) |l(S )|be the energy (grand canonical potential) of the problem, where surface tension, a positive constant |S|, |l(S )| denote the area resp.volume of

S , l(

S ),

w(S ) = v (S ) F (x) dx ,is the disjoining pressure potential, where

F (x) = c s dy|x y| p .Here c is a negative constant, p > 4 a positive constant ( p = 6 for nitrogen)and x R 3 \ s , where s denotes the closure of s , i. e., the union of s with its boundary s . The disjoining pressure potential prevents theinterface S to meet the container wall.Suppose that S 0 denes a local minimum of the energy functional, then

2H + F = 0 on S 0 ,where H is the mean curvature of S 0.


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1.3 Capillary interfaces

Consider a container partially lled with a bounded amount of liquid, seeFigure 1.3. Suppose that the associate energy functional is given by

)

l

v

s

x3

g

liquid(

vapour

(solid)

S

N

N

S

)

(

Figure 1.3: Liquid in a container

E (S ) = |S| |W (S )|+

l (S )

F (x) dx,

where F = Y , andY potential energy per unit mass, for example Y = gx3, g = const. 0, local density of the liquid, surface tension, = const. > 0, (relative) adhesion coefficient between the uid and the container wall,

W wetted part of the container wall,l domain occupied by the liquid.

Additionally we have for given volume V of the liquid the constraint

|l(

S )

|= V.

It turns out that a minimizer S 0 of the energy functional under the volumeconstraint satises, see [21],2H = + F on S 0cos = on S 0,


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1.3. CAPILLARY INTERFACES 13

where H is the mean curvature of

S 0, a (constant) Lagrange parameter

and is the angle between the surface S 0 and the container wall at S 0.Remark. The angle between two surfaces at a point of intersection is bydenition the angle between the associated normals at this point.

Remark. The term |W|in the above energy functional is called wetting energy .

In the case that the capillary surface S is a graph over R 2, i. e.,x3 = u(x1, x2) denes the capillary surface, and if the container wall is thecylinder surface R , then the related boundary value problem is

div T u = g

u +

in

T u = cos on ,where is the exterior unit normal at , and

T u := u

1 + |u|2,

div Tu is twice the mean curvature of the surface dened by x3 = u(x1, x2),see an exercise. The above problem describes the ascent of a liquid, water

Figure 1.4: Ascent of liquid in a wedge

for example, in a vertical cylinder with constant cross section . It is as-sumed that gravity is directed downwards in the direction of the negative


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x3-axis. Figure 1.4 shows that liquid can rise along a vertical wedge which

is a consequence of the strong nonlinearity of the underlying equations, seeFinn [21].

Remark. Liquid can pilled up on a glass, see Figure 1.5. Here the capillarysurface S satises a variational inequality at S where S meets the containerwall along an edge, see [48]. Such an edge is called wetting barrier .

The photos Figure 1.4 and Figure 1.5 were taken from [50].

Figure 1.5: Piled up of liquid

1.4 Floating bodies

Figure 1.6: A oating paper clip

Consider a particle on a surface of a liquid which lls partially a container.It is assumed that gravity g is directed downwards in the direction of the


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1.4. FLOATING BODIES 15

negative x3-axis. Bodies of density exceeding that of the liquid can oat on

a liquid. Examples are paper clips, see Figure 1.6, needles or razor blades.Even for small particles or if the gravity is small, surface tension is thedominating force.

Some methods of measurement of surface tension are based on the equi-librium conditions for oating bodies. Assume a cylindrical homogeneouslybody = D [0, L] with constant cross section D hangs vertically in a liq-uid, see Figure 1.7, where F is the force directed upwards to keep the bodyin an equilibrium. We suppose that the body can move in the x3-direction

0

F

S1 02

liquid

g

Figure 1.7: Device for measurement of surface tension

only and that no rotation is possible. Let z = v(x), x = ( x1, x2), denes thefree surface S 0. Then the equilibrium conditions are given by

2H 0 + g1v(x) + 0 = 0 on S 0cos 1 = 1 on 1S 0cos 2 = 2 on 2S 0

2 S 0 0, e3 ds + g1 W 2 (S 0 ) y3(v) N 0 (v), e3 dA + g2||

+ 0 W 2 (S 0 ) N 0 (v), e3 dA = F ,where e3 is the unit vector directed in the positive x3-direction, W 2(S 0)is the wetted part of = , N is the normal at directed into the


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body , 1 is the density of the liquid, 2 the density of the body and

y(v) = ( y1(v), y2(v), y3(v)) is the parameter representation of the boundaryof the oating body in an equilibrium.

Remark. a, b or a b denotes the scalar product ni=1 a i bi for vectors

a = ( a1, . . . , a n ), b = ( b1, . . . , bn ) in R n .


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1.5. PROBLEMS 17

1.5 Problems

1. Prove the basic lemma in the calculus of variations: let R n be adomain and f C () such that

f (x)h(x) dx = 0for all hC

10 (). Then f 0 in .

2. Show that 2 H 2 K 0 holds for every regular surface.3. Prove that Q(x) is orthogonal to

F c, see Section 1.1.

4. Show that d iv Tu is twice the mean curvature of S , dened by x3 =u(x1, x2).

5. Write the capillary problem

div Tu = g

u +

in

T u = cos on as a boundary value problem for a quasilinear equation of second order.

6. Prove that a minimizer in C 2() of

J (v) = F (x,v,v) dx g(v, v) ds,is a solution of the boundary value problem

n

i=1

x i

F u x i = F u in

n

i=1

F u x i i = gu on ,

where = ( 1, . . . , n ) is the exterior unit normal at the boundary .

7. Find an energy functional E such that its extremals satisfydiv Tu =

g

u +

in

T u = cos on .


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Chapter 2

Liquid layers

The classical macroscopic theory of capillarity is usually applied to describethe location and the behaviour of the uid in a container. The interfacebetween the liquid and the vapour makes a constant contact angle withthe container wall if the container is made from homogeneous material, ac-cording the general theory of Gauss [29]. The associated strongly nonlinearboundary value problem which describes the interface arose from studies of Young [81], Laplace [39] and Gauss [29].

A rened mesoscopic description incorporates the disjoining pressurewhich results from the long range forces between the uid and the solidsubstrates. The theory of disjoining pressure of liquid layers was initiatedby Derjaguin [15, 16]. In Schiller et al. [63], in particular, a single wedge isconsidered as a special geometry of solid material.

Let s R 3 be a domain occupied by homogeneous solid material, lthe liquid layer, v the domain lled with vapour and S is the capillarysurface which is the interface between liquid and vapour, see Figure 1.2. Let

E (S , ) = |S|+ w(S ) |l(S )| (2.1)be the energy (grand canonical potential) of the problem, where surface tension, |S|, |l(S )| denote the area resp. volume of S , l(S ),

w(S ) = v (S ) F (x) dx , (2.2)is the disjoining pressure potential, where

F (x) = c s dy|x y| p . (2.3)19


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20 CHAPTER 2. LIQUID LAYERS

Here c is a negative constant, p > 4 a positive constant ( p = 6 for liquid

nitrogen) and x R 3 \ s , where s denotes the closure of s , i. e., theunion of S with its boundary s . Finally, set = kT ln(X ) ,

where density of the liquid,k Boltzmann constant,T absolute temperature,X reduced (constant) vapour pressure, 0 < X < 1.

More precisely, is the difference between the number densities of theliquid and the vapour phase. However, since in most practical cases thevapour density is rather small, can be replaced by the density of the liquidphase.

The above negative constant is given by c = H/ 2, where H is theHamaker constant, see [34], p. 177. For a liquid nitrogen lm on quartz onehas about H = 10 20Nm .Ansatz (2.1) can be justied by using density functional theory for uidsin combination with the sharp-kink approximation, see Bieker and Diet-rich [6]. In our approach we neglect a small curvature correction of thesurface tension. This correction is negligibly small if curvature radii aremuch larger than a molecular diameter (0.3 nm). Formulas and asymptoticexpansions of F (x) for plane, spherical, cylindrical and wedge geometriesare published by Philip [56].

Let S ( ), | | < 0, be the family of comparison surfaces given byz(u, ) = x(u) + (u)N (u),

where x(u), , u U R 2, u = ( u1, u2) = ( , ), is a parameter representa-tion of the surface S 0 = S (0) such that the unit normal on S 0N =

xx

|xx |is directed out of the uid into vapour. It is assumed that

S 0 and the scalar

function (u) are sufficiently regular.In the case of a capillary surface with boundary the boundary of the

admissible comparison surfaces S ( ) must lie on the boundary of the soliddomain, see Finn [21], p. 7, for the construction of such surfaces. In ourchoice of energy the liquid covers completely the solid surface since thefunction F dened by (2.3) becomes unbounded near the solid surface.


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2.1. GOVERNING FORMULAS 21

2.1 Governing formulas

A necessary condition such that S 0 denes a local minimum of the energyfunctional E (S , 0) is

dd E (S ( ), 0) =0 = 0 ,

which leads to the equation, see the corollaries to Lemma A.2.1A.2.3 of thethe appendix to this chapter,

2

S 0

H dA +

S 0

F dA

S 0

dA = 0

for all , where H is the mean curvature of S 0, see Section 1.1. An interfaceS 0 is said to be an equilibrium surface if this equation holds. From thisequation we get

Theorem 2.1. Suppose that S 0 denes an equilibrium, then on S 02H + F = 0 . (2.4)

An existing equilibrium state S 0 is said to be stable by denition if d2d 2 E (S ( ), 0) =0 > 0

for all not identically zero. From the corollaries to Lemma A.2.4A.2.6 of the appendix to this chapter, integration by parts and by using the equilib-rium condition (2.4) we nd

Theorem 2.2. An equilibrium interface S 0 is stable if S 0 | |2 2(2H 2 K ) 2 + F, N 2 dA > 0 (2.5)

for all W 1,2(S 0) \ {0}.Here K is the Gauss curvature of the capillary surface S 0, see Section 1.1.Concerning the denition of on S 0 see for example [7], p. 113. TheSobolev space W 1,2(S 0) may be replaced by the space C 1(S 0) of continu-ously differentiable functions. Concerning the denition of Sobolev spaces


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see [2], for instance. If

S 0 is unbounded, then we assume that has compact

support, i. e., 0 on S 0 \ M , where M is a compact set of S 0.Thus, since there is no side condition on , S 0 is stable if

2(2H 2 K ) + 1

F, N > 0 on S 0. (2.6)If the left hand side of (2.6) is constant on S 0, which is satised in allexamples considered in the following, then (2.6) is also necessary for stability.

If inequality (2.5) is satised, then S 0 denes at least a weak local min-imum of the energy functional (2.1), i. e., E (S 0, 0) E (S , 0) for all S ina C 1-neighbourhood of S 0. This follows from the Taylor expansion of theassociated energy functional, see [48], where capillary interfaces are consid-ered.

Remark. In the volume constraint case |l(S )| = const. the equilibriumcondition is2H + F = 0 on S 0 ,

where is the (constant) Lagrange multiplier. An equilibrium is stable inthe volume constrained case if inequality (2.5) is satised for all not iden-tically zero which satisfy the side condition S 0 dA = 0.

2.2 Explicit solutionsIn this section we consider examples of pores with simple geometry where weknow equilibria, i. e., surfaces S 0 which satisfy the necessary condition (2.4).Then we ask whether or not such a surface is stable in the sense that thecondition (2.6) is satised. In general, for given solid surface the equilibriaare not known explicitly.

2.2.1 Adsorption in slit cavities

Assume that two parallel plates have a distance 2 d from each other, and theregions y3 > d and y3 d y21 + y22 + ( x3 y3)2 p/ 2 dy+ c y3 < d y21 + y22 + ( x3 y3)2 p/ 2 dy .


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2.2. EXPLICIT SOLUTIONS 23

s

s

y3d

d

dh

solid

vapour

solid

d+h

liquid

liquid

Figure 2.1: Slit cavity

The plane surfaces S 1 : y3 = d h and S 2 : y3 = d + h dene stable equi-libria if F = 0 on S i (equilibrium condition) and F/N F, N > 0on S i (stability condition).Let x be on S 1 and set q = h/d , then

F (x) = cd3 p (1(q ) + 2(q )) ,

where

1(q ) = y3 > 1 y21 + y22 + (1 q y3)2 p/ 2 dy2(q ) = y3 < 1 y21 + y22 + (1 q y3)2 p/ 2 dy.

Integration is elementary and yields

1(q ) = 4

( p 3)( p 2)q p+3

2(q ) = 4

( p 3)( p 2)(2 q ) p+3 .

The normal derivative F/N := F, N of F on S 1 isF N

= cd2 p 1(q ) + 2(q ) .

We recall that the normal is always directed into the vapour and the constantc is negative. Set

f (q ) = cd3 p (1(q ) + 2(q )) ,


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then f (q ) > 0, f (q ) > 0 on 0 < q < 2, and lim q0 f (q ) =

, limq2 f (q ) =

. Let 0 be the zero of f (q ) = 0, i. e. 0 = 1 because of the symmetry of f (q ) with respect to q = 1. Set

X 0 = exp 1kT

f (1)

and assume the constant X , 0 < X < X 0, is given, then there are exactlytwo pairs of planes which are equilibria. These planes are dened by thetwo zeros 1, 2 of

kT ln(X ) = f (q ) , 0 < 1 < 1 < 2 < 2 ,see Figure 2.2, and the layer with thickness h = 1d is stable. We repeat

q

11 2

Figure 2.2: Layer with thickness h = 1d is stable

the same considerations on S 2. Let X , 0 < X < X 0 be given, then the liquid layers with thickness h = 1d are stable.

Remark. The above considerations show that the cavity lls up in a stablemanner in contrast to cylinder or ball cavities, where not all curvaturesdisappear, see the following subsections. This contradicts experiments. Amodied energy ansatz which takes into account the interaction of moleculesin the vapour domain yields probably the right behaviour of liquid layers inslit cavities.


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From above it follows a result for a single plate: For any 0 < X < 1

there is exactly one plane which is an equilibrium, and this equilibrium is stable. These results for plates and the consideration in Schiller et al. [63],p. 2230, suggest a conjecture: if two plates are tilded, i. e., a wedge isoccupied by solid material, then there is at least one stable equlibrium whichis asymptotically a plane lm far away from the edge of the wedge.

2.2.2 Adsorption in cylinder cavities

Let the cross section of the tube be a disk with radius R. The solid domainis

s = {(y1, y2, y3)R 3 : y21 + y22 > R 2, < y 3 < } .We claim that there are stable liquid layers given by r2 < x 21 + x22 < R 2,< x 3 < , where r > 0, see Figure 2.3. The potential (2.3) is

N

solidliquid

vapour

Rr

S

Figure 2.3: Cylinder cavity

F (x) = cR3 p(q ), q = r/R ,

where r 2 = x21 + x22, and

(q ) =

(y1 q )2 + y22 + y23 p/ 2 dy ,

the domain of integration is here y21 + y22 > 1, < y 3 < .For stability considerations we need some properties of (q ). The func-tion (q ) is, see Philip [56],

(q ) = 3/ 2 12 ( p 3)

12 p F

12

( p 3), 12

( p 1);1; q 2 ,


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where F denotes hypergeometric functions. From integral representation of

hypergeometric functions, see for example [1], p. 558, Eq. 15.3.1, it follows (q ) > 0 and (q ) > 0 on 0 < q < 1. The denition of (q ) implies that(q ) > 0, limq1 (q ) = , (0) > 0 and (0) = 0.

The cylinder surface

S = {(x1, x2, x3)R 3 : x21 + x22 = r 2, < x 3 < }denes an equilibrium if

= r cR3 p

rR

since the mean curvature of S is H = (2 r ) 1

. Such a surface is stable if

r 2 cR2 p

rR

> 0

since the Gauss curvature K of a cylinder surface is zero.Set

f (r ) = r cR3 p

rR

.

From the above properties of it follows that f (r ) is positive, strictly convexon 0 < r < R , and lim r 0 f (r ) = , limr R f (r ) = , see Figure 2.4 for atypical graph of f (r ).

1 2 3 4

1

2

3

4

5

Figure 2.4: Graph of f (r ), layer in a pore

Let r 0 be the zero of f (r ) = 0 and set

X 0 := exp 1kT

f (r 0) .

Then


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(i) for given 0 < X < X 0 there are exactly two cylinders which are equilibria.

These spheres are dened by the zeros r1, r2, r1 < r 0 < r 2, of

kT ln(X ) = f (r ) .The liquid layer associated to the larger radius r 2 is stable and the other one is not stable. If X 0 < X 1, then there is no equilibrium.(ii) limR 0 X 0(R) = 0 .

The behaviour (ii) follows from denition of X 0 since r0 0, and itshows for given (even small) reduced vapour pressure that there are noliquid layers if the pore is sufficiently small.

2.2.3 Adsorption on a cylinder

The solid domain is

s = {(y1, y2, y3)R 3 : y21 + y22 < R 2, < y 3 < }and the potential (2.3) is given by

F (x) = cR3 p(q ), q = r/R ,

where r 2 = x21 + x22, r > R , and

(q ) = (y1 q )2 + y22 + y23 p/ 2

dy ,

the domain of integration is here y21 + y22 < 1, < y 3 < .The cylinder surface

S = {(x1, x2, x3)R 3 : x21 + x22 = r 2, < x 3 < }denes an equilibrium if

= r cR3 p

rR

since the mean curvature of S is H = (2r ) 1. Such a surface is stable if r 2 + cR2 p rR > 0.

From properties of , we omit the details, it follows that for given 0 R 2},R > 0, and the potential (2.3) is

F (x) = cR3 p(q ),

where(q ) := y

21 + y

22 + ( y3 q )2

p/ 2 dy.

with q = r/R , and r 2 = x21 + x22 + x23, 0 < r < R . The domain of integrationis y21 + y22 + y23 > 1. An elementary calculation yields, if p > 4,

(q ) = 2

( p 2)( p 3)1q

1 p 4

(1 q ) p+4 (1 + q ) p+4

+ (1 q ) p+3 (1 + q ) p+3 .It follows that (q ) > 0, (q ) > 0, 0 < q < 1, since (q ), 0 < q < 1,can be written as a convergent power series expansion (q ) = k=0 a2kq

2k ,where all coefficients a2k are positive.

A sphere S : x21 + x22 + x23 = r 2 denes an equilibrium if

= 2r cR3 p rR =: f (r )since the mean curvature of S is H = r 1. Such a surface is stable if

2r 2 cR2 p

rR

> 0

since the Gauss curvature of S is K = r 2.Thus the same assertions (i) and (ii) of the previous subsection 2.2 hold for the case of spheres, where f and are taken from above.

Remark. In contrast to the above result each layer is stable in the volumeconstrained case. This follows since S 0 : |x| = r , 0 < r < R , satises theequilibrium condition

2H 2|c|R3 prR = 0 ,


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where H = r 1, and is the constant such that this equation holds for

given r . The stability criterion (2.5) is satised for all W 1,2(S 0) \ {0}which satisfy the side condition S 0 dA = 0, sinceF N

= 2|c|R2 prR

is positive and

S 0 | |2 2r 2 2 dA = B 1 (0) | |2 2 2 dA 0for all W 1,2(B 1(0)) which satisfy the side condition

B 1 (0) dA = 0,

where B 1(0) denotes the boundary of the ball B1(0), see [78]. The proof of this inequality is based on methods from [77].

2.2.5 Adsorption on a sphere

The domain occupied by solid material is

s = {(y1, y2, y3)R 3 : y21 + y22 + y23 < R 2},R > 0, and the potential (2.3)

F (x) = cR3 p(q ),

where(q ) = y

21 + y

22 + ( y3 q )2

p/ 2 dy

with q = r/R , and r 2 = x21 + x22 + x23, 0 < R < r . The domain of integrationis y21 + y22 + y23 < 1. An elementary calculation yields, if p > 4,

(q ) = 2

( p 2)( p 3)1q

1 p 4

(1 + q ) p+4 (q 1) p+4

+ (q 1) p+3 + (1 + q ) p+3 .A sphere

S : x21 + x22 + x23 = r 2, r > R , denes an equilibrium if

= 2r

+ |c|R3 prR

since the mean curvature of S is H = r 1. Such a surface is stable if

2r 2 |c|R2 p

rR

> 0


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since the Gauss curvature of

S is K = r 2.

We obtain qualitatively the same result as in the case of a layer on acylinder. For given 0 < X < 1 there is exactly one sphere which is inequilibrium, and this sphere is stable, see Figure 2.5 for a typical graph of g(x).

15 20 25 30

0.1

0.2

0.3

Figure 2.5: Graph of g(x), Adsorbtion on a sphere

2.3 Pores of general geometry

In the case of general geometry of a pore equilibria are not known explicitely.A proposal is to start with an equilibrium surface S (0) for a large |0|calculated by a numerical method, then a numerical continuation procedureyields an equilibrium S () for increasing parameter . Simultaneously onechecks whether or not S () is stable. If S () satises the stability criterion(2.5) for < c and S (c) violates this condition, then c is said to be thestability bound of this continuation procedure.

Such a numerical continuation was applied to a unilateral problem for therectangular plate, see [51]. In this example the parameter of continuation isan eigenvalue.

2.4 Stability

Here we consider the question whether or not equilibria dene also minimiz-ers of the associated energy functional. The considerations in this section


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2.4. STABILITY 31

were suggested by a paper of Wente [79] concerning capillary tubes of general

cross sections.

2.4.1 Bounded surfaces

Here we suppose that the single cavity R 3 \ s is bounded. An example isa spherical cavity.

Theorem 2.3. (Bounded cavities.) Let R 3 \ s be a bounded single pore.Suppose that the free interface S 0 of the liquid layer is sufficiently regular,satises the necessary condition (2.4), and the second variation is positive for all nonzero variations, see (2.5). Then there exists a > 0 such that

E (S 0, 0) < E (S 0, 0) for all surfaces S different from S 0, and located in a -neighbourhood of S 0.We suppose that S is sufficiently regular such that the divergence theorem holds.Proof. The proof is based on a method of H. A. Schwarz [67] for minimalsurfaces. We will show that an equilibrium surface S 0 can be embeddedin a foliation, provided that the second variation is positive for all nonzerovariations. Dene for a small > 0 the -neighbouhood of S 0 by

D(

S 0) =

{y

R 3 : y = x(u) + s N (u),

< s <

}.

A family S (), (0 , 0 + ), which covers D simply is called foliation ,and a surface S 0 is called embedded in this family if S 0 = S (0).Lemma 2.1. (Existence of an embedding family). Let S 0 be a solution of

2H + F 0 = 0 on S 0,and suppose that the second variation of E (S , 0) at S 0 is positive on W 1,2(S 0)\{0}. Then there exists a foliation S () of D(S 0), where > 0 is sufficiently small, and each element of this foliation solves

2H + F

= 0 on

S ().

Proof. We will show that the embedding family is given by x + (u, )N ,where the scalar function is the solution of

2H (x + N ) + F (x + N ) (0 + ) = 0 .


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The constant is from the interval (

, ). Dene the mapping

M (, ) : C 2, (S 0) ( , ) C (S 0),where 0 < < 1 is a constant H older exponent, by

M (, ) = 2H (x + N ) + F (x + N ) (0 + ).Since M (0, 0) = 0 , and M (0, 0), dened by

M (0, 0) h = h 2(2H 2 K )h + F N

h,

see [7], p. 186, for the formula of the rst variation of H , is a regular mappingfrom C 2, (S 0) C (S 0), it follows from an implicit function theorem thatthere is a unique solution

= (u, ) = v(u) + r (u, )

of M (, ) = 0, where r (u, ) C 2, (S 0) ||r ( )||C 2 , (S 0 ) = o( ) as 0, r exists, r C 2, (S 0) and lim 0 ||r (u, )||C 2 , (S 0 ) = 0. Moreover, v is thesolution of the Jacobi equation

v 2(2H 2 K )v + F N

v 1 = 0 (2.7)

on S 0. From this equation and from the fact that the second variation ispositive for all nonzero variations, we obtain that v > 0 on S 0 as follows.Set

(u) = max {v(u), 0}and

q = 2(2H 2 K ) + 1

F N

.

Then

0 = S 0 ( v + qv 1 ) dA= S 0 (v + qv 1 ) dA= { v 0} (| |2 + q 2 + 1 ) dA= S 0 (| |2 + q 2 + 1 ) dA.


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2.4. STABILITY 33

Combining this equation and inequality (2.5), we nd that

0, i. e., v

0

on S 0. This inequality and equation v + q v =

1

on S 0imply that v > 0 on S 0. If not, then there is a P S 0 where v(P ) = 0.Then v = 1 / and v 0 at P , which is a contradiction. Theprevious argument is called touching principle . Finally, for given y = x(u)+s N (u) D(S 0), > 0 sufficiently small, there exists a unique solution(s) of (u, ) = s since (u, 0) = v(u) > 0.

Let xD(S 0) and consider the associated surface S ((x)) from the familyS (). We recall that (x) is constant on S ((x)), and

2H + F (x) (x) = 0 on S ((x)) .Let n be the inward normal on the surface S ((x)) at x, see Figure 2.6, then

div n = 2H.Combining both equations, we have at xD (S 0) the equation

n

S

S

x

(x))S (

0

n

=

Figure 2.6: Proof of Theorem 2.1

div n + 1

(F (x) (x)) = 0 .Let S be any other regular surface sitting inside of D(S 0). Here we assumethat S is inside of the domain dened by S 0, see Figure 2.6. For the general


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case see a remark in [18], p. 81 (at the end of the proof of Lemma 1). Let

T 0,S be the domain between S 0 S (0) and S . Then0 = T 0 , S div n + 1 (F (x) (x)) dx

= T 0 , S n dA + 1 T 0 , S (F (x) (x)) dx,where is the exterior unit normal on T 0,S . Since = n on S 0, it follows

|S||S 0| = S (1 n ) dA 1 T 0 , S (F (x) (x)) dx. (2.8)From the denition (2.1) of the energy we get

E (S , 0) E (S 0, 0) = (|S||S 0|) + T 0 , S F (x) dx 0|T 0,S |. (2.9)Combining (2.9) and (2.8), we nd that

E (S , 0) E (S 0, 0) = S (1 n ) dA + T 0 , S ((x) 0) dx.Since (x) > 0 if x

T 0,S , the theorem is shown.

Remark. Liquid layers in equilibrium on and in spherical cavities studiedabove which are stable in the sense that the second variation is positive forall nonzero variations are stable in the sense of the previous theorem.

Remark. In all explicit examples above this lemma is superuous sincean embedding family is dened in a natural way, in the cases of ball orcylinder cavities by varying the radius r . The resulting surfaces are planes,concentric spheres or coaxial cylinders, resp.

2.4.2 Unbounded surfaces

Here we suppose that the single cavity R 3 \ s is unbounded. Examples areslit and cylinder cavities. In fact, there are no such cavities. On the otherhand, such objects serve as an approximation of long cylinder cavities, forexample. In general, the denition of the energy by (2.1) makes no senseif S is unbounded. But the difference (2.9) is well dened if S 0 is replced


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2.4. STABILITY 35

by a sufficiently regular bounded subdomain

MS 0. For given sufficiently

small positive dene a -neighbourhood of M byD M (S 0) = {y R 3 : y = x + s N, x M, < s < }.

In this section we consider the example of a cylinder type cavity, see Fig-ure 2.7. We will nd an embedding family x+ ( )N of M, where = (u, )

M

MS

l

v

s

0

M

Figure 2.7: Cylinder type cavity

is the solution of the boundary value problem

2H (x + N ) + F (x + N ) (0 + ) = 0 on M = on M.

Replacing the lateral surfaces of D(M) by surfaces 0 D M (S 0) closeto such that n = 0 on 0, where is the normal on 0 and n is thenormal on the surfaces of the embedding family. The resulting domain isdenoted by D0 (M), see Figure 2.8. Set M0 = S 0 D 0 (M) and supposethat a sufficiently regular surface S is in D0 (M), see Figure 2.8. Let T 0,S be the domain between M0 and S . Then the considerations of the proof of the theorem in the previous subsection leads toTheorem 2.4 (Unbounded cavities.) Suppose that the free interface S 0 of the liquid layer is sufficiently regular, satises the necessary condition (2.4)and the second variation, see (2.5), is positive for all nonzero variations with compact support. Then there exists a > 0 such that

E (S, M0, 0) := (|S||M0|) + T 0 , S F (x) dx 0|T 0,S | > 0 for all surfaces S different from M0, and which are located in D0 (M).


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S

00

00

D0 (M)

Figure 2.8: Proof of Theorem 3.1

Remark. In the case of a cylinder cavity, see Section 2.2.2, the lateralsurfaces are already perpendicular on the surfaces of the embedding family,i. e., we can set = 0.

Remark. Liquid layers in equilibrium on and in cylindrical cavities studiedin Section 2.2.2 which are stable in the sense that the second variation ispositive for all nonzero variations with compact support are stable in thesense of the previous Theorem 2.4.

2.5 AppendixConsider a family S ( ), | | < 0, of surfaces given by

z(u, ) = x(u) + (u)N (u),

where is a given sufficiently regular function on the parameter domainU R 2 where the support is a compact subset of U . Set

W (u, ) = E (u, )G(u, ) F 2(u, ),where E (u, ), G(u, ) F (u, ) are the coefficients of the rst fundamental

form of S ( ), and N (u, ), H (u, ) are the normal and the mean curvatureassociated to the surface S ( ), resp.Lemma A.2.1.

dd |S ( )| = 2 U H (u, ) N (u, ), z (u, ) W (u, ) du.


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2.5. APPENDIX 37

Proof. Set u = ( , ), then

dd |S ( )| = U 1W (u, ) E (u, ) z (u, ), z (u, )

F (u, )[ z (u, ), z (u, ) + z (u, ), z (u, ) ]+ G(u, ) z (u, ), z (u, ) du.

The formula of the lemma follows by integration by parts, see [18, 19], p.45, p. 44, resp., and by using the formula z = 2HN , see [18, 19], p. 71,p. 72, resp., for a proof of this formula.

Corollary. dd |S ( )| =0 = 2 S 0 H 0 dA

where S 0 = S (0) , and H 0 denotes the mean curvature of S 0.Let

F (x) = c s dy|x y| p ,where c is a positive constant, and p 4 is a constant. Set

w(S ) = v (S ) F (x) dx,here v denotes the domain lled with vapour.

Lemma A.2.2.dd

w(S ( )) = U F (z(u, )) N (u, ), zt (u, ) W (u, ) du.Proof.

v (S ( )) F (x) dx = v (S 0 ) F (x) dx

0 U F (z(u, t ))det z(u, t ) (u, t ) dudt.This formula holds also if S ( ) is not completely located in the vapour regionassociated to S 0, see a remark in [18], p.81.

Sincez (u, t ) (u, t )

= z (u, t )z (u, t ), zt (u, t )= N (u, t ), zt (u, t ) W (u, t ),


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it follows that

dd v (S ( )) F (x) dx = U F (z(u, )) N (u, ), zt (u, ) W (u, ) du.

Corollary. dd

w(S ( )) =0 = S 0 F (x(u)) dA.Lemma A.2.3.

dd |l(S ( )) | = U N (u, ), z (u, ) W (u, ) du.Proof. If S ( ) is completely located in the vapour region associated to S 0,then l(S ( )) = l(S 0)1, where 1 = {z(u, t ) : u U, 0 < t < }. Thenal formula holds for the general case, see a remark in [18], p.81. Then

|l(S ( )) | = |l(S 0)|+

0 U N (u, t ), zt (u, t ) W (u, t ) dudt.Corollary.

dd |l(S ( )) | =0 = S 0 dA.

Second derivatives of the integrals with respect to are used for stabilityconsiderations.

SetD(u, ) = N (u, ), z (u, ) W (u, ),

and let be the Laplace-Beltrami operator on S 0, and H 0 = H (u, 0),K 0 = K (u, 0) the mean and Gauss curvature of

S 0, resp.

Lemma A.2.4.

d2

d 2 |S ( )| =0 = 2 S 0 + 2(2 H 20 K 0) dA2 U H 0(u)D (u, 0) du.


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2.5. APPENDIX 39

Proof. From Lemma A.2.1 we get

d2

d 2 |S ( )| =0 = 2 S 0 H (u, 0) dA 2 U H (u, 0)D (u, 0) du.Then the formula of the lemma follows since

2H (u, 0) = + 2(2 H 20 K 0) see for example [7] or the appendix to the chapter on capillary interfaces.

Finally we obtain from Lemma A.2.2 and Lemma A.2.3

Lemma A.2.5. d2

d 2w(S ( )) =0 = S 0 F, N 2dA + U F (x(u))D (u, 0) du.

Lemma A.2.6.

d2

d 2 |l(S ( )) =0 = U D (u, 0) du.

Remark. In the above integrals over

S 0 we can assume that

C 1(

S 0) or

C 2(S 0), resp., if S 0 is bounded. This follows by partition of unity andsince S 0 has no boundary.Remark. We did not need an explicit formula for D (u, 0) since it wasassumed that the interface S 0 is in an equilibrium, i. e., equation (2.4) issatised. On the other hand there is an explicit formula

D (u, 0) = 2H (u, 0) 2W (u, 0), (2.10)see an exercise.


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2.6 Problems

1. Assume 2(H 2 K ) + F, N / is constant at the equilibrium stateS 0. Show that 2(H 2 K ) + F, N / 0 is a necessary conditionfor stability.Hint: := const.

2. Show that (q ), dened in Section 2.2.2 (adsorption in cylinder cav-ities) satises (q ) > 0, (q ) > 0, (q ) > 0 on 0 < q < 1, andlimq1 (q ) = , limq0+0 (q ) > 0, (0) = 0.

3. Set

f (r ) = r cR3 p

rR ,

see Section 2.2.2. Prove that f (r ) is positive, strictly convex on 0 0, (relative) adhesion coefficient between the uid and the container wall,

W wetted part of the container wall,l domain occupied by the liquid.Additionally we have for given volume V of the liquid the constraint

|l(S )| = V. (3.2)By |S|, |l(S )| we denote the area resp. volume of S , l(S ).

43


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44 CHAPTER 3. CAPILLARY INTERFACES

The problem is to nd minimizers of

E in an appropriate class of comparison

surfaces S .Let S 0 be a given regular surface. Then we will derive necessary andsufficient conditions such that this surface denes a minimizer of the asso-ciated energy functional subject to a given family of comparison surfaces.Assume x : u U R 3, u = ( u1, u2) or u = ( , ), denes the regularsurface S 0, where U R 2 is a domain such that x(u) if u U , andx(u) R 3 \ {s } if u U . We choose the parameters such that thenormal

N S 0 = x x

|x x |is directed out of the liquid, and that the normal N at the container wallis directed out of the solid material s of the container, see Figure 1.3.

3.2 Equilibrium conditions

Following Finn [21], Chapter 1, we dene for a given conguration S 0, 0,where 0 = 0, a one parameter family of admissible comparison surfaceswhich are in general not yet volume preserving. Set

z0(u, ) = x(u) + (u) + r (u, ),

| | < 0, the remainder r is continuously differentiable with respect to all

arguments, such that r = O( ) as 0, z0(u, ) R 3 \ (s0) if u U ,z0(u, ) if uU , z0(u, ) 0 if u 2U , and (u) = (u)N S 0 (u) + (u)T S 0 (u)

is a given vector eld. Here N S 0 denotes the unit normal to S 0 pointed tothe exterior of the liquid, and T S 0 is a unit tangent eld dened on a stripU of U of width , such that on S 0, 0 := T S 0 is orthogonal to S 0 andpoints to the exterior of the liquid, see Figure 3.1. In generalization to thenormal at the boundary of a two dimensional domain the vector 0 at S 0is called conormal . We assume that and are sufficiently regular on U ,supp U , and 2 + 2 1. Dene on S 0 = {x(u) : u U } the angle [0, ] by cos = N N S 0 . This angle depends on P S 0. Later wewill see that this angle is a constant. Since , N = 0 at S 0 and N , N S 0 ,T S 0 are in a common plane, it follows

(u)cos (u)sin = 0 , (3.3)


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3.2. EQUILIBRIUM CONDITIONS 45

0

TS

S

S 0

0

Figure 3.1: T S 0 at S 0

where uU . The family z0(u, ), where satises equation (4.3), is calleda family of admissible comparison surfaces. Such a family exists providedthe boundary of the container is sufficiently smooth. The proof exploits the

implicit function theorem, we omit the details. Set

V = { = N S 0 + T S 0 : (, ) satises (3 .3)}.Suppose that for given V the family z(u, ) is volume preserving, i. e.,|(S ( )| = V for all | | < 0. It follows that

S 0 dA = 0 (3.4)since, see the corollary to Lemma A.3.4 of the appendix to this chapter,

dd |(S ( )| =0 = S 0 dA.On the other hand, let V satises the side condition (3.4), then there

exists a volume preserving family of admissible comparison surfaces givenby z(u, ) = x(u) + + O( 2) with a sufficiently regular remainder O( 2),see Lemma A.3.1 of the appendix to this chapter. Suppose that S 0 is en-ergy minimizing subject to such a family of comparison surfaces S ( ), i. e.,E (S ( )) E (S 0). Then

dd E (S ( )) =0 = 0 . (3.5)

A sufficiently regular surface S 0 dened by x(u) is said to be an extremal if equation (3.5) is satised, where S ( ) is a volume preserving family of comparison surfaces dened above. Set

V 0 = { V : S 0 dA = 0}. (3.6)


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Let

f (x), = dd E (S ( )) =0

g (x), = dd |(S ( )| =0 ,

whereg (x), = S 0 dA

and, see the corollaries to Lemma A.3.2A.3.4 of the appendix to this chap-ter,

f (x), = 2 S 0 H 0 dA + S 0 ds S 0 ( sin + cos ) ds + S 0 F (x) dA.

From a Lagrange multiplier rule, see Chapter 8, it follows that there existsa real constant such that

f (x), + g (x), = 0 (3.7)

for all V .Theorem 3.1. Suppose that S is an extremal, then

2H + F (x) + = 0 on S 0cos = on S 0.

Proof. See Finn [21], p. 10. From the above formula (3.7) we get

0 =

S 0

(2H + F (x) + ) dA+ S 0 [ sin + (1 cos )] ds

for all ( (u), (u)) satisfying 2 + 2 1 in U and (3.3). Thus

2H + F (x) + = 0 on S 0


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3.2. EQUILIBRIUM CONDITIONS 47

and

S 0 [ sin + (1 cos )] ds = 0for all ( (, ), (, )), ( , )U , satisfying (3.3).Set

= sin , = cos ,

where = (, ), | | 1, C (U ). Then

S 0 ( cos ) ds = 0for all these . It follows that cos = on S 0.Remark. If Y is the gravitational potential Y = gx3, g gravity, directeddownwards into the negative x3-axis, see Figure 1.3, then

2H = 0 + g

x3 on S 0where 0 = / . If is a constant (incompressible uid), then the constant

= g

is called capillarity constant .

3.2.1 The capillary tube

Consider a capillary tube with constant cross section R 2. We assumethat the cross section is perpendicular to the x3-axis, see Figure 3.2. Supposethat the capillary surface S is a graph over and that the potential Y isthe gravitational potential Y = gx3, where g denotes the gravity directeddownwards into the negative x3-axis. Set 0 = / , then the boundaryvalue problem of Theorem 3.1 reads as

div Tu = u + 0 in (3.8) T u = cos on , (3.9)where is the exterior unit normal at (where it exists), and

T u = u

1 + |u|2.


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x3

g

liquid

S

Figure 3.2: Capillary tube

If > 0, then the following mapping eliminates the Lagrange parameterin (3.8). Set

u = v 0

, (3.10)

then we arrive at

div Tv = v in T v = cos on .If is bounded and sufficiently regular, then the Lagrange parameter isdened through the data of the problem. To see this, we integrate equa-tion (3.8) over , use the boundary condition (3.9) and obtain

0 = | |cos V ||

,

where V denotes the volume of the liquid. In this calculation we supposethat there are no dry spots on the bottom . Consequently let v be asolution of the previous boundary value problem, then the solution u of theproblem (3.8) is given by (3.10).

Scaling. In many applications one considers tubes with a cross section of a small size. It is often convenient to have such a small parameter in theequation instead in the domain. An important example is a disk Ba (0)


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with radius a and center at the origin. The boundary value problem in

consideration is

div Tu = u in (a) T u = cos on (a)

where x (a), u = u(x), and a is a positive parameter, a representativelength. The mapping (similarity transformation)

y = xa

, v(y) = 1a

u(ay)

transforms this problem into

div T v = Bv in T v = cos on ,

whereB = a 2

is called Bond number . Here we assume that the mapping y = x/a maps(a) onto a xed domain , independent of a .

3.3 Explicit solutions

There are only few examples where explicit solutions are known. We will seein Chapter 6 that such solutions can serve as a leading term for asymptoticexpansions of solutions to problems where no explicit solutions are known.

3.3.1 Ascent of a liquid at a vertical wall

Let = {(x1, x2) R 2 : x1 > 0} be the right half plane, and consider theboundary value problemdiv Tu = u in

T u = cos on ,

where is a positive constant. Now we assume additionally that there existsa solution u which depends on x = x1 only such that lim x u(x) = 0 andlimx u (x) = 0. In the following we will nd such a solution. From amaximum principle of Finn and Hwang [25], see Section 6.1, it follows thatthis solution is the only one of the original boundary value problem without


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the additionally assumptions . Thus we have to consider a boundary value

problem for an ordinary differential equation.

u (x)

1 + u (x)2= u(x), 0 < x < (3.11)

limx0, x> 0

u (x)

1 + u (x)2 = cos (3.12)

limx

u(x) = 0 , limx

u (x) = 0 . (3.13)

From the identities

gh

x

S

liquid

u

Figure 3.3: Ascent of liquid at a vertical wall

u (x)

1 + u (x)2=

u (x)(1 + u (x)2)3/ 2

(3.14)

= 1

u (x) 1

1 + u (x)2(3.15)

it follows from (3.11) that

1

u (x) 1

1 + u (x)2= u

1

1 + u (x)2=

12

(u2) .


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Consequently we have

1

1 + u (x)2 =

12

u 2 + C,

where C is a constant. From the boundary conditions (3.13) we see thatC = 1. Then

1

1 + u (x)2 = 1

12

u 2. (3.16)

Combining this formula with the boundary condition (3.12), we get for theascent h = u(0) of the liquid at the wall, see Figure 3.3 for the case 0 / 2,

h = 2 1 sin .Instead of u = u(x) we consider the inverse function x = x(u). Here we

assume that x (u) = 0. It turns out that this inequality is satised for thesolution which we will calculate in the following. Then we obtain from (3.16)equation

x (u)

1 + x (u)2 = 1

12

u 2. (3.17)

We have used the additional assumption that x (u) < 0, which is satisedfor the calculated solution. We suppose here that 0 < / 2. From (3.17)it follows

x (u) = u2/ 2 1

1 (u 2/ 2 1)2= u2/ 1u/ 2 u2

,

where := 2 / . Using the substitution

= 2 u2,we nd thatx(u) =

u2/ 2

1

1 (u 2/ 2 1)2 du + C = + d 2 2 + C = 2 u2 +

2 2 ln

2 + 2 u2 2 2 u2+ C.


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The constant follows from the condition x(u(0)) = 0, where u(0) = h, and

the nal formula is

x(u) = 2 h2 2 u2 + 2

ln( 2 2 h2)u( 2 2 u2)h

,

where = 2 / .

3.3.2 Ascent of a liquid between two parallel plates

Let (d) = {(x1, x2) R 2 : d/ 2 < x 1 < d/ 2, d > 0, and consider theboundary value problem

div Tu = u in T u = cos 1 on 1 T u = cos 2 on 2,

where is a positive constant, and 1, 2 are the two walls, see Figure 3.4.Now we assume additionally that there exists a solution u which depends onx = x1 only, see Figure 3.4. In the following we will nd such a solution. As

d/2

u

g

liquid 21

S

1

2

xd/2

Figure 3.4: Ascent of liquid between two vertical walls

in the previous example, from a maximum principle of Finn and Hwang [25],see Section 6.1, it follows that this solution is the only one of the originalboundary value problem without the additionally assumption . Thus we have


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to consider a boundary value problem for an ordinary differential equation.

After scaling y = x/d , v(y) := u(dy)/d , we arrive at the following boundaryvalue problem, where B = d2.

v (y)

1 + v (y)2= Bv(y), 1/ 2 < y < 1/ 2 (3.18)

limy( 1/ 2)+0

v (y)

1 + v (y)2 = cos 1 (3.19)

limy(1/ 2) 0

v (y)

1 + v (y)2 = cos 2. (3.20)

To simplify the consideration here, we consider the case that the contactangles on 1 and 2 are the same: = 1 = 2. Then we seek a solution of

v (y)

1 + v (y)2= Bv(y), 0 < y < 1/ 2

limy(1/ 2) 0

v (y)

1 + v (y)2 = cos

v (0) = 0 .

As in the case of the ascent on a vertical wall we see that the inverse functiony(v) satises

y (v)

1 + y (v)2 = C v2

a2 , (3.21)

where C is a constant and a = 2/B . Suppose that 0 / 2, andassume v (y) > 0, 0 < y 1/ 2, if < / 2. The previous assumption issatised for the solution which we will derive in the following. Then

y (v) = C v2/a 2

1 (C v2/a 2)2y(v) =

v

v0

C t2/a 2

1 (C t2/a 2)2 dt, v0 v v1,

where

v0 = v(0) = 2B C 1, v1 = v(1/ 2) = 2B C sin ,see (3.21). The substitution

cos = C t2

a2, 0

2

,


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where denotes the angle of inclination of the curve dened by v(y), leads

to the elliptic integral

X ( ) = a2

0

cos C cos

d.

Then we nd the constant C 1 as the solution of 12

= a2

(/ 2)

0

cos C cos

d.

It follows that C if a = 2/B . We recall that B = d2.Consequently B = 2 C

(/ 2)

0cos d + O

1C 3/ 2

= 2 C cos + O

1C 3/ 2

as C , which implies thatC =

2B

cos2 + O( B )as B 0, see an exercise. Then

v0, v1 = 2B

cos + O(1)

as B 0.Since u(x) = dv(y), we obtain nally

u0, u1 = 2d

cos + O(d)

as d 0.3.3.3 Zero gravity solutions

Here we consider domains R 2 such that there are solutions u(x) of

div T u = 2H in (3.22) T u = cos on , (3.23)

where H is a constant.


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Constant circular cross section

Let (a) = Ba (0) be a disk with radius a and center at the origin. Weassume that the solution u(x) in consideration is rotationally symmetricwith center at the origin, i. e., u(x) = v(r ), where r = x21 + x22, then

1r

rv (r )

1 + v (r )2= 2 H (3.24)

limr a 0

v (r )

1 + v (r )2 = cos . (3.25)

From a comparison principle, see Section 6.1, it follows that every solution

must be rotationally symmetric. Integrating (3.22) over ( a) and using theboundary condition, we get

2H = 2cos

a . (3.26)

Equation (3.24) implies

rv (r )

1 + v (r )2 = Hr 2 + C.

Set r = 0, it follows that the constant C is zero. Then, if 0 / 2,

v(r ) = 1H 2 r 2 + C with another constant C . Thus v(r ) denes a circle with radius 1 / |H | andcenter at (0 , C ). Suppose that the liquid of given volume V occupies adomain as shown in Figure 3.5a, i. e., we consider a tube closed by a atbottom, and suppose that there is enough liquid such there is no dry spotat the bottom. Then we nd from

V = 2 a

0rv (r ) dr

thatC =

V + 23 a 2

a 2 .

Assume there is no bottom, and let S + and S be dened by

v+ = 1H 2 r 2 + C + , v = 1H 2 r 2 + C ,


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(a) (b)

liquid

liquid

S

S

+

Figure 3.5: Liquid in a circular tube

resp., with constants C + , C , see Figure 3.5b. Then

C + C = V + 43 a

2

a 2 .

Here we assume that there is enough liquid such that S + is above of S .

Liquid between two coaxial cylinders

Consider problem (3.22), (3.23) over an annulus domain 0 < a . For example, N (u) :=

N S 0 (u)cos T S 0 (u)sin is an extension of N (u), uU to U . We canassume that the remainder O( 3) is sufficiently regular, which follows froman implicit function theorem, we omit the details. The function q ( ) satisesq (0) = 0 and q (0) = 0, see Lemma A.3.1. We recall that

V = { = N S 0 + T S 0 : (, ) satises (3 .3)}and

V 0 = { V : S 0 dA = 0}.Suppose that S 0 = S (0) is an equilibrium interface, i. e., x(u) satises thenecessary conditions of Theorem 3.1. Set

L(S , ) = E (S ) + (|(S )| V ),then, if V 0,

E (S ( )) E (S 0) = L(S ( ), ) L(S 0, )=

dd L(S ( ), ) =0 +

2

2 d2

d 2 L(S ( ), ) =0 + O(3).

Since

d

d L(S ( ), ) =0 = 0for all V , see the Lagrange equation (3.7), in particular for V 0. Thus

d2

d 2 L(S ( ), ) =0 0is a necessary condition such that the inequality

E (S ( )) E (S 0)holds for all | | < 0.

Next we will calculate d2

d 2 L(S ( ), ) =0under the assumption that S 0 denes an equilibrium, i. e., it satises theequations of Theorem 3.1.


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From Lemma A.3.2A.3.4 of the appendix to this chapter we see that

dd L(S ( ), ) = 2 U H (u, ) N (u, ), z (u, ) W (u, ) du

+ S ( ) (u, ), z (u, ) ds( ) U Z 0, (, ), (u, ) |Z 0, (, )| d + U F (z(u, ) N (u, ), z (u, ) W (u, ) du+

U

N (u, ), z (u, ) W (u, ) du, (3.28)

where ds( ) = |Z (, )|d , Z (, ) = z(u( ), ), Z 0(, t ) = z0(u( ), t ) andu( ) is a regular parameter representation of U .Set

D(u, ) = N (u, ), z (u, ) W (u, ),

then

d2

d 2 L(S ( ), ) =0 = 2 U H (u, 0)D (u, 0) du2 U H (u, 0)D (u, 0) du+ U (u, ), z (u, ) =0 |Z (, 0)| d + U (u, 0), z (u, 0) |Z (, )| =0 d U (u, ), Z 0, (, ) =0 |Z 0, (, )| d 1 U (u, 0), Z 0, (, 0) |Z 0, (, )| =0 d + U F (x(u)) , z (u, 0) D(u, 0) du+ U F (x(u))D (u, 0) du.

From the differential equation 2H 0(u) + F (x(u)) + = 0 on S 0 and theformula, see Lemma A.3.8 of the appendix to this chapter,

2H (u, 0) = (u) + 2(2 H 20 (u) K 0(u)) (u) + 2 H 0, (u) (u),


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3.4. STABILITY 63

where H 0, K 0 are the mean and Gauss curvature of

S 0, resp., and

are dened through = N 0 + x , x = 0 on S 0, and , satisfy cos sin on U , we get d2

d 2 L(S ( ), ) =0 = S 0 + 2(2 H 20 K 0) dA+ S 0 F, N 0 2 dA+ U ( (u, ), z (u, ) (u, ), Z 0, (, ) ) =0

|Z 0, (, 0)| d + U ( (u, 0), z (u, 0) (u, 0), Z 0, (, 0) )

|Z 0, (, )| =0 d .Here we have used that D (u, 0) = W (u, 0), Z (, 0) = Z 0, (, 0),

z (u, 0) = z0, (u, 0) = = N 0 + x ,

and

2H 0, (u) + F (x(u)) , x (u) = 0on

S 0.

Next we show that the integrands of the last two terms in the previousformula for the second derivative of L vanish. We have

(u, 0), z (u, 0) (u, 0), Z 0, (, 0) = 0 , since z (u, 0) = and Z 0, (, 0) = on S . At S 0 the vectors 0, 0, N are in a common plane, and 0 and N , N 0 , resp., are orthogonal. Thuswe have at S 0 with scalar functions a, b the formula 0 = a 0 + bN . Usingthe boundary condition cos = on S 0 it follows

0 0 cos = N sin . (3.29)Then the brackets vanish since N at S 0. This is a consequence of formula

, 0 0 = (cos ),where is here a continuous function dened on U , see the proof of The-orem 3.1.


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For the brackets in the integrands of the remaining boundary integrals

we get from formula (3.29) above that

( (u, ), z (u, ) (u, ), Z 0, (, ) ) =0= 0, z (u, 0) 0, Z 0, (, 0) + , z (u, 0) (u, 0), = K | |2N , 0 k 0 + , (u, 0) (u, 0)= K | |2 sin + , (u, 0) (u, 0) .

Finally, we obtain

, (u, 0) (u, 0) = N 0 + 0, (u, 0) , (u, 0)= N 0, (u, 0)=

0

+ K S 0 .

Here we have used that (u, 0), (u, 0) = 0, = a 0 with a scalar functiona, and the formula, see Lemma A.3.6 of the appendix to this chapter,

N 0, (u, 0) = 0

+ K S 0 ,

where K S 0

is the curvature at P

S 0 of the plane curve dened through

the intersection of S 0 and the plane through P which is spanned by N S 0 atP and 0 := (0) at P . The curvature K S 0 is considered as nonnegative if the curve in consideration bends in the direction of N S 0 at P .

Summarizing, we obtain after integration by parts, see [18, 19], p. 45, p.44, resp.,

Lemma 3.1. Suppose that x(u) which denes S 0 satises the equations of Theorem 3.1. Then

d2

d 2 L(S ( ), ) =0 =

S 0 | |

2 2(2H 20 K 0) 2 dA+ S 0 F, N 0 2 dA+ 1 S 0 K S 0 | |2K sin ds .


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3.4. STABILITY 65

Corollary. Suppose that k are different from 0 or , then

d2

d 2 L(S ( ), ) =0 = S 0 | |2 2(2H 20 K 0) 2 dA+ S 0 F, N 0 2 dA+ S 0 cos sin K S 0 1sin K 2 ds.

Proof. The formula follows since

| |2 = N 0 + 0, N 0 + 0= 2 + 2

on S , and cos sin = 0on S 0. Remark. We recall that 2 H 20 K 0 0 holds for every regular surface.Remark. The second variation formula of Lemma 3.1 was derived byWente [75] under the assumption that is different from 0 or . In thecase of constant mean curvature, i. e. if F = 0, a proof was given by Rosand Souam [61], also for contact angles different from 0 or . Our proof includes the borderline cases = 0 and = as well as the case of anonvanishing F .

At some points and notations our proof is close to that of Ros andSuam [61]. Moreover, we do not assume the existence of a smooth continu-ation of the given surface S 0 accross its boundary.

An extremal is often called stable by denition if the second variation,given by the formula of the above lemma, is positive for all nonvanishing .

3.4.1 Strong minimizersLet S 0 be a sufficiently regular bounded capillary interface in equilibrium,i. e., S 0 satises the equilibrium conditions of Theorem 3.1. Dene theLagrange functional L(S , ) for sufficiently regular admissible interfaces S by

L(S , ) = E (S ) + (|l(S )| V 0),


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where

E is given by (3.1),

R and V 0 =

|l(

S 0)

|.

Admissible means that S and S R 3 \ s . If S is volume preserv-ing, i. e. |l(S )| = V 0, thenL(S , ) = E (S ).

Dene for a small > 0 a -neighbourhood D of S 0 byD = {xR 3 \ s : dist( x, S 0) < }.

Let S ( ), | | < , be a family of sufficiently regular and admissible surfacesdened by z(u, )R 3, uU , where U R 2 is a xed parameter domain.A family S ( ) which covers D simply is called a foliation , and the surface

S 0 is called embedded in this family if

S 0 is dened by z(u, 0).

Remark. The method of foliation was used by Wente [79] to prove a resultconcerning capillary tubes of nonconstant circular cross sections.

Assumptions. (i)There exists an embedding foliation of S 0 dened by 2H = ( ) + F (z(u, )) in S ( )

N S ( ) N = cos on S ( ),where H is the mean curvature of S ( ) at z(u, ).(ii) |(S ( )) | is increasing and ( ) is decreasing with growing .Theorem 3.3. Suppose that S 0 satises the previous assumptions (i)(ii).Then E (S ) E (S 0) for all admissible volume preserving comparison surfaces S D .Proof. Let x D and consider the associated surface S ( (x)) from thefamily S ( ). We recall that (x) is constant on S ( (x)) and

2H = ( (x)) + F in S ( (x)) (3.30)N S ( (x)) N = cos on S ( (x)) . (3.31)

Let n be the normal directed out of the liquid, see Figure 3.9, on the surface

S ( (x)) at x, then

div n = 2H (3.32)at x, where H is the mean curvature of S ( (x)) at x. For a proof of for-mula (3.32) see [18], pp. 77, for example. Combining (3.30) and (3.32), weget

div n + 1

(( (x)) + F ) = 0


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3.4. STABILITY 67

at x

D . Let T + and T be the domains enclosed by

S 0 and

S which are

nT

T

S

S

S

S

S( )

+

+

_

_

_

+

S( )

>0

+ _0

0


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and

L(S , 0) L(S 0, 0) = (|S||S 0| (|W (S )||W (S 0)|))+ l (S ) F dx l (S 0 ) F dx + 0 (|l(S )| |l(S 0)|) ,

it follows that

L(S , 0) L(S 0, 0) = S + (1 n ) dA + S (n + 1) dA+ T (F + ( (x))) dx T + (F + ( (x))) dx+

l (S )

F dx

l (S 0 )

F dx + 0 (

|l(

S )

| |l(

S 0)

|) .

Thus

L(S , 0) L(S 0, 0) = S + (1 n ) dA + S (n + 1) dA+ T (( (x)) 0) dx + T + (0 ( (x))) dx. (3.33)

Since S is volume preserving by assumption we haveL(S , 0) L(S 0, 0) = E (S ) E (S 0).

We get from (3.33) that E (S ) E (S 0). Example. As an example we consider the capillary tube with boundedcross section R 2. Assume that the capillary interface S 0 is given by agraph z = u(x) over the x-plane, x = ( x1, x2). Then

div Tu = u + 0 in T u = cos on ,

where T u = u/ 1 + |u|2, is the exterior unit normal at and > 0is the capillary constant. Let V 0 be the given volume. A foliation is explicitlygiven by

S ( ) : z(x, ) = u(x) + ,

where R , | | < . Then V ( ) = V 0 + || is the volume of liquid underS ( ), and the Lagrange parameter is

( ) = 1

||(| |cos V ( )) .

All assumptions (i)(ii) are satised.


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3.4. STABILITY 69

3.4.2 On the existence of an embedding foliation

Let S 0 be a sufficiently regular surface dened by x = x(u) satisfying theequilibrium equations of Theorem 3.1. Let be a real parameter in a neigh-bourhood of 0. Let S ( ) be an admissible family of surfaces, dened by

z(u, ) = x(u) + (u, ),

where (u, ) = (u, )N S 0 + (u, )T S 0 ,

with supp (u, )U ( 0, 0) for a 0 > 0, and U is a closed boundarystrip of U as introduced in Section 3.3. Admissible means, see Section 3.1,that z(u, ) is in the interior of the container if u

U and on the containerwall if u

U .

Assume S ( ) is sufficiently regular and satises

2H = ( ) + F (z(u, )) in S ( ) (3.34)N S ( ) N = cos on S ( ) (3.35)|l(S ( )) | = V 0 + , (3.36)

where H denotes the mean curvature of S ( ) at z(u, ), N is the normalon the container wall directed as shown in Figure 1.3, and V 0 = |l(S 0)) |is the given volume.

Suppose that

(u, ) = 1(u) + 12

2(u) 2 + O( 3)

(u, ) = 1(u) + 12

2(u) 2 + O( 3)

( ) = 0 + 1 + O( 2),

where the coefficients and the remainders are sufficiently regular.Set

l(u) = l(u)N S 0 + l(u)T S 0 , l = 1 , 2,

then

z(u, ) = x(u) + 1(u) + 12 2(u)

2

+ O( 3

).By assumption, z(u, ) is on the container wall if uU . Consequently wehave at U :

1(u)cos 1(u)sin = 0 (3.37) 2(u) = K | 1|2N . (3.38)


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Then we get from equation (3.34) and Lemma A.3.8 of the appendix (rst

variation of mean curvature) that on S 0( 1 + 2(2 H 2 K ) 1 + 2 H , w ) = 1 +

F N

1 + x F, T 1,

where T = w x , , and H = H (u) is the mean curvature of S 0 at x(u). Usingthe rst order necessary condition2H = 0 + F (x(u)) on S 0,

the above equation reduces to

( 1 + 2(2 H 2

K )

1) =

1 +

F

N

1.

From (3.37) we obtain that 1 = 0 on U if = 0 or = . In the case0 < < , we nd from the above boundary condition (3.35), equations(3.37), (3.38) and the corollary to Lemma A.3.9 of the appendix to thischapter that

1

+cos sin

K S 1sin

K 1 = 0

on S 0. Finally, equation (3.36) implies that

S

0

1 dA = 1 .

Set

p = 2(2H 2 K ) + 1

F N

q = cos

sin K S

1sin

K .

Consequently ( , ) := ( 1, 1) is a solution of

+ p +

= 0 in S 0 (3.39) + q = 0 on S 0 if 0 < < (3.40)

= 0 on S 0 if = 0 , or = (3.41)

S 0 dA = 1 , (3.42)provided there exists a sufficiently regular family of admissible surfaces.


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3.4. STABILITY 71

On the other hand, there exists a unique solution ( , ) of the system

(3.39)(3.42) if the second variation is positive. Set

Q(, ) = S 0 ( + p ) dA + S 0 q ds if 0 < < Q(, ) = S 0 ( + p ) dA if = 0 , or = .

Lemma 3.2. Suppose that 0 < < / 2. Then there exists a unique solu-tion (, ) of (3.39)(3.42), provided the second variation Q(, ) is positive for all W 1,2(S 0) \ {0} satisfying

S 0 dA = 0 if 0 < < / 2. If = 0

or = we have to replace W 1,2 through W 1,20 .

Proof. For the convenience of the reader we will give a proof. Set J () =Q(, )/ 2 and consider the minimum problem min J (), where the mini-mum is taken over all W 1,2(S 0) or W 1,20 (S 0), resp., satisfying theside condition S 0 dA = 1. Take a sufficiently regular 0 which satisesthe previous side condition, and set = + 0. Then the above minimumproblem changes to min J ( + 0), where the minimum is taken over all W 1,2(S 0) or W 1,20 (S 0), resp., satisfying the side condition

S 0

dA = 0 .

The existence of a solution 0 follows from the assumption concerning thesecond variation. Consequently there exists a Lagrange multiplier suchthat ( 0 + 0) is a solution of (3.39)(3.42). The uniqueness is also a conse-quence of the positivity of the second variation. Assume ( 1, 1) and ( 2, 2)are solutions of (3.39)(3.42), then := 1 2 satises (3.39)(3.41), with = 1 2, and the homogeneous side condition S 0 dA = 0. It followsthat Q(, ) = 0. Consequently = 0, which implies 1 = 2, see the abovedifferential equation (3.39) with = 1 2. Lemma 3.3. Assume 1 < 0 and that the second variation Q(, ) is pos-itive for all W 1,2(S 0) \ {0} or W 1,2(S 0) \ {0}, resp., satisfying S 0 dA = 0 . Then 1(u) > 0 on U in the case that 0 < < , and 1(u) > 0 on the open set U if = 0 or = .Proof. Set = 1 and = 1/ in this proof. Let = + + , where + = max {, 0}, = min {, 0}, and assume < 0 on a subset of U of


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positive measure. Then

S 0 dA < 0.Dene 0 = + + , where

= 1 1

S 0 dA.

Then > 1 and S 0 0 dA = 0. We have, see (3.39)(3.40),Q(, ) + S 0 dA = 0 (3.43)

for all W 1,2(S 0). Inserting = 0 und = 0 + (1 ) , we see that

0 = Q( 0, 0) + (1 )Q( , 0)= Q( 0, 0) + (1 )Q ( , )

since 0 = + + . Hence Q( , ) > 0 because of 0 0 and > 1. Onthe other hand, if we insert = in equation (3.43), we get

Q( , ) + S 0 dA = 0 ,which is a contradiction if

0. Thus we have shown (u)

0 in U ,

provided that 0. If < 0, then it follows from the differential equa-tion (3.39) and the strong maximum principle that (u) > 0 in U . Nextwe show that (u) > 0 on U , provided that 0 < < . Let (P ) = 0 atP U . Then we will show that

> 0 (3.44)

at P , which is a contradiction to the boundary condition (3.40). In fact,the Hopf boundary point lemma says that /n > 0, where n = ( n1, n 2)is the exterior normal at U . We recall that U is sufficiently smoothby assumption. Since /n =

,n and / =

, , where =

gg n , see [18], p. 44. Then n = gg n n > 0

since (g ) is positive denite. This inequality says that = ( 1, 2) con-sidered as a vector at P in R 2 makes an angle with n which is less than .


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3.5. APPENDIX 73

Thus at P we have = a n + b t, where a > 0. Here t denotes a unit tangent

vector at U at P . Then / = a /n + b/t = a /n at P , whichproves inequality (3.44).

The above two lemmas suggest the following conjecture.

Conjecture. Suppose that 1 < 0 and that the second variation is positive for variations, not identically zero, satisfying the side condition S 0 dA =0. Then there exists an embedding foliation of admissible surfaces.3.5 Appendix

Consider a family of admissible comparison surfaces given by

z0(u, ) = x(u) + (u) + r (u, ),

| | < 0, the remainder r is continuously differentiable with respect to allarguments, such that r = O( ) as 0, z0(u, ) R 3 \ (s0) if u U ,z0(u, ) if uU , and

(u) = (u)N S 0 (u) + (u)T S 0 (u)

is a given vector eld. Here N S 0 denotes the unit normal to S 0 pointed to theexterior of the liquid, and T S 0 is a unit tangent eld dened on closed stripU of U of width , such that on S 0, 0 := T S 0 is orthogonal to S 0 andpoints to the exterior of the liquid, see Figure 3.1. We assume that and are sufficiently regular on U , supp U , 2 + 2 1, and , N = 0, i. e., satisfy (3.3). By (u, ), resp. (u, ), we denote the exterior normal to S ( ) in S ( ), resp., in the container wall , see Figure 3.10. At P S ( )the vectors N , N (u, ), (u, ) and (u, ) are all in the same plane since S ( ) is a curve on S ( ) as well as on the container wall . Here and in thefollowing we set N (u, ) = N S ( ) at uU .

To get a volume preserving admissible family of comparison surfaces wereplace z0(u, ) through

z(u, ,q ) := z0(u, ) + q 0(u)N 0(u),

where q R , N 0 = N S 0 , supp 0 U , and

S 0 0(u) dA = 1 .


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S

()

()N()

N

()

Figure 3.10: Normals in consideration

q)

S0

S* ( ,0)

S* ( ,

Figure 3.11: Volume preserving comparison surfaces

Then S ( , q ), dened by z(u, ,q ), is a family of admissible comparisonsurfaces, in general not yet volume preserving, see Figure 3.11.

We recall that V is the set of all N S 0 + T S 0 , where (, ) satises theboundary condition (u)cos (u)sin = 0 on S 0, and V 0 is the subsetof V such that

S 0 dA = 0.

Lemma A.3.1. For given V satisfying the side condition

S 0 dA = 0 ,there exists a regular function q = q ( ) such that the family of admissi-ble congurations given by S ( ), where S ( ) is dened through z(u, ) =


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3.5. APPENDIX 75

z(u, ,q ( )) , is volume preserving for all

R 2,

| |< 0. The function q ( )

satises q (0) = 0 and q (0) = 0 .

Proof. We have, see Figure 3.11,

l(S ( , q )) = 12l(S 0),where

1 = {z(u,t, 0) : uU, 0 < t < },2 = {z(u, , ) : uU, 0 < < q }.

Here we assume for simplicity that S

( , q ) is above of S 0 and of S

( ,0) .The nal formula is valid for every S ( , q ), see a remark in [18], p. 81.Set

f ( , q ) = |l(S ( , q )) |.From

det z(u,t, 0)

(u, t ) = z z , zt

= N (u,t, 0), zt (u,t, 0) W (u,t, 0),det

z(u, , ) (u, )

= z z , z = N (u, , ), z (u, , ) W (u, , ),

we nd that

f ( , q ) = U

0N (u,t, 0), zt (u,t, 0) W (u,t, 0) dudt

+ U q

0N (u, , ), z (u, , ) W (u, , ) dud

+ |l(S 0)|.The assertion of the lemma follows since

f q(0, 0) = S 0 0 dA = 1f (0, 0) = S 0 dA.


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Lemma A.3.2.

dd |S ( )| = 2 U H (u, ) N (u, ), z (u, ) W (u, ) du

+ U z (u, ), (u, ) ds( ),where H (u, ) denotes the mean curvature of S ( ) at uU , and

ds( ) = |Z (, c)|d, Z (, c) := z(u( ), c),here u( ) is a regular parameter representation of U .

Proof. Set u = ( , ), then

|S ( )| = U E (u, )G(u, ) F 2(u, ) duand

dd |S ( )| = U 1W (u, ) E (u, ) z (u, ), z (u, )

F (u, )[ z (u, ), z (u, ) + z (u, ), z (u, ) ]+ G(u, ) z (u, ), z (u, ) du.

The formula of the lemma follows by integration by parts, see [18, 19], p.45, p. 44, resp., and by using the formula

z = 2HN , see [18, 19], p. 71,

p. 72, resp., for a proof of this formula.

Corollary.

dd |S ( )| =0 = 2 S 0 H (u, 0) (u) dA + S 0 (u) ds.

Proof. The corollary follows since

z (u, 0) = ,

and = (u)N S 0 (u) + (u)T S 0 .

Lemma A.3.3.dd |W (S ( )) | = U Z 0, (, ), (u, ) |Z 0, (, )| d,


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Lemma A.3.4.

dd l (S ( )) F (x) dx = U F (z(u, )) N (u, ), z (u, ) W (u, ) du.

Proof. Here we assume for simplicity that S ( ) is above of S (0). The nalformula is valid for every S ( ), see a remark in [18], p. 81. Thenl(S ( ), ( )) = 1l(S 0),

with1 = {z(u, t ) : uU, 0 < t < }.

Consequently

l (S ( )) F (x) dx = U

0F (z(u, t ))det

z(u,,t ) (u, t )

dudt.

Since

det z(u, t )

(u, t ) = zz , zt

= N (u, t ), zt (u, t ) W (u, t ),we obtain the formula of the lemma.

Corollary.

dd l (S ( )) F (x) dx =0 = S 0 F (x(u)) (u) dA.

Set H 0 = H (u, 0), 0 = (, 0), 0 = (, 0), N 0 = N (u, 0), which is N S 0 atuU .Let N (u, ) be the unit normal to S ( ) pointed to the exterior of theliquid, where S ( ) is given by z(u, ) = x(u) + (u) + O( 2). For notationsand formulas used in the following see [18], Part I. To simplify the notationswe omit the variable u and set N (0) = N (u, 0).

Lemma A.3.5.N (0) = g N (0)) , , x , .


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3.5. APPENDIX 79

Proof. We have

N (0) = c x + cN (0) .Since N ( ), N ( ) = 1, it follows c = 0. Consequently

N (0) = c x , (3.45)

andN (0) , x , = c g . (3.46)

The identity N ( ), z, ( ) = 0 implies thatN (0) , x , + N (0) , , = 0 .

With (3.46), we obtain

N (0) ,

,= c g

,

which implies thatc = g N (0) , , .

Together with (3.45) we get the formula of the lemma. .

Let ( ) be the exterior normal to S ( ) in S ( ), see Figure 4.4, and K S 0 isthe curvature at P S 0 of the plane curve dened through the intersectionof S 0 and the plane through P which is spanned by N S 0 at P and 0 := (0)at P . The curvature K S 0 is considered as nonnegative if the curve bends inthe direction of N S 0 at P .

Lemma A.3.6.N (0) , (0) =

0

+ K S 0 ,

where , dene the variation z0(u, ).

Proof. Since N ( ), ( ) = 0 we getN (0) , (0) = N (0) , (0) .

From Lemma A.3.5. and 0 = 0 x , it follows

N (0) , 0 = g N (0) , , x , , 0 x , =

g g N (0) , (N (0)

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