Liquid Layers, Capillary Interfaces, and Floating Bodies

Transcription

1 Liquid Layers, Capillary Interfaces, and Floating Bodies Lecture Notes Erich Miersemann Mathematisches Institut Universität Leipzig Version March, 2013

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3 Contents 1 Introduction Mean curvature, Gauss curvature Liquid layers Capillary interfaces Floating bodies Problems Liquid layers Governing formulas Explicit solutions Adsorption in slit cavities Adsorption in cylinder cavities Adsorption on a cylinder Adsorption in spherical cavities Adsorption on a sphere Pores of general geometry Stability Bounded surfaces Unbounded surfaces Appendix Problems Capillary interfaces Governing energy Equilibrium conditions The capillary tube Explicit solutions Ascent of a liquid at a vertical wall Ascent of a liquid between two parallel plates

4 4 CONTENTS Zero gravity solutions Stability Strong minimizers On the existence of an embedding foliation Appendix Problems Floating bodies Governing energy Equilibrium conditions Restricted movements Explicit solutions Stability The floating ball Appendix Problems Wetting barriers Equilibrium conditions Stability Problems Asymptotic formulas Comparison principles Applications An interior estimate Narrow circular tube, interior ascent Wide circular tube Ascent at a needle Narrow tube of general bounded cross section, interior ascent Ascent in a wedge A numerical method Problems Surface tension Wilhelmy s plate method A rod method Padday s method A new method

5 CONTENTS Problems Lagrange multiplier rules Equations Inequalities Problems

6 6 CONTENTS

7 Preface In these notes we study liquid layers, capillary interfaces and floating 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 or contact angle occur. Thus the admissible variations which yield necessary or sufficient conditions are much easier than the variations which we will use for the floating body problem where we have to take into account contact angles between the capillary surface and the solid materials as well as rigid motions of the body. The chapter on liquid layers is based on a cooperation with Peter Schiller and Hans-Jörg Mögel from the Institute of Physical Chemistry at the TU Bergakademie Freiberg. Concerning capillary interfaces we recommend the inspiring book Equilibrium Capillary Surfaces of Robert Finn [21]. The recent interest concerning floating bodies was probably initiated through the paper Convex particles at interfaces of E. Raphaël, 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 (liquid layers), Chapter 3 (capillary interfaces) and Chapter 4 (floating bodies) independently from each other with the consequence that some formulas used in Chapter 2 and Chapter 3 follow from more general formulas derived in Chapter 4. Each of the Chapters 2 4 contains an appendix with the lemmas, including proofs, which are used in that chapter. Some of the exercises in the sections Problems are just exercises and others 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 conversations over many years, and I am thankful one of the authors of the article [57] for sending me a reprint years ago. Part I of the book Minimal Surfaces I of U. Dierkes, S. Hildebrandt, A. Küster and O. Wohlrab [18] (second edition of U. Dierkes, S. Hildebrandt, F. Sauvigny [19]) was an essential help for me to calculate some variational formulas. Moreover I thank 7

8 8 CONTENTS Stefan Ackermann for his kindly and permanent help to produce source files for these and other lecture notes.

9 Chapter 1 Introduction 1.1 Mean curvature, Gauss curvature Assume x : u U R 3, u = (u 1, u 2 ) or u = (α, β), defines a regular surface S, where U R 2 is a domain in R 2. The normal N = N(u) on S is defined through N = x u 1 x u 2 x u 1 x u 2. The coefficients of the first fundamental form are given by E = x α, x α, F = x α, x β, G = x β, x β, and the coefficients of the second fundamental form are l = N α, x α, m = N α, x β = N β, x α, n = N β, x β. The mean curvature H = H(u) is defined 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 R 1, R 2, see [7], pp. 57, one has H = 1 ( ), K = 1. 2 R 1 R 2 R 1 R 2 9

10 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 or not a given equilibrium surface defines a strong minimizer of the associated energy functional. Consider a family {F c } of regular surfaces which can be described as its level surfaces Suppose that s(x) 0, then F c = {x R 3 : s(x) = c}. Q(x) := s(x) s(x) is orthogonal to all surfaces F c, see Figure 1.1, and div Q(x) = 2H(x), (1.2) where H(x) is the mean curvature at x of the surface of the family which contains x, see [18], p.77, for a proof of this formula, where x = x(u) is a regular C 2 -parametrization of F c such that N(u) = Q(x(u)). F c 1 Q(x) x F c 2 Figure 1.1: Level curves 1.2 Liquid layers Porous materials have a large amount of cavities different in size and geometry. Such materials swell and shrink in dependence on air humidity. Here we consider an isolated cavity, see [64] for some cavities of special geometry.

11 1.2. LIQUID LAYERS 11 Let Ω s R 3 be a domain occupied by homogeneous solid material. The question is whether or not liquid layers Ω l on Ω s are stable, where Ω v is the domain filled with vapour and S is the capillary surface which is the interface between liquid and vapour, see Figure 1.2. solid Ω s Ω l liquid N S Ω v vapour Figure 1.2: Liquid layer in a pore Let E(S, µ) = σ S + w(s) µ Ω l (S) be the energy (grand canonical potential) of the problem, where σ surface tension, µ a positive constant S, Ω l (S) denote the area resp. volume of S, Ω l (S), w(s) = F(x) dx, Ω v(s) is the disjoining pressure potential, where F(x) = c Ω s dy x y p. Here c is a negative constant, p > 4 a positive constant (p = 6 for nitrogen) and x R 3 \ Ω s, where Ω s denotes the closure of Ω s, i. e., the union of Ω s with its boundary Ω s. The disjoining pressure potential prevents the interface S to meet the container wall. Suppose that S 0 defines a local minimum of the energy functional, then 2σH + F µ = 0 on S 0, where H is the mean curvature of S 0.

12 12 CHAPTER 1. INTRODUCTION 1.3 Capillary interfaces Consider a container partially filled with a bounded amount of liquid, see Figure 1.3. Suppose that the associate energy functional is given by x 3 Ω v ( vapour) S N S N Γ Ω s (solid) Ω l ( liquid) g Figure 1.3: Liquid in a container E(S) = σ S σβ W(S) + Ω l (S) F(x) dx, where F = Y ρ, and Y potential energy per unit mass, for example Y = gx 3, g = const. 0, ρ local density of the liquid, σ surface tension, σ = const. > 0, β (relative) adhesion coefficient between the fluid and the container wall, W wetted part of the container wall, Ω l domain occupied by the liquid. Additionally we have for given volume V of the liquid the constraint Ω l (S) = V. It turns out that a minimizer S 0 of the energy functional under the volume constraint satisfies, see [21], 2σH = λ + F on S 0 cos γ = β on S 0,

13 1.3. CAPILLARY INTERFACES 13 where H is the mean curvature of S 0, λ a (constant) Lagrange parameter and γ 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 by definition the angle between the associated normals at this point. Remark. The term σβ W in the above energy functional is called wetting energy. In the case that the capillary surface S is a graph over Ω R 2, i. e., x 3 = u(x 1, x 2 ) defines the capillary surface, and if the container wall is the cylinder surface Ω R, then the related boundary value problem is div Tu = gρ σ u + λ σ in Ω ν Tu = cos γ on Ω, where ν is the exterior unit normal at Ω, and Tu := u 1 + u 2, div Tu is twice the mean curvature of the surface defined by x 3 = u(x 1, x 2 ), 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 Ω. It is assumed that gravity is directed downwards in the direction of the negative

14 14 CHAPTER 1. INTRODUCTION x 3 -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, see Finn [21]. Remark. Liquid can pilled up on a glass, see Figure 1.5. Here the capillary surface S satisfies a variational inequality at S where S meets the container wall 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 floating paper clip Consider a particle on a surface of a liquid which fills partially a container. It is assumed that gravity g is directed downwards in the direction of the

15 1.4. FLOATING BODIES 15 negative x 3 -axis. Bodies of density exceeding that of the liquid can float 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 the dominating force. Some methods of measurement of surface tension are based on the equilibrium conditions for floating bodies. Assume a cylindrical homogeneously body Ω = D [0, L] with constant cross section D hangs vertically in a liquid, see Figure 1.7, where F is the force directed upwards to keep the body in an equilibrium. We suppose that the body can move in the x 3 -direction F g Ω ν γ 0 γ S liquid Figure 1.7: Device for measurement of surface tension only and that no rotation is possible. Let z = v(x), x = (x 1, x 2 ), defines the free surface S 0. Then the equilibrium conditions are given by 2σH 0 + gρ 1 v(x) + λ 0 = 0 on S 0 cos γ 1 = β 1 on 1 S 0 cos γ 2 = β 2 on 2 S 0 σ ν 0, e 3 ds + gρ 1 y 3 (v) N Σ0 (v), e 3 da + gρ 2 Ω 2 S 0 W 2 (S 0 ) +λ 0 N Σ0 (v), e 3 da = F, W 2 (S 0 ) where e 3 is the unit vector directed in the positive x 3 -direction, W 2 (S 0 ) is the wetted part of Σ = Ω, N Σ is the normal at Σ directed into the

16 16 CHAPTER 1. INTRODUCTION body Ω, ρ 1 is the density of the liquid, ρ 2 the density of the body and y(v) = (y 1 (v), y 2 (v), y 3 (v)) is the parameter representation of the boundary of the floating body in an equilibrium. Remark. a, b or a b denotes the scalar product n i=1 a ib i for vectors a = (a 1,...,a n ), b = (b 1,...,b n ) in R n.

17 1.5. PROBLEMS Problems 1. Prove the basic lemma in the calculus of variations: let Ω R n be a domain and f C(Ω) such that f(x)h(x) dx = 0 for all h C0 1 (Ω). Then f 0 in Ω. Ω 2. Show that 2H 2 K 0 holds for every regular surface. 3. Prove that Q(x) is orthogonal to F c, see Section Show that div Tu is twice the mean curvature of S, defined by x 3 = u(x 1, x 2 ). 5. Write the capillary problem div Tu = gρ σ u + λ σ in Ω ν Tu = cos γ on Ω as a boundary value problem for a quasilinear equation of second order. 6. Prove that a minimizer in C 2 (Ω) of J(v) = F(x, v, v) dx is a solution of the boundary value problem Ω n F uxi x i n F uxi ν i i=1 i=1 Ω = F u in Ω = g u on Ω, g(v, v) ds, where ν = (ν 1,...,ν n ) is the exterior unit normal at the boundary Ω. 7. Find an energy functional E such that its extremals satisfy div Tu = gρ σ u + λ σ in Ω ν Tu = cos γ on Ω.

18 18 CHAPTER 1. INTRODUCTION

19 Chapter 2 Liquid layers The classical macroscopic theory of capillarity is usually applied to describe the location and the behaviour of the fluid in a container. The interface between the liquid and the vapour makes a constant contact angle with the container wall if the container is made from homogeneous material, according the general theory of Gauss [29]. The associated strongly nonlinear boundary value problem which describes the interface arose from studies of Young [81], Laplace [39] and Gauss [29]. A refined mesoscopic description incorporates the disjoining pressure which results from the long range forces between the fluid and the solid substrates. The theory of disjoining pressure of liquid layers was initiated by Derjaguin [15, 16]. In Schiller et al. [63], in particular, a single wedge is considered as a special geometry of solid material. Let Ω s R 3 be a domain occupied by homogeneous solid material, Ω l the liquid layer, Ω v the domain filled with vapour and S is the capillary surface which is the interface between liquid and vapour, see Figure 1.2. Let E(S, µ) = σ S + w(s) µ Ω l (S) (2.1) be the energy (grand canonical potential) of the problem, where σ surface tension, S, Ω l (S) denote the area resp. volume of S, Ω l (S), w(s) = F(x) dx, (2.2) Ω v(s) is the disjoining pressure potential, where F(x) = c Ω s 19 dy x y p. (2.3)

20 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 \ Ω s, where Ω s denotes the closure of Ω s, i. e., the union of Ω S with its boundary Ω s. Finally, set µ = ρkt ln(x), where ρ density of the liquid, k Boltzmann constant, T absolute temperature, X reduced (constant) vapour pressure, 0 < X < 1. More precisely, ρ is the difference between the number densities of the liquid and the vapour phase. However, since in most practical cases the vapour density is rather small, ρ can be replaced by the density of the liquid phase. The above negative constant is given by c = H/π 2, where H is the Hamaker constant, see [34], p For a liquid nitrogen film on quartz one has about H = Nm. Ansatz (2.1) can be justified by using density functional theory for fluids in combination with the sharp-kink approximation, see Bieker and Dietrich [6]. In our approach we neglect a small curvature correction of the surface tension. This correction is negligibly small if curvature radii are much larger than a molecular diameter (0.3 nm). Formulas and asymptotic expansions of F(x) for plane, spherical, cylindrical and wedge geometries are published by Philip [56]. Let S(ɛ), ɛ < ɛ 0, be the family of comparison surfaces given by z(u, ɛ) = x(u) + ɛξ(u)n(u), where x(u),, u U R 2, u = (u 1, u 2 ) = (α, β), is a parameter representation of the surface S 0 = S(0) such that the unit normal on S 0 N = x α x β x α x β is directed out of the fluid into vapour. It is assumed that S 0 and the scalar function ξ(u) are sufficiently regular. In the case of a capillary surface with boundary the boundary of the admissible comparison surfaces S(ɛ) must lie on the boundary of the solid domain, see Finn [21], p. 7, for the construction of such surfaces. In our choice of energy the liquid covers completely the solid surface since the function F defined by (2.3) becomes unbounded near the solid surface.

21 2.1. GOVERNING FORMULAS Governing formulas A necessary condition such that S 0 defines a local minimum of the energy functional E(S, µ 0 ) is [ ] d dɛ E(S(ɛ), µ 0) = 0, ɛ=0 which leads to the equation, see the corollaries to Lemma A.2.1 A.2.3 of the the appendix to this chapter, 2σ Hξ da + Fξ da µ ξ da = 0 S 0 S 0 S 0 for all ξ, where H is the mean curvature of S 0, see Section 1.1. An interface S 0 is said to be an equilibrium surface if this equation holds. From this equation we get Theorem 2.1. Suppose that S 0 defines an equilibrium, then on S 0 2σH + F µ = 0. (2.4) An existing equilibrium state S 0 is said to be stable by definition if [ ] d 2 dɛ 2 E(S(ɛ), µ 0) > 0 for all ξ not identically zero. From the corollaries to Lemma A.2.4 A.2.6 of the appendix to this chapter, integration by parts and by using the equilibrium condition (2.4) we find Theorem 2.2. An equilibrium interface S 0 is stable if ( σ ξ 2 2(2H 2 K)ξ 2 + F, N ξ 2) da > 0 (2.5) S 0 for all ξ W 1,2 (S 0 ) \ {0}. Here K is the Gauss curvature of the capillary surface S 0, see Section 1.1. Concerning the definition of ξ on S 0 see for example [7], p The Sobolev space W 1,2 (S 0 ) may be replaced by the space C 1 (S 0 ) of continuously differentiable functions. Concerning the definition of Sobolev spaces ɛ=0

22 22 CHAPTER 2. LIQUID LAYERS see [2], for instance. If S 0 is unbounded, then we assume that ξ has compact support, i. e., ξ 0 on S 0 \ M, where M is a compact set of S 0. Thus, since there is no side condition on ξ, S 0 is stable if 2(2H 2 K) + 1 σ F, N > 0 on S 0. (2.6) If the left hand side of (2.6) is constant on S 0, which is satisfied in all examples considered in the following, then (2.6) is also necessary for stability. If inequality (2.5) is satisfied, then S 0 defines at least a weak local minimum of the energy functional (2.1), i. e., E(S 0, µ 0 ) E(S, µ 0 ) for all S in a C 1 -neighbourhood of S 0. This follows from the Taylor expansion of the associated energy functional, see [48], where capillary interfaces are considered. Remark. In the volume constraint case Ω l (S) = const. the equilibrium condition is 2σH + F λ = 0 on S 0, where λ is the (constant) Lagrange multiplier. An equilibrium is stable in the volume constrained case if inequality (2.5) is satisfied for all ξ not identically zero which satisfy the side condition S 0 ξ da = Explicit solutions In this section we consider examples of pores with simple geometry where we know 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 the condition (2.6) is satisfied. In general, for given solid surface the equilibria are not known explicitly Adsorption in slit cavities Assume that two parallel plates have a distance 2d from each other, and the regions y 3 > d and y 3 < d are occupied by the solid, see Figure 2.1. The potential (2.3) is given by ( F(x) = c y y2 2 + (x 3 y 3 ) 2) p/2 dy y 3 >d ( +c y y2 2 + (x 3 y 3 ) 2) p/2 dy. y 3 < d

23 2.2. EXPLICIT SOLUTIONS 23 solid liquid vapour liquid solid y 3 d d h d+h d Ω s Ω s Figure 2.1: Slit cavity The plane surfaces S 1 : y 3 = d h and S 2 : y 3 = d + h define stable equilibria if F µ = 0 on S i (equilibrium condition) and F/ N F, N > 0 on S i (stability condition). Let x be on S 1 and set q = h/d, then F(x) = cd 3 p (ψ 1 (q) + ψ 2 (q)), where ψ 1 (q) = ψ 2 (q) = ( y y2 2 + (1 q y 3 ) 2) p/2 dy y 3 >1 y 3 < 1 ( y y (1 q y 3 ) 2) p/2 dy. Integration is elementary and yields ψ 1 (q) = ψ 2 (q) = 4π (p 3)(p 2) q p+3 4π (p 3)(p 2) (2 q) p+3. The normal derivative F/ N := F, N of F on S 1 is F N = cd2 p ( ψ 1(q) + ψ 2(q) ). We recall that the normal is always directed into the vapour and the constant c is negative. Set f(q) = cd 3 p (ψ 1 (q) + ψ 2 (q)),

24 24 CHAPTER 2. LIQUID LAYERS then f(q) > 0, f (q) > 0 on 0 < q < 2, and lim q 0 f(q) =, lim q 2 f(q) =. Let κ 0 be the zero of f (q) = 0, i. e. κ 0 = 1 because of the symmetry of f(q) with respect to q = 1. Set ( X 0 = exp 1 ) ρkt f(1) and assume the constant X, 0 < X < X 0, is given, then there are exactly two pairs of planes which are equilibria. These planes are defined by the two zeros κ 1, κ 2 of ρkt ln(x) = f(q), 0 < κ 1 < 1 < κ 2 < 2, see Figure 2.2, and the layer with thickness h = κ 1 d is stable. We repeat µ κ q Figure 2.2: Layer with thickness h = κ 1 d is stable the same considerations on S 2. Let X, 0 < X < X 0 be given, then the liquid layers with thickness h = κ 1 d are stable. Remark. The above considerations show that the cavity fills up in a stable manner in contrast to cylinder or ball cavities, where not all curvatures disappear, see the following subsections. This contradicts experiments. A modified energy ansatz which takes into account the interaction of molecules in the vapour domain yields probably the right behaviour of liquid layers in slit cavities.

25 2.2. EXPLICIT SOLUTIONS 25 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 is occupied by solid material, then there is at least one stable equlibrium which is asymptotically a plane film far away from the edge of the wedge Adsorption in cylinder cavities Let the cross section of the tube be a disk with radius R. The solid domain is Ω s = {(y 1, y 2, y 3 ) R 3 : y y 2 2 > R 2, < y 3 < }. We claim that there are stable liquid layers given by r 2 < x x2 2 < R2, < x 3 <, where r > 0, see Figure 2.3. The potential (2.3) is solid liquid N vapour S r R Figure 2.3: Cylinder cavity where r 2 = x x2 2, and ψ(q) = F(x) = cr 3 p ψ(q), q = r/r, ((y1 q) 2 + y y 2 3) p/2 dy, the domain of integration is here y1 2 + y2 2 > 1, < y 3 <. For stability considerations we need some properties of ψ(q). The function ψ(q) is, see Philip [56], ψ(q) = π 3/2Γ( 1 [ 2 (p 3)) 1 Γ ( 1 2 p) F 2 (p 3), 1 ] (p 1); 1;q2, 2

26 26 CHAPTER 2. LIQUID LAYERS where F denotes hypergeometric functions. From integral representation of hypergeometric functions, see for example [1], p. 558, Eq , it follows ψ (q) > 0 and ψ (q) > 0 on 0 < q < 1. The definition of ψ(q) implies that ψ(q) > 0, lim q 1 ψ(q) =, ψ(0) > 0 and ψ (0) = 0. The cylinder surface S = {(x 1, x 2, x 3 ) R 3 : x x 2 2 = r 2, < x 3 < } defines an equilibrium if µ = σ ( r r cr3 p ψ R) since the mean curvature of S is H = (2r) 1. Such a surface is stable if σ ( r 2 cr2 p ψ r ) > 0 R since the Gauss curvature K of a cylinder surface is zero. Set f(r) = σ ( r r cr3 p ψ. R) From the above properties of ψ it follows that f(r) is positive, strictly convex on 0 < r < R, and lim r 0 f(r) =, lim r R f(r) =, see Figure 2.4 for a typical graph of f(r) 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 1 ) ρkt f(r 0). Then

27 2.2. EXPLICIT SOLUTIONS 27 (i) for given 0 < X < X 0 there are exactly two cylinders which are equilibria. These spheres are defined by the zeros r 1, r 2, r 1 < r 0 < r 2, of ρkt 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) lim R 0 X 0 (R) = 0. The behaviour (ii) follows from definition of X 0 since r 0 0, and it shows for given (even small) reduced vapour pressure that there are no liquid layers if the pore is sufficiently small Adsorption on a cylinder The solid domain is Ω s = {(y 1, y 2, y 3 ) R 3 : y y 2 2 < R 2, < y 3 < } and the potential (2.3) is given by F(x) = cr 3 p ψ(q), q = r/r, where r 2 = x x2 2, r > R, and ((y1 ψ(q) = q) 2 + y2 2 + y3) 2 p/2 dy, the domain of integration is here y y2 2 < 1, < y 3 <. The cylinder surface S = {(x 1, x 2, x 3 ) R 3 : x x 2 2 = r 2, < x 3 < } defines an equilibrium if µ = σ ( r r cr3 p ψ R) since the mean curvature of S is H = (2r) 1. Such a surface is stable if σ ( r 2 + cr2 p ψ r ) > 0. R From properties of ψ, we omit the details, it follows that for given 0 < X < 1 there is exactly one cylinder surface which is an equilibrium, and this is stable, see Figure 2.5 for a typical graph of σ/r + c R 3 p ψ(r/r). We recall that the constant c is negative.

28 28 CHAPTER 2. LIQUID LAYERS Adsorption in spherical cavities The domain occupied by solid material is here Ω s = {(y 1, y 2, y 3 ) R 3 : y1 2 + y2 2 + y3 2 > R 2 }, R > 0, and the potential (2.3) is F(x) = cr 3 p ψ(q), where ψ(q) := (y y (y 3 q) 2) p/2 dy. with q = r/r, and r 2 = x 2 1 +x2 2 +x2 3, 0 < r < R. The domain of integration is y1 2 + y2 2 + y2 3 > 1. An elementary calculation yields, if p > 4, 2π 1 ( 1 [ ψ(q) = (1 q) p+4 (1 + q) p+4] (p 2)(p 3) q p 4 + [ (1 q) p+3 (1 + q) p+3]). It follows that ψ (q) > 0, ψ (q) > 0, 0 < q < 1, since ψ(q), 0 < q < 1, can be written as a convergent power series expansion ψ(q) = k=0 a 2kq 2k, where all coefficients a 2k are positive. A sphere S : x x2 2 + x2 3 = r2 defines an equilibrium if µ = 2σ ( r r cr3 p ψ =: f(r) R) since the mean curvature of S is H = r 1. Such a surface is stable if 2σ ( r 2 cr2 p ψ r ) > 0 R 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 ψ are taken from above. Remark. In contrast to the above result each layer is stable in the volume constrained case. This follows since S 0 : x = r, 0 < r < R, satisfies the equilibrium condition ( r 2σH 2π c R 3 p ψ λ = 0, R)

29 2.2. EXPLICIT SOLUTIONS 29 where H = r 1, and λ is the constant such that this equation holds for given r. The stability criterion (2.5) is satisfied for all ξ W 1,2 (S 0 ) \ {0} which satisfy the side condition S 0 ξ da = 0, since S 0 F ( N = 2π c R2 p ψ r ) R is positive and ( ξ 2 2r ) 2ξ2 da = B 1 (0) ( ξ 2 2ξ 2) da 0 for all ξ W 1,2 ( B 1 (0)) which satisfy the side condition B 1 (0) ξ da = 0, where B 1 (0) denotes the boundary of the ball B 1 (0), see [78]. The proof of this inequality is based on methods from [77] Adsorption on a sphere The domain occupied by solid material is R > 0, and the potential (2.3) Ω s = {(y 1, y 2, y 3 ) R 3 : y y y 2 3 < R 2 }, F(x) = cr 3 p ψ(q), where ψ(q) = (y y (y 3 q) 2) p/2 dy with q = r/r, and r 2 = x 2 1 +x2 2 +x2 3, 0 < R < r. The domain of integration is y1 2 + y2 2 + y2 3 < 1. An elementary calculation yields, if p > 4, 2π 1 ( 1 [ ψ(q) = (1 + q) p+4 (q 1) p+4] (p 2)(p 3) q p 4 + [ (q 1) p+3 + (1 + q) p+3]). A sphere S : x x2 2 + x2 3 = r2, r > R, defines an equilibrium if µ = 2σ ( r r + c R3 p ψ R) since the mean curvature of S is H = r 1. Such a surface is stable if 2σ ( r 2 c R2 p ψ r ) > 0 R

30 30 CHAPTER 2. LIQUID LAYERS since the Gauss curvature of S is K = r 2. We obtain qualitatively the same result as in the case of a layer on a cylinder. For given 0 < X < 1 there is exactly one sphere which is in equilibrium, and this sphere is stable, see Figure 2.5 for a typical graph of g(x) 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(µ 0 ) for a large µ 0 calculated by a numerical method, then a numerical continuation procedure yields an equilibrium S(µ) for increasing parameter µ. Simultaneously one checks whether or not S(µ) is stable. If S(µ) satisfies the stability criterion (2.5) for µ < µ c and S(µ c ) violates this condition, then µ c is said to be the stability bound of this continuation procedure. Such a numerical continuation was applied to a unilateral problem for the rectangular plate, see [51]. In this example the parameter of continuation is an eigenvalue. 2.4 Stability Here we consider the question whether or not equilibria define also minimizers of the associated energy functional. The considerations in this section

31 2.4. STABILITY 31 were suggested by a paper of Wente [79] concerning capillary tubes of general cross sections Bounded surfaces Here we suppose that the single cavity R 3 \ Ω s is bounded. An example is a spherical cavity. Theorem 2.3. (Bounded cavities.) Let R 3 \ Ω s be a bounded single pore. Suppose that the free interface S 0 of the liquid layer is sufficiently regular, satisfies the necessary condition (2.4), and the second variation is positive for all nonzero variations, see (2.5). Then there exists a δ > 0 such that E(S 0, µ 0 ) < E(S 0, µ 0 ) for all surfaces S different from S 0, and located in a δ-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 minimal surfaces. We will show that an equilibrium surface S 0 can be embedded in a foliation, provided that the second variation is positive for all nonzero variations. Define for a small δ > 0 the δ-neighbouhood of S 0 by D δ (S 0 ) = {y R 3 : y = x(u) + s N(u), δ < s < δ}. A family S(µ), µ (µ 0 ɛ, µ 0 +ɛ), which covers D δ simply is called foliation, and a surface S 0 is called embedded in this family if S 0 = S(µ 0 ). Lemma 2.1. (Existence of an embedding family). Let S 0 be a solution of 2σH + F µ 0 = 0 on S 0, and suppose that the second variation of E(S, µ 0 ) at S 0 is positive on W 1,2 (S 0 )\ {0}. Then there exists a foliation S(µ) of D δ (S 0 ), where δ > 0 is sufficiently small, and each element of this foliation solves 2σH + F µ = 0 on S(µ). Proof. We will show that the embedding family is given by x + ξ(u, τ)n, where the scalar function ξ is the solution of 2σH(x + ξn) + F(x + ξn) (µ 0 + τ) = 0.

32 32 CHAPTER 2. LIQUID LAYERS The constant τ is from the interval ( ɛ, ɛ). Define the mapping M(ξ, τ) : C 2,κ (S 0 ) ( ɛ, ɛ) C κ (S 0 ), where 0 < κ < 1 is a constant Hölder exponent, by M(ξ, τ) = 2σH(x + ξn) + F(x + ξn) (µ 0 + τ). Since M(0, 0) = 0, and M ξ (0, 0), defined by M ξ (0, 0) h = σ h 2σ(2H 2 K)h + F N h, see [7], p. 186, for the formula of the first variation of H, is a regular mapping from C 2,κ (S 0 ) C κ (S 0 ), it follows from an implicit function theorem that there is a unique solution ξ = ξ(u, τ) = τv(u) + r(u, τ) of M(ξ, τ) = 0, where r(u, τ) C 2,κ (S 0 ) r(τ) C 2,κ (S 0 ) = o(τ) as τ 0, r τ exists, r τ C 2,κ (S 0 ) and lim τ 0 r τ (u, τ) C 2,κ (S 0 ) = 0. Moreover, v is the solution of the Jacobi equation σ v 2σ(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 is positive for all nonzero variations, we obtain that v > 0 on S 0 as follows. Set ζ(u) = max{ v(u), 0} and Then q = 2(2H 2 K) + 1 σ F N. 0 = ( v ζ + qv ζ 1 ζ) da S 0 σ = ( v ζ + qv ζ 1 ζ) da S 0 σ = ( ζ 2 + qζ ζ) da { v 0} σ = ( ζ 2 + qζ ζ) da. S 0 σ

33 2.4. STABILITY 33 Combining this equation and inequality (2.5), we find that ζ 0, i. e., v 0 on S 0. This inequality and equation v + q v = 1 σ on S 0 imply that v > 0 on S 0. If not, then there is a P S 0 where v(p) = 0. Then v = 1/σ and v 0 at P, which is a contradiction. The previous argument is called touching principle. Finally, for given y = x(u) + s N(u) D δ (S 0 ), δ > 0 sufficiently small, there exists a unique solution µ(s) of ξ(u, µ) = s since ξ µ (u, 0) = v(u) > 0. Let x D δ (S 0 ) and consider the associated surface S(µ(x)) from the family S(µ). We recall that µ(x) is constant on S(µ(x)), and 2σH + F(x) µ(x) = 0 on S(µ(x)). Let n be the inward normal on the surface S(µ(x)) at x, see Figure 2.6, then div n = 2H. Combining both equations, we have at x D δ (S 0 ) the equation ν= n n S ( µ (x)) x ν S S 0 Figure 2.6: Proof of Theorem 2.1 div n + 1 (F(x) µ(x)) = 0. σ Let S be any other regular surface sitting inside of D δ (S 0 ). Here we assume that S is inside of the domain defined by S 0, see Figure 2.6. For the general

34 34 CHAPTER 2. LIQUID LAYERS 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(µ 0 ) and S. Then 0 = = T 0,S T 0,S (div n + 1σ ) (F(x) µ(x)) dx n ν da + 1 σ T 0,S (F(x) µ(x)) dx, where ν is the exterior unit normal on T 0,S. Since ν = n on S 0, it follows S S 0 = S (1 n ν) da 1 σ T 0,S (F(x) µ(x)) dx. (2.8) From the definition (2.1) of the energy we get E(S, µ 0 ) E(S 0, µ 0 ) = σ( S S 0 ) + F(x) dx µ 0 T 0,S. (2.9) T 0,S Combining (2.9) and (2.8), we find that E(S, µ 0 ) E(S 0, µ 0 ) = σ (1 n ν) da + (µ(x) µ 0 ) dx. S T 0,S Since µ(x) > µ 0 if x T 0,S, the theorem is shown. Remark. Liquid layers in equilibrium on and in spherical cavities studied above which are stable in the sense that the second variation is positive for all nonzero variations are stable in the sense of the previous theorem. Remark. In all explicit examples above this lemma is superfluous since an embedding family is defined in a natural way, in the cases of ball or cylinder cavities by varying the radius r. The resulting surfaces are planes, concentric spheres or coaxial cylinders, resp Unbounded surfaces Here we suppose that the single cavity R 3 \ Ω s is unbounded. Examples are slit and cylinder cavities. In fact, there are no such cavities. On the other hand, such objects serve as an approximation of long cylinder cavities, for example. In general, the definition of the energy by (2.1) makes no sense if S is unbounded. But the difference (2.9) is well defined if S 0 is replced

35 2.4. STABILITY 35 by a sufficiently regular bounded subdomain M S 0. For given sufficiently small positive δ define a δ-neighbourhood of M by D M δ (S 0) = {y R 3 : y = x + s N, x M, δ < s < δ}. In this section we consider the example of a cylinder type cavity, see Figure 2.7. We will find an embedding family x+ξ(τ)n of M, where ξ = ξ(u, τ) Ω s Ω l δm S 0 δm M Ω v Figure 2.7: Cylinder type cavity is the solution of the boundary value problem 2σH(x + ξn) + F(x + ξn) (µ 0 + τ) = 0 on M ξ = τ on M. Replacing the lateral surfaces Σ of D δ (M) by surfaces Σ 0 Dδ M(S 0) close to Σ such that ν n = 0 on Σ 0, where ν is the normal on Σ 0 and n is the normal on the surfaces of the embedding family. The resulting domain is denoted by Dδ 0(M), see Figure 2.8. Set M 0 = S 0 Dδ 0 (M) and suppose that a sufficiently regular surface S is in Dδ 0(M), see Figure 2.8. Let T 0,S be the domain between M 0 and S. Then the considerations of the proof of the theorem in the previous subsection leads to Theorem 2.4 (Unbounded cavities.) Suppose that the free interface S 0 of the liquid layer is sufficiently regular, satisfies 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 δ > 0 such that E(S, M 0, µ 0 ) := σ( S M 0 ) + F(x) dx µ 0 T 0,S > 0 T 0,S for all surfaces S different from M 0, and which are located in D 0 δ (M).

36 36 CHAPTER 2. LIQUID LAYERS Σ 0 Σ 0 S D δ 0 (M) Σ 0 Σ 0 Figure 2.8: Proof of Theorem 3.1 Remark. In the case of a cylinder cavity, see Section 2.2.2, the lateral surfaces are already perpendicular on the surfaces of the embedding family, i. e., we can set Σ = Σ 0. Remark. Liquid layers in equilibrium on and in cylindrical cavities studied in Section which are stable in the sense that the second variation is positive for all nonzero variations with compact support are stable in the sense of the previous Theorem Appendix Consider a family S(ɛ), ɛ < ɛ 0, of surfaces given by z(u, ɛ) = x(u) + ɛξ(u)n(u), where ξ is a given sufficiently regular function on the parameter domain U R 2 where the support is a compact subset of U. Set W(u, ɛ) = E(u, ɛ)g(u, ɛ) F 2 (u, ɛ), where E(u, ɛ), G(u, ɛ) F(u, ɛ) are the coefficients of the first fundamental form of S(ɛ), and N(u, ɛ), H(u, ɛ) are the normal and the mean curvature associated to the surface S(ɛ), resp. Lemma A.2.1. d S(ɛ) = 2 dɛ U H(u, ɛ) N(u, ɛ), z ɛ (u, ɛ) W(u, ɛ) du.

37 2.5. APPENDIX 37 Proof. Set u = (α, β), then d dɛ S(ɛ) = 1 ( E(u, ɛ) zβ (u, ɛ), z βɛ (u, ɛ) U W(u, ɛ) F(u, ɛ)[ z α (u, ɛ), z βɛ (u, ɛ) + z β (u, ɛ), z αɛ (u, ɛ) ] +G(u, ɛ) z α (u, ɛ), z αɛ (u, ɛ) ) 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. [ ] d dɛ S(ɛ) = 2 H 0 ξ da ɛ=0 S 0 where S 0 = S(0), and H 0 denotes the mean curvature of S 0. Let dy F(x) = c Ω s x y p, where c is a positive constant, and p 4 is a constant. Set w(s) = F(x) dx, Ω v(s) here Ω v denotes the domain filled with vapour. Lemma A.2.2. d dɛ w(s(ɛ)) = U F(z(u, ɛ)) N(u, ɛ), z t (u, ɛ) W(u, ɛ) du. Proof. Ω v(s(ɛ)) F(x) dx = Ω v(s 0 ) F(x) dx ɛ 0 U F(z(u, t))det z(u, t) (u, t) dudt. This formula holds also if S(ɛ) is not completely located in the vapour region associated to S 0, see a remark in [18], p.81. Since z(u, t) (u, t) = z α (u, t) z β (u, t), z t (u, t) = N(u, t), z t (u, t) W(u, t),

38 38 CHAPTER 2. LIQUID LAYERS it follows that d dɛ Ω v(s(ɛ)) F(x) dx = U F(z(u, ɛ)) N(u, ɛ), z t (u, ɛ) W(u, ɛ) du. Corollary. [ ] d dɛ w(s(ɛ)) = F(x(u))ξ da. ɛ=0 S 0 Lemma A.2.3. d dɛ Ω l(s(ɛ)) = U N(u, ɛ), z ɛ (u, ɛ) W(u, ɛ) du. Proof. If S(ɛ) is completely located in the vapour region associated to S 0, then Ω l (S(ɛ)) = Ω l (S 0 ) Ω 1, where Ω 1 = {z(u, t) : u U, 0 < t < ɛ}. The final formula holds for the general case, see a remark in [18], p.81. Then ɛ Ω l (S(ɛ)) = Ω l (S 0 ) + N(u, t), z t (u, t) W(u, t) dudt. 0 U Corollary. [ ] d dɛ Ω l(s(ɛ)) = ξ da. ɛ=0 S 0 Second derivatives of the integrals with respect to ɛ are used for stability considerations. Set D(u, ɛ) = N(u, ɛ), z ɛ (u, ɛ) W(u, ɛ), 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. [ ] d 2 dɛ 2 S(ɛ) = 2 ξ ( ξ + 2(2H0 2 K 0 )ξ ) da 2 S 0 ɛ=0 U H 0 (u)d ɛ (u, 0) du.

39 2.5. APPENDIX 39 Proof. From Lemma A.2.1 we get [ ] d 2 dɛ 2 S(ɛ) = 2 ξh ɛ (u, 0) da 2 S 0 U ɛ=0 Then the formula of the lemma follows since H(u, 0)D ɛ (u, 0) du. 2H ɛ (u, 0) = ξ + 2(2H 2 0 K 0 )ξ 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. [ d 2 dɛ 2w(S(ɛ)) ] ɛ=0 = F, N ξ 2 da + S 0 U F(x(u))D ɛ (u, 0) du. Lemma A.2.6. [ d 2 dɛ 2 Ω l(s(ɛ)) = ɛ=0 U D ɛ (u, 0) du. Remark. In the above integrals over S 0 we can assume that ξ C 1 (S 0 ) or ξ C 2 (S 0 ), resp., if S 0 is bounded. This follows by partition of unity and since S 0 has no boundary. Remark. We did not need an explicit formula for D ɛ (u, 0) since it was assumed that the interface S 0 is in an equilibrium, i. e., equation (2.4) is satisfied. On the other hand there is an explicit formula see an exercise. D ɛ (u, 0) = 2H(u, 0)ξ 2 W(u, 0), (2.10)

40 40 CHAPTER 2. LIQUID LAYERS 2.6 Problems 1. Assume 2(H 2 K)+ F, N /σ is constant at the equilibrium state S 0. Show that 2(H 2 K) + F, N /σ 0 is a necessary condition for stability. Hint: ξ := const. 2. Show that ψ(q), defined in Section (adsorption in cylinder cavities) satisfies ψ(q) > 0, ψ (q) > 0, ψ (q) > 0 on 0 < q < 1, and lim q 1 ψ(q) =, lim q 0+0 ψ(q) > 0, ψ (0) = Set f(r) = σ ( r r cr3 p ψ, R) see Section Prove that f(r) is positive, strictly convex on 0 < r < R, lim r 0 f(r) = and lim r R f(r) =. 4. Consider the adsorption on a cylinder, see Section Show that for given 0 < X < 1 there is exactly one cylinder surface which is an equilibrium, and this equilibrium is stable. 5. Prove formula (2.10). Hint. Show that, see [7], p.184, W(ɛ) = W 0 (1 2ɛHξ + ɛ 2 Kξ ) 2 ɛ2 ξ 2 + O(ɛ 3 ) by using the following formulas. E(ɛ) = x u1 + ɛn u1 ζ + ɛnζ u1, x u1 + ɛn u1 ζ + ɛnξ u1 = E 0 2ɛlξ + ɛ 2 N u1, N u1 ξ 2 + ɛ 2 ξ 2 u 1 F(ɛ) = x u1 + ɛn u1 ξ + ɛnξ u1, x u2 + ɛn u2 ξ + ɛnξ u2 = F 0 2ɛmξ + ɛ 2 N u1, N u2 ξ 2 + ɛ 2 ξ u1 ξ u2 G(ɛ) = x u2 + ɛn u2 ξ + ɛnξ u2, x u2 + ɛn u2 ξ + ɛnξ u2 = G 0 2ɛnξ + ɛ 2 N u2, N u2 ζ 2 + ɛ 2 ζ 2 u 2, where l, m, n are the coefficients of the second fundamental form defined by l = x u1, N u1 2m = x u1, N u2 x u2, N u1 n = x u2, N u2.

41 2.6. PROBLEMS 41 The coefficients e, f, g of the third fundamental form defined by e = N u1, N u1, f = N u1, N u2, g = N u2, N u2 satisfy the relations, see [7], p. 69, for example, KE 2Hl + e = 0 KF 2Hm + f = 0 KG 2Hn + g = 0. It follows that Define E(ɛ) = E 0 2ɛlξ + ɛ 2 ξ 2 (2Hl KE) + ɛ 2 ξ 2 u 1 F(ɛ) = F 0 2ɛmξ + ɛ 2 ξ 2 (2Hm KF) + ɛ 2 ξ u1 ξ u2 G(ɛ) = G 0 2ɛnξ + ɛ 2 ξ 2 (2Hn KG) + ɛ 2 ξ 2 u 2. ξ 2 = E 0ξu 2 2 2F 0 ξ u1 ξ u2 + G 0 ξu 2 1 W0 2. Then the above expansion of W(ɛ) follows. 6. Prove the corollary to Lemma A.2.1. Hint. Partition of unity, and S 0 has no boundary.

42 42 CHAPTER 2. LIQUID LAYERS

43 Chapter 3 Capillary interfaces 3.1 Governing energy Let Ω s R 3 be a domain occupied by homogeneous solid material. This domain Ω s is called container. By Ω l = Ω l (S) we denote the domain filled with liquid, where S denotes the interface between the liquid and the vapour which filles the domain Ω v, see Figure 1.3. Let W(S) = Ω s Ω l (S) be the wetted part of the container wall Ω s. In this chapter we assume that the container wall is a sufficiently smooth surface, and that the energy of the problem is given by E(S) = σ S σβ W(S) + F(x) dx, (3.1) Ω l (S) where F = Y ρ, and Y potential energy per unit mass, for example Y = gx 3, g = const. gravitational force, ρ local density of the liquid, σ surface tension, σ = const. > 0, β (relative) adhesion coefficient between the fluid and the container wall, W wetted part of the container wall, Ω l domain occupied by the liquid. Additionally we have for given volume V of the liquid the constraint Ω l (S) = V. (3.2) By S, Ω l (S) we denote the area resp. volume of S, Ω l (S). 43

44 44 CHAPTER 3. CAPILLARY INTERFACES The problem is to find minimizers of E in an appropriate class of comparison surfaces S. Let S 0 be a given regular surface. Then we will derive necessary and sufficient conditions such that this surface defines a minimizer of the associated energy functional subject to a given family of comparison surfaces. Assume x : u U R 3, u = (u 1, u 2 ) or u = (α, β), defines the regular surface S 0, where U R 2 is a domain such that x(u) Γ if u U, and x(u) R 3 \ {Ω s Ω} if u U. We choose the parameters such that the normal N S0 = x α x β x α x β is directed out of the liquid, and that the normal N Γ at the container wall is directed out of the solid material Ω s of the container, see Figure Equilibrium conditions Following Finn [21], Chapter 1, we define for a given configuration S 0, Σ 0, where Σ 0 = Ω 0, a one parameter family of admissible comparison surfaces which are in general not yet volume preserving. Set z 0 (u, ɛ) = x(u) + ɛζ(u) + r(u, ɛ), ɛ < ɛ 0, the remainder r is continuously differentiable with respect to all arguments, such that r = O(ɛ) as ɛ 0, z 0 (u, ɛ) R 3 \ (Ω s Ω 0 ) if u U, z 0 (u, ɛ) Γ if u U, z 0 (u, ɛ) Σ 0 if u 2 U, and ζ(u) = ξ(u)n S0 (u) + η(u)t S0 (u) is a given vector field. Here N S0 denotes the unit normal to S 0 pointed to the exterior of the liquid, and T S0 is a unit tangent field defined on a strip U δ of U of width δ, such that on S 0, ν 0 := T S0 is orthogonal to S 0 and points to the exterior of the liquid, see Figure 3.1. In generalization to the normal at the boundary of a two dimensional domain the vector ν 0 at S 0 is called conormal. We assume that ξ and η are sufficiently regular on U, supp η U δ, and ξ 2 + η 2 1. Define on S 0 = {x(u) : u U} the angle γ [0, π] by cos γ = N Γ N S0. This angle depends on P S 0. Later we will see that this angle is a constant. Since ζ, N Γ = 0 at S 0 and N Γ, N S0, T S0 are in a common plane, it follows ξ(u)cos γ η(u)sin γ = 0, (3.3)

45 3.2. EQUILIBRIUM CONDITIONS 45 S 0 T δs S 0 0 Figure 3.1: T S0 at S 0 where u U. The family z 0 (u, ɛ), where ζ satisfies equation (4.3), is called a family of admissible comparison surfaces. Such a family exists provided the boundary of the container is sufficiently smooth. The proof exploits the implicit function theorem, we omit the details. Set V = {ζ = ξn S0 + ηt S0 : (ξ, η) satisfies (3.3)}. Suppose that for given ζ V the family z(u, ɛ) is volume preserving, i. e., Ω(S(ɛ) = V for all ɛ < ɛ 0. It follows that S 0 ξ da = 0 (3.4) since, see the corollary to Lemma A.3.4 of the appendix to this chapter, [ ] d dɛ Ω(S(ɛ) = ξ da. S 0 ɛ=0 On the other hand, let ζ V satisfies the side condition (3.4), then there exists a volume preserving family of admissible comparison surfaces given by z(u, ɛ) = x(u) + ɛζ + O(ɛ 2 ) with a sufficiently regular remainder O(ɛ 2 ), see Lemma A.3.1 of the appendix to this chapter. Suppose that S 0 is energy minimizing subject to such a family of comparison surfaces S(ɛ), i. e., E(S(ɛ)) E(S 0 ). Then [ ] d dɛ E(S(ɛ)) = 0. (3.5) ɛ=0 A sufficiently regular surface S 0 defined by x(u) is said to be an extremal if equation (3.5) is satisfied, where S(ɛ) is a volume preserving family of comparison surfaces defined above. Set V 0 = {ζ V : ξ da = 0}. (3.6) S 0

46 46 CHAPTER 3. CAPILLARY INTERFACES Let where f (x), ζ = g (x), ζ = [ ] d dɛ E(S(ɛ)) [ ] d dɛ Ω(S(ɛ) g (x), ζ = ξ da S 0 ɛ=0, ɛ=0 and, see the corollaries to Lemma A.3.2 A.3.4 of the appendix to this chapter, f (x), ζ = 2σ ξh 0 da + σ η ds S 0 S 0 σβ (ξ sinγ + η cos γ) ds + ξf(x) da. S 0 S 0 From a Lagrange multiplier rule, see Chapter 8, it follows that there exists a real constant λ such that for all ζ V. f (x), ζ + λ g (x), ζ = 0 (3.7) Theorem 3.1. Suppose that S is an extremal, then 2σH + F(x) + λ = 0 on S 0 cos γ = β on S 0. Proof. See Finn [21], p. 10. From the above formula (3.7) we get 0 = ( 2σH + F(x) + λ)ξ da S 0 +σ [ βξ sin γ + η(1 β cos γ)] ds S 0 for all (ξ(u), η(u)) satisfying ξ 2 + η 2 1 in U and (3.3). Thus 2σH + F(x) + λ = 0 on S 0

47 3.2. EQUILIBRIUM CONDITIONS 47 and [ βξ sinγ + η(1 β cos γ)] ds = 0 S 0 for all (ξ(α, β), η(α, β)), (α, β) U, satisfying (3.3). Set ξ = τ sinγ, η = τ cos γ, where τ = τ(α, β), τ 1, τ C( U). Then S 0 τ(β cos γ) ds = 0 for all these τ. It follows that cosγ = β on S 0. Remark. If Y is the gravitational potential Y = gx 3, g gravity, directed downwards into the negative x 3 -axis, see Figure 1.3, then 2H = λ 0 + gρ σ x 3 on S 0 where λ 0 = λ/σ. If ρ is a constant (incompressible fluid), then the constant is called capillarity constant The capillary tube κ = ρg σ Consider a capillary tube with constant cross section Ω R 2. We assume that the cross section is perpendicular to the x 3 -axis, see Figure 3.2. Suppose that the capillary surface S is a graph over Ω and that the potential Y is the gravitational potential Y = gx 3, where g denotes the gravity directed downwards into the negative x 3 -axis. Set λ 0 = λ/σ, then the boundary value problem of Theorem 3.1 reads as div Tu = κu + λ 0 in Ω (3.8) ν Tu = cos γ on Ω, (3.9) where ν is the exterior unit normal at Ω (where it exists), and Tu = u 1 + u 2.

48 48 CHAPTER 3. CAPILLARY INTERFACES x 3 g S liquid γ Ω Figure 3.2: Capillary tube If κ > 0, then the following mapping eliminates the Lagrange parameter in (3.8). Set then we arrive at u = v λ 0 κ, (3.10) div T v = κv ν Tv = cos γ in Ω on Ω. If Ω is bounded and sufficiently regular, then the Lagrange parameter is defined through the data of the problem. To see this, we integrate equation (3.8) over Ω, use the boundary condition (3.9) and obtain λ 0 = Ω cos γ κv Ω, where V denotes the volume of the liquid. In this calculation we suppose that there are no dry spots on the bottom Ω. Consequently let v be a solution of the previous boundary value problem, then the solution u of the problem (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 the equation instead in the domain. An important example is a disk B a (0)

49 3.3. EXPLICIT SOLUTIONS 49 with radius a and center at the origin. The boundary value problem in consideration is div T u = κu ν Tu = cos γ in Ω(a) on Ω(a) where x Ω(a), u = u(x), and a is a positive parameter, a representative length. The mapping (similarity transformation) transforms this problem into y = x a, v(y) = 1 a u(ay) div T v = Bv ν Tv = cos γ in Ω on Ω, where B = κa 2 is called Bond number. Here we assume that the mapping y = x/a maps Ω(a) onto a fixed domain Ω, independent of a. 3.3 Explicit solutions There are only few examples where explicit solutions are known. We will see in Chapter 6 that such solutions can serve as a leading term for asymptotic expansions of solutions to problems where no explicit solutions are known Ascent of a liquid at a vertical wall Let Ω = {(x 1, x 2 ) R 2 : x 1 > 0} be the right half plane, and consider the boundary value problem div T u = κu in Ω ν Tu = cos γ on Ω, where κ is a positive constant. Now we assume additionally that there exists a solution u which depends on x = x 1 only such that lim x u(x) = 0 and lim x u (x) = 0. In the following we will find such a solution. From a maximum principle of Finn and Hwang [25], see Section 6.1, it follows that this solution is the only one of the original boundary value problem without

50 50 CHAPTER 3. CAPILLARY INTERFACES the additionally assumptions. Thus we have to consider a boundary value problem for an ordinary differential equation. ( u (x) 1 + u (x) 2 ) = κu(x), 0 < x < (3.11) lim x 0, x>0 From the identities u u (x) 1 + u (x) 2 = cos γ (3.12) lim u(x) = 0, lim x x u (x) = 0. (3.13) h γ S g liquid x 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) ( ) u (x) 2 (3.15) = 1 u (x) it follows from (3.11) that ( ) 1 1 u (x) 1 + u (x) 2 ( = κu u (x) 2 ) = 1 2 κ(u2 ).

51 3.3. EXPLICIT SOLUTIONS 51 Consequently we have u (x) 2 = 1 2 κu2 + C, where C is a constant. From the boundary conditions (3.13) we see that C = 1. Then u (x) = κu2. (3.16) Combining this formula with the boundary condition (3.12), we get for the ascent h = u(0) of the liquid at the wall, see Figure 3.3 for the case 0 γ π/2, 2 h = 1 sinγ. κ 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 satisfied for the solution which we will calculate in the following. Then we obtain from (3.16) equation x (u) 1 + x (u) = κu2. (3.17) We have used the additional assumption that x (u) < 0, which is satisfied for the calculated solution. We suppose here that 0 γ < π/2. From (3.17) it follows x (u) = where α := 2/κ. Using the substitution we find that x(u) = = τ + α = κu 2 /2 1 1 (κu 2 /2 1) 2 u 2 /α 1 u/α 2α u 2, τ = 2α u 2, κu 2 /2 1 1 (κu 2 /2 1) 2 du + C = 2α u 2 + α 2 2α ln dτ 2α τ 2 + C ( ) 2α + 2α u 2 + C. 2α 2α u 2

52 52 CHAPTER 3. CAPILLARY INTERFACES The constant follows from the condition x(u(0)) = 0, where u(0) = h, and the final formula is x(u) = 2α h 2 ( α 2α u ln ( 2α ) 2α h 2 )u ( 2α, 2α u 2 )h where α = 2/κ Ascent of a liquid between two parallel plates Let Ω(d) = {(x 1, x 2 ) R 2 : d/2 < x 1 < d/2, d > 0, and consider the boundary value problem div T u = κu in Ω ν Tu = cos γ 1 on Γ 1 ν Tu = cos γ 2 on Γ 2, where κ is a positive constant, and Γ 1, Γ 2 are the two walls, see Figure 3.4. Now we assume additionally that there exists a solution u which depends on x = x 1 only, see Figure 3.4. In the following we will find such a solution. As u g γ 2 d/2 S d/2 x Γ 1 γ 1 liquid Γ 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 original boundary value problem without the additionally assumption. Thus we have

53 3.3. EXPLICIT SOLUTIONS 53 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 boundary value problem, where B = κd 2. ( v (y) 1 + v (y) 2 ) = Bv(y), 1/2 < y < 1/2 (3.18) lim y ( 1/2)+0 lim y (1/2) 0 v (y) 1 + v (y) 2 = cos γ 1 (3.19) v (y) 1 + v (y) 2 = cos γ 2. (3.20) To simplify the consideration here, we consider the case that the contact angles on Γ 1 and Γ 2 are the same: γ = γ 1 = γ 2. Then we seek a solution of ( v (y) 1 + v (y) 2 ) = Bv(y), 0 < y < 1/2 lim y (1/2) 0 v (y) 1 + v (y) 2 = cosγ v (0) = 0. As in the case of the ascent on a vertical wall we see that the inverse function y(v) satisfies y (v) 1 + y (v) = C v2 2 a2, (3.21) where C is a constant and a = 2/B. Suppose that 0 γ π/2, and assume v (y) > 0, 0 < y 1/2, if γ < π/2. The previous assumption is satisfied for the solution which we will derive in the following. Then where y (v) = y(v) = v 0 = v(0) = C v 2 /a 2 1 (C v 2 /a 2 ) 2 v see (3.21). The substitution C t 2 /a 2 v 0 1 (C t 2 /a 2 ) dt, v 2 0 v v 1, 2 2 C 1, v1 = v(1/2) = C sinγ, B B cos ξ = C t2 a 2, 0 ξ π 2,

54 54 CHAPTER 3. CAPILLARY INTERFACES where ξ denotes the angle of inclination of the curve defined by v(y), leads to the elliptic integral X(ξ) = a 2 ξ 0 cos τ C cos τ dτ. Then we find the constant C 1 as the solution of 1 2 = a (π/2) γ 2 0 cos τ C cos τ dτ. It follows that C if a = 2/B. We recall that B = κd 2. Consequently B = 2 C (π/2) γ = as C, which implies that 0 2 C cos γ + O as B 0, see an exercise. Then as B 0. Since u(x) = dv(y), we obtain finally ( 1 cos τ dτ + O ( ) 1 C 3/2 C = 2 B cos2 γ + O( B) v 0, v 1 = 2 cos γ + O(1) B C 3/2 ) as d 0. u 0, u 1 = 2 cos γ + O(d) κd Zero gravity solutions Here we consider domains Ω R 2 such that there are solutions u(x) of div T u = 2H in Ω (3.22) ν Tu = cos γ on Ω, (3.23) where H is a constant.

55 3.3. EXPLICIT SOLUTIONS 55 Constant circular cross section Let Ω(a) = B a (0) be a disk with radius a and center at the origin. We assume that the solution u(x) in consideration is rotationally symmetric with center at the origin, i. e., u(x) = v(r), where r = x x2 2, then ( ) 1 rv (r) = 2H (3.24) r 1 + v (r) 2 lim r a 0 v (r) 1 + v (r) 2 = cos γ. (3.25) From a comparison principle, see Section 6.1, it follows that every solution must be rotationally symmetric. Integrating (3.22) over Ω(a) and using the boundary condition, we get Equation (3.24) implies 2H = 2 cos γ. (3.26) a rv (r) 1 + v (r) 2 = Hr2 + C. Set r = 0, it follows that the constant C is zero. Then, if 0 γ π/2, 1 v(r) = H 2 r2 + C with another constant C. Thus v(r) defines a circle with radius 1/ H and center at (0, C). Suppose that the liquid of given volume V occupies a domain as shown in Figure 3.5a, i. e., we consider a tube closed by a flat bottom, and suppose that there is enough liquid such there is no dry spot at the bottom. Then we find from that V = 2π a 0 rv(r) dr C = V πa2 πa 2. Assume there is no bottom, and let S + and S be defined by 1 1 v + = H 2 r2 + C +, v = H 2 r2 + C,

56 56 CHAPTER 3. CAPILLARY INTERFACES + S liquid liquid S (a) (b) Figure 3.5: Liquid in a circular tube resp., with constants C +, C, see Figure 3.5b. Then C + C = V πa2 πa 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 < x < b < and assume that there exists a solution which is defined as a graph over the x-plane, then div Tu = 2H ν Tu = cos γ 1 ν Tu = cos γ 2 in a < x < b on x = a on x = b Scaling x = by, v(y) := u(by)/b leads to div Tv = 2A in q < y < 1 ν Tv = cos γ 1 on y = q ν Tv = cos γ 2 on y = 1, where q = a/b. By integration of the differential equation we obtain for the constant A = cos γ 2 + q cos γ 1 1 q 2.

57 3.3. EXPLICIT SOLUTIONS 57 Suppose the solution in consideration is rotationally symmetric. From a maximum principle, see Section 6.1, it follows that every solution must be rotationally symmetric. Set w(r) = v(y), r = y1 2 + y2 2, see Figure 3.6. Then the boundary value problem is + S γ 2 liquid γ 1 liquid S (a) (b) Figure 3.6: Liquid between two coaxial cylinders ( ) 1 rw (r) = 2A, q < r < 1 r 1 + w (r) 2 lim r q+0 lim r 1 0 w (r) 1 + w (r) 2 = cos γ 1 w (r) 1 + w (r) 2 = cosγ 2. Elementary calculations lead to the elliptic integral w(r) = r q f(s) 1 f 2 (s) ds + const., where f(s) = 1 s ( q cos γ1 + A(s 2 q) ), see an exercice. In Figure 3.6b the upper surface S + is given by bw( y )+B, and the lower surface S by bw( y ) + C with a constant C.

58 58 CHAPTER 3. CAPILLARY INTERFACES Cross section is a regular n-gon Consider problem (3.22), (3.23), where 0 γ π/2. Let Ω be a regular n-gon, where the corners lie on the unit circle. It is easily seen that the lower hemisphere given by with v 0 (x; γ) = 1 H 1 H 2 x 2, H = cos γ cos(π/n), is a solution of (3.22), (3.23) if H 1, see Figure 3.7 for the case that Ω is a square. The previous inequality implies that π/n γ, or, equivalently, α + γ π/2, where 2α denotes the interior angle at the corners. From a maximum principle, see Section 6.1, it follows that every solution of (3.22), (3.23) is given by v 0 + const., provided that π/n γ holds. The following S Ω Figure 3.7: Cross section is a square proposition shows that there is no solution of (3.22), (3.22) if π/n > γ. Let Ω R 2 be a domain with a corner located at the origin and assume that this corner with opening angle 0 < 2α < π is locally defined by straight lines, see Figure 3.8. Theorem 3.2 (Concus and Finn [12]). Suppose that u C 2 (Ω a ) C 1 (Ω a \ {0}) is a solution of div Tu = f(x, u) in Ω a ν Tu = cos γ on Σ (1) a Σ (2) a,

59 3.3. EXPLICIT SOLUTIONS 59 where f(x, u) is given function satisfying sup x Ω a f(x, u(x)) <. Then α + γ π 2. x 2 ν (1) Σ a ν Ω a α ν α (2) Σ a a Γ a x 1 Figure 3.8: Corner domain Proof. Integrating the differential equation over Ω c, 0 < c a, we find that Σ (1) c Σ (2) c ν Tu ds + ν Tu ds = Γ c 2 Σ c cos γ + Since ν Tu 1, it follows Γ c ν Tu ds = O( Ω c ). 2 Σ c cos γ Γ c + O(c 2 ) 2c cos γ cos α 2c tan α + O(c2 ), Ω c f(x, u(x)) dx which implies that cosγ sinα, and the theorem is shown. We recall that 0 γ π/2.

60 60 CHAPTER 3. CAPILLARY INTERFACES Corollary. If there is a corner of Ω with interior angle 2α defined locally through straight lines, then there is no bounded solution of div Tu = f(x, u) ν Tu = cos γ in Ω on Ω, where f(x, u) is given function satisfying if α + γ < π/2 holds. sup f(x, u) = m(c) < x Ω, u C Thus one expects that liquid, oil for example, flows out of a glass with edges of interior angles less than π 2γ. On the other hand an upper edge can serve as a wetting barrier which prevents the liquid to flow out of the glass, see Chapter Stability We will show that a capillary interface in equilibrium defines a strong energy minimizer if there exists an embedding foliation of capillary surfaces and if the Lagrange multiplier decreases with increasing volume of liquid in the container. We also give a proof of Wente s second variation formula which includes the borderline cases that the contact angle is 0 or π. Finally, we discuss the question of the existence of an embedding foliation for a given capillary interface in equilibrium. Let K Γ be the curvature at P S 0 of the plane curve defined by the intersection of Γ with the plane through P and the linear span N S0 (at P) and ν 0. This curvature is considered as nonnegative if the curve bends in the direction of N Γ. Extending an idea of Finn, we consider a class of volume preserving admissible comparison surfaces S(ɛ) defined by z(u, ɛ) = x(u) + ɛζ(u) + q(ɛ)ξ 0 (u)n 0 (u) + ɛ2 2 ζ(u) 2 r(u) + O(ɛ 3 ), (3.27) where ζ V 0 is given, r(u) = K Γ (u)n Γ (u)ρ(u). Here K Γ (u) and N Γ (u) denote smooth extensions of K Γ and N Γ, which are defined on S 0, to U δ, and ρ(u) is a given smooth function which satisfies

61 3.4. STABILITY 61 ρ(u) = 1 if u U and ρ(u) = 0 if dist (u, U) > δ. For example, N Γ (u) := N S0 (u)cos γ T S0 (u)sin γ is an extension of N Γ (u), u U to U δ. We can assume that the remainder O(ɛ 3 ) is sufficiently regular, which follows from an implicit function theorem, we omit the details. The function q(ɛ) satisfies q(0) = 0 and q (0) = 0, see Lemma A.3.1. We recall that and V = {ζ = ξn S0 + ηt S0 : (ξ, η) satisfies (3.3)} V 0 = {ζ V : S 0 ξ da = 0}. Suppose that S 0 = S(0) is an equilibrium interface, i. e., x(u) satisfies the necessary conditions of Theorem 3.1. Set then, if ζ V 0, L(S, λ) = E(S) + λ( Ω(S) V ), E(S(ɛ)) E(S 0 ) = L(S(ɛ), λ) L(S 0, λ) [ ] [ ] d = ɛ L(S(ɛ), λ) + ɛ2 d 2 L(S(ɛ), λ) dɛ ɛ=0 2 dɛ2 Since [ ] d L(S(ɛ), λ) = 0 dɛ ɛ=0 ɛ=0 + O(ɛ 3 ). for all ζ V, see the Lagrange equation (3.7), in particular for ζ V 0. Thus [ ] d 2 L(S(ɛ), λ) dɛ2 ɛ=0 0 is a necessary condition such that the inequality holds for all ɛ < ɛ 0. Next we will calculate E(S(ɛ)) E(S 0 ) [ ] d 2 L(S(ɛ), λ) dɛ2 under the assumption that S 0 defines an equilibrium, i. e., it satisfies the equations of Theorem 3.1. ɛ=0

62 62 CHAPTER 3. CAPILLARY INTERFACES From Lemma A.3.2 A.3.4 of the appendix to this chapter we see that d L(S(ɛ), λ) = 2σ H(u, ɛ) N(u, ɛ), z ɛ (u, ɛ) W(u, ɛ) du dɛ U +σ ν(u, ɛ), z ɛ (u, ɛ) ds(ɛ) S(ɛ) σβ Z 0,ɛ (τ, ɛ), ν(u, ɛ) Z 0,τ (τ, ɛ) dτ U + F(z(u, ɛ) N(u, ɛ), z ɛ (u, ɛ) W(u, ɛ) du U +λ N(u, ɛ), z ɛ (u, ɛ) W(u, ɛ) du, (3.28) U where ds(ɛ) = Z τ (τ, ɛ) dτ, Z(τ, ɛ) = z(u(τ), ɛ), Z 0 (τ, t) = z 0 (u(τ), t) and u(τ) is a regular parameter representation of U. Set D(u, ɛ) = N(u, ɛ), z ɛ (u, ɛ) W(u, ɛ), then [ ] d 2 L(S(ɛ), λ) = 2σ H dɛ2 ɛ (u, 0)D(u, 0) du ɛ=0 U 2σ H(u, 0)D ɛ (u, 0) du U [ ] +σ U ɛ ν(u, ɛ), z ɛ(u, ɛ) Z τ (τ, 0) dτ ɛ=0 [ ] +σ ν(u, 0), z ɛ (u, 0) U ɛ Z τ(τ, ɛ) dτ ɛ=0 [ ] σβ U ɛ ν(u, ɛ), Z 0,ɛ(τ, ɛ) Z 0,τ (τ, ɛ) dτ ɛ=0 [ ] σβ ν(u, 0), Z 0,ɛ (τ, 0) 1 U ɛ Z 0,τ(τ, ɛ) dτ ɛ=0 + F(x(u)), z ɛ (u, 0) D(u, 0) du U +λ F(x(u))D ɛ (u, 0) du. U From the differential equation 2σH 0 (u) + F(x(u)) + λ = 0 on S 0 and the formula, see Lemma A.3.8 of the appendix to this chapter, 2H ɛ (u, 0) = ξ(u) + 2(2H 2 0(u) K 0 (u))ξ(u) + 2H 0,α (u)η α (u),

63 3.4. STABILITY 63 where H 0, K 0 are the mean and Gauss curvature of S 0, resp., ξ and η α are defined through ζ = ξn 0 + η α x α, η α x α = ην 0 on S 0, and ξ, η satisfy ξ cos γ η sinγ on U, we get [ ] d 2 ( L(S(ɛ), λ) = σ dɛ2 ξ + 2(2H 2 0 K 0 )ξ ) ξ da ɛ=0 S 0 + F, N 0 ξ 2 da S 0 [ ] +σ ɛ ( ν(u, ɛ), z ɛ(u, ɛ) β ν(u, ɛ), Z 0,ɛ (τ, ɛ) ) U +σ U Z 0,τ (τ, 0) dτ ( ν(u, 0), z ɛ (u, 0) β ν(u, 0), Z 0,ɛ (τ, 0) ) [ ] ɛ Z 0,τ(τ, ɛ) Here we have used that D(u, 0) = ξw(u, 0), Z τ (τ, 0) = Z 0,τ (τ, 0), and z ɛ (u, 0) = z 0,ɛ (u, 0) = ζ = ξn 0 + η α x α, 2σH 0,α (u) + F(x(u)), x α (u) = 0 on S 0. Next we show that the integrands of the last two terms in the previous formula for the second derivative of L vanish. We have ɛ=0 ν(u, 0), z ɛ (u, 0) β ν(u, 0), Z 0,ɛ (τ, 0) = ν 0 βν, ζ since z ɛ (u, 0) = ζ and Z 0,ɛ (τ, 0) = ζ on S. At S 0 the vectors ν 0, ν 0, N Γ are in a common plane, and ν 0 and N Γ, N Σ0, resp., are orthogonal. Thus we have at S 0 with scalar functions a, b the formula ν 0 = aν 0 +bn Γ. Using the boundary condition cosγ = β on S 0 it follows ɛ=0 dτ. ν 0 ν 0 cos γ = N Γ sinγ. (3.29) Then the brackets vanish since ζ N Γ at S 0. This is a consequence of formula ζ, ν 0 βν 0 = τ(cos γ β), where τ is here a continuous function defined on U, see the proof of Theorem 3.1.

64 64 CHAPTER 3. CAPILLARY INTERFACES For the brackets in the integrands of the remaining boundary integrals we get from formula (3.29) above that [ ] ɛ ( ν(u, ɛ), z ɛ(u, ɛ) β ν(u, ɛ), Z 0,ɛ (τ, ɛ) ) ɛ=0 = ν 0, z ɛɛ (u, 0) β ν 0, Z 0,ɛɛ (τ, 0) + ν ɛ, z ɛ (u, 0) β ν ɛ (u, 0), ζ = K Γ ζ 2 N Γ, ν 0 β k ν 0 + ζ, ν ɛ (u, 0) βν ɛ (u, 0) = K Γ ζ 2 sinγ + ζ, ν ɛ (u, 0) βν ɛ (u, 0). Finally, we obtain ζ, ν ɛ (u, 0) βν ɛ (u, 0) = ξn 0 + ην 0, ν ɛ (u, 0) β ζ,ν ɛ (u, 0) = ξ N 0, ν ɛ (u, 0) ( ) ξ = ξ + K S0 η. ν 0 Here we have used that ν ɛ (u, 0), ν(u, 0) = 0, ζ = aν 0 with a scalar function a, and the formula, see Lemma A.3.6 of the appendix to this chapter, N 0, ν ɛ (u, 0) = ξ ν 0 + K S0 η, where K S0 is the curvature at P S 0 of the plane curve defined through the intersection of S 0 and the plane through P which is spanned by N S0 at P and ν 0 := ν(0) at P. The curvature K S0 is considered as nonnegative if the curve in consideration bends in the direction of N S0 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 defines S 0 satisfies the equations of Theorem 3.1. Then [ ] d 2 L(S(ɛ), λ) dɛ2 ɛ=0 = σ ( ξ 2 2(2H0 2 K 0 )ξ 2) da S 0 + F, N 0 ξ 2 da S 0 ( +σ ξηks0 ζ 2 K Γ sinγ ) ds. 1 S 0

65 3.4. STABILITY 65 Corollary. Suppose that γ k are different from 0 or π, then [ ] d 2 ( L(S(ɛ), λ) = σ dɛ2 ξ 2 2(2H0 2 K 0 )ξ 2) da ɛ=0 S 0 + F, N 0 ξ 2 da S 0 ( cos γ +σ sinγ K S 0 1 ) sin γ K Γ ξ 2 ds. S 0 Proof. The formula follows since ζ 2 = ξn 0 + ην 0, ξn 0 + ην 0 = ξ 2 + η 2 on S, and on S 0. ξ cos γ η sinγ = 0 Remark. We recall that 2H 2 0 K 0 0 holds for every regular surface. Remark. The second variation formula of Lemma 3.1 was derived by Wente [75] under the assumption that γ is different from 0 or π. In the case of constant mean curvature, i. e. if F = 0, a proof was given by Ros and Souam [61], also for contact angles different from 0 or π. Our proof includes the borderline cases γ = 0 and γ = π as well as the case of a nonvanishing F. At some points and notations our proof is close to that of Ros and Suam [61]. Moreover, we do not assume the existence of a smooth continuation of the given surface S 0 accross its boundary. An extremal is often called stable by definition if the second variation, given by the formula of the above lemma, is positive for all nonvanishing ξ Strong minimizers Let S 0 be a sufficiently regular bounded capillary interface in equilibrium, i. e., S 0 satisfies the equilibrium conditions of Theorem 3.1. Define the Lagrange functional L(S, λ) for sufficiently regular admissible interfaces S by L(S, λ) = E(S) + λ( Ω l (S) V 0 ),

66 66 CHAPTER 3. CAPILLARY INTERFACES where E is given by (3.1), λ R and V 0 = Ω l (S 0 ). Admissible means that S Γ and S R 3 \ Ω s. If S is volume preserving, i. e. Ω l (S) = V 0, then L(S, λ) = E(S). Define for a small δ > 0 a δ-neighbourhood D δ of S 0 by D δ = {x R 3 \ Ω s : dist(x, S 0 ) < δ}. Let S(τ), τ < ɛ, be a family of sufficiently regular and admissible surfaces defined by z(u, τ) R 3, u U, where U R 2 is a fixed parameter domain. A family S(τ) which covers D δ simply is called a foliation, and the surface S 0 is called embedded in this family if S 0 is defined by z(u, 0). Remark. The method of foliation was used by Wente [79] to prove a result concerning capillary tubes of nonconstant circular cross sections. Assumptions. (i)there exists an embedding foliation of S 0 defined by N S(τ) N Γ 2σH = λ(τ) + F(z(u, τ)) in S(τ) = cosγ on S(τ), where H is the mean curvature of S(τ) at z(u, τ). (ii) Ω(S(τ)) is increasing and λ(τ) is decreasing with growing τ. Theorem 3.3. Suppose that S 0 satisfies the previous assumptions (i) (ii). Then E(S) E(S 0 ) for all admissible volume preserving comparison surfaces S D δ. Proof. Let x D δ and consider the associated surface S(τ(x)) from the family S(τ). We recall that τ(x) is constant on S(τ(x)) and 2σH = λ(τ(x)) + F in S(τ(x)) (3.30) N S(τ(x)) N Γ = cos γ on S(τ(x)). (3.31) Let n be the normal directed out of the liquid, see Figure 3.9, on the surface S(τ(x)) at x, then div n = 2H (3.32) at x, where H is the mean curvature of S(τ(x)) at x. For a proof of formula (3.32) see [18], pp. 77, for example. Combining (3.30) and (3.32), we get div n + 1 (λ(τ(x)) + F) = 0 σ

67 3.4. STABILITY 67 at x D δ. Let T + and T be the domains enclosed by S 0 and S which are + T + S n ν S( τ ) τ>0 _ Σ + S _ 0 Σ S+ 0 _ T S( τ ) τ <0 _ S Figure 3.9: Proof of Theorem 2.1 above or below, resp., of S 0. See Figure 3.9 for further notations used in the following. Integrating over T + and T, resp., we obtain 0 = Σ + cos γ S n ν da + 1 (F + λ(τ(x))) dx S + σ T + = Σ + cos γ S S+ + (n ν 1) da S (F + λ(τ(x))) dx σ T + and 0 = Σ cos γ + S0 S + (n ν +1) da+ 1 (F + λ(τ(x))) dx S σ T Consequently we have S S 0 ( Σ + Σ )cos γ = (1 n ν)da + (n ν + 1) da S + S + 1 (F + λ(τ(x))) dx 1 (F + λ(τ(x))) dx. σ T σ T + Since W(S) W(S 0 ) = Σ + Σ

68 68 CHAPTER 3. CAPILLARY INTERFACES and L(S, λ 0 ) L(S 0, λ 0 ) = σ ( S S 0 β( W(S) W(S 0 ) )) + F dx F dx + λ 0 ( Ω l (S) Ω l (S 0 ) ), Ω l (S) Ω l (S 0 ) it follows that ( ) L(S, λ 0 ) L(S 0, λ 0 ) = σ (1 n ν) da + (n ν + 1)dA S + S + (F + λ(τ(x))) dx (F + λ(τ(x))) dx T T + + F dx F dx + λ 0 ( Ω l (S) Ω l (S 0 ) ). Thus Ω l (S) Ω l (S 0 ) ( L(S, λ 0 ) L(S 0, λ 0 ) = σ (1 n ν) da + S + (n ν + 1) da S ) + (λ(τ(x)) λ 0 )dx + T (λ 0 λ(τ(x))) dx. T + (3.33) Since S is volume preserving by assumption we have L(S, λ 0 ) L(S 0, λ 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 bounded cross section Ω R 2. Assume that the capillary interface S 0 is given by a graph z = u(x) over the x-plane, x = (x 1, x 2 ). Then div Tu = κ u + λ 0 in Ω ν Tu = cos γ on Ω, where Tu = u/ 1 + u 2, ν is the exterior unit normal at Ω and κ > 0 is the capillary constant. Let V 0 be the given volume. A foliation is explicitly given by S(τ) : z(x, τ) = u(x) + τ, where τ R, τ < ɛ. Then V (τ) = V 0 + τ Ω is the volume of liquid under S(τ), and the Lagrange parameter is All assumptions (i) (ii) are satisfied. λ(τ) = 1 ( Ω cos γ κv (τ)). Ω

69 3.4. STABILITY On the existence of an embedding foliation Let S 0 be a sufficiently regular surface defined by x = x(u) satisfying the equilibrium equations of Theorem 3.1. Let τ be a real parameter in a neighbourhood of 0. Let S(τ) be an admissible family of surfaces, defined by where z(u, τ) = x(u) + ζ(u, τ), ζ(u, τ) = ξ(u, τ)n S0 + η(u, τ)t S0, with supp η(u, τ) U δ ( τ 0, τ 0 ) for a τ 0 > 0, and U δ is a closed boundary strip of U as introduced in Section 3.3. Admissible means, see Section 3.1, that z(u, τ) is in the interior of the container if u U and on the container wall if u U. Assume S(τ) is sufficiently regular and satisfies 2σH = λ(τ) + F(z(u, τ)) in S(τ) (3.34) N S(τ) N Γ = cos γ on S(τ) (3.35) Ω l (S(τ)) = V 0 + τ, (3.36) where H denotes the mean curvature of S(τ) at z(u, τ), N Γ is the normal on the container wall Γ directed as shown in Figure 1.3, and V 0 = Ω l (S 0 )) is the given volume. Suppose that ξ(u, τ) = ξ 1 (u)τ ξ 2(u)τ 2 + O(τ 3 ) η(u, τ) = η 1 (u)τ η 2(u)τ 2 + O(τ 3 ) λ(τ) = λ 0 + λ 1 τ + O(τ 2 ), where the coefficients and the remainders are sufficiently regular. Set ζ l (u) = ξ l (u)n S0 + η l (u)t S0, l = 1, 2, then z(u, τ) = x(u) + ζ 1 (u)τ ζ 2(u)τ 2 + O(τ 3 ). By assumption, z(u, τ) is on the container wall if u U. Consequently we have at U: ξ 1 (u)cos γ η 1 (u)sin γ = 0 (3.37) ζ 2 (u) = K Γ ζ 1 2 N Γ. (3.38)

70 70 CHAPTER 3. CAPILLARY INTERFACES Then we get from equation (3.34) and Lemma A.3.8 of the appendix (first variation of mean curvature) that on S 0 σ( ξ 1 + 2(2H 2 K)ξ 1 + 2H,α ηw α ) = λ 1 + F N ξ 1 + x F, T η 1, where T = w α x,α, and H = H(u) is the mean curvature of S 0 at x(u). Using the first order necessary condition the above equation reduces to 2σH = λ 0 + F(x(u)) on S 0, σ( ξ 1 + 2(2H 2 K)ξ 1 ) = λ 1 + F N ξ 1. From (3.37) we obtain that ξ 1 = 0 on U if γ = 0 or γ = π. In the case 0 < γ < π, we find from the above boundary condition (3.35), equations (3.37), (3.38) and the corollary to Lemma A.3.9 of the appendix to this chapter that ( ξ 1 cos γ ν + sinγ K S 1 ) sin γ K Γ ξ 1 = 0 on S 0. Finally, equation (3.36) implies that S 0 ξ 1 da = 1. Set p = 2(2H 2 K) + 1 F σ N q = cos γ sinγ K S 1 sinγ K Γ. Consequently (ξ, λ) := (ξ 1, λ 1 ) is a solution of ξ + p ξ + λ σ = 0 in S 0 (3.39) ξ ν + q ξ = 0 on S 0 if 0 < γ < π (3.40) ξ = 0 on S 0 if γ = 0, or γ = π (3.41) ξ da = 1, S 0 (3.42) provided there exists a sufficiently regular family of admissible surfaces.

71 3.4. STABILITY 71 On the other hand, there exists a unique solution (ξ, λ) of the system (3.39) (3.42) if the second variation is positive. Set Q(φ, ψ) = ( φ ψ + p φψ) da + q φψ ds if 0 < γ < π S 0 S 0 Q(φ, ψ) = ( φ ψ + p φψ) da if γ = 0, or γ = π. S 0 Lemma 3.2. Suppose that 0 < γ < π/2. Then there exists a unique solution (ξ, λ) of (3.39) (3.42), provided the second variation Q(φ, φ) is positive for all φ W 1,2 (S 0 ) \ {0} satisfying S 0 φ da = 0 if 0 < γ < π/2. If γ = 0 or γ = π we have to replace W 1,2 through W 1,2 0. Proof. For the convenience of the reader we will give a proof. Set J(φ) = Q(φ, φ)/2 and consider the minimum problem min J(φ), where the minimum is taken over all φ W 1,2 (S 0 ) or φ W 1,2 0 (S 0 ), resp., satisfying the side condition S 0 φ da = 1. Take a sufficiently regular φ 0 which satisfies the previous side condition, and set φ = ψ + φ 0. Then the above minimum problem changes to minj(ψ + φ 0 ), where the minimum is taken over all ψ W 1,2 (S 0 ) or ψ W 1,2 0 (S 0 ), resp., satisfying the side condition S 0 ψ da = 0. The existence of a solution ψ 0 follows from the assumption concerning the second variation. Consequently there exists a Lagrange multiplier λ such that (ψ 0 +φ 0 ) is a solution of (3.39) (3.42). The uniqueness is also a consequence of the positivity of the second variation. Assume (ξ 1, λ 1 ) and (ξ 2, λ 2 ) are solutions of (3.39) (3.42), then ξ := ξ 1 ξ 2 satisfies (3.39) (3.41), with λ = λ 1 λ 2, and the homogeneous side condition S 0 ξ da = 0. It follows that Q(ξ, ξ) = 0. Consequently ξ = 0, which implies λ 1 = λ 2, see the above differential equation (3.39) with λ = λ 1 λ 2. Lemma 3.3. Assume λ 1 < 0 and that the second variation Q(φ, φ) is positive for all φ W 1,2 (S 0 ) \ {0} or φ W 1,2 (S 0 ) \ {0}, resp., satisfying S 0 φ da = 0. Then ξ 1 (u) > 0 on U in the case that 0 < γ < π, and ξ 1 (u) > 0 on the open set U if γ = 0 or γ = π. Proof. Set ξ = ξ 1 and λ = λ 1 /σ in this proof. Let ξ = ξ + + ξ, where ξ + = max{ξ,0}, ξ = min{ξ,0}, and assume ξ < 0 on a subset of U of

72 72 CHAPTER 3. CAPILLARY INTERFACES positive measure. Then Define ξ 0 = ξ + + αξ, where S 0 ξ da < 0. α = 1 1 S 0 ξ da. Then α > 1 and S 0 ξ 0 da = 0. We have, see (3.39) (3.40), Q(ξ, φ) + λ Φ da = 0 S 0 (3.43) for all φ W 1,2 (S 0 ). Inserting φ = ξ 0 und ξ = ξ 0 + (1 α)ξ, we see that 0 = Q(ξ 0, ξ 0 ) + (1 α)q(ξ, ξ 0 ) = Q(ξ 0, ξ 0 ) + (1 α)αq(ξ, ξ ) since ξ 0 = ξ + +αξ. Hence Q(ξ, ξ ) > 0 because of ξ 0 0 and α > 1. On the other hand, if we insert φ = ξ in equation (3.43), we get Q(ξ, ξ ) + λ ξ da = 0, S 0 which is a contradiction if λ 0. Thus we have shown ξ(u) 0 in U, provided that λ 0. If λ < 0, then it follows from the differential equation (3.39) and the strong maximum principle that ξ(u) > 0 in U. Next we show that ξ(u) > 0 on U, provided that 0 < γ < π. Let ξ(p) = 0 at P U. Then we will show that ξ ν > 0 (3.44) at P, which is a contradiction to the boundary condition (3.40). In fact, the Hopf boundary point lemma says that ξ/ n > 0, where n = (n 1, n 2 ) is the exterior normal at U. We recall that U is sufficiently smooth by assumption. Since ξ/ n = ξ,α n α and ξ/ ν = ξ,α ν α, where ν α = gg αβ n β, see [18], p. 44. Then ν α n α = gg αβ n α n β > 0 since (g αβ ) is positive definite. This inequality says that ν = (ν 1, ν 2 ) considered as a vector at P in R 2 makes an angle with n which is less than π.

73 3.5. APPENDIX 73 Thus at P we have ν = a n+bt, where a > 0. Here t denotes a unit tangent vector at U at P. Then ξ/ ν = a ξ/ n+b ξ/ t = a ξ/ n at P, which proves inequality (3.44). The above two lemmas suggest the following conjecture. Conjecture. Suppose that λ 1 < 0 and that the second variation is positive for variations, not identically zero, satisfying the side condition S 0 φ da = 0. Then there exists an embedding foliation of admissible surfaces. 3.5 Appendix Consider a family of admissible comparison surfaces given by z 0 (u, ɛ) = x(u) + ɛζ(u) + r(u, ɛ), ɛ < ɛ 0, the remainder r is continuously differentiable with respect to all arguments, such that r = O(ɛ) as ɛ 0, z 0 (u, ɛ) R 3 \ (Ω s Ω 0 ) if u U, z 0 (u, ɛ) Γ if u U, and ζ(u) = ξ(u)n S0 (u) + η(u)t S0 (u) is a given vector field. Here N S0 denotes the unit normal to S 0 pointed to the exterior of the liquid, and T S0 is a unit tangent field defined on closed strip U δ of U of width δ, such that on S 0, ν 0 := T S0 is orthogonal to S 0 and points to the exterior of the liquid, see Figure 3.1. We assume that ξ and η are sufficiently regular on U, supp η U δ, ξ 2 +η 2 1, and ζ, N Γ = 0, i. e. ξ, η satisfy (3.3). By ν(u, ɛ), resp. ν(u, ɛ), we denote the exterior normal to S(ɛ) in S(ɛ), resp., in the container wall Γ, see Figure At P S(ɛ) the vectors N Γ, N(u, ɛ), ν(u, ɛ) and ν(u, ɛ) are all in the same plane since S(ɛ) is a curve on S(ɛ) as well as on the container wall Γ. Here and in the following we set N(u, ɛ) = N S(ɛ) at u U. To get a volume preserving admissible family of comparison surfaces we replace z 0 (u, ɛ) through z (u, ɛ, q) := z 0 (u, ɛ) + qξ 0 (u)n 0 (u), where q R, N 0 = N S0, supp ξ 0 U, and S 0 ξ 0 (u) da = 1.

74 74 CHAPTER 3. CAPILLARY INTERFACES N(ε) ν (ε) N Γ ν (ε) S (ε) Γ Figure 3.10: Normals in consideration S* ( ε,0) S* ( ε, q) S 0 Figure 3.11: Volume preserving comparison surfaces Then S (ɛ, q), defined by z (u, ɛ, q), is a family of admissible comparison surfaces, in general not yet volume preserving, see Figure We recall that V is the set of all ξn S0 + ηt S0, where (ξ, η) satisfies the boundary condition ξ(u)cos γ η(u)sin γ = 0 on S 0, and V 0 is the subset of V such that S 0 ξ da = 0. Lemma A.3.1. For given ζ V satisfying the side condition S 0 ξ da = 0, there exists a regular function q = q(ɛ) such that the family of admissible configurations given by S(ɛ), where S(ɛ) is defined through z(u, ɛ) =

75 3.5. APPENDIX 75 z (u, ɛ, q(ɛ)), is volume preserving for all ɛ R 2, ɛ < ɛ 0. The function q(ɛ) satisfies q(0) = 0 and q (0) = 0. Proof. We have, see Figure 3.11, where Ω l (S (ɛ, q)) = Ω 1 Ω 2 Ω l (S 0 ), Ω 1 Ω 2 = {z (u, t, 0) : u U, 0 < t < ɛ}, = {z (u, ɛ, τ) : u U, 0 < τ < q}. Here we assume for simplicity that S (ɛ, q) is above of S 0 and of S (ɛ, 0). The final formula is valid for every S (ɛ, q), see a remark in [18], p. 81. Set f(ɛ, q) = Ω l (S (ɛ, q)). From det z (u, t, 0) (u, t) det z (u, ɛ, τ) (u, τ) = zα zβ, z t = N (u, t, 0), zt (u, t, 0) W(u, t, 0), = zα zβ, z τ = N (u, ɛ, τ), zτ(u, ɛ, τ) W(u, ɛ, τ), we find that f(ɛ, q) = + ɛ U 0 q U 0 + Ω l (S 0 ). N (u, t, 0), z t (u, t, 0) W(u, t, 0) dudt N (u, ɛ, τ), z τ(u, ɛ, τ) W(u, ɛ, τ) dudτ The assertion of the lemma follows since f q (0, 0) = ξ 0 da = 1 S 0 f ɛ (0, 0) = ξ da. S 0

76 76 CHAPTER 3. CAPILLARY INTERFACES Lemma A.3.2. d dɛ S(ɛ) = 2 H(u, ɛ) N(u, ɛ), z ɛ (u, ɛ) W(u, ɛ) du U + z ɛ (u, ɛ), ν(u, ɛ) ds(ɛ), U where H(u, ɛ) denotes the mean curvature of S(ɛ) at u U, and ds(ɛ) = Z τ (τ, c) dτ, Z(τ, c) := z(u(τ), c), here u(τ) is a regular parameter representation of U. Proof. Set u = (α, β), then S(ɛ) = and d dɛ S(ɛ) = U E(u, ɛ)g(u, ɛ) F 2 (u, ɛ) du 1 ( E(u, ɛ) zβ (u, ɛ), z βɛ (u, ɛ) U W(u, ɛ) F(u, ɛ)[ z α (u, ɛ), z βɛ (u, ɛ) + z β (u, ɛ), z αɛ (u, ɛ) ] +G(u, ɛ) z α (u, ɛ), z αɛ (u, ɛ) ) 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. [ ] d dɛ S(ɛ) = 2 H(u, 0)ξ(u) da + η(u) ds. ɛ=0 S 0 S 0 Proof. The corollary follows since z ɛ (u, 0) = ζ, and ζ = ξ(u)n S0 (u) + η(u)t S0. Lemma A.3.3. d dɛ W(S(ɛ)) = U Z 0,ɛ (τ, ɛ), ν(u, ɛ) Z 0,τ (τ, ɛ) dτ,

77 3.5. APPENDIX 77 where Z 0 (τ, t) = z 0 (u(τ), t) and u(τ) is a regular parameter representation of U. Proof. Let u(τ) be a regular parameter representation of U. Then W(S(ɛ)) = U ɛ 0 Z 0,τ (τ, t) Z 0,t (τ, t) dτdt + W(S(0)). Here we assume again that S(ɛ) is above of S(0). The final formula is valid for every 1 S(ɛ), see a remark in [18], p. 81. The assertion of the lemma follows since the tangential plane on the container wall at P S(t) is spanned by the orthogonal vectors ν(u(τ), t) and t(τ, t) = Z 0,τ (τ, t)/ Z 0,τ (τ, t). We recall that ν(u(τ), t) and t(τ, t) are orthogonal vectors. Then Z 0,t (τ, t) = aν(u(τ), t) + bt(τ, t) with scalar functions a and b depending on τ and t. Using the Lagrange identity, we find that Z 0,τ (τ, t) Z 0,t (τ, t) 2 = Z 0,τ (τ, t) 2 Z 0,t (τ, t) 2 Z 0,τ (τ, t), Z 0,t (τ, t) 2 Corollary. = Z 0,τ (τ, t) 2 (a 2 + b 2 ) Z 0,τ (τ, t), aν(u(τ), t) + bt(τ, t) 2 = Z 0,τ (τ, t) 2 (a 2 + b 2 ) b 2 Z 0,τ (τ, t) 2 = a 2 Z 0,τ (τ, t) 2 = Z 0,t (τ, t), ν(u(τ), t) Z 0,τ (τ, t) 2 [ ] d dɛ W(S(ɛ)) = (ξ(u)sin γ + η(u)cos γ) ds. ɛ=0 S 0 Proof. The formula follows since Z 0,τ (τ, 0) dτ = ds and Z 0,ɛ (τ, 0), ν(u(τ), 0) = ζ(u(τ)), ν(u(τ), 0) = ξ(u(τ)) N(u(τ), 0), ν(u(τ), 0) +η(u(τ)) ν(u(τ), 0), ν(u(τ), 0) = ξ(u(τ))sin γ + η(u(τ))cos γ.

78 78 CHAPTER 3. CAPILLARY INTERFACES Lemma A.3.4. d dɛ Ω l (S(ɛ)) F(x) dx = U F(z(u, ɛ)) N(u, ɛ), z ɛ (u, ɛ) W(u, ɛ) du. Proof. Here we assume for simplicity that S(ɛ) is above of S(0). The final formula is valid for every S(ɛ), see a remark in [18], p. 81. Then Ω l (S(ɛ), Σ(ɛ)) = Ω 1 Ω l (S 0 ), with Consequently Ω l (S(ɛ)) Ω 1 = {z(u, t) : u U, 0 < t < ɛ}. F(x) dx = U ɛ 0 F(z(u, t))det z(u, µ, t) (u, t) dudt. Since det z(u, t) (u, t) = z α z β, z t = N(u, t), z t (u, t) W(u, t), we obtain the formula of the lemma. Corollary. [ d dɛ Ω l (S(ɛ)) ] F(x) dx = F(x(u))ξ(u) da. S 0 ɛ=0 Set H 0 = H(u, 0), ν 0 = ν(τ, 0), ν 0 = ν(τ, 0), N 0 = N(u, 0), which is N S0 at u U. Let N(u, ɛ) be the unit normal to S(ɛ) pointed to the exterior of the liquid, where S(ɛ) is given by z(u, ɛ) = x(u) + ɛζ(u) + O(ɛ 2 ). For notations and formulas used in the following see [18], Part I. To simplify the notations we omit the variable u and set N (0) = N ɛ (u, 0). Lemma A.3.5. N (0) = g αβ N(0)), ζ,β x,α.

79 3.5. APPENDIX 79 Proof. We have N ɛ (0) = c α x α + cn(0). Since N(ɛ), N(ɛ) = 1, it follows c = 0. Consequently and The identity N(ɛ), z,α (ɛ) = 0 implies that With (3.46), we obtain which implies that N (0) = c α x,α (3.45) N (0), x,β = c α g αβ. (3.46) N (0), x,α + N(0), ζ,α = 0. N(0), ζ,α = c α g αβ, c α = g αβ N(0), ζ,β. Together with (3.45) we get the formula of the lemma.. Let ν(ɛ) be the exterior normal to S(ɛ) in S(ɛ), see Figure 4.4, and K S0 is the curvature at P S 0 of the plane curve defined through the intersection of S 0 and the plane through P which is spanned by N S0 at P and ν 0 := ν(0) at P. The curvature K S0 is considered as nonnegative if the curve bends in the direction of N S0 at P. Lemma A.3.6. N(0), ν (0) = ξ + K S0 η, ν 0 where ξ, η define the variation z 0 (u, ɛ). Proof. Since N(ɛ), ν(ɛ) = 0 we get N(0), ν (0) = N (0), ν(0). From Lemma A.3.5. and ν 0 = ν γ 0 x,γ it follows N (0), ν 0 = g αβ N(0), ζ,β x,α, ν γ 0 x,γ = g αβ g αγ N(0), (ξn(0) + η k x,k ),β ν γ 0 = N(0), (ξn(0) + η k x,k ),β ν β 0 = N(0), ξ,β N(0) + ξn(0),β + η k,β x,k + η k x,kβ ν β 0 = ξ,β ν β 0 N(0), ηk x kβ ν β 0 = ξ + N(0), ν 0 η. ν 0 ν 0

80 80 CHAPTER 3. CAPILLARY INTERFACES The previous equation holds since we get from N(0), x,k = 0 that thus, since η k = ην k 0, N(0), x,kβ = N(0),β, x,k. N(0), x,kβ η k ν β 0 = η N(0),β, x,k ν k 0ν β 0 = η N(0) ν 0, ν 0 The assertion of the lemma follows from the formula at S 0. N(0), ν 0 = K S0 ν 0 For the convenience of the reader we will give the proof of the previous formula. Lemma A.3.7. at S 0. N(0), ν 0 = K S0 ν 0 Proof. Let P S 0, and consider the intersection of the plane Π governed by N(0) and ν 0 at P with the surface S 0. Denote by X(s) = x(u(s)) the representation of this curve, parametrized by its arc length, and let N(s) = N(u(s)) be the normal on S 0 at X(s). The curvature K(s) of the plane curve X(s) is defined by K(s) = X (s). Since X (s), X (s) = 1 it follows that X (s) is perpendicular on X (s). Thus X (s) = K(s)N Π (s), where N Π (s) denotes the normal at X(s) in the plane Π. Since N Π (s 0 ) = N(s 0 ) = N 0 at P = X(s 0 ) it follows that at S 0. Since at S 0 X (s 0 ) = K S0 N 0 (3.47) we get ν(s) = X (s) = x,α (u(s))(u α (s)) = x,α (u(s))ν α (s) N (s) = N,α (u(s))(u α (s)) = N,α (u(s))ν α (s) = N(s) ν(s) ν, N(s) ν(s) = X (s), N (s).

81 3.5. APPENDIX 81 Hence, since X (s), N(s) = 0, we find that X (s 0 ), N(s 0 ) = ν 0, N(0) ν 0. The assertion of the lemma follows from formula (3.47). Corollary. If S 0 is rotationally symmetric with respect to the x 3 -axis, then on S 0. N(0) ν 0 = K S0 ν 0 Proof. The vector N(0)/ ν 0 is a linear combination of ν 0 and N(0). Since N(0)/ ν 0, N(0) = 0, the formula follows from the lemma above. Suppose that S 0 is a sufficiently regular surface defined bx x : U R 3, where U R 2 is a fixed sufficiently regular parameter domain. Let S(ɛ) be a perturbed surface given by z(u, ɛ) = x(u) + ɛ[ξ(u)n S0 (u) + η α (u)x,α (u)] + O(ɛ 2 ), where ξ, η α and the remainder are sufficiently regular. The normal N S0 is defined by N(u) = (x,1 x,2 )/ x,1 x,2. Let H(u, ɛ) be the mean curvature of the perturbed surface S(ɛ) at z(u, ɛ). If there is no tangent component η α x,α, then we have the well known formula 2H ɛ (u, 0) = ξ +2(2H 2 K)ξ, see [7], p. 186, for example. Here H = H(u) and K = K(u) denote the mean curvature and the Gauss curvature, resp., of the given surface S 0 at x(u). In the general case, a formula for 2H ɛ (u, 0) does not seem to appear in the literature. Lemma A H ɛ (u, 0) = ξ + 2(2H 2 K)ξ + 2H,α (u, 0)η α. Proof. The following proof was suggested through an unidentified referee of the second proof below. Since H ɛ (u, 0) is a linear operator, we have H ɛ (u, 0) = H 1,ɛ (u, 0) + H 2,ɛ (u, 0), where H 1 (u, ɛ) is the mean curvature at z 1 (u, ɛ) of the surface defined by z 1 (u, ɛ) = x(u) + ɛξ(u)n S0 (u) + O(ɛ 2 ),

82 82 CHAPTER 3. CAPILLARY INTERFACES and H 2 (u, ɛ) is the mean curvature at z 2 (u, ɛ) of the surface defined by z 2 (u, ɛ) = x(u) + ɛη α (u)x,α (u). Since the normal variation formula is known, see [7], p. 186, for example, we have 2H 1,ɛ (u, 0) = ξ + 2(2H 2 K)ξ. Fix u in the open parameter set U. Since x(u+ɛη(u)) = x(u)+ɛη α (u)x,α (u)+ O(ɛ 2 ) and H(u + ɛη(u)) is the mean curvature of S 0 at x(u + ɛη(u)), we get H(u + ɛη(u)) = H 2 (u, ɛ) + O(ɛ 2 ). Consequently H 2,ɛ (u, 0) = H,α (u)η α (u). Another proof is by straightforward calculation based on the formula 2H(u, ɛ) = b αβ (u, ɛ)g αβ (u, ɛ) and using the Gauss, Weingarten and Codazzi equations. Moreover this proof includes a proof of the normal variation formula. Concerning formulas used in the following see [18, 19], Part I. Let N(ɛ) be the normal associated to the surface S(ɛ), and set ζ = λn(0) + η α x,α. Here and in the following N(0), g αβ (0) etc. denote vectors or functions which are related to the given surface S(0). Since 2H(ɛ) = b αβ (ɛ)g αβ (ɛ) we have 2H (0) = b αβ (0)gαβ (0) + b αβ (0)(g αβ ) (0). (3.48) In the following we set b αβ = b αβ (0), g αβ = g αβ (0), g αβ = g αβ (0), N = N(0), H = H(0) and K = K(0). We have z,α (ɛ) = x,α + ɛ[η β,αx,β + η β x,αβ + λ,α N + λn,α ] + O(ɛ 2 ). Together with the Gauss equations and the Weingarten equations x,αβ = Γ γ αβ x,γ + b αβ N (3.49) N,α = b β αx,β, (3.50)

83 3.5. APPENDIX 83 where it follows b β α := b αγ g βγ, (3.51) z,α = x,α + ɛ[ξ γ αx,γ + ν α N] + O(ɛ 2 ), (3.52) with ξα γ : = η,α γ + Γ γ αβ ηβ b αβ g βγ λ, (3.53) ν α : = b αβ η β + λ,α. (3.54) Next we will show that (g αβ ) (0) = ξγg β αγ ξδ α gδβ. (3.55) With (3.52) we obtain g αβ (0) = z,α(0), x,β + x,α, z,β (0) = ξαx γ,γ + ν α N, x,β + x,α, ξ γ β x,γ + ν β N. Thus From it follows that g αβ (0) = ξγ αg γβ + ξ γ β g αγ. (3.56) (g αβ (ɛ)) 1 = (g αβ (ɛ)) (g αβ ) (0) = g αγ g γδ (0)gδβ. Then, see (3.56), Next we prove that (g αβ ) (0) = g αγ [ξ τ γg τδ + ξ τ δ g γτ]g δβ = ξ β γg αγ ξ α δ gδβ. b αβ (0) = (gγτ ν τ ),α g γβ + g γτ ν τ Γ ρ γαg ρβ + b ρ αξ τ β g ρτ. (3.57) Since we have b αβ (ɛ) = N,α (ɛ), z,β (ɛ), b αβ (0) = N,α(0), x,β N,α, z,β (0).

84 84 CHAPTER 3. CAPILLARY INTERFACES From Lemma we have Then N (0) = g αβ ν β x,α. N,α(0), x,β = (g γτ ν τ x,γ ),α, x,β = (g γτ ν τ ),α g γβ g γτ ν τ x,γα, x,β = (g γτ ν τ ),α g γβ g γτ ν τ Γ ρ γαg ρβ, where we have used Gauss equations again. Together with N,α, z,β (0) = bρ αx,ρ, ξ τ β x,τ + ν β N = b ρ αξ τ β g ρτ we obtain formula (3.57) for b αβ (0). Combining this formula with formula (3.55), we find, see (3.48), (3.51) and definitions (3.53), (3.54) of ξα γ and ν α, that where with 2H (0) = (g ατ ν τ ),α + g γτ ν τ Γ α γα + b αγ ξ γ β gαβ b αβ (ξ β γg αγ + ξ α δ gδβ ) = (g ατ ν τ ),α + g γτ ν τ Γ α γα b ατ ξ α δ gδτ = l 1 (λ) + l 2 (η), l 1 (λ) : = (g ατ λ,τ ),α + g γτ Γ α γαλ,τ + b γ β bβ γλ l 2 (η) : = (g ατ b τκ η κ ),α + g γτ b τκ Γ α γαη κ b κτ g δτ η κ,δ b ατg δτ Γ α δκ ηκ = h κ η κ, Since, see [18], p. 55 and p. 19, h κ := (g ατ b τκ ),α + g γτ Γ α γαb τκ b αβ g δβ Γ α δκ. Γ α γα = 1 W W,γ and it follows 2H = b b 2 2, K = det (b β α), l 1 (λ) = (g ατ λ,τ ),α + 1 W W,γg γτ λ,τ + 2(2H 2 K)λ = λ + 2(2H 2 K)λ,

85 3.5. APPENDIX 85 see the definition of λ in [18], p. 43. This is the well known formula for the normal variation of 2H. Using formula (g ατ ),κ = g ρα Γρκ τ g ρτ Γρκ α, (3.58) see [18], p. 27, we obtain for the coefficients h κ in the definition of l 2 (η) h κ = ( g ρα Γρα τ g ρτ Γρα)b α τκ + g ατ b τκ,α + g γτ b τκ Γγα α b αβ g δβ Γδκ α = g ατ b τκ,α g ρα b τκ Γρα τ b αβ g δβ Γδκ α. From the Codazzi equations, see [18], p. 29, we get b τκ,α = b τα,κ + Γ ɛ ταb ɛκ Γ ɛ τκb ɛα, where we have used the symmetry properties b τκ = b κτ and Γ ɛ τκ = Γ ɛ κτ. With formula (3.58) we find finally h κ = g ατ b ατ,κ + g ατ (Γταb ɛ ɛκ Γτκb ɛ ɛα ) g ρα b τκ Γρα τ b αβ g δβ Γδκ α = g ατ b ατ,κ g ατ b ɛα Γτκ ɛ b αβ g δβ Γδκ α = (g ατ b ατ ),κ b ατ (g ατ ),κ g ατ b ɛα Γτκ ɛ b αβ g δβ Γδκ α = (g ατ b ατ ),κ = 2H,κ. We assume that admissible comparison surfaces are defined through x(u) + ɛζ(u) + ɛ2 2 ζ(u) 2 r(u) + O(ɛ 3 ), where ζ(u) = ξ(u)n 0 (u) + η(u)t S0 (u), for all ɛ < ɛ 0, see Section 3.4 for the definition of r(u). We recall that this implies the boundary condition ξ(u)cos γ η(u)sin γ = 0 on U. Denote by N(ɛ), N(ɛ) the normals on S(ɛ) and Γ, resp., at S(ɛ). In the following lemma we have ν, ζ/ ζ = ±1 since ν = ±ζ/ ζ. If ζ = 0, then we define ν, ζ/ ζ by a fixed constant, say by 0. Lemma A.3.9. [ ] d N(ɛ), N(ɛ) dɛ ɛ=0 = ξ ν sinγ + K Sη sinγ K Γ ζ ν, ζ/ ζ sin γ.

86 86 CHAPTER 3. CAPILLARY INTERFACES Proof. From Lemma A.3.5 it follows N (0), N(0) = g αβ N(0), ζ,α x,α, N(0). Since we have N(0) = N(0)cos γ ν sinγ, x,α, N(0) = x,α, ν sin γ = x,α, x,β ν β sin γ = g ατ ν τ sinγ. Then on S 0, Next we show that N (0), N(0) = g αβ N(0), ζ,β g ατ ν τ sinγ = N(0), ζ,β ν β sin γ = ξ,β ν β sinγ + N(0), T,β η ν β sin γ = ξ ν sin γ N,β(0), T η ν β sinγ = ξ ν sin γ N, ν η sinγ ν = ξ ν sin γ + K Sη sin γ. N(0), N (0) = K Γ ζ ν, ζ/ ζ sin γ. A parameter representation of a neighbourhood on Γ of S 0 is z(ɛ, τ) = x(u(τ)) + ɛζ(u(τ)) + ɛ2 2 ζ(u) 2 r(u(τ)) + O(ɛ 3 ), where u(τ) is a parameter representation of U, and ɛ < ɛ 0. Set z,1 = z,ɛ, z,2 = z,τ, then N (0) = c α z,α (0, τ) + cn(0) since N(ɛ), N(ɛ) = 1. Consequently we have = c α z,α (0, τ), (3.59) N (0), z,β (0, τ) = c α g αβ (0, τ),

87 3.5. APPENDIX 87 where a bar indicates quantities which are related to Γ. Since N(ɛ), z,β (ɛ, τ) = 0, β = 1, 2, it follows N (0), z,β (0, τ) + N(0), z,βɛ (0, τ) = 0 Thus Then, see (3.59), c α = g αβ (0, τ) N(0), z,βɛ (0, τ). Combining this equation with we get finally N (0) = g αβ (0, τ) N(0), z,βɛ (0, τ) z,α (0, τ). z,α (0, τ), N(0) = z,α (0, τ), N(0)cos γ + ν sinγ = z,α (0, τ), ν sinγ = z,α (0, τ), z,ρ (0, τ) ν ρ sinγ = g αρ ν ρ sinγ, N (0), N(0) = N(0), z,βɛ (0, τ) ν β sin γ Here we have used that = N(0), ζ(u(τ)) 2 r(u(τ)) ν 1 sinγ N(0), ζ(u(τ)),τ ν 2 sinγ = N(0), ζ(u) 2 K Γ N(0) ν 1 sinγ = K Γ ζ 2 ν 1 sinγ = K Γ ζ ν, ζ/ ζ sin γ. ν = ± ζ ζ = ν1 ζ ν 1 z,1 (0, τ), which implies that ν 2 = 0. Corollary. If γ satisfies 0 < γ < π, then [ ] [ ( d ξ cos γ N(ɛ), N(ɛ) = dɛ ν + sin γ K S 1 sinγ K Γ ɛ=0 ) ] ξ sinγ.

88 88 CHAPTER 3. CAPILLARY INTERFACES Proof. We have on U and, see (3.3), ζ 2 = ξ 2 + η 2 η = cos γ sin γ ξ.

89 3.6. PROBLEMS Problems 1. Show that there is a correction term r(α, β; ɛ) such that S(ɛ) Ω s. 2. Show that N Γ, N S and T S are in a common plane, provided that the container wall is sufficiently regular. Hint: Denote by g the tangent at S in P S, then all vectors N Γ, N S and T S are perpendicular on g. 3. Show that det(x α, x β, N S ) = E 0 G 0 F 2 0. Hint: where Then det(x α, x β, N S ) = 3 N i D 1, i=1 x α x β = (D 1, D 2, D 3 ), N S = (N 1, N 2, N 3 ). 3 N i D 1 = i=1 Finally, use the Lagrange identity D D2 2 + D2 3 = x α x β. (a b) (c d) = (a c)(b d) (b c)(a d). 4. Let S be given by x 3 = u(x 1, x 2 ). Show that 2H = div Tu. 5. Suppose that S is a capillary surface over Ω R 2. Show that N S N Γ = ν Tu = cosγ, where Γ = Ω (, ). 6. Derive the final formula for the ascent of a liquid at a vertical wall, see Section Let Ω be the half plane Ω = {(x 1, x 2 ) : x 1 > 0}. Suppose that u C 2 (Ω) satisfies div Tu = λ, where λ is constant. Prove that λ = 0, i. e., u is a solution of the minimal surface equation. The same result holds if Ω is a sector in the plane. 8. Let u(x) = v(r) be a rotationally symetric surface with respect to the x 3 -axis, r = x x2 2. Show that ( ) div Tu = 1 rv (r). r 1 + v (r) 2

90 90 CHAPTER 3. CAPILLARY INTERFACES 9. See the problem of the ascent of liquid at two parallel vertical plates. Prove the asymptotic formula for the constant C. 10. See the problem of the ascent of liquid at two parallel vertical plates. Consider the case of different (constant) contact angles on each plate and express the the solution of the capillary problem through elliptic integrals. 11. See the problem of the ascent of liquid at two vertical walls. Find asymptotic formulas for the heights u 0, u 1 which are uniform in 0 γ π/ See the problem of the ascent of liquid at two vertical walls. Find asymptotic formulas in the case of different contact angles at each plate. 13. Show that there exists a zero gravity capillary interface between two coaxial cylinders for any given contact angles γ 1, γ 2 [0, π], see Section 3.3 Hint: Show that f(s) 1 and f = 1 if and only if γ 1, γ 2 = 0 or π and s = a or s = R. 14. See Figure 3.5. Find a lower bound V 0 such that there is no dry spot at the bottom or that S + is above of S, resp., if V > V Let 0 < α π. Set 16. Set Ω α = {(r, θ) : r > 0, 0 < θ π}, where (r, θ) are polar coordinates with center at the origin. and consider the linear boundary value problem u = 0 in Ω α, u/ ν = cos γ, where 0 γ < 2π. Show that there is a solution u C 2 (Ω α \ {0}). (Thus, there is a striking difference between the original problem (3.22), (3.23) and the linearized problem.) where D(ɛ) = z (ɛ), N(ɛ) W(ɛ), z(ɛ) = x(u) + ɛ[ξ(u)n S0 (u) + η α x,α (u)] + O(ɛ 2 ) and the remainder is differentiable. Show that D(0) = ξw(0).

91 3.6. PROBLEMS Consider the same family of surfaces z(ɛ) as in the previous exercise. Set W(ɛ) = E(ɛ)G(ɛ) F 2 (ɛ) det(g αβ (ɛ)). Show that W (0) = 2HW(0)ξ + (η α W(0)),α. Hint. Use Gauss and Weingarten equations, see [18], p Consider the family of surfaces of second order defined at the beginning of Section Find a formula for D (0). 19. Consider pure normal perturbations defined by z(ɛ) = x(u) + ɛξ(u)n(u). Recover the formula [ ] d 2 dɛ 2 S(ɛ) ( = 2K0 ξ 2 + ξ 2) da, ɛ=0 S 0 see [7], p. 184, from Lemma A.3.2. Hint. 2H ɛ (u, 0) = ξ + 2(2H(u, 0) K(u, 0))ξ D ɛ (u, 0) = 2H(u, 0)ξ 2 W(u, 0) N(u, 0), ν ɛ (u, 0) = ξ ν Consider pure normal perturbations defined by z(ɛ) = x(u) + ɛξ(u)n(u). Set D(ɛ) = z (ɛ), N(ɛ) W(ɛ). Show that D(ɛ) = det(z α, z β, z ɛ ) and, by using the following hint, D (0) = 2Hζ E 2 0 G 0 F0 2.

92 92 CHAPTER 3. CAPILLARY INTERFACES Hint: Consider (D(ɛ)) 2, where and use the formula (D(ɛ)) 2 = det D(ɛ) = det(z α, z β, z ɛ ), z α z α z α z β z α z ɛ z β z α z β z β z β z ɛ z ɛ z α z ɛ z β z ɛ z ɛ 21. Discuss the conjecture at the end of Section

93 Chapter 4 Floating bodies 4.1 Governing energy The case of convex particles in two dimensions was considered in the article of E. Raphaël et al [57]. Partial results concerning equilibrium criteria were achieved in [41]. Floating criteria for a long cylinder were given in [22], and in three dimensions and for convex particles in [28]. We will derive equilibrium conditions by constructing a two parameter family of admissible comparison configurations. In contrast to the classical problem in capillarity we have to take into account translations and rotations of the particle additionally. Let Ω be the solid particle which can float and is N S Σ N Σ N Γ γ 2 Ω (particle) S γ 1 Γ g Ω l (liquid) Ω s (solid) Figure 4.1: A floating particle, notations away from the boundary of the container. Let Ω s be the fixed solid container. 93

94 94 CHAPTER 4. FLOATING BODIES Set Σ = Ω and Γ = Ω s. In this chapter we suppose that the boundaries Σ and Γ are sufficiently smooth. In Chapter 5 we consider particles with edges. Let E(S, Σ) = σ S σβ 1 W 1 (S) σβ 2 W 2 (S) (4.1) + F 1 (x) dx + F 2 (x) dx Ω l (S,Σ) be the energy of the problem, where σ surface tension (a positive constant), S area of the capillary surface S, Ω l (S, Σ) the liquid domain, β 1 (relative) adhesion coefficient between the fluid and the container wall, β 2 (relative) adhesion coefficient between the fluid and the particle, W 1 (S) wetted part of the container, W 2 (S) wetted part of the particle, F 1 = gρ 1 x 3, where ρ 1 is the density of the liquid, F 2 = gρ 2 x 3, where ρ 2 is the density of the particle. To simplify the presentation, we suppose that β 1, β 2 and ρ 1, ρ 2 are constants. For given volume C of the liquid we have additionally the side condition Ω Ω l (S, Σ) = C, (4.2) where Ω l (S, Σ) denotes the volume of the liquid domain. Let S 0, Σ 0 be a given configuration. Then we will derive necessary and sufficient conditions such that this configuration defines a minimizer of the associated energy functional subject to a given family of comparison configurations. Assume x : u U R 3, u = (u 1, u 2 ) or u = (α, β), defines the regular surface S 0, where U R 2 is a double connected domain such that x(u) Γ if u 1 U, x(u) Σ if u 2 U, and x(u) R 3 \ {Ω s Ω} if u U. Let y : V R 3, v = (v 1, v 2 ), be a regular parameter representation of Σ, the boundary of the particle, where V R 2. We choose the parameters such that the normal N S = x u 1 x u 2 x u 1 x u 2 is directed out of the liquid, and that the normal N Σ = y v 1 y v 2 y v 1 y v 2 at Ω is directed into the particle. Moreover we assume the normal N Γ at the container wall is directed out of the solid material Ω s of the container, see Figure 4.1.

95 4.2. EQUILIBRIUM CONDITIONS 95 To simplify the presentation we do not consider particles where vapour is enclosed between the capillary surface and the particle. 4.2 Equilibrium conditions Following Finn [21], Chapter 1, we define for a given configuration S 0, Σ 0, where Σ 0 = Ω 0, a one parameter family of admissible comparison surfaces which are in general not yet volume preserving. Set z 0 (u, ɛ) = x(u) + ɛζ(u) + r(u, ɛ), ɛ < ɛ 0, the remainder r is continuously differentiable with respect to all arguments, such that r = O(ɛ) as ɛ 0, z 0 (u, ɛ) R 3 \ (Ω s Ω 0 ) if u U, z 0 (u, ɛ) Γ if u 1 U, z 0 (u, ɛ) Σ 0 if u 2 U, and ζ(u) = ξ(u)n S0 (u) + η(u)t S0 (u) is a given vector field. Here N S0 denotes the unit normal to S 0 pointed to the exterior of the liquid, and T S0 is a unit tangent field defined on closed strips U i,δ of i U of width δ, such that on S 0, ν 0 := T S0 is orthogonal to 1 S 0, 2 S 0, resp., and points to the exterior of the liquid, see Figure 4.2. We assume that ξ and η are sufficiently regular on U, supp η U 1,δ U 2,δ, and ξ 2 + η 2 1. Define on 1 S 0 = {x(u) : u 1 U} the angle γ 1 [0, π] δ 1 S 0 δ particle 2 S 0 ν 0 ν 0 liquid Figure 4.2: Normals ν 0, notations by cos γ 1 = N Γ N S0 and on 2 S 0 = {x(u) : u 2 U} the angle γ 2 [0, π] by cos γ 2 = N Σ0 N S0. We recall that N Σ0 is ditected into the particle. These angles depend on P S 0. Later we will see that these angles are constants. Since ζ, N Γ = 0 at S 0 it follows ξ(u)cos γ 1 η(u)sin γ 1 = 0, u 1 U, (4.3) ξ(u)cos γ 2 η(u)sin γ 2 = 0, u 2 U. (4.4)

96 96 CHAPTER 4. FLOATING BODIES Thus for unchanged Σ 0 the family z 0 (u, ɛ), where ζ satisfies equations (4.3) and (4.4), defines a family of admissible comparison configurations. Such a family exists provided the boundaries of the container and of the particle are sufficiently regular. The proof exploits the implicit function theorem, we omit the details. To incorporate rigid motions of the particle we define a family of admissible comparison configurations, in general not yet volume preserving, as follows. Let R(ω) = R 1 (ω 1 )R 2 (ω 3 )R 3 (ω 3 ) be a rotation matrix, where R i (ω i ) defines a rotation around the x i -axis with angle ω i. These rotations are given by R 1 (τ) = R 2 (τ) = R 3 (τ) = cos τ sinτ 0 sin τ cos τ cos τ 0 sin τ sinτ 0 cos τ cos τ sinτ 0 sinτ cos τ Let χ(u), u U, be a sufficiently regular function such that χ(u) = { 0 : u 1 U 1 : u 2 U. Set V 0 = {v V : y(v) W 2 (S 0 )}, where W 2 (S 0 ) is the wetted part of the unperturbed particle. For given ζ and α, a R 3 we define a two parameter family of admissible configurations, not yet volume preserving in general, (S (c), Σ(c)), c = (µ, ɛ), where the capillary surface S (c) = S (µ, ɛ) is given by z (u, c) = (1 χ(u))z 0 (u, ɛ) + χ(u)(r(µα)z 0 (u, ɛ) + µa), u U, and the wetted part Σ(c) of the moved particle is defined by z Σ (v, c) = R(µα)y(v) + µa, v V (ɛ), where V (ɛ) = {v V : y(v) W 2 (S (0, ɛ)}

97 4.2. EQUILIBRIUM CONDITIONS 97 is the parameter domain of Σ which defines the wetted part of the unchanged particle Ω 0 subject to S(0, ɛ). The moved particle is given by Ω(µ) = {y R 3 : y = R(µα)x + µa, x Ω 0 }. To get a volume preserving admissible family of configurations we replace z (u, c) through z (u, c, q) := z (u, c) + qξ 0 (u)n 0 (u), where q R, N 0 = N S0, supp ξ 0 U, supp ξ 0 supp χ = and S 0 ξ 0 (u) da = 1. Then (S (c, q), Σ(c)), where S (c, q) is defined by z (u, c, q), provides a family of admissible configurations, in general not yet volume preserving. Let X = {(ζ, α, a) C 1 (U) C 0 (U) R 3 R 3 : ζ satisfies (4.3) and (4.4)} and let X 0 be the subspace where w satisfies the side conditions S 0 ξ da = 0 (4.5) and χ(u) N S0 (u), (α i R αi (0))x(u) + a da (4.6) S 0 + N Σ0 (v), (α i R αi (0))y(v) + a da = 0. W 2 (S 0 ) From Lemma A.4.1 of the appendix to this chapter we have that for given w X 0 there exists a regular function q = q(c) such that the family of admissible configurations given by (S(c), Σ(c)), where S(c) is defined through z(u, c) = z (u, c, q(c)), is volume preserving for all c R 2, c < c 0. The function q(c) satisfies q(0, 0) = 0, q µ (0, 0) = 0 and q ɛ (0, 0) = 0. In the following we assume for ζ and the remainder r(u, ɛ) in the definition of z 0 (u, ɛ) that all integrals below exist. Suppose that for fixed volume C

98 98 CHAPTER 4. FLOATING BODIES Ω(µ) S * * ( µ,ε, q) liquid S * * (µ,ε, 0) Ω(0) S * * ( µ, 0,0) S * * (0,ε,0) liquid S * * (0,0,0) S * * (0,ε, q) Figure 4.3: Volume preserving motion of the particle of the liquid the above family (S(c), Σ(c)) of volume preserving admissible configurations satisfies for all c, c < c 0, It follows E(S(c), Σ(c)) E(S(0), Σ(0)). [ c E(S(c), Σ(c))] c=0 = 0, (4.7) [ c Ω l (S(c), Σ(c) ] c=0 = 0. (4.8) Since the left hand side of (4.8) defines two bounded linear functionals on X, equation (4.8) defines the subspace X 0. Set L(S, Σ, λ) = E(S, Σ) + λ( Ω l (S, Σ) C). From a Lagrange multiplier rule, see Chapter 8, it follows that there exists a real λ 0 such that the family (S (c), Σ(c)), in general not volume preserving, where w X instead of w from the subspace X 0, satisfies [ c L(S (c), Σ(c), λ 0 )] c=0 = 0,

99 4.2. EQUILIBRIUM CONDITIONS 99 where L(S (c), Σ(c), λ 0 ) = σ S (c) σβ 1 W 1 (S (c)) σβ 2 W 2 (S (c)) + F 1 (x) dx + F 2 (x) dx Ω l (S (c),σ(c)) +λ 0 ( Ω l (S (c), Σ(c)) C). The argument c indicates that the function or vector in consideration is related to the perturbed surface S (c). Set W(c) = E(c)G(c) F 2 (c), where E, F, G are the coefficients of the first fundamental form of S (c) defined by z (u, c). To simplify the notation, we omit the star in the following Ω(µ) Σ(c) ν (c) N Γ N(c) ν (c) ν (c) S (c) ν (c) liquid Γ Figure 4.4: Normals in consideration of this section, and we write S(c), z(u, c) instead of S (c), z (u, c), resp. By ν(u, c), resp. ν(u, c), we denote the exterior normal to S(c) in S(c), resp., at the container wall Γ or at the boundary Ω(µ) of the particle at P S(c), see Figure 4.4. If P S(c), then the vectors N Γ, N(u, c), ν(u, c) and ν(u, c) are all in the same plane since S(c) is a curve on S(c) as well as on the container wall Γ or on Ω(µ). Here and in the following we set N(u, c) = N S(c) at u U. Let H 0 = H(u, 0), ν 0 = ν(u, 0), ν 0 = ν(τ, 0), where u U, N 0 = N(u, 0), which is N S0 at u U, and N Σ0 = N Σ (v,0). We recall that S 0 = S(0), where S 0, Σ 0 is the given configuration under consideration.

100 100 CHAPTER 4. FLOATING BODIES Equation L ɛ (S 0, Σ 0, λ 0 ) = 0 implies the well known equilibrium conditions of a liquid in a fixed container, see Finn [21], p. 10. Since L(S(c), Σ(c), λ 0 ) = σ S(c) σ 1 W 1 (S(c) σβ 2 W 2 (S(c)) (4.9) + F 1 (x) dx + F 2 (x) dx + λ 0 ( Ω l (S(c), Σ(c)) C), Ω l (S(c),Σ(c)) Ω(µ) we obtain from the corollaries to Lemma A.4.2 A.4.5 the formula [L ɛ (S(c), Σ(c), λ 0 )] c=0 = 2σ H 0 ξ da + σ ζ, ν 0 β 1 ν 0 ds S 0 1 S 0 + σ ζ, ν 0 β 2 ν 0 ds + F 1 (x(u))ξ da 2 S 0 S 0 + λ 0 S 0 ξ da. Equilibrium conditions follow from L ɛ (S 0, Σ 0, λ 0 ) = 0 for all ζ = ξn 0 +ηt S0, where T S0 = ν 0 on S 0, and ξ cos γ k η sinγ k = 0 on k S 0. Consider a subset of these ζ s where ζ = 0 on S 0, we get the equation on S 0. Then we have 2σH 0 + F 1 + λ 0 = 0 (4.10) k S 0 ζ, ν 0 β k ν 0 ds = 0 for all ζ defined above. This equation implies the boundary conditions on k S 0 since on k S 0 cos γ k = β k. ζ, ν 0 β k ν 0 = ξ N 0, ν 0 β k ν 0 + η ν 0, ν 0 β k ν 0 = β k ξ N 0, ν 0 + η(1 β k ν, ν 0 ) = ξβ k sinγ k + η(1 β k cos γ k ). Following Finn [21], p. 10, we set ξ = τ k sinγ k and η = τ k cos γ k, where τ k C( k U). Then ξ cos γ k η sin γ k = 0 is satisfied for all τ k. Then ζ, ν 0 β k ν 0 = τ k (cos γ k β k ). (4.11) In contrast to the classical capillary problem with a fixed container we get additional conditions from the fact that the particle can move in the

101 4.2. EQUILIBRIUM CONDITIONS 101 liquid. These conditions are given by equation L µ (S 0, Σ 0, λ 0 ) = 0. We have, see the corollaries to Lemma A.4.2 A.4.5, [L µ (S(c), Σ(c), λ 0 )] c=0 = 2σ H 0 (u) N 0 (u), z µ (u, 0, 0) da S 0 +σ ν 0 (u), z µ (u, 0, 0) ds + σ ν 0 (u), z µ (u, 0, 0) ds 1 S 0 2 S 0 + F 1 (x(u)) N 0 (u), z µ (u, 0, 0) da S 0 + F 1 (y(v)) N Σ0 (v), z Σ,µ (v,0) da Since W 2 (S 0 ) + F 2 (y), (α i R αi (0))y + a dy Ω ( 0 +λ 0 N 0 (u), z µ (u, 0, 0) da + S 0 W 2 (S 0 ) z Σ,µ (v,0) = (α i R αi (0))y(v) + a, z µ (u, 0, 0) = χ(u)((α i R αi (0))y(v) + a), N Σ0 (v), z Σ,µ (v,0) da where χ(u) = 1 on 2 U and χ(u) = 0 on 1 U, we find from L µ (S 0, Σ 0, λ 0 ) = 0 and by using equation (4.10) that σ ν 0 (u), (α i R αi (0))x(u) + a ds 2 S 0 + F 1 (y(v)) N Σ0 (v), (α i R αi (0))y(v) + a da for all α, a R 3. W 2 (S 0 ) + F 2 (y), (α i R αi (0))y + a dy Ω 0 +λ 0 N Σ0 (v), (α i R αi (0))y(v) + a da = 0 (4.12) W 2 (S 0 ) Let S(c), Σ(c) be the family of volume preserving admissible configurations defined above. Summarizing the previous formulas, we have shown Theorem 4.1. Suppose that the configuration S 0, Σ 0 defines a local minimum of E(S(c)), Σ(c)) under the side condition Ω l (S(c), Σ(c)) = C, C > 0 )

102 102 CHAPTER 4. FLOATING BODIES is a given constant. Then there exists a real constant λ 0 such that 2σH 0 + F 1 + λ 0 = 0 on S 0 (4.13) cos γ 1 = β 1 at 1 S 0 (4.14) cos γ 2 = β 2 at 2 S 0, (4.15) σ ν 0 (u) ds + 2 S 0 + F 1 (y(v))n Σ0 (v) da (4.16) W 2 (S 0 ) F 2 (y) dy + λ 0 N Σ0 (v) da = 0, Ω 0 W 2 (S 0 ) σ R αi (0)x(u), ν 0 (u) ds (4.17) 2 S 0 + F 1 (y(v)) R αi (0)y(v), N Σ0 (v) da + F 2 (y), R αi (0)y dy W 2 (S 0 ) Ω 0 + λ 0 R αi (0)y(v), N Σ0 (v) da = 0, W 2 (S 0 ) where i = 1, 2, 3. Remark. In general, if the body is not in an equilibrium, then the force F = (F 1, F 2, F 3 ) and the torques (moments of force) M k with respect to the x k -axis keep the body in an equilibrium if the following equations are satisfied: F k = σ ν 0 (u), e k ds + F 1 (y(v)) N Σ0 (v), e k da 2 S 0 W 2 (S 0 ) + F 2 (y), e k dy + λ 0 N Σ0 (v), e k da Ω 0 W 2 (S 0 ) and M k = σ R αk (0)x(u), ν 0 (u) ds 2 S 0 + F 1 (y(v)) R αk (0)y(v), N Σ0 (v) da + W 2 (S 0 ) Ω 0 F 2 (y), R αk (0)y dy + λ 0 W 2 (S 0 ) R αk (0)y(v), N Σ0 (v) da. Here we denote by e k the unit vectors of the standard basis in R 3.

103 4.2. EQUILIBRIUM CONDITIONS Restricted movements Here we consider some of possible cases of restricted motions of the body. Vertical movement. Suppose that the particle can move in the x 3 -direction only and no rotation is allowed, see Figure 4.5, then it follows from the above remark that the vertical force F 3 directed upwards keeps the particle in an equilibrium if (4.13) (4.15) and F 3 = σ ν 0 (u), e 3 ds + F 1 (y(v)) N Σ0 (v), e 3 da 2 S 0 W 2 (S 0 ) + F 2 (y), e 3 dy + λ 0 N Σ0 (v), e 3 da Ω 0 W 2 (S 0 ) are satisfied. If the body can rotate around the x 3 -axis additionally, then the torque M 3 vanishes, see the above remark where the torques are defined. x 3 F 3 g liquid Figure 4.5: Vertical movement of the particle Remark. The buoyancy B of the body is given by B = F 1 (y(v)) N Σ0 (v), e 3 da. W 2 (S 0 )

104 104 CHAPTER 4. FLOATING BODIES We recall that that N Σ0 denotes the interior normal at the boundary of the body. Horizontal movement. Assume the body can move in the x 1 -direction only and no rotation is allowed, see Figure 4.6. Then the horizontal force F 1 g x 1 F 1 liquid Figure 4.6: Horizontal movement of the particle directed in the positive x 1 -axis keeps the body in an equilibrium if equations (4.13) (4.15) and F 1 = σ ν 0 (u), e 1 ds + gρ 1 y 3 (v) N Σ0 (v), e 1 da 2 S 0 W 2 (S 0 ) + λ 0 N Σ0 (v), e 1 da W 2 (S 0 ) are satisfied, where y(v) = (y 1 (v), y 2 (v), y 3 (v)), v V, defines the boundary of the particle. Finn [23] discovered that parallel plates which are parallel to the gravity force can attract or repel each other depending on the contact angles on the plates. Here we consider a plate parallel to a fixed vertical wall, and we assume that the plate can move in the x 1 -direction only, see Figure 4.7. Then, see the above formulas, the force F 1 is given by F 1 = σ ν 0 (v), e 1 ds + gρ 1 y 3 (v) N Σ0 (v), e 1 da, 2 S 0 W 2 (S 0 ) where N Σ0 is the interior normal, where it exist, at the boundary of the plate, and F 1 is positive if the wall attracts the plate and negative if the

105 4.2. EQUILIBRIUM CONDITIONS 105 z 2 g wall general level u l γ l γ r u r S 0 S 0 x F x z 1 γ w d plate liquid Figure 4.7: Floating plate wall repels the plate. Consider the case that the plate is infinitely long in both y-directions, and denote by F1 L the force acting on a part of length L in the x 2 -direction. Then F1 L = Lσ(sinγ l sinγ r ) gρ 1L(u 2 l u2 r), where u l is the ascent of the liquid at the left part of the plate and u r is the ascent at the right plane, both measured from the general level of the liquid which is the line given by z = 0. We assume that the plate is sufficiently heigh and deep such that z 1 < u l, u r < z 2 is satisfied. Since, see Section 3.3.1, u 2 r = 2σgρ 1 (1 sinγ r ), we get ( Fx L = L σ(sin γ l 1) + gρ ) 1 2 u2 l, where u l = u l (d, γ w, γ l ). In particular, if γ w = γ l =: γ, then, see Section 3.3.2, u l = 2 cos γ + O(d) κd

106 106 CHAPTER 4. FLOATING BODIES as d 0. Consequently ( ) 2σ Fx L 2 = L gρ 1 d 2 cos2 γ + O(1) as d 0. Thus the plate is attracted if γ 0 and if the distance d to the wall is sufficiently small. Rotation around the x 3 -axis. Suppose that the particle can rotate around the x 3 -axis only and no translation is allowed, see Figure 4.8, then the particle is in an equlibrium if equations (4.13) (4.15) are satisfied and if the torque M 3 vanishes, i. e., σ R α3 (0)x(u), ν 0 (u) ds + gρ 1 y 3 (v) R α3 (0)y(v), N Σ0 (v) da 2 S 0 W 2 (S 0 ) +λ 0 R α3 (0)y(v), N Σ0 (v) da = 0, W 2 (S 0 ) where R α3 (0) = Remark. If the volume is infinite, then λ 0 = 0 in the above formula Explicit solutions Here we consider examples which lead to explicit results. Horizontal capillary surfaces Consider a floating ball in a cylindrical tube, not necessarily with circular cross section. Assume the liquid makes a contact angle of π/2 with the container wall. Let ρ 1 be the (constant) density of the liquid and ρ 2 the (constant) density of the ball. We suppose that 0 < ρ 2 < ρ 1. In general, for given γ 2 (0, π) the capillary surface is not a subset of a plane. On the other hand, we have Proposition 4.1. Suppose that the gravity g is positive and directed downwards in the negative x 3 -axis, see Figure 4.1. Then there exists a unique contact angle γ 2 such that the a horizontal capillary surface S 0 satisfies the equilibrium criteria of the theorem. Moreover we have 0 < γ 2 < π/2 if

107 4.2. EQUILIBRIUM CONDITIONS 107 liquid Figure 4.8: Rotation around the x 3 axis 2ρ 2 > ρ 1, γ 2 = π/2 if 2ρ 2 = ρ 1 and π/2 < γ 2 < π if 2ρ 2 < ρ 1. Every ball moved horizontally and stays away from the boundary of the container defines an equilibrium. Proof. Suppose that S 0 is a subset of the plane x 3 = z 0, see Figure 4.9 for notations, and makes a contact angle γ with the ball. From the equilibrium condition (4.13) we get λ 0 = g 1 ρ 1 z 0. Since 2 S 0 ν 0 ds = 0, equilibrium condition (4.16) reduces to gρ 1 W 2 (S 0 ) y 3 (v)n Σ0 (v) da+gρ 2 Ω dx+λ 0 W 2 (S 0 ) N Σ0 (v) da = 0. This equation is satisfied if γ is a zero of f(γ) defined by f(γ) = ρ 1 ( 1 3 cos3 γ cos γ 2 3 ) ρ 2. The assertions of the proposition follow since f(0) = 4/3 (ρ 1 ρ 2 ), f(π) =

108 108 CHAPTER 4. FLOATING BODIES g Ω 0 γ N Σ 0 ν 0 z 0 liquid Figure 4.9: Horizontal capillary surface 4/3 ρ 2, f(π/2) = 0 : 2ρ 2 = ρ 1 > 0 : 2ρ 2 > 2ρ 1. < 0 : 2ρ 2 < 2ρ 1 Since f (γ) = ρ 1 sin 3 γ we get uniqueness of the contact angle. The equilibrium criterion (4.17) follows from the symmetry properties of the particle or by formal calculations. In fact, in the case of a ball as a particle the equilibrium condition (4.17) is superfluous because we can set α = 0 in the definition of the admissible configurations. Zero gravity Consider a floating ball in the zero gravity case. Then for every given contact angle γ 2 (0, π) there is a horizontal capillary surface since the equilibrium condition (4.13) shows that λ 0 = 0, and condition (4.16) is satisfied. This behaviour for floating balls under zero gravity was firstly discovered by Finn [26]. An important consequence is a further discovery of Finn that Young s surface tension diagram fails for floating particles, see [27]. Suppose the particle contains a rotationally part and that all angles between this part and planes which intersect the axis of rotation perpendicular are different from each other. Assume the contact angle between the particle and the liquid is in this set of angles, then there is a horizontal surface in equilibrium, see Figure One expects that one can keep such a particle vertically in zero gravity provided that a tilted particle from the axis of symmetry of the symmetric part of the particle makes a non constant contact

109 4.3. STABILITY 109 γ 2 liquid Figure 4.10: Horizontal capillary surface, zero gravity angle with planes perpendicular to the axis of symmetry. For example, this assumption is satisfied for a rotationally ellipsoid different from a ball. Floating ball and rotationally symmetric capillary surfaces Here we consider a floating ball in a cylindrical tube with a disk as cross section where the center of the ball is on the axis of the cylinder and the gravity force is directed downwards, see Figure We seek rotationally capillary surfaces which satisfy the equilibrium conditions (4.13) (4.16). We recall that in the case of a ball the equilibrium condition (4.17) is superfluous because we can set α = 0 in the definition of the admissible configurations. Kitchen sink experiments show that a ping-pong ball does not stay in the middle of the tube in contrast to the case where the capillary surface hangs on the interior edge of the tube and makes a contact angle larger than π/2, see Figure Stability Consider an admissible configuration (S 0, Σ 0 ) which satisfies the equilibrium conditions of Theorem 4.1, and let (S(c), Σ(c)) be the volume preserving admissible family of configurations defined above. We recall that S(c) is given by z(u, c) = (1 χ(u))z 0 (u, ɛ) + χ(u)(r(µα)z 0 (u, ɛ) + µa) + q(c)ξ 0 (u)n 0 (u),

110 110 CHAPTER 4. FLOATING BODIES g Ω 0 γ 1 γ 2 N Σ 0 liquid Figure 4.11: Floating ball, rotationally symmetric capillary surface u U, and Σ(c) is defined through z Σ (v, c) = R(µα)y(v) + µa, v V (ɛ), where V (ɛ) = {v V : y(v) W 2 (S(0, ɛ))}. The function q(c) is defined in Lemma A.4.1. Since we need second derivatives with respect to ɛ and µ, we have to expand of z 0 (u, ɛ) up to second order. Let K Γ be the curvature at P S 0 of the plane curve defined by the intersection of Γ with the plane through P and the linear span N S0 (at P) and ν 0. This curvature is considered as nonnegative if the curve bends in the direction of N Γ. Extending an idea of Finn, we consider a class of admissible comparison surfaces S(ɛ) defined by z 0 (u, ɛ) = x(u) + ɛζ(u) + ɛ2 2 ζ(u) 2 r(u) + O(ɛ 3 ), where r(u) = K Γ (u)n Γ (u)ρ(u). Here K Γ (u) and N Γ (u) denote smooth extensions of K Γ and N Γ, which are defined on S 0, to U δ, and ρ(u) is a given smooth function which satisfies ρ(u) = 1 if u U and ρ(u) = 0 if dist (u, U) > δ. For example, N Γ (u) := N S0 (u)cos γ T S0 (u)sin γ is an extension of N Γ (u), u U to U δ. We can assume that the remainder O(ɛ 3 ) is sufficiently regular, which follows from an implicit function theorem, we omit the details.

111 4.3. STABILITY 111 g Ω 0 γ 1 γ 2 N Σ 0 liquid Figure 4.12: Floating ball, wetting barrier For the derivatives of z and z Σ, resp., we have z µ (u, 0) = χ(u)((α i R αi (0))x(u) + a) (4.18) z ɛ (u, 0) = ζ (4.19) z µµ (u, 0) = χ(u) ( α i α j R αi α j (0) ) x(u) + q µµ (0)ξ 0 (u)n 0 (u) (4.20) z µɛ (u, 0) = χ(u))((α i R αi (0))ζ(u)) + q µɛ (0)ξ 0 (u)n 0 (u) (4.21) z ɛɛ (u, 0) = ζ(u) 2 r(u) + q ɛɛ (0)ξ 0 (u)n 0 (u) (4.22) z Σ,µ (v,0) = (α i R αi (0))y(v) + a (4.23) z Σ,ɛ (v,0) = 0 (4.24) z Σ,µµ (v,0) = (α i α j R αi α j (0))y(v) (4.25) z Σ,µɛ (v,0) = 0 (4.26) z Σ,ɛɛ (v,0) = 0. (4.27) We recall that q(0) = 0 and c q(0) = 0. Formulas for the second derivatives of q(c) follow from the proof of Lemma A.4.1. A necessary condition of second order that the equilibrium configuration (S 0, Σ 0 ) defines a local minimum of the energy functional is [L ck c l (S(c), Σ(c), λ 0 )] c=0 c k c l 0 for every c R 2 and for all (ζ, α, a) X 0, where X 0 is the subset of X satisfying the side conditions (4.5) and (4.6), see above. To keep the overview in the calculations we consider a two parameter family with parameters µ and ɛ. Replacing µ by ɛ we get a one parameter

112 112 CHAPTER 4. FLOATING BODIES family of configurations S(ɛ, ɛ), Σ(ɛ, ɛ). Thus a necessary condition of second order that an equilibrium configuration (S 0, Σ 0 ) defines a local minimizer of the energy functional is { [ [ ] L µµ(s(c), Σ(c), λ 0 )]µ=ɛ + 2 L µɛ (S(c), Σ(c), λ 0 ) (4.28) µ=ɛ [ ] } + L ɛɛ (S(c), Σ(c), λ 0 ) 0 µ=ɛ for all (ξ, α, a) which satisfy the side condition ξ da + χ(u) N S0 (u), (α i R αi (0))x(u) + a da (4.29) S 0 S 0 + N Σ0 (v), (α i R αi (0))y(v) + a da = 0. W 2 (S 0 ) The previous formula is a consequence of the corollary to Lemma A.4.4 of the appendix to this chapter and of { d [ ] } Ω l ((S(µ, ɛ), Σ(µ, ɛ)) dɛ µ=ɛ = 0. ɛ=0 An extremal is often called stable by definition if the strict inequality is satisfied in inequality (4.28) for nonvanishing (ξ, α, a) satisfying the side condition (4.29). To answer the question whether a given extremal defines a strong minimizer of the associated energy functional requires the construction of an embedding family of configurations, see [67, 79, 52] for related problems. We do not address this open problem here. From formula (4.9) and Lemma A.4.2 Lemma A.4.5 we get L ɛ (S(c), Σ(c), λ 0 ) = 2σ H(u, c) N(u, c), z ɛ (u, c) W(u, c) du U +σ ν(u, c), z ɛ (u, c) ds(c) + σ ν(u, c), z ɛ (u, c) ds(c) 1 S(c) 2 S(c) σβ 1 Z 0,ɛ (τ, ɛ), ν(u, 0, ɛ) Z 0,τ (τ, ɛ) dτ 1 U σβ 2 Z 0,ɛ (τ, ɛ), ν(u, 0, ɛ) Z 0,τ (τ, ɛ) dτ 2 U + F 1 (z(u, c) N(u, c), z ɛ (u, c) W(u, c) du U +λ 0 N(u, c), z ɛ (u, c) W(u, c) du, (4.30) U ɛ=0

113 4.3. STABILITY 113 where ds(c) = Z τ (τ, c) dτ, Z(τ, c) = z(u(τ), c) and Z 0 (τ, t) = z 0 (u(τ), t). Set D(u, c) = N(u, c), z ɛ (u, c) W(u, c), then L ɛɛ (S 0, Σ 0, λ 0 ) = 2σ H ɛ (u, 0)D(u, 0) du U 2σ H 0 (u)d ɛ (u, 0) du U [ ] +σ 1 U ɛ ν(u, c), z ɛ(u, c) Z τ (τ, 0) dτ c=0 [ ] +σ ν(u, 0), z ɛ (u, 0) 1 U ɛ Z τ(τ, c) dτ c=0 [ ] +σ 2 U ɛ ν(u, c), z ɛ(u, c) Z τ (τ, 0) dτ c=0 [ ] +σ ν(u, 0), z ɛ (u, 0) 2 U ɛ Z τ(τ, c) dτ c=0 [ ] σβ 1 1 U ɛ ν(u, 0, ɛ), Z 0,ɛ(τ, ɛ) Z 0,τ (τ, ɛ) dτ ɛ=0 [ ] σβ 1 ν(u, 0, 0), Z 0,ɛ (τ, 0) 1 U ɛ Z 0,τ(τ, ɛ) dτ ɛ=0 [ ] σβ 2 2 U ɛ ν(u, 0, ɛ), Z 0,ɛ(τ, ɛ) Z 0,τ (τ, ɛ) dτ ɛ=0 [ ] σβ 2 ν(u, 0, 0), Z 0,ɛ (τ, 0) 2 U ɛ Z 0,τ(τ, ɛ) dτ ɛ=0 + F 1 (x(u)), z ɛ (u, 0) D(u, 0) du U +λ 0 F 1 (x(u))d ɛ (u, 0) du. U From the differential equation 2σH 0 (u) + F 1 (x(u)) + λ 0 = 0 on S 0 and formula, see Lemma A.3.8 of the appendix to Chapter 3, 2H ɛ (u, 0) = ξ(u) + 2(2H 2 0(u) K 0 (u))ξ(u) + 2H 0,α (u)η α (u), where H 0, K 0 are the mean and Gauss curvature of S 0, resp., ξ and η α are defined through ζ = ξn 0 + η α x α, η α x α = ην 0 on S 0, and ξ, η satisfy

114 114 CHAPTER 4. FLOATING BODIES equations (4.3), (4.4) on U 1, U 2, resp., we get ( L ɛɛ (S 0, Σ 0, λ 0 ) = σ ξ + 2(2H 2 0 K 0 )ξ ) ξ da S 0 F 1 + ξ 2 da S 0 N 0 [ ] +σ ɛ ( ν(u, c), z ɛ(u, c) β 1 ν(u, 0, ɛ), Z 0,ɛ (τ, ɛ) ) 1 U Z 0,τ (τ, 0) dτ [ ] +σ 2 U ɛ ( ν(u, c), z ɛ(u, c) β 2 ν(u, 0, ɛ), Z 0,ɛ (τ, ɛ) ) Z 0,τ (τ, 0) dτ +σ ( ν(u, 0), z ɛ (u, 0) β 1 ν(u, 0), Z 0,ɛ (τ, 0) ) 1 U [ ] ɛ Z 0,τ(τ, ɛ) ɛ=0 +σ ( ν(u, 0), z ɛ (u, 0) β 2 ν(u, 0), Z 0,ɛ (τ, 0) ) 2 U [ ] ɛ Z 0,τ(τ, ɛ) ɛ=0 Here we have used that D(u, 0) = ξw(u, 0), Z τ (τ, 0) = Z 0,τ (τ, 0), and the formula z ɛ (u, 0) = z 0,ɛ (u, 0) = ζ = ξn 0 + η α x α, 2σH 0,α (u) + F 1 (x(u)), x α (u) = 0 on S 0. Next we show that the integrands of the last two terms in the previous formula vanish. We have for k = 1, 2 that dτ dτ. ( ν(u, 0), z ɛ (u, 0) β k ν(u, 0), Z 0,ɛ (τ, 0) )) = ν 0 β k ν, ζ since z ɛ (u, 0) = ζ and Z 0,ɛ (τ, 0) = ζ on S. At 1 S 0 the vectors ν 0, ν 0, N Γ and at 2 S 0 the vectors ν 0, ν 0, N Σ0 are in a common plane, resp., and ν 0 and N Γ, N Σ0, resp., are orthogonal, we have at k S 0 with scalar functions a k, b k the formula ν 0 = a k ν 0 + b k N Γk, where Γ 1 = Γ and Γ 2 = Σ 0. Using the boundary conditions (4.14), (4.15) it follows c=0 c=0 ν 0 ν 0 cos γ k = N Γk sinγ k. (4.31)

115 4.3. STABILITY 115 Thus the brackets vanish since ζ N Γk at S 0. We recall that N Γ2 = N Σ0, where N Σ0 is the interior normal at the boundary of the particle. For the brackets in the integrands of the remaining boundary integrals we get from formula (4.31) above that [ ] ɛ ( ν(u, c), z ɛ(u, c) β k ν(u, 0, ɛ), Z 0,ɛ (τ, ɛ) ) c=0 = ν 0, z ɛɛ (u, 0) β k ν 0, Z 0,ɛɛ (τ, 0) + ν ɛ, z ɛ (u, 0) β k ν ɛ (u, 0), ζ = K Γk ζ 2 N Γk, ν 0 β k ν 0 + ζ, ν ɛ (u, 0) β k ν ɛ (u, 0) = K Γk ζ 2 sinγ k + ζ, ν ɛ (u, 0) β k ν ɛ (u, 0). Finally we find that ζ, ν ɛ (u, 0) β k ν ɛ (u, 0) = ξn 0 + ην 0, ν ɛ (u, 0) β k ζ,ν ɛ (u, 0) = ξ N 0, ν ɛ (u, 0) ( ) ξ = ξ + K S0 η. ν 0 Here we have used that ν ɛ (u, 0, 0), ν(u, 0, 0 = 0, ζ = aν 0 with a scalar function a, and the formula, see the Lemma A.3.6 of the appendix to Chapter 3, N 0, ν ɛ (u, 0) = ξ ν 0 + K S0 η, where K S0 is the curvature at P S 0 of the plane curve defined through the intersection of S 0 and the plane through P which is spanned by N S0 at P and ν 0 := ν(0) at P. The curvature K S0 is considered as nonnegative if the curve bends in the direction of N S0 at P. Summarizing, we obtain after integration by parts, see [18, 19], p. 45, p. 44, resp., Lemma 4.1. ( L ɛɛ (S 0, Σ 0, λ 0 ) = σ ξ 2 2(2H0 2 K 0 )ξ 2) da S 0 F 1 + ξ 2 da S 0 N 0 ( +σ ξηks0 ζ 2 ) K Γ sinγ 1 ds 1 S 0 ( +σ ξηks0 ζ 2 ) K Σ0 sinγ 2 ds 2 S 0

116 116 CHAPTER 4. FLOATING BODIES Corollary. Suppose that γ k are different from 0 or π, then ( L ɛɛ (S 0, Σ 0, λ 0 ) = σ ξ 2 2(2H0 2 K 0 )ξ 2) da S 0 F 1 + ξ 2 da S 0 N 0 ( cos γ1 +σ +σ 1 S 0 2 S 0 K S0 1 ) K Γ ξ 2 ds sin γ 1 sinγ 1 ( cos γ2 K S0 1 ) K Σ0 ξ 2 ds. sin γ 2 sinγ 2 Proof. The formula follows since ζ 2 = ξn 0 + ην 0, ξn 0 + ην 0 = ξ 2 + η 2 on S, and ξ cos γ k η sinγ k = 0 on k S 0. Next we will calculate L µµ (S 0, Σ 0, λ 0 ). From the formula (4.9) for L and

117 4.3. STABILITY 117 Lemma A.4.2 A.4.5 we see that L µ (S(c), Σ(c), λ 0 ) = 2σ H(u, c) N(u, c), z µ (u, c) W(u, c) du U +σ ν(u, c), z µ (u, c) Z τ (τ, c) dτ 2 U ɛ [ ] + F 1 (z(u, µ, t)) N(u, µ, t), z t (u, µ, t) W(u, µ, t) dudt U 0 µ + F 1 (z(u, µ,0)) N(u, µ,0), z µ (u, µ,0) W(u, µ,0) du U + F 1 (z Σ (v, µ)) N Σ (v, µ), z Σ,µ (v, µ) W Σ (v, µ) dv V 0 + F 2 (R(µα)y + µa), (α i R αi (µα))y + a dy Ω ( 0 ɛ [ ] +λ 0 N(u, µ, t), z t (u, µ, t) W(u, µ, t) dudt U 0 µ + N(u, µ,0), z µ (u, µ,0) W(u, µ,0) du + U N Σ (v, µ), z Σ,µ (v, µ) W Σ (v, µ) dv V 0 ). Here we have used that z µ (u, c) = 0 on 1 U. We recall that V 0 = {v V : y(v) W 2 (S 0 )}. Set Q(u, c) = N(u, c), z µ (u, c) W(u, c). Using the differential equation (4.13), we get L µµ (S 0, Σ 0, λ 0 ) = 2σ H µ (u, 0)Q(u, 0) du U [ ] +σ 2 U µ ( ν(u, c), z µ(u, c) Z τ (τ, c) ) dτ c=0 + F 1 (x(u)), z µ (u, 0) Q(u, 0) du U [ ] d + V 0 dµ (F 1(z Σ (v, µ)) N Σ (v, µ), z Σ,µ (v, µ) W Σ (v, µ)) [ ] d + Ω 0 dµ F 2(R(µα)y + µa), (α i R αi (µα))y + a dy µ=0 ( [ ] ) d +λ 0 dµ ( N Σ(v, µ), z Σ,µ (v, µ) W Σ (v, µ)) dv. V 0 µ=0 µ=0 dv

118 118 CHAPTER 4. FLOATING BODIES The surface S(µ,0) is defined through where Then z(u, µ,0) = x(u) + µχ((α i R αi (0))x(u) + a) + O(µ 2 ) = x(u) + µχ(ξ 1 N 0 + η α 1 x,α ) + O(µ 2 ), ξ 1 = (α i R αi (0))x(u) + a, N 0 (4.32) η α 1 = g αβ (α i R αi (0))x(u) + a, x,β. (4.33) 2H µ (u, 0) = (χξ 1 ) + 2(2H 2 0 K 0 )(χξ 1 ) + 2H 0,α (u, 0)(χη α 1 ) (4.34) and, since Q(u, 0) = N 0, z µ (u, 0) W(u, 0) = χ N 0, α i R αi (0))x(u) + a W(u, 0) = χ N 0, ξ 1 N 0 + η α x,α W(u, 0) = χξ 1 W(u, 0), we get 2σ H µ (u, 0)Q(u, 0) du (4.35) U ( = σ (χξ1 ) + 2(2H0 2 K 0 )(χξ 1 ) + 2H 0,α (u, 0)(χη1 α ) ) (χξ 1 )da. S 0 Since Σ(c) is defined by a rigid motion of Σ(0, ɛ), we have Z τ (τ, c) = Z 0,τ (τ, ɛ) W Σ (v, µ) = W Σ (v,0) N Σ (v, µ) = R(µα)N Σ (v,0) ν(u, µ, ɛ) = R(µα)ν(u, 0, ɛ). Using formulas (4.18), (4.19), we find for the integrand of the second integral of the above formula for L µµ that [ ] µ ( ν(u, c), z µ(u, c) Z τ (τ, c) ) c=0 [ ] = µ ν(u, c), z µ(u, c) Z 0,τ (τ, 0) c=0 = ( ν 0, z µµ (u, 0) + ν µ (u, 0), z µ (u, 0) ) Z 0,τ (τ, 0) = ( ν 0, (α i α j R αi α j (0))x(u) + (α i R αi (0))ν 0, (α i R αi (0))x(u) + a ) Z 0,τ (τ, 0). (4.36)

119 4.3. STABILITY 119 We recall that ds = Z 0,τ (τ, 0) dτ on 2 S. Concerning the third integral we have F 1 (x(u)), z µ (u, 0) = F 1 (x(u)), ξ 1 N 0 + η α 1 x,α From the differential equation on S 0 we find that Then = ξ 1 F 1 N 0 + F 1 (x(u)), η α 1 x,α. (4.37) 2σH 0 (u) + F 1 (x(u)) + λ 0 = 0 2σH 0,α (u) + F 1 (x(u)), x,α = 0. 2σH 0,α (u)(χη α 1 ) + F 1 (x(u)), χη α 1 x,α = 0. (4.38) For the integrand of the forth integral of the formula for L µµ we get [ ] d dµ (F 1(z Σ (v, µ)) N Σ (v, µ), z Σ,µ (v, µ) W Σ (v, µ)) µ=0 [ ] d = dµ (F 1(z Σ (v, µ)) N Σ (v, µ), z Σ,µ (v, µ) ) W Σ (v,0) µ=0 = F 1 (z Σ (c,0)), z Σ,µ (v,0) N Σ (v,0), z Σ,µ (v,0) W Σ (v,0) [ ] d +F 1 (y(v)) dµ N Σ(v, µ), z Σ,µ (v, µ) W Σ (v,0) µ=0 = F 1 (y(v)), (α i R αi (0))y(v) + a N Σ0, (α i R αi (0))y(v) + a W Σ (v,0) +F 1 (y(v)) ( (α i R αi (0))N Σ0, (α i R αi (0))y(v) + a + N Σ0, (α i α j R αi,α j (0))y(v) ) W Σ (v,0). (4.39) Finally we have [ ] d dµ F 2(R(µα)y + µa), (α i R αi (µα))y + a µ=0 = 2 F 2 (y)((α i R αi (0))y + a),(α i R αi (0))y + a + F 2 (y), (α i α j R αi,α j (0))y (4.40) Combining the above equations (4.35) (4.40) we arrive at, after integration by parts,

120 120 CHAPTER 4. FLOATING BODIES Lemma 4.2. L µµ (S 0, Σ 0, λ 0 ) ( = σ (χξ1 ) 2 2(2H0 2 K 0 )(χξ 1 ) 2) da S 0 F 1 + (χξ 1 ) 2 ξ 1 da σ ξ 1 ds S 0 N 0 2 S 0 ν 0 ( +σ ν0, (α i α j R αi α j (0))x(u) 2 S 0 + (α i R αi (0))ν 0, (α i R αi (0))x(u) + a ) ds ( + F1 (y(v)), (α i R αi (0))y(v) + a N Σ0, (α i R αi (0))y(v) + a W 2 (S 0 ) +F 1 (y(v)) ( (α i R αi (0))N Σ0, (α i R αi (0))y(v) + a + N Σ0, (α i α j R αi α j (0))y(v) )) da ( + 2 F 2 (y)((α i R αi (0))y + a), (α i R αi (0))y + a Ω 0 F 2 (y), (α i α j R αi α j (0))y ) dy +λ 0 ( W 2 (S 0 ) ( (αi R αi (0))N Σ0, (α i R αi (0))y(v) + a + N Σ0, (α i α j R αi α j (0))y(v) )) ) da, where ξ 1 = (α i R αi (0)x(u) + a, N 0 and ξ 1 ν 0 = (α i R αi (0)ν 0, N 0 + (α i R αi (0)x(u) + a, N 0 ν 0. Concerning the previous formula we recall that x/ ν 0 = ν α 0 x,α = ν 0. Next we calculate L ɛµ (S 0, Σ 0, λ 0 ). From formula (4.30) for L ɛ (S(c), Σ(c), λ 0 ) we

121 4.3. STABILITY 121 obtain (L ɛ ) µ (S 0, Σ 0, λ 0 ) = 2σ H µ (u, 0)D(u, 0) du 2σ H(u, 0)D µ (u, 0) du U U [ ] +σ 2 U µ z ɛ(u, c), ν(u, c) Z 0,τ (τ, 0) dτ c=0 [ ] + U µ F 1(z(u, c)) D(u, 0) du + F 1 (z(u, c))d µ (u, 0) du c=0 U +λ 0 D µ (u, 0) du. We recall that U D(u, c) = N(u, c), z ɛ (u, c) W(u, c). Using the differential equation (4.13) we have (L ɛ ) µ (S 0, Σ 0, λ 0 ) = 2σ H µ (u, 0)D(u, 0) du U [ ] +σ 2 U µ z ɛ(u, c), ν(u, c) Z 0,τ (τ, 0) dτ c=0 + F 1 (x(u)), z µ (u, 0) D(u, 0) du. From equation (4.34) for H µ (u, 0), U D(u, 0) = ξ 0 W(u, 0) ν(u, c) = R(µα)ν(u, 0, ɛ) and since on 2 S 0, see formulas (4.19), (4.18), (4.21), z ɛ (u, 0) = ζ = ξ 0 N 0 + η 0 ν 0 z µ (u, 0) = (α i R αi (0))x(u) + a = ξ 1 N 0 + η α 1 x,α z ɛµ (u, 0) = (α i R αi (0))ζ we find after integration by parts = (α i R αi (0))(ξ 0 N 0 + η 0 ν 0 ),

122 122 CHAPTER 4. FLOATING BODIES Lemma 4.3. (L ɛ ) µ (S 0, Σ 0, λ 0 ) ( = σ (χξ1 ), ξ 0 2(2H 2 ) 0 K 0 )(χξ 1 )ξ 0 da S 0 ξ 1 F 1 σ ξ 0 ds + (χξ 1 )ξ 0 da 2 S 0 ν 0 S 0 N 0 ( +σ ξ0 ( (α i R αi (0))N 0, ν 0 + N 0, (α i R αi (0))ν 0 ) 2 S 0 +2η 0 (α i R αi (0))ν 0, ν 0 ) ds, where ξ 0, η 0 satisfy ξ 0 cos γ 2 η 0 sinγ 2 = 0 on 2 U, ξ 1 = (α i R αi (0)x(u) + a, N 0 and ξ 1 ν 0 = (α i R αi (0)ν 0, N 0 + (α i R αi (0)x(u) + a, N 0 ν 0. Remark. The second variation formula of Lemma 4.1 was derived in Chapter 3 already since [ ] d 2 L ɛɛ (S 0, Σ 0, λ 0 ) = dɛ 2 L(S(ɛ), Σ 0, λ 0 ), ɛ=0 where S(ɛ) is the volume preserving admissible family of comparison surfaces defined in Chapter The floating ball In the case of a floating ball we can assume that α = 0. Suppose that the gravity is positive and is directed downwards in the negative x 3 -axis. Then F 1 = gρ 1 x 3, F 2 = gρ 2 y 3, and the equilibrium conditions of Theorem 4.1 reduce to equations (4.13) (4.16). Assume that S 0, Σ 0 defines an equilibrium configuration. Since ξ 1 = a, N 0 and ξ 1 ξ 0 ν 0 = a, N 0 ξ 0 ν 0 ξ 1 ξ 1 ν 0 = a, N 0 a, N 0, ν 0

123 4.3. STABILITY 123 we get for the second derivatives of L, here we assume that 0 < γ i < π, ( L ɛɛ (S 0, Σ 0, λ 0 ) = σ ξ0 2 2(2H0 2 K 0 )ξ 2 ) 0 da S 0 F 1 + ξ0 2 da S 0 N 0 ( cos γ1 +σ K S0 1 ) K Γ ξ0 2 ds 1 S 0 sinγ 1 sinγ 1 ( cos γ2 +σ K S0 1 ) K Σ0 ξ0 2 ds 2 S 0 sinγ 2 sinγ 2 ( L ɛµ (S 0, Σ 0, λ 0 ) = σ (χξ1 ), ξ 0 2(2H 2 ) 0 K 0 )(χξ 1 )ξ 0 da S 0 F 1 + (χξ 1 )ξ 0 da σ a, N 0 ξ 0 ds S 0 N 0 2 S 0 ν 0 ( L µµ (S 0, Σ 0, λ 0 ) = σ (χξ1 ) 2 2(2H0 2 K 0 )(χξ 1 ) 2) da S 0 F 1 + (χξ 1 ) 2 da σ a, N 0 a, N 0 ds S 0 N 0 2 S 0 ν 0 +gρ 1 a 3 N Σ0, a da, W 2 (S 0 ) where ξ 1 = a, N 0. We recall that an equilibrium configuration S 0, Σ 0 is said to be stable by definition if L ɛɛ + 2L µɛ + L µµ 0 for all (ξ 0, a) W 1,2 (S 0 ) R 3 satisfying the side condition ξ 0 da + χ N 0, a da + N Σ0, a da = 0. S 0 S 0 W 2 (S 0 ) Remark. Consider a floating ball in a cylindrical tube with constant circular cross section, and suppose that the gravity is positive and is directed downwards in the negative x 3 -axis, the direction of the cylinder axis. Assume that S 0, Σ 0 defines an equilibrium configuration such that S 0 is rotationally symmetric with respect to the x 3 -axis, see Figure Then ξ 1 ξ 0 ν 0 = ξ 0 a, ν 0 K S0 ξ 1 ξ 1 ν 0 = a, N 0 a, ν 0 K S0,

124 124 CHAPTER 4. FLOATING BODIES where K S0 is constant along S 0, see the corollary to Lemma A.3.7. Horizontal capillary surfaces Consider the equilibrium configurations of Proposition 4.1, where the capillary surface is horizontally, then ξ 1 = a 3, K S0 = 0, K Γ = 0 and K Σ0 = 1/R, where R is the radius of the ball. Consequently we have L ɛɛ + 2L µɛ + L µµ = σ ξ 0 2 da + gρ 1 ξ 2 σ 0 da + ξ0 2 ds S 0 S 0 R sinγ 2 2 S 0 +2σa 3 χ, ξ 0 da + 2gρ 1 a 3 χξ 0 da S 0 S 0 +σa 2 3 χ 2 da + gρ 1 a 2 3 χ 2 da + gρ 1 πr 2 a 2 3 sin 2 γ 2. S 0 S 0 From Schwarz inequality it follows that σ L ɛɛ + 2L µɛ + L µµ ξ0 2 ds + gρ 1 πr 2 a 2 3 sin 2 γ 2 R sin γ 2 2 S 0 for all (ξ 0, a) W 1,2 (S 0 ) R 3, and in particular for (ξ 0, a) satisfying the side condition (4.29) with α = 0. If a 3 = 0, then L ɛɛ + 2L µɛ + L µµ = L ɛɛ which is non negative and zero if and only if ξ 0. Thus we have shown Proposition 4.2. The equilibrium configurations of Proposition 4.1 are stable in the sense that L ɛɛ + 2L µɛ + L µµ 0, and equality holds if and only if ξ 0 = 0 and a 3 = Appendix Let X be the set of all w = (ζ, α, a) C 1 (U) C 0 (U) R 3 R 3 such that ζ satisfies the side conditions (4.3) and (4.4) on U, and let X 0 be the subset such that w satisfies the side conditions S 0 ξ da = 0 and χ(u) N S0 (u), (α i R αi (0))x(u) + a da S 0 + N Σ0 (v), (α i R αi (0))y(v) + a da = 0. W 2 (S 0 )

125 4.4. APPENDIX 125 Lemma A.4.1. For given w X 0 there exists a regular function q = q(c) such that the family of admissible configurations given by (S(c), Σ(c)), where S(c) is defined through z(u, c) = z (u, c, q(c)), is volume preserving for all c R 2, c < c 0. The function q(c) satisfies q(0, 0) = 0, q µ (0, 0) = 0 and q ɛ (0, 0) = 0. Proof. We have, see Figure 4.3, where Ω l (S (µ, ɛ, q), Σ(µ, ɛ)) = Ω 1 Ω 2 Ω 3 Ω 4 Ω l (S 0, Σ 0 ), Ω 1 = {z (u, µ, t,0) : u U, 0 < t < ɛ}, Ω 2 = {z (u, ρ, 0, 0) : u U, 0 < ρ < µ}, Ω 3 = {z Σ (v, ρ) : v V 0, 0 < ρ < µ}, Ω 4 = {z (u, µ, ɛ, τ) : u U, 0 < τ < q}. Here we assume for simplicity that S (µ, ɛ, q) and Σ(µ, ɛ) are above of S 0 and of Σ 0, resp. The final formula is valid for every S (µ, ɛ, q) and Σ(µ, ɛ) we omit the details. For the general case see a remark in [18], p. 81. We point out that the dashed surfaces in Figure 4.3 are identically since supp χ supp ξ 0 =. Set f(µ, ɛ, q) = Ω l (S (µ, ɛ, q), Σ(µ, ɛ)). From det z (u, µ, t,0) (u, t) det z (u, ρ, 0, 0) (u, ρ) det z Σ(v, ρ) (v, ρ) det z (u, µ, ɛ, τ) (u, τ) = z α z β, z t = N (u, µ, t,0), z t (u, µ, t,0) W(u, µ, t,0), = z α z β, z ρ = N (u, ρ, 0, 0), z t (u, ρ, 0, 0) W(u, ρ, 0, 0), = z Σ,v1 z Σ,v2, z Σ,ρ = N Σ (v, ρ), z Σ,ρ (v, ρ) W Σ (v, ρ) = z α z β, z τ = N (u, µ, ɛ, τ), z τ (u, µ, ɛ, τ) W(u, µ, ɛ, τ),

126 126 CHAPTER 4. FLOATING BODIES we find that f(µ, ɛ, q) = ɛ U 0 µ U 0 q U V 0 0 µ 0 + Ω l (S 0, Σ 0 ). N (u, µ, t,0), z t (u, µ, t,0) W(u, µ, t,0) dudt N (u, ρ, 0, 0), z ρ (u, ρ, 0, 0) W(u, ρ, 0, 0) dudρ N (u, µ, ɛ, τ), z τ (u, µ, ɛ, τ) W(u, µ, ɛ, τ) dudτ N Σ (v, ρ), z Σ,ρ (v, ρ) W Σ (v, ρ) dvdρ The assertion of the lemma follows since f q (0, 0, 0) = f ɛ (0, 0, 0) = ξ 0 da = 1 S 0 ξ da S 0 and f µ (0, 0, 0) = + χ(u) N S0 (u), (α i R αi (0))x(u) + a da S 0 N Σ0 (v), (α i R αi (0))y(v) + a da. W(S 0 ) On the other hand we have Corollary. Suppose that for given (ζ, α, a) X and a given regular scalar function q(c), where q(0, 0) = 0, q µ (0, 0) = 0, q ɛ (0, 0) = 0 the configuration (S(c), Σ(c)) defined through z(u, c) = z (u, c) + q(c)ξ 0 (u)n 0 (u), u U, and z Σ (v, µ), v V (ɛ), is volume preserving, then (ζ, α, a) is in the subspace X 0. Proof. From the formulas in the proof above we see that f µ (0, 0, 0) + f q (0, 0, 0)q µ (0, 0, 0) = 0 and f ɛ (0, 0, 0) + f q (0, 0, 0)q ɛ (0, 0, 0) = 0.

127 4.4. APPENDIX 127 Lemma A.4.2. c k S(c) = 2 + U 1 S(c) + 2 S(c) H(u, c) N(u, c), z ck (u, c) W(u, c) du z ck (u, c), ν(u, c) ds(c) z ck (u, c), ν(u, c) ds(c), where H(u, c) denotes the mean curvature of S(c) at u U, and ds(c) = Z τ (τ, c) dτ, Z(τ, c) := z(u(τ), c), here u(τ) is a regular parameter representation of 1 U or of 2 U, resp. Proof. Set u = (α, β), then S(c) c k = 1 ( E(u, c) zβ (u, c), z β,ck (u, c) U W(u, c) F(u, c)[ z α (u, c), z β,ck (u, c) + z β (u, c), z α,ck (u, c) ] +G(u, c) z α (u, c), z α,ck (u, c) ) du. The formula of Lemma 2.2 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. [ ] µ S(c) c=0 = 2 H(u, 0) N(u, 0), χ(u)((α i R αi (0))x(u) + a) da S 0 + α i R αi x(u) + a, ν(u, 0) ds. 2 S 0 [ ] ɛ S(c) = 2 H(u, 0)ξ(u) da + η(u) ds + η(u) ds. c=0 S 0 1 S 0 2 S 0 Proof. The first formula of the corollary is a consequence of formula z µ (u, 0) = χ(u)((α i R αi (0))x(u) + a)).

128 128 CHAPTER 4. FLOATING BODIES We recall that χ(u) = 0 if u 1 U and χ(u) = 1 if u 2 U. The second formula of the previous corollary follows since z ɛ (u, 0) = ζ, and ζ = ξ(u)n S0 (u) + η(u)t S0 (u). Lemma A.4.3. ɛ W k(s(c)) = k U Z 0,ɛ (τ, ɛ), ν(u, 0, ɛ) Z 0,τ (τ, ɛ) dτ, k = 1, 2, where Z 0 (τ, t) = z 0 (u(τ), t) and u(τ) = u k (τ) is a regular parameter representation of 1 U or 2 U, resp, and µ W k(s(c)) = 0, k = 1, 2. Proof. Since z(u, µ, ɛ) = z(u, 0, ɛ) on 1 U and W 2 (S(0, ɛ) does not change under rigid motion we get W k (S(µ, ɛ)) = W k (S(0, ɛ)). Let u(τ) be a regular parameter representation of 1 U or 2 U. Then ɛ W k (S(0, ɛ)) = Z 0,τ (τ, t) Z 0,t (τ, t) dτdt + W k (S(0, 0)). k U 0 Here we assume for simplicity that 1 S(0, ɛ) is above of 1 S(0, 0). The final formula is valid for every 1 S(0, ɛ), we omit the details. For the general case see a remark in [18], p. 81. The assertion of the lemma follows since the tangential plane on the container wall at P 1 S(0, t) is spanned by the orthogonal vectors ν(u(τ), 0, t) and t(τ, t) = Z 0,τ (τ, t)/ Z 0,τ (τ, t). We recall that ν(u(τ), 0, t) and t(τ, t) are orthogonal vectors. Then Z 0,t (τ, t) = aν(u(τ), 0, t) + bt(τ, t) with scalar functions a and b depending on τ and t. Using the Lagrange identity, we find that Z 0,τ (τ, t) Z 0,t (τ, t) 2 = Z 0,τ (τ, t) 2 Z 0,t (τ, t) 2 Z 0,τ (τ, t), Z 0,t (τ, t) 2 = Z 0,τ (τ, t) 2 (a 2 + b 2 ) Z 0,τ (τ, t), aν(u(τ), 0, t) + bt(τ, t) 2 = Z 0,τ (τ, t) 2 (a 2 + b 2 ) b 2 Z 0,τ (τ, t) 2 = a 2 Z 0,τ (τ, t) 2 = Z 0,t (τ, t), ν(u(τ), 0, t) Z 0,τ (τ, t) 2

129 4.4. APPENDIX 129 Corollary. [ ] ɛ W k(s(c)) = (ξ(u)sin γ k + η(u)cos γ k ) ds. c=0 k S 0 Proof. The formula follows since Z 0,τ (τ, 0) dτ = ds and Z 0,ɛ (τ, 0), ν(u(τ), 0, 0) = ζ(u(τ)), ν(u(τ), 0, 0) = ξ(u(τ)) N(u(τ), 0, 0), ν(u(τ), 0, 0) +η(u(τ)) ν(u(τ), 0, 0), ν(u(τ), 0, 0) = ξ(u(τ))sin γ k + η(u(τ))cos γ k. Lemma A.4.4. ɛ µ Ω l (S(c),Σ(c)) Ω l (S(c),Σ(c)) ɛ U 0 µ F 1 (x) dx = F 1 (x) dx = U F 1 (z(u, c)) N(u, c), z ɛ (u, c) W(u, c) du, [ F 1 (z(u, µ, t)) N(u, µ, t), z t (u, µ, t) W(u, µ, t) + F 1 (z(u, µ,0)) N(u, µ,0), z µ (u, µ,0) W(u, µ,0) du U + F 1 (z Σ (v, µ)) N Σ (v, µ), z Σ,µ (v, µ) W Σ (v, µ) dv. V 0 ] dudt Proof. Here we assume for simplicity that S(c) and Σ(c) are above of S(0) and of Σ(c), resp. The final formula is valid for every S(c) and Σ(c), we omit the details. For the general case see a remark in [18], p. 81. Then, see Figure 4.13, with Ω l (S(c), Σ(c)) = Ω 1 Ω 2 Ω 3 Ω l (S(0), Σ(0)), Ω 1 = {z(u, µ, t) : u U, 0 < t < ɛ}, Ω 2 = {z(u, ρ, 0) : u U, 0 < ρ < µ}, Ω 3 = {z Σ (v, ρ) : v V 0, 0 < ρ < µ}.

130 130 CHAPTER 4. FLOATING BODIES S (µ,ε) Ω(µ) S (µ,0) Σ(µ) Ω(0) S (0,ε) S (0,0) Σ(0) Ω ( S 0, Σ ) l 0 Figure 4.13: Changing of the liquid region Consequently Ω l (S(c),Σ(c)) F 1 (x) dx = U ɛ 0 µ U V 0 F 1 (z(u, µ, t))det 0 µ 0 Ω l (S(0),Σ(0)) F 1 (z(u, ρ, 0))det z(u, µ, t) (u, t) z(u, ρ, 0) (u, ρ) F 1 (z Σ (v, ρ))det z Σ(v, ρ) (v, ρ) F 1 (x) dx. dudt dudρ dvdρ

131 4.4. APPENDIX 131 Since det det z(u, µ, t) (u, t) z(u, ρ, 0) (u, ρ) det z Σ(v, ρ) (v, ρ) = z α z β, z t = N(u, µ, t), z t (u, µ, t) W(u, µ, t), = z α z β, z ρ = N(u, ρ, 0), z ρ (u, ρ, 0) W(u, ρ, 0), = z Σ,v1 z Σ,v2, z Σ,ρ = N Σ (v, ρ), z Σ,ρ (v, ρ) W Σ (v, ρ), we obtain the formulas of the lemma. Remark. Since the particle is moved by a rigid motion, we have W Σ (v, µ) = W Σ (v,0) and N Σ (v, µ) = R(µα)N Σ (v,0). Corollary. [ ɛ [ µ where Ω l (S(c),Σ(c)) Ω l (S(c),Σ(c)) F 1 (x) dx F 1 (x) dx ] ] c=0 c=0 = = + F 1 (x(u))ξ(u) da S 0 F 1 (x(u)) N(u, 0, 0), z µ (u, 0, 0) da S 0 F 1 (y(v)) N Σ0 (v,0), z Σ,µ (v,0) da, W 2 (S 0 ) z µ (u, 0, 0) = χ(u)((α i R αi (0))x(u) + a), z Σ,µ (v,0) = (α i R αi (0))y(v) + a. Lemma A.4.5. d F 2 (y) dy = dµ Ω(µ) Ω(0) F 2 (R(µα)y + µa), (α i R αi (µα))y + a dy. Proof. Since Ω(µ) = {q R 3 : q = R(µα)y + µa, where y Ω 0 },

132 132 CHAPTER 4. FLOATING BODIES we get Ω(µ) F 2 (q) dq = = Ω(0) Ω(0) F 2 (R(µα)y + µa) q det y dy F 2 (R(µα)y + µa) dy. Corollary. [ ] d F 2 (y) dy = F 2 (y), (α i R αi (0))y + a dy. dµ Ω(µ) Ω(0) c=0

133 4.5. PROBLEMS Problems 1. Consider a floating body in an equilibrium which floats in a liquid of infinite volume and suppose that the capillary interface is horizontal, see Figure Show that B=weight of displaced fluid, where B vapour g general level liquid displaced fluid Figure 4.14: Displaced liquid=buoyancy denotes the buoyancy of the body. 2. Consider a cylinder of constant circular cross section which hangs in a fluid of infinite volume. Let γ be the constant contact angle between the liquid and the body and suppose the cylinder can move vertically only, see Figure Show that vapour liquid γ g displaced fluid general level Figure 4.15: Capillarity and buoyancy < weight of displaced fluid if γ < π/2 B = > weight of displaced fluid if γ > π/2 = weight of displaced fluid if γ = π/2.

134 134 CHAPTER 4. FLOATING BODIES The displaced fluid is sketched in Figure Determine lim d F L x (d, γ w, γ l ), see Section (attraction/repelling of a plate). 4. Express F L x (d, γ w, γ l ) in terms of elliptic integrals. 5. Discuss the sign of F L x (d, γ w, γ l ), 6. Discuss the rotation of a cylindrical body, which has a constant square as cross section, around the middle axis which coincides with the middle axis of the container with constant square as cross section, see the figure. 7. Find an embedding foliation to the problem of floating bodies.

135 Chapter 5 Wetting barriers Consider a container or a floating particle with edges or lines where different materials meet each other, see [45] for the case of of wetting barriers on the container wall. Γ + ω γ S Γ liquid Figure 5.1: A container with an edge, γ γ γ + ω + π Suppose that the capillary surface hangs on such a line of the container wall or the particle, see Figure 5.1 and Figure 5.2. A typical example of such a floating particle is a floating razor blade. Along the wetting barrier the contact angle is not determined, see a remark in [31], p Suppose that the configuration is in an equilibrium, see the definition below, then it turns out that the contact angle is in an interval defined by the data. We will see that all the equilibrium conditions of Theorem 4.1 hold with the only exception that we have to replace the equation which defines the contact angle by inequalities. 135

136 136 CHAPTER 5. WETTING BARRIERS + Γ S γ 1 ω particle γ Γ liquid Figure 5.2: A particle with an edge, γ γ γ + ω + π 5.1 Equilibrium conditions In the following we consider the case of a floating particle with an edge. The case of a container with an edge is contained in this case if one keeps the particle fixed, i. e., a = 0 and α = 0, and if the boundary of the particle is replaced by the container wall. In contrast to the case of smooth particles, we define two families S 0 ± (ɛ) of admissible comparison surfaces by z ± 0 (u, ɛ) = x(u) + ɛζ± (u) + O(ɛ 2 ), where 0 < ɛ < ɛ 0, and ζ ± = ξ ± (u)n S0 (u)+η ± (u)t S0 (u) are given. We recall that S 0 ± (ɛ) is called admissible if it meets Ω 0 at its boundary Σ 0 only. More precisely, we assume that z 0 ± (u, ɛ) R3 \ Ω for all u U and 0 < ɛ ɛ 0, and z 0 ± (u, ɛ) Σ± 0 if u U, see Figure 5.4 for notations. Suppose that the interior open angle ω, here we use another notation than in [45], of the edge satisfies 0 < ω < π, then it can be shown, we omit the details, that z0 (u, ɛ) defines an admissible family of comparison surfaces if 0 < γ < π, and z 0 + (u, ɛ) is admissible if π ω < γ < 2π ω, see Figure 5.3 (a). If π < ω < 2π, see Figure 5.3 (b), then z 0 + (u, ɛ) and z 0 (u, ɛ) define admissible comparison surfaces. Since ζ + N Σ +, ζ N 0 Σ, where 0 u U 2, see Figure 5.4 for notations, then ζ + satisfies ξ + cos(γ + ω) η + sin(γ + ω) = 0 (5.1) at U 2, and in the case of the second family we have ξ cos γ + η sinγ = 0 (5.2) at 2 U. Let S ± (c) be the comparison surface defined by z ± (u, c) = z ± 0 (u, ɛ) + χ(u)( R(µα)z ± 0 (u, ɛ) + µa z± 0 (u, ɛ)),

137 5.1. EQUILIBRIUM CONDITIONS 137 ν + particle γ ω ν + γ ω Ω liquid ν Ω liquid ν particle (a) (b) Figure 5.3: Convex and reentrant edge where c = (µ, ɛ), ɛ 0 and µ < µ 0. Suppose that E(S ± (c), Σ(c)) E(S(0), Σ(0)), Ω l (S ± (c), Σ(c) = C. In contrast to Chapter 4 one of the resulting equations has to be replaced by an inequality: [ ] [ ] ɛ E(S± (c), Σ(c)) 0, c=0 µ E(S± (c), Σ(c)) = 0, (5.3) c=0 [ c Ω l (S ± (c), Σ(c)) ] = 0. (5.4) c=0 We call a capillary surface in an equilibrium if (5.3) and (5.4) are satisfied Let X ± be the set of all (ζ ±, α, a) C 1 (U) C 0 (U) R 3 R 3 satisfying (5.1), (5.2), resp, and ζ ± = a(u)ν ± on U 2, where a(u) is nonnegative and continuous on U 2. Thus X + and X define convex cones with the vertex at zero. From a Lagrange multiplier rule of Chapter 8 it follows that there is a real λ 0 such that [ ] ɛ L(S± (c), Σ(c), λ 0 ) 0, (5.5) c=0 ] [ µ L(S± (c), Σ(c), λ 0 ) c=0 = 0 (5.6) for all (ζ, α, a) X + or in X, resp. Following the proof of Theorem 3.1, we obtain the differential equation (4.13), the boundary condition (4.14) on

138 138 CHAPTER 5. WETTING BARRIERS S 0 liquid N S 0 ν + 0 γ ω ν ν 0 0 vapor Σ + 0 Σ 0 particle N Σ + 0 N Σ 0 Figure 5.4: Particle with an edge, notations the container wall and, see the proof of Theorem 3.1, [ ] ɛ L(S+ (c), Σ(c), λ 0 ) = ζ +, ν 0 β 2 ν 0 + ds c=0 S 0 = a(u) ν 0 +, ν 0 β 2 ν 0 + ds S 0 = a(u) ( cos(γ + ω) β +) ds 0 S 0 for all a(u) which are continuous and nonnegative at U 2. Thus, since β + = cos γ +, we have the inequality cos γ + cos(γ + ω π). (5.7) For the comparison surfaces S we get [ ] ɛ L(S (c), Σ(c), λ 0 ) = ζ, ν 0 + β 2 ν0 ds c=0 S 0 = a(u) ν0, ν 0 + β 2 ν0 ds S 0 = a(u) ( cos γ + β ) ds 0 S 0 for all a(u) which are continuous and nonnegative at U 2. Thus cos γ cos γ. (5.8)

139 5.1. EQUILIBRIUM CONDITIONS 139 Then we have the following necessary conditions: Case 0 ω < π. If 0 γ π and π ω < γ < 2π ω, then there are admissible families of comparison surfaces S + (c) and S (c) and, see (5.8) and (5.7), γ γ γ + ω + π. If 0 < γ < π ω, then there exists an admissible family S (c) and (5.8) implies that γ γ. If π < γ 2π ω, then there is an admissible family S + (c) which implies that γ γ + ω + π. Case π < ω < 2π. In this case, there exist admissible families of comparison surfaces S + (ɛ) and S (ɛ). Then Summarizing, we get γ γ γ + ω + π. Theorem 5.1. Let γ be the angle between Γ and the capillary surface in equilibrium, then γ γ γ + ω + π at 2 S 0. Additionally, in any case we have that γ, γ + [0, π], ω [0, 2π) and 0 γ 2π ω. Equations (4.16) and (4.17) follow again from equation (5.6). Remark. A capillary surface which hangs at a wetting barrier is sometimes called pinned at this line. A mathematical founded proof for this behaviour of a liquid does not seem to appear in the literature with the exception of the paper [45]. If the strict inequalities γ < γ < γ + ω + π hold and if the admissible variations do not coincide with the equilibrium surface at the barrier line then this variation is positive. Moreover, an equilibrium defines a strict weak local minimizer of the energy provided the strict inequalities hold and if an additional eigenvalue criterion is satisfied, see [45], p. 246.

140 140 CHAPTER 5. WETTING BARRIERS 5.2 Stability In the following we consider the case of a container with wetting barriers in the absence of a floating particle. Let S ± (ɛ) be the family of volume preserving admissible comparison surfaces defined above through z ± (u, ɛ) where a = 0 and α = 0 and suppose that S 0 = S(0) is an equilibrium interface. Then E(S ± (ɛ)) E(S 0 ) = L(S ± (ɛ), λ) L(S 0, λ) [ ] d = ɛ dɛ L± (S(ɛ), λ) + ɛ2 ɛ=0 2 +O(ɛ 3 ). [ ] d 2 dɛ 2 L± (S(ɛ), λ) ɛ=0 We have [ ] d dɛ L+ (S(ɛ), λ) [ ] d dɛ L (S(ɛ), λ) ɛ=0 ɛ=0 = = a(u)(cos(γ + ω π) cos γ + ) ds S 0 a(u)( cos γ + cosγ ) ds. S 0 Since ζ ± = ξ ± N S0 + η ± T S0 and a > 0 on a part of U if and only if ξ ± > 0 on that part of U provided ξ ± C 0 ( U) and γ is different from 0 or 2π ω. Consequently there is a positive constant c such that E(S ± (ɛ)) E(S 0 ) 0 for all 0 < ɛ < ɛ 0 (ζ) if ξ + > 0, ξ > 0, resp., on a part of U and if the strict inequalities γ < γ < γ + ω + π are satisfied. Thus, see the corollary to Lemma 3.1 of Chapter 3, a necessary condition that the extremal S defines a (weak or strong) local minimizer of the associated energy is that ( σ ξ 2 2(2H0 2 K 0 )ξ 2) da + F 1, N 0 ξ 2 da 0 S 0 S 0 for all ξ W 1,2 0 (S 0 ) satisfying the side condition S 0 ξ da = 0. To answer the question whether a given extremal defines a strong minimizer of the associated energy functional requires the construction of an embedding family of configurations, see Chapter 2 and Chapter 3 for the case of liquid layers and capillary interfaces.

141 5.3. PROBLEMS Problems 1. Consider a liquid which hangs at the upper edge of a cylindrical container with constant circular cross section, see Figure 5.1. Suppose that ω = π/2 and that the gravity vanishes. Show that the following stability criterion holds: S 0 ( ξ 2 2(2H 2 0 K 0 )ξ 2) da > 0 for all ξ W 1,2 0 (S 0 ) \ {0} satisfying the side condition S 0 ξ da = 0. Hint: S 0 is a spherical cap, and see [77]. 2. Consider a floating razor blade and find the necessary equilibrium conditions for γ. 3. Consider a liquid trapped in a groove as shown in Figure 5.5. S 0,L L Γ + γ ω 2 Γ ω 1 Figure 5.5: Liquid trapped in a groove (i) Find the necessary equilibrium conditions for γ. (ii) Consider the case of zero gravity and check whether the following stability criterion holds: S 0,L ( ξ 2 2(2H 2 0 K 0 )ξ 2) da > 0

142 142 CHAPTER 5. WETTING BARRIERS for all ξ W 1,2 0 (S 0 ) \ {0}. Here S 0,L denotes a section of length L of S 0. Hint: S 0 is a cylindrical cap. 4. Find an embedding foliation to the problem of wetting barriers (open problem). 5. Does an equilibrium, which satisfies the above stability condition, define a strong minimizer of the associated energy (open problem)? 6. Extend the above stability considerations to the case of a floating particle with wetting barriers. 7. Find the volume of the liquid which rests at the lower end of the tube, see Figure 5.6 which is taken from [50]. Figure 5.6: Capillary tube, constant cross section 8. Find the volume of the liquid which rests at the lower end of the tube, see Figure 5.7 which is taken from [50]. 9. Discuss the regularity (conjecture C 1,α ) of the trace of the liquid which hangs at the upper edge on a wall, see Figure 5.8 which is taken from [50]. 10. Find the stability bound of the radius of the dry spot of the drop with a hole, see Figure 5.9 which is taken from [50].

143 5.3. PROBLEMS 143 Figure 5.7: Capillary tube, variable cross section Figure 5.8: Liquid hangs partially at the upper edge 11. Find the capillary surface in a tube if its upper edge is below of the rise of the liquid of the same tube where the upper edge is above of the rise of the liquid, see Figure Discuss the case of a rounded upper edge of a tube with constant circular cross section, see [48], pp , and Figure 5.11, where Γ ɛ is a circular arc of radius ɛ.

144 144 CHAPTER 5. WETTING BARRIERS Figure 5.9: Drop with a hole g? liquid Figure 5.10: Upper edge of the tube is below of the rise of liquid ε γ γ S ε Γ ε ε γ Figure 5.11: Rounded edge

145 Chapter 6 Asymptotic formulas Here we consider capillarity problems depending on a small or large real parameter. In general, there are no explicit solutions with the exception of few examples, see previous sections. Therefore asymptotic formulas are of interest if the related parameter is small or large, resp. The main tools to get asymptotic formulas are comparison principles. 6.1 Comparison principles The following comparison principles exploit the special strong nonlinearity of the governing equation. There are no counterparts for linear elliptic equations, see some examples in this section. Set Tu = u 1 + u 2 and Nu = div Tu κu, where κ is a nonnegative constant. We recall that an open set is said to be connected if it is not a union of two nonempty open sets with empty intersection. Theorem 6.1 (Concus and Finn[12, 21]). Let Ω R 2 be a bounded connected domain and assume the boundary Σ = Ω admits a decomposition where Σ = Σ α Σ β Σ 0, (a) Σ 0 can be covered, for any ε > 0, by a countable number of disks B δi of radius δ i, such that δ i < ε, i. e. H 1 (Σ 0 ) = 0, where H 1 is the onedimensional Hausdorff measure, 145

146 146 CHAPTER 6. ASYMPTOTIC FORMULAS (b) Σ α is Lipschitz and Σ β C 1. Let u, v C 2 (Ω) with the properties (i) Nu Nv in Ω, (ii) lim sup(u v) 0 for any approach to Σ α from within Ω, (iii) (Tu Tv) ν 0 almost everywhere (with respect to L 1 (Σ β )) on Σ β as a limit from points of Ω, where ν denotes the exterior unit normal at Σ β. Then: (A) If κ > 0 or if Σ α, then v u in Ω. If v(x 0 ) = u(x 0 ) at any point x 0 Ω, then v(x) u(x). (B) If κ = 0, Σ α =, then v(x) u(x) + const. in Ω. Proof. See Finn [21], pp Define for a positive constant M Ω M 0 = {x Ω : 0 < u(x) v(x) < M} Ω = {x Ω : u(x) v(x) 0} Ω + = {x Ω : u(x) v(x) M} Λ(ɛ) = Ω ( B δi ) Ω(ɛ) = Ω \ B δi. Assume u > v on a subset of Ω. Then there exists a positive constant M such that Ω M 0 Ω(ε) for all 0 < ε < ε 0, ε 0 sufficiently small. Define for x Ω the function w = 0 : x Ω u v : x Ω M 0. M : x Ω + The function w belongs to the Sobolev class W 1,2 (Ω) L (Ω), see for example [30], pp Since w 0 and Nu Nv 0, it follows from the

147 6.1. COMPARISON PRINCIPLES 147 definition of N and after integration by parts that 0 w(nu Nv) dx = w (Tu Tv) dx where Ω(ε) κ +M Ω(ε) Σ + (ε) Ω(ε) w(u v) dx + w(tu Tv) ν ds Λ(ε) (Tu Tv) ν ds + w(tu Tv) ν ds, Σ M 0 (ε) Σ M 0 (ε) = (Ω M 0 \ B δi ) Ω Σ + (ε) = (Ω + \ B δi ) Ω Σ (ε) = (Ω \ B δi ) Ω, see Figure 6.1. Thus, assigning symbols to the integrals on the right in order B δ i + Σ (ε) _ Σ (ε) u v > M = u v < 0 = 0<u v<m M 0 Σ (ε) Figure 6.1: Proof of Theorem 1 of appearance, 0 Q(ε) W(ε) + J Λ (ε) + J + (ε) + J M 0 (ε). We observe that W(ε) 0 and W(ε) = 0 if and only if κ = 0. Since Σ + (ε) Σ β it follows J + (ε) 0 and since all points from Σ M 0 (ε) where

148 148 CHAPTER 6. ASYMPTOTIC FORMULAS w 0 are contained in Σ β we have J0 M (ε) 0. From the definition of w we see that Q(ε) = (u v) (Tu Tv) dx, Ω M 0 (ε) where Ω M 0 (ε) = ΩM 0 \ ( B δ i ). Thus, Q(ε) 0 and Q(ε) = 0 if and only if u v in Ω M 0 (ε). This follows since the vector field A(p) := p/ 1 + p 2, p R n, satisfies A(p) A(q), p q c p q 2, where c = c(p, q) is a positive constant, see an exercise. Finally, the inequality J Λ (ε) 4πMε follows from the fact that Tf 1 for all f C 1 (Ω). Consequently lim(q(ε) + W(ε)) 0. ε 0 This implies, since Q(ε) and W(ε) are nonnegative and increasing with decreasing ε, that Q(ε) = 0 and W(ε) = 0 for each ε, 0 < ε ε 0. We conclude from W(ε) = 0 that in the case κ > 0 a construction of Ω M 0 is not possible for each ε and M. Thus v u in Ω. Let κ = 0. If there is a subset where v(x) < u(x), then u v c, where c is a constant and 0 < c < M, on each connected set of Ω M 0 (ε) since u v on these sets, according to the above considerations. The constants may differ from set to set. Since u v = 0 or u v = M on the boundary of these components if the boundary point in consideration is in Ω and since u v is continuous in Ω by assumption, it follows that the union of these connected sets is Ω, see Figure 6.2. Then we have u v = const. > 0 in Ω. Case Σ. The above inequality is a contradiction to the assumption lim sup(u v) 0 from any approach to Σ α from within Ω, consequently v u in Ω. Case Σ =. In this case an arbitrary constant can be added to v without changing the hypotheses. Thus there is always a domain in Ω where 0 < u (v C) < M. From above it follows that u = v C + const. in Ω. It remains to show that from v(x) u(x) in Ω and u(x 0 ) = v(x 0 ) for at least one x 0 Ω it follows that u v in Ω. This is a consequence of the

149 6.1. COMPARISON PRINCIPLES 149 u v = > M u v=c u v = < 0 0<c<M Figure 6.2: Proof of Theorem 1 strong maximum principle 1 since Nu Nv = 2 a ij (x)w xi x j (x) + i,j=1 2 b i (x)w xi (x) κw, where w = u v, a ij = a ji, a ij, b i C(Ω), and the right hand side is elliptic in Ω. Remark. The previous result fails in the case of linear elliptic equations. In that case we can not take away in the boundary condition one point from the the boundary as the following example shows. Let Ω R 2 be the domain Ω = {x B 1 (0) : x 2 > 0}, i=1 1 Strong maximum principle. Suppose that Ω R n is a connected domain. Let Lu = n n a ij(x)w xi x j (x) + b i(x)w xi (x) + c(x)w, i,j=1 where a i,j = a ji, a ij, b i, c C(Ω), c 0 in Ω R n, and the right hand side is elliptic in Ω. Assume w C 2 (Ω) satisfies Lw 0 (Lw 0) in Ω. Then if w achieves its positive supremum (negative infimum) in Ω, w is a constant. i=1

150 150 CHAPTER 6. ASYMPTOTIC FORMULAS x 2 1 x 1 Figure 6.3: Counterexample for linear problems Assume u C 2 (Ω) C(Ω \ {0}) is a solution of u = 0 in Ω u = 0 on Ω \ {0}. This problem has solutions u 0 and u = Im(z + z 1 ), where z = x 1 + ix 2. For another example see an exercise. In contrast to this behaviour of the Laplace equation, one has uniqueness according to the above theorem if u = 0 is replaced by the minimal surface equation x 1 ( u x1 1 + u 2 ) ( + x 2 u x2 1 + u 2 ) = 0. The following theorem concerns in particular unbounded domains. Theorem 6.2 (Finn and Hwang [25]). Assume that Ω R 2, not necessarily bounded, κ > 0 and that the assumptions (a) (b) and (i) (iii) of Theorem 6.1 are satisfied, then u v in Ω, and if u(x) = v(x) at any x Ω, then u(x) v(x) in Ω. Proof. If u > v at some point of Ω, then there are positive constants m 1, m 2 and a set Ω 12 Ω of positive measure where 0 < m 1 < u v < m 2 <. Set 0 : u v m 1 w(x) = u v m 1 : x Ω 12 m 2 m 1 : u v m 2.

151 6.1. COMPARISON PRINCIPLES 151 For any R > 0 set B R = {x R n : x < R} and Ω R = Ω B R, Γ R = Ω B R, Σ α,r = Σ α B R, Σ β,r = Σ β B R, see Figure 6.4 for an example. If R is sufficiently large, then the measure of Ω R Ω 12 is positive. Integration by parts gives x 2 Σ β Ω 12 Σ 0 Ω R x 1 Γ R Σ α Figure 6.4: Proof of Theorem 6.2 w n 1 (div Tu div Tv) dx Ω R = w n 1 (Tu Tv) ν ds + w n 1 (Tu Tv) ν ds Γ R Σ R,β + w n 1 (Tu Tv) ν ds (n 1) w n 2 (Tu Tv) w dx. Σ R,α Ω R Since (Tu Tv) ν 2, (Tu Tv) ν 0 on Σ β, w = 0 on Σ α and (Tu Tv) w 0, we find that κ w n 1 (u v) dx 2 Ω R Γ R w n 1 ds.

152 152 CHAPTER 6. ASYMPTOTIC FORMULAS We have 2 Thus we get finally w n dx w n 1 dx. Ω R Ω R κ w n dx 2 Ω R Γ R w n 1 ds. Set Then Q(R) Q(R) = w n dx. Ω R 2 w n 1 ds κ Γ R 2 ( ) (n 1)/n ( ) 1/n w n ds ds. κ Γ R Γ R Consequently 1 ρ Q(ρ)n/(n 1) ( ) 2 n/(n 1) ( 1 w n ds κ ρ Γ ρ Γ ρ ds ) 1/(n 1). Set 2 J(R) = R R 1 1 ρ Q(ρ)n/(n 1) dρ. Ω R {u v m 1 } w n 1 (u v) dx = 0 (u v) dx + (u v m 1) n 1 (u v) dx + = +m 1 {m 1 <u v<m 2 } {u v m 1 } w n dx + {m 1 <u v<m 2 } {m 1 <u v<m 2 } (u v m 1) n 1 dx + {u v m 2 } (m 2 m 1) n 1 (u v) dx (u v m 1) n 1 (u v m 1) dx {u v m 2 } + (m 2 m 1) n 1 ((u v) (m 2 m 1)) dx {u v m 2 } (m 2 m 1) n 1 (m 2 m 1) dx Ω R w n dx.

153 6.1. COMPARISON PRINCIPLES 153 Let 0 < R 1 < R <, then, since ds = ρ n 1 dω, we obtain ( ) 2 n/(n 1) R J(R) 1 ( ) 1/(n 1) w n ds ds dρ κ R 1 ρ Γ ρ Γ ρ ( ) 2 n/(n 1) R ( ) 1/(n 1) = w n ds dω dρ κ R 1 Γ ρ Γ ρ/ρ ( ) 2 n/(n 1) R ω n 1/(n 1) w n ds dρ κ = ω n 1/(n 1) ( 2 κ) n/(n 1) (Q(R) Q(R 1 )), R 1 Γ ρ where ω n denotes the n-dimensional unit sphere. Set C 1 = ω n 1/(n 1) ( 2 κ) n/(n 1), then, if R 1 is sufficiently small, it follows that for all R > R 1 J (R) (J(R)) n/(n 1) C 2 R, (Q(R)) n/(n 1) /R [C 1 (Q(R) Q(R 1 ))] n/(n 1) where C 2 is a positive constant which does not depend on R. Let R 1 A < B <. Then or B A J (ρ) (J(ρ)) n/(n 1) dρ C 2(lnB lna) [ (n 1) (J(A)) 1/(n 1) (J(B)) 1/(n 1)] C 2 (lnb lna), which is a contradiction since 0 < J(A) < J(B). The remainder of the proof follows from the strong maximum principle. Remark. The previous theorem does not cover the case κ = 0. Consider the minimal surface equation over R 2, then every linear function is a solution. Bernstein [5] has shown that every solution must be a linear function.

154 154 CHAPTER 6. ASYMPTOTIC FORMULAS For another example consider the strip Ω = {(x, y) : y < cos γ}, and the boundary value problem div Tu = 2 in Ω, ν Tu = cos γ on Ω. Solutions are given by the cylindrical surfaces u(x, y) = xtanα 1 y 2 cos 1 α + β, for arbitrary α, β with α < π/2. Tam [70] has proved that this class of surfaces yields the totality of the solutions of this boundary value problem. Remark. The previous Theorem 6.2 fails in the case of linear elliptic equations. Let Ω = R 2 \ {0} and consider the problem div Tu = κu in Ω. From the theorem follows that u 0 in Ω is the unique solution. Another consequence of the comparison principle is that every solution must be rotationally symmetric. Let u(x) = v(r), 0 < r <, be a rotationally symmetric solution. Then ( ) 1 rv (r) = κv(r), 0 < r <. r 1 + (v (r)) 2 The associated linearized problem is Every solution is given by 1 r (rv (r)) = κv(r), 0 < r <. v(r) = C 1 I 0 ( κr) + C 2 K 0 ( κr) with constants C 1, C 2. Here I 0, K 0 denote modified Bessel functions of second kind and of order zero. Remark. The crucial points are that the previous theorems hold without growth hypotheses at the exceptional set Σ 0 or, see Theorem 6.2 at infinity. Moreover, the comparison principle holds also when the boundary contact angle γ is 0 or π, where u becomes necessarily unbounded near Σ β. 6.2 Applications As applications of the above comparison principles we will derive an a priori estimate, some asymptotic formulas and an effective numerical method.

155 6.2. APPLICATIONS An interior estimate Let u C 2 (Ω) be a solution of the capillary equation div Tu = κu in Ω R 2, where the capillary constant κ is positive. Proposition 6.1. (Concus and Finn [13]). Let B δ be an open disk with radius δ such that B δ Ω. Then sup B δ u 2 κδ + δ. Proof. Let B δ, 0 < δ < δ, be concentric to B δ and denote by v (x) the lower hemisphere with radius δ over B δ with lower point z = v 0 := 2 κδ. Then div Tv = 2 δ = κv 0 κv in B δ that and ν Tv = 1 on B δ. The comparison principle Theorem 4.1 says u(x) v (x) < 2 κδ + δ in B δ. Letting δ δ, the proof is finished. Remark. There is no assumption on the boundary behaviour of u in this proposition Narrow circular tube, interior ascent Here we consider the problem div TU = κu in B a (0) ν TU = cos γ on B a (0), where κ > 0 is the capillary constant and B a (0) is a disk with radius a and center at the origin.

156 156 CHAPTER 6. ASYMPTOTIC FORMULAS Set x = ay and V (y) = U(x)/a, then V = V (y) is a solution of where B = κa 2 is the Bond number. div TV = BV in B 1 (0) (6.1) ν TV = cos γ on B 1 (0), (6.2) Proposition 6.2. ([45]). For every nonnegative integer n there exist n + 1 radially symmetric functions φ l (y; γ), l = 0,...n, analytical in B 1 (0) and bounded on B 1 (0), such that V (y; γ, B) = 2 cos γ B n + φ l (y; γ)b l + O(B n+1 ) l=0 uniformly in B 1 (0) and γ, 0 γ π, as B tends to zero. The function φ 0 is a solution of a nonlinear boundary value problem and the functions φ l, l 1 are solutions of linear boundary value problems. In particular we find that φ 0 (y; γ) = 2 1 sin 3 γ 3 cos r γ cos γ 2 cos 2 γ, where r = y y2 2, and φ 1 (y; γ) = (1 sinγ)2 (1 + 3 sin 2 γ) 6 cos 7 γ + (1 sinγ)sin2 γ 3 cos 5 γ 1 1 r 2 cos 2 γ cos 2 γ ln ( r 2 cos 2 γ 1 2 cos 3 ln(1 + sinγ) γ ). Remark. In the case n = 0 we obtain the Laplace formula V (y; γ, B) = 2 cos γ B + φ 0(y; γ) + O(B), see Laplace [39] for a formal proof and D. Siegel [68] for a proof. Then the center height of U is given by U(0) = 2 cos γ κa ( 1 cos γ sin 3 ) γ cos 3 a + O(κa 3 ). γ

157 6.2. APPLICATIONS 157 Set β = cos γ. Lemma 6.1. For given N N {0} there exist functions φ l (r; β), β 1, which are in C 1 [0, 1] and analytical in 0 s < 1 with respect to s and satisfy sup φ l (r; β) c l < uniformly with respect to β 1 and 0 s 1, such that satisfies V N := 2 cos γ B N + φ l B l l=0 div TV N BV N = O(B N+1 ) uniformly in Ω = B 1 (0) and β 1, and lim V N s V N 2 = β, V N(0) = 0, where the derivatives are taken with respect to r. Proof. We have ( ) div Tv N Bv N = 1 v N Bv r 1 + (v N ) 2 N. The formal ansatz V N = C N B + φ k B k, where C is a constant, and the conditions ( ) div TV N BV N = 1 V N BV r 1 + (V N = O(B N+1 ) (6.3) N )2 V lim N = β (6.4) r (V N )2 lead to a nonlinear boundary value problem which determines φ 0 and to linear boundary value problems for φ k, k 1. Let γ [0, π/2]. Then lim r 1 0 k=0 φ φ 2 0 = cos γ,

158 158 CHAPTER 6. ASYMPTOTIC FORMULAS (1 ɛ 0 )cos γ 1 r 2 cos 2 γ φ 0(r) cos γ 1 r 2 cos 2 γ for an ɛ 0 (0, 1) and for all r (ɛ 0, 1), and The functions φ l satisfy 0 (1 r 2 cos 2 γ)φ 0(r) cos γ. φ l (r) (1 + φ 0 (r) 2 = O(1 r) ) 3/2 as r 1 and uniformly in γ [0, π/2], and sup φ l (r; γ) <, ) sup ((1 r cos γ) 1/2 φ l (r; γ) <, ( ) sup (1 r cos γ) 3/2 φ l (r; γ) <, where the supremum is taken over 0 < r < 1 and 0 γ π/2. From the behaviour of φ 0 it follows the asserted behaviour of the functions φ k, k 1, using the strong nonlinearity of the problem, see [45] for details. Since V N = C B + φ 0 + N φ k B k, k=1 we have where Q = 1 + V 2 N = (1 + φ 2 0 )(1 + Q), 2φ 0 N k=1 φ k Bk + ( N k=1 φ k B k 1 + φ 2 0 ) 2, (6.5) and (1 + v 2 ) 1/2 = (1 + φ 2 0 ) 1/2 1 + ( ) 1/2 µ=1 µ Q µ.

159 6.2. APPLICATIONS 159 Then v (1 + v 2 ) 1/2 = (φ 0 + N φ k B k)(1 + φ 2 k=1 = (1 + φ 2 0 ) 1/2( φ N k=1 µ=1 k=1 0 ) 1/2 ( ) 1/2 φ k Bk Q µ). µ 1 + ( ) 1/2 µ=1 N ( 1/2 φ k B k + µ µ=1 µ Q µ ) φ 0Q µ For simplicity we consider here the case N = 1. For the general case N 1 see [45]. Set v = C B + φ 0 + φ 1 B, then v (1 + v 2 ) 1/2 = (1 + φ 2 0 ) 1/2 ( φ 0 + φ φ 2 0 ) B + O(B 2 ), and the boundary condition (6.4) implies From lim r 1 0 lim r 1 0 div Tv Bv = = 1 r φ 0 (1 + φ 2 0 )1/2 = β, φ 0(0) = 0 φ 1 (1 + φ 2 0 )3/2 = 0, φ 1(0) = 0. ( rv 1 + v 2) Bv 1 r ( rφ φ 2 0 ) C { 1 ( +B rφ r 1(1 + φ 2 0 ) 3/2) } φ0 + O(B 2 ) and the requirement div Tv Bv = O(B 2 )

160 160 CHAPTER 6. ASYMPTOTIC FORMULAS as B 0 we find the boundary value problems ( ) 1 rφ 0 = C on (0, 1) (6.6) r 1 + φ 2 0 and lim r 1 0 φ φ 2 0 = β, φ 0(0) = 0 (6.7) ( 1 rφ 1 r (1 + φ 2 0 )3/2 lim r 1 0 ) = φ 0 on (0, 1) (6.8) φ 1 (1 + φ 2 0 )3/2 = 0, φ 1(0) = 0. (6.9) From the boundary value problem (6.6), (6.7) we obtain φ 0 (r) = Φ 0 (r) + K, where Φ 0 (r) = 1 β 1 β 2 r 2, and K is a constant which is determined through the boundary value problem (6.8), (6.9) which defines the next term φ 1, up to a constant, in the asymptotic expansion. An easy calculation shows that K = sin 3 β cos 3 β. The function φ 1 is defined by (6.8), (6.9) up to an unknown constant. This constant is fixed by the boundary value problem for the next term φ 2. In general, the quotient Q defined by (6.5) can be written as Q = N a k (γ; r)b k + O(B N+1 ), k=1 where the functions a k and the remainder are uniformly bounded with respect to r (0, 1) and γ [0, π], see [45], p In particular, it follows that div TV N BV N = O(B N+1 ) in B 1 (0) ν TV N = cos γ on B 1 (0),

161 6.2. APPLICATIONS 161 where the remainder O(B N+1 ) is uniformly bounded with respect to r (0, 1) and γ [0, π]. Lemma 6.2. For any given nonnegative integer n the solution V of the boundary value problem (6.1), (6.2) satisfies V = V n 1 + O(B n ) uniformly with respect to β 1 and 0 < r < 1, where V 1 = 2 cos γ/b. Proof. We recal that β = cos γ. Set V + = V n + AB n, where A is a positive constant which will be determinated later. Then div TV + BV + = div TV n BV n AB n+1 = O(B n+1 ) AB n+1 0, provided A is large enough. Then, since lim r 1 0 (V + ) = β, 1 + (V + ) 2 it follows from the comparison principle (Theorem 6.1) that V V + on (0, 1). The same reason yields a lower bound if V := V n AB n. Thus V = V n + O(B n ) = V n 1 + O(B n ) Wide circular tube Here we consider the boundary value problem ( rw 1 + w 2) = κrw 0 < r < R lim r R 0 w 1 + w 2 = cos γ w (0) = 0,

162 162 CHAPTER 6. ASYMPTOTIC FORMULAS where 0 γ π and the capillary constant κ is positive. In contrast to the above narrow tube we are here interested in the case that R is large. A formal asymptotic solution of this boundary value problem for large κr was calculated by Concus [11] by using a boundary layer technique which goes back to Laplace [39]. The idea is to assume that there is a central core region covering most of the base domain in which w is small, and a boundary layer region near the wall in which w increases rapidly to its given boundary value. Matching the core and the boundary layer solutions in the transition circle determines the thickness of the boundary layer. This method was used by Perko [55] to prove that a certain boundary layer approximation is asymptotically correct. More precisely, it is shown that for each given boundary contact angle γ away from the critical angles γ = 0 or γ = π the relative error in in the ordinate and slope of this boundary-layer approximation is uniform of order 1/R ln(1/r) as R for 0 < r < R. A formal second order boundary layer approximation was calculated by Rayleigh [58]. Away from the boundary, for example in 0 < r < R (5/2)lnR, the slope and the ordinate decrease exponentially as R, see Siegel [68]. We will prove an explicit asymptotic formula when κr tends to infinity. It turns out the expected result that the leading term defines the capillary surface over the half plane with the given boundary contact angle. More precisely, let w(r; R, γ, κ) be the solution of the above boundary value problem and v(s; γ, κ) the solution of the following boundary value problem ( v = κv on 0 < s < 1 + v 2) lim s +0 v 1 + v 2 = cos γ, which defines the capillary surface over a half plane with the same boundary contact angles, see Section Proposition 6.3. w(r; R, γ, κ) v(r r; γ, κ) 2.1 π 2 γ 1 κr uniformly in r [0, R] and γ [0, π], provided that κr 6.4. In particular, 2 w 1 1 sinγ 2.1 π κ 2 γ 1 κr,

163 6.2. APPLICATIONS 163 where w 1 = w(r; R, γ, κ) is the ascent of the capillary surface at the boundary of the tube. Proof. See [47]. The proof is based on the comparison principle and on a mapping which brings the right parameter onto the right place of the equation. Instead of the above boundary value problem we consider the normalized problem ( ) ρu (ρ) = ρu(ρ), 0 < ρ < M 1 + u (ρ) 2 where lim ρ M 0 u (ρ) 1 + u (ρ) 2 = cos γ, u (0) = 0, M = κr, ρ = κr, u(ρ) = κw(r). Then v(s) = u(m s; M, γ) solves the boundary value problem ( ) 1 (M s)v (s) = v(s), 0 < s < M M s 1 + v (s) 2 lim s +0 v (s) 1 + v (s) 2 = cos γ, v (M) = 0. This boundary value problem becomes singular if s = M. Then we find an approximate solution v n (s; M, γ) in powers of 1/M k, k 0, such that satisfies, see [47], Using the comparison function U n (x) := v n (M x ; M, γ) div TU n U n c n+1 M n+1 in B M (0) lim ν TU n = cos γ on B M (0). x M 0 we obtain U + = U n + c n+1 M n+1, div TU + U + 0 in B M (0) lim ν x M 0 TU+ = cos γ on B M (0).

164 164 CHAPTER 6. ASYMPTOTIC FORMULAS Since the function U(x) = u( x ; M, γ) satisfies div TU U = 0 in B M (0) lim ν TU x M 0 = cos γ on B M(0), the comparison principle (Theorem 6.1) implies that Analogously the comparison function U(x) U + (x) in B M (0). U = U n c n+1 M n+1 yields a lower bound of U(x) in B M (0). Remark. In fact it can be shown, see [47], that there exists a complete asymptotic expansion in powers of 1/M k, k 0. More precisely, there is an approximate solution of the type v n (s; M, γ) = n k=0 φ k (s; M, γ) 1 M k, where ψ k, k 1, are defined through linear boundary value problems, see [47]. Then for given integer n 0 there exists a constant c n+1 such that u(ρ; M, γ) v n (M ρ; M, γ) c n+1 M n+1 uniformly in ρ [0, M] and γ [0, π] where u(ρ; M, γ) is defined in the previous sketch of proof Ascent at a needle We seek a surface S: U = U(x), x = (x 1, x 2 ), defined over the base domain Ω := R 2 \B a (0), where B a (0) is a disk with (small) radius a and the center at x = 0, such that U satisfies the nonlinear elliptic boundary value problem. div TU = κ U in Ω, ν TU = cos θ on Ω, where TU = U 1 + U 2,

165 6.2. APPLICATIONS 165 κ = ρg/σ (ρ = density change across free surface, g = gravitational acceleration, σ = surface tension) is the (positive) capillary constant. It is assumed that the gravity is positive and directed downwards. Further, θ in 0 θ π denotes in this subsection the constant contact angle between the capillary surface and the cylinder with cross section B a (0). The vector ν is the exterior unit normal on Ω, that is, the interior normal on B a (0). No explicit solution of the above boundary value problem is known. It was shown by Johnson and Perko [36] that there exists a radially symmetric solution. From a maximum principle of Finn and Hwang (Theorem 6.2) for unbounded domains it follows that this symmetric solution is the only one. Set u(r) = U(x), r = x x2 2. Then we have, see [49], Proposition 6.4. Let B = κa 2 and γ = Euler s constant. Then the ascent u(a) of a liquid on a circular needle with radius a satisfies u(a) a = cos θ as B 0, uniformly in θ [0, π]. ( ) 1 2 lnb + γ 2 ln2 + ln(1 + sinθ) + O(B 1 5 ln 2 B) Uniformly means that the remainder satisfies O(B 1 5 ln 2 B) cb 1 5 ln 2 B for all 0 < B B 0, B 0 sufficiently small, where the constant c depends on B 0 only and not on the contact angle θ. It is noteworthy that the special nonlinearity of the problem implies that the expansion is uniform with respect to θ [0, π] although Du tends to infinity as θ 0 or θ π and therefore the differential equation will be singular on Ω. Moreover, we do not need any growth assumption at infinity, which is a further consequence of the strong nonlinearity of the problem. In the case of complete wetting, i. e. if θ = 0, the formula u(a) a ( ) 1 lnb as a 0 has been derived formally by Derjaguin [17] by matching of expansions. We recall that B = κa 2. Higher order approximations where obtained formally by James [35] and Lo [40], also by matching arguments. Turkington [73] has proved that u(a) 1 2 cos θ a lnb as a 0.

166 166 CHAPTER 6. ASYMPTOTIC FORMULAS The idea of the proof of Proposition 6.4 is as follows. Since the solution is radially symmetric, we have the boundary value problem ( ) 1 ru (r) = κu(r) in a < r <, r 1 + (u (r)) 2 lim r a+0 u (r) 1 + (u (r)) 2 = cos θ. Set r = as, v(s) = 1 a u(as), B = κa2, then the above problem is changed to a problem where the unknown function is defined on a fixed interval: ( ) 1 sv (s) = Bv(s) in 1 < s <, s 1 + (v (s)) 2 lim s 1+0 v (s) 1 + (v (s)) 2 = cos θ. Then we can find an upper and a lower C 1 -solution of this boundary value problem. We obtain the lower and the upper solution by gluing together a boundary layer expansion near the boundary s = 1 with a second expansion which is valid far from the boundary such that the resulting function is in C 1. This method of composing of functions defined on different annular domains was used in [46], where a numerical method for the circular tube was proposed, see Section Narrow tube of general bounded cross section, interior ascent Here we consider capillary tubes with general cross section which is constant along the cylinder. In contrast to the case of circular cross section the situation is more delicate. Let Ω R 2 be a bounded domain with a piecewise C 2,λ boundary and u a solution of the boundary value problem div T u = Bu in Ω (6.10) ν Tu = cos γ (6.11) on the smooth parts of Ω, where Tu = u/ 1 + u 2, B is a positive constant (Bond number), γ, 0 γ < π/2, denotes the constant contact

167 6.2. APPLICATIONS 167 angle between the capillary surface and the container wall, and the vector ν is the exterior unit normal on the smooth parts of Ω. A formal ansatz u = C B + w 0(x, γ) + w 1 (x, γ)b +... in powers of small B, where C is a constant, leads to div Tw 0 = C in Ω (6.12) ν Tw 0 = cos γ (6.13) on the smooth parts of Ω. Suppose a sufficiently regular w 0 is a solution of (6.12), (6.13) then we find after integration of equation (6.12) over Ω that C = Ω Ω cos γ. (6.14) A solution of (6.12), (6.13) where C satisfies (6.14) is said to be a zero gravity solution. Under the assumption that Ω C 2,λ and that there is a zero gravity solution Siegel [69] has shown that u(x, γ, B) = Ω cos γ Ω B + w 0(x, γ) + O(B) for each fixed γ, 0 < γ < π/2, as B tends to zero. Here w 0 (x, γ) denotes the zero gravity solution which satisfies the side condition Ω w 0 (x, γ) dx = 0. In the following we suppose that Ω is a bounded domain with a C 2,λ boundary or with a boundary consisting of a finite number of C 2,λ curves Σ 1,...,Σ M and that Σ l and Σ l+1 meet at an interior angle 2α l such that 0 < α l < π/2, α l + γ > π/2. Set k l = sinα l / cos γ, define ω l, 0 < ω l < π/2, through tan ω l = 1 tan α l /k 2 l, and suppose that 2ω l < π/2. Then we get, see [44],

168 168 CHAPTER 6. ASYMPTOTIC FORMULAS Proposition 6.5. Suppose there exists a zero gravity C 2,λ (Ω) solution. Then there are n + 1 functions w l C 2,λ (Ω), l = 0,...,n, analytical in Ω and depending on γ, such that u(x, γ, B) = Ω cos γ Ω B + w 0(x, γ) + n w l (x, γ)b l + O(B n+1 ) l=1 uniformly in Ω and for each fixed γ, 0 < γ < π/2, as B tends to zero. Remark. In the case that there is no classical zero gravity solution the asymptotic behaviour as B 0 is more complicated, see [21], Chapter 6, [71] and [66]. In the following two examples there exists a zero gravity solution which can be written in terms of elementary functions, elliptic integrals or elliptic functions Annulus Let Ω = {x R 2 : q < x < 1} for a fixed q, 0 < q < 1. From the comparison principle Theorem 6.1 it follows that each solution is radially symmetric and uniquely determined. A radially solution exists, see [36]. Set r = x, then r w0(r) = c q s 2 q s 2 c 2 (s 2 q) 2 ds, where c = cos γ/(1 q), is a zero gravity solution. This integral can be expressed in terms of elliptic integrals, see [9], formulas and The asymptotic expansion is given by u(x, γ, B) = 2 cos γ n (1 q)b + w l (x, γ)b l + O(B n+1 ) l=0 as B 0, where w l (x, γ) are radially symmetric functions and w 0 is given by w 0 (x, γ) = w 0 (r) + C 0, where C 0 = q 2 q rw 0(r) dr. This integral can be expressed, after integration by parts, by elliptic integrals and Jacobian elliptic functions, see [9], formulas and

169 6.2. APPLICATIONS 169 Regular n-gon Let Ω be a regular n-gon, where the corners lie on the unit circle. It is easily seen that the lower hemisphere defined by w 0(x, γ) = 1 H 1 H 2 x 2, where H = cos γ/cos(π/n), is a solution of the zero gravity problem, which is analytical on Ω, provided that H < 1, or equivalently, that α l + γ > π/2 is satisfied. The assumption 2ω l < π/2 holds in the cases n = 3 and n = 4 for each γ, 0 < γ < π/2. If n 5, then 2ω l < π/2 is a condition on the boundary contact angle γ. Then, according to Proposition 6.5, we get the asymptotic formula u(x, γ, B) = 2 cos γ n B cos(π/n) + w l (x, γ)b l + O(B n+1 ), l=0 where w 0 (x, γ) = w0 (x, γ) + C 0, with a constant C 0 such that w 0 (x, γ) dx = 0. After some calculation we find that ( 2π C 0 = 3n sin(π/n)cos 3 1 n γ π Ω π/n 0 ( ) 3/2 1 cos2 γ dθ) cos 2. θ This integral can be expressed in terms of elementary functions Ascent in a wedge In 1712 Brooke Taylor [72] conjectured that the trace of the capillary interface on the walls of a narrow wedge is a hyperbola, see also Figure 1.4 of the introduction. Here we consider the boundary value problem div T u = κu in Ω (6.15) ν Tu = cosγ (6.16) on the smooth parts of Ω,where κ = const. > 0 and ν is the exterior unit normal on the smooth parts of Ω. Let x = 0 be a corner of Ω with the interior angle 2α satisfying 0 < 2α < π. For simplicity we assume that the

170 170 CHAPTER 6. ASYMPTOTIC FORMULAS corner is bounded by lines near x = 0 and that Ω R = Ω B R coincides with the circular sector {x R 2 : x 1 tan α x 2 <} B R for a sufficiently small R, where B R denotes the disk with center at the origin and radius R, see Figure 6.5. Furthermore, we assume that the contact x 2 ν α Ω R ν α R x 1 Figure 6.5: Corner in consideration angle satisfies 0 γ < π/2. Concus and Finn [12, 13] have shown that every solution of (6.15), (6.16) is bounded near the corner if and only if α + γ π/2 is satisfied and that in the unbounded case α + γ < π/2 u(x; α, γ) = h 1(θ; α, γ) r + O(1) as r 0, where r, θ are polar coordinates centered at x = 0, and h 1 is defined by h 1 (θ; α, γ) = cos θ k 2 sin 2 θ κk with k = sin α/cos γ. In fact, there exists an asymptotic expansion of u in powers of r 4l 1, l = 0, 1,..., in particular as r 0. u(x; α, γ) = h 1(θ; α, γ) r + O(r 3 )

171 6.2. APPLICATIONS 171 Proposition 6.6 ([43]). For a given nonnegative integer m there exist positive constants r 0, A and m + 1 functions h 4l 1 (θ; α, γ), l = 0,...,m, analytical on (α, α) and bounded on [α, α], such that m u(x) h 4l 1 (θ; α, γ)r 4l 1 A r 4m+3 l=0 in Ω r0. Moreover, the constants r 0, A and the functions h 4l 1 (θ; α, γ) do not depend on the solution considered. Sketch of the proof. The proof is by induction. A function, we omit the arguments γ and α in the following, v n (x) := n h 4l 1 (θ)r 4l 1 l=0 is said to be an approximate solution if div Tv n κv n = O(r 4n+3 ) in Ω R ν Tv n cos γ = O(r 4n+3 ) on Σ R, as r 0, where Σ R = ( Ω Ω) \ {0}. (i) v 0 := h 1 (θ)r 1 is an approximate solution. This follows from a calculation and by exploiting the strong nonlinearity of the problem. In contrast to this estimate one gets v 0 r 3 as r 0. (ii) Assume v n is an approximate solution, then there exist positive constants A, r 0, independent on the solution considered, such that in Ω r0. u(x) v n (x) A r 4n+3 This estimate is a consequence of the comparison principle of Concus and Finn (Theorem 6.1). The comparison function used here is w = v n + Aq(θ)r 4n+3 with an appropriate function q(θ). (iii) Suppose that v m (x) is an approximate solution. Then there exist functions h 4m+3 (θ), analytical in ( α, α) and bounded on [ α, α], such that v m+1 (x) = v m (x) + h 4m+3 (θ)r 4m+3

172 172 CHAPTER 6. ASYMPTOTIC FORMULAS is an approximate solution. We seek a function q(θ) such that w = v m + q(θ)r 4m+3 is an approximate solution where n = m + 1. After some calculation one finds that q(θ) is a solution of an inhomogeneous linear boundary value problem for a second order ordinary differential equation. Then one shows that the associated homogeneous boundary value problem has the zero solution only. Thus there exists a unique solution of the inhomogeneous boundary value problem. To see that the homogeneous boundary value problem has the zero solution only we argue as follows. Suppose that there is a solution q 0 (θ) of the homogeneous boundary value problem, then we find another asymptotic formula u(x) = v m (x) + q 0 (θ)r λ lnr + O(r 4m+3 ) as r 0. Since also holds, we get q 0 (θ) = 0 on [ α, α]. u(x) = v m (x) + O(r 4m+3 ) The formula u = h 1 (θ; α, γ)/r 1 + O(1) as r 0 implies a formula for the area of the capillary surface near the corner. Set Ω r,r0 = Ω r0 \Ω r, 0 < r < r 0. Proposition 6.7. Let 0 < 2α < π, 0 γ < π/2 and α + γ < π/2. Then 1 + u 2 dx = 2 ( π ) Ω r,r0 κ 2 α γ ln r + O(1) as r 0. The remainder O(1) is independent on the solution considered and on γ, 0 < γ γ 0, where α + γ 0 < π/2. Proof. From the differential equation we get u div Tu dx = κ u 2 dx. Ω r,r0 Ω r,r0 Integration by parts implies u Ωr,r0 2 dx + u ν Tu ds = κ u 1 + u 2 Ω 2 dx. r,r0 Ω r,r0 Then Ω r,r0 1 + u 2 dx = J 1 + J 2 + J 3,

173 6.2. APPLICATIONS 173 where dx J 1 = Ω r,r0 1 + u 2 J 2 = u ν Tu ds Ω r,r0 J 3 = κ u 2 dx. Ω r,r0 Set h(θ) := h 1 (θ; α, γ), then we find from the boundary condition and the asymptotic formula u(x) = h(θ)r 1 + O(1) as r 0 that The asymptotic formula implies Summarizing, we obtain J 2 = 2h(α)(ln r 0 lnr)cos γ = O(1). J 3 = κ(lnr lnr 0 ) Ω r,r0 1 + u 2 dx = α α h 2 (θ) dθ + O(1). ( α ) 2h(α)cos γ κ h 2 (θ) dθ ln r + O(1). α After an easy calculation we arrive at the formula of the proposition. From an exercise it follows that the remainder is uniformly bounded if γ 0. Remark. Concerning the existence of a solution see [20] when α +γ > π/2 and [24] if α + γ π/2. The case of different contact angles on the walls of the wedge was discussed in [10]. See also Lancaster [37] where a conjecture of Concus and Finn was confirmed. This conjecture says that solutions are discontinuous near the corner if the contact angles γ 1, γ 2, 0 < γ i < π, satisfy 2α < γ 2 γ 1 < π. The case of the ascent of liquid along wedges over cusps was firstly considered in Scholz [65, 66] A numerical method Here we consider rotationally symmetric capillary problems in the presence of gravity. By composing asymptotic expansions defined on different annular subdomains one obtains a numerical procedure for the calculation the capillary surface in a circular tube or between two coaxial circular tubes. Explicit error estimates uniform with respect to the boundary contact angle

174 174 CHAPTER 6. ASYMPTOTIC FORMULAS are given. The proof of this estimate is based on the comparison principle of Concus and Finn (Theorem 6.1). The main idea of the method can be extended to the capillary tube problem with general cross section. An open problem is here to find error estimates. Let Ω be the annular domain B = {x R 2 : r 0 < x x2 2 < 1}, where r 0 0. The solution is radially symmetric and satisfies the onedimensional boundary value problem ( ) ru (r) = Bru(r) in r 0 < r < 1 (6.17) 1 + u (r) 2 lim r 1 0 u (r) 1 + u (r) 2 = cos γ (6.18) u (r) lim = cos γ, (6.19) r r u (r) 2 where γ and γ are the given constant contact angles at the container walls. In the case of a capillary tube, i. e. if r 0 = 0, then γ = π/2, or the boundary condition at r 0 has to be replaced by u (0) = 0. Even in this radially symmetric case no explicit solution is known except u 0 in the case γ = π/2. For an integer n 2 let 0 r 0 < r 1 <... < r n 1 < r n = 1 be a subdivision of the interval r 0 < r < 1. On each subinterval I k : r k 1 < r < r k, 1 k n, we consider the boundary value problem ( ) ru (r) = Bru(r) in r k 1 < r < r k, (6.20) 1 + u (r) 2 u (r) 1 + u (r) 2 = { cos γk 1 if r = r k 1 cos γ k if r = r k (6.21) for angles 0 γ k π, 0 k n, where γ 0 = π γ, see Figure 6.6. For given angles γ k 1, γ k there exists a unique solution of the problem (6.20), (6.21). Concerning the question of existence of rotationally symmetric solutions see [36]. Set a k = cos γ k and let u k (r) = u(r k 1, r k, a k 1, a k, r) be the solution of (6.20), (6.21). By the same reasoning as in [68, 69] one obtains an asymptotic formula or even a complete asymptotic expansion of the solution u k (r) with respect to h k = r k r k 1 as h k 0, see [44, 45].

175 6.2. APPLICATIONS 175 γ γ k 1 γ k γ 0 γ r r 0 k 1 r k r n =1 Figure 6.6: Subdivision, notation Moreover, for fixed r k 1, r k this expansion is uniform with respect to the angle coordinates a k 1, a k from the closed interval [ 1, 1]. This follows as in [45]. The finite asymptotic sum is given by u (m,h,b) (r k 1, r k, a k 1, a k, r) = C 1(r k 1, r k, a k 1, a k ) h k B m + φ l (r k 1, r k, a k 1, a k, r r k 1 )B l h 2j+1 h k, k l=0 where the constant C 1 and the functions φ l are defined by a recurrent system of boundary value problems, see [44]. In particular, C 1 and φ 0 are given by the following formulas. Set v = r k 1, w = r k, x = a k 1 and y = a k, then Let C 1 (v, w, x, y) = 2v v + w x + 2w v + w y. (6.22) f(v, w, x, y, ζ) = vx + ( vζ + (w v)ζ 2 /2 ) C 1 (v, w, x, y) v + (w v)ζ f(v, w, x, y, ζ) q 0 (v, w, x, y, ζ) = 1 f 2 (v, w, x, y, ζ),

176 176 CHAPTER 6. ASYMPTOTIC FORMULAS then φ 0 (v, w, x, y, ζ) = ζ 0 q 0 (v, w, x, y, τ) (1 ) 2vτ + (w v)τ2 v + w q 0 (v, w, x, y, τ) dτ. (6.23) This sum defines an approximate solution of the boundary value problem (6.20) (6.21) in the sense that the sum satisfies the boundary condition (6.21) and solves the differential equation (6.20) up to a function which can be estimated by a positive power of Bh 2 k, see [44, 45]. We define the composed function U (m,h,b) (r, a, r) by U (m,h,b) (r, a, r) = u (m,h,b) (r k 1, r k, a k 1, a k, r), where r I k, r = (r 0, r 1,...,r n 1, r n ) and a = (a 0, a 1,...,a n 1, a n ) with a 0 = cos γ, a n = cos γ. This function U (m,h,b) (r, a, r) is in C 1 [0, 1) if and only if the following system of n 1 nonlinear equations, k = 1, 2,...,n 1, is satisfied. C 1 (r k 1, r k, a k 1, a k ) h k B + m j=0 = C 1(r k, r k+1, a k, a k+1 ) h k+1 B + dτ φ j (r k 1, r k, a k 1, a k, 1)B j h 2j+1 k (6.24) m j=0 φ j (r k, r k+1, a k, a k+1, 0)B j h 2j+1 k+1. From now on we will consider the case m = 0, and equidistant subdivisions r k = hk, where h = 1/n. Under these assumptions there exists a solution of the system (6.24) if, roughly speaking, a smallness condition is satisfied for the product Bh. More precisely, set η(h, cos γ) = 1 1 cos γ + 1 cos γ + h(1 h)cos γ, (6.25) then one concludes from formulas (6.22) and (6.23) which define C 1 and φ 0, that there exists a solution 0 a 1 a 2... a n 1 cos γ of the system (6.24) if 6Bη(h, cos γ)h 2 1 (6.26) holds. Set Tu = u/ 1 + u 2. The function Û(0,h,B)(a, x) = U (0,h,B) (r, a, r), where r = x r2 2, satisfies on the union of the annular domains {r : (k 1)h < r < kh}, k = 1,...,n,

177 6.2. APPLICATIONS 177 the inequality div T Û (0,h,B) BÛ(0,h,B) c 1 Bh, (6.27) where c 1 = 4η(h, cos γ). This estimate follows from the special properties of C 1 and φ 0, defined by (6.22) and (6.23). If there exists a solution of the nonlinear system (6.24) of the n 1 equations then Û(0,h,B) C 1 in the open annular domain Ω. Under the assumption Û(0,h,B) C 1 (B 1 ) one concludes by the same reasoning as in [44, 45] from (6.27) and the comparison principle of Concus and Finn in its weak form (Theorem 7.1 in Finn [21]), an error estimate. We apply this comparison principle to the comparison functions U ± = Û(0,h,B)±c 1 h. Since div TU + BU + 0 in Ω and lim x 1 ν TU + = cos γ on Ω. Then u U + holds in Ω, where u denotes the solution of the nonlinear boundary value problem (6.17) (6.18). By the same reasoning, the comparison function U yields a lower bound for w. Thus the above considerations are a sketch of Proposition 6.8 ([46]). Suppose that 6Bη(h, cos γ)h 2 1, where η is defined by (6.26). Then there exists a solution 0 a 1 a 2... a n 1 cos γ of the system (6.24), when m = 0 and r k 1 r k = h = 1/n, such that max U (0,h,B) (a, r) u(r) 4η(h, cos γ)h holds. 0 r 1 The crucial point here is that this estimate yields an explicit error estimate of order O( h) uniform with respect to the boundary contact angle γ in 0 γ π/2 despite the fact that u becomes unbounded if γ 0 or γ π. The reason for this behaviour is the strong nonlinearity of the problem. There is no counterpart for linear problems. Remark. It is worthy to note that already Runge [62] proposed a scheme for the calculation of the solution of (6.17)-(6.19) in Another method was presented by Concus and Pereyra [14]. Indeed, no error estimate uniform in γ was known. Numerical experiments. In the case of first order approximation, i. e., if m = 0, we solved numerically the nonlinear system (6.24) with Newton iteration by using standard software packages, see below. Let r = (r 0, r 1,...,r n ) with a = (a 0, a 1,...,a n 1, a n ) with a 0 = cos γ and a n = cosγ. Set z = (a 1,...,a n 1 ) and let m = 0 in (6.24). Then the system (6.24) can be

178 178 CHAPTER 6. ASYMPTOTIC FORMULAS written as Mz = Be(r, a)+p, where B = Bond number, M is the tridiagonal (n 1, n 1) matrix with entries m kl, k, l = 1, 2,...,n 1, defined by m ii = 2r i(r i 1 r i ) r i 1 + r i + 2r i(r i r i 1 ) r i + r i+1, m i,i+1 = 2r i+1(r i r i 1 ) r i + r i+1, m i+1,i = 2r i(r i+2 r i+1 ) r i + r i+1, m kl = 0 else. The (n 1) vector e is given by the coordinates e k (r, a) = (r k 1 r k ) 2 (r k r k 1 )φ 0 (r k, r k 1, a k, a k 1, 0) (r k r k 1 ) 2 (r k+1 r k )φ 0 (r k 1, r k, a k 1, a k, 1), k = 1,...,n 1. The coordinates of the (n 1) vector p are p n 1 = 2 cos γ(r n 1 r n 2 )/(r n 1 +1) and p k = 0 if k = 1,...,n 2. The associated Newton iteration scheme is defined by ( ) M BDe(r, a (n) ) z (n+1) = p + B (e(r, a (n) ) De(r, a (n) ) z (n)) with an initial vector a (0) defined by a (0) k = (k/n)cos γ, k = 0, 1,...,n, for example. Here De denotes the Jacobian matrix of e(r, a) with the entries e k / a l, k, l = 1,...,n 1, De is a tridiagonal matrix. Replacing the matrix De(r, a (n) ) by De(r, a (0) ), one obtains a simplified Newton scheme which runs faster. Figures 6.7 show composed functions in the case of a capillary tube with 10 equidistant subdivisions of the interval (0, 1), B = 13.4 and cos γ = 0.99, after one and two iterations, resp. Mathematica program The following Mathematica program concerns the above numerical procedure. It solves numerically the boundary value problem, see Figure 6.8 for

179 6.2. APPLICATIONS Figure 6.7: Composed functions after one and two Newton iterations notations, ( ) ru (r) 1 + (u (r)) 2 u (r) 1 + (u (r)) 2 u (r) 1 + (u (r)) 2 = κ r u(r) in rin < r < rout = cos γ 1, at r = rin = cos γ 0, at r = rout. Some notations concerning the program: kappa: capillary constant, ar0:= cosθ 1, θ 1 contact angle at the interior cylinder, arn:=cosθ 0, θ 0 contact angle at the exterior cylinder, rin: radius of the interior cylinder, rout: radius of the exterior cylinder, run: number of Newton iterations. Hints how to use the program: loading the program in Mathematica, then type alg. After running type hc, for example, and one gets the ascent at the interior cylinder, heighto gives the ascent at the exterior cylinder. Remark. The program includes the case of a capillary surface in a circular cylinder. In this case one has rin = 0 and θ 1 = π/2.

180 180 CHAPTER 6. ASYMPTOTIC FORMULAS g θ 1 θ 0 rin rout r Figure 6.8: Capillary interface between two coaxial cylinders b:=kappa rout 2 kappa:=13.4 arn:=0 ar0:= run:=5 n1:=100 n2:=50 n:=n1 + n2 rin:=0.1 rout:=0.3 rout0:=0.3 r0:=rin/rout k0:=0.8 a1[i ]:= 2 (r[[i + 1]] r[[i]]) r[[i + 2]]/(r[[i + 1]] + r[[i + 2]]) b1[i ]:=2 (r[[i + 2]] r[[i + 1]]) r[[i + 1]]/(r[[i]] + r[[i + 1]])+ 2 (r[[i + 1]] r[[i]]) r[[i + 1]]/(r[[i + 1]] + r[[i + 2]]) c1[i ]:= 2 (r[[i + 3]] r[[i + 2]]) r[[i + 1]]/(r[[i + 1]] + r[[i + 2]]) m:=table[switch[i j, 1, a1[i], 0, b1[i], 1, c1[i 1],, 0], {i, n 1}, {j, n 1}] p:=join[{ar0 r[[1]] 2 (r[[3]] r[[2]])/(r[[1]] + r[[2]])}, Table[0, {n 3}], {arn 2 (r[[n]] r[[n 1]]) r[[n + 1]]/(r[[n]] + r[[n + 1]])}]

181 6.2. APPLICATIONS 181 c[u, v, x, y ]:=2 ( u x + v y)/(u + v) f[u, v, x, y, t ]:=(u x + c[u, v, x, y] (u t + 1/2 (v u) t 2))/(u + (v u) t) q[u, v, x, y, t ]:=f[u, v, x, y, t]/sqrt[1 f[u, v, x, y, t] 2] w0[u, v, x, y ]:=NIntegrate[q[u, v, x, y, s] ( 1 +(2 u s + (v u) s 2)/(u + v)), {s,0, 1}, AccuracyGoal->10] w1[u, v, x, y ]:=NIntegrate[q[u, v, x, y, s] (2 u s + (v u) s 2)/(u + v), {s, 0, 1}, AccuracyGoal->10] e[r, a ]:=Table[(r[[k + 2]] r[[k + 1]]) 2 (r[[k + 1]] r[[k]]) w0[r[[k + 1]], r[[k + 2]], a[[k + 1]], a[[k + 2]]] (r[[k + 1]] r[[k]]) 2 (r[[k + 2]] r[[k + 1]]) w1[r[[k]], r[[k + 1]], a[[k]], a[[k + 1]]], {k, n 1}] dw0[l, u, v, x, y ]:=NIntegrate[dq[l, u, v, x, y, s] ( 1 + (2 u s+ (v u) s 2)/(u + v)), {s,0, 1}, AccuracyGoal->10] dw1[l, u, v, x, y ]:=NIntegrate[dq[l, u, v, x, y, s] (2 u s +(v u) s 2)/(u + v), {s,0, 1}, AccuracyGoal->10] dq[1, u, v, x, y, t ]:=(1 f[u, v, x, y, t] 2) ( 3/2) (u 2 u (u t + 1/2 (v u) t 2)/(u + v))/(u + (v u) t) dq[2, u, v, x, y, t ]:=(1 f[u, v, x, y, t] 2) ( 3/2) 2 v (u t + 1/2 (v u) t 2)/((u + v) (u + (v u) t)) a2[r, a, i ]:=(r[[i + 2]] r[[i + 1]]) 2 (r[[i + 1]] r[[i]]) dw0[2, r[[i + 1]], r[[i + 2]], a[[i + 1]], a[[i + 2]]] b2[r, a, i ]:=(r[[i + 2]] r[[i + 1]]) 2(r[[i + 1]] r[[i]]) dw0[1, r[[i + 1]], r[[i + 2]], a[[i + 1]], a[[i + 2]]] (r[[i + 1]] r[[i]]) 2(r[[i + 2]] r[[i + 1]]) dw1[2, r[[i]], r[[i + 1]], a[[i]], a[[i + 1]]]

182 182 CHAPTER 6. ASYMPTOTIC FORMULAS c2[r, a, i ]:= (r[[i + 2]] r[[i + 1]]) 2 (r[[i + 3]] r[[i + 2]]) dw1[1, r[[i + 1]], r[[i + 2]], a[[i + 1]], a[[i + 2]]] de[r, a ]:=Table[Switch[i j, 1, a2[r, a, i], 0, b2[r, a, i], 1, c2[r, a, i 1],, 0], {i, n 1}, {j, n 1}] a0:=table[r[[k]] ar0, {k,1, n + 1}] r1:=table[n[r0 + (k/n1)(k0 Sqrt[r0] r0)], {k,0, n1}] r2:=table[n[k0sqrt[r0] + (k/n2)(1 k0 Sqrt[r0])], {k, n2}] r:=join[r1, r2] z:=take[a, {2, n}] alg:={a = a0; Do[Print[a]; a = Flatten[Join[{ar0}, LinearSolve[m b de[r, a], p + b (e[r, a] de[r, a].z)], {arn}]], {run}]} wab[u, v, x, y, t ]:=NIntegrate[q[u, v, x, y, s], {s,0, t}]+ NIntegrate[q[u, v, x, y, s] ( 1 + (2 u s + (v u) s 2)/(u + v)), {s,0, 1}] fab[u, v, x, y, t ]:=1/b c[u, v, x, y]/(v u) + (v u) wab[u, v, x, y, t] fun[l, r, a, t ]:=Which[r[[l]]<=t < r[[l + 1]], fab[r[[l]], r[[l + 1]], a[[l]], a[[l + 1]], (t r[[l]])/(r[[l + 1]] r[[l]])], True, 0] func[r, a, t ]:=Sum[fun[l, r, a, t], {l, n 1}] + Which[r[[n]]<=t<=1, fab[r[[n]], r[[n + 1]], a[[n]], a[[n + 1]], (t r[[n]])/(r[[n + 1]] r[[n]])], True, 0] func1[rout, r, a, s ]:=rout func[r, a, s/rout] hc:=func1[rout, r, a, rin] heighto:=func1[rout, r, a, rout] ho0:=func1[rout, r, a, rout0] heightm:=hc heighto fig1:=plot[func1[rout, r, a, s], {s, rin, rout}, AxesOrigin->{rin, 0},

183 6.2. APPLICATIONS 183 PlotRange->{{rin, 0.4}, {0, 0.4}}, AspectRatio->1.1] General cross section In the presence of gravity, the equilibrium free surface S : u = u(x 1, x 2 ) of a liquid inside a tube satisfies the nonlinear elliptic Neumann problem div ν u 1 + u 2 u 1 + u 2 = Bu in Ω, (6.28) = cos γ on Ω, where Ω R 2 is the cross section of the tube, B the Bond number which is a positive constant, and γ in 0 γ π denotes the constant contact angle between the capillary surface and the cylinder wall. The vector ν is the exterior unit normal at Ω. Thus the problem of finding a capillary surface is a geometric one: to find a surface whose mean curvature is a prescribed function of position and which meets prescribed boundary walls in a prescribed angle γ, see a remark in Finn [21], p.12. We recall that div Tu 2H(x), where H(x) is the mean curvature of the surface S: z = u(x) at (x, u(x)). This observation leads to the following scheme for the numerical calculation of the capillary surface. Consider a triangulation of a two dimensional domain Ω with, for simplicity, a polygonal boundary. We assume that adjoining triangles join either a single corner or an entire side. Let z k (x) be a hemisphere over a ball B rk with radius r k. By ν k we denote the outer unit normal at the boundary Ω k \{corners} of the triangle Ω k. Suppose that (i) Ω k B rk for each triangle Ω k. (ii) ν k Tz k = ν l Tz l on Ω k Ω l when this set is a common side of the two triangles Ω k, Ω l. (iii) 2/r k = Bv k (x (k) ) for one point x (k) Ω k. (iv) If Ω k Ω l is a common side, then z k (x (k) ) = z l (x (k) ) for an x (k) Ω k Ω l, for example, for the center of the common side. (v) ν k Tz k = cos γ on Ω Ω k when this set is a side of Ω k.

184 184 CHAPTER 6. ASYMPTOTIC FORMULAS The assumptions (ii) (v) define a nonlinear system of equations where the unknowns are the boundary contact angles of the spheres with the plane perpendicular to the (x 1, x 2 ) plane which contains the common side of two triangles. From equation (6.28), (i) and (iii) one concludes that div Tv k Bv k B 2 diam(ω k) in Ω k. A capillary surface over a quadrangle was calculated with this method (200 triangles, 280 equations) on a microcomputer, see [46]. Remark. Probably the first scheme for the calculation of capillary surfaces over general domains was proposed in [33]. For schemes with software packages see [4, 8].

185 6.3. PROBLEMS Problems 1. Show that φ 0 (y; γ) = 2 1 sin 3 γ 3 cos r γ cos γ 2 cos 2 γ, which is defined in the theorem of Section 6.2.2, is uniformly bounded with respect to 0 < r < 1 and 0 γ π. 2. Show that the angle 2ω l, which is defined in Section 6.2.5, is the angle at the corner of the capillary surface about the related corner in the base domain. 3. Conjecture: Proposition 6.5 holds if γ = 0, provided the boundary of Ω is sufficiently smooth. 4. Write w0, defined in Section in the example of an annulus, in terms of elliptic integrals. 5. Write C 0, defined in Section in the example of an annulus, in terms of elliptic integrals and Jacobian elliptic functions. 6. Show that the infinite series l=0 h 4l 1(θ; α, γ)r 4l 1, see Proposition 6.6, can not converge pointwise to u(x). 7. Show that the asymptotic formula of Proposition 6.6 is uniform with respect to γ 0. More precisely, suppose that γ 0 satisfies α + γ 0 < π/2, then the constants A and r 0 are independent on γ, 0 < γ γ 0. Hint: Let h(θ; γ) := h 1 (θ; α, γ) and q(θ, γ) := h 4l 1 (θ; α, γ), l 1. Then sup q(θ, γ) < q θ (θ, γ) sup (h 2 (θ; γ) + h 2 θ (θ; γ))1/2 < sup q θθ (θ, γ) (h 2 (θ; γ) + h 2 θ (θ; γ))3/2 <, where the supremum is taken over θ ( α, α) and 0 < γ γ 0. Then follow calculations in [43].

186 186 CHAPTER 6. ASYMPTOTIC FORMULAS 8. Determine the solution of the ascent between two parallel plates, see Section 3.3.2, by using a numerical method adapted from Section 6.2.7, and discus the sign of F L x (d, γ w, γ l ), see Section (attraction/repelling of a plate). 9. Discuss the problem of surface structure and contact angle, see [50], pp. 15.

187 Chapter 7 Surface tension Some methods of measurement of surface tension are based on the equilibrium conditions for floating bodies. Assume a cylindrical homogeneous body Ω = D [0, L] with constant cross section D hangs vertically in a liquid which fills partly a container, see Figure 7.1, where F is the force directed upwards to keep the body in an equilibrium. We suppose that the F g Ω γ γ S general level h liquid Figure 7.1: Device for measurement of surface tension body can move in the x 3 -direction only and that no rotation is possible. Let z = v(x), x = (x 1, x 2 ), define the free surface S 0. Then, see the remark 187

188 188 CHAPTER 7. SURFACE TENSION below of Theorem 4.1, the equilibrium conditions are given by 2σH 0 + gρ 1 v(x) + λ 0 = 0 on S 0 cos γ 1 = β 1 on 1 S 0 cos γ 2 = β 2 on 2 S 0 σ ν 0, e 3 ds + gρ 1 y 3 (v)(x) N Σ0, e 3 da + gρ 2 Ω 2 S 0 W 2 (S 0 ) +λ 0 N Σ0, e 3 da = F 3. W 2 (S 0 ) The constant λ 0 can be calculated explicitly. It depends on the volume of the liquid and on the location of the body Ω. Set v = u λ 0 gρ 1, then u satisfies the equilibrium conditions 2σH 0 + gρ 1 u(x) = 0 on S 0 cos γ 1 = β 1 on 1 S 0 cos γ 2 = β 2 on 2 S 0 σ ν 0, e 3 ds + gρ 1 y 3 (v) N Σ0, e 3 da + gρ 2 Ω = F 3. (7.1) 2 S 0 W 2 (S 0 ) Here u denotes the ascent of the liquid from the general level which is approximately a horizontal plane in the case that body is sufficiently far away from the boundary of the container. Above of the (x 1, x 2 )-plane the ascent u(x) is close to zero, see estimates derived in [68]. Numerical calculations for rotationally capillary surfaces suggest that the error is much smaller. From formula (7.1) we get σ 2 S 0 cos γ 2 + gρ 1 h D + gρ 2 Ω = F 3, where h denotes the distance of the bottom of Ω from the general line which is positive if the bottom is above of the general line and negative if the bottom is below. Then σ = F 3 gρ 1 h D gρ 2 Ω 2 S 0 cos γ 2. (7.2) The force F = (0, 0, F 3 ) and h are known from measurements. In general, the contact angle γ 2 and 2 S 0 are unknown.

189 7.1. WILHELMY S PLATE METHOD Wilhelmy s plate method Here Ω is a thin plate where the cross section is a thin rectangle, see Figure 7.2. In the Wilhelmy method, see Wilhelmy [80], it is assumed that the F l L g h d Figure 7.2: Wilhelmy device for measurement of surface tension contact angle γ 2 is zero, and 2 S 0 is replaced by 2l+2d. Thus the Wilhelmy method yields an upper bound σ for the surface tension σ, where. σ = F 3 gρ 1 hld gρ 2 Ω. 2l + 2d 7.2 A rod method Here the plate is replaced by a rod with constant circular cross section, and the rod is placed in the middle of a container with constant circular cross section. Since S 0 is rotationally symmetric, we have 2 S 0 = 2πR, where R is the radius of the cross section of the rod. Suppose again that the contact angle γ 2 is zero, then. σ = F gρ 1hπR 2 gρ 2 πlr 2. 2πR

190 190 CHAPTER 7. SURFACE TENSION 7.3 Padday s method Here the capillary surface hangs on the lower edge of a rod with constant circular cross section, see Figure 7.3. F g Ω γ + ν 0 γ h γ 2 Figure 7.3: Padday s device for measurement of surface tension From formula (7.2) we see that σ = F 3 gρ 1 hπr 2 gρ 2 πlr 2 2πR sinγ 2. The contact angle γ 2 is not determined. Assume the capillary surface is in an equilibrium, then, see Theorem 5.1, γ 2 satisfies the inequalities γ γ 2 γ + + π 2, where γ, γ + denote the contact angle associated to the bottom and to the cylindrical surface, resp., of the rod. In [54] it is proposed to determine the surface tension from the maximum pull on the rod. 7.4 A new method In the following we consider a new method 1 how to find the surface tension without any assumption on the contact angle, see [53]. This parameter identification method is based on the measurement of the ascents e and i at the exterior and interior at a capillary tube, see Figure Patent application, January 19, 2011

191 7.4. A NEW METHOD 191 Consider a transparent tube with a constant circular cross section dipped into the middle of a container with a circular cross section which is filled with liquid. For the following notations see Figure 7.4. Let z S i g γ i i γe S e e γ R general level r i r e R 0 R r Figure 7.4: Ascents of liquid at a tube r e exterior radius of the tube, r i interior radius of the tube, R interior radius of the circular container, γ e contact angle at the exterior of the tube, γ i contact angle at the interior of the tube, γ R contact angle at the container wall. Suppose that the contact angles are constant and are in the closed interval [0, π/2]. The interior capillary surface S i above of the general level is defined by z = u i (r; γ i, κ), where u i is the solution of the Laplace boundary

192 192 CHAPTER 7. SURFACE TENSION value problem ( ) ru (r) = κ r u(r) in 0 < r < r i (7.3) 1 + (u (r)) 2 u (r) lim = cos γ i (7.4) r r i (u (r)) 2 u (0) = 0. (7.5) And the exterior surface S e over r e < x < R 0 is defined by w e = w e (r; γ e, κ), where w e is the solution of ( rw (r) 1 + (w (r)) 2 ) = κ r w(r) in r e < r < R 0 (7.6) lim r r e+0 w (r) 1 + (w (r)) 2 = cos γ e (7.7) w (R 0 ) = 0. (7.8) We fill the container with liquid up to the top such that the capillary surface hangs on the upper edge of the container and makes a contact angle γ R a little larger than π/2 with the interior container wall. The liquid remains in the container since the upper interior edge of the container is a wetting barrier, see Chapter 5. Let R 0 satisfies r e < R 0 < R. In examples we will pick an R 0 below of the horizontal part of the capillary surface, see Figure 7.4. Let i and e are the observed ascents measured from the general level, see Figure 7.4. Up to small errors which we can neclect, see [53] for details, we have i = u i (r i ; γ i, κ) e = w e (r e ; γ e, κ). These equations define functions κ i (γ i ) and κ e (γ e ). Assume that the interior and the exterior contact angle are the same, then we get from equation κ i (γ) = κ e (γ) both, the contact angle and the capillary constant. Here we have used that the surface tension of the capillary surface in the tube is equal to the surface tension of the capillary surface outside the tube since the same liquid is in and outside of the tube.

193 7.4. A NEW METHOD 193 Remark. Asymptotic formulas for the ascent in the interior of the tube, see [39, 68] and for the ascent at the exterior of the tube, see [40, 49], support the conjecture that the functions κ i (γ) is strictly concave from below and κ e (γ) is strictly convex from below. We solve the above boundary value problems (7.3) (7.5) and (7.6) (7.8) numerically with a method proposed in [46], see Section This procedure is a finite element method where the elements are capillary surfaces between narrow coaxially circular cylinders. This method admits an explicit error estimate of order O(h) uniformly in γ i, γ e, γ R [0, π], provided that κh 2 c with a known constant c, where h is the step size of the method. Remark. Asymptotic formulas for the ascent in the interior of the tube, see [39, 68] and for the ascent at the exterior of the tube, see [40, 49] are too rough and yield unrealistic results for the surface tension. Thus we have to solve the related boundary value problems numerically. Experiments with a glass tube show that we get realistic results. Figure 7.5 shows our device for measurements of the exterior and the interior ascents 2. We have used a glass tube of interior radius r i = cm and exterior radius r e = cm. The interior radius of the container is R = 5.5 cm. In the numerical calculations we set R 0 = 4.5 cm. Assume ρ = 10 3 kg/cm 3 for the density of the water which we use, and g = 9.81 m/s 2 for the local gravity. The temperature of the liquid and of the environment is about 20 o Celsius. After some time the surfaces are in a stationary position which is described by the Laplace boundary value problem. Distilled water. In this experiment we have used distilled water from a public market. We see from a photograph that approximately e = 0.25 cm and i = 1.1 cm. From numerical calculations we obtain that 13 cm 2 < κ < 14 cm 2 and 5 o < γ < 10 o. This result is in a good agreement with measurements with a Wilhelmy tensiometer which gives σ = 73.8 mn/m for the surface tension. Then we find for the capillary constant κ = gρ/σ = 13.3 cm 2 Distilled water with some surfactant. In this experiment we have used distilled water from a public market mixed with some drops of a washing-up liquid. We see from a photograph that approximately e = 0.19 cm and i = 0.49 cm. From numerical calculations we obtain that 34 cm 2 < κ < 2 This foto was taken from [53]

194 194 CHAPTER 7. SURFACE TENSION Figure 7.5: A device for measurement of surface tension 35 cm 2 and that the contact angle is about zero. Thus we have for the surface tension 28 mn/m < σ < 29 mn/m which is in agreement with measurements using a Wilhelmy plate tensiometer. Remark. Once one knows the surface tension then we find the contact angle on a wire or a rod with constant circular cross section from the ascent of liquid at the exterior of the related cylinder, see [53], p. 5, for an example.