Capillary surfaces and floating bodies

Abstract

We investigate capillary surfaces with boundary components on a floating body. The unknowns of this problem, the free surface and the position and orientation of the floating body are determined by minimizing the corresponding energy; besides gravitation and cohesion forces, we consider also adhesion between the fluid and both the wall of the container and the floating body. Existence of a solution is shown in the class of Caccioppoli sets under suitable restrictions on the data, as well as for surfaces that are graphs of real functions.

1 Introduction

We consider a cylindrical container $G:= \Omega \times \mathbb{R }^+$, where $\Omega$ is a bounded, simply connected domain in $\mathbb{R }^2$; G is partially filled with some liquid of density $\rho$. A rigid body $\mathcal{B }$ whose density ${\rho }_0$ is smaller than $\rho$ is floating on the liquid, and it is assumed that the interface between the liquid and the air above is governed by surface tension. Besides the gravitational force, there is adhesion between the fluid and both the outer wall of the container and the floating body. We then look for configurations in which these forces are in equilibrium; the unknowns are the position of the floating body and the capillary surface, and they are determined by a variational problem for the energy.

A special version of this problem has been studied in [1]; as [1] is part of an investigation of free-boundary problems for the Navier–Stokes equations, we allowed only for contact angles of $\pi /2$ because in domains with edges of this type solutions to the Navier–Stokes equations are regular up to the boundary. Such a contact angle occurs only if there is no adhesion energy. In the present paper, we prove the existence of capillary surfaces under the standard assumptions, in particular with non-vanishing adhesion energy.

In Sect. 2, we study the energy functional for the case that the region occupied by the fluid is a Caccioppoli set. Let $\mathcal{B }$ denote the rigid body, and let $\mathcal{B }(c,R):\, = \{y = c + Rx; \, x \in \mathcal{B }\}$ where $c \in \mathbb{R }^3$ denotes a translation and $R = R (d,\alpha ) \in SO(3)$ describes a rotation with respect to some axis with direction $d,\Vert d\Vert = 1$, about some angle $\alpha$. The quantities $c$ and $R$ are restricted by requiring that the floating body is contained in $G$, i.e $\mathcal{B }(c,R) \subseteq G; \mathcal{B }(c,R)$ is assumed to be a closed set. If we denote the domain occupied by the fluid by $E$, we have $E \subseteq G \setminus \mathcal{B }(c,R)$ and ${\fancyscript{L}}^3(E) = V_0$ where $V_0$ is the volume of the Lebesgue measurable set $E$. The energy functional then reads

$$\begin{aligned} \mathcal{F }(c,R;E):&= \sigma \int \limits _{G \setminus \mathcal{B }(c,R)} |D \varphi _E | + \kappa \int \limits _{\partial G} {\varphi }_E \, \text{ d} \sigma \, +\, \kappa _0 \int \limits _{\partial {\mathcal{B }}(c,R)} {\varphi }_E \, \text{ d} \sigma \nonumber \\&+\rho g \int \limits _{G \setminus \mathcal{B }(c,R)} x_3 {\varphi }_E \, \text{ d}x\, +\, {\rho }_0 g \int \limits _{\mathcal{B }(c,R)} x_3 \text{ d} x, \end{aligned}$$

(1)

where

$$\begin{aligned} \int \limits _{G \setminus \mathcal{B }(c,R)} \left|D\varphi _E\right|\, :=\, \sup \, \left\{ \int \limits _{G \setminus \mathcal{B }(c,R)} \varphi _E\, \mathrm div \, g\ \text{ d}x:\, g \in C^1_c\, \left(G \setminus \mathcal{B }(c,R);\ \mathbb R ^3\right),\ \Vert g\Vert _{C^0} \le 1 \right\} \end{aligned}$$

denotes the total variation of the characteristic function ${\varphi }_E$ of $E$; $\sigma$ is the coefficient of surface tension. The integrals $\int _{\partial G} {\varphi }_E \, \text{ d} \sigma$ and $\int _{\partial \mathcal{B }(c,R)} \varphi _E \, \text{ d} \sigma$ denote the area of the wetted part of the container’s boundary $\partial G$ and the boundary $\partial \mathcal{B }(c,R)$ of the floating body, respectively; the corresponding coefficients of the adhesion energy are $\kappa$ and $\kappa _0$; their relative size with respect to $\sigma$ determines the contact angles. The last two integrals in (1) represent the gravitational energy of the fluid and of the floating body; $g$ is the gravitational constant. We prove that there is a minimizing configuration $(\mathcal{B }(c,R),E)$ to (1); concerning the data $\sigma ,\kappa$ and $\kappa _0$, we make no further restrictions than in the classical capillary problem without floating bodies. The functional $\mathcal{F }(c,R;E)$ is to be minimized in the class

$$\begin{aligned} \mathcal{C }&= \left\{ \,(c,R;E)\,:\, c \in \mathbb R ^3,\, R \in SO(3)\ \text{ such} \text{ that}\ \mathcal{B }(c,R) \subseteq G;\right. \nonumber \\&\qquad \qquad \left. E \subset G \backslash \mathcal{B } (c,R) \text{ measurable} \text{ with} \, \mathcal{L }^3(E)\, =\, V_0\right\} . \end{aligned}$$

(2)

In Sect. 3, we formulate the variational problem in the class of graphs. The body $\mathcal{B }$ is assumed to be strictly convex such that the non-wetted part of its boundary can be described by a graph; also the capillary surface is assumed to be a graph which is certainly appropriate in our case because of the influence of the gravitational force. Archimedes’ principle characterizes the position of the floating body by an equilibrium condition. In the presence of capillary and adhesion forces, the first variation of the energy leads to an equilibrium of forces that act on $\mathcal{B }$, where the resultant of the surface forces is a vector field on the contact line that is tangential to the capillary surface. This result was first proved by McCuan [4] in the general case of parametric surfaces and deformable bodies.

The existence theorem requires more restrictions on the data compared to the variational problem in the class of Caccioppoli sets. This more special solution allows, however, to prove regularity of the minimizer as well as geometrical properties, in particular that the upper part of $\partial B$ that is not wetted is a simply connected set. These results are contained in Sect. 4.

2 Existence of solutions: the general case

We first prove that the infimum of the energy functional $\mathcal{F }(c,R;E)$ from (1) is attained for some configuration $(c_0,R_0;E_0)$ from $\mathcal{C }$, which is defined in (2). In [1], we adopted the variational method for Caccioppoli sets to the case that a floating body is involved. If also adhesion energy is part of the functional it is well known that only the sum of surface and adhesion energy, $\sigma \int _G |D\varphi _E| + \kappa \int _{\partial G} \varphi _E \, \text{ d} \sigma$, is lower semi-continuous with respect to $L^1$-convergence; the integral over $\partial G$ is generally not lower semi-continuous by itself. To prove semi-continuity, one uses the estimate of Emmer [2]

$$\begin{aligned} \int \limits _{\partial G} u \, \text{ d}\, \sigma \le \sqrt{1 + L^2} \ \int \limits _{G (\varepsilon )} |Du| + C_{\varepsilon } \int \limits _{G (\varepsilon )} |u|, \end{aligned}$$

(3)

which holds for functions $u \in BV(G). G(\varepsilon )$ is a strip near the boundary, and $L$ denotes the Lipschitz constant of $\partial G$. In our problem, we consider $G \setminus \mathcal{B }(c,R)$ instead of $G$; then, the boundary need not be Lipschitz continuous even if $\partial G$ and $\partial \mathcal{B }(c,R)$ are smooth because the rigid body $\mathcal{B }$ might touch the boundary $\partial G$ of the container. Hence, we must prove Emmer’s lemma (3) for this case.

In the analog of (3), we require $\partial \Omega$ to be of class $C^2$ and $\mathcal{B }(c,R)$ to have a projection $P(c,R)$ into $\Omega$ such that the curvature of $\partial P(c,R)$ is larger than the curvature of $\partial \Omega$:

$$\begin{aligned} \min _{R \in SO{(3)}}\min _{x^{\prime }\in \,\partial P(c,R)}\ K (\partial P(c,R),x^{\prime }) \ >\ \max _{x^{\prime }\in \,\partial \Omega }\ K(\partial \Omega , x^{\prime }), \end{aligned}$$

where $K(\gamma , x^{\prime })$ denotes the curvature of a curve $\gamma$ at some point $x^{\prime }\!\in \gamma$. If then $\mathcal{B }(c,R)$ touches $\partial G$ it does so in a single point $y_0$. In a neighborhood $U_{\varepsilon _0}$ of $y_0$, we can describe the boundaries $\partial G$ and $\partial \mathcal{B }(c,R)$ by the graphs of the functions

$$\begin{aligned} y_3 = \omega \left(y_1,y_2\right) \text{ and} \, y_3 = \beta \left(y_1,y_2\right),\ \left(y_1,y_2\right) \in A_{\varepsilon _0} \subseteq E \end{aligned}$$

where $(y_1,y_2) \equiv y^{\prime }$ are cartesian coordinates with center $y^{\prime }_0 = (0,0)$ in the tangent plane at $y_0$. We have $\beta (y^{\prime }) > \omega (y^{\prime })$ for all $y^{\prime }$ in the domain $A_{\varepsilon _0}$ where $\sigma$ and $\omega$ are defined except for $y^{\prime } = (0,0)$.

Local strips between $\partial \mathcal{B }(c,R)$ and $\partial G$ around $y_0$ can be defined by

$$\begin{aligned} \mathcal{B }^{*}_{\varepsilon }(c,R):=\!\bigcup _{0\, <\,\delta \, <\, \varepsilon } \beta ^{*}(\delta ) \text{ and} \,G^{*}_{\varepsilon }:=\!\bigcup _{0\, <\,\delta \, <\, \varepsilon } \omega ^{*}(\delta )\, \end{aligned}$$

where

$$\begin{aligned} \beta ^{*}(\delta )&= \left\{ \,y = \left(y^{\prime },y_3\right)\!:\ y_3 =\, \beta (y^{\prime })\, -\, \delta \tau (y^{\prime }),\ y^{\prime } \in \, A_{\varepsilon _0}\right\} \\ \omega ^{*}(\delta )&= \left\{ \,y = \left(y^{\prime },y_3\right)\!:\ y_3 =\, \omega (y^{\prime })\, +\, \delta \tau (y^{\prime }),\ y^{\prime } \in \, A_{\varepsilon _0}\right\} \end{aligned}$$

with

$$\begin{aligned} \tau (y^{\prime }):=\, \left\{ \! \begin{array}{c@{}c@{}l}\displaystyle \frac{\beta (y^{\prime }) - \omega (y^{\prime })}{3 \, \varepsilon }&\text{ if}&\quad \beta (y^{\prime })\, -\, \omega (y^{\prime })\, \le \, 3 \, \varepsilon \\ 1&\text{ if}&\quad \beta (y^{\prime })\, -\, \omega (y^{\prime })\, \ge 3\, \varepsilon \end{array} \right. \end{aligned}$$

These sets replace the neighborhoods $G(\varepsilon )$ of constant thickness in Emmer’s proof of (3).

Lemma 2.1

Let $\partial G$ and $\partial \mathcal{B } (c,R)$ touch in a single point $y_0$ as described above. Then, there holds for $u \in BV (G \backslash \mathcal{B } (c,R))$.

$$\begin{aligned} \int \limits _{\partial G\, \cap \, U_{\varepsilon _0}} u \, \text{ d}\sigma \,&\le \, \sqrt{1 + L^2}\ \int \limits _{G^{*}_{\varepsilon }} |Du| + C_{\varepsilon }\ \int \limits _{G^{*}_{\varepsilon }} |u|\, \mathrm{d}x\end{aligned}$$

(4)

$$\begin{aligned} \int \limits _{\partial \mathcal{B }(c,R)\, \cap \, U_{\varepsilon _0}} u \, \text{ d}\sigma \,&\le \sqrt{1 + L^2}\ \int \limits _{\mathcal{B }^{*}_{\varepsilon }(c,R)}\, |Du| + C_{\varepsilon } \int \limits _{\mathcal{B }^{*}_{\varepsilon }(c,R)}\, |u|\, \mathrm{d}x \end{aligned}$$

(5)

Here, $L$ is an upper bound for $|D\omega |$ and $|D \beta |$ in $A_{\varepsilon _0}$.

Remark 2.2

(i) :

As we assume that $\partial \mathcal{B }$ and $\partial G$ are regular surfaces, we can make $L$ small by covering $\partial G$ and $\partial \mathcal{B }$ by small open sets. Hence, the value of $L$ is no extra restriction for $\sigma$ and $\kappa$ or $\kappa _0$ in the existence theorem.

(ii) :

Away from the point of contact, the original proof of Emmer can be applied because there $G \setminus \partial B(c,R)$ has a smooth boundary.

Proof

To $u \in BV(G \setminus \mathcal{B }(c,R))$ we consider the trace $u_{\delta }$ on the surface $\omega ^{*}(\delta )$ for some $\delta \in (0,\varepsilon )$. Then,

$$\begin{aligned} \int \limits _{\Gamma _{\varepsilon _0}} u \, \text{ d} \sigma \, \le \int \limits _{\Gamma _{\varepsilon _0}} |u - u_{\delta }| \, \text{ d}\sigma \, + \int \limits _{\Gamma _{\varepsilon _0}} |u_{\delta }|\, \text{ d}\sigma \end{aligned}$$

(*)

with $\Gamma _{\!\varepsilon _0} = \partial G \cap U_{\varepsilon _0}$, and we estimate the difference $|u-u_{\delta }|$ by $|Du|$:

$$\begin{aligned} \int \limits _{\Gamma _{\varepsilon _0}} |u-u_{\delta }| \, \text{ d}\sigma&= \displaystyle \int \limits _{A_{\varepsilon _0}} \left|u\left(y^{\prime }, \omega (y^{\prime })\right) - u \left(y^{\prime }, \omega (y^{\prime }) + \delta \tau (y^{\prime })\right)\right| \cdot \sqrt{1 + \left|D\omega \left(y^{\prime }\right)\right|^2}\, \text{ d}y^{\prime }\\&\le \displaystyle \sqrt{1 + L^2}\ \int \limits _{A_{\varepsilon _0}} \int \limits ^{\omega (y^{\prime })\ +\ \delta \tau (y^{\prime })}_{\omega (y^{\prime })}\, \left|\,\frac{\partial u}{\partial y_3}\, (y^{\prime },t)\right|\, \text{ d}t\, \text{ d}y^{\prime } \\&\le \displaystyle \sqrt{1 + L^2}\ \int \limits _{G^{*}_{\varepsilon }}|Du|\\ \int \limits _{\Gamma _{\epsilon _0}} \left|u_{\delta }(y)\right| \, \text{ d}\sigma \,&= \, \int \limits _{A_{\varepsilon _0}} \left|u\left(y^{\prime }, \omega (y^{\prime })\, +\, \delta \tau (y^{\prime })\right)\right|\, \sqrt{1 + \left|D\left(\omega (y^{\prime })\, +\, \delta \tau (y^{\prime })\right)\right|^2}\, \text{ d}y^\prime \\&+ \int \limits _{A_{\varepsilon _0}} \left|u\left(y^{\prime }, \omega (y^{\prime })\, +\, \delta \tau (y^{\prime })\right)\right|\ \left[ \sqrt{1 + |D\omega (y^\prime )|^2}\right. \\&\left.- \sqrt{1 + \left|D(\omega (y^\prime ) + \delta \tau (y^\prime ))\right|^2}\,\right] \, \text{ d}y^{\prime } \end{aligned}$$

Integration with respect to $\delta$ gives $\int \nolimits _{G_\varepsilon ^*} |u| \, \text{ d} y$ for the first integral and $C \int _{G^{*}_{\varepsilon }} |u| \, \text{ d} y$ for the second term. We now integrate $({*})$ with respect to $\delta \in (0,{\varepsilon })$ and obtain

$$\begin{aligned} \varepsilon \int \limits _{\Gamma _{\varepsilon _0}} u \, \text{ d}\sigma \, \le \, \varepsilon \ \sqrt{1 + L^2} \int \limits _{G^{*}_{\varepsilon }} |Du|\, +\, C \int \limits _{G^{*}_{\varepsilon }} |u|\, \text{ d}x, \end{aligned}$$

which proves (4). The case (5) can be handled in the same way. $\square$

Theorem 2.3

Assume that $\partial \mathcal{B } (c,R)$ and $\partial G$ touch in at most one point, and let $\sigma , \rho , \rho _0, \kappa , \kappa _0$ be constants with $\rho > \rho _0 > 0, \sigma > 0$ and $\sigma - |\kappa |\sqrt{1-L^2} < 0, \sigma - |\kappa _0|\sqrt{1+L^2} > 0$. Then, there exists an element $(c_0, R_0; E_0)\, \in \, \mathcal{C }$, such that

$$\begin{aligned} \mathcal{F }\left(c_0, R_0, E_0\right)\, \le \, \mathcal{F }(c,R;E) \text{ for} \text{ all}(c,R;E) \in \mathcal{C }. \end{aligned}$$

(6)

Proof

We first show that $\mathcal{F }(c,R;E)$ is bounded from below on $\mathcal{C }$. The gravitational energies are clearly positive; for the adhesion terms we have

$$\begin{aligned} \kappa _0 \int \limits _{\partial \mathcal{B }(c,R)} \varphi _E \, \text{ d}\sigma , \ge \, -\, |\kappa _0|\; |\partial \mathcal{B }|\, >\, -\infty \end{aligned}$$

and, if $\partial \mathcal{B }(c,R)$ and $\partial G$ touch in a point $y_0$,

$$\begin{aligned} \kappa \int \limits _{\partial G \cap U_{\varepsilon _0}} \varphi _E \, \text{ d}\sigma \, \ge \, -\, |\kappa | \sqrt{1 + L^2} \int \limits _{G^{*}_{\varepsilon }} \left|D \varphi _E\right|\, -\, |\kappa |C_{\varepsilon } \int \limits _{G^{*}_{\varepsilon }} \varphi _E\, \text{ d}x, \end{aligned}$$

which follows from Lemma 2.1.

For the part of the boundary that lies outside of $U_{\varepsilon _0}$ (and that may be all of $\partial G$ if $\partial \mathcal{B }(c,R)$ and $\partial G$ do not touch at all), we have from Sect. 3

$$\begin{aligned} \kappa \int \limits _{\Gamma \backslash U_{\varepsilon _0}} \varphi _E\, \text{ d} \sigma \ge -|\kappa |\sqrt{1+L^2} \int \limits _{G(\varepsilon ) \backslash U_{\varepsilon _0}} |D\varphi _E|-C^{\prime }_{\varepsilon } \int \limits _{G(\varepsilon ) \backslash U_{\varepsilon _0}} \varphi _E\, \mathrm{d}x \end{aligned}$$

with

$$\begin{aligned} G(\varepsilon )=\{x\in G: \mathrm{dist} (x,\Gamma )<\varepsilon \}. \end{aligned}$$

The first term is majorized by $\sigma \int _{G \setminus \mathcal{B }(c,R)} |D \varphi _E|$ because of the assumption on the data, and the second one remains finite because of $\int _{G^{*}_{\varepsilon }} \varphi _E\, \text{ d}x \le \, \mathcal{L }^3(E) = V_0$.

Therefore, there exists a constant $c_0$ such that

$$\begin{aligned} \mathcal{F }(c,R;E)\, \ge \, c_0\, >\, -\infty \,\,\forall \, (c,R;E) \in \mathcal{C }, \end{aligned}$$

and a minimizing sequence $\{ (c_n, R_n; E_n)\}^{\infty }_{n=1}$ from $\mathcal{C }$:

$$\begin{aligned} \lim _{n \rightarrow \infty } \mathcal{F }\left(c_n,R_n;E_n\right)\, =\, m_0\, \equiv \, \inf _{(c,R;E)\, \in \, \mathcal{C }}\, \mathcal{F }(c,R;E). \end{aligned}$$

We claim that this sequence is bounded:

$$\begin{aligned} |c_n|\, +\, |R_n|\, +\, \parallel \varphi _{E_n} \parallel _{BV}\ \le \, C\,\,\quad \forall \, n \in \mathbb{N }. \end{aligned}$$

We may assume that

$$\begin{aligned} \mathcal{F }\left(c_n,R_n;E_n\right)\, \le \, m_0\, +\, 1\quad \forall \, n \in \mathbb{N }, \end{aligned}$$

and because of $\rho g \int \nolimits _{G} x_3\, \varphi _{E_n}\ \text{ d}x \ge 0,\ \rho _0\, g\, \int \nolimits _{\mathcal{B }(c_n,R_n)} x_3\ \text{ d}x \ge 0$ and $|\,\kappa _0 \int \nolimits _{\partial \mathcal{B }(c_n,R_n)} \varphi _{E_n}\ \text{ d}\sigma | \le \,c_1$ we have

$$\begin{aligned} \sigma \int \limits _{G \setminus \mathcal{B }(c_n,R_n)} |D\varphi _{E_n}|\, +\, \kappa \int \limits _{\partial G} \varphi _{E_n}\, \text{ d}\sigma \, \le \, m_0\, +\, 1\, +\, c_1. \end{aligned}$$

Using (4) again we get

$$\begin{aligned} \left(\sigma - |\kappa |\sqrt{1 + L^2}\right)\, \int \limits _{G \setminus \mathcal{B }(c_n,R_n)}\, |D\varphi _{E_n}|\, \le \, m_0\, +\, 1\, +\, c_1\, +\, C_{\varepsilon } \cdot V_0. \end{aligned}$$

Because of the assumption on $\sigma$ and $\kappa$ and the volume constraint this gives

$$\begin{aligned} \parallel \varphi _{E_n}\parallel _{L^1(G)}\, +\, \int \limits _{G \setminus \mathcal{B }(c_n,R_n)} |D\varphi _{E_n}|\, \le \, C. \end{aligned}$$

The area integral can be written in the form

$$\begin{aligned} \int \limits _{G} \left|D\varphi _{E_n}\right|\, -\,\int \limits _{\partial \mathcal{B }(c_n,R_n)} \varphi _{E_n} \, \text{ d}\sigma , \end{aligned}$$

which gives

$$\begin{aligned} \parallel \varphi _{E_n}\parallel _{L^1(G)}\, +\, \int \limits _{G} \left|D\varphi _{E_n}\right|\, \le \, C. \end{aligned}$$

The values of $R_n$ belong to a compact set, hence $R_n$ is bounded independently of $n \in \mathbb{N }$. The $x_1$- and the $x_2$ component of $c_n$ is bounded by diam $\Omega$, and the $x_3$-component is bounded because of

$$\begin{aligned} \rho _0\, g \int \limits _{\mathcal{B }(c_n,R_n)} x_3\, \text{ d}x\, \le \, m_0\, +\, 1, \end{aligned}$$

and, thus, we have shown that the minimizing sequence $\{(c_n,R_n;E_n)\}^\infty _{n=1}$ is bounded:

$$\begin{aligned} |\,c_n|\, +\, |R_n|\ + \parallel \!\varphi _{E_n}\!\parallel _{BV(G)}\ \le \, C \,\,\quad \forall \, n \in {\mathbb{N }}. \end{aligned}$$

This implies that there is a subsequence, which we again denote by $\{(c_n,R_n,E_n)\}^\infty _{n=1}$ that converges for $n \rightarrow \infty$:

$$\begin{aligned} c_n \rightarrow c_0,\ R_n \rightarrow R_0,\ \varphi _{E_n} \rightarrow \varphi _{E_0}\text{ in}L^1(G). \end{aligned}$$

In the last steps of the proof, we show that $\mathcal{F }$ is lower semi-continuous with respect to this minimizing sequence.

The integral $\rho _0\, g \int _{\mathcal{B }(c_n,R_n)} x_3\ \text{ d}x$ is continuous because $\mathcal{B }(c_n,R_n)$ converges uniformly to $\mathcal{B }(c_0,R_0)$ for $n \rightarrow \infty$. The integrand $x_3 \varphi _{E_n} (x)$ in the term that represents the gravitational energy of the fluid is non-negative, and therefore the integral is lower semi-continuous according to Fatou’s lemma.

In order to compare the traces of $\varphi _{E_n}$ and $\varphi _{E_0}$ on the boundary of the rigid body, we set

$$\begin{aligned} \widetilde{E}_n:=\, T_n\, E_n, \end{aligned}$$

where $T_n$ is the rigid motion that maps $\mathcal{B }(c_n,R_n)$ onto $\mathcal{B }(c_0,R_0)$; $\widetilde{E}_n$ is generally not an element of $\mathcal{C }$, but $(\varphi _{\widetilde{E}_n} - \varphi _{E_0})$ restricted to $\partial \mathcal{B }(c_0,R_0)$ can be estimated by (5). For the area energy

$$\begin{aligned} {\fancyscript{A}}(c,R;E):=\, \sigma \int \limits _{G \setminus \mathcal{B }(c,R)} |D\varphi _E|\, +\, \kappa \int \limits _{\partial G} \varphi _E\ \text{ d}\sigma \, +\, \kappa _0 \int \limits _{\partial \mathcal{B }(c,R)} \varphi _E\ \text{ d}\sigma \end{aligned}$$

we get using

$$\begin{aligned} \kappa _0 \int \limits _{\partial \mathcal{B }(c_0,R_0)} \varphi _{E_0}\ \text{ d}\sigma \, -\, \kappa _0 \int \limits _{\partial \mathcal{B }(c_n,R_n)} \varphi _{E_n}\ \text{ d}\sigma \, =\, \kappa _0 \int \limits _{\partial \mathcal{B }(c_0,R_0)} \varphi _{E_0}\, -\, \varphi _{\widetilde{E}_n}\, \text{ d}\sigma \end{aligned}$$

and then (5) again:

$$\begin{aligned}&{\fancyscript{A}}(c_0,R_0;R_0) - {\fancyscript{A}}(c_n,R_n;E_n) \\&\quad \le \, \sigma \int \limits _{G \setminus ({\mathcal{B }}^{*}_{\varepsilon }(c_{0},R_{0}) \, \cup \, G^{*}_{\varepsilon })}\, |D\varphi _{E_0}| \,\, + \; \sigma \int \limits _{\mathcal{B }^{*}_{\varepsilon }(c_{0},R_{0})} |D \varphi _{E_0}|\; + \, \sigma \int \limits _{G^{*}_{\varepsilon }} |D \varphi _{E_0}|\\&\quad - \, \sigma \int \limits _{G \setminus (\mathcal{B }^{*}_{\varepsilon }(c_n,R_n)\, \cup \, G^{*}_{\varepsilon })}\, |D\varphi _{E_n}|- \, \sigma \int \limits _{\mathcal{B }^{*}_{\varepsilon }(c_n,R_n)}\, \left|D\varphi _{E_n}\right|\, -\,\sigma \int \limits _{G^{*}_{\varepsilon }}\, \left|D\varphi _{E_n}\right|\\&\quad + \, |\kappa | \sqrt{1 + L^2}\ \int \limits _{G^{*}_{\varepsilon }}\, |D(\varphi _{E_0}\, -\, \varphi _{E_n})| + \, |\kappa |\ C_{\varepsilon } \int \limits _{G^{*}_{\varepsilon }}\, |\varphi _{E_0}\, -\, \varphi _{E_n}|\, \text{ d}x \\&\quad + \, |\kappa _0| \sqrt{1 + L^2} \int \limits _{\mathcal{B }^{*}_{\varepsilon }(c_0,R_0)}\, \left|D(\varphi _{E_0}\, -\, \varphi _{\widetilde{E}_n})\right| + \, |\kappa _0|\ C_{\varepsilon } \int \limits _{\mathcal{B }^{*}_{\varepsilon }(c_0,R_0)}\, \left|\varphi _{E_0}\, -\, \varphi _{\widetilde{E}_n}\right|\, \text{ d}x. \end{aligned}$$

Now we have

(i) :

$\left(|\kappa |\,\sqrt{1 + L^2}\, -\, \sigma \right)\, \int \nolimits _{G^{*}_{\varepsilon }} |D \varphi _{E_n}|\, \le \, 0 \forall \, n \in {\mathbb{N }}$ because of the assumptions on the data.

(ii) :

$|\kappa | C_{\varepsilon }\, \int \nolimits _{G^{*}_{\varepsilon }}|\varphi _{E_0} - \varphi _{E_n}|\text{ d}x \rightarrow 0, n \rightarrow \infty$, for every $\varepsilon > 0$.

(iii) :

$\int \nolimits _\mathcal{B ^{*}_{\varepsilon }(c_0,R_0)}|\varphi _{E_0} - \varphi _{\widetilde{E}_n}|\text{ d}x\, \le \int \nolimits _{\mathcal{B }^{*}_{\varepsilon }(c_0,R_0)}|\varphi _{E_0} - \varphi _{E_n}|\text{ d}x\, + \int \nolimits _{\mathcal{B }^{*}_{\varepsilon }(c_0,R_0)}|\varphi _{E_n} - \varphi _{\widetilde{E}_n}|\text{ d}x$, and both integrals converge to zero for $n \rightarrow \infty$.

(iv) :

$\int \nolimits _{\mathcal{B }^{*}_{\varepsilon }} |D \varphi _{\widetilde{E}_n}|\, =\, \mathcal{H }^2 (\partial \tilde{E}_n \cap \mathcal{B }^{*}(c_0,R_0))$; the Hausdorff measure $\mathcal H ^2$ is invariant under rigid motions, and therefore we have $\mathcal H ^2 (T_n (\partial E_n) \cap \mathcal{B }^{*}_{\varepsilon }(c_0,R_0)) =$ $\mathcal H ^2 (T^{-1}_{n} ( T_n (\partial E_n) \cap \mathcal{B }^{*}_{\varepsilon }(c_0,R_0))) = \mathcal{H }^2 (\partial E_n \cap \mathcal{B }^{*}_{\varepsilon } (c_n,R_n))$ which gives $|\kappa _0| \sqrt{1 + L^2} \int \nolimits _{\mathcal{B }^{*}_{\varepsilon }(c_0,R_0)}|D\varphi _ {\widetilde{E}_n}|\, -$ $\sigma \int \nolimits _{\mathcal{B }^{*}_{\varepsilon }(c_n,R_n)}|D \varphi _{E_n}| \le 0$.

(v) :

The integrals $\sigma \int \limits _\mathcal{B ^{*}_{\varepsilon }(c_0,R_0)}|D \varphi _{E_0}|\quad \text{ and} \,\,\sigma \int \limits _{G^{*}_{\varepsilon }}|D \varphi _{E_0}|$  are of order $O(\varepsilon )$.

(vi) :

$\int \nolimits _{G \setminus \mathcal{B }^{*}_{\varepsilon } (c_0,R_0)}|D \varphi _{E_0}| \le \liminf _{n \rightarrow \infty } \int \nolimits _{G \setminus \mathcal{B }^{*}_{\varepsilon } (c_n,R_n)}|D \varphi _{E_n}|$  has been proved in [1].

Using (i)–(vi) in the inequality for ${\fancyscript{A}}(c,R;E)$, we obtain the lower semi-continuity of $\mathcal{F }$, and this concludes the proof of Theorem 2.3. $\square$

3 Existence of solutions in the class of graphs

We now investigate capillary surfaces $\Sigma$ in the presence of floating bodies under the assumption that both $\Sigma$ and the upper part of $\partial \mathcal{B }(c,R)$ are graphs of real functions $u$ and $h$, respectively; we use the same notation as in [1]. If we assume that $\mathcal{B } = \mathcal{B }(0,R)$ is strictly convex, there is a smallest number $\alpha = \alpha (R)$ such that

$$\begin{aligned} \partial \mathcal{B }_{\alpha } \equiv \partial \mathcal{B } (0,R) \cap \left\{ x: x_3 > \alpha (R) \right\} \end{aligned}$$

can be written as the graph of a real function $\overline{h}: B (\alpha ) \rightarrow \mathbb R$, where $B (\alpha )$ is the projection of $\partial \mathcal{B }_{\alpha }$ onto the $x^{\prime }$-plane. We assume that there is another number $h_0 (R) > \alpha (R)$, such that

$$\begin{aligned} vol \ \left(\mathcal{B } (0,R) \cap \left\{ x:x_3>h_0(R)\right\} \right)=\frac{1}{3} \ vol \ (\mathcal{B }) \end{aligned}$$

and define $B(h_0):= \{x^{\prime }\!\in B(\alpha )\!:\,\overline{h}(x^{\prime })>h_0\},\, C(h_0):= \partial B (h_0)$.

Then, we can define the function $h\!:\, \Omega \rightarrow \mathbb R$ by

$$\begin{aligned} h(x^{\prime }):= \left\{ \begin{array}{l@{}l@{}l} \overline{h}(x^{\prime }) ,&x^{\prime }\!\in B (h_0) \\ h_0 ,&x^{\prime }\!\in \Omega \setminus \overline{B (h_0)}. \end{array} \right. \end{aligned}$$

(7)

This function enters the variational problem as an obstacle. The surface $\Sigma$ is assumed to be the graph of a real function $\overline{u}\!: \Omega \setminus B(h)\rightarrow \mathbb R$ where $B(h) \subseteq B(h_0)$ is the projection of the dry part of $\partial \mathcal{B }_{\alpha }$ into the $x^{\prime }$-plane, which means that $\Gamma := \text{ graph}\, h (\gamma )$, is the contact line where $h(x^{\prime }) = u(x^{\prime }),\ x^{\prime } \in \gamma$, holds. In order to work with functions that are defined on the whole domain $\Omega$, we set

$$\begin{aligned} u(x^{\prime }) :\, = \left\{ \begin{array}{l@{}l@{}l} \overline{u}(x^{\prime }) ,&x^{\prime }\!\in \Omega \setminus \overline{B(h)} \\ \overline{h}(x^{\prime }) ,&x^{\prime }\!\in B(h). \end{array} \right. \end{aligned}$$

(8)

With these quantities, the energy (1) is of the form

$$\begin{aligned} \mathcal{F } (c,R;u)&= \sigma \int \limits _{\Omega } \sqrt{1 + |Du|^2} + \sigma \int \limits _{B(h_0)\cap \{u > h\}} \sqrt{1 + |Dh|^2} \text{ d}x^{\prime }\!-\! \sigma \int \limits _{B(h_0)} \sqrt{1 + |Dh|^2} \text{ d}x^{\prime }\nonumber \\&+ \kappa _0 \left\{ \left|\partial \mathcal{B }\,\right| + \displaystyle \int \limits _{\mathcal{B }(h_0)\,\cap \,\{u > h\}} \sqrt{1 + |Dh|^2} \text{ d}x^{\prime }\, - \int \limits _{B(h_0)} \sqrt{1 + |Dh|^2} \text{ d}x^{\prime }\right\} \nonumber \\&+ \displaystyle \kappa \int \limits _{\partial \Omega } u\ \text{ d}\sigma + \displaystyle \frac{\rho g}{2} \int \limits _{\Omega } u^2\ \text{ d}x^{\prime } - \left(\rho - \rho _0\right)g \int \limits _{\mathcal{B }(c,R)} x_3\ \text{ d}x \end{aligned}$$

(9)

where $h$ and $h_0$ depend on $c$ and $R$. The first integral is defined for $u \in BV(\Omega )$, which means

$$\begin{aligned} \int \limits _{\Omega } \sqrt{1 + |Du|^2}:\,&= \, \sup \left\{ \,\int \limits _{\Omega } g_3\, +\, u\left(\frac{\partial g_1}{\partial x_2} + \frac{\partial g_2}{\partial x_3}\right)\text{ d}x^{\prime }:\right.\\&\qquad \qquad \left. g=(g_1,g_2,g_3) \in C^1_c\left(\Omega ;\mathbb R ^3\right),\, \parallel g \parallel _{C^0(\Omega )}\ \le \ 1 \right\} . \end{aligned}$$

If $u$ is a smooth function with $u > h_0$, the first three integrals equal $\int \nolimits _{\Omega \,\cap \,\{u > h\}} \sqrt{1 + |Du|^2} \text{ d}x^{\prime }$ and this is the area of the capillary surface; the quantities in brackets equal the area of the wetted part $\Sigma _{\mathcal{B }}$ of $\partial \mathcal{B } (c,R)$. We look for minimizers in the class

$$\begin{aligned} \mathcal{C }\, :&= \, \left\{ (c,R;u) \in \mathbb R ^3 \times SO(3) \times BV(\Omega )\!:\, \mathcal{B }(c,R)\, \subseteq \, G;\right.\nonumber \\&\qquad \qquad \left.\int \limits _{\Omega } u(x^{\prime })\ \text{ d}x^{\prime } = V_0 + vol(\mathcal{B }),\, u(x^{\prime })\, \ge \, h(x^{\prime }) \text{ a.e.} \text{ in}\,\, \Omega \right\} . \end{aligned}$$

(10)

In this setup, we can calculate the first variation of $\mathcal{F }$; variations of $u$ lead to the Euler–Lagrange equations

$$\begin{aligned} \upsigma \text{ div} \left( \frac{Du}{\sqrt{1 + |Du|^2}}\right)\, =\, \rho gu\, +\, \lambda ~ \text{ in} \Omega \setminus \overline{B(h)} \end{aligned}$$

(11)

and the boundary conditions

$$\begin{aligned} \frac{Du \cdot n}{\sqrt{1 + |Du|^2}}\,&= \, -\, \frac{\kappa }{\sigma } \quad \ \text{ on} \partial \Omega \end{aligned}$$

(12)

$$\begin{aligned} \frac{1 + Du \cdot Dh}{\sqrt{1 + |Du|^2} \cdot \sqrt{1 + |Dh|^2}}\,&= \, -\, \frac{\kappa _0}{\sigma } \quad \text{ on} \gamma . \end{aligned}$$

(13)

Remark 3.1

For the proof of (13), we assume that $u$ is smooth up to the contact line, and therefore the surface energy and the adhesion energy can be written in the form

$$\begin{aligned} I_1(u)\, +\, I_2(u)\, =\, \sigma \int \limits _{\{u\,>\,h\}} \sqrt{1 + |Du|^2}\text{ d}x^{\prime }\, +\, \kappa _0 \int \limits _{\{u\,>\,h\}} \sqrt{1 + |Dh|^2}\mathrm{d}x^{\prime }. \end{aligned}$$

we consider only these two integrals because the rest of the energy terms will not be affected by the variations that determine the contact angle. For the contact line $\Gamma \, =\, \{(x^{\prime },x_3) \in \partial \mathcal{B }(c,R):\, x_3 =\, u(x^{\prime })\}$, we have $u(x^{\prime }) = h(x^{\prime })\forall \, x^{\prime } \in \gamma$, and hence the normal to $\gamma$ is

$$\begin{aligned} n(x^{\prime }_0)\, =\, -\, \frac{D(u - h) (x_0^{\prime })}{\left|D(u - h) (x^{\prime }_0)\right|}. \end{aligned}$$

Curves that lie in a neighborhood of $\gamma$ can be described as $\gamma _{\delta } =\, \{x^{\prime }\!\in \Omega \!:\, x^{\prime } =\, x^{\prime }_0 + \delta n(x^{\prime }_0),\, x_0 \in \gamma \}$. A variation $u(x^{\prime }) +\, \varepsilon \,\varphi (x^{\prime })$ of $u$ meets $\partial \mathcal{B } (c,R)$ in some curve near $\Gamma$, and it is given by $u(x^{\prime }) + \varepsilon \,\varphi (x^{\prime }) = h(x^{\prime })\forall \, x^{\prime }\!\in \gamma _{\delta }$, where $\delta = \delta (\varepsilon ,x^{\prime }_0)$ will be determined by $u(x^{\prime }_0 + \delta (\varepsilon ,x^{\prime }_0)\, n\, (x^{\prime }_0)) +\, \varepsilon \varphi (x^{\prime }_0 + \delta (\varepsilon ,x^{\prime }_0)\, n(x^{\prime }_0)) = h(x^{\prime }_0 + \delta (\varepsilon ,x^{\prime }_0)\, n(x^{\prime }_0))$ which leads to

$$\begin{aligned} u(x^{\prime }_0)\, + \, \delta (\varepsilon , x^{\prime }_0) Du(x^{\prime }_0) \cdot n(x^{\prime }_0) + \varepsilon \varphi (x^{\prime }_0)\, =\, h(x^{\prime }_0)\, +\, \delta (\varepsilon ,x^{\prime }_0) Dh(x^{\prime }_0) \cdot n(x^{\prime }_0) + \text{ o} (\varepsilon ) \end{aligned}$$

for $|\varepsilon |$ small; therefore, we get

$$\begin{aligned} \delta (\varepsilon ,x^{\prime }_0)\, =\, -\, \frac{\varepsilon \cdot \varphi (x^{\prime }_0)}{D(u - h)\,(x^{\prime }_0) \cdot n(x^{\prime }_0)}. \end{aligned}$$

The first variation of $I_1$ is

$$\begin{aligned} \delta \, I_1(u,\varphi )\, =\, \lim \limits _{\varepsilon \rightarrow 0}\frac{\sigma }{\varepsilon }\left\{ \int \limits _{\{u\,+\,\varepsilon \varphi \,>\,h\}} \sqrt{1 + \left|D(u + \varepsilon \varphi )\right|^2}\, \text{ d}x^{\prime }\, -\, \int \limits _{\{u\,>\,h\}} \sqrt{1 + |Du|^2}\, \mathrm{d}x^{\prime } \right\} \end{aligned}$$

with

$$\begin{aligned}&\frac{1}{\varepsilon }\, \int \limits _{\{u\,>\,h\}} \sqrt{1 + \left|D(u + \varepsilon \varphi )\right|^2}\, -\, \sqrt{1 + |Du|^2}\mathrm{d}x^{\prime }\, \longrightarrow \, \int \limits _{\{u\,>\,h\}} \frac{Du \cdot D\varphi }{\sqrt{1 + |Du|^2}}\mathrm{d}x^{\prime }\\&\quad =\, -\, \int \limits _{\{u\,>\,h\}} \text{ div} \left(\frac{Du}{\sqrt{1 + |Du|^2}}\right)\, \varphi \mathrm{d}x^{\prime }\, +\, \oint \limits _{\gamma }\, \frac{Du \cdot n}{\sqrt{1 + |Du|^2}}\, \varphi \, \mathrm{d}s \end{aligned}$$

and

$$\begin{aligned}&\displaystyle \frac{1}{\varepsilon }\, \left(\int \limits _{\{u\,+\,\varepsilon \varphi \,>\,h\}} - \int \limits _{\{u\,>\,h\}}\right)\, \sqrt{1 + \left|D(u + \varepsilon \varphi )\right|^2}\, \mathrm{d}x^{\prime }\\&\quad = \displaystyle \frac{1}{\varepsilon }\, \oint \limits _{\gamma }\!\int \limits _{0}^{\delta (\varepsilon ,x^{\prime }_0)}\! \sqrt{1 + \left|D(u + \varepsilon \varphi )\,(x^{\prime }_0\,+\,tn(x^{\prime }_0)\right|^2}\, \mathrm{d}t\, \mathrm{d}s \\&\quad \longrightarrow \displaystyle \oint \limits _{\gamma } \sqrt{1 + \left|D(u + \varepsilon \varphi )\,(x^{\prime }_0\,+\,\delta (\varepsilon ,x^{\prime }_0)\, n(x^{\prime }_0)\right|^2}_{|_{\varepsilon =0}}\!\!\!\! \cdot \ \delta ^{\prime }(\varepsilon ,x^{\prime }_0)_{|_{\varepsilon =0}}\text{ d}s\\&\quad = \displaystyle \oint \limits _{\gamma } \sqrt{1 + |Du|^2}\, \cdot \left(\,\frac{-\,\varphi }{D(u - h) \cdot n}\,\right) \mathrm{d}s. \end{aligned}$$

In the same way, we get

$$\begin{aligned} \delta I_2(u,\varphi )\, =\, \kappa _0\, \oint \limits _{\gamma }\, \sqrt{1 + |Dh|^2}\, \cdot \left(\,\frac{-\,\varphi }{D(u - h) \cdot n}\,\right)\, \mathrm{d}s. \end{aligned}$$

Finally

$$\begin{aligned}&\sigma \, \oint \limits _{\gamma }\, \frac{Du \cdot n}{\sqrt{1 + |Du|^2}}\, \cdot \varphi \, \mathrm{d}s + \displaystyle \sigma \, \oint \limits _{\gamma }\, \sqrt{1 + |Du|^2}\, \cdot \left(-\, \frac{\varphi }{D(u - h) \cdot n}\right)\, \mathrm{d}s\\&\quad \displaystyle +\, \kappa _0\, \oint \limits _{\gamma }\, \sqrt{1 + |Dh|^2}\, \cdot \left(-\, \frac{\varphi }{D(u - h) \cdot n}\right)\, \mathrm{d}s = 0 \end{aligned}$$

for all $\varphi \in C^0(\gamma )$ gives the equation for the contact angle:

$$\begin{aligned} \sigma \, \left\{ \, \frac{Du \cdot n}{\sqrt{1 + |Du|^2}}\, -\, \frac{\sqrt{1 + |Du|^2}}{D(u - h) \cdot n}\,\right\}&+ \displaystyle \kappa _0\ \frac{-\,\sqrt{1 + |Dh|^2}}{D(u - h) \cdot n} = 0 \end{aligned}$$

which implies

$$\begin{aligned} \displaystyle \sigma \, \left\{ -\, |Du|^2 +\, Du \cdot Dh\, +\, 1\, -\, |Du|^2\right\} +\displaystyle \kappa _0\ \sqrt{1 + |Du|^2}\ \sqrt{1 + |Dh|^2}\ =\ 0 \end{aligned}$$

and thus

$$\begin{aligned} \displaystyle \frac{(Du,-1)}{\sqrt{1 + |Du|^2}}\, \cdot \, \frac{(Dh,-1)}{\sqrt{1 + |Dh|^2}}\, =\, - \frac{\kappa _0}{\sigma } \end{aligned}$$

which means that $-\, \frac{\kappa _0}{\sigma } = \cos \vartheta$ gives the contact angle between the capillary surface and the floating body.

The contact angle from (13) is restricted to some interval $[0, \vartheta _0]$ with $\vartheta _0 < \pi$. This means that the tangent of graph $u$ ranges between being parallel to the tangent of graph $h$ (because of $u \ge h)$ and pointing in the $x_3$-direction (because $\Sigma$ is a graph). Therefore, $\kappa _0$ must be restricted, too:

$$\begin{aligned} -\, \frac{\kappa _0}{\sigma }\, \equiv \, \cos \alpha _0\, \ge \, -\, \frac{\left|Dh(R;x^{\prime })\right|}{\sqrt{1 + \left|Dh(R;x^{\prime })\right|^2}}\quad \forall \ x^{\prime }\in B(h_0),\, R \in SO(3); \end{aligned}$$

(14)

with this condition on the data, we can perform variations $T(x) = e + d\,\wedge \,x$ of the position and orientation of $\mathcal{B }(c,R)$, that is of $h(c,R;x^{\prime })$, and get from [1] Theorem 3.2 the equilibrium condition

$$\begin{aligned} \oint \limits _{\Gamma } E \cdot N\, \text{ d}s + pg \int \limits _{{\Sigma }_{\mathcal{B }}} -\, E \cdot N_{\mathcal{B }}\, x_3\ \text{ d}\sigma - \rho _0 g\, \left(e+d \wedge x_s\right)_3\ vol(\mathcal{B }) = 0 \end{aligned}$$

(15)

with $E \cdot N$, the component of the variation $E$ that is normal to the contact line $\Gamma$ and tangent to graph $(u),N_\mathcal{B }$ the normal $\partial \mathcal{B }$ and $x_s$ the center of gravity of $\mathcal{B }(c,R)$.

Theorem 3.2

For $G,h = h(c,R)$, and $\sigma ,\rho _0,\rho ,\kappa _0,\kappa$ as above with $\sigma > |\kappa |$ the variational problem

$$\begin{aligned} \mathcal{F }(c,R;u) \rightarrow \min \quad \text{ in} ~\mathcal{C } \end{aligned}$$

(16)

has a solution $(c_0,R_0;u_0) \in \mathcal{C }$.

Proof

(i) :

$\mathcal{F }(c,R;u)$ is bounded from below on $\mathcal{C }$, because $\kappa \int _{\partial G} u\, \text{ d}\sigma$ can be estimated by Emmer’s Lemma (3) and the condition on $\kappa$. As $|\partial \mathcal{B }|$ is finite, the adhesion term $\,\kappa _0 \{\,|\partial \mathcal{B }|\, - \int _{B(h_0)\,\cap \,\{u > h\}} \sqrt{1 + |Dh|^2}\, \text{ d}x^{\prime } + \int _{B(h_0)} \sqrt{1 + |Dh|^2}\, \text{ d}x^{\prime }\}$ can be estimated for any $\kappa _0$.

(ii) :

Any minimizing sequence $\{(c_n,R_n;u_n)\}^{\infty }_{n=1}\,\subseteq \,\mathcal{C }$ satisfies $\displaystyle \lim _{n \rightarrow \infty }\mathcal{F }(c_n,R_n;u_n) = m_0 := \inf \{\mathcal{F }(c,R;u)\!: (c,R;u) \in \mathcal{C }\}$. It is bounded: $|c_n| + |R_n|\, + ||u_n||_{BV(\Omega )}\, \le C \forall \, n \in {\mathbb{N }}$. This is known for $\mathcal{E }(c,R;u)$; the adhesion energy can be treated as in (i).

(iii) :

A sequence $\{u_n\}^{\infty }_{n=1}$ that is bounded in $BV(\Omega )$ has a subsequence which converges in $L^1(\Omega )$. Then, $\mathcal{F }$ is lower semi-continuous for a minimizing sequence $\{(c_n,R_n;u_n)\}$ with $u_n \rightarrow u_0$ in $L^1(\Omega ), n \rightarrow \infty$, and this proves the existence of a minimizer. $\square$

Remark 3.3

Also in this setup, the floating body can touch the wall of the container. But as we are working with real functions $u$ that satisfy $u \ge h_0$ on $\partial \Omega$ and a possible point of contact lies below $\{x_3=h_0\}$ it does not affect $u$ at all. Additional considerations in Emmer’s Lemma are therefore not necessary.

4 Boundedness, regularity, and geometrical properties of the solution

We first prove that a minimizer $u$ to the variational problem (16) is essentially bounded. In order to show regularity of the solutions, we have to make sure that $u$ does not touch the obstacle $h_0$, that is, $u(x^{\prime }) > h_0$ a.e. in $\Omega _0 := \Omega \setminus \overline{B(h_0)}$, because then the usual variations of $u$ lie in $\mathcal{C }$, and therefore, the standard methods from regularity theory for the mean curvature equation apply. Finally we show that the set $\{x^{\prime }\!\!\in \Omega \!: u(x^{\prime }) = h(x^{\prime })\}$ is simply connected.

Lemma 4.1

Let $(c,R;u)$ be a solution to the variational problem

$$\begin{aligned} \mathcal{F } (c,R;u) \rightarrow \min \quad \text{ in}\,\, \mathcal{C }. \end{aligned}$$

(17)

Then, there exists a constant $C$, such that

$$\begin{aligned} u(x^{\prime })\, \le \, C ~\text{ a.e.} \text{ in}\,\, \Omega . \end{aligned}$$

Proof For $\kappa = \kappa _0 = 0$, the result was proved in [1], Lemma 5.1 by using the comparison function

$$\begin{aligned} v_{t,\varepsilon }(x^{\prime })\, = \left\{ \begin{aligned}&\min \ \left(u (x^{\prime }),t\right) +\, \varepsilon&\text{ for}&x^{\prime }\!\in \{u > h\} \\&u(x^{\prime })&\text{ for}&x^{\prime }\!\in \Omega \setminus \{u > h\}, \end{aligned} \right. \end{aligned}$$

where $\varepsilon$ was chosen to be

$$\begin{aligned} \varepsilon \, =\, \frac{1}{\left|\{u>h\}\right|}\ \int \limits _{A(t)} u - t\, \text{ d}x^{\prime },A(t):\, = \left\{ x^{\prime }\!\in \Omega \!:\, u(x^{\prime })\, >\, t\right\} . \end{aligned}$$

Because of this $v_{t,\varepsilon }$ satisfies the volume constraint and therefore belongs to $\mathcal{C }$. This implies

$$\begin{aligned} \mathcal{F }(c,R;u) \le \mathcal{F }\left(c,R;v_{t,\varepsilon }\right), \end{aligned}$$

(18)

in particular for large values of $t$, and from this inequality it was deduced that there is a $t^{*} > 0$ such that the measure of $A(t^{*})$ vanishes, which means that $u$ is essentially bounded.

The adhesion energy of the body $\mathcal{B }$ is

$$\begin{aligned} \kappa _0 \left\{ \left|\partial \mathcal{B }\right|\, + \int \limits _{B(h_0)\,\cap \{u\,>\,h\}} \sqrt{1 + |Dh|^2}\text{ d}x^{\prime } - \int \limits _{B(h_0)} \sqrt{1 + |Dh|^2}\text{ d}x^{\prime }\right\} , \end{aligned}$$

and this quantity remains unchanged if we replace $u$ by $v_{t,\varepsilon }$. Hence, this term does not affect inequality (18).

To control the adhesion energy of the walls of the container, we have to compare $\kappa \!\int \limits _{\partial \Omega } u\, \text{ d}\sigma$ and $\kappa \!\int \limits _{\partial \Omega } v_{t,\varepsilon }\, \text{ d}\sigma$. Then, we get

$$\begin{aligned} \int \limits _{\partial \Omega } v_{t,\varepsilon }\, \text{ d}\sigma&= \displaystyle \int \limits _{\partial \Omega \,\setminus \,\overline{\{u\,>\,h_0\}}} v_{t,\varepsilon }\, \text{ d}\sigma + \displaystyle \int \limits _{\partial \Omega \,\cap \,\{u\,>\,h_0\}} v_{t,\varepsilon }\, \text{ d}\sigma \\&= \displaystyle \int \limits _{\partial \Omega \,\setminus \,\overline{\{u\,>\,h_0\}}} u\, \text{ d}\sigma + \displaystyle \int \limits _{\partial \Omega \,\cap \,\{u\,>\,h_0\}} \min (u,t) + \varepsilon \, \text{ d}\sigma \\&= \displaystyle \int \limits _{\partial \Omega \,\setminus \,\overline{\{u\,>\,h_0\}}} u\, \text{ d}\sigma +\displaystyle \int \limits _{\partial \Omega \,\cap \,\{u\,>\,h_0\}\,\cap \,A(t)} t + \varepsilon \, \text{ d}\sigma + \displaystyle \int \limits _{[\,\partial \Omega \,\cap \,\{u\,>\,h_0\}\,]\,\setminus \,A(t)} u + \varepsilon \, \text{ d}\sigma . \end{aligned}$$

Recalling the choice of $\varepsilon$, we obtain:

$$\begin{aligned} \int \limits _{\partial \Omega } u\, \text{ d}\sigma \, - \int \limits _{\partial \Omega } v_{t,\varepsilon }\, \text{ d}\sigma&= - \varepsilon \left[\,\left|\partial \Omega \cap \!\left\{ u > h_0\right\} \cap A(t)\right|\right.\\&\left.+\, \left|\,\left[\partial \Omega \cap \!\{u > h_0\}\,\right] \setminus \,A(t)\right|\,\right] + \displaystyle \int \limits _{\partial \Omega \,\cap \, \{u > h_0\}\,\cap A(t)} u - t\, \text{ d}\sigma \\&= -\displaystyle \frac{\left|\partial \Omega \,\cap \,\{u > h_0\}\right|}{\left|\{u > h\}\right|} \int \limits _{A(t)} u - t\, \text{ d}x^{\prime } + \displaystyle \int \limits _{\partial \Omega \,\cap \,A(t)} u - t\, \text{ d}\sigma . \end{aligned}$$

We now apply Emmer’s lemma to the second term and get

$$\begin{aligned} \int \limits _{\partial \Omega } u\, \text{ d}\sigma \, - \int \limits _{\partial \Omega } v_{t,\varepsilon }\, \text{ d}\sigma \, \ge \, - \, \displaystyle \frac{\left|\partial \Omega \cap \{u > h_0\}\right|}{\left|\{u\,>\,h\}\right|}\, \int \limits _{A(t)} u - t\, \text{ d}x^{\prime } \\ - \, \displaystyle \sqrt{1 + L^2}\, \int \limits _{A(t)} |Du| - C \int \limits _{A(t)} (u-t)\, \text{ d}x^{\prime }. \end{aligned}$$

If we use this inequality in (18) and proceed with the other terms as in the proof of Lemma 5.1 from [1], we arrive at the same inequality (33) from [1], only with a different constant. The rest of the proof follows then again from [3], pp. 211–213. $\square$

Lemma 4.2

Let $(c,R;u)$ be a solution to the variational problem

$$\begin{aligned} \mathcal{F }(c,R;u) \rightarrow \min \text{ in} \,\,\mathcal{C }. \end{aligned}$$

(19)

For the coefficients in the energy, we assume

$$\begin{aligned} |\kappa |,\, |\kappa _0|\, <\, \sigma \ll \rho g \end{aligned}$$

in case $\kappa$ or $\kappa _0$ are negative. Then, there holds

$$\begin{aligned} u(x^{\prime })\, >\, h_0 \text{ a.e.} \text{ on}\,\, \Omega _0. \end{aligned}$$

Proof

For $\kappa$ and $\kappa _0 \ge 0$ the proof from [1], Lemma 5.2, applies because if the graph of $u$ is cutoff at some height $t$, the parts of $\partial \Omega$ and $\partial \mathcal{B }$ that are wetted according to the function $\min (u,t)$ have smaller area than before and hence the adhesion energy decreases because of $\kappa ,\kappa _0 \ge 0$. If $\kappa$ or $\kappa _0$ is negative, we therefore cut off $u$ only locally such that the adhesion energy is not affected. As this cannot be done in such a way that the area of $u$ decreases, we must compensate a possible increase in the surface energy by an appropriate change in the gravitational energy. And this can be achieved if $\sigma /\rho g$ is sufficiently small. We now assume the proposition of the lemma not to be true, which means that there is a set $A \subseteq \Omega _0$ of positive measure such that $u(x^{\prime })=h_0\; \forall \, x^{\prime }\!\in A$. As in [1], we distinguish several cases.

(i) :

There is a $\delta >0$, such that

$$\begin{aligned} u(x^{\prime })\, >\, h_0\, +\, \delta \end{aligned}$$

for all $x^{\prime }$ of some subset of $\partial \Omega$ that has positive measure. Then, the boundary value problem

$$\begin{aligned} \left\{ \! \begin{array}{c@{}c@{}l}\displaystyle \text{ div} \left( \frac{Dw}{\sqrt{1+|Dw|^2}} \right)\, =\, 0 \text{ in} \quad \Omega _0\\ w = h_0 \quad \text{ on} \,\,C(h_0),\, w\, =\, \min \left(u,h_0 + \delta \right) \quad \text{ on}\,\, \partial \Omega \end{array}\right. \end{aligned}$$

admits a unique regular solution $w_\delta$, if $\delta$ is small enough. This solution satisfies the strong maximum principle

$$\begin{aligned} h_0\, <\, w_\delta (x^{\prime })\, <\, h_0\, +\, \delta \forall \ x^{\prime } \in \Omega _0, \end{aligned}$$

and therefore,

$$\begin{aligned} u_\delta \, :=\, \min \left(u,w_\delta \right) \end{aligned}$$

satisfies $u_\delta (x^{\prime }) > h_0$, a.e., on $\Omega _0$. Because $w_\delta$ minimizes the area locally we have

$$\begin{aligned} \int \limits _{\{u\,<\,w_\delta \}} \sqrt{1 + |Du_\delta |^2} \ \text{ d}x\, < \int \limits _{\{u\,<\,w_\delta \}} \sqrt{1 + |Du|^2}. \end{aligned}$$

We now change $u_\delta$ such that it includes the same volume as $u$ and hence belongs to $\mathcal{C }$. Let $s$ be a real number such that

$$\begin{aligned} \int \limits _{A(s)} u(x^{\prime })-s \, \text{ d}x^{\prime }=2\cdot V_\delta , \end{aligned}$$

where $V_\delta :\, =\int _{\{u<w_\delta \}} w_\delta - u \, \text{ d}x^{\prime }$.

Then, we choose $\varphi \in C^1_c(A(s))$ such that $\int _{A(s)} \varphi (x^{\prime }) \, \text{ d}x^{\prime }=1$. We have $u_{\delta ,t}(x^{\prime }) \equiv u_\delta (x^{\prime }) - t\varphi (x^{\prime }) = u(x^{\prime })$ for all $x^{\prime }\!\in \partial \Omega \,\cup \,C(h)$, and the adhesion energy remains the same when we consider $u - t\varphi$ instead of $u$. With $t = V_\delta$ we have $\int _{\Omega } u_\delta - t\varphi \, \text{ d}x^{\prime } = \int _{\Omega } u\, \text{ d}x^{\prime }$, and for the area we get

$$\begin{aligned}&\int \limits _{\Omega \,\cap \,\{u\,>\,h\}} \sqrt{1 + \left|\,D(u - t \varphi )\,\right|^2}\, \text{ d}x^{\prime } \\&\quad = \int \limits _{\Omega \,\cap \,\{u\,>\,h\}} \sqrt{1 + |Du|^2} \, \text{ d}x^{\prime } - t \int \limits _{A(t)} \frac{Du \cdot D\varphi }{\sqrt{1 + |Du|^2}}\ \text{ d}x^{\prime } +\, \text{ O}\left(t^2\right) \end{aligned}$$

hence,

$$\begin{aligned} \int \limits _{\Omega \,\cap \,\{u\,>\,h\}} \sqrt{1 + \left|D(u - t\varphi )\right|^2}\text{ d}x^{\prime }\, \le \, \int \limits _{\Omega \,\cap \,\{u\,>\,h\}} \sqrt{1 + |Du|^2}\text{ d}x^{\prime } +\, V_\delta \int \limits _{A(s)} |D\varphi |. \end{aligned}$$

The change in the gravitational energy is, up to the constant $\frac{\rho g}{2}$:

$$\begin{aligned}&\int \limits _{\Omega \,\cap \,\{u\,>\,h\}} u^2 - h^2\, \text{ d}x^{\prime } - \int \limits _{\Omega \,\cap \,\{u\,>\,h\}}(u - t\varphi )^2 - h^2\, \text{ d}x^{\prime } \\&\quad =\displaystyle \int \limits _{A(s)}u^2 - (u-t\varphi )^2\, \text{ d}x^{\prime } + \int \limits _{\Omega _0\,\cap \,\{u\,<\,w_\delta \}}u^2 - w^2_\delta \, \text{ d}x^{\prime }. \end{aligned}$$

We have

$$\begin{aligned} \int \limits _{\Omega _0\,\cap \,\{u\,<\,w_\delta \}} u^2 - w^2_\delta \, \text{ d}x^{\prime }&= \int \limits _{\Omega _0\,\cap \,\{u\,<\,w_\delta \}} (u + w_\delta )\,(u - w_\delta )\, \text{ d}x^{\prime }\, \\&\ge \min _{\Omega _0\,\cap \,\{u\,<\,w_\delta \}} (u + w_\delta )\ \cdot \int \limits _{\Omega _0\,\cap \,\{u\,<\,w_\delta \}} u - w_\delta \, \text{ d}x^{\prime }\, \ge \, -\ 2h_0\, V_\delta , \end{aligned}$$

and in the same way

$$\begin{aligned} \int \limits _{A(s)} u^2 - (u - t\varphi )^2\, \text{ d}x^{\prime }\, \ge \, 2s \cdot V_\delta , \end{aligned}$$

such that we get for the total energy

$$\begin{aligned} \mathcal{F }(R,c;u)\, -\, \mathcal{F }\left(R,c;u_{\delta ,t}\right)\, \ge \, -\, \sigma \cdot V_\delta \cdot C\, +\, \frac{\rho g}{2} \left(2s - 2h_0\right)V_\delta \, >\, 0, \end{aligned}$$

and this is a contradiction to $u$ being a minimizer of $\mathcal{F }$.

(ii) :

There is a $\delta > 0$, such that

$$\begin{aligned} u(x^{\prime }) > h_0 + \delta \end{aligned}$$

for all $x^{\prime }$ of some subset of $C(h_0)$ that has positive measure. As in [1], Lemma 5.2, we construct a function $w_\delta$ that minimizes $\int _{\Omega _0} \sqrt{1 + |Dw|^2}$ among all functions in $BV(\Omega _0)$ with boundary data $w = u$ on $\partial \Omega _0$. We then correct the comparison function such that it satisfies the volume constraint; this can be done as in (i) above.

(iii)If $u(x^{\prime }) > h_0 + \delta$ holds on subsets of positive measure of $C(h_0)$ and of $\partial \Omega$, then we can argue as in (i). Therefore, it remains to consider the case that $u(x^{\prime }) = h_0$ on all of $C(h_0)$ and $\partial \Omega$.

(iv)Now assume that $u(x^{\prime }) \equiv h_0$ on $\Omega _0 \cup C (h_0) \cup \partial \Omega$. Then, $u$ cannot be minimizing because the function $u_{\alpha ,t}$ has less energy, where $u_{\alpha ,t}$ is constructed in the following way. We choose $\alpha \ll 1$ and set $u_\alpha := \min (u,h_0+\alpha )$. Then, $u_\alpha$ is not an element of $\mathcal{C }$ because its volume exceeds that of $u$ by $V_\alpha = \int _{\{u_\alpha > u\}} u_\alpha - u\,\text{ d}x^{\prime }$. Hence, we correct $u_\alpha$ as in part (i) of the proof, and then, $\mathcal{F }(u_{\alpha ,t}) < \mathcal{F }(u)$ follows as before.

(v)If $u = h_0$ on $C(h_0) \cup \partial \Omega$, but not identically $h_0$ on $\Omega _0$, we can use the fact that $u$ is a minimizer with respect to variations that have compact support in $\Omega _0$. Then, the adhesion energy is not affected, and we can therefore proceed as in [1], Lemma 5.2,  (iv). $\square$

Lemma 4.3

Let $(c,R;u)$ be a solution of

$$\begin{aligned} \mathcal{F }(c,R;u) \rightarrow \min \text{ in}\,\, \mathcal{C }. \end{aligned}$$

(20)

(i)If $\kappa _0 \!\le \! 0$, and if $\mathcal{B }$ is not completely wetted, that is, the set $\{ x^{\prime }\!\!\in \!\! B (h_0)\!:\! u(x^{\prime }) \!=\! h(x^{\prime })\}$ has positive measure, then $u$ is regular: $u \in C^{2 + \alpha }(\Omega \,\cap \,\{u > h\})\, \cap \, C^{1 + \alpha } (\overline{\Omega \cap \{u > h\}})$. Furthermore, $u$ meets the body $\mathcal{B }$ under a constant angle $\vartheta$ with $\cos \vartheta = -\kappa _0{\big /}\sigma$, and the boundary $\gamma$ of $\Omega \,\cap \,\{u > h\}$ is locally a $C^{1 + \alpha }$-curve.

(ii)If $|Du(x^{\prime })| < \infty$ for all $x^{\prime }\!\in \gamma$ then the set $\Omega \,\cap \{u > h\}$ is connected, or equivalently the non-wetted part of $\partial \mathcal{B }$ is simply connected.

Proof

The condition on the coincidence set implies that the capillary surface meets the body $\mathcal{B }$. Then, the boundary $\gamma$ is locally a $C^{1+\alpha }$-curve according to the regularity theorem of Taylor [5]. This result is proved for any $\kappa _0$ with $|\kappa _0|<\sigma$ in the context of geometric measure theory, and hence the set $E$ that is occupied by the fluid may be any Caccioppoli set $E$ and not necessarily the subgraph $U := \{x \in G\!:\, h(x^{\prime }) < x_3 < u(x^{\prime })\}$ of some function $u \in BV(\Omega )$.

We now restrict the adhesion coefficient $\kappa _0$ in such a way that the corresponding contact angle $\vartheta$ with $\cos \vartheta = -\, \kappa _0{\big /}\sigma$ can be realized in the class of graphs. If we have a priori no knowledge about the curve $\Gamma$ in which $\partial \mathcal{B }$ and $\Sigma$ meet, the vertex of $\partial \mathcal{B }$ might be a point of $\Gamma$ and consequently the normal to $\partial \mathcal{B }$ at that element of $\Gamma$ points into the $x_3$-direction. Then, $\Sigma$ can only be a graph if $\kappa _0 \le 0$ because then $\vartheta$ ranges between $0$ and $\pi /2$. If it is known that the normal vectors in points of $\Gamma$ make an angle $\varphi$ with the $x_3$-direction that is larger or equal to some $\varphi _0 > 0$ then one can allow for a contact angle between $0$ and $\frac{\pi }{2} + \varphi _0$, and consequently the lemma holds also for the corresponding positive values of $\kappa _0$.

If (ii) were not true, there would be an open set $M\!\subset \subset \!B (h_0)$ with $u(x^{\prime }) > h(x^{\prime })\; \forall \ x^{\prime } \in M$ and $u(x^{\prime }) = h(x^{\prime })\; \forall \ x \in \partial M$. Then, there would exist at least one point $x^{\prime }_0 \in M$ such that

$$\begin{aligned} h\left(x^{\prime }_0\right)\, \le \, h\left(x^{\prime }\right)\,\, \forall x^{\prime } \in \overline{M}. \end{aligned}$$

(*)

Clearly $x^{\prime }_0$ must lie on the boundary $\partial M$, because for $x^{\prime }_0 \in M$ there would be some point $x^{\prime }_1 = x^{\prime }_0 - \varepsilon \, Dh(x^{\prime }_0)$ which lies in $M$ for $\varepsilon$ small enough and for which $h(x^{\prime }_1) < h(x^{\prime }_0)$ holds. This follows from the fact that $h$ is strictly concave and therefore $Dh(x^{\prime }_0) \ne 0$. In such a point $x^{\prime }_0 \in \partial M \equiv \gamma$ the tangent vector $\tau$ to the curve $\Gamma = \text{ graph}\ h (\gamma )$ satisfies $\tau _3 (x^{\prime }_0) = 0$. Let $x^{\prime } = x^{\prime }(t), |t| < \varepsilon$, be a representation of the curve $\gamma$ with $x^{\prime }(0) = x^{\prime }_0$. Then, $x(t) = \left(x^{\prime }(t),\, h\left(x^{\prime }(t)\right)\right),\, |t| < \varepsilon$, describes $\Gamma$ locally around $x_0 = \left(x^{\prime }_0, h(x^{\prime }_0)\right)$. Because of $(*)$ we have in particular

$$\begin{aligned} h\left(x^{\prime }(0)\right)\, \le \, h\left(x^{\prime }(t)\right) \forall t \in (-\varepsilon ,\varepsilon ), \end{aligned}$$

hence

$$\begin{aligned} 0\, =\, \frac{d}{\text{ d}t} h\left(x^{\prime }(t)\right)_{|_{t=0}}&= \displaystyle Dh\left(x^{\prime }_0\right) \cdot \frac{d}{\text{ d}t} x^{\prime }(t)_{|_{t=0}}\\&\equiv \tau _{\varepsilon }(x_0). \end{aligned}$$

Now let $n = n(x^{\prime }_0)$ be the direction perpendicular to $\gamma$ and pointing into $M$. Then, $h_n(x^{\prime }_0)\, :=\, Dh(x^{\prime }_0) \cdot n(x^{\prime }_0) > 0$ and because of $u_\tau (x^{\prime }_0) = h_\tau (x^{\prime }_0) = 0$ we have

$$\begin{aligned} u_n\left(x^{\prime }_0\right)&= \displaystyle -\, \frac{1}{h_n(x^{\prime }_0)} < 0. \end{aligned}$$

This is, however, a contradiction to $u(x^{\prime }) > h(x^{\prime })\forall \ x^{\prime } \,\,\in M$, and the lemma is proved. $\square$