Sep 30, 2007 - Neel and Siegel, but the general existence problem for arbitrary drop ...... David Siegel, Equilibrium configurations for a floating drop, J. Math.
Floating drops and functions of bounded variation Alan Elcrat
∗
Ray Treinen
†
September 30, 2007
Abstract A variational problem for three fluids in which gravitational and surface tension forces are in equilibrium is studied using sets of finite perimeter and functions of bounded variation. Existence theorems are proven which imply the existence of an axisymmetric floating drop. This problem has been studied previously as a free boundary problem for axisymmetric capillary surfaces by Elcrat, Neel and Siegel, but the general existence problem for arbitrary drop volumes was left open. The results presented here settle a version of this question.
1
Introduction
We consider here an equilibrium problem for capillary surfaces in which there are three fluids and, in general, three bounding surfaces. In configurations with rotationally symmetric containers, symmetry may be expected, and possible solutions to the energy minimization problem can be given by curves in the meridian plane. Minimization of the potential energy for this equilibrium problem then leads formally to a free boundary problem for axisymmetric capillary surfaces in which a sessile drop, an interior capillary surface, and an exterior capillary surface meet at a point. This differential equation problem has been studied extensively both theoretically and numerically by each of us [2], [3], [11]. A typical configuration is shown in Figure 1.
0.6
0.4
height(u)
0.2
0
−0.2
−0.4
−0.6
−0.8 −0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
radius(r)
Figure 1: A typical floating drop. 1 Department 2 Department
of Mathematics and Statistics, Wichita State University, Wichita, KS, 67218 of Mathematics, Kansas State University, Manhattan, KS, 66502
1
There are difficulties, however, in establishing existence in full generality using these methods. We reconsider the basic problem here using variational methods in spaces of functions of bounded variation. See for example, Ambrosio, Fusco, and Pallara [1], Evans and Gariepy [5], Giusti [7], Massari and Miranda [10], or Ziemer [14] for background on functions of bounded variation. See also Emmer [4], Gonz´alez [8] and Vogel [12], [13] for examples of just some of the many applications to capillary surfaces. The general setup follows. Let Ω ⊆ R3 be connected and open. The fluids are represented by the sets E0 , E1 , E2 , where Ω = ∪Ei and Ei ∩ Ej = ∅ for i 6= j. Let ρ0 , ρ1 , ρ2 be densities and such that ρ0 ≤ ρ1 ≤ ρ2 . Let σ01 , σ02 , σ12 be surface tensions. Also, all the multiple indices are symmetric: eg. σij = σji . In order that the fluids remain separate we assume σ01 , σ02 , σ12 > 0. The surface tensions will determine the contact angles with the wall and at any point where all three fluids meet. In order to be able to achieve force balance at any point of contact of three fluids we assume for i, j = 0, 1, 2 each σij is less than the sum of the other two [2]. Here Sij := ||∂Ei ∩ ∂Ej ∩ Ω||H2 are the surface areas of each surface and g is the acceleration due to gravity. Now Z 3 X Sij . (1) |DχEi | = Ω
j=0 j6=i
Define α0
=
α1
=
α2
=
1 (σ01 + σ02 − σ12 ) 2 1 (σ01 + σ12 − σ02 ) 2 1 (σ02 + σ12 − σ01 ) 2
that is, σ01 σ12
= =
α0 + α1 α1 + α2
σ02
=
α0 + α2 .
Formally define the energy functional to be FΩ (E0 , E1 , E2 ) =
Z Z 2 X αi |DχEi | + ρi g Ω
i=0
Ei
z dV
.
(2)
We will sometimes refer to this as F when clear, or as F(E0 , E1 , E2 ; Ω) when the subscript notation becomes cumbersome. Note that wetting energy terms could be included on ∂Ω, however this generality is not needed in the solution of our problem. Given a Borel set E, Giusti [7, Proposition 3.1] states that we may alter E on a set of measure zero so that 0 < |E ∩ B(x, ρ)| < ω3 ρ3 (3) for all x ∈ ∂E and all ρ > 0 where ω3 is the measure of the unit ball in R3 . We will assume that all the sets in this paper are normalized in this way.
2
The floating drop
If (2) is computed for the configurations in [2], the value will not be finite because E0 and E2 have infinite volume. We will approach this problem by considering minimizers over a certain class of admissible sets 2
in bounded subsets of R3 and using a limiting procedure we will show that there is a minimizer over a certain class of admissible sets to (2) restricted to arbitrary compact sets. Prescribe the volume v1 of the “drop”, that is the volume of the set E1 . Definition 2.1 The triple (E0 , E1 , E2 ) is radially symmetric if E1 = {r < r(z)}, E0 = {r ∈ ∪I0 (z)}, and E2 = {r ∈ ∪I2 (z)} for some intervals I0 (z) and I2 (z). Theorem 2.2 If (E0 , E1 , E2 ) is the limit of radially symmetric triples, then (E0 , E1 , E2 ) is radially symmetric. Proof. First consider E1 . Denote by E1n a convergent sequence converging to E1 , where E1n is part of a radially symmetric triple. Assume that rn (z) > 0 does not converge to r(z) > 0 for z in a set of positive measure. That is |rn − rm | ≥ δ > 0 for a set of positive z-measure where rn , rm > 0, for simplicity say rn − rm ≥ δ > 0 and Z Z rn Z π r2 dr dz ≥ δC > 0 dx = π (4) |E1n − E1m | = n m 3 rm E1 −E1 for some positive constant C. This contradicts E1n → E1 in L1 , so rn (z) → r(z) a.e. in z. Then for E0 consider (an (z), bn (z)) where 0 < an , bn < ∞. If E0n (part of a radially symmetric triple) is converging in L1 and an (z), bn (z) are not converging, then we may assume |an (z) − am (z)| ≥ δ > 0 and |bn (z) − bm (z)| ≥ δ > 0 in a set of positive measure (passing to a subsequence if necessary). Then Z dV |E0n − E0m | = E0n −E0m
= π =
π 3
Z Z
Z
bn
2
r dr −
an
Z
bm
2
r dr
am
!
dz
π b3n − a3n − b3m + a3m dz ≥ δC 3
(5)
for some positive constant C. This contradicts E0n → E0 in L1 , so an (z) → a(z) and bn (z) → b(z) a.e. in z. The case where the interval is semi-infinite is completely similar to the case for E1 . The proof that E2 is radially symmetric is the same as for E0 . Definition 2.3 A minimizer over radially symmetric, compact perturbations is the radially symmetric triple (E0 , E1 , E2 ), where each Ei is a set of locally finite perimeter, if for a given compact set K the following conditions hold: R 1. Ω χE1 = v1 ,
2. RFor any radially symmetric (A0 , A1 , A2 ), where each Ai is a set of locally finite perimeter and Ω χA1 = v1 , such that Ei △ Ai ⊂ K, then (E0 , E1 , E2 ) satisfies FK (E0 , E1 , E2 ) ≤ FK (A0 , A1 , A2 ).
Here △ is the symmetric difference: E △ A = (E − A) ∪ (A − E). We also require that the solutions are non-trivial, in that they have three fluids, and E0 , E2 are both semi-infinite. We will need the following special case of a result proved by Massari [9]: Theorem 2.4 Let Ω be bounded, αi ≥ 0 and σij > 0. Then there exists (E0 , E1 , E2 ), where Ei are sets of finite perimeter, that minimizes F over K
:=
{(E0 , E1 , E2 )| ∪ Ei = Ω, Ei ∩ Ej = ∅ for i 6= j P |Ei | = vi , prescribed volumes, where vi = |Ω|, and Ei are sets of finite perimeter}. 3
(6)
The following corollary follows immediately from Massari’s proof and Theorem 2.2. Corollary 2.5 Let the hypotheses of Theorem 2.4 hold, and additionally assume that Ω is symmetric about the vertical axis. There exists radially symmetric (E0 , E1 , E2 ), where Ei are sets of finite perimeter, that minimizes F over K′ := K ∩ {(E0 , E1 , E2 ) is radially symmetric}. We will need the following: Theorem 2.6 The minimum for the functional F with all σij = 0 in a bounded, connected domain Ω over K is achieved by a configuration consisting of flat interfaces between the fluids, which are ordered decreasing with higher density. Proof. Define the height of the interface between E0 and E1 to be z1 = sup{z : (x, y, z) ∈ E1 } and define the height of the interface between E1 and E2 to be z2 = sup{z : (x, y, z) ∈ E2 }. Assume that there exists (A0 , A1 , A2 ) such that F(A0 , A1 , A2 ) < F(E0 , E1 , E2 ) − δ
(7)
for some δ > 0. This means that (A0 , A1 , A2 ) must differ from (E0 , E1 , E2 ) on a set of positive measure. Recall that we are asuming that the sets are normalized as in (3). Thus not all of the following are true: 1. A0 = E0 , 2. A1 = E1 , 3. A2 = E2 , where we are identifying sets equal up to measure zero. First, assume that 1. is not true. Then A˜0 = {z < z1 } ∩ A0
(8)
has positive measure and must displace some fluid below z = z1 . Then the volume constraints force one or both of 2. or 3. to be false. Thus some of A1 or A2 or both must be above z = z1 . Label A˜1 A˜2
= =
{z > z1 } ∩ A1 {z > z1 } ∩ A2 .
(9) (10)
One of A˜1 or A˜2 must have positive measure. By volume considerations |A˜1 | ≤ |A˜0 | and |A˜2 | = |A˜0 |−|A˜1 |. Select A ⊂ A˜0 such that |A| = |A˜1 | and B ⊂ A˜0 − A such that |B| = |A˜2 |. Then switch: identify A with A1 and A˜1 with A0 ; identify B with A2 and A˜2 with A0 . This lowers the value of the energy functional. Note that 1. is now true. Volume considerations now imply that 2. is false if and only if 3. is false. Now label A˜1 A˜2
=
{z < z2 } ∩ A1
(11)
=
{z > z2 } ∩ A2 .
(12)
We deduce that |A˜1 | = |A˜2 | and we switch by identifying A˜1 with A2 and A˜2 with A1 . This lowers the value of the energy functional and supplies a contradiction as our new sets A0 , A1 , A2 are equal to E0 , E1 , E2 respectively. The need for Ω to be connected is so that the fluids can move freely under perturbations. The main result of the paper is 4
Theorem 2.7 For a given volume v1 there exists radially symmetric (E0 , E1 , E2 ) that minimizes FK over compact symmetric perturbations in Ω = R2 × (−T, T ) for any compact set K. Proof. Consider the cylinder C(n), with n large enough that v1 < |C(n)|. Set v0n = 21 (|C(n)| − v1 ) and v2n = 21 (|C(n)| − v1 ) and consider the minimizer of the functional F over K′ in the bounded domain C(n). We denote this minimizer in this cylinder as (E0n , E1n , E2n ) and note the triple is radial, as follows from Corollary 2.5. Our first goal is to find a convergent subsequence of E1n as n → ∞. We will need to following two lemmas to show this. Lemma 2.8 E1n is uniformly bounded in Ω for all n. Proof. To show that E1n is uniformly bounded in {−T < z < T } we first assume otherwise. That is for every a > 0 there is an n large enough that rn > a where rn is the maximal radius of E1n . Now ! Z Z 2 X αi χEin zdV |DχEin | + ρi g i=0
>
C(n)
C(n)
α1 πrn2 + πσ02 n2 +
2 X
ρi g
C(n)
i=0
>
=
Z
χEin zdV
(13)
( 1 ρ0 ρ2 α1 πrn2 + πσ02 n2 + gπ 2 n4 2 (2v0 T − πn2 ) + 2 (πn2 − 2v2 T ) + 2 v0 v2 ) ρ1 2 2 2 2T (v0 − v2 )v0 v2 + πn (v2 − v0 ) v02 v22 2 gπ 2 n4 πn 2 2 α1 πrn + πσ02 n + (ρ2 − ρ0 ) − 2T 2v2 v2
(14) (15)
as we replace the surface area terms of the drop with the area of a disc and the the remaining surface area terms with the area of a washer. Clearly this reduces the surface area. The gravity terms are reduced by replacing them with value of F from the sets from Theorem 2.6 with v0 = v2 . This is no longer an admissible configuration, but it is a lower bound. Next we see that ! Z Z 2 X αi χEin zdV |DχEin | + ρi g i=0
≤ =
C(n)
C(n)
πr12 (σ01 + σ12 ) + 2πr1 α1 πr12 + πσ02 n2 +
1 v1 2 πr12
(σ01 + σ12 ) + πσ02 (n2 − r12 ) +
2 X i=0
v1 π (σ01 + σ12 ) + (ρ2 − ρ0 )g (r12 h2 − 4T 2 n2 ) r1 8
ρi g
Z
C(n)
χAni zdV
(16) (17)
where A1 = An1 is the cylinder of radius r1 and height following from v1 = πr12 h that is divided evenly by the xy-plane, An2 = (C(n) − A1 ) ∩ {z < 0} and An0 = C(n) − (A1 ∪ An2 ). The sets An0 , A1 , An2 form an admissible configuration for C(n) and so have energy no less than that of the minimizer. The gravity terms reduce here due to the symmetry of the configuration about the xy-plane and integration. Combining these two inequalities we obtain 2 πn gπ 2 n4 2 2 (ρ2 − ρ0 ) − 2T α1 πrn + πσ02 n + 2v2 v2 v1 π < α1 πr12 + πσ02 n2 + (σ01 + σ12 ) + (ρ2 − ρ0 )g (r12 h2 − 4T 2 n2 ) (18) r1 8 5
T
−n
−R
R
E n1
n
−T
Figure 2: The set E1n bounded in the cylinder of radius R, height 2T , independently of C(n).
which, as πn2 v2 = 2
πn2 2T − , v1
(19)
we can rewrite as α1 π(rn2 − r12 ) −
v1 (σ01 + σ12 ) r1
2 π gπ 2 n4 πn < (ρ2 − ρ0 )g (r12 h2 − 4T 2 n2 ) − (ρ2 − ρ0 ) − 2T 8 2v2 v2 " !# 2 πn π 2 2 π 2 2 2 r h − n T − = (ρ2 − ρ0 )g 2 2 − 2T 8 1 2 2T − πn 2T − πn v1 v1
(20) (21)
which goes to −∞ as n → ∞. This contradicts the assumption that rn → ∞. Thus we know that there is a large enough radius R where E1n ⊂ C(R) for all n. See Figure 2.
Lemma 2.9 supn
R
C(R)
|DχE1n | < ∞.
Proof. Note that FC(n) (E0n , E1n , E2n ) ≤ FC(n) (An0 , An1 , An2 ) for all admissible A’s. In particular we can set A1 = An1 to be the cylinder of height 2T with radius r1 that satisfies v1 = 2πr12 T . Since E1n ⊂ C(R) we may set An2 = E2n and An0 = E0n outside of C(R) and within C(R) − C(r1 ) set An2 to be everything below a height hn which is determined to get the volumes correct. 6
From the inequality for the energy we obtain, primarily by bounding the trace on ∂C(R): α1
Z
|DχE1n |
C(R)
≤
2 X
αi
i=0
− α0
Z
Z
C(n)
|DχAni | + ρi g
|DχE0n | − α2
C(n)
< α1 4πr1 T + α0
Z
Z
C(n)
Z
χAni zdV − ρi g
C(n)
χEin zdV
|DχE2n |
(22)
C(n) 2
8πT (r1 + R) + 2π(R −
r12 )
−
Z
!
|DχE0n |
C(R)
+ α2
!
2
8πT (r1 + R) + 2π(R −
r12 )
−
Z
!
|DχE2n |
C(R)
+
2 X i=0
< 4πα1 r1 T + 2π(α0 + α2 )(R + r1 )(4T + R − r1 ) + 2 (Σρi ) g
Z
ρi g
Z
C(R)
zχAni − zχEin dV (23)
zdV.
(24)
C(R)
The right side does not depend on n, thus taking supremum with respect to n from both sides implies R supn C(R) |DχE1n | < ∞.
Set Ω = C(R). As E1n ⊂ Ω which is bounded and open, the general compactness theorem for BV n functions applies and this implies the existence of a limit to a subsequence of E1n . Denote E1 = lim E1 j . Recall that (E0n , E1n , E2n ) is the minimizer on the cylinder C(n). Next we show that E0n and E2n have convergent subsequences with limits E0 , E2 such that ∪Ei = Ω. p First extend each E0n to include p the set A˜n0 = {(x, y, z) : n < x2 + y 2 , 0 < z < T }. Similarly extend E2n to include A˜n2 = {(x, y, z) : n < x2 + y 2 , −T < z < 0}. Now each of E0n and E2n are defined outside of C(n), and can be components of an admissible configuration on larger cylinders. Consider a cylinder C(k) of radius k. Denote restrictions of E0n to C(k) by E0n (k) := E0n ∩ C(k) and similarly for E2n .
Lemma 2.10 The sequences of sets E0n (k), E2n (k) have convergent subsequences on the cylinder C(k), denoting the limiting sets by E0 (k) and E2 (k) respectively.
Proof. Now F(E0n , E1n , E2n ; C(n))
= +
F(E0n , E1n , E2n ; C(k)) + F(E0n , E1n , E2n ; C(n) − C(k)) Z 2 X αi (T + χEin − T − χEin )dH2 . i=0
(25)
∂C(k)∩Ω
Observing (25) becomes an inequality < if the sum is replaced with 2πkT (α0 + α1 + α2 ), and if k > R then we may use 2πkT (α0 + α2 ), then as (E0n , E1n , E2n ) is a minimizer, for sets (E0k , E1k , E2k ) that minimize F on C(k), we have F(E0n , E1n , E2n ; C(n)) ≤ F(E0k , E1k , E2k ; C(n)). Then writing everything out explicitly: 7
2 X
αi
C(k)
i=0
2 X
+
Z
ρi g
0, we rearrange (26):
|Dχ
E0n
C(k)
| + α2
Z
|Dχ
E2n
|
