Master part 3
This is the third part of my series of posts regarding my master thesis. Here I follow the article of Massari [1] regarding the formulation of the problem and the proof of the existence result in the case of two fluids. Although the ideas are the same, the proof presented here is different from the one given in the referred article.
Suppose we have two fluids , with the considerations above. Then the energy functional has the following form
but since almost everywhere on , we see that the position of is uniquely determined by the position of , so it is obvious that the problem
has a solution if and only if the problem
This formulation can be proven to be equivalent to the initial one, by replacing in . Notice that the term is in fact equal to , since we have only two fluids, and the measure interface between them, which is inside , is equal to their relative perimeter with respect to . Also, notice that . Then we have
where . Notice that this has the same form as (2) with and except for the constant , which does not affect the existence of the minimizer.
There are a few simple conditions which are needed for the existence of the minimizer for (2). First note that the term is bounded by and the last term is bounded by . If we have then is not bounded from below, and the minimizer does not exist. Therefore, a necessary condition for the existence of a minimum is . This is a reasonable condition from a physical point of view, because it simply says that , which is true. In fact , or else the two fluids would mix, which is not the case here. Hence, in the following we will use the condition .
Therefore, if we know that is bounded from below. Suppose is a minimizing sequence. Then is bounded above by a constant , and therefore
Therefore is bounded and since , by the compactness property it follows that the sequence has a convergent subsequence in the topology. Since the perimeter is lower semicontinuous for the topology, it follows that the limit point of the convergent subsequence of is also a set with finite perimeter. If we could prove that is lower semicontinuous with respect to the topology, then the existence of a minimizer would be proved.
First note that if then the functional is not lower semicontinuous.
Example 1 Take (a cylinder with two halfspheres glued at its ends). Note that has boundary and the interior sphere property: there exists such that for any there exists a ball of radius such that .
Take the open unit ball centered in origin, and denote the translation of with vector , where .
Note that for we have and converges to in as . First, let’s notice that for and , where is the gravity term, and it is continuous.
On the other hand , and therefore . This limit is strictly positive for and it follows that is not necessarily lower semicontinuous if .
Example 2 Consider the cube and the unit open ball. Define . Then is open and has boundary and the interior sphere property.
Let’s consider now sets such that and is the complement in of a compact subset . In particular, this means that , so the functional has the following form
where is the gravity term, and it is continuous. Consider sets of the form such that . The volume of can explicitly be calculated in terms of , because can be split in parts which form a cube of side , six boxes of size , three cylinders with radius and height and a ball of radius ,
note that for every such that we can find such that . Denote the positive solution of this equation and note that as we must have . Therefore, the diameter of the figure in the direction of the axes , which is equal to tends to which is the diameter of the sphere of volume .
Pick such that the sphere of volume centered at the origin doesn’t fit inside (pick the diameter of the sphere equal to where is small enough. By the continuity of we can deduce that there is an such that is equal to , and since the ball of radius centered at the origin does not fit in . Moreover, by implicit differentiation, we see that is increasing. Consider now the limit
Notice now, that if then and the result of the above limit is strictly positive. This proves that is not necessarily lower semicontinuous if .
In the following, for a function we denote by its trace on .
Definition 1 Suppose is an open set with Lipshitz boundary and . For every we define
As noted in [2] if has boundary of class then and for open sets with Lipschitz boundary we have where is the Lipschitz constant of . The following trace inequality is proved in [2].
Proposition 2 Suppose is a bounded open set with Lipschitz boundary and . Then for every there exists a constant such that for every we have
With the aid of the above result we will prove an inequality which is of great importance in the proofs of the theorems in this chapter.
Theorem 3 Suppose is an open set with Lipschitz boundary, and as in Definition 1. Define . Then for every there exists a constant such that for every we have
Proof: We will use Proposition 2 for the function where if and .
We have
and note that we can take . \hfill
Italo Tamanini proves in [3] a similar inequality for open sets with the interior sphere condition, i.e. there exists some such that for any there exists a ball of radius such that . Equivalently, the curvature of is bounded from above. This inequality states that for every set with finite perimeter we have
where is a constant which depends on and . This is a stronger inequality than (4), but its proof uses in an essential way the interior sphere property of , and cannot be directly generalized to sets with Lipschitz boundary. This inequality is used in [1] in the proof of the existence result, but here we will use the inequality (4), as the proof still works with this weaker inequality.
Theorem 4 Suppose is a bounded open set with boundary. If and then the functional is lower semicontinuous for the convergence of characteristic functions and therefore the problem (2) has a solution.
Proof: Let’s consider cases separately. If then the perimeter and boundary terms give which is lower semicontinuous. If then the perimeter and boundary terms differ by a constant of which is also lower semicontinuous. Therefore these cases do not pose any problem, and from now on we suppose that .
Consider a sequence of sets of finite perimeter such that in the convergence. Consider small enough and note that for considered in the hypothesis we have . We denote with the gravity term associated to and note that is continuous with respect to the convergence. Denote . Then applying inequality (4) we have
By choosing small enough such that , using the fact that is continuous and the total variation is lower semicontinuous we obtain that
Since is arbitrary small, we can deduce that
which means that is lower semicontinuous for the convergence of characteristic functions.
The boundedness condition (3) together with the compactness property now lead to the existence of a minimizer for problem (2). \hfill
Since the fact that has boundary is a bit restricting (for examples, squares, rectangles, cubes and cylinders are not included in this class), we can state another variant of the result for domains with Lipschitz continuous boundary. Note that for with Lipschitz continuous boundary we have where is the Lipschitz constant of . It is easy to see that if we impose the stronger hypothesis then the proof of the next result works exactly as the proof of the above theorem.
Theorem 5 Suppose is a bounded open set with Lipschitz continuous boundary. If and then the functional is lower semicontinuous for the convergence of characteristic functions and therefore the problem (2) has a solution.
[1] Massari, U., The parametric problem of capillarity: The case of two and three fluids
[2] Anzellotti, G. and Giaquinta, M., BV functions and traces
[3] Tamanini, Italo, Il problema della capillarita su domini non regolari

March 1, 2013 at 11:59 amMaster 4  Beni Bogoşel's blog

March 1, 2013 at 12:46 pmMaster 5  Beni Bogoşel's blog