PHD 4324 (Indiana)

Proof: If we assume that then there exists a subsequence denoted also such that . Then is bounded in and is reflexive there exists a subsequence converges weakly to in …

The problem of finding a minimal resistance body (due to Newton) consists of finding the shape of a body which travels in a straight line through a fluid when we are given a certain fixed section of it, orthogonal to the flow of the fluid. The classical problem presented here makes some assumptions about the fluid and about the movement, which are not really accurate, taking into account the physics of the fluids, but which turns to be a good approximation in the case the liquid is rare, such as the movement of the airplanes. The assumptions made are:

• the single shock property: every particle which hits the body is reflected and it doesn’t influence the behavior of other particles in the fluid, moreover, if a particle hits the body, it never touches the body after that moment.
• the part of the body below the fixed orthogonal section is neglected, which means that it is considered that its resistance is zero.

Considering a point ( denotes the fixed orthogonal section), the force with which a particle hitting the body in the point holds the body back can be calculated decomposing the normal force as in the figure to be proportional to , where is the angle made by the normal to the surface of the body with the direction of the flow.  Using the equality and the fact that the slope of the tangent line to the body we are motivated to chose the resistance functional by  taking the mean of all such local resistances over , resulting in the formula .

Our problem is minimizing on some class of admissible functions . Choosing the right class of admissible functions is important, because of the following:

• if we do not impose a boundedness condition on , then we can take to be the function which gives us a very long cone. For this cone, the slope of its generator is very large, and that slope is counted in . Since this is in the denominator, the resistance gets very small, as we take a higher and higher cone. Then the infimum of the resistance is zero, but is never attained, yielding no solution for the problem.
• taking as an admissible class the functions bounded above by some constant , we still don’t have a solution. This is because we can chose a shape with many zig-zag’s, i.e. imagine the picture rotated about a vertical axis. This produces a shape which comes again from a function with gradient of great size (the steeper the slopes of the zig-zag’s, the greater the size). Therefore, in this case the infimum is zero and is never achieved. At this point, one may argue that in reality, the shapes considered are obviously not optimal, because the more zig-zag’s we put, the larger the resistance will be. This strange fact is a consequence of the property of ‘single-shock’, considered in our assumptions, property which is does not hold in reality.
• 

Moreover, there may be some problems due to the functional , which is small as . This means that our functional is not coercive, and we cannot apply the direct methods of the calculus of variation. The admissible class which assures us of the existence of the solutions is .

Compactness lemma: For every and every the class is compact with respect to the strong topology of .

Proof:  Let be a sequence of elements of ; since all are concave, they are locally Lipschitz continuous on , that is , where is a suitable constant. From the fact that , the constants can be chosen independent of ; we can take . Therefore is equi-Lipschitz continuous and equi-bounded on every subset which is relatively compact in . By theorem, is compact relative to the uniform convergence in . By a diagonal argument, we can construct a subsequence of , denoted for simplicity such that uniformly on all compact subsets of for a suitable .

Since the gradients are equibounded on every , by the Lebesgue dominated convergence theorem, in order to conclude the proof it is enough to show that for a.e. .

To do this, fix an integer and a point where all are differentiable (almost all points of are of this kind). The functions are concave, so that we get for every ,

, where is the -th vector in the standard basis in . Passing to the limit for in the last inequality we obtain

.

Taking finalizes the proof:

, which is the desired result.

The result above proves the existence of a solution for the Newton problem in the class . Here are a few facts about the solution of the Newton problems. If is one dimensional then the solution is a triangle for large enough (such that the slope of the side is greater than ) or a trapezoid with lateral sides of slope . For a great period of time, the solutions of the Newton problem in the real case and is a disk, were thought to be radially symmetric. Recently, a few results and approaches prove that this is not the case. Here are some of them:

• P. Guasoni calculated the value of the functional on a shape of the form of a screwdriver (convex hull of the disk and a segment parallel to the plane of the disk) and it turns out that this is smaller than the resistance of the computed rotationally symmetric solution;
• One necessary condition of optimality for a solution of the Newton problem states that if in an open set the function is of class and does not touch the upper bound , then is . This relation is not satisfied by the rotationally symmetric solution; (for a proof see Bucur-Buttazzo, Variational Methods in Shape Optimization Problems, Theorem 2.2.6)
• See Theorem 2.23 from the same book as above for another proof;

Again, one necessary condition of optimality is that the set is non-void, for if then we can use a dilation , which decreases the resistance functional. There are some interesting things which can be seen in numerical computations of the optimal solution about the set . All the sets seem to be regular polygons with sides, and the number increases as increases. If we consider all sets of the form and denote with this subclass of , it can be shown that is also compact for the same topologies, and therefore, the Newton problem has a solution in this class. All such solutions are proved to be more optimal than any radially symmetric solution. Even if we know many things about the optimal solution, there a full characterization is not available even if is a disk.

Another well known shape optimization problem is the Optimal Mixing of Two Materials. The problem is formulated like this: We have a region and we must fill the region with two materials with conductivities such that contains the first material and contains the second material. We search for an optimal configuration (i.e. an optimal shape ) which is the most performant, with respect to a given cost functional. The volume of each material can also be prescribed. Denoting , the combined conductivity of the two materials, we get the state equation

 where is the given source density, and we denote by the unique solution of this equation. It is well known that if we consider an arbitrary cost functional of the form then a general optimal configuration does not exist. However, if we add an additional perimeter penalization like , where , then the optimization problem has a solution.

It is important that this definition would agree with the classical definition for sets, namely, if is a bounded open set of class then , where represents the surface element on .

We say that a set of finite Lebesgue measure is a set of finite perimeter in if its characteristic function belongs to . This means that the distributional gradient is a vector valued measure with finite total variation. The total variation is called the perimeter of .

In the same way we can define the perimeter of a Lebesgue measurable set relative to an open set . We say that is a set of finite perimeter relative to if the characteristic function belongs to the space .

The gradient is be seen as a distribution due to the following relations:

This definition of the perimeter wouldn’t be of any good if it didn’t agree with the usual formula for the perimeter of bounded open sets of class . You can find a proof of this in this post following the lines from Henrot, Pierre, Variation et Optimization des Formes.

Having defined the perimeter of a Lebesgue measurable set in a general way, which agrees with the classical definition, we may now approach the proof of existence for the isoperimetric type problems presented below.

Let us first define a class of admissible domains for our problem. As it can be seen in this post, if the set where the admissible domains belong is not bounded, then the shape optimization problem may have no solution, therefore we impose that the admissible domains to be contained in a compact set . Instead of fixing the Lebesgue measure, we can impose the more general constraint where is a given constant and . When is the constant function we get the usual volume constraint on . The isoperimetric problem can be formulated as follows:

Given a compact subset of and a function find the subset of whose perimeter is minimal among all subsets of for which , where is given.

Denoting and we have the following result:

Theorem: With the notations above, if is a compact set and if is nonempty, then the minimization problem admits at least a solution.

Proof: Since the class of admissible sets is not empty, there exists a minimizing sequence with the property that . Since , it follows that the measures of form a bounded sequence, and the perimeters form also a bounded sequence. This means that is a bounded sequence in , where is a ball containing . Then we can extract a subsequence (we denote it by the same ) which converges weakly* to some function in the sense that

The function has to be of the form for some set with finite perimeter; moreover (up to a set of measure zero) and , which shows that . This domain achieves the minimum for the functional , since the perimeter is weakly* lower semicontinuous function on .

The same approach can be made to prove the existence in the case that , the relative perimeter with respect to an open set . This proves the existence result for the Dido problem presented here.

The above proof follows the lines from Dorin Bucur, Giuseppe Buttazzo, Variational Methods in Shape Optimization Problems.

Day 1. Speaker – Giuseppe Buttazzo

Optimization problems have the following form:   where is a functional (sometimes called cost) and is the set of admissible objects. A Shape Optimization Problem has the following form: , where again is a functional (e.g. area or perimeter) and is a class of admissible domains (e.g. bounded area, convex, connected). There are a few aspects of a shape optimization problem, each important in its own way:

1. Existence of a solution. This is not a trivial question, because sometimes optimal forms do not exist. A general method is to provide a topology for such that the map is lower semicontinuous and the sublevels of are compact. (i.e. is compact for every ). This is not easy in general, because the two facts are in contradiction. For the compacity we need fewer open sets, but for the lower continuity of we need more open sets. The key is to find a balance between the two. There is not a general topology for ; changing the functional we may need to change the topology we use, or the class of admissible domains.
2. Uniqueness. This is not generally the case for shape optimization problems, because sometimes, if we have a solution, its translates or rigid motions of the shape are are also a solution.
3. Regularity. In some problems, we may get existence, and we may wonder if the shapes we found are regularly enough (e.g of class , etc).
4. Necessary conditions of optimality. These are conditions for which we have the following implication: is optimal implies satisfies . Maybe sometimes not all objects which satisfy are optimal.
5. Numerical approximation. This is is an important tool, since in many cases it turns out that the optimal shape is not what we would expect. Numerical approximation can give us some idea of what we are looking for and what should we try and prove theoretically. See for example the discussion on the Newton problem, where many people tried to prove that the optimal solution in case of a disk is radial. After seeing numerically that this is not the case, the theoretical proof of the existence of a better non-radial solution appeared.

Some of the problems of Shape Optimization can be written in the form

, for which classical results in the Calculus of Variations provide the existence of a solution given that the two following conditions hold:

• (H1) Convexity: is convex for every ;
• (H2) Coercivity: there exist and such that .

The notation simply means that and . (this roughly means that on )

A theorem in Calculus of variation simply states that under conditions (H1) and (H2) the problem admits a minimizing solution .

Isoperimetric Problems

This type of problems is known back to the ancient Greeks. The Queen Dido problem is quite known. The problem is to encompass the maximum area along the coast having a very long rope at your disposal(the coast line does not count). If you consider the coast to be straight, then the problem can be solved using a symmetrization about the line of the coast and the Isoperimetric Problem, which says that the maximum area of a region with fixed perimeter is the circle. Altering the coast line can lead to problems. If the coast is not bounded then there may be no solution, but it can be proved that if the coast is contained in a fixed compact set, then the Dido problem has a solution even if the coast is not straight. For the existence you may take a look at this paper, or you can find it in the book of Henrot, Pierre, Variation et Optimisation des Formes, une Analyse Geometrique. Assuming the existence, it can be proved that the shape of the rope must be a portion of a circle. The proof goes pretty easy using the Isoperimetric Inequality.

In the image on the right side there is an example of an unbounded coast for which the dido problem has no solution. The points are so that each segment is the half of the preceding one, and the points are chosen such that each of the triangles has area 1.  The considered problem is . Consider a segment of length sliding along the line towards . As the segment gets closer and closer to it covers more and more small segments, and that means that the area of grows to . Therefore any position can be improved and the minimum does not exist. The same picture can give a non-existence case for the problem .

For the existence of solution for Isoperimetric Problems in higher dimensions, take a look at the following posts: Post1, Post2.

Best packing

Assume you have a bounded object and you want to pack it using some expensive paper. Therefore, you want to minimize the cost of the packing. Since the cost depends on the perimeter(area) of the packing, the problem has the form .

In two dimensions the result is easy to guess and it is the convex hull of the figure . In three dimensions things are not that simple anymore, since the convex hull may not be the best packing. Take for example a dumbbell.

An easy calculation can show that it is sometimes more economic to pack the dumbbell as it is and not use its convex hull.

Best aerodinamical shape

This is attributed to Newton, who first thought of this problem. The problem consists of finding the best 3D aerodinamical shape which is above an open set . This is equivalent to finding a function for which the resistance is minimal. This problem is the main subject of the second day of the course, and will be presented in the next post.

satisfying .

Applications:

1. Prove that for every there exists satisfying.
2. Pick . Prove that for every there exists such that .

Source: Haim Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, 2011

Proof: For the initial lemma, just argue by contradiction. The two application are more or less immediate after using the given lemma.

1) does not have Lipschitz boundary (i.e. its boundary is not locally the graph of Lipschitz functions).

2) .

3) For every ball which contains , there is no function in which extends .

This problem gives a counter example which states that if doesn’t have Lipschitz boundary, then there may be no extension to some functions in to greater Sobolev spaces.

Proof: 1) We use the fact that a set has Lipschitz boundary if and only if it has the -cone property (that is, for every there exists a unit vector such that for every the cone is contained in ). Pick which is on the boundary of , and since the graph of is tangent to in the origin, if we pick some close to in , its -cone will not be contained in . Therefore does not have Lipschitz boundary.

Another approach would be to argue by contradiction. Suppose has Lipschitz boundary, so around the origin   is the graph of a Lipschitz function. That is not possible, since the boundary of around is not the graph of any function, no matter what coordinates we establish centered in .

2) Notice that the partial derivatives with respect to are zero, and is smooth with respect to in . So the partial derivatives exist and we must check that they and belong to . To do this, note that . The partial derivative with respect to is zero, and therefore belongs to . The partial derivative with respect to is . To check that this belongs to , see that which is finite.

3) Suppose that admits an extension to . Since is a ball and therefore it is bounded and of class , it follows by the Rellich-Kondrachov Theorem (See Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations (2011) Thm 9.16) that for all with compact injection. Anyway, this means that the extension of is in for . But then , which for large enough it is not finite since becomes negative. Contradiction.

• every subsequence has a subsequence converging to .

Then .

Proof: Suppose that does not converge to . Then there exists an open neighborhood of and a subsequence such that . But the hypothesis states that there exists a subsequence of the latter which converges to ; this means that from a point on, all but finitely many terms of this subsequence are in . This contradicts the fact that .

Therefore, the assumption made was false, and does indeed converge to .

Proof: Pick a continuous linear functional on . Then is a continuous linear functional on , and . Therefore, since was an arbitrary linear functional, it follows that weakly in .

Since is weakly convergent, it follows that it is bounded, and therefore, by compactness of , for any subsequence the sequence has a convergent subsequence (strongly) in . Since , that convergent subsequence must converge to . Therefore, we have the following property for :

Every subsequence of contains a convergent subsequence with limit equal to .

This is enough to conclude that strongly in .

Note that this generalizes the usual Poincare inequality, which says that the above inequality holds for some on the space , a space which does not contain the non-zero constant functions.

Proof: Suppose that the given inequality is not true. Then for every there exists such that . Consider the sequence defined by . From the previous inequality it follows that , therefore as .

On the other hand we have , which implies that is bounded in . Then there is a subsequence denoted without loss of generality which converges weakly to . Since is closed, it follows that . Since the inclusion is compact, it follows that has a subsequence which converges strongly in to the same . Without loss of generality, denote this sequence by . Then, using the weak-sequential lower semicontinuity of we have

.

From the above, we get that and therefore is constant. Moreover, , and we have found a non-zero constant . This contradicts the definition of .