Home > shape optimization > Shape Optimization Course – Day 1

Shape Optimization Course – Day 1

November 4, 2011 Leave a comment Go to comments

The main speakers of the course were Giuseppe Buttazzo and Edouard Oudet. See more details in the Shape Optimization page.

Day 1. Speaker – Giuseppe Buttazzo

Optimization problems have the following form:   \min \{ F(x) : x \in A\} where F is a functional (sometimes called cost) and A is the set of admissible objects. A Shape Optimization Problem has the following form: \min\{ F(\Omega) : \Omega \in \mathcal{A}\}, where again F is a functional (e.g. area or perimeter) and \mathcal{A} 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 \mathcal{A} such that the map F is lower semicontinuous and the sublevels of F are compact. (i.e. \{ F(\Omega) \leq t\} is compact for every t). 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 F we need more open sets. The key is to find a balance between the two. There is not a general topology for \mathcal{A}; changing the functional F 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 C^1,C_2, etc).
  4. Necessary conditions of optimality. These are conditions (C) for which we have the following implication: \Omega is optimal implies \Omega satisfies (C). Maybe sometimes not all objects which satisfy (C) 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

(P) \ \ \displaystyle \inf \left\{ I(u) = \int_\Omega f(x,u(x),\nabla u(x))dx : u \in u_0+W_1^{1,p}(\\Omega)\right\}, for which classical results in the Calculus of Variations provide the existence of a solution given that the two following conditions hold:

  • (H1) Convexity: \xi \mapsto f(x,u,\xi) is convex for every (x,u) \in\overline{\Omega}\times \Bbb{R};
  • (H2) Coercivity: there exist p>q\geq 1 and \alpha_1>0, \alpha_2,\alpha_3 \in \Bbb{R} such that \displaystyle f(x,u,\xi) \geq \alpha_1 |\xi|^p+\alpha_2 | u|^q+\alpha_3,\ \forall (x,u,\xi) \in \overline{\Omega} \times \Bbb{R} \times \Bbb{R}^n.

The notation u \in u_0 +W_0^{1,p}(\Omega) simply means that u,u_0 \in W^{1,p}(\Omega) and u-u_0 \in W_0^{1,p}(\Omega). (this roughly means that u=u_0 on \partial \Omega)

A theorem in Calculus of variation simply states that under conditions (H1) and (H2) the problem (P) admits a minimizing solution \overline{u} \in u_0 + W_0^{1,p} (\Omega).

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 A,B,D,E,... are so that each segment is the half of the preceding one, and the points C,G,H are chosen such that each of the triangles ABC,DBG,DEH has area 1.  The considered problem is \max\{ |\Omega| : Per(\Omega)=a\}. Consider a segment of length a sliding along the line AF towards F. As the segment gets closer and closer to F it covers more and more small segments, and that means that the area of \Omega grows to \infty. Therefore any position can be improved and the minimum does not exist. The same picture can give a non-existence case for the problem \min\{ Per(\Omega) : |\Omega|=|\Delta ABC|\}.

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 K 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 \min \{ Per(\Omega) : K \subset \Omega\}.

In two dimensions the result is easy to guess and it is the convex hull of the figure K. 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 \Omega \subset \Bbb{R}^2. This is equivalent to finding a function u: \Omega \to \Bbb{R}_+ for which the resistance R(u) is minimal. This problem is the main subject of the second day of the course, and will be presented in the next post.

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