Home > shape optimization > Shape Optimization Course – Day 2

Shape Optimization Course – Day 2

Speaker – Giuseppe Buttazzo

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 $x \in \Omega$ ($\Omega$ denotes the fixed orthogonal section), the force with which a particle hitting the body in the point $u(x)$ holds the body back can be calculated decomposing the normal force as in the figure to be proportional to $\cos^2 \theta(x)$, where $\theta(x)$ is the angle made by the normal to the surface of the body with the direction of the flow.  Using the equality $\cos^2 \theta(x)=\displaystyle \frac{1}{1+\tan^2 \theta (x)}$ 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 $\Omega$, resulting in the formula $\displaystyle R(u)=\int_\Omega \frac{1}{1+|\nabla u|^2}dx$.

Our problem is minimizing $R(u)$ on some class of admissible functions $u: \Omega \to \Bbb{R}$. Choosing the right class of admissible functions is important, because of the following:

• if we do not impose a boundedness condition on $u$, then we can take $u$ 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 $|\nabla u|$. 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 $u$ bounded above by some constant $M$, 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 $R(u)$, which is small as $|\nabla u| \to \infty$. 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 $C_M=\{u \text{ concave on }\Omega : 0\leq u\leq M\}$.

Compactness lemma: For every $M>0$ and every $p<\infty$ the class $C_M$ is compact with respect to the strong topology of $W_{loc}^{1,p}(\Omega)$.

Proof:  Let $(u_n)$ be a sequence of elements of $C_M$; since all $u_n$ are concave, they are locally Lipschitz continuous on $\Omega$, that is $\forall K \subset \subset \Omega,\ \forall x,y \in K,\ \ |u_n(x)-u_n(y)| \leq C_{n,K} |x-y|$, where $C_{n,K}$ is a suitable constant. From the fact that $0 \leq u_n \leq M$, the constants $C_{n,K}$ can be chosen independent of $n$; we can take $C_{n,K}=2M/d(K,\partial \Omega)$. Therefore $(u_n)$ is equi-Lipschitz continuous and equi-bounded on every subset $K$ which is relatively compact in $\Omega$. By $Ascoli-Arzela$ theorem, $(u_n)$ is compact relative to the uniform convergence in $K$. By a diagonal argument, we can construct a subsequence of $(u_n)$, denoted for simplicity $(u_n)$ such that $u_n \to u$ uniformly on all compact subsets of $\Omega$ for a suitable $u \in C_M$.

Since the gradients $\nabla u_n$ are equibounded on every $K \subset \subset D$, by the Lebesgue dominated convergence theorem, in order to conclude the proof it is enough to show that $\nabla u_n(x) \to \nabla u(x)$ for a.e. $x \in D$.

To do this, fix an integer $k \in [1,N]$ and a point $x \in D$ where all $u_n$ are differentiable (almost all points of $D$ are of this kind). The functions $t \mapsto u_m(x+te_k)$ are concave, so that we get for every $\varepsilon >0$,

$\displaystyle \frac{u_n(x+\varepsilon e_k)-u_n(x)}{\varepsilon} \leq \nabla_k u_n(x) \leq \frac{u_n(x-\varepsilon e_k)-u_n(x)}{-\varepsilon}$, where $e_k$ is the $k$-th vector in the standard basis in $\Bbb{R}^N$. Passing to the limit for $n \to \infty$ in the last inequality we obtain

$\displaystyle\frac{u(x+\varepsilon e_k)-u(x)}{\varepsilon}\leq\lim\inf \nabla_k u_n(x) \leq \lim\sup \nabla_k u_n(x)\leq\frac{u(x-\varepsilon e_k)-u(x)}{-\varepsilon}$.

Taking $\varepsilon \to 0$ finalizes the proof:

$\nabla_k u(x) \leq \lim\inf \nabla_k u_n(x) \leq \lim\sup \nabla_k u_n(x) \leq \nabla_k u(x)$, which is the desired result.

The result above proves the existence of a solution for the Newton problem in the class $C_M$. Here are a few facts about the solution of the Newton problems. If $\Omega$ is one dimensional then the solution is a triangle for $M$ large enough (such that the slope of the side is greater than $1$) or a trapezoid with lateral sides of slope $1$. For a great period of time, the solutions of the Newton problem in the real case $N=2$ and $\Omega$ 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 $\omega$ the function $u$ is of class $C^2$ and does not touch the upper bound $M$, then $\det \nabla^2 u \equiv 0$ is $\omega$. 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 $u=M$ is non-void, for if $u then we can use a dilation $u=\lambda u(x)$, which decreases the resistance functional. There are some interesting things which can be seen in numerical computations of the optimal solution about the set $K_M= \overline{\{ x \in \Omega: u(x)=M\}}$. All the sets $K_M$ seem to be regular polygons with $n_M\geq 3$ sides, and the number $n_M$ increases as $M$ increases. If we consider all sets of the form $convexhull((K \times \{M\} \cup \Omega\times \{0\})$ and denote with $P_M$ this subclass of $C_M$, it can be shown that $P_M$ is also compact for the same topologies, and therefore, the Newton problem has a solution $w_M$ in this class. All such solutions $w_M$ 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 $\Omega$ 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 $D$ and we must fill the region with two materials with conductivities $\alpha, \beta$ such that $\Omega$ contains the first material and $D\setminus \Omega$ contains the second material. We search for an optimal configuration (i.e. an optimal shape $\Omega$) which is the most performant, with respect to a given cost functional. The volume of each material can also be prescribed. Denoting $a_\Omega(x)=\alpha \chi_{\Omega}(x)+\beta \chi_{D\setminus \Omega}$, the combined conductivity of the two materials, we get the state equation

$\displaystyle\begin{cases}-\text{div}(a_{\Omega}(x)\nabla u)=f & \text{in } D,\\ u=0 & \text{on }\partial D\end{cases}$  where $f$ is the given source density, and we denote by $u_{\Omega}$ the unique solution of this equation. It is well known that if we consider an arbitrary cost functional of the form $\int_D j(x,\chi_\Omega,u_\Omega,\nabla u_\Omega)dx$ then a general optimal configuration does not exist. However, if we add an additional perimeter penalization like $J(u,\Omega)=\int_D j(x,\chi_\Omega, u,\nabla u)dx +\sigma \text{Per}_D(A)$, where $\sigma >0$, then the optimization problem $\min\{J(u,\Omega) : \Omega \subset D,\ u \text{ satisfies the above equation}\}$ has a solution.