Calculus of Variations and Geometric Measure Theory
home | mail | papers | authors | news | seminars | events | open positions | login

Besicovitch's magic method and problems of minimal resistance

Alexander Plakhov

created by dicastro on 09 Dec 2013
modified on 12 Dec 2013

18 dec 2013 -- 16:00   [open in google calendar]

Aula Seminari - Department of Mathematics, University of Pisa

Abstract.

We consider piecewise smooth functions $u : \bar{\Omega} \rightarrow \mathbb{R}$ defined on the closure of a bounded domain $\Omega \subset \mathbb{R}^2$ satisfying the conditions $u(x)<0$ for $x\in \Omega $ and $u(x)=0$ for $x\in\partial \Omega$ (in other words, the graph of $u$ forms a "dimple" on the plane).

We also consider a flow of particles that fall on the graph of $u$ vertically down and reflect from it in the perfectly elastic manner. It is assumed that u satisfies the so-called "single impact condition" (SIC): each particle reflected at a non-singular point of the graph, further moves freely above the graph until it leaves the dimple. This condition can be stated analytically as follows: for any regular point $x\in\Omega$ and any $t > 0$ such that $x - t \nabla u ( x ) \in \bar{\Omega}$, \[ \frac{u(x − t\nabla u(x)) − u(x)}{t} \leq \frac 1 2(1 −
\nabla u(x)
^2). \tag{1} \] The force of resistance of the dimple to the flow (more precisely, the vertical projection of this force) equals $2\rho
\Omega
R(u; \Omega)$, where $ρ$ is the flow density, $
\Omega
$ is the area of $\Omega$, and \[ R(u;\Omega) = \frac{1}{
\Omega
} \int_\Omega \frac{dx}{1+
\nabla u(x)
^2}. \] This formula is true provided that the SIC (1) is fulfilled.

The problem is: minimize the value of "specific resistance" $R(u; \Omega)$. It has two versions which are eventually equivalent: (a) $\inf_{u,\Omega} R(u; \Omega)$ (b) $\inf_{u} R(u;\Omega)$ for a given $\Omega$. Obviously, $\sup R(u; \Omega) = 1$ and $\inf R(u; \Omega) \geq 1/2$. The main question is to find if $\inf R(u; \Omega) > 1/2$ or $\inf R(u; \Omega) = 1/2$. I will prove that the latter is true. This result is somewhat counterintuitive: one needs to provide a sequence of functions with the slope of the graph being "almost" $45^\circ$ in the most part of the region $\Omega$. That is, most part of reflected particles move "almost" horizontally and do not meet obstacles on the way. A part of the construction is borrowed from Besicovitch’s solution of the Kakeya problem: what is the minimum area of a plane region in which a unit line segment can be rotated continuously through $360^\circ$.

Credits | Cookie policy | HTML 5 | CSS 2.1