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 $360^\circ$.

http://cvgmt.sns.it/seminar/331/