OPhO 2025, Финален кръг · част I — Задача 2. Theory Question 2

Efficient Lighting - Solution

Ivan's solution

(A.1) The problem might be reduced to considering the room $S$ as a sequence of chambers, each containing one light source, such that all of the light rays emitted from the source stay confined within the corresponding chamber. One may be inspired by the fact that, considering a circle with a light source at its centre, all light rays arrive at the centre after exactly one reflection off the circular wall. The radius of the circle being irrelevant regarding reflections, one can select specific parts of the wall and prolong said parts, obtaining the room illustrated by Figure (1). Note that the light sources do not necessarily need to be collinear.

Another possible solution is to consider the reflective properties of parabolas. Construct chambers around each source made of parabolic walls such that one is wider than the other. The gaps between them allow only strictly vertical light rays to pass, enabling the construction of the entire room $S$ by prolonging these corridors in any way. Note the $45^{\circ}$ angle for corridors that need to bend to reach another light source.

Another interesting room to consider is the following one, which uses elliptic and hyperbolic walls. However, in this case, the light sources need to be strictly collinear.

(A.2) / (A.3) Analogous to part (a), this part is concerned with controlling all light rays emitted from within some region $R$ to return to $R$ after a finite number of reflections. This is easily achieved by considering the reflective property of an arbitrary ellipse, so that one may attempt the following construction of $S$:

Note that the regions $R_{\mathrm{i}}$ are of arbitrary shape, hence one can build any corridor/canal towards a specific region-like source. The same can be obtained by considering parabolas:

Hubert's solution: Formal proof for the existence of configurations for both parts follows.

(A.1) Suppose we have a finite set of point sources $S = \{x_i, i \in [1, n]\}$. We claim that it is possible to choose an ordering of $i$ such that the line segments $(x_i, x_{i+1})$ form disjoint sets for all $i$ (i.e. they do not intersect). This can be achieved by choosing an ordering such that the x-coordinates of $x_i$ are arranged in an increasing order. This is achievable by the total orderedness of the reals. Should two points have the same x-coordinate we order points in order of increasing y-coordinate. Then, any two line segments have disjoint support on the x-axis and are therefore non-intersecting. For line segments which are vertical the same argument can be applied for support on the y-axis.

Given such an ordering and such a construction of line segments $l_i = (x_i, x_{i+1})$ we make the following construction: each point $x_i$ is surrounded by the epsilon-ball $B_\epsilon(x_i)$. Then, along each line segment $l_i$ for $i \in [1, n)$, construct two sectors which has a radius of $\frac{|x_i - x_{i+1}|}{2} + \delta_1$ (i.e. the sector reaches just beyond the midpoint of the line segment connecting the two points) centred on $x_i$ and $x_{i+1}$ respectively. The sectors should extend an angle $\delta_2$ counterclockwise. We claim that the room consisting of the union of the epsilon-ball and all such sectors satisfies the requirements of the problem for some $\epsilon, \delta_1, \delta_2 > 0$.

Since a circle reflects all rays from its centre back towards the centre, it is clear that each union of the epsilon-ball and all sectors centred at $x_i$ is fully illuminated by the source, and that all rays are parallel to any intersections of the boundaries with any source, so the only requirement is that these sets do not self intersect. In other words, if we consider that $S_i$ is the union of $B_\epsilon(x_i)$ and the two sectors centred at $x_i$, we want to show that there exist $\epsilon, \delta_1, \delta_2 > 0$ such that $(S_i \backslash \partial S_i) \cap (S_j \backslash \partial S_j) = \emptyset$.

Firstly, we know the epsilon-balls do not intersect for some $\epsilon > 0$, since any finite set of points is nowhere dense. Suppose without loss of generality that the x-coordinates of $x_i$ and $x_{i+1}$ were different. (If not, the axes can be rotated infinitesimally anticlockwise.) Then we are able to choose $\delta_2$ sufficiently small such that no point in the sector pointing towards $x_{i+1}$ has an x-coordinate less than that of $x_i$ by choosing it to be less than the slope of $l_i$. Assume further without loss of generality that $x_{i-1}$ has a different x-coordinate with $x_i$ (this is achievable for any configuration of three points). Then by choosing $\epsilon$ to be less than the finite difference of the x-coordinates of $x_i$ and $x_{i-1}$ we guarantee the non-intersection between the sector and the epsilon-ball around $B_\epsilon(x_{i-1})$. The intersection between two sectors occurs only on the boundaries by construction. Hence the condition is satisfied.

It is also easy to show that $\cup_i S_i$ is simply connected by observing that the curve consisting of the line segments $l_i$ has no return path except through the same sequence of epsilon-balls and sectors, which can necessarily be contracted to a point. This completes the proof.

(A.2) / (A.3) Similar to the previous idea, we construct the room to be bounded by the boundary of each source, with a small corridor along the straight line connecting each source. The difference is that the incident angle of the light is arbitrary, so one needs to use an ellipse or parabola to refocus the light back towards the source, as shown in the illustration above. The $\epsilon\delta$ arguments will not be repeated as they are similar. The key point is to prove that, for any given finite disjoint sequence of simply connected sources $S_i$, we can choose an ordering for $i$ and points $x_{i,2} \in \partial S_i$, $x_{i+1,1} \in \partial S_{i+1}$ such that the line segments $l_i = (x_{i,2}, x_{i+1,1})$ are pairwise disjoint with both $S_j\ \forall j$ and $l_j, j \neq i$.

This can be proven by construction via a greedy algorithm. For any pair of sets $S_i, S_j$ we can define a minimum distance between the two sets, $D_{ij} = \min_{x_1 \in S_i, x_2 \in S_j} |x_1 - x_2|$. We find $\{i, j\} = \arg\min_{i \neq j} D_{ij}$. Then we find $\{x_1, x_2\} = \arg\min_{x_1 \in S_i, x_2 \in S_j} |x_1 - x_2|$ and construct the line segment $(x_1, x_2)$. We then consider $S_i \cup S_j$ to be one set and reiterate the process until all sets are connected. The lines necessarily do not intersect with any set by the minimum distance property, and do not intersect with any other line as if they did, you can swap the pairs of points to obtain shorter lines, as in a minimum spanning tree. This proves the construction works for any arrangement of sources.

Gas - solution

We consider the gas cloud of density $\rho(\mathbf{x})$ and velocity field $\mathbf{v}(\mathbf{x})$. The total energy of this cloud is

where we have neglected the constant factor, $r = |\mathbf{x}|$, and $K$ is a constant relating to the mass of the central object. As viscous forces are always dissipative, we expect to minimise this functional. There are, however, several physics constraints, including the continuity equation:

as well as conservation of angular momentum, where we choose the units such that the magnitude is 1 and choose the z-axis to be along its direction (this is possible as all forces are internal):

We also note that the functional $E$ is time-varying. We want to study the steady state, so we consider the time-average over an infinite horizon, which, as the limit tends to infinity, will be dominated by the steady state:

We minimise this functional. With the two constraints, we introduce the Lagrange multipliers $\lambda(\mathbf{x}, t)$ and $\mu(t)$ for the continuity equation (which applies at every x and t) and the angular momentum conservation (which applies at every t). The full functional is therefore

Now, we vary this with respect to $\rho, \mathbf{v}, \lambda$ and $\mu$, taking fixed $T$ and taking the limit later. We also assume the boundary conditions that all quantities have finite support in $\mathbf{x}$ so boundary terms vanish (which is a physical assumption). The latter two return the constraints. Varying with respect to $\mathbf{v}$ gives

while varying over $\rho$ gives

where in both cases we have implicitly taken the limit $\lambda \to 0$ as $t \to \infty$ as the steady state is unconstrained. Simplifying produces

We take the dot product of both sides of (8) with $\mathbf{v}$ which gives

Substituting into (9) we get

which is impossible as $K > 0$, and $\partial_t\lambda$ must be sometimes positive for any periodic solution. Hence, by contradiction, there does not exist a steady state.

Addendum While the above solution is mathematically rigorous, we received a submission from the team ”rotting crustaceans” which in our opinion is also rigorous while being more intuitive. We attach a modified sketch of their solution below.

For the viscous fluid the viscous dissipation function is defined as $\Phi = \tau_{ij}\partial_i u_j$, where Einstein notation is used and $\tau_{ij}$ is the stress tensor

At any sort of energy minimum we must have $\Phi$ zero everywhere, which implies rigid body motion. Moreover, for a given fixed density distribution with fixed total angular momentum, the minimum energy configuration is rigid body. This can be proved by considering the KE:

while the angular momentum is

where we have defined $\mathbf{L}$ to be along the z-axis. Then we have

by Cauchy-Schwarz inequality, and therefore

completing the proof. This therefore implies that if there existed an energy minimum which is not rigid body, one could keep the density distribution the same and instead rotate the body rigidly. (This also eliminates any time-varying solutions.)

Now, for the rigid body solutions, we can show that they cannot be an energy minimum by considering the fact that, for any given particle orbiting around a central mass of mass $M$ at a distance $r$ and with angular momentum $l$, it has energy

This is negative by the boundedness of the mass, and therefore if the mass is moved inwards (decreasing $r$) while conserving angular momentum, the energy will decrease. We can therefore imagine a situation where, if there were a given rigidly rotating density distribution which is a local energy minimum, one can move a parcel of mass inwards while conserving the angular momentum to further decrease the energy. We can then redistribute the angular momentum so that the motion is again rigid body - as there is a slight decrease in moment of inertia, $\omega$ will increase slightly and the angular momentum is transferred outwards - and so on and so forth, until all the mass is concentrated at the central mass and an infinitesimal mass is orbiting infinitely far away and/or at infinite angular velocity. This is clearly unphysical and therefore there does not exist any configuration which is an energy minimum and stable.