Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 1 / 2D (Methods)

For positive differentiable speed $c(y)$ and a smooth ray written as a graph $y(x)$, the Fermat principle minimizes travel timeA variation vanishing at the endpoints gives the Euler-Lagrange equation for $F=\sqrt{1+y^{\prime 2}}/c$. Its derivatives are $F_y=-c^{\prime }\sqrt{1+y^{\prime 2}}/c^2$ and $F_{y^{\prime }}=y^{\prime }/(c\sqrt{1+y^{\prime 2}})$. Thusand multiplying by $c^2(1+y^{\prime 2}{)}^{3/2}$ givesThe displayed differential equation requires differentiability, beyond mere continuity of $c$. For constant positive speed it reduces to $y^{\prime \prime }=0$, so rays are straight lines; a straight segment minimizes travel time because it has the shortest Euclidean length.

For $c(y)=y$ in the positive-height region, the Beltrami identity gives $F-y^{\prime }{F}_{y^{\prime }}=1/[y\sqrt{1+y^{\prime 2}}]=1/R$. Hence $y^2(1+y^{\prime 2})=R^2$, whose nonvertical solutions are $(x-h{)}^2+y^2=R^2$. Formally imposing the given boundary points fixes $h=1/2$, $R=1/2$, and gives the ray in a linear-speed mediumIt is the upper semicircle of diameter one, with vertical limiting tangents at its endpoints.

There is a necessary endpoint qualification. The speed is zero at the two stated endpoints. For any rectifiable path entering positive height, $ds/y\ge |dy|/y$, and its travel time diverges logarithmically on leaving or approaching height zero. Thus the semicircle is the formal interior ray, but there is no finite-time minimizer between the zero-speed endpoints. For endpoints at height $ϵ>0$ the circular ray has radius $\sqrt{1/4+{ϵ}^2}$ and finite travel time $2\mathrm{arsinh}(1/(2ϵ))$, which tends to infinity as $ϵ\downarrow 0$. This explains the intended semicircle as a limiting hyperbolic geodesic, without asserting finite travel time where $c=0$.