Catenary light ray in a linear refractive index ($x(y)=-1/\beta +(c/\beta )\cosh [\beta (y-y_0)/c]$)

For height $x$ and horizontal position $y$, Fermat principle in a positive index profile $n(x)=1+\beta x$, $\beta >0$, conserves $n(x)y^{\prime }/\sqrt{1+(y^{\prime }{)}^2}$ on a graph $y(x)$. For positive magnitude $c$ of the horizontal optical momentum, integration gives the displayed catenary. Its minimum height is $x_0=(c-1)/\beta$, at $y=y_0$. Thus $c>1$ produces a smooth turning ray wholly above a road at $x=0$. The two graph branches have opposite signs of $dy/dx$; $x(y)$ continues smoothly through their common horizontal tangent. For a signed negative graph constant, use $|c|$ in the prefactor; $c=0$ gives a vertical ray separately.

For speed $c(y)=1/y$ in $y>0$, Fermat principle gives the elapsed-time integrand $y\sqrt{1+(y^{\prime }{)}^2}$. Its symmetric stationary light rays through $(\pm d,H)$ are the displayed catenaries. Set $t=d/a$. The endpoint equation is $\cosh t/t=H/d$. This function decreases and then increases, with unique minimum where $t\tanh t=1$. Thus above the minimum there are two stationary rays, at the minimum one, and below it none. The travel time on either ray is $T=ad+(a^2/2)\sinh (2d/a)$, and its minimum height is $a$.

For $d=1$ and $H=\cosh 1$, one solution is $a=1$, with $T=1+\sinh 2/2$. A second has $a\simeq 0.7021026952$, with $T\simeq 2.8224491447$. Hence these endpoints do not uniquely specify the ray. Stationarity is distinct from global minimization: paths descending almost to $y=0$ have limiting travel time $H^2$, below either stationary value in this example.