Two stationary catenary rays in an inverse-height speed profile ($y=a\cosh (x/a),a\cosh (d/a)=H$)

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.