 Solution

For fixed endpoint values, stationarity requires the Euler-Lagrange equation $f_y-d(f_{y^{\prime }})/dx=0$. Differentiate $y^{\prime }{f}_{y^{\prime }}-f$ along an extremal: its derivative is $y^{\prime }[d(f_{y^{\prime }})/dx-f_y]-f_x$. Thus when $f_x=0$ the Beltrami identity makes that quantity constant.

Take $y$ as depth measured downward. Conservation of energy from rest gives $m{v}^2/2-mgy=0$, hence speed $v=\sqrt{2gy}$. With arc length $ds=\sqrt{1+y^{\prime 2}}dx$, the travel time is $T=(2g{)}^{-1/2}{\int }_0^{x_B}\sqrt{(1+y^{\prime 2})/y}dx$. For $f=\sqrt{(1+y^{\prime 2})/y}$, the Beltrami quantity is $-1/\sqrt{y(1+y^{\prime 2})}$. Thus $y(1+y^{\prime 2})=c>0$.

Put $y=c{\sin }^2\theta$ and take $y^{\prime }=\cot \theta$ along the first arch. Since $dy/d\theta =2c\sin \theta \cos \theta$, one obtains $dx/d\theta =2c{\sin }^2\theta$. Integrating from the initial cusp gives the cycloid solving the brachistochrone problem:For the equal-height target, the first complete arch has ${\theta }_B=\pi$, so $c=l/\pi$. Also $ds/d\theta =2c\sin \theta$ and $v=\sqrt{2gc}\sin \theta$ for $0<\theta <\pi$. ThereforeThe endpoint zeros of speed cancel the zeros of $ds/d\theta$, leaving finite travel time. Global minimality follows from the maximum-depth bound for a brachistochrone: any competing track with maximum depth $h$ satisfies $T\ge [l/\sqrt{h}+\pi \sqrt{h}]/\sqrt{2g}\ge \sqrt{2\pi l/g}$. The displayed cycloid attains both inequalities with $h=l/\pi$.