 Solution

For a variation $y+\epsilon \eta$ with $\eta (a)=0$ but arbitrary $\eta (b)$, differentiation and integration by parts giveFirst choose variations supported inside the interval. The fundamental lemma of the calculus of variations gives the Euler-Lagrange equation. Then allow arbitrary endpoint variation to obtain the natural boundary conditions for a free endpoint:Because $F$ has no explicit dependence on $x$,Thus the Beltrami identity is $F-y^{\prime }{F}_{y^{\prime }}=k$, constant along an extremal.

For the brachistochrone problem, put $u=-y>0$, so conservation of energy gives the bead speed $v=\sqrt{2gu}$. Its travel time isThe Beltrami identity now gives $u(1+u^{\prime 2})=2R$ for a positive constant $R$. A descending cycloid parametrizes this relation:The free endpoint has $F_{u^{\prime }}=0$, hence $u^{\prime }(x_0)=0$. On the first descending arc this occurs at $\theta =\pi$. ThereforeAlong this arc $ds=2R\sin (\theta /2)d\theta$ and $v=2\sqrt{gR}\sin (\theta /2)$, soThe initial vertical tangent causes no divergent travel time: the quotient $ds/v$ has the finite constant limit $\sqrt{R/g}d\theta$.

To verify global minimality, take any admissible wire graph of finite travel time below the release level, and let $H>0$ be its maximum depth. For $0<u\le H$, the Cauchy-Schwarz inequality givesIndeed the two coefficients on the right, after multiplication by $\sqrt{u}$, form a unit vector. Since the path starts at depth zero and reaches depth $H$, the weighted total variation is at least the integral over all depths from $0$ to $H$. ConsequentlyThe last inequality is minimized at $H=2x_0/\pi$. Our cycloid attains equality throughout: its depth increases to $H$ and $u^{\prime }=\sqrt{H/u-1}$. Thus it achieves the genuine minimum over the wire graphs, not merely a stationary time. Reflection handles a negative horizontal endpoint coordinate by replacing $x_0$ with $|x_0|$.