= Fixed-length brachistochrone between equal-height endpoints
{title2=$L>1,\quad b<1,\quad c>0$}
For horizontal separation one, let $y>0$ be downward depth and constrain the <arc length> to $L$. <Conservation of energy> makes the <travel time> proportional to $\int\sqrt{1+y'^2}/\sqrt y\,dx$. An <isoperimetric constraint> adds a <Lagrange multiplier> $\lambda$, and the <Beltrami identity> gives $(y^{-1/2}+\lambda)/\sqrt{1+y'^2}=H>0$. With $b=\lambda/H$, $c=H^2$ and $s=\sin(\phi/2)$, the stationary arch is $y=s^2/[c(1-bs)^2]$, $dx/d\phi=s^2/[c(1-bs)^3]$, for $0\le\phi\le2\pi$. The endpoint and length integrals fix $b,c$ uniquely for $L>1$; symmetry gives $x(2\pi-\phi)=1-x(\phi)$. At $L=1$ the only curve is horizontal and a particle released from rest never completes the journey. The finite-time stationary arches have vertical endpoint tangents, so the graph is smooth only on the open interval. These first-variation equations alone do not establish global minimality in a larger admissible class.
Back to article page
