Fixed-length brachistochrone between equal-height endpoints ($L>1,b<1,c>0$)

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 $\int \sqrt{1+y^{\prime 2}}/\sqrt{y}dx$. An isoperimetric constraint adds a Lagrange multiplier $\lambda$, and the Beltrami identity gives $(y^{-1/2}+\lambda )/\sqrt{1+y^{\prime 2}}=H>0$. With $b=\lambda /H$, $c=H^2$ and $s=\sin (\varphi /2)$, the stationary arch is $y=s^2/[c(1-bs{)}^2]$, $dx/d\varphi =s^2/[c(1-bs{)}^3]$, for $0\le \varphi \le 2\pi$. The endpoint and length integrals fix $b,c$ uniquely for $L>1$; symmetry gives $x(2\pi -\varphi )=1-x(\varphi )$. 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.