Brachistochrone problem

The brachistochrone problem asks for the curve along which a particle released from rest reaches a specified target fastest under a uniform gravitational field, without friction. With downward depth $u$, conservation of energy gives speed $\sqrt{2gu}$ and travel time $\int \sqrt{(1+u^{\prime 2})/(2gu)}dx$. The Beltrami identity makes $u(1+u^{\prime 2})$ constant, leading to a cycloid. If only the horizontal endpoint coordinate $x_0>0$ is fixed, the natural boundary conditions for a free endpoint select the half arch ending at its deepest point, with depth $2x_0/\pi$ and minimum time $\sqrt{\pi x_0/g}$.

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.

Measure polar angle $\theta$ from the downward vertical on the lower hemisphere of radius $R$. A particle released from rest at the rim and sliding without energy loss has speed $\sqrt{2gR\cos \theta }$. The spherical arc length along $\varphi =\varphi (\theta )$ is $R\sqrt{1+{\sin }^2\theta {\varphi }^{\prime 2}}|d\theta |$. Thus its travel time has integrand proportional to $\sqrt{(1+{\sin }^2\theta {\varphi }^{\prime 2})/\cos \theta }$. Since the azimuth is a cyclic coordinate, the Euler-Lagrange equation givesSolving for ${\varphi }^{\prime 2}$ gives the displayed first integral on a segment where $\theta$ is a valid path parameter. For finite $C$, approaching the rim $\theta \uparrow \pi /2$ gives ${\varphi }^{\prime }\to 0$. Thus a regular stationary path begins along the meridian toward the bottom. This is an initial direction condition; a path joining other rim points can turn before reaching the bottom. Monotone segments use positive $|d\theta |$ in the time integral, and a turning point requires another path parameter or a split into segments.

Consider a frictionless track from rest between equal-height endpoints separated horizontally by $l>0$. Let depth $y$ be positive downward and let the maximum depth be $h>0$. Conservation of energy gives speed $\sqrt{2gy}$. The Cauchy-Schwarz inequality, applied to the tangent displacement, givesThe squared norm of the coefficient vector is $1/h+(h-y)/(hy)=1/y$. The track goes from depth zero to $h$ and returns to zero, so the second term integrates to at least $2{\int }_0^h\sqrt{(h-y)/(hy)}dy=\pi \sqrt{h}$. Integrating the first term gives $l/\sqrt{h}$, even if the track backtracks horizontally. Division by $\sqrt{2g}$ proves the first bound. The arithmetic-geometric mean inequality proves the second.

Equality requires $h=l/\pi$ and tangent slope $|dy/dx|=\sqrt{(h-y)/y}$, on the descending and ascending halves respectively. Substitution $y=h{\sin }^2\theta$ integrates to the cycloid $x=h(\theta -\frac{1}{2}\sin 2\theta )$, $0\le \theta \le \pi$. It attains the bound, proving global minimality of the single-arch brachistochrone problem between these endpoints. Endpoint singularities are handled as limits of the integrals; the cycloid has finite travel time.

Measure $\theta \in [-\pi /2,\pi /2]$ from the vertically downward direction on the lower half of a cylinder of radius $R$, and let $z$ be axial distance. From rest at an edge, conservation of energy gives speed $\sqrt{2gR\cos \theta }$. The travel time is $T=(2gR{)}^{-1/2}\int \sqrt{R^2+z^{\prime 2}}/\sqrt{\cos \theta }d\theta$. Since $z$ is a cyclic coordinate, its conjugate quantity $z^{\prime }/[\sqrt{\cos \theta }\sqrt{R^2+z^{\prime 2}}]=K$ is constant, giving the displayed formula. For finite positive axial separation, $0<K<1$ is fixed uniquely by integrating $z^{\prime }$ between the edges: the separation grows strictly from zero to infinity as $K$ increases from zero to one. The time integrand is strictly convex in $z^{\prime }$, so its supporting-line inequality proves global minimality for the fixed endpoints. As the starting edge is approached, $z^{\prime }\to 0$: the initial direction is directly down the curved surface, with no axial component. This differs from the constant-pitch shortest geodesic on the unrolled cylinder.