Tautochrone problem

Remarks

The problem is solved: the answer is an arc of a cycloid with its cusps pointing up. For the cycloid traced by a circle of radius $a$, the descent time from every starting point is $\pi \sqrt{a / g}$. Huygens gave a geometric solution and used it for the design of pendulum clocks, in his book Horologium Oscillatorium of 1673.

In the nineteenth century Abel treated the problem through an integral equation. Let $y$ be the height above the lowest point and $s(y)$ the arc length from the lowest point, and let $\varphi(y) = s'(y)$. By conservation of energy, the speed at height $y$ of a bead released at height $h$ is $\sqrt{2 g (h - y)}$, so the descent time is $$T(h) = \frac{1}{\sqrt{2g}} \int_0^h \frac{\varphi(y)}{\sqrt{h - y}} \, dy.$$ Asking that $T(h) = T$ for all $h$ is Abel's integral equation with constant right side $\sqrt{2g} \, T$. The Abel inversion formula gives $\varphi(y) = \frac{\sqrt{2g} \, T}{\pi \sqrt{y}}$, and the curves with $s'(y)$ proportional to $y^{-1/2}$ are exactly the cycloids, with $a = g T^2 / \pi^2$. Check: on the cycloid $s^2 = 8 a y$, energy conservation gives $s'' = -\frac{g}{4a} s$, a harmonic oscillation whose quarter period $\frac{\pi}{2} \sqrt{4a/g} = \pi \sqrt{a/g}$ does not depend on the amplitude.