The following is a specific instance of the brachistochrone problem, which I first encountered in grad school, and I have occasionally used as hw problem in teaching CM.

A particle is started from rest at the origin and constrained to fall under gravity along a path $y(x)$ which passes through the point $x=5$, $y=-1$ (in arbitrary units, for example meters). We will assume that the gravitational potential is linear, $V=mgz$.

a) Determine the path that minimizes the time taken. Make a plot of that path.

b) Is there another path that makes the time taken stationary? If yes, make a plot of that path and explain whether this path is a minimum, a maximum, or a saddle point.

The solution to the brachistochrone problem is of course very well known, so this assignment is really about finding a specific cycloid that satisfies the boundary conditions. As part b indicates, there is more than one: The standard cycloid, and two cycloids that ``bounce''.

enter image description here

Now it is clear that the simple cycloid is the absolute minimum, because traversal time is proportional to the angle traced out. But what about the other two? Naively they should be saddles, but the second variation of the action functional is manifestly positive, indicating that they are local minima. But that can't be right, unless there is something funny about the topology of the space of paths. Are the higher cycloids saddle points or minima?

P.S.: To see that the higher cycloids cannot easily be dismissed as solutions, consider this plot of the velocity components $(v_x,v_y)$ as a function of time for the second cycloid.

enter image description here

The corresponding components of the acceleration are:

enter image description here

Clearly, the acceleration (and the forces of constraint) are perfectly smooth.

  • $\begingroup$ There are many paths for which the traversal time is infinite (or maybe negative), right? I think there should be no local maxima, only singularities. $\endgroup$
    – Daniel
    Commented Aug 19, 2020 at 16:58
  • $\begingroup$ @Daniel Yes, there are obviously arbitrarily shallow, infinite time, path. However, it would seem to me that you can continuously deform one of the cycloids into the others without encountering an infinite time path $\endgroup$
    – Thomas
    Commented Aug 19, 2020 at 20:00
  • $\begingroup$ Good point. A different thought: Isn't traversal time undefined for the two solutions that bounce? See en.wikipedia.org/wiki/Norton%27s_dome $\endgroup$
    – Daniel
    Commented Aug 19, 2020 at 20:51
  • $\begingroup$ @Daniel In principle the bounce solutions seem fine -- finite travel time (clear from t-symmetry), continuous forces, etc. However, in terms of the variational problem these paths are indeed singular. There is an infinitesimal $\delta z(x)$ that makes the travel time infinite. $\endgroup$
    – Thomas
    Commented Aug 19, 2020 at 23:52
  • $\begingroup$ I agree that there exist finite-travel-time solutions, but I don't agree that these are solutions are unique. Along the path, force is proportional to $\frac{dy}{dx}$. This derivative is undefined at a cusp, so the force is also undefined at a cusp. Usually this doesn't matter, since it's only one point, but the particle also arrives at the cusp with zero velocity. It seems reasonable to define the force at the cusp to be zero. Then a consistent (but not unique!) solution to Newton's laws is a particle that rolls up to the cusp, stops, and sits there forever. $\endgroup$
    – Daniel
    Commented Aug 20, 2020 at 0:46

1 Answer 1


TL;DR: A path built piecewise from more than 1 cycloid (each with possibly different energy $E$, see below), and with cusps at the $x$-axis, is not stationary.

Sketched proof:

  1. Recall that the action (=spent time) of the brachistochrone problem is $$S~=~\int_0^a\! \mathrm{d}x~L,\qquad L~=~\sqrt{\frac{1+y^{\prime 2}}{y}},\qquad y~\geq~ 0,\tag{1}$$ with boundary conditions $y(0)=0$ and $y(a)=b$. (Here the $y$-axis is pointing downwards and we chose for simplicity units of time and space such that $2g=1$.)

  2. Physically, we demand that the path $x\mapsto y(x)$ is at least continuous. Mathematically, the integrand should just be Lebesgue integrable. To be as simple as possible but also incorporate OP's examples we will strike a convenient compromise and assume that the path $x\mapsto y(x)$ is piecewise continuously differentiable, although we will allow the derivative $y^{\prime}\equiv \frac{dy}{dx}$ to become singular at the points between the pieces as long as the integrand remains Lebesgue integrable.

  3. It follows that a stationary path necessarily satisfies Euler-Lagrange (EL) equation within the interior of each piece. Additional conditions may arise at the points between the pieces.

  4. Since the Lagrangian $L$ has no explicit $x$-dependence the corresponding notion of energy (within a piece) is conserved: $$E~=~ y^{\prime} \frac{\partial L}{\partial y^{\prime}}-L~\stackrel{(1)}{=}~-\frac{1}{\sqrt{y(1+y^{\prime 2})}}~<~0.\tag{2}$$

  5. The piece solution is a cycloid: $$\begin{align} 2E^2x~=~&\theta-\sin\theta~\approx~\frac{\theta^3}{6},\cr 2E^2y~=~&1-\cos\theta~\approx~\frac{\theta^2}{2},\end{align}\tag{3}$$ where the approximation is valid close to the cusp. The cusp-equation becomes $$ y~\stackrel{(3)}{\propto}~ x^{2/3}.\tag{4}$$ Near the cusp, the particle is performing a free falling motion, which is smooth as a function of time $t$.

  6. The idea is now to truncate the cusp at some horizontal the level $y=\epsilon\ll 1$, i.e. at some $x~\propto~ y^{3/2}~=~\epsilon^{3/2}$. (We consider for simplicity just the right branch of the cusp -- the left branch is similar.) The action of the cusp is $$L~\stackrel{(1)+(2)}{=}~\frac{1}{|E|y}~\stackrel{(4)}{\propto}~ x^{-2/3}\qquad\Rightarrow\qquad S~\propto~x^{1/3} ~\propto~\epsilon^{1/2}.\tag{5}$$ For comparison, the action of the horizontal path is as expected faster: $$L~\stackrel{(1)}{=}~\frac{1}{\sqrt{y}}~=~ \frac{1}{\sqrt{\epsilon}}\qquad\Rightarrow\qquad S~\propto~\frac{x}{\sqrt{\epsilon}} ~\stackrel{(4)}{\propto}~\epsilon.\tag{6}$$ This shows that we can change the action to first order in $\epsilon$, and hence the path is not stationary. $\Box$

  • $\begingroup$ Thanks, that's a very nice observation. Two quick questions: 1) If there is a term of order $\sqrt{\epsilon}$, does that not imply that the variational derivative $\delta S/\delta x$ is ill defined (infinite)? 2) Am I allowed to use $E=const$ in carrying out the variation ($E=const$ only for stationary paths)? $\endgroup$
    – Thomas
    Commented Sep 6, 2020 at 19:30
  • 1
    $\begingroup$ 1) It is certainly not a stationary path. 2) $E=const$ is true for cycloid segments, but not for generic paths. $\endgroup$
    – Qmechanic
    Commented Sep 6, 2020 at 19:46

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