Just for fun, I am working through some problems in Mathematics of Classical and Quantum Physics by Byron and Fuller. Problem 2.13 reads:

Prove that a particle moving under gravity in a plane from a fixed point $P$ to a vertical line $L$ will reach the line in minimum time by following the cycloid from $P$ to $L$ that intersects $L$ at right angles.

Solving the Brachistochrone is more or less straightforward calculus of variations, taking $P$ to be the origin wlog. The action to be minimized is given by (ignoring the optimization irrelevant leading factor of $\frac{1}{\sqrt{2g}}$):

$$ f(x, y, y') = \sqrt{\frac{1 + {\left(\frac{dy}{dx}\right)}^2}{y}} $$

And the derivation of the resulting cycloid, which can be found literally everywhere, gives the parametric function:

\begin{align*} x &= a(\theta - \sin \theta) \\ y &= a(1 - \cos \theta) \end{align*}

Where $a$ is a free parameter allowed to vary. (Note: gravity is taken to be in the y-positive direction to eliminate a pair of redundant minus signs.) If $L$ is taken to be the line $x = x_b$, then in theory it ought to be possible to minimize the cost function with respect to $a$.

If the cycloid intersects the line at a right angle, then it follows that $\frac{dy}{d\theta} = 0$ and $\frac{dx}{d\theta} \ne 0$ at the point of intersection.

\begin{align*} \frac{dy}{d\theta} = 0 &= a \sin(\theta) \\ \theta &= n\pi : n \in \mathbb{Z} \\ \frac{dx}{d\theta} \ne 0 &= a(1 - \cos \theta) \\ \theta &\ne 2n\pi : n \in \mathbb{Z} \end{align*}

Because only the first solution could possibly be the curve of fastest descent, the value of $\theta$ must be $\pi$, which gives $a = x_b / \pi$. So, if we take the derivative of the cost function with respect to $a$ and set it equal to zero, it ought to be the case that we find the equal yields this value. Unfortunately, this is not what I find. Instead, I think I might be being too careless with what is a function of what.

The cost function is given by:

$$ I = \int_0^{x_b} \sqrt{\frac{x'^2 + y'^2}{y}} dx $$

Where $x$ and $y$ are functions of $\theta$, and their primed derivatives are with respect to $\theta$.

The most problematic part is going to be finding the upper limit ($\theta_b$), which we only know implicitly:

$$ x_b = a(\theta_b - \sin \theta_b) $$

Because $\theta_b$ is also a function of $a$, it is necessarily to make this explicit. So let $\theta_b(a)$ be the unique solution to the aforementioned implicit equation. With some algebraic manipulation $I$ becomes:

\begin{align*} I &= \int_0^{x_b} \sqrt{\frac{{x'(\theta)}^2 + {y'(\theta)}^2}{y(\theta)}} dx \\ &= \int_0^{\theta_b(a)} a(1 - \cos \theta) \sqrt{\frac{a^2{(1 - \cos \theta)}^2 + a^2{(\sin \theta)}^2}{a(1-\cos \theta)}} d\theta \\ &= \int_0^{\theta_b(a)} a(1 - \cos \theta) \sqrt{2a} d\theta \\ &= {\left[\sqrt{2} a^{3/2} (\theta - \sin \theta)\right]}_{0}^{\theta_b(a)} \\ &= \sqrt{2} a^{3/2} (\theta_b(a) - \sin \theta_b(a)) \\ \end{align*}

Next, take the derivative and set to 0.

\begin{align*} \frac{dI}{da} = 0 &= \frac1{\sqrt{2a}} (3a(\theta_b(a) - \sin \theta_b(a)) + 2a^2 \theta_b'(a) (1 - \cos \theta_b(a))) \\ &= \frac1{\sqrt{2a}}(3x_b + 2a^2 \theta_b'(a) (1 - \cos \theta_b(a))) \end{align*}

From here, we have to take care to find $\theta_b'(a)$.

\begin{align*} x_b &= a(\theta_b(a) - \sin \theta_b(a)) \\ \frac{d}{da} x_b &= \frac{d}{da} (a(\theta_b(a) - \sin \theta_b(a))) \\ 0 &= \theta_b(a) - \sin \theta_b(a) + a \theta_b'(a) (1 - \cos \theta_b(a)) \\ \theta_b'(a) &= - \frac{\theta_b(a) - \sin \theta_b(a)}{a (1 - \cos \theta_b(a))} \\ \end{align*}

Now to continue solving:

\begin{align*} 0 &= \frac1{\sqrt{2a}}(3x_b + 2a^2 \theta_b'(a) (1 - \cos \theta_b(a))) \\ &= \frac1{\sqrt{2a}}(3x_b - 2x_b) \\ &= \frac{x_b}{\sqrt{2a}} \end{align*}

Which seems to imply that the critical point lies at $a = +\infty$, which is clearly wrong (given this would imply an infinite distance to travel), and also not equal to the previously found value $a = x_b/\pi$. Which begs the question, where is my error? And is there a better approach for this problem?


2 Answers 2


Hint :

You must start from the time to travel between $\theta_1$ and $\theta_2$ on a cycloid :

\begin{equation} t_{2}-t_{1}=\sqrt{\dfrac{\,R\,}{g}}\, \int\limits_{\theta_{1}}^{\theta_{2}}\mathrm{d}\theta=\sqrt{\dfrac{\,R\,}{g}}\, \left(\theta_{2}-\theta_{1}\right) \tag{01}\label{01} \end{equation} If the motion of the particle starts at $\left(x_{1},y_{1}\right)=\left(0,0\right)$, so $\:\theta_{1}=0\:$, then the time $\:t\:$ needed to reach at $\left(x_{2},y_{2}\right)=\left(x,y\right)$, with $\:\theta_{2}=\theta \:$, is \begin{equation} t=\sqrt{\dfrac{\,R\,}{g}}\, \theta = \dfrac{\theta}{\omega} \tag{02}\label{02} \end{equation} where \begin{equation} \omega = \dfrac{\,\theta \,}{t}=\dfrac{\mathrm{d}\theta }{\mathrm{d} t}=\sqrt{\dfrac{\,g\,}{R}}=\text{constant} \tag{03}\label{03} \end{equation}

Note : The variable $\,R\,$ is yours $\,a$.

Then think of a $\,R-$parametric family of cycloids intersecting the vertical line with $\theta(R)$. The time for the $\,R-$cycloid would be \begin{equation} t(R)=\sqrt{\dfrac{\,R\,}{g}}\, \theta(R) \tag{04}\label{04} \end{equation} and for the minimum $\,t(R)\,$ find the roots of \begin{equation} \dfrac{\mathrm d t(R)}{\mathrm d R}=0 \tag{05}\label{05} \end{equation} The angle $\theta(R)$ is the following implicit function of $\,R$(1) \begin{equation} \theta(R)-\sin\left[\theta(R)\right]=L/R \tag{06}\label{06} \end{equation} where $\,L\,$ the horizontal distance of the vertical line from point $\texttt P$.

We must remind that the slope $\,\mathrm d y(\theta)/\mathrm d x(\theta)\,$ on a cycloid and the angle $\,\theta\,$ are related as follows \begin{equation} \dfrac{\mathrm d y}{\mathrm d x}=\dfrac{\sin\theta}{1-\cos\theta}=\cot\left(\dfrac{\theta}{2}\right) \tag{07}\label{07} \end{equation}

enter image description here


(1) The angle $\,\theta(L/R)\,$ as function of the dimensionless variable $\,L/R\,$ would have a graph like that in Figure-02 below. Note that for $\,L/R=k\pi\,(k=1,2,3\cdots)\,$ we have $\,\theta\left(L/R\right)=k\pi\,$ while the graph is repeat of a segment of width $\,2\pi$ : $\,\theta\left(L/R+2\pi\right)=\theta\left(L/R\right)+2\pi\,$. enter image description here


\begin{equation} \sqrt{g}\,\dfrac{\mathrm d t(R)}{\mathrm d R}\boldsymbol{=}\dfrac{\mathrm d }{\mathrm d R}\left(\sqrt{R}\,\theta\right)\boldsymbol{=}\dfrac{1}{2\sqrt{R}}\,\theta\boldsymbol{+}\sqrt{R}\,\dfrac{\mathrm d \theta}{\mathrm d R} \tag{08}\label{08} \end{equation} \begin{equation} \theta\boldsymbol{=}\sin\theta\boldsymbol{+}\dfrac{L}{R} \boldsymbol{\implies} \tag{09}\label{09} \end{equation} \begin{equation} \dfrac{\mathrm d \theta}{\mathrm d R}\boldsymbol{=}\cos\theta\dfrac{\mathrm d \theta}{\mathrm d R}\boldsymbol{-}\dfrac{L}{R^2} \boldsymbol{\implies} \nonumber \end{equation} \begin{equation} \dfrac{\mathrm d \theta}{\mathrm d R}\boldsymbol{=}\boldsymbol{-}\dfrac{L}{(1\boldsymbol{-}\cos\theta)R^2} \tag{10}\label{10} \end{equation} \eqref{08},\eqref{09},\eqref{10} $\:\boldsymbol{\implies}$ \begin{equation} \sqrt{g}\,\dfrac{\mathrm d t(R)}{\mathrm d R}\boldsymbol{=}\dfrac{1}{2\sqrt{R}}\left(\sin\theta\boldsymbol{+}\dfrac{L}{R}\right)\boldsymbol{-}\sqrt{R}\dfrac{L}{(1\boldsymbol{-}\cos\theta)R^2} \tag{11}\label{11} \end{equation} \begin{equation} \dfrac{\mathrm d t(R)}{\mathrm d R}\boldsymbol{=}0 \quad \boldsymbol{\implies}\quad \cos\left(\dfrac{\theta}{2}\right)\cdot \left[\sin^3\left(\dfrac{\theta}{2}\right)\boldsymbol{-}\dfrac{L}{2R}\cos\left(\dfrac{\theta}{2}\right)\right]\boldsymbol{=}0 \tag{12}\label{12} \end{equation} etc

  • 1
    $\begingroup$ Thanks so much. I managed to figure out what was wrong with my approach with the help of your excellent notes on the topic that you linked from one of the other answers. I would write a self-answer, but as the question is closed I am unable to. In summary, my approach was flawed because I was finding the derivative along the wrong curve. I should have been going across curves of equal $x_b$ rather than along constant $R$ or $\theta$. Once I realized that I also realized I should be using Lagrangian multipliers to perform this constraint optimization. It was straightforward from there. $\endgroup$ Aug 3, 2021 at 3:56


  1. To have a well-defined variational problem, we have to impose adequate boundary conditions (BC). Recall in particular that BCs are necessary$^1$ for the proof of the Euler-Lagrange (EL) equation.

  2. The initial BC is an essential/Dirichlet BC $y(x\!=\!x_i)=y_i$. However, the final BC is clearly not an essential/Dirichlet BC. It must then for consistency reasons be a natural BC $p(x\!=\!x_f)=0$. This turns out to be a Neumann BC $y^{\prime}(x\!=\!x_f)=0$, i.e. the final tangent is horizontal/forms a right angle with the vertical line.

  3. Since the EL equation of the modified brachistochrone problem is unchanged, the solution is still a cycloid.


$^1$ Conversely, if we haven't fixed the BCs, we cannot assume the EL equation.


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