Remark 3.4.1.
The next paragraph is going to perform several dirty tricks. I wrote a little blog post about why they make sense.
Section 3.4 Application to the brachistochrone problem
For the special case of the brachistochrone problem the integrand \(f(y, y')\) and its partial derivatives are readily found to be
\begin{equation}
f=\sqrt{\frac{1+(y')^2}{y-\alpha}}, \quad
f_{y}=-\frac{1}{2} \sqrt{\frac{1+(y')^2}{(y-a)^{3}}}, \quad
f_{y^{\prime}}=\frac{y^{\prime}}{\sqrt{(y-\alpha)\left(1+(y')^2\right)}}\tag{3.4.1}
\end{equation}
and the condition (3.3.4) of the preceding section, which must hold along every minimizing arc \(E_{12}\text{,}\) can therefore be expressed in the form
\begin{equation*}
\frac{y^{\prime}}{\sqrt{1+(y')^2}}=\sqrt{y-\alpha}\left\{c-\frac{1}{2} \int_{x_{1}}^{x} \sqrt{\frac{1+(y')^2}{(y-\alpha)^{3}}}\,dx\right\}
\end{equation*}
The second member of this equation, which we may denote by \(\phi(x)\text{,}\) is a continuous function of \(x\) since both \((y-\alpha)\) and the integral in it are continuous along \(E_{12}\text{.}\) By solving the equation for \(y'\) we find that
\begin{equation*}
y'=\frac{\phi}{\sqrt{1-\phi^2}}
\end{equation*}
is also continuous. Turning again to the next to last equation with this result we see now that the second member \(\phi(x)\) of that equation has also a continuous derivative, since when \(y\) is continuous and has a continuous derivative the same is true of both \((y-\alpha)\) and the integral occurring in it. Hence \(y'\) in the last equation must have a continuous derivative and we find the following result:
For the brachistochrone problem a minimizing arc \(E_{12}\) lying entirely below the line \(y = \alpha\) can have no corners and must have continuous curvature. Analytically stated, this means that the derivatives \(y'(x)\text{,}\) \(y''(x)\) exist and are continuous along \(E_{12}\text{.}\)
Aside
Since we now know that the second derivative \(y''(x)\) exists along the minimizing arc we can be sure that the equation \(f=y'f_{y'} = \text{constant}\) deduced in the preceding section also holds along it. When the values of \(f\) and its derivative \(f_{y'}\) for the brachistochrone problem are substituted from (3.4.1) this equation becomes
\begin{equation}
f-y'f_{y'} = \frac{1}{\sqrt{y-\alpha}\sqrt{1+(y')^2}} = \frac{1}{\sqrt{2b}},\tag{3.4.2}
\end{equation}
the value of the constant being chosen for convenience in the form \(1/\sqrt{2b}\text{.}\)
The curves which satisfy the differential equation (3.4.2) may be found in the customary manner by solving the equation for \(y' = \frac{dy}{dx}\) and separating the variables, but we shall find them more easily if we profit by the experience of others and introduce a new variable \(u\) defined by the equation
\begin{equation}
y' = -\tan \frac{u}{2} = -\frac{\sin u}{1+\cos u}.\tag{3.4.3}
\end{equation}
From the differential equation (3.4.2) it follows then, with the help of some simple trigonometry, that along a minimizing arc \(E_{12}\) we must have
\begin{gather*}
y-\alpha=\frac{2 b}{1+(y')^{2}}=2 b \cos ^{2} \frac{u}{2}=b(1+\cos u),\\
\frac{d x}{d u}=\frac{d x}{d y} \frac{d y}{d u}=2 b \cos ^{2} \frac{u}{2}=b(1+\cos u),\\
x=a+b(u+\sin u),
\end{gather*}
where the last equation is found from the next to last by an integration and \(a\) is the new constant so introduced. It will be shown in the next section that curves which satisfy the first and third of these equations are the cycloids described in the following theorem:
Theorem 3.4.2.
A curve down which a particle, started with the initial velocity \(v_1\) at the point 1, will fall in the shortest time to a second point 2 is necessarily an arc having equations of the form
\begin{equation}
x-a = b(u + \sin u), \quad y-\alpha = b(1+\cos u).\tag{3.4.4}
\end{equation}
These represent the locus of a point fixed on the circumference of a circle of radius \(b\) as the circle rolls on the lower side of the line
\begin{equation*}
y = \alpha = y_1 - \frac{v_1^2}{2g}.
\end{equation*}
Such a curve is called a cycloid.
