Background: Equation of Motion

Okay. First I want to see if my "Newtonian Mechanics" lens of the problem is correct.

Brachistochrone Problem Visualization

Let the particle's path be given by $\vec{r}(t) = (x(t), y(t))$ and just as in figure 1, WLOG, assume that the path lies in the first quadrant; according to figure 1, we identify the following forces on the point mass:

$$ \vec{w} = (0, mg) $$

$$ \vec{N} = (N_x, -N_y) $$

$$ \text{where $m, g, N_x, N_y \in \mathbb{R^{+}} := [0, \infty)$} $$

Thus, in Cartesian coordinates $ \vec{F} = m\vec{a} = \vec{w} + \vec{N} \implies (\ddot{x}, \ddot{y}) = (N_x, mg - N_y) $

Let $v^2 := \dot{x}^2 + \dot{y}^2$; then, utilizing the fact that the tangential $\mathbf{\hat{T}}$ and normal $\mathbf{\hat{N}}$ unit vectors to the curve are

$$ \mathbf{\hat{T}} = \frac{1}{v} (\dot{x}, \dot{y}) $$

$$ \mathbf{\hat{N}} = \frac{1}{v} (-\dot{y}, \dot{x}) $$

Let $N^2 := {N_x}^2 + {N_y}^2$; we project the force vectors onto the unit tangent and unit normal via

$$ \vec{w} = (\vec{w}\cdot\mathbf{\hat{T}})\mathbf{\hat{T}} + (\vec{w}\cdot\mathbf{\hat{N}})\mathbf{\hat{N}} $$

$$ \vec{N} = 0 \mathbf{\hat{T}} - N\mathbf{\hat{N}} $$

since the particle is constrained to move along the curve, we apply the normal condition

$$ -N\mathbf{\hat{N}} = (\vec{w}\cdot\mathbf{\hat{N}})\mathbf{\hat{N}} $$

it follows that

$$ \vec{w} = \frac{mg}{v} (\dot{y} \mathbf{\hat{T}} + \dot{x} \mathbf{\hat{N}})\tag{1} $$

$$ \vec{N} = 0 \mathbf{\hat{T}} - \frac{mg}{v}\dot{x}\mathbf{\hat{N}}\tag{2} $$

Using the fact that the acceleration of the particle is given by

$$ \ddot{\vec{r}} = \dot{v}\mathbf{\hat{T}} + \frac{\ddot{x}\ddot{y} - \dot{y}\ddot{x}}{v} \mathbf{\hat{N}} $$

we find that by the normal condition, the normal component is zero so we are left with only tangential acceleration and hence:

$$ m\dot{v} = \frac{mg\dot{y}}{v}\tag{3} $$

where by the ODE given by equation 3, using separation of variables, we get:

$$ v = \sqrt{2gy} $$

Let $s := ||{\vec{r}}||$, and suppose $y$ is paremetrised by $x$, so that now $\vec{r} = (x, y(x))$. Then, taking into account that $v = ds/dt$ where we denote $T$ as the time it takes for the particle to reach point $x(T) = p$ from the origin:

$$ T = \int_{0}^{T} dt = \int_{0}^{p} \sqrt{\frac{1 + y'(x)^2}{2gy(x)}} dx\tag{4} $$

Calculus of Variations Stuff

This process is well documented so I will be omitting a lot of steps here. Using the integrand from equation 4, define:

$$ f(y, y', x) = \sqrt{\frac{1 + y'(x)^2}{2gy(x)}}\tag{5} $$

and by the Euler-Lagrange equation, we end up with the following differential equation:

$$ y'(x) = \sqrt{\frac{c - y(x)}{y(x)}} $$

Here, $c$ is a constant where, according to how we aligned our coordinate system and how we assumed the particle's path (i.e. $x > 0$ and $y > 0$), it necessarily follows that $c > 0$.

Using the paremetrization $y = c \sin^2\big(\tilde{t}\big)$ where it necessarily follows that $\tilde{t} \in [0, \pi / 2)$, we obtain the following solutions:

$$ x(\tilde{t}) = c(t - \frac{1}{2} \sin(2\tilde{t})) $$

$$ y(\tilde{t}) = c\bigg(\frac{1}{2} - \frac{1}{2} \cos(2\tilde{t})\bigg) $$


Now, in order to check the solution, let's express the forces given by eq 1 and eq 2 in cartesian coordinates and write out the corresponding equation of motion; working out all the details yields:

$$ \vec{F} = \vec{N} + \vec{w} $$

$$ \Rightarrow m(\ddot{x}, \ddot{y}) = \frac{mg\dot{y}}{v^2} \big(\dot{x} , \dot{y}) $$

Perhaps this was where I made a mistake: I assumed that we can treat $\tilde{t}$ as $t$, so we can check if the solution given by (5) satisfies eq (6). However, if we assume that, then the RHS of (6) ends up as:

$$ mg\bigg(\frac{1}{2} \sin(2t), \frac{1}{2} + \frac{1}{2}\cos(2t)\bigg) $$

whereas the LHS of (6) yields:

$$ m(2c \sin(2t), 2c \cos(2t)) $$

Looking at only the first entries of (7) and (8), it follows that

$$ c = \frac{g}{4} $$

HOWEVER, if we assume this value of $c$, then, using the second entries of (7) and (8), it necessarily follows that:

$$ \boxed{\frac{g}{2}\cos(2t) = \frac{g}{2} + \frac{g}{2} \cos(2t)} $$

Now, if you ask me, this is a contradiction. However, if we are generous with our estimates and consider the y coordinate acceleration yielding a discrepancy of $\frac{g}{2} \approx 4.9 \ \frac{m}{s^2}$ negligible, then I guess we don't have a contradiction..?