Study of Brachistochrone Problem.

This blog will deal with one of the most elegant topics in mathematics. The famous Brachistochrone curve. Vsuace did a great job at explaining it in this video. This looks good to me, but it doesn’t feel good; because of course the video does not contain tons of fancy mathematical vocabulary and millions of lines of $\LaTeX$.

To get started, we will just put forward a small fact that, this topic is from a larger field of mathematics called the Calculus of Variations.

Before we get into the Analysis of Brachistochrone Curve (which happens to be the heart of this post), let us brush-up on some concepts.

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Partial Differentiation (advanced preliminary):

Let us look at how we define the derivative of a given function, say we have $g(x)$ and we have,
$$\boxed{\frac{\mathrm{d}}{\mathrm{d}x}g(x)=g'(x)=\lim_{c \to 0}\frac{f(x+ c)-f(x)}{c}}$$
Provided this limit exists.

Now, for a function of many variables it is not easy to compute the total derivative (the usual derivative). Therefore, we calculate the Partial Derivative. You might have already guessed it by now, we define partial derivative as, say we have a function $h(x,y,z)$, then,
$$\boxed{\frac {\partial}{\partial x}h(x,y,z) = \lim_{c \to 0} \frac{h(x+c, y,z)-f(x,y,z)}{c}}$$
Let us look into a small example now, say we have,
$$f(x,y,z) = x^2y^3+x+y+z+\sin(x) \cos(y)$$
Now,
$$\frac{\partial}{\partial x}f(x,y,z) = 2xy^3+1+0+0+\cos(x) \cos(y)$$
similarly,
$$\frac{\partial}{\partial y}f(x,y,z) = 3x^2y^2+0+1+0-\sin(x) \sin(y)$$
and,
$$\frac{\partial}{\partial z}f(x,y,z) = 0 +0+0+1+0$$

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Idea of Speed and Time (basic preliminary):

Let us consider, the following scenario, (Made using Geogebra)

We have, in the figure we have denoted the path ACB as $\pi_1$ and the path ADB as $\pi_2$.
Consider that, The time taken for a body to move from A to B using path $\pi_1$ is $t_{AB_{\pi_1}}$ similarly, the time taken for a body to move from A to B using path $\pi_2$ is $t_{AB_{\pi_2}}$.
Also, consider that the distance for $\pi_1$ be $l_1$ and the distance for $\pi_2$ be $l_2$.

Then,
Speed ($\pi_1$) =$v_1$= $\frac{l_1}{t_{AB_{\pi_1}}}$ and Speed ($\pi_2$) =$v_2$ =$\frac{l_2}{t_{AB_{\pi_2}}}$.

3. Idea of Distance(basic preliminary):

Distance between two points $(a,b)$ and $(c,d)$ is given by,
$$l = \sqrt{(a-c)^2+(b-d)^2}$$
In case, the distance is measured from origin ($O(0,0)$) to some point $A(x,y)$, we get,
$$l = \sqrt{(x-0)^2+(y-0)^2} = \sqrt{x^2 + y^2}$$

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Now that the notion is clear, we shall proceed into the analysis.
[NOTE : Some topics which are new to me (or you) will be explained as and when it is required].

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Time needed for a body to travel from $A$ to $B$ is given by,
$t_{AB}=\frac{l}{v}$ for a linear path.

For a curve or a non-linear path, we consider piece-wise linear distance ($\mathrm{d}l$) and speed ($v$). We can define, $t_{AB}$ as,
$$\boxed{t_{AB} = \int_{A}^{B}\frac{\mathrm{d}l}{v}} \tag*{(1)}$$
Now, we must understand a fact that, all distance is composed of $x$ and $y$ coordinates. We consider the point $A$ as the origin ($O(0,0)$) and $B$ as $(x,y)$. Hence,
$$l = \sqrt{x^2 + y^2}\\ \implies \mathrm{d}l = \sqrt{\mathrm{d}x^2 + \mathrm{d}y^2}\\ \implies \mathrm{d}l = \sqrt{(1 + \frac{\mathrm{d}y^2}{\mathrm{d}x^2})\mathrm{d}x^2}$$
$$\boxed{\implies \mathrm{d}l = \sqrt{(1 + \frac{\mathrm{d}y^2}{\mathrm{d}x^2})}\mathrm{d}x} \tag*{(2)}$$

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Also, since the translation is both in the $x$ and $y$ axes, we can say that the speed is gained by $KE$ and $PE$ equality,
$$KE = PE\\ \implies \frac{1}{2}mv^2 = mgy \\ $$
$$\implies \boxed{v = \sqrt{2gy}} \tag*{(3)}$$

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Use $(2)$ and $(3)$ in $(1)$, we have,
$$\boxed{t_{AB} = \int_{A}^{B}\frac{\sqrt{(1 + \frac{\mathrm{d}y^2}{\mathrm{d}x^2})}\mathrm{d}x}{\sqrt{2gy}}} \tag*{(4)}$$
We write
$$\frac{\mathrm{d}y}{\mathrm{d}x}=y'$$
Now, $(4)$ becomes,
$$t_{AB} = \int_{A}^{B}\frac{\sqrt{(1 +y'^2 )}}{\sqrt{2gy}}\mathrm{d}x \tag*{(4.1)}$$
Therefore, the function is,
$$\theta = \frac{\sqrt{(1 +y'^2 )}}{\sqrt{2gy}}\\ \boxed{\theta = (1 +y'^2 )^{1/2}(2gy)^{-1/2}} \tag*{(5)}$$

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Please notice that, the integral $t_{AB}$ can be written as,
$$t_{AB} = \int_{A}^{B} f(y, y', x) \mathrm{d}x \tag*{(6)}$$
In our case, $f(y, y', x)$ is $\theta$.

Equation$(6)$ closely follows the The Euler-Lagrange equation.

It is pretty simple to define,

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The Euler-Lagrange equation

For any $I$ such that,
$$I = \int_{a}^{b}f(y,y',x)\mathrm{d}x$$
Where, $y' = \frac{\mathrm{d}y}{\mathrm{d}x}$, then $I$ has an stationary point, if the Euler-Lagrange Equation, given by,
$$\frac{\partial f}{\partial y}-\frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial f}{\partial y'}=0$$
is satisfied.

Therefore, for our analysis,
$$\boxed{\frac{\partial \theta}{\partial y} - \frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial \theta}{\partial y'}=0} \tag*{(7)}$$

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Stationary value

This is a value at the stationary point.
A stationary point $a$ is the point at which the first derivative of a function becomes zero,
$$f'(x)|_{x=a} = 0 ; a = \text{Stationary point}$$
Side note :
Find the stationary points of $f(x) = x^3-6x^2+9x$.

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Beltrami Identity

Our $\theta$ (from $(5)$) is such a cool expression, because $x$ does not appear in that; therefore, of course $\boxed{\frac{\partial \theta}{\partial x} = 0}$; this again leads us to another beautiful form, Beltrami Identity, which is given by,
$$\boxed{\theta - y' \frac{\partial \theta}{\partial y'}=C ; C \text{ is a constant}}.\tag*{(8)}$$

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We now find, $\frac{\partial \theta}{\partial y'}$,
$$\frac{\partial \theta}{\partial y'}=\frac{\partial}{\partial y'}[(1 +y'^2 )^{1/2}(2gy)^{-1/2}] \\ \implies \frac{\partial \theta}{\partial y'} =(2gy)^{-1/2} \frac{1}{2}(1 +y'^2 )^{-1/2}\cdot 2y'$$
$$\therefore \boxed{\frac{\partial \theta}{\partial y'} = (2gy)^{-1/2} (1 +y'^2 )^{-1/2}\cdot y'} \tag*{(9)}$$

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Use $\theta$ and $\frac{\partial \theta}{\partial y'}$ in equation $(8)$, we get,
$$(1 +y'^2 )^{1/2}(2gy)^{-1/2}-y' \cdot (2gy)^{-1/2} (1 +y'^2 )^{-1/2}\cdot y' = C$$
$$\implies \frac{\sqrt{(1 +y'^2 )}}{\sqrt{2gy}}-\frac{y'^2}{\sqrt{2gy}\sqrt{1+y'^2}} = C$$
$$\implies \frac{1+y'^2-y'^2}{\sqrt{2gy}\sqrt{1+y'^2}}=C$$
$$\implies \boxed{\frac{1}{\sqrt{2gy}\sqrt{1+y'^2}}=C} \tag*{(10)}$$
Rearranging a little gives,
$$\sqrt{1+y'^2}=\frac{1}{\sqrt{2gy}C}$$
$$\implies \boxed{(1+\frac{\mathrm{d}y}{\mathrm{d}x})y = \frac{1}{2gC^2} = M^2} \tag*{(11)}$$
This is the equation of the cycloid as per [4], I am yet to figure out how this happened.
The solution of $(11)$ can be found using a parametric equation,
$$\boxed{x(t) = \frac{1}{2}M^2(t-\sin t)\\ y(t)=\frac{1}{2}M^2(1-\cos t)} \tag*{(12)}$$
To derive the above equation, please refer Math-stackexchange|Derive the parametric equations of a cycloid.

I have even plotted the solutions at $(12)$, we can see that it is a cycloid,

If you wish to play with the visualisation please go to Cycloid | Parametric @ Desmos by Pragyaditya

Cheers!

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References:
1. Stationary Points
2. Euler Lagrange Equation
3. Derivation of Beltrami Identity
4. Introduction to calculus of variations
5. Brachistochrone @ Wolfram