# Lagrangian Derivation of Brachistocrone? [closed]

I've been trying to derive the functional description of the brachistocrone. To do this, I used the Lagrangian to describe motion along a functional path. given a function $f(x)$ where $0<y<a$ for all values of x . The lagrangian of this function is $L = KE - PE = \frac{1}{2} m (1+f'(x)^2)\dot x^2 - mgf(x)$.

This simplifies to $\ddot x = \frac {-gf'(x)}{1+f'(x)^2}$.

Now, in order to solve for which function f(x) is the shortest path in time and space between two points greater than 0, we would have to solve our acceleration for displacement. This mathematics is beyond me, does anyone else understand how to finish the problem and derive the brachistocrone?

• you have a $y$ in there but nothing depends on it. Do you mean $0<x<a$? – ZeroTheHero Nov 15 '17 at 21:27
• – Frobenius Nov 10 at 22:33

From what I am looking at, it appears you are applying Euler-Lagrange's equations to the Lagrangian of the system of a particle constrained on a certain curve of shape f(x). However, by doing this, you are minimising the action of this particle, and thus you are getting the equations of motion of the particle. Granted, if you where able to solve them in the general case, you would then be able to compute the time it takes for the particle to reach the bottom, and thus minimise by varying f(x), finding the form of the brachistochrone.

However, this seems a little bit overkill to me. Let's see if there is a better functional to consider other than the action of your system.

Just as a side note, the equation you derived seem to be wrong, the correct equation should be $$\ddot{x}(1+f'^2)=-f'f''\dot{x}^2-gf'$$

Anyway, since Lagrangian formalism is all about minimisation, let's consider, what do we want to minimize here ? It's very straightforward, and it is the time it takes for the particle to fall down the ramp. Now, parametrising the form of the ramp as you did, how can we compute the time for the ball to reach the bottom ?

To cross the infinitesimal distance $ds = \sqrt{1+f'^2}dx$ at speed v takes a time dt given by : $$dt = \frac{ds}{v}$$ We're basically done ! The total time for the ball to reach the bottom will then simply be $T = \int dt = \int \frac{ds}{v} = \int \frac{\sqrt{1+f'2}}{v}dx$.

As we can see, we still miss an element to be able to start doing some lagrangian mechanics. Indeed, we have the speed $v$ that we would like to express as a function of $x$, so that we have an integral on $x$. To do this, we exploit the conservation of energy which tells us that $\frac{1}{2}mv^2+mgh = E$. If we set our y-axis to be directed downwards for convenience (so here, $0<f(x)<-a$) as well as set the Energy to be 0 at $x=0$, we obtain $v(x) = \sqrt{2gf(x)}$

Thus, we have the functional that we would like to minimize : $$T = \int \sqrt{\frac{1+f'(x)^2}{2gf(x)}}dx$$. We can strip away the constant $2g$ which plays no role (we just showed that the brachistochrone has the same shape here or on the moon, by the way !), and the problem can be formulated as a function of a Lagrangian :

$$L = \sqrt{\frac{1+f'(x)^2}{f(x)}}$$

Note that here x plays the role of the time in the usual lagrangian, in the sense that the Euler equations are now : $$\frac{d}{dx}\frac{\partial L}{\partial f'} = \frac{\partial L}{\partial f}$$ Just for the record, $f(x)$ is usually renamed $y(x)$.

See if you have more luck working on this formulation of the problem, if not, tell me and I'll complete my answer !

• In the final equation ( $\frac{d}{dx} \frac{\partial L}{\partial f'} = \frac{\partial L}{\partial f}$), I could solve the equation for x, then find the minimum of the function? – BooleanDesigns Nov 16 '17 at 21:07
• No, rather this equation would give you directly the sought ou function $f(x)$ ! You solve it for $f(x)$ ! – Frotaur Nov 16 '17 at 21:53
• Yeah, would you mind finishing your solution, I managed to follow you through to the step when you changed $T = \int { \frac{ds}{v}} dx$ into the Lagrangian and dropped the integral. This must be me missing something fundamental about Lagrangian mechanics. After that, I tried to solve the Lagrangian, but either I'm not seeing the trick, or the math is simply beyond me. Thanks for any insight you can give. – BooleanDesigns Nov 17 '17 at 4:48
• I'll do this tonight if I have some time ! – Frotaur Nov 17 '17 at 8:18

Under the assumption that your $f(x)=y(x)$, note that $$T=\int dt = \int \frac{ds}{v}$$ with $ds=\sqrt{dx^2+dy^2}$. The speed $v$ can be inferred from conservation of energy. If your particle is starting from rest at height $y_0$ then $$E=mgy_0=mgy+\frac{1}{2}mv^2 \quad \Rightarrow\quad v=\sqrt{2g(y_0-y)}$$ so $$T=\int\sqrt{\frac{dx^2+dy^2}{2g(y_0-y)}}$$ The sweet trick is to realize that, by making $dy$ your integration variable, i.e. $$T=\int\sqrt{\frac{(dx/dy)^2+1}{2g(y_0-y)}}dy$$ you have a Lagrangian $L=\sqrt{\frac{(dx/dy)^2+1}{2g(y_0-y)}}$ where $x$ is ignorable, so that the Euler-Lagrange equation reads $$\frac{d}{dy}\frac{\partial L}{\partial (dx/dy)}-\frac{\partial L}{\partial x}=0$$ and you get a first integral for free: \begin{align} \frac{\partial L}{\partial (dx/dy)}=C \end{align} where $C$ is a constant. From that point it's not terribly hard to reorganize the terms to get an equation that is separable in $x$ and $y$. This well-known problem is to find the curve joining two points, along which a particle falling from rest under the influence of gravity travels from the higher to the lower point in the least time.

It's supposed that the motion on the curve is frictionless.

The time to travel from point 1 to point 2 is $$\begin{equation} t_{12}\boldsymbol{=}\int\limits_{1}^{2}\mathrm{d}t\boldsymbol{=}\int\limits_{1}^{2} \dfrac{\mathrm{d}s}{v} \tag{01}\label{01} \end{equation}$$ where $$\:s\:$$ the arc-length parameter and $$\:v\:$$ the speed $$\begin{equation} \mathrm{d}s\boldsymbol{=}\sqrt{\mathrm{d}x^{2}\boldsymbol{+}\mathrm{d}y^{2}}\boldsymbol{=}\sqrt{1+y'^{\,2}}\:\mathrm{d}x \tag{02}\label{02} \end{equation}$$ In equation \eqref{02} $$\begin{equation} y'\boldsymbol{\equiv}\dfrac{\mathrm{d}y}{\mathrm{d}x} \tag{03}\label{03} \end{equation}$$ while from energy conservation $$\begin{equation} v\boldsymbol{=}\sqrt{2gy} \tag{04}\label{04} \end{equation}$$ So $$\begin{equation} t_{12}\boldsymbol{=}\dfrac{1}{\sqrt{2g}}\int\limits_{1}^{2}\sqrt{\dfrac{1\boldsymbol{+}y'^{\,2}}{y}}\,\mathrm{d}x \tag{05}\label{05} \end{equation}$$ Least time means $$\begin{equation} \sqrt{2g}\;t_{12}\boldsymbol{=}\int\limits_{1}^{2}L\left(y,y',x\right)\mathrm{d}x\boldsymbol{=}\text{extremum} \tag{06}\label{06} \end{equation}$$ where $$\begin{equation} L\left(y,y',x\right) \boldsymbol{\equiv} \sqrt{\dfrac{1\boldsymbol{+}y'^{\,2}}{y}}\boldsymbol{=}\left(\dfrac{1\boldsymbol{+}y'^{\,2}}{y}\right)^{\frac{1}{2}} \tag{07}\label{07} \end{equation}$$ is the Lagrangian of the problem.

The Euler-Lagrange equation $$\begin{equation} \dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{\partial L}{\partial y'}\right)\boldsymbol{-}\dfrac{\partial L}{\partial y}\boldsymbol{=}0 \tag{08}\label{08} \end{equation}$$ gives after lengthy differentiation or using the Beltrami Identity the following differential equation for the curve
$$\begin{equation} y'^{\,2}\boldsymbol{+}2y y''\boldsymbol{+}1\boldsymbol{=}0 \tag{09}\label{09} \end{equation}$$  