1

from

https://mathworld.wolfram.com/BrachistochroneProblem.html

I found the EL equation (29) and the parametric solution equations $~(32)~,(33)~$.

Eq. (29) \begin{align*} &{~(1+y'^2)\,(1+\mu\,y')+2(y-\mu\,x)\,y''=0}\tag 1 \end{align*} Eq. (32),(33) \begin{align*} &{x=k \left( \theta-\sin \left( \theta \right) +\mu\, \left( 1-\cos \left( \theta \right) \right) \right)}\\ &{y={k} \left( 1-\cos \left( \theta \right) +\mu\, \left( \theta+\sin \left( \theta \right) \right) \right)} \end{align*} with:

$$\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$$

and

$$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)= \frac{\frac{d}{d\theta}\left(\frac{dy}{dx}\right)}{\frac{dx}{d\theta}}$$

in Eq. (1) , the result is not equal zero , why ??

for solution of Brachistochrone Problem with out friction $~\mu=0~$ I obtain that Eq (1) is qual zero.

Edit

those equations are the solutions of equation (1)

$$~x(\theta)=k\,(~\theta+\theta_0+\sin(\theta+\theta_0)-\mu\,\cos(\theta+\theta_0))\\ ~y(\theta)=k\,[1+\cos(\theta+\theta_0)+\mu^2+\mu(\theta+\theta_0+\sin(\theta+\theta_0))]$$

where $~\theta_0~$ is a constant to adapt the initial condition.

for example $~\mu=0~,\theta_0=-\pi~$ you obtain

enter image description here

so which solutions are the correct one ??

Eli
  • 413

1 Answers1

1

starting with \begin{align*} &\mathcal{L}=\sqrt{\frac{1+y'^2}{y-\mu\,x}} \quad\text{with}\\ & y=z+\mu\,x~,y'=z'+\mu\\ &\mathcal{L}=\sqrt{\frac{1+(z'+\mu)^2}{z}} \end{align*} and because $~\mathcal{L}~$ is independent of x the energy $~E~$ is constant \begin{align*} &E(z'~,z)= z'\,\frac{\partial\mathcal{L}}{\partial z'}-\mathcal{L}=C\quad\Rightarrow\\ &\frac{1}{E^2}\bigg|_{(z'=y'-\mu~,z=y-\mu\,x)}=\frac{(1+y'^2)\,(y-\mu\,x)}{(1+y'^2)^2}=c \end{align*}

to addapt the initial condition $~y(0)=0~$ we transfer $~y\mapsto\,y-y_0~$ thus

\begin{align*} &c=\frac{(1+y'^2)\,(y-y_0-\mu\,x)}{(1+y'^2)^2}\tag 1 \end{align*}

solve equation (1) with \begin{align*} &y'=\cot\left(\frac{1}{2}\theta\right)\quad,c=2\\ &\boxed{x=k \left( \theta-\sin \left( \theta \right) +\mu\, \left( 1-\cos \left( \theta \right) \right) \right)}\qquad\qquad (32) \end{align*} for $~y~$ we obtain \begin{align*} &y=\left( 2\,{\mu}^{2}+\mu\, \left( \theta+\sin \left( \theta \right) \right) -\cos \left( \theta \right) +{y_0}+1 \right) ~,\text{and with}~, y_0=-{2\,\mu^2}{k}\quad\Rightarrow\\ &\boxed{y={k} \left( 1-\cos \left( \theta \right) +\mu\, \left( \theta+\sin \left( \theta \right) \right) \right)~}\qquad\qquad (33)\\ &y(0)=0 \end{align*} the EL equation (29) with $~y\mapsto y+y_0~$ is: \begin{align*} &\boxed{~(1+y'^2)\,(1+\mu\,y')+2(y-y_0-\mu\,x)\,y''=0}\tag 2 \end{align*} equations $~(32)~(33)~$ are also the solution of equation (2)

so we have to modified the EL equation to get it equal zero.

Eli
  • 413