Why do I need the Beltrami identity to solve the brachistochrone problem?

Question

Brachistochrone problem

The time to travel from point $p_1$ to $p_2$ is given by this integral $$t_{12}=\int_{p_1}^{p_2}\frac{ds}{v}.$$ With $ds=\sqrt{dx^2+dy^2}=\sqrt{1+y'^2}\,dx$ and $v=\sqrt{2g\,y}$, we obtain $$t_{12}=\int_{p_1}^{p_2}\underbrace{\sqrt{\frac{1+y'^2}{2g\,y(x)}}}_{f(y,y')}\,dx,$$ and because the function $f$ is Independent of $x$, we can use the Beltrami identity $$f-y'\frac{\partial f}{\partial y'}=c\tag 1$$ to obtain the solution.

Instead, I want to use a different approach. Taking $ds=\sqrt{\dot x^2+\dot y^2}\,dt$ and $v=\sqrt{2g\,y}$ gives $$ t_{12}\mapsto\int_{p_1}^{p_2}\frac{\sqrt{\dot x^2+\dot y^2}\,dt}{\sqrt{2g\,y}},$$ which implies $$f(y,\dot{y},\dot{x})=\sqrt{\frac{\dot{x}^2+\dot{y}^2}{2g\,y}}\equiv\mathcal L(y,\dot{y},\dot{x}).$$ So the first integral is $$\left(\frac{\partial \mathcal L}{\partial{\dot{x}}}\right)^2=c.\tag 2$$ From equation (2), I obtain the equation $$\frac{\dot x^2}{(\dot x^2+\dot y^2)y}=\tilde k^2,$$ implying $$\frac{(\dot x^2+\dot y^2)y}{\dot x^2}=k^2,$$ which reduces to $$(1+y'^2)y(x)=k^{2},$$ to which the solution is $$x(t)=\frac{k^2}{2}[t-\sin(t)],\quad y(t)=\frac{k^2}{2}[1-\cos(t)],$$ which is indeed the solution of brachistochrone problem.

So what is wrong with this approach?

Answer 1

Your equation $$ \frac{\dot x^2}{(\dot x^2+\dot y^2)\,y}=\tilde k^2\quad\Rightarrow \frac{(\dot x^2+\dot y^2)\,y}{\dot x^2}=k^2 \tag{1} $$ is correct. This is effectively a consequence of the Euler-Lagrange equation $$ \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}} \right) = \frac{\partial \mathcal{L}}{\partial x}, $$ noting that $\partial \mathcal{L}/\partial x = 0$. In general, though, this would not be sufficient to identify the solutions; you would also need to take into account the equation of motion for $y$, $$ \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{y}} \right) = \frac{\partial \mathcal{L}}{\partial y}. $$ If you are given an arbitrary Lagrangian depending on two functions, there is no guarantee that a solution that works for one of the equations of motion will necessarily work for the other. So it's not a valid move to take one of the equations of motion, find a solution for it, and claim that it's the solution that extremizes the action.

Now, it happens that you can get away with only solving the $x$ equation in this particular case. For example, here's another solution to (1): $$ \tag{2} x(t) = k^2 \sin^{-1} \left( \frac{\sqrt{t}}{k}\right) - \sqrt{ t(k^2 - t)} \qquad y = t. $$ In fact, you can generate any number of solutions to (1) by just setting $y(t)$ to be any function you want, solving the resulting equation for $\dot{x}$, and integrating.

But it turns out if you do this you'll still end up with the same curve, just reparametrized in a weird way. (It is left as an exercise to the reader to show what the relationship is between the $t$ in (2) and the $t$ in the OP's original solution.) This is because your Lagrangian has reparametrization invariance; your integral for the time of travel works out to have exactly the same form even if you replace $t$ in your integral with an arbitary parameter $\lambda(t)$. This "hidden symmetry" effectively means that you can set either $x(t)$ or $y(t)$ to be whatever you want it to be; it just corresponds to a change of parameter, and a solution to one EOM will necessarily solve the other as well. (I think.)

Answer 2

what is the relation between the first integral and the Betrami Identity

assume that the Lagrangian is a function $~y~,\dot{y}~,\dot{x}~$ but not a function of x $~L=L(~y~,\dot{y}~,\dot{x}~)~$ from here it follows that \begin{align*} & \frac{dL}{dt}= \frac{\partial L}{\partial\dot{x}}\ddot{x}+ \frac{\partial L}{\partial y}\dot{y}+ \frac{\partial L}{\partial \dot{y}}\ddot{y} \tag 1 \end{align*}

and with Euler Lagrange equations \begin{align*} &\frac{\partial L}{\partial y}=\frac{d}{dt}\frac{\partial L}{\partial \dot{y}}\quad, \frac{\partial L}{\partial x}=\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}=0 \quad\Rightarrow \frac{\partial L}{\partial\dot{x}}=c\quad,\text{(this is the first integral)} \end{align*} substitute to equation (1) \begin{align*} & \frac{dL}{dt}=\frac{\partial L}{\partial\dot{x}}\ddot{x}+ \left(\frac{d}{dt}\frac{\partial L}{\partial \dot{y}}\right)\dot{y}+ \frac{\partial L}{\partial \dot{y}}\ddot{y}=\frac{\partial L}{\partial\dot{x}}\,\frac{d}{dt}\dot{x}+ \frac{d}{dt}\left(\,\dot{y}\,\frac{\partial L}{\partial \dot{y}}\right) \end{align*} and integrate \begin{align*} &L=\frac{\partial L}{\partial\dot{x}}\,\dot{x}+\,\dot{y}\,\frac{\partial L}{\partial \dot{y}}\quad,\text{or} \end{align*} \begin{align*} & \frac{\partial L}{\partial\dot{x}}\,\dot{x}=\underbrace{L-\dot{y}\,\frac{\partial L}{\partial \dot{y}}}_{\text{Betrami identity}}=k \end{align*} with $~\dot{y}=y'\,\dot{x}~$ \begin{align*} & \frac{\partial L}{\partial\dot{x}}\,\dot{x}={L-y'\,\dot{x}\,\frac{\partial L}{\partial\dot{y}}}\bigg|_{\dot{y}=y'\,\dot x}=k\quad,\Rightarrow \end{align*}

\begin{align*} &\boxed{~~\frac{\partial L}{\partial\dot{x}}\bigg|_{\dot{y}=y'\,\dot x}= L(y,y')-y'\,\frac{\partial L}{\partial y'}=k~~} \end{align*}

Example

\begin{align*} &L=\sqrt{\frac{1+y'^2}{y}}\quad,\text{with} y'=\frac{\dot{y}}{\dot{x}}\quad, L= \sqrt{\frac{\dot{x}^2+\dot{y}^2}{y}} \end{align*}

\begin{align*} & \frac{\partial L}{\partial\dot{x}}=\frac{\dot{x}}{\sqrt{(\dot{x}^2+\dot{y}^2)}\,\sqrt{y}} \overset{\text{with}~~ \dot{y}=y'\,\dot{x}}{=} \frac{1}{\sqrt{1+y'^2}\,\sqrt{y}} \end{align*}

Conclusion

we can use the First integral to obtain $~y(x)~$