Euler-Lagrange formula

Question

Let $y:[-1,1]\to [2-1,2+1]$ be a $C^1$-smooth function, and $F(y,y'):=y\sqrt{1+y'^2}$.

Suppose $y(x)$ satisfy the Euler-Lagrange equation, i.e. $$\frac{\partial F}{\partial y}-\frac{d}{dx}\frac{\partial F}{\partial y'}=0.$$

Q: Why $F-y'\frac{\partial F}{\partial y'}\equiv Const.$ holds ?

Answer 1

As an aside, note that as $y'$ is not presumed to be classically differentiable, $y''$ (and therefore $\frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial F}{\partial y'}$) may only exist in the sense of a weak derivative.

Now, consider that for $F$ a function of only $y$ and $y'$, $$\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\left[F-y'\frac{\partial F}{\partial y'}\right] &= y'\frac{\partial F}{\partial y}+y''\frac{\partial F}{\partial y'}-\left(y''\frac{\partial F}{\partial y'}+y'y''\frac{\partial^2F}{\partial y'^2}+y'^2\frac{\partial^2F}{\partial y\,\partial y'}\right) \\ &= y'\left(\frac{\partial F}{\partial y}-y'\frac{\partial^2F}{\partial y\,\partial y'}-y''\frac{\partial^2F}{\partial y'^2}\right) \\ &= y'\left(\frac{\partial F}{\partial y}-\frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial F}{\partial y'}\right) \end{align*}$$ Therefore, if $\frac{\partial F}{\partial y}-\frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial F}{\partial y'}\equiv 0$, we have that $F-y'\frac{\partial F}{\partial y'}\equiv c$ for some constant $c$.

Answer 2

The key observation is the absence of $x$ in $F(y,y') =y \sqrt{1+y'^2}$.

Let $F(y,y')$ be any $C^1$ smooth function that doesn't depend explicitly on $x$.

Hint 1. Multiply the Euler-Lagrange equation by $y'$ and use the chain rule to get $$y' \frac{\partial F}{\partial y} + y'' \frac{\partial F}{\partial y'} = \frac{d}{dx}(y' \frac{\partial F}{\partial y'}) \tag{1}$$

Hint 2. The total derivative of $F$ is just the LHS of (1), i.e. $\frac{dF}{dx} = y' \frac{\partial F}{\partial y} + y'' \frac{\partial F}{\partial y'}$, see why?

Hence, $\frac{d}{dx}(F - y' \frac{\partial F}{\partial y'}) = 0$, and so $F - y' \frac{\partial F}{\partial y'} = \text{Constant}$.

Answer 3

This is just the Beltrami identity. It can be viewed as a consequence of (i) the fact that the $F$-function has no explicit $x$-dependence, i.e. it possesses a symmetry; and (ii) Noether's theorem.