2
$\begingroup$

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 ?

$\endgroup$
0
$\begingroup$

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 $c$.

$\endgroup$
1
$\begingroup$

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 $F - y' \frac{\partial F}{\partial y'} = \text{Constant}$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.