Beltrami Identity

Let $$y'\equiv \frac{dy}{dx}.$$ Let $J$ be the line integral of the function $f(x,y,y')$ between $x_1$ and $x_2$ $$J=\int^{x_2}_{x_1}f(x,y,y')dx.$$ When $J$ is at extremum while $x_1,y_1$ and $x_2,y_2$ are held fixed, then $$\begin{align} \delta J &= \int^{x_2}_{x_1}\left(\frac{\partial f}{\partial y}\delta y + \left(\frac{\partial f}{\partial y'}\right)\delta y'\right)dx\\ &= \int^{x_2}_{x_1}\left(\frac{\partial f}{\partial y}\delta y + \left(\frac{\partial f}{\partial y'}\right)\frac{d(\delta y)}{dx}\right)dx\\ &= \left[\delta y \frac{\partial f}{\partial y'}\right]^{x_2}_{x_1} + \int^{x_2}_{x_1}\left(\frac{\partial f}{\partial y} - \frac{d}{dx}\frac{\partial f}{\partial y'} \right)\delta y dx.\\ \end{align}$$ As we are finding extremum with fixed points $x_1,y_1$ and $x_2,y_2$, $\delta y$ at $x_1$ and $x_2$ are both zero. Then we have $$\frac{\partial f}{\partial y} - \frac{d}{dx}\frac{\partial f}{\partial y'} = 0.$$

In particular, if $$\frac{\partial f}{\partial x} = 0,$$ then $$\begin{align} \frac{df}{dx} &= \frac{\partial f}{\partial x} + y'\frac{\partial f}{\partial y} + y''\frac{\partial f}{\partial y'}\\ &= y'\frac{\partial f}{\partial y} + y''\frac{\partial f}{\partial y'}\\ &= y'\left(\frac{\partial f}{\partial y} - \frac{d}{dx}\frac{\partial f}{\partial y'}\right) + y'\frac{d}{dx}\frac{\partial f}{\partial y'} + y''\frac{\partial f}{\partial y'}\\ &= y' (0) + \frac{d}{dx}\left(y'\frac{\partial f}{\partial y'}\right). \end{align}$$ So, $$\frac{d}{dx}\left(f-y'\frac{\partial f}{\partial y'}\right)=0.$$ Then, we have $$f-y'\frac{\partial f}{\partial y'} = C,$$ for some constant $C$. This is known as the Beltrami Identity.