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.