Calculus of Variations (Gelfand & Fomin): Proof of Functional Dependence Identity

I'm on page 40 of the book (Section 2.10 - Variational Problems in Parametric Form). It states that for

$$\int_{t_0}^{t_1}F\left(x,y,\frac{\dot{y}}{\dot{x}}\right)\dot{x}dt=\int_{t_0}^{t_1}\Phi\left(x,y,\dot{x},\dot{y}\right)dt$$

where the Euler equation for the left-hand side (LHS) is

$$F_y-\frac{d}{dx}F_{y'}=0$$

and the system of Euler equations for the right-hand side (RHS) are

$$\Phi_x-\frac{d}{dt}\Phi_{\dot{x}}=0$$

$$\Phi_y-\frac{d}{dt}\Phi_{\dot{y}}=0$$

that the system of Euler equations are dependent and related by

$$\dot{x}\left(\Phi_x-\frac{d}{dt}\Phi_{\dot{x}}\right)+\dot{y}\left(\Phi_y-\frac{d}{dt}\Phi_{\dot{y}}\right)=0$$.

Can someone prove this? I'm lost.

Thoughts

1. David Widder's "Advanced Calculus" tells us that two functions are dependent if their Jacobian vanishes (i.e. is identically zero)

2. Formulating the equation as

$$\frac{\dot{y}}{\dot{x}}=-\frac{\left(\Phi_x-\frac{d}{dt}\Phi_{\dot{x}}\right)}{\left(\Phi_y-\frac{d}{dt}\Phi_{\dot{y}}\right)}$$

looks eerily similar to implicit differentiation.

Calling

$$\Phi(x,y,\dot x, \dot y)=F\left(x,y,\frac{\dot y }{\dot x}\right)\dot x$$

and performing

$$\dot{x}\left(\Phi_x-\frac{d}{dt}\Phi_{\dot{x}}\right)+\dot{y}\left(\Phi_y-\frac{d}{dt}\Phi_{\dot{y}}\right)$$

we get after some algebra

$$\dot x\left((\dot x-\dot x)F_x+(\dot y-\dot y)F_y\right) = 0$$

NOTE

$$\dot y \left(F^{(0,1,0)} \dot x-F^{(1,0,1)} \dot x+F^{(0,0,2)} \left(\frac{\ddot x \dot y}{\left(\dot x\right)^2}-\frac{\ddot y}{\dot x}\right)-F^{(0,1,1)} \dot y\right)+\dot x \left(\frac{F^{(0,0,1)} \ddot y}{\dot x}-\frac{F^{(0,0,1)} \ddot x \dot y}{\left(\dot x\right)^2}-F^{(0,0,1)} \left(\frac{\ddot y}{\dot x}-\frac{\ddot x \dot y}{\left(\dot x\right)^2}\right)+\frac{\dot y \left(F^{(1,0,1)} \dot x+F^{(0,0,2)} \left(\frac{\ddot y}{\dot x}-\frac{\ddot x \dot y}{\left(\dot x\right)^2}\right)+F^{(0,1,1)} \dot y\right)}{\dot x}-F^{(0,1,0)} \dot y\right) = 0$$

NOTE

Considering

$$I = \int_{x_1}^{x_2} F(x,y,y')dx$$

we have

$$y' = \frac{dy}{dx}= \frac{\dot y}{\dot x}\to dx = \dot x dt$$

and then

$$I = \int_{t_1}^{t_2}F\left(x,y,\frac{\dot y}{\dot x}\right)\dot x dt$$

then

$$\Phi_x = F_x \dot x\\ \Phi_{\dot x} = F-\dot x F_{y'}\frac{\dot y}{(\dot x)^2} = F-y' F_{y'}$$

etc. Having in mind the identity for $$g(x,y,y')$$

$$\frac{d}{dx}\left(y'g_{y'}-g\right) = y'\frac{d}{dx}g_{y'}-g_x-g_y y'=-y'\left(g_y-\frac{d}{dx}g_{y'}\right)-g_x$$

we have

$$\frac{d}{dt}\Phi_{\dot x} = \dot x\frac{d}{dx}\left(F-y' F_{y'}\right) = \dot x\left\{y'\left(F_y-\frac{d}{dx}F_{y'}\right)+F_x\right\}$$

and also

$$\Phi_x = F_x\dot x,\ \ \ \Phi_{\dot y} = \dot x F_{y'}\frac{1}{\dot x} = F_{y'}$$

So concluding with $$\frac{d}{dt}\Phi_{\dot y} = \dot x\frac{d}{dx}F_{y'}$$

$$\Phi_x-\frac{d}{dt}\Phi_{\dot x} = -\dot y\left(F_y-\frac{d}{dx}F_{y'}\right)\\ \Phi_y-\frac{d}{dt}\Phi_{\dot y} = \dot x\left(F_y-\frac{d}{dx}F_{y'}\right)$$

• Can you please show the algebra? – A. Hendry Feb 28 at 23:59
• @A.Hendry See note attached. – Cesareo Mar 1 at 0:24
• Was this from Wolfram? – A. Hendry Mar 1 at 0:43
• Also, I still don't understand how the original identity was derived. The book just says it can be "easily verified". Do you know how it was derived? – A. Hendry Mar 1 at 0:44
• Thank you, by the way, for your help! – A. Hendry Mar 1 at 0:47