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
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David Widder's "Advanced Calculus" tells us that two functions are dependent if their Jacobian vanishes (i.e. is identically zero)
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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.
Best Answer
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 $\frac{d}{dt}\Phi_{\dot y} = \dot x\frac{d}{dx}F_{y'}$
$$ \Phi_x-\frac{d}{dt}\Phi_{\dot x} = -\dot y\left(F_y-\frac{d}{dx}F_{y'}\right)\\ \Phi_y-\frac{d}{dt}\Phi_{\dot y} = \dot x\left(F_y-\frac{d}{dx}F_{y'}\right) $$