Euler's Equation for Vanishing Variation in Canonical Variables

Theorem
Consider the following system of differential equations:


 * $\begin{cases} F_{y_i} - \dfrac \d {\d x} F_{y_i'} = 0\\

\dfrac{\d {y_i} }{\d x} = y_i'\end{cases}$

where $i \in \set{1, \ldots, n}$.

Let the coordinates $\paren{x, \family {y_i}_{1 \mathop \le i \mathop \le n},\family {y_i'}_{1 \mathop \le i \mathop \le n}, F}$ be transformed to canonical variables:
 * $\paren{x,\family {y_i}_{1 \mathop \le i \mathop \le n},\family {p_i}_{1 \mathop \le i \mathop \le n} ,H}$

Then the aforementioned system of differential equations is transformed into:


 * $ \begin{cases}

\dfrac {\d y_i} {\d x} = \dfrac {\partial H} {\partial p_i} \\ \dfrac {\d p_i} {\d x} = - \dfrac {\partial H} {\partial y_i} \end{cases}$

Proof
Find the full differential of Hamiltonian:

By equating coefficients of differentials in last two equations we find that:


 * $\dfrac {\partial H} {\partial x} = - \dfrac {\partial F} {\partial x},\quad \dfrac {\partial H} {\partial y_i} = - \dfrac {\partial F} {\partial y_i}, \quad \dfrac {\partial H} {\partial p_i} = y_i'$

From the third identity it follows that:


 * $\paren{\dfrac {\d y_i} {\d x} = y_i} \implies \paren{\dfrac {\d y_i} {\d x} = \dfrac {\partial H} {\partial p_i} }$

while the second identity together with the definition of $p_i$ assures that:


 * $\paren{\dfrac {\partial F} {\partial y_i} - \dfrac \d {\d x} \dfrac {\partial F} {\partial y_i} = 0} \implies \paren{\dfrac {\d p_i} {\d x} = - \dfrac {\partial H} {\partial y_i} }$