Euler's Equation for Vanishing Variation in Canonical Variables

Theorem
Consider the following system of differential equations:


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

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

where $i = \left({1, \ldots, n }\right)$.

Let the coordinates $\left({x, \langle y_i \rangle_{1 \mathop \le i \mathop \le n}, \langle y_i' \rangle_{1 \mathop \le i \mathop \le n}, F}\right)$ be transformed to canonical variables:
 * $\left({ x, \langle y_i \rangle_{1 \mathop \le i \mathop \le n}, \langle p_i \rangle_{1 \mathop \le i \mathop \le n}, H }\right)$

Then the aforementioned system of differential equations is transformed into:


 * $ \begin{cases}

\dfrac {\rd y_i} {\rd x} = \dfrac {\partial H} {\partial p_i} \\ \dfrac {\rd p_i} {\rd 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:


 * $\left({\dfrac {\rd y_i} {\rd x} = y_i}\right) \implies \left({\dfrac {\rd y_i} {d x} = \dfrac {\partial H} {\partial p_i} }\right)$

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


 * $\left({\dfrac {\partial F} {\partial y_i} - \dfrac \rd {\rd x} \dfrac {\partial F} {\partial y_i} = 0}\right) \implies \left({\dfrac {\rd p_i} {\rd x} = -\dfrac {\partial H} {\partial y_i} }\right)$