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 $\paren{x,\langle y_i\rangle_{1\mathop\le i\mathop\le n},\langle y_i' \rangle_{1\mathop\le i\mathop\le n},F}$ be transformed to canonical variables:
 * $\paren{x,\langle y_i\rangle_{1\mathop\le i\mathop\le n},\langle p_i\rangle_{1\mathop\le i\mathop\le n},H}$

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:


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