Euler's Equation for Vanishing Variation in Canonical Variables

Theorem
Suppose we have the following system of differential equations:


 * $ \begin{cases}

\displaystyle F_{ y_i} - \frac{ \mathrm d }{ \mathrm d x } F_{ y_i'}=0 \\ \displaystyle \frac{ \mathrm d {y_i} }{ \mathrm d x }= y_i' \end{cases} $

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

Suppose the coordinates $ \left({ x, \langle y_i \rangle_{1 \le i \le n}, \langle y_i' \rangle_{1 \le i \le n}, F } \right)$ are transformed to canonical variables $\left({ x, \langle y_i \rangle_{1 \le i \le n}, \langle p_i \rangle_{1 \le i \le n}, H }\right)$.

Then the aforementioned system of differential equations is transformed into


 * $ \begin{cases}

\displaystyle \frac{ \mathrm d y_i }{ d x }=\frac{ \partial H }{ \partial p_i } \\ \displaystyle \frac{ \mathrm d {p_i} }{ \mathrm d x }= - \frac{ \partial H }{ \partial y_i } \end{cases} $

Proof
Find the full differential of Hailtonian:

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


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

From the third identity it follows that


 * $\displaystyle \left({ \frac{ \mathrm d y_i }{ d x }= y_i } \right) \implies \left({ \frac{ \mathrm d y_i }{ d x }= \frac{ \partial H }{ \partial p_i } } \right)$

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


 * $\displaystyle \left({ \frac{ \partial F }{ \partial y_i } - \frac{ \mathrm d }{ \mathrm d x } \frac{ \partial F }{ \partial y_i }= 0 } \right) \implies \left({ \frac{ \mathrm d p_i }{ \mathrm d x }= -\frac{ \partial H }{ \partial y_i } } \right)$