Euler's Equation for Vanishing Variation in Canonical Variables

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

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}

\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 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:


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

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


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