Jacobi's Theorem/Proof 2

Proof
Consider canonical Euler's equations:


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

Apply a canonical transformation:
 * $\tuple {x, \mathbf y, \mathbf p, H} \to \tuple {x, \boldsymbol \alpha, \boldsymbol \beta, H^*}$

where $\Phi = S$.

By Conditions for Transformation to be Canonical:


 * $p_i = \dfrac {\partial S} {\partial y_i}$
 * $\beta_i = \dfrac {\partial S} {\partial \alpha_i}$
 * $H^* = H + \dfrac {\partial S} {\partial x}$

Because $S$ is a solution to the Hamilton-Jacobi equation:
 * $H^* = 0$

In these new coordinates canonical Euler's equations are:


 * $\dfrac {\d\alpha_i} {\d x} = \dfrac {\partial H^*} {\partial \beta_i}$


 * $\dfrac {\d \beta_i} {\d x} = -\dfrac {\partial H^*} {\partial \alpha_i}$

By $H^* = 0$:


 * $\dfrac {\d \alpha_i} {\d x} = 0$
 * $\dfrac {\d\beta_i} {\d x} = 0$

which imply that $ \alpha_i$ and $\beta_i$ are constant along each extremal.

$\beta_i$ constancy provides with $n$ first integrals:


 * $\dfrac {\partial S} {\partial \alpha_i} = \beta_i$

Because $S = \map S {x, \mathbf y, \boldsymbol \alpha}$, the aforementioned set of first integrals is also a system of equations for functions $y_i$.

Thus, functions $y_i$ can be found.

Functions $p_i$ are found by the results of Conditions for Transformation to be Canonical:


 * $p_i = \dfrac {\partial} {\partial y_i} \map S {x, \mathbf y, \boldsymbol \alpha}$

Then:


 * $\map {y_i} {x, \boldsymbol \alpha, \boldsymbol \beta}$


 * $\map {p_i} {x, \boldsymbol \alpha, \boldsymbol \beta}$

are solutions to canonical Euler's equations.