Jacobi's Theorem

Theorem
Let $ \mathbf y= \langle y_i \rangle_{1 \le i \le n}$, $ \boldsymbol \alpha = \langle \alpha_i \rangle_{1 \le i \le n}$, $ \boldsymbol \beta= \langle \beta_i \rangle_{1 \le i \le n}$ be vectors, where $\alpha_i$ and $ \beta_i$ are parameters.

Let $ S= S \left({ x, \mathbf y, \boldsymbol \alpha } \right)$ be a complete solution of Hamilton-Jacobi equation.

Let


 * $ \begin{vmatrix} \displaystyle

\frac{ \partial^2 S}{ \partial \alpha_i \partial y_k} \end{vmatrix} \ne 0$

Let


 * $ \displaystyle \frac{ \partial S}{ \partial \alpha_i}= \beta_i$

Then


 * $ \displaystyle p_i= \frac{ \partial S}{ \partial y_i} \left({ x, \mathbf y, \boldsymbol \alpha } \right)$


 * $ \displaystyle y_i= y_i \left({ x, \boldsymbol \alpha, \boldsymbol \beta } \right)$

constitute a general solution of the canonical Euler's equations.

Proof 1
Consider the total derivative of $ \displaystyle \frac{ \partial S}{ \partial \alpha_i}$ wrt $x$:

Next, consider the total derivative of $p_i$ wrt $x$:

On the other hand, partial derivative of Hamilton-Jacobi equation yields

Comparison of this and previous expressions leads to the conclusion that


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

Proof 2
Consider canonical Euler's equations:


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

By Conditions for Transformation to be Canonical, where $ \Phi = S$, where $ \boldsymbol \alpha$ are the new momenta, and $ \boldsymbol \beta$ are the new coordinates:


 * $ \displaystyle p_i= \frac{ \partial S}{ \partial y_i}, \quad \beta_i = \frac{ \partial S}{ \partial \alpha_i}, \quad H^*= H+ \frac{ \partial S}{ \partial x}$

However, $S$ satisfies Hamilton-Jacobi equation.

Hence, in these new coordinates Euler's equations become:


 * $ \displaystyle \frac{ \mathrm d \alpha_i}{ \mathrm d x}= 0, \quad \frac{ \mathrm d \beta}{ \mathrm d x}= 0$

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


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

Use this to determine functions $ y_i$

Set


 * $ \displaystyle p_i= \frac{ \partial}{ \partial y_i} S \left({ x, \mathbf y, \boldsymbol \alpha } \right)$

Then solutions are


 * $ y_i \left({ x, \boldsymbol \alpha, \boldsymbol \beta } \right)$


 * $ p_i \left({ x, \boldsymbol \alpha, \boldsymbol \beta } \right)$