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

By comparison of this and previous expressions:


 * $ \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}$

Apply a canonical transformation $ \left({ x, \mathbf y, \mathbf p, H } \right) \to \left({ x, \boldsymbol \alpha, \boldsymbol \beta, H^*  } \right) $, where $ \Phi = S$.

By Conditions for Transformation to be Canonical:


 * $ \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}$

Since $S$ satisfies Hamilton-Jacobi equation, $ H^*=0$.

In these new coordinates canonical Euler's equations are:


 * $ \displaystyle \frac{ \mathrm d \alpha_i}{ \mathrm d x}= \frac{ \partial H^*}{ \partial \beta_i}$


 * $ \displaystyle \frac{ \mathrm d \beta_i}{ \mathrm d x}= -\frac{ \partial H^*}{ \partial \alpha_i}$

By $ H^*=0$:


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

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

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


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

Because $ S= S \left({x, \mathbf y, \boldsymbol \alpha } \right)$, 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:


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

Then


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


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

are solutions to canonical Euler's equations.