Jacobi's Theorem/Proof 1

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Theorem

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

Let $S = \map S {x, \mathbf y, \boldsymbol \alpha}$ be a a complete solution of the Hamilton-Jacobi equation.


Let:

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

where $\begin {vmatrix} \cdot \end{vmatrix}$ is a determinant.


Let:

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


Then:

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

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


Proof

Consider the total derivative of $\dfrac {\partial S} {\partial \alpha_i}$ with respect to $x$:

\(\ds \frac \d {\d x} \frac {\partial S} {\partial \alpha_i}\) \(=\) \(\ds \frac {\partial^2 S} {\partial x \partial \alpha_i} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial \alpha_i} \frac {\d y_j} {\d x} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial \alpha_j \partial \alpha_i} \frac {\d \alpha_j} {\d x}\)
\(\ds \) \(=\) \(\ds \frac {\partial^2 S} {\partial x \partial \alpha_i} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial \alpha_i} \frac {\d y_j} {\d x}\) $\alpha_j$ is a parameter, independent of $x$
\(\ds \) \(=\) \(\ds -\frac {\partial H} {\partial \alpha_i} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial \alpha_i} \frac {\d y_j} {\d x}\) $S$ is a Solution to Hamilton-Jacobi Equation
\(\ds \) \(=\) \(\ds -\sum_{j \mathop = 1}^n \frac {\partial H} {\partial p_j} \frac {\partial p_j} {\partial \alpha_i} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial \alpha_i} \frac {\d y_j} {\d x}\)
\(\ds \) \(=\) \(\ds -\sum_{j \mathop = 1}^n \frac {\partial H} {\partial p_j} \frac {\partial^2 S} {\partial \alpha_i \partial y_j} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial \alpha_i} \frac {\d y_j} {\d x}\) Derivation of Hamilton-Jacobi Equation
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 1}^n \map {\frac {\partial^2 S} {\partial x \partial \alpha_i} } {\frac {\d y_j} {\d x} - \frac {\partial H} {\partial p_j} }\)
\(\ds \) \(=\) \(\ds 0\) as $\dfrac {\d \beta_i} {\d x} = 0$
\(\ds \leadsto \ \ \) \(\ds \frac {\d y_j} {\d x}\) \(=\) \(\ds \frac {\partial H} {\partial p_j}\)


Next, consider the total derivative of $p_i$ with respect to $x$:

\(\ds \frac {\d p_i} {\d x}\) \(=\) \(\ds \frac \d {\d x} \frac {\partial S} {\partial y_i}\)
\(\ds \) \(=\) \(\ds \frac {\partial^2 S} {\partial x \partial y_i} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial y_i} \frac {\d y_j} {\d x} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial \alpha_j \partial y_i} \frac {\d \alpha_j} {\d x}\)
\(\ds \) \(=\) \(\ds \frac {\partial^2 S} {\partial x \partial y_i} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial y_i} \frac {\d y_j} {\d x}\) $\alpha_j$ is a parameter, independent of $x$
\(\ds \) \(=\) \(\ds \frac {\partial^2 S} {\partial x \partial y_i} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial y_i} \frac {\partial H} {\partial p_j}\) as $\dfrac {\d y_j} {\d x} = \dfrac {\partial H} {\partial p_j}$


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

\(\ds \frac {\partial^2 S} {\partial x \partial y_i}\) \(=\) \(\ds -\frac {\partial H} {\partial y_i} - \sum_{j \mathop = 1}^n \frac {\partial H} {\partial p_j} \frac {\partial p_j} {\partial y_i}\)
\(\ds \) \(=\) \(\ds -\frac {\partial H} {\partial y_i} - \sum_{j \mathop = 1}^n \frac {\partial H} {\partial p_j} \frac {\partial^2 S} {\partial y_i \partial y_j}\) Derivation of Hamilton-Jacobi Equation
\(\ds \leadsto \ \ \) \(\ds -\frac {\partial H} {\partial y_i}\) \(=\) \(\ds \frac {\partial^2 S} {\partial x \partial y_i} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_i \partial y_j} \frac {\partial H} {\partial p_j}\)

By comparison of this and previous expressions:

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

$\blacksquare$


Source of Name

This entry was named for Carl Gustav Jacob Jacobi.


Sources

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