Partial Derivatives of Solution of Hamilton-Jacobi Equation are First Integrals of Euler's Equations

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Theorem

Let $\mathbf y = \sequence {y_i}_{1 \mathop \le i \mathop \le n}$, $\bsalpha = \sequence {\alpha_i}_{1 \mathop \le i \mathop \le m}$ be vectors, where $m \le n$.

Let $S = \map S {x, \mathbf y, \bsalpha}$ be a solution of the Hamilton-Jacobi equation, where $\bsalpha$ are parameters.


Then each partial derivative:

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

is a first integral of canonical Euler's equations.


Proof

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

\(\ds \frac \d {\d x} \frac {\partial S} {\partial \alpha_i}\) \(=\) \(\ds \frac {\partial^2 S} {\partial x \partial\alpha_i} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial \alpha_i} \frac {\d y_j} {\d x} + \sum_{j \mathop = 1}^m + \frac {\partial^2 S} {\partial \alpha_j \partial \alpha_i} \frac {\d \alpha_j} {\d x}\) Total derivative of $S$ with respect to $x$
\(\ds \) \(=\) \(\ds \frac {\partial^2 S} {\partial x \partial \alpha_i} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial \alpha_i} \frac {\d y_j} {\d x}\) $ \alpha_i$ is parameter, independent of $x$
\(\ds \) \(=\) \(\ds -\frac {\partial H} {\partial \alpha_i} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial \alpha_i} \frac {\d y_j} {\d x}\) $S$ satisfies Hamilton-Jacobi equation
\(\ds \) \(=\) \(\ds -\frac {\partial x} {\partial \alpha_i} \frac {\partial H} {\partial x} - \sum_{j \mathop = 1}^n \frac {\partial y_j} {\partial \alpha_i} \frac {\partial H} {\partial y_j} - \sum_{j \mathop = 1}^n \frac {\partial p_j} {\partial \alpha_i} \frac {\partial H} {\partial p_j} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial \alpha_i} \frac {\d y_j} {\d x}\) Partial derivative of multivariate composite function $\map H {x, \mathbf y, \mathbf p}$
\(\ds \) \(=\) \(\ds -\sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial \alpha_i} \frac {\partial H} {\partial p_j} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial \alpha_i} \frac {\partial y_j} {\partial x}\) $x$, $y_j$ independent of $\alpha_i$; $S$ satisfies Hamilton-Jacobi equation, thus $p_j = \dfrac {\partial S} {\partial y_j}$
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 1}^n \map {\frac {\partial^2 S} {\partial y_j \partial \alpha_i} } {\frac {\d y_j} {\d x} - \frac {\partial H} {\partial p_j} }\)

If Euler's equations are satisfied, the right hand side vanishes.

Hence

$\dfrac \d {\d x} \dfrac {\partial S} {\partial \alpha_i} = 0$

or:

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

where $C_i$ is a constant.

$\blacksquare$


Sources

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