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

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Theorem

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

Let $S = \map S {x, \mathbf y, \boldsymbol \alpha}$ be a solution of Hamilton-Jacobi quation, where $\boldsymbol \alpha$ are parameters.


Then each 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 $x$:

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

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.

$\blacksquare$


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