Conditions for Function to be First Integral of Euler's Equations for Vanishing Variation

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Theorem

Let $\Phi = \map {\Phi} {x, \family {y_i}_{1 \mathop \le i \mathop \le n}, \family {p_i}_{1 \mathop \le i \mathop \le n} }$ be a real function.

Let $H$ be Hamiltonian.

Then a necessary and sufficient condition for $\Phi$ to be the first integral of Euler's Equations is

$\dfrac {\partial \Phi} {\partial x} + \sqbrk{\Phi, H} = 0$


Corollary 1

Let $\dfrac {\partial\Phi} {\partial x}=0$.


Then $\Phi$ is the first integral if its Poisson Bracket with the Hamiltonian vanishes.


Corollary 2

Let $\Phi = H$.

Let $\dfrac {\partial H} {\partial x} = 0$.


Then $H$ is the first integral of Euler's Equations.


Proof

\(\displaystyle \dfrac {\d \Phi} {\d x}\) \(=\) \(\displaystyle \frac {\partial\Phi} {\partial x} + \sum_{i \mathop = 1}^n \frac {\partial \Phi} {\partial y_i} \frac {\partial y_i} {\partial x} + \sum_{i \mathop = 1}^n \frac {\partial \Phi} {\partial p_i} \frac{\partial p_i} {\partial x}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\partial \Phi} {\partial x} + \sum_{i \mathop = 1}^n \frac {\partial \Phi} {\partial y_i} \frac {\partial H} {\partial p_i} - \sum_{i \mathop = 1}^n \frac {\partial \Phi} {\partial p_i} \frac{\partial H} {\partial y_i}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\partial \Phi} {\partial x} + \sqbrk{\Phi, H}\)

For $\Phi$ to be the first integral:

$\dfrac {\d \Phi} {\d x} = 0$

Hence the result.

$\blacksquare$



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