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

From ProofWiki
Jump to: navigation, search


Let $\Phi=\map {\Phi} {x,\langle y_i\rangle_{1\le i\le n},\langle p_i\rangle_{1\le i\le n} }$ 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.


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

For $\Phi$ to be the first integral:

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

Hence the result.