Conditions for Functional to be Extremum of Two-variable Functional over Canonical Variable p

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Theorem

Let $y=\map y x$ and $\map F {x,y,y'}$ be real functions.

Let $\displaystyle\frac{\partial^2 F}{\partial {y'}^2}\ne 0$.

Let $\displaystyle J\sqbrk y=\int_a^b \map F {x,y,y'}\rd x$

Let $J\sqbrk{y,p}=\int_a^b\paren{-\map H {x,y,p}+py'}\rd x$, where $H$ is the Hamiltonian of $J\sqbrk y$.


Then $\displaystyle J\sqbrk y=J\sqbrk{y,p}\big\vert_{\frac{\delta J\sqbrk{y,p} }{\delta p}=0}$

Proof

Euler's equation for $J\sqbrk{y,p}$:

\(\displaystyle \frac{\delta J\sqbrk{y,p} }{\delta p}\) \(=\) \(\displaystyle \frac{\partial}{\partial p}\paren{-\map H {x,y,p}+py'}\) $\quad$ Depends only on $p$ and not its derivatives $\quad$
\(\displaystyle \) \(=\) \(\displaystyle -\frac{\partial H}{\partial p}+y'\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 0\) $\quad$ $\quad$
\(\displaystyle \) \(\implies\) \(\displaystyle y'=\frac{\partial H}{\partial p}\) $\quad$ $\quad$


Substitute this result back into the functional $J\sqbrk {y,p}$:

\(\displaystyle J\sqbrk{y,p}\big\vert_{\frac{\delta J\sqbrk{y,p} }{\delta p}=0}\) \(=\) \(\displaystyle \int_a^b\paren{-H+p\frac{\partial H}{\partial p} }\rd x\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b\paren{\map F {x,y,y'}-py'+p\frac{\partial\paren{-\map F {x,y,y'}+py'} }{\partial p} } \rd x\) $\quad$ Definition of Hamiltonian $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b\paren{\map F {x,y,y'}-py'+py'}\rd x\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b \map F {x,y,y'}\rd x\) $\quad$ $\quad$


$\blacksquare$

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