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


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

Proof

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

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


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

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


$\blacksquare$

Sources