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

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

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

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 $\displaystyle J \sqbrk y = \bigvalueat {J \sqbrk {y, p} } {\frac {\delta J \sqbrk{y, p} } {\delta p} \mathop = 0}$


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
\(\displaystyle \) \(=\) \(\displaystyle -\frac {\partial H} {\partial p} + y'\)
\(\displaystyle \) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle y'\) \(=\) \(\displaystyle \frac {\partial H} {\partial p}\)

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

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