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

## 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

- 1963: I.M. Gelfand and S.V. Fomin:
*Calculus of Variations*... (previous) ... (next): $\S 4.18$: The Legendre Tranformation