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 $\dfrac {\partial^2 F} {\partial {y'}^2} \ne 0$.
Let $\ds J \sqbrk y = \int_a^b \map F {x, y, y'} \rd x$
Let $\ds J \sqbrk {y, p} = \int_a^b \paren {-\map H {x, y, p} + p y'} \rd x$, where $H$ is the Hamiltonian of $J \sqbrk y$.
Then $\ds J \sqbrk y = \bigvalueat {J \sqbrk {y, p} } {\frac {\delta J \sqbrk{y, p} } {\delta p} \mathop = 0}$
Proof
Euler's equation for $J \sqbrk {y, p}$:
\(\ds \frac {\delta J \sqbrk{y, p} } {\delta p}\) | \(=\) | \(\ds \frac {\partial} {\partial p} \paren {-\map H {x, y, p} + p y'}\) | Depends only on $p$ and not its derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac {\partial H} {\partial p} + y'\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds y'\) | \(=\) | \(\ds \frac {\partial H} {\partial p}\) |
Substitute this result back into the functional $J \sqbrk {y, p}$:
\(\ds \bigvalueat {J \sqbrk {y, p} } {\frac {\delta J \sqbrk {y, p} } {\delta p} \mathop = 0}\) | \(=\) | \(\ds \int_a^b \paren {-H + p \frac{\partial H}{\partial p} } \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_a^b \paren {\map F {x, y, y'} - p y' + p \frac {\partial \paren {-\map F {x, y, y'} + p y'} } {\partial p} } \rd x\) | Definition of Hamiltonian | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_a^b \paren {\map F {x, y, y'} - p y' + p y'} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_a^b \map F {x, y, y'} \rd x\) |
$\blacksquare$
Sources
- 1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations ... (previous) ... (next): $\S 4.18$: The Legendre Tranformation