Euler's Equation/Independent of y

Theorem

Let $y$ be a mapping

Let $J$ be a functional such that

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

Then the corresponding Euler's equation can be reduced to:

$F_{y'} = C$

where $C$ is an arbitrary constant.

Proof

Assume that:

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

Euler's equation for $J$ is:

$\dfrac \d {\d x} F_{y'} = 0$

Integration yields:

$F_{y'} = C$

$\blacksquare$

Sources

• 1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations ... (previous) ... (next): $\S 1.4$: The Simplest Variational Problem. Euler's Equation