Euler–Lagrange equation and standard functionals

The equation

For $J[y] = \int_{a}$ᵇ F(x, y, y′) dx with fixed endpoints: $**\partial{}F/\partial{}y - d/$dx$(\partial{}F/\partial{}y') = 0.**$

Two first integrals

• F independent of $y \Rightarrow \partial{}F/\partial{}y' =$ const (momentum conservation).
• F independent of x ⇒ **Beltrami$**: F - y' \partial{}F/\partial{}y' =$ const (energy conservation).

Standard functionals

J[y]	Euler–Lagrange	Extremals
$\int(y')^{2}$	y″ = 0	straight lines
$\int\sqrt{1 + (y')^{2}} ($arc length)	y″ = 0	straight lines
$\int\sqrt{1 + (y')^{2}}/\sqrt{y} ($brachistochrone)	—	cycloid
$\int y\sqrt{1 + (y')^{2}} ($surface of revolution)	—	catenary
$\int[(y')^{2} - y^{2}]$	y″ + y = 0	sines
$\int(1 + x^{3}y')y'$	d/dx$(1 + 2x^{3}y') = 0$	$y = d - c/(2x^{2})$

Null Lagrangians — the trap

A term that is a total derivative d/dx G(x, y) contributes only boundary terms and does not affect the Euler–Lagrange equation. In $\int[(y')^{2} - y$|y|y′ + xy] dx the middle term is −d/dx(|y|$^{3}/3)$, so the equation reduces to 2y″ = x.

Several unknowns and constraints

• Vector case: one Euler–Lagrange equation per component. For $\int[(y')^{2} + (z')^{2} + y'z']: 2y'' + z'' = 0$ and 2z″ + y″ = 0 ⇒ y″ = z″ = 0.
• **Isoperimetric constraint$** \int{}G = c$: extremise $F + \lambda{}G$; the multiplier is fixed by the constraint.
• Higher derivatives: $\partial{}F/\partial{}y - d/$dx$(\partial{}F/\partial{}y') + d^{2}/$dx$^{2}(\partial{}F/\partial{}y'') = 0$.

Weak vs strong extrema

A weak minimum compares y′ too $(C^{1}-$neighbourhood); a strong minimum only compares $y (C^{0}-$neighbourhood$). \int(1 - (y')^{2})y^{2}$ has y ≡ 0 as a weak minimum but not a strong one — in $a C^{0}-$ball, y′ can be huge and make J negative.

Key takeaways

• No x ⇒ Beltrami; no y ⇒ momentum constant.
• Exact-derivative terms drop out of the Euler–Lagrange equation.
• Weak vs strong extremum is decided by which norm the neighbourhood uses.