Calculus of Variations

The equation

For $J[y]={\int }_a$ᵇ F(x, y, y′) dx with fixed endpoints: $\ast \ast \partial F/\partial y-d/$dx$(\partial F/\partial y^{\prime })=0.\ast \ast$

Two first integrals

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

Standard functionals

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 $\int [(y^{\prime }{)}^2-y$|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 $\int [(y^{\prime }{)}^2+(z^{\prime }{)}^2+y^{\prime }{z}^{\prime }]:2y^{\prime \prime }+z^{\prime \prime }=0$ and 2z″ + y″ = 0 ⇒ y″ = z″ = 0.
• **Isoperimetric constraint$\ast \ast \int G=c$: extremise $F+\lambda G$; the multiplier is fixed by the constraint.
• Higher derivatives: $\partial F/\partial y-d/$dx$(\partial F/\partial y^{\prime })+d^2/$dx${}^2(\partial F/\partial y^{\prime \prime })=0$.

Weak vs strong extrema

A weak minimum compares y′ too $(C^1-$neighbourhood); a strong minimum only compares $y(C^0-$neighbourhood$).\int (1-(y^{\prime }{)}^2)y^2$ 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.