Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 2 / 13D (Variational Principles)

For fixed endpoint values, the first variation of the functionalvanishes exactly when the Euler-Lagrange equationholds. If $f$ has no explicit $x$ dependence, differentiation along an extremal gives the Beltrami identityFor a further constraint $J[z]=J_0$, introduce a constant Lagrange multiplier $\lambda$ and apply the Euler-Lagrange equation to $I+\lambda J$; $\lambda$ is then chosen so that the constraint holds.

Enforce the fixed arc length with a multiplier. Apart from an irrelevant common factor, the augmented integrand isThe Beltrami identity givesAfter absorbing constants into $z_0$ and a positive scale $B$, integration yields the catenaryThe endpoint conditions and prescribed length giveThe second equation determines $B>0$ implicitly because $L>2a$, and the first then determines $z_0$.

For the lower semicircleFor a general potential, the Beltrami first integral isSubstitution of the semicircle therefore requiresThus, up to an arbitrary additive constant and a multiplicative strength,The additive constant is absorbed by the length multiplier. The semicircle has length $\pi a$, as required.