Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 9

For fixed initial and final coordinates, the two Euler-Lagrange equations areIndeed the first variation of the functional isIntegration by parts moves the time derivatives from the variations; their endpoint values are zero. The fundamental lemma of the calculus of variations then gives the two equations above.

The total derivative contributes only the boundary termThis is fixed by the endpoint data, so its first variation vanishes. It cannot change the Euler-Lagrange equations or which path minimizes $F$.

For a direct check, let $h=g_x\dot{x}+g_y\dot{y}$. ThenThe equality uses commuting mixed partial derivatives. Thus $\boxed{dg/dt\text{ contributes zero to both Euler-Lagrange equations}}$, as in total time derivatives preserve fixed-endpoint Euler-Lagrange equations.

This is an isoperimetric constraint, imposed on the whole path rather than separately at each time. Introduce a constant Lagrange multiplier $\lambda$ and replace the variational integrand by $\tilde{f}=f+\lambda u$. The modified Euler-Lagrange equations areSolve these together with the fixed endpoint conditions, treating $\lambda$ as an unknown constant, then choose its value so thatThe multiplier is constant because there is one scalar integral constraint. A condition imposed at every time would instead call for a time-dependent multiplier. As usual, this multiplier procedure assumes the constraint is regular, so its first variation does not vanish identically on all admissible variations.

The needed partial derivatives areSubstitution into the Euler-Lagrange equations givesThe ${\gamma }_1$ terms cancel in the first equation, leavingThese are accompanied by $x(0)=y(0)=0$ and $x(T)=y(T)=S$. The cancellation is consistent with part (b): the term proportional to ${\gamma }_1$ is the total derivative of $\alpha {\gamma }_1(x+\beta x^2/2)$.

Impose $y=x$ before varying. Writing $\delta ={\delta }_1+{\delta }_2$, the restricted integrand isIts middle term is the total derivative of $\alpha ({\gamma }_1+{\gamma }_2)(x+\beta x^2/2)$, so it can be dropped for fixed endpoints by part (b). The remaining Euler-Lagrange equation is $\delta \ddot{x}=\alpha \beta$. Integrating twice and imposing $x(0)=0$, $x(T)=S$ givesThis is the unique global minimizer among paths on the specified line. To prove it, add any admissible variation $\eta$ with $\eta (0)=\eta (T)=0$. The change in fuel, after dropping the fixed boundary term, isThe first integral vanishes by integration by parts and the equation for $x$. Equality forces $\eta =0$. No monotonicity or restriction to the segment between endpoints was imposed in the question; the minimizing coordinate can initially decrease for sufficiently large $T$.