Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 4 / 28I / b / Solution

The fixed terminal height makes $-y(T)+T=T-1$, so this is a minimum-time problem. An optimal path can be taken entirely in $y\ge 1/2$. To justify the positive-$y$ calculation despite the absolute value in the dynamics, the time of a spatial path is its length in the metric $ds/\sqrt{|y|}$. Reflection to $|y|$ preserves this time; clipping the reflected height below $1/2$ to $1/2$ leaves the endpoints unchanged and cannot increase time, since it decreases vertical length and increases the speed available. Any excursion below $1/2$ strictly wastes time. Thus an optimum does not meet $y=0$.

Use $H=\sqrt{y}(pu+qv)-1$. Its maximum is $\sqrt{y}\sqrt{p^2+q^2}-1$, attained atThe costate equations and free-terminal-time transversality condition giveThe last identity holds throughout because the maximized autonomous Hamiltonian is constant. An abnormal extremal would instead have maximized Hamiltonian $\sqrt{y}\sqrt{p^2+q^2}=0$, forcing both costates to vanish and contradicting their nontriviality. Thus the normal choice is valid. Also $p\ne 0$, since otherwise $x$ cannot reach one; in fact $p>0$ because $\dot{x}$ has its constant sign.

Now express both requested derivatives using the costates:Consequently the optimal trajectory satisfiesThe maximizing controls above and the fixed endpoints determine the relevant extremal; the displayed equation is the requested necessary trajectory equation.