Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 4 / 28I (Optimization and Control)

Pontryagin maximum principle leaves the initial costate unknown. Terminal conditions determine these unknown constants and, when terminal time is free, the time. In a maximization convention with terminal reward $\Phi$ and running cost $L$, use $H=p\cdot f-L$. A free terminal coordinate gives $p_i(T)={\partial }_i\Phi$; a fixed terminal coordinate instead allows a constraint multiplier. More generally $p(T)-\nabla \Phi$ is normal to the permitted terminal manifold. The free-terminal-time transversality condition is $H(T)+{\Phi }_T=0$. These conditions convert the adjoint and state equations into a boundary-value problem. Reversing the costate sign gives the equivalent minimization convention.

Maximize $y(1)$. Along the positive-$y$ trajectory use Hamiltonian of an optimal-control problem $H=\sqrt{y}(pu+qv)$. The terminal costate is $(p(1),q(1))=(0,1)$, andThus $p=0$. Maximizing over $u^2+v^2=1$ gives $u=0,v=1$ as long as $q>0$. The state equation integrates to $\sqrt{y(t)}=1/\sqrt{2}+t/2$, and the adjoint equation gives $q(t)=\sqrt{y(1)/y(t)}>0$, validating the maximization. HenceThis is globally optimal, not just an extremal: whenever $y>0$, $d\sqrt{y}/dt=v/2\le 1/2$. If a competitor reaches $y=0$ and later becomes positive, apply that bound after its last zero; it has even less remaining time to attain a positive terminal height. A negative terminal height cannot improve this positive maximum. Therefore the minimum objective is $\boxed{-y(1)=-(1/\sqrt{2}+1/2{)}^2}$.

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.