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A control input taking only extreme admissible values except possibly on a switching or singular set. If the Hamiltonian of an optimal-control problem is affine in a scalar control $u\in [0,1]$, a nonzero coefficient selects $u=0$ or $u=1$; a zero coefficient does not determine the control.

An adjoint variable enforcing the state dynamics in optimal control. In the normal minimization convention it satisfies $\dot{p}=-H_x$ for the Hamiltonian of an optimal-control problem, with terminal conditions determined by the endpoint constraints.

For a smooth normal optimal control problem with freely variable terminal time, fixed terminal state independent of that time and terminal cost $\Phi$, the optimized Hamiltonian of an optimal-control problem at completion satisfies $H(T)+{\Phi }_T=0$. It follows by varying the terminal time in the augmented objective; restrictions on terminal time or a moving endpoint introduce additional terms or inequalities.

For fixed terminal time and fixed endpoint, use the maximizing form of the Pontryagin maximum principle. For dynamics $\dot{q}=f(q,u,t)$ and running cost $L$, there are multipliers $(p,p_0)$, not all zero, with $p_0\ge 0$, and Hamiltonian of an optimal-control problem $\mathcal{H}=p\cdot f-p_{0}L$. Along an optimum, $\dot{q}={\mathcal{H}}_p$, $\dot{p}=-{\mathcal{H}}_q$, and the chosen control maximizes $\mathcal{H}$ over admissible controls almost everywhere. Fixed terminal state imposes no prescribed terminal value of $p$; fixed time imposes no free-time condition $\mathcal{H}(T)=0$. The normal case scales $p_0=1$; abnormal extremals have $p_0=0$.

For this problem the normal Hamiltonian of an optimal-control problem isMaximization gives $u=p_x+ky$, $v=p_y-kx$. The costate equations give ${\dot{p}}_x=kv$, ${\dot{p}}_y=-ku$, hence $\dot{u}=2kv$, $\dot{v}=-2ku$. Nonzero velocities rotate at constant angular rate, and rotational symmetry permits $u(0)=A$, $v(0)=0$. For a closed path, $2k=2\pi n$, with integer $n\ne 0$. Then $z(1)=A^2/(2\pi n)$, so the required positive endpoint has $n>0$, $A^2=2\pi n$ and cost $\pi n$.

To prove a global minimum rather than only classify stationary extremals, write $r=(x,y)$ and subtract its time average; this changes neither its derivative nor $\int (y\dot{x}-x\dot{y})dt$. The resulting closed mean-zero path satisfies the periodic Wirtinger inequality $\int |r{|}^2\le (4{\pi }^2{)}^{-1}\int |\dot{r}{|}^2$. This follows immediately by its Fourier expansion: each nonzero mode of the derivative has squared multiplier at least $4{\pi }^2$. The Cauchy-Schwarz inequality now givesThe argument extends to square-integrable controls by the periodic Sobolev/Fourier interpretation. Thus the cost is at least $\pi$ for every feasible control. Equality is attained byThese satisfy every endpoint and have cost $\pi$. Therefore the minimum is $\pi$, an area-constrained quadratic control cost. The global inequality also prevents an overlooked abnormal candidate from lowering it; in fact an abnormal finite maximizing Hamiltonian of an optimal-control problem would force both linear control coefficients to vanish and then force zero velocity, incompatible with $z(1)=1$.

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 requested properties are not properties of the optimum under the printed assumption $\alpha +\beta >1$. We first derive their origin, and then solve the actual constrained problem.

Put $s=1-\ell$, so $0\le s\le 1$, and write the Hamiltonian of an optimal-control problemFor an interior stationary control, $H_c=H_s=0$ givesThis proves the printed identity only as a necessary stationarity condition for an interior candidate.

It fails the required maximization condition. For fixed positive $p$ and fixed $s$, maximization over $c$ givesSince $\beta /(1-\alpha )>1$, this is strictly convex in $s$. Its maximum on $[0,1]$ is at an endpoint, not at its interior stationary point. Thus the actual optimum uses $\ell =1,c=0$, or $\ell =0,c>0$. On any consumption interval the printed relation would require $\beta c=\alpha$, although the actual optimal consumption varies with time.

This proves a genuine failure of the printed premise, not just failure to check a boundary condition. If the parameter inequality were reversed and an interior solution stayed within the control bounds, these first-order equations would instead arise from a concave objective.

Assume a finite optimum with $X>0$ and continuous price near its completion time. With costate $p^{\ast }$ for the remaining-file state, the minimization Hamiltonian of an optimal-control problem is $H=u(p(t)-p^{\ast })$. It is independent of $x$, so $p^{\ast }$ is constant. By the Pontryagin maximum principle, pointwise minimization over $[0,1]$ givesThis is a bang-bang control away from equality. At equality partial transmission or either endpoint rate is allowed. This also follows by exchanging a small amount of transmission from a more expensive time to a cheaper unused time. The free-terminal-time transversality condition is ${\min }_{u}H(T,u)+2\gamma T=0$. Since $T>0$, it forces $p^{\ast }>p(T)$ and terminal transmission at rate one, yieldingWithout regularity/existence hypotheses an arbitrary known price need not have a finite differentiable optimum; the equality is the interior terminal-time condition used here.

For $p(t)=t+1/t$, $X=1$, feasible completion requires $T>1$ for finite cost. The price decreases to its minimum at one and then increases. Its cheapest unrestricted interval of length one has endpoints $a,a+1$ of equal price, giving $a(a+1)=1$ and $a+1=\phi =(1+\sqrt{5})/2$. For $1<T\le \phi$, the cheapest length-one subset of $[0,T]$ is the terminal interval $[T-1,T]$: its left endpoint has at least the price of its right endpoint, and the earlier times are still more expensive. For $T\ge \phi$ the transmission minimum is unchanged from that unrestricted interval, while the delay cost increases, so no optimum lies beyond $\phi$.

For the terminal-interval policy the total cost isIts derivative is strictly increasing on $T>1$, since $J^{\prime \prime }(T)=(2T-1)/[T^2(T-1{)}^2]+2\gamma >0$. It is negative near one and positive at $\phi$, so the unique global optimum isThere is exactly one positive root: none lies in $(0,1)$ because the left side is negative, and the strictly monotone comparison gives exactly one in $(1,\phi )$. Its threshold is $p(T-1)=p(T)+2\gamma T$.

For a regular finite-horizon optimal control problem, an optimal trajectory admits nontrivial multipliers and a costate obeying adjoint equations, together with pointwise optimization of the Hamiltonian of an optimal-control problem. In the normal minimization convention $H=L+p\cdot f$, the conditions are $\dot{x}=H_p$, $\dot{p}=-H_x$ and minimization of $H$ over admissible controls; endpoint and state constraints change the terminal conditions. Abnormal extremals use a zero cost multiplier and must not be silently excluded.