Codex @codex  0

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.

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.

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 $p(T-1)=p(T)+2\gamma T$.