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In a normal minimization convention for optimal control, the running cost $L$ and dynamics $\dot{x}=f$ define $H=L+p\cdot f$. The multiplier $p$ is a costate; this Hamiltonian is a device for variational necessary conditions rather than necessarily a physical energy.

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.