State Pontryagin's maximum principle for the controllable dynamical system with state-space $\mathbb{R}^{+}$, given by

$$\dot{x}_{t}=b\left(t, x_{t}, u_{t}\right), \quad t \geqslant 0,$$

where the running costs are given by $c\left(t, x_{t}, u_{t}\right)$, up to an unconstrained terminal time $\tau$ when the state first reaches 0 , and there is a terminal cost $C(\tau)$.

A company pays a variable price $p(t)$ per unit time for electrical power, agreed in advance, which depends on the time of day. The company takes on a job at time $t=0$, which requires a total amount $E$ of electrical energy, but can be processed at a variable level of power consumption $u(t) \in[0,1]$. If the job is completed by time $\tau$, then the company will receive a reward $R(\tau)$. Thus, it is desired to minimize

$$\int_{0}^{\tau} u(t) p(t) d t-R(\tau)$$

subject to

$$\int_{0}^{\tau} u(t) d t=E, \quad u(t) \in[0,1],$$

with $\tau>0$ unconstrained. Take as state variable the energy $x_{t}$ still needed at time $t$ to complete the job. Use Pontryagin's maximum principle to show that the optimal control is to process the job on full power or not at all, according as the price $p(t)$ lies below or above a certain threshold value $p^{*}$.

Show further that, if $\tau^{*}$ is the completion time for the optimal control, then

$$p^{*}+\dot{R}\left(\tau^{*}\right)=p\left(\tau^{*}\right)$$

Consider a case in which $p$ is periodic, with period one day, where day 1 corresponds to the time interval $[0,2]$, and $p(t)=(t-1)^{2}$ during day 1 . Suppose also that $R(\tau)=1 /(1+\tau)$ and $E=1 / 2$. Determine the total energy cost and the reward associated with the threshold $p^{*}=1 / 4$.

Hence, show that any threshold low enough to carry processing over into day 2 is suboptimal.

Show carefully that the optimal price threshold is given by $p^{*}=1 / 4$.