Editing Optimal control (section)

===Finite time===
{{confusing|section|reason=the law of evolution mentioned in the example is not mentioned in the article and is probably not the same as [[evolution]]|date=October 2018}}
Consider the problem of a mine owner who must decide at what rate to extract ore from their mine. They own rights to the ore from date <math>0</math> to date <math>T</math>. At date <math>0</math> there is <math>x_0</math> ore in the ground, and the time-dependent amount of ore <math>x(t)</math> left in the ground declines at the rate of <math>u(t)</math> that the mine owner extracts it. The mine owner extracts ore at cost <math>u(t)^2/x(t)</math> (the cost of extraction increasing with the square of the extraction speed and the inverse of the amount of ore left) and sells ore at a constant price <math>p</math>. Any ore left in the ground at time <math>T</math> cannot be sold and has no value (there is no "scrap value"). The owner chooses the rate of extraction varying with time <math>u(t)</math> to maximize profits over the period of ownership with no time discounting.

{{ordered list
| 1 = Discrete-time version
{{pb}}
The manager maximizes profit <math>\Pi</math>:
<math display="block">\Pi = \sum_{t=0}^{T-1} \left[ pu_t - \frac{u_t^2}{x_t} \right] </math>
subject to the law of motion for the state variable <math>x_t</math>
<math display="block">x_{t+1} - x_t = - u_t</math>

Form the Hamiltonian and differentiate:
<math display="block">\begin{align}
H &= pu_t - \frac{u_t^2}{x_t} - \lambda_{t+1} u_t \\
\frac{\partial H}{\partial u_t} &= p - \lambda_{t+1} - 2\frac{u_t}{x_t} = 0 \\
\lambda_{t+1} - \lambda_t &= -\frac{\partial H}{\partial x_t} = -\left( \frac{u_t}{x_t} \right)^2
\end{align}</math>

As the mine owner does not value the ore remaining at time <math>T</math>,
<math display="block">\lambda_T = 0</math>

Using the above equations, it is easy to solve for the <math>x_t</math> and <math>\lambda_t</math> series
<math display="block">\begin{align}
\lambda_t &= \lambda_{t+1} + \frac{\left(p-\lambda_{t+1}\right)^2}{4} \\
x_{t+1} &= x_t \frac{2 - p + \lambda_{t+1}}{2}
\end{align}</math>
and using the initial and turn-T conditions, the <math>x_t</math> series can be solved explicitly, giving <math>u_t</math>.

| 2 = Continuous-time version
{{pb}}
The manager maximizes profit <math>\Pi</math>:
<math display="block">\Pi = \int_0^T \left[ pu(t) - \frac{u(t)^2}{x(t)} \right] dt </math>
where the state variable <math>x(t)</math> evolves as follows:
<math display="block"> \dot x(t) = - u(t) </math>

Form the Hamiltonian and differentiate:
<math display="block">\begin{align}
H &= pu(t) - \frac{u(t)^2}{x(t)} - \lambda(t) u(t) \\
\frac{\partial H}{\partial u} &= p - \lambda(t) - 2\frac{u(t)}{x(t)} = 0 \\
\dot\lambda(t) &= -\frac{\partial H}{\partial x} = -\left( \frac{u(t)}{x(t)} \right)^2
\end{align}</math>

As the mine owner does not value the ore remaining at time <math>T</math>,
<math display="block">\lambda(T) = 0</math>

Using the above equations, it is easy to solve for the differential equations governing <math>u(t)</math> and <math>\lambda(t)</math>
<math display="block">\begin{align}
\dot\lambda(t) &= -\frac{(p-\lambda(t))^2}{4} \\
u(t) &= x(t) \frac{p- \lambda(t)}{2}
\end{align}</math>
and using the initial and turn-T conditions, the functions can be solved to yield
<math display="block">x(t) = \frac{\left(4-pt+pT\right)^2}{\left(4+pT\right)^2} x_0  </math>
}}