For what objective function is PID control the optimal control? I am wondering if it's possible to demonstrate that PID control is optimal control for some set of problems. For what objective function is PID control the optimal control?
The reason I ask is that I'm interested in the question of what modification of PID control is appropriate for problems in which some data are missing. From a decision-theoretic perspective, the right way to go about that is to integrate the objective function over the missing variables, and optimize the result. So in order to get started, it seems an objective function is needed, hence my question.
 A: A PD/PID controller can be reformulated as a state-feedback controller.
Consider the following differential equation (spring mass damper system with force per mass as input).
$$\ddot{x}+a\dot{x}+bx = u$$
This can be formulated in state-space as
$$\dfrac{d}{dt}\begin{bmatrix}x_1\\x_2 \end{bmatrix}=\begin{bmatrix}0 & 1\\-b& -a \end{bmatrix}\begin{bmatrix}x_1\\x_2 \end{bmatrix}+\begin{bmatrix}0\\1 \end{bmatrix}u$$
Now, use the full-state feedback $u=-[k_\text{P},k_\text{D}][x_1,x_2]^T=-k_\text{P}x_1-k_\text{D}x_2=-k_\text{P}x-k_\text{D}\dot{x}$ as a control input to drive the system to the origin. You will see that this is actually a PD-controller.
You could use a quadratic cost with an infinite horizon as the cost function. We know that for a linear-time invariant system with quadratic cost the full-state feedback controller is an optimal solution.
A similar procedure is possible with a PID controller. 
$$\ddot{x}+a\dot{x}+bx = -k_\text{P}x-k_\text{D}\dot{x}-k_\text{I}\int_{t=t_0}^{t}x(\tau)~d\tau$$
Differentiate 
$$\dddot{x}+a\ddot{x}+b\dot{x}=-k_\text{P}\dot{x}-k_\text{D}\ddot{x}-k_\text{I}x$$
The new input is $$\tilde{u}=-k_\text{P}\dot{x}-k_\text{D}\ddot{x}-k_\text{I}x=-[k_\text{I},k_\text{P},k_\text{D}][x_1,x_2,x_3]^T$$
for the system
$$\dfrac{d}{dt}\begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix}=\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\0 & -b & -a \end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix}+\begin{bmatrix}0\\0\\1 \end{bmatrix}\tilde{u}.$$
