So I'm studying a control subject and the teacher asked a question about optimization that I got really interested. We studied that for some performance index
$$J=\int_{0}^{\infty}[\vec{x}^TQ\vec{x}+\vec{u}^TR\vec{u}]dt$$
We can give some weight for the $\vec{x}$ and $\vec{u}$ term giving different values to $Q$ and $R$. Taking that $R=I$, we can minimize the position value by using $$Q=\begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix} $$ Which gives $J=\int_{0}^{\infty}[x^2+u^2]dt$, or the velocity by $$Q=\begin{pmatrix} 0 & 0 \\ 0 & 1 \\ \end{pmatrix} $$
Which gives $J=\int_{0}^{\infty}[\dot{x}^2+u^2]dt$. Until now it's fine, but we were asked what value of $Q$ we would need if we wanted to minimize the acceleration $\ddot{x}$ so we get $J=\int_{0}^{\infty}[\ddot{x}^2+u^2]dt$.
My guess was analyzing it for a mass, spring and dumper system such as: $$\dot{\begin{pmatrix} x \\ \dot{x} \\ \end{pmatrix}}= \begin{pmatrix} 0 & 1\\ \omega_n^2 & 2\zeta\omega_n \\ \end{pmatrix}\begin{pmatrix} x \\ \dot{x} \\ \end{pmatrix}+\begin{pmatrix} 0 \\ \frac{1}{m} \\ \end{pmatrix}u $$
So, knowing that $\ddot{x}=\omega_n^2x+2\zeta\omega_n\dot{x}+u/m$, then $$\ddot{x}^2=\omega_n^4x^2+4\zeta^2\omega_n^2\dot{x}^2+u^2/m^2+4\omega_n^3\zeta\dot{x}x+4\omega_n\zeta\dot{x}u/m+2\omega_n^2ux/m$$
But in this expression there are crossed terms with $u,x$ and $u,\dot{x}$, so I thought that the only way would be to add another term to the performance index such as $$J=\int_{0}^{\infty}[\vec{x}^TQ\vec{x}+\vec{u}^TR\vec{u}+\vec{x}^TT\vec{u}]dt$$
or
$$J=\int_{0}^{\infty}[\vec{x}^TQ\vec{x}+\vec{u}^TR\vec{u}+\vec{u}^TT\vec{x}]dt$$
Would this be correct? or is done by another way? (I'm a bit new to this)
