stability of dynamic system, find control Our system is $\dot{w} = Aw+Bv$ where $A = \begin{pmatrix}1 & 0 & 2 \\ 1 & 7 & 3\\1 & 2 & 0\end{pmatrix}, B = \begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix}$
and our goal is to asymptotically stabilize this system. $v$ however is not our controller.
$v$ is $x_1$ where $x_1$ is part of the solution to another system
$\begin{cases}\dot{x_1}=\cos(t)+1-x_2\\\dot{x_2}=\cos(3+x_3)+(2+\cos(x_3))u\\\dot{x_3}=\sin(2x_1+x_3+4t)-x_3+u\end{cases}$
Where $u$ is our controller. So we impact $v$ by changing $u$, which we can change freely.
Choose $u$ that stabilizes this system
What I tried:
Let $c = \begin{pmatrix}7.58686705 & -68.15812197 & -24.11610062\end{pmatrix}$
Using a computer, it is possible to check that the matrix $A+Bc$ has eigenvalues with strictly negative real part.
Hence if we choose $v = cw$, then our original system is $\dot{w} = (A+Bc)w$ which is asymptotically stable, because eigenvalues have negative real part.
Thus, it would be great if we could choose $u$ such that the solution $x_1$ to the other system is $cw$.
What is the correct $u$?
 A: You probably do not only want to find a $u$ such that $v$ goes to a desired value, you probably also want that $u$ ($x_1$, $x_2$ and $x_3$ as well) would remain bounded. For this you can try to find a $u$ such that $x_1=v$ goes to a desired value. However, instead of trying to dictate $v$ I would propose input-output linearization and use state feedback for a system with an extended state dimension. For this you have to differentiate $v$ until it is a function of $u$, so
$$
\frac{d^n}{dt^n}v = f(t,x,u).
$$
In order to do input-output linearization you want to be able to set $f(t,x,u)$ equal to a new virtual input $\nu$, such that $u$ can be found by solving $f(t,x,u) = \nu$. A stabilizing control law for $\nu$ can now for example be chosen using linear control method like pole placement or LQR, since the resulting input-output linearized system is linear time invariant with respect to $\nu$. For example when $n=3$ the extended system with input-output linearization becomes
$$
\dot{z} = 
\begin{bmatrix}
A & B & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix} z +
\begin{bmatrix}
0 \\
0 \\
0 \\
1
\end{bmatrix} \nu,
$$
with $z = \begin{bmatrix}w^\top & v & \dot{v} & \ddot{v}\end{bmatrix}^\top$, which can be shown to be controllable so there exists a $K$ such that $\nu = K\,z$ would drive $z$ exponentially to zero.
In order to guarantee that $u$ remains bounded requires you to show that solving $f(t,x,u) = \nu$ for $u$ doesn't have any singularities. Bounds on $x_1$, $x_2$ and $x_3$ might also be required in order for $u$ to be bounded, for example if you would have $u=\nu+x_3$. In order to show that $x_1$, $x_2$ and $x_3$ are bounded one might be able to use that $v$ and its derivatives should go to zero due to the control law for $\nu$.
You will learn more by trying to solve this yourself so I will not post an actual solution, but feel free to ask more questions if these hints aren't clear enough.
