I'm having trouble proving the following:
For $A\in\mathbb{R}^{3x3}$, $B\in\mathbb{R}^{3x1}$, $U\subset\mathbb{R}$, consider the restricted control problem:
$\dot x=Ax+Bu$, $x(0)=x_{0}$, $u(t)\in U$
Let $A=\begin{bmatrix} 0&0&1\\0&1&0\\0&1&0\\\end{bmatrix}$ and $B=\begin{bmatrix} 0\\1\\0\end{bmatrix}$
Let $U=[-1,1]$.
Prove that the controllable set $C\subset \{x_0:|[x_0]_2|<1\}$.
The fixed time t controllable set is defined as follows: $C(t)= \{x_0\in\mathbb{R}^n \ \exists u\in\mathbb{U} : x(t)=0\}$ and $\mathbb{U}=\{u:[0,\infty)\to U\subset\mathbb{R}^m | u(.) \text{is measureable}\}$ is the set of admissible controls.
Then the controllable set is $C=\cup_{t>0}C(t)$.
I have absolutely no idea how to do this. Any help will be very much appreciated.
The general solution is$$x(t) = e^{tA}x_0 + \int_0^te^{(t-s)A}Bu(s)\,ds.$$Since $A^2=A^3=\ldots$, one has $$ e^{tA} = \sum_{n=0}^\infty\frac{t^nA^n}{n!} = I_3 + tA + A^2\sum_{n=2}^\infty\frac{t^n}{n!} = I_3 + tA + (e^t - t - 1)A^2. $$ Now, if $x_0\in C$, there exist $t>0$ and $u\in\mathbb U$ such that $x(t) = 0$, i.e., $$ e^{tA}x_0 = -\int_0^te^{(t-s)A}Bu(s)\,ds. $$ Applying $e^{-tA}$ from the left gives \begin{align*} x_0 &= -\int_0^tu(s)e^{-sA}B\,ds = -\int_0^tu(s)(I - sA + (e^{-s}+s-1)A^2)B\,ds\\ &= -\int_0^tu(s)\begin{pmatrix}e^{-s}+s-1\\ e^{-s}\\ e^{-s}-1\end{pmatrix}\,ds. \end{align*} Hence, for the second entry of $x_0$ we have $$ |[x_0]_2| = \left|\int_0^tu(s)e^{-s}\,ds\right|\le\int_0^t|u(s)|e^{-s}\,ds\le\int_0^te^{-s}\,ds = \left[-e^{-s}\right]_0^t = 1-e^{-t}\,<\,1. $$