The question asks me to solve the positioning problem where:
$$ \dot{x_1} = x_2 $$ $$ \dot{x_2} = u_1 \in U_{bb} $$ $$ x_1(0) = - \text{X} (<0) $$ $$ x_2 (0) = 0 $$ $$ x_1(t_1) = 0$$ $$ x_2(t_1) = 0$$
I'm having trouble making the A matrix that I understand is required for the problem to be solved. i.e. the $$ \dot{x} = Ax + Bu$$ part of the control problem.
I think I may also need a bit of an explanation, too. When the lecturer teaches this I have a little trouble grasping the movements that lead to the answer.
Some help would be great. All the literature I've found on the topic turns out to be a little denser than I can handle.
Edit:
When I try this is how I see the formula:
$$ \begin{pmatrix} 1 \\ 1 \\ \end{pmatrix} \dot{x} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix} x + \begin{pmatrix} 0 \\ 1 \\ \end{pmatrix} u $$
Is this correct? If so, what happens from here?
The equation you've written down for the state space form is almost correct. On the left-hand side, typically the vector $x$ is defined as \begin{equation} x = \left(\begin{array}{c} x_1 \\ x_2 \end{array}\right) \end{equation} so that \begin{equation} \dot{x} = \left(\begin{array}{c} \dot{x}_1 \\ \dot{x}_2 \end{array}\right) \end{equation} which means you don't need the extra $(1\,\,\,\, 1)^T$ vector on the left-hand side (and indeed, virtually always in linear systems the left-hand side only contains $\dot{x}$ -- I can't recall seeing any other form of this equation). Then the corrected form of this equation is \begin{equation} \dot{x} = \left(\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right)x + \left(\begin{array}{c} 0 \\ 1\end{array}\right)u. \end{equation} To see that this makes sense, multiplying $Ax$ through lets us recover the original differential equation that we started with.