Solve the problem of optimal control to the origin for the system

\displaystyle \dot{x_1} = - 4x_1 + 2x_2

\displaystyle \dot{x_2} = 2x_1 - 4x_2 + 2u

where \displaystyle |u| \leq 2

I have done a bit, but Im not really sure where to go next....

\displaystyle J[x] = \int_{t_0}^{t_1} 1 dt

Let
\displaystyle f_0 (x_1, x_2, u) = 0

\displaystyle f_1 (x_1, x_2, u) = -4x_1 + 2x_2

\displaystyle f_2 (x_1, x_2, u) = 2x_1 - 4x_2 + 2u where u \leq |2|

Then

\displaystyle \vec{\dot{x}} = Ax + \left( \begin{array}{ccc} 0 \\ 3 \end{array} \right) u where A = \left( \begin{array}{ccc} -4 & 2 \\ 2 & -4 \end{array} \right)

\vec{\dot{x}} = \left( \begin{array}{ccc} \dot{x_1} \\ \dot{x_2} \end{array} \right) , \vec{x} = \left( \begin{array}{ccc} x_1 \\ x_2 \end{array} \right) , |u| \leq 2

Then

H = \psi_0 f_0 + \psi_1 f_1 + \psi_2 f_2 where \psi= -1

= -1 + \psi_1(- 4x_1 + 2x_2) + \psi_2(2x_1 - 4x_2) + 2 \psi_2 u

The costate equations are

\displaystyle \dot{\psi_1} = - \frac{\partial H}{\partial x_1} = 4 \psi_1 - 2 \psi_2

\displaystyle \dot{\psi_2} = - \frac{\partial H}{\partial x_2} = 4 \psi_2 - 2 \psi_1

\displaystyle \vec{\dot{\psi}} = -A^T \vec{\psi}

\displaystyle = \left( \begin{array}{ccc} 4 & -2 \\ -2 & 4 \end{array} \right)\vec{\psi}

so u^* maximizes H as a function of u. As u is linear in H and |u| \leq 2 then H is maximized such that u^* = 2 when 2 \psi_2 > 0 and u^* = -2when 2 \psi_2 < 0

u* = sgn (2 \psi_2)
The eigenvalues are found when det (A - \lambda I) = 0

when solving we have two eigenvalues

\lambda = -4 \pm 2 \sqrt{7}.

I'm not sure if what I have done is correct, and I don't know where to go from here.