Time-optimal control to the origin for two first order ODES - Trying to take control as we speak!

I want to find the time optimal control to the origin of the system:

$$\dot{x}_1 = 3x_1+ x_2$$ $$\dot{x}_2 = 4x_1 + 3x_2 + u$$ where $|u|\leq 1$

I ran straight into the problem full strength, hit it with all I have got:

$\begin{pmatrix} \dot{x}_1 \\ \dot{x}_2\end{pmatrix} = \begin{pmatrix}3 & 1 \\ 4 & 3\end{pmatrix}\begin{pmatrix}x_1 \\ x_2\end{pmatrix}+ \begin{pmatrix}0\\u \end{pmatrix}$

So we get eigenvalues from $(3-\lambda)^2-4=\lambda^2-6\lambda +5=(\lambda-1)(\lambda-5),\lambda=1,5$

And bam, I got hit by that... Hit straight headfirst with two eigenvalues - same sign - no complex component. But oh no... That means... No it couldn't be?? It is repulsive, the control is going to be hard.

I am running out of time my friends, the oxygen is low and I need to land at base to refill it, but the planets is pulsating with a magenetic field of opposite polarity to the ship at the time being(positive eigenvalues)! How do I get back in the optimal time using my rockets?

Question: How do I setup the optimal control so I can land before my oxygen depletes? I don't know where to go from here. Is the information decoded from the transmission entitled 'Committing to a name' correct?

Editing in my current knowledge as an answer as I try it again.

I can have a shot, I think I've finished the problem sufficiently.

So, you should be able to get to the following;

$$H = -1 + \psi_1(3x_1 + x_2) + \psi_2(4x_1 + 3x_2) + u \psi_2$$

Now, as a direct consequence of PMP, we can basically set $S = \psi_2$, so with that, we get $u^* = sgn(S) = sgn(\psi_2) = \pm 1$. Now, what this basically means is that $u^*$ will switch depending on whether $\psi_2$ is positive or negative. We do eventually work it out, but we need to make sure that $\psi_2$ has, at most, one zero.

Now, again, you should be able to work out the appropriate eigenvalues and eigenvectors, which should come out to be: $$\lambda_1 = 5, v_1 = \left( \begin{array}{c} 1\\ 2\\ \end{array} \right)$$

$$\lambda_2 = 1, v_2 = \left( \begin{array}{c} 1\\ -2\\ \end{array} \right)$$

Then, this gives you an expression for $x$:

$$x(t) = \alpha_1 \left( \begin{array}{c} 1\\ 2\\ \end{array} \right)e^{5t} + \alpha_2 \left( \begin{array}{c} 1\\ 2\\ \end{array} \right)e^{t} $$

Using a similar process, you should be able to work out the appropriate eigenvalues and vectors for the costate equations (the eigenvalues are just the negatives of $1$ and $5$, the vectors differ a little bit, but not too much).

You should get something along the lines of;

$$\psi(t) = \beta_1 \left( \begin{array}{c} 2\\ 1\\ \end{array} \right)e^{-5t} + \beta_2 \left( \begin{array}{c} -2\\ 1\\ \end{array} \right) e^{-t}$$

Now, this means that $\psi_2 = \beta_1 e^{-5t} + \beta_2 e^{-t}$, which clearly can have at most one zero.

Now, you can sub $u^* = \pm 1$ into your initial state equations, to work out your $P$ and $Q$ start points respectively. If you take $u^* = 1$, you should get the $C^+$ paths; $$x(t) = \alpha_1 \left( \begin{array}{c} 1\\ 2\\ \end{array} \right)e^{5t} + \alpha_2 \left( \begin{array}{c} 1\\ 2\\ \end{array} \right)e^{t} +\left( \begin{array}{c} \frac{1}{5}\\ \frac{-3}{5}\\ \end{array} \right) $$

And with $u^* = -1$, the following $C^-$ paths;

$$x(t) = \alpha_1 \left( \begin{array}{c} 1\\ 2\\ \end{array} \right)e^{5t} + \alpha_2 \left( \begin{array}{c} 1\\ 2\\ \end{array} \right)e^{t} +\left( \begin{array}{c} \frac{-1}{5}\\ \frac{3}{5}\\ \end{array} \right) $$

From there, it should be pretty easy to draw out your solution. If that gives you some trouble, I'll be in the 2:00pm tutorial tomorrow, so you can have a look at my diagram if you'd like. I'm the kid with the stupid long hair and blue headphones!!