# Control Theorie: Drawing Transition Matrix Trajectories

Given a transition matrix like e.g.

$$\phi(t)=\begin{bmatrix} e^{-3t} & 0 \\ 0 & e^{-3t} \end{bmatrix}$$

What is the analytical way of drawing x(t) for two given starting points? I only received this solution, which I can somehow understand in this case:

Since one x2 is at zero for the right point and one therm is decaying, it hast to go to zero. The left point somehow goes on a line to zero because both x1(t) and x2(t) got the same factor. The hint which was given confirms this assumption: $$x_1(t) = e^{-3t} \cdot x_1(0)$$ $$x_2(t) = e^{-3t} \cdot x_2(0)$$ $$x_1(t) = C \cdot x_2(t)$$

The reason I'm questioning my assumption is a second task which was given: $$\phi(t)=\begin{bmatrix} e^{2t} & 0 \\ 0 & e^{-2t} \end{bmatrix}$$ With those starting points (and solution in blue) To solve this exercise I tried to find a connection (like the C before). The equations for $x_1$ and $x_2$ both have difference coefficients. Besides the starting point they have a factor of $e^{4}$ in difference.

The hint in the solution which was given confused me more. As before, it can be written: $$x_1(t) = e^{2t} \cdot x_1(0)$$ $$x_2(t) = e^{-2t} \cdot x_2(0)$$ but then $$x_1(t) \cdot x_2(t) = 1 \cdot x_1(0)\cdot x_2(0) \rightarrow hyperbole$$

Of course the math seems legit, but the same solution could have applied in the exercise before, which would have led to something different. Is there a mistake, or why is this the proper way?

• How could the same solution obtained in the first case? $x_1(t) \cdot x_2(t) = e^{-6t} \cdot x_1(0)\cdot x_2(0)$ – Chu Aug 20 '18 at 8:32
• @Chu first I thought so too. But that is wrong. It's not a general solution to multiply them. It's all about removing the time variable as far as I understood. In the first case this can be achieved by dividing them. – Mr.Sh4nnon Aug 20 '18 at 8:45
• Yes of course, that's correct. – Chu Aug 20 '18 at 8:50
• So I'm not sure, if you were suggesting a solution or asking a question. – Mr.Sh4nnon Aug 20 '18 at 8:57