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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: enter image description here

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) enter image description here 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?

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    \$\begingroup\$ 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) \$ \$\endgroup\$ – Chu Aug 20 '18 at 8:32
  • \$\begingroup\$ @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. \$\endgroup\$ – Mr.Sh4nnon Aug 20 '18 at 8:45
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    \$\begingroup\$ Yes of course, that's correct. \$\endgroup\$ – Chu Aug 20 '18 at 8:50
  • \$\begingroup\$ So I'm not sure, if you were suggesting a solution or asking a question. \$\endgroup\$ – Mr.Sh4nnon Aug 20 '18 at 8:57
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Correct assumption. For example, see the plot below representing two state space trajectories starting from t = 0 with initial states (-0.5; 0.3) and (0.5; 0.3) and identical eigenvalues you provided:

Simu Dirceu

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  • \$\begingroup\$ okay thank you :) so both are correct. but how comes that in the second example those two get multiplied? i don't see a pattern or generic scheme which i could apply. \$\endgroup\$ – Mr.Sh4nnon Aug 19 '18 at 18:50
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    \$\begingroup\$ Any procedure that leads to elimination of the variable time (remember it is practical to sketch phase portraits for only two variables). The procedure of dividing one equation by other other is quite general. Note that if, in your example, the roots were -1 and -2 (distinct but with same sign), the phase portrait would be composed of parabolas - confirm by yourself. \$\endgroup\$ – Dirceu Rodrigues Jr Aug 19 '18 at 20:22
  • \$\begingroup\$ Ou now I get it! Removing time variable nails it! Thank you. \$\endgroup\$ – Mr.Sh4nnon Aug 19 '18 at 20:37

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