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Suppose I have the linear system $\dot{x}=Ax$, with $A=\left[ \begin{array}{cc} -1 & 0 \\ 0 & 2\\ \end{array}\right]$. I know that the phase portrait of the linear system has a saddle in $(0,0)$, so how am I supposed to know where a particle will flow given an initial point in a small neighborhood $N_\epsilon (x_0)$ of radius $\epsilon=0.3$, given $x_0=(-3,0)$?

I know that if $x_0$ has it's second component $x_2>0$, given $x=(x_1,x_2)$, it the particle will flow upwards towards $(0,+\infty)$, and if $x_2<0$ it will go downwards to $(0,-\infty)$, but given the saddle in $(0,0)$, should the particle just stop at $(0,0)$, or bifucate itself to go to $(0,\pm\infty)$? I am quite confused to see what is the mathematical and physical explanation ...

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  • $\begingroup$ "should the particle just stop at (0,0)?" No. Starting from any (x,0) with x nonzero, the particle moves closer and closer to (0,0), without ever reaching it. So there is no "stop" here. $\endgroup$ – Did Mar 17 '14 at 7:45
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Here is phase portrait showing the phase lines for multiple points (blue trajectories), the initial point $(-3,0)$ in green and $N_\epsilon (x_0)$ of radius $\epsilon=0.3$ in red.

enter image description here

Some observations:

  • Notice what happens to initial points inside the circle? If they are slightly above the y-axis, they shoot off to positive infinity, if they are slightly below the y-axis, the go to negative infinity.
  • Notice what happens to the point $(-3,0)$? It stays on the horizontal axis. Why? If we write $\dfrac{y'}{x'} = \dfrac{2y}{x}$, what happens when $y$ is zero, regardless of the value of $x \ne 0$? We have a constant slope of $0$ and the trajectory stays there.
  • From the previous statement, we have $\dfrac{y'}{x'} = \dfrac{2y}{x}$. What happens to the slope when $x = 0$? It is infinite (vertical).
  • If we solve this system, $x' = -x, y' = 2y, x(0) = -3$, we have:

$$x(t) = -3e^{-t}, y(t) = 0$$

What happens if you do a parametric plot of that?

  • I would say it is safe to state that the trajectory just continues toward the origin but never quite reaches it.
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  • $\begingroup$ Ok, I already have the plot, and I knew how it went, but my question is what happens to the point if it starts exactly at $(-3,0)$? I know it will go to $(0,0)$. Does it stop there? $\endgroup$ – Arturo Mar 17 '14 at 0:06
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    $\begingroup$ As soon as $x$ becomes $0$, what happens to the slope? Physically, what does this represent, maybe nothing. It depends on the system I suppose. If this represents number of animals, that is okay (population expires), if this represent a voltage or current, the system could breakdown or explode. It depends on how the system is defined to handle those cases where the model just breaks down. In this case, the slope becomes infinite when $x = 0$. $\endgroup$ – Amzoti Mar 17 '14 at 0:09
  • $\begingroup$ Great, this makes sense, awesome! $\endgroup$ – Arturo Mar 17 '14 at 0:11
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    $\begingroup$ "the trajectory stops at the origin" Actually it does not. Starting from any (x,0) with x nonzero, the particle moves closer and closer to (0,0), without ever reaching it. So there is no "stop" here. $\endgroup$ – Did Mar 17 '14 at 7:45
  • $\begingroup$ Point taken. Thanks. $\endgroup$ – Amzoti Mar 17 '14 at 12:28

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