I have the following dynamical system

$${dp \over dt} = p(1-p-q)$$ $${dq \over dt} = q(p-{1 \over 2}-q)$$

and I have to show that the first quadrant ( $p, q \ge 0$ ) is an invariant set. I know what this means but I'm having trouble coming up with a strategy to show it.

Would it be sufficient to show that each axis is an invariant set and by virtue of a trajectory not being able to cross an invariant set then each quadrant must also be invariant? Or is there a better way?

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  • $\begingroup$ That's a good idea, yes. $\endgroup$ – Hans Lundmark Jun 7 '15 at 9:31
  • $\begingroup$ Actually the two axes (set of points $(p,q)$ such that $p=0$ or set of points $(p,q)$ such that $q=0$) are invariant sets. If you get there you stay there. Thus, the trajectories starting from $p,q\geq 0$ cannot cross the axes in order to change quadrant. $\endgroup$ – RTJ Jun 7 '15 at 19:45
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    $\begingroup$ And one cannot get to one of the axes p=0, q>0, and p>0, q=0, starting from p>0, q>0, except possibly in infinite time. $\endgroup$ – Did Jun 12 '15 at 19:02

A diagram to show that the quadrant $(p>0,q>0)$ is invariant:

enter image description here

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