Consider the system $$\begin{cases} \dot x = x\left(5-\frac{5x}{12}-\frac{y}{1+x}\right)\\ \dot y = y\left(1-\frac{y}{5x}\right) . \end{cases}$$ I want to show that there is a positively invariant set such that the closures of all orbits with positive initial conditions (i.e. $x(0),y(0) > 0$) inside this set are contained within. This is complicated by the fact that the system is not defined on $x = 0$.
Considering the rectangle $[0,12]\times[0,60]$, three of the sides are easy to show:
- if $x > 12$ then $\dot x < 0$,
- if $y > 60$ and $x \le 12$ then $\dot y < 0$; but the latter will hold in finite time by the preceding observation (i.e. we might have to "go around" first),
- if $y = 0$ then $\dot y = 0$ and the trajectories go along the y-axis nullcline but do not cross, so this side poses no problem.
But one cannot directly check the system at $x=0$.
Indeed, starting from the interior of the first quadrant one cannot reach the axes in finite time. But this does not immediately preclude that some solution could tend to zero in the limit and thus the closure of the orbit would not be defined in the positively invariant set. How does one show that the closure of every orbit is contained in at least $(x,y) \in (0,12]\times[0,60]$?