I have found the phase plane for a "predator-prey model" system of differential equations: $$\dot{x}=x^2-x^3-xy \qquad \dot{y}=yx-ya$$ where $x$ represents the population of prey and $y$ represents the population of predator. I want to interpret this for $a \geq 1$. It looks like this (I have used $a=1.1$):
I have used the Jacobian and nullclines to find equilibrium points at $(0,0)$, $(1,0)$ and $(a,a-a^2)$ and found that when $a\geq 1$,
$(0,0)$ is a saddle (one eigenvalue is 0, the other is negative) - (although pplane didn't seem to recognise this as a stable point, any ideas why?),
$(1,0)$ is a stable sink (both eigenvalues are negative), and
$(a,a-a^2)$ is a saddle (the eigenvalues are $\frac{a-2a^2\pm a \sqrt{4a^2-3}}{2}$ so the $\frac{a-2a^2- a \sqrt{4a^2-3}}{2}$ is negative and $\frac{a-2a^2+ a \sqrt{4a^2-3}}{2}$ is positive at a=1.1)
I want to use this portrait to show that the predators go extinct. How to I interpret this output this way?
The positive quadrant is stable by the dynamics hence the only relevant equilibrium points are $(0,0)$ and $(1,0)$. After that, start from $(x(0),y(0))$ with $x(0)\gt0$ and $y(0)\gt0$ and follow the arrows...
Since $x'\lt0$ when $x\gt1$, $x\leqslant1$ eventually, then $y'\leqslant-(a-1)y$ hence $y\to0$ and, since $(0,0)$ is a saddle point only reachable through the line $x=0$ and $x(t)\gt0$ for every $t$, $(x(t),y(t))\to(1,0)$.