= Solution
The positive quadrant is invariant. The non-axis <nullclines> are $u+v=1$ and $u=1/\beta$. Below $u+v=1$, $u$ increases; above it, $u$ decreases. To the left of $u=1/\beta$, $v$ increases; to the right, it decreases.
For $\beta>1$, the <stable manifold> of the interior saddle separates trajectories tending to $(1,0)$ from those for which $u\to0$ and $v\to\infty$. Exactly on that manifold, the trajectory tends to the saddle. For the illustrated $r=1,\beta=2$, the separating manifold is exactly $v=u$, since the vector field preserves that diagonal.
For $\beta<1$, every strictly positive initial state ultimately has $u\to0$ and $v\to\infty$. Indeed $\dot u\leq ru(1-u)$ implies $\limsup u\leq1$, so eventually $1-\beta u$ is bounded below by a positive constant. Then $v$ grows exponentially, forcing $u$ down. At $\beta=1$, a positive $v$ eventually drives $u$ below one; thereafter $v$ increases. A finite positive limit for $v$ would force $u\to1$, inconsistent with $\dot u\to-rv<0$, so the same exclusion outcome holds.
There are no interior <periodic orbits>: the <Bendixson-Dulac criterion> with multiplier $1/(uv)$ gives
$$
\frac\partial{\partial u}\frac{r(1-u-v)}v+\frac\partial{\partial v}\frac{1-\beta u}u=-\frac rv<0.
$$
The <phase portraits> in the preceding figure show the threshold geometry and the absence of a bounded attracting coexistence state. Initial states on the axes retain the corresponding one-species dynamics.
Back to article page
