 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 $u\to 0$ 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 $u\to 0$ and $v\to \infty$. Indeed $\dot{u}\le ru(1-u)$ implies $\mathrm{limsup}u\le 1$, 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 $u\to 1$, 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 $1/(uv)$ givesThe 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.