= Solution
The prey have per-capita growth $\mu_1$, loss through encounters with predators at rate $\alpha_1v$, and density-dependent competition $\delta u$. Predators suffer per-capita mortality $\mu_2$, gain from consuming prey at rate $\alpha_2u$, and undergo constant harvesting or removal $\epsilon$. Constant harvesting makes the equations biologically valid only while the predator population remains nonnegative.
At an equilibrium with positive populations,
$$
u=\frac{\mu_1-\alpha_1v}{\delta},\qquad v(\alpha_2u-\mu_2)=\epsilon.
$$
Writing $A=\alpha_2\mu_1-\delta\mu_2>0$ gives
$$
\alpha_1\alpha_2v^2-Av+\delta\epsilon=0,\qquad \boxed{v_\pm=\frac{A\pm\sqrt{A^2-4\alpha_1\alpha_2\delta\epsilon}}{2\alpha_1\alpha_2}.}
$$
For $0<\epsilon<A^2/(4\alpha_1\alpha_2\delta)$, both roots are positive and give $u>\mu_2/\alpha_2>0$ via the second equilibrium equation. Thus there are two positive equilibria.
Using the equilibrium identities in the Jacobian gives
$$
J=\begin{pmatrix}-\delta u&-\alpha_1u\\ \alpha_2v&\epsilon/v\end{pmatrix},\qquad \boxed{\operatorname{tr}J=-\delta u+\frac{\epsilon}{v},\quad \det J=u\left(\alpha_1\alpha_2v-\frac{\delta\epsilon}{v}\right).}
$$
The smaller root satisfies $v_-v_+=\delta\epsilon/(\alpha_1\alpha_2)$ and $v_-<v_+$, so $v_-^2<\delta\epsilon/(\alpha_1\alpha_2)$. Hence its <determinant> is negative: \b[it is always a saddle and unstable]. For small $\delta,\epsilon$, this is the equilibrium with $u\sim\mu_1/\delta$ and $v\sim\delta\epsilon/(\alpha_2\mu_1)$, as required. The <determinant> argument is stronger than merely taking the small-parameter limit.
Back to article page
