So I'm facing a modified predator-prey model:
$$ \dfrac{dU}{dt} = aU\left(1 - \dfrac{U}{K_U}\right) - cUV,$$
$$ \dfrac{dV}{dt} = ceUV - bV,$$ where $a, b, c, e > 0$.
To determine the stability, I found out the equilibria at $(0, 0)$, $(K_U, 0)$ and $(b/(ce), a/c - (ab)/(ce^2 K_U))$. The 1st, trivial one should be unstable like the original (without the $1 - U/K_U$ factor), so my concerns lie on the other 2.
If I understand correctly, the stability of the 2nd equilibrium should depend on whether $ceK_U$ or $b$ is larger. If the former $\leq$ latter, it is still stable. Did I misunderstand anything?
For the last one, I'm very confused about how to handle the eigenvalue equation $$\lambda(ab + ceK_U \lambda) + (b/e)(ae - (ab)/(cK_U)) = 0.$$ The usual quadratic formula gives another quadratic equation about $ceK_U$ within the square root (let's denote it as $\sigma$). It is still okay when $\sigma$ is real -- when $\sigma \leq a^2 b^2$, the node is stable. Since the square root looks complicated, (a) should I not use quadratic formula to determine the eigenvalues? (b) When $\sigma$ is complex, what happens to the equation system? For example, does the solution attract?
Thanks in advance.
No. When $ceK_U=b$, the equation gives $\dot V=(U-K_U)V$ so for a small time there will be exponential growth in number of predators if $U>K_U$. So the equilibrium is not stable.
Only the real part matters. The imaginary part of eigenvalue measures the rate of trajectories rotating as it spirals in (or out) near the equilibrium so doesn't affect stability. If $ceK_U>b$ this equilibrium is stable since it is either a pair of complex conjugate eigenvalues with negative real part $-ab/(2ceK_U)$, or two negative reals. If $ceK_U<b$ then there is a negative and a positive eigenvalue, so unstable. Equality means you will need to analyse what happens along the degenerate direction.