# Predator-Prey Model

• Feb 13th 2011, 07:09 PM
Truthbetold
Predator-Prey Model
From the following equations, set $H1$ at a fixed positive value. Show that as $H2 --> c^{-}$, the equilibrium point inside the population quadrant approaches the point $(0, \frac{a+H1}{b})$ on the y-axis and that if
H2 = c, all points on the y-axis are equilibrium points of system (9).

$x' = (-a - H1 + by)x$
$y' = (c - H2 - kx)y$

I decided to let H1 = 1.

It would seem that as $H2 --> c^{-}$ , the equation would reduce to $-kxy$.

I don't know how to solve for the equilibrium point, however.
The only mention in my book I find is the following excerpt:

"If the harvest coefficients in the above 2 equations are too large, the internal equilibrium point $\frac{c - H2}{k}, \frac{a+H1}{b}$ crosses the positive y-axis, and one (or both) species becomes extinct, as we see in the next example."

Where should I start in proving the equilibrium point?

Thanks!
• Feb 14th 2011, 02:50 PM
TheEmptySet
The equilibrium points occur when $\displaystyle \frac{dy}{dt}=0, \text{ and } \frac{dx}{dt}=0$

If you set

$\frac{dx}{dt}=(-a-H_1+by)x=0$ and
$\frac{dy}{dt}=-kyx=0$

Since the population of both is positive we reject $(x,y)=(0,0)$

And solve

$-a-H_1+by=0$ and
$-kyx=0$

This gives the desired conclusion.