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Math Help - Predator-Prey Model

  1. #1
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    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!
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  2. #2
    Behold, the power of SARDINES!
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    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.
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