48
3. Predator-Prey Dynamics
FIGURE 3.4
(7)
and
stock and then run the model. You will see the numeric display showing
the values of each stock at each point in time and displaying the final value
when the model stops . That value should be very close to 200 for Wand
very close to 150 for 5.
You can analytically solve the model for those steady-state conditions. In
steady-state, each of the two populations must have reached a constant
level and not change anymore. Consequently,
dW =w=P * A * W * 5 * (K - W) - Q* W =0
(6)
dt
K
d5
-
.
=5 =R * 5 - A * W * 5 =O.
dt
Let us solve equations (6) and (7) for steady-state values of W>O and
5>0. Divide equation (7) by 5 and rearrange the terms to get
W=R .
A
(8)
For R = .8 and A = .004, the steady-state humpback whale population size
will be W = 200.
Next, solve equation (6) for S. After dividing both sides by Wand rearranging terms, you should get
5 - -.lL * (---.!S.-)
-P*A
K-W '
(9)
Insert into equation (7) Q = .03, P= .1, A = .004 and K= 400, as well as
the steady-state value for lv, and you will get 5 = 150 as the steady-state
sand lance population size.
The differences of the modeled steady-state values and those derived analytically are due to errors associated with the numeric solution of our
model. Change for subsequent runs the DT and integration method, and