MATH 2280-002
Lecture Notes: 01/17/2013
Math 2280 Section 002 [SPRING 2013]
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An Example of A Logistic Model
Last time we talked about how and when logistic models are used. Today we’ll look at a specific
example involving a logistic DE and consider the implications of this model.
Example. Scientists are studying the problem of overpopulation in a colony of lab mice. Initially, there are 500 mice, but since the population exceeds the resources available, the population
decreases. The scientists are using the logistic DE
dP
= (900 − 6P )P
dt
to model the population. Find P (t).
We begin by recognizing that logistic differential equations are separable (and non-linear). This
suggests that we use the separation of variables technique that we reviewed earlier. Begin by
separating P and t.
dP
= dt
(900 − 6P )P
Now integrate both sides.
Z
dP
=
(900 − 6P )P
Z
dt = t + C1
The left-hand side of the equation is a little harder. We need partial fractions in order to evaluate
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the integral. Decompose (900−6P
)P into partial fractions.
1
A
B
=
+
(900 − 6P )P
900 − 6P
P
Since the denominators 900 − 6P and P are degree 1 polynomials, the numerators should be
degree 0 (that is, constant). To solve for A and B, clear denominators by multiplying both sides
by (900 − 6P )P .
1 = AP + B(900 − 6P )
This identity has to hold for any choice of P , so we choose a P that make one of two summands
vanish. If P = 0,
1 = 0 + B(900 − 0),
so B must be
1
900 .
Similarly, if P =
900
6 ,
1=
so A must be
6
900 .
900
A − 0,
6
We found the decomposition
1
6
1
1 1
=
+
.
(900 − 6P )P
900 900 − 6P
900 P
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MATH 2280-002
Lecture Notes: 01/17/2013
When we integrate, we get
Z
Z
1
6
1 1
−1
−1 900 − 6P dP +
dP =
[ln |600 − 6P | − ln |P |] =
ln
900 900 − 6P
900 P
900
900 P
Remember that we were calculating the left-hand side of an equation where the right-hand side
was equal to t + C1 .
−1 900 − 6P ln
= t + C1
900 P
900 − 6P = −900t + C2
ln P
900 − 6P
= C3 e−900t ,
P
where C3 is either ±eC2 . (In class, we argued that, in fact, C3 = −eC2 by looking at the slope
field.) Solve for C3 by using the initial condition P (0) = 500.
900 − 6(500)
= C3 e 0
(500)
This tells us that A = − 21
5 .
900 − 6P
21
= − e−900t
P
5
21 −900t
900
−6=− e
P
5
900
21
= 6 − e−900t
P
5
900
P =
−900t
6 − 21
5 e
1500
P =
10 − 7e−900t
Great, we found P (t), but what does this equation really mean? What sort of behavior does it
predict over time? As t → ∞, e−900t → 0, so
1500
1500
=
= 150.
−900t
t→∞ 10 − 7e
10
lim P (t) = lim
t→∞
According to our model, the colony population should approach the carrying capacity (also called
the limiting population) of 150 mice as time progresses.
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MATH 2280-002
2
Lecture Notes: 01/17/2013
Logistic Models and Critical Points
Let’s try to tie the above example together with our understanding of critical points. The equation
dP
dt = (900 − 6P )P is autonomous and has critical points at P = 0, 150.
Exercise. Construct a phase diagram for the DE dP
dt = (900 − 6P )P to show that the critical point
P = 0 is unstable, while the critical point P = 150 is stable.
Solution curves to dP
dt = (900 − 6P )P which start at P -values greater than 150 will approach the
line P = 150 asymptotically from above, and solution curves which start at values between 0 and
150 will approach the line P = 150 asymptotically from below. Notice that this agrees with our
findings for P (t) in the specific case that P (0) = 500 > 150. According to our model, our prediction
should be quite robust, so even if we add or remove a few mice, it shouldn’t matter in the long run.
Next time, we’ll contrast this with a logistic DE possessing an unstable critical point which causes
very different behavior.
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