Owen Lewis
Math 5110
Homework #2
9/29/2015
A Read Chapter 3 in the textbook.
Just read. This one isn’t hard. If you haven’t done it already, please do in the future. It
will provide valuable supplemental information to what was covered in lecture.
⌅
1 In 1973 Varley, Gradwell and Hassell proposed the following model of a population
with a density dependent reproduction rate
p n +1 =
which can be rewritten as
p n +1 =
✓
l 1 b
p ,
a n
(1)
◆
⌘
1 b ⇣
p
lpn .
a n
(2)
(a) Interpret the two terms in equation 2 in your own words.
(b) Find the equilibrium population. Under what conditions is it stable? Sketch the
bifurcation diagram.
(c) What happens to the population when the stability conditions are violated? Include a cobweb diagram of an illustrative case.
(a) The second term corresponds to a linear growth model with a growth rate of l. However, this is modulated by the first term, which corresponds to a population dependent
change in the growth rate. Assuming that b > 0, this term will penalize large populations
by decreasing the growth rate. When the population is pn = 1, this reduces to a linear
model with growth rate l/a.
⌅
(b) By inspection, we can see that p⇤ = 0 is an equilibrium as long as we are not dividing
by zero (i.e. as long as 1 b > 0, or 1 > b). We find other equilibria by setting
p⇤ =
Multiplying both sides by pb⇤
pb⇤
1
l 1
p
a ⇤
b
.
gives us
l
= , and therefore p⇤ =
a
1
✓ ◆1/b
l
.
a
To determine the stability of these equilibria, we linearize the system.
f ( p) =
l 1
p
a
b
l
b) p
a
) f 0 ( p ) = (1
b
.
Notice that f 0 (0) is not even a well defined value. However it is true that as long as b > 0,
then
lim f 0 ( p) = ±•,
p !0
So we may regard this equilibrium as unstable. Finally, we evaluate the derivative at the
non-trivial equilibrium
0
f ( p⇤ ) = f
0
✓ ◆1/b !
l
= (1
a
Clearly this equilibrium is stable if |1
l
b)
a
✓ ◆1/b !
l
a
b
=1
b.
b| < 1, or
1<1 b<1
1>b 1>1
2 > b > 0.
In Figure 1, we show two bifurcation diagrams. One is for when the quantity l/a > 1
and one is for l/a < 1. The nature of the bifurcation is the same, the only difference is
the plot of the curve.
0.9
4
0.8
3.5
3
p∗ = 0.5(1/b)
p∗ = 1.5(1/b)
0.7
2.5
2
0.6
0.5
0.4
0.3
0.2
1.5
0.1
1
0.5
1
1.5
2
2.5
3
3.5
0
4
b
0
0.5
1
1.5
2
2.5
3
3.5
4
b
Figure 1: Bifurcation curves for l/a = 1.5 and l/a = 0.5 respectively.
⌅
(c) Once b > 0, the equilibrium is unstable and the eigenvalue of the linearized system
is negative. This leads to increasing oscillations. The population will experience “boom”
2
and “bust” cycles which increase in magnitude. Eventually oscillating to “infinity”. This
is hard to see in the cobweb plot of Figure 2. It appears that a 2 cycle has emerged. However, if we run the simulation longer and plot it as a function of n, then you clearly see
the oscillation becoming larger and larger, not settling on a cycle. These plots were made
with b = 2.003.
Cobweb plot of system
Solution sequence
7
8
7
7
6
6
x 10
5
Output
Output
5
4
4
3
3
2
2
1
1
0
0.5
1
1.5
2
2.5
3
0
3.5
0
200
Input
400
600
800
1000
1200
n
Figure 2: b = 2.003, p0 = 0.5.
⌅
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Input
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