Lab 1.5: Modeling the extinction of the Passenger Pigeon
Differential Equations, Fall 2002, TESC
Text = Differential Equations (2002, ed.2) by Blanchard, Devaney, and Hall (pp.143-145)
Sources: handwritten notes in notebook zD; Mathematica calculations in Lab15.nb
10.Nov.2002- E.J. Zita (with Don Middendorf)
_______________________________________________________________________________
We investigate the population changes of the passenger pigeon for four cases:
(1) logistic growth with constant harvesting (hunting)
(2) the same case, but with harvesting at 10% over the critical rate
(3) rebuilding the population after hunting stops
(4) logistic growth with constant harvesting plus a geometric nest-loss term.
Short answers:
(1) Equilbrium population values depend on hunting rate. Critical hunting rate Cmax = NK/4.
For the reference case of growth rate k=0.25 (1/year) and carrying capacity N = 50
million birds, we find Cmax = 3125000 (birds/year).
(2) For hunting at a rate of 1.1 Cmax , 50 million pigeons drop to ten thousand in less than 32
years, and the pigeons go extinct shortly thereafter.
(3) If hunting stops when there are only ten thousand pigeons left, they recover to one
million in about 18.5 years.
(4) If we include a term r P representing nest loss due to pigeons overcrowding, we find
that even if hunting stops, the pigeon population keeps declining if r  k P . For ten
thousand birds in our reference case, the critical r=25.
Methods: We derived critical points, phase plots, and solutions to the differential equations
analytically by hand wherever possible. Then we used Mathematica to plot relations, check our math,
and perform the harder calculations. We reference that source explicitly whenever we use results
obtained from our Mathematica notebook "lab15.nb". Handwritten results from EJZ's Diffeq
notebook are scanned into an attached pdf file, with pages numbered zD.#.
_______________________________________________________________________________
Part 1: While the passenger pigeons are being hunted, the differential equation for their changing
population P in time t can be modeled as logistic growth with constant harvesting (p.144)
dP
P

 kP 1    C
dt
 N
(1)
where the linear growth rate k = 1/4 and the carrying capacity N = 5 x 107.
The equilibrium points (stable population) are where the population is steady: dP/dt=0.
Solving the resultant quadratic equation on the RHS of (1), we find that this occurs for population

values P0  12 N 1± 1 

4C 
 which depend on the hunting rate C.
NK 
If C=NK/4, there is a single equilibrium population value at P=N/2, half the carrying capacity.
This is the largest sustainable hunting rate: Cmax = NK/4 = 3125000 (birds/year).
At higher hunting rates, the population decreases. At lower hunting rates, the solution bifurcates with
two equilibrium population values, similar to the example on p.100 (but with signs reversed).

4C 
P  12 N 1+ 1 
 is a source (or a stable equilibrium point: dP/dt<0 above and >0 below) and
NK 


4C 
P  12 N 1- 1 
 is a sink (or unstable equilibrium point: dP/dt > 0 above and <0 below).
NK 

In sum, for hunting rates below Cmax,
populations starting out above (P > P+) will drop to P+ and remain steady,
populations starting out between (P- < P < P+) will increase to P+ and remain steady, and
populations starting out below P- will go extinct.
Note that the point C = Cmax and P=N/2 satisfies the mathematical conditions for a bifurcation point:
d2P/dt2=0 (node) and dP/dt=0 (equilibrium) at this point.
See attached handwritten notes (zD.34) for the bifurcation diagram.
With some work, you can find the solutions in Mathematica:
P
H
L
Cmax
7 ´ 10
6 ´ 10
5 ´ 10
4 ´ 10
3 ´ 10
2 ´ 10
1 ´ 10
2
7
Cmax
6 ´ 10
7
7
4 ´ 10
7
7
2 ´ 10
7
7
7
7
2
HL
2
P
HL
P
7
4
6
8
10
12
1.1 Cmax
6 ´ 10
4 ´ 10
2 ´ 10
4
10
12
t
P
HL
2 Cmax
7
5 ´ 10
7
2.5 ´ 10
7
7
7
2
-4 ´ 10
8
t -2 ´ 10 7
2
-2 ´ 10
6
7
7
4
6
8
10
12
t -2.5 ´ 10 7
-5 ´ 10
-7.5 ´ 10
7
7
4
6
8
t
Part 2: If the harvest rate was 110% of Cmax, how long would it take for the population to decline to
P1 = 10,000, assuming that the initial population was near the carrying capacity?
We can linearize about an equilibrium point (as in 1.9 # 22) in the handwritten notes (zD.35) to obtain
an analytic approximation. Far from equilibrium points, however (e.g. at P ~ N or P ~ P1), the
linearization solution does not hold. (For example, the solution involves a log with a negative
argument near P1.)
@Dik y
{
Assume that P(0) = N = 50,000,000 = 5 x 107 and investigate solutions with Mathematica (see
f1
C_, P_
P2
:= k P -
- C
N
attached file Lab1.5.nb). We plot
3125000
rate C in terms of the critical hunting rate Cmax = kN/4 =

Red: Cmax /2; Green: Cmax; Blue: 2 Cmax
C ranges from Cmax /4 to Cmax
Equilibrium
f = dP dt
4 ´ 10
1 ´ 10
-2 ´ 10
7
2 ´ 10
7
3 ´ 10
7
4 ´ 10
7
-6 ´ 10
P values
7
P
5 ´ 10
7
3 ´ 10
7
6
2 ´ 10
-4 ´ 10
for various values of the hunting
6
1 ´ 10
7
7
6
1.5 ´ 10
6
2 ´ 10
6
2.5 ´ 10
6
3 ´ 10
6
C
Above is an example of a phase plot at C < Cmax. (There are no equilibrium values for C = 1.1 Cmax.)
8
<

 A
E
The general solution to equation (1) is:
P(t) =
N
N-
4 C - k N Tan
k
P pigeons
4 C- k N t
2
4 ´ 10
N
k
(2)
At a hunting rate of 1.1 Cmax, the population
drops to P1 = 10,000 in 31.9886 years,
2 ´ 10
7
7
5
-2 ´ 10
10
15
20
25
30
35
HL
t years
7
and all pigeons disappear within t=32 years.
(Population does not go negative - it stops at 0.)
You can get a very rough solution to the problem above by simply inserting numbers. If the


population starts out near the carrying capacity, then 1 
P
  0 and the diffeq becomes
N
(3) dP/dt = - C, which is simply linear decay, with a solution P = a - Ct. Using the initial condition
P(0) = N, we determine the integration constant a = N and find the general solution to (3):
(4) P(t) = N - Ct. Then the time it takes for the population to decline to P1 in this approximation is:
t1 
N  P1 5 107  104

 14.5 years
C
1.1 165 107
The population doesn't really decay linearly - it actually takes about 32 years to drop to P1, as seen
above.
Part 3: Once the population reaches P1 = 104, how long would it take to rebuild to P2 = 106 if
harvesting were completely halted (C=0)? An analytic solution is tractable in this case. The diffeq is
simple logistic growth:
dP
P

 kP 1  
dt
 N
(3)
This is a separable equation that can be integrated with the same method we used in Lab 1.4 part 2.
dP
k
 P( N  P)   N dt
Use partial fractions to write
 1
1  1
1

 (kt   )


N  PN
N ( N  P )  N
N P
  ln P  ln( N  P )   ln 
   kt  
 P 
N P
 kt

  Be
P


N
P (t ) 
1  Be  kt
1
1
1


P( N  P) PN N ( N  P)
cancel (1/N) terms and multiply by (-1)
Use the initial condition P(0) = P1 to find
B = (N - P1) / P1= 4999
(4)
This is the general solution to (3) above.
We can solve P(t2) = P2 = 106 analytically to find that the time to return the population to 100,000
pigeons is 18.5 years (after t1.)
N P
 e  kt
PB
N P
PB
 kt  ln
  ln
PB
N P
5
1
PB
1 10 *5 103
t  ln
 ln
k N  P 1 4 5  103  105
t  4 ln (102)  18.5 years
Pigeons recover after hunting stops
analytic
N
P = €€€€€€€€€€€€€€€€
€€€€€€€€
€€€€
- k
t€€
7 1 + B* E
5 ´ 10
4 ´ 10
3 ´ 10
2 ´ 10
1 ´ 10
Zoom in to see that P=100,000 after 18.5 years
Pigeons
1.4 ´ 10
1.2 ´ 10
7
7
1 ´ 10
800000
7
600000
7
10
20
30
40
50
HL
recovering
6
6
6
400000
200000
t years
5
10
15
20
HL
t years
Part 4:
If the pigeons cluster in trees so densely that the branches come tumbling down, nestlings and all,
the new diffeq could include a loss term proportional to the size of a circular cluster of birds,
where r is a geometric factor:


dP
P

 kP  r P 1    C
dt
 N
(5)
If there is no hunting (C=0), there are equilibrium points (dP/dt=0) when P=0, P=N, and P 
k2
.
r2
So the population is steady if there are no pigeons, if they are at carrying capacity, or when the nestloss geometric factor has the critical value r0  k P . If the population starts at P = P1 = 104 and the
growth rate is k = 1/4, then the critical value is r0 = 102/4 = 25. Assuming P<N, so the second term in
parenthesis in (5) is positive, we can see that for higher values (r > r0), the population declines, and
for (r< r0) the population grows.
H
L
H
L
Red: Cmax /2; Green: Cmax; Blue: 1.1 Cmax;
Black: C=0 (all for r=r0)
f = dP dt
3 ´ 10
2 ´ 10
1 ´ 10
-1 ´ 10
-2 ´ 10
-3 ´ 10
Red: r0; Green: 20 r0; Blue: 40 r0
(all for C=0: just shift down if C>0)
f = dP dt
c
6
3 ´ 10
r
6
6
2 ´ 10
6
6
1 ´ 10
7
2 ´ 10
7
3 ´ 10
7
4 ´ 10
7
P
5 ´ 10
7
1 ´ 10
6
6
6
1 ´ 10
7
2 ´ 10
7
3 ´ 10
7
4 ´ 10
7
P
5 ´ 10
7
6
-1 ´ 10
6
H L
Red: r=0~ r0; Green: 20 r0; Blue: 40 r0
f = dP dt
1.1 Cmax
1 ´ 10
-1 ´ 10
-2 ´ 10
-3 ´ 10
-4 ´ 10
7
2 ´ 10
7
3 ´ 10
7
4 ´ 10
7
P
5 ´ 10
7
6
6
6
6
There are no equilibrium solutions (for r= r0) for the hunting rate in part 2, C=1.1 Cmax.
In this case, dP/dt < 0 always, so the population is guaranteed to decline to zero no matter how high it
starts. Extinction is inevitable unless the hunting stops or the birds stop losing nests to overcrowding.