On Wednesday night we began discussing a new population model, Logistic
Growth. This model explains what happens to a population’s growth rate as it
approaches, and exceeds, its habitat’s carrying capacity. The carrying capacity is
essentially a cap on the size of a population that can be adequately sustained in the
habitat. The carrying capacity, however, is not a maximum population number – there
is room for a population size beyond that of the carrying capacity, but the population
will no longer experience growth, but instead, decline.
𝑥𝑛 :population at time 𝑛
The following variables are used for the logistic growth model: 𝑘:carrying capacity and
𝑟:growth rate
the model is: 𝑥𝑛+1 = 𝑥𝑛 + 𝑟 (1 −
𝑥𝑛
) 𝑥𝑛 .
𝑘
So, the population at the next time step includes
the entire previous population and an added portion of the previous population based
on the growth rate and the carrying capacity. From this model it can be seen that if
𝑥
𝑥𝑛 < 𝑘, then 1 − 𝑘𝑛 > 0 and the population will grow, on the other hand, if 𝑥𝑛 > 𝑘, then
1−
𝑥𝑛
𝑘
< 0 and the population will decay. Also of interest, if 𝑥𝑛 is very large or
significantly greater than the carrying capacity, then it is possible for the population at
the next time step to become less than 0 and then the population will be extinct.
With a specific example it is easier to see how the carrying capacity affects the
1
1
𝑥
𝑛
growth rate: Let k = 1000 and 𝑟 = 10, then the model is: 𝑥𝑛+1 = 𝑥𝑛 + 10 (1 − 1000
) 𝑥𝑛
or 𝑥𝑛+1 = 1.1𝑥𝑛 − .0001𝑥𝑛 2 , where the 1.1𝑥𝑛 represents a population growth of 10% and
−.0001𝑥𝑛 2 will slow down the growth as the population approaches the carrying
capacity and will reverse the growth (population decay) if 𝑥𝑛 goes beyond the carrying
capacity. This can be seen in the table below:
𝑥𝑛
100
𝑥𝑛+1
109 = 1.1(100) − .0001(100)2
growth
109
9% =
= 1.09
200
500
800
900
216
525
816
909
1000
1000
1500
1425
8%
5%
2%
1%
growth is slowed as the population approaches
carrying capacity
0%
at carrying capacity so no growth
-5%
beyond the carrying capacity so population decay
100
As we continued to observe what was going on with both the growth rate and
the scatter plot of data points in Excel, we began to notice some obvious max and min
peaks that were forming when the population was significantly large and that the
population would oscillate between these peaks. We then looked at a bifurcation
diagram which shows branching points (bifurcation points) that related to the peaks we
saw and, in turn, the oscillation values. Before we further investigated the oscillations,
we performed a change of variable on our earlier model, 𝑥𝑛+1 = 𝑥𝑛 + 𝑟 (1 −
let 𝑥𝑛 = 𝑘 (1
1
+ 𝑟 ) 𝑢𝑛 ,
𝑥𝑛
) 𝑥𝑛 .
𝑘
We
and through several steps of algebraic manipulation, we rewrote
our model as: 𝑢𝑛+1 = 𝜌𝑢𝑛 (1 − 𝑢𝑛 ) and as an iteration function: 𝑔(𝑢) = 𝜌𝑢(1 − 𝑢) where
𝑟
𝑥
𝜌 = 1 + 𝑟 and 𝑢 =
∙ .
𝑟+1 𝑘
After the change of variable and then replacing the u with x for simplicity in
function notation, we have the iteration function: 𝑔(𝑥) = 𝜌𝑥(1 − 𝑥) where 0 < 𝑥0 < 1 and
𝜌 > 0. The graph of this function is the family of parabolas which open down and have
x-intercepts of 0 and 1. The fixed points are where the parabola and the line y = x
intersect which will always occur at x = 0 and the other point will depend on the value
of 𝜌. More specifically, the fixed points are where 𝑥 = 𝜌𝑥(1 − 𝑥)
or 1 = 𝜌(1 − 𝑥)
1
so, 𝑥 = 0 and 𝑥 = 1 − 𝜌 are the two fixed points.
Further, we can determine if these fixed points are either attracting or repelling
by finding the derivative at each point:
𝑔′ (0) = 𝜌 which will be attracting when |𝑝| < 1 which implies |1 + 𝑟| < 1 or −2 < 𝑟 < 0 this
means r is negative and the population has a negative growth rate – this
situation is not of interest.
1
1
𝑔′ (1 − 𝜌) = 𝜌 − 2𝜌 (1 − 𝜌) = 2 − 𝜌 which will be attracting when |2 − 𝜌| < 1 or 1 < 𝜌 < 3
To better explain the Excel observations of the branching oscillation points and
total break down, we tested several 𝜌 values which are summarized in the table on the
next page.
We found that when 𝜌 > 3 the cobwebs begin cycling through two
accumulation points and as 𝜌 increases, there are more accumulation points which
eventually lead to total break down once 𝜌 > 4. These cycles occur because every
other term in the iteration function converges to one of the two accumulation points
(and then 1 of 4 and then 1 of 8 and so on until total breakdown) What happens is the
iteration function contains subsequences which have converging fixed points and then
the main sequence has its own additional converging fixed points – all these attracting
fixed points are the accumulation points that create the cycles and eventually, 𝜌 > 4,
there are too many.
𝜌 – value
Iteration
function
𝜌=1
𝜌=2
𝑔(𝑥) = 𝑥(1 − 𝑥)
𝑔(𝑥) = 2𝑥(1 − 𝑥)
𝜌 = 2.5
𝑔(𝑥) = 2.5𝑥(1 − 𝑥)
𝜌=3
𝑔(𝑥) = 3𝑥(1 − 𝑥)
𝜌>3
𝜌>4
𝑔(𝑥) = 𝜌𝑥(1 − 𝑥)
𝑔(𝑥)
= 𝜌𝑥(1 − 𝑥)
Graph w/
y=x
Fixed point
1−
1
=0
𝜌
1−
1 1
=
𝜌 2
Cobweb
description
Short bounces
between parabola
and y = x, eventually
goes to 0
Fixed point is the
vertex of the
parabola, so a
cobweb will move
quickly towards the
fixed point
Converge?
Yes, but converges
very slowly
Yes, quickly
1−
1 3
=
𝜌 5
1−
1 2
=
𝜌 3
Fixed point is to the
left of the vertex –
the cobweb will
spiral between
parabola and
y = x moving
towards the fixed
point
Fixed point is where
the slope of the
tangent line is
perpendicular to the
slope of y = x.
Cobweb will still spiral
towards the fixed
point
Yes
Yes, but converges
very slowly
1−
1 2
>
𝜌 3
Cobweb will
cycle
between two
values
meaning the
sequence
cannot
escape – it is
bounded
between two
accumulation
points
No – cannot
converge to
a single point
1−
1 3
>
𝜌 4
Cobweb
will leave
the
bounded
area and
go to
negative
infinity
No – total
breakdown
Questions I have:
1. I did not follow the purpose for changing the variable. Why was this step
necessary to our further discussion of the cycles? Was it so we could form an
iteration function with a simple quadratic equation?
2. How does someone determine that the way to change the variable is by letting
1
𝑥𝑛 = 𝑘 (1 + 𝑟) 𝑢𝑛 ?
3. Back to harvesting: I was wondering about the following situation (thinking of
controlling the deer population)and wondered if some version of the harvesting
model could be used:
If there is no control over the starting values of a population, but the
current population numbers can be determined (I suppose these could
be used as the starting values) can one find the harvesting amount to
bring the population to the steady state?