1. Population Modeling Mathematical Biology Lecture 2 James A. Glazier
(Partially Based on Brittain Chapter 1)

2. Population Models Simple and a Good Introduction to Methods Two Types
Continuum [Britton, Chapter 1]
Discrete [Britton, Chapter 2]

3. Continuum Population Models
Given a Population, N0 of an animal, cell, bacterium,… at time t=0, What is the Population N(t) at time t? Assume that the population is large so treat N as a continuous variable.
Naively:
Continuum Models are Generally More Stable than discrete models (no chaos or oscillations)

4. Malthusian Model (Exponential Growth)
For a Fertility Rate, b, a Death Rate, d, and no Migration: 
In reality have a saturation: limited food, disease, predation, reduced birth-rate from crowding…

5. Density Dependent Effects
How to Introduce Density Dependent Effects?
Decide on Essential Characteristics of Data.
Write Simplest form of f(N) which Gives these Characteristics.
Choose Model parameters to Fit Data
Generally, Growth is Sigmoidal, i.e. small for small and large populations 
f(0) = f(K) = 0, where K is the Carrying Capacity and f(N) has a unique maximum for some value of N, Nmax
The Simplest Possible Solution is the Verhulst or Logistic Equation

6. Verhulst or Logistic Equation
A Key Equation—Will Use Repeatedly
Assume Death Rate  N, or that Birth Rate Declines with Increasing N, Reaching 0 at the Carrying Capacity, K:
The Logistic Equation has a Closed-Form Solution:
No Chaos in Continuum Logistic Equation
Click for Solution Details

7. Solving the Logistic Equation
2) Now let:
1) Start with Logistic Equation:
3) Substitute:
4) Solve for N(t):

8. General Issues in Modeling
Not a model unless we can explain why the death rate d~N/K.
Can always improve fit using more parameters.
Meaningless unless we can justify them.
Logistic Map has only three parameters N0, K, r – doesn't fit real populations, very well. But we are not just curve fitting.
Don't introduce parameters unless we know they describe a real mechanism in biology.
Fitting changes in response to different parameters is much more useful than fitting a curve with a single set of parameters.

9. Idea: Steady State or Fixed Point
For a Differential Equation of Form
is a Fixed Point 
So the Logistic Equation has Two Fixed Points, N=0 and N=K
Fixed Points are also often designated x*

10. Idea: Stability Is the Fixed Point Stable?
I.e. if you move a small distance e away from x0 does x(t) return to x0?
If so x0 is Stable, if not, x0 is Unstable.

11. Calculating Stability: Linear Stability Analysis
Consider a Fixed Point x0 and a Perturbation e. Assume that:
Taylor Expand f around x0:
So, if
Response Timescale, t, for disturbance to grow or shrink by a factor of e is:

12. Example Logistic Equation
Start with the Logistic Eqn.
Fixed Points at N0=0 and N0=K
For N0=0
unstable
For N0=K
stable

13. Phase Portraits Idea: Describe Stability Behavior Graphically
Arrows show direction of Flow
Generally:

14. Solution of the Logistic Equation
For N0>K, N(t) decreases exponentially to K
For N0<2, K/N(t) increases sigmoidally to K
For K/2<N0<K, N(t) increases exponentially to K

15. Example Stability in Population Competition
Consider two species, N1 and N2, with growth rates r1 and r2 and carrying capacities K1 and K2, competing for the same resource. Both obey Logistic Equation.
If one species has both bigger carrying capacity and faster growth rate, it will displace the other.
What if one species has faster growth rate and the other a greater carrying capacity?
An example of a serious evolutionary/ecological question answerable with simple mathematics.

16. Population Competition—Contd.
Start with all N1 and no N2. Represent population as a vector (N1, N2) Steady state is (K1,0). What if we introduce a few N2?
In two dimensions we need to look at the eigenvalues of the Jacobian Matrix evaluated at the fixed point.
Evaluate at (K1,0).

17. Stability in Two Dimensions
Cases:
1) Both Eigenvalues Positive—Unstable
2) One Eigenvalue Positive, One Negative—Unstable
1) Both Eigenvalues Negative—Stable

18. Population Competition—Contd.
Eigenvalues are solutions of
-r1 always < 0 so fixed point is stable  r2(1-K1/K2)<0 i.e. if K1>K2. Fixed Point Unstable (i.e. species 2 Invades Successfully)  K2>K1
Independent of r2! So high carrying capacity wins out over high fertility (called K-selection in evolutionary biology).
A surprising result. The opposite of what is generally observed in nature.