Population Modeling

Mathematical Biology Lecture 2

James A. Glazier

(Partially Based on Brittain Chapter 1)

Population Models

Simple and a Good Introduction to Methods

Two Types

• Continuum [Britton, Chapter 1]

• Discrete [Britton, Chapter 2]

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)

Deaths -n Immigratio Births d

)(d

t

tN

Malthusian Model (Exponential Growth)

• For a Fertility Rate, b, a Death Rate, d, and no Migration:

NdbdNbNt

tN)(

d

)(d

tdbNtN )(

0 e )(

In reality have a saturation: limited food, disease, predation, reduced birth-rate from crowding…

Density Dependent Effects• How to Introduce Density Dependent Effects?

1) Decide on Essential Characteristics of Data.

2) Write Simplest form of f(N) which Gives these Characteristics.

3) 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

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:

KNrN

t

tN 1

d

)(d

rt

rt

eNNK

eKNtN

00

0

No Chaos in

Continuum Logistic Equation

Click for Solution Details

Solving the Logistic Equation

N

N

N

N

NKN

KdN

rt

KNrN

dNt

KNrN

dNdt

KNrNdt

dN

0

0

1

1

1

1

dNNKN

NdxdN

NKN

K

dNNKN

NdN

NKN

Kdx

dNNKN

NKdx

NKNx

2

2

2

log

NKN

NKN

rNK

Nr

NK

NKN

rt

NKxr

t

NK

dNdx

rt

KNrN

dNt

N

N

N

N

N

N

NKN

NKN

N

N

NKN

NKN

N

N

0

0

log

log

log

log

log1

log1

2log

1

log21

21

1

00

000

000

0

2) Now let:

3) Substitute:

1) Start with Logistic Equation:

4) Solve for N(t):

rt

rt

rtrt

rtrt

eNNK

eKNtN

eNNKNeKN

NKNeNKNNKN

NKNe

00

0

000

000

0

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.

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*

xfx

0x 00 xf

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.

Calculating Stability: Linear Stability Analysis

• Consider a Fixed Point x0 and a Perturbation . Assume that: 0x

0

00

0

302

22

000

0

neglect 0

!2

xdx

dft

xdx

df

dt

d

dt

xd

dt

dx

xdx

df

xdx

fdx

dx

dfxfxf

et

STABLE MARGINALLY sorder termhigher on depends then 0

STABLE llyexponentia shrinks then 0

UNSTABLElly exponentia grows then 00

t

t

txdx

df

• Taylor Expand f around x0:

• So, if

Response Timescale, , for disturbance to grow or shrink by a factor of e is:

Example Logistic Equation

Nf

KNrN

t

tN 1

d

)(dStart with the Logistic Eqn.

Fixed Points at N0=0 and N0=K

For N0=0

For N0=K stable

unstable

Phase Portraits • Idea: Describe Stability Behavior Graphically

Arrows show direction of Flow

Generally:

Solution of the Logistic EquationSolution of the Logistic Equation:

KeNK

eKNtN

trt

rt

lim0

0

1

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

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.

Population Competition—Contd.

NfNf

KNNNr

t

tNK

NNNrt

tN

11

2

2122

2

1

2111

1 1 d

)(d and 1

d

)(d

2

22

2

212

2

22

1

11

1

11

1

211

2

2

1

2

2

1

1

1

1

1

K

Nr

K

NNr

K

Nr

K

Nr

K

Nr

K

NNr

N

f

N

fN

f

N

f

J

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).

2

12

11

2

12

1

11

1

1

1

11

1010

1

K

Kr

rr

K

Kr

K

Kr

K

K

K

Kr

J

Stability in Two Dimensions

N1

N2

N1

N2

1) Both Eigenvalues Positive—Unstable

Cases:

2) One Eigenvalue Positive, One Negative—Unstable

1) Both Eigenvalues Negative—Stable

N1

N2

Population Competition—Contd.

0-det IJ

2

121

2

121

2

12

11

1,

110

det-det

K

Krr

K

Krr

K

Kr

rr

IJ

-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.

Eigenvalues are solutions of