Population Modeling Mathematical Biology Lecture 2 James A. Glazier (Partially Based on Brittain...
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Transcript of Population Modeling Mathematical Biology Lecture 2 James A. Glazier (Partially Based on Brittain...
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