Transcript P548/M548 Mathematical Biology

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: d N (t )  Births  Immigratio n - Deaths
dt
• Continuum Models are Generally More Stable
than discrete models (no chaos or oscillations)
Malthusian Model (Exponential
Growth)
• For a Fertility Rate, b,
a Death Rate, d, and
no Migration: d N (t )
dt

 bN  dN  (b  d ) N
N (t )  N 0 e (b  d )t
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:

d N (t )
 rN 1  N
K
dt

• The Logistic Equation has a Closed-Form
Solution:
KN0e rt
No Chaos in
N t  
Click for Solution Details
K  N 0  N 0e rt
Continuum
Logistic Equation
Solving the Logistic Equation
1) Start with Logistic
dN
Equation:
 rN 1  N K 
2) Now let:
x  log  N K  N 
K  2N
 dx 
dN
N K  N 
K
2N
 dx 
dN 
dN
N K  N 
N K  N 
K
2N

dN  dx 
dN
N K  N 
N K  N 
dt
 dt 
dN
rN 1  N K 
N
t 
dN
 rN 1  N K 
N0
N
t 
1
KdN

r N 0 N K  N 
4) Solve for N(t):
3) Substitute:
N
t
dN
 rN 1  N K 
e rt 
N0
log N  K  N 
N
1 
2dN 
t  
dx


r log N 0 K  N 0  N0 K  N 

1  log N  K  N 

 t  x
 2 log  K  N  
N0 
r  log N 0  K  N 0 
N K  N  
1
1
  1 log N  K  N 0  
N
 t  log

log


2
r
r   K  N
r 
N 0  K  N  
K

N


 
N0
 
N0
N0 K  N 
 N 0  K  N  e rt  N  K  N 0 
 KN 0 e rt  N K  N 0  N 0 e rt
N
N
N K  N0 
N
KN 0 e rt
 N t  
K  N 0  N 0 e rt

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
x  f x 
•
x0
is a Fixed Point 
f x0   0
• So the Logistic Equation has Two Fixed Points,
N=0 and N=K
• Fixed Points are also often designated x*
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 e.
Assume that: e   x0
• Taylor Expand f around x0:
• So, if
df
e2 d2 f
x0  
x0    e 3 
f x0  e   f x0   e
2
dx
2! dx
df
x0   neglect
0 e
dx
dx d x0  e  de
df
x0 



e
dt
dt
dt
dx
df
t  x0 
 e t  e e dx
0
 

df
x0   0 then e t  grows exponentia lly  UNSTABLE
dx
 0 then e t  shrinks exponentia lly  STABLE
 0 then e t  depends on higher order term s  MARGINALLY STABLE
Response Timescale, t, for disturbance to grow or shrink by a factor of e is:
Example Logistic Equation
Start with the
Logistic Eqn.


d N (t )
 rN 1  N
K
dt

f N 
Fixed Points at N0=0 and N0=K
For N0=0
unstable
For N0=K
stable
Phase Portraits
• Idea: Describe Stability Behavior Graphically
Arrows show direction of Flow
Generally:
Solution of the Logistic Equation
Solution of the Logistic Equation:
KN 0 e rt
N t  
 K
rt
K  N 0 e  1 limt 
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.
d N1 (t )
N  N 2   and d N 2 (t )  r N 1  N1  N 2  
 r1 N1 1  1
2 2
K1 
K2 

dt
d
t




f1  N 
f1  N 
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.
 f1

N
J  1
 f 2
 N
 1
N
f1    N1  N 2  
 r1 1
 r1 1 

K1
K1
N 2   


f 2 
N2


r

2
N 2  
K2


N1

K1

 N  N 2  
N 
  r2 2 
r2 1  1
K2
K 2 


Evaluate at (K1,0).
  K1 K1 
K1 
 r1 1 



r
 r1
  r1

1

K1  
  K1 K1 
 K1  
J 




0 r2 1 
 K 


0
r2 1  1   
 K2  

K
2 


 r1
Stability in Two Dimensions
Cases:
N2
1) Both Eigenvalues Positive—Unstable
N1
N2
2) One Eigenvalue Positive, One
Negative—Unstable
N1
N2
1) Both Eigenvalues Negative—Stable
N1
Population Competition—Contd.
Eigenvalues are solutions of
det J - I   det
 r1  
0
det J - I   0
 r1
  K 

 K1 
  r1    r2 1  1    
  
r2 1 
  K2 

K
2 


 K 
    r1 , r2 1  1 
 K 2 

-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.