Lecture 2
Modelling the dynamics of
populations. First order growth and
decay. The logistics equation.
What is a model?

  




dV    v .n  A   n dA 

 t CV
surface
 ( So  Si )
Mathematical models improve with age.....
BEST – IST, 2006
Princípio de conservação
The rate of
accumulation inside
a control volume

t
What flows in
minus what
flows out
=
 dV    
+
  
Production minus
destruction



 v .n  A   n dA  ( S o  S i )
CV
surface
Closed (isolated) Volume

t
 dV  ( S
o
 Si )
CV
• If β is uniform inside the volume:

Vol
 ( So  Si )
t
• Writting the source/sink terms per unit of volume
one gets:
c
 so  si
t
Population Dynamics
c
n
 kc
t
In case (n=1) => 1st order :
The analytical solution is:
c  c0 e
(n=0)
=> zero orde decay/growth
(linear evolution )
(n=1) => 1st order (exponential)
……..
c
K>0
kt
c0
If (n=1) => 1st order :
K >0 implies exponential growth
K<0 asymptotic decay towards zero.
K<0
t
1st order decay
• Usually it is assumed that:
– faecal bacterian mortality is a 1st order decay
accounted using the T90,
– Pesticides have a 1st order decay accounted by the
half-life time.
• How to compute k?
c  c0 e
kt
=>
0.1c0  c0 e
kT90
=>
ln 0.1
k
T90
T90=1 hour=> k=-6.4E-4 s-1.
In case of the half-life time the calculation is identical, using ln(0.5)
The “Logistic”solution
•The solution so called "Logistic “ solution admits that the
exponential growth is not sustainable. Admitting that there
is a maximum population and consequently K must be
variable.
c
c
n
 kc
t
k  k0 cmax  c  / cmax
Cmax
C0
t
Numerical solution(explicit)
c
n
 kc
t
k  k0 cmax  c  / cmax
c t  t  c t
*
 kc
t
Using the explicit method:
c
t  t
 1  kt c
Discretizing the time derivative
one gets:
t
Comparison of the numerical and
analytical solutions
See the Excel workbook “dinâmica de populações”
Numerical solution (explicit)
In the explicit solution we got:
c t  t  1  kt c t
If k<o then the parentheses can be negative if
the time step is high. In this case the new
concentration would be negative and the
method would be unstable. The stability
condition is:
k 0
1  kt   0  t  1 / k
In this passage the inequality sign
change when we divide by k<0
Numerical Solution (implicit)
c
n
 kc
t
k  k0 cmax  c  / cmax
In this case one gets:
t  t
c
 kc*
t
c t  t  c t / 1  kt 
c
t
Now instabillities can appear if k>0:
k 0
1  kt   0  t  1 / k
Criteria for Stability
• When we have mortality, if the method is explicit, the number
of individuals who dies is a function of the value that we had
at the beginning of the time step. This implies that the
mortality value is overestimated. If the step time is too large
we can eliminate more individuals than existing ones and we
get a negative value (the same can be said for the
concentration).
• When we have growth problem arises in the implicit method
because physically the number of children is in proportion to
the number of parents and the calculation must be explicit.
The implicit calculation would be equivalent to saying that
"the kids would be born bringing children on their lap".
Generalizing one can say that:
• The sources must be calculated explicitly and sinks
must be calculated implicitly for model stability.
• If the model is stable which time step should be
used? A time step small enough to assure that the
numerical solution does not move away from the
analytical solution.
c
K>0
implícito
explícito
c0
K<0
t
Final Considerations
• The models based on first order decays can be
realistic for properties that do not arise in the
natural environment and consequently when
they are released they decay continuously.
• The models based on first-order growth are
unrealistic. The logistic equation can give them
some realism.
• Models must reproduce the processes of
production and decay. The Lotka-Volterra model
is the simplest that attempts to achieve this goal.
Prey-Predator Model (Lotka-Volterra)
•In the equation: c  kcn
t
k  k0 cmax  c  / cmax
only logistics could limit grow. In the real world
there is always predator that also contributes to
limit the prey.
c p
 k p c p  k g c p cz
t
cz
 eg k g c p cz  k mz cz
t
Lotka-Volterra Equations
Lotka Volterra Model Limitations
• It does not conserve the total mass. A very simple
nature would require at least 3 state variables:
dc p
 k p c p  k g c p cz
dt
dc z
 e g k g c p c z  k mz c z
dt
dc D
 k p c p  1  e g k g c p c z  k mz c z
dt
• Note: The derivatives are now total derivatives to describe the case of a
material system in motion.
• Could kp beconstant? Is it reasobable that the prey is
fed with detritus? We need extra variables...
Generic shape of the evolution
equations
 c p

 vj


 k p c p  k g c p cz
dt
t
x j x j x j
dc p
c p
c p
dc z c z
c z
 c z

 vj


 e g k g c p c z  k mz c z
dt
t
x j x j x j
dc D c D
c D
 c D

 vj


 k p c p  1  e g k g c p c z  k mz c z
dt
t
x j x j x j
In these equations have added the difusive transport.