Transcript The Chemostat

Homework 4, Problem 3
The Allee Effect
Homework 4, Problem 4a
The Ricker Model
Homework 4, Problem 4a
The Beverton-Holt Model
The Chemostat
Group Model Building Exercise
The Chemostat
• An apparatus for the continuous culture of
bacterial populations in a steady state.
• Nutrients are supplied continuously to the
culture vessel.
• The cells in the vessel grow continuously on
these nutrients.
• Residual nutrients and cells are removed
from the vessel at the same rate by an
overflow, thus maintaining the culture at a
constant volume.
The Chemostat
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Your Task
• Derive a mathematical model of
bacteria growth in a chemostat
• Use the model to design the system so
that
– The flow rate will not be so great as to
wash out the entire culture
– The nutrient replenishment is sufficiently
rapid so the culture is able to grow
normally
Building The Model
• Variables
Nutrient Concentration
in growth chamber
C(t)
mass/vol
Bacteria Density
in growth chamber
N(t)
#/vol
Building The Model
• Parameters
Stock Nutrient
Concentration
C0
mass/vol
Volume of growth
Chamber
V
vol.
Inflow/Outflow Rate
F
vol/time
Bacteria growth rate
k
1/time
Yield constant
1/
#/mass
Assumptions
• Growth chamber is well mixed
– No spatial variations
• There is a single, growth limiting
nutrient
• Bacteria growth rate depend on nutrient
availability, that is k = k(C)
• Nutrient depletion occurs continuously
as result of reproduction
The Model
dN
F
 k(C)N  N
dt
V
reproduction

dC
F
F
 k(C)N  C  C0
dt
V
V
consumption

outflow
outflow
inflow
What Should k(C) Be?
• Last time, when deriving logistic growth,
we assumed k(C) = C
– Bacteria growth increases linearly with
nutrient concentration
• Jacques Monod found that following
function provided the best fit to data for
bacteria feeding on a single nutrient:
rC
k(C) 
aC
Monod Function
The Monod Function
• This growth function
is monotonically
increasing with limit
r as C → infinity.
• The parameter r is
the maximum
growth rate
• The parameter a is
called that halfsaturation constant
k(C) 
r
rC
aC

r/2
a
C
Sample Parameter Values
• For E.coli grown on glucose at 30 degrees
Celsius
r = 1.35 per hour
a = 0.004 g/liter
1/ = 0.23
If glucose is plentiful so that the specific
growth rate is maintained at r = 1.35, the
model predicts a doubling time of
approximately half and hour.
Bacteria Growth Without
Maintenance
dN
rC

N
dt a  C
dC
rC
 
N
dt
aC
• The steady state for the bacteria
depends on the initial conditions.
Back to the Chemostat
dN
rC

N  dN
dt a  C

dC
rC
 
N  dC  dC0
dt
aC
• d = F/V is the dilution rate (1/time)
• If 
the chemostat were filled it would take 1/d
hours to empty it. For this reason, 1/d is the
mean residence time of bacteria cells in the
chemostat
Nondimensionalize
Choose Arbitrary Scales:
N
n
A
C
c
B
  Tt
Substitute in Model:
dn
rBc

An  dAn
AT
dt a Bc



dc
rBc
BT
 
An  dBc  dC0
dt
a  Bc
Nondimensionalize
N
n
A
Simplify

Choices

C
c
B
  Tt
dn r c
d

n n
dt T a 
T

c

B
dc
Ar c
d
dC0

n c
dt
BT a  c
T
BT
B
Tr
Ba
Td
dC0
B
T
BT
A
r
Nondimensionalize
N
n
ad
r
C
c
a
  dt
dn r 
c
 dt  d 1 c n  n

Rename 

dc
c
C0

n c 
dt
1 c
a
r
a1 
d
C0
a2 
a
Only 2 parameter
groupings govern
the dynamics
Analysis of The Model
dn
c
 a1
nn
dt
1 c
dc
c

n  c  a2
dt
1 c

Steady States
0  a1
c
1 c
nn
c
0
n  c  a2
1 c
n1  0
c1  a2

1  c  1
n 2  a1a2 
 2
a1 1
a

1 1

Analysis of The Model
dn
c
 a1
nn
dt
1 c

dc
c

n  c  a2
dt
1 c
Steady States

Exists iff
1
a2 
a1 1
n1  0

1 
n 2  a1a2 

a

1 1

c1  a2
Exists iff
1
c2 
a1 1

a1 1
Stability
Let

c
g(n,c)  
n  c  a2
1 c
c
f (n,c)  a1
nn
1 c
Compute

Let

f
a11  (n e ,c e
)
x
f
a12  (n e ,c e )
y
g
a21  (ne ,c e )
x
a22 

  a11a22  a12a21
  a11  a22
1,2 
g
(n e ,c e )
y

   2  4
2
Stability of Continuous Models
• In a continuous model, a steady state
will be stable provided that eigenvalues
are both negative (if real) or have
negative real part (if complex).
• As with discrete models complex
eigenvalues are associated with
oscillatory solutions.
Necessary and Sufficient
Conditions
• For a system of two equations, a
steady state will be stable if:
  a11  a22  0
  a11a22  a12a21  0


Stability of Chemostat
• For n1 = 0, c1 = a2
f
a1a2
a11  (0,a2 ) 
1
x
1 a2


g
a2
a21  (0,a2 )  
0
x
1 a2

a1a2

2
1 a2
a12 
f
(0,a2 )  0
y
g
a22  (0,a1 )  1
y
 a1a2

1
   
1 a2 

Stability of Chemostat
• For n1 = 0, c1 = a2
 a1a2
?
  
1 0
1 a2 
 > 0 if:
1
a2 
,a1  1
1 a1
?
a1a2

2 0
1 a2
 < 0 if  > 0

So the elimination state is stable whenever the nontrivial
steady state does not exist.
Stability of the Chemostat
• For n2, c2


f
a11  (n2 ,c 2 )  0
x
f
a1n2
a12  (n2 ,c 2 ) 
0
2
y
(1 c 2 )
g
1
a21  (0,a2 )    0
x
a
1
g
n2
a22  (0,a1)  
1  0
2
y
(1 c 2 )
 n 2

  
 1 0
2
(1 c 2 )  
n2

0
2
(1 c 2 )
Therefore the nontrivial steady state is stable whenever it exists.
Stability of the Chemostat
• Check for oscillations: 1,2 
Is
   2  4
  4   0?
2
 n 2

  
 1
0
2
(1 c 2 )

n2

0
2
(1 c 2 )
 
n2

2
2
2





n2
n2
2
  4  
 1  4

1  0
2
2  
2
(1 c 2 )


 (1 c 2 )  (1 c 2 )
Therefore no oscillations are possible.
Conclusions
• It is always possible to design the
chemostat so that at steady state there
will be bacteria populating the growth
chamber.
• For the nontrivial steady state to exist
we must have:
1
a2 
,a1  1
a1 1
Interpretation
What do these results mean in term of the
original parameters?
That is what should the flow rate, the
volume, and the stock nutrient
concentration be in order to ensure
continuous culture?
This is part of HW 5
Phase Portraits
• A graphical picture that summaries the
behavior of a system of two ODEs.
• Example
dx
 xy  y
dt
dy
 xy  x
dt
Nullclines
• Nullclines are curves of zero slope
– That is curves for which
dx
dy
0
0
and
dt
dt
– Therefore steady states are located at the
intersection nullclines


x-Nullclines
dx
 xy  y  0
dt
y(x 1)  0
y 0
x 1

• On these lines, the slope (velocity) in the xdirection is zero


• The only movement can be in the y-direction
(ie up or down)
y-Nullclines
dy
 xy  x  0
dt
x(y 1)  0
y 1
x 0

• On these curves, the slope (velocity) in the ydirection is zero
• The only movement

 can be in the x-direction
(ie left or right)
Graph the Nullclines
y 1
x-nullclines
y-nullclines
y
x 0
1
x 1

1

x
• Label the Steady States

• Mark the Direction of Motion
y 0
Direction of Motion On y = 0
dx
 xy  y
dt
y
1
dy
 xy  x
dt


x
1
On y = 0:
dx
0
dt
dy
 x  0
dt
So movement is in the negative y-direction (down)
Direction of Motion On x = 1
dx
 xy  y
dt
y
1
dy
 xy  x
dt


x
1
On x = 1:
dx
0
dt
dy
 y 1
dt
So movement is down for y < 1 and up for y > 1
Direction of Motion On x = 0
dx
 xy  y
dt
y
1
dy
 xy  x
dt


x
1
On x = 0:
dx
 y
dt
dy
0
dt
So movement is in the negative x-direction (left)
Direction of Motion On y = 1
dx
 xy  y
dt
y
1
dy
 xy  x
dt


x
1
On y = 1:
dx
 x 1
dt
dy
0
dt
So movement is left for x < 1 and right for x > 1
Fill in the Direction Field
dx
 xy  y
dt
y
1
dy
 xy  x
dt


1
x
All trajectories move away from the non-trivial steady state, therefore
x = 1, y = 1 is unstable.
Some trajectories move towards the origin, but some move away;
therefore x = 0, y=0 is unstable.
Note that arrows change direction across a steady state.
Actual Phase Portrait
Another Example
dx
 x2  y
dt
dy
xy
dt
• Determine nullclines


– x-nullclines:
y  x2
• On this curve, slope (velocity) in the x-direction is zero
• Trajectories can only move off of this curve in the ydirection (up or down)

– y-nullclines:
yx
• On this curve, slope (velocity) in the x-direction is zero
• Trajectories can only move off of this curve in the ydirection (up or down)
Graph the Nullclines
dx
 x2  y
dt
dy
xy
dt
y
x-nullclines
y  x2


y-nullclines
yx
1

1
 x
• Plot nullclines, label steady states, mark direction of
motion
Direction of Motion On y = x
y  x2
y
1
yx


1
On y = x:
dx
 x2  y
dt
dy
xy
dt
dy
0
dt


x
dx
 x 2  x  x(x 1)
dt
So movement is left if x < 1 and right if x > 1
Direction of Motion On y =
y  x2
y
1
yx
dx
 x2  y
dt
dy
xy
dt


1
On y = x2:
2
x
dx
0
dt


x
dy
 x  x 2  x(1 x)
dt
So movement is up if x < 1 and down if x > 1
Fill in the Direction Field
y  x2
y
1
yx
dx
 x2  y
dt
dy
xy
dt


1


x
Some trajectories move towards the nontrivial steady state, but others
move away; therefore x = 1, y = 1 is unstable.
Trajectories seem to move toward the origin, therefore the origin could be
stable.
Note that arrows change direction across a steady state.
Actual Phase Portrait
Classifying Steady States
• Unstable Node (Source)
– Both eigenvalues are real
and positive
 0
 2  4
 0
• Saddle Point (unstable)

– Eigenvalues have opposite


sign
 0
Classifying Steady States
• Unstable Spiral (Source)
– Complex eigenvalues with
positive real part
 0
 2  4
• Neutral Center

– Complex eigenvalues with zero
realpart
 0
 2  4
Classifying Steady States
• Stable Spiral (Sink)
– Complex eigenvalues with
negative real part
 2  4
 0
• Stable Node (Sink)

– Eigenvalues are real and

negative
 0
 0
Summary


 2  4



1,2 
   2  4
2
• The local stability properties of steady states of a nonlinear
system of two equations can be ascertained by

determining b and g and noting the region
of parameter
space in which they lie.
Global Behavior From Local
Information
• For systems of 2 equations, local
stability properties of steady states can
be used to determine global behavior.
• There are a limited number of ways that
trajectories can flow in the phase plane
(due to continuity)
Properties of Trajectories in
the Phase Plane
• Solution curves can only intersect at
steady states
• If a solution curve is a closed loop, it
must enclose at least one steady state.
– That steady state cannot be a saddle point
Asymptotic Behavior of
Trajectories
• Trajectories in the phase plane can
– Approach a steady steady
– Approach infinity
– Approach a closed loop (a limit cycle)
• A trajectory itself may be a closed loop or else it may
approach or recede from one
– Be a heteroclinic trajectory
• Connects two different steady states
– Be a homoclinic trajectory
• Returns to the the same steady state