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Dynamic simulation of microbial growth (Linearized stability analysis based on eigenvalues.)
Instructor: Nam Sun Wang
Constitutive relations:
μ( )s
0.3 s
50 s
... Monod specific growth rate
Y( )s 0.
0.001 s ... yield coefficient
Dynamic Equations:
dxdt x s D s , , , f
( μ( )s D )x
dsdt x s D s , , , f
D s f
s
μ( )s
Y( )s
x
Steady-states: (calculated by setting d/dt=0 with initial guesses: x 1 s 0
Note that with an initial guess of x=0, computer gives the washout steady-state, which leads to
one positive and one negative eigenvalues. Thus, the washout steady-state is unstable. With an
initial guess of x=1, computer gives the non-washout steady-state. If the non-washout
steady-state is also unstable, we have a limit cycle.
Given dxdt x s D s , , , f
dsdt x s D s , , , f
ss D s , f
Find( x s, )
The 0th answer is the biomass steady-state: x ss
D s , f
ss D s , f
0
The 1st answer is the substrate steady-state: s ss
D s , f
ss D s , f
1
An example: ss( 0.1 200, )=
Form the Jacobian matrix such that dX/dt=AX, where X is the deviation variable. (Evaluate d/dx and
d/ds with |Symbolic|Differentiate on Variable| on dynamic equation.)
A x s D s , , , f
μ( )s D
μ( )s
Y( )s
d
d s
μ( )s x
D
d
d s
μ( )s
Y( )s
x
μ( )s
Y( )s
2
x
d
d s
Y( )s
Find the eigenvalue of matrix A, evaluated at steady-state points
eigenvals
μ( )s D
μ( )s
Y( )s
d
d s
μ( )s x
D
d
d s
μ( )s
Y( )s
x
μ( )s
Y( )s
2
x
d
d s
Y( )s
Evaluate the above expression symbolically gives the following long expressions for eigenvalues:
2 Y( )s
2
μ( )s Y( )s
2
.. 2 D Y( )s
2 d
d s
μ( )s x Y( )s
μ( )s x
d
d s
Y( )s
μ( )s
2
Y( )s
4
... 2 μ( )s Y( )s
3 d
d s
μ( )s x
2 μ( )s
2
2 Y( )s
2
μ( )s Y( )s
2
.. 2 D Y( )s
2 d
d s
μ( )s x Y( )s
μ( )s x
d
d s
Y( )s
μ( )s
2
Y( )s
4
... 2 μ( )s Y( )s
3 d
d s
μ( )s x
2 μ( )s
2
Steady-state stability changes from unstable to stable when the real part (i.e., the part preceeding
the square root sign) crosses 0. ==> stable(D,sf)=0.
stable x s D s , , , f
μ( )s Y( )s
2
.. 2 D Y( )s
2 d
d s
μ( )s x Y( )s
μ( )s x
d
d s
Y( )s
D 0.1 s f
stable_D s f
root stable x , , , , ss
D s , f
s ss
D s , f
D s f
D stable_D( 50 )=0.
stable_sf( D ) root stable x , , , , ss
D s , f
s ss
D s , f
D s f
s f
stable_sf( 0.1 )=2.
Approach to steady-state changes from exponential to oscillatory when the imaginary part (i.e., the
part within the square root sign) crosses 0. ==> oscillation(D,sf)=
oscillation x s D s , , , f
μ( )s
2
Y( )s
4
... 2 μ( )s Y( )s
3 d
d s
μ( )s x
2 μ( )s
2
Y( )s
2
x
d
d s
Y( )s
d
d s
μ( )s
2
x
2
Y( )s
2 . 2
d
d s
μ( )s x
2
Y( )s μ( )s
d
d s
Y( )s
μ( )s
2
x
2 d
d s
Y( )s
2
oscillation_D s f
root oscillation x , , , , ss
D s , f
s ss
D s , f
D s f
D oscillation_D( 200 )=0.
oscillation_sf( D ) root oscillation x , , , , ss
D s , f
s ss
D s , f
D s f
s f
oscillation_sf( 0.1 )=15.
An alternate approach: Eigenvalue:
λ D s , f
eigenvals A x , , , ss
D s , f
s ss
D s , f
D s f
An example: Possitive real part shows that the
non-washout steady-state is unstable, and
the imaginary part shows spiral.
λ( 0.1 200, ) =
0.01839 + 0.21524i
0.01839 0.21524i
s f
200 ... initial guess
stable_sf( D ) root Re λ D s , , f
0
s f
stable_sf( 0.1 )=152.
D 0.07 0.08, ..0.
0 0.1 0.
0
200
stable_sf( D)
D
Y( )s
2
x
d
d s
Y( )s
d
d s
μ( )s
2
x
2
Y( )s
2 . 2
d
d s
μ( )s x
2
Y( )s μ( )s
d
d s
Y( )s
μ( )s
2
x
2 d
d s
Y( )s
2
Y( )s
2
x
d
d s
Y( )s
d
d s
μ( )s
2
x
2
Y( )s
2 . 2
d
d s
μ( )s x
2
Y( )s μ( )s
d
d s
Y( )s
μ( )s
2
x
2 d
d s
Y( )s
2
