Design of Fermentation Processes
389
(9.102)
This was of course just what was done in Example 6.5 when the two non-linear rates
r,
and
r2
were linearized around a given state (.
x°,s°
).
The solution of the two linear differential equations with constant coefficients is a weighted sum of
exponentials. If both exponentials have negative arguments then the perturbation will die out, and
the steady state will be exponentially approached.
The
eigenvalues!^
of the
Jacobian Matrix
J determine the exponentials of the solution.
Matrix J of dimension (N x N) has N eigenvalues. These are determined as the N zeros of the
polynomial obtained by calculating the determinant of the matrix J - AI.
For our case of N = 2 the determinant is a polynomial of degree 2 in
X.
Det
p-D -X
. ~ YxsM
YX
stisx ~ D ~ !
(9.103)
jis
is defined as j ~ j
and the subscript ss stands for the steady state that is investigated.
p = D
(while
p
must be different from
D
during the transient since
s
is
different from
s0).
With this simplification the following equation is obtained for
X:
X(Yxsp,x + D + X) + YxsxDps =0
x =
-D
~ Y*Ws
(9.104)
There is always one negative eigenvalue = -
D,
and if
ft,
is positive for the steady state the other
eigenvalue is also negative.
For Monod kinetics p increases monotonically with
s,
and
ps
is positive for any steady state.
For substrate inhibition kinetics, Eq (9.28), all steady states to the left of the maximum
of p
(sj at
s = (Kj Ks)'a
have
fts
> 0, while
0 to the right of the maximum
The stability of the steady state is determined by the following
If all eigenvalues have a negative real part then the steady state is
asymptotically stable.