HIERARCHAL INDUCTIVE PROCESS MODELING AND ANALYSIS ...

For the Zooplankton not to go to zero as t goes to infinity, (a 12 a 13 − a 11 − a 16 )
would have to be greater than zero. This may be a clue to refining the constraints on
the parameter selection process, so that it is strictly positive, insuring a zooplankton
concentration not going to zero for this model structure.
This result is then used to further our analysis by looking at the Phytoplankton
(1) equation as knowing P(t) will help us find bounds for the other entities. The
phytoplankton differential equation like those of Nitrate and Iron is composed of a
minimum function M(t), not often found in differential equations. In order to find
bounds for P(t) we must first find bounds for M(t). Recall,
{
F
M(t) = min
(F + a 5 ) , N
(N + a 4 ) , E P UR (t)
(e− a 2 )(1−e −E P UR (t)(1+a 3 e(E P UR (t)e1.089−2.12log 10 (a 1 ) ) )
a 1 ))
}
M(t) being a minimum function it will always pick the smallest value of the 3
functions stated above, thus using (15) we can safely estimate the range of M(t) to
be:
0 ≤ M(t) ≤ F upperbound
F upperbound + a 5
= a 21
a 21 + a 5
= 0.53262. (18)
Using the lower bound of (16), we are trying to find an upper bound for P(t) since
Z(t) is subtracted we used its small value (i.e. lower bound), and Lemma 2 we have,
[
dP [ ]
dt = (1 − E ice (t))a 0 e (0.06933∗E T H 2 O(t))
M(t)(1 − a 6 ) − a 9 − a 17
]P
(
)
− a 13 (1 − e (−a 14P ) )Z
≤
[ [
(1 − E ice (t))a 0 e (0.06933∗E T H 2 O(t))
]
M(t)(1 − a 6 ) − a 9 − a 17
]P
44