HIERARCHAL INDUCTIVE PROCESS MODELING AND ANALYSIS ...

That being said let’s continue the analysis of Model C with Detritus.
dD
dt = (
+
)
(1 − a 9 )(a 8 P + a 10 Z 2 )
(
(1 − a 9 )(1 − a 11 )
Using (35) and (39) we get,
(a 12 P 2 )
Z
Z 2 + a 12 a 13 P
} {{ 2
}
Z Grazing Rate
)
− D(a 14 + a 17 )
dD
(
)
dt ≤ (1 − a 9 )(a 8 P 0 e αut + a 10 K 2 ) +
((1 − a 9 )(1 − a 11 ) K )
− D(a 14 + a 17 )
a 13
Then solving for the upper bound,
dD u
dt
+ D u (a 14 + a 17 ) =(1 − a 9 )a 8 P 0 e αut +
(
a 10 K + (1 − a )
11)
(1 − a 9 )K
a 13
Using Proposition 1,
(
a 10 K + (1−a 11)
D u a 13
)(1 − a 9 )K
(t) =
+ (1 − a 9)a 8
P 0 e αut
(a 14 + a 17 ) α
} {{ } u + (a 14 + a 17 )
} {{ }
β u1 β u2
)
(D 0 − β u1 − β u2 P 0 e −(a 14+a 17 )t
where β u1 = 1.721033 and β u2 = 1.7508e −04
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