AMATH/BIOL 382 Problem Set 1
Solutions
1. One can begin by observing that equation two factors as
0 = 2
a
ss
(2

b
ss
)
.
The case
a
ss
= 0 can be ruled out, since this cannot satisfy the first equation. Thus
b
ss
= 2.
Substituting into the first equation gives
0 = 3

4
a
ss

4
a
ss
so
a
ss
= 3
/
8. Substituting into the other equations gives
c
ss
=
2
·
2
·
(3
/
8)
2
.
5
=
3
5
d
ss
=
2
·
2
·
(3
/
8) =
3
2
.
as indicated.
2.
Closed Network
a) The system is described by
d
dt
a
(
t
)
=

k
1
a
(
t
)
b
(
t
)
d
dt
b
(
t
)
=

k
1
a
(
t
)
b
(
t
) +
k
2
d
(
t
)
d
dt
c
(
t
)
=
k
1
a
(
t
)
b
(
t
)

k
3
c
(
t
)
d
dt
d
(
t
)
=
k
1
a
(
t
)
b
(
t
)

k
2
d
(
t
)
d
dt
e
(
t
)
=
k
3
c
(
t
)
d
dt
f
(
t
)
=
k
3
c
(
t
)
b) We note that the following combinations of species each satisfy a conservation:
b
(
t
) +
d
(
t
)
=
T
1
a
(
t
) +
c
(
t
) +
e
(
t
)
=
T
2
e
(
t
)

f
(
t
)
=
T
3
,
for constants
T
1
,
T
2
,
T
3
whose values are set by the initial conditions. These can be verified
by considering flow in the network or by considering combinations of rates of change. (This
set of three conservations is not unique, e.g.
a
(
t
) +
c
(
t
) +
f
(
t
) is also constant.)
1
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Setting
d
(
t
)
=
T
1

b
(
t
)
e
(
t
)
=
T
2

a
(
t
)

c
(
t
)
f
(
t
)
=

T
3
+
e
(
t
)
and substituting into the DEs for the remaining three variables, we have
d
dt
a
(
t
)
=

k
1
a
(
t
)
b
(
t
)
d
dt
b
(
t
)
=

k
1
a
(
t
)
b
(
t
) +
k
2
(
T
1

b
(
t
))
d
dt
c
(
t
)
=
k
1
a
(
t
)
b
(
t
)

k
3
c
(
t
)
.
c) Steady state values of the concentrations must satisfy
0
=

k
1
a
ss
b
ss
0
=

k
1
a
ss
b
ss
+
k
2
(
T
1

b
ss
)
0
=
k
1
a
ss
b
ss

k
3
c
ss
.
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 Spring '10
 Leroy
 Steady State, Constant of integration, London Buses route K3, Daewoo K3, k3*a

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