Homework 9 Solutions
Math 5110/6830
1. Recall the system from class:
dc
dt
dx0
dt
dx2
dt
dp
dt
= k−1 x2 − k1 x0 c2
= k−1 x2 + k2 x2 − k1 x0 c2
= k1 x0 c2 − k−1 x2 − k2 x2
= k2 x2
With the conditions:
d
(x0 + x2 ) = 0
dt
x0 + x2 = r
Then we can reduce the system to (using x0 = r − x2 ):
dc
dt
dx2
dt
= k−1 x2 − k1 (r − x2 )c2
= k1 (r − x2 )c2 − k−1 x2 − k2 x2
Now, using the quasi-steady state assumption:
dx2
dt
x2
Plugging this into
dx2
dt
=0
= k1 (r − x2 )c2 − k−1 x2 − k2 x2 = 0
=
k1 rc2
k1 c2 + k−1 + k2
dc
dt :
dc
dt
k1 rc2
(k−1 + k1 c2 )
− k1 rc2
2
k1 c + k−1 + k2
k−1 + k1 c2
=
−
1
(k1 rc2 )
k1 c2 + k−1 + k2
k−1 + k1 c2 − (k1 c2 + k−1 + k2 )
=
(k1 rc2 )
k1 c2 + k−1 + k2
−k2 k1 rc2
=
k1 c2 + k−1 + k2
=
Let
kmax
kn
= k2 r
k−1 + k2
=
k1
Then,
dc
dt
=
−kmax c2
kn + c2
2. (a) c-nullcline:
dc
dt
=
x1
=
=
−k1 rc + (k−1 + k1 c)x1 = 0
k1 rc
k−1 + k1 c
Kc
a+c
where
k−1
k1
K=r
and
a=
dx1
dt
=
−k1 rc + (k−1 + k2 + k1 c)x1
x1
=
x1 -nullcline:
=
k1 rc
k−1 + k2 + k1 c
Kc
b+c
where
K=r
and
b=
k−1 + k2
k1
Note that b > a.
(b) There is only one point that the nullclines intersect. Phase-plane:
(c) See above graph.
(d) When x1 = 0 and c = c0 :
(e) If the receptors begin begin unbound, then the amount of nutrient will decrease. This can be seen
from the above graph.
3. (a)
dX
= k1 AX − k2 XY
dt
dY
= k2 XY − k3 Y
dt
(b) Equilibria points are (0, 0) and kk32 , kk21 A
J(X ∗ , Y ∗ )
=
k1 A − k2 Y ∗
k2 Y ∗
−k2 X ∗
k2 X ∗ − k3
For (0,0) we have:
J(0, 0)
=
k1 A
0
0
−k3
Since the evals
of this are λ1 = k1 A and λ2 = −k3 , then (0, 0) is a saddle point.
k3 k 1
For k2 , k2 A we have:
J
k3 k1
, A
k2 k2
=
0
−k3
k1 A
0
Evals here are imaginary with zero real parts. Therefore, we’ll get circles so there will be oscillations
around this point.