1. BIOL 364 - Assignment 3
N Li - 20343046
11/01/13
Question 1: Chemostat Analysis:
Given:
dN
dt
= α
C
1 + C
N − N
dC
dt
= −
C
1 + C
N − C + β
Perform appropriate analyses to determine conditions for continual insulin
production.
Analysis
Equilibria
Our equilibrium equations are:
0 = N α
C
1 + C
− 1 [∗], 0 = −
C
1 + C
N − C + β [∗∗] (1)
From [∗], we obtain: N*
= 0, C*
= 1
α−1
From [∗∗], we obtain: N*
= (−C+β)(1+C)
C
, C*
One can show that C*
is the solution to the following quadratic equation:
C2
+ C (N + 1 − β) − β = 0 (2)
1

2. note: For the rest of this paper, we will consider only the + C*
root.
Our equilibria pairs are given by:
N*
= 0, C*
, N*
=
(−C + β) (1 + C)
C
, C*
=
1
α − 1
(3)
Jacobian Matrix
The Jacobian is given as:
J =




df1
dx
df1
dy
df2
dx
df2
dy



 (4)
where:
f1 = α
C
1 + C
N − N, f2 = −
C
1 + C
N − C + β (5)
In our case, the Jacobian is:
J =



α C
1+C − 1 α 1
(1+C)
2 N
− C
1+C −N 1
(1+C)
2 − 1


 (6)
Evaluation
J @ N*
= 0, C*
J =



α C*
1+C* − 1 0
− C*
1+C* −1


 (7)
Since J is lower triangular, the eigenvalues are easily obtained as:
λ1 = α
C*
1 + C*
− 1, λ2 = −1 (8)
2

3. For the first pair of equilibrium points to be stable, we require that:
C*
<
1
α − 1
(9)
note: λ2 is always < 0
J @ N*
, C*
= 1
α−1
J =



0 (α−1)
2
α N*
−1
α − N* (α−1)
2
α2 − 1


 (10)
A direct calculation of the eigenvalues is complicated, so instead we use the
Routh-Hurwitz stability conditions:
Tr(J) < 0 and det(J) > 0
Tr(J) < 0: This condition gives:
−N* (α − 1)2
α2
− 1 < 0 (11)
Which leads to:
N*
>
α2
(α − 1)2 (12)
det(J) > 0: This condition gives:
(α − 1)2
α2
N*
> 0 (13)
Which leads to:
N*
> 0 (14)
It follows naturally that (Tr(J) < 0) ∩ (det(J) > 0) yields:
N*
>
α2
(α − 1)2 , α > 0
3

4. If we substitute the value of N*
into the above relation, we find that an
equivalent expression is:
β >
2α − 1
(α − 1)2 (15)
Interpretation
We’ve calculated the conditions necessary for stability for both eq’m pairs,
but biologically speaking we only care about the second equilibrium pair.
The first pair tells us that N*
= 0 will be stable. Since N is the amount of
insulin produced, the 0 eq’m violates our biological condition.
Conclusion
Given the above reasoning, the required condition for the continual
production of insulin is:
β >
2α − 1
(α − 1)2
In layman’s terms, there has to be enough nutrients flowing into the system
to support production.
Question 2:
Given:
dx
dt
= 1 − (r + 1) x + x2
y
dy
dt
= x (r − xy) , r > 0
(1): To show that ∃ a positive equilibrium, we begin by calculating the
system’s isoclines.
4

5. Table 1: Isocline Combinations
function x nullcline y nullcline
f1 1 − (r + 1) x + x2
y
−1+xr+x
x2
f2 0, r
y
r
x
To find the eq’m points, compute:
−1 + xr + x
x2
= 0 =⇒ x*
=
1
r + 1
(2.1)
Next:
eval
−1 + xr + x
x2
, x*
=
1
r + 1
=⇒ y*
= 0 (2.2)
This gives us the first eq’m point:
1
r + 1
, 0 (2.3)
To find the second eq’m point, compute:
−1 + r
y
r + r
y
r
y
2 = y =⇒ y*
= r (2.4)
This gives us the second eq’m point:
(1, r) (2.5)
note: Convince yourself that (2.3) and (2.5) are the only 2 eq’m points.
From the above, we see that (2.5) is a strictly positive eq’m, (since r > 0).
5

6. Thus, we have proven the existence of a positive equilibrium.
(2): We concern ourselves with only the strictly positive eq’m, then the
Jacobian is given by:
J @ x*
= 1, y*
= r
J =


r − 1 1
−r −1

 (2.6)
We need to determine a bound on r. So, we’ll use the Routh-Hurwitz
conditions.
The reader can verify the following results:
Tr(J) < 0 =⇒ 0 < r < 2
det(J) > 0 =⇒ 1 > 0 , always true
The interval obtained for r will be important in our asymptotic stability
consideration later.
We also need the eigenvalues of the system. They are given by:
λ1 = −1 +
1
2
r +
1
2
√
−4r + r2, λ2 = −1 +
1
2
r −
1
2
√
−4r + r2 (2.7)
To evaluate asymptotic stability, we need to recall the solution to the
system.
Mainly:
If (λ1, v1) and (λ2, v2) are 2 eigen-pairs, then the solution to the system is
given by:
Y = c1eλ1t
v1 + c2eλ2t
v2
With this in mind, we see that there are 3 possible cases to consider:
1. 0 < r < 2
6

7. 2. r = 2
3. r > 2
For illustration purposes, let’s pick a trial ro value s.t. 0 < ro < 2 for case 1.
ro = 1:
λ1 = −1
2
+ 1
2
I
√
3 and λ2 = −1
2
− 1
2
I
√
3
Notice that Re(λ1, λ2) are negative. This implies as t → ∞ the solution
decays and is asymptotically stable. This phenomenon is universal as long
as 0 < r < 2.
Let’s now consider case 2. Again, we pick a trial ro value, but this time
making sure that ro = 2.
ro = 2:
λ1 = I and λ2 = −I
In this case, both eigenvalues are imaginary. This implies as t → ∞ the
solution orbits in a stable trajectory. The system is neutrally stable. This
phenomenon only occurs for r = 2.
Finally, let’s consider case 3. Pick ro = 3 for example.
ro = 3:
λ1 = 1
2
+ 1
2
I
√
3 and λ2 = 1
2
− 1
2
I
√
3
Here, Re(λ1, λ2) are positive. This implies as t → ∞ the solution grows
unbounded. This means that the system is asymptotically unstable. This
phenomenon is universal for r > 2.
(3): Heuristically, we already know that case 2 will generate a Hopf
Bifurcation. Why is this?
Consider the definition of a Hopf Bifurcation.
7

8. Definition: Hopf Bifurcation
The appearance or the disappearance of a periodic orbit through a local
change in the stability properties of a steady point.
With this definition, it becomes clear exactly what we’re looking for. Any
value of r that produces a periodic orbit will generate a Hopf Bifurcation.
Below, the phase portrait of our system is plotted for an r value of 2:
The code which produces this is given:
with(DEtools)
DE1 := diff(x(t), t) = 1 − (2 + 1) ∗ x(t) + x(t)2
∗ y(t);
DE2 := diff(y(t), t) = x(t) ∗ (2 − x(t) ∗ y(t));
phaseportrait([DE1, DE2], [x, y], t = 0..20, [[x(0) = 1, y(0) = 0], [x(0) = 0, y(0) =
2], [x(0) = 0, y(0) = −2], [x(0) = 3, y(0) = 0], [x(0) = 4, y(0) = 0], [x(0) = 5, y(0) = 0]], x =
0..5, y = −3..3, color = aquamarine, linecolor = blue)
8