Math 5110
Midterm 3
Problems:
1. We will analyze a simple model of a chemical oscillator. This system is known as the “Brusselator”
and is named after the scientist who proposed it. It is often used as a very simple “proxy” for a
physiological calcium oscillator in cellular dynamics. In dimensionless form, the equations are
dx
= 1 − (b + 1)x + x2 y,
dt
dy
= bx − x2 y.
dt
(1)
Here, x and y are the dimensionless concentrations of two chemical species. The single parameter b
is assumed to be non-negative.
(a) Find all equilibria of the system.
(b) Sketch the null clines and then use this information to help construct a “trapping region” in
the phase plane. Make sure you also use the equations to justify that your region is a trapping
region.
(c) Now determine the stability of all fixed points that you found. Identify any parameter values
at which a bifurcation occurs.
(d) Identify the types of any bifurcations that you found in the previous section. How can you tell?
(e) Putting all this information together, what are all possible behaviors of the Brusselator that
may be observed depending on b?
Solution:
1.
(a) First, we will find the equilibria. Setting dy/dt = 0, we immediately see that 0 = bx − x2 y.
Substituting this into the equation for the x-cline, we have 1 − x = 0. This gives us x∗ = 1.
Now, we substitute this back into the equation for the y-cline to get 0 = b − y, or y ∗ = b. This
is the only equilibrium of the system.
(b) The information requested by the problem is all contained in Figure 1. The phase plane was
generated for b = 0.5, but qualitatively is similar for other values. The x-cline is shown in
pink, and given by the equation (1 + b)x − x2 y = 1. The orange lines are the y-clines and are
given by the equations x = 0 and y = b/x.
We construct the trapping region out of the 5 line segments shown in black. Line 1 is vertical
and lies entirely to the left/above the pink x-cline. This means that everywhere on line 1, dx/dt
is positive. Therefore the vector field points into the region on this line. Line 2 is perfectly
horizontal and lies entirely to the right/above the orange y-cline given by y = b/x. In this
region, dy/dt is negative, and therefore points into the trapping region everywhere on line 2.
Line 3 is also horizontal and lies entirely to the right/below the same y-cline. Therefore dy/dt
is positive here, and thus the vector field points into the region. Line 4 is perfectly vertical and
lies to the right/below the pink x-cline, therefore dx/dt is negative and the vector field points
into the region along this line.
Finally, we consider line 5. In the figure, it is not drawn to scale. However, consider a line
which lies in the region where dx/dt is positive (left/above the pink cline) and dy/dt is negative
(above/right of the orange y = b/x cline). Suppose that this line has slope −1. We will show
that along this line segment, dy/dt has greater magnitude than dx/dt. This will imply that the
vector field points down and to the right with a slope less than −1 and thus into the region.
Consider the quantity
bx − x2 y
dy/dt
=
.
dx/dt
1 − x − bx + x2 y
Because of where we chose to draw line 5 relative to the y-cline we know the quantity bx − x2 y
is negative, and thus −bx + x2 y > 0 . As long as line 2 is long enough and meets line 5 at
a point to the right of x = 1, we know that 1 − x < 0. Letting −bx + x2 y = a > 0 and
x − 1 = b > 0, we have
−a
dy/dt
=
.
dx/dt
a−b
Again, because of how we chose to draw line 5 relative to the x cline, we know the denominator
is positive, and therefore a > b. This implies that
dy/dt
−a
=
< −1.
dx/dt
a−b
This implies that the vector field points downward steeper than line 5 (which has a slope of
−1), and thus into the region.
2
5
1
3
4
Figure 1: Phase plane of the Brusselator with b = 0.5. The x-cline is shown in pink, while the y-cline is
shown in orange. The trapping region is indicated with black and is composed of 5 straight line segments.
(c) To evaluate the stability of the fixed point, we first calculate the Jacobian:
−(b + 1) + 2xy x2
J=
.
b − 2xy
−x2
We then evaluate the Jacobian at the only fixed point (x∗ , y ∗ ) = (1, b):
−(b + 1) + 2b 1
b−1 1
J(1, b) =
=
.
b − 2b
−1
−b −1
A quick calculation shows that the trace of the Jacobian is τ = b − 2 and the determinant is
∆ = 1. Because the determinant is positive, the stability of the fixed point is totally given by
the trace. The fixed point is stable if τ < 0, which is equivalent to b < 2. It is unstable if
τ > 0, which is equivalent to b > 2. This means that a bifurcation must occur at b = 2.
(d) Because the only equilibrium in the model switched stability without creating or destroying
any new ones, we have a strong suspicion that a Hopf bifurcation occurred. To show that this
is true, consider the trace and determinant of the Jacobian “near” the bifurcation (i.e. when
b ≈ 2. In this situation τ ≈ 0, but ∆ = 1 because ∆ does not depend on the parameter b. This
means that the eigenvalues of the Jacobian are
√
√
0± 0−4
τ ± τ 2 − 4∆
λ=
≈
= ±i.
2
2
Clearly, at the bifurcation value b = 2, we have a complex pair of eigenvalues that transition
from negative real part to positive real part (they cross the imaginary axis of the complex
plane). This is the definition of a Hopf bifurcation.
(e) Now, we can use the trapping region from before to determine what happens when the equilibrium is unstable. When b > 2, we can remove a small circle around the fixed point from our
trapping region. Because it is an unstable fixed point, we know that the vector field must point
away from the equilibrium and thus into our trapping region. Because our newly modified
trapping region contains no equilibria, it must contain a stable limit cycle.
This completes our characterization of the system:
• When b < 2, there is a single fixed point and it is stable. We expect all trajectories of
the system to approach the concentrations x∗ = 1 and y ∗ = b. Figure 2 demonstrates this
behavior.
• When b > 2, there is still a single fixed point at x∗ = 1 and y ∗ = b, but it is unstable.
The fixed point lies within a stable limit cycle. Because it is the only stable structure in
the system, we expect all solutions to approach this limit cycle and oscillate behavior.
Figure 3 demonstrate this behavior.
x ’ = 1 − (1 + b) x + x2 y
y ’ = b x − x2 y
b = 1.5
x ’ = 1 − (1 + b) x + x2 y
y ’ = b x − x2 y
4
3.5
4
3
3.5
x
y
3
2
x and y
y
2.5
b = 1.5
1.5
2.5
2
1
1.5
0.5
1
0
0
0.5
1
1.5
2
x
2.5
3
3.5
4
0.5
0
2
4
6
8
10
12
14
t
Figure 2: Phase plane of the Brusselator with b = 1.5. The left pane shows the phase plane and a single
trajectory which settles to the stable equilibrium. The left panel shows the x and y concentrations over
time settling down the the steady state concentrations x∗ = 1 and y ∗ = b = 1.5.
2
x ’ = 1 − (1 + b) x + x y
y ’ = b x − x2 y
b = 2.5
2
x ’ = 1 − (1 + b) x + x y
y ’ = b x − x2 y
4
b = 2.5
x
y
3.5
3.5
3
3
2.5
2
x and y
y
2.5
1.5
2
1
1.5
0.5
1
0
0
0.5
1
1.5
2
x
2.5
3
3.5
4
0.5
0
5
10
15
20
t
25
30
35
Figure 3: Phase plane of the Brusselator with b = 2.5. The left pane shows the phase plane and a single
trajectory which starts near the untable equilibrium. The left panel shows the x and y concentrations over
time oscillating around the steady state concentrations x∗ = 1 and y ∗ = b = 2.5.