Download WORKSHEET: ONE-PARAMETER BIFURCATIONS Here we discuss

WORKSHEET: ONE-PARAMETER BIFURCATIONS
Here we discuss (briefly) the study of differential equations that depend on a single parameter.
Here are two examples of differential equations with a single parameter µ:
dy
y
dy
= µy,
= µy 1 −
dt
dt
N
Many physical systems depend on a parameter. The question is whether or not the value of the
parameter affects the qualitative behavior of the solution. This can be visualized by creating a
bifurcation diagram. In this worksheet, all differential equations are first order and autonomous
with a single parameter. This type of equation can be described as:
dy
= fµ (y)
dt
0.1. Bifurcation diagrams. We’ll start with an example:
dy
= y2 − y + µ
dt
For any parameter µ, the equilibrium solutions are found by looking at y 2 − y + µ = 0. We can look
at the equilibrium solutions for all parameter values by plotting y 2 − y + µ = 0 on the µy-plane.
Problem: Plot this on the axes below.
2
y
1
K4
K3
K2
K1
x
0
1
K1
Now, any vertical line µ = µ0 , on this plane corresponds to a phase line . Why is that? If you
plot a vertical line, it will intersect the the graph y 2 − y + µ = 0 in several points. The points of
intersection are the equilibrium solutions of y 2 − y + µ0 .
Problem: For example, look at the vertical line µ = −2. Find the intersection points with the
line µ = −2 and y 2 − y + µ = 0. Convince yourself that these really are equilibrium solutions to
y 2 − y + µ0 = 0. Plot the line µ = −2 on the axes above and put arrows on it so that it is the phase
line for this differential equation. Do this also for the lines µ = −4, µ = 0, µ = 1/4, and µ = 1.
When you are done, you will have drawn a bifurcation diagram.
Definition: A bifurcation diagram of the differential equation dy/dt = fµ (y) is the graph of the
equation fµ (y) = 0 in the µy-plane along with several vertical lines µ = µ1 , µ2 , . . . , µn , which are
embellished with arrows so that each line µ = µi becomes a phase line for the autonomous equation
dy/dt = fµi (y).
1
2
WORKSHEET: ONE-PARAMETER BIFURCATIONS
0.2. Bifurcation points. Let’s continue with our example. Imagine moving a vertical line along
the µ axis (i.e. sliding the vertical line to the left or to the right). For µ < 1/4, all of the phase
lines look pretty much the same. In fact, they all have 2 equilibrium solutions. One of the solutions
is a sink and one is a source. The µ = 1/4 differential equation has exactly 1 equilibrium solution,
y = 1/2. For µ > 1/4, there are NO equilibrium solutions. The value µ = 1/4 is called a bifurcation
value of the one parameter equation dy/dt = fµ (y).
Definition: A bifurcation value for a differential equation dy/dt = fµ (y) is a value µ0 for µ where
the phase lines for µ near µ0 differ for µ > µ0 and µ < µ0 .
Problem: Find the bifurcation points for the differential equation with one parameter dy/dt =
y(y − 1)(y − 2) + µ. Hint: Plot z = f (y) = y(y − 1)(y − 2) + µ on the yz-plane. For different values
of µ, f (y) has different roots. The values of µ at which the number of roots are changing must be
the bifurcation points. In the picture below, I have graphed fµ (y) for three different values of µ. On
the second graph, we have a bifurcation diagram of this differential equation (although it doesn’t
have any vertical lines). Use calculus and these graphs to find the bifurcation values. In particular,
you will need to find the z values of the local maximum and local minimum when µ = 0.
2
3
1
y
2
1
K
0
0.5
K
1
K
0.5
1.0
y
1.5
2.0
2.5
K10
K5
0
5
m
10
K1
2
0.3. Meaning of Bifurcations. By definition, a bifurcation point is a value of the parameter at
which the qualitative behavior of the differential equation is changing. Bifurcations can be bad news.
A small change in the value of the parameter can drastically affect the nature of the solution.
Problem: Consider a population of fish in a lake which has logistic growth if otherwise undisturbed. The parameters k > 0 and N > 0 are fixed for the remainder of this problem:
P
dP
= kP 1 −
dt
N
Now suppose that the fish are harvested at a rate C > 0. In other words, fish are removed from the
lake at a constant rate. The differential equation is:
P
dP
= kP 1 −
−C
dt
N
This is a single parameter differential equation. Find the bifurcation point of this one parameter
system and interpret the results in terms of the situation.