Download Bifurcations in Ordinary Differential Equations: Identifying and Determining Values - Prof and more Assignments Differential Equations in PDF only on Docsity!
Math 320: Ordinary Differential Equations Section 1.7: Bifurcations February 4, 2003
I. One-Parameter Families of Differential Equations and Bifurcations
What are parameters, and why are they important to the study of differential equations? What models have we looked at that use parameters?
We’ll indicate that a given differential equation is dependent upon a parameter (we’re assuming a single parameter dependence here) by writing:
dy dt
= fμ(t, y)
where μ is the parameter used in the equation.
So, what is a bifurcation? Although parameters are fixed (constant) for each specific case, as you change parameter values to meet the situation being modelled, the solution to the differential equation changes. When a small change in a parameter value leads to a drastic change in the nature and long term behavior of the solution of the differential equation, we say a bifurcation has occurred (or that the differential equation bifurcates).
Here we’ll focus on single parameter, autonomous differential equations, i.e. ODE’s of the form:
II. Identifying Bifurcations in One-Parameter Autonomous ODE’s
Consider the example dy dt = y^2 + 4y + μ
i.e. where fμ(y) = y^2 + 4y + μ. We want to get a quantitative idea of the solutions graph to the ODE using (NOTE: we can do this only because the differential equation is ).
Calculate the equilibrium points and sketch the phase line for μ = -1, 1, 3, 5 and 7, then put the phase lines next to each other in increasing order of μ:
μ = −1 μ = 1 μ = 3 μ = 5 μ = 7
−
2
0
y−vals
Homework: Read Section 1.7: Do p 107, # 1, 3
III. Determining Bifurcation Values From here on out, we will assume that the differential equation dydt = fμ(y) depends smoothly on the parameter, i.e. that the partials of fμ with respect to y and μ exist and are continuous. So, changing the value of μ by only a small amount results in changing the graph of fμ(y) only a little. This will be a key idea.
Suppose you had an parameterized autonomous ODE dydt = fμ(y) with an equilibrium value y 0 when μ = μ 0 , and f (^) μ′ 0 (y 0 ) < 0. The linearization theorem tells us that this this equilibrium point is a sink. What can we say will happen for y-values near this sink, y 0 , if we change the parameter just a little bit? Use a graph to help.
Thus, no bifurcation occurs near y 0. The same would hold true if f (^) μ′ 0 (y 0 ) > 0, so y 0 was a source. In fact, we can now make the following observation:
Bifurcations can only occur at values of μ = μ 0 such that for an equilibrium point y = y 0 ,
- fμ 0 (y 0 ) = 0 (i.e. y 0 is an equilibrium point of dydt = fμ 0 (y)
- f (^) μ′ 0 (y 0 ) = 0 (i.e. the linearization test fails and the graph of fμ 0 (y) is tangent to the y-axis at the point y 0 )
Determine the bifurcation values of dy dt
= fμ(y) = y(1 − y^2 ) + μ
Homework: Read Section 1.7: Do p 107, # 5, 9, 15, 20-
