For each of the following questions, draw the phase portrait as function of the control parameter μ. classify the bifurcations that occur as μ varies, and find all the bifurcation values of u. θ' = μsin θ-sin 2θ

Part a) Fixed points are integral multiples of π.

−2π −π 0 π 2π

Vector Field

Part b) When r > 1, the absolute value of x is always greater than the absolute value of

sin x unless x = 0, which is the only fixed point. The derivative of (rx − sin x) is r − 1 at

point x = 0, since it’s positive x = 0 is unstable.

Vector Field

Part c) As r decreases, the graph of y = rx has more intersections with the graph of

y = sin x, i.e. more fixed points are created. At an intersection point x = c, if (y = sin x)

crosses (y = rx) from the below, then x = c is unstable, and vice versa. When (y = rx)

touches (y = sin x) at a new point, a bifurcation occurs, and after the bifurcation the smaller

fixed point will be unstable.

Notice that 0 is always a fixed point and it changes from unstable to stable as r passes 1.

We conclude that when r decreases from ∞ to 0, there is a subcritical pitchfork bifurcation

at r = 1 and saddle-node bifurcations when 0 < r < 1.

Part d) When r ≪ 1, y = rx touches y = sin x at approximately the peaks of its graph,

i.e. x =

π

2 + 2kπ, where k is a positive integer. Therefore bifurcations occur near r =

2π

4k+1 .

Part e) When r further decreases, two loci of fixed points will merge and vanish, which

is clear if you stare the figure in part c for a while. These are also saddle-node bifurcations,

shown below.

3.6.2 (Page 86)

When h = 0 this system is the same as the one in Section 3.2. As h varies, the curves in

the following pictures move vertically.

The full details about the answer is attached.

Detailed solution of the question is given in the attached files.

Step-by-step explanation: