Answer:
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.
