CHAPTER 3, Period Three Points and Sharkovsky’s Theorem.

3.1 The Converse of Sharkovsky’s Theorem and Tent Maps.

The following function is one of the maps used for the converse of Sharkovsky’s Theorem. It was shown to have a 5-cycle, namely {1,3,4,2,5}, but no 3-cycle. We can check that it has no 3-cycle by graphing the third iterate:

f[x_]:=Piecewise[{{2x+1,1<x<2},{-x+7,2<x<3},{-2x+10,3<x<4},{-x+6,4<x<5}}]

Plot[{f[x],x},{x,0,6},AspectRatio->Automatic]

Plot[{f[f[f[x]]],x},{x,0,6},AspectRatio->Automatic]

Plot[{f[f[x]],x},{x,0,6},AspectRatio->Automatic]

The third iterate has a single fixed point, which must be a fixed point of f(x), and not a period-3 point. The second iterate has three fixed points, two of which must be a 2-cycle. The existence of a 2-cycle follows from Sharkovsky’s Theorem, since there is a 5-cycle (in fact we there exists points of all periods less than 5 in the Sharkovsky ordering).

Another map of interest in this section is the truncated tent map. This map can be used to prove the converse of Sharkovsky’s Theorem. It is defined on [0,1] by (Subscript[T, h])(x)=min{h,T(x)}, where T(x) is the standard tent map. Here is the map for h=2/3.

T[x_]:=Piecewise[{{2x,x<1/3},{2/3,1/3<x<2/3},{2(1-x),2/3<x}}]

Plot[T[x],{x,0,1}, Ticks->None]

Again using the Piecewise command, we can define  and plot (Subscript[T, h])(x) for arbitrary h.

T[x_,h_]:=Piecewise[{{2x,x<h/2},{h,h/2<x<1-h/2},{2(1-x),1-h/2<x}}]

Manipulate[Plot[{x,T[x,h]},{x,0,1},PlotRange->{0,1}],{h,0,1}]

Manipulate[Plot[{x,T[T[x,h],h]},{x,0,1},PlotRange->{0,1}],{h,0,1}]

Notice that (Subscript[T, h])(x) has a single fixed point for h≤2/3, and no 2-cycle. A 2-cycle is created when h>2/3.

Manipulate[Plot[{x,T[T[T[x,h],h],h]},{x,0,1},PlotRange->{0,1}],{h,0,1}]

3.1 Exercises.

1. Use Mathematica to find the values of h for which (Subscript[T, h])(x) has a 4-cycle. Find where the 3-cycle is created.

2. Use Mathematica to check that the map defined piecewise linear, so that f:[1,7]-> [1,7] with f(1)=4, f(2)=7, f(3)=6, f(4)=5, f(5)=3, f(6)=2, f(7)=1 has no 5-cycle, but clearly {1,4,5,3,6,2,7} is a 7-cycle.

3. Use Mathematica to determine if the map defined piecewise linearly, so that f:[1,4]-> [1,4] with f(1)=3, f(2)=1, f(3)=4, f(4)=2 has a 3-cycle, and show that the fixed points are {5/3, 5/2, 10/3 }. What about the map  f:[1,4]-> [1,4] with f(1)=4, f(2)=3, f(3)=1, f(4)=2 (see Exercise 3.2 # 10)?

4. Use a ManipulatePlot  to show that   (Subscript[L, μ])(x)=μ x(1-x), has a 3-cycle when μ=1-√8.

5. Plot f:[-1,1]-> [-1,1] defined by f(x)=x+1, -1≤x≤0, f(x)=cos(x), 0≤x≤1, and find a 3-cycle. If (Subscript[g, a])(x)=a f(x), 0≤a≤1, find the approximate value of a where the 3-cycle first arises.