Math 4/5027 - Nonlinear Dynamics and Chaos
Solution Set #3 - Spring 1999

Problems are 5 points each. Undergraduate total = 20. Graduate total = 35.

Problem T3.2. Setting a=2.5 in the logistic map g, we find that the fixed point tex2html_wrap_inline62 is repelling and the fixed point tex2html_wrap_inline64 is attracting for tex2html_wrap_inline66 . By Theorem 3.4, the (bounded) orbits that originate in tex2html_wrap_inline66 and converge to the periodic orbit tex2html_wrap_inline70 have the same Lyapunov number and exponent as that periodic orbit. The Lyapunov number of this periodic orbit is simply L(0.6)=|g'(0.6)|=-0.5 and the Lyapunov exponent is tex2html_wrap_inline74 . The reason the problem asks about ``most bounded orbits'' is that tex2html_wrap_inline76 , so the Lyapunov exponent is not defined for orbits that include tex2html_wrap_inline78 . Thus for most orbits that originate in tex2html_wrap_inline66 , we have tex2html_wrap_inline82 .

Problem T3.6. From the example in the book for tex2html_wrap_inline84 , we know that tex2html_wrap_inline86 . We now take tex2html_wrap_inline88 and compute

displaymath42

Thus C(x) satisfies the conjugacy condition.

Problem T3.7. It seems I worked most of this problem in class! If x is a fixed point of T, then G(C(x)) = C(T(x)) = C(x), which says that C(x) is a fixed point of G. Similarly, if x is a period-k point of T, then note that tex2html_wrap_inline108 and

displaymath43

Thus C(x) is a fixed point of tex2html_wrap_inline112 or a period-k point of G. It also follows that if x is a genuine period-k point of T (no lower order period), then C(x) is a genuine period-k point of G.

Problem 3.1. The map tex2html_wrap_inline130 has fixed points at the roots of tex2html_wrap_inline132 or at

displaymath44

a. We see that there are two fixed points for tex2html_wrap_inline134 , one fixed point ( tex2html_wrap_inline136 ) for tex2html_wrap_inline138 , and no fixed points for tex2html_wrap_inline140 .

b. By graphical iteration for tex2html_wrap_inline142 , we see that tex2html_wrap_inline144 as tex2html_wrap_inline146 .

c. Analyzing the stability of the two fixed points for tex2html_wrap_inline148 , we find that tex2html_wrap_inline150 and

displaymath45

(This means that tex2html_wrap_inline152 takes the - branch of the square root and vice versa.) The condition that tex2html_wrap_inline156 implies that

displaymath46

This says that tex2html_wrap_inline158 is unstable and tex2html_wrap_inline152 is stable for tex2html_wrap_inline162 .

d. The period-2 points satisfy

displaymath47

It helps to recognize that the fixed point are roots of the polynomial. This leaves the period-2 points as

displaymath48

Not surprisingly, these fixed point exist for tex2html_wrap_inline164 which is where the period-1 points lose stability. Recall that tex2html_wrap_inline166 . A little algebra shows that tex2html_wrap_inline168 . Thus we have stability of the period-2 pair when tex2html_wrap_inline170 or when -1<4(1-a)<1 or when tex2html_wrap_inline174 .

e. For a=2, the fixed points are unstable. The period-2 pair tex2html_wrap_inline178 has tex2html_wrap_inline180 and is unstable. Given the similarity of this map to the logistic map, it should not be surprising that it admits period points of all periods and has chaotic orbits. This will be confirmed in the problem 3.3 below.

Problem 3.2. The circle map tex2html_wrap_inline182 has two natural bins: tex2html_wrap_inline184 and tex2html_wrap_inline186 . A sketch of the map shows that f(L) = [0,1] and f(R)=[0,1]. This means the transition graph is fully connected: both nodes, L and R, are connected to both nodes. This means that period points of all orders exist. Note that as the intervals R and L are split, they maintain their order (unlike the logistic or tent map); that is, at level k an interval of the form tex2html_wrap_inline202 (where tex2html_wrap_inline204 is R or L) splits into tex2html_wrap_inline210 on the left and tex2html_wrap_inline212 on the right. All subintervals on the kth level have length tex2html_wrap_inline216 .

Problem 3.3. The goal is to find a conjugate map between G(x)=4x(1-x) and tex2html_wrap_inline220 . The conjugate map must satisfy G(C(x))=C(g(x)) which leads to the functional equation

displaymath49

A few preliminary observations will help. T maps tex2html_wrap_inline226 onto itself and G maps [0,1] onto itself. Therefore, C must map tex2html_wrap_inline234 (because C works on values of T and G works on values of C). At some point, a bit of trial and error is needed. An obvious place to begin is with a linear conjugate map C(x)=ax+b, where a and b must be determined. Substituting this trial solution into the functional equation above and matching coefficients of 1, x, and tex2html_wrap_inline252 yields three consistent equations for a and b. I find that tex2html_wrap_inline258 and tex2html_wrap_inline260 , so that a candidate conjugate map is C(x) = (x+2)/4. If you reversed the the role of g and G in the relation G(C(x))=C(g(x)), you would have found the inverse of C, or tex2html_wrap_inline272 . It's important to check the properties of C. First, it does satisfy G(C(x))=C(T(x)), and C(-2)=0 and C(2)=1. Note also that T(0) is the maximum of T and tex2html_wrap_inline286 is the maximum of G. Note also that x=1 is a fixed point of T and tex2html_wrap_inline294 is a fixed point of G.

We can appeal to the conjugacy (just as we did for the tent map and the logistic map) to conclude that because G has chaotic orbits g also has chaotic orbits.

Problem 3.9. The interval of interest can be split into three bins tex2html_wrap_inline302 , tex2html_wrap_inline304 , and tex2html_wrap_inline306 . With the given information, the map must look as shown below. We can see that tex2html_wrap_inline308 , tex2html_wrap_inline310 , tex2html_wrap_inline312 , tex2html_wrap_inline314 , tex2html_wrap_inline316 , tex2html_wrap_inline318 , and tex2html_wrap_inline320 . The transition graph is also shown below.

We can find period-1 and period-2 orbits easily. The path tex2html_wrap_inline322 can be chosen to have any length tex2html_wrap_inline324 that we wish. Therefore the map has periodic points of all possible periods.