Math 4/5027 - Nonlinear Dynamics and Chaos
Solution Set #1 - Spring 1999
Problems are worth 5 points each. Undergraduate total = 35. Graduate total = 55.
Problem T1.5. The fixed points of 
satisfy
f(x)=x and are x=0 and x=3. You can show that 
. The fixed points of 
satisfy 
which is a quartic polynomial. It is essential to note that the fixed
points of are also fixed points of f. Therefore, it is possible
to factor x(x-3) out of , leaving a quadratic polynomial for the
genuine period-2 points. This polynomial is 
, which has
roots, 
. Thus 
is the
period-2 orbit.
Problem T1.6. This problem was done in class, so I'll be brief.
The fixed points of g(x) = ax(1-x) are x=0 and x= (a-1)/a. We
showed in class that x=0 is stable for 0<a<1 and x= (a-1)/a is
stable for 1<a<3. To find the period-2 points, the trick is to factor
these first two fixed points out of the quartic polynomial 
.
This leaves a quadratic,
to be solved for the period-2 points. The roots are
The stability of the period-2 points is determined by
Note this means that 
has the same value at both fixed points.
The condition for stability is
For the values of a of interest, this stability condition reduces to
.
Problem T1.8. We argued rather rigorously in class that the map
G(x)=4x(1-x) has the property that 
has 
points at which
(maximum points) and 
points at which 
(zeros). Therefore has 
fixed points, some of which are of
lower period. Assume for the moment that among the fixed points
are points of all lower periods (which could never happen because
k cannot be divisible by all integers less than k). If we subtract
out all of the periodic points with periods 
, we
subtract out
lower order periodic points, which is (two) less that the total number of period-k points. So there must be points whose minimum period is k. Thus there is a period-k orbit. (This argument can be made tighter by noting that if k is even the largest lower order period is k/2 and if k is odd the largest lower order period is k/3.)
Problem T1.9. (a) The fixed points of G(x)=4x(1-x) are x=0 and
x=3/4. It is easily shown that G'(0)=4 and G'(3/4)=-2, so both
fixed points are sources. The period-2 points of G are 
. Recall the important calculation for the derivative of
:
Clearly, 
, so the period-2 points are also
sources.
(b) Here is Table 1.3 through k=10.
Problem 1.2a. Given that 
, we see that x=0 is the
only fixed point. However f'(0)=1, which means the usual stabilty test
is inconclusive. Therefore, it helps to graph f carefully near x=0.
You will see that f is concave down at x=0 and lies below the
identity line y=x. A few graphical iterations shows that the fixed
point x=0 attracts points with x>0 and repels points with x<0.
Such a point is often called semi-stable.
Problem 1.4.
This problem was meant to be done analytically and is most easily done
this way! With the graphs of G and 
in front of you (page 22), it
is evident that 
and 
. The remaining six points must be
grouped into two period-3 orbits.Note that 
for k=2,4,8
and 
for k=3,5,7. Recall that the points of a period k
orbit must have the same value of . Therefore, 
must be a period-3 orbit and 
must be the other
period-3 orbit.
CORRECTION: All references to G^3 or G^k in the previous solution should be to their derivatives.
Problem 1.8. This counting exercise was (accidentally!) done in class. Briefly, there are no asymptotically periodic points because all periodic points are sources. There is a countable number of periodic points and each periodic points has a countable number of preimages. Therefore (using an argument analogous to the Cantor diagonalization idea), the number of periodic or eventually periodic points is countable. Because the number of points on [0,1] is uncountable, there are points that neither periodic nor eventually periodic.
Problem 1.15. Remember that this is a verification exercise: show
that the given explicit formula works for all 
. Just to warm
up, for n=0, we see that
For the remaining cases, it helps to let 
or
. Therefore for n=1, we have
which is the definition of 
according to the map. We now assume
that the formula is true at the nth step. It's easiest to note that
and do a direct
verification:
which confirms the formula for the n+1 step.
Sleep Model. The second order difference equation that follows from the assumptions is
Notice that if 
(you get less sleep on night n than on
night n-1), then your sleep time goes up on night n+1 compared to
night n, and vice versa. As described in class, trial solutions of the
form 
lead to the characteristic poynomial 
.
The roots are p=1 and p=-a. Using 
and 
as arbitrary
constants, the general solution is
Given two initial values for 
and , the arbitrary constants
can be found to be
Here are the cases of interest:
is constant for all n.
as 
,
and the sequence approaches a steady state value of which is
a weighted average of and . and
.
as .