Homework: Dynamical Systems Homework
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Dynamical Systems Homework
Solutions Manual

SOLUTIONS ARE IN BOLD

Questions from Part 2.

Verify that the quadratic family f(x)=x^2+c has fixed points of the form
.5 (1+ sqrt(1-4c)) and .5(1-sqrt(1-4c)).
Fixed points of f are points x such that f(x) = x^2+c = x. This is equivalent to the equation x^2-x+c=0, which has the above roots using the quadratic formula.
For a variety of values of c, compare the graphs of functions in the quadratic family to the identity map. Use this to determine where these functions have fixed points. For each c, how many fixed points are there? Are there the same number of fixed points for every c? Relate your findings to the previous problem.
As c increases, the parabola f_c moves upwards. From the following picture, we can see that for c < 1/4, there are two fixed points, and for c > 1/4, there are no fixed points. This agrees with the fact that the roots we found above are not real for c > 1/4.
Note that at c=1/4 (in green), f is tangent to the identity function.
Use Chaos Lab #2 to graph the second iterate of the quadratic family. To do this, change the value in the Iterate column from 1 to 2. For parameter c=-1, check graphically the number of points of period two. You may wish to simultaneously graph the first and second iterates to see which points are least period two.
For c=-1, there are four points of intersection of the second iterate of f_c (in green) and the identity function. Two of these are fixed points, so there are two points of period two. See the picture below.

Questions from Part 3.
Draw a qualitative diagram for the eventual behavior of all points under the quadratic map for c=-0.4. Your diagram does not need to have exact values of fixed and periodic points. The idea is to compare pictures qualitatively.
The behavior is qualitatively the same as for c=0. Thus the same qualitative picture holds. See picture below.
Draw a similar diagram for c=-1.1. As you investigate, magnify the region near the smaller fixed point. What has happened? What is the period of the attracting set? In some way, indicate the period of all the periodic points. Also remember to indicate what happens to large negative values of x.
For c=-1.1, there is an attracting period two orbit. Although there are still two fixed points, they are now repelling.
Draw a similar diagram for c=-1.3. What is the period of the attracting orbit here? Indicate in which order the points in the orbit occur.
For c=-1.3, a period four orbit is now the attracting set. There are still two fixed points, as well as a period two orbit, but they are all repelling.
What happens for c=-2?
Points -1 and 2 are fixed points. Points less than -2 or greater than 2 go off to infinity. All other points have orbits with no periodicity and are unpredictable in the long term (although stay bounded). This is an example of the behavior commonly known as chaos.
Does every value c give different dynamics? Look at some intermediate values between c=-0.4 and c=-1.1. Is the dynamics at each intermediate value qualitatively distinct from the dynamics you have already studied? If not, what happens?
Dynamics of the quadratic family do not change with every distinct c value; there are regions of c values for which the family has the same dynamics. The dynamics remains the same in the region 0 > c > -0.75 and in the region -0.75 > c > -1.1. The behavior seen for values c=-0.4, -1.1, and -1.3 represents all distinct dynamical behavior occuring for 0 > c > -1.3.
Between c=-0.4 and c=-1.1, at approximately what c value does a bifurcation occur?
From the pictures below, it appears that the fixed point changes from attracting to repelling at the value c=-0.75. In the next section, you verify this analytically.

In the first picture below, c>-0.75, and there is an attracting fixed point.
In the second picture, c<-0.75, and the attracting set is a period two orbit.

Questions from Part 4.
For a map in the quadratic family, what is the slope of its graph at the fixed point?
The derivative of f_c(x) is f_c'(x)=2x. This evaluated at the fixed points gives the slope of the tangent line at each of the fixed points.
At the larger fixed point (1+SquareRoot[1-4c])/2, the slope is 1+SquareRoot[1-4c]
At the smaller fixed point (1-SquareRoot[1-4c])/2, the slope is 1-SquareRoot[1-4c].
Iteration of a linear map with this slope is similar to iteration of the quadratic function near the fixed point. For what values of c should fixed points for the quadratic family be attracting? Repelling? Neutral? Test your values using the Chaos Lab #2.
The larger fixed point is always repelling.
The smaller fixed point is attracting for c > -0.75, repelling for c < -0.75, and neutral for c=-0.75.
This relies on the answer to the previous problem combined with the following facts about a fixed point. A fixed point for a function is:
Explanation:
The linear function g(x)=a x has fixed point 0. g^n(x)= a ^n x, and if |a| < 1, then a^n becomes arbitrarily small as n becomes large. If |a| > 1, a^n becomes arbitrarily large. Thus the fixed point 0 is attracting for |a| < 1 and repelling for |a| > 1.
Since the iteration of a nonlinear function near a fixed point is similar to iteration of a linear function with same slope, we get the above fact.
State the relationship between slope of the graph at the fixed point and whether the fixed point is attracting, repelling, or neutral.
See the answer to the above question.
Recall that in order to find period two points, you graphed the second iterate of the map. Can you find a relationship between the slope of the graph of the second iterate of the function and whether the period two points are attracting, repelling, or neutral?
The slope of the graph of the second iterate at the period two points determines whether the period two points are attracting, repelling, or neutral. If the absolute value of the slope is:
Describe geometrically how your relationship might generalize to period n points.
The slope of the graph of the nth iterate at the period n points determines whether they are attracting, repelling, or neutral. The condition is the same as before.

Questions from Part 5.
In Chaos Lab #3, between -0.4 and -1.1, the orbit diagram curve splits into two branches. How does this relate to what you saw in Chaos Lab #2? Why are there two branches now? What do they correspond to? (Hint: What was the eventual behavior of points in this region?)
The curve splits at the bifurcation point, which you verified previously as -0.75.
For c>-0.75, the stable set is a fixed point. Thus there is only one visible curve for these c values. For c<-0.75, a period two orbit becomes the attracting set. The two branches you see correspond to this attracting period two orbit.
Investigate further bifurcations as c decreases. Describe the sequence of bifurcations. In other words, how does the period of the attracting set change at each bifurcation?
As c decreases, the period two orbit ceases to be attracting, and an attracting period four orbit appears. In turn, this stops being attracting, and an attracting period eight orbit appears. In general, each bifurcation indicates a doubling in the period of the attracting set. For this reason, such a sequence of bifurcations is sometimes referred to as the period doubling sequence.
How would you describe the orbit diagram near c=-2?
As in the iteration diagram, the orbit diagram near c=-2 is a big mess. It displays behavior sometimes referred to as chaos.
General dynamical systems questions
The line x=0 divides the plane in half. Label the half "+" and "-", corresponding to positive and negative x values. For c=-2, pick a point, iterate four times. Write down a series of +'s and -'s corresponding to whether the x-value for the point and the iterates are positive or negative. This ordered list of +'s and -'s is called the itinerary for the point.
Using Chaos Lab #2 for c=-2, this is a matter of observing whether iterates are in the left or right half of the picture.
Find a point which starts in the + region, and with first iterate in the + region, second iterate in the - region. How close do you have to be to this point to stay in these same regions under iteration? Try the same thing for a point with itinerary +,+,-,+. Again, how close do you have to be to this point to have the same itinerary? Try again for the itinerary +,+,-,+,-.
Using Chaos Lab #2 for c=-2, experiment with a variety of starting points, writing down the first four iterates, until you find a point which has the appropriate itinerary. x=1.6 will do. As the length of the itinerary becomes longer, the size of the region of points with the same itinerary decreases. The size decreases by a constant scale factor for each extra + or - in the itinerary.
Based on the previous two questions on itinerary, for a physical system modelled by the quadratic family, make a statement about your ability to predict the future.
Your ability to predict the future will depend on two things:
If there is any error in the initial conditions, in future predictions it will be exponentially compounded. The further into the future you wish to predict, the more iterations involved. Since a model for the physical world has inherent inaccuracies, predicting only a short time into the future may be good, but at some time, predictions will lose all accuracy.
The pattern of bifurcations you have concentrated on here is called the period doubling sequence. Can you see any other patterns of bifurcations? To help think about this, look at the strange way of ordering the integers described in the Geometry Forum article Sharkovskii's Theorem.
You should be able to see a region with an attracting period three orbit. A slightly less obvious region is of period six. There are regions of other periods. From the article above, you know that in the region in which there is an attracting period n, there are points of every period ">" n. (They are repelling so you do not see them.)
The picture below shows the period 3 region magnified.