Math 4/5027 - Nonlinear Dynamics and Chaos
Solution Set #2 - Spring 1999

Problems = 5 points each (T2.7 is 10 points). Undergraduate total = 25. Graduate total = 40.

Problem T2.2. If we view the map in components as (f(x,y),g(x,y)), then we must solve

simultaneously for the fixed points. The first fixed point (0,0) is evident; but don't forget the second fixed point (-0.6,-0.6).

*Problem T2.5. To find the period two points, we must solve the equations or

Solving the second equation for y and substituting into the first equation, we have (eventually)

Notice that we have factored out the known period-1 points, leaving a quadratic for the period-2 points. Solving the quadratic, the period-2 points are

The condition for existence of period-2 points is or .

*Problem T2.7. The real goal of this problem is to determine, not just verify, the given intervals in a.

(a) The fixed points of the Henon map are the solutions of the system

Setting y=x in the first equation, the first component satisfies which has roots

The problem considers the case b=0.4, which gives the fixed points where

Notice that the fixed points exist only for . Also note that for all , while for and for a >0. Furthermore, as . The diagram, below and left, shows how the fixed points vary with the parameter a.

Now we need to investigate the stability of these fixed points. The Jacobian is given by

A quick calculation shows that the eigenvalues satisfy which has roots

Some sleuthing is now required. Let's consider first. Note that for all values of . Therefore

For the behavior of , notice that when a=-0.09, we have and . As a increases, decreases, and approaches 0. Therefore, for , and , making the fixed point a saddle for all .

Now consider . It helps to determine for what values of a and we have . A little calculation shows that when a=-0.09, , and . Similarly, when a=0.27, , and . As a increases, increases, and decreases from 1 to 0, while decreases from -0.4 to (see figure above and right). We see that the fixed point changes from a sink to a source as passes through -1, which occurs at a=0.27.

(b) As noted above, when a=0.27, and the eigenvalues of the Jacobian are and 0.4.

(c) We found the period-2 points of the Henon map in T2.5 above. They are the roots of the quadratic

With b=0.4, the period two points are and (recall that x=y for all of the periodic points), where

Clearly, these period-2 points exist only for a>0.27. To determine the stability of these points, we need to compute , the Jacobian of , which is given by . Using the expression for the Jacobian given above, we see that

The eigenvalues satisfy the quadratic equation

which has roots

The discriminant in this expression is negative when or when 0.36<a<0.76. Thus for values of a in this interval, the eigenvalues are complex. Furthermore, for values of a in this interval, the eigenvalues have a constant absolute value of 0.4. So for 0.36<a<0.76, the period-2 pair is a sink. But there is a larger interval in which the period-2 pair is a sink.

(d) As described above, when a=0.85, we have .

Problem 2.1a. To find the eigenvalues, we must solve the equation

The roots are and , which says that (0,0) is a source (because both eigenvalues are greater than 1 in absolute value).

Problem 2.1c. To find the eigenvalues, we must solve the equation

The roots are and , which says that (0,0) is a sink (because both eigenvalues are less than 1 in absolute value).

Problem 2.3. The fixed points satisfy

Solving this system, we find two fixed points, (0,0) and (3,9). The Jacobian calculation gives us

The eigenvalues corresponding to the point (0,0) are , meaning that (0,0) is a saddle. The eigenvalues corresponding to the point (3,9) are , which says (3,9) is a source (disagreeing with the book).

Problem 2.8a. Following the procedure outlined in Theorem 2.24 of the book, we must compute the eigenvalues and eigenvectors of . We find that

The eigenvalues are and , with corresponding eigenvectors (1,1) and (1,-1). Thus the image of the unit disk is an ellipse with a semi-major axis in the (1,1) direction of length and a semi-minor axis in the (1,-1) direction of length . The area of the image ellipse is .

Problem 2.8b. We must compute the eigenvalues and eigenvectors of

The eigenvalues are , with corresponding eigenvectors and . This says that the image of the unit disk is an ellipse with a semi-major axis in the (1,-2) direction of length and a semi-minor axis in the (2,1) direction of length . The area of the image ellipse is .