When we substitue (0,0,0,0) into the generalized Jacobian, we get:

The eigenvalues of this matrix are:

We have three cases of possible values of lambda to examine to find the possible values of the first two eigenvalues:

If lambda > 0, then the eigenvalues in question are complex conjugates with a positive real part.

If lambda = 0, then the eigenvalues are pure imaginary.

If lambda < 0, then the eigenvalues are complex conjugates with negative real parts

We have 5 cases of possible values to examine for c1 and the resulting
values of the third and fourth eigenvalues.

When c1 >= 2, the last two eigenvalues are negative and real.

When 0 < c1 < 2, the eigenvalues are complex conjugates with negative real parts

When c1 = 0, the eigenvalues are pure imaginary.

When -2 < c1 < 0, the eigenvalues are complex conjugates with positive real parts.

Fianlly, when c1 <= -2, the eigenvalues in question are negative real numbers.

Thus, if lambda < 0 and c1 > 0, all four eigenvalues have negative
real parts and the origin acts as a sink. If lambda > 0 and c1 <
0, all four eigenvalues have positive real parts and the origin acts
as a source for points close by. If lambda and c1 have the same sign,
then we have a four dimensional saddle with the stable eigenvectors
along the x and y axes (w and z must both equal 0 for the graph to
collapse into the origin) and unstable eigevectors along the w and z
axes. If either lambda or c1 equal 0, then we cannot determine the
stability of the origin since we have a purely imaginary
eigenvalue.

c1 = .75, lambda = -.1

c1 = 1.5, lambda = -.1

c1 = .1, lambda = -.75

c1 = .1, lambda = -1.5

c1 = 1.5, lambda = -1.5

Here, we see that the solutions for these restrictions on lambda and c1 always decay into the origin as time increases. As c1 increases and lambda is fixed, the solution decays more quickly. This makes sense since as c1 increases, the damping of the oscillator is increasing. When lambda increases and c1 is fixed constant, the oscillator also decays faster into a spiral, but it remains in a spiral longer than for smaller values of lambda.

c1 = .75, lambda = .1

c1 = 1.5, lambda = .1

c1 = .1, lambda = .75

c1 = .1, lambda = 1.5

c1 = .1, lambda = 3.0

c1 = .75, lambda = 1.5

c1 = 1.25, lambda = 3.0

In these situations, we notice some interesting behavior. The solution converges to a limit cycle as t increases for all values of c1 and lambda. However, in each case, the limit cycle is different. For small values of lambda, the limit cycle is period one. As lambda increases, the limit cycle becomes larger and more bow-tie shaped. It is interesting to note that every limit cycle is either symmetrical or by applying kappa, we can make a symmetrical figure. For certain values of lambda, the period of the limit cycle increases. For instance, at about lambda = 1.28, the limit cycle becomes period 2. The jump between different period limit cycles becomes shorter as t increases. In this way, the results reminded us of the bifurcation trees so often found in the study of chaos. These trees start out with one branch which splits into two after a while. After a shorter time, these branches split again. This continues until the whole interval seems to be full of branches. (See any good textbook on fractals and chaos for more on bifurcation trees).

As c1 increases, more interesting behavior occurs. For small values of lambda, an increase in the damping constant causes a smaller and more-elliptical limit cycle. For larger values of lambda, an increase in c1 will cause a decrease in the period of the limit cycle.

Since the origin should act as a saddle values of lambda and c1 with
the same size, we should be able to find some initial conditions that
cause the solution to decay to the origin. These initial conditions
are of the form (x,y,0,0).

c1 = .1, lambda = .1, I.C. = (1,1,0,0)

c1 = .1, lambda = .1, I.C. = (1,1,0,.1)

Here we see, that both w(0) and z(0) must equal 0 for the solution to decay to 0, and not just one of them.

c1 = -1, lambda = .1

c1 = -.1, lambda = 1.0

c1 = -.1, lambda = 2.0

In these cases, the solutions appear to be similar to most of the phase portraits for the saddles. Increasing c1 and lambda has the same effect as in the above section.

The key difference between this quadrant in the c1-lambda plane and
the quadrant where lambda and c1 have the same signs is that there is
no stable eigenvectors along the x and y axes.

c1 = -1, lambda = 1, I.C.> = (.0001,.0001,0,0)

Created by Jason Sardell (hyena@wpi.edu) & Thomas Szymkiewicz (mrtom@wpi.edu)