1. Each of the following functions undergoes a bifurcation of fixed points at the given parameter value. In each case, use algebraic or graphical methods to identify this bifurcation as either a saddle-node or period-doubling bifurcation, or neither of these. In each case, sketch the phase portrait for typical parameter values below, at, and above the bifurcation value.

- b. F(x)=x + x^2 + A, A=-1

To find the fixed points of F, solve F(x)= x for x. to get the solutions x=sqrt(-A) and x=-sqrt(-A). In order to determine the behavior of these fixed points, we need to evaluate the equation F'(x)=2x+1 at sqrt(-A) and -sqrt(-A). For A between -1 and 0, it is easy to see that F'(-sqrt(-A)) falls between -1 and 1, which implies -sqrt(-A) is an attracting fixed point. On the other hand, sqrt(-A) is a repelling fixed point for all A less than 0, since F'(sqrt(-A)) is greater than 1. At A=-1, the attracting fixed point -sqrt(-(-1))=-1 becomes neutral, with F'(-1)=2(-1) + 1=-1. This implies that F(x)=x + x^2 +A undergoes a period-doubling bifurcation at A=1. - c. G(x)=Mx+x^3, M=-1

First set G(x)=x and solve for x. The solutions are x=0, x=sqrt(1-M) and x=-sqrt(1-M). For M greater than or equal to 1, there is only one real-valued fixed point at the origin, and there are three for M<1. From the equation G'(x)=M+3x^2, we see that G'(0)=M, which implies that the fixed point at the origin is attracting for M between -1 and 1. Evaluating G'(x) at the other two fixed points, sqrt(1-M) and -sqrt(1-M), yields the same response of 3-2M. This implies that both fixed points are attracting for M between 1 and 2. So, when M=-1, we have G'(0)=-1 and the derivative at the other two points is greater than one, indicating a period doubling bifurcation at the origin with the other two fixed points repelling. - e. S(x)=M sin x, M=1

S(x) has a fixed point at 0 for all values of M. For M between -1 and 1, this is the only fixed point of the function. S'(x)=Mcosx indicates that at the origin we have S'(x)=M. This implies that for values of M between -1 and 1, the origin is repelling. The origin becomes neutral at M=1, and is repelling for values of M greater than 1. At the bifurcation point, the function gives rise to two other fixed points, one less than 0 and one greater than 0. This can easily seen by graphing the function. These two fixed points will be attracting for M>1.

- i. E(x)=L(e^x - 1), L=1

E(x) has a fixed point at the origin for all values of L. Since E'(x)=Le^x, E'(0)=L. Therefore, when the absolute value of L is less than one, the origin is attracting. At the bifurcation point of L=1, the fixed point is neutral. And when the absolute value of L is greater than 1, 0 is repelling. Also, for values of L greater than one, another fixed point appears.

The next three exercises apply to the family Q_c(x) = x^2 +c.

3. Prove that the cycle of period 2 given by q_+ and q_- is attracting for c between -5/4 and -3/4.

First note that (Q_c)'(x)=2x. It can easily be verified that the derivative of the second iterate of Q_c(x) is equal to (Q_c)'(q_+)(Q_c)'(q_-)=4c+4 for each value of c. To find the parameter values for which the two cycle is attracting we must solve for which c values 4c+4 is between -1 and 1. These are simply the values of c between -5/4 and -3/4.

4. Prove that this cycle is neutral for c=-5/4.

From the previous solution, we see that in order to show that the 2-cycle is neutral for c=-5/4 all we have to do is plug c=-5/4 into 4c+4, which yields -1.

5. Prove that this cycle is repelling for c<-5/4.

Again, refering to exercise 3, when c<-5/4, 4c+4<-1, which implies that the 2-cycle is repelling.

Exercises 8-11 deal with the logistic family of functions given by F_L(x)=Lx(1-x).

8. Describe the bifurcation that occurs when L=1.

For L not equal to 0, F_L has two fixed points at 0 and (L-1)/L. Using the fact that (F_L)'(x)=L(1-2x), we have (F_L)'(0)=L, which implies that for L between -1 and 1, the origin is attracting, then becomes neutral at L=1, and repelling for values of L greater than 1. Evaluating the derivative at the other fixed point, (L-1)/L, yields the value of 2-L. This indicates that the fixed point is repelling for L less than 1, neutral at L=1, and attracting for values of L between 1 and 3. Therefore, as L increases through the bifurcation point at L=1 there is an exchange of attractiveness from the origin to the other fixed point at (L-1)/L.

9. Sketch the phase portrait and bifurcation diagram near L=1.

To see a piture of the bifurcation diagram click here.

10. Describe the bifurcation that occurs when L=3.

From exercise 8, we know that the fixed point (L-1)/L is attracting for L between 1 and 3, and repelling for values of L greater then 3. The derivative of F_3 at the fixed point 2/3 is 2-3=-1, indicating a period-doubling bifurcation at L=3.

11. Sketch the phase portrait and bifurcation diagram near L=3.

Just beyond the value of L=3, we can see the begining of the period-doubling process as expected from exercise 10.

To see a piture of the bifurcation diagram click here.

For notation simplification purposes in #12-14 lambda=**L**.

12. Describe the bifurcation that occurs when **L**=-1.

Recall that F has two fixed points one at x=0 and one at
x=(**L**-1)/**L**

This does not change as **L** passes through -1. To see what does change
we'll consider each fixed point separately.

Starting with x=0.

When **L** is slightly greater than -1, |F'(0)|<1 which implies x=0 is
attracting.

When **L** is slightly less than -1, |F'(0)|>1 which implies x=0 is
repelling.

At **L**= -1, F'(0)=-1 which indicates that the fixed point is neutral and
that a period doubling bifurcation occurs.

(ie. There is an attracting
two cycle born at **L**=-1.)

For the other fixed pt, x=(**L**-1)/**L**;

When **L** is slightly greater than -1, |F'((**L**-1)/**L**)|>1
which implies x=(**L**-1)/**L** is repelling.

When **L** is slightly less than -1, |F'((**L**-1)/**L**)|>1 which
implies x=(**L**-1)/**L** is repelling.

And when **L**= -1, |F'((**L**-1)/**L**)|=|F'(2)|=3>1 which
implies x=(**L**-1)/**L** is repelling.

So there is no bifurcation at the fixed point x=(**L**-1)/**L** when
**L**=-1.

13. Sketch the phase portrait and bifurcation diagram near
**L**=-1.

Not here yet. Sorry.

14. Compute and explicit formula for the periodic points of period 2
for F.

First compute the second iterate of F:

F^2(x)

=**L**(**L**x(1-x))(1-**L**x(1-x))

=**L**(**L**x-**L**x^2)(1-**L**x+**L**x^2)

=**L**(**L**x-**L**^2x^2+**L**^2x^3-**L**x^2+**L**^2x^3-**L**^2x^4)

=**L**(**L**x-(**L**+**L**^2)x^2+2**L**^2x^3-**L**^2x^4)

=**L**^2x-**L**^2(1+**L**)x^2+2**L**^3x^3-**L**^3x^4.

To find the fixed points of F^2 (ie. the period 2 points of F), set
F^2(x)=x, rearrange terms , and get

(**L**^2-1)x-**L**^2(1+**L**)x^2+2**L**^3x^3-**L**^3x^4=0 **

This 4th-degree polynomial may be factored into a pair of quadratics.

One of these factors *must* be (**L**-1)x-**L**x^2 since the
fixed points of F are also fixed points of F^2.

Long dividing out this
2nd-degree polynomial
from (**), we find that

[(**L**^2-1)x-**L**^2(1+**L**)x^2+2**L**^3x^3-**L**^3x^4**]
/ [**(**L**-1)x-**L**x^2**]**

=**L**^2x^2-**L**(**L**+1)x+(**L**+1)

and so

[(**L**-1)x-**L**x^2][**L**^2x^2-**L**(**L**+1)x+(**L**+1)
]=0

This new factor may be solved using the quadratic formula and the
simplified result of that is:

x=[(**L**+1) + Sqrt[(**L**+1)(**L**+3)] ] / 2**L**

and

x=[(**L**+1) - Sqrt[(**L**+1)(**L**+3)] ] / 2**L**

Hence, this 2-cycle exist when **L**>3 or **L**<-1 which agrees with the
results of exercises 10 and 12.

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