## Solutions to Homework Assignment #3

Page 67: Nos. 1b,c,e,i; 3, 4, 5, 8, 9, 10, 11, 12, 13, 14

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.

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.

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(Lx(1-x))(1-Lx(1-x))
=L(Lx-Lx^2)(1-Lx+Lx^2)
=L(Lx-L^2x^2+L^2x^3-Lx^2+L^2x^3-L^2x^4)
=L(Lx-(L+L^2)x^2+2L^2x^3-L^2x^4)
=L^2x-L^2(1+L)x^2+2L^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+2L^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-Lx^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+2L^3x^3-L^3x^4] / [(L-1)x-Lx^2]
=L^2x^2-L(L+1)x+(L+1)
and so
[(L-1)x-Lx^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)] ] / 2L
and
x=[(L+1) - Sqrt[(L+1)(L+3)] ] / 2L

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