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