Part II. A study of the quadratic map Q(x) = x^2 + b.
Part II. A study of the quadratic map Q(x) = x^2 + b.
1) Compute the fixed points of Q(x).
Input := Q[x_] = x^2 + b;
Input := Solve[Q[x] == x, x]
Output =
1 - Sqrt[1 - 4 b] 1 + Sqrt[1 - 4 b]
{{x -> -----------------}, {x -> -----------------}}
2 2
Input := fp1 = x /. %[[1]]
Output =
1 - Sqrt[1 - 4 b]
-----------------
2
Input := fp2 = x /. %%[[2]]
Output =
1 + Sqrt[1 - 4 b]
-----------------
2
2) Compute the slope of Q(x) at each fixed point.
Input := s1 = Q'[fp1]
Output = 1 - Sqrt[1 - 4 b]
Input := s2 = Q'[fp2]
Output = 1 + Sqrt[1 - 4 b]
3) Verify that for all -2 <= b <= 1/4, if |x0| > (1 + Sqrt[1 - 4 b])/2 then the orbit of x0 will diverge to positive infinity.
Since Q is an even function, Q(-x) = Q(x). So, any negative input has the same dynamics as the corresponding positive input. Thus, it suffices to consider the orbits corresponding to x0 > (1 + Sqrt[1 - 4 b])/2. Noting that Q'(x) > 1 for all x > fp2, we must have Q(x) > x for all x > fp2. Thus, the orbits diverge to positive infinity.
4) A bifurcation occurs at b = 1/4. Describe the dynamics of ALL orbits (consider all real numbers x0) for b close to and on both sides of 1/4.
b > 1/4, all points diverge to positive infinity.
b = 1/4, if |x0| > 1/2 then diverge to infinity otherwise attracted to the fixed point 1/2 (x0=1/2 is already fixed, and x0=-1/2 is "eventually fixed" after 1 iteration.)
b < 1/4, if |x0| > (1 + Sqrt[1 - 4 b])/2 then diverge to infinity
if x0 is a fixed point, then the orbit is fixed
if x0 is between the fixed points |x0| < (1 + Sqrt[1 - 4 b])/2 the orbit is attracted to fp1.
5) Let p+ > p- be the two fixed points which exist for all b < 1/4. Find the b-values (for b < 1/4) where each fixed point switches stability (if any.)
The fixed point p+ = fp2 doesn't change stability since the slope s2 > 1 for all b < 1/4. However, we see that the fixed point p- = fp1 switches stability when
Input := Solve[s1 == -1, b]
Output =
3
{{b -> -(-)}}
4
Input := b2 = b /. %[[1]]
Output = 3 -(-) 4
6) Graph Q^2(x) = Q(Q(x)) along with x for various values of b between -2 and 1/4. Can you tell for what b-value two period-2 points of Q(x) emerge?
Input := q2[x_] = Q[Q[x]]
Output =
2 2
b + (b + x )
Input := q21 = q2[x] /. b -> 1/4; q22 = q2[x] /. b -> 0; q23 = q2[x] /. b -> -3/4; q24 = q2[x] /. b -> -3/2; q25 = q2[x] /. b -> -2;
Input :=
Plot[{q21, q22, q23, q24, q25, x}, {x, -2, 2},
AspectRatio -> Automatic,
PlotRange -> {-2, 2}]
Output = -Graphics-
Period-2 points appear to emerge as b decreases through b = -3/4.
7) Determine formulas for the period-2 points of Q(x) (i.e., the new fixed points of Q^2(x) which are not fixed points of Q(x).)
Input := per2 = Solve[q2[x]==x, x]
Output =
1 - Sqrt[1 - 4 b] 1 + Sqrt[1 - 4 b]
{{x -> -----------------}, {x -> -----------------},
2 2
-1 - Sqrt[1 - 4 (1 + b)] -1 + Sqrt[1 - 4 (1 + b)]
{x -> ------------------------}, {x -> ------------------------}}
2 2
Input := p21 = x /. per2[[3]] p22 = x /. per2[[4]]
Output =
-1 - Sqrt[1 - 4 (1 + b)]
------------------------
2
Output =
-1 + Sqrt[1 - 4 (1 + b)]
------------------------
2
8) Find the b-value(s) for which the period-2 points switch(s) from attracting to repelling.
Let's first note that both points have the same stability property.
Input := slope = Simplify[q2'[p21]]
Output = 4 + 4 b
Input := Simplify[q2'[p22]]
Output = 4 + 4 b
Now, we want to know when the slope goes through 1 in magnitude. It's clear from the graphs in number 6 that we are particularly interested in when the slope is negative 1.
Input := per4 = Solve[slope == -1, b]
Output =
5
{{b -> -(-)}}
4
Input := b4 = b /. per4[[1]]
Output = 5 -(-) 4
9) Let b=-1.3 and plot Q(x), Q^2(x), Q^3(x), and Q^4(x) along with the replacement line y = x. List how many periodic points Q(x) has of every period less than or equal to 4. Do you think that Q(x) has any higher periodic orbits (when b = -1.3?) Why or why not?
Input :=
b = -1.3;
Plot[{Q[x], q2[x], q2[Q[x]], q2[q2[x]], x}, {x, -2, 2},
AspectRatio -> Automatic,
PlotRange -> {-2, 2}]
Output = -Graphics-
The period-4 map does appear to have unique fixed points which cross the replacement line with a slope of less than one in magnitude. Thus, there must be 4 attracting period-4 points (all on the same orbit), 2 repelling period-2 points (on the same orbit), and two repelling fixed points. There are no periodic points of any other period.