We review a simple derivation of
the onset and bifurcation points the 3-cycle
of the logistic map.
The logistic map is defined as
zn+1 = F(zn) = rzn (1 − zn).
We want to find the r values that permit a stable 3-cycle,
i.e., a solution of zn+3 = zn,
which can sustain a small deviation in the initial z0.
The two boundaries r0 and r1
of the stable window (r0, r1)
are called the onset and bifurcation points, respectively.
By a change of variables
xn = r (zn − 1/2), R = (r2 − 2 r)/4,
(1)
we get a simplified map
xn+1 = f(xn) = R − xn2.
The onset and bifurcation points will be solved
in terms of R, and the corresponding r values
are obtained by (1).
Let a, b and c be the three points in the 3-cycle:
b=f(a), c=f(b), a=f(c).
(2)
We define the cyclic polynomials
X = a+b+c,
Xk = ak+bk+ck (for k = 2,3),
Y = ab+bc+ca, and Z = abc.
The equations of X, Xk, Y, and Z
are more helpful than those of a, b and c
in determining the two points of the 3-cycle.
Onset point.
From (2), we have
c−b = f(b) − f(a) = −(b+a)(b−a);
similar,
a−c = −(c+b)(c−b)=(c+b)(b+a)(b−a).
Now (b−a) + (c−b) + (a−c) = 0 means
1 − (b+a) + (c+b)(b+a) = 0.
Cycling variables
a→ b,
b→ c,
c→ a
twice,
and adding up the three versions yields
3 − 2X + X2 + Y = 0.
(3)
On the other hand, the sum of (2) gives
X = 3R − X2 = 3R − X2 + 2Y (for X2 = X2 + 2 Y).
Using (3) for Y yields
X2 − X + 2 − R = 0,
(4)
This equation has a real root only if 2 − R ≤ 1/4.
Thus, R = 7/4 (so r = 1+√{1+4R} = 1+ √8)
is the onset point of the only real 3-cycle
[1,2,3,4,5].
Bifurcation point.
Let us express cyclic polynomials as linear functions of X.
Using (3) and (4), we get
Y = X − R − 1,
(5)
and
X2 = X2 − 2Y = 3R−X,
(6)
For X3,
we have a3 = Ra − ab from b = f(a),
summing over cyclic versions yields
The composite map f(f(f(x))) is marginally stable
at the onset (bifurcation) point, so
[(d)/(da)]f(f(f(a))) = f′(c)f′(b)f′(a) = +1 (−1)
[1].
As f′(x) = −2x, we get
whose real solution R = 7/4 agrees with the previous result.
At the bifurcation point Z = 1/8, we have
R3 − 2 R2 + 9R/8 − 81/64 = 0.
The only real solution is
R = [2/3] +[1/4] 3√{ [1915/54] + [5/2] √{201}} +[1/4] 3√{ [1915/54] − [5/2] √{201}}
[3,4].
Since a, b and c are the distinct roots of
where we have used (5) and (8),
and eliminated R by (4).
Thus, as long as X is real
[which is true after the onset R ≥ 7/4 by (4)],
we have ∆ > 0, and the cycle points are real.
(Thank Beiye Feng for suggesting this point).