(Fµ ) (x) =

Fµ (Fµ (x))Fµ (x)3 +2Fµ (Fµ (x))Fµ (x)Fµ (x)+Fµ (Fµ (x))Fµ (x)Fµ (x)+Fµ (Fµ (x))Fµ (x).

At (x, µ) = (0, 0) this simpli¬es to

‚3F ‚ 2F

’ 2 3 (0, 0) + 3 2 (0, 0) . (2.6)

‚x ‚x

Proposition 2.2.2 (Period doubling bifurcation). Suppose that F is C 3 ,

that

‚ 3F ‚2F

d»

(d)F0 (0) = ’1 (e) (0) > 0, and (f) 2 3 (0, 0) + 3 2 (0, 0) > 0.

dµ ‚x ‚x

Then there are non-empty intervals (µ1 , 0) and (0, µ2 ) and > 0 so that

(i) If µ ∈ (µ1 , 0) then Fµ has one repelling ¬xed point and one attracting

orbit of period two in (’ , )

—¦2

If µ ∈ (0, µ2 ) then Fµ has a single ¬xed point in (’ , ) which is in

(ii)

fact an attracting ¬xed point of Fµ .

The statement of the theorem is summarized in Figure 2.7:

Proof. Let

H(x, µ) := F —¦2 (x, µ) ’ x.

Then by the remarks before the proposition, H vanishes at the origin together

with its ¬rst two partial derivatives with respect to x. The expression (2.6)

(which used condition (d)) together with conditions (f) gives

‚3H

(0, 0) < 0.

‚x3

One of the zeros of H corresponds to the ¬xed point, let us factor this out:

De¬ne P (x, µ) by

H(x, µ) = (x ’ x(µ))P (x, µ). (2.7)

Then

‚H ‚P

= P + (x ’ µ)

‚x ‚x

2

‚2P

‚H ‚P

+ (x ’ x(µ)) 2

=2

‚x2 ‚x ‚x

‚3H 2

‚3H

‚P

= 3 2 + (x ’ x(µ)) 3 .

‚x3 ‚x ‚x

41

2.2. LOCAL BIFURCATIONS.

1

0.8

← attracting double point

0.6

0.4

0.2

“ attracting fixed point

repelling fixed point“

0

-0.2

-0.4

-0.6

-0.8

-1

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

µ

Figure 2.7: Period doubling bifurcation.

42 CHAPTER 2. BIFURCATIONS

So P vanishes at the origin together with its ¬rst partial derivative with respect

to x, while

‚3H ‚ 2P

(0, 0) = 3 2 (0, 0)

‚x3 ‚x

so

‚2P

(0, 0) < 0. (2.8)

‚x2

We claim that

‚P

(0, 0) < 0, (2.9)

‚µ

so that we can apply the implicit function theorem to P (x, µ) = 0 to solve for

—¦2

µ as a function of x. This will allow us to determine the ¬xed points of Fµ

which are not ¬xed points of Fµ , i.e. the points of period two. To prove (2.9)

we compute ‚H both from its de¬nition H(x, µ) = F —¦2 (x, µ) ’ x and from (2.7)

‚x

to obtain

‚H ‚F ‚F

(x, µ) ’ 1

= (F (x, µ), µ)

‚x ‚x ‚x

‚P

= P (x, µ) + (x ’ x(µ)) (x, µ).

‚x

‚F

Recall that x(µ) is the ¬xed point of Fµ and that »(µ) = ‚x (x(µ), µ). So

substituting x = x(µ) into the preceding equation gives

»(µ)2 ’ 1 = P (x, µ).

Di¬erentiating with respect to µ and setting µ = 0 gives

‚P

(0, 0) = 2»(0)» (0) = ’2» (0)

‚µ

which is < 0 by (e).

By the implicit function theorem, (2.9) implies that there is a C 2 function

ν(x) de¬ned near zero as the unique solution of P (x, ν(x)) ≡ 0. Recall that P

and its ¬rst derivative with respect to x vanish at (0, 0). We now repeat the

arguments of the last subsection: We have

‚P/‚x

ν (x) = ’

‚P/‚µ

so

ν (0) = 0

and