Solution to Homework #5Roy Malka

1. Questions 2,3,4 of Homework #5 of M. Cross class.

Bifurcations: (M. Cross) Consider the bifurcations of the stationarysolutions of a particle undergoing damped one dimensional motion ina potential V (x) described by the equation of motion

ηx = −dV (x)

dx

Solution:

(a) the potential V1(x) = −rx2 + bx4 with b positive;

ηx = −(−2rx + 4bx3) = x(2r − 4bx2) → ηy = y(r′ − y2)

where y := 2√

bx, and r′ := 2r. Thus we have here a supercriticalpitchfork bifurcation. We have FP at the origin for all values of r′, andthe creation of two FPs at y∗ = ±√r′, for r′ > 0. The stability is clearwhen we notice that DF

∣∣y=0

= (r′− y2)− 2y2∣∣y=0

= r′ thus, the origin

is stable for r′ < 0 and unstable for r′ > 0, thus the two other solutionsthat exist for r′ > 0 are stable.

Geometrically, we can see that the potential is double-well, for positiver′. The minima corresponds to the stable FPs, while the maximumcorresponds to the unstable FP at the origin. For negative r′ there isonly one minima at the origin.

(b) the potential V2(x) = −rx2 + bx3 + cx4 with b and c positive;

Geometrically, we see that the potential has the same qualitative prop-erties (number of extrema and stability) but the symmetry of the stableFPs is broken, in this case it means that transcritical bifurcation occurin addition to the saddle-node (the pitchfork is combined out of tran-scritical - change of stability- and saddle node - creation of a pair ofFPs in the same point):

1

−1 −0.5 0 0.5 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

x

r

Figure 1: Bifurcation diagram of supercritical pitchfork

−1 0 1

−0.3

−0.2

−0.1

0

0.1

0.2

r=0.1

−1 0 1

−0.02

−0.01

0

0.01

0.02

0.03

0.04

0.05

r=−0.1

−2 0 20

0.5

1

1.5

2

2.5

3

3.5r=−0.5

Figure 2: The potential for different r values (V1:solid/blue, V2:dash/red andV3:dash-dot/green)

ηx = −(−2rx + 3bx2 + 4cx3) = x(2r − 3bx − 4cx2).

Thus, x = 0 is a sol. and there are two sol. for 0 < ∆ = 9b2 + 32cr ⇒r > −9b2

32c. While DF

∣∣x=0

= 2r, thus the FP x = 0 still changes sta-bility at r = 0. This reveals that the saddle-node and the transcriticalbifurcations are not at the same point.

Since the saddle-node generates a negative sol. (r = −9b2

32c⇒ x = −3b

8c)

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we get the following bifurcation diagram:

−1 −0.5 0 0.5 1−1.5

−1

−0.5

0

0.5

1

x

r

Figure 3: tanscritical and saddle-node bifurcation but not at the same point

Geometrically, the parabola extremum does not lie on the x = 0 solu-tion.

(c) the potential V3(x) = −rx2 − bx4 + cx6 with b and c positive

This model has both properties of pitchfork bifurcation: x = 0 is al-ways a solution and the model has even symmetry. Locally, for smallneighborhood around the bifurcation point (x, r) = (0, 0) we can ne-glect the last term, thus we get: x(2r+4bx2), which is the normal formof subcritical pitchfork bifurcation (see Fig. 4), thus we should havetwo unstable non-zero FP, for small negative r. However, for large xthe neglected term (x6) is the most dominant, which contributes twoadditional stable FP further from zero.

Analytically:

x = 2x(r + 2bx2 − 3cx4)

we can analyze the right term in the product as a quadratic equationfor y := x2, we get that for (2b)2 + 12rc > 0 ⇒ r > − b2

3cthere are two

solutions y1,2 = b3c± 1

2

√(2b)2 + 12cr which are positive for rc < 0 thus,

(when c assumed to be positive) for r ∈ (− b2

3c, 0) there are 4 non-zero

3

−1 −0.5 0 0.5 1−1.5

−1

−0.5

0

0.5

1

1.5

x

r

Figure 4: pitchfork bifurcation with dual-saddle-node (duality ensues fromsymmetry)

FPs (x1,2 = ±√y1, x3,4 = ±√y2) for r < − b2

3conly the origin is a FP,

and for positive r, two non-zero FPs x1,2 = ±√y1 (when y1 > 0).

Type I Intermittency:(M. Cross)solution:(a) verify that yn+1 = −ε+yn +y2

n has a saddle-node bifurcation point:

FP: −ε + y2 = 0 → y = ±√ε which become a non-hyperbolic pointfor ε = 0 (DF

∣∣±√ε

= 1± 2√

ε)

(b) For samll ε we can use the following continuum eq.

dy

dt= y2 − ε,

We want to estimate the time to pass y = 0 through the ”gap”,but the gap exists only for ε < 0, thus, y2 − ε = y2 + |ε|. Thesolution is

∫dt =

∫dy

y2+|ε| = 1√|ε| tan−1

(y√|ε|

)+ C, for y À

√|ε|,

| tan−1(

y√|ε|

)| ' π2

thus when we go back to the discrete model we

get that n ∝ 1√|ε| .

(c)For a=3.8282 I found the following patterns of period-3-chaotic dy-namics: 12(p3),22(c) 24 (p3) 4 (c) 66 (p3) 56 (c) 30 (p3).

(d)Dependence of ”period-3” residence time on r:We get that after the quasi-period-3 there is window of period-3 and

4

0.505 0.51 0.515 0.52 0.525 0.53 0.535 0.54 0.545 0.55

0.5

0.52

0.54

0.56

0.58

0.6

x(n)

x(n+

4)

logistic map

Figure 5: Type I Intermittency

then there is period-3 doubling (period 6,12) with the increase in a.

2. Construct numerically, by iterating initial conditions and leaving outthe transients, the bifurcation diagram for:

(a) The quadratic map: xn+1 = rxn(1−xn) xn ∈ [0, 1] for r ∈ [0, 4]

(b) The sine map: xn+1 = r sin πxn xn ∈ [0, 1] for r ∈ [0, 1]

(c) Let rn denote the nth period doubling bifurcation in the dou-bling bifurcation sequence. For both maps, find numerically, foras large an n as you can, the ratio: δn = rn−rn−1

rn+1−rn. Can you see

convergence to the Universal Feigenbaum constant δ = 4.669201..?(bonus(15pts): derive more sophisticated ways to find δn).Solution:using the above formula for δn, rn was obstructed from simula-tions, and got δ7 = 4.5 for the logistic map. δ5 = 4.6667 for thesin map. (n is smaller due to the scale of r).

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0 0.5 1 1.5 2 2.5 3 3.5 40

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1logistic map

r

x

Figure 6: Bifurcation of logistic map

Figure 7: Bifurcation diagram of sin map

More sophisticated way will relay on the normalization theory.

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3. Let ΣN consist of all sequences of natural numbers {0, 1, 2, .., N − 1}.Let σ denote the shift map on these sequences. ΣN is a metric space,with the metric d(s, t) =

∑∞i=1

|si−ti|N i

(a) Find CardPerk(σ) : the number of the periodic points of σ ofperiod k.

Solution:By definition of Perk each periodic trajectory has k elements, andeach element is one out of N elements, thus, we have Nk differ-ent k periodic trajectories (including the relevant period-k′, fork′ < k).

(b) Show that σ has a dense orbit.

Solution:Let s∗ ∈ ΣN which consists of all blocks of size 1, all blocks of size2, that is all finite blocks, (N’:=N-1)

s∗ = (012 . . . N ′︸ ︷︷ ︸ |′

00 01 02 . . . 0N ′ 10 11 . . . N ′N︸ ︷︷ ︸ | 000 001 . . . N ′N ′N ′︸ ︷︷ ︸ | . . .︸︷︷︸)

Claim: O = {σk(s∗) : k ∈ N} is a dense orbit.

Proof: Let ε > 0 and arbitrary s ∈ ΣN . ∀ε, by construction of s∗,there exist k ∈ N s.t. d(σk(s∗), s) < ε, therefore the orbit is dense.

(c) Consider the map: xn+1 = 3xn mod 1. Prove that the map ischaotic (hint: use the symbolic dynamics on Σ3). Prove that themiddle-third Cantor set Λ is invariant under the map and thatthe map has a dense orbit on Λ (hint: use the subset of Σ3 ofsequences containing only the symbols {0, 2}).Solution:

The map is chaotic if it satisfy:

i. F has sensitive dependence to initial conditions (SDIC).

ii. F is topological transitive (TP).

iii. F has dense periodic orbit (DP).

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Direct:

Notice that the given map is a ”triple-cover” that is its a map ,with I defined as a union of disjoint sets I = I1 ∪ I2 ∪ I3, s.t.,F (I1) = F (I2) = F (I3) = F (I) (Ii = [(i − 1)/3, i/3) i = 1, 2, 3.).Thus, any U ⊂ I will cover I for some k0, (∃k0 : k ≥ k0F

k(U) =I).

Let F : I → I, for all x ∈ I and all neighborhoods of x: U ⊂I, s.t. x ∈ U , then ∃n s.t. F n(U) = I. Therefore F is sensitiveto initial conditions (SDIC). In addition, since F n(U) = I, for allopen V ⊂ I, F n(U)∩V 6= ∅, thus F is also topologically transitive(TP).

Notice: 3x mod 1 where x ∈ I (unit interval) is conjugate to3θ, θ ∈ S1 (circle).

Let F n(θ) = 3nθ so θ is periodic of period n iff 3nθ = θ + 2πk →θ = 2πk

3n−1for k integer: 0 ≤ k < 3n − 1

Using Symbolic dynamics:

Let x ∈ I, we represent it by a ternary basis, that is, x =∑∞i=1

xi

3i xi ∈ {0, 1, 2}. if x ∈ Ikm = [k−1

3m , k3m ) ↔ xm = k, and

{xk}∞k=1 ∈ Σ3. Using this representation, the map F is conjugateto the shift map over Σ3.

TP: Let O ⊂ Σ3, open, then there exist an open ball Un ⊂ O, s.t.,∀t, s ∈ Un si = ti for i = 1...n (an homeomorphic set of Ik

n in Σ3).Since σn(Uε) = Σ3, σn(Uε) ∩ V 6= ∅ and TP follows.

SDIC: From the same property for each x ∈ U we can find U 3y 6= x s.t. |σn(x)− σn(y)| > δ ≥ 1/3 since y 6= x ⇒ ∃i0 xi0 6= yi0

thus, for |σi0(x)− σi0(y)| ≥ 1/3.

DP: Was proven in previous section for arbitrary N.

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We want to representant the Cantor set by Λ = {{xk}∞k=1|xk ∈{0, 2}}, thus we need the following representation: I1

1 = [0, 13]),

I21 = (1

3, 2

3), and I3

1 = [23, 1], where the rest should follow accord-

ingly1. Now the Cantor set is represented by sequences havingonly {0, 2} alphabet.

Λ = {x = {xi}∞i=1 s.t. xi ∈ {0, 2}}, for all k ∈ Z+, σkx ∈ Λ henceσkΛ = Λ.

Claim: O = {σks∗|k ∈ Z+, s∗ ∈ Λ} ⊂ Λ and dense in Λ.

Proof: Construct s∗ as before, all finite blocks, but using thealphabet {0, 2}. Hence, the trajectory d(σk(s∗), s) < ε for alls ∈ Λ, hence O is dense in Λ.

1This representation is equivalent to the identification of the seq. 0.02222222... =0.10000..., and 0.011111... = 0.200000 and all similar endings, where we choose the versionwithout ”1” in the representation.

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