Solution to Homework #5 Roy Malka. Questions 2,,4 of Homework #5 of M. Cross class. Bifurcations: (M. Cross) Consider the bifurcations of the stationary solutions of a particle undergoing damped one dimensional motion in a potential V () described by the equation of motion dv () ηẋ = d (a) the potential V () = r 2 + b 4 with b positive; ηẋ = ( 2r + 4b ) = (2r 4b 2 ) ηẏ = y(r y 2 ) where y := 2 b, and r := 2r. Thus we have here a supercritical pitchfork bifurcation. We have FP at the origin for all values of r, and the creation of two FPs at y = ± r, for r >. The stability is clear when we notice that DF y= = (r y 2 ) 2y 2 y= = r thus, the origin is stable for r < and unstable for r >, thus the two other solutions that eist for r > are stable. Geometrically, we can see that the potential is double-well, for positive r. The minima corresponds to the stable FPs, while the maimum corresponds to the unstable FP at the origin. For negative r there is only one minima at the origin. (b) the potential V 2 () = r 2 + b + c 4 with b and c positive; Geometrically, we see that the potential has the same qualitative properties (number of etrema and stability) but the symmetry of the stable FPs is broken, in this case it means that transcritical bifurcation occur in addition to the saddle-node (the pitchfork is combined out of transcritical - change of stability- and saddle node - creation of a pair of FPs in the same point):
.8.6.4.2.2.4.6.8.5.5 r Figure : Bifurcation diagram of supercritical pitchfork.2 r=..5 r=..5 r=.5..4. 2.5.2 2...5.2...2.5 2 2 Figure 2: The potential for different r values (V :solid/blue, V 2 :dash/red and V :dash-dot/green) ηẋ = ( 2r + b 2 + 4c ) = (2r b 4c 2 ). Thus, = is a sol. and there are two sol. for < = 9b 2 + 2cr r > 9b2. While DF 2c = = 2r, thus the FP = still changes stability at r =. This reveals that the saddle-node and the transcritical bifurcations are not at the same point. Since the saddle-node generates a negative sol. (r = 9b2 2c = b 8c ) 2
we get the following bifurcation diagram:.5.5.5.5.5 r Figure : tanscritical and saddle-node bifurcation but not at the same point Geometrically, the parabola etremum does not lie on the = solution. (c) the potential V () = r 2 b 4 + c 6 with b and c positive This model has both properties of pitchfork bifurcation: = is always a solution and the model has even symmetry. Locally, for small neighborhood around the bifurcation point (, r) = (, ) we can neglect the last term, thus we get: (2r +4b 2 ), which is the normal form of subcritical pitchfork bifurcation (see Fig. 4), thus we should have two unstable non-zero FP, for small negative r. However, for large the neglected term ( 6 ) is the most dominant, which contributes two additional stable FP further from zero. Analytically: ẋ = 2(r + 2b 2 c 4 ) we can analyze the right term in the product as a quadratic equation for y := 2, we get that for (2b) 2 + 2rc > r > b2 there are two c solutions y,2 = b ± c 2 (2b)2 + 2cr which are positive for rc < thus, (when c assumed to be positive) for r ( b2, ) there are 4 non-zero c
.5.5.5.5.5.5 r Figure 4: pitchfork bifurcation with dual-saddle-node (duality ensues from symmetry) FPs (,2 = ± y,,4 = ± y 2 ) for r < b2 only the origin is a FP, c and for positive r, two non-zero FPs,2 = ± y (when y > ). Type I Intermittency:(M. Cross) solution: (a) verify that y n+ = ɛ+y n +y 2 n has a saddle-node bifurcation point: FP: ɛ + y 2 = y = ± ɛ which become a non-hyperbolic point for ɛ = (DF ± ɛ = ± 2 ɛ) (b) For samll ɛ we can use the following continuum eq. dy dt = y2 ɛ, We want to estimate the time to pass y = through the gap, but the gap eists only for ɛ <, thus, y 2 ɛ = y 2 + ɛ. The solution is dt = tan ( ) y + C, for y ɛ, ɛ ɛ tan ( ) y π ɛ 2 get that n. ɛ dy y 2 + ɛ = thus when we go back to the discrete model we (c)for a=.8282 I found the following patterns of period--chaotic dynamics: 2(p),22(c) 24 (p) 4 (c) 66 (p) 56 (c) (p). (d)dependence of period- residence time on r: We get that after the quasi-period- there is window of period- and 4
logistic map.6.58 (n+4).56.54.52.5.55.5.55.52.525.5.55.54.545.55 (n) Figure 5: Type I Intermittency then there is period- doubling (period 6,2) with the increase in a. 2. Construct numerically, by iterating initial conditions and leaving out the transients, the bifurcation diagram for: (a) The quadratic map: n+ = r n ( n ) n [, ] for r [, 4] (b) The sine map: n+ = r sin π n n [, ] for r [, ] (c) Let r n denote the nth period doubling bifurcation in the doubling bifurcation sequence. For both maps, find numerically, for as large an n as you can, the ratio: δ n = r n r n r n+ r n. Can you see convergence to the Universal Feigenbaum constant δ = 4.6692..? (bonus(5pts): derive more sophisticated ways to find δ n ). using the above formula for δ n, r n was obstructed from simulations, and got δ 7 = 4.5 for the logistic map. δ 5 = 4.6667 for the sin map. (n is smaller due to the scale of r). 5
logistic map.9.8.7.6.5.4..2..5.5 2 r 2.5.5 4 Figure 6: Bifurcation of logistic map Figure 7: Bifurcation diagram of sin map More sophisticated way will relay on the normalization theory. 6
. Let Σ N consist of all sequences of natural numbers {,, 2,.., N }. Let σ denote the shift map on these sequences. Σ N is a metric space, with the metric d(s, t) = i= s i t i N i (a) Find CardP er k (σ) : the number of the periodic points of σ of period k. By definition of P er k each periodic trajectory has k elements, and each element is one out of N elements, thus, we have N k different k periodic trajectories (including the relevant period-k, for k < k). (b) Show that σ has a dense orbit. Let s Σ N which consists of all blocks of size, all blocks of size 2, that is all finite blocks, (N :=N-) s = (2 }{{... N } } 2... N {{... N N} }. {{.. N N N } }{{}... ) Claim: O = {σ k (s ) : k N} is a dense orbit. Proof: Let ɛ > and arbitrary s Σ N. ɛ, by construction of s, there eist k N s.t. d(σ k (s ), s) < ɛ, therefore the orbit is dense. (c) Consider the map: n+ = n mod. Prove that the map is chaotic (hint: use the symbolic dynamics on Σ ). Prove that the middle-third Cantor set Λ is invariant under the map and that the map has a dense orbit on Λ (hint: use the subset of Σ of sequences containing only the symbols {, 2}). 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). 7
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 = I I 2 I, s.t., F (I ) = F (I 2 ) = F (I ) = F (I) (I i = [(i )/, i/) i =, 2,.). Thus, any U I will cover I for some k, ( k : k k F k (U) = I). Let F : I I, for all I and all neighborhoods of : U I, s.t. U, then n s.t. F n (U) = I. Therefore F is sensitive to initial conditions (SDIC). In addition, since F n (U) = I, for all open V I, F n (U) V, thus F is also topologically transitive (TP). Notice: mod where I (unit interval) is conjugate to θ, θ S (circle). Let F n (θ) = n θ so θ is periodic of period n iff n θ = θ + 2πk θ = 2πk n for k integer: k < n Using Symbolic dynamics: Let I, we represent it by a ternary basis, that is, = i i= i i {,, 2}. if Im k = [ k k, ) m m m = k, and { k } k= Σ. Using this representation, the map F is conjugate to the shift map over Σ. TP: Let O Σ, open, then there eist an open ball U n O, s.t., t, s U n s i = t i for i =...n (an homeomorphic set of I k n in Σ ). Since σ n (U ɛ ) = Σ, σ n (U ɛ ) V and TP follows. SDIC: From the same property for each U we can find U y s.t. σ n () σ n (y) > δ / since y i i y i thus, for σ i () σ i (y) /. DP: Was proven in previous section for arbitrary N. 8
We want to representant the Cantor set by Λ = {{ k } k= k {, 2}}, thus we need the following representation: I = [, ]), I 2 = (, 2), and I = [ 2, ], where the rest should follow accordingly. Now the Cantor set is represented by sequences having only {, 2} alphabet. Λ = { = { i } i= s.t. i {, 2}}, for all k Z +, σ k Λ hence σ k Λ = Λ. Claim: O = {σ k s k Z +, s Λ} Λ and dense in Λ. Proof: Construct s as before, all finite blocks, but using the alphabet {, 2}. Hence, the trajectory d(σ k (s ), s) < ɛ for all s Λ, hence O is dense in Λ. This representation is equivalent to the identification of the seq..2222222... =..., and... =.2 and all similar endings, where we choose the version without in the representation. 9