advertisement

MATH 345: Assignment 6 Solutions Q1 Choose three different initial values x0 = 0, x0 = 1/2 and x0 = 2. If we take these initial conditions √ and apply the map xn+1 = xn , that we get the following results (rounded to 4 decimal places). x1 x2 x3 x0 0 0 0 0 0.5 0.7071 0.8409 0.9170 2 1.4142 1.1892 1.0905 If we were to continue this table we would find that the iterates beginning with x0 6= 0 converge to 1. (i) From our calculator experiment hypothesize that the fixed points are x∗ = 0 and x∗ = 1. To show √ this we must solve the equation: x∗ = x∗ , x∗ ≥ 0. Solving this yields the two solutions x∗ = 0, 1 as expected. (ii) Linear stability analysis: evaluate f 0 (x) = µ = f 0 (0) is undefined 1 µ = f 0 (1) = , |µ| < 1 2 d √ x dx = 1 √ 2 x at the fixed points. ⇒ standard analysis fails here ⇒ x∗ = 1 is a stable fixed point (iii), (iv) Figure 1: Left: Staircase diagram. Right: Phase portrait (v) The basin of attraction for the fixed point x∗ = 1 is (0, ∞). 1 Q2 (a) 26 iterations required. After 32 iterations: x32 = 0.567143281, the last few digits are still changing. (b) 3 iterations. After 5 iterations: x5 = 0.567143290. Result here in (b) should be more accurate since the map is superstable and that after 5 iterations all the digits remain unchanged, indicating that all the digits that show up are accurate. Q3 xn+1 = rxn (1 − xn ) (1) (a) Description of iterations Iterates appear to converge monotonically to a fixed value, x ≈ 0.4736842 Iterates appear to converge in an alternating fashion to a fixed value, x ≈ 0.6428571 Iterates appear to converge to a 2-cycle with points: x ≈ 0.479427 → 0.8236033 → (repeat). Iterates appear to converge to a 4-cycle with points, x ≈ 0.8749973 → 0.3828197 → 0.8269407 → 0.5008842 → (repeat). 3.6 Iterates appear to be choatic; do not appear to converge to anything. 3.835 Iterates appear to converge to a 3-cycle with points: x ≈ 0.9586346 → 0.1520743 → 0.4945144 → (repeat) 4.0 Iterates appear to be choatic; do not appear to converge to anything. r 1.9 2.8 3.3 3.5 (b) To get the fixed point, x∗ , solve the equation x∗ = rx∗ (1 − x∗ ). 1 ∗ ∗ ∗ x ((1 − r) + rx ) = 0 ⇒ x = 0, 1 − r 1 Therefore when r = 1.9, x∗ = 1 − 1.9 ≈ 0.47368421 and when r = 2.8, x∗ = 1 − 0.642857142. There is agreement with the results from XPP and the theory. (c) From the lectures, the Lyapunov exponent for a fixed point x∗ is λ = ln |f 0 (x∗ )|. f (x) = rx(1 − x), ⇒ f 0 (x) = r − 2rx = r(1 − 2x), ⇒ λ = ln |r(1 − 2x)| The Lyapunov exponents are as follows for the fixed points 1 , r = 1.9, x∗ = 1 − 1.9 1 r = 2.8, x∗ = 1 − , 2.8 1 ≈ −2.302585 λ = ln 1.9 1 − 2 1 − 1.9 1 ≈ −0.223144 λ = ln 2.8 1 − 2 1 − 2.8 2 1 2.8 ≈ (d) 1 , XPP computes the Lyapunov exponent to be λ = −2.32187 For r = 1.9, x∗ = 1 − 1.9 1 ∗ For r = 2.8, x = 1 − 2.8 , XPP computes the Lyapunov exponent to be λ = −0.22315 The results from XPP are reasonably accurate, within 1%. (e) For r = 3.3, compare XPP results for stable 2-cycle. {p, q}, where p, q are given by: p r + 1 ± (r − 3)(r + 1) = 0.4794270, 0.8236033 p, q = 2r (This solution is obtained by using the fact that p = f (f (p)) for a 2-cycle, leading to a quartic equation and realizing that x = 0, 1 − 1r are solutions to that equation and can be factored out, since they are fixed points) Lyapunov exponent: For a period 2-cycle, {p, q}, if x0 = p then x1 = q, x2 = p, x3 = q, ... n−1 X ln |f 0 (xi )| = ln |f 0 (p)| + ln |f 0 (q)| + ln |f 0 (p)| + ln |f 0 (q)| + ... i=0 ( = n ln |f 0 (p)| + n2 ln |f 0 (q)| 2 n−1 ln |f 0 (p)| + n−1 ln |f 0 (q)| 2 2 + ln |f 0 (p)| if n is even if n is odd Dividing by n and taking the limit as n → ∞, regardless of whether n is even or odd, or the initial point is x0 = p or x0 = q, we get: n−1 1X 1 1 λ = lim ln |f 0 (xi )| = ln |f 0 (p)| + ln |f 0 (q)| n→∞ n 2 2 i=0 1 1 ln |f 0 (0.4794270)| + ln |f 0 (0.8236033)| 2 2 = −0.6189372 = Using XPP to compute the Lyapunov exponent gives λ = −0.619539, which is reasonably accurate. (f ) A superstable fixed point, x∗ , for the iteration our case we have: 1 0 0 f (x) = f 1 − =r 1−2+ r xn+1 = f (xn ) is one that satisfies f 0 (x∗ ) = 0. In 2 r = 2 − r = 0 iff r = 2 ⇒ x∗ = 1 2 Using XPP, convergence to x∗ = 12 = 0.5000000 when r = 2 does seem somewhat faster than 1 to x∗ = 1 − 1.9 = 0.4736842 when r = 1.9 (e.g. starting a distance 0.1 away from x∗ , convergence takes about 8 iterations when r = 1.9, but only 4 when r = 2.) Using Maple with 20 digits, starting at x0 = x∗ − 0.1, convergence take about 20 iterations when r = 1.9, only 5 when r = 2. 3 (g) For r = 4 and x0 = 0, we have x1 = x2 = x3 = 0, ..., so λ = ln |f 0 (0)| = ln 4. x0 = 0 is not a typical initial value, since λ(x0 = 0) 6= ln 2. x0 = 1 is not a typical initial value, since x1 = 0, x2 = 0, ... x0 = 21 is not a typical initial value, since x1 = 1, x2 = 0, x3 = 0, ... Any orbit that eventually contains xn = 0 for some n is not typical, since all further iterations of that point will be zero and λ = ln 4 6= ln 2 Starting with x0 = 0.1, 0.2, 0.3, XPP computes λ = 0.700871, 0.699009, 0.659690 respectively, all of which are reasonably close to ln 2 ≈ 0.693 Q4 2 (a) The fixed points are obtained by solving x = f (x) √ = x(r − x ). We find immediately that the fixed point xs = 0 exists for all r and xs = ± r − 1 exist for r > 1. (b) The stability is determined by the magnitude of multiplier λ = f 0 (xs ) = r − 3x2s . Since |f 0 (0)| = |r|, the fixed point xs = 0 is table for |r| < 1 or −1 < r < 1. √ For the two fixed points xs = ± r − 1, |f 0 (xs )| = |r − 3(r − 1)| = |3 − 2r|. Thus, they are stable for 1 < r < 2. For r > 2, they are both unstable. (c) The period-2 cycles are roots of x = f 2 (x) = x(r − x2 )[r − x2 (r − x2 )2 ] ⇒ x[r2 − 1 − r(r2 + 1)x2 + 3r2 x4 − 3rx6 + x8 ] = 0. On the other hand, if we expand the lhs of the equation x(x2 −r+1)(x2 −r−1)(x4 −rx2 +1) = 0 we also obtain x[((x2 −r)2 −1)(x4 −rx2 +1)] = (x[r2 −1−r(r2 +1)x2 +3r2 x4 −3rx6 +x8 ] = 0. This is identical to the expression obtained from x = f 2 (x). √ Therefore, the roots that differ from the fixed s = 0, ± r − 1 expressed in pairs of √ points x√ p, q (f (p) = q, f (q) = p) are: (i) {p, q} = { r + 1, − r + 1} thatsexists for r > −1sand bir r r r2 r r2 furcates from the critical point at (xs , rc ) = (0, −1), (ii) {p, q} = { + − 1, − − 1 }, 2 4 2 4 s s r r r r2 r r2 and (iii) {p, q} = { − + − 1, − − − 1 }. (ii) and (iii) exist for r > 2 and 2 4 2 4 both are bifurcated from the bifurcation points (xs rc ) = (±1, 2). They are symmetrical and should have identical stability property (i.e. they become unstable at the same value of r). (d) The stability of the period-2 cycles are determined by the magnitude of multiplier λ = f 0 (p)f 0 (q). The the period cycle (i), λ = f 0 (p)f 0 (q) = (r − 3p2 )(r − 3q 2 ) = (r − 3(r + 1))2 = (2r + 3)2 which is larger than 1 for all r > −1. Therefore, (i) is unstable for the whole domain of 4 values of r in which it exists. The stability for (ii) and (iii) are identical. Thus, we only need r to study one of rthem, say 2 r r r2 r − 1][− − 3 − 1] = (ii). In this case, λ = f 0 (p)f 0 (q) = (r − 3p2 )(r − 3q 2 ) = [− + 3 2 4 2 4 √ r2 r2 − 9( − 1) = −2r2 + 9. Note that |9 − 2r2 | < 1 for 2 < r < 5, (ii) and (iii) are stable. 4 4√ For r > 5, they both become unstable. (e) See Fig. 4e. (Fig. 4e) x 1 r −2 −1 1 2 5 3 −1 Blue = fixed point branches Brown = period 2 cycle branches stable unstable (f) See Fig. 4f. 1 0.5 0 -0.5 -1 -1 -0.5 0 0.5 1 1.5 (Fig. 4f) (g) See Fig. 4g. 5 2 2.5 3 1 0.5 0 -0.5 -1 2 2.2 2.4 2.6 2.8 (Fig. 4g) (h) See Fig. 4h. 0.61 0.6 0.59 0.58 0.57 0.56 0.55 2.836 2.838 2.84 2.842 2.844 (Fig. 4h) (i) Results using x0 = 1.175 and x0 = 3 are given below. 1 1 x_0=3 x_0=1.175 0.5 x 0.5 x 0 -0.5 0 -0.5 -1 -1 -1 -0.5 0 0.5 1 1.5 2 2.5 3 -1 r -0.5 0 0.5 1 r 6 1.5 2 2.5 3 The bifurcation √ in (e) shows that there exists a pair of unstable 2-cycle solution: √ diagram {p, q} = {− r + 1, r + 1} that is unstable for all values of r > −1.√Notice also that the crossing points between f (x) = rx − x3 and y = −x are identically ± r + 1. When r = 3, √ ± r + 1 = ±2 which are exactly the same as the local max and local min values of the function f (x) = rx − x3 at r = 3. √ √ Therefore, for r ∈ [−1, 3] the interval Ir = [− r + 1, r + 1] defines an invariant interval for the map at parameter r in the sense that f (Ir ) ⊆ Ir . While it is safe to use initial points x0 ∈ Ir to guarantee that the iteration does not diverge outside of Ir . For initial conditions outside of Ir , convergence is not guaranteed. For r = 3, this interval is I3 = [−2, 2]. x0 = 3 is outside of this interval. 7