Bounds for (generalised) Lyapunov exponents for deterministic and random products of shears

Transcription

1 Bounds for (generalised) Lyapunov exponents for deterministic and random products of shears Rob Sturman School of Mathematics University of Leeds Applied & Computational Mathematics seminar, 15 March 2017 University of Edinburgh Joint work with James Springham (Leeds) Jean-Luc Thiffeault (University of Wisconsin-Madison) Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

2 Periodic, deterministic, random Motivation fluid mixing Motivation Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

3 Periodic, deterministic, random Motivation fluid mixing Motivation Basic model is a combination of shears: ( ) ( ) G = & F = Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

4 Periodic, deterministic, random Periodic The Arnold Cat Map compose F and G H = G F (a toral automorphism) ( ) ( ) = ( ) is hyperbolic + Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

5 Periodic, deterministic, random Periodic The Arnold Cat Map compose F and G H = G F (a toral automorphism) ( ) ( ) = Lyapunov exponents ( ) is hyperbolic + 1 lim n n log DHn x v = log(3 + 5)/2 Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

6 Periodic, deterministic, random Periodic The Arnold Cat Map compose F and G H = G F (a toral automorphism) ( ) ( ) = Lyapunov exponents ( ) is hyperbolic + 1 lim n n log DHn x v = log(3 + 5)/2 2 The Cat Map is strong mixing: A, B, lim n µ(hn (A) B) = µ(a)µ(b) Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

7 Periodic, deterministic, random Periodic The Arnold Cat Map compose F and G H = G F (a toral automorphism) ( ) ( ) = Lyapunov exponents ( ) is hyperbolic + 1 lim n n log DHn x v = log(3 + 5)/2 2 The Cat Map is strong mixing: A, B, lim n µ(hn (A) B) = µ(a)µ(b) 3 Exponential decay of correlations (φ H n )ψdµ φdµ ψdµ = O(θ n ) Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

8 Periodic, deterministic, random Deterministic Linked twist maps on the torus Id P = {(x, y) y 1/2}, Q = {(x, y) x 1/2} Id and define F(x, y) = G(x, y) = { (x + 2y, y) if (x, y) P (x, y) { (x, y + 2x) (x, y) if x / P if (x, y) Q if x / Q S And finally H(x, y) = G F(x, y) Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

9 Periodic, deterministic, random Deterministic LTM S Id Id A trajectory always landing in overlap S is uniformly hyperbolic Almost all trajectories land outside S The closer to a boundary, the longer the return to S Arbitrarily long sequences of Fs and Gs But these are rare enough that Lyapunov exponent λ is non-zero No uniform growth = algebraic decay of correlations (polynomial, not exponential, mixing) Computing value of λ is hard Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

10 Periodic, deterministic, random LTM return time partition Deterministic Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

11 Periodic, deterministic, random Random matrix products Random G = ( ) & F = ( ) At each iterate, choose F or G with probability 1/2. Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

12 Periodic, deterministic, random Random Random braiding of particle paths Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

13 Periodic, deterministic, random Tangled magnetic fields Random Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

14 Periodic, deterministic, random Lyapunov exponents Definition of Lyapunov exponents Diffeomorphisms: 1 λ(x, v) = lim n n log D xh n v Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

15 Periodic, deterministic, random Lyapunov exponents Definition of Lyapunov exponents Diffeomorphisms: 1 λ(x, v) = lim n n log D xh n v For products of random matrices M N = N k=1 A i k : 1 λ = lim N N E log M 1 N = lim N N E log X N almost surely, where X N = A in X N 1. Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

16 Periodic, deterministic, random Lyapunov exponents Definition of Lyapunov exponents Diffeomorphisms: 1 λ(x, v) = lim n n log D xh n v For products of random matrices M N = N k=1 A i k : 1 λ = lim N N E log M 1 N = lim N N E log X N almost surely, where X N = A in X N 1. Convergence is given by the celebrated Furstenberg-Kesten theorem (1960). Equivalence of the two definitions is given by the Oseledets multiplicative ergodic theorem (1968) λ is independent of choice of norm Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

17 Periodic, deterministic, random Hard to compute analytically... Lyapunov exponents There is a paucity of exact results concerning Lyapunov exponents for random matrices, as famously lamented by Kingman (1973): Pride of place among the unsolved problems of subadditive ergodic theory must go to the calculation of [the Lyapunov exponent]... In none of the applications described here is there an obvious mechanism for obtaining an exact numerical value, and indeed this usually seems to be a problem of some depth. Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

18 Periodic, deterministic, random Existing work... but easy computationally The submultiplicativity of matrix norms provides the most popular upper bound in the literature : E k = 2 k E {log M 2 k }, (for products drawn from 2 possible matrices) The E k decrease monotonically to λ (for any matrix norm) as k This is easy, if not efficient the number of matrix product calculations required increases exponentially with k Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

19 Periodic, deterministic, random Existing work Many (very good) algorithms and approximations... Mannion, Products of 2 2 random matrices (1993) Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

20 Periodic, deterministic, random Existing work Many (very good) algorithms and approximations... Mannion, Products of 2 2 random matrices (1993) Pollicott, Inventiones mathematicae, 181, 1 (2010): Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

21 Periodic, deterministic, random Existing work Many (very good) algorithms and approximations... Mannion, Products of 2 2 random matrices (1993) Pollicott, Inventiones mathematicae, 181, 1 (2010): Protasov & Jungers, Lower and upper bounds for the largest Lyapunov exponent of matrices (2013) Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

22 Periodic, deterministic, random Our contribution Why is this different? Rigorous, explicit (elementary) bounds (Not algorithmic, no complex functions, no Fourier decomposition) Positive and negative entries in matrices (but see also D. Viswanath, Random Fibonacci sequences and the number Mathematics of Computation of the American Mathematical Society (2000)) Generalised Lyapunov exponents Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

23 Periodic, deterministic, random Generalised Lyapunov exponents Generalised Lyapunov exponents Often want more than knowledge of the infinite time limit of matrix product growth Study deviations from the long-time average Growth rate of the qth moment of the matrix product norm Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

24 Periodic, deterministic, random Generalised Lyapunov exponents Generalised Lyapunov exponents Often want more than knowledge of the infinite time limit of matrix product growth Study deviations from the long-time average Growth rate of the qth moment of the matrix product norm Define the generalized Lyapunov exponents as 1 l(q) = lim N N log E M N q, Note that l (0) is the usual Lyapunov exponent l(1) measures topological entropy (in the sense of growth of material lines) Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

25 Periodic, deterministic, random Generalised Lyapunov exponents Generalised Lyapunov exponents are hard to compute J. Vanneste, Estimating generalized Lyapunov exponents for products of random matrices. Physical Review E 81.3 (2010) For large q, l(q) is controlled by exceedingly rare realisations of the matrix products, and hence it is difficult to estimate reliably using Monte Carlo numerical methods, even with importance sampling. Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

26 Bounding Lyapunov exponents Express the problem differently Group the matrices together W.l.o.g. assume that the first matrix in the product is F Group F s and G s together into J blocks e.g. (FFGGG)(FG)(FGG)(FFFFGG)(FGG)... so the random product is J M J = F a j G b j, a j + b j = n j, j=1 J n j = N, j=1 with 1 a i, b i < n i, so n i 2. Now the a i and b i are the i.i.d. random variables, with identical probability distribution P(x) = 2 x, x 1. Hence, the joint distribution P(a, b) = P(a) P(b) = 2 (a+b). We have the expected values Ea = Eb = 2, so En = 4 Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

27 Bounding Lyapunov exponents Express the problem differently Group the matrices together Recall 1 λ = lim N N E log X N almost surely, where X N = FX N 1 or GX N 1. Then where 1 λ = lim J 4J E log X J X J = F a J G b J X J 1 X J 1 F a J 1G b J 1X J 2 X J 2 F a 1G b 1X 0. X 0 We will compute bounds on each term in the RHS product, by bounding the orientation of the vectors X i, and recalling that the choice of norm is arbitrary. Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

28 Shears Bounding Lyapunov exponents The simple case We take the specific matrices ( ) 1 0 F =, G = 1 1 and let ( ) 1 1, 0 1 ( ) M ab = F a G b 1 b =. a 1 + ab A more general problem takes the matrices F = ( ) 1 0, G = α 1 ( ) ( ) 1 β, M 0 1 ab = F a G b 1 bβ =. aα 1 + aαbβ Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

29 Bounding Lyapunov exponents Invariant cones Simple case (single, positive shears) Take v u/v = 1 F = ( ) 1 0, G = 1 1 ( ) u Lemma The cone C = {0 u/v 1} in tangent space is invariant under M ab = F a G b for all a, b 1. Moreover, this is the minimal such cone. Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

30 Norm bounds I. Bounding Lyapunov exponents Invariant cones Lemma (L -norm) The norm M ab X for a vector X C satisfies the bounds 1 + ab M abx X 1 + a + ab Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

31 Norm bounds I. Bounding Lyapunov exponents Invariant cones Lemma (L -norm) The norm M ab X for a vector X C satisfies the bounds 1 + ab M abx X 1 + a + ab Lemma (L 1 -norm) The norm M ab X 1 for a vector X C satisfies the bounds (a + b + ab) M abx 1 X b + ab Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

32 Norm bounds II. Bounding Lyapunov exponents Invariant cones Lemma (L 2 -norm) The norm M ab X 2 for a vector X C satisfies the bounds M ab X 2 2 X ( 2 + C ab + ) C ab (C ab + 4), where and M ab X 2 2 X 2 2 C ab = (a + b) 2 + a 2 b 2. { (1 + ab) 2 + b 2 min ( (1 + a + ab) 2 + (1 + b) 2) 1 2 Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

33 Putting it together Bounding Lyapunov exponents Invariant cones 1 λ = lim J 4J E log X J 1 = lim J 4J E log F aj G bj X J 1 X J 1 F a 1G b 1X 0 X 0 Using the geometric probability distribution for the a s and b s, we then have, for the L -norm: a,b=1 2 a b log(1 + ab) 4λ a,b=1 and similar expressions for the other two norms. 2 a b log(1 + a + ab) Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

34 Accuracy Accuracy in the simple case Simple case The lowest upper bound (U 2 ) and largest lower bound (L 1 ) differ by about 8%. The true value (from explicit calculation via a standard algorithm) is Norm Lower bound Upper bound L 1 L 1 (1, 1) = U 1 (1, 1) = L 2 L 2 (1, 1) = U 2 (1, 1) = L L (1, 1) = U (1, 1) = Table: Bounds for the maximal Lyapunov exponent for the matrix product in the case α = β = 1. Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

35 Accuracy Positive shears of α and β General shears The invariant cone is now C = {0 u/v 1/α} Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

36 Accuracy Positive shears of α and β General shears The invariant cone is now Theorem C = {0 u/v 1/α} The Lyapunov exponent λ(α, β) for the product M N satisfies max L k(α, β) 4λ(α, β) min U k(α, β) k {1,2, } k {1,2, } where L k (α, β) = U k (α, β) = 2 a b log φ k (a, b, α, β) a,b=1 a,b=1 2 a b log ψ k (a, b, α, β), Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

37 Accuracy General shears φ 1 (a, b, α, β) = 1 + α 1+α (a + bβ + aαbβ) ((1 + aαbβ) 2 + b 2 β 2 ) 1/2 φ 2 (a, b, α, β) = min ( 1 ( 1+α α 2 (1 + a + aαbβ) 2 + (1 + αbβ) 2)) 1/2 { aαbβ φ (a, b, α, β) = min {max(1 + aαbβ, bβ), max(α(1 + a + aαbβ), 1 + ψ 1 (a, b, α, β) = 1 + bβ + aαbβ ( ) 1/2 ψ 2 (a, b, α, β) = 2 + C aαbβ + C aαbβ (C aαbβ + 4) { ψ (a, b, α, β) = (1 + a + aαbβ) α 1 max(1 + aα + aαbβ, 1 + bβ) α < 1 where C aαbβ = (aα + bβ) 2 + (aαbβ) 2. Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

38 Accuracy of the bounds Accuracy General shears λ α = β Linf upper Linf lower L1 upper L1 lower L2 upper L2 lower standard Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

39 Accuracy Asymptotic explicit bounds General shears Corollary The Lyapunov exponent λ(α, β) for the product M N for α, β 1 satisfies K + log αβ 4λ K + log( αβ + 1 ) + 1 log(1 + αβ), αβ 2 where K = 2 a b log ab a,b=1 Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

40 Accuracy of the bounds Accuracy General shears Bounds λ α = β Linf upper L2 upper L1 upper Linf lower L2 lower L1 lower Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

41 Accuracy of the bounds Accuracy General shears Upper bound - Lower bound Linf L1 L2 Combined α = β Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

42 Accuracy Improving the bounds Comparing two independent geometric distributions Lemma When a and b are both drawn from i.i.d. geometric distributions with parameter 1/2, we have P(a = b) = P(a > b) = P(b > a) = 1/3. Proof. We have P(a = b) = P(a = i b = i) = i=1 i=1 2 2i = 1/4 1 1/4 = 1 3. The remaining two equalities follow by symmetry. Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

43 Accuracy Improving the bounds The cone can be made smaller Lemma The cone C = {0 u v 1 α } is mapped into the following cones, in the following cases: 1 When a = b, M ab (C) = {0 u v 1+αβ α(2+αβ) }; 2 when a > b, M ab (C) = {0 u v 1+αβ α(3+2αβ) }; 3 when a < b, M ab (C) = C. and hence expressions like 4λ a,b=1 2 a b log 1 (ψ(a = b) + ψ(a > b) + ψ(a < b)) 3 Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

44 Accuracy Improving the bounds Accuracy of the bounds, improved Upper bound - Lower bound Linf L1 L2 Linf improved L1 improved L2 improved Combined Combined, improved α = β Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

45 Accuracy Improving the bounds Generalised Lyapunov exponents Theorem We have 4l(q, α, β) 4l(q, α, β) max k {1,2, } max k {1,2, } log log with φ k, ψ k and defined as above. a,b=1 a,b=1 2 a b (φ k (a, b, α, β)) q 2 a b (ψ k (a, b, α, β)) q Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

46 Accuracy Generalised Lyapunov exponent Improving the bounds l(q) q Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

47 Accuracy Generalised Lyapunov exponents Generalised Lyapunov exponents When q is a positive integer we can evaluate this easily by expanding the power q. Since ( ) ( ) 2 a b a n b m = 2 a a n 2 b b m a,b=1 a=1 b=1 we require values of the polylogarithm Li q ( 1 2 ), defined by z k Li s (z) = k s. k=1 For integer q = s we have special values 2 a a n = 1, 2, 6, 26, 150, 1082, 9366,... for n = 0, 1, 2, 3, 4, 5, 6,... a=1 and so the L norm, for example, gives... Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

48 Accuracy Generalised Lyapunov exponents Generalised Lyapunov exponents Corollary Generalised Lyapunov exponents in the case α = β = 1 are bounded by: 1 4 log 5 l(1) 1 4 log log 45 l(2) 1 4 log log 797 l(3) 1 4 log log l(4) 1 4 log log l(5) 1 4 log Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

49 Accuracy Switch one of the shears Reversing the twist Take F = ( ) 1 0, G = α 1 ( ) 1 β 0 1 u/v = 1 v Now ( ) M ab = F a G b 1 bβ =. aα 1 aαbβ u is only hyperbolic a, b when αβ > 4. We ll assume α, β > 2 for ease. Lemma The cone C = { 1 u/v 0} in tangent space is invariant under M ab = F a G b for all a, b 1. Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

50 Accuracy of the bounds Accuracy Reversing the twist 0.20 Ũ 1 L Ũ 2 L 2 Ũ L Ũ 1 ˆL Ũ 2 ˆL 2 Ũ ˆL Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

51 Conclusions Conclusions Alternating shears in (trivially) uniformly hyperbolic LTMs are more interesting non-zero Lyapunov exponents We can derive pretty accurate, explicit, upper and lower bounds for Lyapunov exponents for random products of shears of varying strength Generalised Lyapunov exponents can also be bounded (which are also hard to compute) Our results, unusually, include matrices with negative entries Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

52 Conclusions Next?... Linked twist maps also produce a sequence FFGFFGFGGGFFGFGGGF... = J F a j G b j but now the a j, b j are no longer random we don t have a probability distribution. But a block (F a G b ) represents the return map to the hyperbolic overlap regions... and such a block still possess the invariant cone and bounds on the growth of the vector norm... and Kac s lemma states that given a set Q with measure µ(q), the expected return time is 1/µ(Q)... j=1 so this leads us back to the return time partition. Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

53 Next?... Conclusions x 1 = y 1 = 1/2 Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44

54 Next?... Conclusions x 1 = y 1 = 1/2 x 1 = y 1 = 3/4 Rob Sturman (University of Leeds) Mixing for products of shears 15 March / 44