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

Transcription

1 Bounds for (generalised) Lyapunov exponents for random products of shears Rob Sturman Department of Mathematics University of Leeds LAND Seminar, 13 June 2016 Leeds Joint work with Jean-Luc Thiffeault (Madison-Wisconsin)

2 Motivation Fluid Mixing Fluid mixing is essentially shearing in two transverse directions, usually periodically.

3 Motivation Random braiding of particle paths

4 Motivation Tangled magnetic fields

5 Lyapunov exponents Definition Diffeomorphisms: λ(x, v) = lim n 1 n log D xf n v

6 Lyapunov exponents Definition Diffeomorphisms: 1 λ(x, v) = lim n n log D xf 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 = M N X N 1.

7 Lyapunov exponents Definition Diffeomorphisms: 1 λ(x, v) = lim n n log D xf 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 = M N 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

8 Lyapunov exponents Hard to compute... 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.

9 Existing work A standard bound The submultiplicativity of matrix norms provides the most popular upper bound in the literature : E k = 2 k E {log M 2 k }, 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

10 Existing work Many (very good) algorithms and approximations... Mannion, Products of 2 2 random matrices (1993)

11 Existing work Many (very good) algorithms and approximations... Mannion, Products of 2 2 random matrices (1993) Pollicott, Inventiones mathematicae, 181, 1 (2010):

12 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)

13 Our contribution Why is this different? Rigorous, explicit (elementary) bounds Positive and negative entries in matrices Generalised Lyapunov exponents

14 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

15 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 J 4J log E M J q, Note that l (0) is the usual Lyapunov exponent l(1) measures topological entropy (in the sense of growth of material lines)

16 Generalised Lyapunov exponents Generalised Lyapunov exponents are hard Vanneste, Estimating generalised Lyapunov exponents for products of random matrices: Monte Carlo method; numerical solution to an eigenvalue problem; asymptotic results

17 Express the problem differently Group the matrices together W.l.o.g. assume that the first matrix in the product is A 1 = A (and we write A 2 = B). Group A s and B s together into J blocks the random product is J M J = A a j B 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

18 Express the problem differently Group the matrices together Recall 1 λ = lim N N E log X N almost surely, where X N = M N X N 1. Then where 1 λ = lim J 4J E log X J X J = M a J b J X J 1 X J 1 M aj 1 b J 1 X J 2 X J 2 M a 1 b 1 X 0. X 0 We will compute bounds on each term in the RHS product, by bounding the orientation on the vectors X i, and recalling that the choice of norm is arbitrary.

19 The simple case Shears We take the specific matrices ( ) 1 0 A =, B = 1 1 and let ( ) 1 1, 0 1 ( ) M ab = A a B b 1 b =. a 1 + ab A more general problem takes the matrices A = ( ) 1 0, B = α 1 ( ) ( ) 1 β, M 0 1 ab = A a B b 1 bβ = aα 1 + aαbβ

20 Invariant cones Simple case (single, positive shears) Take v u/v = 1 A = ( ) 1 1, B = 0 1 ( ) u Lemma The cone C = {0 u/v 1} in tangent space is invariant under M ab = A a B b for all a, b 1. Moreover, this is the minimal such cone.

21 Invariant cones Norm bounds I. Lemma (L -norm) The norm M ab X for a vector X C satisfies the bounds 1 + ab M abx X 1 + a + ab

22 Invariant cones Norm bounds I. 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

23 Invariant cones Norm bounds II. Lemma (L 2 -norm) The norm M ab X 2 for a vector X C satisfies the bounds M ab X ( 2 1 X C ab + ) C ab (C ab + 4), 2 where and M ab X 2 X 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

24 Invariant cones Putting it together 1 λ = lim J 4J E log X J 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)

25 Simple case Accuracy in the 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.

26 General shears Positive shears of α and β The invariant cone is now C = {0 u/v 1/α}

27 General shears Positive shears of α and β 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 2 a b log ψ k (a, b, α, β), a,b=1

28 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/ { aαbβ φ (a, b, α, β) = min {max(1 + aαbβ, bβ), max(α(1 + a + aαbβ), 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.

29 General shears Accuracy of the bounds λ Linf upper Linf lower L1 upper L1 lower L2 upper L2 lower standard α = β

30 General shears Accuracy of the bounds Bounds λ Linf upper L2 upper L1 upper Linf lower L2 lower L1 lower α = β

31 General shears Accuracy of the bounds Upper bound - Lower bound Linf L1 L2 Combined α = β

32 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.

33 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.

34 Improving the bounds 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.

35 Improving the bounds Accuracy of the bounds Upper bound - Lower bound Linf L1 L2 Linf improved L1 improved L2 improved Combined Combined, improved α = β

36 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 a,b=1 a,b=1 with φ k, ψ k and defined as above. 2 a b (φ k (a, b, α, β)) q 2 a b (ψ k (a, b, α, β)) q

37 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...

38 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, 1) 1 4 log log 45 l(2, 1, 1) 1 4 log log 797 l(3, 1, 1) 1 4 log log l(4, 1, 1) 1 4 log log l(5, 1, 1) 1 4 log

39 Generalised Lyapunov exponents Generalised Lyapunov exponent l(q) q

40 Reversing the twist Switch one of the shears Take A = ( ) 1 α, B = 0 1 ( ) 1 0 β 1 u/v = 1 v Now ( ) M ab = A a B b 1 bβ =. aα 1 aαbβ u is only hyperbolic 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 = A a B b for all a, b 1.

41 Accuracy Accuracy of the bounds λ Linf upper Linf lower L1 upper L1 lower L2 upper L2 lower α = β

42 Accuracy Accuracy of the bounds λ Linf upper Linf lower Linf lower improved L1 upper L1 lower L1 lower improved L2 upper L2 lower L2 lower improved α = β

43 Accuracy Accuracy of the bounds Bounds λ Linf upper L2 upper L1 upper Linf lower L2 lower L1 lower L1 lower L2 lower Linf lower α = β

44 Accuracy Accuracy of the bounds Upper bound - Lower bound Linf L1 L2 L1 improved L2 improved Linf improved Combined Combined, improved α = β

45 Accuracy Conclusions 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