Pattern size in Gaussian fields from spinodal decomposition

Transcription

1 Pattern size in Gaussian fields from spinodal decomposition Luigi Amedeo Bianchi (joint work with D. Blömker and P. Düren) 8th Random Dynamical Systems Workshop Bielefeld, 6th November 2015 Research partially supported by the DFG Grant BL-535/9-2

2 Sit-rep Introduction The problem Ingredients of the proof Remarks Generalisations 1/35

3 Cahn-Hilliard-Cook equation The equation t u = (ε 2 u + f (u)) f derivative of a double well potential, (e.g. f (u) = u u 3 ) A model for relative concentration of an alloy after quenching. Dynamics dominated by strongly unstable space. Initial condition: constant homogeneous concentration u = m. 2/35

4 Cahn-Hilliard-Cook equation The equation t u = (ε 2 u + f (u)) + t W, f derivative of a double well potential, (e.g. f (u) = u u 3 ) t W derivative of a Q-Wiener process. A model for relative concentration of an alloy after quenching. Dynamics dominated by strongly unstable space. Initial condition: constant homogeneous concentration u = m. Perturbation causes spinodal decomposition. 2/35

5 Spinodal decomposition 3/35

6 Linearisation Good approximation by linearisation in m: t u = (ε f (m))u + t W on [0, 1] 2. L 2 -basis: e k,l (x) = C cos(kπx) cos(lπy). Assuming Qe k,l = αk,l 2 e k,l, the solution is the stochastic convolution t α k,l e (t s)λ k,l db k,l (s)e k,l. k,l N 0 The strong subspace is Rε γ := {(k, l) N 2 : λ k,l > γλ max }, γ (0, 1). Projecting there we have that at fixed time t the solution is well approximated by u(x, y) α k,l c k,l cos(kπx) cos(lπy), c k,l Gaussians. (k,l) R ε 4/35

7 Cosine series 5/35

8 Some comments We have R ε = {(k, l) N 2 α < (kε) 2 + (lε) 2 < α }, (growing and moving with ε 0), with γ 1 1 γ α = 2π 2 and α = 2π 2 with γ (0, 1). We restrict to f (x, y) = c k,l cos(kπx) cos(lπy), x, y [0, 1] 2 c k,l N(0, 1) i.i.d. (k,l) R ε Note that f is neither stationary nor isotropic; its law might change under translation or rotation (as a function extended to R 2 ). 6/35

9 Snake like pattern Patterns like this one appear in other models (reaction-diffusions,... ) 7/35

10 Snake like pattern 7/35

11 Sit-rep Introduction The problem Ingredients of the proof Remarks Generalisations 8/35

12 The question Question What is the characteristic thickness of the pattern (i.e. snake-like structures) in our model f (x, y), on the unit square (x, y) [0, 1] 2? 9/35

13 The question Question What is the characteristic thickness of the pattern (i.e. snake-like structures) in our model f (x, y), on the unit square (x, y) [0, 1] 2? Candidate answer In the previous picture (and similar simulations) the average thickness appears to be 2πε. 9/35

14 The question Question What is the characteristic thickness of the pattern (i.e. snake-like structures) in our model f (x, y), on the unit square (x, y) [0, 1] 2? Candidate answer In the previous picture (and similar simulations) the average thickness appears to be 2πε. Strategy To address the question, we draw a straight line across the unit square and we count the (average) number of zeros of f on that segment, then we divide the length of the segment by the number of zeros,. 9/35

15 Strategy Count number of zeros on lines and divide length of segment by it. 10/35

16 How many zeros of a random polynomial? Definition (Equator of a point and a curve) Let P S n. P is the hyperplane to PO in O, intersected with S n. Let γ(t) rectifiable, then γ = {P P γ}. 11/35

17 How many zeros of a random polynomial? Definition (Equator of a point and a curve) Let P S n. P is the hyperplane to PO in O, intersected with S n. Let γ(t) rectifiable, then γ = {P P γ}. Definition (Multiplicity and area swept out) Multiplicity of Q γ is #{t R q γ(t) } (equators containing Q). γ is the integral of multiplicity over γ (area swept out by equators). 11/35

18 How many zeros of a random polynomial? Definition (Equator of a point and a curve) Let P S n. P is the hyperplane to PO in O, intersected with S n. Let γ(t) rectifiable, then γ = {P P γ}. Definition (Multiplicity and area swept out) Multiplicity of Q γ is #{t R q γ(t) } (equators containing Q). γ is the integral of multiplicity over γ (area swept out by equators). Lemma If γ is a rectifiable curve, γ its length, γ area S n = γ π. 11/35

19 Where are the random polynomials? 12/35

20 Where are the random polynomials? Let p(x) = a n x n + + a 1 x + a 0, and consider a = (a 0, a 1, a 2,..., a n ), v(t) = (1, t, t 2,..., t n ). 12/35

21 Where are the random polynomials? Let p(x) = a n x n + + a 1 x + a 0, and consider a = (a 0, a 1, a 2,..., a n ), v(t) = (1, t, t 2,..., t n ). t is a zero of p(x) iff a v(t). Project w(t) = v(t)/ v(t) (and a to ā). 12/35

22 Where are the random polynomials? Let p(x) = a n x n + + a 1 x + a 0, and consider a = (a 0, a 1, a 2,..., a n ), v(t) = (1, t, t 2,..., t n ). t is a zero of p(x) iff a v(t). Project w(t) = v(t)/ v(t) (and a to ā). Number of 0s of p(x) is number of times ā is swept by an equator, which is the multiplicity of ā in w. 12/35

23 Where are the random polynomials? Let p(x) = a n x n + + a 1 x + a 0, and consider a = (a 0, a 1, a 2,..., a n ), v(t) = (1, t, t 2,..., t n ). t is a zero of p(x) iff a v(t). Project w(t) = v(t)/ v(t) (and a to ā). Number of 0s of p(x) is number of times ā is swept by an equator, which is the multiplicity of ā in w. Assume a i N(0, 1) iid, then ā is uniform on S n, as the joint density is a function of the radius only. 12/35

24 Where are the random polynomials? Let p(x) = a n x n + + a 1 x + a 0, and consider a = (a 0, a 1, a 2,..., a n ), v(t) = (1, t, t 2,..., t n ). t is a zero of p(x) iff a v(t). Project w(t) = v(t)/ v(t) (and a to ā). Number of 0s of p(x) is number of times ā is swept by an equator, which is the multiplicity of ā in w. Assume a i N(0, 1) iid, then ā is uniform on S n, as the joint density is a function of the radius only. Let now E n be the expected number of zeros. It is the portion of the surface swept out (with multiplicity), so E n = w area S n = w π. 12/35

25 Where are the random polynomials? Let p(x) = a n x n + + a 1 x + a 0, and consider a = (a 0, a 1, a 2,..., a n ), v(t) = (1, t, t 2,..., t n ). t is a zero of p(x) iff a v(t). Project w(t) = v(t)/ v(t) (and a to ā). Number of 0s of p(x) is number of times ā is swept by an equator, which is the multiplicity of ā in w. Assume a i N(0, 1) iid, then ā is uniform on S n, as the joint density is a function of the radius only. Let now E n be the expected number of zeros. It is the portion of the surface swept out (with multiplicity), so E n = w area S n = w π. Now we only need to compute the length of w, using Kac s formula. 12/35

26 A theorem by Edelman and Kostlan Theorem (Edelman&Kostlan, 1995) Let v(x) = (g 0 (x),..., g n (x)) T be any collection of differentiable functions and c 0,..., c n be independent and identically distributed Gaussians centred in 0. Given the function h(x) = n c k g k (x), the density of real zeros of h on an interval I is δ(x) = 1 d π dx w(x), where w(x) = v(x). v(x) R n R n k=0 The expected number of real zeros of h on I is then δ(x) dx. I 13/35

27 Main result We need to introduce, in the spirit of the previous theorem, the following notation: w t (x) = cos(kπx) cos(lπt) cos2 (mπx) cos 2 (nπt) m,n Rε ( ) d W t (x) = dx w t(x) (k,l) R ε 2 = S 3 S 1 (k,l) R ε ( ) 2 S2 where we have S 1 = cos 2 (mπx) cos 2 (nπt) m,n R ε S 2 = mπ cos(mπx) sin(mπx) cos 2 (nπt) m,n R ε S 3 = m 2 π 2 sin 2 (mπx) cos 2 (nπt). m,n R ε S 1 14/35

28 Main result (cont.) Let L t = {(x, t) : x [0, 1]} for t [0, 1]. Theorem For any γ (0, 1) and any horizontal line L t for x, t (0, 1) the function W t (x) defined on L t behaves asymptotically as (2ε) 2 for ε 0. This means that the average number of zeros is (2πε) 1, so the mean pattern size is 2πε. 15/35

29 Dependency on γ Remark The result is independent of γ, even if the number of Fourier modes involved is much smaller for γ 1 than for γ 0. As we can see, while the average asymptotic pattern size along lines remains the same, the domain with higher γ looks more organized. The pattern seems to be more regular in some sense. (γ = 0.1 and γ = 0.9) 16/35

30 Sit-rep Introduction The problem Ingredients of the proof Remarks Generalisations 17/35

31 Ergodic Theory Theorem (Birkhoff ergodic theorem) Let (X, µ) be a probability space. If T is µ-invariant and ergodic and g is integrable, then for a.e. z X 1 lim N N N g ( T k (z) ) = k=1 X g(ζ) dµ(ζ). Moreover if T is continuous and uniquely ergodic with measure µ and if g is continuous, then the limit holds for all z X (instead of a.e.). Example The map z z + α on the unit circle is uniquely ergodic if and only if α is irrational. In this case the unique ergodic measure is the Lebesgue measure. 18/35

32 Weighted averaging condition Requirement (Weighted averaging condition) Let ([0, 1] d, λ) be the probability space with the Lebesgue-measure λ. We say that (f, (a m )) with f : [0, 1] d R continuous and extended by periodicity to R d and a m R fulfils the weighted averaging condition, if for every x 0 [0, 1] d, every α N d and Q L = d [1,..., α i L] N d, i=1 the following holds: 1 a m f (m 1 x1 0, m 2 x2 0,..., m d xd 0 ) m Q L a m L m Q L [0,1] d f (x) dx. For any open set M R d + we define the (scaled) projection M L = (L M) N d. 19/35

33 Weighted averaging condition (cont.) Lemma Let (f, (a m )) fulfil the weighted averaging condition with the weights a generating a measure. Then for any open measurable set S R d + 1 a m f (m 1 x1 0, m 2 x2 0,..., m d xd 0 ) L f (x) dx. S L a m S [0,1] d L 20/35

34 Weighted averaging condition (cont.) Lemma Let (f, (a m )) fulfil the weighted averaging condition with the weights a generating a measure. Then for any open measurable set S R d + 1 a m f (m 1 x1 0, m 2 x2 0,..., m d xd 0 ) L f (x) dx. S L a m S [0,1] d L Q 2 L R L Q 3 L Q 1 L 20/35

35 Weights generating measures Requirement (Generation of measures) We require that the weights a = (a m ) generate a measure λ a on R d + which is equivalent to the Lebesgue measure λ, i.e. there exists an α > 0 such that for each set with open interior M R d +, L α M L a L λ a (M). Example Consider a k,l = k 2 in dimension d = 2. L 4 M L a = L 4 (k,l) L M k 2 = k 2 L 2 L (k,l) M 1 L N2 M ξ 2 d(ξ, η). The measure λ (k2,1) has a Lebesgue-density (ξ, η) ξ 2. The density is a. e. strictly positive, so the measures λ (k2,1) and λ are equivalent. 21/35

36 Key Lemma Lemma For x, t (0, 1), lim ε 0 1 R ε S 1 = 1 4, lim ε 0 ε2 S3 S 1 = 1 4, lim ε S2 = 0, ε 0 S 1 where the S i are S 1 = cos 2 (mπx) cos 2 (nπt) m,n R ε S 2 = mπ cos(mπx) sin(mπx) cos 2 (nπt) m,n R ε S 3 = m 2 π 2 sin 2 (mπx) cos 2 (nπt). m,n R ε 22/35

37 Proof of key lemma (sketch) We consider only the case x, t / Q. First limit. We can use Birkhoff s ergodic theorem. Define T x (z) = z + x: this is a measure-preserving and uniquely ergodic transformation (since x Q). Then N 1 1 cos 2 (πkx) = 1 N N k=0 N 1 k=0 cos 2 ( πt k x (0) ) N 1 0 cos 2 (πx) dx = 1 2. All the coefficients a m are 1 in this case and the function is multiplicative. Then the result follows immediately from previous lemma and the fact that [0,1] 2 cos 2 (πx 1 ) cos 2 (πx 2 ) d(x 1, x 2 ) = /35

38 Proof of key lemma (sketch) Second limit. The asymptotic behaviour is (by previous limit) By previous example ε2 ε 2 S 3 4π 2 S 1 R ε lim ε 0 ε2 R ε a R ε = lim ε 0 ε 2 where the rescaled domain is k,l R ε k 2 sin 2 (kπx) cos 2 (lπt). 1 R ε k 2 = λ a(r) λ(r) = 1 4π 2, k,l R ε R = {(η, ξ) R 2 α < ξ 2 + η 2 < α }, We can check on rectangles and then on the quarter ring that ε 2 S3 S 1 4π 2 ε 2 R ε a R ε 1 R ε a k 2 sin 2 (kπx) cos 2 1 (lπt) ε 0 4. k,l R ε 24/35

39 Proof of theorem Proof of main theorem. From key lemma: W t (x) = S 3 S 1 ( S2 S 1 ) 2 1 4ε 2 as ε 0. By E-K s theorem, the number of expected zeros on the horizontal line L t of length 1 is, for a given ε, N = π 4ε 2 dx = 1 2π ε. 0 This is the same as saying that the average pattern size is 1 N = 2πε. 25/35

40 Sit-rep Introduction The problem Ingredients of the proof Remarks Generalisations 26/35

41 Fast convergence To get an idea of the speed of convergence, let s consider the rescaled density of zeros εδ(x) ε = ε = ε = /35

42 Asymptotic result Now, fix x 0 (0, 1) and look at εδ(x 0 ) as ε 0. 1e+0 1e-1 ε δ(x0) 1e-2 x0 = 10 1 x0 = 10 2 x0 = e n, where ε = 10 n 28/35

43 Sit-rep Introduction The problem Ingredients of the proof Remarks Generalisations 29/35

44 Other Fourier domains Lemma Let D ε = ε 1 D N 2 be a scaled domain in Fourier space. Then δ(x) 1 2πε λ (k2,1)(d), δ(x) 1 λ(d) 2πε λ (1,l 2 )(D), λ(d) respectively on horizontal and vertical lines. Proof. Analogous to the one of the key lemma. We obtain the following asymptotic equivalences δ(x) 2 S 3 1 S 1 4 D ε (k2,1) 1 D ε 4ε 2 λ(k2,1)(d). λ(d) 30/35

45 Examples of Fourier domains l l k 2 k 2 k 1 R γ ε Q 1 ε k 1 k 2 k k 2 k l l k 2 Q 2 ε k 1 k 1 + k 2 Q 3 ε 2 k 1 k 1 k 2 k k 1 k 2 k 31/35

46 Corresponding patterns y R Q y x x y Q Q y x x 32/35

47 Some numbers Domain Corr. coeff. Avg. # 0s Avg. # 0s (sampled) R ( 1) Q ( 1.016) Q ( 1.375) Q Q (hor.) ( 1.375) (ver.) ( 2.318) The correction coefficient depends on γ (the quarter ring is a special case!). 33/35

48 Further generalisations Relax iid and equal variance assumption on coefficients. (Free, using the fact that E-K result holds for a more general family of gaussian random coefficients). 34/35

49 Further generalisations Relax iid and equal variance assumption on coefficients. (Free, using the fact that E-K result holds for a more general family of gaussian random coefficients). Spaces of higher dimension? 34/35

50 EOF Thank you for your attention! Bibliography A. Edelman, and E. Kostlan. How many zeros of a random polynomial are real? Bulletin of the AMS, 32(1):1 37, L. A. Bianchi, D. Blömker, and P. Düren Pattern size in Gaussian fields from spinodal decomposition. arxiv: v2, 2015