Stochastic homogenization 1

Transcription

1 Stochastic homogenization 1 Tuomo Kuusi University of Oulu August 13-17, 2018 Jyväskylä Summer School 1 Course material: S. Armstrong & T. Kuusi & J.-C. Mourrat : Quantitative stochastic homogenization and large-scale regularity, arxiv:

2 Scope The goal is to introduce basic concepts in stochastic homogenization for linear, uniformly elliptic equations of the form (a ε (x) u ε (x)) = 0 in U R d, ε > 0, d 2, where a ε (x) = a( x ) and the diffusion matrix a( ) satisfies ε ξ 2 a(x)ξ ξ Λ ξ 2 for some Λ 1 and for all ξ R d and for almost every x R d. If you are not familiar with the following concepts, please go through Appendixes from the course book. You should recall Basic knowledge about Sobolev spaces Knowledge about basic a priori estimates for elliptic equations is useful, but I will also discuss them during the course

3 Homogenization: Paradigm Homogenization means that, in an appropriate way, the original equation (a ε (x) u ε (x)) = 0 in U R d, ε > 0, d 2, homogenizes to an effective equation (a u) = 0 with constant coefficients a such that u ε is close to u Two basic questions are: When can one expect homogenization? (Qualitative theory) How fast is homogenization happening? (Quantitative theory)

4 When can one expect homogenization? Model assumptions for coefficients are that a is periodic quasi-periodic almost periodic stationary random fields. Let us first take a look of the easiest case, namely the periodic setting. We assume that a(x + z) = a(x) for every x Z d and a.e. x R d.

5 Periodic 1D Let ε = 1, k N, and solve an ODE k The unique solution is u(x) = ( which can equivalently be written as u ε (x) = x + ( 0 1 { (aε (x)(u ε ) (x)) = 0 u ε (0) = 0, u ε (1) = a(t) dt) ε 1 1 a ε (t) dt) 0 0 x/ε x/ε x 1 a ε (t) dt, ( 1 a(t) dt) dt a(t)

6 Periodic 1D Set now, for x R, φ(x) = ( a(t) dt) 0 x x ( 1 a(t) The solution u ε can be written by means of φ as u(x) = x + εφ ( x ε ) dt) dt a(t) Observe that u has two parts. Homogeneous solution u(x) = x and the small wiggles εφ ( ) coming from the anisotropic nature of the problem ε

7 Correctors Following the analogue suggested by 1D-example, we define in the periodic first-order corrector. Denote the the periodic Sobolev space as H 1 per([0, 1] d ) = {u H 1 loc(r d ) u(x + z) = u(x) for z Z d and a.e x R d }. One can identify this space as the completion of smooth Z d -periodic functions w.r.t. to the norm of H 1 ([0, 1] d ). This space is actually a Hilbert space. First-order corrector φ e H 1 per([0, 1] d ), e B 1, is the unique weak solution satisfying [0,1] d φ e (x) dx = 0 to the equation (a(x) (e + φ e (x))) = 0 in [0, 1] d. Notice that since a( ) is assumed to be Z d -periodic, the above equation is actually satisfied in the whole space. Exercise. Prove that there exists a unique solution φ e H 1 per([0, 1] d ) modulo a constant satisfying the above equation. Hint: One way is to find a suitable version of Lax-Millgram Lemma from the literature, state it, prove it, and apply it to obtain the existence.

8 Two-scale expansion Let φ j H 1 per([0, 1] d ), j {1,..., d}, be the solution of (a ( φ j + e j )) = 0 in [0, 1] d, [0,1] d φ j (x) dx = 0, where e j is the unit vector parallel to x j -axis, and choose the constant so that φ j has zero mean. Denote by φ ε and φ ε the vector and the matrix, respectively, having the components, for i, j {1,..., d}, (φ ε (x)) j = φ j ( x ε ) and ( φε (x)) ij = xi φ j ( x ε ). We can test the equation of φ j by itself, using the periodicity, and obtain by the Poincaré inequality that φ j L2 ([0,1] d ) C φ j L2 ([0,1] d ) C.

9 Two-scale expansion Suppose that we have the heterogenous solution u ε H 1 (U) and homogenized solution u W 2, (U) solving (a ( x ε ) uε ) = f in U, u ε = g on U, (a u) = f in U, { u = g on U. The very basic two-scale expansion around u is defined as w ε (x) = u(x) + εφ ε (x) u(x) We will show that using this it is possible to deduce estimates how close u ε and u are in L 2, and u ε and w ε in H 1

10 Two-scale expansion: The basic computation We still want to tinker the definition of the two-scale expansion. Namely, we choose a smooth cut-off function η ε C0 (U) so that ηε = 1 in ε away from U and η ε L (U) Cε 1. Then w ε and u have the same boundary values. Set w ε (x) = u(x) + εη ε (x)φ ε (x) u(x) = u(x) + εη ε (x) φ ε k ( x ε ) x k u(x) d Our goal is to show that (a ε ( w ε u ε ) H 1 (U) Cε u W 2, (U). Since u ε w ε H0 1 (U), this, in turn, implies u ε u H1 (U) + uε w ε H1 (U) Cε u W 2, (U).

11 Two-scale expansion: The basic computation Having compute where w ε = u + εη ε φ ε u = u + εη ε (x) φ ε k xk u, d d w ε = (e k + φ ε k) xk u + G ε, d G ε = (η ε 1) φ ε k xk u + ε φ ε k (η ε xk u) and then, using the equation of φ k, (a ε w ε d ) = xk u (a ε (e k + φ ε k)) d =0 d + a ε (e k + φ ε k) xk u + (a ε G ε ).

12 Two-scale expansion: The basic computation We have thus found the following formula: and d (a ε w ε ) = a ε (e k + φ ε k) xk u + (a ε G ε ). a ε G ε L2 (U) Cε u W 2, (U) This suggests to define the effective matrix as a = a(x) (I d + φ(x)) dx [0,1] d so that the above formula can be rewritten as d (a ε w ε a u) = (a ε (e k + φ ε k) ae k ) xk u + (a ε G ε ). Notice, indeed, that since a is a constant matrix, a u(x) = a 2 u(x). Recall that G ε = (η ε 1) φ ε u + ε (η ε u) φ ε.

13 Two-scale expansion: The basic computation The formula (a ε w ε a u) = (a ε (I d + φ ε ) a) 2 u + (a ε G ε ), tells now two sources of errors. Since a ε G ε L2 (U) Cε u W 2, (U), we have that (a ε G ε ) H 1 (U) Cε u W 2, (U). We are thus left to establish (a ε (I d + φ ε ) a) 2 u H 1 (U) Cε u W 2, (U).

14 Two-scale expansion: The basic computation Set now F e (x) = a(x) (e + φ e (x)) ae By the equation of φ e, F e is solenoidal (that is, F e = 0), Z d -periodic, and it has a zero mean. Therefore, one finds the following Helmholtz projection for F e : F e (x) = S e, and the matrix S e is Z d -periodic, zero mean, and skew-symmetric: S e,ij = S e,ji. Moreover, by Poincaré s inequality, S e L2 ([0,1] d ) C S e L2 ([0,1] d ) C.

15 Two-scale expansion: The basic computation Using this computation, notice that (a ε (I d + φ ε ) a) 2 u = F ε e k xk u d d = ε S ε e k xk u = ε (S ε e k xk u) Here we used the following consequence of the skew-symmetry of S e : S ε e k xk u = i,j xi (S ε e k ) ij xj x k u = i,j d xi ((S ε e k ) ij xj x k u) i,j (S ε e k ) ij xi x j x k u =0

16 Example: Periodic setting We have now found the formula (a ε w ε a u) = (ε S ε e k xk u + a ε G ε ), d and Since this yields d ε S ε e k xk u + a ε G ε L 2 (U) (a u) = (a ε u ε ), Cε u W 2, (U). (a ε ( w ε u ε ) H 1 (U) Cε u W 2, (U), as desired.

17 Scope: Towards stochastic homogenization Before we were taking a look of periodic setting. However, Real applications are more accurately modelled by random models

18 Scope: Towards stochastic homogenization Microprocessor (scale 180nm)! Virus surface! Battery structure! Solar cell structure!

19 Assumptions on coefficients Reasonable assumptions for coefficients are as follows. Ω = {a a symmetric matrix R d d, λ(a) [1, Λ]}, Λ 1. For U R d, F U denotes the σ-algebra generated by a U a(x)φ(x) dx, φ C c (U). Write F = F(R d ). P is a probability measure on (Ω, F R d ) Assumptions on P. (P1) Stationarity P T z = P for z Z d, where T z is a translation T zf (x) = f (x + z) (P2) Unit range dependency If dist(u, V ) 1, then F U and F V are P-independent We denote the expectation with respect to P by E. That is, if X Ω R is an F-measurable random variable, we write E[X ] = Ω X (a) dp(a)

20 In practice The probability distribution can be obtained by analyzing the material. Microprocessor (scale 180nm)! Virus surface! Battery structure! Solar cell structure!

21 Another concrete example A piece of the random checkerboard. The conductivity matrix equals identity matrix I in the white region, and 4I in the black region. Probability measure P is a product measure so that at each cube a fair coin is tossed to decide the value I or 4I. It can be shown by so-called Dykhne formula that a = 2I. Notice that P [a 4I in macroscopic cube] = 2 #(small cubes).

22 Correctors It turns out that so-called correctors play a central role in the theory. First-order corrector φ e H 1 loc (Rd ), e R d, is the unique weak solution, modulo an additive constant, to the equation (a(x) (e + φ e (x, a( )))) = 0 in R d. Observe that φ e depends on coefficients in the whole space. Now, whenever we are referring to a solution, it is actually a function of both x and a. This is to say that in reality the solution lives possibly in an infinite dimensional space. However, it is usually convenient to suppress a from the notation for u since the PDE is cast in the physical space. Following the reasoning from the periodic setting, one can show that the effective (homogenized) elliptic, symmetric and deterministic matrix a is defined via ae = E [ a(x) (e + φ e (x)) dx] e R d. [0,1] d

23 Two-scale expansion Suppose that we have the heterogenous solution u ε and homogenized solution u (a ( x ε ) u ε) = f in U, u ε = g on U, (a u) = f in U, { u = g on U. Then, defining so-called two-scale expansion w ε (x) = u(x) + εφ ε (x) u(x), one of the main goals is to prove that with a stochastic constant X. u ε u L2 (U) X ε, u ε w ε L2 (U) X ε 1 2

24 Back to the checkerboard example As noticed before, P [a 4I in macroscopic cube] = 2 #(small cubes) and a = 2I. Consider thus the problem from before in this very unlikely event that a 4I in B 1. We have { 4 uε = 1 in B 1, u ε = 0 on B 1, { 2 u = 1 in B 1, u = 0 on B 1. We can find an explicit solutions 2u ε (x) = u(x) = 1 2d (1 x 2 ) and thus we have that u ε u L2 (B = 1 1) 4d u L 2 (B 1) = c(d) C(d)ε. This shows that the stochastic constant X in the error estimate is necessary, and the task is then to show that it is integrable in probability space stemming to P [a 4I in macroscopic cube] = 2 #(small cubes).

25 Correctors Let us now go back to correctors, which satisfy the equation (a(x) (e + φ e )) = 0 in R d. How do the correctors look like, for example, in the case of checkerboard?

26 Corrector Approximation of the graph of φ e1 solving div(a(x)(e 1 + φ e1 (x))) = 0 in R 2

27 Corrector Approximation of the graph of φ e2 solving div(a(x)(e 2 + φ e2 (x))) = 0 in R 2