Densities for the Navier Stokes equations with noise

Transcription

1 Densities for the Navier Stokes equations with noise Marco Romito Università di Pisa Universitat de Barcelona March 25, 2015

2 Summary 1 Introduction & motivations 2 Malliavin calculus 3 Besov bounds 4 Other functionals 5 Time regularity 6 The Fokker Planck equations approach 7 Open problems MR: Densities for 3DNSe with noise 2 / 21

3 The equations Consider the Navier Stokes equations, u + (u )u + p = ν u + η, div u = 0, u(0) = x, either on the torus T 3 with periodic boundary conditions (and zero mean) or on a bounded domain with Dirichlet boundary conditions. Here η is Gaussian noise which is white in time and coloured in space. For instance assume periodic boundary conditions, η(t, y) = S dw = k Z 3 σ k β k (t)e k (y), and e k (y) = e ik y are the Fourier exponential. MR: Densities for 3DNSe with noise 3 / 21

4 What we look for and why Theorem Under suitable assumptions on the covariance of the driving noise, suitable finite dimensional projections of any martingale weak solution of the 3D Navier Stokes equations have a density with respect to the Lebesgue measure. Why densities? probabilistic form of regularity (hot issue for NSE), better understanding of the evolution of the flow, ideas related to uniqueness in law, quantifying uncertainty. Why do we consider finite dimensional projections: there is a reference measure, easier: curse of regularity solved, ideas related to long time behaviour [MatPar2006] [HaiMat2006] MR: Densities for 3DNSe with noise 4 / 21

5 Existence of densities We focus on the following problem dx = b(x) dt + db How to prove the existence of a density: Girsanov transformation, the Fokker-Planck equation, Malliavin calculus,????? MR: Densities for 3DNSe with noise 5 / 21

6 A standard method The idea is to use integration by parts. Having a smooth density would give, [ φ ] E h (x t) = So, if we know that by duality there is a density. φ h (y)f t(y) dy = [ φ ] E h (x t) c h φ φ(y) f t (y) dy. h How to prove this? MR: Densities for 3DNSe with noise 6 / 21

7 Integration by parts Assume that we can find a variation of the noise H that compensates the variation h, that is h = D H x t. Then [ φ ] E h (x t) = E[Dφ(x t )h] = E[Dφ(x t )D H x t ] = [ = E[D H φ(x t )] = E φ(x t ) ] H dw by the chain rule and integration by parts. How to find H?: essentially one needs to (pseudo)invert the map u(t, x) u(t, x) + h h ΨH = t 0 H(s)D s x t ds that is H = Ψ M 1 t h. x MR: Densities for 3DNSe with noise 7 / 21

8 Malliavin calculus and Navier-Stokes We are dealing with a process u such that du ν u dt + (u )u dt + p = S dw, whose Malliavin derivative D H u satisfies NO! MAYBE d dt D Hu ν D H u + (u )D H u + (D H u )u + q = SH, { Trying to prove invertibility of the Malliavin matrix forces to use the dynamics of the linearisation. A decent estimate on the Malliavin derivative is the same as an estimate on the difference of two solutions! the dynamics is well defined and good for short times, an invertible covariance allows to prove invertibility without using the dynamics. It falls in the class of equations with non-regular coefficients, Fournier-Printemps (starting point for what we ll explain here), Bally and Caramellino (based on interpolation), Kohatsu-Higa and co-workers. MR: Densities for 3DNSe with noise 8 / 21

9 Set Primer on Besov spaces ( 1 h f)(x) = f(x + h) f(x), ( n h f)(x) = 1 h ( n 1 h f)(x) and given any integer n > s, f B s p,q = f L p + ( h 1 n h f q L p h sq Some nice properties of Besov spaces dh h n ) 1 q [+ sup h 1 ] n h f L p h s different (more general) definition in terms of Littlewood-Paley decomposition, B s p,p = W s,p, B s, = C s, s 0 non integer, (B s p,q) = B s p,q, q <, (I ) : B s p,q B s 2 p,q isomorphism, MR: Densities for 3DNSe with noise 9 / 21

10 Smoothing lemma Theorem (deterministic fractional integration by parts) If µ is a finite measure on R d and there are s > 0, an integer m > s, and α (0, 1) such that for every φ C c (R d ) and every h R d, R d m h φ(x) µ(dx) c h s φ C α, b then µ has a density f B r 1, for every r < s α. We use the above lemma to prove existence of a density for the solution x of our simple model problem at time t = 1 We look for an estimate dx = b(x) dt + db. E[φ(x 1 + h) φ(x 1 )] h s φ C α b MR: Densities for 3DNSe with noise 10 / 21

11 Fact 1: exploiting the short times Recall: wish to estimate Consider E[φ(x 1 + h) φ(x 1 )] h s φ C α b dx ɛ = η ɛ (t)b(x ɛ ) + db. η ɛ { dx 1 ɛ = b(x ) + db, 1 time dx ɛ = b(x ɛ ) + db, [fournier printemps (2010)] MR: Densities for 3DNSe with noise 11 / 21

12 Fact 1: exploiting the short times Recall: wish to estimate Consider E[φ(x 1 + h) φ(x 1 )] h s φ C α b dx ɛ = η ɛ (t)b(x ɛ ) + db. η ɛ { dx 1 ɛ = b(x ) + db, 1 time dx ɛ = b(x ɛ ) + db, [fournier printemps (2010)] MR: Densities for 3DNSe with noise 11 / 21

13 Hence Decomposition of the error s 1 ɛ: x ɛ (s) = x(s), s 1 ɛ: x ɛ (s) is the one step explicit Euler approximation of x, starting at time 1 ɛ with time step ɛ. and E[φ(x 1 + h)] E[φ(x 1 )] = E[φ(x 1 + h)] E[φ(x ɛ + h)] 1 + E[φ(x ɛ + h)] E[φ(x ɛ )] E[φ(x ɛ )] E[φ(x 1 1 )] where numerical error probabilistic error MR: Densities for 3DNSe with noise 12 / 21

14 The numerical error There is not much that can be done here, E[φ(x ɛ 1 )] E[φ(x 1 )] [φ] α E [ x ɛ 1 x 1 α] and so that x 1 x ɛ = ɛ b(x s ) ds b ɛ, numerical error ɛ α How to improve it? The numerical error can be made smaller using a higher order numerical method. More regularity of the drift b needed. MR: Densities for 3DNSe with noise 13 / 21

15 Fact 2: inverting the covariance (a.k.a. the Absorb the variation in the noise probabilistic error) x ɛ 1 (t) + h = i. c ɛ db s + h 1 = i. c. + (db s + H ds) 1 ɛ with an obvious value of H h ɛ and by the Girsanov formula (actually Cameron Martin here!) prob. error E [( ) G 1 1 φ(x ɛ ) ] h φ 1 ɛ How to improve it? Look for higher order increments n h φ(x 1) that yield the same numerical error but prob. error ( h ɛ ) n. MR: Densities for 3DNSe with noise 14 / 21

16 A balance of regularity Putting all together, total error probabilistic + numeric h ɛ φ + [φ] α ɛ α and by optimizing in ɛ, E[φ(x 1 + h)] E[φ(x 1 )] h 2α 2α+1 φ C α α derivatives given for the estimate, 2α derivatives obtained 2α+1 use the smoothing lemma with n = 1, α α and s = 2α 2α+1. MR: Densities for 3DNSe with noise 15 / 21

17 Other functionals The same idea can be used on quantities that derive from an equation (although might not solve one), such as in Navier Stokes the balance of energy: E t = 1 u(t, x) 2 dx + ν 2 To keep it simple, let us think again at x R d, t 0 u(s, x) 2 dx ds dx = b(x) dt + db, and E t = i (x i t) 2 It is immediate to check that the corresponding energy computed for x ɛ has a generalized χ 2 distribution. Apparently odd: it turns out that the regularity depends on the number of degrees of freedom of the χ 2 (here d), and only in infinite dimension the unconditioned result is recovered. MR: Densities for 3DNSe with noise 16 / 21

18 Time regularity Let f be the density of the solution of dx = b(x) dt + db Theorem f(t) f(s) B α 1, t s β/2, for all α + β < 1 f(t) f(s) L 1 t s 1 2, more challenging, through Girsanov transformation, [f(t) f(s)] B α 1, t s β/2 easier: { n n h h(f(t) f(s)) L 1 f(t) L 1 + n h f(s) L 1 f(t) f(s) L 1 h t s, t s h MR: Densities for 3DNSe with noise 17 / 21

19 The associated Fokker Planck equation Consider again dx = b(x) dt + db, x(0) = x 0 with b L. We know that the density f B 1 1,. In fact we can show that f B 1,. The density solves { t f = 1 f (bf), 2 f(0) = δ x0, that is f(t, x) = p t (x x 0 ) + where p t is the heat kernel. t 0 p t s (bf)(x) dt, MR: Densities for 3DNSe with noise 18 / 21

20 Besov bounds with the Fokker Planck equation f(t, x) = p t (x x 0 ) + hence t 2 h f(t, x) L 1 = 2 h p t (x x 0 ) + 0 p t s (bf)(x) dt, t 2 h p t L 1 + b f L 1 and conclude with estimates for the heat kernel 0 ( 2 h p t s) (bf)(x) dt t 0 2 h p t s L 1 ds L 1 MR: Densities for 3DNSe with noise 19 / 21

21 Hölder bounds with the Fokker Planck equation To see the Hölder bound, consider for simplicity the equation for the stationary solution (density of the invariant measure): 1 2 f (bf) = 0, that is, if g is the Poisson kernel (g(x) = x 2 d, d 3), f(x) = g (bf) = ( g1 Bɛ (0)) (bf) + ( g1 Bɛ (0) c) (bf) b f ɛ d + ɛ 1 d b f L 1 and choose ɛ = (2 b ) 1/d. The Hölder norm follows likewise. Watch out! Needs to know that already f <. MR: Densities for 3DNSe with noise 20 / 21

22 Some open problems Local estimate (see the problem with the energy). Probabilistic proof of the Fokker Planck approach. Degenerate noise completely open (at least in this way). MR: Densities for 3DNSe with noise 21 / 21