A Sufficient Condition for the Kolmogorov 4/5 Law for Stationary Martingale Solutions to the 3D Navier-Stokes Equations

Transcription

1 A Sufficient Condition for the Kolmogorov 4/5 Law for Stationary Martingale Solutions to the 3D Navier-Stokes Equations Franziska Weber Joint work with: Jacob Bedrossian, Michele Coti Zelati, Samuel Punshon-Smith October 24, 2018

2 2 Stochastic Navier-Stokes equations t u + (u )u + p ν u = t W, div u = 0, on T 3 (NSE) ν inverse of Reynolds number, ν = 1 Re W white-in-time, colored-in-space Gaussian process t W (t, x) = σ k e k (x)dβ k (t), k=1 {β k (t)} k independent one-dimensional Brownian motions {e k } k orthonormal eigenfunctions of the Stokes operator on T 3 {σk } k fixed constants satisfying coloring condition ε = 1 2 σ k 2 < k=1

3 3 Symmetries of Navier-Stokes equations (on R 3 ) If u(t, x) is a solution of (NSE), then Space translations: u(t, x + z) for z R 3 is a solution as well Time translations: u(t + τ, x), τ R solution Galilean transformations: U 0 + u(t, x U 0 t), U 0 R 3 solution Parity: u(t, x) solution Rotations: Ru(t, R x), R SO(R 3 ) solution Scaling: λ s u(λ 1 s t, λx) solution for λ R +, s = 1.

4 4 But... Figure: Homogeneous turbulence behind a grid (picture from Frisch (1995))

5 Figure: One second of a signal recorded by a hot-wire (sampled at 5kHz) in wind tunnel (picture from Frisch (1995)) 5

6 Figure: One second of a signal recorded by a hot-wire (sampled at 5kHz) in wind tunnel four seconds later (picture from Frisch (1995)) 6

7 Figure: Histogram for previous figure sampled 5000 times over time-span of 150s (picture from Frisch (1995)) 7

8 Figure: Same histogram a few minutes later (picture from Frisch (1995)) 8

9 9 Kolmogorov 1941 (K41) Probabilistic description of turbulence At high Reynolds numbers the symmetries of (NSE) are restored in a statistical sense, flow is statistically stationary and locally homogeneous and isotropic: Statistically stationary: u(t + τ, ) law = u(t, ), τ > 0 Velocity increment δ h u(x) = u(x + h) u(x), h R 3 Local homogeneity: δh u(x + y) law = δ h u(x) Local isotropy: RδR hu(x) law = δ h u(x), R SO(3) Kolmogorov formulates hypotheses that should hold for such flows based on experimental observations and derives additional predicitions about these

10 10 K41 First hypothesis of similarity: For locally homogeneous and isotropic turbulence, the laws of δ h u are uniquely determined by the kinematic viscosity ν and the mean energy dissipation rate ε ν u 2 L 2 x. Second hypothesis of similarity: Let λ K = ν 3/4 ε 1/4 the Kolmogorov scale. If h is large in comparison to λ K, then the laws of the velocity increments δ h u are uniquely defined by the mean energy dissipation rate ε and do not depend on ν.

11 K41 Intermittency 2nd hypothesis is debated, it would lead to scale invariance: There exists s R such that δ λh u(x) law = λ s δ h u(x) for all λ R +, h λ K For example this would imply, E (δ h u) 2 = Cε 2/3 h 2/3, where C is a universal constant (Since units (δ h u) 2 [L] 2 /[T ] 2, ε [L] 2 /[T ] 3 and by scale invariance (δ h u) 2 l 2s only possible exponent is s = 1/3 ) Physical experiments indicate scale invariance might not be true/c is not universal Intermittency corrections Theory of intermittency is largely based on empirical considerations, no direct derivation from fluid dynamics equations/mathematical theory, relates to higher regularity/smoothness of solutions 11

12 Intermittency: Experimental observations Figure: From Anselmet, Gagne, Hopfinger, Antonio (1983) 12

13 13 4/5 law For h [l D, l I ], (dissipation scale l D ε 1/4 ν 3/4, integral scale l I ), ( ) h 3 E δ h u 4 h 5 ε h where δ h u(x) = u(x + h) u(x). Balance of dissipation due to nonlinear effects with energy input 4/5 law should be independent of intermittency corrections/higher order regularity of solutions! Second hypothesis of similarity should not be needed for derivation 4/5 law should be deducible directly from fluid mechanics equations without further assumptions!

14 14 Statistically stationary solutions Kolmogorov derived his laws of turbulence for this case Statistically stationary means: u( + τ) has the same distribution as u( ), i.e. u( + τ) law = u( ) for every τ > 0 EF (u( + τ)) = EF (u( )), F continuous function Energy balance: νe u 2 L 2 x = ε, where ε = 1 2 k=1 σ2 k <

15 15 Energy balance for statistically stationary solutions Itô s formula: Stochastic process u = (u (1), u (2),... ), du (i) = µ (i) dt + k v (i) k dβ k f = f (t, u) satisfies df = f t dt + i = f t since i,j,k f u (i) du(i) i,j 2 f u (i) u (i) v (j) k v (j) 2 f u (i) u (j) d[u(i), u (j) ](t) k + f u d[β l, β m ](t) = δ lm dt, d[u (i), u (j) ](t) = 1 2 k (i) µ(i) v (i) k dt + i,k v (j) k dt, f (i) v u (i) k dβ k,

16 Energy balance for statistically stationary solutions For f (t, u) = 1 2 u 2, du = ( (u )u + ν u p)dt + k σ k e k dβ k : ( d( u 2 ) = 2 ) σk e 2 k u u 2 + ν u u p u dt k + k σ k e k u dβ k Integrate over torus T 3 : ( ) ( 1 2 d u 2 1 ) dx = σk 2 ν u 2 dx dt + 2 k k Integrate in time, take expectation: 1 2 E u 2 (t)dx 1 2 E u 0 2 dx = εt νe t Assume u is stationary ε = νe u 2 dx = νe u 2 L 2 x 0 σ k e k u dx dβ k u 2 dxds 16

17 17 Martingale solutions Stochastic equivalent of Leray-Hopf weak solutions Stochastic basis (Ω, (F t ) t [0,T ], P, {β k } k ), progressively measurable stochastic process u : [0, T ] Ω L 2 div sample paths of u in C t (H α x ) L t L 2 div L2 t H 1 div, α < 0 P-a.s. ω Ω u(ω) is a weak solution of the stochastic Navier-Stokes equations du + (u )udt + pdt = ν udt + dw t, Energy inequality: a.e. t 2 > t t2 2 E u(t 2) 2 L 2 1 x 2 E u(t 1) 2 L 2+νE u(s) 2 x L 2 ds ε(t 2 t x 1 ) t 1 k σ2 k energy input ε = 1 2 Such solutions exist (Bensoussan, Temam (1973))

18 18 Martingale solutions Stationary martingale solutions: Same as before but paths u( + τ) have same law as u( ) for each τ 0. Such solutions exist (Flandoli, Gatarek (1995)) Stationary energy inequality νe u 2 L 2 x ε Homogeneous forcing homogeneous stationary solution exists

19 Weak anomalous dissipation Stochastic heat equation: t v ν v = t W Same energy balance for stationary solutions: ε = νe v 2 L 2 x But no energy cascade! Energy balance doens t say anything about nonlinear effects nor contains enough information to derive 4/5 law Poincaré inequality (domain bounded)/fourier transform: Weak anomalous dissipation if ν u ν 2 L 2 x Cν uν 2 L 2 x ε lim ν 0 νe uν 2 L 2 x = 0. (WAD) Energy in low frequencies gets transferred to high frequencies (WAD) holds for passive scalar t f + v f ν f = t W, where v is weakly mixing (Bedrossian, Coti Zelati, Glatt-Holtz (2016)) or solution of stochastic 2D (NSE) or hyperviscous 3D (NSE) (Bedrossian, Blumenthal, Punshon-Smith (2018)) 19

20 20 What we prove Velocity increment: δ h u(x) := u(x + h) u(x), h R 3 Averaged 3rd order longitudinal structure function: S (l) = 1 4π E (δ lˆn u ˆn) 3 dxds(ˆn) S 2 T 3 Theorem (4/5 law) Let {u ν } ν>0 be a sequence of stationary martingale solutions to (NSE), which satisfy (WAD), and let S be the third order structure function. Then, there exists l D = l D (ν) with lim ν 0 l D = 0 such that lim lim sup l I 0 ν 0 sup l [l D,l I ] S (l) + 4 l 5 ε = 0. (1)

21 21 Remarks Only assumption is weak anomalous dissipation: νe u ν ν 0 L 2 x 0 No smoothness/regularity of solutions required Rigorous derivation under assumption of regularity/integrability that weak martingale solutions satisfy ( stochastic equivalent of Leray-Hopf solutions) No energy balance needed, inequality νe u 2 2 ε is enough Assuming symmetries (homogeneity, isotropy), same results holds without averaging. Proof ideas from Frisch (1995), Monin, Yaglom (1965), Eyink (2003), Nie, Tanveer (1995), but mathematically rigorous and with weaker assumptions (only (WAD)) Lagrangian version of 4/5 law: Drivas (2018)

22 22 4/3 law Velocity increment: δ h u(x) := u(x + h) u(x), h R 3 Averaged 3rd order structure function: S 0 (l) = 1 4π E δ lˆn u 2 δ lˆn u ˆn dxds(ˆn) S 2 T 3 Theorem (4/3 law) Let {u ν } ν>0 be a sequence of stationary martingale solutions to (NSE), which satisfy (WAD), and let S be the third order structure function. Then, there exists l D = l D (ν) with lim ν 0 l D = 0 such that lim lim sup l I 0 ν 0 sup l [l D,l I ] S 0 (l) l ε = 0. (2)

23 23 What we prove Velocity increment: δ h u(x) := u(x + h) u(x), h R 3 Averaged 3rd order longitudinal structure function: S (l) = 1 4π E (δ lˆn u ˆn) 3 dxds(ˆn) S 2 T 3 Theorem (4/5 law) Let {u ν } ν>0 be a sequence of stationary martingale solutions to (NSE), which satisfy (WAD), and let S be the third order structure function. Then, there exists l D = l D (ν) with lim ν 0 l D = 0 such that lim lim sup l I 0 ν 0 sup l [l D,l I ] S (l) + 4 l 5 ε = 0. (1)

24 Kármán-Howarth-Monin relation Define Γ(t, h) = E u(t, x) u(t, x + h)dx, T 3 D k (t, h) = E (δ h u δ h u)δ h u (k) dx, T 3 a(h) = 1 σk 2 e k (x) e k (x + h)dx, 2 T 3 k Let η(h) smooth, isotropic, compactly supported test function of the form η(h) = φ( h )I + ϕ( h )ĥ ĥ, ĥ = h h, φ(l) and ϕ(l) smooth and compactly supported on (0, ). Then, η(h) : Γ(T, h) dh η(h) : Γ(0, h) dh = 2T η(h) : a(h)dh R 3 R 3 R 3 1 T T k η(h) : D k (t, h) dhdt + 2ν η(h) : Γ(t, h) dhdt. 2 0 R 3 0 R 3 24

25 Kármán-Howarth-Monin relation If u is stationary, 1 T k η(h) : D k (t, h) dhdt 2 0 R 3 = 2T η(h) : a(h)dh + 2ν R 3 where T 0 R 3 η(h) : Γ(t, h) dhdt, Γ(t, h) = E u(t, x) u(t, x + h)dx, T 3 D k (h) = E (δ h u δ h u)δ h u (k) dx, T 3 a(h) = 1 σk 2 e k (x) e k (x + h)dx, 2 T 3 k η(h) = φ( h )I + ϕ( h )ĥ ĥ, ĥ = h h. 25

26 26 Proof idea Mollify equations for u(t, x) and u(t, x + h) For each x T 3 f κ = (f ) κ := γ κ f. du κ (t, x) + div x (u u) κ (t, x)dt + p κ (t, x)dt = ν u κ (t, x)dt + dw κ (t, x), Compute evolution of u κ (t, x) u κ (t, x + h) using stochastic product rule Integrate against isotropic test function η(h), estimate all terms and pass κ 0. Pressure term vanishes because of isotropic test function

27 27 4/3 law Choose η(h) = φ( h )I in KHM-relation and average Γ and a over sphere Γ(l) = 1 I : Γ(lˆn)dS(ˆn), a(l) = 1 tr(a(lˆn))ds(ˆn) 4π S 2 4π S 2 Flux term (D k (h) : I ) k φ(h)dh = T 3 k T 3 E T 3 δ h u 2 δ h u φ(h) dx dh, KHM-relation becomes ( 1 S 0 (l)l 2 φ (l)dl = l 2 φ(l) ν Γ + ν 2 ) 4 R + R + l Γ + ā(l) dl That s an ODE (in the sense of distributions): ( l l 3 S ) ( 0 = l 2 S ) ( l l S 0 = 4l 2 ν Γ + ν 2 ) l Γ + ā.

28 28 Integrate ODE S 0 (l) l = 1 l 3 l Estimate terms 4 l 3 l 0 0 τ (4ν Γ 2 (τ) + 4ν 2 ) τ Γ (τ) + 4ā(τ) dτ τ 2 ā(l)dτ = 4 3ā(0) + 4 l l 3 τ 2 (ā(τ) ā(0)) dτ Continuity of a 2nd term is o l 0 (1) Integrate by parts once more 0 We obtain 1 l 3 l 0 [ τ 2 Γ (τ) + 2τ Γ (τ) ] dτ = Γ (l) l S 0 (l) l = 4ν Γ (l) l 4 3 ε + o l 0(1)

29 29 Remains to estimate ν l Γ (l) ν l ( ) 1/2 ( ) 1/2 E u 2 L 2 E u 2 (εν) 1/2 ( ) 1/2 x L 2 x E u 2 L l 2 x Weak anomalous dissipation condition can choose l D (ν) 0 ( ) 1/2 such that νe u 2 L 2 x = o(ld ): lim sup ν 0 l (l D,l I ) ν l Γ (l) = 0 Hence lim lim sup l I 0 ν 0 sup l [l D,l I ] S 0 (l) l ε = 0.

30 30 Remarks l I depends only of continuity of a. l D needs to satisfy l D ε 1/2 ν 1/2 (E u 2 2 )1/2. Taylor micro scale: λ T (ν) ε 1/2 ν 1/2 (E u 2 ) 1/2 Kolmogorov dissipation scale: λ K (ν) ε 1/4 ν 3/4. Different test function η and a few more terms: Obtain 4/5 law under same assumptions

31 31 Necessary conditions I Theorem {u} ν>0 sequence of stationary martingale solutions to the Navier-Stokes equations satisfying Energy balance: For ν sufficiently small νe u 2 L 2 x = ε, Regularity: There exists C > 0 and s > 1 such that sup νe s u 2 L 2 C, x ν (0,1) then lim sup l 0 ν (0,1) ( ) S (l) l + S 0 (l) l = 0

32 32 Necessary conditions II Corollary Let {u} ν>0 be a sequence of stationary martingale solutions to (NSE) such that for some c 0 and some l D (ν) with lim l D = 0 ν 0 there holds lim lim l I 0 ν 0 Then, for all s > 5/4, sup l [l D,l I ] S 0 (l) l + c = 0. lim inf ν 0 Eν s u 2 L 2 x =. If inf ν Eν u 2 L 2 x > 0, then (WAD) implies that lim inf ν 0 Eν s u 2 L 2 x =, s > 1.