THE CASCADE OF EDDIES IN TURBULENCE. David Ruelle IHES CIRM, July - - PowerPoint PPT Presentation

the cascade of eddies in turbulence
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THE CASCADE OF EDDIES IN TURBULENCE. David Ruelle IHES CIRM, July - - PowerPoint PPT Presentation

THE CASCADE OF EDDIES IN TURBULENCE. David Ruelle IHES CIRM, July 2019 A theory of hydrodynamic turbulence based on non-equilibrium statistical mechanics. J. Statist. Phys. 169 ,1039-1044(1917). (arXiv:1707.02567). Incompressible


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SLIDE 1

THE CASCADE OF EDDIES IN TURBULENCE.

David Ruelle

IHES

CIRM, July 2019

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SLIDE 2

“A theory of hydrodynamic turbulence based on non-equilibrium statistical mechanics.” J. Statist. Phys. 169,1039-1044(1917). (arXiv:1707.02567).

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SLIDE 3

Incompressible Navier-Stokes equation.

∂v ∂t + v · ∇v = −∇p ρ + ν∇2v + f , ∇ · u = 0 where v = velocity field p = pressure ρ = density ν = kinematic viscosity f = external force

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SLIDE 4

Mathematicians who have studied turbulence.

Jean LERAY: “turbulent” solutions of Navier-Stokes [Caffarelli-Kohn-Nirenberg theorem]. Andrey N. KOLMOGOROV: using dimensional analysis to study energy cascade in 3-D “inertial range” assuming homogeneous turbulence. [comparison with 2-D]

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SLIDE 5

Overview.

  • Study intermittency exponents ζp such that

|∆v|p ∼ ℓζp where ∆v is contribution to fluid velocity at small scale ℓ. [ Claim: ζp = p 3 − 1 ln κ ln Γ(p 3 + 1) experimentally (ln κ)−1 = 0.32 , i.e., κ ≈ 20 or 25 ].

  • Distribution of radial velocity increment and relation with

Kolmogorov-Obukhov.

  • Reynolds number ≈ 100 at onset of turbulence.
  • Problem: study decomposition of eddy into daughter eddies.
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SLIDE 6

References:

  • F. Anselmet, Y. Gagne, E.J. Hopfinger, and R.A. Antonia.

“High-order velocity structure functions in turbulent shear flows.”

  • J. Fluid Mech. 140,63-89(1984).

A.N. Kolmogorov. “A refinement of previous hypotheses concerning the local structure of turbulence in a viscous incompressible fluid at high Reynolds number.” J. Fluid Mech. 13,82-85(1962).

  • D. Ruelle. “Hydrodynamic turbulence as a problem in

nonequilibrium statistical mechanics.” PNAS 109,20344-20346(2012).

  • D. Ruelle. “Non-equilibrium statistical mechanics of turbulence.”
  • J. Statist. Phys. 157,205-218(2014).
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SLIDE 7
  • J. Schumacher, J. Scheel, D. Krasnov, D. Donzis, K. Sreenivasan,

and V. Yakhot. “Small-scale universality in turbulence.” PNAS 111,10961-10965(2014). also contributions by G. Gallavotti, and P. Garrido, and Ruelle to

  • Chr. Skiadas (editor) The foundations of chaos revisited: from

Poincar´ e to recent advancements. Springer, Heidelberg, 2016.

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SLIDE 8
  • 1. Obtaining the basic probability distribution.
  • Kinetic energy goes down from large spatial scale ℓ0 to small

scales through a cascade of eddies of increasing order n so that v =

  • n≥0

vn with viscous cutoff. Eddy of order n − 1 in ball R(n−1)i decomposes after time T(n−1)i into eddies of order n contained in balls Rnj ⊂ R(n−1)i. Balls Rnj form a partition of 3-space into roughly spherical polyhedra of linear size ℓnj, lifetime Tnj.

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  • Assume that the dynamics of each eddy is universal, up to

scaling of space and time, and independent of other eddies. Conservation of kinetic energy E yields

  • j

E(Rnj) Tnj = E(R(n−1)i) T(n−1)i Universality of dynamics and inviscid scaling give for initial eddy velocities vn ℓnj = T(n−1)i Tnj · vn−1 ℓn−1 hence

  • j
  • Rnj

|vn|3 ℓnj =

  • R(n−1)i

|vn−1|3 ℓ(n−1)i (implies intermittency).

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SLIDE 10
  • For simplicity assume size ℓnj depends only on n: ℓ(n−1)i/ℓnj = κ.

Then κ

  • j
  • Rnj

|vn|3 =

  • R(n−1)i

|vn−1|3

  • Assume that the distribution of the vn between different Rnj

maximizes entropy: microcanonical distribution → canonical distribution: ∼ exp[−β|vn|3] d3vn Integrating over angular variables: ∼ exp[−β|vn|3]|vn|2 d|vn| = 1 3 exp[−β|vn|3] d|vn|3 hence Vn = |v|3 has distribution β exp[−βVn] dVn

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  • Finally since the average value β−1 of Vn is Vn−1/κ, Vn is

distributed according to κ Vn−1 exp

  • − κVn

Vn−1

  • dVn

Starting from a given value of V0 the distribution of Vn is given by κ dV1 V0 e−κV1/V0 · · · κ dVn Vn−1 e−κVn/Vn−1 (∗) The validity of (∗) is limited by dissipation due to the viscosity ν: we must have V 1/3

n

ℓn > ν

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  • 2. Calculating ζp.
  • To compute the mean value of |vn|p = V p/3

n

we note that κ Vn−1

  • exp
  • − κVn

Vn−1

  • .V p/3

n

dVn = Vn−1 κ p/3 exp[−w].wp/3 dw = κ−p/3V p/3

n−1Γ

p 3 + 1

  • hence, using induction and ℓn/ℓ0 = κ−n,

V p/3

n

= κ V0

  • exp
  • −κV1

V0

  • dV1 · · ·

κ Vn−1

  • exp
  • − κVn

Vn−1

  • .V p/3

n

dVn = κ−np/3V p/3 Γ p 3 + 1 n = V p/3 ℓn ℓ0 p/3 Γ p 3 + 1 n

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  • Therefore

ln|vn|p = lnV p/3

n

= ln V p/3 + p 3 ln ℓn ℓ0

  • − ln(ℓn/ℓ0)

ln κ ln Γ p 3 +1

  • = ln V p/3

+ ln ℓn ℓ0

  • .

p 3 − 1 ln κ ln Γ p 3 + 1

  • = ln
  • V p/3

ℓn ℓ0 ζp where ζp = p 3 − 1 ln κ ln Γ p 3 + 1

  • r

|vn|p = V p/3 ℓn ℓ0 ζp ∼ ℓζp

n

as announced.

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  • 3. Estimating the probability distribution F(u) of the radial

velocity increment u. Relation with Kolmogorov-Obukhov.

  • If r ≈ ℓn we have u ≈ un ≈ radial component of vn

⇒ rough estimate of the probability distribution of u: F(u) =

  • n
  • k=1

∞ κ dVk Vk−1 e−κVk/Vk−1

  • 1

2V 1/3

n

χ[−V 1/3

n

,V 1/3

n

](u)

= 1 2(κn V0 )1/3

  • · · ·
  • w1···wn>(κn/V0)|u|3

n

  • k=1

dwk e−wk w1/3

k

  • The distribution Gn(y) of y = (κn/V0)1/3|u| is given by

Gn(y) =

  • · · ·
  • w1···wn>y3

n

  • k=1

dwk e−wk w1/3

k

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SLIDE 15
  • This satisfies

etGn(et) = (φ∗(n−1) ∗ ψ)(t) (∗∗) with φ(t) = 3 exp(3t − e3t) , ψ(t) = et ∞

t

e−sφ(s) ds [ ⇒ Gn(y) is a decreasing function of y].

  • For small u, Gn gives a good description of the distribution of u,

with normalized |u|2 (see Schumacher et al.).

  • (∗∗) suggests a lognormal distribution with respect to u in

agreement with Kolmogorov-Obukhov, but this fails because φ, ψ tend to 0 only exponentially at −∞.

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SLIDE 16
  • 4. The onset of turbulence.
  • We may estimate the Reynolds number Re = |v0|ℓ0/ν for the
  • nset of turbulence by taking

1 ≈

  • ν

|v1|ℓ1

  • =
  • ν

V 1/3

1

κ−1ℓ0

  • = Re−1

κ4/3 V0 κV1 1/3 [Relation to dissipation is dictated by dimensional arguments] ⇒ Re ≈ κ4/3 ∞ κV1 V0 −1/3 κ dV1 V0 e−κV1/V0 = κ4/3 ∞ α−1/3 dα e−α = κ4/3Γ 2 3

  • Taking 1/ ln κ = .32 hence κ4/3 = 64.5, with Γ(2/3) ≈ 1.354 gives

Re ≈ 87 agreeing with Re ≈ 100 as found in Schumacher et al.

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SLIDE 17
  • 5. Problem.

Study numerically the statistics of the decomposition of one eddy

  • f order n − 1 into κ3 eddies of order n.