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

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]

Overview. • Study intermittency exponents ζ p such that �| ∆ v | p � ∼ ℓ ζ p where ∆ v is contribution to fluid velocity at small scale ℓ . [ Claim: 1 ζ p = p ln κ ln Γ( p 3 − 3 + 1) (ln κ ) − 1 = 0 . 32 experimentally , 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.

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).

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.

1. Obtaining the basic probability distribution. • Kinetic energy goes down from large spatial scale ℓ 0 to small scales through a cascade of eddies of increasing order n so that � v = v n n ≥ 0 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 R nj ⊂ R ( n − 1) i . Balls R nj form a partition of 3-space into roughly spherical polyhedra of linear size ℓ nj , lifetime T nj .

• 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 = E ( R ( n − 1) i ) E ( R nj ) � T nj T ( n − 1) i j Universality of dynamics and inviscid scaling give for initial eddy velocities = T ( n − 1) i v n · v n − 1 ℓ nj T nj ℓ n − 1 hence | v n | 3 | v n − 1 | 3 � � � = ℓ nj ℓ ( n − 1) i R nj R ( n − 1) i j (implies intermittency).

• For simplicity assume size ℓ nj depends only on n : ℓ ( n − 1) i /ℓ nj = κ . Then � � | v n | 3 = � | v n − 1 | 3 κ R nj R ( n − 1) i j • Assume that the distribution of the v n between different R nj maximizes entropy: microcanonical distribution → canonical distribution: ∼ exp[ − β | v n | 3 ] d 3 v n Integrating over angular variables: ∼ exp[ − β | v n | 3 ] | v n | 2 d | v n | = 1 3 exp[ − β | v n | 3 ] d | v n | 3 hence V n = | v | 3 has distribution β exp[ − β V n ] dV n

• Finally since the average value β − 1 of V n is V n − 1 /κ , V n is distributed according to κ − κ V n � � exp dV n V n − 1 V n − 1 Starting from a given value of V 0 the distribution of V n is given by κ dV 1 e − κ V 1 / V 0 · · · κ dV n e − κ V n / V n − 1 ( ∗ ) V 0 V n − 1 The validity of ( ∗ ) is limited by dissipation due to the viscosity ν : we must have V 1 / 3 ℓ n > ν n

2. Calculating ζ p . • To compute the mean value of | v n | p = V p / 3 we note that n κ � − κ V n � V n − 1 � p / 3 � � � exp[ − w ] . w p / 3 dw . V p / 3 exp dV n = n V n − 1 V n − 1 κ � p � = κ − p / 3 V p / 3 n − 1 Γ 3 + 1 hence, using induction and ℓ n /ℓ 0 = κ − n , � = κ � − κ V 1 κ � − κ V n � V p / 3 � � � � . V p / 3 exp dV 1 · · · exp dV n n n V 0 V 0 V n − 1 V n − 1 � p � n � ℓ n � p / 3 � p � n = κ − np / 3 V p / 3 = V p / 3 Γ 3 + 1 Γ 3 + 1 0 0 ℓ 0

• Therefore � ℓ n − ln( ℓ n /ℓ 0 ) + p � p ln �| v n | p � = ln � V p / 3 � = ln V p / 3 � � 3 ln ln Γ 3 +1 n 0 ℓ 0 ln κ � ℓ n � p 1 � p � ℓ n � ζ p � = ln V p / 3 � �� � V p / 3 + ln . 3 − ln κ ln Γ 3 + 1 = ln 0 0 ℓ 0 ℓ 0 where 1 ζ p = p � p � 3 − ln κ ln Γ 3 + 1 or � ζ p ∼ ℓ ζ p � ℓ n �| v n | p � = V p / 3 n 0 ℓ 0 as announced.

3. Estimating the probability distribution F ( u ) of the radial velocity increment u . Relation with Kolmogorov-Obukhov. • If r ≈ ℓ n we have u ≈ u n ≈ radial component of v n ⇒ rough estimate of the probability distribution of u : � ∞ n κ dV k 1 � � � e − κ V k / V k − 1 F ( u ) = χ [ − V 1 / 3 ] ( u ) , V 1 / 3 2 V 1 / 3 V k − 1 n n 0 n k =1 n dw k e − w k 2( κ n = 1 � � ) 1 / 3 � · · · w 1 / 3 V 0 w 1 ··· w n > ( κ n / V 0 ) | u | 3 k =1 k • The distribution G n ( y ) of y = ( κ n / V 0 ) 1 / 3 | u | is given by n dw k e − w k � � � G n ( y ) = · · · w 1 / 3 w 1 ··· w n > y 3 k =1 k

• This satisfies e t G n ( e t ) = ( φ ∗ ( n − 1) ∗ ψ )( t ) ( ∗∗ ) with � ∞ φ ( t ) = 3 exp(3 t − e 3 t ) ψ ( t ) = e t e − s φ ( s ) ds , t [ ⇒ G n ( y ) is a decreasing function of y ]. • For small u , G n 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 −∞ .

4. The onset of turbulence. • We may estimate the Reynolds number R e = | v 0 | ℓ 0 /ν for the onset of turbulence by taking κ 4 / 3 � V 0 ν ν � 1 / 3 � � � � � = R e − 1 � 1 ≈ = | v 1 | ℓ 1 V 1 / 3 κ V 1 κ − 1 ℓ 0 1 [Relation to dissipation is dictated by dimensional arguments] ⇒ � ∞ � − 1 / 3 κ dV 1 � κ V 1 R e ≈ κ 4 / 3 e − κ V 1 / V 0 V 0 V 0 0 � ∞ � 2 α − 1 / 3 d α e − α = κ 4 / 3 Γ � = κ 4 / 3 3 0 Taking 1 / ln κ = . 32 hence κ 4 / 3 = 64 . 5, with Γ(2 / 3) ≈ 1 . 354 gives R e ≈ 87 agreeing with R e ≈ 100 as found in Schumacher et al.

5. Problem. Study numerically the statistics of the decomposition of one eddy of order n − 1 into κ 3 eddies of order n .