turbulent weak solutions of the euler equationstpantev/seminarslides/vicol.upenn.pdf · turbulent...

Turbulent weak solutions of the Eulerequations

Vlad Vicol(Princeton University)

Colloquium, U PennFebruary 8, 2017

The Navier-Stokes equations

I consider an incompressible fluid continuum ⇢ R3, of constantdensity (⇢0 = 1)

I conservation of mass/incompressibility reads

rx · u = @x1u1 + @x2u2 + @x3u3 = 0

I conservation of momentum (Newton’s second law of motion) +assumptions on the internal stress tensor of the fluid yield

@t u + (u · rx)u = � 1⇢0

rxp + ⌫�xu + f

I unknowns: velocity field u(x , t) and internal pressure field p(x , t)

I ⌫ > 0 is the kinematic viscosityI f (x , t) is an external body forceI supplement with initial data u0 for the Cauchy problemI for simplicity: periodic boundary conditions T3

= [�⇡,⇡]3

The Navier-Stokes equations

I consider an incompressible fluid continuum ⇢ R3, of constantdensity (⇢0 = 1)

I conservation of mass/incompressibility reads

rx · u = @x1u1 + @x2u2 + @x3u3 = 0

I conservation of momentum (Newton’s second law of motion) +assumptions on the internal stress tensor of the fluid yield

@t u + (u · rx)u = � 1⇢0

rxp + ⌫�xu + f

I unknowns: velocity field u(x , t) and internal pressure field p(x , t)I ⌫ > 0 is the kinematic viscosityI f (x , t) is an external body forceI supplement with initial data u0 for the Cauchy problemI for simplicity: periodic boundary conditions T3

= [�⇡,⇡]3

History

@t u + (u · r)u = � 1⇢0

rp + ⌫�u + f

r · u = 0

I the Euler equations, formally obtained by setting ⌫ = 0 in theNavier-Stokes equations, were derived by L. Euler (1757)

I the introduction of internal viscous friction forces is due toC.L. Navier (1822) and G. Stokes (1845)

I to solve d’Alambert paradox: birds cannot fly in potential Euler flowI under the assumption that the shear stress is proportional to the

gradient of the velocity

Non-dimensionalizationI physical laws should hold independently of units:

I let U be the characteristic velocity in the flowI let L be the characteristic length scale in the flowI then T = U/L is the characteristic time scale in the flowI all the terms in Navier-Stokes/Euler have units LT�2

I x 0= xL�1; t 0 = ⌫tL�2; u0

= uL⌫�1; p0= pL2⌫�2⇢�1

0I drop the “primes”

I the dimensionally independent Navier-Stokes equations become

@t u + u · ru = �rp +

1Re

�u + f

r · u = 0

I the non-dimensional control parameter that measures thecomplexity of the flow is the Reynolds number

Re =

UL⌫

=

inertial forcesviscous forces

Control parameter: Reynolds number

I bacteria Re ⇡ 10�5

I blood flow in aorta Re ⇡ 102

I pitch in major league baseball Re ⇡ 105

I in the wake of a blue whale Re ⇡ 108

I in the wake of a Boeing 747 Re ⇡ 1012

Turbulent behavior in wake of cylinder

Figure: Von Karman vortex street behind circular cylinder at Re = 105. VanDyke (’82).

Figure: Wake behind two identical circular cylinder at Re = 240. Frisch (’95).

Homogenous Turbulence

Figure: Wake behind two identical circular cylinder at Re = 1800. Frisch (’95).

Figure: Homogenous Turbulence behind a grid at Re = 2300. Frisch (’95).

Experimentally observed statistical features

I in real flows turbulence is generated at boundariesI hydrodynamic turbulence deals with the universal statistical

features, which are expected to hold away from boundaries, andat small scales

I given the complexity observed in turbulent fluid flows it isunreasonable to predict pathwise behavior of solutions of theCauchy problem, instead...

I for any observable F of the solution u, let hF (u)i be a suitableaverage of the observed quantity

I in practice: long-time average hF (u)i = limT!11T

´ T0 F (u(t))dt

I in theory: ensemble average at statistical equilibriumhF (u)i =

´L2 F (u)dµRe(u), where µRe is the unique ergodic

invariant measure induced by the dynamicsI if the strong law of large numbers holds, we can drop the expected

value from the ergodic theorem, and the two concepts agree

Observables: Kinetic energyI In 3D Navier-Stokes (and Euler), the only a-priori controlled

quantity which is positive definite is the kinetic energy:

E(t) =12

T3

|u(t , x)|2dx .

I Proof of energy balance:I Taking the inner product of Navier-Stokes with u we obtain:

@t|u|2

2+r ·

✓u✓|u|2

2+ p

◆◆= f · u � 1

Re

|ru|2 + 1Re

�

|u|2

2

I Integrating over T3, we arrive at the energy balance:

dEdt

= � 1Re

T3

|ru|2 dx +

T3

f · u dx

I In Navier-Stokes: justified for any Leray-Hopf weak solution (i.e.u 2 L1

t L2x \ L2

t˙H1

x ), which obeys either u 2 L4t,x [Shinbrot ’74].

I Need not be a smooth solution, and need not be unique either !

Dissipation anomaly

I Recall the energy balance:

dEdt

= � 1Re

T3

|ru|2 dx +

T3

f · u dx

I The energy dissipation rate per unit mass is:

"Re =1Re

⌧ T3

|ru(·, x)|2dx�

I Experimentally:" = lim inf

Re!1"Re > 0

I Non-zero energy dissipation in inviscid limit: if all parameters arekept same, except the Reynolds number, which is sent to infinity,the mean energy dissipation per unit mass does not vanish.

Figure: The drag coefficient Cd behaves as a constant for Re � 1. Kineticenergy dissipated per unit mass �dE/dt is CdU3/L.

Structure functions

I Fix p � 1, ` > 0.I The longitudinal pth order structure function, is defined as

Sp(`) =

⌧ S2

T3

�� (u(x + `ˆz) � u(x)) · ˆz��pdxdˆz

�

I Units: the quantity ("`)p/3 has exactly the same units as Sp(`)!I Formally, this is the fundamental observation for Kolmogorov’s

1941 theory of turbulence.

The 4/5-lawI Four-fifths law: for ` in the inertial range (⌧ size of box, but �

molecular diffusion scale), and uniformly as Re ! 1 one sees:

S3(`) ⇠ �45"`

VOLUME 77, NUMBER 8 P HY S I CA L REV I EW LE T T ER S 19 AUGUST 1996

number k , 2.5. The total energy of all modes in eachof the first two shells was maintained constant in time.Simulations were carried out for ten eddy turnover times.The velocity initial conditions were prescribed to have aGaussian phase distribution with compact spectral supportat low wave numbers. The microscale Reynolds numberRl ≠ 220. Taylor’s hypothesis was not necessary.The microscale Reynolds number is only moderately

large in both experiment and simulations, and a criticalquestion concerns the scaling range. A traditional way—see for example, Ref. [9]—is to obtain the scaling regionfrom the flat part of kDu

3r

lyr versus r [3]. Unfortunately,it is not known if this procedure is valid exactly in thepresence of strong anisotropies such as occur [10] in pipeflows, or if some nontrivial correction is needed [11]. Wehave examined the extent of scaling in the energy spectraldensity, considered the so-called extended scale similarity(ESS) [12], and the notion of relative scaling [13] and,in general, the sensitivity of the results to the scalingregion used.Figure 1 shows a plot of the compensated spectral

density for the velocity data from the experiment; inaccordance with Taylor’s hypothesis, spectral frequency istreated as wave number. Scaling exists over a decade orso. We shall indicate this as the K range. Figure 2 plotsthe ratio kDu

3r

lyr against r for both the experiment andsimulations. The two flows are at comparable Reynoldsnumbers, yet the scaling region (to be called the Rrange) is substantially smaller for the experiment than forsimulations; it is definitely smaller than the K range. Thecutoff at the small-scale end is roughly the same in allcases, but the R range in the homogeneous simulationas well as the K range in the experiment extend tomuch larger scales (or lower frequencies) than doesthe experimental R range. That the scaling in one-

FIG. 1. The spectral density of u multiplied by f

5y3, where fis the frequency, plotted to show the flat region. Scaling occursover a decade (the K range). There is no perceptible differenceeven when a power-law exponent slightly different from 5y3 isused to compensate for the frequency rolloff.

dimensional longitudinal spectrum extends to smallerwave numbers than one should expect has been discussedin Monin and Yaglom [14], p. 357, but it has notbeen noted before that different manners of forcing andconsequent anisotropies can change the extent of thescaling so drastically. This matter will be discussedelsewhere in more detail. We have examined manystructure function plots and consistently used least-squarefits to the R range of Fig. 2 to obtain the numbers tobe quoted below, and verified that the relative trends arerobust even for the K range as well as for the ESS method.One noteworthy feature of the plus/minus structure

functions is shown in Fig. 3, which plots the logarithmof the ratio S

2q

yS

1q

against log10 r for various valuesof q. It can be seen readily that the ratio S

2q

yS

1q

isgreater than unity for all r , L whenever q . 1 andsmaller than unity whenever q , 1. Here L is the so-called integral scale of turbulence characteristic of thelarge scale turbulence. By definition, the ratio should beexactly unity for q ≠ 1. For one-dimensional data suchas those considered here, it follows from the definition ofgeneralized dimensions D

q

that the ratio of the minus toplus structure functions scales as

sryLdsq21dsD2q

2D

1q

d.

For consistency with the observation that S

2q

yS

1q

isgreater than unity for q . 1 and less for q , 1, oneshould have

D

2q

, D

1q

FIG. 2. The quantity kDu

3r

lyr as a function of r. Squares,experiment; circles, simulations; dots indicate Kolmogorov’s45 th law. It is believed that the slight bump in the leftpart of the experimental data is the bottleneck effect [see G.Falkovich, Phys. Fluids 6, 1411 (1994); D. Lohse and A.Mueller-Groeling, Phys. Rev. Lett. 74, 1747 (1995)]. Whilethe bottleneck effects discussed in these two papers referespecially to second-order structure functions (or to energyspectrum), a similar effect is likely to exist for the third-order aswell. This is typical of most measurements [see, for example,Y. Gagne, Docteur ès-Sciences Physiques Thèse, Université deGrenoble, France (1987)].

1489

Figure: The quantity �S3(r)/(r") as a function of r . Squares denoteexperimental observations of centerline in pipe flow at Re = 230000. Circlesindicate data from a 5123 DNS of homogenous turbulence at Re = 220. Dotsindicate Kolmogorov’s 4/5 law. [K. R. Sreenivasan et. al. – PRL ’96]

The 2/3-law

I Two-thirds law: for ` in the inertial range, uniformly as Re ! 1:

S2(`) ⇠ C2"2/3`2/3

Figure: log� log plot of S2(`) from the S1 wind tunnel of ONERA.Re ⇡ 3 ⇥ 107. [Frisch ’95]

Kolmogorov’s 5/3 power spectrumI Energy per unit volume in eddies of size �1: for in the inertial

range, uniformly as Re ! 1:

E() =12

dd

⌧ T3

|Pu|2dx�

⇡ "2/3�5/3

the particular shape of E(k) depends on the particular physical regime and the mechanism ofturbulence generation. At intermediate scales (i.e., in the inertial range) the spectrum shape is (14)and is assumed to be universal (H1). At scales smaller than the dissipation scale Ld (i.e., in thedissipation range), the spectrum falls off due to the disappearance of kinetic energy into the thermalreservoir of molecular collisions (i.e., “heat death”), and its shape is steeper than any power law,often taken to have a shape

E(k) � k�e�ck/kd

as k ! 1, where numerical calculations indicate that ↵ ⇡ 3.3 and c ⇡ 7.1 (Chen et al., 1993).The dissipation range spectrum is also universal if the inertial range is. Note that the total energyE is not universal; it depends on the particular dynamics of the energy-containing range and alsodepends weakly on the scale separation between Lo and Ld when Re is finite.

There is considerable experimental and observational evidence in support of the Kolmogorovspectrum shape (14), e.g., Fig. 2 from various experiments. There is also good experimental sup-port for an approximately universal shape for the dissipation range (Fig. 3).

Figure 2: Normalized energy spectra from nine different turbulent flows with Re� values rangingfrom 130 to 13,000, plotted in log-log coordinates. The wavenumber and energy spectrum havebeen divided by ln[Re�/Re�] with Re� = 75, and the resulting curves have been shifted to givethe best possible superposition. (Gagne and Castaing, 1991; also see Grant et al., 1961)

Another important length scale in turbulence is the Taylor microscale �, defined as the squareroot of the ratio of the variances of the velocity and the velocity gradient. We can estimate theformer as ⇠ V 2

o and the latter as ⇠ �/� from the definition of dissipation in the energy budget;hence,

� =

�V 2

0

��

�1/2

=

��Lo

Vo

�1/2

, (15)

6

Figure: Normalized energy spectra from nine different turbulent flows withRe� values ranging from 130 to 13, 000, plotted in log� log coordinates. Thewavenumber and energy spectrum have been divided by log[Re� /Re⇤], withRe⇤ = 75, and the resulting curves have been shifted to give the bestpossible superposition. [Gagne and Castaing, ’91].

Weak solutions of the Euler equations

I The Euler equations may be written as

@t u + r · (u ⌦ u + pI) = 0, r · u = 0.

where we have ignored the body force (f ⌘ 0).I Onsager [1949] connected the nonvanishing energy dissipation

rate in the Re ! 1 limit (" > 0) to the roughness of solutions tothe Euler equations...

I ... i.e., the solutions are “weak”I Weak solutions: the equation holds in the sense of distributions:

ˆR

ˆT3

u · @t�+ (u ⌦ u) : r� dxdt = 0

for any � 2 C10 (R ⇥ T3

), such that r · � = 0.I This makes sense as soon as u 2 L2

t,x .I Pressure is obtained by solving ��p = r · (r · (u ⌦ u)).

I Recall: for any C1 smooth incompressible vector field v , we haveˆT3(v · rv) · v dx =

ˆT3

v j@j|v |2

2dx = �

ˆT3(@j v j

)

|v |2

2dx = 0

or equivalentlyˆT3(v ⌦ v) : rv dx =

ˆT3(v i v j

)(@i v j) dx= 0.

I Thus, if u 2 C0t C1

x is a weak solution of the Euler equations, thenthe kinetic energy is conserved

ddt

E(t) =ddt

✓12

ˆT3

|u(x , t)|2dx◆= 0.

I Is the energy conserved for weak solutions?

I Let u 2 L1t L2

x be a weak solution of the Euler equations. Then:

E(t) = lim

!1E(t) = lim

!1

T3

|Pu(t , x)|2

2dx

I Taking the inner product of the Euler equations with P2u, and

integrating over space-time, we obtain:

E(T ) � E(0) =ˆ T

0

T3

@tPu · Pu dxdt

= �ˆ T

0

T3

P(u · ru) · Pu + rPp · Pu dxdt

=

ˆ T

0

T3

P(u ⌦ u) : rPu dxdt

=

ˆ T

0

T3

⇣P(u ⌦ u) � (Pu ⌦ Pu)

⌘: rPu dxdt

=:

ˆ T

0�⇧dt

Onsager’s theory of anomalous dissipation

I Fundamental object: energy flux through frequencies of size :⇧ =

�T3

⇣(Pu ⌦ Pu) � P(u ⌦ u)

⌘: rPu dx .

I Onsager ’49: E(t) is a constant function of time on [0,T ] iff

lim

!1

ˆ T

0⇧(t)dt = 0

“in principle, turbulent dissipation as described could take placejust as readily without the final assistance of viscosity. In theabsence of viscosity the standard proof of conservation ofenergy does not apply, because the velocity field does notremain differentiable! ”

I Moreover, the Onsager and Kolmogorov theories are connected:

lim

`!0�5S3(`)

4`=

Dlim

!1⇧

E= " = lim

Re!1"Re

assuming that the sequence of statistically steady weak solutionsof the Navier-Stokes equations, converge in L2

t,x to a statisticallysteady weak solution of the Euler equations. See [Eyink ’94].

Onsager conjecture

I Fix a Banach scale X↵, measuring the regularity of the weaksolution to 3D Euler. Examples:

I Onsager: X↵= C0

t C↵x . This is L1-based.

I 4/5th law: X↵= L3

t B↵3,1,x . This is L3-based.

I Kolmogorov spectrum: X↵= C0

t H↵x . This is L2-based.

I Given one of these Banach scales: the Onsager conjecture is todetermine a number ↵O , such that:

Rigidity For ↵ > ↵O , and any weak solution u 2 X↵ of the 3D Eulerequations, E is a constant function of time.

Flexibility For ↵ < ↵O , there exists a weak solution u 2 X↵ of the 3D Eulerequations, such that E is not constant in time.

I In the flexible regime, one may also ask for other softness of thePDE: e.g. can attain any given non-negative energy profile?

Does the value of the value of ↵O depend on X↵?

I Onsager stated his conjecture on the Holder-scale, andconjectured that ↵O = 1/3.

I The same value of ↵O = 1/3 is expected to hold on the L3-scale,in view of the 4/5th-law.

I If Kolmogorov’s predictions for the second order structurefunctions are robust, then we may expect that ↵O = 1/3 also onthe L2-scale.

I Or on any Lp scale for that matter... Sp(`) ⇡ ("`)p/3

I Except... [Landau-Lifshitz ’59]: the rate of energy dissipation isintermittent, i.e., spatially inhomogenous, and cannot be treatedas a constant. Thus ⇣p may deviate from the conjectured p/3, assoon as p 6= 3.

Deviations of ⇣p from p/3

Figure: Data: inverted white triangles: [Van Atta-Park ’72]; black circles, whitesquares, black triangles: [Anselmet, Gagne, Hopfinger, Antonia ’84] atRe = 515, 536, 852; + signs: from S1 ONERA wind tunnel. [Frisch ’95]

I lognormal model of [Kolmogorov ’62]:⇣p = p/3 � (µ/18)p(p � 3), with µ = 0.25. Here ⇣2 = 0.694444.

I �-model [Frisch-Sulem-Nelkin ’78]: ⇣p = p/3 + (3 � D)(1 � p/3),with D = 2.8. Here ⇣2 = 0.733333.

I log-Poisson model of [She-Leveque ’94]:⇣p = p/9 + 2(1 � (2/3)p/3). Here ⇣2 = 0.695937.

I mean-field theory of [Yakhot ’01]: ⇣p = ap/(b � cp) witha = 0.185; b = 0.473 and c = 0.0275. Here ⇣2 = 0.700758.

I Is there anything ”universal” for p 6= 2?

Rigid side of the Onsager conjectureI [Eyink ’94]: requires a bit more than L3

t C↵x , for ↵ > 1/3

I [Constantin-E-Titi ’94]: proveˆ T

0|⇧|dt . 1�3↵kuk3

L3t B↵

3,1

so that u 2 L3t B↵

3,1, with ↵ > 1/3 implies energy conservationI [Duchon-Robert ’00]

I [Cheskidov-Constantin-Friedlander-Shvydkoy ’08]: proveˆ T

0|⇧2j |dt .

1X

i=1

2�2/3|j�i|2ikP⇡2i uk3L3

so that u 2 L3t B1/3

3,c0implies

lim

j!1

ˆ T

0|⇧2j |dt = 0

and thus energy conservation. (Is sharp for Burgers.)

Rigid side of the Onsager conjectureI [Eyink ’94]: requires a bit more than L3

t C↵x , for ↵ > 1/3

I [Constantin-E-Titi ’94]: proveˆ T

0|⇧|dt . 1�3↵kuk3

L3t B↵

3,1

so that u 2 L3t B↵

3,1, with ↵ > 1/3 implies energy conservationI [Duchon-Robert ’00]I [Cheskidov-Constantin-Friedlander-Shvydkoy ’08]: prove

ˆ T

0|⇧2j |dt .

1X

i=1

2�2/3|j�i|2ikP⇡2i uk3L3

so that u 2 L3t B1/3

3,c0implies

lim

j!1

ˆ T

0|⇧2j |dt = 0

and thus energy conservation. (Is sharp for Burgers.)

Flexible side of the Onsager conjectureI [Scheffer ’93]: L2

t,x ; [Shnirelman ’00]: L1t L2

x with E 0 < 0.I These weak solutions have compact support in space and time.

I [De Lellis-Szekelyhidi ’09-’11]: L1t,x with E 0 < 0.

I Moreover, given any e(t) � 0 of compact support, there exists aweak solution u 2 L1

t,x such that E(t) = e(t).I These examples fit in a long tradition of counterintuitive examples

in geometry, such as the [Nash ’54]- [Kuiper ’55] isometricembedding: for any ✏ > 0 there is a C1 isometric embeddingu : Sn�1 ! B"(0) in Rn.

I A primary example of [Gromov ’86] h-principles: 8✏ > 0 and a C1

short embedding v , 9 a C1 isometric embedding u, s.t.ku � vkC0 ✏.

I Analogy between Dv in short embeddings, and subsolutions u ofthe Euler equation.

I Proof builds on [Muller -Sverak ’99]: there are critical points ofstrongly elliptic variational functionals F (u) =

´f (Du), with smooth

integrands f , which are W 1,1, but nowhere C1.I Solution built by adding up highly oscillatory plane waves, and the

convergence holds in w⇤ � L1.I This method faces serious difficulties for convergence in C0.






t,x such that E(t) = e(t).

I These examples fit in a long tradition of counterintuitive examplesin geometry, such as the [Nash ’54]- [Kuiper ’55] isometricembedding: for any ✏ > 0 there is a C1 isometric embeddingu : Sn�1 ! B"(0) in Rn.













t,x such that E(t) = e(t).I These examples fit in a long tradition of counterintuitive examples

in geometry, such as the [Nash ’54]- [Kuiper ’55] isometricembedding: for any ✏ > 0 there is a C1 isometric embeddingu : Sn�1 ! B"(0) in Rn.








I [De Lellis-Szekelyhidi ’12]: C1/10�t,x

I [Isett ’13]; [Buckmaster-De Lellis-Szekelyhidi ’13]: C1/5�t,x

I [Buckmaster ’13]: C1/3�x a.e. in t and C1/5�

t,x ;[Buckmaster-De Lellis-Szekelyhidi ’14]: L1

t C1/3�x

I [Buckmaster-Masmoudi-V. ’16]: C0t H1/3�

x

I [Isett ’16]: C0t C1/3�

x with compact support in time.

I [De Lellis - Buckmaster - Szekelyhidi - V. ’17]: C0t C1/3�

x whichcan attain any positive energy profile, and h-principles hold.

Main result

Theorem (Buckmaster - Masmoudi - V. ’17)Fix any ↵ < 5/14. There exist infinitely many weak solutions

u 2 C0t H↵

x

of the 3D Euler equations which have compact support in time.

I First result with regularity above 1/3!I First analytical confirmation that S2(`) may deviate from `2/3.

Main result


u 2 C0t H↵

x


I First result with regularity above 1/3!

I First analytical confirmation that S2(`) may deviate from `2/3.

Main result


u 2 C0t H↵

x


I First result with regularity above 1/3!I First analytical confirmation that S2(`) may deviate from `2/3.

Main result


u 2 C0t H↵

x


I Main new ideas:I explore spatial intermittency of turbulent flowI use Beltrami waves to build Dirichlet-kernel-like objects, as the

building blocks in convex integrationI implement higher order Reynolds stress corrections

I 5/14 is probably not sharp. [Frisch-Sulem ’75]: for ↵ > 5/6energy is conserved.

I Result does not work in 2D: there are not enough points withinteger coordinates on growing circles.

Thank you!

Figure: The level sets of large vorticity in a numerical simulation of fullydeveloped turbulence display isotropic small scale coherent features.

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