The Navier-Stokes equations I consider an incompressible fluid continuum R 3, of constant density ( 0 = 1) I conservation of mass/incompressibility re

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1 Turbulent weak solutions of the Euler equations Vlad Vicol (Princeton University) Colloquium, U Penn February 8, 2017

2 The Navier-Stokes equations I consider an incompressible fluid continuum R 3, of constant density ( 0 = 1) I conservation of mass/incompressibility reads r x u x1 u 1 x2 u 2 x3 u 3 = 0 I conservation of momentum (Newton s second law of motion) + assumptions on the internal stress tensor of the fluid t u +(u r x )u = 1 0 r x p + x u + f I unknowns: velocity field u(x, t) and internal pressure field p(x, t)

3 The Navier-Stokes equations I consider an incompressible fluid continuum R 3, of constant density ( 0 = 1) I conservation of mass/incompressibility reads r x u x1 u 1 x2 u 2 x3 u 3 = 0 I conservation of momentum (Newton s second law of motion) + assumptions on the internal stress tensor of the fluid t u +(u r x )u = 1 0 r x p + x u + f I unknowns: velocity field u(x, t) and internal pressure field p(x, t) I >0is the kinematic viscosity I f (x, t) is an external body force I supplement with initial data u 0 for the Cauchy problem I for simplicity: periodic boundary conditions T 3 =[, ] 3

4 t u +(u r)u = 1 0 rp + u + f r u = 0 I the Euler equations, formally obtained by setting = 0 in the Navier-Stokes equations, were derived by L. Euler (1757) I the introduction of internal viscous friction forces is due to C.L. Navier (1822) and G. Stokes (1845) I to solve d Alambert paradox: birds cannot fly in potential Euler flow I under the assumption that the shear stress is proportional to the gradient of the velocity

5 Non-dimensionalization I physical laws should hold independently of units: I let U be the characteristic velocity in the flow I let L be the characteristic length scale in the flow I then T = U/L is the characteristic time scale in the flow I all the terms in Navier-Stokes/Euler have units LT 2 I x 0 = xl 1 ; t 0 = tl 2 ; u 0 = ul 1 ; p 0 = pl I drop the primes I the dimensionally independent Navier-Stokes equations t u + u ru = r u = 0 rp + 1 Re u + f

6 Non-dimensionalization I physical laws should hold independently of units: I let U be the characteristic velocity in the flow I let L be the characteristic length scale in the flow I then T = U/L is the characteristic time scale in the flow I all the terms in Navier-Stokes/Euler have units LT 2 I x 0 = xl 1 ; t 0 = tl 2 ; u 0 = ul 1 ; p 0 = pl I drop the primes I the dimensionally independent Navier-Stokes equations t u + u ru = r u = 0 rp + 1 Re u + f I the non-dimensional control parameter that measures the complexity of the flow is the Reynolds number Re = UL = inertial forces viscous forces

7 Control parameter: Reynolds number I bacteria Re 10 5 I blood flow in aorta Re 10 2 I pitch in major league baseball Re 10 5 I in the wake of a blue whale Re 10 8 I in the wake of a Boeing 747 Re 10 12

8 Turbulent behavior in wake of cylinder Figure: Von Karman vortex street behind circular cylinder at Re = 105. Van Dyke ( 82). Figure: Wake behind two identical circular cylinder at Re = 240. Frisch ( 95).

9 Homogenous Turbulence Figure: Wake behind two identical circular cylinder at Re = Frisch ( 95). Figure: Homogenous Turbulence behind a grid at Re = Frisch ( 95).

10 Experimentally observed statistical features I in real flows turbulence is generated at boundaries I hydrodynamic turbulence deals with the universal statistical features, which are expected to hold away from boundaries, and at small scales

11 Experimentally observed statistical features I in real flows turbulence is generated at boundaries I hydrodynamic turbulence deals with the universal statistical features, which are expected to hold away from boundaries, and at small scales I given the complexity observed in turbulent fluid flows it is unreasonable to predict pathwise behavior of solutions of the Cauchy problem, instead...

12 Experimentally observed statistical features I in real flows turbulence is generated at boundaries I hydrodynamic turbulence deals with the universal statistical features, which are expected to hold away from boundaries, and at small scales I given the complexity observed in turbulent fluid flows it is unreasonable to predict pathwise behavior of solutions of the Cauchy problem, instead... I for any observable F of the solution u, let hf(u)i be a suitable average of the observed quantity I in practice: long-time average hf(u)i 1 =limt!1 F(u(t))dt T 0 I in theory: ensemble average at statistical equilibrium hf(u)i = L F(u)dµ 2 Re (u), where µ Re is the unique ergodic invariant measure induced by the dynamics I if the strong law of large numbers holds, we can drop the expected value from the ergodic theorem, and the two concepts agree T

13 Observables: Kinetic energy I In 3D Navier-Stokes (and Euler), the only a-priori controlled quantity which is positive definite is the kinetic energy: E(t) = 1 u(t, x) 2 dx. 2 T 3

14 Observables: Kinetic energy I In 3D Navier-Stokes (and Euler), the only a-priori controlled quantity which is positive definite is the kinetic energy: E(t) = 1 u(t, x) 2 dx. 2 T 3 I Proof of energy balance: I Taking the inner product of Navier-Stokes with u we obtain: u 2 u t 2 + r u 2 + p = f u 1 Re ru Re u 2 2

15 Observables: Kinetic energy I In 3D Navier-Stokes (and Euler), the only a-priori controlled quantity which is positive definite is the kinetic energy: E(t) = 1 u(t, x) 2 dx. 2 T 3 I Proof of energy balance: I Taking the inner product of Navier-Stokes with u we obtain: u 2 u t 2 + r u 2 + p = f u 1 Re ru Re u 2 2 I Integrating over T 3, we arrive at the energy balance: de dt = 1 ru 2 dx + Re T 3 T 3 f u dx

16 Observables: Kinetic energy I In 3D Navier-Stokes (and Euler), the only a-priori controlled quantity which is positive definite is the kinetic energy: E(t) = 1 u(t, x) 2 dx. 2 T 3 I Proof of energy balance: I Taking the inner product of Navier-Stokes with u we obtain: u 2 u t 2 + r u 2 + p = f u 1 Re ru Re u 2 2 I Integrating over T 3, we arrive at the energy balance: de dt = 1 ru 2 dx + Re T 3 T 3 f u dx I In Navier-Stokes: justified for any Leray-Hopf weak solution (i.e. u 2 L 1 t L 2 x \ L 2 t Ḣ1 x ), which obeys either u 2 L 4 t,x [Shinbrot 74]. I Need not be a smooth solution, and need not be unique either!

17 Dissipation anomaly I Recall the energy balance: de dt = 1 ru 2 dx + Re T 3 I The energy dissipation rate per unit mass is: " Re = 1 ru(, x) 2 dx Re T 3 T 3 f u dx

18 Dissipation anomaly I Recall the energy balance: de dt = 1 ru 2 dx + Re T 3 I The energy dissipation rate per unit mass is: " Re = 1 ru(, x) 2 dx Re T 3 T 3 f u dx I Experimentally: " =liminf Re!1 " Re > 0 I Non-zero energy dissipation in inviscid limit: if all parameters are kept same, except the Reynolds number, which is sent to infinity, the mean energy dissipation per unit mass does not vanish.

19 Figure: The drag coefficient Cd behaves as a constant for Re energy dissipated per unit mass de/dt is CdU 3 /L. 1. Kinetic

20 Structure functions I Fix p 1, `>0. I The longitudinal p th order structure function, is defined as S p (`) = (u(x + `ẑ) u(x)) ẑ p dxdẑ S 2 T 3

21 Structure functions I Fix p 1, `>0. I The longitudinal p th order structure function, is defined as S p (`) = (u(x + `ẑ) u(x)) ẑ p dxdẑ S 2 T 3 I Units: the quantity ("`) p/3 has exactly the same units as S p (`)! I Formally, this is the fundamental observation for Kolmogorov s 1941 theory of turbulence.

22 The 4/5-law I Four-fifths law: for ` in the inertial range ( size of box, but molecular diffusion scale), and uniformly as Re!1one sees: S 3 (`) 4 5 "` Figure: The quantity S 3 (r)/(r") as a function of r. Squares denote experimental observations of centerline in pipe flow at Re = Circles indicate data from a DNS of homogenous turbulence at Re = 220. Dots indicate Kolmogorov s 4/5 law. [K. R. Sreenivasan et. al. PRL 96]

23 The 2/3-law I Two-thirds law: for ` in the inertial range, uniformly as Re!1: S 2 (`) C 2 " 2/3`2/3 Figure: log log plot of S 2 (`) from the S1 wind tunnel of ONERA. Re [Frisch 95]

24 Kolmogorov s 5/3 power spectrum I Energy per unit volume in eddies of size apple 1 : for apple in the inertial range, uniformly as Re!1: E(apple) = 1 d P appleapple u 2 dx " 2/3 apple 5/3 2 dapple T 3 Figure: Normalized energy spectra from nine different turbulent flows with Re values ranging from 130 to 13, 000, plotted in log log coordinates. The wavenumber and energy spectrum have been divided by log[re / Re ], with Re = 75, and the resulting curves have been shifted to give the best possible superposition. [Gagne and Castaing, 91].

25 Weak solutions of the Euler equations I The Euler equations may be written 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!1limit (" >0) to the roughness of solutions to the Euler equations...

26 Weak solutions of the Euler equations I The Euler equations may be written 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!1limit (" >0) to the roughness of solutions to the Euler equations... I... i.e., the solutions are weak I Weak solutions: the equation holds in the sense of distributions: ˆ ˆ t +(u u): r dxdt = 0 R T 3 for any 2 C 1 0 (R T3 ), such that r = 0. I This makes sense as soon as u 2 L 2 t,x. I Pressure is obtained by solving p = r (r (u u)).

27 I Recall: for any C 1 smooth incompressible vector field v, we have ˆ T 3 (v rv) v dx = ˆT3 v j v 2 2 dx = ˆT3 (@ j v j ) v 2 2 dx = 0 or equivalently ˆ ˆ (v v): rv dx = (v i v j )(@ i v j ) dx= 0. T 3 T 3 I Thus, if u 2 Ct 0C1 x is a weak solution of the Euler equations, then the kinetic energy is conserved d dt E(t) = d ˆ 1 u(x, t) 2 dx = 0. dt 2 T 3 I Is the energy conserved for weak solutions?

28 I Let u 2 L 1 t L 2 x be a weak solution of the Euler equations. Then: E(t) = lim apple!1 E apple(t) = lim apple!1 T 3 P appleapple u(t, x) 2 dx 2

29 I Let u 2 L 1 t L 2 x be a weak solution of the Euler equations. Then: E(t) = lim apple!1 E apple(t) = lim apple!1 T 3 P appleapple u(t, x) 2 dx 2 I Taking the inner product of the Euler equations with Pappleapple 2 u, and integrating over space-time, we obtain: E apple (T ) E apple (0) = ˆ T 0 T t P appleapple u P appleapple u dxdt

30 I Let u 2 L 1 t L 2 x be a weak solution of the Euler equations. Then: E(t) = lim apple!1 E apple(t) = lim apple!1 T 3 P appleapple u(t, x) 2 dx 2 I Taking the inner product of the Euler equations with Pappleapple 2 u, and integrating over space-time, we obtain: E apple (T ) E apple (0) = = ˆ T 0 T t P appleapple u P appleapple u dxdt ˆ T 0 T 3 P appleapple (u ru) P appleapple u + rp appleapple p P appleapple u dxdt

31 I Let u 2 L 1 t L 2 x be a weak solution of the Euler equations. Then: E(t) = lim apple!1 E apple(t) = lim apple!1 T 3 P appleapple u(t, x) 2 dx 2 I Taking the inner product of the Euler equations with Pappleapple 2 u, and integrating over space-time, we obtain: E apple (T ) E apple (0) = = = ˆ T 0 T t P appleapple u P appleapple u dxdt ˆ T ˆ T 0 T 3 P appleapple (u ru) P appleapple u + rp appleapple p P appleapple u dxdt 0 T 3 P appleapple (u u): rp appleapple u dxdt

32 I Let u 2 L 1 t L 2 x be a weak solution of the Euler equations. Then: E(t) = lim apple!1 E apple(t) = lim apple!1 T 3 P appleapple u(t, x) 2 dx 2 I Taking the inner product of the Euler equations with Pappleapple 2 u, and integrating over space-time, we obtain: E apple (T ) E apple (0) = = = = ˆ T 0 T t P appleapple u P appleapple u dxdt ˆ T ˆ T 0 T 3 P appleapple (u ru) P appleapple u + rp appleapple p P appleapple u dxdt P appleapple (u u): rp appleapple u dxdt 0 T 3 P appleapple (u u) (P appleapple u P appleapple u) : rp appleapple u dxdt ˆ T 0 T 3

33 I Let u 2 L 1 t L 2 x be a weak solution of the Euler equations. Then: E(t) = lim apple!1 E apple(t) = lim apple!1 T 3 P appleapple u(t, x) 2 dx 2 I Taking the inner product of the Euler equations with Pappleapple 2 u, and integrating over space-time, we obtain: E apple (T ) E apple (0) = = = = =: ˆ T 0 T t P appleapple u P appleapple u dxdt ˆ T ˆ T 0 T 3 P appleapple (u ru) P appleapple u + rp appleapple p P appleapple u dxdt P appleapple (u u): rp appleapple u dxdt 0 T 3 P appleapple (u u) (P appleapple u P appleapple u) : rp appleapple u dxdt ˆ T 0 T ˆ 3 T 0 apple dt

34 Onsager s theory of anomalous dissipation I Fundamental object: energy flux through frequencies of size apple: apple = T 3 (P appleapple u P appleapple u) P appleapple (u u) : rp appleapple u dx. I Onsager 49: E(t) is a constant function of time on [0, T ] iff ˆ T lim apple (t)dt = 0 apple!1 0 in principle, turbulent dissipation as described could take place just as readily without the final assistance of viscosity. In the absence of viscosity the standard proof of conservation of energy does not apply, because the velocity field does not remain differentiable!

35 I Moreover, the Onsager and Kolmogorov theories are connected: lim `!0 5S 3 (`) 4` D E = lim apple apple!1 = " = lim Re!1 " Re assuming that the sequence of statistically steady weak solutions of the Navier-Stokes equations, converge in L 2 t,x to a statistically steady weak solution of the Euler equations. See [Eyink 94].

36 Onsager conjecture I Fix a Banach scale X, measuring the regularity of the weak solution to 3D Euler. Examples: I Onsager: X = Ct 0 Cx. This is L 1 -based. I 4/5th law: X = L 3 t B3,1,x. This is L 3 -based. I Kolmogorov spectrum: X = Ct 0 Hx. This is L 2 -based.

37 Onsager conjecture I Fix a Banach scale X, measuring the regularity of the weak solution to 3D Euler. Examples: I Onsager: X = Ct 0 Cx. This is L 1 -based. I 4/5th law: X = L 3 t B3,1,x. This is L 3 -based. I Kolmogorov spectrum: X = Ct 0 Hx. This is L 2 -based. I Given one of these Banach scales: the Onsager conjecture is to determine a number O, such that: Rigidity For > O, and any weak solution u 2 X of the 3D Euler equations, E is a constant function of time. Flexibility For < O, there exists a weak solution u 2 X of the 3D Euler equations, such that E is not constant in time.

38 Onsager conjecture I Fix a Banach scale X, measuring the regularity of the weak solution to 3D Euler. Examples: I Onsager: X = Ct 0 Cx. This is L 1 -based. I 4/5th law: X = L 3 t B3,1,x. This is L 3 -based. I Kolmogorov spectrum: X = Ct 0 Hx. This is L 2 -based. I Given one of these Banach scales: the Onsager conjecture is to determine a number O, such that: Rigidity For > O, and any weak solution u 2 X of the 3D Euler equations, E is a constant function of time. Flexibility For < O, there exists a weak solution u 2 X of the 3D Euler equations, such that E is not constant in time. I In the flexible regime, one may also ask for other softness of the PDE: e.g. can attain any given non-negative energy profile?

39 Does the value of the value of O depend on X? I Onsager stated his conjecture on the Hölder-scale, and conjectured that O = 1/3. I The same value of O = 1/3 is expected to hold on the L 3 -scale, in view of the 4/5th-law.

40 Does the value of the value of O depend on X? I Onsager stated his conjecture on the Hölder-scale, and conjectured that O = 1/3. I The same value of O = 1/3 is expected to hold on the L 3 -scale, in view of the 4/5th-law. I If Kolmogorov s predictions for the second order structure functions are robust, then we may expect that O = 1/3 also on the L 2 -scale. I Or on any L p scale for that matter... S p (`) ("`) p/3

41 Does the value of the value of O depend on X? I Onsager stated his conjecture on the Hölder-scale, and conjectured that O = 1/3. I The same value of O = 1/3 is expected to hold on the L 3 -scale, in view of the 4/5th-law. I If Kolmogorov s predictions for the second order structure functions are robust, then we may expect that O = 1/3 also on the L 2 -scale. I Or on any L p scale for that matter... S p (`) ("`) p/3 I Except... [Landau-Lifshitz 59]: the rate of energy dissipation is intermittent, i.e., spatially inhomogenous, and cannot be treated as a constant. Thus p may deviate from the conjectured p/3, as soon as p 6= 3.

42 Deviations of p from p/3 Figure: Data: inverted white triangles: [Van Atta-Park 72]; black circles, white squares, black triangles: [Anselmet, Gagne, Hopfinger, Antonia 84] at Re = 515, 536, 852; + signs: from S1 ONERA wind tunnel. [Frisch 95]

43 I lognormal model of [Kolmogorov 62]: p = p/3 (µ/18)p(p 3), with µ = Here 2 = I -model [Frisch-Sulem-Nelkin 78]: p = p/3 +(3 D)(1 p/3), with D = 2.8. Here 2 = I log-poisson model of [She-Leveque 94]: p = p/9 + 2(1 (2/3)p/3). Here 2 = I mean-field theory of [Yakhot 01]: p = ap/(b cp) with a = 0.185; b = and c = Here 2 = I Is there anything universal for p 6= 2?

44 Rigid side of the Onsager conjecture I [Eyink 94]: requires a bit more than L 3 t C x, for >1/3 I [Constantin-E-Titi 94]: prove ˆ T 0 apple dt. apple 1 3 kuk 3 L 3 t B 3,1 so that u 2 L 3 t B 3,1, with >1/3implies energy conservation I [Duchon-Robert 00]

45 Rigid side of the Onsager conjecture I [Eyink 94]: requires a bit more than L 3 t C x, for >1/3 I [Constantin-E-Titi 94]: prove ˆ T 0 apple dt. apple 1 3 kuk 3 L 3 t B 3,1 so that u 2 L 3 t B 3,1, with >1/3implies energy conservation I [Duchon-Robert 00] I [Cheskidov-Constantin-Friedlander-Shvydkoy 08]: prove ˆ T 0 2 j dt. 1X 2 2/3 j i 2 i kp 2 i uk 3 L 3 i=1 so that u 2 L 3 t B1/3 3,c 0 implies ˆ T lim 2 j dt = 0 j!1 0 and thus energy conservation. (Is sharp for Burgers.)

46 Flexible side of the Onsager conjecture I [Scheffer 93]: L 2 t,x ; [Shnirelman 00]: L1 t L 2 x with E 0 < 0. I These weak solutions have compact support in space and time.

47 Flexible side of the Onsager conjecture I [Scheffer 93]: L 2 t,x ; [Shnirelman 00]: L1 t L 2 x with E 0 < 0. I These weak solutions have compact support in space and time. I [De Lellis-Szekelyhidi 09-11]: L 1 t,x with E 0 < 0. I Moreover, given any e(t) 0 of compact support, there exists a weak solution u 2 L 1 t,x such that E(t) =e(t).

48 Flexible side of the Onsager conjecture I [Scheffer 93]: L 2 t,x ; [Shnirelman 00]: L1 t L 2 x with E 0 < 0. I These weak solutions have compact support in space and time. I [De Lellis-Szekelyhidi 09-11]: L 1 t,x with E 0 < 0. I Moreover, given any e(t) 0 of compact support, there exists a weak solution u 2 L 1 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] isometric embedding: for any >0there is a C 1 isometric embedding u : S n 1! B "(0) in R n. I A primary example of [Gromov 86] h-principles: 8 >0 and a C 1 short embedding v, 9 a C 1 isometric embedding u, s.t. ku vk C 0 apple. I Analogy between Dv in short embeddings, and subsolutions u of the Euler equation. I Proof builds on [Muller -Sverak 99]: there are critical points of strongly elliptic variational functionals F(u) = f (Du), with smooth integrands f, which are W 1,1, but nowhere C 1. I Solution built by adding up highly oscillatory plane waves, and the convergence holds in w L 1. I This method faces serious difficulties for convergence in C 0.

49 I [De Lellis-Szekelyhidi 12]: C 1/10 t,x I [Isett 13]; [Buckmaster-De Lellis-Szekelyhidi 13]: C 1/5 t,x I [Buckmaster 13]: C 1/3 x a.e. in t and C 1/5 t,x ; [Buckmaster-De Lellis-Szekelyhidi 14]: L 1 t C1/3 x

50 I [De Lellis-Szekelyhidi 12]: C 1/10 t,x I [Isett 13]; [Buckmaster-De Lellis-Szekelyhidi 13]: C 1/5 t,x I [Buckmaster 13]: C 1/3 x a.e. in t and C 1/5 t,x ; [Buckmaster-De Lellis-Szekelyhidi 14]: L 1 t C1/3 x I [Buckmaster-Masmoudi-V. 16]: C 0 t H1/3 x

51 I [De Lellis-Szekelyhidi 12]: C 1/10 t,x I [Isett 13]; [Buckmaster-De Lellis-Szekelyhidi 13]: C 1/5 t,x I [Buckmaster 13]: C 1/3 x a.e. in t and C 1/5 t,x ; [Buckmaster-De Lellis-Szekelyhidi 14]: L 1 t C1/3 x I [Buckmaster-Masmoudi-V. 16]: C 0 t H1/3 x I [Isett 16]: C 0 t C1/3 x with compact support in time.

52 I [De Lellis-Szekelyhidi 12]: C 1/10 t,x I [Isett 13]; [Buckmaster-De Lellis-Szekelyhidi 13]: C 1/5 t,x I [Buckmaster 13]: C 1/3 x a.e. in t and C 1/5 t,x ; [Buckmaster-De Lellis-Szekelyhidi 14]: L 1 t C1/3 x I [Buckmaster-Masmoudi-V. 16]: C 0 t H1/3 x I [Isett 16]: C 0 t C1/3 x with compact support in time. I [De Lellis - Buckmaster - Szekelyhidi - V. 17]: Ct 0C1/3 x which can attain any positive energy profile, and h-principles hold.

53 Main result Theorem (Buckmaster - Masmoudi - V. 17) Fix any <5/14. There exist infinitely many weak solutions u 2 C 0 t H x of the 3D Euler equations which have compact support in time.

54 Main result Theorem (Buckmaster - Masmoudi - V. 17) Fix any <5/14. There exist infinitely many weak solutions u 2 C 0 t H x of the 3D Euler equations which have compact support in time. I First result with regularity above 1/3!

55 Main result Theorem (Buckmaster - Masmoudi - V. 17) Fix any <5/14. There exist infinitely many weak solutions u 2 Ct 0 Hx of the 3D Euler equations which have compact support in time. I First result with regularity above 1/3! I First analytical confirmation that S 2 (`) may deviate from `2/3.

56 Main result Theorem (Buckmaster - Masmoudi - V. 17) Fix any <5/14. There exist infinitely many weak solutions u 2 Ct 0 Hx of the 3D Euler equations which have compact support in time. I Main new ideas: I explore spatial intermittency of turbulent flow I use Beltrami waves to build Dirichlet-kernel-like objects, as the building blocks in convex integration I implement higher order Reynolds stress corrections

57 Main result Theorem (Buckmaster - Masmoudi - V. 17) Fix any <5/14. There exist infinitely many weak solutions u 2 C 0 t H x of the 3D Euler equations which have compact support in time. I Main new ideas: I explore spatial intermittency of turbulent flow I use Beltrami waves to build Dirichlet-kernel-like objects, as the building blocks in convex integration I implement higher order Reynolds stress corrections I 5/14 is probably not sharp. [Frisch-Sulem 75]: for >5/6 energy is conserved.

58 Main result Theorem (Buckmaster - Masmoudi - V. 17) Fix any <5/14. There exist infinitely many weak solutions u 2 C 0 t H x of the 3D Euler equations which have compact support in time. I Main new ideas: I explore spatial intermittency of turbulent flow I use Beltrami waves to build Dirichlet-kernel-like objects, as the building blocks in convex integration I implement higher order Reynolds stress corrections I 5/14 is probably not sharp. [Frisch-Sulem 75]: for >5/6 energy is conserved. I Result does not work in 2D: there are not enough points with integer coordinates on growing circles.

59 Thank you! Figure: The level sets of large vorticity in a numerical simulation of fully developed turbulence display isotropic small scale coherent features.

Nonuniqueness of weak solutions to the Navier-Stokes equation

Nonuniqueness of weak solutions to the Navier-Stokes equation Tristan Buckmaster (joint work with Vlad Vicol) Princeton University November 29, 2017 Tristan Buckmaster (Princeton University) Nonuniqueness