A Proof of Onsager s Conjecture for the Incompressible Euler Equations

Transcription

1 A Proof of Onsager s Conjecture for the Incompressible Euler Equations Philip Isett UT Austin September 27, 2017

2 Outline Motivation Weak Solutions to the Euler Equations Onsager s Conjecture and Turbulence Brief Survey of Previous Results A Proof of Onsager s Conjecture

3 Motivation: Weak Solutions to the Euler equations The incompressible Euler equations for a homogeneous fluid: t v l + j (v j v l ) + l p = 0 (1) j v j = 0 (2)

4 Motivation: Weak Solutions to the Euler equations The incompressible Euler equations for a homogeneous fluid: t v l + j (v j v l ) + l p = 0 (1) j v j = 0 (2) make sense in integral form for continuous (v, p):

5 Motivation: Weak Solutions to the Euler equations The incompressible Euler equations for a homogeneous fluid: t v l + j (v j v l ) + l p = 0 (1) j v j = 0 (2) make sense in integral form for continuous (v, p): d v l (t, x)dx = p(t, x)n l dσ + v l (t, x)(v n)dσ (3) dt Ω Ω Ω Ω (v n)(t, x)dσ(x) = 0 (4) for all Ω with smooth boundary Ω and interior unit normal n l.

6 Motivation: Sufficiently smooth solutions conserve energy

7 Motivation: Sufficiently smooth solutions conserve energy Take the dot product of the Euler equations with v l v l t v l + v l j (v j v l ) + v l l p = 0 j v j = 0

8 Motivation: Sufficiently smooth solutions conserve energy Take the dot product of the Euler equations with v l v l t v l + v l j (v j v l ) + v l l p = 0 j v j = 0 Then, use the divergence free condition div v = l v l = 0, and integrate d Rn v 2 ] (t, x)dx = div [( v 2 dt 2 R n 2 + p)v dx = 0

9 Motivation: Onsager s Conjecture (1949) 1. Solutions (v, p) to Euler obeying a Hölder estimate t v l + j (v j v l ) + l p = 0 (5) j v j = 0 v(t, x + x) v(t, x) C x α (6) for some α > 1/3 must conserve energy.

10 Motivation: Onsager s Conjecture (1949) 1. Solutions (v, p) to Euler obeying a Hölder estimate t v l + j (v j v l ) + l p = 0 (5) j v j = 0 v(t, x + x) v(t, x) C x α (6) for some α > 1/3 must conserve energy. 2. If the α in (6) is less than 1/3, then v may fail to conserve energy

11 Motivation: Hydrodynamic turbulence Kolmogorov (1941): As ν 0 for solutions to 3D Navier-Stokes: { t v l + j (v j v l ) + l p = ν v l j v j (7) = 0 the energy dissipation rate remains strictly positive as ν 0 ε = lim d vν 2 (t, x)dx > 0. ν 0 dt 2

12 Motivation: Hydrodynamic turbulence Kolmogorov (1941): As ν 0 for solutions to 3D Navier-Stokes: { t v l + j (v j v l ) + l p = ν v l j v j (7) = 0 the energy dissipation rate remains strictly positive as ν 0 ε = lim d vν 2 (t, x)dx > 0. ν 0 dt 2 The velocity fluctuations on average obey a law v(x + x) v(x) p 1/p ε 1/3 x 1/3 for x ( ν 3 /ε ) 1/4

13 Motivation: Hydrodynamic turbulence Kolmogorov (1941): As ν 0 for solutions to 3D Navier-Stokes: { t v l + j (v j v l ) + l p = ν v l j v j (7) = 0 the energy dissipation rate remains strictly positive as ν 0 ε = lim d vν 2 (t, x)dx > 0. ν 0 dt 2 The velocity fluctuations on average obey a law v(x + x) v(x) p 1/p ε 1/3 x 1/3 for x ( ν 3 /ε ) 1/4 Onsager considered the case ν = 0; argued that frequency cascades may lead to energy dissipation in the absence of viscosity.

14 Onsager and Ideal Turbulence Onsager considered the Euler equations in Fourier series form (which converges for v L 2 ) v(x, t) = a k (t)e ik x k da k dt = i [ a k m k a m + (a ] m k)k k 2 m He argued that energy can cascade from low wavenumbers to high wavenumbers, and the cascade can happen so rapidly that part of the energy k a k 2 escapes to infinite frequency (i.e. vanishes to small spatial scales) in finite time. However, only low regularity solutions could behave this way, and he stated that solutions in C α with α > 1/3 must conserve energy.

15 Onsager and Ideal Turbulence Onsager considered the Euler equations in Fourier series form (which converges for v L 2 ) v(x, t) = a k (t)e ik x k da k dt = i [ a k m k a m + (a ] m k)k k 2 m He argued that energy can cascade from low wavenumbers to high wavenumbers, and the cascade can happen so rapidly that part of the energy k a k 2 escapes to infinite frequency (i.e. vanishes to small spatial scales) in finite time. By a statistical physics argument, a typical turbulent flow should have: λ 2 k 2λ a k 2 λ 2/3 (hence regularity exactly 1/3).

16 Motivation: Onsager s Conjecture (1949) 1. Solutions (v, p) to Euler obeying a Hölder estimate t v l + j (v j v l ) + l p = 0 (8) j v j = 0 v(t, x + x) v(t, x) C x α (9) for some α > 1/3 must conserve energy. 2. If the α in (9) is less than 1/3, then v may fail to conserve energy

17 Motivation: Onsager s Conjecture (1949) 1. Solutions (v, p) to Euler obeying a Hölder estimate t v l + j (v j v l ) + l p = 0 (8) j v j = 0 v(t, x + x) v(t, x) C x α (9) for some α > 1/3 must conserve energy. 2. If the α in (9) is less than 1/3, then v may fail to conserve energy Part 1 is known: (Eyink, 94), (Constantin-E-Titi, 94)

18 Motivation: Onsager s Conjecture (1949) 1. Solutions (v, p) to Euler obeying a Hölder estimate t v l + j (v j v l ) + l p = 0 (8) j v j = 0 v(t, x + x) v(t, x) C x α (9) for some α > 1/3 must conserve energy. 2. If the α in (9) is less than 1/3, then v may fail to conserve energy Part 1 is known: (Eyink, 94), (Constantin-E-Titi, 94) Refinements: (Duchon-Robert 00), (Cheskidov-Constantin- Shvydkoy-Friedlander 08): v L 3 t B 1/3 3,c(N), but not L t B 1/3 3,.

19 Motivation: Onsager s Conjecture (1949) 1. Solutions (v, p) to Euler obeying a Hölder estimate t v l + j (v j v l ) + l p = 0 (8) j v j = 0 v(t, x + x) v(t, x) C x α (9) for some α > 1/3 must conserve energy. 2. If the α in (9) is less than 1/3, then v may fail to conserve energy Part 1 is known: (Eyink, 94), (Constantin-E-Titi, 94) Refinements: (Duchon-Robert 00), (Cheskidov-Constantin- Shvydkoy-Friedlander 08): v L 3 t B 1/3 3,c(N), but not L t B 1/3 3,. Different proof on compact manifolds: (I.-Oh, 13).

20 Motivation: Onsager s Conjecture (1949) 1. Solutions (v, p) to Euler obeying a Hölder estimate t v l + j (v j v l ) + l p = 0 (8) j v j = 0 v(t, x + x) v(t, x) C x α (9) for some α > 1/3 must conserve energy. 2. If the α in (9) is less than 1/3, then v may fail to conserve energy Part 1 is known: (Eyink, 94), (Constantin-E-Titi, 94) Refinements: (Duchon-Robert 00), (Cheskidov-Constantin- Shvydkoy-Friedlander 08): v L 3 t B 1/3 3,c(N), but not L t B 1/3 3,. Different proof on compact manifolds: (I.-Oh, 13).

21 Motivation: Onsager s Conjecture (1949) 1. Solutions (v, p) to Euler obeying a Hölder estimate t v l + j (v j v l ) + l p = 0 (10) j v j = 0 v(t, x + x) v(t, x) C x α (11) for some α > 1/3 must conserve energy. 2. If the α in (11) is less than 1/3, then v may fail to conserve energy 3. Energy-dissipating solutions to Euler with Onsager critical regularity arise in the 0 viscosity limit of Navier-Stokes Shell models and continuous model equations: (Cheskidov-Friedlander-Pavlović 06, Ches.-Fried. 08, Ches.-Fried.-Shvydkoy 11, Friedlander-Glatt Holtz-Vicol 14)

22 Motivation: Onsager s Conjecture (1949) 1. Solutions (v, p) to Euler obeying a Hölder estimate t v l + j (v j v l ) + l p = 0 (10) j v j = 0 v(t, x + x) v(t, x) C x α (11) for some α > 1/3 must conserve energy. 2. If the α in (11) is less than 1/3, then v may fail to conserve energy 3. Energy-dissipating solutions to Euler with Onsager critical regularity arise in the 0 viscosity limit of Navier-Stokes Shell models and continuous model equations: (Cheskidov-Friedlander-Pavlović 06, Ches.-Fried. 08, Ches.-Fried.-Shvydkoy 11, Friedlander-Glatt Holtz-Vicol 14)

23 Motivation: Onsager s Conjecture (1949) 1. Solutions (v, p) to Euler obeying a Hölder estimate t v l + j (v j v l ) + l p = 0 (10) j v j = 0 v(t, x + x) v(t, x) C x α (11) for some α > 1/3 must conserve energy. 2. If the α in (11) is less than 1/3, then v may fail to conserve energy 3. Energy-dissipating solutions to Euler with Onsager critical regularity arise in the 0 viscosity limit of Navier-Stokes Shell models and continuous model equations: (Cheskidov-Friedlander-Pavlović 06, Ches.-Fried. 08, Ches.-Fried.-Shvydkoy 11, Friedlander-Glatt Holtz-Vicol 14)

24 Weak solutions that fail to conserve energy

25 Weak solutions that fail to conserve energy Weak solutions in L 2 t,x(r R 2 ) with compact support in space and time (Scheffer, 93) Weak solutions in L 2 t,x(r T 2 ) (Shnirelman, 97) Dissipative solutions in L t L 2 x(r T 3 ) (Shnirelman, 00 )

26 Weak solutions that fail to conserve energy Weak solutions in L 2 t,x(r R 2 ) with compact support in space and time (Scheffer, 93) Weak solutions in L 2 t,x(r T 2 ) (Shnirelman, 97) Dissipative solutions in L t L 2 x(r T 3 ) (Shnirelman, 00 ) Solutions in L t,x C t L 2 x(r R n ) with any energy density v 2 (De Lellis, Székelyhidi, 07) 2 = e(t, x)

27 Convex Integration and Isometric Embeddings (Nash, 54) Constructs surprising, C 1 isometric embeddings in very low codimension. (Borisov, 65, 04) Irregular C 1,α isometric embeddings for analytic metric (Gromov, 86) Generalizes Nash s idea to the method of convex integration in topology and geometry (Müller-Sverak, 04) Elliptic systems with Lipschitz but nowhere C 1 solutions (i.e. u L, but u / C 0 ).

28 Convex Integration and Isometric Embeddings (Nash, 54) Constructs surprising, C 1 isometric embeddings in very low codimension. (Borisov, 65, 04) Irregular C 1,α isometric embeddings for analytic metric (Gromov, 86) Generalizes Nash s idea to the method of convex integration in topology and geometry (Müller-Sverak, 04) Elliptic systems with Lipschitz but nowhere C 1 solutions (i.e. u L, but u / C 0 ). (De Lellis-Székelyhidi, 09) Simpler proofs and extensions of Borisov s results on C 1,α isometric embeddings

29 Continuous weak solutions that fail to conserve energy Theorem (De Lellis, Székelyhidi, 12) For every α < 1/10, solutions (v, p) Ct,x α Ct,x(R 2α T 3 ) that can realize any smooth energy profile v 2 (t, x)dx = e(t) c > 0 2

30 Continuous weak solutions that fail to conserve energy Theorem (De Lellis, Székelyhidi, 12) For every α < 1/10, solutions (v, p) Ct,x α Ct,x(R 2α T 3 ) that can realize any smooth energy profile v 2 (t, x)dx = e(t) c > 0 2 Extension to R T 2 (De Lellis, Székelyhidi 12, Choffrut 12)

31 Continuous weak solutions that fail to conserve energy Theorem (De Lellis, Székelyhidi, 12) For every α < 1/10, solutions (v, p) Ct,x α Ct,x(R 2α T 3 ) that can realize any smooth energy profile v 2 (t, x)dx = e(t) c > 0 2 Extension to R T 2 (De Lellis, Székelyhidi 12, Choffrut 12)

32 Improved regularity of energy non-conserving solutions Theorem (I., 12) For every α < 1/5 there exist nontrivial weak solutions to the incompressible Euler equations on R T 3 in the class v C α t,x p C 2α t,x that are identically 0 outside of a bounded time interval.

33 Improved regularity of energy non-conserving solutions Theorem (I., 12) For every α < 1/5 there exist nontrivial weak solutions to the incompressible Euler equations on R T 3 in the class v C α t,x p C 2α t,x that are identically 0 outside of a bounded time interval. Shorter proof, solutions with arbitrary smooth e(t) = v 2 (t, x)dx c > 0 (Buckmaster-De Lellis-Székelyhidi, 13) Solutions with compact support in Ω R R 3 (I.-Oh, 14)

34 Main Ideas New ideas for 1/10 (DeL, Sze) Euler Reynolds system Nonstationary phase Transport term vs. oscillatory term Beltrami flows (= special stationary solutions to 3D Euler)

35 Main Ideas Euler Reynolds system Nonstationary phase New ideas for 1/10 (DeL, Sze) Transport term vs. oscillatory term Beltrami flows (= special stationary solutions to 3D Euler) New ideas for 1/5 (I.) Frequency Energy Levels used to measure Hölder regularity (= sharp estimates) Nonlinear phase functions and transport of high frequency fluctuations along the coarse scale flow Improved bounds for D t := t + v

36 Main Ideas Euler Reynolds system Nonstationary phase New ideas for 1/10 (DeL, Sze) Transport term vs. oscillatory term Beltrami flows (= special stationary solutions to 3D Euler) New ideas for 1/5 (I.) Frequency Energy Levels used to measure Hölder regularity (= sharp estimates) Nonlinear phase functions and transport of high frequency fluctuations along the coarse scale flow Improved bounds for D t := t + v

37 Onsager s Conjecture in Weaker Topologies Can the exponent 1/5 be improved if we weaken the topology? (Buckmaster, 13) Solutions in v C 1/5 ɛ t,x v(t, ) C 1/3 ɛ for a.e. t with (Buckmaster-De Lellis-Székelyhidi, 14) C 0 solutions in v L 1 t C 1/3 ɛ x (Buckmaster-Masmoudi-Vicol, 16) Solutions with v C t H 1/3 ɛ x Note: The improvement in regularity is in an averaged sense (in L 1 or L 2 ), but achieves the Onsager critical exponent 1/3. To compare: energy conservation requires L 3 t B 1/3 3,c 0 (N).

38 Main Theorem: A Proof of Onsager s Conjecture Theorem (I. 16) For every α < 1/3 there exists a weak solution in the class v C α t,x, p C 2α t,x, (t, x) R T 3 such that v has nonempty, compact support in time.

39 Main Theorem: A Proof of Onsager s Conjecture Theorem (I. 16) For every α < 1/3 there exists a weak solution in the class v C α t,x, p C 2α t,x, (t, x) R T 3 such that v has nonempty, compact support in time. New Ideas: Using Mikado Flows instead of Beltrami Flows to perform the convex integration method (Daneri-Székelyhidi) Difficulty controlling interactions between distinct Mikado flows

40 Main Theorem: A Proof of Onsager s Conjecture Theorem (I. 16) For every α < 1/3 there exists a weak solution in the class v C α t,x, p C 2α t,x, (t, x) R T 3 such that v has nonempty, compact support in time. New Ideas: Using Mikado Flows instead of Beltrami Flows to perform the convex integration method (Daneri-Székelyhidi) Difficulty controlling interactions between distinct Mikado flows Gluing technique Hidden special structure in the linearization of the Euler equations to estimate main terms

41 Main Theorem: A Proof of Onsager s Conjecture Theorem (I. 16) For every α < 1/3 there exists a weak solution in the class v C α t,x, p C 2α t,x, (t, x) R T 3 such that v has nonempty, compact support in time.

42 Main Theorem: A Proof of Onsager s Conjecture Theorem (I. 16) For every α < 1/3 there exists a weak solution in the class v C α t,x, p C 2α t,x, (t, x) R T 3 such that v has nonempty, compact support in time. (Buck., De L., Szé., Vicol, 17) Solutions in v C 1/3 ɛ t,x any, smooth energy profile T v 2 (t, x)dx = e(t) > 0. 3 with

43 Main Theorem: A Proof of Onsager s Conjecture Theorem (I. 16) For every α < 1/3 there exists a weak solution in the class v C α t,x, p C 2α t,x, (t, x) R T 3 such that v has nonempty, compact support in time. (Buck., De L., Szé., Vicol, 17) Solutions in v C 1/3 ɛ t,x any, smooth energy profile T v 2 (t, x)dx = e(t) > 0. 3 with (I., 17) Solutions with borderline endpoint regularity 1 3 v(t, x + x) v(t, x) x B log log x 1 log x 1, B = 4/3 +.

44 Outline Convex Integration for Euler (General Strategy): The Euler Reynolds Equations (= Approximate solutions) Nonstationary Phase Lemma Acceptable errors (high frequency slowly varying)

45 Outline Convex Integration for Euler (General Strategy): The Euler Reynolds Equations (= Approximate solutions) Nonstationary Phase Lemma Acceptable errors (high frequency slowly varying) Mikado flows (Daneri-Székelyhidi) Convex integration using Mikado flows The difficulty with Mikado flows for Onsager s conjecture The Gluing technique Deriving the Gluing equations Dangerous terms Special structure in the equations

46 Continuous Solutions: The Euler-Reynolds Equations (De Lellis, Székelyhidi): Consider the Euler-Reynolds equations t v l + j (v j v l ) + l p = j R jl j v j = 0 (ER)

47 Continuous Solutions: The Euler-Reynolds Equations (De Lellis, Székelyhidi): Consider the Euler-Reynolds equations t v l + j (v j v l ) + l p = j R jl j v j = 0 (ER) The symmetric tensor R jl measures the error from solving Euler.

48 Continuous Solutions: The Euler-Reynolds Equations (De Lellis, Székelyhidi): Consider the Euler-Reynolds equations t v l + j (v j v l ) + l p = j R jl j v j = 0 (ER) The symmetric tensor R jl measures the error from solving Euler. Examples: If (v, p) solves the Euler equations then (v ɛ, p ɛ, Rɛ jl ), Rɛ jl = vɛv j ɛ l (v j v l ) ɛ, vɛ l = η ɛ v l Corollary: Every continuous incompressible Euler flow (v, p) is the uniform limit of a sequence of Euler-Reynolds flows (v q, p q, R q ) with R q C 0 0 as q

49 Continuous Solutions: The Euler-Reynolds Equations (De Lellis, Székelyhidi): Consider the Euler-Reynolds equations t v l + j (v j v l ) + l p = j R jl j v j = 0 (ER) The symmetric tensor R jl measures the error from solving Euler. Examples: Any v l that is incompressible and conveserves momentum t v l + j (v j v l ) = U l T 3 U l (t, x)dx = 0 j R jl = U l

50 Continuous Solutions: Convex Integration for Euler We construct a sequence (v q, p q, R q ) indexed by q solving t v l q + j (v j qv l q) + l p q = j R jl q j v j q = 0 (ERq) where v q+1 = v q + V q, p q+1 = p q + P q solve (ERq+1) with much smaller R q+1 R q 1+δ

51 Continuous Solutions: Convex Integration for Euler We construct a sequence (v q, p q, R q ) indexed by q solving t v l q + j (v j qv l q) + l p q = j R jl q j v j q = 0 (ERq) where v q+1 = v q + V q, p q+1 = p q + P q solve (ERq+1) with much smaller R q+1 R q 1+δ In the limit as q, we get continuous solutions R q C 0 0, V q R q 1/2, P q R q

52 Continuous Solutions: Convex Integration for Euler Start with any smooth solution to Euler-Reynolds on R T 3 t v l + j (v j v l ) + l p = j R jl j v j = 0 and add high-frequency corrections v 1 = v + V, p 1 = p + P, which are designed to get rid of R jl.

53 Continuous Solutions: Convex Integration for Euler Get new solutions v 1 = v + V, p 1 = p + P to Euler-Reynolds t v l 1 + j (v j 1 vl 1) + l p 1 = j R jl 1 j v j 1 = 0 with R 1 C 0 t,x much smaller than R C 0 t,x.

54 Continuous Solutions: Convex Integration for Euler The corrected v 1 = v + V, p 1 = p + P satisfy t v l 1 + j (v j 1 vl 1) + l p 1 = t V l j (V j V l + P δ jl + R jl ) j v j 1 = 0 = not in the form j R jl 1

55 Continuous Solutions: Convex Integration for Euler The corrected v 1 = v + V, p 1 = p + P satisfy t v l 1 + j (v j 1 vl 1) + l p 1 = t V l j (V j V l + P δ jl + R jl ) j v j 1 = 0 = not in the form j R jl 1 so we will have to solve a divergence equation: j R jl 1 = tv l + j (v j V l ) + j (V j v l ) + j (V j V l + P δ jl + R jl ) to define R 1.

56 Continuous Solutions: Convex Integration for Euler The corrected v 1 = v + V, p 1 = p + P satisfy t v l 1 + j (v j 1 vl 1) + l p 1 = t V l j (V j V l + P δ jl + R jl ) j v j 1 = 0 = not in the form j R jl 1 so we will have to solve a divergence equation: j R jl 1 = tv l + j (v j V l ) + j (V j v l ) + j (V j V l + P δ jl + R jl ) to define R 1. The new error R 1 C 0 will only be small when V and P are very oscillatory and are designed carefully depending on the given v l and R jl.

57 The Error terms Let (v, p, R) be a smooth solution to Euler-Reynolds. t v l + j (v j v l ) + l p = j R jl Then v 1 = v + V and p 1 = p + P satisfy t v l 1 + j (v j 1 vl 1) + l p 1 = t V l + j (v j V l ) + j (V j v l ) want = j R jl 1 with R 1 C 0 λ 1 where V l oscillates at large frequency λ. + j (V j V l + P δ jl + R jl )

58 The Error terms We name the terms as follows Transport term: j R jl T = tv l + j (v j V l ) + j (V j v l ) Stress term: j R jl S = LFreq[ j(v j V l + P δ jl + R jl )] High-Frequency Interference terms: j R jl H = HFreq[ j(v j V l + P δ jl )] Each one of R T, R S and R H must be R 1 C 0 λ 1.

59 The Error terms We name the terms as follows Transport term: j R jl T = tv l + j (v j ɛv l ) + j (V j v l ɛ) Stress term: j R jl S = LFreq[ j(v j V l + P δ jl + R jl ɛ )] High-Frequency Interference terms: j R jl H = HFreq[ j(v j V l + P δ jl )] Each one of R T, R S and R H must be R 1 C 0 λ 1. (There is also another term involving errors from mollifying v v ɛ and R R ɛ that we are neglecting here.)

60 The High-Frequency Correction The correction V l is a high-frequency, divergence free wave. For example, in (I., 12), it has the form V l = I e iλξ I v l I + δv l l V l = 0 (by choice of small δv l ) ( t + v j ɛ j )ξ I = 0 ( nonlinear phase functions) (e iλξ I v I ) λe iλξ I v I (by taking (i ξ I ) v I v I )

61 The High-Frequency Correction The correction V l is a high-frequency, divergence free wave. For example, in (I., 12), it has the form V l = I e iλξ I v l I + δv l l V l = 0 (by choice of small δv l ) ( t + v j ɛ j )ξ I = 0 ( nonlinear phase functions) (e iλξ I v I ) λe iλξ I v I (by taking (i ξ I ) v I v I )

62 The High-Frequency Correction The correction V l is a high-frequency, divergence free wave. For example, in (I., 12), it has the form V l = I e iλξ I v l I + δv l l V l = 0 (by choice of small δv l ) ( t + v j ɛ j )ξ I = 0 ( nonlinear phase functions) (e iλξ I v I ) λe iλξ I v I (by taking (i ξ I ) v I v I ) The last condition makes V l approximate a Beltrami flow ( V λv ), which are special stationary solutions to 3D Euler. It is used to control j R jl H = HFreq[ j(v j V l + P δ jl )]

63 The High-Frequency Correction The correction V l is a high-frequency, divergence free wave. For example, in (I., 12), it has the form V l = I e iλξ I v l I + δv l l V l = 0 (by choice of small δv l ) ( t + v j ɛ j )ξ I = 0 ( nonlinear phase functions) (e iλξ I v I ) λe iλξ I v I (by taking (i ξ I ) v I v I ) The last condition makes V l approximate a Beltrami flow ( V λv ), which are special stationary solutions to 3D Euler. It is used to control j R jl H = HFreq[ j(v j V l + P δ jl )]

64 The Error terms again Each one of R T, R S and R H must have size R 1 C 0 λ 1, and requires solving a divergence equation: Transport term: j R jl T = tv l + j (v j ɛv l ) + j (V j v l ɛ) Stress term: j R jl S = LFreq[ j(v j V l + P δ jl + R jl ɛ )] High-Frequency Interference terms: j R jl H = HFreq[ j(v j V l + P δ jl )]

65 Nonstationary phase Lemma (Nonstationary Phase Lemma) Suppose u l (x) and ξ(x) are smooth functions on T 3 and U l (x; λ) = e iλξ(x) u l (x) ξ 1 C 0 A, U l (x)dx = 0 T 3 Then U l is very small in C 1. That is, we can solve j R jl = e iλξ(x) u l (x) R jl C 0 λ 1 The implicit constant depends on A and C k norms of ξ, u l

66 Nonstationary phase Lemma (Nonstationary Phase Lemma) Suppose u l (x) and ξ(x) are smooth functions on T 3 and U l (x; λ) = e iλξ(x) u l (x) ξ 1 C 0 A, U l (x)dx = 0 T 3 Then U l is very small in C 1. That is, we can solve j R jl = e iλξ(x) u l (x) R jl C 0 λ 1 The implicit constant depends on A and C k norms of ξ, u l

67 Nonstationary phase Lemma (Nonstationary Phase Lemma) Suppose u l (x) and ξ(x) are smooth functions on T 3 and U l (x; λ) = e iλξ(x) u l (x) ξ 1 C 0 A, U l (x)dx = 0 T 3 Then U l is very small in C 1. That is, we can solve j R jl = e iλξ(x) u l (x) R jl C 0 λ 1 The implicit constant depends on A and C k norms of ξ, u l

68 Nonstationary phase Lemma: Cartoon Proof In 1D, we want to solve div R(x) = dr dx = eiλξ(x) u(x) = u(x)eiλξ(x) iλ ξ(x) R(x) = = x 0 x X=x 1 X=0 iλ e iλξ(x) u(x)dx 1 d iλξ (X) dx (eiλξ(x) )u(x)dx ( ) d 1 dx ξ(x) u(x) dx 0 x e iλξ(x) 0 Using ξ 1 C 0 A, the solution has size R C 0 λ 1.

69 Nonstationary phase Lemma: Cartoon Proof In 1D, we want to solve div R(x) = dr dx = eiλξ(x) u(x) = u(x)eiλξ(x) iλ ξ(x) R(x) = = x 0 x X=x 1 X=0 iλ e iλξ(x) u(x)dx 1 d iλξ (X) dx (eiλξ(x) )u(x)dx ( ) d 1 dx ξ(x) u(x) dx 0 x e iλξ(x) 0 Using ξ 1 C 0 A, the solution has size R C 0 λ 1.

70 Nonstationary phase: Proof Proof: To solve j R jl = e iλξ(x) u l (x), write ( ) 1 e iλξ(x) u l (x) = j λ eiλξ(x) q jl (x) + j Ř jl i j ξq jl (x) = u l (x), q jl C (T 3 ; R 3 R 3 ) (12) j Ř jl = 1 λ eiλξ(x) j q jl (x) (13) Equation (12) is solved pointwise and leads to a bound q jl C 0 ξ 1 C 0 u l C 0

71 Nonstationary phase: Proof Proof: To solve j R jl = e iλξ(x) u l (x), write ( ) 1 e iλξ(x) u l (x) = j λ eiλξ(x) q jl (x) + j Ř jl i j ξq jl (x) = u l (x), q jl C (T 3 ; R 3 R 3 ) (12) j Ř jl = 1 λ eiλξ(x) j q jl (x) (13) Equation (12) is solved pointwise and leads to a bound q jl C 0 ξ 1 C 0 u l C 0 We can solve (13) because T 3 e iλξ(x) u l (x)dx = 0. The solution satisfies Řjl C 0 λ 1.

72 Nonstationary phase 2 In order to use Mikado flows: Lemma (Generalized Nonstationary Phase, Daneri-Székelyhidi) Suppose u l (x) and ω(x) C (T 3 ) and Γ C (T 3 ; T 3 ) U l (x; λ) = ω(λγ(x))u l (x) ( Γ) 1 C 0 A, U l (x)dx = 0 T 3 ω(x)dx = 0 T 3 Then U l is very small in C 1. That is, we can solve j R jl = ω(λγ(x)) u l (x) }{{}}{{} fast slow R jl C 0 λ 1

73 Nonstationary phase 2 In order to use Mikado flows: Lemma (Generalized Nonstationary Phase, Daneri-Székelyhidi) Suppose u l (x) and ω(x) C (T 3 ) and Γ C (T 3 ; T 3 ) U l (x; λ) = ω(λγ(x))u l (x) ( Γ) 1 C 0 A, U l (x)dx = 0 T 3 ω(x)dx = 0 T 3 Then U l is very small in C 1. That is, we can solve j R jl = ω(λγ(x)) u l (x) }{{}}{{} fast slow R jl C 0 λ 1

74 Nonstationary phase 2 In order to use Mikado flows: Lemma (Generalized Nonstationary Phase, Daneri-Székelyhidi) Suppose u l (x) and ω(x) C (T 3 ) and Γ C (T 3 ; T 3 ) U l (x; λ) = ω(λγ(x))u l (x) ( Γ) 1 C 0 A, U l (x)dx = 0 T 3 ω(x)dx = 0 T 3 Then U l is very small in C 1. That is, we can solve j R jl = ω(λγ(x)) u l (x) }{{}}{{} fast slow R jl C 0 λ 1

75 Nonstationary phase 2: Proof outline To solve j R jl = ω(λγ(x))u l (x), write (using T 3 ω(x)dx = 0) ω(λγ(x))u l (x) = m 0 ˆω(m)e iλm Γ(x) u l (x) (14)

76 Nonstationary phase 2: Proof outline To solve j R jl = ω(λγ(x))u l (x), write (using T 3 ω(x)dx = 0) ω(λγ(x))u l (x) = m 0 ˆω(m)e iλm Γ(x) u l (x) (14) Can apply the previous Lemma if we have nonstationary phase functions, which requires ( Γ) 1 C 0 A (m Γ) 1 A m 1 Applying the Nonstationary Phase Lemma gives a solution with R jl C 0 λ 1

77 Motivation for Mikado flows Theorem (Daneri-Székelyhidi, 16) For every smooth Euler-Reynolds flow ( v, p, R) with R jl cδ jl, c > 0, (15) there exist weak solutions to Euler in v (k) C 1/5 ɛ t,x such that v l (k) vl in L t,x (16) v j (k) vl (k) vj v l R jl in L t,x as k (17)

78 Motivation for Mikado flows Theorem (Daneri-Székelyhidi, 16) For every smooth Euler-Reynolds flow ( v, p, R) with R jl cδ jl, c > 0, (15) there exist weak solutions to Euler in v (k) C 1/5 ɛ t,x such that v l (k) vl in L t,x (16) v j (k) vl (k) vj v l R jl in L t,x as k (17) With Beltrami flows, would require R jl = a(t, x)(δ jl + small). To overcome this restriction, they introduce a different family of stationary solutions to Euler ( Mikado flows ) that provide more algebraic flexibility to achieve an arbitrary stress R jl.

79 Elementary Mikado flows on T 3 Fix a constant integer vector f l Z 3 and define for X T 3 W l (X) = f l ψ f (X), ψ f C (T 3 )

80 Elementary Mikado flows on T 3 Fix a constant integer vector f l Z 3 and define for X T 3 W l (X) = f l ψ f (X), ψ f C (T 3 ) We choose ψ f whose level surfaces are concentric cylinders with an axis pointed in the f l direction. With this choice we have l ψ f (X)f l = 0 orthogonality Then W l (X) is a stationary Euler flow with 0 pressure: l W l (X) = 0 j (W j W l (X)) = j (ψ 2 f (X)f j f l )

81 Elementary Mikado flows on T 3 Fix a constant integer vector f l Z 3 and define for X T 3 W l (X) = f l ψ f (X), ψ f C (T 3 ) We choose ψ f whose level surfaces are concentric cylinders with an axis pointed in the f l direction. With this choice we have l ψ f (X)f l = 0 orthogonality Then W l (X) is a stationary Euler flow with 0 pressure: l W l (X) = 0 j (W j W l (X)) = 2ψ f (X) j ψ f (X)f j f l

82 Elementary Mikado flows on T 3 Fix a constant integer vector f l Z 3 and define for X T 3 W l (X) = f l ψ f (X), ψ f C (T 3 ) We choose ψ f whose level surfaces are concentric cylinders with an axis pointed in the f l direction. With this choice we have l ψ f (X)f l = 0 orthogonality Then W l (X) is a stationary Euler flow with 0 pressure: l W l (X) = 0 j (W j W l (X)) = 0

83 Elementary Mikado flows on T 3 Fix a constant integer vector f l Z 3 and define for X T 3 W l (X) = f l ψ f (X), ψ f C (T 3 ) In addition to j ψ f (X)f j = 0, we require that ψ f (X)dX = 0, T 3 ψf 2 (X)dX = 1 T 3 (18) With these choices, we have: W l (X)dX = 0, T } 3 {{} Oscillation T 3 W j W l (X)dX = f j f l }{{} Nontrivial low-frequency part (19)

84 Elementary Mikado flows on T 3 More generally, fix a finite set F Z 3 and coefficients γ f and set W l (X) = f F γ f ψ f (X)f l supp ψ f supp ψ f =, if f f F We still have l W l = 0, j (W j W l ) = 0 and T W l (X)dX = 0, 3 but now W j W l (X)dX = γf 2 f j f l T 3 f F can be an arbitrary, positive definite tensor.

85 Designing a wave with Mikado flows Using these flows, we design our high-frequency wave V l as follows. At time t = 0 it looks like V l (0, x) = γ f (0, x) f l ψ f (λx) + }{{}}{{}}{{} δv l f F slow fast small

86 Designing a wave with Mikado flows Using these flows, we design our high-frequency wave V l as follows. At time t = 0 it looks like V l (0, x) = γ f (0, x) f l ψ f (λx) + }{{}}{{}}{{} δv l f F slow fast small At nonzero times, it has the form: V l (t, x) = γ f (t, x) f l (t, x) ψ }{{} f (λ Γ(t, x) ) + δv l }{{} f F slow slow ( t +v ɛ )Γ(t, x) = 0, Γ(0, x) = x

87 Designing a wave with Mikado flows The vector field f l satisfies V l (t, x) = γ f (t, x) f l (t, x)ψ f (λγ(t, x)) + δv l f F f l = ( Γ 1 ) l af a f l l [ψ f (λγ(t, x))] = 0, since f a a ψ f = 0

88 Designing a wave with Mikado flows The vector field f l satisfies V l (t, x) = γ f (t, x) f l (t, x)ψ f (λγ(t, x)) + δv l f F f l = ( Γ 1 ) l af a f l l [ψ f (λγ(t, x))] = 0, since f a a ψ f = 0 We can then make V l divergence free by solving l V l = 0 = l [γ f (t, x) f l (t, x)] ψ f (λγ(t, x)) + l δv l }{{}}{{} f F slow fast δv l C 0 λ 1 (starting now we will neglect this term...)

89 Designing a wave with Mikado flows The vector field f l satisfies V l (t, x) = γ f (t, x) f l (t, x) ψ f (λγ(t, x)) +δv l }{{}}{{} f F slow fast f l = ( Γ 1 ) l af a f l l [ψ f (λγ(t, x))] = 0, since f a a ψ f = 0 We can then make V l divergence free by solving l V l = 0 = l [γ f (t, x) f l (t, x)] ψ f (λγ(t, x)) + l δv l }{{}}{{} f F slow fast δv l C 0 λ 1 (starting now we will neglect this term...)

90 Recalling the Error terms again Each one of R T, R S and R H must have size R 1 C 0 λ 1, and requires solving a divergence equation: Transport term: j R jl T = tv l + j (v j ɛv l ) + j (V j v l ɛ) Stress term: j R jl S = LFreq[ j(v j V l + P δ jl + R jl ɛ )] High-Frequency Interference terms: j R jl H = HFreq[ j(v j V l + P δ jl )]

91 Recalling the Error terms again Each one of R T, R S and R H must have size R 1 C 0 λ 1, and requires solving a divergence equation: Transport term: j R jl T = tv l + j (v j ɛv l ) + j (V j v l ɛ) Stress term: j R jl S = LFreq[ j(v j V l + P δ jl + R jl ɛ )] High-Frequency Interference terms: j R jl H = HFreq[ j(v j V l + P δ jl )]

92 The Main Error Terms With this Ansatz the Transport term is under control: Letting D t := ( t + v j ɛ j ) be the advective derivative we have t V l + j (v j ɛv l ) + j (V j v l ɛ) = ( t + v j ɛ j )V l + V j j v l ɛ = f F D t [γ f f l ψ f (λγ(t, x))] + γ f f j ψ f (λγ(t, x)) j v l ɛ j R jl T = (D t [γ f f l ] + γ f f j j vɛ) l ψ f (λγ(t, x)) }{{}}{{} f F slow fast

93 The Main Error Terms With this Ansatz the Transport term is under control: Letting D t := ( t + v j ɛ j ) be the advective derivative we have t V l + j (v j ɛv l ) + j (V j v l ɛ) = ( t + v j ɛ j )V l + V j j v l ɛ = f F D t [γ f f l ψ f (λγ(t, x))] + γ f f j ψ f (λγ(t, x)) j v l ɛ j R jl T = (D t [γ f f l ] + γ f f j j vɛ) l ψ f (λγ(t, x)) }{{}}{{} f F slow fast (Used j V j = 0.)

94 The Main Error Terms The Stress term is controlled as follows: LFreq[ j (V j V l + P δ jl + R jl ɛ )] ( = LFreq [ j f 1,f 2 F f F disjoint if {}}{ γ f1 γ f2 ψ f1 ψ f2 (λγ) f j f )] 1 2 l + P δ jl + R jl ( = LFreq [ j γf 2 ψ2 f (λγ) f )] j f l + P δ jl + Rɛ jl := j [ f F = j [0] = 0 γ 2 f (t, x) f j f l + P (t, x)δ jl + R jl ɛ Here we solve for the γ 2 f (t, x) at each point using that the ( f j f l ) f F span the space of symmetric tensors. ] ɛ

95 The Main Error Terms The remaining High-Frequency Interference term is controlled as follows using the orthogonality f j j [ψ 2 f (λγ)] = 0 HFreq[ j (V j V l )] = j γ 2 f f j f l (ψf 2 (λγ(t, x)) 1) f F j R jl H = j [γ 2 f f j f l ](ψf 2 (λγ(t, x)) 1) f F

96 The Main Error Terms The remaining High-Frequency Interference term is controlled as follows using the orthogonality f j j [ψ 2 f (λγ)] = 0 HFreq[ j (V j V l )] = j γ 2 f f j f l (ψf 2 (λγ(t, x)) 1) f F j R jl H = j [γ 2 f f j f l ] }{{} f F slow (ψf 2 (λγ(t, x)) 1) } {{ } fast := ω(λγ(t,x)) The last term is fast-oscillating since T 3 (ψ 2 f (X) 1)dX = 0. (Using Beltrami flows, the corresponding term is under control only for a very short period of time.)

97 Can we use Mikado flows for Onsager s conjecture? All the error terms discussed above appear sufficiently small for the method of convex integration to yield regularity 1/3 ɛ. However, there is a substantial difficulty standing in the way of using Mikado flows to prove Onsager s conjecture, namely:

98 Can we use Mikado flows for Onsager s conjecture? All the error terms discussed above appear sufficiently small for the method of convex integration to yield regularity 1/3 ɛ. However, there is a substantial difficulty standing in the way of using Mikado flows to prove Onsager s conjecture, namely: Problem: To iterate the previous construction again and again (i.e. perform convex integration) we need to use multiple waves (see next slide). The difficulty comes in dealing with the interactions of distinct Mikado flows that start from different times.

99 Why we Need Multiple Waves A crucial assumption we are using is the bound ( Γ 1 ) C 0 A for the solution to ( t + v j ɛ j )Γ(t, x) = 0, Γ(0, x) = x We can see that this assumption holds only for times of the order t v 1 from the PDE: C 0 ( t + v j ɛ j )( Γ 1 ) a b = jv a ɛ ( Γ 1 ) j b ( Γ 1 ) a b = Ida b at t = 0 Since v C 0 as v converges to a C 1/3 ɛ vector field, we need to use more and more waves starting at different times!

100 Difficulty with Mikado Flows It seems very difficult to control the interactions between two Mikado flow based waves. Suppose we have two such waves V l 1 = f F 1 γ f,1 f j f l ψ f (λγ 1 ), V l 0 = f F 0 γ f,0 f j f l ψ f (λγ 0 ) where Γ 1 and Γ 0 both solve ( t + v ɛ )Γ I = 0, but start as the identity at different times t 1 t 0 v 1 C 0.

101 Difficulty with Mikado Flows It seems very difficult to control the interactions between two Mikado flow based waves. Suppose we have two such waves V l 1 = f F 1 γ f,1 f j f l ψ f (λγ 1 ), V l 0 = f F 0 γ f,0 f j f l ψ f (λγ 0 ) where Γ 1 and Γ 0 both solve ( t + v ɛ )Γ I = 0, but start as the identity at different times t 1 t 0 v 1 C 0. Then the supports of the ψ f (λγ I ) (which are unions of long, λ 1 -thin, λ 1 -separated cylinders deformed by the flow) will in general overlap and we will lose control over the interference term j [V j 1 V l 0 + V j 0 V l 1 ]

102 Strategy to Fix the Problem Idea: Find a new stress error R that is supported in disjoint time intervals of width θ v 1 supp t R [t(i) θ, t(i) + θ] I so that the new velocity field is a perturbation of the old one v ṽ = v + y and R obeys the same estimates as the original R.

103 Strategy to Fix the Problem Idea: More precisely, starting with (v, p, R), find a new Euler-Reynolds flow (ṽ, p, R) with ṽ close to v such that t ṽ l + j (ṽ j ṽ l ) + l p = j Rjl, R = I Z supp R I [t(i) θ, t(i) + θ], t(i) t(i ) 4θ, I I R I θ v 1

104 Strategy to Fix the Problem Idea: More precisely, starting with (v, p, R), find a new Euler-Reynolds flow (ṽ, p, R) with ṽ close to v such that t ṽ l + j (ṽ j ṽ l ) + l p = j Rjl, R = I Z supp R I [t(i) θ, t(i) + θ], t(i) t(i ) 4θ, I I R I θ v 1 Rules: (ṽ, p, R) must obey the same C k estimates as (v, p, R). In particular, the new error R cannot be much larger than the previous error R! ( R C 0 R C 0 is OK.) Also, we require ṽ to be close to v: v ṽ C 0 R 1/2 C 0

105 Constructing the new (ṽ, p, R) We introduce the velocity increment y l and pressure increment p, which satisfy ṽ l = v l + y l, p = p + p and t y l + v j j y l + y j j v l + j (y j y l ) + l p = j Rjl j R jl j y j = 0

106 Constructing the new (ṽ, p, R) We introduce the velocity increment y l and pressure increment p, which satisfy ṽ l = v l + y l, p = p + p and t y l + v j j y l + y j j v l + j (y j y l ) + l p = j Rjl j R jl j y j = 0 Need R = I R I where supp t R I [t(i) θ, t(i) + θ], θ v 1 C 0. Also need y C 0 e 1/2 R R 1/2 C 0 and R C 0 e R R C 0

107 The Gluing Technique Want the new error R = I R I supported in disjoint intervals: supp t R I [t(i) θ, t(i) + θ] R 0 outside of [t(i) θ, t(i) + θ] I

108 The Gluing Technique Want the new error R = I R I supported in disjoint intervals: supp t R I [t(i) θ, t(i) + θ] R 0 outside of [t(i) θ, t(i) + θ] I So the new ṽ l should solve the Euler equations exactly in the gaps between the intervals [t(i) θ, t(i) + θ] and [t(i + 1) θ, t(i + 1) + θ] Also, ṽ l needs to be a close approximation to v l.

109 The Gluing Technique Let u l I = vl + yi l be the unique, smooth solution to Euler starting at the middle of the Ith gap t 0 (I) with initial data u l I(t 0 (I), x) = v l (t 0 (I), x), yi(t l 0 (I), x) = 0

110 The Gluing Technique Let u l I = vl + yi l be the unique, smooth solution to Euler starting at the middle of the Ith gap t 0 (I) with initial data u l I(t 0 (I), x) = v l (t 0 (I), x), yi(t l 0 (I), x) = 0 Then set y l = I η IyI l, ṽl = I η Iu l I with a partition of unity

111 The Gluing Technique Let u l I = vl + yi l be the unique, smooth solution to Euler starting at the middle of the Ith gap t 0 (I) with initial data u l I(t 0 (I), x) = v l (t 0 (I), x), y l I(t 0 (I), x) = 0 Then set y l = I η IyI l, ṽl = I η Iu l I with a partition of unity Theorem (Classical Existence Result) There exists a unique open interval JI containing t 0 (I) such that u I is smooth on J I T 3 and for all T J I endpoints of J I, lim sup u I (t) C 0 = t T (We will have to prove that supp t η I J I to know the formula is well-defined).

112 The New Stress With y l I and yl = I η Iy l I as above, the new R jl is a solution to j Rjl = I η I(t)y l I + I η I η I+1 j (y j I yl I+1 + y j I+1 yl I) + I (η 2 I η I ) j (y j I yl I), where each yi l = ul I vl solves t yi l + v j j yi l + y j I jv l + j (y j I yl I) + l p I = j R jl j y j I = 0 yi(t l 0 (I), x) = 0

113 The New Stress With y l I and yl = I η Iy l I as above, the new R jl is a solution to j Rjl = I η I(t)y l I + I η I η I+1 j (y j I yl I+1 + y j I+1 yl I) + I (η 2 I η I ) j (y j I yl I), Choosing r jl I such that j r jl I = yi l, the new stress will be R jl = I η I(t)r jl I + I η I η I+1 (y j I yl I+1 + y j I+1 yl I) + I (η 2 I η I )y j I yl I Note that supp t R I supp t η I I [t(i) θ, t(i) + θ].

114 The New Stress With y l I and yl = I η Iy l I as above, the new R jl is a solution to j Rjl = I η I(t)y l I + I η I η I+1 j (y j I yl I+1 + y j I+1 yl I) + I (η 2 I η I ) j (y j I yl I), Choosing r jl I such that j r jl I = yi l, the new stress will be R jl = I η I(t)r jl I + I η I η I+1 (y j I yl I+1 + y j I+1 yl I) + I (η 2 I η I )y j I yl I Note that supp t R I supp t η I I [t(i) θ, t(i) + θ].

115 Finding a good Anti-Divergence: Attempt 1 Problem: we get bad estimates from solving j r jl I = y l I. (20) Suppose that e R is the size of the error ( R C 0 e R ) and suppose (optimistically) that yi l C 0 e1/2 R obeys the bound we desire for y l = ṽ l v l. Then our new error has size R C 0 = η I(t)r I +... C 0 θ 1 r I C 0 θ 1 y I C 0 R C 0 θ 1 e 1/2 R +... Our goal was e R. Having e 1/2 R is already too big, and having θ 1 makes this bound diverge to!

116 Finding a good Anti-Divergence: Attempt 2 We can find a better solution to j r jl I = y l I using the equation t y l I = v j j y l I y j I jv l j (y j I yl I) l p I j R jl

117 Finding a good Anti-Divergence: Attempt 2 We can find a better solution to j r jl I = y l I using the equation t y l I = j (v j y l I + y j I vl + y j I yl I + p I δ jl + R jl )

118 Finding a good Anti-Divergence: Attempt 2 We can find a better solution to j r jl I = y l I using the equation t y l I = j (v j y l I + y j I vl + y j I yl I + p I δ jl + R jl ) y l I(t, ) = j = r jl I (t, ) {}}{ t 0 (v j y l I(τ, ) + y j I vl (τ, ) R jl (τ, ))dτ

119 Finding a good Anti-Divergence: Attempt 2 We can find a better solution to j r jl I = y l I using the equation t y l I = j (v j y l I + y j I vl + y j I yl I + p I δ jl + R jl ) y l I(t, ) = j = r jl I (t, ) {}}{ t 0 (v j y l I(τ, ) + y j I vl (τ, ) R jl (τ, ))dτ R C 0 θ 1 r I C v C 0 y I C R C 0 e 1/2 R +... Still not the desired R C 0 e R.

120 Finding a good Anti-Divergence: Attempt 3 Idea: Set r jl I (t 0(I), x) = 0 and solve a transport equation ( t + v i i )[ j r jl I ] = ( t + v i i )y l I, ( t + v i i )[ j r jl I ] = yj I jv l j (y j I yl I) l p I j R jl (Motivation: integration over trajectories is more natural than integrating in time at fixed x.)

121 Finding a good Anti-Divergence: Attempt 3 Idea: Set r jl I (t 0(I), x) = 0 and solve a transport equation ( t + v i i )[ j r jl I ] = ( t + v i i )y l I, ( t + v i i )[ j r jl I ] = yj I jv l j (y j I yl I) l p I j R jl Setting r jl I = ρ jl I + zjl I, we can solve away the last few terms: ( t + v j j )z jl I = y j I yl I p I δ jl R jl (21) Then z I C 0 looks good if we have y I C 0 e 1/2 R, p I C 0 e R

122 Finding a good Anti-Divergence: Attempt 3 Idea: Set r jl I (t 0(I), x) = 0 and solve a transport equation ( t + v i i )[ j r jl I ] = ( t + v i i )y l I, ( t + v i i )[ j r jl I ] = yj I jv l j (y j I yl I) l p I j R jl To handle the linear term, let r jl I = ρ jl I + zjl I where j [( t + v i i )ρ jl I ] = jv i i r jl I yj I jv l (Obtained by commuting j and ( t + v i i ).)

123 Finding a good Anti-Divergence: Attempt 3 To handle the linear term, let r jl I = ρ jl I + zjl I where j [( t + v i i )ρ jl I ] = jv i i r jl I yi I i v l (22) Equation (22) can only be solved if we can invert the divergence on both sides. We need to know the right hand side has integral 0: j v i i r jl I Here we use that i v i = i y i I = 0. yj I jv l = i [ j v i r jl I yi Iv l ] We now invert the divergence to obtain an equation for ρ I.

124 Finding a good Anti-Divergence: Attempt 3 We let ρ jl I solve a transport-elliptic equation: ( t + v i i )ρ jl I = Rjl [ a v i i (ρ ab I + z ab I ) y i I i v b ] where R jl = div 1 is an order 1 operator that inverts divergence. This type of equation can be solved as in (I. 12) as long as y I and z I are smooth. Question: Are the estimates good enough? (e.g. Do we have R C 0 e R?)

125 Finding a good Anti-Divergence: Attempt 3 We let ρ jl I solve a transport-elliptic equation: ( t + v i i )ρ jl I = Rjl [ a v i i (ρ ab I + z ab I ) y i I i v b ] The corresponding estimate for R jl = η I (t)r I +... is: R C 0 θ 1 ρ I C ( t + v )ρ I C R jl [y i I i v b ] C 0 + other terms e 1/2 R θ

126 Finding a good Anti-Divergence: Attempt 3 We let ρ jl I solve a transport-elliptic equation: ( t + v i i )ρ jl I = Rjl [ a v i i (ρ ab I + z ab I ) y i I i v b ] The corresponding estimate for R jl = η I (t)r I +... is: R C 0 θ 1 ρ I C ( t + v )ρ I C R jl [y i I i v b ] C 0 + other terms e 1/2 R θ

127 Finding a good Anti-Divergence: Attempt 3 We let ρ jl I solve a transport-elliptic equation: ( t + v i i )ρ jl I = Rjl [ a v i i (ρ ab I + z ab I ) y i I i v b ] The corresponding estimate for R jl = η I (t)r I +... is: R C 0 θ 1 ρ I C ( t + v )ρ I C R jl i [y i Iv b ] C 0 + other terms e 1/2 R (if we pretend R = div 1 is bounded on C 0 ) But that is still not good enough for R C 0 e R...

128 Finding a good Anti-Divergence: Attempt 3 We let ρ jl I solve a transport-elliptic equation: ( t + v i i )ρ jl I = Rjl [ a v i i (ρ ab I + z ab I ) y i I i v b ] The corresponding estimate for R jl = η I (t)r I +... is: R C 0 θ 1 ρ I C ( t + v )ρ I C R jl i [y i Iv b ] C 0 + other terms e 1/2 R (if we pretend R = div 1 is bounded on C 0 ) But that is still not good enough for R C 0 e R... Key point: We can actually prove R jl [y i I iv b ] C 0 e R! (almost)

129 The Pressure Has a Similar Bad Term The pressure increment has a similar bad term p I = 2 1 l [y j I jv l ] 1 l j [y j I yl I + R jl ] Note that the highlighted operator is of order 1, similar to R jl. Let us show how to (almost) estimate this term by 1 l [y j I jv l ] C 0 e R

130 The Pressure Has a Similar Bad Term Notation: We define the Littlewood-Paley projections P q u l (x) = u l (x h)η q (h)dh R 3 supp ˆη q (ξ) {2 q 2 ξ 2 q+2 } η q (h) = 2 3q η 0 (2 q h) u l (x) = Π 0 u l + P q u l (x), x T 3 q=0

131 The Pressure Has a Similar Bad Term Choose Ξ such that θ 1 v C 0 that 2ˆq 1 Ξ < 2ˆq. Then Ξe 1/2 R and choose ˆq Z such 1 l [y j I jv l ] = 1 l P ˆq [y j I jv l ] + q>ˆq 1 l P q [y j I jv l ] = p I,L + p I,H

132 The Pressure Has a Similar Bad Term Choose Ξ such that θ 1 v C 0 that 2ˆq 1 Ξ < 2ˆq. Then Ξe 1/2 R and choose ˆq Z such 1 l [y j I jv l ] = 1 l P ˆq [y j I jv l ] + q>ˆq 1 l P q [y j I jv l ] = p I,L + p I,H The high frequency term is bounded by p I,H C 0 q>ˆq 1 l P q [y j I jv l ] C 0 q>ˆq 1 l P q }{{} (C 0 C 0 )norm y j I jv l C 0 (Note the operator convolves with an L 1 Schwartz kernel.)

133 The Pressure Has a Similar Bad Term Choose Ξ such that θ 1 v C 0 that 2ˆq 1 Ξ < 2ˆq. Then Ξe 1/2 R and choose ˆq Z such 1 l [y j I jv l ] = 1 l P ˆq [y j I jv l ] + q>ˆq 1 l P q [y j I jv l ] = p I,L + p I,H The high frequency term is bounded by p I,H C 0 q>ˆq 1 l P q y j I jv l C 0 q>ˆq 2 q y j I jv l C 0 Ξ 1 e 1/2 R (θ 1 ) = e R It now remains to bound the low frequency term.

134 The Low Frequency Term The low frequency term has the form p I,L = 1 l P ˆq [y j I jv l ] = ˆq q=0 1 l P q [y j I jv l ] In this case, we do not gain smallness from bounding 1 l P q 2 q 1

135 The Low Frequency Term Step 2: Decompose v into high and low frequencies p I,L = p I,LL + p I,LH p I,LL = 1 l P ˆq [y j I jp ˆq v l ] p I,LH = 1 l P ˆq [y j I jp q v l ] q>ˆq

136 The Low Frequency Term Step 2: Decompose v into high and low frequencies p I,L = p I,LL + p I,LH p I,LL = 1 l P ˆq [y j I jp ˆq v l ] p I,LH = q>ˆq 1 l P ˆq [y j I jp q v l ] And bound the LH term using j y j I = 0: p I,H C 0 q>ˆq 1 l j P ˆq y I C 0 P q v C 0 q>ˆq log Ξ e 1/2 R (2 q v C 0) log Ξ e 1/2 Ξ R 1 ( Ξe 1/2 R ) log Ξ e R

137 Remaining Problematic Term The remaining problematic term is p I,LL = 1 l P ˆq [y j I jp ˆq v l ] or p I,LL = 1 l P ˆq [P ˆq+3 y j I jp ˆq v l ] using that high frequencies of y I do not contribute.

138 Remaining Problematic Term The remaining problematic term is p I,LL = 1 l P ˆq [y j I jp ˆq v l ] We treat this term by decomposing into frequency increments p I,LL = ˆq q= 1 δ q p I,LL δ q p I,LL = 1 l P q+1 [y j I jp q+1 v l ] 1 l P q [y j I jp q v l ] Note: Starting now, 2 q is in the low to medium range of frequencies.

139 Frequency Increments The frequency increment can either fall on the operator or on v: δ q p I,LL = 1 l P q+1 [y j I jp q+1 v l ] + 1 l P q [y j I jp q+1 v l ]

140 Frequency Increments The frequency increment can either fall on the operator or on v: δ q p I,LL = 1 l P q+1 [y j I jp q+1 v l ] + 1 l P q [y j I jp q+1 v l ] Consider the second term. Using j y j I = 0, we have 1 l P q [y j I jp q+1 v l ] = 1 l j P q [y j I P q+1v l ] = 1 l j P q [P q+6 y j I P q+1v l ]

141 Frequency Increments The frequency increment can either fall on the operator or on v: δ q p I,LL = 1 l P q+1 [y j I jp q+1 v l ] + 1 l P q [y j I jp q+1 v l ] Consider the second term. Using j y j I = 0, we have 1 l P q [y j I jp q+1 v l ] = 1 l j P q [y j I P q+1v l ] = 1 l j P q [P q+6 y j I P q+1v l ] In the last line, we observe that frequencies of y l I above 2q+4 do not contribute to the product by the frequency localization.

142 Frequency Increments Now use that we can solve y j I = ir ij I to write 1 l j P q [P q+6 y j I P q+1v l ] = 1 l j P q [P q+6 i r ij I P q+1v l ] C 0 1 l j P q P q+6 i r I C 0[2 q v C 0] (2 + q)2 q r I C 02 q Ξe 1/2 R Note how the 2 q and 2 q cancel out.

143 Frequency Increments Now use that we can solve y j I = ir ij I to write 1 l j P q [P q+6 y j I P q+1v l ] = 1 l j P q [P q+6 i r ij I P q+1v l ] C 0 1 l j P q P q+6 i r I C 0[2 q v C 0] (2 + q) r I C 0( Ξe 1/2 R ) Almost closes if there exists r I such that r I C 0 Ξ e 1/2 R

144 Frequency Increments Now use that we can solve y j I = ir ij I to write 1 l j P q [P q+6 y j I P q+1v l ] = 1 l j P q [P q+6 i r ij I P q+1v l ] C 0 1 l j P q P q+6 i r I C 0[2 q v C 0] (2 + q) r I C 0( Ξe 1/2 R ) Idea: impose a bootstrap assumption on ρ I and z I that implies Ξ r I C 0 e 1/2 R Then summing over q ˆq log Ξ leads to R I C 0 (log Ξ) 2 e R, which is the correct estimate (except for the (log Ξ) 2 )!

145 Loss of Derivatives It turns out that (if one furthermore shrinks the time scale θ by a logarithmic factor) it is possible to close the argument implying the above estimates by using certain weighted C 3,α norms. But there is a catch...

146 Loss of Derivatives It turns out that (if one furthermore shrinks the time scale θ by a logarithmic factor) it is possible to close the argument implying the above estimates by using certain weighted C 3,α norms. But there is a catch, namely this gluing construction loses derivatives. E.g., v and R both enter in the equation for y I t y l I + v j j y l I + y j I jv l + j (y j I yl I) + l p I = j R jl Similarly, bounds on 2 v and 2 R are required to estimate y I and so on...

147 Loss of Derivatives To fully close the argument, we first regularize the Euler-Reynolds flow (v, p, R) (v ɛ, p ɛ, R ɛ ) using a mollifier η ɛ t v l + j (v j v l ) + l p = j R jl t v l ɛ + j (v j ɛv l ɛ) + l p ɛ = j [v j ɛv l ɛ (v j v l ) ɛ + η ɛ R jl ] We apply the Constantin-E-Titi commutator estimate to bound the resulting Stress for ɛ Ξ 1 not too small. This regularization gains derivatives (with acceptable bounds on higher, borrowed derivatives), and allows the whole scheme (i.e. Regularize Gluing Convex integration with Mikado flows repeat) to close.

148 Thank you!