THE ONSAGER CONJECTURE

R. SHVYDKOY

The Euler system governing evolution of an incompressible ideal homogeneous (ρ ≡ 1)fluid is given by

(E)

ut + u · ∇u+∇p = f, f smooth∇ · u = 0, x ∈ Tn or Rn

u(0) = u0.

Let E(t) = 12

∫|u(x, t)|2 dx. For classical solutions, we have the energy balance relation

(EB) E(t) = E0 +

∫ t

0

f · u ds.︸ ︷︷ ︸total work of force

The reason is that Flux = Π =∫u · ∇u · u dx = 0.

Question: How much regularity is needed to still prove (EB)? If we want to approach thisas in the classical sense, then we first need to learn how to control the flux. Heuristically,

Π ∼∫∇

13u · ∇

13u · ∇

13u dx =

∫(∇

13u)3 dx.

So u has to be 13-smooth in the proper sense.

Lars Onsager (version 1949):

(A) If u has smoothness better than 13

(loosely for now), then (EB) holds.

(B) There exists a weak solution to (E) which is “13-smooth” and which violates (EB).

We have

E(t) =

∫ T

0

Π dt︸ ︷︷ ︸total energy flux

+

∫ T

0

f · u dt.︸ ︷︷ ︸work

If [u] = U , [t] = T , [x] = L, then the units of total flux are TU3Ln−1. A function space Xwith [X]3 = TU3Ln−1 will be called Onsager-critical. For example, the following spaces areOnsager-critical:

L3L6 in 2D,

L3L9/2 and L3H56 in 3D,

L3W13,3, L3B

133,p, 1 ≤ p ≤ ∞ in any D

Exercise: Any Onsager-critical space which is Lp-based in time must be L3t .

Reference: Gregory L. Eyink and Katepalli R. Sreenivasan, Onsager and the theory ofhydrodynamic turbulence, Rev. Modern Phys. 78 (2006), no. 1, 87–135. Contains a historicoverview of Onsager’s papers, etc.

1

2 R. SHVYDKOY

Evidence from turbulence.

Kolmogorov proposed a theory in 1941 (K41) that postulates non-vanishing energy dissi-pation rate as a driving mechanism of turbulence. If uν satisfies

∂tuν + uν · ∇uν +∇pν = ν∆uν + f,

then we define the mean energy dissipation rate per unit mass by

εν = ν〈|∇uν |2〉.

Here 〈·〉 denotes ensemble average. Isotropic, homogeneous turbulence postulates that distri-bution laws of velocity increments (displacements), δu(`) = u(r+ `)−u(r), are independentof r and the direction of `. The longitudinal structure functions Sp(`) = 〈(δu(`) · `|`|)

p〉 can

be measured experimentally.Elements of K41 Theory:

• Kolmogorov’s 0th law of turbulence: εν → ε > 0 as ν → 0 (keeping L and U undercontrol, i.e. Re→∞).We also have the following “provable” laws (assuming all formal manipulations arejustified):• Kolmogorov 4

5-law (K-4

5): S3(`) = −4

5ε|`|.

• Karman-Howarth-Monin relation (KHM): Π = −14

div`〈δu(`)|δu(`)|2〉∣∣`=0

.

The K-45

law clearly suggests that turbulence is generically 13-regular. (It also suggests that

regularity needs to be understood in an averaged sense, hence the need for Besov spacesbelow.) The KHM relation also implies that Π 6= 0 =⇒ u is at most 1

3-regular. Π and ε

anomalies can be reconciled on a formal level.Reference: Uriel Frisch, Turbulence, Cambridge University Press, Cambridge, 1995. The

legacy of A. N. Kolmogorov.Suppose 〈·〉 is also stationary and f = f<k0 , k0 is an integral wavenumber, and f<k0 is the

“filtered” field, i.e., the projection on the Fourier side. Let u solve the NSE. Then test withu<k:

1

2∂t〈|u<k|2〉︸ ︷︷ ︸

=0 (stationary)

+ 〈(u · ∇u)<k · u<k〉︸ ︷︷ ︸=Πν(k)

= −ν〈|∇u<k|2〉+ 〈f · u<k〉.︸ ︷︷ ︸=〈f ·u〉,

since f=f<k0

So Πν(k) = −ν〈|∇u<k|2〉+ 〈f · u〉. But

ν〈|∇u<k|2〉 ≤ ν〈|u|2〉k2 ∼ νU2k2 ν→0−→ 0.

On the other hand, testing the NSE with the full u:

∂t〈|u|2〉︸ ︷︷ ︸=0 (stationary)

+ 〈u · ∇u · u〉︸ ︷︷ ︸=0 (regularity)

= −ν〈|∇u|2〉︸ ︷︷ ︸=−εν

+〈f · u〉,

and εν = 〈f · uν〉 ν→0−→ ε. So Π(k) = ε, for all k > k0. Letting k →∞, conclude Π = ε.This naıve computation also shows energy cascade: energy being fed by f dissipates from

large to small scales with constant flux. See Figure 1.The energy spectrum E(k) = 1

2ddk〈|u<k|2〉 distributes according to a scaling law.

THE ONSAGER CONJECTURE 3

Figure 1.

• K41: All quantities depend only on ε and `. Dimensional analysis implies that

E(k) = k0ε

23

k53

.

This famous law has been confirmed experimentally in many situations.• Self-similarity: δu(`) ∼ `h. K-4

5implies h = 1

3, which in turn implies Sp(`) = Cpε

p3 `

p3 .

Deviations are observed but unexplained (see intermittency below).• Locality of flux: Π(k) is generated mostly by scales close to 1

k.

Full confirmation of Onsager Conjecture for Burgers’ Equation:Motivation for use of Besov Spaces.

Burgers’ equation is

(BE) ut + uux = 0.

Suppose u0 has a negative slope somewhere, so that a shock forms. (BE) represents the evo-lution of compressible fluid. After the shock, the solution becomes a weak solution (entropysolution), which describes the motion of particles after an inelastic collision (particles sticktogether and move with the same velocity).

Figure 2.

Consider temporarily the collision of two particles, with velocities u+ and u−, respectively.Momentum is conserved: u+ +u− = u++u−

2+ u++u−

2before and after the collision. But Eb =

Energy before = (u+)2+(u−)2

2, Ea = Energy after = 1

2

(u++u−

2

)2

+ 12

(u++u−

2

)2

=(u++u−

2

)2

.

4 R. SHVYDKOY

So Eb > Ea: energy is lost! The intuition from this simple calculation is reflected in theenergy associated to the Burgers’ equation. Indeed,

dE

dt= − 1

12(u+ − u−)3 < 0.

The proof of this statement is left as an exercise.What regularity does a shock have? Certainly not Holder! How do we measure smooth-

ness?

Holder: u ∈ Cs (= Bs∞,∞) if sup

y 6=0supx

|u(x+ y)− u(x)||y|s

<∞.

[Replace L∞x by Lpx]

Besov: u ∈ Bsp,∞ if u ∈ Lp and sup

y 6=0

∫Rn

|u(x+ y)− u(x)|p

|y|psdx <∞.

The third index “∞” corresponds to L∞y and can be replaced by p Lpy in a proper sense(not used here). Also,

u ∈ Bsp,c0

if u ∈ Bsp,∞ and lim

y→0

∫R

|u(x+ y)− u(x)|p

|y|psdx = 0.

Besov functions have good approximation properties:

Lemma. u ∈ Bsp,∞ (1 ≤ p ≤ ∞, 0 < s < 1) iff ‖uδ − u‖Lp ≤ Cδs as δ → 0, where uδ is a

standard mollification of u.

Lemma. A shock is in the class B1pp,∞ for all 1 ≤ p ≤ ∞.

(Proof left as an exercise. Hint: Check the Heaviside function.)So, generic solutions to Burgers’ equation start dissipating energy in finite time while

remaining in the class B133,∞. On the other hand, we will show that in any dimension, if

u ∈ B133,c0

, then (EB) holds. This answers the Onsager conjecture for Burgers’ equationcompletely.

Littlewood-Paley Theory.

Let χ = ϕ−1 be a decreasing radial C∞ function supported in the unit ball, such thatχ ≡ 1 in |ξ| ≤ 1

2. Define ϕ0 = χ( ξ

2)−χ(ξ) and ϕq = ϕ0( ξ

2q) for q ≥ 1. Then

∑∞q=−1 ϕq = 1

pointwise. Let uq = (ϕqu)∨, so that u =∑∞

q=−1 uq in the sense of distributions. Let u<q

denote the sum u<q =∑q−1

p=−1 up, and let u≥q = u− u<q, etc.

Theorem (Littlewood-Paley). u ∈ Lp iff(∑∞

−1 |uq|2) 1

2 = S(u) ∈ Lp, and ‖u‖Lp ∼ ‖S(u)‖Lp.

Here S(u) is called the “square function.”We will actually work with the dimensional dyadic wavenumber λq = 2q

L.

Lemma. (i) u ∈ Bsp,∞ iff supq λ

sq‖uq‖p <∞. (ii) u ∈ Bs

p,c0iff limq→0 supq λ

sq‖uq‖p = 0.

The proof is left as an exercise.

We will be using time-averages (c.f. ensemble average) 〈f〉 = 1T

∫ T0f dt.

THE ONSAGER CONJECTURE 5

Figure 3.

Onsager Conjecture: Part (A)

Theorem. (Frisch-Sulem ‘75: H56

+ε; Eyink ‘94: C13

+ε; Constantin, E, Titi: B13

+ε

3,∞ ; Duchon-Robert 2000: dissipation distribution, CET, local KMH; Cheskidov, Constantin, Friedlander,RS 2008: present version.) Suppose u ∈ CwL2

x is a weak solution to the Euler equations suchthat

(∗) limq→∞〈λq‖uq‖3

3〉 = 0 on [0, T ].

Then the energy law (EB) holds on the same interval.

Remark. If u ∈ L3B133,∞, then (∗) holds. (Exercise. Hint: use Fatou’s Lemma.)

Proof. The equation ∂tu+ div(u⊗u) +∇p = 0 holds distributionally in D′. Let us test with(u<q)<q. We obtain

1

2

d

dt‖u<q‖2

L2 =

∫(u⊗ u)<q : ∇u<q dx = Πq.

Here A : B = aijbij. We have the identity

(u⊗ u)<q = rq(u, u)− u≥q ⊗ u≥q + u<q ⊗ u<q,

where rq(u, u)(x) =∫hq(y) (u(x+ y)− u(x)) ⊗ (u(x+ y)− u(x)) dy, hq = χ∨, hq(y) =

λnqh(yλq). (Proof: Exercise.) So we are left with

Πq =

∫rq(u, u) : ∇u<q dx︸ ︷︷ ︸

=:A

+

∫u≥q ⊗ u≥q : ∇u<q dx.︸ ︷︷ ︸

=:B

We need the differential Bernstein inequality:

‖∇uq‖p ∼ λq‖uq‖p, 1 ≤ p ≤ ∞.

6 R. SHVYDKOY

Let us first estimate B. We have

|B| ≤ ‖u≥q‖23‖∇u<q‖3 ≤

(∑p≥q

‖up‖3

)2(∑p<q

λp‖up‖3

)

=

(∑p≥q

(λqλp

) 13

λ13p ‖up‖3

)2(∑p<q

(λpλq

) 23

λ13p ‖up‖3

)

=

(∑p

K|p−q|λ13p ‖up‖3

)3

(here Kr = λ− 1

3

|r| )

≤∑p

K|p−q|λp‖up‖33.

Now, let us estimate the term A. First, we have

|A| ≤ ‖∇u<q‖3‖rq(u, u)‖ 32.

As before, ‖∇u<q‖3 ≤∑

p<q λp‖up‖3. Next, we need the Minkowski inequality:Suppose f : Ω→ X is a strongly measurable, Bochner integrable function. Then

∥∥∥∥∫Ω

f dµ

∥∥∥∥X

≤∫

Ω

‖f‖X dµ.

Continuing with the prove of the theorem, we have

‖rq(u, u)‖ 32

=

∥∥∥∥∥∥∥∫

(u(·+ y)− u(·))⊗ (u(·+ y)− u(·))hq(y) dy︸ ︷︷ ︸dµ

∥∥∥∥∥∥∥L

32

≤∫‖u(·+ y)− u(·)‖2

L3hq(y) dy

≤∫‖u<q(·+ y)− u<q(·)‖2

L3hq(y) dy +

∫‖u≥q‖2

L3hq(y) dy

≤∫ 1

0

∫‖∇u<q‖2

L3θ2|y|2λnqh(λqy) dy dθ +

∫ (∑p≥q

‖up‖L3

)2

hq(y) dy

≤

(∑p<q

λp‖up‖3

)21

λ2q

∫λnq (λq|y|)2h(λqy) dy︸ ︷︷ ︸

= const. independent of q!

+

(∑p≥q

‖up‖3

)2

.

So

THE ONSAGER CONJECTURE 7

|A| ≤

(∑p<q

λp‖up‖3

)21

λ2q

+

(∑p<q

λp‖up‖3

)(∑p≥q

‖up‖3

)2

=

(∑p<q

(λpλq

) 23

λ13p ‖up‖3

)3

+

(∑p<q

(λpλq

) 23

λ13p ‖up‖3

)(∑p≥q

(λqλp

) 13

λ13p ‖up‖3

)2

≤ [same as bound for B].

So 〈Πq〉 ≤ C∑

p∈NKp−q〈λp‖up‖33〉. This proves the theorem since 〈Πq〉 → 0 and

‖u<q(t)‖2L2 − ‖u<q(0)‖2

L2 = 〈Πq〉+ 〈f · u<q〉=⇒ E(t)− E(0) = 〈f · u〉.

It also provides a justification for locality of the energy flux in frequency space.

Local Version

The local energy equality is a more detailed (in space) version of the law (EB) which isglobal in nature. Test (E) with ϕu, where ϕ ∈ C∞0 is a scalar function. Then

(LEB)

∫Rn|u(x, t)|2ϕ(x, t) dx−

∫Rn|u(x, t0)|2ϕ(x, t0) dx−

∫ t

t0

∫Rn|u(x, s)|2∂sϕ(x, s) dx ds

=

∫ t

t0

∫Rn

(|u|2 + 2p)u · ∇ϕdx ds+

∫ t

t0

∫Rn

(f · u)ϕdx ds.

We define the global Onsager class R0 by

u ∈ R0 ⇐⇒ (∗) holds,

and for an open set Ω ⊂ Rn × [0, T ], we say that u ∈ R0(Ω) iff ϕu ∈ R0 for all ϕ ∈ C∞0 (Ω).

Theorem. If u ∈ R0(Ω), then (LEB) holds for all ϕ ∈ C∞0 (Ω).

Sister Conjectures

(1) Helicity∫R3 u · ω dx requires u ∈ B

233,c0∩H 1

2 . Helicity flux is given by (exercise)

d

dt

∫u<q · ω<q dx =

∫(u⊗ u)<q : ∇ω<q dx+

∫(u× ω)<q · ∇u<q dx.

(2) Vorticity in 2D: ‖ω‖L2 = enstrophy conserved. A similar Littlewood-Paley argumentrequires u ∈ B1

3,c0or ω ∈ B0

3,∞ ∩ L2. But ωt + u · ∇ω = 0 is a transport equation!

If we just assume ω ∈ L2 (which we do in order to make sense of the enstrophy),then u ∈ W 1,2 → W 1,1 which puts the solution into the framework of “renormalized”solutions in the sense of DiPerna-Lions. As a consequence, measω > a is conserved,for all real a. Hence, ‖ω‖2

L2 = 2∫∞

0λmeasω > λ dλ is conserved. Lesson: The

Onsager conjecture may still hold in Onsager-supercritical classes if there is extrastructure of the equation (or solution) known.

(3) Similar conjectures can be stated for other advective systems, such as SQG, porousmedium, etc.

8 R. SHVYDKOY

Onsager Conjecture: Part (B)

In view of the result on the energy conservation, we know now that B133,∞ is critical. Let

us now state the 2015 version of the Onsager Conjecture, part (B): Construct a pair (u, f)such that

(1) u ∈ L∞t L2x, f ∈ L2, supq〈λq‖uq‖3

3〉 <∞.(2) In the sense of distributions,

∂tu+ u · ∇u+∇p = f∇ · u = 0.

(3) There exists t < T such that E(t) 6= E(0) +∫ t

0f · u ds.

Remark. (1) [L3B133,∞]3 = TU3Ln−1 in any dimension, including n = 1! (2) Note that f ∈ L2

is the minimal assumption under which the pairing∫f · u dx can be understood classically.

The predual to B133,∞ is B

− 13

32,1

, which is weaker but is less physically reasonable for f to be in.

Timeline of attempts to solve part (B) of the Onsager Conjecture.

• Scheffer 1998: u ∈ L2tL

2x has compact support in time.

• Shnirelman 2000: same result, much shorter proof. Both constructions use backwardcascade ideas.• Convex integration era:

DeLellis & Szekelyhidi 2007 (Annals): u ∈ L∞t L∞x even more striking; given anyΩ ∈ Rn × (0, 1), we can have |u| = XΩ. There are extensions in the contexts of theactive scalar equation, MHD, compressible, Perron-Malik, etc.

DeLellis & Szekelyhidi, u ∈ C110−ε

t,x

P. Isett, u ∈ C15−ε

t,x

Buckmaster, DeLellis, Szekelyhidi, u ∈ L1tC

13−ε

x .

P. Isett, u ∈ C1/3−εt,x , September 2016.

All these examples also demonstrate ill-posedness of the Euler system in the spaces wherethe examples live. The penultimate result of Isett shows that criticality of 1/3 from the lower”harder” side of the conjecture. The result represent the triumph of the convex integrationapproach in the Onsager supercritical regularity up to , but not yet including the criticalone 1/3. The description of the technique is out of the scope of these lectures.

It is interesting to compare this with regularity exponents in Holder and Besov spaces:

• C. Bardos, E. Titi ‘09 use an example of solution from DePerna-Majda ‘87 (Lions‘98), namely u(x, t) = 〈u1(x2), 0, u3(x1 − tu1(x2))〉, and show that there exist u1, u2

such that u(0) ∈ Cα, 0 < α < 1, but u(t) /∈ Cα2+ε for any ε > 0.• Cheskidov, RS ‘09: There exists u0 ∈ Bs

p,∞ for s > 0, p > 2 or s > n(2p−1), 1 ≤ p ≤ 2,

such that any weak solution to (E) satisfies

limt→0+‖u(t)− u0‖Bsp,∞ ≥ δ0 > 0.

In particular, in B133,∞, the trajectories are not continuous.

THE ONSAGER CONJECTURE 9

Example of a Non-zero Flux

There exists a field u ∈ B133,∞ such that lim supq→∞Πq > 0.

Figure 4.

Let u1 = uq + uq−1. Also,

Πq−1 =

∫(uq ⊗ uq)≤q−1 : ∇(uq)≤q−1 dx

=

∫(uq + uq−1) · ∇uq−1 · (uq + uq−1) dx

=

∫uq · ∇uq−1 · uq dx

= i∑ξ

∑η1+η2=ξ

(uq(η1) · ξ)(uq−1(ξ) · uq(η2))

= (by hand) = c0 > 0.

Let u =∑∞

k=0 uqk , where |qk − qk−1| = eek. Then Πq−1 = Πq−1 + Res, where Res c0 due

to outer locality estimates on the flux. So, lim supq→∞Πq ≥ c02

.

Looking ahead towards obtaining better regularity, can we improve s = 13

by sacrificingp = 3 to a lower index? Tool: Sobolev embedding or Bernstein’s inequalities. Note that in2D,

curlu = ω ∈ L32 ⇐⇒ u ∈ W 1, 3

2 → W13,3 → B

133,c0

=⇒ Πq = 0.

So, in 2D we can only improve ω to Lp<32 .

Saturating Bernstein’s Inequalities. Bernstein’s inequality is the endpoint Sobolev in-

equality for a given dyadic shell: if | supp u| ⊂ [λk−1, λk+1], then ‖u‖q ≤ λn( 1p− 1q

)

k ‖u‖p for all1 ≤ p < q ≤ ∞.

Proof. Let hk = ϕ∨k−1 + ϕ∨k + ϕ∨k+1. Then u = u ∗ hk. By Young’s inequality,

‖u‖q = ‖u ∗ hk‖q ≤ ‖u‖p‖hk‖r ≤ C‖u‖pλn(1− 1

r)=n( 1

p− 1q

)

k ,

where 1q

+ 1 = 1p

+ 1r, so that ‖hk‖rr =

∫λnrk |ϕ∨0 (λkx)|r dx = Cλnr−nk .

10 R. SHVYDKOY

So on Tn we have the two inequalities ‖u‖p ≤ ‖u‖q ≤ λn( 1p− 1q

)

k ‖u‖p.The lower bound prevails when u is “uniformly” distributed in x and “concentrated”

to a point in ξ. The upper bound prevails when u is “uniformly” distributed in ξ and“concentrated” in x.

Example: u(x) = cos(λkx), all Lp-norms are similar; in complex variables, an even morevivid example is u(x) = eiλk·x. On the Fourier side, u has only one mode.

Example: The Dirichlet kernel DN(x) =∑N

k=−N eikx has modes which are uniformly

distributed on the Fourier side, and DN is concentrated near x = 0 on the x-side. Recallthat

‖DN‖Lp ∼=N1− 1

p , 1 < p ≤ ∞logN, p = 1.

Set u(x) = eiλk·x1Dλk10

(x1) · . . . · eiλk·xnDλk10

(xn).

Figure 5.

We say in this case that the Bernstein inequality has been saturated. Let us now “saturate”our flux construction.

Figure 6.

Each block is scaled by 1λαq

. What should α be? We have

λ13q ‖uq‖3 = λ

13−α

q ‖Dλq‖23 = λ

43

+ 13−α=0

q

if α = 53. But then for all p < 3

2, ‖∇uq‖p = λq‖uq‖p ∼= λ

53− 2p

q ‖uq‖3∼= λ

43− 2p

q , and 43− 2

p≤ 0 iff

p ≤ 32! So u ∈ B

133,∞ ∩B1

32,∞ =⇒ ω ∈ Lp for all p < 3

2.

THE ONSAGER CONJECTURE 11

Mechanisms for Energy Balance Restoration in Onsager Critical andSupercritical Spaces

Transport.

(1) The DiPerna-Majda example u = 〈u1(x2), 0, u3(x1 − tu1(x2))〉 satisfies the Eulerequation without pressure: ∂tu+ u · ∇u = 0. So each component of u is transportedalong a divergence-free field. For this solution, we clearly have E(t) = E(0).

(2) ω ∈ L2 uses transport nature of vorticity in 2D to show enstrophy conservation. Moreprecisely, if ‖ω‖L2 is bounded, and ω is a weak solution to the 2D Euler

ωt + u · ∇ω = 0,

then it is automatically renormalized since u ∈ W 1,2. Hence distribution function|ω| > λ is conserved, hence the enstrophy.

(3) Exploring transport and the maximum principle for the Navier Stokes equations in2D, we can obtain (EB) via the vanishing viscosity limit.

Theorem. (H. Nussenzveig-Lopes, M. Lopes, A. Cheskidov, RS 2015.) Suppose u0 ∈ W 1,p,1 < p <∞ and hence ω0 ∈ Lp(T2). Let us solve the NSE with initial datum ω0:

∂tων + uν · ∇ων = ν∆ων

ων(0) = ω0.

There is a global in time solution with ων ∈ L∞t Lpx.

Then any weak limit u of the sequence uν would solve the Euler equation and u wouldconserve energy, while being Onsager supercritical. Note again that in our flux example,ω ∈ Lp for all p < 3

2. So to achieve the energy conservation, we have to avoid any attempt

to bound the flux. Instead, we will show that the Kolmogorov 0th law fails.

Proof. We know that NSE in 2D is globally well-posed and that it regularized u instantly.This is more than enough to pass to the limit:

uνt + uν · ∇uν = −ν∆uν + ∆pν .

Take the dot product with v divergence-free to obtain∫uν(t) · v dx−

∫u0 · v dx−

∫ t

0

∫uν ⊗ uν︸ ︷︷ ︸→u⊗ustrongly.

: ∇v dx dt = −ν∫ t

0

∫uν∆v dx dt.

Now test NSE for vorticities with ων :d

dt‖ων‖2

L2 = −2ν ‖∇ων‖2L2︸ ︷︷ ︸

enthalpy

.

Let us assume that 1 < p < 2, for otherwise our general part (A) result covers it. By theGagliardo-Nirenberg-Sobolev inequality,

‖ων‖L2 ≤ ‖∇ων‖1− p2

L2 ‖ω‖p2Lp

=⇒ − 2ν‖∇ων‖2L2 ≤ −2ν‖ων‖

42−pL2 ‖ω‖

− 2p2−p

Lp

≤ −2νC0‖ων‖4

2−pL2 .

Here the last inequality is justified by the maximum principle: ‖ων(t)‖Lp ≤ ‖ων0‖Lp .

12 R. SHVYDKOY

Let y(t) = ‖ων(t)‖2L2 . We have the differential inequality

y′ ≤ −C0νy2

2−p =⇒ dy

y2

2−p≤ −c0ν.

Integrate over [δ, t] to obtain

y(t)−p

2−p − y(δ)−p

2−p

− p2−p

≤ −c0ν(t− δ)

=⇒ 1

y(t)p

2−p− 1

y(δ)p

2−p≥ cpν(t− δ).

We have assumed that ων0 ∈ Lp but not in L2 =⇒ y(δ)→∞ as δ → 0. Hence,

‖ων(t)‖2L2 = y(t) ≤

[C

νt

] 2−pp

= C(νt)p−2p .

Recall the energy inequality for solutions to the NSE:

‖uν(t)‖22 = ‖u0‖2

2 − 2ν

∫ t

0

‖ων(s)‖22 ds

‖u0‖22 − Cν

1− 2−pp t1−

2−pp = ‖u0‖2

2 − C(νt)2p−2p → ‖u0‖2

2.

So in the limit as ν → 0, ‖u(t)‖22 ≥ ‖u0‖2

2. But also,

‖u∨(t)‖22 + 2ν

∫ t

0

‖ω∨(s)‖22 ds = ‖u0‖2

2

=⇒ ‖u(t)‖22 ≤ ‖u0‖2

2.

Frozen Singularities. Vortex sheets in 2D or 3D are exactly Onsager-critical solutions.

Figure 7.

In Figure 7, u has a jump across S(t), and (uν)+ = (uν)− by the divergence-free condition.Also, ω = γδS(t), γ = u+

2 − u−2 . S(t) is “frozen” into the flow, that is, it is transported by

the average U = u++u−

2. This is dictated by the weak formulation of the Euler equations.

Π(t) =

∫S(t)

(|u+|2 − |u−|2) dµt,

where dµt = ( ∂tH|∇H| + U · n) dσt, and dσt is surface measure, S(t) = H = 0. But dµt = 0

is the condition of being “frozen.” The difference between Burgers’ shock and the shock of

THE ONSAGER CONJECTURE 13

the vortex sheet is that particles in the Burgers’ case are allowed to “jump” into the shock,while in the vortex sheet they are not. So there is no exchange of energy between the twosides of the sheet.

Symmetries (Hamiltonian, scaling, etc). The Euler equation has a two-parameter fam-ily of scaling symmetries: if (u, p) solves (E), then the new pair

uλ = λαu(λβx, λα+βt),

pλ = λ2αp(λβx, λα+βt)

solves the same equation (E). Solutions that are invariant under one of these symmetries arecalled self-similar ; among them are stationary solutions. They are necessarily homogeneous:

u =U(x)

|x|αp =

P (x)

|x|2α.

It is an interesting area of research to classify all such solutions (V. Sverak for 3D NSE,RS & Xue Luo for 2D Euler, RS for 3D Euler (arXiv)). In general, for α = n−1

3, where

n is the dimension of the space, the regularity of u at the origin becomes exactly Onsager-

critical: B133,∞. Quick Check: Since u is (−α)-homogeneous, its Fourier transform is (−n+α)-

homogeneous, i.e. u(ξ) = Ω(ξ)|ξ|n−α . In the dyadic shell λq−1 ≤ |ξ| ≤ λq+1 all |ξ|s are comparable,

and yet Ω is radially uniformly distributed. Also, Ω is smooth, so it changes smoothly onthe sphere Sn−1. This creates saturation of Bernstein’s inequalities:

‖uq‖3 ∼1

λn−α‖Dαn

λq ‖3 =1

λn−αq

λn(1− 1

3)

q = λα−n

3q = λ

n3− 1

3−n

3q =

1

λ13q

.

So indeed, u ∈ B133,∞ (dot means “homogeneous”). In order to place u into L3 we need to

truncate it at ∞, which creates extra forcing in the equation. This can be done in 2D and3D using modification of the stream-function and stream-field respectively. For α < n

2the

solutions have locally finite energy. We assume U, P ∈ C1(Sn−1). Truncation at ∞ yieldssolutions to forced Euler

u · ∇u+∇p = F,

with F ∈ C∞loc(Rn) and in addition F ∈ W k,p for all p > 1.For all α < 1 in 2D we have a complete classification of homogeneous solutions (Xue Luo,

RS, 2014). And in 3D a partial classification is given (most are excluded RS, 2016). Inthe case of stationary solutions we are studying not energy conservation but energy balancerelation, which states that the work of force is 0. So, for us anomalous energy flux =anomalous work of force:

Π =

∫RnV · F dx,

it concentrates at the origin and can be expressed by

Π =

∫Sn−1

(U · ~n)3dσ,

(exercise).

14 R. SHVYDKOY

0.0 0.5 1.0 1.5 2.0 2.5 3.0

-2

-1

0

1

2

Figure 8. Here α = 13. Elliptic case with B = −1

0.1. 2D case. Let u = ∇⊥Ψ where Ψ = r1−αψ(θ), and p = r−2αP . Then the equations read

−2αP = α(ψ′)2 + (1− α)2ψ2 + (1− α)ψ′′ψ,

ψ(0) = ψ(2π).

This is equivalent to a Monge-Ampere equation

detD2Ψ = −∇p.Daskalopoulos, Savin, 2009: only two kinds of radial behavior. The ODE has a conservedquantity (coming from conservation of the Bernoulli function along particle lines),

B = (2p+ (1− α)2ψ2 + (ψ′)2)ψ2α1−α .

With this law the system becomes Hamiltonian in phase variables (x, y) = (ψ, ψ′):x′ = y

y′ = −(1− α)2x+α

α− 1Bx

α+1α−1 .

with the pressure

P = −y2

2− (1− α)2

2x2 +

B

2x

2αα−1

being the Hamiltonian. Figures below illustrate two instances of different phase portraits –elliptic and hyperbolic types in the case α = 1/3, the Onsager homogeneity.

Classification result excludes all elliptic solutions for α = 1/3, but there exist plenty ofhyperbolic solutions. We have U · ~n = ψ′. So,

Π =

∫ 2π

0

(ψ′(θ))3dθ =

∫ 2π

0

y3dθ.

Since the pressure p(x, y) is even in y, this means that y is odd relative to critical points.Hence

Π = 0.

THE ONSAGER CONJECTURE 15

0.0 0.5 1.0 1.5 2.0 2.5 3.0

-2

-1

0

1

2

Figure 9. Here α = 13. Hyperbolic case with B = 1

0.2. 3D case.

Theorem 0.1 (RS, 2016). In 3D we also have Π = 0.

Mechanism is unknown. Proof is based on analysis of the reduced system on the sphere.We present the proof below.

The Euler system of equations for homogenous solutions reduces to the following systemon S2:

(2− α)f + div v = 0(1a)

v∇f = |v|2 + αf 2 + 2αp(1b)

(1− α)fv + v∇v = −∇p,(1c)

where v is the tangential vector component of U on the sphere, and f is the normal scalarcomponent. All the notation is intrinsic on the manifold. One important quantity thatcomes into consideration is the Bernoulli pressure: H = |v|2 + f 2 + 2p. One can write theequation for H use the system above:

(2) v∇H = 2αfH.

One can also check directly that the flux is given by

(3) Π = −1

2

∫S2fHdσ.

Multiplying (2) with Hn−1, n ∈ N and integrating over the sphere we obtain∫fHndσ = 0,

for all n ∈ N provided α is not of the form 21+2n0

, for some n0 ∈ N. In this case the result

is proved. If, however, α = 21+2n0

, for some n0 ∈ N, then the identity still needs to be

proved for n = n0. We argue as follows. We have∫fHn0Hkdσ = 0, for all k = 1, 2, . . ..

Consequently,∫fHn0G(H)dσ = 0, for all real analytic functions G with G(0) = 0. Letting

G(x) = 1 − e−x2/ε and letting ε → 0 we obtain

∫H 6=0

fHn0dσ = 0. However, on the set

H = 0 the integral vanishes trivially.

16 R. SHVYDKOY

Low dimensionality of singular set. In the Euler case there is no apriori bound on thedimension of the singularity set of a weak solution (or strong at the first critical time). Forthe Navier-Stokes the celebrated Caffarelli-Kohn-Nirenberg theorem states that the parabolicdimension of singularity for suitable weak solutions is no higher than d = 1 with vanishing1-D measure of the set. This can be used to improve upon classical energy equality criteria.

The best general result belongs to Lions: u ∈ L4tL

4x. Onsager space u ∈ L3B

1/33,∞ gives

another condition with scales like L3tL

9/2x , but as stated not in terms of Lebesgue spaces is

not directly comparable to Lions”. However Lions follows directly from Onsager. To see this

simply interpolate between L4L4 and the enstrophy L2B12,2. This implies that u ∈ L3B

1/33,3 ,

hence energy conservation.To illustrate how low dimensionality helps we present results in the case when singularity

occurs at the first time of a hypothetical blow-up on the torus T3.

0.3. Setup. Suppose (u, p) is a suitable weak solution to the Navier-Stokes system on thedomain T3 × [0, T ) with initial condition (u0, p0). For any function φ ∈ C∞(T3 × [0, T ))supported outside the singularity set S we have the local energy equality, for any 0 < t < T :

(4)

∫T3×t

|u|2φ−∫T3×0

|u0|2φ−∫T3×(0,t)

|u|2∂tφ

=

∫T3×(0,t)

|u|2u · ∇φ+ 2

∫T3×(0,t)

p u · ∇φ− 2ν

∫T3×(0,t)

|∇u|2φ− 2ν

∫T3×(0,t)

u⊗∇φ : ∇u.

Let us write the obtained identity using labels for respective terms in the same order as theyappear in (4):

(5) A−B − C = D + 2P − 2νE − 2νF.

Since u ∈ L∞L2 ∩ L2H1, clearly the terms A,B,E converge to their natural limits∫T3×t

|u|2,∫T3×0

|u|2,∫T3×(0,t)

|∇u|2,

respectively, if |φ| ≤ 1 and φ → 1 pointwise a.e. Therefore, to restore the energy equalitywe have to choose a proper sequence of φ’s for which all the other terms in (4) converge tozero.

0.4. 1-Slice Singularities (T. Leslie, RS, 2016). In the case of a singularity that occursat a single time slice, we replace the interval [0, T ) with [−1, 0) for convenience and assumethat S ⊂ T3×0. We assume that S has Hausdorff dimension d; by CKN, we may assumed ∈ [0, 1]. For convenience, we will identify S with its spacial slice at time t = 0. We denoteby Hd(S) the d-dimensional Hausdorff measure of S and assume that Hd(S) <∞.

Let us denote by Br(x) the open ball y ∈ T3 : |y − x| < r. Choose δ ∈ (0, 1),then choose finitely many xi ∈ T3, ri ∈ (0, δ) for all i, such that S ⊂

⋃iBri(xi), and∑∞

i=1 rdi . 2Hd(S). Denote Ii = (−rαi , rαi ) (where α is determined below); let Qi denote the

cylinder Qi = Bri(xi)×Ii, and put Q =⋃iQi, I =

⋃i 2Ii. Let ψ(s) be the usual (symmetric,

radially decreasing) cut-off function on the line with ψ(s) = 1 on |s| < 1.1 and ψ(s) vanishingon |s| > 1.9. Let φi(x, t) = ψ(|x−xi|/ri)ψ(t/rαi ). Define φ = 1− supi φi. Clearly, φ vanisheson an open neighborhood of Q, while any partial derivative ∂φ is supported within the unionof the double-dilated cylinders, let us denote it Q∗. Note that the Lebesgue measure of the

THE ONSAGER CONJECTURE 17

sequence of Q∗’s vanishes as δ → 0; the same is true of the measure of the sequence of Is.Also note that φ is differentiable a.e. and

|∂φ(x, t)| ≤ supi|∂φi(x, t)|, a.e., see [?, Theorem 4.13].

Therefore, for any a > 0, we have the following bounds, which hold for a.e. t:∫T3

|∂tφ(x, t)|a dx ≤∑i

∫T3

|∂tφi(x, t)|a dx ≤∑i

r−αa+3i χIi(t),(6a) ∫

T3

|∇xφ(x, t)|a dx ≤∑i

∫T3

|∇xφi(x, t)|a dx ≤∑i

r−a+3i χIi(t).(6b)

Let us bound term C first. By Holder’s inequality, we have that for all p, q ≥ 2,

|C| ≤ ‖u‖2LqLp

(∫ 0

−1

(∫T3

|∂tφ(x, t)|pp−2 dx

) p−2p

qq−2

dt

) q−2q

≤ ‖u‖2LqLp

∫ 0

−1

(∑i

r− αpp−2

+3

i χIi(t)

) p−2p

qq−2

dt

q−2q

.

(7)

Here and below, we often write LqLp for Lq(I;Lp(T3)). In particular, if q < ∞ we have‖u‖LqLp → 0. So, in order to conclude C → 0, for q <∞, it suffices to prove that the termin parentheses is bounded as δ → 0; the latter need not vanish. The energy equality for thecase q =∞ will follow by interpolation.

The viscous term F is bounded by

(8) |F | ≤∫Q∗|u|2|∇φ|2 dx dt+

∫Q∗|∇u|2 dx dt.

Clearly, the second integral on the right vanishes as δ → 0. For the first integral we have abound similar to C,

(9)

∫Q∗|u|2|∇φ|2 dx dt ≤ ‖u‖2

LqLp

∫ 0

−1

(∑i

r− 2pp−2

+3

i χIi(t)

) p−2p

qq−2

dt

q−2q

.

Before we proceed with estimates for D and P , let us produce conditions on p, q that warrantvanishing of the integral on the right hand sides of (7) and (9). The following lemma willassist us:

Lemma 0.2. Let d, δ, ri, Ii be as above, and let σ, s be positive numbers. Let∑

i rdi = H.

The inequality

(10)

∫ (∑i

r−σi χIi(t)

)s

dt . Hs

holds whenever s ≥ 1 and s(σ + d) ≤ α, or s < 1 and s(σ + d) < α. When d = 0, the aboveholds (trivially) under the non-strict assumption sσ ≤ α.

18 R. SHVYDKOY

Proof. Case 1, s ≥ 1. By Holder, we have(∑i

r−σi χIi(t)

)s

=

(∑i

rdi r−σ−di χIi(t)

)s

≤

(∑i

rdi

)s−1∑i

rd−(σ+d)si χIi(t)

= Hs−1∑i

rd−(σ+d)si χIi(t).

(11)

Integrating in time we obtain∫ (∑i

r−σi χIi(t)

)s

dt . Hs−1∑i

rd−(σ+d)s+αi .

The sum is ≤ H whenever the condition stated in the Lemma is satisfied.Case 2, s < 1. For each j ∈ Z, define Rj := ri : ri ∈ [2−j, 2−j+1), and let Nj

denote the cardinality of Rj. Clearly Nj . 2jdH and Nj = 0 for j ≤ 0. Also denoteJj = [−2(−j+1)α, 2(−j+1)α]. So if ri ∈ Rj, then r−σi χIi(t) ≤ 2jσχJj(t).∫ (∑

i

r−σi χIi(t)

)s

dt ≤∫ (∑

j

Nj2jσχJj(t)

)s

dt . Hs

∫ (∑j

2j(σ+d)χJj(t)

)s

dt

≤ Hs

∫ ∞∑j=1

2j(σ+d)sχJj(t) dt . Hs

∞∑j=1

2j((σ+d)s−α)

The final sum converges to an adimensional number due to the assumption of the Lemma.

With the use of Lemma 0.2 we will now provide estimates on the C, D, P , and F termsfor various values of p, q ≥ 2 and arbitrary α > 0. Then we will optimize over α to providethe largest region in which all terms vanish and hence the energy equality holds.

First, note that p < q ⇐⇒ p−2p

qq−2

< 1, and p ≥ q ⇐⇒ p−2p

qq−2≥ 1. Applying

Lemma 0.2 with σ = αpp−2− 3 and s = p−2

pqq−2

, we see that C vanishes whenever

(12)3− dp

+α

q≤ 3− d

2, p ≥ q ≥ 2;

3− dp

+α

q<

3− d2

, 2 ≤ p < q.

Reasoning similarly, we have F → 0 whenever

(13)3− dp

+α

q≤ 3− d+ α− 2

2, p ≥ q ≥ 2;

3− dp

+α

q<

3− d+ α− 2

2, 2 ≤ p < q.

In both cases the conditions are non-strict if d = 0.The estimates we use to bound the terms D and P depend on whether p ≥ 3 or p < 3;

we consider each case in turn. First, let p ≥ 3. Using Holder’s inequality, together with thebound ‖up‖Lq/3Lp/3 . ‖u‖3

LqLp , we have for all q, p ≥ 3,

(14) |D|+ |P | ≤ ‖u‖3LqLp

∫ 0

−1

(∑i

r− pp−3

+3

i χIi(t)

) p−3p

qq−3

dt

q−3q

.

Arguing as before with the use of Lemma 0.2, we see that |D|+ |P | → 0 whenever

(15)3− dp

+α

q≤ 2 + α− d

3, p ≥ q ≥ 3;

3− dp

+α

q<

2 + α− d3

, 3 ≤ p < q,

THE ONSAGER CONJECTURE 19

with the non-strict inequality in both cases if d = 0.In the case p < 3 we can no longer use straight Holder to bound the D + P term, instead

we will use interpolation with the enstrophy norm. So, when p < 3, we have

(16) |D|+ |P | ≤ ‖u‖3βL2H1‖u‖3(1−β)

LqLp ‖∇φ‖LσL∞ ,where

(17)1

3=β

6+

1− βp

=⇒ β =6− 2p

6− p;

1

σ= 1− 3β

2− 3(1− β)

q=

2pq − 3p− 3q

(6− p)q.

Now

‖∇φ‖σLσL∞ =

∫supir−σi χIi(t) dt ≤

∫supj

2jσχJj(t) dt .∑j

(2α−σ)−j,

and the sum on the right is bounded whenever σ < α. Substituting in for σ and simplifying,we obtain

(18)2 + α

p+α

q<

1 + 2α

3, p < 3 ≤ q.

Let us discuss optimal values of α now. First, regarding the case p ≥ 3. In this case wehave the three constraints (12), (13), and (15), representing a triple of parallel lines. TheC-line pivots around the energy space L∞L2 and rotates counterclockwise (toward a more

stringent condition) as α increases. The DP -line pivots around L3L9−3d2−d and also rotates

counterclockwise (but toward a more relaxed condition) as α increases. The F -line pivots

around L2L6−2d1−d counterclockwise, relaxing as α increases. (Note that some exponents can

become negative for larger d’s; however, the region beyond p, q = ∞ can be disregarded atthis moment.) So, the conditions become optimal when the two lower lines coincide. Simplelinear algebra shows that for d < 1 the C- and DP -lines are lower, for d > 1 the C- andF -lines are lower, and at d = 1 all three lines coincide at their optimal tilt. So, in the cased ≤ 1, setting C- and DP -lines equal we find α = 5−d

2(clearly in this case α ≥ 2 and so the

F -condition is more relaxed). Also notice that the restriction q ≥ 3 is no longer necessary

since beyond that point the C-line can be extended up to the space L5−d3−dL∞ by interpolation

with the energy space. In the case d > 1 we set C- and F -lines equal which results in thevalue of α = 2 as being most optimal. We thus obtained the following conditions

2(3− d)

p+

5− dq≤ 3− d, 5− d

3− d≤ q ≤ p, d ≤ 1(19)

2(3− d)

p+

5− dq

< 3− d, 3 ≤ p < q, d ≤ 1(20)

3− dp

+2

q<

3− d2

, 3 ≤ p < q, 1 < d < 3.(21)

Let us note that in the case d > 1 the optimal line drops below Lions’ space L4L4, thereforeonly the case p < q yields new results. The line intersects the segment [L4L4, L∞L3] at the

space L6+2d3−d L

6+2d1+d . And again, the inequality (20) becomes non-strict in the case d = 0.

For the values 0 ≤ d ≤ 1, the point on the bisectrice separating the open and closed

regions is L11−3d3−d L

11−3d3−d . When d = 1, it becomes the classical Lions space L4L4.

Let us now address the case p < 3. In this case we use (18) to replace (15) in the previousargument, while (12) and (13) are understood under the lighter restriction p, q ≥ 2. Also,

20 R. SHVYDKOY

L∞L3 L∞L2

L2L∞

L5/3L∞

new

L2L6

L3L9/2

L4L4L5L3

x = 1p

y = 1q

1

Figure 10. d = 0

L∞L3 L∞L2

L2L∞

L5−d3−dL∞

new

L2L6

L4L4 L15−3d3−d L3

x = 1p

y = 1q

1

Figure 11. 0 < d < 1

note that region now only lies in the cone q > p. The new DP -line now pivots aroundthe enstrophy point L2L6 counterclockwise as α increases. For d ≤ 1, the non-trivial newregion appears as α increases beyond α = 5−d

2. The F -line is less restrictive than C-line, so

we can disregard it. At α = 5−d2

the C- and DP -lines intersect at L15−3d3−d L3. The point of

intersection reaches its final state at the energy space L∞L2 when α = 4. In the process ittraverses the curve given by

(22) (18− 6d)x2 + (6− 6d)xy − (21− 7d)x− (7− 3d)y + 6− 2d = 0,

in coordinates x = p−1 and y = q−1. Notice that the curve in fact contains both L2L6 andL∞L2 for all values of d, as we expect it to (indeed, these points are the two axes of rotationfor our lines). However, since we are restricted to the case when p < 3, the part of the curve

that we can use is limited to that connecting L15−3d3−d L3 and L∞L2. The curve is a part of

the hyperbola as seen from the negative Hessian (except when d = 1, when the curve is aparabola).

For d > 1, the two lines C and F coincide when α = 2; at this value of α, the new DP -line

cuts through the C-line at space L6+6d3−d L

6+6d1+3d , which is already inside the strip p < 3. It does

not make sense to decrease α since doing so would move the F - and DP -lines clockwise insidethe already discovered region. Increasing α above 2 makes the F -line more relaxed, and theintersection point of C- and DP -lines falls on the same curve (22). This time, however, the

curve begins further to the right, at the space L6+6d3−d L

6+6d1+3d , and ends at L∞L2, corresponding

to the fixed range 2 ≤ α ≤ 4.Finally, recall that in all the arguments above we have assumed q < ∞ in order to

ensure that the vanishing of the terms comes from the norm Lq(I;Lp(T3)), and not from theHausdorff measure of S. However, the case q = ∞ follows automatically by interpolationwith the Leray-Hopf line, which lends the solution strictly inside the quadrant q, p <∞.

The figures above illustrate the new regions uncovered in each case.

THE ONSAGER CONJECTURE 21

L∞L3 L∞L2

L2L∞

new

L2L6

L4L4L6L3

x = 1p

y = 1q

1

Figure 12. d = 1

L∞L3 L∞L2

L2L∞

new

L2L6

L4L4

L6+2d3−d L

6+2d1+d

L6+6d3−d L

6+6d1+3d

x = 1p

y = 1q

1

Figure 13. 1 < d < 3