Source
Constantin, E and Titi (1994), Communications in Mathematical Physics 165:207–209, LIT-1248, read in full in NOTE-982 from the authors’ manuscript. The theorem and the proof are the paper’s; the derivation below follows it and fills in what it leaves out.
What was actually shown
Setting. D = [0, 1]³ with periodic boundary conditions; ‖·‖_p is the L^p(D) norm. u is divergence-free and satisfies the weak Euler equations (1): for every smooth ψ with compact support in time,
−∫∫u_j ∂_tψ_j − ∫u_j(x, 0)ψ_j(x, 0) − ∫∫∂_iψ_j u_iu_j − ∫∫∂_iψ_i p = 0.
Hypothesis. u ∈ C([0, T]; L²) and u ∈ L³(0, T; B^α_{3,∞}) with 1/3 < α < 1, where ‖u‖{B^α{3,∞}} = ‖u‖₃ + sup_y |y|^(−α)‖u(·+y) − u‖₃. (For α ≥ 1 the field lies in B^β_{3,∞} for every β < 1, so take β in (1/3, 1).) φ ≥ 0 is smooth, supported in the unit ball, ∫φ = 1; φ_ε(x) = ε^(−3)φ(x/ε) and v^ε = φ_ε ∗ v.
Step 1. The Besov facts. Write δ_yu(x) = u(x − y) − u(x).
- (6) ‖δ_yu‖₃ ≤ |y|^α‖u‖{B^α{3,∞}}: the definition.
- (7) u^ε(x) − u(x) = ∫φ_ε(y)δ_yu(x) dy, since ∫φ_ε = 1. By Minkowski’s inequality, ‖u^ε − u‖₃ ≤ ∫φ_ε(y)‖δ_yu‖₃ dy ≤ ε^α‖u‖, because |y| ≤ ε on the support.
- (8) ∇u^ε(x) = ∫∇φ_ε(y)u(x − y) dy = ∫∇φ_ε(y)δ_yu(x) dy, since ∫∇φ_ε = 0. So ‖∇u^ε‖₃ ≤ ∫|∇φ_ε(y)| |y|^α dy ‖u‖ ≤ ε^(α−1)‖∇φ‖₁‖u‖, using ∇φ_ε(y) = ε
Step 2. The commutator identity (10). Define r_ε(x) = ∫φ_ε(y) δ_yu(x)⊗δ_yu(x) dy. Expanding the product,
r_ε = (u⊗u)^ε − u^ε⊗u − u⊗u^ε + u⊗u,
using ∫φ_ε = 1 for the last term. Adding u^ε⊗u^ε and subtracting (u − u^ε)⊗(u − u^ε) = u⊗u − u⊗u^ε − u^ε⊗u + u^ε⊗u^ε leaves (u⊗u)^ε. So, pointwise,
(u⊗u)^ε = u^ε⊗u^ε + r_ε − (u − u^ε)⊗(u − u
and the subgrid stress τ_ε = (u⊗u)^ε − u^ε⊗u^ε equals r_ε − (u − u^ε)⊗(u − u^ε): quadratic in increments over distances below ε.
Step 3. The bound (12) on r_ε. By Minkowski in L^{3/2}, and ‖a⊗a‖{3/2} ≤ ‖a‖₃², ‖r_ε‖{3/2} ≤ ∫φ_ε(y)‖δ_yu‖₃² dy ≤ ε^(2α)‖u‖². Likewise ‖(u − u^ε)⊗(u − u^ε)‖{3/2} ≤ ‖u − u^ε‖₃² ≤ ε^(2α)‖u‖², by (7). So ‖τ_ε‖{3/2} ≤ 2ε
Step 4. The mollified energy balance (11). Take ψ(z, t) = θ(t)φ_ε(x − z)e_j in (1), for fixed x and θ smooth with compact support in (0, T). This gives, in the sense of distributions in t,
∂_t u^ε_j(x, t) + ∂_i(u_iu_j)^ε(x, t) + ∂_j p^ε(x, t) = 0.
The pressure: taking ψ = ∇χ with χ smooth in x gives −Δp = ∂_i∂_j(u_iu_j) up to a function of t alone, which has no gradient, so ∇p is determined and lies in W^{−1,3/2} in space; then ∇p^ε is smooth in x and in L^{3/2} in t. So ∂_t u^ε ∈ L^{3/2}(0, T; L²), u^ε is absolutely continuous in time with values in L², and
(d/dt) ½∫|u^ε|² = ∫u^ε·∂_tu^ε = −∫u^ε_j ∂_i(u_iu_j)^ε − ∫u^ε·∇p
The pressure term is ∫p^ε ∇·u^ε = 0, since mollification commutes with divergence. Integrating the other by parts on the torus,
(d/dt) ½∫|u^ε|² = ∫(u_iu_j)^ε ∂_iu
which is the paper’s ∫Tr((u⊗u)^ε ∇u^ε), the stress being symmetric. Integrated from 0 to t, this is (11). The paper’s u^ε(·, 0) is φ_ε ∗ u(·, 0), the initial datum, because u ∈ C([0, T]; L²).
Step 5. The resolved part carries no flux. Since ∇·u^ε = 0, ∫u^ε_iu^ε_j ∂_iu^ε_j = ∫u^ε_i ∂_i(½|u^ε|²) = −∫(∇·u^ε)½|u^ε|² = 0. So by Step 2 the right side of Step 4 is ∫τ_ε : ∇u^ε, the flux through the subgrid stress.
Step 6. Hölder. |∫τ_ε : ∇u^ε| ≤ ‖τ_ε‖{3/2}‖∇u^ε‖₃ ≤ 2ε^(2α)‖u‖² · Cε^(α−1)‖u‖ = Cε^(3α−1)‖u(t)‖³{B^α_{3,∞}}. Integrating in time,
|∫|u^ε(t)|² − ∫|u^ε(0)|²| ≤ Cε^(3α−1)∫₀ᵗ‖u(τ)‖³_{B^α_{3,∞}} dτ,
finite because u ∈ L³(0, T; B^α_{3,∞}). The manuscript’s middle line, with exponents 2/3 and 1/3, is a misprint: it is not homogeneous in u (degree 5/3 against the cubic left side) and would give ε^((5α−1)/3). The line written here is what Hölder gives, and it gives the paper’s last line.
Step 7. The limit. For each t, u(t) ∈ L², so u^ε(t) → u(t) in L² as ε → 0, and ∫|u^ε(t)|² → ∫|u(t)|². The right side of Step 6 tends to 0 exactly when 3α − 1 > 0. Hence ∫|u(t)|² = ∫|u(0)|² for every t in [0, T).
What could have come out otherwise. The exponent 3α − 1 is fixed by counting: the flux is cubic in u, two factors are increments over distance ε (each ε^α) and one is a gradient at scale ε (ε^(α−1)). Had the resolved part u^ε⊗u^ε not dropped out in Step 5, it would contribute a term with no small factor, and nothing would follow. It drops out because transport of resolved energy by a divergence-free resolved field is a divergence. At α = 1/3 the bound is of order one at every ε, which is where Kolmogorov’s scaling (4) puts turbulence.
Checked numerically. On a synthetic divergence-free periodic field in two dimensions (512², spectrum chosen for α ≈ 1/2), with a smooth bump mollifier at ε = 2 to 32 grid points: identity (10) held to a relative error of 10⁻¹⁴; ∫u^ε⊗u^ε : ∇u^ε was zero to rounding; and the fitted exponents of ‖r_ε‖_{3/2}, ‖u − u^ε‖₃ and ‖∇u^ε‖₃ in ε were 1.06, 0.59 and −0.52, against 2α = 1, α = 0.5 and α − 1 = −0.5. The flux itself was far below the bound, as a field with random phases should give. This checks the algebra and the estimates, not the theorem.
What this does not say
- That rough solutions dissipate. The theorem is one-directional. Below 1/3 it gives nothing: neither conservation nor dissipation. Some rough solutions conserve energy. The shear flow u = (f(x₂), 0, 0) is a stationary weak solution for any f in L², however rough, and its flux is zero at every ε, since the only non-zero entry of (u⊗u)^ε, (f²)^ε in position (1, 1), meets ∂₁u^ε₁ = 0. The existence of energy-dissipating weak solutions below 1/3, the negative half, is Isett’s (2018) theorem for Hölder exponents below 1/3, after De Lellis and Székelyhidi, with Buckmaster, De Lellis, Székelyhidi and Vicol (2019) for admissible solutions. None of these is held or read by the record. So “a coarse-grained flux can survive as the scale goes to zero for rough fields” is true, but on that later evidence, not on this paper.
- That turbulence dissipates energy, or does not. The theorem is about Euler weak solutions. The paper’s “physical interpretation”, that ⟨|δu(r)|³⟩^(1/3) ≤ Cr^α with α > 1/3 forces ε̄ = 0, moves to viscous Navier–Stokes and to ensemble averages without a proof. Read the other way, as Onsager did: a non-zero dissipation in the limit of vanishing viscosity requires the limiting field to be no smoother than 1/3 in this sense. That reading also needs the passage from Navier–Stokes to Euler.
- The endpoint α = 1/3. The bound is constant there and decides nothing. Conservation slightly below B^{1/3}_{3,∞} is a later result.
- Where energy is lost. Only the total energy is controlled. A local balance with a dissipation measure is Duchon and Robert’s (2000).
- Anything about the scales that carry the flux. The bound is on the whole flux across ε. Which scales above and below ε contribute it is THEORY-754’s question, under a hypothesis of the same form.
- Anything about units or organization. One field, one balance law, a scale ε chosen by the analyst.
Connections
- THEORY-754. Its hypothesis ‖δu(r)‖₃ ≤ C₀u_rms(r/L)^σ is the B^σ_{3,∞} condition, and its three norm bounds are Steps 1 and 3 here. Its prefactor (u_rms³/L)(ℓ/L)^(3σ−1) is this theorem’s ε^(3α−1): read as ℓ → 0 at a fixed field, it bounds the whole flux and vanishes for σ > 1/3. So the non-zero mean flux that Eyink and Aluie assume for locality to be read as a fraction is possible only at σ₃ ≤ 1/3, and Kolmogorov’s 1/3 is the largest exponent that allows it. Together the two say: regularity above 1/3 removes the flux; at 1/3 it can persist, and is then local in scale. Neither shows that it does persist.
- QUESTION-219. A3 cited this work for “the regularity of a field decides whether the unresolved scales vanish in the limit”. This THEORY holds the half of that the paper proves. It does not answer the question, whether a pattern of transfer picks out a unit; it bears on A5’s precondition, that a flux is defined by a balance law for a specified quantity, and on which fields can carry a flux to the smallest scales at all.
- LIT-1252. Eyink (2005) refines estimates of this kind into locality bounds; not read.
- THEORY-661 and CLAIM-157. Unrelated: no blocks, no gap, no aggregation.
- The negative half (2026-10-20). THEORY-763 (Isett, Proposed): below 1/3, Hölder solutions exist whose flux across vanishing scales survives the limit, in both directions. So the threshold here is sharp in that sense. It is not sharp for one-way, dissipative transfer, which needs admissible solutions.