Introduction to the classical DeGiorgi method and its application in the regularity theory of incompressible Navier-Stokes equations

Introduction to the classical DeGiorgi method and its application in the regularity theory of

incompressible Navier-Stokes equations

Yi-Hsuan Lin

Abstract

The note is mainly for personal record, if you want to read it, please be careful. This note is given by Prof. Chi Hin Chan’s lecture during the summer course in NCTS, 2015. We will introduce the De-Giorgi’s method in the classical setting of elliptic PDE’s in the divergence form with rough diffusive coefficients. It was in this setting that De-Giorgi used his method to establish the regularity of solutions to general linear elliptic PDE’s in the framework of Holder’s continuity.

After this preparation in the classical elliptic setting, we go ahead to discuss a piece of recent work of Alexis Vasseur in which he applied the De-Giorgi’s method to yield an alternative proof of the key proposition in the famous partial regularity theory of Caffarelli, Kohn, Nirenberg.

Further, we will introduce notions such as the Serrin’s regularity criteria, and the notion of suitable weak solutions in the theory of weak solutions to the incompressible Navier-Stokes equations.

1 Introduction to DeGiorgi’s ideas

Let B(1) ⊂ R^N (N ≥ 2). We consider a function u ∈ H¹(B(1)) which is a weak solution to

∂_α{a^αβ∂_βu} = 0 in B(1). (1) Here (a^αβ) : B(1) → R^N2 satisfies the following constraint:

1

Λ|ξ|²≤ a^αβ(x)ξ_αξ_β≤ Λ|ξ|², ∀x ∈ B(1), ξ ⊂ R^N. Or regard ξ ∈ T_x^∗(B(1)). Relation with Riemannian geometry,

g = g_αβdx^α⊗ dx^β and

∆f = 1

√G∂α{g^αβ√ G∂βf }, where g^αβ= d(dx^α, dx^β).

In this note, we focus on the DeGiorgi method, 1957-1960. Compare to the Schauder’s estimate, we need the coefficients lying in C^α(Ω), then the solution will lie in C^2,α(Ω). The idea is that if we rescale uλ(x) = u(λx) and a^αβ_λ (x) = a^αβ(λx), if we make λ smaller and smaller, then a^αβ_λ will behave like a piecewise

constant function. This scaling just “flatten” the graph of solutions, it will not change the height of functions. Therefore, the regularity will similar as the Laplacian case.

1.1 First DeGiorgi lemma

The first step is from the L² to L^∞. If the L²energy of the solution u in B(1) is small enough, then

kuk_L∞(B(¹₂))≤ 1.

Lemma 1.1. There exists an absolute constant ₀ = ₀(N, Λ) > 0 such that the following statement holds for any weak solution u ∈ H¹(B(1)) to (1). If ku₊k²_L2(B(1))< ₀, then

u+≤ 1 in B(1 2).

Remark 1.2. For the whole information of u, just multiply the negative sign and consider the previous lemma again. What if kuk_L2(B(1)) is not small ? Rescale

V = 0

2kukL²(B(1))

u, then kV k_L2(B(1))< ₀. This will imply

kuk_L∞(B(¹₂))≤ 2

₀kuk_L2(B(1)). Set up the Game: Let Rk= 1

2(1 + 1

2^k), k ≥ 0. Let ϕk ∈ C₀^∞(B(Rk)) with

ϕk=

(1 in B(R_k+1), 0 outside B(Rk),

and |∇ϕ_k| ≤ 2 Rk− Rk+1

χ_B(Rk)= 2^k+3χ_B(Rk). Cutting level. Let V₀ = u₊, V_k = [u − (1 − 1

2^k)]₊ with V_k ≤ Vk−1 and

∇V_k = ∇u · χ_{vk>0}. U_k=´

B(Rk)|V_k|². Rough idea: Build up the following

Uk ≤ C₀^kU_k−1^β , β > 1. (2) If U0 = ku+k²_L2(B(1)) < 0, then Uk → 0 as k → ∞. ´

B(¹₂)[u − 1]+ = limk→∞Uk= 0 which implies u ≤ 1 in B(1

2).

Test against (1) by ϕ²_kVk, Vk∈ H/ ₀¹(B(1)) but ϕ²_kVk ∈ H₀¹(B(1)), which will be the test function. Then

0 = ˆ

B(1)

a^αβ∂βu∂α{ϕ²_kVk}

= ˆ

B(1)

a^αβ∂βu · 2ϕk· ∂αϕk· Vk+ ˆ

B(1)

a^αβ∂βu · ϕ²_k∂αVk

= ˆ

B(1)

a^αβ∂βVk· 2ϕk· ∂αϕk· Vk+ ˆ

B(1)

a^αβ∂βVkϕ²_k∂αVk,

and let g_x(A, B) = a^αβ(x) · A_αB_β, ˆ

B(1)

a^αβ∂_αV_k∂_βV_kϕ²_k= −2 ˆ

B(1)

a^αβ∂_βV_k· ϕ_k· ∂_αϕ_k· V_k

= −2 ˆ

B(1)

gx(ϕk∇Vk, Vk∇ϕk)|xdx

≤  ˆ

B(1)

g_x(ϕ_k∇Vk, ϕ_k∇Vk)|_xdx

+1

 ˆ

B(1)

gx(Vk∇ϕk, Vk∇ϕk)|xdx.

This implies ˆ

B(1)

a^αβ∂αVk∂βVkϕ²_k ≤ 4 ˆ

B(1)

a^αβ∂αϕk∂βϕk· V_k²

≤ 4Λ ˆ

B(1)

|∇ϕ_k|²V_k²

≤ 4Λ2^2k+6 ˆ

B(Rk)

V_k²

= 128Λ4^kU_k and use ellipticity condition, we have

ˆ

B(1)

|∇(ϕkVk)|²≤ 2 ˆ

B(1)

|∇ϕk|²V_k²+ 2 ˆ

B(1)

ϕ²_k|∇Vk|²

≤ 256 · 4^k ˆ

B(R_k)

|Vk|²+ 256Λ4^kUk

= 256(1 + Λ)4^kU_k. For N ≥ 3, use the Sobolev embedding theorem,

kϕkVkk

L^{N −22N} (B(1))≤ C0( ˆ

B(1)

|∇(ϕkVk)|²)^1/2.

For N = 2, use the Ladyzhenskaya inequality,

kϕkVkk_L4(B(1))≤ kϕkVkk^1/2_L2(B(1))· k∇(ϕkVk)k^1/2_L2(B(1)).

Note that, {x ∈ B(1) : Vk > 0} ⊂ {x ∈ B(1) : Vk−1(x) > 1

2^k}, χ_{Vk>0} ≤ (2^kVk−1)^{N −24} , and χB(R_k)≤ ϕ

2N N −2

k−1, Uk =

ˆ

B(Rk)

V_k²χ_{Vk>0}≤ ˆ

B(Rk)

V_k²(2^{N −24} )^kV

4 N −2

k−1

≤ (2^{N −24} )^k ˆ

B(R_k)

V

2N N −2

k−1 ≤ (2^{N −24} )^kkϕk−1Vk−1k

2N N −2

L^{N −22N} (B(1))

,

which raised up the index. Finally, Uk≤ (2^{N −24} )^kC0( ˆ

B(1)

|∇(ϕk−1Vk−1)|²)^{N −2N}

≤ (2^{N −24} )^kC0(256(1 + Λ)4^kUk)^{N −2N}

≤ [C0(N, Λ)]^kU

N N −2

k−1 , k ≥ 1.

Thinking: {U_k}^∞_k=1 ⊂ [0, ∞), U_k+1 ≤ (C₀)^k+1U_k^β, k ≥ 0, C₀ > 1, β > 1.

Ul+1∼ (C₀^k+1U_k^β−1)Ukand Uk+1∼ 1

AUk, where A > 1. It means C₀^k+1U_k^β∼ 1 A, then U_k ∼ 1

(AC₀^k+1)^β−11

and U_k+1 ∼ 1

AU_k ∼ 1 A

1 (AC₀^k+1)^β−11

,..., A ∼ C

1 β−1

0 . Want C₀^k+1U^β−1≤ 1

C

1 β−1

0

for all k ≥ 0. ku₊k²_L2(B(1)) ≤ 1 (C0C

1 β−1

0 )^β−11 .

1.2 A detour: DeGiorgi’s isoperimetric inequality

Take u ∈ H¹(B(1)) ∩ C^∞(B(1)), A = {x ∈ B(1) : u(x) ≤ 0}, B = {x ∈ B(1) : u(x) ≥ 1}, C = {x ∈ B(1) : 0 < u(x) < 1}.

Proposition 1.3. The following estimate holds:

|A| · |B| ≤ CNk∇u+kL²(B(1))|C|¹².

Takeu = max{0, inf{1, u}}, x ∈ A, y ∈ B, (Trick: Think of |∇e eu| = |∇eu|χ_C as a function on R^N.)

eu(y) −u(x) = 1 =e ˆ |y−x|

0

d

dτ{u(x + τe y − x

|y − x|)}dτ

= ˆ |y−x|

0

∇u|e_(x+τ y−x

|y−x|)· y − x

|y − x|dτ.

Then 1 ≤

ˆ |y−x|

0

|∇u(x + τe y − x

|y − x|)|dτ ≤ ˆ ∞

0

|∇eu(x + τ y − x

|y − x|)|dτ, and since |x| ≤ 1, if y ∈ B, |y − x| ≤ 2,

|B| ≤ ˆ

y∈B

ˆ ∞ 0

|∇eu(x + τ y − x

|y − x|)|dτ dy

≤ ˆ

|y−x|≤2

ˆ ∞ 0

|∇u(x + τe y − x

|y − x|)|dτ

≤ ˆ 2

0

r^{N −1} ˆ

S^{N −1}

ˆ ∞ 0

|∇u(x + τ ω)|dτ dS(ω)dre

= 2^N N

ˆ ∞ 0

ˆ

S^{N −1}

τ^{N −1}|∇eu(x + τ ω)|dS(ω)dτ

= 2^N N

ˆ

R^N

|∇eu(x + y)

|y|^{N −1} dy = 2^N N

ˆ

C

|∇u(y)|e

|x − y|^{N −1}dy,

then

|A| · |B| ≤ 2^N N

ˆ

y∈C

ˆ

x∈A

|∇u(y)|e

|x − y|^{N −1}dxdy

=2^N N

ˆ

y∈C

|∇eu(y)| · ˆ

x∈A

1

|x − y|^{N −1}dxdy, x ∈ A and |x − y| ≤ 2. A ⊂ {x ∈ |x − y| ≤ 2} provided |y| ≤ 1,

ˆ

A

1

|x − y|^{N −1}dy ≤ ˆ

|x−y|≤2

1

|x − y|^{N −1}dy = CN. Therefore,

|A| · |B| ≤ C_N ˆ

y∈C

|∇eu(y)|dy.

2 Oscillating lemma

Suppose we have a solution u : B(2) → R to (1) which satisfies 1. u ≤ 2 in B(2),

2. |{x ∈ B(1) : u(x) ≤ 0}| ≥ |B(1)|

2 . The second DeGiorgi lemma shows that

Lemma 2.1. ∃ another absolute constant λ = λ(Λ, N ) ∈ (0, 1) if the solution u : B(2) → R satisfies 1,2, then

u ≤ 2 − λ in B(1 2).

Is it possible that ˆ

B(1)

(2(u − 1)₊)²< ₀,

we care since u1 = 2(u − 1) is another solution to (1). If this happens, which means´

B(1)(u1)²< 0, by the first DeGiorgi lemma, we have u₁≤ 1 in B(1

2), equivalently, u ≤ 3 2.

This may be false (do successive cuttings...) and treat u1 just in the same manner as u and let u2= 2(u1− 1), and iterate these steps until we find some uk≤ 3

2. Note that u2= 2(u1− 1) = 2²u − 2²⁻², then u_k= 2^ku − (2^k+ 2^k−1+ · · · + 2)

= 2^k{u − 2(1 − 1 2^k)}.

Look at u = u0 and so on, set uk ∈ H¹(B(1)), Ak = {x ∈ B(1) : uk(x) ≤ 0}, Bk = {x ∈ B(1) : uk(x) ≥ 1}, Ck = {x ∈ B(1) : 0 < uk(x) < 1}. |A1| =

|A₀| + |C₀|, |A_k| = |A_k−1| + |C_k−1| and property 2 gives that |A_k| ≥ |B(1)|

2 .

Remark 2.2. The structure of (1) allows us to obtain: For each solution u_kto (1) in B(2),

ˆ

B(1)

|∇(uk)+|²≤ CΛ,N

ˆ

B(2)

|(uk)+|²≤ 4CΛ,N|B(2)|,

i.e.,

k∇(u_k)₊k_L2(B(1))≤ C_Λ,N.

|A_k| · |B_k| ≤ C_Nk∇(u_k)₊k_L2(B(1))|C_k|¹² ≤ C_Λ,N|C_k|¹².

Remember: We want |B_k| to be small eventually. We want 4|B_k| < ₀, i.e., ˆ

B(1)

(uk+1)²₊= 4 ˆ

B(1)

(uk− 1)²₊χ_Bk≤ 4|Bk| < 0.

If in the worst case

|Bk| ≤ 0

4 holds for all k, then

|Ck| ≥ (|B(1)|

2

0

4 1 CΛ,N

)²:= αN,Λ, ∀k ≥ 0.

But |B(1)| ≥ |A_k| ≥ kα_N,Λ+|B(1)|

2 or k ≤ |B(1)|

2αN,Λ

, choose K₀:= |B(1)|

2αN,Λ

 + 1.

In other words, the propagation on the property |Bk| ≥ 0

4 would stop at some 0 ≤ k ≤ K0, i.e., |BK₀| < 0

4 will imply´

B(1)(uK₀+1)²₊ < 0 with uK₀+1 solves (1). Apply the first DeGiorgi lemma to uK₀+1 and get uK₀+1≤ 1 in B(1

2), i.e., {u − 2(1 − 1

2^K0+1)} ≤ 1

2^K0+1 in B(1 2), then u ≤ 2 − 1

2^K0+1.

Remark 2.3. You can always choose d1with |d1| ≤ λ

2 such that |u − d1| ≤ 2 −λ 2. Let

W⁽¹⁾= 2 2 −λ

2 [u(1

2x) − d1] be a rescaled solution. Finally, we will see

osc_B(1

2k)u ≤ (2 −^λ₂

2 )^kosc_B(1)u or (|u| ≤ 2)

osc_B(rk)u ≤ 4δ^k. Thinking: |x| ∼ r^kwhich means k ∼ ln |x|

ln r , then δ^k ∼ δ^{ln |x|ln r} = (δ^logδ|x|)^{ln r1 logδ e1} =

|x|^{ln r1 logδ e1} ∼ |x|^γ. Then |u(x) − u(0)| ∼ |x|^γ.

3 Introduction to the elements of incompress- ible Navier-Stokes (NS) equations

3.1 Historical view

Starting point: The linear theory. Let Ω be a bounded domain with smooth boundary in R^N, N = 2, 3. Consider

(∂tu − ∆u + ∇P = 0

∇ · u = 0 in [0, T ] × Ω. (3)

To write down the weak formulation, we need the right functional space for the

“test-vector fields” on Ω. Let

Λ¹_c,σ(Ω) = the space of al smooth, compactly supported divergence free v.f on Ω, V (Ω) = Λ¹_c,σ^{k·kH1 (Ω)}, and HΩ= Λ¹_c,σ^L

2(Ω)

. If Ω is bounded, kϕk_H1

0(Ω)= kϕkV = k∇ϕk_L2(Ω)since kϕk_L2(Ω)≤ C(Ω)k∇ϕk_L2(Ω). If Ω = R^N, kϕkV = kϕk_L2(R^N)+ k∇ϕk_L2(R^N) .

Let eV (Ω) = {u ∈ H₀¹(Ω)|∇ · u = 0}, then V (Ω) ⊂ eV (Ω). Are they the same ? If Ω is either R^N, R^N+ or any bounded domain with smooth boundary in R^N for N = 2, 3 (usual domains), then V (Ω) = eV (Ω). The example when V (Ω) eV (Ω) is to take

Ω = R³− {this (x1, x3)-plane with a disc removed}.

Heywood (1970) proved eV (Ω) = finite dimesional space V^Ω, which has some- thing to do non-uniqueness, in general situation.

Theorem 3.1. Let Ω be a “usual domain” in R^N(N = 2, 3). There exists a unique element u ∈ C⁰([0, T ]; HΩ) ∩ L²(0, T ; VΩ) which satisfies

A. ∂tv ∈ L²(0, T ; V_Ω⁰)

B. For a.e. t ∈ [0, T ], and ϕ ∈ VΩ, we have h∂tv|t, ϕi_V0

Ω⊗VΩ+ ˆ

Ω

∇v|t: ∇ϕ = (f (t), ϕ)_V0

Ω⊗VΩ (weak formulation of (3)), C. u(0) = a.

The following properties are fundamental fact to further build the theory of weak solutions to NS equations. Why do we care about f ? Back to the original NS equations

(∂tu − ∆u + u · ∇u + ∇P = 0

∇ · u = 0 in [0, T ] × Ω (4)

Weak formulation for (4). For a.e. t ∈ [0, T ], ∀ϕ ∈ VΩ, we have h∂tu, ϕi_V0

Ω⊗V_Ω+ ˆ

Ω

∇u : ∇ϕ = − ˆ

Ω

(u · ∇u) · ϕ

= ˆ

Ω

(u ⊗ u) : ∇ϕ = ˆ

Ω

u_αu_β∂_αϕ_β.

Reference: Seregin, Lecture notes on Regularity theory to NS equations, Oxford (Ch5) and Leray-Schauder fixed point method Twice. Big Glory (Leray-Hopf, 1930-1940).

Theorem 3.2. Take a ∈ H_RN. There exists u ∈ L^∞(0, T ; H_RN) ∩ L²(0, T ; VΩ) which satisfies

D. For any ϕ ∈ L²(R^N), the function t → ˆ

R^N

u(t)ϕ is continuous, E. ∂tu ∈ L¹(0, T ; V⁰

R^N)

F. For a.e. t ∈ [0, T ], and any ϕ ∈ V_RN, we have h∂_tu, ϕi +

ˆ

R^N

∇u : ∇ϕ = ˆ

R^N

(u ⊗ u) : ∇ϕ, G.ku(t) − ak_L2(R^N)→ 0 as t → 0.

H.For a.e. t ∈ [0, T ], we have 1

2ku(t)k²_L2(R^N)+ ˆ T

0

ˆ

R^N

|∇u(t, x)|²dxdt ≤ 1

2kak²_L2(R^N).

The result is not easy by any standard method. Teman (abstract): either bounded or R^N. Seregin (concrete): bounded domain with smooth boundary.

Moreover, in 2-dimesional space, the condition H will be an equality, and note that we only discuss the partial regularity properties for higher dimensional (N ≥ 3) spaces.

Recall in the Evans’ book, we have ∂tf ∈ L²(0, T ; H⁻¹(Ω)), f ∈ L²(0, T ; H₀¹(Ω)), d

dt|f |² = 1

2h∂_tf (t), f (t)i_H−1(Ω)⊗H₀¹(Ω). If we have ∂_tu ∈ L²(0, T ; V¹

R^N), u ∈ L²(0, T ; V_RN) (u ∈ C⁰([0, T ]; H_RN)), V_RN ⊂ H_RN ⊂ V⁰

R^N and ku(t)k²_L2(R^N)− ku(0)k²_L2(R^N)=

ˆ T 0

h∂tu|_T, u|_Ti_V0

RN⊗V_RNdτ.

Look at (4) and define a projection P : L²(R^N) → H_RN, L²(R^N) = H_RN  {dF ∈ L²(R^N) : F ∈ W_loc^1,2(R^N)}.

| h−∆u, ϕi | = | ˆ

R^N

∇u : ∇ϕ| ≤ k∇ukL²(R^N)kϕkV

RN, ∀ϕ ∈ V_RN, which implies

k(−∆)u|tkV⁰

RN ≤ k∇u|tk_L2(R^N)

and

k(−∆)uk_L2(0,T ;V⁰

RN)≤ kuk_L2(0,T ;V

RN)< ∞.

Moreover, in 2D, we have

| hu · ∇u|t, ϕi = | ˆ

R²

uαuβ∂αϕβ|

≤ ˆ

R²

|u|²|∇ϕ| ≤ ( ˆ

R²

|u|⁴)¹²kϕkV

RN.

Note that we have used the Ladyzhenskaya’s inequality, which states ∀f ∈ H₀¹(R²),

kf k_L4(R²)≤ C0kf k_L¹²2(R²)k∇f k_L¹²2(R²), so we have

ku · ∇ukV⁰

RN ≤ ku(t)k²_L4(R²)≤ C0ku(t)kL²(R²)k∇u(t)kL²(R²),

and ˆ T

0

ku · ∇uk_V⁰

RNdt ≤ C0

ˆ T 0

ku(t)k_L2(R²)k∇u(t)k_L2(R²).

3.2 More weak solutions for the NS equations

In 1980’s, suitable weak solutions which enable us to “localize” the NS equation.

Definition 3.3. (Version 1) A suitable weak solution to (NS) which arises from a ∈ H¹

R³ is an element u ∈ L^∞(0, T ; H_R3)∩L²(0, T ; V_R3) which satisfies condition D to H, and for a.e. t ∈ [0, T ], ϕ ∈ C_c^∞((0, T ] × R³), we have

1 2 ˆ

R³

ϕ(t, x)|u(t, x)|²+ ˆ t

0

ˆ

R³

|∇u(t, x)|²ϕ(t, x)dxdy

≤1 2

ˆ t 0

ˆ

R³

(∂tϕ + ∆ϕ)|u|²dxdt + ˆ t

0

ˆ

R³

∇ϕ · u(1

2|u|²+ P )dxdt.

We do not know if each Leray-Hopf weak solutions will satisfy this definition.

Remark 3.4. It is a fact that for each initial data a ∈ H_R3, there is at least one suitable weak solution to (4) which arises from a.

Key lemma of the Partial Regularity theory.

Lemma 3.5. (Vasseur) For each p > 1, there exists 0 = 0(p) ∈ (0, 1) such that for any suitable weak solution (u, P ) in [−1, 1]×B(1), we have the following implication: If

sup

t∈[−1,1]

ku(t)k²_L2(B(1))+ ˆ 1

−1

ˆ

B(1)

|∇u|²dxdt + ( ˆ 1

−1

[ ˆ

B(1)

|P |dx]^pdt)^2p < 0,

then

|u| ≤ 1 in [−1

2, 1] × B(1 2).

Remark 3.6. For CKN-version: If kuk_L3([−1,1]×B(1))+ kP k

L³2([−1,1]×(B(1))< 0, then

|u| ≤ 1 in [−1

2, 1] × B(1).

From the (4), take divergence on both sides, we have

−∆P = RαRβ{uαuβ},

where R_` stands for the Riesz transformation. When u ∈ L³, i.e., |u|² ∈ L³²([−1, 0] × R³), then P ∈ L³²([−1, 0] × R³).

Now, (u, P ) satisfies the following: ∀ϕ ∈ C_c^∞([−1, 1]×B(1)), a.e. t ∈ (−1, 1], then

1 2

ˆ

B(1)

ϕ(t, ·)|u(t, ·)|²+ ˆ t

−1

ˆ

B(1)

|∇u|²ϕ

≤1 2

ˆ t

−1

(∂_tϕ + ∆ϕ)|u|²+ ˆ t

−1

ˆ

B(1)

u · ∇ϕ{1

2|u|²+ P }. (5)

3.3 Do truncation to |u|

vk = [|u|−(1− 1

2^k)]+, Bk = B(1+ 1

2^3k), B_k−1

3 = B(1+ 2

2^3k), B_k−2

3 = B(1+ 2² 2^3k).

Take a bump function ηk ∈ C_c^∞(B_k−1

3) such that χ_Bk ≤ ηk ≤ χB_{k− 1}

3

,

|∇ηk| ≤ 2 · 2^3kχB

k− 13

,

|∇²ηk| ≤ 4 · 2^6kχB_{k− 1}

3

.

Find the inequality satisfied by vk, and (5) can be abstractly phrased as follows:

∂t

|u|²

2 + |∇u|²− ∆(|u|² 2 ) +1

2∇ · {u|u|²} + u · ∇P ≤ 0.

Note that ∂_tv_k²= vk

|u|u · ∂_tu. You know somehow v_k should satisfy 1

2∂_tv²_k+ D²_k− ∆(v²_k 2 ) +1

2∇ · {u · v_k²} + (vk

|u|− 1)u · ∇P + u · ∇P ≤ 0, but this is not the right way to obtain the inequality. The problem shows up on the big part of vk.

3.4 Look at ( v

|u| − 1)u

1. If vk> 0, then (vk

|u|− 1) = −(1 −₂¹k)

|u| |u|.

2. If v_k(x) = 0, |u| ≤ 1 − 1 2^k(vk

|u| − 1)u ∈ H¹(B(1)) ∩ L^∞(B(1)) and |vk

|u|− 1| · |u| ≤ 1 − 1

2^k < 1.

Note that the idea is vk

|u| and u themselves maybe worse, but combine them together and after canceling out, the term (vk

|u|− 1)u · ηk will be nice. We use for each t ∈ [−1, 1], (vk

|u|− 1)u · ηk to test against (4) 0 =

(∂_tu − ∆u + u · ∇u + ∇P )|_t, (v_k

|u|− 1)u · η_k

H⁻¹(B(1))⊗H¹₀(B(1))

,

the weak formulation of (4) matters. The idea is let 1

2∇ · {u|u|²} handle the large part of |vk|² for us.

u · ∇u, (vk

|u|− 1)uηk

= ˆ

B(1)

uα∂αuβ(vk

|u|− 1)uβ· ηk

= ˆ

B(1)

uα· ∂∂{v_k²− |u|² 2 } · ηk

= ˆ

B(1)

∇ · {u

2 · (v_k²− |u|²)}ηk, since

∂α{v²_k− |u|²

2 } = vk∂αvk− u · ∇u = vk∂α|u| − u · ∇u

= vk

u

|u|∇u − u · ∇u = (vk

|u|− 1)u · ∇u,

|∇(v_k²− |u|²

2 ) ≤ |∇u| and v_k²− |u|²

2 ∈ W^1,2(B(1)).

Now, look at ∆u:

−∆u, (vk

|u|− 1)u · ηk

H¹(B(1))⊗H¹₀(B(1))

= ˆ

B(1)

∇u : ∇{(vk

|u|− 1)u · ηk}

= ˆ

B(1)

∇u : ∇((vk

|u|− 1)u) · η_k+ ˆ

B(1)

∇u : (vk

|u|− 1)u∇η_k:= I₁+ I₂. For I₂,

I₂= ˆ

B(1)

∂_αu_β(vk

|u|− 1)u_β∂_αη_k = ˆ

B(1)

1

2∂_α(v_k²− |u|²)∂_αη_k

= −

∆(v_k²− |u|² 2 ), ηk

 . For I1, note that ∇|u|²= 2u · ∇u = 2|u|∇|u|

I₁= ˆ

B(1)

∇u : ∇((v_k

|u|− 1)u) · ηk= ˆ

B(1)

∇u : ∇(v_k

|u|u) − |∇u|²

= ˆ

B(1)

∇u : ∇(vk

|u|) · u + ∇u : vk

|u|∇u − |∇u|²

= ∇u∇(vk

|u|) · u + (vk

|u|− 1)|∇u|²

= |u|∇|u|²· 1 −₂¹_k

|u|² ∇|u|χ_{vk>0}+ (v_k

|u|− 1)|∇u|²

= [1 − ₂¹_k

|u| ∇|u|²χ_{vk>0}+ (v_k

|u|− 1)|∇u|²].

Note that vk = [|u| − (1 − 1

2^k)]+ and D_k² = vk

|u||∇|u||²χ_{vk>0} + vk

|u||∇u|². u = (1 − v_k

|u|)u + v_k

|u|u and |u|v_k²

2 ≤ (1 − 1 2^k)v_k²

2 +v³_k 2 .

3.5 Short summary

∂_t(v²_k− |u|²

2 ) + ∇ · (u

2(v_k²− |u|²)) + D²_k− |∇u|²+ (vk

|u|− 1)u · ∇P ≡ 0 (6) hold in D⁰, whichever

∂_t(|u|²

2 ) + |∇u|²−1

2|∆(|u|²) + ∇ · (1

2u|u|²) + u · ∇P ≤ 0 (7) which is (5) to test against nonnegative test functions. From (6) and (7), we obtain

∂_t(v_k²

2) + D²_k− ∆(v²_k

2 ) + ∇ · (u

2 − v²_k) + (vk

|u|− 1)u · ∇P + u · ∇P ≤ 0. (8) Define

Uk := sup

Tk≤t≤1

kvk(t)k²_L2(B_k)+ ˆ 1

Tk

ˆ

Bk

D_k²,

by choosing Tk = −1 2(1 + 1

2^k) and derive U_k ≤ C₀^k[1 + kP k_L^p

tL¹_x]U_k−1^β to imply the first DeGiorgi lemma holding.

For ϕ ≥ 0 ∈ C_c^∞((−1, 1) × B(1)) works for (7) but not for (8). For t > 0, 1

2 ˆ

B(1)

v²_k(t, ·)ϕ(t, ·) + ˆ 1

−1

ˆ

B(1)

d²_kϕ

≤ ˆ 1

−1

ˆ

B(1)

(∂tϕ + ∆ϕ)v_k²+ ˆ 1

−1

ˆ

B(1)

u 2v_k²· ∇ϕ +

ˆ 1

−1

ˆ

B(1)

[(vk

|u|− 1)u · ∇P + u∇P ] · ϕdt.

In the final step, let ψ_(t) be a bump function with support in (σ − , s + ) (σ < 0, s > 0) and ψ_(t) = 1 in (σ + , s − ), and connect 1 and 0 by straight lines. Plug the test function ϕ = ψ(t) · ηk(x) into previous calculations, then we have

ˆ σ+

σ−

ˆ

B(1)

ψ(t)ηk(x)D_k²(t, x)dxd

≤ ˆ 1

−1

ψ⁰_(t) · ˆ

B(1)

η_k(x)v_k²(t, x)dxdy + ˆ 1

−1

ˆ

B(1)

ψ_(t)∆η_k(x)v_k²(t, x)dxdt

+ ˆ σ+

σ−

ψ(t) ˆ

Bk− 13

u

2v²_k∇ηk+ ˆ s+

σ−

ˆ

Bk− 13

[(vk

|u|− 1)u · ∇P + u · ∇P ]ψ(t)ηk(x),

and ˆ 1

−1

ψ⁰_(t) · ˆ

B(1)

η_k(x)v_k²(t, x)dxdt

= − 1 2

ˆ s+

s−

ˆ

B(1)

ηk(x)u²_k(t, x)dxdt + 1 2

ˆ σ+

σ−

ˆ

B(1)

ηk(x)v²_kdxdt.

Until now, we have

∂t

v_k²

2 + D²_k− ∆(v_k²

2 ) + ∇ · (uv²_k 2 ) + (vk

|u|− 1)u · ∇P + ∇ · (uP ) ≤ 0 in (−1, 1) × B(1).

Select any −1 < σ < t < 1, ˆ

B(1)

ηk|vk(t, ·)|²+ ˆ t

σ

ˆ

B(1)

D_k²· ηkdxdt

≤ ˆ

B(1)

η_k|vk(σ, ·)|²+ ˆ t

σ

v²_k 2 ∆η_k+

ˆ t σ

ˆ

B(1)

u ·v²_k 2 ∇ηk

+ ˆ t

σ

| ˆ

B(1)

ηk{(vk

|u|− 1)u · ∇P + ∇ · (uP )}dt.

Fix t ∈ [Tk, 1], take average of the above inequality over σ ∈ [Tk−1, Tk], ˆ

B(1)

ηk|vk(t)|²+ ˆ t

Tk

ˆ

B(1)

D_k²ηk≤ Aerage of LHS ≤ Average of LHS

≤ 1

Tk− Tk−1

ˆ T_k Tk−1

ηk|vk(σ, ·)|²dσ + ˆ t

Tk−1

ˆ

B(1)

v²_k

2 |∆ηk| + |u|v²_k 2 |∇ηk| +

ˆ t T_k−1

| ˆ

B(1)

η_k{(v_k

|u|− 1)u · ∇P + ∇ · (uP )|dτ.

Use |∇ηk| ≤ 2 · 2^3k and |∇²uk| ≤ 42^6k, then 1

2U_k≤ sup

T_k≤t≤1

ˆ

B(1)

η_k|vk(t)|²

2 +

ˆ 1 T_k

ˆ

B(1)

D²_kη_k

≤ 2^k ˆ 1

Tk

ˆ

B_{K− 1}

3

|vk|²+ ˆ

Qk−1

(4 · 2^σk)[|vk|²+ |u|v_k² 2 ] +

ˆ t Tk−1

| ˆ

B(1)

ηk{(vk

|u|− 1)u · ∇P + ∇ · (uP )|dτ.

u ∈ L^∞(−1, 1; L²(B(1)) ∩ L²(−1, 1; H¹(B(1)), vk ∈ L^∞(Tk, 1; L²(Bk)). Note that L²(Tk, 1; H¹(B(1)) ⊂ L²(Tk, 1; L⁶(B(1)).

kvkk_L6(Bk)≤ C0{kvkk_L2(Bk)+ k∇vkk_L2(Bk)}, B(1

2) ⊂ Bk⊂ B(1).

For θ ∈ (0, 1), 1 p= θ

2+1 − θ

∞ = θ

6+1 − θ

2 , implying p = 10

3 . Therefore, kvkk

L¹⁰3 (Q_k)≤ kvkk^1−θ_L∞(T_k,1;L²(B_k)· kvkk^θ_L2(T_k,1;H¹(B_k))≤ U_k¹².

4 Deal with the pressure

Motivation: For Leray Hopf weak solution in R³, −∆P = ∂_α∂_β{u_αu_β}, u ∈ L^∞(−1, 1; L²(R³)), |u|(t, ·) ∈ L¹(R³) and |u|(t, ·) ∈ L⁶(R³).

Lemma 4.1. Suppose we have P ∈ L^p(T_k−1, 1; L²(B_k−1)) and some symmetric matrix (G_αβ) ∈ L^∞(T_k−1, 1; L¹(B_k−1)) such that (−∆)P = ∂_α∂_β{G_αβ} (Not the whole space setting). Then

P |B

k− 13

= Pk1+ Pk2,

where P_k1 is the nonlocal part and P_k2 = (−∆)⁻¹∂_α∂_β[G_αβ· ϕ_k] is the local part. Moreover,

kPk1k_Lp(T_k−1,1;L^∞(B_{k− 1}

3

))+ k∇Pk1k_Lp(T_k−1,1;L^∞(B_{k− 1}

3

)). Observe that

−∆(ϕkP ) = −∆ϕk· P − 2∇ϕk· ∇P − ∆P · ϕk

= −P ∆ϕ_k− 2∇ · {P ∇ϕ_k} + 2ϕ_k∆P − ϕ_k∆P

= −P ∆ϕk− 2∇ · {P ∇ϕk} + ϕk∆P and

−ϕk∆P = ϕk∂α∂β{Gαβ}

= ∂α{ϕk∂βGαβ} − ∂αϕk· ∂βGαβ

= ∂_α∂_β[ϕ_kG_αβ] − ∂_α{∂βϕ_k· Gαβ} − ∂αϕ_k∂_βG_αβ

= (−∆)Pk2− ∂αβϕk· Gαβ− 2∂βϕk· ∂αGαβ, or

(−∆)[ϕkP ] = (−∆)Pk2− ∂αβϕk· Gαβ− 2∂βϕk· ∂αGαβ

− ∆ϕ_k· P − 2∇ · {P · ∇ϕ_k}.

Further,

(−∆)Pk1= 1 4π

1

|x|∗ {−∂αβϕk· Gαβ− 2∂βϕk∂αGαβ

− ∆ϕk· P − 2∇ · {P ∇ϕk}.

Since supp∇ϕk, supp∇²ϕk⊂ Bk−1\B_k−2

3, (−∆)Pk1= 0 in B_k−2

3, we estimate kPk1k_L∞(B_{k− 1}

3

) and k∇Pk1k_L∞(B_{k− 1}

3

). For x ∈ B_k−¹

3, ˆ

R³

1

|x − y|∇ · {P ∇ϕk}|ydy = ˆ

y∈B_k−B_{k− 2}

3

1

|x − y|∇y· {P ∇ϕk}dy

= − ˆ

B_k−1\B_{k− 2}

3

1

|y − x|²( y − x

|y − x|)P (y)∇ϕk|ydy

≤ 2 · 2^6k· 2^3k ˆ

B_k−1

|P |.

The conclusion is that

k∇Pk1(t, ·)k_L∞(B_{k− 1}

3

)+ kPk1(t, ·)k_L∞(B_{k− 1}

3

)

≤C₀2^12k ˆ

B(1)

{|{P (t, ·)| + |G|}.

Lemma 4.2. Let p > 1, consider P ∈ L^p(T_k−1, 1; L¹(B_k−1)) and (G_αβ) ∈ L^∞(T_k−1, 1; L¹(B_k−1)) such that (−∆P ) = ∂_αβG_αβ. Then

P |B

k− 13

= Pk1+ Pk2,

where Pk2= (−∆)∂αβ(Gαβϕk) and Pk1 satisfies the following (−∆)Pk1= 0 in [Tk−1, 1] × B_k−2

3

and

kPk1kL^p(T_k−1,1;L^∞(B

k− 13

))+ k∇Pk1kL^p(T_k−1,1;L^∞(B

k− 13

))

≤C02^24k{kP k_Lp(Tk−1,1;L¹(Bk−1))+ kGk_Lp(Tk−1,1;L¹(Bk−1))}.

The details of proof, we refer to Seregin’s big paper and Sverak’s work.