Regularity, blow up, and small scale creation in fluids Sasha - - PowerPoint PPT Presentation

Introduction 2D Euler equation background and statements 3D Euler equation motivation 2d Euler sketch of the proof 1D models

Regularity, blow up, and small scale creation in fluids

Sasha Kiselev (Rice University)

16 September, 2015 Mathflows, Porquerolles

Introduction 2D Euler equation background and statements 3D Euler equation motivation 2d Euler sketch of the proof 1D models

Incompressible Euler equation

Derived by Euler in 1755. Let D be domain with smooth boundary in Rd, d = 2 or 3. ∂tu + (u · ∇)u = ∇p, ∇ · u = 0, u · n|D = 0, where u is fluid velocity and p is pressure. Local/global existence of regular solutions? Stories are very different in two and three dimensions. Let vorticity ω = curlu. In the vorticity form, Euler equation reads as ∂tω + (u · ∇)ω = (ω · ∇)u, u = curl(−∆)−1

D ω, ω(x, 0) = ω0(x).

In dimension two, (ω · ∇)u ≡ 0.

Introduction 2D Euler equation background and statements 3D Euler equation motivation 2d Euler sketch of the proof 1D models

2D global regularity and small scale creation

Theorem (Wolibner; H¨

• lder 1930s)

Suppose D ⊂ R2 is a compact domain with smooth boundary. Assume that the initial data ω0 ∈ C1(D). Then there exists a unique global solution of the 2D Euler equation ω(x, t) ∈ C1(D). Moreover, ∇ω(·, t)L∞ ≤ (1 + ∇ω0L∞)C exp(Cω0L∞t) for some C > 0 and every t ≥ 0. Define trajectories d dt Φt(x) = u(Φt(x), t), Φ0(x) = x. The proof has three ingredients. We have ω(x, t) = ω0(Φ−1

t

(x)), so 1.∇ω(·, t)L∞ ≤ ∇ω0L∞supx,y |Φ−1

t

(x) − Φ−1

t

(y)| |x − y| .

Introduction 2D Euler equation background and statements 3D Euler equation motivation 2d Euler sketch of the proof 1D models

• 2. Purely kinematic bound

exp

• −

t ∇uL∞ ds

• ≤ |Φt(x) − Φt(y)|

|x − y| ≤ exp t ∇uL∞ ds

• .
• 3. Kato’s estimate

∇uL∞ ≤ CωL∞ 1 + log+ ∇ωL∞ . Reason for double exponential: Fourier-analytic, lack of L∞ bound for Riesz transform: ∇u ∼ Rijω. How sharp is the double exponential bound?

Introduction 2D Euler equation background and statements 3D Euler equation motivation 2d Euler sketch of the proof 1D models

• 2. Purely kinematic bound

exp

• −

t ∇uL∞ ds

• ≤ |Φt(x) − Φt(y)|

|x − y| ≤ exp t ∇uL∞ ds

• .
• 3. Kato’s estimate

∇uL∞ ≤ CωL∞ 1 + log+ ∇ωL∞ . Reason for double exponential: Fourier-analytic, lack of L∞ bound for Riesz transform: ∇u ∼ Rijω. How sharp is the double exponential bound? Yudovich, 1960s - examples with some infinite growth of ∇ω at the

• boundary. Later (2000s) - Morgulis-Shnirelman-Yudovich, Koch.

Nadirashvili, 1990s - linear in time growth of ∇ω, D is an annulus. Bahouri-Chemin, 1990s - stationary ”singular cross” solution; leads to u1(x1, 0) ∼ x1 log x1 near the origin.

Introduction 2D Euler equation background and statements 3D Euler equation motivation 2d Euler sketch of the proof 1D models

Denisov, 2010s - superlinear growth of ∇ω on a torus; double exponential growth on finite time intervals; double exponential rate of convergence for forced patches.

Introduction 2D Euler equation background and statements 3D Euler equation motivation 2d Euler sketch of the proof 1D models

Denisov, 2010s - superlinear growth of ∇ω on a torus; double exponential growth on finite time intervals; double exponential rate of convergence for forced patches.

Theorem (K-ˇ Sver´ ak, 2014)

Let D be a unit disk in R2. There exist ω0 ∈ C∞(¯ D), ∇ω0L∞ > 1, such that for the corresponding solution ω(x, t) we have ∇ω(·, t)L∞ ≥ ∇ω0c exp(cω0L∞t) for some c > 0 and every t ≥ 0. The growth in our example happens at the boundary. The geometry is that of a hyperbolic point similar to Bahouri-Chemin cross.

Introduction 2D Euler equation background and statements 3D Euler equation motivation 2d Euler sketch of the proof 1D models

Denisov, 2010s - superlinear growth of ∇ω on a torus; double exponential growth on finite time intervals; double exponential rate of convergence for forced patches.

Theorem (K-ˇ Sver´ ak, 2014)

Let D be a unit disk in R2. There exist ω0 ∈ C∞(¯ D), ∇ω0L∞ > 1, such that for the corresponding solution ω(x, t) we have ∇ω(·, t)L∞ ≥ ∇ω0c exp(cω0L∞t) for some c > 0 and every t ≥ 0. The growth in our example happens at the boundary. The geometry is that of a hyperbolic point similar to Bahouri-Chemin cross. Zlatos - used a similar scenario to build example of a smooth solution

• n a torus with exponential in time growth for second derivatives of ω.

Xu - generalized construction to an arbitrary smooth domain with a symmetry axis. Our construction has been motivated by the numerical simulations of Luo and Hou who proposed a new blow up scenario for 3D Euler equation.

Introduction 2D Euler equation background and statements 3D Euler equation motivation 2d Euler sketch of the proof 1D models

3D axi-symmetric Euler

Denote ˜ Dt = ∂t + ur∂r + uz∂z. Axi-symmetric 3D Euler can be written as ˜ Dt (ruθ) = 0, ˜ Dt

• r −1ωθ
• = r −4∂z
• (ruθ)2

along with Biot-Savart law rur = ∂zψ, ruz = −∂rψ, Lψ = r −1ωθ, L = 1 r ∂r 1 r ∂r

• + 1

r 2 ∂2

z.

A good proxy for 3D axi-symmetric Euler, at least away from the axis, is the 2D inviscid Boussinesq system. Dtθ = 0, Dtω = ∂x1θ, Dt = ∂t + u1∂x1 + u2∂x2, along with the usual 2D Biot-Savart law u = (∂x2(−∆)−1ω, −∂x1(−∆)−1ω).

Introduction 2D Euler equation background and statements 3D Euler equation motivation 2d Euler sketch of the proof 1D models

The Hou-Luo scenario is set on a cylinder, with periodic condition along z axis.

z u periodic in z symmetry plane for the reflection (x, y, z) → (x, y, −z) u · n = 0 at ∂Ω u u “secondary flow” possible finite-time singularity? x z

The initial data for uθ is odd with respect to z, ωθ = 0 initially. Such flow spontaneously creates ”secondary flow” leading to a vortex in each z, r section for fixed θ (this was observed in particular already by Einstein in 1928). Very fast growth of vorticity is observed near the circle of hyperbolic points at the boundary.

Introduction 2D Euler equation background and statements 3D Euler equation motivation 2d Euler sketch of the proof 1D models

In the context of Boussinesq system, the engine of motion is buoyancy of warmer fluid.

y x higher θ lower θ ∂Ω Ω 2D Boussinesq gravity lower θ warmer fluid possible finite-time singularity?

The 2D Euler double exponential growth example is constructed in a similar hyperbolic point scenario. A key role in the construction is played by the certain asymptotic representation of the Biot-Savart law for velocity u1 near the hyperbolic point O.

Introduction 2D Euler equation background and statements 3D Euler equation motivation 2d Euler sketch of the proof 1D models

2D Euler sketch of the proof

Set the origin O at the lowest point of the disk D. Let D+ = {x ∈ D |x1 ≥ 0}, Dγ

1 = {x ∈ D+

π

2 − γ ≥ φ ≥ 0}, and

Dγ

2 = {x ∈ D+

π

2 ≥ φ ≥ γ }.

The initial data ω0 will be odd with respect to x1, ω0(x) ≥ 0 for x ∈ D+. Finally, set Q(x1, x2) = {y ∈ D+ |y1 ≥ x1, y2 ≥ x2 }.

Introduction 2D Euler equation background and statements 3D Euler equation motivation 2d Euler sketch of the proof 1D models

2D Euler sketch of the proof

Set the origin O at the lowest point of the disk D. Let D+ = {x ∈ D |x1 ≥ 0}, Dγ

1 = {x ∈ D+

π

2 − γ ≥ φ ≥ 0}, and

Dγ

2 = {x ∈ D+

π

2 ≥ φ ≥ γ }.

The initial data ω0 will be odd with respect to x1, ω0(x) ≥ 0 for x ∈ D+. Finally, set Q(x1, x2) = {y ∈ D+ |y1 ≥ x1, y2 ≥ x2 }. Main Lemma Fix a small γ > 0. There exists a small δ > 0 such that for every |x| ≤ δ, x ∈ Dγ

1 , then

u1(x1, x2) = −x1 4 π

• Q(x1,x2)

y1y2 |y|4 ω(y1, y2) dy2dy1 + x1ωL∞B1(x1, x2), where |B1| ≤ C(γ). Similarly, for every |x| ≤ δ, x ∈ Dγ

2 ,

u2(x1, x2) = x2 4 π

• Q(x1,x2)

y1y2 |y|4 ω(y1, y2) dy2dy1 + x2ωL∞B2(x1, x2), where |B2| ≤ C(γ).

Introduction 2D Euler equation background and statements 3D Euler equation motivation 2d Euler sketch of the proof 1D models

Corollary Exponential growth is easy. Take ω0 = 1 everywhere on D+ except a thin strip of width δ along the x2 axis, where ω0. Due to incompressibility, the measure of the set in D+ where ω(x, t) < 1 does not exceed Cδ for any t ≥ 0. Then the main term coefficient Ω(x1, x2, t) := 4 π

• Q(x1,x2)

y1y2 |y|4 ω(y1, y2) dy2dy1 ≥ c log δ−1 (1) for all t. Pick δ small enough so that c log δ−1 dominates the error term coefficient B1. Then for a characteristic along the boundary, we have x′

1(t) ≤ −Cx1, leading to exponential in time convergence (and so growth in

∇ω if we choose ω0 = 0 on ∂D.

Introduction 2D Euler equation background and statements 3D Euler equation motivation 2d Euler sketch of the proof 1D models

Corollary Exponential growth is easy. Take ω0 = 1 everywhere on D+ except a thin strip of width δ along the x2 axis, where ω0. Due to incompressibility, the measure of the set in D+ where ω(x, t) < 1 does not exceed Cδ for any t ≥ 0. Then the main term coefficient Ω(x1, x2, t) := 4 π

• Q(x1,x2)

y1y2 |y|4 ω(y1, y2) dy2dy1 ≥ c log δ−1 (1) for all t. Pick δ small enough so that c log δ−1 dominates the error term coefficient B1. Then for a characteristic along the boundary, we have x′

1(t) ≤ −Cx1, leading to exponential in time convergence (and so growth in

∇ω if we choose ω0 = 0 on ∂D. Double exponential growth is more subtle; one needs to get a nonlinear enhancement through growth of Ω. The monotonicity present in the integration domain of integral in (1) plays an important role.

Introduction 2D Euler equation background and statements 3D Euler equation motivation 2d Euler sketch of the proof 1D models

Define the sets Fx′

1,x′′ 1 =

• (x1, x2) ∈ D+ |x′

1 ≤ x1 ≤ x′′ 1 , x2 ≤ x1

• .

Take the initial data as in the Corollary, but in addition fix ǫ << δ and set ω0(x) = 1 on the set Fǫ10,ǫ. We can choose ω0 so that ∇ω0L∞ ǫ−10. Denote u1(x1, t) = min

(x1,x2)∈D+ , x2<x1

u1(x1, x2, t) u1(x1, t) = max

(x1,x2)∈D+ , x2<x1

u1(x1, x2, t) . Define a(t), b(t) by a′(t) = u1(a, t) , a(0) = ǫ10 b′(t) = u1(b, t) , b(0) = ǫ. Denote F(t) := F(a(t), b(t)).

Introduction 2D Euler equation background and statements 3D Euler equation motivation 2d Euler sketch of the proof 1D models

• Claim. ω(x, t) = 1 if x ∈ F(t), for all t.

The reason is that fluid particles from D+ \ F(0) cannot enter F(t) for any time. They can’t enter through vertical boundaries due to the definition of a(t), b(t) or through ∂D due to the boundary condition. Near the diagonal x1 = x2, due to the Main lemma, we have u1(x1, x2) = −x1Ω(x1, x2, t) + O(x1) and u2(x1, x2) = x2Ω(x1, x2, t) + O(x2). So x1(log δ−1 − C) x2(log δ−1 + C) ≤ −u1(x1, x2) u2(x1, x2) ≤ x1(log δ−1 + C) x2(log δ−1 − C). Thus u points outside F(t) and the trajectories cannot enter F(t) through the diagonal either. Now by Main Lemma and a simple estimate, u1(b(t), t) ≥ −b(t)Ω(b(t), b(t)) − Cb(t) Similarly, u1(b(t), t) ≤ −a(t)Ω(a(t), 0) + Ca(t).

Introduction 2D Euler equation background and statements 3D Euler equation motivation 2d Euler sketch of the proof 1D models

But Ω(a(t), 0) ≥ 4 π

• F(t)

y1y2 |y|4 ω(y, t) dy1dy2 + Ω(b(t), b(t)) ≥ 1 2π (− log a(t) + log b(t)) − C + Ω(b(t), b(t)). We get d dt log b(t) ≥ −Ω(b(t), b(t)) − C d dt log a(t) ≤ 1 2π (log a(t) − log b(t)) − Ω(b(t), b(t)) + C. Then d dt (log b(t) − log a(t)) ≥ 1 2π (log b(t) − log a(t)) − 2C. This leads to log b(t) − log a(t) ≥ (9 log ǫ−1 − C) exp(t/2π). Choose ǫ so that ǫ−1 >> C. Then a(t) approaches zero at a double exponential rate, and hence ∇ω(·, t) ǫ−8 exp(t/2π).

Introduction 2D Euler equation background and statements 3D Euler equation motivation 2d Euler sketch of the proof 1D models

1D models

In the 2D Boussinesq case growth in ω makes is difficult to obtain a similar velocity representation. Analysis of simpler models? Two 1D models have been proposed and analyzed.

• 1. Hou-Luo model, periodic setting.

∂tω + u∂xω = ∂xθ, ∂tθ + u∂xθ = 0, ux = Hω, where H is the Hilbert transform. This model can be obtained from 2D Boussinesq by making an assumption that vorticity is constant in a thin layer along the boundary and is zero otherwise.

• 2. Choi-K-Yao (CKY) model. Set on [0, 1],

∂tω + u∂xω = ∂xθ, ∂tθ + u∂xθ = 0, u(x) = −x 1

x

ω(y) y dy. Here initially ω and θ are smooth, compactly supported in [0, 1]. The analog of Biot-Savart law is inspired by the 2D Euler velocity representation formula.

Introduction 2D Euler equation background and statements 3D Euler equation motivation 2d Euler sketch of the proof 1D models

Theorem (Choi-K-Yao, 2013)

Let ω0, θ0 ∈ C∞

0 [0, 1]. Then there exists a unique local smooth

solution of the CKY model. There exist initial data for which the solution blows up in finite time; in particular, we must have that t ω(·, s)L∞ ds → ∞ as t → T for some T < ∞.

Theorem (Choi-Hou-K-Luo-ˇ Sver´ ak-Yao, 2014)

Let ω0, θ0 ∈ C∞ be periodic with period 2. Then there exists a unique local smooth solution of the HL model. There exist initial data such that the solution blows up in finite time; in particular, we must have that t Hω(·, s)L∞ ds → ∞ as t → T for some T < ∞.

Introduction 2D Euler equation background and statements 3D Euler equation motivation 2d Euler sketch of the proof 1D models

1/2 1 M ρ0 ω0 2

ρ0(x) x1 = 1

3

1

1 2

x∞ x2

3 4

· · · . . .

∂tω+u∂xω = ∂xθ, ∂tθ+u∂xθ = 0, u(x) = −x 1

x

ω(y) y dy ≡ −xΩ(x, t). Choose xn so that θ0(xn) = 1

2 + 2−n. Define Φn(t) the trajectory

starting at xn, Φ′

n(t) = u(Φn(t), t). Let ψn(t) = − log Φn(t).

Then ψ′

n(t) = Ω(Φn(t), t). On the other hand,

d dt Ω(Φn(t), t) Φn−1(t)

Φn(t)

∂yθ y dy 2−n Φn−1 = 2−neψn−1(t). This leads to ψ′′

n (t) 2−neψn−1(t). This recursive bound is sufficient to

show ψ∞(t) → ∞ in finite time.

Introduction 2D Euler equation background and statements 3D Euler equation motivation 2d Euler sketch of the proof 1D models

The Biot-Savart law in CKY model is ”almost local” in a sense that (u(x)/x)′ = ω(x)/x. This makes analysis easier. Hou-Liu (2014): detailed, self-similar blow up picture in CKY model. The Biot-Savart law in the HL model is essentially nonlocal, and its analysis is more challenging. The proof of finite time blow up for the HL model is based on control

• f a Lyapunov-like functional

I(t) = L/2 ω(x, t) cot π Lx

• dx.

Here the period of the solution is L, and the solution is odd. We show I′(t) ≥ t I(s)2 ds, I(0) > 1. for a certain class of initial data. This leads to finite time blow up, implying that ω cannot remain regular at the origin. The kinetic energy analog

• |u|2 dx remains finite at the time of blow

up.

Introduction 2D Euler equation background and statements 3D Euler equation motivation 2d Euler sketch of the proof 1D models

Thank you!