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 R d , d = 2 or 3 . ∂ t u + ( 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 = curl u . In the vorticity form, Euler equation reads as ∂ t ω + ( u · ∇ ) ω = ( ω · ∇ ) u , u = curl ( − ∆) − 1 D ω, ω ( x , 0 ) = ω 0 ( 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¨ older 1930s) Suppose D ⊂ R 2 is a compact domain with smooth boundary. Assume that the initial data ω 0 ∈ C 1 ( D ) . Then there exists a unique global solution of the 2D Euler equation ω ( x , t ) ∈ C 1 ( D ) . Moreover, �∇ ω ( · , t ) � L ∞ ≤ ( 1 + �∇ ω 0 � L ∞ ) C exp ( C � ω 0 � L ∞ t ) for some C > 0 and every t ≥ 0 . Define trajectories d dt Φ t ( x ) = u (Φ t ( x ) , t ) , Φ 0 ( x ) = x . The proof has three ingredients. We have ω ( x , t ) = ω 0 (Φ − 1 ( x )) , so t | Φ − 1 ( x ) − Φ − 1 ( y ) | t t 1 . �∇ ω ( · , t ) � L ∞ ≤ �∇ ω 0 � L ∞ sup x , 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 � t �� t � � � ≤ | Φ t ( x ) − Φ t ( y ) | exp − �∇ u � L ∞ ds ≤ exp �∇ u � L ∞ ds . | x − y | 0 0 3. Kato’s estimate �∇ u � L ∞ ≤ C � ω � L ∞ � 1 + log + �∇ ω � L ∞ � . Reason for double exponential: Fourier-analytic, lack of L ∞ bound for Riesz transform: ∇ u ∼ R ij ω. 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 � t �� t � � � ≤ | Φ t ( x ) − Φ t ( y ) | exp − �∇ u � L ∞ ds ≤ exp �∇ u � L ∞ ds . | x − y | 0 0 3. Kato’s estimate �∇ u � L ∞ ≤ C � ω � L ∞ � 1 + log + �∇ ω � L ∞ � . Reason for double exponential: Fourier-analytic, lack of L ∞ bound for Riesz transform: ∇ u ∼ R ij ω. 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 u 1 ( x 1 , 0 ) ∼ x 1 log x 1 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 S ver´ ak, 2014) Let D be a unit disk in R 2 . There exist ω 0 ∈ C ∞ (¯ D ) , �∇ ω 0 � L ∞ > 1 , such that for the corresponding solution ω ( x , t ) we have �∇ ω ( · , t ) � L ∞ ≥ �∇ ω 0 � c exp ( c � ω 0 � L ∞ 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 S ver´ ak, 2014) Let D be a unit disk in R 2 . There exist ω 0 ∈ C ∞ (¯ D ) , �∇ ω 0 � L ∞ > 1 , such that for the corresponding solution ω ( x , t ) we have �∇ ω ( · , t ) � L ∞ ≥ �∇ ω 0 � c exp ( c � ω 0 � L ∞ 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 on 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 ˜ D t = ∂ t + u r ∂ r + u z ∂ z . Axi-symmetric 3D Euler can be written as D t ( ru θ ) = 0 , ˜ ˜ r − 1 ω θ = r − 4 ∂ z ( ru θ ) 2 � � � � D t along with Biot-Savart law ru r = ∂ z ψ, ru z = − ∂ r ψ, L ψ = r − 1 ω θ , � 1 � L = 1 + 1 r 2 ∂ 2 r ∂ r r ∂ r z . A good proxy for 3D axi-symmetric Euler, at least away from the axis, is the 2D inviscid Boussinesq system. D t θ = 0 , D t ω = ∂ x 1 θ, D t = ∂ t + u 1 ∂ x 1 + u 2 ∂ x 2 , along with the usual 2D Biot-Savart law u = ( ∂ x 2 ( − ∆) − 1 ω, − ∂ x 1 ( − ∆) − 1 ω ) .
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 z u periodic in z possible “secondary u · n = 0 at ∂ Ω finite-time flow” u singularity? symmetry plane for the reflection ( x, y, z ) → ( x, y, − z ) x u The initial data for u θ is odd with respect to z , ω θ = 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. The 2D Euler double y exponential growth 2D Boussinesq example is constructed in gravity a similar hyperbolic point scenario. warmer fluid A key role in the Ω construction is played by the certain asymptotic representation of the higher θ lower θ lower θ Biot-Savart law for x ∂ Ω velocity u 1 near the hyperbolic point O . possible finite-time singularity?
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 | x 1 ≥ 0 } , D γ 1 = { x ∈ D + � � π 2 − γ ≥ φ ≥ 0 } , and D γ 2 = { x ∈ D + � � π 2 ≥ φ ≥ γ } . The initial data ω 0 will be odd with respect to x 1 , ω 0 ( x ) ≥ 0 for x ∈ D + . Finally, set Q ( x 1 , x 2 ) = { y ∈ D + | y 1 ≥ x 1 , y 2 ≥ x 2 } .
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 | x 1 ≥ 0 } , D γ 1 = { x ∈ D + � � π 2 − γ ≥ φ ≥ 0 } , and D γ 2 = { x ∈ D + � � π 2 ≥ φ ≥ γ } . The initial data ω 0 will be odd with respect to x 1 , ω 0 ( x ) ≥ 0 for x ∈ D + . Finally, set Q ( x 1 , x 2 ) = { y ∈ D + | y 1 ≥ x 1 , y 2 ≥ x 2 } . Main Lemma Fix a small γ > 0 . There exists a small δ > 0 such that for every | x | ≤ δ, x ∈ D γ 1 , then 4 � y 1 y 2 u 1 ( x 1 , x 2 ) = − x 1 | y | 4 ω ( y 1 , y 2 ) dy 2 dy 1 + x 1 � ω � L ∞ B 1 ( x 1 , x 2 ) , π Q ( x 1 , x 2 ) where | B 1 | ≤ C ( γ ) . Similarly, for every | x | ≤ δ, x ∈ D γ 2 , 4 � y 1 y 2 u 2 ( x 1 , x 2 ) = x 2 | y | 4 ω ( y 1 , y 2 ) dy 2 dy 1 + x 2 � ω � L ∞ B 2 ( x 1 , x 2 ) , π Q ( x 1 , x 2 ) where | B 2 | ≤ C ( γ ) .
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