Diego Córdoba. Interface dynamics for incompressible flows in 2D

Transcription

1 Interface dynamics for incompressible flows in 2D Diego Córdoba

2 Equation ρ t + u ρ = 0, u = 0, { ρ ρ(x 1, x 2, t) = 1, x Ω 1 (t) ρ 2, x Ω 2 (t), with ρ(x, t) an active scalar, (x, t) R 2 R +, ρ 1 ρ 2 are constants, and Ω 2 (t) = R 2 Ω 1 (t). 1 SQG sharp front: ρ temperature, u = ( R 2 ρ, R 1 ρ), Rj = iξ j / ξ. 2 Muskat: ρ density, µ viscosity µ u = p (0, gρ), κ Darcy s law. 3 Water wave: ρ density, ρ(u t + (u )u) = p (0, gρ), Euler.

3 The SQG equation Equation θ t + u θ = 0, u = ψ, θ = ( ) 1/2 ψ, with θ(x, t) the temperature, (x, t) R 2 R +. Constantin, Majda, and Tabak (1994)

4 Sharp fronts We consider weak solutions given by { θ1, Ω(t) θ(x 1, x 2, t) = θ 2, R 2 Ω(t).

5 The 2-D vortex patch problem Contour equation w t + u w = 0, u = ψ, w = ψ, where the vorticity is given by { w0, Ω(t) w(x 1, x 2, t) = 0, R 2 Ω(t). Chemin (1993) Bertozzi and Constantin (1993)

6 The 2-D patch problem for SQG Rodrigo (2005), for a periodic C front, i.e. { θ1, {f (x θ(x 1, x 2, t) = 1, t) > x 2 } θ 2, {f (x 1, t) x 2 }.

7 The α-patch model Córdoba, Fontelos, Mancho and Rodrigo (2005) Contour equation θ t + u θ = 0, u = ψ, θ = ( ) 1 α/2 ψ, 0 < α 1, where the active scalar θ(x, t) satisfies { θ1, Ω(t) θ(x 1, x 2, t) = θ 2, R 2 Ω(t).

8 The contour equation Ω(t) = {x(γ, t) = (x 1 (γ, t), x 2 (γ, t)) : γ [ π, π] = T}, where x(γ, t) is one to one. We have and it gives θ = (θ 1 θ 2 ) γ x(γ, t) δ(x x(γ, t)), u(x, t) = Θ α 2π u = ( ) α/2 1 θ, T γ x(γ η, t) x x(γ η, t) α dη.

9 The normal velocity of the systems reads u(x(γ, t), t) γ x(γ, t) = Θ α 2π T γ x(γ η, t) γ x(γ, t) x(γ, t) x(γ η, t) α dη. The tangential velocity does not change the shape of the boundary, so that we fix the contour α-patch equations as follows: Contour equation x t (γ, t) = Θ α 2π x(γ, 0) = x 0 (γ). T γ x(γ, t) γ x(γ η, t) dη, 0 < α 1, x(γ, t) x(γ η, t) α

10 Numerical simulations Close caption of the corner region at t= for α = 0.5 (a) and t= for α = 1 (b). Observe that the singularity is point-like in both cases.

11 Numerical simulations Evidence of finite time singularity. (a) Evolution of the inverse of the maximum curvature with time. (b) Evolution of the minimum distance between patches with time. (Insets) Here, we represent the latest stage of the evolution together with its linear interpolation.

12 Numerical simulations Rescaled profiles at 20 different times in the interval (3.46,4.46) for the case α = 1.

13 Local well-posedness in H s for 0 < α 1 We define F(x)(γ, η, t) = η x(γ, t) x(γ η, t) γ, η [ π, π], with F(x)(γ, 0, t) = γ x(γ, t) 1. Gancedo (2008) Theorem Let x 0 (γ) H k (T) for k 3 with F(x 0 )(γ, η) <. Then there exists a time T > 0 so that there is a unique solution to the α-patch model for 0 < α 1 in C 1 ([0, T ]; H k (T)), with x(γ, 0) = x 0 (γ).

14 Existence for α = 1; the SQG sharp front We modify the equation as follows: Contour equation x t (γ, t) = T γ x(γ, t) γ x(γ η, t) dη + λ(γ, t) γ x(γ, t), (1) x(γ, t) x(γ η, t) with λ(γ, t) = γ+π γ x(γ, t) ( 2π T γ x(γ, t) 2 γ γ γ x(η, t) ( γ x(η, t) 2 η π T T γ x(γ, t) γ x(γ η, t) ) dη dγ x(γ, t) x(γ η, t) ) γ x(η, t) γ x(η ξ, t) dξ x(η, t) x(η ξ, t) dη. (2)

15 We get γ x(γ, t) 2 = A(t). Hou, Lowengrub and Shelley (1997) Extra cancellations: γ x(γ, t) 2 γx(γ, t) = 0, and γ x(γ, t) 3 γx(γ, t) = 2 γx(γ, t) 2.

16 Darcy s law Law µ v = p (0, 0, g ρ), κ v velocity, p pressure, µ viscosity, κ permeability, ρ density, and g acceleration due to gravity. Muskat (1937) Saffman and Taylor (1958) Equation (Hele Shaw) 12µ v = p (0, g ρ), b2 b distance between the plates.

17 Smooth initial data with µ = const. Equation ρ t + v ρ = 0 v = p (0, ρ) div v = 0 Two-dimensional mass balance equation in porous media (2DPM) v(x) = 1 2π PV ( 2 y 1y 2 R 2 y 4, y 1 2 y 2 2 y 4 ) ρ(x y)dy 1 (0, ρ(x)), 2 ( t + v ) ρ = ( v) ρ.

18 Local existence. Singularities with infinite energy, with a stream function given by ψ(x 1, x 2, t) = x 2 f (x 1, t) + g(x 1, t). Blow-up T 0 ρ BMO (t) dt =. Geometric constraints: η = ρ ρ. Numerical simulations: ρ L (t) e t.

19 Cordoba-Gancedo (2007) We consider the case where the fluid has different densities, that is ρ is represented by { ρ1, {x ρ(x 1, x 2, t) = 2 > f (x 1, t)} ρ 2, {x 2 < f (x 1, t)} being f the interface. Then we have Equation df dt (x, t) = ρ 2 ρ 1 PV 2π R f (x, 0) = f 0 (x); x R. ( x f (x, t) x f (x α, t))α α 2 + (f (x, t) f (x α, t)) 2 dα ρ t + v ρ = 0 v = p (0, ρ) div v = 0 2DPM contour equation

20 The linearized equation If we neglect the terms of order greater than one f t = ρ 1 ρ 2 (H x f ) = ρ 1 ρ 2 Λf, 2 2 f (x, 0) = f 0 (x). Applying the Fourier transform we get ˆf (ξ) = ˆf0 (ξ)e ρ 1 ρ 2 2 ξ t. Problem ρ 1 < ρ 2 stable case, ρ 1 > ρ 2 unstable case.

21 Local well-posedness for the stable case (ρ 2 > ρ 1 ) The following theorem follows Theorem Let f 0 (x) H k for k 3 and ρ 2 > ρ 1. Then there exists a time T > 0 so that there is a unique solution to 2DPM contour equation in C 1 ([0, T ]; H k ) with f (x, 0) = f 0 (x).

22 Global existence? L decays. f (x, t) dx = f 0 (x) dx. Global existence for small initial data: ξ ˆf (ξ) << 1.

23 Ill-posedness for the unstable case (ρ 1 > ρ 2 ) We define g b = ĝ(ξ) e b ξ b 0. If g b <, g is analytic in Iz < b. We get global solutions for the 2-D stable case f (x, t), x R, with x f a (t) C(ε) exp((2σa ρ 2 ρ 1 t)/4), Λ 1+γ f 0 0 < C, and for γ, ζ > 0. Λ 1+γ+ζ f 0 0 =,

24 We take f λ (x 1, t) = λ 1 f (λx 1, λt + λ 1/2 ), {f λ } λ>0 a family of solutions to the unstable case. Then f λ H s(0) = λ s 3 2 f H s(λ 1 2 ) C λ s 3 2 f 1 (λ 1 2 ) C λ s 3 2 e λ 1/2, and f λ H s(λ 1/2 ) = λ s 3 2 f H s(0) λ s 3 2 C Λ 1+γ+ζ f 0 0 =, for s > 3/2 and γ, ζ small enough. Theorem Let s > 3/2, then for any ε > 0 there exists a solution f of 2DPM contour equation with ρ 1 > ρ 2 and 0 < δ < ε such that f H s(0) ε and f H s(δ) =.

25 The Muskat problem Equation ρ t + u ρ = 0, { u = 0, (µ (µ, ρ)(x, t) = 1, ρ 1 ), x Ω 1 (t) µ κ u = p (0, gρ), (µ 2, ρ 2 ), x Ω 2 (t), Siegel, Caflish & Howison (2004) Ambrose (2004)

26 We shall consider z(α + 2πk, t) = z(α, t) + (2πk, 0). Darcy s law yields Biot-Savart and taking limits ω(x, t) = ϖ(α, t)δ(x z(α, t)), u(x, t) = 1 2π PV u j (z(α, t), t) = BR(z, ϖ)(α, t) where BR(z, ϖ)(α, t) = 1 2π PV (x z(β, t) ϖ(β, t)dβ, x z(β, t) 2 ϖ(α, t) 2 α z(α, t) 2 αz(α, t) j = 1, 2, (z(α, t) z(β, t) ϖ(β, t)dβ. z(α, t) z(β, t) 2

27 Closing the equation: In search of a relationship between z and ϖ Beginning with Darcy s law: µ j κ uj = p j (0, g ρ j ) First we substitute u j by its formula involving Birkhoff-Rott. Next, for each j, we write its scalar product with α z(α), and finally we subtract both expressions to get µ 2 + µ 1 ϖ(α, t) + µ2 µ 1 BR(z, ϖ)(α, t) α z(α, t) 2κ κ = ( p 2 (z(α, t), t) p 1 (z 1 (α, t), t)) α z(α, t) g(ρ 2 ρ 1 ) α z 2 (α, t) = α (p 2 (z(α, t), t) p 1 (z(α, t), t)) g(ρ 2 ρ 1 ) α z 2 (α, t) Since α (p 2 (z(α, t), t) p 1 (z(α, t), t)) = 0

28 Darcy s law implies therefore p(x, t) = div ( where Π(α, t) is given by µ(x, t) u(x, t)) g x2 ρ(x, t), κ p(x, t) = Π(α, t)δ(x z(α, t)), Π(α, t) = ( µ2 µ 1 u(z(α, t), t) α z(α, t) + g(ρ 2 ρ 1 ) α z 1 (α, t)). κ It follows that: p(x, t) = 1 2π T ln ( cosh(x 2 z 2 (α, t)) cos(x 1 z 1 (α, t)) ) Π(α, t)dα, for x z(α, t), implying the important identity p 2 (z(α, t), t) = p 1 (z(α, t), t), (3) which is just a mathematical consequence of Darcy s law, making unnecessary to impose it as a physical assumption.

29 Then ϖ(α, t) + A µ T (ϖ)(α, t) = 2gκC(ρ, µ) α z 2 (α, t) where T (ϖ) = 2BR(z, ϖ) α z, A µ = µ2 µ 1 µ 2 + µ 1, C(ρ, µ) = ρ2 ρ 1 µ 2 + µ 1. Baker, Meiron & Orszag (1982) λ < 1 for (I + A µ T ). Contour equation z t (α, t) = BR(z, ϖ)(α, t) + c(α, t) α z(α, t), ϖ(α, t) = (I + A µ T ) 1 ( 2gκC(ρ, µ) α z 2 )(α, t).

30 Cordoba, Cordoba & Gancedo (2008). Theorem Let z 0 (γ) H k (T) for k 3, F(z 0 )(α, β) < and σ 0 (α) = ( p 2 (z 0 (α), 0) p 1 (z 0 (α), 0)) α z 0 (α) > 0. Then there exists a time T > 0 so that there is a solution to the contour equation in C 1 ([0, T ]; H k (T)) with z(α, 0) = z 0 (α). where σ(α) = ( p 2 (z(α)) p 1 (z(α))) α z(α) = µ2 µ 1 κ BR(z, ϖ)(α) α z(α) + g(ρ 2 ρ 1 ) α z 1 (α).

31 Energy estimates: z 2 = z 2 H + F(z) 2 3 L, (I + A µt ) e C z 2, H 2 H 2 We can estimate as follows: ϖ H 2 e C z 2.

32 By energy estimates we obtain d dt z 2 (t) K σ(α, t) αz(α, 3 t) Λ( αz)(α, 3 t)dα + e C( z 2). T Pointwise inequality: f Λ(f ) 1 2 Λ(f 2 ). αz Λ( Tσ 3 αz)dα αz TσΛ( 3 2 )dα = 1 2 as long as σ(α, t) > 0. T Λ(σ) 3 αz 2 dα

33 Sketch of the proof Estimates on the arc-chord condition Lemma d dt F(z) 2 L (t) exp C( F(z) 2 L (t) + z 2 H 3 (t)) 17

34 Sketch of the proof Estimate for the evolution of the minimum of the difference of the gradients of the pressure in the normal direction. This quantity is given by σ(α) = µ2 µ 1 BR(z, ϖ)(α) α z(α) + g(ρ 2 ρ 1 ) α z 1 (α). k Lemma Let z(α) be a solution of the system with z(α, t) C 1 ([0, T ]; H 3 ), and m(t) = mín σ(α, t). α T Then m(t) m(0) t 0 exp C( F(z) 2 L (s) + z 2 H 3 (s))ds. 18

35 Sketch of the proof Regularizing the system Let z ε (α, t) be a solution of the following system: z ε,δ t (α, t) = BR δ (z ε,δ, ϖ ε,δ )(α, t) + c ε,δ (α, t) α z ε,δ (α, t), z ε,δ (α, 0) = z 0 (α), where BR δ (z, ϖ)(α, t) = ( 1 4π 1 4π T T ϖ(β, t) tanh(z 2(α,t) z 2 (β,t) 2 )(1 + tan 2 ( z 1(α,t) z 1 (β,t) 2 )) tan 2 ( z 1(α,t) z 1 (β,t) 2 ) + tanh 2 ( z 2(α,t) z 2 (β,t) 2 ) + δ dβ, ϖ(β, t) tan(z 1(α,t) z 1 (β,t) 2 )(1 tanh 2 ( z 2(α,t) z 2 (β,t) 2 )) tan 2 ( z 1(α,t) z 1 (β,t) 2 ) + tanh 2 ( z 2(α,t) z 2 (β,t) 2 ) + δ dβ ), 19

36 ϖ ε,δ (α, t) = µ2 µ 1 µ 2 + µ 1 φ ε φ ε (2BR(z ε,δ, ϖ ε,δ ) α z ε,δ )(α) 2kg ρ2 ρ 1 µ 2 + µ 1φ ε φ ε ( α z ε,δ 2 )(α), c ε,δ (α, t) = α + π α z ε,δ (α, t) 2π T α z ε,δ (α, t) αbr δ (z ε,δ, ϖ ε,δ )(α, t)dα 2 α α z ε,δ (β, t) α z ε,δ (α, t) βbr δ (z ε,δ, ϖ ε,δ )(β, t)dβ, 2 π φ C c (R), φ(α) 0, φ( α) = φ(α), for ε > 0 and δ > 0. R φ(α)dα = 1, φ ε (α) = φ(α/ε)/ε, 20

37 Sketch of the proof The next step is to integrate the system during a time T with T independent of ε. If z 0 (α) H n, we have the solution z ε C 1 ([0, T ε ]; H n ) and that m ε (t) m(0) t 0 exp C( F(z ε ) 2 L (s) + zε 2 H 3 (s))ds, for m ε (t) = mín α T σε (α, t), and t T ε. If initially σ(α, 0) > 0, there is a time depending on ε, denoted by T ε again, in which σ ε (α, t) > 0. 21

38 Sketch of the proof For t T ε d dt zε 2 H n(t) I + exp C( F(zε ) 2 L (t) + zε 2 H n(t)), for k σ ε (α, t) I = 2π(µ 1 +µ 2 ) T α z ε (α, t) 2φ ε ( αz n ε )(α, t) Λ(φ ε ( αz n ε ))(α, t)dα k σ ε (α, t) 1 2π(µ 1 +µ 2 )A ε (t) T 2 Λ( φ ε ( αz n ε ) 2 )(α, t)dα. k = Λ(σ ε )(α, t) 1 2π(µ 1 +µ 2 )A ε (t) 2 φ ε ( αz n ε ) 2 (α, t)dα, we obtain T I C F(z ε ) L Λ(σ ε ) L n αz ε 2 L 2 exp C( F(z ε ) 2 L + zε 2 H 3 ). 22

39 Finally, for t T ε we have and Sketch of the proof d dt zε 2 H n(t) C exp C( F(zε ) 2 L (t) + zε 2 H n(t)). d dt F(zε ) 2 L (t) C exp C( F(zε ) 2 L (t) + zε 2 H 3 (t)), Therefore d dt ( zε 2 H n(t) + F(zε ) 2 L (t)) C exp C( zε 2 H n(t) + F(zε ) 2 L (t)) for t T ε. Integrating together with q.e.d. m ε (t) m(0) t 0 exp C( F(z ε ) 2 L (s) + zε 2 H 3 (s))ds, 23

40 The fluid interface problem Equation ρ t + u ρ = 0, u = 0, ρ(u t + (u )u) = p (0, gρ), ρ(x, t) = { ρ 1, x Ω 1 (t) ρ 2, x Ω 2 (t),

41 Previous works Rayleigh (1879) Taylor (1950) Craig (1985) Ebin (1987) Beale, Hou & Lowengrub (1993) Wu (1997) Christodoulou & Lindblad (2000) Lindblad (2005) Ambrose & Masmoudi (2005) Coutand & Shkoller (2007) Shatah & Zeng (2008) Zhang & Zhang (2008)

42 Contour equation We consider Ω j (t) = {z(α, t) = (z 1 (α, t), z 2 (α, t)) : α R}: z(α + 2πk, t) = z(α, t) + (2πk, 0), z(α + 2πk, t) = z(α, t). We study the irrotational case Biot-Savart yields for x z(α, t). ω(x, t) = ϖ(α, t)δ(x z(α, t)). u(x, t) = 1 (x z(β, t)) 2π PV ϖ(β, t)dβ, x z(β, t) 2

43 Contour equation We consider Ω j (t) = {z(α, t) = (z 1 (α, t), z 2 (α, t)) : α R}: z(α + 2πk, t) = z(α, t) + (2πk, 0), z(α + 2πk, t) = z(α, t). We study the irrotational case Biot-Savart yields for x z(α, t). ω(x, t) = ϖ(α, t)δ(x z(α, t)). u(x, t) = 1 (x z(β, t)) 2π PV ϖ(β, t)dβ, x z(β, t) 2

44 Contour equation We consider Ω j (t) = {z(α, t) = (z 1 (α, t), z 2 (α, t)) : α R}: z(α + 2πk, t) = z(α, t) + (2πk, 0), z(α + 2πk, t) = z(α, t). We study the irrotational case Biot-Savart yields for x z(α, t). ω(x, t) = ϖ(α, t)δ(x z(α, t)). u(x, t) = 1 (x z(β, t)) 2π PV ϖ(β, t)dβ, x z(β, t) 2

45 Then ( Bernoulli s law) Contour equation ϖ 2 ϖ t = 2A ρ t BR(z, ϖ) α z A ρ α ( 4 α z 2 ) + α(c ϖ) + 2A ρ c α BR(z, ϖ) α z(α, t) 2A ρ g α z 2, (4) for A ρ = (ρ 2 ρ 1 )/(ρ 2 + ρ 1 ) and c parametrization freedom. Contour equation z t = BR(z, ϖ) + c α z. (5) Equations (4) and (5) give weak solutions of Euler. We can choose c to get α z(α, t) 2 = A(t).

46 Then ( Bernoulli s law) Contour equation ϖ 2 ϖ t = 2A ρ t BR(z, ϖ) α z A ρ α ( 4 α z 2 ) + α(c ϖ) + 2A ρ c α BR(z, ϖ) α z(α, t) 2A ρ g α z 2, (4) for A ρ = (ρ 2 ρ 1 )/(ρ 2 + ρ 1 ) and c parametrization freedom. Contour equation z t = BR(z, ϖ) + c α z. (5) Equations (4) and (5) give weak solutions of Euler. We can choose c to get α z(α, t) 2 = A(t).

47 Then ( Bernoulli s law) Contour equation ϖ 2 ϖ t = 2A ρ t BR(z, ϖ) α z A ρ α ( 4 α z 2 ) + α(c ϖ) + 2A ρ c α BR(z, ϖ) α z(α, t) 2A ρ g α z 2, (4) for A ρ = (ρ 2 ρ 1 )/(ρ 2 + ρ 1 ) and c parametrization freedom. Contour equation z t = BR(z, ϖ) + c α z. (5) Equations (4) and (5) give weak solutions of Euler. We can choose c to get α z(α, t) 2 = A(t).

48 Then ( Bernoulli s law) Contour equation ϖ 2 ϖ t = 2A ρ t BR(z, ϖ) α z A ρ α ( 4 α z 2 ) + α(c ϖ) + 2A ρ c α BR(z, ϖ) α z(α, t) 2A ρ g α z 2, (4) for A ρ = (ρ 2 ρ 1 )/(ρ 2 + ρ 1 ) and c parametrization freedom. Contour equation z t = BR(z, ϖ) + c α z. (5) Equations (4) and (5) give weak solutions of Euler. We can choose c to get α z(α, t) 2 = A(t).

49 Conditions Arc-chord: F(z)(α, β, t) = β z(α, t) z(α β, t) α, β ( π, π), and F(z)(α, 0, t) = 1 α z(α, t). Rayleigh-Taylor: σ(α, t) = ( p 2 (z(α, t), t) p 1 (z(α, t), t)) α z(α, t) > 0.

50 Conditions Arc-chord: F(z)(α, β, t) = β z(α, t) z(α β, t) α, β ( π, π), and F(z)(α, 0, t) = 1 α z(α, t). Rayleigh-Taylor: σ(α, t) = ( p 2 (z(α, t), t) p 1 (z(α, t), t)) α z(α, t) > 0.

51 Local-existence Cordoba, Cordoba & Gancedo (2008) Theorem Let z 0 (α) H k and ϖ 0 (α) H k 1 for k 4, F(z 0 )(α, β) <, and σ(α, 0) > 0. Then there exists a time T > 0 so that we have a solution to (4) and (5) in the case ρ 1 = 0, where z(α, t) C 1 ([0, T]; H k ) and ϖ(α, t) C 1 ([0, T]; H k 1 ) with z(α, 0) = z 0 (α) and ϖ(α, 0) = ϖ 0 (α).

52 The linearized equation f tt (α, t) = σλ(f )(α, t), It can be written as follows: σ < 0 e σξ 1 2 t, e σξ 1 2 t σ > 0 cos( σξ 1 2 t), sin( σξ 1 2 t) f t (α, t) = H(g)(α, t), g t (α, t) = σ α f (α, t), and therefore E L (t) = (σ α f 2 + Λ 1 2 g 2 )dα.

53 The linearized equation f tt (α, t) = σλ(f )(α, t), It can be written as follows: σ < 0 e σξ 1 2 t, e σξ 1 2 t σ > 0 cos( σξ 1 2 t), sin( σξ 1 2 t) f t (α, t) = H(g)(α, t), g t (α, t) = σ α f (α, t), and therefore E L (t) = (σ α f 2 + Λ 1 2 g 2 )dα.

54 The linearized equation f tt (α, t) = σλ(f )(α, t), It can be written as follows: σ < 0 e σξ 1 2 t, e σξ 1 2 t σ > 0 cos( σξ 1 2 t), sin( σξ 1 2 t) f t (α, t) = H(g)(α, t), g t (α, t) = σ α f (α, t), and therefore E L (t) = (σ α f 2 + Λ 1 2 g 2 )dα.

55 Energy of the system Case z H 4 : E(t) = z 2 H 3 (t) + π π σ(α, t) ρ 2 α z(α, t) 2 4 αz(α, t) 2 dα + F(z) 2 L (t) + ϖ 2 H (t) + ϕ 2 (t), 2 H for σ(α, t) > 0 and ϕ(α, t) given by ϕ(α, t) = ϖ(α, t) 2 α z(α, t) c(α, t) αz(α, t). Beale, Hou & Lowengrub (1993) Ambrose & Masmoudi (2005)

56 Equation in terms of ϕ(α, t) Case A ρ = 0 (vortex sheet problem): ϖ t (α, t) = α (c ϖ)(α, t). Kelvin-Helmholtz instability arises. The linear formulation is given by f tt (α, t) = 2 αf (α, t) e ξ t, e ξ t Case A ρ = 1 (water wave): ϕ t = α(ϕ 2 ) 2 α z B(t) ϕ α z tbr(z, ϖ) α z g αz 2 α z t(c α z ).

57 Equation in terms of ϕ(α, t) Case A ρ = 0 (vortex sheet problem): ϖ t (α, t) = α (c ϖ)(α, t). Kelvin-Helmholtz instability arises. The linear formulation is given by f tt (α, t) = 2 αf (α, t) e ξ t, e ξ t Case A ρ = 1 (water wave): ϕ t = α(ϕ 2 ) 2 α z B(t) ϕ α z tbr(z, ϖ) α z g αz 2 α z t(c α z ).

58 Equation in terms of ϕ(α, t) Case A ρ = 0 (vortex sheet problem): ϖ t (α, t) = α (c ϖ)(α, t). Kelvin-Helmholtz instability arises. The linear formulation is given by f tt (α, t) = 2 αf (α, t) e ξ t, e ξ t Case A ρ = 1 (water wave): ϕ t = α(ϕ 2 ) 2 α z B(t) ϕ α z tbr(z, ϖ) α z g αz 2 α z t(c α z ).

59 Estimates on the inverse operator (I + T ) 1 Let T be defined by T (u)(α, t) = 2BR(z, u)(α) α z(α), then T : L 2 H 1 and T L 2 H 1 F(z) 4 L z 4. C 2,δ ϖ t (α, t) = (I + T ) 1 (R(z, ϖ))(α, t). Baker, Meiron & Orszag (1982). Córdoba, Córdoba & Gancedo (2008): (I + T ) 1 L 2 L 2 ec( z 2 H 3 + F (z) 2 L ), (conformal mappings, Hopf lemma and Dahlberg-Harnack).

60 Estimates on the inverse operator (I + T ) 1 Let T be defined by T (u)(α, t) = 2BR(z, u)(α) α z(α), then T : L 2 H 1 and T L 2 H 1 F(z) 4 L z 4. C 2,δ ϖ t (α, t) = (I + T ) 1 (R(z, ϖ))(α, t). Baker, Meiron & Orszag (1982). Córdoba, Córdoba & Gancedo (2008): (I + T ) 1 L 2 L 2 ec( z 2 H 3 + F (z) 2 L ), (conformal mappings, Hopf lemma and Dahlberg-Harnack).

61 Estimates on the inverse operator (I + T ) 1 Let T be defined by T (u)(α, t) = 2BR(z, u)(α) α z(α), then T : L 2 H 1 and T L 2 H 1 F(z) 4 L z 4. C 2,δ ϖ t (α, t) = (I + T ) 1 (R(z, ϖ))(α, t). Baker, Meiron & Orszag (1982). Córdoba, Córdoba & Gancedo (2008): (I + T ) 1 L 2 L 2 ec( z 2 H 3 + F (z) 2 L ), (conformal mappings, Hopf lemma and Dahlberg-Harnack).

62 Estimates on the inverse operator (I + T ) 1 Let T be defined by T (u)(α, t) = 2BR(z, u)(α) α z(α), then T : L 2 H 1 and T L 2 H 1 F(z) 4 L z 4. C 2,δ ϖ t (α, t) = (I + T ) 1 (R(z, ϖ))(α, t). Baker, Meiron & Orszag (1982). Córdoba, Córdoba & Gancedo (2008): (I + T ) 1 L 2 L 2 ec( z 2 H 3 + F (z) 2 L ), (conformal mappings, Hopf lemma and Dahlberg-Harnack).

63 Estimates on the inverse operator (I + T ) 1 Let T be defined by T (u)(α, t) = 2BR(z, u)(α) α z(α), then T : L 2 H 1 and T L 2 H 1 F(z) 4 L z 4. C 2,δ ϖ t (α, t) = (I + T ) 1 (R(z, ϖ))(α, t). Baker, Meiron & Orszag (1982). Córdoba, Córdoba & Gancedo (2008): (I + T ) 1 L 2 L 2 ec( z 2 H 3 + F (z) 2 L ), (conformal mappings, Hopf lemma and Dahlberg-Harnack).

64 A priori energy estimates π E(t) = z 2 σ(α, t) H (t) + 3 π ρ 2 α z(α, t) 2 4 αz(α, t) 2 dα + F(z) 2 L (t) + ϖ 2 H (t) + ϕ 2 (t). 2 H Lemma Let z(α, t) and ϖ(α, t) be a solution of (4) and (5). Then, the following a priori estimate holds: d dt E(t) C m q (t) exp(ce(t)), for m(t) = min σ(α, t) = σ(α t, t) > 0 and C and q some α [ π,π] universal constants.

65 The Rayleigh-Taylor condition The true energy: E RT (t) = E(t) + 1 m ε (t). One gets σ ε (α, t) C 1 ([0, T ε ] [ π, π]), and therefore m ε (t) is Lipschitz. d dt (mε )(t) = σ ε t (α t, t), d dt ( 1 m ε )(t) = σε t (α t, t) (m ε (t)) 2, for almost every t.it yields d dt E RT (t) C exp(ce RT (t)), almost everywhere and therefore E RT (t) 1 C ln(exp( CE RT (0)) C 2 t).

66 The Rayleigh-Taylor condition The true energy: E RT (t) = E(t) + 1 m ε (t). One gets σ ε (α, t) C 1 ([0, T ε ] [ π, π]), and therefore m ε (t) is Lipschitz. d dt (mε )(t) = σ ε t (α t, t), d dt ( 1 m ε )(t) = σε t (α t, t) (m ε (t)) 2, for almost every t.it yields d dt E RT (t) C exp(ce RT (t)), almost everywhere and therefore E RT (t) 1 C ln(exp( CE RT (0)) C 2 t).

67 The Rayleigh-Taylor condition The true energy: E RT (t) = E(t) + 1 m ε (t). One gets σ ε (α, t) C 1 ([0, T ε ] [ π, π]), and therefore m ε (t) is Lipschitz. d dt (mε )(t) = σ ε t (α t, t), d dt ( 1 m ε )(t) = σε t (α t, t) (m ε (t)) 2, for almost every t.it yields d dt E RT (t) C exp(ce RT (t)), almost everywhere and therefore E RT (t) 1 C ln(exp( CE RT (0)) C 2 t).

68 The Rayleigh-Taylor condition The true energy: E RT (t) = E(t) + 1 m ε (t). One gets σ ε (α, t) C 1 ([0, T ε ] [ π, π]), and therefore m ε (t) is Lipschitz. d dt (mε )(t) = σ ε t (α t, t), d dt ( 1 m ε )(t) = σε t (α t, t) (m ε (t)) 2, for almost every t.it yields d dt E RT (t) C exp(ce RT (t)), almost everywhere and therefore E RT (t) 1 C ln(exp( CE RT (0)) C 2 t).

69 Vortex Sheets Equation u t + (u )u = p, u = 0, ω(x, t) = ϖ(α, t)δ(x z(α, t)). Contour equation z t = BR(z, ϖ) + c α z, ϖ t = α (c ϖ) (6)

70 Vortex Sheets Equation u t + (u )u = p, u = 0, ω(x, t) = ϖ(α, t)δ(x z(α, t)). Contour equation z t = BR(z, ϖ) + c α z, ϖ t = α (c ϖ) (6)

71 The Parametrization Linearizing ϖ(α, t) = 1 e ξ t, e ξ t. We study the case: ϖ(α, 0)dα = 0. We take a parametrization given by c = 1 2H(ϖ).Let us point out that for z(α, t) = (α, 0) it follows: For the amplitude one finds BR(z, ϖ)(α, t) = 1 H(ϖ)(α, t)(0, 1). 2 ϖ t 1 2 α(ϖhϖ) = 0. Castro, Córdoba & Gancedo (2009)

72 The Parametrization Linearizing ϖ(α, t) = 1 e ξ t, e ξ t. We study the case: ϖ(α, 0)dα = 0. We take a parametrization given by c = 1 2H(ϖ).Let us point out that for z(α, t) = (α, 0) it follows: For the amplitude one finds BR(z, ϖ)(α, t) = 1 H(ϖ)(α, t)(0, 1). 2 ϖ t 1 2 α(ϖhϖ) = 0. Castro, Córdoba & Gancedo (2009)

73 The Parametrization Linearizing ϖ(α, t) = 1 e ξ t, e ξ t. We study the case: ϖ(α, 0)dα = 0. We take a parametrization given by c = 1 2H(ϖ).Let us point out that for z(α, t) = (α, 0) it follows: For the amplitude one finds BR(z, ϖ)(α, t) = 1 H(ϖ)(α, t)(0, 1). 2 ϖ t 1 2 α(ϖhϖ) = 0. Castro, Córdoba & Gancedo (2009)

74 The Parametrization Linearizing ϖ(α, t) = 1 e ξ t, e ξ t. We study the case: ϖ(α, 0)dα = 0. We take a parametrization given by c = 1 2H(ϖ).Let us point out that for z(α, t) = (α, 0) it follows: For the amplitude one finds BR(z, ϖ)(α, t) = 1 H(ϖ)(α, t)(0, 1). 2 ϖ t 1 2 α(ϖhϖ) = 0. Castro, Córdoba & Gancedo (2009)

75 The Parametrization Linearizing ϖ(α, t) = 1 e ξ t, e ξ t. We study the case: ϖ(α, 0)dα = 0. We take a parametrization given by c = 1 2H(ϖ).Let us point out that for z(α, t) = (α, 0) it follows: For the amplitude one finds BR(z, ϖ)(α, t) = 1 H(ϖ)(α, t)(0, 1). 2 ϖ t 1 2 α(ϖhϖ) = 0. Castro, Córdoba & Gancedo (2009)

76 Ill-posedness for the amplitude equation If we take Ω(α, t) = Hϖ(α, t) iϖ(α, t), by using the Green s function, Ω can be extended analytically on the unit ball and satisfies Ω t (u, t) 1 2 iuω(u, t)ω u(u, t) = 0, with u = re iα. Complex trajectories: X t (u, t) = 1 ix(u, t)ω(x(u, t), t), with X(u, 0) = u. 2 Therefore Ω(X(u, t), t) = Ω(u, 0) and X(u, t) = ue 1 2 iω(u,0)t.

77 Ill-posedness for the amplitude equation If we take Ω(α, t) = Hϖ(α, t) iϖ(α, t), by using the Green s function, Ω can be extended analytically on the unit ball and satisfies Ω t (u, t) 1 2 iuω(u, t)ω u(u, t) = 0, with u = re iα. Complex trajectories: X t (u, t) = 1 ix(u, t)ω(x(u, t), t), with X(u, 0) = u. 2 Therefore Ω(X(u, t), t) = Ω(u, 0) and X(u, t) = ue 1 2 iω(u,0)t.

78 Ill-posedness for the amplitude equation If we take Ω(α, t) = Hϖ(α, t) iϖ(α, t), by using the Green s function, Ω can be extended analytically on the unit ball and satisfies Ω t (u, t) 1 2 iuω(u, t)ω u(u, t) = 0, with u = re iα. Complex trajectories: X t (u, t) = 1 ix(u, t)ω(x(u, t), t), with X(u, 0) = u. 2 Therefore Ω(X(u, t), t) = Ω(u, 0) and X(u, t) = ue 1 2 iω(u,0)t.

79 Ill-posedness for the amplitude equation If we take Ω(α, t) = Hϖ(α, t) iϖ(α, t), by using the Green s function, Ω can be extended analytically on the unit ball and satisfies Ω t (u, t) 1 2 iuω(u, t)ω u(u, t) = 0, with u = re iα. Complex trajectories: X t (u, t) = 1 ix(u, t)ω(x(u, t), t), with X(u, 0) = u. 2 Therefore Ω(X(u, t), t) = Ω(u, 0) and X(u, t) = ue 1 2 iω(u,0)t.

80 The formula X(u, t) = ue 1 2 iω(u,0)t yields X(e iα 0, t) < 1 for ϖ(α 0, 0) > 0 and t > 0. If we denote X(e iα 0, t) = R(α 0, t)e iθ(α 0,t), Ω satisfies d n Ω dθ n (X(eiα 0, t), t) = d n Ω dα n (e iα 0, 0) (1 1 2 Ω α(e iα 0, 0)t) n+1 + l.o.t..

81 The formula X(u, t) = ue 1 2 iω(u,0)t yields X(e iα 0, t) < 1 for ϖ(α 0, 0) > 0 and t > 0. If we denote X(e iα 0, t) = R(α 0, t)e iθ(α 0,t), Ω satisfies d n Ω dθ n (X(eiα 0, t), t) = d n Ω dα n (e iα 0, 0) (1 1 2 Ω α(e iα 0, 0)t) n+1 + l.o.t..