Interface dynamics for incompressible flows in...

Interface dynamics for incompressible flowsin 2D

Diego Córdoba

Interface dynamics for incompressible flows in 2D

Equation

ρt + u · ∇ρ = 0,∇ · u = 0,

ρ(x1, x2, t) =

ρ1, x ∈ Ω1(t)ρ2, x ∈ Ω2(t),

with ρ(x , t) an active scalar, (x , t) ∈ R2 × R+, ρ1 6= ρ2 areconstants, and Ω2(t) = R2 r Ω1(t).

1 SQG sharp front: ρ temperature,

u = (−R2ρ,R1ρ), Rj = iξj/|ξ|.

2 Muskat: ρ density, µ viscosity

κu = −∇p − (0, gρ), Darcy’s law.

3 Water wave: ρ density,

ρ(ut + (u · ∇)u) = −∇p − (0, gρ), Euler.

The SQG equation

Equation

θt + u · ∇θ = 0,

u = ∇⊥ψ, θ = −(−∆)1/2ψ,

with θ(x , t) the temperature, (x , t) ∈ R2 × R+.

Constantin, Majda, and Tabak (1994)

Sharp fronts

We consider weak solutions given by

θ(x1, x2, t) =

θ1, Ω(t)θ2, R2 r Ω(t).

The 2-D vortex patch problem

Contour equation

wt + u · ∇w = 0,

u = ∇⊥ψ, w = ∆ψ,

where the vorticity is given by

w(x1, x2, t) =

w0, Ω(t)0, R2 r Ω(t).

Chemin (1993)Bertozzi and Constantin (1993)

The 2-D patch problem for SQG

Rodrigo (2005), for a periodic C∞ front, i.e.

θ(x1, x2, t) =

θ1, f (x1, t) > x2θ2, f (x1, t) ≤ x2.

The α-patch model

Córdoba, Fontelos, Mancho and Rodrigo (2005)

Contour equation

θt + u · ∇θ = 0,

u = ∇⊥ψ, θ = −(−∆)1−α/2ψ, 0 < α ≤ 1,

where the active scalar θ(x , t) satisfies

θ(x1, x2, t) =

θ1, Ω(t)θ2, R2 r Ω(t).

The contour equation

∂Ω(t) = x(γ, t) = (x1(γ, t), x2(γ, t)) : γ ∈ [−π, π] = T,

where x(γ, t) is one to one. We have

∇⊥θ = (θ1 − θ2) ∂γx(γ, t) δ(x − x(γ, t)),

u = −(−∆)α/2−1∇⊥θ,

and it gives

u(x , t) = − Θα

∂γx(γ − η, t)|x − x(γ − η, t)|α

The normal velocity of the systems reads

u(x(γ, t), t) · ∂⊥γ x(γ, t) = − Θα

∂γx(γ − η, t) · ∂⊥γ x(γ, t)|x(γ, t)− x(γ − η, t)|α

The tangential velocity does not change the shape of theboundary, so that we fix the contour α-patch equations asfollows:

Contour equation

xt (γ, t) =Θα

∂γx(γ, t)− ∂γx(γ − η, t)|x(γ, t)− x(γ − η, t)|α

dη, 0 < α ≤ 1,

x(γ,0) = x0(γ).

Numerical simulations

Close caption of the corner region at t=16.515 for α = 0.5 (a)and t= 4.464 for α = 1 (b). Observe that the singularity ispoint-like in both cases.

Evidence of finite time singularity. (a) Evolution of the inverse ofthe maximum curvature with time. (b) Evolution of the minimumdistance between patches with time. (Insets) Here, werepresent the latest stage of the evolution together with itslinear interpolation.

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

Local well-posedness in Hs for 0 < α ≤ 1

We define

F (x)(γ, η, t) =|η|

|x(γ, t)− x(γ − η, t)|∀ γ, η ∈ [−π, π],

withF (x)(γ,0, t) = |∂γx(γ, t)|−1.

Gancedo (2008)

Theorem

Let x0(γ) ∈ Hk (T) for k ≥ 3 with F (x0)(γ, η) <∞. Then thereexists a time T > 0 so that there is a unique solution to theα-patch model for 0 < α ≤ 1 in C1([0,T ]; Hk (T)), withx(γ,0) = x0(γ).

Existence for α = 1; the SQG sharp front

We modify the equation as follows:

Contour equation

xt (γ, t) =

∂γx(γ, t)− ∂γx(γ − η, t)|x(γ, t)− x(γ − η, t)|

dη + λ(γ, t)∂γx(γ, t), (1)

λ(γ, t) =γ+π

∂γx(γ, t)|∂γx(γ, t)|2

·∂γ(∫

∂γx(γ, t)−∂γx(γ−η, t)|x(γ, t)−x(γ−η, t)|

dη)dγ

−∫ γ

∂γx(η, t)|∂γx(η, t)|2

·∂η(∫

∂γx(η, t)−∂γx(η−ξ, t)|x(η, t)−x(η−ξ, t)|

dξ)dη.

We get|∂γx(γ, t)|2 = A(t).

Hou, Lowengrub and Shelley (1997)

Extra cancellations:

∂γx(γ, t) · ∂2γx(γ, t) = 0,

and∂γx(γ, t) · ∂3

γx(γ, t) = −|∂2γx(γ, t)|2.

Darcy’s law

κv = −∇p − (0,0, g ρ),

v velocity, p pressure, µ viscosity, κ permeability, ρ density, andg acceleration due to gravity.

Muskat (1937)Saffman and Taylor (1958)

Equation (Hele–Shaw)

12µb2 v = −∇p − (0, g ρ),

b distance between the plates.

Smooth initial data with µ = const .

Equation

ρt + v · ∇ρ = 0v = −∇p − (0, ρ)div v = 0

Two-dimensional mass balanceequation in porous media (2DPM)

v(x) =1

(−2y1y2

|y |4,y2

1 − y22

|y |4) ρ(x − y)dy − 1

2(0, ρ(x)) ,

(∂t + v · ∇)∇⊥ρ = (∇v)∇⊥ρ.

Local existence.Singularities with infinite energy, with a stream functiongiven by

ψ(x1, x2, t) = x2f (x1, t) + g(x1, t).

Blow-up ⇔∫ T

0‖∇ρ‖BMO(t) dt =∞.

Geometric constraints: η =∇⊥ρ|∇⊥ρ|

Numerical simulations: ‖∇ρ‖L∞(t) ∼ et .

Cordoba-Gancedo (2007)

We consider the case where the fluid has different densities,that is ρ is represented by

ρ(x1, x2, t) =

ρ1, x2 > f (x1, t)ρ2, x2 < f (x1, t)

being f the interface. Then we have

Equation

(x , t) =ρ2 − ρ1

(∂x f (x , t)− ∂x f (x − α, t))α

α2 + (f (x , t)− f (x − α, t))2 dα

f (x ,0) = f0(x); x ∈ R.

ρt + v · ∇ρ = 0v = −∇p − (0, ρ)div v = 0

⇔ 2DPM contour equation

The linearized equation

If we neglect the terms of order greater than one

ft =ρ1 − ρ2

2(H∂x f ) =

ρ1 − ρ2

2Λf ,

f (x ,0) = f0(x).

Applying the Fourier transform we get

f (ξ) = f0(ξ)eρ1−ρ2

2 |ξ|t .

Problemρ1 < ρ2 stable case,ρ1 > ρ2 unstable case.

Local well-posedness for the stable case (ρ2 > ρ1)

The following theorem follows

Theorem

Let f0(x) ∈ Hk for k ≥ 3 and ρ2 > ρ1. Then there exists a timeT > 0 so that there is a unique solution to 2DPM contourequation in C1([0,T ]; Hk ) with f (x ,0) = f0(x).

Global existence?

L∞ decays.∫f (x , t) dx =

∫f0(x) dx .

Global existence for small initial data:∑|ξ||f (ξ)| << 1.

Ill-posedness for the unstable case (ρ1 > ρ2)

We define‖g‖b =

∑|g(ξ)|eb|ξ| b ≥ 0.

If ‖g‖b <∞, g is analytic in |=z| < 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− |ρ2 − ρ1|t)/4),

‖Λ1+γ f0‖0 < C,

and‖Λ1+γ+ζ f0‖0 =∞,

for γ, ζ > 0.

We takefλ(x1, t) = λ−1f (λx1,−λt + λ1/2),

fλλ>0 a family of solutions to the unstable case. Then

‖fλ‖Hs (0) = |λ|s−32 ‖f‖Hs (λ

12 ) ≤ C|λ|s−

32 ‖f‖1(λ

12 ) ≤ C|λ|s−

32 e−λ

‖fλ‖Hs (λ−1/2) = |λ|s−32 ‖f‖Hs (0) ≥ |λ|s−

32 C‖Λ1+γ+ζ f0‖0 =∞,

for s > 3/2 and γ, ζ small enough.

TheoremLet s > 3/2, then for any ε > 0 there exists a solution f of2DPM contour equation with ρ1 > ρ2 and 0 < δ < ε such that‖f‖Hs (0) ≤ ε and ‖f‖Hs (δ) =∞.

The Muskat problem

Equation

ρt + u · ∇ρ = 0,∇ · u = 0,µ

κu = −∇p − (0, gρ),

(µ, ρ)(x , t) =

(µ1, ρ1), x ∈ Ω1(t)(µ2, ρ2), x ∈ Ω2(t),

Siegel, Caflish & Howison (2004)Ambrose (2004)

We shall consider z(α + 2πk , t) = z(α, t) + (2πk ,0).Darcy’s law yields

ω(x , t) = $(α, t)δ(x − z(α, t)),

Biot-Savart

u(x , t) =1

∫(x − z(β, t)⊥

|x − z(β, t)|2$(β, t)dβ,

and taking limits

uj(z(α, t), t) = BR(z, $)(α, t)∓ $(α, t)2|∂αz(α, t)|2

∂αz(α, t) j = 1,2,

BR(z, $)(α, t) =1

∫(z(α, t)− z(β, t)⊥

|z(α, t)− z(β, t)|2$(β, t)dβ.

Closing the equation: In search of a relationship between z and$ Beginning with Darcy’s law:

κuj = −∇pj − (0, g ρj)

First we substitute uj by its formula involving Birkhoff-Rott.Next, for each j, we write its scalar product with ∂αz(α), andfinally we subtract both expressions to get

µ2 + µ1

2κ$(α, t) +

µ2 − µ1

κBR(z, $)(α, t) · ∂αz(α, t)

= −(∇p2(z(α, t), t)−∇p1(z1(α, t), t))·∂αz(α, t)−g(ρ2−ρ1) ∂αz2(α, t)

= −∂α(p2(z(α, t), t)− p1(z(α, t), t))− g(ρ2 − ρ1) ∂αz2(α, t)

∂α(p2(z(α, t), t)− p1(z(α, t), t)) = 0

Darcy’s law implies

∆p(x , t) = −div (µ(x , t)κ

u(x , t))− g ∂x2ρ(x , t),

therefore∆p(x , t) = Π(α, t)δ(x − z(α, t)),

where Π(α, t) is given by

Π(α, t) = (µ2 − µ1

κu(z(α, t), t) ·∂⊥α z(α, t) + g(ρ2−ρ1)∂αz1(α, t)).

It follows that:

p(x , t) = − 12π

cosh(x2−z2(α, t))−cos(x1−z1(α, t)))Π(α, t)dα,

for x 6= z(α, t), implying the important identity

p2(z(α, t), t) = p1(z(α, t), t), (3)

which is just a mathematical consequence of Darcy’s law,making unnecessary to impose it as a physical assumption.

$(α, t) + AµT ($)(α, t) = −2gκC(ρ, µ)∂αz2(α, t)

T ($) = 2BR(z, $) · ∂αz, Aµ =µ2 − µ1

µ2 + µ1 , C(ρ, µ) =ρ2 − ρ1

µ2 + µ1 .

Baker, Meiron & Orszag (1982) |λ| < 1 for (I + AµT ).

Contour equation

zt (α, t) = BR(z, $)(α, t) + c(α, t)∂αz(α, t),$(α, t) = (I + AµT )−1(−2gκC(ρ, µ)∂αz2)(α, t).

Cordoba, Cordoba & Gancedo (2008).

Theorem

Let z0(γ) ∈ Hk (T) for k ≥ 3, F (z0)(α, β) <∞ and

σ0(α) = −(∇p2(z0(α),0)−∇p1(z0(α),0)) · ∂⊥α z0(α) > 0.

Then there exists a time T > 0 so that there is a solution to thecontour equation in C1([0,T ]; Hk (T)) with z(α,0) = z0(α).

σ(α) = −(∇p2(z(α))−∇p1(z(α))) · ∂⊥α z(α)

= µ2−µ1

κ BR(z, $)(α) · ∂⊥α z(α) + g(ρ2 − ρ1)∂αz1(α).

Energy estimates:

|||z|||2 = ‖z‖2H3 + ‖F (z)‖2L∞ , ‖(I + AµT )−1‖H

12→H

12≤ eC|||z|||2 ,

We can estimate as follows:

‖$‖H2 ≤ eC|||z|||2 .

By energy estimates we obtain

ddt|||z|||2(t) ≤ −K

∫Tσ(α, t)∂3

αz(α, t) · Λ(∂3αz)(α, t)dα + eC(|||z|||2).

Pointwise inequality:

f Λ(f ) ≥ 12

Λ(f 2).

−∫

Tσ∂3

αz·Λ(∂3αz)dα ≤ −1

∫TσΛ(|∂3

αz|2)dα = −12

Λ(σ)|∂3αz|2dα

as long asσ(α, t) > 0.

Sketch of the proof

Estimates on the arc-chord condition

Lemmad

dt‖F(z)‖2

L∞(t) ≤ expC(‖F(z)‖2L∞(t) + ‖z‖2

H3(t))

Sketch of the proof

Estimate for the evolution of the minimum of the difference of the gradientsof the pressure in the normal direction. This quantity is given by

σ(α) =µ2 − µ1

kBR(z,$)(α) · ∂⊥α z(α) + g(ρ2 − ρ1)∂αz1(α).

Lemma Let z(α) be a solution of the system with z(α, t) ∈ C1([0, T ];H3),and

m(t) = mınα∈T

σ(α, t).

m(t) ≥ m(0)−∫ t

expC(‖F(z)‖2L∞(s) + ‖z‖2

H3(s))ds.

Sketch of the proof

Regularizing the system

Let zε(α, t) be a solution of the following system:

zε,δt (α, t) = BRδ(zε,δ, $ε,δ)(α, t) + cε,δ(α, t)∂αzε,δ(α, t),

zε,δ(α, 0) = z0(α),

BRδ(z,$)(α, t) =(− 1

∫T$(β, t)

tanh(z2(α,t)−z2(β,t)2 )(1 + tan2(z1(α,t)−z1(β,t)2 ))

tan2(z1(α,t)−z1(β,t)2 ) + tanh2(z2(α,t)−z2(β,t)2 ) + δdβ,

∫T$(β, t)

tan(z1(α,t)−z1(β,t)2 )(1− tanh2(z2(α,t)−z2(β,t)2 ))

tan2(z1(α,t)−z1(β,t)2 ) + tanh2(z2(α,t)−z2(β,t)2 ) + δdβ

$ε,δ(α, t) = −µ2 − µ1

µ2 + µ1φε ∗ φε ∗ (2BR(zε,δ, $ε,δ) · ∂αzε,δ)(α)

−2kgρ2 − ρ1

µ2 + µ1φε ∗ φε ∗ (∂αz

ε,δ2 )(α),

cε,δ(α, t) =α + π

∂αzε,δ(α, t)

|∂αzε,δ(α, t)|2· ∂αBRδ(zε,δ, $ε,δ)(α, t)dα

−∫ α

∂αzε,δ(β, t)

|∂αzε,δ(α, t)|2· ∂βBRδ(zε,δ, $ε,δ)(β, t)dβ,

φ ∈ C∞c (R), φ(α) ≥ 0, φ(−α) = φ(α),

∫Rφ(α)dα = 1, φε(α) = φ(α/ε)/ε,

for ε > 0 and δ > 0.

Sketch of the proof

The next step is to integrate the system during a time T with T independentof ε. If z0(α) ∈ Hn, we have the solution zε ∈ C1([0, T ε];Hn) and that

mε(t) ≥ m(0)−∫ t

expC(‖F(zε)‖2L∞(s) + ‖zε‖2

H3(s))ds,

formε(t) = mın

α∈Tσε(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 σε(α, t) > 0.

Sketch of the proof

For t ≤ T ε

dt‖zε‖2

Hn(t) ≤ I + expC(‖F(zε)‖2L∞(t) + ‖zε‖2

Hn(t)),

I = − k

2π(µ1+µ2)

σε(α, t)

|∂αzε(α, t)|2φε ∗ (∂nαz

ε)(α, t) · Λ(φε ∗ (∂nαzε))(α, t)dα

≤ − k

2π(µ1+µ2)Aε(t)

∫Tσε(α, t)

2Λ(|φε ∗ (∂nαz

ε)|2)(α, t)dα.

= − k

2π(µ1+µ2)Aε(t)

Λ(σε)(α, t)1

2|φε ∗ (∂nαz

ε)|2(α, t)dα,

we obtain

I ≤ C‖F(zε)‖L∞‖Λ(σε)‖L∞‖∂nαzε‖2L2 ≤ expC(‖F(zε)‖2

L∞ + ‖zε‖2H3).

Sketch of the proof

Finally, for t ≤ T ε we have

dt‖zε‖2

Hn(t) ≤ C expC(‖F(zε)‖2L∞(t) + ‖zε‖2

Hn(t)).

dt‖F(zε)‖2

L∞(t) ≤ C expC(‖F(zε)‖2L∞(t) + ‖zε‖2

H3(t)),

Therefore

dt(‖zε‖2

Hn(t) + ‖F(zε)‖2L∞(t)) ≤ C expC(‖zε‖2

Hn(t) + ‖F(zε)‖2L∞(t))

for t ≤ T ε.Integrating together with

mε(t) ≥ m(0)−∫ t

expC(‖F(zε)‖2L∞(s) + ‖zε‖2

H3(s))ds,

q.e.d.

The fluid interface problem

Equation

ρt + u · ∇ρ = 0,∇ · u = 0,ρ(ut + (u · ∇)u) = −∇p − (0, gρ),

ρ(x , t) =

ρ1, x ∈ Ω1(t)ρ2, x ∈ Ω2(t),

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)

Contour equation

We consider ∂Ωj(t) = z(α, t) = (z1(α, t), z2(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

ω(x , t) = $(α, t)δ(x − z(α, t)).

Biot-Savart yields

u(x , t) =1

∫(x − z(β, t))⊥

|x − z(β, t)|2$(β, t)dβ,

for x 6= z(α, t).

Contour equation

z(α + 2πk , t) = z(α, t).

ω(x , t) = $(α, t)δ(x − z(α, t)).

Biot-Savart yields

u(x , t) =1

∫(x − z(β, t))⊥

|x − z(β, t)|2$(β, t)dβ,

for x 6= z(α, t).

Contour equation

z(α + 2πk , t) = z(α, t).

ω(x , t) = $(α, t)δ(x − z(α, t)).

Biot-Savart yields

u(x , t) =1

∫(x − z(β, t))⊥

|x − z(β, t)|2$(β, t)dβ,

for x 6= z(α, t).

Then ( Bernoulli’s law)

Contour equation

$t = −2Aρ∂tBR(z, $) · ∂αz − Aρ∂α(|$|2

4|∂αz|2) + ∂α(c$)

+ 2Aρc ∂αBR(z, $) · ∂αz(α, t)− 2Aρg∂αz2,

for Aρ = (ρ2 − ρ1)/(ρ2 + ρ1) and c parametrization freedom.

Contour equation

zt = BR(z, $) + c ∂αz. (5)

Equations (4) and (5) give weak solutions of Euler.

We can choose c to get |∂αz(α, t)|2 = A(t).

Contour equation

$t = −2Aρ∂tBR(z, $) · ∂αz − Aρ∂α(|$|2

4|∂αz|2) + ∂α(c$)

Contour equation

zt = BR(z, $) + c ∂αz. (5)

Contour equation

$t = −2Aρ∂tBR(z, $) · ∂αz − Aρ∂α(|$|2

4|∂αz|2) + ∂α(c$)

Contour equation

zt = BR(z, $) + c ∂αz. (5)

Contour equation

$t = −2Aρ∂tBR(z, $) · ∂αz − Aρ∂α(|$|2

4|∂αz|2) + ∂α(c$)

Contour equation

zt = BR(z, $) + c ∂αz. (5)

Conditions

Arc-chord:

F(z)(α, β, t) =|β|

|z(α, t)− z(α− β, t)|∀α, β ∈ (−π, π),

andF(z)(α,0, t) =

1|∂αz(α, t)|

Rayleigh-Taylor:

σ(α, t) = −(∇p2(z(α, t), t)−∇p1(z(α, t), t)) ·∂⊥α z(α, t) > 0.

Conditions

Arc-chord:

F(z)(α, β, t) =|β|

|z(α, t)− z(α− β, t)|∀α, β ∈ (−π, π),

andF(z)(α,0, t) =

1|∂αz(α, t)|

Rayleigh-Taylor:

σ(α, t) = −(∇p2(z(α, t), t)−∇p1(z(α, t), t)) ·∂⊥α z(α, t) > 0.

Local-existence

Cordoba, Cordoba & Gancedo (2008)

Theorem

Let z0(α) ∈ Hk and $0(α) ∈ Hk−1 for k ≥ 4,

F(z0)(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 ρ1 = 0, where z(α, t) ∈ C1([0,T]; Hk ) and$(α, t) ∈ C1([0,T]; Hk−1) with z(α,0) = z0(α) and$(α,0) = $0(α).

ftt (α, t) = −σΛ(f )(α, t),

σ < 0⇒ e|σξ|

12 t , e−|σξ|

σ > 0⇒ cos(|σξ|12 t), sin(|σξ|

It can be written as follows:

ft (α, t) = H(g)(α, t),gt (α, t) = −σ∂αf (α, t),

and therefore

EL(t) =

∫(σ|∂αf |2 + |Λ

12 g|2)dα.

ftt (α, t) = −σΛ(f )(α, t),

σ < 0⇒ e|σξ|

12 t , e−|σξ|

and therefore

EL(t) =

∫(σ|∂αf |2 + |Λ

12 g|2)dα.

ftt (α, t) = −σΛ(f )(α, t),

σ < 0⇒ e|σξ|

12 t , e−|σξ|

and therefore

EL(t) =

∫(σ|∂αf |2 + |Λ

12 g|2)dα.

Energy of the system

Case z ∈ H4:

E(t) = ‖z‖2H3(t) +

∫ π

σ(α, t)ρ2|∂αz(α, t)|2

|∂4αz(α, t)|2dα

+ ‖F(z)‖2L∞(t) + ‖$‖2H2(t) + ‖ϕ‖2H4− 1

for σ(α, t) > 0 and ϕ(α, t) given by

ϕ(α, t) =$(α, t)

2|∂αz(α, t)|− c(α, t)|∂αz(α, t)|.

Beale, Hou & Lowengrub (1993)Ambrose & Masmoudi (2005)

Equation in terms of ϕ(α, t)

Case Aρ = 0 (vortex sheet problem):

$t (α, t) = ∂α(c$)(α, t).

Kelvin-Helmholtz instability arises.

The linear formulation is given by

ftt (α, t) = −∂2αf (α, t) ⇒ e|ξ|t , e−|ξ|t

Case Aρ = 1 (water wave):

ϕt = −∂α(ϕ2)

2|∂αz|− B(t)ϕ− ∂tBR(z, $) · ∂αz

|∂αz|

− g∂αz2

|∂αz|− ∂t (c|∂αz|).

$t (α, t) = ∂α(c$)(α, t).

ϕt = −∂α(ϕ2)

2|∂αz|− B(t)ϕ− ∂tBR(z, $) · ∂αz

|∂αz|

− g∂αz2

|∂αz|− ∂t (c|∂αz|).

$t (α, t) = ∂α(c$)(α, t).

ϕt = −∂α(ϕ2)

2|∂αz|− B(t)ϕ− ∂tBR(z, $) · ∂αz

|∂αz|

− g∂αz2

|∂αz|− ∂t (c|∂αz|).

Estimates on the inverse operator (I + T )−1

Let T be defined by T (u)(α, t) = 2BR(z,u)(α) · ∂αz(α),

then T : L2 → H1 and ‖T‖L2→H1 ≤ ‖F(z)‖4L∞‖z‖4C2,δ .

$t (α, t) = (I + T )−1(R(z, $))(α, t).

Baker, Meiron & Orszag (1982).Córdoba, Córdoba & Gancedo (2008):