on self-similar solutions to the incompressible euler...

On Self-similar Solutions to the Incompressible

Euler Equations

Alberto Bressan(∗) and Ryan Murray(∗∗)

(∗) Department of Mathematics, Penn State University,University Park, Pa. 16802, USA.

(∗∗) Department of Mathematics, North Carolina State University,Raleigh, NC 27695, USA.

e-mails: [email protected], [email protected]

March 2, 2020

Abstract

Recent numerical simulations have shown the existence of multiple self-similar solutionsto the Cauchy problem for the 2-dimensional incompressible Euler equation, with initialvorticity in Lploc(R2), 1 ≤ p < +∞. Toward a rigorous validation of these computations, inthis paper we analytically construct self-similar solutions (i) on an outer domain of the form{|x| > R}, and (ii) in a neighborhood of the points where the solution exhibits a spiralingvortex singularity. The outer solution is obtained as the fixed point of a contractivetransformation, based on the Biot-Savart formula and integration along characteristics.The inner solution is constructed using a system of adapted coordinates, following theapproach of V. Elling (Comm. Math. Phys., 2016).

1 Introduction

The classical Euler equations, describing the motion of a non-viscous, homogeneous, incom-pressible fluid, take the form {

ut + (u · ∇)u = −∇p ,

divu = 0 .(1.1)

Here u = u(t, x) is the fluid velocity while p is the pressure. For a given initial condition

u(0, x) = ū(x) , x ∈ Rd, (1.2)

the well-posedness of the Cauchy problem, in various functional spaces, has been an outstand-ing mathematical problem [2, 3, 6, 11, 21, 24]. In recent years, a new approach based on convexintegration [10, 13, 14, 15] has led to the construction of a large family of weak solutions, cul-minating with the proof of the famous Onsager’s conjecture [5, 20]. This approach, relying on

1

a Baire category argument, yields infinitely many solutions with the same initial data. Similarresults have been obtained also for multi-dimensional inviscid compressible flow [8]. In theseconstructions, the usual admissibility criteria based on energy or entropy dissipation fail toselect a unique solution. Yet, one may still hope that some other admissibility criterium canbe found, selecting unique solutions which depend continuously on the initial data.

We believe that this is not the case. Indeed, as soon as the vorticity is only required be inLploc, for some p < +∞, the ill-posedness of the Euler equations appears to be “incurable”.To support this claim, in the present paper we study a simple example of 2-dimensionalincompressible Euler flow where numerical simulations show the existence of two distinctsolutions for the same initial data. In polar coordinates, x = (x1, x2) = (r cos θ, r sin θ), theinitial vorticity ω = curl ū has the form

ω(x) = r− 1µ Ω(θ), (1.3)

where 12 < µ < +∞. As shown in Figure 1, left, we assume that Ω ∈ C∞(R) is a non-negative,

smooth, periodic function which satisfies

Ω(θ) = Ω(π + θ), Ω(θ) = 0 if θ ∈[π

4, π]. (1.4)

Figure 1: The supports of the initial vorticity ω, ωε1, and ωε2, considered at (1.3)-(1.6).

We then consider two sequences of initial data, with vorticity ωε1, ωε2 ∈ L∞(R2), as shown in

Figure 1, center and right. Namely

ωε1(x).=

ω(x) if |x| > ε,ε− 1µ if |x| ≤ ε,

(1.5)

ωε2(x).=

{ω(x) if |x| > ε,

0 if |x| ≤ ε.(1.6)

By Yudovich’ theorem [24], for every ε > 0 the Cauchy problem for the incompressible Eulerequation (1.1) with initial data (1.5) or (1.6) has a unique solution. Numerical simulations ofthese solutions, performed by Wen Shen [23], are shown in Figures 2 and 3, respectively. Asε→ 0, the initial data ωε1, ωε2 converge to the same limit ω in L

ploc. However, at times t > 0 the

corresponding solutions converge to two different limits. Both of these limits yield solutionsto the Euler equations (1.1) with the same initial data (1.3).

At an intuitive level, this behavior is easy to explain. For the initial data ω in (1.3)-(1.4), thevorticity is supported inside two wedges. In the approximations (1.5), the presence of a dense

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Figure 2: The vorticity distribution at time t = 1, for a solution to (1.1) with initial vorticity (1.5).

Figure 3: The vorticity distribution at time t = 1, for a solution to (1.1) with initial vorticity (1.6).

patch of vorticity around the origin stirs these two wedges together, forming a single spiralingvortex (Fig. 2). On the other hand, in the approximations (1.6) each of the two wedges curlsup on itself, and two separate vortex spirals are generated (Fig. 3).

The above simulations provide numerical evidence of the existence of multiple self-similarsolutions to the Cauchy problem (1.1)–(1.3). However, proving the existence of two distinctexact solutions, close to the computed ones, is a nontrivial task. Toward this goal, the maindifficulties to overcome are:

(i) The exact solution is a function defined on the whole plane R2, while a numerical ap-proximation is computed only on a bounded domain D.

(ii) The exact solution is not smooth, but contains a singularity at the center of each spiral.In a neighborhood of such point, a-posteriori error estimates will break down.

We thus propose an approach based on a domain decomposition:

R2 = Do ∪ Dm ∪ Di. (1.7)

As shown in Fig. 4, here Do is an outer domain, Dm is an intermediate, bounded domainwhere the solution remains smooth, and Di is an inner domain consisting of one or two discsaround the spirals’ centers.

In the present paper we focus on the analytical construction of solutions on the two domainsDo and Di. Once this is accomplished, the existence of a globally defined solution is reduced

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to a problem that can be resolved by a numerical computation. Indeed, if a numerical solutionis available on Dm, which matches sufficiently well with the analytical solutions on the innerand outer domains, then a version of the Newton-Kantorovich theorem [1, 12] will yield theexistence of an exact solution defined on the entire plane R2. We remark that our solutionsare C∞ on the intermediate domain Dm. The analysis of a numerical algorithm, together withposteriori error estimates on an intermediate domain, has been carried out in [4]. Appropriatematching conditions will be studied in a future paper.

The remainder of this paper is organized as follows. Section 2 reviews the vorticity formulationof the Euler equations, deriving the system (2.9) satisfied by self-similar solutions. In Section 3we construct self-similar solutions on an outer domain Do = {x ∈ R2 ; |x| > R}, with givenasymptotic conditions as |x| → +∞. The existence and uniqueness of these solutions isestablished in Theorem 3.1.

The remaining sections carry out the construction of solutions in a small neighborhood of aspiral’s center. Here the analysis heavily relies on the ground-breaking work of Volker Elling[16, 17, 18]. Self-similar solutions are obtained within an adapted system of coordinates whichfollow the pseudo-streamlines. As shown in Elling’s papers, this approach is successful aslong as our solutions remain sufficiently close to radially symmetric ones. Section 4 describesthe construction of adapted coordinates, where the characteristic curves for the linear, firstorder PDE in (2.9) are taken as coordinate lines. Sections 5 and 6 analyze radially symmetricsolutions and the linearized system of equations satisfied by a small perturbation of suchsolutions. In terms of a Fourier series, these take the form Y (β, φ) =

∑k Yk(β)e

ikφ.

We point out the major difference between our analysis and the results in [16, 17]. To constructa globally defined solution of (2.9) in terms adapted coordinates, one needs to assume thatsuch solution is close to radially symmetric on the entire space R2. In [16, 17], this wasguaranteed by the assumption that all Fourier components of the vorticity Yk(β) vanish, for0 < |k| < N , with N large enough.

On the other hand, our construction allows all components of the vorticity to be nonzero,including k = ±1 and k = ±2. In our case, however, the solution is constructed not on theentire space R2 but only in a neighborhood of the spiral’s center. We remark that the analysisof the Fourier components |k| = 2 and |k| = 1 is essential in order to understand the behavior ofthe solutions shown respectively in Fig 2 and in Fig. 3, near the spirals’ centers. Additionally,the boundary data for the stream function Ψ is allowed to be any function which is closeenough to the radially symmetric one, with the restriction that one cannot assign the k = ±1Fourier coefficients. This restriction reflects the fact that we need to construct solutions ona ball that is centered precisely at the singularity of the vortex, as being off center will yieldnon-zero coefficients for Ψ on the k = ±1 modes.

Section 7 contains the analysis of the nonlinear problem, in adapted coordinates. The main re-sult, stated in Theorem 7.1, yields the existence of a solution operator, mapping the boundarydata into the corresponding solution, on a suitably small neighborhood of the origin.

For solutions as in Figures 2, 3, we expect that the stream function will become close toradially symmetric, as a spiral’s center is approached. On the other hand, the vorticity onlysatisfies a linear transport equation, which does not provide any smoothing. Hence, if it is notclose to radially symmetric as |x| → ∞, the vorticity will never approach radial symmetry.To address this issue, Proposition 7.3 gives a further refinement of Theorem 7.1. Here the

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existence of a solution is proved also for large vorticity, far from radially symmetric, but on asuitably small neighborhood of the spiral’s center.

Finally, in Section 8 we analyze how the solution on the inner domain varies, depending onthe boundary data. This is a basic step, toward the eventual goal of matching the solution onthe inner domain Di with a solution computed on the intermediate domain Dm.

i

mD

D

D

D

D

D

o

m

i

o

Figure 4: The decomposition (1.7) of the plane into an outer, a middle, and an inner domain. Left:the case of the single spiraling vortices, as in Fig. 2. Right: the case of two spiraling vortices, as inFig. 3.

2 Self-similar solutions

Let u = u(t, x) be a solution to the Euler equations (1.1). Calling ω = curl u = (−u1,x2 +u2,x1)and taking the curl of both sides of (1.1) one obtains the linear transport equation

ωt + u · ∇ω = 0. (2.1)

We recall that the velocity u can be recovered from the vorticity by the Biot-Savart formula

u(x) =1

2π

∫R2

(x− y)⊥

|x− y|2ω(y) dy. (2.2)

The assumption that div u = 0 implies the existence of a stream function ψ such that

u = ∇⊥ψ, (u1, u2) = (−ψx2 , ψx1). (2.3)

The Euler equations (1.1) can be reformulated as follows: find two scalar functions ω, ψ definedfor (t, x) ∈ R+ × R2, such that ωt +∇

⊥ψ · ∇ω = 0,

∆ψ = ω.(2.4)

We seek solutions which are self-similar, so that

u(t, x) = tαU( xtµ

), p(t, x) = tβP

( xtµ

), (2.5)

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for some exponents α, β, µ. Notice that, taking t = 1, one finds

U(x) = u(1, x), P (x) = p(1, x). (2.6)

Therefore, the value of the solution at t = 1 determines its values at all positive times.

Inserting (2.5) into (1.1) one obtains

αtα−1U − µtα−1 xtµ· ∇U + tαU · tα−µ∇U = − tβ−µ∇P. (2.7)

This forces α− 1 = 2α− µ = β − µ, hence α = µ− 1 and β = 2α. Setting

y =x

tµ

for some constant µ, these self-similar solutions will be described using the notationu(t, x) = tµ−1U

(xtµ

),

ω(t, x) = t−1Ω(xtµ

),

ψ(t, x) = t2µ−1Ψ(xtµ

).

(2.8)

Inserting (2.8) in (2.3)-(2.4), one finds

tµ−1U(y) = (u1, u2)(t, x) = (−ψx2 , ψx1)(t, x) = t2µ−1(−t−µΨy2 , t−µΨy1) = tµ−1∇⊥y Ψ(y),

Ω = t ω = t∆xψ = ∆yΨ,

0 = ωt +∇⊥x ψ · ∇xω = − t−2Ω(y)− t−2µx

tµ· ∇yΩ(y) + t−2 (−Ψy2Ωy1 + Ψy1Ωy2) .

This yields the equations (∇⊥Ψ− µy

)· ∇Ω = Ω ,

∆Ψ = Ω ,(2.9)

while the velocity is recovered byU = ∇⊥Ψ. (2.10)

The vector fieldq(y)

.= U(y)− µy (2.11)

is called the pseudo-velocity. Its integral curves are the pseudo-streamlines.

Notice that the system (2.9) is equivalent to the third order PDE for the stream function(∇⊥Ψ− µy

)· ∇(∆Ψ) = ∆Ψ . (2.12)

Throughout this paper, we assume that the rescaling parameter µ satisfies

µ >1

2, µ, 2µ /∈ N. (2.13)

To construct solutions of (2.9), one can proceed as follows.

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• Given Ψ, solve the first equation for Ω. This is a linear, first order equation that can besolved explicitly, integrating along characteristics, with suitable asymptotic conditionsat |y| → ∞.

• Insert Ω in the second equation, solve for Ψ.

This will yield a map Ψ 7→ Ω 7→ Ψ. Working in a suitable function space, one needs to find afixed point of this map, which will in turn provide a self-similar solution of (1.1).

Our eventual goal is to construct solutions of (2.9) on the entire plane R2 with prescribedasymptotic behavior as |y| → +∞. Our solutions will be everywhere smooth, except for oneor two points where a spiraling singularity occurs. Solutions will be constructed separately onthree domains:

• An outer domain, of the form Do = {x ∈ R2 ; |x| > R}, for some large radius R. Herethe fluid velocity will be small, and the solution will be obtained as a small perturbationof the asymptotic data.

• A bounded intermediate domain Dm, where the solution is still smooth and can beaccurately computed by numerical approximations.

• An inner domain Di, consisting of one or two small discs centered at points wherethe spiraling singularities occur. On these domains, the solution will be constructedanalytically, using the adapted coordinates introduced in [16, 17].

By suitably patching together these three solutions defined on Do,Dm,Di, a globally definedsolution of (2.9) can be obtained. The present paper will focus on the construction of solutionson the outer domain Do and on the inner domain Di.

3 Solutions on an outer domain

In this section we construct solutions to the system(∇⊥Ψ− µx

)· ∇Ω = Ω ,

∆Ψ = Ω ,(3.1)

on an outer domain of the form

DR.= {x ∈ R2 ; |x| ≥ R}. (3.2)

We recall that the velocity is recovered by U(x) = ∇⊥Ψ(x). We seek a solution (Ω,Ψ) of(3.1) with suitable conditions given on the circumference {|x| = R}, and asymptotic values as|x| → +∞. It will be convenient to use the polar coordinate notation

x(r, θ).= (r cos θ, r sin θ) ∈ R2, r = |x|, θ = ]x, (3.3)

Ω(r, θ).= Ω(r cos θ, r sin θ), Ψ(r, θ)

.= Ψ(r cos θ, r sin θ).

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Moreover, we denote by T the unit circumference. Namely: T = [0, 2π] with endpointsidentified. Any map φ : T 7→ R can thus be identified with a 2π-periodic function of a realvariable.

Given a function ω : T 7→ R, we impose the asymptotic condition

limr→+∞

r1µ Ω(r, θ)

.= ω(θ), (3.4)

Together with ω we also consider the 2π-periodic function ψ : T 7→ R, defined as the solutionto the linear second order ODE

ψθθ(θ) +

(2− 1

µ

)2ψ(θ) = ω(θ). (3.5)

Notice that the homogeneous equation

ψ′′ + λ2ψ = 0

has nonzero 2π-periodic solutions only if λ is an integer. Since we always assume that µsatisfies (2.13), this guarantees that (3.5) has a unique 2π-periodic solution.

Remark 3.1 If λ is not an integer, the unique periodic solution to

ψ′′(θ) + λ2ψ(θ) = f(θ)

is given by

ψ(θ) =1

2λ sinλπ

∫ π−π

f(θ + π − α) cos(λα) dα. (3.6)

Notice that for this periodic problem the Green’s function has constant sign if 0 < λ < 1, but changes

sign if λ > 1.

As r → +∞, on the stream function Ψ we impose the asymptotic condition

limr→+∞

r1µ−2

Ψ(r, θ).= ψ(θ). (3.7)

Remark 3.2 To motivate (3.7), consider two functions of the form

Ω(r, θ) = r−1µ ω(θ), Ψ(r, θ) = r2−

1µ ψ(θ). (3.8)

Assume that

Ω = ∆Ψ = Ψrr +Ψrr

+Ψθθr2

. (3.9)

Inserting (3.8) into (3.9) one obtains

r−1µ ω(θ) =

(2− 1

µ

)(1− 1

µ

)r−

1µψ(θ) +

(2− 1

µ

)r−

1µψ(θ) + r−

1µψθθ(θ). (3.10)

Dividing both sides by r−1/µ we obtain (3.5).

Finally, along the inner boundary SR.= {x ∈ R2 ; |x| = R}, we impose a Dirichlet condition

Ψ(R, θ) = ψ0(θ), (3.11)

for some given function ψ0 : T 7→ R. Notice that no boundary value is imposed on Ω(R, θ).

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3.1 Rescaling of coordinates.

It will be convenient to rescale coordinates, introducing the variables

z =x

R, Ω̃(z) = Ω(Rz), Ψ̃(z) = R−2Ψ(Rz). (3.12)

By the relations

∇zΩ̃(z) = R∇xΩ(x), ∇zΨ̃(z) = R−1∇xΨ(x), ∆zΨ̃(z) = ∆xΨ(x),

in these rescaled variables the equations (3.1) take exactly the same form, namely(R∇⊥z Ψ̃− µRz

)·R−1∇zΩ̃ = Ω̃ ,

∆zΨ̃ = Ω̃ .(3.13)

However, we now have to solve (3.13) on the outer domain

D1 = {z ∈ R2 ; |z| ≥ 1} . (3.14)

Moreover, the asymptotic conditions (3.4), (3.7) and the boundary condition (3.11) now be-come

limr→+∞

r1µ Ω̃(y(r, θ)) = R

− 1µω(θ),

limr→+∞

r1µ−2

Ψ̃(y(r, θ)) = R− 1µψ(θ),

Ψ̃(y(1, θ)) = R−2ψ0(θ).

(3.15)

In other words, solving the original system on the exterior domainDR withR large is equivalentto solving it on D1, but with smaller boundary and asymptotic data.

Throughout the sequel, we shall thus seek a solution to the system (3.1) on the outer domainD1, with asymptotic and boundary conditions

limr→+∞

r1µ Ω(r, θ) = ω(θ),

limr→+∞

r1µ−2

Ψ(r, θ) = ψ(θ),

Ψ(1, θ) = ψ0(θ),

(3.16)

assuming that (3.5) holds and that the norms ‖ω‖C2(T), ‖ψ0‖C2(T) are sufficiently small. Ouranalysis has two main goals:

(i) Prove the existence and uniqueness of a solution to the problem (3.1), (3.16), for allsufficiently small boundary and and asymptotic data.

(ii) Given a radius r1 > 1, study the differential of the map

Λ1 : ψ0 7→ (ω1, ψ1), (3.17)

where ω1, ψ1 : T 7→ R denote the restrictions of the functions Ω,Ψ to the circumference∂Br1

.= {x ∈ R2 ; |x| = r1}.

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3.2 Statement of the main result.

We always assume that the rescaling parameter µ satisfies (2.13). Moreover, we let N0, N1 ∈ Nbe the smallest integers such that

N0 + 3

µ> 2, 1 +

N1 + 3

µ> 2 , (3.18)

and define � > 0 by the identity

2 + �.= min

{N0 + 3

µ, +

N1 + 3

µ

}. (3.19)

Theorem 3.1 Assume that µ satisfies (2.13) and let the integers N0, N1 ≥ 1 be as in (3.18).Consider any functions ω, ψ, ψ0 : T 7→ R, with ψ satisfying (3.5). If the norms

‖ω‖CN0+N1 (T) , ‖ψ0‖C2(T) ,

are sufficiently small, then the problem (3.1) on the exterior of the unit disc, with boundary andasymptotic conditions (3.16), has a unique solution (Ω,Ψ). Here Ω ∈ C1(D1) and Ψ ∈ C2(D1).

A proof will be worked out in the remainder of this section.

3.3 Review of the Poisson equation on an outer domain.

A Green’s function for the Laplace operator on the outer domain D1 = {x ∈ R2 ; |x| > 1} is

G(x, y) =1

2πln

(1

|y|· |y − x||y∗ − x|

), y∗

.=

y

|y|2. (3.20)

This provides a solution to

∆xG(x, y) = δy , G(x, y) = 0 for |x| = 1,

where δy denotes the Dirac measure concentrating a unit mass at the point y. Assuming thatg is continuous while f satisfies the bound

f(x) = O(1) · |x|−2−1µ , (3.21)

the solution to the boundary value problem{∆u(x) = f(x) |x| > 1,

u(x) = g(x) |x| = 1,(3.22)

can be written as

u(x) =

∫|y|>1

G(x, y) f(y) dy +

∫|y|=1

∇yG(x, y) · n(y) g(y) dσ

=1

2π

∫|y|>1

ln

(1

|y|· |y − x||y∗ − x|

)f(y) dy +

|x|2 − 12π

∫|y|=1

g(y)

|y − x|2dσ.

(3.23)

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Here n(y) = −y is the unit outer normal to the domain D1 at the boundary point y ∈ ∂D1.Moreover, dσ denotes integration w.r.t. arc-length, along the unit circumference.

Observe that, for every fixed x, the first integral on the right hand side of (3.23) is absolutelyconvergent provided that

f(y) = O(1) · |y|−α for some α > 2. (3.24)

Lemma 3.1 Assume that g ∈ C1(T) and (3.24) holds. Then the integrals in (3.23) areabsolutely convergent. Moreover, there exist a constant C0 independent of f, g such that

|∇u(x)| = C0 |x|−1 ·(∥∥ |x|αf∥∥

L∞+ ‖g‖C1

). (3.25)

Proof. 1. If (3.24) holds, differentiating w.r.t. x, one obtains

∇u(x) = 12π

∫|y|>1

(x− y|x− y|2

− x− y∗

|x− y∗|2

)f(y) dy

+2x

2π

∫|y|=1

g(y)

|x− y|2dσ − |x|

2 − 12π

∫|y|=1

x− y|x− y|4

g(y) dσ.

(3.26)

We observe that∫|y|>1

|y|−α

|y − x|dy =

(∫|y|>1, |y−x|< |x|

2

+

∫|y|>1, |y−x|> |x|

2

)|y|−α

|y − x|dy

≤∣∣∣x2

∣∣∣−α ∫|z|≤ |x|

2

1

|z|dz +

2

|x|

∫|y|>1

|y|−α dy

= 2απ · |x|1−α + 4πα− 2

|x|−1.

Since we are assuming α > 2, the first integral in (3.26) has size O(1) · |x|−1∥∥ |x|αf∥∥

L∞.

2. We now consider the second and third integrals in (3.26). For |x| ≥ 2, these integrals havesize O(1) · |x|−1‖g‖C0 . In the case 1 < |x| < 2, using polar coordinates we write the secondintegral in (3.23) as

v(r, θ) =r2 − 1

2π

∫ 2π0

g(θ + α)

|r − cosα|2 + | sinα|2dα.

Observing that vr(r, θ) and vθ(r, θ) both have size O(1) · ‖g‖C1 , we obtain the bound ∇v(x) =O(1) · |x|−1‖g‖C1 , also for 1 < |x| < 2.

3.4 A linear, first order PDE.

Motivated by (3.1), given a vector field v = (v1, v2), in this subsection we study the moregeneral equation

(v(x)− µx) · ∇u = u. (3.27)

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We seek a solution to (3.27) on the outer domain D1 = {x ∈ R2 ; |x| ≥ 1}, with asymptoticbehavior given by

limr→+∞

r1/µu(r, θ) = ω(θ). (3.28)

The linear, first order equation (3.27) can be solved by the classical method of characteristics[19, 22]. This yields the system {

ẋ = v(x)− µx ,

U̇ = U .(3.29)

Throughout the following, we shall assume

|v(x)| ≤ ε1|x|1−1µ , |∇v(x)| ≤ ε1|x|−

1µ , (3.30)

for some ε1 > 0 sufficiently small. It will be convenient to rewrite the above equations in polarcoordinates x = (r cos θ, r sin θ), using r as independent variable. At any point x, denoteby a(x), b(x) the components of v(x) in the radial and angular direction. More preciselyintroducing the unit vector e

.= x/|x| = (cos θ, sin θ), let

v(x) = a(x)e + b(x)e⊥. (3.31)

Using polar coordinates, from (3.29) one obtainsṙ = a(r, θ)− µr,

θ̇ = b(r, θ) r−1,

U̇ = U,

dθ

dr= − b(r, θ)r

−1

µr − a(r, θ),

dU

dr= − U

µr − a(r, θ).

(3.32)

The last two equations in (3.32) can be written in the equivalent formdθ

dr= − r−1−

1µA(r, θ),

dU

dr= − U

µr

(1 + r−1/µB(r, θ)

),

(3.33)

where we defined

A(r, θ).= r

−1+ 1µ

b(r, θ)

µ− r−1a(r, θ), B(r, θ) = r

−1+ 1µ

a(r, θ)

µ− r−1a(r, θ). (3.34)

Notice that the bounds in (3.30) imply{A(r, θ) = O(1) · ε1 ,

B(r, θ) = O(1) · ε1 ,

{Ar(r, θ) = O(1) · ε1 r−1,

Br(r, θ) = O(1) · ε1 r−1,

{Aθ(r, θ) = O(1) · ε1,

Bθ(r, θ) = O(1) · ε1.(3.35)

Fix an angle Θ ∈ [0, 2π]. Given the asymptotic data

limr→+∞

θ(r) = Θ , limr→+∞

r1µU(r) = ω(Θ), (3.36)

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the solution to (3.33) can be obtained as the fixed point of a contractive map. Indeed, givenθ ∈ C0([1,+∞[), consider the Picard operator

(Pθ)(r) .= Θ +∫ +∞r

s−1− 1

µA(s, θ(s)) ds . (3.37)

Consider any two functions θ, θ̃ ∈ C0([1,+∞[) and set δ .= ‖θ− θ̃‖C0 . Using the bound on Aθgiven in (3.35), for every r ≥ 1 we obtain∣∣(Pθ)(r)− (P θ̃)(r)∣∣ = O(1) · ε1 ∫ +∞

rs−1− 1

µ δ ds = O(1) · ε1µ · δ .

Therefore, for ε1 > 0 small enough, the map (3.37) is a strict contraction in the space C0, witha unique fixed point θ(·). By (3.37), this solution satisfies

|θ(r)−Θ| ≤ µr−1/µ ‖A‖C0 . (3.38)

To solve the second equation in (3.33) we set

Z(r).= r1/µU(r).

This yields

dZ

dr=

1

µr−1+ 1

µU − r1µU

µr

(1 + r−1/µB(r, θ(r))

)= r

−1− 1µB(r, θ(r))

µ· Z . (3.39)

Integrating, one obtains∫ Z(+∞)Z(r)

dZ

Z= −

∫ +∞r

s−1− 1

µB(s, θ(s))

µds ,

Z(r) = Z(+∞) · exp{∫ +∞

rs−1− 1

µB(s, θ(s))

µds

}, (3.40)

and finally

U(r) = exp

{∫ +∞r

s−1− 1

µB(s, θ(s))

µds

}· r−1/µ ω(Θ). (3.41)

Next, we recall that the components of the gradient p = (p1, p2) = (ux1 , ux2) satisfy thecharacteristic equations

ṗj = −∑i

vi,xjpi + (1 + µ)pj .

Calling η = |p|, using r as independent variable, and integrating along a characteristic, wenow obtain

−dη(r)dr

≤ |Dv(r)|+ 1 + µµr − a(r, θ)

η(r).

Hence, for any r2 > r1 ≥ 1, one has

η(r2)

η(r1)≥ exp

{∫ r2r1

1 + µ+O(1) · ε1r−1/µ

µr − ε1r1−1µ

dr

}= exp

{1 + µ

µ

∫ r2r1

(r−1 +O(1)ε1r−1−

1µ

)dr

}.

(3.42)

13

η(r1) ≤[η(r2)r

1+ 1µ

2

]· r−1− 1

µ

1 ·(

1 +O(1) · ε1r−1/µ1).

Setting r = r1 and letting r2 → +∞, by the asymptotic condition (3.28) we obtain

|p(r)| = η(r) = O(1) · ‖ω‖C1 r−1− 1

µ . (3.43)

The next lemma collects the main properties of the solution u of (3.27)-(3.28). Moreover,given a second vector field ṽ, we estimate the difference between u and the solution ũ of

(ṽ(x)− µx) · ∇ũ = ũ, (3.44)

with the same asymptotic condition (3.28)

Lemma 3.2 (asymptotic estimates). Let v, ṽ : D1 7→ R2 be C1 vector fields satisfying thebounds

|v(x)|, |ṽ(x)| ≤ ε1 |x|1−1µ , |Dv(x)|, |Dṽ(x)| ≤ ε1|x|−

1µ , (3.45)

|v(x)− ṽ(x)| ≤ ε2 |x|−1, (3.46)

for some constants µ > 12 and ε1, ε2 > 0, and for all |x| ≥ 1. Moreover, assume that ω ∈ C1(T).

(i) If ε1 > 0 is sufficiently small, then the equation (3.27) has a unique solution withasymptotic behavior (3.28). This solution satisfies the bounds

|u(x)| ≤ 2 ‖ω‖C0 · |x|− 1µ , (3.47)

|∇u(x)| ≤ 3 ‖ω‖C1 · |x|−1− 1

µ . (3.48)

(ii) If ũ is a solution to (3.44) satisfying the same asymptotic conditions (3.28), then (3.46)implies

|u(x)− ũ(x)| ≤ O(1) · ε2 ‖ω‖C1 · |x|−2− 1

µ . (3.49)

Proof. 1. For each Θ ∈ T, call

r 7→ θ(r,Θ), r 7→ U(r,Θ)

the corresponding solution to the characteristic system of ODEs (3.33) with asymptotic con-ditions (3.36). By (3.37) we have

∂

∂Θθ(r,Θ) = 1 +O(1) · ε1 r−1/µ.

Hence, if ε1 > 0 is small enough, for every r ≥ 1 the map Θ 7→ θ(r,Θ) is a bijection from Tinto itself. In particular, for every angle ϑ ∈ T there exists a unique Θ = Θ(r, ϑ) such thatθ(r,Θ) = ϑ. The desired solution, obtained by the method of characteristics, is thus

u(r, ϑ) = U(r,Θ(r, ϑ)).

14

From (3.41) and the bound on B in (3.35) it follows

|U(r,Θ)| =(1 +O(1) · ε1 r−1/µ

)r−1/µω(Θ). (3.50)

This immediately implies (3.47). In turn, (3.48) is a consequence of (3.43).

2. To prove (3.49), we first estimate the difference between the corresponding solutions of thecharacteristic system of ODEs. The assumption (3.46) implies

|a(r, θ)− ã(r, θ)| ≤ ε2r−1 , |b(r, θ)− b̃(r, θ)| ≤ ε2r−1,

hence|A(r, θ)− Ã(r, θ)|+ |B(r, θ)− B̃(r, θ)| = O(1) · ε2r−2+

1µ .

Using this bound in (3.37) and (3.41), we obtain

|θ(r,Θ)− θ̃(r,Θ)| = O(1) · ε2 r−2, (3.51)

|U(r,Θ)− Ũ(r,Θ)| = O(1) · ε2 r−2−1µ ‖ω‖C0 . (3.52)

In addition to (3.52), we now use the bound (3.48) on the gradient ∇u together with (3.51),and obtain

|u(r, ϑ)− ũ(r, ϑ)| = O(1) · ε2 r−2−1µ ‖ω‖C1 .

This establishes the inequality (3.49).

The following lemma provides an error estimate, bounding the difference between an approx-imation u] and the exact solution u of (3.27)-(3.28).

Lemma 3.3 (error estimates). Let v : D1 7→ R2 be a C1 vector field satisfying the assump-tions in (3.45), for some ε > 0 sufficiently small. Let u] ∈ C0(D1) be a function which satisfiesthe asymptotic condition (3.28) and the approximate equation

(v − µx) · ∇u] = u] + e(x), |e(x)| ≤ ε4 |x|−α (3.53)

for some ε4, α > 0. Then the difference between u] and the exact solution u of (3.27) can be

estimated as|u(x)− u](x)| = O(1) · ε4 |x|−α−

2µ . (3.54)

Proof. For a fixed Θ ∈ T, let r 7→ U(r) = u(t, θ(r)) be the functions considered in (3.41). Todetermine the corresponding function U ](r) = u](r, θ(r)), we define Z](r)

.= r1/µU ](r). The

equations in (3.39)–(3.41) are now replaced by

U̇ ] = U ] + e,dU ]

dr= −U

] + e

µr

(1 + r−1/µB(r, θ)

),

dZ]

dr=

1

µr−1+ 1

µU ]−r1µU ] + e

µr

(1 + r−1/µB(r, θ(r))

)= r

−1− 1µB(r, θ(r))

µ·Z]+e](r), (3.55)

wheree](r) = r

−1− 1µ

(1 + r−1/µB(r, θ(r))

) eµ

= O(1) · ε4 r−1−α−2µ .

15

Integrating, one obtains

Z](r) = Z](+∞) · exp{∫ +∞

rs−1− 1

µB(s, θ(s))

µds

}+O(1) · ε4 r−α−

1µ , (3.56)

and finally

U ](r) = exp

{∫ +∞r

s−1− 1

µB(s, θ(s))

µds

}· r−1/µ ω(Θ) +O(1) · ε4 r−α−

2µ . (3.57)

Comparing (3.57) with (3.41) we obtain (3.54).

3.5 An approximating sequence.

A solution to the system (3.1) on the domain D1, with asymptotic and boundary conditions(3.16), will be obtained as the limit of a sequence of approximations (Ωn,Ψn). Assume thatan initial guess (Ω0,Ψ0) is available, which satisfies the conditions (3.16) and provides anapproximate solution to (3.1), namely∣∣∣(∇⊥Ψ0 − µx) · ∇Ω0 − Ω0∣∣∣ = O(1) · |x|−2− 1µ , ∆Ψ0 = Ω0 . (3.58)We then set v0 = ∇⊥Ψ0 and construct a sequence of approximations (Ωn,Ψn,vn)n≥1 suchthat

(vn−1 − µx) · ∇Ωn = Ωn ,

∆Ψn = Ωn ,

vn = ∇⊥Ψn .

(3.59)

At each step, the above equations for Ωn,Ψn are solved with the same asymptotic and bound-ary conditions as in (3.16).

Inductive estimates on the functions Ωn will be obtained by applying Lemma 3.2 to the vectorfields vn. In turn, the functions Ψn can be recovered by

Ψn(x) = Ψn−1(x) +1

2π

∫|y|>1

ln

(1

|y|· |y − x||y∗ − x|

)(Ωn(y)− Ωn−1(y)) dy. (3.60)

Assuming that the asymptotic and boundary data ω, ψ, and ψ0 are sufficiently small, we willshow that the above sequence of approximations is convergent.

3.6 A leading order approximation.

To start the inductive procedure (3.59), it would be natural to choose

Ω0(r, θ) = r− 1µ ω(θ), Ψ0 = r

2− 1µ ψ(θ), (3.61)

so that

−µx · ∇Ω0 = Ω0 , ∆Ψ0 = Ω0 , |v0| = |∇⊥Ψ0| = O(1) · r1−1µ . (3.62)

16

By (3.61)-(3.62) it follows

(v0 − µx) · ∇Ω0 − Ω0 ≤ |∇Ψ0| |∇Ω0| = O(1) · |x|−2µ . (3.63)

As in (3.59), let Ω1 be the solution to

(v0 − µx) · ∇Ω1 = Ω1 ,

with asymptotic condition given at (3.16). In view of (3.63), the error estimate in Lemma 3.3with α = 2/µ yields ∣∣Ω1(x)− Ω0(x)∣∣ = O(1) · |x|− 4µ . (3.64)If µ < 2, then for n = 1 the integral in (3.60) is absolutely convergent, and the inductiveprocedure can begin.

However, this is not the case if µ > 2. To ensure that, when n = 1, the integral in (3.60)is absolutely convergent, we thus need to find a better approximation (Ω0,Ψ0). Namely, weshall require that {

(∇⊥Ψ0 − µx) · ∇Ω0 = Ω0 +O(1) · |x|−α,

∆Ψ0 = Ω0 .(3.65)

for some α > 0 such that α+ 2µ > 2. To construct such an approximation, we use separationof variables and try with a finite sum of the form

Ω0(r, θ) =

N0∑j=1

r− jµωj(θ) +

N1∑k=1

r−1− k

µ ω̃k(θ), (3.66)

Ψ̂0(r, θ) =

N0∑j=1

r2− j

µψj(θ) +

N1∑k=1

r1− k

µ ψ̃k(θ), (3.67)

choosingN0, N1 ∈ N to be the smallest integers such that (3.18) holds. Setting e.= (cos θ, sin θ),

we compute∇(rαω(θ)) = rα−1[αω(θ) e + ω′(θ) e⊥] ,

∇⊥(rαψ(θ)) = rα−1[−ψ′(θ) e + αψ(θ) e⊥] ,

∆(rαψ(θ)) = rα−2[α2ψ(θ) + ψ′′(θ)] .

(3.68)

Using the identities (3.68), from (3.66)-(3.67) we thus obtain

∇Ω0(r, θ) =∑j≥1

r−1− j

µ

[− jµωj(θ)e + ω

′j(θ)e

⊥]

+∑k≥1

r−2− k

µ

[(− 1− k

µ

)ω̃k(θ)e + ω̃

′k(θ)e

⊥],

(3.69)

−µx · ∇Ω0 =∑j≥1

r− jµ · j ωj(θ) +

∑k≥1

r−1− k

µ (µ+ k)ω̃k(θ), (3.70)

∇⊥Ψ̂0(r, θ) =∑j≥1

r1− j

µ

[ψ′j(θ)e +

(2− j

µ

)ψj(θ)e

⊥]

+∑k≥1

r− kµ

[ψ̃′k(θ)e +

(1− k

µ

)ψ̃k(θ)e

⊥],

(3.71)

17

∆Ψ̂0(r, θ) =∑j≥1

r− jµ

[(2− j

µ

)2ψj(θ) + ψ

′′j (θ)

]+∑k≥1

r−1− k

µ

[(1− j

µ

)2ψ̃j(θ) + ψ̃

′′j (θ)

].

(3.72)

As r → +∞, the asymptotic conditions (3.16) imply

ω1 = ω, ψ1 = ψ.

We need to determine all further terms in the expansions (3.66)-(3.67). Notice that, if ωj , ω̃kare known, then the equation ∆Ψ0 = Ω0 uniquely determines the functions ψj , ψ̃k. Indeed,

ψ′′j (θ) +(

2− jµ)2ψj(θ) = ωj(θ),

ψ̃′′k(θ) +(

1− jµ)2ψ̃k(θ) = ω̃k(θ).

(3.73)

We assume here that µ, 2µ /∈ N, hence all the above equations have a unique solution.

The functions ωj , ω̃k will be determined by an inductive procedure, based on the first identityin (3.65). Assume that ωj , ω̃k have already been determined for all integers j < j

∗ and k < k∗,in such a way that (3.73) holds and moreover

(∇⊥Ψ̂0 − µx) · ∇Ω0 − Ω0 = o(|x|−γ

)for all γ < min

{j∗

µ, 1 +

k∗

µ

}.

Two cases must be considered.

CASE 1: j∗

µ < 1 +k∗

µ .

We can then uniquely determine the function ωj∗ by imposing that

(∇⊥Ψ̂0 − µx) · ∇Ω0 − Ω0 = o(|x|−j∗/µ). (3.74)

Indeed, inserting (3.66), (3.70),(3.71) in (3.74) and collecting terms of order r−j∗/µ, we obtain

(j∗ − 1)ωj∗(θ) = Fj∗(θ), (3.75)

where the function Fj∗(θ) is computed in terms of the previous functions ωj , ω̃k and their firstderivatives ω′j , ω̃

′k , with j < j

∗ and k < k∗.

CASE 2: 1 + k∗

µ <j∗

µ .

In this case we can uniquely determine the function ω̃k∗ by imposing that

(∇⊥Ψ̂0 − µx) · ∇Ω0 − Ω0 = o(|x|1−(k∗/µ)). (3.76)

Indeed, inserting (3.66), (3.70),(3.71) in (3.74) and collecting terms of order r1−(k∗/µ), we

obtain(µ+ k∗ − 1) ω̃k∗(θ) = Gk∗(θ), (3.77)

where the function Gk∗(θ) is computed in terms of the previous functions ωj , ω̃k, ω′j , ω̃′k, with

j < j∗ and k < k∗.

By induction, all functions ωj , ω̃k can thus be uniquely determined in a finite number of steps.

18

We observe that, at each step, the formula for ωj∗ or ω̃k∗ contains the first order derivativesof the previous functions ωj or ω̃k with j < j

∗ and k < k∗. We thus have an estimate of theform

max1≤j≤N0

‖ωj‖C1(T) + max1≤k≤N1

‖ω̃k‖C1(T) ≤ C · ‖ω‖CN (T) , N.= N0 +N1 . (3.78)

Example 3.1 Assume 2 < µ < 52 . By (3.18) we can take N0 = 2, N1 = 0. The approximation thistakes the form

Ω0(r, θ) = r− 1µω1(θ) + r

− 2µω2(θ), Ψ̂0(r, θ) = r2− 1µψ1(θ) + r

2− 2µψ2(θ).

Given the function ω : T 7→ R, we begin by taking ω1 = ω, ψ1 = ψ, where ψ is uniquely determined by(3.5). In turn, ω2 is obtained by imposing the condition

(∇⊥Ψ̂0 − µx) · ∇Ω0 − Ω0 = o(|x|−2/µ).

In view of (3.69)–(3.71), this yields(r1−

1µψ′1 − µr

)[− 1µr−1−

1µω1 −

2

µr−1−

2µω2

]+ r1−

1µ

(2− 1

µ

)ψ1

[r−1−

1µω′1 + r

−1− 2µω′2

]−(r−

1µω1 + r

− 2µω2

)= o(|x|−2/µ).

Notice that terms of order O(1) ·r−1/µ already cancel. Collecting terms of order O(1) ·r−2/µ we obtain

ψ′1ω1 + 2ω2 +

(2− 1

µ

)ψ1ω

′1 − ω2 = 0,

ω2 = − ψ′1ω1 −(

2− 1µ

)ψ1ω

′1 .

Finally, ψ2 is determined by the first equation in (3.73), with j = 2.

Having determined all coefficients in the finite sums (3.66)-(3.67), it now remains to adjustthe boundary condition (3.16) at r = 1. Since in general Ψ̂0(1, θ) 6= ψ0(θ), a correction termmust be added. Recalling (3.23), we thus consider the function

Ψ0(x) = Ψ̂0(x) + Ψ]0(x), Ψ

]0(x)

.=|x|2 − 1

2π

∫ 2π0

ψ0(θ)− Ψ̂0(1, θ)|x1 − cos θ|2 + |x2 − sin θ|2

dθ.

(3.79)

Notice that the above definition implies ∇⊥Ψ]0(x) = O(1) · |x|−1. Hence

∇⊥Ψ]0(x) · ∇Ω0(x) = O(1) · |x|−2− 1

µ . (3.80)

3.7 Convergence of the approximating sequence.

Having determined the functions Ω0,Ψ0, Ψ̂0 as in (3.66)-(3.67) and (3.79), we now define

v0(x).= ∇⊥Ψ0(x). (3.81)

We then construct the sequence (Ωn,Ψn,vn)n≥1 according to (3.59), always with asymptoticand boundary conditions (3.16). The following analysis will establish

19

• a priori bounds in a strong norm,

• convergence estimates in a weaker norm.

Consider the spaces

X.={

Ω ∈ C0(D1) ; ‖Ω‖X.= sup

x|x|2+

1µ |Ω(x)| < +∞

},

Y.={

v ∈ C0(D1;R2) ; ‖v‖Y.= sup

x|x| |v(x)| < +∞

}.

Recalling (3.78), by induction we will establish the bounds

|Ωn(x)| = O(1) · |x|−1/µ(‖ψ0‖C1 + ‖ω‖CN

), (3.82)

|∇Ωn(x)| = O(1) · |x|−1−1µ(‖ψ0‖C1 + ‖ω‖CN

), (3.83)

‖Ωn+1 − Ωn‖X = O(1) · ‖vn − vn−1‖Y · ‖ω‖CN , (3.84)

|vn(x)| = O(1) · |x|1−1µ(‖ψ0‖C1 + ‖ω‖CN

), (3.85)

|Dvn(x)| = O(1) · |x|−1/µ(‖ψ0‖C1 + ‖ω‖CN

), (3.86)

‖vn − vn−1‖Y = O(1) · ‖Ωn − Ωn−1‖X . (3.87)

1. We begin by observing that, by the construction of Ω0,Ψ0,v0 at (3.66)-(3.67), (3.79), and(3.81), the above estimates (3.82), (3.83), (3.85), and (3.86) hold true when n = 0.

2. To estimate the difference Ω1 −Ω0, we observe that Ω0 satisfies the approximate equation

(v0 − µx) · ∇Ω0 = Ω0 +O(1) · |x|−2−�(‖ψ0‖C1 + ‖ω‖CN

),

where � is the constant in (3.19). We compare Ω0 with the exact solution Ω1 of

(v0 − µx) · ∇Ω1 = Ω1 ,

always with the same asymptotic condition (3.16). The error estimate proved in Lemma 3.3yields

Ω1(x)− Ω0(x) = O(1) · |x|−2−�(‖ψ0‖C1 + ‖ω‖CN

), (3.88)

for all x ∈ D1.

In turn, the estimate (3.25) in Lemma 3.1 yields∣∣v1(x)− v0(x)∣∣ = ∣∣∇Ψ1(x)−∇Ψ0(x)∣∣= O(1) · |x|−1 ·

∥∥ |x|2+ε(Ω1 − Ω0)∥∥L∞ = O(1) · |x|−1(‖ψ0‖C1 + ‖ω‖CN) . (3.89)

3. The remainder of the proof is achieved by induction. By (3.89), the estimate (3.49) inLemma 3.2 yields∣∣Ω2(x)−Ω1(x)∣∣ = O(1)·|x|−2− 1µ ·∥∥ |x|(v2−v1)∥∥L∞ ≤ C1 |x|−2− 1µ(‖ψ0‖C1 +‖ω‖CN) , (3.90)

20

for some constant C1. In turn, the bound (3.25) in Lemma 3.1 yields∣∣∇Ψ2(x)−∇Ψ1(x)∣∣ = O(1) · |x|−1 · ∥∥ |x|2+ε(Ω1−Ω0)∥∥L∞ = O(1) · |x|−1(‖ψ0‖C1 + ‖ω‖CN) .(3.91)

Assuming that Ωn,Ψn satisfy uniform bounds for all n ≥ 2, the bound (3.84) follows from(3.49), while (3.87) is a consequence of (3.25). Notice that, by taking ‖ω‖CN small enough,this implies

‖Ωn+1 − Ωn‖X ≤1

2‖Ωn − Ωn−1‖X . (3.92)

Hence the the sequence (Ωn)n≥1 is absolutely convergent.

4. Assuming that the vector field vn−1 satisfies the uniform bounds as in (3.85)-(3.86), theestimates (3.82)-(3.83) on the solution Ωn of

(vn−1 − µx) · ∇Ωn = Ωn , limr→∞

r1/µΩn(r, θ) = ω(θ),

follow from (3.47)-(3.48) in Lemma 3.2. It thus remains to give a proof of (3.85)-(3.86).

5. The bound (3.85) follows from Lemma 3.1 and the formula

|vn(x)| ≤ |v1(x)|+1

2π

∫|y|>1

{1

|x− y|+

1

|x− y∗|

} ∣∣Ωn(y)− Ω0(y)∣∣ dy. (3.93)Indeed, by (3.90) and (3.92) one has∣∣Ωn(x)− Ω1(x)∣∣ ≤ 2C1 |x|−2− 1µ(‖ψ0‖C1 + ‖ω‖CN) . (3.94)6. To prove (3.86), we write vn = v1 + wn, where

wn(x).= ∇⊥

(1

2π

∫|y|>1

ln

(1

|y|· |y − x||y∗ − x|

)(Ωn(y)− Ω0(y)

)dy

). (3.95)

The Jacobian matrix Dwn(x) can be estimated in terms of the Hessian matrix of secondderivatives D2φ, where

φ(x) = Ψn −Ψ0 =1

2π

∫|y|>1

ln

(1

|y|· |y − x||y∗ − x|

)(Ωn(y)− Ω0(y)

)dy (3.96)

is the solution to {∆φ = Ωn − Ω0 , |x| > 1,

φ(x) = 0, |x| = 1.(3.97)

A bound on Dwn(x) will be obtained using Schauder’s regularity estimates on φ.

For any ρ ≥ 3 consider the annular region

Γρ.={x ∈ R2 ; ρ

3≤ |x| ≤ 3ρ

}.

21

Using the rescaled variable y = ρx, and setting φ̂(y) = φ(ρy), one obtains an equation on thefixed domain Γ1, namely

∆φ̂(y) = ρ2(Ωn(ρy)− Ω0(ρy)

) .= f(y). (3.98)

We observe that, by (3.66) and (3.83), the right hand side of (3.98) satisfies the estimates

|f(y)| = O(1) · ρ2∣∣Ωn(ρy)− Ω0(ρy)∣∣ = O(1) · ρ2 · ρ−2− 1µ = O(1) · ρ−1/µ. (3.99)

In addition, by (3.83), the gradient of Ωn − Ω0 over Γρ can be bounded as

supx∈Γρ

|∇Ωn(x)−∇Ω0(x)| = O(1) · ρ−1−1µ .

In turn, this implies

|∇yf(y)| = O(1) · ρ3 · ρ−1−1µ . (3.100)

Together, (3.99) and (3.100) imply that, for any y, y′ ∈ Γ1,

|f(y)− f(y′)| ≤ O(1) ·min{|y − y′| · ρ2−

1µ , ρ−1/µ

}= O(1) · |y − y′|

12µ . (3.101)

By (3.99) and (3.100), the right hand side of (3.98) is bounded and Hölder continuous, uni-formly w.r.t. ρ. Moreover, the function φ̂ satisfies uniform bounds for |y| = 1/3 and |y| = 3.Schauder’s interior regularity estimates yield

|D2yφ̂(y)| ≤ κ4 ,1

2< |y| < 2, (3.102)

for some constant κ4 independent of ρ. Returning to the original coordinates, for any ρ ≥ 3we obtain

|D2φ(x)| ≤ κ4|x|−2 ,ρ

2< |x| < 2ρ. (3.103)

To complete the proof, it now remains to prove that (3.103) holds also for 1 < |x| < 2/3,possibly with a larger constant κ4. But this is an immediate consequence of Schauder’sboundary regularity estimates, because φ(x) = 0 for |x| = 1.

7. By the previous estimates, we have the convergence ‖Ωn − Ω‖X → 0 and ‖vn − v‖Y → 0,as n→∞. In particular, this implies the uniform convergence of the functions Ωn and vn onthe outer domain D1. Since Ωn is recovered from vn−1 by solving the PDE in (3.59), whileΨn − Ψ0 is recovered by (3.96), we conclude that (Ω,Ψ) provide a solution to the problem(3.1), (3.16). This completes the proof of Theorem 3.1.

Remark 3.3 (Dependence on the boundary data). It is of interest to see how the solution(Ω,Ψ) depends on the data ψ0 in (3.16). In the special case where the asymptotic data ω ≡ 0, accordingto (3.23) the explicit solution is

Ψ(x) =|x|2 − 1

2π

∫|y|=1

ψ0(y)

|y − x|2dσ, Ω(x) = 0. (3.104)

In general, for ω small, a good approximation to the differential of the solution map ψ0 7→ (Ω,Ψ) isprovided by differential of the first iteration of our scheme (3.59).

22

4 Solutions on an inner domain

In the remainder of this paper we consider the problem of constructing solutions to (2.9) ona small disc

BR.= {x ∈ R2 ; |x| < R},

with boundary data {Ω(R, θ) = ω(θ),

Ψ(R, θ) = ψ(θ).(4.1)

Remark 4.1 Throughout the following, we assume that the integral curves of the pseudo-velocityq(y) = ∇⊥Ψ(y)− µy are spirals winding around the origin. We notice that, by adding an appropriatelinear function to Ψ in (2.9), one may shift the centers of these spirals to an arbitrary point in the

plane. More generally, this shift can be achieved by adding a function which is linear on a large disc,

and satisfies a suitable decay estimate as |x| → +∞. In the sequel, we will thus restrict our attentionto the case of spirals centered at zero. Solutions on more general inner domains, as shown in Fig. 4,

right, can be constructed in a similar way.

q

φ

β

(β,φ)

1R0

y

y

y( )_

φq

Figure 5: Construction of the adapted coordinate system. When β = 0, one starts by definingy(0, φ) = ȳ(φ)

.= (R cosφ,R sinφ). For β > 0, the point y(β, φ) lies at the intersection of the pseudo-

streamline through ȳ(φ) with the ray making an angle φ+ β with the y1-axis.

4.1 Adapted coordinates.

We regard the first equation in (2.9) as a linear, first order equation for Ω. The characteristiccurves, i.e. the pseudo-streamlines, are the integral curves of the vector field q = ∇⊥Ψ− µy.Following [16, 17], it will be convenient to make a change of variables (y1, y2) 7→ (β, φ), suchthat (see Fig. 5)

• Pseudo-streamlines have equation φ = constant.

• β = θ − φ, where r = |y|, θ = ] y are the standard polar coordinates on R2.

23

• For fixed φ we have

limβ→0+

r(β, φ) = R , limβ→0+

θ(β, φ) = φ . (4.2)

We calculate the left hand side of (2.9) in terms of the new variables (β, φ).

(∇⊥Ψ(y)− µy) · ∇Ω = (−Ψy2 − µy1)Ωy1 + (Ψy1 − µy2)Ωy2

=[− (Ψββy2 + Ψφφy2)− µy1)

](Ωββy1 + Ωφφy1)

+[(Ψββy1 + Ψφφy1)− µy2

](Ωββy2 + Ωφφy2)

=[(βy1φy2 − βy2φy1)Ψβ − µ(y1φy1 + y2φy2)

]Ωφ

+[(βy2φy1 − βy1φy2)Ψφ − µ(y1βy1 + y2βy2)

]Ωβ .

(4.3)

If {φ = constant} is the equation of a pseudo-streamline, then in the above expression thecoefficient of Ωφ should vanish. Namely

0 = (βy1φy2 − βy2φy1)Ψβ − µ(y1φy1 + y2φy2). (4.4)

We now observe that the 2× 2 Jacobian matrix of the variable transformation satisfies[y1,β y1,φy2,β y2,φ

]=

[βy1 βy2φy1 φy2

]−1=

1

βy1φy2 − βy2φy1·[

φy2 −βy2−φy1 βy1

]. (4.5)

In view of (4.5), from (4.4) it follows

Ψβ + µ y × yβ = Ψβ + µ(y1y2,β − y2y1,β) = 0. (4.6)

Using the variablesθ(β, φ) = β + φ, y = y(r, θ), (4.7)

we can writeyβ = yrrβ + yθθβ ,

y × yβ = (y × yr)rβ + (y × yθ)θβ = |y|2. (4.8)

Together with (4.6), this yieldsΨβ = − µ |y|2. (4.9)

Using polar coordinates, the map (β, φ) 7→ (r, θ) thus takes the form{r = (−Ψβ/µ)1/2,

θ = β + φ.(4.10)

It will be convenient to use the auxiliary variable ` = ln r, so that ` =12 ln(−Ψβ/µ) ,

θ = β + φ.(4.11)

24

For future use, we also introduce the differential operator

∂ϕ = ∂φ − ∂β, (4.12)

describing a directional derivative in the radial direction, i.e. with θ = constant.

We now compute the Jacobian matrix of the transformation (4.11) and its inverse.

J.=

(`β `φ

θβ θφ

)=

Ψββ2Ψβ Ψβφ2Ψβ1 1

, (4.13)J−1 =

(β` βθ

φ` φθ

)=

1

`ϕ

(−1 `φ

1 −`β

). (4.14)

Notice that the Jacobian determinant is

det J = `β − `φ =Ψββ −Ψβφ

2Ψβ

.= −

Ψβϕ2Ψβ

= − `ϕ . (4.15)

Moreover, by (4.10)-(4.11), the determinant of the coordinate transformation (β, φ) 7→ (y1, y2)is computed by

det

(y1,β y1,φ

y2,β y2,φ

)= r det

(rβ rφ

θβ θφ

)= r r` det

(`β `φ

θβ θφ

)

= r2Ψβφ −Ψββ−2Ψβ

=Ψβϕ2µ

.

(4.16)

For future use we observe that, for any function f , one has

`ϕ fθ = `ϕfφ − `φfϕ = (`ϕf)φ − (`φf)ϕ . (4.17)

We also observe that

∂θ = βθ ∂β + φθ ∂φ =1

`ϕ

(`φ∂β − `β∂φ

)=

1

`ϕ

(`φ∂φ− `φ∂ϕ− `β∂φ

)= ∂φ−

`φ`ϕ∂ϕ . (4.18)

Therefore, partial derivatives w.r.t. `, θ can be expressed as∂` =

1

`ϕ∂ϕ ,

∂θ = ∂φ −`φ`ϕ∂ϕ .

(4.19)

4.2 The equation in the new coordinates.

Adopting a notation from differential geometry, we denote by t 7→ exp(tq)(y0) the solution to

ẏ = q(y), y(0) = y0 . (4.20)

25

Here q(y) = ∇⊥Ψ(y) − µy is the pseudo-velocity, introduced at (2.11). By construction, thepoint y(β, φ) lies on the integral curve of q starting from ȳ(φ) = (R cosφ,R sinφ). Hence

y(β, φ) = exp(τq)(ȳ(φ)) (4.21)

for some τ = τ(β). The first equation in (2.9) thus leads to

Ω(β, φ) = eτ(β) Ω(0, φ). (4.22)

The time τ(β) can now be estimated by comparing the area elements at the two points ȳ andy. Namely, with reference to Fig. 6, we claim that

e−2µτ(β) =

(yφ × (∇⊥Ψ− µy)

)(y(β, φ))(

yφ × (∇⊥Ψ− µy))

(ȳ(φ)). (4.23)

To prove (4.23), we rely on the identities

ẏ = yβ β̇ = ∇⊥Ψ− µy, (4.24)

div (∇⊥Ψ− µy) ≡ − 2µ , (4.25)d

dt

(yφ × yββ̇

)= − 2µ

(yφ × yββ̇

). (4.26)

Integrating (4.26) over the time interval [0, τ(β)] and using (4.24) and (4.21), we obtain (4.23).

Observing thatyφ ×∇⊥Ψ = y1,φΨy1 + y2,φΨy2 = Ψφ

and using (4.22)-(4.23), one obtains

Ω(β, φ) = eτ(β) Ω(0, φ) =(e−2µτ(β)

)− 12µ ω(φ)

=

(

Ψφ + µ y × yφ)

(β, φ)(Ψφ + µ y × yφ

)(0, φ)

−12µ

ω(φ) ,

(4.27)

where ω is the boundary value prescribed at (4.1). When β = 0, by (4.1) one has(Ψφ + µ y × yφ

)(0, φ) = ψ

′(φ) + µR2. (4.28)

Using the relations

y × yφ = r2 θφ = −Ψβµ,

we can write (4.27) in the form

Ω(β, φ) = (Ψφ −Ψβ)−12µ ω∗(φ) = Ψ

− 12µ

ϕ ω∗(φ). (4.29)

Here, in view of (4.28), we defined

ω∗(φ).=(ψ′(φ) + µR2

) 12µω(φ). (4.30)

26

(τ) yy_

φ = constant

= constantβ

Figure 6: By (4.25), along any trajectory ẏ = ∇⊥Ψ(y) − µy, the area element decreases at anexponential rate. Namely, dA(y(τ)) = e−2µτdA(ȳ). This motivates the formula (4.27).

It remains to write the second equation in (2.9) in terms of the coordinates β, φ. Writing theLaplace operator in polar coordinates, one obtains

Ψrr +Ψrr

+Ψθθr2

= Ω. (4.31)

Recalling that ` = log r, this can be written as

`ϕ(Ψ`` + Ψθθ) = r2 `ϕ Ω. (4.32)

Using (4.19)-(4.17) we compute

Ψ`` + Ψθθ = ∂`

[1

`ϕΨϕ

]+ ∂θ

[Ψφ −

`φ`ϕ

Ψϕ

], (4.33)

`ϕ · (Ψ`` + Ψθθ) =(

Ψϕ`ϕ

)ϕ

+

[`ϕ

(Ψφ −

`φ`ϕ

Ψϕ

)]φ

−[`φ

(Ψφ −

`φ`ϕ

Ψϕ

)]ϕ

=

[(1 + (`φ

)2)Ψϕ`ϕ− `φΨφ

]ϕ

+[`ϕΨφ − `φΨϕ

]φ.

(4.34)

Using (4.34), (4.29), and the first identity in (4.10), from (4.32) one obtains[(1 + (`φ

)2)Ψϕ`ϕ− `φΨφ

]ϕ

+[`ϕΨφ − `φΨϕ

]φ

= −Ψβµ`ϕΨ

− 12µ

ϕ ω∗(φ) . (4.35)

Replacing the expressions `φ, `ϕ by their values in terms of the partial derivatives of Ψ, givenat (4.13), as in [16, 17] one eventually obtains the non-linear PDE[(

1 +(Ψβφ

2Ψβ

)2) 2ΨβΨϕΨβϕ

−ΨβφΨφ

2Ψβ

]ϕ

+

[ΨβϕΨφ −ΨβφΨϕ

2Ψβ

]φ

= −Ψβϕ2µ

Ψ− 1

2µϕ ω

∗(φ) . (4.36)

Since the variable φ represents an angle, a solution of (4.36) will be defined for β > 0, φ ∈ R.It will be 2π-periodic w.r.t. the variable φ.

In view of (4.1) and (4.9), the third order equation (4.36) should be supplemented by the twoboundary conditions

Ψ(0, φ) = ψ(φ),

Ψβ(0, φ) = − µR2.(4.37)

Notice that the first boundary condition in (4.1) does not show up in (4.37). Instead, it is usedin (4.30) to obtain the explicit formula (4.29) for the vorticity in the adapted coordinates.

27

ϕ

φβ

φ

β

2π

0

Figure 7: The coordinates β, φ and the vector field ∂ϕ.= ∂φ − ∂β .

4.3 Recovering the solution in the original coordinates.

As soon as the stream function Ψ = Ψ(β, φ) of (4.36) has been constructed as a function of theauxiliary variables β, φ, one can recover Ψ in terms of the polar coordinates r, θ by inverting(4.10). Namely,

Ψ(r(β, φ), θ(β, φ)

)= Ψ(β, φ). (4.38)

Here one has to check that (4.2) holds and that the determinant of the coordinate transfor-mation at (4.16) does not vanish. In our context, this will be justified by the conclusion ofTheorem 7.1.

y1

2y

θ

Figure 8: The images of the coordinate lines. The blue spiral corresponds to {φ = constant}, whileclosed red curves correspond to {β = constant}. Finally, the ray θ = constant corresponds to the line{ϕ .= φ− β = constant}.

Remark 4.2 At first sight, the equation (4.36) looks much harder to solve than the third orderequation (2.12) or the equivalent system (2.9). The key advantage offered by (4.36) is that, at least insome cases, it allows us to resolve the singularity near the spiral’s center. In other words, while thesolution Ψ(y) loses regularity near the center of the vortex spiral, in terms of the coordinates (β, φ) weexpect that the same function Ψ will remain uniformly smooth.

Due to its complexity, the equation (4.36) will be solved by a perturbation argument. Starting from a

radially symmetric solution, we first solve the linearized equation describing a first order perturbation.

28

In a second step, using the implicit function theorem, we obtain a solution to the original nonlinear

problem. As shown in [16, 17], this approach can be successful as long as the solutions remain sufficiently

close to a radially symmetric one.

5 Radially symmetric solutions

Radially symmetric solutions of (2.9), defined for all r ≥ 0, can be easily constructed. Asin [16, 17], such solutions play a fundamental role in our analysis. Indeed, the solutionswe construct on the inner domain are obtained by suitably small perturbations of a radiallysymmetric solution.

In the radially symmetric case where Ω = Ω(r), the first equation in (2.9) reduces to

−µrΩr = Ω. (5.1)

Hence the vorticity has the form

Ω(r) = c0 r− 1µ (5.2)

for some constant c0. In turn the second equation yields

Ψrr +Ψrr

= c0r− 1µ , (5.3)

(rΨr)r = c0r1− 1

µ ,

rΨr = c0

(2− 1

µ

)−1r

2− 1µ .

Assuming µ > 1/2, the stream function is thus computed as

Ψ(r) = c0

(2− 1

µ

)−2r

2− 1µ . (5.4)

Finally, the identity U = ∇⊥Ψ implies that the velocity satisfies

|U | = |Ψr| = c0(

2− 1µ

)−1r

1− 1µ , (5.5)

hence

U(x) = c0

(2− 1

µ

)−1|x|−

1µ x⊥. (5.6)

We rewrite this solution in terms of the adapted coordinates β, φ. By radial symmetry itfollows β = β(r). From the identity (4.9) we obtain

(βr)−1Ψr = − µr2,

hence

βr = −Ψrµr2

= − 1µr2

c0

(2− 1

µ

)−1r

1− 1µ = − c0

2µ− 1r−1− 1

µ .

29

This yields

β =c0 µ

2µ− 1r− 1µ . (5.7)

Next, since the angular variable is θ = β+φ, the curves {φ = constant} correspond to algebraicspirals (see Fig. 9):

θ =c0 µ

2µ− 1r− 1µ + φ . (5.8)

Remark 5.1 The present analysis refers to radially symmetric solutions defined over the entire planeR2. For solutions defined on the disc BR = {x ∈ R2 ; |x| < R}, the adapted coordinates β, φ definedin Section 4.1 were chosen so that

r = R =⇒ β = 0, φ = θ.

This only produces a shift in these coordinates, replacing (5.7)-(5.8) with

β =c0 µ

2µ− 1(r−

1µ −R−

1µ), θ =

c0 µ

2µ− 1(r−

1µ −R−

1µ)

+ φ . (5.9)

For convenience, throughout the following we shall use the coordinates (5.7)-(5.8). In particular, onthe inner domain where |y| ≤ R, we have

β ∈ [b,+∞[ , b = c0µ2µ− 1

R−1/µ. (5.10)

The boundary conditions (4.37) become{Ψ(b, φ) = ψ(φ),

Ψβ(b, φ) = − µR2.(5.11)

It is of interest to compute the various quantities appearing in the equations (4.36), in thisradially symmetric case. Setting

κ.=

(c0 µ

2µ− 1

)µ,

one finds

r = κβ−µ,

rϕ = − rβ = µκβ−1−µ,

Ψφ = Ωφ = rφ = 0,

Ω = Ω0 β,

Ψ = µκ2

2µ−1 β1−2µ,

Ψβ = − µr2 = − µκ2 β−2µ,

Ψββ = 2µ2κ2 β−1−2µ.

(5.12)

Remark 5.2 Inserting

Ψβ = − µκ2β−2µ, Ψββ = 2µ2κ2β−1−2µ, Ψφ = 0, Ψϕ = −Ψβ ,

30

in (4.36), one obtains

−[

2(Ψβ)2

Ψββ

]β

+

[1

2µ− 1(−Ψβ)1−

12µΩ

]β

= 0,

−[

2µ2κ4β−4µ

2µ2κ2β−1−2µ

]β

+

[1

2µ− 1(µκ2β−2µ)1−

12µ

]β

Ω0 = 0,

−κ ·[β1−2µ

]β

+(µκ2)1−

12µ

2µ− 1[β1−2µ

]β

Ω0 = 0.

This yields

Ω0 = (2µ− 1)µ12µ−1κ

1µ−1. (5.13)

Remark 5.3 Consider a fluid in steady motion, with velocity U = ∇⊥Ψ independent of time, asin (5.6). Consider all fluid particles that at time t = 0 lie on the positive x1-axis, which In radialcoordinates is

{(r, θ) ; θ = 0, r > 0}.

At time t = 1, these moving particles are located along the spiral{(r, θ) ; θ = |U(r)|/r , r > 0

}.

By (5.5) and (5.8), one checks that this spiral coincides with the spiral{(r(β, φ), θ(β, φ)

); β > 0, φ = 0

}.

Figure 9: The algebraic spiral θ = 32r−4/3 obtained by taking κ = 1, µ = 3/4 in (5.8).

5.1 Rescaled equations

Motivated by (5.12), we expect that our solutions will have size Ψ = O(1) · β1−2µ. In order toconstruct solutions on a domain where β >> 1, again following [16, 17] it is thus convenientto introduce the rescaled variables

Ψ̃ = β2µ−1Ψ . (5.14)

Defining the operatorsA

.= −β∂ϕ , B0

.= β∂β , (5.15)

∂ϕ.= β∂ϕ + 2µ− 1 = −A+ 2µ− 1,

∂β.= β∂β + 1− 2µ = B0 + 1− 2µ ,

(5.16)

31

from (5.14) one obtains

Ψβ = β1−2µΨ̃β + (1− 2µ)β−2µΨ̃

.= β−2µ∂βΨ̃ ,

Ψφ = β1−2µΨ̃φ ,

Ψϕ = β1−2µΨ̃ϕ − (1− 2µ)β−2µΨ̃

.= β−2µ∂ϕΨ̃ ,

Ψβϕ = β−2µ(βΨ̃βϕ + (1− 2µ)(Ψ̃ϕ − Ψ̃β) + (1− 2µ)2µβ−1Ψ̃

).

(5.17)

Using (5.17), we can write (4.36) in the equivalent form

0 = ∂ϕT(ϕ) + ∂φT

(φ) + T (0)ω∗, (5.18)

where

T (ϕ).= 2

1 +(∂β∂φΨ̃2∂βΨ̃

)2× ∂βΨ̃ ∂ϕΨ̃(∂ϕ + 1)∂βΨ̃

−∂φ∂βΨ̃ · ∂φΨ̃

2∂βΨ̃, (5.19)

T (φ).=

(∂ϕ + 1)∂βΨ̃ · ∂φΨ̃− ∂φ∂βΨ̃ · ∂ϕΨ̃2∂βΨ̃

, (5.20)

T (0).=

1

2µ(∂ϕ + 1)∂βΨ̃ · (∂ϕΨ̃)−1/2µ . (5.21)

More explicitly, substituting (5.17) in the third order equation (4.36) and dividing both sidesby β−2µ, we eventually obtain

0 =ω∗

2µ

((1 − 2µ)βΨ̃ϕ + β2Ψ̃βϕ − (1 − 2µ)βΨ̃β + (1 − 2µ)2µΨ̃

)(βΨ̃ϕ − (1 − 2µ)Ψ̃

)−1/2µ+

(((1 − 2µ)βΨ̃ϕ + β2Ψ̃βϕ − (1 − 2µ)βΨ̃β + (1 − 2µ)2µΨ̃

)Ψ̃φ −

((1 − 2µ)Ψ̃φ + βΨ̃βφ

)(βΨ̃ϕ − (1 − 2µ)Ψ̃

)2((1 − 2µ)Ψ̃ + βΨ̃β)

)φ

+(β∂ϕ − (1 − 2µ)

)[2((1 − 2µ)Ψ̃ + βΨ̃β

)(1 +

((1 − 2µ)Ψ̃φ + βΨ̃βφ2((1 − 2µ)Ψ̃ + βΨ̃β)

)2)

× βΨ̃ϕ − (1 − 2µ)Ψ̃(1 − 2µ)βΨ̃ϕ + β2Ψ̃βϕ − (1 − 2µ)βΨ̃β + (1 − 2µ)2µΨ̃

− Ψ̃φ((1 − 2µ)Ψ̃φ + βΨ̃φβ)2((1 − 2µ)Ψ̃ + βΨ̃β)

].

(5.22)

It remains to identify the appropriate boundary conditions for the rescaled equations. Recall-ing (5.10) and (5.14), from (5.11) one obtains Ψ̃(b, φ) = b

2µ−1Ψ(b, φ) = b2µ−1ψ(φ).= ψ̃(φ) ,

−µκ2b−2µ = − µR2 = Ψβ(b, φ) = (1− 2µ)b−2µΨ̃(b, φ) + b1−2µΨ̃β(b, φ).(5.23)

In these boundary conditions, we must guarantee that the second equation is satisfied, as thisis intimately related to the change of variables for the adapted coordinates. The validity ofthe second equation will also be a major concern later on, when we construct our functionspaces. As a consequence, this will require us to give up some flexibility in the choice of ψ̃.

32

We remark that, after the rescaling, all the expressions (5.19)–(5.21) involve only the differ-ential operators ∂φ, ∂β and ∂ϕ. A crucial part of the analysis in Section 6 is that the lasttwo operators have bounded inverses within the space of bounded continuous functions (seePropositions 6.1 and Lemma 7.5). This is because the unbounded part of these operators,namely β∂β and β∂ϕ, couple differentiation with multiplication by β. This will help facilitatethe construction of appropriate function spaces in which solutions will exist.

6 Linearizing around the radially symmetric solution

Always following [16, 17], we now linearize the rescaled equation (5.22) around the constantsolution Ψ̃ ≡ 1. Consider a family of perturbed solutions of the form

Ψ̃� = 1 + � Y (φ, β) + o(�),

Ω� =(2µ− 1)1+

12µ

µ+ � ω(φ) + o(�).

(6.1)

Inserting (6.1) into (5.22) and matching terms of order �, we obtain the linearized equation

0 =ω

2µ

((1− 2µ)2µ

)(2µ− 1)−1/2µ

+(2µ− 1)1+1/2µ

2µ2

(((1− 2µ)(β∂ϕ − β∂β + 2µ)Y + β2Yβϕ

)(2µ− 1)−1/2µ

+(2µ(1− 2µ)

)−12µ

(2µ− 1)−1−1/2µ(βYϕ + (2µ− 1)Y

))

+(1− 2µ)2µYφφ

2(1− 2µ)+

(1− 2µ)((1− 2µ)Yφφ + βYβφφ

)2(1− 2µ)

+2(β∂ϕ − (1− 2µ)

)[ 2µ− 1(1− 2µ)2µ

(β∂β + (1− 2µ))Y +

(1− 2µ)2µ(1− 2µ)

(β∂β − (1− 2µ)

)Y

+(1− 2µ)2 1((1− 2µ)2µ

)2((1− 2µ)2µ+ (1− 2µ)β∂ϕ + β2∂βϕ − (1− 2µ)β∂β)Y].

(6.2)Using the fact that β2Yβϕ = (β∂ϕ + 1)β∂βY , setting

ω̃ = 2µ2(2µ− 1)1−1/2µ ω ,

and performing some simple cancellations, we obtain

ω̃ = (2µ− 1)(β∂ϕ + (1− 2µ)β∂ϕ + (β∂ϕ + 1)β∂β − (1− 2µ)β∂β − (2µ− 1)2

)Y

+µ2(β∂β + 1)Yφφ

+(β∂ϕ − (1− 2µ)

)(2µ((β∂ϕ − (1− 2µ)

)−(β∂β + (1− 2µ)

)+(

(1− 2µ)2µ+ (1− 2µ)β∂ϕ + (β∂ϕ + 1)β∂β − (1− 2µ)β∂β))

Y.

(6.3)

33

We can factor the above operators as

ω̃ = (2µ− 1)(βYϕ + βYβ + (β∂ϕ − (1− 2µ))(β∂β + (1− 2µ))

)Y

+µ2(β∂β + 1)Yφφ

+(β∂ϕ − (1− 2µ)

)((β∂ϕ − (1− 2µ)

)(β∂β + 1)− (2µ− 1)

(β∂β + (1− 2µ)

))Y.

(6.4)Combining terms, one obtains

ω̃ = (2µ− 1)(β∂ϕ + β∂β)Y + µ2∂2φ(β∂β + 1)Y + (β∂ϕ + (2µ− 1))2(β∂β + 1)Y. (6.5)

We thus recover the linearized equation in the same form derived in [16, 17], namely

ω̃ = (2µ− 1)βYφ +(β∂ϕ + (2µ− 1) + iµ∂φ

)(β∂ϕ + (2µ− 1)− iµ∂φ

)(β∂β + 1)Y. (6.6)

Remark 6.1 The equation for the rescaled perturbation is independent of the constant κ in (5.12).

For convenience, in the above we chose κ =(

2µ−1µ

)1/2.

6.1 Fourier transform in φ

We shall compute the solution of (6.6), and later also of the original equation (5.18), in termsof a Fourier series in φ. Writing

Y (β, φ) =∑k∈Z

Yk(β) eikφ, ω̃(φ) =

∑k∈Z

ωk eikφ, (6.7)

from (6.6) we obtain a countable family of decoupled ODEs for the coefficients Yk, namely

A−k A+k B Yk + (2µ− 1)ikβ Yk = ωk . (6.8)

Here A+k , A−k , and B denote first order differential operators:

A−k = β(∂β − ik)− (2µ− 1)− kµ,

A+k = β(∂β − ik)− (2µ− 1) + kµ,

B = β∂β + 1.

(6.9)

We remark that, when k = 0, the equation (6.8) has the simple solution

Y0(β) = (2µ− 1)−2ω0 + c0β−1 , (6.10)

Notice that here we neglect the other two homogeneous solutions, namely β2µ−1 and log(β)β2µ−1,because they are unbounded as β →∞. We observe that, when we invert the rescaling (5.14)and go back to the original variable Ψ, adding these two homogeneous solutions to Ψ̃ amountsto adding constants to Ψ and to the Green’s function of the Laplace operator in R2. Forconvenience, we choose to re-introduce the first of these homogeneous solutions only at thevery end of our analysis.

34

Toward the analysis of the nonlinear problem, we shall need to consider the equation withk = 0 also in the presence of a non constant right hand side. This will be studied togetherwith the cases k 6= 0.

When k 6= 0, a more careful analysis is needed. In the following, we focus on the case k ≥ 1.The analysis for k ≤ −1 is entirely analogous.

Looking at (6.8), it would seem natural to treat the term (2µ− 1)ikβ as a relatively boundedperturbation of the third order operator A−k A

+k B. As shown in [17, 18], this approach works

well for large values of k. However, when k is small, this strategy is not effective. For thisreason, in the following we write the left hand side of (6.8) in a different way, which is effectivefor all µ, k on the domain β ∈ [b,+∞[ , with b large. More precisely, for k 6= 0, we considerthe equivalent linear differential equation

Â+k Â−k BYk + (2µ− 1)(ikβ −B)Yk = ωk , (6.11)

where we let

Â+k.= β(∂β − ik)− a+k , (6.12)

Â−k.= β(∂β − ik)− a−k , (6.13)

a+k.= (2µ− 1) +

√k2µ2 − (2µ− 1), a−k

.= (2µ− 1)−

√k2µ2 − (2µ− 1). (6.14)

Since we are always assuming µ > 1/2, a direct computation yields

0 < a−1 < a+1 , −1 < a

−2 < 0 < a

+2 , (6.15)

a−k+1 < a−k < −1 < 0 < a

+k < a

+k+1 for k ≥ 3. (6.16)

The advantage of this factorization will become apparent in the proof of Proposition 6.2, whereit will achieve a cancellation of certain terms after integration by parts. In the special casek = 0, instead of (6.12)-(6.13) we define

Â+0.= Â−0

.= β ∂β − (2µ− 1).

Notice that, by (6.9), these operators coincide with A−0 and A+0 .

6.2 Solutions to the linearized equation.

The main goal of this section is to establish the solvability of (6.11), within an appropriatefunction space, on the half line β ∈ [b,+∞[ , for some b > 0 suitably large and with suitableboundary conditions. We begin by introducing some appropriate function spaces in which thethree operators Â+k , Â

−k , B, possess a right inverse.

Definition 6.1 For a fixed δ > 0, we define the spaces of continuous functions g : [b,+∞[ 7→ Rby setting

Hδ.={g ∈ C0([b,+∞[ ; ‖g‖δ

.= sup

β∈[b,∞]

∣∣βδg(β)∣∣ < ∞}. (6.17)H+δ

.={g ∈ C0([b,+∞[ ; there exists the limit g∞ = lim

β→∞g(β),

and moreover ‖g‖+δ.= |g∞|+ ‖g − g∞‖δ < ∞

}.

(6.18)

35

In other words, Hδ is the space of all continuous functions which converge to zero at rateg(β) = O(1) · β−δ, while H+δ contains all functions converging to some limit g∞ at rateg(β)− g∞ = O(1) ·β−δ. The form of (6.10) requires that we use H+δ in constructing solutionsfor the k = 0 mode. On the other hand, we shall see that for k 6= 0 we may restrict ourattention to Hδ.

In the same spirit as Proposition 11 in [17], we now establish the right invertibility of theabove operators.

Proposition 6.1 Let 0 < δ < −a−2 . Then the operators defined by[(Â+k )

†g](β) = eikββa

+k

∫ ∞β

s−a+k −1e−iksg(s) ds for k ≥ 1,

[(Â−k )

†g](β) =

eikββa

−k

∫ βbs−a

−k −1e−iksg(s) ds if k ≥ 2,

eikββa−k

∫ ∞β

s−a−k −1e−iksg(s) ds if k = 1,[

B†g](β) = β−1

∫ βbg(s) ds,

are injective operators on Hδ, providing the right inverses of Â+k , Â

−k , and B, respectively.

Moreover, they are bounded linear operators on Hδ, with norms

‖(Â+k )†‖L(Hδ) ≤

1

a+k + δ, (6.19)

‖(Â−k )†‖L(Hδ) ≤

1

−a−k − δif k ≥ 2,

1

a−k + δif k = 1,

(6.20)

‖B†‖L(Hδ) ≤1

1− δ. (6.21)

Furthermore, for any � > 0 and k 6= 0, one has that

‖(Â+k )†‖L(H�) ≤

1

a+k + �, ‖(Â+k )

†1‖H� ≤C

kb−1+�, (6.22)

where C does not depend on k, and 1 denotes the constant function.

Proof: We first show boundedness of the operators. In the case of B†, it is straightforwardto check that

‖B†g‖δ ≤ supββδ−1

∫ βb|g(s)| ds ≤ ‖g‖δ · sup

ββδ−1

∫ βbs−δ ds ≤ ‖g‖δ

1− δ,

where in the second inequality we used the fact that 0 < δ < 1.

Concerning (Â−k )†, for k ≥ 2, recalling that 0 < δ < −a−2 < −a

−k we estimate

‖(Â−k )†g‖δ ≤ sup

β∈[b,∞)βδ+a

−k

∫ βbs−a

−k −1|g(s)| ds

≤ ‖g‖δ supβ∈[b,∞)

βδ+a−k

∫ βbs−a

−k −1−δ ds ≤ ‖g‖δ

−a−k − δ.

36

The same estimate also works for (A+k )† and (A−k )

† (when k = 1), with the additional observa-tion that in this case any choice of 0 < δ is possible. On the other hand, a direct computationshows that these are right inverses, and hence must be invertible.

The last estimate in (6.22), concerning the action of (Â+k )† on constant functions, is achieved

by an integration by parts. This concludes the proof.

Remark 6.2 The previous estimates are valid on any domain [b,+∞[ . Under the above assumptionson δ, in the cases where b shows up in the expressions (namely in (Â−k )

† and in B†), we use the fact

that, after integrating, we obtain a positive power s > 0. Therefore, we can bound |βs − bs| with βs.

We notice that the previous proposition, along with the equation (6.10), directly establishesthe solvability of (6.11) within H+δ .

Next, we define the functionsE0k(β)

.= β−1 for k ∈ Z,

E−k (β).= b−a

−k · β−1

∫ βb e

ikssa−k ds for k ∈ Z, k 6= 0,

E+k (β).= b−a

+k · β−1

∫ βb e

ikssa+k ds for k ∈ Z, k 6= 0.

(6.23)

Remark 6.3 It is straightforward to verify that each of these functions is in the kernel of Â+k Â−k B.

Moreover, E+k is unbounded as long as a+k > 1. Since this is the case for many choices of µ and k, we

will not include E+k as a part of our function space. It is also easy to check that E−k decays faster than

β−1, for all |k| ≥ 2. Later on, we shall identify situations where we can include E−k in our functionspaces, in order to satisfy the boundary conditions. Here the factors b−a

−k and b−a

+k provide a natural

normalization, see Proposition 6.3.

With these definitions at hand, we are prepared to define the appropriate function spaces onwhich we will construct solutions to these linearized equations. It is important to observethat the choice of these function spaces, which leads to an intermediate Banach algebra, laterin Section 7, will allow us to handle non-linear terms as well. For convenience, we use thenotation

〈k〉 .= (1 + k2)1/2

Definition 6.2 Recalling (6.23), we introduce the spaces

Ek.=

B†(Â−k )

†Hδ ⊕ CE0k ⊕ CE−k for |k| ≥ 2,

B†(Â−k )†Hδ ⊕ CE0k for |k| = 1,

B†(Â−k )†H+δ ⊕ CE

0k for k = 0,

(6.24)

Fk.=

{Â+kHδ if k 6= 0,Â+kH

+δ if k = 0,

(6.25)

37

with norms‖gk + a1E0k + a2E

−k ‖Ek = ‖A

−k Bgk‖δ + 〈k〉|a1|+ 〈k〉|a2| for |k| ≥ 2,

‖gk + a1E0k‖Ek = ‖A−k Bgk‖δ + 〈k〉|a1| for |k| = 1,

‖gk + a1E0k‖Ek = ‖A−k Bgk‖

+δ + 〈k〉|a1| for k = 0,

(6.26)

‖Â+k gk‖Fk = ‖gk‖δ, for k 6= 0,

‖Â+k gk‖Fk = ‖gk‖+δ , for k = 0.

(6.27)

Here the elements of Fk are understood in the sense of distributions. We notice that, if gk ∈ Fkis a bounded continuous function, then (for k 6= 0) we have ‖gk‖Fk = ‖(Â

+k )†gk‖δ.

Notice that we have not included E−k as part of our function spaces when |k| = 1. This isbecause, in these cases, the term AE−k (which shows up in the expression for T

(ϕ)) may notbe in H+δ . See Lemma 7.4 for more details.

Our first main goal is to prove the solvability of the linear equation (6.11) within Ek, for aright hand side in Fk. For this purpose, we show that the perturbation term in (6.11) doesnot spoil the invertibility of the equation within these spaces, provided we work on a domainβ ∈ [b,+∞[ with b large enough. Notice that here we do not yet prescribe boundary conditionsat β = b. This will be done at a later stage.

Proposition 6.2 Let µ > 1/2 and 0 < δ < −a−2 be given. Then there exists a constant C > 0so that for all k ∈ Z, k 6= 0,

‖(2µ− 1)(ikβ −B)‖L(B†(Â−k )

†Hδ;Fk) < Cb−1. (6.28)

By choosing b large enough, one can construct a linear, bounded (uniformly in k) solutionoperator for (6.11), mapping a right hand side ωk ∈ Fk to a solution Yk ∈ B†(Â−k )

†Hδ.

This solution operator can be written as

Yk =[B†(Â−k )

†(Â+k )† +Qk

]ωk ,

where Qk satisfies‖Qk‖L(Fk;B†(Â−k )†Hδ) ≤ Cb

−1.

Proof. We need to bound the operator norm∥∥(Â+k )†((2µ− 1)ikβ − (2µ− 1)B)B†(Â−k )†∥∥L(Hδ).Recalling the formula for (Â−k )

†, writing (Â−k )†g(s) = eiksh(s) and integrating by parts, one

38

obtains

((2µ− 1)ikβ − (2µ− 1)B)B†(Â−k )†g

= (2µ− 1)(ik

∫ βbeiksh(s) ds− eikβh(β)

)= − (2µ− 1)

(−eikbh(b)−

∫ βbeiksh′(s) ds

)= − (2µ− 1)((Â−k )

†g)(b)

+

−(2µ− 1)

∫ βb

(s−1g(s) + a−k e

ikssa−k −1

∫ sbe−iktt−a

−k −1g(t) dt

)ds for k > 2,

−(2µ− 1)∫ βb

(s−1g(s) + a−k e

ikssa−k −1

∫ ∞s

e−iktt−a−k −1g(t) dt

)ds for k = 1, 2.

By changing the order of integration and integrating by parts, the double integral in each ofthe above expressions yields only a term of order β−1−δ + b−1−δ. Applying (Â+k )

† and usingthe bound (6.22), we thus obtain terms which are bounded in the Hδ norm by C(b

−2 + b−1),with a constant C independent of k. Notice that the first term in this expression, namely(2µ− 1)((Â−k )

†g)(b) (which does not depend on β), after applying (Â+k )† for the same reason

is bounded by Cb−1 in the Hδ norm.

Concerning the single integral, by applying (Â−k )† one obtains

eikββa+k

∫ ∞β

s−a+k −1e−iks

∫ sbt−1g(t) dt ds = eikββa

+k

∫ ∞b

t−1g(t)

∫ ∞t∨β

s−a+k −1e−iks ds dt.

Integrating by parts, we find that this expression is bounded by b−1 in the Hδ norm. Thisestablishes the estimate (6.28). In turn, for b large enough, we obtain the solvability of (6.11).The bounds on the solution operator follow by a standard perturbation argument.

Remark 6.4 By setting β = b and computing the composed operator B†(Â−k )†(Â+k )

†, onechecks that the solutions constructed in Proposition 6.2 satisfy Yk(b) = 0.

Next, we will show that, by including the functions E0k , E−k in the space Ek, we can solve

(6.11) with some prescribed boundary conditions. As a first step, we introduce an operatorcorresponding to our rescaled boundary condition:

H(Y ) .= (1− 2µ)b−2µY + b1−2µ∂βY. (6.29)

Proposition 6.3 Fix µ > 1/2 and 0 < δ < −a−2 . Let b be sufficiently large, so that theconclusion of Proposition 6.2 holds. Then the functions E0k and (when |k| ≥ 2) E

−k are not

elements of B†(Â−k )†Hδ (respectively, of B

†(Â−k )†H+δ when k = 0).

However, there exist functions Ẽ0k = E0k + Z

0k and, for |k| ≥ 2, Ẽ

−k = E

−k + Z

−k , such that

A−k A+k BE

0k + (2µ− 1)ikβ Ẽ0k = 0 A−k A

+k BE

−k + (2µ− 1)ikβ Ẽ

−k = 0. (6.30)

39

Here the additional terms Z0k , Z−k satisfy

Z0k = B†(Â−k )

†(Â+k )†(2µ− 1)ik +W 0k , with ‖W 0k ‖Ek ≤ Cb

−2+δ,

Z−k = B†(Â−k )

†(Â+k )†(−(2µ− 1)eikb − (2µ− 1)b−a

−k

∫ βbeiksa−k s

a−k −1 ds

)+W−k ,

with ‖W−k ‖Ek ≤ Cb−2+δ , where C > 0 is a constant independent of k, not necessarily the

same as in Proposition 6.2.

Furthermore, there exists a unique solution (in Ek) to the homogeneous problem with boundaryconditions

A−k A+k B Yk + (2µ− 1)ikβ Yk = 0,

{H(Yk)(b) = bk, Yk(b) = dk for k ≥ 3,H(Yk)(b) = bk for k = 0, 1, 2.

Such solution can be written in the form

Yk =

{c0kẼ

0k + c

−k Ẽ−k for k ≥ 3,

c0kẼ0k for k = 0, 1, 2,

with

c0k = O(1) · bdk , c−k = O(1) ·(b−2µdk + bk

b−2µ

)when k ≥ 3,

c0k = O(1) · b2µ+1bk when k = 0, 1, 2.

Proof. 1. The case k = 0 is resolved by a direct computation, as no perturbation term ispresent.

In the case k ≥ 1, since by construction E0k and E−k lie in the kernel of Â

−k B, they are not

included in the space B†(Â−k )†Hδ. Moreover, the definition (6.23) implies E

−k (b) = 0.

2. We now construct Z0k and Z−k so that the equation (6.30) is satisfied. We observe that(

Â+k Â−k B − (2µ− 1)B + (2µ− 1)ikβ

)E0k = (2µ− 1)ik. (6.31)

SinceÂ†k(2µ− 1)ik = (2µ− 1)β

−1 +O(1) · β−2 ,

the Hδ norm of the above quantity has magnitude O(1) · bδ−1. Thus by taking the constantfunction ωk(β) = (2µ − 1)ik for all β ∈ [b,+∞[ , in (6.11), by applying Proposition 6.2 weobtain the existence of Z0k with the desired properties.

On the other hand, to study E−k for |k| ≥ 2, integrating by parts we obtain(Â+k Â

−k B − (2µ− 1)B + (2µ− 1)ikβ

)E−k = (2µ− 1)b

−a−k

(−eikββa

−k + ik

∫ βbeikssa

−k ds

)= − (2µ− 1)eikb − (2µ− 1)b−a

−k

∫ βbeiksa−k s

a−k −1 ds.

40

We recall that, for |k| ≥ 2, one has a−k < 0. Applying (Â+k )† to the previous expression one

obtains a function inHδ whose norm has sizeO(1)·b−1+δ. Once again, applying Proposition 6.2we obtain the existence of Z−k satisfying the desired bounds.

3. Having constructed Ẽ0k and Ẽ−k , we now show that they allow us to solve the ODE (6.11)

with the prescribed boundary conditions, satisfying the required bounds.

In the case |k| = 1, it suffices to take

bk = H(c0kẼ0k)(b) = c0k(1− 2µ)b−2µ(E0k(b) + Z0k(b)) + c0kb1−2µ((E0k)′(b) + (Z0k)′(b)).

By a direct computation one checks that E0k(b) = b−1 and (E0k)

′(b) = −b−2. Hence

H(E0k)(b) = − 2µb−2µ−1.

Furthermore, since Z0k ∈ Ek, we have that Z0k(b) = 0. Another direct computation yields(Z0k −W 0k )′(b) = O(1) · b−3.

On the other hand, by using the fact that ‖W 0k ‖Ek ≤ b−2+δ, then taking a derivative of theexpression for B and recalling that Â−k is bounded on Hδ, we obtain

(W 0k )′(b) = O(1) · b−3+δ.

Combining these facts, always in the case k = 1, we conclude

c0k = O(1) · b2µ−1bk .

On the other hand, when |k| ≥ 2, we have Ẽ−k (b) = 0, because E−k (b) = 0 and Z

−k is an

element of Ek. The form of E0k , along with the construction of Z0k and Remark 6.4, impliesthat, for |k| ≥ 2, one has c0k = bdk. Hence

bk = H(c0kẼ0k + c−k Ẽ−k )(b) = O(1) · dkb

−2µ + c−k H(Ẽ−k ).

We now compute

H(E−k )(b) = eikbb−2µ,

{H(Z−k )(b) = O(1) · b

−2µ−2+δ,

H(Z−k −W−k )(b) = O(1) · b

−2µ−2.

In turn this implies

c−k = O(1) ·(b−2µdk + bk

b−2µ

).

Define the spaces of admissible boundary values

Bk =

{C2 for |k| ≥ 2,C for |k| = 0, 1,

{ ∥∥(bk, dk)∥∥Bk .= 〈k〉(|bk|+ |dk|) for |k| ≥ 2,‖bk‖Bk

.= 〈k〉|bk| for |k| = 0, 1.

(6.32)Notice that we assign two boundary values for |k| ≥ 2, but only one for |k| = 0, 1. Thefollowing proposition summarizes the results of this subsection.

41

Proposition 6.4 Consider the linear boundary value problem for (6.8), namely

A−k A+k B Yk + (2µ− 1)ikβ Yk = ωk,

{H(Yk)(b) = bk, Yk(b) = dk , for |k| ≥ 2,H(Yk)(b) = bk , for |k| = 0, 1.

Under the same assumptions as in Proposition 6.2, this equation admits a linear, boundedsolution operator, mapping the source term and initial data (ωk, bk, dk) ∈ Fk × Bk to thesolution Yk ∈ Ek (or, in the case |k| = 0, 1, mapping (ωk, bk) to Yk). The norm of this linearsolution operator is bounded uniformly w.r.t. k.

7 Solution of the non-linear problem

In this section we construct a solution of the original non-linear problem (5.18). This will beachieved using an implicit function theorem, again following the analysis in [17].

For convenience, we recall the form of the equation:

0 = ∂ϕT(ϕ) + ∂φT

(φ) + T (0)ω∗,

where

T (ϕ).= 2

1 +(∂β∂φΨ̃2∂βΨ̃

)2× ∂βΨ̃ ∂ϕΨ̃(∂ϕ + 1)∂βΨ̃

−∂φ∂βΨ̃ · ∂φΨ̃

2∂βΨ̃, (7.33)

T (φ).=

(∂ϕ + 1)∂βΨ̃ · ∂φΨ̃− ∂φ∂βΨ̃ · ∂ϕΨ̃2∂βΨ̃

, (7.34)

T (0).=

1

2µ(∂ϕ + 1)∂βΨ̃ · (∂ϕΨ̃)−1/2µ . (7.35)

To help the reader, we outline here the main steps of the analysis.

1. In Section 7.1 we begin by defining an intermediate function space G−, which is a Banachalgebra obtained by summing functions which are periodic in φ and which (in the k 6= 0modes) decay rapidly enough as β → +∞. In a similar way, we also define an outputspace F , obtained by multiplying elements of Fk by eikφ, and a space E , obtained bymultiplying elements of Ek by eikφ.

2. In Section 7.2 we show that the outside derivatives appearing in (5.18) (namely ∂ϕ, ∂φ)are bounded as operators from G− into F . This is indeed the primary reason for definingthe space Fk in (6.25) in terms of the operators Â−k . This will allow us to more easilyhandle the derivatives that are on the outside of the non-linearities.

3. In Section 7.

on self-similar solutions to the incompressible euler...

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