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Peking Mathematical Journal (2019) 2:155–238https://doi.org/10.1007/s42543-019-00011-4

1 3

ORIGINAL ARTICLE

Global Steady Prandtl Expansion over a Moving Boundary I

Sameer Iyer1

Received: 7 November 2017 / Accepted: 26 October 2018 / Published online: 8 March 2019 © Peking University 2019

AbstractThis is the first of three papers in which we prove that steady, incompressible Navier–Stokes flows posed over the moving boundary, y = 0 , can be decomposed into Euler and Prandtl flows in the inviscid limit globally in [1,∞) × [0,∞) , assum-ing a sufficiently small velocity mismatch. In this part, sharp decay rates and self-similar asymptotics are extracted for both Prandtl and Eulerian layers.

Keywords Boundary layers · Navier–Stokes equations · Inviscid limit

1 Introduction

We consider the steady, incompressible Navier–Stokes equations in two dimensions:

in the domain,

The boundary Y = 0 is moving with velocity ub > 0 . The no-slip boundary condi-tions are placed on this portion of the boundary:

The boundary conditions at X = 1 will be prescribed explicitly in the text. We take X = 1 for convenience (this enables us to replace weights of (1 + x)k with xk ). Throughout this paper, we assume that prescribed Euler flow is the shear flow:

(1.1)UNSUNSX

+ VNSUNSY

+ PNSX

= �ΔUNS,

(1.2)UNSVNSX

+ VNSVNSY

+ PNSY

= �ΔVNS,

(1.3)UNSX

+ VNSY

= 0,

(1.4)Ω = [1,∞) ×ℝ+.

(1.5)UNS(X, 0) = ub = 1 − �, VNS(X, 0) = 0.

This research was completed under partial support by NSF Grant 1209437.

* Sameer Iyer [email protected]

1 Fine Hall, Department of Mathematics, Princeton University, Princeton, NJ 08544, USA

http://crossmark.crossref.org/dialog/?doi=10.1007/s42543-019-00011-4&domain=pdf

156 S. Iyer

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We are interested in the limit as � → 0 . Formally, one expects that the solutions to Navier–Stokes equations in (1.1)–(1.3) converges to the Euler shear flow in (1.6). This does not happen, however, due to the mismatch at the boundary Y = 0 , between the no-slip condition enforced for Navier–Stokes, (1.5), and UE(X, Y = 0) = 1.

To account for the mismatch at the boundary, Prandtl in 1904 proposed a thin fluid boundary layer which connects the velocity of ub to the Euler velocity of 1. The Prandtl hypothesis is that the Navier–Stokes solutions can be decomposed, up to leading order in � , as the sum of the prescribed Euler flow and a boundary layer, the latter of which corrects the disparity at the boundary between Euler and Navier–Stokes:

The contribution of this sequence of papers is to validate the boundary layer theory, Eqs. (1.7), in the domain Ω , which in particular implies that the tangential variable can be taken in [1,∞) if the mismatch is sufficiently small:

Boundary Layer Expansion

We will work with scaled, boundary layer variables:

The scaled Navier–Stokes unknowns are then given by:

These unknowns satisfy the following system:

Note that the steady Prandtl system is obtained by considering the leading order in � of the above system (1.11)–(1.13). We start with the following asymptotic expansion:

(1.6)(UE,VE) = (1, 0).

(1.7)UNS = 1 + u0p+ (�), VNS = 0 +

√�v0

p+√�v1

e+ (�).

(1.8)UE − UNS|Y=0 = 1 − ub = 1 − (1 − 𝛿) = 𝛿 ≪ 1.

(1.9)x = X, y =Y√�.

(1.10)U�(x, y) = UNS(X, Y), V�(x, y) =VNS(X, Y)

√�

, P�(x, y) = PNS(X, Y).

(1.11)U�U�x+ V�U�

y+ P�

x= U�

yy+ �U�

xx,

(1.12)U�V�x+ V�V�

y+

P�y

�= V�

yy+ �V�

xx,

(1.13)U�x+ V�

y= 0.

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Here, � ∈ [0,1

4) . We will use the word “profiles” to refer to the terms which appear

in the expansions (1.14)–(1.15), excluding the remainders, [u, v, P]. All of the pro-files with subscript-e are functions of Eulerian variables, (x, Y), whereas all terms with subscript-p are functions of boundary layer variables, (x, y). Here, [ui

p, vi

p] are

boundary layers to be constructed. The number of intermediate layers, n, is depend-ent on universal constants. The pressures Pi

e,Pi

p are the pressures associated with

the i’th Euler and Prandtl layers, respectively. We will show that Pip= 0 , that is, the

leading-order pressure in the boundary layers is zero. The pressures Pi,a

P,P1,a

e are

auxiliary pressures, which are higher order, whose purpose is to capitalize on the gradient structure of our problem (see (5.31)). After these layers are constructed, the Navier–Stokes remainders [u, v, P] are then constructed. Let us now designate the names for the partial expansions:

For the pressure expansion:

We insert the expansions (1.14)–(1.16) into (1.11)–(1.13) and collect a hierarchy of equations in powers of � . Doing so yields the linearized Prandtl equations:

(1.14)U�(x, y) = 1 + u0p+

n∑

i=1

�i

2 uie+ �

i

2 uip+ �

n

2+�u(x, y),

(1.15)V�(x, y) =

n−1∑

i=0

�i

2 vip+ �

i

2 vi+1e

+ �n

2 vnp+ �

n

2+�v(x, y),

(1.16)P�(x, y) =

n∑

i=1

�i

2Pie+ �

i

2Pip+ �iPi,a

e+ �

i+1

2 Pi,ap+ �

n

2+�P(x, y).

(1.17)u(i)s∶= 1 +

i−1∑

j=0

𝜖j

2 ujp+

i∑

j=1

𝜖j

2 uje, u(i)

s∶= 1 +

i∑

j=0

𝜖j

2 ujp+

i∑

j=1

𝜖j

2 uje,

(1.18)v(i)s∶=

i−1∑

j=0

𝜖j

2 vjp+

i∑

j=1

𝜖j

2−

1

2 vje, v(i)

s∶=

i∑

j=0

𝜖j

2 vjp+

i∑

j=1

𝜖j

2−

1

2 vje.

(1.19)P(i)s∶=

i−1∑

j=1

�j

2Pjp+

i−1∑

j=1

�j+1

2 Pj,ap+

i∑

j=1

�j

2Pje+

i∑

j=1

�jPj,ae,

(1.20)P(i)s∶=

i∑

j=1

𝜖j

2Pjp+

i∑

j=1

𝜖j

2Pje+

i∑

j=1

𝜖jPj,ae+

i∑

j=1

𝜖j+1

2 Pj,ap.

158 S. Iyer

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and the Euler equations:

The forcing term f (i) will be defined precisely in (5.45). These equations are derived rigorously in the analysis leading up to Eqs. (3.5), (4.21), (5.17), and (5.46).

Boundary Data

The no-slip boundary condition at the boundary {y = 0} is the most important, and must be enforced at each order in � , which gives:

The in-flow ( x = 1 ) boundary conditions for the leading order boundary layer, u0p , is:

We assume the rapid decay of the profile:

We will in addition assume the following smallness condition:

The in-flow ( x = 1 ) boundary conditions for the boundary-layer profiles are:

Here, Ui(y) , for i ≤ 1 ≤ n − 1 , will be prescribed to be rapidly decaying in y, so:

For the final Prandtl layer, unp , which occurs at order �

n

2 , the in-flow data is deter-mined through the analysis and is not explicitly prescribed. This is due to retaining that [un

p, vn

p] are divergence free, while cutting off vn

p for large values of y. The reader

is invited to turn to Eq. (5.125) and corresponding discussion for details regarding this matter. We will enforce ‖⟨y⟩mUn(y)‖L∞ ≤ C(m) . However, Un is an auxiliary

(1.21)(1 + u0p)ui

px+ u(i)

sxuip+ v(i)

suipy+ u0

py

(vip− vi

p(x, 0)

)+ Pi

px= ui

pyy+ f (i),

(1.22)uip(x, 0) = −ui

e(x, 0), lim

y→∞uip(x, y) = 0, ui

p(1, y) = Ui(y),

(1.23)uiex+ Pi

ex= 0, vi

ex+ Pi

eY= 0, ui

ex+ vi

eY= 0.

(1.24)u0p(x, 0) = −�, ui

e(x, 0) + ui

p(x, 0) = 0 for i ≥ 1, u(x, 0) = 0,

(1.25)vi−1p

(x, 0) + vie(x, 0) = 0 for i ≥ 1, vn

p(x, 0) = 0, v(x, 0) = 0.

(1.26)u0p(1, y) = U0(y), U0(0) = −�, lim

y→∞U0(y) = 0.

(1.27)‖⟨y⟩m�jyU0(y)‖L∞ ≤ C(m, j) for any m, j ≥ 0.

(1.28)‖⟨y⟩m�jyU0(y)‖L∞ ≤ (�;j,m) for any m ≥ 0, j = 0, 1, 2.

(1.29)uip(1, y) = Ui(y), u(1, y) = 0, v(1, y) = 0 for 1 ≤ i ≤ n − 1.

(1.30)‖⟨y⟩muip(1, y)‖L∞ = ‖⟨y⟩mUi(y)‖L∞ ≤ C(i,m) for 1 ≤ i ≤ n − 1.

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in-flow, which is used to construct [unp, vn

p] . That is: un

p(1, y) ≠ Un(y) , and the in-flow

velocity, unp(1, y) , is given by an implicit, bounded profile which decays as y → ∞ .

We will need several compatibility conditions on the in-flow data for the Prandtl lay-ers. The first of these is:

However, we shall also need higher-order compatibility conditions on the Ui(y) at y = 0 (see for instance Remarks 2.6, 4.11), which we refrain from depicting explic-itly here. Finally, the boundary conditions of the boundary-layer profiles as y → ∞ are:

These boundary conditions are known as the “matching condition”, and physically correspond to the Navier–Stokes flow matching the outer Euler flow away from the boundary, y = 0 . According to our construction, we will have rapid matching up to order �

n−1

2 :

At the highest order in � , we enforce the matching condition:

Let us now turn to the Euler flows. At leading order, we have the prescription: [u0

e, v0

e] = [1, 0] . The higher-order Euler flows will be described starting in Sects. 3

and 5.1. The higher-order Euler flows are obtained as suitable Poisson extensions of the Y = 0 boundary data, vi−1

p(x, 0) (see (1.24)), which depend on the constructed

Prandtl layers. For these higher-order Euler flows, we do not prescribe the in-flow data, ui

e(1, Y) . Rather, the in-flow conditions are obtained through the analysis, so we

state:

Main ResultTo state the main result of the sequence of three papers, we need to introduce

the norm Z in which we control the remainder solutions, [u, v]:

Definition 1.1 The norm Z is defined through:

(1.31)U0(0) = −�, �yyU0(y) = 0, Ui(0) = −uie(1, 0).

(1.32)limy→∞

[uip(x), vi

p(x)] = lim

y→∞[u(x), v(x)] = 0 for all x ≥ 1 and all 0 ≤ i ≤ n.

(1.33)limy→∞

[⟨y⟩Nuip(x), ⟨y⟩Nvi

p(x)] = 0 for all x ≥ 1, for 1 ≤ i ≤ n − 1.

(1.34)limy→∞

�n

2 [unp(x, y), vn

p(x, y)] = lim

y→∞�

n

2+�[u(x, y), v(x, y)] = 0 for all x ≥ 1.

(1.35)Eulerian in-flow =1 +

n∑

i=1

�i

2 uie(1, ⋅).

160 S. Iyer

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Here, Ni are large, universal numbers which can remain unspecified for now. The parameter n from (1.14)–(1.15) will be taken much larger than any of the Ni . The norms ‖ ⋅ ‖Xi

are energy norms, while the norms ‖ ⋅ ‖Yi are elliptic norms. For the purposes of stating the main result, we can refrain from being too specific with regard to the definitions of these norms. The essential point that we will record con-cerns the uniform component:

The main result of the full sequence of papers, the present paper together with [28, 29] is:

Theorem 1.2 Suppose the outer Euler flow is prescribed with u0e= 1 . Suppose the

boundary and in-flow data are specified satisfying the conditions outlined in (1.24)–(1.34). Then there exists an n depending on only universal constants such that the asymptotic expansions in (1.14)–(1.16) are valid globally on the domain Ω , for 0 ≤ 𝛾 <

1

4 , so long as the mismatch between the Eulerian boundary trace and the

motion of the boundary, � , and the viscosity, � , are taken sufficiently small relative to universal constants, and 𝜖 ≪ 𝛿 . The remainders, [u, v], in the expansions (1.14)–(1.15) are uniquely determined in the space Z:

where � is any fixed constant such that 𝛾 + 𝜅 <1

4.

Because n2 is large relative to N4 in (1.37), we immediately find:

Corollary 1.3 (Inviscid L∞ Convergence) Under the hypothesis of Theorem 1.2, there exists a unique Navier–Stokes solution [UNS,VNS,PNS] on Ω such that:

(1.36)

‖u, v‖Z ∶= ‖u, v‖X1∩X2∩X3+ �N2‖u, v‖Y2

+ �N3‖u, v‖Y3 + �N4‖ux1

4 ,√�vx

1

2 ‖L∞

+ �N5 supx≥20

‖√�vxx

3

2 , uxx5

4 ‖L∞ + �N6 supx≥20

‖uyx1

2 ‖L2y

+ �N7

�

�∞

20

x4‖√�vxx‖2L∞

y

dx

� 1

2

.

(1.37)�N4‖ux1

4 ,√�vx

1

2 ‖L∞ ≤ ‖u, v‖Z .

(1.38)‖u, v‖Z ≲ 𝜖1

4−𝛾−𝜅 ,

(1.39)sup(X,Y)∈Ω

|||UNS(X, Y) − 1 − u0

p(X, y)

|||X1

4 ≲ 𝜖1

2 ,

(1.40)sup(X,Y)∈Ω

��VNS(X, Y) −

√𝜖v0

p(X, y) −

√𝜖v1

e(X, Y)

��X1

2 ≲ 𝜖.

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Existing Literature

Let us first discuss the issue of establishing wellposedness of the Prandtl equation, which becomes an issue in the unsteady setting (in contrast to the steady setting of the present paper). This program was initiated in the classic works [40, 41], in which, under the monotonicity assumption U𝜖

y(t = 0) > 0 , globally regular solutions

are constructed on the [0, L] ×ℝ+ , where L is sufficiently small, and local solu-tions are constructed for arbitrary, but finite L. This was extended in [50], in which global weak solutions were constructed for arbitrary L, under both monotonicity and favorable outer-Euler pressure ( �xPE(t, x) ≤ 0 for t ≥ 0 ) assumptions.

From a physical standpoint, the monotonicity and favorable pressure assump-tions mentioned above are stabilizing and in particular prevent boundary layer separation. This phenomena was known to Prandtl; see Figure 2 in [44]. More recently, it was announced in [6] that a proof of boundary layer separation in the steady setting has been obtained.

The main tool used both in [40] and [50] is the Crocco transform. Still under monotonicity hypothesis, local wellposedness was obtained in [2] and [37], nei-ther works using the Crocco transform. [2] uses energy methods coupled with a Nash–Moser iteration, and [37] uses energy methods applied to a good unknown which enjoys crucial cancellation properties. Generalizing to multiple monoto-nicity regions, [32] has shown the Prandtl equation is locally well-posed, if an analyticity assumption is made on the complement of the monotonicity regions.

Indeed, when the assumption of monotonicity is removed, the wellposedness results are largely in the analytic or Gevrey setting. The reader should consult [26, 31, 35, 45, 46], and [14] for some results in this direction. In the Sobolev setting without monotonicity, the equations are linearly and nonlinearly ill-posed (see [13] and [15]). A finite-time blowup result was obtained in [10] when the outer Euler flow is taken to be zero, in [33] for a particular, periodic outer Euler flow, and in [25] for both the inviscid and viscous Prandtl equations. The above discussion is not comprehensive: we refer the reader to the review articles, [9, 17] and references therein for a more thorough review of the wellposedness theory.

The question with which we are concerned is the validity of the asymptotic expansion (1.14)–(1.16) in the inviscid limit. Let us first discuss unsteady flows. Local-in-time convergence is established in [45, 46] in the analyticity framework, in [16] in the Gevrey setting, and in [36] when the initial vorticity distribution is supported away from the boundary. The reader should see also [3, 38] for related results. Despite the boundary layer classically having thickness

√� , an interesting

criteria was given in [30] which points to phenomena occurring in a sub-layer of size � . There are also several linear and nonlinear instability results (for instance, [17–23]) which show the invalidity of Prandtl’s expansion generically in Sobolev spaces in the unsteady setting.

For steady flows, there are very few validity results. [24] is the first result in this direction, establishing the validity of the boundary layer expansion for steady-state flows in a rectangular domain over a moving boundary. Geometric effects of the boundary were subsequently considered in [27]. The crucial idea in [24] was the use of a positivity estimate, which is coupled with energy estimates and elliptic

162 S. Iyer

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estimates. Both of these results are local in the tangential variable. In the present work, we prove the validity of the boundary layer expansion globally in the tangen-tial variable, x, in the setting of small data.

One preliminary piece of our analysis is to obtain the asymptotics of the Prandtl layer, u0

p . This has nontrivial dynamics due to the mismatched boundary conditions,

u0p(y = 0) = −� , while limy→∞ u0

p(y) = 0 . These asymptotics were first studied in

[48, pg. 493, Inequality 5] using maximum principle techniques and are valid for large data. The result in [48] gives that the difference between u0

p and a Gaussian

“front” solution to the heat equation (call it w) is o(1) in x, uniformly in y. Under the hypothesis of small data, we sharpen these asymptotics in the following sense: first, w is shown to belong to a higher-order Sobolev space Hk(m) , where this weight is in the self-similar variable z = y√

x . Second, we obtain the rates of decay of w in x in

various norms.

Notation and Important Parameters

There are three important parameters in this paper: � , � , and n (see (1.14)–(1.16)). The notation A ≲ B means A ≤ CB , where C is some constant which is independent of small �, � , and large n. Constants denoted by (�) or (�) satisfy (�),(�) → 0 as �, � → 0 , respectively. Given any parameter, say p, constants denoted by C(p) mean those constants which depend (perhaps poorly) on large values of p. Given two parameters, � and � , for instance, we will write (�;�) to denote a constant which depends on � and � , but such that for fixed � , � can be made small to make the constant small, for instance � × � . We define ‖f‖p

Lpy

∶= ∫ f (x, y)pdy . When unspeci-fied, ‖ ⋅ ‖Lp means the Lp norm of two variables. We define here the differential operators:

The variable z will denote a self-similar variable, so typically z = y√x or z = �√

x .

Finally, the word “profiles” refers to terms in the expansion (1.14)–(1.15), excluding the remainders [u, v].

1.1 Introduction: Sharp Decay of Profiles

The key issue that we must capture in our analysis is the decay as x → ∞ of various quantities. More specifically, a central difficulty is to control contributions from the nonlinearity V�U�

y . Let us now introduce the equations for the remainders, [u, v, P]:

Here, the terms Su, Sv contain the linearizations of [u, v] around the previously con-structed profiles, [u(n)

s, v(n)

s] , while f and g can be regarded as an abstract forcing term

for now. To organize this discussion, let us record the heuristic:

(1.41)Δ�u ∶= (�xx + ��yy)u for any profile u; Δ[uie, vi

e] ∶= (�xx + �YY )[u

ie, vi

e].

(1.42)−Δ�u + Su + Px = f , −Δ�v + Sv +Py

�= g, ux + vy = 0.

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First, let us discuss u(n)syv from (1.43), which will motivate the crucial decay rates

appearing in (1.49). Applying the scaled multiplier of (u, �v) to the system (1.42) requires controlling the large convective term, ∫ ∫ u0

pyuv . We do not have the ability

to create a derivative through the Poincare inequality, and so we trade factors of x and y in the following manner:

The first crucial observation we make is the identification of a self-similar front,�∗(

y√x) , which bridges the boundary conditions: u0

p(x, 0) = −� , u0

p(x,∞) = 0 .

Then, temporarily identifying u0p≈ �∗(

y√x) , (1.44) will be satisfied:

Requirement 1.4 (Self-Similarity of Prandtl Profiles)

Summarizing the energy estimate that we obtain:

Above, the key point is the loss of weight, x1

2 . The next ingredient is recovering this weight in the positivity estimate, (see Proposition 3.4 of [28]), which is summarized:

To prove (1.47), we must apply the weighted multiplier vyx . Referring to the final two terms in (1.43), this gives (temporarily ignoring factors of �):

The latter two L2 quantities are controlled by the left-hand sides of (1.47)–(1.48). From this, we obtain the requirements:

Requirement 1.5 (Uniform Decay)

We emphasize that this requirement is inflexible, and we cannot sacrifice even a logarithmic factor of x here. The main contribution of this paper is the construction of

(1.43)Difficult Contributions from V𝜖U𝜖y= u(n)

syv + v(n)

suy + 𝜖

n

2+𝛾vuy.

(1.44)

�� u0pyuv��≤ ‖u0

pyy2x

−1

2 ‖2L∞

��u

y

��L2

��vx

1

2

y

��L2≤ (�)‖uy‖L2‖vyx 1

2 ‖L2 .

(1.45)‖y2x−1

2 u0py‖L∞ =

��y2x−1��

∗

�y√x

��L∞= ‖z2��

∗(z)‖L∞ ≤ (�).

(1.46)‖uy‖2L2 ≤ (�)��√

�vx, vy

�x

1

2

��

2

L2+ forcing terms.

(1.47)��{√𝜖vx, vy}x

1

2��2

L2≲ ‖uy‖2L2 + forcing terms.

(1.48)�� v(n)

suy ⋅ vyx

��+�� vuy ⋅ vyx

��≤ ‖{v(n)

s, v}x

1

2 ‖L∞‖uy‖L2‖vyx1

2 ‖L2 .

(1.49)v(n)s

+ v ∼ x−

1

2 , as x → ∞.

164 S. Iyer

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profiles uip, vi

p, ui

e, vi

e which satisfy the requirement of v(n)

s in (1.49). The most difficult

task is to obtain the estimate (1.49) for the Eulerian profiles, v1e , as these are solutions to

elliptic boundary value problems in which the boundary condition exhibits exactly the required decay rate, |v1

e(x, 0)| ≤ x

−1

2 . We refer the reader to the crucial Proposition 3.2 in which we introduce novel techniques centered around the explicit integral represen-tation of v1

e , enabling us to prove the required decay, |v1

e| ≲ x

−1

2.Note that we construct the expansions, (1.14)–(1.16) for any n ∈ ℕ . Our ability to do

this relies on the Cauchy–Riemann structure of the Eulerian profiles, which we use in (5.31).

1.2 Presentation of Results

The purpose of this paper is to construct each of the profiles appearing in the expan-sions (1.14)–(1.16), with the exception of the final terms, [u, v, P]. Let us first introduce the following notation for u(n)

s, v(n)

s:

We shall also have occasion to further split uPR to distinguish the final layer via:

Inserting the expansion (1.14)–(1.16) into the scaled NS equations, (1.11)–(1.13), motivates the following definition:

Definition 1.6 The n’th remainder is denoted by:

The main result of this paper is:

Theorem 1.7 Let n ≥ 2 ∈ ℕ . Let �, � be sufficiently small relative to universal con-stants, and 𝜖 ≪ 𝛿 . Let the boundary and in-flow data from (1.24)–(1.35) be pre-scribed. Then there exist Prandtl profiles [ujp, v

jp,P

jp] for j = 1,… , n , Euler profiles

[uje, v

je,P

je] for j = 1,… , n , and auxiliary pressures [Pj,a

p ,Pj,ae ] for j = 1,… , n such

(1.50)uPR∶=

n∑

j=0

𝜖j

2 ujp, uE

R∶= 1 +

n∑

j=1

𝜖j

2 uje, uR ∶= u(n)

s= uP

R+ uE

R,

(1.51)vPR∶=

n∑

j=0

𝜖j

2 vjp, vE

R∶=

n∑

j=1

𝜖j

2−

1

2 vje, vR ∶= v(n)

s= vP

R+ vE

R.

(1.52)uP,n−1

R=

n∑

j=0

�j

2 ujp, so that uP

R= u

P,n−1

R+ �

n

2 unp.

(1.53)Ru,n ∶= −Δ𝜖 u(n)s

+ u(n)su(n)sx

+ v(n)su(n)sy

+ P(n)sx,

(1.54)Rv,n ∶= −Δ𝜖 v(n)s

+ u(n)sv(n)sx

+ v(n)sv(n)sy

+𝜕y

𝜖P(n)s.

165

1 3


that for Ru,n,Rv,n as defined in (1.53)–(1.54), and for any � ∈ [0,1

4) , n ≥ 2 , and for

�n =1

10 000 , 𝜅 > 0 arbitrarily small, the following remainder estimate holds for any

k ≥ 0:

The following bounds hold on [uR, vR] by construction, for any [k, j,m] ≥ 0 , so long as n is sufficiently large relative to m.

(1.55)�−

n

2−� ��

kxRu,n +

√��k

xRv,n�� ≤ C(n, �)�

1

4−�−�

x−k−

3

2+2�n ,

(1.56)�−

n

2−��

√��k

xRu,n,

√��k

xRv,n��L2

y

≤ C(n, �)�1

4−�−�

x−k−

5

4+2�n+� .

(1.57)‖�kx�jyvPRzmx

k+j

2+

1

2 ‖L∞ ≤ C(k, j,m) if k ≥ 1,

(1.58)‖�jyvPRzmx

j

2+

1

2 ‖L∞ ≤ C(j,m) if j ≥ 2,

(1.59)‖�jyvPRzmx

j

2+

1

2 ‖L∞ ≤ (�;m, j) if j = 0, 1.

(1.60)‖𝜕kx𝜕jyuPRzmx

k+j

2 ‖L∞ ≤ C(k, j,m) for k > 1, j ≥ 0,

(1.61)‖�xuPRzmx‖L∞ ≤ (�;m),

(1.62)‖�x�jyuPRzmx‖L∞ ≤ C(m, j) for j ≥ 1,

(1.63)‖�jyuP,n−1

Ryjzm‖L∞ ≤ (�;m, j) for 0 ≤ j ≤ 2,

(1.64)‖𝜕jyuP,n−1

Ryjzm‖L∞ ≤ C(m, j) for j > 2,

(1.65)‖�jyunpyjx

1

2−�n‖L∞ ≤ C(n, j) for all j ≥ 0,

(1.66)‖𝜕kx𝜕j

YvERxk+j+

1

2 ‖L∞ ≤ C(k, j) for k + j > 0,

(1.67)‖𝜕kx𝜕j

YuERxk+j+

1

2 ‖L∞ ≤ √𝜖C(k, j) for k + j > 0,

(1.68)‖�kxvERxk−

1

2 Y‖L∞ ≤ C(k, j) for k ≥ 1,

166 S. Iyer

1 3

The profiles uR, vR from (1.50)–(1.51) arise as coefficients in the linearized prob-lem for the Navier–Stokes remainders [u, v, P], which is to be analyzed in [28] (i.e., Global steady Prandtl expansion over a moving boundary II). The estimates obtained in (1.57)–(1.67) are therefore essential to the analysis of [28]. In particular, we invite the reader to compare the requirements discussed in (1.45) and (1.49) with the estimates we prove in (1.57)–(1.69).

The structure of this paper is as follows:

(Step 1) Construction of the zeroth-order Prandtl layers, [u0p, v0

p] (Sect. 2): the distin-

guishing features of u0p are the mismatched boundary conditions, as seen from

(2.5). As shown in Proposition 2.2, this contributes a “front”-profile which looks similar to e� , defined in (2.11). The decay rates for Prandtl profiles from estimates (1.57)–(1.67) are dictated by this front profile. For the zeroth layer, this is formalized in Proposition 2.9, Corollaries 2.11 and 2.12. It is essential that we obtain estimates weighted in the self-similar variable, z = y√

x , as can

be seen from (1.57)–(1.69) above. (Step 2) Construction of Euler-1 layers, [u1

e, v1

e] (Sect. 3): given the boundary condi-

tions, which are known to satisfy the progressive estimate: |𝜕kxv1e(x, 0)| ≲ x

−1

2−k ,

we must obtain the sharp uniform estimate |𝜕kxv1e(x, Y)| ≲ x

−1

2−k . This is a deli-

cate matter, as these profiles are elliptic, and as indicated in (1.49), the required decay rates must be obtained exactly. We introduce a method to obtain the required pointwise estimates using directly the Poisson integral formulation, which is carried out in Proposition 3.2. This section is a major contribution of this paper.

(Step 3) Construction of Prandtl-1 layer, [u1p, v1

p] (Sect. 4): this profile is controlled

using coupled energy and positivity estimates, given in Lemma 4.6 and Lemma 4.8.

(Step 4) Construction of intermediate Euler and Prandtl layers (Sect. 5): the essential mechanism here is as follows: as one considers linearizations for [ujp, v

jp] for

j > 1 , one encounters terms which scale poorly in z = y√x , due to Euler–Euler

interactions. However, due to the Cauchy–Riemann structure present in the Euler profiles (see (3.7)), we may introduce auxiliary pressures Pi,a

p,P

i,a

E which

creates cancellations of all terms that are “purely-Eulerian”. This is seen in (5.30), (5.31).

(Step 5) Construction of the final Prandtl layer, [unp, vn

p] (Sect. 5.3): the final Prandtl layer

satisfies the boundary condition vnp(x, 0) = 0 , and so has a contribution as y ↑ ∞ .

We cut off this layer in the region z ≤ 1√� , which honors the parabolic scaling of

the Prandtl layers. This is a generalization of the cutoff used in [24]. It is delicate to ensure that these cutoff layers obey desirable estimates, which is done in Lemma 5.28. It is also delicate to ensure that this process contributes an error that satisfies the estimates (1.55)–(1.56) above. This is proven in Lemma 5.27.

(1.69)‖{uER− 1, vE

R}x

1

2 , vERYx

3

2 ‖L∞ ≤ (�).

167

1 3


2 Asymptotics of Prandtl Layer, u0

p

Our starting point is the leading order terms from (1.21), which yields the following system for the Prandtl layer, [u0

p, v0

p]:

As shown in [24, pg. 9], taking [u0p, v0

p] to solve the system (2.1)–(2.2) creates an

error:

This Ru,0 contribution is higher order in � , and so will be accounted for as a forcing term in the construction of the next Prandtl layer, [u1

p, v1

p] (see (4.23)). After intro-

ducing the von Mises coordinates,

the equation for u0p becomes parabolic, with x being the time-like variable:

Via the maximum principle, as in [24],

For the analysis in Sect. 2, it is convenient to introduce the shifted unknown

The shifted unknown then satisfies the IBVP:

2.1 Existence of the Front Profile

The dynamics of the solution to Eqs. (2.1)–(2.2), or equivalently, (2.8), are gov-erned by a self-similar “front”, in the sense of [5]. This is due to the mismatch in boundary conditions at y = 0 and y = ∞ , as seen from (2.8). Inserting a self-simi-lar anzatz �∗(z) = �∗(

�√x) into Eq. (2.8) gives the following ODE:

(2.1)(1 + u0

p

)u0px+

(v0p+ v1

e(x, 0)

)u0py= u0

pyy, u0

px+ v0

py= 0,

(2.2)u0p(x, 0) = −�, v0

p(x, 0) = −v1

e(x, 0), u0

p(1, y) = U0(y).

(2.3)Ru,0 ∶= �u0pxx

+√�yv1

eYu0py+ �u0

py ∫y

0 ∫y

y�v1eYY

dy��dy�.

(2.4)� = ∫y

0

(1 + u0

p

)dy�,

(2.5)u0px= ��((1 + u0

p)u0

p�), u0

p(x, 0) = −�, u0

p(x,∞) = 0, u0

p(1, �) = U0.

(2.6)1 + u0p≥ 1 − �.

(2.7)q = u0p+ �.

(2.8)qx = ��

((1 − � + q)q�

), q(x, 0) = 0, q(x,∞) = �, q(1, �) = u0

p(1, �) + �.

(2.9)(1 − � + �∗)�

��∗+ |��

∗|2 + z

2��∗= 0, �∗(0) = q(x, 0) = 0, �∗(∞) = q(x,∞) = �.

168 S. Iyer

1 3

Here, “ ′ ” denotes �z , where z is the self-similar variable for �∗ . As the profile �∗ that we seek is a nonlinear variant of the Gaussian error function, we study:

where e� is the Gaussian front profile with value � at +∞:

Let us record the following:

Lemma 2.1 With e� being defined as (2.11), e�(�√x) is an explicit solution of the heat

equation, which bridges two distinct boundary conditions at y = 0 and y = ∞:

Proof We first record the identities:

Differentiating (2.11) gives the identities:

One then checks that: �xe�(�√x) = ��e�(

�√x) is equivalent to (2.14). The boundary

condition at 0 is trivial from (2.11), and the boundary condition at ∞ arises from: e�(∞) =

�√�∫ ∞

0e−

t2

4 dt = � . The lemma is proven. □

Using the heat equation for e� , coupled with (2.10), we obtain:

The first task is to obtain the existence of a solution, � , to the above boundary value problem in a suitable Sobolev space.1

(2.10)� = �∗ − e� ,

(2.11)e�(z) =�√� ∫

z

0

e−

t2

4 dt.

(2.12)��x − ��

�e�

��√x

�= 0, e�(0) = 0, e�(∞) = �.

(2.13)�z

��=

1√x,

�z

�x= −

z

2x.

(2.14)e��(z) =

�√�e−

z2

4 , e��(z) = −

z

2x

�√�e−

z2

4 = −z

2xe��(z).

(2.15)(1 − � + �)� �� + |� �|2 + z

2� � = −2� �e�

�− |e�

�|2 − �e��

�− e��

�� − e�e��,

�(0) = �(∞) = 0.

1 It is clear that by rescaling z → (1 − �)1

2 z , we can replace the factor of 1 − � in front of � ′′ by simply 1. This rescaling would change the main linear operator, (1 − �)� �� +

z

2� � to � �� +

z

2� � . For notational ease,

then, we work simply with the � ′′ instead of (1 − �)� �� . The actual self-similar variable, then, is really (1 − �)

�√x , but as (1 − �) is near 1, this causes no confusion in the analysis to follow.

169

1 3


Proposition 2.2 For � sufficiently small, there exists a unique solution to Eq. (2.15) in H2

w(ℝ+) satisfying ‖𝜓‖H2

w≲ 𝛿 , where H2

w is a weighted variant of H2 which is for-

mally defined in (2.24).

Proof For this argument, fix r > 0 . We will eventually let r → ∞ . Our starting point is to establish the existence of solutions to the linear operator

We recall the identity given in [5]:

which in turn implies

if and only if

The subscripts are included to emphasize the domain, (0, r) ⊂ ℝ . Consider:

B is clearly bounded and coercive on H10(0, r) . One obtains the existence of a unique

H1 weak solution to the system (2.19), and correspondingly to (2.18) from the Lax–Milgram Lemma. Via elliptic regularity, this implies H2 regularity which, in original unknowns, translates to the existence of a unique �(r) ∈ H2(0, r) for each f ∈ L2(0, r) such that:

Let us rewrite the nonlinear problem (2.15) as a fixed point to:

(2.16)(−�2

z−

z

2�z

).

(2.17)ez2

8

(−�2

z−

z

2�z

)e−

z2

8 � =

(−�2

z+

z2

16+

1

4

)� ,

(2.18)(−�2

z−

z

2�z

)� = f , �(0) = �(r) = 0

(2.19)

(−𝜕2

z+

z2

16+

1

4

)��(r) = f , ��(r)(0) = ��(r)(r) = 0, ��(r) = e

z2

8 𝜓(r), f = ez2

8 f .

(2.20)

B[��(r), v] ∶= ∫r

0

𝜕z��(r) ⋅ 𝜕zv + ∫r

0

(z2

16+

1

4

)��(r)v, H1

0(0, r) × H1

0(0, r) → ℝ.

(2.21)‖�(r)‖H2(0,r) ≤ C(r)‖f‖L2(0,r), �(r)(0) = �(r)(r) = 0.

(2.22)−� ��(r)

−z

2� �(r)

= f (� (r)) + (e� + �)��

(r), �(r)(0) = �(r)(r) = 0,

170 S. Iyer

1 3

where

Define now the norm:

and the parameter:

Via a standard integration by parts argument applied to (2.22), we obtain the stabi-lizing estimate:

which proves that if ‖� (r)‖H2w≤ R(�) , then

Selecting � small enough then ensures 3R(𝛿)4 < R(𝛿)2 , ensuring that

The second step is to prove that the nonlinear map N is a contraction map on BR(𝛿) ⊂ H2

w . As such, label the pairs:

Taking differences and performing a standard integration by parts argument give:

(2.23)f (� (r)) = � (r)��

(r)+|||�

�

(r)

|||2

+ 2e��

(r)+ |e�

�|2 + e��

�� (r) + e�e

��.

(2.24)‖�(r)‖2H2w

∶= ∫ �� (r)�2 + ∫ (1 + z2)��(r),�

�(r)�2,

(2.25)R(�) = max�‖e�‖L∞ , ‖e��‖L∞ , ‖e

��‖L∞ , ‖(1 + z2)

1

2 e��‖L2 , ‖e��(1 + z2)

1

2 ‖L2�.

(2.26)‖�(r)‖2H2w

≤ R(�)4 + R(�)2‖� (r)‖2H2w

+ ‖� (r)‖4H2w

,

(2.27)‖�(r)‖2H2w

≤ R(�)4 + 2R(�)2‖� (r)‖2H2w

.

(2.28)𝜓(r) = N(𝜓(r)) ∈ BR(𝛿) ⊂ H2w

whenever 𝜓(r) ∈ BR(𝛿) ⊂ H2w.

(2.29)−� ��

(i,r)−

z

2� �(i,r)

= f (�(i,r)) +(e� + �

)�

��

(r), �(i,r)(0) = �(i,r)(r) = 0, i = 1, 2.

(2.30)‖�1,r − �2,r‖2H2w

≤ R(�)2‖�1,r − �2,r‖2H2w

,

171

1 3


By the contraction mapping theorem, there exists a unique solution to the nonlinear problem (2.22)–(2.23). Moreover, as this solution lies in the ball BR(�) , it obeys the estimate

uniformly in r. Therefore, we may let r → ∞ to obtain a solution to the problem (2.15). □

We may repeat the procedure in the above theorem with any weight, giving:

By further differentiating Eq. (2.15), we can obtain

Our front from here on will be denoted as �∗ = � + e� . We will abuse the notation and depict

We have the following corollary to the preceding analysis:

Corollary 2.3 The front �∗ obeys the following bounds, for any m, k, l ≥ 0:

Proof This follows immediately from writing �∗ − � = � + (e� − �) , the definition of e� in (2.11), the bounds in (2.33), and then the chain rule together with the identi-ties (2.13). □

Let us now record the following fact:

Corollary 2.4 A solution �∗ to the system (2.9) must satisfy 𝜙�∗(0) > 0.

Proof Define uf = �∗, vf = ��∗ . By (2.35), |uf | ≤ c0� for some constant c0 . The sys-

tem (2.9) is then: u�f= vf , (1 − � + uf )v

�f+ |vf |2 +

z

2vf = 0 . The invariant set

Γ = {uf = C, vf = 0} , for −c0� ≤ C ≤ c0� contains equilibria. The solution �∗ cor-responds to an orbit with initial condition (uf (0) = 0, vf (0)) and final condition (uf (∞) = �, vf (∞)) . By ODE uniqueness, the trajectory cannot cross the set Γ unless at (�, 0) . Thus, if vf (0) ≤ 0 , vf (z) ≤ 0 for all z, which violates uf (∞) = � . Thus, we must have vf (0) > 0 . □

2.2 Zeroth Prandtl Layer, u0p

We define the remainder, w, via:

(2.31)‖�(r)‖H2w(0,r) ≤ R(�),

(2.32)‖⟨z⟩M�‖H2 ≤ (�;M).

(2.33)‖⟨z⟩M�‖Hk ≤ (�;M, k).

(2.34)�∗(z) = �∗(x, �).

(2.35)‖zm�l��kx(�∗ − �)‖Lp� ≤ (�)x−k− l

2+

1

2p .

(2.36)w = q − �∗(⋅), g ∶= w(1, ⋅) = q(1, ⋅) − �∗(⋅) = U0(1, ⋅) + � − �∗(⋅).

172 S. Iyer

1 3

The initial data in (2.36) are clearly rapidly decaying after recalling (1.26)–(1.28), via the relation: g = U0 − (e� − �) − � . Moreover, the following smallness is obtained, again by recalling the assumptions (1.26)–(1.28):

The equation satisfied by w is the following (after substituting q = w + �∗):

where

Let us first make the following basic observation regarding our initial data:

Lemma 2.5 (Compatibility of Initial Data) Suppose the compatibility conditions in (1.31) are enforced. Then the initial data of wx respects the boundary condition wx(1, 0) = 0.

Proof This follows upon evaluating (2.1) at y = 0 and (2.8) at � = 0, x = 1.

Remark 2.6 (Higher Order Compatibility) This same calculation for higher-order compatibility conditions at x = 1, y = 0 can be made. We omit displaying them explicitly, but will feel free to assume higher-order compatibility of the initial data at y = 0 as needed.

Let us now define a series of norms in which we control the remainder, w:

We shall need the general, differentiated, and weighted variant of the above norms:

Remark 2.7 One should take note of several points. First, each application of �x to the system (2.38) adds enhanced decay of x−1 , which is reflected in the norms above. This is because of the behavior of the profiles �∗ under the application of �x ; see estimate (2.35).

Second, upon controlling w in the Q-norms in (2.41), the dynamics of q = w + �∗ will be dominated by those of the front �∗ (so long as the parameter 𝜎0 <

1

4 , which

(2.37)‖⟨y⟩m�jyg‖L∞ ≤ (�;j,m) for any m, and j = 0, 1, 2.

(2.38)wx = (1 − �)w�� + ��(qw� + w��

∗)= (1 − �)w�� + K,

(2.39)w(x, 0) = 0, w(x,∞) = 0, w(1, �) = g(�),

(2.40)K = w2�+ 2�∗

�w� + ww�� + �∗w�� + w�∗

��.

(2.41)‖w‖2Q(0,0)

∶= supx≥1 � w2,w2

�x,w2

��x2 + �

∞

1 � w2�,w2

��x,w2

��x2.

(2.42)

‖w‖2Q(�0,k)

∶= supx≥1 � ��k

xw�2x2k−2�0 , ��k

xw��2x2k+1−2�0 , ��kxw��2x2k+2−2�0

+ �∞

1 � ��kxw��2x2k−2�0 , ��kxw��2x2k+1−2�0 , ��kxw��2x2k+2−2�0 .

173

1 3


we will enforce). The small parameter �0 arises for technical reasons when perform-ing weighted estimates, but should be ignored for the unweighted estimates.

The final point is that upon heuristically identifying wx ≈ w�� through Eq. (2.38), one sees that the norms Q(�0, k) have an iterative structure:

the quantity on the left-hand side above being in ‖w‖Q(0,k+1) and the quantity on the right-hand side above being in ‖w‖Q(0,k) (upon taking sup in x). The reason for this structure is that Eq. (2.38) is quasilinear.

Through standard Sobolev interpolation, it is clear that:

Lemma 2.8 For any �0 ≥ 0 , and k ≥ 0,

Proof As the profile w decays at � → ∞ for each fixed x, we have:

A similar computation works for the �kxw� term in (2.44). □

We now give the energy estimates for w:

Proposition 2.9 For any 𝜎0 > 0 , and m ≥ 0 , w satisfies the following estimate:

Remark 2.10 Notice that the weights, z = �√x , honor the parabolic scaling of (2.38),

and are propagated by the linear flow.

Proof The proof proceeds in several steps. □

(2.43)∫ |�k+1x

w|2x2(k+1) d� ≈ ∫ |�kxw��|2x2k+2 d�,

(2.44)supx

‖𝜕kxwx

1

4−𝜎0+k‖L∞

𝜂+ sup

x

‖𝜕kxw𝜂x

3

4−𝜎0+k‖L∞

𝜂≲ ‖w‖Q(𝜎0,k).

(2.45)

�𝜕kxw�2x

1

2−2𝜎0+2k =

��2∫

∞

𝜂

𝜕kxw𝜕k

xw𝜂x

1

2−2𝜎0+2kd𝜂�

��≲ ‖wxk−𝜎0‖L2

𝜂‖w𝜂x

k−𝜎0+1

2 ‖L2𝜂

≲ ‖w‖2Q(𝜎0,k)

.

(2.46)‖zmw‖Q(�0,0) ≤ (�;m, �0),(2.47)‖zmw‖Q(𝜎0,k) ≤ C(k,m, 𝜎0), for k > 0.

174 S. Iyer

1 3

Step 1: Multiplier M = w

Applying the multiplier M = w to the system (2.38) generates the following positive terms:

Next, we must treat the nonlinear and linearized terms in K. The boundary condition w(x, 0) = 0 enables us to integrate by parts the quasilinear term:

Above, we have used Corollary 2.3 to absorb two factors of � into �∗��

. We have also used the Hardy inequality, which is available as w(x, 0) = 0 . Summarizing, we have:

Step 2: Multiplier M = −w��x

Next, we will apply the multiplier M = −w��x . This generates the positive terms:

Let us now turn to the terms in K:

Summarizing, we have:

(2.48)∫ (wx − (1 − �)w��) ⋅ w =�x

2 ∫ w2 + (1 − �)∫ w2�.

(2.49)��∫ w2

𝜂⋅ w + ww𝜂𝜂 ⋅ w

��=��∫ ww2

𝜂

��≲ ‖w‖L∞

𝜂

�

∫ w2𝜂

�,

(2.50)

�� ∗w�� ⋅ w + � �∗��w ⋅ w + � 2�∗

�w� ⋅ w

��=�� ∗

��w2 + � �∗w2

�

��

≤ ‖�∗��2‖L∞

� �w2

�2+ ‖�∗‖L∞‖w�‖2L2

�

≤ (�)� w2�.

(2.51)𝜕x � w2 + � w2𝜂≲

�(𝛿) + ‖w‖L∞𝜂

�� w2

𝜂.

(2.52)−∫(�x − (1 − �)��

)w ⋅ w��x =

�x

2 ∫ w2�x − ∫ w2

�+ (1 − �)∫ w2

��x.

(2.53)�� w2

�w��x + � ww2

��x��≤ ‖w,w�x

1

2 ‖L∞�

�‖w�‖2L2

�

+ ‖x1

2w��‖2L2�

�,

(2.54)

�� ∗�w�w��x + � �∗w2

��x + � �∗

��ww��x

��≤ ‖�∗, �x

1

2�∗��, x

1

2�∗�‖L∞

�‖w�‖2L2

�

+ ‖w��x1

2 ‖2L2�

�

≤ (�)�‖w�‖2L2�

+ ‖w��x1

2 ‖2L2�

�.

(2.55)𝜕x ∫ w2

𝜂x + ∫ w2

𝜂𝜂x ≲

�1 + ‖w𝜂x

1

2 ‖L∞𝜂

�∫ w2

𝜂+ ‖w,w𝜂x

1

2 ‖L∞𝜂 ∫ w2

𝜂𝜂x.

175

1 3


Step 3: Multiplier wxx2

The third step is to take �x of Eq. (2.38). Doing so gives the system:

First, we will record that ‖⟨𝜂⟩mwx(1, 𝜂)‖L∞ ≲ (𝛿;m) , for any m ≥ 0 , by (2.57) and (1.26)–(1.28). Let us now apply the multiplier M = wxx

2 to the system (2.56). Again, we will remain cognizant of the boundary condition wx(x, 0) = 0 when integrating by parts. Doing so gives the positive terms:

Next, we come to the nonlinearity in Kx , integrating by parts as necessary:

(2.56)(�x − (1 − �)��

)wx = �xK, wx(x, 0) = 0,

(2.57)wx(1, ⋅) = (1 − �)g�� + g2�+ 2�∗

�g� + gg�� + �∗g�� + �∗

��g.

(2.58)∫(�x − (1 − �)��

)wx ⋅ wxx

2 = �x ∫ w2xx2 − ∫ w2

xx + (1 − �)∫ w2

�xx2.

(2.59)

�� w�w�xwxx2 + wxw��wxx

2 + ww��xwxx2��

≤ ‖w�x1

2 ‖L∞�‖w�xx‖L2

�‖wxx

1

2 ‖L2�+ ‖w‖L∞‖w�xx‖2L2

�

,

(2.60)

�� ∗�xw�wxx

2 + � �∗�w�xwxx

2��

≤ ‖�∗�xx�‖L∞

�‖w�‖L2

�‖wx�x‖L2

�+ ‖�∗

��‖L∞

�‖w�xx‖2L2

�

≤ (�)‖w�‖L2�‖wx�x‖L2

�+ (�)‖w�xx‖2L2

�

,

(2.61)

�� ∗xw��wxx

2 + � �∗w��xwxx2��

≤ ‖�∗xx‖L∞

�‖w��x

1

2 ‖L2�‖wxx

1

2 ‖L2�+ ‖�∗,�∗

��‖L∞

�‖w�xx‖2L2

�

≤ (�)‖w��x1

2 ‖L2�‖wxx

1

2 ‖L2�+ (�)‖w�xx‖2L2

�

,

(2.62)

�� ∗��x

wwxx2 + � �∗

��w2xx2��

≤ ‖�∗��x

�x3

2 ‖L∞�‖wxx

1

2 ‖L2�‖w�‖L2

�+ ‖�∗

��x‖L∞‖wxx

1

2 ‖2L2

≤ (�)‖wxx1

2 ‖L2�‖w�‖L2

�+ (�)‖wxx

1

2 ‖2L2�

.

176 S. Iyer

1 3

Summarizing this piece:

With these estimates in hand, we may now apply a standard continuous induc-tion argument to conclude the global existence of w satisfying (2.46) with k = m = �0 = 0.

Step 4: Weighted Estimates

We now apply the weighted multiplier M = x−2�0zmw to the system (2.38). First, we have the following positive terms:

The last term above has been estimated inductively at the (m − 1)’th iterate. Next, we address the terms in K:

Summarizing:

By integrating in x:

The next step is to apply the multiplier M = −w��x1−2�0z2m . First, this gives:

(2.63)𝜕x � w2

xx2 + � w2

𝜂xx2 ≲

�(𝛿) + ‖w,w𝜂x1

2 ‖L∞𝜂

�� w2

𝜂xx2

+

�1 + (𝛿) + ‖w,w𝜂x

1

2 ‖L∞𝜂

�� w2

xx + (𝛿)� w2

𝜂,w2

𝜂𝜂x.

(2.64)

∫(wx − w��

)⋅ wz2mx−2�0 =

�x

2 ∫ w2z2mx−2�0 +(m

2+ �0

)∫ w2z2mx−1−2�0

+ ∫ w2�z2mx−2�0 − m(2m − 1)∫ w2z2m−2x−1−2�0 .

(2.65)��∫

�w2𝜂+ ww𝜂𝜂

�⋅ wz2mx−𝜎0

��≲ ‖w‖L∞

𝜂‖w𝜂z

mx−𝜎0‖2L2𝜂

,

(2.66)

��∫�2𝜙∗

𝜂w𝜂 + 𝜙∗w𝜂𝜂 + w𝜙∗

𝜂𝜂

�⋅ z2mx−2𝜎0w

��≲ ‖𝜂𝜙∗

𝜂, 𝜂2𝜙∗

𝜂𝜂‖L∞

𝜂‖w𝜂z

mx−𝜎0‖2L2𝜂

.

(2.67)

𝜕x ∫ w2z2mx−2𝜎0 + ∫ w2𝜂z2mx−2𝜎0 + ∫ w2z2mx−1−2𝜎0 ≲ ∫ w2z2m−2x−1−2𝜎0 .

(2.68)supx≥1 � w2z2mx−2𝜎0 + � � w2

𝜂z2mx−2𝜎0 + � � w2z2mx−1−2𝜎0

≲ �x=1

w2y2m + � � w2z2m−2x−1−2𝜎0 ≤ (𝛿,m, 𝜎0).

177

1 3


Next, let us turn to the nonlinear terms in K:

and the linearized terms:

Summarizing,

Integrating in x:

The final step is to apply the multiplier M = wxx2−2�0z2m to the system (2.56). This

generates the following terms:

(2.69)

− ∫(�x − ��

)w ⋅ w��z

2mx1−2�0

=�x

2 ∫ w2�z2mx1−2�0 + ∫ w2

��z2mx1−2�0 +

m

2 ∫ w2�z2mx−2�0

−1 − 2�0

2 ∫ w2�z2mx−2�0 + 2m∫ w�wxz

2m−1x1

2−2�0 .

(2.70)

��w2�+ ww��

�⋅ w��z

2mx1−2�0��≤ ‖w,w�x

1

2 ‖L∞�

�‖w�z

mx−�0‖2L2�

×‖w��zmx

1

2−�0‖2

L2�

�,

(2.71)

��∗�w� + �∗w�� + �∗

��w�⋅ w��z

2mx1−2��

≤ ‖x1

2�∗�, x

1

2 ��∗��,�∗‖L∞‖w�x

−�0zm‖L2�‖w��z

mx1

2−�0‖L2

�.

(2.72)

𝜕x ∫ w2𝜂z2mx1−2𝜎0 + ∫ w2

𝜂𝜂z2mx1−2𝜎0 + ∫ w2

𝜂z2mx−2𝜎0

≲ ∫ w𝜂wxz2m−1x

1

2−2𝜎0

≲1

10 000 ∫ w2𝜂z2mx−2𝜎0 + C ∫ w2

xx1−2𝜎0z2m−2.

(2.73)supx≥1 � w2

𝜂z2mx1−2𝜎0 + �

∞

1 � w2𝜂𝜂z2mx1−2𝜎0 + �

∞

1 � w2𝜂z2mx−2𝜎0

≲ �x=1

w2𝜂y2m + � � w2

xx1−2𝜎0z2m−2 ≤ (𝛿;𝜎0,m).

178 S. Iyer

1 3

We turn to the nonlinearities contained in Kx , for which we integrate by parts as needed:

Next, we come to the linearizations around the front profile, �∗:

Summarizing the results of this multiplier:

(2.74)∫

(�x − ��

)wx ⋅ wxx

2−2�0z2m =�x

2 ∫ w2xx2−2�0z2m + ∫ w2

�xx2−2�0z2m

+

(m

2− 1 + �0

)∫ w2

xx1−2�0z2m − m(2m − 1)∫ w2

xx1−2�0z2m−2.

(2.75)

��∫�w𝜂w𝜂x + w2

xw𝜂𝜂 + ww𝜂𝜂xw

�⋅ wxx

2−2𝜎0z2m��

≲ ‖w,w𝜂x1

2 ‖L∞�‖wxx

1

2−𝜎0zm‖2

L2𝜂

+ ‖wxx1

2−𝜎0zm−1‖2

L2𝜂

+ ‖wxx1

2−𝜎0zm‖2

L2𝜂

�.

(2.76)

�� ∗�xw�wxx

2−2�0z2m + � �∗�w�xwxx

2−2�0z2m��

≤ ‖�∗�x�xz2m,�∗

��z2m‖L∞

�‖w�‖2L2

�

+ ‖wx�x‖2L2�

�,

(2.77)�� ∗

xw��wxx

2−2�0z2m��≤ ‖z2mx�∗

x‖L∞

�‖w��x

1

2−�0‖2

L2�

+ ‖wxx1

2−�0‖2

L2�

�,

(2.78)

��∫ 𝜙∗wx𝜂𝜂wxx2−2𝜎0z2m

��=��∫ 𝜙∗

𝜂wxwx𝜂x

2−2𝜎0z2m

+2m∫ 𝜙∗wxwx𝜂x3

2−2𝜎0z2m−1

��≲ ‖𝜂𝜙∗

𝜂z2m,𝜙∗‖L∞

�‖xwx𝜂‖2L2

𝜂

+ ‖x1−𝜎0wx𝜂zm‖2

L2𝜂

+‖zm−1wxx1

2−𝜎0‖2

L2𝜂

�,

(2.79)

��∫ 𝜙∗𝜂𝜂x

wwxx2−2𝜎0z2m + ∫ 𝜙∗

𝜂𝜂w2xx2−2𝜎0z2m

��≲ ‖z2mx𝜂2𝜙∗

𝜂𝜂x, z2m𝜂2𝜙∗

𝜂𝜂‖L∞

�‖w𝜂‖2L2

𝜂

+ ‖wx𝜂x‖2L2𝜂

�.

(2.80)

𝜕x

2 � w2xx2−2𝜎0z2m + � w2

𝜂xx2−2𝜎0z2m

≲ � w2xx1−2𝜎0z2m + � w2

xx1−2𝜎0z2m−2 + (𝛿)� w2

𝜂,w2

𝜂𝜂x,w2

x𝜂x2.

179

1 3


Again, we may integrate in x to obtain:

The third and fourth terms on the right-hand side are controlled by ‖w‖2Q , and the

first term is controlled by the initial data. For the second term, we must use Eq. (2.38):

As the majorizing terms above have been controlled, we can conclude:

By subsequently relating wx to w�� and w�x to w�� , we have shown that ‖zmw‖Q(�0,0) ≤ (�,m, �0) , which is the k = 0 case of (2.46). We may upgrade the previous set of estimates to higher-order x derivatives by iterating Steps 2, 3, and 4. The mechanism for doing so is that each application of �x to the profile terms �∗ pro-duces an extra factor of x−1 , and similarly each time �x hits z, one extra factor of x−1 is produced by (2.13). As this is similar to the previous steps, we omit the details. We remark that controlling higher-order derivatives does not require smallness of the initial datum (hence, the smallness assumption (1.28) is only for j = 0, 1, 2 ). □

We now give the following corollaries:

Corollary 2.11 (u0p Asymptotics) For any m ≥ 0 , 2 ≤ p ≤ ∞,

(2.81)

supx≥1 � w2

xx2−2𝜎0z2m + �

∞

1 � w2𝜂xx2−2𝜎0z2m

≲ �x=1

wx(1)2y2m + �

∞

1 � w2xx1−2𝜎0z2m + �

∞

1 � w2xx1−2𝜎0z2m−2

+ (𝛿)�∞

1 � w2𝜂,w2

𝜂𝜂x

1

2 ,w2x𝜂x.

(2.82)

‖wxx1

2−𝜎0zm‖L2 ≤ ‖w𝜂𝜂x

1

2−𝜎0zm‖L2

+��

�w2𝜂+ 2𝜙∗

𝜂w𝜂 + ww𝜂𝜂 + 𝜙∗w𝜂𝜂 + w𝜙∗

𝜂𝜂

�x

1

2−𝜎0zm

��L2≤ ‖w𝜂𝜂x

1

2−𝜎0zm‖L2‖w𝜂x

1

2 ,𝜙∗𝜂x

1

2 ,w,

𝜙∗,𝜙∗𝜂𝜂𝜂x

1

2 ‖L∞‖w𝜂x−𝜎0zm,w𝜂𝜂x

1

2−𝜎0zm‖L2

≲ ‖w𝜂𝜂x1

2−𝜎0zm,w𝜂x

−𝜎0zm‖L2 .

(2.83)supx≥1 � w2

xx2−2�0z2m + �

∞

1 � w2�xx2−2�0z2m ≤ (�).

(2.84)‖zm𝜕ly𝜕kxu0p‖Lpy ≤ C(m, l, k)x

−k−l

2+

1

2p for 2k + l > 2,

(2.85)‖zm�ly�kxu0p‖Lpy ≤ (�;m, l, k)x−k− l

2+

1

2p for 2k + l ≤ 2.

180 S. Iyer

1 3

Proof We start with the representation of u0p via (2.36):

The first quantity on the right-hand side above is controlled by Corollary 2.3. For the second quantity on the right-hand side, we simply use (2.46) with 0 < 𝜎0 <

1

4 , cou-

pled with the standard Sobolev interpolation. Recall now: � = ∫ y

01 + u0

p(x, �)d� ,

from which we have the equivalence: (1 − �)y ≤ � ≤ (1 + �)y . Moreover, we have that the Jacobians of the change of coordinates are given by: ��(y) = 1 + u0

p, y�(�) =

1

1+u0p

, both of which are bounded and bounded away from

zero. Therefore, we can transfer (2.86) to (x, y) coordinates. □

We may give bounds for v0p:

Corollary 2.12 (Estimates for v0p ) We have the following bounds for v, for any m ≥ 0:

Proof First, we give L2 via the Hardy inequality, which is available for m ≥ 0 as �kxv(x, y = ∞) = 0:

We have used estimate (2.84) for k > 0 , and one obtains the smallness of (�) for k = 0 by applying (2.85). The L∞ estimate then follows via standard Sobolev inter-polation. Indeed, as limy→∞ v0

p(x, y) = 0 , with rapid decay for fixed x, one writes:

Dividing both sides by xm then gives:

(2.86)‖zm�l��kxu0p‖Lp� = ‖zm�l

��kx

�� + e� − �

�‖Lp� + ‖zm�l

��kxw‖Lp� .

(2.87)‖zm�kxv0p(x, ⋅)‖L2

y≤ C(k,m)x−k−

1

4 for k ≥ 1,

(2.88)‖zmv0p(x, ⋅)‖L2

y≤ (�;m)x− 1

4 ,

(2.89)‖zm�kxv0p(x, ⋅)‖L∞

y≤ C(k,m)x−k−

1

2 for k ≥ 1,

(2.90)‖zmv0p(x, ⋅)‖L∞

y≤ (�;m)x− 1

2 .

(2.91)‖zm𝜕kxv0p(x, ⋅)‖L2

y≤ ‖zmy𝜕k

xv0py(x, ⋅)‖L2

y= ‖zmy𝜕k

xu0px(x, ⋅)‖L2

y≲ x

−k−1

4 .

(2.92)|ym�k

xv0p(x, y)|2 = ∫

∞

y

2(ym�kxv0p)�y(y

m�kxv0p) dy�

= ∫∞

y

2ym�kxv0pmym−1�k

xv0pdy� + ∫

∞

y

2(ym�kxv0p)ym�k

xv0pydy�.

181

1 3


from which the desired results follow upon consultation with (2.91) and (2.84)–(2.85). □

3 Euler‑1 Layer

3.1 Derivation of Equations

In this section, we construct the next order in the expansion, [u1e, v1

e,P1

e] . The analysis

will be taking place in Euler coordinates, that is (x, Y), where Y =√�y as in (1.9).

We make the notational convention that differential operators applied to [u1e, v1

e,P1

e]

will always be in (x, Y) coordinates, so for example Δu1e∶= �2

xu1e+ �2

Yu1e . Let us now

expand the partial expansion to find the lowest-order terms which we then take to solve the Euler-1 equation. The first equation is:

Removing now the terms found in Eq. (2.1), and retaining the lowest-order purely Euler terms gives:

The remaining terms from the above expansion are placed into the next order, which is discussed in detail in Eqs. (4.1)–(4.8). Let us now turn to the second equation:

Taking the order-1 purely Eulerian terms gives:

The remaining terms are contributed to the next-order error in (4.1)–(4.8). Putting (3.2)–(3.4) together with the divergence-free condition yields the following system for the Euler-1 layer:

(2.93)‖zm𝜕kxv0p(x, ⋅)‖L∞

y≲ ‖zm𝜕k

xv0p(x)‖

1

2

L2y

‖zm𝜕kxv0py(x)‖

1

2

L2y

+ x−

1

4 ‖zm−1

2 𝜕kxv0p‖L2

y,

(3.1)

− Δ𝜖u(1)s

+ u(1)su(1)sx

+ v(1)su(1)sy

+ P(1)sx

= −Δ𝜖u0p− 𝜖

3

2Δu1e+

�u0s+√𝜖u1

e

��u0px+√𝜖u1

ex

�

+ (v0p+ v1

e)(u0

py+ 𝜖u1

eY) +

√𝜖P1

ex+ 𝜖P1,a

ex.

(3.2)√�[u1

ex+ P1

ex] = 0.

(3.3)

−Δ𝜖v(1)s

+ u(1)sv(1)sx

+ v(1)sv(1)sy

+𝜕y

𝜖P(1)s

= −Δ𝜖v0p− 𝜖Δv1

e+

�u0s+√𝜖u1

e

�(v0

px+ v1

ex)

+ (v0p+ v1

e)

�v0py+√𝜖v1

eY

�+ P1

eY+√𝜖P

1,a

eY.

(3.4)v1ex+ P1

eY= 0.

182 S. Iyer

1 3

with prescribed boundary data

Without loss of generality, taking P1e= −u1

e , gives the div-curl system

Notice that these equations are the Cauchy–Riemann equations for v1e= Re(f ) ,

u1e= Im(f ) , f holomorphic on Ω . Then by taking the curl of the equation,

The following boundary decay rates follow from (2.89)–(2.90):

3.2 Uniform Decay Estimates

Motivated by (3.9), let us define for this section f ∶= �x≥−10E(m)

ℝ+

v0p , where E(m)

ℝ+

is the m’th order Sobolev extension operator on the half-line ℝ+ (see [1, Chapter 5]). Here, m is selected as large as required by the remainder of our analysis. Then, f agrees with v0

p on [0,∞) and is cut off past x ≤ −10 . Then by standard Sobolev extension

theory, ‖f‖Hm ≤ C(m)‖v0p‖Hm . Our setup for this subsection is:

Using the Poisson kernel, we have:

We will need the following pointwise estimates on the Poisson kernel:

Lemma 3.1 For x > 0 , k ≥ 0 , and 0 ≤ s ≤ k , we have:

Proof First, let us address the k = 0 case. Indeed,

For k ≥ 1 , we record the basic identities:

(3.5)u1ex+ P1

ex= 0, v1

ex+ P1

eY= 0, u1

ex+ v1

eY= 0, in Ω,

(3.6)v1e(x, 0) = −v0

p(x, 0).

(3.7)v1ex− u1

eY= 0, u1

ex+ v1

eY= 0.

(3.8)Δv1e= 0 in Ω, v1

e(x, 0) = −v0

p(x, 0).

(3.9)|�kxv0p(x, 0)| ≤ C(k)x

−1

2−k, |v0

p(x, 0)| ≤ (�)x− 1

2 .

(3.10)Δv = 0 on ℍ, v(x, 0) = f (x), |�kxf (x)| ≤ x

−k−1

2 , |f (x)| ≤ (�)x− 1

2 .

(3.11)v(x, Y) = PY ∗ f = ∫∞

−∞

PY (x − t)f (t)dt = ∫∞

−∞

Y

Y2 + (x − t)2f (t)dt.

(3.12)|𝜕kxPY (x)| ≲ x−s−1Y−(k−s).

(3.13)xPY (x) =Yx

Y2 + x2≤ Y2 + x2

Y2 + x2≤ 1.

183

1 3


For the k even case, we multiply (3.14) by xk+1 , use the binomial formula, and then apply Young’s inequality for products in the following manner:

where the Young’s conjugates are:

Performing the same calculation using (3.15), we have:

(3.14)�kxPY (x) =

k

2∑

j=0

cj,kYx2j(Y2 + x2)

−(k

2+1)−j if k even,

(3.15)�kxPY (x) =

k−1

2∑

j=0

cj,kYx2j+1(Y2 + x2)

−k+1

2−1−j if k odd.

(3.16)

|xs+1Yk−s𝜕kxPY (x)| =

|||||||

k

2∑

j=0

cj,kY1+k−sx2j+s+1(Y2 + x2)

−(k

2+1)−j

|||||||

≲

k

2∑

j=0

cj,kY1+k−sx2j+s+1

Yk+2+2j + xk+2+2j

≲

k

2∑

j=0

Y(1+k−s)(

k+2+2j

1+k−s)+ x

(s+1+2j)(k+2+2j

s+1+2j)

Yk+2+2j + xk+2+2j≲ 1,

(3.17)1 =1 + k − s

k + 2 + 2j+

s + 1 + 2j

k + 2 + 2j.

(3.18)

|xs+1Yk−s𝜕kxPY (x)| =

|||||||

k−1

2∑

j=0

cj,kY1+k−sxs+2j+2(Y2 + x2)

−k+1

2−1−j

|||||||

≲

k−1

2∑

j=0

cj,kY1+k−sxs+2j+2

Yk+3+2j + xk+3+2j

≲

k−1

2∑

j=0

cj,kY(1+k−s)(

k+2j+3

1+k−s)+ x

(s+2j+2)(k+2j+3

s+2j+2)

Yk+3+2j + xk+3+2j≲ 1.

184 S. Iyer

1 3

Here, the Young’s conjugates are:

This proves (3.12). □

We now prove the following uniform estimates for v1e(x, Y) , which now drop

superscripts, and denote by v for notational ease:

Proposition 3.2 Let v be the Poisson extension defined via (3.11). Then v satisfies the following bounds:

Proof The Y = 0 case is clear by (3.9), and so we must treat the case when Y > 0 . In this regime, several qualitative facts are available for use. In particular, PY is smooth, has unit mass, and is positive everywhere. As is standard, we shall exploit these qualitative facts to extract the quantitative bounds independent of Y > 0 . Fix-ing any 𝛼 > 0 , we shall split the integral (3.11) into three pieces:

For the sake of concreteness, fix: � =1

10 000 . � and � will be taken small relative to

this universal constant, � . Let us first turn to I3:

(3.19)1 =1 + k − s

k + 2j + 3+

s + 2j + 2

k + 2j + 3.

(3.20)supY≥0

|�kxv(x, Y)| ≤ C(k)x

−k−1

2 , supY≥0

|v(x, Y)| ≤ (�)x− 1

2 .

(3.21)

v(x, Y) = ∫−

x

1+�

−∞

PY (x − t)f (t)dt + ∫x

1+�

−x

1+�

PY (x − t)f (t)dt

+ ∫∞

x

1+�

PY (x − t)f (t)dt

= I1 + I2 + I3.

x1

2 |I3| =||||||�

∞

x

1+�

PY (x − t)x1

2 f (t)dt

||||||

≤||||||�

∞

x

1+�

PY (x − t){x

1

2 − |t|1

2

}f (t)dt

||||||+

||||||�

∞

x

1+�

PY (x − t)|t|1

2 f (t)dt

||||||

185

1 3


where we have used for each fixed Y > 0 , PY ≥ 0 and ∫ PY = 1 . By symmetry, I1 works the same way. Therefore, we are left with I2 , where the main mechanisms are pointwise estimates on PY . According to (3.12), with k = s = 0,

as x − t > 0 into the region − x

1+�≤ t ≤ x

1+� . Thus, we have:

Combining (3.22)–(3.24), we may conclude that

We now need to establish this procedure for higher-order x derivatives for v. We continue with the I1, I2, I3 splitting used above. Upon differentiating (3.21) with respect to x, let us treat each term Ii individually.

(3.22)

≤ �∞

x

1+�

PY (x − t)

��x

�t�

� 1

2

− 1

��t�

1

2 f (t)dt + ‖fx1

2 ‖L∞ �∞

−∞

PY (x − t)dt

≤ supt≥ x

1+�

��

�x

�t�

� 1

2

− 1

��‖fx

1

2 ‖L∞ �∞

−∞

PY (x − t)dt + (�) ≤ (�),

(3.23)sup−

x

1+𝛼≤t≤ x

1+𝛼

PY (x − t) ≲1

x,

(3.24)|I2| ≤ �

x

1+𝛼

−x

1+𝛼

PY (x − t)|f (t)|dt ≲ 1

x �x

1+𝛼

−x

1+𝛼

|f (t)|dt ≤ (𝛿)x �

x

1+𝛼

−x

1+𝛼

|1 + t|−1

2 dt

≤ (𝛿)x− 1

2 .

(3.25)|||x1

2 v||| ≤ (�).

(3.26)

�xI1 =−1

1 + �PY

(x +

x

1 + �

)f(−

x

1 + �

)+ ∫

−x

1+�

−∞

�xPY (x − t)f (t)dt

=−1

1 + �PY

(x +

x

1 + �

)f(−

x

1 + �

)− ∫

−x

1+�

−∞

�tPY (x − t)f (t)dt

= B1 + ∫−

x

1+�

−∞

PY (x − t)f �(t)dt.

186 S. Iyer

1 3

Here,

Note now in a similar manner to (3.12),

Using the same method as evaluating (3.22):

Next, we turn to I2 , which after differentiating gives:

Via (3.12) with k = 1, s = 0,

(3.27)

B1 =−1

1 + �PY

(x +

x

1 + �

)f(−

x

1 + �

)− PY

(x +

x

1 + �

)f(

−x

1 + �

).

(3.28)|B1| ≲ PY

(2 + 𝛼

1 + 𝛼x)||||f(−

x

1 + 𝛼

)||||≲

1

x(𝛿)x− 1

2 ≲ x−

3

2 .

(3.29)

x3

2

��−

x

1+𝛼

−∞

PY (x − t)f �(t)dt��

≤ �−

x

1+𝛼

−∞

PY (x − t)�t�3

2 �f �(t)�dt + �−

x

1+𝛼

−∞

PY (x − t)��

�x

3

2 − �t�3

2

��f �(t)�dt

≤ �∞

−∞

PY (x − t)�t�3

2 �f �(t)�dt + �−

x

1+𝛼

−∞

PY (x − t)��

�x

3

2 − �t�3

2

��f �(t)�dt

≤ ‖x3

2 f �‖L∞ + �−

x

1+𝛼

−∞

PY (x − t)

��

��x

�t�

� 3

2

− 1

��t�

3

2 �f �(t)�dt

≲ ‖x3

2 f �‖L∞ + supt≤− x

1+𝛼

��

��x

�t�

� 3

2

− 1

��‖x

3

2 f �‖L∞ �∞

−∞

PY (x − t)dt ≤ C.

(3.30)

�xI2 =1

1 + �PY

(x −

x

1 + �

)f(

x

1 + �

)

+1

1 + �PY

(x +

x

1 + �

)f(−

x

1 + �

)

+ ∫x

1+�

−x

1+�

�xPY (x − t)f (t)dt.

187

1 3


For I2 , this then facilitates the bound:

For the �xI3 term, we must integrate by parts in the following manner:

Here, B3 contains the boundary terms:

Again, via direct calculation using the definition of PY , it is clear that:

We may estimate the integral in (3.33) in identical fashion to (3.29). This procedure can be iterated for higher-order derivatives. To see this, let us start with the splitting given in (3.21). We will apply �k

x , where k ≥ 2:

Our starting point is the expression (3.26), from which it becomes clear that apply-ing �k

x to I1 , for generic constants c𝛼 > 0 (which can change from term to term), we

have:

(3.31)supY>0,t∈[−

x

1+𝛼,

x

1+𝛼]

|𝜕xPY (x − t)| ≤ supY>0,t∈[−

x

1+𝛼,

x

1+𝛼]

1

(x − t)2≲

1

x2.

(3.32)||||||∫

x

1+𝛼

−x

1+𝛼

𝜕xPY (x − t)f (t)dt

||||||≲

1

x2 ∫x

1+𝛼

−x

1+𝛼

(1 + t)−

1

2 dt ≲ x−

3

2 .

(3.33)

�xI3 =1

1 + �P(x −

x

1 + �

)f(

x

1 + �

)+ ∫

∞

x

1+�

�xPY (x − t)f (t)dt

= B3 + ∫∞

x

1+�

PY (x − t)f �(t)dt.

(3.34)B3 =

1

1 + �PY

(x −

x

1 + �

)f(

x

1 + �

)− PY

(x −

x

1 + �

)f(

x

1 + �

)

=1

1 + �PY

(�

1 + �x)f(

x

1 + �

)− PY

(�

1 + �x)f(

x

1 + �

).

(3.35)|B3| ≲ x−

3

2 .

(3.36)�kxv(x, Y) = �k

xI1 + �k

xI2 + �k

xI3.

188 S. Iyer

1 3

Using (3.12), we estimate:

For the integration term in (3.37), we follow (3.29) to give:

By symmetry, the same estimate applies to I3 , and so we turn to I2 . Referring to (3.30), one sees the following expression:

(3.37)�kxI1 =

k−1∑

j=0

cj�jxPY (c�x)�

k−1−jx

f (c�x) + ∫−

x

1+�

−∞

PY (x − t)�ktf (t)dt.

(3.38)|𝜕jxPY (c𝛼x)𝜕

k−1−jx

f (c𝛼x)| ≤ c𝛼x−j−1x

−(k−1−j)−1

2 ≲ x−k−

1

2 .

(3.39)

xk+

1

2

��−

x

1+𝛼

−∞

PY (x − t)𝜕ktf (t)dt

��

≤ �−

x

1+𝛼

−∞

PY (x − t)�t�k+1

2��𝜕

ktf (t)

��dt

+ �−

x

1+𝛼

−∞

PY (x − t)��

�xk+

1

2 − �t�k+1

2

��𝜕

ktf (t)

��dt

≤ �∞

−∞

PY (x − t)�t�k+1

2��𝜕

ktf (t)

��dt

+ �−

x

1+𝛼

−∞

PY (x − t)��

�xk+

1

2 − �t�k+1

2

��𝜕

ktf (t)

��dt

≤ ‖xk+1

2 𝜕kxf‖L∞ + �

−x

1+𝛼

−∞

PY (x − t)

��

��x

�t�

�k+1

2

− 1

��t�k+

1

2��𝜕

ktf (t)

��dt

≲ ‖xk+1

2 𝜕kxf‖L∞ + sup

t≤− x

1+𝛼

��

��x

�t�

�k+1

2

− 1

��‖xk+

1

2 𝜕kxf‖L∞

× �∞

−∞

PY (x − t)dt ≤ C.

(3.40)�kxI2 =

k−1∑

j=0

cj�jxPY (c�x)�

k−1−jx

f (c�x) + ∫x

1+�

−x

1+�

�kxPY (x − t)f (t) dt,

189

1 3


where c𝛼 > 0 denotes generic constants which depend on � , not on �, � , and are strictly positive. Referring to (3.38), it remains to estimate the integral term above:

This then proves the desired result. □

By repeating the above proof, we actually have the following:

Corollary 3.3 Consider boundary values g(x) satisfying:

Then the Poisson extension satisfies:

We now give an estimate more suitable to obtaining L2 in Y estimates.

Lemma 3.4 For any k ≥ 1 , and 0 ≤ k − s ≤ 1 , we have:

Proof By Lemma 3.2, the claim is only meaningful for Y ≥ x , so make this assump-tion to begin. Again, we start with the splitting (3.21). By symmetry it suffices to consider I2, I3 . Applying �k

x , we have the expression (for generic constants c𝛼 > 0):

First,

(3.41)

||||||�

x

1+𝛼

−x

1+𝛼

𝜕kxPY (x − t)f (t) dt

||||||≤ �

x

1+𝛼

−x

1+𝛼

|𝜕kxPY (x − t)||f (t)| dt

≲ sup−

x

1+𝛼≤t≤ x

1+𝛼

|𝜕kxPY (x − t)|�

x

1+𝛼

−x

1+𝛼

|(1 + t)|1

2 dt

≲ x−k−1x1

2 = x−k−

1

2 .

(3.42)|�kxg(x)| ≤ x−k−w, where w ∈ (0, 1).

(3.43)supY

|𝜕kx{PY ∗ g}| ≲ x−k−w.

(3.44)|𝜕kxv(x, Y)| ≲ x

−s−1

2 Y−(k−s).

(3.45)�kxI2 =

k−1∑

j=0

cj,k,��jxPY (c�x)�

k−1−jx

f (c�x) + ∫x

1+�

−x

1+�

�kxPY (x − t)f (t)dt.

(3.46)|𝜕j

xPY (c𝛼x)𝜕

k−1−jx

f (c𝛼x)| ≲ Y−(k−s)x−j−1+(k−s)x−(k−1−j)−

1

2

≲ Y−(k−s)x−s−

1

2 .

190 S. Iyer

1 3

Above, we have used (3.12) for the case j ≥ 1 . For the j = 0 case, we use the assumption that Y ≥ x , and the following estimate on PY , which is valid so long as k − s ≤ 1:

For the integral term in (3.45), via (3.12),

From here, the estimate on the integral in (3.45) follows:

For the I3 contribution, we apply �kx and integrate by parts in t; it reads:

First, as shown in (3.46),

Turning to the integration in (3.50), our point of view on the kernel is:

Then in the regime Y ≥ x , we have (since k − s ≤ 1):

(3.47)PY (x − t) =Y

Y2 + (x − t)2≤ Y

Y2=

1

Y≤ Y−(k−s)x−1+(k−s).

(3.48)sup

Y≥0, t∈[− x

1+𝛼,

x

1+𝛼

] |𝜕kxPY (x − t)| ≲ Y−(k−s)x−s−1.

(3.49)||||||∫

x

1+𝛼

−x

1+𝛼

𝜕xPY (x − t)f (t)dt

||||||≲ Y−(k−s)x−s−1 ∫

x

1+𝛼

−x

1+𝛼

(1 + t)−

1

2 dt ≲ Y−(k−s)x−s−

1

2 .

(3.50)�xI3 =

k−1∑

j=0

cj,k,��jxPY (c�x)�

k−1−jx

f (c�x) + ∫∞

x

1+�

PY (x − t)�ktf (t)dt.

(3.51)|𝜕jxPY (c𝛼x)𝜕

k−1−jx

f (c𝛼x)| ≲ Y−(k−s)x−s−

1

2 .

(3.52)PY (x − t) =Y

Y2 + (x − t)2≤ Y

Y2=

1

Y.

(3.53)

||||||�

∞

x

1+�

PY (x − t)�ktf (t)dt

||||||≤ 1

Y �∞

x

1+�

t−k−

1

2 dt ≤ 1

Yx−k+

1

2

≤ 1

Yx−k+

1

2

(Y

x

)1−(k−s) ≤ Y−(k−s)x−s−

1

2 .

191

1 3


This establishes the desired result. □

Lemma 3.5 (Interpolation Estimates) We have the following decay estimates:

Proof We treat the case when k = 1 , with the extension for higher k values being obvious. This follows via standard interpolation arguments: fix an x, and consider vY (x, ⋅) as a function of the Y variable, defined on the half-space Y ≥ 0 . Then via Gagliardo–Nirenberg interpolation estimates, and via harmonicity vxx = −vYY,

Importantly, via Remark 4, Page 125 in [39], we need not include a lower-order term on the right-hand side of (3.55), which is due to the fact that our domain, for each fixed x, is the half-space Y ≥ 0 . Taking the supremum over x and applying (3.20) yield the desired result. The smallness of (�) for k = 1 case is guaranteed due to the v term in (3.55). □

Proposition 3.6 (Euler Correctors) Suppose [u1e, v1

e] solves the boundary value

problem:

Then the following bounds hold for any k, j ≥ 0:

Proof We shall take v1e= v from the above lemmas. For the u1

e profile, we define for

x ≥ 1:

From here it is clear that the Cauchy–Riemann equations hold:

We have used that v1ex(x, Y) → 0 as x → ∞ . From Lemma 3.5, it is then clear that:

(3.54)xk+

1

2 ‖�kYv(x, ⋅)‖L∞

Y≤ C, x

3

2 ‖�Yv(x, ⋅)‖L∞Y≤ (�).

(3.55)x3

2 ‖vY (x, ⋅)‖L∞Y≲

�x

1

2 ‖v(x, ⋅)‖L∞Y

� 1

2�x

5

2 ‖vxx(x, ⋅)‖L∞Y

� 1

2

.

(3.56)Δv1e= 0, v1

e(x, 0) = −v0

p(x, 0), u1

e= �

∞

x

v1eY(x�, Y)dx�, for x ≥ 1.

(3.57)supY≥0

|�kx�j

Yv1e|xk+j+

1

2 ≤ C, supY≥0

[|u1

e, v1

e|x

1

2 + |u1ex, v1

eY|x

3

2

] ≤ (�).

(3.58)u1e(x, Y) ∶= ∫

∞

x

v1eY(x�, Y)dx�.

(3.59)u1ex= −v1

eY, u1

eY= ∫

∞

x

v1eYY

= −∫∞

x

v1exx

= v1ex.

192 S. Iyer

1 3

□

Remark 3.7 (In-Flow Conditions) Note that we do not solve for the Euler correctors given an arbitrary in-flow condition at x = 1 . Rather, we take [u1

e, v1

e,P1

e] to be the

explicit profiles obtained by applying the Poisson extension to f. In this sense, the setup we consider here is distinct from the construction of Euler profiles in [24].

4 Prandtl Layer 1

4.1 Derivation of Linearized Prandtl Equations

In this section, we will construct the Prandtl correctors, [u1p, v1

p,P1

p] . Let us now

obtain the equations that they will satisfy. We do this in a manner which can be eas-ily generalized in the next section. We expand the nonlinear terms:

Let us now denote:

(3.60)|u1e| =

||||�∞

x

u1exdx�

||||=||||�

∞

x

v1eYdx�

||||≤ (�)�

∞

x

|x�|−3

2 dx� ≤ (�)x− 1

2 .

(4.1)

u(1)su(1)sx

=

�u(0)s

+√𝜖u1

e+√𝜖u1

p

��u(0)sx

+√𝜖u1

ex+√𝜖u1

px

�

= u(0)su(0)sx

+√𝜖u(1)

sxu1p+√𝜖u(1)

su1px+ 𝜖u1

pu1px+√𝜖u1

exu0p+√𝜖u1

eu0px

+√𝜖u1

ex+ 𝜖u1

eu1ex,

(4.2)v(1)su(1)sy

=

�v0p+ v1

e+√𝜖v1

p

��u0py+ 𝜖u1

eY+√𝜖u1

py

�

= v(1)su0py+√𝜖u(1)

syv1p+√𝜖v(1)

su1py+ 𝜖v1

pu1py+ 𝜖v0

pu1eY

+ 𝜖v1eu1eY,

(4.3)

u(1)sv(1)sx

=

�1 + u0

p+√𝜖u1

e+√𝜖u1

p

��v0px+ v1

ex+√𝜖v1

px

�

= u(0)sv0px+√𝜖v1

pxu(1)s

+√𝜖v(1)

sxu1p+ 𝜖u1

pv1px+ u0

pv1ex+√𝜖u1

ev0px

+ v1ex+√𝜖u1

ev1ex,

(4.4)

v(1)sv(1)sy

=

�v0p+ v1

e+√𝜖v1

p

��v0py+√𝜖v1

eY+√𝜖v1

py

�

= v0pv0py+√𝜖v1

pyv(1)s

+√𝜖v(1)

syv1p+ 𝜖v1

pv1py

+√𝜖v0

pv1eY

+ v1ev0py+√𝜖v1

ev1eY.

193

1 3


The purpose of the terms above is to separate the Euler–Prandtl terms and the pure-Euler terms. The pure-Euler terms are harmful to our analysis, because they scale differently from the Euler–Prandtl terms. From a practical point of view, their pres-ence prevents the application of weights of the form z = y√

x , and therefore obstructs

self-similarity. It turns out that all pure-Euler terms are of “gradient-type”; see (5.30). Thus, by introducing appropriate potential functions in the pressure expan-sion, we can force these terms to vanish identically. With this notation, we may write the Navier–Stokes expansion as:

The normal equation is expanded via:

In the above expressions, [u(0)s, v(0)

s] , [u(1)

s, v(1)

s] are known at this stage, and u1

p, v1

p are

unknowns, to be constructed in this step. Similarly for the pressures, P1e is known,

but the auxiliary pressures P1,a

E,P1,a

p are to be defined in this section. Finally, as can

be seen from the definitions in (4.5), (4.6), Ru,1

E,R

v,1

E,E

u,1

E,E

v,1

E are all knowns. Let us

now simplify the above expressions. First, via the construction of [u0p, v0

p] we may

write:

(4.5)Ru,1

E∶= v1

eu0py+√�u1

exu0p+√�u1

eu0px+ �v0

pu1eY, E

u,1

E∶= �v1

eu1eY

+ �u1eu1ex,

(4.6)Rv,1

E= u0

pv1ex+√�u1

ev0px+√�v0

pv1eY

+ v1ev0py, E

v,1

E∶=

√�u1

ev1ex+√�v1

ev1eY.

(4.7)

− Δ𝜖 u(1)s

+ u(1)su(1)sx

+ v(1)su(1)sy

+ P(1)x

= −Δ𝜖 u(0)s

+ u(0)su(0)sx

+ v(0)su(0)sy

+ P(0)x

+√𝜖

�−Δ𝜖u

1p+ u(1)

su1px+ u(1)

sxu1p+ u(1)

syv1p+ v(1)

su1py+√𝜖u1

pu1px

+√𝜖v1

pu1py+ P1

px

�+ 𝜖P1,a

px+ R

u,1

E+ E

u,1

E− 𝜖

3

2Δu1e+ 𝜖P

1,a

Ex+√𝜖(u1

ex+ P1

ex).

(4.8)

− Δ𝜖 v(1)s

+ u(1)sv(1)sx

+ v(1)sv(1)sy

+P(1)y

𝜖= −Δ𝜖 v

(0)s

+ u(0)sv(0)sx

+ v(0)sv(0)sy

+P(0)y

𝜖

+√𝜖

�− Δ𝜖v

1p+ u(1)

sv1px+ v(1)

sxu1p+ v(1)

sv1py+ v(1)

syv1p+√𝜖u1

pv1px

+√𝜖v1

pv1py+

P1py

𝜖

�+

𝜕y

𝜖𝜖P1,a

p+ R

v,1

E+ E

v,1

E

− 𝜖Δv1e+

𝜕y

𝜖𝜖P

1,a

E+

�𝜕y

𝜖

√𝜖P1

e+ v1

ex

�.

(4.9)

Ru,0 ∶= −Δ𝜖 u(0)s

+ u(0)su(0)sx

+ v(0)su(0)sy

+ P(0)x

+ v1eu0py

= −𝜖u0pxx

+√𝜖yv1

eYu0py+ 𝜖u0

py ∫y

0 ∫y�

y

v1eYY

dy��dy�.

194 S. Iyer

1 3

This then accounts for the lowest-order term from Ru,1

E in (4.5), allowing us to

redefine:

Let us now define the auxiliary Euler pressure:

so that, combined with the equations we have taken for [u1e, v1

e] , we have:

Lemma 4.1 With P1,a

E as in (4.11), and Eu,1

E,E

v,1

E as in (4.5)–(4.6),

Remark 4.2 Denoting by ∇� ∶= (�x,�Y√�) , the above lemma reads:

Thus, the purely Eulerian terms are of gradient structure, which is then exploited with the introduction of our auxiliary pressure.

Proof The proof follows by direct calculation and an appeal to (3.5):

Similarly,

The claim has been proven. □

Corollary 4.3 All “purely Eulerian” terms from the expansions (4.7), (4.8) vanish identically. That is:

Proof First, by harmonicity of [u1e, v1

e] , we have:

(4.10)Ru,1

E=√�u1

exu0p+√�u1

eu0px+ �v0

pu1eY.

(4.11)P1,a

E∶= −

1

2(|u1

e|2 + |v1

e|2),

(4.12)�x�P1,a

E+ E

u,1

E=

�y

��P

1,a

E+ E

v,1

E= 0.

(4.13)∇��P1,a

E+ (E

u,1

E,E

v,1

E) = 0.

(4.14)��xP1,a

E= −�u1

eu1ex− �v1

ev1ex= −�u1

eu1ex− �v1

eu1eY

= −Eu,1

E.

(4.15)�Y√��P

1,a

E= −

√�u1

eu1eY

− �v1ev1eY

= −√�u1

ev1ex− �v1

ev1eY

= −Ev,1

E.

(4.16)Eu,1

E− �

3

2Δu1e+ �P

1,a

Ex+√�(u1

ex+ P1

ex) = 0,

(4.17)Ev,1

E− �Δv1

e+

�y

��P

1,a

E+

��y

�

√�P1

e+ v1

ex

�= 0.

195

1 3


Next, regarding the final terms in both (4.7) and (4.8), by the Eqs. (3.2) and (3.4), we see that these terms drop out:

Coupled with (4.12), this establishes the desired claim. □

In the above calculation, we use crucially the Cauchy–Riemann structure of [u1e, v1

e]

in (3.7); harmonicity alone does not suffice here. Next, define the auxiliary Prandtl pressure:

Combining all this, we take our Prandtl-1 equation to be:

together with the divergence-free condition u1px+ v1

py= 0 , and the boundary

conditions:

Here, the forcing term is defined as:

The boundary contribution of v1p(x, 0) in (4.21) again arises from the calculation:

Consolidating (4.18), (4.19), (4.12), (4.20), (4.21), (4.24) with the expressions (4.7) and (4.8) then shows that the following remainder is contributed:

(4.18)�3

2Δu1e= �Δv1

e= 0.

(4.19)u1ex+ P1

ex= 0, v1

ex+ P1

eY= 0.

(4.20)P1,ap

∶= ∫∞

y

[−Δ𝜖 v

(0)s

+ u(0)sv(0)sx

+ v(0)sv(0)sy

]+ ∫

∞

y

Rv,1

E.

(4.21)−u1pyy

+ (1 + u0p)u1

px+ u(1)

sxu1p+ u0

py(v1

p− v1

p(x, 0)) + v(1)

su1py+ P1

px= f (1),

(4.22)u1p(x, 0) = 0, lim

y→∞[u1

p(x, y), v1

p(x, y)] = 0, v1

p(x, 0) = −v2

e(x, 0), u1

p(1, y) = U1(y).

(4.23)f (1) ∶= �−

1

2 [Ru,0 + Ru,1

E+ �P1,a

px].

(4.24)v2e(x, Y)u0

py= v2

e(x, 0)u0

py+√�yu0

pyv2eY

+ �u0py ∫

y

0 ∫y�

y

v2eYY

= −v1p(x, 0)u0

py+√�yu0

pyv2eY

+ �u0py ∫

y

0 ∫y�

y

v2eYY

.

(4.25)

Ru,1 =√�

�−�u1

pxx+√�yv2

eYu0py+ �u0

py ∫y

0 ∫y�

y

v2eYY

+

��u1

eY+√�u1

py

�v1p

+√�(u1

e+ u1

p)u1

px

�,

196 S. Iyer

1 3

Let us emphasize the boundary condition at y → ∞ for v1p means that we will define:

We may evaluate Eq. (4.21) at y = ∞ to see that the pressure term drops out. That is, the pressure in the Prandtl layer is constant, so we may WLOG take P1

p= 0.

4.2 Global in x Existence and Decay

The first step is to homogenize the boundary conditions by introducing the new unknowns:

where �(y) is a cutoff function satisfying:

The mean-zero condition is meant to ensure that v(0) = 0 . The homogenized profiles now satisfy the following system:

along with the divergence-free condition ux + vy = 0 . Here:

We note that v(x, y) does not vanish as y → ∞ due to the definition in (4.28). The essential feature of v(x, y) that will be in use is that v(x, 0) = 0 . Examining (4.31), one observes that v is always accompanied by u0

p , which decays rapidly in y for each

fixed x. Let us now define the norm in which we shall control the Prandtl solutions:

(4.26)Rv,1 =

√�

�−Δ�v

1p+ u(1)

sv1px+ v(1)

sxu1p+ v(1)

sv1py+ v(1)

syv1p+√�u1

pv1px+√�v1

pv1py

�.

(4.27)v1p(x, y) = ∫

∞

y

u1px(x, y�)dy�.

(4.28)

u = u1p+ �(y)u1

e(x, 0), v = v1

p− v1

p(x, 0) + u1

ex(x, 0)I� (y), I� (y) = ∫

∞

y

�(�)d�,

(4.29)�(0) = 1, �ky�(0) = 0, ∫

∞

0

�(y)dy = 0.

(4.30)(1 + u0

p)ux − uyy + (u, v) = f (1) + ; u(x, 0) = v(x, 0) = 0, lim

y→∞u(x, y) = 0,

(4.31) ∶= u(1)sxu + u0

pyv + v(1)

suy,

(4.32) ∶= u(1)

s�(y)u1

ex(x, 0) − � ��u1

e(x, 0) − �(y)u(1)

sxu1e(x, 0) − u0

pyI� (y)u

1ex(x, 0)

+ v(1)s� �(y)u1

e(x, 0).

(4.33)

‖u‖2P(X1,�)

∶= sup1≤x≤X1

�

� u2x−2� + u2yx1−2�

�+ �

X1

1 ��u2x−1−2� + u2

yx−2� + u2

xx1−2�

�.

197

1 3


We shall also need the following, differentiated version, of the above Prandtl-layer norm:

Through the H1y↪ L∞

y embedding, it is clear that:

Finally, we shall write the global norms as:

Remark 4.4 A comparison of the norms Pk with the estimates valid for [u0p, v0

p] in

(2.85), (2.90) shows that u1p is roughly “ x−

1

4-better” than u0p . There is no front pro-

file as in the case of [u0p, v0

p] , which is because the present boundary condition,

u1p(x, 0) = −u1

e(x, 0) decays as x → ∞ , due to (3.60), in contrast with the boundary

condition for u0p(x, 0) = −� . The secondary reason is because f (1) in (4.23) contains

derivative and nonlinear terms from the previous layers, which enhances the decay in x.

The first step is to give the following estimates on the forcing terms, which capitalize on the structure of these terms which either have many derivatives (thereby enhancing decay in x) or are of product form (also enhancing decay in x):

Lemma 4.5 (Forcing Estimates) For any k,m ≥ 0 , and arbitrary N > 0,

Proof The estimate for the terms is direct, thanks to Proposition 3.6; estimate (3.57). For f (1) , in consultation with the definition (4.23), we start with Ru,0 , defined in (4.9):

(4.34)

‖u‖2Pk(X1,�)

∶= sup1≤x≤X1

�

� ��kxu�2x2k−2� + ��k

xuy�2x2k+1−2�

�

+ �X1

1 ��k

xu�2x2k−1−2� + ��k

xuy�2x2k−2� + ��k+1

xu�2x2k+1−2�

�.

(4.35)sup1≤x≤X1

xk+

1

4−�‖�k

xu‖L∞ ≤ ‖u‖Pk(X1,�)

.

(4.36)‖u‖P(�) ∶= ‖u‖P(∞,�), ‖u‖Pk(�)∶= ‖u‖Pk(∞,�).

(4.37)�zm�kx � ≤ C(k,m)⟨y⟩−Nx−k−

1

2 ,

(4.38)‖zm�kxf (1)‖L2

y≤ C(k,m)⟨x⟩−k−

5

4 .

(4.39)√𝜖‖u0

pxx‖L2

y≲√𝜖x

−7

4 ,

(4.40)‖yu0pyv1eY‖L2

y≤ ‖x−

1

4 yu0py‖L2

ysupy

�v1eY�x

1

4 ≲ x−

5

4 ,

198 S. Iyer

1 3

Second, we control Ru,1

E via:

Next, according to (4.20), we have:

For each of these terms, the ability to trade x3

4+

�

2 for y3

2+� is in constant use:

Above, we use the established estimates in (2.85), (2.90) for the Prandtl-0 profiles, and (3.57) for the Euler-1 profiles. The desired estimates are proven for m = 0 . For general m the estimates work in an identical manner, after noticing that powers of z play no role when accompanied by [u0

p, v0

p] , which appear in every term above. □

The above lemma relies crucially on Corollary 4.3 to apply the weight zm . We now give the following energy estimate:

Lemma 4.6 Let 𝜎 > 0 and fix any X1 > 1 . Then:

(4.41)√𝜖��u0py �

y

0 �y�

y

v1exxdy��dy�

��L2y

≤ √𝜖‖y2u0

py‖L2

y‖v1

exx‖L∞

y≲√𝜖x

−7

4 .

(4.42)𝜖−

1

2 ‖Ru,1

E‖L2

y≤ ‖u1

ex‖L∞

y‖u0

p‖L2

y+ ‖u1

e‖L∞

y‖u0

px‖L2

y+√𝜖‖v0

p‖L2

y‖u1

eY‖L∞

y

≲ x−

5

4 .

(4.43)

‖P1,apx‖L2

y≤ ��

⟨y⟩3

2+𝜅𝜕x

�−Δ𝜖 v

(0)s

+ u(0)sv(0)sx

+ v(0)sv(0)sy

��L∞y

+ ‖⟨y⟩3

2+𝜅Rv,1

Ex‖L∞

y.

(4.44)

‖⟨y⟩3

2+𝜅Δ𝜖v

0px‖L∞

y+ ‖⟨y⟩

3

2+𝜅𝜕x{u

(0)sv(0)sx}‖L∞

y+ ‖⟨y⟩

3

2+𝜅𝜕x{v

(0)sv(0)sy}‖L∞

y

+ ‖⟨y⟩3

2+𝜅𝜕x{u

0pv1ex}‖L∞

y+��⟨y⟩

3

2+𝜅𝜕x

�√𝜖u1

ev0px

��L∞y

+��⟨y⟩

3

2+𝜅𝜕x

�√𝜖v0

pv1eY

��L∞y

+ ‖⟨y⟩3

2+𝜅𝜕x{v

1ev0py}‖L∞

y≲ x

−3

2 .

(4.45)sup

x∈[1,X1]� x−2𝜎u2 + �

X1

1 � x−1−2𝜎u2 + �X1

1 � x−2𝜎u2y

≲ (𝛿;𝜎)�X1

1 � v2yx1−2𝜎 + C(𝜎).

199

1 3


The constant above depends poorly on small �.

Remark 4.7 The need for this 𝜎 > 0 is to avoid certain x-integrations being critical. We make the notational convention that we will not rename different values for � (for instance 2� ), as we can always redefine � to be smaller. However, within a sin-gle calculation (for instance the upcoming proof) we fix a 𝜎 > 0.

Proof Applying the multiplier M = ux−2� to the system (4.30) gives the following terms:

The constant in the above estimate depends poorly on � , as there is a factor of � accompanying the ∫ x−1−2�u2 term. The final term above then gets placed into the contributions from , to which we now turn (see the definition in (4.31)):

Above, we have used the Prandtl-0 bounds in (2.85), which crucially provides the smallness of {u0

px, u0

py} in terms of (�) . No smallness of Eulerian profiles is required

due to the extra factor of √� . The key estimate which forces a loss of �x derivative is

the following:

The structure of the key estimate above, (4.48), is omnipresent in our analysis: we use the y-absorption of u0

py to produce a vy term via Hardy’s inequality (which is

valid as v(x, 0) = 0 ). This then forces a loss of x1

2−� in the decay, which must then be

regained in the next lemma. Again, the estimate ‖yu0py‖L∞ ≤ (�) arises from (2.85).

Next, we arrive at the forcing terms. First, via (4.37):

(4.46)

∫((1 + u0

p)𝜕x − 𝜕yy

)u ⋅ ux−2𝜎 ≳ 𝜕x ∫ (1 + u0

p)x−2𝜎u2

+ ∫ x−1−2𝜎u2 + ∫ x−2𝜎u2y− ∫ u0

pxu2x−2𝜎 .

(4.47)

�� ⋅ ux−2��≤ ‖xu(1)

sx, yu0

py, v(1)

syx‖∞

�

� u2x−1−2� + � v2yx1−2�

�

≤ (�)�

� u2x−1−2� + � v2yx1−2�

�.

(4.48)

�� u0pyv ⋅ ux−2�

��≤ ‖u0

pyy‖L∞ �

��v

yux−2�

��≤ (�)‖vyx 1

2−�‖L2

y‖ux−

1

2−�‖L2

y.

(4.49)

�� ⋅ ux−2𝜎��≲ � ⟨y⟩−Nx−

1

2 �u�x−2𝜎 ≲ ‖x−1

2−𝜎⟨y⟩−

N

2 ‖L2y

��x−𝜎

u

⟨y⟩N

2

��L2y

≤ 1

100 000‖uyx−𝜎‖2L2

y

+ Cx−1−2𝜎 .

200 S. Iyer

1 3

Upon taking an integration in dx , the majorizing terms above are finite. Next, according to (4.38), via Young’s inequality:

Placing these estimates together:

Integrating above from x = 1 to x = X1 yields:

Finally, we reason as follows: the second and third terms on the left-hand side above are positive, which then gives for this X1:

For any X2 ∈ [1,X1] , the same estimate holds, namely:

Above, we have used the monotonicity of the right-hand side as X2 < X1 . This allows us to replace in (4.50) the first term on the left-hand side with sup1≤x≤X1

∫ x−2�u2dy . This then gives the desired estimate in (4.45). □

We now recover the vy term on the right-hand side of (4.45) via:

Lemma 4.8 Let 𝜎 > 0 and fix any X1 > 1 . Then:

�� f (1) ⋅ ux−2��≤ ‖f (1)x

1

2−�‖L2

y‖ux−

1

2−�‖L2

y≤ Cx

−3

2 +1

100 000 � u2x−1−2�dy.

𝜕x � x−2𝜎u2 + � x−1−2𝜎u2 + � x−2𝜎u2y≲ (𝛿)� u2

xx1−2𝜎 + x−1−2𝜎 .

(4.50)� X−2𝜎

1u2(X1)dy + �

X1

1 � x−1−2𝜎u2 + �X1

1 � x−2𝜎u2y

≲ (𝛿)�X1

1 � v2yx1−2𝜎 + C.

(4.51)� X−2𝜎1

u2(X1)dy ≲ (𝛿)�X1

1 � v2yx1−2𝜎 + C.

(4.52)� X−2𝜎

2u2(X2)dy ≲ (𝛿)�

X2

1 � v2yx1−2𝜎 + C ≲ (𝛿)�

X1

1 � v2yx1−2𝜎 + C.

(4.53)sup

1≤x≤X1� u2

yx1−2𝜎 + �

X1

1 � u2xx1−2𝜎 ≲ C + (𝛿)�

X1

1 � u2x−1−2𝜎

+ �X1

1 � u2yx−2𝜎 .

201

1 3


Proof We now apply the multiplier M = uxx1−2� to the system in (4.30). Doing so

yields the following positive terms:

Note crucially that the middle term in the above estimate, ‖uyx−�‖2L2y

, has been esti-mated in (4.45) upon taking an x-integration. To close this sequence of estimates, therefore, it is crucial that this small parameter (�) is attached to the ‖vyx

1

2−�‖2

L2

term in (4.45). Next, we come to the profile terms, (see (4.31) for the definition):

We have used estimates (2.85), (2.90), and (3.57), which provide the smallness of (�) . Next, we come to f (1) , for which we use the estimate in (4.38) (with k = 0):

Piecing the above estimates together gives:

Now, we take an integration from x = 1 to x = X1:

The last step is to come to the terms in , from (4.32). The most delicate term here requires successive integration by parts:

(4.54)∫

((1 + u0

p)𝜕x − 𝜕yy

)u ⋅ uxx

1−2𝜎 ≳ 𝜕x ∫ u2yx1−2𝜎 − ∫ u2

yx−2𝜎 + ∫ u2

xx1−2𝜎 .

(4.55)

�� ⋅ uxx1−2�

��≤ ‖xu(1)

sx, yu0

py, v(1)

sx

1

2 ‖L∞�‖ux−

1

2−�‖2

L2y

+ ‖uxx1

2−�‖2

L2y

�

≤ (�)�‖ux− 1

2−�‖2

L2y

+ ‖uxx1

2−�‖2

L2y

�.

(4.56)

�� f (1)uxx1−2�

��≤ ‖x

1

2−�f (1)‖L2

y‖uxx

1

2−�‖L2

y≤ Cx

−3

2 +1

100 000‖uxx

1

2−�‖2

L2y

.

(4.57)�x � u2

yx1−2� + � u2

xx1−2�

≤ � u2yx−2� + (�)� u2x−1−2� + Cx−1−� + � ⋅ uxx

1−2� .

(4.58)� u2

y(X1)X

1−2𝜎1

+ �X1

1 � u2xx1−2𝜎

≲ �X1

1 � u2yx−2𝜎 + (𝛿)�

X1

1 � u2x−1−2𝜎 + C + �X1

1 � ⋅ uxx1−2𝜎 .

202 S. Iyer

1 3

We have used |u1e(X1, 0)| ≲ X

−1

2

1 and |u1

ex(x, 0)| ≲ x

3

2 , according to (3.57). Finally, we have absorbed the boundary contribution at x = X1 , ∫ u2

yX1−2�1

into the left-hand side of (4.58). A consultation with (4.32) shows that we can perform a similar cal-culation with the remaining terms in because these terms either have one extra x-derivative, or are accompanied by v(1)

s , which contributes additional decay of x−

1

2 . This then gives:

Using a similar line of reasoning as in Lemma 4.6, we can replace ∫ u2y(X1)X

1−2�1

with the supremum over all x ∈ [1,X1] , thereby yielding the desired result. □

Consolidating the results of the previous two lemmas and applying contraction mapping:

Corollary 4.9 For �, � sufficiently small relative to � , a solution to the Prandtl system in (4.30) satisfies the following a priori estimate in the space P:

For �, � sufficiently small, there exists a unique solution to (4.30), satisfying ‖u‖P(X1)

≤ C(�) . The constant above in (4.60) is independent of X1 , and so we can immediately send X1 → ∞ , thereby yielding a global solution on x ∈ [1,∞) satisfy-ing: ‖u‖P ≤ C(�).

(4.59)

�X1

1 � 𝜒 ��(y)x1−2𝜎u1e(x, 0)ux

= −�X1

1 � 𝜒 ��(y)u𝜕x{u1e(x, 0)x1−2𝜎} − �x=1

u1e(1, 0)u𝜒 ��(y)dy

+ �x=X1

X1−2𝜎1

u1e(X1, 0)𝜒

��(y)u(X1, y)dy

= −�X1

1 � 𝜒 ��(y)u𝜕x{u1e(x, 0)x1−2𝜎} + C

− �x=X1

X1−2𝜎1

u1e(X1, 0)𝜒

�(y)uy(X1, y)dy

≤ �X1

1 � 𝜒 �(y)uy𝜕x{u1e(x, 0)x1−2𝜎} + C +

1

100 000 �x=X1

X1−2𝜎1

u2y(X1)

≲ ‖uyx−𝜎‖L2‖𝜒 �(y)x−

1

2−𝜎‖L2 + C ≤ C +

1

100 000‖uyx−𝜎‖2L2 .

� u2y(X1)X

1−2𝜎1

+ �X1

1 � u2xx1−2𝜎 ≲ �

X1

1 � u2yx−2𝜎 + (𝛿)�

X1

1 � u2x−1−2𝜎 + C.

(4.60)‖u‖2P(X1,𝜎)

≲ C(𝜎).

203

1 3


It is possible to successively differentiate the system in �x , and re-apply the previ-ous estimates, noting that the added x-derivative adds a factor of x−1 to each term above, enabling us to enhance the multiplier to [xk−2��k

xu, x1+k−2��k+1

xu] . This is the

reason for the differentiated version of the Prandtl-norm in (4.34). To do so, we sim-ply need:

Lemma 4.10 (Initial Conditions) For each k ≥ 0 , the initial data �kxu(1, y) is of order

� and decays rapidly in y.

Proof This follows from using Eq. (4.21) to write:

We may now multiply by (1 + y)n , use the inequality ‖ v(1,y)

y‖L∞ ≲ ‖vy(1, y)‖L∞ for

functions v satisfying v(x, 0) = 0 (here we take v = v1p(1, y) − v1

p(1, 0) ), and use

‖⟨y⟩nU1(y)u0px(1, y)‖L∞ ≤ (�)‖u0

px(1, y)‖L∞ , to obtain: ‖⟨y⟩nu1

px(1, y)‖L∞

y≤ C . It is

clear that the same procedure may be applied to higher-order x-derivatives. □

Remark 4.11 (Higher-Order Compatibility) To apply the procedure described, we require high-order compatibility condition such that the initial data of u1

px(1, 0) = 0 ,

thereby honoring the boundary condition. These condition can be ascertained induc-tively from (4.61). For instance:

We suppose these compatibility conditions for large k.

Repeating the previous set of estimates after applying the self-similar weight zM , and upon applying �k

x to the system, we arrive at the following:

Lemma 4.12 Given any 𝜎 > 0 , let �, � be sufficiently small relative to � . Consider the system given in (4.21), together with the boundary conditions (4.22). Let all deriva-tives of the prescribed data U1(y) be exponentially decaying in its argument. Then there exists a unique, global in x solution [up, vp] , satisfying:

(4.61)u1px(1, y) = U1yy(y) + u(1)

sx(1, y)U1(y) + yu0

py(1, y)

v1p(1, y) − v1

p(1, 0)

y

+ v(1)s(1, y)U1y(y) + f (1)(1, y).

(4.62)0 = u1px(1, 0) = U1yy(0) + u(1)

sx(1, 0)U1(0) + v(1)

sU1y(0) + f (1)(1, 0).

(4.63)‖zM�kxu‖Pk(�)

≤ C(M, k, �).

204 S. Iyer

1 3

It is now our task to extract similar estimates for the profiles u1p, v1

p from (4.63).

First,

Lemma 4.13 For any m, k ≥ 0, 𝜎 > 0,

Proof This follows by writing u1p= u − �(y)u1

e(x, 0) and using |𝜕k

xu1e(x, 0)| ≲ x

−k−1

2 . □

Next, we may give the following uniform decay estimate for v1p:

Corollary 4.14 (Uniform Estimates for v1p ) For any m ≥ 0,

Proof First, according to (4.27), we have the rapid decay v1p→ 0 as y → ∞ . By the

divergence-free condition, and the trace inequality, and for any small 𝜅 > 0,

We have used the 𝜅 > 0 to avoid the critical Hardy inequality. The Hardy inequal-ity we have used (with power y−

1

2+�v1p ) relies on the vanishing of v1

p at y = ∞ . From

here the desired bound follows for m = 0, k = 0 , and k,m ≥ 1 works analogously. □

The final ingredient we will need is to understand the connection between the norms we have controlled, Pk , and the quantities u1

py, u1

pyy . This is the content of the

following:

Lemma 4.15

Proof First, let us record:

(4.64)supx≥1 � z2m|�k

xu1p|2x2k−2� ≤ C(M,m, k, �).

(4.65)‖zm𝜕kxv1p‖L∞

y≲ x

−k−3

4+𝜎C(𝜎,m, k).

(4.66)

�v1p�2 ≲

��∞

y

v1pvpy

��≤ �

∞

0

��

v1p

y1

2−𝜅

y1

2−𝜅v1py

��≤��

v1p

y1

2−𝜅

��L2y

‖y1

2−𝜅v1py‖L2

y

≤ ‖y1

2+𝜅v1py‖L2

y‖y

1

2−𝜅v1py‖L2

y= ‖x

1

4+𝜅z

1

2+𝜅v1py‖L2

y‖x

1

4−𝜅z

1

2−𝜅v1py‖L2

y

≤ x1

2 x−1+𝜎x−1+𝜎 = x−

3

2+2𝜎 .

(4.67)x3

4−𝜎‖zmu1

py‖L∞

y+ x

1

2−𝜎‖zmu1

py‖L2

y+ x1−𝜎‖zmu1

pyy‖L2

y≤ C < ∞.

(4.68)x1−2� � z2m�u1

py�2 ≤ x1−2� � z2mu2

y+ x1−2� � z2m�2�u1

e�2(x, 0)

≤ ‖zmu‖P(�) + C.

205

1 3


Via Eq. (4.21), we have:

We use the decay rates established for u0p in (2.85), and the pointwise decay of the

Euler profiles established in (3.44) and the relations in (4.64). Let us give in detail,

For the final line we have used (4.68). Next, via Sobolev interpolation in the y direction,

Coupled with the u1pyy

estimate in (4.68), this then establishes the desired bounds. The weighted estimate in z follows analogously. □

We will select now,

Summarizing the results of this section:

Proposition 4.16 (Prandtl-1 Layer Bounds) Given any n ∈ ℕ , let �1 be as in (4.74), and let �, � be sufficiently small relative to n and �1 . Consider the system given in (4.21), together with the boundary conditions (4.22). Let all derivatives of the pre-scribed data U0(y) be exponentially decaying in its argument. Then there exists a unique, global in x solution [u1

p, v1

p] , satisfying:

(4.69)x1−𝜎‖u1

pyy‖L2

y≤ x1−𝜎‖u1

px‖L2

y+ x1−𝜎‖‖L2

y+ x1−𝜎‖f (1)‖L2

y

≲ ‖u‖P1(𝜎)+ ‖u‖P(𝜎) + C.

(4.70)x1−�‖u(1)sxu1p‖L2

y≤ ‖xu(1)

sx‖L∞x−�‖u1p‖L2y ≤ (�)‖u‖P + (�),

(4.71)x1−�‖u0

py(v1

p(x, y) − v1

p(x, 0))‖L2

y≤ x1−�‖yu0

py‖L∞

��

v1p(x, y) − v1

p(x, 0)

y

��L2y

≤ (�)x1−�‖u1px‖L2

y≤ (�)‖u‖P1

,

(4.72)x1−�‖v(1)su1py‖L2

y≤ ‖x

1

2 v(1)s‖L∞x

1

2−�‖u1

py‖L2

y≤ (�)‖u‖P + (�).

(4.73)

x3

4−�‖u1

py‖L∞

y≤ �

x1−�‖u1pyy

‖L2y

� 1

2�x

1

2−�‖u1

py‖L2

y

� 1

2 ≤ C‖u‖P +1

100x1−�‖u1

pyy‖L2

y.

(4.74)� = �1 =3

3n × 10 000.

(4.75)‖zM�kxu1p‖Pk(�1)

+ xk+

3

4−�1‖zm�k

xv1p‖L∞

y≤ C(M, k, n), for any k,m ≥ 0.

206 S. Iyer

1 3

5 Intermediate Layers

We now construct intermediate layers i = 2 through i = n − 1 . This is achieved induc-tively, starting with the construction of the Euler layer, ui

e, vi

e . Let us fix the parameters

for the sake of concreteness:

5.1 Construction of Euler Layer, [uie, vi

e]

For this step in the construction, we suppose that [u(i−1)s

, v(i−1)s

] have already been con-structed. The inductive hypothesis on the 1,… , i − 1 Prandtl profiles are as follows:

where [ujp, vjp] satisfy the system given in (5.46) for 2 ≤ j ≤ i − 1 , and for i = 2 that

[u1p, v1

p] satisfy (4.21). To obtain the equations for [ui

e, vi

e] , we expand the nonlinear

terms including the new Euler terms:

We will now define several terms:

Definition 5.1 The (i − 1)’th remainder is denoted by:

(5.1)�i =3i

3n × 10 000for i = 2,… , n.

(5.2)‖zm�kxujp‖Pk(�j)

≤ C(k,m, n), for j = 1,… , i − 1,

(5.3)u(i)su(i)sx=

(u(i−1)s

+ 𝜖i

2 uie

)(u(i−1)sx

+ 𝜖i

2 uiex

)

= u(i−1)s

u(i−1)sx

+ 𝜖i

2 u(i−1)sx

uie+ 𝜖

i

2 u(i−1)s

uiex+ 𝜖iui

euiex,

(5.4)v(i)su(i)sy=

�v(i−1)s

+ 𝜖i−1

2 vie

��u(i−1)sy

+√𝜖𝜖

i

2 uieY

�

= v(i−1)s

u(i−1)sy

+ 𝜖i−1

2 u(i−1)sy

vie+√𝜖𝜖

i

2 v(i−1)s

uieY

+ 𝜖ivieuieY.

(5.5)u(i)sv(i)sx=

(u(i−1)s

+ 𝜖i

2 uie

)(v(i)sx+ 𝜖

i−1

2 viex

)

= u(i−1)s

v(i−1)sx

+ 𝜖i

2 v(i−1)sx

uie+ 𝜖

i−1

2 u(i−1)s

viex+ 𝜖

i−1

2 uieviex,

(5.6)v(i)sv(i)sy=

(v(i−1)s

+ 𝜖i−1

2 vie

)(v(i−1)sy

+ 𝜖i

2 vieY

)

= v(i−1)s

v(i−1)sy

+ 𝜖i−1

2 v(i−1)sy

vie+ 𝜖

i

2 v(i−1)s

vieY

+ 𝜖i−

1

2 vievieY.

(5.7)Ru,i−1 ∶= −Δ𝜖 u(i−1)s

+ u(i−1)s

u(i−1)sx

+ v(i−1)s

u(i−1)sy

+ P(i−1)sx

+ 𝜖i−1

2 vieu0py,

207

1 3


We will also split the Euler–Euler interaction terms and the Euler–Prandtl terms via:

Definition 5.2

Remark 5.3 The natural definition of the remainder term, Ru,i−1, should be:

and the natural definition of Ru,i

E would contain the lowest-order term, �

i−1

2 vieu0py

. However, for i < n , the quantity of interest in (5.20) is the sum, Ru,i−1 + R

u,i

E , that is

contributed to the next order (see the definition of the forcing in (5.29)). Thus, for

(5.8)Rv,i−1 ∶= −Δ𝜖 v(i−1)s

+ u(i−1)s

v(i−1)sx

+ v(i−1)s

v(i−1)sy

+𝜕y

𝜖P(i−1)s

.

(5.9)

Ru,i

E∶= �

i

2 uiex

�i−1�

j=0

�j

2 ujp

�+ �

i

2 uie

�i−1�

j=0

�j

2 ujpx

�

+ �i

2

√�ui

eY

�i−1�

j=0

�j

2 vjp

�+ �

i−1

2 vie

�i−1�

j=1

�j

2 ujpy

�,

(5.10)

Rv,i

E∶= �

i−1

2 viex

i−1�

j=0

�j

2 ujp+ �

i

2 uie

i−1�

j=0

�j

2 vjpx

+ �i−1

2 vie

i−1�

j=0

�j

2 vjpy+√��

i−1

2 vieY

i−1�

j=0

�j

2 vjp,

(5.11)

Eu,ie

∶= �i�uieuiex+ vi

euieY

�+ �

i

2 uiex

i−1�

j=1

�j

2 uje+ �

i

2 uie

i−1�

j=1

�j

2 ujex

+√��

i

2 uieY

�i−1�

j=1

�j−1

2 vje

�+ �

i−1

2 vie

�i−1�

j=1

�1

2 �j

2 uj

eY

�,

(5.12)

Ev,ie

∶= �i−

1

2

�uieviex+ vi

evieY

�+ �

i

2 uie

�i−1�

j=1

�j−1

2 vjex

�+ �

i−1

2 viex

�i−1�

j=1

�j

2 uje

�

+√��

i−1

2 vie

i−1�

j=1

�j−1

2 vj

eY+√��

i−1

2 vieY

i−1�

j=1

�j−1

2 vje.

(5.13)Ru,i−1“ = ” − Δ𝜖 u(i−1)s

+ u(i−1)s

u(i−1)sx

+ v(i−1)s

u(i−1)sy

+ P(i−1)sx

,

208 S. Iyer

1 3

convenience (see the calculations (5.50) and (5.51)), we add and subtract one factor �

i−1

2 vieu0py

, which explains the definitions of (5.7) and (5.9). This, however, will not be done for i = n.

The Navier–Stokes expansion reads:

For the normal equation, the expansions read:

From here, we simply read off the highest-order terms that are “purely-Eulerian”. All of the remaining terms will be treated in the next subsection. In (5.14), these are at order �

i

2 , and in (5.15), these are at order �i−1

2 :

When paired with the divergence-free condition, we arrive at the equations that are taken for the [ui

e, vi

e] , which are the Cauchy–Riemann equations:

The boundary conditions for the i’th Euler layer is vie(x, 0) = −vi−1

p(x, 0) . According

to the inductive hypothesis, the decay rate of this boundary condition is:

(5.14)

− Δ𝜖u(i)s+ u(i)

su(i)sx+ v(i)

su(i)sy+ P(i)

sx= −Δ𝜖 u

(i−1)s

+ u(i−1)s

u(i−1)sx

+ v(i−1)s

u(i−1)sy

+ P(i−1)sx

− 𝜖i

2+1Δui

e+ 𝜖

i

2 u(i−1)sx

uie+ 𝜖

i

2 u(i−1)s

uiex+ 𝜖iui

euiex+ 𝜖

i−1

2 u(i−1)sy

vie

+√𝜖𝜖

i

2 v(i−1)s

uieY

+ 𝜖ivieuieY

+ 𝜖i

2Piex+ 𝜖iP1,a

ex

= 𝜖i

2

�uiex+ Pi

ex

�+ 𝜖

i

2+1Δui

e+ Ru,i−1 + R

u,i

E+ Eu,i

e+ 𝜖iPi,a

ex.

(5.15)

− Δ𝜖v(i)s+ u

(i)sv(i)sx+ v

(i)sv(i)sy+

𝜕y

𝜖P(i)s

= −Δ𝜖 v(i−1)s

+ u(i−1)s

v(i−1)sx

+ v(i−1)s

v(i−1)sy

+𝜕y

𝜖P(i−1)s

− 𝜖i+1

2 Δvie+ 𝜖

i

2 v(i−1)sx

ui

e+ 𝜖

i−1

2 u(i−1)s

vi

ex+ 𝜖

i−1

2 ui

evi

ex+ 𝜖

i−1

2 v(i−1)sy

vi

e

+ 𝜖i

2 v(i−1)s

vi

eY+ 𝜖

i−1

2 vi

evi

eY+ 𝜖

i−1

2 Pi

eY+ 𝜖

i−1

2P1,a

eY.

(5.16)�i

2

[uiex+ Pi

ex

]= 0, �

i−1

2

[viex+ Pi

eY

]= 0.

(5.17)uiex+ Pi

ex= 0, vi

ex+ Pi

eY= 0, ui

ex+ vi

eY= 0.

(5.18)

�vie(x, 0)� = �vi−1

p(x, 0)� ≤ ‖vi−1

p(x, y)‖

1

2

L2y

‖vi−1py

(x, y)‖1

2

L2y

≤ ‖yvi−1py

(x, y)‖1

2

L2y

‖vi−1py

(x, y)‖1

2

L2y

≤ C(�, n)x−3

4+�i−1 , for i ≥ 2.

209

1 3


A comparison of (5.18) to (3.9) shows that the decay rate of the boundary condi-tion has improved, enabling us to improve the Euler decay rates. Indeed, as the system (5.17) is an identical system to the first Euler layer (and is in particular the Cauchy–Riemann equations), we may simply repeat the analysis given there to conclude:

Proposition 5.4 (Euler-i Layer) Let i ≥ 2 . The i’th Euler layer, defined by the Cauchy–Riemann equations (5.17) taken with boundary conditions (5.18) satisfy the enhanced decay rates:

Proof This follows by repeating the arguments in Sect. 3 with the enhanced bound-ary condition (5.18). □

Remark 5.5 It is not possible to enhance the rates (5.19) much more (for instance, past x−1 for k = m = 0 . This is due to the restriction of w ∈ (0, 1) in Corollary 3.3.

We have the simplified expression for (5.14), (5.15).

Lemma 5.6 According to definitions in (5.7)–(5.12), one has

Proof By construction in Proposition 5.4, uie, vi

e are harmonic, and so Δui

e,Δvi

e terms

vanish identically. This then accounts for all of the terms in (5.14). □

5.2 Construction of Prandtl Layer, [uip, vi

p]

For this step in the construction, we suppose that [u(i)s, v(i)

s] have been constructed. The

inductive hypothesis on these profiles are that the following remainders (according to the definitions in (5.7) for Ru,i−1 and (5.8) for Rv,i−1 ) have been accumulated:

(5.19)xk+m+

3

4−�i−1 |||�

kx�mYvie

||| + x3

4−�i−1 |ui

e| ≤ C(k,m, n).

(5.20)−Δ�u(i)s+ u(i)

su(i)sx+ v(i)

su(i)sy+ P(i)

sx= Ru,i−1 + R

u,i

E+ Eu,i

e+ �iPi,a

ex,

(5.21)−Δ�v(i)s+ u(i)

sv(i)sx+ v(i)

sv(i)sy+

�y

�P(i)s= Rv,i−1 + R

v,i

E+ Ev,i

e+ �

i−1

2PieY.

(5.22)

Ru,i−1 = �i−1

2

��ui−1

pxx+√�yu0

pyvieY

+ ui−1px

i−1�

j=1

�j

2 {uje+ uj

p}

+�u0py ∫

y

0 ∫y�

y

vieYY

dy��dy� + vi−1p

�i−1�

j=1

�j

2

�ujpy+√�u

j

eY

��,

210 S. Iyer

1 3

The induction will start at i = 2 , and so (5.22), (5.23) should be compared to (4.26) for this case and to (5.48) for the general i case. The relevant profile estimates, according to Propositions 3.6, 5.4 and 4.16 which hold inductively are:

By writing u(i)s= u(i)

s+ 𝜖

i

2 u(i)p

, v(i)s= v(i)

s+ 𝜖

i−1

2 v(i)p

, and expanding the Navier–Stokes equations, we obtain the two expansions:

and:

Define the forcing term to be those terms from (5.27) which do not appear in the bracket:

where we have used (5.20) to simplify the expression. The first step, here, is to intro-duce a potential pressure which eliminates the “purely” Eulerian terms from above:

(5.23)Rv,i−1 = �

i−1

2

[−Δ�v

i−1p

+ u(i−1)s

vi−1px

+ v(i−1)sx

ui−1p

+ v(i−1)s

vi−1py

+ v(i−1)sy

vi−1p

+�i−1

2 ui−1p

vi−1px

+ �i−1

2 vi−1p

vi−1py

].

(5.24)xk+m+

1

2 |�kx�mYv1e| + x

1

2 |u1e| ≤ C(k,m, n),

(5.25)xk+m+

3

4−�j−1 |||�

kx�mYvje

||| + x3

4−�j−1 |uj

e| ≤ C(k,m, n) for j = 2,… , i,

(5.26)‖zm�kxujp‖Pk(�j)

≤ C(k,m, n) for j = 1,… , i − 1.

(5.27)

− Δ𝜖 u(i)s+ u(i)

su(i)sx+ v(i)

su(i)sy+ P(i)

sx= −Δ𝜖u

(i)s+ u(i)

su(i)sx+ v(i)

su(i)sy+ P(i)

sx+ 𝜖

i+1

2 Pi,apx

+ 𝜖i

2

[−Δ𝜖u

ip+ u(i)

sxuip+ u(i)

suipx+ u(i)

syvip+ v(i)

suipy

+𝜖i

2 uipuipx+ 𝜖

i

2 vipuipy+ Pi

px

],

(5.28)

− Δ𝜖 v(1)s

+ u(i)sv(i)sx+ v

(i)sv(i)sy+

𝜕y

𝜖P(i)s= −Δ𝜖v

(i)s+ u

(i)sv(i)sx+ v

(i)sv(i)sy+

P(i)sy

𝜖+ 𝜖

i−1

2 Pi,a

py

+ 𝜖i

2

[− Δ𝜖v

i

p+ u

(1)svi

px+ v

(i)sxui

p+ v

(i)syvi

p+ v

(i)svi

py+ 𝜖

i

2 ui

pvi

px

+𝜖i

2 vi

pvi

py+

Pi

py

𝜖

].

(5.29)−�

i

2 f (i) ∶= −Δ�u(i)s+ u(i)

su(i)sx+ v(i)

su(i)sy+ P(i)

sx+ �

i+1

2 Pi,apx

= Ru,i−1 + Ru,i

E+ Eu,i

e+ �iP1,a

ex+ �

i+1

2 Pi,apx,

211

1 3


Definition 5.7 The i’th auxiliary Euler pressure, P1,ae

, is defined by:

We may now check that:

Lemma 5.8 With the definition above (5.30), P1,ae

serves as a gradient potential to eliminate the purely Eulerian terms, [Eu,i

e,Ev,i

e] from the expansion

Proof By scaling, we will write:

We will go term by term through the definition in (5.30), starting with:

Next,

Third,

Comparing these expressions, (5.33)–(5.35) to the expression (5.11), one observes the exact cancellation:

Next, we will move to the �Y√� terms:

(5.30)P1,ae

∶= −

i−1∑

j=1

�j−i

2 vievje−

i−1∑

j=1

�j−i

2 uieuje−

1

2|vi

e|2 − 1

2|ui

e|2.

(5.31)(�x,

�y

�

)�iP1,a

e+(Eu,ie,Ev,i

e

)= 0.

(5.32)��x,

�y

�

��iP1,a

e=

��x,

�Y√�

��iP1,a

e.

(5.33)

�i�x

(−

i−1∑

j=1

�j−i

2 vievje

)= −

i−1∑

j=1

�j+i

2

(viexvje+ vi

evjex

)= −

i−1∑

j=1

�j+i

2

(uieYvje+ vi

euj

eY

).

(5.34)�i�x

(−

i−1∑

j=1

�j−i

2 uieuje

)= −

i−1∑

j=1

�j+i

2

(uiexuje+ ui

eujex

).

(5.35)�i�x

(−1

2|vi

e|2 − 1

2|ui

e|2)= −�ivi

eviex− �iui

euiex= −�ivi

euieY

− �iuieuiex.

(5.36)�x�iP1,a

e+ E

u,i

E= 0.

212 S. Iyer

1 3

Next,

Third,

Comparing these expressions, (5.37)–(5.39) to the expression (5.11), one observes the exact cancellation:

This establishes the desired result, (5.31). □

Remark 5.9 One should notice the essential role played by the Cauchy–Riemann equations, uj

eY= v

jex in the equalities above.

Let us now turn to the terms outside of the bracket in (5.28), which we also sim-plify via (5.21) and subsequently via (5.31):

Motivated by this, define the auxiliary pressure via:

Definition 5.10 The i’th auxiliary Prandtl pressure, Pi,a

P, is defined via:

(5.37)−�Y√��i

�i−1�

j=1

�j−i

2 vievje

�= −

i−1�

j=1

�i+j

2−

1

2

�vieYvje+ vi

evj

eY

�.

(5.38)

−�Y√��i

�i−1�

j=1

�j−i

2 uieuje

�= −

i−1�

j=1

�i+j

2−

1

2

�uieYuje+ ui

euj

eY

�

= −

i−1�

j=1

�i+j

2−

1

2

�viexuje+ ui

evjex

�.

(5.39)

�Y√��i�−�vi

e�2 − �ui

e�2�= −�

i−1

2 vievieY

− �i−

1

2 uieuieY

= −�i−

1

2 vievieY

− �i−

1

2 uieviex.

(5.40)�Y√��iP1,a

e+ E

v,i

E= 0.

(5.41)

− Δ�v(i)s+ u(i)

sv(i)sx+ v(i)

sv(i)sy+

P(i)sy

�+ �

i−1

2 Pi,apy

= Rv,i−1 + Rv,i

E+ Ev,i

e+ �

i−1

2P1,a

eY+ �

i−1

2 Pi,apy

= Rv,i−1 + Rv,i

E+ �

i−1

2 Pi,apy.

(5.42)�i+1

2 Pi,a

P∶= � ∫

∞

y

Rv,i−1 + Rv,i

E.

213

1 3


Immediately from this definition, we have:

Lemma 5.11 According to (5.42), the following identity holds:

Proof By direct calculation from (5.42),

Combined with (5.41) then implies the desired result. □

We are now able to rewrite the forcing for our equation, which was defined in (5.29):

The �−i

2 factor arises as we are in the i’th order of the construction. Let us briefly comment on the orders of the three terms on the right-hand side of (5.45). Accord-ing to (5.22), it is evident that Ru,i−1 is of order �

i

2 . According to (5.9), it is clear that Ru,i

E is of order �

i

2 . According to (5.42) coupled with (5.23) and (5.10), it is clear that Pi,a

P is of order 1. Thus, all of the terms in (5.45) are of order 1 or higher. Reading off

from (5.27), (5.28), the system we will now consider is:

Again, by evaluating the equation at y = ∞ gives Pipx= 0 . Coupled with the normal

equation Pipy= 0 then shows that Pi

p= 0 . After this construction, Ru,i,Rv,i contain

the terms from (5.27), (5.28) which were omitted in the construction of [uip, vi

p]:

Lemma 5.12 For each i, with [Ru,i,Rv,i] defined as in (5.7)–(5.8), with [uip, vi

p] taken

to solve the system (5.46), the following identities hold:

(5.43)−Δ�v(i)s+ u(i)

sv(i)sx+ v(i)

sv(i)sy+

P(i)sy

�+ �

i−1

2 Pi,apy

= 0.

(5.44)�i−1

2 Pi,a

P= −Rv,i−1 − R

v,i

E.

(5.45)f (i) = −�−

i

2

[Ru,i−1 + R

u,i

E+ �

i+1

2 Pi,apx

].

(5.46)(1 + u0p)ui

px+ u(i)

sxuip+ v(i)

suipy+ u0

py

(vip− vi

p(x, 0)

)+ Pi

px= ui

pyy+ f (i),

(5.47)uip(x, 0) = −ui

e(x, 0), lim

y→∞uip(x, y) = 0, ui

p(1, y) = Ui(y), Pi

py= 0.

(5.48)

Ru,i = �i

2

��ui

pxx+√�yu0

pyvi+1eY

+ �u0py ∫

y

0 ∫y�

y

vi+1eYY

dy��dy�

+vip

i�

j=1

�j

2

�ujpy+√�u

j

eY

�+ ui

px

i�

j=1

�j

2 {uje+ uj

p}

�,

214 S. Iyer

1 3

Proof Starting with the definition in Eq. (5.7), we have:(5.49)

Rv,i = �i

2

[−Δ�v

ip+ u(i)

svipx+ v(i)

sxuip+ v(i)

svipy+ v(i)

syv(i)p+ �

i

2 uipvipx+ �

i

2 vipvipy

].

(5.50)

Ru,i ∶= −Δ𝜖 u(i)s+ u(i)

su(i)sx+ v(i)

su(i)sy+ P(i)

sx+ 𝜖

i

2 vi+1e

u0py

= −Δ𝜖u(i)s+ u(i)

su(i)sx+ v(i)

su(i)sy+ P(i)

sx+ 𝜖

i+1

2 Pi,apx

+ 𝜖i

2

�−Δ𝜖u

ip+ u(i)

sxuip+ u(i)

suipx+ u(i)

syvip+ v(i)

suipy

+𝜖i

2 uipuipx+ 𝜖

i

2 vipuipy+ Pi

px

�+ 𝜖

i

2 vi+1e

u0py

= −𝜖i

2 f (i) + 𝜖i

2

�−Δ𝜖u

ip+ u(i)

sxuip+ u(i)

suipx+ u(i)

syvip+ v(i)

suipy

+𝜖i

2 uipuipx+ 𝜖

i

2 vipuipy+ Pi

px

�+ 𝜖

i

2 vi+1e

u0py

= 𝜖i

2 f (i) + 𝜖i

2

�− ui

pyy+ u(i)

sxuip+ (1 + u0

p)ui

px+ u0

pyvip+ v(i)

suipy

+ 𝜖i

2 uipuipx+ 𝜖

i

2 vipuipy+ Pi

px− 𝜖ui

pxx+

i�

j=1

𝜖j

2 {uje+ uj

p}ui

px

+

i�

j=1

𝜖j

2

�ujpy+√𝜖u

j

eY

�vip

�+ 𝜖

i

2 vi+1e

u0py

= −𝜖i

2 f (i) + 𝜖i

2

�− ui

pyy+ u(i)

sxuip+ (1 + u0

p)ui

px+ u0

pyvip+ v(i)

suipy

+ 𝜖i

2 uipuipx+ 𝜖

i

2 vipuipy+ Pi

px− 𝜖ui

pxx+

i�

j=1

𝜖j

2 {uje+ uj

p}ui

px

+

i�

j=1

𝜖j

2

�ujpy+√𝜖u

j

eY

�vip

�+ 𝜖

i

2

�− vi

p(x, 0)u0

py+√𝜖yu0

pyvi+1eY

+𝜖u0py ∫

y

0 ∫y�

y

vi+1eYY

dy�� dy�

�

= 𝜖i

2

�−𝜖ui

pxx+

i�

j=1

𝜖j

2

�uje+ uj

p

�uipx+

i�

j=1

𝜖j

2

�ujpy+√𝜖u

j

eY

�vip+√𝜖yu0

pyvi+1eY

+𝜖u0py ∫

y

0 ∫y�

y

vi+1eYY

dy�� dy�

�.

215

1 3


For the calculation in (5.50), we have used the expansion (5.27), (5.29), and have simply rearranged the terms and used:

Finally, for the calculation in (5.50), we used (5.46). The identity (5.48) then fol-lows. For the Rv,i contribution, we combine (5.43) with the expression (5.28), and finally with the condition that Pi

p= 0 from (5.47).

A comparison with (5.22), (5.23) shows that this closes the construction. We now give estimates on the forcing term based on the inductively assumed decay rates and regularity in (5.2), together with the [u0

p, v0

p] bounds in (2.85) and the

[u1e, v1

e] bounds in (3.57).

Lemma 5.13 (Forcing Estimates) Let i ≥ 2 . For any m, k ≥ 0 , with f (i) as in (5.45),

Proof We start with the Ru,i

E terms, which are defined in (5.9):

In (5.56), we have used the enhanced Eulerian decay rate in (5.19). Next, we have the Prandtl contributions from Ru,i−1 , according to (5.22):

(5.51)

�i

2 vi+1e

u0py= �

i

2

�−vi

p(x, 0)u0

py+√�yu0

pyvi+1eY

+ �u0py ∫

y

0 ∫y�

y

vi+1eYY

dy�� dy�

�.

(5.52)‖zm�kxf (i)‖L2

y≤ C(k,m, �, n)x−k−

5

4+2�i−1 .

(5.53)i−1�

j=0

𝜖j

2 ‖uiexujp‖L2

y≤

i−1�

j=0

𝜖j

2 ‖uiexx

3

2 ‖L∞x−3

2 ‖ujp‖L2

y≲ x

−5

4 ,

(5.54)i−1�

j=0

‖uie𝜖

j

2 ujpx‖L2

y≲

i−1�

j=0

𝜖j

2 ‖uiex

1

2 ‖L∞‖x−1

2 ujpx‖L2

y≤ x

−5

4 ,

(5.55)i−1�

j=0

𝜖j+1

2 ‖uieYvjp‖L2

y≲

i−1�

j=0

𝜖j+1

2 ‖uieYx

3

2 ‖L∞x−3

2 ‖zmvjp‖L2

y≲ x

−7

4 ,

(5.56)𝜖i−1

2

i−1�

j=1

𝜖j

2 ‖vieujpy‖L2

y≲

i−1�

j=1

‖viex

3

4−𝜎i−1‖L∞‖x−

3

4+𝜎i−1uj

py‖L2

y≤ x

−5

4+2𝜎i−1 .

(5.57)√�‖ui−1

pxx‖L2

y≤ √

�x−

7

4+�i−1 ,

216 S. Iyer

1 3

We now move to the term Pi,apx

. For this we note:

We now turn to evaluating the right-hand side of (5.63). Let us comment that it is essential at this stage that the pressure P1,a

e was introduced to eliminate all of the

purely Eulerian contributions, as those terms cannot handle the weight of y3

2+� . Each

term below has at least one Prandtl ( [uip, vi

p] ) factor, and so we may exchange the

weight of y3

2+� for decay, x

3

4+

�

2 . For the calculations below, we simply pair this with the decay rates in (5.2):

(5.58)‖yu0pyvieY‖L2

y≤ ‖x−

1

4 yu0py‖L2

yx

1

4 ‖vieY‖L∞

y≲ x

−5

4 ,

(5.59)

√��u0py �

y

0 �y�

y

vieYY

dy�dy��L2

y

≤ √�‖y2x−

3

4 u0py‖L2

yx

3

4 ‖vieYY

‖L∞y≤ √

�x−

7

4 ,

(5.60)�j−1

2 ‖vi−1p

ujpy‖L2

y≤ �

j−1

2 x−

5

4+2�i−1 , j ≥ 1,

(5.61)�j

2 ‖vi−1p

uj

eY‖L2

y≤ �

j

2 x−2+2�i−1 , j ≥ 1,

(5.62)𝜖j

2 ‖ui−1px

{uje+ uj

p}‖L2

y≲√𝜖x

−5

4+2𝜎i−1 , j ≥ 1.

(5.63)�−

i

2 �i+1

2 ‖Pi,apx‖L2

y≤ √

��−

i−1

2 ‖⟨y⟩3

2+��x(R

v,i−1 + Rv,i

E)‖L∞

y.

(5.64)‖y3

2+𝜅𝜕x(v

iexujp)‖L∞

y+ ‖y

3

2+𝜅𝜕x(u

ievjpx)‖L∞

y≲ x

−3

2+2𝜎i−1 ,

(5.65)‖y3

2+𝜅𝜕x(v

ievjpy)‖L∞

y≲ x

−3

2+2𝜎i−1 ,

(5.66)‖y3

2+𝜅𝜕x(v

ieYvjp)‖L∞

y≲ x−2+2𝜎i−1 ,

(5.67)‖y3

2+𝜅𝜕x(Δ𝜖v

i−1p

)‖L∞y≲ x

−5

4+2𝜎i−1 ,

(5.68)‖y3

2+𝜅𝜕x(u

(i−1)s

vi−1px

)‖L∞y+ ‖y

3

2+𝜅𝜕x(v

(i−1)sx

ui−1p

)‖L∞y≲ x

−3

2+2𝜎i−1 ,

217

1 3


This proves the claim for k = 0,m = 0 . The general case can be obtained in an iden-tical fashion, upon noticing that powers of z will not influence the estimates when accompanied by Prandtl profiles, which are present in each term above. □

Indeed, the boundary condition in (5.46) has improved, referring to (5.19):

according to (5.19). We can expect enhanced decay due to the enhanced decay of the boundary data. This is the content of the following:

Proposition 5.14 (Construction of i’th Prandtl Layer) For i ≥ 2 , there exists a solu-tion [ui

p, vi

p] to the system (5.46), satisfying for any m, k ≥ 0:

This will be proven in several steps. We homogenize the boundary conditions by defining:

where � is a localized cutoff function selected to have mean zero. These profiles satisfy:

where

We will record the estimate on , which follows directly from (5.71):

We introduce the stream function,

(5.69)‖y3

2+𝜅𝜕x(v

(i−1)s

vi−1py

)‖L∞y+ ‖y

3

2+𝜅𝜕x(v

(i−1)sy

vi−1p

)‖L∞y≲ x

−3

2+2𝜎i−1 ,

(5.70)‖y3

2+𝜅𝜕x(𝜖

i−1

2 ui−1p

vi−1px

+ vi−1p

vi−1py

)‖L∞y≲ 𝜖

i−1

2 x−

5

4+2𝜎i−1 .

(5.71)|||uie(x, 0)

||| ≤ x−

3

4+�i−1 ,

(5.72)‖zmx1

4 �kxuip‖Pk(�i)

+ xk+1−�i‖zm�kxvip‖L∞

y≤ C(k,m, n).

u = up + �(y)uie(x, 0), v = vp − vp(x, 0) + un

ex(x, 0)I� (y), I� (y) = ∫

∞

y

�(�)d�,

(5.73)(1 + u0p)ux − uyy + (u, v) = f (i) + ,

(5.74) ∶= u(i)

s�(y)ui

ex(x, 0) − � ��ui

e(x, 0) − �(y)u(i)

sxune(x, 0)

− u0pyI� (y)u

iex(x, 0) + v(i)

s� �(y)ui

e(x, 0).

(5.75)� � ≲ ⟨y⟩−Nx−3

4+𝜎i−1 .

(5.76)�(x, y) ∶= ∫∞

y

u(x, y�)dy�, �y = −u, �x = v − v(x,∞).

218 S. Iyer

1 3

The stream function satisfies the following system:

For technical reasons (due to certain integrals being critical), it is necessary to start with the stream function formulation to obtain the desired, enhanced decay.

Lemma 5.15 The stream function � solving the system (5.77) satisfies:

Proof Applying the multiplier M = x−

1

2−2�i� to the above system yields the follow-

ing terms on the left-hand side:

For the first term on the right-hand side of (5.77), we will now give the bound via Hardy’s inequality, and (2.85):

Next, let us address the profile terms in . For the first term from , we will split:

(5.77)(�x − �yy

)� = �

∞

y

u0pux + �

∞

y

+ �∞

y

f (n) + �∞

y

,

(5.78)�y(x, 0) = u(x, 0) = 0, �(x,∞) = 0, �(1, y) = ∫∞

y

Un(y�)dy�.

(5.79)supx � x

−1

2−2𝜎i𝜓2 + � � x

−3

2−2𝜎i𝜓2 + � � x

−1

2−2𝜎i u2

≲ C + (𝛿)� � v2yx

3

2−2𝜎i .

(5.80)

𝜕x ∫ x−

1

2−2𝜎i𝜓2 + ∫ x

−3

2−2𝜎i𝜓2 + ∫ x

−1

2−2𝜎i𝜓2

y≲ ∫

(𝜕x − 𝜕yy

)𝜓 ⋅ 𝜓x

−1

2−2𝜎i .

(5.81)

� x−

1

2−2�i� �

∞

y

u0puxdy

�dy ≤ ‖x−3

4−�i�‖L2

y

��x

1

4 �∞

y

u0puxdy

��L2

y

≤ ‖x−3

4−�i�‖L2

y‖x

1

4−�i yu0

pux‖L2

y

≤ (�)‖x− 3

4−�i�‖L2

y‖x

3

4−�i ux‖L2

y.

(5.82)∫ x

−1

2−2�i� ∫

∞

y

u(i)sxu dy� dy =

i−1∑

j=0∫ x

−1

2−2�i� ∫

∞

y

�j

2 ujpxu dy� dy

+

i∑

j=1∫ x

−1

2−2�i� ∫

∞

y

�j

2 ujexu dy� dy.

219

1 3


For the first term above, we apply the Hardy inequality in y (it suffices to consider j = 0):

We have used the smallness which is guaranteed by (2.85). For the second term in (5.82), we will integrate by parts in y and recall �

y→∞

��→ 0 , (again it suffices to con-sider j = 1):

The first term in (5.84):

For the second term in (5.84), we apply Hardy in y and appeal to estimate (3.44) with k = 2, s = 1:

Let us now move to the second term in . One integration by parts in y produces:

(5.83)

�� x−

1

2−2�i� �

∞

y

u0pxu dy� dy

��≤ ‖x−

3

4−�i�‖L2

y

��x

1

4−�i �

∞

y

u0pxu dy�

��L2y

≤ ‖x−3

4−�i�‖L2

y‖x

1

2 yu0px‖L∞‖x−

1

4−�i u‖L2

y

≤ (�)‖x− 3

4−�i�‖2

L2y

+ (�)‖x− 1

4−�i u‖2

L2y

.

(5.84)

∫ x−

1

2−2�i� ∫

∞

y

√�u1

ex�y� = ∫ x

−1

2−2�i

√�u1

ex�2 + ∫ x

−1

2−2�i� ∫

∞

y

�u1exY

� .

(5.85)�� x

−1

2−2𝜎i

√𝜖u1

ex𝜓2

��≤ √

𝜖‖u1exx‖L∞‖x−

3

4−𝜎i𝜓‖2

L2y

≲√𝜖‖x−

3

4−𝜎i𝜓‖2

L2y

.

(5.86)

�� x−

1

2−2𝜎i𝜓 �

∞

y

𝜖u1exY

𝜓��≤ √

𝜖‖u1exY

Yx‖L∞‖x−3

4−𝜎i𝜓‖2

L2y

=√𝜖‖v1

exxYx‖L∞‖x−

3

4−𝜎i𝜓‖2

L2y

≲√𝜖‖x−

3

4−𝜎i𝜓‖2

L2y

.

(5.87)

∫ x−

1

2−2�i� ∫

∞

y

v(i)suydy

�dy = ∫ x−

1

2−2�i v(i)

s��y + ∫ x

−1

2−2�i� ∫

∞

y

u(i)sxudy�dy.

220 S. Iyer

1 3

For the second term in (5.87), one notices this is exactly the term on the left-hand side of (5.82). For the first term in (5.87), we estimate:

We have used the smallness guaranteed by (2.90). The final term from is:

Consolidating the previous estimates, taking � small enough relative to �i (and con-sequently relative to large n, according to (5.1)) and applying Young’s inequality give:

For the forcing terms, we appeal to the bounds in (5.52), coupled with (5.75):

(5.88)

�� x−

1

2−2�i v(i)

s��y

��≤ ‖v(i)

sx

1

2 ‖L∞�‖x−

3

4−�i�‖2

L2y

+ ‖x−1

4−�i�y‖2L2

y

�

≤ (�)�‖x− 3

4−�i�‖2

L2y

+ ‖x−1

4−�i�y‖2L2

y

�.

(5.89)

�� x−

1

2−2�i� �

∞

y

u0pyvdy�dy

��≤ ‖x−

3

4−�i�‖L2

y

��x

1

4−�i �

∞

y

u0pyv��L2

y

≤ ‖x−3

4−�i�‖L2

yx

1

4−�i‖yu0

pyv‖L2

y

≤ ‖y2x−1

2 u0py‖L∞‖x−

3

4−�i�‖L2

y‖vyx

3

4−�i‖L2

y

≤ (�)‖x− 3

4−�i�‖L2

y‖vyx

3

4−�i‖L2

y.

(5.90)𝜕x � x

−1

2−2𝜎i𝜓2 + � x

−3

2−2𝜎i𝜓2 + � x

−1

2−2𝜎i𝜓2

y

≲ (𝛿)� v2yx

3

2−2𝜎i + � x

−1

2−2𝜎i𝜓 �

∞

y

[f (i) + ]

.

(5.91)

��x−

1

2−2�i� �

∞

y

f (i) dy��L2

y

≤ ‖x−3

4−�i�‖L2

y

��x

1

4−�i �

∞

y

f (i) dy��L2

y

≤ ‖x−3

4−�i�‖L2

y‖x

1

4−�i yf (i)‖L2

y

≤ ‖x−3

4−�i�‖L2

y‖x

3

4−�i zf (i)‖L2

y

≤ ‖x−3

4−�i�‖L2

yx−

1

2−�i

≤ 1

100 000‖x−

3

4−�i�‖2

L2y

+ Cx−1−��

,

221

1 3


where 𝜎′ > 0 . We have used (5.1) via: �i − 2�i−1 = 3�i−1 − 2�i−1 = �i−1 , to calculate:

Next, through Young’s inequality,

where 𝜎′ > 0 . Inserting these into (5.90), relabeling �y = u , and integrating in x, one obtains the desired result in (5.79). □

From here, we may repeat the calculations in Lemmas 4.6–4.8:

Lemma 5.16 The solutions [u, v] to the system (5.46) satisfy the following inequality:

Proof We multiply both sides of Eq. (5.73) by ux1

2−2�i and integrate by parts. This

gives on the left-hand side:

One now sees that it is crucial to have controlled the final term in (5.95), which is the purpose of (5.90). Moving next to the sequence of terms in (u, v):

(5.92)‖x3

4−𝜎i zf (i)‖L2

y≲ x

3

4−𝜎i x

−5

4+2𝜎i−1 = Cx

−1

2−𝜎i+2𝜎i−1 = Cx

−1

2−𝜎i−1 .

(5.93)

�� x−

1

2−2𝜎i𝜓 �

∞

y

dy��≲ � x

−1

2−2𝜎i𝜓⟨y⟩−Nx−

3

4+𝜎i−1

≲1

100 000‖x−

3

4−𝜎i𝜓‖2

L2y

+ Cx−1−𝜎�

,

(5.94)

supx � u2x

1

2−2𝜎i + � � u2

yx

1

2−2𝜎i ≲ C + (𝛿)� � v2

yx

3

2−2𝜎i + � � u2x

−1

2−2𝜎i .

(5.95)

∫((1 + u0

p)𝜕x − 𝜕yy

)u ⋅ ux

1

2−2𝜎i ≳ 𝜕x ∫ (1 + u0

p)u2x

1

2−2𝜎i + ∫ u2

yx

1

2−2𝜎i

− ∫ u0pxu2x

1

2−2𝜎i − ∫ (1 + u0

p)u2x

−1

2−2𝜎i .

(5.96)�� u(i)

sxu2x

1

2−2�i

��≤ ‖u(i)

sxx‖L∞ � u2x

−1

2−2�i ≤ (�)‖ux− 1

4−�i‖2

L2y

,

222 S. Iyer

1 3

We can summarize the above terms by writing:

Note that the smallness in the above estimates is guaranteed by (2.85), (2.90), and (3.57). We now move to the forcing terms, f and :

Next, we use the Hardy inequality:

Combining the above estimates together, one obtains, for some 𝜎′ > 0 small,

(5.97)

�� v(i)suyux

1

2−2𝜎i

��=

��

v(i)sy

2u2x

1

2−2𝜎i

��≲ ‖u(i)

sxx‖L∞‖ux−

1

4−𝜎i‖2

L2y

≤ (𝛿)‖ux− 1

4−𝜎i‖2

L2y

,

(5.98)

�� u0pyuvx

1

2−2𝜎i

��≲ ‖u0

pyy‖L∞

��v

yx

3

4−𝜎i

��L2y

‖ux−1

4−𝜎i‖L2

y

≲ (𝛿)‖vyx 3

4−𝜎i‖L2

y‖ux−

1

4−𝜎i‖L2

y.

(5.99)�� ⋅ ux

1

2−�i

��≤ (�)‖vyx 3

4−�i‖2

L2y

+ (�)‖ux− 1

4−�i‖2

L2y

.

(5.100)

�� f (i) ⋅ ux1

2−2𝜎i

��≤ ‖f (i)x

3

4−𝜎i‖L2

y‖ux−

1

4−𝜎i‖L2

y≲ x

−1

2−𝜎�‖ux−

1

4−𝜎i‖L2

y

≤ Cx−1−𝜎�

+1

100 000‖ux−

1

4−𝜎i‖2

L2y

.

(5.101)

�� ⋅ ux1

2−2𝜎i

��≲ � ⟨y⟩−Nx−

3

4+𝜎i−1 �u�x

1

2−2𝜎i

≲ ‖⟨y⟩−N+1‖L2yx−

1

2−𝜎��

u

y

��L2y

x1

4−𝜎i

≤ Cx−1−𝜎�

+1

100 000‖uyx

1

4−𝜎i‖2

L2y

.

(5.102)

𝜕x � (1 + u0p)u2x

1

2−2𝜎i + � u2

yx

1

2−2𝜎i ≲ Cx−1−𝜎

�

+ (𝛿)� v2yx

3

2−2𝜎i + � u2x

−1

2−2𝜎i ,

223

1 3


and so integrating in x gives the desired result. □

The third step in establishing the desired bounds is:

Lemma 5.17 For solutions [u, v] to the system in (5.73), one has the following posi-tivity estimate:

Proof We apply the multiplier uxx3

2−2�i to the system (5.73). This gives on the left-

hand side:

Let us now move to the terms in :

We can summarize this contribution via:

(5.103)

supx � u2

yx

3

2−2𝜎i + � � u2

xx

3

2−2𝜎i ≲ C + � � u2

yx

1

2−2𝜎i + (𝛿)� � u2x

−1

2−2𝜎i .

(5.104)∫

((1 + u0

p)𝜕x − 𝜕yy

)u ⋅ uxx

3

2−2𝜎i ≳ ∫ (1 + u0

p)u2

xx

3

2−2𝜎i

+ 𝜕x ∫ u2yx

3

2−2𝜎i − ∫ u2

yx

1

2−2𝜎i .

(5.105)

�� u(i)sxuuxx

3

2−2𝜎i

��≤ ‖u(i)

sxx‖L∞‖ux−

1

4−𝜎i‖L2

y‖uxx

3

4−𝜎i‖L2

y

≲ (𝛿)‖ux− 1

4−𝜎i‖L2

y‖uxx

3

4−𝜎i‖L2

y,

(5.106)

�� v(i)suyuxx

3

2−2�i

��≤ ‖v(i)

sx

1

2 ‖L∞‖uyx1

4−�‖L2

y‖uxx

3

4−�i‖L2

y

≤ (�)‖uyx 1

4−�i‖L2

y‖uxx

3

4−�i‖L2

y,

(5.107)

�� u0pyuxvx

3

2−2𝜎i

��≲ ‖u0

pyy‖L∞

��v

yx

3

4−𝜎i

��L2y

‖uxx3

4−𝜎i‖L2

y

≲ (𝛿)‖vyx 3

4−𝜎i‖2

L2y

.

(5.108)

�� ⋅ uxx3

2−2�i

��≤ (�)‖uxx 3

4−�i‖2

L2y

+ (�)‖uyx 1

4−�i‖2

L2y

+ (�)‖ux− 1

4−�i‖2

L2y

.

224 S. Iyer

1 3

Again, the smallness is guaranteed by (2.85), (2.90), and (3.57). Next, the forcing is given in the same way, due to the choice of �i relative to �i−1:

We now integrate in x up to some fixed point X1:

The last part is to control the term above:

We have absorbed the x boundary contribution, ∫ u2yX

3

2−2�i

1 into the left-hand side

of (5.110). For the other terms in , we can perform a similar calculation. As X1 is arbitrary, this then implies the desired result, (5.103). □

(5.109)

�� f (i) ⋅ uxx3

2−2𝜎i

��≤ ‖f (i)x

3

4−𝜎i‖L2

y‖uxx

3

4−𝜎i‖L2

y

≲ x−

1

2−𝜎�‖uxx

3

4−𝜎i‖L2

y≲ x−1−𝜎

�

+1

100 000‖uxx

3

4−𝜎i‖2

L2y

.

(5.110)

� u2y(X1)X

3

2−2𝜎i

1+ �

X1

1 � u2xx

3

2−2𝜎i

≲ C + �X1

1 � u2yx

1

2−2𝜎i + (𝛿)�

X1

1 � u2x−

1

2−2𝜎i

+ �X1

1 � ⋅ uxx3

2−2𝜎i .

(5.111)

�X1

1 � � ��(y)x3

2−2�i ui

e(x, 0)ux

= −�X1

1 � � ��(y)u�x{uie(x, 0)x

3

2−2�i} − �x=1

uie(0, 0)u� ��(y)dy

+ �x=X1

X3

2−2�i

1uie(X1, 0)�

��(y)u(X1, y)dy

= −�X1

1 � � ��(y)u�x{uie(x, 0)x

3

2−2�i} + C

− C �x=X1

X3

2−2�i

1uie(X1, 0)�

�(y)uy(X1, y)dy

≤ �X1

1 � � �(y)uy�x

�uie(x, 0)x

3

2−2�i

�+ C +

1

100 000 �x=X1

X3

2−2�i

1u2y(X1)

≤ C‖uyx1

4−�i‖L2‖� �(y)x

−1

2−�i‖L2 + C ≤ C +

1

100 000‖uyx

1

4−�i‖2

L2.

225

1 3


Proof of Proposition 5.14 Consolidating estimates the three estimates (5.79), (5.94), and (5.103), and subsequently taking � small enough, one obtains:

It is a standard matter now to obtain weighted in z estimates, and to also succes-sively differentiate the system in x and repeat the previously established bounds. This procedure establishes the desired result for ui

p in (5.72). To estimate vi

p , we take:

and repeat the procedure as in estimate (4.65). □

5.3 Final Prandtl Layer

We now construct the final Prandtl layer, [unp, vn

p] . This final, n’th layer is slightly

different because vnp will be taken to satisfy the boundary condition: vn

p(x, 0) = 0 .

According to (5.46), the system for the n’th layer is:

As usual from the Prandtl layers, by evaluating the equation at y = ∞ , it is clear that the leading order Prandtl pressure is constant, that is Pn

p= 0 . Once up, vp are

constructed to solve (5.114), we shall cut them off, thereby defining [unp, vn

p] . The rel-

evant remainders are given in (1.53)–(1.54), which we recall here for convenience:

Definition 5.18 The n’th remainder is denoted by:

(5.112)

supx � 𝜓2x

−1

2−2𝜎i + u2x

1

2−2𝜎i + u2

yx

3

2−2𝜎i

+ � � 𝜓2x−

3

2−2𝜎i + u2x

−1

2−2𝜎i + u2

yx

1

2−2𝜎i

+ u2xx

3

2−2𝜎i ≲ (𝛿;n).

(5.113)vip= ∫

∞

y

uipxdy�,

(5.114)(1 + u0p)upx + u(n)

sxup + v(n)

supy + u0

pyvp + Pn

px= upyy + f (n),

(5.115)[up(x, 0), vp(x, 0)] = [−un

e(x, 0), 0], lim

y→∞up(x, y) = 0, up(1, y) = Un(y),

(5.116)Pnpy= 0, upx + vpy = 0.

(5.117)Ru,n ∶= −Δ𝜖 u(n)s

+ u(n)su(n)sx

+ v(n)su(n)sy

+ P(n)sx,

(5.118)Rv,n ∶= −Δ𝜖 v(n)s

+ u(n)sv(n)sx

+ v(n)sv(n)sy

+𝜕y

𝜖P(n)s.

226 S. Iyer

1 3

Remark 5.19 We refer the reader to Remark 5.3. In this case, there is no (n + 1)’th Euler construction, which explains the definition in (5.117).

Using similar arguments to (5.48), (5.49), we have the following errors:

Here (n) = (n)(up, vp) is an error term created by cutting off these layers, which will be defined in (5.137). One can repeat the procedure used to construct the previ-ous Prandtl layers to conclude:

Proposition 5.20 Let �n be defined according to (5.1). The Prandtl layer, up con-structed to satisfy Eq. (5.114) together with the boundary conditions in (5.115) satisfies:

Let us summarize the decay rates which result from this construction, by recall-ing the definition of our Prandtl norms, Pk , given in (4.33)–(4.34):

Corollary 5.21 For any k, j,M ≥ 0 , the profiles up satisfy the following decay rates:

To satisfy vp(x, 0) = 0 , vp is obtained from up via: vp(x, y) = − ∫ y

0upx(x, y

�) dy� . This is distinct from (4.27) and (5.112), and distinguishes the final, n- th Prandtl layer from the previous layers. Then,

Corollary 5.22 vp obeys the following uniform decay estimate:

Proof Using the boundary condition vp(x, 0) = 0 , one can write:

(5.119)

Ru,n = �n

2

��un

pxx+ vn

p

�n�

j=1

�j

2

�ujpy+√�u

j

eY

��+ un

px

n�

j=1

�j

2

�uje+ uj

p

�+ (n)

�,

(5.120)Rv,n = �

n

2

[−Δ�v

np+ u(n)

svnpx+ v(n)

sxunp+ v(n)

svnpy+ v(n)

syvnp+ �

n

2 unpvnpx+ �

n

2 vnpvnpy

].

(5.121)‖x1

4 zm�kxup‖Pk(�n)

≤ C(k,m, n).

(5.122)xk+

j

2+

1

2−�n‖zM�k

x�jyup‖L∞

y+ x

k+j

2+

1

4−�n‖zM�k

x�jyup‖L2

y≤ C(k, j,M, n).

(5.123)xk+1−�n‖�kxvp‖L∞

y≤ C(k, n).

(5.124)

�vp�2 ≲��

y

0

vpvpy��≤ �

∞

0

��

vp

y1

2+𝜅

y1

2+𝜅vpy

��≤��

vp

y1

2+𝜅

��L2y

‖y1

2+𝜅vpy‖L2

y

≤ ‖y1

2−𝜅vpy‖L2

y‖y

1

2+𝜅vpy‖L2

y= ‖x

1

4−𝜅z

1

2−𝜅vpy‖L2

y‖x

1

4+𝜅z

1

2+𝜅vpy‖L2

y

≤ x1

2 x−1+𝜎nx−1+𝜎n = x−

3

2+2𝜎n .

227

1 3


We have used the 𝜅 > 0 to avoid the critical Hardy inequality. The Hardy inequality we have used (with power y−

1

2−�v1p ) relies on the vanishing of vn

p at y = 0.

Next, we introduce a cutoff function which honors the scaling, z = y√x:

Here, we make the notational convention for integrands: z� = y√x�

or z� = y�√x where

“ ′ ” denotes the integration variable. It is clear, then, that �xunp + �yvnp= 0 . The fol-

lowing boundary conditions are also clear:

Remark 5.23 This cutoff honors the parabolic scaling of the Prandtl layers for all x > 0 . The cutoff introduced in [24] was in the region y ≥ 1√

� . Locally in x, z is

equivalent to y, so these cutoffs agree locally in x.

We must now record two properties about the cutoff layers. First, the uniform estimates of the cutoff layers remain unchanged:

Lemma 5.24 The following pointwise decay bounds hold for the cutoff layers, for any k ≥ 0:

Proof First,

Next,

(5.125)vnp∶= �(

√�z)vp, un

p∶= ∫

∞

x

�y

��

�√�

y√x�

�vp(x

�, y)

�dx�.

(5.126)[unp, vn

p] → 0 as y → ∞, [un

p, vn

p]|y=0 = [up, vp]|y=0.

(5.127)|||𝜕

kxvnp

||| ≲ C(k, n)x−k−1+𝜎n ,

(5.128)|||𝜕kxvnpy

||| ≲ C(k, n)x−k−3

2+𝜎n ,

(5.129)|||𝜕

kxunpy

||| ≲ C(k, n)x−k−1+𝜎n ,

(5.130)|||𝜕kxunp

||| ≲ C(k, n)x−k−1

2+𝜎n .

(5.131)�vnpy� ≤

√𝜖

√x𝜒 ��vp� + 𝜒�vpy� ≲ x

−3

2+𝜎n .

228 S. Iyer

1 3

Finally, for unp,

By differentiating successively in x, the result for k ≥ 1 follows in the same manner. □

Lemma 5.25 (L2y Estimates) The following L2

y decay bounds hold for the cutoff lay-

ers, for any k ≥ 0:

Proof To compute the L2 norms, we must trade in the following way:

where o(�) means � or any derivative of � , the essential feature being that z ≤ 1√� .

Using this:

For unpy

, again according to the expression in (5.132), and the decay rates in (5.122):

It is now clear that we can repeat these calculations for higher x and y derivatives, thereby obtaining the desired result. □

(5.132)

�unpy� =

��∞

x

𝜕yy

�𝜒(

√𝜖z�)vp

�dx�

��≤ ��

∞

x

𝜖

x�𝜒 ��vp

��

+

��2�

∞

x

√𝜖

√x�vpy

��+��

∞

x

𝜒vpyy��

≲ x−1+𝜎n .

(5.133)�unp� ≤ �

∞

x

√�

1√x�� vp� + �

∞

x

�vpy� ≤ x−

1

2+�n .

(5.134)‖𝜕kx𝜕jyunp‖L2

y≲ C(k, 𝜎, n)x−k−

j

2−

1

4+𝜎n .

(5.135)o(𝜒)��𝜖

1

4+

𝜅

2 ⟨y⟩1

2+𝜅x−

1

4−

𝜅

2�� ≲ 1,

(5.136)‖unp‖L2

y≤ �

∞

x

��

√𝜖

√x�𝜒 �vp

��L2y

+ �∞

x

‖𝜒vpy‖L2y≲ x

−1

4+𝜎n .

(5.137)

‖unpy‖L2

y≲ ∫

∞

x

��𝜖𝜒 ��

x�vp��L2

y

+ ∫∞

x

��

√𝜖

√x�vpy

��L2y

+ ∫∞

x

‖vpyy‖L2y≲ x

−3

4+𝜎n .

229

1 3


The next task is to control the error made by cutting off the layers. This error is obtained by inserting the new, cutoff layers into Eq. (5.114):

We will proceed to expand (5.138), term by term:

Let us provide justification to the above expressions, first turning to (5.140). Apply-ing the definition in (5.125) yields:

the final equality following from an integration by parts in x. Similarly, for (5.141):

(5.138) (n) ∶= (1 + u0p)un

px+ u(n)

sxunp+ v(n)

sunpy+ u0

pyvnp− un

pyy− f (n).

(5.139)(1 + u0p)un

px= �(1 + u0

p)upx − (1 + u0

p)

√�

√x� �vp,

(5.140)u(n)sxunp= �u(n)

sxup + u(n)

sx ∫∞

x

√�

√x�� vp + � �

√�z

x�up,

(5.141)v(n)sunpy= �v(n)

supy + v(n)

s ∫∞

x

�

x�� vp + 2

√�

√x�� vpy −

√�

√x�� zupy,

(5.142)u0pyvnp= �u0

pyvp,

(5.143)

unpyy

= �upyy − ∫∞

x

√�

x�� z

2upyy + C1

�

x�� vpy + C2�

�

√�

√x�vpyy

+�

3

2

(x�)3

2

� ��vp.

(5.144)

u(n)sxunp= u(n)

sx ∫∞

x

√�

√x�� vp + u(n)

sx ∫∞

x

�vpy

= u(n)sx ∫

∞

x

√�

√x�� vp − u(n)

sx ∫∞

x

�upxdx�

= u(n)sx ∫

∞

x

� √�

√x�� vp + u(n)

sx

z√�

x�� up

�dx� + �u(n)

sxup,

230 S. Iyer

1 3

This same computation is performed for (5.143):

Summing the above terms together, we arrive at our expression:

(5.145)

v(n)sunpy= v(n)

s ∫∞

x

�yy(�vp)dx�

= v(n)s ∫

∞

x

��

x�� vp + 2� �

√�

√x�vpy + �vpyy

�dx�

= v(n)s ∫

∞

x

�

x�� vp + 2� �

√�

x�vpy − v(n)

s ∫∞

x

�upxydx�

= v(n)s ∫

∞

x

��

x�� vp + 2� �

√�

x�vpy −

√��

x�zupy

�dx� + v(n)

s�upy.

(5.146)

unpyy

= ∫∞

x

�3y(�vp)dx

�

= ∫∞

x

��

3

2

(x�)3

2

� ��vp + C1

�

x�� vpy + C2

√�

√x�vpyy�

�

�+ ∫

∞

x

�vpyyy

= ∫∞

x

��

3

2

(x�)3

2

� ��vp + C1

�

x�� vpy + C2

√�

√x�vpyy�

�

�− ∫

∞

x

�upxyy

= ∫∞

x

��

3

2

(x�)3

2

� ��vp + C1

�

x�� vpy + C2

√�

√x�vpyy�

�

�− ∫

∞

x

�� upyy

z

2x�

√�

�dx�

+ �upyy.

(5.147)

(n) = −(1 + u0p)

√�

√x� �vp + u(n)

sx �∞

x

√�

√x�� vp + u(n)

sx �∞

x

� �

√�z

x�up

+ v(n)s �

∞

x

�

x�� vp + 2

√�

√x�� vpy −

√�

x�� zupy − �

∞

x

√�

x�� z

2upyy

+ �∞

x

�C2�

�

√�

√x�vpyy + C1

�

x�� vpy +

�3

2

(x�)3

2

� ��vp

�+ (1 − �)f (n).

231

1 3


Lemma 5.26 For 𝜅 > 0 arbitrarily small, the error, (n) obeys the following bound:

Proof We will proceed in order from (5.147), and we will focus on the L2y estimates,

as the uniform estimates are straightforward. First,

The essential characteristic in the above calculation is the ability to pay a factor of �

1

4+

�

2 x−

1

4−

�

2 to obtain an L2y quantity due to the presence of the cutoff function � ′

(5.135). In a similar manner, we have:

The third term on the right-hand side of (5.147) can be estimated analogously. Com-ing now to the first term on the second line of (5.147),

The second term on the second line of (5.147):

(5.148)� (n)� ≤ C(n, �)�1

4−�x−

3

2+�n+� , ‖ (n)‖L2

y≤ C(n, �)�

1

4−�x−

5

4+�n+� .

(5.149)

��(1 + u0

p)

√𝜖

√x𝜒 �(z)vp

��L2y

≲√𝜖

��⟨y⟩

1

2+𝜅⟨y⟩−

1

2−𝜅𝜒 �vp

1√x

��L2y

≲√𝜖‖⟨y⟩

1

2+𝜅𝜒 �x

−1

2 vp‖L∞y≲√𝜖‖z

1

2+𝜅𝜒 �x

−1

4+

𝜅

2 vp‖L∞y

≲ 𝜖1

4−𝜅x−

5

4+

𝜅

2+𝜎n .

(5.150)

��u(n)sx �

∞

x

√𝜖

√x�𝜒 �vpdx

�

��L2y

≤ ‖u(n)sx‖L∞ �

∞

x

��

√𝜖

√x�𝜒 �vp

��L∞y

‖⟨y⟩−1

2−𝜅‖L2

ydx�

≲ x−1 �∞

x

(x�)−1+𝜎n𝜖1

4−

𝜅

2 x−

1

4+

𝜅

2 ‖𝜖1

4+

𝜅

2 x−

1

4−

𝜅

2 ⟨y⟩1

2+𝜅𝜒 �‖L∞

≲ 𝜖1

4+𝜅x−

5

4+𝜎n+

𝜅

2 .

(5.151)

��v(n)s �

∞

x

𝜖

x𝜒 ��vp

��L2y

≤ ‖v(n)s‖L∞ �

∞

x

𝜖3

4−

𝜅

2 (x�)−

3

4+

𝜅

2 (x�)−1+𝜎n‖𝜖1

4+

𝜅

2 x−

1

4−

𝜅

2 ⟨y⟩1

2+𝜅𝜒 ��‖L∞

≲ 𝜖3

4−

𝜅

2 x−

5

4+

𝜅

2+𝜎n .

(5.152)

��v(n)s �

∞

x

√𝜖

√x�𝜒 �vpy

��L2y

≤ ‖v(n)s‖L∞ �

∞

x

𝜖1

4−

𝜅

2 (x�)−

1

4+

𝜅

2 (x�)−

3

2+𝜎ndx�

≲ 𝜖1

4−

𝜅

2 x−

5

4+𝜎n+

𝜅

2 .

232 S. Iyer

1 3

Moving to the third, it is slightly different due to the weight z, but this causes no harm as zupy is known to be in L∞

y according to (5.121):

Next, we move to the unpyy

contributions:

according to the rates, ‖upyyzx5

4−�n , vpyyx

7

4−�n , vpyx

5

4−�n‖L2

y≤ C from (5.122). Finally,

for the vp term on the third line of (5.147), one must use the trade-off in (5.135), to obtain:

The f (n) term can then be estimated upon observing that f (n) is exponentially small in the support of 1 − �:

for any N large, according to estimate (5.52). This now concludes the proof of (5.148). □

By construction, one has the following:

Lemma 5.27 (Remainder Estimates) For Ru,n,Rv,n as defined in (5.120), and for any � ∈ [0,

1

4) , n ≥ 2 , and for �n as in (5.1), 𝜅 > 0 arbitrarily small,

(5.153)

��v(n)s ∫

∞

x

√𝜖

x�𝜒 �zupy

��L2y

≲ x−

1

2 ∫∞

x

√𝜖

x�(x�)

−3

4+𝜎ndx� ≲

√𝜖x

−5

4−𝜎n .

(5.154)∫∞

x

��C1

𝜖

x�𝜒 ��vpy + C2

√𝜖

√x�vpyy𝜒

� − 𝜒 �upyyz

2x�

√𝜖

��L2y

≲√𝜖x

−5

4+𝜎n ,

(5.155)

∫∞

x

��

𝜖3

2

(x�)3

2

𝜒 ��vp

��L2y

dx� ≲ 𝜖5

4−

𝜅

2 ∫∞

x

(x�)1

4+

𝜅

2 (x�)−

3

2 ‖vp‖L∞y

≲ 𝜖5

4−

𝜅

2 ∫∞

x

(x�)1

4+

𝜅

2 (x�)−

3

2 (x�)−1+𝜎n

≲ 𝜖5

4−

𝜅

2 x−

5

4+

𝜅

2+𝜎n .

(5.156)‖(1 − 𝜒)f (n)‖L2

y= ‖(1 − 𝜒)zNz−Nf (n)‖L2

y

≲ 𝜖N

2 ‖zNf (n)‖L2y≲ 𝜖

N

2 x−

5

4+𝜎n−1 ≲ 𝜖

N

2 x−

5

4+𝜎n ,

(5.157)𝜖−

n

2−𝛾 ��𝜕

kxRu,n +

√𝜖𝜕k

xRv,n�� ≲ C(n, 𝜅)𝜖

1

4−𝛾−𝜅

x−k−

3

2+2𝜎n ,

233

1 3


Proof This follows from (5.148) and those bounds established in (5.121). We shall proceed term by term from (5.120). First,

First,

It is clear that (5.159.1), (5.159.2), and (5.159.3) are identical, and so we move to:

and finally:

We next move to the second term in Ru,n , for which we immediately use (5.135)

where we have used (5.127), (5.129). Similarly, for j ≥ 1 , one has:

(5.158)𝜖−

n

2−𝛾‖

√𝜖𝜕k

xRu,n,

√𝜖𝜕k

xRv,n‖L2

y≲ C(n, 𝜅)𝜖

1

4−𝛾−𝜅

x−k−

5

4+2𝜎n+𝜅 .

(5.159)

�1−�‖unpxx

‖L2y= �1−�‖�x(�vp)y‖L2

y= �1−�

��x

��

√x

√�vp + �vyp

��L2y

= �1−��

x3

2

zvp +� �√�

x3

2

vp +� �

√x

√�vpx +

√��

xzvpy + �vpxy

��L2y

= (5.159.1) +⋯ + (5.159.5).

(5.160)

�(5.159.1)� ≲ 𝜖1−𝛾��√𝜖z ⋅

�⟨y⟩

1

2+𝜅x−

1

4−

𝜅

2 𝜖1

4+

𝜅

2

�𝜒 ��

��L∞y

𝜖1

4−

𝜅

2 ‖vp‖L∞x−5

4+

𝜅

2

≲ 𝜖5

4−𝛾−

𝜅

2 x−

9

4+

𝜅

2+𝜎n .

(5.161)�(5.159.4)� ≲ 𝜖1−𝛾

√𝜖

x‖zvpy‖L2

y≲ 𝜖1−𝛾

√𝜖

xx−

5

4+𝜎n ,

(5.162)�(5.159.5)� = 𝜖1−𝛾‖𝜒vpxy‖L2y≲ 𝜖1−𝛾x

−9

4+𝜎n .

(5.163)𝜖

n

2−𝛾‖vn

punpy‖L2

y≤ 𝜖

n

2−𝛾−

1

4−

𝜅

2 x1

4+

𝜅

2 ‖vnp‖L∞

y‖un

py‖L∞

y

≲ 𝜖n

2−𝛾−

1

4−

𝜅

2 x1

4+

𝜅

2 x−2+2𝜎n = 𝜖n

2−𝛾−

1

4−

𝜅

2 x−

7

4+

𝜅

2+2𝜎n .

(5.164)𝜖

1

2−𝛾‖vn

pujpy‖L2

y≤ 𝜖

1

2−𝛾‖vn

p‖L∞‖ujpy‖L2y ≲ 𝜖

1

2−𝛾x−1+𝜎nx

−1

2+𝜎n ≲ 𝜖

1

2−𝛾x−

3

2+2𝜎n

234 S. Iyer

1 3

where we have used the specification of the norm ‖ ⋅ ‖P given in (4.33). For the term vnpuj

eY , we calculate, for 𝜅 > 0:

The last estimate follows from applying L2Y to the Eulerian self-similar estimate,

(3.44):

The next term in Ru,n is, for j ≥ 1 , for which we use (4.75) for the ujp term, (5.19) for the uje term, and (5.134) for the u(n)

px term in L2:

The final term in Ru,n is the error term, (n) , which has been controlled in (5.148). We may now move to Rv,n . In so doing, we first note that �vn

pxx can be estimated iden-

tically to (5.168) below. Second, using (5.134), we have:

For the L2 bound on the vnpx

term in (5.120), we employ (5.135):

The next two terms are:

(5.165)𝜖

1

2+

j

2−𝛾‖vn

puj

eY‖L2

y≤ 𝜖

1

4+

j

2−𝛾‖vn

p‖L∞

y‖uj

eY‖L2

Y≲ C(n)𝜖

1

4+

j

2−𝛾x−1+𝜎nx−1+𝜅 .

(5.166)|u1eY| = |v1

ex| ≤ C(�)x−1+�Y

−1

2−� .

(5.167)‖𝜖j

2 u(n)px(uj

e+ uj

p)‖L2

y≲ 𝜖

j

2 x−

1

4+𝜎n‖un

px‖L2

y≲ 𝜖

1

4−

𝜅

2 x−

1

4+𝜎nx

−5

4+𝜎n .

𝜖1

2−𝛾‖vn

pyy‖L2

y≲ 𝜖

1

2−𝛾x−

7

4+𝜎n .

(5.168)

𝜖1

2−𝛾‖u(n)

svnpx‖L2

y≲ 𝜖

1

2−𝛾‖vn

px‖L2

y≤ 𝜖

1

2−𝛾‖1

z≤ 1√𝜖

vnpx‖L2

y

≤ 𝜖1

2−𝛾‖1

z≤ 1√𝜖

x−2+𝜎‖L2y= 𝜖

1

2−𝛾‖1

z≤ 1√𝜖

x−2+𝜎n (1 + y)1

2+𝜅(1 + y)

−1

2−𝜅‖L2

y

≤ 𝜖1

2−𝛾‖1

z≤ 1√𝜖

(1 + y)1

2+𝜅x−

1

4−

𝜅

2 ‖L∞yx−

7

4+𝜎n+

𝜅

2 ‖(1 + y)−

1

2−𝜅‖L2

y

≤ 𝜖1

2−𝛾‖1

z≤ 1√𝜖

(1 + z)1

2+𝜅‖L∞

yx−

7

4+𝜎n+

𝜅

2 ≤ 𝜖1

4−𝛾−𝜅

x−

7

4+𝜎n+

𝜅

2 .

(5.169)𝜖1

2−𝛾‖v(n)

sxunp‖L2

y≤ √

𝜖‖v(n)sx‖L∞‖unp‖L2y ≲

√𝜖x

−3

2 x−

1

4 ,

(5.170)�1

2−�‖v(n)

svnpy‖L2

y≤ �

1

2−�x−

7

4+2�n .

235

1 3


Now, by using the trade-off in (5.135):

We will now move to:

where we have used (5.127) and (5.134). This completes all of the terms in (5.120), thereby establishing (5.158). The uniform estimates follow in an analogous manner. □

We will need to retain the self-similarity of unp (the ability to absorb factors of

z). As the profiles unp, vn

p are higher order in � , we will be happy to pay factors of �

to retain this ability:

Lemma 5.28 The following point-wise decay estimates hold for any m ≥ 0,

Proof Via the definition:

Next, applying �y yields:

(5.171)𝜖1

2−𝛾‖v(n)

syvnp‖L2

y≤ 𝜖

1

4−𝛾−𝜅‖v(n)

sy‖L∞

y‖vn

p‖L∞

yx

1

4+

𝜅

2 ≲ 𝜖1

4−𝛾−𝜅

x−

7

4+2𝜎n+

𝜅

2 .

(5.172)𝜖n

2−𝛾√𝜖‖un

pvnpx‖L2

y≲ 𝜖

n

2−𝛾+

1

2 ‖unp‖L2

y‖vn

px‖L∞

y≲ 𝜖

n

2−𝛾+

1

2 x−

9

4+2𝜎n ,

(5.173)𝜖n

2−𝛾√𝜖‖vn

pvnpy‖L2

y≲ 𝜖

n

2−𝛾+

1

2 x−

9

4+2𝜎n ,

(5.174)|||z

m𝜕kxvnp

||| ≲ 𝜖−

m

2 C(k, n,m)x−k−1+𝜎n ,

(5.175)|||zm𝜕k

xvnpy

||| ≲ 𝜖−

m

2 C(k, n,m)x−k−3

2+𝜎n ,

(5.176)|||yj𝜕j

yunp

||| ≲ C(k, n,m)x−1

2+𝜎n ,

(5.177)|||z

m𝜕kxunpy

||| ≲ 𝜖−

m

2 C(k, n,m)x−1+𝜎n−k for k ≥ 1.

(5.178)zm�vnp� ≤ zm�𝜒(

√𝜖z)vp� ≤ 𝜖

−m

2 �vp� ≲ 𝜖−

m

2 x−1+𝜎n .

(5.179)zm�vnpy� = zm

√𝜖

��𝜒 �

vp√x

��+ 𝜒zm�vpy� ≲ 𝜖

−m

2 x−

3

2+𝜎n .

236 S. Iyer

1 3

The estimate for unpy

is more complicated. The m = 0 case was treated in (5.130). Suppose m = 1 . Then,

We have:

Next,

Lastly,

Piecing these estimates together gives:

as desired. Higher y-derivatives of unp follow an identical calculation. Let us now

move to x-derivatives of unpy

:

From here, one can estimate:

□

(5.180)

yunpy= y∫

∞

x

�yy(�vp)dx� = ∫

∞

x

�

x�� yvp + ∫x

√�

√x�� yvpy + ∫

∞

x

�yvpyy

= I1 + I2 + I3.

(5.181)

�I1� ≤ ‖𝜒 ��z‖L∞ �∞

x

𝜖�vp�√x�dx� ≤ √

𝜖 �∞

x

(x�)−

3

2+𝜎dx� ≲

√𝜖x

−1

2+𝜎n .

(5.182)�I2� ≤ �∞

x

√𝜖𝜒 ��zvpy�dx� ≲ �

∞

x

√𝜖(x�)

−3

2 dx� ≲√𝜖x

−1

2+𝜎n .

(5.183)�∞

x

|y𝜒vpyy|dx� ≤ �∞

x

(x�)1

2 zvpyydx� ≲ x

−1

2+𝜎n .

(5.184)y|unpy| ≲ x

−1

2+𝜎n ,

(5.185)unpyx

= �yy(�vp) =�

x� ��vp +

√�

√x� �vpy + �vpyy.

(5.186)zm�un

pxy� ≲ 𝜖

x�𝜒 ��zm�vp� +

√𝜖

√x�𝜒 ��zm�vpy� + 𝜒zm�vpyy�

≤ 𝜖−

m

2 x−2+𝜎n .

237

1 3


We are now ready to prove Theorem 1.7.

Proof of Theorem 1.7 Consolidating Corollaries 2.11–2.12, Propositions 3.6, 4.16, 5.4, 5.14, Lemma 5.27, and Lemma 5.28 gives Theorem 1.7. □

Acknowledgements The author thanks Yan Guo for many valuable discussions regarding this research. The author also thanks Bjorn Sandstede for introducing him to the paper [5].

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