Regularity and Expansion for Steady Prandtl Equations

Abstract

Due to degeneracy near the boundary, the question of high regularity for solutions to the steady Prandtl equations has been a longstanding open question since the celebrated work of Oleinik. We settle this open question in affirmative in the absence of an external pressure. Our method is based on energy estimates for the quotient, $q = \frac{v}{\bar{u}}$, $\bar{u}$ being the classical Prandtl solution, via the linear derivative Prandtl (LDP) equation. As a consequence, our regularity result leads to the construction of Prandtl layer expansion up to any order.

1 Introduction

1.1 Steady Prandtl equation

The 2D, steady, Prandtl equations are given by

$$\begin{aligned} \left. \begin{aligned}&\bar{u} \bar{u}_x + \bar{v} \bar{u}_y - \bar{u}_{yy} = - \partial _x p_E(x, 0) \\&\bar{u}_x + \bar{v}_y = 0 \end{aligned} \right\} \text { on } (x, y) \in (0, L) \times \mathbb {R}_+. \end{aligned}$$

(1)

The quantity $p_E(x, 0)$ above is considered prescribed, and $\partial _x p_E(x, 0)$ evidently acts as a forcing term to the Prandtl equations. In this paper, we are concerned with the homogeneous Prandtl equations, that is $p_E = 0$.

The Prandtl Eq. (1) are thought of as an evolution equation, with x being a time-like variable, and y being space-like. This can be formally seen through a crude derivative count $\bar{u} \partial _x \approx \partial _{yy}$, which indicates (1) is a quasilinear, parabolic equation. As a result, the Eq. (1) are supplemented with boundary conditions at $y = 0, y \rightarrow \infty$, and an initial condition at $x = 0$. The quantity L appearing in (1) is thought of as the time over which we are considering the evolution. To summarize the datum that is prescribed is

$$\begin{aligned} \bar{u}|_{y = 0} = \bar{v}|_{y = 0} = 0, \qquad \bar{u}|_{y \uparrow \infty } = u_E(\infty ), \qquad \bar{u}|_{x = 0} = \bar{U}_0(y). \end{aligned}$$

(2)

The conditions at $y = 0$ are called the no-slip condition, which is the classical boundary condition for viscous flows. The condition at $y = \infty$ is classically known as the Euler matching condition, and physically describes the horizontal component, u, matching up with some ambient Euler flow at $\infty$. Thus, the quantity $u_E(\infty )$ is a prescribed constant. We can, without loss of generality, assume this constant is 1.

The Prandtl system (1)–(2) is a celebrated system of PDEs that arise naturally when considering the physically important problem of describing the vanishing viscosity limit of Navier-Stokes flows on domains with boundaries. The precise link to the inviscid limit problem will be made clear later in the introduction, in Sect. 1.2. In addition, the Prandtl system by itself poses significant mathematical challenges that have attracted substantial attention over the years, beginning with the seminal work of Oleinik, [47].

The following local regularity result for the Prandtl system, (1)–(2), is classical ([47], P. 21, Theorem 2.1.1):

Theorem 1

(Oleinik). Assume $\bar{U}_0 \in C^\infty (\mathbb {R}_+)$ satisfies the following

$$\begin{aligned} \begin{aligned}&\bar{U}_0(y)> 0 \text { for } y> 0, \qquad \bar{U}_0'(0) > 0, \qquad \bar{U}_0''(0) = \bar{U}_0'''(0) = 0, \end{aligned} \end{aligned}$$

(3)

Then for some $L > 0$, there exists a solution, $[\bar{u}, \bar{v}]$ to (1)–(2) satisfying, for some $y_0, m_0 > 0$,

$$\begin{aligned}&\sup _{x \in (0,L)} \sup _{y \in (0, y_0)} |\bar{u}, \bar{v}, \bar{u}_{y}, \bar{u}_{yy}, \bar{u}_{x}| \lesssim C(\bar{U}_0), \end{aligned}$$

(4)

$$\begin{aligned}&\sup _{x \in (0,L)} \sup _{y \in (0, y_0)} \bar{u}_{y}> m_0 > 0. \end{aligned}$$

(5)

The method employed by Oleinik to prove Theorem 1 is to pass to the following change of coordinates, known as the von-Mise transform:

$$\begin{aligned}&(x,\psi ) = (x,\int _0^y \bar{u}(x,y') \,\mathrm {d}y'), \tilde{u}(x,\psi ) := \bar{u}(x,y), \partial _x \tilde{u} = \partial _\psi \{ \tilde{u} \tilde{u}_\psi \}. \end{aligned}$$

(6)

The equation above is quasilinear, degenerate diffusion equation. Estimates (4)–(5) are subsequently proven using maximum principle techniques.

Despite this, establishing higher regularity has been an important open problem. One cannot simply differentiate the Eq. (6) and repeat Oleinik’s argument because the commutators that arise from this process are uncontrollable near the boundary $\{\psi = 0\} = \{y = 0\}$. One contribution of the present paper is to resolve this problem by introducing a novel set of energy estimates enabling us to proving higher regularity estimates.

Theorem 2

Assume the data $\bar{U}_0(y)$ is provided satisfying conditions (3). Assume also generic compatibility conditions at the corner (0, 0) (as defined explicitly in Definition 1) up to order $M_0$. Fix $N > 0$ sufficiently large relative to universal constants, and assume the exponential decay of derivatives:

$$\begin{aligned} \Vert \partial _y^{j} \bar{U}_0 e^{Ny} \Vert _{L^\infty } \le C_0 \text { for } j = 1,\ldots ,M_0. \end{aligned}$$

(7)

Then there exists an $0< L<< 1$ depending on $C_0$ so that on $0 \le x \le L$, Oleinik’s solutions guaranteed by Theorem 1 obey the following estimates for $0 < 2\alpha + \beta \le M_0$ and for $\gamma \le \frac{M_0}{2} - 1$:

$$\begin{aligned} \Vert \partial _x^\alpha \partial _y^\beta \bar{u} e^{Ny} \Vert _{L^\infty } + \Vert \partial _x^{\gamma } \bar{v} \Vert _{L^\infty } \lesssim C_0. \end{aligned}$$

The system governing the behavior of the higher order derivatives of $[\bar{u}, \bar{v}]$ is, naturally, the linearized Prandtl equations. We now state precisely the linearized Prandtl system we consider, where we will work with some more generality than is strictly required for the proof of Theorem 2. Indeed, we consider the following problem for the unknown vector field, $[u_p, v_p]$:

$$\begin{aligned} \begin{aligned}&\bar{u} \partial _x u_p + u_p \partial _x \bar{u} + \bar{v} \partial _y u_p + v_p \partial _y \bar{u} - \partial _{yy} u_p = f, \qquad (x, y) \in (0, L) \times \mathbb {R}_+ \\&\partial _x u_p + \partial _y v_p = 0, \qquad (x, y) \in (0, L) \times \mathbb {R}_+ \\&\left[ u_p, v_p\right] |_{y = 0} = \left[ u_0(x), 0\right] , \qquad \lim _{y \rightarrow \infty } u_p(x, y) = 0, \qquad v_p|_{x = 0} = V_0(y). \end{aligned} \end{aligned}$$

(8)

The boundary condition $u_0(x)$ is a prescribed, smooth function (which certainly is allowed to be zero). We note that in our prescription of (8), the initial datum is now prescribed at the level of $v_p|_{x = 0}$ as opposed to $u_p$ as in (1).

The method that we use to obtain Theorem 2 is sufficiently robust also to obtain estimates for (8). This is an important distinction between our technique and that of Oleinik. Indeed, the proof of Theorem 1 relies on a very specific use of the maximum principle which cannot apply to the more general, linearized problem. We now state our main result regarding the linearized Prandtl system, (8).

Theorem 3

Fix any $M_0$ and an N large relative to universal constants. Assume compatibility conditions on the initial datum, $V_0$, as specified in Definitions 1 and 2. Assume also that

$$\begin{aligned} \Vert \partial _y^{k} V_0 e^{Ny} \Vert _{L^2\left( \mathbb {R}_+\right) } \le C_0 < \infty \text { for } k = 1,\ldots ,2M_0 + 3. \end{aligned}$$

(9)

Assume the forcing, f, satisfies the bounds

$$\begin{aligned} \Vert \partial _{x}^{k+1} \partial _y f e^{Ny} \Vert _{L^2} + \Vert \partial _x^{k + 2} \partial _y f \langle y \rangle \Vert _{L^2} \le C_1 < \infty \text { for } k = 0,\ldots ,M_0. \end{aligned}$$

(10)

Then the following estimates are valid

$$\begin{aligned}&\Vert \partial _x^\alpha v_p \Vert _{L^\infty } \lesssim C(C_0, C_1) \text { for } 0 \le \alpha \le M_0 + 1, \end{aligned}$$

(11)

$$\begin{aligned}&\Vert \partial _x^{\alpha } \partial _y^{\beta } v_p e^{Ny} \Vert _{L^\infty } \lesssim C(C_0, C_1) \text { for } 0 < \alpha + \frac{\beta }{2} \le M_0 + 1, \end{aligned}$$

(12)

$$\begin{aligned}&\Vert \partial _x^{\alpha } \partial _y^\beta u_p e^{Ny} \Vert _{L^\infty } \lesssim C(C_0, C_1) \text { for } \alpha + \frac{\beta + 1}{2} \le M_0 + 1. \end{aligned}$$

(13)

1.2 Inviscid limit of Navier–Stokes

We now describe the link of Theorems 2 and 3 with the related and very important problem of describing the vanishing viscosity limit of Navier–Stokes flows. Consider the steady, incompressible Navier–Stokes equations with viscosity $\epsilon$ posed on the domain

$$\begin{aligned} \varOmega := (0, L) \times \mathbb {R}_+, \text { with coordinates } x \in (0, L) \text { and } Y \in \mathbb {R}_+. \end{aligned}$$

The equations for the velocity field $\varvec{u}^{NS} := [u^{NS}, v^{NS}]$ and pressure $P^{NS}$ read

$$\begin{aligned}&\varvec{u}^{NS} \cdot \nabla \varvec{u}^{NS} + \nabla P^{NS} = \epsilon \varDelta \varvec{u}^{NS}, \text {div}(\varvec{u}^{NS}) = 0 \text { on } \varOmega \end{aligned}$$

with the following boundary conditions

$$\begin{aligned}&\varvec{u}^{NS}|_{Y = 0} = 0, \text { (no-slip condition)}, \\&\varvec{u}^{NS}|_{Y \uparrow \infty } = [u^0_e(Y), 0], \text { (convergence to an Euler flow) }, \\&\varvec{u}^{NS}|_{x = 0, L} = \text { in-flow and out-flow provided through asymptotic expansion.} \end{aligned}$$

The shear Euler profile $u^0_e(Y)$ is given data. The in-flow and out-flow data are prescribed individually through the expansion (14).

A fundamental question in fluid mechanics is to characterize the limit of $\varvec{u}^{NS}$ as $\epsilon \downarrow 0$. Due to the mismatch of the no-slip boundary condition satisfied by $\varvec{u}^{NS}$ for $\epsilon > 0$ and the typical no-penetration condition satisfied by a generic Euler flow, (which is $v^0_e|_{Y = 0} = 0$), one cannot expect $L^\infty$ convergence of $\varvec{u}^{NS}$ to an Euler flow $[u^0_e(Y), 0]$. Instead, one expects convergence to what is now known as the Prandtl boundary layer.

The first step in quantifying the asymptotic in $\epsilon$ behavior of $\varvec{u}^{NS}$ is to rescale the normal variable, Y, and the normal component of the velocity, $v^{NS}$ according to

$$\begin{aligned}&(x,y) = \left( x, \frac{Y}{\sqrt{\epsilon }}\right) , \\&\left[ U^\epsilon (x,y), V^\epsilon (x,y), P^\epsilon (x,y)\right] = \left[ U^{NS}(x,Y), \frac{v^{NS}(x,Y)}{\sqrt{\epsilon }}, P^{NS}(x,Y)\right] . \end{aligned}$$

The original, unscaled, variables (x, Y) are known as “Euler variables”, whereas the new, scaled variables (x, y) are known as “Prandtl variables”.

One subsequently asymptotically expands $[U^\epsilon , V^\epsilon , P^\epsilon ]$ in the following manner:

$$\begin{aligned} \begin{aligned} U^\epsilon \, =\,&u^0_e + u^0_p + \sum _{i = 1}^n \sqrt{\epsilon }^i \left( u^i_e + u^i_p\right) + \epsilon ^{N_0} u^{(\epsilon )} := u_s + \epsilon ^{N_0} u^{(\epsilon )}, \\ V^\epsilon \, =\,&v^0_p + v^1_e + \sum _{i = 1}^{n-1} \sqrt{\epsilon }^i \left( v^i_p + v^{i+1}_e\right) + \sqrt{\epsilon }^n v^n_p + \epsilon ^{N_0} v^{(\epsilon )} := v_s + \epsilon ^{N_0} v^{(\epsilon )}, \\ P^\epsilon \, =\,&P^0_e + P^0_p + \sum _{i = 1}^n \sqrt{\epsilon }^i \left( P^i_e + P^i_p\right) + \epsilon ^{N_0} P^{(\epsilon )} := P_s + \epsilon ^{N_0} P^{(\epsilon )}, \end{aligned} \end{aligned}$$

(14)

where the coefficients are independent of $\epsilon$. First, the Euler flow $[u^0_e(Y), 0]$ is a given shear flow satisfying the assumptions delineated in Theorem 4. Next, $[u^i_e, v^i_e]$ are Euler correctors, which satisfy elliptic equations in the Euler variables (x, Y). The terms $[u^i_p, v^i_p]$ are Prandtl correctors, which satisfy parabolic equations in Prandtl variables, (x, y). x behaves as a time-like variable whereas y behaves as a space-like variable.

Let us also introduce the following notation:

$$\begin{aligned} \begin{aligned} \bar{u}^i_p := u^i_p - u^i_p|_{y = 0}, \bar{v}^i_p := v^i_p - v^i_p|_{y = 0}, \bar{v}^i_e := v^i_e - v^i_e|_{Y = 0}. \end{aligned} \end{aligned}$$

(15)

The profile $\bar{u}^0_p, \bar{v}^0_p$ from (15) is classically known as the “boundary layer”; one sees from (14) that it is the leading order approximation to the Navier-Stokes velocity, $U^\epsilon$. We sometimes use the notation $[\bar{u}, \bar{v}] = [\bar{u}^0_p, \bar{v}^0_p]$ due to the distinguished nature of the leading order boundary layer. Indeed, the quantity $[\bar{u}^0_p, \bar{v}^0_p]$ will solve the Prandtl Eq. (1). As a matter of notation, we adopt the convention that for a given function, f(x, y), the quantity $\bar{f} := f - f(x,0)$.

The final layer,

$$\begin{aligned} \left[ u^{(\epsilon )}, v^{(\epsilon )}, P^{(\epsilon )}\right] = \left[ \varvec{u}^{(\epsilon )}, P^{(\epsilon )}\right] . \end{aligned}$$

are called the “remainders” and importantly, they depend on $\epsilon$.

An important first step in understanding the asymptotic behavior of $(U^\epsilon , V^\epsilon , P^\epsilon )$ is to understand the approximate terms in the expansion (14), that is, all the terms aside from the remainders. One outcome of the present paper is that our analysis to establish Theorems 2, 3 will also help us to construct each term in the approximate solution and prove regularity estimates on these terms. Informally, our result below in this direction, Theorem 4, says given reasonably well-behaved boundary data on the sides $x = 0, L$, each of the terms in $[u_s, v_s, P_s]$ (see (14)) can be constructed and are sufficiently regular. In order to state such a result precisely, we must introduce the equations satisfied by each term in $[u_s, v_s, P_s]$.

We now list the equations to be satisfied by the sub-leading order terms from (14). For $1 \le i \le n$, the i’th Euler layer satisfies the following system:

$$\begin{aligned} \left. \begin{aligned}&u^0_e \partial _x u^i_e + \partial _Y u^0_e v^i_e + \partial _x P^i_e =: f^i_{E,1}, \\&u^0_e \partial _x v^i_e + \partial _Y P^i_e =: f^i_{E,2}, \\&\partial _x u^i_e + \partial _Y v^i_e = 0, \\&v^i_e|_{Y = 0} = - v^0_p|_{y = 0}, v^i_e|_{x = 0, L} = V_{E, \{0, L\}}^i u^i_e|_{x = 0} = U^i_{E}. \end{aligned} \right\} \end{aligned}$$

(16)

In this case, since Eq. (16) is elliptic in (x, Y) we provide boundary data on both sides, $x = 0, L$. The given data for this problem is therefore the three functions $U^i_E(Y), V^i_{E,0}(Y), V^i_{E,L}(Y)$. The forcing terms $f^i_{E,1}, f^i_{E,2}$ are specifically given in Definition 5, and should be regarded as given for the purposes of stating the result.

For $1 \le i \le n-1$, the i’th Prandtl layer satisfies

$$\begin{aligned} \left. \begin{aligned}&\bar{u} \partial _x u^i_p + u^i_p \partial _x \bar{u} + \partial _y \bar{u} [v^i_p - v^i_p|_{y = 0}] + \bar{v} \partial _y u^i_p + \partial _x P^i_p - \partial _{yy} u^i_p := f^{(i)}, \\&\partial _x u^i_p + \partial _y v^i_p = 0, \partial _y P^i_p = 0\\&u^i_p|_{y = 0} = -u^i_e|_{y = 0}, [u^i_p, v^i_p]_{y \rightarrow \infty } = 0, v^i_p|_{x = 0} = V^i_P. \end{aligned} \right\} \end{aligned}$$

(17)

In this case, Eq. (17) is parabolic, with x controlling the evolution. As a result, the given data is the function $V^i_P(y)$. The forcing term, $f^{(i)}$, is defined in Definition 5. It should be regarded as given for the present discussion. For $i = n$, the difference is that $v^n_p$ will be defined so as to satisfy $v^n_p|_{y = 0} = 0$, unlike $v^i_p$ in (17), and subsequently cut-off as $y \uparrow \infty$. The key point is that system (17) is of the form considered in (8) and so we can apply Theorem 3.

We summarize now the given data in order to construct $[u_s; v_s]$. The parameters $n,N_0$ as in the expansions of equation (14) can be considered part of the given data. Next, the outer Euler shear flow $u^0_e(Y)$, initial datum for the leading order Prandtl layer, $\bar{U}^0_P(y)$, initial datum for the intermediate Prandtl layers, $\bar{V}^i_P(y)$, and data on the sides for the Euler profiles: $V^i_{E,0}(Y); V^i_{E,L}(Y); U^i_{E,0}(Y)"$ are all given functions.

Theorem 4

(Construction of Approximate Navier–Stokes Solution). Fix any $N_0 > 1$ and any $n \ge 2$. Assume the shear flow $u^0_e(Y) \in C^\infty$, whose derivatives decay rapidly, and which is bounded above and below: $c_0 \le u^0_e \le C_0$ for some universal constants $0< c_0< C_0 < \infty$. Assume (3) regarding $\bar{U}^0_P$, and the conditions

$$\begin{aligned}&\bar{V}^i_{Pyyy}(0) = \partial _x g^{(i)}|_{x = 0, y = 0}, \end{aligned}$$

(18)

$$\begin{aligned}&\bar{V}^i_P|'''(0) = \partial _{xy}g^{(i)}|_{x = 0, y =0}, \end{aligned}$$

(19)

$$\begin{aligned}&\bar{U}^0_{P y}(0) U^{i}_{E,0}(0) - \int _0^\infty \bar{U}^0_{P} e^{-\int _1^y \bar{v}^0_{p}} \left\{ f^{(i)}(y) - r^{(i)}(y) \right\} \,\mathrm {d}y = 0, \end{aligned}$$

(20)

where $r^{(i)}(y) := \bar{V}^i_P \bar{U}^0_{P y} - \bar{U}^0_{P} \bar{V}^i_{Py}$, and $g^{(i)} = g^{(i)}(f^{(i)}, u^i_e|_{Y = 0})$ is an explicit quantity depending on the forcing $f^{(i)}$ and the boundary data $u^i_e|_{Y = 0}$, and is defined in (32). We assume on the data standard elliptic compatibility conditions at the corners (0, 0) and (L, 0). Define the weight $w_N(Y)$ to be either $\langle Y \rangle ^N$ or $e^{NY}$.

$$\begin{aligned}&\arrowvert \partial ^k_YV^i_{E,0}\mid +\arrowvert \partial ^k_YV^i_{E,L}\mid +\arrowvert \partial ^k_YU^i_{E,0}\mid \lesssim w^{-1}_N \,\, \mathrm{for}\,\, 0 \le k \le K \nonumber \\&\,\, \mathrm{and} \,\, 1\le i \le n, \\&\arrowvert \partial ^k_y \bar{V}^i_P\mid +\arrowvert \partial ^k_y(\bar{U}^i_p-\bar{U}^i_p(\infty ))\mid \lesssim e^{-Ny} \,\,\mathrm{for} \,\, 0 \le k \le K \nonumber \\&\,\, \mathrm{and} \,\, 1\le i \le n. \end{aligned}$$

(21)

Then, for $L<< 1$, all profiles in $[u_s, v_s]$ exist and are smooth on $\varOmega$. The following estimates hold:

$$\begin{aligned} \begin{aligned}&\bar{u}^0_p \ge 0, \bar{u}^0_{py}|_{y = 0} > 0, \bar{u}^0_{pyy}|_{y = 0} = \bar{u}^0_{pyyy}|_{y = 0} = 0, \\&\Vert \nabla ^k \{ u^0_p, v^0_p \} e^{\frac{N}{2}y} \Vert _{\infty } \lesssim 1 \text { for } 0 \le k \le \frac{K}{2}, \\&\Vert u^i_p\Vert _{\infty } + \Vert \partial _x^k v^i_p \Vert _{\infty } + \Vert \nabla ^k u^i_p e^{\frac{N}{2}y} \Vert _{\infty } \lesssim \text { for } 0 \le k \le \frac{K}{2} \text { and } 1 \le i \le n, \\&\Vert \nabla ^k \{ u^i_e, v^i_e \} w_{\frac{N}{4}} \Vert _{\infty } \lesssim 1 \text { for } 0 \le k \le \frac{K}{2} \text { and for } 1 \le i \le n.\\ \end{aligned} \end{aligned}$$

(22)

In addition the following estimate on the remainder forcing holds:

$$\begin{aligned} \Vert F_R w_{\frac{N}{4}}\Vert _{H^1} \lesssim \varepsilon ^\frac{{n-1-2N_0}}{2}, \end{aligned}$$

(23)

where $F_R$ is the quantity defined in (126).

These constructions and regularity theory of the approximate solution to Navier–Stokes are of intrinsic interest. We also remark that another motivation for establishing our result is that these leading order constructions are important from the point of view of applications to the validity theory. Indeed, the estimates that we establish in this paper are in use in the works [25, 26].

1.3 Notations

We adopt several notation conventions throughout the paper. First, for functions f(x, y), we define the norm $\Vert f \Vert := \Vert f \Vert _{L^2} = (\int _0^L \int _0^\infty f(x, y)^2 \,\mathrm {d}y \,\mathrm {d}x)^{\frac{1}{2}}$. Next, we define the antiderivative via $I_x[f]:= \int _0^x f(x',y) \,\mathrm {d}x'$. We define (f, g) := $\int fgdy$ to be the inner product in the y variable.

1.4 Main ideas

The key ingredient in the analysis of (36) is a quotient estimate for $q_x$. We illustrate the main estimates for the simplified equation:

$$\begin{aligned} - \partial _{xy} \{ \bar{u}^2 q_y \} + v_{yyyy} = 0. \end{aligned}$$

(24)

By taking $\partial _x$ of the main part of (36), taking the inner-product with $q_x$ (Lemma 10), and rearranging the main contributions, we obtain (upon omitting at each line below easy to estimate terms)

$$\begin{aligned}&- \left( \partial _{xy} \left\{ \bar{u}^2 q_{xy} \right\} , q_x\right) + \left( v_{xyyyy}, q_x\right) \nonumber \\&\quad \sim \frac{\partial _x}{2} \int \bar{u}^2 q_{xy}^2 + \left( v_{xyy}, q_{xy}\right) _{y = 0} + \left( v_{xyy}, q_{xyy}\right) \nonumber \\&\quad \sim \frac{\partial _x}{2} \int \bar{u}^2 q_{xy}^2 + (\bar{u} q_{xyy} + 2 \bar{u}_y q_{xy} + \bar{u}_{yy}q, q_{xy})_{y = 0} \nonumber \\&\qquad + \left( \bar{u} q_{xyy} + 2 \bar{u}_y q_{xy} + 2 \bar{u}_{xy} q_y, q_{xyy}\right) \nonumber \\&\quad \sim \frac{\partial _x}{2} \int \bar{u}^2 q_{xy}^2 + \int \bar{u} q_{xyy}^2 + \left( 2 \bar{u}_y q_{xy}, q_{xy}\right) _{y = 0} - \left( \bar{u}_y q_{xy}, q_{xy}\right) _{y = 0} \nonumber \\&\qquad - \left( \bar{u}_{xy} q_y, q_{xyy}\right) \nonumber \\&\quad \sim \frac{\partial _x}{2} \int \bar{u}^2 q_{xy}^2 + \int \bar{u} q_{xyy}^2 + \left( \bar{u}_y q_{xy}, q_{xy}\right) _{y = 0} + \left( \bar{u}_{xyy}q_y, q_{xy}\right) \nonumber \\&\qquad + \left( \bar{u}_{xy} q_{yy}, q_{xy}\right) - \left( \bar{u}_{xy} q_y, q_{xy}\right) _{y = 0} \end{aligned}$$

(25)

The key positive boundary contribution $(\bar{u}_y q_{xy}, q_{xy})_{y = 0}$ holds naturally for Prandtl solutions as $\bar{u}_y|_{y = 0} > 0$.

Next, for the term $(\bar{u}_{xy} q_{yy}, q_{xy})$ the key point is that $\bar{u}_{xy}|_{y = 0} \ne 0$. It is thus important to avoid the degeneracy at $y = 0$ in the term $\Vert \bar{u} q_{xy} \Vert$ by invoking a higher order norm term:

$$\begin{aligned} \Vert q_{yy} \Vert \lesssim \Vert v_{yyy} \Vert \text { for } y<< 1. \end{aligned}$$

(26)

In order to close, we must further estimate $v_{yyyy}$ from the Eq. (25), which gives (Lemma 11

$$\begin{aligned} \Vert v_{yyyy} \Vert \le \Vert \partial _{xy} \{ \bar{u}^2 q_y \} \Vert . \end{aligned}$$

(27)

Finally, we establish decay in y in Lemma 12.

We introduce

$$\begin{aligned} \bar{u}^\theta := \bar{u} + \theta > 0 \end{aligned}$$

(28)

in our construction of approximate solutions and regard LDP (36) as an initial value problem for given $v|_{x = 0}$ or $q|_{x = 0}$. We finally recover $u^0 = u|_{x = 0}$ and solve the original system, (8), via a necessary integrability condition (20).

The first three positive quantities in (26) form our basic energy norm, $\Vert \cdot \Vert _{\mathcal {E}}$ in (40). The remainder of our analysis is centered around propagating control over a slightly stronger norm, X, and upgrading to higher $\partial _x$ derivatives. We note here that similar quotient estimates to those developed in this paper have played a role in establishing the validity of the Prandtl layer expansion, see for instance [25, 26, 35].

2 Regularity for the Prandtl and Linearized Prandtl Equations

Our objective in this section is to establish Theorems 2 and 3, both of which will follow from the study of the system (8).

2.1 Linearized derivative Prandtl equation (LDP)

We homogenize the system (8) to remove the boundary condition $u|_{y = 0} = u_0(x)$ via:

$$\begin{aligned} u = u_p - u_0(x) \psi (y), v = v_p + u_{0x}(x) I_\psi (y), I_\psi (y) := \int _y^\infty \psi (\theta ) \,\mathrm {d}\theta . \end{aligned}$$

(29)

Here, we select $\psi$ to be a $C^\infty$ function satisfying the following:

$$\begin{aligned} \psi (0) = 1, \int _0^\infty \psi = 0, \psi \text { decays as } y \uparrow \infty , \psi ^{(k)}(0) = 0 \text { for } k \ge 1. \end{aligned}$$

(30)

According to (8), the homogenized unknowns [u, v] satisfy the system:

$$\begin{aligned} \begin{aligned}&\bar{u} \partial _x u + u \partial _x \bar{u} + \bar{v} \partial _y u + v \partial _y \bar{u} - \partial _{yy} u = : g_1, \\&\quad u_x + v_y = 0, \\&\quad [u, v]|_{y = 0} = 0, \lim _{y \rightarrow \infty } u(x, y) = 0, v|_{x = 0} = \bar{V}_0(y), \end{aligned} \end{aligned}$$

(31)

where the following quantities have been introduced:

$$\begin{aligned} \begin{aligned}&g_1 := f + G, \\&\quad G := - \bar{u} \psi u_{0x} - \bar{u}_{x} \psi u_0 - \bar{v} \psi ' u_0 - \bar{u}_{y} u_{0x} I_\psi + \psi '' u_0, \\&\quad \bar{V}_0(y) := V_0 -u_0(0) I_\psi (y). \end{aligned} \end{aligned}$$

(32)

By applying $\partial _y$, we obtain the system:

$$\begin{aligned} - \bar{u} v_{yy} + v \bar{u}_{yy} - u \bar{v}_{yy} + \bar{v} u_{yy} - u_{yyy} = \partial _y g_1. \end{aligned}$$

(33)

We rewrite the Rayleigh operator as

$$\begin{aligned} - \bar{u} v_{yy} + v \bar{u}_{yy} = - \partial _y \{ \bar{u}^2 q_y \}, q := \frac{v}{\bar{u}}. \end{aligned}$$

(34)

By further taking $\partial _x$ we derive the following linear Derivative Prandtl Equation (LDP) for the quotient, q, which is the main focus of our paper:

$$\begin{aligned} \begin{aligned}&- \partial _{xy} \{ \bar{u}^2 q_y \} + \partial _y^4 v + \kappa \varLambda + \kappa U = \partial _{xy} g_1, \\&\quad q|_{y = 0} = 0, q|_{x = 0} = \frac{1}{\bar{u}}|_{x= 0}(y) \bar{V}_0(y) := f_0(y). \end{aligned} \end{aligned}$$

(35)

We have introduced the artificial parameter $\kappa$ above. The system of interest is $\kappa = 1$. The reason for including this artificial parameter is, in Proposition 2 below, we also treat the $\kappa = 0$ case of nonlinear Prandtl, (1). We have defined the following operators:

$$\begin{aligned}&\varLambda (v) := \bar{v}_{xyy}I_x\left[ v_y\right] + \bar{v}_{yy} v_y - \bar{v}_{x}I_x\left[ v_{yyy}\right] - \bar{v} v_{yyy}, \end{aligned}$$

(36)

$$\begin{aligned}&U(u^0) := - \bar{v}_{xyy}u^0 + \bar{v}_{x}u^0_{yy}. \end{aligned}$$

(37)

We now state our basic proposition for the homogenized system. To do so, we need to specify norms in which we measure the quantity, q, as well as the forcing, f. Let $\chi$ denote the following cut-off function:

$$\begin{aligned} \chi (y) = {\left\{ \begin{array}{ll} 1 \text { on } 0 \le y \le 1 \\ 0 \text { on } y \ge 2\end{array}\right. } \chi '(y) \le 0 \text { for all }y. \end{aligned}$$

(38)

Fix $w \,=\, e^{Ny}$ for the N from Theorem 2. Denote by $q^{(k)} \,:=\, \partial _x^k q$. We will now define several norms:

$$\begin{aligned} \begin{aligned} \Vert q \Vert _X \,:= \,&\sup _{0 \le x \le L} \left( \Vert \bar{u} q_{xy} \Vert _{L^2_y} + \Vert q_{yyy} w (1 - \chi ) \Vert _{L^2_y} \right) + \Vert \sqrt{\bar{u}} q_{xyy} w \Vert + \Vert v_{yyyy} w \Vert \\&+ \Vert \sqrt{\bar{u}_y} q_{xy} \Vert _{y = 0}, \\ \Vert q \Vert _{\mathcal {E}} :=&\sup _{0 \le x \le L} \Vert \bar{u} q_{xy} \Vert _{L^2_y} + \Vert \sqrt{\bar{u}} q_{xyy} \Vert + \Vert \sqrt{\bar{u}_y} q_{xy} \Vert _{y = 0}\\ \Vert q \Vert _{\mathcal {H}} :=&\sup _{0 \le x \le L} \Vert q_{yyy} w\{1 - \chi \}\Vert _{L^2_y} + \Vert v_{yyyy} w \{1 - \chi \} \Vert + \Vert q_{xyy} w \{1 - \chi \} \Vert \\ \Vert q \Vert _{X_k}\, :=\,&\Vert q^{(k)} \Vert _X, \Vert q \Vert _{\mathcal {E}_k}\, :=\, \Vert q^{(k)} \Vert _{\mathcal {E}}, \Vert q \Vert _{\mathcal {H}_k} \,:=\, \Vert q^{(k)} \Vert _{\mathcal {H}}, \\ \Vert q \Vert _{X_{\langle k \rangle }} =&\sum _{i = 0}^k \Vert q \Vert _{X_{i}}. \end{aligned} \end{aligned}$$

(39)

We also define the following norm in which we control the forcing term:

$$\begin{aligned} \Vert f \Vert _{\tilde{X}_{\langle k \rangle }} = \Vert e^{Ny} \partial _x^{k} \partial _{xy}f \Vert + \Vert \langle y \rangle \partial _x \partial _x^k \partial _{xy} f \Vert . \end{aligned}$$

(40)

Proposition 1

Let the initial data, $V_0(y)$, the forcing, f, and the boundary data, $u_0(x)$, satisfy the compatibility conditions (48) and (49). Assume also the integral condition, (53). Then there exists a unique solution to (8) satisfying the following estimate:

$$\begin{aligned} \Vert q \Vert _{X_{\langle k \rangle }} \lesssim \Vert f \Vert _{\tilde{X}_{\langle k \rangle }} + C\left( V_0, u_0\right) . \end{aligned}$$

Theorem 2 is a corollary of Proposition 1 upon differentiating the Prandtl Eq. (1), in x. Theorem 3 is also an immediate corollary of Proposition 1.

2.2 Compatibility conditions and reconstruction of $u^0$

It is possible and natural to solve (8) for a given initial condition $u^0 = u|_{x = 0}$, instead of a given $v^0$. In such a formulation, no integrability condition, (53), is needed, but $v^0$ must be determined from solving

$$\begin{aligned} -\bar{u} v_y + \bar{v} u^0_y + \bar{u}_y v + \bar{u}_x u^0 - u^0_{yy} = g_1 \text { at } x= 0. \end{aligned}$$

(41)

In this case, the compatibility conditions are $\partial _x^k v^0(0) = \partial _x^k v^0_y(0) = 0$, where

$$\begin{aligned} \partial _x^k v^0 :=&\partial _x^k \left\{ - \bar{u} \int _0^y \frac{1}{\bar{u}} \left\{ \bar{v} u^0_y - u^0 \bar{u}_x + u^0_{yy} + g_1 \right\} \right\} , \end{aligned}$$

(42)

$$\begin{aligned} \partial _x^k v^0_y :=&\partial _x^k \left\{ - \bar{u} \int _0^y \frac{1}{\bar{u}} \left\{ \bar{v} u^0_y - u^0 \bar{u}_x + u^0_{yy} + g_1 \right\} \right\} _y \end{aligned}$$

(43)

However, in our v-formulation, it is interesting to reconstruct $u^0$ via a given initial condition $V_0$. The purpose of this subsection is to reconstruct $u^0$ from $V_0$ under a necessary integrability condition on $V_0$, (53).

Our aim now is to derive compatibility conditions for the initial data. By computing $\partial _x$ of (32) and evaluating at $y = 0$, we obtain the condition:

$$\begin{aligned} v_{yyy}|_{y = 0}(x) = \partial _x g_1|_{y = 0}(x), \qquad x \in (0,L). \end{aligned}$$

The first order compatibility condition that arises from this entails matching now the initial datum evaluated at $y = 0$ with the boundary datum evaluated at $x = 0$ in the standard manner for initial-boundary value problems:

$$\begin{aligned} V_0'''(0) = \bar{V}_0'''(0) = \partial _x g_1|_{y = 0}(0). \end{aligned}$$

(44)

The first equality above follows from the property of $\psi '''(0) = 0$, whereas the second equality above is the natural outcome of the aforementioned matching process.

We also require the second-order compatibility which can be obtained as follows. Taking $\partial _x$ of (34):

$$\begin{aligned} - \partial _x \{ - \bar{u} v_{yy}+ v \bar{u}_{yy} + \bar{v} u_{yy} - u \bar{v}_{yy} \} + \partial _y^4 v = \partial _{xy}g_1. \end{aligned}$$

Evaluating at $y = 0$ gives the identity:

$$\begin{aligned} \partial _y^4 v|_{y = 0}(x) = \partial _{xy} g_1|_{y = 0}(x) \text { for } x \in (0,L). \end{aligned}$$

This then gives our second order compatibility condition of

$$\begin{aligned} \partial _y^4 V_0(0) = \partial _y^4 \bar{V}_0(0) = \partial _{xy}g_1|_{y = 0}(0). \end{aligned}$$

(45)

This derivation motivates our definition of compatibility conditions, which we state formally now.

Definition 1

(Compatibility at Corners). The first and second order compatibility conditions are given by

$$\begin{aligned} \partial _y^3 V_0(0) = \partial _x g_1|_{y = 0}(0), \qquad \partial _y^4 V_0(0) = \partial _{xy}g_1|_{y = 0}(0). \end{aligned}$$

(46)

Higher order compatibility conditions can be derived in the same manner, and are stated implicitly as follows

$$\begin{aligned} \partial _y^3 \left( \partial _x^k v|_{x = 0}\right) (0) =&\partial _x^{k+1} g_1|_{y = 0}(0) \end{aligned}$$

(47)

$$\begin{aligned} \partial _y^4 \left( \partial _x^k v|_{x = 0}\right) (0) =&\partial _x^{k+1} \partial _y g_1|_{y =0}(0). \end{aligned}$$

(48)

Starting from the q formulation in (36), we will further distribute on the Rayleigh term:

$$\begin{aligned} - \partial _y \left\{ \bar{u}^2 q_{xy} \right\} - \partial _y \left\{ 2 \bar{u} \bar{u}_{x} q_y \right\} + \partial _y \partial _x \left\{ \bar{v} u_y - u\bar{v}_{y} \right\} + \partial _y \partial _y^3 v = \partial _y \partial _x g_1. \end{aligned}$$

We now compute at $\{x = 0\}$:

$$\begin{aligned} \bar{u}^2 q_{xy} =&- \int _y^\infty \partial _y\left\{ \bar{u}^2 q_{xy} \right\} \,\mathrm {d}y' \nonumber \\ =&\int _y^\infty \partial _y \left\{ \partial _x g_1 - \partial _y^3 v + 2 \bar{u} \bar{u}_{x} q_y - \bar{v}_{x} u^0_y + \bar{v} v_{yy} - v_y \bar{v}_{y} + u^0 \bar{v}_{xy} \right\} \,\mathrm {d}y' \nonumber \\ =&- \left\{ \partial _x g_1 - \partial _y^3 v + 2 \bar{u} \bar{u}_{x} q_y - \bar{v}_{x}u^0_y + \bar{v} v_{yy} - v_y \bar{v}_{y} + u^0 \bar{v}_{xy} \right\} . \end{aligned}$$

(49)

It is clear that all quantities are vanishing at $y = 0$. We thus have that $\bar{u} q_{xy}|_{x = 0} \in L^2(\mathbb {R}_+)$. A computation of $\partial _y$ shows:

$$\begin{aligned}&\partial _{xy} g_1+ \partial _y^4 v + \partial _y \left\{ 2 \bar{u}_{x}\bar{u} q_y - \bar{v}_{x} u^0_y + \bar{v} v_{yy} - v_y \bar{v}_{y} + u^0 \bar{v}_{yy} \right\} |_{y = 0} \\&\quad = \partial _{xy}g_1(0,0) + \partial _y^4v|_{x = 0}(0) = 0. \end{aligned}$$

Thus $q_{xy}$ itself is in $L^2$. Using this we may easily bootstrap to higher order in $\partial _y$ compatibility conditions for $v^0$ which we refrain from writing. These conditions in turn assure that:

Lemma 1

Assume the compatibility conditions on $V_0$ given in Definition 1. Assume exponential decay on $\partial _y^k V_0$ for $k \ge 1$. Then there exist functions $f_k(y) \in L^2_w(\mathbb {R}_+) \cap C^\infty (\mathbb {R}_+)$ for exponential weight w such that

$$\begin{aligned} \partial _x^k q_y|_{x = 0} = f_k(y) \in L^2_w\left( \mathbb {R}_+\right) \text { for } k \ge 1. \end{aligned}$$

(50)

Moreover, $f_k$ depend only on the given profile $V_0$ and the forcing term $g_1$.

Our task now is to establish criteria on the initial data, $V_0$ so that $u^0(y) := u_p|_{x = 0}$ can be bounded. We evaluate the velocity Eq. (8) at $x = 0$ to obtain the equation:

$$\begin{aligned} \mathcal {L}_{\parallel } u^0 = f(y) - r(y), u^0(0) = u_0(0), \qquad r(y) := V_0 \bar{u}_{y} - \bar{u} V_0', \end{aligned}$$

(51)

where we have defined the operators

$$\begin{aligned} \mathcal {L}_{\parallel } u^0 := - u^0_{yy} + \bar{v} u^0_{y} - u^0 \bar{v}_{y}, \qquad \mathcal {L}_{\parallel , y} u^0 := - u^0_{yyy} + \bar{v} u^0_{yy} - u^0 \bar{v}_{yy}, \end{aligned}$$

To invert Eq. (52) for $u^0(y)$, we assume the following condition on the datum $V_0(y)$:

$$\begin{aligned} -\bar{u}_{y}(0, 0) u_0(0) - \int _0^\infty \bar{u} e^{-\int _1^y \bar{v}} \left\{ f(y) - r(y) \right\} \,\mathrm {d}y = 0. \end{aligned}$$

This, then, motivates the following definition upon replacing notationally $\bar{u}(0, y) = \bar{U}_0(y)$:

Definition 2

(Compatibility Integral Condition). We say the datum, $V_0(y)$ satisfies the “compatibility integral condition” if

$$\begin{aligned}&\int _0^\infty \bar{U}_0 e^{- \int _1^y \bar{v}(0, z) \,\mathrm {d}z} (V_0 \bar{U}_0' - \bar{U}_0 V_0') \,\mathrm {d}y \nonumber \\&\quad = \bar{U}_0'(0) u_0(0) + \int _0^\infty \bar{U}_0 e^{- \int _1^y \bar{v}(0, z) \,\mathrm {d}z} f(y) \,\mathrm {d}y. \end{aligned}$$

(52)

Lemma 2

Elements of the three dimensional kernel of $\mathcal {L}_{\parallel , y}$ can be written as the following linear combination: $c_1 \bar{u} + c_2 \tilde{u}_s + c u_R$, where $c_1, c_2, c \in \mathbb {R}$. Here:

$$\begin{aligned}&\tilde{u}_s := \bar{u} \int _1^y \frac{|\bar{u}(1)|^2}{|\bar{u}|^2} \exp \left[ \int _1^z \bar{v} \,\mathrm {d}w \right] \,\mathrm {d}z, \\&u_R := \tilde{u}_s \int _0^y \bar{u} \exp \left[ -\int _1^z \bar{v} \right] -\bar{u} \int _0^y \tilde{u}_s \exp \left[ - \int _1^z \bar{v} \right] . \end{aligned}$$

Proof

Note $\mathcal {L}_{\parallel , y} u = 0$ if and only if $\mathcal {L}_{\parallel } u = c$ for a constant c. One solution to the homogeneous equation, $\mathcal {L}_{\parallel } u = 0$, is $\bar{u}$. By supposing the second spanning solution is of the form $\tilde{u}_s := \bar{u} a(y)$, we may derive the equation: $a''(y) = \Big [ \bar{v} - 2\frac{\bar{u}_{y}}{\bar{u}} \Big ] a'(y)$. Solving this equation gives one solution:

$$\begin{aligned} a'(y) = \frac{|\bar{u}(1)|^2}{|\bar{u}|^2} \exp \left[ \int _1^y \bar{v} \right] , a(y) = \int _1^y \frac{|\bar{u}(1)|^2}{|\bar{u}|^2} \exp \left[ \int _1^z \bar{v} \,\mathrm {d}w\right] \,\mathrm {d}z. \end{aligned}$$

(53)

$\square$

We shall need asymptotic information about $\tilde{u}_s$:

Lemma 3

As defined in Lemma 2, $\tilde{u}_s$ satisfies the following asymptotics:

$$\begin{aligned} \begin{aligned}&\tilde{u}_s|_{y = 0} \sim -1 \text { and }\tilde{u}_{sy} \sim 1 \text { as } y \downarrow 0, \\&\quad \tilde{u}_{sy}, \tilde{u}_{sy}, \tilde{u}_s \sim \exp [\bar{v}(\infty ) y] \text { as } y \uparrow \infty . \end{aligned} \end{aligned}$$

(54)

Proof

For convenience, denote $g(y) = \exp ( \int _1^y \bar{v} )$. By rewriting $\bar{v} = \bar{v}(\infty ) + (\bar{v} - \bar{v}(\infty ))$, and using that the latter difference decays rapidly, we obtain the basic asymptotics $g \sim \exp ( \bar{v}(\infty ) y)$ as $y \uparrow \infty$. An expansion of a, given in (54), near $y = 0$ gives $a(y) \approx \int _1^y \frac{1}{z^2} \,\mathrm {d}z \sim -\frac{1}{z}|_{1}^y = 1-\frac{1}{y}$. Thus: $\tilde{u}_s|_{y = 0} \sim \bar{u} (1 - \frac{1}{y} ) \sim -1$. At $y = \infty$, we have the asymptotics:

$$\begin{aligned} \tilde{u}_s = \bar{u} \int _1^y \frac{|\bar{u}(1)|^2}{|\bar{u}|^2} g(z) \sim \int _1^y \exp \left[ \bar{v}(\infty ) z\right] \,\mathrm {d}z \sim \exp \left[ \bar{v}(\infty ) y\right] . \end{aligned}$$

We now differentiate to obtain

$$\begin{aligned} \tilde{u}_{sy} = \bar{u}_{y}a + \bar{u} a'(y) = \bar{u}_{y}a(y) + \frac{|\bar{u} (1)|^2}{\bar{u}} g(y) \sim \exp [\bar{v}(\infty ) y]. \end{aligned}$$

To evaluate $\tilde{u}_{sy}$ at $y = 0$, we need more precision. Expansions give:

$$\begin{aligned}&\bar{u} a'(y) = \frac{|\bar{u}(1)|^2}{\bar{u}} g(y) \sim \frac{|\bar{u}(1)|^2}{\bar{u}_{y}(0)y} g(y) \text { for }y \sim 0, \text { and } \\&\quad \bar{u}_{ y}a(y) \sim \bar{u}_{y}(0) |\bar{u}(1)|^2 \int _1^y \frac{1}{|\bar{u}|^2} g(z) \sim \frac{|\bar{u}(1)|^2}{\bar{u}_{y}(0)} \int _1^y \frac{g(z)}{z^2} \,\mathrm {d}z. \end{aligned}$$

We have used the fact that $\frac{1}{|\bar{u}|^2}$ does not contribute a factor of $\frac{1}{z}$ following the singularity of $\frac{1}{z^2}$. Indeed, Taylor expanding, using that $\bar{u}^0_{ yy}(0) = \bar{u}_{yyy}(0) = 0$ (see the first identity in (23)), and the elementary identity for any $a, b \in \mathbb {R}$, $\frac{1}{a-b} - \frac{1}{a} = \frac{b}{a(a-b)}$, one obtains:

$$\begin{aligned} \frac{1}{|\bar{u}(z)|^2} = \frac{1}{\bar{u}_{y}(0)^2 z^2} + \mathcal {O}(z). \end{aligned}$$

It remains to show $\int _1^y \frac{g(z)}{z^2} \,\mathrm {d}z \sim -\frac{g(y)}{y}$. We decompose the integral into region $[1, y_*]$ and $[y_*, y]$ for $0 < y \le z \le y_*$. The $[1, y_*]$ integral contributes an $\mathcal {O}(1)$ constant. In the $[y_*, y]$ region, the Taylor expansion is valid:

$$\begin{aligned} \int _{y_*}^y \frac{g(z)}{z^2} \,\mathrm {d}z \sim \int _{y_*}^y \frac{g(y)}{z^2} \,\mathrm {d}z + g'(y)\int _{y_*}^y \frac{z-y}{z^2} \,\mathrm {d}z \sim g(y) \left( \frac{1}{y_*} - \frac{1}{y} \right) + g'(y) \phi (y), \end{aligned}$$

where $| \phi (y)| \lesssim |\log y|$. We now use that $\bar{v}(0) = \bar{v}_{p}(0) = 0$ and $g'(y) = \bar{v}(y) g(y)$ to show that $g'(y) \sim y^2$. Thus, $g'(y) \phi (y)$ vanishes as $y \rightarrow 0$. We thus have verified that $I(y) \sim -\frac{g(y)}{y}$.

We now compute two derivatives:

$$\begin{aligned} \tilde{u}_{syy}&= \bar{u}_{yy}a + 2\bar{u}_{y}a'(y) + \bar{u} a''(y) \\&\sim a''(y) \sim \partial _{y} \left\{ \frac{1}{|\bar{u}|^2} \exp \left[ \bar{v}(\infty ) y\right] \right\} \sim \exp \left[ \bar{v}(\infty ) y\right] \text { as } y \uparrow \infty . \end{aligned}$$

$\square$

Lemma 4

Assume the integral condition, (53) is satisfied by the initial data $v_0$. Then the solution $u^0$ to (52) exists and satisfies:

$$\begin{aligned}&|\partial _y^k u^0 e^{My}|_\infty \le C_{K,M} (V^0, g_1) \text { for }k \ge 1, \end{aligned}$$

(55)

$$\begin{aligned}&u^0(0) = u_0(0) \text { and } \lim _{y \uparrow \infty } u^0(y) = 0. \end{aligned}$$

(56)

Proof

First, we compute the Wronskian of $\bar{u}^0_{p}$ and $\tilde{u}_s$:

$$\begin{aligned} W = \bar{u} \tilde{u}_{sy} - \tilde{u}_s \bar{u}_{y} = \bar{u}(1)^2 \exp \left[ \int _1^y \bar{v} \right] . \end{aligned}$$

Next, we express the solution to (52) in the following manner:

$$\begin{aligned} u^0 \,=\,&u_0(0) \frac{\tilde{u}_s}{\tilde{u}_s(0)} + c_1 \bar{u} - \frac{1}{|\bar{u}(1)|^2} \tilde{u}_s \int _0^y \bar{u} e^{- \int _1^z \bar{v}} \left( f(z) - r(z) \right) \,\mathrm {d}z \nonumber \\&+ \frac{1}{\bar{u}(1)^2} \bar{u} \int _0^y \tilde{u}_s e^{-\int _1^z \bar{v}} ( f(z) - r(z) ) \,\mathrm {d}z. \end{aligned}$$

(57)

We now compute $\tilde{u}_s(0) = - \frac{|\bar{u}(1)|^2}{\bar{u}_{y}(0)} e^{\int _1^0 \bar{v}}$. Using this, we now evaluate at $y = \infty$ and observe that the terms with a $\tilde{u}_s$ prefactor vanish according to the integral condition, (53).

$$\begin{aligned} -\frac{\bar{u}_{ y}(0)}{\bar{u}(1)^2} u_0(0) e^{-\int _1^0 \bar{v}} - \frac{1}{|\bar{u}(1)|^2} \int _0^\infty \bar{u} e^{- \int _1^y \bar{v}} (f - r(y)) \,\mathrm {d}y = 0. \end{aligned}$$

This proves that $u^0$ as defined in (58) is bounded as $y \uparrow \infty$. We next notice that the derivative of $\int _0^y \bar{u} e^{-\int _1^\infty \bar{v}} (f - r(z) )$ is the integrand itself, which decays fast enough to eliminate $\tilde{u}_s$ at $\infty$. Therefore we also see that $\partial _y^k u^0$ for $k \ge 1$ decays rapidly.

Finally, we need to ensure that $u^0 \rightarrow 0$ as $y \uparrow \infty$. It is clear that $\mathcal {L}_{\parallel , y} u^0 = 0$, and so we are free to modify $u^0$ by factors of $\bar{u}$. Thus we modify (58) by subtracting off a factor of $c \bar{u}$, for c appropriately selected so as to ensure $u^0(\infty ) = 0$. This concludes the proof. $\quad \square$

Summarizing the above,

Lemma 5

Assume smooth data, $V_0$, are prescribed that satisfies the compatibility conditions (48), (49), as well as higher order compatibility conditions. Assume also that $V_0$ satisfies the integral condition (53). Let $q = \frac{v}{\bar{u}}$ solve (36) and $u^0$ be constructed from v via (52). Then $[u = u^0 - \int _0^x v_y, v]$ solve (32). Further, let $[u_p, v_p]$ be reconstructed from [u, v] using (30). Then $[u_p, v_p]$ are solutions to (8). Moreover, the initial datum $u^0 = u|_{x = 0}$ satisfies the decay estimates (56).

2.3 Construction of approximate solution

First, we recall the definition $\bar{u}_\theta = \bar{u} + \theta$ from (29). We also define

$$\begin{aligned} q^{(\theta )} := \frac{v^{(\theta )}}{\bar{u}}. \end{aligned}$$

(58)

Definition 3

($\theta$-Approximant)

$$\begin{aligned}&- \partial _{xy}\left( \bar{u}_\theta ^2 q^{(\theta )}_y\right) + v^{(\theta )}_{yyyy} + \kappa \varLambda (v^{(\theta )}) + \kappa U(u^0) = F^{(\theta )}, \end{aligned}$$

(59)

$$\begin{aligned}&\quad q^{(\theta )}|_{y = 0} = 0, \qquad q^{(\theta )}|_{x = 0} = Q_\theta (y), \end{aligned}$$

(60)

where the forcing $F^{(\theta )} := \partial _{xy} g_1$. We also introduce the notation $V_\theta (y) = \bar{u} Q_\theta (y)$.

Definition 4

($\theta$-Dependent Compatibility Conditions). We now define the compatibility required on the initial conditions we take, $Q_{\theta }(y)$. We assume that $u^{0}(y)$ is given. We need to enforce standard parabolic compatibility conditions on $Q_\theta (y)$ at $y = 0$. These can be read off from the Eq. (60) in the usual manner. For instance, for the first order compatibility condition, we obtain

$$\begin{aligned} Q_{\theta , 1}(y) := q^{(\theta )}_x(0, y) = \int _y^\infty \left( \frac{\mathcal {G}_0}{(\bar{u}_\theta ^2)} + 2 \frac{\bar{u}_{\theta x}}{\bar{u}_\theta } Q_{\theta }' - \frac{V_{\theta }'''}{\bar{u}_\theta ^2}\right) \,\mathrm {d}y'. \end{aligned}$$

(61)

where

$$\begin{aligned} \mathcal {G}_0(y) := \left( F^{(\theta )} - \kappa U\left( u^0\right) - \kappa \varLambda \left( v^{(\theta )} \right) |_{x = 0}\right. \end{aligned}$$

(62)

We thus need to ensure that $Q_{\theta , 1}(0) = 0$, which is the “first order compatibility condition”. Higher order compatibility conditions are obtained in a similar manner.

To construct solutions to (60), we will introduce an artificial truncation at $y = N$ and then send $N \rightarrow \infty$. Moreover, will make $\bar{u}_\theta (y)$ constant for $y > \frac{N}{2}$. To do this, we define $\bar{u}_\theta ^{(N)} \in C^\infty$ satisfying the following properties

$$\begin{aligned} \bar{u}_{\theta }^{(N)}(x, y) = {\left\{ \begin{array}{ll} \bar{u}_\theta (x, y) \text { for } 0 \le y \le \frac{N}{4} \\ \bar{u}_\theta (x, \frac{N}{2}) \text { for } \frac{N}{2} \le y \le N \end{array}\right. }. \end{aligned}$$

(63)

We shall now consider the problem on $(x, y) \in (0, L) \times (0, N)$,

$$\begin{aligned} \begin{aligned}&- \partial _{y}\left( \left( \bar{u}_{\theta }^{(N)}\right) ^2 q^{(\theta , N)}_{xy}\right) + v^{(\theta , N)}_{yyyy}+ \kappa \varLambda (v^{(\theta , N)}) + \kappa U\left( u^0\right) = F^{(\theta )} \\&\quad q^{\left( \theta , N\right) }|_{y = 0, N} = 0, \qquad q^{(\theta , N)}_{yy}|_{y = 0, N} = 0, \qquad q^{(\theta , N)}|_{x = 0} = f_0(y). \end{aligned} \end{aligned}$$

(64)

We fix a Fourier sine basis adapted to the interval [0, N], by letting

$$\begin{aligned} e_i(y) := \sin ( i \frac{2\pi }{N} y ). \end{aligned}$$

(65)

For each $n < \infty$, we search for solutions, $q^{(n,N,\theta )} \in \text {span}\{e_i\}$ for $i = 1,\ldots ,n$, which satisfy, for $e_i, i = 1,\ldots ,n$,

$$\begin{aligned}&\left( \partial _x \left( \left( \bar{u}_\theta ^{(N)}\right) ^2 q^{\left( N, n,\theta \right) }_y \right) , e_i'\right) + \left( \partial _{yy} \left( \bar{u} q^{\left( N, n, \theta \right) } \right) , e_i''\right) + 2 \bar{u}_y q^{\left( N, n,\theta \right) }_y e_i'(0)\nonumber \\&\quad + \kappa \left( \bar{v}_{xyy}I_x\left[ v^{(N, n,\theta )}_y\right] , e_i\right) + \kappa \left( \bar{v}_{yy} v^{(N, n,\theta )}_y, e_i\right) + \kappa \left( I_x\left[ v^{\left( N, n,\theta \right) }_{yy}\right] , \bar{v}_{xy}e_i + \bar{v}_x e_i'\right) \nonumber \\&\quad + \kappa \left( v^{(N, n,\theta )}_{yy}, \bar{v}_y e_i + \bar{v} e_i'\right) + \kappa \left( U\left( u^0\right) , e_i\right) = \left( F^{(\theta )}, e_i\right) \end{aligned}$$

(66)

We now propose $q^{(N, n,\theta )} = \sum _{l = 1}^n b^{(n)}_l(x) e_l$. The first claim is that there exist coefficients, $b_i^{(n)}(x)$ such that $q^{(n,\theta )}$ satisfies (67). Inserting this expansion into (67), we obtain

$$\begin{aligned}&\left( \partial _x \left\{ \left( \bar{u}_\theta ^{(N)}\right) ^2 \sum _{l = 1}^n b^{(n)}_l(x) e_l' \right\} , e_i'\right) + \left( \partial _{yy} \{ \bar{u} \sum _{l = 1}^n b^{(n)}_l(x) e_l \}, e_i''\right) \nonumber \\&\quad + 2\left( \bar{u}_y \sum _{l = 1}^n b^{(n)}_l(x) e_l'(0), e_i'(0)\right) + \kappa \sum _{l = 1}^n I_x\left[ b^{(n)}_l(x)\right] \left( \bar{v}_{xyy} \partial _y \left( \bar{u} e_l\right) , e_i\right) \nonumber \\&\quad + b^{(n)}_l(x) \left( \bar{v}_{yy} \partial _y \left( \bar{u} e_l\right) , e_i\right) + \kappa ( I_x[ \partial _{yy} ( \bar{u} \sum _{l = 1}^n b^{(n)}_l(x) e_l(y)) ], \bar{v}_{xy} e_i + \bar{v}_x e_i' ) \nonumber \\&\quad + \kappa \left( \partial _{yy} \left( \bar{u} \sum _{l = 1}^n b^{(n)}_l(x) e_l \right) , \bar{v}_y e_i + \bar{v} e_i'\right) +\left( \kappa U(u^0), e_i\right) \nonumber \\&\quad - 2\left( \bar{u}_\theta ^{(N)} \partial _x \bar{u}_\theta ^{(N)} \sum _{l = 1}^n b^{(n)}_l(x) e_l' , e_i'\right) = \left( F^{(\theta )}, e_i\right) . \end{aligned}$$

(67)

We thus obtain an $n \times n$ system of ODE’s for $b_l^{(n)}$ $l = 1,\ldots ,n$, which can be solved using standard ODE theory. Indeed, for $i = 1,\ldots ,n$, we have the following system of ODEs:

$$\begin{aligned} \varGamma ^{\ell }_i \dot{b}^{(n)}_\ell (x) + \sum _{l = 1}^{n} A^l_i(x) b_l^{(n)}(x) + \sum _{l = 1}^n K^l_i(x) I_x\left[ b^{(n)}_l(x)\right] = F_i(x), \text { for } i = 1,\dots ,n \end{aligned}$$

(68)

where

$$\begin{aligned} A^l_i(x)\, :=\,&\left( \partial _{yy}\left( \bar{u} e_l(y)\right) , e_i''(y) \right) + 2 \bar{u}_y(x,0) e_l'(0) e_i'(0) + \kappa \left( \bar{v}_{yy} \partial _y \left( \bar{u} e_l\right) , e_i\right) \nonumber \\&+ \kappa \left( \partial _{yy}\left( \bar{u} e_l\right) , \bar{v}_y e_i + \bar{v} e_i' \right) , \end{aligned}$$

(69)

$$\begin{aligned} K^l_i(x) :=&\kappa \left( \bar{v}_{xyy} \partial _y \left( \bar{u} e_l\right) , e_i\right) + \kappa \left( \partial _{yy}(\bar{u} e_l), \bar{v}_{xy} e_i + \bar{v}_x e_i' \right) \end{aligned}$$

(70)

$$\begin{aligned} \varGamma ^\ell _i :=&\left( \bar{u}_\theta ^2 e_\ell ', e_i' \right) \end{aligned}$$

(71)

$$\begin{aligned} F_i(x) :=&\left( F^{(\theta )}, e_i\right) . \end{aligned}$$

(72)

This can be solved for each fixed $n \in \mathbb {N}$, as the coefficients $A^l_i(x), K^l_i(x), F_i(x)$ are smooth. Moreover, the coefficients $\varGamma$ form a symmetric, positive definite matrix. To establish positive definiteness, we fix a vector $\mathbf {b} =( b_j)_{j = 1}^n$, Multiplying $\mathbf {b}^T \varGamma \mathbf {b} = \Vert \bar{u}_\theta \eta \Vert _{L^2_y}^2$, where $\eta := \sum _j b_j e_j'$. In turn, this implies that $\eta = 0$, which also implies that $b_j = 0$.

We now would like to send $n, N \rightarrow \infty$, for which we need to rely upon uniform bounds. By linearity of (68), we can replace the $e_i(y)$ by $q^{(n, N, \theta )}$, which gives

$$\begin{aligned}&\frac{\partial _x}{2}\left( |\bar{u}_\theta ^{(N)}|^2 q^{(n, N, \theta )}_y, q^{(n, N, \theta )}_y\right) + \left( \partial _{yy}\left( \bar{u} q^{(n, N, \theta )}\right) , q^{\left( n, N, \theta \right) }_{yy}\right) + 2 \bar{u}_y |q^{(n, N, \theta )}_y(0)|^2\nonumber \\&\quad + \kappa \left( \bar{v}_{xyy} I_x\left[ v^{(n, N, \theta )}_y\right] , q^{(n, N, \theta )}\right) + \kappa (\bar{v}_{yy} v^{(n, N, \theta )}_y, q^{(n, N,\theta )}) \nonumber \\&\quad + \kappa \left( I_x\left[ v^{(n, N, \theta )}_{yy}\right] , \bar{v}_{xy}q^{(n, N, \theta )} + \bar{v}_x q^{(n, N, \theta )}_y\right) + \kappa \left( v^{(n, N, \theta )}_{yy}, \bar{v}_y q^{(n, N, \theta )} + \bar{v} q^{(n, N, \theta )}_y\right) \nonumber \\&\quad + \kappa \left( U(u^0), q^{(n, N, \theta )}\right) - \left( \bar{u}_\theta ^{(N)} \partial _x \bar{u}_\theta ^{(N)} q^{\left( n, N, \theta \right) }_y, q^{\left( n ,N, \theta \right) }_y\right) = \left( F^{(\theta )}, q^{\left( n, N, \theta \right) }\right) . \end{aligned}$$

(73)

We now introduce some notation to clarify the next two lemmas:

$$\begin{aligned} \mathcal {E}_{(k)}(x)\, :=\,&\Vert \bar{u}_\theta ^{(N)} \partial _x^{k-1} q^{(n, \theta , N)}_{yy} \Vert _{L^2_y} + \Vert \bar{u}_\theta ^{(N)} \partial _x^k q^{(n, \theta , N)}_{y} \Vert _{L^2_y} \\ \mathcal {D}_{(k)}(x) \, :=\,&\Vert \sqrt{\bar{u}} \partial _x^{k-1} q^{(n, \theta , N)}_{yyy} \Vert _{L^2_y} + \Vert \sqrt{\bar{u}} \partial _x^k q^{(n, \theta , N)}_{yy} \Vert _{L^2_y} \end{aligned}$$

Lemma 6

The function $q^{(n, N, \theta )}$ satisfies the following inequality for $k \ge 0$,

$$\begin{aligned}&\sup _{x} \Vert \bar{u}_\theta ^{(N)} \partial _x^k q^{(n, N, \theta )}_y \Vert _{L^2_y}^2 + \Vert \sqrt{\bar{u}} \partial _x^k q^{(n, N, \theta )}_{yy} \Vert _{L^2_y}^2 + \bar{u}_y(0) |\partial _x^k q^{(n, N, \theta )}_y(0)|^2 \nonumber \\&\quad \le C_\theta \Vert \partial _x^k F^{(\theta )} \langle y \rangle \Vert ^2 + \Vert u^0 \Vert _{H^1_y}^2 + \Vert \bar{u}_\theta ^{(N)} \partial _x^k q^{(n, N, \theta )}_y|_{x = 0} \Vert _{L^2_y}^2. \end{aligned}$$

(74)

Proof

Our starting point is (74). By expanding the second term above using $v^{(n, \theta , N)} = \bar{u} q^{(n, \theta , N)}$, we obtain

$$\begin{aligned}&\left( \partial _{yy}(\bar{u} q^{(n, N, \theta )}), q^{(n, N, \theta )}_{yy}\right) = \Vert \sqrt{\bar{u}} q^{(n, N, \theta )}_{yy} \Vert _{L^2_y}^2 - \bar{u}_y(0) |q^{(n, N, \theta )}_y(0)|^2 \nonumber \\&\quad - 2\left( \bar{u}_{yy} q^{(n, N, \theta )}_y, q^{(n, N, \theta )}_y\right) + \frac{1}{4} \left( \bar{u}_{yyyy} q^{(n, N, \theta )}, q^{(n, N, \theta )}\right) . \end{aligned}$$

(75)

The contribution at $y = 0$ is absorbed into the third term from (74) upon using the prefactor of 2.

We now estimate

$$\begin{aligned} |2\left( \bar{u}_{yy} q^{(n, N, \theta )}_y, q^{(n, N, \theta )}_y\right) - \frac{1}{4} (\bar{u}_{yyyy} q^{(n, N, \theta )}, q^{(n, N, \theta )})| \le C_\theta \Vert \bar{u}_\theta ^{(0)} q^{(n, N, \theta )}_y \Vert _{L^2_y}^2. \end{aligned}$$

We next, estimate the fourth through eighth terms from (74) above by

$\kappa C_\theta \mathcal {E}_0(x)^{\frac{1}{2}} (\mathcal {D}_0(x)^{\frac{1}{2}} + \Vert u^0 \Vert _{H^1_y})$, and the ninth term from (74) by $o_L(1) C_\theta \sup _x \mathcal {E}_0(x)$, upon invoking that $\theta > 0$.

From here, we achieve our basic energy identity

$$\begin{aligned} \sup _{x} \Vert \bar{u}_\theta ^{(N)} q^{(n, N, \theta )}_y \Vert _{L^2_y}^2 + \Vert \sqrt{\bar{u}} q^{(n, N, \theta )}_{yy} \Vert _{L^2_y}^2 + \bar{u}_y(0) |q^{(n, N, \theta )}_y(0)|^2 \nonumber \\ \le C_\theta \Vert \hat{F} \langle y \rangle \Vert ^2 + o_L(1) \Vert u^0 \Vert _{H^1_y}^2 + o_L(1) \sup _x \Vert \bar{u}_\theta ^{(0)} q^{(n, N, \theta )}_y \Vert _{L^2_y}^2 + \Vert \bar{u}_\theta ^{(0)} q^{(n, N, \theta )}_y|_{x = 0} \Vert _{L^2_y}^2, \end{aligned}$$

which, for L sufficiently small, ensures the estimate (75). This concludes the proof of estimate (75)

Estimate (76) follows in essentially an identical manner to (75) upon differentiating k times in x. The only exception is we need to treat

$$\begin{aligned} \left( \partial _x^k v^{(n, N, \theta )}_{yy}, \partial _x^k q^{(n, N, \theta )}_{yy}\right) =&\Vert \sqrt{\bar{u}} q^{(n, N, \theta )}_{yy} \Vert _{L^2_y}^2 + 2( \bar{u}_y \partial _x^k q^{(n, N, \theta )}_y, \partial _x^k q^{(n, N, \theta )}_{yy}) \\&+ \left( \bar{u}_{yy} \partial _x^k q^{(n, \theta , N)}, \partial _x^k q^{(n, \theta , N)}_{yy}\right) + \mathcal {C}_2, \end{aligned}$$

where the commutators in $\mathcal {C}_2$ are defined by

$$\begin{aligned} \mathcal {C}_2 :=&\sum _{j = 1}^k \left( {\begin{array}{c}k\\ j\end{array}}\right) \left( \left( \partial _x^j \bar{u}_{yy} \partial _x^{k-j} q^{(n, \theta , N)}, \partial _x^k q^{\left( n, \theta , N\right) }_{yy}\right) + 2\left( \partial _x^j \bar{u}_y \partial _x^{k-j} q^{(n, \theta , N)}_y,\partial _x^k q^{(n, \theta , N)}_{yy} \right) \right. \nonumber \\&\left. + \left( \partial _x^j \bar{u} \partial _x^{k-j} q^{(n, \theta , N)}_{yy},\partial _x^k q^{(n, \theta , N)}_{yy}\right) \right) . \end{aligned}$$

(76)

We can easily estimate now

$$\begin{aligned} |\mathcal {C}_2| \le C_\theta \mathcal {E}_{(k)}(x)^{\frac{1}{2}} \mathcal {D}_{(k)}(x)^{\frac{1}{2}}, \end{aligned}$$

(77)

which concludes the proof of the lemma. $\quad \square$

Lemma 7

The function $q^{(n, N, \theta )}$ satisfies the following inequality for $k \ge 1$:

$$\begin{aligned}&\sup _x \Vert \bar{u}_\theta ^{(N)} \partial _x^{k-1} q^{(n, \theta , N)}_{yy} \Vert _{L^2_y}^2 + \Vert \sqrt{\bar{u}} \partial _x^{k-1} q^{(n, \theta , N)}_{yyy} \Vert ^2\nonumber \\&\quad \le C_\theta \Vert \partial _x^k F^{(\theta )} \langle y \rangle \Vert ^2 + \Vert u^0 \Vert _{H^1_y}^2 + \Vert \bar{u}_\theta ^{(N)} \partial _x^k q^{(n, N, \theta )}_y|_{x = 0} \Vert _{L^2_y}^2. \end{aligned}$$

(78)

Proof

We start from the weak formulation, (68), which, together with $e_i'' = (\frac{2\pi i}{N})^2 e_i$, immediately implies the validity of the following identity

$$\begin{aligned}&-\frac{\partial _x}{2}\left( |\bar{u}_\theta ^{(N)}|^2 q^{\left( n, N, \theta \right) }_y, q^{\left( n, N, \theta \right) }_{yyy}\right) - \left( \partial _{yy}\left( \bar{u} q^{(n, N, \theta )}\right) , q^{(n, N, \theta )}_{yyyy}\right) + 2 \bar{u}_y |q^{(n, N, \theta )}_y(0)|^2 \nonumber \\&\quad - \kappa \left( \bar{v}_{xyy} I_x\left[ v^{(n, N, \theta )}_y\right] , q_{yy}^{(n, N, \theta )}\right) - \kappa (\bar{v}_{yy} v^{(n, N, \theta )}_y, q_{yy}^{(n, N, \theta )}) \nonumber \\&\quad - \kappa (I_x\left[ v^{\left( n, N, \theta \right) }_{yy}\right] , \bar{v}_{xy}q_{yy}^{(n, N, \theta )} + \bar{v}_x q^{(n, N, \theta )}_{yyy}) - \kappa \left( v^{(n, N, \theta )}_{yy}, \bar{v}_y q_{yy}^{(n, N, \theta )} - \bar{v} q^{(n, N, \theta )}_{yyy}\right) \nonumber \\&\quad - \kappa (U(u^0), q_{yy}^{(n, N, \theta )}) + \left( \bar{u}_\theta ^{(N)} \partial _x \bar{u}_\theta ^{(N)} q^{(n, N, \theta )}_y, q^{(n ,N, \theta )}_{yy}\right) = - \left( F^{(\theta )}, q_{yy}^{(n, N, \theta )}\right) . \end{aligned}$$

(79)

From here, we can integrate by parts both terms using also that for each fixed $n < \infty$, $q_{yy}|_{y = 0} = q_{yy}|_{y = N} = 0$, which produces for the first two terms above

$$\begin{aligned} \sum _{j = 1}^2 (80.j) =&\frac{\partial _x}{2} \Vert \bar{u}_\theta ^{(N)} q^{(n, \theta , N)}_{yy} \Vert _{L^2_y}^2 + \left( \partial _y \bar{u}_\theta ^{(N)} q^{(n, \theta , N)}_{xy}, q^{(n, \theta , N)}_{yy}\right) + \left( v^{(n, \theta , N)}_{yyy}, q^{(n, \theta , N)}_{yyy}\right) . \end{aligned}$$

Expanding out $v^{(n, \theta , N)}_{yyy} = \partial _y^3(\bar{u} q^{(n, \theta , N)})$, we obtain furthermore the following identity

$$\begin{aligned} \sum _{j = 1}^2 (80.j) =&\frac{\partial _x}{2} \Vert \bar{u}_\theta ^{(N)} q^{(n, \theta , N)}_{yy} \Vert _{L^2_y}^2 + \Vert \sqrt{\bar{u}} q^{(n, N, \theta )}_{yyy} \Vert _{L^2_y}^2 + \mathcal {R}_1, \end{aligned}$$

where we have defined

$$\begin{aligned} \mathcal {R}_1 :=&- \frac{9}{2} \left( \bar{u}_{yy} q^{(n, \theta , N)}_{yy}, q^{(n, \theta , N)}_{yy}\right) - 4 \left( \bar{u}_{yyy} q^{(n, \theta , N)}_y, q^{(n, \theta , N)}_{yy} \right) \nonumber \\&- \left( \bar{u}_{yyyy} q^{(n, \theta , N)}, q^{(n, \theta , N)}_{yy}\right) + \left( \partial _y \bar{u}_\theta ^{(N)} q^{(n, \theta , N)}_{xy}, q^{(n, \theta , N)}_{yy}\right) . \end{aligned}$$

(80)

We can see that we may easily estimate

$$\begin{aligned} |\mathcal {R}_1| \lesssim \Vert q_{yy}^{(n, \theta , N)} \Vert _{L^2_y}( \Vert q_{yy}^{(n, \theta , N)} \Vert _{L^2_y} + \Vert q_{xy}^{(n, \theta , N)} \Vert _{L^2_y} ). \end{aligned}$$

(81)

We next estimate in a straightforward manner using that $\theta > 0$,

$$\begin{aligned} |\sum _{j = 4}^{9} (80.j)| \le C_\theta \mathcal {E}_1(x)^{\frac{1}{2}} \left( \mathcal {D}_1(x)^{\frac{1}{2}} + \Vert u^0 \Vert _{H^2_y}\right) . \end{aligned}$$

This concludes the proof of the lemma for $k = 1$. The $k > 1$ case works in a largely identical manner. $\quad \square$

We note that our compatibility conditions from Definition 1 imply that higher order x derivatives of the Eq. (60) holds at $\{x = 0\}$, and therefore we can evaluate the ODE (69) to obtain the initial condition for $\dot{b}^n_i$ (and higher order derivatives thereof). Combining (75), (79), and (75), we may thus send $n \rightarrow \infty$ and $N \rightarrow \infty$ to obtain a distributional solution to (60), which can then be upgraded to a strong solution in the usual manner. We state this now as a lemma:

Lemma 8

Fix any $\theta > 0$. Assume compatibility conditions on $q^{(\theta )}|_{x= 0}$ according to Definition 1. Assume $F^{(\theta )}$ satisfies $\sum _{j = 0}^{M_0}\Vert \partial _x^j F^{(\theta )} \langle y \rangle \Vert < \infty$. Then there exists an $L = L(\theta ) > 0$ such that there exists a unique solution to (60) on the interval (0, L).

2.4 Uniform estimates in $\theta$

In this section, we provide uniform in $\theta$ estimates for the problem (60). As such, we shall define the $\theta$-dependent version of our basic X norm

$$\begin{aligned} \Vert q \Vert _{X_\theta } :=&\sup _{0 \le x_0 \le L} \left( \Vert \bar{u}_\theta q_{xy} \Vert _{x = x_0} + \Vert q_{yyy} w (1 - \chi ) \Vert _{x= x_0} \right) \nonumber \\&+ \Vert \sqrt{\bar{u}} q_{xyy} w \Vert + \Vert v_{yyyy} w \Vert , \end{aligned}$$

(82)

$$\begin{aligned} \Vert q \Vert _{X_{k, \theta }} :=&\sum _{j = 0}^k \Vert q^{(j)} \Vert _{X_{\theta }}. \end{aligned}$$

(83)

We adopt the notation that

$$\begin{aligned} p_{k} := p\left( \Vert \bar{q} \Vert _{X_k}\right) , \qquad p_{\langle k \rangle } := p\left( \Vert \bar{q} \Vert _{X_{\langle k \rangle }}\right) , \end{aligned}$$

(84)

where p is an inhomogeneous polynomial of one variable of unspecified power. In general, we will suppress those constants which depend on $\Vert \bar{q} \Vert _{X_{\langle k \rangle }}$, and only display those which depend on $\Vert \bar{q} \Vert _{X_{\langle k + 1 \rangle }}$.

Lemma 9

The following inequalities are valid:

$$\begin{aligned}&\Vert q^{(k)}_{xy} \Vert + \Vert \{q^{(k)}_y, q^{(k)}_{yy} \} w \Vert + \Vert q^{(k)} \langle y \rangle ^{-1} \Vert \le o_L(1)(1 + \Vert q \Vert _{X_{k, \theta }}) \end{aligned}$$

(85)

$$\begin{aligned}&\Vert v^{(k)} \langle y \rangle ^{-1} \Vert + \Vert \{v^{(k)}_{y}, v^{(k)}_{yy}, v^{(k)}_{yyy}, v^{(k)}_{xy} \} w \Vert \le o_L(1) (1 + \Vert q \Vert _{X_{k, \theta }}) \end{aligned}$$

(86a)

$$\begin{aligned}&\Vert q^{(k)}_{xy} w \Vert + \Vert v^{(k)}_{xyy} w \Vert \lesssim 1 + \Vert q \Vert _{X_{k, \theta }} \end{aligned}$$

(86b)

$$\begin{aligned}&\Vert q^{(k)} \Vert _\infty + \Vert v^{(k)}_y \Vert _\infty + \Vert q^{(k)}_y \Vert _{\infty } \lesssim 1 + o_L(1) \Vert q \Vert _{X_{\langle k, \theta \rangle }} \end{aligned}$$

(86c)

$$\begin{aligned} \\&\Vert v^{(k)}_{yy}, v^{(k)}_{yyy} \Vert _\infty \le 1 + o_L(1) \Vert q \Vert _{X_{\langle k + 1, \theta \rangle }}. \end{aligned}$$

(86d)

Proof

The first step is to obtain control over $\Vert q^{(k)}_{xy} \Vert$ via interpolation.

$$\begin{aligned} \Vert q^{(k)}_{xy} \{1 - \chi \left( \frac{y}{\delta }\right) \} \Vert \le \delta ^{-1} \Vert \bar{u}_\theta q^{(k)}_{xy} \Vert \le L \delta ^{-1} \sup _x \Vert \bar{u}_\theta q^{(k)}_{xy}\Vert _{L^2_y} \le L \delta ^{-1} \Vert q \Vert _{X_{k, \theta }}. \end{aligned}$$

Near the $\{y = 0\}$ boundary, one interpolates:

$$\begin{aligned} |\left( \chi \left( \frac{y}{\delta }\right) \partial _y \{ y \}, |q^{(k)}_{xy}|^2\right) | \lesssim&\Vert \chi y q^{(k)}_{xyy} \Vert ^2 + \left( \frac{y}{\delta }\chi '\left( \frac{y}{\delta }\right) , |q^{(k)}_{xy}|^2\right) \\ \lesssim&\delta \Vert q^{(k)} \Vert _{X_\theta }^2 + L^2 \delta ^{-2} \Vert q^{(k)} \Vert _{X_\theta }^2. \end{aligned}$$

Optimizing $\sqrt{\delta } + L \delta ^{-1}$, one obtains $\delta = L^{2/3}$. Thus, $\Vert q^{(k)}_{xy} \Vert \lesssim L^{1/3} \Vert q \Vert _{X_{k, \theta }}$. From here, a basic Poincare inequality gives:

$$\begin{aligned} \Vert q^{(k)}_y \Vert = \Vert q^{(k)}_y|_{x = 0} + \int _0^x q^{(k)}_{xy} \Vert \lesssim \sqrt{L} |q^{(k)}_y|_{x = 0}\Vert + L \Vert q^{(k)}_{xy} \Vert \end{aligned}$$

From here, Hardy inequality gives immediately $\Vert q^{(k)} \langle y \rangle ^{-1} \Vert \le \Vert q^{(k)}_y \Vert$.

The next step is to establish the uniform bound via straightforward Sobolev embedding:

$$\begin{aligned} |q^{(k)}|^2 \lesssim \sup _x |q^{(k)}_y \langle y \rangle \Vert ^2 \lesssim |q^{(k)}_y|_{x = 0} \langle y \rangle \Vert ^2 + L \Vert q^{(k)}_{xy} \langle y \rangle \Vert ^2 \lesssim 1 + o_L(1) \Vert q \Vert _{X_{k, \theta }}. \end{aligned}$$

A Hardy computation gives:

$$\begin{aligned} \Vert q^{(k)}_{xy} w (1 - \chi )\, \Vert \lesssim \,&\Vert q^{(k)}_{xy} \Vert _{2,loc} + \Vert q^{(k)}_{xyy} w (1 - \chi )\Vert \lesssim \Vert q \Vert _{X_{k, \theta }}. \end{aligned}$$

We record the following expansions which follow from the product rule upon recalling that $v = \bar{u} q$:

$$\begin{aligned} \begin{aligned} |v^{(k)}| \lesssim&\sum _{j = 0}^k |\bar{u}^j q^{(k-j)}| \\ |v^{(k)}_y| \lesssim&\sum _{j = 0}^k |\bar{u}^j_{y} q^{(k-j)}| + |\bar{u}^j q^{(k-j)}_y| \\ |v^{(k)}_{yy}| \lesssim&\sum _{j = 0}^k |\bar{u}^j_{yy} q^{(k-j)}| + |\bar{u}^j_{y} q^{(k-j)}_{y}| + |\bar{u}^j q^{(k-j)}_{yy}| \\ |v^{(k)}_{yyy}| \lesssim&\sum _{j = 0}^k |\bar{u}^j_{yyy} q^{(k-j)}| + |\bar{u}^j_{yy} q^{(k-j)}_y| + |\bar{u}^j_{y} q^{(k-j)}_{yy}|+ |\bar{u}^j q^{(k-j)}_{yyy}| \\ |v^{(k)}_{xy}| \lesssim&\sum _{j = 0}^k |\bar{v}^{j}_{yy} q^{(k-j)}| + |\bar{u}^j_{y} q^{(k-j)}_x| + |\bar{v}^j_{y} q^{(k-j)}_y| + |\bar{u}^j q^{(k-j)}_{xy}| \\ |v^{(k)}_{xyy}| \lesssim&\sum _{j = 0}^k |\bar{v}^j_{yyy} q^{(k-j)}| + |\bar{u}^j_{yy} q^{(k-j)}_x| + |\bar{v}^j_{yy} q^{(k-j)}_y| + |\bar{u}^j_{y} q^{(k-j)}_{xy}| \\&\quad + |\bar{v}^j_{y} q^{(k-j)}_{yy}| + |\bar{u}^j q^{(k-j)}_{xyy}|. \end{aligned} \end{aligned}$$

(86e)

We will restrict to $k = 0$ for the remainder of the proof, as the argument works for general k in a straightforward way. From (87), $\Vert v_y\Vert$ follows obviously. Next,

$$\begin{aligned} \Vert v_{yy} \Vert \lesssim&\Vert \bar{u}_{yy} q \Vert + \Vert \bar{u}_{y} q_y \Vert + \Vert \bar{u} q_{yy} \Vert \lesssim \sqrt{L} + o_L(1) \Vert q \Vert _{X_\theta }. \end{aligned}$$

(86f)

From here, $\Vert v_{yyy} \Vert _{loc}$ can be interpolated in the following way:

$$\begin{aligned}&\left( v_{yyy}, v_{yyy} \chi \left( \frac{y}{\delta }\right) \right) = \left( \partial _y \{ y \} \chi \left( \frac{y}{\delta }\right) , |v_{yyy}|^2\right) \\&\quad = - \left( y \chi \left( \frac{y}{\delta }\right) v_{yyy}, \partial _x^{k-1} v_{yyyy}\right) - \left( y \delta ^{-1} \chi '\left( \frac{y}{\delta }\right) , |v_{yyy}|^2\right) \\&\quad \lesssim \delta ^2 \Vert v_{yyyy} \Vert ^2 + \Vert \psi _\delta v_{yyy} \Vert ^2. \end{aligned}$$

For the far-field component, we may majorize via:

$$\begin{aligned} |\left( v_{yyy}, v_{yyy} \{1 - \chi (\frac{y}{\delta }) \} \right) | \lesssim \Vert \psi _\delta v_{yyy} \Vert ^2. \end{aligned}$$

Here $\psi _\delta = 1 - \chi (\frac{10 y}{\delta })$, the key point being that both $\{1 - \chi (\frac{y}{\delta })\}$ and $\chi '(\frac{y}{\delta })$ are supported in the region where $\psi _\delta = 1$. To estimate this term, we may integrate by parts:

$$\begin{aligned} \left( \psi v_{yyy}, v_{yyy}\right) =&- \left( \psi v_{yy}, v_{yyyy}\right) - \left( \delta ^{-1} \psi ' v_{yy}, v_{yyy}\right) \\ =&- \left( \psi v_{yy}, v_{yyyy}\right) + \left( \delta ^{-2} \frac{\psi ''}{2}, |v_{yy}|^2\right) \\ \lesssim&\delta ^2 \Vert v_{yyyy} \Vert ^2 + N_\delta \Vert v_{yy} \Vert ^2. \end{aligned}$$

Thus,

$$\begin{aligned} \Vert v_{yyy} \Vert \le&\delta \Vert v_{yyyy} \Vert + N_\delta \Vert v_{yy} \Vert . \end{aligned}$$

(87)

We combine the above with (88) to select $\delta = L^{0+}$ to achieve control over $\Vert \partial _y^j v^{(k)}\Vert$ for $j = 1,2,3$.

We now use the elementary formula $\frac{1}{a+b} = \frac{1}{a} - \frac{b}{a+b}$ to write:

$$\begin{aligned} q = \frac{v}{u_s} = \frac{v}{u_{sy}(0) y + [u_s - u_{sy}(0)y]} = \frac{1}{u_{sy}(0)} \frac{v}{y} - v \frac{u_s - u_{sy}(0)y}{y u_{sy}(0) u_s}. \end{aligned}$$

Using the estimates $u_s \gtrsim y$ as $y \downarrow 0$ and $|u_s - u_{sy}(0)y| \lesssim y^2$ as $y \downarrow 0$, it is easy to see that the second quotient above is bounded and in fact $\mathcal {C}^\infty$. We may thus limit our study to $q_0 := \frac{v}{y}$. We let $k_1 + k_2 = 3$ and differentiate the formula:

$$\begin{aligned} q_0(x,y) = \frac{1}{y} \int _0^y v_y(x,y') \,\mathrm {d}y' = \int _0^1 v_y(x,ty) \,\mathrm {d}t, \end{aligned}$$

where we changed variables via $ty = y'$. From here, we can obtain:

$$\begin{aligned} \Vert q_{yy} \chi (y) \Vert \lesssim \Vert \int _0^1 t^2 v_{yyy}(x, ty) \,\mathrm {d}t \Vert _{loc} \lesssim \Vert v_{yyy} \Vert . \end{aligned}$$

(88)

Away from the $\{y = 0\}$ boundary, we estimate trivially:

$$\begin{aligned} \Vert q_{yy}\left\{ 1 - \chi \left( \frac{y}{\delta }\right) \right\} w \Vert&\lesssim \Vert \bar{u} q_{yy} w \{1 - \chi \left( \frac{y}{\delta }\right) \} \Vert _{2} \\&\lesssim \sqrt{L} \Vert \bar{u} q_{yy} w\Vert _{x = 0} + L \Vert \sqrt{\bar{u}} q_{xyy} w \Vert . \end{aligned}$$

From here, obtaining $\Vert q_y w \Vert$ follows from Hardy. We now turn our attention to the weighted estimates for $v_y, v_{yy}, v_{xy}, v_{xyy}$, which follow from (87), whereas for $v_{yyy}$, we use the Prandtl equation to produce the identity:

$$\begin{aligned} v_{yyy} =&\bar{u}_{yyy} q + 3 \bar{u}_{yy} q_y + 3 \bar{u}_{y} q_{yy} + \bar{u} q_{yyy} \\ =&\left( - \bar{u} \bar{v}_{yy} + \bar{v} \bar{u}_{yy} \right) q + 3 \bar{u}_{yy} q_y + 3 \bar{u}_{y} q_{yy} + \bar{u} q_{yyy} \end{aligned}$$

The uniform estimates subsequently follow from straightforward Sobolev embeddings. $\square$

We now perform our first energy estimate.

Lemma 10

($\mathcal {E}_k$ Estimate). Assume q solves (36). Then the following inequality is valid:

$$\begin{aligned} \begin{aligned} \Vert q \Vert _{\mathcal {E}_{k, \theta }}^2 \lesssim&\Vert \bar{u} \partial _x^k q_{xy}|_{x = 0}\Vert ^2 + o_L(1)\left( p_{\langle k \rangle } + \kappa p_{\langle k + 1 \rangle }\right) \left( 1 + \Vert q \Vert _{X_{\langle k, \theta \rangle }}^2 \right) \\&+ o_L(1) C(u^0) + o_L(1) \Vert \partial _x^{k+1} \partial _{xy}g_1 \langle y \rangle \Vert ^2. \end{aligned} \end{aligned}$$

(89)

Proof

We have the following energy inequality,

$$\begin{aligned} \frac{\partial _x}{2} \Vert \bar{u}_\theta \partial _x^{k+1} q^{(\theta )}_y \Vert _x^2 + \Vert \sqrt{\bar{u}_\theta } \partial _x^{k+1} q_{yy}^{(\theta )} \Vert ^2 + 2 \bar{u}_{y}(0) \left( \partial _x^{k+1} q^{(\theta )}_y(0) \right) ^2 \le |\mathcal {I}_{k+1}^{(\theta )}|, \end{aligned}$$

(90)

where the integral $\mathcal {I}^{(\theta )}_{k+1}$ is defined

$$\begin{aligned} \mathcal {I}_{k+1}^{(\theta )} :=&\sum _{j = 1}^{k+1} \left( {\begin{array}{c}k+2\\ j\end{array}}\right) \left( \left( \partial _x^j \{ \bar{u}_\theta ^2 \} \partial _x^{k+2-j} q^{(\theta )}_y, \partial _x^{k+1} q^{(\theta )}_y \right) + \left( \partial _x^j \bar{u} \partial _x^{k+1-j} q^{(\theta )}_{yy}, \partial _x^{k+1} q^{(\theta )}_{yy}\right) \right) \nonumber \\&+ \sum _{j = 0}^{k+1} \left( {\begin{array}{c}k+1\\ j\end{array}}\right) \left( 2 \partial _x^j \partial _y \bar{u} \partial _x^{k+1-j} q^{(\theta )}_{y}, \partial _x^{k+1} q^{(\theta )}_{yy}\right) \nonumber \\&+ \sum _{j = 0}^{k+1} \left( {\begin{array}{c}k+1\\ j\end{array}}\right) \left( \partial _x^j \partial _y^2 \bar{u} \partial _x^{k+1-j} q^{(\theta )}, \partial _x^{k+1} q^{(\theta )}_{yy}\right) + \kappa \left( \partial _x^{k+1} U, \partial _x^{k+1} q^{(\theta )}\right) \nonumber \\&+\kappa \sum _{j = 0}^{k+1} \left( {\begin{array}{c}k+1\\ j\end{array}}\right) \left( \partial _x^j \bar{v}_{xyy} I_x\left[ \partial _x^{k+1-j} v^{(\theta )}_y\right] + \partial _x^j \bar{v}_{yy} \partial _x^{k+1-j} v^{(\theta )}_y, \partial _x^{k+1} q^{(\theta )}\right) \nonumber \\&+ \kappa \sum _{j = 0}^{k+1} \left( {\begin{array}{c}k+1\\ j\end{array}}\right) \left( \left( I_x\left[ \partial _x^{k+1-j} v^{(\theta )}_{yy} \right] , \partial _y \left\{ \partial _x^j \bar{v}_x \partial _x^{k+1} q^{(\theta )} \right\} \right) \right. \nonumber \\&\left. + \left( \partial _x^{k+1-j} v^{(\theta )}_{yy}, \partial _y \left\{ \partial _x^j \bar{v} \partial _x^{k+1} q^{(\theta )} \right\} \right) \right) + 2\sum _{j =1}^{k+1} \left( {\begin{array}{c}k+1\\ j\end{array}}\right) \partial _x^j \bar{u}_{y} \partial _x^{k+1-j} q^{(\theta )}_y(0) \partial _x^{k+1} q^{(\theta )}_y(0) \nonumber \\&+ \left( \partial _x^{k+1} F^{(\theta )}, \partial _x^{k+1} q^{(\theta )}\right) = \sum _{\ell = 1}^{11} \mathcal {I}^{(\theta )}_{k+1,\ell }. \end{aligned}$$

(91)

For this estimate, we further integrate over $x \in [0, x_0]$, where $0< x_0 < L$. This produces

$$\begin{aligned} \sup _x \Vert \bar{u} \partial _x^{k+1} q_y \Vert _x^2 + \Vert \sqrt{\bar{u}} \partial _x^{k+1} q_{yy} \Vert ^2 + 2 \Vert \sqrt{\bar{u}_{y}} \partial _x^{k+1} q_y \Vert _{y = 0}^2 \le \int _0^L |\mathcal {I}_{k+1}^{(\theta )}|, \end{aligned}$$

(92)

We recall now the various contributions into $\mathcal {I}_{k+1}^{(\theta )}$ from (93). We start with the first of these, via

$$\begin{aligned} \int _0^{L} |\mathcal {I}_{k+1, 1}^{(\theta )}|= & {} \int _0^L \Big | \sum _{j = 1}^{k+2} \left( {\begin{array}{c}k+2\\ j\end{array}}\right) \sum _{j_1 = 0}^j \left( {\begin{array}{c}j\\ j_1\end{array}}\right) \left( \partial _x^{j_1} \bar{u}_\theta \partial _x^{j - j_1} \bar{u}_\theta \partial _x^{k+2-j} q_y, \partial _x^{k+1} q_y\right) \Big | \\\lesssim & {} \Big \Vert \frac{\partial _x^{\langle \frac{k+1}{2} \rangle } \bar{u}_\theta }{\bar{u}_\theta } \Big \Vert _\infty ^2 \Vert \bar{u}_\theta \partial _x^{\langle k + 1 \rangle } q_y \Vert ^2 \\&+ \Big \Vert \frac{\partial _x^{\langle \frac{k+2}{2} \rangle }\bar{u}_\theta }{\bar{u}_\theta } \Big \Vert _\infty \Vert \partial _x^{\langle \frac{k+2}{2} \rangle } q_y \Vert _{L^\infty _x L^2_y} \Vert \partial _x^{k+2} \bar{u}_\theta \Vert _{L^2_x L^\infty _y} \Vert \bar{u}_\theta \partial _x^{k+1} q_y \Vert \\\lesssim & {} o_L(1) p\left( \Vert \bar{q} \Vert _{X_{\langle k \rangle }}\right) \Vert q \Vert _{X_{\langle k \rangle }}^2. \end{aligned}$$

We have used the fact that $\frac{\partial _x^{j} \bar{u}_\theta }{\bar{u}_\theta } \sim \partial _x^{j-1} \partial _{yy} \bar{v}$ for any j, in particular with $j = \langle \frac{k+2}{2} \rangle$ and then appealed to (86d). We also use (86a) to bound $\Vert \partial _x^j q_y\Vert$ for $j = \frac{k+2}{2}$.

We now move to $\partial _y^4$ term from the specification of $\mathcal {I}^{(\theta )}_{k+1}$, the first of which reads

$$\begin{aligned} \int _0^L | \mathcal {I}_{k+1,2}^{(\theta )} | \le \int _0^L | \sum _{j = 1}^{k+1} \left( {\begin{array}{c}k+1\\ j\end{array}}\right) \left( \partial _x^j \bar{u} \partial _x^{k+1-j} q_{yy}, \partial _x^k q_{yy}\right) |. \end{aligned}$$

(93)

The $j = k+1$ case from above contributes

$$\begin{aligned} |\int _0^{x_0} \left( \partial _x^{k+1} \bar{u} q_{yy}, \partial _x^{k+1} q_{yy}\right) | \,\le \,&\Big \Vert \frac{\partial _x^{k+1} \bar{u}}{\bar{u}} \Big \Vert _{L^2_x L^\infty _y} \Vert \sqrt{\bar{u}} q_{yy} \Vert _{L^\infty _x L^2_y} \Vert \sqrt{\bar{u}} \partial _x^{k+1} q_{yy} \Vert \\ \lesssim \,&\left( 1 + o_L(1) \Vert \bar{q} \Vert _{X_{\langle k \rangle }}\right) \left( 1 + \Vert q \Vert _{X_{\langle k \rangle }}\right) \Vert \sqrt{\bar{u}} \partial _x^{k} q_{xyy} \Vert . \end{aligned}$$

Above, we have used the Hardy and Agmon inequalities, as well as (86b), which gives

$$\begin{aligned} \Big \Vert \frac{\partial _x^{k+1} \bar{u}}{\bar{u}} \Big \Vert _{L^2_x L^\infty _y} =&\Big \Vert \frac{\partial _x^{k} \bar{v}_y}{\bar{u}} \Big \Vert _{L^2_x L^\infty _y} \lesssim \Vert v^{(k)}_{yy} \Vert _{L^2_x L^\infty _y} + \Vert v^{(k)}_y \Vert _{L^2_x L^\infty _y} \\ \lesssim \,&\Vert v^{(k)}_{yyy} \Vert + \Vert v^{(k)}_{yy} \Vert + \Vert v^{(k)}_y \Vert \lesssim o_L(1) \left( 1 + \Vert \bar{q} \Vert _{X_{\langle k \rangle }}\right) . \end{aligned}$$

The intermediate cases, $j = 1,\ldots ,k$ can be bounded above by

$$\begin{aligned} \Big \Vert \frac{\partial _x^{\langle k \rangle } \bar{u}}{\bar{u}} \Big \Vert _{\infty } \Vert \sqrt{\bar{u}} \partial _x^{\langle k \rangle } q_{yy} \Vert \Vert \sqrt{\bar{u}} \partial _x^k q_{xyy} \Vert \lesssim \left( 1 + o_L(1) p_{k}\right) \Vert q \Vert _{X_k} \Vert q \Vert _{X_{\langle k \rangle }}. \end{aligned}$$

The next term is

$$\begin{aligned} \int _0^{L} |\mathcal {I}_{k+1, 3}^{(\theta )}| =&\int _0^{L} |\sum _{j = 0}^{k+1} \left( {\begin{array}{c}k+1\\ j\end{array}}\right) \left( 2 \partial _x^j \partial _y \bar{u} \partial _x^{k+1-j} q_y, \partial _x^{k+1} q_{yy}\right) |\nonumber \\ =&\int _0^{L} | \left( 2 \bar{u}_{y} \partial _x^{k+1} q_y, \partial _x^{k+1} q_{yy} \right) | + \sum _{1 \le j \le \frac{k+1}{2}} \int _0^{L} |\left( \partial _x^j \bar{u}_{y} \partial _x^{k+1 - j} q_y, \partial _x^{k+1} q_{yy} \right) | \nonumber \\&+ \sum _{\frac{k+1}{2} \le j \le k+1} \int _0^{L} |\left( \partial _x^j \bar{u}_{y} \partial _x^{k+1-j} q_y, \partial _x^{k+1} q_{yy} \right) |. \end{aligned}$$

(94)

First, using $\bar{u}_{yy} = \bar{U}_0''(y) - I_x[\bar{v}_{yyy}]$ and also integration by parts gives:

$$\begin{aligned} (96.1) \le&\int _0^{L} |( \bar{u}_{yy} \partial _x^{k+1} q_y, \partial _x^{k+1} q_y )| + \int _0^{L} \bar{u}_{y}(0) |\partial _x^{k+1} q_y(0)|^2 \\ \le&o_L(1)(1 + p_{\langle 2 \rangle }) \Vert q \Vert _{X_k}^2 + \Vert \sqrt{\bar{u}_y} |\partial _x^{k+1}q_y \Vert _{y = 0}^2. \end{aligned}$$

We now absorb that the crucial boundary term above can be absorbed into the left-hand side of (94) due to the factor of 2 in (94) as compared with the factor of 1 above.

We now treat, where we have used again that $\bar{u}_x = - \bar{v}_y$ and subsequently (86a)–(86b) to bound the $\bar{u}$ contributions:

$$\begin{aligned} (96.2) \le \,&\int _0^{L} |\left( \partial _x^{k+1} q_y, \partial _x^j \bar{u}_{yy} \partial _x^{k+1-j} q_y \right) | + \int _0^{L} |\left( \partial _x^{k+1} q_y, \partial _x^j \bar{u}_{y} \partial _x^{k+1 -j} q_{yy} \right) | \\&\, + \int _0^{L}| \partial _x^j \bar{u}_{y} \partial _x^{k-j+1} q_y \partial _x^{k+1} q_y(0)| \\ \le&\, \int _0^{L} | \left( \partial _x^{k+1} q_y, \partial _x^j \bar{u}_{yy} \partial _x^{k+1-j} q_y \right) | + \int _0^{L}| \left( \partial _x^{k+1} q_y, \partial _x^j \bar{u}_{y} \partial _x^{k+1 -j} q_{yy} \right) |\\&\, + \int _0^{L}| \partial _x^j \bar{u}_{y} \partial _x^{k-j+1} q_y \partial _x^{k+1} q_y(0)| \\ \lesssim \,&\Vert \partial _x^{\langle \frac{k+1}{2}\rangle } \bar{u}_{yy} \Vert _\infty \Vert \partial _x^{\langle k \rangle } q_y \Vert \Vert \partial _x^{k+1} q_y \Vert + \Vert \partial _x^{\langle \frac{k+1}{2} \rangle } \bar{u}_{y} \Vert _\infty \Vert \partial _x^{\langle k \rangle } q_{yy} \Vert \Vert \partial _x^{k+1} q_{y} \Vert \\&\, + o_L(1) \Vert q \Vert _{X_{\langle k \rangle }}^2 \\ \lesssim \,&o_L(1) p_{\langle k \rangle } \left( \Vert q \Vert _{X_{\langle k \rangle }}^2 + 1 \right) , \end{aligned}$$

Similarly, we obtain by invoking the definition of $\Vert \bar{q} \Vert _{X_k}$, as well as (86a)

$$\begin{aligned} (96.3) \le \,&\int _0^{L} |\left( \partial _x^{k+1} \bar{u}_{yy} q_y, \partial _x^{k+1} q_y\right) | + \int _0^{L}| (\partial _x^{k+1} \bar{u}_y q_{yy}, \partial _x^{k+1}q_y)|\\&\, + \int _0^{L}| \sum _{j = \frac{k+1}{2}}^{k} \left( \partial _x^j \bar{u}_{yy} \partial _x^{k+1-j} q_y, \partial _x^{k+1}q_y\right) | + \int _0^{L}| \sum _{j = \frac{k+1}{2}}^{k}\left( \partial _x^j \bar{u}_y \partial _x^{k+1-j} q_{yy}, \partial _x^{k+1}q_y\right) | \\ \lesssim&\Vert \partial _x^k \bar{v}_{yyy} \Vert _{L^2_x L^\infty _y} \Vert q_y \Vert _{L^\infty _x L^2_y} \Vert \partial _x^{k+1}q_y \Vert + \Vert \partial _x^k \bar{v}_{yy} \Vert _{L^2_x L^\infty _y} \Vert q_{yy} \Vert _{L^\infty _x L^2_y} \Vert \partial _x^{k+1} q_{yy} \Vert \\&\, + \Vert \partial _x^{\langle k - 1 \rangle } \bar{v}_{yyy} \Vert _{L^\infty } \Vert \partial _x^{\langle \frac{k+1}{2} \rangle } q_y \Vert \Vert \partial _x^{k+1} q_y \Vert + \Vert \partial _x^{\langle k - 1 \rangle } \bar{v}_{yy} \Vert _{L^\infty } \Vert \partial _x^{\langle \frac{k+1}{2} \rangle } q_{yy} \Vert \Vert \partial _x^{k+1} q_y \Vert \\ \lesssim \,&o_L(1) \left( 1 + \Vert \bar{q} \Vert _{X_{\langle k \rangle }}\right) \left( 1 + \Vert q \Vert _{X_{\langle k \rangle }} \right) ^2 \end{aligned}$$

Next, we move to:

$$\begin{aligned} \int _0^L |\mathcal {I}_{k+1, 4}^{(\theta )}| \le \sum _{j = 0}^{k+1} \left( {\begin{array}{c}k+1\\ j\end{array}}\right) \int _0^{L} |\left( \partial _x^j \bar{u}_{yy} \partial _x^{k+1 - j} q, \partial _x^{k+1} q_{yy}\right) |. \end{aligned}$$

(95)

We split the above term into several cases. First, let us handle the $j = 0$ case for which (4) gives us the required bound by Hardy’s inequality :

$$\begin{aligned} |(97)[j=0]| \lesssim&\Vert \frac{1}{\bar{u}} \bar{u}_{yy} y \Vert _\infty \Vert \partial _x^{k} q_{xy} \Vert \Vert \sqrt{\bar{u}} \partial _x^k q_{xyy} \Vert \lesssim o_L(1) p_{\langle k \rangle } \Vert q \Vert _{X_{k}}^2. \end{aligned}$$

We now handle the case of $1 \le j \le k/2$, which requires a localization using $\chi$ as defined in (39). We invoke (86b) for the $\bar{v}$ contribution and (86d):

$$\begin{aligned} |(97)[1-\chi ]| \lesssim&\Vert \partial _x^{\frac{k}{2}} \bar{v}_{yyy} \Vert _{L^\infty _x L^2_y} \Vert \partial _x^{\langle k \rangle } q \Vert _{L^2_x L^\infty _y} \Vert q \Vert _{X_k} \lesssim o_L(1) p_{\langle k \rangle } \Vert q \Vert _{X_{\langle k \rangle }}^2. \end{aligned}$$

For the localized component, we integrate by parts in y and invoking Lemma 9:

$$\begin{aligned} (97)[\chi ] =&\sum _{j = 0}^{k+1} \int _{0}^{x_0} ( \partial _x^k q_y \chi ', \partial _x^j \bar{u}_{yy} \partial _x^{k+1-j}q ) + \sum _{j = 0}^{k+1} \int _0^{x_0} \left( \partial _x^k q_y \chi , \partial _x^j \bar{u}_{yyy} \partial _x^{k+1-j} q \right) \nonumber \\&+ \sum _{j = 0}^{k+1} \int _0^{x_0} \langle \partial _x^k q_y \chi , \partial _x^j \bar{u}_{yy} \partial _x^{k+1-j} q_y \rangle \nonumber \\ \lesssim&\Vert \partial _x^{k-1} q_{xy} \Vert \Vert \partial _x^{\frac{k}{2}} \bar{v}_{yyy} \Vert \Vert \partial _x^{\langle k \rangle } q \Vert _{L^\infty _{loc}} + \Vert \partial _x^{k-1} q_{xy} \Vert \Vert \partial _x^{\frac{k}{2}} \bar{v}_{yyyy} \Vert \Vert \partial _x^{\langle k \rangle } q \Vert _{L^\infty _{loc}} \nonumber \\&+ \Vert \partial _x^{k-1} q_{xy} \Vert \Vert \partial _x^{\frac{k}{2}} \bar{v}_{yyy} \Vert _\infty \Vert \partial _x^{\langle k \rangle } q_{xy} \Vert \nonumber \\ \lesssim&o_L(1)p_{\langle k \rangle } \Vert q \Vert _{X_{\langle k \rangle }}^2. \end{aligned}$$

(96)

We now treat the case in which $k/2 \le j \le k$, which still requires localization and (86d) and (86b)

$$\begin{aligned} |(97)[\chi _{\ge 1}]| \lesssim \,&\Vert \partial _x^k q_{xyy} \sqrt{\bar{u}} \Vert \Vert \partial _x^{\frac{k}{2}} q \Vert _{L^2_x L^\infty _y} \Vert \partial _x^{\langle k - 1 \rangle } \bar{v}_{yyy} \Vert _{L^\infty _x L^2_y} \\ \lesssim \,&o_L(1) p_{\langle k \rangle } \Vert q \Vert _{X_{\langle k \rangle }}^2. \end{aligned}$$

Finally, we deal with the case when $y \lesssim 1$ for $k/2 \le j \le k$, which again requires integration by parts in y as in (98), upon invoking (86a), (86b)

$$\begin{aligned} |(97)| \lesssim&\Vert \partial _x^{k-1} q_{xy} \Vert \Vert \partial _x^{\langle k - 1 \rangle } \bar{v}_{yyy} \Vert \Vert \partial _x^{\frac{k}{2}} q \Vert _{L^\infty _{loc}} \\&\, + \Vert \partial _x^{k-1} q_{xy} \Vert \Vert \partial _x^{\langle k - 1 \rangle } \bar{v}_{yyyy} \Vert \Vert \partial _x^{\frac{k}{2}} q \Vert _{L^\infty _{loc}} \\&\, + \Vert \partial _x^{k-1} q_{xy} \Vert \Vert \partial _x^{\langle k - 1 \rangle }\bar{v}_{yyy} \Vert _\infty \Vert \partial _x^{\langle k/2 \rangle } q_{xy} \Vert \\ \lesssim \,&o_L(1)p_{\langle k \rangle }\left( 1 + \Vert q \Vert _{X_{\langle k \rangle }}^2\right) . \end{aligned}$$

We now move to the $\varLambda$ terms. By Hardy’s inequality in y, we obtain

$$\begin{aligned} \kappa \int _0^L |\mathcal {I}_{k+1,6}^{(\theta )}| \le&\kappa \sum _{j = 0}^{k+1} \left( {\begin{array}{c}k+1\\ j\end{array}}\right) \int _0^{x_0}\left( \partial _x^{j} \bar{v}_{xyy} \partial _x^{k+1-j}I_x[v_y] , \partial _x^{k+1} q\right) \\ \lesssim \,&\kappa \Vert \partial _x^{\langle \frac{k}{2} \rangle } \bar{v}_{xyy} \langle y \rangle \Vert _\infty \Vert \partial _x^{\langle k+1 \rangle } I_x[v_y] \Vert \Vert \partial _x^{k+1} q_y \Vert \\&+ \kappa \Vert \partial _x^{\langle k+1 \rangle } \bar{v}_{xyy} \Vert \Vert \partial _x^{\langle \frac{k+1}{2} \rangle } I_x[v_y] \langle y \rangle \Vert _\infty \Vert \partial _x^{k+1} q_y \Vert . \end{aligned}$$

Next, we similarly have

$$\begin{aligned} \kappa \int _0^L |\mathcal {I}_{k+1,7}^{(\theta )}| \le \,&\kappa \sum _{j = 0}^{k+1} \left( {\begin{array}{c}k+1\\ j\end{array}}\right) \int _0^{L}| \left( \partial _x^j \bar{v}_{yy} \partial _x^{k+1-j} v_y, \partial _x^{k+1} q \right) | \\ \lesssim \,&\kappa \, \Vert \partial _x^{\langle \frac{k+1}{2} \rangle } \bar{v}_{yy} \langle y \rangle \Vert _\infty \Vert \partial _x^{\langle k+1 \rangle } v_y \langle y \rangle \Vert \Vert \partial _x^{k+1} q_y \Vert . \end{aligned}$$

Next, we have

$$\begin{aligned} \kappa \int _0^L |\mathcal {I}_{k+1,8}^{(\theta )}| \le&\kappa \sum _{j = 0}^{k+1} \left( {\begin{array}{c}k+1\\ j\end{array}}\right) \int _0^{L}| \left( \partial _x^j \bar{v}_{x} \partial _x^{k+1-j} I_x\left[ \partial _y^3 v\right] , \partial _x^{k+1} q \right) | \\ \le&\kappa \,\sum _{j = 0}^{k+1} \left( {\begin{array}{c}k+1\\ j\end{array}}\right) \int _0^{L}| \left( \partial _x^j \bar{v}_{xy} \partial _x^{k+1-j} I_x[\partial _y^2 v], \partial _x^{k+1} q \right) | \\&\, +\kappa \sum _{j = 0}^{k+1} \left( {\begin{array}{c}k+1\\ j\end{array}}\right) \int _0^{L} |\left( \partial _x^j \bar{v}_{x} \partial _x^{k+1-j} I_x[\partial _y^2 v], \partial _x^{k+1} q_y \right) |\\ \lesssim&\kappa \, \Vert \partial _x^{\langle \frac{k+1}{2} \rangle } \bar{v}_{xy} \langle y \rangle \Vert _\infty \Vert \partial _x^{\langle k+1 \rangle } I_x\left[ v_{yy}\right] \Vert \Vert \partial _x^{k+1} q_y \Vert \\&+\, \kappa \Vert \partial _x^{\langle \frac{k+1}{2}\rangle } \bar{v}_{x} \Vert _\infty \Vert \partial _x^{\langle k + 1 \rangle } I_x\left[ v_{yy}\right] \Vert \Vert \partial _x^{k+1} q_{y} \Vert \\&+\, \kappa \Vert \partial _x^{\langle \frac{k+1}{2} \rangle } I_x\left[ v_{yy}\right] \langle y \rangle \Vert _\infty \Vert \partial _x^{\langle k + 1 \rangle } \bar{v}_{xy} \Vert \Vert \partial _x^{k+1} q_y\Vert \\&+\kappa \, \Vert \partial _x^{\langle k +1 \rangle } \bar{v}_{xy} \Vert \Vert \partial _x^{\langle \frac{k+1}{2} \rangle } I_x\left[ v_{yy}\right] \langle y \rangle \Vert _\infty \Vert \partial _x^{k+1} q_{y} \Vert .\\ \kappa \int _0^L |\mathcal {I}_{k+1,9}^{(\theta )}| \le&\kappa \, \sum _{j = 0}^{k+1} \left( {\begin{array}{c}k+1\\ j\end{array}}\right) \int _0^{L} |\left( \partial _x^j \bar{v} \partial _x^{k+1 - j} v_{yyy}, \partial _x^{k+1} q \right) | \\ \le&\kappa \, \sum _{j = 0}^{k+1} \left( {\begin{array}{c}k+1\\ j\end{array}}\right) \left( \int _0^{L} | \left( \partial _x^{k+1 - j} v_{yy} \partial _x^j \bar{v}_{y}, \partial _x^{k+1} q \right) |\right. \\&\left. + \int _0^{L}| \left( \partial _x^{k+1 - j} v_{yy} \partial _x^j \bar{v}, \partial _x^{k+1} q_y \right) | \right) \\ \lesssim \,&\kappa \Vert \partial _x^{\langle \frac{k+1}{2} \rangle } \bar{v}_{y} \langle y \rangle \Vert _\infty \Vert \partial _x^{\langle k + 1 \rangle } v_{yy} \Vert \Vert \partial _x^{k+1} q_y \Vert \\&+ \kappa \Vert \partial _x^{\langle \frac{k+1}{2} \rangle } \bar{v}_{} \Vert _\infty \Vert \partial _x^{\langle k + 1 \rangle } v_{yy} \Vert \Vert \partial _x^{k+1} q_{y} \Vert \\&+\kappa \Vert \partial _x^{\langle \frac{k+1}{2} \rangle } v_{yy} \langle y \rangle \Vert _\infty \Vert \partial _x^{\langle k+1 \rangle } \bar{v}_{y} \Vert \Vert \partial _x^{k+1} q_y \Vert \\&+ \kappa \Vert \partial _x^{\langle \frac{k+1}{2} \rangle } \bar{v} \Vert _\infty \Vert \partial _x^{\langle k + 1 \rangle } v_{yy} \Vert \Vert \partial _x^{k} q_{xy} \Vert . \end{aligned}$$

Summarizing the $\varLambda$ contributions by repeated use of Lemma 9 with $k+1$ for the $\bar{v}$ contribution gives:

$$\begin{aligned} \sup _{x_0 \le L} \int _0^{x_0} \left( \partial _x \partial _x^k \varLambda , \partial _x \partial _x^k q \right) \lesssim \kappa o_L(1)p_{\langle k + 1 \rangle } \left( 1 + \Vert q \Vert _{X_{\langle k \rangle }}^2 \right) . \end{aligned}$$

Next, we have the contributions at $y = 0$:

$$\begin{aligned} \int _0^L |\mathcal {I}_{k+1,10}^{(\theta )}| \le&\sum _{j = 1}^{k+1} \left( {\begin{array}{c}k+1\\ j\end{array}}\right) \int _0^{L} |\partial _x^j \bar{u}_y \partial _x^{k+1-j} q_y(0) \partial _x^{k+1} q_y(0) | \\ \le&o_L(1)\left( 1 + p_{\langle k \rangle }\right) \Vert \partial _x^{k+1-j} q_y(0) \Vert _{L^2_x} \Vert \partial _x^k q_y(0) \Vert _{L^2_x} \le o_L(1) \left( 1 + p_{\langle k \rangle }\right) \Vert q \Vert _{X_{\langle k \rangle }}^2. \end{aligned}$$

Above, we have used that $\partial _x^j \bar{u}_y(x, 0) = - \partial _x^{j-1} \bar{v}_{yy}(x, 0)$, and then (86b).

Finally, we have the $u^0$ contributions for which we rely upon (86b) for $k + 1$ for the $\bar{v}$ term:

$$\begin{aligned} \kappa \int _0^L |\mathcal {I}_{k+1,5}^{(\theta )}|&=\kappa \int _0^{L} |\left( \partial _x^{k+1} U, \partial _x^{k} q_x \right) | =\kappa \int _0^{L} \left( u^0 \partial _x^{k+1} \bar{v}_{xyy} - u^0_{yy} \partial _x^{k+1} \bar{v}_{x}, \partial _x^{k}q_x \right) \\&\quad \lesssim \kappa o_L(1) \Vert u^0, u^0_{yy} \cdot \langle y \rangle \Vert _\infty \Vert \partial _x^{k+1} \bar{v}_{xyy} \Vert \Vert \partial _x^{k} q_{xy} \Vert \\&\quad \lesssim \kappa o_L(1) p_{\langle k + 1 \rangle } \Vert q \Vert _{X_{\langle k \rangle }}. \end{aligned}$$

This concludes the proof of the lemma. $\square$

Lemma 11

($\partial _x^k$ $\partial _y^4$ Estimate). Assume v is a solution to (36). Then the following estimate holds:

$$\begin{aligned} \begin{aligned} \Vert \partial _x^k v_{yyyy} \Vert _{2,loc} \lesssim&\Vert q \Vert _{\mathcal {E}_{\langle k \rangle }} + o_L(1)\left( p_{\langle k \rangle } + \kappa p_{\langle k + 1 \rangle }\right) \left( 1 + \Vert q \Vert _{X_{\langle k \rangle }} \right) \\&+\kappa o_L(1) C\left( u^0\right) + \Vert \partial _{xy} \partial _x^k g_1 \Vert _{2,loc}. \end{aligned} \end{aligned}$$

(97)

Proof

We apply $\partial _x^k$ to the Eq. (36) to obtain the following pointwise inequality:

$$\begin{aligned} \begin{aligned} |v^{(k)}_{yyyy}| \lesssim&|\bar{u}^{j_1} \bar{u}^{j_2} q^{(k-j)}_{xyy}| + |\bar{u}_{x}^{j_1} \bar{u}_{y}^{j_2} q_y^{k-j}| + |\bar{u}^{j_1} \bar{u}_{xy}^{j_2} q_y^{k-j}| + |\bar{u}^{j_1} \bar{u}_{y}^{j_2} q_{xy}^{k-j}| \\&+ |\bar{u}^{j_1} \bar{u}_{x}^{j_2} q^{(k-j)}_{yy}| + \kappa |\bar{v}^j_{xyy} I_x\left[ v^{(k-j)}_y\right] | + \kappa |\bar{v}^j_{yy}v^{(k-j)}_y| + \kappa |v^{j}_{sx} I_x\left[ v^{(k-j)}_{yyy}\right] | \\&+ |\bar{v}^j v^{(k-j)}_{yyy} | + \kappa |v^{(k)}_{sxyy} u^0| + \kappa |v^{(k)}_{sx} u^0_{yy}| + |\partial _{xy}g_1^k|. \end{aligned} \end{aligned}$$

(98)

Placing the terms on the right-hand side above in $L^2_{loc}$ gives the desired result by applying Lemma 9 with k and $k+1$:

$$\begin{aligned}&\Vert (100.1)\Vert \lesssim \Vert \bar{v}_{y}^{\langle k - 1 \rangle }, \bar{v}_{yy}^{\langle k - 1 \rangle } \Vert _\infty ^2 \Vert \bar{u} q_{xyy}^{\langle k \rangle } \Vert , \\&\quad \Vert (100.2) \Vert \lesssim \Vert \bar{v}_{y}^{\langle k - 1 \rangle } \Vert _\infty \Vert \bar{v}_{yy}^{\langle k - 1 \rangle } \Vert _\infty \Vert q^{\langle k \rangle }_y \Vert \\&\quad \Vert (100.3) \Vert \lesssim \Vert \bar{v}_{y}^{\langle k - 1 \rangle } \Vert _\infty \Vert \bar{v}_{yy}^{\langle k - 1 \rangle } \Vert _\infty \Vert q^{\langle k \rangle }_y \Vert \\&\quad \Vert (100.4) \Vert \lesssim \Vert \bar{v}^{\langle k - 1 \rangle }_{y} \Vert _\infty \Vert \bar{v}^{\langle k - 1 \rangle }_{yy} \Vert _\infty \Vert q^{\langle k \rangle }_{xy} \Vert \\&\quad \Vert (100.5) \Vert \lesssim \Vert \bar{v}_{y}^{\langle k - 1 \rangle } \Vert _\infty \Vert \bar{v}^{\langle k \rangle }_{y} \Vert _\infty \Vert q^{\langle k \rangle }_{yy} \Vert \\&\quad \Vert (100.6, 7) \Vert \lesssim \kappa \Vert \bar{v}^{\langle k + 1 \rangle }_{yy} \Vert \Vert v_y^{\langle k \rangle } \Vert _\infty \\&\quad \Vert (100.8, 9) \Vert \lesssim \left( \Vert \bar{v}^{\langle k \rangle } \Vert _\infty + \kappa \Vert \bar{v}^{\langle k + 1 \rangle } \Vert _\infty \right) \Vert v^{\langle k \rangle }_{yyy} \Vert \\&\quad \Vert (100.10, 11) \Vert \lesssim \kappa \Vert u^0, u^0_{yy} \langle y \rangle ^2 \Vert _\infty \Vert \bar{v}^{k+1}_{yy} \Vert \end{aligned}$$

These estimates immediately establish the proof of the lemma. $\square$

We now move to a $\Vert \cdot \Vert _{\mathcal {H}_k}$ estimate, for which we first recall the definition in (40).

Lemma 12

(Weighted $\partial _x^k H^4$). Assume q solves (36). Then the following estimate is valid:

$$\begin{aligned} \begin{aligned} \Vert q \Vert _{\mathcal {H}_k}^2 \lesssim&\Vert \partial _x^k \partial _{xy} g_1 \cdot w \chi \Vert ^2 + C_k(q_0) + \kappa o_L(1) C\left( u^0\right) \\&+\left( o_L(1) p_{\langle k \rangle } + \kappa o_L(1) p_{\langle k + 1 \rangle }\right) \left( 1+ \Vert q \Vert _{X_{\langle k \rangle }}^2 \right) . \end{aligned} \end{aligned}$$

(99)

Proof

We take $\partial _x^k$ of Eq. (60), which produces:

$$\begin{aligned} \begin{aligned}&- \partial _{xy} \left( \bar{u}_\theta ^2 q^{(k)}_y \right) + v^{(k)}_{yyyy} = \partial _x^k \partial _{xy}g_1 - \kappa \partial _x^k \varLambda (v) - \kappa \partial _x^k U \\&\quad - \partial _{xy} \left( \sum _{j = 1}^k \sum _{j_1 + j_2 = j} c_{j_1, j_2, j} \partial _x^{j_1}\bar{u}_\theta \partial _x^{j_2} \bar{u}_\theta q^{(k-j)}_y \right) . \end{aligned} \end{aligned}$$

(100)

We now square and integrate both sides in y, for frozen $x = x_0$, against the weight $w (1 - \chi )$, which produces the inequality

$$\begin{aligned}&\Vert \left( - \partial _{xy} \left( \bar{u}_\theta ^2 q^{(k)}_y \right) + v^{(k)}_{yyyy} \right) w (1 - \chi ) \Vert _{L^2_y}^2 \nonumber \\&\quad \le \Vert \partial _x^k \partial _{xy}g_1 w (1-\chi ) \Vert _{L^2_y}^2 + \kappa \Vert \partial _x^k \varLambda (v) w (1-\chi ) \Vert _{L^2_y}^2 +\kappa \Vert \partial _x^k U w (1-\chi )\Vert _{L^2_y}^2 \nonumber \\&\qquad + \Vert \partial _{xy} \left( \sum _{j = 1}^k \sum _{j_1 + j_2 = j} c_{j_1, j_2, j} \partial _x^{j_1}\bar{u}_\theta \partial _x^{j_2} \bar{u}_\theta q^{(k-j)}_y \right) w (1-\chi )\Vert _{L^2_y}^2. \end{aligned}$$

(101)

We start by expanding out the terms on the left-hand side, which gives

$$\begin{aligned}&\Vert \left[ \partial _{xy} \{\bar{u}_\theta ^2 q^{(k)}_y \} - v^{(k)}_{yyyy}\right] \cdot w \{1 - \chi \}\Vert _{x = x_0}^2 \\&\quad \ge \Vert v^{(k)}_{yyyy} \{1 - \chi \} w \Vert _{x = x_0}^2 + \Vert [ \bar{u}_\theta ^2 q^{(k)}_{xyy} \{1 - \chi \} w \Vert _{x = x_0}^2 - 4 \Vert \bar{u}_\theta \bar{u}_{\theta x} q^{(k)}_{yy} w(1-\chi ) \Vert _{L^2_y}^2 \\&\qquad - 4 \Vert \bar{u}_{\theta y} \bar{u}_{\theta x} q^{(k)}_y w(1-\chi ) \Vert _{L^2_y}^2 - 4 \Vert \bar{u} \bar{u}_{\theta x y} q^{(k)}_y w (1 - \chi ) \Vert _{L^2_y}^2 - 4 \Vert \bar{u}_\theta \bar{u}_{\theta y} q^{(k)}_{xy} w(1-\chi ) \Vert _{L^2_y}^2 \\&\qquad - \left( 2 v^{(k)}_{yyyy}, \bar{u}_\theta ^2 q^{(k)}_{xyy} w^2 (1 - \chi )^2\right) - 4\left( v^{(k)}_{yyyy}, \bar{u}_\theta \bar{u}_{\theta x} q^{(k)}_{yy} w^2 (1 - \chi )^2\right) \\&\qquad - 4(v^{(k)}_{yyyy}, \bar{u}_{\theta y} \bar{u}_{\theta x} q^{(k)}_y w^2 (1 - \chi )^2) - 4 (v^{(k)}_{yyyy}, \bar{u}_\theta \bar{u}_{\theta x y} q^{(k)}_y w^2 (1-\chi )^2)\\&\qquad - 4 (v^{(k)}_{yyyy}, \bar{u}_\theta \bar{u}_{\theta y} q^{(k)}_{xy} w^2 (1 - \chi )^2) =: \mathcal {P}_1 + ... + \mathcal {P}_{11}. \end{aligned}$$

All terms are estimated in a straightforward manner except for $\mathcal {P}_7$, upon invoking $\bar{u}_x + \bar{v}_y = 0$, so we begin with:

$$\begin{aligned}&|\mathcal {P}_3| \lesssim \Vert \bar{u}_{x} \Vert _\infty ^2 \Vert q^{(k)}_{yy} \cdot w \Vert _{x = x_0}^2 \\&\quad |\mathcal {P}_4| \lesssim \Vert \bar{u}_{y} \Vert _\infty ^2 \Vert \bar{v}_{y} \Vert _\infty ^2 \Vert q^{(k)}_y w \Vert _{x = x_0}^2 \\&\quad |\mathcal {P}_5| \lesssim \Vert \bar{u} \bar{v}_{yy} \Vert _\infty ^2 \Vert q^{(k)}_y w \Vert _{x = x_0}^2, \\&\quad |\mathcal {P}_6| \lesssim \Vert \bar{u} \bar{u}_{y} w \Vert _\infty ^2 \Vert q^{(k)}_{xy} \Vert _{x = x_0}^2, \\&\quad |\mathcal {P}_8| \lesssim \Vert \bar{u}_{x} \Vert _\infty | v^{(k)}_{yyyy} w \{1 -\chi \} \Vert _{x = x_0} \Vert q^{(k)}_{yy} w \{1 -\chi \}\Vert _{x = x_0} \\&\quad |\mathcal {P}_9| \lesssim \Vert \bar{u}_{y} \bar{v}_{y} \Vert _\infty |q^{(k)}_y w \{1 -\chi \}\Vert _{x = x_0} \Vert v^{(k)}_{yyyy} w \{1 -\chi \}\Vert _{x = x_0}, \\&\quad |\mathcal {P}_{10}| \lesssim \Vert \bar{u} \bar{v}_{yy} \Vert _\infty |q^{(k)}_y w \{1 -\chi \}\Vert _{x = x_0} \Vert v^{(k)}_{yyyy}w \{1 -\chi \}\Vert _{x = x_0} \\&\quad |\mathcal {P}_{11}| \lesssim \Vert \bar{u} \bar{u}_{y} w \Vert _\infty \Vert q^{(k)}_{xy} \Vert _{x = x_0} \Vert v^{(k)}_{yyyy} w \{1 -\chi \}\Vert _{x = x_0}. \end{aligned}$$

Upon integrating in x and applying Lemma 9, we may summarize the above estimates via:

$$\begin{aligned} |\mathcal {P}_3| +\cdots + |\mathcal {P}_6| + |\mathcal {P}_8| +\cdots + |\mathcal {P}_{11}| \lesssim o_L(1) p\left( \Vert \bar{q} \Vert _{X_{\langle 1 \rangle }}\right) \left( 1 + \Vert q \Vert _{X_k} \right) . \end{aligned}$$

We thus move to $\mathcal {P}_7$ for which we integrate by parts once in y to obtain

$$\begin{aligned} \mathcal {P}_7 \,=\,&2\left( v^{(k)}_{yyy}, \bar{u}_\theta ^2 q^{(k)}_{xyyy} w^2(1 - \chi )^2\right) + 4\left( v^{(k)}_{yyy}, \bar{u}_\theta \bar{u}_{\theta y} q^{(k)}_{xyy} w^2 (1 - \chi )^2\right) \\&+ 4\left( v^{(k)}_{yyy}, \bar{u}_\theta ^2 w w_y (1 - \chi )^2\right) - 4\left( v^{(k)}_{yyy}, \bar{u}_\theta ^2 w^2 (1 - \chi ) \chi '\right) =: \mathcal {P}_{7,1} + \dots + \mathcal {P}_{7,4}. \end{aligned}$$

We now expand using that $v = \bar{u} q$,

$$\begin{aligned} v^{(k)}_{yyy} \,:=\,&\partial _x^k \{ \bar{u} q \}_{yyy} = \partial _x^k \left\{ \bar{u}_{yyy}q + 3 \bar{u}_{yy} q_y + 3 \bar{u}_y q_{yy} + \bar{u} q_{yyy} \right\} \\ =&\sum _{j = 0}^k c_j \partial _x^j \bar{u}_{yyy} \partial _x^{k-j} q + 3 c_j \partial _x^j \bar{u}_{yy} \partial _x^{k-j} q_y + 3 c_j \partial _x^j \bar{u}_y \partial _x^{k-j} q_{yy} + c_j \partial _x^j \bar{u} \partial _x^{k-j} q_{yyy}. \end{aligned}$$

First, upon integrating by parts once in y (ignoring commutator terms, which are dealt with below), let us highlight the main positive contribution from the last term above, for $j = 0$:

$$\begin{aligned}&2\left( \bar{u} \partial _x^k q_{yyy}, \bar{u}_\theta ^2 q^{(k)}_{xyyy} \{1 - \chi \}^2 w^2\right) _{x = x_0} = \left( 2 \bar{u} \bar{u}_\theta ^2 q^{(k)}_{yyy}, q^{(k)}_{xyyy} \left\{ 1 - \chi \right\} ^2 w^2\right) _{x = x_0} \\&\quad = \partial _x \Vert |\bar{u}|^{\frac{1}{2}} \bar{u}_\theta q^{(k)}_{yyy} \{1 - \chi \} w \Vert _{x =x_0}^2 - \left( \left( \bar{u} \bar{u}_\theta ^2\right) _x q^{(k)}_{yyy}, q^{(k)}_{yyy} \left\{ 1 - \chi \right\} ^2 w^2\right) _{x = x_0}. \end{aligned}$$

Hence:

$$\begin{aligned}&- 2\left( v^{(k)}_{yyyy}, \bar{u}_\theta ^2 q^{(k)}_{xyy} w^2 \{1 - \chi \}^2\right) _{x = x_0} \nonumber \\&\quad = \partial _x \Vert q^{(k)}_{yyy} |\bar{u}|^{\frac{3}{2}} w \{1 - \chi \} \Vert _{x = x_0}^2 - \left( |q^{(k)}_{yyy}|^2, \left( \bar{u} \bar{u}_\theta ^2\right) _{x} w^2 \{1 - \chi \}^2\right) _{x = x_0} \nonumber \\&\qquad - \left( q_{xyy}^k, \partial _y \left\{ \left[ \bar{u}_{yyy}^j q^{(k-j)} + 3 \bar{u}_{yy}^j q_y^{k-j} + 3 \bar{u}_{y}^j q_{yy}^{k-j} \right] \right. \right. \nonumber \\&\qquad \left. \left. \times \bar{u}_\theta ^2 w^2 \{1 - \chi \}^2 \right\} \right) + \left( 2 v^{(k)}_{yyy}, q^{(k)}_{xyy} \partial _y\left\{ \bar{u}_\theta ^2 w^2 \left\{ 1 - \chi \right\} ^2\right\} \right) _{x = x_0} \nonumber \\&\qquad + \sum _{j = 1}^k c_j (\partial _x^j \bar{u} \partial _x^{k-j} q_{yyy}, \bar{u}_\theta ^2 q^{(k)}_{xyyy} w^2 \{1 - \chi \}^2)_{x = x_0} \nonumber \\&\quad = \partial _x \Vert q^{(k)}_{yyy} |\bar{u}|^{\frac{3}{2}} w \{1 - \chi \} \Vert _{x = x_0}^2 + \varUpsilon _1 + \dots + \varUpsilon _6. \end{aligned}$$

(102)

First, we estimate upon integrating from $x = 0$ to $x = x_0$,

$$\begin{aligned} |\int _0^{x_0} \varUpsilon _1(x) \,\mathrm {d}x| \lesssim o_L(1) \sup _x \Vert |\bar{u}|^{\frac{3}{2}} q^{(k)}_{yyy} w\{1- \chi \} \Vert _{x = x_0}^2 \lesssim o_L(1) \Vert q \Vert _{X_k}^2. \end{aligned}$$

Next,

$$\begin{aligned} \varUpsilon _2 =&- \left( q^{(k)}_{xyy} \bar{u}^j_{yyyy}, q^{(k-j)} \bar{u}_\theta ^2 w^2 \{1 - \chi \}^2\right) _{x = x_0} - ( q^{(k)}_{xyy} \bar{u}^j_{yyy}, q^{(k-j)} \bar{u}_\theta \bar{u}_{\theta y} w^2 \{1 - \chi \}^2)_{x = x_0} \\&- \left( q^{(k)}_{xyy} \bar{u}^j_{yyy}, q^{(k-j)}_y \bar{u}_\theta ^2 w^2 \{1 - \chi \}^2\right) _{x = x_0} - ( q^{(k)}_{xyy} \bar{u}^j_{yyy}, q^{(k-j)} \bar{u}_\theta ^2 2 ww_{y} \{1 - \chi \}^2)_{x = x_0} \\&- \left( q^{(k)}_{xyy} \bar{u}^j_{yyy}, q^{(k-j)} \bar{u}_\theta ^2 w^2 \{1 - \chi \} \chi '\right) _{x = x_0} =: \varUpsilon _{2,1} + \dots + \varUpsilon _{2,5}. \end{aligned}$$

We will estimate each term above with the help of the Prandtl identities, which follow from (1), for $\bar{u}$:

$$\begin{aligned} \begin{aligned}&|\bar{u}^j_{yyy}| \lesssim |\partial _x^j \left\{ \bar{u} \bar{v}_{yy} + \bar{v} \bar{u}_{yy} \right\} |, \\&\quad |\bar{u}^j_{yyyy}| \lesssim |\partial _x^j \left\{ \bar{u}_{y} \bar{v}_{yy} + \bar{u} \bar{v}_{yyy} + \bar{v}_{y} \bar{u}_{yy} + \bar{u} \bar{v} \bar{v}_{yy} + \bar{v}^2 \bar{u}_{yy} \right\} | \end{aligned} \end{aligned}$$

(103)

Inserting this expansion into term $\varUpsilon _{2,1}$, gives

$$\begin{aligned} |\varUpsilon _{2,1}| \,\lesssim \,&|\left( \partial _x^j(\bar{u}_y \bar{v}_{yy}) q^{(k)}_{xyy} , q^{(k-j)} \bar{u}_\theta ^2 w^2 \{1 - \chi \}^2\right) | + |( \partial _x^j(\bar{u} \bar{v}_{yyy}) q^{(k)}_{xyy} , q^{(k-j)} \bar{u}_\theta ^2 w^2 \{1 - \chi \}^2)| \\&+\, |\left( \partial _x^j(\bar{u}_{yy} \bar{v}_{y}) q^{(k)}_{xyy} , q^{(k-j)} \bar{u}_\theta ^2 w^2 \{1 - \chi \}^2\right) | + |\left( \partial _x^j(\bar{u} \bar{v} \bar{v}_{yy}) q^{(k)}_{xyy} , q^{(k-j)} \bar{u}_\theta ^2 w^2 \{1 - \chi \}^2\right) |\\&+ |\left( \partial _x^j(\bar{v}^2 \bar{u}_{yy}) q^{(k)}_{xyy} , q^{(k-j)} \bar{u}_\theta ^2 w^2 \{1 - \chi \}^2\right) | = : \varUpsilon _{2,1,1} + \cdots + \varUpsilon _{2,1,5}. \end{aligned}$$

First, upon integration in x and using (86c), (86d)

$$\begin{aligned} \int _0^{L}| \varUpsilon _{2,1,1} | \lesssim&\int _0^{L} | (q^{(k)}_{xyy}, q^{\langle k \rangle } \bar{u}_\theta ^2 w^2 \{1 - \chi \}^2 \bar{u}^{\langle k \rangle }_{y} \bar{v}^{\langle k \rangle }_{yy})| \\ \lesssim \,&\int _0^{L} \Vert q^{(k)}_{xyy} \{1 - \chi \} w\Vert _{x = x_0} \Vert q^{\langle k \rangle }\Vert _\infty \Vert \bar{v}^{\langle k -1 \rangle }_{yy} \Vert _\infty \Vert \bar{v}^{\langle k \rangle }_{yy} w \{1 - \chi \}\Vert _{x = x_0}, \\ \lesssim&\, \left( 1 + o_L(1) p_k\right) ^2 \Vert q \Vert _{X_k} o_L(1) \Vert 1 + \Vert q \Vert _{X_k}). \end{aligned}$$

Next, upon invoking (86b) and (86d),

$$\begin{aligned} \int _0^{L}| \varUpsilon _{2,1,2}| \lesssim&|\int _0^{L}( q^{(k)}_{xyy}, q^{\langle k \rangle } \bar{u}_\theta ^2 w^2 \{1 - \chi \}^2 \partial _x^{\langle k \rangle } \bar{u} \partial _x^{\langle k \rangle } \bar{v}_{yyy})| \\ \lesssim&\Vert q^{\langle k \rangle }\Vert _\infty \Vert \partial _x^{\langle k -1 \rangle } \bar{v}_{y} \Vert _\infty \Vert q^{(k)}_{xyy} \{1 - \chi \} w\Vert \Vert \partial _x^{\langle k \rangle } \bar{v}_{yyy} \{1 - \chi \} w\Vert \\ \lesssim&\left( 1 + o_L(1) p_k\right) ^2 \Vert q \Vert _{X_k} o_L(1) \left( 1 + \Vert q \Vert _{X_k}\right) . \end{aligned}$$

Next, upon invoking (86b), (86d), (86f), we have

$$\begin{aligned} \int _0^{L} |\varUpsilon _{2,1,3} | \lesssim&\int _0^{x_0}|\left( q^{(k)}_{xyy}, q^{\langle k \rangle } \partial _x^{\langle k \rangle } \bar{v}_{y} \partial _x^{\langle k \rangle } \bar{u}_{yy} \{1 - \chi \}^2 w^2 \bar{u}_\theta ^2\right) | \\ \lesssim&\Vert q^{(k)}_{xyy} \{1 - \chi \} w\Vert \Vert q^{\langle k \rangle }\Vert _\infty \Vert \partial _x^{\langle k - 1 \rangle } \bar{v}_{yyy} \langle y \rangle ^2\Vert _\infty \Vert \partial _x^{\langle k \rangle } \bar{v}_{y} w \{1 - \chi \}\Vert , \\ \lesssim&\left( 1 + o_L(1) p_k\right) ^2 \Vert q \Vert _{X_k} o_L(1) \left( 1 + \Vert q \Vert _{X_k}\right) . \end{aligned}$$

Next, upon invoking (86b), (86d), we have

$$\begin{aligned} \int _0^{L} | \varUpsilon _{2,1,4} | \lesssim&\int _0^{L} |\left( q^{(k)}_{xyy}, q^{\langle k \rangle } \bar{u}_\theta ^2 w^2 \{1 - \chi \}^2 \partial _x^{\langle k - 1 \rangle } \bar{v}_{y} \partial _x^{\langle k \rangle } \bar{v} \partial _x^{\langle k \rangle } \bar{v}_{yy}\right) | \\ \lesssim&\Vert q^{(k)}_{xyy} \{1 - \chi \} w\Vert \Vert q^{\langle k \rangle }\Vert _\infty \Vert \partial _x^{\langle k \rangle }\bar{v}\Vert _\infty \Vert \partial _x^{\langle k - 1 \rangle } \bar{v}_{y}\Vert _\infty \Vert \partial _x^{\langle k \rangle } \bar{v}_{yy} w \{1 - \chi \}\Vert , \\ \lesssim&\left( 1+ o_L(1) p(\Vert q \Vert _{X_k}) \right) \left( 1 + o_L(1) p_k\right) ^2 \Vert q \Vert _{X_k} o_L(1) \left( 1+ \Vert \bar{q} \Vert _{X_k}\right) . \end{aligned}$$

Next, upon invoking (86d)

$$\begin{aligned} \int _0^{L} |\varUpsilon _{2,1,5}| \lesssim&|\int _0^{L} \left( q^{(k)}_{xyy}, q^{\langle k \rangle } \partial _x^{\langle k \rangle } \bar{v} \partial _x^{\langle k \rangle }\bar{v} \partial _x^{\langle k - 1 \rangle } \bar{v}_{yyy} \bar{u}_\theta ^2 \{1 - \chi \}^2 w^2\right) | \\ \lesssim&\Vert q^{(k)}_{xyy} \{1 - \chi \} w\Vert \Vert q^{\langle k \rangle }\Vert _\infty \Vert \partial _x^{\langle k \rangle } \bar{v}\Vert _\infty ^2 \Vert \partial _x^{\langle k - 1 \rangle }\bar{v}_{yyy} \{1 - \chi \} w \Vert \\ \lesssim&\Vert q \Vert _{X_k} \left( 1 + o_L(1) p\left( \Vert q \Vert _{X_k}\right) \right) \left( 1 + o_L(1) p_k\right) o_L(1) \left( 1 + p_k\right) \\ \lesssim&o_L(1) + o_L(1) p_{\langle k \rangle } + o_L(1) \Vert q \Vert _{X_{\langle k \rangle }}^2. \end{aligned}$$

We now move to:

$$\begin{aligned} \int _0^{L} | \varUpsilon _{2,2}| \lesssim&\, \int _0^{x_0} | \left( q^{(k)}_{xyy}, q^{\langle k \rangle } w^2 \bar{u} \bar{u}_{y} \{1 - \chi \}^2 \left[ \bar{u}^{\langle k \rangle } \bar{v}^{\langle k \rangle }_{yy} + v^{\langle k \rangle }_s \bar{u}^{\langle k \rangle }_{yy}\right] \right) | \\ \lesssim&\, \Vert q^{(k)}_{xyy} w \{1 - \chi \}\Vert \Vert q^{\langle k \rangle }\Vert _\infty \left[ \Vert \bar{v}^{\langle k - 1 \rangle }_{y}\Vert _\infty \Vert \bar{v}^{\langle k \rangle }_{yy} w \{1 - \chi \}\Vert \right. \\&\left. + \Vert \bar{v}^{\langle k \rangle }\Vert _\infty \Vert \bar{u}^{\langle k \rangle }_{yy} w \{1 - \chi \}\Vert \right] \Vert \Vert \bar{u}_{y} \langle y \rangle \Vert _\infty \\ \lesssim \,&\, \Vert q \Vert _{X_k} \left( 1 + o_L(1) \Vert q \Vert _{X_k}\right) \left( 1 + o_L(1) p_{k-1}\right) o_L(1) \left( 1 + \Vert \bar{q} \Vert _{X_k}\right) . \end{aligned}$$

Above, we have invoked (86b) and (86d).

Next, we use (105) and Lemma 9 to estimate,

$$\begin{aligned} \int _0^{x_0} |\varUpsilon _{2,3}| \,\le \,&\, \sum _{j = 0}^{k} \left( {\begin{array}{c}k\\ j\end{array}}\right) \int _x |(q^{(k)}_{xyy} \bar{u}^{j}_{yyy}, q^{(k-j)}_{y} \bar{u}^2 w^2 \{1 - \chi \}^2 )_{x = x_0} \,\mathrm {d}x_0 \\ \lesssim \,&\, \int _x |(q^{(k)}_{xyy} \bar{u}_{yyy}, q^{(k)}_y \bar{u}^2 w^2 \{1 - \chi \}^2)_{x = x_0} \,\mathrm {d}x_0 \\&+\, \int _x |(q^{(k)}_{xyy} \bar{u}^{(k)}_{yyy}, q_y w^2 \{1 - \chi \}^2)_{x = x_0} \,\mathrm {d}x_0| \\&+\, \sum _{j= 1}^{k-1} \int _x |(q^{(k)}_{xyy} \bar{u}^{(j)}_{yyy}, q^{(k-j)}_y \bar{u}^2 w^2 \{1 - \chi \}^2)_{x = x_0}| \\ \lesssim \,&\, \Vert \bar{u}_{yyy} \Vert _\infty \Vert \bar{u} q^{(k)}_{xyy} w \{1 - \chi \} \Vert \Vert q^{(k)}_y w \{1 - \chi \} \Vert \\&+\, \Vert q^{(k)}_{xyy} w \{1 - \chi \} \Vert \Vert \bar{u}^{(k)}_{yyy} w \Vert \Vert q_y \Vert _\infty \\&+ \,\Vert \bar{v}_{yyyy}^{\langle k - 2 \rangle } w \Vert _{L^\infty _x L^2_y} \Vert q^{(k)}_{xyy} w \Vert \Vert q^{\langle k - 1 \rangle }_y \Vert _{L^2_x L^\infty _y} \\ \lesssim \,\,&\, o_L(1) (1 + o_L(1) p_k) \Vert q \Vert _{X_k}. \end{aligned}$$

Next, again by invoking Lemma 9, we estimate

$$\begin{aligned} \int _x |\varUpsilon _{2,4}| \lesssim&|\left( q^{(k)}_{xyy}, q^{\langle k \rangle } w^2 \{1 - \chi \}^2 \bar{u}^2 \left[ \bar{u}^{\langle k \rangle } \bar{v}^{\langle k \rangle }_{yy} + \bar{v}^{\langle k \rangle } \bar{u}^{\langle k \rangle }_{yy} \right] \right) | \\ \lesssim&\Vert q^{(k)}_{xyy} w \{1 - \chi \}\Vert _{x = x_0} \Vert q^{\langle k \rangle }\Vert _\infty \left[ \Vert \bar{v}^{\langle k \rangle }_{yy} w \{1 - \chi \}\Vert _{x = x_0} \Vert \bar{v}^{\langle k - 1 \rangle }_{y} \Vert _\infty \right. \\&\left. + \Vert \bar{v}^{\langle k \rangle }\Vert _\infty \times \Vert \bar{v}^{\langle k - 1 \rangle }_{yyy} w \{1 - \chi \}\Vert _{x = x_0} \right] \\ \lesssim&o_L(1) p_k \Vert q \Vert _{X_{\langle k \rangle }}^2, \\ \int _x |\varUpsilon _{2,5}| \lesssim&|\left( q^{(k)}_{xyy}, q^{\langle k \rangle } \{1 - \chi \} \{1 - \chi \}' \left[ \bar{u}^{\langle k \rangle } \bar{v}^{\langle k \rangle }_{yy} + \bar{v}^{\langle k \rangle } \bar{u}^{\langle k \rangle }_{yy} \right] \right) | \\ \lesssim&\, \Vert q^{(k)}_{xyy}\Vert _{loc} \Vert q^{\langle k \rangle }\Vert _{\infty , loc} \left[ \Vert \bar{v}_{y}^{\langle k - 1\rangle }\Vert _{\infty , loc} \Vert \bar{v}^{\langle k \rangle }_{yy}\Vert _{loc}\right. \\&\left. +\, \Vert \bar{v}^{\langle k \rangle }\Vert _{loc} \Vert \bar{v}^{\langle k - 1 \rangle }_{yyy}\Vert _{\infty , loc} \right] \\ \lesssim&\,\, o_L(1) p_k \Vert q \Vert _{X_{\langle k \rangle }}^2. \end{aligned}$$

We now estimate the following:

$$\begin{aligned} \int _x |\varUpsilon _3| \lesssim&\, |\left( q^{(k)}_{xyy} \bar{u}^j_{yyy}, q^{(k-j)}_y \bar{u}^2 w^2 \{1 - \chi \}^2\right) | + |\left( q^{(k)}_{xyy} \bar{u}^j_{yy}, q^{(k-j)}_{yy} \bar{u}^2 w^2 \{1 - \chi \}^2\right) | \\&+\, |( q^{(k)}_{xyy}, \bar{u}^j_{yy} q^{(k-j)}_y \bar{u} \bar{u}_{y} w^2 \{1 - \chi \}^2)| + |\left( q^{(k)}_{xyy}, \bar{u}^j_{yy} q^{(k-j)}_y \bar{u}^2 ww_{y} \{1 - \chi \}^2\right) | \\&+\, |\left( q^{(k)}_{xyy}, \bar{u}^j_{yy} q^{(k-j)}_y \bar{u}^2 w^2 \{1 - \chi \} \{1 - \chi \}' \right) | \\ \lesssim&o_L(1) p_{k} \Vert q \Vert _{X_{\langle k \rangle }}^2. \end{aligned}$$

Above, we have used the estimates which proceed below, with the use of the identities (105) and Lemma 9:

$$\begin{aligned} \int _x |\varUpsilon _{3,1}| \lesssim&|\left( q^{(k)}_{xyy}, q^{\langle k \rangle }_y \bar{u}^2 w^2 \{1 - \chi \}^2 \left[ \bar{u}^{\langle k \rangle } \bar{v}^{\langle k \rangle }_{yy} + \bar{v}^{\langle k \rangle } \bar{u}^{\langle k \rangle }_{yy} \right] \right) | \\ \lesssim&\, \Vert q^{(k)}_{xyy} w \{1 - \chi \}\Vert _{x = x_0} |q^{\langle k \rangle }_y w\{1 - \chi \}\Vert _{x = x_0} \left[ \Vert \bar{v}_{y}^{\langle k - 1 \rangle }\Vert _\infty \Vert \bar{v}^{\langle k - 1 \rangle }_{yy} \Vert _\infty \right. \\&\left. +\, \Vert \bar{v}^{\langle k \rangle } \Vert _\infty \Vert \bar{v}_{yy}^{\langle k- 1\rangle } \Vert _\infty \right] \\ \int _x |\varUpsilon _{3,2}| \lesssim&\Vert q^{(k)}_{xyy} w \{1 - \chi \}\Vert _{x = x_0} |q^{\langle k \rangle } \{1 - \chi \} w \langle y \rangle ^{-1}\Vert _{x = x_0} \Vert \bar{v}_{yyy}^{\langle k - 1 \rangle } \langle y \rangle \Vert _\infty \\ \int _x |\varUpsilon _{3,3}| \lesssim&\Vert q^{(k)}_{xyy} w \{1 - \chi \}\Vert _{x = x_0} |q^{\langle k \rangle } \{1 - \chi \} w \langle y \rangle ^{-2}\Vert _{x = x_0} \Vert \bar{v}_{yyy}^{\langle k - 1 \rangle } \Vert _\infty \Vert \bar{u}_{y} \langle y \rangle ^2 \Vert _\infty \\ \int _x |\varUpsilon _{3,4}|\lesssim&\Vert q^{(k)}_{xyy} w \{1 - \chi \}\Vert _{x = x_0} \Vert \bar{v}_{yyy}^{\langle k - 1 \rangle } \Vert _\infty |q_y^{\langle k \rangle } w \{1 - \chi \}\Vert _{x = x_0} \\ \int _x |\varUpsilon _{3,5}| \lesssim&\Vert q^{(k)}_{xyy}\Vert _{x = x_0, loc} \Vert q_y^{\langle k \rangle }\Vert _{x = x_0,loc} \Vert \bar{v}_{yyy}^{\langle k - 1 \rangle } \Vert _{\infty , loc}. \end{aligned}$$

We now move to, using Lemma 9 to control the $\bar{v}$ contribution:

$$\begin{aligned} \int _x |\varUpsilon _{4}|\lesssim&\, \int _x \left( q^{(k)}_{xyy}, \bar{u}^j_{yy} q_{yy}^{k-j} \bar{u}^2 w^2 \{1 - \chi \}^2\right) + \int _x \left( q^{(k)}_{xyy}, \bar{u}_{y}^j q_{yyy}^{k-j} \bar{u}^2 w^2 \{1 - \chi \}^2\right) \\&+\, | \int _x \left( q^{(k)}_{xyy}, \bar{u}_{y}^j q_{yy}^{k-j} \bar{u} \bar{u}_{y} w^2 \{1 - \chi \}^2\right) + \int _x \left( q^{(k)}_{xyy}, \bar{u}_{y}^j q_{yy}^{k-j} \bar{u}^2 ww_{y} \{1 - \chi \}^2\right) \\&+ \, \int _x \left( q^{(k)}_{xyy}, \bar{u}_{y}^j q_{yy}^{k-j} \bar{u}^2 w^2 \{1 - \chi \} \chi '\right) \\ \lesssim&\int _x \left[ \Vert q^{(k)}_{xyy} w \{1 - \chi \}\Vert |q_{yy}^{\langle k \rangle } w \{1 - \chi \} \Vert _{x = x_0} \Vert \bar{v}^{\langle k - 1 \rangle }_{yy} \Vert _\infty \right. \\&+ \Vert q^{(k)}_{xyy} w \{1 - \chi \}\Vert |q_{yyy}^{\langle k \rangle } w \{1 - \chi \}\Vert _{x = x_0} \Vert \bar{v}^{\langle k- 1 \rangle }_{yy} \Vert _\infty \\&+\, \Vert q^{(k)}_{xyy} w \{1 - \chi \}\Vert _{x = x_0} \Vert q_{yy}^{\langle k \rangle } w \{1 - \chi \}\Vert _{x = x_0} \Vert \bar{v}_{yy}^{\langle k - 1 \rangle } \Vert _\infty \Vert \bar{u}_{y} \Vert _\infty \\&+ \,\Vert q^{(k)}_{xyy} w \{1 - \chi \}\Vert _{x = x_0} \Vert q_{yy}^{\langle k \rangle } w \{1 - \chi \}\Vert _{x = x_0} \Vert \bar{v}_{yy}^{\langle k - 1 \rangle } \Vert _\infty \\&\left. + \Vert q^{(k)}_{xyy}\Vert _{x = x_0,loc} \Vert q^{\langle k \rangle } \Vert _{x = x_0,loc} \Vert \bar{v}_{yy}^{\langle k - 1 \rangle } \Vert _{\infty , loc} \right] \\ \lesssim&o_L(1) p_k \Vert q \Vert _{X_{\langle k \rangle }}^2. \end{aligned}$$

Next, since $\bar{u}_y$ is bounded by Theorem 1,

$$\begin{aligned} \int _x |\varUpsilon _{5}| \lesssim&\, | \int _x \left( v^{(k)}_{yyy}, q^{(k)}_{xyy} \left\{ \bar{u} \bar{u}_{y} w^2 \{1 - \chi \}^2 + \bar{u}^2 ww_{y} \{1 - \chi \}^2 + \bar{u}^2 w^2 \{1 - \chi \} \chi ' \right\} \right) | \\ \lesssim&\int _{x} \left[ \Vert q^{(k)}_{xyy} w \{1 - \chi \}\Vert _{x = x_0} \Vert v^{(k)}_{yyy} w \{1 - \chi \} \Vert _{x = x_0} \Vert \bar{u}_{y}\Vert _\infty \right. \\&+ \,\Vert q^{(k)}_{xyy} w \{1 - \chi \}\Vert _{x = x_0} \Vert v^{(k)}_{yyy} w \{1 - \chi \}\Vert _{x = x_0} \\&\left. +\, \Vert q^{(k)}_{xyy} w|_{x = x_0,loc} \Vert v^{(k)}_{yyy} \{1 - \chi \}|_{x = x_0,loc} \right] \\ \lesssim&\, o_L(1) \Vert q \Vert _{X_{\langle k \rangle }}^2. \end{aligned}$$

To conclude, we have

$$\begin{aligned} \int _x |\varUpsilon _{6}| \,=\,&|- \sum _{j =1}^k \int _x |\left( {\begin{array}{c}k\\ j\end{array}}\right) \left( q^{(k)}_{xyy}, \partial _y \left\{ \bar{u}^2 \partial _x^j \bar{u} \partial _x^{k-j} q_{yyy} w^2 \{1 - \chi \}^2 \right\} \right) _{x = x_0}| \\ \lesssim&\, \int _x \Vert q^{(k)}_{xyy} w \{1 - \chi \}\Vert _{x = x_0} \Vert \bar{v}_{yy}^{\langle k - 1 \rangle } \Vert _\infty \Vert q^{\langle k - 1 \rangle }_{yyyy} \{1 - \chi \} w \Vert _{x = x_0} \\ \lesssim \,&\, o_L(1) p_k \Vert q \Vert _{X_{\langle k \rangle }}^2, \end{aligned}$$

all of which are acceptable contributions due to the cut-off $\{1 - \chi \}$.

This now concludes our treatment (104) and consequently the left-hand side of (103). We now move to the terms from the right-hand side of (103), upon using Lemma 9 for the $\bar{v}$ contribution,

$$\begin{aligned}&\,\int _x \Vert \partial _{xy} \{ \partial _x^{j_1}\bar{u} \partial _x^{j_2} \bar{u} \partial _x^{k-j} q_y \} w \{1 - \chi \}(x) \Vert _{x = x_0}^2 \\&\quad \lesssim \,\int _x \Vert \left[ \partial _x^{\langle k + 1 \rangle }\bar{u}_{y} \partial _x^{\langle k \rangle }\bar{u} \partial _x^{\langle k -1 \rangle } q_y + \partial _x^{\langle k +1 \rangle } \bar{u} \partial _x^{\langle k \rangle }\bar{u}_{y} \partial _x^{\langle k - 1 \rangle } q_y \right. \\&\qquad +\, \partial _x^{\langle k \rangle } \bar{u} \partial _x^{\langle k \rangle } \bar{u} \partial _x^{\langle k -1 \rangle } q_{yy} + \partial _x^{\langle k \rangle } \bar{u} \partial _x^{\langle k \rangle } \bar{u}_{y} \partial _x^{\langle k \rangle } q_y \\&\qquad \left. +\, |\partial _x^{\langle k \rangle } \bar{u}|^2 \partial _x^{\langle k \rangle } q_{yy} \right] w \{1 - \chi \}(x) \Vert _{x = x_0}^2 \\&\quad \lesssim \,\int _x \left[ \Vert \partial _x^{\langle k - 1 \rangle } \bar{v}_{y} \Vert _\infty \Vert \partial _x^{\langle k \rangle } \bar{v}_{yy} w \{1 - \chi \} \Vert _{x = x_0} \Vert \partial _x^{\langle k - 1 \rangle } q_y w \{1 - \chi \} \Vert _{\infty }^2 \right. \\&\qquad +\, \Vert \partial _x^{\langle k - 1 \rangle } \bar{v}_{yy} \Vert _\infty ^2 \Vert \partial _x^{\langle k \rangle } \bar{v}_{y} \Vert _\infty ^2 \Vert \partial _x^{\langle k - 1 \rangle } q_y w \{1 - \chi \} \Vert _{x = x_0}^2 \\&\qquad +\, \Vert \partial _x^{\langle k - 1 \rangle } \bar{v}_{y} \Vert _\infty ^2 \Vert \partial _x^{\langle k - 1 \rangle } q_{yy} w \{1 - \chi \}\Vert _{x = x_0}^2\\&\qquad + \Vert \partial _x^{\langle k-1 \rangle } \bar{v}_{y} \Vert _\infty \Vert \partial _x^{\langle k - 1 \rangle } \bar{v}_{yy} \Vert _\infty ^2 \Vert \partial _x^{\langle k \rangle } q_y w \{1 - \chi \} \Vert _{x = x_0}^2 \\&\qquad \left. +\, \Vert \partial _x^{\langle k - 1 \rangle } \bar{v}_{y} \Vert _\infty ^2 |\partial _x^{\langle k \rangle } q_{yy} w \{1 - \chi \} \Vert _{x = x_0}^2 \right] \\&\quad \lesssim \, o_L(1) p_k \Vert q \Vert _{X_{\langle k \rangle }}^2. \end{aligned}$$

We now move to the $\varLambda$ terms:

$$\begin{aligned} \partial _x^k \varLambda \,=\, \sum _{j = 0}^k \bar{v}^j_{xyy} I_x\left[ v^{(k-j)}_y\right] + \bar{v}^j_{yy} v^{(k-j)}_{y} - \bar{v}^j_{x} I_x\left[ v^{(k-j)}_{yyy}\right] - \bar{v}^j v^{(k-j)}_{yyy} \end{aligned}$$

We estimate directly:

$$\begin{aligned}&\Vert \bar{v}^j_{xyy} I_x\left[ v^{(k-j)}_y\right] w \{1 - \chi \}(x) \Vert _{x = x_0}^2 \lesssim \,\Vert \bar{v}^{\langle k \rangle }_{xyy} w \{1 - \chi \}(x)\Vert _{x = x_0}^2 \Vert v_y^{\langle k \rangle } \Vert _\infty ^2 \\&\quad \,\Vert \bar{v}^j_{yy} v^{(k-j)}_y w \{1 - \chi \}(x) \Vert _{x = x_0}^2 \lesssim \Vert \bar{v}^{\langle k \rangle }_{xyy} w \{1 - \chi \}(x) \Vert _{x = x_0}^2 \Vert v_y^{\langle k \rangle } \Vert _\infty ^2\\&\quad \Vert |\bar{v}^j_{x} I_x[v^{(k-j)}_{yyy} \{1 - \chi \} w (x) \Vert _{x = x_0}^2 \lesssim \Vert \bar{v}^{\langle k +1 \rangle } \Vert _\infty \Vert v^{\langle k \rangle }_{yyy} w \{1 - \chi \} (x) \Vert _{x = x_0}^2 \\&\quad \Vert \bar{v}^j |v^{(k-j)}_{yyy} \{1 - \chi \} w(x) \Vert _{x = x_0}^2 \lesssim \Vert \bar{v}^{\langle k +1 \rangle } \Vert _\infty \Vert v^{\langle k \rangle }_{yyy} w \{1 - \chi \} (x) \Vert _{x = x_0}^2. \end{aligned}$$

Upon integrating in x, the above terms are majorized by $o_L(1) p(\Vert \bar{q} \Vert _{X_{\langle k+ 1 \rangle }}) (1 + \Vert q \Vert _{X_k} )$, upon applying Lemma 9 for $k+1$. We now move to the $U(u^0)$ terms:

$$\begin{aligned} \int |\partial _x^k U(u^0)|^2 w^2 \{1 - \chi \}^2 \le&\int \left[ |\bar{v}^k_{xyy}|^2 |u^0|^2 + |\bar{v}^k_{x}|^2 |u^0_{yy}|^2 \right] w^2 \{1 - \chi \}^2 \\ \le&\Vert u^0 \Vert _\infty ^2 \Vert \bar{v}^k_{xyy} w \{1 - \chi \}\Vert _{x = x_0}^2 + \Vert u^0_{yy} w \{1 - \chi \}\Vert ^2 \Vert \bar{v}^k_{x}\Vert _\infty ^2. \end{aligned}$$

Integrating, the above is majorized by $C(u^0) o_L(1) p(\Vert \bar{q} \Vert _{X_{\langle k +1 \rangle }})$, upon applying Lemma 9 for $k+1$. Similarly, the g contributions are clearly estimated via $\Vert \partial _{xy} g_1^{k} \{1 - \chi \} w \Vert ^2$. $\square$

Proposition 2

For $k \ge 0$, and let q solve (36). Then:

$$\begin{aligned} \begin{aligned} \Vert q \Vert _{X_{ k }} \lesssim&C(q_0) + \Vert \partial _x^k \partial _{xy}g_1 w \Vert ^2 + o_L(1) \Vert \partial _x^k \partial _x \partial _{xy}g_1 \langle y \rangle \Vert ^2 + \kappa o_L(1) C(u^0) \\&+ o_L(1)\left( p\left( \Vert \bar{q} \Vert _{X_{\langle k \rangle }}\right) + \kappa p\left( \Vert \bar{q} \Vert _{X_{\langle k + 1 \rangle }}\right) \right) \left( C(q_0) + \Vert q \Vert _{X_{k}}^2 \right) . \end{aligned} \end{aligned}$$

(104)

Proof

We add together (91), a small multiple of (99) and (101). On the left-hand side, this produces

$$\begin{aligned} \begin{aligned}&\sup \left[ |q^{(k)}_{yyy}w \{1-\chi \}|^2 + |\bar{u} q^{(k)}_{xy}|^2\right] +\Vert v^{(k)}_{yyyy} w \{1-\chi \} \Vert ^2 \\&\quad + \Vert v^{(k)}_{yyyy} \Vert _{loc} + \Vert q^{(k)}_{xyy} w \{1-\chi \} \Vert ^2 + \Vert q^{(k)}_{xyy} \sqrt{\bar{u}} \Vert ^2, \end{aligned} \end{aligned}$$

(105)

which can clearly be combined to majorize $\Vert q^{(k)} \Vert _X$. On the right-hand side

$$\begin{aligned} \begin{aligned}&\Vert \partial _{xy}\partial _x^k g_1 w \{1-\chi \} \Vert ^2 + C(q_0) + \kappa o_L(1) C(u^0) + o_L(1)(p\left( \Vert \bar{q} \Vert _{X_k}\right) \\&\quad + \kappa p\left( \Vert \bar{q} \Vert _{X_{\langle k+ 1 \rangle }}\right) \left( C\left( q_0\right) + \Vert q^{(k)} \Vert _X^2\right) + o(1) \Vert q^{(k)} \Vert _{\mathcal {E}} \\&+ o_L(1) \Vert \partial _{xxy}\partial _x^k g_1 \langle y \rangle \Vert ^2 + |\bar{u} q_{xy}^k(0,\cdot )|^2 \end{aligned} \end{aligned}$$

(106)

Of these, the $o(1) \Vert q^{(k)} \Vert _{\mathcal {E}}$ term is absorbed to the left-hand side. Finally, the initial value $|\bar{u} q^{(k)}_{xy}(0,\cdot )|^2$ is obtained through (51). $\square$

We can upgrade to higher y regularity by using the equation. In this direction, we establish the following lemma:

Lemma 13

Let q solve (36). Then the following inequality is valid for any $k \ge 0$,

$$\begin{aligned} \Vert \partial _y^{j} v \Vert \lesssim \Vert \partial _y^{j - 4} F \Vert + p_{\langle k \rangle } \Vert q \Vert _{X_{\langle k \rangle }} + C(u^0), \qquad 0 \le j \le 4 + 2k. \end{aligned}$$

(107)

Proof

We will first address the case of $k = 1, j = 5$. We take $\partial _y$ of Eq. (60) to obtain by using simply the definitions (37)–(38),

$$\begin{aligned} \Vert \partial _y^5 v \Vert \le&\Vert F_y \Vert + \kappa o_L(1) \Vert \partial _y U\left( u^0\right) \Vert _{L^2_y} + \kappa \Vert \partial _y \varLambda (v) \Vert + \Vert \partial _{xyy} \left( \bar{u}^2 q_y\right) \Vert \nonumber \\ \lesssim&\Vert F_y \Vert + \kappa o_L(1) C\left( u^0\right) + \kappa p_{\langle 1 \rangle } \Vert q \Vert _{X_{\langle 1 \rangle }} + \Vert \partial _{xyy}\left( \bar{u}^2 q_y\right) \Vert . \end{aligned}$$

(108)

We next expand using the product rule and estimate using Lemma 9,

$$\begin{aligned} \Vert \partial _{xyy}\left( \bar{u}^2 q_y\right) \Vert \lesssim&\Vert 2 \bar{u} \bar{u}_x q_{yyy} \Vert + \Vert \bar{u}^2 q_{xyyy} \Vert + \Vert \partial _y^2\left( \bar{u}^2\right) q_{xy} \Vert \nonumber \\&+ \Vert \partial _{xyy}\left( \bar{u}^2\right) q_y \Vert + \Vert \left( \bar{u} \bar{u}_y\right) _x q_{yy} \Vert + \Vert \bar{u} \bar{u}_y q_{xyy} \Vert \nonumber \\ \lesssim&\Vert q \Vert _{X_{\langle 1 \rangle }} \end{aligned}$$

(109)

It is clear that we can upgrade to higher y regularity by iterating the above. $\square$

We now come to the proofs of two of our main results.

Proof of Proposition 1

Proposition 1 is a direct consequence of Proposition 2 and Lemma 13. $\quad \square$

Proof of Theorem 2

We begin by reformulating the Prandtl Eq. (1) into the D-Prandtl system, analogous to (36), which produces ($U = \varLambda = f = 0$)

$$\begin{aligned} - \partial _{xy} \left\{ \bar{u}^2 q_y \right\} + \partial _y^4 v = 0, \quad q := \frac{v}{\bar{u}}. \end{aligned}$$

(110)

From here, Proposition 2 is applied with $g_1 = 0, \kappa = 0$, and $q = \bar{q}$ to give

$$\begin{aligned} \Vert q \Vert _{X_k} = \Vert \bar{q} \Vert _{X_k} \lesssim C\left( q_0\right) \lesssim C\left( \bar{U}_0\right) . \end{aligned}$$

Above, we have used that the constant $C(q_0)$ depends on $\Vert \bar{u} q_{xy} \Vert _{x = 0}$. From (112), we obtain

$$\begin{aligned} - \bar{u} q_{xy} = - \frac{v_{yyy}}{\bar{u}} + 2 \bar{u}_x q_y, \end{aligned}$$

(111)

from which

$$\begin{aligned} \Vert \bar{u} q_{xy} \Vert _{x = 0} \le \Vert \frac{1}{\bar{u}} v_{yyy} \Vert _{x = 0} + \Vert 2 \bar{u}_x q_{y} \Vert _{x = 0} \le C(\bar{U}_0). \end{aligned}$$

(112)

It is important to note that $v_{yyy}|_{x = 0}(0) = 0$ from $\bar{u}_{yy}|_{y = 0} = 0$ according to the Prandtl Eq. (1). Above, we use that the quantities $\partial _y^N v|_{x= 0}$ for any $N \ge 0$ is determined according to the initial data, $u^0_p|_{x = 0}$. Hence, (112) becomes

$$\begin{aligned} \Vert q \Vert _{X_0} = \Vert \bar{q} \Vert _{X_0} \le C(u^0). \end{aligned}$$

This concludes the proof. $\square$

3 Construction of Approximate Navier–Stokes Solution

3.1 Specification of equations

We will assume the expansions:

$$\begin{aligned} U^\epsilon = \tilde{u}^n_s + \epsilon ^{N_0} u, \quad V^\epsilon = \tilde{v}^n_s + \epsilon ^{N_0} v, \quad P^\epsilon = \tilde{P}^n_s + \epsilon ^{N_0} P. \end{aligned}$$

(113)

We will denote the partial expansions:

$$\begin{aligned} u_s^i =&\sum _{j = 0}^i \sqrt{\epsilon }^j u^j_e + \sum _{j = 0}^{i-1} \sqrt{\epsilon }^j u^j_p, \qquad \tilde{u}_s^i = u_s^i + \sqrt{\epsilon }^i u^i_p, \end{aligned}$$

(114)

$$\begin{aligned} v_s^i =&\sum _{j = 1}^i \sqrt{\epsilon }^{j-1} v^j_e + \sum _{j = 0}^{i-1} \sqrt{\epsilon }^j v^j_p, \qquad \tilde{v}_s^i = v_s^i + \sqrt{\epsilon }^i v^i_p, \end{aligned}$$

(115)

$$\begin{aligned} P^i_s =&\sum _{j = 0}^i \sqrt{\epsilon }^j P^j_e, \qquad \tilde{P}_s^i = P_s^i + \sqrt{\epsilon }^i \left\{ P^i_p + \sqrt{\epsilon } P^{i,a}_p \right\} . \end{aligned}$$

(116)

We will also define $u^{E,i}_s = \sum _{j = 0}^i \sqrt{\epsilon }^j u^j_e$ to be the “Euler” components of the partial sum. Similar notation will be used for $u^{P,i}_s, v^{E,i}_s, v^{P,i}_s$. The following will also be convenient:

$$\begin{aligned} \begin{aligned} u_s^E :=&\sum _{i =0}^n \sqrt{\epsilon }^i u^i_e, \quad v_s^E := \sum _{i = 1}^n \sqrt{\epsilon }^{i-1} v^i_e, \\ u_s^P :=&\sum _{i = 0}^n \sqrt{\epsilon }^i u^i_p, \quad v_s^P := \sum _{i = 0}^n \sqrt{\epsilon }^i v^i_p, \\ u_s =&u_s^P + u_s^E, \quad v_s = v_s^P + v_s^E. \end{aligned} \end{aligned}$$

(117)

The $P^{i,a}_p$ terms are “auxiliary Pressures” in the same sense as those introduced in [27] and [31] and are for convenience. We will also introduce the notation:

$$\begin{aligned} \bar{u}^i_p := u^i_p - u^i_p|_{y = 0}, \qquad \bar{v}^i_p := v^i_p - v^i_p(x,0), \qquad \bar{v}^i_e = v^i_e - v^i_e|_{Y = 0}. \end{aligned}$$

(118)

We first record the properties of the leading order $(i = 0)$ layers. For the outer Euler flow, we will take a shear flow, $[u^0_e(Y), 0, 0]$. The derivatives of $u^0_e$ decay rapidly in Y and that is bounded below, $|u^0_e| \gtrsim 1$.

For the leading order Prandtl boundary layer, the equations are given in (1), for the i’th Euler layer, $i \ge 1$, the equations are given by (16), whereas for the i’th Prandtl layer the equations are given by (17).

The relevant definitions of the forcing terms in those equations are given below. Note that as a matter of convention, summations that end with a negative number are empty sums.

Definition 5

(Forcing Terms).

$$\begin{aligned}&-f^i_{E,1} := u^{i-1}_{ex} \sum _{j = 1}^{i-2} \sqrt{\epsilon }^{j-1} \{u^j_e + u^j_p(x,\infty ) + u^{i-1}_e \sum _{j = 1}^{i-2} \sqrt{\epsilon }^{j-1} u^j_{ex} \\&\quad + \sqrt{\epsilon }^{i-2}[ \{u^{i-1}_{e} + u^{i-1}_p(x,\infty ) \} u^{i-1}_{ex} + v^{i-1}_e u^{i-1}_{eY}] \\&\quad + u^{i-1}_{eY} \sum _{j = 1}^{i-2} \sqrt{\epsilon }^{j-1} v^j_e + v^{i-1}_e \sum _{j = 1}^{i-2} \sqrt{\epsilon }^{j-1} u^j_{eY} - \sqrt{\epsilon } \varDelta u^{i-1}_e, \\&-f^i_{E,2} := v^{i-1}_{eY} \sum _{j = 1}^{i-2} \sqrt{\epsilon }^{j-1} v^j_e + v_e^{i-1} \sum _{j = 1}^{i-2} \sqrt{\epsilon }^{j-1} v^j_{eY} + \sqrt{\epsilon }^{i-2}\left[ v^{i-1}_e v^{i-1}_{eY} + u^{i-1}_e v^{i-1}_{ex}\right] \\&\quad + \left\{ u_e^{i-1} + u^{i-1}_p(x,\infty )\right\} \sum _{j =1}^{i-2} \sqrt{\epsilon }^{j-1} v^j_{ex} + v^{i-1}_{ex} \sum _{j = 1}^{i-2} \sqrt{\epsilon }^{j-1} \left\{ u^j_e + u^j_p(x,\infty ) \right\} \\&\quad - \sqrt{\epsilon } \varDelta v^{i-1}_e, \\&-f^{(i)} := \sqrt{\epsilon } u^{i-1}_{pxx} + \epsilon ^{-\frac{1}{2}} \{ v^i_e - v^i_e(x,0) \} u^0_{py} + \epsilon ^{-\frac{1}{2}} \{ u^0_e - u^0_e(0) \} u^{i-1}_{px} + \epsilon ^{-\frac{1}{2}} \{ u^{P, i-1}_{sx} \\&\quad - \bar{u}^0_{sx} \} u^{i-1}_p + \epsilon ^{-\frac{1}{2}} \{ u^{E, i-1}_{sx} - \bar{u}^0_{sx} \} \{u^{i-1}_p - u^{i-1}_p(x,\infty ) \} + \epsilon ^{-\frac{1}{2}} v^{i-1}_p \{ \bar{u}^{i-1}_{sy} \\&\quad - u^0_{py} \} + u^{i-1}_{px} \sum _{j =1 }^{i-1} \sqrt{\epsilon }^{j-1}(u^j_e + u^j_p) + \epsilon ^{-\frac{1}{2}} (v_s^{i-1} - v_s^1) u^{i-1}_{py} + \epsilon ^{-\frac{1}{2}} (v^1_e \\&\quad - v^1_e(x,0)) u^{i-1}_{py} + \sqrt{\epsilon } u^i_{eY} \sum _{j = 0}^{i-1} \sqrt{\epsilon }^j v^j_p + v^i_e \sum _{j = 1}^{i-1} \sqrt{\epsilon }^{j-1} u^j_{py} + u^i_{ex} \sum _{j = 0}^{i-1} \sqrt{\epsilon }^j \{u^j_p \\&\quad - u^j_p(x,\infty ) \} + u^i_e \sum _{j = 0}^{i-1} \sqrt{\epsilon }^j u^j_{px} + \int _y^\infty \partial _x \{ \sqrt{\epsilon }^2 u^i_e \sum _{j = 0}^{i-1} \sqrt{\epsilon }^j v^j_{px} + \sqrt{\epsilon } v^i_{ex} \\&\quad \times \sum _{j = 0}^{i-1} \sqrt{\epsilon }^j \{u^j_p - u^j_p(x,\infty )\} + \sqrt{\epsilon }^2 v^i_{eY} \sum _{j = 0}^{i-1} \sqrt{\epsilon }^j v^j_p + \sqrt{\epsilon } v^i_e \sum _{j = 0}^{i-1} \sqrt{\epsilon }^j v^j_{py} \\&\quad + \sqrt{\epsilon } v^{i-1}_s v^{i-1}_{py} + \sqrt{\epsilon } v_{sy}^{i-1} v^{i-1}_p + \sqrt{\epsilon } v^{E,i-1}_{sx} \{u^{i-1}_p - u^{i-1}_p(x,\infty )\} \\&\quad + \sqrt{\epsilon } v^{P,i-1}_{sx} u^{i-1}_{p} + \sqrt{\epsilon } u_s^{i-1} v^{i-1}_{px} + \sqrt{\epsilon } \varDelta _\epsilon v^{i-1}_p + \sqrt{\epsilon }^i \{ u^{i-1}_p v^{i-1}_{px} + v^{i-1}_p v^{i-1}_{py} \} \} \,\mathrm {d}z. \end{aligned}$$

For $i = 1$ only, we make the following modifications. The aim is to retain only the required order $\sqrt{\epsilon }$ terms into $f^{(1)}$. $f^{(2)}$ will then be adjusted by including the superfluous terms. Define:

$$\begin{aligned} \begin{aligned} f^{(1)} :=&- u^0_p u^1_{ex}|_{Y = 0} - u^0_{px} u^1_e|_{Y = 0} - \bar{u}^0_{eY}(0) y u^0_{px} - v^0_p u^0_{eY} - v^1_{eY}(0) y u^0_{py}. \end{aligned} \end{aligned}$$

(119)

For the final Prandtl layer, we must enforce the boundary condition $v^n_p|_{y = 0} = 0$. Define the quantities $[u_p, v_p, P_p]$ to solve

$$\begin{aligned} \left. \begin{aligned}&\bar{u} \partial _x u_p + u_p \partial _x \bar{u} + \partial _y \bar{u} v_p + \bar{v} \partial _y u_p + \partial _x P_p - \partial _{yy} u_p := f^{(n)}, \\&\partial _x u_p + \partial _y v_p = 0, \qquad \partial _y P^i_p = 0\\&\left[ u_p, v_p\right] |_{y = 0} = \left[ -u^n_e, 0\right] |_{y = 0}, \qquad u_p|_{y \rightarrow \infty } = 0 \qquad v_p|_{x = 0} = V_P^n. \end{aligned} \right\} \end{aligned}$$

(120)

Note the change in boundary condition of $v_{p}|_{y = 0} = 0$ which contrasts the $i = 1,\ldots ,n-1$ case. This implies that $v_p = \int _0^y u_{px} \,\mathrm {d}y'$. For this reason, we must cut-off the Prandtl layers:

$$\begin{aligned} u^n_p :=&\chi \left( \sqrt{\epsilon }y\right) u_p + \sqrt{\epsilon } \chi '\left( \sqrt{\epsilon }y\right) \int _0^y u_p(x, y') \,\mathrm {d}y', \\ v^n_p :=&\chi \left( \sqrt{\epsilon }y\right) v_p. \end{aligned}$$

Here $\mathcal {E}^n$ is the error contributed by the cut-off:

$$\begin{aligned} \mathcal {E}^{(n)} := \bar{u} \partial _x u^{n}_{p} + u^n_p \partial _x \bar{u} +\bar{v} \partial _y u^n_{p} + v^n_p \partial _y \bar{u} - u^n_{pyy} - f^{(n)}. \end{aligned}$$

Computing explicitly:

$$\begin{aligned} \mathcal {E}^{(n)} :=&(1-\chi ) f^{(n)} + \bar{u} \sqrt{\epsilon } \chi '\left( \sqrt{\epsilon }y\right) v_p(x,y) + \bar{u}_{x} \sqrt{\epsilon } \chi ' \int _0^y u_p \nonumber \\&+ \bar{v} \sqrt{\epsilon } \chi ' u_p + \epsilon \bar{v} \chi '' \int _0^y u_p + \sqrt{\epsilon } \chi ' u_p \nonumber \\&+ \epsilon ^{\frac{3}{2}} \chi ''' \int _0^y u_p + 2\epsilon \chi '' u_p + \sqrt{\epsilon } \chi ' u_{py}. \end{aligned}$$

(121)

We will now define the contributions into the next order, which will serve as the forcing for the remainder term:

$$\begin{aligned}&{f}^{(n+1)} := \sqrt{\epsilon }^n \left[ \epsilon u^n_{pxx} + v^n_p\left\{ \bar{u}^n_{sy} - u^0_{py} \right\} + \left\{ u^0_e - u^0_e(0) \right\} u^n_{px} \right. \nonumber \\&\quad + u^n_{px} \sum _{j = 1}^n \sqrt{\epsilon }^j \left( u^j_e + u^j_p\right) + \left\{ u^n_{sx} - \bar{u}^0_{sx} \right\} u^n_p + \left( v^n_s - v^1_s\right) u^n_{py} \nonumber \\&\quad \left. + \left\{ v^1_e - v^1_e(x,0) \right\} u^n_{py} \right] + \sqrt{\epsilon }^n \mathcal {E}^{(n)} + \sqrt{\epsilon }^{n+2} \varDelta u^n_e \nonumber \\&\quad + \sqrt{\epsilon }^n u^n_{ex} \sum _{j = 1}^{n-1} \sqrt{\epsilon }^j u^j_e + \sqrt{\epsilon }^n u^n_e \sum _{j = 1}^{n-1} \sqrt{\epsilon }^j u^j_{ex} + \sqrt{\epsilon }^{2n} [ u^n_e u^n_{ex} \nonumber \\&\quad + v^n_e u^n_{eY}] + \sqrt{\epsilon }^{n+1} u^n_{eY} \sum _{j= 1}^{n-1} \sqrt{\epsilon }^{j-1} v^j_e + \sqrt{\epsilon }^{n-1}v^n_e \sum _{j = 1}^{n-1} \sqrt{\epsilon }^{j+1} u^j_{eY}, \end{aligned}$$

(122)

and

$$\begin{aligned}&{g}^{(n+1)} := \sqrt{\epsilon }^n \left[ v_s^n \partial _y v^n_p + \partial _y v_s^n v^n_p + \partial _x v^n_s u^n_p + u^n_s \partial _x v^n_p - \varDelta _\epsilon v^n_p \right. \nonumber \\&\quad \left. + \sqrt{\epsilon }^n \left( u^n_p \partial _x v^n_p + v^n_p \partial _y v^n_p \right) \right] + \left( \sqrt{\epsilon }\right) ^n \partial _Y v^n_e \sum _{j = 1}^{n-1} (\sqrt{\epsilon })^{j-1} v^j_e \nonumber \\&\quad + \sqrt{\epsilon }^{n-1} v^n_e \sum _{j = 1}^{i-1} \sqrt{\epsilon }^j \partial _Y v^j_e + \sqrt{\epsilon }^{2n-1} \left[ v^n_e v^n_{eY} + u^n_e \partial _x v^n_e\right] \nonumber \\&\quad + \sqrt{\epsilon }^n u^n_e \sum _{j =1}^{n-1} \left( \sqrt{\epsilon }\right) ^{j-1}\partial _x v^j_e + \sqrt{\epsilon }^{n-1} \partial _x v^n_e \sum _{j = 0}^{n-1} \sqrt{\epsilon }^j u^j_e + \sqrt{\epsilon }^{n+1} \varDelta v^n_e. \end{aligned}$$

(123)

Finally, we have

$$\begin{aligned} F_R := \epsilon ^{-N_0} \left( \partial _y {f}^{(n+1)} - \epsilon \partial _x {g}^{(n+1)}\right) . \end{aligned}$$

(124)

3.2 Construction of Euler layers

Our starting point is the system (16). Going to vorticity yields the system we will analyze:

$$\begin{aligned} \begin{aligned}&u^0_e \varDelta v^i_e + u^0_{eYY} v^i_e = F^{(i)} := \partial _Y f^i_{E,1} - \partial _x f^i_{E,2}, \\&\quad v^i_e|_{Y = 0} = - v^{i-1}_p|_{y = 0}, \quad v^i_e|_{x = 0,L} = V^i_{E, \{0, L\}}, \quad u^i_e|_{x = 0} = U^i_{E,0}. \end{aligned} \end{aligned}$$

(125)

The data for $u^i_e|_{x = 0}$ is required because $u^i_e = u^i_e|_{x = 0} - \int _0^x v^i_{eY}$ will be recovered through the divergence free condition upon constructing $v^i_e$.

We will quantify the decay rates as $Y \uparrow \infty$ for the quantities $V^i_{E,{0, L}}$ and $F^{(i)}$.

Definition 6

In the case of $i = 1$, define $w_{m_1} = Y^{m_1}$ if $|v^1_{e}|_{x = 0}| \lesssim Y^{-m_1}$ or $w_{m_1} = e^{m_1 Y}$ if $|v^1_e|_{x = 0}| \lesssim e^{-m_1 Y}$ as $Y \uparrow \infty$. This now fixes whether or not $w_{m}$ will refer to polynomial or exponential growth rates. For other layers, we will assume:

$$\begin{aligned} \begin{aligned}&|V^i_{E,\{0, L\}}| \lesssim w_{m_i}^{-1} \text { for } m_i>> m_1 \\&\quad |F^{(i)}| \lesssim w_{l_i}^{-1} \text { for some } l_i>> 0. \end{aligned} \end{aligned}$$

(126)

Finally, let $m_i' := \min \{ m_i, l_i \}$.

Define:

$$\begin{aligned} S(x,Y) = \left( 1 - \frac{x}{L}\right) \frac{V_{i,0}(Y)}{v^{i-1}_p(0,0)} v^{i-1}_p(x,0) + \frac{x}{L}\frac{V_{i,L}(Y)}{v^{i-1}_p(L,0)}v^{i-1}_p(x,0), \end{aligned}$$

(127)

and consider the new unknown:

$$\begin{aligned} \bar{v} := v^i_e - S, \end{aligned}$$

which satisfies the Dirichlet problem:

$$\begin{aligned} -u^0_e \varDelta \bar{v} + u^0_{eYY} \bar{v} = F^{(i)} + \varDelta S, \qquad \bar{v}|_{\partial \varOmega } = 0. \end{aligned}$$

(128)

From here, we have for any $m < m_i' - n_0$ for some fixed $n_0$, perhaps large,

$$\begin{aligned} ||v \cdot w_m||_{H^1} \lesssim 1. \end{aligned}$$

(129)

To go to higher-order estimates, we must invoke that the data are well-prepared in the following sense: taking two $\partial _Y^2$ to the system yields:

$$\begin{aligned} \partial _Y^2 v^i_e(0,Y) =&\partial _Y^2 V_{i,0}(Y), \end{aligned}$$

(130)

$$\begin{aligned} \partial _Y^2 v^i_e(L,Y) =&\partial _Y^2 V_{i,L}(Y), \end{aligned}$$

(131)

$$\begin{aligned} \partial _Y^2 v^i_e(x,0) =&\frac{1}{u^0_e(0)} \left\{ v^{i-1}_{pxx}(x,0) + u^0_{eYY}(0) v^{i-1}_p(x,0) + F^{(i)}(x,0) \right\} . \end{aligned}$$

(132)

Our assumption on the data, which are compatibility conditions, ensure:

$$\begin{aligned} \partial _Y^2 V_{i,0}(0) =&\frac{1}{u^0_e(0)} \left\{ v^{i-1}_{pxx}(0,0) + u^0_{eYY}(0) v^{i-1}_p(0,0) + F^{(i)}(0,0) \right\} , \end{aligned}$$

(133)

$$\begin{aligned} \partial _Y^2 V_{i,0}(L) =&\frac{1}{u^0_e(0)} \left\{ v^{i-1}_{pxx}(L,0) + u^0_{eYY}(0) v^{i-1}_p(L,0) + F^{(i)}(L,0) \right\} . \end{aligned}$$

(134)

It is natural at this point to introduce the following definition:

Definition 7

(Well-Prepared Boundary Data) Consider the corner (0, 0). There exists a value of $\Big (\partial _Y^2 v^i_e|_{Y = 0}\Big )|_{x = 0}$ which is obtained by evaluating (134) at $x = 0$. There exists a value of $\Big (\partial _Y^2 v^i_e|_{x = 0} \Big )|_{Y = 0}$ which is obtained by evaluating (132) at $Y = 0$. These two values should coincide. The analogous statement should also hold for the corner (L, 0). In this case, we say that the boundary data are “well-prepared to order 2”. The data are “well-prepared to order 2k” if we can repeat the procedure for $\partial _Y^{2k}$.

We thus have the following system:

$$\begin{aligned}&-u^0_e \varDelta v^1_{eYY} + u^0_{eYY} v^1_{eYY} + \partial _Y^4 u^0_e v^1_e + 2 \partial _Y^3 u^0_e v^1_{eY} \nonumber \\&\quad - 2 u^0_{eY} \varDelta v^1_{eY} - u^0_{eYY} \varDelta v^i_e = \partial _{YY} F^{(i)}. \end{aligned}$$

(135)

We can define another homogenization in the same way:

$$\begin{aligned} S_{(2)}(x,Y) = \left( 1 - \frac{x}{L}\right) \frac{V''_{i,0}(Y)}{\partial _Y^2 v^i_{e}(x,0)} \partial _Y^2 v^{i-1}_p(x,0) + \frac{x}{L}\frac{V''_{i,L}(Y)}{\partial _Y^2 v^{i-1}_p(L,0)}\partial _Y^2 v^{i-1}_p(x,0), \end{aligned}$$

(136)

which is smooth and rapidly decaying by the assumption that the data are well-prepared. Let us consider the system for $\bar{v} := v^1_{eYY} - B_{(2)}$. The first step is to rewrite:

$$\begin{aligned} v^i_{exx} = -v^i_{eYY} + \frac{u^0_{eYY}}{u^0_e} v^i_e + F^i = - \bar{v} + S_{(2)} + F^i. \end{aligned}$$

We can now rewrite the system (137) in terms of $\bar{v}$:

$$\begin{aligned}&-u^0_e \varDelta \bar{v} + u^0_{eYY} \bar{v} + \partial _Y^4 u^0_e v^1_e + 2\partial _Y^3 u^0_e v^1_{eY} \nonumber \\&\quad -2u^0_{eY} \left( \partial _Y\left\{ \bar{v} + S_2 \right\} + \partial _Y \left\{ \bar{v} + S_2 + F^i \right\} \right) \nonumber \\&\quad - u^0_{eYY} F^i = u^0_e \varDelta S_2 + u^0_{eYY} S_2 + \partial _{YY}F^i. \end{aligned}$$

(137)

Obtaining estimates for $\bar{v}$ yields for any $m < m_i' - n_0$:

$$\begin{aligned} ||\bar{v} \cdot w_m||_{H^1} \lesssim 1. \end{aligned}$$

(138)

Translating to the original unknown gives:

$$\begin{aligned} ||v^i_{eYY}, v^1_{eYYx}, v^i_{eYYY} \cdot w_M||_{L^2} \lesssim 1. \end{aligned}$$

(139)

Using the equation and Hardy in Y, we can obtain:

$$\begin{aligned} ||v^i_{exx}, v^i_{exxx}, v^1_{exY}, v^i_{exxY} \cdot w_M||_{L^2} \lesssim 1. \end{aligned}$$

(140)

Thus, we have the full $H^3$ estimate. $u^1_e$ can be recovered through the divergence free condition:

$$\begin{aligned} u^i_e(x,Y) := u^i_e(0,Y) - \int _0^x \partial _Y v^i_e(x', Y) \,\mathrm {d}x'. \end{aligned}$$

(141)

The compatibility conditions can be assumed to arbitrary order by iterating this process, and thus we can obtain:

Proposition 3

There exists a unique solution $v^i_e$ satisfying (127). With $u^i_e$ defined through (143), the tuple $[u^i_e, v^i_e]$ satisfy the system (16). For any $k \ge 0$ and $M \le m_i' - n_0$ for some fixed value $n_0 > 0$:

$$\begin{aligned} ||\{u^i_e, v^i_e \} w_M ||_{H^k} \le C_{k,M}. \end{aligned}$$

(142)

Proof

The existence follows from Lax-Milgram, whereas the estimates follow from continuing the procedure resulting in (141)–(142). $\quad \square$

Corollary 1

Assume $m_i>> m_1$ for $i = 2,\ldots ,n$. Then:

$$\begin{aligned} \Vert \{ u^i_e, v^i_e \} w_{\frac{m_1}{2}} \Vert _{H^k} \lesssim 1. \end{aligned}$$

(143)

In particular, if $m_i = N>> 1$, then (145) is valid with $m_1 = \frac{N}{2}$.

Proof

This follows from two points. First, for the $i = 1$ case, the forcing is absent and therefore the parameter $l_1$ can be taken arbitrarily large. In particular this implies that $m_1' = m_1$. Second, a subsequent application of the above proposition shows that the i-th layer quantities decay like $m_1 - n_0$. An examination of the forcing terms $f^i_{E,1}, f^i_{E,2}$ shows that these quantities decay as $w_{m_1 - n_0}^{-1}$. Thus, for $i \ge 2$, we can take the parameter $l_i = m_1 - n_0 = m_i'$. Therefore, if $m_1$ is sufficiently large, $\frac{m_1}{2}<< m_1 - 10 n_0$. $\quad \square$

Recall the definition of $m_i$ from Definition 6. The main estimate here is:

Lemma 14

Let $v^i_e$ be a solution to (127). For any $m' < m_i' := \min \{m_i, l_i\}$. Then, the following estimate holds:

$$\begin{aligned} \Vert v^i_e w_{m'} \Vert _{W^{k,\infty }} \lesssim 1. \end{aligned}$$

(144)

Proof

We first homogenize $v^i_e$ by introducing $\bar{v}_e := v^i_e - S$, where S was defined in (129). Recall the definition of $\chi$ in (39). We will localize using $1 -\chi (\frac{Y}{N})$ for some large, fixed $N > 1$. A direct computation produces the following:

$$\begin{aligned} \varDelta \left( \left\{ 1 - \chi \left( \frac{Y}{N}\right) \right\} \bar{v}_e\right) =&-\left\{ 1 - \chi \left( \frac{Y}{N}\right) \right\} \frac{u^0_{eYY}}{u^0_e} v^1_e - \left\{ 1 - \chi \left( \frac{Y}{N}\right) \right\} \varDelta S \\&+ 2 \partial _Y \left\{ 1 - \chi \left( \frac{Y}{N}\right) \right\} \bar{v}_{eY} + \partial _{YY} \left\{ 1 - \chi \left( \frac{Y}{N}\right) \right\} \bar{v}_e \\&+ \left( 1 - \chi \left( \frac{Y}{N}\right) \right) \frac{F^{(i)}}{u^0_e} := R. \end{aligned}$$

Let $w = w_{m'}^{-1}$. Now we define the quotient $q^\delta = \frac{\{ 1 - \chi (\frac{Y}{N}) \} \bar{v}_e}{w(Y) + \delta }$, which satisfies:

$$\begin{aligned} \underbrace{\varDelta q^\delta + 2 \frac{w_Y}{w + \delta } q^\delta _Y}_{:= \mathcal {T}_\delta } + \frac{w_{YY}}{w+\delta } q^\delta = \frac{R}{w+\delta }. \end{aligned}$$

(145)

Case 1: $w_{m_i}$ are polynomials in Y. The following inequalities hold, independent of small $\delta$:

$$\begin{aligned} |\frac{R}{w+\delta }| \lesssim 1, \qquad |\frac{w_{YY}}{w+\delta }\{ 1 - \chi (\frac{Y}{N}) \}| \le o(1). \end{aligned}$$

(146)

The second inequality above holds because $|w_{YY}| \lesssim Y^{-2} |w|$ for polynomial decay, so by taking N large, we can majorize the desired quantity by o(1). To apply the maximum principle to $q^\delta$, we introduce the following barrier, for m large and fixed and for $f = f(x) := 10 - (1+x)^2$ (which we note satisfies $f''(x) < - 1$):

$$\begin{aligned} q_{\pm }^\delta := q^\delta \pm f(mx) \left( \sup |\frac{R}{w+\delta }| + \sup |\frac{w_{YY}}{w} q^\delta | \right) . \end{aligned}$$

Immediate computations gives $\mathcal {T}_\delta [q_-^\delta ] \ge 0$ and $\mathcal {T}_\delta [q_+^\delta ] \le 0$. Due to the homogenization, $\bar{v}_e|_{x = 0} = \bar{v}_e|_{x = L} = 0$, which immediately implies that $q^\delta _-|_{x = 0} < 0$ and $q^\delta _-|_{x = L} < 0$, and $q^\delta _+|_{x = 0} > 0$, $q^\delta _+|_{x = L} > 0$. Due to the presence of the cut-off, $q^\delta _{\pm }|_{Y = \frac{N}{2}} = 0$. Thus, applying the maximum principle to both $q_-^\delta , q_+^\delta$ on the domain $(x, Y) \in (0, L) \times (\frac{N}{2}, \infty )$, gives:

$$\begin{aligned} \Vert q^\delta \Vert _\infty \lesssim \sup |\frac{R}{w+\delta }| + \sup |\frac{w_{YY}}{w}q^\delta |. \end{aligned}$$

Applying both estimates in (148) gives $\Vert q^\delta \Vert _\infty \lesssim 1$ uniformly in $\delta > 0$. Due to the cutoff $\{1 - \chi (\frac{Y}{N})\}$, all quantities are supported away from $Y = 0$, we may differentiate the Eq. (147), in Y to obtain the new system:

$$\begin{aligned} \varDelta q^\delta _Y + 2 \frac{w_Y}{w+\delta } q^\delta _{YY} + \left[ \frac{w_{YY}}{w+\delta } + 2 \partial _Y\left\{ \frac{w_Y}{w+\delta } \right\} \right] q^\delta _Y = \partial _Y\left\{ \frac{R}{w+\delta } \right\} - \partial _Y\left\{ \frac{w_{YY}}{w+\delta } \right\} q^\delta . \end{aligned}$$

Clearly, we may repeat the above argument for the unknown $q^\delta _Y$. Bootstrapping further to $q^{\delta }_{YY}$ and using the equation, we establish:

$$\begin{aligned} \Vert q^\delta _{xx} \Vert _\infty \lesssim 1. \end{aligned}$$

For each fixed Y, $q^\delta _x(x_*, y) = 0$ for some $x_*= x_*(Y) \in [0,L]$ since $q^\delta (0,Y) = q^\delta (L,Y) = 0$. Thus, using the Fundamental Theorem of Calculus $\Vert q_x^\delta \Vert _\infty \lesssim 1$. Finally, we use the pointwise in Y equality:

$$\begin{aligned} |q^\delta _x - q_x| = \delta | \frac{\chi \bar{v}_{ex}}{w(w+\delta )} | \rightarrow 0 \text { as } \delta \downarrow 0, \text { pw in } Y. \end{aligned}$$

Thus, for each fixed Y, there exists a $\delta _*= \delta _*(Y) > 0$ such that for $0< \delta < \delta _*(Y)$, $|q^\delta _x - q_x| \le 1/2$. Thus, $|q_x(Y)| \lesssim 1$. This is true for all Y. Thus, $\Vert q_x \Vert _\infty \lesssim 1$.

Case 2: $w_{m_i}$ are exponential in Y. In this case, we start with (147), and perform $H^k$ energy estimates. We replace (148) with:

$$\begin{aligned} \Vert \frac{R}{w+\delta } \langle Y \rangle ^M \Vert < \infty \text { for large } M. \end{aligned}$$

(147)

From here, straightforward energy estimates show $\Vert q^\delta \Vert _{H^k} \lesssim 1$ for any k as in estimate (144). This is achieved by repeatedly differentiating in Y and using that the cutoff $\{1 - \chi (\frac{Y}{N}) \}$ localizes away from the boundary $\{Y = 0\}$. We thus conclude $\Vert q^\delta _{xx} \Vert _{\infty } \lesssim \Vert q^\delta \Vert _{H^4} \lesssim 1$ using Sobolev embedding. The proof then concludes as in the polynomial case. $\quad \square$

Proof of Theorem 4

Theorem 4 is a consequence of Theorem 1, Proposition 3, Corollary 1, and Lemmas 13, 14. $\quad \square$