The Feynman–Lagerstrom Criterion for Boundary Layers

Abstract

We study the boundary layer theory for slightly viscous stationary flows forced by an imposed slip velocity at the boundary. According to the theory of Prandtl (in: International mathematical congress, Heidelberg, 1904; see Gesammelte Abhandlungen II, 1961) and Batchelor (J Fluid Mech 1:177–190, 1956), any Euler solution arising in this limit and consisting of a single “eddy” must have constant vorticity. Feynman and Lagerstrom (in: Proceedings of IX international congress on applied mechanics, 1956) gave a procedure to select the value of this vorticity by demanding a necessary condition for the existence of a periodic Prandtl boundary layer description. In the case of the disc, the choice—known to Batchelor (1956) and Wood (J Fluid Mech 2:77–87, 1957)—is explicit in terms of the slip forcing. For domains with non-constant curvature, Feynman and Lagerstrom give an approximate formula for the choice which is in fact only implicitly defined and must be determined together with the boundary layer profile. We show that this condition is also sufficient for the existence of a periodic boundary layer described by the Prandtl equations. Due to the quasilinear coupling between the solution and the selected vorticity, we devise a delicate iteration scheme coupled with a high-order energy method that captures and controls the implicit selection mechanism.

1 Introduction

Let $M\subset \mathbb {R}^2$ be a bounded, simply connected domain. Consider the Navier–Stokes equations

$$\begin{aligned} \partial _t u^\nu +u^\nu \cdot \nabla u^\nu&= -\nabla p^\nu + \nu \Delta u^\nu \qquad \text {in} \ M, \end{aligned}$$

(1)

$$\begin{aligned} \nabla \cdot u^\nu&=0 \qquad \qquad \qquad \ \ \quad \text {in} \ M. \end{aligned}$$

(2)

Motion is excited through the boundary, where stick boundary conditions are supplied

$$\begin{aligned} u^\nu \cdot \hat{n}&= 0 \qquad \text {on}\ \partial M, \end{aligned}$$

(3)

$$\begin{aligned} u^\nu \cdot \hat{\tau }&= f \qquad \text {on}\ \partial M, \end{aligned}$$

(4)

where $\hat{n}$ is the unit outer normal vector field on the boundary and $\hat{\tau }=\hat{n}^\perp$, the unit tangent field. In the above, there is a given autonomous slip velocity $f:\partial M \rightarrow \mathbb {R}$, which should be thought of as being generated by motion of the boundary (via so-called “stick" or “no slip" boundary conditions—the fluid velocity on the boundary matches its speed), and is responsible for the generation of complex fluid motions in the bulk. If the viscosity is large relative to the forcing, then it is easy to see that all solutions converge to a unique steady state as $t\rightarrow \infty$ [38]. However, as viscosity is decreased, one generally expects solutions to develop and retain non-trivial variation in time, perhaps even forever harboring turbulent behavior. There one general exception to this expectation in a special setting, proved in Sect. 3:

Theorem 1

(Absence of turbulence) Let $M=\mathbb {D}$ be the disk of radius $\textsf{R}$ and $f = \frac{1}{2} {\omega _0}\textsf{R}$ for any given ${\omega _0}\in \mathbb {R}$ be a constant slip on the boundary. For any distributionally divergence-free $u_0\in L^2$, the unique Leray-Hopf weak solution converges at long time to solid body rotation $u_\textsf{sb}=\frac{1}{2}{\omega _0}x^\perp$ having vorticity $\omega _0$. In fact,

$$\begin{aligned} \Vert u(t)- u_\textsf{sb}\Vert _{L^2}\le \Vert u_0- u_\textsf{sb}\Vert _{L^2} e^{-\lambda _1 \nu t} \end{aligned}$$

where $\lambda _1$ is the first positive eigenvalue of $-\Delta$ with Dirichlet boundary conditions on $\mathbb {D}$.

Remark

The forcing (slip velocity) in Theorem 1 can be arbitrarily large and yet for any viscosity Navier–Stokes has a one-point attractor. This is the analogue of Marchioro’s results on the absence of turbulence on the torus with ‘gravest mode’ body forcing [31, 32].

Theorem 1 highlights a peculiarity of solid body rotation on the disk: if you center a circular basin on fluid on a record player, all motion will eventually be solid body. A question arises:

Question 1

What if the imposed slip is non-constant, or the domain is not a disk?

Fig. 1

Cartoon of streamlines of a steady Navier–Stokes solutions on the ellipse forced by an imposed slip $u^\nu \cdot \tau = f$ on the left. Streamlines of constant vorticity Euler solution $\psi _* = \frac{a^2}{2(1+a^2)}(\frac{x^2}{a^2}+y^2)$ shown on the right

As mentioned above, one might expect that if either of these conditions is violated, time dependence generally survives. However, if the boundary forcing is special this need not be the case. For instance, consider the velocity field with constant (unit) vorticity on any M

$$\begin{aligned} u_{*} = K_M [1], \qquad K_M:= \nabla ^\perp \Delta ^{-1}_{ D}. \end{aligned}$$

(5)

Any such velocity field satisfies both the Euler and Navier–Stokes equations in the bulk. As such, it is a stationary solution of Euler, and also of Navier–Stokes provided $u_*$ is taken as initial data and it is forced consistently on the boundary: $f_* = u_{*}\cdot \tau .$ Thus, for any domain there is a family of non-trivial time-independent solutions uniformly in $\nu \ge 0$.

For force sufficiently close to that generated by a stationary Euler solution, asymptotic stability may occur but is a delicate issue. To begin to understanding these issues, we are interested in the question of the existence of sequence of stationary Navier–Stokes solutions approximating an Euler flow in this setting. The Prandtl–Batchelor theory [1, 35] provides a restriction on the type of stationary Euler solutions that can arises as inviscid limits. Namely, it stipulates that they have constant vorticity within closed streamlines, so-called “eddies". See also Childress [4, 5] and Kim [22, 23]. The result, proved in Sect. 4, is:

Theorem 2

(Prandtl–Batchelor Theorem) Let $M\subset \mathbb {R}^2$ be a simply connected domain with smooth boundary. Let $\psi :M\rightarrow \mathbb {R}$ be a $W^{4,1+}(M)$ streamfunction of a steady, non-penetrating solution of Euler $u_e$ having a single stagnation point which is non-degenereate in a sense that the period of revolution of a particle is a differentiable function of the streamline. Suppose $\{u^\nu \}_{\nu >0}$ is a family satisfying (1), (2) together with

$$\begin{aligned} \lim _{\nu \rightarrow 0} \Vert u ^\nu -u_e \Vert _{H^{5/2+}(U)} \rightarrow 0, \end{aligned}$$

(6)

for all interior open subsets $U\subset M$. Then $u_e= {\omega _0}u_*$ for a constant ${\omega _0}\in \mathbb {R}$ and $u_*$ is (5).

See Fig. 1. for a cartoon of this convergence. In the above theorem, M can be thought of as a streamline of an Euler solution occupying some larger spatial domain. If the limiting Euler solution consists of multiple eddies, the above shows that, within each eddy, the vorticity tends to become constant. The vorticity of the resulting solution would be a staircase landscape separated, perhaps, by vortex sheets. Such a picture is consistent with the general expectation of the emergence of weak solutions in the inviscid limit on bounded domains [7, 8, 12]. We remark that similar selection principles to Theorem 2 appear also in two-dimensional passive scalar problems [33, 36], and in steady heat distribution in three-dimensional integrable magnetic fields [10].

If the boundary data is a sufficiently small perturbation of the corresponding slip of an unit vorticity Euler flow $u_*$ on whole vessel M,

$$\begin{aligned} f = u_* \cdot \tau + {\varepsilon }g. \end{aligned}$$

(7)

then the inviscid limit of steady Navier–Stokes solutions might be expected to consist of just a single eddy having constant vorticity (5), that is, for some appropriate constant ${\omega _0}\in \mathbb {R}$,

$$\begin{aligned} u^\nu \rightarrow {\omega _0}u_* \qquad \text {as} \qquad \nu \rightarrow 0. \end{aligned}$$

(8)

See Conjecture 1. This naturally leads to the following question:

Question 2

Given boundary data (3), (4), (7), how is the limiting vorticity ${\omega _0}$ (8) selected?

This question was discussed by Batchelor (1956) [1] and Wood (1957) [40] for disk domains, and the resulting prediction is called the Batchelor–Wood formula. This analysis was done independently by Feynman–Lagerstrom (1956) [14] who also generalized this formula to domains with non-constant curvature. See also [29, 30]. The idea is: Navier–Stokes with small viscosity should approximate Euler (constant vorticity) in the bulk of the domain, and interpolate to the given boundary conditions across a layer of width $\sqrt{\nu }$. In this layer, predicted by Prandtl [35], the leading order behavior of the fluid is captured by a simpler boundary layer equation which is supplied with data at infinity from Euler and at zero (the boundary), from Navier–Stokes. The complication is that the Euler solution is known only up to a constant multiple. The specific vorticity value ${\omega _0}$ is then fixed by demanding the corresponding Prandtl equation for the boundary layer admits a periodic solution.

We note that the (Prandtl–Batchelor or Feynman–Lagerstrom) theory was developed to address the specific phenomenon which only arises for stationary flows on closed domains. For the much different setting of unsteady flows, the question of which Euler flow is achieved in the inviscid limit is essentially completely determined by the initial data. For stationary flows on non-closed domains, for example on $[0, L] \times \mathbb {R}_+$, there is an analogue of data prescribed on the sides $\{x = 0\}, \{x = L\}$ which similarly fixes the inviscid Euler flow (see for instance, results of [15, 21] for results in this direction). In contrast, for closed domains, the only prescribed data is the slip boundary data, $f(\theta )$. Therefore, the selection mechanism is less obvious to uncover, and historically motivated investigations of Prandtl–Batchelor or Feynman–Lagerstrom, and from a rigorous standpoint, the results in this paper.

On the disk $M= \mathbb {D}$ of radius $\textsf{R}$, this amounts to:

(9)

This picture has been rigorously justified by Kim [24, 25] for the boundary layer and recently by Fei, Gao, Lin and Tao [13] for Navier–Stokes. The latter constructs a sequence of steady Navier–Stokes solutions on the disk forced by (7) converging towards this predicted end state. In the case of a general domain, Feynman–Lagerstrom argued that selecting ${\omega _0}$ to ensure a certain periodicity is a necessary condition for the existence of such a layer and therefore for convergence, but did not speak to its sufficiency. We now review their theory.

Fig. 2

Boundary layer geometry, depicted (abusing the required asymptotic $\nu \rightarrow 0$ for the sake of illustration) in von Mises coordinates. The level $\psi \in [0,\infty )$ denotes the distance from the boundary in a layer of size $\sqrt{\nu }$, upon rescaling so that this extends indefinetely. See Sect. 2. The unique vorticity value $\omega _0$ so that such a boundary layer exists is selected nonlinearly via (12). It is given approximately by the Feynman-Lagerstrom formulae (13) or (14)

Recall Prandtl’s boundary layer equations written in von Mises coordinates $(s,\overline{\psi })$, where s is the periodic coordinate on the boundary and $\bar{\psi }:=\psi /\sqrt{\nu }$ is the rescaled streamline coordinate. From hereon, we denote $\bar{\psi }=\psi$ and

$$\begin{aligned} q_e (s):= u_*\cdot \hat{\tau }(\gamma (s)) \end{aligned}$$

where $\gamma : [0,|\partial M|)\rightarrow \partial M$ is the arc-length parametrization of the boundary, is the tangential slip along the boundary of unit vorticity Euler solutions (5). The Prandtl equations—which determine an unknown function $q:[0,L)\times \mathbb {R}^+\rightarrow \mathbb {R}$ which serves as an approximation of the tangential Navier–Stokes velocity $u^\nu \cdot \hat{\tau }$ in an $O(\sqrt{\nu })$ boundary layer—are (see [34]):

$$\begin{aligned} \partial _s Q - q\partial _\psi ^2 Q= 0, \qquad Q = q^2-\omega _0^2 q_e ^2, \end{aligned}$$

(10)

which is to be satisfied on $(\psi ,s)\in [0,\infty )\times [0,L)$ where $L=|\partial M|$ is the length of the boundary. For completeness, we derive these equations in Sect. 2. The solution q must connect Navier–Stokes to Euler: at the boundary ($\psi =0$), the solution q takes the Navier–Stokes data and away from the boundary ($\psi =\infty$), the solution q assumes the Eulerian behavior:

$$\begin{aligned} q(0,s)= f(s), \qquad q(\infty ,s) = {\omega _0}q_e (s) \end{aligned}$$

which, for Q, translates to the data

$$\begin{aligned} Q(0,s)&=f^2(s) - \omega _0^2 q_e ^2(s), \qquad Q(\infty ,s) =0. \end{aligned}$$

The solution q (or, equivalently, Q) of Eq. (10) must be periodic in the s variable (so that the boundary layer closes). See Fig. 2. Feynman–Lagerstrom noted that this leads to a self-consistency condition on ${\omega _0}$. We will enforce this in the following way. First, we rewrite (10) as

$$\begin{aligned} \partial _s Q -{\omega _0}q_e \partial _\psi ^2 Q = \Big (1- \frac{{\omega _0}q_e }{\sqrt{\omega _0^2 q_e ^2+ Q }}\Big )\partial _s Q. \end{aligned}$$

Integrating the above equation over the boundary, we obtain

$$\begin{aligned} \partial _\psi ^2 \int _0^L q_e (s) Q(\psi ,s) \textrm{d}s = \mathcal {N}^{\omega _0}[Q] \end{aligned}$$

where the nonlinearity is explicitly

$$\begin{aligned} \mathcal {N}^{\omega _0}[Q](\psi ):= -\frac{1}{{\omega _0}} \int _0^L \left( 1- \frac{{\omega _0}q_e (s)}{\sqrt{\omega _0^2 q_e ^2(s)+ Q }}\right) \partial _s Q \textrm{d}s. \end{aligned}$$

(11)

Then, for some scalars A and B, we obtain the identity

$$\begin{aligned} \int _0^L q_e (s)Q(\psi ,s) \textrm{d}s = A+ B \psi - \int _\psi ^\infty (\psi -y) \mathcal {N}^{\omega _0}[Q](y) \textrm{d}y. \end{aligned}$$

From the boundary conditions, we have that $A=B=0$. We thus obtain the nonlinear condition that at each $\psi \in \mathbb {R}_+$, we have

$$\begin{aligned} \int _0^L q_e (s)Q(\psi ,s) \textrm{d}s = \int _\psi ^\infty (y-\psi ) \mathcal {N}^{\omega _0}[Q](y) \textrm{d}y. \end{aligned}$$

Evaluating at $\psi =0$, we find a nonlinear, nonlocal condition determining the constant $\omega _0$:

$$\begin{aligned} \int _0^L q_e (s)(f^2(s) - \omega _0^2 q_e ^2(s)) \textrm{d}s = \int _0^\infty y \mathcal {N}^{\omega _0}[Q](y) \textrm{d}y. \end{aligned}$$

(12)

Remark

(Feynman–Lagerstrom formulae) Letting $1- \omega _0^2 =:{\varepsilon }\overline{\omega }_0$ with $\overline{\omega }_0=O(1)$, we anticipate $Q = O({\varepsilon })$. Since $\frac{1}{\sqrt{ \omega _0^2+x}}-\frac{1}{\sqrt{ \omega _0^2}} = - \frac{x}{2\omega _0^3} + O (x^2)$, we see that $\mathcal {N}^{\omega _0}[Q](\psi )=O(Q^2) = O({\varepsilon }^2)$. Moreover $\mathcal {N}^{\omega _0}[Q](y)$ is trivial in the case of the boundary having constant curvature $\kappa :=\hat{\tau }\cdot \nabla \hat{n} \cdot \hat{\tau }$. Indeed, in this case of M being a disk, it is readily seen that the integrand in (11) is a total derivative in s and hence $\mathcal {N}^{\omega _0}[Q](\psi )\equiv 0$), see the next Remark. Thus, as pointed out by Feynman and Lagerstrom [14], to leading order, condition (12) is

$$\begin{aligned} \omega _0^2 = \frac{\int _0^L q_e (s)f^2(s) \textrm{d}s}{\int _0^L q_e ^3(s) \textrm{d}s} + o(|\partial _s \kappa |). \end{aligned}$$

(13)

This formula is exact (having $\partial _s \kappa =0$) when M is the disk, and generally only for the disk.¹ Recalling $f(s) = q_e (s) + {\varepsilon }g(s)$ so that $f^2(s) - \omega _0^2 q_e ^2(s)= (1- \omega _0^2 ) q_e ^2(s) + 2{\varepsilon }g(s)q_e (s) + {\varepsilon }^2\,g^2(s)$, to leading order in ${\varepsilon }$ (the deviation of NS data from unit vorticity Euler slip) we have

$$\begin{aligned} \omega _0^2 = 1+ 2 {\varepsilon }\frac{\int _0^L q_e ^2(s)g(s) \textrm{d}s}{\int _0^L q_e ^3(s) \textrm{d}s} + O({\varepsilon }^2). \end{aligned}$$

(14)

Remark

(Wood’s formula when $M=\mathbb {D}$) On the disk of radius $\textsf{R}$, the constant vorticity solution (5) is solid body rotation $u_*(x)=\frac{1}{2} x^\perp$, so that $q_e =\frac{\textsf{R}}{2}$ is a constant. In fact, by Serrin’s theorem [37] constraining a domain admitting a solution of $\Delta \psi = 1$ with constant Neumann and Dirichlet data, the disk is the unique domain for which the solid body Euler solution has constant boundary slip velocity $q_e$. See also [26]. On the disk, (13) without an error term is exact and, with the circumference $L=2\pi \textsf{R}$, agrees with (9) of Wood [40].

In this paper, we rigorously establish this prediction by constructing a periodic boundary layer verifying the Feynman–Lagerstrom condition if the constant vorticity Euler solution $u_*$ on M defined by (5) has no stagnation points on the boundary. We have

Theorem 3

(Existence of a periodic Prandtl boundary layer) Let M be a simply connected domain with $L=|\partial M|$. Denote $\mathbb {T}_L= [0,L)$. Let $q_e:\mathbb {T}_L\rightarrow \mathbb {R}$ be a smooth, non-vanishing function. Let $f(s):= q_e (s)+ {\varepsilon }g(s)$ with smooth $g: \mathbb {T}_L \rightarrow \mathbb {R}$. For all ${\varepsilon }$ sufficiently small, depending only the data $(q_e,g)$, there exists a unique constant ${\omega _0}\in \mathbb {R}$ and function $q:\mathbb {T}_L\times \mathbb {R}^+\rightarrow \mathbb {R}$ such that the pair $(\omega _0, q)$ solves the Prandtl equations on $\mathbb {T}_L\times \mathbb {R}^+$:

$$\begin{aligned} \partial _s Q - q\partial _\psi ^2 Q&= 0, \qquad Q = q^2-\omega _0^2 q_e ^2, \nonumber \\ Q(s,0)&=f^2(s) - \omega _0^2 q_e ^2(s),\nonumber \\ Q(s,\infty )&=0. \end{aligned}$$

(15)

Moreover, the solution Q lies in the space $X_{2,50}$ defined by (25) and enjoys $\Vert Q\Vert _{X_{2,50}} \lesssim {\varepsilon }$. The selected vorticity $\omega _0$ can be expressed as follows: there exists a constant $C>0$ so that

$$\begin{aligned} \omega _0^2 = \frac{\int _0^L q_e (s)f^2(s) \textrm{d}s}{\int _0^L q_e ^3(s) \textrm{d}s} + {\omega }_{\textrm{Err}}\qquad \text {where} \qquad |{\omega }_{\textrm{Err}}| \le C {\varepsilon }^2. \end{aligned}$$

(16)

The sign of $\omega _0$ agrees with that of the background $q_e$ which, in this case, is positive.

This theorem, proved in Sect. 5, provides the first rigorous confirmation of the Feynman–Lagerstrom formula, and justifies their claim that for $|f-q_e |\ll |f|$ (translating to ${\varepsilon }\ll 1$), the leading term in (13) serves as a good approximation for the selected vorticity.

The constant ${\omega _0}$ satisfies (12), and has an explicit component, determined by $q_e (s)$ and f(s), as well as an implicit component $\overline{\omega }_{\textrm{Err}}$ which is smaller amplitude and for which we obtain bounds. We emphasize that the $\omega _0$ appearing in (15) is nonlinearity selected as soon as the domain, M, is no longer a disk (for example, M is an ellipse). This requires, at an analytical level, a delicate coupling between the choice of constant, $\omega _0$, and the control of the solution Q in an appropriately chosen norm, which is the main innovation of our work.²

We anticipate the Prandtl system, which we analyze in this paper, to be stable in the inviscid limit for the full Navier–Stokes system. Indeed, this is what is proved in [13] when M is a disk and $u_e = x^\perp$. However, as discussed above, in that very special setting the constant ${\omega _0}$ can be explicitly determined (9). In general, this is not the case and ${\omega _0}$ is only implicitly determined by the condition (12) described above, making the inviscid limit more delicate. Nevertheless, we believe that the nonlinearly determined constant $\omega _0$ will describe, to leading order in viscosity, the selection principle. That is, we believe that Navier–Stokes vorticity $\omega ^{\nu }$ should obey an asymptotic expansion

$$\begin{aligned} \omega ^{\nu } = \omega _0 + O(\sqrt{\nu }) \end{aligned}$$

(17)

in the interior of the domain. In fact, we issue the following

Conjecture 1

Let $M\subset \mathbb {R}^2$ be any simply connected domain such that the constant vorticity Euler solution $u_*$ on M defined by (5) has a single eddy (streamfunction has a single, non-degenerate, critical point). Suppose that Navier–Stokes is forced by a slip of the form (7), e.g. $f = u_*\cdot \tau + {\varepsilon }g$ for some smooth function $g:\partial M \rightarrow \mathbb {R}$. Then, there exists an ${\varepsilon }_*:={\varepsilon }_*(M, g)$ such that for all ${\varepsilon }<{\varepsilon }_*$ we have weak convergence in $L^2(M)$

$$\begin{aligned} u^{\nu } \rightharpoonup \omega _0 u_* \qquad \text {as} \qquad \nu \rightarrow 0 \end{aligned}$$

along a sequence of steady Navier–Stokes solutions, where $\omega _0:=\omega _0(M,g)$ is (16) of Thm 3.

Of course, stronger convergence can be expected, along with a boundary layer description such as that established by [13] on the disk. The fact that moving boundaries can stabilize the inviscid limit is a well known phenomenon from the work of Guo and Nguyen [16] and Iyer [19, 20]. Verifying the expansion (17) to prove the above conjecture will require substantially new ideas. In the context of elliptical domains M, this is work in preparation.

Finally we remark that the failure of a boundary layer to exist is indicative of the existence of multiple eddies: constant vorticity regions are separated by internal layers which can be thought of as free boundaries. This can happen either if the constant vorticity solution on that domain has multiple eddies, or if the given slip data is far from that of a constant vorticity slip (according to our Theorem 3). Kim [26] showed that if the Navier–Stokes boundary slip is only slightly negative in places, the Prandtl–Batchelor theory still applies to good approximation in the bulk. For the situation of being far from compatible slip data, see Kim and Childress [28] for an analytical investigation on a rectangle, Greengard and Kropinski [17] for a numerical investigation on disk domains, and Henderson, Lopez and Stewart [18] for laboratory experiments.

2 Derivation of the Prandtl Boundary Layer Equation

In this section, we derive the Prandtl equations for any simply connected domain M. Assume that $s:[0,L]\rightarrow \partial M$ be the arc-length parametrization of the boundary $\partial M$. For $s\in \mathbb T_L$, let $\tau (s)$ and n(s) be unit the tangential vector to the boundary $\partial M$. There exists $\delta >0$ such that for any $x\in M$ such that $\textrm{dist}(x,\partial M)<\delta$, there exists a unique $s\in \mathbb T_L$ and $x(s)\in \partial M$ such that

$$\begin{aligned} \textrm{dist}(x,\partial M)=|x-x(s)|. \end{aligned}$$

Moreover, one has the representation

$$\begin{aligned} x(s,z)=x(s)+zn(s), \end{aligned}$$

where $x(s)\in \partial M$ and $z=\textrm{dist}(x,\partial M)$, see [3]. The map

$$\begin{aligned} \begin{aligned} \{x\in M:\quad 0<\textrm{dist}(x,\partial M)<\delta \}&\rightarrow \mathbb T_L\times (0,\delta )\\ x&\rightarrow (z,s) \end{aligned} \end{aligned}$$

is a diffeomorphism. We also define the following quantities for the domain M:

$$\begin{aligned} \gamma (s)=x_1''(s)x_2'(s)-x_1'(s)x_2''(s),\quad J(z,s)=1+z\gamma (s)>0, \end{aligned}$$

where $\gamma$ represents the boundary curvature, and J is a Jacobian for a near-wall mapping used to derived the following form of Navier–Stokes, see [3] and Appendix 6.

Now for $x=x(s,z)$, we denote $\tau (s)$ and n(s) to be the tangential and normal vector at $x(s)\in \partial M$ on the boundary. Consider the steady Navier–Stokes equations

$$\begin{aligned}\begin{aligned} u^\nu \cdot \nabla u^\nu +\nabla p^\nu&=\nu \Delta u^\nu ,\\ \nabla \cdot u^\nu&=0, \end{aligned} \end{aligned}$$

written in the region $\textrm{dist}(x,\partial M)<\delta$. We define

$$\begin{aligned}\begin{aligned} u_\tau (s,z)&=u^\nu (x)\cdot \tau (s)=u^\nu (x(s,z))\cdot \tau (s),\\ u_n(s,z)&=u^\nu (x) \cdot n(s)=u^\nu (x(s,z))\cdot n(s). \end{aligned} \end{aligned}$$

By direct calculation, provided in Appendix 6, the Navier–Stokes equations become

$$\begin{aligned} \begin{aligned}&\frac{u_\tau }{J}\partial _s u_\tau +u_n\partial _z u_\tau -\frac{\gamma }{J}u_\tau u_n+\frac{1}{J} \partial _s p\\&\quad =\nu \left\{ \frac{1}{J} \partial _z (J\partial _z u_\tau )+\frac{1}{J} \partial _s\left( \frac{1}{J} \partial _s u_\tau \right) -\frac{1}{J} \partial _s\left( \frac{\gamma u_n}{J} \right) -\frac{\gamma }{J}\left( \gamma u_\tau +\partial _s u_n \right) \right\} ,\\&\frac{u_\tau }{J} \partial _s u_n+u_n\partial _z u_n -\frac{\gamma }{J}u_\tau ^2+\partial _z p\\&\quad =\nu \left\{ \frac{1}{J} \partial _z\left( J\partial _z u_n \right) +\frac{1}{J} \partial _s\left( \frac{1}{J} \partial _s u_n \right) -\frac{1}{J} \partial _s\left( \frac{\gamma u_\tau }{J} \right) -\frac{\gamma }{J}\left( \partial _s u_\tau -\gamma u_n \right) \right\} ,\\&\partial _z u_n+\frac{1}{J} \partial _s u_\tau -\frac{\gamma }{J} u_n=0. \end{aligned} \end{aligned}$$

Remark

On the disk with the usual polar coordinates $(\theta ,r)\in \mathbb T\times [0,1]$, we have

$$\begin{aligned} \gamma (\theta )=-1,\quad J(z,\theta )=\frac{1}{r},\qquad u_\tau =u_\theta ,\quad u_n=u_r. \end{aligned}$$

Near the boundary, in a layer of width $\sqrt{\nu }$, we anticipate that Navier–Stokes velocity field $(u_\tau , u_n)$ will look like a small boundary layer correction $(u_\tau ^P, v_n^P)$, to a constant vorticity $({\omega }_0q_e,0)$ Euler flow, as discussed in the introduction. That is,

$$\begin{aligned} u_\tau (s,z)\sim {\omega }_0q_e (s)+u_\tau ^P\left(s,Z\right),\qquad u_n(s,z)\sim \sqrt{\nu }v_n^P\left(s,Z\right),\qquad Z=\frac{z}{\sqrt{\nu }}, \end{aligned}$$

where $\lim _{Z\rightarrow \infty }u_\tau ^P(s,Z)=0$. See discussion in Oleinik and Samokhin [34]. Plugging in this ansatz into the Navier–Stokes equations near the boundary, and using the approximation

$$\begin{aligned} \tfrac{1}{J}=\tfrac{1}{1+z\gamma (s)}=\tfrac{1}{1+\sqrt{\nu }Z\gamma (s)}\sim 1-\sqrt{\nu }Z \gamma (s)+O(\nu ), \end{aligned}$$

we obtain the equations

$$\begin{aligned} \begin{aligned}&({\omega }_0q_e (s)+u_\tau ^P)\partial _s({\omega }_0q_e (s)+u_\tau ^P)+v_n^P\partial _Z (q_e (s)+u_\tau ^P)+\partial _s p-\partial _Z^2 u_\tau ^P=0,\\&\partial _Z p=0 \end{aligned} \end{aligned}$$

(18)

along with the divergence free condition

$$\begin{aligned} \partial _s({\omega }_0 q_e (s)+u_\tau ^P)+\partial _Z v_n^P=0 \end{aligned}$$

Taking $Z\rightarrow \infty$ in the Eq. (18), we obtain

$$\begin{aligned} \partial _s p=-{\omega }_0^2 q_e (s)q_e '(s) \end{aligned}$$

Replacing the pressure by the above into the equation (18), we obtain the Prandtl equations:

$$\begin{aligned}&({\omega }_0 q_e (s)+u_\tau ^P)\partial _s({\omega }_0 q_e (s)+u_\tau ^P)\\&\qquad +v_n^P\partial _Z ({\omega }_0q_e (s)+u_\tau ^P)-{\omega }_0^2 q_e (s)q_e'(s)-\partial _Z^2 u_\tau ^P=0 \\&\partial _s({\omega }_0q_e (s)+u_\tau ^P)+\partial _Z v_n^P=0. \end{aligned}$$

Define the von Mises variables $(s,\psi )$ such that

$$\begin{aligned} \partial _Z \psi (s,Z)={\omega }_0 q_e (s)+u_\tau ^P(s,Z),\qquad -\partial _s \psi (s,Z)=v_n^P(s,Z). \end{aligned}$$

Let $q=q(s,\psi )={\omega }_0 q_e (s)+u_\tau ^P$, the Prandtl equation becomes

$$\begin{aligned} q(\partial _s q-v_n^P\partial _\psi q)+qv_n^P \partial _\psi q-\frac{1}{2} \partial _s\left({\omega }_0^2 q_e (s)^2 \right)-q\partial _\psi (q\partial _\psi q)=0 \end{aligned}$$

which reduces to

$$\begin{aligned} \partial _s q^2-{\omega }_0^2\partial _s q_e ^2 -q\partial _\psi ^2 q^2 =0. \end{aligned}$$

Letting $Q=q^2-{\omega }_0^2 q_e ^2$, the above equation becomes (10), namely (See Fig. 3 for a visualization)

$$\begin{aligned} \partial _s Q-q\partial _\psi ^2 Q=0. \end{aligned}$$

Fig. 3

Numerical solution Q of (10) for some representative data $q_e$ and f. Thinking of s at “time", the equation is nonlinear heat sourced at the wall $\psi =0$. Decay away from the boundary is rapid (exponential) in $\psi$

3 Proof of Theorem 1: Absence of Turbulence

Let $v^\nu = u^\nu - u_e$ be the difference of solutions of Euler and Navier–Stokes, where Navier–Stokes is forced by Euler’s slip velocity. On general domains M, it satisfies

$$\begin{aligned} \begin{array}{ll} \partial _t v^\nu + (v^\nu + u_e)\cdot \nabla v^\nu + v^\nu \cdot \nabla u_e = -\nabla q +\nu \Delta v^\nu + \nu \Delta u_e &{} \quad \text {in }\ M, \\ \nabla \cdot v^\nu =0 &{}\quad \text {in }\ M,\\ v^\nu \cdot \hat{n} = 0 &{}\quad \text {on}\ \partial M,\\ v^\nu \cdot \hat{\tau } = 0 &{}\quad \text {on}\ \partial M. \end{array} \end{aligned}$$

Whence the error energy (which holds for Leray-Hopf solutions in dimension two) satisfies

$$\begin{aligned} \frac{1}{2}\frac{\textrm{d}}{\textrm{d}t} \Vert v^\nu \Vert _{L^2}^2&\le - \int _M v^\nu \cdot \nabla u_e \cdot v^\nu \textrm{d}x - \nu \Vert \nabla v^\nu \Vert _{L^2}^2 + \nu \int v^\nu \cdot \Delta u_e. \end{aligned}$$

In general, we may bound

$$\begin{aligned} \frac{1}{2}\frac{\textrm{d}}{\textrm{d}t} \Vert v^\nu \Vert _{L^2}^2&\le \left( \Vert \nabla u_e\Vert _{L^\infty } - \frac{\nu \lambda _1}{2}\right) \Vert v^\nu \Vert _{L^2}^2 + \frac{2\nu }{\lambda _1} \Vert \Delta u_e\Vert _{L^2}^2, \end{aligned}$$

since $v^\nu |_{\partial \mathbb {D}}=0$ so we may apply the Poincaré inequality $\lambda _1 \Vert v^\nu \Vert _{L^2} \le \Vert \nabla v^\nu \Vert _{L^2}^2$ where $\lambda _1$ is the first positive eigenvalue of $-\Delta _D$ on M. We remark, using the results of [38, Chapter 7] (which establish uniform bounds on the steady states), a similar energy identity can be used to prove global attraction of the unique steady state for Navier–Stokes forced by imposed slip on any domain, provided viscosity is large enough.

On the disk $M=\mathbb {D}$, if $u_e=u_\textsf{sb}= {\omega _0}x^\perp$ so that $\nabla u_\textsf{sb}= {\omega _0}\begin{pmatrix} 0&{} -1\\ 1&{} 0 \end{pmatrix}$ and $\Delta u_e =0$, we have

$$\begin{aligned} \int _\mathbb {D} v^\nu \cdot \nabla u_\textsf{sb}\cdot v^\nu \textrm{d}x =\int _\mathbb {D} v^\nu \cdot (v^\nu )^\perp \textrm{d}x=0. \end{aligned}$$

On the disk of radius $\textsf{R}$, this is $\lambda _1= (j_0/\textsf{R})^2$ where $j_0$ is the first zero of $J_0$ the Bessel function of the first kind and order zero). We thus have the stated result.

Remark

On the ellipse $u_\textsf{sb}= {\omega _0}(-y, \alpha x)$ so that $\nabla u_\textsf{sb}= {\omega _0}\begin{pmatrix} 0&{} -1\\ \alpha &{} 0\end{pmatrix}$ and $v \cdot \nabla u_\textsf{sb}\cdot v= {\omega _0}(\alpha -1)v_1v_2$. It follows that provided

$$\begin{aligned} \nu >\nu _*:={\omega _0}\lambda _1^{-1}(1-\alpha ), \end{aligned}$$

then the solid body rotation solution is the global attractor. In particular, as the eccentricity of the elliptical domain goes to zero, $\alpha \rightarrow 1$ and the critical viscosity $\nu _*$ goes to zero. Curiously, all flows in this elliptical family are isochronal [41], meaning that the period of revolution of a particle does not depend on the particular streamline. As such, the form examples of cut points in group of area preserving diffeomorphisms of those domains, see discussion in [9, 11]. The lack of differential rotation in the Euler solution may have important consequences for the asymptotic stability and realizability in the inviscid limit.

4 Proof of Theorem 2: Prandtl–Batchelor Theory

First, by [6, Lemma 5], under the stated assumptions we have that

$$\begin{aligned} \begin{array}{ll} \Delta \psi = F(\psi ) &{} \quad \text {on} \ M,\\ \psi = c_* &{} \quad \text {on} \ \partial M \end{array} \end{aligned}$$

for some $C^1$ function $F:\mathbb {R}\rightarrow \mathbb {R}$ and constant $c_*\in \mathbb {R}$. Suppose without loss of generality that $\{\psi = 0\}$ is the unique critical point in M, so that $\textrm{rang} (\psi ) = [0,c_*]$. By the assumption (6), we have the convergence $\psi ^\nu \rightarrow \psi$ in $H^{7/2+}(U)$ and thus in $C^1(U)$ for all interior open subsets $U\subset M$. It follows that we have convergence of the streamlines (level sets of $\psi ^\nu$). Specifically, for any $c\in \textrm{rang} (\psi )$, the set $\{\psi ^\nu = c\}$ is a closed streamline (at least for sufficiently small $\nu :=\nu (c)$) converging to $\{\psi = c\}$. In what follows, for fixed c we assume $\nu$ is sufficiently small for the above to hold.

Integrating the Navier–Stokes vorticity balance in the sublevel set $\{\psi ^\nu \le c\}$

$$\begin{aligned} 0&=\int _{\{\psi ^\nu \le c\}}\bigg [ u^\nu \cdot \nabla \omega ^\nu - \nu \Delta \omega ^\nu \bigg ] \textrm{d}x\\&=\int _{\{\psi ^\nu = c\}}\bigg [ (u^\nu \cdot \hat{n}^\nu ) \omega ^\nu - \nu \hat{n}^\nu \cdot \nabla \omega ^\nu \bigg ]\textrm{d}\ell =- \nu \int _{\{\psi ^\nu = c\}} \hat{n}^\nu \cdot \nabla \omega ^\nu \textrm{d}\ell , \end{aligned}$$

where $\hat{n}^\nu = \nabla \psi ^\nu /|\nabla \psi ^\nu |$ is the unit normal to streamlines $\{\psi ^\nu = c\}$. Thus

$$\begin{aligned} \int _{\{\psi = c\}} \hat{n}\cdot \nabla \omega \textrm{d}\ell&= \int _{\{\psi = c\}} \hat{n}\cdot \nabla \omega \textrm{d}\ell - \int _{\{\psi ^\nu = c\}} \hat{n}^\nu \cdot \nabla \omega ^\nu \textrm{d}\ell \\&= \int _{\{\psi ^\nu = c\}} \hat{n}^\nu \cdot \nabla (\omega -\omega ^\nu ) \textrm{d}\ell - \int _\Omega \Delta F(\psi ) \textbf{1}_{{\{\psi ^\nu = c\}}\Delta {\{\psi = c\}} } \textrm{d}x \end{aligned}$$

where $A \Delta B$ denotes the symmetric difference between two sets. Under our assumptions, there exists an open set $O\subset M$ containing the streamline $\{\psi ^\nu = c\}$ uniformly in $\nu$. By the trace theorem,

$$\begin{aligned}{} & {} \int _{\{\psi ^\nu = c\}} n^\nu \cdot \nabla (\omega ^\nu -\omega ) \textrm{d}\ell \lesssim \Vert n^\nu \Vert _{H^{1/2+}(O)} \Vert \omega ^\nu -\omega \Vert _{H^{3/2+}(O)}\\{} & {} \qquad \lesssim \Vert \omega ^\nu \Vert _{H^{3/2}(O)} \Vert \omega ^\nu -\omega \Vert _{H^{3/2+}(O)}. \end{aligned}$$

Combined with the fact that $\omega = F(\psi )$, we find

$$\begin{aligned} F'(c) \int _{\{\psi = c\}} u \cdot \textrm{d}\ell&= \int _{\{\psi ^\nu = c\}} \Delta (\omega ^\nu - F(\psi )) \textrm{d}x - \int \Delta F(\psi ) \textbf{1}_{{\{\psi ^\nu = c\}}\Delta {\{\psi = c\}} } \textrm{d}x. \end{aligned}$$

Thus, for any $\delta >0$, we have the bound

$$\begin{aligned} \left| F'(c) \int _{\Gamma (c)} u \cdot \textrm{d}s \right|&\lesssim \ \Vert \omega ^\nu \Vert _{H^{3/2}(O)} \Vert \omega ^\nu -\omega \Vert _{H^{3/2+}(O)} \\&\qquad + \Vert \Delta F(\psi ) \Vert _{L^{1+\delta } (M)} \left( \textrm{Area} \left( {\{\psi ^\nu = c\}}\Delta {\{\psi = c\}} \right) \right) ^{1/\delta }. \end{aligned}$$

Consequently, using (6) and taking the limit of the upper bound, we have

$$\begin{aligned} F'(c) \int _{\Gamma (c)} u \cdot \textrm{d}s =0. \end{aligned}$$

By our hypotheses that $\psi$ has a single stagnation point $\{\psi =0\}$ in M, the circulation $\oint _{\{\psi = c\}} u \cdot \textrm{d}\ell \ne 0$ for all $c\ne 0$. Thus, since $F'$ is continuous, we must have that $F'(c)=0$ for all $c\in \textrm{rang} (\psi )$ so that $F= {\omega _0}$ for some ${\omega _0}\in \mathbb {R}$.

5 Proof of Theorem 3

5.1 Iteration and Bootstraps

Here we produce a unique solution $(Q,{\omega _0})$ of

$$\begin{aligned} \partial _s Q - q\partial _\psi ^2 Q&= 0, \nonumber \\ Q&:= q^2-\omega _0^2 q_e ^2, \nonumber \\ Q(s,0)&= {\varepsilon }\overline{{\omega _0}} q_e ^2(s) + 2{\varepsilon }g(s)q_e (s) + {\varepsilon }^2 g^2(s),\nonumber \\ Q(s,\infty )&=0, \end{aligned}$$

(19a)

on $(s,\psi )\in \mathbb {T} \times \mathbb {R}^+$, for arbitrary $g:\mathbb {T}\rightarrow \mathbb {R}$ and sufficiently small ${\varepsilon }:= {\varepsilon }(g;q_e )$. Here, ${\omega _0}$ is to be determined together with Q, and we introduced $\overline{{\omega _0}}\in \mathbb {R}$ (anticipated to be an O(1) quantity as it depends on ${\varepsilon }$) defined by

$$\begin{aligned} 1-\omega _0^2 = {\varepsilon }\overline{{\omega _0}}. \end{aligned}$$

To prove this result, it is convenient to rewrite (19) as

$$\begin{aligned} \partial _s Q -{\omega _0}q_e \partial _\psi ^2 Q = \left( 1- \tfrac{{\omega _0}q_e }{\sqrt{\omega _0^2 q_e ^2+ Q }}\right) \partial _s Q. \end{aligned}$$

(20)

We will study of following iteration scheme

$$\begin{aligned} \partial _s Q_n -{\omega _0}_{n-1} q_e \partial _\psi ^2 Q_n&= \left( 1- \tfrac{{\omega _0}_{n-1} q_e }{\sqrt{{\omega _0}_{n-1} ^2 q_e ^2+ Q_{n-1} }}\right) \partial _s Q_{n-1}, \\ Q_n(s,0)&= (1- {\omega _0}_n^2) q_e ^2(s) + 2{\varepsilon }g(s)q_e (s) + {\varepsilon }^2 g^2(s)\nonumber \\&= {\varepsilon }\overline{{\omega _0}}_n q_e ^2(s) + 2{\varepsilon }g(s)q_e (s) + {\varepsilon }^2 g^2(s), \nonumber \\ Q_n(s,\infty )&=0,\nonumber \end{aligned}$$

(21)

with $Q_{-1} = 0$, ${\omega _0}_{-1}= 1$. Schematically, we think that $({\omega _0}_{n-1}, Q_{n-1}) \mapsto {\omega _0}_n \mapsto Q_n$, that is ${\omega _0}_n$ is determined on the onset by a compatibility condition for the linear problem which depends on the prior iterate, and $Q_n$ is subsequently solved for ${\omega _0}_{n-1}$. Let

$$\begin{aligned} f(Q, s;{\omega _0})&:=\left( 1- \tfrac{{\omega _0}q_e }{\sqrt{\omega _0^2 q_e ^2+ Q }}\right) \partial _s Q= \left( 1 - \sqrt{1 - \frac{Q}{\omega _0^2 q_e ^2 + Q}}\right) \partial _s Q. \end{aligned}$$

(22)

In this system ${\omega _0}_n$ is chosen to enforce that

(23)

Conceptually, it is clearer to separate out the explicit component of $\overline{{\omega _0}}_n$, which is O(1) and independent of n, and the smaller amplitude implicit component of $\overline{{\omega _0}}_n$ as follows

(24)

With this, we can solve the above equation for $Q_n$. For $\psi \ge 0$, we define

$$\begin{aligned} \langle \psi \rangle =1+\psi \end{aligned}$$

For a function $f=f(s,\psi )$ defined on $\mathbb T_L\times \mathbb R_+$, we define

$$\begin{aligned} \Vert f\Vert _{X_{k,m}}^2&=\sum _{k'=0}^k\sum _{m'=0}^m\left\rbrace \Vert \langle \psi \rangle ^{m'}\partial _s^{k'}f\Vert _{L^2(\mathbb T_L\times \mathbb R_+)}^2+\Vert \langle \psi \rangle ^{m'}\partial _s^{k'+1}f\Vert _{L^2(\mathbb T_L\times \mathbb R_+)}^2\right\lbrace \nonumber \\&\quad +\sum _{k'=0}^k\sum _{m'=0}^m\left\rbrace \Vert \langle \psi \rangle ^{m'}\partial _\psi \partial _s^{k'}f\Vert _{L^2(\mathbb T_L\times \mathbb R_+)}^2+\Vert \langle \psi \rangle ^{m'}\partial _\psi ^2\partial _s^{k'}f \Vert _{L^2(\mathbb T_L\times \mathbb R_+)}^2 \right\lbrace . \end{aligned}$$

(25)

We will construct the unique solution of the Eq. (20) in the space $X_{2,50}$. By the standard Sobolev embedding, we also have

$$\begin{aligned} \Vert f\Vert _{L^\infty }\lesssim \Vert f\Vert _{L^2}+\Vert \partial _sf\Vert _{L^2}+\Vert \partial _\psi f\Vert _{L^2}+\Vert \partial _s\partial _\psi f\Vert _{L^2}\lesssim \Vert f\Vert _{X_{1,0}}. \end{aligned}$$

Remark

(Exponential decay of Q for $\psi \gg 1$) In fact, one can prove existence in a space encoding exponential decay in $\psi$, as should be expect for a (nonlinear) heat equation with data at $\psi =0$. For simplicity of presentation, we prove only algebraic decay but the requisite modifications involving exponential weights are standard.

Indeed, we have the following result that tells us this iteration is well-defined.

Lemma 4

Let $n \ge 0$. Assume that $Q_{n-1} \in X_{2,50}$, and that (27)–(28) are valid until index $n -1$. Assume further that $\overline{{\omega _0}}_{n}$ is defined according to (23). Then, there exists a unique solution, $Q_n$ to the system (21).

Proof

We first of all write the system (21) as follows

$$\begin{aligned} \partial _s Q_n -{\omega _0}_{n-1} q_e \partial _\psi ^2 Q_n&= F_{n-1}, \\ Q_n(s,0)&= b_{n-1}(s),\\ Q_n(s,\infty )&=0, \end{aligned}$$

We introduce the variable

$$\begin{aligned} t = J(s), \qquad \frac{\textrm{d}t}{\textrm{d}s} = J'(s) = {\omega _0}_{n-1} q_e (s). \end{aligned}$$

We notice that $\frac{\textrm{d}t}{\textrm{d}s}$ is bounded above and below and hence determines an invertible transformation due to the fact that ${\omega _0}_{n-1} q_e (s) > 0$. We also note that by writing ${\omega _0}_{n-1} q_e (s) = \langle {\omega _0}_{n-1} q_e \rangle + ({\omega _0}_{n-1} q_e - \langle {\omega _0}_{n-1} q_e \rangle )$, we have

$$\begin{aligned} t = \langle {\omega _0}_{n-1} q_e \rangle s + \int _0^s {\omega _0}_{n-1} (q_e - \langle q_e \rangle ) \textrm{d}s', \end{aligned}$$

which maps $\mathbb {T}_L$ into $\mathbb {T}_{{\omega _0}_{n-1} \langle q_e \rangle L}$. We next introduce

$$\begin{aligned} V_n(t, \psi ) = V_n(J(s), \psi ) = Q_n(s, \psi ). \end{aligned}$$

This object satisfies the system

$$\begin{aligned} \partial _t V_n - \partial _\psi ^2 V_n = \frac{F_{n-1}}{{\omega _0}_{n-1} q_e } =: G_{n-1}. \end{aligned}$$

We expand the solution of $V_n$ in a Fourier basis in the t variable as follows

$$\begin{aligned} ik \widehat{V}^k_n - \partial _\psi ^2 \widehat{V}^k_n = \widehat{G}_{n-1}^{k}. \end{aligned}$$

The zero mode equation is exactly the Feynman-Lagerstrom formula, (23). For the k’th mode, where $k \ne 0$, we write the explicit formula:

$$\begin{aligned} \hat{V}_n^k=e^{-\sqrt{ik}\psi }\hat{b}_{n-1}^k+\frac{1}{2\sqrt{ik}}\int _0^\infty \left(e^{-\sqrt{ik}(\psi +\psi ')}-e^{-\sqrt{ik}|\psi -\psi '|}\right)\hat{G}_{n-1}^kd\psi ', \end{aligned}$$

where $\sqrt{ik}$ is the complex square root of ik with positive real part. We now observe that for $G_{n-1} \in X_{2,50}$, the above integrals converge to zero as $\psi \rightarrow \infty$ (when $k \ne 0$). This completes the proof.

We define the differences $\triangle \overline{{\omega _0}}_n$ and $\triangle Q_n$ to be

Differences in Q obey:

$$\begin{aligned}&\partial _s (Q_n-Q_{n-1}) -{\omega _0}_{n-1} q_e \partial _\psi ^2 (Q_n- Q_{n-1}) \nonumber \\&\quad = \left( 1- \tfrac{{\omega _0}_{n-1} q_e }{\sqrt{{\omega _0}_{n-1} ^2 q_e ^2+ Q_{n-1} }}\right) \partial _s Q_{n-1}- \left( 1- \tfrac{{\omega _0}_{n-2} q_e }{\sqrt{{\omega _0}_{n-2} ^2 q_e ^2+ Q_{n-2} }}\right) \partial _s Q_{n-2}, \nonumber \\&\qquad + ( {\omega _0}_{n-1} -{\omega _0}_{n-2}) \partial _\psi ^2 Q_{n-1}, \nonumber \\&(Q_n-Q_{n-1}) (s,0) = {\varepsilon }( \overline{{\omega _0}}_{n}- \overline{{\omega _0}}_{n-1}) q_e ^2(s),\nonumber \\&Q_n(s,\infty ) =0. \end{aligned}$$

(26)

Remark

(Obtaining the sharper bounds stated in Theorem 3) In what follows, we will bootstrap bounds of ${\varepsilon }^{1-}$, specifically ${\varepsilon }^{0.97}$ for $|\overline{{\omega _0}}_{\textrm{Err}, n}|$ and ${\varepsilon }^{0.99}$ for $\Vert Q_{n} \Vert _{X_{2, 50}}$, although any power less than 1 would suffice by the same argument given below. This is not essential, it is to avoid keeping track of large constants for simplicity of the bootstrap argument. In fact, from these bounds one can deduce a posteriori sharper estimates of the form $|\overline{{\omega _0}}_{\textrm{Err}, n}| \le C_1 {\varepsilon }$ and $\Vert Q_{n} \Vert _{X_{2, 50}} \le C_2 {\varepsilon }$ for some, possibly large, constants $C_1,C_2>0$ by taking the proved bounds on ${\omega _0}$ and Q, returning to the equation, and performing the estimate again.

We will establish the following bounds

$$\begin{aligned} |\overline{{\omega _0}}_{\textrm{Err}, n}|&\le {\varepsilon }^{0.97} , \end{aligned}$$

(27)

$$\begin{aligned} \Vert Q_{n} \Vert _{X_{2, 50}}&\le {\varepsilon }^{0.99}, \end{aligned}$$

(28)

$$\begin{aligned} | \Delta \overline{{\omega _0}}_{n}|&\le {\varepsilon }^{1.97} |\Delta \overline{{\omega _0}}_{n-1}| + {\varepsilon }^{-0.02} \Vert \Delta Q_{n-1} \Vert _{X_{1,4}},\end{aligned}$$

(29)

$$\begin{aligned} \Vert \Delta Q_n \Vert _{X_{2, 50}}&\le {\varepsilon }^{\frac{3}{4}} \Vert \Delta Q_{n-1} \Vert _{X_{2,50}} + {\varepsilon }^{\frac{1}{2}} |\Delta \overline{{\omega _0}}_n| + {\varepsilon }^{\frac{3}{2}} |\Delta \overline{{\omega _0}}_{n-1}|, \end{aligned}$$

(30)

which immediately imply the main result. The bounds (27)–(28) show that $(\overline{{\omega _0}}_{\textrm{Err},n}, Q_n) \in B_{{\varepsilon }^{1.99}, {\varepsilon }^{0.99}} \subset \mathbb {R} \times X_{2, 50}$, whereas the bounds (29)–(30) show that iteration converges to a unique fixed point. A standard fixed point result imply that these bootstrap bounds give the main theorem:

Proof

We insert the bound (29) into the second term on the right-hand side of (30) in order to get the following

$$\begin{aligned} | \Delta \overline{{\omega _0}}_{n}|&\le {\varepsilon }^{1.97} |\Delta \overline{{\omega _0}}_{n-1}| + {\varepsilon }^{-0.02} \Vert \Delta Q_{n-1} \Vert _{X_{2,50}} ,\\ \Vert \Delta Q_n \Vert _{X_{2, 50}}&\le {\varepsilon }^{\frac{1}{4}} \Vert \Delta Q_{n-1} \Vert _{X_{2,50}} + {\varepsilon }^{\frac{3}{2}} |\Delta \overline{{\omega _0}}_{n-1}|. \end{aligned}$$

Define $\textbf{Y}_n:= (\Delta Q_n, {\varepsilon }\Delta \overline{{\omega _0}}_n) \in X_{2,50} \times \mathbb {R}$, endowed with the product norm. Then

$$\begin{aligned} \Vert \textbf{Y}_n \Vert _{X_{2,50} \times \mathbb {R}} \le {\varepsilon }^{\frac{1}{5}}\Vert \textbf{Y}_{n-1} \Vert _{X_{2,50} \times \mathbb {R}}. \end{aligned}$$

It is therefore clear that

$$\begin{aligned} \Vert \textbf{Y}_n \Vert _{X_{2,50} \times \mathbb {R}}&\le {\varepsilon }^{\frac{n}{5}}\Vert \textbf{Y}_{0} \Vert _{X_{2,50} \times \mathbb {R}} \\&\le {\varepsilon }^{\frac{n}{5}} ( \Vert Q_{-1} \Vert _{X_{2,50}} + \Vert Q_0 \Vert _{X_{2,50}} + {\varepsilon }|\overline{{\omega _0}}_{-1}| + {\varepsilon }|\overline{{\omega _0}}_0|) \\&\le C{\varepsilon }^{\frac{n}{5}}. \end{aligned}$$

This then implies that $(Q_n, \overline{{\omega _0}}_{n})$ is a Cauchy sequence in $X_{2,50} \times \mathbb {R}$, and hence converges to a limit, $(Q_{\infty }, \overline{{\omega _0}}_{\infty })$. We can therefore pass to the limit in Eq. (21) as well as in (23) to conclude that $(Q_\infty , \overline{{\omega _0}}_{\infty })$ satisfy the system (19).

We now prove uniqueness. We assume that $(Q_{1}, {\omega _0}_{,1})$ and $(Q_2, {\omega _0}_{,2})$ are two solutions to (19) in the space $X_{2,50} \times \mathbb {R}$. We may therefore write an analogous Eq. (26) on $Q_1 - Q_2$ (without the iteration), which reads:

$$\begin{aligned}&\partial _s (Q_1-Q_{2}) -{\omega _0}_{,1} q_e \partial _\psi ^2 (Q_1- Q_{2}) \\&\quad = \left( 1- \tfrac{{\omega _0}_{,1} q_e }{\sqrt{{\omega _0}_{,1} ^2 q_e ^2+ Q_{1} }}\right) \partial _s Q_{1}- \left( 1- \tfrac{{\omega _0}_{,2} q_e }{\sqrt{{\omega _0}_{,2} ^2 q_e ^2+ Q_{2} }}\right) \partial _s Q_{2}, \\&\qquad + ( {\omega _0}_{,1} -{\omega _0}_{,2}) \partial _\psi ^2 Q_{2}, \\&(Q_1-Q_{2}) (s,0) = {\varepsilon }( \overline{{\omega _0}}_{,1}- \overline{{\omega _0}}_{,2}) q_e ^2(s),\\&(Q_1 - Q_2)(s,\infty ) =0, \end{aligned}$$

as well as the analogue of expression (26) (again without the iteration)

Re-applying the a-priori estimates on these systems results in the following bounds:

$$\begin{aligned} | \overline{{\omega _0}}_{,1} - \overline{{\omega _0}}_{,2}|&\le \epsilon ^{1.97} | \overline{{\omega _0}}_{,1} - \overline{{\omega _0}}_{,2}| + \epsilon ^{-0.02} \Vert Q_1 - Q_2 \Vert _{X_{2,50}}, \\ \Vert Q_1 - Q_2 \Vert _{X_{2,50}}&\le {\varepsilon }^{\frac{3}{4}} \Vert Q_1 - Q_2 \Vert _{X_{2,50}} + {\varepsilon }^{\frac{1}{2}} | \overline{{\omega _0}}_{,1} - \overline{{\omega _0}}_{,2}|, \end{aligned}$$

which are the analogues of (29)–(30). The two bounds above clearly imply that $\overline{{\omega _0}}_{,1} = \overline{{\omega _0}}_{,2}$ and $Q_1 = Q_2$. This proves uniqueness.

5.2 $\overline{{\omega _0}}_{\textrm{Err}, n}$ Estimates

Here, we will establish the bootstrap bound (27). Indeed,

Lemma 5

Assume (27)–(30) are valid until the index $n - 1$. Then $\overline{{\omega _0}}_{\textrm{Err},n}$ satisfies:

$$\begin{aligned} |\overline{{\omega _0}}_{\textrm{Err}, n}| \le {\varepsilon }^{0.97}. \end{aligned}$$

Proof

Recall the expression (24), after which we estimate as follows

$$\begin{aligned} |\overline{{\omega _0}}_{\textrm{Err}, n}|&\le \frac{1}{{\varepsilon }} \left| \int _0^\infty y \int _0^{L} f(Q_{n-1}(y,s), s;{\omega _0}_{n-1}) \textrm{d}s\textrm{d}y\right| \\&\lesssim \frac{1}{{\varepsilon }} \left\| \left( 1- \tfrac{{\omega _0}_{n-1} q_e }{\sqrt{{\omega _0}_{n-1}^2 q_e ^2+ Q_{n-1} }}\right) \partial _s Q_{n-1} \langle \psi \rangle ^4 \right\| _{L^2} \\&\lesssim \frac{1}{{\varepsilon }} \left\| \left( 1- \tfrac{{\omega _0}_{n-1} q_e }{\sqrt{{\omega _0}_{n-1}^2 q_e ^2+ Q_{n-1} }}\right) \Vert _{L^\infty } \Vert \partial _s Q_{n-1} \langle \psi \rangle ^4 \right\| _{L^2} \\&= \frac{1}{{\varepsilon }} \left\| 1 - \sqrt{1 - \frac{Q_{n-1}}{{\omega _0}_{n-1}^2 q_e ^2 + Q_{n-1}}}\right\| _{L^\infty } \left\| \partial _s Q_{n-1} \langle \psi \rangle ^4 \right\| _{L^2} \\&\lesssim \frac{1}{{\varepsilon }} \left\| \frac{Q_{n-1}}{{\omega _0}_{n-1}^2 q_e ^2 + Q_{n-1}}\right\| _{L^\infty } \left\| \partial _s Q_{n-1} \langle \psi \rangle ^4 \right\| _{L^2} \lesssim \frac{1}{{\varepsilon }} {\varepsilon }^{0.99} {\varepsilon }^{0.99} < {\varepsilon }^{0.97}, \end{aligned}$$

where we have invoked the bootstrap bound (28) in the final step, as well as the $L^\infty$ estimate

$$\begin{aligned} \left\| \frac{Q_{n-1}}{{\omega _0}_{n-1}^2 q_e ^2 + Q_{n-1}}\right\| _{L^\infty }&\lesssim \frac{1}{{\omega _0}_{n-1}^2 q_e ^2 - \Vert Q_{n-1} \Vert _{L^\infty }} \Vert Q_{n-1} \Vert _{L^\infty } \\&\lesssim \frac{1}{{\omega _0}_{n-1}^2 q_e ^2 - \Vert Q_{n-1} \Vert _{X_{1,0}}} \Vert Q_{n-1} \Vert _{X_{1,0}} \\&\lesssim \frac{1}{{\omega _0}_{n-1}^2 q_e ^2 - C {\varepsilon }^{0.99}} {\varepsilon }^{0.99} \lesssim {\varepsilon }^{0.99}. \end{aligned}$$

Above, we have used the following Sobolev inequality on $\mathbb {T} \times \mathbb {R}_+$, which reads $\Vert f \Vert _{L^\infty } \lesssim \Vert f \Vert _{X_{1,0}}$. This Sobolev embedding will be used repeatedly to estimate nonlinear terms.

5.3 $\Delta \overline{{\omega _0}}_{n}$ Estimates

Here we prove the following lemma.

Lemma 6

Assume (27)–(30) are valid until the index $n - 1$. Assume (27) and (28) are valid until index n. Then the following bound holds

$$\begin{aligned} |\Delta \overline{{\omega _0}}_n| \le \epsilon ^{1.97} |\Delta \overline{{\omega _0}}_{n-1}| + {\varepsilon }^{-0.02} \Vert \Delta Q_{n-1} \Vert _{X_{1,4}}. \end{aligned}$$

Proof

We use the expression

To estimate $I_1$, we have

To estimate $I_2$, we need to use the identity (22) to estimate

$$\begin{aligned} \begin{aligned} f(Q, {\omega _0}) - f(\overline{Q}, \overline{{\omega _0}})&= \left( 1 - \sqrt{1 - \frac{Q}{\omega _0^2 q_e ^2 + Q}}\right) \partial _s Q \\&\qquad - \left( 1 - \sqrt{1 - \frac{\overline{Q}}{\overline{{\omega _0}}^2 q_e ^2 + \overline{Q}}}\right) \partial _s \overline{Q} \\&= \left( 1 - \sqrt{1 - \frac{Q}{\omega _0^2 q_e ^2 + Q}}\right) \partial _s Q \\&\qquad - \left( 1 - \sqrt{1 - \frac{Q}{\omega _0^2 q_e ^2 + Q}}\right) \partial _s \overline{Q} \\&\quad + \left( 1 - \sqrt{1 - \frac{Q}{\omega _0^2 q_e ^2 + Q}}\right) \partial _s \overline{Q}\\&\qquad - \left( 1 - \sqrt{1 - \frac{\overline{Q}}{\overline{{\omega _0}}^2 q_e ^2 + \overline{Q}}}\right) \partial _s \overline{Q} \\&= BD_1 + BD_2. \end{aligned} \end{aligned}$$

Clearly, using the inequality $|1 - \sqrt{1-x}| \le |x|$ for $x\le 1$, we have

$$\begin{aligned} \Vert BD_1 \langle \psi \rangle ^4 \Vert _{L^2}&\lesssim \left\Vert 1 - \sqrt{1 - \frac{Q}{\omega _0^2 q_e ^2 + Q}}\right\Vert_{L^\infty } \Vert \partial _s (Q - \overline{Q}) \langle \psi \rangle ^4 \Vert _{L^2} \\&\lesssim \Vert Q \Vert _{X_{1,0}} \Vert Q - \overline{Q} \Vert _{X_{0,4}}. \end{aligned}$$

and, using the inequality $|\sqrt{1-x} - \sqrt{1-y}| \lesssim |x - y|$ for $x, y \ll 1$, we have

$$\begin{aligned} \Vert BD_2 \langle \psi \rangle ^4 \Vert _{L^2}&\lesssim \Vert \partial _s \overline{Q} \langle \psi \rangle ^4 \Vert _{L^2} \left\Vert \frac{Q}{\omega _0^2 q_e ^2 + Q} - \frac{\overline{Q}}{\overline{{\omega _0}}^2 q_e ^2 + \overline{Q}}\right\Vert_{L^\infty } \\&\lesssim \Vert \partial _s \overline{Q} \langle \psi \rangle ^4 \Vert _{L^2} \left\Vert\frac{\overline{{\omega _0}}^2 Q -\omega _0^2 \overline{Q}}{(\omega _0^2 q_e ^2 + Q)(\overline{{\omega _0}}^2 q_e ^2 + \overline{Q})}\right\Vert_{L^\infty } \\&\lesssim \Vert \partial _s \overline{Q} \langle \psi \rangle ^4 \Vert _{L^2}\left\Vert Q - \overline{Q} \right\Vert_{L^\infty } + \Vert \partial _s \overline{Q} \langle \psi \rangle ^4 \Vert _{L^2} \Vert Q \Vert _{L^\infty } |\omega _0^2 - \overline{{\omega _0}}^2| \\&\lesssim \Vert \overline{Q} \Vert _{X_{0,4}} \Vert Q - \overline{Q} \Vert _{X_{1,0}} + \Vert \overline{Q} \Vert _{X_{0,4}} \Vert Q \Vert _{X_{1,0}} |\omega _0^2 - \overline{{\omega _0}}^2|. \end{aligned}$$

Therefore, we have

$$\begin{aligned} |I_2|&\lesssim \frac{1}{\epsilon } \Vert Q_{n-1} \Vert _{X_{1,0}} \Vert Q_{n-1} - Q_{n-2} \Vert _{X_{0,4}} + \frac{1}{\epsilon } \Vert Q_{n-2} \Vert _{X_{0,4}} \Vert Q_{n-1} - Q_{n-2} \Vert _{X_{1,0}} \\&\quad + \frac{1}{\epsilon } \Vert Q_{n-2} \Vert _{X_{0,4}} \Vert Q_{n-1} \Vert _{X_{1,0}} | {\omega _0}_{n-1}^2 - {\omega _0}_{n-2}^2| \\&\lesssim \frac{1}{{\varepsilon }} \epsilon ^{0.99} \Vert \Delta Q_{n-1} \Vert _{X_{1,4}} + \frac{1}{{\varepsilon }} \epsilon ^{0.99} \epsilon ^{0.99} \epsilon |\overline{{\omega _0}}_{n-1} - \overline{{\omega _0}}_{n-2}| \\&\le {\varepsilon }^{-0.02} \Vert \Delta Q_{n-1} \Vert _{X_{1,4}} + \frac{{\varepsilon }^{1.97}}{2} |\overline{{\omega _0}}_{n-1} - \overline{{\omega _0}}_{n-2}|. \end{aligned}$$

Pairing these bounds together, we get the desired result.

5.4 Abstract Q Estimates

For future use, it turns out we will have a need to develop our estimates on a slightly more abstract system. Therefore, we consider

$$\begin{aligned} \partial _s Q - {\omega _0}q_e \partial _\psi ^2 Q&= F + \partial _{\psi }^2 G,\\ Q(s, 0)&= b(s), \nonumber \\ Q(s, \infty )&= 0.\nonumber \end{aligned}$$

(31)

We develop a high-order energy method to treat equation (31). We commute $\partial _s^k$ to obtain

$$\begin{aligned} \partial _s Q^{(k)} - {\omega _0}q_e \partial _\psi ^2 Q^{(k)}&= F^{(k)} + \partial _{\psi }^2 G^{(k)} + A_{\text {comm},k}, \\ Q^{(k)}(s, 0)&= b^{(k)}(s), \nonumber \\ Q^{(k)}(s, \infty )&= 0,\nonumber \end{aligned}$$

(32)

where the commutator term

$$\begin{aligned} A_{\text {comm},k} := \sum _{k' = 0}^{k-1} \left( {\begin{array}{c}k\\ k'\end{array}}\right) \partial _s^{k-k'} q_e \partial _\psi ^2 Q^{(k')}, \end{aligned}$$

and where we adopt the short-hand $f^{(k)}(s, \psi ):= \partial _s^k f(s, \psi )$ for an abstract function $f(s, \psi )$.

Proposition 7

Assume that the boundary condition b(s) and the source term $F + \partial _{\psi }^2\,G$ satisfy the Feynman–Lagerstrom compatibilty condition (23). Then the solution Q to (31) obeys the following inequality:

$$\begin{aligned} \Vert Q \Vert _{X_{k,m}} \lesssim \sum _{k' = 0}^k \Vert \partial _s^{k'} F \langle \psi \rangle ^{m+4} \Vert _{L^2} +\sum _{k' = 0}^k \sum _{j= 0}^2 \Vert \partial _s^{k'} \partial _{\psi }^j G \langle \psi \rangle ^m \Vert _{L^2} +\Vert b\Vert _{H^{k+1}_s} . \end{aligned}$$

(33)

The first task is we lift the boundary condition b(s) by considering the lift function

$$\begin{aligned} L(s, \psi ) := L[b](s, \psi ) := e^{-\psi } b(s) \end{aligned}$$

and consequently

$$\begin{aligned} \mathring{Q} := Q - L[b], \end{aligned}$$

which satisfies the following system

$$\begin{aligned} \partial _s \mathring{Q} - {\omega _0}q_e \partial _\psi ^2 \mathring{Q}&= F + \partial _{\psi }^2 G + G_{\text {Lift}},\\ \mathring{Q}(s, 0)&= 0, \nonumber \\ \mathring{Q}(s, \infty )&= 0\nonumber \end{aligned}$$

(34)

where

$$\begin{aligned} G_{\text {Lift}}:= e^{-\psi } ( {\omega _0}q_e b(s) - b'(s)). \end{aligned}$$

(35)

We will need to work in higher order norms. Therefore, we present the equations upon commuting $\partial _s^k$ to (34), which yield

$$\begin{aligned} \partial _s \mathring{Q}^{(k)} - {\omega _0}q_e \partial _\psi ^2 \mathring{Q}^{(k)} =&F^{(k)} + \partial _{\psi }^2 G^{(k)} +G_{\text {Lift}}^{(k)}+ C_{\text {comm},k},\nonumber \\ \mathring{Q}^{(k)}(s, 0) =&0, \nonumber \\ \mathring{Q}^{(k)}(s, \infty ) =&0. \end{aligned}$$

(36)

Above, we define the commutator term as follows:

$$\begin{aligned} C_{\text {comm},k}:= {\omega _0}\textbf{1}_{k \ge 1} \sum _{k' =0}^{k-1} \left( {\begin{array}{c}k\\ k'\end{array}}\right) \partial _s^{k-k'} q_e \partial _\psi ^2 \mathring{Q}^{(k')}. \end{aligned}$$

(37)

Lemma 8

For any $\delta >0$ the following bounds hold (where the constant $C_\delta \uparrow \infty$ as $\delta \downarrow 0$):

$$\begin{aligned} \Vert \partial _{\psi } Q^{(k)} \langle \psi \rangle ^m \Vert _{L^2}^2&\le C_{\delta } \Vert F^{(k)} \langle \psi \rangle ^m \Vert _{L^2}^2 +C_{\delta } \Vert \partial _{\psi }^2 G^{(k)} \langle \psi \rangle ^m \Vert _{L^2}^2 + C_{\delta } \Vert b \Vert _{H^{k+1}_s}^2 \nonumber \\&\quad + \delta \Vert Q^{(k)} \langle \psi \rangle ^m \Vert _{L^2}^2 + C_\delta \textbf{1}_{k \ge 1} \sum _{k' = 0}^{k-1} \Vert \partial _\psi Q^{(k')} \langle \psi \rangle ^m \Vert _{L^2}^2\nonumber \\&\quad + \textbf{1}_{m \ge 1}\Vert Q^{(k)} \langle \psi \rangle ^{m-1} \Vert _{L^2}^2. \end{aligned}$$

(38)

Proof

We multiply (36) by $\mathring{Q}^{(k)}\langle \psi \rangle ^{2m}$ and integrate by parts to get the identity

$$\begin{aligned} \begin{aligned}&\frac{\partial _s}{2} \int _{\mathbb {R}_+} |\mathring{Q}^{(k)}|^2 \langle \psi \rangle ^{2m}\textrm{d}\psi + {\omega _0}q_e (s) \int _{\mathbb {R}_+} |\partial _{\psi }\mathring{Q}^{(k)}|^2 \langle \psi \rangle ^{2m} \textrm{d}\psi \\&\quad = \int (F^{(k)} + \partial _{\psi }^2G^{(k)})\mathring{ Q}^{(k)} \langle \psi \rangle ^{2m} \textrm{d}\psi + \int G^{(k)}_{\text {Lift}} \mathring{Q}^{(k)} \langle \psi \rangle ^{2m} \textrm{d}\psi \\&\qquad + \frac{2m(2m-1)}{2} {\omega _0}q_e (s) \int |\mathring{Q}^{(k)}|^2 \langle \psi \rangle ^{2m-2} \textrm{d}\psi \\&\qquad - \int {\omega _0}\textbf{1}_{k \ge 1} \sum _{k' =0}^{k-1} \left( {\begin{array}{c}k\\ k'\end{array}}\right) \partial _s^{k-k'} q_e \partial _\psi \mathring{Q}^{(k')} \partial _\psi \mathring{Q}^{(k)} \langle \psi \rangle ^{2m} \textrm{d}\psi \\&\qquad -\int {\omega }_0\textbf{1}_{k\ge 1}\sum _{k'=0}^{k-1}\left( {\begin{array}{c}k\\ k'\end{array}}\right) \partial _s^{k-k'}q_e\partial _\psi \mathring{Q}^{(k')}\mathring{Q}^{(k)} 2m\langle \psi \rangle ^{2m-1}. \end{aligned} \end{aligned}$$

We now integrate in $s \in \mathbb {T}$, and the $\partial _s$ term drops out due to periodicity. This implies

$$\begin{aligned} \begin{aligned} \left\Vert\partial _\psi \mathring{Q}^{(k)}\langle \psi \rangle ^m\right\Vert_{L^2}^2&\lesssim \left|\int \int (F^{(k)} + \partial _{\psi }^2 G^{(k)}) \mathring{Q}^{(k)} \langle \psi \rangle ^{2m} \textrm{d}\psi \textrm{d}s \right|\\&\quad +\left|\int \int G^{(k)}_{\text {Lift}} \mathring{Q}^{(k)} \langle \psi \rangle ^{2m} \textrm{d}\psi \textrm{d}s \right|\\&\quad +\left\Vert\mathring{Q}^{(k)}\langle \psi \rangle ^{m-1} \right\Vert_{L^2}^2\\&\quad +\left|\int \int {\omega _0}\textbf{1}_{k \ge 1} \sum _{k' =0}^{k-1} \left( {\begin{array}{c}k\\ k'\end{array}}\right) \partial _s^{k-k'} q_e \partial _\psi \mathring{Q}^{(k')} \partial _\psi \mathring{Q}^{(k)} \langle \psi \rangle ^{2m} \textrm{d}\psi \textrm{d}s\right|\\&\quad +\left|\int \int {\omega }_0\textbf{1}_{k\ge 1}\sum _{k'=0}^{k-1}\left( {\begin{array}{c}k\\ k'\end{array}}\right) \partial _s^{k-k'}q_e\partial _\psi \mathring{ Q}^{(k')}\mathring{Q}^{(k)}\langle \psi \rangle ^{2m-1}\textrm{d}\psi \textrm{d}s\right|. \end{aligned} \end{aligned}$$

This implies

$$\begin{aligned} \begin{aligned} \left\| \partial _\psi \mathring{Q}^{(k)}\langle \psi \rangle ^m\right\| _{L^2}^2&\le \delta \left\| \mathring{Q}^{(k)}\langle \psi \rangle ^m \right\| _{L^2}^2+C_\delta \left\| F^{(k)}\langle \psi \rangle ^m \right\| _{L^2}^2+C_\delta \left\| \partial _\psi ^2 G^{(k)}\langle \psi \rangle ^m \right\| _{L^2}^2\\&\quad +C_\delta \Vert b\Vert _{H^{k+1}_s}+ C_0\textbf{1}_{\{m \ge 1\}} \left\| \mathring{Q}^{(k)}\langle \psi \rangle ^{m-1}\right\| _{L^2}^2\\&\quad +C_\delta \textbf{1}_{k \ge 1}\sum _{k'=0}^{k-1}\left\| \partial _\psi \mathring{Q}^{(k')}\langle \psi \rangle ^m \right\| _{L^2}^2+\delta \left\| \partial _\psi \mathring{Q}^{(k)}\langle \psi \rangle ^m \right\| _{L^2}^2, \end{aligned} \end{aligned}$$

where $\delta >0$ is small and $C_\delta \sim \delta ^{-1}$. The result follows immediately, using the fact that

$$\begin{aligned} \mathring{Q}^{(k)}=Q^{(k)}-e^{-\psi }b^{(k)}(s). \end{aligned}$$

This concludes the proof of the lemma.

Lemma 9

Let $k \ge 0, m \ge 0$. The solution $Q^{(k)}$ to (32) satisfies the following estimate:

$$\begin{aligned} \Vert \partial _s Q^{(k)} \langle \psi \rangle ^m \Vert _{L^2}^2&\lesssim \Vert F^{(k)} \langle \psi \rangle ^m \Vert _{L^2}^2 + \Vert \partial _{\psi }^2 G^{(k)} \langle \psi \rangle ^m \Vert _{L^2}^2 + \Vert b \Vert _{H^{k+1}_s}^2 \nonumber \\&\quad + \textbf{1}_{m \ge 1} \Vert \partial _\psi Q^{(k)} \langle \psi \rangle ^{m-1} \Vert _{L^2}^2 + \textbf{1}_{k \ge 1} \sum _{k' =0}^{k-1} \Vert \partial _\psi ^2 Q^{(k')} \langle \psi \rangle ^m \Vert _{L^2}^2. \end{aligned}$$

(39)

Proof

We multiply (36) by $\frac{1}{q_e (s)} \partial _s \mathring{Q}^{(k)} \langle \psi \rangle ^{2\,m}$ and integrate by parts to produce

$$\begin{aligned}&\int \frac{1}{q_e (s)} (F^{(k)} + \partial _{\psi }^2 G^{(k)}) \partial _s \mathring{Q}^{(k)} \langle \psi \rangle ^{2m} \textrm{d}\psi + \int \frac{1}{q_e (s)} G^{(k)}_{\text {Lift}} \partial _s \mathring{Q}^{(k)} \langle \psi \rangle ^{2m} \textrm{d}\psi \\&\qquad + \int \frac{1}{q_e (s)}C_{\text {comm},k} \partial _s \mathring{Q}^{(k)} \langle \psi \rangle ^{2m} \textrm{d}\psi \\&\quad = \int \frac{1}{q_e (s)} |\partial _s \mathring{Q}^{(k)}|^2 \langle \psi \rangle ^{2m} \textrm{d}\psi + {\omega _0}\int \partial _{\psi }^2 \mathring{Q}^{(k)} \partial _s \mathring{Q}^{(k)} \langle \psi \rangle ^{2m} \textrm{d}\psi \\&\quad = \int \frac{1}{q_e (s)} |\partial _s \mathring{Q}|^2 \langle \psi \rangle ^{2m}\textrm{d}\psi - {\omega _0}\frac{\partial _s}{2} \int |\partial _\psi \mathring{Q}|^2 \langle \psi \rangle ^{2m}\textrm{d}\psi \\&\qquad - {\omega _0}2m \int \partial _\psi \mathring{Q}^{(k)} \partial _s \mathring{Q}^{(k)} \langle \psi \rangle ^{2m-1} \textrm{d}\psi . \end{aligned}$$

Above, we have used the homogeneous boundary condition for $\mathring{Q}$ to integrate by parts. We now integrate over $s \in \mathbb {T}_L$ and use periodicity to eliminate the second term on the right-hand side above, which results in

$$\begin{aligned}\begin{aligned}&\int \int \frac{1}{q_e(s)}\left| \partial _s\mathring{Q}^{(k)}\right| ^2\langle \psi \rangle ^{2m}\textrm{d}\psi \textrm{d}s\\&\quad \le C_\delta \left\| F^{(k)}\langle \psi \rangle ^m\right\| _{L^2}^2+C_\delta \left\| \partial _\psi ^2 G^{(k)}\langle \psi \rangle ^m\right\| _{L^2}^2\\&\qquad +C_\delta \left\| G_{\text {Lift}}^{(k)}\langle \psi \rangle ^m \right\| _{L^2}^2+C_\delta \left\| C_{\text {comm},k} \langle \psi \rangle ^m \right\| _{L^2}^2\\&\qquad +\delta \left\| \partial _s\mathring{Q}^{(k)}\langle \psi \rangle ^m \right\| _{L^2}^2+C_\delta \left\| \partial _\psi \mathring{Q}^{(k)}\langle \psi \rangle ^{m-1} \right\| _{L^2}^2. \end{aligned} \end{aligned}$$

Recalling (37), (35) and absorbing the last term on the right-hand side to the left, we get

$$\begin{aligned} \begin{aligned} \left\| \partial _s\mathring{Q}^{(k)}\langle \psi \rangle ^m \right\| _{L^2}&\le C_\delta \left( \left\| F^{(k)}\langle \psi \rangle ^m\right\| _{L^2}^2+\left\| \partial _\psi ^2 G^{(k)}\langle \psi \rangle ^m\right\| _{L^2}^2+\Vert b\Vert _{H^{k+1}_s}^2\right) \\&\quad +C_\delta \textbf{1}_{k \ge 1} \sum _{k' =0}^{k-1} \left\| \partial _\psi ^2\mathring{Q}^{(k')}\langle \psi \rangle ^m\right\| _{L^2}^2. \end{aligned} \end{aligned}$$

We conclude the proof of the lemma, upon using the fact that $\mathring{Q}^{(k)}=Q^{(k)}-e^{-\psi }b^{(k)}(s).$

We now need to estimate the zero mode of $Q^{(k)}$. Clearly, this is nontrivial only for $k = 0$ (when $k\ge 1$ there is no zero mode).

Lemma 10

The zero mode, $Q^{(=0)}$, to the solution of (31), satisfies the following bound:

$$\begin{aligned} \Vert Q^{(= 0)} \langle \psi \rangle ^m\Vert _{L^2_\psi }^2 \lesssim \&\Vert Q^{(\ne 0)} \langle \psi \rangle ^m \Vert _{L^2_\psi }^2 + \Vert F \langle \psi \rangle ^{m + 4} \Vert _{L^2}^2 + \Vert G \langle \psi \rangle ^m\Vert _{L^2}^2. \end{aligned}$$

(40)

Proof

We integrate Eq. (31) to generate the identity for each $\psi \in \mathbb {R}_+$:

$$\begin{aligned} \partial _{\psi }^2 \int _0^L q_e (s) Q(s, \psi ) \textrm{d}s = - \frac{1}{{\omega _0}}\int _0^L (F + \partial _\psi ^2 G) \textrm{d}s , \end{aligned}$$

after which we integrate twice from $\infty$ to get

$$\begin{aligned} \int _0^L q_e (s) Q(s, \psi ) \textrm{d}s =&- \frac{1}{{\omega _0}} \int _{\psi }^\infty \int _{\psi '}^\infty \int _0^L(F + \partial _\psi ^2 G ) \textrm{d}s \textrm{d}\psi '' \psi ' \\ =&- \frac{1}{{\omega _0}} \int _{\psi }^\infty \int _{\psi '}^\infty \int _0^L F \textrm{d}s \textrm{d}\psi '' \psi ' - \frac{1}{{\omega _0}} \int _0^L G \textrm{d}s. \end{aligned}$$

We now separate out the left-hand side

$$\begin{aligned} \int _0^L q_e (s) Q(s, \psi ) \textrm{d}s&= \langle q_e \rangle Q^{(= 0)}(\psi ) + \int _0^L (q_e (s) - \langle q_e \rangle ) Q(s, \psi ) \textrm{d}s \\&=\langle q_e \rangle Q^{(=0)}(\psi )+\int _0^L \left(q_e -\langle q_e \rangle \right)\left(Q^{(=0)}(\psi )+Q^{(\ne 0)}(s,\psi ) \right)\textrm{d}s\\&=\langle q_e \rangle Q^{(=0)}(\psi )+\int _0^L q_e ^{(\ne 0)}(s)Q^{(\ne 0)}(s,\psi )\textrm{d}s. \end{aligned}$$

This implies

$$\begin{aligned} \begin{aligned} Q^{(=0)}(\psi )&=-\frac{1}{{\omega }_0\langle q_e \rangle }\int _\psi ^\infty \int _{\psi '}^\infty \int _0^L F \textrm{d}s\textrm{d}\psi ''\textrm{d}\psi '-\frac{1}{{\omega }_0\langle q_e \rangle }\int _0^L G\textrm{d}s\\&\quad -\frac{1}{\langle q_e \rangle }\int _0^L q_e ^{(\ne 0)}(s)Q^{(\ne 0)}(s,\psi )\textrm{d}s. \end{aligned} \end{aligned}$$

We therefore obtain

$$\begin{aligned} \Vert Q^{(= 0)} \langle \psi \rangle ^m\Vert _{L^2_\psi }^2 \lesssim&\int _{\mathbb {R}_+} \Big ( \int _0^L q_e ^{(\ne 0)}(s) Q^{(\ne 0)}(s, \psi ) \textrm{d}s \Big )^2 \langle \psi \rangle ^{2m} \textrm{d}\psi \\&+ \int _{\mathbb {R}_+} \langle \psi \rangle ^{2m} \Big ( \int _{\psi }^\infty \int _{\psi '}^\infty \int _0^L F \Big )^2 \textrm{d}\psi + \Vert G^{(=0)} \langle \psi \rangle ^m\Vert _{L^2_\psi }^2 \\ =:&\ \mathcal {I}_1 + \mathcal {I}_2 + \mathcal {I}_3 . \end{aligned}$$

Clearly, $\mathcal {I}_3$ is majorized by the last term on the right-hand side of (40). We will estimate the first term above, which we call $\mathcal {I}_1$. Using Hölder’s inequality, we get

$$\begin{aligned} \mathcal {I}_1 :=&\int _{\mathbb {R}_+} \langle \psi \rangle ^{2m} \Big ( \int _0^L q_e ^{(\ne 0)}(s)Q^{(\ne 0)}(s, \psi ) \textrm{d}s \Big )^2 \textrm{d}\psi \\ \lesssim&\ \Vert Q^{( \ne 0)}\langle \psi \rangle ^m \Vert _{L^2}^2 . \end{aligned}$$

To estimate $\mathcal {I}_2$, we need to pay weights as follows using Cauchy-Schwartz:

$$\begin{aligned} \Big |\int _{\psi '}^\infty \int _0^L F\Big | \lesssim \Vert F \langle \psi \rangle ^{m + 4} \Vert _{L^2} \langle \psi ' \rangle ^{-m-\frac{7}{2}}, \end{aligned}$$

which therefore implies that $|\mathcal {I}_2| \lesssim \Vert F \langle \psi \rangle ^{m + 4} \Vert _{L^2}$. This completes the proof.

Lemma 11

Let $k \ge 0, m \ge 0$. The solution $Q^{(k)}$ to (32) satisfies the following estimate:

$$\begin{aligned} \Vert \partial _\psi ^2 Q^{(k)} \langle \psi \rangle ^m \Vert _{L^2}^2 \lesssim&\ \Vert \partial _s Q^{(k)} \langle \psi \rangle ^m \Vert _{L^2}^2 + \Vert F^{(k)} \langle \psi \rangle ^m \Vert _{L^2}^2 + \Vert \partial _\psi ^2 G^{(k)} \langle \psi \rangle ^m \Vert _{L^2}^2\nonumber \\&+ \textbf{1}_{k \ge 1} \sum _{k' = 0}^{k-1} \Vert \partial _\psi ^2 Q^{(k')} \langle \psi \rangle ^m \Vert _{L^2}^2. \end{aligned}$$

(41)

Proof

We simply rearrange Eq. (32) and apply $L^2$ norm to both sides.

Proof of Proposition 7

Consolidating the bounds (38), (39), (40), (41), we proved (33).

5.5 $Q_n$ Estimates

Lemma 12

Assume (27) is valid up to index n and (28) is valid up to index $n-1$. Then

$$\begin{aligned} \Vert Q_n \Vert _{X_{2, 50}} \le \epsilon ^{0.99}. \end{aligned}$$

Proof

For this bound, motivated by Eq. (21), we set

$$\begin{aligned} F&:= \left( 1- \tfrac{{\omega _0}_{n-1} q_e }{\sqrt{{\omega _0}_{n-1} ^2 q_e ^2+ Q_{n-1} }}\right) \partial _s Q_{n-1},\\ G&:= 0, \\ b&:= {\varepsilon }\overline{{\omega _0}}_n q_e ^2(s) + 2{\varepsilon }g(s)q_e (s) + {\varepsilon }^2 g^2(s). \end{aligned}$$

According to (33), we fix $k = 2, m = 50$, which results in

$$\begin{aligned} \Vert Q_n \Vert _{X_{2,50}}&\lesssim \sum _{k' = 0}^{2} \Vert \partial _s^{k'} F \langle \psi \rangle ^{54} \Vert _{L^2} +\Vert b \Vert _{H^{3}_s}. \end{aligned}$$

(42)

We therefore estimate the two quantities appearing on the right-hand side above. To make notation simpler, we define

$$\begin{aligned} U_{n-1} := \frac{Q_{n-1}}{{\omega _0}_{n-1}^2 q_e ^2 + Q_{n-1}}, \end{aligned}$$

(43)

then

$$\begin{aligned} F=\left(1-\sqrt{1-U_{n-1}}\right)\partial _s Q_{n-1}. \end{aligned}$$

By a direct calculation, we have the following identities

$$\begin{aligned} \begin{aligned} \partial _s U_{n-1}&= \frac{\partial _s Q_{n-1}}{{\omega _0}_{n-1}^2 q_e ^2 + Q_{n-1}} - \frac{Q_{n-1}}{({\omega _0}_{n-1}^2 q_e ^2 + Q_{n-1})^2} ({\omega _0}_{n-1}^2 \partial _s q_e ^2 + \partial _s Q_{n-1}), \\ \partial _s^2 U_{n-1}&= \frac{\partial _s^2 Q_{n-1}}{{\omega _0}_{n-1}^2 q_e ^2 + Q_{n-1}} - 2\frac{\partial _s Q_{n-1}}{({\omega _0}_{n-1}^2 q_e ^2 + Q_{n-1})^2} ({\omega _0}_{n-1}^2 \partial _s q_e ^2 + \partial _s Q_{n-1}) \\&\quad + 2 \frac{Q_{n-1}}{({\omega _0}_{n-1}^2 q_e ^2 + Q_{n-1})^3} ({\omega _0}_{n-1}^2 \partial _s q_e ^2 + \partial _s Q_{n-1})^2 \\&\quad - \frac{Q_{n-1}}{({\omega _0}_{n-1}^2 q_e ^2 + Q_{n-1})^2} ({\omega _0}_{n-1}^2 \partial _s^2 q_e ^2 + \partial _s^2 Q_{n-1}).\\ \end{aligned} \end{aligned}$$

First we estimate $\Vert F\langle \psi \rangle ^{54}\Vert _{L^2}$. We have

$$\begin{aligned} \Vert F \langle \psi \rangle ^{54} \Vert _{L^2}&\lesssim \Vert U_{n-1} \langle \psi \rangle ^4 \Vert _{L^\infty } \Vert \partial _s Q_{n-1} \langle \psi \rangle ^{50} \Vert _{L^2} \\&\lesssim \frac{1}{1 - \Vert Q_{n-1} \Vert _{L^\infty }} \Vert Q_{n-1} \langle \psi \rangle ^4 \Vert _{L^\infty } \Vert \partial _s Q_{n-1} \langle \psi \rangle ^{50} \Vert _{L^2} \\&\lesssim \frac{1}{1 - {\varepsilon }^{0.99}}\Vert Q_{n-1} \langle \psi \rangle ^4 \Vert _{X_{1,4}}\Vert \partial _s Q_{n-1} \langle \psi \rangle ^{50} \Vert _{L^2} \\&\lesssim \Vert Q_{n-1} \Vert _{X_{1,4}} \Vert Q_{n-1} \Vert _{X_{0,50}} \\&\lesssim \ {\varepsilon }^{0.99} {\varepsilon }^{0.99}. \end{aligned}$$

We now show that $\left\Vert\partial _s F\langle \psi \rangle ^{54}\right\Vert_{L^2}\lesssim \epsilon ^{0.99}\epsilon ^{0.99}$. We have

$$\begin{aligned}\begin{aligned} \partial _s F&= (1 - \sqrt{1-U_{n-1}}) \partial _s^2 Q_{n-1} + \frac{1}{2} (1 - U_{n-1})^{-\frac{1}{2}} \partial _s U_{n-1} \partial _s Q_{n-1} \\&=: A_{1} + A_2.\\ \end{aligned} \end{aligned}$$

We first bound $A_1$. We have

$$\begin{aligned} \Vert A_1 \langle \psi \rangle ^{54} \Vert _{L^2}&\lesssim \Vert U_{n-1} \langle \psi \rangle ^4 \Vert _{L^\infty } \Vert \partial _s^2 Q_{n-1} \langle \psi \rangle ^{50} \Vert _{L^2} \\&\lesssim \frac{1}{1 - \Vert Q_{n-1} \Vert _{L^\infty }} \Vert Q_{n-1} \langle \psi \rangle ^4 \Vert _{L^\infty } \Vert \partial _s^2 Q_{n-1} \langle \psi \rangle ^{50} \Vert _{L^2} \\&\lesssim \frac{1}{1 - {\varepsilon }^{0.99}}\Vert Q_{n-1} \Vert _{X_{1,4}}\Vert \partial _s^2 Q_{n-1} \langle \psi \rangle ^{50} \Vert _{L^2} \\&\lesssim \Vert Q_{n-1} \Vert _{X_{1,4}} \Vert Q_{n-1} \Vert _{X_{1,50}} \\&\lesssim \ {\varepsilon }^{0.99} {\varepsilon }^{0.99}. \end{aligned}$$

We now estimate $A_2$. We have

$$\begin{aligned} \Vert A_2 \langle \psi \rangle ^{54} \Vert _{L^2}&\lesssim \Vert \partial _s U_{n-1} \langle \psi \rangle ^{4} \Vert _{L^\infty } \Vert \partial _s Q_{n-1} \langle \psi \rangle ^{50} \Vert _{L^2} \\&\lesssim \Vert Q_{n-1} \Vert _{X_{2,4}} \Vert Q_{n-1} \Vert _{X_{0,50}} \\&\lesssim {\varepsilon }^{0.99} {\varepsilon }^{0.99}. \end{aligned}$$

We now show that

$$\begin{aligned} \left\Vert\partial _s^2 F\langle \psi \rangle ^{54} \right\Vert_{L^2}\lesssim \epsilon ^{0.99}\epsilon ^{0.99}. \end{aligned}$$

By a direct calculation, we get

$$\begin{aligned} \begin{aligned} \partial _s^2 F&= (1 - \sqrt{1-U_{n-1}}) \partial _s^3 Q_{n-1} + (1 - U_{n-1})^{-\frac{1}{2}} \partial _s U_{n-1} \partial _s^2 Q_{n-1} \\&\quad + (1 - U_{n-1})^{-\frac{1}{2}} \partial _s^2 U_{n-1} \partial _s Q_{n-1} + (1 - U_{n-1})^{-\frac{3}{2}}|\partial _s U_{n-1} |^2 \partial _s^2 Q_{n-1} \\&=: B_1 + B_2 + B_3 + B_4. \end{aligned} \end{aligned}$$

We first establish the following bounds on the auxiliary quantities $U_{n-1}$. We have

$$\begin{aligned} \Vert \partial _s U_{n-1} \langle \psi \rangle ^{m} \Vert _{L^\infty }&\lesssim \Vert \partial _s Q_{n-1} \langle \psi \rangle ^{m} \Vert _{L^\infty } + \Vert Q_{n-1} \langle \psi \rangle ^{m} \Vert _{L^\infty } (1 + \Vert \partial _s Q_{n-1} \Vert _{L^\infty }) \\&\lesssim \Vert Q_{n-1}\Vert _{X_{2,m}} + \Vert Q_{n-1} \Vert _{X_{2,m}} (1 + \Vert Q_{n-1} \Vert _{X_{2,0}}) \\&\lesssim \Vert Q_{n-1}\Vert _{X_{2,m}}. \end{aligned}$$

Similarly, we have

$$\begin{aligned} \Vert \partial _s^2 U_{n-1} \langle \psi \rangle ^{m} \Vert _{L^2}&\lesssim \Vert \partial _s^2 Q_{n-1} \langle \psi \rangle ^{m} \Vert _{L^2} + \Vert \partial _s Q_{n-1} \langle \psi \rangle ^{m} \Vert _{L^2} (1 + \Vert \partial _s Q_{n-1} \Vert _{L^\infty }) \\&\quad + \Vert Q_{n-1} \langle \psi \rangle ^m \Vert _{L^2} (1 + \Vert \partial _s Q_{n-1} \Vert _{L^\infty })^2 \\&\quad + \Vert Q_{n-1} \langle \psi \rangle ^m \Vert _{L^\infty } (1 + \Vert \partial _s^2 Q_{n-1} \Vert _{L^2}) \\&\lesssim \Vert Q_{n-1} \Vert _{X_{2,m}}. \end{aligned}$$

We can now estimate of $\left\Vert\partial _s^2 F\langle \psi \rangle ^{54}\right\Vert_{L^2}$. We first bound $B_1$. We have

$$\begin{aligned} \Vert B_1 \langle \psi \rangle ^{54} \Vert _{L^2}&\lesssim \Vert U_{n-1} \langle \psi \rangle ^4 \Vert _{L^\infty } \Vert \partial _s^3 Q_{n-1} \langle \psi \rangle ^{50} \Vert _{L^2} \\&\lesssim \frac{1}{1 - \Vert Q_{n-1} \Vert _{L^\infty }} \Vert Q_{n-1} \langle \psi \rangle ^4 \Vert _{L^\infty } \Vert \partial _s^3 Q_{n-1} \langle \psi \rangle ^{50} \Vert _{L^2} \\&\lesssim \frac{1}{1 - {\varepsilon }^{0.99}} \Vert Q_{n-1} \Vert _{X_{1,4}} \Vert Q_{n-1} \Vert _{X_{2,50}} \\&\lesssim \Vert Q_{n-1} \Vert _{X_{1,4}} \Vert Q_{n-1} \Vert _{X_{2,50}} \\&\lesssim \ {\varepsilon }^{0.99} {\varepsilon }^{0.99}. \end{aligned}$$

We next move to $B_2$, for which we have

$$\begin{aligned} \Vert B_2 \langle \psi \rangle ^{54} \Vert _{L^2}&\lesssim \Vert \partial _s U_{n-1} \langle \psi \rangle ^{4} \Vert _{L^\infty } \Vert \partial _s^2 Q_{n-1} \langle \psi \rangle ^{50} \Vert _{L^2} \\&\lesssim \Vert Q_{n-1} \Vert _{X_{2,4}} \Vert Q_{n-1} \Vert _{X_{1,50}} \\&\lesssim {\varepsilon }^{0.99} {\varepsilon }^{0.99}. \end{aligned}$$

As for $B_3$, we have

$$\begin{aligned} \Vert B_3 \langle \psi \rangle ^{54} \Vert _{L^2}&\lesssim \Vert \partial _s^2 U_{n-1} \langle \psi \rangle ^{50} \Vert _{L^2} \Vert \partial _s Q_{n-1} \langle \psi \rangle ^{4} \Vert _{L^\infty } \\&\lesssim \Vert \partial _s^2 U_{n-1} \langle \psi \rangle ^{50} \Vert _{L^2} \Vert \partial _s Q_{n-1} \langle \psi \rangle ^{4} \Vert _{H^1_s H^1_\psi } \\&\lesssim \Vert Q_{n-1} \Vert _{X_{2,50}} \Vert Q_{n-1} \Vert _{X_{2,4}} \\&\lesssim \ {\varepsilon }^{0.99} {\varepsilon }^{0.99}. \end{aligned}$$

We finally conclude with an estimate on $B_4$, for which we have

$$\begin{aligned} \Vert B_4 \langle \psi \rangle ^{54} \Vert _{L^2}&\lesssim \Vert \partial _s U_{n-1} \langle \psi \rangle ^2 \Vert _{L^\infty }^2 \Vert \partial _s^2 Q_{n-1} \langle \psi \rangle ^{50} \Vert _{L^2} \\&\lesssim \Vert Q_{n-1} \Vert _{X_{2,2}}^2 \Vert Q_{n-1} \Vert _{X_{1,50}} \\&\lesssim {\varepsilon }^{0.99} {\varepsilon }^{0.99} {\varepsilon }^{0.99}. \end{aligned}$$

To conclude the proof of lemma, we need to estimate the $H^3_s$ norm of b,

$$\begin{aligned} \Vert b \Vert _{H^3_s} \lesssim {\varepsilon }|{\omega _0}_{n}| \Vert q_e \Vert _{H^3_s} + {\varepsilon }\Vert g \Vert _{H^3_s} \Vert q_e \Vert _{H^3_s} + {\varepsilon }^2 \Vert g \Vert _{H^3_s}^2 \lesssim {\varepsilon }. \end{aligned}$$

Therefore, according to (4260), the lemma is proven.

5.6 $\Delta Q_n$ Estimates

Our main objective in this section is to close the final bootstrap bound, (30). We begin with a lemma which allows us to control our auxiliary quantity, $U_{n-1}$, introduced in (43).

Lemma 13

Let $0 \le m \le 50$. Assume (27) - (30) are valid until the index $n - 1$. Assume (27) - (29) are valid until the index n. The quantities $U_{n-1}, U_{n-2}$ satisfy:

$$\begin{aligned} \sum _{j = 0}^1 \Vert ( \partial _s^j U_{n-1} - \partial _s^j U_{n-2}) \langle \psi \rangle ^m \Vert _{L^\infty }&\lesssim \Vert \Delta Q_{n-1} \Vert _{X_{2,m}} + \epsilon ^{1.99} |\Delta \overline{{\omega _0}}_{n-1}| , \end{aligned}$$

(44)

$$\begin{aligned} \sum _{j = 0}^2 \Vert ( \partial _s^j U_{n-1} - \partial _s^j U_{n-2}) \langle \psi \rangle ^m \Vert _{L^2}&\lesssim \Vert \Delta Q_{n-1} \Vert _{X_{2,m}} + \epsilon ^{1.99} |\Delta \overline{{\omega _0}}_{n-1}|. \end{aligned}$$

(45)

Proof

Recalling (43), we have

$$\begin{aligned} U_{n-1} - U_{n-2}&= \Big (\frac{Q_{n-1}}{{\omega _0}_{n-1}^2 q_e ^2 + Q_{n-1}} - \frac{Q_{n-2}}{{\omega _0}_{n-1}^2 q_e ^2 + Q_{n-1}}\Big )\nonumber \\&\quad + Q_{n-2} \Big ( \frac{1}{{\omega _0}_{n-1}^2 q_e ^2 + Q_{n-1}} - \frac{1}{{\omega _0}_{n-2}^2 q_e ^2 + Q_{n-2}} \Big ) \nonumber \\&= \ \frac{\Delta Q_{n-1}}{{\omega _0}_{n-1}^2 q_e ^2 + Q_{n-1}} + \frac{Q_{n-2}}{({\omega _0}_{n-1}^2 q_e ^2 + Q_{n-1})({\omega _0}_{n-2}^2 q_e ^2 + Q_{n-2})} \Delta Q_{n-1} \nonumber \\&\quad + \frac{Q_{n-2}}{({\omega _0}_{n-1}^2 q_e ^2 + Q_{n-1})({\omega _0}_{n-2}^2 q_e ^2 + Q_{n-2})} q_e ^2 ( {\omega _0}_{n-2}^2 - {\omega _0}_{n-1}^2) \nonumber \\&= \alpha \Delta Q_{n-1} + \gamma {\varepsilon }|\Delta \overline{{\omega _0}}_{n-1}|, \end{aligned}$$

(46)

where the coefficients are defined by

$$\begin{aligned} \alpha&:=\frac{1}{{\omega _0}_{n-1}^2 q_e ^2 + Q_{n-1}} + \frac{Q_{n-2}}{({\omega _0}_{n-1}^2 q_e ^2 + Q_{n-1})({\omega _0}_{n-2}^2 q_e ^2 + Q_{n-2})}, \\ \gamma&:= \frac{Q_{n-2}}{({\omega _0}_{n-1}^2 q_e ^2 + Q_{n-1})({\omega _0}_{n-2}^2 q_e ^2 + Q_{n-2})} q_e ^2. \end{aligned}$$

According to our bootstraps, we claim the following bounds. There exists a decomposition of $\partial _s^2 \alpha = \alpha _A + \alpha _B$ such that

$$\begin{aligned} \Vert \alpha \Vert _{L^\infty } + \Vert \partial _s \alpha \Vert _{L^\infty }&\lesssim 1, \end{aligned}$$

(47)

$$\begin{aligned} \Vert \alpha _A \Vert _{L^\infty } + \Vert \alpha _B \Vert _{L^2}&\lesssim 1, \end{aligned}$$

(48)

$$\begin{aligned} \Vert \gamma \langle \psi \rangle ^m \Vert _{L^\infty } + \Vert \partial _s \gamma \langle \psi \rangle ^m \Vert _{L^\infty }&\lesssim {\varepsilon }^{0.99} ,\end{aligned}$$

(49)

$$\begin{aligned} \Vert \partial _s^2 \gamma \langle \psi \rangle ^m \Vert _{L^2}&\lesssim {\varepsilon }^{0.99}, \end{aligned}$$

(50)

$$\begin{aligned} \Vert \gamma \langle \psi \rangle ^m \Vert _{L^2} + \Vert \partial _s \gamma \langle \psi \rangle ^m \Vert _{L^2}&\lesssim {\varepsilon }^{0.99}. \end{aligned}$$

(51)

We will prove these bounds as follows. First, we define

$$\begin{aligned} \alpha ^{(1)}&:= \frac{1}{{\omega _0}_{n-1}^2 q_e ^2 + Q_{n-1}} = \frac{1}{D_{n-1}},\\ \alpha ^{(2)}&:= \frac{Q_{n-2}}{({\omega _0}_{n-1}^2 q_e ^2 + Q_{n-1})({\omega _0}_{n-2}^2 q_e ^2 + Q_{n-2})} = \frac{Q_{n-2}}{D_{n-1} D_{n-2}}, \\ D_{n}&:= {\omega _0}_{n}^2 q_e ^2 + Q_{n}. \end{aligned}$$

after which the following identities are valid:

$$\begin{aligned} \alpha = \alpha ^{(1)} + \alpha ^{(2)}, \qquad \gamma = q_e ^2 \alpha ^{(2)}. \end{aligned}$$

(52)

We will henceforth prove the following bounds. We claim there exists a decomposition of $\partial _s^2 \alpha ^{(1)} = \alpha ^{(1)}_A + \alpha ^{(1)}_B$, where

$$\begin{aligned} \Vert \alpha ^{(1)} \Vert _{L^\infty } + \Vert \partial _s \alpha ^{(1)} \Vert _{L^\infty }&\lesssim 1 , \end{aligned}$$

(53)

$$\begin{aligned} \Vert \alpha ^{(1)}_A \Vert _{L^\infty } + \Vert \alpha ^{(1)}_B \Vert _{L^2}&\lesssim 1, \end{aligned}$$

(54)

$$\begin{aligned} \Vert \alpha ^{(2)} \langle \psi \rangle ^m \Vert _{L^\infty } + \Vert \partial _s \alpha ^{(2)} \langle \psi \rangle ^m \Vert _{L^\infty }&\lesssim {\varepsilon }^{0.99} ,\end{aligned}$$

(55)

$$\begin{aligned} \Vert \partial _s^2 \alpha ^{(2)} \langle \psi \rangle ^m \Vert _{L^2}&\lesssim {\varepsilon }^{0.99}, \end{aligned}$$

(56)

$$\begin{aligned} \Vert \alpha ^{(2)} \langle \psi \rangle ^m \Vert _{L^2} + \Vert \partial _s \alpha ^{(2)} \langle \psi \rangle ^m \Vert _{L^2}&\lesssim {\varepsilon }^{0.99}, \end{aligned}$$

(57)

upon which using (52), we obtain (47), (48), (49), (50), and (51).

Proof of (53)

Clearly, we have

$$\begin{aligned} \Vert \alpha ^{(1)} \Vert _{L^\infty }&\lesssim \frac{1}{\inf |D_{n-1}|} \lesssim \frac{1}{{\omega _0}_{n-1}^2 q_e ^2 - \Vert Q_{n-1} \Vert _{L^\infty } } \lesssim \frac{1}{1- {\varepsilon }^{0.99}} \lesssim 1. \end{aligned}$$

Next, we have the identity $\partial _s \alpha ^{(1)} = \frac{\partial _s D_{n-1}}{D_{n-1}^2}$. Since we have already established a lower bound on $D_{n-1}$, it suffices to estimate $\partial _s D_{n-1}$:

$$\begin{aligned} \Vert \partial _s \alpha ^{(1)} \Vert _{L^\infty } \lesssim&\Vert \partial _s D_{n-1} \Vert _{L^\infty } \lesssim \Vert {\omega _0}_{n-1}^2 \partial _s\{ q_e ^2 \} \Vert _{L^\infty } + \Vert \partial _s Q_{n-1} \Vert _{L^\infty } \lesssim 1 + {\varepsilon }^{0.99} \lesssim 1, \end{aligned}$$

where we have invoked the bootstraps (27) and (28). This proves the bound (53).

Proof of (54)

For this bound, we differentiate once more to find the identity

$$\begin{aligned} \partial _s^2 \alpha ^{(1)}&= \frac{\partial _s^2 D_{n-1}}{D_{n-1}^2} -2 \frac{|\partial _s D_{n-1}|^2}{D_{n-1}^3} \\&= [ \frac{1}{D_{n-1}^2} {\omega _0}_{n-1}^2 \partial _s^2 \{ q_e ^2 \} + \frac{\partial _sD_{n-1}}{D_{n-1}^3} {\omega _0}_{n-1}^2 \partial _s \{ q_e ^2 \} ] \\&\qquad + [ \frac{1}{D_{n-1}^2}\partial _s^2 Q_{n-1} + \frac{\partial _sD_{n-1}}{D_{n-1}^3} \partial _s Q_{n-1} ] \\&=: \alpha ^{(1)}_A + \alpha ^{(1)}_B. \end{aligned}$$

We estimate

$$\begin{aligned} \Vert \alpha ^{(1)}_A \Vert _{L^\infty } \lesssim |{\omega _0}_{n-1}|^2 + \Vert \partial _s D_{n-1} \Vert _{L^\infty } |{\omega _0}_{n-1}|^2 \lesssim 1, \end{aligned}$$

and

$$\begin{aligned} \Vert \alpha ^{(1)}_B \Vert _{L^2} \lesssim \Vert \partial _s^2 Q_{n-1} \Vert _{L^2} + \Vert \partial _s D_{n-1} \Vert _{L^\infty } \Vert \partial _s Q_{n-1} \Vert _{L^2} \lesssim {\varepsilon }^{0.99}. \end{aligned}$$

This proves the bound (54).

Proof of (55)

We turn now to the definition of $\alpha ^{(2)}$. We will use freely the bounds $|D_{n-1}| + |D_{n-2}| > rsim 1$ and $\Vert \partial _s D_{n-1} \Vert _{L^\infty } + \Vert \partial _s D_{n-2} \Vert _{L^\infty } \lesssim 1$, which have already been established. First, we have

$$\begin{aligned} \Vert \alpha ^{(2)} \langle \psi \rangle ^m \Vert _{L^\infty } \lesssim \Vert Q_{n-2} \langle \psi \rangle ^m \Vert _{L^\infty } \lesssim \Vert Q_{n-2} \Vert _{X_{2,m}} \lesssim {\varepsilon }^{0.99}. \end{aligned}$$

Next, we have the identity

$$\begin{aligned} \partial _s \alpha ^{(2)} = \frac{\partial _s Q_{n-2}}{D_{n-1} D_{n-2}} - \frac{Q_{n-2}\partial _s \{ D_{n-1} D_{n-2} \}}{D_{n-1}^2 D_{n-2}^2}, \end{aligned}$$

(58)

from which we obtain

$$\begin{aligned} \Vert \partial _s \alpha ^{(2)} \langle \psi \rangle ^m \Vert _{L^\infty }&\lesssim \Vert \partial _s Q_{n-2} \langle \psi \rangle ^m \Vert _{L^\infty } + \Vert Q_{n-2} \langle \psi \rangle ^m \Vert _{L^\infty } ( \Vert D_{n-1} \Vert _{L^\infty } \Vert \partial _s D_{n-2} \Vert _{L^\infty } \\&\quad + \Vert D_{n-2} \Vert _{L^\infty } \Vert \partial _s D_{n-1} \Vert _{L^\infty }) \\&\lesssim \Vert \ Q_{n-2} \Vert _{X_{2,m}} + \Vert Q_{n-2} \Vert _{X_{1,m}} ( \Vert D_{n-1} \Vert _{L^\infty } \Vert \partial _s D_{n-2} \Vert _{L^\infty } \\&\quad + \Vert D_{n-2} \Vert _{L^\infty } \Vert \partial _s D_{n-1} \Vert _{L^\infty }) \\&\lesssim {\varepsilon }^{0.99}. \end{aligned}$$

This proves the bound (55).

Proof of (56)

We differentiate (58) again to obtain the identity

$$\begin{aligned} \partial _s^2 \alpha ^{(2)}&= \frac{\partial _s^2 Q_{n-2}}{D_{n-1} D_{n-2}} -2 \frac{\partial _s Q_{n-2}\partial _s \{ D_{n-1} D_{n-2} \}}{D_{n-1}^2 D_{n-2}^2} + 2\frac{Q_{n-2}|\partial _s \{ D_{n-1} D_{n-2} \}|^2 }{D_{n-1}^3 D_{n-2}^3}, \end{aligned}$$

after which we obtain the bound

$$\begin{aligned} \Vert \partial _s^2 \alpha ^{(2)} \langle \psi \rangle ^m \Vert _{L^2}&\lesssim \Vert \partial _s^2 Q_{n-2} \langle \psi \rangle ^m \Vert _{L^2} + \Vert \partial _s Q_{n-2} \langle \psi \rangle ^m \Vert _{L^2} \Vert \partial _s \{ D_{n-1} D_{n-2} \} \Vert _{L^\infty } \\&\quad + \Vert Q_{n-2} \langle \psi \rangle ^m \Vert _{L^2} \Vert \partial _s \{ D_{n-1} D_{n-2} \} \Vert _{L^\infty }^2 \\&\lesssim \Vert Q_{n-2} \Vert _{X_{2,m}} \\&\lesssim {\varepsilon }^{0.99}. \end{aligned}$$

This proves the bound (56).

Proof of (57)

We have

$$\begin{aligned} \Vert \alpha ^{(2)} \langle \psi \rangle ^m \Vert _{L^2} \lesssim \Vert Q_{n-2} \langle \psi \rangle ^m \Vert _{L^2} \lesssim \Vert Q_{n-2} \Vert _{X_{2,m}} \lesssim {\varepsilon }^{0.99}, \end{aligned}$$

and upon using (58), we have

$$\begin{aligned} \Vert \partial _s \alpha ^{(2)} \langle \psi \rangle ^m \Vert _{L^2}&\lesssim \ \Vert \partial _s Q_{n-2} \langle \psi \rangle ^m \Vert _{L^2} + \Vert Q_{n-2} \langle \psi \rangle ^m \Vert _{L^2} ( \Vert D_{n-1} \Vert _{L^\infty } \Vert \partial _s D_{n-2} \Vert _{L^\infty } \\&\quad + \Vert D_{n-2} \Vert _{L^\infty } \Vert \partial _s D_{n-1} \Vert _{L^\infty }) \\&\lesssim \ \Vert Q_{n-2} \Vert _{X_{2,m}} + \Vert Q_{n-2} \Vert _{X_{1,m}} ( \Vert D_{n-1} \Vert _{L^\infty } \Vert \partial _s D_{n-2} \Vert _{L^\infty } \\&\quad + \Vert D_{n-2} \Vert _{L^\infty } \Vert \partial _s D_{n-1} \Vert _{L^\infty }) \\&\lesssim {\varepsilon }^{0.99}. \end{aligned}$$

We have therefore established (53)–(57) and hence (47)–(51). From here, the desired estimates, (44)–(45), follow from an application of the product rule applied to the identity (46). Indeed, we have:

$$\begin{aligned} \Vert (U_{n-1} - U_{n-2} ) \langle \psi \rangle ^m \Vert _{L^\infty }&\lesssim \Vert \alpha \Vert _{L^\infty } \Vert \Delta Q_{n-1} \langle \psi \rangle ^m \Vert _{L^\infty } + \Vert \gamma \langle \psi \rangle ^m \Vert _{L^\infty } {\varepsilon }|\Delta \overline{{\omega _0}}_{n-1}| \\&\lesssim \Vert \Delta Q_{n-1} \Vert _{X_{2,m}} + {\varepsilon }^{1.99} |\Delta \overline{{\omega _0}}_{n-1}|, \end{aligned}$$

where we have used the bounds (47) and (49). In $L^2$, we similarly have

$$\begin{aligned} \Vert (U_{n-1} - U_{n-2} ) \langle \psi \rangle ^m \Vert _{L^2}&\lesssim \Vert \alpha \Vert _{L^\infty } \Vert \Delta Q_{n-1} \langle \psi \rangle ^m \Vert _{L^2} + \Vert \gamma \langle \psi \rangle ^m \Vert _{L^2} {\varepsilon }|\Delta \overline{{\omega _0}}_{n-1}| \\&\lesssim \Vert \Delta Q_{n-1} \Vert _{X_{2,m}} + {\varepsilon }^{1.99} |\Delta \overline{{\omega _0}}_{n-1}|, \end{aligned}$$

where we have used the bounds (47) and (51).

Next, we have upon differentiating (46), the identity

$$\begin{aligned} \partial _s \{ U_{n-1} - U_{n-2} \} = \alpha \partial _s \Delta Q_{n-1} + \Delta Q_{n-1} \partial _s \alpha + {\varepsilon }|\Delta {\omega _0}_{n-1}| \partial _s \gamma , \end{aligned}$$

after which we have the following $L^\infty$ bound:

$$\begin{aligned} \Vert \partial _s \{ U_{n-1} - U_{n-2} \} \langle \psi \rangle ^m \Vert _{L^\infty }&\lesssim \Vert \alpha \Vert _{L^\infty } \Vert \partial _s \Delta Q_{n-1} \langle \psi \rangle ^m \Vert _{L^\infty }\\&\quad + \Vert \partial _s \alpha \Vert _{L^\infty } \Vert \Delta Q_{n-1} \langle \psi \rangle ^m \Vert _{L^\infty } \\&\quad + {\varepsilon }\Vert \partial _s \gamma \langle \psi \rangle ^m \Vert _{L^\infty } |\Delta {\omega _0}_{n-1}| \\&\lesssim \Vert \Delta Q_{n-1} \Vert _{X_{2,m}} + {\varepsilon }^{1.99} |\Delta {\omega _0}_{n-1}|, \end{aligned}$$

where we have invoked (47) and (49). In $L^2$, we similarly have

$$\begin{aligned} \Vert \partial _s \{ U_{n-1} - U_{n-2} \} \langle \psi \rangle ^m \Vert _{L^2}&\lesssim \Vert \alpha \Vert _{L^\infty } \Vert \partial _s \Delta Q_{n-1} \langle \psi \rangle ^m \Vert _{L^2} \\&\quad + \Vert \partial _s \alpha \Vert _{L^\infty } \Vert \Delta Q_{n-1} \langle \psi \rangle ^m \Vert _{L^2} \\&\quad + {\varepsilon }\Vert \partial _s \gamma \langle \psi \rangle ^m \Vert _{L^2} |\Delta {\omega _0}_{n-1}| \\&\lesssim \Vert \Delta Q_{n-1} \Vert _{X_{2,m}} + {\varepsilon }^{1.99} |\Delta {\omega _0}_{n-1}|, \end{aligned}$$

where we have used the bounds (47) and (51).

Differentiating (46) twice in s, we obtain the identity

$$\begin{aligned} \partial _s^2 \{ U_{n-1} - U_{n-2} \}&= \alpha \partial _s^2 \Delta Q_{n-1} + \Delta Q_{n-1} \partial _s^2 \alpha + 2 \partial _s \Delta Q_{n-1} \partial _s \alpha + {\varepsilon }|\Delta {\omega _0}_{n-1}| \partial _s^2 \gamma \\&= \alpha \partial _s^2 \Delta Q_{n-1} + \Delta Q_{n-1} \alpha _A + \Delta Q_{n-1} \alpha _B + 2 \partial _s \Delta Q_{n-1} \partial _s \alpha \\&\quad + {\varepsilon }|\Delta {\omega _0}_{n-1}| \partial _s^2 \gamma , \end{aligned}$$

where we use the decomposition $\partial _s^2 \alpha = \alpha _A + \alpha _B$. We now estimate the $L^2$ norm as follows:

$$\begin{aligned}&\Vert \partial _s^2 \{ U_{n-1} - U_{n-2} \} \langle \psi \rangle ^m \Vert _{L^2}\\&\quad \lesssim \Vert \alpha \Vert _{L^\infty } \Vert \partial _s^2 \Delta Q_{n-1} \langle \psi \rangle ^m \Vert _{L^2} + \Vert \alpha _A \Vert _{L^\infty } \Vert \Delta Q_{n-1} \langle \psi \rangle ^m \Vert _{L^2} \\&\qquad + \Vert \Delta Q_{n-1} \langle \psi \rangle ^m \Vert _{L^\infty } \Vert \alpha _B \Vert _{L^2} + \Vert \partial _s \Delta Q_{n-1} \langle \psi \rangle ^m \Vert _{L^2} \Vert \partial _s \alpha \Vert _{L^\infty } \\&\qquad + {\varepsilon }\Vert \partial _s^2 \gamma \langle \psi \rangle ^m \Vert _{L^2} |\Delta {\omega _0}_{n-1}| \\&\quad \lesssim \Vert \partial _s^2 \Delta Q_{n-1} \langle \psi \rangle ^m \Vert _{L^2} + \Vert \Delta Q_{n-1} \langle \psi \rangle ^m \Vert _{L^2} + \Vert \Delta Q_{n-1} \langle \psi \rangle ^m \Vert _{L^\infty } \\&\qquad + \Vert \partial _s \Delta Q_{n-1} \langle \psi \rangle ^m \Vert _{L^2} + {\varepsilon }^{1.99} |\Delta {\omega _0}_{n-1}| \\&\quad \lesssim \Vert \Delta Q_{n-1} \Vert _{X_{2,m}} + {\varepsilon }^{1.99} |\Delta {\omega _0}_{n-1}|, \end{aligned}$$

where we have used the bounds (47)–(50). We have therefore established the bounds (44)–(45), and this concludes the proof of the lemma.

Lemma 14

Assume (27)–(30) are valid until the index $n - 1$. Assume (27)–(29) are valid until the index n. Then

$$\begin{aligned} \Vert \Delta Q_n \Vert _{X_{2, 50}} \le&{\varepsilon }^{\frac{3}{4}} \Vert \Delta Q_{n-1} \Vert _{X_{2,50}} + {\varepsilon }^{\frac{1}{2}} |\Delta \overline{{\omega _0}}_n| + {\varepsilon }^{\frac{3}{2}} |\Delta \overline{{\omega _0}}_{n-1}|. \end{aligned}$$

(59)

Proof

For this estimate, motivated by (26), we set

$$\begin{aligned} F&:= \left( 1- \tfrac{{\omega _0}_{n-1} q_e }{\sqrt{{\omega _0}_{n-1} ^2 q_e ^2+ Q_{n-1} }}\right) \partial _s Q_{n-1}- \left( 1- \tfrac{{\omega _0}_{n-2} q_e }{\sqrt{{\omega _0}_{n-2} ^2 q_e ^2+ Q_{n-2} }}\right) \partial _s Q_{n-2}, \\ G&:= ( {\omega _0}_{n-1} -{\omega _0}_{n-2}) Q_{n-1} ,\\ b&:= {\varepsilon }( \overline{{\omega _0}}_{n}- \overline{{\omega _0}}_{n-1}) q_e ^2(s). \end{aligned}$$

According to (33), we fix $k = 2, m = 50$, which results in

$$\begin{aligned} \Vert \Delta Q_n \Vert _{X_{2, 50}}&\lesssim \sum _{k' = 0}^{2} \Vert \partial _s^{k'} F \langle \psi \rangle ^{54} \Vert _{L^2} + \sum _{k' = 0}^{2} \sum _{j = 0}^2 \Vert \partial _s^{k'} \partial _\psi ^j G \langle \psi \rangle ^{50} \Vert _{L^2} +\Vert b \Vert _{H^{3}_s}. \end{aligned}$$

(60)

We therefore estimate the two quantities appearing on the right-hand side above. We first address the term F, which we rewrite as follows

$$\begin{aligned} F&= (1 - \sqrt{1 - U_{n-1}}) \partial _s Q_{n-1} - (1 - \sqrt{1 - U_{n-2}}) \partial _s Q_{n-2} \\&= (1 - \sqrt{1 - U_{n-1}}) \partial _s \Delta Q_{n-1} + \partial _s Q_{n-2}(\sqrt{1 - U_{n-1}} - \sqrt{1 - U_{n-2}}) := F_1 + F_2. \end{aligned}$$

An identical calculation to the estimate of the forcing, F, in Lemma 12 results in the bound

$$\begin{aligned} \Vert F_1 \langle \psi \rangle ^{54} \Vert _{H^2_s L^2_\psi } \lesssim {\varepsilon }^{0.99} \Vert \Delta Q_{n-1} \Vert _{X_{2,50}}. \end{aligned}$$

We develop the following identities

$$\begin{aligned} \partial _s F_2&= \partial _s^2 Q_{n-2}\left( \sqrt{1 - U_{n-1}} - \sqrt{1 - U_{n-2}}\right) \\&\qquad - \frac{1}{2} \partial _s Q_{n-2}\Big ( \frac{\partial _s U_{n-1}}{\sqrt{1 - U_{n-1}}} - \frac{\partial _s U_{n-2}}{\sqrt{1-U_{n-2}}} \Big ) \\&= \partial _s^2 Q_{n-2}\left( \sqrt{1 - U_{n-1}} - \sqrt{1 - U_{n-2}}\right) \\&\qquad - \frac{1}{2} \partial _s Q_{n-2}\frac{\partial _s U_{n-1} - \partial _s U_{n-2}}{\sqrt{1 - U_{n-1}}} \\&\quad - \frac{1}{2} \partial _s Q_{n-2} \Big ( \frac{\partial _s U_{n-2}}{\sqrt{1 - U_{n-1}}} - \frac{\partial _s U_{n-2}}{\sqrt{1 - U_{n-2}}} \Big )\\&= C_1 + C_2 + C_3. \end{aligned}$$

We estimate $\partial _s F_2$ as follows. First,

$$\begin{aligned} \Vert C_1 \langle \psi \rangle ^{54} \Vert _{L^2}&\lesssim \Vert (\sqrt{1 - U_{n-1}} - \sqrt{1 - U_{n-2}} ) \langle \psi \rangle ^{4} \Vert _{L^\infty } \Vert \partial _s^2 Q_{n-2} \langle \psi \rangle ^{50} \Vert _{L^2} \\&\lesssim \Vert (U_{n-1} - U_{n-2}) \langle \psi \rangle ^{4} \Vert _{L^\infty } \Vert \partial _s^2 Q_{n-2} \langle \psi \rangle ^{50} \Vert _{L^2} \\&\lesssim \Vert (U_{n-1} - U_{n-2}) \langle \psi \rangle ^{4} \Vert _{L^\infty } \Vert Q_{n-2} \Vert _{X_{1,50}} \\&\lesssim \ {\varepsilon }^{0.99} \Vert (U_{n-1} - U_{n-2}) \langle \psi \rangle ^{4} \Vert _{L^\infty }. \end{aligned}$$

Next, to estimate $C_2$, we have

$$\begin{aligned} \Vert C_2 \langle \psi \rangle ^{54} \Vert _{L^2}&\lesssim \Vert \partial _s Q_{n-2} \langle \psi \rangle ^{50} \Vert _{L^\infty } \Vert ( \partial _s U_{n-1} - \partial _s U_{n-2}) \langle \psi \rangle ^{4} \Vert _{L^2} \\&\lesssim \Vert \partial _s Q_{n-2} \Vert _{X_{2,50}} \Vert ( \partial _s U_{n-1} - \partial _s U_{n-2}) \langle \psi \rangle ^{4} \Vert _{L^2} \\&\lesssim \ {\varepsilon }^{0.99}\Vert ( \partial _s U_{n-1} - \partial _s U_{n-2}) \langle \psi \rangle ^{4} \Vert _{L^2}. \end{aligned}$$

Finally, to estimate $C_3$, we have

$$\begin{aligned} \Vert C_3 \langle \psi \rangle ^{54} \Vert _{L^2}&\lesssim \left\Vert \partial _s Q_{n-2} \langle \psi \rangle ^{50} \right\Vert_{L^2} \Vert \partial _s U_{n-2} \langle \psi \rangle ^{4} \Vert _{L^\infty } \Vert U_{n-1} - U_{n-2} \Vert _{L^\infty } \\&\lesssim \Vert Q_{n-2} \Vert _{X_{0,50}} \Vert Q_{n-2} \Vert _{X_{2,4}}\Vert U_{n-1} - U_{n-2} \Vert _{L^\infty } \\&\lesssim {\varepsilon }^{0.99} {\varepsilon }^{0.99} \Vert U_{n-1} - U_{n-2} \Vert _{L^\infty }. \end{aligned}$$

We now move to the second derivative, $\partial _s^2 F_2$, which we will treat as follows:

$$\begin{aligned} \partial _s^2 F_2 = \partial _s C_1 + \partial _s C_2 + \partial _s C_3. \end{aligned}$$

We have

$$\begin{aligned} \partial _s C_1&= \partial _s^3 Q_{n-2}(\sqrt{1 - U_{n-1}} - \sqrt{1 - U_{n-2}})\\&\qquad -\frac{1}{2} \partial _s^2 Q_{n-2} \Big ( \frac{\partial _s U_{n-1}}{\sqrt{1- U_{n-1}}} - \frac{\partial _s U_{n-2}}{\sqrt{1 - U_{n-2}}} \Big ) \\&= \partial _s^3 Q_{n-2}(\sqrt{1 - U_{n-1}} - \sqrt{1 - U_{n-2}})-\frac{1}{2} \partial _s^2 Q_{n-2} \Big ( \frac{\partial _s U_{n-1} - \partial _s U_{n-2}}{\sqrt{1- U_{n-1}}} \Big ) \\&\quad - \frac{1}{2} \partial _s^2 Q_{n-2} \Big ( \frac{\partial _s U_{n-2}}{\sqrt{1 - U_{n-1}}} - \frac{\partial _s U_{n-2}}{\sqrt{1 - U_{n-2}}}\Big ) \\&= : C_{1,1} + C_{1,2} + C_{1,3}. \end{aligned}$$

First, we estimate

$$\begin{aligned} \Vert C_{1,1} \langle \psi \rangle ^{54} \Vert _{L^2}&\lesssim \left\Vert (\sqrt{1 - U_{n-1}} - \sqrt{1 - U_{n-2}} ) \langle \psi \rangle ^{4}\right\Vert_{L^\infty } \left\Vert \partial _s^3 Q_{n-2} \langle \psi \rangle ^{50} \right\Vert_{L^2} \\&\lesssim \left\Vert (U_{n-1} - U_{n-2}) \langle \psi \rangle ^{4} \right\Vert_{L^\infty } \Vert \partial _s^3 Q_{n-2} \langle \psi \rangle ^{50} \Vert _{L^2} \\&\lesssim \left\Vert (U_{n-1} - U_{n-2}) \langle \psi \rangle ^{4}\right\Vert_{L^\infty } \Vert Q_{n-2} \Vert _{X_{2,50}} \\&\lesssim \ {\varepsilon }^{0.99} \Vert (U_{n-1} - U_{n-2}) \langle \psi \rangle ^{4} \Vert _{L^\infty }. \end{aligned}$$

Next, to estimate $C_{1,2}$, we have

$$\begin{aligned} \Vert C_{1,2} \langle \psi \rangle ^{54} \Vert _{L^2}&\lesssim \Vert \partial _s^2 Q_{n-2} \langle \psi \rangle ^{50} \Vert _{L^2} \Vert ( \partial _s U_{n-1} - \partial _s U_{n-2}) \langle \psi \rangle ^{4} \Vert _{L^\infty } \\&\lesssim \Vert Q_{n-2} \Vert _{X_{1,50}} \Vert ( \partial _s U_{n-1} - \partial _s U_{n-2}) \langle \psi \rangle ^{4} \Vert _{L^\infty } \\&\lesssim \ {\varepsilon }^{0.99}\Vert ( \partial _s U_{n-1} - \partial _s U_{n-2}) \langle \psi \rangle ^{4} \Vert _{L^\infty }. \end{aligned}$$

Next, to estimate $C_{1,3}$, we have

$$\begin{aligned} \Vert C_{1,3} \langle \psi \rangle ^{54} \Vert _{L^2}&\lesssim \Vert \partial _s^2 Q_{n-2} \langle \psi \rangle ^{50} \Vert _{L^2} \Vert \partial _s U_{n-2} \langle \psi \rangle ^{4} \Vert _{L^\infty } \Vert U_{n-1} - U_{n-2} \Vert _{L^\infty } \\&\lesssim \Vert Q_{n-2} \Vert _{X_{1,50}} \Vert Q_{n-2} \Vert _{X_{2,4}}\Vert U_{n-1} - U_{n-2} \Vert _{L^\infty } \\&\lesssim \ {\varepsilon }^{0.99} {\varepsilon }^{0.99} \Vert U_{n-1} - U_{n-2} \Vert _{L^\infty }. \end{aligned}$$

We next move to the $\partial _s C_2$ contributions, for which we record the identity

$$\begin{aligned} \partial _s C_2&= - \frac{1}{2} \partial _s^2 Q_{n-2}\frac{\partial _s U_{n-1} - \partial _s U_{n-2}}{\sqrt{1 - U_{n-1}}} - \frac{1}{2} \partial _s Q_{n-2}\frac{\partial _s^2 U_{n-1} - \partial _s^2 U_{n-2}}{\sqrt{1 - U_{n-1}}} \\&\quad + \frac{1}{4} \partial _s Q_{n-2} \partial _s U_{n-1} (1 - U_{n-1})^{-\frac{3}{2}}(\partial _s U_{n-1} - \partial _s U_{n-2}) \\&=: C_{2,1}+C_{2,2} + C_{2,3}. \end{aligned}$$

We first estimate $C_{2,1}$ for which we have

$$\begin{aligned} \Vert C_{2,1} \langle \psi \rangle ^{54} \Vert _{L^2}&\lesssim \Vert \partial _s^2 Q_{n-2} \langle \psi \rangle ^{50} \Vert _{L^2} \Vert (\partial _s U_{n-1} - \partial _s U_{n-2}) \langle \psi \rangle ^{4} \Vert _{L^\infty } \\&\lesssim \Vert Q_{n-2} \Vert _{X_{1,50}} \Vert (\partial _s U_{n-1} - \partial _s U_{n-2}) \langle \psi \rangle ^{4} \Vert _{L^\infty }\\&\lesssim \ {\varepsilon }^{0.99} \Vert (\partial _s U_{n-1} - \partial _s U_{n-2}) \langle \psi \rangle ^{4} \Vert _{L^\infty }. \end{aligned}$$

Next, we have

$$\begin{aligned} \Vert C_{2,2} \langle \psi \rangle ^{54} \Vert _{L^2}&\lesssim \Vert \partial _s Q_{n-2} \langle \psi \rangle ^{50} \Vert _{L^\infty } \Vert ( \partial _s^2 U_{n-1} - \partial _s^2 U_{n-2}) \langle \psi \rangle ^{4} \Vert _{L^2} \\&\lesssim \Vert \partial _s Q_{n-2} \Vert _{X_{2,50}} \Vert ( \partial _s^2 U_{n-1} - \partial _s^2 U_{n-2}) \langle \psi \rangle ^{4} \Vert _{L^2} \\&\lesssim \ {\varepsilon }^{0.99}\Vert ( \partial _s^2 U_{n-1} - \partial _s^2 U_{n-2}) \langle \psi \rangle ^{4} \Vert _{L^2}. \end{aligned}$$

Finally, we have the $C_{2,3}$ contribution for which we estimate

$$\begin{aligned} \Vert C_{2,3} \langle \psi \rangle ^{54} \Vert _{L^2}&\lesssim \Vert \partial _s Q_{n-2} \langle \psi \rangle ^{50} \Vert _{L^\infty } \Vert \partial _s U_{n-1} \langle \psi \rangle ^4 \Vert _{L^\infty } \Vert (\partial _s U_{n-1} - \partial _s U_{n-2}) \Vert _{L^2} \\&\lesssim \Vert Q_{n-2} \Vert _{X_{2,50}} \Vert Q_{n-1} \Vert _{X_{2,4}} \Vert (\partial _s U_{n-1} - \partial _s U_{n-2}) \Vert _{L^2} \\&\lesssim \ {\varepsilon }^{0.99} {\varepsilon }^{0.99} \Vert (\partial _s U_{n-1} - \partial _s U_{n-2}) \Vert _{L^2}. \end{aligned}$$

We next compute $\partial _s C_3$, which results in the following identity,

$$\begin{aligned} \partial _s C_3&= \frac{1}{2} \partial _s^2 Q_{n-2} \Big ( \frac{\partial _s U_{n-2}}{\sqrt{1 - U_{n-1}}} - \frac{\partial _s U_{n-2}}{\sqrt{1 - U_{n-2}}} \Big ) \\&\quad + \frac{1}{2} \partial _s Q_{n-2} \Big ( \frac{\partial _s^2 U_{n-2}}{\sqrt{1 - U_{n-1}}} - \frac{\partial _s^2 U_{n-2}}{\sqrt{1 - U_{n-2}}} \Big ) \\&\quad - \frac{1}{4} \partial _s Q_{n-2} \partial _s U_{n-2} \frac{1}{(1 - U_{n-1})^{\frac{3}{2}}} (\partial _s U_{n-1} - \partial _s U_{n-2}) \\&\quad - \frac{1}{4} \partial _s Q_{n-2} |\partial _s U_{n-2}|^2\left( \frac{1}{(1- U_{n-1})^{\frac{3}{2}}} - \frac{1}{(1- U_{n-2})^{\frac{3}{2}}} \right) =: \sum _{i = 1}^4 C_{3,i}. \end{aligned}$$

We estimate first $C_{3,1}$ as follows

$$\begin{aligned} \Vert C_{3,1} \langle \psi \rangle ^{54} \Vert _{L^2}&\lesssim \Vert \partial _s^2 Q_{n-2} \langle \psi \rangle ^{50} \Vert _{L^2} \Vert \partial _s U_{n-2} \langle \psi \rangle ^{4} \Vert _{L^\infty } \Vert U_{n-1} - U_{n-2} \Vert _{L^\infty } \\&\lesssim \Vert Q_{n-2}\Vert _{X_{1,50}} \Vert Q_{n-2} \Vert _{X_{2,4}} \Vert U_{n-1} - U_{n-2} \Vert _{L^\infty } \\&\lesssim \ {\varepsilon }^{0.99} {\varepsilon }^{0.99} \Vert U_{n-1} - U_{n-2} \Vert _{L^\infty }. \end{aligned}$$

Next, we estimate $C_{3,2}$ as follows

$$\begin{aligned} \Vert C_{3,2} \langle \psi \rangle ^{54} \Vert _{L^2}&\lesssim \Vert \partial _s Q_{n-2} \langle \psi \rangle ^{50} \Vert _{L^\infty } \Vert \partial _s^2 U_{n-2} \langle \psi \rangle ^4 \Vert _{L^2} \Vert U_{n-1} - U_{n-2} \Vert _{L^\infty } \\&\lesssim \Vert Q_{n-2} \Vert _{X_{2,50}} \Vert \partial _s^2 U_{n-2} \langle \psi \rangle ^4 \Vert _{L^2} \Vert U_{n-1} - U_{n-2} \Vert _{L^\infty } \\&\lesssim \Vert Q_{n-2} \Vert _{X_{2,50}} \Vert Q_{n-2} \Vert _{X_{2,4}} \Vert U_{n-1} - U_{n-2} \Vert _{L^\infty } \\&\lesssim \ {\varepsilon }^{0.99} {\varepsilon }^{0.99} \Vert U_{n-1} - U_{n-2} \Vert _{L^\infty }. \end{aligned}$$

Next, we estimate $C_{3,3}$ as follows

$$\begin{aligned} \Vert C_{3,3} \langle \psi \rangle ^{54} \Vert _{L^2}&\lesssim \Vert \partial _s Q_{n-2} \langle \psi \rangle ^{50} \Vert _{L^\infty } \Vert \partial _s U_{n-2} \langle \psi \rangle ^{4} \Vert _{L^\infty } \Vert \partial _s U_{n-1} - \partial _s U_{n-2} \Vert _{L^2} \\&\lesssim \Vert Q_{n-2} \Vert _{X_{2,50}} \Vert Q_{n-2} \Vert _{X_{2,4}} \Vert \partial _s U_{n-1} - \partial _s U_{n-2} \Vert _{L^2} \\&\lesssim \ {\varepsilon }^{0.99} {\varepsilon }^{0.99} \Vert \partial _s U_{n-1} - \partial _s U_{n-2} \Vert _{L^2}. \end{aligned}$$

Finally, we estimate $C_{3,4}$ as follows

$$\begin{aligned} \Vert C_{3,4} \langle \psi \rangle ^{54} \Vert _{L^2}&\lesssim \Vert \partial _s Q_{n-2} \langle \psi \rangle ^{50} \Vert _{L^\infty } \Vert \partial _s U_{n-2} \langle \psi \rangle ^{2} \Vert _{L^\infty }^2 \Vert U_{n-1} - U_{n-2} \Vert _{L^2} \\&\lesssim \Vert Q_{n-2} \Vert _{X_{2,50}} \Vert Q_{n-2} \Vert _{X_{2,2}}^2 \Vert U_{n-1} - U_{n-2} \Vert _{L^2} \\&\lesssim \ {\varepsilon }^{0.99} {\varepsilon }^{0.99} {\varepsilon }^{0.99} \Vert U_{n-1} - U_{n-2} \Vert _{L^2}. \end{aligned}$$

Now, upon invoking (44)–(45), the above estimates give

$$\begin{aligned}&\Vert F_2 \langle \psi \rangle ^{54} \Vert _{H^2_s L^2_\psi } \\&\quad \lesssim {\varepsilon }^{0.99} \Big ( \sum _{j = 0}^1 \Vert ( \partial _s^j U_{n-1} - \partial _s^j U_{n-2}) \langle \psi \rangle ^4 \Vert _{L^\infty } \\&\qquad + \sum _{j = 0}^2 \Vert ( \partial _s^j U_{n-1} - \partial _s^j U_{n-2}) \langle \psi \rangle ^4 \Vert _{L^2}\Big ) \\&\quad \lesssim \ {\varepsilon }^{0.99}\left(\Vert \Delta Q_{n-1} \Vert _{X_{2,4}} + \epsilon ^{1.99} |\Delta \overline{c}_{n-1}|\right). \end{aligned}$$

Next, we clearly have

$$\begin{aligned}&\sum _{k' = 0}^{2} \sum _{j = 0}^2 \Vert \partial _s^{k'} \partial _\psi ^j G \langle \psi \rangle ^{50} \Vert _{L^2} \le |{\omega _0}_{n-1} - {\omega _0}_{n-2}| \sum _{k' = 0}^{2} \sum _{j = 0}^2\left\Vert \partial _s^{k'} \partial _\psi ^j Q_{n-1} \langle \psi \rangle ^{50} \right\Vert_{L^2} \\&\quad \lesssim \frac{1}{\left|{\omega _0}_{n-1} + {\omega _0}_{n-2}\right|} |{\omega _0}_{n-1}^2 - {\omega _0}_{n-2}^2| \Vert Q_{n-1} \Vert _{X_{2,50}} \\&\quad \lesssim \ {\varepsilon }{\varepsilon }^{0.99} |\overline{{\omega _0}}_{n-1} - \overline{{\omega _0}}_{n-2}|. \end{aligned}$$

Finally, we have the boundary condition

$$\begin{aligned} \Vert b \Vert _{H^3_s} \lesssim {\varepsilon }\Vert q_e \Vert _{H^3}^3 |\Delta \overline{{\omega _0}}_n|. \end{aligned}$$

Consolidating all the above bounds with estimate (59) concludes the proof of the lemma.

Appendix A. Derivation of Near-Boundary Navier–Stokes Equations

In this section, we give the detailed calculations for the Navier–Stokes equations claimed in Sect. 2. We recall the standard identities, which will be used in the next lemmas:

$$\begin{aligned} n'(s)=\gamma (s)\tau (s),&\qquad \tau '(s)=\gamma (s)n(s). \end{aligned}$$

We recall that the map

$$\begin{aligned} \begin{aligned} \{x\in M:\quad 0<\textrm{dist}(x,\partial M)<\delta \}&\rightarrow \mathbb T_L\times (0,\delta )\\ x&\rightarrow (s,z)=\left(s(x_1,x_2),z(x_1,x_2) \right) \end{aligned} \end{aligned}$$

is a diffeomorphism. In this transformation, we have

$$\begin{aligned} \nabla _x s=\frac{\tau }{J},\qquad \nabla _x z=n(s). \end{aligned}$$

For a vector field $u:M\rightarrow \mathbb R^2$, we also have

$$\begin{aligned} \begin{aligned}&u_1=n_2u_\tau -\tau _2u_n=\tau _1u_\tau -\tau _2u_n,\\&u_2=-n_1u_\tau +\tau _1u_n=\tau _2u_\tau +\tau _1u_n. \end{aligned} \end{aligned}$$

Lemma 15

For any vector field $u:M\rightarrow \mathbb R^2$ and scalar function $f:M\rightarrow \mathbb R$ supported near the boundary $\partial M$ there holds

$$\begin{aligned} \begin{aligned}&u\cdot \nabla f=\frac{u_\tau }{J}\partial _s f+u_n\partial _z f.\\ \end{aligned} \end{aligned}$$

In particular, by choosing $u=\tau$ and $u=n$ respectively, there hold

$$\begin{aligned} \begin{aligned} \tau \cdot \nabla f=\frac{1}{J}\partial _s f,\qquad n\cdot \nabla f=\partial _z f.\\ \end{aligned} \end{aligned}$$

Proof

This follows by direct calculation. We have

$$\begin{aligned} u\cdot \nabla f&=u\cdot \nabla _x f(s,z)=u_1\left(\partial _sf\partial _{x_1}s+\partial _z f\partial _{x_1}z\right)+u_2\left(\partial _sf\partial _{x_2}s+\partial _z f\partial _{x_2}z\right)\\&=u_1\left(\partial _s f\frac{\tau _1}{J}+\partial _z f n_1\right)+u_2(\partial _s f\cdot \frac{\tau _2}{J}+\partial _z f n_2)\\&=\frac{u\cdot \tau }{J}\partial _s f+(u\cdot n)\partial _z f. \end{aligned}$$

Lemma 16

The following identities holds for any given vector field $u:M\rightarrow \mathbb R^2$:

$$\begin{aligned} \begin{aligned} (u\cdot \nabla u)\cdot \tau&=\left(\frac{u_\tau }{J},u_n\right)\cdot \nabla _{s,z}u_\tau -\frac{\gamma (s)}{J}u_\tau u_n,\\ (u\cdot \nabla u)\cdot n&=\left(\frac{u_\tau }{J},u_n\right)\cdot \nabla _{s,z}u_n-\frac{\gamma (s)}{J}u_\tau ^2,\\ \Delta u\cdot \tau&=\frac{1}{J}\partial _z(J\partial _zu_\tau )+\frac{1}{J}\partial _s\left(\frac{1}{J}\partial _s u_\tau \right)-\frac{1}{J}\partial _s\left(\frac{\gamma u_n}{J} \right)-\frac{\gamma }{J}\left(\gamma u_\tau +\partial _s u_n \right),\\ \Delta u\cdot n&=\frac{1}{J}\partial _z(J\partial _zu_n)+\frac{1}{J}\partial _s\left(\frac{1}{J}\partial _s u_n \right)-\frac{1}{J}\partial _s \left(\frac{\gamma u_\tau }{J} \right)-\frac{\gamma }{J}\left(\partial _s u_\tau -\gamma u_n \right),\\ \nabla \cdot u&=\frac{1}{J}\left(\partial _s u_\tau -\gamma (s)u_n \right)+\partial _z u_n,\\ \nabla ^\perp \cdot u&=\partial _1u_2-\partial _2 u_1=\frac{\gamma }{J}u_\tau -\partial _zu_\tau +\partial _s u_n. \end{aligned} \end{aligned}$$

Proof

We check the first, the third and the fifth identities only, and the proofs for other identities are similar. We have

$$\begin{aligned} \begin{aligned} (u\cdot \nabla u)\cdot \tau&=(u\cdot \nabla _x u_1)\tau _1+(u\cdot \nabla _x u_2)\tau _2\\&=\left(\frac{u_\tau }{J},u_n\right) \cdot \nabla _{s,z} u_1 \tau _1+\left(\frac{u_\tau }{J},u_n\right)\cdot \nabla _{s,z} u_2 \tau _2\\&=\left(\frac{u_\tau }{J},u_n \right)\cdot \nabla _{s,z}u_\tau -\frac{u_\tau }{J} \left(\tau _1'(s)u_1+\tau _2'(s)u_2\right). \end{aligned} \end{aligned}$$

We note that

$$\begin{aligned} \begin{aligned} \tau _1'u_1+\tau _2'u_2&=\gamma n_1(\tau _1u_\tau -\tau _2u_n)+\gamma n_2(\tau _2u_\tau +\tau _1u_n)\\&=\gamma (-\tau _2)(\tau _1u_\tau -\tau _2u_n)+\gamma \tau _1(\tau _2u_\tau +\tau _1u_n)\\&=-\gamma \tau _1\tau _2u_\tau +\gamma \tau _2^2 u_n+\gamma \tau _1\tau _2u_\tau +\gamma \tau _1^2 u_n=\gamma u_n. \end{aligned} \end{aligned}$$

Combining the above with the previous calculation, we obtain

$$\begin{aligned} (u\cdot \nabla u)\cdot \tau =\left(\frac{u_\tau }{J},u_n\right)\cdot \nabla _{s,z}u_\tau -\frac{\gamma (s)}{J}u_\tau u_n. \end{aligned}$$

Now we show the third identity. We have

$$\begin{aligned} \begin{aligned} \Delta u\cdot \tau&=\Delta u_1\cdot \tau _1+\Delta u_2\cdot \tau _2=\sum _{i=1}^2\frac{1}{J}\left(\partial _z(J\partial _z u_i)+\partial _s\left(\frac{1}{J}\partial _s u_i\right) \right)\tau _i\\&=\frac{1}{J}\partial _z(J\partial _zu_\tau )+\frac{1}{J}\partial _\theta \left(\frac{1}{J}\partial _\theta (u\cdot \tau ) \right)-\frac{1}{J}\partial _s\left(\frac{1}{J}u\cdot \partial _s \tau \right)-\sum _i \frac{1}{J}\partial _s u_i\tau _i'\\&=\frac{1}{J}\partial _z(J\partial _zu_\tau )+\frac{1}{J}\partial _\theta \left(\frac{1}{J}\partial _s u_\tau \right)-\frac{1}{J}\partial _s\left(\frac{\gamma u_n}{J} \right)-\frac{\gamma }{J}\left(\gamma u_\tau +\partial _s u_n \right). \end{aligned} \end{aligned}$$

For incompressibility, we find

$$\begin{aligned} \begin{aligned} \nabla \cdot u&=\partial _{x_1}u_1+\partial _{x_2}u_2=\sum _{i}\partial _{x_i}s \partial _s u_i+\partial _{x_i}z\partial _z u_i=\sum _i \frac{\tau _i}{J}\partial _s u_i+n_i\partial _z u_i\\&=\frac{1}{J}\left(\partial _s u_\tau -\gamma (s)u_n \right)+\partial _z u_n. \end{aligned} \end{aligned}$$

The proof is complete.