text:
\begin{document}

\title{A Priori Estimates for Two-Dimensional Water Waves with Angled Crests}
\author{Rafe H.~Kinsey\thanks{R.H.~Kinsey is supported in part by a University of Michigan Rackham Regents Fellowship and by NSF grants DMS-0800194 and DMS-1101434.} \and  Sijue Wu\thanks{S.~Wu is supported in part by NSF grant  DMS-1101434.}}
\newcommand{\Addresses}{{\bigskip\footnotesize

R.H.~Kinsey, \textsc{Department of Mathematics, University of Michigan, Ann Arbor, MI 48109}
\par\nopagebreak
  \textit{E-mail address}: \texttt{rkinsey@umich.edu  }

  \medskip

  S.~Wu (Corresponding author), \textsc{ Department of Mathematics, University of Michigan, Ann Arbor, MI 48109   }\par\nopagebreak
  \textit{E-mail address}: \texttt{sijue@umich.edu}
  }}

\date{}

\maketitle

\begin{abstract}
We consider the two-dimensional water wave problem in the case where the free interface of the fluid meets a vertical wall at a possibly non-right angle; our problem also covers interfaces with angled crests.  We assume that the fluid is inviscid, incompressible, and irrotational, with no surface tension and with air density zero.  We construct a low-regularity energy and prove a closed energy estimate for this problem. Our work differs from earlier work in that, in our case, only a degenerate Taylor stability criterion holds, with $-\f{\partial P}{\partial \vec{n}} \ge 0$, instead of the strong Taylor stability criterion $-\f{\partial P}{\partial \vec{n}} \ge c > 0$.
\end{abstract}

\tableofcontents

\section{Introduction}\label{sec:intro}
\subsection{Water wave problems}\label{sec:1}
A class of water wave problems concerns the dynamics of the free surface separating a zero-density region (air) from an  incompressible fluid (water), under the influence of gravity.

Let $\Omega(t)$ be the fluid region, $\Sigma(t)$ be the free surface between the fluid and the air, and $\Upsilon$, if it exists, be the fixed rigid boundary of $\Omega(t)$  (e.g., the ocean bottom or the coast), for time $t \ge 0$; thus $\partial(\Omega(t)) = \Sigma(t) \cup \Upsilon$.  We will henceforth assume that the fluid is not only incompressible but also irrotational, and we will neglect surface tension and viscosity.
Assume that the fluid density is $1$. If the gravity field is $-\vec{j}$, the governing equations of motion are
\begin{align}
  \label{eq:1}  \vec{v}_t + \vec{v} \cdot \nabla \vec{v} &= - \vec{j} - \nabla P \quad \text{ on } \Omega(t)\\
 \label{eq:2} \text{div } \vec{v}& = 0 \quad \text{ on } \Omega(t) \\
 \label{eq:3} \text{curl } \vec{v} &= 0 \quad \text{ on } \Omega(t) \\
\label{eq:4} P &= 0 \quad \text{ on } \Sigma(t) \\
  \label{eq:5} (1,\vec{v}) \text{ is tangent}&\text{ to the free surface }(t,\Sigma(t))\\
 \label{eq:6}   \vec{v} \text{ is tangent to}&\text{ the fixed boundary } \Upsilon \text{ (if it exists)},
\end{align}
where $\vec{v}$ is the fluid velocity and $P$ is the fluid pressure.  We also assume that
\begin{equation}
  \label{eq:620}
  \vec{v}(x,y,t) \to 0 \text{ as } y \to -\infty.
\end{equation}
An important quantity governing the stability of these problems is $-\f{\partial P}{\partial \vec{n}}$, where $\vec{n}$ is the outward-facing unit normal vector to $\Sigma(t)$.

The study of water waves dates back centuries; early mathematical works include those by Stokes \cite{stokes}, Levi-Civita \cite{levi-civita} and G.I.~Taylor \cite{taylor}.  \cite{nalimov}, \cite{yosihara}, and \cite{craig} obtained early local-in-time existence results for the two-dimensional water wave problem for small data.  In 1997, \cite{wu1997} proved, for the infinite depth two-dimensional water wave problem   \eqref{eq:1}-\eqref{eq:5} and \eqref{eq:620} with $\Upsilon = \emptyset$,
that the Taylor stability criterion $-\f{\partial P}{\partial \vec{n}} \ge c > 0$ always holds and that the problem is locally well-posed in Sobolev spaces $H^s, s > 4$, for arbitrary data.  \cite{wu1999} proved a similar result in three dimensions. Since then, there have been numerous results on local well-posedness in both two and three dimensions, for the water wave equations with nonzero vorticity \cite{christodoulou-lindblad} \cite{lindblad}  \cite{coutand-shkoller}, with a fixed bottom \cite{lannes}, and with nonzero surface tension \cite{ambrose-masmoudi} \cite{coutand-shkoller}. \cite{alazard-b-z} proved local well-posedness of the problem for interfaces in $C^{3/2 + \e}$.\footnote{A recent preprint, \cite{alazard-b-z-strichartz}, improves on these results using Strichartz estimates, reducing the required smoothness of the interface to a fraction under $C^{3/2}$.  \cite{hunter-ifrim-tataru} also has a low-regularity result that improves on \cite{alazard-b-z} for 2-D water waves. Both \cite{alazard-b-z-strichartz} and \cite{hunter-ifrim-tataru} work in the setting where the strong Taylor stability criterion $-\frac{\partial P}{\partial \vec{n}}\ge c>0$ holds.}  All of these works assume the strong Taylor stability criterion $-\f{\partial P}{\partial \vec{n}} \ge c > 0$. In addition, in the past several years \cite{wu2009}, \cite{wu2011}, \cite{germain-masmoudi-shatah}, \cite{pusateri-ionescu}, \cite{alazard-delort} and \cite{hunter-ifrim-tataru} have proved results showing almost global or  global well-posedness in two and three dimensions for sufficiently small and smooth initial data.

All of these results assume either no fixed boundary or else a fixed boundary that is a positive distance away from the free interface.  In actual oceans, of course, the free surface does intersect with the rigid boundary, e.g., on the coast.  It is important, therefore, to study the water wave problem in this setting.  In this paper, we will begin such a program, by considering the two-dimensional water wave equations in the presence of a rigid vertical wall that interacts with the free interface.

In addition to addressing the interaction of the free surface with the rigid vertical wall, our research  covers singularities in water waves away from the rigid boundary. Indeed, by Schwarz reflection, a model of waves making a non-right angle with a rigid vertical wall corresponds to a symmetric angled crest in the middle of the ocean; see Figure \ref{fig:a}.
More generally, our work applies to water waves with very low regularity, including those with angled crests.\footnote{We note that the angled crests here don't have to be symmetric.}  Our work differs from the low-regularity result of \cite{alazard-b-z}, which assumes that the strong Taylor stability criterion $-\f{\partial P}{\partial \vec{n}} \ge c > 0$ holds and considers surfaces in  $C^{3/2+ \e}$.  In our case, we allow surfaces that are not even in $C^1$ where only a degenerate Taylor stability criterion $-\f{\partial P}{\partial \vec{n}} \ge 0$ holds.\footnote{In \cite{alazard-b-z}, Alazard et al.~consider the water wave problem in the presence of a bottom bounded away from the interface. In this framework, the strong Taylor stability condition doesn't necessarily hold, so it is assumed. In our case, the only rigid boundary is the vertical wall and we can prove that the degenerate Taylor stability condition always holds. The degeneracy occurs at the singularity in the interface and when the water wave meets the wall with a non-right angle.} Our research relates to recent work of Wu \cite{wu:self}, who constructed a class of convection-dominated self-similar solutions of the two-dimensional water wave equation with such angled crests. Our work indicates that the water wave equations \eqref{eq:1}-\eqref{eq:620} admit solutions with this type of singularity.
\begin{figure}[H]
\centering
\includegraphics[width=2in]{reflected-a.png}
\caption[Symmetric Angled Crests]{Under a Schwarz reflection, a non-right angle at a vertical wall corresponds to a symmetric angled crest in the middle of surface.}
\label{fig:a}
\end{figure}

For simplicity, we consider a rigid boundary consisting of two vertical walls, with water of infinite depth in between the walls. Working in two dimensions $(x,y) \in \R^2$, we assume that our fixed walls are at $x = 0, 1$, with the fluid region $\Omega_0(t) \subset [0,1] \times (-\infty, c)$ for some $c < \infty$. We assume that the water surface makes a right angle with the wall $x=0$, but we allow a non-right (i.e., not 90$^\circ$) angle at $x=1$.\footnote{Although in our problem there is a vacuum-liquid-solid triple point, one should not confuse it with the dynamic contact angle problem, where typically surface tension is very important: our problem and the contact angle problem are two different problems.} \footnote{We started studying our problem by focusing only on resolving the difficulties when the free surface makes a non-right angle $\ang$ at the wall $x=1$. The result we obtain, however, covers cases where the free surface has angled crests; in particular, the angle of the water wave at $x=0$ doesn't have to be $90^\circ$.}

This angle will be of fundamental importance for the remainder of the paper.\footnote{We will see in \S \ref{sec:27}  that the angle $\ang$ must be no more than $\f{\pi}{2}$. Therefore a non-right angle $\ang$ is necessarily $<90^\circ$. }
 We will use $\ang$ to denote this angle; see Figure \ref{fig:b}. If there is a right angle at $x=1$, then existing techniques ought to apply, so long as there are no singularities in the middle of the free surface. What's novel about our work is that it extends to the non-right angle regime, and allows singularities in the middle of the free surface.

We begin by symmetrizing the problem, using Schwarz reflection to expand the domain across the $y$-axis, using the fact that \eqref{eq:6} implies that $v_1(0,y, t) \equiv 0$. Precisely, for $\vec{v} = (v_1, v_2)$ and $x \in [0,1]$, we define
\begin{equation}
  \label{eq:366}
  \vec{v}(-x,y, t) = (-v_1(x,y, t), v_2(x,y, t)).
\end{equation}
It is easy to check that equations \eqref{eq:1} through \eqref{eq:620} continue to hold in the expanded domain. Henceforth, we shall work exclusively in this reflected domain $\Omega(t)$. $\Omega(t) \subset [-1,1] \times (-\infty, c)$ is symmetric with respect to the $y$-axis, with  solid vertical walls $\Upsilon$ at $x = -1, 1$. When we say the {\em wall} we will mean $x \equiv \pm 1$, and when we say the {\em corners}, we will mean the corner of the water surface $\Sigma(t)$ at $x = \pm 1$. See Figure \ref{fig:b}.
\begin{figure}[H]
\centering
\includegraphics[width=2in]{reflected-b.png}
\caption[The Geometric Framework]{The fluid domain $\Omega_0(t)$ has solid walls at $x = 0, 1$, with a possibly non-right angle $\ang$ at $x = 1$ and a right angle at $x = 0$.  The figure shows the reflected domain $\Omega(t)$ under a Schwarz reflection across $x=0$, with a symmetric free surface $\Sigma(t)$.}
\label{fig:b}
\end{figure}

Allowing non-right angles at the wall and singularities in the surface introduces significant challenges.

In \cite{wu1997}, a key result was proving that the Taylor sign condition $-\f{\partial P}{\partial \vec{n}} \ge c > 0$ always holds as long as the surface is smooth and non-self-intersecting. In the situation where $-\f{\partial P}{\partial \vec{n}} < 0$, \cite{ebin} has shown that the free boundary problem is not well-posed, a fact which is not surprising, since it is the positive sign of $-\f{\partial P}{\partial \vec{n}}$ that gives the hyperbolicity of the equation.  In our situation, however, we will have that $-\f{\partial P}{\partial \vec{n}} = 0$ at the wall when there is a non-right angle, and at the points on the surface where there are singularities (e.g., angled crests).\footnote{Precisely, $\nabla P=0$ at the wall when $\ang\ne\frac\pi 2$ and at the points on the free surface where there are angled crests.}  In this case, the water wave equation is degenerate hyperbolic, and it is harder to construct an energy that can be closed.

A second challenge comes from the potential asymmetry of certain quantities at the left and right boundaries of the domain. Thanks to the Schwarz reflection, we can now treat our problem as being periodic.  The possibly non-right angle of the free surface with the wall introduces a fundamental asymmetry: if the surface is sloping downwards near $x = -1$, it is sloping upwards near $x=1$.  Therefore, the values of certain quantities at $x = -1$ will not agree with their values at $x = 1$.  This leads to problems both with integration by parts and with handling estimates of various commutators involving the (periodic) Hilbert transform.  This challenge actually proves to be a blessing in disguise, because it directs us to precisely the quantities that {\em are} well-behaved as periodic functions.

A third challenge also proves to have a silver lining.  \cite{wu1997} relied on the Riemann mapping to flatten out the fluid domain.  When the angle of the free surface with the wall is not $90^\circ$, or when there is an angled crest in the middle of the free surface,  the Riemann mapping has a singularity.  Instead of avoiding the Riemann mapping, we embrace it, since it directly captures the geometry of the interface,  and gives us heuristics to determine which quantities we might or might not be able to control. We used these heuristics, and more importantly the self-similar solution constructed in \cite{wu:self}, to guide us in the construction of our energy.

In this paper, we prove an a priori estimate  for solutions of the water wave equation in this framework.
 We follow the general approach of Wu's earlier papers \cite{wu1997} and \cite{wu1999}, in reducing the water wave problem to an equation on the free surface, differentiating the equation with respect to time, and then using the fact that the velocity is antiholomorphic to express the nonlinearities in terms of commutators involving the Hilbert transform.  The novelty is that we introduce a new energy that relies on two different singular weights and controls some natural holomorphic quantities, and that our estimates {\em do not} depend on any positive lower bound for $-\f{\partial P}{\partial \vec{n}}$.\footnote{We will show that in our framework $-\f{\partial P}{\partial \vec{n}} \ge 0$. In fact, a positive lower bound on $-\f{\partial P}{\partial \vec{n}}$ is roughly equivalent to a non-singular Riemann mapping or, equivalently, a non-singular surface; see \S \ref{sec:27}.}   Substantial technical difficulties exist due to the very low regularity involved in the energy and the lack of a positive lower bound for  $-\f{\partial P}{\partial \vec{n}}$.\footnote{In particular, we were led to an energy consisting only of a limited and specific collection of weighted quantities rather than the usual assortment of quantities in more traditional energies (e.g., squares of classical Sobolev space norms). See the characterization of the energy at \eqref{eq:532} and \eqref{eq:575} in \S \ref{sec:36d}.}  Nevertheless, we have overcome these challenges.

Our energy inequality is a crucial step towards proving local existence in this framework. This will be the focus of an upcoming paper.

We mention that our energy is finite for the self-similar water waves constructed in \cite{wu:self}.

\subsection{Outline of the paper}\label{sec:4}
In the next subsection, \S \ref{sec:5}, we present some of the notations and conventions
and introduce the function spaces and norms to be used in the paper. Then, in \S \ref{sec:6}, we introduce the Lagrangian framework for the water wave problem,  following \cite{wu1997} and \cite{wu2009}.  In particular, in \S \ref{sec:10}, we discuss some of the boundary challenges mentioned above, and introduce a special  derivative that we will use in our energies. In \S \ref{sec:11}, we introduce the Riemann mapping and use it to derive a form of the equation in ``Riemannian coordinates.'' Finally, in \S \ref{sec:energy}, we define the energy (in \S \ref{sec:35}) and state our main result, the a priori inequality (in \S \ref{sec:38}).  We offer some discussion on our choice of energy in \S \ref{sec:36}.   We begin the proof in \S \ref{sec:39}, and then in \S \ref{sec:42} we outline the remainder of the proof, which takes up sections \S \ref{sec:43} through \S \ref{controlE1}.

In \S \ref{sec:36e} we give a characterization of the energy in terms of the  velocity and position of the free surface, and we discuss the types of singularities allowed when our energy is finite.

The derivation of the free surface equations and the proof of the main result rely on understanding the boundary behaviors, the holomorphicity, and the means of various quantities; we leave these, as well as some basic identities and inequalities used in the proof of the main result, to appendices \S\ref{sec:hmb}-\S \ref{sec:tech}. The reader may want to read these appendices before certain sections in the main text. We have two additional appendices that might be useful to the reader. In \S \ref{notation}, we provide an overview of the notation used in the paper, with cross references to where everything was initially defined.  In \S \ref{quantities}, we list various quantities controlled by the energy, again with cross references.

\subsection{Notations, conventions and function spaces}\label{sec:5}
We will define most of our notations throughout the text, as we introduce our various quantities.  Here we only list some general conventions and notations.

Since we are in two dimensions, we will often work in complex coordinates $(x,y) = x + iy$.  We will use $\Re z:=x$ and $\Im z:=y$  to represent the real and imaginary parts, respectively, of  $z=x+iy$.

Compositions are always in terms of the spatial variable. For example, for $f=f(\alpha, t)$, $g=g(\alpha,t)$, we define $f\circ g=f\circ g(\alpha,t):=f(g(\alpha,t),t)$. An expression $f_{x}(x,y)$ means $\partial_xf(x,y)$; we occasionally use the notation $f'$, which is always the spatial derivative in whatever coordinates we are using.

Once we have reduced the water wave equations to an equation in one spatial dimension, we will primarily be working with the spatial domain $I := [-1,1]$. We will often refer to the ``boundary''; this refers to what happens at $\pm 1$. We write $\bound{f} := f(1) - f(-1)$. We will use
\begin{equation}
  \label{eq:660}
  \avg_I f := \f{1}{\abs{I}} \int_I f(x) dx = \f{1}{2} \int_{-1}^1 f(x) dx
\end{equation}
for the mean of a function $f$. Here, and elsewhere for other integrals,
when there is no risk of ambiguity, we will often drop the subscript $I$.

We will often use commutators.  We define
\begin{equation}
  \label{eq:659}
  [A,B] := AB - BA.
\end{equation}
We will use the following notation as an abbreviation for a type of higher-order Calderon commutator:
\begin{equation}
  \label{eq:624}
  [f,g;h](\aa) := \frac\pi{4i}\int \f{f(\aa) -f(\bb)}{\sin(\f{\pi}{2}(\aa-\bb))} \f{g(\aa) - g(\bb)}{\sin(\f{\pi}{2}(\aa-\bb))} h(\bb) d\bb.
\end{equation}

We will use $C$ as a placeholder to refer to a universal constant, possibly varying from line to line. We will also often use the notation $f \lec g$, which means that there exists some universal constant $C$ such that $f \le C g$.

We will at several points have long series of identities or inequalities.  When we say ``on the RHS'' of an equation block with a string of multiple equalities or inequalities, we mean all the terms on the right hand side of the last equality or inequality sign in the string. Similarly, when we say ``on the LHS,'' we mean all the terms to the left of the very first equality or inequality sign in  the string of equalities and inequalities.  We have tried to avoid saying ``on the $n$th line'' when any of the mathematical formulas splits into more than one typographic line, but if we have, ``line'' refers to the mathematical, not typographic, line.

We have tried to give extensive cross references for each time we use a result or estimate.  We tend to refer to equation numbers, rather than propositions, since it seems that these will be easier to find as cross references. When we refer to an equation number as part of a proposition, we are of course referring to the whole proposition, including any conditions assumed.

When we are deriving estimates, we sometimes use the cross references within our equations, e.g.:
\begin{equation}
  \label{eq:661}
  \begin{aligned}
    f  & \le g
    \\ & \le h
  \end{aligned}
\end{equation}
and
\begin{equation}
  \label{eq:662}
  \begin{aligned}
    h & \lec j + f
    \\ & \lec j + \eqref{eq:661}
    \\ & \lec j + h.
  \end{aligned}
\end{equation}
This means \eqref{eq:661} is used to obtain \eqref{eq:662}. We hope this will help the reader locate the previous estimate or estimates.

In several of our more complicated estimates, we will split terms up $f = I + II$ and then $I = I_1 + I_2$, $I_1 = I_{11} + I_{12}$, etc.  Such notation will be {\em local to each section}. There is an ambiguity between the use of $I$ as a placeholder, its use as the identity operator, and its use as $I:= [-1,1]$. It should be clear from the context which one is being used.

We now introduce  the function spaces and norms we will use.   We will work with functions $f(\cdot,t)$ defined on $I=[-1,1]$. Except when necessary to avoid ambiguity, we neglect to write the time variable; when it is not specified,  function spaces and norms are in terms of the spatial variable.

We say $f\in C^k(J)$, $J=(-1,1)$ or $[-1,1]$,  if for every $0 \le l \le k$, $\partial_x^l f$ is a continuous function on the interval $J$.
We say $f \in C^k(S^1)$ (i.e., periodic $C^k$) if for every $0 \le l \le k$,
$\partial_x^l f\in C^0[-1,1]$ and
$\partial_x^lf(1) = \partial_x^lf(-1)$. ($\partial_x^lf$ at the endpoints $1$ or $-1$ is the derivative from the left or right, respectively.) Note in particular that saying $f \in C^0(S^1)$ implies that $\bound{f}=0$.

For $1 \le p < \infty$, we define our $L^p$ spaces by the norms
\begin{equation}
  \label{eq:395}
  \nm{f}_{L^p} := \nm{f}_{L^p(I)}:=\paren{\int_{[-1,1]} \abs{f}^p}^{1/p},
\end{equation}
and we define $L^\infty$ analogously. We will sometimes deal with weighted $L^p$ spaces. We write
\begin{equation}
  \label{eq:428}
  \nm{f}_{L^p( \omega)} = \nm{f}_{L^p(\omega dx)} := \paren{\int_I \abs{f(x)}^p \omega(x) dx}^{1/p}
\end{equation}
for weights $\omega \ge 0$.
Whenever we write $L^p$, we will be referring to $L^p(I)$, in the spatial variable. For weighted $L^p$ spaces, we always write $L^p(\omega)$ or $L^p(\omega dx)$,   where $\omega$ is the weight function.

Note that if $f \in C^0(S^1)$, with $f' \in L^p$, then if we extend $f$ periodically to $(-3,3)$,  the weak derivative of the extended function is the periodic extension of $f'$ to $(-3,3)$ and is in $L^p(-3,3)$.

We define the homogeneous half-derivative space $\dot{H}^{1/2}$ by the norm
\begin{equation}
  \label{eq:23}
  \nm{f}_{\dot{H}^{1/2}} := \paren{\frac\pi 8 \iint_{I \times I} \f{\abs{f(\aa) - f(\bb)}^2}{\sin^2(\f{\pi}{2}(\aa - \bb))} d\aa d\bb}^{1/2}.
\end{equation}

Through the remainder of the paper, when we say the boundary value of a function $G$ defined on the fluid region $\Omega(t)$ (resp., on $P^-:=I\times (-\infty, 0]$), we mean the value of  $G$ on the free surface (resp., on $I\times \{0\}$); we do not include the value on vertical walls $x=\pm1$. Except when there's a risk of confusion, we will slightly abuse notation and say that a function $f$ on the free surface (resp., on $I\times \{0\}$) is ``holomorphic'' (or ``antiholomorphic''); what we mean, precisely, is that it is the boundary value of a function that is holomorphic (or antiholomorphic) in the fluid region $\Omega(t)$ (resp., on $P^-$).

\section{The surface equation in the Lagrangian framework}\label{sec:6}

Let $z(\alpha, t)=x(\a,t) + i y(\a,t)$, $\a\in I=[-1,1]$  be a parametrization  of the free surface $\Sigma(t)$ in  the {\em Lagrangian} variable $\a$, i.e., $z_t(\a,t) = \vec{v}(z(\a,t),t)$ is the velocity and $z_{tt}$ is the acceleration of the particle occupying position $z(\a,t)$ at time $t$. We choose the initial parametrization to be in arc-length coordinates: $|z_\a(\a,0)| \equiv 1$.\footnote{\label{fn:za}We choose this initial parametrization for the sake of convenience in discussion. Our a priori estimate is a purely geometric one and does not depend on this parametrization. On the other hand, if $\abs{z_\a(\cdot, 0)}\equiv 1$, then $\abs{z_\a(\cdot, t)}$ is bounded away from $0$ and infinity so long as our  energy $E(t)$ remains finite.  This follows from  $\partial_t |z_\a|\le |z_\a||D_\a z_t|$,  $\partial_t |1/{z_\a}|\le |1/{z_\a}||D_\a z_t|$, $\|D_\a z_t\|_{L^\infty}\le C(E(t))$ (see \S \ref{sec:quants}), and the Gronwall inequality.} Along the free surface,  the Euler equation \eqref{eq:1} is $z_{tt} + i = -\nabla P$. By equation \eqref{eq:4},  we know $\nabla P$ is orthogonal to the free surface.  Since $iz_\a$ is normal to the free surface,  we can rewrite our main equation as
\begin{equation}
  \label{eq:8}
  z_{tt} + i = i \mathfrak{a} z_\a,
\end{equation}
where
\begin{equation}
  \label{eq:28}
  \mathfrak{a} = - \f{\partial P}{\partial \vec{n}} \f{1}{\abs{z_\a}} \in \R
\end{equation}
for $\f{\partial P}{\partial \vec{n}}:=\vec{n}\cdot \nabla P$ the outward-facing normal derivative. The incompressibility and irrotationality conditions \eqref{eq:2} and \eqref{eq:3} imply that the conjugate velocity is holomorphic; therefore $\bar{z}_t$ is the boundary value of a holomorphic function in the fluid region.

\subsection{The quasilinear equation}\label{sec:9}
We henceforth focus on the equations on the free surface.\footnote{We may solve for the velocity on $\Omega(t)$ from its boundary values (including the condition that it goes to zero as $y \to -\infty$), and then solve for the pressure from the velocity.} As in \cite{wu1997} and following works, we differentiate  \eqref{eq:8} with respect to time and take conjugates, turning it into the quasilinear equation\footnote{We call it ``quasilinear'' because in the classical situation \cite{wu1997}, this equation is quasilinear with the RHS the lower-order term. However, in our setting, due to the degeneracy of $-\f{\partial P}{\partial \vec{n}}$ we do not know a priori that this is still the case; only by our proof do we show that the RHS is, indeed, lower-order and that \eqref{eq:11} is in fact quasilinear. All references to \eqref{eq:11} and related equations being ``quasilinear'' should be interpreted with this in mind.}
\begin{equation}
  \label{eq:11}
  \bar{z}_{ttt} + i \af \bar{z}_{t\a} = -i \af_t \bar{z}_\a,
\end{equation}
where we continue to have $\bar{z}_t$  the boundary value of a holomorphic function. This is the basic equation we will work with throughout the paper.

The holomorphicity of $\bar{z}_t$ implies that $\af \partial_\a\bar{z}_t$ can be written as $\af \partial_\a \H\bar{z}_t$, where $\H$ is the Hilbert transform with respect to the free surface $\Sigma(t)$. Because $i  \partial_\a \H$ is a nonnegative operator, a natural positive energy can be defined.

In \cite{wu1999} and \cite{wu2009}, coordinate-independent formulas for the RHS were derived, using the holomorphicity of $\bar{z}_t$ and the invertibility of the double-layer potential.  We will instead follow the original approach of \cite{wu1997}, relying on the Riemann mapping version of the equation to derive the RHS.  We do so in \S \ref{sec:28}.

\subsection{The degeneracy of the Taylor stability criterion}

Recall that in \cite{wu1997} it was essential to show the {\em strong Taylor stability criterion}  $-\frac{\partial P}{\partial\vec{n}} \ge c_0 > 0$. Here we are not so lucky, since we only have $-\frac{\partial P}{\partial\vec{n}}  \ge 0$.  We will prove this fully in \S \ref{sec:27a} and \S \ref{sec:27}. For now, we make a quick observation: $\nabla P$ is degenerate when the free surface makes a non-right angle at the corner.\footnote{This observation has also been made by \cite{alazard-b-z-2} in \S 6.1.} Indeed, from \eqref{eq:8}, we have
\begin{equation}
  \label{eq:9}
  \f{x_\a} {-y_\a}= \f{y_{tt} + 1}{x_{tt}}.
\end{equation}
Along the wall, we know that $x_t \equiv 0$ by \eqref{eq:6}, so we have $x_{tt} \equiv 0$. Observe that the angle $\ang$ between the water surface and the wall  $x = -1$ is given by
\begin{equation}
  \label{eq:365}
\tan \ang =\f{ x_\a}{-y_\a}.
\end{equation}
If $(y_{tt} + i)/x_{tt}$ is infinite, then the angle must be $90^\circ$. Therefore, the only possible way for the angle of the free surface with the wall to be other than a right angle is if the numerator $y_{tt} + 1$ is zero at the corner. This implies that $\nabla P = 0$ at the corner, hence $-\frac{\partial P}{\partial \vec{n}}:=-\vec{n}\cdot \nabla P=0$ at the corner.

\subsection{Boundary behavior and a special derivative}\label{sec:10}
We discuss here the boundary behavior of our basic quantity $\bar z_t$ and define a weighted spatial derivative that will be used in constructing higher-order
energies. For this paper we will also need to understand the boundary properties of some other quantities. We will leave those to \S\ref{sec:19} in the appendices.

In our framework, the conjugate velocity $\bar{\vec{v}}$ has real part $\Re \bar{\vec{v}}$ odd and  imaginary part $\Im \bar{\vec{v}}$ even, and  $\Re\vec{v}(\pm 1,y)=0$ by \eqref{eq:6}. Therefore, $\bar{\vec{v}}$ is a periodic holomorphic function in $\Omega(t)$, with $\bar{\vec{v}}(-1,y)=\bar{\vec{v}}(1,y)$, and our basic function $\bar z_t$ satisfies
\begin{equation}
  \label{eq:1010}
  \bound{\bar{z}_t} = 0.
\end{equation}

We would like the quantities we work with to have such periodic  boundary behavior, since we often integrate by parts and since periodicity will be essential for certain results involving the periodic Hilbert transform that we will use. Periodicity is not preserved by the spatial derivative $\partial_\a$; for example, we do not necessarily have $\bound{\partial_\a \bar{z}_t}=0$.
Notice, however, that for holomorphic functions, $\partial_z = -i \partial_y$ and $\partial_y$ preserves periodicity. We therefore look for the derivative on the surface $\Sigma(t)$ that corresponds to $\partial_z$ inside the domain. This derivative is
\begin{equation}
  \label{eq:367}
  D_\a := \f{1}{z_\a}\partial_\a.
\end{equation}
Indeed, if $g(\a,t) = G(z(\a,t),t)$, and $G$ is holomorphic, then
 $ \partial_\a g = (G_z \circ z) z_\a$
so
 $ D_\a g=( \partial_zG)\circ z=- i (\partial_y  G)\circ z$.
Thus $D_\a^k g$ is the boundary value of holomorphic function $\partial_z^k G$, provided $G$ is holomorphic. $D_\a^k g$ is in addition periodic for any $k \ge 1$, so long as $G$ is periodic and holomorphic.

We may therefore conclude that  $D_\a^k \bar{z}_t$ is the boundary value of holomorphic function $\partial_z^k \bar {\vec{v}}$ in $\Omega(t)$, and
\begin{equation}
  \label{eq:434}
  \bound{D_\a^k \bar{z}_t} = 0, \text{ for any } k \ge 0.
\end{equation}

We will use $D_\a$ as the spatial derivative in constructing higher-order
energies.  In addition to preserving   holomorphicity and periodic boundary behavior, it transforms well under the Riemann mapping, to be discussed in the next section.

\section{The Riemann mapping version}\label{sec:11}
We now introduce a version of  water wave equations using the Riemann mapping to flatten out the curved free interface.\footnote{The idea of using Riemann mapping to study the wellposedness of 2-d water waves dates back to \cite{wu1997}.} The Riemann mapping version of the equations offers a key advantage. Because we are working on a flat domain, our Hilbert transform is now
the periodic Hilbert transform $\HH$ defined by
\begin{equation}
  \label{eq:174a}
  \HH f(\aa) := \f{1}{\hdenomconst} \pv \int_I \cot(\f{\pi}{2}(\aa - \bb)) f(\bb) d\bb.
\end{equation}
Since $\HH f \in i \R$ for $f$ real-valued, this allows us to invert the operator $(I-\HH)$ on purely real (resp., purely imaginary) functions by taking real (resp., imaginary) parts.

The problem with using the Riemann mapping in our framework is that the possibly non-right angle at the corner and singularities in the middle of the interface can create singularities in the Riemann mapping. This differs from previous work (e.g., \cite{wu1997}), where there were no such singularities.  On the other hand, the Riemann mapping is helpful for us in that it gives us heuristic information about the nature of our singularities,  which leads us to the right energy functional.

\subsection{The Riemannian coordinates and notations}\label{sec:12a}

Let $\Omega_0(t) \subset [0,1] \times (-\infty,c)$ be the unreflected fluid domain, and let
\begin{equation}
  \label{eq:16}
  \Phi_0: {\Omega}_0(t) \to P_0^- := \set{(x,y)}{x \in [0,1], y \le 0}
\end{equation}
be the unique Riemann mapping that takes the two upper corners to $(0,0)$ and $(1,0)$ and $\infty$ to $\infty$.  Let $\Phi$ be the Schwarz reflection of $\Phi_0$,
\begin{equation}
  \label{eq:414}
\Phi: {\Omega(t)} \to P^- := \set{(x,y)}{x \in [-1,1], y \le 0}.
\end{equation}
We know  $\Phi$ is a biholomorphic map that takes the free surface $\Sigma(t)$ to $I\times \{0\}$, and  wall to  wall. Let
\begin{equation}
  \label{eq:175}
\aa=  h(\a,t) := \Phi(z(\a,t),t) : I \to I
\end{equation}
be the change of coordinates taking the Lagrangian variable $\a$ to the Riemannian variable $\aa$, and let $h\i$ be the spatial inverse of $h$, defined by $h(h\i(\aa,t),t) = \aa$. We define
\begin{equation}
  \label{eq:107}
  Z(\aa,t) := z \circ h\i (\aa,t) = z(h\i(\aa,t),t).
\end{equation}
$Z=Z(\aa,t)$ is a parametrization of the free surface $\Sigma(t)$ in Riemannian variable $\aa$.
We  write
\begin{equation}\label{eq:370}
\begin{aligned}
  & Z_t := z_t \circ h\i; \quad Z_{tt} := z_{tt} \circ h\i;\\ Z_{,\aa} :=& \partial_\aa Z;  \quad  Z_{t,\aa}: =\partial_\aa Z_t;  \quad  Z_{tt,\aa}: =\partial_\aa Z_{tt}; \text{ etc.}
  \end{aligned}
\end{equation}
and
\begin{equation}
  \label{eq:632}
\AA := (\af h_\a) \circ h\i; \quad   \AAt := (\af_t h_\a) \circ h\i.
\end{equation}

Observe that $Z = z \circ h\i = \Phi\i$. Therefore
\begin{equation}
  \label{eq:63}
 Z_{,\aa}(\aa,t) = \partial_\aa (\Phi\i(\aa,t)), \quad\text{ and }\quad \f{1}{Z_{,\aa}} = \Phi_z \circ Z.
\end{equation}

Observe also that under the change of variables
$(D_\a f) \circ h\i = \f{1}{Z_{,\aa}} \partial_\aa (f \circ h\i)$.
We therefore define
\begin{equation}
  \label{eq:124}
  D_\aa := \f{1}{Z_{,\aa}} \partial_\aa.
\end{equation}

We note that for holomorphic functions on $P^-$, $\partial_\aa=\partial_{z'}$,  so $\partial_\aa$  also preserves holomorphicity. However, it has two drawbacks: it does not preserve periodicity and, for non-right angles or singular surfaces, it is more singular than $D_\aa$. We will give more details in \S \ref{sec:27} after deriving some necessary formulas.

\subsection{The water wave equations in the Riemann mapping framework}\label{sec:26}
We now derive  the water wave equations in Riemann mapping framework. We follow the approach of \cite{wu1997}, although we work with the real and imaginary parts together instead of separating $Z_t = X_t + iY_t$ into real and imaginary parts.

In this section, we will present a full derivation of the RHS of the quasilinear equation in Riemannian coordinates.  This won't end up be the precise way we estimate these quantities for higher-order energies; instead, we will have to do those by hand, in later sections.\footnote{Because we are using weighted derivatives $D_\aa$ (to ensure periodic boundary behavior and prevent singularity), and because our energy involves only a selected collection of weighted quantities,  we have to take more care in applying derivatives to the RHS than \cite{wu1997} does. Another source of difficulty is that our derivative $D_\aa$ is not purely real, so inverting $(I-\HH)$ is more subtle.} Rather, the derivation in this section proves useful because it provides several key formulas, especially one that allows us to estimate the crucial quantity $\f{\af_t}{\af}$.    Still, it seems most natural to motivate the following calculations by writing as if our goal were to derive the full quasilinear Riemann mapping version of the equations, rather than just to get a technical formula for $\f{\af_t}{\af}$.

Beginning with the conjugated form of our equation \eqref{eq:8} and with \eqref{eq:11},
 we precompose both sides with $h\i$ to get the surface equations in the flattened Riemann mapping coordinate:
\begin{equation}
  \label{eq:179}
  \bar{Z}_{tt} - i = - i \AA \bar{Z}_{,\aa};
\end{equation}
\begin{equation}
  \label{eq:178}
   \bar Z_{ttt} + i \AA \bar Z_{t,\aa} = -i \AAt  \bar Z_{,\aa},
\end{equation}
where $\bar Z_t$ is the boundary value of the periodic holomorphic function $\bar{\vec{v}}\circ \Phi^{-1}$, with $\bound{{\bar Z_t}}=0$.

The following proposition gives a characterization of the boundary value of  a periodic holomorphic function on $P^-$.
\begin{proposition}\label{prop:hilbe}
a.  Let $g \in L^p$ for some $p > 1$.
  Then $g$ is the boundary value of a holomorphic function $G$ on $P^-$ satisfying $G(-1+iy) = G(1+iy)$ for all $y < 0$ and $G(x+iy) \to c_0$ as $y \to -\infty$ if and only if
  \begin{equation}
    \label{eq:1571}
    (I-\HH) g = c_0.
  \end{equation}
Moreover, $c_0 = \avg_I g$.

b. Let $ f \in L^p$ for some $p > 1$. Then $\mathbb P_H  f:=\frac12(I+\mathbb H)  f$ is the boundary value of a periodic holomorphic function $\mathcal F$ on $P^-$, with $\mathcal F(x+iy)\to \frac12 \avg_I  f$ as $y\to -\infty$.

\end{proposition}
Proposition~\ref{prop:hilbe} is a classical result, which can be proved by the Cauchy integral formula; see \cite{Journe}. We note that our basic quantity $D_\aa^k \bar Z_t$ is the boundary value of the periodic holomorphic function $\partial_{z'}^k \paren{\bar {\vec{v}}\circ \Phi^{-1}}$ and satisfies
\begin{equation}
\label{eq:1026}
(I-\mathbb H)D_\aa^k \bar Z_t=0,\quad \bound{D_\aa^k \bar Z_t}=0;\qquad\text{for }k\ge 0
\end{equation}
 by  \eqref{eq:315} and \eqref{eq:259}; see discussions therein.

We define the following projection operators:
\begin{equation}
  \label{eq:1581}
   \P_H f := \f{(I + \HH)}{2} f; \quad \P_A f := \f{(I-\HH)}{2} f.
\end{equation}
We will refer to  $\P_H$ as the ``holomorphic projection'' and $\P_A$ as the ``antiholomorphic projection.''  These operators are, indeed, proper projections when interpreted modulo a constant.

We now seek formulas for $\AA$ and  $\AAt$. For derivations here and hereafter,
we will rely on the technicalities given in appendices \S \ref{sec:hmb}-\S \ref{sec:tech}. We will work  under the assumption that all the regularity properties required in appendices \S \ref{sec:hmb}-\S \ref{sec:tech} are met by our solutions.\footnote{Note that a key step in the proof of an existence result is having  a priori estimates for sufficiently regular solutions.}

\subsubsection{\texorpdfstring{$\AA$}{A}  and the quantity  \texorpdfstring{$A_1$}{A1} }\label{sec:27a}
We first derive a formula for $\AA$.   Historically, in \cite{wu1997}, this derivation showed that the strong Taylor stability criterion would automatically hold. There is now \cite{wu1999} a direct proof of this via basic elliptic theory without using the Riemann mapping. For our purposes, though, this original derivation here will be crucial because it introduces a quantity, $A_1$, that compares the degeneracy of $-\f{\partial P}{\partial \vec{n}}$ directly with that of the Riemann mapping and therefore the geometry of the surface.

We begin with \eqref{eq:179}. The key observation is that $\bar{Z}_{tt} - i$ is in some sense close to holomorphic, since $\bar{Z}_{t}$ is holomorphic. Therefore, if we apply $(I-\HH)$ to this, it should be ``small'' in some sense.  If the RHS  were purely imaginary, we would be able to invert $(I-\HH)$ by taking imaginary parts, and thus get a formula for the RHS.  The RHS is not purely real or imaginary, but if we multiply both sides by the holomorphic $Z_{,\aa}$, it is:
\begin{equation}
  \label{eq:180}
  Z_{,\aa}(\bar{Z}_{tt} - i) =- i \AA \abs{Z_{,\aa}}^2.
\end{equation}

  We will apply $(I-\HH)$ to both sides of \eqref{eq:180}. Before we do so, we must expand out $\bar{Z}_{tt}$. Let
\begin{equation}
  \label{eq:376}
  F(z(\a,t),t) := \bar{z}_t(\a,t).
\end{equation}
Note that $F= \bar{\vec{v}}$, a periodic holomorphic function in $\Omega(t)$.  We will use this expansion several times in the sequel, always with this definition of $F$. By the chain rule,
\begin{equation}
  \label{eq:181}
  \bar{z}_{tt} = \f{d}{dt} F(z(\a,t),t) = (F_z \circ z) z_t + (F_t \circ z).
\end{equation}
Recall from \S \ref{sec:10} that $\partial_z = D_\a$ for  holomorphic functions. Therefore, $  F_z \circ z = \f{\bar{z}_{t\a}}{z_\a}$,
and thus $
  \bar{z}_{tt} = \f{\bar{z}_{t\a}}{z_\a} z_t + F_t\circ z$.
We precompose with $h\i$:
\begin{equation}
  \label{eq:184}
  \bar{Z}_{tt} = \paren{\f{\bar{Z}_{t,\aa}}{Z_{,\aa}}} Z_t + F_t \circ Z.
\end{equation}
We can now write our equation \eqref{eq:180} as
\begin{equation}
  \label{eq:185}
  \bar{Z}_{t,\aa} Z_t + Z_{,\aa} (F_t \circ Z) - i Z_{,\aa} = -i \AA \abs{Z_{,\aa}}^2.
\end{equation}
We apply $(I-\HH)$ to both sides. By \eqref{eq:474}, \eqref{eq:186} and \eqref{eq:516}, writing $(I-\HH)( Z_t \bar{Z}_{t,\aa}) = [Z_t,\HH] \bar{Z}_{t,\aa}$, we get
\begin{equation}
  \label{eq:174}
  [Z_t,\HH] \bar{Z}_{t,\aa} -i = (I-\HH) \paren{- i \AA \abs{Z_{,\aa}}^2}.
\end{equation}
We now take imaginary parts of both sides. This gives us the new quantity $A_1$:
\begin{equation}
  \label{eq:208}
  A_1 : = \AA \abs{Z_{,\aa}}^2
  = \Im\paren{- [Z_t,\HH] \bar{Z}_{t,\aa}} + 1.
\end{equation}
This is the same $A_1$ as that in \cite{wu1997}. It's easy to see that $\Im \paren{-[Z_t,\HH] \bar{Z}_{t,\aa}}$ is non-negative, by integration by parts. Indeed, if $Z_t = X_t + iY_t$, then
\begin{equation}
  \label{eq:132}
  \begin{aligned}
   \Im \paren{-[Z_t,\HH] \bar{Z}_{t,\aa}} & = - \Im \f{1}{\hdenomconst} \int (Z_t(\aa) - Z_t(\bb)) \cot(\f{\pi}{2}(\aa -\bb)) \bar{Z}_{t,\bb} d\bb
 \\ & =  \f{1}{\hdenomconstreal} \int \f{1}{2} \braces{- \partial_\bb \bracket{(X_t(\aa) - X_t(\bb))^2 + (Y_t(\aa) - Y_t(\bb))^2}} \cot(\f{\pi}{2}(\aa - \bb)) d\bb
\\ & = \f{\pi}{ 8} \int  \f{(X_t(\aa) - X_t(\bb))^2 + (Y_t(\aa) - Y_t(\bb))^2}{\sin^2(\f{\pi}{2}(\aa -\bb))} d\bb
\\&  \ge 0.
\end{aligned}
\end{equation}
Here, there is no boundary term in the integration by parts because $\bound{\bar Z_t }=0$.
Therefore
\begin{equation}
  \label{eq:394}
  A_1 \ge 1.
\end{equation}

\subsubsection{Degenerate Taylor stability criterion and the singularity of the surface}\label{sec:27}

We can draw a few important conclusions from the derivations in \S \ref{sec:27a}. For the sake of exposition, we will in this section focus on the angle $\ang$ at the wall, and we will move the corner from $\pm 1$ to $0$; angled crests and other singularities in the middle of the surface can be handled similarly.  Let $\ang$ be the angle at the corner.   Then the Riemann mapping $\Phi(z)\approx z^r$ at the corner, where $r \ang = \f{\pi}{2}$. By \eqref{eq:63}, ${Z_{,\aa}} = (\Phi^{-1})_{z'}$. Therefore,
\begin{equation}\label{eq;1010}
Z_{,\aa}=\partial_\aa\Phi^{-1}\approx (\aa)^{1/r-1}
\end{equation}
 at the corner.

We first note that by \eqref{eq:180} and \eqref{eq:208}
\begin{equation}
  \label{eq:251}
  \f{1}{Z_{,\aa}} = i \f{\bar{Z}_{tt} - i}{A_1}.
\end{equation}
Therefore in the regime where the acceleration $|Z_{tt}|<\infty$, $Z_{,\aa}\ne0$, by \eqref{eq:394}.  This implies $r\ge 1$ and $\ang\le \pi/2$. Similarly, this implies that angled crests have an angle $\le \pi$.\footnote{In fact, the angle $\ang$ at the corner is no more than $90^\circ$ always, even if $\abs{Z_{tt}} = \infty$.  Indeed, if $|Z_{tt}|=\infty$ at the corner, then $y_{tt}$ must be infinity at the corner, because $x_{tt}=0$ at the corner. This implies, by \eqref{eq:9} and \eqref{eq:365}, that $\ang=\frac\pi2$. Note that this argument applies only at the corner; for angled crests in the middle, we need $\abs{Z_{tt}} < \infty$, which will hold in our energy regime.}

Now, because
\begin{equation}
  \label{eq:209}
  A_1 \circ h = \f{\af \abs{z_\a}^2}{h_\a}
\end{equation}
and
\begin{equation}
  \label{eq:50}
  \f{-\partial P}{\partial \vec{n}} = \abs{z_\a}\af = \f{A_1 \circ h}{\abs{Z_{,\aa} \circ h}},
\end{equation}
$\f{- \partial P}{\partial \vec{n}}\ge 0$ always holds.
In the setup where the free surface is $C^{1,\gamma}$ and makes a right angle at the corner ($\ang=\pi/2$),
$0<c_0\le\abs{ (\Phi^{-1})_{z'}}\le C_0<\infty$.   This together with the estimate $A_1 \ge 1$  gave \cite{wu1997} a strictly positive lower bound for the Taylor coefficient $-\f{\partial P}{\partial \vec{n}}$.  In our situation,   $\f{1}{Z_{,\aa}} \to 0$ at the corner if $\ang<\frac \pi 2$; similarly, $\f{1}{Z_{,\aa}} = 0$ at an angled crest if the interior angle is $<\pi$. If $A_1$ is in addition bounded from above---which will be true when our energy is finite---we know that the degeneracy of $-\f{\partial P}{\partial \vec{n}}$ corresponds precisely to the degeneracy of the Riemann mapping $\Phi$. Because the degeneracy of the Riemann mapping corresponds precisely to the singularity of the free surface, we can conclude that the degeneracy of the Taylor coefficient $-\f{\partial P}{\partial \vec{n}}$ corresponds precisely to the singularities of the free surface, whether they are angled crests in the middle or non-right angles at the walls.

We note also that $\af\ge 0$; that,  by \eqref{eq:50}, $\af$ degenerates alongside $-\frac{\partial P}{\partial \vec{n}}$; and that $\af\approx h_\a\approx \frac{h_\a}{A_1\circ h}$ by \eqref{eq:209}, provided $A_1$ and $|z_\a|$ are bounded away from 0 and $\infty$.

Finally, we note that our spatial derivative $D_\aa = \f{1}{Z_{,\aa}} \partial_\aa$ is more regular, in a sense, than $\partial_\aa$. We know $\f{1}{Z_{,\aa}} \to 0$ at singularities, which indicates that the weight function $\f{1}{Z_{,\aa}}$ has some ``regularizing'' effect. Indeed, we have $D_\aa \bar Z_t, \ D_\aa \bar Z_{tt} \in L^\infty$, but $\partial_\aa \bar Z_t,\ \partial_\aa \bar Z_{tt}$ are only in $L^2$ in our energy space, for example; see \S \ref{sec:quants}.

\subsubsection{\texorpdfstring{$\AAt$}{At}}\label{sec:28}
Now we seek a formula for the nonlinearity on the RHS of \eqref{eq:178}, $\AAt$. As in \S \ref{sec:27a}, we will hope that the LHS is close to holomorphic and apply $(I-\HH)$ to both sides and then invert.  Here we start by multiplying both sides of \eqref{eq:178} by $Z_{,\aa}$, since we want the RHS to be purely imaginary:
\begin{equation}
  \label{eq:194}
  Z_{,\aa}(\bar{Z}_{ttt} + i \AA \bar{Z}_{t,\aa}) = - i \AAt \abs{Z_{,\aa}}^2.
\end{equation}
We once again carefully expand the LHS. As before, let $F(z(\a,t),t) = \bar{z}_t(\a,t)$. Again, we have
\begin{equation}
  \label{eq:380}
  \bar{z}_{tt} = (F_z\circ z) z_t + F_t\circ z,
\end{equation}
so
\begin{equation}
  \label{eq:195}

  \bar{z}_{ttt}  = (F_{zz} \circ z) z_t^2 + 2 (F_{tz} \circ z) z_t + (F_z\circ z) z_{tt} + F_{tt}\circ z.

\end{equation}
We now solve for $F_z \circ z,$ $F_{zz} \circ z$ and $F_{tz} \circ z$. Since $\partial_z = D_\a$ on  holomorphic functions,
\begin{equation}
  \label{eq:199}
  F_z \circ z= D_\a \bar{z}_t,\quad F_{zz} \circ z= D_\a^2 \bar{z}_t.
\end{equation}
We solve for $F_{tz} \circ z$ by applying $\partial_z = D_\a$ to \eqref{eq:380}:
\begin{equation}
  \label{eq:198}
  F_{tz} \circ z = D_\a \paren{\bar{z}_{tt} - (D_\a \bar{z}_t) z_t}.
\end{equation}
Therefore, by substituting \eqref{eq:199} and \eqref{eq:198} into \eqref{eq:195}, we get
\begin{equation}
  \label{eq:200}
  \bar{z}_{ttt} = (D_\a^2 \bar{z}_t) z_t^2 + 2 z_t D_\a \paren{\bar{z}_{tt} - (D_\a \bar{z}_t) z_t} + (D_\a \bar{z}_t) z_{tt} + F_{tt} \circ z.
\end{equation}
Precomposing with $h\i$, we have
\begin{equation}
  \label{eq:202}
  \bar{Z}_{ttt} = (D_\aa^2 \bar{Z}_t) Z_t^2 + 2 Z_t D_\aa(\bar{Z}_{tt} - (D_\aa \bar{Z}_t) Z_t) + (D_\aa \bar{Z}_t) Z_{tt} + F_{tt} \circ Z.
\end{equation}
We now go back to \eqref{eq:194}, substituting in \eqref{eq:202} to get
\begin{equation}
  \label{eq:203}
  Z_{,\aa} \paren{(D_\aa^2 \bar{Z}_t) Z_t^2 + 2 Z_t D_\aa(\bar{Z}_{tt} - (D_\aa \bar{Z}_t) Z_t) + (D_\aa \bar{Z}_t) Z_{tt} + F_{tt} \circ Z + i \AA \bar{Z}_{t,\aa}} = -i \AAt \abs{Z_{,\aa}}^2.
\end{equation}
We simplify, distributing the $Z_{,\aa}$ and then using the identity $Z_{tt} + i = i \AA Z_{,\aa}$ on the last term:
\begin{equation}
  \label{eq:204}
  (\partial_\aa D_\aa \bar{Z}_t) Z_t^2 + 2 Z_t \partial_\aa(\bar{Z}_{tt} - (D_\aa \bar{Z}_t)Z_t) + 2\bar{Z}_{t,\aa} Z_{tt} + Z_{,\aa} (F_{tt} \circ Z) + i \bar{Z}_{t,\aa} = - i \AAt \abs{Z_{,\aa}}^2.
\end{equation}
We now apply $(I-\HH)$ to both sides.  Various terms will disappear on the LHS and others will turn into commutators, due to holomorphicity; specifically, we use \eqref{eq:474}, \eqref{eq:222}, and \eqref{eq:504}. We get
\begin{equation}
  \label{eq:205}
  \begin{split}
      [Z_t^2,\HH]\partial_\aa D_\aa \bar{Z}_t + 2[Z_t,\HH] \partial_\aa (\bar{Z}_{tt} - (D_\aa \bar{Z}_t) Z_t) + 2[Z_{tt},\HH] \bar{Z}_{t,\a}
\\ = (I-\HH) \braces{-i \AAt \abs{Z_{,\aa}}^2}.
\end{split}
\end{equation}
We could continue working with this equation, but two integrations by parts will give us a nicer equation to work with.  We take the first term and the second part of the second term and integrate by parts both terms, noting that we have no boundary terms. We get
\begin{equation}
  \label{eq:265}
  \bracket{Z_t^2,\HH} \partial_\aa D_\aa \bar{Z}_t - 2[Z_t,\HH]\partial_\aa\paren{(D_\aa \bar{Z}_t) Z_t} = - \f{\pi}{4i} \int \f{(Z_t(\aa) - Z_t(\bb))^2}{\sin^2(\f{\pi}{2}(\aa - \bb))} D_\bb \bar{Z}_t(\bb) d\bb.
\end{equation}
This is a type of higher-order Calderon commutator, which we write as $-[Z_t,Z_t; D_\aa \bar{Z}_t]$ (see \eqref{eq:624}). We therefore can rewrite \eqref{eq:205} as
\begin{equation}
  \label{eq:266}
  - i (I-\HH)  \braces{\AAt \abs{Z_{,\aa}}^2} =   2[Z_t,\HH] \bar{Z}_{tt,\aa} + 2[Z_{tt},\HH] \bar{Z}_{t,\a} - [Z_t, Z_t; D_\aa \bar{Z}_t].
\end{equation}
Taking imaginary parts, we get
\begin{equation}
  \label{eq:207}
   \AAt \abs{Z_{,\aa}}^2 = - \Im \paren{2[Z_t,\HH] \bar{Z}_{tt,\aa} + 2[Z_{tt},\HH] \bar{Z}_{t,\a} - [Z_t, Z_t; D_\aa \bar{Z}_t]}.
\end{equation}

\subsubsection{Formulas for \texorpdfstring{$\f{\af_t}{\af}$}{at/a} and  \texorpdfstring{$\f{\AAt}{\AA}$}{At/A}}\label{sec:29}
We could now plug \eqref{eq:207} into \eqref{eq:178} to get the quasilinear equation in Riemannian coordinates.  Instead, though, we will focus on a different quantity. Observe that dividing \eqref{eq:207} by \eqref{eq:208} we have
\begin{equation}
  \label{eq:210}
  \f{\af_t}{\af} \circ h\i = \f{\AAt}{\AA} = \f{\AAt \abs{Z_{,\aa}}^2}{\AA \abs{Z_{,\aa}}^2} = \f{- \Im \paren{2[Z_t,\HH] \bar{Z}_{tt,\aa} + 2[Z_{tt},\HH] \bar{Z}_{t,\aa} - [Z_t, Z_t; D_\aa \bar{Z}_t]}}{A_1}.
\end{equation}
If we want to control $\nm{\f{\af_t}{\af}}_{L^\infty}$, then, because $A_1 \ge 1$ by \eqref{eq:394}, it suffices to control the numerator of the RHS of \eqref{eq:210}, which we will be able to do.

\section{The main result}\label{sec:energy}
\subsection{Definition of the energy}\label{sec:35}
We now introduce the energy for which we will prove an a priori inequality.  We consider a general equation of the form
\begin{equation}
  \label{eq:171}
  (\partial_t^2 + i \mathfrak{a} \partial_\a) \th = G_\th
\end{equation}
with the constraint that $\theta$ is the boundary value of a periodic holomorphic function on $\Omega(t)$.
Our base case is \eqref{eq:11}, written in this form as
\begin{equation}
  \label{eq:362}
  (\partial_t^2 + i \mathfrak{a} \partial_\a) \bar{z}_t = -i \af_t \bar{z}_{\a},
\end{equation}
with $\th = \bar{z}_t$ and $G_{\bar{z}_t} = - i \af_t \bar{z}_\a$.  Higher-order cases will come from applying $D_\a^k$ to \eqref{eq:11}; in those cases, $\th  =  D_\a^k \bar{z}_t$ with
\begin{equation}
  \label{eq:30}
G_\th =D_\a^k(-i\mathfrak{a}_t \bar{z}_\a) + [\partial_t^2 + i \mathfrak{a}\partial_\a,D_\a^k] \bar{z}_t.
\end{equation}

A natural energy for \eqref{eq:171} would be
\begin{equation}
  \label{eq:384}
  \int \abs{\th_t}^2 d\a + \Re \int i \paren{\mathfrak{a} \partial_\a \th} \bar{\th} d\a.
\end{equation}
Here, the fact that $\th$ is holomorphic allows the second term to be rewritten as $\int i \paren{\mathfrak{a} \H \partial_\a \th} \bar{\th} d\a$, where $\H$ is the Hilbert transform for the curved domain $\Omega$. Because $i \H \partial_\a$ is a positive operator---in flattened coordinates, it corresponds to $\abs{D} = \sqrt{-\Delta}$---the second term will be positive up to an error term (due to the coefficient $\af$).

We will, instead, use two variants of this energy, differing primarily in that we multiply or divide by the roughly equivalent singular weights $\af$ and $\f{h_\a}{A_1 \circ h}$. In earlier works, where there was a strict bound $\mathfrak{a} \ge c_0 > 0$  (in addition to the upper bound we continue to have),  all of these energies were essentially equivalent, but here the singularity $\mathfrak{a} = 0$ makes an important difference, and so the choice of the weights is critical.  We will discuss this further in \S \ref{sec:36} below.

We note that we are phrasing everything here in Lagrangian coordinates. With a change of variables, we can easily switch to Riemannian coordinates. We will often express our basic quantities in Lagrangian coordinates when we need to take a time derivative, but use Riemannian coordinates when we need to estimate terms, since that gives us access to the easily invertible $(I-\HH)$ operator.

Our first energy differs from \eqref{eq:384} primarily by a multiplicative factor of $h_\a$:
\begin{equation}
  \label{eq:328}
  E_{a,\th}(t) := \int_I \abs{\th_t}^2 \f{h_\a}{A_1 \circ h} d\a + \Re \int_I (i \af \partial_\a \th) \bar{\th} \f{h_\a}{A_1 \circ h} d\a + \Re \int_I \paren{i \f{1}{\bar{z}_\a}\partial_\a \f{h_\a}{z_\a}} \abs{\th}^2 h_\a d\a +  \int_I \abs{\th}^2 \f{h_\a}{A_1 \circ h} d\a.
\end{equation}
Here, the $A_1 \circ h$ appears for technical reasons, which will be explained below in \S \ref{sec:36}.  The first two terms are the main terms.  The third term is a correction to the second term, to be explained below, and the fourth term is lower-order.

For our second energy, $E_b$, we divide through by $\mathfrak{a}$ in the main terms:
\begin{equation}
  \label{eq:172}
  E_{b,\th}(t) := \int_I \f{1}{\mathfrak{a}} \abs{\th_t}^2 d\a + \int_I (i\partial_\a \th) \bar{\th} d\a + \int_I \f{(A_1 \circ h)}{\mathfrak{a}} \abs{\th}^2 d\a.
\end{equation}
Here, the first two integrals are the primary terms; the last is lower-order. It's easy to check, by integration by parts, that the second term is purely real.

Our total energy consists primarily of $E_{a,\th}$ using two $D_\a$ derivatives and $E_{b,\th}$ using one $D_\a$ derivative. In addition, we will include one other, lower-order term in our total energy:  $\abs{\bar{z}_{tt}(\a_0,t) - i}$ for some fixed $\a_0 \in I$. Our total energy therefore is
\begin{equation}
  \label{eq:49}
  E=E(t) := E_{a,D_\a^2\bar{z}_t}(t) + E_{b,D_\a\bar{z}_t} (t) + \abs{\bar{z}_{tt}(\a_0,t)-i}.
\end{equation}
Even though we have now specialized to the case where $\th = D_\a^2 \bar{z}_t$ and $\th = D_\a \bar{z}_t$ for $E_{a,\th}$ and $E_{b,\th}$, respectively, we will in the following present some of the proofs about these energies in broader generality.  We will use $E_{a,\th}$ and $E_{b,\th}$ to refer to the generic first- and second-type energies, and use $E_a := E_{a,D_\a^2 \bar{z}_t}$ and $E_b := E_{b,D_\a\bar{z}_t}$ for the specific energies.

\subsection{Discussion}\label{sec:36}
We offer a few comments on our energy.

We begin with $E_a$. On changing variables to Riemannian coordinates, the first and the fourth terms of $E_a$ become $\nm{\th_t \circ h\i}_{L^2(1/A_1)}^2$ and $\nm{\th \circ h\i}_{L^2(1/A_1)}^2$. Since we will be able to control $\nm{A_1}_{L^\infty}$ by our energy, these two terms therefore are equivalent to $\nm{\th_t \circ h\i}_{L^2}$ and $\nm{\th \circ h\i}_{L^2}$.

The second term and third terms in $E_{a,\th}$ treated together equal $\nm{\f{1}{Z_{,\aa}} (\th \circ h\i)}_{\dot{H}^{1/2}}^2$. Indeed, when we use $A_1 \circ h = \f{\af \abs{z_\a}^2}{h_\a}$ \eqref{eq:209} and then change variables, these two  terms in $E_{a,\th}$ become
\begin{equation}
  \label{eq:331}
  \begin{aligned}
    \Re \int (i \af \partial_\a \th) \bar{\th} \f{h_\a}{A_1 \circ h} d\a + \Re \int_I \paren{i \f{1}{\bar{z}_\a}\partial_\a \f{h_\a}{z_\a}} \abs{\th}^2 h_\a d\a  & =  \Re \int i \f{ h_\a^2}{ \abs{z_\a}^2} (\partial_\a \th) \bar{\th} d\a + \Re \int_I \paren{i \f{1}{\bar{z}_\a}\partial_\a \f{h_\a}{z_\a}} \abs{\th}^2 h_\a d\a
    \\ & = \Re \int i  \paren{\partial_\a \paren{\th \f{h_\a}{z_\a}}} \paren{\bar{\th} \f{h_\a}{\bar{z}_\a}} d\a
\\ & = \Re \int i \paren{\partial_\aa \paren{\f{1}{Z_{,\aa}} (\th \circ h\i)}}\paren{\f{1}{\bar{Z}_{,\aa}} (\bar{\th} \circ h\i)} d\aa.
\end{aligned}
\end{equation}
This equals the square of the $\dot{H}^{1/2}$ norm of $\f{1}{Z_{,\aa}}(\th \circ h\i)  $; see \S \ref{sec:half}.  (Note that $\f{1}{Z_{,\aa}}(\th \circ h\i)$ is holomorphic.) This explains why we divided through by $A_1 \circ h$ in our definition of $E_{a,\th}$. This $\dot{H}^{1/2}$ control will be crucial in allowing us to close our energy inequality.

We now specialize to $E_a = E_{a,D_\a^2 \bar{z}_t}$. By the above discussions,
\begin{equation}
  \label{eq:400}
  E_a = \nm{(\partial_t D_\a^2 \bar{z}_t) \circ h\i}_{L^2(1/A_1)}^2 +  \nm{\f{1}{Z_{,\aa}}D_\aa^2 \bar{Z}_t}_{\dot{H}^{1/2}}^2 + \nm{D_\aa^2 \bar{Z}_t}_{L^2(1/A_1)}^2.
\end{equation}
We will be able to control the commutator $[\partial_t,D_\a^2]$, so the first term corresponds roughly to $\nm{D_\aa^2 \bar{Z}_{tt}}_{L^2(1/A_1)}$ or $\nm{D_\aa^2 \bar{Z}_{tt}}_{L^2}$; we make this more precise in \S \ref{sec:quants}.

Similarly, by using the half-derivative and the fact that $\nm{D_\a \bar{z}_t}_{L^2(\f{(A_1 \circ h)}{\af})}  = \nm{\bar{Z}_{t,\aa}}_{L^2}$ (by \eqref{eq:209}),
\begin{equation}
  \label{eq:408}
  E_b = \nm{\partial_t D_\a \bar{z}_t}_{L^2(\f{1}{\af})}^2 +  \nm{D_\aa \bar{Z}_t}_{\dot{H}^{1/2}}^2 + \nm{\bar{Z}_{t,\aa}}_{L^2}.
 \end{equation}
Again, on changing variables to Riemannian coordinates,  we see that,  modulo a commutator $[\partial_t,D_\a]$ and a factor of $A_1$,
$  \nm{\partial_t D_\a \bar{z}_t}_{L^2(\f{1}{\af})} \approx \nm{\bar{Z}_{tt,\aa}}_{L^2}$;
we make this more precise, again, in \S \ref{sec:quants}. We will see in \eqref{eq:361} that $\nm{A_1}_{L^\infty}$ is bounded by $\nm{\bar{Z}_{t,\aa}}_{L^2}^2+1$. This explains our choice of the lower-order term in \eqref{eq:408}.
We note that both $\nm{A_1}_{L^\infty}$ and $\nm{Z_{t,\aa}}_{L^2}$ are invariant with respect to the scaling $\aa \sim t^2$ of the water wave equations \eqref{eq:1}-\eqref{eq:5}, \eqref{eq:620} (with $\Upsilon=\emptyset$).

The combination of $E_a$ and $E_b$, with their roughly inverse singular weights $\f{h_\a}{A_1 \circ h}$ and $\frac1{\af}$, will allow us to control the $L^\infty$ norms of various useful quantities (via the weighted Sobolev inequality \eqref{eq:396}).

Finally, we also include $\abs{\bar{z}_{tt}(\a_0,t) - i}$ in our energy since it, along with control of $\nm{\bar{Z}_{tt,\aa}}_{L^2}$, will allow us to control the lower-order term $\nm{\bar{Z}_{tt}-i}_{L^\infty}$. We note that $\nm{\bar{Z}_{tt}-i}_{L^\infty}$ is invariant with respect to the scaling $\aa \sim t^2$, and we remark that our estimate does not depend on control of $\nm{Z_t}_{L^\infty}$.

In \S \ref{sec:36e} we give a characterization of our energy in terms of only $\bar{Z}_t$ and $\f{1}{Z_{,\aa}}$ and their spatial derivatives, and we discuss what types of singularities are allowed by a finite energy.

\subsection{The main result}\label{sec:38}
We now state our main result. We assume that the solution is suitably regular so that  all the calculations in our proof make sense.

\begin{theorem}\label{maintheorem}[A priori estimate]
There exists a polynomial $p=p(x)$ with universal coefficients such that for any solution of water wave equations \eqref{eq:1}-\eqref{eq:620} with  $E(t) < \infty$ for all $t \in [0,T]$,
\begin{equation}
  \label{eq:690}
  \f{d}{dt} E(t) \le  p\paren{E(t)}
\end{equation}
for all $t \in [0,T]$.
\end{theorem}
Here  the water wave equations \eqref{eq:1}-\eqref{eq:620} are those defined on the symmetric domain $\Omega(t)$,  with fixed boundary $\Upsilon=\{x=\pm1\}$, as specified in \S \ref{sec:1}.
We do not state  precise polynomial $p(x)$, but it can be calculated by carefully combining the estimates we make in the proof. Estimates for the individual parts of the energy $E_a$, $E_b$, and $\abs{\bar{z}_{tt}(\a_0)-i}$ are listed in \S \ref{sec:39}. Although we state the a priori inequality for the original equations \eqref{eq:1}-\eqref{eq:620} on $\Omega(t)$, our proof relies exclusively on the Lagrangian reduction to the equation \eqref{eq:8} on the free surface $\Sigma(t)$, along with the accompanying statements about $\bar{z}_t$'s holomorphicity and periodicity.

\subsection{The proof}\label{sec:39}
We begin the proof by differentiating each component of $E(t)$ in time. For the two main energies, $E_a$ and $E_b$, we then integrate by parts to arrive at a term $\partial_t^2\theta+i\af \partial_\alpha \theta$ and use the basic equation $\partial_t^2\theta+i\af \partial_\a \theta=G_\theta$ to replace it with $G_\theta$.  What remain to be estimated will be $G_\theta$, along with several ancillary terms.  We control those quantities  in  \S \ref{sec:43} through \S \ref{controlE1} in terms of a polynomial of the energy.

\subsubsection{The estimate for \texorpdfstring{$E_a$}{Ea}}\label{sec:40a}
We begin by differentiating $E_a$ with respect to $t$.

We will work initially with general $\th$ satisfying $\bound{\th} = 0$, $(I-\HH) (\th \circ h\i) = 0$, and the basic equation \eqref{eq:171}, and then we will specialize to the $\th = D_\a^2 \bar{z}_t$\footnote{See \eqref{eq:1026}.} in our energies.  The periodicity ensures that there is no boundary term in the integration by parts.

We use the fact that $\f{\af h_\a}{(A_1 \circ h)} = \f{h_\a^2}{\abs{z_\a}^2}$ (by \eqref{eq:209}) in the following calculation.
\begin{equation}
  \label{eq:385}
  \begin{aligned}
  \f{d}{dt} E_{a,\th}(t) & = \int (\th_{tt} \bar{\th}_t + \th_t \bar{\th}_{tt}) \f{h_\a}{A_1 \circ h} d\a  + \int \abs{\th_t}^2 \f{h_{t\a}}{A_1 \circ h} d\a - \int \abs{\th_t}^2 \f{h_\a}{A_1 \circ h} \f{(A_1 \circ h)_t}{A_1 \circ h} d\a
\\ & + \Re \int i \paren{\f{h_\a^2}{\abs{z_\a}^2}}_t  \th_\a \bar{\th} d\a +\underbrace{\Re \int i \f{h_\a^2}{\abs{z_\a}^2} \th_{t\a} \bar{\th} d\a}_{-\Re \int i \paren{\f{h_\a^2}{\abs{z_\a}^2}}_\a \th_t \bar{\th} d\a+ \Re \int \f{h_\a^2}{\abs{z_\a}^2} \th_t \bar{i\th_\a}d\a} + \Re \int i \f{h_\a^2}{\abs{z_\a}^2} \th_\a \bar{\th}_t  d\a
\\ & + \Re \int i \paren{\f{1}{\bar{z}_\a} \partial_\a \f{h_\a}{z_\a}}_t \abs{\th}^2 h_\a d\a + \Re \int i \paren{\f{1}{\bar{z}_\a} \partial_\a \f{h_\a}{z_\a}} (\th_t \bar{\th} + \th \bar{\th}_t) h_\a d\a + \Re \int i \paren{\f{1}{\bar{z}_\a} \partial_\a \f{h_\a}{z_\a}} \abs{\th}^2 h_{t\a} d\a
\\ & +  \int (\th_t \bar{\th} + \bar{\th}_t \th) \f{h_\a}{A_1 \circ h} +  \int \abs{\th}^2 \f{h_{t\a}}{A_1 \circ h} d\a -  \int \abs{\th}^2 \f{h_\a}{A_1 \circ h} \f{(A_1 \circ h)_t}{A_1 \circ h} d\a
\\ &= \int 2 \Re \paren{(\th_{tt} + i\af \th_\a)\bar{\th}_t} \f{h_\a}{A_1 \circ h} d\a
 + \int \paren{\abs{\th_t}^2+\abs{\th}^2}   \paren{\f{h_{t\a}}{h_\a} - \f{(A_1 \circ h)_t}{A_1 \circ h}} \f{h_\a}{A_1 \circ h} d\a
\\ & + \Re \int i \paren{\f{1}{\bar{z}_\a} \partial_\a \f{h_\a}{z_\a}}_t \abs{\th}^2 h_\a d\a
 + \Re \int i \paren{\f{1}{\bar{z}_\a} \partial_\a \f{h_\a}{z_\a}} (\th_t \bar{\th} + \th \bar{\th}_t) h_\a d\a
 \\&+ \Re \int i \paren{\f{1}{\bar{z}_\a} \partial_\a \f{h_\a}{z_\a}} \abs{\th}^2 \f{h_{t\a}}{h_\a} h_\a d\a
+   2 \Re \int \th_t \bar{\th} \f{h_\a}{A_1 \circ h} d\a
\\& - \Re \int i \paren{\f{h_\a^2}{\abs{z_\a}^2}}_\a \th_t \bar{\th} d\a
 + \Re  \int i \paren{\f{h_\a^2}{\abs{z_\a}^2}}_t  \th_\a \bar{\th} d\a.
\end{aligned}
\end{equation}

Now we show how we control each of these terms.

For the first, we replace $\th_{tt} + i \af \th_\a$ with the RHS $G_\th$ by the main equation \eqref{eq:171} and then use Cauchy-Schwarz:
\begin{equation}
  \label{eq:43}
  \int 2 \Re \paren{(\th_{tt} + i\af \th_\a)\bar{\th}_t} \f{h_\a}{A_1 \circ h} d\a
\\ \lec  \paren{\int  \abs{G_\th}^2\f{h_\a}{A_1 \circ h}d\a}^{1/2} \paren{\int \abs{\bar{\th}_t}^2 \f{h_\a}{A_1 \circ h}d\a}^{1/2}.
\end{equation}
The first factor,  which involves the RHS of the basic equation,
is the main term to control. For $\th= D_\a^2 \bar{z}_t$,
\begin{equation}
  \label{eq:391}
  G_\th = D_\a^2(-i\af_t \bar{z}_\a) + [\partial_t^2 +i\af \partial_\a,D_\a^2] \bar{z}_t.
\end{equation}
We estimate these terms in \S \ref{controlE1}.

In \S \ref{sec:98a}, we will control
\begin{equation}
  \label{eq:189}
  \abs{\Re \int i \paren{\f{1}{\bar{z}_\a} \partial_\a \f{h_\a}{z_\a}}_t \abs{\th}^2 h_\a d\a} \lec \eqref{eq:368}.
\end{equation}
Because $\f{1}{\bar{z}_\a} \partial_\a \f{h_\a}{z_\a}= D_\a \f{h_\a}{z_\a}=\paren{D_\aa \f{1}{Z_{,\aa}}}\circ h$, we estimate
\begin{equation}
  \label{eq:190}
    \abs{\Re \int i \paren{\f{1}{\bar{z}_\a} \partial_\a \f{h_\a}{z_\a}} (\th_t \bar{\th} + \th \bar{\th}_t) h_\a d\a}     \lec \nm{D_\aa \f{1}{Z_{,\aa}}}_{L^\infty} \nm{A_1}_{L^\infty} E_{a,\th}.
 \end{equation}
Similarly, we estimate
\begin{equation}
  \label{eq:191}

    \abs{\Re \int i \paren{\f{1}{\bar{z}_\a} \partial_\a \f{h_\a}{z_\a}} \abs{\th}^2 \f{h_{t\a}}{h_\a} h_\a d\a}
    \lec \nm{D_\aa \f{1}{Z_{,\aa}}}_{L^\infty} \nm{\f{h_{t\a}}{h_\a}}_{L^\infty} \nm{A_1}_{L^\infty} E_{a,\th}.

\end{equation}
We observe that
\begin{equation}
  \label{eq:388}
  \nm{\f{\paren{\f{h_\a^2}{\abs{z_\a}^2}}_\a}{h_\a}}_{L^\infty} = \nm{\partial_\aa \f{1}{\abs{Z_{,\aa}}^2}}_{L^\infty} \le 2 \nm{D_\aa \f{1}{Z_{,\aa}}}_{L^\infty},
\end{equation}
so, using Cauchy-Schwarz, we have
\begin{equation}
  \label{eq:389}
  \abs{- \Re \int i \paren{\f{h_\a^2}{\abs{z_\a}^2}}_\a \th_t\bar{\th}d\a} \lec \nm{A_1}_{L^\infty}  \nm{D_\aa \f{1}{Z_{,\aa}}}_{L^\infty} E_{a,\th}.
\end{equation}
In \S \ref{sec:48} we control
\begin{equation}
  \label{eq:193}
  \Re  \int i \paren{\f{h_\a^2}{\abs{z_\a}^2}}_t  \th_\a \bar{\th} d\a  \lec \eqref{eq:552}.
\end{equation}
We estimate  the remaining two terms of \eqref{eq:385} by Cauchy-Schwarz and H\"older.

We now combine these estimates and specialize to $\th = D_\a^2 \bar{z}_t$. Each of the remaining factors we will control in \S \ref{sec:43}; we list the location of the final estimate for each quantity of the following in the subscripts. We get
\begin{equation}
  \label{eq:66}
  \begin{aligned}
    \abs{\f{d}{dt} E_{a}} & \lec \underbrace{\nm{G_{D_\a^2 \bar{z}_t}}_{L^2(\f{h_\a}{A_1 \circ h})}}_{\lec \eqref{eq:1600}} E_a^{1/2}
     + \underbrace{\nm{\f{h_{t\a}}{h_\a}}_{L^\infty}}_{\lec \eqref{eq:451}}E_a + \underbrace{\nm{\f{(A_1 \circ h)_t}{A_1 \circ h}}_{L^\infty}}_{\lec \eqref{eq:455}}E_a
\\ & + \underbrace{\paren{1 + \nm{\f{h_{t\a}}{h_\a}}_{L^\infty}}}_{\lec 1+ \eqref{eq:451}} \nm{A_1}_{L^\infty} \underbrace{\nm{D_\aa \f{1}{Z_{,\aa}}}_{L^\infty}}_{\lec \eqref{eq:1023}} E_a
 + E_a
\\ & + \underbrace{\Re \int i \paren{\f{1}{\bar{z}_\a} \partial_\a \f{h_\a}{z_\a}}_t \abs{\th}^2 h_\a d\a}_{\lec \eqref{eq:368}}
 + \underbrace{\Re  \int i \paren{\f{h_\a^2}{\abs{z_\a}^2}}_t  \th_\a \bar{\th} d\a}_{\lec \eqref{eq:552}}.
  \end{aligned}
\end{equation}

\subsubsection{The estimate for \texorpdfstring{$E_b$}{Eb}}\label{sec:40}
Now we consider our second energy. Once again, we work first with general $\th$ satisfying $\bound{\th} = 0$, $(I-\HH)(\th \circ h\i) = 0$, and the main equation \eqref{eq:171}.  Then we specialize to $\th = D_\a \bar{z}_t$.\footnote{See \eqref{eq:1026}.} The periodicity ensures there is no boundary term when we integrate by parts.
\begin{equation}
  \label{eq:224}
  \begin{aligned}
    \f{d}{dt} E_{b,\th}(t) & = \int \f{1}{\mathfrak{a}} (\th_{tt} \bar{\th}_t + \th_t \bar{\th}_{tt}) d\a - \int \f{\mathfrak{a}_t}{\mathfrak{a}} \f{1}{\mathfrak{a}} \abs{\th_t}^2 d\a
+ \underbrace{\int i \th_{t\a} \bar{\th} d\a}_{= {\int \bar{i \th_\a} \th_t d\a}} + \int i \th_\a \bar{\th}_t d\a
\\ & + \int \f{(A_1 \circ h)}{\mathfrak{a}} (\th_t \bar{\th} + \th \bar{\th}_t) d\a + \int \f{(A_1 \circ h)_t}{\af} \abs{\th}^2 d\a - \int \f{\mathfrak{a}_t}{\mathfrak{a}} \f{(A_1 \circ h)}{\mathfrak{a}} \abs{\th}^2 d\a
\\ &
= 2 \Re \int \f{G_\th}{\mathfrak{a}} \bar{\th}_t d\a  - \int \f{\mathfrak{a}_t}{\mathfrak{a}} \f{1}{\mathfrak{a}} \abs{\th_t}^2 d\a
\\ & + \int \f{A_1\circ h}{\mathfrak{a}} (\th_t \bar{\th} + \th \bar{\th}_t) d\a + \int \paren{\f{(A_1 \circ h)_t}{(A_1 \circ h)} - \f{\mathfrak{a}_t}{\mathfrak{a}}} \f{(A_1 \circ h)}{\mathfrak{a}} \abs{\th}^2 d\a.
\end{aligned}
\end{equation}

By H\"older and Cauchy-Schwarz, we conclude that
\begin{equation}
  \label{eq:173}
  \abs{\f{d}{dt} E_{b,\th}(t)} \lec  \nm{\f{G_\th}{\sqrt{\mathfrak{a}}}}_{L^2} E_{b,\th}^{1/2}  + \paren{\|A_1\|_{L^\infty}^{1/2} + \nm{\f{\mathfrak{a}_t}{\mathfrak{a}}}_{L^\infty} + \nm{\f{(A_1 \circ h)_t}{(A_1 \circ h)}}_{L^\infty}} E_{b,\th}.
\end{equation}
For $\th = D_\a \bar{z}_t$, we control $\nm{\f{G_\th}{\sqrt{\mathfrak{a}}}}_{L^2}$ in \S \ref{sec:E2}, at  \eqref{eq:311}. We control $\|A_1\|_{L^\infty}$ at \eqref{eq:361},  $\nm{\f{\af_t}{\af}}_{L^\infty}$ at \eqref{eq:424} and  $\nm{\f{(A_1 \circ h)_t}{(A_1 \circ h)}}_{L^\infty}$ at \eqref{eq:455}.

\subsubsection{The proof for \texorpdfstring{$\abs{z_{tt}(\a_0,t)-i}$}{| ztt(a0,t) - i |}}
Finally, we show that we can control $\frac d{dt}\abs{\bar{z}_{tt}(\a_0,t)-i}$. By differentiating with respect to $t$, we have, by the basic quasilinear equation \eqref{eq:11},
\begin{equation}
  \label{eq:664}
  \begin{aligned}
    \f{d}{dt}\abs{\bar{z}_{tt}(\a_0) - i} & \le \abs{\bar{z}_{ttt}(\a_0)}
     = \abs{i\af_t(\a_0) \bar{z}_\a(\a_0) + i \af(\a_0) \bar{z}_{t\a}(\a_0)}
    \\ & \lec \paren{\nm{\f{\af_t}{\af}}_{L^\infty} + \nm{D_\a \bar{z}_t}_{L^\infty}} \abs{\bar{z}_{tt}(\a_0) - i}.
  \end{aligned}
\end{equation}
We control $\nm{\f{\af_t}{\af}}_{L^\infty}$ below at \eqref{eq:424} and $\nm{D_\a \bar{z}_t}_{L^\infty}$ at \eqref{eq:15}.

\subsection{Outline of the remainder of the proof}\label{sec:42}
In  sections \S \ref{sec:43} through \S \ref{controlE1}, we complete the proof of the a priori inequality \eqref{eq:690}.

In \S \ref{sec:43}, we control various quantities that are necessary for our proof. In \S \ref{sec:quants}, we carefully list the basic quantities controlled by our energy. In \S \ref{sec:at-a}-\S \ref{sec:47}, we estimate various other quantities in terms of quantities already controlled in previous subsections of \S \ref{sec:43}. These quantities include many of the quantities to be controlled in \S \ref{sec:39} above, and are used in the remainder of the proof.  In appendix \S \ref{quantities}, we list and give references to all the quantities controlled in \S \ref{sec:43}, which we then use, sometimes without citation, in \S \ref{sec:98a} through \S \ref{controlE1}.

In \S \ref{sec:98a} and \S \ref{sec:48} we estimate the terms from \eqref{eq:189} and \eqref{eq:193} in the estimate of $\f{d}{dt}E_a$ above. Finally, in \S \ref{sec:E2} and \S \ref{controlE1} we conclude the estimates for $\f{d}{dt}E_b$ and $\f{d}{dt}E_a$, respectively, by controlling the $G_\th$ terms, completing the proof.

The basic approach for many of the estimates is to try and use the fact that certain quantities are purely real-valued and others are holomorphic to express problematic terms as commutators involving the Hilbert transform, and then use the commutator estimates from \S \ref{sec:31} to avoid loss of derivatives. The estimates are very tight, and sometimes convoluted. Among the reasons for the complexity are:  we have very little regularity to work with and our energy has direct control of only a selected collection of (weighted) quantities; we are working with a weighted derivative $D_\aa$ that has to be commuted with $\HH$ and whose complex-valued weight makes inverting $(I-\HH)$ on real-valued functions more difficult; we have no positive lower bound for the Taylor coefficient $-\frac{\partial P}{\partial \vec{n}}=\af\abs{z_\a}$; because our estimates are tight, we have to take care in using different estimates for different terms, including treating certain terms as commutators while keeping others in $(I-\HH)$ form; we must take care that the quantities in our commutators have appropriate periodic boundary behavior; and we sometimes have to carefully decompose certain quantities into their holomorphic and antiholomorphic projections. We will give enough details to facilitate reading.

We use $C(E)$ to indicate a universal polynomial of $E$,  which may differ from line to line.

Throughout the remaining derivations, we will repeatedly rely on the identity \eqref{eq:209}
\begin{equation}
 \label{eq:492}
  A_1 \circ h = \f{\af \abs{z_\a}^2}{h_\a}.
\end{equation}

\section{Quantities controlled by our energy}\label{sec:43}
Here we collect together many quantities that are controlled by our energy. In appendix \S \ref{quantities}, we give a list of all the quantities controlled in this section that are quoted without citation in future sections or subsections.

\subsection{Basic quantities controlled by the energy}\label{sec:quants}

In this section, we present a list of basic quantities controlled by our energy. Because conjugations and commutations of $\partial_t$ with $D_\a$ add complexity, we take care to list some of those estimates as well. We list all of the basic terms controlled here at \eqref{eq:1050} below.

We start with \eqref{eq:400} and \eqref{eq:408}. $E_a$ and $E_b$ directly control
  \begin{align}
   \label{eq:498}
 \nm{D_\a^2 \bar{z}_t}_{L^2(\f{h_\a}{A_1 \circ h} d\a)},\quad\nm{D_\aa^2 \bar{Z}_t}_{L^2(\f{1}{A_1} d\aa)}, &\quad  \nm{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t}_{\dot{H}^{1/2}}  \le  E_a^{1/2}\\
 \label{eq:239}
   \paren{ \int \abs{D_\a \bar{z}_t}^2 \f{d\a}{\af}}^{1/2}\le \nm{\bar{Z}_{t,\aa}}_{L^2} &\le E_b^{1/2},
\end{align}
where we used $A_1\ge 1$ \eqref{eq:394} in \eqref{eq:239}. By \eqref{eq:208}, using  the commutator estimate \eqref{eq:32}, and then \eqref{eq:239},  we have
\begin{equation}
  \label{eq:361}

    \nm{A_1}_{L^\infty}
    \lec \nm{Z_{t,\aa}}_{L^2}^2 + 1
     \lec E_b + 1.

\end{equation}
Thanks  to \eqref{eq:361}, we can now control $\nm{D_\a^2 \bar{z}_t}_{L^2(h_\a d\a)}=\nm{D_\aa^2 \bar{Z}_t}_{L^2}$ by a polynomial of our energy $E$.

Now we control $\nm{D_\a \bar{z}_t}_{L^\infty} = \nm{D_\aa \bar{Z}_t}_{L^\infty}$. We work in Riemannian coordinates and use the weighted Sobolev inequality \eqref{eq:638} with weight $\omega =\f{1}{\abs{Z_{,\aa}}^2}$ (and $\e = 1$). Note that $\avg (D_\aa \bar{Z}_t)^2 = 0$ by \eqref{eq:48}. This gives
\begin{equation}
  \label{eq:206}
  \begin{aligned}
 \nm{D_\aa \bar{Z}_t}_{L^\infty} & \lec \paren{\int \abs{D_\aa \bar{Z}_t}^2 \abs{Z_{,\aa}}^2 d\aa}^{1/2} + \paren{\int \abs{\partial_\aa D_\aa \bar{Z}_t}^2 \f{1}{\abs{Z_{,\aa}}^2} d\aa}^{1/2}
 \\ & = \nm{\bar{Z}_{t,\aa}}_{L^2} + \nm{D_\aa^2 \bar{Z}_t}_{L^2}
 \le E_b^{1/2}+ \nm{A_1 }_{L^\infty}^{1/2} E_a^{1/2}
\\ & \lec C(E).
\end{aligned}
\end{equation}
 We conclude that
\begin{equation}
  \label{eq:15}
  \nm{D_\a z_t}_{L^\infty} = \nm{D_\aa Z_t}_{L^\infty} = \nm{D_\a \bar{z}_t}_{L^\infty} = \nm{D_\aa \bar{Z}_t}_{L^\infty}  \lec C(E).
\end{equation}

Now we use the commutator identity \eqref{eq:120} to move the $\partial_t$ inside the first term in $E_b$:
\begin{equation}
  \label{eq:403}
\begin{aligned}
  \paren{\int \abs{D_\a \bar{z}_{tt}}^2 \f{d\a}{\af}}^{1/2} & \le \paren{\int \abs{\partial_t D_\a \bar{z}_t}^2 \f{d\a}{\af}}^{1/2} + \paren{\int \abs{[\partial_t,D_\a] \bar{z}_t}^2 \f{d\a}{\af}}^{1/2}
\\ & \le E_b^{1/2} + \nm{D_\a z_t}_{L^\infty} \paren{ \int \abs{D_\a \bar{z}_t}^2 \f{d\a}{\af}}^{1/2}
\\ & \lec C(E)
\end{aligned}
\end{equation}
by \eqref{eq:15} and \eqref{eq:239}.  By changing variables and by \eqref{eq:492}, we conclude from \eqref{eq:403} and \eqref{eq:361} that
\begin{equation}
  \label{eq:406}
\begin{aligned}
   \nm{\bar{Z}_{tt,\aa}}_{L^2} & \lec \nm{A_1}_{L^\infty}^{1/2}  \paren{\int \abs{D_\a \bar{z}_{tt}}^2 \f{d\a}{\af}}^{1/2}
\\ & \lec C(E).
\end{aligned}
\end{equation}

Note that  $\abs{D_\a^2 f}\ne \abs{D_\a^2 \bar{f}}$. Nevertheless, for generic $f$, we can control $D_\a^2 f$ by $D_\a^2 \bar{f}$ in $L^2({h_\a} d\a)$ norm, at the expense of some lower-order terms.    For notational convenience, we define here
\begin{equation}
  \label{eq:401}
  \abs{D_\a} := \f{1}{\abs{z_\a}} \partial_\a = \f{z_\a}{\abs{z_\a}} D_\a.
  \end{equation}
  We expand:
\begin{equation}
  \label{eq:65aa}
  \begin{aligned}
   D_\a^2 f & = \paren{\f{\abs{z_\a}}{z_\a}}^2 \abs{D_\a}^2 f + \f{\abs{z_\a}}{z_\a} \paren{\abs{D_\a} \f{\abs{z_\a}}{z_\a}} \abs{D_\a} f\\
   D_\a^2 \bar f & = \paren{\f{\abs{z_\a}}{z_\a}}^2 \abs{D_\a}^2 \bar f + \f{\abs{z_\a}}{z_\a} \paren{\abs{D_\a} \f{\abs{z_\a}}{z_\a}} \abs{D_\a} \bar f.
   \end{aligned}
   \end{equation}
Therefore,
   \begin{equation}\label{eq:65a}
   \abs{D_\a^2 f }\le
\abs{D_\a^2 \bar{f}} + 2 \abs{\abs{D_\a} \f{\abs{z_\a}}{z_\a}} \abs{D_\a \bar{f}}
\end{equation}
and so
\begin{equation}
  \label{eq:283}

    \paren{\int \abs{D_\a^2 f}^2 {h_\a d\a}}^{1/2}    \le  \paren{\int \abs{D_\a^2 \bar{f}}^2 {h_\a d\a}}^{1/2} + 2 \nm{D_\a \bar{f}}_{L^\infty} \paren{\int \abs{D_\a \f{\abs{z_\a}}{z_\a}}^2 {h_\a d\a}}^{1/2}.

\end{equation}
By \eqref{eq:279} and then \eqref{eq:100}, \eqref{eq:8}, \eqref{eq:492}, and the fact $A_1 \ge 1$,
\begin{equation}
  \label{eq:284}
    \abs{D_\a \f{\abs{z_\a}}{z_\a}}^2 h_\a   =  \abs{D_\a \f{\bar{z}_{tt} - i}{\abs{\bar{z}_{tt} -i}}}^2 h_\a
 \le  \abs{D_\a \bar{z}_{tt}}^2 \f{h_\a}{\abs{\bar{z}_{tt} - i}^2 }
  =  \abs{D_\a \bar{z}_{tt}}^2 \f{h_\a}{\af^2 \abs{z_\a}^2 }
\le  \abs{D_\a \bar{z}_{tt}}^2 \f{1}{\af }.
\end{equation}
Plugging this into \eqref{eq:283}, and using \eqref{eq:403}, we get
\begin{equation}
  \label{eq:404}
  \paren{\int \abs{D_\a^2 f}^2 h_\a d\a}^{1/2}
\lec  \paren{\int \abs{D_\a^2 \bar{f}}^2 h_\a d\a}^{1/2}
  + \nm{D_\a \bar{f}}_{L^\infty} C(E).
\end{equation}

We now apply \eqref{eq:404} to $f = z_t$, using \eqref{eq:15} to control $\nm{D_\a \bar{z}_t}_{L^\infty}$ and \eqref{eq:498} and \eqref{eq:361} to control $\nm{D_\aa^2 \bar Z_t}_{L^2}$:
\begin{equation}
  \label{eq:405}

 \nm{D_\aa^2 Z_t}_{L^2}  = \paren{\int \abs{D_\a^2 z_t}^2 h_\a d\a}^{1/2}
 \lec C(E).

\end{equation}

We now control $\nm{D_\a^2 \bar{z}_{tt}}_{L^2(h_\a d\a)}$. We use the commutator identity \eqref{eq:121} to get
\begin{equation}
  \label{eq:122}
  \begin{aligned}
    \nm{D_\aa^2 \bar{Z}_{tt}}_{L^2} & = \nm{D_\a^2 \bar{z}_{tt}}_{L^2({h_\a d\a})}
\\ & \le \nm{\partial_t D_\a^2 \bar{z}_{t}}_{L^2({h_\a d\a})} + 2 \nm{(D_\a z_t)D_\a^2 \bar{z}_t}_{L^2({h_\a d\a})} + \nm{(D_\a^2 z_t) D_\a \bar{z}_t}_{L^2({h_\a d\a})}
\\ & \le\nm{A_1}_{L^\infty} E_a^{1/2} + 2\nm{D_\a z_t}_{L^\infty} \nm{D_\a^2 \bar{z}_t}_{L^2({h_\a d\a})} + \nm{D_\a^2 z_t}_{L^2({h_\a d\a})} \nm{D_\a \bar{z}_t}_{L^\infty}
\\ & \lec C(E).
\end{aligned}
\end{equation}
We will also need to control $D_\aa^2 Z_{tt}$; we delay doing this until later, after we control $\nm{D_\a \bar{z}_{tt}}_{L^\infty}$.

We will also at one point need to control
\begin{equation}
  \label{eq:419}
  \begin{aligned}
      \nm{D_\a \partial_t D_\a \bar{z}_t}_{L^2({h_\a}d\a)} & \le \nm{\partial_t D_\a D_\a \bar{z}_t}_{L^2({h_\a}d\a)} + \nm{[\partial_t,D_\a] D_\a \bar{z}_{t}}_{L^2({h_\a}d\a)}
      \\ & \le \|A_1\|_{L^\infty}E_a^{1/2} + \nm{D_\a z_t}_{L^\infty} \nm{D_\a^2 \bar{z}_{t}}_{L^2({h_\a}d\a)}
      \\ & \le C(E).
\end{aligned}
\end{equation}

We now control $\nm{\bar{z}_{tt} - i}_{L^\infty} = \nm{\bar{Z}_{tt} - i}_{L^\infty}$. Recall from our definition of the energy \eqref{eq:49} that the energy includes $\abs{\bar{z}_{tt}(\a_0,t)-i}$ for some fixed $\a_0 \in I$. Let $\aao = h(\a_0, t)$. Then, by the fundamental theorem of calculus, for arbitrary $\aa \in I$,
\begin{equation}
  \label{eq:29}
  \begin{aligned}
    \abs{\bar{Z}_{tt}(\aa,t)- i}     & \lec \abs{\bar{Z}_{tt}(\aao,t)- i} + \nm{\bar{Z}_{tt,\aa}}_{L^1(I)}
    \\ & \lec \abs{\bar{Z}_{tt}(\aao,t)- i} + \eqref{eq:406}.
  \end{aligned}
\end{equation}
We conclude that
\begin{equation}
  \label{eq:378}
  \begin{aligned}
    \nm{z_{tt} + i}_{L^\infty} & = \nm{Z_{tt}+i}_{L^\infty} = \nm{\bar{z}_{tt}-i}_{L^\infty} = \nm{\bar{Z}_{tt}-i}_{L^\infty}
  \\ & \lec C(E).
\end{aligned}
\end{equation}
Because of \eqref{eq:251} and \eqref{eq:394}, we can also conclude that
\begin{equation}
  \label{eq:357}
  \nm{\f{1}{Z_{,\aa}}}_{L^\infty} \lec C(E).
\end{equation}

We use this to control $\nm{D_\a \bar{z}_{tt}}_{L^\infty}$, using the weighted Sobolev inequality \eqref{eq:396} in Riemannian coordinates with weight $\omega = \f{1}{\abs{Z_{,\aa}}^2}$ (and $\e = 1$):\footnote{Note that unlike our proof for $\nm{D_\aa \bar{Z}_t}_{L^\infty}$ at \eqref{eq:206} above, we don't necessarily have that $\avg (D_\aa \bar{Z}_{tt})^2$ is zero, so we get a third term in the Sobolev inequality.}
\begin{equation}
  \label{eq:17}
  \begin{aligned}
     \nm{D_\a z_{tt}}_{L^\infty} & = \nm{D_\aa Z_{tt}}_{L^\infty} = \nm{D_\a \bar{z}_{tt}}_{L^\infty} = \nm{D_\aa \bar{Z}_{tt}}_{L^\infty}
\\ & \lec \nm{\bar{Z}_{tt,\aa}}_{L^2} + \nm{D_\aa^2 \bar{Z}_{tt}}_{L^2} + \paren{\int \abs{D_\aa \bar{Z}_{tt}}^2 d\aa}^{1/2}
\\ & \lec \paren{1 + \nm{1/Z_{,\aa}}_{L^\infty}} \nm{\bar{Z}_{tt,\aa}}_{L^2} + \nm{D_\aa^2 \bar{Z}_{tt}}_{L^2}
\\ & \lec \paren{1 +\eqref{eq:357}} \eqref{eq:406} +  \eqref{eq:122}\\&\lec C(E).
\end{aligned}
\end{equation}

Finally, we use \eqref{eq:404}, \eqref{eq:122}, and \eqref{eq:17} to control $D_\a^2 z_{tt}$ and $D_\aa^2 Z_{tt}$:
\begin{equation}
  \label{eq:418}
  \begin{aligned}
    \nm{D_\aa^2 Z_{tt}}_{L^2} &=  \nm{D_\a^2 z_{tt}}_{L^2({h_\a d\a})}
\\ & \lec \nm{D_\aa^2 \bar{z}_{tt}}_{L^2(h_\a d\a)}+  \nm{D_\aa \bar{z}_{tt}}_{L^\infty}C(E)\\ & \lec \eqref{eq:122} + \eqref{eq:17} C(E)\lec C(E).
  \end{aligned}
\end{equation}

To sum up, we have the following quantities and their counterparts in Lagrangian coordinates controlled by universal polynomials of the energy $E$:
\begin{equation}\label{eq:1050}
\begin{aligned}
&\nm{D_\aa^2 \bar{Z}_{tt}}_{L^2}, \nm{D_\aa^2 {Z}_{tt}}_{L^2},  \nm{D_\aa^2 \bar{Z}_t}_{L^2}, \nm{D_\aa^2 {Z}_t}_{L^2}, \nm{D_\a \partial_t D_\a \bar{z}_t}_{L^2({h_\a}d\a)},\\& \nm{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t}_{\dot{H}^{1/2}}, \nm{D_\aa \bar{Z}_{tt}}_{L^\infty}, \nm{D_\aa {Z}_{tt}}_{L^\infty},  \nm{D_\aa \bar{Z}_t}_{L^\infty}, \nm{D_\aa {Z}_t}_{L^\infty},\\&  \nm{ \bar{Z}_{tt, \aa}}_{L^2},   \nm{ \bar{Z}_{t, \aa}}_{L^2},  \int \abs{D_\a \bar{z}_t}^2 \f{d\a}{\af},\  \int \abs{D_\a \bar{z}_{tt}}^2 \f{d\a}{\af},  \ \nm{\f{1}{Z_{,\aa}}}_{L^\infty}, \nm{Z_{tt}+i}_{L^\infty} , \nm{A_1}_{L^\infty} \\&
\lec C(E).
\end{aligned}
\end{equation}

\subsection{Controlling \texorpdfstring{$\nm{\f{\af_t}{\af}}_{L^\infty}$}{|| at/a || L infinity}} \label{sec:at-a}

We now show that we can control $\nm{\f{\af_t}{\af}}_{L^\infty}$, using \eqref{eq:210}.  Because $A_1 \ge 1$ \eqref{eq:394}, it suffices to control
\begin{equation}
  \label{eq:267}
  \nm{2[Z_t,\HH] \bar{Z}_{tt,\aa} + 2[Z_{tt},\HH] \bar{Z}_{t,\aa} - [Z_t, Z_t; D_\aa \bar{Z}_t]}_{L^\infty}.
\end{equation}
We control the first two terms by \eqref{eq:32}, and the last term by H\"older and then Hardy's inequality \eqref{eq:77}.
We have
\begin{equation}
  \label{eq:424}
  \nm{\f{\af_t}{\af}}_{L^\infty} \lec  \nm{Z_{t,\aa}}_{L^2} \nm{\bar{Z}_{tt,\aa}}_{L^2}+ \nm{Z_{t,\aa}}_{L^2}^2 \nm{D_\aa \bar{Z}_t}_{L^\infty}\lec C(E).
\end{equation}

\subsection{Controlling \texorpdfstring{$\nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}$}{|| partial a' 1/Z,a' || L2}}\label{sec:44}
Recall from \eqref{eq:251} that

 $ \f{1}{Z_{,\aa}} = i \f{\bar{Z}_{tt} - i}{A_1}$.

Therefore,
\begin{equation}
  \label{eq:252}
  \partial_\aa \f{1}{Z_{,\aa}} = i \f{\bar{Z}_{tt,\aa}}{A_1} - i \f{\bar{Z}_{tt} -i }{A_1^2} \partial_\aa A_1.
\end{equation}
Because $A_1 \ge 1$ \eqref{eq:394}, we can control the first term by $\nm{\bar{Z}_{tt,\aa}}_{L^2}$.  Now we address the second term.

We recall that  $ A_1 = \Im\paren{ - [Z_t,\HH] \bar{Z}_{t,\aa}} + 1$ \eqref{eq:208}. Therefore,
\begin{equation}
  \label{eq:254}
  \begin{aligned}
      \partial_\aa A_1 & = \partial_\aa \Im \f{-1}{\hdenomconst} \int (Z_t(\aa) - Z_t(\bb)) \cot(\f{\pi}{2}(\aa -\bb)) \bar{Z}_{t,\bb} d\bb
\\ & = - \underbrace{\Im Z_{t,\aa} \HH \bar{Z}_{t,\aa}}_{0} + \Im \f{1}{\hdenomconst} \int \f{\pi}{2} \f{(Z_t(\aa) - Z_t(\bb))}{\sin^2(\f{\pi}{2}(\aa - \bb))} \bar{Z}_{t,\bb}(\bb) d\b
\\ & =   \Im \f{1}{\hdenomconst} \int \f{\pi}{2} \f{(Z_t(\aa) - Z_t(\bb))}{\sin^2(\f{\pi}{2}(\aa - \bb))} \bar{Z}_{t,\bb}(\bb) d\b,
\end{aligned}
\end{equation}
where the first term disappears because $\HH \bar{Z}_{t,\aa} = \bar{Z}_{t,\aa}$ \eqref{eq:474} and so $Z_{t,\aa} \HH \bar{Z}_{t,\aa}$ is purely real. Therefore, multiplying \eqref{eq:254} by $\abs{\bar{Z}_{tt}(\a) - i}$
 and splitting into two parts, we have
\begin{equation}
  \label{eq:255}
  \begin{aligned}
      \abs{\bar{Z}_{tt} - i} \partial_\aa A_1 & = \Im \f{1}{\hdenomconst} \int \f{\pi}{2} \f{(Z_t(\aa) - Z_t(\bb))}{\sin^2(\f{\pi}{2}(\aa - \bb))} (\abs{\bar{Z}_{tt}(\aa)-i} - \abs{\bar Z_{tt}(\bb)-i}) \bar{Z}_{t,\bb}(\bb) d\b
\\&+ \Im \f{1}{\hdenomconst} \int \f{\pi}{2} \f{(Z_t(\aa) - Z_t(\bb))}{\sin^2(\f{\pi}{2}(\aa - \bb))}\abs{\bar{Z}_{tt}(\bb) - i} \bar{Z}_{t,\bb}(\bb) d\b
\\ & = I + II.
\end{aligned}
\end{equation}
We need to control $\nm{I}_{L^2}$ and $\nm{II}_{L^2}$.
By \eqref{eq:113},
\begin{equation}
  \label{eq:257}
  \nm{I}_{L^2} \lec\nm{Z_{t,\a}}_{L^2}  \nm{\bar{Z}_{tt,\a}}_{L^2} \nm{\bar{Z}_{t,\aa}}_{L^2} = \nm{\bar{Z}_{t,\aa}}_{L^2}^2 \nm{\bar{Z}_{tt,\aa}}_{L^2}.
\end{equation}
For $II$, we replace  $\abs{\bar Z_{tt}(\bb)-i}$ by $\abs{\f{(-i A_1(\bb))}{Z_{,\bb}}}$ \eqref{eq:251} and use estimate \eqref{eq:258}, noticing that $\f{1}{Z_{,\bb}}\bar{Z}_{t,\bb}=D_\aa \bar{Z}_t$:
\begin{equation}
  \label{eq:427}
  \nm{II}_{L^2} \lec \nm{Z_{t,\aa}}_{L^2} \nm{A_1 D_\aa \bar{Z}_{t}}_{L^\infty} \le \nm{Z_{t,\aa}}_{L^2} \nm{D_\aa \bar{Z}_t}_{L^\infty} \nm{A_1}_{L^\infty}.
\end{equation}

We conclude that
\begin{equation}
  \label{eq:1018}
  \begin{aligned}
  \nm{ \paren{\bar{Z}_{tt} - i} \partial_\aa A_1  }_{L^2} &\lec \nm{\bar{Z}_{tt,\aa}}_{L^2}\nm{Z_{t,\aa}}_{L^2}^2 + \nm{Z_{t,\aa}}_{L^2} \nm{D_\aa \bar{Z}_t}_{L^\infty} \nm{A_1}_{L^\infty}\\& \lec C(E)
  \end{aligned}
\end{equation}
and
\begin{equation}
  \label{eq:425}
  \begin{aligned}
  \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2} &\lec \nm{\bar{Z}_{tt,\aa}}_{L^2}+  \nm{ \paren{\bar{Z}_{tt} - i} \partial_\aa A_1  }_{L^2}
    \\& \lec C(E).
  \end{aligned}
\end{equation}

\subsection{Controlling \texorpdfstring{$\nm{\f{h_{t\a}}{h_\a}}_{L^\infty}$}{|| hta/ha || L infinity}}\label{hta-ha}
We recall \eqref{eq:175}:  $ h(\a,t) = \Phi(z(\a,t),t) = \Phi \circ z$. Therefore
$ h_\a = (\Phi_z \circ z) z_\a$,
and
\begin{equation}
  \label{eq:244}
  h_t = (\Phi_t \circ z)+ (\Phi_z \circ z) z_t = (\Phi_t \circ z) + \f{h_\a}{z_\a} z_t.
\end{equation}
We precompose with $h^{-1}$:
\begin{equation}
  \label{eq:246}
  (h_t \circ h\i) (\aa,t) = \Phi_t \circ Z + \f{1}{Z_{,\aa}} Z_t.
\end{equation}
Differentiating with respect to $\aa$ gives
\begin{equation}
  \label{eq:248}
  \begin{aligned}
  (h_t \circ h\i)_\aa &= \partial_\aa(\Phi_t \circ Z) + D_\aa Z_t + Z_t \partial_\aa \f{1}{Z_{,\aa}}\\&
  = \partial_\aa(\Phi_t \circ Z) +2 \Re D_\aa Z_t - \f{\bar{Z}_{t,\aa}}{\bar{Z}_{,\aa}} + Z_t \partial_\aa \f{1}{Z_{,\aa}}.
\end{aligned}
\end{equation}
We replaced  $D_\aa Z_t$ by $ 2 \Re D_\aa Z_t - \f{\bar{Z}_{t,\aa}}{\bar{Z}_{,\aa}}$, since we can turn $(I-\mathbb H)( \f{\bar{Z}_{t,\aa}}{\bar{Z}_{,\aa}})$ into a controllable commutator.
Observe that the LHS and the second term on the RHS are purely real. Therefore, if we apply $\Re(I-\HH)$ to both sides, the LHS and the second term on the RHS remain unchanged, the term $\Re(I-\HH) \partial_\aa(\Phi_t \circ Z)$ on the RHS disappears by \eqref{eq:312}, and the remaining terms become commutators by \eqref{eq:474} and \eqref{eq:242}:
\begin{equation}
      \label{eq:249}
  \begin{aligned}
  (h_t \circ h\i)_\aa &= 2 \Re D_\aa Z_t + \Re \braces{-(I-\HH) \paren{\f{1}{\bar{Z}_{,\aa}} \bar{Z}_{t,\aa}} + (I-\HH) Z_t \partial_\aa \f{1}{Z_{,\aa}}}
\\ & =2 \Re D_\aa Z_t + \Re \braces{ -\bracket{\f{1}{\bar{Z}_{,\aa}},\HH} \bar{Z}_{t,\aa} + [Z_t,\HH] \partial_\aa \f{1}{Z_{,\aa}}}.
\end{aligned}
\end{equation}
We use \eqref{eq:32} to control these:
\begin{equation}
  \label{eq:449}
  \nm{\bracket{\f{1}{\bar{Z}_{,\aa}},\HH} \bar{Z}_{t,\aa}}_{L^\infty} \lec \nm{\partial_\aa \f{1}{\bar{Z}_{,\aa}}}_{L^2} \nm{\bar{Z}_{t,\aa}}_{L^2}
\end{equation}
\begin{equation}
  \label{eq:450}
   \nm{\bracket{Z_t,\HH} \partial_\aa \f{1}{Z_{,\aa}}}_{L^\infty} \lec \nm{Z_{t,\aa}}_{L^2} \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}.
\end{equation}
Observe that
\begin{equation}
  \label{eq:452}
  \f{h_{t\a}}{h_\a} \circ h\i = \partial_\aa (h_t \circ h\i).
\end{equation}
We therefore conclude that
\begin{equation}
  \label{eq:451}
  \begin{aligned}
  \nm{\f{h_{t\a}}{h_\a}}_{L^\infty} &= \nm{\partial_\aa(h_t \circ h\i)}_{L^\infty} \lec \nm{D_\aa Z_t}_{L^\infty} + \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2} \nm{\bar{Z}_{t,\aa}}_{L^2}\\&\lec C(E).
  \end{aligned}
\end{equation}

\subsection{Controlling \texorpdfstring{$\nm{\f{(A_1 \circ h)_t}{(A_1 \circ h)}}_{L^\infty}$}{|| A1t/A1 || L infinity}}\label{sec:41}
Recall that $ A_1 \circ h = \f{\af \abs{z_\a}^2}{h_\a}$ \eqref{eq:209}.
Therefore,
\begin{equation}
  \label{eq:215}
  \f{\f{d}{dt} (A_1 \circ h)}{A_1 \circ h} = \f{\af_t}{\af} - \f{h_{t\a}}{h_\a} + 2 \Re D_\a z_t.
\end{equation}
We have controlled each of the terms on the RHS in $L^\infty$ in the previous sections. We conclude that
\begin{equation}
  \label{eq:455}
  \begin{aligned}
    \nm{\f{\f{d}{dt}(A_1 \circ h)}{A_1 \circ h}}_{L^\infty} & \lec \nm{\f{\af_t}{\af}}_{L^\infty} + \nm{\f{h_{t\a}}{h_\a}}_{L^\infty} + \nm{D_\a z_t}_{L^\infty}\lec C(E).
\end{aligned}
\end{equation}

\subsection{Controlling \texorpdfstring{$\nm{D_\aa \f{1}{Z_{,\aa}}}_{L^\infty}$}{|| Da' 1/Z,a' || L infinity} and \texorpdfstring{$\nm{(\bar Z_{tt}-i)\partial_\aa \f{1}{Z_{,\aa}}}_{L^\infty}$}{|| bar Ztt - i partial a' 1/Z,a' || L infinity}} \label{sec:46}
Recall from \eqref{eq:251} that
$ \f{1}{Z_{,\aa}} = i \f{\bar{Z}_{tt} - i}{A_1}$.
Therefore,
\begin{equation}
\label{eq:1015}
D_\aa \f{1}{Z_{,\aa}}= i \f{D_\aa\bar{Z}_{tt} }{A_1}-i\f{\bar{Z}_{tt} - i}{A_1^2}{D_\aa A_1}=i \f{D_\aa\bar{Z}_{tt} }{A_1}+ \f{(\bar{Z}_{tt} - i)^2}{A_1^3}{\partial_\aa A_1}.
\end{equation}
 Because we can control the first term on the RHS by $\nm{D_\aa Z_{tt}}_{L^\infty}$, it suffices to focus on the second term.
 We start from \eqref{eq:254}, and then use a similar idea to \eqref{eq:255}:
\begin{equation}
  \label{eq:1014}
  \begin{aligned}
      \abs{\bar{Z}_{tt} - i}^2 \partial_\aa A_1 & = \Im \f{1}{\hdenomconst} \int \f{\pi}{2} \f{(Z_t(\aa) - Z_t(\bb))}{\sin^2(\f{\pi}{2}(\aa - \bb))} (\abs{\bar{Z}_{tt}(\aa)-i}^2 - \abs{\bar Z_{tt}(\bb)-i}^2) \bar{Z}_{t,\bb}(\bb) d\b
\\&+ \Im \f{1}{\hdenomconst} \int \f{\pi}{2} \f{(Z_t(\aa) - Z_t(\bb))}{\sin^2(\f{\pi}{2}(\aa - \bb))}\abs{\bar{Z}_{tt}(\bb) - i}^2 \bar{Z}_{t,\bb}(\bb) d\b
\\ & = I + II.
\end{aligned}
\end{equation}
To control $\nm{I}_{L^\infty}$, we use the mean value theorem and the periodicity of $Z_{tt}$ \eqref{eq:437} to estimate
\begin{equation}
  \label{eq:7}
  \abs{\f{\abs{\bar{Z}_{tt}(\aa)-i}^2 - \abs{\bar Z_{tt}(\bb)-i}^2} {\sin(\f{\pi}{2}(\aa - \bb))}  }\lec \nm{(\bar{Z}_{tt}(\aa)-i)\partial_\aa \bar{Z}_{tt}(\aa) }_{L^\infty}\lec \|A_1\|_{L^\infty} \nm{D_\aa \bar Z_{tt}}_{L^\infty}.
\end{equation}
From Cauchy-Schwarz and Hardy's inequality \eqref{eq:77}, we get
\begin{equation}\label{eq:1016}
\nm{I}_{L^\infty}\lec \nm{Z_{t,\aa}}_{L^2}^2\|A_1\|_{L^\infty} \nm{D_\aa \bar Z_{tt}}_{L^\infty}.
\end{equation}
For $II$,
observe that $\abs{\bar{Z}_{tt}(\aa) - i}^2 \bar{Z}_{t,\aa}(\aa)=\f{A_1^2}{\bar Z_{,\aa}}D_\aa \bar Z_t$
and $\bound{  \f{A_1^2}{\bar Z_{,\aa}}D_\aa \bar Z_t}=0$.\footnote{By \eqref{eq:259}, \eqref{eq:438}, and the evenness of $A_1$.} We integrate by parts as in \eqref{eq:98}:
\begin{equation}\label{eq:1017}
\begin{split}
 II=\Im\f{1}{\hdenomconst} \int \f{\pi}{2} \f{(Z_t(\aa) - Z_t(\bb))}{\sin^2(\f{\pi}{2}(\aa - \bb))}\abs{\bar{Z}_{tt}(\bb) - i}^2 \bar{Z}_{t,\bb}(\bb) d\b \\=\Im\mathbb H \paren{Z_{t,\aa} \f{A_1^2}{\bar Z_{,\aa}}D_\aa \bar Z_t}-\Im[Z_t,\mathbb H] \partial_\aa \paren{\f{A_1^2}{\bar Z_{,\aa}}D_\aa \bar Z_t}.
 \end{split}
\end{equation}
We estimate the second term by \eqref{eq:32}:
\begin{equation}\label{eq:1019}
\nm{[Z_t,\mathbb H] \partial_\aa \paren{\f{A_1^2}{\bar Z_{,\aa}}D_\aa \bar Z_t}}_{L^\infty}\lec
\|Z_{t,\aa}\|_{L^2}\nm{ \partial_\aa \paren{\f{A_1^2}{\bar Z_{,\aa}}D_\aa \bar Z_t}}_{L^2}.
\end{equation}
We expand the first term, using the conjugate of \eqref{eq:474}, $\mathbb H Z_{t,\aa} =-Z_{t,\aa}$, and  noticing that $\Im \paren{Z_{t,\aa} \f{A_1^2}{\bar Z_{,\aa}}D_\aa \bar Z_t}=0$:
\begin{equation}\label{eq:1021}
\Im \mathbb H \paren{Z_{t,\aa} \f{A_1^2}{\bar Z_{,\aa}}D_\aa \bar Z_t}=-\Im \paren{Z_{t,\aa} \f{A_1^2}{\bar Z_{,\aa}}D_\aa \bar Z_t+\bracket{\f{A_1^2}{\bar Z_{,\aa}}D_\aa \bar Z_t,\mathbb H}Z_{t,\aa}}=-\Im\bracket{\f{A_1^2}{\bar Z_{,\aa}}D_\aa \bar Z_t,\mathbb H}Z_{t,\aa}.
\end{equation}
We estimate the RHS by  \eqref{eq:32}:
\begin{equation}\label{eq:1022}
\nm{\bracket{\f{A_1^2}{\bar Z_{,\aa}}D_\aa \bar Z_t,\mathbb H}Z_{t,\aa}}_{L^\infty}\lec \|Z_{t,\aa}\|_{L^2}\nm{ \partial_\aa \paren{\f{A_1^2}{\bar Z_{,\aa}}D_\aa \bar Z_t}}_{L^2}.
\end{equation}
Now,
\begin{equation}\label{eq:1020}
\begin{aligned}
\nm{ \partial_\aa \paren{\f{A_1^2}{\bar Z_{,\aa}}D_\aa \bar Z_t}}_{L^2}
& \lec \nm{A_1}_{L^\infty}^2\nm{D^2_\aa \bar Z_t}_{L^2}\\&+\nm{A_1}_{L^\infty}^2\nm{\partial_\aa\f{1}{Z_{,\aa}}}_{L^2}\nm{D_\aa \bar Z_t}_{L^\infty}+\nm{ \f{A_1}{\bar Z_{,\aa}}\partial_\aa A_1  }_{L^2}\nm{D_\aa \bar Z_t}_{L^\infty}.
\end{aligned}
\end{equation}

Combining the above calculations and using \eqref{eq:1018}, \eqref{eq:425}, the estimates in \S \ref{sec:quants}, and the fact $A_1\ge 1$, we conclude
\begin{equation}\label{eq:1023}
\begin{aligned}
\nm{D_\aa \f{1}{Z_{,\aa}}}_{L^\infty}&\le \nm{D_\aa\bar{Z}_{tt} }_{L^\infty}+\nm{(\bar{Z}_{tt} - i)^2\partial_\aa A_1 }_{L^\infty}\\&\lec \nm{D_\aa\bar{Z}_{tt} }_{L^\infty}+\nm{Z_{t,\aa}}_{L^2}^2\|A_1\|_{L^\infty} \nm{D_\aa \bar Z_{tt}}_{L^\infty}+\|Z_{t,\aa}\|_{L^2}\nm{ \partial_\aa (\f{A_1^2}{\bar Z_{,\aa}}D_\aa \bar Z_t)}_{L^2}\\&\lec C(E).
\end{aligned}
\end{equation}
We  record here the estimate for a related quantity, which we will use in later sections:
\begin{equation}\label{eq:1024}
\nm{(\bar Z_{tt}-i)\partial_\aa \f{1}{Z_{,\aa}}}_{L^\infty}\le \nm{A_1}_{L^\infty}\nm{D_\aa \f{1}{Z_{,\aa}}}_{L^\infty}\lec C(E).
\end{equation}

\subsection{Controlling \texorpdfstring{$\nm{(I+\HH) D_\aa Z_{t}}_{L^\infty}$}{|| (I+H) Da' Zt || L infinity}}\label{sec:45c}
We note that this is the conjugate of the second term on the RHS of \eqref{eq:249}. We give details nevertheless. By taking a conjugate, using \eqref{eq:474} to get a commutator, and using commutator estimate \eqref{eq:32}, we have
\begin{equation}
  \label{eq:170}
  \begin{aligned}
    \nm{(I+\HH) D_\aa Z_t}_{L^\infty} & = \nm{(I-\HH) \f{1}{\bar{Z}_{,\aa}} \bar{Z}_{t,\aa}}_{L^\infty}
     = \nm{\bracket{\f{1}{\bar{Z}_{,\aa}},\HH} \bar{Z}_{t,\aa}}
   \\&  \lec \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2} \nm{\bar{Z}_{t,\aa}}_{L^2}\lec C(E).
  \end{aligned}
\end{equation}

\subsection{Controlling \texorpdfstring{$\nm{\partial_\aa (I-\HH) \f{Z_t}{Z_{,\aa}}}_{L^\infty}$}{|| partial a' (I-H) Zt/Z,a' || L infinity} and   \texorpdfstring{$\nm{(I-\HH) \braces{Z_t \partial_\aa \f{1}{Z_{,\aa}}}}_{L^\infty}$}{|| (I-H) Zt partial a' 1/Z,a' || L infinity}}\label{sec:47}
In this section, we estimate $\nm{\partial_\aa (I-\HH) \f{Z_t}{Z_{,\aa}}}_{L^\infty}$  and $ \nm{(I-\HH) \braces{Z_t \partial_\aa \f{1}{Z_{,\aa}}}}_{L^\infty}$, which we will use in \S \ref{controlE1}.

We know $\bound{\f{Z_t}{Z_{,\aa}}} = 0$ by \eqref{eq:437} and \eqref{eq:438}. By \eqref{eq:176} we may commute in the derivative, so
\begin{equation}
  \label{eq:127}
  \begin{aligned}
    \partial_\aa (I-\HH) \f{Z_t}{Z_{,\aa}} & = (I-\HH) \partial_\aa \f{Z_t}{Z_{,\aa}}
\\ & =   (I-\HH) D_\aa Z_t + (I-\HH) \braces{Z_t \partial_\aa \f{1}{Z_{,\aa}}}.
\end{aligned}
\end{equation}
First, we control the first term. Rewriting $(I-\HH) = 2I - (I+\HH)$, we have
\begin{equation}
  \label{eq:154}
  \begin{aligned}
    \nm{(I-\HH) D_\aa Z_t}_{L^\infty} & \lec  \nm{D_\aa Z_t}_{L^\infty} + \nm{(I+\HH) D_\aa Z_t}_{L^\infty},
  \end{aligned}
\end{equation}
where the second term was estimated at \eqref{eq:170} above and the first term in \S \ref{sec:quants}.

We write the second term of \eqref{eq:127} as a commutator by \eqref{eq:242}, and control this by \eqref{eq:32}:
\begin{equation}
  \label{eq:151}
  \begin{aligned}
    \nm{(I-\HH) \braces{Z_t \partial_\aa \f{1}{Z_{,\aa}}}}_{L^\infty} & = \nm{[Z_t,\HH] \partial_\aa \f{1}{Z_{,\aa}}}_{L^\infty}
  \\ & \lec \nm{Z_{t,\aa}}_{L^2} \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}\lec C(E).
\end{aligned}
\end{equation}
We conclude from \eqref{eq:154}, \eqref{eq:151}, and \eqref{eq:170} that
\begin{equation}
  \label{eq:513}
  \nm{\partial_\aa (I-\HH) \f{Z_t}{Z_{,\aa}}}_{L^\infty} \lec \nm{D_\aa Z_t}_{L^\infty} + \nm{Z_{t,\aa}}_{L^2} \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}\lec C(E).
\end{equation}

\section{Controlling \texorpdfstring{$\Re \int i \paren{\f{1}{\bar{z}_\a} \partial_\a \f{h_\a}{z_\a}}_t \abs{\th}^2 h_\a d\a$}{Re int i 1/za partial a (ha/za)t |theta|2 ha da}}\label{sec:98a}
In this section, we control from \eqref{eq:66} the term
\begin{equation}
  \label{eq:229}
  \begin{aligned}
    \Re \int i \paren{\f{1}{\bar{z}_\a} \partial_\a \f{h_\a}{z_\a}}_t \abs{\th}^2 h_\a d\a & = - \Re \int i \f{\bar{z}_{t\a}}{\bar{z}_\a} \paren{\f{1}{\bar{z}_\a} \partial_\a \f{h_\a}{z_\a}} \abs{\th}^2 h_\a d\a
\\ & +  \Re \int i \paren{\f{1}{\bar{z}_\a} \partial_\a \partial_t \f{h_\a}{z_\a}} \abs{\th}^2 h_\a d\a
  \end{aligned}
\end{equation}
for $\th = D_\a^2 \bar{z}_t$.
We can control the first of these terms by
\begin{equation}
  \label{eq:169}
  \begin{aligned}
    \abs{-\Re \int i \f{\bar{z}_{t\a}}{\bar{z}_\a} \paren{\f{1}{\bar{z}_\a} \partial_\a \f{h_\a}{z_\a}} \abs{\th}^2 h_\a d\a} & \lec \nm{D_\a \bar{z}_t}_{L^\infty} \nm{D_\aa \f{1}{Z_{,\aa}}}_{L^\infty} \nm{D_\aa^2 \bar{Z}_t}_{L^2}^2.
  \end{aligned}
\end{equation}
Therefore, it suffices to focus on the second term on the RHS of \eqref{eq:229}. We expand it out:
\begin{equation}
  \label{eq:232}
  \begin{aligned}
    \Re \int i &\paren{\f{1}{\bar{z}_\a} \partial_\a \partial_t \f{h_\a}{z_\a}} \abs{\th}^2 h_\a d\a  = \Re \int i \paren{\f{1}{\bar{z}_\a} \partial_\a \paren{\f{h_\a}{z_\a} \paren{ \f{h_{t\a}}{h_\a} - \f{z_{t\a}}{z_\a}}}} \abs{\th}^2 h_\a d\a
\\ & = \Re \int i \paren{\f{1}{\bar{z}_\a} \partial_\a \f{h_\a}{z_\a}} \paren{ \f{h_{t\a}}{h_\a} - \f{z_{t\a}}{z_\a}} \abs{\th}^2 h_\a d\a
 + \Re \int i \paren{ \f{h_\a}{|z_\a|^2} \partial_\a \paren{ \f{h_{t\a}}{h_\a} - \f{z_{t\a}}{z_\a}}} \abs{\th}^2 h_\a d\a.
  \end{aligned}
\end{equation}
We can estimate the first term on the RHS by
\begin{equation}
  \label{eq:237}
  \abs{\Re \int i \paren{\f{1}{\bar{z}_\a} \partial_\a \f{h_\a}{z_\a}} \paren{ \f{h_{t\a}}{h_\a} - \f{z_{t\a}}{z_\a}} \abs{\th}^2 h_\a d\a} \lec \nm{D_\aa \f{1}{Z_{,\aa}}}_{L^\infty} \paren{\nm{\f{h_{t\a}}{h_\a}}_{L^\infty} + \nm{D_\a z_t}_{L^\infty}} \nm{D_\aa^2 \bar{Z}_t}_{L^2}.
\end{equation}
Therefore, it suffices to focus on the second term on the RHS of \eqref{eq:232}. Observe that because $h$ is real-valued,
\begin{equation}
  \label{eq:238}
  \Re \int i \paren{ \f{h_\a}{|z_\a|^2} \partial_\a \paren{ \f{h_{t\a}}{h_\a} - \f{z_{t\a}}{z_\a}}} \abs{\th}^2 h_\a d\a = - \Re \int i \paren{ \f{h_\a}{|z_\a|^2} \partial_\a \paren{\f{z_{t\a}}{z_\a}}} \abs{\th}^2 h_\a d\a.
\end{equation}
We now drop $\Re$ and the $i$, write $D_\a^2 \bar{z}_t = \th$, and switch to Riemannian coordinates. For consistency with the quantities we've controlled elsewhere, we will take a conjugate. We have
\begin{equation}
  \label{eq:292}
  \int  \paren{\f{1}{\abs{Z_{,\aa}}^2} \partial_\aa \paren{\f{\bar{Z}_{t,\aa}}{\bar{Z}_{,\aa}}}} \abs{D_\aa^2 \bar{Z}_t}^2 d\aa.
\end{equation}

We want to take advantage of the holomorphicity and antiholomorphicity of various of these factors.  To do this, we first replace the $\f{1}{\bar{Z}_{,\aa}}$ with a $\f{1}{Z_{,\aa}}$ inside the first factor to make it closer to holomorphic:
\begin{equation}
  \label{eq:294}
  \begin{aligned}
     & \int  \paren{\f{1}{\abs{Z_{,\aa}}^2} \partial_\aa \paren{\f{\bar{Z}_{t,\aa}}{\bar{Z}_{,\aa}}}} \abs{D_\aa^2 \bar{Z}_t}^2 d\aa \\& = \int  \paren{\f{1}{\abs{Z_{,\aa}}^2} \f{Z_{,\aa}}{\bar{Z}_{,\aa}} \partial_\aa \paren{\f{\bar{Z}_{t,\aa}}{Z_{,\aa}}}} \abs{D_\aa^2 \bar{Z}_t}^2 d\aa
 + \int  \paren{\f{1}{\abs{Z_{,\aa}}^2} \paren{\partial_\aa \f{Z_{,\aa}}{\bar{Z}_{,\aa}}}  \paren{\f{\bar{Z}_{t,\aa}}{Z_{,\aa}}}} \abs{D_\aa^2 \bar{Z}_t}^2 d\aa
\\ & =\int  \paren{\f{1}{\bar{Z}_{,\aa}^2} \partial_\aa D_\aa \bar{Z}_t} \abs{D_\aa^2 \bar{Z}_t}^2 d\aa
 - \int  \paren{\f{1}{\abs{Z_{,\aa}}^2}  \paren{\partial_\aa \f{Z_{tt} + i}{\bar{Z}_{tt} - i}}  D_\aa \bar{Z}_t} \abs{D_\aa^2 \bar{Z}_t}^2 d\aa,
\end{aligned}
\end{equation}
where we have used \eqref{eq:179} to replace the $\f{Z_{,\aa}}{\bar{Z}_{,\aa}}$ with $- \f{Z_{tt} + i}{\bar{Z}_{tt} - i}$. We can estimate the second term by
\begin{equation}
  \label{eq:304}
      \abs{\int  \paren{\f{1}{\abs{Z_{,\aa}}^2}  \paren{\partial_\aa \f{Z_{tt} + i}{\bar{Z}_{tt} - i}}  D_\aa \bar{Z}_t} \abs{D_\aa^2 \bar{Z}_t}^2 d\aa}
 \lec \nm{D_\aa Z_{tt}}_{L^\infty} \nm{D_\aa \bar{Z}_t}_{L^\infty} \nm{D_\aa^2 \bar{Z}_t}_{L^2}^2,
\end{equation}
where we have used $\f{1}{A_1} = \abs{\f{1}{Z_{,\aa}(Z_{tt} + i)}}$ \eqref{eq:251}, $A_1 \ge 1$ \eqref{eq:394}, and \eqref{eq:100}.

It therefore remains only to control the first term on the RHS of \eqref{eq:294}.  Now we take advantage of holomorphicity. We rewrite this as
\begin{equation}
  \label{eq:306}
     \int  \paren{\f{1}{\bar{Z}_{,\aa}^2} \partial_\aa D_\aa \bar{Z}_t} \abs{D_\aa^2 \bar{Z}_t}^2 d\aa
      = \int \paren{(\P_A + \P_H) \paren{\f{1}{\bar{Z}_{,\aa}} \partial_\aa D_\aa \bar{Z}_t} }\paren{\bar{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t}} \P_H D_\aa^2\bar{Z}_t d\aa,
\end{equation}
where we have used \eqref{eq:315} to insert $\P_H$ in front of the $D_\aa^2 \bar{Z}_t$ and decomposed the first factor into the holomorphic and antiholomorphic projections. Now we use the adjoint property \eqref{eq:471} to turn the $\P_H$ into a $\P_A$ on the opposite factors, and control using Cauchy-Schwarz:
\begin{equation}
  \label{eq:308}
  \begin{aligned}
     &\abs{\int \paren{(\P_A + \P_H) \paren{\f{1}{\bar{Z}_{,\aa}} \partial_\aa D_\aa \bar{Z}_t} }\paren{\bar{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t}}  \P_H D_\aa^2\bar{Z}_t d\aa}
\\ & \lec \nm{ \P_A\braces{ \paren{(\P_A + \P_H)\paren{ \f{1}{\bar{Z}_{,\aa}} \partial_\aa D_\aa \bar{Z}_t} }\paren{\bar{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t}}}}_{L^2} \nm{D_\aa^2 \bar{Z}_t}_{L^2}.
  \end{aligned}
\end{equation}
It now remains only to control this first factor.

First we consider the term with the $\P_H$. In this case, we can rewrite this as a commutator:
\begin{equation}
  \label{eq:353}
  \begin{aligned}
   \P_A\braces{ \paren{\P_H \paren{\f{1}{\bar{Z}_{,\aa}} \partial_\aa D_\aa \bar{Z}_t}} \paren{\bar{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t}}} & = \frac12 \bracket{\bar{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t},\HH}\P_H \paren{\f{1}{\bar{Z}_{,\aa}} \partial_\aa D_\aa \bar{Z}_t }
\\ & + \f{1}{4} \bar{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t} \paren{\avg \f{1}{\bar{Z}_{,\aa}} \partial_\aa D_\aa \bar{Z}_t}.
  \end{aligned}
\end{equation}
Here, the mean term appears because of \eqref{eq:159}.  We now use commutator estimate \eqref{eq:96} for the first term and H\"older for the second term, to conclude that
\begin{equation}
  \label{eq:355}
  \begin{split}
   & \nm{\P_A\braces{ \paren{\P_H \paren{\f{1}{\bar{Z}_{,\aa}} \partial_\aa D_\aa \bar{Z}_t}} \paren{\bar{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t}}}}_{L^2} \\& \lec  \nm{\bar{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t}}_{\dot{H}^{1/2}} \nm{\P_H \paren{\f{1}{\bar{Z}_{,\aa}} \partial_\aa D_\aa \bar{Z}_t}}_{L^2}
 + \nm{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t}_{L^2} \nm{\f{1}{\bar{Z}_{,\aa}} \partial_\aa D_\aa \bar{Z}_t}_{L^1}
\\ & \lec  \nm{D_\aa^2 \bar{Z}_t}_{L^2} \paren{ \nm{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t}_{\dot{H}^{1/2}} + \nm{\f{1}{Z_{,\aa}}}_{L^\infty} \nm{D_\aa^2 \bar{Z}_t}_{L^2}}.
\end{split}
\end{equation}

Finally, we consider the $\P_A$ term in the first factor on the RHS of \eqref{eq:308}. By the $L^2$ boundedness of $\P_A$,
it suffices to control
\begin{equation}
  \label{eq:360}
  \begin{aligned}
    &\nm{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t(I-\HH)\paren{ \f{1}{\bar{Z}_{,\aa}} \partial_\aa D_\aa \bar{Z}_t}}_{L^2} \\ & \lec \nm{ D_\a^2 \bar{Z}_t \bracket{\f{1}{Z_{,\aa}},\HH} \f{1}{\bar{Z}_{,\aa}} \partial_\aa D_\aa \bar{Z}_t}_{L^2}
 + \nm{D_\aa^2 \bar{Z}_t (I-\HH) \f{1}{\bar{Z}_{,\aa}} D_\aa^2 \bar{Z}_t}_{L^2}
\\ & \lec \nm{D_\a^2 \bar{Z}_t}_{L^2} \nm{ \bracket{\f{1}{Z_{,\aa}},\HH} \f{1}{\bar{Z}_{,\aa}} \partial_\aa D_\aa \bar{Z}_t}_{L^\infty}
 + \nm{D_\a^2 \bar{Z}_t}_{L^2} \nm{ \bracket{\f{1}{\bar{Z}_{,\aa}},\HH} D_\aa^2 \bar{Z}_t}_{L^\infty}
\\ & \lec \nm{D_\a^2 \bar{Z}_t}_{L^2}^2 \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2},
\end{aligned}
\end{equation}
where we've used \eqref{eq:315} to get the second commutator and used commutator estimate \eqref{eq:32}.

We now combine our estimates, concluding that
\begin{equation}
  \label{eq:368}
  \begin{aligned}
    \Re \int i &\paren{\f{1}{\bar{z}_\a} \partial_\a \f{h_\a}{z_\a}}_t \abs{\th}^2 h_\a d\a \ \lec \eqref{eq:169} + \eqref{eq:232}
\\ & \lec \eqref{eq:169}  + \eqref{eq:237} + \eqref{eq:304}
 + \nm{D_\aa^2 \bar{Z}_t}_{L^2} \cdot (\eqref{eq:355} + \eqref{eq:360})
\\ & \lec C(E).
  \end{aligned}
\end{equation}

\section{Controlling \texorpdfstring{$\Re  \int i \paren{\f{h_\a^2}{\abs{z_\a}^2}}_t  \th_\a \bar{\th} d\a$}{Re int i (ha2/za2)t theta a bar theta da}}
\label{sec:48}
We now show that we can control the following term from the RHS of \eqref{eq:66}:
\begin{equation}
  \label{eq:78}
  \Re  \int i \paren{\f{h_\a^2}{\abs{z_\a}^2}}_t  \th_\a \bar{\th} d\a = \Re \int i \paren{2 \f{h_{t\a}}{h_\a} - 2 \Re D_\a z_t} \f{h_\a^2}{\abs{z_{\a}}^2} \th_\a \bar{\th} d\a.
\end{equation}
Here, all results will be expressed in terms of general energy $E_{a,\th}$ for $\th$ satisfying $(I-\HH) (\th \circ h\i) = 0$ and $\bound{\th} = 0$, rather than specifying $\th = D_\a^2 \bar{z}_t$.\footnote{$D_\a^2\bar z_t$ satisfies $(I-\HH)D_\a^2 \bar{z}_t\circ h^{-1}=0, \bound{D_\a^2 \bar{z}_t}=0$; see \eqref{eq:1026}.}

We begin by rewriting this as
\begin{equation}
  \label{eq:80}
  \begin{aligned}
     \Re \int i \paren{2 \f{h_{t\a}}{h_\a} - 2 \Re D_\a z_t} \f{h_\a^2}{\abs{z_{\a}}^2} \th_\a \bar{\th} d\a & = \Re \int 2 i\paren{\f{h_{t\a}}{h_\a} - \Re D_\a z_t}\paren{\partial_\a \paren{\th \f{h_\a}{z_\a}}} \bar{\th} \f{h_\a}{\bar{z}_\a} d\a
\\ & - \Re \int 2i \paren{\f{h_{t\a}}{h_\a} - \Re D_\a z_t} \paren{\f{h_\a}{\bar{z}_\a} \partial_\a \f{h_\a}{z_\a}} \th \bar{\th} d\a
\\ & = I + II.
\end{aligned}
\end{equation}

II is easy to control, via H\"older and change of variables to Riemannian coordinates:
\begin{equation}
  \label{eq:82}
  \begin{aligned}
    \abs{II} & \lec \paren{\nm{\f{h_{t\a}}{h_\a}}_{L^\infty} + \nm{D_\a z_t}_{L^\infty}}\nm{D_\aa \f{1}{Z_{,\aa}}}_{L^\infty}  \nm{A_1}_{L^\infty} E_{a,\th}.
  \end{aligned}
\end{equation}
Therefore, we can focus on $I$ from \eqref{eq:80}.
We introduce the following notations:
\begin{equation}
  \label{eq:83}
  \psi := \paren{\f{h_\a}{z_\a} \th} \circ h\i; \quad \Th := \th \circ h\i; \quad B := \paren{\f{h_{t\a}}{h_\a} - \Re D_\a z_t} \circ h\i=(h_t\circ h^{-1})_\aa-\Re D_\aa Z_t.
\end{equation}
We know $(I-\HH) \Th = 0$, $\bound{\Th}=0$, and
\begin{equation}
 \label{eq:555}
  \bound{B} = 0, \quad  \bound{\psi} = 0\\
\end{equation}
\begin{equation}
  \label{eq:558}
  \quad (I-\HH) \psi = 0
 \end{equation}
\begin{equation}
  \label{eq:565}
 \nm{\psi}_{\dot{H}^{1/2}}+ \nm{A_1}_{L^\infty}^{-1/2}\nm{\Th}_{L^2} \lec E_{a,\th}^{1/2},
\end{equation}
where \eqref{eq:555} follows from \eqref{eq:567}, \eqref{eq:570}, $\bound{\th}=0$, and \eqref{eq:438};
\eqref{eq:558} follows from \eqref{eq:516},  $(I-\HH) \Th = 0$ and  principle no.2~in \S \ref{sec:20} (and for $\th=D_\a^2\bar z_t$ specifically from \eqref{eq:456});
and \eqref{eq:565} is immediate from the definition of $E_{a, \theta}$ and change of variables. Upon changing variables, we can write $I= \Re \int 2i B (\partial_\aa \psi) \bar{\psi} d\aa $.

{\bf Step 1.} {\it{Green's identity.}}
We now show that we can control $I$ from \eqref{eq:80}. The main idea is to use Green's identity
to move the derivative from $\psi := \paren{\f{h_\a}{z_\a} \th} \circ h\i$ onto $B$. We note that $i\partial_\aa \psi = i \partial_\aa \HH \psi$ by \eqref{eq:558}, and that the operator $i \partial_\aa \HH$ corresponds to the Dirichlet-Neumann operator $\nabla_n$.\footnote{Recall that the Dirichlet-Neumann operator is defined by $\nabla_n f :=\nabla_n f^\hbar$,  the outward-facing normal derivative of $f^\hbar$, where $f^\hbar$ is the extension of $f$ that is harmonic and periodic in $P^-$ and tending to a constant at infinity.  For $f$ real-valued, we can derive this  by noting that $(I+\HH)f$ is holomorphic, so $i\partial_\aa(I+\HH)f = \nabla_n (I+\HH)f$. Taking real parts gives the identity.}
Letting  $\psi^\hbar$ and $B^\hbar$ be the (periodic) harmonic extension of $\psi$ and $B$ to $P^-$ respectively, we have
\begin{equation}

  \label{eq:86}
  I
   = \Re \int 2i B (\partial_\aa \psi) \bar{\psi} d\aa  = \Re \int 2 B (\nabla_n \psi) \bar{\psi} d\aa

= \int B \nabla_n (\abs{\psi^\hbar}^2) d\aa.

\end{equation}
By Green's identity,\footnote{Here, to justify Green's identity, we can map (biholomorphically) the space $P^- $ to the unit disk minus the slit, and then use the periodicity of all of the functions involved to consider the harmonic extensions of these functions to the whole unit disk.}
\begin{equation}
  \label{eq:550}
  \begin{aligned}
    \int B \nabla_n (\abs{\psi^\hbar}^2) d\aa & = \int (\nabla_n B) \abs{\psi}^2 d\aa + \int_{P^-} B^\hbar \Delta (\abs{\psi^\hbar}^2) dv
\\ & = I_1 + I_2.
  \end{aligned}
\end{equation}
We control the second term, $I_2$, by
\begin{equation}
  \label{eq:87}
  \begin{aligned}
       \abs{I_2} & = \abs{\int_{P^-} B^\hbar \Delta (\abs{\psi^\hbar}^2) dv}
 =   2\abs{ \int_{P^-} B^\hbar \abs{\nabla \psi^\hbar}^2 dv}
\\ &  \le 2\nm{B^\hbar}_{L^\infty} \int_{P^-} \abs{\nabla\psi^\hbar}^2 dv
 = \nm{B^\hbar}_{L^\infty}  \int_{P^-} \Delta (\abs{\psi^\hbar}^2)dv
\\ & = \nm{B}_{L^\infty} \int \nabla_n (\abs{\psi^\hbar}^2) d\aa
= 2\nm{B}_{L^\infty} \Re \int i (\partial_\aa \psi) \bar{\psi} d\aa,
\\ & =2 \nm{B}_{L^\infty} \nm{\psi}_{\dot{H}^{1/2}}\lec \paren{\nm{\f{h_{t\a}}{h_\a}}_{L^\infty} + \nm{D_\aa Z_t}_{L^\infty}}\nm{\psi}_{\dot{H}^{1/2}}
     \end{aligned}
   \end{equation}
by the maximum principle and another application of Green's identity.

{\bf Step 2.} {\it Controlling $I_1$.}
We are left from Step 1 with controlling
\begin{equation}
  \label{eq:89}
  \begin{aligned}
   I_1 & =  \int (\nabla_n B) \abs{\psi}^2 d\aa
   = \Re \int (i  \partial_\aa \HH    B) \abs{\psi}^2 d\aa
    \\ & = \Re \int (i  \HH\partial_\aa    B) \abs{\psi}^2 d\aa
      = \Re \int \f{1}{Z_{,\aa}} (i\HH \partial_\aa B) \Th \bar{\psi} d\aa,
\end{aligned}
\end{equation}
where we have commuted $\partial_\aa$ outside the $\HH$ by \eqref{eq:176} since $\bound{B} = 0$ \eqref{eq:555}.

We commute the $\f{1}{Z_{,\aa}}$ factor inside the $\HH$, and then apply the adjoint property \eqref{eq:471}:
\begin{equation}
  \label{eq:90}
  \begin{aligned}
    I_{1} & = \Re \int i\paren{\bracket{\f{1}{Z_{,\aa}}, \HH}\partial_\aa B} \Th \bar{\psi} d\aa + \Re \int i \paren{\HH \paren{\f{1}{Z_{,\aa}} \partial_\aa B}} \Th \bar{\psi} d\aa
\\ & = \Re \int i\paren{\bracket{\f{1}{Z_{,\aa}}, \HH}\partial_\aa B} \Th \bar{\psi} d\aa - \Re \int i \paren{\f{1}{Z_{,\aa}} \partial_\aa B} \HH \paren{\Th \bar{\psi}} d\aa
\\ & = \Re \int i\paren{\bracket{\f{1}{Z_{,\aa}}, \HH}\partial_\aa B} \Th \bar{\psi} d\aa + \Re \int i \paren{\f{1}{Z_{,\aa}} \partial_\aa B} \bracket{\bar{\psi},\HH} \Th d\aa - \Re \int i \paren{\f{1}{Z_{,\aa}} \partial_\aa B}  \bar{\psi} \HH \Th d\aa
\\ & = I_{11} + I_{12} + I_{13}.
\end{aligned}
\end{equation}

Observe that because $\HH \Th = \Th$,
\begin{equation}
  \label{eq:559}

    I_{13}  =  -\Re \int i \paren{\f{1}{Z_{,\aa}} \partial_\aa B}  \bar{\psi} \Th d\aa
      = -\Re \int i \paren{\partial_\aa B} \abs{\psi}^2 d\aa
     = 0,

\end{equation}
since $B \in \R$. It remains to control $I_{11}$ and $I_{12}$.

We use Cauchy-Schwarz and then the  $\dot{H}^{1/2} \times L^2$ commutator estimate \eqref{eq:96} to control $I_{12}$:
\begin{equation}
  \label{eq:92}

    \abs{ I_{12}} \le \nm{D_\aa B}_{L^2} \nm{[\bar{\psi},\HH] \Th}_{L^2}
     \lec \nm{D_\aa B}_{L^2}  \nm{\bar{\psi}}_{\dot{H}^{1/2}} \nm{\Th}_{L^2}.

\end{equation}
We have controlled $\nm{\bar{\psi}}_{\dot{H}^{1/2}}$ and $\nm{\Th}_{L^2}$ at \eqref{eq:565}, and we will control $\nm{D_\aa B}_{L^2}$ by \eqref{eq:546} in Step 3 below.

It remains to control $I_{11}$ from \eqref{eq:90}.  Here we use Proposition \ref{prop:iden}, identity \eqref{eq:19}.  Because $(I-\HH) \f{1}{Z_{,\aa}} = \avg \f{1}{Z_{,\aa}}$ \eqref{eq:516} and $\avg \partial_\aa B=0$ by \eqref{eq:555}, we can rewrite
\begin{equation}
  \label{eq:562}

    \bracket{\f{1}{Z_{,\aa}}, \HH}\partial_\aa B =
    \P_H  \paren{\bracket{\f{1}{Z_{,\aa}}, \HH}\partial_\aa B} - \f{1}{2}\avg D_\aa B.

\end{equation}
We plug \eqref{eq:562} into  $I_{11}$, and then use adjoint property \eqref{eq:471}:
\begin{equation}
  \label{eq:563}
  \begin{aligned}
    I_{11}
    & = \Re \int i \braces{\P_H  \paren{\bracket{\f{1}{Z_{,\aa}}, \HH}\partial_\aa B}}\Th \bar{\psi} d\aa - \Re \braces{ \paren{ \f{1}{2}\avg D_\aa B} \int i \Th \bar{\psi} d\aa}
\\ & = \Re \int i \paren{\bracket{\f{1}{Z_{,\aa}}, \HH}\partial_\aa B}\P_A \paren{\Th \bar{\psi}} d\aa - \Re\braces{ \paren{ \f{1}{2}\avg D_\aa B} \int i \Th \bar{\psi} d\aa}.

  \end{aligned}
\end{equation}
To control the first term, we use Cauchy-Schwarz, and then control the first factor with the $L^2 \times L^\infty$ estimate
\eqref{eq:228} and control the second factor by rewriting it as a commutator by \eqref{eq:558} and then using the $\dot{H}^{1/2} \times L^2$ estimate
\eqref{eq:96}. We use \eqref{eq:251} to rewrite $\f{1}{\bar{Z}_{,\aa}} = -i\f{Z_{tt} + i}{A_1}$ in the second term.
\begin{equation}
  \label{eq:564}
  \begin{aligned}
    \abs{I_{11}} & \le \nm{\bracket{\f{1}{Z_{,\aa}}, \HH}\partial_\aa B}_{L^2} \nm{\P_A \paren{\Th \bar{\psi}}}_{L^2}+\abs{\paren{ \f{1}{2}\avg D_\aa B} \int i \Th \bar{\psi} d\aa}
\\ & \le \nm{\bracket{\f{1}{Z_{,\aa}}, \HH}\partial_\aa B}_{L^2} \nm{\f{1}{2}\bracket{\bar{\psi},\HH} \Th}_{L^2}
+ \nm{D_\aa B}_{L^2} \int \abs{\f{Z_{tt} + i}{A_1} \abs{\Th}^2} d\aa\\ & \lec \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}\nm{B}_{L^\infty} \nm{\bar{\psi}}_{\dot{H}^{1/2}} \nm{\Th}_{L^2}+\nm{D_\aa B}_{L^2} \nm{Z_{tt} + i}_{L^\infty} E_{a,\th}.
  \end{aligned}
\end{equation}
We have controlled all the quantities on the last line, except for $\nm{D_\aa B}_{L^2}$.

{\bf Step 3.} {\it Controlling $\nm{D_\aa B}_{L^2}$.}
We must control $\nm{D_\aa B}_{L^2}$, where $B$ is as defined by \eqref{eq:83}. By \eqref{eq:249},
\begin{equation}
  \label{eq:530}
  B=  (h_t \circ h\i)_\aa - \Re D_\aa Z_t =
   \Re D_\aa Z_t + \Re \braces{ -\bracket{\f{1}{\bar{Z}_{,\aa}},\HH} \bar{Z}_{t,\aa} + [Z_t,\HH] \partial_\aa \f{1}{Z_{,\aa}}}.
\end{equation}
Therefore, noting that $\abs{\partial_\aa \Re f} \le \abs{\partial_\aa f}$ and so $\abs{D_\aa \Re f} \le \abs{D_\aa f}$,
\begin{equation}
  \label{eq:541}

  \nm{D_\aa B}_{L^2} \le
  \nm{D_\aa^2 Z_t}_{L^2} + \nm{D_\aa   \bracket{\f{1}{\bar{Z}_{,\aa}},\HH}   \bar{Z}_{t,\aa}}_{L^2} + \nm{D_\aa[ Z_t, \HH]\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}.

\end{equation}
We've controlled $\nm{D_\aa^2 Z_t}_{L^2}$, so it suffices to focus on the second and third terms. In what follows, we work on $D_\aa [f,\HH] \partial_\aa g$ for general functions $f$ and $g$ satisfying $\bound{f}=\bound{g}=0$. Once we have an appropriate estimate, we will apply it to $f=\f{1}{\bar{Z}_{,\aa}}$, $g=\bar{Z}_{t}$ for the second term, and $f= Z_t$, $g=\f{1}{Z_{,\aa}}$ for the third term.

We know
\begin{equation}
  \label{eq:1081}
  \begin{aligned}
      D_\aa [f,\HH] \partial_\aa g
     & = \f{1}{Z_{,\aa}} \partial_\aa \f{1}{\hdenomconst} \int (f(\aa ) - f(\bb)) \cot(\f{\pi}{2}(\aa -\bb)) \partial_\bb g(\bb) d\bb
    \\ & = \f{1}{Z_{,\aa}} (\partial_\aa f) \HH \partial_\aa g - \f{1}{Z_{,\aa}} \f{1}{\hdenomconst} \int \f{\pi}{2} \f{f(\aa) - f(\bb)}{ \sin^2(\f{\pi}{2}(\aa - \bb))} \partial_\bb g(\bb)d\bb.
  \end{aligned}
\end{equation}
Via the boundedness of the Hilbert transform, we control the first of these terms by $\nm{D_\aa f}_{L^\infty} \nm{\partial_\aa g}_{L^2}$. Therefore, it suffices to focus on the second term.  We commute the $\f{1}{Z_{,\aa}}$ inside, getting
  \begin{equation}
    \label{eq:1091}
    -\f{\pi}{4i} \int \f{f(\aa) - f(\bb)}{\sin^2(\f{\pi}{2}(\aa - \bb))} D_\bb g(\bb) d\bb -\f{\pi}{4i} \int \f{(f(\aa) - f(\bb))}{\sin(\f{\pi}{2}(\aa - \bb))}\f{\paren{\f{1}{Z_{,\aa}}(\aa) - \f{1}{Z_{,\bb}}(\bb)}}{\sin(\f{\pi}{2}(\aa - \bb))} \partial_\bb g(\bb) d\bb.
  \end{equation}
We control the first term by \eqref{eq:258}:
\begin{equation}
  \label{eq:1101}
 \nm{ \f{\pi}{4i}\int \f{f(\aa) - f(\bb)}{\sin^2(\f{\pi}{2}(\aa - \bb))} D_\bb g(\bb) d\bb}_{L^2} \lec \nm{\partial_{\aa}f}_{L^2} \nm{D_\bb g}_{L^\infty}.
\end{equation}
We control the second term by \eqref{eq:113}:
\begin{equation}
  \label{eq:1111}
  \nm{\f{\pi}{4i}\int \f{(f(\aa) - f(\bb))}{ \sin(\f{\pi}{2}(\aa - \bb))}\f{\paren{\f{1}{Z_{,\aa}}(\aa) - \f{1}{Z_{,\bb}}(\bb)}}{\sin(\f{\pi}{2}(\aa - \bb))} \partial_\bb g(\bb) d\bb}_{L^2} \lec \nm{\partial_{\aa}f}_{L^2} \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}\nm{\partial_{\aa}g}_{L^2}.
\end{equation}
We conclude that
\begin{equation}
  \label{eq:5481}
  \begin{aligned}
    \nm{D_\aa [f, \HH] \partial_\aa g}_{L^2} & \lec \nm{D_\aa f}_{L^\infty} \nm{\partial_\aa g}_{L^2}+ \nm{D_\aa g}_{L^\infty} \nm{\partial_\aa f}_{L^2}+\nm{\partial_{\aa}f}_{L^2} \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}\nm{\partial_{\aa}g}_{L^2}.
\end{aligned}
\end{equation}

We can conclude from \eqref{eq:541} and \eqref{eq:5481} that
\begin{equation}
  \label{eq:546}

    \nm{D_\aa B}_{L^2}
 \lec \nm{D_\aa^2 Z_t}_{L^2} + \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2} \nm{D_\aa \bar{Z}_t}_{L^\infty} + \nm{Z_{t,\aa}}_{L^2}\paren{ \nm{D_\aa \f{1}{Z_{,\aa}}}_{L^\infty} + \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}^2}.

\end{equation}

{\bf Step 4.} {\it Conclusion.}
We now combine our various estimates. We have
\begin{equation}
  \label{eq:552}
  \begin{aligned}
    \abs{ \Re  \int i \paren{\f{h_\a^2}{\abs{z_\a}^2}}_t  \th_\a \bar{\th} d\a} & \le \abs{I} + \abs{II} \le \abs{I_1} +\abs{I_2}+\eqref{eq:82}
    \\ &\le \abs{I_{11}} +\abs{I_{12}}  +\eqref{eq:87}+\eqref{eq:82}
    \\ & \le \eqref{eq:564}+\eqref{eq:92}+\eqref{eq:87}+\eqref{eq:82},
 \end{aligned}
\end{equation}
where we use \eqref{eq:546} to control $\nm{D_\aa B}_{L^2}$.

In particular, by specifying $\th=D_\a^2\bar z_t$, we have
\begin{equation}\label{eq:1027}
\abs{ \Re  \int i \paren{\f{h_\a^2}{\abs{z_\a}^2}}_t  \th_\a \bar{\th} d\a}\lec C(E).
\end{equation}

\section{Controlling \texorpdfstring{$G_\th$}{G theta} of \texorpdfstring{$E_b$}{Eb}}\label{sec:E2}
By \eqref{eq:173}, we must control
\begin{equation}
  \label{eq:421}
  \paren{\int \f{1}{\af} \abs{ D_\a(-i \af_t \bar{z}_\a) + [\partial_t^2 + i \af \partial_\a, D_\a] \bar{z}_t}^2 d\a}^{1/2}.
\end{equation}

We control the commutator via \eqref{eq:53}:
\begin{equation}
  \label{eq:422}

  \paren{\int \f{1}{\af} \abs{[\partial_t^2 + i \af \partial_\a, D_\a] \bar{z}_t}^2 d\a}^{1/2}  \lec (\nm{D_\a z_{tt}}_{L^\infty} + \nm{D_\a z_t}_{L^\infty}^2) \nm{D_\a \bar{z}_t}_{L^2(\f{1}{\af} d\a)},
\end{equation}
where we have controlled all the quantities on the RHS in \S \ref{sec:quants}. We are left with the term $\paren{\int \f{1}{\af} \abs{D_\a(-i\af_t \bar{z}_\a)}^2 d\a}^{1/2}$.
Since ${\af}\abs{z_\a}^2=(A_1 \circ h) {h_\a}$ \eqref{eq:209}, $A_1\ge 1$  \eqref{eq:394}, and $(\af_t \bar{z}_\a)\circ h^{-1}=\AA_t \bar{Z}_\aa$ we have
\begin{equation}\label{eq:1033}
\paren{\int \f{1}{\af} \abs{D_\a(-i\af_t \bar{z}_\a)}^2 d\a}^{1/2}\le \paren{\int \f1{h_\a} \abs{\partial_\a(-i\af_t \bar{z}_\a)}^2 d\a}^{1/2}=\paren{\int \abs{\partial_\aa(-i\AA_t \bar{Z}_\aa)}^2 d\aa}^{1/2},
\end{equation}
 where in the second step we changed to  Riemannian coordinates. We write $-i\AA_t \bar{Z}_\aa$ as $\f{\AA_t}{\AA} (-i\AA \bar{Z}_\a)$ and apply $\partial_\aa$. Since $\bar{Z}_{tt}-i =-i \AA \bar{Z}_{,\aa}$ \eqref{eq:179}, we have
 \begin{equation}
  \label{eq:218}
 \partial_\aa \paren{-i\AAt \bar{Z}_{,\aa}}=  (-i \AA \bar{Z}_{,\aa})\partial_\aa \paren{\f{\AAt}{\AA}} + \f{\AAt}{\AA} \bar{Z}_{tt,\aa}.
\end{equation}
Therefore,
\begin{equation}\label{eq:214}
\paren{\int \abs{\partial_\aa(-i\AA_t \bar{Z}_\aa)}^2 d\aa}^{1/2}\le \paren{\int  \abs{\mathcal A \bar{Z}_\aa \partial_\aa \paren{\f{\mathcal A_t}{\mathcal A}}}^2 d\aa}^{1/2}+\nm{\f{\AA_t}{\AA}}_{L^\infty}\nm{
\bar{Z}_{tt,\aa}}_{L^2}.
\end{equation}
We controlled the factors in the second term on the RHS in \eqref{eq:406} and \eqref{eq:424}. We can therefore concentrate on the first term.

We  seek a way of writing $ \mathcal A \bar{Z}_\aa \partial_\aa \paren{\f{\AAt}{\AA}}$.  We will go through a slightly convoluted derivation that enables us to express this in terms of a commutator structure; we have to take care that we can invert $(I-\HH)$ and we want to make sure that the advantageous weight $\mathcal A \bar{Z}_\aa$ is placed appropriately.

Starting from \eqref{eq:218}, we replace the LHS by the derivative of the LHS of our quasilinear equation \eqref{eq:178}, and then apply $(I-\HH)$ to the equation. We get
\begin{equation}
  \label{eq:61}
  (I-\HH) \braces{(-i \AA \bar{Z}_{,\aa}) \partial_\aa \f{\AAt}{\AA}} = (I-\HH) \partial_\aa \paren{\bar{Z}_{ttt} + i \AA \bar{Z}_{t,\aa}} - (I-\HH) \braces{\f{\AAt}{\AA} \partial_\aa \bar{Z}_{tt}}.
\end{equation}
We want to move the factor  $\AA \bar{Z}_{,\aa}$ outside on the LHS, so we do this:
\begin{equation}
  \label{eq:69}
  (-i\AA \bar{Z}_{,\aa}) (I-\HH) \partial_\aa \f{\AAt}{\AA} = (I-\HH) \partial_\aa \paren{\bar{Z}_{ttt} + i \AA \bar{Z}_{t,\aa}} - (I-\HH)  \braces{\f{\AAt}{\AA} \partial_\aa \bar{Z}_{tt}} + [i\AA \bar{Z}_{,\aa},\HH] \partial_\aa \f{\AAt}{\AA}.
\end{equation}
Now $\f{\AAt}{\AA}$ is real and $\HH$ is purely imaginary, so $\abs{ \partial_\aa \f{\AAt}{\AA} }\le\abs{  (I-\HH) \partial_\aa \f{\AAt}{\AA} }$.
Taking absolute value on both sides,   we have
\begin{equation}
  \label{eq:482}
  \abs{\AA \bar{Z}_{,\aa} \partial_\aa \f{\AAt}{\AA}} \le \abs{(I-\HH) \partial_\aa \paren{\bar{Z}_{ttt} + i \AA \bar{Z}_{t,\aa}} - (I-\HH)  \braces{\f{\AAt}{\AA} \partial_\aa \bar{Z}_{tt}} + [i\AA \bar{Z}_{,\aa},\HH] \partial_\aa \f{\AAt}{\AA}}.
\end{equation}

We can easily control the $L^2$ norm of the second and third terms. By the $L^2$ boundedness of $\HH$ and H\"older for the second term and estimate \eqref{eq:228} for the third term, and since $i \AA \bar{Z}_{,\aa} = -(\bar{Z}_{tt} - i)$,
\begin{equation}
  \label{eq:227}
  \nm{ - (I-\HH)  \braces{\f{\AAt}{\AA} \bar{Z}_{tt,\aa}} - [i\AA \bar{Z}_{,\aa},\HH] \partial_\aa \f{\AAt}{\AA} }_{L^2} \lec \nm{Z_{tt,\aa}}_{L^2} \nm{\f{\AAt}{\AA}}_{L^\infty}.
\end{equation}

We can now focus on controlling
\begin{equation}
  \label{eq:74}
  \nm{(I-\HH) \partial_\aa \paren{\bar{Z}_{ttt} + i \AA \bar{Z}_{t,\aa}}}_{L^2}=\nm{\partial_\aa  (I-\HH) \paren{\bar{Z}_{ttt} + i \AA \bar{Z}_{t,\aa}}}_{L^2},
\end{equation}
where we used $\bound{\paren{\bar{Z}_{ttt} + i \AA \bar{Z}_{t,\aa}}} = 0$ \eqref{eq:483} to  commute $\partial_\aa$ outside $(I-\HH)$ by \eqref{eq:176}.

Recall from \eqref{eq:202} and $i \AA = \f{Z_{tt} + i}{Z_{,\aa}}$ that
\begin{equation}
  \label{eq:51}
  \bar{Z}_{ttt} + i \AA \bar{Z}_{t,\aa} = (D_\aa^2 \bar{Z}_t) Z_t^2 + 2 Z_t D_\aa(\bar{Z}_{tt} - (D_\aa \bar{Z}_t) Z_t) + 2 (D_\aa \bar{Z}_t) Z_{tt} + F_{tt} \circ Z + i  D_\aa \bar{Z}_t,
\end{equation}
where $F(z(\a,t),t) := \bar{z}_t(\a,t)$.
Under $(I-\HH)$, by \eqref{eq:315}, \eqref{eq:313} and \eqref{eq:504} the last two terms disappear and the rest turn into commutators:
\begin{equation}
  \label{eq:250}
   (I-\HH) \paren{\bar{Z}_{ttt} + i \AA \bar{Z}_{t,\aa} } = [Z_t^2,\HH] D_\aa^2 \bar{Z}_t + 2 [Z_t, \HH] D_\aa (\bar{Z}_{tt} - (D_\aa \bar{Z}_t) Z_t) + 2 [Z_{tt},\HH] D_\aa \bar{Z}_t.
\end{equation}
Therefore, by \eqref{eq:74} and \eqref{eq:250}, we have to control the $L^2$ norm of
\begin{equation}
  \label{eq:163}
  \partial_\aa [Z_t^2,\HH] D_\aa^2 \bar{Z}_t + 2 \partial_\aa [Z_t, \HH] D_\aa (\bar{Z}_{tt} - (D_\aa \bar{Z}_t) Z_t) + 2 \partial_\aa [Z_{tt},\HH] D_\aa \bar{Z}_t.
\end{equation}
We use the identity
\begin{equation}
  \label{eq:269}
  \partial_\aa [f,\HH] g = f' \HH g - \f{1}{\hdenomconst} \int \f{\pi}{2} \f{f(\aa) - f(\bb)}{\sin^2(\f{\pi}{2}(\aa -\bb))} g(\bb) d\bb
\end{equation}
to expand out each term in \eqref{eq:163}, and use \eqref{eq:315} and \eqref{eq:313} to remove the  $\HH$s from the RHS. We get
\begin{equation}
  \label{eq:36}
\begin{aligned}
\eqref{eq:163}&=
   2Z_t Z_{t,\aa} D_\aa^2 \bar Z_t+2 Z_{t,\aa} D_\aa(\bar{Z}_{tt} - (D_\aa \bar{Z}_t) Z_t) + 2 Z_{tt,\a}  D_\aa \bar{Z}_t
   - \f{\pi}{4i} \int  \f{Z_t^2(\aa) - Z_t^2(\bb)}{\sin^2(\f{\pi}{2}(\aa - \bb))} D_\bb^2 \bar Z_t d\bb
\\& - \f{\pi}{2i} \int  \f{Z_t(\aa) - Z_t(\bb)}{\sin^2(\f{\pi}{2}(\aa - \bb))} D_\bb (\bar{Z}_{tt} - (D_\bb \bar{Z}_t) Z_t) d\bb
 - \f{\pi}{2i} \int  \f{Z_{tt}(\aa) - Z_{tt}(\bb)}{\sin^2(\f{\pi}{2}(\aa -\bb))} D_\bb \bar{Z}_t d\bb.
\end{aligned}
\end{equation}
We expand out the RHS. We note that certain terms cancel out with others, and we further observe the following identity:
\begin{equation}
  \label{eq:235}
     \int \f {f^2(\aa) - f^2(\bb) } {\sin^2(\f{\pi}{2}(\aa - \bb)) }g(\bb) d\bb -2\int \f {(f(\aa) - f(\bb)) f(\bb) } {\sin^2(\f{\pi}{2}(\aa - \bb)) }g(\bb) d\bb
 =  \int \f{(f(\aa) - f(\bb))^2 }{\sin^2(\f{\pi}{2}(\aa - \bb))} g(\bb) d\bb.
\end{equation}
We have
\begin{equation}\label{eq:1032}
\begin{aligned}
\eqref{eq:163}&= 2 Z_{t,\aa} (D_\aa \bar{Z}_{tt} - (D_\aa \bar{Z}_t) D_\aa Z_t) + 2 Z_{tt,\a}  D_\aa \bar{Z}_t
   - \f{\pi}{4i} \int  \f{(Z_t(\aa) - Z_t(\bb))^2}{\sin^2(\f{\pi}{2}(\aa - \bb))} D_\bb^2 \bar Z_t d\bb
\\& - \f{\pi}{2i} \int \f{Z_t(\aa) - Z_t(\bb)}{\sin^2(\f{\pi}{2}(\aa - \bb))} (D_\bb \bar{Z}_{tt} - (D_\bb \bar{Z}_t) D_\bb Z_t) d\bb
 - \f{\pi}{2i} \int  \f{Z_{tt}(\aa) - Z_{tt}(\bb)}{\sin^2(\f{\pi}{2}(\aa -\bb))} D_\bb \bar{Z}_t d\bb.
\end{aligned}
\end{equation}
We now apply H\"older to the first two terms, \eqref{eq:113} to the third term, and \eqref{eq:258} to the last two terms.  We get
\begin{equation}
  \label{eq:310}
  \begin{aligned}
   &\nm{(I-\HH) \partial_\aa \paren{\bar{Z}_{ttt} + i \AA \bar{Z}_{t,\aa}}}_{L^2} =\nm{\eqref{eq:163}}_{L^2} \\&\lec \nm{Z_{t,\aa}}_{L^2} (\nm{D_\aa \bar{Z}_{tt}}_{L^\infty} + \nm{D_\aa \bar{Z}_t}_{L^\infty}^2) + \nm{Z_{tt,\aa}}_{L^2} \nm{D_\aa \bar{Z}_t}_{L^\infty}
 +  \nm{Z_{t,\aa}}_{L^2}^2 \nm{D_\aa^2 \bar{Z}_t}_{L^2}.
 \end{aligned}
\end{equation}

We now combine our various estimates. We have
\begin{equation}\label{eq:327}
\nm{\AA \bar{Z}_{,\aa} \partial_\aa \f{\AAt}{\AA}}_{L^2}\le \eqref{eq:227}+\eqref{eq:310}\lec C(E)
\end{equation}
and
\begin{equation}\label{eq:311}
\begin{aligned}
 &\paren{\int \f{1}{\af} \abs{ D_\a(-i \af_t \bar{z}_\a) + [\partial_t^2 + i \af \partial_\a, D_\a] \bar{z}_t}^2 d\a}^{1/2}\le \eqref{eq:422}+\eqref{eq:214}\\ & \le \eqref{eq:422}+\eqref{eq:327}+\nm{\f{\AA_t}{\AA}}_{L^\infty} \nm{ \bar{Z}_{tt,\aa}}_{L^2}\lec C(E).
 \end{aligned}
 \end{equation}

We can now conclude that $\f d{dt} E_b$ is bounded by a polynomial of $E$.

We record here the estimate
\begin{equation}\label{eq:1034}
\nm{D_\aa \f{\AAt}{\AA}}_{L^2}\le \nm{\AA \bar{Z}_{,\aa} \partial_\aa \f{\AAt}{\AA}}_{L^2} \lec C(E),
\end{equation}
which holds because $\abs{\AA\bar{Z}_{,\aa}}=\f{A_1}{\abs{{Z}_{,\aa}}}\ge \f{1}{\abs{{Z}_{,\aa}}}$; we will use this in \S \ref{controlE1}.

\section{Controlling \texorpdfstring{$G_\th$}{G theta} of \texorpdfstring{$E_a$}{Ea}}\label{controlE1}
From \eqref{eq:66}, we must control
\begin{equation}
  \label{eq:399}
  \paren{\int  \abs{D_\a^2 (-i \af_t \bar{z}_\a) + [\partial_t^2 + i \af \partial_\a, D_\a^2]\bar{z}_t}^2\f{h_\a}{A_1 \circ h}d\a}^{1/2}.
\end{equation}

Recall that $A_1\ge 1$ \eqref{eq:394}. We control the commutator via  \eqref{eq:52} and H\"older:
\begin{equation}
  \label{eq:415}
  \begin{aligned}
 & \nm{[\partial_t^2 + i \af \partial_\a, D_\a^2]\bar{z}_t}_{L^2(\f{h_\a}{A_1 \circ h}d\a)}\le  \nm{[\partial_t^2 + i \af \partial_\a, D_\a^2]\bar{z}_t}_{L^2({h_\a}d\a)}\\& \lec  \nm{D_\a z_{tt}}_{L^\infty} \nm{D_\a^2 \bar{z}_t}_{L^2({h_\a}d\a)} + \nm{D_\a z_t}_{L^\infty}^2 \nm{D_\a^2 \bar{z}_t}_{L^2({h_\a}d\a)} + \nm{D_\a z_t}_{L^\infty}\nm{D_\a \partial_t D_\a \bar{z}_t}_{L^2({h_\a}d\a)} \\&+ \nm{D_\a^2 z_{tt}}_{L^2({h_\a}d\a)} \nm{D_\a \bar{z}_t}_{L^\infty} + \nm{D_\a z_t}_{L^\infty}^2 \nm{D_\a^2 z_t}_{L^2({h_\a}d\a)} + \nm{D_\a^2 z_t}_{L^2({h_\a}d\a)} \nm{D_\a \bar{z}_{tt}}_{L^\infty} \\& + \nm{D_\a z_t}_{L^\infty} \nm{D_\a^2 \bar{z}_{tt}}_{L^2({h_\a}d\a)}.
 \end{aligned}
\end{equation}
We have controlled all quantities on the RHS in \S \ref{sec:quants}. We are left with the term $\paren{\int \abs{D_\a^2 \paren{\af_t \bar{z}_\a}}^2\f{h_\a}{A_1 \circ h} d\a}^{1/2}$.

We know
\begin{equation}
  \label{eq:134a}
  \paren{\int \abs{D_\a^2 \paren{\af_t \bar{z}_\a}}^2\f{h_\a}{A_1 \circ h} d\a}^{1/2}\le \paren{\int \abs{D_\a^2 \paren{\af_t \bar{z}_\a}}^2{h_\a} d\a}^{1/2}=\paren{\int \abs{D_\aa^2 \paren{\AA_t \bar{Z}_\a}}^2 d\aa}^{1/2},
\end{equation}
where we  changed to Riemannian coordinate in the second step. We will now  focus on estimating
\begin{equation}
  \label{eq:10}
  \paren{\int \abs{D_\aa^2 \paren{\AA_t \bar{Z}_\a}}^2 d\aa}^{1/2}.
\end{equation}

Our plan will be to rewrite $D_\aa^2 \paren{\AA_t \bar{Z}_\a}$ in a form which we can control via our energy. We will take advantage of the fact that $\f{\AA_t}{\AA}$ is  real-valued and use $(I-\HH)$ to turn the main part of $D_\aa^2 \paren{\AA_t \bar{Z}_\a}$  into a series of commutators, as we did in \S \ref{sec:28}.  Getting to this point will require a long and somewhat convoluted series of calculations.

We begin by writing $\AA_t \bar{Z}_\a = \paren{\f{\AA_t}{\AA}} \AA \bar{Z}_\a$.  By the product rule
\begin{equation}
  \label{eq:46}
  D_\aa^2 \paren{\AA_t \bar{Z}_\a} = \paren{D_\aa^2 \paren{\f{\AA_t}{\AA}}} \AA \bar{Z}_\a + 2 D_\aa \paren{\f{\AA_t}{\AA}} D_\aa(\AA \bar{Z}_\a) + \f{\AA_t}{\AA} D_\aa^2 (\AA \bar{Z}_\a).
\end{equation}
We can handle the second and third terms directly, using $\AA \bar{Z}_\a = i(\bar{Z}_{tt} - i)$ \eqref{eq:179}:
\begin{equation}
  \label{eq:316}
    \nm{2D_\aa \paren{\f{\AA_t}{\AA}}D_\aa(\AA \bar{Z}_\a)+\f{\AA_t}{\AA} D_\a^2 (\AA \bar{Z}_\a)}_{L^2}     \le
    2\nm{D_\aa \paren{\f{\AA_t}{\AA}}}_{L^2} \nm{D_\a \bar{Z}_{tt}}_{L^\infty}+\nm{\f{\AA_t}{\AA}}_{L^\infty} \nm{D_\a^2 \bar{Z}_{tt}}_{L^2},
\end{equation}
where we have controlled all the quantities on the RHS in \S \ref{sec:quants} and in \eqref{eq:424} and \eqref{eq:1034}.   It therefore suffices to focus on the first term on the RHS of \eqref{eq:46}, $ \paren{D_\aa^2 \paren{\f{\AA_t}{\AA}}} \AA \bar{Z}_\a=i(\bar Z_{tt}-i) D_\aa^2 \paren{\f{\AA_t}{\AA}} $.

{\bf Step 1.} {\it Controlling $\paren{D_\aa^2 \paren{\f{\AA_t}{\AA}}} \AA \bar{Z}_\a$ or $i(\bar Z_{tt}-i) D_\aa^2 \paren{\f{\AA_t}{\AA}}$.}
We now rearrange this term so that we can apply $(I-\HH)$ in a way so that we will be able to invert the operator by taking real parts.  Note that $\f{\AA_t}{\AA}$ is purely real. Unfortunately, our derivative $D_\aa = \f{1}{Z_{,\aa}} \partial_\aa$ is not purely real.  To get around this, we factor the derivative into a real derivative and a complex modulus-one weight.  Recall our notation

 $ \abs{D_\aa} = \f{1}{\abs{Z_{,\aa}}} \partial_\aa$.

Since $D_\aa = \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}} \abs{D_\aa}$, we rewrite
\begin{equation}
  \label{eq:320}
  D_\aa^2 = \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}^2 \abs{D_\aa}^2 + \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}\paren{\abs{D_\aa} \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}} \abs{D_\aa}.
\end{equation}
Therefore,
\begin{equation}
  \label{eq:321}
i(\bar Z_{tt}-i)    D_\aa^2 \f{\AA_t}{\AA} = i(\bar Z_{tt}-i)  \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}^2 \abs{D_\aa}^2 \f{\AA_t}{\AA} + i(\bar Z_{tt}-i)  \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}\paren{\abs{D_\aa} \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}} \abs{D_\aa}\f{\AA_t}{\AA}.
\end{equation}
It will turn out that we'll want to replace $\abs{D_\aa}^2$ in the first term of the RHS of \eqref{eq:321} with $\partial_\aa \paren{\f{1}{\abs{Z_{,\aa}}} \abs{D_\aa}}$, since this will give us the proper commutator estimate,\footnote{See footnote \ref{fn-expl} below for a further explanation.} and it further turns out that we want to make the switch now rather than later, when it would be in commutator form. Doing this, we get
\begin{equation}
  \label{eq:149}
  \begin{split}
    i(\bar{Z}_{tt} - i) D_\aa^2 \f{\AAt}{\AA} = i(\bar{Z}_{tt} - i) \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}^2 \partial_\aa \paren{\f{1}{\abs{Z_{,\aa}}} \abs{D_\aa} \f{\AAt}{\AA}}  \\ \underbrace{- i(\bar{Z}_{tt} - i) \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}^2 \paren{\partial_\aa \f{1}{\abs{Z_{,\aa}}}} \abs{D_\aa} \f{\AAt}{\AA}  + i(\bar{Z}_{tt} - i) \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}\paren{\abs{D_\aa} \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}} \abs{D_\aa}\f{\AAt}{\AA}}_{e},
  \end{split}
\end{equation}
where we will use
\begin{equation}
  \label{eq:324}
  e := -i(\bar{Z}_{tt} - i) \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}^2 \paren{\partial_\aa \f{1}{\abs{Z_{,\aa}}}} \abs{D_\aa} \f{\AAt}{\AA}  +i(\bar{Z}_{tt} - i) \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}\paren{\abs{D_\aa} \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}} \abs{D_\aa}\f{\AAt}{\AA}
\end{equation}
as an abbreviation to save space and de-emphasize the less central error terms, which we will control directly, below at \eqref{eq:325}. We now apply $(I-\HH)$ to both sides of \eqref{eq:149}:
\begin{equation}
  \label{eq:55}
  (I-\HH) \braces{i(\bar{Z}_{tt} - i) D_\aa^2 \f{\AAt}{\AA}} = (I-\HH) \braces{i(\bar{Z}_{tt} - i) \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}^2 \partial_\aa \paren{\f{1}{\abs{Z_{,\aa}}} \abs{D_\aa} \f{\AAt}{\AA}}} + (I-\HH) e.
\end{equation}
Observe that the first term on the RHS is purely real, except for the controllable factor $-i(\bar{Z}_{tt} - i) \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}^2$. We commute that part outside the $(I-\HH)$.  We get
\begin{equation}
  \label{eq:56}
  \begin{aligned}
      (I-\HH) \braces{i(\bar{Z}_{tt} - i) D_\aa^2 \f{\AAt}{\AA}} & = i(\bar{Z}_{tt} - i) \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}^2  (I-\HH) \partial_\aa \paren{\f{1}{\abs{Z_{,\aa}}} \abs{D_\aa} \f{\AAt}{\AA}}
\\ & + \bracket{i(\bar{Z}_{tt} - i) \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}^2,\HH}  \partial_\aa \paren{\f{1}{\abs{Z_{,\aa}}} \abs{D_\aa} \f{\AAt}{\AA}} + (I-\HH) e.
  \end{aligned}
\end{equation}
Because $\HH$ is purely imaginary,  $\abs{\partial_\aa \paren{\f{1}{\abs{Z_{,\aa}}} \abs{D_\aa} \f{\AAt}{\AA}}}\le \abs{(I-\HH) \partial_\aa \paren{\f{1}{\abs{Z_{,\aa}}} \abs{D_\aa} \f{\AAt}{\AA}} }$. By taking absolute values, we have
\begin{multline}
  \label{eq:58}
  \abs{i(\bar{Z}_{tt} - i) \partial_\aa \paren{\f{1}{\abs{Z_{,\aa}}} \abs{D_\aa} \f{\AAt}{\AA}}}
 \\ \le \abs{(I-\HH) \braces{i(\bar{Z}_{tt} - i) D_\aa^2 \f{\AAt}{\AA}} - \bracket{i(\bar{Z}_{tt} - i) \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}^2,\HH}  \partial_\aa \paren{\f{1}{\abs{Z_{,\aa}}} \abs{D_\aa} \f{\AAt}{\AA}} - (I-\HH) e}.
\end{multline}

Now we may begin controlling these terms. Recall that what we needed to control was the $L^2$ norm of  \eqref{eq:149}. We can estimate this by
\begin{equation}
  \label{eq:71}
  \begin{aligned}
    \nm{-i(\bar{Z}_{tt} - i) D_\aa^2 \f{\AAt}{\AA} }_{L^2} &   \lec \nm{i(\bar{Z}_{tt} - i) \partial_\aa \paren{\f{1}{\abs{Z_{,\aa}}} \abs{D_\aa} \f{\AAt}{\AA}}}_{L^2} + \nm{e}_{L^2}
 \lec \nm{\eqref{eq:58}}_{L^2} + \nm{e}_{L^2}
\\ & \lec \nm{(I-\HH) \braces{i(\bar{Z}_{tt} - i) D_\aa^2 \f{\AAt}{\AA}}}_{L^2}
\\ & + \nm{\bracket{i(\bar{Z}_{tt} - i) \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}^2,\HH}  \partial_\aa \paren{\f{1}{\abs{Z_{,\aa}}} \abs{D_\aa} \f{\AAt}{\AA}}}_{L^2} + \nm{e}_{L^2}.
\end{aligned}
\end{equation}
Thus, it suffices to focus on these three terms.  The first term will be the main term that will eventually give us the commutator structure that we want.  We first control the remaining terms.

First we check the error term, $e$ \eqref{eq:324}. We control
\begin{equation}
  \label{eq:325}
  \begin{aligned}
  \nm{e}_{L^2} &\le \paren{\nm{(\bar{Z}_{tt} - i)\partial_\aa \f{1}{\abs{Z_{,\aa}}}}_{L^\infty} +\nm{(\bar{Z}_{tt} - i) \paren{\abs{D_\aa} \f{\abs{Z_{,\aa}}}{Z_{,\aa}}}}_{L^\infty}}\nm{D_\aa \f{\AAt}{\AA}}_{L^2}\\&\lec \nm{(\bar{Z}_{tt} - i)\partial_\aa \f{1}{Z_{,\aa}}}_{L^\infty} \nm{D_\aa \f{\AAt}{\AA}}_{L^2},
  \end{aligned}
\end{equation}
where in the second step we used  $\abs{\partial_\aa \abs{f}} \le \abs{\partial_\aa f}$ \eqref{eq:100}, and $\abs{\partial_\aa \f{f}{\abs{f}}} \le \abs{\f{f'}{\abs{f}}}$ \eqref{eq:100}. We have controlled both factors on the RHS in \eqref{eq:1024} and \eqref{eq:1034}.

Now we estimate the second term on the RHS of \eqref{eq:71}. We use $L^2 \times L^\infty$ commutator estimate \eqref{eq:228}.
Observe that because either (i) the angle $\ang=\frac\pi 2$, in which case $\bound{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}} = 0$, or else (ii) the angle  $\ang\ne\frac\pi 2$ and so $\bar{Z}_{tt} - i = 0$ at the corners (see \eqref{eq:626}) this commutator does satisfy the boundary condition we need for \eqref{eq:228}.\footnote{\label{fn-expl} Note that this estimate for $[f,\HH] \partial_\aa g$, unlike the $L^\infty \times L^2$ estimate, does {\em not} require that $\bound{g} = 0$. This explains why we moved from $\abs{D_\aa}^2 \f{\AAt}{\AA}$ to $\partial_\aa \paren{\f{1}{\abs{Z_{,\aa}}} \abs{D_\aa} \f{\AAt}{\AA}}$; had we tried to estimate
    \begin{equation}
      \label{eq:337}
      \nm{\bracket{\f{1}{\abs{Z_{,\aa}}} \paren{-i(\bar{Z}_{tt} - i) \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}^2},\HH}  \partial_\aa \abs{D_\aa} \f{\AAt}{\AA}}_{L^2}
    \end{equation}
by $L^\infty \times L^2$ commutator estimate \eqref{eq:305}, we would have been able to control each of the terms (they would be similar to the estimates for $e$ in \eqref{eq:325} above), but we don't know that $\bound{\abs{D_\aa} \f{\AAt}{\AA}}$ is zero.} We have
\begin{equation}
  \label{eq:336}
  \begin{aligned}
  &\nm{\bracket{-i(\bar{Z}_{tt} - i) \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}^2,\HH}  \partial_\aa \paren{\f{1}{\abs{Z_{,\aa}}} \abs{D_\aa} \f{\AAt}{\AA}}}_{L^2}\\& \lec \nm{\partial_\aa \paren{-i(\bar{Z}_{tt} - i) \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}^2}}_{L^2} \nm{\f{1}{Z_{,\aa}} D_\aa \f{\AAt}{\AA}}_{L^\infty}
  \lec  \nm{Z_{tt,\aa}}_{L^2}\nm{\f{1}{Z_{,\aa}} D_\aa \f{\AAt}{\AA}}_{L^\infty},
  \end{aligned}
\end{equation}
where in the second step we replaced $\f{\abs{Z_{,\aa}}}{Z_{,\aa}}$ with $\f{\bar{Z}_{tt} - i}{\abs{\bar{Z}_{tt} - i}}$ \eqref{eq:279} to obtain the last inequality.
We have controlled the first factor, $\nm{Z_{tt,\aa}}_{L^2}$, in \S \ref{sec:quants}.
What remains to be done is to show that we can control $\nm{\f{1}{Z_{,\aa}} D_\aa \f{\AAt}{\AA}}_{L^\infty}$. We defer doing so until the end, in Step 7.

We're left with the first, main term of the RHS of \eqref{eq:71}. Observe that by \eqref{eq:46}, our main equation $\bar Z_{ttt}+i\AA \bar Z_{t,\aa}=-i\AA_t\bar Z_{,\aa}$ \eqref{eq:178}, and the $L^2$ boundedness of $\HH$,
\begin{equation}
  \label{eq:341}
  \begin{aligned}
     \nm{(I-\HH) \braces{i(\bar{Z}_{tt} - i) D_\aa^2 \f{\AAt}{\AA}}}_{L^2}  & \lec \nm{(I-\HH) \paren{D_\aa^2 (\AA_t \bar{Z}_{,\aa})}}_{L^2}
 + \eqref{eq:316}\\&= \nm{(I-\HH) D_\aa^2(\bar{Z}_{ttt} + i \AA \bar{Z}_{t,\aa})}_{L^2}+ \eqref{eq:316}.
 \end{aligned}
\end{equation}
We have therefore reduced things (except for the one term  we're deferring to the end) to controlling  $ \nm{(I-\HH) D_\aa^2(\bar{Z}_{ttt} + i \AA \bar{Z}_{t,\aa})}_{L^2}$, which will allow us to use desirable commutators.

{\bf Step 2.} {\it Controlling $\nm{(I-\HH) \paren{D_\aa^2 (\bar{Z}_{ttt} + i \AA \bar{Z}_{t,\aa}) } }_{L^2} $. }\label{sec:100}
We now expand out  $(I-\HH) D_\aa^2 (\bar{Z}_{ttt} + i \AA \bar{Z}_{t,\aa})$, as we did in \S \ref{sec:28}. From \eqref{eq:202}, we have
\begin{equation}
  \label{eq:300}
  \bar{Z}_{ttt} = (D_\aa^2 \bar{Z}_t) Z_t^2 + 2 Z_t D_\aa(\bar{Z}_{tt} - (D_\aa \bar{Z}_t) Z_t) + (D_\aa \bar{Z}_t) Z_{tt} + F_{tt} \circ Z,
\end{equation}
where $F(z(\a,t),t) := \bar{z}_t(\a,t)$. By $Z_{tt} + i = i \AA Z_{,\aa}$ \eqref{eq:179},
$ i \AA \bar{Z}_{t,\aa} = (Z_{tt} + i) D_\aa \bar{Z}_{t}$.
Putting these together and applying $(I-\HH) D_\aa^2$, we have
\begin{multline}
  \label{eq:302}
  (I-\HH) D_\aa^2  (\bar{Z}_{ttt} + i \AA \bar{Z}_{t,\aa})
\\ = (I-\HH) D_\aa^2 \paren{(D_\aa^2 \bar{Z}_t) Z_t^2 + 2 Z_t D_\aa(\bar{Z}_{tt} - (D_\aa \bar{Z}_t) Z_t) + (D_\aa \bar{Z}_t) (2 Z_{tt} + i) + F_{tt} \circ Z}.
\end{multline}
Observe that each of these terms has holomorphic factors, which will (hopefully) give us controllable commutators or disappear under $(I-\HH)$.

First we address the last two terms, $(I-\HH) D_\aa^2 \paren{(D_\aa \bar{Z}_t) (2 Z_{tt} + i) + F_{tt} \circ Z}$.  We may rewrite
\begin{equation}
  \label{eq:303}
  \begin{aligned}
    (I-\HH) D_\aa^2 &\paren{(D_\aa \bar{Z}_t) (2 Z_{tt} + i)+ F_{tt} \circ Z}

    \\  &  =  2(I-\HH) \paren{(D_\aa^3 \bar{Z}_t) (Z_{tt} + i) + 2 (D_\aa^2 \bar{Z}_t)(D_\aa Z_{tt}) + (D_\aa \bar{Z}_t) (D_\aa^2 Z_{tt})},
  \end{aligned}
\end{equation}
where we have used \eqref{eq:504} and \eqref{eq:315}.

By H\"older, we can control the second and third terms of the RHS of \eqref{eq:303}:
\begin{equation}
  \label{eq:506}
\begin{aligned}
  \nm{(I-\HH) \paren{2(D_\aa^2 \bar{Z}_t)(D_\aa Z_{tt}) + (D_\aa \bar{Z}_t) (D_\aa^2 Z_{tt})}}_{L^2} &
\\ \lec \nm{D_\aa^2 \bar{Z}_t}_{L^2} &\nm{D_\aa Z_{tt}}_{L^\infty} + \nm{D_\aa \bar{Z}_t}_{L^\infty} \nm{D_\aa^2 Z_{tt}}_{L^2}.
\end{aligned}
\end{equation}
To control the first term on the RHS of \eqref{eq:303}, we use \eqref{eq:474} to write
\begin{equation}
  \label{eq:508}
  \begin{aligned}
    \nm{(I-\HH)(D_\aa^3 \bar{Z}_t) (Z_{tt} + i)}_{L^2} & =  \nm{\bracket{\f{Z_{tt} + i}{Z_{,\aa}},\HH} \partial_\aa D_\aa^2 \bar{Z}_t}_{L^2}
    \\ & \lec \nm{\partial_\aa \f{Z_{tt} + i}{Z_{,\aa}}}_{L^\infty} \nm{D_\aa^2 \bar{Z}_t}_{L^2}
    \\ & \lec \paren{\nm{D_\aa Z_{tt}}_{L^\infty} + \nm{(Z_{tt} + i)\partial_\aa \f{1}{Z_{,\aa}}}_{L^\infty}}\nm{D_\aa^2 \bar{Z}_t}_{L^2},
  \end{aligned}
\end{equation}
where we have used the $L^\infty \times L^2$ commutator estimate \eqref{eq:305}.

We conclude from \eqref{eq:303}, \eqref{eq:506} and \eqref{eq:508} that we can control
\begin{equation}
  \label{eq:509}

    \nm{(I-\HH) D_\aa^2 \paren{(D_\aa \bar{Z}_t) (2 Z_{tt} + i) + F_{tt} \circ Z}}_{L^2}\lec \eqref{eq:506} + \eqref{eq:508}\lec C(E).

\end{equation}

We can now focus on what remains from \eqref{eq:302},
\begin{equation}
  \label{eq:70}
  \begin{aligned}
    (I-\HH) & D_\aa^2 \braces{(D_\aa^2 \bar{Z}_t) Z_t^2 + 2 Z_t D_\aa(\bar{Z}_{tt} - (D_\aa \bar{Z}_t) Z_t)}
\\ & = (I-\HH) \paren{(D_\aa^2 (Z_t^2)) (D_\aa^2 \bar{Z}_t) + 2 (D_\aa^2 Z_t) D_\aa(\bar{Z}_{tt} - (D_\aa \bar{Z}_t) Z_t)}
\\ & + (I-\HH) \paren{2 (D_\aa (Z_t^2)) (D_\aa^3 \bar{Z}_t) + 4 (D_\aa Z_t) D_\aa^2(\bar{Z}_{tt} - (D_\aa \bar{Z}_t) Z_t)}
\\ & + (I-\HH) \paren{Z_t^2 (D_\aa^4 \bar{Z}_t) + 2 Z_t D_\aa^3 (\bar{Z}_{tt} - (D_\aa \bar{Z}_t) Z_t)}
\\ & = i + ii + iii.
\end{aligned}
\end{equation}
Terms $i$ and $ii$ are easy to control. The details are uninteresting and distract from our main derivation, so we defer them to Step 6,

 below.

We may therefore focus on $iii$.

{\bf Step 3.} {\it Controlling $iii$ from \eqref{eq:70}.}\label{sec:693}
We now focus on controlling
\begin{equation}
  \label{eq:307}
 iii = (I-\HH) Z_t^2 (D_\aa^4 \bar{Z}_t) + (I-\HH)2 Z_t D_\aa^3 (\bar{Z}_{tt} - (D_\aa \bar{Z}_t) Z_t) = I + II.
\end{equation}
These terms appear to require too much regularity to close the estimate: we have up to four derivatives here, when our energy controls only two derivatives. Our strategy is to expand the equation and then exploit the holomorphicity of certain of the factors involved; other terms can be controlled directly, and the rest of the terms cancel out with each other.

We begin by expanding $II$ from \eqref{eq:307}. Rewriting the identity as $\P_A + \P_H$, we have
\begin{equation}
  \label{eq:79}
  \begin{aligned}
    II
    & = 2 (I-\HH)\braces{\paren{(\P_A + \P_H) \f{Z_t}{Z_{,\aa}} }\partial_\aa D_\aa^2 (\bar{Z}_{tt} - (D_\aa \bar{Z}_t) Z_t)}
\\ & = 2\bracket{\paren{\P_A \f{Z_t}{Z_{,\aa}}},\HH} \partial_\aa D_\aa^2 (\bar{Z}_{tt} - (D_\aa \bar{Z}_t) Z_t) + \underbrace{ (I-\HH)\braces{ \paren{\P_H \f{Z_t}{Z_{,\aa}}} \partial_\aa D_\aa^2 (\bar{Z}_{tt} - (D_\aa \bar{Z}_t) Z_t)}}_{0 \text{ by }\eqref{eq:518}},
\end{aligned}
\end{equation}
where we've used \eqref{eq:222} to get the first commutator. We now expand what remains from \eqref{eq:79}:
\begin{equation}
  \label{eq:95}
  \begin{aligned}
    II & = 2\bracket{\paren{\P_A \f{Z_t}{Z_{,\aa}}},\HH} \partial_\aa (D_\aa^2 \bar{Z}_{tt} - (\partial_\aa D_\aa^2 \bar{Z}_t) \f{Z_t}{Z_{,\aa}} - 2 (D_\aa^2 \bar{Z}_t) D_\aa Z_t - (D_\aa \bar{Z}_t) D_\aa^2 Z_t)
\\ & = II_1 + II_2 + II_3 + II_4.
\end{aligned}
\end{equation}
We control everything but $II_2$ directly by the $L^\infty \times L^2$ commutator estimate \eqref{eq:305}. To apply this, we need both parts to have periodic boundary behavior. We know $\bound{\P_A \f{Z_t}{Z_{,\aa}}} = 0$ \eqref{eq:161}.  The reason that
\begin{equation}
  \label{eq:521}
  \bound{(D_\aa^2 \bar{Z}_{tt} - 2 (D_\aa^2 \bar{Z}_t) D_\aa Z_t - (D_\aa \bar{Z}_t) D_\aa^2 Z_t)} = 0
\end{equation}
is that it equals $D_\aa (F_{tz} \circ Z) + (\partial_\aa D_\aa^2 \bar{Z}_t) \f{Z_t}{Z_{,\aa}}$, both of which have  periodic boundary behavior by \eqref{eq:437}, \eqref{eq:259}, and \eqref{eq:67}. We get
\begin{equation}
  \label{eq:520}
  \begin{aligned}
    &\nm{II_1 + II_3 + II_4}_{L^2}
      \\ & \lec \nm{\partial_\aa \P_A \f{Z_t}{Z_{,\aa}}}_{L^\infty} \paren{\nm{D_\a^2 \bar{Z}_{tt}}_{L^2} + \nm{D_\aa^2 \bar{Z}_t}_{L^2} \nm{D_\aa Z_t}_{L^\infty} + \nm{D_\aa^2 Z_t}_{L^2}\nm{D_\aa \bar{Z}_t}_{L^\infty}}.
  \end{aligned}
\end{equation}

We are now left with $II_2$ from \eqref{eq:95}. We switch back from the commutator form, commuting the $\partial_\aa$ in the second term outside $(I-\HH)$, which we can do because  $\bound{ (\partial_\aa D_\aa^2 \bar{Z}_t)  \f{Z_t}{Z_{,\aa}}}=0$ \eqref{eq:437} \eqref{eq:259}, and then we expand the factor $\f{Z_t}{Z_{,\aa}}$ by rewriting the identity as $\P_A + \P_H$:
\begin{equation}
  \label{eq:119}
  \begin{aligned}
    II_2 & = -2 (I-\HH)\braces{\P_A \f{Z_t}{Z_{,\aa}} \,\partial_\aa \paren{(\partial_\aa D_\aa^2 \bar{Z}_t)  \f{Z_t}{Z_{,\aa}}}}+2 \P_A \f{Z_t}{Z_{,\aa}}\partial_\aa(I-\HH) \paren{ (\partial_\aa D_\aa^2 \bar{Z}_t)  \f{Z_t}{Z_{,\aa}}}
\\ &= -2 (I-\HH)\braces{\P_A \f{Z_t}{Z_{,\aa}} \,\partial_\aa \paren{(\partial_\aa D_\aa^2 \bar{Z}_t)  \P_A \f{Z_t}{Z_{,\aa}}}}\\&-2 (I-\HH)\braces{\P_A \f{Z_t}{Z_{,\aa}} \,\partial_\aa \paren{(\partial_\aa D_\aa^2 \bar{Z}_t) \P_H \f{Z_t}{Z_{,\aa}}}}+2 \P_A \f{Z_t}{Z_{,\aa}}\partial_\aa(I-\HH) \braces{ (\partial_\aa D_\aa^2 \bar{Z}_t) \P_A \f{Z_t}{Z_{,\aa}}}\\&=II_{21}+II_{22}+II_{23},
\end{aligned}
\end{equation}
where  in the penultimate line we used  $(I-\HH) \braces{ (\partial_\aa D_\aa^2 \bar{Z}_t) \P_H \f{Z_t}{Z_{,\aa}}}=0$ by \eqref{eq:522}.

Next, we expand $I$ from \eqref{eq:307}:
\begin{equation}
  \label{eq:112}
  I = (I-\HH) \f{Z_t^2}{Z_{,\aa}^2} \partial_\aa^2 D_\aa^2 \bar{Z}_t + (I-\HH) Z_t^2 \paren{\f{1}{Z_{,\aa}} \partial_\aa \f{1}{Z_{,\aa}}} \partial_\aa D_\aa^2 \bar{Z}_t = I_1 + I_2.
\end{equation}
We once again expand the identity operator into the sum of the projections $\P_A + \P_H$:
\begin{equation}
  \label{eq:118}
  \begin{aligned}
    I_1 &   = (I-\HH) \braces{\paren{\P_A \f{Z_t}{Z_{,\aa}}}^2 \partial_\aa^2 D_\aa^2 \bar{Z}_t} + (I-\HH) \braces{2\paren{\P_A \f{Z_t}{Z_{,\aa}}} \paren{\P_H \f{Z_t}{Z_{,\aa}}} \partial_\aa^2 D_\aa^2 \bar{Z}_t}
\\&+ (I-\HH) \braces{\paren{\P_H \f{Z_t}{Z_{,\aa}}}^2 \partial_\aa^2 D_\aa^2 \bar{Z}_t}
\\ & = I_{11} + I_{12} + I_{13}.
\end{aligned}
\end{equation}

We now carefully inspect  \eqref{eq:119}, \eqref{eq:112} and \eqref{eq:118} and combine terms.

First, we combine $II_{21}, I_{11}$, and $I_{23}$, using \eqref{eq:474} to rewrite $II_{23}$ in commutator form:
\begin{equation}\label{eq:1052}
II_{21}+I_{11}+II_{23}=-(I-\HH) \partial_\aa\braces{\paren{\P_A \f{Z_t}{Z_{,\aa}}}^2 \partial_\aa D_\aa^2 \bar{Z}_t}+ 2 \P_A \f{Z_t}{Z_{,\aa}}\partial_\aa\bracket{\P_A \f{Z_t}{Z_{,\aa}}, \HH}  (\partial_\aa D_\aa^2 \bar{Z}_t) .
\end{equation}
Commuting the $\partial_\aa$ in the first term outside $(I-\HH)$, then rewriting it in commutator form by \eqref{eq:474}, we have
\begin{equation}\label{eq:1030}
\begin{aligned}
II_{21}&+I_{11} +II_{23} =-\partial_\aa\bracket{\paren{\P_A \f{Z_t}{Z_{,\aa}}}^2, \HH} \braces{ \partial_\aa D_\aa^2 \bar{Z}_t}+ 2 \P_A \f{Z_t}{Z_{,\aa}}\partial_\aa\bracket{\P_A \f{Z_t}{Z_{,\aa}}, \HH}  (\partial_\aa D_\aa^2 \bar{Z}_t)\\ &\  \  \  -\bracket{ \partial_\aa,\HH}\braces{\paren{\P_A \f{Z_t}{Z_{,\aa}}}^2 \partial_\aa D_\aa^2 \bar{Z}_t}\\&
=-\frac1{2i}\int \frac{\pi}2\frac{\paren{\P_A \f{Z_t}{Z_{,\aa}}(\aa)-\P_A \f{Z_t}{Z_{,\bb}}(\bb)}^2}{\sin^2(\frac\pi2(\aa-\bb))}\partial_\bb D_\bb^2 \bar{Z}_t\,d\bb-\bracket{ \partial_\aa,\HH}\braces{\paren{\P_A \f{Z_t}{Z_{,\aa}}}^2 \partial_\aa D_\aa^2 \bar{Z}_t}\\ &=:-\bracket{ \P_A \f{Z_t}{Z_{,\aa}}, \P_A \f{Z_t}{Z_{,\aa}}; \partial_\aa D_\aa^2 \bar{Z}_t}+R_1,
\end{aligned}
\end{equation}
where we arrived at the second equality by applying the derivatives and combining the first two terms. We can control the first term on the RHS via \eqref{eq:6a}, relying on boundary condition \eqref{eq:161} (and \eqref{eq:259}):
\begin{equation}
  \label{eq:156}
 \begin{aligned}
    \nm{\bracket{\P_A \f{Z_t}{Z_{,\a}}, \P_A \f{Z_t}{Z_{,\aa}}; \partial_\b D_\bb^2 \bar{Z}_t}}_{L^2} & \le \nm{\partial_\aa \paren{\P_A \f{Z_t}{Z_{,\aa}}}}_{L^\infty}^2 \nm{D_\aa^2 \bar{Z}_t}_{L^2},
 \end{aligned}
\end{equation}
where we have controlled the first factor at \eqref{eq:513} and the second factor in \S \ref{sec:quants}.
We will handle the remainder $R_1$ a little later, after we combine the next three terms  in $II_2$ and $I$.

We next combine $II_{22}$, $I_2$ and $I_{12}$. We note that after applying the derivative in $II_{22}$ by the product rule,  one of the terms cancels out with  $I_{12}$. We have
\begin{equation}\label{eq:1031}
II_{22}+I_2+I_{12} =  (I-\HH) \braces{-2\paren{\P_A \f{Z_t}{Z_{,\aa}}} \paren{(\partial_\aa D_\aa^2 \bar{Z}_t) \P_H \partial_\aa\f{Z_t}{Z_{,\aa}}}+ Z_t^2 \paren{\f{1}{Z_{,\aa}} \partial_\aa \f{1}{Z_{,\aa}}} \partial_\aa D_\aa^2 \bar{Z}_t },
\end{equation}
where we've used $\bound{\f{Z_t}{Z_{,\aa}}} = 0$ \eqref{eq:524} to commute the $\partial_\aa$ in the first term inside $\P_H$ by \eqref{eq:176}. We rewrite

\begin{equation}
  \label{eq:129}
  \begin{aligned}
II_{22}+I_2+I_{12}
 & = (I-\HH) \braces{(\partial_\aa D_\aa^2 \bar{Z}_t) \braces{\f{Z_t}{Z_{,\aa}} Z_t \partial_\aa \f{1}{Z_{,\aa}}-2 \paren{\P_A \f{Z_t}{Z_{,\aa}}}\paren{\P_H \partial_\aa \f{Z_t}{Z_{,\aa}}}}}\\
 & = (I-\HH) \braces{(\partial_\aa D_\aa^2 \bar{Z}_t) \braces{\f{Z_t}{Z_{,\aa}} Z_t \partial_\aa \f{1}{Z_{,\aa}}-2 \paren{ (\P_A + \P_H) \f{Z_t}{Z_{,\aa}}}\paren{\P_H \partial_\aa \f{Z_t}{Z_{,\aa}}}}}
\\ & + 2 (I-\HH) \braces{(\partial_\aa D_\aa^2 \bar{Z}_t) \braces{\paren{\P_H \f{Z_t}{Z_{,\aa}}}\paren{\partial_\aa\P_H  \f{Z_t}{Z_{,\aa}}}}}
\\ & =: A + R_2.
\end{aligned}
\end{equation}
We begin by considering the second term, $R_2$,
which we claim will cancel out with $I_{13}$ from \eqref{eq:118} and $R_1$ from \eqref{eq:1030}. Indeed, these three terms are
\begin{equation}
  \label{eq:231}
  \begin{aligned}
R_1+I_{13} + R_2 & = -\bracket{ \partial_\aa,\HH}\braces{\paren{\P_A \f{Z_t}{Z_{,\aa}}}^2 \partial_\aa D_\aa^2 \bar{Z}_t}+(I-\HH) \partial_\aa \braces{ (\partial_\aa D_\aa^2 \bar{Z}_t)\paren{\P_H \f{Z_t}{Z_{,\aa}}}^2}\\ &
= -\bracket{ \partial_\aa,\HH}\braces{\paren{\P_A \f{Z_t}{Z_{,\aa}}}^2 \partial_\aa D_\aa^2 \bar{Z}_t}+\partial_\aa \underbrace{(I-\HH) \braces{(\partial_\aa D_\aa^2 \bar{Z}_t)\paren{\P_H \f{Z_t}{Z_{,\aa}}}^2}}_{0 \text{ by }\eqref{eq:522}}\\& + [\partial_\aa,\HH] \braces{(\partial_\aa D_\aa^2 \bar{Z}_t)\paren{\P_H \f{Z_t}{Z_{,\aa}}}^2}
\\&= -[\partial_\aa,\HH]\braces{  \paren{(\P_A - \P_H)\f{Z_t}{Z_{,\aa}}}\paren{(\P_A + \P_H) \f{Z_t}{Z_{,\aa}}}  \paren{\partial_\aa D_\aa^2 \bar{Z}_t}}
\\ & =  [\partial_\aa,\HH]\braces{ \paren{ {\HH}\f{Z_t}{Z_{,\aa}}}\paren{ \f{Z_t}{Z_{,\aa}}}  \paren{\partial_\aa D_\aa^2 \bar{Z}_t}}
= 0.
\end{aligned}
\end{equation}
This is zero because the first factor has periodic boundary behavior by \eqref{eq:161}, the $Z_t$ has periodic boundary behavior by \eqref{eq:437}, and the $\f{1}{Z_{,\aa}}$ joins with the $\partial_\aa D_\aa^2 \bar{Z}_t$ to ensure periodic boundary behavior by \eqref{eq:259}.

Therefore
\begin{equation}\label{eq:1054}
iii=I+II= -\bracket{ \P_A \f{Z_t}{Z_{,\aa}}, \P_A \f{Z_t}{Z_{,\aa}}; \partial_\aa D_\aa^2 \bar{Z}_t}+A+(II_1+II_3+II_4).
\end{equation}
Since we have estimated the first and third sets of terms at \eqref{eq:156} and \eqref{eq:520} above, all that remains is to control
  \begin{equation}
    \label{eq:130}
    A = (I-\HH) \braces{(\partial_\aa D_\aa^2 \bar{Z}_t) \f{Z_t}{Z_{,\aa}} \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}} - 2 \P_H \partial_\aa \f{Z_t}{Z_{,\aa}}}}.
  \end{equation}
Let $(*)= Z_t \partial_\aa \f{1}{Z_{,\aa}} - 2 \P_H \partial_\aa \f{Z_t}{Z_{,\aa}}$.  We expand $(*)$ as follows:
\begin{equation}
  \label{eq:131}
  \begin{aligned}
    (*) &= Z_t \partial_\aa \f{1}{Z_{,\aa}} - (I+\HH) D_\aa Z_t - (I+\HH) \braces{Z_t \partial_\aa \f{1}{Z_{,\aa}}}
\\ &= -\HH \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}} - (I+\HH) D_\aa Z_t.
\end{aligned}
\end{equation}
Therefore, we need to control
\begin{equation}
  \label{eq:133}
  (I-\HH)\braces{ (\partial_\aa D_\aa^2 \bar{Z}_t) \f{Z_t}{Z_{,\aa}} \HH\paren{ Z_t \partial_\aa \f{1}{Z_{,\aa}}}}
\end{equation}
and
\begin{equation}
  \label{eq:135}
  (I-\HH)\braces{ (\partial_\aa D_\aa^2 \bar{Z}_t) \f{Z_t}{Z_{,\aa}} (I+\HH) D_\aa Z_t}.
\end{equation}
We will control \eqref{eq:133} in the next step, Step 4, and \eqref{eq:135} in Step 5.

{\bf Step 4.} {\it Controlling \eqref{eq:133}.}\label{sec:83}
We begin with \eqref{eq:133}.  Our goal is to take advantage of the fact that we can control $\nm{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t}_{\dot{H}^{1/2}}$.  Therefore, we rewrite \eqref{eq:133} so that we can isolate $\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t$ for a commutator estimate:
\begin{equation}
  \label{eq:136}
  \begin{aligned}
    \eqref{eq:133} & = (I-\HH)\braces{ \paren{\f{1}{Z_{,\aa}}\partial_\aa D_\aa^2 \bar{Z}_t} Z_t \HH\paren{ Z_t \partial_\aa \f{1}{Z_{,\aa}}}}
\\ & =   (I-\HH) \braces{\paren{ Z_t \HH \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}} \partial_\aa \paren{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t}}
\\ & - (I-\HH) \braces{\paren{ Z_t \HH \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}} \paren{\partial_\aa \f{1}{Z_{,\aa}}} D_\aa^2 \bar{Z}_t}
\\ & = (a) + (b).
\end{aligned}
\end{equation}
We expand $(a)$ using $I = \P_A + \P_H$:
\begin{equation}
  \label{eq:137}
  \begin{aligned}
    (a) & =   (I-\HH) \braces{\paren{(\P_A + \P_H) \paren{Z_t \HH \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}}} \partial_\aa \paren{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t}}
    \\ & =(I-\HH) \braces{\paren{\P_A \paren{Z_t \HH \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}}} \partial_\aa \paren{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t} }
    \\ & + \underbrace{(I-\HH) \braces{\paren{\P_H \paren{Z_t \HH \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}}} \partial_\aa \paren{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t}}}_{0 \text{ by } \eqref{eq:526}}
    \\ & = \bracket{\paren{\P_A \paren{Z_t \HH \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}}},\HH} \partial_\aa\paren{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t}
  \end{aligned}
\end{equation}
by \eqref{eq:525} and \eqref{eq:526}. We now use \eqref{eq:21} to estimate the RHS of \eqref{eq:137}, using the $\dot{H}^{1/2}$ control:
\begin{equation}
  \label{eq:138}
  \nm{(a)}_{L^2} \lec \nm{\partial_\aa \P_A \paren{Z_t \HH \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}}}_{L^2}\nm{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t}_{\dot{H}^{1/2}}.
\end{equation}
To apply \eqref{eq:21}, we require
\begin{equation}
  \label{eq:392}
  \bound{\P_A \paren{Z_t \HH \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}}}=0.
\end{equation}
(Note that the second required boundary condition, $\bound{\paren{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t}}=0,$ holds by \eqref{eq:259} and \eqref{eq:438}.) It isn't obvious why this first boundary condition should hold. We will show below at \eqref{eq:275} that it does, in fact, hold. Assuming that, because we've controlled $\nm{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t}_{\dot{H}^{1/2}}$ at \eqref{eq:498}, we've reduced things to controlling $\nm{\partial_\aa \paren{\P_A \paren{Z_t \HH \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}}}}_{L^2}$.

We first find a useful way of rewriting $Z_t \HH \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}$. Observe that
\begin{equation}
  \label{eq:139}

      \bracket{Z_t,[Z_t,\HH]}
      = Z_t^2 \HH - 2 Z_t \HH Z_t + \HH Z_t^2.

  \end{equation}
Therefore,
\begin{equation}
  \label{eq:140}
  \begin{aligned}
    \bracket{Z_t,[Z_t,\HH]} \partial_\aa \f{1}{Z_{,\aa}} & = Z_t^2 \HH \partial_\aa \f{1}{Z_{,\aa}} - 2 Z_t \HH \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}} + \HH \paren{ Z_t^2 \partial_\aa \f{1}{Z_{,\aa}}}
\\ & = (I+\HH) \paren{Z_t^2 \partial_\aa \f{1}{Z_{,\aa}}} - 2 Z_t \HH \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}},
\end{aligned}
\end{equation}
where $\HH \partial_\aa \f{1}{Z_{,\aa}}= \partial_\aa \f{1}{Z_{,\aa}}$   by \eqref{eq:242}. Therefore,
\begin{equation}
  \label{eq:141}
  2Z_t\HH \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}} =  (I+\HH) \paren{Z_t^2 \partial_\aa \f{1}{Z_{,\aa}}}- [Z_t,[Z_t,\HH]] \partial_\aa \f{1}{Z_{,\aa}}.
\end{equation}
Now we apply $\P_A$ to this, getting
\begin{equation}
  \label{eq:262}
 2\P_A \paren{ Z_t \HH \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}} =  \P_A (I+\HH) \paren{Z_t^2 \partial_\aa \f{1}{Z_{,\aa}}} -  \P_A [Z_t,[Z_t,\HH]] \partial_\aa \f{1}{Z_{,\aa}}.
\end{equation}
By \eqref{eq:159}, the first term on the RHS is just a mean term. Because this mean will eventually disappear under the derivative, we denote it simply by $c$. We are left with
\begin{equation}
  \label{eq:264}
    2\P_A \paren{ Z_t\HH \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}} = c - \P_A [Z_t,[Z_t,\HH]] \partial_\aa \f{1}{Z_{,\aa}}.
\end{equation}
Now we will show, as promised, that $\bound{\P_A  \paren{Z_t\HH \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}}}=0$.  We expand
\begin{equation}
  \label{eq:273}
     \bracket{Z_t,[Z_t,\HH]} \partial_\aa \f{1}{Z_{,\aa}}  = \f{1}{\hdenomconst} \int \paren{Z_t(\aa) -Z_t(\bb)}^2 \cot(\f{\pi}{2}(\aa -\bb)) \partial_\bb \f{1}{Z_{,\bb}} d\bb.
\end{equation}
Because $Z_t\big|_\partial=0$, we see, by the fundamental theorem of calculus and H\"older, that the kernel function $\paren{Z_t(\aa) -Z_t(\bb)}^2 \cot(\f{\pi}{2}(\aa -\bb))$ is continuous for  $(\alpha', \beta')\in I\times I\setminus\{\abs{\aa-\bb} \not\in \{0,2\}\}$, periodic in $\alpha'$,  and satisfies
\begin{equation}\label{eq:12}
  \big|\paren{Z_t(\aa) -Z_t(\bb)}^2 \cot(\f{\pi}{2}(\aa -\bb))\big|\lec  \int|Z_{t,\alpha'}|^2\,d\alpha'.
\end{equation}
Therefore by the dominated convergence theorem,
\begin{equation}
 \label{eq:529}
  \bound{[Z_t,[Z_t,\HH]] \partial_\aa \f{1}{Z_{,\aa}}} = 0.
\end{equation}
Furthermore we can estimate
\begin{equation}
  \label{eq:649}
  \nm{[Z_t,[Z_t,\HH]] \partial_\aa \f{1}{Z_{,\aa}}}_{L^\infty} \lec  \nm{Z_{t,\aa}}_{L^2}^2 \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}
\end{equation}
by H\"older and Hardy's inequality \eqref{eq:77}, and estimate
\begin{equation}
  \label{eq:650}
  \nm{\partial_\aa [Z_t,[Z_t,\HH]] \partial_\aa \f{1}{Z_{,\aa}}}_{L^2} \lec \nm{Z_{t,\aa}}_{L^2}^2 \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}
\end{equation}
by \eqref{eq:144}. We may then conclude by \eqref{eq:264} and \eqref{eq:439} that
\begin{equation}
    \label{eq:275}
  \bound{\P_A [Z_t,[Z_t,\HH]] \partial_\aa \f{1}{Z_{,\aa}}} = \bound{\P_A \paren{ Z_t\HH \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}}}= 0.
\end{equation}

Now we return to controlling $\nm{\partial_\aa \paren{\P_A \paren{Z_t \HH \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}}}}_{L^2}$.  By \eqref{eq:264},
\begin{equation}
  \label{eq:276}
  \begin{aligned}
    &\nm{\partial_\aa \P_A \paren{Z_t \HH \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}}}_{L^2}  = \f{1}{2} \nm{\partial_\aa \P_A [Z_t,[Z_t,\HH]] \partial_\aa \f{1}{Z_{,\aa}}}_{L^2}.
    \\ & =  \f{1}{2}\nm{\P_A \partial_\aa[Z_t,[Z_t,\HH]] \partial_\aa \f{1}{Z_{,\aa}}}_{L^2}
    \lec \nm{\partial_\aa[Z_t,[Z_t,\HH]] \partial_\aa \f{1}{Z_{,\aa}}}_{L^2}
    \\ & \lec  \nm{Z_{t,\aa}}_{L^2}^2 \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2},
  \end{aligned}
\end{equation}
where we have used \eqref{eq:176} and \eqref{eq:529} to commute the $\partial_\aa$ inside the $\P_A$  in the second line; then used the $L^2$ boundedness of $\P_A$; and finally used estimate \eqref{eq:650} in the last line.

Now we return to controlling part $(b)$ from \eqref{eq:136}.  We begin by decomposing two of the factors into their holomorphic and antiholomorphic projections:
\begin{equation}
  \label{eq:145}
  \begin{aligned}
    (b) & =- (I-\HH) \braces{\paren{\HH \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}} \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}} D_\aa^2 \bar{Z}_t}
\\ &  = - (I-\HH) \braces{\paren{ ( -\P_A + \P_H)  \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}}\paren{(\P_A + \P_H) \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}}  D_\aa^2 \bar{Z}_t}.
\end{aligned}
\end{equation}
Observe that $\HH =  -\P_A + \P_H$. The cross terms cancel, and we're left with
\begin{equation}
  \label{eq:146}
  (I-\HH) \braces{\paren{\P_A \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}}^2 D_\aa^2 \bar{Z}_t} - (I-\HH) \braces{\paren{\P_H \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}}^2 D_\aa^2 \bar{Z}_t}.
\end{equation}
The second of these terms disappears by \eqref{eq:531}. We control the first term of \eqref{eq:146} by the boundedness of $\HH$ and  H\"older's inequality, and conclude that
\begin{equation}
  \label{eq:150}
  \nm{(b)}_{L^2} \lec \nm{D_\aa^2 \bar{Z}_t}_{L^2} \nm{\P_A\paren{ Z_t \partial_\aa \f{1}{Z_{,\aa}}}}_{L^\infty}^2,
\end{equation}
whose second factor we controlled in \eqref{eq:151}.

We conclude from \eqref{eq:136} that
  \begin{equation}
    \label{eq:538}
    \begin{aligned}
      \nm{\eqref{eq:133}}_{L^2} & \lec \eqref{eq:138} + \eqref{eq:150}
      \\ & \le  \nm{Z_{t,\aa}}_{L^2}^2 \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2} \nm{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t}_{\dot{H}^{1/2}} + \nm{D_\aa^2 \bar{Z}_t}_{L^2}\nm{\P_A\paren{ Z_t \partial_\aa \f{1}{Z_{,\aa}}}}_{L^\infty}^2.
    \end{aligned}
  \end{equation}

{\bf Step 5.} {\it Controlling \eqref{eq:135}.}\label{sec:84}
Now we are left with controlling \eqref{eq:135}, which we expand using $I = \P_A + \P_H$:
\begin{equation}
  \label{eq:152}
  \begin{aligned}
    \eqref{eq:135} & =(I-\HH)\braces{ (\partial_\aa D_\aa^2 \bar{Z}_t) \f{Z_t}{Z_{,\aa}} (I+\HH) D_\aa Z_t}\\
    &  = (I-\HH) \braces{(\partial_\aa D_\aa^2 \bar{Z}_t) \paren{(\P_A + \P_H)\f{Z_t}{Z_{,\aa}}} (I+\HH) D_\aa Z_t}
\\ & =  (I-\HH) \braces{(\partial_\aa D_\aa^2 \bar{Z}_t) \paren{\P_A \f{Z_t}{Z_{,\aa}}} (I+\HH) D_\aa Z_t},
\end{aligned}
\end{equation}
where the $\P_H$ term disappears by \eqref{eq:535}.  We expand again $I = \P_A + \P_H$ with what remains:
\begin{equation}
  \label{eq:117}
   \eqref{eq:135} =  (I-\HH)\braces{ \paren{(I+\HH) D_\aa Z_t}  (\P_A + \P_H) \braces{(\partial_\aa D_\aa^2 \bar{Z}_t) \paren{\P_A \f{Z_t}{Z_{,\aa}}} }}.
\end{equation}
The $\P_H$ part would be zero by holomorphicity, except that there is a mean term; by \eqref{eq:536}, what we're left with from that part is, in absolute value,
\begin{equation}
  \label{eq:260}
  \begin{aligned}
    \abs{\f{1}{2} \paren{\avg D_\aa Z_t}\paren{\avg (\partial_\aa D_\aa^2 \bar{Z}_t) \paren{\P_A \f{Z_t}{Z_{,\aa}}}}} & = \abs{\f{1}{2} \paren{\avg D_\aa Z_t}\paren{\avg (D_\aa^2 \bar{Z}_t) \paren{\partial_\aa \P_A \f{Z_t}{Z_{,\aa}}}}}
    \\ & \lec \nm{D_\aa Z_t}_{L^\infty} \nm{D_\aa^2 \bar{Z}_t}_{L^2} \nm{\partial_\aa \P_A \f{Z_t}{Z_{,\aa}}}_{L^\infty}.
  \end{aligned}
\end{equation}
By \eqref{eq:161} and \eqref{eq:259} there is no boundary term in the integration by parts.

What's left from \eqref{eq:117} to control is the $\P_A$ part,
\begin{equation}
  \label{eq:147}
  (I-\HH) \braces{\paren{(I+\HH) D_\aa Z_t} \P_A \paren{(\partial_\aa D_\aa^2 \bar{Z}_t)\paren{\P_A \f{Z_t}{Z_{,\aa}}}}}.
\end{equation}
Now we use the boundedness of $\HH$ and H\"older to control this in $L^2$ by
\begin{equation}
  \label{eq:153}
  \nm{(I+\HH) D_\aa Z_t}_{L^\infty} \nm{\P_A \braces{(\partial_\aa D_\aa^2 \bar{Z}_t)\paren{\P_A \f{Z_t}{Z_{,\aa}}}}}_{L^2}.
\end{equation}
We rewrite the second factor as a commutator, using \eqref{eq:474}:
\begin{equation}
  \label{eq:155}
  \begin{aligned}
    \nm{\P_A \braces{(\partial_\aa D_\aa^2 \bar{Z}_t)\paren{\P_A \f{Z_t}{Z_{,\aa}}}}}_{L^2} & = \nm{\f{1}{2}\bracket{\paren{\P_A \f{Z_t}{Z_{,\aa}}},\HH}\partial_\aa D_\aa^2 \bar{Z}_t}_{L^2}
\\ &  \lec \nm{\partial_\aa (I-\HH) \f{Z_t}{Z_{,\aa}}}_{L^\infty} \nm{D_\aa^2 \bar{Z}_t}_{L^2},
\end{aligned}
\end{equation}
where we've used the $L^\infty \times L^2$ commutator estimate \eqref{eq:305}.

We conclude from \eqref{eq:117}, \eqref{eq:260}, \eqref{eq:153} and \eqref{eq:155} that
\begin{equation}
  \label{eq:537}
  \nm{\eqref{eq:135}}_{L^2} \lec \paren{ \nm{D_\aa Z_t}_{L^\infty} + \nm{(I+\HH) D_\aa Z_t}_{L^\infty}} \nm{\partial_\aa (I-\HH) \f{Z_t}{Z_{,\aa}}}_{L^\infty} \nm{D_\aa^2 \bar{Z}_t}_{L^2} .
\end{equation}

{\bf Step 6.} {\it Concluding the estimate for $\nm{(I-\HH) \paren{D_\aa^2 (\AA_t \bar Z_{,\aa})}}_{L^2}$ from \eqref{eq:341}.}\label{sec:638}
We start from \eqref{eq:1054}. We have, from \eqref{eq:520}, \eqref{eq:156}, \eqref{eq:130}, \eqref{eq:538} and \eqref{eq:537} that
\begin{equation}
  \label{eq:702}

    \nm{iii}_{L^2}   \lec \eqref{eq:156} + \eqref{eq:520} + \eqref{eq:538} + \eqref{eq:537}\lec C(E).
\end{equation}

We must estimate $i$ and $ii$ from \eqref{eq:70}.  When we expand this out carefully, all the problematic terms cancel and what remains can be controlled easily using H\"older and the $L^2$ boundedness of $\HH$.

We expand $i$:
\begin{equation}
    \label{eq:329}
       i = (I-\HH) \paren{2 (D_\aa Z_t)^2 (D_\aa^2 \bar{Z}_t) + (2 D_\aa^2 Z_t) (D_\aa \bar{Z}_{tt}) - 2 (D_\aa^2 Z_t) (D_\aa Z_t) (D_\aa \bar{Z}_t)}.
\end{equation}
We can therefore estimate
\begin{equation}
  \label{eq:510}
  \nm{i}_{L^2} \lec \nm{D_\aa Z_t}_{L^\infty}^2 \nm{D_\aa^2 \bar{Z}_t}_{L^2} + \nm{D_\aa^2 Z_t}_{L^2} (\nm{D_\aa \bar{Z}_{tt}}_{L^\infty} + \nm{D_\aa Z_t}_{L^\infty}^2).
\end{equation}

We expand $ii$:
\begin{equation}
  \label{eq:330}
      ii
= (I-\HH) \paren{4 (D_\aa Z_t) (D_\aa^2 \bar{Z}_{tt}) - 8 (D_\aa Z_t)^2 (D_\aa^2 \bar{Z}_t) - 4 (D_\aa Z_t)(D_\aa \bar{Z}_t) (D_\aa^2 Z_t)},
\end{equation}
and we estimate
\begin{equation}
  \label{eq:511}
  \nm{ii}_{L^2} \lec \nm{D_\aa Z_t}_{L^\infty} \nm{D_\aa^2 \bar{Z}_{tt}}_{L^2} +  \nm{D_\aa Z_t}_{L^\infty}^2 \nm{D_\aa^2 \bar{Z}_t}_{L^2} + \nm{D_\aa Z_t}_{L^\infty}^2 \nm{D_\aa^2 Z_t}_{L^2}.
\end{equation}

Combining our various estimates, we have, by \eqref{eq:302}, that
\begin{equation}
  \label{eq:323}
    \nm{(I-\HH) \paren{D_\aa^2 (\AA_t \bar Z_{,\aa})}}_{L^2}  \lec \eqref{eq:509} +\nm{i}_{L^2} +
    \nm{ii}_{L^2} + \nm{iii}_{L^2} \lec \eqref{eq:509} +\eqref{eq:510} + \eqref{eq:511} + \eqref{eq:702}\lec C(E).
\end{equation}

{\bf Step 7.} {\it Concluding the estimate for $G_\th$ of $E_a$.}\label{combining}
Recall that we still needed to control $\nm{\f1{Z_{,\aa}}D_\aa \f{\AA_t}{\AA}}_{L^\infty}$. Here, we use the Peter-Paul trick Proposition \ref{peter-paul}.  By \eqref{eq:71}, \eqref{eq:325}, \eqref{eq:336}, \eqref{eq:341}, \eqref{eq:323},  $\f1{Z_{,\aa}}=i\f{\bar Z_{tt}-i}{A_1}$ \eqref{eq:251}, and $A_1\ge 1$ \eqref{eq:394},
we now have an estimate
\begin{equation}
  \label{eq:340}
  \nm{\f1{Z_{,\aa}} D^2_\aa\f{\AA_t}{\AA}}_{L^2}\lec C(E)+\nm{ Z_{tt,\aa}}_{L^2} \nm{ \f1{Z_{,\aa}}D_\aa \f{\AA_t}{\AA}     }_{L^\infty}.
\end{equation}
This implies
\begin{equation}
  \label{eq:1035}
  \begin{aligned}
  \nm{D_\aa\paren{\f1{Z_{,\aa}} D_\aa\f{\AA_t}{\AA}}}_{L^2}&\lec \nm{D_\aa \f1{Z_{,\aa}}}_{L^\infty}\nm{D_\aa\f{\AA_t}{\AA}}_{L^2}+C(E)+\nm{ Z_{tt,\aa}}_{L^2} \nm{ \f1{Z_{,\aa}}D_\aa \f{\AA_t}{\AA}     }_{L^\infty}
  \\&\lec  C(E)+\nm{ Z_{tt,\aa}}_{L^2} \nm{ \f1{Z_{,\aa}}D_\aa \f{\AA_t}{\AA}     }_{L^\infty},
  \end{aligned}
\end{equation}
by \eqref{eq:1023} and \eqref{eq:1034}.
We may now apply Proposition \ref{peter-paul}. We have
\begin{equation}\label{eq:1036}
   \nm{ \f1{Z_{,\aa}}D_\aa \f{\AA_t}{\AA} }_{L^\infty}\lec (\nm{ Z_{tt,\aa}}_{L^2}+1)\paren{\nm{D_\aa \f{\AA_t}{\AA}}_{L^2}+C(E)+
  \nm{\f1{Z_{,\aa}} D_\aa \f{\AA_t}{\AA}}_{L^2}}\lec C(E),
\end{equation}
where we have controlled $\nm{D_\aa \f{\AA_t}{\AA}}_{L^2}$ in \eqref{eq:1034} and $\nm{\f1{Z_{,\aa}}}_{L^\infty}$ and $\nm{ Z_{tt,\aa}}_{L^2}$ in \S \ref{sec:quants}.

Combining \eqref{eq:46}, \eqref{eq:316}, \eqref{eq:71}, \eqref{eq:325}, \eqref{eq:336}, \eqref{eq:341}, \eqref{eq:323} and \eqref{eq:1036}, we have
\begin{equation}\label{eq:1037}
\nm{D^2_\aa\paren{ {\AA_t}{\bar Z_{,\aa}}}}_{L^2}\lec C(E).
\end{equation}

We conclude by \eqref{eq:399}, \eqref{eq:415}, \eqref{eq:134a}, \eqref{eq:1050} and \eqref{eq:1037} that
\begin{equation}\label{eq:1600}
\nm{G_{D^2_\a \bar z_t}}_{L^2(\frac{h_\a}{A_1\circ h})}\le \eqref{eq:415} +\eqref{eq:134a}\le \eqref{eq:415} +\eqref{eq:1037}\lec C(E).
\end{equation}

We have now shown that $\frac d{dt}E_a$ is bounded by a polynomial of $E$. This concludes the proof of Theorem~\ref{maintheorem}. \qed

\section{A characterization of the energy}\label{sec:36e}
Our energy is expressed in terms of not only the free surface $Z$, the velocity $Z_t$, and their spatial derivatives, but also time derivatives of these quantities. In this section, we give a characterization of our energy in terms of the free surface $Z$, the velocity $Z_t$, and their spatial derivatives. We do so in the Riemannian variable, as it directly captures the geometry of the interface $Z$. We also, in \S \ref{sec:13a}, offer a heuristic discussion of the singularities inherent in the problem and the crest angles allowed by our energy, as indicated by the Riemann mapping.

\subsection{A characterization of the energy  in terms of position and velocity}\label{sec:36d}
  In this section, we translate the terms of our energy involving time derivatives into terms depending only on the free surface $Z$, the velocity $Z_t$, and their spatial derivatives. We do this using the basic identity \eqref{eq:251}
\begin{equation}
  \label{eq:528}
  \f{1}{Z_{,\aa}} = i \f{\bar{Z}_{tt} - i}{A_1},
\end{equation}
\eqref{eq:208}, and the holomorphicity of $\bar Z_t$ and various other quantities discussed in \S \ref{sec:20}. These basic water wave equations allow us to show that quantities involving $\bar{Z}_{tt}$ can be controlled by analogous quantities involving $\f{1}{Z_{,\aa}}$, along with various lower-order terms.\footnote{We remark that for these estimates we {\em do not} ever rely on (high order) $\dot{H}^{1/2}$ parts of the energies.}

The estimate we prove is
\begin{multline}
  \label{eq:532}
  E(t) \le C\bigg(\nm{\bar{Z}_{t,\aa}}_{L^2}, \nm{D_\aa^2 \bar{Z}_t}_{L^2}, \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2},
  \\ \nm{D_\aa^2 \f{1}{Z_{,\aa}}}_{L^2}, \nm{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t}_{\dot{H}^{1/2}}, \nm{D_\aa \bar{Z}_t}_{\dot{H}^{1/2}}, \nm{\f{1}{Z_{,\aa}}}_{L^\infty}\bigg),
\end{multline}
where the constant depends polynomially on its terms. We remark that this inequality can be reversed: each of the factors on the RHS of \eqref{eq:532} is controlled by the energy. That is,
\begin{multline}
  \label{eq:575}
  \nm{\bar{Z}_{t,\aa}}_{L^2}, \nm{D_\aa^2 \bar{Z}_t}_{L^2}, \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}, \nm{D_\aa^2 \f{1}{Z_{,\aa}}}_{L^2}, \nm{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t}_{\dot{H}^{1/2}}, \\ \nm{D_\aa \bar{Z}_t}_{\dot{H}^{1/2}},   \nm{\f{1}{Z_{,\aa}}}_{L^\infty}   \lec C(E(t)).
\end{multline}
Therefore, these quantities fully characterize our energy. In the proof of our a priori estimate, we have shown \eqref{eq:575} for every term except $\nm{D_\aa^2 \f{1}{Z_{,\aa}}}_{L^2}$, which we never had a need to control. One can adapt the argument in \S \ref{sec:573} below to show that $\nm{D_\aa^2 \f{1}{Z_{,\aa}}}_{L^2}$ can be controlled by the energy.\footnote{To do this, it comes down once again to estimating $\nm{\f{1}{Z_{,\aa}} D_\aa^2 A_1}_{L^2}$, except this time we need to do this without the dependence on $\nm{D_\aa^2 \f{1}{Z_{,\aa}}}_{L^2}$. That dependence comes from estimate \eqref{eq:411}. (It also comes from using $\nm{D_\aa^2 \f{1}{Z_{,\aa}}}_{L^2}$ in the Sobolev inequality for $\nm{D_\aa \f{1}{Z_{,\aa}}}_{L^\infty}$; this is not a problem, since $\nm{D_\aa \f{1}{Z_{,\aa}}}_{L^\infty}$ is controlled by the energy.) To handle \eqref{eq:411}, we take advantage of the fact that $(I-\HH) \braces{\partial_\aa D_\aa \f{1}{Z_{,\aa}}} = 0$ (this is due to \eqref{eq:516}, \eqref{eq:242} and the second principle in \S \ref{sec:20}) to rewrite the problematic term as a commutator and then use commutator estimate \eqref{eq:228}:
  \begin{equation}
    \label{eq:574}

      \nm{(I-\HH) \braces{\f{A_1}{Z_{,\aa}} \partial_\aa D_\aa \f{1}{Z_{,\aa}}}}_{L^2}  = \nm{\bracket{\bar{Z}_{tt},\HH} \partial_\aa D_\aa \f{1}{Z_{,\aa}}}_{L^2}
  \lec \nm{\bar{Z}_{tt,\aa}}_{L^2} \nm{D_\aa \f{1}{Z_{,\aa}}}_{L^\infty},

  \end{equation}
both of which are controlled by the energy. }

We remark because both $\bar{Z}_t$ and $\f{1}{Z_{,\aa}}$ are the boundary values of periodic holomorphic functions, the weighted derivative $D_\aa$ corresponds to the complex derivative $\partial_z$, or the gradient of the corresponding quantities in the spatial domain $P^-$. We also note that ${Z_{,\aa}} = (\Phi^{-1})_z$ is a natural geometric quantity well-suited to this problem: it captures the geometry of the free surface directly through the Riemann mapping $\Phi^{-1}: P^-\to \Omega(t)$.

\subsubsection{The proof}
Throughout the following proof we will rely on the fact that $A_1 \ge 1$ \eqref{eq:394}, the estimate \eqref{eq:361}
\begin{equation}
  \label{eq:533}
  \nm{A_1}_{L^\infty} \lec 1 + \nm{\bar{Z}_{t,\aa}}_{L^2}^2,
\end{equation}
 the Sobolev estimate \eqref{eq:206}
\begin{equation}
  \label{eq:502}
  \nm{D_\aa \bar{Z}_t}_{L^\infty} \lec \nm{\bar{Z}_{t,\aa}}_{L^2} + \nm{D_\aa^2 \bar{Z}_t}_{L^2},
\end{equation}
and the estimate
\begin{equation}
  \label{eq:453}
  \nm{D_\aa \f{1}{Z_{,\aa}}}_{L^\infty} \lec \nm{D_\aa^2 \f{1}{Z_{,\aa}}}_{L^2} + \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2},
\end{equation}
which holds by Sobolev inequality \eqref{eq:638}.\footnote{Note that $\avg \paren{D_\aa \f{1}{Z_{,\aa}}}^2 = 0$ by the same argument that was used at \eqref{eq:48} to show $\avg (D_\aa \bar{Z}_t)^2 = 0$.}

We begin by noting that it suffices to control only the first terms of $E_a$ and $E_b$, since the remaining terms of the energy are (up to a factor of $A_1$) already on the RHS of \eqref{eq:532}.

For the first term of $E_a$, by the commutator identity \eqref{eq:121},
\begin{equation}
  \label{eq:24}
  \begin{aligned}
    \int \abs{\partial_t D_\a^2 \bar{z}_t}^2 \f{h_\a}{A_1 \circ h} d\a & \lec \int \abs{D_\a^2 \bar{z}_{tt}}^2 \f{h_\a}{A_1 \circ h} d\a + \int \abs{[\partial_t,D_\a^2] \bar{z}_t}^2 \f{h_\a}{A_1 \circ h} d\a
\\ & \lec \nm{D_\aa^2 \bar{Z}_{tt}}_{L^2}^2 + \int \abs{(D_\a z_t)D_\a^2 \bar{z}_t + (D_\a^2 z_t) D_\a \bar{z}_t}^2 \f{h_\a}{A_1 \circ h} d\a
\\ & \lec \nm{D_\aa^2 \bar{Z}_{tt}}_{L^2}^2 + \nm{D_\a z_t}_{L^\infty}^2(\nm{D_\aa^2 \bar{Z}_{t}}_{L^2}^2 + \nm{D_\aa^2 Z_t}_{L^2}^2).
  \end{aligned}
\end{equation}
By \eqref{eq:65a} and \eqref{eq:100},
\begin{equation}
  \label{eq:64}

      \nm{D_\aa^2 Z_t}_{L^2} \lec \nm{D_\aa^2 \bar{Z}_t}_{L^2} + \nm{D_\aa \bar{Z}_t}_{L^\infty} \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}.

\end{equation}
We conclude that
\begin{equation}
  \label{eq:65}
  \begin{aligned}
      \int \abs{\partial_t D_\a^2 \bar{z}_t}^2 \f{h_\a}{A_1 \circ h} d\a &\lec   C\paren{\nm{D_\aa^2 \bar{Z}_{tt}}_{L^2}, \nm{\bar{Z}_{t,\aa}}_{L^2}, \nm{D_\aa^2 \bar{Z}_t}_{L^2}, \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}}.
\end{aligned}
\end{equation}

For the first term of $E_b$, we use the commutator identity \eqref{eq:120} to get
\begin{equation}
  \label{eq:412}
  \begin{aligned}
    \int &\abs{\partial_t D_\a \bar{z}_t}^2 \f{1}{\af} d\a  \lec \int \abs{D_\a \bar{z}_{tt}}^2 \f{1}{\af} d\a + \int \abs{[\partial_t,D_\a] \bar{z}_t}^2 \f{1}{\af} d\a
\\ & \lec \int \abs{D_\a \bar{z}_{tt}}^2 \f{(A_1 \circ h)}{\af} d\a + \int \abs{(D_\a z_t)D_\a \bar{z}_t}^2 \f{(A_1 \circ h)}{\af} d\a
\\ & \lec \nm{\bar{Z}_{tt,\aa}}_{L^2}^2 + \nm{D_\a z_t}_{L^\infty}^2 \nm{\bar{Z}_{t,\aa}}_{L^2}^2
 \le C(\nm{\bar{Z}_{tt,\aa}}_{L^2}, \nm{\bar{Z}_{t,\aa}}_{L^2}, \nm{D_\aa^2 \bar{Z}_t}_{L^2}).
  \end{aligned}
\end{equation}

All that remains to do from \eqref{eq:65} and \eqref{eq:412} is to estimate $\nm{\bar{Z}_{tt,\aa}}_{L^2}$ and $\nm{D_\aa^2 \bar{Z}_{tt}}_{L^2}$ in terms of $Z_t$ and $\f{1}{Z_{,\aa}}$, which we now do, in \S \ref{sec:534} and \S \ref{sec:573}.

\subsubsection{Controlling \texorpdfstring{$\nm{\bar{Z}_{tt,\aa}}_{L^2}$}{|| bar Ztt,a' || L2}}\label{sec:534}

Using \eqref{eq:528}, we estimate
\begin{equation}
  \label{eq:534}
    \nm{\partial_\aa \bar{Z}_{tt}}_{L^2}  \lec \nm{A_1}_{L^\infty} \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2} + \nm{D_\aa A_1}_{L^2}.
\end{equation}
To control $\nm{D_\aa A_1}_{L^2}$, we follow a similar procedure to what we did in \eqref{eq:254}-\eqref{eq:255}, except instead of using $\bar{Z}_{tt} - i$, we use $\f{1}{Z_{,\aa}}$ and estimate things in terms of $\f{1}{Z_{,\aa}}$. We get
\begin{equation}
  \label{eq:496}
    \nm{D_\aa A_1}_{L^2} \lec \nm{\bar{Z}_{t,\aa}}_{L^2}^2 \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2} + \nm{Z_{t,\aa}}_{L^2} \nm{D_\aa \bar{Z}_t}_{L^\infty}.
\end{equation}
Combining \eqref{eq:534} and \eqref{eq:496} we conclude that
\begin{equation}
  \label{eq:545}
  \begin{aligned}
    \nm{\partial_\aa \bar{Z}_{tt}}_{L^2} \le C\paren{\nm{\bar{Z}_{t,\aa}}_{L^2}, \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}, \nm{D_\aa^2 \bar{Z}_t}_{L^2}}.
  \end{aligned}
\end{equation}

\subsubsection{Controlling \texorpdfstring{$\nm{D_\aa^2 \bar{Z}_{tt}}_{L^2}$}{|| Da' 2 bar Ztt || L2}}\label{sec:573}
From \eqref{eq:528}, we have
\begin{equation}\label{eq:1000}
iD_\aa^2 \bar{Z}_{tt}= \underbrace{A_1 D_\aa^2 \f{1}{Z_{,\aa}}+2(D_\aa A_1) D_\aa \f{1}{Z_{,\aa}}}_{e_1} + \f{1}{Z_{,\aa}} D_\aa^2 A_1.
\end{equation}
We estimate $\nm{D_\aa^2 \bar{Z}_{tt}}_{L^2}$ through  the following procedure. First we note that the only challenging term to control on the RHS of \eqref{eq:1000} is the last one, $\f{1}{Z_{,\aa}} D_\aa^2 A_1$. We observe that this is almost real, modulo factors of $\f{1}{Z_{,\aa}}$ and its derivatives. Therefore, we will be able to use the $\Re (I-\HH)$ trick and, through a series of commutators,  reduce the estimate for $\f{1}{Z_{,\aa}} D_\aa^2 A_1$ to an estimate
of $(I-\HH) (D_\aa^2 \f{A_1}{Z_{,\aa}} )=(I-\HH)(iD_\aa^2 \bar{Z}_{tt})$. Since $\bar{Z}_t$ is holomorphic, we will be able to rewrite $(I-\HH)(iD_\aa^2 \bar{Z}_{tt})$ in terms of commutators and obtain favorable estimates. We now give the details.

We first estimate the error term $e_1$ in \eqref{eq:1000}:
\begin{equation}
  \label{eq:573}
  \begin{aligned}
    \nm{e_1}_{L^2} & \lec \nm{A_1}_{L^\infty} \nm{D_\aa^2 \f{1}{Z_{,\aa}}}_{L^2} + \nm{D_\aa A_1}_{L^2} \nm{D_\aa \f{1}{Z_{,\aa}}}_{L^\infty}
\\ & \lec (1 + \nm{\bar{Z}_{t,\aa}}_{L^2}^2) \nm{D_\aa^2 \f{1}{Z_{,\aa}}}_{L^2} + \eqref{eq:496} \paren{\nm{D_\aa^2 \f{1}{Z_{,\aa}}}_{L^2} + \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}}.
  \end{aligned}
\end{equation}

It remains to control $\nm{\f{1}{Z_{,\aa}} D_\aa^2 A_1}_{L^2}$. We want to use $(I-\HH)$ to turn our quantity into commutators, but to do so we need to factor $D_\aa$ into a real-weighted derivative $\abs{D_\aa} := \f{1}{\abs{Z_{,\aa}}} \partial_\aa$ so that we may invert $(I-\HH)$. From \eqref{eq:320}, we have
\begin{equation}
  \label{eq:168}
  \f{1}{Z_{,\aa}} D_\aa^2 A_1 = \f{1}{Z_{,\aa}} \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}^2 \abs{D_\aa}^2 A_1 + \underbrace{\f{1}{Z_{,\aa}} \f{\abs{Z_{,\aa}}}{Z_{,\aa}}\paren{\abs{D_\aa} \f{\abs{Z_{,\aa}}}{Z_{,\aa}}} \abs{D_\aa} A_1}_{e_2}.
\end{equation}
We multiply both sides by $\paren{\f{Z_{,\aa}}{\abs{Z_{,\aa}}}}^3$ so that the first term on the RHS is purely real:
\begin{equation}
  \label{eq:274}
  \paren{\f{Z_{,\aa}}{\abs{Z_{,\aa}}}}^3\f{1}{Z_{,\aa}} D_\aa^2 A_1 = \paren{\f{Z_{,\aa}}{\abs{Z_{,\aa}}}}^3\f{1}{Z_{,\aa}} \paren{\f{\abs{Z_{,\aa}}}{Z_{,\aa}}}^2 \abs{D_\aa}^2 A_1 + \paren{\f{Z_{,\aa}}{\abs{Z_{,\aa}}}}^3 e_2.
\end{equation}
Now we apply $\Re(I-\HH)$ to each side, and conclude from the fact that $A_1 \in \R$ that
\begin{equation}
  \label{eq:369}
  \abs{\f{1}{Z_{,\aa}} \abs{D_\aa}^2 A_1} \lec \abs{(I-\HH) \braces{\paren{\f{Z_{,\aa}}{\abs{Z_{,\aa}}}}^3\f{1}{Z_{,\aa}} D_\aa^2 A_1}} + \abs{(I-\HH) \braces{\paren{\f{Z_{,\aa}}{\abs{Z_{,\aa}}}}^3 e_2}}.
\end{equation}
We conclude from \eqref{eq:168} and \eqref{eq:369} that
\begin{equation}
  \label{eq:372}
  \nm{\f{1}{Z_{,\aa}} D_\aa^2 A_1}_{L^2} \lec \nm{e_2}_{L^2} + \nm{(I-\HH) \braces{\paren{\f{Z_{,\aa}}{\abs{Z_{,\aa}}}}^3\f{1}{Z_{,\aa}} D_\aa^2 A_1}}_{L^2}.
\end{equation}
By \eqref{eq:100} and \eqref{eq:496} we estimate
\begin{equation}
  \label{eq:373}
  \nm{e_2}_{L^2} \lec \nm{D_\aa \f{1}{Z_{,\aa}}}_{L^\infty} \nm{ D_\aa A_1}_{L^2}\lec \eqref{eq:453}\eqref{eq:496}.
\end{equation}
It remains to estimate $\nm{(I-\HH) \braces{\paren{\f{Z_{,\aa}}{\abs{Z_{,\aa}}}}^3\f{1}{Z_{,\aa}} D_\aa^2 A_1}}_{L^2}$. To get the right commutator estimate, we first rewrite this as
\begin{equation}
  \label{eq:374}
  \begin{aligned}
      \nm{(I-\HH) \braces{\paren{\f{Z_{,\aa}}{\abs{Z_{,\aa}}}}^3\f{1}{Z_{,\aa}} D_\aa^2 A_1}}_{L^2} & \lec \nm{(I-\HH) \braces{\paren{\f{Z_{,\aa}}{\abs{Z_{,\aa}}}}^3\f{1}{Z_{,\aa}} \partial_\aa \paren{\f{1}{Z_{,\aa}} D_\aa A_1}}}_{L^2}
\\ & + \nm{(I-\HH) \braces{\paren{\f{Z_{,\aa}}{\abs{Z_{,\aa}}}}^3\f{1}{Z_{,\aa}} \paren{\partial_\aa \f{1}{Z_{,\aa}}} D_\aa A_1}}_{L^2}.
    \end{aligned}
  \end{equation}
We estimate the second term on the RHS of \eqref{eq:374} directly:
\begin{equation}
  \label{eq:379}
  \begin{aligned}
    \nm{(I-\HH) \braces{\paren{\f{Z_{,\aa}}{\abs{Z_{,\aa}}}}^3\f{1}{Z_{,\aa}} \paren{\partial_\aa \f{1}{Z_{,\aa}}} D_\aa A_1}}_{L^2} & \lec \nm{D_\aa \f{1}{Z_{,\aa}}}_{L^\infty} \nm{ D_\aa A_1}_{L^2}
\\ & \lec \eqref{eq:453}\eqref{eq:496}.
\end{aligned}
\end{equation}
For the first term on the RHS of \eqref{eq:374}, we commute the  factor $\paren{\f{Z_{,\aa}}{\abs{Z_{,\aa}}}}^3$ outside the $(I-\HH)$, bringing along $\f{1}{Z_{,\aa}}$ to ensure that the commutator is controllable, and then bringing the $\f{1}{Z_{,\aa}}$ back inside:
\begin{equation}
  \label{eq:383}
  \begin{aligned}
    & \nm{(I-\HH) \braces{\paren{\f{Z_{,\aa}}{\abs{Z_{,\aa}}}}^3\f{1}{Z_{,\aa}} \partial_\aa \paren{\f{1}{Z_{,\aa}} D_\aa A_1}}}_{L^2}
  \lec \nm{\bracket{\paren{\f{Z_{,\aa}}{\abs{Z_{,\aa}}}}^3\f{1}{Z_{,\aa}},\HH}\partial_\aa \paren{\f{1}{Z_{,\aa}} D_\aa A_1}}_{L^2}
 \\ & + \nm{\bracket{\f{1}{Z_{,\aa}},\HH}\partial_\aa \paren{\f{1}{Z_{,\aa}} D_\aa A_1}}_{L^2} + \nm{(I-\HH)\braces{\f{1}{Z_{,\aa}} \partial_\aa \paren{\f{1}{Z_{,\aa}} D_\aa A_1}}}_{L^2}.
  \end{aligned}
\end{equation}
We estimate the first two terms on the RHS of \eqref{eq:383} using commutator estimate \eqref{eq:228}\footnote{Note that this does not require that $\f{1}{Z_{,\aa}} D_\aa A_1$ has periodic boundary behavior; this explains why we chose this commutator and moved $\f{1}{Z_{,\aa}}$ around instead of using the $L^\infty \times L^2$ estimate. Cf.~footnote \ref{fn-expl}.}:
\begin{equation}
  \label{eq:397}
  \begin{aligned}
    \nm{\bracket{\paren{\f{Z_{,\aa}}{\abs{Z_{,\aa}}}}^3\f{1}{Z_{,\aa}},\HH}\partial_\aa \paren{\f{1}{Z_{,\aa}} D_\aa A_1}}_{L^2} & + \nm{\bracket{\f{1}{Z_{,\aa}},\HH}\partial_\aa \paren{\f{1}{Z_{,\aa}} D_\aa A_1}}_{L^2}
\\ & \lec \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2} \nm{\f{1}{Z_{,\aa}} D_\aa A_1}_{L^\infty}.
  \end{aligned}
\end{equation}
We will postpone estimating $ \nm{\f{1}{Z_{,\aa}} D_\aa A_1}_{L^\infty}$ until the end of this long series of calculations. For the moment, we take the last term from the RHS of \eqref{eq:383}:
\begin{equation}
  \label{eq:407}
  \begin{aligned}
    &\nm{(I-\HH)\braces{\f{1}{Z_{,\aa}} \partial_\aa \paren{\f{1}{Z_{,\aa}} D_\aa A_1}}}_{L^2}\\&  \lec \nm{(I-\HH)D_\aa^2\paren{\f{1}{Z_{,\aa}}A_1}}_{L^2}
 + \nm{(I-\HH) \braces{D_\aa \paren{A_1 D_\aa \f{1}{Z_{,\aa} } }}}_{L^2}.
  \end{aligned}
\end{equation}
We estimate the second term by
\begin{equation}
  \label{eq:411}
  \begin{aligned}
    \nm{(I-\HH) \braces{ D_\aa \paren{A_1 D_\aa \f{1}{Z_{,\aa} } }}  }_{L^2} & \lec \nm{D_\aa^2 \f{1}{Z_{,\aa}}}_{L^2} \nm{A_1}_{L^\infty} +\nm{D_\aa \f{1}{Z_{,\aa}}}_{L^\infty} \nm{ D_\aa A_1}_{L^2}
\\ & \lec \nm{D_\aa^2 \f{1}{Z_{,\aa}}}_{L^2} (1 + \nm{\bar{Z}_{t,\aa}}_{L^2}^2) +\eqref{eq:453}\eqref{eq:496}.  \end{aligned}
\end{equation}
Finally, for the first term on the RHS of \eqref{eq:407}, we recall from \eqref{eq:184} that
\begin{equation}
  \label{eq:413}
  \f{1}{Z_{,\aa}}A_1 = i (\bar{Z}_{tt} - i) = i (Z_t D_\aa \bar{Z}_t) + i F_{t} \circ Z + 1
\end{equation}
for $F(z(\a,t),t) := \bar{z}_t(\a,t)$. We apply $(I-\HH) D_\aa^2$ to this, and the last two terms disappear. We get
\begin{equation}
  \label{eq:417}
  \begin{aligned}
    &\nm{(I-\HH)D_\aa^2\paren{\f{1}{Z_{,\aa}}A_1}}_{L^2}  = \nm{(I-\HH) D_\aa^2 (Z_t D_\aa \bar{Z}_t)}_{L^2}
\\ & \lec \nm{(I-\HH) \braces{(D_\aa^2 Z_t)D_\aa \bar{Z}_t}}_{L^2}  + \nm{(I-\HH) \braces{(D_\aa Z_t)(D_\aa^2 \bar{Z}_t)}}_{L^2}
\\ & + \nm{(I-\HH) \braces{\f{Z_t}{Z_{,\aa}}  \partial_\aa  D_\aa^2 \bar{Z}_t}}_{L^2}.
\end{aligned}
\end{equation}
We estimate the first two terms directly by
\begin{equation}
  \label{eq:420}
  \begin{aligned}
      &\nm{(I-\HH) \braces{(D_\aa^2 Z_t)D_\aa \bar{Z}_t}}_{L^2}  + \nm{(I-\HH) \braces{(D_\aa Z_t)(D_\aa^2 \bar{Z}_t)}}_{L^2}
\\ & \lec \nm{D_\aa \bar{Z}_t}_{L^\infty}(\nm{D_\aa^2 Z_t}_{L^2} + \nm{D_\aa^2 \bar{Z}_t}_{L^2})
 \lec \eqref{eq:502} (\nm{D_\aa^2 \bar{Z}_t}_{L^2} + \eqref{eq:64}).
\end{aligned}
\end{equation}
We are left with the last term on the RHS of \eqref{eq:417}. We first decompose $\f{Z_t}{Z_{,\aa}}$ into its holomorphic and antiholomorphic projections. The term with the holomorphic projection disappears by \eqref{eq:522}.  With what remains, we use \eqref{eq:474} to get a commutator, which we control by commutator estimate \eqref{eq:305}:
\begin{equation}
  \label{eq:429}
  \begin{aligned}
&\nm{(I-\HH) \braces{\f{Z_t}{Z_{,\aa}} \partial_\aa  D_\aa^2 \bar{Z}_t}}_{L^2}  =  \nm{(I-\HH) \braces{\paren{\P_A \f{Z_t}{Z_{,\aa}}} \partial_\aa D_\aa^2 \bar{Z}_t}}_{L^2}
\\ & = \nm{\bracket{\paren{\P_A \f{Z_t}{Z_{,\aa}}},\HH}\partial_\aa D_\aa^2 \bar{Z}_t}_{L^2}
 \lec \nm{\partial_\aa \P_A \f{Z_t}{Z_{,\aa}}}_{L^\infty} \nm{ D_\aa^2 \bar{Z}_t}_{L^2}

\\ & \lec \paren{\nm{D_\aa^2 \bar{Z}_t}_{L^2} + \nm{Z_{t,\aa}}_{L^2} \paren{1 +\nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}}}  \nm{D_\aa^2 \bar{Z}_t}_{L^2}
  \end{aligned}
\end{equation}
by \eqref{eq:513} and \eqref{eq:502}.

We now give the estimate for $\nm{\f{1}{Z_{,\aa}} D_\aa A_1}_{L^\infty}=\nm{\f{1}{\abs{Z_{,\aa}}^2} \partial_\aa A_1}_{L^\infty}$ in \eqref{eq:397}. We do so using \eqref{eq:254}. We have
\begin{equation}\label{eq:1001}
\begin{aligned}
 \frac1{\abs{Z_{,\aa}}^2} \partial_\aa A_1 & = \Im \f{1}{\hdenomconst} \int \f{\pi}{2} \f{(Z_t(\aa) - Z_t(\bb))}{\sin^2(\f{\pi}{2}(\aa - \bb))} \paren{\frac1{\abs{Z_{,\aa}}^2} -\frac1{\abs{Z_{,\bb}}^2}    } \bar{Z}_{t,\bb}(\bb) d\b
\\&+ \Im \f{1}{\hdenomconst} \int \f{\pi}{2} \f{(Z_t(\aa) - Z_t(\bb))}{\sin^2(\f{\pi}{2}(\aa - \bb))} \frac1{\bar Z_{,\bb}}D_\bb \bar{Z}_{t}(\bb) d\b
\\ & = I + II.
\end{aligned}
\end{equation}
From H\"older's inequality, Hardy's inequality \eqref{eq:77} and the mean value theorem,\footnote{Note that  $\f{1}{\abs{Z_{,\aa}}^2}$ is periodic.}  we have
\begin{equation}\label{eq:1002}
\nm{I}_{L^\infty}\lec \nm{Z_{t,\aa}}_{L^2}^2\nm{D_\aa\frac1{Z_{,\aa}}}_{L^\infty}\lec \nm{Z_{t,\aa}}_{L^2}^2\eqref{eq:453}.
\end{equation}
We rewrite $II$ using integration-by-parts identity \eqref{eq:98}:
\begin{equation}
\label{eq:1003}
\begin{aligned}
 \f{1}{\hdenomconst}& \int \f{\pi}{2} \f{(Z_t(\aa) - Z_t(\bb))}{\sin^2(\f{\pi}{2}(\aa - \bb))} \frac1{\bar Z_{,\bb}}D_\bb \bar{Z}_{t}(\bb) d\b
\\ & =-[Z_t, \HH]\partial_\aa \paren{\frac1{\bar Z_{,\aa}}D_\aa \bar{Z}_{t}}+ \HH \paren{Z_{t,\aa}\frac1{\bar Z_{,\aa}}D_\aa \bar{Z}_{t}}
\\&= -[Z_t, \HH]\partial_\aa \paren{\frac1{\bar Z_{,\aa}}D_\aa \bar{Z}_{t}}- \bracket{\f{1}{Z_{,\aa}}\bar{D_\aa \bar{Z}_{t} }, \HH}\bar{Z}_{t,\aa}+\abs{ D_\aa \bar{Z}_{t}}^2.
\end{aligned}
\end{equation}
Using \eqref{eq:32} on the first two terms on the RHS of \eqref{eq:1003}, we get
 \begin{equation}\label{eq:1004}
 \begin{aligned}
 \nm{II}_{L^\infty} & \lec \nm{Z_{t,\aa}}_{L^2}\nm{\partial_\aa \paren{\frac1{\bar Z_{,\aa}}D_\aa \bar{Z}_{t}}}_{L^2}+\nm{ D_\aa \bar{Z}_{t}}_{L^\infty}^2
\\ & \lec \nm{Z_{t,\aa}}_{L^2}\paren{ \nm {D^2_\aa \bar{Z}_{t}}_{L^2}+ \nm{\partial_\aa \frac1{Z_{,\aa}}}_{L^2}\nm{D_\aa \bar{Z}_{t}}_{L^\infty}}+\nm{ D_\aa \bar{Z}_{t}}_{L^\infty}^2.
\end{aligned}
\end{equation}
Combining \eqref{eq:1001}, \eqref{eq:1002}, \eqref{eq:1004}, \eqref{eq:453}, and \eqref{eq:502}, we have
\begin{equation}
\label{eq:1005}
\begin{aligned}
\nm{\frac1{\abs{Z_{,\aa}}^2} \partial_\aa A_1}_{L^\infty} & \le \nm{I}_{L^\infty}+\nm{II}_{L^\infty}
\\ & \le C\paren{ \nm{\bar{Z}_{t,\aa}}_{L^2}, \nm{D_\aa^2 \bar{Z}_t}_{L^2}, \nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}, \nm{D_\aa^2\f{1}{Z_{,\aa}}   }_{L^2}}.
\end{aligned}
\end{equation}
We now sum up these estimates. From \eqref{eq:372}, and using  \eqref{eq:1005} to estimate $\nm{\f{1}{Z_{,\aa}} D_\aa A_1}_{L^\infty}$, we have
\begin{equation}
  \label{eq:489}
  \begin{aligned}
    \nm{\f{1}{Z_{,\aa}}D_\aa^2 A_1}_{L^2} & \lec \eqref{eq:373} + \eqref{eq:379} + \eqref{eq:397} + \eqref{eq:411} + \eqref{eq:420} + \eqref{eq:429}
\\ & \lec
 C\paren{\nm{D_\aa^2 \f{1}{Z_{,\aa}}}_{L^2},\nm{\bar{Z}_{t,\aa}}_{L^2},\nm{D_\aa^2 \bar{Z}_t}_{L^2},\nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}}.
  \end{aligned}
\end{equation}

Combining \eqref{eq:573} and \eqref{eq:489} we conclude that
\begin{equation}
  \label{eq:454}
  \nm{D_\aa^2 \bar{Z}_{tt}}_{L^2} \lec C\paren{\nm{D_\aa^2 \f{1}{Z_{,\aa}}}_{L^2},\nm{\bar{Z}_{t,\aa}}_{L^2},\nm{D_\aa^2 \bar{Z}_t}_{L^2},\nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}}.
\end{equation}

\subsection{Singularities and the angle of the crest}\label{sec:13a}
In \S \ref{sec:36d} at \eqref{eq:532} and \eqref{eq:575} we characterized our energy in terms of various $L^2$ and $\dot{H}^{1/2}$ norms of quantities in Riemannian coordinates (as well as a single quantity, $\f{1}{Z_{,\aa}}$, in $L^\infty$).  When there is a non-right angle $\ang$ at the corner, or when there is a singularity in the middle of the free surface, the Riemann mapping will have a singularity. In this section, we discuss this singularity and what it suggests about the angle $\ang$ or the interior angle of an angled crest in the middle of the free surface, continuing the discussion from \S \ref{sec:27}.  We emphasize that unlike the rest of the paper, which is fully rigorous, the discussion in this section is heuristic and is under various unstated assumptions.

As in \S \ref{sec:27}, we will phrase our discussion in terms of a singularity at the corner, but it applies more broadly to singularities in the middle of the free surface.\footnote{The angled crests in the middle of the free surface don't have to be symmetric.} We thus henceforth focus on the angle $\ang$ at the corner. Of course, our energy is finite in the regime when $\ang = \f{\pi}{2}$, so we can focus on the case where $\ang < \f{\pi}{2}$.\footnote{Recall from the discussion in \S \ref{sec:27}  that we cannot have $\ang > \f{\pi}{2}$ in our energy regime.}

Throughout this section, we will abuse notation and say, e.g., $\Phi(z) \approx z^r$ or $h(\a) \approx \a^r$ at the corner, when in fact the corners are at $z, \a = \pm 1$, not $0$.

If $\ang$ is the  angle the water wave makes with the wall, the Riemann mapping $\Phi(z)$ should behave like $z^r$ at the corner, where $r \ang = \f{\pi}{2}.$ For $\ang < \f{\pi}{2},$ we have $r > 1$.

Recall from \eqref{eq:532} and \eqref{eq:575} that among the quantities that  characterize the energy, $\nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}$, $\nm{D_\aa^2 \f{1}{Z_{,\aa}}}_{L^2}$ and $\nm{\f{1}{Z_{,\aa}}}_{L^\infty}$  are the terms purely related to the surface. We will see through the following calculation what non-right angles $\ang$ are allowed if $\nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}$, $\nm{D_\aa^2 \f{1}{Z_{,\aa}}}_{L^2}$ and $\nm{\f{1}{Z_{,\aa}}}_{L^\infty}$ are finite.

Note that
 $ Z(\aa) = \Phi\i(\aa) \approx (\aa)^{1/r}$
so $ Z_{,\aa} = \partial_\aa(\Phi\i) \approx (\aa)^{1/r - 1}$,
and
\begin{equation}
  \label{eq:672}
  \f{1}{Z_{,\aa}} \approx (\aa)^{1 - 1/r},  \quad
 \quad
\partial_\aa \f{1}{Z_{,\aa}} \approx (\aa)^{-1/r} \quad (r \not= 1).
\end{equation}
Therefore, assuming $r > 1$, $\partial_\aa \f{1}{Z_{,\aa}} \in L^2$ if and only if $r > 2$ if and only if $\ang < \f{\pi}{4}$.  Similarly, $D_\aa^2 \f{1}{Z_{,\aa}} \in L^2$ so long as $r > 2$.\footnote{We remark that, even though $E_a$ (which roughly includes $\nm{D_\aa^2 \f{1}{Z_{,\aa}}}_{L^2}$) is higher-order than $E_b$ (which roughly includes $\nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}$) in terms of the number of derivatives, the two energies are comparable in the sense that they allow precisely the same angles.}

We conclude from this discussion that our energy will be finite only when $\ang < \f{\pi}{4}$ (or when $\ang = \f{\pi}{2}$). This coincides precisely with the angles in the self-similar solutions of \cite{wu:self}.\footnote{We recall that our energy is finite for these solutions.}  For singularities in the middle of the free surface, this suggests that the interior angle  must be less than $\f{\pi}{2}$.\footnote{We note that our energy does not apply to Stokes waves of maximum height (interior angle $ = \f{2\pi}{3}$).}

In the following discussion, we work under the assumption that $0 < c_1 \le \abs{z_\a} \le c_2 < \infty$ for all time in our timeframe, so the function $z(\a)$  doesn't affect the singularity (to first order).\footnote{This holds so long as our energy is finite; see footnote \ref{fn:za}.}

Since $z \approx \a$, $|z_\a|\approx 1$, we have
\begin{equation}
  \label{eq:295}
 h(\a) = \Phi(z(\a)) \approx \a^r,\qquad  h\i(\aa) \approx (\aa)^{1/r}.
\end{equation}
On differentiating \eqref{eq:295}, we get that
\begin{equation}
  \label{eq:350}
  h_\a(\a) \approx \a^{r-1},\qquad
  h_\a \circ h\i (\aa) \approx (\aa)^{\f{1}{r}(r -1)} = (\aa)^{1 - \f{1}{r}}.
\end{equation}

The behavior of the angle $\ang$ over time is of significant interest. This angle is determined by $z_\a$ at the corner.  Therefore, the behavior of $z_{t\a}$ and $z_{tt\a}$ at the corner should determine how the angle changes.  Since $Z_{t,\aa} = \f{z_{t\a}}{h_\a} \circ h\i$ and $\f{1}{h_\a} \circ h\i \approx (\aa)^{1/r - 1}$, we must have $z_{t\a} \to 0$ at the corner if $Z_{t,\aa} \in L^2$, as our energy assumes. A similar argument holds for $z_{tt\a}$. This suggests that if initially $\ang<\frac\pi 4$,  the angle would not change while the energy remained finite. This holds true also for interior angles at the angled crests. Careful analysis of such behavior at the corner would be an avenue for future research.

\begin{appendices}

\section{Holomorphicity, mean and boundary behavior}\label{sec:hmb}
\subsection{The Hilbert transform \texorpdfstring{$\HH$}{H}}\label{sec:18}
Recall that in \S \ref{sec:11} we introduced the Hilbert transform $\HH$:
\begin{equation}
  \label{eq:440}
  \HH f(\aa) := \f{1}{2 i} \pv \int_I \cot(\f{\pi}{2}(\aa - \bb)) f(\bb) d\bb,\qquad\text{for }\aa \in [-1,1].
\end{equation}
We know from Proposition~\ref{prop:hilbe} that a function $g$ is the boundary value of a periodic holomorphic function on $P^-$ if and only if  $(I-\HH) g=\avg_I g$, and that for any function $f$, $(I+\HH)f$ is the boundary value of a periodic holomorphic function on $P^-$, with $(I-\HH)(I+\HH)f=\avg f$.  Recall also that we defined at \eqref{eq:1581} the holomorphic and antiholomorphic projection operators
\begin{equation}
  \label{eq:158}
  \P_A f := \f{(I-\HH)}{2} f;\qquad \P_H f := \f{(I + \HH)}{2} f.
\end{equation}
Here we gather some basic properties of the Hilbert transform $\HH$ that will be used in this paper:

\begin{proposition}\label{prop:1}

a.  Let $1 < p < \infty$. Then there exists $C_p < \infty$ such that for all $f \in L^p$
  \begin{equation}
    \label{eq:441}
    \nm{\HH f}_{L^p} \le C_p \nm{f}_{L^p}.
  \end{equation}

 b.  Let $f \in L^p$ for some $p > 1$. Then
  \begin{equation} \label{eq:159}
     \HH^2 f  = f - \avg f;\qquad  \P_A \P_H f  =  \P_H \P_A f =\f{1}{4} \avg f.
\end{equation}

 c.  Let $f \in L^p$, $g \in L^q$, where $\f{1}{p} + \f{1}{q} = 1$, $1 < p, q < \infty$. Then
  \begin{equation}
    \label{eq:471}
    \int f(\HH g)   = - \int (\HH f) g; \qquad
  \int (\P_A f) g  = \int f (\P_H g).
\end{equation}

d. Let $f \in L^p, g \in L^q$ for $p>1$, $q>1$, $\f{1}{p} + \f{1}{q} < 1$.  Then
\begin{equation}
  \label{eq:160}
  \P_A \braces{(\P_H f) (\P_H g)}  = \f{1}{8} \paren{\avg f} \paren{\avg g}.
\end{equation}
\end{proposition}

Parts a., b., and c.~are classical results. We know the product of periodic holomorphic functions is periodic holomorphic. Part d.~is an easy consequence of Proposition~\ref{prop:hilbe}, parts b.~and c., and the fact that $(\P_H f) (\P_H g)$ is the boundary value of a periodic holomorphic function on $P^-$.

We also note the following commutator:
\begin{proposition}
  Let $f \in C^0[-1,1]$ with $f' \in L^p$ for some $p > 1$.  If $ \bound{f} = 0$, then
\begin{equation}
  \label{eq:176}
[\partial_\aa,\HH] f = 0.
\end{equation}
\end{proposition}
This proposition can be verified easily by integration by parts.

Finally, we will need the following proposition.
\begin{proposition}\label{prop:iden}
  Let $f \in L^\infty$, $g\in L^p$ for some $p> 1$.  Suppose $(I-\HH) f = \avg f$. Then
  \begin{equation}
    \label{eq:19}
    [f,\HH]g = \f12(I+\HH) \paren{[f,\HH]g} - \frac12\avg (fg) + \frac12\paren{\avg f} \paren{\avg g}.
  \end{equation}
\end{proposition}
\begin{proof}
We begin by observing that
\begin{equation}
  \label{eq:634}
 \bracket{f,\HH} g
     = f(I+\HH)g - (I+\HH) (fg).
  \end{equation}
  Because $(I-\HH) f = \avg f$, by Proposition~\ref{prop:hilbe}, $[f,\HH]g$ is the boundary value of a periodic holomorphic function on $P^-$.  Therefore,
\begin{equation}
  \label{eq:635}
  \begin{aligned}
    (I-\HH)([f,\HH]g) &=\avg [f,\HH]g  =   \avg f(I+\HH)g-\avg fg \\& =  \avg ((I-\HH) f) g-\avg( fg)    = \paren{\avg f} \paren{\avg g} - \avg (fg)
  \end{aligned}
\end{equation}
by \eqref{eq:471}.
\end{proof}

\subsection{Boundary properties}\label{sec:19}
Because we are working on a compact domain, we have to worry about boundary terms when integrating by parts. Furthermore, various results involving the periodic Hilbert transform require that certain functions be periodic. Therefore, it is critical that various potentially problematic boundary terms either disappear or can be controlled.  For this reason, we need to take care in specifying what the boundary properties of our various functions are.

Our hope is that we are dealing with functions $f(\a,t)$ that satisfy
\begin{equation}
  \label{eq:352}
 \bound{f} := f(1,t) - f(-1,t) = 0.
\end{equation}
Recall that in \S \ref{sec:10} we have shown that our basic function $\bar{z}_t$ satisfies $\bound{\bar{z}_t} = 0$ \eqref{eq:1010}.  Because this boundary difference is constant over time, and doesn't change under conjugation,
 \begin{equation}
   \label{eq:430}
 \bound{\bar{z}_t} = 0, \quad \bound{\bar{z}_{tt}} = 0, \quad \bound{z_t} = 0,  \quad \bound{z_{tt}} = 0.
 \end{equation}
  Recall also that $D_\a$ preserves holomorphicity and periodic boundary behavior of holomorphic functions, so $\bound{D^k_\a \bar z_t}=0$ for any $k\ge 0$ \eqref{eq:434}. For the same reason, we  have
  \begin{equation}
  \label{eq:435}
   \bound{\partial_t D_\a^k \bar{z}_t} = 0,\quad \bound{\bar{D_\a^k \bar{z}_t}} = 0, \quad\bound{\bar{\partial_t D_\a^k \bar{z}_t}} = 0,\qquad k\ge 0.
\end{equation}

Now by applying change of coordinates to \eqref{eq:430} and \eqref{eq:435}, we have
\begin{equation}
  \label{eq:437}
  \bound{\bar{Z}_{t}} = \bound{\bar{Z}_{tt}} = \bound{Z_t} = \bound{Z_{tt}} = 0
\end{equation}
and
\begin{equation}
  \label{eq:259}
  \bound{D_\aa^k \bar{Z}_t} = \bound{\bar{D_\aa^k \bar{Z}_t}} =0 \text{ for }k \ge 0.
\end{equation}
We caution that this periodic boundary behavior does {\em not} necessarily hold for $D_\aa^k \bar{Z}_{tt}$ or $D_\aa^k Z_t$, because the holomorphicity is crucial and disappears under differentiation by $t$ or conjugation.

We do, however, have
\begin{equation}
  \label{eq:567}
  \bound{\Re D_\aa Z_t} = 0.
\end{equation}
This is used in \S \ref{sec:48}. Here, the $\Re$ is the savior. Note that $\Re D_\aa \bar{Z}_t$ is even and $\Im D_\aa \bar{Z}_t$ is odd; this is because $D_\aa$ flips the parity of the real and imaginary parts, and $\Re \bar{Z}_t$ is odd and $\Im \bar{Z}_t$ is even. On the other hand, we know
$\Re \f{\bar{Z}_{,\aa}}{Z_{,\aa}}$ is even and $\Im \f{\bar{Z}_{,\aa}}{Z_{,\aa}}$ is odd.  We write
\begin{equation}
  \label{eq:568}

    \Re (D_\aa Z_t)  = \Re \paren{\f{\bar{Z}_{,\aa}}{Z_{,\aa}} \bar{ D_\aa \bar Z_t }}
     = \paren{ \Re \f{\bar{Z}_{,\aa}}{Z_{,\aa}}}\paren{\Re D_\aa \bar Z_t} + \paren{\Im \f{\bar{Z}_{,\aa}}{Z_{,\aa}}} \paren{\Im D_\aa \bar Z_t }.

\end{equation}
Each of the terms on the RHS of \eqref{eq:568} are even. This concludes the proof of \eqref{eq:567}.

We list a few other boundary properties that we need in the following.  We  have:
\begin{equation}
  \label{eq:438}
  \bound{\f{1}{Z_{,\aa}}} = 0;
\end{equation}
this follows from \eqref{eq:63} and the symmetry of the Riemann Mapping $\Phi$.

We also can state the following:
\begin{equation}
  \label{eq:626}
  \f{1}{Z_{,\aa}}(\pm 1) \equiv 0 \text{ or } \braces{\text{ the angle $\ang$ is $90^\circ$ and } \bound{\f{z_\a}{\abs{z_\a}}} = 0}.
\end{equation}
This holds because $\f{1}{Z_{,\aa}} = 0$ when $\ang < \f{\pi}{2}$; otherwise, $\ang = \f{\pi}{2}$ and so $y_\a(\pm 1) \equiv 0$, and thus $z_\a|_\partial=0$.

We at one point need the fact that
\begin{equation}
  \label{eq:483}
\bound{\paren{\bar{Z}_{ttt} + i \AA \bar{Z}_{t,\aa}}} = \bound{\bar{Z}_{ttt} + (Z_{tt} + i) D_\aa \bar{Z}_t} = 0.
\end{equation}
This follows from the fact that each of the factors $\bar{Z}_{ttt}, (Z_{tt} + i),$ and $D_\aa \bar{Z}_t$ satisfies the periodic boundary conditions.

We also at one point need the fact that
\begin{equation}
  \label{eq:570}
  \bound{\f{h_{t\a}}{h_\a} \circ h\i} = 0.
\end{equation}
This follows because $h$ and therefore $h_t$ are odd, and $\partial_\a$ flips parity.

 We know periodic boundary behavior is preserved by the Hilbert transform $\HH$:

\begin{proposition}\label{prop-good-boundary}
Let $f \in C^0[-1,1]$, with $f'\in L^2[-1,1]$, and $\bound{f} = 0$.  Then
\begin{equation}
  \label{eq:439}
  \bound{(\HH f)} = 0.
\end{equation}
\end{proposition}

We note that functions $f$ satisfying $\bound{f} = 0$ form an algebra; any product of them retain this behavior.  In particular, we often use
\begin{equation}
  \label{eq:524}
  \bound{\f{Z_t}{Z_{,\aa}}} = 0,
\end{equation}
which follows from \eqref{eq:437} and \eqref{eq:438}.  From Proposition~\ref{prop-good-boundary}, we can conclude that
\begin{equation}
  \label{eq:161}
  \bound{\HH \f{Z_t}{Z_{,\aa}}},\quad \bound{\P_A \f{Z_t}{Z_{,\aa}}}, \quad\bound{\P_H \f{Z_t}{Z_{,\aa}}} = 0.
\end{equation}
To apply the proposition, we rely on the fact that $\partial_\aa \f{Z_t}{Z_{,\aa}} \in L^2$, which is  true in our energy regime since $D_\a Z_t\in L^\infty$,  $Z_{t,\aa} \in L^2,\  \partial_\aa \f{1}{Z_{,\aa}}\in L^2$, $\frac1{Z,_\alpha}\in L^\infty$,\footnote{ $\frac1{Z,_\alpha}\in C^0$ because $\frac1{Z,_\alpha}\in L^\infty$ and $\partial_\aa \f{1}{Z_{,\aa}}\in L^2$. } along with a priori assumption $Z_t \in C^0$ and the boundary condition \eqref{eq:524}.

Finally,  let $F=\bar {\vec{v}}$  be the conjugate of the velocity, as defined in \S \ref{sec:27a}. We know that $F$ is periodic holomorphic on $\Omega(t)$, therefore so are $F_t$, $F_{tz}$,  $F_{tt}$, $F_{tzz}$, etc. We show in  \S \ref{sec:28} that
\begin{equation}
  \label{eq:514}
 F_{t} \circ z= \bar{z}_{tt} - (D_\a \bar{z}_t) z_t,\qquad F_{tz} \circ z= D_\a(\bar{z}_{tt} - (D_\a \bar{z}_t) z_t).
 \end{equation}
 Therefore
\begin{equation}
  \label{eq:611}
  \bound{F_t \circ Z}  = 0;\quad
  \bound{F_{tz}\circ z} = \bound{D_\a((\bar{z}_{tt} - (D_\a \bar{z}_t) z_t))} = 0;
  \quad\bound{F_{ttz} \circ Z} = 0;
\end{equation}

\begin{equation}
  \label{eq:67}
  \bound{D_\aa (F_{tz} \circ Z)} = \bound{F_{tzz} \circ Z} = 0.
\end{equation}

\subsection{Holomorphic functions and what disappears under \texorpdfstring{$(I-\HH)$}{(I-H)}}\label{sec:20}
In this section, we note which of the functions we are dealing with are the boundary values of periodic holomorphic functions---and which, further, have mean zero and thus disappear under $(I-\HH)$. From Proposition~\ref{prop:hilbe} we know that, to show that $(I-\HH)$ of various functions disappears, it suffices to show that they are boundary values of periodic holomorphic functions and to show that their means are zero.

We rely fundamentally on the following facts:

First, we have that the conjugate velocity is holomorphic, and goes to zero as $y \to -\infty$ by \eqref{eq:620}, so
\begin{equation}
  \label{eq:571}
  (I-\HH) \bar{Z}_t = 0.
\end{equation}

Then we have three identities about the Riemann mapping. Recall that
\begin{equation}
  \label{eq:627}
  Z_{,\aa} = \partial_\aa \Phi\i(\aa,t);\qquad \f{1}{Z_{,\aa}} = \Phi_z \circ Z.
\end{equation}
Both of these are clearly holomorphic. Therefore, we have
\begin{equation}
  \label{eq:516}
  (I-\HH) \f{1}{Z_{,\aa}} = \avg \f{1}{Z_{,\aa}},\qquad (I-\HH) Z_{,\aa} = 1.
\end{equation}
The mean $\avg Z_{,\aa} = 1$ by the fundamental theorem of calculus, since $Z(1,t) = 1, Z(-1,t) =-1$ for all time.

Finally, we have
\begin{equation}
  \label{eq:312a}
  (I-\HH) \braces{\Phi_t \circ Z} = \avg_I \Phi_t \circ Z.
\end{equation}
Here, we have $\Phi_t \circ \Phi\i$ is holomorphic because it is the limit of holomorphic functions, and we know that $\Phi_t$ is periodic by the Schwarz reflection.

From these facts, we will be able to deduce everything else that we need, from the following principles:
\begin{itemize}
\item If $(I-\HH)f = c$ and $\bound{f} = 0$ then $(I-\HH) \partial_\aa f = 0$, since $[\partial_\aa,\HH]f = 0$ by \eqref{eq:176}.
\item If $(I-\HH) f = 0$ and $(I-\HH) g = c(t)$ then $(I-\HH) (fg) = 0$, since $fg$ is the boundary value of the product of two periodic holomorphic functions, one of which goes to $0$ as $y \to -\infty$, and therefore the product goes to $0$  as $y\to-\infty$.
\item If $(I-\HH) f = 0$ and $\bound{f} = 0$, then $(I-\HH) D_\aa f = 0$, since $(I-\HH) \partial_\aa f = 0$ and $(I-\HH) \f{1}{Z_{,\aa}} = c$.
\item  Therefore, if $G(z,t)$ is a periodic holomorphic function on $\Omega(t)$ going to zero as $y \to -\infty$, that is,
  \begin{equation}
    \label{eq:381}
    (I-\HH) G(z(h\i(\aa,t),t),t) = (I-\HH) (G \circ Z) = 0,
  \end{equation}
then $(I-\HH) G_z(z(h\i(\aa,t),t),t) = 0$, since $G_z \circ Z = D_\aa (G \circ Z)$.
\item If $G(z,t)$ is a periodic holomorphic function on $\Omega(t)$ going to zero as $y \to -\infty$, so $(I-\HH) (G \circ Z) = 0,$ then $(I-\HH) G_t\circ Z = 0$.\footnote{Note that this argument does not apply to $\Phi_t \circ Z$ itself, because $\Phi$ is not periodic.} $G_t \circ Z$ is periodic holomorphic, since it is the limit of periodic holomorphic functions.  It remains to show that $\avg G_t \circ Z = 0$. Note that $\avg G \circ Z = 0$ for all time. Also, since $\Phi(\Phi^{-1}(\aa, t), t)=\a'$, we have $(\Phi_z\circ Z) \cdot(\Phi^{-1})_t + \Phi_t\circ Z=0$, and therefore
\begin{equation}\label{eq:588}
(\Phi^{-1})_t= (-Z_{,\aa})  \Phi_t\circ Z.
\end{equation}
Therefore, using \eqref{eq:588} and the fact $Z=\Phi^{-1}$, we have
  \begin{equation}
    \label{eq:57}
    \begin{aligned}
      0 & = \f{d}{dt} \avg G(Z(\aa,t),t)
       = \avg G_t(Z(\aa,t),t) +\avg G_z \circ Z\, (\Phi^{-1})_t(\aa,t)
      \\ & = \avg G_t \circ Z - \avg (\partial_\aa (G \circ Z)) \Phi_t\circ \Phi^{-1}.
    \end{aligned}
  \end{equation}
It therefore suffices to show that the second integral is zero. To do this, we use $(I-\HH) \partial_\aa (G \circ Z) = 0$, integration by parts, the adjoint property \eqref{eq:471} and \eqref{eq:312a},
\begin{equation}
  \label{eq:60}
  \begin{aligned}
    \avg (\partial_\aa (G \circ Z)) &\Phi_t\circ \Phi^{-1}  = \avg \braces{\P_H (\partial_\aa (G \circ Z))} \Phi_t\circ \Phi^{-1}
    \\ & = \avg (\partial_\aa (G \circ Z)) \P_A \Phi_t\circ \Phi^{-1}
     = - \avg (G \circ Z) \partial_\aa \paren{\P_A \Phi_t\circ \Phi^{-1}}
    \\&=0.
  \end{aligned}
\end{equation}
\end{itemize}

\subsubsection{Identities}\label{sec:21a}
We now present the various identities. These all follow from the above principles and the fact that the quantities have periodic boundary behavior. In general, they can be derived by an inductive argument, and later identities follow from the earlier ones.

\begin{equation}
  \label{eq:242}
  (I-\HH) \partial_\aa \f{1}{Z_{,\aa}} = 0
\end{equation}
\begin{equation}
  \label{eq:315}
  (I-\HH) D_\aa^k \bar{Z}_t = 0,\qquad \text{for } 0\le k\le 3
\end{equation}
\begin{equation}
  \label{eq:474}
 (I-\HH) \partial_\aa D^k_\aa \bar{Z}_t = 0,\qquad \text{for } 0\le k\le 2
\end{equation}
\begin{equation}
  \label{eq:456}
  (I-\HH) \braces{\f{1}{Z_{,\aa}} D_\aa^k \bar{Z}_t} = 0, \qquad \text{for } 0\le k \le 2
\end{equation}

Let $F=\bar{\vec{v}}$ be the conjugate velocity. $F$ is a periodic holomorphic function on $\Omega(t)$ going to $0$ as $y\to -\infty$, so  $ (I-\HH) \braces{F \circ Z} = 0$.
We have the following statements about $F$:
\begin{equation}
  \label{eq:186}
  (I-\HH) \braces{(Z_{,\aa})^jD_\aa^k (F_t \circ Z)} = 0, \qquad\text{for } j=0,1,\ k=0,1
\end{equation}
\begin{equation}
  \label{eq:504}
  (I-\HH) (Z_{,\aa})^j D_\aa^k (F_{tt} \circ Z) = 0, \qquad\text{for } j=0,1,\ k=0,2.
\end{equation}

Recall that we showed in \S \ref{sec:28} that $F_{t} \circ z= \bar{z}_{tt} - (D_\a \bar{z}_t) z_t$ and $F_{tz} \circ z= D_\a(\bar{z}_{tt} - (D_\a \bar{z}_t) z_t)$.
We begin with the basic identity
\begin{equation}
  \label{eq:313}
  (I-\HH) D_\aa^k  (\bar{Z}_{tt} - (D_\aa \bar{Z}_t) Z_t) = 0,\quad \text{for } 0\le k\le 3.
\end{equation}
This gives
\begin{equation}
    \label{eq:222}
    (I-\HH) \partial_\aa D_\aa^k (\bar{Z}_{tt} - (D_\aa \bar{Z}_t) Z_t) = 0,\quad\text{for } 0\le k\le 2.
\end{equation}
We also have
\begin{equation}
  \label{eq:518}
  (I-\HH) \braces{\paren{\P_H \f{Z_t}{Z_{,\aa}}}\partial_\aa D_\aa^2 (\bar{Z}_{tt} - (D_\aa \bar{Z}_t) Z_t)} = 0
\end{equation}
\begin{equation}
  \label{eq:312}
  (I-\HH) \partial_\aa (\Phi_t \circ Z) = 0
\end{equation}
\begin{equation}
  \label{eq:522}
  (I-\HH) \braces{(\partial_\aa D_\aa^2 \bar{Z}_t)\paren{\P_H \f{Z_t}{Z_{,\aa}}}^k} = 0,\quad\text{for }k=1,2
\end{equation}
\begin{equation}
  \label{eq:525}
  (I-\HH) \partial_\aa \paren{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t} = 0
\end{equation}
\begin{equation}
  \label{eq:526}
  (I-\HH) \braces{\paren{\P_H \paren{Z_t \HH \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}}} \partial_\aa \paren{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t}} = 0
\end{equation}
\begin{equation}
  \label{eq:531}
  (I-\HH) \braces{\paren{\P_H \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}}^2 D_\aa^2 \bar{Z}_t} = 0
\end{equation}
\begin{equation}
  \label{eq:535}
  (I-\HH) \braces{(\partial_\aa D_\aa^2 \bar{Z}_t) \paren{\P_H \f{Z_t}{Z_{,\aa}}} (I+\HH) D_\aa Z_t} = 0
\end{equation}
Finally, we have one, rare, equation where the mean is not zero.\footnote{By adjointness, we can write the second mean as $\avg (\P_H \partial_\aa D_\aa^2 \bar{Z}_t) \f{Z_t}{Z_{,\aa}}= \avg Z_t D_\aa^3 \bar{Z}_t$.} Using \eqref{eq:160},
\begin{equation}
  \label{eq:536}
  \begin{split}
    (I-\HH) \braces{\paren{(I+\HH) D_\aa Z_t} \P_H \bracket{(\partial_\aa D_\aa^2 \bar{Z}_t) \paren{\P_A \f{Z_t}{Z_{,\aa}}} }}
\\ = \f{1}{2} \paren{\avg D_\aa Z_t}\paren{\avg (\partial_\aa D_\aa^2 \bar{Z}_t) \paren{\P_A \f{Z_t}{Z_{,\aa}}}}.
\end{split}
\end{equation}

\subsubsection{Mean conditions}
We have implicitly in the preceding section shown that various quantities are mean-zero, but we don't use that fact other than in those identities. We will at one point, though, require an explicit mean-zero condition, to use a variant of the Sobolev inequality.  This is that
\begin{equation}
  \label{eq:48}
  \avg (D_\aa \bar{Z}_t)^2 d\aa = 0.
\end{equation}
We note that $\avg (D_\aa \bar{Z}_t) = 0$ because $(I-\HH) D_\aa \bar{Z}_t = 0$.  Therefore, $D_\aa \bar{Z}_t$ is the boundary value of a periodic holomorphic function going to $0$ as $y \to -\infty$, so its square will also be the boundary value of a periodic holomorphic function going to $0$ as $y \to -\infty$.

\section{Useful inequalities and identities}\label{sec:tech}
We present here assorted inequalities and identities that we need in our paper. None of the results here are original, so we omit the proofs in most cases. We refer the reader to \cite{kinsey} for details of the proofs.

\subsection{Sobolev inequalities and the Peter-Paul trick}\label{sec:sob}
We present here the one-dimensional Sobolev inequality we  use in our proof.

\begin{proposition}[Weighted Sobolev Inequality with $\e$]\label{sobolev-with-weight}
  Let $\e > 0$. Then for all $f \in C^1(-1,1)\cap L^2(-1,1)$,
  \begin{equation}
    \label{eq:396}
    \nm{f}_{L^\infty} \lec \f{1}{\e} \nm{f}_{L^2(\f{1}{\omega})} + \e \nm{f'}_{L^2(\omega)} + \nm{f}_{L^2}
  \end{equation}
for any weight $\omega \ge 0$.

Furthermore
\begin{equation}
  \label{eq:638}
  \avg f^2 = 0 \To \nm{f}_{L^\infty} \lec \f{1}{\e} \nm{f}_{L^2(\f{1}{\omega})} + \e \nm{f'}_{L^2(\omega)}.
\end{equation}
\end{proposition}
The second version, \eqref{eq:638}, will give us slightly more flexibility; the condition $\avg f^2 = 0$ will actually follow from $\avg f = 0$ for $f$ the boundary value of a periodic holomorphic function on $[-1,1]\times (-\infty, 0)$ (see the discussion for \eqref{eq:48}), and so is not so constraining. We omit the proof since it is fairly standard.

We will once use the following ``Peter-Paul'' trick.\footnote{So-called because one takes from Peter to give to Paul.}
\begin{proposition}[Peter-Paul Trick]\label{peter-paul}
Let $f \in C^1(-1,1)$. Suppose
\begin{equation}
  \label{eq:296}
  \nm{D_\aa f}_{L^2} \le c_1 + c_2 \nm{f}_{L^\infty},
\end{equation}
where  $D_\aa := \f{1}{Z_{,\aa}}\partial_\aa$. Suppose further that $\nm{f}_{L^\infty} < \infty$. Then
\begin{equation}
  \label{eq:297}
  \nm{f}_{L^\infty} \lec  (c_2+1)( c_2 \nm{f}_{L^2({\abs{Z_\aa}^2})} + c_1+ \nm{f}_{L^2}),
\end{equation}
where the constant implicit in $\lec$ is universal.
\end{proposition}
\begin{proof}
By the weighted Sobolev inequality \eqref{eq:396} with $\omega = \f{1}{\abs{Z_{,\aa}}^2}$ and then \eqref{eq:296},
\begin{equation}
    \label{eq:298}
    \begin{aligned}
    \nm{f}_{L^\infty} & \le C\paren{\f{1}{\e} \nm{f}_{L^2({\abs{Z_{,\aa}}^2})} + \e \nm{D_\aa f}_{L^2} + \nm{f}_{L^2}}

    \\ & \le C\paren{\f{1}{\e}   \nm{f}_{L^2({\abs{Z_{,\aa}}^2})  } + \e (c_1 + c_2 \nm{f}_{L^\infty}) + \nm{f}_{L^2}}.
  \end{aligned}
\end{equation}
We choose $\e = \min(\f{1}{2C c_2},1)$, so $C\e c_2 \le \f{1}{2}$. Subtracting $Cc_2 \e \nm{f}_{L^\infty}$ from both sides, we get
\begin{equation}
  \label{eq:240}
  \f{1}{2} \nm{f}_{L^\infty} \le C\paren{\f{1}{\e}  \nm{f}_{L^2({\abs{Z_{,\aa}}^2})   } + \e c_1 + \nm{f}_{L^2}},
\end{equation}
which gives desired inequality.
\end{proof}

\subsection{Derivatives and complex-valued functions}\label{sec:comp}

Because our functions will be complex-valued, and we will often be looking at derivatives of angular and modular parts of these functions, we note here a few elementary facts about such functions.

 Let $f(\alpha) = r(\alpha) e^{i\th(\alpha)}$, where $r$, $\theta$ are real-valued functions. Then
\begin{equation}
  \label{eq:100}
 \abs{\partial_\a \abs{f}} \le \abs{f'};\qquad \quad
  \abs{\partial_\a \f{f}{\abs{f}}} \le \abs{\f{f'}{\abs{f}}}.
\end{equation}

From $-\frac{\partial P}{\partial\vec{n}}\ge 0$, we have $\af\ge 0$ and
\begin{equation}
  \label{eq:279}
    \f{z_{tt} + i}{\abs{z_{tt} + i}} = i \f{z_\a}{\abs{z_\a}}.
\end{equation}
We will use this fact in this paper to replace the angular part of the spatial derivative $z_\a$ with the angular part of the time derivative $z_{tt} + i$.

\subsection{Hardy's inequality and commutator estimates }\label{sec:31}
We present here the basic estimates we will rely on for this paper.  Several of these estimates control quantities of the form $[f,\HH] g'$ by something involving $f'$ and $g$; they thus reduce the amount of regularity required on $g$, at the expense of further regularity on $f$.

For many of these estimates, we must pay close attention to the boundary conditions. Recall that $f \in C^0(S^1)$ implies that $\bound{f} = 0$.  Many of these estimates do not hold if this periodic boundary condition is removed. To save space, we have not always explicitly cited these boundary conditions when we quote these estimates, but they are always met, by the results in \S \ref{sec:19}.

\begin{proposition}[Hardy's Inequality]
\label{hardy-inequality}
Let $f \in C^0(S^1) \cap C^1(-1,1)$,
(and so $\bound{f} = 0$),
with $f' \in L^2$. Then there exists $C > 0$ independent of $f$ such that for any $\aa \in I$,
\begin{equation}
  \label{eq:77}
\abs{\int_I \f{(f(\aa) - f(\bb))^2}{\sin^2(\f{\pi}{2}(\aa-\bb))} d\bb} \le C \nm{f'}_{L^2}^2.
\end{equation}
\end{proposition}
\begin{proposition}[$L^2 \times L^\infty$ Estimate]\label{prop-l2-linfty-hh}
 There exists a constant $C > 0$ such that for any $f \in C^0(S^1) \cap C^1(-1,1)$ with $f' \in L^2$, $g \in C^0[-1,1]$ with $g' \in L^p$ for some $p > 1$\footnote{
 We require this only to ensure that $[f,\HH] g'$ is well-defined.} (so $\bound{f} = 0$, though possibly $\bound{g} \not= 0$),
\begin{equation}
  \label{eq:228}
  \nm{[f,\HH]\partial_\aa g}_{L^2} \le C \nm{f'}_{L^2} \nm{g}_{L^\infty}.
\end{equation}
\end{proposition}

\begin{proof}
This result is the periodic modification of a result from \cite{wu2009}, which in turn is a consequence of the $T(b)$ theorem \cite{david-journe-semmes}.

We begin by integrating by parts,
\begin{equation}
  \label{eq:98}
[f,\HH] \partial_\aa g = \HH (f'g) - \f{1}{2i} \int \f{\pi}{2}\f{f(\aa) - f(\bb)}{\sin^2(\f{\pi}{2}(\aa - \bb))} g(\bb) d\b + \bound{\f{1}{2i} (f(\aa) - f(\bb))\cot(\f{\pi}{2}(\aa - \bb)) g(\bb)},
\end{equation}
where we  have a boundary term because we didn't place periodic boundary assumption on $g$.
We control the first term by the $L^2$ boundedness of $\HH$ and H\"older. We control the last term by Hardy's inequality \eqref{eq:77}.

We handle the second term by using the identity
\begin{equation}\label{eq:286}
\paren{\f\pi 2}^2\f1{\sin^2(\f \pi 2 \aa)}=\sum_{l\in \mathbb Z}\f 1{(\aa+2l)^2}
\end{equation}
 and reducing it to  the real line version, Proposition 3.2 in \cite{wu2009}. For details, see \cite{kinsey}.

\end{proof}

\begin{proposition}[$L^2 \times L^\infty$ Estimate Variant]
 There exists a constant $C > 0$ such that for any $f \in C^0(S^1) \cap C^1(-1,1)$ with $f' \in L^2$, $g\in L^\infty[-1,1]$ (and so  $\bound{f} = 0$ but possibly $\bound{g} \not=0$),
  \begin{equation}
    \label{eq:258}
    \nm{ \int \f{(f(\aa) - f(\bb))}{\sin^2(\f{\pi}{2}(\aa - \bb))} g(\bb) d\bb}_{L^2} \le C \nm{f'}_{L^2} \nm{g}_{L^\infty}.
  \end{equation}
\end{proposition}

This is just the second term on the RHS of \eqref{eq:98}.

\begin{proposition}[$L^\infty \times L^2$ Estimate \cite{calderon}]\label{prop-linfty-l2-hh}
 There exists a constant $C > 0$ such that for any $f \in C^1[-1,1] \cap C^0(S^1)$, $g \in C^0(S^1) \cap C^1(-1,1)$ with $g' \in L^p$ for some $p > 1$\footnote{We assume $g' \in L^p$ only to ensure $[f,\HH] g'$ is well-defined.} (and so $\bound{f} = \bound{g} = 0$),
\begin{equation}
  \label{eq:305}
  \nm{[f,\HH] \partial_\aa g}_{L^2} \le C \nm{f'}_{L^\infty} \nm{g}_{L^2}.
\end{equation}
\end{proposition}
To prove Proposition~\ref{prop-linfty-l2-hh}, we begin with the integration-by-part formula \eqref{eq:98}, noting that the third term, the boundary term, is zero since $\bound{f}=\bound{g}=0$. We handle the first term by the $L^2$ boundedness of $\HH$ and H\"older, and the second term by identity \eqref{eq:286}, reducing it to
  the classical result on $\R$ by \cite{calderon}.\footnote{The result was later extended by \cite{coifman-meyer}.}

\begin{proposition}\label{l2-half-deriv-smoothing}
 There exists a constant $C > 0$ such that for any $f, g \in C^1(-1,1) \cap C^0(S^1)$ with $f' \in L^2$ and $g' \in L^p$ for some $p>1$ (and so $\bound{f} = \bound{g} = 0$),
\begin{equation}
  \label{eq:21}
  \nm{[f,\HH] \partial_\aa g}_{L^2} \le C \nm{f'}_{L^2} \nm{g}_{\dot{H}^{1/2}}.
\end{equation}
\end{proposition}

\begin{proof}
  We integrate by parts, first rewriting $\partial_\bb g(\bb) = \partial_\bb (g(\bb) - g(\aa))$:
  \begin{equation}
    \label{eq:13}
    [f,\HH] \partial_\aa g = \f{1}{\hdenomconst} \int f_\bb(\bb)  \cot(\f{\pi}{2}(\aa - \bb))  (g(\bb) - g(\aa)) d\bb - \f{1}{\hdenomconst} \int \f{\pi}{2}\f{f(\aa) - f(\bb)}{\sin^2(\f{\pi}{2}(\aa -\bb))} (g(\bb) - g(\aa)) d\bb,
  \end{equation}
where there is no boundary term because of the periodic boundary conditions.

For the first term, we apply Cauchy-Schwarz:
\begin{equation}
  \label{eq:14}
  \abs{\int f'(\bb) \cot(\f{\pi}{2}(\aa - \bb)) (g(\bb) - g(\aa)) d\bb} \le \nm{f'}_{L^2} \paren{\int \abs{g(\aa) - g(\bb)}^2 \abs{\cot^2(\f{\pi}{2}(\aa - \bb))} d\bb}^{1/2}.
\end{equation}
Taking $L^2$ of this in $\aa$ and using the boundedness of cosine to replace $\cot^2$ with $\f{1}{\sin^2}$, we get the needed estimate.

For the second term, we use Cauchy-Schwarz:
\begin{equation}
  \label{eq:22}
  \nm{ \int \f{f(\aa) - f(\bb)}{\sin^2(\f{\pi}{2}(\aa - \bb))} (g(\bb) - g(\aa)) d\bb}_{L^2_\aa} \le \paren{\int \int \f{\abs{f(\aa) - f(\bb)}^2}{\sin^2(\f{\pi}{2}(\aa - \bb))} d\bb \int \f{\abs{g(\aa) - g(\bb)}^2}{\sin^2(\f{\pi}{2}(\aa - \bb))} d\bb d\aa}^{1/2},
\end{equation}
and then we use Hardy's inequality \eqref{eq:77} to $f$ to get our inequality \eqref{eq:21}.
\end{proof}

\begin{proposition}
  There exists a constant $C > 0$ such that for any  $f \in \dot{H}^{1/2},  g \in L^2$ (and so $\bound{f} = 0$),
\begin{equation}
  \label{eq:96}
  \nm{[f,\HH]g}_{L^2} \le C \nm{f}_{\dot{H}^{1/2}} \nm{g}_{L^2}.
\end{equation}
\end{proposition}
\begin{proof}
  This is immediate by Cauchy-Schwarz and the boundedness of cosine.
\end{proof}
\begin{proposition}
 There exists a constant $C > 0$ such that for any $f, g \in C^1(-1,1) \cap C^0(S^1)$ with $f', g' \in L^2$ and $h \in L^2$ (and so $\bound{f}, \bound{g} = 0$),
  \begin{equation}
    \label{eq:113}
    \nm{[f,g;h]}_{L^2} := \nm{\frac\pi{4i}\int \f{f(\aa) - f(\bb)}{\sin(\f{\pi}{2}(\aa - \bb))} \f{g(\aa) - g(\bb)}{\sin(\f{\pi}{2}(\aa - \bb))} h(\bb) d\bb}_{L^2} \le C \nm{f'}_{L^2} \nm{g'}_{L^2} \nm{h}_{L^2}.
  \end{equation}
\end{proposition}

\begin{proof}
By Cauchy-Schwarz,
  \begin{multline}
    \label{eq:114}
    \abs{\int \f{f(\aa) - f(\bb)}{\sin(\f{\pi}{2}(\aa - \bb))} \f{g(\aa) - g(\bb)}{\sin(\f{\pi}{2}(\aa - \bb))} h(\bb) d\bb} \\\le \paren{\int \abs{\f{f(\aa) - f(\bb)}{\sin(\f{\pi}{2}(\aa - \bb))}}^2 d\bb}^{1/2} \paren{ \int  \abs{\f{g(\aa) - g(\bb)}{\sin(\f{\pi}{2}(\aa - \bb))} h(\bb)}^{2} d\bb}^{1/2}.
  \end{multline}
Now we take the $L^2$ of this in the $\alpha'$ variable. By Hardy's inequality \eqref{eq:77}, we control the $f$ factor by $\nm{f'}_{L^2}$, and are left with
\begin{equation}
  \label{eq:115}
  \paren{\int  \int  \abs{\f{g(\aa) - g(\bb)}{\sin(\f{\pi}{2}(\aa - \bb))} h(\bb)}^{2} d\bb d\aa}^{1/2}.
\end{equation}
Applying Fubini and then using Hardy's inequality \eqref{eq:77} once more gives the result.
\end{proof}

\begin{proposition}
 There exists a constant $C > 0$ such that for any $f \in C^1(-1,1) \cap C^0(S^1)$ with $f' \in L^2$, $g \in L^2$ (and so $\bound{f} = 0$),
  \begin{equation}
    \label{eq:32}
    \nm{[f,\HH] g}_{L^\infty} \le C \nm{f'}_{L^2} \nm{g}_{L^2}.
  \end{equation}
 \end{proposition}
\begin{proof}
  Estimate \eqref{eq:32} holds by Cauchy-Schwarz and Hardy's inequality \eqref{eq:77}.
\end{proof}

\begin{proposition}
 There exists a constant $C > 0$ such that for any $f, g \in C^1(-1,1) \cap C^0(S^1)$ with $f', g' \in L^2$, and $h \in L^2$ (and so $\bound{f}, \bound{g} = 0$),
\begin{equation}
  \label{eq:144}
  \nm{\partial_\aa [f,[g,\HH]] h}_{L^2} \lec \nm{f'}_{L^2} \nm{g'}_{L^2} \nm{h}_{L^2}.
\end{equation}
\end{proposition}

\begin{proof}
We differentiate:
\begin{multline}
  \label{eq:363}
  \partial_\aa \pv \f{1}{\hdenomconst} \int (f(\aa) - f(\bb)) (g(\aa) - g(\bb)) \cot(\f{\pi}{2}(\aa -\bb)) h(\bb) d\bb
\\ = f' [g,\HH] h + g'[f,\HH] h  - \pv \f{1}{\hdenomconst} \int \f{\pi}{2}\f{(f(\aa) - f(\bb))(g(\aa) - g(\bb))}{\sin^2(\f{\pi}{2}(\aa -\bb))} h(\bb) d\bb.
\end{multline}
We control the first two terms by H\"older and then \eqref{eq:32}. We control the last term by \eqref{eq:113}.
\end{proof}

\begin{proposition}[Higher-Order Calderon Commutator \cite{coifman-meyer}]
There exists a constant $C > 0$ such that for any $f \in C^1(-1,1) \cap C^0(S^1)$ with $f' \in L^\infty$, $h \in C^0(S^1) \cap C^1(-1,1)$ with $h' \in L^p$ for some $p > 1$ (and so $\bound{f}, \bound{h} = 0$)
  \begin{equation}
  \label{eq:6a}
  \nm{[f,f; \partial_\aa h]}_{L^2} \le C \nm{f'}_{L^\infty}^2 \nm{h}_{L^2}.
\end{equation}
\end{proposition}
\begin{proof}
The proof is entirely analogous to the proof of \eqref{eq:305}, and now follows from the work of \cite{coifman-meyer}, which extends the original result of \cite{calderon} used for \eqref{eq:305} to allow two difference quotient factors, instead of one. To move from $\R$ to our compact domain, we do the same infinite summation argument, with the only difference being that instead of \eqref{eq:286},  we use $(\f\pi 2)^2\partial_\aa \f{1}{\sin^2(\f{\pi}{2}\aa)} =  -\sum \f{1}{(\aa + 2l)^3}$.
\end{proof}

\subsection{The \texorpdfstring{$\dot{H}^{1/2}$}{dot H 1/2} norm}\label{sec:half}
We present here the following proposition.
\begin{proposition}\label{prop:half}
  Let $f \in C^1(-1,1) \cap C^0(S^1)$ with $f' \in L^2$. Then
\begin{equation}
  \label{eq:549}
  (I-\HH)f = \avg f \To \nm{f}_{\dot{H}^{1/2}}^2 =  \int i (\partial_\aa f) \bar{f} d\aa.
\end{equation}
\end{proposition}

\eqref{eq:549} holds because for $f$ satisfying the assumption of Proposition~\ref{prop:half},
\begin{equation}
  \label{eq:26}
  \int i (\partial_\aa f) \bar{f} d\aa= \int i (\partial_\aa \HH f) \bar{f} d\aa,
\end{equation}
and $i \,\partial_\aa \HH f=|D|f$, where $|D|$ is the positive operator satisfying $|D|^2=-\partial^2_\aa$.
It is easy to see that $\int i (\partial_\aa f) \bar{f} d\aa$ is real-valued by integration by parts.

\subsection{Commutator identities}\label{sec:33}
We include here for reference the various commutator identities that are necessary.
\begin{align}
  \label{eq:120}
  [\partial_t,D_\a] &= - (D_\a z_t) D_\a;\\
  \label{eq:121}
    \bracket{\partial_t,D_\a^2} &   = [\partial_t,D_\a] D_\a + D_\a [\partial_t,D_\a]
        = -2(D_\a z_t) D_\a^2 - (D_\a^2 z_t) D_\a;\\
  \label{eq:212}
     \bracket{\partial_t^2,D_\a} &=\partial_t\bracket{\partial_t,D_\a} + \bracket{\partial_t,D_\a}\partial_t= (-D_\a z_{tt}) D_\a + 2(D_\a z_t)^2 D_\a - 2(D_\a z_t) D_\a \partial_t.
  \end{align}
To calculate $[i\mathfrak{a}\partial_\a,D_\a]$, we use $i\mathfrak{a} z_\a = z_{tt} + i$ \eqref{eq:8} to rewrite $i\mathfrak{a} \partial_\a = i \mathfrak{a} z_\a D_\a = (z_{tt} + i) D_\a$. Therefore
\begin{equation}
  \label{eq:47}
  [i\mathfrak{a}\partial_\a,D_\a] = [(z_{tt}+i)D_\a,D_\a] = -(D_\a z_{tt}) D_\a.
\end{equation}
Adding \eqref{eq:212} and \eqref{eq:47}, we conclude that
\begin{equation}
  \label{eq:53}
   \bracket{\partial_t^2 + i \mathfrak{a} \partial_\a, D_\a} =  (-2D_\a z_{tt}) D_\a + 2(D_\a z_t)^2 D_\a - 2(D_\a z_t) D_\a \partial_t.
\end{equation}

Because $[(\partial_t^2 + i \af \partial_\a),D_\a^2] = [(\partial_t^2 + i \af \partial_\a),D_\a] D_\a + D_\a[(\partial_t^2 + i \af \partial_\a),D_\a],$ we have
\begin{equation}
  \label{eq:52}
  \begin{aligned}
    \bracket{(\partial_t^2 + i \af \partial_\a),D_\a^2} & =

(-4D_\a z_{tt}) D_\a^2 + 4(D_\a z_t)^2 D_\a^2 - 2(D_\a z_t) D_\a \partial_t D_\a - (2D_\a^2 z_{tt}) D_\a
\\ & + 4(D_\a z_t) (D_\a^2 z_t) D_\a - 2(D_\a^2 z_t) D_\a \partial_t - 2(D_\a z_t) D_\a^2 \partial_t.
\end{aligned}
\end{equation}

\section{Summary of notation}\label{notation}
We list here the various notations we've introduced in the paper. See also \S \ref{sec:5} for a discussion of the conventions used.

\begin{itemize}
\item $\bound{f} := f(1,t) - f(-1,t)$.
\item  $\ang$ is the angle the water wave makes with the wall. See Figure~\ref{fig:b}.
\item $I := [-1,1]$ (except when it's used for the identity or as an abbreviation for a quantity to be controlled).
\item $\Re z$, $\Im z$ are the real and imaginary parts of a complex number $z$.
\item  Function spaces $C^k(-1,1), C^k[-1,1], C^k(S^1)$,  $L^p$, and $\dot{H}^{1/2}$    are defined in \S \ref{sec:5}. We define $\nm{f}_{\dot{H}^{1/2}} := \paren{\frac\pi 8\iint \f{\abs{f(\aa) - f(\bb)}^2}{\sin^2(\f{\pi}{2}(\aa -\bb))} d\aa d\bb}^{1/2}$.
\item $\avg f = \avg_I f  := \f{1}{2} \int_I f(\bb) d\bb$.
\item $z(\a,t)$ is the Lagrangian parametrization, and $z_t(\a,t) = \vec{v}(z(\a,t),t)$ is the velocity. $z_{tt} = \partial_t z_t$, etc.
\item $\af = \f{\abs{\bar{z}_{tt} - i}}{\abs{z_\a}} = \f{-\partial P}{\partial \vec{n}} \f{1}{\abs{z_\a}}$.
We refer to $-\f{\partial P}{\partial \vec{n}}$ as the Taylor coefficient; $\vec{n}$ is the outward-facing normal to $\Sigma(t)$.
\item $h: \a \mapsto \aa$ is defined by $h(\a) = \Phi(z(\a,t),t)$ and gives the Riemannian coordinates, where $\Phi$ is the Riemann mapping defined in \S \ref{sec:12a}.  $\partial_\aa (f \circ h\i) = \f{\partial_\a f}{h_\a} \circ h\i$.  $d\aa = h_\a d\a$. Full details are in \S \ref{sec:12a}.
\item $\alpha$ and $\beta$ are our variables in Lagrangian coordinates; $\aa$ and $\bb$ are our variables in Riemannian coordinates.
\item $\HH$ is the Hilbert transform in Riemannian coordinates, defined by
  \begin{equation}
    \label{eq:665}
    \HH f(\aa) := \f{1}{\hdenomconst} \pv \int_I \cot(\f{\pi}{2}(\aa-\bb)) f(\bb) d\bb.
  \end{equation}
We use $\H$ to refer to the Hilbert transform for the curved domain in Lagrangian coordinates.
\item We define $\P_A := \f{(I-\HH)}{2}$ and $\P_H := \f{(I+\HH)}{2}$ as the antiholomorphic and holomorphic projections.
\item $[f,g;h](\aa) := \frac\pi{4i}\int \f{(f(\aa) - f(\bb))(g(\aa) - g(\bb))}{\sin^2(\f{\pi}{2}(\aa - \bb))} h(\bb) d\bb$, the higher-order Calderon commutator.
\item We use $F(z(\a,t),t) := \bar{z}_t(\a,t)$ at several points (and do not to use $F$ for any other purpose).
\item $Z := z \circ h\i, Z_t := z_t \circ h\i, Z_{tt} := z_{tt} \circ h\i$. $Z_{,\aa} = \partial_\aa(z \circ h\i), Z_{t,\aa} = \partial_\aa(z_t \circ h\i)$, etc.
\item Compositions and inverses are always with respect to the spatial variable.
\item $\AA := (\af h_\a) \circ h\i$, $\AAt := (\af_t h_\a) \circ h\i$.
 \item $D_\a := \f{1}{z_\a} \partial_\a$. $\abs{D_\a} := \f{1}{\abs{z_\a}} \partial_\a$; $D_\aa := \f{1}{\abs{Z_{,\aa}}} \partial_\aa$. $\abs{D_\aa} := \f{1}{\abs{Z_{,\aa}}} \partial_\aa$.
\item $A_1 = \AA \abs{Z_{,\aa}}^2 = i Z_{,\aa}(\bar{Z}_{tt} - i) \in \R$ \eqref{eq:208}. On changing variables, we have
  \begin{equation}
    \label{eq:663}
    A_1 \circ h = \f{\af \abs{z_\a}^2}{h_\a},
  \end{equation}
  originally derived at \eqref{eq:209}; we use this repeatedly without citation.  We  often use $A_1 \ge 1$ \eqref{eq:394}, and also use $\f{1}{Z_{,\aa}} = i \f{\bar{Z}_{tt} - i}{A_1}$ \eqref{eq:251}.
\item We define our energies in \S \ref{sec:35}. We define generic energies $E_{a,\th}$ and $E_{b,\th}$, and then specialize to $E_a := E_{a,D_\a^2 \bar{z}_t}$ and $E_b := E_{b,D_\a \bar{z}_t}$.  We use $G_\th$ to describe the RHS of the equation $(\partial_t^2 + i \af \partial_\a)\th  = G_\th$. For $\th = D_\a^k \bar{z}_t$, $G_\th = D_\a^k(-i\mathfrak{a}_t \bar{z}_\a) + [\partial_t^2 + i \mathfrak{a}\partial_\a,D_\a^k] \bar{z}_t.$
\item $\Th := \th \circ h\i$;  $B := \paren{\f{h_{t\a}}{h_\a} - \Re D_\aa z_t} \circ h\i$; $\psi := \paren{\f{h_\a}{z_\a} \th} \circ h\i$ \eqref{eq:83}.
\item See \S \ref{sec:5} for a discussion of how broadly $I, II, I_1, I_{12}$, etc., are defined. In short, they are unambiguous within each section, but ambiguous between sections. \item We use $C(E)$ to represent a polynomial of the energy $E$.
\end{itemize}

\section{Main quantities controlled} \label{quantities}
We list here the various quantities that are controlled by our energy, for ease of reference.  We don't list every single quantity we have controlled, but we do include any quantities that we give at the end of a concluding inequality without further explanation.

\begin{itemize}
\item
In \S \ref{sec:quants}, we controlled
\begin{equation}\label{eq:1550}
\begin{aligned}
&\nm{D_\aa^2 \bar{Z}_{tt}}_{L^2}, \nm{D_\aa^2 {Z}_{tt}}_{L^2},  \nm{D_\aa^2 \bar{Z}_t}_{L^2}, \nm{D_\aa^2 {Z}_t}_{L^2}, \nm{D_\a \partial_t D_\a \bar{z}_t}_{L^2({h_\a}d\a)},\\& \nm{\f{1}{Z_{,\aa}} D_\aa^2 \bar{Z}_t}_{\dot{H}^{1/2}}, \nm{D_\aa \bar{Z}_{tt}}_{L^\infty}, \nm{D_\aa {Z}_{tt}}_{L^\infty},  \nm{D_\aa \bar{Z}_t}_{L^\infty}, \nm{D_\aa {Z}_t}_{L^\infty},\\&  \nm{ \bar{Z}_{tt, \aa}}_{L^2},   \nm{ \bar{Z}_{t, \aa}}_{L^2},  \int \abs{D_\a \bar{z}_t}^2 \f{d\a}{\af},\  \int \abs{D_\a \bar{z}_{tt}}^2 \f{d\a}{\af},  \ \nm{\f{1}{Z_{,\aa}}}_{L^\infty}, \nm{Z_{tt}+i}_{L^\infty} , \nm{A_1}_{L^\infty}.
\end{aligned}
\end{equation}

\item $\nm{\f{\af_t}{\af}}_{L^\infty} = \nm{\f{\AAt}{\AA}}_{L^\infty}$ is controlled at \eqref{eq:424} in \S \ref{sec:at-a}.
\item $\nm{\partial_\aa \f{1}{Z_{,\aa}}}_{L^2}$ is controlled at \eqref{eq:425} in \S \ref{sec:44}.
\item $\nm{\f{h_{t\a}}{h_\a}}_{L^\infty}$ is controlled at \eqref{eq:451} in \S \ref{hta-ha}.
\item $\nm{(I+\HH) D_\aa Z_t}_{L^\infty}$ is controlled at \eqref{eq:170} in \S \ref{sec:45c}.
\item $\nm{D_\aa \f{1}{Z_{,\aa}}}_{L^\infty}$ is controlled at \eqref{eq:1023} in \S \ref{sec:46}. The related term $\nm{(Z_{tt} + i) \partial_\aa \f{1}{Z_{,\aa}}}_{L^\infty}$ is also estimated there.
\item $\nm{\partial_\aa \P_A \f{Z_t}{Z_{,\aa}}}_{L^\infty}$ is controlled at \eqref{eq:513} in \S \ref{sec:47}. The related term $\nm{\P_A \paren{Z_t \partial_\aa \f{1}{Z_{,\aa}}}}_{L^\infty}$ we estimated at \eqref{eq:151}.
\end{itemize}

\end{appendices}

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