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Navier-Stokes, exterior flow, three dimensions, stationary solutions, asymptotic behavior, laminar wake
---------------0310291149206
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@mastersthesis{haldi001,
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                                         Analyse du cas asymétrique à deux dimensions}
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\begin{document}

\title{Leading order down-stream asymptotics of non-symmetric stationary
Navier-Stokes flows in three dimensions}
\author{Peter Wittwer\thanks{Supported in part by the Fonds National Suisse.}\\{\small D\'{e}partement de Physique Th\'{e}orique}\\{\small Universit\'{e} de Gen\`{e}ve, Switzerland}\\{\small peter.wittwer@physics.unige.ch}}
\maketitle

\begin{abstract}
We consider stationary solutions of the incompressible Navier-Stokes equations
in three dimensions. We give a detailed description of the fluid flow in a
half-space through the construction of an inertial manifold for the dynamical
system that one obtains when using the coordinate along the flow as a time.

\end{abstract}

\section{Introduction}

In this paper we generalize the techniques introduced in \cite{pw003} from two
to three dimensions. This generalization is in many ways straightforward, so
that the results follow as in \cite{pw003}, \textit{mutatis mutandis}. New are
the presence of a branch of zero-modes for the velocity field and the
nontrivial algebraic structure of the equations for the velocity and vorticity
components transverse to the flow.

\subsection{Statement of the problem}

We consider, in $d=3$ dimensions, the time independent incompressible
Navier-Stokes equations
\begin{align}
-(\mathbf{u}\cdot\mathbf{\nabla})\mathbf{u}+\Delta\mathbf{u}-\mathbf{\nabla}p
&  =\mathbf{0}~,\label{navier}\\
\mathbf{\nabla}\cdot\mathbf{u} &  =0~,\label{div0}%
\end{align}
in a half-space $\Omega=\{(x,\mathbf{y})\in\mathbf{R}^{3}\left\vert {}\right.
x\geq1\}.$ We are interested in modeling a situation where fluid enters the
half-space $\Omega$ through the surface $\Sigma=\{(x,\mathbf{y})\in
\mathbf{R}^{3}\left\vert {}\right.  x=1\}$ and where the fluid flows at
infinity parallel to the $x$-axis at a nonzero constant speed $\mathbf{u}%
_{\mathbf{\infty}}\equiv(1,\mathbf{0})$. We therefore impose the boundary
conditions
\begin{align}
\lim_{\substack{x^{2}+\left\vert \mathbf{y}\right\vert ^{2}\rightarrow
\infty\\x\geq1}}\mathbf{u}(x,\mathbf{y})  & =\mathbf{u}_{\mathbf{\infty}%
}~,\label{bc001}\\
\left.  \mathbf{u}\right\vert _{\Sigma}  & =\mathbf{u}_{\mathbf{\infty}%
}+\mathbf{u}_{\ast}~,\label{bc002}%
\end{align}
with $\mathbf{u}_{\ast}$ in a certain set of vector fields satisfying
$\lim_{\left\vert \mathbf{y}\right\vert \rightarrow\infty}\mathbf{u}_{\ast
}(\mathbf{y})=\mathbf{0}.$

\bigskip

The following theorem is our main result.

\begin{theorem}
\label{theorem0}Let $\Sigma$ and $\Omega$ be as defined above. Then, for each
$\mathbf{u}_{\ast}=(u_{\ast},\mathbf{v}_{\ast})$ in a certain set of vector
fields $\mathcal{S}$ to be defined later on, there exist a vector field
$\mathbf{u}=\mathbf{u}_{\infty}+(u,\mathbf{v})$ and a function $p$ satisfying
the Navier-Stokes equations (1) and (2) in $\Omega$ subject to the boundary
conditions (3) and (4). Furthermore,
\begin{align}
\lim\limits_{x\rightarrow\infty}x\left(  \sup\limits_{\mathbf{y}\in
\mathbf{R}^{2}}\left\vert \left(  u-u_{as}\right)  (x,\mathbf{y})\right\vert
\right)   &  =0~,\label{th2}\\
\lim\limits_{x\rightarrow\infty}x^{3/2}\left(  \sup\limits_{\mathbf{y}%
\in\mathbf{R}^{2}}\left\vert \left(  \mathbf{v}_{1}-\mathbf{v}_{1,as}\right)
(x,\mathbf{y})\right\vert \right)   &  =0~,\label{th3}\\
\lim\limits_{x\rightarrow\infty}x\left(  \sup\limits_{\mathbf{y}\in
\mathbf{R}^{2}}\left\vert \left(  \mathbf{v}_{2}-\mathbf{v}_{2,as}\right)
(x,\mathbf{y})\right\vert \right)   &  =0~,\label{th3a}%
\end{align}
where $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ are the irrotational and
divergence free parts of $\mathbf{v}$, respectively, where
\begin{align}
u_{as}(x,\mathbf{y}) &  =\frac{1}{4\pi x}\theta(x)e^{-\frac{y^{2}}{4x}%
}~c+\frac{1}{2\pi}\frac{x}{\left(  x^{2}+y^{2}\right)  ^{\frac{3}{2}}}%
~d+\frac{1}{2\pi}\frac{\mathbf{y}\cdot\mathbf{b}}{\left(  x^{2}+y^{2}\right)
^{\frac{3}{2}}}~,\label{tuas}\\
\mathbf{v}_{1,as}(x,\mathbf{y}) &  =\frac{\mathbf{y}}{8\pi x^{2}}%
\theta(x)e^{-\frac{y^{2}}{4x}}~c+\frac{1}{2\pi}\frac{\mathbf{y}}{\left(
x^{2}+y^{2}\right)  ^{\frac{3}{2}}}~d\nonumber\\
&  -\frac{1}{2\pi}\frac{1}{\sqrt{x^{2}+y^{2}}}\frac{\mathrm{sign}(x)}%
{\sqrt{x^{2}+y^{2}}+\left\vert x\right\vert }\left(  \mathbf{1}-\frac{1}%
{\sqrt{x^{2}+y^{2}}}\left(  \frac{1}{\sqrt{x^{2}+y^{2}}}+\frac{1}{\sqrt
{x^{2}+y^{2}}+\left\vert x\right\vert }\right)  \mathbf{yy}^{T}\right)
\mathbf{b}~,\label{tv1as}\\
\mathbf{v}_{2,as}(x,\mathbf{y}) &  =\frac{1}{4\pi x}\theta(x)e^{-\frac{y^{2}%
}{4x}}~\mathbf{a}\nonumber\\
&  +\frac{1}{2\pi}\theta(x)\left(  \frac{1}{y^{2}}\left(  e^{-\frac{y^{2}}%
{4x}}-1\right)  \mathbf{1}-2\frac{1}{y^{4}}\left(  e^{-\frac{y^{2}}{4x}%
}-1+\frac{y^{2}}{4x}e^{-\frac{y^{2}}{4x}}\right)  \mathbf{yy}^{T}\right)
\mathbf{a}~,\label{tvas}%
\end{align}
with $\theta$ the Heaviside function, with $y=\sqrt{y_{1}^{2}+y_{2}^{2}}$,
where $(y_{1},y_{2})=\mathbf{y}$, with $\mathbf{1}$ the unit $2\times2$
matrix, with $\mathbf{yy}^{T}$ the $2\times2$ matrix with entries $\left(
\mathbf{yy}^{T}\right)  _{ij}=y_{i}y_{j}$, and where the numbers $c$ and $d$
and the vectors $\mathbf{a}$ and $\mathbf{b}$ are related to the initial
conditions $u_{\ast}$ and $\mathbf{v}_{\ast}$ as follows,%
\begin{align}
d  & =\left\langle -i\mathbf{e}^{T}\lim_{k\rightarrow0}\widehat{\mathbf{v}%
_{\ast,1}}(k\mathbf{e})\right\rangle ~,\label{thddd}\\
c  & =-d+\left\langle \lim_{k\rightarrow0}\widehat{u_{\ast}}(k\mathbf{e}%
)\right\rangle ~,\label{thccc}\\
\mathbf{b}  & =\left\langle -i\mathbf{e}\lim_{k\rightarrow0}\widehat{u_{\ast}%
}(k\mathbf{e})\right\rangle ~,\label{thbbb}\\
\mathbf{a}  & =-\mathbf{b}+\left\langle 2\lim_{k\rightarrow0}\widehat
{\mathbf{v}_{\ast,2}}(k\mathbf{e})\right\rangle -\left\langle 2\lim
_{k\rightarrow0}\widehat{\mathbf{v}_{\ast,1}}(k\mathbf{e})\right\rangle
~,\label{thaaa}%
\end{align}
where $~\widehat{}~$ denotes Fourier transform, where $\mathbf{e\equiv
e}(\vartheta)=(\cos(\vartheta),\sin(\vartheta))$ and where $\mathbf{v}%
_{\ast,1}$ and $\mathbf{v}_{\ast,2}$ are the irrotational and divergence free
parts of $\mathbf{v}_{\ast}$, respectively. The average $\left\langle
~.~\right\rangle $ is defined by the equation%
\begin{equation}
\left\langle ~.~\right\rangle =\frac{1}{2\pi}\int_{0}^{2\pi}~.~~d\vartheta
~.\label{average}%
\end{equation}

\end{theorem}

\bigskip

A proof of this theorem is given in Section 9.

\bigskip

The set $\mathcal{S}$\ in Theorem \ref{theorem0} will be specified in Section
9, once appropriate function spaces have been introduced.\textit{\ }For an
interpretation of the results see \cite{pw001}, \cite{pw002}. For related
results see for example \cite{gald98a}, \cite{gald98b}, \cite{Deuring001},
\cite{Galdi0001}, \cite{maria001}, \cite{Farwig001} and references therein.
For an application of related two-dimensional results for an numerical
implementation of two-dimensional stationary exterior flow problems see
\cite{vincent001}.

\subsection{Organization of the paper}

The rest of this paper is organized as follows. In Section 2, Section 3 and
Section 4 we rewrite equation (\ref{navier}) and Section (\ref{div0}) as a
dynamical system with the coordinate parallel to the flow playing the role of
time. The discussion will be formal. At the end of the discussion we get a set
of integral equations. In Sections 5 and 6 we then prove that these integral
equations admit a solution. This solution is analyzed in some detail in
Section 7 and Section 8. In Section 9 we finally prove Theorem \ref{theorem0}
by using the results from Sections 5-8.

\section{The dynamical system}

Define, for given $\mathbf{u}$ and $p$, the vector field $\mathbf{P}$ by the
equation
\begin{equation}
\mathbf{P}=-(\mathbf{u\cdot\nabla})\mathbf{u}+\Delta\mathbf{u}-\mathbf{\nabla
}p~.\label{pp}%
\end{equation}
With this notation, the Navier-Stokes equations (\ref{navier}), (\ref{div0})
are
\begin{align}
\mathbf{P}  & =\mathbf{0}~,\label{p0}\\
\mathbf{\nabla\cdot u}  & =0~.\label{nsp}%
\end{align}
Let $\mathbf{W}=\mathbf{\nabla\times P}$ be the vorticity of the vector field
$\mathbf{P}$. After some vector algebra (or see \textit{e.g.} \cite{batchelor}%
), we find from (\ref{pp}) that
\begin{equation}
\mathbf{W}=\mathbf{\nabla\times}\left(  \mathbf{u\times\omega}\right)
+\Delta\mathbf{\omega}~,\label{ww0}%
\end{equation}
with%
\begin{equation}
\mathbf{\omega=\nabla\times u}\label{omega}%
\end{equation}
the vorticity of the fluid. Note that%
\begin{align}
\mathbf{\nabla\cdot\omega}  & =0~,\label{dviw0}\\
\mathbf{\nabla\cdot W}  & =0~.\label{dviW0}%
\end{align}
The Navier-Stokes equation (\ref{nsp}) can be solved by first determining
$\mathbf{\omega}$ and $\mathbf{u}$ from the vorticity equations%
\begin{equation}
\mathbf{W}=\mathbf{0}~,\label{ww}%
\end{equation}
together with (\ref{omega}) and (\ref{nsp}), and then the pressure $p$ by
solving the Poisson equation
\[
\Delta p=\mathbf{\nabla\cdot}\left(  (\mathbf{u\cdot\nabla})\mathbf{u}%
-\Delta\mathbf{u}\right)
\]
in $\Omega$, subject to the boundary condition $\mathbf{P}=\mathbf{0}$ normal
to the boundary $\Sigma$.

As in \cite{pw003} we consider now the coordinate parallel to the flow as a
time coordinate, and rewrite the equations (\ref{ww}) as a dynamical system.
We first introduce some notation. Let $\mathbf{x}=(x,\mathbf{y})$ with
$\mathbf{y}=(y_{1},y_{2})$, $\mathbf{u}=(1,\mathbf{0})+(u,\mathbf{v})$ with
$\mathbf{v}=(v_{1},v_{2})$, $\mathbf{\omega}=(w,\mathbf{\tau})$ with
$\mathbf{\tau}=(\tau_{1},\tau_{2})$, $\mathbf{\nabla}=(\partial_{x}%
,\mathbf{\nabla}^{\mathbf{\bot}})$ with $\mathbf{\nabla}^{\mathbf{\bot}%
}=(\partial_{y_{1}},\partial_{y_{2}})$, $\Delta^{\mathbf{\bot}}%
=\mathbf{\partial}_{y_{1}}^{2}+\mathbf{\partial}_{y_{2}}^{2}$, and let
$\mathbf{W}=(W,\mathbf{T)}$. Let furthermore $\mathbf{\sigma}_{2}$ be the
second Pauli matrix, \textit{i.e.},%
\[
\mathbf{\sigma}_{2}=\left(
\begin{array}
[c]{cc}%
0 & -i\\
i & 0
\end{array}
\right)  ~.
\]
Then, we find for $\mathbf{\omega}$ as defined in (\ref{omega})
\begin{equation}
\mathbf{\omega=}%
\begin{pmatrix}
\mathbf{\nabla}^{\mathbf{\bot}}\cdot i\mathbf{\sigma}_{2}\mathbf{v}\\
-i\mathbf{\sigma}_{2}\partial_{x}\mathbf{v}+i\mathbf{\sigma}_{2}%
\mathbf{\nabla}^{\mathbf{\bot}}u
\end{pmatrix}
~,\label{nxu}%
\end{equation}
and similarly,
\begin{equation}
\mathbf{u\times\omega=}%
\begin{pmatrix}
\mathbf{v}\cdot i\mathbf{\sigma}_{2}\mathbf{\tau}\\
-i\mathbf{\sigma}_{2}\mathbf{\tau}-i\mathbf{\sigma}_{2}\left(  u\mathbf{\tau
-}\omega\mathbf{v}\right)
\end{pmatrix}
~.\label{uw}%
\end{equation}
Therefore,%
\begin{equation}
\mathbf{\nabla\times}\left(  \mathbf{u\times\omega}\right)  =%
\begin{pmatrix}
\mathbf{\nabla}^{\mathbf{\bot}}\cdot\mathbf{\tau}\\
-\partial_{x}\mathbf{\tau}%
\end{pmatrix}
+%
\begin{pmatrix}
\mathbf{\nabla}^{\mathbf{\bot}}\cdot\mathbf{q}_{0}\\
-\mathbf{q}%
\end{pmatrix}
~,\label{nonl}%
\end{equation}
where%
\begin{equation}
\mathbf{q=}\partial_{x}\mathbf{q}_{0}-i\mathbf{\sigma}_{2}\mathbf{\nabla
}^{\mathbf{\bot}}q_{1}~,\label{q}%
\end{equation}
and%
\begin{align}
\mathbf{q}_{0}  & =u\mathbf{\tau-}\omega\mathbf{v}~,\label{q0}\\
q_{1}  & =\mathbf{v}\cdot i\mathbf{\sigma}_{2}\mathbf{\tau}~.\label{q1}%
\end{align}
Using (\ref{nxu}) and (\ref{nonl}), we find that the equations (\ref{ww}),
(\ref{omega}) and (\ref{nsp}) are in component form equal to%
\begin{align}
\omega & =\mathbf{\nabla}^{\mathbf{\bot}}\cdot i\mathbf{\sigma}_{2}%
\mathbf{v}~,\label{ee1}\\
\mathbf{\tau}  & \mathbf{=}-i\mathbf{\sigma}_{2}\partial_{x}\mathbf{v}%
+i\mathbf{\sigma}_{2}\mathbf{\nabla}^{\mathbf{\bot}}u~,\label{ee2}\\
0  & =\partial_{x}u+\mathbf{\nabla}^{\mathbf{\bot}}\cdot\mathbf{v}%
~,\label{ee3}\\
W  & =\mathbf{\nabla}^{\mathbf{\bot}}\cdot\mathbf{\tau}+\mathbf{\nabla
}^{\mathbf{\bot}}\cdot\mathbf{q}_{0}+\partial_{x}^{2}\omega+\Delta
^{\mathbf{\bot}}\omega=0~,\label{ee4}\\
\mathbf{T}  & =-\partial_{x}\mathbf{\tau}-\mathbf{q}+\partial_{x}%
^{2}\mathbf{\tau}+\Delta\mathbf{^{\mathbf{\bot}}\tau}=\mathbf{0}~.\label{ee5}%
\end{align}


\subsection{The dynamics}

Equations (\ref{ee2}), (\ref{ee3}) and (\ref{ee5}) are equivalent to
\begin{align}
\partial_{x}\mathbf{\tau}  & =\mathbf{\eta}~,\label{ds1}\\
\partial_{x}\mathbf{\eta}  & =\mathbf{\eta}-\Delta\mathbf{^{\mathbf{\bot}}%
\tau}+\mathbf{q}~,\label{ds2}\\
\partial_{x}u  & =-\mathbf{\nabla}^{\mathbf{\bot}}\cdot\mathbf{v}%
~,\label{ds3}\\
\mathbf{\partial}_{x}\mathbf{v}  & =\mathbf{\nabla}^{\mathbf{\bot}%
}u+i\mathbf{\sigma}_{2}\mathbf{\tau}~.\label{ds4}%
\end{align}
The equations (\ref{ds1})-(\ref{ds4}) are very similar to the equations (16)
in \cite{pw003}, and are the dynamical system which we will study below.

\subsection{The constraints}

The remaining two equations (\ref{ee1}) and (\ref{ee4}) have no
two-dimensional analogue. They are related to the fact that the vector fields
$\mathbf{\omega}$ and $\mathbf{W}$ have to be divergence free.

\subsubsection{Equation (\ref{ee1})}

Equation (\ref{ee1}) provides an explicit way of computing the first component
$\omega$ of the vorticity in terms of the transverse components $\mathbf{v}$
of the velocity field, at any \textquotedblleft time\textquotedblright\ $x$.
We will also need expressions for the \textquotedblright
time\textquotedblright-derivatives $\partial_{x}\omega$ and $\partial_{x}%
^{2}\omega$ of $\omega$. Differentiating (\ref{ee1}) with respect to $x$ and
using (\ref{ds4}) shows that%
\begin{equation}
\partial_{x}\omega=-\mathbf{\nabla}^{\mathbf{\bot}}\cdot\mathbf{\tau
}~,\label{divomega}%
\end{equation}
which is nothing else than $\mathbf{\nabla\cdot\omega}=0$ written in component
form, and differentiating (\ref{divomega}) with respect to $x$ and using
(\ref{ds1}) leads to%
\begin{equation}
\partial_{x}^{2}\omega=-\mathbf{\nabla}^{\mathbf{\bot}}\cdot\mathbf{\eta
}~.\label{dxxomega}%
\end{equation}


\subsubsection{Equation (\ref{ee4})}

To motivate what follows we first note that on the linear level the
\textit{r.h.s.} in (\ref{ds1})-(\ref{ds4}) depends only on the irrotational
part of $\mathbf{v}$. This is the reason for the appearance of a branch of
zero modes in our dynamical system (see below), and is related to the fact
that $W$ is a (nonlinear) invariant of the dynamical system. Namely,
differentiating (\ref{ee4}) with respect to $x$ and using (\ref{ds1}%
)-(\ref{ds4}) and (\ref{ee1}) we find that
\begin{equation}
\partial_{x}W=-\mathbf{\nabla}^{\mathbf{\bot}}\mathbf{T}~,\label{W}%
\end{equation}
which is nothing else than $\mathbf{\nabla\cdot W}=0$ written in component
form. From (\ref{W}) it follows that $\partial_{x}W=0$ if $\mathbf{T}=0$,
which means that the function $W$ is independent of $x$ if the equations
(\ref{ds1})-(\ref{ds4}) and (\ref{ee1}) are satisfied, and it is therefore
sufficient to require $W$ to be zero at $x=1$ (or any other convenient value
of $x$) in order to satisfy equation (\ref{ee4}). Therefore, and provided we
can rewrite $W$ in such a way that it does not contain derivatives with
respect to $x$ anymore, equation (\ref{ee4}) can be solved by choosing
appropriate \textquotedblleft initial conditions\textquotedblright\ for our
dynamical system. Indeed, using (\ref{dxxomega}) we get from (\ref{ee4}) that%
\begin{equation}
W=\Delta^{\mathbf{\bot}}\omega+\mathbf{\nabla}^{\mathbf{\bot}}\cdot
(\mathbf{\tau-\eta)}+\mathbf{\nabla}^{\mathbf{\bot}}\cdot\mathbf{q}%
_{0}~,\label{cond1a}%
\end{equation}
with $\omega$ given by (\ref{ee1}), and this is an expression for $W$
containing only the \textquotedblleft dynamical variables\textquotedblright%
\ $\mathbf{\tau}$, $\mathbf{\eta}$, $u$, and $\mathbf{v}$, and is free of $x$-derivatives.

\subsection{Conclusion}

We conclude that the system of equations (\ref{ee1})-(\ref{ee5}) is equivalent
to the dynamical system (\ref{ds1})-(\ref{ds4}) with the nonlinear term as
$\mathbf{q}$ defined in (\ref{q})-(\ref{q1}), with $\omega$ as defined in
(\ref{ee1}) and with initial conditions chosen such that $W$ as defined in
(\ref{cond1a}) equals zero.

\section{Fourier transforms}

Following the ideas in \cite{pw003}, we now Fourier transform equation
(\ref{ds1})-(\ref{ds4}) in the transverse directions. Throughout this and
subsequent sections, vectors will be treated notation-wise as $2\times1$
matrices, an upper script $^{T}$ meaning matrix transposition. Let
\[
\mathbf{\tau}(x,\mathbf{y})=\left(  \frac{1}{2\pi}\right)  ^{2}\int
_{\mathbf{R}^{2}}e^{-i\mathbf{k\cdot y}}~\mathbf{\hat{\tau}}(\mathbf{k}%
,x)~d^{2}\mathbf{k}~,
\]
with $\mathbf{k}=(k_{1},k_{2})$, and accordingly for the other functions. For
(\ref{ds1})-(\ref{ds4}) we then get (for simplicity we drop the hats and use
in Fourier space $t$ instead of $x$ for the \textquotedblleft
time\textquotedblright-variable),%
\begin{align}
\mathbf{\dot{\tau}}  & =\mathbf{\eta}~,\nonumber\\
\mathbf{\dot{\eta}}  & =\mathbf{\eta}+k^{2}\mathbf{\tau}+\mathbf{q}%
~,\nonumber\\
\dot{u}  & =i\mathbf{k}^{T}\mathbf{v}~,\nonumber\\
\mathbf{\dot{v}}  & =-i\mathbf{k}u+i\mathbf{\sigma}_{2}\mathbf{\tau
}~,\label{dd}%
\end{align}
the dot meaning derivative with respect to $t$ and $k=\sqrt{k_{1}^{2}%
+k_{2}^{2}}$. From (\ref{q}), (\ref{q0}) and (\ref{q1}) we get that%
\begin{equation}
\mathbf{q=\partial}_{t}\mathbf{\mathbf{q}}_{0}\mathbf{-\sigma}_{2}%
\mathbf{\mathbf{k}}q_{1}~,\label{qq}%
\end{equation}
where%
\begin{align}
\mathbf{q}_{0}  & =\left(  \frac{1}{2\pi}\right)  ^{2}(u\ast\mathbf{\tau
-}\omega\ast\mathbf{\mathbf{v)}}~,\label{qq0}\\
q_{1}  & =\left(  \frac{1}{2\pi}\right)  ^{2}\mathbf{v^{T}\ast}\left(
i\mathbf{\mathbf{\sigma}}_{2}\mathbf{\tau}\right)  ~,\label{qq1}%
\end{align}
and $\ast$\ being the convolution product. Equation (\ref{ee1}) becomes%
\begin{equation}
\omega=\mathbf{k}^{T}\mathbf{\sigma}_{2}\mathbf{v}~,\label{ok}%
\end{equation}
and from (\ref{cond1a}) we find that%
\begin{equation}
W=\mathbf{-}k^{2}\omega-i\mathbf{k}^{T}(\mathbf{\tau-\eta)}-i\mathbf{k}%
^{T}\mathbf{q}_{0}~.\label{wk}%
\end{equation}


\subsection{Stable and unstable modes}

The equations (\ref{dd}) are of the form $\mathbf{\dot{z}}=L\mathbf{z}%
+\mathbf{\chi}$, with $\mathbf{z}=(\mathbf{\tau},\mathbf{\eta},u,\mathbf{v})$,
$\mathbf{\chi}=(\mathbf{0},\mathbf{q},0,\mathbf{0})$ and with $L$ a matrix
with the following block structure
\begin{equation}
L=\left(
\begin{array}
[c]{cc}%
L_{1} & 0\\
L_{3} & L_{2}%
\end{array}
\right)  ~,\label{l}%
\end{equation}
with $L_{1}$ a $4\times4$ matrix $L_{2}$ a $3\times3$ matrix, $L_{3}$ a
$3\times4$ matrix and $0$ the $4\times3$ zero matrix. For $L_{1}$ we have
\begin{equation}
L_{1}(\mathbf{k})=\left(
\begin{array}
[c]{cc}%
0 & 1\\
k^{2} & 1
\end{array}
\right)  ~,\label{l1}%
\end{equation}
the matrix entries being $2\times2$ matrices. For $L_{2}$ we have
\begin{equation}
L_{2}(\mathbf{k})=\left(
\begin{array}
[c]{cc}%
0 & i\mathbf{k}^{T}\\
-i\mathbf{k} & 0
\end{array}
\right)  ~,\label{l2}%
\end{equation}
the first line of the matrix consisting of a $1\times1$ and a $1\times2$
matrix and the second line of a $2\times1$ and a $2\times2$ matrix, and for
$L_{3}$ we have
\begin{equation}
L_{3}(\mathbf{k})=\left(
\begin{array}
[c]{cc}%
0 & 0\\
i\mathbf{\sigma}_{2} & 0
\end{array}
\right)  ~,\label{l3}%
\end{equation}
the first line of the matrix consisting of two $1\times2$ matrices and the
second line of two $2\times2$ matrices. The matrix $L(\mathbf{k})$ can be
diagonalized (see Appendix I for details). Namely, let
\begin{align*}
\Lambda_{0}(k)  & =\sqrt{1+4k^{2}}~,\\
\Lambda_{+}(k)  & =\frac{1+\Lambda_{0}(k)}{2}~,\\
\Lambda_{-}(k)  & =\frac{1-\Lambda_{0}(k)}{2}~,
\end{align*}
and let $\mathbf{z}=S\mathbf{\zeta}$, with $S$ a matrix with the same block
structure as $L$,
\begin{equation}
S=\left(
\begin{array}
[c]{cc}%
S_{1} & 0\\
S_{3} & S_{2}%
\end{array}
\right)  ~,\label{s}%
\end{equation}
with
\begin{align}
S_{1}(\mathbf{k})  & =\left(
\begin{array}
[c]{cc}%
1 & 1\\
\Lambda_{+} & \Lambda_{-}%
\end{array}
\right)  ~,\label{s1}\\
& \nonumber\\
& \nonumber\\
S_{2}(\mathbf{k})  & =\left(
\begin{array}
[c]{ccc}%
0 & 1 & 1\\
\frac{1}{k}\mathbf{\sigma}_{2}\mathbf{k} & -\frac{i}{k}\mathbf{k} & \frac
{i}{k}\mathbf{k}%
\end{array}
\right)  ~,\label{s2}\\
& \nonumber\\
& \nonumber\\
S_{3}(\mathbf{k})  & =\left(
\begin{array}
[c]{cc}%
\frac{-1}{\Lambda_{+}}\mathbf{k}^{T}\mathbf{\sigma}_{2} & \frac{-1}%
{\Lambda_{-}}\mathbf{k}^{T}\mathbf{\sigma}_{2}\\
& \\
i\left(  \frac{1}{\Lambda_{+}}P_{2}(\mathbf{k})+P_{1}(\mathbf{k})\right)
\mathbf{\sigma}_{2} & i\left(  \frac{1}{\Lambda_{-}}P_{2}(\mathbf{k}%
)+P_{1}(\mathbf{k})\right)  \mathbf{\sigma}_{2}%
\end{array}
\right)  ~,\label{s3}%
\end{align}
and where%
\begin{align*}
P_{1}(\mathbf{k})  & =\frac{\mathbf{kk}^{T}}{k^{2}}~,\\
P_{2}(\mathbf{k})  & =\mathbf{\sigma}_{2}\frac{\mathbf{kk}^{T}}{k^{2}%
}\mathbf{\sigma}_{2}~,
\end{align*}
are the projection operators on the irrotational and divergence free part of a
vector field, respectively. Then, $\mathbf{\dot{\zeta}=}D\mathbf{\zeta+}%
S^{-1}\mathbf{\chi}$, with
\begin{equation}
S^{-1}=\left(
\begin{array}
[c]{cc}%
S_{1}^{-1} & 0\\
\left(  S^{-1}\right)  _{3} & S_{2}^{-1}%
\end{array}
\right) \label{ss1}%
\end{equation}
again a matrix with the same block structure as $L$, with
\begin{align}
S_{1}^{-1}(\mathbf{k})  & =\left(
\begin{array}
[c]{cc}%
-\frac{\Lambda_{-}}{\Lambda_{0}} & \frac{1}{\Lambda_{0}}\\
& \\
\frac{\Lambda_{+}}{\Lambda_{0}} & -\frac{1}{\Lambda_{0}}%
\end{array}
\right)  ~,\label{sm1}\\
& \nonumber\\
& \nonumber\\
S_{2}^{-1}(\mathbf{k})  & =\left(
\begin{array}
[c]{cc}%
0 & \frac{1}{k}\mathbf{k}^{T}\mathbf{\sigma}_{2}\\
& \\
\frac{1}{2} & \frac{i}{2k}\mathbf{k}^{T}\\
& \\
\frac{1}{2} & \frac{-i}{2k}\mathbf{k}^{T}%
\end{array}
\right)  ~,\label{sm2}\\
& \nonumber\\
& \nonumber\\
\left(  S^{-1}\right)  _{3}(\mathbf{k})  & =\left(
\begin{array}
[c]{cc}%
\frac{i}{k^{3}}\mathbf{k}^{T} & -\frac{i}{k^{3}}\mathbf{k}^{T}\\
& \\
(k-1)\frac{1}{2k^{2}}\mathbf{k}^{T}\mathbf{\sigma}_{2} & \frac{1}{2k^{2}%
}\mathbf{k}^{T}\mathbf{\sigma}_{2}\\
& \\
(k+1)\frac{1}{2k^{2}}\mathbf{k}^{T}\mathbf{\sigma}_{2} & \frac{1}{2k^{2}%
}\mathbf{k}^{T}\mathbf{\sigma}_{2}%
\end{array}
\right)  ~,\label{sm3}%
\end{align}
and with $D=S^{-1}LS$ a diagonal matrix with diagonal entries $\Lambda_{+}$,
$\Lambda_{+}$, $\Lambda_{-}$, $\Lambda_{-}$, $0$, $k$, and $-k$. Note that
$\Lambda_{+}(k)\geq1$ and $\Lambda_{-}(k)\leq0$ and that $\Lambda
_{-}(k)\approx-k^{2}$ for small values of $k$. We also have the identities
$\Lambda_{+}+\Lambda_{-}=1$, $\Lambda_{+}-\Lambda_{-}=\Lambda_{0}$, and
$\Lambda_{+}\Lambda_{-}=-k^{2}$, which will be routinely used below. Let
$\mathbf{\zeta=(\tau}_{+},\mathbf{\tau}_{-},v_{0},v_{+},v_{-})$. Using the
definitions we find that the equations (\ref{ds1})-(\ref{ds4}) are equivalent
to
\begin{align}
\mathbf{\dot{\tau}}_{+}  & =\Lambda_{+}\mathbf{\tau}_{+}+\frac{1}{\Lambda_{0}%
}\mathbf{q}~,\nonumber\\
\mathbf{\dot{\tau}}_{-}  & =\Lambda_{-}\tau_{-}-\frac{1}{\Lambda_{0}%
}\mathbf{q}~,\nonumber\\
\dot{v}_{0}  & =\frac{-i}{k^{3}}\mathbf{k}^{T}\mathbf{q}~,\nonumber\\
\dot{v}_{+}  & =k~v_{+}+\frac{1}{2k^{2}}\mathbf{k}^{T}\mathbf{\sigma}%
_{2}\mathbf{q}~,\nonumber\\
\dot{v}_{-}  & =-k~v_{-}+\frac{1}{2k^{2}}\mathbf{k}^{T}\mathbf{\sigma}%
_{2}\mathbf{q}~,\label{ddpm}%
\end{align}
with $\mathbf{q}$\textbf{\ }as defined in (\ref{qq})-(\ref{qq1}). For
convenience later on we write $\mathbf{z=}S\mathbf{\zeta}$ in component form.
Namely,
\begin{align}
\mathbf{\tau}  & =\mathbf{\tau}_{+}+\mathbf{\tau}_{-}~,\label{tau}\\
\mathbf{\eta}  & =\Lambda_{+}\mathbf{\tau}_{+}+\Lambda_{-}\mathbf{\tau}%
_{-}~,\label{eta}\\
u  & =-\mathbf{k}^{T}\mathbf{\sigma}_{2}\mathbf{\bar{\tau}}+v_{+}%
+v_{-}~,\label{u1}\\
\mathbf{v}  & \mathbf{=}\mathbf{v}_{1}+\mathbf{v}_{2}~,\label{uv}%
\end{align}
where%
\begin{equation}
\mathbf{\bar{\tau}}=\frac{1}{\Lambda_{+}}\mathbf{\tau}_{+}+\frac{1}%
{\Lambda_{-}}\mathbf{\tau}_{-}~,\label{tauh}%
\end{equation}%
\begin{align}
\mathbf{v}_{1}  & =P_{1}i\mathbf{\sigma}_{2}\mathbf{\tau}-\frac{i}%
{k}\mathbf{k}~(v_{+}-v_{-})~,\label{v1}\\
\mathbf{v}_{2}  & =P_{2}i\mathbf{\sigma}_{2}\mathbf{\bar{\tau}}+\frac{1}%
{k}\mathbf{\sigma}_{2}\mathbf{k}~v_{0}~.\label{v2}%
\end{align}
Using that $\mathbf{k}^{T}\mathbf{\sigma}_{2}P_{1}=0$, we get using the
definition (\ref{ok}) the following expression for $\omega$,%
\begin{equation}
\omega=\mathbf{k}^{T}\mathbf{\sigma}_{2}\mathbf{v}_{2}~,\label{ok0}%
\end{equation}
and from (\ref{v2}) we therefore get that%
\begin{equation}
\omega=i\mathbf{k}^{T}\mathbf{\bar{\tau}}+k~v_{0}~.\label{ok1}%
\end{equation}
Therefore, we find from (\ref{wk}) that%
\begin{equation}
W=\mathbf{-}k^{3}v_{0}-i\mathbf{k}^{T}\mathbf{q}_{0}~.\label{wkts}%
\end{equation}


We conclude that the system of equations (\ref{ee1})-(\ref{ee5}) is equivalent
to the dynamical system (\ref{ddpm}) with the nonlinear term $\mathbf{q}$ as
defined in (\ref{qq}), (\ref{qq0}), (\ref{qq1}), with $\omega$ as defined in
(\ref{ok0}), with $\mathbf{\tau}$, $\mathbf{\eta}$, $u$, $\mathbf{v}$ as
defined in (\ref{tau}), (\ref{eta}), (\ref{u1}) and (\ref{uv}), and with
initial conditions chosen such that $W$ as defined in (\ref{wkts}) equals zero.

\section{The integral equations}

To solve (\ref{ddpm}) we convert it into an integral equation. The $+$-modes
are unstable (remember that $\Lambda_{+}(k)\geq1)$ and we therefore have to
integrate these modes backwards in time starting with $\mathbf{\tau}%
_{+}(\mathbf{k},\infty)\equiv u_{+}(\mathbf{k},\infty)\equiv0$. We get
\begin{align}
\mathbf{\tau}_{+}(\mathbf{k},t)  & =-\frac{1}{\Lambda_{0}}\int_{t}^{\infty
}e^{\Lambda_{+}(t-s)}\mathbf{q}(\mathbf{k},s)~ds~,\label{inteq1}\\
\mathbf{\tau}_{-}(\mathbf{k},t)  & =\mathbf{\tilde{\tau}}_{-}^{\ast
}(\mathbf{k})e^{\Lambda_{-}(t-1)}-\frac{1}{\Lambda_{0}}\int_{1}^{t}%
e^{\Lambda_{-}(t-s)}\mathbf{q}(\mathbf{k},s)~ds~,\label{inteq2}\\
v_{0}(\mathbf{k},t)  & =v_{0}^{\ast}(\mathbf{k})+\frac{i}{k^{3}}\mathbf{k}%
^{T}\int_{t}^{\infty}\mathbf{q(k},s)~ds~,\label{inteq3}\\
v_{+}(\mathbf{k},t)  & =\frac{-1}{2k^{2}}\mathbf{k}^{T}\mathbf{\sigma}_{2}%
\int_{t}^{\infty}e^{k(t-s)}\mathbf{q}(\mathbf{k},s)~ds~,\label{inteq4}\\
v_{-}(\mathbf{k},t)  & =\tilde{v}_{-}^{\ast}(\mathbf{k})e^{-k(t-1)}+\frac
{1}{2k^{2}}\mathbf{k}^{T}\mathbf{\sigma}_{2}\int_{1}^{t}e^{-k(t-s)}%
\mathbf{q}(\mathbf{k},s)~ds~.\label{inteq5}%
\end{align}
Equation (\ref{qq}) implies that $\mathbf{k}^{T}\mathbf{q(k},t)=\partial
_{t}\mathbf{k}^{T}\mathbf{q}_{0}\mathbf{(k},t)$, and therefore (\ref{inteq3})
is equivalent to%
\begin{equation}
v_{0}(\mathbf{k},t)=v_{0}^{\ast}(\mathbf{k})-\frac{i}{k^{3}}\mathbf{k}%
^{T}\mathbf{q}_{0}\mathbf{(k},t)~,\label{inteq3p}%
\end{equation}
and we therefore get from (\ref{wkts}) that
\begin{equation}
W(k,t)=\mathbf{-}k^{3}v_{0}^{\ast}(\mathbf{k})~,\label{wkts1}%
\end{equation}
for $t\geq1$, which shows that the equation $W=0$ is equivalent to choosing
the initial condition
\begin{equation}
v_{0}^{\ast}=0~.\label{v00}%
\end{equation}


To overcome the problem related to the singular behavior of various
expressions in (\ref{inteq1})-(\ref{inteq5}) at $k=0$, we now proceed exactly
as in \cite{pw003}. Namely, we substitute the integral equations
(\ref{inteq1})-(\ref{inteq5}) into the change of coordinates (\ref{tau}%
)-(\ref{uv}), and integrate by parts the time derivatives acting on
$\mathbf{q}_{0}$. This leads to the following integral equations for
$\mathbf{\tau}$, $u$, $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$:%

\begin{align}
\mathbf{\tau}(\mathbf{k},t)  & =\left(  \mathbf{\tilde{\tau}}_{-}^{\ast
}(\mathbf{k})+\frac{1}{\Lambda_{0}}\mathbf{\mathbf{q}}_{0}(\mathbf{k}%
,1)\right)  e^{\Lambda_{-}(t-1)}\nonumber\\
& +\frac{\mathbf{\mathbf{\sigma}_{2}\mathbf{k}}}{\Lambda_{0}}\int_{1}%
^{t}e^{\Lambda_{-}(t-s)}q_{1}(\mathbf{k},s)~ds\nonumber\\
& +\frac{\mathbf{\mathbf{\sigma}_{2}\mathbf{k}}}{\Lambda_{0}}\int_{t}^{\infty
}e^{\Lambda_{+}(t-s)}q_{1}(\mathbf{k},s)~ds\nonumber\\
& -\frac{\Lambda_{-}}{\Lambda_{0}}\int_{1}^{t}e^{\Lambda_{-}(t-s)}%
\mathbf{\mathbf{q}}_{0}(\mathbf{k},s)~ds\nonumber\\
& -\frac{\Lambda_{+}}{\Lambda_{0}}\int_{t}^{\infty}e^{\Lambda_{+}%
(t-s)}\mathbf{\mathbf{q}}_{0}(\mathbf{k},s)~ds~,\label{itau}%
\end{align}%
\begin{align}
u(\mathbf{k},t)  & =-\mathbf{k}^{T}\mathbf{\sigma}_{2}\frac{1}{\Lambda_{-}%
}\left(  \mathbf{\tilde{\tau}}_{-}^{\ast}(\mathbf{k})+\frac{1}{\Lambda_{0}%
}\mathbf{\mathbf{q}}_{0}(\mathbf{k},1)\right)  e^{\Lambda_{-}(t-1)}\nonumber\\
& +\left(  \tilde{v}_{-}^{\ast}(\mathbf{k})-\frac{1}{2k^{2}}\mathbf{k}%
^{T}\mathbf{\sigma}_{2}\mathbf{\mathbf{q}}_{0}(\mathbf{k},1)\right)
e^{-k(t-1)}\nonumber\\
& +\frac{\Lambda_{+}}{\Lambda_{0}}\int_{1}^{t}e^{\Lambda_{-}(t-s)}%
q_{1}(\mathbf{k},s)~ds\nonumber\\
& -\frac{1}{2}\int_{1}^{t}e^{-k(t-s)}q_{1}(\mathbf{k},s)~ds\nonumber\\
& +\frac{1}{2}\int_{t}^{\infty}e^{k(t-s)}q_{1}(\mathbf{k},s)~ds\nonumber\\
& +\frac{\Lambda_{-}}{\Lambda_{0}}\int_{t}^{\infty}e^{\Lambda_{+}(t-s)}%
q_{1}(\mathbf{k},s)~ds\nonumber\\
& -\frac{1}{2k}\mathbf{k}^{T}\mathbf{\sigma}_{2}\int_{1}^{t}e^{-k(t-s)}%
\mathbf{\mathbf{q}}_{0}(\mathbf{k},s)~ds\nonumber\\
& -\frac{1}{2k}\mathbf{k}^{T}\mathbf{\sigma}_{2}\int_{t}^{\infty}%
e^{k(t-s)}\mathbf{\mathbf{q}}_{0}(\mathbf{k},s)~ds\nonumber\\
& +\frac{1}{\Lambda_{0}}\int_{1}^{t}e^{\Lambda_{-}(t-s)}\mathbf{k}%
^{T}\mathbf{\sigma}_{2}\mathbf{\mathbf{q}}_{0}(\mathbf{k},s)~ds\nonumber\\
& +\frac{1}{\Lambda_{0}}\int_{t}^{\infty}e^{\Lambda_{+}(t-s)}\mathbf{k}%
^{T}\mathbf{\sigma}_{2}\mathbf{\mathbf{q}}_{0}(\mathbf{k},s)~ds~,\label{iu}%
\end{align}%
\begin{align}
\mathbf{v}_{1}(\mathbf{k},t)  & =P_{1}i\mathbf{\sigma}_{2}\mathbf{\tau
}(\mathbf{k},t)\nonumber\\
& +\frac{i}{k}\mathbf{k}\left(  \tilde{v}_{-}^{\ast}(\mathbf{k})-\frac
{1}{2k^{2}}\mathbf{k}^{T}\mathbf{\sigma}_{2}\mathbf{\mathbf{q}}_{0}%
(\mathbf{k},1)\right)  e^{-k(t-1)}\nonumber\\
& -\frac{1}{2}\frac{i}{k}\mathbf{k}\int_{1}^{t}e^{-k(t-s)}q_{1}(\mathbf{k}%
,s)~ds\nonumber\\
& -\frac{1}{2}\frac{i}{k}\mathbf{k}\int_{t}^{\infty}e^{k(t-s)}q_{1}%
(\mathbf{k},s)~ds\nonumber\\
& -\frac{1}{2}\int_{1}^{t}e^{-k(t-s)}P_{1}i\mathbf{\sigma}_{2}%
\mathbf{\mathbf{q}}_{0}(\mathbf{k},s)~ds\nonumber\\
& +\frac{1}{2}\int_{t}^{\infty}e^{k(t-s)}P_{1}i\mathbf{\sigma}_{2}%
\mathbf{\mathbf{q}}_{0}(\mathbf{k},s)~ds~,\label{iv1}%
\end{align}%
\begin{align}
\mathbf{v}_{2}(\mathbf{k},t)  & =\frac{1}{\Lambda_{-}}P_{2}i\mathbf{\sigma
}_{2}\left(  \mathbf{\tilde{\tau}}_{-}^{\ast}(\mathbf{k})+\frac{1}{\Lambda
_{0}}\mathbf{\mathbf{q}}_{0}(\mathbf{k},1)\right)  e^{\Lambda_{-}%
(t-1)}\nonumber\\
& -\frac{1}{\Lambda_{0}}\int_{1}^{t}e^{\Lambda_{-}(t-s)}P_{2}i\mathbf{\sigma
}_{2}\mathbf{\mathbf{q}}_{0}(\mathbf{k},s)~ds\nonumber\\
& -\frac{1}{\Lambda_{0}}\int_{t}^{\infty}e^{\Lambda_{+}(t-s)}P_{2}%
i\mathbf{\sigma}_{2}\mathbf{\mathbf{q}}_{0}(\mathbf{k},s)~ds~,\label{iv2}%
\end{align}
with $\omega$ given by (\ref{ok0}) and with $\mathbf{q}_{0}$ and $q_{1}$ given
by (\ref{qq0}) and (\ref{qq1}), respectively. Note that the function
$\mathbf{\eta}$ does not need to be constructed since it does not appear in
the nonlinearities $\mathbf{q}_{0}$ and $q_{1}$.

\subsection{Choice of initial conditions}

A closer look at (\ref{itau})-(\ref{iv2}) reveals, that the problem concerning
the decision by $k$ in the equations (\ref{inteq1})-(\ref{inteq5}) has not
disappeared. However, in this new representation, the invariance properties of
the equations have become manifest, and we see that the problem can be
eliminated by a proper choice of initial conditions, \textit{i.e.},
$\mathbf{\tau}$, $u$ and $\mathbf{v}$ are either regular or singular for all
times. In particular, as we will see, if we set%
\begin{align}
\mathbf{\tilde{\tau}}_{-}^{\ast}(\mathbf{k})  & =\mathbf{\tau}_{-}^{\ast
}(\mathbf{k})-\frac{1}{\Lambda_{0}}\mathbf{\mathbf{q}}_{0}(\mathbf{k}%
,1)~,\label{tm}\\
\tilde{v}_{-}^{\ast}(\mathbf{k})  & =v_{-}^{\ast}(\mathbf{k})+\frac{1}{2}%
\frac{1}{k^{2}}\mathbf{k}^{T}\mathbf{\sigma}_{2}\mathbf{\mathbf{q}}%
_{0}(\mathbf{k},1)~,\label{um}%
\end{align}
with%
\begin{align}
\mathbf{\tau}_{-}^{\ast}(\mathbf{k})  & =-\mathbf{\sigma}_{2}\mathbf{k}%
~\tau_{-,1}^{\ast}(\mathbf{k})+k^{2}P_{1}\mathbf{\tau}_{-,2}^{\ast}%
(\mathbf{k})~,\label{at}\\
v_{-}^{\ast}(\mathbf{k})  & =v_{-,1}^{\ast}(\mathbf{k})-\frac{\mathbf{k}^{T}%
}{k}\mathbf{\sigma}_{2}\mathbf{v}_{-,2}^{\ast}(\mathbf{k})~,\label{av}%
\end{align}
with $\tau_{-,1}^{\ast}$, $\mathbf{\tau}_{-,2}^{\ast}$, $v_{-,1}^{\ast}$ and
$\mathbf{v}_{-,2}^{\ast}$ smooth, then $\omega$, $\mathbf{\tau}$,
$\mathbf{q}_{0}$ and $q_{1}$ are smooth, and $u$, $\mathbf{v}_{1}$ and
$\mathbf{v}_{2}$ are smooth modulo certain explicit discontinuities at
$\mathbf{k}=0$. This corresponds to choosing initial conditions exactly as
singular as dictated by the nonlinearity. We expect this choice to be general
enough to cover all cases of stationary exterior flows.

Our choice of $\mathbf{\tau}_{-}^{\ast}$ in (\ref{at}) implies that
$\mathbf{k}^{T}\mathbf{\tau}_{-}^{\ast}(\mathbf{k})/k^{2}=\mathbf{k}%
^{T}\mathbf{\tau}_{-,2}^{\ast}(\mathbf{k})$, and therefore $\lim
_{\mathbf{k}\rightarrow0}\mathbf{k}^{T}\mathbf{\tau}_{-}^{\ast}(\mathbf{k}%
)/k^{2}=0$. This will translate below into $\omega(0)=0$, which means that the
longitudinal vorticity when averaged over transversal planes equals zero. This
is dictated by the fact that, due to the divergence-freeness of the vorticity,
this average is independent of the choice of the transversal plane, and should
therefore be chosen to be equal to zero in our case (since there should be no
nonzero vorticity average far ahead of an obstacle; see \cite{pw001} and
\cite{pw002} for the physical interpretation of the problem).

Below, we will prove existence of solutions to (\ref{itau})-(\ref{iv2}) for
certain classes of continuous complex valued functions $\tau_{-,1}^{\ast}$,
$v_{-,1}^{\ast}$ and continuous maps $\mathbf{\tau}_{-,2}^{\ast}$,
$\mathbf{v}_{-,2}^{\ast}$ with values in $\mathbf{C}^{2}$.

Once the existence of solutions has been established, we will restrict
attention to maps $\tau_{-,1}^{\ast}$, $\mathbf{\tau}_{-,2}^{\ast}$,
$v_{-,1}^{\ast}$ and $\mathbf{v}_{-,2}^{\ast}$ with even and odd real and
imaginary parts, respectively. This corresponds to the restriction to real
valued solutions of (\ref{ds1})-(\ref{ds4}).

Note that in the parametrization (\ref{at}) and (\ref{av}) is not unique,
\textit{i.e.}, several choices of $\tau_{-,1}^{\ast}$, $\mathbf{\tau}%
_{-,2}^{\ast}$, $v_{-}^{\ast}$ and $\mathbf{v}_{-,2}^{\ast}$ lead to the same
$\mathbf{\tau}_{-}^{\ast}$ and $v_{-}^{\ast}$. The parametrization has been
chosen as given for convenience later on.

It turns out that, in contrast to the two-dimensional case, the decomposition
of the nonlinearity $\mathbf{q}$ into $\mathbf{q}_{0}$ and $q_{1}$ is detailed
enough to prove the existence of a solution to (\ref{itau})-(\ref{iv2}). This
is due to the fact that, because of the dimension-dependence of power
counting, the same nonlinearity has a smaller amplitude in three dimensions
than in two dimensions (see for example \cite{wayne001} for the basics).

\section{Function spaces}

In order to prove the existence of a solution for (\ref{itau})-(\ref{iv2}) we
will apply, for fixed initial conditions $r^{\ast}=(\tau_{-,1}^{\ast
},\mathbf{\tau}_{-,2}^{\ast},v_{-,1}^{\ast}\mathbf{v}_{-,2}^{\ast})$, the
contraction mapping principle to the map $(\mathbf{\tilde{q}}_{0,}\tilde
{q}_{1})=\mathcal{N}(\mathbf{q}_{0},q_{1})$ that is formally defined by first
computing $\mathbf{\tau}$, $u$, $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ using
(\ref{itau})-(\ref{iv2}), then $\omega$ and $\mathbf{v}$ using (\ref{ok0}) and
(\ref{uv}) and then $\mathbf{q}_{0}$ and $q_{1}$ by using (\ref{qq0}) and
(\ref{qq1}). We now define the functions spaces that will be used below:

Let, for $\alpha$, $p\geq0$ and $k\geq0$,%
\begin{equation}
\mu_{\alpha}^{p}(k,t)=\frac{1}{1+\left(  kt^{p}\right)  ^{\alpha}}~.\label{mu}%
\end{equation}
Let furthermore%
\begin{align*}
\mu_{\alpha}(k,t)  & =\mu_{a}^{1/2}(k,t)~,\\
\bar{\mu}_{\alpha}(k,t)  & =\mu_{a}^{1}(k,t)~.
\end{align*}
We then consider, for fixed $\alpha\geq0$ and $\nu\in\mathbf{N}$, the Banach
spaces $\mathcal{V}_{\alpha}^{\nu}$ of continuous complex valued maps
$\mathbf{f}\equiv(f_{1},\dots,f_{\nu})\in\mathcal{C}(\mathbf{R}^{2}%
\mathbf{,C}^{\nu})$ equipped with the norm
\[
\left\Vert \mathbf{f}\right\Vert _{\alpha}=\sup_{\mathbf{k\in R}^{2}}%
\frac{\left\vert \mathbf{f}(\mathbf{k})\right\vert }{\mu_{\alpha}(\left\vert
\mathbf{k}\right\vert ,1)}~,
\]
where,%
\[
\left\vert \mathbf{f}(\mathbf{k})\right\vert =\sum_{i=1,\dots,\nu}\left\vert
f_{i}(\mathbf{k})\right\vert ~,
\]
and the Banach space $\mathcal{B}_{\alpha,\beta}^{\nu}$ of continuous maps
$\mathbf{f}$ from $[1,\infty)$ to $\mathcal{V}_{\alpha}^{\nu}$ equipped the
norm
\[
\left\Vert \mathbf{f}\right\Vert _{\alpha,\beta}=\sup_{t\geq1}t^{\beta
}||\mathbf{f}(t^{-1/2}.~,t)||_{\alpha}~.
\]
Finally, we define the Banach space%
\[
\mathcal{V}_{\alpha}=\mathcal{V}_{\alpha}^{1}\oplus\mathcal{V}_{\alpha+1}%
^{2}\oplus\mathcal{V}_{\alpha}^{1}\oplus\mathcal{V}_{\alpha}^{2}~,
\]
equipped with the norm%
\[
\left\Vert (f_{1},f_{2},f_{3},f_{4})\right\Vert _{\alpha}=\left\Vert
f_{1}\right\Vert _{\alpha}+\left\Vert f_{2}\right\Vert _{\alpha+1}+\left\Vert
f_{3}\right\Vert _{\alpha}+\left\Vert f_{4}\right\Vert _{\alpha}~,
\]
and the Banach space $\mathcal{B}_{\alpha}$,%
\[
\mathcal{B}_{\alpha}=\mathcal{B}_{\alpha,3/2}^{2}\oplus\mathcal{B}%
_{\alpha,3/2}^{1}~,
\]
equipped with the norm%
\[
\left\Vert (f_{1},f_{2})\right\Vert _{\alpha}=\left\Vert f_{1}\right\Vert
_{\alpha,3/2}+\left\Vert f_{2}\right\Vert _{\alpha,3/2}~.
\]


\begin{theorem}
\label{theorem1}Let $\alpha>2$. Let $r^{\ast}=(\tau_{-,1}^{\ast},\mathbf{\tau
}_{-,2}^{\ast},v_{-,1}^{\ast},\mathbf{v}_{-,2}^{\ast})\in\mathcal{V}%
_{\alpha+1}^{1}$ with $\varepsilon_{0}=\left\Vert r^{\ast}\right\Vert
_{\alpha+1}$. Then, $\mathcal{N}$ is well defined as a map from $\mathcal{B}%
_{\alpha}$ to $\mathcal{B}_{\alpha}$ and contracts, for $\varepsilon_{0}$
sufficiently small, the ball $B_{\alpha}(\varepsilon_{0})=\left\{  \rho
\in\mathcal{B}_{\alpha}\left\vert {}\right.  \left\Vert \rho\right\Vert
_{\alpha}\leq\varepsilon_{0}\right\}  $ into itself.
\end{theorem}

Theorem \ref{theorem1} implies that for $\varepsilon_{0}$ small enough
$\mathcal{N}$ has a unique fixed point in $B_{\alpha}(\varepsilon_{0})$,
\textit{i.e.}, the integral equations (\ref{itau})-(\ref{iv2}) have a solution.

\section{Proof of Theorem \ref{theorem1}}

The proof is organized as follows: we first show that $\mathcal{N}$ is well
defined and maps, for small enough initial conditions $r^{\ast}=(\tau
_{-,1}^{\ast},\mathbf{\tau}_{-,2}^{\ast},v_{-,1}^{\ast},\mathbf{v}_{-,2}%
^{\ast})$, a ball in $\mathcal{B}_{\alpha}$ into itself. Then, we show that
$\mathcal{N}$ is a contraction on this ball.

Let $\varepsilon_{0}$ be as in Theorem \ref{theorem1}. Throughout all proofs
we then denote by $\varepsilon$ a constant multiple of $\varepsilon_{0}$,
\textit{i.e.}, $\varepsilon=\mathrm{const.}~\varepsilon_{0}$ with a constant
that may be different from instance to instance. Also, to simplify notation,
we will write throughout all proofs $k$ instead of $\left\vert \mathbf{k}%
\right\vert $.

\subsection{$\mathcal{N}$ is well defined}

We first prove bounds on $\mathbf{\tau}$, $u$, $\mathbf{v}_{1}$ and
$\mathbf{v}_{2}$:

\begin{proposition}
\label{pp11}Let $\alpha>0$. Let $r^{\ast}=(\tau_{-,1}^{\ast},\mathbf{\tau
}_{-,2}^{\ast},v_{-,1}^{\ast},\mathbf{v}_{-,2}^{\ast})\in\mathcal{V}%
_{\alpha+1}^{1}$ with $\varepsilon_{0}=\left\Vert \mathbf{r}^{\ast}\right\Vert
_{\alpha+1}$, and let $(\mathbf{q}_{0},q_{1})\in B_{\alpha}(\varepsilon)$.
Then, $\omega$ and $\mathbf{\tau}$ as defined by (\ref{ok0}) and (\ref{itau})
are a continuous maps from $\mathbf{R}^{2}\times\lbrack1,\infty)$ to
$\mathbf{C}^{2}$ and $\mathbf{C}$, respectively, and $u$, $\mathbf{v}_{1}$ and
$\mathbf{v}_{2}$ as defined by (\ref{iu}), (\ref{iv1}) and (\ref{iv2}) are of
the form%
\begin{align}
u(\mathbf{k},t)  & =u_{E}(\mathbf{k},t)+\frac{1}{k}i\mathbf{k}^{T}%
\mathbf{u}_{O}(\mathbf{k},t)~,\label{dec1}\\
\mathbf{v}_{1}(\mathbf{k},t)  & =\mathbf{v}_{1,C}(\mathbf{k},t)+P_{1}%
\mathbf{v}_{1,E}(\mathbf{k},t)+\frac{i\mathbf{k}^{T}}{k}v_{1,O}(\mathbf{k}%
,t)~,\label{dec2}\\
\mathbf{v}_{2}(\mathbf{k},t)  & =P_{2}\mathbf{v}_{2,E}(\mathbf{k}%
,t)~,\label{dec3}%
\end{align}
with $u_{E}$, $\mathbf{u}_{O}$, $\mathbf{v}_{1,C}$, $\mathbf{v}_{1,E}$,
$v_{1,O}$ and $\mathbf{v}_{2,E}$ continuous maps from $\mathbf{R}^{2}%
\times\lbrack1,\infty)$ to $\mathbf{C}$ and $\mathbf{C}^{2}$, respectively.
Furthermore, we have the bounds
\begin{align}
\left\vert \omega(\mathbf{k},t)\right\vert  & \leq\frac{\varepsilon}{t^{1/2}%
}\mu_{\alpha}(k,t)~,\label{bpm0}\\
\left\vert \mathbf{\tau}(\mathbf{k},t)\right\vert  & \leq\frac{\varepsilon
}{t^{1/2}}\mu_{\alpha}(k,t)~,\label{bpm1}\\
\left\vert u(\mathbf{k},t)\right\vert  & \leq\varepsilon\mu_{\alpha
}(k,t)~,\label{bpm3}\\
\left\vert \mathbf{v}_{1}(\mathbf{k},t)\right\vert  & \leq\varepsilon\bar{\mu
}_{\alpha+1}(k,t)+\frac{\varepsilon}{t^{1/2}}\mu_{\alpha}(k,t)~,\label{bpm4}\\
\left\vert \mathbf{v}_{2}(\mathbf{k},t)\right\vert  & \leq\varepsilon\mu
_{a+1}(k,t)~,\label{bpm5}%
\end{align}
uniformly in $\mathbf{k}\in\mathbf{R}^{2}$ and $t\geq1$.
\end{proposition}

See Appendix II for a proof.

\bigskip

Now we prove bounds on $\mathbf{q}_{0}$ and $q_{1}$:

\begin{proposition}
\label{pp12}Let $\alpha>2$. Let $\omega$ and $\mathbf{\tau}$ be continuous
maps from $\mathbf{R}^{2}\times\lbrack1,\infty)$ to $\mathbf{C} $ and
$\mathbf{C}^{2}$ satisfying the bounds (\ref{bpm0}) and (\ref{bpm1}),
respectively, and let $u$, $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ by continuous
maps from $\mathbf{R}^{2}\setminus\{0\}\times\lbrack1,\infty)$ to $\mathbf{C}$
and $\mathbf{C}^{2}$, respectively, satisfying the bounds (\ref{bpm3}%
)-(\ref{bpm5}). Then, $\mathbf{q}_{0}$ and $q_{1}$ as defined by (\ref{qq0})
and (\ref{qq1}) are continuous maps from $\mathbf{R}^{2}\times\lbrack
0,\infty)$ to $\mathbf{C}^{2}$ and $\mathbf{C}$, respectively, and we have the
bounds
\begin{align}
\left\vert \mathbf{q}_{0}(\mathbf{k},t)\right\vert  & \leq\frac{\varepsilon
^{2}}{t^{3/2}}\mu_{a}(k,t)~,\label{nqb1}\\
\left\vert q_{1}(\mathbf{k},t)\right\vert  & \leq\frac{\varepsilon^{2}%
}{t^{3/2}}\mu_{a}(k,t)~,\label{nqb3}%
\end{align}
uniformly in $\mathbf{k}\in\mathbf{R}^{2}$ and $t\geq1$, and therefore
$\left\Vert (\mathbf{q}_{0},q_{1})\right\Vert _{\alpha}\leq\varepsilon^{2}$.
\end{proposition}

See Appendix III for a proof.

\bigskip

Proposition \ref{pp11} together with Proposition \ref{pp12} imply that, for
$\rho\in B_{\alpha}(\varepsilon)$, $\left\Vert \mathcal{N}(\rho)\right\Vert
_{\alpha}\leq\varepsilon^{2}$. Therefore, $\mathcal{N}$ is well defined as a
map from $\mathcal{B}_{\alpha}$ to $\mathcal{B}_{\alpha}$. Furthermore, since
$\varepsilon^{2}=\mathrm{const.}$ $\varepsilon_{0}^{2} $, it follows that
$\mathcal{N}$ maps $B_{\alpha}(\varepsilon_{0}) $ into itself for
$\varepsilon_{0}$ small enough.

\subsection{$\mathcal{N}$ is Lipschitz}

In order to complete the proof of Theorem \ref{theorem1} it remains to be
shown that $\mathcal{N}$ is Lipschitz:

\begin{proposition}
\label{prop12}Let $\alpha>2$. Let $r^{\ast}=(\tau_{-,1}^{\ast},\mathbf{\tau
}_{-,2}^{\ast},v_{-,1}^{\ast},\mathbf{v}_{-,2}^{\ast})\in\mathcal{V}%
_{\alpha+1}$ with $\varepsilon_{0}=\left\Vert \mathbf{r}^{\ast}\right\Vert
_{\alpha+1}$, and let $\rho$, $\tilde{\rho}\in B_{\alpha}(\varepsilon_{0})$.
Then
\begin{equation}
\left\Vert \mathcal{N}(\rho)-\mathcal{N}(\tilde{\rho})\right\Vert _{\alpha
}\leq\varepsilon\left\Vert \rho-\tilde{\rho}\right\Vert _{\alpha
}~.\label{lipschitz}%
\end{equation}

\end{proposition}

See Appendix IV for a proof.

\bigskip

Proposition \ref{pp11} together with Proposition \ref{pp12} show that, for
$\alpha>2$, $\mathcal{N}$ maps the ball $B_{\alpha}(\varepsilon_{0})$ into
itself for $\varepsilon_{0}$ small enough, and Proposition \ref{prop12}
therefore shows that $\mathcal{N}$ is a contraction of $B_{\alpha}%
(\varepsilon_{0})$ into itself for $\varepsilon_{0}$ small enough. This
completes the proof of Theorem \ref{theorem1}.

\section{Invariant quantities}

We now restrict attention to maps $\tau_{-,1}^{\ast}$, $\mathbf{\tau}%
_{-,2}^{\ast}$, $v_{-,1}^{\ast}$ and $\mathbf{v}_{-,2}^{\ast}$ with even and
odd real and imaginary parts, respectively.

\begin{proposition}
Let $\mathbf{k}=k\mathbf{e}$ with $\mathbf{e}$ a unit vector, let%
\begin{align}
\mathbf{a}  & =-i\mathbf{\sigma}_{2}\left(  \mathbf{\tau}_{-,2}^{\ast}%
(0)+\int_{1}^{\infty}\mathbf{\mathbf{q}}_{0}(0,s)~ds\right)  ~,\label{aaa}\\
\mathbf{b}  & =i\mathbf{\sigma}_{2}\left(  \mathbf{v}_{-,2}^{\ast}(0)+\frac
{1}{2}\int_{1}^{\infty}\mathbf{\mathbf{q}}_{0}(0,s)~ds\right)  ~,\label{bbb}\\
c  & =-\tau_{-,1}^{\ast}(0)+\int_{1}^{\infty}q_{1}(0,s)~ds~,\label{ccc}\\
d  & =v_{-,1}^{\ast}(0)-\frac{1}{2}\int_{1}^{\infty}q_{1}(0,s)~ds~,\label{ddd}%
\end{align}
and let%
\begin{equation}
\mathbf{r}(t)=i\mathbf{\sigma}_{2}\int_{t}^{\infty}(1-e^{t-s}%
)\mathbf{\mathbf{q}}_{0}(0,s)~ds~.\label{rt}%
\end{equation}
Then, in the limit $k\rightarrow0$, the equations (\ref{iu}), (\ref{iv1}) and
(\ref{iv2}) reduce to%
\begin{align}
\lim_{k\rightarrow0}u(k\mathbf{e},t)  & =c+d+i\mathbf{e}^{T}\mathbf{b}%
~,\label{upm}\\
\lim_{k\rightarrow0}\mathbf{v}_{1}(k\mathbf{e},t)  & =-P_{1}\mathbf{b}%
+P_{1}\mathbf{r}(t)+i\mathbf{e}d~,\label{v1pm}\\
\lim_{k\rightarrow0}\mathbf{v}_{2}(k\mathbf{e},t)  & =P_{2}\mathbf{a}%
+P_{2}\mathbf{r}(t)~.\label{v2pm}%
\end{align}


\begin{proof}
This follows immediately using Proposition \ref{pp11}.
\end{proof}
\end{proposition}

From (\ref{upm})-(\ref{v2pm}) we can extract the time independent (real)
quantities $\mathbf{a}$, $\mathbf{b}$, $c$ and $d$. Namely, let
$\mathbf{e\equiv e}(\vartheta)=(\cos(\vartheta),\sin(\vartheta))$, and let the
average $\left\langle ~.~\right\rangle $ be defined in (\ref{average}). Then,
we see from (\ref{upm}), (\ref{v1pm}) and (\ref{v2pm}) that%
\begin{align*}
\left\langle \lim_{k\rightarrow0}\mathbf{v}_{1}(k\mathbf{e},t)\right\rangle  &
=-\frac{1}{2}\mathbf{b+}\frac{1}{2}\mathbf{r}(t)~,\\
\left\langle \lim_{k\rightarrow0}\mathbf{v}_{2}(k\mathbf{e},t)\right\rangle  &
=\frac{1}{2}\mathbf{a+}\frac{1}{2}\mathbf{r}(t)~,
\end{align*}
and therefore we have, for any $t\geq1$,
\begin{align*}
d  & =\left\langle -i\mathbf{e}^{T}\lim_{k\rightarrow0}\mathbf{v}%
_{1}(k\mathbf{e},t)\right\rangle ~,\\
c+d  & =\left\langle \lim_{k\rightarrow0}u(k\mathbf{e},t)\right\rangle ~,\\
\mathbf{b}  & =\left\langle -i\mathbf{e}\lim_{k\rightarrow0}u(k\mathbf{e}%
,t)\right\rangle ~,\\
\mathbf{a}+\mathbf{b}  & =\left\langle 2\lim_{k\rightarrow0}\mathbf{v}%
_{2}(k\mathbf{e},t)\right\rangle -\left\langle 2\lim_{k\rightarrow0}%
\mathbf{v}_{1}(k\mathbf{e},t)\right\rangle ~.
\end{align*}


\subsection{Interpretation of $\phi$ and $\psi$}

We note that (\ref{aaa})-(\ref{ddd}) imply that the quantities $\phi$ and
$\psi$
\begin{align}
\phi & =c+2d=-\omega_{-}^{\ast}(0)+2u_{-,2}^{\ast}(0)~,\label{zero_flux}\\
\psi & =\mathbf{a}+2\mathbf{b}=-\mathbf{\tau}_{-,2}^{\ast}(0)+2\mathbf{v}%
_{-,2}^{\ast}(0)~,\label{zero_whatever}%
\end{align}
are directly given in terms of the initial conditions. For the downstream
asymptotics of a stationary flow around an arbitrary body we have that
$\phi=0$ (zero flux at infinity) and that $\psi=0$ (matching at $x=0$; see
Appendix VIII for some more details). The quantities $c=-2d$ and
$\mathbf{a}=-2\mathbf{b}$ are in these cases related to the drag and lift
exerted on the body. For a similar interpretation in the two-dimensional case
see \cite{pw001}, \cite{pw002} and \cite{vincent001}.

\section{Asymptotic behavior}

The following theorem provides the leading order behavior of solutions whose
existence has been shown in Theorem \ref{theorem1}. We again restrict
attention to maps $\tau_{-,1}^{\ast}$, $\mathbf{\tau}_{-,2}^{\ast}$,
$v_{-,1}^{\ast}$ and $\mathbf{v}_{-,2}^{\ast}$ with even and odd real and
imaginary parts, respectively.

\bigskip

\begin{theorem}
\label{th4}Let $\alpha>2$. Let $r^{\ast}=(\tau_{-,1}^{\ast},\mathbf{\tau
}_{-,2}^{\ast},v_{-,1}^{\ast},\mathbf{v}_{-,2}^{\ast})\in\mathcal{V}%
_{\alpha+1}$ with $\varepsilon_{0}=\left\Vert \mathbf{r}^{\ast}\right\Vert
_{\alpha+1}$ sufficiently small. Then, the equations (\ref{itau})-(\ref{iv2})
have a solution and%
\begin{align}
\lim_{t\rightarrow\infty}t\int_{\mathbf{R}}\left\vert u(\mathbf{k}%
,t)-u_{as}(\mathbf{k},t)\right\vert ~d^{2}\mathbf{k}~  & =0~,\label{uas}\\
\lim_{t\rightarrow\infty}t^{3/2}\int_{\mathbf{R}}\left\vert \mathbf{v}%
_{1}(\mathbf{k},t)-\mathbf{v}_{1,as}(\mathbf{k},t)\right\vert ~d^{2}%
\mathbf{k}~  & =0~,\label{vas}\\
\lim_{t\rightarrow\infty}t\int_{\mathbf{R}}\left\vert \mathbf{v}%
_{2}(\mathbf{k},t)-\mathbf{v}_{2,as}(\mathbf{k},t)\right\vert ~d^{2}%
\mathbf{k}~  & =0~,\label{vas2}%
\end{align}
where
\begin{align}
u_{as}(\mathbf{k},t)  & =e^{-k^{2}t}~c+e^{-kt}~d+\frac{i\mathbf{k}^{T}}%
{k}e^{-kt}~\mathbf{b}~,\label{uuas}\\
\mathbf{v}_{1,as}(\mathbf{k},t)  & =i\mathbf{k}e^{-k^{2}t}~c+\frac{i}%
{k}\mathbf{k}e^{-kt}~d-P_{1}e^{-kt}~\mathbf{b}~,\label{vv1as}\\
\mathbf{v}_{2,as}(\mathbf{k},t)  & =P_{2}e^{-k^{2}t}~\mathbf{a}=e^{-k^{2}%
t}~\mathbf{a}-P_{1}e^{-k^{2}t}~\mathbf{a}~\label{vv2as}%
\end{align}
with $\mathbf{a}$, $\mathbf{b}$, $c$ and $d$ as defined in (\ref{aaa}%
)-(\ref{ddd}).
\end{theorem}

The existence of a solution follows from Theorem \ref{theorem1}. A proof of
(\ref{uas}), (\ref{vas}) and (\ref{vas2}) can be found in Appendix V.

\section{Proof of Theorem \ref{theorem0}}

We again restrict attention to maps $\tau_{-,1}^{\ast}$, $\mathbf{\tau}%
_{-,2}^{\ast}$, $v_{-,1}^{\ast}$ and $\mathbf{v}_{-,2}^{\ast}$ with even and
odd real and imaginary parts, respectively.

For $\alpha>2$ we have proved in Section 6 the existence of a solution of the
equations (\ref{itau})-(\ref{iv2}) or respectively (\ref{dd}), satisfying (to
avoid confusion we now write the hats for the Fourier transforms)
\begin{align}
\left\vert \hat{u}(\mathbf{k},t)\right\vert  & \leq\varepsilon\mu_{\alpha
+1}(k,t)~,\label{uubound}\\
\left\vert \mathbf{\hat{v}}_{1}(\mathbf{k},t)\right\vert  & \leq
\varepsilon\bar{\mu}_{\alpha+1}(k,t)+\frac{\varepsilon}{t^{1/2}}\mu_{\alpha
}(k,t)~,\label{uvbound}\\
\left\vert \mathbf{\hat{v}}_{2}(\mathbf{k},t)\right\vert  & \leq\varepsilon
\mu_{\alpha+1}(k,t)~.\label{uv2bound}%
\end{align}
Since, for $\alpha>2$, the real and imaginary parts of the functions
$\mathbf{k}\mapsto\hat{u}(\mathbf{k},t)$, $\mathbf{k}\mapsto\mathbf{\hat{v}%
}_{1}(\mathbf{k},t)$ and $\mathbf{k}\mapsto\mathbf{\hat{v}}_{2}(\mathbf{k},t)
$ are, respectively, even and odd functions in $L^{1}(\mathbf{R}^{2}%
,d^{2}\mathbf{k})$ for all $t\geq1$, their Fourier transforms%
\begin{align*}
u(x,\mathbf{y})  & =\left(  \frac{1}{2\pi}\right)  ^{2}\int_{\mathbf{R}^{2}%
}e^{-i\mathbf{k\cdot y}}~\hat{u}(\mathbf{k},x)~d^{2}\mathbf{k}~,\\
\mathbf{v}_{1}(x,\mathbf{y})  & =\left(  \frac{1}{2\pi}\right)  ^{2}%
\int_{\mathbf{R}^{2}}e^{-i\mathbf{k\cdot y}}~\mathbf{\hat{v}}_{1}%
(\mathbf{k},x)~d^{2}\mathbf{k}~,\\
\mathbf{v}_{2}(x,\mathbf{y})  & =\left(  \frac{1}{2\pi}\right)  ^{2}%
\int_{\mathbf{R}^{2}}e^{-i\mathbf{k\cdot y}}~\mathbf{\hat{v}}_{2}%
(\mathbf{k},x)~d^{2}\mathbf{k}~,
\end{align*}
are by the Riemann-Lebesgue lemma real valued continuous maps of $\mathbf{y}$
and vanish as $\left\vert \mathbf{y}\right\vert \rightarrow$ $\infty$ for each
$x\geq1$. Moreover, using (\ref{uubound}), (\ref{uvbound}) and (\ref{uv2bound}%
), we find that, for $x\geq1$,%
\begin{align}
\sup_{\mathbf{y}\in\mathbf{R}^{2}}\left\vert u(x,\mathbf{y})\right\vert  &
\leq\frac{\varepsilon}{\left\vert x\right\vert }~,\label{ubound1}\\
\sup_{\mathbf{y}\in\mathbf{R}^{2}}\left\vert \mathbf{v}_{1}(x,\mathbf{y}%
)\right\vert  & \leq\frac{\varepsilon}{\left\vert x\right\vert ^{3/2}%
}~,\label{vbound1}\\
\sup_{\mathbf{y}\in\mathbf{R}^{2}}\left\vert \mathbf{v}_{2}(x,\mathbf{y}%
)\right\vert  & \leq\frac{\varepsilon}{\left\vert x\right\vert }%
~.\label{v2bound1}%
\end{align}
As a consequence, $u$ and $\mathbf{v=v}_{1}+\mathbf{v}_{2}$ converge to zero
whenever $\left\vert x\right\vert +\left\vert \mathbf{y}\right\vert
\rightarrow\infty$ in $\Omega$ (see Section 5 of \cite{pw001} for details),
and satisfy therefore not only (\ref{ww}) but also the boundary conditions
(\ref{bc001}), (\ref{bc002}). The reconstruction of the pressure from $u$ and
$\mathbf{v}$ is standard. For $\alpha>4$ second derivatives of $u$ and
$\mathbf{v}$ are continuous in direct space, and one easily verifies using the
definitions that the triple $(u,\mathbf{v},p)$ satisfies the Navier-Stokes
equations (\ref{navier}). The set $\mathcal{S}$ in Theorem \ref{theorem0} is
by definition the set of all vector fields $(u,\mathbf{v})$ obtained this way,
restricted to $\Sigma$. Finally, equations (\ref{th2})-(\ref{th3a}) are a
direct consequence of Theorem \ref{th4} (see Appendix VIII for the computation
of the Fourier transforms). This completes the proof of Theorem \ref{theorem0}.

\section{Appendix I}

In this appendix we construct a matrix $S$, with the same block structure as
$L$,
\begin{equation}
S=\left(
\begin{array}
[c]{cc}%
S_{1} & 0\\
S_{3} & S_{2}%
\end{array}
\right)  ~,
\end{equation}
such that
\begin{equation}
S^{-1}LS=D=\left(
\begin{array}
[c]{cc}%
D_{1} & 0\\
0 & D_{2}%
\end{array}
\right)  ~,
\end{equation}
with $D_{1}$ a diagonal $4\times4$ matrix with diagonal entries $\Lambda_{+}$,
$\Lambda_{+}$, $\Lambda_{-}$, $\Lambda_{-}$ and with $D_{2}$ diagonal
$3\times3$ matrix with diagonal entries $0$, $k$, and $-k.$ (Note the branch
of zero modes which is not present in the two-dimensional case.) The matrix
$S_{1}$ diagonalizes $L_{1}$. Namely, $D_{1}=S_{1}^{-1}L_{1}S_{1}$, where
$S_{1}$ is given by (\ref{s1}), the entries being $2\times2$ matrices. The
inverse of $S_{1}$ is given by (\ref{sm1}), the entries again being $2\times2$
matrices. The matrix $S_{2}$ diagonalizes $L_{2}$, namely $D_{2}=S_{2}%
^{-1}L_{2}S_{2}$, where $S_{2}$ is given by (\ref{s2}), the first line being
$1\times1$ matrices and the second line $2\times1$ matrices. The inverse of
$S_{2}$ is given by (\ref{sm2}), the first column being $1\times1$ matrices
and the second column $1\times2$ matrices.

We now compute $S_{3}$. Since $S$ has to satisfy $LS=SD$, we find for $S_{3}$
the equation $L_{3}S_{1}+L_{2}S_{3}=S_{3}D_{1}$, which can be solved as
follows. Let $S_{3}=S_{2}Y$. Then, using that $L_{2}=S_{2}D_{2}S_{2}^{-1}$, we
get the following equation for the matrix $Y$,%
\[
S_{2}^{-1}L_{3}S_{1}=-D_{2}Y+YD_{1}~,
\]
which can be solved for $Y$ entry by entry,\textit{\ i.e.},
\[
Y_{ij}=\frac{1}{-\left(  D_{2}\right)  _{ii}+\left(  D_{1}\right)  _{jj}%
}\left(  S_{2}^{-1}L_{3}S_{1}\right)  _{ij}~,
\]
for $i=1,\dots,3$, $j=1,\dots,4$. Explicitly, we have the $3\times4$ matrix%
\[
L_{3}S_{1}(\mathbf{k})=\left(
\begin{array}
[c]{cc}%
0 & 0\\
i\mathbf{\sigma}_{2} & i\mathbf{\sigma}_{2}%
\end{array}
\right)  ~,
\]
and therefore%
\[
S_{2}^{-1}L_{3}S_{1}(\mathbf{k})=\left(
\begin{array}
[c]{cc}%
\frac{i}{k}\mathbf{k}^{T} & \frac{i}{k}\mathbf{k}^{T}\\
& \\
\frac{-1}{2k}\mathbf{k}^{T}\mathbf{\sigma}_{2} & \frac{-1}{2k}\mathbf{k}%
^{T}\mathbf{\sigma}_{2}\\
& \\
\frac{1}{2k}\mathbf{k}^{T}\mathbf{\sigma}_{2} & \frac{1}{2k}\mathbf{k}%
^{T}\mathbf{\sigma}_{2}%
\end{array}
\right)  ~,
\]
which leads to%
\[
Y(\mathbf{k})=\left(
\begin{array}
[c]{cc}%
\frac{1}{\Lambda_{+}}\frac{i}{k}\mathbf{k}^{T} & \frac{1}{\Lambda_{-}}\frac
{i}{k}\mathbf{k}^{T}\\
& \\
\frac{1}{\Lambda_{+}-k}\frac{-1}{2k}\mathbf{k}^{T}\mathbf{\sigma}_{2} &
\frac{1}{\Lambda_{-}-k}\frac{-1}{2k}\mathbf{k}^{T}\mathbf{\sigma}_{2}\\
& \\
\frac{1}{\Lambda_{+}+k}\frac{1}{2k}\mathbf{k}^{T}\mathbf{\sigma}_{2} &
\frac{1}{\Lambda_{-}+k}\frac{1}{2k}\mathbf{k}^{T}\mathbf{\sigma}_{2}%
\end{array}
\right)  ~.
\]
Using moreover the identities
\begin{align*}
\frac{1}{\Lambda_{+}-k}-\frac{1}{\Lambda_{+}+k}  & =\frac{2k}{\Lambda_{+}}~,\\
\frac{1}{\Lambda_{-}-k}-\frac{1}{\Lambda_{-}+k}  & =\frac{2k}{\Lambda_{-}}~,\\
\frac{1}{\Lambda_{+}-k}+\frac{1}{\Lambda_{+}+k}  & =2~,\\
\frac{1}{\Lambda_{-}-k}+\frac{1}{\Lambda_{-}+k}  & =2~,
\end{align*}
we finally get for $S_{3}$ the matrix (\ref{s3}). We also need $S^{-1}$. We
find that
\[
S^{-1}=\left(
\begin{array}
[c]{cc}%
S_{1}^{-1} & 0\\
(S^{-1})_{3} & S_{2}^{-1}%
\end{array}
\right)  ~,
\]
with $(S^{-1})_{3}=-S_{2}^{-1}S_{3}S_{1}^{-1}=-YS_{1}^{-1}$, for which we find
(\ref{sm3}).

\section{Appendix II}

In this appendix we prove Proposition \ref{pp11}. We first prove the
continuity, then the bounds. Throughout this and subsequent sections we make
extensive use of Proposition \ref{basic} (see Appendix VII).

We note that the maps $u$, $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ as defined in
(\ref{iu}), (\ref{iv1}) and (\ref{iv2}) are explicitly of the form indicated
in (\ref{dec1})-(\ref{dec3}). The continuity of the maps $u_{E}$,
$\mathbf{u}_{O}$, $\mathbf{v}_{1,C}$, $\mathbf{v}_{1,E}$, $v_{1,O}$ and
$\mathbf{v}_{2,E}$ is elementary, since all the integrals converge uniformly
in $k$. Namely, we have that $\left\vert \mathbf{q}_{0}(\mathbf{k}%
,s)\right\vert \leq\varepsilon/s^{3/2}$ and $\left\vert q_{1}(\mathbf{k}%
,s)\right\vert \leq\varepsilon/s^{3/2}$ uniformly in $k\geq0$, and that
$1/s^{3/2}$ is integrable at infinity.

\subsection{Bound on $\omega$}

The bound (\ref{bpm0}) follows immediately from (\ref{bpm5}) using the
definition (\ref{ok0}) of $\omega$.

\subsection{Bound on $\mathbf{\tau}$}

We write $\mathbf{\tau}=\sum_{i=1}^{5}\mathbf{\tau}_{i}$, with $\mathbf{\tau
}_{i}$ the $i$-th term in (\ref{itau}) and we bound each of the terms
individually. The inequality (\ref{bpm1}) then follows using the triangle inequality.

\begin{proposition}
\label{tbounds}For all $\alpha\geq0$ we have the bounds%
\begin{align}
\left\vert \mathbf{\tau}_{1}(\mathbf{k},t)\right\vert  & \leq\frac
{\varepsilon}{t^{1/2}}\mu_{a}(k,t)~,\label{tb1}\\
\left\vert \mathbf{\tau}_{2}(\mathbf{k},t)\right\vert  & \leq\frac
{\varepsilon}{t^{1/2}}\mu_{a}(k,t)~,\label{tb2}\\
\left\vert \mathbf{\tau}_{3}(\mathbf{k},t)\right\vert  & \leq\frac
{\varepsilon}{t^{3/2}}\mu_{a}(k,t)~,\label{tb3}\\
\left\vert \mathbf{\tau}_{4}(\mathbf{k},t)\right\vert  & \leq\frac
{\varepsilon}{t}\mu_{a}(k,t)~,\label{tb4}\\
\left\vert \mathbf{\tau}_{5}(\mathbf{k},t)\right\vert  & \leq\frac
{\varepsilon}{t^{3/2}}\mu_{a}(k,t)~,\label{tb5}%
\end{align}
uniformly in $t\geq1$ and $\mathbf{k}^{2}\in\mathbf{R}^{2}$.
\end{proposition}

For $\mathbf{\tau}_{1}$ we have, using that $k^{2}=-\Lambda_{-}\Lambda_{+} $,
that $\Lambda_{+}^{1/2}\mu_{\alpha+1}(k,1)\leq\mathrm{const.}\mu_{\alpha
+1/2}(k,1)$ and using Proposition \ref{basic} (see Appendix VII), that%
\begin{align*}
& \left\vert \left(  -\mathbf{\sigma}_{2}\mathbf{k}~\tau_{-,1}^{\ast
}(\mathbf{k})+k^{2}P_{1}\mathbf{\tau}_{-,2}^{\ast}(\mathbf{k})\right)
e^{\Lambda_{-}(t-1)}\right\vert \\
& \leq\varepsilon\left(  \mu_{a+1/2}(k,1)\left\vert \Lambda_{-}\right\vert
^{1/2}+\mu_{a+1}(k,1)\left\vert \Lambda_{-}\right\vert \right)  e^{\Lambda
_{-}(t-1)}\\
& \leq\frac{\varepsilon}{t^{1/2}}\mu_{a}(k,t)+\frac{\varepsilon}{t}\mu
_{a}(k,t)~,
\end{align*}
and (\ref{tb1}) follows. Next, splitting the integral in the definition of
$\mathbf{\tau}_{2}$ into two parts we find that%
\begin{align*}
\left\vert \frac{1}{\Lambda_{0}}\int_{1}^{\frac{t+1}{2}}e^{\Lambda_{-}%
(t-s)}q_{1}(\mathbf{k},s)~ds\right\vert  & \leq\varepsilon\mu_{a+1}%
(k,1)e^{\Lambda_{-}\frac{t-1}{2}}\int_{1}^{\frac{t+1}{2}}\frac{1}{s^{3/2}}ds\\
& \leq\varepsilon\mu_{a+1}(k,1)e^{\Lambda_{-}\frac{t-1}{2}}\left(  \frac
{t-1}{t}\right) \\
& \leq\varepsilon\mu_{a+1}(k,t)~,
\end{align*}
and that%
\begin{align*}
\left\vert \frac{1}{\Lambda_{0}}\int_{\frac{t+1}{2}}^{t}e^{\Lambda_{-}%
(t-s)}q_{1}(\mathbf{k},s)~ds\right\vert  & \leq\varepsilon\frac{1}{\Lambda
_{0}}\mu_{a}(k,t)\int_{\frac{t+1}{2}}^{t}\frac{1}{s^{3/2}}ds\\
& \leq\frac{\varepsilon}{t^{1/2}}\frac{1}{\Lambda_{0}}\mu_{a}(k,t)\\
& \leq\varepsilon\mu_{a+1}(k,t)~.
\end{align*}
Therefore we get, using the triangle inequality, that%
\begin{equation}
\left\vert \frac{1}{\Lambda_{0}}\int_{1}^{t}e^{\Lambda_{-}(t-s)}%
q_{1}(\mathbf{k},s)~ds\right\vert \leq\varepsilon\mu_{a+1}(k,t)~.\label{i003}%
\end{equation}
The bound (\ref{tb2}) now follows using that $\varepsilon k\mu_{a+1}%
(k,t)\leq\varepsilon\mu_{a}(k,t)/t^{1/2}$. For $\mathbf{\tau}_{3}$ we have
that%
\begin{align*}
\left\vert \frac{\mathbf{\sigma}_{2}\mathbf{\mathbf{k}}}{\Lambda_{0}}\int
_{t}^{\infty}e^{\Lambda_{+}(t-s)}q_{1}(\mathbf{k},s)~ds\right\vert  &
\leq\frac{\varepsilon}{t^{3/2}}\frac{\left\vert k\right\vert }{\Lambda_{0}}%
\mu_{a}(k,t)\int_{t}^{\infty}e^{\Lambda_{+}(t-s)}ds\\
& \leq\frac{\varepsilon}{t^{3/2}}\frac{1}{\Lambda_{+}}\frac{\left\vert
k\right\vert }{\Lambda_{0}}\mu_{a}(k,t)\\
& \leq\frac{\varepsilon}{t^{3/2}}\mu_{a}(k,t)~,
\end{align*}
which proves (\ref{tb3}). The integral defining $\mathbf{\tau}_{4}$ we split
into two parts. We have that
\begin{align*}
\left\vert \frac{\Lambda_{-}}{\Lambda_{0}}\int_{1}^{\frac{t+1}{2}}%
e^{\Lambda_{-}(t-s)}\mathbf{q}_{0}(\mathbf{k},s)~ds\right\vert  &
\leq\varepsilon\mu_{a+1}(k,1)e^{\Lambda_{-}\frac{t-1}{2}}\left\vert
\Lambda_{-}\right\vert \int_{1}^{\frac{t+1}{2}}\frac{1}{s^{3/2}}ds\\
& \leq\varepsilon\mu_{a+1}(k,1)e^{\Lambda_{-}\frac{t-1}{2}}\left\vert
\Lambda_{-}\right\vert \left(  \frac{t-1}{t}\right) \\
& \leq\frac{\varepsilon}{t}\mu_{a+1}(k,t)~,
\end{align*}
and that%
\begin{align*}
\left\vert \frac{\Lambda_{-}}{\Lambda_{0}}\int_{\frac{t+1}{2}}^{t}%
e^{\Lambda_{-}(t-s)}\mathbf{q}_{0}(\mathbf{k},s)~ds\right\vert  & \leq
\frac{\varepsilon}{t}\frac{1}{\Lambda_{0}}\mu_{a}(k,t)\int_{\frac{t+1}{2}}%
^{t}e^{\Lambda_{-}(t-s)}\left\vert \Lambda_{-}\right\vert ds\\
& \leq\frac{\varepsilon}{t}\frac{1}{\Lambda_{0}}\mu_{a}(k,t)~,
\end{align*}
and (\ref{tb4}) follows using the triangle inequality. For $\mathbf{\tau}_{5}$
we finally have that%
\begin{align*}
\left\vert \frac{\Lambda_{+}}{\Lambda_{0}}\int_{t}^{\infty}e^{\Lambda
_{+}(t-s)}\mathbf{\mathbf{q}}_{0}(\mathbf{k},s)~ds\right\vert  & \leq
\frac{\varepsilon}{t^{3/2}}\mu_{a}(k,t)\int_{t}^{\infty}e^{\Lambda_{+}%
(t-s)}ds\\
& \leq\frac{\varepsilon}{t^{3/2}}\frac{1}{\Lambda_{+}}\mu_{a}(k,t)\\
& \leq\frac{\varepsilon}{t^{3/2}}\mu_{a}(k,t)~,
\end{align*}
which proves (\ref{tb5}).

\subsection{Bound on $u$}

We write $u=\sum_{i=1}^{10}u_{i}$, with $u_{i}$ the $i$-th term in (\ref{iu}),
and we bound each of the terms individually. The inequality (\ref{bpm3}) then
follows using the triangle inequality.

\begin{proposition}
\label{p11}For all $\alpha\geq0$ we have the bounds%
\begin{align}
\left\vert u_{1}(\mathbf{k},t)\right\vert  & \leq\varepsilon\mu_{a}%
(k,t)~,\label{u01b1}\\
\left\vert u_{2}(\mathbf{k},t)\right\vert  & \leq\varepsilon\bar{\mu}%
_{a+1}(k,t)~,\label{u01b2}\\
\left\vert u_{3}(\mathbf{k},t)\right\vert  & \leq\varepsilon\mu_{a}%
(k,t)~,\label{u01b3}\\
\left\vert u_{4}(\mathbf{k},t)\right\vert  & \leq\varepsilon\bar{\mu}%
_{a+1}(k,t)+\frac{\varepsilon}{t^{1/2}}\mu_{a}(k,t)~,\label{u01b4}\\
\left\vert u_{5}(\mathbf{k},t)\right\vert  & \leq\frac{\varepsilon}{t^{1/2}%
}\mu_{a}(k,t)~,\label{u01b5}\\
\left\vert u_{6}(\mathbf{k},t)\right\vert  & \leq\frac{\varepsilon}{t^{3/2}%
}\mu_{a}(k,t)~,\label{u01b6}\\
\left\vert u_{7}(\mathbf{k},t)\right\vert  & \leq\varepsilon\bar{\mu}%
_{a+1}(k,t)+\frac{\varepsilon}{t^{1/2}}\mu_{a}(k,t)~,\label{u01b7}\\
\left\vert u_{8}(\mathbf{k},t)\right\vert  & \leq\frac{\varepsilon}{t^{1/2}%
}\mu_{a}(k,t)~,\label{u01b8}\\
\left\vert u_{9}(\mathbf{k},t)\right\vert  & \leq\frac{\varepsilon}{t^{1/2}%
}\mu_{a}(k,t)~,\label{u01b9}\\
\left\vert u_{10}(\mathbf{k},t)\right\vert  & \leq\frac{\varepsilon}{t^{3/2}%
}\mu_{a}(k,t)~,\label{u01b10}%
\end{align}
uniformly in $t\geq1$ and $\mathbf{k}\in\mathbf{R}^{2}$.
\end{proposition}

For $u_{1}$ we have%
\begin{align*}
\left\vert -\mathbf{k}^{T}\mathbf{\sigma}_{2}\frac{1}{\Lambda_{-}}\left(
-\mathbf{\sigma}_{2}\mathbf{k}~\tau_{-,1}^{\ast}(\mathbf{k})\right)
e^{\Lambda_{-}(t-1)}\right\vert  & \leq\left\vert \Lambda_{+}\tau_{-,1}^{\ast
}(\mathbf{k})e^{\Lambda_{-}(t-1)}\right\vert \\
& \leq\varepsilon\mu_{a}(k,1)e^{\Lambda_{-}(t-1)}~,
\end{align*}
and (\ref{u01b1}) follows using Proposition \ref{basic}. For $u_{2}$ we have,
\[
\left\vert \left(  v_{-,1}^{\ast}(\mathbf{k})-\frac{\mathbf{k}^{T}}%
{k}\mathbf{\sigma}_{2}\mathbf{v}_{-,2}^{\ast}(\mathbf{k})\right)
e^{-k(t-1)}\right\vert \leq\mu_{a+1}(k,1)e^{-k(t-1)}%
\]
and (\ref{u01b2}) follows using Proposition \ref{basic}. For $u_{3}$ we use
(\ref{i003}) and get
\[
\left\vert \frac{\Lambda_{+}}{\Lambda_{0}}\int_{1}^{t}e^{\Lambda_{-}%
(t-s)}q_{1}(\mathbf{k},s)~ds\right\vert \leq\varepsilon\Lambda_{+}\mu
_{\alpha+1}(k,t)~,
\]
and (\ref{u01b3}) follows. The integral defining $u_{4}$ we split into two
parts. We have that
\begin{align*}
\left\vert \frac{1}{2}\int_{1}^{\frac{t+1}{2}}e^{-k(t-s)}q_{1}(\mathbf{k}%
,s)~ds\right\vert  & \leq\varepsilon e^{-k\frac{t-1}{2}}\mu_{\alpha}%
(k,1)\int_{1}^{\frac{t+1}{2}}\frac{1}{s^{3/2}}~ds\\
& \leq\varepsilon\mu_{\alpha}(k,1)e^{-k\frac{t-1}{2}}\left(  \frac{t-1}%
{t}\right) \\
& \leq\varepsilon\bar{\mu}_{\alpha+1}(k,t)~,
\end{align*}
and that
\begin{align*}
\left\vert \frac{1}{2}\int_{\frac{t+1}{2}}^{t}e^{-k(t-s)}q_{1}(\mathbf{k}%
,s)~ds\right\vert  & \leq\varepsilon\mu_{\alpha}(k,t)\int_{\frac{t+1}{2}}%
^{t}\frac{1}{s^{3/2}}~ds\\
& \leq\frac{\varepsilon}{t^{1/2}}\mu_{\alpha}(k,t)~,
\end{align*}
and (\ref{u01b4}) follows. Next, to bound $u_{5}$, we use that
\begin{align*}
\left\vert \frac{1}{2}\int_{t}^{\infty}e^{k(t-s)}q_{1}(\mathbf{k}%
,s)~ds\right\vert  & \leq\varepsilon\mu_{\alpha}(k,t)\int_{t}^{\infty}\frac
{1}{s^{3/2}}~ds\\
& \leq\frac{\varepsilon}{t^{1/2}}\mu_{\alpha}(k,t)~,
\end{align*}
and (\ref{u01b5}) follows. Next, for $u_{6}$ we have that
\begin{align*}
\left\vert \frac{\Lambda_{-}}{\Lambda_{0}}\int_{t}^{\infty}e^{\Lambda
_{+}(t-s)}q_{1}(\mathbf{k},s)~ds\right\vert  & \leq\frac{\varepsilon}{t^{3/2}%
}\frac{\left\vert \Lambda_{-}\right\vert }{\Lambda_{0}}\mu_{\alpha}%
(k,t)\int_{t}^{\infty}e^{\Lambda_{+}(t-s)}~ds\\
& \leq\frac{\varepsilon}{t^{3/2}}\frac{\left\vert \Lambda_{-}\right\vert
}{\Lambda_{0}\Lambda_{+}}\mu_{\alpha}(k,t)~,
\end{align*}
and (\ref{u01b6}) follows. The bounds (\ref{u01b7}) and (\ref{u01b8}) on
$u_{7}$ and $u_{8}$ are obtained exactly as the bounds on $u_{4}$ and $u_{5}$.
To bound on $u_{9}$ we split the corresponding integral into two parts. We
have that%
\begin{align*}
\left\vert \frac{1}{\Lambda_{0}}\int_{1}^{\frac{t+1}{2}}e^{\Lambda_{-}%
(t-s)}\mathbf{\mathbf{q}}_{0}(\mathbf{k},s)~ds\right\vert  & \leq\frac
{1}{\Lambda_{0}}\int_{1}^{\frac{t+1}{2}}e^{\Lambda_{-}(t-s)}\frac{\varepsilon
}{s^{3/2}}\mu_{a}(k,s)~ds\\
& \leq\varepsilon\frac{1}{\Lambda_{0}}e^{\Lambda_{-}\frac{t-1}{2}}\mu
_{a}(k,1)\int_{1}^{\frac{t+1}{2}}\frac{1}{s^{3/2}}~ds\\
& \leq\varepsilon\mu_{a+1}(k,t)~,
\end{align*}
and that
\begin{align*}
\left\vert \frac{1}{\Lambda_{0}}\int_{\frac{t+1}{2}}^{t}e^{\Lambda_{-}%
(t-s)}\mathbf{\mathbf{q}}_{0}(\mathbf{k},s)~ds\right\vert  & \leq\frac
{1}{\Lambda_{0}}\int_{\frac{t+1}{2}}^{t}e^{\Lambda_{-}(t-s)}\frac{\varepsilon
}{s^{3/2}}\mu_{a}(k,s)~ds\\
& \leq\varepsilon\frac{1}{\Lambda_{0}}\mu_{a}(k,t)\int_{\frac{t+1}{2}}%
^{t}\frac{ds}{s^{3/2}}\\
& \leq\frac{\varepsilon}{t^{1/2}}\frac{1}{\Lambda_{0}}\mu_{a}(k,t)\\
& \leq\varepsilon\mu_{a+1}(k,t)~,
\end{align*}
and therefore we find using the triangle inequality that%
\begin{equation}
\left\vert \frac{1}{\Lambda_{0}}\int_{1}^{t}e^{\Lambda_{-}(t-s)}%
\mathbf{\mathbf{q}}_{0}(\mathbf{k},s)~ds\right\vert \leq\varepsilon\mu
_{a+1}(k,t)~.\label{i004}%
\end{equation}
The bound (\ref{u01b9}) now follows, since%
\[
\left\vert u_{9}(\mathbf{k},t)\right\vert \leq\varepsilon k\mu_{a+1}%
(k,t)\leq\frac{\varepsilon}{t^{1/2}}\mu_{a}(k,t)~.
\]
Finally, since $u_{10}=i\mathbf{k}^{T}/\Lambda_{+}~\mathbf{\tau}_{5}$ and
since $\left\vert i\mathbf{k}^{T}/\Lambda_{+}\right\vert \leq\mathrm{const.}$,
(\ref{u01b10}) follows from (\ref{tb5})\textrm{.}

\subsection{Bound on $\mathbf{v}_{1}$}

We write $\mathbf{v}_{1}=\sum_{i=1}^{6}\mathbf{v}_{1,i}$, with $\mathbf{v}%
_{1,i}$ the $i$-th term in (\ref{iv1}), and we bound each of the terms
individually. The inequality (\ref{bpm4}) then follows using the triangle inequality.

\begin{proposition}
\label{v1bounds}For all $\alpha\geq0$ we have the bounds%
\begin{align}
\left\vert \mathbf{v}_{1,1}(\mathbf{k},t)\right\vert  & \leq\frac{\varepsilon
}{t^{1/2}}\mu_{a}(k,t)~,\label{vb1}\\
\left\vert \mathbf{v}_{1,2}(\mathbf{k},t)\right\vert  & \leq\varepsilon
\bar{\mu}_{\alpha+1}(k,t)~,\label{vb2}\\
\left\vert \mathbf{v}_{1,3}(\mathbf{k},t)\right\vert  & \leq\varepsilon
\bar{\mu}_{\alpha+1}(k,t)+\frac{\varepsilon}{t^{1/2}}\mu_{\alpha
}(k,t)~,\label{vb3}\\
\left\vert \mathbf{v}_{1,4}(\mathbf{k},t)\right\vert  & \leq\frac{\varepsilon
}{t^{1/2}}\mu_{a}(k,t)~,\label{vb4}\\
\left\vert \mathbf{v}_{1,5}(\mathbf{k},t)\right\vert  & \leq\varepsilon
\bar{\mu}_{a+1}(k,t)+\frac{\varepsilon}{t^{1/2}}\mu_{a}(k,t)~,\label{vb5}\\
\left\vert \mathbf{v}_{1,6}(\mathbf{k},t)\right\vert  & \leq\frac{\varepsilon
}{t^{1/2}}\mu_{a}(k,t)~,\label{vb6}%
\end{align}
uniformly in $\mathbf{k}\in\mathbf{R}^{2}$ and $t\geq1$.
\end{proposition}

Inequality (\ref{vb1}) follows from (\ref{bpm1}). Next, since $\mathbf{v}%
_{1,2}(\mathbf{k},t)=u_{1}(\mathbf{k},t)i\mathbf{k}/k$ we find that
\[
\left\vert \mathbf{v}_{1,2}(\mathbf{k},t)\right\vert \leq\left\vert
u_{1}(\mathbf{k},t)\right\vert ~,
\]
and therefore (\ref{vb2}) follows from (\ref{u01b1}). Similarly, (\ref{vb3}),
(\ref{vb4}), (\ref{vb5}) and (\ref{vb6}) follow from (\ref{u01b4}),
(\ref{u01b5}), (\ref{u01b7}) and (\ref{u01b8}), since
\begin{align*}
\mathbf{v}_{1,3}(\mathbf{k},t)  & =i\frac{\mathbf{k}}{k}~u_{4}(\mathbf{k}%
,t)~,\\
\mathbf{v}_{1,4}(\mathbf{k},t)  & =-i\frac{\mathbf{k}}{k}~u_{5}(\mathbf{k}%
,t)~,\\
\mathbf{v}_{1,5}(\mathbf{k},t)  & =-i\frac{\mathbf{k}}{k}~u_{7}(\mathbf{k}%
,t)~,\\
\mathbf{v}_{1,6}(\mathbf{k},t)  & =i\frac{\mathbf{k}}{k}~u_{8}(\mathbf{k},t)~.
\end{align*}


\subsection{Bound on $\mathbf{v}_{2}$}

We write $\mathbf{v}_{2}=\sum_{i=1}^{3}\mathbf{v}_{2,i}$, with $\mathbf{v}%
_{2,i}$ the $i$-th term in (\ref{iv2}), and we bound each of the terms
individually. The inequality (\ref{bpm5}) then follows using the triangle inequality.

\begin{proposition}
For all $\alpha\geq0$ we have the bounds%
\begin{align}
\left\vert \mathbf{v}_{2,1}(\mathbf{k},t)\right\vert  & \leq\varepsilon
\mu_{a+1}(k,t)~,\label{v2b1}\\
\left\vert \mathbf{v}_{2,2}(\mathbf{k},t)\right\vert  & \leq\varepsilon
\mu_{a+1}(k,t)~,\label{v2b2}\\
\left\vert \mathbf{v}_{2,3}(\mathbf{k},t)\right\vert  & \leq\frac{\varepsilon
}{t}\mu_{a+1}(k,t)~,\label{v2b3}%
\end{align}
uniformly in $\mathbf{k}\in\mathbf{R}^{2}$ and $t\geq1$.
\end{proposition}

For $\mathbf{v}_{2,1}$ we have, using that $k^{2}=-\Lambda_{-}\Lambda_{+}$ and
using Proposition \ref{basic} (see Appendix VII), that%
\begin{align*}
\left\vert \frac{1}{\Lambda_{-}}P_{2}i\mathbf{\sigma}_{2}\left(  k^{2}%
P_{1}\mathbf{\tau}_{-,2}^{\ast}(\mathbf{k})\right)  e^{\Lambda_{-}%
(t-1)}\right\vert  & \leq\varepsilon\mu_{a+2}(k,1)\Lambda_{+}e^{\Lambda
_{-}(t-1)}\\
& \leq\varepsilon\mu_{a+1}(k,t)~,
\end{align*}
and (\ref{v2b1}) follows. The bound (\ref{v2b2}) follows from (\ref{i004}),
since%
\[
\left\vert \mathbf{v}_{2,2}(\mathbf{k},t)\right\vert \leq\left\vert
P_{2}i\sigma_{2}\right\vert \varepsilon\mu_{a+1}(k,t)\leq\varepsilon\mu
_{a+1}(k,t)~.
\]
Similarly, since $\mathbf{v}_{2,3}=P_{2}i\sigma_{2}\mathbf{\tau}_{5}%
/\Lambda_{+}$, we have that%
\begin{align*}
\left\vert \mathbf{v}_{2,3}(\mathbf{k},t)\right\vert  & \leq\frac{1}%
{\Lambda_{+}}\frac{\varepsilon}{t^{3/2}}\mu_{a}(k,t)\\
& \leq\frac{\varepsilon}{t}\mu_{a+1}(k,t)~,
\end{align*}
and the bound (\ref{v2b3}) follows.

\section{Appendix III}

In this appendix we give a proof of Proposition \ref{pp12}. Note that the
bounds (\ref{bpm4}) and (\ref{bpm5}) imply that%
\begin{equation}
\left\vert \mathbf{v}(\mathbf{k},t)\right\vert \leq\varepsilon\mu_{\alpha
}(k,t)~.\label{f001}%
\end{equation}


We first prove the bound on $\mathbf{q}_{0}$. Namely, using Proposition
\ref{c1} (see Appendix VII), we find from (\ref{bpm0}, (\ref{bpm1}),
(\ref{bpm3}) and (\ref{f001}) that%
\begin{align*}
\left\vert \left(  u\ast\mathbf{\tau}\right)  (\mathbf{k},t)\right\vert  &
\leq\frac{\varepsilon^{2}}{t^{3/2}}\mu_{a}(k,t)~,\\
\left\vert \left(  \omega\ast\mathbf{v}\right)  (\mathbf{k},t)\right\vert  &
\leq\frac{\varepsilon^{2}}{t^{3/2}}\mu_{a}(k,t)~,
\end{align*}
and (\ref{nqb1}) follows using the triangle inequality. Similarly we have for
$q_{1}$, using (\ref{f001}) and (\ref{bpm1}) that
\[
\left\vert \left(  \mathbf{v}\ast(i\sigma_{2}\mathbf{\tau}\right)
(\mathbf{k},t)\right\vert \leq\frac{\varepsilon^{2}}{t^{3/2}}\mu_{a}(k,t)~,
\]
which proves (\ref{nqb3}). This completes the proof of Proposition \ref{pp12}.

\section{Appendix IV}

In this appendix we prove Proposition \ref{prop12}. Let $\rho^{1}\equiv
(\rho_{1}^{1},\rho_{2}^{1})$, $\rho^{2}\equiv(\rho_{1}^{2},\rho_{2}^{2})\in
B_{\alpha}(\varepsilon_{0})$. Then, by Proposition \ref{pp11} and Proposition
\ref{pp12}, $\rho\equiv\mathcal{N}(\rho^{1})-\mathcal{N}(\rho^{2})$ is well
defined and $\rho\in\mathcal{B}_{\alpha}$. Let $\rho\equiv(\rho_{1},\rho_{2}%
)$, and let $\omega^{i}$, $\mathbf{\tau}^{i}$, $u^{i}$, $\mathbf{v}^{i}$,
$i=1,2$, be the quantities (\ref{ok0}), (\ref{itau}), (\ref{iu}) and
(\ref{uv}) computed from $\rho^{1}$ and $\rho^{2}$, respectively. Using the
identity $ab-\tilde{a}\tilde{b}=(a-\tilde{a})b+\tilde{a}(b-\tilde{b})$
(distributive law) we find that%
\begin{align}
\rho_{1}  & =\left(  \frac{1}{2\pi}\right)  ^{2}\left(  u^{1}\ast\mathbf{\tau
}^{1}-\omega^{1}\ast\mathbf{v}^{1}\right) \nonumber\\
& -\left(  \frac{1}{2\pi}\right)  ^{2}\left(  u^{2}\ast\mathbf{\tau}%
^{2}-\omega^{2}\ast\mathbf{v}^{2}\right) \nonumber\\
& =\left(  \frac{1}{2\pi}\right)  ^{2}\left[  \left(  u^{1}-u^{2}\right)
\ast\mathbf{\tau}^{1}+u^{2}\ast\left(  \mathbf{\tau}^{1}-\mathbf{\tau}%
^{2}\right)  \right] \nonumber\\
& -\left(  \frac{1}{2\pi}\right)  ^{2}\left[  \left(  \omega^{1}-\omega
^{2}\right)  \ast\mathbf{v}^{1}+\omega^{2}\ast\left(  \mathbf{v}%
^{1}-\mathbf{v}^{2}\right)  \right]  ~,\nonumber
\end{align}
and similarly that%
\[
\rho_{2}=-\frac{1}{4\pi}\left[  \left(  \mathbf{v}^{1}-\mathbf{v}^{2}\right)
\ast(i\sigma_{2}\mathbf{\tau}^{1})+\mathbf{v}^{2}\ast\left(  i\sigma
_{2}(\mathbf{\tau}^{1}-\mathbf{\tau}^{2})\right)  \right]  ~.
\]
Therefore, and since the quantities $\omega^{i}$, $\mathbf{\tau}^{i}$, $u^{i}%
$, $\mathbf{v}^{i}$, $i=1,2$ are linear (respectively affine) in $\rho^{1}$
and $\rho^{2}$, the bound (\ref{lipschitz}) now follows \textit{mutatis
mutandis} as in the proof of Proposition \ref{pp11} and Proposition \ref{pp12}.

\section{Appendix V}

In this appendix we prove (\ref{uas}), (\ref{vas}) and (\ref{vas2}).

\subsection{Asymptotic behavior of $u$}

Let%
\[
U(\mathbf{k},t)=\Lambda_{+}\left(  \mathbf{-}\tau_{-,1}^{\ast}(\mathbf{k}%
)+\frac{1}{\Lambda_{0}}\int_{1}^{t}q_{1}(\mathbf{k},s)~ds\right)
e^{\Lambda_{-}(t-1)}~.
\]
Using the triangle inequality we get that%
\begin{align}
\left\vert u(\mathbf{k},t)-u_{as}(k,t)\right\vert  & \leq\left\vert
U(\mathbf{k},t)-c~e^{-k^{2}t}\right\vert \nonumber\\
& +\left\vert c~e^{-k^{2}t}-u_{as}(\mathbf{k},t)\right\vert \nonumber\\
& +\left\vert u(k,t)-U(k,t)\right\vert ~.\label{utt}%
\end{align}
We bound each term in (\ref{utt}) separately. First, we have that%
\begin{equation}
\lim_{t\rightarrow\infty}U(\frac{\mathbf{k}}{t^{1/2}},t)=c~e^{-k^{2}%
}~,\label{ld1}%
\end{equation}
and furthermore that%
\begin{align*}
\left\vert U(\mathbf{k},t)\right\vert  & \leq\left(  \varepsilon\mu_{\alpha
}(k,1)+\varepsilon\mu_{\alpha}(k,1)\int_{1}^{\infty}\frac{1}{s^{3/2}%
}ds\right)  e^{\Lambda_{-}(t-1)}\\
& \leq\varepsilon\mu_{\alpha}(k,t)~,
\end{align*}
so that%
\begin{equation}
\left\vert U(\frac{\mathbf{k}}{t^{1/2}},t)\right\vert \leq\varepsilon
\mu_{\alpha}(k,1)~.\label{ld2}%
\end{equation}
From (\ref{ld1}) and (\ref{ld2}) it follows by the Lebesgue dominated
convergence theorem that%
\begin{align*}
& \lim_{t\rightarrow\infty}t\int_{\mathbf{R}^{2}}\left\vert U(\mathbf{k}%
,t)-c~e^{-k^{2}t}\right\vert ~d^{2}\mathbf{k}\\
& =\lim_{t\rightarrow\infty}\int_{\mathbf{R}^{2}}\left\vert U(\frac
{\mathbf{k}}{t^{1/2}},t)-c~e^{-k^{2}}\right\vert ~d^{2}\mathbf{k}\\
& =0~,
\end{align*}
as required. Next%
\begin{align*}
\lim_{t\rightarrow\infty}t\int_{\mathbf{R}^{2}}\left\vert u_{as}%
(\mathbf{k},t)-c~e^{-k^{2}t}\right\vert ~d^{2}\mathbf{k}~  & \leq
\lim_{t\rightarrow\infty}t\int_{\mathbf{R}^{2}}\left(  ~d+\frac{i\mathbf{k}%
^{T}}{k}~\mathbf{b}\right)  e^{-kt}d^{2}\mathbf{k}\\
& \leq\lim_{t\rightarrow\infty}t\int_{\mathbf{R}^{2}}\left(  \left\vert
d\right\vert +\left\vert \mathbf{b}\right\vert \right)  e^{-kt}~d^{2}%
\mathbf{k}\\
& =\lim_{t\rightarrow\infty}t~\left(  \left\vert d\right\vert +\left\vert
\mathbf{b}\right\vert \right)  \frac{2\pi}{t^{2}}=0~,
\end{align*}
as required. For the last term in (\ref{utt}) we have, writing as in Appendix
II $u=\sum_{i=1}^{10}u_{i}$, with $u_{i}$ the $i$-th term in (\ref{iu}),%
\begin{align}
u(\mathbf{k},t)-U(\mathbf{k},t)  & =u_{2}(\mathbf{k},t)\nonumber\\
& +\frac{\Lambda_{+}}{\Lambda_{0}}\int_{1}^{t}\left(  e^{\Lambda_{-}%
(t-s)}-e^{\Lambda_{-}(t-1)}\right)  q_{1}(\mathbf{k},s)~ds\nonumber\\
& \sum_{i=4\dots10}u_{i}(\mathbf{k},t)~.\label{utt3}%
\end{align}
Since%
\begin{align}
\int_{1}^{\frac{t+1}{2}}e^{\Lambda_{-}(t-s)}\left\vert \Lambda_{-}\right\vert
(s-1)\frac{\varepsilon}{s^{3/2}}\mu_{\alpha}(k,s)~ds  & \leq\varepsilon
\mu_{\alpha}(k,1)e^{\Lambda_{-}\frac{t-1}{2}}\left\vert \Lambda_{-}\right\vert
\left(  \frac{t-1}{t}\right)  ^{2}t^{1/2}\nonumber\\
& \leq\frac{\varepsilon}{t^{1/2}}\mu_{\alpha}(k,t)~,\label{ii1}%
\end{align}
and
\begin{align}
\int_{\frac{t+1}{2}}^{t}e^{\Lambda_{-}(t-s)}\left\vert \Lambda_{-}\right\vert
(s-1)\frac{\varepsilon}{s^{3/2}}\mu_{\alpha}(k,s)~ds  & \leq\frac{\varepsilon
}{t^{1/2}}\mu_{\alpha}(k,t)\int_{\frac{t+1}{2}}^{t}e^{\Lambda_{-}%
(t-s)}\left\vert \Lambda_{-}\right\vert ~ds\nonumber\\
& \leq\frac{\varepsilon}{t^{1/2}}\mu_{\alpha}(k,t)~,\label{ii2}%
\end{align}
we find, using that $1-e^{x}\leq-x$ for all $x\leq0$, that%
\begin{align}
\left\vert \frac{\Lambda_{+}}{\Lambda_{0}}\int_{1}^{t}\left(  e^{\Lambda
_{-}(t-s)}-e^{\Lambda_{-}(t-1)}\right)  q_{1}(\mathbf{k},s)~ds\right\vert  &
\leq\int_{1}^{t}e^{\Lambda_{-}(t-s)}\left(  1-e^{\Lambda_{-}(s-1)}\right)
\frac{\varepsilon}{s^{3/2}}\mu_{\alpha}(k,s)~ds\nonumber\\
& \leq-\int_{1}^{t}e^{\Lambda_{-}(t-s)}\Lambda_{-}(s-1)\frac{\varepsilon
}{s^{3/2}}\mu_{\alpha}(k,s)~ds\nonumber\\
& \leq\frac{\varepsilon}{t^{1/2}}\mu_{\alpha}(k,t)~,\label{ii3}%
\end{align}
and therefore we find from (\ref{utt3}) using the triangle inequality and
using Proposition \ref{p11}, that%
\begin{align*}
\left\vert u(\mathbf{k},t)-U(\mathbf{k},t))\right\vert  & \leq\left\vert
u_{2}(\mathbf{k},t)\right\vert +\frac{\varepsilon}{t^{1/2}}\mu_{a}%
(k,t)+\sum_{i=4\dots10}\left\vert u_{i}(\mathbf{k},t)\right\vert \\
& \leq\varepsilon\bar{\mu}_{a+1}(k,t)+\frac{\varepsilon}{t^{1/2}}\mu
_{a}(k,t)~,
\end{align*}
from which it follows that%
\begin{align*}
\lim_{t\rightarrow\infty}t\int_{\mathbf{R}^{2}}\left\vert u(\mathbf{k}%
,t)-U(\mathbf{k},t)\right\vert ~dk  & \leq\lim_{t\rightarrow\infty}%
t\int_{\mathbf{R}^{2}}\left(  \varepsilon\bar{\mu}_{a+1}(k,t)+\frac
{\varepsilon}{t^{1/2}}\mu_{a}(k,t)\right)  ~d^{2}\mathbf{k}\\
& \leq\lim_{t\rightarrow\infty}t\left(  \frac{\varepsilon}{t^{2}}%
+\frac{\varepsilon}{t^{3/2}}\right)  =0~,
\end{align*}
as required. This completes the proof of (\ref{uas}).

\subsection{Asymptotic behavior of $v_{1}$}

Let%
\[
\mathbf{V}_{1}(\mathbf{k},t)=\left(  \tau_{-,1}^{\ast}(\mathbf{k})-\frac
{1}{\Lambda_{0}}\int_{1}^{t}q_{1}(\mathbf{k},s)~ds\right)  (-i\mathbf{k}%
)e^{\Lambda_{-}(t-1)}~.
\]
Using the triangle inequality we get that%
\begin{align}
\left\vert \mathbf{v}_{1}(\mathbf{k},t)-\mathbf{v}_{1,as}(\mathbf{k}%
,t)\right\vert  & \leq\left\vert \mathbf{V}_{1}(\mathbf{k},t)-c~i\mathbf{k}%
e^{-k^{2}t}\right\vert \nonumber\\
& +\left\vert c~i\mathbf{k}e^{-k^{2}t}-\mathbf{v}_{1,as}(\mathbf{k}%
,t)\right\vert \nonumber\\
& +\left\vert \mathbf{v}_{1}(\mathbf{k},t)-\mathbf{V}_{1}(\mathbf{k}%
,t)\right\vert ~.\label{vt}%
\end{align}
We bound each term in (\ref{vt}) separately. First, we have that%
\begin{equation}
\lim_{t\rightarrow\infty}t^{1/2}\mathbf{V}_{1}(\frac{\mathbf{k}}{t^{1/2}%
},t)=c~(-i\mathbf{k})e^{-k^{2}}~,\label{ldv1}%
\end{equation}
and furthermore that%
\begin{align*}
\left\vert \mathbf{V}_{1}(\mathbf{k},t)\right\vert  & \leq\left(
\varepsilon\mu_{\alpha+1}(k,1)+\varepsilon\mu_{\alpha+1}(k,1)\int_{1}^{\infty
}\frac{1}{s^{3/2}}ds\right)  ke^{\Lambda_{-}(t-1)}\\
& \leq\varepsilon k\mu_{\alpha+1}(k,t)~,
\end{align*}
so that%
\begin{equation}
\left\vert t^{1/2}\mathbf{V}_{1}(\frac{\mathbf{k}}{t^{1/2}},t)\right\vert
\leq\varepsilon k\mu_{\alpha+1}(k,1)~.\label{ldv2}%
\end{equation}
From (\ref{ldv1}) and (\ref{ldv2}) it follows by the Lebesgue dominated
convergence theorem that%
\begin{align*}
& \lim_{t\rightarrow\infty}t^{3/2}\int_{\mathbf{R}^{2}}\left\vert
\mathbf{V}_{1}(\mathbf{k},t)-c~(-i\mathbf{k})e^{-k^{2}t}\right\vert
~d^{2}\mathbf{k}\\
& =\lim_{t\rightarrow\infty}\int_{\mathbf{R}^{2}}\left\vert t^{1/2}%
\mathbf{V}_{1}(\frac{\mathbf{k}}{t^{1/2}},t)-c~(-i\mathbf{k})e^{-k^{2}%
}\right\vert ~d^{2}\mathbf{k}\\
& =0~,
\end{align*}
as required. Next%
\begin{align*}
\lim_{t\rightarrow\infty}t^{3/2}\int_{\mathbf{R}^{2}}\left\vert u_{as}%
(\mathbf{k},t)-c~e^{-k^{2}t}\right\vert ~d^{2}\mathbf{k}~  & \leq
\lim_{t\rightarrow\infty}t^{3/2}\int_{\mathbf{R}^{2}}\left(  \frac{i}%
{k}\mathbf{k}e^{-kt}~d-P_{1}e^{-kt}~\mathbf{b}\right)  e^{-kt}d^{2}%
\mathbf{k}\\
& \leq\lim_{t\rightarrow\infty}t^{3/2}\int_{\mathbf{R}^{2}}\left(  \left\vert
d\right\vert +\left\vert \mathbf{b}\right\vert \right)  e^{-kt}~d^{2}%
\mathbf{k}\\
& =\lim_{t\rightarrow\infty}t^{3/2}\left(  \left\vert d\right\vert +\left\vert
\mathbf{b}\right\vert \right)  \frac{2\pi}{t^{2}}=0~,
\end{align*}
as required. Finally, for the last term in (\ref{vt}) we have the following Proposition:

\begin{proposition}
\label{last}Let $\mathbf{v}_{1}$ and $\mathbf{V}_{1}$ be as defined above.
Then,%
\begin{align}
\left\vert \mathbf{v}_{1}(\mathbf{k},t)-\mathbf{V}_{1}(\mathbf{k}%
,t)\right\vert  & \leq\frac{\varepsilon}{t^{2/3}}\mu_{\alpha}(k,t)+\frac
{\varepsilon}{t^{1/3}}\mu_{\alpha}^{5/6}(k,t)\nonumber\\
& +\frac{\varepsilon}{t^{1/2}}\mu_{\alpha}^{3/4}(k,t)+\varepsilon\bar{\mu
}_{\alpha+1}(k,t)~.\label{vvp}%
\end{align}

\end{proposition}

See Appendix VI for a proof.

\bigskip

From Proposition \ref{last} it follows that%
\begin{align*}
& \lim_{t\rightarrow\infty}t^{3/2}\int_{\mathbf{R}^{2}}\left\vert
\mathbf{v}_{1}(\mathbf{k},t)-\mathbf{V}_{1}(\mathbf{k},t)\right\vert
~d^{2}\mathbf{k}\\
& \leq\lim_{t\rightarrow\infty}t^{3/2}\int_{\mathbf{R}^{2}}\left(
\frac{\varepsilon}{t^{2/3}}\mu_{\alpha}(k,t)+\frac{\varepsilon}{t^{1/3}}%
\mu_{\alpha}^{5/6}(k,t)+\frac{\varepsilon}{t^{1/2}}\mu_{\alpha}^{3/4}%
(k,t)+\varepsilon\bar{\mu}_{\alpha+1}(k,t)\right)  ~d^{2}\mathbf{k}\\
& \leq\lim_{t\rightarrow\infty}t^{3/2}\left(  \frac{\varepsilon}{t^{5/3}%
}+\frac{\varepsilon}{t^{2}}\right)  =0~,
\end{align*}
as required. This completes the proof of (\ref{vas}).

\subsection{Asymptotic behavior of $v_{2}$}

Let%
\[
\mathbf{V}_{2}(\mathbf{k},t)=P_{2}i\mathbf{\sigma}_{2}\left(  -\Lambda
_{+}\mathbf{\tau}_{-,2}^{\ast}(\mathbf{k})-\frac{1}{\Lambda_{0}}\int_{1}%
^{t}\mathbf{\mathbf{q}}_{0}(\mathbf{k},s)~ds\right)  e^{\Lambda_{-}(t-1)}~.
\]
Using the triangle inequality we get that%
\begin{align}
\left\vert \mathbf{v}_{2}(\mathbf{k},t)-\mathbf{v}_{2,as}(\mathbf{k}%
,t)\right\vert  & \leq\left\vert \mathbf{V}_{2}(\mathbf{k},t)-P_{2}e^{-k^{2}%
t}~\mathbf{a}\right\vert \nonumber\\
& +\left\vert \mathbf{v}_{2}(\mathbf{k},t)-\mathbf{V}_{2}(\mathbf{k}%
,t)\right\vert ~.\label{vt2}%
\end{align}
We bound each term in (\ref{vt2}) separately. First, we have that%
\begin{equation}
\lim_{t\rightarrow\infty}\mathbf{V}_{2}(\frac{\mathbf{k}}{t^{1/2}}%
,t)=P_{2}e^{-k^{2}}\mathbf{a}~,\label{ldv11}%
\end{equation}
and furthermore that%
\begin{align*}
\left\vert \mathbf{V}_{2}(\mathbf{k},t)\right\vert  & \leq\left(
\varepsilon\mu_{\alpha+1}(k,1)+\varepsilon\mu_{\alpha+1}(k,1)\int_{1}^{\infty
}\frac{1}{s^{3/2}}ds\right)  e^{\Lambda_{-}(t-1)}\\
& \leq\varepsilon\mu_{\alpha+1}(k,t)~,
\end{align*}
so that%
\begin{equation}
\left\vert \mathbf{V}_{2}(\frac{\mathbf{k}}{t^{1/2}},t)\right\vert
\leq\varepsilon\mu_{\alpha+1}(k,1)~.\label{ldv22}%
\end{equation}
From (\ref{ldv11}) and (\ref{ldv22}) it follows by the Lebesgue dominated
convergence theorem that%
\begin{align*}
& \lim_{t\rightarrow\infty}t\int_{\mathbf{R}^{2}}\left\vert \mathbf{V}%
_{2}(\mathbf{k},t)-P_{2}e^{-k^{2}t}\mathbf{a}\right\vert ~d^{2}\mathbf{k}\\
& =\lim_{t\rightarrow\infty}\int_{\mathbf{R}^{2}}\left\vert \mathbf{V}%
_{2}(\frac{\mathbf{k}}{t^{1/2}},t)-P_{2}e^{-k^{2}}\mathbf{a}\right\vert
~d^{2}\mathbf{k}\\
& =0~,
\end{align*}
as required. For the second term in (\ref{vt2}) we have, writing as in
Appendix IV $\mathbf{v}_{2}=\sum_{i=1}^{3}\mathbf{v}_{2,i}$, with
$\mathbf{v}_{2,i}$ the $i$-th term in (\ref{iv2}),%
\begin{align}
\mathbf{v}_{2}(\mathbf{k},t)-\mathbf{V}_{2}(\mathbf{k},t)  & =-\frac
{1}{\Lambda_{0}}\int_{1}^{t}\left(  e^{\Lambda_{-}(t-s)}-e^{\Lambda_{-}%
(t-1)}\right)  P_{2}i\mathbf{\sigma}_{2}\mathbf{\mathbf{q}}_{0}(\mathbf{k}%
,s)~ds\nonumber\\
& +\mathbf{v}_{2,3}(\mathbf{k},t)~.\label{v2diff}%
\end{align}
Using (\ref{ii1}) and (\ref{ii2}) and that $1-e^{x}\leq-x$ for all $x\leq0$,
we find as in (\ref{ii3}) that%
\begin{align*}
\left\vert \frac{1}{\Lambda_{0}}\int_{1}^{t}\left(  e^{\Lambda_{-}%
(t-s)}-e^{\Lambda_{-}(t-1)}\right)  P_{2}i\mathbf{\sigma}_{2}%
\mathbf{\mathbf{q}}_{0}(\mathbf{k},s)~ds\right\vert  & \leq\frac{1}%
{\Lambda_{0}}\int_{1}^{t}e^{\Lambda_{-}(t-s)}\left(  1-e^{\Lambda_{-}%
(s-1)}\right)  \frac{\varepsilon}{s^{3/2}}\mu_{\alpha}(k,s)~ds\\
& \leq\frac{\varepsilon}{t^{1/2}}\frac{1}{\Lambda_{0}}\mu_{\alpha}(k,t)~,
\end{align*}
and therefore we find from (\ref{v2diff}) using the triangle inequality and
using (\ref{v2b3}), that%
\begin{align*}
\left\vert \mathbf{v}_{2}(\mathbf{k},t)-\mathbf{V}_{2}(\mathbf{k}%
,t)\right\vert  & \leq\frac{\varepsilon}{t^{1/2}}\mu_{a}(k,t)+\left\vert
\mathbf{v}_{2,3}(\mathbf{k},t)\right\vert \\
& \leq\frac{\varepsilon}{t^{1/2}}\mu_{a}(k,t)~,
\end{align*}
from which it follows that%
\begin{align*}
\lim_{t\rightarrow\infty}t\int_{\mathbf{R}^{2}}\left\vert \mathbf{v}%
_{2}(\mathbf{k},t)-\mathbf{V}_{2}(\mathbf{k},t)\right\vert ~d^{2}\mathbf{k}  &
\leq\lim_{t\rightarrow\infty}t\int_{\mathbf{R}^{2}}\frac{\varepsilon}{t^{1/2}%
}\mu_{a}(k,t)~d^{2}\mathbf{k}\\
& \leq\lim_{t\rightarrow\infty}t\frac{\varepsilon}{t^{3/2}}=0~,
\end{align*}
as required. This completes the proof of (\ref{vas2}).

\section{Appendix VI}

In this appendix we prove Proposition \ref{last}. The proof is rather lengthy
and we therefore split it in several pieces. We start by proving some general bounds.

\subsection{Three inequalities}

\begin{proposition}
\label{ppp}Let $\alpha\geq0$. Then,%
\begin{align}
\int_{1}^{t}\left(  e^{-k(t-s)}-e^{-k(t-1)}\right)  ~\frac{1}{s^{3/2}}%
\mu_{\alpha}(k,s)~ds  & \leq\mathrm{const.}~\left(  \frac{1}{t^{2/3}}%
\mu_{\alpha}(k,t)+\frac{1}{t^{1/3}}\mu_{\alpha}^{5/6}(k,t)\right)
~,\label{ppp1}\\
\int_{t}^{\infty}e^{k(t-s)}~\frac{1}{s^{3/2}}\mu_{\alpha}(k,s)~ds  &
\leq\mathrm{const.}~\left(  \frac{1}{t^{2/3}}\mu_{\alpha}(k,t)+\frac
{1}{t^{1/2}}\mu_{\alpha}^{3/4}(k,t)\right)  ~,\label{ppp2}\\
\frac{k}{\Lambda_{0}}\int_{1}^{t}\left(  e^{\Lambda_{-}(t-s)}-e^{\Lambda
_{-}(t-1)}\right)  \frac{1}{s^{3/2}}\mu_{\alpha}(k,s)~ds  & \leq
\mathrm{const.}~\frac{1}{t}\mu_{\alpha}(k,t)~,\label{ppp3}%
\end{align}
uniformly in $\mathbf{k}\in\mathbf{R}^{2}$ and $t\geq1$.
\end{proposition}

We first prove (\ref{ppp1}) for $1\leq t\leq2$. We have%
\begin{align*}
\int_{1}^{t}\left(  e^{-k(t-s)}-e^{-k(t-1)}\right)  \frac{1}{s^{3/2}}%
\mu_{\alpha}(k,s)~ds  & \leq\varepsilon\mu_{\alpha}(k,1)\int_{1}^{t}\frac
{ds}{s^{3/2}}\\
& \leq\varepsilon\mu_{\alpha}(k,1)\leq\frac{\varepsilon}{t^{2/3}}\mu_{\alpha
}(k,t)~,
\end{align*}
as required. For $t>2$ we split the integral in (\ref{ppp1}) into two. For the
first part we have%
\begin{align*}
& \int_{1}^{t-(t-1)^{5/6}}\left(  e^{-k(t-s)}-e^{-k(t-1)}\right)  \frac
{1}{s^{3/2}}\mu_{\alpha}(k,s)~ds\\
& \leq\varepsilon\mu_{\alpha}(k,1)\int_{1}^{t-(t-1)^{5/6}}\left(
e^{-k(t-s)}-e^{-k(t-1)}\right)  \frac{1}{s^{3/2}}~ds\\
& \leq\varepsilon\mu_{\alpha}(k,1)\int_{1}^{t-(t-1)^{5/6}}e^{-k(t-s)}\left(
1-e^{k(s-1)}\right)  \frac{1}{s^{3/2}}~ds\\
& \leq\varepsilon\mu_{\alpha}(k,1)e^{-k(t-1)^{5/6}}\int_{1}^{t-(t-1)^{5/6}%
}(s-1)k\frac{1}{s^{3/2}}~ds\\
& \leq\varepsilon t^{1/2}\mu_{\alpha}(k,1)e^{-k(t-1)^{5/6}}k\left(  \frac
{t-1}{t}\right)  ^{2}\\
& \leq\frac{\varepsilon}{t^{1/3}}\mu_{\alpha}^{5/6}(k,t)~,
\end{align*}
as required, and for the other part we get,%
\begin{align*}
& \int_{t-(t-1)^{5/6}}^{t}\left(  e^{-k(t-s)}-e^{-k(t-1)}\right)  \frac
{1}{s^{3/2}}\mu_{\alpha}(k,s)~ds\\
& \leq\frac{\varepsilon}{t^{3/2}}\mu_{\alpha}(k,t)\int_{t-(t-1)^{5/6}}^{t}ds\\
& \leq\frac{\varepsilon}{t^{3/2}}\mu_{\alpha}(k,t)t^{5/6}\leq\frac
{\varepsilon}{t^{2/3}}\mu_{\alpha}(k,t)~,
\end{align*}
as required. We now prove (\ref{ppp2}). Namely,%
\begin{align*}
& \int_{t}^{\infty}e^{k(t-s)}\frac{1}{s^{3/2}}\mu_{\alpha}(k,s)~ds\\
& \leq\varepsilon\mu_{\alpha}(k,t)\left(  \int_{t}^{t+t^{3/4}}e^{k(t-s)}%
\frac{1}{s^{3/2}}~ds+\int_{t+t^{3/4}}^{\infty}e^{k(t-s)}\frac{1}{s^{3/2}%
}~ds\right) \\
& \leq\varepsilon\mu_{\alpha}(k,t)\left(  \int_{t}^{t+t^{3/4}}\frac{1}%
{s^{3/2}}~ds+e^{-kt^{3/4}}\int_{t+t^{3/4}}^{\infty}\frac{1}{s^{3/2}}~ds\right)
\\
& \leq\varepsilon\mu_{\alpha}(k,t)\left(  \frac{1}{t^{3/4}}+\frac{1}{t^{1/2}%
}e^{-kt^{3/4}}\right) \\
& \leq\frac{\varepsilon}{t^{3/4}}\mu_{\alpha}(k,t)+\frac{\varepsilon}{t^{1/2}%
}\mu_{\alpha}^{3/4}(k,t)\\
& \leq\frac{\varepsilon}{t^{2/3}}\mu_{\alpha}(k,t)+\frac{\varepsilon}{t^{1/2}%
}\mu_{\alpha}^{3/4}(k,t)~,
\end{align*}
as required. We finally prove (\ref{ppp3}). We have that%
\begin{align*}
& \frac{k}{\Lambda_{0}}\int_{1}^{t}\left(  e^{\Lambda_{-}(t-s)}-e^{\Lambda
_{-}(t-1)}\right)  \frac{1}{s^{3/2}}\mu_{\alpha}(k,s)~ds\\
& \leq\varepsilon\mu_{\alpha+1/2}(k,1)\int_{1}^{\frac{t+1}{2}}\left(
e^{\Lambda_{-}(t-s)}-e^{\Lambda_{-}(t-1)}\right)  \left\vert \Lambda
_{-}\right\vert ^{1/2}\frac{1}{s^{3/2}}~ds\\
& \leq\varepsilon\mu_{\alpha+1/2}(k,1)\int_{1}^{\frac{t+1}{2}}e^{\Lambda
_{-}(t-s)}\left(  1-e^{\Lambda_{-}(s-1)}\right)  \left\vert \Lambda
_{-}\right\vert ^{1/2}\frac{1}{s^{3/2}}~ds\\
& \leq\varepsilon\mu_{\alpha+1/2}(k,1)e^{\Lambda_{-}\frac{t-1}{2}}\int
_{1}^{\frac{t+1}{2}}(s-1)\left\vert \Lambda_{-}\right\vert ^{3/2}\frac
{1}{s^{3/2}}~ds\\
& \leq\varepsilon t^{1/2}\mu_{\alpha+1/2}(k,1)e^{\Lambda_{-}\frac{t-1}{2}%
}\left\vert \Lambda_{-}\right\vert ^{3/2}\left(  \frac{t-1}{t}\right)  ^{2}\\
& \leq\frac{\varepsilon}{t}\mu_{\alpha+1/2}(k,t)~,
\end{align*}
and that%
\begin{align*}
& \frac{k}{\Lambda_{0}}\int_{\frac{t+1}{2}}^{t}\left(  e^{\Lambda_{-}%
(t-s)}-e^{\Lambda_{-}(t-1)}\right)  \frac{1}{s^{3/2}}\mu_{\alpha}(k,s)~ds\\
& \leq\varepsilon\frac{\Lambda_{+}^{1/2}}{\Lambda_{0}}\mu_{\alpha}%
(k,t)\int_{\frac{t+1}{2}}^{t}e^{\Lambda_{-}(t-s)}\left\vert \Lambda
_{-}\right\vert ^{1/2}\frac{1}{s^{3/2}}~ds\\
& +\varepsilon\frac{\Lambda_{+}^{1/2}}{\Lambda_{0}}\mu_{\alpha}(k,t)e^{\Lambda
_{-}(t-1)}\left\vert \Lambda_{-}\right\vert ^{1/2}\int_{\frac{t+1}{2}}%
^{t}\frac{1}{s^{3/2}}~ds\\
& \leq\frac{\varepsilon}{t^{3/2}}\mu_{\alpha}(k,t)\int_{\frac{t+1}{2}}%
^{t}e^{\Lambda_{-}(t-s)}\left\vert \Lambda_{-}\right\vert ^{1/2}~ds\\
& +\frac{\varepsilon}{t^{1/2}}\mu_{\alpha}(k,t)e^{\Lambda_{-}(t-1)}\left\vert
\Lambda_{-}\right\vert ^{1/2}\left(  \frac{t-1}{t}\right) \\
& \leq\frac{\varepsilon}{t^{3/2}}\mu_{\alpha}(k,t)\int_{\frac{t+1}{2}}%
^{t}\frac{1}{\sqrt{t-s}}~ds+\frac{\varepsilon}{t}\mu_{\alpha}(k,t)\\
& \leq\frac{\varepsilon}{t}\mu_{\alpha}(k,t)~,
\end{align*}
and (\ref{ppp3}) now follows using the triangle inequality. This completes the
proof of Proposition \ref{ppp}.

\subsection{Proof of Proposition \ref{last}}

Let $\mathbf{v}_{D}=\mathbf{v}_{1}-\mathbf{V}_{1}$. Using the definitions and
writing as in Appendix IV $\mathbf{v}_{1}=\sum_{i=1}^{6}\mathbf{v}_{1,i}$,
with $\mathbf{v}_{1,i}$ the $i$-th term in (\ref{iv1}) and $\mathbf{\tau}%
=\sum_{i=1}^{5}\mathbf{\tau}_{i}$, with $\mathbf{\tau}_{i}$ the $i$-th term in
(\ref{itau}), we find that
\begin{align}
\mathbf{v}_{D}(\mathbf{k},t)  & =\frac{i\mathbf{\mathbf{k}}}{\Lambda_{0}}%
\int_{1}^{t}\left(  e^{\Lambda_{-}(t-s)}-e^{\Lambda_{-}(t-1)}\right)
q_{1}(\mathbf{k},s)~ds\nonumber\\
& +\sum_{i=3}^{5}P_{1}i\mathbf{\sigma}_{2}\mathbf{\tau}_{i}(\mathbf{k}%
,t)\nonumber\\
& +\sum_{i=2}^{6}\mathbf{v}_{1,i}(\mathbf{k},t)~.\label{vtt3}%
\end{align}
We write $\mathbf{v}_{D}=\sum_{i=1}^{3}\mathbf{v}_{D,i}$, with $\mathbf{v}%
_{D,i}$ the $i$-th of the three terms in (\ref{vtt3}), and we now bound each
term individually. The inequality (\ref{vvp}) then follows using the triangle inequality.

First, using (\ref{ppp3}) we find for $\mathbf{v}_{D,1}$ that%
\begin{align*}
& \left\vert \frac{i\mathbf{\mathbf{k}}}{\Lambda_{0}}\int_{1}^{t}\left(
e^{\Lambda_{-}(t-s)}-e^{\Lambda_{-}(t-1)}\right)  q_{1}(\mathbf{k}%
,s)~ds\right\vert \\
& \leq\frac{k}{\Lambda_{0}}\int_{1}^{t}\left(  e^{\Lambda_{-}(t-s)}%
-e^{\Lambda_{-}(t-1)}\right)  \frac{1}{s^{3/2}}\mu_{\alpha}(k,s)~ds\\
& \leq\frac{\varepsilon}{t}\mu_{\alpha}(k,t)~,
\end{align*}
as required. Next using (\ref{tb3})-(\ref{tb5}) of Proposition \ref{tbounds}
we find for $\mathbf{v}_{D,2}$that%
\[
\left\vert \sum_{i=3}^{5}P_{1}i\mathbf{\sigma}_{2}\mathbf{\tau}_{i}%
(\mathbf{k},t)\right\vert \leq\sum_{i=3}^{5}\left\vert \mathbf{\tau}%
_{i}(\mathbf{k},t)\right\vert \leq\frac{\varepsilon}{t}\mu_{\alpha}(k,t)~,
\]
as required. This leaves us with proving an improved version of Proposition
\ref{v1bounds}.

\begin{proposition}
For all $\alpha\geq0$ we have the bounds%
\begin{align}
\left\vert \mathbf{v}_{1,2}(\mathbf{k},t)\right\vert  & \leq\varepsilon
\bar{\mu}_{\alpha+1}(k,t)~,\label{111}\\
\left\vert \mathbf{v}_{1,3}(\mathbf{k},t)\right\vert  & \leq\varepsilon
\bar{\mu}_{\alpha+1}(k,t)+\frac{\varepsilon}{t^{2/3}}\mu_{\alpha}%
(k,t)+\frac{\varepsilon}{t^{1/3}}\mu_{\alpha}^{5/6}(k,t)~,\label{222}\\
\left\vert \mathbf{v}_{1,4}(\mathbf{k},t)\right\vert  & \leq\frac{\varepsilon
}{t^{2/3}}\mu_{\alpha}(k,t)+\frac{\varepsilon}{t^{1/2}}\mu_{\alpha}%
^{3/4}(k,t)\label{333}\\
\left\vert \mathbf{v}_{1,5}(\mathbf{k},t)\right\vert  & \leq\bar{\mu}%
_{\alpha+1}(k,t)+\frac{\varepsilon}{t^{2/3}}\mu_{\alpha}(k,t)+\frac
{\varepsilon}{t^{1/3}}\mu_{\alpha}^{5/6}(k,t)~,\label{444}\\
\left\vert \mathbf{v}_{1,6}(\mathbf{k},t)\right\vert  & \leq\frac{\varepsilon
}{t^{2/3}}\mu_{\alpha}(k,t)+\frac{\varepsilon}{t^{1/2}}\mu_{\alpha}%
^{3/4}(k,t)\label{555}%
\end{align}
uniformly in $\mathbf{k}\in\mathbf{R}^{2}$ and $t\geq1$.
\end{proposition}

The bound (\ref{111}) is identical to the bound (\ref{vb2}) in Proposition
\ref{v1bounds}. We now prove the bounds (\ref{222}) and (\ref{444}). We have
that%
\begin{align*}
\left\vert -\frac{1}{2}\frac{i}{k}\mathbf{k}\int_{1}^{t}e^{-k(t-s)}%
q_{1}(\mathbf{k},s)~ds\right\vert  & \leq\varepsilon\int_{1}^{t}%
e^{-k(t-s)}\frac{1}{s^{3/2}}\mu_{\alpha}(k,s)~ds~,\\
\left\vert -\frac{1}{2}\int_{1}^{t}e^{-k(t-s)}P_{1}i\mathbf{\sigma}%
_{2}\mathbf{\mathbf{q}}_{0}(\mathbf{k},s)~ds\right\vert  & \leq\varepsilon
\int_{1}^{t}e^{-k(t-s)}\frac{1}{s^{3/2}}\mu_{\alpha}(k,s)~ds~,
\end{align*}
which proves the bounds, since by Proposition \ref{basic} (see Appendix VII)%
\begin{align*}
\varepsilon e^{-k(t-1)}\int_{1}^{t}\frac{1}{s^{3/2}}\mu_{\alpha}(k,s)~ds  &
\leq\varepsilon e^{-k(t-1)}\mu_{\alpha}(k,1)\left(  \frac{t-1}{t}\right) \\
& \leq\varepsilon\bar{\mu}_{\alpha+1}(k,t)~,
\end{align*}
and therefore, and using (\ref{ppp1}),%
\begin{align*}
& \varepsilon\int_{1}^{t}e^{-k(t-s)}\frac{1}{s^{3/2}}\mu_{\alpha}(k,s)~ds\\
& \leq\varepsilon e^{-k(t-1)}\int_{1}^{t}\frac{1}{s^{3/2}}\mu_{\alpha
}(k,s)~ds+\varepsilon\int_{1}^{t}\left(  e^{-k(t-s)}-e^{-k(t-1)}\right)
\frac{1}{s^{3/2}}\mu_{\alpha}(k,s)~ds\\
& \leq\varepsilon\bar{\mu}_{\alpha+1}(k,t)+\frac{\varepsilon}{t^{2/3}}%
\mu_{\alpha}(k,t)+\frac{\varepsilon}{t^{1/3}}\mu_{\alpha}^{5/6}(k,t)~,
\end{align*}
as required. The bounds (\ref{333}) and (\ref{555}), finally, are an immediate
consequence of (\ref{ppp2}), since%
\begin{align*}
\left\vert -\frac{1}{2}\frac{i}{k}\mathbf{k}\int_{t}^{\infty}e^{k(t-s)}%
q_{1}(\mathbf{k},s)~ds\right\vert  & \leq\varepsilon\int_{t}^{\infty
}e^{k(t-s)}\frac{1}{s^{3/2}}\mu_{\alpha}(k,s)~ds~,\\
\left\vert \frac{1}{2}\int_{t}^{\infty}e^{k(t-s)}P_{1}i\mathbf{\sigma}%
_{2}\mathbf{\mathbf{q}}_{0}(\mathbf{k},s)~ds\right\vert  & \leq\varepsilon
\int_{t}^{\infty}e^{k(t-s)}\frac{1}{s^{3/2}}\mu_{\alpha}(k,s)~ds~.
\end{align*}
This completes the proof of Proposition \ref{last}

\section{Appendix VII}

\subsection{Main technical Lemma}

\begin{proposition}
\label{basic}Let $\alpha^{\prime}\geq\beta^{\prime}\geq\gamma^{\prime}\geq0$
and $\mu>0$. Then, we have the bound%
\begin{equation}
\frac{1}{1+k^{\alpha^{\prime}}}e^{\mu\Lambda_{-}\left(  t-1\right)
}\left\vert \Lambda_{-}\right\vert ^{\beta^{\prime}}\left(  \dfrac{t-1}%
{t}\right)  ^{\gamma^{\prime}}\leq\mathrm{const.}~\frac{1}{t^{\beta^{\prime}}%
}\frac{1}{1+\left(  kt^{1/2}\right)  ^{\alpha^{\prime}-\beta^{\prime}%
+\gamma^{\prime}}}~,\label{decay1}%
\end{equation}
uniformly in $\mathbf{k}\in\mathbf{R}^{2}$ and $t\geq1$. Similarly, for
positive $\alpha^{\prime}$, $\beta^{\prime}$, $\gamma^{\prime}$with
$\alpha^{\prime}-\beta^{\prime}+\gamma^{\prime}\geq0$ and $\mu>0$ we have the
bound%
\begin{equation}
\frac{1}{1+k^{\alpha^{\prime}}}e^{-\mu k\left(  t-1\right)  }k^{\beta^{\prime
}}\left(  \dfrac{t-1}{t}\right)  ^{\gamma^{\prime}}\leq\mathrm{const.}%
~\frac{1}{t^{\beta^{\prime}}}\frac{1}{1+\left(  kt\right)  ^{\alpha^{\prime
}-\beta^{\prime}+\gamma^{\prime}}}~,\label{decay2}%
\end{equation}
uniformly in $\mathbf{k}\in\mathbf{R}^{2}$ and $t\geq1$.
\end{proposition}

\begin{proof}
We first prove (\ref{decay1}). For $1\leq t\leq2$ we have that%
\begin{align*}
\frac{1}{1+k^{\alpha^{\prime}}}e^{\mu\Lambda_{-}(t-1)}\left\vert \Lambda
_{-}\right\vert ^{\beta^{\prime}}\left(  \dfrac{t-1}{t}\right)  ^{\gamma
^{\prime}} &  \leq\mathrm{const.}~\frac{1}{1+k^{\alpha^{\prime}}}e^{\mu
\Lambda_{-}(t-1)}\left\vert \Lambda_{-}\left(  t-1\right)  \right\vert
^{\gamma^{\prime}}~\left\vert \Lambda_{-}\right\vert ^{\beta^{\prime}%
-\gamma^{\prime}}\\
&  \leq\mathrm{const.}~\frac{1}{1+k^{\alpha^{\prime}}}~\left\vert \Lambda
_{-}\right\vert ^{\beta^{\prime}-\gamma^{\prime}}\\
&  \leq\mathrm{const.}~\frac{1}{1+k^{\alpha^{\prime}-\beta^{\prime}%
+\gamma^{\prime}}}\\
&  \leq\mathrm{const.}~\frac{1}{t^{\beta^{\prime}}}\frac{1}{1+\left(
kt^{1/2}\right)  ^{\alpha^{\prime}-\beta^{\prime}+\gamma^{\prime}}}~,
\end{align*}
as claimed, and for $t>2$ we use that
\begin{align*}
&  \left(  1+\left(  kt^{1/2}\right)  ^{\alpha^{\prime}-\beta^{\prime}%
+\gamma^{\prime}}\right)  e^{\mu\Lambda_{-}(t-1)}\left\vert \Lambda
_{-}t\right\vert ^{\beta^{\prime}}\left(  \dfrac{t-1}{t}\right)
^{\gamma^{\prime}}\\
&  \leq\mathrm{const.}\left(  1+\left(  kt^{1/2}\right)  ^{\alpha^{\prime}%
}\right)  e^{\frac{1}{2}\mu\Lambda_{-}t}\left\vert \Lambda_{-}t\right\vert
^{\beta^{\prime}}\\
&  \leq\mathrm{const.}\left(  1+\frac{k^{\alpha^{\prime}}}{\left\vert
\Lambda_{-}\right\vert ^{\alpha^{\prime}/2}}\left\vert \Lambda_{-}t\right\vert
^{\alpha^{\prime}/2}\left\vert \Lambda_{-}t\right\vert ^{\beta^{\prime}%
}e^{\frac{1}{2}\mu\Lambda_{-}t}\right) \\
&  \leq\mathrm{const.}\left(  1+\frac{k^{\alpha^{\prime}}}{\left\vert
\Lambda_{-}\right\vert ^{\alpha^{\prime}/2}}\right) \\
&  \leq\mathrm{const.}\left(  1+k^{\alpha^{\prime}/2}\right)  \leq
\mathrm{const.}\left(  1+k^{\alpha^{\prime}}\right)  ~,
\end{align*}
and (\ref{decay1}) follows. We now prove (\ref{decay2}). For $1\leq t\leq2$
and $k\leq1$ we have that%
\begin{align*}
\frac{1}{1+k^{\alpha^{\prime}}}e^{-\mu k\left(  t-1\right)  }k^{\beta^{\prime
}}\left(  \dfrac{t-1}{t}\right)  ^{\gamma^{\prime}} &  \leq\mathrm{const.}\\
&  \leq\mathrm{const.}~\frac{1}{t^{\beta^{\prime}}}\frac{1}{1+\left(
kt\right)  ^{\alpha^{\prime}-\beta^{\prime}+\gamma^{\prime}}}~,
\end{align*}
and for $1\leq t\leq2$ and $k>1$ we have that%
\begin{align*}
\frac{1}{1+k^{\alpha^{\prime}}}e^{-\mu k\left(  t-1\right)  }k^{\beta^{\prime
}}\left(  \dfrac{t-1}{t}\right)  ^{\gamma^{\prime}} &  \leq\mathrm{const.}%
~\frac{1}{1+k^{\alpha^{\prime}}}e^{-\mu k\left(  t-1\right)  }\left(  k\left(
t-1\right)  \right)  ^{\gamma^{\prime}}~k^{\beta^{\prime}-\gamma^{\prime}}\\
&  \leq\mathrm{const.}~\frac{1}{1+k^{\alpha^{\prime}}}~k^{\beta^{\prime
}-\gamma^{\prime}}\\
&  \leq\mathrm{const.}~\frac{1}{1+k^{\alpha^{\prime}-\beta^{\prime}%
+\gamma^{\prime}}}\\
&  \leq\mathrm{const.}~\frac{1}{t^{\beta^{\prime}}}\frac{1}{1+\left(
kt\right)  ^{\alpha^{\prime}-\beta^{\prime}+\gamma^{\prime}}}~.
\end{align*}
Finally, for $t>2$ we use that
\begin{align*}
&  \left(  1+\left(  kt\right)  ^{\alpha^{\prime}-\beta^{\prime}%
+\gamma^{\prime}}\right)  e^{-\mu k\left(  t-1\right)  }\left(  kt\right)
^{\beta^{\prime}}\left(  \dfrac{t-1}{t}\right)  ^{\gamma^{\prime}}\\
&  \leq\mathrm{const.}\left(  1+\left(  kt\right)  ^{\alpha^{\prime}%
-\beta^{\prime}+\gamma^{\prime}}\right)  e^{-\frac{1}{2}\mu kt}\left(
kt\right)  ^{\beta^{\prime}}\\
&  \leq\mathrm{const.}\leq\mathrm{const.}\left(  1+k^{\alpha^{\prime}}\right)
~,
\end{align*}
and (\ref{decay2}) follows.
\end{proof}

\subsection{Bound on convolution}

\begin{proposition}
\label{c1}Let $\alpha>2$, and let $a_{1}$ be a piecewise continuous, and
$a_{2}$ be a continuous function from $\mathbf{R}^{2}\times\lbrack1,\infty) $
to $\mathbf{C}$ satisfying the bounds,%
\[
\left\vert a_{i}(\mathbf{k},t)\right\vert \leq\mu_{\alpha}(k,t)~,
\]
$i=1,2$. Then, the convolution $a_{1}\ast a_{2}$ is a continuous function from
$\mathbf{R}^{2}\times\lbrack1,\infty)$ to $\mathbf{C}$ and we have the bound%
\begin{equation}
\left\vert \left(  a_{1}\ast a_{2}\right)  (\mathbf{k},t)\right\vert
\leq\mathrm{const.}\frac{1}{t}\mu_{\alpha}(k,t)~,\label{c1b11}%
\end{equation}
uniformly in $t\geq1$, $\mathbf{k}\in\mathbf{R}^{2}$.
\end{proposition}

\begin{proof}
Continuity is elementary. We now prove (\ref{c1b11}). Let%
\[
D(\mathbf{k})=\left\{  \mathbf{\kappa}\in\mathbf{R}^{2}\left\vert {}\right.
\left\vert \mathbf{k}-\mathbf{\kappa}\right\vert \leq k/2\right\}  ~.
\]
For $\mathbf{k}^{\prime}\in D(\mathbf{k})$ we have that%
\[
k^{\prime}\geq k-\left\vert \mathbf{k}-\mathbf{k}^{\prime}\right\vert
\geq\frac{1}{2}k~.
\]
Therefore we have for $a_{1}\ast a_{2}$,%
\begin{align*}
\left\vert \left(  a_{1}\ast a_{2}\right)  (\mathbf{k},t)\right\vert  &
=\int_{\mathbf{R}^{2}\setminus D(\mathbf{k})}\mu(k^{\prime},t)\mu(\left\vert
\mathbf{k}-\mathbf{k}^{\prime}\right\vert ,t)~d^{2}\mathbf{k}^{\prime}\\
& +\int_{D(\mathbf{k})}\mu(k^{\prime},t)\mu(\left\vert \mathbf{k}%
-\mathbf{k}^{\prime}\right\vert ,t)~d^{2}\mathbf{k}^{\prime}\\
& \leq\left(  \sup_{\mathbf{k}^{\prime}\mathbf{\in R}^{2}\setminus
D(\mathbf{k})}\mu(\left\vert \mathbf{k}-\mathbf{k}^{\prime}\right\vert
,t)\right)  \int_{\mathbf{R}^{2}\setminus D(\mathbf{k})}\mu(k^{\prime
},t)~d^{2}\mathbf{k}^{\prime}\\
& +\left(  \sup_{\mathbf{k}^{\prime}\mathbf{\in}D(\mathbf{k})}\mu(k^{\prime
},t)\right)  \int_{D(\mathbf{k})}\mu(\left\vert \mathbf{k}-\mathbf{k}^{\prime
}\right\vert ,t)~d^{2}\mathbf{k}^{\prime}\\
& \leq\mathrm{const.}\mu(k/2,t)\int_{\mathbf{R}^{2}}\mu(k^{\prime}%
,t)~d^{2}\mathbf{k}^{\prime}\\
& +\mathrm{const.}\mu(k/2,t)\int_{\mathbf{R}^{2}}\mu(\left\vert \mathbf{k}%
-\mathbf{k}^{\prime}\right\vert ,t)~d^{2}\mathbf{k}^{\prime}\\
& \leq\mathrm{const.}\frac{1}{t}\mu_{\alpha}(k/2,t)\leq\mathrm{const.}\frac
{1}{t}\mu_{\alpha}(k,t)~,
\end{align*}
and (\ref{c1b11}) follows. This completes the proof of Proposition \ref{c1}.
\end{proof}

\section{Appendix VIII}

For the convenience of the reader we recollect in this appendix some
expressions for Fourier transforms. Let $\mathbf{x}=(x,\mathbf{y)}$, with
$\mathbf{y}=(y_{1},y_{2})$, with $y=\sqrt{y_{1}^{2}+y_{2}^{2}}$, let
$\mathbf{k}=(k_{1},k_{2})$, with $k=\sqrt{k_{1}^{2}+k_{2}^{2}}$ and define $G$
by the equation%
\begin{align*}
G(x,\mathbf{y)}  & \mathbf{=}\mathbf{-}\frac{1}{4\pi}\frac{1}{\left\vert
\mathbf{x}\right\vert }\\
& \equiv\mathbf{-}\frac{1}{4\pi}\frac{1}{\sqrt{x^{2}+y^{2}}}~.
\end{align*}
The function $G$ is the Greens function of the Laplacean, \textit{i.e.}, we
have%
\[
\Delta G(\mathbf{x)}=\delta(\mathbf{x)}~,
\]
and therefore%
\begin{equation}
G(x,\mathbf{y})=\left(  \frac{1}{2\pi}\right)  ^{2}\int_{\mathbf{R}^{2}%
}e^{-i\mathbf{k\cdot y}}~\widehat{G}(\mathbf{k},x)~d^{2}\mathbf{k}%
~,\label{green}%
\end{equation}
where%
\begin{align}
\widehat{G}(\mathbf{k},x)  & =-\frac{1}{2\pi}\int_{\mathbf{R}}e^{-i\mathbf{k}%
_{0}x}\frac{1}{k_{0}^{2}+k^{2}}~dk_{0}\nonumber\\
& =-\frac{1}{k}\frac{1}{2\pi}\int_{\mathbf{R}}e^{-i\mathbf{k}_{0}\left(
kx\right)  }\frac{1}{k_{0}^{2}+1}~dk_{0}\nonumber\\
& =-\frac{1}{2}\frac{1}{k}e^{-k\left\vert x\right\vert }~.\label{gk}%
\end{align}
The vector field $\mathbf{u}_{S}$ of a point source is%
\begin{equation}
\mathbf{u}_{S}(x,\mathbf{\mathbf{y}})=\mathbf{\nabla}G(x,\mathbf{y})=\left\{
\begin{array}
[c]{l}%
\dfrac{1}{4\pi}\dfrac{x}{\left(  x^{2}+y^{2}\right)  ^{\frac{3}{2}}}\\
\\
\dfrac{1}{4\pi}\dfrac{\mathbf{y}}{\left(  x^{2}+y^{2}\right)  ^{\frac{3}{2}}%
}~,
\end{array}
\right. \label{us}%
\end{equation}
or in Fourier space,
\begin{equation}
\widehat{\mathbf{u}_{S}}(\mathbf{\mathbf{k}},x)=\binom{\partial_{x}%
}{-i\mathbf{k}}\widehat{G}(\mathbf{k},x)=\left\{
\begin{array}
[c]{l}%
\dfrac{1}{2}\mathrm{sign}(x)e^{-k\left\vert x\right\vert }\\
\\
\dfrac{1}{2}\dfrac{i\mathbf{k}}{k}e^{-k\left\vert x\right\vert }~.
\end{array}
\right. \label{usk}%
\end{equation}
The vector fields (\ref{us}) and (\ref{usk}), multiplied by $2d$, are one term
in the asymptotic expressions (\ref{tuas})-(\ref{tvas}) and (\ref{uuas}%
)-(\ref{vv2as}), respectively. Next, let $\mathbf{e}$ be unit vector in
$\mathbf{R}^{2}$. Define $G_{1}$ by the equation%
\begin{align*}
G_{1}(\mathbf{e,}x,\mathbf{y})  & =\int_{x}^{\mathrm{sign}(x)\infty
}\mathbf{\nabla}^{\bot}G(\xi,\mathbf{y})\cdot\mathbf{e}~d\xi\\
& =\frac{1}{4\pi}\int_{x}^{\mathrm{sign}(x)\infty}\frac{\mathbf{y}%
^{T}\mathbf{e}}{\left(  \xi^{2}+y^{2}\right)  ^{\frac{3}{2}}}~d\xi\\
& =\frac{1}{4\pi}\frac{\mathbf{y}^{T}\mathbf{e}}{\sqrt{x^{2}+y^{2}}}%
~\frac{\mathrm{sign}(x)}{\sqrt{x^{2}+y^{2}}+\left\vert x\right\vert }~,
\end{align*}
and define, for $x\neq0$, the vector field $\mathbf{u}_{C}$ by the equation%
\begin{align}
& \mathbf{u}_{C}(\mathbf{e},x,\mathbf{\mathbf{y}})\nonumber\\
& =\mathbf{\nabla}G_{1}(\mathbf{e,}x,\mathbf{y})\nonumber\\
& =\left\{
\begin{array}
[c]{l}%
-\dfrac{1}{4\pi}\dfrac{\mathbf{y}^{T}\mathbf{e}}{\left(  x^{2}+y^{2}\right)
^{\frac{3}{2}}}\\
\\
\dfrac{1}{4\pi}\dfrac{1}{\sqrt{x^{2}+y^{2}}}\dfrac{\mathrm{sign}(x)}%
{\sqrt{x^{2}+y^{2}}+\left\vert x\right\vert }\left[  \mathbf{1}-\dfrac
{1}{\sqrt{x^{2}+y^{2}}}\left(  \dfrac{1}{\sqrt{x^{2}+y^{2}}}+\dfrac{1}%
{\sqrt{x^{2}+y^{2}}+\left\vert x\right\vert }\right)  \mathbf{yy}^{T}\right]
\mathbf{e}~.
\end{array}
\right. \label{uc}%
\end{align}
We have the following limits,%
\begin{equation}
\lim_{x\rightarrow0_{\pm}}\mathbf{u}_{C}(\mathbf{e},x,\mathbf{\mathbf{y}%
})=\left\{
\begin{array}
[c]{l}%
-\dfrac{1}{4\pi}\dfrac{\mathbf{y}^{T}\mathbf{e}}{y^{3}}\\
\\
\pm\dfrac{1}{4\pi}\dfrac{1}{y^{2}}\left(  \mathbf{1}-2\dfrac{\mathbf{yy}^{T}%
}{y^{2}}\right)  \mathbf{e}~.
\end{array}
\right. \label{limit1}%
\end{equation}
Using that%
\begin{align*}
\widehat{G_{1}}(\mathbf{e,k},x)  & =-i\mathbf{k}^{T}\mathbf{e}\int
_{x}^{\mathrm{sign}(x)\infty}\widehat{G}(\mathbf{k},\xi)~d\xi\\
& =\frac{1}{2}i\mathbf{k}^{T}\mathbf{e}~\frac{1}{k}\int_{x}^{\mathrm{sign}%
(x)\infty}e^{-k\left\vert \xi\right\vert }d\xi\\
& =\frac{1}{2}i\mathbf{k}^{T}\mathbf{e}~\frac{\mathrm{sign}(x)}{k^{2}%
}e^{-k\left\vert x\right\vert }~,
\end{align*}
we get in Fourier space, for $x\in\mathbf{R}\setminus\{0\}$, that%
\begin{equation}
\widehat{\mathbf{u}_{C}}(\mathbf{e},\mathbf{\mathbf{k}},x)=\binom{\partial
_{x}}{-i\mathbf{k}}\widehat{G}_{1}(\mathbf{k},x)=\left\{
\begin{array}
[c]{l}%
-\dfrac{i}{2}\mathbf{k}^{T}\dfrac{1}{k}e^{-k\left\vert x\right\vert
}~\mathbf{e}\\
\\
\dfrac{1}{2}P_{1}~\mathrm{sign}(x)e^{-k\left\vert x\right\vert }~\mathbf{e}~.
\end{array}
\right. \label{uck}%
\end{equation}
The vector fields (\ref{uc}) and (\ref{uck}), multiplied by $-2\mathbf{b}$,
are one term in the asymptotic expressions (\ref{tuas})-(\ref{tvas}) and
(\ref{uuas})-(\ref{vv2as}), respectively. Next, define for $x>0$ the function
$H$ by the equation%
\[
H(\mathbf{k},x)=\theta(x)e^{-k^{2}x}~,
\]
with $\theta$ the Heaviside function. For $x>0$ this is nothing else than the
heat Kernel in Fourier space and therefore, for $x>0$,
\begin{align*}
H(x,\mathbf{y})  & =\left(  \frac{1}{2\pi}\right)  ^{2}\int_{\mathbf{R}^{2}%
}e^{-i\mathbf{k\cdot y}}~H(\mathbf{k},x)~d^{2}\mathbf{k}\\
& =\frac{1}{4\pi x}e^{-\frac{y^{2}}{4x}}~.
\end{align*}
The vector field%
\begin{equation}
\widehat{\mathbf{u}_{W}}(\mathbf{\mathbf{k}},x)=\left\{
\begin{array}
[c]{l}%
H(\mathbf{k},x)\\
\\
i\mathbf{k}H(\mathbf{k},x)
\end{array}
\right. \label{uWk}%
\end{equation}
is divergence free, and for its inverse Fourier transform we have, for $x>0$,%
\begin{equation}
\widehat{\mathbf{u}_{W}}(x,\mathbf{y})=\left\{
\begin{array}
[c]{l}%
\dfrac{1}{4\pi x}e^{-\frac{y^{2}}{4x}}\\
\\
\dfrac{\mathbf{y}}{8\pi x^{2}}e^{-\frac{y^{2}}{4x}}~.
\end{array}
\right. \label{uW}%
\end{equation}
The vector fields (\ref{uW}) and (\ref{uWk}), multiplied by $c$, are one term
in the asymptotic expressions (\ref{tuas})-(\ref{tvas}) and (\ref{uuas}%
)-(\ref{vv2as}), respectively. Finally, for $x>0$, let%
\begin{equation}
\widehat{\mathbf{u}_{V}}(\mathbf{e},\mathbf{\mathbf{k}},x)=\left\{
\begin{array}
[c]{l}%
0\\
\\
-P_{1}H(\mathbf{k},x)\mathbf{e}~.
\end{array}
\right. \label{uak}%
\end{equation}
To compute the Fourier transform we define%
\begin{align*}
H_{1}(\mathbf{e},\mathbf{k},x)  & =\frac{-i\mathbf{k}^{T}\mathbf{e}}{k^{2}%
}H(\mathbf{k},x)\\
& =\frac{-i\mathbf{k}^{T}\mathbf{e}}{k^{2}}e^{-k^{2}x}=\int_{x}^{\infty
}\left(  -i\mathbf{k}^{T}\mathbf{e}\right)  ~e^{-k^{2}t}dt~,
\end{align*}
which becomes in direct space%
\[
H_{1}(\mathbf{e},x,\mathbf{y})=-\frac{\mathbf{y}^{T}\mathbf{e}}{8\pi}\int
_{x}^{\infty}\frac{1}{\xi^{2}}e^{-\frac{y^{2}}{4\xi}}~d\xi=\frac
{\mathbf{y}^{T}\mathbf{e}}{2\pi}\frac{1}{y^{2}}\left(  e^{-\frac{y^{2}}{4x}%
}-1\right)  ~.
\]
Therefore,%
\begin{equation}
\mathbf{u}_{V}(\mathbf{e},x,\mathbf{y})=\left\{
\begin{array}
[c]{l}%
0\\
\\
\dfrac{1}{2\pi}\left[  \dfrac{1}{y^{2}}\left(  e^{-\frac{y^{2}}{4x}}-1\right)
\mathbf{1}-2\left(  \dfrac{1}{y^{2}}\left(  e^{-\frac{y^{2}}{4x}}-1\right)
+\dfrac{1}{4x}e^{-\frac{y^{2}}{4x}}\right)  \dfrac{\mathbf{yy}^{T}}{y^{2}%
}\right]  \mathbf{e}~,
\end{array}
\right. \label{ua}%
\end{equation}
and we have the limit,%
\begin{equation}
\lim_{x\rightarrow0_{+}}\mathbf{u}_{V}(\mathbf{e},x,\mathbf{\mathbf{y}%
})=\left\{
\begin{array}
[c]{l}%
0\\
\\
-\dfrac{1}{2\pi}\dfrac{1}{y^{2}}\left(  \mathbf{1}-2\dfrac{\mathbf{yy}^{T}%
}{y^{2}}\right)  \mathbf{e}~.
\end{array}
\right. \label{limit2}%
\end{equation}
The vector fields (\ref{ua}) and (\ref{uak}), multiplied by $\mathbf{a}$, are
one term in the asymptotic expressions (\ref{tuas})-(\ref{tvas}) and
(\ref{uuas})-(\ref{vv2as}), respectively.

\subsection{Matching at $x=0$}

Since $\mathbf{u}_{V}=0$ for $x<0$, we see that the asymptotic expressions
(\ref{tuas})-(\ref{tvas}) are continuous at $x=0$, provided%
\[
\lim_{x\rightarrow0_{-}}\mathbf{u}_{C}(-2\mathbf{b},x,\mathbf{\mathbf{y}%
})=\lim_{x\rightarrow0_{+}}\mathbf{u}_{V}(\mathbf{a},x,\mathbf{\mathbf{y}%
})\mathbf{+}\lim_{x\rightarrow0_{+}}\mathbf{u}_{C}(-2\mathbf{b}%
,x,\mathbf{\mathbf{y}})~,
\]
and from (\ref{limit1}) and (\ref{limit2}) we see that this is only the case
if $\psi=\mathbf{a}+2\mathbf{b}=0$.

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