\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 59, pp. 1--35.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/59\hfil Well-posedness]
{Well-posedness of one-dimensional Korteweg models}
\author[S. Benzoni-Gavage, R. Danchin, S. Descombes\hfil EJDE-2006/59\hfilneg]
{Sylvie Benzoni-Gavage,  Rapha\"el Danchin, St\'ephane Descombes}  

\address{Sylvie Benzoni-Gavage \newline
Institut Camille Jordan, Universit\'e Claude Bernard Lyon I\\
21, avenue Claude Bernard,
F-69622 Villeurbanne cedex, France}
\email{benzoni@math.univ-lyon1.fr}

\address{Rapha\"el Danchin \newline
Centre de Math\'ematiques,  Universit\'e Paris XII Val de Marne\\
61, avenue du G\'en\'eral de Gaulle,
F-94010 Cr\'eteil cedex, France}
\email{danchin@univ-paris12.fr}

\address{St\'ephane Descombes \newline
UMPA, ENS Lyon \\
46, all\'ee d'Italie,
F-69364  Lyon cedex 07, France}
\email{sdescomb@umpa.ens-lyon.fr}


\date{}
\thanks{Submitted June 14, 2004. Published May 2, 2006.}
\subjclass[2000]{76N10, 76T10}
\keywords{Capillarity; Korteweg stress;  local well-posedness;
\hfill\break\indent Schr\"odinger equation}

\begin{abstract}
  We investigate the initial-value problem
  for one-dimensional compressible  fluids endowed with internal
  capillarity. We focus on the   isothermal inviscid case with
  variable capillarity. The resulting equations
  for the density and the velocity, consisting of
  the mass conservation law and the momentum conservation
  with Korteweg stress,
  are a  system of third order nonlinear dispersive
  partial differential equations. Additionally, this
  system is Hamiltonian  and admits travelling solutions,
  representing propagating phase boundaries with internal structure.
  By change of unknown, it roughly
  reduces to a  quasilinear Schr\"odinger equation.
  This new formulation enables us to prove
  local well-posedness for smooth perturbations
  of travelling profiles
  and almost-global existence for small enough perturbations.
  A blow-up criterion is also derived.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}[theorem]{Definition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}


\def\norm#1#2{\|{#2\|}_{#1}}
\def\Lp#1#2{\|{#2}\|_{L^{#1}}}


\section*{Introduction}

We are concerned with compressible
fluids endowed with internal capillarity. The models we consider are
originated from the XIXth century work by van der Waals \cite{vdW}
and Korteweg \cite{Korteweg} and were actually derived in their
modern form in the 1980s using the second gradient theory,
see for instance \cite{TN,JLCD}.
They result in dispersive systems of Partial Differential
Equations. In fact, special cases of these models also arise
in other contexts, {\em e.g.} quantum mechanics.
Our main motivation is about fluids though, especially liquid-vapor
mixtures with phase changes. Indeed, Korteweg models allow phase
``boundaries'' of nonzero thickness that are often called
{\em diffuse interfaces} -- by contrast with sharp interfaces in
the Laplace-Young's theory. The
interest for diffuse interfaces has been renewed in the late 1990s
also for numerical purposes, see \cite{AFW} for a nice
review.

The mathematical analysis of Korteweg models is rather recent. One
may quote only a few papers \cite{BDL,DD,HL2}, in which
nonzero viscosity and its regularizing effects play
a fundamental role. One  should also quote
the recent work of Li and Marcati \cite{LM}, which
concerns a similar model in QHD (Quantum HydroDynamics),
with weaker dissipation -- due to relaxation --included.
Here, we concentrate on
purely dispersive models, which are still physically meaningful.
Although very different from dissipative smoothing, dispersive
smoothing is known to exist for various equations,
see for instance the seminal work by Kato \cite{Kato2} on the
Korteweg-de Vries equation, \cite{CKS,KS} on generalized Schr\"odinger
equations, etc. See also the work by Bedjaoui and Sainsaulieu
on a dispersive two-phase flow model \cite{BedjaouiSainsaulieu}.
However, up to our knowledge, no result of this kind
is known on the most general models we are interested in.
This is a direction of research in progress, in connection with
the recent work of Kenig, Ponce and Vega \cite{KPV}.
The positive counterpart of neglecting dissipation phenomena is that
like most other purely dispersive models
({\em e.g.}
Korteweg-de Vries, Boussinesq, standard Schr\"odinger equations), the models
we consider can be viewed, to some extent, as Hamiltonian systems.
The Hamiltonian structure of non-dissipative Korteweg models has been
discussed in a companion paper \cite{BDDJ}.

We address here the local-in-time well-posedness of the non-dissipative
Korteweg models.
On the one hand, for monotone pressure laws,
it has been pointed out by Gavrilyuk and Gouin \cite{GG}
that these models admit a symmetric form, in the classical sense
of hyperbolic systems of conservation laws,
at least for their first order part and with
 non-dissipative -- in a sense that
 we make precise below -- higher order terms.  Even though this kind
of systems enjoy natural $L^2$ estimates, it is not clear
how to show their well-posedness (this is done in
\cite{BedjaouiSainsaulieu} by an artificial viscosity method).
On the other hand, when concentrating on
models with {\em constant capillarity},
Korteweg models -- at least some of them --
can be dealt with by Kato's theory of
abstract evolution equations \cite{Kato}, disregarding
the monotonicity of the pressure. For more general
capillarities though, because of the nonlinearity in higher order terms,
they are {\em not} amenable to Kato's theory.

Restricting to one space dimension, where
we can use  Lagrangian coordinates,
we have been able to deal with
both a nonmonotone pressure and
a nonconstant capillarity. For this, we have introduced
an additional unknown that
gives rise to a system coupling a transport
equation with a variable coefficients
Schr\"odinger equation. Taking advantage of
symmetry properties of this system, inspired from
a work by Coquel on the numerics of Korteweg models,
and introducing suitable
gauge functions, inspired from a work by Lim and Ponce \cite{LP},
we obtain higher order energy estimates {\em without loss of
derivatives}, and eventually
prove the local-in-time well-posedness
in  Sobolev spaces of the one dimensional Korteweg models.
As a matter of fact, our main
theorem is slightly more general
since it also states
existence for data pertaining to $H^k$ perturbations
of any smooth reference solution
whose derivatives have sufficient decay at infinity.
Our main motivation
for proving this is to investigate
the stability of travelling
wave solutions for the one-dimensional model.
Indeed, typical travelling solutions
fail to belong to Sobolev spaces
since they have different endstates
at $-\infty$ and $+\infty$ (see our companion paper \cite{BDDJ}
for more details).


We finally derive a blow-up criterion involving
the Lipschitz norm of the solution
and get a lower bound
for the existence time
in terms of the norm of the data
which entails almost global existence for small
perturbations of a
global solution (e.g a capillary profile).



\section{Derivation of Korteweg models}\label{s:intro}

\subsection{A general model in Eulerian coordinates}
Korteweg-type models are based on an extended version of
nonequilibrium thermodynamics,
which assumes that the energy of the fluid not only depends on
standard variables but on the gradient of the
density. In terms of the
free energy for instance, this principle takes the form of a
generalized Gibbs relation
$$
{\rm d}F=-S\,{\rm d} T+g\,{\rm d}\rho+ \mathbf{\phi}^{*}\cdot{\rm d}{\bf w}\,,
$$
where $F$ denotes the free energy per unit volume,
$S$ the entropy per unit volume\footnote{By convention, extensive
quantities per unit volume are denoted by upper case letters
and their specific counterparts will be denoted by the same, lower
case, letters.}, $T$ the temperature,
$g$ the chemical potential and, in the additional term,
${\bf w}$ stands for $\nabla \rho$. The potential $\mathbf{\phi}$ is most
often assumed of the form
$$
\mathbf{\phi}=K {\bf w},
$$
where $K$ is called the capillarity coefficient,
which may depend on both $\rho$ and $T$.
In this case, $F$ decomposes into a standard part $F_{0}$
and an additional term due to gradients of density,
$$
F(\rho,T,\nabla\rho)=F_{0}(\rho,T)+
\frac{1}{2} K(\rho,T) \|\nabla\rho\|^{2}\,,
$$
and similar decompositions hold for $S$ and $g$.
We shall use this special form in our subsequent analysis.
For the moment we keep the abstract potential $\mathbf{\phi}$ and
we define  the Korteweg tensor as
$$
{\bf K}:= (\rho  \mathop{\rm div} \mathbf{\phi}) {\bf I} -\mathbf{\phi}  {\bf w}^{*}\,.
$$
Neglecting dissipation phenomena,
the conservation of mass, momentum and energy read
\begin{gather*}
\partial_t \rho+\mathop{\rm div}(\rho \mathbf{u})= 0\,,\\
\partial_t(\rho \mathbf{u}^*)+
\mathop{\rm div}(\rho \mathbf{u} \mathbf{u}^{*}+p{\bf I} )= \mathop{\rm div}  {\bf K}\,,\\
\partial_t(E+\tfrac{1}{2} \rho |\mathbf{u}|^2)+
\mathop{\rm div}(( E+\tfrac{1}{2} \rho |\mathbf{u}|^2+p ) \mathbf{u} )
=\mathop{\rm div} ( {\bf K}  \mathbf{u}+{\bf W} )\,,
\end{gather*}
where $p=\rho g-F$ is the (extended) pressure,
$E=F+T S$ is the internal energy per unit volume,
and
$$
{\bf W}:=( \partial_{t}\rho+\mathbf{u}^*\cdot \nabla\rho )
\mathbf{\phi}=-( \rho \mathop{\rm div} \mathbf{u} )  \mathbf{\phi}
$$
is the interstitial working that was first
introduced by Dunn and Serrin \cite{DS}.
This supplementary term ensures that the entropy $S$
satisfies the conservation law
$$
\partial_{t}S+\mathop{\rm div}(S \mathbf{u})=0\,.
$$
(This is obtained through formal computation, for presumably smooth
solutions.)
There is also an alternate form of the momentum equation (still for smooth
solutions).
Using the mass conservation law and the relation
$$
{\rm d} g=-s \,{\rm d} T+ v \,{\rm d} p+
 v \mathbf{\phi}^*\cdot {\rm d} {\bf w}\,,
$$
with $s$ the specific entropy and $ v:=1/\rho$ the specific volume,
we arrive at the equation
$$
\partial_t\mathbf{u}+
( \mathbf{u}^*\cdot \nabla ) \mathbf{u}= \nabla( \mathop{\rm div} \mathbf{\phi}-g )-s \nabla T\,.
$$
The resulting evolution system for $(\rho,\mathbf{u},S)$ is
\begin{equation}
    \label{eq:Korteweg-general}
\begin{gathered}
\partial_t \rho+\mathop{\rm div}(\rho \mathbf{u})= 0\,,\\
\partial_t\mathbf{u}+( \mathbf{u}^*\cdot \nabla ) \mathbf{u}
 =\nabla( \mathop{\rm div} \mathbf{\phi}-g )-s \nabla T \,,\\
\partial_{t}S+\mathop{\rm div}(S \mathbf{u}) =0\,.
\end{gathered}
\end{equation}
The ``Hamiltonian structure'' of \eqref{eq:Korteweg-general}
is discussed in \cite{BDDJ}. In particular, in the isothermal case this
system reduces to
\begin{equation}
    \label{eq:Korteweg-isotherm}
\begin{gathered}
\partial_t \rho+\mathop{\rm div}(\rho \mathbf{u})= 0\,,\\
\partial_t\mathbf{u}+( \mathbf{u}^*\cdot \nabla ) \mathbf{u}
 =\nabla( \mathop{\rm div} \mathbf{\phi}-g )\,.\\
\end{gathered}
\end{equation}

We shall see that for proving well-posedness,
${\bf w}$, or a similar quantity, must be considered at first
as an independent unknown.
To do so, one also needs an evolution equation for ${\bf w}$,
which is easily obtained by
differentiating the mass conservation law. We get
$$
\partial_{t}{\bf w}^*+
\mathop{\rm div}(\mathbf{u}  {\bf w}^{*})
+\mathop{\rm div}(\rho  \,{\rm D} \mathbf{u})= 0\,,
$$
where $\,{\rm D}\mathbf{u}=(\nabla \mathbf{u})^{*}$ is by definition the matrix
of coefficient $\partial_{j}u_{i}$ on the $i$-th row and
$j$-th column. Therefore, one may also look at the equations of
motion as
\begin{equation} \label{eq:lccappur}
\begin{gathered}
\partial_t \rho+\mathop{\rm div}(\rho \mathbf{u})= 0\,,\\
\partial_t(\rho \mathbf{u}^*)+
\mathop{\rm div}(\rho \mathbf{u}  \mathbf{u}^*+p {\bf I}-{\bf K} ) =0\,,\\
\partial_{t}S+\mathop{\rm div}(S \mathbf{u})= 0\,,\\
\partial_{t}{\bf w}^*+
\mathop{\rm div}(\mathbf{u}   {\bf w}^{*})+\mathop{\rm div}(\rho  D \mathbf{u})= 0\,.
\end{gathered}
\end{equation}
It has been pointed out by Gavrilyuk and Gouin \cite{GG} that,
if the total energy $H:=E+\tfrac{1}{2} \rho \|\mathbf{u}\|^{2}$
is a convex function of the conservative
variable $(\rho,\mathbf{m},S,\mathbf{w})$ -- with
$\mathbf{m}:=\rho \mathbf{u}$ --, then the system \eqref{eq:lccappur} admits
a symmetric form similar to Friedrichs symmetric hyperbolic
systems of conservation laws. This symmetrization procedure naturally
involves the Legendre transform $\Pi$ of $H$, which is
a function of the dual variables $(q:=
g-\frac{1}{2} \|\mathbf{u}\|^2,\mathbf{u},T,\mathbf{\phi})$.
Indeed, one easily finds that
$$
{\rm d} H={\rm d} E-\frac{1}{2} \|{\bf u}\|^2\,{\rm d} \rho+
\mathbf{u}^*\cdot {\rm d} \mathbf{m}=
(g-\frac{1}{2} \|\mathbf{u}\|^2)\,{\rm d}\rho+\mathbf{u}^*\cdot {\rm d}
\mathbf{m}+T\,{\rm d} S+\mathbf{\phi}^*
\cdot d{\bf w}\,.
$$
The Legendre transform $\Pi$ is by definition such that
$$
{\rm d}{\Pi}=\rho\,{\rm d} q+{\bf m}^*\cdot{\rm d} \mathbf{u}
+S\,{\rm d} T+{\bf w}^* \cdot{\rm d} \mathbf{\phi}\,,
$$
and
$$
{\Pi}=\rho q+{\bf m}^*\cdot {\bf u}+S T+{\bf w}^*
\cdot \mathbf{\phi}-H =p+{\bf w}^*\cdot \mathbf{\phi}\,.
$$
Then it is not difficult to see that \eqref{eq:lccappur} also reads
\begin{equation} \label{eq:lccappursym}
\begin{gathered}
\partial_t \Big(\frac{\partial {\Pi}}{\partial q}\Big)+
\mathop{\rm div}\Big(\frac{\partial ({\Pi} {\bf u})}{\partial q}\Big)
= 0\,,\\
\partial_{t}\Big(\frac{\partial {\Pi}}{\partial {\bf u}}\Big)+
\mathop{\rm div}\Big(\frac{\partial ({\Pi} {\bf u})}
{\partial {\bf u}}\Big)-
\mathop{\rm div}\Big(\frac{\partial {\Pi}}{\partial q}\;
D \mathbf{\phi}\Big)
= 0\,,\\
\partial_{t}\Big(\frac{\partial {\Pi}}{\partial T}\Big)+
\mathop{\rm div}\Big(\frac{\partial ({\Pi} {\bf u})}{\partial T}\Big)
= 0\,,\\
\partial_{t}\Big(\frac{\partial {\Pi}}
{\partial \mathbf{\phi}}\Big)+
\mathop{\rm div}\Big(\frac{\partial ({\Pi} {\bf u})}
{\partial \mathbf{\phi}}\Big)+
\mathop{\rm div}\Big(\frac{\partial {\Pi}}{\partial q}\;
D {\bf u}\Big)= 0\,.\\
\end{gathered}
\end{equation}
The first-order part in \eqref{eq:lccappursym} is exactly of the
Friedrichs symmetric type.  An abstract form for this system is
$$
\partial_{t}\mathbf{U}+
\sum_{\alpha=1}^{d} {\bf A}_{\alpha}(\mathbf{U})
\partial_{\alpha}\mathbf{U}+\sum_{\alpha=1}^{d} \partial_{\alpha}
 \Bigl( \sum_{\beta=1}^{d} {\bf B}_{\alpha,\beta}(\mathbf{U})\;
\partial_{\beta}\mathbf{U} \Bigr)=0\,,
$$
with the property that for
some positive definite symmetric
matrix $\Sigma(\mathbf{U})$ all matrices
$\Sigma(\mathbf{U})\;{\bf A}_{\alpha}(\mathbf{U})$ are symmetric and
$$
\sum_{\alpha}\sum_{\beta\;}\;(\mathbf{X}^\alpha)^{*}\;
\Sigma(\mathbf{U})\;{\bf B}_{\alpha,\beta}(\mathbf{U})\;\mathbf{X}^\beta=0
$$
for all vectors $\mathbf{X}^1$, \dots, $\mathbf{X}^d$ in $\mathbb{R}^{2d+2}$.
This is what we mean by non-dissipative second order part of the
system, consistently with the usual terminology for
parabolic systems of second order conservation laws.
One may observe that a
similar system (in nonconservative form) is studied in
\cite{BedjaouiSainsaulieu}.
Here
$$
\mathbf{U}=\begin{pmatrix}\rho\\
\mathbf{m}\\
S\\
\mathbf{w}\end{pmatrix},\quad
\Sigma={\rm D}^{2}H\,,\quad
{\bf A}_{\alpha}={\rm D}^{2}({\Pi}   u_{\alpha})\; \Sigma\,,
$$
where the Hessian ${\rm D}^{2}({\Pi}   u_{\alpha})$
of $\Pi   u_{\alpha}$ is taken with
respect to the dual variables
$$\mathbf{Q}=\begin{pmatrix}q\\
\mathbf{u}\\
T\\
\mathbf{\phi}\end{pmatrix}\,.
$$
The actual expression of ${\bf B}_{\alpha,\beta}$ also involves the
Hessian matrix ${\rm D}^{2}H$, and more precisely
$$
{\bf B}_{\alpha,\beta} \frac{\partial}{\partial U_{k}}=
-\rho \frac{\partial^2 H}{\partial w_{\alpha} \partial U_{k}}\;
\frac{\partial}{\partial m_{\beta}}+
\rho \frac{\partial^2 H}{\partial m_{\alpha} \partial U_{k}}\;
\frac{\partial}{\partial w_{\beta}}
$$
for all $k\in\{1,2d+2\}$ (with $U_{1}=\rho$, $U_{2}=m_{1}$,
\dots, $U_{d+1}=m_{d}$, $U_{d+2}=S$,
$U_{d+3}=w_{1}$,\dots, $U_{2d+2}=w_{d}$).
The non-dissipativeness of the
corresponding second order terms in \eqref{eq:lccappursym}
is a calculus exercise, and is equivalent to the fact
that these terms formally cancel out in the computation of
$$
\frac{{\rm d}}{{\rm d} t} \int \mathbf{U}^{*} \Sigma(\mathbf{U}) 
\mathbf{U}\,{\rm d}\mathbf{x}=
2 \int \mathbf{Q} \partial_{t}\mathbf{U}\,{\rm d}\mathbf{x}
$$
along solutions.

Unfortunately, this interesting formulation is limited to
monotone pressure laws, since it makes use of the convexity
of the total energy.


\subsection{Eulerian capillarity models}
 For the rest of this article, we assume that
$\mathbf{\phi}=K {\bf w}$.
Then we can write
$$
g=g_{0}+ \tfrac{1}{2} K'_{\rho} \|\nabla\rho\|^{2}\,,
$$
where $g_0$ is independent of $\nabla\rho$.
In particular, the isothermal model reduces to
\begin{equation}
    \label{eq:multidE}
\begin{gathered}
\partial_t \rho+\mathop{\rm div}(\rho \mathbf{u})=0\,,\\
\partial_t\mathbf{u}+ ( \mathbf{u}^*\cdot \nabla ) \mathbf{u}=
\nabla( K \Delta\rho+\tfrac{1}{2} K'_{\rho} |\nabla\rho|^{2}
-g_{0} )\,,
\end{gathered}
\end{equation}
where $g_{0}$ and $K$ are given, smooth functions of $\rho$
(with $K > 0$). One may also write this system in conservative form,
noting that
$$
p=p_{0}+ \tfrac{1}{2} ( \rho K'_{\rho}-K ) \|\nabla\rho\|^{2}\,,\quad
p_{0}=\rho g_{0}-F_{0}\,,
$$
hence the (complicated) momentum equation
\begin{align*}
&\partial_t(\rho \mathbf{u}^*)+ \mathop{\rm div}(\rho \mathbf{u} \mathbf{u}^{*})+\nabla p_{0}\\
&=\nabla (\rho K \Delta\rho+\tfrac{1}{2}
( K+\rho K'_{\rho} ) \|\nabla\rho\|^{2} )
-\mathop{\rm div} ( K \nabla\rho\otimes\nabla\rho ).
\end{align*}

In one space dimension, our isothermal model reduces to
\begin{equation}    \label{eq:1dE}
\begin{gathered}
\partial_t \rho+\partial_x (\rho  u)=0\,,\\
\partial_t u+ u \partial_x  u=
\partial_x ( K \partial_{xx}^2 \rho+\tfrac{1}{2} K'_{\rho}\;(\partial_x \rho)^{2}
-g_{0}(\rho) )\,.
\end{gathered}
\end{equation}
We point out that models of this kind actually arise in various
other contexts.
In the special case
$$ K(\rho)=\frac{1}{4\rho}\,,
$$
the system \eqref{eq:multidE} is equivalent -- for irrotational
flows -- to a nonlinear Schr\"odinger equation known as the
Gross-Pitaevskii equation
$$
i \partial_{t}\psi+\frac{1}{2} \Delta\psi
= g_{0}(|\psi|^{2}) \psi
$$
for $\psi=\sqrt{\rho} \mbox{e}^{i \varphi}$, $\nabla \varphi=\mathbf{u}$.
See for instance \cite{BOS}, where
$g_{0}(\rho)=\tfrac{1}{4} \rho^{2}$. It is also the case
$K$ proportional to $1/\rho$ that is considered in \cite{LM}, with
almost no restriction on $g_{0}$.
One may also observe that, in one dimension with
$$
g_{0}(\rho)=\tfrac{\rho}{4} \,,\quad K(\rho)=\frac{1}{4\rho}\,,
$$
Equations in \eqref{eq:1dE} appear as an equivalent form of
the filament equation, see \cite{AK} p. 353.

    \subsection{Lagrangian capillarity models}
The one-dimensional isothermal model becomes even simpler
in Lagrangian formulation. Introducing $y$ the mass Lagrangian coordinate
so that\footnote{
This change of variable may be justified rigorously for, say, $C^1$ functions
$(\rho,u)$ and $\rho$ bounded away from zero.}
$\,{\rm d} y=\rho\,{\rm d} x-\rho  u\,{\rm d} t$
 we obtain with a little piece of
calculus the -- at least formally -- equivalent system
\begin{equation}\label{eq:1dL}
\begin{gathered}
\partial_t   v-\partial_y   u = 0\,,\\
\partial_t   u+ \partial_y  p_{0}=
-\partial_y  \left( \kappa\;\partial_{yy}^2 v+\tfrac{1}{2}
\kappa'_{ v} (\partial_y   v)^{2} \right)\,,\\
\end{gathered}
\end{equation}
with
$v:=1/\rho$ and $\kappa( v):=K(1/{ v}) (1/ v)^{5}$.
In the special case $\kappa=$ constant, {\em i.e}.
$K=\mbox{cst} \rho^{-5}$,
the  system \eqref{eq:1dL} is formally equivalent to the (good) Boussinesq
equation and is amenable to the theory of Kato \cite{Kato}, see
 \cite{Bona} for more details.
Our aim here is to deal with general capillarities, motivated by
physical reasons - since there is no reason why $K$ should be proportional
to $\rho^{5}$ - as well as by the various analogies mentioned above.

Following an idea of  Coquel \cite{Coquel},
we rewrite the velocity equation as
$$
\partial_t   u+ \partial_y  p_{0}-\partial_y  (\alpha \partial_y  w )=0\,, \quad
\alpha=\sqrt{\kappa}\,,\quad w=-\alpha \partial_y   v.
$$
Applying the differential
operator $-\partial_y (\alpha \cdot )$ to the first line of
\eqref{eq:1dL}, we  find that $w$ satisfies the equation
$$
\partial_t  w+\partial_y  (\alpha \partial_y   u )=0\,.
$$
Considering $w$ as an additional unknown, we are led to the system
\begin{equation}\label{eq:1dLaug}
\begin{gathered}
\partial_t   v-\partial_y   u = 0\,,\\
\partial_t   u+ \partial_y  p_{0}-\partial_y  (\alpha \partial_y  w )=0\,,\\
\partial_t  w+\partial_y  (\alpha \partial_y   u )=0\,.
\end{gathered}
\end{equation}
Alternatively,
because of the constraint $w=-\alpha \partial_y   v$ we
may rewrite \eqref{eq:1dLaug} as
\begin{equation}\label{eq:1dLaugb}
\begin{gathered}
\partial_t   v-\partial_y   u = 0\,,\\
\partial_t   u-\partial_y  (\alpha( v) \partial_y  w )=q( v) w\,,\\
\partial_t  w+\partial_y  (\alpha( v) \partial_y   u )=0
\end{gathered}
\end{equation}
with $q( v):={p_{0}'( v)}/{\alpha( v)}$.


\section{Well-posedness results}\label{s:wp}

We are interested in the well-posedness of  \eqref{eq:1dE} and
\eqref{eq:1dL} for data with finite specific volumes ({\em i.e.}
away from vacuum) and finite densities.
More precisely, as both systems are known to admit global smooth
solutions (like of course constant states, but also travelling wave
solutions, see for instance \cite{Benzoni,BDDJ}), our main purpose is
to show the well-posedness of  \eqref{eq:1dE} and \eqref{eq:1dL} in
affine spaces  about such reference solutions.

Our strategy is to first prove the well-posedness of the
system \eqref{eq:1dLaugb}.

\subsection{Semigroup method}

The special case where $\alpha$ is constant is much easier, and
enters the framework of Kato \cite{Kato}. As a matter of fact,
\eqref{eq:1dLaugb} can always be put in the abstract form
$$
\partial_t  \mathbf{v}-\mathcal{A}( v)\cdot\mathbf{v}= \mathcal{F}( v,w)\,,
$$
with
\begin{gather*}
\mathbf{v}:=\begin{pmatrix} v \\
 u\\
w \end{pmatrix},\quad
\mathcal{A}( v):=\begin{pmatrix} 0 & \partial_y  & 0 \\
\partial_y  & 0  & \partial_y   \alpha( v) \partial_y \\
0 & -\partial_y   \alpha( v) \partial_y  & 0
\end{pmatrix}, \\
\mathcal{F}( v,w):=
\begin{pmatrix}0
\\ \frac{p'_{0}( v)+1}{\alpha} w \\ 0
\end{pmatrix}.
\end{gather*}
(In fact, we have used again the
expected relation $w=-\alpha \partial_y   v$ to
obtain a right hand side $\mathcal{F}$ of order $0$.)
When $\alpha$ is constant,
the antisymmetric operator $\mathcal{A}( v)$ of course has constant
coefficients, which makes a big difference.
Indeed, because $\mathcal{A}$ is the infinitesimal generator of
a group of unitary operators on $L^{2}(\mathbb{R})^{3}$
(this is due to Stone's
theorem, see \cite{Pazy}, p. 41), it is possible to
apply (a slight adaptation of) Theorem 6 in \cite{Kato} (p. 36) using
the operator
$S=(1-\partial^2_{yy}) I_{3}$ and
the spaces
$Y=H^{2}(\mathbb{R})^{3}$, $X=L^{2}(\mathbb{R})^{3}$,
and thus show the following for any $p_{0}$ a smooth function of $ v$.

\begin{theorem}\label{thm:capcst}
    If $\underline{\mathbf{v}}=(\underline{ v},\underline{ u},
    \underline{w})^t$ is smooth and exponentially decaying to
    $( v_{\pm}, u_{\pm},0)$ when $x\to\pm\infty$, then
    there is a  $T\geq 0$ so that the
 Cauchy problem associated with
$$
\partial_t  \mathbf{v}-\mathcal{A} \mathbf{v}=\mathcal{F}( v,w)
$$
and initial data in $\underline{\mathbf{v}}(0)+H^{2}(\mathbb{R})^{3}$
admits a unique solution such that
$\mathbf{v}-\underline{\mathbf{v}}$ belongs to
$\mathcal{C}(0,T;H^{2}(\mathbb{R})^{3})\cap \mathcal{C}^{1}(0,T;L^{2}(\mathbb{R})^{3})$.
\end{theorem}

This theorem is essentially the same  as Theorem 1 in \cite{Bona},
extended to solutions
having nonzero, and possibly different, limits at infinity. Our motivation for this
extension is the stability of diffuse interfaces.

For a general function $\alpha$, the semi-group approach breaks down
because the operator $\mathcal{A}( v)$ involves
the derivative $\partial_y  v$. Thus the Lipschitz estimate requested by
Kato's theorem,
$$
\| (\mathcal{A}( v_{1})-\mathcal{A}( v_{2}))\cdot \mathbf{v}\|_{X}
 \lesssim \|\mathbf{v}_{1}-\mathbf{v}_{2}\|_{X} \|\mathbf{v}\|_{Y}
$$
means that the norm of vectors in $X$
should control the first derivative of their first components.
This goal does not seem possible to achieve without augmenting also
the regularity index of the other components (to keep a semigroup generated
by $\mathcal{A}( v)$ in $X$). For this reason the Lipschitz
estimate above is quite unlikely.

\subsection{Alternative method}

We introduce the complex valued function
$z=u+i w$ and regard equivalently the system \eqref{eq:1dLaugb}
as the coupling of the variable coefficient Schr\"odinger equation
\begin{equation}\label{eq:1dLaugc}
\partial_t  z+i \partial_y  (\alpha( v) \partial_y  z )=q( v) \mathop{\rm Im} z
\end{equation}
with the compatibility equation
$$
\partial_t   v-\partial_y  \mathop{\rm Re} z = 0\,.
$$
 Let $J_v$ be an open interval of $(0,+\infty)$ and
$k\geq2$ be an integer. Our main assumption is
\begin{itemize}
\item[(H1)] Both $p_{0}$ and $\alpha$
belong to $W^{k+2,\infty}_{\rm loc}(J_v)$
and $\alpha$ is positive on $J_v$.
\end{itemize}
For a nonconvex pressure law $ v\mapsto p_0( v)$, it is easy to
show that  \eqref{eq:1dLaugb} does admit non-constant, bounded, global,
smooth solutions, which are in fact travelling wave solutions.
This is because of  the Hamiltonian structure of $\eqref{eq:1dLaugb}$,
which implies that the governing equations for the travelling waves
are also Hamiltonian, see \cite{Benzoni} or our companion paper
\cite{BDDJ} for more details.
In what follows, we assume the existence of a reference global smooth
solution, regardless of the convexity properties of $p$.

\begin{theorem}\label{existence}
Under assumption {\rm  (H1)} with $k\geq2$,
let $(\overline u,\overline v)$ be a given global classical solution of system
$\eqref{eq:1dLaugb}$ with $\overline v(\mathbb{R}^2)\subset J_v$ and
$$
\partial_y \overline u\in C(\mathbb{R};H^{k+1}),\quad
\partial_y \overline v\in C(\mathbb{R};H^{k+2}).
$$
Let $u_0\in \overline u(0)+H^k$ and $v_0\in\overline v(0)+H^{k+1}$ be  such that
 $v_0(\mathbb{R})\subset\subset J_v$.
There exists a positive $T$ such that the Cauchy problem
associated with the system $(\ref{eq:1dLaugb})$ and initial data
$(u_{0},v_{0})$ has a unique
solution $(u,v)$ with $v([-T,T]\times\mathbb{R})
\subset\subset J_v$
and, denoting $\widetilde  u:=u-\overline u$ and $\widetilde v:=v-\overline v$,
\begin{equation}\label{espace}
(\widetilde u,\widetilde v)\in\mathcal{C}([-T,T];H^k\times H^{k+1})\cap \mathcal{C}^1([-T,T];H^{k-2}\times H^{k-1}).
\end{equation}
 \end{theorem}

\begin{remark} \label{rmk1} \rm
Constants are obviously global classical solutions of \eqref{eq:1dLaugb}.
If $(\overline u,\overline v)$ is constant then  the theorem can be slightly
improved, see Section \ref{ss:constant}.
\end{remark}

 \begin{remark}\label{three} \rm
One can find an explicit bound by below for $T$ (see \eqref{time1}).
Besides, one can show that
for $H^k$ data, the time of existence
of a $H^k$ solution is the same as for a
$H^2$ solution (see section \ref{blowupcriterion}).

In the case where
the Sobolev norms of $\partial_y \overline u$ and $\partial_y \overline v$ are independent of the time,  there exists a constant $C=C(\alpha,q,\overline u,\overline v,v_0(\mathbb{R}))$
such that  $T$ may be chosen such that
\begin{equation}\label{lowerbound}
T\geq\frac1C
\log\left(1+\frac1{\norm{H^2}{\widetilde u_0}
+\norm{H^2}{\partial_y \widetilde  v_0}}\right).
\end{equation}
Hence, if $\norm{H^2}{\widetilde u_0}
+\norm{H^2}{\partial_y  \widetilde v_0}\sim\varepsilon$ for a small $\varepsilon$, the life span
 is greater than $C\log\varepsilon^{-1}$.
\end{remark}

 For simplicity, we shall restrict ourselves to the evolution
for {\it positive} times.
As system $(\ref{eq:1dLaugb})$  is time reversible,
adapting our proof to {\it negative} is straightforward.

Our approach is quite classical. It
consists in deriving first energy estimates
{\em without loss of derivatives} -- in sufficiently high
order Sobolev spaces -- for a linearized version of \eqref{eq:1dLaugc},
and then solve the nonlinear problem through an iterative scheme.
There are some difficulties in both steps that will be pointed out
along the detailed proofs.


\section{A variable coefficient linear Schr\"odinger equation}

To show the (local) well-posedness of \eqref{eq:1dLaugb} by means
of an iterative scheme,
we shall need the resolution of the linear Schr\"odinger equation
\begin{equation}\label{eq:linear}
\partial_t  z+i \partial_y  (a(y,t) \partial_y  z )=f(y,t)\,.
\end{equation}
 Whatever the function $a$ (smooth enough and {\it real}-valued), the
operator $i \partial_y  a \partial_y $ is obviously antisymmetric on $L^{2}(\mathbb{R})$.
Under suitable assumptions on the asymptotic behavior of $a$ it is
not difficult to show that $i \partial_y  a \partial_y $ is also skewadjoint
\footnote{It suffices to check that
the range of $\partial_y  a \partial_y -\lambda$ is dense in $L^{2}(\mathbb{R})$
for all $\lambda\notin \mathbb{R}$.} in $L^{2}(\mathbb{R})$.
Therefore $i \partial_y  a \partial_y $ is
the infinitesimal generator of a group of unitary operators,
and the standard semigroup theory (see \cite{Pazy}, p. 145--147)
enables us to prove the following.


\begin{theorem}     \label{thm:lSchr}
Assuming that $a$ is real valued, belongs to
$\mathcal{C}^{1}([0,T];W^{1,\infty}(\mathbb{R}))$ and that $a(t)$ has finite
limits at $x=\pm\infty$ for all $t\in [0,T]$,
the Cauchy problem
\begin{gather*}
\partial_t  z+i \partial_y  (a(y,t) \partial_y  z )=f(y,t)\,,\\
z(0)=z_{0}
\end{gather*}
with $z_{0}\in H^{2}(\mathbb{R})$ and
$f\in L^{1}([0,T];H^{2}(\mathbb{R}))\cap\mathcal{C}([0,T];L^2(\mathbb{R}))$
admits a unique (classical) solution
$z\in \mathcal{C}([0,T];H^{2}(\mathbb{R}))\cap \mathcal{C}^{1}([0,T];L^{2}(\mathbb{R}))$.
If $z_0\in L^2$ and $f\in L^{1}([0,T];L^{2}(\mathbb{R}))$,  we get a
mild solution $u\in \mathcal{C}([0,T];L^{2}(\mathbb{R}))$ given
by Duhamel's formula:
$$
z(t)=\mathcal{S}(t,0) z_{0}+\int_{0}^t
\mathcal{S}(t,s) f(s)\,{\rm d} s
$$
with $\mathcal{S}(t,s)$ the solution operator of the homogeneous equation.
\end{theorem}


\subsection{A priori estimates}

Our aim is to obtain a more precise result, and especially {\em a
priori} estimates in view of our iterative scheme for the nonlinear problem.
Of course we immediately have from Duhamel's formula the estimate
$$
\|z(t)\|_{L^2} \leq \|z(0)\|_{L^2}+
\int_{0}^{t} \|f(s)\|_{L^{2}}\,{\rm d} s\,,
$$
which can also be derived directly from the equation \eqref{eq:linear}.
Indeed, multiplying \eqref{eq:linear} by $\overline z$
and integrating over $\mathbb{R}$ we get
\begin{equation}
\label{dtorder0}
\frac 12\frac {\,{\rm d}}{ \,{\rm dt}}\Lp 2z^2-\Im\int\partial_y
(a\partial_y  z) \overline z\,{\rm d} y=
 \Re\int f\overline z\,{\rm
d} y
 \end{equation}
and an integration by parts shows that
the integral  in the left-hand side has
no imaginary part since $a$ is {\it real}. So it
follows from Cauchy-Schwarz inequality that (at least formally)
$$
\frac{\,{\rm d}}{ \,{\rm d} t}\Lp 2z \leq
\Lp 2f
$$
hence the result.

However, this $L^2$ estimate does not provide enough information to solve
the non-linear system \eqref{eq:1dLaugb}, which
equivalently reads
\begin{equation}\label{eq:1dLaugcsyst}
\begin{gathered}
\partial_t   v-\partial_y  \mathop{\rm Re} z = 0\,,\\
\partial_t  z+i \partial_y  (\alpha( v) \partial_y  z )=q( v) \mathop{\rm Im} z\,.
\end{gathered}
\end{equation}
The second-order term $\partial_y (\alpha(v)\partial_y  z)$
induces us to prove a priori estimates for  equation (\ref{eq:linear})
in higher  order Sobolev spaces, a matter which is not obvious
despite the linearity of the equation, neither on the Duhamel formula
 -- because differentiation operators do not commute with
 the solution operator $\mathcal{S}(t,s)$ -- nor in the
 direct fashion described in $L^2$. Indeed,
let  $z^{(k)}$ (resp. $a^{(k)}$ and $f^{(k)}$)
denote the $k$-th order derivative of $z$ (resp. $a$ and $f$)
with respect to $y$.
By Leibniz formula, we have
\begin{align*}
\partial_y ^k\Bigl(\partial_y (a\partial_y  z)\Bigr)
&= \partial_y  \Bigl(\partial_y ^k(a\partial_y  z)\Bigr)\\
&= \partial_y (a\partial_y  z^{(k)})+ \sum_{j=0}^{k-1}
{\textstyle\binom kj}\;
\partial_y (z^{(j+1)}a^{(k-j)})\\
&= \partial_y (a\partial_y  z^{(k)})+k\partial_y  a  z^{(k+1)}+
 \sum_{\ell=1}^{k} {\textstyle\binom {k+1}{\ell-1}}\;
z^{(\ell)}a^{(k+2-\ell)}
\end{align*}
so we get the following equation for $z^{(k)}$:
\begin{equation}\label{eq:ordrek}
\partial_t  z^{(k)}+i\partial_y (a\partial_y  z^{(k)})=f^{(k)}-
ik(\partial_y  a) \partial_y  z^{(k)}-i\sum_{j=1}^k {\textstyle \binom {k+1}{j-1}}
z^{(j)}a^{(k+2-j)}.
\end{equation}
This can be rewritten using Duhamel's formula as
\begin{align*}
z^{(k)}(t)
&=\mathcal{S}(t,0) z_{0}^{(k)}\\
&\quad +\int_{0}^t \mathcal{S}(t,s) \Big(
f^{(k)}- ik(\partial_y  a) \partial_y  z^{(k)}-i\sum_{j=1}^k
{\textstyle \binom {k+1}{j-1}} z^{(j)}a^{(k+2-j)}\Big)(s)\,{\rm d} s\,.
\end{align*}
We see that the time integral involves the derivative $\partial_y  z^{(k)}$.
This derivative cannot disappear through a direct {\em a priori}
estimate either,
because the real part of \eqref{eq:ordrek} multiplied by $\overline z^{(k)}$
contains the term
$$
\mathop{\rm Im} \int (\partial_y  a) \partial_y  z^{(k)} \overline{z^{(k)}}\,
{\rm d}y,
$$
which cannot be rewritten without a derivative of $z^{(k)}$, unless
$a$ is constant or $z^{(k)}$ is real.

This is a well identified problem for {\it variable}
coefficients  Schr\"odinger equations. In \cite{LP} for instance,
W. Lim and G. Ponce  overcome the difficulty
by introducing appropriate weighted Sobolev spaces.
We are going to show how weights, also called gauges, can
help to compensate the loss of derivative in our context.


Let $\phi_k$ denote the $k$-th order gauge (to be determined),
and multiply the equation (\ref{eq:ordrek}) by $\phi_k$.
Since
$$
\phi_k\partial_y (a\partial_y  z^{(k)})=\partial_y (a\partial_y (\phi_kz^{(k)}))-z^{(k)}\partial_y (a\partial_y \phi_k)
-2a\partial_y \phi_k\partial_y  z^{(k)},
$$
we find the equation for $\phi_kz^{(k)}$,
\begin{equation}\label{ordrek0}
\begin{aligned}
&\partial_t (\phi_kz^{(k)})+i \partial_y (a\partial_y (\phi_kz^{(k)}))\\
&=\phi_k f^{(k)} +z^{(k)}\partial_t \phi_k-i\sum_{j=1}^{k-1}
{\textstyle \binom {k+1}{j-1}} \phi_kz^{(j)}a^{(k+2-j)}\\
&+iz^{(k)}\bigl(\partial_y (a\partial_y \phi_k)-\frac{k(k+1)}2\partial_{yy}^2 a\;\phi_k\bigr)
+i\partial_y  z^{(k)}
\Bigl(2a\partial_y \phi_k-k\phi_k\partial_y  a\Bigr).
\end{aligned}
\end{equation}
 From the above equality, it is now clear  that the loss of derivative will
be avoided if and only if $\phi_k$ satisfies
$$
2a\partial_y \phi_k-k\phi_k\partial_y  a=0.
$$
Choose $\phi_k:=a^{k/2}$ so that the last term in \eqref{ordrek0}
vanishes. Then multiply (\ref{ordrek0})  by
$\phi_k\overline{z^{(k)}}$ and integrate in time.
If we keep only the real part of the equation,
the second order term in the left-hand side vanishes, as well as
the last remaining term of the right-hand side,
hence the identity
  \begin{equation}\label{eq:jaugek}
\begin{aligned}
\frac12\frac{\,{\rm d}}{ \,{\rm d} t}\Lp2{\phi_kz^{(k)}}^2
&=  \Re\int\phi_k^2f^{(k)}\overline{z^{(k)}}
\,{\rm d} y   + \int\phi_k\partial_t \phi_k|z^{(k)}|^2\,{\rm d} y \\
&\quad +\sum_{j=1}^{k-1} {\textstyle \binom{k+1}{j-1}}
\Im\int a^{(k+2-j)}\phi_k^2
z^{(j)}\overline{z^{(k)}}\,{\rm d} y.
\end{aligned}
\end{equation}
We can now state estimates in $H^k$ for the variable coefficient
Schr\"odinger equation.

\begin{proposition}\label{estimateslinear}
Let $z$ be a solution of \eqref{eq:linear} on $\mathbb{R}\times[0,T]$.
Assume in addition that $a$ is bounded away from zero
by $\underline a$ and is bounded by $\widetilde{a}$.
Then for all $t\in[0,T]$, we have
\begin{equation}\label{order0}
\Lp 2{z(t)}\leq \Lp2{z_0}+\int_0^t \Lp2{f(\tau)}\,d\tau.
\end{equation}
Besides, denoting for  $k\in\mathbb{N}^*$,
$$
Z_k(t):=\Big( \sum_{j=1}^{k}\Lp2{(a^{\frac j2}z^{(j)})(t)}^2\Big)^{1/2}
\quad\mbox{and}\quad
F_k(t):=\Big( \sum_{j=1}^{k}\Lp2{(a^{\frac j2}f^{(j)})(t)}^2\Big)^{1/2},
$$
 we have
\begin{equation}\label{orderk}
Z_k(t)\leq {\rm e}^{A_k(t)}\Big(Z_k(0)
+\int_0^t{\rm e}^{-A_k(\tau)}F_k(\tau)\,d\tau\Big),
\end{equation}
with $A_k(t):=\int_0^t\Bigl(\frac k2\Lp{\infty}{\partial_t \log a(\tau)}+C_k
\bigl(\norm{H^{k-2}}{\partial_y ^3 a(\tau)}+\Lp{\infty}{\partial_y ^2a(\tau)}\bigr)\Bigr)\,d\tau$
 for some positive
constant $C_k$ depending only on $k$, $\underline a$ and $\widetilde{a}$ if $k\geq 2$,
and $C_1=0$.
  \end{proposition}

\begin{proof}
The $L^2$ estimate was already pointed out.
In order to prove the $k$-th order estimate, we sum equalities
\eqref{eq:jaugek} for $j=1,\dots,k$. This implies
by Cauchy-Schwarz inequality,
\begin{equation}\label{orderk1}
\frac12\frac{\,{\rm d}}{ \,{\rm d} t} Z_k^2 \leq
  Z_k\Big(F_k+
\frac k2\Lp\infty{\partial_t\log a}\;Z_k
+\sum_{j=2}^k\sum_{\ell=1}^{j-1} {\textstyle \binom{j+1}{\ell-1}}
\widetilde a^{\frac j2}
\Lp2{a^{(j+2-\ell)}z^{(\ell)}}\Big).
\end{equation}
Note that in the case $k=1$, the last term vanishes  so
that straightforward calculations yield \eqref{orderk}.
Let us assume from now on that $k\geq2$.
Since
$ a^{(j+2-\ell)}z^{(\ell)}
=\bigl(\partial_y ^{(j-1)-(\ell-1)}{a''}\bigr)
\bigl(\partial_y ^{\ell-1}{z'}\bigr)$,
inequality \eqref{tame1} in the appendix ensures that
$$
\Lp2{a^{(j+2-\ell)}z^{(\ell)}}
\leq C_{j,\ell}\Big(\Lp{\infty}{a''}\norm{H^{j-1}}{z'}
+\Lp{\infty}{z'}\norm{L^2}{a^{(j+1)}}\Big).
$$
Now, because $$  \Lp{\infty}{z'}^{2}\leq
\Lp2{z'}\Lp2{z''}
\leq \underline a^{-\frac32}\Lp2{a^{1/2}z'}\Lp2{az''},$$
we conclude that\footnote{remark that we
always have $j+1\geq3$.}
whenever $2\leq j\leq k$ and $1\leq\ell\leq j-1$,
$$
\Lp2{a^{(j+2-\ell)}z^{(\ell)}}
\leq C_{j,\ell}\left(\Lp{\infty}{a''}+\norm{H^{k-2}}{a^{(3)}}\right)Z_k,
$$
for some constant $C_{j,\ell}$ depending only on $j$, $\ell$ and $\underline a$.
Plugging this latter inequality in \eqref{orderk1},
we end up with
$$
\frac12\frac{\,{\rm d}}{ \,{\rm d} t} Z_k^2\leq
  F_kZ_k+
\Big(\frac k2\Lp\infty{\partial_t \log a}+
 C_{k}\bigl(\Lp{\infty}{a''}+\norm{H^{k-2}}{a^{(3)}}\bigr)\Big)Z_k^2.
$$
Then Gronwall lemma entails  inequality (\ref{orderk}).
 \end{proof}

\subsection{Existence of regular solutions}

This section is devoted to the regularity of
solutions of \eqref{eq:linear}, namely we want to
 prove the following theorem.

\begin{theorem}     \label{thm:lSchrreg}
Let $k$ be an integer such that $k\geq 2$, Let $a=a(y,t)$ be
bounded by $\widetilde a$ and bounded away from zero by $\underline a>0$
on $\mathbb{R}\times[0,T]$, and satisfy $\partial_y ^2 a\in L^1(0,T;H^{k-1}(\mathbb{R}))$ and
$\partial_t  a\in L^1(0,T;L^\infty(\mathbb{R}))$.
Let  $z_{0}$ be in $H^{k}(\mathbb{R})$ and
$f$ in $L^{1}(0,T;H^{k}(\mathbb{R}))$.
Then the  Cauchy problem
\begin{equation}
  \label{regularit}
\begin{gathered}
\partial_t  z+i \partial_y  (a(y,t) \partial_y  z )=f(y,t)\,,\\
z(0)=z_{0}
\end{gathered}
\end{equation}
admits a unique  solution
$z\in \mathcal{C}([0,T];H^{k}(\mathbb{R}))$.
If, besides, $f\in  \mathcal{C}([0,T];H^{k-2}(\mathbb{R}))$
then $z$ also belongs to $\mathcal{C}^{1}([0,T];H^{k-2}(\mathbb{R}))$.
\end{theorem}

\begin{proof}
The proof is based on the following fourth-order  regularization of
equation \eqref{regularit}:
\begin{equation} \label{regular}
\begin{gathered}
\partial_t  z_{\varepsilon}+i \partial_y  (a(y,t) \partial_y  z_{\varepsilon} )
+\varepsilon \partial^4_{y} z_{\varepsilon}
=f\,,\\
z_{\varepsilon}(0)=z_{0},
\end{gathered}
\end{equation}
where $\varepsilon >0$ stands for a positive parameter
bound to go to zero.

 We shall see in the proof however that
this regularization does not enable us to  pass to the limit
in the very space $\mathcal{C}([0,T];H^k)$.
For doing so, we  shall adapt the method by J. Bona and  R. Smith in \cite{BS}
which amounts to
smoothing out conveniently  the  data and the variable coefficients.

Let us briefly describe the main steps of the proof:
\begin{enumerate}
\item  Getting  bounds for $z_\varepsilon$ similar to
\eqref{orderk} and independent of $\varepsilon$.
\item
Stating well-posedness  in $H^k(\mathbb{R})$
for the problem \eqref{regular} with  $\varepsilon>0$.
\item
Showing that the  family of solutions $(z_\varepsilon)$
to   \eqref{regular} with regularized data and coefficient
converges to some $z$ in the space $\mathcal{C}([0,T];H^k(\mathbb{R}))$
when $\varepsilon$ approaches $0$.
 \end{enumerate}

\subsubsection*{Step 1: uniform a priori estimates}
Let $p\in\mathbb{N}$.
We assume that we are given a solution
$z\in\mathcal{C}([0,T];H^p(\mathbb{R}))$
to \eqref{regular}\footnote{For
notational convenience, we drop the indices $\varepsilon$ in this step}
corresponding to data
$z_{0}\in H^{p}(\mathbb{R})$ and $f\in L^{1}([0,T];H^{p}(\mathbb{R}))$.
We claim that   $z$ satisfies
an estimate similar to \eqref{orderk}
with a constant $C$ {\it independent} of $\varepsilon$.

Let $\ell$ be an integer such that $0 \leq \ell \leq p$ and
$\phi_{\ell}=a^{\frac{\ell}{2}}$ be the $\ell$-th order gauge introduced
previously.  Applying Leibniz formula
to $\partial_y ^4(\alpha\beta)$ with $\alpha:=1/\phi_\ell$ and $\beta=\phi_\ell z^{(\ell)}$,
one obtains the following expansion:
\begin{align*}
\phi_{\ell} \partial^4_y z^{(\ell)}
&=\partial^4_y(\phi_{\ell} z^{(\ell)})
+4 \phi_\ell \partial_y (\phi_\ell^{-1}) \partial_y ^3(\phi_{\ell} z^{(\ell)})
+6 \phi_\ell \partial_y ^2(\phi_\ell^{-1}) \partial^2_y(\phi_{\ell} z^{(\ell)})
\\
&\quad +4 \phi_\ell \partial_y ^3(\phi_\ell^{-1}) \partial_y (\phi_{\ell} z^{(\ell)})
+\phi_\ell \partial_y ^4(\phi_\ell^{-1}) (\phi_{\ell} z^{(\ell)}).
\end{align*}
Hence, applying the operator $\phi_\ell\partial_y ^\ell$ to the evolution equation
in \eqref{regular} and denoting  $\widetilde z_\ell:=a^{\ell/2}z^{(\ell)}$,
 we get
\begin{equation}\label{ordrek0epsilon}
\begin{aligned}
&\partial_t \widetilde z_\ell +i \partial_y (a\partial_y \widetilde z_{\ell})
+\varepsilon \partial^4_{y}\widetilde z_\ell \\
&=a^{\frac{\ell}{2}} f^{(\ell)}
+z^{(\ell)}\partial_t  a^{\frac{\ell}{2}}
+ i \widetilde z_\ell\Bigl(a^{-\frac{\ell}{2}}\partial_y (a\partial_y  a^{\frac{\ell}{2}})
-\frac{\ell(\ell+1)}2\partial_y ^2 a\Bigr)\\
&\quad -i{  \sum_{j=1}^{\ell-1}}
 {\textstyle \binom {\ell+1}{j-1}}
a^{\frac{\ell}{2}}z^{(j)}a^{(\ell+2-j)}
-\varepsilon a^{\ell/2}
\Bigl(4 \partial_y (a^{-\ell/2}) \partial_y ^3\widetilde z_\ell\\
&\quad +6 \partial_y ^2(a^{-\ell/2}) \partial^2_y\widetilde z_\ell
+4 \partial_y ^3(a^{-\ell/2}) \partial_y \widetilde z_\ell
+\partial_y ^4(a^{-\ell/2}) \widetilde z_\ell
\Bigr).
\end{aligned}
\end{equation}
Now, multiplying \eqref{ordrek0epsilon} by $\overline{\widetilde z_\ell}$
and performing a time integration, we deduce that
\begin{equation}\label{eq:jaugekepsilon}
\begin{aligned}
\frac12\frac{\,{\rm d}}{ \,{\rm d} t} \Lp2{\widetilde z_\ell}^2
+\varepsilon \Lp2{\partial^2_y \widetilde z_\ell}^2
&= \Re\int
a^{\ell} f^{(\ell)}\overline{z^{(\ell)}}
\,{\rm d} y   + \frac\ell2\int
\partial_t \log a\;|\widetilde z_\ell|^2 \,{\rm d} y \\
&\quad  +\sum_{j=1}^{\ell-1} {\textstyle \binom{\ell+1}{j-1}}
\Im\int a^{(\ell+2-j)}a^{\ell}
z^{(j)}\overline{z^{(\ell)}}\,{\rm d} y \\
&\quad -\varepsilon \Re\bigl(4R_1+6R_2+4R_3+R_4)
\end{aligned}
\end{equation}
with
\begin{gather*}
R_1:=\int a^{\ell/2}\partial_y (a^{-\ell/2})\;(\partial_y ^3\widetilde z_\ell)\;
\overline{\widetilde z_\ell}\,{\rm d} y, \quad
R_2:=\int a^{\ell/2} \partial_y ^2(a^{-\ell/2})
(\partial^2_y \widetilde z_\ell)\;
\overline{\widetilde z_\ell}\,{\rm d} y,\\
R_3:=\int a^{\ell/2}\partial_y ^3(a^{-\ell/2})\;(\partial_y \widetilde z_\ell)\;
\overline{\widetilde z_\ell}\,{\rm d} y, \quad
R_4:= \int a^{\ell/2}\partial_y ^4(a^{-\ell/2})
 |\widetilde z_\ell|^2\,{\rm d} y.
\end{gather*}
 To bound $R_1$, we perform an integration by parts:
\begin{align*}
R_1&=-\int\bigl(\partial_y (a^{\ell/2})\partial_y (a^{-\ell/2})
+a^{\ell/2}\partial_y ^2(a^{-\ell/2})\bigr) (\partial_y ^2\widetilde z_\ell)
\overline{\widetilde z_\ell}\,{\rm d} y\\
&-\int a^{\ell/2}\partial_y (a^{-\ell/2})
(\partial_y ^2\widetilde z_\ell)
\partial_y \overline{\widetilde z_\ell}\,{\rm d} y\,.
\end{align*}
Using H\"older inequalities and  applying the interpolation inequality
$$
\Lp{\infty}{F'}^2\leq2\Lp{\infty}{F}\Lp{\infty}{F''},
$$
to $F=a^{\pm\ell/2}$, we get by Young's inequality
$$
|4R_1|\leq \frac18\Lp2{\partial_y ^2\widetilde z_\ell}^2
+C\Lp{\infty}{\partial_y ^2a}^2\Lp2{\widetilde z_\ell}^2
$$
for some constant $C$ depending only on
$\widetilde a \geq \|a\|_{L^\infty}$ and $\underline a^{-1}
\geq  \|a^{-1}\|_{L^\infty}$.
We get a similar bound for $R_2$ using only H\"older and
Young inequalities.

We claim that the same bound holds true for $R_3$ and $R_4$.
This is only a matter of using suitable integration by parts
and standard inequalities like above.
Indeed, we may rewrite
\begin{align*}
R_3&=-\int\big(\partial_y (a^{\ell/2})
\partial_y ^2(a^{-\ell/2}) (\partial_y \widetilde z_\ell)
\overline{\widetilde z_\ell} + a^{\ell/2}
\partial_y ^2(a^{-\ell/2})|\partial_y \widetilde z_\ell|^2\\
&\quad + a^{\ell/2}
\partial_y ^2(a^{-\ell/2}) (\partial_y ^2\widetilde z_\ell)
\overline{\widetilde z_\ell}\big)\,{\rm d} y
\end{align*}
and
\begin{align*}
R_4&=-\int\big(\partial_y ^2(a^{\ell/2})\partial_y ^2(a^{-\ell/2})
|\widetilde z_\ell|^2
+2 \partial_y (a^{\ell/2})\partial_y ^2(a^{-\ell/2})
\partial_y \bigl(|\widetilde z_\ell|^2\bigr) \\
&\quad + a^{\ell/2}\partial_y ^2(a^{-\ell/2})
\partial_y ^2\bigl(|\widetilde z_\ell|^2\bigr)\big)\,{\rm d} y.
\end{align*}
Therefore, we  eventually find a constant
$C$ depending only on $\widetilde a$
and $\underline a$ so that
$$
\Bigl|4R_1+6R_2+4R_3+R_4\Bigr| \leq  \frac12\Lp2{\partial_y ^2\widetilde z_\ell}^2
+C\Lp{\infty}{\partial_y ^2a}^2\Lp2{\widetilde z_\ell}^2.
$$
The other terms appearing in the
right-hand side of \eqref{eq:jaugekepsilon}
may be bounded like in the case $\varepsilon=0$. Hence
\begin{align*}
&\frac12\frac{\,{\rm d}}{ \,{\rm d} t}\Lp2{\widetilde z_\ell}^2
+\frac\varepsilon2 \Lp2{\partial^2_y \widetilde z_\ell}^2\\
&\leq \Lp2{\widetilde z_\ell}\Lp2{a^{\ell/2}f^{(\ell)}}
 +\big(\frac\ell2\Lp{\infty}{\partial_t \log a}+C\varepsilon\Lp{\infty}{\partial_y ^2a}^2
\big) \Lp2{\widetilde z_\ell}^2 \\
&\quad +C\big(\Lp{\infty}{\partial_y ^2a}+\norm{H^{\ell-2}}{\partial_y ^3a}\big)
\norm{H^\ell}{z}^2.
\end{align*}
Summing the above inequalities  for $\ell=0,\dots,p$
and applying the usual  Gronwall type argument,
we deduce that for $t$ in $[0,T]$ and $\varepsilon\geq0$,
we have
\begin{equation}\label{orderkregularized}
\Big( Z_{p}^2(t)
+\varepsilon\sum_{\ell=0}^p\Lp2{\partial_y ^2
\bigl(a^{\ell/2}z^{(\ell)}\bigr)}^2\Big)^{1/2}
\leq {\rm e}^{A_{p,\varepsilon}(t)}
\Big(Z_{p}(0)
+\int_0^tF_p(\tau) {\rm e}^{-A_{p,\varepsilon}(\tau)}\,d\tau\Big)
\end{equation}
with
\begin{gather*}
Z_{p}^2(t)=\sum_{\ell=0}^p \Lp2{a^{\ell/2}\partial_y ^{(\ell)}z(t)}^2,\quad
F_{p}^2(t)=\sum_{\ell=0}^p
\Lp2{a^{\ell/2}\partial_y ^{(\ell)}f(t)}^2,\\
A_{p,\varepsilon}(t)=\int_0^t\big(\frac p2\Lp{\infty}{\partial_t \log a}
+C_p\bigl(\Lp{\infty}{\partial_y ^2 a}+
\norm{H^{p-2}}{\partial_y ^3a}+\varepsilon\Lp{\infty}{\partial_y ^2a}^2\bigr)\big)d\tau
\end{gather*}
where $C_0=C_1=0$ and $C_p$ depends only on $p$, $\underline a$ and $\widetilde a$
for $p\geq 2$.

\subsubsection*{Step 2: solving the regularized equation}
This step is devoted to the proof of the following result.

\begin{proposition}\label{thm:regular}
Let $k$, $a$, $z_0$ and $f$ satisfy the hypotheses of
theorem $\ref{thm:lSchrreg}$.
System $\eqref{regular}$ has
a unique solution $z_\varepsilon$ in the space
$$
\mathcal{C}([0,T];H^k)\cap L^2(0,T;H^{k+2}).
$$
\end{proposition}

\begin{proof}
Denote by $S(t)$ the analytic semi-group generated by  $\partial_y ^4$
and, for $t>0$,
$$
E_t^\varepsilon:=\mathcal{C}([0,t];H^k)\cap L^2(0,t;H^{k+2})
$$
endowed with the norm
$$
\norm{E_t^\varepsilon}z:=
\norm{L^\infty_t(H^k)}z+\varepsilon^{1/2}
\norm{L^2_t(H^{k})}{\partial_y ^2z}.
$$
We claim that for suitably small $t>0$,
 the operator $\Phi$
defined for $\tau\in[0,t]$ by
\begin{equation} \label{fix}
\Phi(z)(\tau)=S(\varepsilon\tau)z_{0}
+ \int_{0}^{\tau}S(\varepsilon(\tau-s))
\Bigl ( f(s)-i\partial_y  (a(y,s) \partial_y  z(y,s) )
\Bigr )   ds
\end{equation}
has a fixed point in $E_t^\varepsilon$.

Obviously
$\Phi$ maps  $E_t^\varepsilon$ in  $E_t^\varepsilon$.
Indeed, on the one hand, using
standard properties of $S(\tau)$,
the terms pertaining to $z_0$ and $f$ belong
to $E_t^\varepsilon$. On the other hand,
 by virtue of \eqref{tame2}, we have
$$
\norm{H^{k+1}}{a\partial_y  z}
\lesssim \Lp{\infty} a\norm{H^{k+1}}{\partial_y  z}+\Lp{\infty}{\partial_y  z}\norm{H^{k-1}}{\partial_y ^2 a}.
$$
Hence, by standard  computations relying on H\"older inequality,
$$
\norm{L^1_t(H^{k+1})}{a\partial_y  z}\leq
C\Bigl(\varepsilon^{-\frac12}\norm{L^2_t(L^\infty)}a
+\norm{L^1_t(L^\infty)}a+\norm{L^1_t(H^{k-1})}{\partial_y ^2 a}\Bigr)
\norm{E_t^\varepsilon}z.
$$
This entails that  $\partial_y (a\partial_y  z)\in L^1(0,t;H^k)$, hence the
Duhamel term pertaining to $\partial_y (a\partial_y  z)$ also
belongs to $E_t^\varepsilon$.


Therefore,  if $z_2$ and $z_1$ both belong to $E_t^\varepsilon$,
we have
for some constant $C$ depending only on $k$,
\begin{align*}
&\norm{E_t^\varepsilon}{\Phi(z_2)-\Phi(z_1)}\\
&\leq C\norm{L^1_t(H^{k+1})}{a\partial_y (z_2-z_1)},\\
&\leq C\Bigl(\varepsilon^{-\frac12}\norm{L^2_t(L^\infty)}a
+\norm{L^1_t(L^\infty)}a+\norm{L^1_t(H^{k-1})}{\partial_y ^2 a}\Bigr)
\norm{E_t^\varepsilon}{z_2-z_1}.
\end{align*}
Choosing $t$ so small as to satisfy
$$
2C\bigl(\varepsilon^{-\frac12}\norm{L^2_t(L^\infty)}a
+\norm{L^1_t(L^\infty)}a+\norm{L^1_t(H^{k-1})}{\partial_y ^2 a}\bigr)\leq 1,
$$
we conclude that $\Phi$ is a contractive map
so that it has a unique fixed point in $E_t^\varepsilon$.
Whence \eqref{regular} has a unique solution
in  $E_t$ on the time interval $[0,t]$.

Obviously, the above proof may be repeated starting from time $t$.
We end up with a solution on the whole interval $[0,T]$.
Uniqueness stems from estimate \eqref{orderkregularized}.
\end{proof}


\subsubsection*{Step 3. Passing to the limit}
In this part, we are given  a nonnegative smooth function  $\theta$
with support in $[-1,1]$ and such that $\int_\mathbb{R}\theta(y)\,dy=1$,
and a smooth radial function $\chi$
whose Fourier transform is supported in $[-1,1]$.
For $\eta>0$,
we denote $\theta_\eta:=\eta^{-1}\theta(\eta^{-1}\cdot)$
and $\chi_\eta:=\eta^{-1}\chi(\eta^{-1}\cdot)$.

We shall make use repeatedly of the following
two lemmas the proof of which is left to the reader.

\begin{lemma}\label{convolution1}
Let  $X$ be a  Banach space and $u\in L^1_{\rm loc}(\mathbb{R};X)$.
Then $\theta_\eta*u$ tends to $u$ in $L^1_{\rm loc}(\mathbb{R};X)$
when $\eta$ approaches $0$. Besides, for all $q\geq1$, $a<b$ and $\eta>0$,
the function  $\theta_\eta*u$ belongs
to $L^q(a,b;X)$ and we have
$$
\norm{L^q(a,b;X)}{\theta_\eta*u}\leq
\eta^{\frac1q-1}\norm{L^1(a-\eta,b+\eta;X)}u.
$$
\end{lemma}

\begin{lemma}\label{convolution2}
Let $u$ be in $H^k(\mathbb{R};\mathbb{C})$. Then $\chi_\eta*u$ belongs
to $H^\infty(\mathbb{R};\mathbb{C})$ and tends to $u$ in $H^k(\mathbb{R};\mathbb{C})$ when
 $\eta$ approaches zero. Besides, there
exists a constant
$C$ depending only on $p$ and $k$ and such that
$$
\norm{H^p}{\chi_\eta*u}\leq C\eta^{k-p}\norm{H^k}u\quad\text{for }
p\geq k.
$$
If $u\in\mathcal{S}'(\mathbb{R};\mathbb{C})$ satisfies $\partial_y ^2u\in H^{k-2}(\mathbb{R};\mathbb{C})$
then $\chi_\eta*u-u\in H^k(\mathbb{R};\mathbb{C})$  and
$$
\norm{H^p}{\chi_\eta*u-u}\leq C\eta^{k-p}\norm{H^{k-2}}{\partial_y ^2
u}\quad\text{for } p\leq k.
$$
\end{lemma}

For $\varepsilon>0$, mollify the data and the coefficient $a$ by setting
$$
z_{0,\varepsilon}:=\chi_{\eta(\varepsilon)}*_yz_0,\quad
f_\varepsilon:=\chi_{\eta(\varepsilon)}*_yf,\quad
a_\varepsilon:=\theta_{\varepsilon}*_t\bigl(\chi_{\eta(\varepsilon)} *_y a\bigr),
$$
where $\eta$ is a positive function tending to $0$ in $0$
to be chosen hereafter, and with the
convention that $a(y,t)=0$ if $t>T$.

According to lemma \ref{convolution2}
and Sobolev embeddings, we have
$\underline a/2\leq a_\varepsilon(y,t)\leq2\widetilde a$
for small enough $\varepsilon$.
This fact will be used repeatedly in the sequel.

Theorem \ref{thm:regular} provides
a solution $z_\varepsilon\in\mathcal{C}([0,T];H^k)\times L^2(0,T;H^{k+2})$
for the system
\begin{equation}
\begin{gathered}
\partial_t  z_{\varepsilon}+i \partial_y  (a_\varepsilon(y,t) \partial_y  z_{\varepsilon} )
+\varepsilon \partial^4_{y} z_{\varepsilon}
=f_\varepsilon\,,\\
z_{\varepsilon}(0)=z_{0,\varepsilon}.
\end{gathered}
\end{equation}
We claim
that $(z_\varepsilon)_{\varepsilon>0}$ is a {\it Cauchy sequence}
in $\mathcal{C}([0,T];H^k)$.
We first notice that  $(z_\varepsilon)$ is uniformly bounded
in $ \mathcal{C}([0,T];H^k)$. Indeed, combining
lemmas \ref{convolution1} and \ref{convolution2}
with estimate \eqref{orderkregularized} leads to
\begin{equation}\label{estk}
\begin{aligned}
\norm{H^k}{z_\varepsilon(t)}&\leq
C\Bigl(\norm{H^k}{z_0}+\norm{L_t^1(H^k)}{f(\tau)}\Bigr)
\exp\Bigl(C\bigl(\norm{L_{t+\varepsilon}^1(L^\infty)}{\partial_y ^2 a}^2\\
&\quad +\norm{L_{t+\varepsilon}^1(L^\infty)}{\partial_t \log a}
+\norm{L_{t+\varepsilon}^1(H^{k-1})}{\partial_y ^2a}\bigr)\Bigr)
\end{aligned}
\end{equation}
for some $C$ depending only on $k$, $\underline a$ and $\widetilde a$.
Let $0<\delta\leq\varepsilon$ and $\zeta_\varepsilon^\delta:=z_\varepsilon-z_\delta$.
Since $\zeta_\varepsilon^\delta$ satisfies
\begin{gather*}
\partial_t \zeta_\varepsilon^\delta+i\partial_y (a_\delta\partial_y \zeta_\varepsilon^\delta)+\delta\partial_y ^4\zeta_\varepsilon^\delta
=\bigl(f_\varepsilon-f_\delta\bigr)+(\delta-\varepsilon)\partial_y ^4z_\varepsilon
+i\partial_y \bigl((a_\delta-a_\varepsilon)\partial_y  z_\varepsilon\bigr),\\
{\zeta_\varepsilon^\delta}_{\mid t=0}=z_{0,\varepsilon}-z_{0,\delta},
\end{gather*}
inequality \eqref{orderkregularized} and lemmas
\ref{convolution1}, \ref{convolution2} ensure that
\begin{equation}\label{eq:cauchy1}
\begin{aligned}
\norm{H^k}{\zeta_\varepsilon^\delta(t)}
&\leq  C \exp\Bigl(C\bigl(\norm{L_{t+\varepsilon}^1(L^\infty)}{\partial_y ^2a}^2
+\norm{L_{t+\varepsilon}^1(L^\infty)}{\partial_t \log a}
+\norm{L_{t+\varepsilon}^1(H^{k-1})}{\partial_y ^2a}\bigr)\Bigr)
\\
&\quad \times \Bigl(\norm{H^k}{z_{0,\varepsilon}-z_{0,\delta}}
+\norm{L_t^1(H^k)}{f_\varepsilon-f_\delta}
+\varepsilon\norm{L_t^1(H^k)}{\partial_y ^4z_\varepsilon}\\
&\quad +\norm{L_t^1(H^{k+1})}{ (a_\delta-a_\varepsilon)\partial_y  z_\varepsilon}\Bigr)
\end{aligned}
\end{equation}
 for some $C$ depending only on $k$, $\underline a$ and $\widetilde a$.
By  lemmas \ref{convolution1}, \ref{convolution2},
the first two terms
of the right-hand side tend to zero as $\varepsilon$ and $\delta$
approach zero.

Let us admit for a while the following lemma.

\begin{lemma}\label{estk+2}
There exists a constant $C$ depending only on $k$,
$\underline a$ and $\widetilde a$ and such that
\begin{align*}
&\norm{H^{k+2}}{z_\varepsilon(t)}+\varepsilon^{1/2}\norm{L_t^2(H^{k+4})}{z_\varepsilon}\\
&\leq C\frac{1}{\eta(\varepsilon)^{2}} \Bigl(\norm{H^k}{z_0}+\norm{L_t^1(H^k)}{f}+
\norm{L_t^\infty(L^\infty)}{\partial_y  z_\varepsilon}
\norm{L_{t+\varepsilon}^1(H^{k-2})}{\partial_y ^3a}\Bigr)\\
& \times \exp\Bigl(C\bigl(\norm{L_{t+\varepsilon}^1(L^\infty)}{\partial_y ^2 a}^2
+\norm{L_{t+\varepsilon}^1(L^\infty)}{\partial_t \log a}
+\norm{L_{t+\varepsilon}^1(L^\infty)}{\partial_y ^2 a}\bigr)\Bigr).
\end{align*}
\end{lemma}
On the one hand, since we assumed that $k\geq2$, Sobolev embedding combined
with inequality \eqref{estk}
implies that $\partial_y  z_\varepsilon$ is uniformly bounded in $L^\infty(0,T;L^\infty)$.
Lemma \ref{estk+2} then
supplies a constant $C_T\geq0$ such that for all $\varepsilon>0$,
we have
\begin{equation}\label{eq:cauchy2}
\varepsilon\int_0^t\norm{H^k}{\partial_y ^4z_\varepsilon}\,d\tau=
C_T[\eta(\varepsilon)]^{-2}\sqrt\varepsilon. \end{equation}
On the other hand, according to inequality \eqref{tame2},
\begin{align*}
&\norm{L^1_t(H^{k+1})}{(a_\delta-a_\varepsilon)\partial_y  z_\varepsilon}\\
&\lesssim\norm{L^1_t(L^\infty)}{a_\delta-a_\varepsilon}
\norm{L^\infty_t(H^{k+1})}{\partial_y  z_\varepsilon}
+\norm{L_t^\infty(L^\infty)}{\partial_y  z_\varepsilon}
\norm{L_t^1(H^{k-1})}{\partial_y ^2(a_\delta-a_\varepsilon)}.
\end{align*}
Since $\partial_y  z_\varepsilon$ is uniformly bounded in $L^\infty(0,T;L^\infty)$
and $\partial_y ^2 a\in L^1(0,T;H^{k-1})$, the
Lebesgue dominated convergence theorem
 entails that
the last term approaches $0$   when $\varepsilon,\delta\to 0$.

The estimation of
$\norm{L_T^1(L^\infty)}
{a_\delta-a_\varepsilon}$ relies on the identity
$$
a_\delta-a_\varepsilon=\chi_{\eta(\varepsilon)}*_y
\bigl[(\theta_\delta-\theta_\varepsilon)*_ta\bigr]
+\theta_\delta*_t\bigl[(\chi_{\eta(\delta)}-\chi_{\eta(\varepsilon)})
*_ya\bigr].
$$ From lemma \ref{convolution1}, we gather
$$
\norm{L_T^1(L^\infty)}{a_\delta-a_\varepsilon}
\leq\norm{L_T^1(L^\infty)}{(\theta_\delta-\theta_\varepsilon)*_ta}
+\norm{L_T^1(L^\infty)}{(\chi_{\eta(\delta)}-\chi_{\eta(\varepsilon)})
*_ya}.
$$
The first term may be bounded by
$$
\alpha(\varepsilon):=\sup_{0<\delta\leq\delta'\leq\varepsilon}
\norm{L_T^1(L^\infty)}{(\theta_\delta-\theta_{\delta'})*_ta}
$$
which, in view of lemma \ref{convolution1}
is a nondecreasing positive function tending  to $0$ in $0$.

To bound the other term, one can argue by
interpolation and write
\begin{align*}
&\Lp{\infty}{(\chi_{\eta(\delta)}-\chi_{\eta(\varepsilon)})*_y a} \\
&\leq \Lp2{(\chi_{\eta(\delta)}-\chi_{\eta(\varepsilon)})*_y a}^{1/2}
\Lp2{(\chi_{\eta(\delta)}-\chi_{\eta(\varepsilon)})
*_y\partial_y  a}^{1/2},\\
&\leq C\Lp2{(\chi_{\eta(\delta)}-\chi_{\eta(\varepsilon)})
*_ya}^{\frac{2k+1}{2k+2}}\norm{H^{k-1}}
{(\chi_{\eta(\delta)}-\chi_{\eta(\varepsilon)})
*_y\partial_y ^2 a}^{\frac1{2k+2}}.
\end{align*}
Hence, using lemma \ref{convolution2},
$$
\Lp{\infty}{(\chi_{\eta(\delta)}-\chi_{\eta(\varepsilon)})
*_ya}\leq
C[\eta(\varepsilon)]^{k+\frac12}\norm{H^{k-1}}{\partial_y ^2 a}.
$$
Combining this with lemma \ref{estk+2}, we conclude that
for some $C_T\geq0$,
$$
\norm{L^1_t(L^\infty)}{a_\delta-a_\varepsilon}
\norm{L^\infty_t(H^{k+1})}{\partial_y  z_\varepsilon}\leq C_T
\Bigl([\eta(\varepsilon)]^{k-\frac32}
+\alpha(\varepsilon)[\eta(\varepsilon)]^{-2}\Bigr).
$$
Of course, with no loss of generality, one can assume
that $\alpha(\varepsilon)\geq\varepsilon$. Now, choosing
$\eta(\varepsilon)=\alpha(\varepsilon)^{\frac16}$
and coming back to \eqref{eq:cauchy2},
we conclude that $(z_\varepsilon)$ has some limit $z$ in
$\mathcal{C}([0,T];H^k)$ when $\varepsilon$ approaches $0$.
\end{proof}

For the proof of Lemma \ref{estk+2},
the starting point is \eqref{eq:jaugekepsilon} with
$0\leq\ell\leq k+2$.
All the terms of the right-hand side are going
to be bounded like in the proof of \eqref{orderkregularized}
except for
$$
\sum_{j=1}^{\ell-1} {\textstyle \binom{\ell+1}{j-1}}
\Im\int a_\varepsilon^{(\ell+2-j)}a_\varepsilon^{\ell}
z_\varepsilon^{(j)}\overline{z^{(\ell)}_{\varepsilon}}\,{\rm d} y
$$
 that we are now going to estimate for $2\leq\ell\leq k+2$.
By  H\"older inequality and \eqref{tame1}, we have
for some $C$ depending only on $k$, $\underline a$ and $\widetilde a$,
\begin{align*}
&\Big|\int a_\varepsilon^{(\ell+2-j)}a_\varepsilon^{\ell}
z^{(j)}\overline{z^{(\ell)}_{\varepsilon}}\,{\rm d} y\Big| \\
&\leq C\Lp2{a_\varepsilon^{\ell/2}z^{(\ell)}_\varepsilon}
\Bigl(\Lp{\infty}{\partial_y ^2
a_\varepsilon}\norm{H^\ell}{z_\varepsilon}+\Lp{\infty}{\partial_y  z_\varepsilon}\norm{H^{\ell-2}}{\partial_y ^3a_\varepsilon}\Bigr).
\end{align*}
Hence, summing inequalities \eqref{eq:jaugekepsilon}
for $\ell=0,\dots,k+2$ and using the above inequality,
\begin{align*}
&\frac12\frac d{dt}\sum_{\ell=0}^{k+2}
\Lp2{a_\varepsilon^{\ell/2}z^{(\ell)}_\varepsilon}^2+\frac\varepsilon2
\Lp2{\partial_y ^2\bigl(a_\varepsilon^{\ell/2}z^{(\ell)}_\varepsilon\bigr)}^2 \\
&\lesssim \Re\sum_{\ell=0}^{k+2}\int a_\varepsilon^\ell z^{(\ell)}_\varepsilon
f^{(\ell)}_\varepsilon\,{\rm d} y +\Lp{\infty}{\partial_y  z_\varepsilon}\norm{H^k}{\partial_y ^3a_\varepsilon}
\norm{H^{k+2}}{z_\varepsilon} \\
&\quad +\bigl(\Lp{\infty}{\partial_y ^2a_\varepsilon}+\varepsilon\Lp{\infty}{\partial_y ^2a_\varepsilon}^2+\Lp{\infty}{\partial_t \log
a_\varepsilon}\bigr)\norm{H^{k+2}}{z_\varepsilon}^2.
\end{align*}
Since $\partial_y ^2\bigl(a_\varepsilon^{\ell/2}z^{(\ell)}_\varepsilon\bigr)=
\partial_y ^2\bigl(a_\varepsilon^{\ell/2}\bigr)z^{(\ell)}_\varepsilon
+2\partial_y \bigl(a_\varepsilon^{\ell/2}\bigr)\partial_y  z^{(\ell)}_\varepsilon
+a_\varepsilon^{\ell/2}\partial_y ^2z^{(\ell)}_\varepsilon$,
 for some $C$ depending only on $k$, $\underline a$ and $\widetilde a$,
 we clearly have
$$
\norm{H^{k+4}}{z_\varepsilon}^2\leq C\Bigl(\Lp{\infty}{\partial_y ^2 a_\varepsilon}^2\sum_{\ell=0}^{k+2}
\Lp2{a_\varepsilon^{\ell/2}z^{(\ell)}_\varepsilon}^2\;+\;\sum_{\ell=0}^{k+2}
\Lp2{\partial_y ^2\bigl(a_\varepsilon^{\ell/2}z^{(\ell)}_\varepsilon\bigr)}^2\Bigr).
$$
Hence, denoting $Z_\varepsilon^2:=\sum_{\ell=0}^{k+2}
\Lp2{a_\varepsilon^{\ell/2}z^{(\ell)}_\varepsilon}^2$ and $\kappa:=C^{-1}$,
we end up with
\begin{align*}
\frac12\frac d{dt}Z_\varepsilon^2+\frac{\kappa\varepsilon}2\norm{H^{k+4}}{z_\varepsilon}^2
&\lesssim Z_\varepsilon\Bigl(\norm{H^{k+2}}{f_\varepsilon}
+\Lp{\infty}{\partial_y  z_\varepsilon}\norm{H^k}{\partial_y ^3a_\varepsilon}\Bigr) \\
&\quad +\bigl(\varepsilon\Lp{\infty}{\partial_y ^2a_\varepsilon}^2+\Lp{\infty}{\partial_y ^2a_\varepsilon}+\Lp{\infty}{\partial_t \log
a_\varepsilon}\bigr)Z_\varepsilon^2.
\end{align*}
According to lemmas \ref{convolution1} and
\ref{convolution2}, we have
\begin{gather*}
\begin{aligned}
&\int_0^t\Bigl(\varepsilon\Lp{\infty}{\partial_y ^2a_\varepsilon}^2+\Lp{\infty}{\partial_y ^2a_\varepsilon}+\Lp{\infty}{\partial_t \log
a_\varepsilon}\Bigr)\,d\tau \\
&\leq C\left(\int_0^{t+\varepsilon}\Lp{\infty}{\partial_y ^2a}^2\,d\tau\right)^2
+ C\int_0^{t+\varepsilon}\Bigl(\Lp{\infty}{\partial_y ^2a}+\Lp{\infty}{\partial_t \log a}\Bigr)\,d\tau,
\end{aligned}
\\
\norm{L_t^1(H^k)}{\partial_y ^3a_\varepsilon}\leq
C[\eta(\varepsilon)]^{-2}\norm{L_{t+\varepsilon}^1(H^{k-2})}{\partial_y ^3a},
\\
Z_\varepsilon(0)\leq
C[\eta(\varepsilon)]^{-2}\norm{H^k}{z_0},\\
\norm{L_t^1(H^{k+2})}{f_\varepsilon}\leq C[\eta(\varepsilon)]^{-2}\norm{L_t^1(H^{k})}{f},
\end{gather*}
so that  Gronwall lemma yields the desired
inequality.


\section{The Korteweg model}

This section is devoted to the proof of
local well-posedness for  (\ref{eq:1dLaugb}).

\subsection{Uniqueness and continuity with respect to the data}
Let us start with the proof of uniqueness in theorem \ref{existence}.
This is a straightforward corollary of the proposition below.

\begin{proposition}\label{uniqueness}
    Under the assumption {\rm  (H1)} with $k=2$, let
 $(u_1,v_1)$ and $(u_2,v_2)$ be two solutions of $(\ref{eq:1dLaugb})$
on $[0,T]\times\mathbb{R}$ with $v_{1}$ and $v_{2}$ both taking values in
a compact subset $K$ of $J_v$.
Let  $\delta u:=u_2-u_1$ and  $\delta v:=v_2-v_1$. Assume that
$\delta u$ belongs to $\mathcal{C}([0,T];L^2)$,
$\delta v$ belongs to $\mathcal{C}([0,T];H^{1})$
 and that in addition, $\partial_y  u_1\in L^1(0,T;H^1)$, $\partial_{yy}^2 v_1\in L^1(0,T;H^1)$
and $\partial_y  v_1\in L^\infty(\mathbb{R}\times [0,T])$.
Denote $z_j=u_j+iw_j$  with $w_j:=-\alpha(v_j)\partial_y  v_j$ for $j=1,2$, and
$\delta z:= z_2-z_1$.
There exists a constant $C$ depending only
on $K$ and on the functions $\alpha$ and $q$,  such that
$$
\Lp2{\delta v(t)}^2+\Lp2{\delta z(t)}^2\leq \Big(\Lp2{\delta v(0)}^2+\Lp2{\delta z(0)}^2\Big)
{\rm e}^{C\int_0^t(1+\Lp{\infty}{\partial_y  v_1})(1+\norm{H^1}{\partial_y 
z_1})\,d\tau}.
$$
\end{proposition}

\begin{proof}
Taking the difference of the systems
satisfied by $(v_{1},z_{1})$ and $(v_{2}, z_{2})$
we easily compute that
\begin{gather}\label{dv}
\partial_t \delta v=\partial_y \delta u,\\  \label{eq:dz}
\partial_t \delta z+i\partial_y (\alpha(v_2)\partial_y \delta z)=f
\end{gather}
with
\begin{align*}
f&=(q(v_2)-q(v_1))\Im z_1+q(v_2)\Im\delta z \\
&\quad +i\partial_y (\alpha(v_1) -\alpha(v_2)) \partial_y  z_1
+i(\alpha(v_1)-\alpha(v_2)) \partial_{yy}^2 z_1.
\end{align*}
Multiplying (\ref{dv}) by $\delta v$, and integrating
by parts the right-hand side term, we readily get
\begin{equation}\label{uniq1}
\begin{aligned}
\frac12\frac{\,{\rm d}}{ \,{\rm d} t}\Lp2{\delta v(t)}^2
&=-\int\delta u \partial_y \delta v\,{\rm d} y \\
&=\int\frac{\delta w}{\alpha(v_2)}\delta u\,{\rm d} y
+\int w_1\Bigl(\frac1{\alpha(v_2)}-\frac1{\alpha(v_1)}\Bigr)\delta u\,{\rm d} y.
\end{aligned}
\end{equation}
On the other hand, Equation \eqref{eq:dz} is of the form
\eqref{eq:linear} with
$z=\delta z$,
$a=\alpha(v_{2})$
so we can apply the identity (\ref{dtorder0}). This yields
\begin{align*}
\frac12\frac d{ \,{\rm d} t}\Lp2{\delta z(t)}^2
&=\int\Bigl(q(v_2)-q(v_1)\Bigr)w_1\delta u\,{\rm d} y +\int
q(v_2)\delta w \delta u\,{\rm d} y\\
&\quad + \Im\int\partial_y (\alpha(v_2)-\alpha(v_1)) \partial_y  z_1 \overline\delta z\,{\rm d} y\\
&\quad + \Im\int(\alpha(v_2)-\alpha(v_1)) \partial_{yy}^2 z_1 \overline\delta z\,{\rm d} y.
\end{align*}
Adding (\ref{uniq1}) to this equality, we get
\begin{align*}
&\frac12\frac{\,{\rm d}}{ \,{\rm d} t}\Bigl(\Lp2{\delta v(t)}^2
+\Lp2{\delta z(t)}^2\Bigr)\\
&\leq \Lp{\infty}{\frac1{\alpha(v_2)}}\Lp2{\delta w}\Lp2{\delta u}
+\Lp{\infty}{w_1}\Lp2{\frac1{\alpha(v_2)}-\frac1{\alpha(v_1)}}\Lp2{\delta u}
 \\
&\quad +\Lp2{q(v_2)-q(v_1)}\Lp{\infty}{w_1}\Lp2{\delta z} +\Lp{\infty}{q(v_2)}\Lp2{\delta z}^2
\\
&\quad +\Lp2{\partial_y (\alpha(v_2)-\alpha(v_1))}\Lp{\infty}{\partial_y  z_1}\Lp2{\delta z}\\
&\quad +\Lp{\infty}{\alpha(v_2)-\alpha(v_1)}\Lp2{\partial_{yy}^2 z_1}\Lp2
{\delta z}.
\end{align*}
 All terms of the type $\Lp2{F(v_2)-F(v_1)}$ can be bounded
 by the mean value theorem.
Furthermore, we have
$$
\partial_y (\alpha(v_2)-\alpha(v_1))=-\frac{\alpha'(v_2)}{\alpha(v_2)} \delta w
-\biggl(\frac{\alpha'(v_2)}{\alpha(v_2)}-
\frac{\alpha'(v_1)}{\alpha(v_1)}\biggr)w_1
$$
so that
$$
\Lp2{\partial_y (\alpha(v_2)-\alpha(v_1))} \leq C ( \|\delta w\|_{L^2}+
\|w_{1}\|_{L^\infty}\;\|\delta v\|_{L^2} )\,.
$$
Using the Sobolev embedding
$H^1  \hookrightarrow  L^\infty$,
we eventually get the inequality
\begin{align*}
&\frac12\frac{\,{\rm d}}{ \,{\rm d} t}\Bigl(\Lp2{\delta v(t)}^2+\Lp2{\delta z(t)}^2\Bigr)\\
&\leq C\Bigl(\Lp2{\delta v(t)}^2+\Lp2{\delta z(t)}^2\Bigr)
\Bigl(1+\Lp{\infty}{w_1}\Bigr) \Bigl(1+\norm{H^1}{\partial_y  z_1}\Bigr)
\end{align*}
for some constant $C$ depending only on the functions $\alpha$ and $q$,
and on $K$.
Applying Gronwall lemma completes the proof.
\end{proof}

\begin{remark} \rm
Let $\overline z$ be a smooth solution to
$\eqref{eq:1dLaugc}$. Combined with the existence theorem, Proposition $\ref{uniqueness}$
shows the Lipschitz continuity of the mapping
$$
\begin{array}{lll}
\overline z(0)+H^2 & \longrightarrow & \overline z+\mathcal{C}([0,T];L^2)\\
z_0 & \longmapsto & z\quad
\text{solution of \eqref{eq:1dLaugc} with data } z_0\,.
\end{array}
$$
Actually, by adapting the method of
\cite{BS}, the above map may be shown to be
continuous from $\overline z(0)+H^k$ to
$\overline z+\mathcal{C}([0,T];H^k)$.
\end{remark}


\subsection{Existence of a local solution}\label{ss:is}

Our aim here is to prove existence in Theorem \ref{existence}.
Uniqueness under condition (\ref{espace})
is given  by Proposition \ref{uniqueness}.
Remark that it actually holds in a class which is much larger  than
the one defined in \eqref{espace}.
The existence proof proceeds in a classical way through the four
main steps:
\begin{itemize}
\item[(a)] Construction of approximate solutions,
\item[(b)] Uniform a priori estimates in large norm,
\item[(c)] Convergence in small norm,
\item[(d)] Continuity results.
\end{itemize}
To simplify the presentation however, we shall
introduce the  auxiliary  function $\lambda:=\Lambda(v)$
where  $\Lambda$ stands for  a primitive of $\alpha$.

We obviously have $w=-\partial_y \lambda$. Besides, observing that $\Lambda$
is by assumption on $\alpha$ monotonically increasing, we may also
use its reciprocal
$\Lambda^{-1}:J_\lambda\longrightarrow J_v$
and  we  have $v=\Lambda^{-1}(\lambda)$.
Therefore, as far as $v$ is valued in $J_v$,
solving system \eqref{eq:1dL} amounts to solving
\begin{equation}\label{eq:1dLbis}
\begin{gathered}
\partial_t \lambda-\alpha_{\sharp}(\lambda)\partial_y   u = 0\,,\\
\partial_t  z+ i\partial_y \left(\alpha_\sharp(\lambda)\partial_y  z\right)
=q_\sharp(\lambda)w,
\end{gathered}
\end{equation}
with $\alpha_{\sharp}=\alpha\circ \Lambda^{-1}$,
$q_{\sharp}:=q\circ \Lambda^{-1}$, $z=u+iw$,
and under the constraint  $w=-\partial_y \lambda$.

In what follows, we  show that
system (\ref{eq:1dLbis}) has a local solution
$(u,\lambda)$  with $\lambda$ valued in $J_\lambda$ and such that
$(\widetilde u:=u-\overline u,\widetilde \lambda:=\lambda-\overline\lambda)$
belongs to the space
$$
E_T^k:=\mathcal{C}\bigl([0,T];H^k\times H^{k+1}\bigr)\cap
\mathcal{C}^1\bigl([0,T];H^{k-2}\times H^{k-1}\bigr).
$$
According to corollaries \ref{comp3} and \ref{comp4},
this gives theorem \ref{existence}.


\subsubsection*{(a) Construction of approximate solutions}
Our approximate scheme will of course take advantage of the linear
estimates in Proposition \ref{estimateslinear}.
The most natural way of computing the iterate $z^{n+1}$ in terms of
$(\lambda^n,z^n=u^n+iw^n)$ is to consider the equation
\begin{equation}
\label{approx0}\partial_t  z^{n+1}+i \partial_y  (\alpha_\sharp(\lambda^n) \partial_y 
z^{n+1} )=q_\sharp(\lambda^n) w^n\,.
\end{equation}
We remark however that since we did not assume that
the data belong to a Sobolev space,  theorem \ref{thm:lSchr}
does not supply a solution for \eqref{approx0}.
Hence we are going to
work with $\widetilde z^n:=z^n-\overline z$
(with the obvious notation $\overline z=\overline u+i\overline w$ and
$\overline w=-\alpha(\overline v)\partial_y \overline v$)
rather than with $z^n$.

Regarding the computation of $\lambda^{n+1}$,
we must keep in mind that
we expect in the limit that
$w=-\partial_y \lambda$. Therefore it will  be suitable to have also
$w^n=-\partial_y \lambda^n$ for all $n\in\mathbb{N}$.
This induces us to set
$$
\lambda^{n+1}(y,t):=\lambda_0(y)+\int_0^t
\Bigl(\alpha_\sharp(\lambda^n)\partial_y u^{n+1}\Bigr)(y,\tau) d\tau
\quad\text{where}\quad\lambda_0(y):=\Lambda(v_0(y)).
$$
Indeed, the term $\alpha_\sharp(\lambda^n)\partial_y u^{n+1}$ will be
continuous in $y$ and $t$ (by Sobolev embeddings)
so that differentiating the above inequality with respect to $t$
yields
\begin{equation}\label{iterative0}
\partial_t \lambda^{n+1}=\alpha_\sharp(\lambda^n)\partial_y u^{n+1}
\end{equation}
in the classical sense.
Then, differentiating with respect to $y$, we get
 $$
\partial_t\partial_y \lambda^{n+1}=\partial_y \partial_t   \lambda^{n+1}=
\partial_y\Bigl(\alpha_\sharp(\lambda^n)\partial_y u^{n+1}\Bigr)
=-\partial_t w^{n+1}
$$
in the weak sense -- in fact, this equality will be true
in $H^{k-2}$ because $\partial_y  \lambda^{n+1}+w^{n+1}$
will belong to  $\mathcal{C}^1([0,T];H^{k-2})$.
So, if the initial data for $w^{n+1}$ is chosen so that
$w^{n+1}(y,0)=w_0=-\alpha(v_0)\partial_y  v_0=-\partial_y \lambda_0$,  we shall have
the identity
\begin{equation}\label{iterative1}
\partial_y  \lambda^{n+1} +w^{n+1} \equiv 0.
\end{equation}


Finally, our scheme is as follows.
For the first term $(\lambda^0,z^{0}=u^{0}+iw^{0})$ of the sequence, we just
set for all $(y,t)\in\mathbb{R}^2$,
$$
\lambda^0(y,t):=\lambda_0(y)\,,\;
u^0(y,t):=u_0(y)\quad\text{and}\quad w^0(y,t):=w_0:=-\partial_y \lambda_0(y,t).
$$
Obviously  $\lambda^0$ is valued in $J_\lambda$   and
$(\widetilde u^0:=u^0-\overline u^0,\widetilde\lambda^0:=\lambda^0-\overline\lambda^0)$
belongs to $ \cap_{T>0}E_T^k$.

 Then we define $(\lambda^n,z^{n}=u^{n}+iw^{n})$ inductively in the
following way. Suppose $(\lambda^n,z^{n})$ has been defined in such a way that
$\lambda^n$ is valued in $J_\lambda$,
$w^n=-\partial_y \lambda^n$ and,  for some $T>0$,   $(\widetilde u^n:=u^n-\overline
u,\widetilde\lambda^n:=\lambda^{n}-\overline\lambda)$ belongs to  $E_T^k$.
  Then inequality \eqref{tame2} and corollaries \ref{comp2}
and \ref{comp4} insure that
the right-hand side of the first line of the following system
 \begin{equation}\label{approx}
\begin{gathered}
\partial_t  \widetilde z^{n+1}+i\partial_y (\alpha_\sharp(\lambda^n)\partial_y 
\widetilde z^{n+1})=F^n_1+F^n_2+F^n_3+F^n_4,\\
 \widetilde z^n_{|t=0}=\widetilde u_0+i\widetilde w_0,
 \end{gathered}
\end{equation}
with
\begin{gather*}
F^n_1:=i\partial_{yy}^2\overline z\bigl(\alpha_\sharp(\overline\lambda)-\alpha_\sharp(\lambda^n)
\bigr), \quad
F^n_2:=i\partial_y \overline\lambda \partial_y \overline
z\bigl(\alpha'_\sharp(\overline\lambda)-\alpha'_\sharp(\lambda^n)\bigr),\\
F^n_3:=\bigl(q_\sharp(\lambda^n)-q_\sharp(\overline\lambda)\bigr)\overline w, \quad
F^n_4:=\bigl(q_\sharp(\lambda^n)
+i\partial_y \overline z \alpha'_\sharp(\lambda^n)\bigr)\widetilde w^n,
\end{gather*}
belongs to $\mathcal{C}([0,T];H^k)\cap \mathcal{C}^1([0,T];H^{k-2})$.

Hence Theorem \ref{thm:lSchrreg} ensures that system \eqref{approx} has
 a unique solution
$\widetilde z^{n+1}$ in $\mathcal{C}([0,T];H^k)\cap
\mathcal{C}^1([0,T];H^{k-2})$.
  Then we set
$\widetilde u^{n+1}:=\Re \widetilde z^{n+1}$,
$u^{n+1}:=\widetilde u^{n+1}+\overline u$,
$\widetilde w^{n+1}:=\Im\widetilde z^{n+1}$,
$w^{n+1}:=\widetilde w^{n+1}+\overline w$  and
$z^{n+1}:=u^{n+1}+iw^{n+1}$ so that
$z^{n+1}$ satisfies \eqref{approx0} as required.
Finally, we set
$$
\lambda^{n+1}(t):=\lambda_0+\int_0^t\alpha_\sharp(\lambda^n)\partial_y
u^{n+1}\,d\tau.
$$
Of course we have to check
that $\lambda^{n+1}$ is valued in $J_\lambda$.
This will be the case for small enough time.


\subsubsection*{(b) Uniform a priori estimates in large norm}
Let  $K_\lambda$ be
a compact subset of $J_\lambda$ containing $\lambda_0(\mathbb{R})$
and such that
$\delta:=d(\lambda_0(\mathbb{R}),{}^cK_\lambda)>0$.
Throughout this section, we shall denote by
$C$ a ``constant''
-- which may change from line to line -- depending only
 on $k$, $q$, $\alpha,$ $\delta$ and $K_\lambda$.
Let $\alpha^n:=\alpha_\sharp(\lambda^n)$,
$ \alpha_0=\alpha_\sharp(\lambda_0)$,
\begin{gather*}
\widetilde X^0(t):=\widetilde X_0:=\Big(\sum_{j=0}^k\Lp2{\alpha_0^{\frac j2}
\partial_y ^j\widetilde z_0}^2\Big)^{1/2},  \\
\widetilde X^{n}(t):=\Big(\sum_{j=0}^k\Lp2{\bigl((\alpha^{n-1})^{\frac j2}
\partial_y ^j\widetilde z^{n}\bigr)(t)}^2\Big)^{1/2}
\end{gather*}
for $n\geq1$. Further define
$ \widetilde Y^n(t):=\Lp2{\widetilde\lambda^n(t)}+\widetilde X^n(t)$
and $\widetilde Y_0:=\Lp2{\widetilde\lambda_0}+\widetilde X_0$.
\smallbreak
We introduce the induction hypothesis
\begin{itemize}
\item[(H$_{n}^T$)]
$\ \widetilde z^n\in \mathcal{C}([0,T];H^k)\cap \mathcal{C}^1([0,T];H^{k-2}) $,
$\widetilde\lambda^n\in \mathcal{C}([0,T];H^{k+1})$ and
$\lambda^n(\mathbb{R}\times[0,T])\subset K_\lambda$
with the inequality
$$
\widetilde Y^n(t)\leq \frac{\widetilde
Y_0{\rm e}^{2C_0\int_0^t\overline Z(\tau)\,d\tau}}{1-\widetilde Y_0
\int_0^t{\rm e}^{2C_0\int_0^\tau\overline Z(\tau')\,d\tau'}\,d\tau}
$$
for some constant
$C_0\geq0$ depending only on  $k$, $\alpha$, $q$
 and $K_\lambda$  to be determined
afterwards and
$$
\overline Z:=\bigl(1+\norm{H^{k+1}}{\partial_y \overline u}+\norm{H^{k+2}}{\overline w}\bigr)^3.
$$
  \end{itemize}
We are going to show
there is a positive time $T>0$
such that the scheme described in \S {\rm a)} yields a
sequence $(\lambda^n,z^n)$ satisfying (H$_{n}^T$) for all $n\in\mathbb{N}$.


Obviously (H$_{0}^T$)  is satisfied for all $T$
and any $C_0\geq0$.
Now we fix $n\in\mathbb{N}$ and assume that (H$_{p}^T$) is true for all
$p\leq n$.
 According to Proposition
\ref{estimateslinear}, we have that for all $t\in\mathbb{R}^+$,
\begin{equation}\label{iterative1a}
\widetilde X^{n+1}(t)\leq \mbox{e}^{A_k^n(t)}\Big(\widetilde X_0
+\int_0^t\mbox{e}^{-A_k^n(\tau)}h^n(\tau)\,d\tau\Big),
\end{equation}
with
\begin{gather*}
A_k^n(t):=C\;\int_0^t\Bigl(
\Lp{\infty}{\partial_t \log\alpha^n(\tau)}+
\norm{H^k}{\partial_y \alpha^n(\tau)}\Bigr)\,d\tau,\\
h^n(t)=C\sum_{j=1}^4\norm{H^k}{F^n_j}.
\end{gather*}
We need a bound for $A_k^n(t)$.
For $n=0$, the first term $\partial_t \log\alpha^0(\tau)$ is zero.
Otherwise, for $n\geq1$, we have
by definition of $v^n$ and $\alpha^n$ and according to (\ref{iterative0}),
$$
\partial_t \log\alpha^n=[\log\alpha_{\sharp}]'(\lambda^n)
\alpha_\sharp(\lambda^{n-1})\partial_y 
u^n=\frac{\alpha'_\sharp(\lambda^n)}{\alpha_\sharp(\lambda^n)}
\;\alpha_\sharp(\lambda^{n-1})\partial_y 
u^n\,.
$$
Since $\Lp{\infty}{\partial_y  \widetilde u^n}\lesssim \norm{H^k}{\widetilde u^n}$,
 using (H$_{n-1}^T$) and (H$_{n}^T$), we have
that for all $t\in[0,T]$,
$$
\Lp{\infty}{\partial_t \log\alpha^n}\leq C\bigl(\widetilde X^n(t) +\Lp{\infty}{\partial_y \overline u(t)}\bigr)
$$
for some $C=C(k,\alpha,\underline\lambda,\widetilde\lambda)$.
The second term in $A_k^n(t)$,
$\partial_y \alpha^n=\alpha'_\sharp(\lambda^n)\partial_y  \lambda^n$ can be bounded
using inequality \eqref{comp2b} in the appendix, which yields
$$
\norm{H^k}{\partial_y \alpha^n} \leq C
\norm{H^k}{w^n}\sum_{j=0}^k\Lp{\infty}{\lambda^n}^j\Lp{\infty}{\alpha_\sharp^{(j+1)}
(\lambda^n)}
\leq C\norm{H^k}{w^n}.
 $$
Hence, we eventually  obtain
\begin{equation}\label{iterative3}
\forall t\in[0,T],\;A_k^n(t)\leq C\int_0^t
\Bigl(\widetilde X^n(\tau)+\Lp{\infty}{\partial_y \overline u(\tau)}+\norm{H^k}{\overline w}\Bigr)\,d\tau.
\end{equation}
Let us now bound the source term $h^n(t)$.
Using the fact that $H^k$ is an algebra and corollary \ref{comp4}, we get
\begin{align*}
\norm{H^k}{F_1^n}&\lesssim \norm{H^k}{\partial_{yy}^2\overline
z}\norm{H^k}{\alpha_\sharp(\overline\lambda)-\alpha_\sharp(\lambda^n)},\\
&\lesssim \norm{H^k}{\partial_{yy}^2\overline
z}\bigl(\norm{H^k}{\widetilde\lambda^n}+\norm{L^\infty}{\widetilde\lambda^n}(\norm{H^{k-1}}{\overline w}
+\norm{H^{k-1}}{\widetilde w^n})\bigr).
\end{align*}
Hence, as $H^k\hookrightarrow L^\infty$,
\begin{equation}\label{F1n}
\norm{H^k}{F_1^n}\lesssim\norm{H^k}{\partial_{yy}^2\overline z}
\bigl(1+\norm{H^{k-1}}{\overline w}\bigr)\norm{H^k}{\widetilde\lambda^n}.
\end{equation}
 From similar computations, we get
\begin{gather}\label{F2n}
\norm{H^k}{F_2^n}\lesssim\norm{H^k}{\partial_y \overline z}\norm{H^k}{\overline w}
\bigl(1+\norm{H^{k-1}}{\overline
w}\bigr)\norm{H^k}{\widetilde\lambda^n},\\
\label{F3n}
\norm{H^k}{F_3^n}\lesssim\norm{H^k}{\overline w}
\bigl(1+\norm{H^{k-1}}{\overline w}\bigr)\norm{H^k}{\widetilde\lambda^n}.
\end{gather}
We remark that here  $\alpha_\sharp$
(resp. $q_\sharp$) has $k+2$ (resp. $k+1$) locally bounded derivatives.
To bound $F_4^n$, we make use of inequality \eqref{comp2c}
which yields
\begin{align*}
\norm{H^k}{q_\sharp(\lambda^n)\widetilde w^n}
&\lesssim \norm{H^k}{\widetilde w^n}+\norm{L^\infty}{\widetilde\lambda^n}\norm{H^k}{w^n},
\\
&\lesssim \norm{H^k}{\widetilde w^n}+\norm{H^k}{\overline w}\norm{H^{k+1}}{\widetilde\lambda^n}.
\end{align*}
Similar computations enable us to handle the term
$\alpha'_\sharp(\lambda^n)\widetilde w^n$
so that we end up with
\begin{equation}\label{F4n}
\norm{H^k}{F_4^n}\lesssim\bigl(1+\norm{H^k}{\partial_y \overline z}\bigr)
\bigl(\norm{H^k}{\widetilde w^n}
+\norm{H^k}{\overline w}\norm{H^{k+1}}{\widetilde\lambda^n}\bigr).
\end{equation}
Plugging inequalities (\ref{iterative3}), \eqref{F1n},   \eqref{F2n},
 \eqref{F3n} and \eqref{F4n} in  (\ref{iterative1a}), we conclude that
\begin{equation}\label{iterative4}
\begin{aligned}
\widetilde X^{n+1}(t)
&\leq \exp\Big(C\int_0^t\bigl(\widetilde X^n+\Lp{\infty}{\partial_y \overline u}
+\norm{H^k}{\overline w}\bigr)\,d\tau\Big) \\
&\quad\times \biggl(\widetilde X_0 +C\int_0^t{\rm e}^{-C\int_0^\tau
\bigl(\widetilde X^n+\Lp{\infty}{\partial_y \overline u}+\norm{H^k}{\overline w}\bigr)\,d\tau'}
\\
&\quad\times
\Bigl(\widetilde X^n+\bigl(1+\norm{H^{k-1}}{\overline w}\bigr)
\bigl(\norm{H^k}{\overline w}\bigl(1+\norm{H^{k-1}}{\overline w} \\
&\quad +\norm{H^k}{\partial_y \overline z}\bigr)
+\norm{H^k}{\partial_{yy}^2\overline z}\bigr)\widetilde Y^n\Bigr)\,d\tau\biggr).
\end{aligned}
\end{equation}
To complete the estimates, we still have
to find a bound for $\Lp2{\widetilde\lambda^n}$.
This is quite straightforward. Indeed, we have
$$
\widetilde\lambda^{n+1}(t)=\widetilde\lambda_0+\int_0^t
\Bigl(\bigl(\alpha_\sharp(\lambda^n)-\alpha_\sharp(\overline\lambda)\bigr)
\partial_y \overline u +\alpha_\sharp(\lambda^n)\partial_y \widetilde u^{n+1}\Bigr)\,d\tau,
$$
whence
$$
\|\widetilde\lambda^{n+1}(t)\|_{L^2}\leq \|{\widetilde\lambda_0}\|_{L^2}
+C\int_0^t\Bigl(
\Lp{\infty}{\partial_y \overline u}\|{\widetilde\lambda^n}\|_{L^2}
+\Lp2{\partial_y \widetilde u^{n+1}}\Bigr)\,d\tau.
$$
Adding this latter inequality to \eqref{iterative4}, we get
\begin{align*}
&\widetilde Y^{n+1}(t)\\
&\leq {\rm e}^{C\int_0^t\bigl(\widetilde Y^n+\overline Z\bigr)\,d\tau}
\biggl(\widetilde Y_0
+C\int_0^t{\rm e}^{-C\int_0^\tau
\bigl(\widetilde Y^n+\overline Z\bigr)\,d\tau'}
\overline Z\widetilde Y^{n}\,d\tau\biggr)
+C\int_0^t\widetilde Y^{n+1}(\tau)\,d\tau,
\end{align*}
with $\overline Z:=\bigl(1+\norm{H^{k+1}}{\partial_y \overline u}
+\norm{H^{k+2}}{\overline w}\bigr)^3$.

Now,  applying Gronwall lemma, we get, up to a change of $C$,
\begin{equation}\label{iterative5}
\widetilde Y^{n+1}(t)\leq
{\rm e}^{C\int_0^t\bigl(\widetilde Y^n+\overline Z\bigr)\,d\tau}
\biggl(\widetilde Y_0
+C\int_0^t{\rm e}^{-C\int_0^\tau
\bigl(\widetilde Y^n+\overline Z\bigr)\,d\tau'}
\overline Z\widetilde Y^{n}\,d\tau\biggr).
\end{equation}
 We choose for $C_0$ the constant $C$ appearing in
the above inequality (note that this choice is {\it independent}
of $n$) and we assume that $T$ satisfies
\begin{equation}
\label{time}
C_0\widetilde Y_0\int_0^T{\rm e}^{2C_0\int_0^t\overline Z(\tau)\,d\tau}\,dt<1.
\end{equation}
Taking advantage of
 (H$_{n}^T$), straightforward calculations yield
for $0\leq s\leq t\leq T$,
$$
{\rm e}^{C_0\int_s^t\widetilde Y^n(\tau)\,d\tau}\leq
\frac{1-C_0\widetilde Y_0\int_0^s
{\rm e}^{2C_0\int_0^{\tau}\overline Z\,d\tau'}\,d\tau}
{1-C_0\widetilde Y_0\int_0^t
{\rm e}^{2C_0\int_0^{\tau}\overline Z\,d\tau'}\,d\tau}.
$$
Inserting the above inequality in \eqref{iterative5}, we get
$$
\widetilde Y^{n+1}(t)\leq
\frac{\widetilde Y_0{\rm e}^{C_0\int_0^{\tau}\overline Z\,d\tau'}\,d\tau}
{1-C_0\widetilde Y_0\int_0^t
{\rm e}^{2C_0\int_0^{\tau}\overline Z\,d\tau'}\,d\tau}
+C_0\widetilde Y_0\int_0^t
\frac{\overline Z {\rm e}^{C_0\int_s^t\overline Z\,d\tau'}
{\rm e}^{2C_0\int_0^{s}\overline Z\,d\tau'}}
{1-C_0\widetilde Y_0\int_0^t
{\rm e}^{2C_0\int_0^{\tau}\overline Z\,d\tau'}\,d\tau}\;ds,
$$
whence
$$
\widetilde Y^{n+1}(t)\leq
\frac{\widetilde Y_0{\rm e}^{2C_0\int_0^{\tau}\overline Z\,d\tau'}\,d\tau}
{1-C_0\widetilde Y_0\int_0^t
{\rm e}^{2C_0\int_0^{\tau}\overline Z\,d\tau'}\,d\tau}
$$
as required.
With our definition of
$\lambda^{n+1}$, we have (up to a change of $C_0$)
\begin{align*}
\norm{L^\infty}{\widetilde\lambda^{n+1}(t)-\widetilde\lambda_0}
&\leq C_0\int_0^t\bigl(\|\widetilde\lambda^n\|_{L^\infty}\Lp{\infty}{\partial_y \overline  u}
+\norm{L^\infty}{\partial_y \widetilde u^{n+1}}\bigr)\,ds,\\
&\leq C_0\int_0^t\bigl(1+\Lp{\infty}{\partial_y \overline u}\bigr)
\bigl(\widetilde Y^ n+\widetilde Y^{n+1}\bigr)\,ds,\\
&\leq \int_0^t\frac{C_0\overline Z\widetilde Y_0{\rm e}^{2C_0\int_0^s\overline Z\,d\tau'}}
{1-C_0\widetilde Y_0\int_0^s
{\rm e}^{2C_0\int_0^{\tau}\overline Z\,d\tau'}\,d\tau}\,ds,\\
&\leq
-\sup_{\tau\in[0,t]}\overline Z(\tau)\;\log\biggl({1-C_0\widetilde Y_0\int_0^t
{\rm e}^{2C_0\int_0^{\tau}\overline Z\,d\tau'}\,d\tau}\biggr).
\end{align*}
Therefore, the condition
\begin{equation}\label{time1}
\int_0^T{\rm e}^{2C_0\int_0^t\overline Z\,d\tau}\,dt\leq
\frac{1-exp(-\delta/\sup_{t\in[0,T]}\overline Z(t))}{C_0\widetilde Y_0}
\end{equation}
which is a stronger condition than \eqref{time} ensures that
$\lambda^{n+1}\subset K_\lambda$ on $[0,T]$.

\subsubsection*{(c) Convergence in small norm}
We aim at  proving
that $(z^n)_{n\in\mathbb{N}}$ is convergent in the affine space
$\overline z+\mathcal{C}([0,T];L^2)$ and that
$(\lambda^n)_{n\in\mathbb{N}}$ is convergent in
the affine space $\overline\lambda+\mathcal{C}([0,T];L^2)$.

Let $\delta z^n:=z^{n+1}-z^n$, $\delta u^n:=u^{n+1}-u^n$,
$\delta w^n:=w^{n+1}-w^n$
and $\delta\lambda^n:=\lambda^{n+1}-\lambda^n$.
On the one hand, we have
\begin{align*}
&\partial_t \delta z^n+i\partial_y (\alpha_{\sharp}(\lambda^n)\partial_y \delta z^n)\\
&= -i\partial_y  z^n\partial_y (\alpha_{\sharp}(\lambda^n)-\alpha_{\sharp}(\lambda^{n-1}))
-i\partial_{yy}^2 z^n(\alpha_{\sharp}(\lambda^n)-\alpha_{\sharp}(\lambda^{n-1})) \\
&\quad +q_{\sharp}(\lambda^n)\delta w^{n-1}+
(q_{\sharp}(\lambda^n)-q_{\sharp}(\lambda^{n-1}))w^{n-1}.
\end{align*}
Hence, using the basic $L^{2}$ estimate
\begin{align*}
\frac12\frac{\,{\rm d}}{ \,{\rm d} t}\Lp2{\delta z^n}^2
&\leq \Lp2{\delta z^n}\Bigl(\Lp{\infty}{\partial_y  z^n}\Lp2{\partial_y (\alpha_{\sharp}
(\lambda^n)-\alpha_{\sharp}(\lambda^{n-1}))} \\
&\quad +\Lp2{\partial_{yy}^2 z^n}\Lp{\infty}{\alpha_{\sharp}(\lambda^n)-\alpha_{\sharp}(\lambda^{n-1})}
+\Lp{\infty}{q_{\sharp}(\lambda^n)}\Lp2{\delta w^{n-1}}\\
&\quad +\Lp2{q_{\sharp}(\lambda^n)
-q_{\sharp}(\lambda^{n-1})}\Lp{\infty}{w^{n-1}}\Bigr).
\end{align*}
By the Sobolev embedding $H^1\hookrightarrow L^\infty$,  this implies
\begin{equation}\label{cauchy1}
\begin{aligned}
&\frac12\frac{\,{\rm d}}{ \,{\rm d} t}\Lp2{\delta z^n}^2\\
&\leq \Lp2{\delta z^n} \Bigl(\norm{H^1}{\partial_y 
z^n}\norm{H^1}{\alpha_{\sharp}(\lambda^n)-\alpha_{\sharp}(\lambda^{n-1})}
\\
&\quad +\Lp{\infty}{q_{\sharp}(\lambda^n)}\Lp2{\delta w^{n-1}}
+\Lp2{q_{\sharp}(\lambda^n)-q_{\sharp}(\lambda^{n-1})}\Lp{\infty}{w^{n-1}}
\Bigr)\,.
\end{aligned}
\end{equation}
On the other hand, by \eqref{iterative0}, we have
$$
\partial_t  \delta\lambda^n=(\alpha_{\sharp}(\lambda^n)-\alpha_{\sharp}(\lambda^{n-1}))\partial_y 
u^{n+1}+\alpha_{\sharp}(\lambda^{n-1})\partial_y \delta u^n
 $$
so that taking the $L^2$-scalar product with  $\delta\lambda^n$ and integrating by parts
in the last term,
\begin{align*}
\frac12\frac
d{ \,{\rm d} t}\Lp2{\delta\lambda^n}^2
&= \int\Bigl((\alpha_{\sharp}(\lambda^n)
-\alpha_{\sharp}(\lambda^{n-1}))\partial_y  u^{n+1}\delta\lambda^n \\
&\quad -\delta u^n\alpha_{\sharp}'(\lambda^{n-1})\partial_y  \lambda^{n-1}\delta\lambda^n
-\delta u^n\alpha_{\sharp}(\lambda^{n-1})\partial_y \delta\lambda^n\Bigr)\,{\rm d} y.
\end{align*}
Now, as $\partial_y  \lambda^{n-1}=-w^{n-1}$,
straightforward computations yield
\begin{equation}\label{cauchy2}
\begin{aligned}
\frac12\frac{\,{\rm d}}{ \,{\rm d} t}\Lp2{\delta\lambda^n}^2
&\leq     \Lp2{\alpha_{\sharp}(\lambda^n)-\alpha_{\sharp}(\lambda^{n-1})}
\Lp{\infty}{\partial_y  u^{n+1}}\Lp2{\delta\lambda^n} \\
&\quad +\Lp2{\delta u^n}\Lp{\infty}{\alpha_{\sharp}'(\lambda^{n-1})}\Lp{\infty}{w^{n-1}}
\Lp2{\delta\lambda^n} \\
&\quad +\Lp2{\delta u^n}\Lp{\infty}{\alpha_{\sharp}(\lambda^{n-1})}\Lp2{\delta w^n}.
\end{aligned}
\end{equation}
Since the sequence $(\lambda^n)_{n\in\mathbb{N}}$ is bounded
in $L^\infty(\mathbb{R}\times 0,T)$, corollary \ref{comp4} ensures that
there is a constant $C$
such that
\begin{gather*}
\max\left(\Lp2{\alpha_{\sharp}(\lambda^n)-\alpha_{\sharp}(\lambda^{n-1})},
\Lp2{q_{\sharp}(\lambda^n)-
q_{\sharp}(\lambda^{n-1})}\right)\leq
C\Lp2{\delta\lambda^{n-1}},\\
\\
\norm{H^1}{\alpha_{\sharp}(\lambda^n)-\alpha_{\sharp}(\lambda^{n-1})}
\leq C\Bigl(\Lp2{w^{n-1}}+\Lp2{w^n}\Bigr)
\Bigl(\Lp2{\delta\lambda^{n-1}}+\Lp2{\delta w^{n-1}}\Bigr).
\end{gather*}
Hence, adding (\ref{cauchy1}) and (\ref{cauchy2}) yields
\begin{align*}
\frac12\frac{\,{\rm d}}{ \,{\rm d} t}(Y^n)^2
&\leq CY^n\Bigl((1+\norm{H^1}{\partial_y 
z^n}+\Lp{\infty}{\partial_y  u^{n+1}}+\Lp{\infty}{w^{n-1}}\\
&\quad +\Lp2{w^{n-1}}+\Lp2{w^n})Y^{n-1}
 +(1+\Lp{\infty}{w^{n-1}})Y^n\Bigr)
\end{align*}
with
$\ Y^n(t):=\big(\Lp2{\delta\lambda^n}^2+\Lp2{\delta z^n}^2)\big)^{1/2}$.

Because $(\partial_y  u^n)_{n\in\mathbb{N}}$
is bounded in $L^\infty(0,T;H^1)$ and  $(w^n)_{n\in\mathbb{N}}$
is bounded in $L^\infty(0,T;H^2)$,  we eventually  get
$$
Y^n(t)\leq C_1\int_0^tY^{n-1}(\tau)\,d\tau+C_2\int_0^tY^n(\tau)\,d\tau,
$$
whence
$$
\mbox{e}^{-C_2t}Y^n(t)\leq C_1\int_0^t\mbox{e}^{-C_2\tau}
Y^{n-1}(\tau)\,d\tau.
$$
A standard induction argument enables us to conclude that
$$
Y^n(t)\leq \frac{C_1^n}{n!}\mbox{e}^{C_2t}Y^0(t)\,.
$$
The series $\sum_{n} C_1^n / n!$ being convergent, this
shows that
$(\widetilde z^n)_{n\in\mathbb{N}}$ and $(\widetilde\lambda^n)_{n\in\mathbb{N}}$ are Cauchy sequences
in the Banach space $\mathcal{C}([0,T];L^2)$.
We conclude that $(z^n)_{n\in\mathbb{N}}$
tends to some function $z$ in $\overline z+\mathcal{C}([0,T];L^2)$,
and that $(\lambda^n)_{n\in\mathbb{N}}$ tends to some function $\lambda$ in
$\overline\lambda+\mathcal{C}([0,T];L^2)$. Passing to the limit in the linear
equation \eqref{iterative1} we readily get
$$
\partial_y  \lambda =-w\,.
$$
Furthermore, we have in the limit
$\lambda\in J_\lambda$. So we can
define $v:=\Lambda^{-1}(\lambda)$, and we have
$(v-\overline v)\in\mathcal{C}([0,T];L^2)$.
Now, by using the uniform bounds of \S b), we have in addition
\begin{equation}\label{reg}
\widetilde z:=z-\overline z\in L^\infty(0,T;H^k)\cap {\rm Lip} (0,T;H^{k-2}).
\end{equation}
An interpolation argument shows that for $\varepsilon>0$,
$\widetilde z^n\to \widetilde z$ in $\mathcal{C}([0,T];H^{k-\varepsilon})$ and
$v^n\to v$ in $\overline v+\mathcal{C}([0,T];H^{k+1-\varepsilon})$.
This suffices to
show that $w=-\partial_y  \Lambda(v)=-\alpha(v)\partial_y  v$ and to
 pass to the limit in \eqref{iterative0} and \eqref{approx}, thus
 obtaining
\begin{gather*}
\partial_t  \Lambda(v)=\alpha(v) \partial_y  u \\
\partial_t  z+i\partial_y (\alpha(v)\partial_y  z)=q(v)w\,,\\
z_{|t=0}=v_0+iw_0,
\end{gather*}
Simplifying by $\alpha(v)$ in the first equation,
we get
$\partial_t  v=\partial_y  u$.


\subsubsection*{(d) Continuity results}
Using (\ref{reg}) and  the
fact that both $z$ and $\widetilde z$
are solutions of \eqref{eq:1dLaugc},
it can be easily shown that
$\partial_t \widetilde z+i\partial_y (\alpha(v)\partial_y \widetilde z)$
belongs to $L^\infty(0,T;H^k)$
(see equation \ref{eq:perturbee} below).
Hence  Theorem \ref{thm:lSchrreg} ensures that
 $\widetilde z$ belongs to  $\mathcal{C}([0,T];H^k)$.
This new result implies that
$\partial_t \widetilde z\in \mathcal{C}([0,T];H^{k-2})$.
Therefore $\widetilde z$ also belongs to
$\mathcal{C}^1([0,T];H^{k-2})$.
 Now, since
$$
\partial_t \widetilde\lambda=\bigl(\alpha_\sharp(\lambda)
-\alpha_\sharp(\overline\lambda)\bigr)\partial_y  \overline u
+\alpha_\sharp(\lambda)\partial_y \widetilde u,
$$
 Corollaries  \ref{comp2} and \ref{comp4} guarantee
that the right-hand side above
belongs to the space $\mathcal{C}([0,T];H^{k-1})$ so  that
$\widetilde\lambda\in\mathcal{C}^1([0,T];H^{k-1})$.
Since moreover $\partial_y \lambda=-w$ and $w\in \mathcal{C}([0,T];H^k)$,
we thus have $\lambda\in\mathcal{C}([0,T];\overline\lambda+H^{k+1})$.
Applying corollary \ref{comp4}, we conclude that
$$
v\in\overline v+\mathcal{C}([0,T];H^{k+1})\cap
\mathcal{C}^1([0,T];H^{k-1}).
$$

\begin{remark} \rm
The lower bound for the existence time
given by \eqref{time1} depends
on  $k$, $\underline b$, $\widetilde b$, $\alpha$, $q$,  $\norm{H^k}{u_0}$
and $\norm{H^k}{\partial_y  v_0}$. In the next subsection,
we shall see that for $H^k$ data ($k\geq2)$, the time of existence
in $H^k$ is the same as in $H^2$. Hence inequality \eqref{time1}
with $k=2$ provides a lower bound.
This proves remark \ref{three}.
\end{remark}


\subsection{Continuation results and life span}\label{blowupcriterion}

This section is devoted to the proof of a continuation
 criterion for $H^k$ solutions to (\ref{eq:1dL}).
Let us first explain what we mean by an $H^k$ solution.

\begin{definition}  \rm
Under assumption {\rm  (H1)} with   $k\geq2$,  assume that $(u,v)$
is a couple of functions of $(y,t)\in\mathbb{R}\times[0,T)$ such that
$v$ is valued in $J_v$.
We shall say that $(u,v)$ is a $H^k$ solution
of $\eqref{eq:1dL}$   on the time interval $[0,T)$
if  $(u,v)$
satisfies $\eqref{eq:1dL}$ on $\mathbb{R}\times[0,T)$ (in the weak sense) and
\begin{equation}\label{smooth}
(u-\overline u,v-\overline v)\in\mathcal{C}([0,T);H^k\times
H^{k+1})\cap \mathcal{C}^1([0,T);H^{k-2}\times H^{k-1})
\end{equation}
where $(\overline u,\overline v)$ stands for a classical solution of
 $\eqref{eq:1dL}$ on $\mathbb{R}\times\mathbb{R}$ such that
for all $(y,t)\in\mathbb{R}^2$,
$$
\overline v(y,t)\in J_v,\quad
\partial_y \overline u\in \mathcal{C}(\mathbb{R};H^{k+1})\quad\mbox{and}\quad
\partial_y \overline v\in \mathcal{C}(\mathbb{R};H^{k+2}).
$$
For given data $(u_0,v_0)$ such that $u_0-\overline u(0)\in H^k$
and $v_0-\overline v(0)\in H^{k+1}$, we define the lifespan of a $H^k$ solution
 as the supremum of all $T$ such that
$\eqref{eq:1dL}$ has a $H^k$ solution on $[0,T)$.
\end{definition}

Our main continuation result  is based on the following lemma.

\begin{lemma}\label{controle}
 Under assumption {\rm  (H1)},
let $(u,v)$ be an $H^k$ solution of $(\ref{eq:linear})$ on
$\mathbb{R}\times[0,T)$ with $v$ valued in $K_v\subset\subset J_v$.
Denote by $(\overline u,\overline v)$ a classical solution
of  $(\ref{eq:linear})$ such that $\eqref{smooth}$ is fulfilled.
Let $\Lambda$ be a primitive of $\alpha$,
$\lambda:=\Lambda(v)$, $w=-\partial_y \lambda$ and $z=u+iw$.
Further define $\overline\lambda:=\Lambda(\overline v)$,
$\overline w=-\partial_y \overline\lambda$, $\overline z=\overline u+i\overline w$,
$\widetilde\lambda:=\lambda-\overline\lambda$, $\widetilde w:=w-\overline w$,
$\widetilde u:=u-\overline u$ and $\widetilde z:=z-\overline z$.
Let
$$
\widetilde Y_k(t):=\Big(\|\widetilde\lambda\|_{L^2}^2+
\sum_{j=0}^{k}\Lp2{(\alpha(v)^{\frac j2}\widetilde
z^{(j)})(t)}^2\Big)^{1/2}.
$$
Then there exists a constant $C$ depending only on $k$, $q$,
$\alpha$  and $K_v$,
 and such  that
$$
\widetilde Y_k(t)\leq \widetilde Y_k(0){\rm e}^{Ct}{\rm e}^{C\int_0^t
\Lp{\infty}{\partial_y  \widetilde z(\tau)}\,d\tau}
{\rm e}^{C\int_0^t\{\norm{H^k}{\partial_{yy}^2\overline z}+\norm{H^k}{\overline w}
(1+\norm{H^{k-1}}{\overline w})(1+\norm{H^k}{\partial_y \overline z})\}\,d\tau}.
$$
\end{lemma}

\begin{proof}
Denoting $\alpha_\sharp=\alpha\circ\Lambda^{-1}$
and $q_\sharp=q\circ\Lambda^{-1}$, one easily find that
$\widetilde z$ solves
\begin{equation}\label{eq:perturbee}
\begin{aligned}
\partial_t  \widetilde z+i\partial_y (\alpha_\sharp(\lambda)\partial_y \widetilde z)
&=
\underbrace{i\partial_{yy}^2\overline z \bigl(\alpha_\sharp(\overline\lambda)
-\alpha_\sharp(\lambda)\bigr)}_{F_1}
+\underbrace{i\partial_y \overline\lambda \partial_y \overline
z \bigl(\alpha'_\sharp(\overline\lambda)-\alpha'_\sharp(\lambda)\bigr)}_{F_2}
\\
&\quad +\underbrace{\bigl(q_\sharp(\lambda)-q_\sharp(\overline\lambda)\bigr)
 \overline w}_{F_3} +\underbrace{\bigl(q_\sharp(\lambda)
+i\partial_y \overline z \alpha'_\sharp(\lambda)\bigr)\widetilde w}_{F_4}.
\end{aligned}
\end{equation}
 Hence, summing  equalities
\eqref{eq:jaugek} for $j=0,\dots,k$, we get
by Cauchy-Schwarz inequality
\begin{equation}\label{blowup1}
\begin{aligned}
\frac12\frac{\,{\rm d}}{ \,{\rm d} t} \widetilde X_k^2
&\leq
\max(1,\widetilde\alpha)^{\frac k2}\widetilde X_k\sum_{j=1}^4\norm{H^k}{F_j}
+\frac k2\Lp{\infty}{\partial_t\log\alpha_\sharp(\lambda)}\widetilde X_k^2 \\
&\quad +\sum_{j=2}^k\sum_{\ell=1}^{j-1} {\textstyle \binom{j+1}{\ell-1}}
\widetilde\alpha^{\frac j2}\Lp2{\alpha^{\frac j2}\partial_y ^j\widetilde z}
\Lp2{\partial_y ^{j+2-\ell}\bigl(\alpha_\sharp(\lambda)\bigr) \partial_y ^{\ell}
\widetilde z}
\end{aligned}
\end{equation}
with
$$
\widetilde X_k(t):=\Big( \sum_{j=0}^{k}\Lp2{(\alpha(v)^{\frac j2}\widetilde
z^{(j)})(t)}^2\Big)^{1/2}.
$$
As $w=-\partial_y \lambda$, $\overline w=-\partial_y \overline\lambda$ and
$\widetilde w=-\partial_y \widetilde\lambda$,
corollary \ref{comp4} combined with the
embedding $H^k\hookrightarrow L^\infty$ yields
for some  constant $C$ depending only on $k$, $\underline b$,
$\widetilde b$  and~$\alpha$,
\begin{equation}\label{F1}
\begin{aligned}
\norm{H^k}{F_1}
&\leq C\norm{H^k}{\partial_{yy}^2\overline z}
\norm{H^k}{\alpha_\sharp(\lambda)-\alpha_\sharp(\overline\lambda)},\\
&\leq C\norm{H^k}{\partial_{yy}^2\overline z}\bigl(1+\norm{H^{k-1}}{\overline w}\bigr)
\norm{H^k}{\widetilde\lambda}.
\end{aligned}
\end{equation}
Similar computations yield
\begin{gather}\label{F2}
\norm{H^k}{F_2} \leq C\norm{H^k}{\partial_y \overline z}\norm{H^k}{\overline w}
\bigl(1+\norm{H^{k-1}}{\overline w}\bigr)
\norm{H^k}{\widetilde\lambda},\\
\label{F3}
\norm{H^k}{F_3} \leq C\norm{H^k}{\overline w}\bigl(1+\norm{H^{k-1}}{\overline w}\bigr)
\norm{H^k}{\widetilde\lambda}.
\end{gather}
Regarding $F_4$, we apply inequality
\eqref{comp2c} to $q_\sharp(\lambda)\partial_y \widetilde\lambda$ and
$\alpha'_\sharp(\lambda)\partial_y \widetilde\lambda$ so
that we  get
\begin{equation}\label{F4}
\norm{H^k}{F_4}\leq C\bigl(1+\norm{H^k}{\partial_y \overline z}\bigr)
\Bigl(\norm{H^k}{\widetilde w}+\norm{H^k}{\overline w}\norm{H^{k+1}}{\widetilde\lambda}\Bigr).
\end{equation}
Because $\partial_t \log\alpha_\sharp(\lambda)=\alpha'_\sharp(\lambda)\partial_y  u$,
we obviously have
\begin{equation}\label{blowup3}
\Lp{\infty}{\partial_t\log\alpha_\sharp(\lambda)}\leq C\Lp{\infty}{\partial_y  u}.
\end{equation}
For bounding the last term in \eqref{blowup1},
we first use
inequality \eqref{tame1} in the appendix which implies
\begin{equation}\label{blowup4}
\Lp2{\partial_y ^{j+2-\ell}\bigl(\alpha_\sharp(\lambda)\bigr) \partial_y ^{\ell}\widetilde z}
\lesssim
\Lp{\infty}{\partial_y \widetilde z}\norm{H^{j-1}}{\partial_{yy}^2\bigl(\alpha_\sharp(\lambda)\bigr)}
+\Lp{\infty}{\partial_{yy}^2\bigl(\alpha_\sharp(\lambda)\bigr)}
\norm{H^j}{\widetilde z}.
\end{equation}
On the one hand,
$$
\norm{H^{j-1}}{\partial_{yy}^2\bigl(\alpha_\sharp(\lambda)\bigr)}^2
=\sum_{m=2}^{j+1}\Lp2{\partial_y ^m\alpha_\sharp(\lambda)}^2,
$$
hence according to lemma \ref{comp1},
$$
\norm{H^{j-1}}{\partial_{yy}^2\bigl(\alpha_\sharp(\lambda)\bigr)}\lesssim\norm{H^j}w,
$$
on the other hand,
$\partial_{yy}^2\bigl(\alpha_\sharp(\lambda)\bigr)=\alpha''_\sharp(\lambda)(\partial_y \lambda)^2
+\alpha'_\sharp(\lambda)\partial_{yy}^2\lambda$
so that, since $\Lp{\infty}{\partial_y \lambda}^2\lesssim\Lp{\infty}\lambda\Lp{\infty}{\partial_{yy}^2\lambda}$, we get
$$\Lp{\infty}{\partial_{yy}^2\bigl(\alpha_\sharp(\lambda)\bigr)}\lesssim\Lp{\infty}{\partial_y  w}.$$
Coming back to \eqref{blowup4}, we end up with
\begin{equation}\label{blowup5}
\Lp2{\partial_y ^{j+2-\ell}\bigl(\alpha_\sharp(\lambda)\bigr) \partial_y ^{\ell}z}
\lesssim\bigl(\Lp{\infty}{\partial_y \widetilde z}
+\norm{H^k}{\overline w}\bigr) \norm{H^k}{\widetilde z}.
\end{equation}
Plugging inequalities \eqref{F1},
\eqref{F2}, \eqref{F3}, \eqref{F4},  \eqref{blowup3} and \eqref{blowup5}
in \eqref{blowup1}, we get
\begin{equation}\label{blowup6}
\begin{aligned}
\frac12\frac{\,{\rm d}}{ \,{\rm d} t} \widetilde X_k^2
&\lesssim \bigl(1+\Lp{\infty}{\partial_y \widetilde z}+\Lp{\infty}{\partial_y \overline u}\bigr)
\widetilde X_k^2\\
&\quad +\bigl(1+\norm{H^{k-1}}{\overline w}\bigr)
\bigl(\norm{H^k}{\partial_{yy}^2\overline z}+\norm{H^k}{\overline w}(1+\norm{H^k}{\partial_y \overline z})\bigr)
\widetilde X_k\widetilde Y_k.
\end{aligned}
\end{equation}
To conclude, we still have to bound
$\|{\widetilde\lambda}\|_{L^2}$.
For doing so, we use the fact that
$$
\partial_t \widetilde\lambda=\bigl(\alpha_\sharp(\lambda)
-\alpha_\sharp(\overline\lambda)\bigr) \partial_y \overline u
+\alpha_\sharp(\lambda)\partial_y \widetilde u,
$$
whence
$$
\frac12\frac{\,{\rm d}}{ \,{\rm d} t}\Lp2{\widetilde\lambda}^2
\lesssim\Lp{\infty}{\partial_y \overline u}\Lp2{\widetilde\lambda}^2+\Lp2{\widetilde\lambda}
\Lp2{\partial_y \widetilde u}.
$$
Adding this last inequality to \eqref{blowup6}, we eventually get
$$
\displaylines{
\frac12\frac{\,{\rm d}}{ \,{\rm d} t} \widetilde Y_k^2 \lesssim
\Bigl(1+\Lp{\infty}{\partial_y \widetilde z}
+\bigl(1+\norm{H^{k-1}}{\overline w}\bigr)
\bigl(\norm{H^k}{\partial_{yy}^2\overline z}+\norm{H^k}{\overline w}(1+\norm{H^k}{\partial_y \overline z})\bigr)\Bigr)
\widetilde Y_k^2.}
$$
Then  Gronwall's lemma completes the proof.
 \end{proof}

One can now state a continuation result
which is very similar to the standard one  for quasi-linear hyperbolic
symmetric systems.

\begin{proposition} Under assumption {\rm  (H1)},
let $(u,v)$ be a $H^k$  solution of $(\ref{eq:linear})$ on $\mathbb{R}\times[0,T)$.
Assume in addition that
\begin{equation}\label{blowup}
\int_0^T\Bigl(\Lp{\infty}{\partial_y  u(\tau)}
+\Lp{\infty}{\partial_{yy}^2 v(\tau)}\Bigr)\,d\tau<\infty
\quad\text{and}\quad
v(\mathbb{R}\times[0,T))\subset\subset J_v.\end{equation}
Then $(u,v)$ may be continued beyond $T$ into a smooth solution
of $\eqref{eq:1dL}$.
\end{proposition}

\begin{proof}
Let $(u,v)$ satisfy the hypotheses of the proposition and denote
 by $(\overline u,\overline v)$ a classical solution of \eqref{eq:linear} such that
$\eqref{smooth}$ is fulfilled.
Introducing $w=-\partial_y \bigl(\Lambda(v)\bigr)$, a
 straightforward interpolation shows that
\begin{align*}
&\int_0^T\Lp{\infty}{\partial_y   w(t)}\,dt\\
&\leq C\int_0^T\Bigl(\Lp{\infty}{\alpha(v)}\Lp{\infty}{\partial_{yy}^2 v}
+\Lp{\infty}{\alpha'(v)}\Lp{\infty} v\Lp{\infty}{\partial_{yy}^2 v}\Bigr)\,dt<\infty.
\end{align*}
Since $\partial_y \overline v \in \mathcal{C}(\mathbb{R};H^{k+2})$ and
$\partial_y \overline u\in \mathcal{C}(\mathbb{R};H^{k+1})$,
$\int_0^T\bigl(\Lp{\infty}{\partial_y   \overline u(t)}+\Lp{\infty}{\partial_y \overline w(t)}\bigr)\,dt<\infty$.
Therefore, lemma \ref{controle} may be applied. From it, we get
(with an obvious notation)
$$
\widetilde u\in L^\infty(0,T;H^k),\quad
\widetilde w\in L^\infty(0,T;H^k)\quad\mbox{and}\quad
\partial_y  \widetilde v\in L^\infty(0,T;H^k).
 $$
Let $\eta$ be a positive time which satisfies
\eqref{time1} with $\norm{L^\infty(0,T;H^k)}{\widetilde z}$
instead of $\widetilde Y_0$.
Theorem \ref{existence} supplies a solution on the time interval
$[0,\eta]$ for \eqref{eq:1dL} with data $(u(T-\frac\eta2),v(T-\frac\eta2))$.
By virtue of uniqueness, this solution is
 a continuation of $(u,v)$ beyond $T$.
 \end{proof}

Because $H^2\hookrightarrow {\rm Lip}$,
we conclude that the $H^k$ regularity is controlled by the $H^2$
regularity so that {\bf the time of existence in $H^2$ is the same
as in $H^k$}.

\subsection{Further comments on the case of a constant
profile}\label{ss:constant}

In this section, we briefly review how theorem \ref{existence}
and blow-up criteria may be improved
if we restrict ourselves to the case of a {\it constant profile}
$(\overline u,\overline v)$.
The main improvement is that
we do not have to suppose that $v$ has a limit at $-\infty$
and $+\infty$. Only assumptions on $\partial_y  v$ are needed.


Going along the lines of the proof of theorem \ref{existence},
one can observe that the scheme reduces to solving
\begin{gather*}
\partial_t  \widetilde z^{n+1}+i\partial_y (\alpha_\sharp(\lambda^n)\partial_y 
\widetilde z^{n+1})=q_\sharp(\lambda^n)\widetilde w^n,\\
 \widetilde z^n_{|t=0}=\widetilde v_0+i\widetilde w_0,
 \end{gather*}
with $\widetilde w_0:=-\partial_y \lambda_0$, $\lambda_0=\Lambda(v_0)$
and $\widetilde u_0=u_0-\overline u$. Then we set
$u^{n+1}:=\overline u+\Re z^{n+1}$, $w^{n+1}:=\Im z^{n+1}$ and
$$
\lambda^{n+1}=\Lambda(v_0)+\int_0^t\alpha(v^n)\partial_y u^{n+1}.
$$
Therefore, most of the terms in $F_1^n$, $F_2^n$, $F_3^n$ and $F_4^n$
vanish and we end up with the inequality
$$
\widetilde X^{n+1}(t)\leq
{\rm e}^{C\int_0^t\widetilde X^n(\tau)\,d\tau}
\Big(\widetilde X_0+C\int_0^t{\rm e}^{-C\int_0^\tau \widetilde X^n(\tau')
\,d\tau'}\widetilde X^n(\tau)\,d\tau\Big)
$$
for some constant $C$ depending only on
$k$, $\underline b$, $\widetilde b$, $\alpha$ and $q$.

Also one has to assume only that $q$ and $\alpha$ have $k+1$
bounded derivatives (instead of $k+2$ in the general case)
and  no control on $\|{\widetilde\lambda^n}\|_{L^2}$
 is needed to close the estimates.
Therefore, we eventually get the following existence theorem.

\begin{theorem}\label{existencebis}
Let $\overline u\in \mathbb{R}$ and  $k\geq2$. Under assumption {\rm  (H1)}
with $k-1$, let $K_v\subset\subset J_v$,
$u_0\in \overline u+H^k$ and $v_0\in L^\infty$ with  $\partial_y  v_0\in H^k$
and $v_0(\mathbb{R})\subset K_v$.
There exists a positive $T$ such that the Cauchy problem
associated with the system $(\ref{eq:1dLaugb})$ and initial data
$(u(0),v(0))=(u_{0},v_{0})$ has a unique
solution $(u,v)$ which satisfies
\begin{equation}\label{espace1}
u-\overline u\,,\;\partial_y  v\;\in\; \mathcal{C}([0,T];H^k)\cap \mathcal{C}^1([0,T];
H^{k-2}) \quad\mbox{and}\quad v(\mathbb{R}\times[0,T])\subset J_v
\end{equation}
with moreover,
\begin{equation}\label{espace2} (v-v_0)\in \mathcal{C}([0,T];H^{k+1})\cap
\mathcal{C}^1([0,T];H^{k-1}).\end{equation}
 Besides, there exists a constant $C=C(\alpha,q,K_v)$
such that $T$ may be chosen such that
$$
T\geq\frac1C\log\big(1+\frac1{\norm{H^2}{u_0-\overline u}+\norm{H^2}{\partial_y  v_0}}\big).
$$
\end{theorem}

\begin{proof}
Under the assumptions of theorem \ref{existencebis},
the estimate of lemma \ref{controle} reduces to
\begin{equation}\label{lifespan}
\widetilde X_k(t)\leq \widetilde X_k(0){\rm e}^{Ct}{\rm e}^{C\int_0^t\Lp{\infty}{\partial_y  z(\tau)}\,d\tau}
\end{equation}
with $C$ depending only on $k$, $q$,
$\alpha$ and $J_v$, $z:=(u-\overline u)+iw$
and
$$
X_k(t):=\Big(\sum_{j=0}^{k}\Lp2{(\alpha(v)^{\frac
j2}z^{(j)})(t)}^2\Big)^{1/2}.
$$
 From the above estimate, we gather  that the blow-up criterion
stated in \eqref{blowup} remains true under the assumptions
of theorem \ref{existencebis} and that the time of existence
in $H^k$ is the same as in $H^2$.
Indeed, from \eqref{lifespan} and Sobolev embeddings
we get
$$
\widetilde X_k(t)\leq \widetilde X_k(0){\rm e}^{Ct}
{\rm e}^{C\int_0^tX_2(\tau)\,d\tau}.
$$
Hence, $\widetilde X_k$ remains bounded
as long as $\widetilde X_2$ does. Thus
the lifespan in $H^k$ is the same as in $H^2$.
Now, the above inequality with $k=2$ yields
$$
\widetilde X_2(t)\leq \frac{\widetilde X_2(0)e^{Ct}}
{1-\widetilde X_2(0)(e^{Ct}-1)}\quad\text{while}\quad
\widetilde X_2(0)(e^{Ct}-1)<1.
$$
This  gives the desired lower bound
for the life span in $H^2$.
\end{proof}



\section{Appendix}

In this section, we state some technical estimates for
products or composition of functions which have been used
repeatedly throughout the paper.

Most of them are based on the following Gagliardo-Nirenberg
 inequality.

\begin{lemma}\label{Gagliardo}
Let $k\in\mathbb{N}$ and $j\in\{0,\dots,k\}$. There exists a constant $C_{j,k}$
depending only on $j$ and $k$ such that
$$
\Lp{\frac{2k}j}{\partial_y ^j v}
\leq C_{j,k}\Lp{\infty} v^{1-\frac jk}
\norm{L^2}{\partial_y ^k v}^{\frac jk}.
$$
\end{lemma}

We can now state  some tame estimates for the product of two functions.

\begin{lemma}
Let $k\in\mathbb{N}$ and  $j\in\{0,\dots,k\}$.
  There exist a constant $C_{j,k}$ depending only on $j$ and on
$k$ and a constant $C_k$ depending only on $k$, and such that
\begin{gather}\label{tame1}
\Lp2{\partial_y ^ju \partial_y ^{k-j}v}
\leq C_{j,k}\Bigl(\Lp{\infty} u\Lp2{\partial_y ^kv}
+\Lp{\infty} v\Lp2{\partial_y ^ku}\Bigr),\\
\label{tame2}
\norm{H^k}{uv}\leq C_{k}\Bigl(\Lp{\infty}{u}\norm{H^k}v+
\Lp{\infty} v\Lp2{\partial_y ^ku}\Bigr).
\end{gather}
\end{lemma}

\begin{proof}
Because
$$
\norm{H^k}{uv}\leq C_k\Bigl(\Lp2{uv}+\Lp2{\partial_y ^k(uv)}\Bigr),
$$
 Leibniz formula entails that
$$
\norm{H^k}{uv}\leq C_k\Bigl(\Lp{\infty} u\Lp2{v}+
\sum_{j=0}^k\Lp2{\partial_y ^ju \partial_y ^{k-j}v}\Bigr)
$$
so that inequality \eqref{tame1} (used repeatedly)
yields \eqref{tame2}.
Let us focus on the proof of \eqref{tame1}.
According to H\"older inequality, we have
$$
\Lp2{\partial_y ^ju \partial_y ^{k-j}v}\leq
\Lp{\frac{2k}j}{\partial_y ^ju}
\Lp{\frac{2k}{k-j}}{\partial_y ^{k-j}v}.
$$
This obviously yields \eqref{tame1} if $j=0$ or $k$. Else,
 using lemma \ref{Gagliardo}, we get
$$
\Lp2{\partial_y ^ju \partial_y ^{k-j}v}\leq C_{j,k}
\Bigl(\Lp{\infty} u\Lp2{\partial_y ^{k}v}\Bigr)^{1-\frac jk}
\Bigl(\Lp{\infty} v\Lp2{\partial_y ^ku}\Bigr)^{j/k},
$$
and Young inequality  leads to $(\ref{tame1})$.
\end{proof}

Let us now state estimates in Sobolev spaces for the composition
of functions.

\begin{lemma}\label{comp1}
Let $k\geq1$ and $F$ be in $W^{k,\infty}_{\rm loc}(\mathbb{R})$.
There  exists a constant $C_{k}$ such that
for all $v\in L^\infty$ such  that $\partial_y ^k v\in
L^2$,   there holds
$$
\norm{L^2}{\partial_y ^k\bigl(F(v)\bigr)}\leq C_{k}\norm{L^2}{\partial_y ^k v}
\sum_{j=0}^{k-1}\Lp{\infty} v^j\Lp{\infty}{F^{(j+1)}(v)}.
$$
\end{lemma}

\begin{proof}
This inequality stems from  Fa\'a-di-Bruno's formula:
$$
\partial_y ^k\bigl(F(v)\bigr)=\sum \frac{k!}{i_1!\dotsm i_k!}\;
\biggl(\frac{\partial_y  v}{1!}\biggr)^{i_1}
\dotsm\biggl(\frac{\partial_y ^kv}{k!}\biggr)^{i_k}F^{(i_1+\dots+i_k)}(v)
$$
where the sum is over all the $(i_1,\dots,i_k)\in\mathbb{N}^k$
such that $i_1+2i_2+\dots+ki_k=k$.
On the one hand, H\"older inequality gives
\begin{equation}\label{tame2a}
\Lp2{\bigl(\partial_y  v\bigr)^{i_1}
\dotsm\bigl(\partial_y ^kv\bigr)^{i_k}F^{(i_1+\dots+i_k)}(v)}
\leq\Lp{\infty}{F^{(i_1+\dots+i_k)}(v)}
\prod_{\ell=1}^k\Lp{\frac{2k}\ell}{\partial_y ^\ell v}^{i_\ell}.
\end{equation}
On the other hand, lemma \ref{Gagliardo}  yields for $1\leq\ell\leq k$,
$$
\Lp{\frac{2k}\ell}{\partial_y ^\ell v}
\leq C_{\ell,k}\Lp{\infty} v^{1-\frac\ell k}
\norm{L^2}{\partial_y ^k v}^{\frac\ell k}.
$$
Inserting this inequality in (\ref{tame2a})
completes the proof of lemma \ref{comp1}.
\end{proof}

\begin{corollary}\label{comp2}
Let $F\in W^{k,\infty}_{\rm loc}$, $k\in\mathbb{N}$.
There exists a constant $C_k$ so that
\begin{gather}\label{comp2a}
\norm{H^k}{F(v)w}\leq C_k\Bigl(\Lp{\infty}{F(v)}\norm{H^k}w
+\Lp{\infty} w\norm{L^2}{\partial_y ^k v}
\sum_{j=0}^{k-1}
\Lp{\infty}{F^{(j+1)}(v)}\Lp{\infty} v^j\Bigr).
\\ \label{comp2b}
\norm{H^k}{F(v)\partial_y  v}\leq C_k\norm{H^{k}}{\partial_y  v}
\sum_{j=0}^{k}
\Lp{\infty}{F^{(j)}(v)}\Lp{\infty} v^j.
\\ \label{comp2c}
\begin{aligned}
&\norm{H^k}{F(v)\partial_y  w}\\
&\leq C_k\Bigl(\Lp{\infty}{F(v)}\norm{H^k}{\partial_y  w}
+\Lp{\infty} w\norm{H^{k}}{\partial_y  v}
\sum_{j=0}^{k}
\Lp{\infty}{F^{(j+1)}(v)}\Lp{\infty} v^j\Bigr).
\end{aligned}
\end{gather}
\end{corollary}

\begin{proof}
The three results are  obvious if $k=0$ so let us assume that $k\geq1$.
Then applying inequality \eqref{tame2} yields
$$
\norm{H^k}{F(v)w}\leq C_k\Bigl(\Lp{\infty}{F(v)}\norm{H^k}w
+\Lp{\infty} w\Lp2{\partial_y ^k\bigl(F(v)\bigr)}\Bigr).
$$
Lemma \ref{comp1} enables us to
bound the last term in the right-hand side. This
yields (\ref{comp2a}).

For proving (\ref{comp2b}), we introduce
$\mathcal{F}$ a primitive of $F$.
We have
$$
\norm{H^k}{F(v)\partial_y  v}\leq C_k\left(\Lp2{F(v)\partial_y  v}
+\Lp2{\partial_y ^{k+1}\bigl(\mathcal{F}(v)\bigr)}\right).
$$
Now, applying lemma \ref{comp1} yields the desired result.

For proving \eqref{comp2c}, we first notice that
by virtue of Leibniz formula, we have
$$
\norm{H^k}{F(v)\partial_y  w}\leq C_k\sum_{0\leq\ell\leq j\leq k}
\Lp2{\partial_y ^\ell\bigl(F(v)\bigr)\partial_y ^{j+1-\ell}w}.
$$
Hence, according to \eqref{tame1},
\begin{align*}
\norm{H^k}{F(v)\partial_y  w}
&\leq C_k\sum_{0\leq j\leq k}
\Bigl(\Lp{\infty}{F(v)}\Lp2{\partial_y ^{j+1} w}
+\Lp{\infty} w\Lp2{\partial_y ^{j+1}\bigl(F(v)\bigr)}\Bigr),\\
&\leq C_k\Bigl(\Lp{\infty}{F(v)}\norm{H^k}{\partial_y  w} +\Lp{\infty}
w\sum_{m=1}^{k+1}\Lp2{\partial_y ^m\bigl(F(v)\bigr)}\Bigr),
\end{align*}
and applying lemma  \ref{comp1} achieves the proof
of \eqref{comp2c}.
\end{proof}

\begin{corollary}\label{comp3}
Let $I$ be a bounded  interval of $\mathbb{R}$ and $F$ be in $W^{k,\infty}(\mathbb{R};I)$
for some $k\in\mathbb{N}$ and satisfy $F(0)=0$.
Let $v\in H^k$ be a $I$-valued function.
There exists a constant $C=C_k$ such that
$$
\norm{H^k}{F(v)}\leq
C_k\Bigl(\norm{L^\infty(I)}{F'}\Lp2 v
+\norm{L^2}{\partial_y ^k v}
\sum_{j=0}^{k-1}\Lp{\infty}{v}^j\norm{L^\infty(I)}{F^{(j+1)}}\Bigr).
$$
\end{corollary}

\begin{proof}
We first use the fact that
$$
\norm{H^k}{F(v)}\leq C_k\Bigl(\Lp2{F(v)}+\Lp2{\partial_y ^k(F(v))}\Bigr).
$$
The last term in the right-hand side may be bounded according to lemma
\ref{comp1}.
For bounding the first one, we take advantage of first order
Taylor's formula
$$
F(v)=v\int_0^1F'(\tau v)\,d\tau.
$$
 which obviously implies $\Lp2{F(v)}\leq\norm{L^\infty(I)}{F'}\Lp2 v$.
\end{proof}

\begin{corollary}\label{comp4}
Let $I$ be a bounded  interval of $\mathbb{R}$ and $F$ be in $W^{{k+1},\infty}(\mathbb{R};I)$
for some $k\in\mathbb{N}$.
Let $v$ and $w$ be two $I$-valued functions such that
$\partial_y  v$ and $\partial_y  w\in H^{k-1}$ and $w-v\in H^k$.
   Then $F(w)-F(v)$ belongs
to $H^{k}$ and there exists a constant
$C=C_k$ such that
\begin{align*}
\norm{H^k}{F(w)-F(v)}
&\leq C_k\Bigl(\norm{L^\infty(I)}{F'}\norm{H^k}{w-v}
+\Lp{\infty}{w-v}\Bigl(\norm{L^2}{\partial_y ^k v}+\norm{L^2}{\partial_y ^k w}\Bigr)\\
&\quad\times \sum_{j=0}^{k-1}\Bigl(\Lp{\infty}{v}+\Lp{\infty}
w\Bigr)^j\norm{L^\infty(I)}{F^{(j+2)}}\Bigr).
\end{align*}
\end{corollary}

\begin{proof}
Arguing by density, it suffices to prove the inequality for smooth
$I$-valued functions.
According to first order Taylor's formula, we have
$$
F(w)-F(v)=\int_0^1(w-v)F'(v+\tau(w-v))\,d\tau.
 $$
Therefore,
$$
\norm{H^k}{F(w)-F(v)}\leq\int_0^1\norm{H^k}{(w-v)F'(v+\tau(w-v))}\,d\tau.
$$
Fix a $\tau\in[0,1]$. From corollary \ref{comp2}, we get
\begin{align*}
&\norm{H^k}{(w-v)F'(v+\tau(w-v))}\\
&\leq C_k\Bigl( \Lp{\infty}{F'(v+\tau(w-v))}\norm{H^k}{w-v}
 +\Lp{\infty}{w-v}\norm{L^2}{\partial_y ^k v+\tau(\partial_y ^k w-\partial_y ^k v)}\\
&\quad\times \sum_{j=0}^{k-1} \norm{L^\infty(I)}{F^{(j+2)}}
\Lp{\infty}{v+\tau(w-v)}^j\Bigr)
\end{align*}
whence the desired inequality follows.
\end{proof}

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\end{document}
