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{ %\large
UNIVERSITY OF OKLAHOMA
\par
\vspace{0.16in}
GRADUATE COLLEGE
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EXISTENCE AND
 STABILITY OF SOLUTIONS
 \par
 \vspace{0.17in}
 TO A MODEL EQUATION FOR
\par
\vspace{0.17in}
 DISPERSION-MANAGED SOLITARY WAVES
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A DISSERTATION
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SUBMITTED TO THE GRADUATE FACULTY
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in partial fulfillment of the requirements for the
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degree of
\par
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DOCTOR OF PHILOSOPHY
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By
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ESTAPRAQ KAHLIL
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%Department of Mathematics
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Norman, Oklahoma
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%Norman, OK 73019 USA
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2015
}
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EXISTENCE AND STABILITY OF SOLUTIONS
\par
TO  A  MODEL EQUATION FOR
\par
DISPERSION-MANAGED SOLITARY WAVES
\par
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A DISSERTATION APPROVED FOR THE
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DEPARTMENT OF MATHEMATICS
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BY
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\  \  \  \  \  \  \  & Dr.~John P.~Albert, Chair\\
 \\ &  \\
  \\ &  \\
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  & Dr.~Tomasz Przebinda \\
\\ & \\
 \\ &  \\
\hline
 & Dr.~Semion Gutman \\
\\   & \\
 \\ &  \\
\hline
 & Dr.~R\"{u}diger Landes \\
 \\ &  \\
  \\ &  \\
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   & Dr.~Georgia Kosmopoulou \\
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}
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\copyright \    Copyright by ESTAPRAQ KAHLIL \ 2015

All Rights Reserved.
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%\emph{\huge{{\fontencoding{T1}\fontfamily{calligra}\selectfont To the memory of my mother.}}}
\emph{\large{ To the memory of my mother.}}
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\chapter*{Acknowledgements}

{\doublespacing

First and foremost I would like to thank my advisor Dr.~John P. Albert. It has
been an honor to be his student. I appreciate all his contributions of time,
ideas, and support of my Ph.D. The joy and enthusiasm he
has for his research motivated me even during the tough times of my journey.

I also would like to express my appreciation to my committee members: Dr.~Semion Gutman, Dr.~R\"{u}diger Landes, Dr.~Tomasz Przebinda, and Dr.~Georgia Kosmopoulou. Thank you for
giving me the opportunity to be a part of the Ph.D program in the Mathematics department.
A special thanks to Dr.~Tomasz Przebinda for his passion, encouraging me from the beginning of my journey.

A big thanks for Yoana Walschap and Dr.~Alan Roche for their kindness, support and
for helping me to be a part of the Ph.D program. Your help means a lot to me.

Thanks for all the faculty members in the Mathematics department, in particular Dr. Ralf Schmidt for
supporting and helping me to pursue my goal. I am also thankful to the Mathematics department staff,
in particular Paula Killian, and Cristin Sloan for making the graduate life easier for me and for all other students.

I would like to thank all my friends who made my graduate life an awesome experience.

I would like to thank all my family members for their love and support.
I am also thankful for my dad, I know your sprit is always with me, thanks for your special love.

Mom, you gave me so much, you were my strength.
I think about you every second that passes.
I wish you were here, but I know that your spirit is always around me.
Thanks for your love and faith in me and for teaching me how I should never give up.

Most of all, I would like to thank my fiance, Darren Jaworski, you mean the world to me, thanks for your unmeasurable love and understanding.


}




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\chapter*{Abstract}
{\doublespacing

We study the existence and stability of solutions of the initial-value problem for the
dispersion-managed nonlinear Schr\"{o}dinger (DMNLS) equation, a model equation for optical
pulses in a dispersion-managed fiber.     One interesting feature of the DMNLS
equation is that the nonlinear term involves the non-local operator  $T(s)=e^{-iD(s)\partial^2_x}$,
where the periodic function $D(s)$ governs the dispersive properties of the fiber.   Another interesting
feature is that even when the average dispersion $\alpha$ is equal to zero, the equation
 is known to have solitary-wave solutions.

For the Cauchy problem for the DMNLS equation with initial data in $H^s$ with
$s\geq 1$, under
weak assumptions on the variable dispersion and nonlinear coefficients, we prove local well-posedness  for all $\alpha\in\textbf{R}$, and  global well-posedness for
$\alpha \ne 0$. We also use a Strichartz estimate on $T(s)$  to establish global well-posedness for initial data in $L^2$
for all $\alpha\in\textbf{R}$, and  local well-posedness for data
in $L^2\cap L^{\infty}$ in the case $\alpha=0$.

We also revisit the proofs of existence and stability of solitary waves due to Zharnitsky et al. in the case
$\alpha >0$ and to Kunze in the case $\alpha=0$.   We show that their arguments, based on a concentration
compactness approach to a variational characterization of solitary waves, continue to be valid under
weak assumptions on the dispersion and nonlinear coefficients.



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\chapter{Introduction}

\section{The dispersion-managed nonlinear Schr\"odinger equation}
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An important model equation for pulses in fiber optics is the one-dimensional nonlinear
Schr\"{o}dinger equation (NLS),
\begin{equation}
\label{onezero}
i u_z + u_{\tau\tau} = |u|^2 u,
\end{equation}
which can be derived from Maxwell's equations under the assumption that the pulse is a slowly
modulated sinusoidal wave \cite{A}. Here $u= u(z, \tau)$ is the complex-valued envelope of the
electromagnetic field. $z$ measures distance along the fiber, and $\tau$ is the time. The role of
the NLS equation in describing the dynamics of such phenomena makes it central to the understanding
and design of long-distance fiber optics communication systems \cite{M}. The NLS equation and its
variants also appear in studies of gravity waves, plasma waves, energy transport along molecular chains,
and many other applications.

In this thesis we study an variant of \eqref{onezero} which was derived in \cite{GT} as a model for optical pulses in a
dispersion-managed fiber: that is, a fiber which has been treated to alter its dispersive properties to
enhance the stability of pulses \cite{A}.  For the reader's information we summarize here the assumptions made in the derivation
of this equation; for a more detailed description of these assumptions and their regimes of physical validity,
a good reference is \cite{M}.

The propagation of light in a one-dimensional fiber is described by an electromagnetic field vector
$\bold E=\bold E(z,t)$, where $z$ measures distance along the fiber and $t$ is time.  Here we assume that
$\bold E$ has small magnitude relative to interatomic electric fields; say $|\bold E|=O(\epsilon)$ with $\epsilon < < 1$.
We also assume that $\bold E$ is, to leading order, polarized in the $z$-direction, so that if $\bold e_z$ is the unit vector
in the $z$-direction, then $|\bold E(z,t)- \epsilon E_1(z,t) \bold e_z|= O(\epsilon^2)$ for some scalar function
$E_1(z,t)$ of order one.  Further we assume that $E_1$ is a nearly monochromatic wave propagating in the $z$ direction, with wave number $k$
and frequency $\omega$, i.e.,
$$E_1(z,t)= u(\epsilon z,\epsilon t)\ e^{i(kz-\omega t)}+ \bar{u}(\epsilon z,\epsilon t)\ e^{-i(kz-\omega t)},$$
where the derivatives of $u$ with respect to its arguments $Z = \epsilon z$ and $\tau = \epsilon t$ are order one.  In other words, $u$
is a slowly varying function of $z$ and $t$.


Under the preceding assumptions, a formal argument starting from Maxwell's equations (see \cite{A}) shows that $u$ satisfies approximately the variable-coefficient
NLS equation
\begin{equation}
\begin{aligned}\label{oneone}
iu_Z + d(Z)u_{\tau\tau} + \gamma(Z) |u|^2u =0,
\end{aligned}
\end{equation}
where $d(Z)$ is determined by the optical properties of the fiber, and affects the dispersion of signals
(the spreading of signals due to the fact that components of different wavelengths travel with different velocities), and $\gamma(Z)$ models the amplification and decay of signals due to the presence
of amplifiers and/or loss in the fiber.

In the 1990's it was suggested that signals in optical fibers might propagate more stably if one constructed the fiber so that segments with large positive dispersion
alternate with segments of large negative dispersion.  The idea was that the large absolute values of the dispersion would suppress undesirable nonlinear effects,
but that if it was arranged that the average value of the dispersion was small (due to cancellation of the negative and positive dispersion with each other), then
undesirable spreading of signals would not occur \cite{Tu}.  Experimental studies (see \cite{Z2} for references) found that this would indeed be the case.   The technique of
constructing fibers with such dispersive properties is known as dispersion management.

At around the same time, Gabitov and Turitsyn in \cite{GT} proposed an interesting model equation for optical waves in a dispersion-managed fiber.  To derive it, one assumes
that the dispersion coefficient function $d(Z)$ in \eqref{oneone} is periodic with small period $\mu$, where $\mu<<1$; and that in absolute value,
$d(Z)$ is large, of size $O(1/\mu)$, and has mean value $\alpha$ which is $O(1)$.  Thus we assume that
$$
d(Z)=\alpha + \frac{1}{\mu}\ \Delta\left(\frac{Z}{\mu}\right),$$
where $\Delta(\zeta)$ is a function of period 1 and mean value zero; i.e.,
$$
\int_0^1\Delta(\zeta)\ d\zeta=0.
$$
We also assume that $\gamma(Z)$ is non-negative and periodic with period $\mu$, which, if $g(Z)$ is not constant, corresponds to the assumption that
the signal is amplified by amplifiers spaced $\mu$ units apart.  Let $g(\zeta)$ be defined by $g(\zeta)=\gamma(\mu \zeta)$, so
that $\gamma(\zeta)$ is periodic with period 1.


Under the preceding assumptions, one can approximate the solutions
of \eqref{oneone} by solutions of the averaged equation
\begin{equation}
\label{onethree}
iu_Z + \alpha u_{\tau\tau}+\int_0^1 \ g(s)T^{-1}(s)\left[|T(s)u|^2 T(s)u\right]\ ds=0.
\end{equation}
(see, e.g., chapter 10 of [A] for a derivation).  In fact, in \cite{Z2} it is shown that solutions of \eqref{onethree} do approximate those of \eqref{oneone} well
when $\mu$ is small (see Theorem 4.1 of \cite{Z2}).

There does not seem to be a generally accepted name for
\eqref{onethree} in the literature.  We will refer to it below as
the dispersion-managed nonlinear Schr\"{o}dinger (DMNLS) equation.

In equation \eqref{onethree}, the operator $T(s)$ is defined as follows.
Define $D(s)$ for $s \in \mathbf R$ by
\begin{equation}
D(s):=\int_0^s \Delta(\zeta)d\zeta.
\label{defDs}
\end{equation}
Notice that since $\displaystyle \int_0^1\Delta(\zeta)\ d\zeta = 0$, then
$D(s)$ is periodic with period $1$.  For $s\in [0,1]$, define $T(s):L^2 \rightarrow L^2$ by
\begin{equation}
T(s)= e^{-iD(s)\partial^2_x},
\end{equation}
or in other words
\begin{equation}
\mathcal{F}(T(s)f)[\omega]=e^{iD(s)\omega^2}\mathcal{F}(f)[\omega],
\label{defTs}
\end{equation}
 where $\mathcal{F}$ is the
Fourier transform.   We describe the properties of $T(s)$ in more detail below, but we mention here
that clearly $T(s)$ is an invertible (in fact unitary) operator on $L^2$
with inverse given by $T^{-1}(s)= e^{iD(s)\partial^2_x}$, so the operator $T^{-1}(s)$ in \eqref{onethree}
is well-defined.


To conform to the usual choice of variables used for NLS and its variants in the literature, from now on we will replace $z$
by $t$ and $\tau$ by $x$, so that \eqref{onethree} becomes
\begin{equation}
iu_t + \alpha u_{xx}+\int_0^1 g(s)T^{-1}(s)\left[|T(s)u|^2 T(s)u\right]\ ds=0 .
\label{onefour}
\end{equation}
In \eqref{onefour}, when $T(s)$ or $T^{-1}(s)$ is applied to a function of two variables $x$ and $t$,
it is understood that $t$ is held fixed and the operators are viewed as acting on functions of the variable $x$.

The results in this thesis hold under rather general assumptions on the functions $\Delta(s)$
(which determines $T(s)$ through \eqref{defDs} and \eqref{defTs}) and $g(s)$.  We assume, unless otherwise stated, only that
$\Delta(s)$ is integrable on $[0,1]$ and that $g(s)$ is measurable on $[0,1]$.


\section{Previous results and statement of main results}

Two of the central topics in the theory of nonlinear dispersive equations are well-posedness of the
initial-value problem, and the existence and stability of important special solutions.

For an initial-value
problem to be (globally) well-posed, its solution $u(x,t)$ should exist for all times $t$, for arbitrary choices of the
initial data $u(x,0)$ in some function class, and the solution should be unique and depend continuously on the initial
data.

When special solutions $u_s(x,t)$ such as solitary waves or bound-state solutions exist, the question arises of
how important these solutions are in the evolution of more general solutions.  An important first step in answering
this question is to study the stability of the special solutions.  To prove stability, one tries to show that if the equation
is solved with general initial data $u(x,0)$ that is sufficiently close (in some appropriate function space)
to $u_s(x,0)$, the initial data for the special solution, then for all times $t$ the solution $u(x,t)$ arising from $u(x,0)$
stays close to the special solution $u_s(x,t)$.  In particular, to prove a stability result one generally
also needs to prove a well-posedness result, to guarantee that $u(x,t)$ exists for all time.

For the NLS equation \eqref{onezero}, global well-posedness of the initial-value problem in $H^s$ for $s \ge 0$ was proved
in \cite{T}. Also, \eqref{onezero} has important special solutions
known as solitary-wave (or bound-state) solutions, which are solutions of the
form $u_s(\tau,z)=e^{i\theta z}\phi(\tau),$ where $\theta$ is a constant and $\phi(\tau) \to 0$ as $|\tau| \to \infty$.
The stability of these solitary waves is a classical result for which it is a little difficult to identify a first
author.  Certainly any proof of stability owes much to the original proof of stability of solitary waves for the Korteweg-de Vries
equation given by
Benjamin \cite{Be} and Bona \cite{B}. For the case of the NLS equation \eqref{onezero} it is probably fair to give much of the
credit to Cazenave and Lions for devising a nice method of
proof of stability of solitary-wave solutions (see \cite{C}, chapter 8).
More recently, in \cite{MP} it has been proved that solitary-wave solutions of \eqref{onezero} are stable in $L^2$.

For the variable-coefficient NLS equation \eqref{oneone}, global well-posedness in $L^2$ was
proved in \cite{ASS}, under the assumptions that $d(Z)$ is periodic and piecewise constant and $g(Z) \equiv 1$.  Global well-posedness
in $H^1$ is also known (see Remark 3.3 of \cite{ASS}).   Of course, solitary waves as defined above could not exist for the variable-coefficient
equation \eqref{oneone}, but it is an interesting question whether \eqref{oneone} has solutions which somehow resemble solitary waves (see, for example,
\cite{PZ} for results on this topic).

For the DMNLS equation \eqref{onefour}, it is noted in \cite{Z2}  that it is straightforward to prove
global well-posedness in $H^1$ in the case $\alpha \ne 0$ and local well-posedness in $H^1$ in the case $\alpha = 0$.  Also,
\cite{Z2} includes a result on existence of solitary waves and their stability in $H^1$ in the case $\alpha > 0$.  The assumptions used in \cite{Z2} were
that $\Delta(s)=D'(s)$ is piecewise constant on $[0,1]$ and that $g(s)=1$ on $[0,1]$.  For the case $\alpha=0$, Kunze \cite{Z1} used an
interesting version of the concentration compactness method to prove the existence of solitary-wave solutions
to \eqref{onefour} when $d(s)$ is piecewise constant and $g(s)\equiv 1$.  We note that the question remains open whether solitary-wave solutions to equation \eqref{onefour} exist in the case
$\alpha < 0$, although numerical evidence suggests that they do not (see \cite{PZ}, p. 749).

The main results of this thesis are as follows.  They all apply to the DMNLS equation \eqref{onefour}.

In Theorem \ref{twothree} below, we include a proof, valid for all $\alpha \in \mathbf R$, of global well-posedness of the initial-value problem in $H^r$ for
all $r \ge 1$.
   As indicated in \cite{Z2}, the proof is a straightforward contraction-mapping argument, but we wanted to include
the details to set the stage for the results which follow.  This result only requires that $\Delta(s)$ be integrable and $g(s)$ be bounded and measurable
on $[0,1]$.

In Theorem \ref{theoremtwonine}, again for all $\alpha \in \mathbf R$, we prove global well-posedness of the initial-value problem in $L^2$.  The proof makes
use of a Strichartz estimate for the family of operators $T(t)$ (Theorem \ref{theoremtwosix}).  Here we have to assume more about $\Delta(s)$, namely that it is
piecewise of one sign, and bounded away from zero (see Assumption D1 below).  Our final well-posedness result, Theorem \ref{theoremtwoten}, is for the case
$\alpha=0$, also under the extra assumption on $\Delta(s)$, and states that the initial-value problem is locally well-posed in $L^2\cap L^\infty$.

In Chapter 3, we consider the existence and stability of solitary-wave solutions to \eqref{onefour}.   In Section \ref{stabalphapos} we consider the case when $\alpha > 0$,
 and show in Theorem \ref{theoremthreetwo} that for every $\lambda > 0$, \eqref{onefour} has a non-empty
 stable set $G_\lambda$ of solitary-wave solutions $e^{i\theta t}\phi(x)$ satisfying $\displaystyle \int_{-\infty}^\infty \phi^2 \ dx = \lambda$.
The proof, which proceeds by showing that $G_\lambda$ is the solution set to a variational problem, is essentially the same as that given in \cite{Z2} for the
case when $\Delta(s)$ is piecewise constant and $g(s)=1$ on $[0,1]$.  We were interested in writing out the details so as to check what are the minimal assumptions
on $\Delta(s)$ and $g(s)$ for the proof to work.  It turns out that all that is required is for $\Delta(s)$ and $g(s)$ to be integrable on $[0,1]$ and for $g(s)$ to be non-negative
and not identically equal to zero.

In Section \ref{stabalphazero} we consider the case $\alpha=0$, and state an existence and stability result for solitary waves in Theorem \ref{theoremthreeninteen}.  Here we
have to add additional assumptions on $\Delta(s)$ and $g(s)$; namely that $\Delta(s)$ and $g(s)$ are piecewise absolutely continuous.
  Since the proof only differs in a couple of places from that given in \cite{K} for the case when $\Delta(s)$ is piecewise constant, we do not give the full proof, but
only indicate how the proof in \cite{K} should be modified in the more general case.

Finally we would like to mention a couple of open problems associated to the results in this thesis.  It is not yet known whether \eqref{onefour} is globally
well-posed in $H^r$ for any $r>1$.  In the case $\alpha = 0$, it is not known whether \eqref{onefour}
is locally well-posed in $H^r$ for any $r > 0$, or whether it is globally well-posed in $L^2 \cap L^\infty$.  Also, we would like to know whether
a well-posedness result can be proved for equation \eqref{onefour} in Strichartz-type spaces $L^q_t L^p_x$,  along the lines of the result given
for equation \eqref{oneone} in \cite{ASS}.

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\section{Preliminaries}
The set of natural numbers $\{1,2,3,\cdots \}$ and  the set of all integers are written
$\textbf{N}$ and $\textbf{Z}$, respectively. The set of all real numbers is denoted by $\textbf{R}$.

For any measurable function $f$ on $\textbf{R}$ and any $p\in[1,{\infty})$, we define
\begin{equation*}
\left\|f\right\|_{L^p}=\left(\inftyint|f(x)|^{p}\ dx\right)^\frac{1}{p} ,
\end{equation*}
and $L^p = L^p (\textbf{R})$ denotes the space of all $f$ for which $\|f\|_p$ is finite.
The space $L^{\infty}$ is defined as the space of all measurable functions $f$ on $\textbf{R}$ such that
\begin{equation*}
\left\|f\right\|_{L^\infty}=\text{ess sup}_{x\in\textbf{R}} \left|f\left(x\right)\right|
\end{equation*}
is finite.

For any measurable function $f(x,t)$ on $\textbf{R}\times \textbf{R}$ and any $p\in[1,{\infty})$ and
$q\in[1,{\infty})$, we define
\begin{equation*}
\left\|f\right\|_{L^q_tL^p_x}=\left(\inftyint\left(\inftyint|f(x,t)|^{p}\ dx\right)^\frac{q}{p}\ dt\right)^\frac{1}{q} ,
\end{equation*}
and $L^q_t L^p_x = L^q_t L^p_x (\textbf{R}\times \textbf{R})$ denotes the space of all $f$ for which
$\|f\|_{L^q_tL^p_x}$ is finite.

If $E$ is a subset of $\textbf{R}$ then $C^{\infty}_{0}\left(E\right)$ denotes the space of
infinitely differentiable functions with compact support in $E$.  A larger space, the Schwartz space
$S(\textbf{R})$, is defined to be the set of all $C^{\infty}$
functions on
$\textbf{R}$ such that for every nonnegative integer $m$ and every multi-index $\alpha$,
\begin{equation*}
\sup_{x\in\textbf{R}}\ (1+|x|^2)^\frac{m}{2}|D^\alpha u(x)|<\infty.
\end{equation*}
If $f$ is any function in $S(\textbf{R})$, the Fourier transform of $f$ is denoted by
$\widehat f$ or $\mathcal{F}(f)$ and is defined by
\begin{equation*}
\mathcal{F}(f)[\omega]:={\widehat f}\left(\omega\right)=\inftyint e^{-i\omega x}f(x)\ dx.
\end{equation*}
The inverse of the Fourier transform is given for $f \in S(\textbf{R})$ by
\begin{equation*}
\mathcal{F}^{-1}(f)[x]=\frac{1}{2\pi}\inftyint e^{i\omega x}f(\omega)\ d\omega,
\end{equation*}
and for all $f$ and $g$ in $S(\textbf{R})$ one has the formula
\begin{equation}
\label{ftconv}
\mathcal{F}(fg)=\mathcal{F}(f) \ast \mathcal{F}(g),
\end{equation}
where the convolution $p \ast q$ of two functions $p$ and $q$ is defined by
$$
p\ast q(\omega) = \frac{1}{2\pi}\inftyint p(\omega-\omega_1)q(\omega_1)\ d\omega_1.
$$


For any tempered distribution $f$ on $\textbf{R}$ whose Fourier transform $\hat{f}$ is a function,
and any $s\in\textbf{R}$, we define
\begin{equation*}
\|f\|^2_{H^s}=\left(\inftyint(1+|\omega|^2)^{s}|{\hat f}(\omega)|^2\ d\omega \right)
\end{equation*}
and $H^s =H^s (\textbf{R}) $ denotes the Sobolev space of all $f$ for which $\|f\|_{H^s}$ is finite.


If $X$ is any Banach space with norm $\|\cdot\|_X$, and $[a,b]\subset \textbf{R}$, we define
$C\left([a,b];\textbf{R}\right)$ to be the Banach space of all continuous maps $u:[a,b]\to X$
with norm $$\|u\|_{C\left([a,b];\textbf{R}\right)}=\sup_{t\in[a,b]}\|u(t)\|_X.$$

In this thesis we will use $C$ to stand for several constants, whose value can vary from line to line.

We recall the following standard results.
\begin{lem}\label{lemmaonefour}
There exists $C>0$ such that for all $f\in H^1$, if $g=|f|$ then
$$\| g \|_{H^1}\leq C\|f\|_{H^1}.$$
\end{lem}

\begin{proof}
See theorem $6.17$ in \cite{LL}.
\end{proof}

\begin{thm}[Riesz-Thorin Interpolation Theorem]\label{lemmaonefive}
Suppose $1\le p_0 < p_1\le \infty$, $1\le q_0 < q_1\le \infty$, and suppose $L$ is a bounded linear operator from $L^{p_0}(X)$ to $L^{q_0}(Y)$ with norm $M_0$ and from $L^{p_1}(X)$ to $L^{q_1}(Y)$ with norm $M_1$. Then for all $\theta\in (0,1),$ $L$ is bounded from $L^{p_{\theta}}(X)$ to $L^{q_{\theta}}(Y)$ with norm $M_{\theta}$ such that
$$M_{\theta}\le M^{1-\theta}_0 M^{\theta}_1 ,$$
where  $\displaystyle \frac{1}{p_\theta}=\frac{1-\theta}{p_0} +\frac{\theta}{p_1}$
and $\displaystyle \frac{1}{q_\theta}=\frac{1-\theta}{q_0} +\frac{\theta}{q_1}$.
\end{thm}

\begin{proof}
See \cite{LP}.
\end{proof}

\begin{thm}[Hardy-Littlewood Inequality]\label{lemmaonesix}
Suppose $\beta\in (0,1)$ and $p>1$. Then for all $f\in L^p$, the integral
\begin{equation*}
I_\beta f(x)=\inftyint \frac{f(y)}{|x-y|^{1-\beta}}\ dy
\end{equation*}
is absolutely convergent for almost every $x\in \textbf{R}$. Moreover, there exists $C>0$ such that for all $f\in L^p$,
\begin{equation*}
\|I_{\beta}f\|_{L^q}\leq C\|f\|_p ,
\end{equation*}
where $\displaystyle \frac{1}{q}= \frac{1}{p}- \beta$.
\end{thm}

\begin{proof}
See Theorem 2.6 in \cite{LP}.
\end{proof}

\begin{thm}[Banach algebra property]\label{lemmaoneseven}
If $u, v \in H^r(\textbf{R}^n)$ for $r>\frac{n}{2}$, then $uv\in H^r(\textbf{R}^n)$ and
$$\|uv\|_{H^r(\textbf{R}^n)}\le C\|u\|_{H^r(\textbf{R}^n)} \|v\|_{H^r(\textbf{R}^n)},$$
The constant $C$ depends only on $r$ and $n$.
\end{thm}

\begin{proof}
See Theorem 3.4 in \cite{LP}.
\end{proof}

\begin{thm}[Minkowski's Integral Inequality]\label{lemmaoneeight}
Let $f$ be a nonnegative function, and let $1\le p< \infty$. Then
$$\left( \inftyint \left( \inftyint f(x,y)\ dy\right)^p dx\right)^{\frac{1}{p}} \le \inftyint \left( \inftyint f(x,y)^p\ dx\right)^{\frac{1}{p}}\ dy.$$
\end{thm}

\begin{proof}
See Theorem 2.4 in \cite{LL}.
\end{proof}
Note that Minkowski's integral inequality can be written as
$$\left\| \inftyint f(\cdot,y)\ dy\right\|_{L^p} \le \inftyint \left\| f(\cdot,y)\right\|_{L^p} dy.$$
In fact, this inequality can be generalized by replacing the $L^p$ norm with other Banach function norms,
such as the $H^1$ norm (see for example \cite{S} for a general result). We will use such generalized
Minkowski's inequalities freely below.

\begin{thm}[Generalized H\"{o}lder's Inequality]\label{lemmaonenine}
Let $p_1$, \dots, $p_m$ be such that $1\le p_j\le \infty$ for $1 \le j \le m$ and $\displaystyle \sum_{j=1}^m \frac{1}{p_j}=1$, and
suppose $f_j\in L^{p_j}$ for $1 \le j \le m$. Then $\displaystyle \prod_{j=1}^m f_j = (f_1 f_2 \cdots f_m) \in L^1$, and
$$\inftyint \left|\prod_{j=1}^m f_j\right|\ dx   \le \prod_{j=1}^m\|f_j\|_{L^{p_j}}.$$
\end{thm}

\begin{proof}
See Theorem 2.3 in \cite{LL} and the remarks following.
\end{proof}

\begin{thm}[Plancherel's Theorem]\label{lemmaoneten}
Let $f\in L^2$. Then $\hat{f} \in L^2$ and
$$\|\hat{f}\|_{L^2}=\|f\|_{L^2}.$$
Also, for all $g\in L^2$, $<\hat{f},\hat{g}>_{L^2}=<f,g>_{L^2}$
\end{thm}

\begin{proof}
See Theorem 1.3 in \cite{LP}.
\end{proof}

\begin{thm}[Gagliardo-Nirenberg Inequality]
\label{theoremoneeleven}
Let $q$, $r$ be any numbers satisfying $1\le q\le \infty$ and $1<r\le \infty$. If $u$ is any function
in $L^q$ such that $\frac{du}{dx}\in L^r$, then
$$\|u\|_{L^p}\le C\left\|\frac{du}{dx}\right\|^{\theta}_{L^r} \|u\|^{1-\theta}_{L^q},$$
where $$\frac{1}{p}=\theta \left(\frac{1}{r}
-1\right)+\left(1-\theta\right)\frac{1}{q},$$ for all $\theta \in
[0,1]$.
\end{thm}

\begin{proof}
See Theorem 9.3, page 29, of  \cite{F}.
\end{proof}

\begin{cor}
\label{LinfHone}
If $u \in H^1$, then $u \in L^\infty$, and
$$
\|u\|_{L^\infty} \le C\|u\|_{H^1}.
$$
\end{cor}

\begin{proof}
This follows from the Gagliardo-Nirenberg Inequality if we take $p=\infty$, $r=q=2$, and $\theta=1/2$.
\end{proof}

We next state a few basic lemmas concerning the operators $T(s)$ defined in \eqref{defTs}.

\begin{lem}\label{lemmaoneone}
For every $u \in L^2$, $\overline{T(s)u}=T^{-1}(s)\overline{u}$.
\end{lem}
\begin{proof}
It is enough to show that the equality holds for $u\in C^{\infty}_0 (\textbf{R})$.\\
For such $u$, we have
\begin{equation*}
\begin{aligned}
\overline{T(s)u}(x)&=\overline{\frac{1}{2\pi} \inftyint e^{ix\omega } \widehat{T(s)u(\omega)}\ d\omega}\\
&=\frac{1}{2\pi} \inftyint e^{-ix\omega } \overline{\widehat{T(s)u(\omega)}}\ d\omega\\
&=\frac{1}{2\pi} \inftyint e^{-ix\omega } \overline{e^{i{\omega}^2 D(s)}\widehat{u(\omega)}}\ d\omega \\
&=\frac{1}{2\pi} \inftyint e^{-ix \omega } {e^{-i{\omega}^2 D(s)}}\overline{\widehat{u(\omega)}}\ d\omega \\
%&=\frac{1}{2\pi} \inftyint e^{-ix\omega } {e^{-i{\omega}^2D(s)}\widehat\overline{u(-\omega)}}\ d\omega , \\
&=\frac{1}{2\pi} \inftyint e^{-ix\omega} e^{-i{\omega}^2 D(s)} \widehat{\overline{u(-\omega)}}\ d\omega \\
&=\frac{1}{2\pi} \inftyint e^{ix\omega} e^{-i{\omega}^2 D(s)} \widehat{\overline{u(\omega)}}\ d\omega \\
&=(T^{-1}(s)\overline{u})(x).
\end{aligned}
\end{equation*}
\end{proof}

\begin{lem}\label{lemmaonetwo}
For $u, v \in L^2$, $$\inftyint u(x)\overline{T^{-1}(s)v(x)}\ dx=\inftyint T(s)u(x)\overline{v(x)}\ dx.$$
\end{lem}
\begin{proof}
By Plancherel's Theorem,
\begin{equation*}
\begin{aligned}
\inftyint u(x)\overline{T^{-1}(s)v(x)}\ dx &=\inftyint \widehat{u}(\omega)\overline{\widehat{T^{-1}(s)v}}(\omega)\ d\omega\\
&=\inftyint \widehat{u}(\omega)\overline{e^{-i\omega^2 D(s)}\widehat{v}}(\omega)\ d\omega\\
&=\inftyint \widehat{u}(\omega) e^{i\omega^2 D(s)}\overline{\widehat{v}}(\omega)\ d\omega\\
&=\inftyint \widehat{T(s)u}(\omega) \overline{\widehat{v}}(\omega)\ d\omega,\\
&=\inftyint T(s)u(x)\overline{v(x)}\ dx.
\end{aligned}
\end{equation*}
\end{proof}

\begin{lem}\label{lemmaonethree}
Let $k\in \textbf{R}$. Then for all $f \in H^k$,
$$\|T(s)f\|_{H^k}=\|f\|_{H^k}.$$
\end{lem}

\begin{proof}
Observe that
\begin{equation*}
\begin{aligned}
\|T(s)f\|^2_{H^k} &= \inftyint (1+|\omega|^2)^k |\widehat{T(s)f}(\omega)|^2 \ d\omega \\
&= \inftyint (1+|\omega|^2)^k |e^{i\omega^2 D(s)}\widehat{f}(\omega)|^2 \ d\omega \\
&= \inftyint (1+|\omega|^2)^k |\widehat{f}(\omega)|^2 \ d\omega \\
&=\|f\|^2_{H^k}.
\end{aligned}
\end{equation*}
\end{proof}

We will also use below the operator $e^{i\partial^2_x t}: L^2 \to L^2$ defined by
$\widehat {e^{i\partial^2_x t}}f [\omega]=e^{-i\omega^2 t}\widehat{f}(\omega)$. The same proof as given in Lemma \ref{lemmaonethree} shows
that $\|e^{i\partial^2_x t}f\|_{H^k}=\|f\|_{H^k}$, for all $f\in H^k$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%------------------------------Begin : Chapter Two---------------------------------%
%---------------------------Well-posedness of the IVP ---------------------------- %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\chapter{Well-posedness of the initial-value problem}
%\renewcommand{\theequation}{\arabic{chapter}.\arabic{equation}}
%\setcounter{chapter}{2} \setcounter{equation}{0}

Rewrite the DMNLS equation \eqref{onethree} as
\begin{equation}
\label{twoone}
i u_t +\alpha u_{xx} + Q(u)=0
\end{equation}
where $\alpha \in \textbf{R}$, and we define $Q(u)$ as
\begin{equation}\label{twotwo}
 Q(u)=\int_0^1  g(s)T^{-1}(s)[ |T(s)u|^2 T(s)u] ds.
\end{equation}
Notice that $Q(u)$ will be well-defined if, for example,
$\left|T(s)u\right|^2T(s)u\in H^r$ for all $s\in[0,1]$ and some $r\in\textbf{R}$.

\begin{definition}
\label{defstrongsoln} Suppose $X \subset H^r$ for some $r \in \mathbf R$, and suppose $u_0 \in X$ and
$M>0$. We say
$u(x,t)\in C([0,M]; X)$ is a strong solution of \eqref{twoone}
with initial data $u_0$ if
\begin{enumerate}[label=(\alph*)]
\item for every $t \in [0,M]$, the function $u= u(t)= u(\cdot,t)$
satisfies $\left|T(s)u\right|^2T(s)u\in H^r$
for almost every $s\in[0,1]$,   so $Q(u)$ is
well-defined,   and
\item for all $t \in [0, M]$,
\begin{equation*}
u=e^{it\alpha\partial^2_x} u_0 + i\int_0^t e^{i\alpha(t-t')\partial^2_x} Q(u(t'))\ dt'.
\end{equation*}
\end{enumerate}
\end{definition}


\section{Conserved quantities}
Define functionals $E: H^1 \to \textbf{R}$ and $P:H^1 \to
\textbf{R}$ by
\begin{equation}
E(u)=\inftyint \int_0^1\left(\alpha \left |u_{x}\right|^2
-\frac12\ g(s)\left|T(s)u\right|^4\right)\ ds\ dx
\label{twothree}
\end{equation}
and
\begin{equation}
P\left(u\right)= \inftyint \left|u\right|^2 dx.
\label{twofour}
\end{equation}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{thm}\label{lemmatwotwo}\
%\begin{enumerate}[label=(\alph*)]
If $u=u\left(x,t\right)\in C([0,M];H^r)$ is a  strong solution of
\eqref{twoone} with $r$ sufficiently large, then
$E\left(u\left(x,t\right)\right)$ and
$P\left(u\left(x,t\right)\right)$ are independent of $t\in[0,M]$.
%\end{enumerate}
\end{thm}

\begin{proof} For $r$ sufficiently large, the assumption that $u=u\left(x,t\right)\in C([0,M];H^r)$
guarantees that $u$ and its derivatives with respect to $x$ are
smooth enough and tend to zero rapidly enough as $|x| \to \infty$,
that the integrations by parts below will be justified.  We assume
in advance that $r$ has been so chosen and $u=u\left(x,t\right)\in
C([0,M];H^r)$ is a strong solution of \eqref{twoone}, so that
\begin{equation}
\label{twofive}
i u_t + \alpha u_{xx}+\int_0^1 g(s) T^{-1}(s) (|T(s) u|^2 T(s) u) ds=0.
\end{equation}
for $t \in [0,M]$.

To prove that $P(u)$ is independent of $t$, first multiply by
$\overline{u}$ and integrate with respect to $x$ over $(-\infty,
\infty)$, to get
\begin{equation}
\label{twosix}
i \inftyint \overline{u} u_t\ dx+ \alpha \inftyint \overline{u} u_{xx}\ dx
+\inftyint \bar{u} \left\{\int_0^1 g(s) T^{-1}(s) (|T(s) u|^2 T(s) u)\ ds\right\}\ dx=0.
\end{equation}
Taking the conjugate gives
\begin{equation}
\label{twoseven}
-i \inftyint \overline{u}_t u\ dx + \alpha \inftyint \overline{u}_{xx} u\ dx
+\inftyint u\int_0^1 g(s)\overline{T^{-1}(s) (|T(s) u|^2 T(s)u)}\ ds\ dx=0,
\end{equation}
and subtracting equation \eqref{twosix} from equation \eqref{twoseven} gives
\begin{equation}
\label{twoeight}
i \inftyint(\overline{u} u_t + \overline{u}_t u)\ dx +\alpha \inftyint(\overline{u} u_{xx} - \overline{u}_{xx} u)\ dx+A-B,
\end{equation}
where
$$A= \inftyint \overline{u}\int_0^1 g(s)T^{-1}(s) (|T(s) u|^2 T(s)u)\ ds\ dx$$
and
$$B= \inftyint u_{xx} \int_0^1 g(s)\overline{T^{-1}(s) (|T(s) u|^2 T(s)u)}\ ds\ dx.$$
Now
\begin{equation*}
\begin{aligned}
A&= \int_0^1 g(s)\inftyint \overline{u} T^{-1}(s) (|T(s) u|^2 T(s)u)\ dx\ ds\\
&= \int_0^1 g(s)\inftyint \overline{u \overline{T^{-1}(s) (|T(s) u|^2 T(s)u)}}\ dx\ ds.\\
\end{aligned}
\end{equation*}
From Lemma \ref{lemmaonetwo} it follows that
\begin{equation*}
\begin{aligned}
A&= \int_0^1 g(s)\inftyint \overline{T(s)u |T(s) u|^2 \overline{T(s)u}}\ dx\ ds\\
&= \int_0^1 g(s)\inftyint \overline{|T(s) u|^4}\ dx\ ds\\
&= \int_0^1 g(s)\inftyint |T(s) u|^4\ dx\ ds.
\end{aligned}
\end{equation*}
Similarly,
\begin{equation*}
\begin{aligned}
B&= \int_0^1 g(s)\inftyint u \overline{T^{-1}(s) (|T(s) u|^2 T(s)u)}\ dx\ ds,
\end{aligned}
\end{equation*}
and from Lemma \ref{lemmaonetwo} it follows that
\begin{equation*}
\begin{aligned}
B&= \int_0^1 g(s)\inftyint T(s)u \overline{(|T(s) u|^2 T(s)u)}\ dx\ ds\\
&= \int_0^1 g(s)\inftyint T(s)u (|T(s) u|^2 \overline{T(s)u})\ dx\ ds\\
&= \int_0^1 g(s)\inftyint |T(s) u|^4\ dx\ ds.
\end{aligned}
\end{equation*}
Therefore, $A-B=0$. Also, an integration by parts shows that $\int \bar{u}_{xx}u
-\bar{u}u_{xx} =0.$
So
\begin{equation*}
\frac{d}{dt}\inftyint |u|^2\ dx=\inftyint(\overline{u} u_t + \overline{u}_t u)\ dx=0,
\end{equation*}
and hence $\inftyint | u|^2\ dx$ is independent of $t$.

To prove $E$ is independent of $t$, write
$$\frac{dE(u)}{dt}=\frac{d}{dt} \inftyint \left\{\alpha|u_x|^2
-\frac{1}{2}\int_0^1 g(s)\left|T(s)u\right|^4\ ds\right\}\ dx$$
$$=\inftyint \left\{\alpha\left(u_x\overline{u_{xt}}+\overline{u_x}u_{xt}\right)
-\frac{1}{2}\int_0^1
g(s)\left(2T(s)uT(s)u_t\overline{\left(T(s)u\right)}^2\right.\right.$$
$$+\left.\left.\left(T(s)u\right)^2 2\overline{\left(T(s)u\right)}\overline{\left(T(s)u_t\right)}\right)\ ds\right\}\ dx,$$
$$=-\alpha\inftyint\overline{u_{xx}} u_t \ dx
-\inftyint\int_0^1 g(s) u_t\overline{T^{-1}(s)\left
\{\left|T(s)u\right|^2 T(s)u\right\}}\ ds\ dx + c.c.,$$
where in the last step we have used Lemma \ref{lemmaonetwo}, and $c.c.$ denotes
the complex conjugate of the first two terms on the right-hand side.

Let $\displystyle{V=\alpha u_{xx}+\int_0^1 g(s)T^{-1}(s)\left\{\left|T(s)u\right|^2 T(s)u\right\}\ ds.}$
Then
$$\frac{dE(u)}{dt}=-\left(\inftyint u_t\overline{V}\ dx+\inftyint \overline{u_t}V\ dx\right).$$
But by equation \eqref{twoone}, $u_t=iV$ and $\overline{u_t}=-i\overline{V}$.
So
$$\frac{dE(u)}{dt}=-\left(\inftyint iV\overline{V}\ dx+\inftyint -i\overline{V}V\ dx\right)=0.$$
 \end{proof}

\begin{lem} Suppose $\alpha \ne 0$.  Then there exists a function $f_\alpha:[0,\infty)\times[0,\infty) \to
[0,\infty)$ such that if $a \ge 0$ and $b \ge 0$, and $u \in H^1$
satisfies $P(u) \le a$ and $|E(u)| \le b$, then
$$
\|u\|_{H^1} \le f_\alpha(a,b).
$$
\label{control_H1}
\end{lem}

\begin{proof} For $u \in H^1$ we have
\begin{equation*}
\begin{aligned}
\left\|u\right\|^2_{H^1}&= \left({\frac{1}{\alpha}}\alpha\right)
\left(\inftyint |u|^2 \ dx +\inftyint |u_x|^2\ dx\right)
\\&= {\frac{1}{\alpha}}\left(\alpha P\left(u\right)+E\left(u\right)+
\frac12\inftyint\int_0^1\left|T(s)u\right|^4\ ds\ d{x}\right)
\\&\leq {\frac{1}{|\alpha|}}\left(|\alpha| a+b+C\int_0^1\left\|T(s)u_{x}\right\|_{L^2}
\left\|T(s)u\right\|^3_{L^2}\ ds\right),
\end{aligned}
\end{equation*}
where in the last step we used the Gagliardo-Nirenberg Inequality,
Theorem \ref{lemmaoneten}, with $p=4$, $r=q=2$, and $\theta=1/4$.
Therefore, by Lemma \ref{onethree},
\begin{equation*}
\begin{aligned}
\left\|u\right\|^2_{H^1}
 &\leq \frac{1}{|\alpha|}
\left(|\alpha| a+ b+C \int_0^1\left\|u_{x}\right\|_{L^2}
\left\|u\right\|^3_{L^2}\ ds\right)
\\&\le \frac{1}{|\alpha|}
\left(|\alpha| a+ b+Ca^3 \|u\|_{H^1}\right).
\end{aligned}
\end{equation*}

This proves there exists $M_1>0$ and $M_2>0$, depending only on
$a$, $b$, and $\alpha$, such that
\begin{equation*}
\left\|u\right\|^2_{H^1}\leq M_1 + M_2 \left\|u\right\|_{H^1},
\end{equation*}
and the conclusion of the Lemma follows easily.
\end{proof}

\section{Well-posedness in $H^1$}
Here is our well-posedness result for the DMNLS equation
\eqref{twoone} in Sobolev spaces $H^r$ of order $r \ge 1$.

\begin{thm}\label{thmtwothree} Suppose $\alpha \in \mathbf R$ and
$r \ge 1$.

\begin{enumerate}
[label=(\alph*)]

 \item  Suppose $\Delta(s)$ and $g(s)$ are integrable on $[0,1]$.  For every $u_0 \in H^r$,
there exists a number $M>0$, which can be chosen to depend only on
$\|u_0\|_{H^r}$, such that equation \eqref{twoone} has a unique
strong solution $u \in C([0,M]; H^r)$ with initial data $u_0$.
Moreover, the solution depends continuously on the initial data;
that is, the map $u_0\mapsto u$ is continuous from $H^r$ to
$C([0,M]; H^r)$.

\item With $M$ as in part $(a)$,  $E(t)$ and $P(t)$ are
independent of $t$ for $t \in [0,M]$.

\item If $\alpha \ne 0$ and $r=1$, then $M$ in part $(a)$ can be taken
arbitrarily large.
\end{enumerate}
\end{thm}

\begin{proof}
Suppose $u_0\in H^r$ is givin. If $M>0$ and $v\in C([0,M]; H^r)$, define
\begin{equation*}
\phi(v)=\exp{\left(it\alpha \partial^2_x\right)} u_0 +\int_0^t
\exp{\left(i\alpha(t-t')\partial^2_x\right)}\left[Q(v)\right]\
dt',
\end{equation*}
where $Q$ is defined in \eqref{twotwo}. Let $|||\cdot|||$ denote
the norm in $C([0, M];H^r)$, so $\displaystyle
|||v|||=\sup_{t\in[0,M]} \|v(.,t)\|_{H^r}$, and let $E(M,a)$ be
the closed ball of radius $a$ in $H^r$, centered at the origin:
$$E(M,a)=\{v\in C([0, M];H^r):|||v|||\leq a\}.$$
We want to show that for every $a>2\|u_0\|_{H^r}$, there exists
$M>0$ such that $\phi$ defines a contraction map on $E(M, a)$.

First we have to show that if $M$ is chosen small enough, then
$\phi$ maps $E(M,a)$ into itself. Suppose $v \in E(M,a)$ and
$a>2\|u_0\|_{H^r}$. Then for all $t\in [0, M]$, by Minkowski's
Integral Inequality,
\begin{equation*}
\|\phi(v)\|_{H^r} \le \|\exp{\left(i\alpha
t\partial^2_x\right)}u_0\|_{H^r} +\int_0^t \| \exp{\left( i\alpha
(t-t')\partial^2_x\right)}Q(v)\|_{H^r}\ dt'.
\end{equation*}
From Lemma \ref{lemmaonethree}, we then have that
\begin{equation*}
\|\phi(v)\|_{H^r}\le \|u_0\|_{H^r} +\int_0^t \|Q(v) \|_{H^r}\ dt'.
\end{equation*}
Now, again by Minkowski's Integral Inequality,
for $v=v(t')$ we have
\begin{equation*}
\begin{aligned}
\|Q(v)\|_{H^r} &=\left\|\int_0^1 g(s)T^{-1}(s)(|T(s)v|^2 T(s)v)\ ds\right\|_{H^r}\\
&\le \beta \int_0^1\|T^{-1}(s)(|T(s)v|^2 T(s)v)\|_{H^r}\ ds;\\
\end{aligned}
\end{equation*}
and hence, by Lemma \ref{lemmaonethree},
\begin{equation*}
\begin{aligned}
\|Q(v)\|_{H^r}&\leq  \beta \int_0^1\| |T(s)v|^2 T(s)v\|_{H^r}\ ds.\\
\end{aligned}
\end{equation*}
From Lemma \ref{lemmaoneseven} it follows that
\begin{equation*}
\|Q(v)\|_{H^r}\le \beta C\int_0^1 \| |T(s)v| \|^2_{H^r}
\|T(s)v\|_{H^r}\ ds,
\end{equation*}
and hence, by Lemma \ref{lemmaonefour},
\begin{equation*}
\|Q(v)\|_{H^r}\le  \beta C\int_0^1 \| T(s)v \|^3_{H^r}\ ds.
\end{equation*}
By Lemma \ref{lemmaonethree},
$$ \|Q(v)\|_{H^r} \le \beta C\int_0^1 \|v\|^3 _{H^r}\ ds=\beta C\|v(t')\|^3 _{H^r}.$$
Since $v\in E(M,a)$, we have $$\|v(t')\|_{H^r}\le a$$ for all $t'
\in[0, M]$. Therefore it follows that
\begin{equation}
\begin{aligned}
\label{twotwelve} \|\phi(v)\|_{H^r}\le \|u_0\|_{H^r} + \int_0^t
\|v(t')\|^3_{H^r}\ dt' \le \|u_0\|_{H^r} +Ca^3 t.
\end{aligned}
\end{equation}
Now choose $M$ such that
\begin{equation}
\begin{aligned}
Ca^3M\le a/2.
\end{aligned}
\label{twothirteen}
\end{equation}
From \eqref{twotwelve} we obtain
\begin{equation*}
\begin{aligned}
\|\phi (v)\|_{H^r} \le a/2 + a/2 =a .
\end{aligned}
\end{equation*}
This proves that $\phi(v)\in E(M,a)$, whenever \eqref{twothirteen} holds and $a>2\|u_0\|_{H^r}$.\\
Also for $u,v \in E(M,a)$, for all $t\in [0,M]$,
\begin{equation*}
\|\phi(u) - \phi(v) \|_{H^r}= \left\|\int_0^t \exp{\left(i\alpha
(t-t')\partial^2_x\right)} [Q(u) - Q(v)]\ dt' \right\|_{H^r}.
\end{equation*}
So, by Minkowski's inequality,
\begin{equation*}
\|\phi(u) - \phi(v)\|_{H^r}\le \int_0^t \|Q(u) - Q(v)\|_{H^r}\
dt'.
\end{equation*}
Now, for each $t'\in [0,t]$, letting $u=u(t')$ and  $v=v(t')$, we have
\begin{equation*}
\begin{aligned}
\|Q(u)-Q(v)\|_{H^r}&=\left\|\int_0^1 \left(g(s)T^{-1}(s)|T(s)u|^2
T(s)u) - g(s) T^{-1}(s)|T(s)v|^2 T(s)v\right)\ ds\right\|_{H^r}
\\&\sup_{s\in[0,1]} \left\|T^{-1}(s)\left(|T(s)u|^2 T(s)u - |T(s)v|^2 T(s)v\right)\right\|_{H^r}\int_0^1 |g(s)|\ ds.
\end{aligned}
\end{equation*}
So, by Lemma \ref{lemmaonethree},
\begin{equation*}
\|Q(u) - Q(v)\|_{H^r}\le \sup_{s\in[0,1]}\left\| |T(s)u|^2 T(s)u -
|T(s)v|^2 T(s)v\right\|_{H^r}\gama,
\end{equation*}
where $\gama=\int_0^1 |g(s)|\ ds<\infty$ by assumption.

Now for all $s\in[0,1]$
\begin{equation*}
\begin{aligned}
&\| |T(s)u|^2T(s)u- |T(s)v|^2T(s) v\|_{H^r}
\\&\le \| |T(s)u|^2 (T(s) u- T(s) v)\\
&\ \ \ \ \ +(T(s) u-T(s) v )\bar{T} (s) u T(s) v+ \bar{(T (s) u- T (s)
v)} (T(s) v)^2 \|_{H^r}
\\&\le \|T(s) u- T(s)v\|_{H^r}\left(\|T(s)v\|^2_{H^r}+\|T(s)u\|^2_{H^r}+\|T(s) v\|_{H^r}\|T(s) u\|_{H^r}\right)
\\&= \|u- v\|_{H^r}\left(\|v\|^2_{H^r}+ \|u\|^2_{H^r} + \|v\|_{H^r} \|u\|_{H^r} \right)
\\&\le 3a^2 \sup_{t\in[0,T]} \|u-v\|_{H^r} = 3 a^2|||u-v|||.
\end{aligned}
\end{equation*}
So for $u,v\in E(M,a)$
\begin{equation*}
\begin{aligned}
|||Q(u)- Q(v)|||
&=\sup_{0\leq t\leq M}\|Q(u)- Q(v)\|_{H^r} \\
&\le \beta \int_0^1 3 a^2 |||u-v|||\ ds\\
&= 3 a^2 \beta |||u-v||| .
\end{aligned}
\end{equation*}
It follows that
\begin{equation*}
\begin{aligned}
|||\phi (u)-\phi (v)|||
&\le \sup_{0\leq t\leq M} \int_0^t 3 a^2\beta |||u-v|||\ dt'\\
&=\sup_{0\leq t\leq M} 3 a^2\beta t |||u-v||| \\
&=3 a^2 \beta C M |||u-v||| .
\end{aligned}
\end{equation*}
If we now choose $M$ such that $3a^2\beta C M < \frac12$, then it
follows that
\begin{equation*}
\begin{aligned}
|||\phi(u)-\phi(v)|||<\frac12 |||u-v|||,
\end{aligned}
\end{equation*}
for all $u,v$ in $E(M,a)$, so $\phi$ is a contraction.

Note that we have shown that if $M$ and $a$ are any positive
numbers such that
\begin{equation}
\begin{aligned}
a &> 2|||u_0|||_{H^r},\\
M &< \frac{1}{2a^2 C},\\
M &< \frac{1}{6a^2\beta C}
\end{aligned}
\label{con_map_conditions}
\end{equation}
all hold, then $\phi:E(M,a) \to E(M,a)$ and
$$
|||\phi(u)-\phi(v)||| \le \frac12 |||u-v|||
$$
for all $u, v \in E(M,a)$.  In particular, from
\eqref{con_map_conditions} we see that $M$ can be chosen to depend
only on $\|u_0\|_{H^r}$, and not otherwise on $u_0$.


By the Banach Contraction Mapping Theorem, $\phi$ has a unique
fixed point $u\in E(M,a)$, which is therefore a strong solution of
\eqref{twoone} in $C\left([0,M]; H^r\right)$.

To prove the continuity of the solution with respect to the
initial data, first fix $u_0 \in H^r$, and define a neighborhood
$\Lambda$ of $u_0$ in $H^r$ by
\begin{equation}
\Lambda = \left\{ \widetilde{u_0} \in H^r: \|u_0 -
\widetilde{u_0}\|_{H^r}< \|u_0\|_{H^r}\right\}.
\end{equation}
Observe that we can extend $\phi$ to a map depending on the
initial data $\widetilde{u_0}$ as a parameter, by defining
$\phi:C\left([0,M];H^r\right)\times \Lambda \to
C\left([0,M];H^r\right)$ as
$$\phi(v, \widetilde{u_0}):=e^{(it\alpha\partial^2_x)}\widetilde{u_0}
+\int_0^t e^{\left(i\alpha(t-t')\partial^2_x\right)}Q(v)\ dt'.$$

Now let
\begin{equation}
a=10\|u_0\|_{H^r}
\end{equation}
and choose $M$ such that the last two inequalities of
\eqref{con_map_conditions} are satisfied.  Then for all
$\widetilde{u_0} \in \Lambda$, we have
$$\|\widetilde{u_0}\|_{H^r}\le 2\|u_0\|_{H^r}$$
and so
\begin{equation}
a\geq 2\|\widetilde{u_0}\|_{H^r}.
\end{equation}
Therefore the inequalities \eqref{con_map_conditions} hold with
$u_0$ replaced by $\widetilde{u_0}$, for all $\widetilde{u_0} \in
\Lambda$.  As noted above after \eqref{con_map_conditions}, it
follows that $\phi:E(M,a) \times \Lambda \to E(M,a)$ and that
$\phi$ is uniformly contractive in the first argument, with
constant $1/2$, for all $\widetilde{u_0}$.  It then follows from
the proof of the Banach Contraction Mapping Theorem (see, for
example, Exercise $A.4$ page $18$ in \cite{GD}), that for each
$\widetilde{u_0} \in \Lambda$, there is a unique fixed point $u$
of $\phi(\cdot,\widetilde{u_0})$ in $E(M,a)$, and the map
$\widetilde{u_0} \mapsto u$ is continuous from $H^r$ to
$C([0,M];H^r)$.  This then completes the proof of part $(a)$ of
the Theorem.

To prove part $(b)$, first approximate $u_0$ in $H^r$ norm by a
sequence $u_{0n}$ of functions in $H^{r_0}$, for $r_0$ large
enough that Theorem \ref{lemmatwotwo} holds with $r=r_0$.  Then by
part $(a)$, for each $n$ there is a strong solution $u_n$ of
\eqref{twoone} in $C([0,M];H^{r_0})$ with initial data $u_0$. Then
from Theorem \ref{lemmatwotwo}, $E(u_n(\cdot,t)) = E(u_{0n})$ and
$P(u_n(\cdot,t)=P(u_{0n})$ for $t \in [0,M]$.  But as shown in
part $(a)$, solutions depend continuously on the initial data in
$H^r$ norm, so since $u_{0n} \to u_{0}$ in $H^r$, then $u_n \to u$
in $C([0,M];H^r)$.  Since $E$ and $P$ are continuous maps from
$H^r$ to $\mathbf R$ for $r \ge 1$, it follows that
$E(u(\cdot,t))=E(u_0)$ and $P(u(\cdot,t))=P(u_0)$ for $t \in
[0,M]$, thus proving part $(b)$ of the Theorem.

To prove part $(c)$, suppose $\alpha \ne 0$ and let $u_0 \in H^1$ be
given.  Let
$$
M_s = \sup \left\{ M > 0:  \text{there exists a strong solution
$u$ of \eqref{twoone} in $C([0,M];H^1)$} \right\}.
$$
We claim that $M_s= \infty$.  To see this, we suppose to the
contrary that $M_s < \infty$ and will get a contradiction.

Let $B=f_\alpha(P(u_0),|E(u_0)|)$, where $f_\alpha$ is the
function defined in Lemma \ref{control_H1}.   By part $(a)$, there
exists a number $M_1 > 0$ such that if $v_0 \in H^1$ and
$\|v_0\|_{H^1} \le B$, then a strong solution $v$ of
\eqref{twoone} with initial data $v_0$ exists in $C([0,M_1];H^1)$.
Let $t_1=M_s - M_1/2$, and let $v_0(x) = u(x,t_1)$, where $u \in
C([0,t_1];H^1)$ is a strong solution of \eqref{twoone} with
initial data $u_0$. By part $(b)$, we have that $P(v_0)=P(u_0)$
and $E(v_0)=E(u_0)$, so by Lemma \ref{control_H1}, $\|v_0\|_{H^1}
\le B$.  Therefore a strong solution $v$ of \eqref{twoone} with
initial data $v_0$ exists in $C([0,M_1];H^1)$.

But it is easy to see that if we define
$$
w(x,t)=\begin{cases} u(x,t) \quad&\text{for $t \in [0,t_1]$}\\
v(x,t-t_1) \quad&\text{for $t \in [t_1,M_1+t_1]$},\end{cases}
$$
then $w$ is a strong solution of \eqref{twoone} in
$C([0,M_1+t_1],H^1)$ with initial data $u_0$.  Since $M_1 + t_1 >
M_s$, this contradicts the definition of $M_s$.  So $(c)$ is
proved.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%            The End of the well-Posedness in H^s here            %
%          Start:  Strichartz and other inequalities (Kunze)       %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\section{Well-posedness in $L^2$ and in $L^2 \cap L^\infty$}

Next we would like to prove a well-posedness result for equation
\eqref{twoone} in $L^2$.  For this purpose we need to put extra assumptions on
$\Delta(s)$; namely that it is piecewise of one sign and bounded away from zero.  So, for the
remainder of Chapter 2, the following assumption will be in force:

\begin{assump}
The function $\Delta(s)$ is integrable on $[0,1]$; and there exist  $\Delta_0 > 0$ and numbers $s_0$, $s_1$, \dots, $s_n$, with
$0=s_0<s_1<s_2<\cdots < s_{n-1}<s_n =1$, such that  for all $j \in \{1,\cdots,n\}$, either
$\Delta(s)\ge \Delta_0$ for almost every $s \in [s_{j-1}, s_j]$, or
$\Delta(s)\le -\Delta_0$ for almost every $s \in [s_{j-1}, s_j]$.
\end{assump}

Recall that $D(s)=\int_0^s \Delta(\zeta)\ d\zeta$.  So it follows from the above assumption that, for all $j \in \{1,\cdots,n\}$, $D(s)$ is absolutely
continuous on $[s_{j-1}, s_j]$ and either $D'(s) \ge \Delta_0$ almost everywhere on $[s_{j-1}, s_j]$, or $D'(s) \le -\Delta_0$ almost everywhere on $[s_{j-1}, s_j]$.


%--------------------Strichartz Estimates-----------------------
\begin{thm}[Strichartz Estimate]\label{lemmatwofour}
Suppose
$2 \le p \le\infty$ and
$$
\frac{2}{q} = \frac12 - \frac{1}{p}\ .
$$
There exists $C>0$ such that for all $f \in L^2$,
%\begin{enumerate}[label=(\alph*)]
%\item
\begin{equation*}
\|e^{it\partial^2_x}f\|_{L^q_tL^p_x}\le  C \|f\|_2.
\end{equation*}
%\item \begin{equation*}
%\left\|\int_{-\infty}^\infty e^{i (t-s)\partial^2_x}g(., s)\ ds\right\|_{L^q_t L^p_x}\le C \|g\|_{L^{q'}_tL^{p'}_x},
%\end{equation*}
%\item  \begin{equation*} \left\|\int_{-\infty}^\infty e^{i s
%\partial^2_x} g(., s)\ ds\right\|_2  \le C \|g\|_{L^{q'}_tL^{p'}_x}.
%\end{equation*}
%\end{enumerate}
\end{thm}

\begin{proof}
See Theorem 4.2 of \cite{LP}.
\end{proof}

\begin{lem}\label{lemmatwofive}
For $\phi \in S(\textbf{R})$ and $t \in \textbf{R}$
 \[\exp(-i\partial^2_x t) \phi (x) = \frac{1}{2\pi} \inftyint \frac{1}{\sqrt{4\pi i t}}\exp\left[-i\left(\frac{(x-y)^2} {4 t}\right)\right]\phi(y)\ dy\]
\end{lem}
\begin{proof}
See \cite{LP}, page 17.
\end{proof}

In the next theorem we verify that the Strichartz estimate in Theorem \ref{lemmatwofour} remains valid if $\displaystyle e^{it\partial^2_x}$ is replaced by $\displaystyle T(t)=e^{iD(t)\partial_x^2}$.
A similar result appears in Lemma 2.5 of \cite{ASS}.

\begin{thm}\label{theoremtwosix}  Suppose $D'(s)=\Delta(s)$ satisfies Assumption D1.
Assume $q\in [4,\infty],$ and define $r\in[2,\infty]$ by
$$
\frac{2}{q}=\frac12-\frac{1}{r}.
$$
Then for all $u \in L^2$,
\begin{equation}
\label{twosixteen} \|T(t)u\|_{{L^q_t}((0,1),
L^r_x)}:=\left(\int_0^1 \|T(t)u\|^q_{L^r}\ dt\right)^{1/q}\leq
C\|u\|_{L^2},
\end{equation}
with $C$ depending only on $q.$
\end{thm}

\begin{proof}
We have
\begin{equation*}
\begin{aligned}
\|T(t)u\|_{{L^q_t}((0,1),
L^r_x)}&=\left(\int_0^1 \|T(t)u\|^q_{L^r}\ dt\right)^{1/q}
\\&=\left(\sum_{j=1}^n\int_{s_{j-1}}^{s_j} \|T(t)u\|^q_{L^r}\ dt\right)^{1/q}
\\&\le C \sum_{j=1}^n\left(\int_{s_{j-1}}^{s_j} \|T(t)u\|^q_{L^r}\ dt\right)^{1/q}
\\&=C \sum_{j=1}^n\|T(t)u\|_{{L^q_t}((s_{j-1},s_j),
L^r_x)},
\end{aligned}
\end{equation*}
where $s_0,\dots,s_n$ are the numbers in Assumption D1.  By Assumption D1 it is enough to prove the estimate \eqref{twosixteen} on an arbitrary finite interval $[a,b]$ instead of $[0,1]$, under the assumption
that $D(t)$ is absolutely continuous on $[a,b]$ and either $D'(t) \ge \Delta_0$ almost everywhere on $[a,b]$ or $D'(t) \le -\Delta_0$ almost everywhere on $[a,b]$.

From $L^p$ duality it follows that
\begin{equation}
\label{twoseventeen}
\begin{aligned}
&\|T(\cdot)u\|_{{L^q_t}((a,b), L^r_x)} =\\
 &\ \ = \sup\left\{\int_a^b
(T(t)u,\phi(t))_{L^2}\ dt: \phi\in L^{q'}\left((a,b),L^{r'}\right)
\text{and}\ \|\phi\|_{{L^{q'}}((a,b),L^{r'})}=1\right\},
\end{aligned}
\end{equation}
where $\displaystyle\frac{1}{q'}+\frac{1}{q}=1$ and
$\displaystyle\frac{1}{r'}+\frac{1}{r}=1$. Now
\begin{equation}
\label{twoeighteen}
\begin{aligned}
\int_a^b \left(T(t)u,\phi(t)\right)_{L^2}\ dt&=\int_a^b \left(u,T^*(t) \phi(t)\right)_{L^2}\ dt\\
&= \left(u, \int_a^b T^*(t) \phi(t)\ dt\right)_{L^2}\\
&\leq \|u\|_{L^2} \left\|\int_a^b T^*(t) \phi(t)\ dt\right\|_{L^2},
\end{aligned}
\end{equation}
where $T^*\left( t\right)$ is the adjoint of $T\left(t\right)$. Notice that since $T(t)$ is unitary by Lemma \ref{lemmaonethree}, then we have that
 $T^*\left(t\right)=T^{-1}\left(t\right).$

 From \eqref{twoseventeen} and \eqref{twoeighteen} it follows that to prove the theorem, it is enough to show that for all $\phi \in L^{q'}\left((a,b),L^{p'}\right),$
\begin{equation}
\label{twoninteen}
\left\|\int_a^b T^*(t)\phi(t)\ dt\right\|_{L^2}\leq C \|\phi\|_{L^{q'}((a,b),L^{r'})}.
\end{equation}

We have
\begin{equation}
\label{twotwenty}
\begin{aligned}
\left\|\int_a^b T^*(t) \phi(t)\ dt\right\|^2_{L^2}&= \left(\int_a^b T^*(t) \phi(t)\ dt,\int_a^b T^*(s) \phi(s)\ ds\right)_{L^2}
\\&= \int_a^b \int_a^b \left( T^*(t)\phi(t),T^*(s) \phi(s)\right)_{L^2}\ dt\ ds
\\&= \int_a^b \left( \phi(t),\int_a^b T(t)T^*(s)\phi(s)\ ds\right)_{L^2}\ dt.
\end{aligned}
\end{equation}
Let
\begin{equation}\label{twotwentyone}
\theta_{\phi}(t) = \int_a^b T(t)T^*(s) \phi(s)\ ds.
\end{equation}
Then by using H\"{o}lder's Inequality, first in the space variable and then in the time variable, we obtain
\begin{equation}
\label{twotwentytwo}
\begin{aligned}
&\int_a^b \left( \phi(t),\int_a^b T(t)T^*(s) \phi(s)\ ds\right)_{L^2}\ dt\leq\int_a^b \|\phi(t)\|_{L^{r'}}\|\theta_{\phi}(t)\|_{L^r}\ dt\\
&\leq\left(\int_a^b\|\phi(t)\|^{q'}_{L^{r'}}\ dt\right)^{1/q'}\left(\int_a^b\|\theta_{\phi}(t)\|^q_{L^{r}}\ dt\right)^{1/q}
=\|\phi\|_{L^{q'}((a,b),L^{r'})}\|\theta_{\phi}\|_{L^{q}((a,b),L^{r})}.
\end{aligned}
\end{equation}
So to prove \eqref{twoninteen}, by \eqref{twotwenty} and \eqref{twotwentytwo}, it is enough to show that
\begin{equation}\label{twotwentyonenew}
\|\theta_{\phi}\|_{L^{q}((a,b),L^{r})}\leq C \|\phi\|_{L^{q'}((a,b),L^{r'})}.
\end{equation}

By Minkowski's Integral Inequality, for each $t\in (0,1)$ we have
\begin{equation*}
\begin{aligned}
\|\theta_{\phi}(t)\|_{L^{r}}= \left\|\int_a^b T(t)T^*(s)\phi(s)\
ds\right\|_{L^r} \leq \int_a^b \|T(t)T^*(s) \phi(s)\|_{L^r}\ ds.
\end{aligned}
\end{equation*}
We have $$\mathcal{F}\left({T(t)T^*(s)\phi(s)}\right)=\exp{\left(i\omega^2(D(t)-D(s))\right)}\ \widehat{\phi}(s),$$
so $$T(t)T^*(s)\phi(s)=\exp{\left(-i\partial^2_x (D(t)-D(s))\right)}\phi(s),$$
and hence by Lemma \ref{lemmatwofive},
\begin{equation*}
\left(T(t)T^*(s)\phi(s)\right)[x]=\frac{1}{2\pi}{\frac{1}{\sqrt{4\pi i (D(t)-D(s))}}}\inftyint\exp{\left(\frac{-i|x-y|^2}{4(D(t)-D(s))}\right)}\ \phi(s,y)dy.
\end{equation*}
Taking the supremum over $x\in \textbf{R}$ gives
\begin{equation}\label{twotwentytwonew}
\|T(t)T^*(s)\phi(s)\|_{L^\infty}\leq
{\frac{C\|\phi(s)\|_{L^1}}{|D(t)-D(s)|^{1/2}}}  .
\end{equation}
On the other hand, since $T$ is unitary, we have
\begin{equation}\label{twotwentythreenew}
\|T(t)T^*(s)\phi(s)\|_{L^2}= \|\phi(s)\|_{L^2}.
\end{equation}
From \eqref{twotwentytwonew}, \eqref{twotwentythreenew} and the
Riesz-Thorin Interpolation Theorem (Theorem \ref{lemmaonefive}),
we get
\begin{equation}
\|T(t)T^*(s)\phi(s)\|_{L^r} \leq
\left(\frac{C}{\sqrt{|D(t)-D(s)|}}\right)^{1-(2/r)} C^{2/r}
\|\phi(s)\|_{L^{r'}},
\end{equation}
for some constant $C$ independent of $\phi$.
So
\begin{equation}
\|\theta_{\phi}(t)\|_{L^r} \leq C\int_a^b{\frac{
\|\phi(s)\|_{L^{r'}}}{|D(t)-D(s)|^{(1/2)-(1/r)}}}\ ds.
\end{equation}

But from our assumption on $D(t)$, it follows that either
\begin{equation*}
D(t)-D(s)=\int_s^t \Delta(\zeta)\ d\zeta \ge \Delta_0 |t-s|
\end{equation*}
or
\begin{equation*}
D(t)-D(s)=\int_s^t \Delta(\zeta)\ d\zeta \le -\Delta_0 |t-s|,
\end{equation*}
for all $s$ and $t$ in $(a,b)$, so in any case
\begin{equation*}
|D(t)-D(s)| \ge\Delta_0 |t-s|.
\end{equation*}
Therefore
\begin{equation}
\label{thetaphilessthang}
\|\theta_{\phi}(t)\|_{L^r} \leq \frac{C}{\Delta_0^{(1/2)-(1/r)}}\ g(t),
\end{equation}
where
\begin{equation*}
g(t):= \int_a^b{\frac{ \|\phi(s)\|_{L^{r'}}}{|t-s|^{
(1/2)-(1/r)}}}\ ds.
\end{equation*}

Now we consider separately the cases when $r > 2$ and $r=2$.

If $r > 2$, we define
$$
\beta=\frac{1}{2}+\frac{1}{r},
$$
and we have that $\beta \in(0,1)$.  Define $f(s)$ for $s \in \mathbf R$ by
$$
f(s):=
\begin{cases}
\|\phi(s)\|_{L^{r'}} & \text{for $s\in[a, b]$},\\
0  & \text{for $s \notin [a, b]$}.
\end{cases}
$$
Then $g(t)$ is the Hardy-Littlewood fractional integral $g(t)=I_\beta f(t)$, in the notation of Theorem \ref{lemmaonesix}.
From the relation $\displaystyle
\frac{2}{q}=\frac{1}{2}-\frac{1}{r}$, we see that $\displaystyle
\frac{1}{q}=\frac{1}{q'}-\beta$. Therefore Theorem \ref{lemmaonesix} gives
$$\|I_\beta f\|_{L^q}\leq C\|f\|_{L^{q'}},$$
and so
\begin{equation}
\label{twotwentyseven}
\begin{aligned}
\|g\|_{L^q[a,b]}&=\|I_{\beta}f\|_{L^q[a,b]}
\leq \|I_{\beta}f\|_{L^q (\textbf{R})}
\\& \leq C\|f\|_{L^{q'}}=C\left(\int_a^{b}\|\phi(s)\|^{q'}_{r'}\ ds\right)^{1/q'}=C \|\phi\|_{L^{q'}((a,b),L^{r'})}.
\end{aligned}
\end{equation}
Together with \eqref{thetaphilessthang}, this proves \eqref{twotwentyonenew} when $r>2$.

In the remaining case, when $r=2$, we have $q=\infty$ and $q' =1$, so
$$\|g\|_{L^q[a,b]}=\sup_{t\in [a,b]} \int_a^b\frac{\|\phi(s)\|_{r'}}{|t-s|^0} \ ds
= \left(\int_a^b\|\phi(s)\|^{q'}_{r'}\ ds\right)^{1/q'},$$
so \eqref{twotwentyseven} again holds, and as before \eqref{twotwentyonenew} then follows from \eqref{thetaphilessthang}.

As remarked above, from \eqref{twotwentyonenew} we then deduce \eqref{twoninteen}, and from there we get \eqref{twosixteen}.
So the proof of the Theorem is complete.
 \end{proof}


\begin{cor}\label{corollarytwoseven}
Suppose $u\in C\left([0,M];L^2\right)$ for some $M>0$. Then for
every $t\in [0, M]$, if $u=u(\cdot,t)$, we have
$\left|T(s)u\right|^2T(s)u\in L^2$ for almost every $s\in [0,1].$
\end{cor}

\begin{proof}
Let $t\in [0,M]$ be given. By Theorem \ref{theoremtwosix} with $r=6$ and $q=6$, we
have that for $u=u(.,t),$
$$\int_0^1 \left\|T(s)u\right\|^6_{L^6}\ ds\leq C\left\|u\right\|_{L^2}^6<\infty.$$
So for almost every $s\in [0,1]$, we have that $T(s)u\in L^6$, and
therefore $\left|T(s)u\right|^2 T(s)u\in L^2.$
\end{proof}

The following Lemma is from \cite{K}, where it is proved for the case where $\Delta(t)$
is piecewise constant.

\begin{lem}\ \label{lemmatwoeightnew}
Suppose $D'(s)=\Delta(s)$ satisfies Assumption $D1$ and $g(s)$ is bounded on $[0,1]$,
\begin{enumerate}[label=(\alph*)]
\item We have $Q(u)\in L^{\infty}$ for $u\in L^2$, and
\begin{equation}
\|Q(u)-Q(v)\|_{L^{\infty}}\le C\left(\|u\|_{L^2}+\|v\|_{L^2}\right)^2 \|u-v\|_{L^2},
\end{equation}
for $u,v \in L^2$.
In particular, $\|Q(u)\|_{L^{\infty}}\leq C \|u\|_{L^2}^3$ for $u \in L^2.$
\item We have $Q(u)\in L^2$ for $u\in L^2$, and
\begin{equation*}
\|Q(u)-Q(v)\|_{L^2}\leq C\left(\|u\|_{L^2}+\|v\|_{L^2}\right)^2 \|u-v\|_{L^2},
\end{equation*}
for $u,v \in L^2$.
In particular, $\|Q(u)\|_{L^2}\leq C \|u\|_{L^2}^3$ for $u \in L^2.$
\end{enumerate}
\end{lem}

\begin{proof}
We have
\begin{equation*}
\begin{aligned}
|Q(u)-Q(v)|&\leq C\int_0^1 |T^*(t)\left\{|T(t)u|^2T(t)u-|T(t)v|^2T(t)v\right\}|\ dt\\
&= C\sum_{i=1}^n \int_{s_{i-1}}^{s_i}\left|T^*(t)\left\{|T(t)u|^2 T(t)u-|T(t)v|^2 T(t)v\right\}\right|\ dt,\\
\end{aligned}
\end{equation*}
where $s_0, \dots, s_n$ are the numbers in Assumption D1.  From  Lemma \ref{lemmatwofive} and the
triangle inequality and using that $g(t)$ is bounded on $[0,1]$, we obtain that
\begin{equation*}
\begin{aligned}
|Q(u)-Q(v)|&\leq C\sum_{i=1}^n \int_{s_{i-1}}^{s_i}\frac{1}{\sqrt{|D(t)|}}(\||T(t)u|^2\left[T(t)u-T(t)v\right]\|_{L^1}\\
\\&+\|T(t)uT(t)v\left[\overline{T(t)u}-\overline{T(t)v}\right]\|_{L^1}+ \|{|T(t)v|^2[T(t)u-T(t)v]\|_{L^1})\ dt,\\
&\leq C\sum_{i=1}^n \int_{s_{i-1}}^{s_i}\frac{1}{\sqrt{|D(t)|}}(\|T(t)u\|^2_{L^3}\|T(t)u- T(t)v\|_{L^3}
\\&+\|T(t)u\|_{L^3} \|T(t)v\|_{L^3}\|T(t)u- T(t)v\|_{L^3}\\
&+ \|T(t)v\|^2_{L^3} \|T(t)u- T(t)v\|_{L^3})\ dt.\\
\end{aligned}
\end{equation*}
Applying the Generalized H\"{o}lder's Inequality with exponents $\frac{4}{3}$, $6$, and $12$, or
$\frac{4}{3}$, $12$ $12$, and $12$, we get
\begin{equation*}
\begin{aligned}
|Q(u)-Q(v)|&\leq C\sum_{i=1}^n \left[\int_{s_{i-1}}^{s_i} |D(t)|^{-2/3}\ dt\right]^{3/4}\\
&\left(\|T(.)u\|^2_{L^{12}_t(L^3_x)}+ \|T(.)u\|_{L^{12}_t(L^3_x)}\|T(.)v\|_{L^{12}_t(L^3_x)}\right) \|T(.)(u-v)\|_{L^{12}_t(L^3_x)},\\
\end{aligned}
\end{equation*}
and so, by using Theorem \ref{theoremtwosix}, we obtain
\begin{equation*}
\begin{aligned}
|Q(u)-Q(v)|&\leq C\sum_{i=1}^n \left[\int_{s_{i-1}}^{s_i} |D(t)|^{-2/3}
\ dt\right]^{3/4}\left(\|u\|^2_{L^2}+\|v\|^2_{L^2}\right)\|u-v\|_{L^2}.
\end{aligned}
\end{equation*}

Now we claim that our assumption D1 on $\Delta(s)$ implies that $\displaystyle D(t)=\int_0^t\Delta(s)\ ds$ satisfies
$\displaystyle\int_{s_{i-1}}^{s_i} |D(t)|^{-2/3}\ dt < \infty$  for each $i=1,2,\cdots,n$.   To see this,
notice that if $D(t_0)=0$ for any $t_0 \in [s_{i-1},s_i]$, then since Assumption D1 implies
$|D(t)-D(t_0)| \ge \Delta_0 |t-t_0|$ for all $t \in [s_{i-1},s_i]$, we have in particular that
$|D(t)|^{-2/3}\le (\Delta_0 |t-t_0|)^{-2/3}$ for $t$ in a neighborhood of $t_0$, and therefore
$|D(t)|^{-2/3}$ is integrable in a neighborhood of $t_0$, proving the claim.

It follows then that we have
\begin{equation*}
 |Q(u)-Q(v)|\leq C \left(\|u\|^2_{L^2}+\|v\|^2_{L^2}\right)\|u-v\|_{L^2}.
\end{equation*}
This proves part $(a)$ of the Lemma.

To prove part $(b)$, first let us define $$Q(u,v,w)=\int_0^1
g(s)T^{-1}(s)\left\{T(s)u\overline{T(s)v}T(s)w\right\}\ ds.$$
 Then by applying the Generalized H\"older's Inequality, first in
 the $x$-variable with exponents 3, 3, and 3; and then in the $s$-variable with exponents 6, 6, 6, and 2; we obtain
\begin{equation*}
\begin{aligned}
&\|Q(u, v, w)\|_{L^2}\leq C\int_0^1\|T^{-1}(s)\left(T(s)u\ \overline{T(s)v}\ T(s)w\right)\|_{L^2}\ ds\\
& \quad= \int_0^1\|\left(T(s)u\ \overline{T(s)v}\ T(s)w\right)\|_{L^2}\ ds\\
&\quad\leq C\int_0^1 \left\|T(s)u\right\|_{L^6} \left\|T(s)v\right\|_{L^6}\left\|T(s)w\right\|_{L^6} \ ds\\
&\quad\leq
C\|T(s)u\|_{L^6\left((0,1),L^6_x\right)}\|T(s)v\|_{L^6\left((0,1),L^6_x\right)}
\|T(s)w\|_{L^6\left((0,1),L^6_x\right)}\left(\int_0^1 1^2\ dx\right)^{1/2}\\
&\quad \le C\|u\|_{L^2}\|v\|_{L^2}\|w\|_{L^2},
\end{aligned}
\end{equation*}
where in the last inequality we have used Theorem
\ref{theoremtwosix}. Taking $u=v=w$ now gives
$$\|Q(u)\|_{L^2}\leq C\|u\|^3_{L^2}.$$

It follows from what we have just proved that
\begin{equation}
\begin{aligned}
\left\|Q(u)-Q(v)\right\|_{L^2}&=\left\|Q(u-v,u,u)-Q(v,v-u,u)+Q(v,v,u-v)\right\|_{L^2}\\
&\leq \left\|Q(u-v,u,u)\right\|_{L^2}+\left\|Q(v,v-u,u)\right\|_{L^2}+\left\|Q(v,v,u-v)\right\|_{L^2}\\
&\leq C\left(\left\|u-v\right\|_{L^2}\left\|u\right\|^2_{L^2}
+\left\|u-v\right\|_{L^2}\left\|u\right\|^2_{L^2}\left\|v\right\|^2_{L^2}
+\left\|u-v\right\|_{L^2}\left\|v\right\|^2_{L^2}\right)\\
&\leq C\left\|u-v\right\|_{L^2} \left(\left\|u\right\|_{L^2}+\left\|v\right\|_{L^2}\right)^2.
\end{aligned}
\end{equation}
Thus $(b)$ has been proved. \end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                            The well-Posedness in L^2 starts here                                      %
%The proof of the well-posedness in H^s ends here and next the proof is about the well-posedness in L^2 %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{thm}\ \label{theoremtwonine}
Suppose $D'(s)=\Delta(s)$ satisfies Assumption $D1$ and $g(s)$ is bounded on $[0,1]$.
\begin{enumerate}[label=(\alph*)]
\item Suppose $\alpha\in \textbf{R}$. For every $u_0 \in L^2$,
there exists $M>0$ such that the DMNLS equation \eqref{twoone} has
a unique strong solution $u \in C([0,M];L^2)$ with initial data
$u_0$. Moreover, the solution depends continuously on the initial
data; that is, the map $u_0\mapsto u$ is continuous from $L^2$ to
$C([0,M];L^2)$.

\item With $M$ as in part $(a)$,  $E(t)$ and $P(t)$ are
independent of $t$ for $t \in [0,M]$.

 \item The number $M$ in $(a)$ can be taken
arbitrarily large.
\end{enumerate}
\end{thm}

\begin{proof}\
As in the proof of Theorem \ref{thmtwothree}, define
\begin{equation*}
\phi(v)=\exp(i\alpha t\partial^2_x) u_0 +i\int_0^t \exp(i\alpha
(t-t')\partial^2_x)[ Q(u)]\ dt',
\end{equation*}
where $Q$ is as defined in \eqref{twotwo}.

By Corollary \ref{corollarytwoseven}, if $u\in C([0,M];L^2)$ then
$u$ satisfies part $(a)$ of Definition \ref{defstrongsoln} for
$r=0$.  Therefore to prove that such a $u$ is a strong solution in
$L^2$, it is enough to show that $\phi(u)=u$, or in other words
that $\phi$ has a fixed point in $C([0,M];L^2)$ for some $M>0$.

Let
\begin{equation*}
\begin{aligned}
E(M,a)=\{v\in C([0, M]; L^2(\textbf{R})) :
|||v|||=\sup_{t\in[0,M]}  \|v(t)\|_{L^2}\leq a\}.
\end{aligned}
\end{equation*}
We want to show that for every $a>2\|u_0\|_{L^2}$ there exists $M>0$ such that $\phi$ defines a contraction
map on $E(M, a)$.

First we have to find $M$ such that $\phi: E(M,a)\rightarrow E(M,a)$.

By using Lemma \ref{lemmaonethree} and Lemma \ref{lemmatwoeightnew}(a), we get
\begin{equation*}
\begin{aligned}
&\|\phi (u)\|_{L^2} \le \|e^{i\alpha \partial^2_x t} u_0\|_{L^2}
+\left\| \int_0^t e^{ i \alpha \partial^2_x (t-t')} Q(u)\ dt'\right\|_{L^2} \\
&\le \|u_0\|_{L^2} +  \int_0^t \|e^{ i\alpha \partial^2_x (t-t')} Q(u) \|_{L^2}dt'  \\
%&\le \|u_0\|_{L^2} +  \int_0^t \|e^{ i\partial^2_x (-t')} Q(u)\|_{L^2}dt'\\
&= \|u_0\|_{L^2} +  \int_0^t \|Q(u)\|_{L^2}dt'\\
&\le\|u_0\|_{L^2}+C M \sup_{0\leq t\leq M}\|u\|^3_{L^2} \\
&\le \frac{a}{2}+C M a^3 \\
\end{aligned}
\end{equation*}

So
\begin{equation*}
|||\phi(u)||| \le \frac{a}{2}+CM a^3.
\end{equation*}
Choose $M$ such that
\begin{equation}
C a^3 M \le a/2,
\label{CT}
\end{equation}
then
\begin{equation*}
|||\phi (u)|||_{M} \le a/2 + a/2 =a .
\end{equation*}
This proves $\phi(u)\in E(M,a)$, whenever \eqref{CT} holds and $a>2\|u_0\|_{L^2}$.

Similarly, by using Lemma \ref{lemmatwoeightnew}(a), we can show for $u,v \in E(M,a)$, for all $t\in [0,M]$,
\begin{equation*}
\begin{aligned}
|||\phi(u)- \phi(v)||| &=\sup_{0\leq t\leq M}\|Q(u)- Q (v)\|_{L^2} \\
&\le\sup_{0\leq t\leq M} a^2 C t |||u-v|||\\
&=a^2 C M |||u-v|||.
\end{aligned}
\end{equation*}

Choose $M$ such that $a^2 C M < \frac{1}{2}$, then
\begin{equation*}
|||Q(u)-Q(v)|||<\frac{1}{2} |||u-v|||
\end{equation*}
on $E(M,a)$, so $\phi$ is a contraction map.
So by the Banach contraction mapping theorem, $\phi$ has a fixed point $u\in E(M,a)$.
Thus $u$ is a strong solution of \eqref{onethree} in $C([0,M];L^2).$ This proves existence of a
solution in $C([0,M];L^2)$ for some $M>0$, completing the proof of $(a)$.

Part $(b)$ of the theorem now follows from exactly the same considerations as used to prove part $(b)$ of Theorem
\ref{thmtwothree}, so we can omit the details here.

The proof of part $(c)$ is even simpler here than in Theorem \ref{thmtwothree}, since here we need only the fact that $P(u)$ is independent of time
for solutions of \eqref{twoone}.  Let
$$
M_s = \sup \left\{ M > 0:  \text{there exists a strong solution
$u$ of \eqref{twoone} in $C([0,M]; L^2)$} \right\}.
$$
We claim that $M_s= \infty$.  To see this, we suppose to the
contrary that $M_s < \infty$ and will get a contradiction.

 Let $B=\|u_0\|_{L^2}$.   By part $(a)$, there
exists a number $M_1 > 0$ such that if $v_0 \in L^2$ and
$\|v_0\|_{H^1} \le B$, then a strong solution $v$ of
\eqref{twoone} with initial data $v_0$ exists in $C([0,M_1]; L^2)$.
Let $t_1=M_s - M_1/2$, and let $v_0(x) = u(x,t_1)$, where $u \in
C([0,t_1];L^2)$ is a strong solution of \eqref{twoone} with
initial data $u_0$. By part $(b)$, we have that $P(v_0)=P(u_0)$, so $\|v_0\|_{L^2}
=\|u_0\|_{L^2}= B$.  Therefore a strong solution $v$ of \eqref{twoone} with
initial data $v_0$ exists in $C([0,M_1];H^1)$. But, as explained in the proof of Theorem \ref{thmtwothree},
it follows that there exists a strong solution $w$ of \eqref{twoone} in
$C([0,M_1+t_1]; L^2)$ with initial data $u_0$,  and since $M_1 + t_1 >
M_s$, this contradicts the definition of $M_s$.  So $(c)$ is
proved.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%           The well-possedness in L^2 \intersects L^\infty when \alpha=0         %                                                  %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{thm}\label{theoremtwoten}
Suppose $\alpha=0$. Let $u_0 \in  L^2\cap L^{\infty}$.  There exists
$M>0$ such that a unique strong solution of
\eqref{twoone} with initial data $u_0$ exists in $C([0, M]; L^2
\cap L^\infty)$. Moreover, the solution depends continuously on
the initial data; that is, the map $u_0\mapsto u$ is continuous
from $L^2\cap L^{\infty}$ to $ C([0, M]; L^2 \cap L^{\infty})$.
\end{thm}

\begin{proof}
According to Definition \ref{defstrongsoln} in the case $\alpha = 0$, what we are looking for is a number
$M > 0$ and a function $u\in C\left([0, M]; L^2 \cap L^\infty\right)$ such that
\begin{equation}
u(t)=u_0 + i \int_0^t Q(u(t'))\ dt'
\end{equation}
for all $t\in [0,M]$.

For a given $M>0$, define $\phi(u)$ for
$u\in C\left([0, M]; L^2 \cap L^{\infty}\right)$ by
\begin{equation}
\phi(u)=u_0 + i \int_0^t Q(u(t'))\ dt'.
\end{equation}
Let
$$E(M,a)=\{u\in C\left([0, M]; L^2 \cap L^\infty \right)  :|||u|||=\sup_{t\in[0,M]}  \|u(.,t)\|_{L^2}+\|u(.,t)\|_{L^\infty}<\infty\}.$$
Suppose $u_0 \in  L^2 \cap L^{\infty}$. We want to show that for every $a>4(\|u_0\|_{L^2}+\|u_0\|_{L^\infty})$ there exists $M>0$ such that $\phi$ defines a contraction map on $E(M, a)$.

First we have to show
$\phi: E(M,a)\rightarrow E(M,a)$. For all $t\in[0,M]$,

\begin{equation*}
\begin{aligned}
\|\phi (u)\|_{L^\infty} &\le \|u_0\|_{L^\infty}+\left\| \int_0^t Q (u)\ dt'\right\|_{L^\infty} \\
&\le a/4 +  M\sup_{0\le t \le M} \|Q(u)\|_{L^\infty}.
\end{aligned}
\end{equation*}
By using Lemma \ref{lemmatwoeightnew}(a), we then get
\begin{equation*}
\begin{aligned}
\|\phi(u)\|_{L^\infty} &\le a/4 + M\sup_{0\le t \le M} \|u(t)\|^3_{L^2} \\
&\le  a/4 +  M a^3 \\
\end{aligned}
\end{equation*}

Now, for fixed $t\in[0, M]$,
\begin{equation*}
\begin{aligned}
&\|\phi(u)\|_{L2} \le \|u_0\|_{L^2} +\left\| \int_0^t Q(u)\ dt'\right\|_{L^2} \\
&\le a/4 + \int_0^t \|Q(u)\|_{L^2} dt' \\
\end{aligned}
\end{equation*}
By using Lemma \ref{lemmatwoeightnew}(b), we get
\begin{equation*}
\begin{aligned}
\|\phi (u)\|_{L2}&\le a/4 + M C  \|u\|^3_{L^2} \\
&\le a/4 + M C a^3 .\\
\end{aligned}
\end{equation*}
So
\begin{equation*}
|||\phi(u)||| \le  a/2 +  2M C a^3 .
\end{equation*}

Choosing $M$ such that
\begin{equation}\label{twoeightyfour}
C a^3 M \le a/4,
\end{equation}
we get
\begin{equation*}
|||\phi (u)||| \le a/2 + a/2 =a .
\end{equation*}
This proves that $\phi(u)\in E(M, a)$, whenever \eqref{twoeightyfour} holds and
$$a>4(\|u_0\|_{L^2}+\|u_0\|_{L^\infty})$$.
Similarly, by using Lemma \ref{lemmatwoeightnew}(a), for $u,v \in E(M, a)$, for all $t\in [0, M]$,
\begin{equation*}
\begin{aligned}
\|\phi(u) - \phi(v)\|_{L^\infty}&=\left\|\int_0^t [Q(u) - Q(v)]\ dt'\right\|_{L^\infty}\\
&\le \int_0^t \|Q(u) - Q(v)\|_{L^\infty}\ dt'\\
&\le \beta C  \sup_{0 \le t \le M} t \left(\|u\|_{L^2} + \|v\|_{L^2} \right)^2 \|u-v\|_{L^2}\\
\end{aligned}
\end{equation*}

\begin{equation*}
\begin{aligned}
&\le \beta C M \sup_{0 \le t \le M} \left(\|u\|_{L^2} + \|v\|_{L^2} \right)^2 \|u-v\|_{L^2}\\
&\le \beta C M a^2  \sup_{0 \le t \le M} \|u-v\|_{L^2}.\\
\end{aligned}
\end{equation*}

By using Lemma \ref{lemmatwoeightnew}(b), for $u,v \in E(M, a)$, for all $t\in [0, M]$,
\begin{equation*}
\begin{aligned}
\|\phi(u) - \phi(v)\|_{L^2}&=\left\|\int_0^t [Q(u) - Q(v)]\ dt'\right\|_{L^2}\\
&\le \int_0^t \|Q(u) - Q(v)\|_{L^2}\ dt'\\
\end{aligned}
\end{equation*}
\begin{equation*}
\begin{aligned}
&\le \beta C \sup_{0 \le t \le M} t \left(\|u\|_{L^2} + \|v\|_{L^2} \right)^2 \|u-v\|_{L^2}\\
&\le \beta C Ma^2 \sup_{0 \le t \le M} \|u-v\|_{L^2}.\\
\end{aligned}
\end{equation*}

Now we have
\begin{equation*}
\begin{aligned}
|||\phi(u)-\phi(v)||| &\le \sup_{0\le t\le M} \|\phi(u)-\phi(v)\|_{L^2} + \|\phi(u) - \phi(v) \|_{L^\infty}\\
&\le 2\beta C M a^2 \sup_{0 \le t \le M} \|u-v\|_{L^2}.
\end{aligned}
\end{equation*}
Choose $M$ such that $2a^2\beta C M < \frac{1}{2}$, then
\begin{equation*}
|||\phi(u)-\phi(v)|||\le \frac{1}{2} |||u-v|||
\end{equation*}
on $E(T,a)$, so $\phi$ is a contraction map. So by the Banach
Contraction Mapping Theorem, $\phi$ has a fixed point $v\in
E(M,a)$. Therefore $u$ is a strong solution of \eqref{twoone} in
$C\left([0,M]; L^2\cap L^{\infty}\right)$ for some $M>0$.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                      The well-Posedness in L^8L^4 starts here                                           %                         %
%The proof of the well-posedness in L^2 ends here and next the proof is about the well-posedness in %L^8L^4 %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{thm}\label{lemma:well-posedness L}
%Let $u_0 \in L^2(\textbf{R})$. There exists $T>0$ and a solution $u \in L^8([-T, T]:L^4)$ of \eqref{a1} with %initial data $u_0$.
%\end{thm}

%......................... the proof of the wellpossedness in L^2 when alpha!=0.
%\begin{proof}
%Assume $\alpha=1$
%\begin{equation}
%v=\exp{it\Delta } v_0 + \int_0^t \exp{i\Delta (t-t')} <\phi>(v,v,v) dt'.
%\end{equation}

% Define
%\begin{equation}
%\phi(v)=\exp{it\Delta } v_0 + \int_0^t \exp{i\Delta (t-t')} <\phi>(v,v,v) dt',
%\end{equation}

%where

%$<\phi>(v,v,v)=\int^0_1 \phi(v,v,v,t) dt$

%and,

%$\phi(v,v,v,t)= T^{-1}(v)[ |T(s)u|^2 T(s)v].$

%let
%$E(T,a)=\{v\in C([0, T]:L^2(\textbf{R})) \cap L^8_t L^4_x :|||v|||=\sup_{t\in[0,T]}  \|v(.,t)\|_{L^2}+\|v(.,t)\|%_{L^8_tL^4_x}\}$
%We want to show for  every $a>2\|v_0\|_{H^1}$ there exists $T>0$ such that $\phi$ define a contraction %map on $E(T, a)$.

%First we have to show  $\phi: E(T,a)\rightarrow E(T,a)$.
%For all $t\in[0,T]$,

%\begin{equation*}
%\begin{aligned}
%&\|\phi (v)\|_{L^4L^8} \le \|\exp{i\Delta t} v_0\|_{L^4L^8} +  \| \int_0^t \exp{ i\Delta (t-t')} <\phi> (v,v,v) dt' \|%_{L^4L^8}  \\
%&\le \|v_0\|_{L^2} +  \| \int_0^t \exp{ i\Delta (t-t')} \int_0^1 \exp{i\Delta D(s)} [\|\exp{-i\Delta D(s)}u\|^2 \exp{- %i\Delta D(s)} u] dsdt' \|_{L^4L^8}  \\
%&\le \|v_0\|_{L^2} +  \| \int_0^t \exp{ i\Delta (t-t')} \int_0^1 f(s,x,t') dsdt' \|_{L^4L^8} \\
%&\le \|v_0\|_{L^2} +  \| \int_0^t \exp{ i\Delta (t-t')}  ∑ f(s_1,x,t') \Delta s dt'ds \|_{L^4L^8} \\
%&\le \|v_0\|_{L^2} +  \| \int_0^1 \int_0^t \exp{ i\Delta (t-t')} f(s,x,t') dt'ds \|_{L^4L^8} \\
%&\le \|v_0\|_{L^2} + \beta \int_0^1 \| \int_0^t  \exp{ i\Delta (t-t')} \exp{i\Delta D(s)} [\|\exp{-i\Delta D(s)}u\|^2 %\exp{-i\Delta D(s)} u]dt' \|_{L^4L^8} ds \\
%&= \|v_0\|_{L^2} + \beta \int_0^1 \| \int_0^t  \exp{ i\Delta (t-(t'-D(s)))} [\|\exp{-i\Delta D(s)}u\|^2 \exp{-i\Delta %D(s)} u]dt' \|_{L^4L^8} ds \\
%&\le \|v_0\|_{L^2} +C\beta \int_0^1 \| |\exp{-i\Delta D(s)}u|^2 \exp{-i\Delta D(s)} udt' \|
%_{L^{\frac{4}{3}}L^{\frac{8}{7}}} ds \\
%&\le \|v_0\|_{L^2} +C\beta \int_0^1 \| |\exp{-i\Delta D(s)}u|^2 \exp{-i\Delta D(s)} udt' \|_{L^{\frac{4} %{3}}L^{\frac{8}{7}}} ds \\
%&\le C\|v_0\|_{L^2} + C\beta\| \exp{ -i\Delta D(s)} u\|^3_{L^4L^{\frac{24}{7}}}ds\\
%&\le \|v_0\|_{L^2} + \beta  \left( \|v\|_{L^4 L^{\frac{24}{7}} } \right)^3 \\
%&\le C a+C\beta T^{\frac{1}{2}} a^3.
%\end{aligned}
%\end{equation*}

%Now, for fix $t\in[0,T]$,

%\begin{equation*}
%\begin{aligned}
%&\|\phi (v)\|_{L2} \le \|\exp{i\Delta t} v_0\|_{L^2} +  \| \int_0^t \exp{ i\Delta (t-t')} <\phi> (v,v,v) dt' \|_{L2} \\
%&\le \|v_0\|_{L^2} +  \| \int_0^t \exp{ i\Delta (t-t')} \int_0^1 \exp{i\Delta D(s)} [\|\exp{-i\Delta D(s)}u\|^2 \exp{- %i\Delta D(s)} u] dsdt' \|_{L^2} \\
%&\le \|v_0\|_{L^2} +  \int_0^1 \| \int_0^t \exp{ i\Delta (t-t')} \exp{i\Delta D(s)} [\|\exp{-i\Delta D(s)}u\|^2 \exp{-  %i\Delta D(s)} u] dsdt' \|_{L^2} \\
%&\le \|v_0\|_{L^2} +C\beta \int_0^1 \| |\int_{-D(s)}^{t-D(s)} \exp{-i\Delta (t-\tilde{t})} [\|\exp{-i\Delta D(s)}u\|^2 %\exp{-i\Delta D(s)} u]dt' \|_{L^2} ds\\
%&\le \|v_0\|_{L^2} +C\beta \int_0^1 \| |\exp{-i\Delta D(s)}u|^2 \exp{-i\Delta D(s)} u]dt' \|_{L^{\frac{4} %{3}}L^{\frac{8}{7}}} ds\\
%5&\le C\|v_0\|_{L^2} + C\beta\| \exp{ -i\Delta D(s)} u\|^3_{L^4L^{\frac{24}{7}}}ds\\
%&\le \|v_0\|_{L^2} + \beta  \left( \|v\|_{L^4 L^{\frac{24}{7}} } \right)^3 \\
%&\le C a+C\beta T^{\frac{1}{2}} a^3.
%\end{aligned}
%\end{equation*}

%So

%\begin{equation*}
%|||\phi(v)|||_T \le C a+C\beta T^{\frac{1}{2}} a^3.
%\end{equation*}

%choose $T$ such that
%\begin{equation}
%C a^3 T \le a/2,
%\label{CT}
%\end{equation}
%then
%\begin{equation*}
%|||\phi (v)|||_{T} \le a/2 + a/2 =a .
%\end{equation*}
%This proves $\phi(v)\in E(T,a)$, whenever \ref{CT} holds and $a>2\|\|v_0\|\|_{L^2}$, and   $T<1/{2Ca^2}$.
%Also for $u,v \in E(T,a)$, for all $t\in [0,T]$,

%\begin{equation*}
%\begin{aligned}
%\|\phi(u) - \phi(v) \|_{L^4L^8}&= \|\int_0^t \exp{i\Delta (t-t')} [<\phi>(u,u,u) - <\phi>(v,v,v)] dt'\|_{L^4L^8}\\
%&\le \int_0^t \|<\phi>(u,u,u) - <\phi>(v,v,v)\|_{L^4L^8} dt'
%\end{aligned}
%\end{equation*}
%and
%\begin{equation*}
%\begin{aligned}
%&\|<\phi>(u,u,u)-<\phi>(v,v,v)\|_{L^4L^8}\\
%&=\|\int_0^1 (g(s) T^{-1}(s)(|T(s)u|^2 T(s)u) - g(s) T^{-1}(s)(|T(s)u|^2 T(s)v)) ds\|_{L^4L^8}\\
%&=\|\int_0^1 g(s) T^{-1}(s)(|T(s)u|^2 T(s)u - |T(s)u|^2 T(s)v) ds\|_{L^4L^8}\\
%&\le \beta \int_0^1 || |T(s)u|^2 T(s)u - |T(s)u|^2 T(s)v ||_{L^4L^8} ds,
%\end{aligned}
%\end{equation*}
%so
%\begin{equation*}
%\begin{aligned}
%&\| |T(s)u|^2T(s)u- |T(s)u|^2T(s) v\|_{L^4L^8} \\
%&\le C\| T(s)u- T(s) v\|_{L^4L^8} (\|T(s)u\|^2_{L^4L^8}\\&-\|T(s)u\|_{L^4L^8} \|T(s) v\|_{L^4L^8} )
%+\|T(s)u- T(s) v\|_{L^4L^8}\|T(s) v\|^2_{L^4L^8}\\
%&= \|u- v\|_{L^4L^8} (\|u\|^2_{L^4L^8}-\|v\|^2_{L^4L^8} +\|u\|_{L^4L^8}\|v\|_{L^4L^8})\\
%&\le \|u- v\|_{L^4L^8} (3a^2) \\
%&\le\sup_{t\in[0,T]} \|u-v\|_{L^4L^8} 3 a^2C\\
%&= 3 a^2C|||u_0-v_0|||_{E(T,a)}.
%\end{aligned}
%\end{equation*}
%In the same way we can prove,
%\begin{equation*}
%\begin{aligned}
%\| |T(s)u|^2T(s)u- |T(s)u|^2T(s) v\|_{L^2} \le C||u_0-v_0||_{L^2}.
%\end{aligned}
%\end{equation*}

%So for $u,v\in E(T,a)$
%\begin{equation*}
%\begin{aligned}
%|||<\phi>(u,u,u)- <\phi>(v,v,v)|||_{E(T,a)}
%&=\sup_{0\leq t\leq T}\|<\phi>(u,u,u)- <\phi> (v,v,v)\|_{L^4L^8} \\
%&\le \beta \int_0^1 3 a^2 C|||u-v|||_{E(T,a)} ds\\
%&= 3 a^2C|||u-v|||_{E(T,a)},
%\end{aligned}
%\end{equation*}
%so
%\begin{equation*}
%\begin{aligned}
%|||<\phi>(u,u,u)-<\phi>(v,v,v)|||_{E(T,a)}
%&\le \sup_{0\leq t\leq T} \int_0^1 3 a^2\beta C|||u-v|||_{E(T,a)} dt'\\
%&=\sup_{0\leq t\leq T} 3 a^2\beta C t |||u-v|||_{E(T,a)}\\
%&=3 a^2 \beta C T |||u-v|||_{E(T,a)}.
%\end{aligned}
%\end{equation*}
%Choose $T$ such that $3a^2\beta C T < \frac{1}{2}$, then
%\begin{equation*}
%|||<\phi>(u,u,u)-<\phi>(v,v,v)|||_{E(T,a)}<\frac{1}{2} |||u-v|||_{E(T,a)}
%\end{equation*}
%on $E(T,a)$, so $\phi$ is a contraction.
%So by Banach contractionn mapping theorem, $\phi$ has a fixed point $v\in E(T,a)$ so $v$ is a strong %solution of $15$ in $L^8([0,T], L^4)$
%\end{proof}
%The end of section two

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                          Begining: Chapter Three               %                                                                                  %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\chapter{Existence and stability of solitary waves}
%\renewcommand{\theequation}{\arabic{chapter}.\arabic{equation}}
%\setcounter{chapter}{3} \setcounter{equation}{0}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                       Begining: Section three one              %                                                                             %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Variational approach to solitary waves}
\begin{definition}  A solution of \eqref{onethree} of the form
\begin{equation}\label{threeone}
u(t,x)=e^{i\theta t}\phi(x),
\end{equation}
where $\theta\in \textbf{R}$ and $\phi\in L^2$, is called a solitary-wave solution.
\end{definition}

Substituting \eqref{threeone} into \eqref{onethree}, we see that \eqref{threeone} defines a
solitary-wave solution of \eqref{onethree} if and only if $\phi(x)$ satisfies the equation
\begin{equation}\label{threetwo}
-\theta \phi(x)+\alpha \phi''(x)+ Q(\phi(x))=0.
\end{equation}
We now observe that \eqref{threetwo} can be viewed as the Euler-Lagrange equation
for the following variational problem.

For $\lambda> 0$, define
\begin{equation*}
I_\lambda=\inf\ \{E(u): u \in H^1\quad \mbox{and}\quad P\left(u\right)=\lambda\}.
\end{equation*}

The set of minimizers for the problem of minimizing $E(\psi)$ subject to $P(\psi)=\lambda$ is
\begin{equation}\label{Glambda}
G_\lambda=\{\phi\in H^1:E\left(\phi\right)=I_\lambda \quad \mbox{and}\quad P\left(\phi\right)=\lambda\}.
\end{equation}

According to the calculus of variations (see, e.g., Theorem $1$ on page $243$ of \cite{LU}), if $\phi$ is any element of $G_\lambda$, then $\phi$ must
satisfy the Euler-Lagrange equation $E'(\phi)+\tilde{\theta}P'(\phi)=0$, where $E'(\phi)$ and $P'(\phi)$ are
the Frechet derivatives of $E$ and $P$ at $\phi$, and $\tilde{\theta}$, the Lagrange multiplier, is a constant.

An elementary computation shows that for $\phi \in H^1$ we have $E'(\phi)=\alpha \phi''+Q(\phi)$ and
$P'(\phi)=2\phi$. Therefore the elements of $G_\lambda$ (if any exist) give rise to solitary waves
through \eqref{threeone}, with $\theta=2\tilde{\theta}$. Such solitary waves are called ground-state solitary waves.

%We are going to give a shorter proof of Theorem \ref{theoremthreeone} in
%the next section (see Theorem \ref{theoremthreeeighteen} below).

%In this chapter, we are going to show the stability of the solution of \eqref{onethree}
%with $\alpha=0$. In this case the energy functional is given by
%\begin{equation}
%E(u)=\frac{-1}{2}\int_0^1 g(s)\inftyint \left|T(s)u\right|^4\ dx\ ds.
%\end{equation}

%We define $G_{\lambda}$ to be the set of minimizers for $I\left(\lambda\right),$
%\begin{equation*}
%G_\lambda=\{\psi\in L^2:E\left(\psi\right)=I\left(\lambda\right)\quad \mbox{and}
%R\quad Q\left(\psi\right)=\lambda\}.
%\end{equation*}
%We assume that there exist numbers $\{t_i\}_{i=0,\cdots,n}$ in $[0,1]$, with
%$0=t_0<t_1<t_2<\cdots < t_{n-1}<t_n =1$ and $\beta>0$ such that $|\Delta(t)|\geq \Delta_0$ for
%$t\neq t_i$ and $\Delta(t)$ is continuous on $(t_{i-1}, t_i)$ for all $i=1,\cdots,n$ and has limits
%from the left and the right at $0, {t_1}, {t_2},\cdots ,{t_n}, 1$
%Then $D(t)$ is differentiable function of $t$ on $[0,{t_1}],[{t_1}, {t_2}],\cdots, [{t_n}, 1]$
%with $D'(t)\neq0$ everywhere.

%\begin{thm}\label{theoremthreethree}
%Suppose $\alpha=0$. Then for every $\lambda>0$ there exists a non-empty set
%$G_\lambda\subseteq L^2\left(\textbf{R}\right)\bigcap L^{\infty}\left(\textbf{R}\right)$
%such that for every  $\phi\in G_\lambda$, $\exp(i\omega z)\phi(\tau)$ is a solitary-wave solution of equation
%\eqref{onethree} with $\inftyint \phi^2\ d\tau=\lambda.$ The set $G_\lambda$ is stable in the following sense.
%For every $\epsilon>0$, there exists $\delta>0$ such that if $u_0 \in L^2$ and $\phi\in G_\lambda$ and
%\begin{equation}
%\left\|{u_0}-\phi\right\|_{L^2}<\delta,
%\end{equation}
%then the solution $u\left(z,\tau\right)$ of equation \eqref{onethree} with
%$u\left(0,\tau\right)=u_0\left(\tau\right)$ satisfies
%\begin{equation}
%\inf_{\psi\in\textbf{S}}\left\|u\left(z,\tau\right)-\psi
%\left(\tau+y\right)\right\|_{L^2\left(d\tau\right)}<\epsilon
%\end{equation}
%for all $z\in\textbf{R}$.
%\end{thm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                          Beginning: Section three two             %                                                                              %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Existence and stability in $H^1$ of solitary waves when $\alpha>0$} \label{stabalphapos}

In this section we prove the following result on the existence and stability of
solitary-wave solutions of the DMNLS equation \eqref{twoone} in the case when $\alpha >0$.

\begin{thm}\label{theoremthreetwo}
Suppose $\alpha>0$. Assume $\Delta(s)$ is an integrable function on $[0,1]$, and $g(s)$ is a non-negative integrable
function on $[0,1]$ such that $\displaystyle \int_0^1 g(s)\ ds>0$.

Then for every $\lambda>0$ there exists a non-empty set
$G_\lambda\subseteq H^1$ such that for every \ $\phi\in
G_\lambda$, there exists $\theta$ such that $\exp(i\theta
t)\phi(x)$ is a solitary-wave solution of \eqref{twoone} with
$\displaystyle \inftyint \phi^2\ dx=\lambda$.

 The set $G_\lambda$ is stable in
the following sense: for every $\epsilon>0$, there exists
$\delta>0$ such that if $u_0 \in H^1$ and $\phi\in G_\lambda$ and
\begin{equation}
\left\|{u_0}-\phi\right\|_{H^1}<\delta,
\end{equation}
then the solution $u\left(x,t\right)$ of \eqref{twoone} with $u\left(x,0\right)
=u_0\left(x\right)$ satisfies
\begin{equation}
\inf_{\psi\in G_\lambda}\left\|u\left(\cdot,t\right)-\psi
\left(x\right)\right\|_{H^1}<\epsilon
\end{equation}
for all $t\geq 0$.
\end{thm}

Theorem \ref{theoremthreetwo} is proved in \cite{Z1} and \cite{Z2}
for the case where $g(s) \equiv 1$ and $\Delta(s)$ is piecewise
constant.  As we show below, it turns out that essentially the
same proof works with the more general assumptions on $g(s)$ and
$\Delta(s)$ stated above.

We will obtain Theorem \ref{theoremthreetwo} as a corollary of the following theorem
about the behavior of arbitrary minimizing sequences for the variational problem.
We define a minimizing sequence for $I_\lambda$ to be any sequence $\{u_n\}$ of
functions in $H^1$ satisfying
\begin{equation}
P\left(u_n\right)=\lambda,
\end{equation}
for all $n$, and
\begin{equation}
\lim_{n\to\infty}E\left(u_n\right)=I_\lambda.
\end{equation}


\begin{thm}\label{theoremthreeseventeen}
The set $G_{\lambda}$ is not empty. Moreover, for every minimizing sequence $\{u_n\}$
for $I_{\lambda}$, the following are true:
\begin{enumerate}
\item
there exists a subsequence $\{u_{n_k}\}$ of $\{u_n\}$ and a sequence $\{y_k\}$ of real numbers and an element $g\in G_{\lambda}$ such
that
$$
\lim_{k \to \infty}\|u_{n_k}(\cdot +y_k)-g\|_{H^1}=0.
$$
\item
$$\lim_{n\to\infty}\inf_{g\in G_{\lambda}, y\in\textbf{R}} \left\|u_n(.+y)-g\right\|_{H^1}=0.$$
\item
$$\lim_{n\to\infty}\inf_{g\in G_{\lambda}} \left\|u_n-g\right\|_{H^1}=0.$$
\end{enumerate}
\end{thm}

The idea behind the proof of Theorem \ref{theoremthreeseventeen} is that, for any given minimizing sequence $\{u_n\}$, we can apply the
Concentration Compactness Principle to the sequence of non-negative functions $\rho_n$
defined by $\rho_n = |u_n|^2$.  This is done as follows. First, we define a sequence of nondecreasing functions
$M_n:\left[0,\infty\right]\to\left[0,\lambda\right]$ by
\begin{equation}\label{defmk}
M_n(r)=\sup_{y\in\textbf{R}}\int_{y-r}^{y+r}\rho_n\ dx.
\end{equation}
From Helley's selection theorem \cite{H}, it follows that any uniformly bounded sequence of
nondecreasing functions on $\left[0,\infty\right)$ must have a subsequence which converges pointwise to a
nondecreasing limit function on $\left[0,\infty\right)$. Hence $\{M_n\}$ has such a subsequence, which
converges to a limit function $M(r)$. Define
\begin{equation}\label{qthreenine}
q=\lim_{r\to\infty}M\left(r\right),
\end{equation}
so that $0\leq q\leq\lambda$.

The Concentration Compactness Principle, as given in \cite{L}, states that we can characterize the behavior of the
sequence $\rho_n$ in useful ways based on the value on $q$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%         The concentration compactness lemma                                   %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{thm}\label{theoremthreefour}
Suppose $\lambda>0$, and let $\{{\rho}_n\}$ be a sequence of
nonnegative functions in $L^1$ satisfying $\displaystyle
\inftyint{{\rho}_n}(x)\ dx=\lambda$ for all $n$.  Let $M_n$ and $M$ be as defined in \eqref{defmk}.
Then there are three possibilities:

\begin{enumerate}

\item{(Vanishing)}: If $q=0$, then there exists a subsequence $\{\rho_{n_k}\}$ of $\{\rho_n\}$ such that for every $R>0$,
$$
\lim_{k\to\infty} \sup_{y\in\textbf{R}}\int_{y-R}^{y+R}\rho_{n_k}(x)\ dx=0.
$$

\item{(Dichotomy)}: If $q \in (0,\lambda)$, then there exists a subsequence $\{\rho_{n_k}\}$ of $\{\rho_n\}$
such that for every $\epsilon>0$,
there exist a number $k_0$ and nonnegative functions $\rho^1_k$ and ${\rho}^2_k$ in  $L^1$, such that for all
$k\geq k_0$,
\begin{equation*}
\begin{aligned}
\left\|{\rho}_{n_k}-\left({\rho}^1_k+{\rho}^2_k \right)\right\|_{L^1}&\leq\epsilon, \\
\left|\inftyint{\rho}^1_{k}\ dx-q\right|&\leq\epsilon,\\
\left|\inftyint{\rho}^2_{k}\ dx-(\lambda-q)\right|&\leq\epsilon,
\end{aligned}
\end{equation*}
$\rho^1_k$ and  $\rho^2_k$ have disjoint support, and
$$
\text{dist}(\text{supp}\ {\rho}^1_k, \text{supp}\ {\rho}^2_k)\to\infty
$$
as $k \to \infty$.

\item{(Compactness)}: If $q = \lambda$, then there is a a subsequence $\{{\rho}_{n_k}\}$ of $\{\rho_n\}$ and a
sequence $\{y_k\}$ of real numbers such that ${\rho}_{n_k}(\cdot-y_k)$
is tight; i.e, for every $\epsilon>0$, there is $R >0$ large enough such that
$$
\int_{- R}^R  {\rho}_{n_k}(x-y_k)\ dx\geq\lambda-\epsilon
$$
for all  $k\in \textbf{N}$.

\end{enumerate}
\end{thm}

 For a nice exposition of the proof of Theorem \ref{theoremthreefour}, the reader may consult Lemma 8.3.8 of
\cite{C}.

The next steps towards a proof of Theorem \ref{theoremthreetwo} are to show that, for every minimizing
sequence $\{u_n\}$ of the variational problem, we must have $q=\lambda$, so that
the ``compactness'' alternative of Theorem \ref{theoremthreefour} holds.   We will do this by showing in the
lemmas which follow that the assumptions that $q=0$ and $q \in (0,\lambda)$ lead to contradictions.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%   I(\lambda) > -\infnty
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{lem}\label{lemmathreefive}
For all $\lambda>0$, one has
$I_\lambda >-\infty.$
\end{lem}

\begin{proof}
Suppose $u\in H^1$ and $P(u)=\lambda$. Let $\beta>0$ be such that
$0\leq g(s)\leq \beta$ for $s\in[0,1]$. Then it follows from Lemma
\ref{onethree} and the Gagliardo-Nirenberg Inequality, Theorem
\ref{lemmaoneten}, with $p=4$, $r=q=2$, and $\theta=1/4$, that for
every $s\in [0,1],$
\begin{equation*}
\begin{aligned}
\inftyint \left|T(s)u\right|^4\ dx&=
\left\|T\left(s\right)u\right\|^4_{L^4} \leq C \left\|T\left(s\right)\frac{du}{dx}\right\|_{L^2} \left\|T\left(s\right)u\right\|^3_{L^2}\\
&= C \left\|\frac{du}{dx}\right\|_{L^2} \left\|u\right\|^3_{L^2} =
C \left\|\frac{du}{dx}\right\|_{L^2} \lambda^{3/2}.
\end{aligned}
\end{equation*}
Therefore we have
\begin{equation}
\begin{aligned}
E\left(u\right)&\geq\alpha\left\|\frac{du}{dx}\right\|^2_{L^2}
-\frac12 \int_0^1 C g(s) \left\|\frac{du}{dx}\right\|_{L^2}
\lambda^{3/2}\ ds
\\& \geq \alpha \left\|\frac{du}{dx}\right\|^2_{L^2} - \frac{C\gamma
\lambda^{3/2}}{2} \left\|\frac{du}{dx}\right\|_{L^2}
\end{aligned}
\end{equation}
where $\displaystyle \gamma = \int_0^1 g(s)\ ds$.
Thus
$$
E(u) \ge f\left(\left\|\frac{du}{dx}\right\|_{L^2}\right),
$$
where $f:\mathbf R \to \mathbf R$ is the quadratic function defined by $\displaystyle f(x):=
\alpha x^2-\frac{C\gamma \lambda^{3/2}}{2}x$.  Since the minimum value of $f(x)$ on $\mathbf R$ is
$$
m:=-\frac{\gamma^2C^2\lambda^3}{16\alpha},
$$ it follows that $E(u) \ge m$ for all $u \in H^1$ such that $P(u) = \lambda$.  This
proves that
$$
I_\lambda \ge m>-\infty.$$
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%% here we show that I(\lambda)<0 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{lem}\label{lemmathreesix}
For each $\lambda > 0,$ we have $I_\lambda <0.$
\end{lem}

\begin{proof}
It is enough to show that for given $\lambda>0$ there exists $u\in H^1$ such that $P(u)=\lambda$ and $E(u)<0.$
We follow the proof on page 62 of \cite{Z1}, which needs only slight modification here to treat
the case where $g(s)$ is nonconstant. Start by defining
\begin{equation}\label{b8}
u(x)=A_0\exp\left(-x^2/2\sigma_0\right),
\end{equation}
where $A_0$ and $\sigma_0$ are positive real numbers, to be chosen later.
Recall that $T(s)[u](x)$ is defined for $s\in\left[0,1\right]$ as
\begin{equation}\label{b9}
T(s)[u](x)=p(s,x),
\end{equation}
where $p$ is the solution of
\begin{equation}\label{b99}
\begin{aligned}
i{\frac{\partial p}{\partial s}}-D'(s){\frac{\partial^2 p}{\partial x^2}}&=0
\\
p\left(0,x\right)&= u(x).\
\end{aligned}
\end{equation}
It is straightforward to verify that the solution of \eqref{b9} is given by
\begin{equation*}
p(s,x)=A(s)\exp\left(-x^2/2\sigma(s)\right),
\end{equation*}
where
\begin{equation}
\sigma(s)=\sigma_0-2iD(s)
\end{equation}
and
\begin{equation}
A(s)=\frac{{A_0}\sqrt{\sigma_0}}{\sqrt{\sigma(s)}}.
\end{equation}
Here $\sqrt{\sigma(s)}$ is taken as the square root
of $\sigma(s)$ which has positive real part. Therefore
\begin{equation*}
T(s)[u](x)=A(s)\exp\left(-x^2/2\sigma(s)\right).
\end{equation*}
Since
\begin{equation*}
\begin{aligned}
P(u)&= {A_0}^2\inftyint{\exp\left(-x^2/\sigma_0\right)}\ dx
\\&= {A_0}^2\sqrt{\sigma_0\pi},
\end{aligned}
\end{equation*}
then to get $P(u)=\lambda,$ for a given $\sigma_0$ we must choose $A_0$ so that
\begin{equation}\label{th3412}
\lambda={A_0}^2 \sqrt{\sigma_0\pi}.
\end{equation}
We also have that
\begin{equation*}
\begin{aligned}
\inftyint\left|\frac{du}{dx}\right|^2\ dx&= {\frac{{A_0}^2}{\sigma^2_0}}\inftyint
{x^2}\exp\left(\frac{-x^2}{\sigma_0}\right)\ dx
\\&= \frac{{A_0}^2}{\sqrt{\sigma_0}}{\frac{\sqrt{\pi}}{2}}\\&= \frac{\lambda}{2\sigma_0}.
\end{aligned}
\end{equation*}
and
\begin{equation*}
\begin{aligned}
&\inftyint\int_{0}^{1}g(s)\left|T\left(s\right)[u]\right|^4\ ds\ dx
\\&= {A_0}^4{\sigma_0}^2\int_{0}^{1}g(s)\left(\inftyint{\frac{1}{\left|\sigma
\left(s\right)\right|^2}}\left|{\exp\left({\frac{-x^2}{2\sigma
\left(s\right)}}\right)}\right|^4dx\right)\ ds
\\&= {A_0}^4{\sigma_0}^2
\int_{0}^{1}{\frac{g(s)}{\left|\sigma\left(s\right)\right|^2}}\inftyint
{\exp\left({\frac{-2x^2 \Re(\sigma(s))}{\left|\sigma\left(s\right)
\right|^2}}\right)}\ dx\ ds
\\&= {A_0}^4{\sigma_0}^{3/2}
{\sqrt{\frac{\pi}{2}}}\int_{0}^{1}{\frac{g(s)}{\left|\sigma\left(s\right)\right|}}\ ds
\\&= \frac{{\lambda}^2{\sqrt{\sigma_0}}}{\sqrt{2\pi}}\int_{0}^{1}
{\frac{g(s)}{\left|\sigma\left(s\right)\right|}}\ ds.
\end{aligned}
\end{equation*}
Hence
\begin{equation}
\label{eofu}
E\left(u\right)=\left(\frac{\lambda\alpha}
{2\sigma_0}\right)\left(1-\frac{\lambda{\sigma_0}^{3/2}}
{\alpha\sqrt{2\pi}}\int_0^1\frac{g(s)\ ds}{\sqrt{{\sigma_0}^2+4 D(s)^2}}\right).
\end{equation}
Now choose $\sigma_0$ so large that $\sigma_0\geq2D(s)$ for all
$s\in\left[0,1\right]$, and $$\sigma_0
\geq\frac{4\pi\alpha^2}{\lambda^2\gamma^2},$$ where $\displaystyle
\gamma=\int_0^1 g(s)\ ds$, and $\gamma >0$ by assumption. Then
\begin{equation*}
\int_0^1\frac{g(s)\
ds}{\sqrt{{\sigma_0}^2+4D^2\left(s\right)}}\geq
{\frac{\gamma}{\sqrt2\ \sigma_0}},
\end{equation*}
so
\begin{equation*}
1-\frac{\lambda{\sigma_0}^{3/2}}{\alpha\sqrt{2\pi}}{\int_0^1\frac{g(s)\
ds} {\sqrt{{\sigma_0}^2+4 D\left(s\right)^2}}} \leq
{1-{\frac{\lambda{\sigma_0}^{3/2}}
{\alpha\sqrt{2\pi}}}{\frac{\gamma}{\sqrt{2} \sigma_0}}} =
1-{\frac{\lambda{\sigma_0}^{1/2}\gamma}{\alpha2\sqrt{\pi}}}< 0,
\end{equation*}
and hence $E(u)<0.$ Thus, for this choice of $\sigma_0$, and with $A_0$ chosen as in \eqref{th3412},
we have that $u$ in \eqref{b8} satisfies both $P(u)=\lambda$ and $E(u)<0$.
\end{proof}

\begin{lem}\label{lemmathreeseven}
\par\medskip\noindent
Suppose $u\in C^{\infty}\left(\textbf{R}\right)$, and for $j\in Z$, define
$Q_j=\left[j-1,j+1\right]$. Then for all $j$,
$$\left\|u\right\|_{L^\infty\left(Q_j\right)}
\leq \frac{1}{2}\left\|u\right\|_{L^1\left(Q_j\right)}+
\left\|\frac{du}{dx}\right\|_{L^1\left(Q_j\right)}.$$
\end{lem}

\begin{proof}
Assume $u\in C^{\infty}\left(\textbf{R}\right).$ Then for all $x_0\in Q_j$ and $y\in Q_j$,
\begin{equation}
u\left(x_0\right)=u\left(y\right)+ \int_y^{x_0} {\frac{du}{dx}}\ dx,
\end{equation}
so
\begin{equation}
\left|u(x_0)\right|\leq\left|u(y)\right|+\left\|\frac{du}{dx}\right\|_{L^1(Q_j)}.
\end{equation}
Integrate both sides with respect to $y$ over $Q_j$ to get
\begin{equation}
\int_{j-1}^{j+1}\left|u(x_0)\right|dy\leq
\int_{j-1}^{j+1}\left|u(y)\right|dy+\int_{j-1}^{j+1}\left\|\frac{du}{dx}\right\|_{L^1\left(Q_j\right)}dy
\end{equation}
or
\begin{equation}
2\left|u(x_0)\right|\leq\left\|u\right\|_{L^1}+ 2\left\|\frac{du}{dx}
\right\|_{L^1\left(Q_j\right)}.
\end{equation}
Hence
\begin{equation}
\left\|u\right\|_{L^\infty\left(Q_j\right)}=\sup_{x_0\in Q_j}
\left|u(x_0)\right|\leq{\frac{1}{2}}\left\|u\right\|_{L^1\left(Q_j\right)}+
\left\|\frac{du}{dx}\right\|_{L^1\left(Q_j\right)}.
\end{equation}
\end{proof}

\begin{lem}\label{lemmathreeeight}
There exists $C>0$ such that, for all $u\in H^1(\textbf{R})$,
\begin{equation}\label{threetwentysix}
\inftyint|u|^4\ dx\leq C\left(\sup_{y\in{\textbf{R}}}\int_{y-1}^{y+1}|u|^2\ dx
\right)\left\|u\right\|^2_{H^1}
\end{equation}
\end{lem}

\begin{proof}
First assume $u\in C^{\infty}_0\left(\textbf{R}\right).$ Applying Lemma \ref{lemmathreeseven} to
$u^2$ instead of $u$, we get
\begin{equation*}
\begin{aligned}
\left\|u^2\right\|_{L^{\infty}\left(Q_j\right)} &\leq {\frac{1}{2}}
\left\|u^2\right\|_{L^1\left(Q_j\right)}+ \left\|\frac{d u^2}{dx}\right\|_{L^1\left(Q_ j\right)}
\\&\leq {\frac{1}{2}}{\left\|u\right\|^2}_{L^2\left(Q_j\right)}
+\int_{Q_j}2\left|{u}\right|\left|{u_x}\right|\ dx
\\&\leq {\frac{1}{2}}{\left\|u\right\|^2}_{L^2\left(Q_j\right)}+
2{\left\|u\right\|}_{L^2\left(Q_j\right)}\left\|\frac{du}{dx}\right\|_{L^2\left(Q_j\right)}
\\&= \left\|u\right\|_{L^2\left(Q_j\right)}\left(\frac{1}{2}\left\|u\right\|_{L^2\left(Q_j\right)}
+2\left\|\frac{du}{dx}\right\|_{L^2\left(Q_j\right)}\right).
\end{aligned}
\end{equation*}
Hence
\begin{equation} \label{c10}
\begin{aligned}
\left\|u\right\|^2_{L^\infty(Q_j)} &\leq C\left\|u\right\|_{L^2\left(Q_j\right)}
\left(\left\|u\right\|_{L^2\left(Q_j\right)}+\left\|\frac{du}{dx}\right\|_{L^2\left(Q_j\right)}\right).
\end{aligned}
\end{equation}
Since
\begin{equation}\label{c11}
\int_{Q_j}\left|u\right|^4\ dx\ \leq\ \left\|u\right\|^2_{L^2\left(Q_j\right)}
\left\|u\right\|^2_{L^\infty\left(Q_j\right)},
\end{equation}
it follows that
\begin{equation*}
\begin{aligned}
\int_{Q_j}|u|^4\ dx&\leq C
\left\|u\right\|^3_{L^2\left(Q_j\right)}
\left(\left\|u\right\|_{L^2\left(Q_j\right)}
+\left\|\frac{du}{dx}\right\|_{L^2\left(Q_j\right)}\right)
\\&\leq C\left(\int_{Q_j}|u|^2\ dx\right)^{3/2}
\left(\int_{Q_j}{\left(|u|^2+\left|\frac{du}{dx}\right|^2\right)\
dx}\right)^{1/2}
\\&\le C\left(\int_{Q_j}|u|^2\ dx\right)\int_{Q_j}
\left(|u|^2+ \left|\frac{du}{dx}\right|^2\right)\ dx.
\end{aligned}
\end{equation*}
Now sum over $j\in \mathbf{Z}$ to get
\begin{equation}\label{threetwentynine}
2\inftyint|u|^4\ dx\leq C\left(\sup_{j \in
\mathbf{Z}}\int_{Q_j}\left|u\right|^2\
dx\right)\left\|u\right\|^2_{H^1}.
\end{equation}

This proves \eqref{threetwentysix} for $u\in C^{\infty}_0(\textbf{R}).$ The result
for general $u$ in $H^1$ follows by approximating $u$ with a sequence
$u_n\in C^{\infty}_0(\textbf{R})$ such that $u_n\to u$ in $H^1$ norm.
Then \eqref{threetwentynine} holds for each $u_n$, and we obtain
\eqref{threetwentysix} by passing to the limit as $n\to \infty.$
%Proving the inequality on P.7363 of Zharnitstay.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Here we are proving vanishing does not occur by using the commutator./we adjust our prove to a better way.%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{lem}\label{lemmathreenine}
Suppose $\{u_n\}$ is a minimizing sequence. Then there exists $B>0$ such that,
for all $n\in \textbf{N}$, $\left\|u_n\right\|_{H^1}\leq B$.
\end{lem}

\begin{proof}
Since $\{E\left(u_n\right)\}$ is a convergent sequence of real
numbers, then it is bounded.  Moreover, $P(u_n) = \lambda$ for all
$n$.  So the conclusion follows immediately from Lemma
\ref{control_H1}.
\end{proof}

\begin{lem}\label{lemmathreeten}
\par\medskip\noindent
Suppose $\{u_n\}$ is a minimizing sequence for $I_\lambda$. Then there exists $C_0 > 0$
(independent of n) such that for all sufficiently large $n\in \textbf{N}$,
there exists $s_n \in [0,1]$ for which
\begin{equation}
\sup_{y\in\textbf{R}}\int_{y-1}^{y+1}\left|T\left(s_n\right)u_n\right|^2\ dx\geq C_0.
\end{equation}
\end{lem}

\begin{proof}
Since $\{u_n\}$ is a minimizing sequence, we have by Lemma \ref{lemmathreesix} that
\begin{equation}
\lim_{n\to \infty}E(u_n)=I_\lambda < 0,
\end{equation}
and hence
\begin{equation}
\liminf_{n\to{\infty}}\int_0^1g(s)\left(\int_{-\infty}^{\infty}\left|T(s)u_n\right|^4\ dx\right)\ ds=C > 0.
\end{equation}
So, by passing to a subsequence if necessary, we can assume that
for every sufficiently large $n$,
\begin{equation*}
\int_0^1 g(s)\inftyint|T(s)u_n|^4\ dx\ ds\geq C,
\end{equation*}
and therefore
\begin{equation*}
0< C \le \gamma \sup_{s \in [0,1]}\inftyint|T(s)u_n|^4\ dx,
\end{equation*}
where as before we define $\displaystyle \gamma = \int_0^1 g(s)\ ds > 0$.
Hence there must exist $s_n \in [0,1]$ such that
\begin{equation}\label{threethirtyone}
\inftyint |T(s_n)u_n|^4\ dx \ge \frac{C}{2\gamma} > 0.
\end{equation}

Since $\left\|u_n\right\|_{H^1}$ is bounded by Lemma
\ref{lemmathreenine}, then $\left\|T(s_n)u_n\right\|_{H^1}$ is
bounded by Lemma \ref{lemmaonethree}. Therefore, from Lemma
\ref{lemmathreeeight} together with \eqref{threethirtyone}, we
obtain that
\begin{equation}
\sup_{y\in\textbf{R}}\int_{y-1}^{y+1}\left|T\left(s_n\right)u_n\right|^2\ dx\geq C_0
\end{equation}
for some $C_0$ which does not depend on $n$.
\end{proof}

Next we will rule out the possibilities that the sequence $\rho_n =|u_n|^2$ satisfies either of the ``vanishing'' or ``dichotomy'' alternatives described in the sense of Theorem \ref{theoremthreefour}.
We follow the argument of \cite{Z2}, slightly modified here to take the variable coefficient $\Delta(s)$ into account.
For each $n$ and $s \in [0,1]$, let us define
$$
\epsilon_n(s)=\sup_{y\in\textbf{R}} \int_{y-1}^{y+1}\left|T(s)u_n(x)\right|^2\ dx.
$$


\begin{lem}\label{theoremsevenone}
Suppose $\{u_n\}$ is a sequence that is bounded in $H^1$, satisfies the constraint $\|u_n\|_{L^2}=\lambda$ for all $n$, and
vanishes in the sense of Theorem \ref{theoremthreefour}, so that $\epsilon_n(0)\to 0$.
Then for each $s \in [0,1]$, the sequence $T(s)u_n$ is also vanishing, and the following estimate holds:
\begin{equation}
\sup_{s \in [0,1]}\epsilon_n(s)\le \frac{32}{3}\left(\epsilon_n(0)+\sqrt{\epsilon^2_n(0)+C\epsilon_n(0)}\right).
\end{equation}
\end{lem}

\begin{proof}
By definition, $T(s)u_n$ satisfies the
linear Schr\"{o}dinger equation
\begin{equation}
\label{linschrodts}
i\frac{\partial}{\partial s}(T(s)u_n)=\Delta(s)\frac{\partial^2}{(\partial x)^2}T(s)u_n
\end{equation}
for $s \in \mathbf R$ and $x \in \mathbf R$. Multiplying equation \eqref{linschrodts} by $\overline{T(s)u_n}$, subtracting the resulting equation
from the original, and integrating with respect to $x$ from $-R$ to $R$, we obtain
\begin{equation} \label{intx}
\begin{aligned}
\frac{d}{ds}\int_{-R}^{+R}\left|T(s)u_n\right|^2\ dx
&=-2 \Delta(s) \left(\Im \left[\overline{T(s)u_n(R)}T(s)u_{nx}(R)\right]\right.\\
&-\Im \left.\left[\overline{T(s)u_n(-R)}T(s)u_{nx}(-R)\right]\right).
\end{aligned}
\end{equation}
Integrating equation \eqref{intx} with respect to $s$ from $0$ to $t$, and recalling that $T(0)u_n=u_n$, we obtain
\begin{equation}\label{thirtythree}
\begin{aligned}
\int_{-R}^{+R}\left|T(t)u_n\right|^2\ dx-\int_{-R}^{+R}\left|u_n\right|^2\ dx
&=-2 \int_0^t\Delta(s) \Im\left[\overline{T(s)u_n(R)}T(s)u_{nx}(R)\right]\ ds\\
&+2 \int_0^t\Delta(s) \Im\left[\overline{T(s)u_n(-R)}T(s)u_{nx}(-R)\right]\ ds.
\end{aligned}
\end{equation}
Assume now that $T(t)u_n(x)$ has been translated in $x$ so that
$$\epsilon_n(t) \le 2\int_{-1}^{1}\left|T(t)u_n(x)\right|^2\ dx.$$
This is always possible since we can shift the initial data using translational invariance.
Using the obvious inequalities
$$
\int_{-1}^{1}\left|T(t)u_n(x)\right|^2\ dx\leq \int_{-R}^{+R}\left|T(t)u_n(x)\right|^2\ dx
$$
 and
$$\int_{-R}^{+R}\left|u_n(x)\right|^2\ dx
\le\left(R+2\right)\epsilon_n(0)\leq 2R\epsilon_n(0)$$
for $R\geq 2$, we can write
\begin{equation}\label{first}
\frac12\epsilon_n(t)-2R\epsilon_n(0)\le \left|\int_{-R}^{+R}\left|T(t)u_n(x)\right|^2\ dx
-\int_{-R}^{+R}\left|u_n(x)\right|^2\ dx\right|.
\end{equation}
On the other hand using \eqref{thirtythree}, we have
\begin{equation}\label{second}
\begin{aligned}
\left|\int_{-R}^{+R}\left|T(t)u_n(x)\right|^2\ dx-\int_{-R}^{+R}\left|u_n(x)\right|^2\ dx\right|
&\le 2\int_0^1 |\Delta(s)|\left|T(s)u_n(R)\right|\left|T(s)u_{nx}(R)\right|\ ds
\\&+2\int_0^1 |\Delta(s)|\left|T(s)u_n(-R)\right|\left|T(s)u_{nx}(-R)\right|\ ds.
\end{aligned}
\end{equation}
Assuming that $\epsilon_n(t)>8\epsilon_n(0)$ (for otherwise we are done), we can find
$R_n\ge 2$ such that $\frac12\epsilon_n(t)-4R_n\epsilon_n(0)=0$.

Combining the inequalities  \eqref{first} and \eqref{second} gives
$$
\begin{aligned}
\frac12\epsilon_n(t)&-2R\epsilon_n(0)\le
2\int_0^1 |\Delta(s)|\left|T(s)u_n(R)\right|\left|T(s)u_{nx}(R)\right|\ ds
\\&+2\int_0^1 |\Delta(s)|\left|T(s)u_n(-R)\right|\left|T(s)u_{nx}(-R)\right|\ ds.
\end{aligned}
$$
Now integrating both sides of the last inequality with respect to $R$, from $R=1$ to $R=R_n$, we obtain
\begin{equation}
\begin{aligned}
\frac{3\epsilon^2_n(t)}{64\epsilon_n(0)}&-\frac12\epsilon_n(t)+\epsilon_n(0)\le
2\int_0^1 |\Delta(s)|\inftyint \left|T(s)u_n(R)\right|\left|T(s)u_{nx}(R)\right|\ dR\ ds
\\&\le 2\int_0^1 |\Delta(s)|\left\|T(s)u_n\right\|_{L^2}
\left\|T(s)u_{nx}\right\|_{L^2}\ ds
\\&= 2\int_0^1 |\Delta(s)|\|u_n\|_{L^2}\left\|u_{nx}\right\|_{L^2}\ ds.
\end{aligned}
\end{equation}
Since $\|u_n\|_{H^1}$ is bounded and $\displaystyle \int_0^1 |\Delta(s)|\ ds<\infty$, this implies
$$
\frac{3\epsilon^2_n(t)}{64\epsilon_n(0)}-\frac12\epsilon_n(t)\le C.
$$
Hence
$$\epsilon_n(t)\le \frac{32}{3}\left(\epsilon_n(0)+\sqrt{\epsilon^2_n(0)+C\epsilon_n(0)}\right),$$
where $C$ is independent of $n$ and $t$.
\end{proof}

Now we can rule out the ``vanishing'' alternative for minimizing sequences.

\begin{lem}
\label{novanish}
If $\{u_n\}$ is a minimizing sequence for $I_\lambda$, then $q \ne 0$.
\end{lem}

\begin{proof} If $q=0$, then for some subsequence of $u_n$, which we still denote by $u_n$, we have
\begin{equation}\label{start}
\lim_{n \to \infty}\epsilon_n(0)=\lim_{n \to \infty}\sup_{y\in \textbf{R}}\int_{y-1}^{y+1}\left|u_n\right|^2 \ dx = 0.
\end{equation}
But then Lemma \ref{theoremsevenone} implies
$\displaystyle \lim_{n \to \infty}\epsilon_n(s_n)= 0$, contradicting Lemma \ref{lemmathreeten}.
\end{proof}

The next lemma is used to describe how minimizing sequences $\{u_n\}$ would behave
in the case when $q \in (0,\lambda)$.  We follow the proof of inequality (32) in \cite{Z2}.

\begin{lem}\label{lemmathreethirteen}  Suppose $\{u_n\}$ is a minimizing sequence for $I_\lambda$ and suppose
$q \in (0,\lambda)$.  Then for some subsequence of $\{u_n\}$, which we continue to denote by $\{u_n\}$, the following is true.
For every  $\epsilon > 0$, there exist a number $N\in \textbf{N}$
and sequences $\{v_N,v_{N+1},\cdots\}$ and $\{w_N, w_{N+1},\cdots\}$
of $H^1$ functions such that for every $n\geq N$,
\begin{enumerate}
\item
$\displaystyle\left|\inftyint\left|v_n(x)\right|^2\ dx-q\right|< \epsilon$
\item
$\displaystyle\left|\inftyint\left|w_n(x)\right|^2\ dx-\left(\lambda-q\right)\right|
<\epsilon$
\item
$E\left(u_n\right)\geq E\left(v_n\right)+E\left(w_n\right)-\epsilon$
\end{enumerate}
\end{lem}

\begin{proof}  Suppose $\epsilon > 0$ is given.  Then by definition of $q$, there exists $r_0$ such that
if $r > r_0$, then $q-\epsilon < M(r) \le q$.  Therefore, after passing to a subsequence of $\{M_n\}$ if necessary, we can say that
 there exist numbers $n_0 \in \mathbf N$ and $r_1 > r_0$ and $r_2 > r_0$ such that
$r_2 - r_1 \ge 6/\epsilon$ and, for all $n \ge n_0$,
$$
q-\epsilon < M_n(r_1) \le M_n(r_2) < q + \epsilon.
$$
It follows that for every $n \ge n_0$, there exists $y_n$ such that
\begin{equation}
\label{integraloverr1}
q-\epsilon < \int_{y_n -r_1}^{y_n+r_1} |u_n|^2\ dx \le \int_{y_n -r_2}^{y_n+r_2} |u_n|^2\ dx < q + \epsilon.
\end{equation}

Now introduce smooth cut-off functions $\rho$ and  $\theta$, defined on $\textbf{R})$ with values in $[0,1]$,
such that  $|\rho_x|<\epsilon$ and $ |\theta_x|<\epsilon$ for all $x \in \mathbf R$;
$\rho(x)=1$ for $|x|\le r_1$; $\rho(x)=0$ for $|x|\ge r_1+2/\epsilon$;
$\theta(x)=1$ for $|x|\geq r_2$; and $\theta(x)=0$ for $|x|\leq r_2-2/\epsilon$.
Define $v_n(x):=\rho(y_n-x)u_n(x)$ and $w_n(x):=\theta(y_n-x)u_n(x)$.  Then it follows
easily from \eqref{integraloverr1} that statements 1 and 2 of the Lemma hold, and it remains only to
prove statement 3.

Now we can write
$$
u_n = v_n + w_n + h_n,
$$
where $\|h_n\|^2_{L^2}<\epsilon$.  Then we have
\begin{equation}
E(u_n)=\inftyint   \int_0^1\left(\alpha |v_{nx} + w_{nx}
+ h_{nx}|^2-\frac12 |T(s)(v_n+w_n+h_n)|^4\right)\ ds\ dx,
\end{equation}
which can be rewritten as
\begin{equation}\label{twentynine}
\begin{aligned}
E(u_n)&=E(v_n)+E(w_n)+ 2\alpha \Re\inftyint  (\overline{v_{nx}}\ w_{nx}
+ \overline{v_{nx}}\ h_{nx}+\overline{w_{nx}} \ h_{nx})\ dx
\\&+\alpha\inftyint |h_{nx}|^2\ dx
- \Re \inftyint \int_0^1 \left(2|T(v_n+w_n)|^2 |Th_n|^2\right.+ \frac12 |Th_n|^4
\\&+2|T(v_n+w_n)|^2T(v_n+w_n)Th_n + (T(v_n+w_n))^2(\overline{Th_n})^2
\\&+2T(v_n+w_n)|Th_n|^2\overline{Th_n}+ 2|Tv_n|^2|Tw_n|^2 + 2|Tv_n|^2Tv_n \overline{Tw_n}
\\&\left.+(Tv_n)^2 (\overline{Tw_n})^2 + 2|Tw_n|^2 Tv_n \overline{Tw_n}\ \right)\ ds\ dx,
\end{aligned}
\end{equation}
where we have used $T$ as an abbreviation for $T(s)$.

To prove statement 3, it is enough to show that each of the integrals on the right-hand side of \eqref{twentynine}
can be bounded below by terms which go to zero with $\epsilon$, uniformly in $n$.

First, notice that
$$
\inftyint  \overline{v_{nx}}\ w_{nx}\ dx = 0,
$$
since $v_n$ and $w_n$ have disjoint supports.


 To estimate the second term in the integral on the first line of \eqref{twentynine}, write
\begin{equation}
\begin{aligned}
\overline{v_{nx}} h_{nx}&= (\rho \overline{u_n})_x(u_n- \rho u_n - \theta u_n)_x
\\&= (\rho \overline{u_n})_x(u_n- \rho u_n)_x =(\rho_x \overline{u_n} + \rho \overline{u_{nx}})(-\rho_x u_n + (1-\rho)u_{nx}),
\end{aligned}
\end{equation}
where we have used $\rho$ as an abbreviation for $\rho(y_n -x)$.  Then we have
\begin{equation}
\begin{aligned}
2\alpha \Re\inftyint \overline{v_{nx}}h_{nx}\ dx&=-2 \alpha\inftyint \rho_x^2|u_n|^2\ dx + 2\alpha \inftyint \rho(1-\rho)|u_{nx}|^2\ dx
\\&-2\alpha \Re\inftyint \rho\rho_x\overline{u_{nx}}u_n\ dx+2\alpha \Re\inftyint(1-\rho)\rho_x\overline{u_{n}}u_{nx}\ dx
\\&\ge -2\alpha\left( \|\rho_x\|_{L^\infty}^2\|u_n\|_{L^2}^2+\|\rho_x\|_{L^\infty}\|u_n\|_{L^2}\|u_{nx}\|_{L^2}\right)
\\& \ge -C\epsilon,
\end{aligned}
\end{equation}
since $\|\rho_x\|_{L^\infty} \le \epsilon$ and $u_n$ is bounded in $H^1$ by Lemma \ref{lemmathreenine}.

A similar argument proves the desired estimate for the third term in the integral on the first line of \eqref{twentynine}.

The first term in the second line of \eqref{twentynine} is non-negative, and so trivially satisfies the desired estimate.

The second term in the second line of \eqref{twentynine} can be estimated by writing
\begin{equation}
\begin{aligned}
\inftyint &\int_0^1 |T(v_n+w_n)|^2|Th_n|^2  \ ds\ dx\\
&\le \int_0^1 \left(\inftyint |T(v_n+w_n)|^4\ dx \inftyint |Th_n|^4\ dx\right)^{1/2}\ ds
\\&\le C\int_0^1 \left(\|T(v_n+w_n)\|_{L^2}\|T(v_n+w_n)_x\|^3_{L^2}\|Th_n\|_{L^2}\|T(h_n)_x\|^3_{L^2}\right)^{1/2}\ ds
\\&=C\int_0^1 \|v_n+w_n\|^{1/2}_{L^2}\|(v_n+w_n)_x\|^{3/2}_{L^2}
\|h_n\|^{1/2}_{L^2}\|h_{nx}\|^{3/2}_{L^2}\ ds
\\&\le C\epsilon^{1/4}.
\end{aligned}
\end{equation}
Here we have used that $\displaystyle \|h_n\|_{L^2}^2 \le \epsilon$, and that $\{w_n\}$, $\{v_n\}$, and $\{h_n\}$ are bounded sequences
in $H^1$, due to the fact that $\{u_n\}$ is bounded in $H^1$.

The remaining terms in the second and third lines of \eqref{twentynine}, along with the first term in the fourth line of \eqref{twentynine},
are estimated similarly.  We have already shown in the preceding paragraph that
$$
\inftyint \int_0^1 |Th_n|^4\ ds\ dx\le C\sqrt{\epsilon},
$$
and we have also that
\begin{equation*}
\begin{aligned}
\left|\int_0^1 \right.&\left.\inftyint |T(v_n+w_n)|^2 T(v_n+w_n)Th_n\ dx\ ds\right|
\\&\le C \int_0^1 \|T(v_n+w_n)\|_{L^4}^3\|Th_n\|_{L^4}\ ds
\\&\le C\|v_n+w_n\|^{3/4}_{L^2}\|(v_n+w_n)_x\|^{9/4}_{L^2}
\|h_n\|^{1/4}_{L^2}\|h_{nx}\|^{3/4}_{L^2}
\\&\le C\epsilon^{1/8},
\end{aligned}
\end{equation*}
\begin{equation*}
\left|\int_0^1 \inftyint (T(v_n+w_n))^2 \overline{Th_n}\ dx\ ds\right|\le C\epsilon^{1/4},
\end{equation*}
and
\begin{equation*}
\left|\int_0^1 \inftyint (T(v_n+w_n))|Th_n|^2\overline{Th_n}\ dx\ dt\right|\le C\epsilon^{3/8}.
\end{equation*}

To estimate the last four terms on the right-hand side of \eqref{twentynine}, which do not involve $h_n$, we first need to establish
the following facts.  Let $\displaystyle r_c=\frac12\left(r_1+r_2\right)$.  We claim that
\begin{equation}
\label{twest}
\int_{y_n-r_c}^{y_n+r_c}\left|T(s)w_n\right|^2 \ dx\le C\epsilon
\end{equation}
and
\begin{equation}
\label{tvest}
\int_{|x-y_n| \ge r_c}\left|T(s)v_n\right|^2 \ dx\le C\epsilon
\end{equation}
for all $n \ge n_0$.

To prove these estimates, we use the argument of Lemma \ref{theoremsevenone}. As in equation \eqref{thirtythree} we obtain, for all $r$
such that $r_c \le r \le r_c + 1/\epsilon$, the relation
$$ \int_{y_n-r}^{y_n+r}\left|T(t)w_n\right|^2 \ dx
=2\int_0^t \Delta(s)\Im \left[\overline{T(s)w_n}\ T(s)w_{nx}\right]_{y_n-r}^{y_n+r}\ ds.$$
Integrating this relation with respect to $r$ over the interval $[r_c, r_c+1/\epsilon]$,  and estimating
the left-hand side from below and the right-hand side from above, we obtain
 \begin{equation}
\begin{aligned}
\frac{1}{\epsilon} \int_{y_n-r_c}^{y_n+r_c}\left|T(t)w_n\right|^2 \ dx
&\leq \int_{r_c}^{r_c+1/\epsilon} \int_{y_n+r}^{y_n-r}\left|T(t)w_n\right|^2\ dx\ dr
\\&=2\left|\int_{r_c}^{r_c+1/\epsilon}\int_0^t \Delta(s) \Im \left[\overline{Tw_n}\ Tw_{nx}\right]_{y_n-r_c}^{y_n+r_c}\ ds\ dr\right|
\\&\le C\int_0^t \Delta \left\|T(s)w_n\right\|_{L^2}\left\|T(s)w_{nx}\right\|_{L^2}\ ds\le C.
\end{aligned}
\end{equation}
This proves \eqref{twest}.  The estimate \eqref{tvest} for $v_n$ is obtained similarly.

Now using \eqref{twest} and \eqref{tvest}, we can write
\begin{equation}
\begin{aligned}
\int_0^1 &\inftyint\left|Tv_n\right|^2 \left|Tw_n\right|^2\ dx\ ds\\
&=\int_0^1 \int_{|x-y_n|\le r_c}\left|Tv_n\right|^2 \left|Tw_n\right|^2\ dx\ ds
 +\int_0^1 \int_{|x-y_n|\geq r_c}\left|Tv_n\right|^2 \left|Tw_n\right|^2\ dx\ ds
\\&\le C\int_0^1\|Tv_n\|_{L^\infty}^2 \int_{|x-y_n|\le r_c}\left|Tw_n\right|^2\ dx\ ds
\\&\ \ \ +C\int_0^1\|Tw_n\|_{L^\infty}^2 \int_{|x-y_n|\le r_c}\left|Tv_n\right|^2\ dx\ ds
\\&\le C\epsilon(\|Tv_n\|_{H^1}^2 + \|Tw_n\|_{H^1}^2) = C\epsilon(\|v_n\|_{H^1}^2 + \|w_n\|_{H^1}^2) \le C\epsilon,
\end{aligned}
\end{equation}
where we have used Corollary \ref{LinfHone}, applied to $Tv_n$ and $Tw_n$.

Similar estimates apply to the remaining terms in \eqref{twentynine}.
\end{proof}



\begin{cor}\label{corollarythreefifteen} Suppose $\{u_n\}$ is a minimizing sequence for $I_\lambda$ and
$0<q<\lambda$. Then
\begin{equation}
I_{\lambda}\ge I_{q}+I_{\lambda-q}.
\end{equation}
\end{cor}

\begin{proof}
First observe that if $v$ is a function such that $\left|P(v)-q\right|<\epsilon$,
then $P(\beta v)=q$, where $\beta=\sqrt{q/P(v)}$ satisfies $\left|\beta-1\right|<A_1\epsilon$
with $A_1$ independent of $g$ and $\epsilon$. Hence
\begin{equation}
I_{\alpha}\leq E({\beta}v) \le E(v)+{A_2}\epsilon ,
\end{equation}
where $A_2$ depends only on $A_1$ and $\left\|v\right\|_{1}$. A similar result holds for functions $w$ such that
\begin{equation}
\left|P(w)-(\lambda-q)\right|<\epsilon.
\end{equation}
From these observations and Lemma \ref{lemmathreethirteen} it follows easily that there exists a subsequence $\{u_{n_k}\}$ of $\{u_n\}$ and corresponding functions $v_{n_k}$ and $w_{n_k}$ such that for all $k$,
\begin{equation}
E\left(v_{n_k}\right)\geq I_{q}-\frac{1}{k},
\end{equation}
\begin{equation}
E\left(w_{n_k}\right)\geq I_{\lambda-\alpha}-\frac{1}{k},
\end{equation}
and
\begin{equation}
E\left(u_{n_k}\right)\geq E\left(v_{n_k}\right)+E\left(w_{n_k}\right)-\frac{1}{k},
\end{equation}
so
\begin{equation}
\label{elim}
E\left(v_{n_k}\right)\geq I_{\alpha}+ I_{\lambda-q}-\frac{1}{k}.
\end{equation}
The desired result is now obtained by taking the limit  as $k \to \infty$ of both sides of the inequality \eqref{elim}.
\end{proof}

% In this part we are proving the subadditivity condition. Proved the Compactness.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%% Subaddivatity %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{lem}\label{lemmathreefourteen}
For all $\lambda_1>0$ and $\lambda_2>0$, one has
\begin{equation}
I_{\left({\lambda_1}+{\lambda_2}\right)}<I_{\lambda_1}+I_{\lambda_2}.
\end{equation}
\end{lem}

\begin{proof}
First we claim that for all $\theta>1$, and $\lambda>0$
\begin{equation}
I_{\theta\lambda}<\theta I_\lambda
\end{equation}
To see this, let $\phi_n$ be such that $\displaystyle \inftyint{\phi_n}^2= \lambda$ and
$E\left(\phi_n\right)\to I_{\lambda},$
and let $\eta$ be such that $\displaystyle\inftyint\left(\eta \phi_n\right)^2=\theta\lambda$,
or $\eta=\sqrt{\theta}$. Then
\begin{equation*}
\begin{aligned}
I_{\theta\lambda}&=\inf\{E\left(u\right):u\in H^1 \ \text{and} \int \left|u\right|^2=\theta\lambda\}
\\&\leq E\left(\eta\phi_n\right)
\\&= \eta^2\left(\alpha\int\left|\phi_{nx}\right|^2\right)
-\eta^4 \inftyint \int_0^1 g(s)\left|T(s)\phi_n\right|^4\ ds\ dx
\\&=\eta^2\alpha\left(\int\left|\phi_{nx}\right|^2\right)
-\eta^2\inftyint\int_0^1 g(s)\left|T\left(s\right)\phi_n\right|^4\ ds\ dx
\\&+\left(\eta^2-\eta^4\right)\inftyint \int_0^1 g(s)\left|T\left(s\right)\phi_n\right|^4\ ds\ dx
\\&=\theta E\left(\phi_n\right)
+\left(\eta^2-\eta^4\right)\inftyint\int_0^1 g(s)\left|T\left(s\right)\phi_n\right|^4\ ds\ dx.
\end{aligned}
\end{equation*}
So, taking $n\to{\infty}$, we get, since $\eta>1$ and therefore $\eta^2-\eta^4 <0$, that
\begin{equation*}
\begin{aligned}
I_{\theta\lambda}&\leq \theta I_\lambda
+ \left(\eta^2-\eta^4\right)\liminf_{n\to{\infty}}\inftyint\int_0^1
g(s)\left|T\left(s\right)\phi_n\right|^4\ ds\ dx.
\end{aligned}
\end{equation*}
Since $I_\lambda<0$ by Lemma \ref{lemmathreesix}, then we have
\begin{equation}
\beta=\liminf_{n\to{\infty}}{\int_{-\infty}^{\infty}\int_0^1
g(s)\left|T\left(s\right)\phi_n\right|^4\ ds\ dx}>0,
\end{equation}
and so
\begin{equation}
I_{\theta\lambda}\leq\theta{I_\lambda}+\left(\eta^2-\eta^4\right)
\beta<\theta{I_\lambda}.
\end{equation}
%Now from the claim and lemma and section $2$ it follows that for all $\lambda_1$,%$\lambda_2>0$
%\begin{equation}
%{I_{\left({\lambda_1}+{\lambda_2}\right)}}<\left
%({\lambda_1}+{\lambda_2}\right){I_1}={I_{\lambda_1}}+{I_{\lambda_2}}
%\end{equation}

Now suppose, without loss of generality, that $\lambda_1\geq\lambda_2$. Then from the claim just proved, it follows that
\begin{equation*}
\begin{aligned}
{I_{\left({\lambda_1}+{\lambda_2}\right)}}&= I_{{\lambda_1}\left(1+{\lambda_2/ \lambda_1}\right)}
<\left(1+{\lambda_2/ \lambda_1}\right)I_{\lambda_1}
\\&\leq I_{\lambda_1}+ {\left(\lambda_2/ \lambda_1\right)}{{\left(\lambda_1/ \lambda_2\right)}I_{\lambda_2}}
\\&= I_{\lambda_1}+ I_{\lambda_2},
\end{aligned}
\end{equation*}
so \[{I_{\left({\lambda_1}+{\lambda_2}\right)}} < I_{\lambda_1}+ I_{\lambda_2}\]
as desired.
\end{proof}

Now we can rule out the ``dichotomy'' alternative of Theorem \ref{theoremthreefour} for minimizing sequences for $I_\lambda$.

\begin{cor}
\label{nodichotomy}
Suppose $\{u_n\}$ is a minimizing sequence for $I_\lambda$.  Then $q \notin (0,\lambda)$.
\end{cor}

\begin{proof}
From Corollary \ref{corollarythreefifteen}, if $q\in\left(0,\lambda\right)$, then by taking $\lambda_1=q$ and $\lambda_2=\lambda-q$ we get
\begin{equation}
{I_{\left({\lambda_1}+{\lambda_2}\right)}}\leq I_{\lambda_1}+I_{\lambda_2}
\end{equation}
which contradicts Lemma \ref{lemmathreefourteen}.
\end{proof}

Finally we examine what happens in the only remaining alternative from Theorem \ref{theoremthreefour}, the case of ``compactness''.

\begin{lem} \label{lemmathreesixteen}
 Suppose $q=\lambda$.  Then there exists a sequence of real numbers $\{y_1,y_2,y_3,\cdots\}$ such that
\begin{enumerate}
\item
for every $z<\lambda$ there exists $r=r(z)>0$ and an integer $N(z)$ such that for all $n > N(z)$,
\begin{equation}
\int_{{y_n}-r}^{{y_n}+r}\left|u_n\right|^2\ dx > z.
\label{intgreaterz}
\end{equation}
\item the sequence
$\{\widetilde{u_{n}}\}$ defined by
$\widetilde{u_n}(x)=u_n(x-y_{n})$ for $x\in\textbf{R}$ has a
subsequence which converges in $H^1$ norm to a function $g\in
G_{\lambda}$.
\end{enumerate}
In particular, $G_{\lambda}$ is nonempty.
\end{lem}

\begin{proof}
Since $q=\lambda$, then there exists $r_0$ such that for all sufficiently large values of $n$ we have
\begin{equation}
M_n\left(r_0\right)= \sup_{y\in\textbf{R}}\int_{y-r_0}^{y+{r_{0}}}\left|u_n\right|^2\ dx > \lambda/2.
\end{equation}
Hence for each sufficiently large $n$ we can find $y_n$ such that
\begin{equation}
\int_{y_n-r_{0}}^{y_n+r_{0}}\left|u_n\right|^2\ dx > \lambda/2.
\end{equation}
Now let $z<\lambda$ be given; clearly we may assume $z>\lambda/2$. Again, since $q =\lambda$ then we
can find $r_0(z)$ and $N(z)$ such that if $n>N(z)$ then
\begin{equation}
\label{smallerintgreaterz}
\int_{y_n(z)-r_{0}(z)}^{y_n(z)+r_{0}(z)}\left|u_n\right|^2\ dx > z
\end{equation}
for some $y_n(z)\in\textbf{R}$. Since
$\displaystyle \int_{-\infty}^\infty\left|f_n\right|^2dx = \lambda$, it follows that
for large $n$ the intervals $[y_{n}-r_0, y_{n}+r_0]$ and
$[y_{n}(z)-r_{0}(z), y_{n}(z)+r_{0}(z)]$ must overlap. Therefore,
defining $r=r(z)=2r_{0}(z)+r_{0}$, we have that $[y_{n}-r,
y_{n}+r]$ contains $[y_{n}(z)-r_{0}(z), y_{n}(z)+r_{0}(z)]$, and
so \eqref{intgreaterz} follows from \eqref{smallerintgreaterz}, for all $n > N(z)$.
This proves statement $1$.

Now statement $1$ implies that for every
$k\in\textbf{N}$, there exists $r_{k}>0$ such that for
all sufficiently large $n$,
\begin{equation}
\int_{-r_k}^{r_k}\left|\widetilde{u_n}\right|^2\ dx > \lambda-\frac{1}{k}.
\end{equation}
By Lemma \ref{lemmathreenine}, the sequence $\{\widetilde{u_n}\}$ is uniformly bounded in $H^1$, and therefore
also in $H^1(-r_k,r_k)$.  Therefore, from Rellch's Lemma, the
compactness of the embedding of $H^1(\Omega)$ into $L^2(\Omega)$ on bounded intervals $\Omega$ it
follows that some subsequence of $\{\widetilde{u_n}\}$ converges in $L^2(-r_k, r_k)$ norm to a limit function
$g\in L^2(-r_k, r_k)$ satisfying
\begin{equation}
\int_{-r_{k}}^{r_{k}}\left|g\right|^2dx \ge \lambda-\frac{1}{k}.
\end{equation}

A Cantor diagonalization argument, together with the fact that
$\displaystyle \int_{-\infty}^\infty\left|\widetilde{u_n}\right|^2\ dx = \lambda$ for all $n$, then shows that
some subsequence of $\{\widetilde{u_n}\}$ converges in $L^2(\textbf R)$ norm to a function $g\in L^2(\textbf{R})$
satisfying $\displaystyle \int_{-\infty}^\infty\left|g\right|^2\ dx = \lambda$.  For ease of notation we continue to denote this
subsequence by $\widetilde{u_n}$. Again using Lemma
\ref{lemmathreenine}, together with the Gagliardo-Nirenberg Inequality, we have
\begin{equation}
\|\widetilde{u_n}-g\|_{L^3}\le C\|\widetilde{u_n}-g\|^{1/6}_{H^1}
\|\widetilde{u_n}-g\|^{5/6}_{L^2}\le C(2B)^{1/6}\|\widetilde{u_n}-g\|^{5/6}_{L^2},
\end{equation}
so $\widetilde{u_n}\to g$ in $L^3$ norm also. Furthermore, by the weak
compactness of the unit sphere and the weak lower continuity of the norm in Hilbert space, we can assume,
by passing to a subsequence again if necessary, that $\widetilde{u_n}$ converges weakly to $g$ in $H^1$, and that
\begin{equation}
\left\|g\right\|_{H^1} \le \liminf_{n\to\infty}\left\|\widetilde{u_n}\right\|_{H^1}.
\end{equation}

It follows then that
\begin{equation}
E(g)\le \lim_{n\to\infty}E(\widetilde{u_n})=I_{\lambda},
\end{equation}
and since $\widetilde{u_n}$ converges in $L^2$ to $g$, we also have that $P(g)=\lim_{n \to \infty}P(\widetilde{u_n})=\lambda$.  From the definition
of $I_\lambda$ we conclude that we must have $E(g)=I_{\lambda}$ and $g\in G_\lambda$. Finally,
$\displaystyle E(g)=\lim_{n\to \infty}E(\widetilde{u_n})$, $\displaystyle \|g\|_{L^3}
=\lim_{n\to\infty}\|\widetilde{u_n}\|_{L^3}$, and $\displaystyle \|g\|_{L^2}
=\lim_{n\to \infty}\|\widetilde{u_n}\|_{L^2}$ together imply that
$\displaystyle\|g\|_{H^1}=\lim_{n\to \infty}\|\widetilde{u_n}\|_{H^1}$,
and from an elementary exercise in Hilbert space theory it then follows
that $\widetilde{u_n}$ converges to $g$ in $H^1$ norm.
\end{proof}

%------------------------------------------------------------------------------------------
%Theorem 2.9 in Albert's paper that we need to prove 2.10

We can now prove Theorem \ref{theoremthreeseventeen}.
\begin{proof}[Proof of Theorem \ref{theoremthreeseventeen}]
Let $q$ be as defined in \eqref{qthreenine}. By Corollary
\ref{nodichotomy},
$q$ cannot be in $(0, \lambda)$, and by Lemma \ref{novanish} we cannot have $q=0$. So it
follows that $q=\lambda$. Hence by Lemma \ref{lemmathreesixteen}, the set $G_{\lambda}$ is nonempty and
statement $1$ of Theorem \ref{theoremthreeseventeen} holds.
Now suppose statement $2$ does not hold; then there exist a subsequence $\{u_{n_k}\}$ of $\{u_n\}$
and a number $\epsilon_0>0$ such that
\begin{equation}
\inf_{g\in G_{\lambda}, y\in\textbf{R}} \left\|u_{n_k}(.+y)-g\right\|_{H^1}\geq \epsilon_0
\end{equation}
for all $k\in\textbf{N}$. But since $\{u_{n_k}\}$ is itself a minimizing sequence for $I_{\lambda}$,
from statement $1$ it follows that there exist a sequence $\{y_k\}$ and $g_0\in G_{\lambda}$ such that
\begin{equation}
\lim_{k\to\infty} \left\|u_{n_k}(.+y_k)-g_{0}\right\|_{H^1}=0.
\end{equation}
This contradiction proves statement $2$.

Since the functionals $E$ and $P$ are invariant under translations,
then $G_{\lambda}$ clearly contains any translate of $g_0$ if it contains $g_0$,
and hence statement $3$ follows immediately from statement $2$.
\end{proof}

%-------------------------------------------------------------------------------------------
% Proof of theorem 2.1 in the paper/ theorem 2.10 in Dr. Albert's paper.

Finally we can complete the proof of Theorem
\ref{theoremthreetwo}.
\begin{proof}[Proof of Theorem \ref{theoremthreetwo}]
We have already seen in Theorem \ref{theoremthreeseventeen} that the set $G_\lambda$
defined in \eqref{Glambda} is non-empty. Also, as explained in Section $3.1$, for every
$\phi\in G_\lambda$ there exists $\theta\in \textbf{R}$ such that $e^{i\theta t}\phi(x)$
is a solitary-wave solution of \eqref{onefour}. Now suppose $G_\lambda$ is not stable.
Then there exist a number $\epsilon_0 >0$,
a sequence $\{\psi_n\}$ of functions in $H^1$, and a sequence of times $\{t_n\}$ such that
\begin{equation}
\inf_{g\in G_{\lambda}} \left\|\psi_n-g\right\|_{H^1} < \frac{1}{n}
\end{equation}
and
\[\inf_{g\in G_{\lambda}} \left\|u_n(\cdot,t_n)-g\right\|_{H^1} \geq \epsilon_0\]
for all $n$, where $u_n(x,t)$ solves \eqref{twoone} with $u_n(x,0)= \psi_n$. Then since
$\psi_n\to G_{\lambda}$ in $H^1$, and $E(g)=I_\lambda$ and $P(g)=\lambda$ for $g\in G_\lambda$, we
have $E(\psi_n)\to I_\lambda$ and $P(\psi_n)\to\lambda$.
Choose $\{\alpha_n\}$ such that $P(\alpha_n \psi_n)=\lambda$ for all $n$; thus $\alpha_n\to 1$.
Hence the sequence $v_n=\alpha_n u_n(.,t_n)$ satisfies $P(v_n)=\lambda$ and
\begin{equation}
\lim_{n\to \infty} E(v_n)= \lim_{n\to\infty}E(u_n(\cdot,t_n))= \lim_{n\to\infty}E(\psi_n)=I_{\lambda},
\end{equation}
and is therefore a minimizing sequence for $I_{\lambda}$. From Theorem \ref{theoremthreeseventeen} it
follows that for all $n$ sufficiently large there exists $g_n \in G_{\lambda}$ such
that $\left\|v_n-g_n\right\|_{H^1} < \epsilon_0/2$. But then, for large $n$,
\begin{equation*}
\begin{aligned}
\epsilon_0\leq\left\|u_n(., t_n)-g_n\right\|_{H^1} &\leq \left\|u_n(.,t_n)-v_n\right\|_{H^1}
+\left\|v_n-g_n\right\|_{H^1}
\\&\leq \left|1-\alpha_n\right|.\left\|u_n(.,t_n)\right\|_{H^1}+\frac{\epsilon_0}{2},
\end{aligned}
\end{equation*}
and taking $n \to \infty$ gives $\epsilon_0\leq \epsilon_0/2,$ a contradiction.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%Section three three%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Existence and stability in $L^2$ of solitary waves when $\alpha=0$}\label{stabalphazero}




The proof of existence and stability of solitary-wave solutions to the DMNLS equation
\eqref{twoone} given in Section \ref{stabalphapos} relied crucially in several places on the assumption that $\alpha > 0$.  In particular,
without the assumption that $\alpha > 0$ one cannot control the $H^1$ norm of minimizing sequences, and so cannot use the
Rellich Lemma as in Lemma \ref{lemmathreesixteen} to prove the convergence of minimizing sequences.  Remarkably, however, Kunze has shown
(\cite{K}, see also \cite{KMZ}) that a variational proof of existence and stability of solitary waves can be given in the case $\alpha = 0$ by using a novel version
of the method of concentration compactness.

In \cite{K,KMZ} it is assumed that $D(s)$ is piecewise constant and that $g(s) \equiv 1$.
In this section we verify that Kunze's arguments carry through to the case when $D(s)$ satisfies the more general assumption
D1 given above in Section 2, together with the further assumption:


\begin{assump}
The numbers $s_0$, \dots, $s_n$ in Assumption D1 can be chosen so that $\Delta(s)$ is absolutely continuous on the interval $(s_{j-1},s_j)$
for each $j = 1, 2, \dots, n$.
\end{assump}

In addition, for the case of variable $g(s)$ considered here, we need an extra regularity
assumption on $g(s)$; namely that $g(s)$ is piecewise absolutely continous on $[0,1]$.

Then the existence and stability result for solitary waves is as follows.

\begin{thm}\label{theoremthreeninteen}
Suppose $\alpha=0$. Assume that $D'(s)=\Delta(s)$ satisfies assumptions D1 and D2,
 and that $g(s)$ is a non-negative
measurable function on $[0,1]$ with $\displaystyle \int_0^1 g(s)\ ds > 0$ which is
piecewise absolutely continous on $[0,1]$.

Then for every $\lambda>0$, there exists a non-empty
set $\widetilde{G_\lambda}\subseteq L^2\bigcap L^{\infty}$
such that for every  $\phi\in \widetilde{G_\lambda}$, there exists $\omega$ such that
$\exp(i\omega t)\phi(x)$ is a solitary-wave solution of \eqref{twoone} with
$\displaystyle \inftyint \phi^2\ dx=\lambda$.

The set $\widetilde{G_\lambda}$ is stable in the following
sense: for every $\epsilon>0$, there exists $\delta>0$ such that if
$u_0 \in L^2$ and
\begin{equation}
\left\|{u_0}-\phi\right\|_{L^2}<\delta
\end{equation}
for some $\phi\in \widetilde{G_\lambda}$,
then the solution $u\left(x,t\right)$ of \eqref{twoone} with $u\left(x,0\right)=u_0\left(x\right)$ satisfies
\begin{equation}
\inf_{\Psi\in \widetilde{G}_{\lambda}}\left\|u\left(x,t\right)-\Psi
\left(x\right)\right\|_{L^2\left(dx\right)}<\epsilon
\end{equation}
for all $t\geq 0$.
\end{thm}

In the case when $\alpha = 0$, the energy functional $E(u)$ is given by
\begin{equation}\label{energy}
E(u)=-\frac12 \inftyint \int_0^1 |T(s)u|^4\ ds\ dx.
\end{equation}
 With $P(u)$ defined as before, we now consider the variational problem of minimizing $E(u)$ over the set of all $u \in L^2$
 satisfying $P(u)= \lambda$.  That is, we define
\begin{equation*}
\widetilde{I_\lambda}=\inf\ \{E(u): u \in L^2 \quad \mbox{and}\quad P\left(u\right)=\lambda\},
\end{equation*}
and
\begin{equation*}
\widetilde{G_\lambda}=\{\phi\in L^2: E\left(\phi\right)=I_\lambda \quad \text{and}\quad P\left(\phi\right)=\lambda\}.
\end{equation*}
The key to the proof of Theorem \ref{theoremthreeninteen} is Kunze's solution of this variational problem, in which he showed that
arbitrary minimizing sequences for $\widetilde{I_\lambda}$ have subsequences which converge in $L^2$ to
the solution set $\widetilde{G_\lambda}$.

Another way to view Kunze's result is to consider the estimate
$$\inftyint \int_0^1 \left|T(s)u\right|^4 \ ds\ dx\leq C\left\|u\right\|^4_{L^2},$$
which, as can be seen from the proof of Corollary \ref{corollarytwoseven} above, is valid for all $u \in L^2$,
for some $C >0$ which is independent of $u$.  Kunze's result shows that the best possible constant
$C$ in this estimate is attained on a non-empty set of functions in $L^2$.

\begin{lem}
\label{negvaralphazero}
For every $\lambda > 0$, we have
$$
0 > \widetilde{I_{\lambda}}  > -\infty.
$$
\end{lem}

\begin{proof}
To see that $\widetilde{I_{\lambda}}  > -\infty$, observe that it follows from Theorem \ref{theoremtwosix} with $r=4$ and $q=8$
that for each $u \in L^2$, we have
\begin{equation*}
\begin{aligned}
-E(u)&=\int_0^1 \|T(s)u\|_{L^4}^4\ ds
\\&  \le \left(\int_0^1 \|T(s)u\|_{L^4}^8\ ds\right)^{1/2}\left(\int_0^1 1\ ds\right)^{1/2}
\\& = \|T(\cdot)u\|_{L^8_t((0,1),L^4_x)}^4 \le C\|u\|_{L^4}^4.
\end{aligned}
\end{equation*}
 It follows that $E(u) \ge - C\lambda^2$ for all $u$ such that $P(u)=\lambda$, and therefore $\widetilde{I_\lambda} \ge - C \lambda^2 > -\infty$.

 To prove that $\widetilde{I_\lambda} < 0$, observe that the proof of Lemma \ref{lemmathreesix} shows that, for $u$ as given in \eqref{b8}, we have
 in place of \eqref{eofu} that
 $$
 E(u)=-\frac{\lambda^2\sqrt{\sigma_0}}{2\sqrt{2\pi}}\int_0^1\frac{g(s)\ ds}{\sqrt{{\sigma_0}^2+4 D(s)^2}}\ .
 $$
 Since $E(u)<0$ when $\sigma_0 > 0$, it follows that $\widetilde{I_\lambda} < 0$.
 \end{proof}


Here is a simple version of the classical Van der Corput lemma for oscillatory integrals (see, e.g., Chapter 1 of [P] for more general
versions).
\begin{lem}
\label{vandercorput}
Suppose $D'(s)=\Delta(s)$ satisfies assumptions D1 and D2, and $g(s)$ satisfies the assumptions of Theorem \ref{theoremthreeninteen}.  Then there exists a constant $C > 0$ such that for all $w \in \mathbf R$,
\begin{equation}
\label{vdcest}
\left|\int_0^1 g(s)e^{i w D(s)}\ ds \right|\le \frac{C}{1+|w|}.
\end{equation}
\end{lem}

\begin{proof}
By, taking the intervals $[s_{j-i},s_j]$ in Assumptions D_1 and D_2 smoller if necessary, we can
assume $g(s)$ is absolutely continous on $[s_{j-i},s_j]$ for each $j=1,\cdots,n$. Since
$g(s)$ is piecewise absolutely continous on $[0,1]$, we have
\begin{equation*}
%\begin{aligned}
\left| \int_0^1 g(s)e^{i w D(s)}\ ds\right| =
\left|\sum_{j=1}^n \int_{s_{j-1}}^{s_j}g(s) e^{i w D(s)}\ ds\right| \le
\sum_{j=1}^n \left|\int_{s_{j-1}}^{s_j} g(s)e^{i w D(s)}\ ds\right|,
%\end{aligned}
\end{equation*}
so it is enough to prove the result with the integral over $[0,1]$ replaced by the integral over an arbitrary interval $[a,b]$,
under the assumption that $D(s)$ is absolutely continuous on $[a,b]$ with $|D'(s)|=|\Delta(s)| \ge \Delta_0$ on $[a,b]$.  Also, since the left-hand side of \eqref{vdcest}
is clearly less than 1 for all $w \in \mathbf R$, we only need to prove that
\begin{equation}
\label{abint}
\left|\int_a^b g(s)e^{i w D(s)}\ ds \right|\le \frac{C}{|w|}
\end{equation}
for all $w$ such that $|w| \ge 1$.

Using integration by parts, we can write
\begin{equation*}
\begin{aligned}&\left|\int_a^bg(s)\exp(iwD(s))\ ds\right|=
\\&\ \ \ =\left|\frac{g(b)\exp(iwD(b))}{iwD'(b)}-\frac{g(a)\exp(iwD(a))}{iwD'(a)}
-\frac{1}{iw}\int_a^b \exp(iwD(s))\frac{d}{ds}\left(\frac{g(s)}{D'(s)}\right)\ ds\right|
\\&\ \ \ \le\frac{1}{|w|}\left(\frac{1}{|D'(b)|}+\frac{1}{|D'(a)|}+\int_a^b\left|\frac{d}{ds}\left(\frac{g(s)}{D'(s)}\right)\right|\ ds\right).
\end{aligned}
\end{equation*}
The assumptions on $\Delta(s)=D'(s)$ and $g(s)$ guarantee that
\begin{equation*}
\begin{aligned}
I=\int_a^b\left|\frac{d}{ds}\left(\frac{g(s)}{D'(s)}\right)\right|\ ds&=\int_a^b\left|\frac{d}{ds}\left(\frac{g(s)}{\Delta(s)}\right)\right|\ ds
\\&\le \int_a^b\frac{|\Delta(s)||g'(s)|+|g(s)||\Delta'(s)|}{\Delta_0^2}\ ds < \infty.
\end{aligned}
\end{equation*}
Also
$$
\frac{1}{|D'(b)|}+\frac{1}{|D'(a)|}\le \frac{2}{\Delta_0}.
$$
So it follows that \eqref{abint} holds with $C = (2/\Delta_0) + I$.
\end{proof}

As in \eqref{twotwo}, let us define
$$
 Q(u):=\int_0^1  g(s)T^{-1}(s)[ |T(s)u|^2 T(s)u] ds.
$$
Recall that if $u \in L^2$, then $Q(u) \in L^2$ also, by Lemma \ref{lemmatwoeightnew}(b).


\begin{lem}\ \label{Qhat}  Suppose $u \in L^2$. Then
the Fourier transform of $Q(u)\in L^2$ is given by
\begin{equation}
\label{qft}
\widehat{Q}(u)(\omega)=\frac{1}{4\pi^2}\inftyint \inftyint  Z(\omega,\omega_1,\omega_2)
\widehat{u}\left(\omega-\omega_1-\omega_2\right)
\widehat{\overline{u}}(\omega_1)\widehat{u}(\omega_2)\ d\omega_1 \ d\omega_2,
\end{equation}
where
$$
Z(\omega,\omega_1,\omega_2)=\int_{0}^{1}g(s) e^{i\alpha(\omega,\omega_1,\omega_2) D(s)}\ ds,
$$
with
$$
\alpha(\omega,\omega_1,\omega_2)=\omega_2+\omega_1^2-\omega_2^2 - (\omega-\omega_1-\omega_2)^2=2(\omega-\omega_2)(\omega_1+\omega_2).
$$
For all $\omega \in \mathbf R$,
\begin{equation}
\label{qftest}
\left|\widehat{Q}(u)(\omega)\right|\le C\inftyint \inftyint
\frac{\left|\widehat{u}\left(\omega-\omega_1-\omega_2\right)
\widehat{\overline{u}}(\omega_1)\widehat{u}(\omega_2)\right|}{1+\left|\alpha(\omega,\omega_1,\omega_2)\right|}\ d\omega_1\omega_2.
\end{equation}
\end{lem}

\begin{proof}
Using \eqref{defTs} and \eqref{ftconv}, we have
\begin{equation}
\begin{aligned}
&\widehat{Q}(u)(\omega)=\int_0^1 g(s)\mathcal{F}\left\{T^{-1}(s)[|T(s)|^2T(s)u] \right\}(\omega)\ ds
\\&=\int_0^1 g(s)e^{-iD(s)\omega^2}\mathcal{F}\left\{T(s)u \cdot\overline{T(s)u}\cdot T(s)u\right\}(\omega)\ ds
\\&=\int_0^1 g(s) e^{-i\omega^2 D(s)}\mathcal{F}\left\{T(s)u\right\}
\ast \mathcal{F}\left\{\overline{T(s)u}\right\}\ast\mathcal{F}\left\{T(s)u\right\}\ ds
%\\&=\int_0^1 e^{i\xi^2 D(s)} \inftyint p(\xi_1) q\ast r(\xi-\xi_1) \ d{\xi}_1 \ ds
%\\&=\int_0^1 e^{i\xi^2 D(s)}\inftyint p(\xi_1) \inftyint r(\xi_2) q(\xi-\xi_1-\xi_2)\ d\xi_2\ d\xi_1\ ds
%\\&=\int_0^1 e^{i\xi^2D(s)}\inftyint e^{i\xi^2_1 D(s)}\overline{\widehat{u_0}}(\xi_1)
%\inftyint e^{-i\xi^2_2 D(s)}\widehat{u_0}(\xi_2)e^{i(\xi-\xi_1-\xi^2)^2 D(s)}
%\widehat{u_0}(\xi-\xi_1-\xi_2)\ d\xi_1\ d\xi_2\ ds
\\&=\inftyint\inftyint\left[\int_{0}^{1}g(s) e^{i\alpha(\omega,\omega_1,\omega_2) D(s)}\ ds\right]
\widehat{u}(\omega-\omega_1-\omega_2)\widehat{\overline u}(\omega_1)\widehat{u}(\omega_2)\ d\omega_1\ d\omega_2,
\end{aligned}
\end{equation}
which proves \eqref{qft}.

The estimate \eqref{qftest} then follows immediately from \eqref{qft} and Lemma \ref{vandercorput}.
\end{proof}

 \begin{proof}[Proof of Theorem \ref{theoremthreeninteen}]
 The same argument as used at the end of Section \ref{stabalphapos} to prove Theorems \ref{theoremthreeseventeen} and \ref{theoremthreetwo}
 shows that the statement of Theorem \ref{theoremthreeninteen} will follow if we can show that for every minimizing sequence $\{u_n\}$
 for $\widetilde(I_\lambda)$ in $L^2$, there exists a subsequence $\{u_{n_k}\}$ and a sequence of numbers $y_k$ such that
 $u_{n_k}(x-y_k)$ converges strongly in $L^2$.

 The proof of the latter statement is given in Kunze's paper \cite{K} for the case when $D(s)$ is piecewise constant and $g(s)\equiv 1$.  It
 is enough to check, then, that Kunze's proof still works under the assumptions on $D(s)$ and $g(s)$ given in Theorem \ref{theoremthreeninteen}.

 Only a few modifications are needed to Kunze's proof, which is rather long.  Therefore rather then repeating all the details of Kunze's proof here,
 we only list the necessary modifications.  We use the notation Lemma $x.y$K to refer to Lemma $x.y$ in \cite{K}.

We already proved above in Theorem \ref{theoremtwosix} and Lemma \ref{lemmatwoeightnew} that Lemma 2.1K and Lemma 2.2K still hold in our case, except
that the formula for $\widehat{Q(u)}$ given in Lemma 2.2K has to be replaced here by the formula in Lemma \ref{Qhat}.

It is easy to see that Lemmas 2.4K and 2.5K hold as well in our case.

Lemma 2.6K is replaced here by Lemma \ref{negvaralphazero}, and Lemmas 2.7K and 2.8K are replaced here by Lemma \ref{theoremsevenone} and the estimates
obtained in the proof of Lemma \ref{lemmathreethirteen}.

Lemma 2.9K and its proof need no modification.  The proof of Lemmas 2.10K and 2.11K are the same in our case, except that in our case to prove the estimates (2.23)
and (2.31) in \cite{K}, one uses our Lemma \ref{Qhat}.

Finally Lemma 2.12K and its proof remain unchanged.

Once the Lemmas in Section 2 of \cite{K} have been proved, the proof given in Sections 3 and 4 of \cite{K} applies almost without change to prove the desired convergence result
for minimizing sequences.  The only modification needed is that in the proof of Lemma 4.1K, the variable coefficient $\Delta(t)$ should be inserted in the formula for $\dot I(t)$,
but the estimate for $\dot I(t)$ still holds because our assumptions imply that $\Delta(t)$ is bounded on $[0,1]$.
 \end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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