\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 279, pp. 1--6.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/279\hfil Solutions to nonlinear Schr\"odinger equations]
{Solutions to nonlinear Schr\"odinger equations
for special initial data}

\author[T. Wada \hfil EJDE-2015/279\hfilneg]
{Takeshi Wada}

\address{Takeshi Wada \newline
Department of Mathematics, Shimane University,
Matsue 690-8504, Japan}
\email{wada@riko.shimane-u.ac.jp}

\thanks{Submitted March 27, 2015. Published November 10, 2015.}
\subjclass[2010]{35Q55}
\keywords{Nonlinear Schr\"odinger Equations; solvability; rough initial data}

\begin{abstract}
 This article concerns the solvability of the nonlinear Schr\"odinger
 equation with gauge invariant power nonlinear term in one space dimension.
 The well-posedness of this equation is known only for $H^s$ with $s\ge 0$.
 Under some assumptions on the nonlinearity, this paper shows that
 this equation is uniquely solvable for special but typical initial data,
 namely the linear combinations of $\delta(x)$ and $\operatorname{p.v.} (1/x)$, which
 belong to $H^{-1/2-0}$. The proof in this article allows $L^2$-perturbations
 on the initial data.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction}

In this article we consider nonlinear Schr\"odinger equations with 
a gauge invariant nonlinear term
\begin{equation}
i\partial_t u+\frac12 \partial_x^2 u =f(u), \label{eq:nls}
\end{equation}
where $u:\mathbb{R}_t\times \mathbb{R}_x\to \mathbb{C}$, $f(u)=|u|^{p-1} u$, $p>1$.
The discussion in this paper is irrelevant to the sign of the nonlinear term, 
so we only treat the defocusing case.
For $1<p<5$, the well-posedness in $L^2$ of the Cauchy problem for \eqref{eq:nls} 
was proved in~\cite{Tsutsumi1987}.
However there are few results on the solvability of \eqref{eq:nls} for initial 
data in negative order Sobolev spaces.
Although Kenig-Ponce-Vega~\cite{KenigPonceVega1996} treated the case where 
$f(u)=u^2,\Bar{u}^2$ or $|u|^2$
and proved the well-posedness in negative order Sobolev spaces, their result 
does not cover the nonlinear term that we treat in this paper.
On the other hand Kita~\cite{Kita2014,KitaPre} solved  \eqref{eq:nls} 
with $u(0)=\delta(x)+L^2$-perturbation for $1<p<3$ by using the fact that 
there exists an exact solution if $u(0)=\delta(x)$.
Namely $u(t)=(2\pi i t)^{-1/2}\exp\{ix^2/2t -(2\pi t)^\alpha/(1-\alpha)\}$ 
satisfies  \eqref{eq:nls}, where $\alpha=(p-1)/2$.

In this article we consider the case where
\begin{equation}\label{eq:ID2}
u(0)=\sqrt{2\pi i}\,\lambda \delta(x)-\sqrt{\frac{2}{\pi i}} \mu \operatorname{p.v.} \frac1x+v_0.
\end{equation}
Here $\lambda,\mu\in \mathbb{C}$,  $v_0\in L^2$, and $\operatorname{p.v.}$ means Cauchy's principle value.
Unlike the case $u(0)=\delta(x)$,  exact solutions for 
\eqref{eq:nls}-\eqref{eq:ID2} are not known.
This makes the problem more difficult.
We introduce a first approximation of the solution to \eqref{eq:nls} and 
determine the difference of the solution itself and the first approximation by 
the contraction mapping principle.
 (Theorems \ref{thm:1} and \ref{thm:2}) We have to assume $1<p<5/2$ if
 $\lambda=0$ and assume $1<p<7/3$ for general case.
Finally we remark on uniqueness of solutions. (Theorem~\ref{thm:3})

\subsection*{Notation}
Throughout this paper we put $\alpha=(p-1)/2$.
$\mathcal{L}=i\partial_t+\partial_x^2/2$.
$U(t)=\exp(it\partial_x^2/2)$ is the free propagator.
$L^r$ denote usual Lebesgue spaces for the space variable.
For $T>0$, $L^q_TL^r$ is the abbreviation of $L^q(0,T;L^r)$.
$X_T=L^\infty_TL^2 \cap L^4_TL^\infty$ and $Y_T=L^1_TL^2+L^{4/3}_TL^1$.


\section{Preliminaries}

\begin{lemma}[Strichartz]\label{lem:Strichartz}
For any $\phi\in L^2$ and for any $F\in Y_T$, the following inequalities 
hold valid:
\begin{gather*}
\| U(t) \phi \|_{X_T}\le C \| \phi \|_2, \\
\| \int_0^t U(t-\tau) F(\tau) d\tau \|_{X_T} \le C \| F\|_{Y_T}.
\end{gather*}
The constants $C$ are independent of $T, \phi$ and $F$.
\end{lemma}

For a proof of the above lemma, see \cite{Strichartz1977,Yajima1987}.
Before proceeding to the nonlinear problem, we consider the linear 
Cauchy problem.

\begin{lemma}\label{lem:asym}
Let $\lambda, \mu \in \mathbb{C}$ and let $u_\mathrm{L}$ be the solution of
\begin{equation}
i\partial_t u_\mathrm{L}+\frac12 \partial_x^2 u_\mathrm{L} =0
\end{equation}
with
\begin{equation}
u_\mathrm{L}(0)=\sqrt{2\pi i}\lambda \delta(x)-\sqrt{\frac{2}{\pi i}}\mu \operatorname{p.v.} \frac1x.
\end{equation}
Then
\[ 
u_\mathrm{L} = U(t) u_\mathrm{L}(0) = \frac{e^{ix^2/2t}}{\sqrt{t}} g(x/\sqrt{t}).
\]
Here
\[ 
g(a)=\lambda+\sqrt{\frac{i}{2\pi}}\mu 
	 \Big[\int_{-\infty}^a e^\frac{-i\eta^2}{2} d\eta 
-\int_a^\infty e^\frac{-i\eta^2}{2} d\eta \Big].
\]
\end{lemma}

We remark that using integration by parts, we can easily show that
\[ 
g(a)=\lambda\pm \mu +O(1/a) \quad\text{as } a\to \pm\infty.
\]


\begin{proof}[Proof of Lemma \ref{lem:asym}]
This is done  by a direct calculation.
It is well-known that
\[ 
\mathcal{F}u_\mathrm{L}(0)= \sqrt{2\pi i}(\lambda + \mu \operatorname{sign}\xi),
\]
where $\mathcal{F}$ is the Fourier transform.
Therefore
\begin{align}
u_\mathrm{L}(t)
&= \sqrt{\frac{i}{2\pi}} \int_{-\infty}^\infty 
e^{ix\xi-it\xi^2/2}(\lambda + \mu \operatorname{sign}\xi)d\xi \\
	&= \sqrt{\frac{i}{2\pi}} e^{ix^2/2t} \int_{-\infty}^\infty 
e^{-it(\xi-x/t)^2/2}(\lambda + \mu \operatorname{sign}\xi)d\xi.
\end{align}
Changing the variable as $-\sqrt{t}(\xi-x/t)=\eta$, we obtain the result.
\end{proof}

\section{Case $\lambda=0$}

In this section we consider the case 
\begin{equation}\label{eq:ID1}
u(0)=-\sqrt{\frac{2}{\pi i}} \mu \operatorname{p.v.} \frac1x  +v_0
\end{equation}
with $v_0\in L^2$.
In this case we put $A(t)=\exp (-i |\mu |^{p-1} t^{1-\alpha}/(1-\alpha))$ and put
$v=u-A u_\mathrm{L}$ where $u$ satisfies \eqref{eq:nls} and $u_\mathrm{L}$
is defined as in Lemma \ref{lem:asym} with $\lambda=0$.
Then $v$ satisfies
\begin{equation}
\mathcal{L}v=R+N
\end{equation}
with $v(0)=v_0$, where
\[ 
R= t^{-\alpha} (-|\mu|^{p-1}+|g(x/\sqrt {t})|^{p-1})A u_\mathrm{L}, \quad
 N=f(A u_\mathrm{L}+v)-f(A u_\mathrm{L}).
\]
By Duhamel's principle, we convert this equation to the integral form
\begin{equation}\label{eq:IE1}
v(t)=U(t)v_0-i \int_0^t U(t-\tau) \{ R(\tau)+N(\tau) \} d\tau.
\end{equation}

\begin{theorem}\label{thm:1}
Let $1<p<5/2$ and let $v_0\in L^2$.
Then there exists $0<T\le 1$ such that \eqref{eq:IE1} has a unique 
solution in $X_T$.
\end{theorem}

\begin{proof}
For simplicity we only prove a priori estimates;
precise proof is done by the contraction mapping principle.
We apply Lemma \ref{lem:Strichartz} to the right-hand side of 
\eqref{eq:IE1} and obtain
\begin{equation}
\| v\|_{X_T} \le C\|v_0\|_2 +C \| R\|_{Y_T}+C \| N\|_{Y_T}.
\label{eq:estv}
\end{equation}
By the remark to Lemma \ref{lem:asym} we have
\begin{equation}
\| R\|_2 \le Ct^{-\alpha -1/2} \| \mu -g(x/\sqrt {t}) \|_2 
\le Ct^{-\alpha -1/2} \| \langle x/\sqrt {t}\rangle^{-1} \|_2 \le Ct^{-\alpha -1/4}.
\end{equation}
Therefore $\| R\|_{Y_T}\le \| R;L^1_TL^2\| \le  CT^{-\alpha+3/4}$.
We proceed to the estimate of the second term in the right-hand side of 
\eqref{eq:estv}.
We can easily show that
\[ 
|N|\le C t^{-\alpha} |v|+C|v|^p. 
\]
Therefore
\[ 
\| N\|_{Y_T} \le C\| t^{-\alpha} |v|;L^1_T L^2\| + \| |v|^p ;L^{q_0'}_TL^{r_0'}\| 
 \]
with $r_0=p+1$ and $q_0 =4(p-1)/(p+1)$ since 
$L^{q_0'}_TL^{r_0'}\hookrightarrow Y_T$.
Applying H\"older's inequality for the time variable we obtain
\begin{align*}
\| N\|_{Y_T} 
&\le CT^{1-\alpha}\| v;L^\infty_T L^2\| 
+ CT^{1-\alpha/2}\| v ;L^{q_0}_TL^{r_0}\|^p\\
&\le CT^{1-\alpha}\|v\|_{X_T} + CT^{1-\alpha/2} \|v\|_{X_T}^p. 
\end{align*}
Therefore, we have proved that
\[ 
\|v\|_{X_T}\le C\|v_0\|_2+ CT^{1-\alpha}\|v\|_{X_T} 
+ CT^{1-\alpha/2} \|v\|_{X_T}^p. 
\]
If $v_1$ and  $v_2$ are solutions to \eqref{eq:IE1}, we similarly obtain
\[ 
\|v_1-v_2 \|_{X_T}\le C(T^{1-\alpha} + T^{1-\alpha/2} \sum_{j=1}^2 
\|v_j\|_{X_T}^{p-1}) \|v_1-v_2 \|_{X_T}. 
\]
By these estimates, we can apply the contraction mapping principle.
\end{proof}

\section{General case}

We first remark that if $|\lambda+\mu|=|\lambda-\mu|$, $\lambda+\mu=0$ or 
$\lambda-\mu=0$, then we can prove analogous results to Theorem \ref{thm:1} 
by the same method.
However in general case the method in the previous section does not work.
Therefore we consider another first approximation $A u_\mathrm{L}$ of the
solution as follows.
Let $\rho \in C^2(\mathbb{R})$ be a real-valued function satisfying
$\rho(a)=|\lambda\pm\mu|^{p-1}$ for $\pm a\ge 1$.
We put
\begin{gather*}
A(t,x)=\exp \Big[ -i\int_0^t \tau^{-\alpha} \rho(x/\tau^\beta)d\tau \Big], \\
v=u-A u_\mathrm{L},
\end{gather*}
where $\beta>0$ is suitably chosen, $u$ is a solution to \eqref{eq:nls} 
with \eqref{eq:ID2}, and $u_\mathrm{L}$ is defined as in Lemma \ref{lem:asym}.
Then $v$ satisfies
\[ \mathcal{L}v=\sum_{j=1}^4 R_j +N, \]
where
\begin{gather*}
R_1=\frac 12 \Big[\int_0^t \tau^{-\alpha-\beta} \rho'(x/\tau^\beta)d\tau \Big]^2 
 Au_\mathrm{L}, \quad
R_2= \frac i2 \int_0^t \tau^{-\alpha-2\beta} \rho''(x/\tau^\beta)d\tau  Au_\mathrm{L}, \\
R_3= i\int_0^t \tau^{-\alpha-\beta} \rho'(x/\tau^\beta)d\tau A \partial_x  u_\mathrm{L}, \quad
R_4=t^{-\alpha} [ |g(x/\sqrt{t})|^{p-1} -\rho(x/t^\beta)] A u_\mathrm{L}
\end{gather*}
Similarly as in the previous section, we convert this equation to the integral form
\begin{equation}\label{eq:IE2}
v(t)=U(t)v_0-i \int_0^t U(t-\tau) \{ \sum_{j=1}^4 R_j(\tau)+N(\tau) \} d\tau.
\end{equation}
We look for the solution to \eqref{eq:IE2} by the contraction mapping principle.

\begin{theorem}\label{thm:2}
Let $1<p<7/3$ and let $(p-2)_+<\beta<(3-p)/2$.
Then there exists $0<T\le 1$ such that \eqref{eq:IE2} has a unique solution 
in $X_T$.
\end{theorem}

\begin{proof}
We only show a priori estimates. By Lemma \ref{lem:Strichartz},
\begin{align*}
\| v\|_{X_T} 
&\le C\|v_0\|_2 +C\sum_{j=1}^4 \| R_j\|_{Y_T}+C \| N\|_{Y_T} \\
&\le  C\|v_0\|_2 +C\| R_1;L_T^{4/3}L^1\|+C\| R_2;L_T^{4/3}L^1\| \\
&\quad+C\| R_3;L_T^1L^2\|+C\| R_4;L_T^1L^2\|+C \| N\|_{Y_T}.
\end{align*}
We estimate the right-hand side term by term.
\begin{align*}
\| R_1\|_1 
&\le Ct^{-1/2} \Big[ \int_0^t \tau^{-\alpha-\beta} 
\|\rho'(x/\tau^\beta)\|_2 d\tau \Big]^2\\
&= Ct^{-1/2} \Big( \int_0^t \tau^{-\alpha-\beta/2} d\tau \Big)^2
= Ct^{3/2-2\alpha -\beta} 
\end{align*}
if $\alpha +\beta/2<1$, and hence 
$\| R_1;L_T^{4/3}L^1\|\le CT^{9/4-2\alpha -\beta}$ under this condition.
Similarly $\| R_2;L^{4/3}L^1\|\le CT^{5/4-\alpha -\beta}$  if $\alpha+\beta<1$.
By the estimate
\[ 
|R_3| \le C\int_0^t \tau^{-\alpha-\beta} |\rho'(x/\tau^\beta)|
d\tau (|x|/t^{3/2}+1/t), 
\]
we obtain
\[ 
\| R_3\|_2 \le C t^{-1/2-\alpha+\beta/2}+Ct^{-\alpha-\beta/2} 
\]
if $\alpha -\beta/2<1$ and $\alpha+\beta/2<1$.
Therefore
\[ 
\| R_3;L_T^1L^2\| \le C T^{1/2-\alpha+\beta/2}+CT^{1-\alpha-\beta/2} 
\]
if  $\alpha -\beta/2<1/2$ and $\alpha+\beta/2<1$.

The estimate of $R_4$ is done as follows.
We can easily show that
\[ 
|R_4|\le 	\begin{cases}
Ct^{-\alpha -1/2}, & |x/\sqrt{t}|\le 1 \text{ or } |x/t^\beta|\le 1, \\[4pt]
Ct^{-\alpha }/|x|,  & |x/\sqrt{t}| \ge 1 \text{ and } |x/t^\beta|\ge 1.
\end{cases}
\]
We first consider the case where $\beta\ge 1/2$.
Then $|x/t^\beta|\ge 1$ follows from $|x/\sqrt{t}| \ge 1$ since we may 
assume $0<t<T<1$.
So we  divide the spatial real-axis into the parts $|x/\sqrt{t}| \le 1$ 
and $|x/\sqrt{t}| \ge 1$ and we denote the corresponding parts of $R_4$ by
$R_{4,<}$ and $R_{4,>}$.
By the estimate above we have $\| R_{4,<}\|_2\le Ct^{-\alpha -1/4}$
 and hence $\| R_{4,<};L^1_T L^2\|\le CT^{-\alpha +3/4}$ if $\alpha <3/4$.
We can similarly estimate $R_{4,>}$ and we obtain 
$\| R_{4};L^1_T L^2\|\le CT^{-\alpha +3/4}$ if $\alpha <3/4$.
If $\beta <1/2$ we divide the spatial real-axis into the parts 
$ |x/t^\beta|\le 1$ and $ |x/t^\beta|\ge 1$ and estimate $R_4$ similarly.
Then we obtain $\| R_{4,<};L^1_T L^2\|\le CT^{-\alpha-\beta/2+1/2}$ 
if $\alpha -\beta/2<1/2$ and
$\| R_{4,>};L^1_T L^2\|\le CT^{-\alpha -\beta/2+1}$ if $\alpha+\beta/2<1$.
On the other hand, the estimate of $N$ is same as in the previous section.
Collecting all the estimates above, we can conclude that
\begin{equation}
\| v\|_{X_T} \le C \| v_0\|_2 +CT^\epsilon 
+CT^{1-\alpha}\|v\|_{X_T} + CT^{1-\alpha/2} \|v\|_{X_T}^p 
\end{equation}
with some $\epsilon>0$, under the conditions that
\[ 
\alpha+\beta<1  \quad \text{and} \quad \alpha-\beta/2<1/2, 
\]
which is possible if $0<\alpha<2/3$ and $(2\alpha-1)_+<\beta<1-\alpha$,
or equivalently the assumption for $p$ and $\beta$.
The estimate for the difference of two solutions is the same as in the previous 
section.
\end{proof}

\section{A remark on the uniqueness}

In the previous two sections we discuss the unique existence of the integral 
equations \eqref{eq:IE1} or \eqref{eq:IE2}.
However the uniqueness of the solution to \eqref{eq:nls} with \eqref{eq:ID1} 
or \eqref{eq:ID2} may fail because different first approximations
derive another solutions. In this section we consider this problem.
Let $\tilde{\rho}\in C^2(\mathbb{R})$ be a different real-valued function from $\rho$
in the previous section but let $\tilde{\rho}$ satisfy 
$\tilde{\rho}(a)=|\lambda\pm\mu|^{p-1}$ for $\pm a\ge 1$.
Let $\tilde{\beta}>0$, and we put
\begin{gather*}
\Tilde{A}(t,x)=\exp \Big[ -i\int_0^t \tau^{-\alpha} 
\Tilde{\rho}(x/\tau^{\Tilde\beta})d\tau \Big], \\
\Tilde v=u-\Tilde A u_\mathrm{L}.
\end{gather*}
As in the previous section, we convert this equation with $\tilde{v}(0)=v_0$ 
into integral form and solve this integral equation by the contraction mapping 
principle.
We want to prove that $v+Au_\mathrm{L}=\tilde v +\Tilde A u_\mathrm{L}$, where $v$ and $A$
are the ones in the previous section.
To this end it is  sufficient to show the following.
\begin{theorem}\label{thm:3}
Let $1<p<5/2$, $\beta>(p-2)_+$ and let $\tilde \beta$ satisfy the same condition.
Then
$Au_\mathrm{L}-\Tilde A u_\mathrm{L} \in X_T$.
\end{theorem}

\begin{proof}
By the estimate
\begin{equation}\label {eq:A-Atilde}
|Au_\mathrm{L}-\Tilde A u_\mathrm{L}|
\le Ct^{-1/2}\int_0^t \tau^{-\alpha} |\rho(x/\tau^\beta)
-\tilde\rho(x/\tau^{\tilde\beta})|d\tau,
\end{equation}
 we can easily show that 
$\|Au_\mathrm{L}-\Tilde A u_\mathrm{L} ; L^4_T L^\infty \| \le CT^{3/4-\alpha}$ if $\alpha<3/4$.
On the other hand, the right-hand side of \eqref{eq:A-Atilde} does not exceed
\[ 
Ct^{-1/2}\int_0^t \tau^{-\alpha} |\rho(x/\tau^\beta)-|g(x/\sqrt{\tau})|^{p-1}| 
d\tau
+Ct^{-1/2}\int_0^t \tau^{-\alpha}|\tilde\rho(x/\tau^{\tilde\beta})
-|g(x/\sqrt{\tau})|d\tau, 
\]
where $g$ is defined in Lemma \ref{lem:asym}, it suffices to show that the 
first integral in the above quantity belongs to $X_T$.
Since
\[  
|\rho(x/\tau^\beta)-|g(x/\sqrt{\tau})|^{p-1}|
\le
\begin{cases}
C , & |x/\sqrt{\tau}|\le 1 \text{ or } |x/\tau^\beta|\le 1, \\[4pt]
C\sqrt{\tau}/|x|,  & |x/\sqrt{\tau}| \ge 1 \text{ and } |x/\tau^\beta|\ge 1,
\end{cases}
\]
we obtain
\[ 
\|t^{-1/2}\int_0^t \tau^{-\alpha} 
|\rho(x/\tau^\beta)-|g(x/\sqrt{\tau})|^{p-1}| d\tau;L^\infty_T L^2\|
\le CT^{-\alpha+\beta/2+1/2}+CT^{-\alpha-\beta/2+1} 
\]
if $\alpha<3/4$ and $\beta>(2\alpha-1)_+$.
Thus we have proved the theorem.
\end{proof}


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\end{document}