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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. ??, pp. 1--16.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/??\hfil Ott-Sudan-Ostrovskiy equations]
{Ott-Sudan-Ostrovskiy equations on a right half-line}

\author[M. P. Arciga-Alejandre, E. I. Kaikina \hfil EJDE-2011/??\hfilneg]
{Mart\'in P. \'Arciga-Alejandre, Elena I. Kaikina}
% in alphabetical order

\address{ 
Instituto de F\'isica y Matem\'aticas \\
UMSNH, Edificio C-3, Ciudad Universitaria, Morelia CP 58089,
Michoac\'an, Mexico}
\email[M. P. Arciga-Alejandre]{mparciga@matmor.unam.mx}
\email[E. I. Kaikina]{ekaikina@matmor.unam.mx}

\thanks{Submitted August 8, 2011. Published November 3, 2011.}
\subjclass[2000]{35Q35}
\keywords{Dispersive nonlinear evolution equation;
 asymptotic behavior; \hfill\break\indent Initial-boundary value problem}

\begin{abstract}
 We consider initial-boundary value problems for the
 Ott-Sudan-Ostrovskiy equation on a right half-line.
 We show the the existence of solutions, global in time,
 and study their asymptotic behavior for large time.
\end{abstract}

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

\section{Introduction} \label{0s}

This article is devoted to the study of the initial-boundary value
problem for the Ott-Sudan-Ostrovskiy equation on the right half-line,
\begin{equation}
\begin{gathered}
u_{t}+uu_x+\alpha u_{xxx}+\int_0^{+\infty}
 \frac{\operatorname{sign}(x-y)u_{y}
(y,t)  }{\sqrt{| x-y| }}dy=0,\quad t>0,\; x>0,\\
u(x,0)=u_0(x),\quad x>0,\\
u(0,t)=0,\quad t>0,
\end{gathered} \label{1.1}
\end{equation}
where $\alpha>0$.

The Ott-Sudan-Ostrovskiy equation is a simple universal model
equation  which appears as the first approximation in the
description of the ion-acoustic waves in plasma \cite{ns1,os}. We
study traditionally important questions in the theory of nonlinear
partial differential equations, such as global in time existence
of solutions to the initial-boundary value problem \eqref{1.1})
and the asymptotic behavior of solutions for large time.

Many publications have dealt with asymptotic representations of
solutions to the Cauchy problem for nonlinear evolution equations
in the previous twenty years. While not attempting to provide a
complete review of these publications, we do list some known
results: \cite{Amick,Biler,Dix2,
Escobedo,GmiraVeron1,bookhkns}, where, in particular,
the optimal time decay estimates and asymptotic formulas of
solutions to different nonlinear local and nonlocal dissipative
equations were obtained. In the case of dispersive equations some
progress in the asymptotic methods was achieved due to the
discovery of the Inverse Scattering Transform method (see books
\cite{AblowitzSegur,Novik}). Some other
functional analytic methods were applied for the study of the
large time asymptotic behavior of solutions to dispersive
equations in \cite{Cazenave1,GinibreOzawa1,kpv2,Strauss1}.

The theory of the initial-boundary value problems on a half-line
for the nonlinear non local equations is relatively young and
traditional questions of a general theory are far from their
conclusion (\cite{lionsMagenes,W}). There are many open natural
questions which we need to study. First of them is how many
boundary data should be posed in the initial-boundary value
problems for it's correct solvability. Another difficulty of the
non local equation on a half-line is due to the influence of the
boundary data. A description of the large time asymptotic behavior
of solutions requires new approach and some reorientation of the
points of view comparing with the Cauchy problem. For example,
comparing with the corresponding case of the Cauchy problem,
solutions can obtain rapid oscillations, can converge to a
self-similar profile, can grow or decay faster, and so on. So
every type of the nonlinearity and boundary data should be studied
individually. For the general theory of nonlinear
pseudodifferential equations on a half-line we refer to the book
\cite{hkbook}. This book is the first attempt to develop
systematically a general theory of the initial-boundary value
problems for nonlinear evolution equations on a half-line, where
pseudodifferential operator ${K}$ on a half-line was
introduced by virtue of the inverse Laplace transformation of the
product of the symbol $K(p)=O(p^{\beta})$ which is analytic in the
right complex half-plane, and the Laplace transform of the
derivative $\partial _x^{[\beta]}u$. The main difficulty in the
boundary value problem \eqref{1.1} is that the operator
\[
Ku=\alpha u_{xxx}+\int
_0^{+\infty}\frac{\operatorname{sign}(x-y)u_{y}(y,t) }{\sqrt {|
x-y| }}dy
\]
in equation \eqref{1.1} has a symbol $K(p)=\alpha
p^3 +| p| ^{1/2}$, which is non analytic. Thus we can not
use methods of the book \cite{hkbook}  directly. Also the order
of the first term of the symbol $K(p)$ is critical, since the
number of the boundary data depends also on the sign of $\alpha$
(see \cite{hkbook}). In paper \cite{JDEH} the initial-boundary
value problem for the nonlinear nonlocal Ott-Sudan-Ostrovskiy
type on the left half-line ($\alpha<0$) was studied. It was proved
that because of the negative sign of the term $u_{xxx}$ two
boundary conditions have to be imposed at $x=0$. In the present
paper we develop the theory of the Ott-Sudan-Ostrovskiy equation
\eqref{1.1} considering the case of the right half-line and$\
\alpha>0$. We will show below that only one boundary value is
necessary and sufficient to pose in the problem \eqref{1.1} for
its solvability and uniqueness. Our approach here is based on the
$ L^{p}$ estimates of the Green function. For constructing
the Green operator in the present paper we follow the idea of
paper \cite{JDE}, reducing the linear problem \eqref{1.1} to the
corresponding Riemann problem. The Laplace transform requires the
boundary data $u(0,t)$, $u_x(0,t)$, $u_{xx}(0,t)$ and so
$u_x(0,t)$ and $u_{xx}(0,t)$ should be determined by the given
data. To achieve this we need to solve the system of nonlinear
singular integro-differential equations with Hilbert kernel.
We believe that the method developed in this paper could be
applicable to a wide class of dissipative nonlinear non local
equations.

To state precisely the results of the present paper we give some
notation. We denote $\langle t\rangle =1+t$,
$\{t\}  =t/\langle t\rangle$. Direct Laplace transformation
$\mathcal{L}_{x\to \xi}$ is
\[
\widehat{u}(\xi)  \equiv\mathcal{L}_{x\to \xi}u=\int
_0^{+\infty}e^{-\xi x}u(x)\,  dx
\]
and the inverse Laplace transformation $\mathcal{L}_{\xi\to  x}^{-1}$
is defined by
\[
u(x)  \equiv\mathcal{L}_{\xi\to  x}^{-1}\widehat
{u}=(2\pi i)^{-1}\int_{\mathbf{-}i\infty}^{i\infty}e^{\xi
x}u(\xi)  d\xi.
\]
Weighted Lebesgue space $L^{q,a}(\mathbb{R}^{+})
=\{\varphi\in\mathcal{S}';\| \varphi\|_{{L}^{q,a}}<\infty\}$, where
\[
\| \varphi\|_{{L}^{q,a}}
=\Big(\int _0^{+\infty}(1+x)^{aq}| \varphi(x)| ^{q}dx\Big)^{1/q}
\]
for $a>0$, $1\leq q<\infty$ and
\[
\| \varphi\|_{{L}^{\infty}}
=\operatorname{ess\,sup}_{x\in\mathbb{R}^{+}}| \varphi(x)  |
\]
By $\Phi^{\pm}$ we denote a left and right limiting values
of sectionally analytic function $\Phi$  given by integral of
Cauchy type
\[
\Phi(z)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{\phi(q)}{q-z}dq
\]
All the integrals are understood in the sense of the Cauchy principal
value.
We introduce $\Lambda(s)\in L^{\infty}(\mathbb{R}^{+})$ by formula
\begin{equation} \label{lambda1}
\begin{split}
\Lambda(s) & =\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}e^{ps-|
p| ^{1/2}}dp
 -\big(\frac{1}{2\pi i}\big)^3
 \int_{-i\infty}^{i\infty}d\xi e^{\xi}
 \int_0^{+\infty}e^{-ps} \\
&\quad \times \frac{\sqrt{p}(p^2+| \xi|^{4})
^{1/8}}{(\sqrt{ip}+\xi)(\sqrt{-ip}+\xi)}dp
\int_0^{\infty} \frac{1}{q+p}\frac{1}{(q^2+| \xi| ^{4})^{1/8}}dq.
\end{split}
\end{equation}
We define the linear operator
\begin{equation}
f(\phi)=\int_0^{+\infty}\phi(y)dy.\label{u1}
\end{equation}
Now we state our main result.

\begin{theorem}\label{thm1}
 Suppose that for small $a>0$ the initial data
$$
u_0 \in{Z}= L^1(\mathbb{R}^{+})\cap L^{1,a}(\mathbb{R}^{+})
 \cap L^{\infty}(\mathbb{R}^{+})
$$
are such that $\| u_0\|_{{Z}}\leq\varepsilon$
is sufficiently small. Then there exists a unique global solution
\[
u\in{C}([0,\infty); L^1(
\mathbb{R}^{+})\cap L^{1,a}(\mathbb{R}^{+})
)\cap{C}((0,\infty); L^{\infty
}(\mathbb{R}^{+}))
\]
to the initial-boundary value problem \eqref{1.1}. Moreover
\begin{equation}
u=A\Lambda(xt^{-2})t^{-2}+O(t^{-2-\gamma}),\label{A1}
\end{equation}
as $t\to \infty$ in $ L^{\infty}$, where $\gamma\in
(0,\min(1,a))$ and
\[
A=f(u_0)-\int_0^{+\infty}f(u_xu))d\tau.
\]
\end{theorem}

\section{Preliminaries} \label{S0}

We consider the following linear initial-boundary value problem on half-line
\begin{equation}
\begin{gathered}
u_{t}+\alpha u_{xxx}+\int_0^{+\infty}\frac{\operatorname{sign}
(x-y)u_{y}(y,t)}{\sqrt{| x-y| }}dy=0,\quad t>0,\; x>0,\\
u(x,0)=u_0(x),\quad x>0,\\
u(0,t)=0,\quad t>0.
\end{gathered}\label{2.1}
\end{equation}
Denote
\begin{equation}
K(q)=\alpha q^3 +| q| ^{1/2},\quad
K_1(q)=\alpha q^3 +q^{1/2},\quad
K_1(k(\xi))=-\xi,\label{ksi}
\end{equation}
where $\operatorname{Re}k(\xi)>0$ for $\operatorname{Re}\xi>0$.
We define
\begin{equation}
\mathcal{G}(t)
\phi=\int_0^{+\infty}G(x,y,t)\phi(y)dy, \label{operator}
\end{equation}
where
\begin{equation}
\begin{split}
&G(x,y,t)\\
&=\big(\frac{1}{2\pi i}\big)^2\int_{-i\infty}^{i\infty}d\xi
e^{\xi t}\int_{-i\infty}^{i\infty}dpe^{px}\frac{Y^{-}(p,\xi)}{K_1(p)+\xi
}(p-k_1)(p-k_2)I^{-}(p,\xi,y)
\end{split}\label{Green}
\end{equation}
for $x>0$, $y>0$, $t>0$. Here and below
\begin{equation}
Y^{\pm}=e^{\Gamma^{\pm}},\label{iks}
\end{equation}
$\Gamma^{+}(p,\xi)$ and $\Gamma^{-}(p,\xi)$ are a left and right
limiting values of sectionally analytic function
\begin{equation}
\Gamma(z,\xi)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{1}{q-z}\ln
\frac{K(q)+\xi}{K_1(q)+\xi}dq,\label{gam}
\end{equation}
and
\begin{equation}
I(z,\xi,y)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{e^{-qy}}{q-z}
\frac{1}{(q-k_2)(q-k_1)}\frac{1}{Y^{+}
(q,\xi)}dq.\label{ksii}
\end{equation}
All the integrals are understood in the sense of the principal
values.

\begin{proposition} \label{P1}
Let the initial data $u_0$ be in $L^1$. Then there
exists a unique solution $u(x,t)$ of the initial-boundary value
problem \eqref{2.1}, which has integral representation
\begin{equation}
u(x,t)=\mathcal{G}(t)u_0.\label{Theorem}
\end{equation}
\end{proposition}

\begin{proof}
To derive an integral representation for the solutions of
\eqref{2.1} we suppose that there exists a solution
$u(x,t)$ of problem \eqref{2.1}, which is continued by zero
outside of $x>0$:
\[
u(x,t)=0\quad \text{for all } x<0.
\]
Let $\phi(p)$ be a function of the complex variable Re$p=0$,
which obeys the H\"{o}lder condition for all finite $p$ and tends
to $0$ as $p\to \pm i\infty$. We define the operator
\[
\mathbb{P}\phi(z)=-\frac{1}{2\pi i}
\int_{-i\infty}^{i\infty}\frac{1}{p-z} \phi(p)dp.
\]
Note that $\mathbb{P}\phi(z)=F(z)$ constitutes a function analytic
n the left and right semi-planes. Here and below these functions
will be denoted $F^{+}(z)$ and $F^{-}(z)$, respectively.
These functions have the limiting values $F^{+}(p)$ and $F^{-}(p)$
at all points of imaginary axis $\operatorname{Re}p=0$, on
approaching the contour from the left and from the right, respectively.
These limiting values are expressed by Sokhotzki-Plemelj formula
\begin{equation}
F^{+}(p)-F^{-}(p)=\phi(p).\label{SP}
\end{equation}
We have for the Laplace transform
\[
\mathcal{L}_{x\to  p}\Big\{
\int_0^{+\infty}\frac{\operatorname{sign} (x-y)u_{y}(y,t)
}{\sqrt{| x-y| }}dy\Big\}
=\mathbb{P}\Big\{  | p|^{1/2}(\mathcal{L}_{x\to  p}\{u\}
-\frac{u(0,t)}{p})\Big\}.
\]
Since $\mathcal{L}_{x\to  q}\{u\}$ is analytic for all
$\operatorname{Re}q>0$ we have
\begin{equation}
\widehat{u}(q,t)=\mathcal{L}_{x\to  q}\{u\}
=\mathbb{P}\{  \widehat{u}(p,t)\} \label{x}
\end{equation}
and
\[
\mathcal{L}_{x\to  q} \{\alpha u_{xxx}\}
 =\mathbb{P} \Big\{  \alpha p^3 \Big(\widehat{u}(p,t)
-\sum_{j=1}^3 \frac{\partial_x^{j-1}u(0,t)}{p^{j}}\Big)\Big\}  .
\]
Therefore, applying the Laplace transform to \eqref{2.1}
with respect to space and time variables we obtain for
$\operatorname{Re}p=0$, $t>0$
\begin{equation}
\begin{gathered}
\widehat{\widehat{u}}(p,\xi)
 =  \frac {1}{K(p)+\xi}\big(\Xi(p,\xi)+\widehat{\Phi}(p,\xi)\big)\\
\Xi(p,\xi)  =  \widehat{u}_0(p)+\alpha
p\widehat{u}_x(0,\xi)+\alpha\widehat{u}_{xx}(0,\xi).
\end{gathered}\label{sol}
\end{equation}
with some function $\widehat{\Phi}(p,\xi)=O(p^{-1/2})$ such that for
all $\operatorname{Re}z>0$
\begin{equation}
\mathbb{P}^{-}\{\widehat{\Phi}\}  (z,\xi)=0\label{ass}
\end{equation}
Here the functions
$\widehat{\widehat{u}}(p,\xi)$,
$\widehat{\Phi}(p,\xi)$, $\widehat{u}_x(0,\xi)$ and
$\widehat{u}_{xx}(0,\xi)$ are the Laplace
transforms for $\widehat{u}(p,t)$, $\Phi(p,t)$, $u_x(0,t)$ and
$u_{xx}(0,t)$ with respect to time, respectively.
 We will find the function $\widehat{\Phi }(p,\xi)$ using the analytic
properties of function $\widehat{\widehat{u}}$ in
the right-half complex planes $\operatorname{Re}p>0$ and
$\operatorname{Re}\xi>0$. For $\operatorname{Re}p=0$, we have
\begin{equation}
\widehat{\widehat{u}}(p,\xi)=-\frac{1}{\pi i}
-\hskip -4mm
\int _{-i\infty}^{i\infty}\frac{1}{q-p}\widehat{\widehat{u}}(q,\xi)dq.
\label{11}
\end{equation}
As in \cite{JDE} we perform the condition \eqref{11} in the
form of nonhomogeneous Riemann problem to find
\[
\widehat{\Phi}(p,\xi)=-Y^{+}(p,\xi)U^{+}(p,\xi),
\]
where
\begin{equation}
U(z,\xi)=\mathbb{P}\big\{  \frac{1}{Y^{+}(p,\xi)}\frac{K_1(p)-K(p)}
{(K(p)+\xi)(K_1(p)+\xi)}\Xi(p,\xi)\big\}  .\label{u11}
\end{equation}

We now return to solution $u(x,t)$ of the problem \eqref{2.1}.
From \eqref{sol} with the help of the integral representation
\eqref{u11} and Sokhotzki-Plemelj formula \eqref{SP} we have for
Laplace transform of solution of the problem \eqref{2.1}
\begin{equation}
\widehat{\widehat{u}}(p,\xi)=\frac{1}{K_1(p)+\xi}(\widehat{u}
_0(p)+\alpha(p\widehat{u}_x(0,\xi)+\widehat{u}_{xx}(0,\xi))
-Y^{-}U^{-}).\label{sol12}
\end{equation}
There exist two roots $k_j(\xi)$ of equation $K_1(z)=-\xi$
such that $\operatorname{Re}k_j(\xi)>0$ for all
$\operatorname{Re}\xi>0$. Therefore in the
expression for the function $\widehat{\widehat{u}}$ the factor
$1/(K_1(z)+\xi)$ has two poles in the point
$z=k_j(\xi)$, $j=1,2$, $\operatorname{Re}z>0$. However the function
$\widehat{\widehat{u}}$ is the limiting value of an analytic
function in $\operatorname{Re}z>0$. Thus in general case the problem
\eqref{2.1} is insolvable. It is soluble only when the functions
$U^{-}(z,\xi)$ satisfies additional conditions. For analyticity of
$\widehat{\widehat{u}}(z,\xi)$ in points $z=k_j(\xi)$ it is
necessary that
\begin{equation}
\operatorname{Res}_{z=k_j(\xi)}\{  \widehat{u}_0(z)
+\alpha(z\widehat{u}_x(0,\xi)+\widehat{u}_{xx}(0,\xi))-Y^{-}U^{-}\}
=0,\quad j=1,2.\label{res}
\end{equation}
Therefore, we obtain for the Laplace transforms of boundary data
$u_x(0,t)$, $u_{xx}(0,t)$ the following system
\begin{equation}
B_1(k_j(\xi),\xi)\widehat{u}_x(0,\xi)+B_2(k_j(\xi),\xi)\widehat
{u}_{xx}(0,\xi)=\frac{1}{\alpha}\{  \mathcal{I}_{3}\widehat{u}
_0\}  (k_j),\quad j=1,2,\label{system}
\end{equation}
where
\begin{gather}
B_1(k_j(\xi),\xi)=k_j(\xi)-Y^{-}(k_j(\xi),\xi)\lim_{y\to
0}\partial_{y}\Upsilon_1(k_j(\xi),\xi,y), \label{v2}
\\
B_2(k_j(\xi),\xi)=1-Y^{-}(k_j(\xi),\xi)\lim_{y\to 0}\Upsilon
_1(k_j(\xi),\xi,y), \label{v21}
\\
\{  \mathcal{I}_{3}\widehat{u}_0\}  (k_j)
=\int_0^{+\infty }dy u_0(y)\Big(Y^{-}(k_j(\xi),\xi)\Upsilon_1(k_j(\xi),\xi
,y)-e^{-k_j(\xi)y}\Big),\label{v22}
\end{gather}
where
\begin{equation}
\Upsilon_1(z,\xi,y)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{1}
{q-z}\frac{1}{Y^{+}(q,\xi)}\frac{K_1(q)-K(q)}{K_1(q)+\xi}e^{-qy}
dq.\label{ya1}
\end{equation}
To solve this system firstly we consider the sectionally
analytic function $\Upsilon_1(z,\xi,y)$ given by Cauchy type integral.
Since, by \eqref{SP},
\begin{equation}
\frac{1}{Y^{+}(q,\xi)}\frac{K(q)+\xi}{K_1(q)+\xi}=\frac{1}{Y^{-}(q,\xi
)},\label{i}
\end{equation}
we have
\begin{equation}
\Upsilon_1(z,\xi,y)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{1}
{q-z}\Big(\frac{1}{Y^{+}(q,\xi)}-\frac{1}{Y^{-}(q,\xi)}\Big)
e^{-qy}dq\label{R1}
\end{equation}
Observe that the function $1/\big(Y^{-}(q,\xi)\big)$ is analytic
for all $\operatorname{Re} q>0$. Therefore, by the Cauchy Theorem
for $y>0$, we find
\begin{equation}
-\lim_{z\to  p,\; \operatorname{Re}z>0}
\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{1}{q-z}
\frac{1}{Y^{-}(q,\xi )}e^{-qy}dq
=\frac{1}{Y^{-}(p,\xi)}e^{-py}.\label{R2}
\end{equation}
Thus from \eqref{R1} and \eqref{R2}, we obtain the
relation
\begin{equation}
\Upsilon_1^{-}(p,\xi,y)=\Psi^{-}(p,\xi,y)+\frac{1}{Y^{-}(p,\xi)}
e^{-py},\label{i4}
\end{equation}
where
\[
\Psi(z,\xi,y)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{1}{q-z}\frac
{1}{Y^{+}(q,\xi)}e^{-qy}dq.
\]
Therefore,
\begin{gather*}
B_1(k_j(\xi),\xi)=Y^{-}(k_j(\xi),\xi)\partial_{y}\Psi(k_j
,\xi,0),\quad j=1,2,\\
B_2(k_j(\xi),\xi)=-Y^{-}(k_j(\xi),\xi)\Psi(k_j,\xi,0),\quad j=1,2
\end{gather*}
and
\begin{align*}
\{  \mathcal{I}_{3}\widehat{u}_0\}  (k_j)
& =\int_0 ^{+\infty}dyu_0(y)
\big[  Y^{-}(k_j(\xi),\xi)\Upsilon_1^{-}(k_j ,
 \xi,y)-e^{-k_j(\xi)y}\big]  \\
& =Y^{-}(k_j(\xi),\xi)\int_0^{+\infty}dyu_0(y)\Psi(k_j(\xi
),\xi,y),\quad j=1,2.
\end{align*}
Here and below
\begin{gather*}
\partial_{y}\Psi(z,\xi,0)  =\lim_{y\to 0}\partial_{y}\frac{1}{2\pi
i}\int_{-i\infty}^{i\infty}\frac{e^{-zy}}{q-z}\frac{1}{Y^{+}(q,\xi)}dq,\\
\Psi(z,\xi,0)  =\lim_{y\to 0}\frac{1}{2\pi i}\int_{-i\infty}
^{i\infty}\frac{e^{-zy}}{q-z}\frac{1}{Y^{+}(q,\xi)}dq,
\end{gather*}
Substituting these formulas in \eqref{system} we obtain
\begin{equation}
\begin{split}
&\widehat{u}_x(0,\xi)\\
&=\frac{1}{\alpha}\int_0^{+\infty}dyu_0(y)\frac
{\Psi(k_1,\xi,0)\Psi(k_2(\xi),\xi,y)-\Psi(k_2,\xi,0)\Psi(k_1(\xi
),\xi,y)}{\partial_{y}\Psi(k_2,\xi,0)-\partial_{y}\Psi(k_1,\xi,0)}
\end{split}\label{value2}
\end{equation}
and
\begin{equation}
\begin{split}
&\widehat{u}_{xx}(0,\xi)\\
&=\frac{1}{\alpha}\int_0^{+\infty}dyu_0
(y)\frac{\partial_{y}\Psi(k_1,\xi,0)\Psi(k_2(\xi),\xi,y)-\partial_{y}
\Psi(k_2,\xi,0)\Psi(k_1(\xi),\xi,y)}{\partial_{y}\Psi(k_2,\xi
,0)-\partial_{y}\Psi(k_1,\xi,0)}.
\end{split}\label{value3}
\end{equation}

Now we return to formula \eqref{sol12}. In accordance with
\eqref{value2}--\eqref{value3}, the function
$\widehat{\widehat{u}}(p,\xi)$ constitutes the limiting value of
an analytic function in $\operatorname{Re}z>0$ and, as a consequence,
its inverse Laplace transform vanish for all $x<0$.

Under assumption $u(0,t)=0$, via definition \eqref{ya1} the
integral representation \eqref{u11} for the function
$U^{-}(p,\xi)$, in $\operatorname{Re}z>0$, takes form
\begin{align*}
& U^{-}(p,\xi)\\
& =\int_0^{+\infty}dyu_0(y)\Upsilon_1^{-}(p,\xi,y)-\alpha\widehat{u}
_x(0,\xi)\partial_{y}\Upsilon_1^{-}(p,\xi,0)+\alpha\Upsilon_1^{-}
(p,\xi,0)\widehat{u}_{xx}(0,\xi),
\end{align*}
where the function $\Upsilon_1(z,\xi,y)$ is  defined by
\eqref{R1}. Using \eqref{i4} from \eqref{sol12} we obtain
\begin{equation}
\begin{split}
\widehat{\widehat{u}}&=\int_0^{\infty}u_0(y)\frac{Y^{-}(p,\xi)}
{K_1(p)+\xi}\Psi^{-}(p,\xi,y)dy\\
&\quad +\alpha\widehat{u}_x(0,\xi)\partial_{y}
\Psi^{-}(p,\xi,0)-\alpha\Psi^{-}(p,\xi,0)\widehat{u}_{xx}(0,\xi)
\end{split} \label{Ss2}
\end{equation}
where $\widehat{u}_x(0,\xi)$ and $\widehat{u}_{xx}(0,\xi)$ are
defined by \eqref{value2}--\eqref{value3}
\[
\Psi(z,\xi,y)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{e^{-qy}}
{q-z}\frac{1}{Y^{+}(q,\xi)}dq.
\]
Applying to \eqref{Ss2} the inverse Laplace transform with respect to
time, and the inverse Fourier transform with respect to space
variables, we obtain
\[
u(x,t)=\mathcal{G}(t)u_0=\int_0^{\infty}G(x,y,t)u_0(y)dy,
\]
where for $x>0$, $y>0$, $t>0$, the  $G(x,y,t)$ was defined by
\begin{equation} \label{Gg}
\begin{split}
& G(x,y,t) \\
& =\frac{1}{2\pi i}\frac{1}{2\pi i}
\int_{-i\infty}^{i\infty}d\xi e^{\xi t}
\int_{-i\infty}^{i\infty}dpe^{px}\frac{Y^{-}}{K_1(p)+\xi}
\Big(\Psi^{-}(p,\xi,y)  \\
& \quad  +\frac{(\partial_{y}\Psi^{-}(p,\xi,0)-\widetilde{\Psi}(k_2
))\Psi(k_1,\xi,y)-(\partial_{y}\Psi^{-}(p,\xi,0)-\widetilde{\Psi}
(k_1))\Psi(k_2,\xi,y)}{\partial_{y}\Psi(k_2,\xi,0)-\partial_{y}
\Psi(k_1,\xi,0)}\Big),
\end{split}
\end{equation}
where
\begin{gather*}
\widetilde{\Psi}(k_j)=\Psi(k_j,\xi,0)\partial_{y}\Psi(k_j,\xi,0),
\quad j=1,2,
\\
\Psi(z,\xi,y)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{e^{-qy}}
{q-z}e^{-\Gamma^{+}(q,\xi)}dq,
\\
\Gamma(z,\xi)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{1}{q-z}\ln
\frac{K(q)+\xi}{K_1(q)+\xi}dq. %\label{gamma}
\end{gather*}
For subsequent considerations it is required to investigate the behavior of
the function $\Gamma(z,\xi)$. Set
\[
\phi(p,\xi)=\ln\frac{K(p)+\xi}{K_1(p)+\xi}\neq0,\quad
\operatorname{Re} p=0,\quad \operatorname{Re} \xi\geq0.
\]
Observe that the function $\phi(p,\xi)$ satisifes
 the H\"{o}lder condition for all finite $p$ and tends to a
definite limit $\phi(\infty,\xi)$ as $p\to \pm i\infty$,
\[
\phi(\infty,\xi)=0.
\]
It can be easily obtained that for large $p$ and some
fixed $\xi$,
\begin{equation}
| \phi(p,\xi)-\phi(\infty,\xi)| \leq C\frac{\xi
}{\langle | p| \rangle ^3 }.\label{uux}
\end{equation}
Therefore,
\begin{equation}
| e^{\Gamma^{\pm}(z,\xi)}| \leq C\label{1e}
\end{equation}
for all $\operatorname{Re}\xi\geq0$. Moreover we have
\[
\frac{1}{q-z}=-\frac{1}{z}-\frac{q}{z^2}+\frac{q^2}{z^2(q-z)}.
\]
and therefore
\begin{equation} \label{g111}
\Gamma(z,\xi)  =\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{1}{q-z}
\ln(\frac{K(q)+\xi}{K_1(q)+\xi})dq
 =A_1(\xi)\frac{1}{z}+O(| z| ^{-2}),
\end{equation}
where
\begin{equation}
A_1(\xi)=\int_{-i\infty}^{i\infty}\ln\big\{\frac{K(q)+\xi}{K_1(q)+\xi
}\big\}  dq.\label{a111}
\end{equation}


In view of \eqref{uux}, for large $|q| $, the integrand
in \eqref{g111} is of order $| q| ^{-3}$;
whence the corresponding integrals with infinite limits are convergent,
in accordance with well-known
criterion for convergence improper integrals.

Using the above estimate, for $| z| >1$, we obtain
\[
e^{-\Gamma^{+}(z,\xi)}-1-A_1(\xi)\frac{1}{z}=e^{-A_1(\xi)/z
+O(| z| ^{-2})}-1+A_1(\xi)\frac{1}{z}=O(|z| ^{-2}).
\]
In view of this fact and the Cauchy Theorem, we obtain
\begin{align*}
\Psi^{-}(p,\xi,0)
& =\frac{1}{2\pi i}\lim_{y\to 0}\lim_{z\to
p,\,\operatorname{Re}z>0}\int_{-i\infty}^{i\infty}
\frac{e^{-qy}}{q-z}\big(\frac {1}{Y^{+}(q,\xi)}-1\big)dq\\
&\quad +\frac{1}{2\pi i}\lim_{y\to 0}\lim_{z\to  p,\,
\operatorname{Re}z>0} \int_{-i\infty}^{i\infty}\frac{e^{-qy}}{q-z}dq\\
& =-1
\end{align*}
and
$\Psi(k_j,\xi,0)=-1$ for $j=1,2$.
In the same way
\begin{align*}
\partial_{y}\Psi^{-}(p,\xi,0)
& =\frac{1}{2\pi i}\lim_{z\to
p,\,\operatorname{Re}z>0}\int_{-i\infty}^{i\infty}
\frac{q}{q-z}\big(\frac{1}
{Y^{+}(q,\xi)}-1+A_1(\xi)\frac{1}{q}\big)dq\\
&\quad -\lim_{y\to 0}\partial_{y}\lim_{z\to p,
 \operatorname{Re}z>0}
\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{e^{-qy}}{q-z}
\big(\frac{A_1(\xi)}{q}-1\big)dq\\
& =-A_1(\xi)+p.
\end{align*}
and
\[
\partial_{y}\Psi(k_j,\xi,0)=-A_1(\xi)+k_j,j=1,2.
\]
Thus after some calculation we attain
\begin{align*}
& [  \partial_{y}\Psi(k_2,\xi,0)-\partial_{y}\Psi(k_1,\xi,0)]
^{-1}\\
& \times(\partial_{y}\Psi^{-}(p,\xi,0)-\Psi(k_2,\xi,0)\partial
_{y}\Psi(k_2,\xi,0))\Psi(k_1,\xi,y) \\
& -(\partial_{y}\Psi^{-}(p,\xi,0)-\Psi(k_1,\xi,0)\partial_{y}
\Psi(k_1,\xi,0))\Psi(k_2,\xi,y))\\
& =\frac{(p-k_2)\Psi(k_1,\xi,y)-(p-k_1)\Psi(k_2,\xi,y)}{k_2
(\xi)-k_1(\xi)}.
\end{align*}
Substituting the above  relation in \eqref{Gg}, we obtain
\eqref{Green}. The proof is complete.
\end{proof}

Now we collect some preliminary estimates of the Green operator
$\mathcal{G}(t)$ defined in \eqref{operator}.

\begin{lemma}\label{Lemma 2.1}
The following estimates are true, provided that the right-hand sides
are finite
\begin{gather}
\| \partial_x^{n}\mathcal{G}(t)\phi\|_{ L^{s,\mu}}
\leq C\widetilde{t}^{-(\frac{1}{r}-\frac{1+\mu}{s}
)-n}\| \phi(\cdot)\|_{ L^{r}}+\widetilde{t}
^{-(\frac{1}{r}-\frac{1}{s})-n}\| \phi(\cdot)\|_{ L^{r,\mu}},\label{L2}
\\
\| (\mathcal{G}(t)\phi-t^{-2}\Lambda
(xt^{-2})f(\phi))\|_{ L^{\infty}}\leq Ct^{-2-2\mu
}\| (\cdot)^{\mu}\phi\|_{ L^1},\label{L3}
\end{gather}
where $\widetilde{t}=\{t\} ^{1/3}\langle
t\rangle ^2$, small $\mu>0$, $1\leq r\leq s\leq\infty$,
$n=0,1$, $\Lambda(s)$ is given by \eqref{lambda1} and $f(\phi)$ is
given by \eqref{u1}.
\end{lemma}

\begin{proof}
By Sokhotzki-Plemelj formula we have
\[
I^{-}(p,\xi,y)=I^{+}(p,\xi,y)-\frac{e^{-py}}{Y^{+}(p,\xi)}.
\]
Inserting this expression in \eqref{Green}, we have
\begin{equation}
G(x,y,t)=J_1(x,y,t)+J_2(x,y,t),\label{1J3}
\end{equation}
where
\begin{equation}
J_1(x,y,t)
=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}e^{px-K(p)t}
e^{-py}dp \label{0I1}
\end{equation}
and
\begin{equation}
\begin{split}
&J_2(x,y,t)\\
&=-(\frac{1}{2\pi i})^2\int_{-i\infty}^{i\infty
}d\xi e^{\xi t}\int_{-i\infty}^{i\infty}e^{px}\frac{Y^{+}(p,\xi)(p-k_1
(\xi))(p-k_2(\xi))}{K(p)+\xi}I^{+}(p,\xi,y)dp.
\end{split}\label{0I2}
\end{equation}
Denote
\[
\mathcal{J}_j(x,t)\phi=\int_0^{+\infty}J_j(x,y,t)\phi
(y)dy
\]
for $x>0$. From  \cite{bookhkns} we have
\[
 \| \partial_x^{n}\mathcal{J}_1(t)\phi\|_{ L^{s,\mu}}
 \leq C\widetilde{t}^{-(\frac{1}{r}-\frac{1+\mu}{s})-n}\| \phi
(\cdot)\|_{ L^{r}}+\widetilde{t}^{-(\frac{1}{r}-\frac{1}
{s})-2n}\| \phi(\cdot)\|_{ L^{r,\mu}}
\]
for $n=0,1,2$, $s\geq r$, $\mu\in(0,1)$. Let the contours
$\mathcal{C}_{i}$ be defined as
\begin{equation}
\mathcal{C}_{i}=\{  p\in(\infty e^{-i(\frac{\pi}{2}-(-1)^{i}
\varepsilon_{i})},0)\cup(0,\infty e^{i(\frac{\pi}
{2}-(-1)^{i}\varepsilon_{i})})\}  ,i=1,2,3,\label{c1}
\end{equation}
where $\varepsilon_j>0$ are small enough, can be chosen such that all
functions under integration are analytic and
$\operatorname{Re}k_j(\xi)>0$, $j=1,2$ for $\xi\in\mathcal{C}_{3}$.
In particular, for example, $K(p)+\xi \neq0$ outside the origin for
all $p\in\mathcal{C}_1$ and $\xi\in \mathcal{C}_{3}$.

Firstly we estimate $\mathcal{J}_1(t)\phi$ . Let
$x-y>0$. Let $x-y<0$. We can rewrite \eqref{0I1} in
the form
\begin{equation}
J_1(x,y,t)=\frac{1}{2\pi i}\int_{\mathcal{C}_1}e^{\xi t}d\xi\frac{1}{2\pi
i}\int_{\mathcal{C}_2}e^{-p(y-x)}\frac{1}{K(p)+\xi}dp\label{2ggg}
\end{equation}
By the choice of contour $\mathcal{C}_2$ we have
$| e^{-p(y-x)}| \leq Ce^{-C| p| | x-y| }$. Therefore, using
obvious estimates:
\begin{equation}
\| e^{-| p| | \cdot|
}\|_{L^{s,\mu}}\leq C| p| ^{-\frac{1+s\mu}{s}
},\label{eee1}
\end{equation}
\[
\| \int_0^{+\infty}e^{-C| p| |
x-y| }\phi(y)dy\|_{L^{s,\mu}}\leq C| p|
^{-1+\frac{1}{r}-\frac{1+s\mu}{s}}\| \phi\|_{L^{r}
}+C| p| ^{-1+\frac{1}{r}-\frac{1}{s}}\|
\phi\|_{L^{r,\mu}}
\]
for all $s,r\geq1$, for $p\in\mathcal{C}_2$, we obtain
\begin{equation} \label{01e}
\begin{split}
 \| \partial_x^{n}\mathcal{J}_1(t)\phi\|_{ L^{s,\mu}}
& \leq C\| \phi\|_{L^{r}}\int_{\mathcal{C}_1
}e^{-C| \xi| t}d\xi\int_{\mathcal{C}_2}\frac
{1}{| K(p)+\xi| }| p| ^{-1+n+\frac{1}
{r}-\frac{1+s\mu}{s}}dp \\
&\quad +C\| \phi\|_{L^{r,\mu}}\int_{\mathcal{C}_1
}e^{-C| \xi| t}d\xi\int_{\mathcal{C}_2}\frac
{1}{| K(p)+\xi| }| p| ^{-1+\frac{1}
{r}+n-\frac{1}{s}}dp \\
& \leq C\widetilde{t}^{-(\frac{1}{r}-\frac{1+\mu}{s})-n}\|
\phi(\cdot)\|_{ L^{r}}+\widetilde{t}^{-(\frac{1}{r}-\frac
{1}{s})-n}\| \phi(\cdot)\|_{ L^{r,\mu}}
\end{split}
\end{equation}
for $n=0,1$, $r>s$, $\mu\in(0,1)$. In the case of $x-y>0$ we use
the fact that for $p\in\mathcal{C}_{3}$, $\operatorname{Re}K(p)>0$ and
\[
\operatorname{Re}K(p)=O(\{p\}^{1/2}\langle p\rangle^3 ).
\]
Therefore, changing in formula \eqref{0I1} contour of integration
by $\mathcal{C}_{3}$ from the book \cite{bookhkns} we obtain estimate
\eqref{01e}.

Now we estimate $\mathcal{J}_2(t)\phi$. For any analytic in
the left half-complex plane function $\phi^{+}(p)$ we have
\begin{align*}
& \int_{-i\infty}^{i\infty}\frac{1}{K(p)+\xi}\phi^{+}(p)dp\\
& =(e^{i\frac{\pi}{4}}-e^{-i\frac{\pi}{4}})\int_0^{+\infty}\phi^{+}
(-p)\frac{\sqrt{p}}{(\sqrt{ip}-p^3 +\xi)(\sqrt{-ip}-p^3 +\xi)}
\end{align*}
where $\operatorname{Re}\xi=0$. Here
\[
K(p)=\begin{cases}
\alpha p^3 +(-ip)^{1/2}, & \operatorname{Im}p>0;\\
\alpha p^3 +(ip)^{1/2}, & \operatorname{Im}p<0.
\end{cases}
\]
Using the above relation, for $J_2(x,y,t)$ we obtain
\begin{equation}
\begin{split}
J_2(x,y,t)
&=-(\frac{1}{2\pi i})^2i\sqrt{2}\int
_{\mathcal{C}_1}d\xi e^{\xi t}\int_0^{\infty}e^{-px}Z^{+}(p,\xi
,y) \\
&\quad\times \frac{\sqrt{p}}{(\sqrt{ip}-p^3 +\xi)(\sqrt{-ip}-p^3 +\xi)}dp.
\end{split} \label{0i2}
\end{equation}
where
\[
Z^{+}(p,\xi,y)=Y^{+}(-p,\xi)(p+k_1(\xi))(p+k_2(\xi))I^{+}(-p,\xi,y).
\]
To estimate $J_2(x,y,t)$ in the case $t<1$ we rewrite $I^{+}$
in the form
\begin{align*}
&I^{+}(p,\xi,y) \\
 & =\frac{1}{2\pi i}
\int_{\mathcal{C}_2}\frac{e^{-qy}}{(q-p)(q-k_1(\xi))(q-k_2(\xi))}(
\frac{1}{Y^{-}(q,\xi)}\frac{K_1(q)+\xi}{K(q)+\xi}-\frac{1}{Y^{-}(q,\xi
)})dq\\
&\quad +\frac{1}{2\pi i}\int_{\mathcal{C}_2}\frac{e^{-qy}}{(q-p)(q-k_1
(\xi))(q-k_2(\xi))}\frac{1}{Y^{-}(q,\xi)}dq
\end{align*}
Therefore, using analytic properties of integrand function in
the second term by the Cauchy Theorem, we obtain
\begin{align*}
I^{+}(p,\xi,y)
& =\frac{1}{(k_2(\xi)-k_1(\xi))}\sum_{j=1}^2(-1)^{j}
\frac{e^{-k_j(\xi)y}}{(p-k_j(\xi))}\frac{1}{Y^{-}(k_j,\xi)}\\
& +\frac{1}{2\pi i}\int_{\mathcal{C}_2}\frac{e^{-qy}}{(q-p)(q-k_1
(\xi))(q-k_2(\xi))}\frac{1}{Y^{-}(q,\xi)}\frac{K_1(q)-K(q)}{K(q)+\xi}dq
\end{align*}
After this observation from the integral representation
\eqref{0I1} by Holder inequality we have arrived at the following
estimate for $r\geq1$, $s\geq 1$, $l^{-1}=1-r^{-1}$, small
$\mu\geq0$, $n=0,1$, $t>1$,
\begin{equation}  \label{J2m1}
\begin{split}
& \| \partial_x^{n}\int_0^{+\infty}J_2(\cdot,y,t)\phi
(y)dy\|_{ L^{s,\mu}} \\
& \leq C\| \phi\|_{ L^{r}}\int e^{-|
\xi| t}d\xi\int_0^{\infty}dp| p| ^{n-\frac
{1+\mu}{s}+\frac{1}{2}}\frac{| p+| \xi| ^{\frac
{1}{3}}| ^2}{| -p^3 +\xi| ^2} \\
&\quad \times\Big(\int_{\mathcal{C}_2}dq\frac{1}{| q-p|
| q| ^{\frac{1}{2}-\frac{1}{r}}}\frac{1}{|
-q^3 +\xi| }+\frac{1}{\langle | \xi|
\rangle ^{\frac{1}{3}+1-\frac{1}{r}}}\Big) \\
& \leq C\| \phi\|_{ L^{r}}
t^{-(\frac{1}{r}-\frac{1+\mu}{s}+n)/3}.
\end{split}
\end{equation}
Here we  used  \eqref{1e} and the estimate
$k_j(\xi)=O(\langle \xi\rangle ^{1/3})$.
Therefore according to \eqref{01e} and \eqref{J2m1}, for $t<1$,
we obtain the estimate \eqref{L2}.

Now we prove the asymptotic formula \eqref{L3}.
For large $t>1$ and $| \xi| >1$, the integrand $e^{\xi t}$
decays as $e^{-Ct}$. However,
in the neighborhood of $\xi=0$, the integrand $e^{\xi t}$ changes
relatively slowly. By this reason we split the integrals
\eqref{0i2} into integrals over sections $| p| <1$, $| \xi| <1$ and
over the rest of the ranges of integration.
In the neighborhood of $p=0$ we have
\[
K(p)+\xi  =\sqrt{| p| }+\xi+O(p^{7/2}),\quad
K_1(p)+\xi  =\sqrt{p}+\xi+O(p^{7/2}).
\]
Also for small $\xi$ by construction
$k_j(\xi)=\alpha^{-2/5}e^{i(-1)^{j} 2\pi/5}+O(\xi)$.
To separate the principal
part of the expansion of
\[
\Gamma^{+}(z,\xi)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{1}{q-z}
\ln\frac{K(q)+\xi}{K_1(q)+\xi}dq
\]
near points $p=0$, $\xi=0$ we introduce the new function
$\widetilde{\Gamma}^{+}(z,\xi)$ by the  relation
\begin{equation} \label{gammanew}
\begin{split}
\widetilde{\Gamma}^{\pm}(z,\xi)
& =\frac{1}{2\pi i}\Big\{  \int_{i|
\xi| ^2}^{i}+\int_{-i}^{-i| \xi| ^2}\Big\}
\frac{1}{q-z}\ln\frac{\sqrt{| q| }}{\sqrt{q}}
dq\\
& =\frac{1}{2\pi i}(-i\frac{\pi}{4})\int_{i| \xi| ^2}
^{i}\big(\frac{1}{q-z}+\frac{1}{q+z}\big)dq \\
& =\frac{1}{8}\ln\frac{z^2+| \xi| ^{4}}{z^2+1}
\end{split}
\end{equation}
It can be proved by a direct calculation that for small
 $|z| <1$, $|\xi|<1$,
\[
\widetilde{\Gamma}^{+}(z,\xi)-\Gamma^{+}(z,\xi)=O(\xi).
\]
Also via \eqref{gammanew}, the functions
\[
w^{\pm}(z,\xi)=e^{\widetilde{\Gamma}^{\pm}(z,\xi)}=(\frac
{z^2+| \xi| ^{4}}{z^2+1})^{1/8}
\]
are analytic in $z\in C$ except for $z\in[  -i,-i| \xi|
^2]  \cup[  i| \xi| ^2,i]  $ and
therefore
\[
w^{+}(z,\xi)=w^{-}(z,\xi)
\]
for all $z\notin[  -i,-i| \xi| ^2]
\cup[ i| \xi| ^2,i]$. Via the last comments
we obtain for small $p>0$, $\xi\in\mathcal{C}_1$
\[
I^{+}(p,\xi,y)=\frac{1}{2\pi i}\frac{1}{k_1(\xi)k_2(\xi)}
\int_{-i\infty }^{i\infty}\frac{1}{q-p}\frac{1}{(q^2+| \xi| ^{4}
)^{1/8}}\frac{K(q)}{K_1(q)}dq(1+O(y^{\mu}p^{\mu})),
\]
where $\mu\in(0,\frac{1}{4})$.

On the basis last relations for points of the contours close the
points $p=0$ and $\xi=0$ integrand of $\mathcal{M}_1(x,y,t)$ has
the following representation
\[
-\big(\frac{1}{2\pi i}\big)^2i\sqrt{2}e^{\xi t}e^{-px}\frac{\sqrt
{p}(p^2+| \xi| ^{4})^{1/8}}{(\sqrt{ip}-p^3
+\xi)(\sqrt{-ip}-p^3 +\xi)}\widetilde{I}^{+}(p,\xi,y).
\]
where
\[
\widetilde{I}^{+}(p,\xi,y)
=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{1}{q-p}
 \frac{1}{(q^2+| \xi| ^{4})^{1/8}}
\frac{K(q)}{K_1(q)}dq.
\]
On integrating this function by extending the limits to $-i\infty$ and
$+i\infty$ we obtain the contributions to $\mathcal{M}_1(x,y,t)$
from the neighborhoods of $\xi=0$ and $p=0$
\begin{align}
& -t^{-2}(\frac{1}{2\pi i})^2\int_{\mathcal{C}_1}d\xi
e^{\xi}\int_0^{+\infty}e^{-pxt^{-2}}\frac{\sqrt{p}(p^2+|
\xi| ^{4})^{1/8}}{(\sqrt{ip}+\xi)(\sqrt{-ip}+\xi
)}dp\label{m}\\
& \times\int_0^{\infty}\frac{1}{q+p}\frac{1}{(q^2+|
\xi| ^{4})^{1/8}}dq.
\end{align}


To estimate the contribution of $R(x,y,t)$ to $G(x,y,t)$ over the
rest of the range of integration we use standard method of
``partition of unity'' and Watson Lemma.
Due to smoothing properties of integrand functions by integrating by
part we obtain
\begin{equation}
R(x,y,t)=y^{\mu}O(t^{-2-2\mu})\label{Iii2}
\end{equation}
For the function $J_1(x,y,t)$ defined by \eqref{0I1}, we have
\begin{equation}
J_1(x,y,t)=\frac{1}{2\pi i}t^{-2}\int_{-i\infty}^{i\infty}e^{pxt^{-2}
-| p| ^{1/2}}dp+y^{\mu}O(t^{-2-2\mu}).\label{Iii3}
\end{equation}
Thus from \eqref{m}, \eqref{Iii2} and \eqref{Iii3}, via
\eqref{1J3} we obtain
\[
\| \mathcal{G}\phi-t^{-2}\Lambda(xt^{-2})f(\phi)\|_{ L^{\infty}}\leq Ct^{-2-2\mu}\| \langle x\rangle
^{\mu}\phi\|_{ L^1}.
\]
By the same argument, it is easy to prove  \eqref{L2} for
$t>1$. The proof is complete.
\end{proof}

\begin{theorem}\label{TL}
Let the initial data $u_0$ belong to
$L^1(\mathbb{R}^{+})\cap L^{1,a}(\mathbb{R}^{+})$, with small
$a\geq0$. Then for some $T>0$ there exists a unique solution
\[
u\in{C}([0,T]  ; L^1(
\mathbb{R}^{+})\cap L^{1,a}(
\mathbb{R}^{+}))\cap {C}((0,T];
{H}_{\infty }^1(\mathbb{R}^{+}))
\]
to the initial-boundary value problem \eqref{1.1}
\end{theorem}

\section{Proof of Theorem \ref{thm1}}

By Proposition \ref{P1} we rewrite the initial-boundary value
problem \eqref{1.1} as the  integral equation
\begin{equation}
u(t)=\mathcal{G}(t)u_0-\int_0^t
\mathcal{G}(t-\tau)\mathcal{N}(u(\tau)
)d\tau\label{Greenformula}
\end{equation}
where $\mathcal{G}$ is the Green operator of the linear problem
\eqref{2.1}. We choose the space
\[
{Z}=\{\phi\in L^1(\mathbb{R}^{+})
\cap L^{1,a}(\mathbb{R}^{+})\}
\]
with $a>0$ is small and the space
\[
{X}=\{\phi\in{C}([0,\infty);{Z})\cap{C}((0,\infty)
;{H}_{\infty}^1(\mathbb{R}^{+})): \| \phi\|_{{X}}<\infty\},
\]
where
\[
\| \phi\|_{{X}}
=\sup_{t\geq0}\Big(\{t\}  ^{-a/3}\langle t\rangle ^{-2a}\|
\phi(t)\|_{ L^{1,a}}+\| \phi(t)\|_{ L^1}
 +\sum_{n=0}^1\{t\}  ^{\frac{n+1}{3}}\langle t\rangle ^{2(n+1)}\|
\phi(t)\|_{ L^{\infty}}\Big)
\]
shows the optimal time decay properties of the solution.
We apply the contraction mapping principle in a ball
${X}_{\rho}=\{  \phi \in{X}: \| \phi\|_{{X}}\leq
\rho\}$ in the space ${X}$ of  radius
\[
\rho=\frac{1}{2C}\| u_0\|_{{Z}}>0.
\]
For $v\in{X}_{\rho}$ we define the mapping $\mathcal{M}(v)$
 by
\begin{equation}
\mathcal{M}(v)=\mathcal{G}(t)u_0-\int_0^t e^{-\tau
}\mathcal{G}(t-\tau)\mathcal{N}(v(\tau)
)d\tau.\label{map}
\end{equation}
We first prove that
$\| \mathcal{M}(v)\|_{{X}}\leq\rho$,
where $\rho>0$ is sufficiently small. By  Lemma \ref{Lemma 2.1}, we
obtain
\begin{gather*}
\ \| \mathcal{G}(t)\phi\|_{ L^1
}\leq C\{t\}  ^{-\frac{n}{3}}\langle t\rangle
^{-2n}\| \phi\|_{ L^1}, \\
 \| \mathcal{G}(t)\phi\|_{ L^{1,a}
}\leq C\{t\}  ^{\frac{a}{3}}\langle t\rangle
^{2a}\| \phi\|_{ L^1}+\| \phi\|_{ L^{1,a}}, \\
 \| \partial_x^{n}\mathcal{G}(t)\phi\|_{ L^{\infty}}
\leq C\{t\}  ^{-(n+1)/3}\langle t\rangle ^{-2(n+1)}
 \| \phi\|_{ L^1}
\end{gather*}
for all $t\geq 0$. Therefore,
\begin{equation}
\| \mathcal{G}u_0\|_{{X}}\leq C\| u_0\|_{{Z}}.\label{est1}
\end{equation}
Also since $v\in{X}_{\rho}$, we obtain
\begin{gather*}
 \| \mathcal{N}(v(\tau))\|_{ L^{1,a}}\ \leq C\| v\|_{ L^{1,a}
}\| v_x\|_{ L^{\infty}}
 \leq C\{  \tau\}  ^{-(2-a)/3}\langle \tau \rangle ^{2a-4}
\| v\|_{{X}}^2,
\\
 \| \mathcal{N}(v(\tau))\|_{ L^1}\leq C\| v\|_{ L^1}\|
v_x\|_{ L^{\infty}}
\leq C\{  \tau\}  ^{-2/3}\langle \tau\rangle
^{-3}\| v\|_{{X}}^2,
\end{gather*}
for all $\tau>0$, and
\[
 \| \mathcal{N}(v(\tau))\|_{ L^{\infty}}\leq C\| v\|_{ L^{\infty}
}\| v_x\|_{ L^{\infty}}
 \leq C\langle \tau\rangle ^{-6}\ \| v\|_{{X}}^2.
\]
for all $\tau>1$. Now by Lemma \ref{Lemma 2.1} we obtain
\begin{align*}
&\| \int_0^t \mathcal{G}(t-\tau)\mathcal{N}(
v(\tau))d\tau\|_{ L^{1,a}}\\
& \leq\int_0^t \{  t-\tau\}  ^{\frac{a}{3}}\langle
t-\tau\rangle ^{2a}\| \mathcal{N}(v(\tau)
)\|_{ L^1}d\tau+\int_0^t \|
\mathcal{N}(v(\tau))\|_{ L
^{1,a}}d\tau\\
& \leq C\| v\|_{{X}}^2\Big[  \int_0^t \{
t-\tau\}  ^{\frac{a}{3}}\langle t-\tau\rangle ^{2a}\{
\tau\}  ^{-2/3}\langle \tau\rangle ^{-4}\ d\tau
+\int_0^t \{  \tau\}  ^{-\frac{2-a}{3}}\langle
\tau\rangle ^{2a-3}\ d\tau\Big] \\
& \leq C\langle t\rangle ^{2a}\{t\}  ^{\frac{a}{3}
}\| v\|_{{X}}^2
\end{align*}
for all $t\geq0$. In the same manner by  Lemma \ref{Lemma 2.1} we
have
\begin{align*}
 \| \int_0^t \mathcal{G}(t-\tau)\mathcal{N}(
v(\tau))d\tau\|_{ L^1}
& \leq\ \int_0^t \| \mathcal{N}(v(\tau))\|_{ L^1}d\tau\\
& \leq C\| v\|_{{X}}^2\int_0^t \{\tau\} ^{-2/3}\langle
\tau\rangle ^{-3}d\tau\\
& \leq C\| v\|_{{X}}^2
\end{align*}
for all $t\geq0$. Also in view of Lemma \ref{Lemma 2.1}, we find
\begin{align*}
& \| \partial_x^{n}\int_0^t \mathcal{G}(t-\tau)
\mathcal{N}(v(\tau))d\tau\|_{ L^{\infty}}\\
& \leq\ \int_0^t \{  t-\tau\}  ^{-(n+1)/3}\langle
t-\tau\rangle ^{-2(n+1)}\| \mathcal{N}(v(
\tau))\|_{ L^1}d\tau d\tau\\
& \leq C\| v\|_{{X}}^2(\int_0^t \{
t-\tau\}  ^{-(n+1)/3}\langle t-\tau\rangle
^{-2(n+1)}\{  \tau\}  ^{-2/3}\langle \tau\rangle
^{-3}d\tau)\\
&\leq C\{t\}  ^{-(n+1)/3}\langle
t\rangle ^{-2(n+1)}\| v\|_{{X}}^2
\end{align*}
for $t\geq0$.  Thus we obtain
\[
\| \int_0^t e^{-\tau}\mathcal{G}(t-\tau)
\mathcal{N}(v(\tau))d\tau\|_{{X}}\leq C\| v\|_{{X}}^2,
\]
hence in view of \eqref{map} and \eqref{est1},
\begin{align*}
\| \mathcal{M}(v)\|_{{X}}
&\leq\| \mathcal{G}u_0\|_{{X}}+\| \int
_0^t \mathcal{G}(t-\tau)\mathcal{N}(v(
\tau))d\tau\|_{{X}}\\
& \leq C\| u_0\|_{{Z}}+C\| v\|_{{X}}^2\\
&\leq\frac{\rho}{2}+C\rho^2<\rho.
\end{align*}
since $\rho>0$ is sufficiently small. Hence the mapping $\mathcal{M}$
transforms a ball ${X}_{\rho}$ into itself. In the same manner we
estimate the difference
\[
\| \mathcal{M}(w)-\mathcal{M}(v)
\|_{{X}}\leq\frac{1}{2}\| w-v\|_{{X}},
\]
which shows that $\mathcal{M}$ is a contraction mapping. Therefore,
 there exists a unique solution
$v\in{C}([0,\infty); L^1(\mathbb{R}^{+})\cap L^{1,a}(
\mathbb{R}^{+}))\cap {C}((0,\infty);{H}_{\infty}^1)$
to the initial-boundary value problem \eqref{1.1}.
Now we can prove the asymptotic formula
\begin{equation}
v(x,t)=A_1\Lambda(xt^{-2})t^{-2}+O(t^{-2-\gamma}),\label{asv}
\end{equation}
where
\[
A_1=f(u_0)-\int_0^{+\infty}e^{-\tau}f(\mathcal{N}(v))d\tau.
\]
Denote
$G_0(t)=t^{-2}\ \Lambda(xt^{-2})$.
From Lemma \ref{Lemma 2.1}, we have
\begin{equation}
t^{2+\gamma}\| \mathcal{G}(t)\phi-G_0(
t)f(\phi)\|_{ L^{\infty}}\leq
C\| \phi\|_{{Z}}\label{General.5.11}
\end{equation}
for all $t>1$. Also in view the definition of the norm ${X}$
we have
\[
| f(\mathcal{N}(v(\tau)))
| \leq\| \mathcal{N}(v(\tau))
\|_{ L^1}\leq C\{  \tau\}  ^{-2/3}
\langle \tau\rangle ^{-4}\| v\|_{{X}}^2.
\]
By a direct calculation, for some small $\gamma_1>0$, $\gamma>0$,
we have
\begin{equation} \label{5.1}
\begin{split}
& \| \int_0^{t/2}| G_0(t-\tau)
-G_0(t)| f(\mathcal{N}(v(\tau)))d\tau\|_{ L^{\infty}} \\
& \leq\langle t\rangle ^{-1}C\| v\|_{{X}
}^2\int_0^{t/2}\| (G_0(t-\tau)
+G_0(t))\|_{ L^{\infty}}\{\tau\}
^{-\gamma_1}\langle \tau\rangle ^{-\gamma_2}
d\tau \\
& \leq C\langle t\rangle ^{-2}\int_0^{t/2}\{
\tau\}  ^{-\gamma_1}\langle \tau\rangle ^{-\gamma_2}
d\tau\leq C\langle t\rangle ^{-\gamma-2}
\end{split}
\end{equation}
and in the same way,
\begin{equation}
\| \langle t\rangle ^{\gamma}G_0(t)
\int_{t/2}^{\infty}f(\mathcal{N}(v(\tau)
))d\tau\|_{ L^{\infty}}\leq C\|v\|_{{X}}^2\label{5.2}
\end{equation}
Also we have
\begin{equation} \label{5.3}
\begin{split}
& \| \int_0^{t/2}(\mathcal{G}(t-\tau)
\mathcal{N}(v(\tau))-G_0(t-\tau)
f(\mathcal{N}(v(\tau)))) d\tau\|_{ L^{\infty}} \\
& +\| \int_{t/2}^t \mathcal{G}(t-\tau)
\mathcal{N}(v(\tau))d\tau\|_{ L^{\infty}} \\
& \leq C\int_0^{t/2}(t-\tau)^{-2}\|
\mathcal{N}(v(\tau))\|_{ L^1}d\tau
 +Ce^{-\frac{t}{2}}\int_{t/2}^t \| \mathcal{N}(
v(\tau))\|_{ L^1}d\tau \\
& \leq Ct^{-2-\gamma}\| v\|_{{X}}^2
\end{split}
\end{equation}
for all $t>1$. By \eqref{Greenformula}, we obtain
\begin{align}
&\langle t\rangle ^{\gamma+2}\| (v(t)-AG_0(t))\|_{{X}} \nonumber\\
&\leq\|(\mathcal{G}(t)u_0-G_0(t)f(u_0)\|_{ L^{\infty}} \nonumber\\
&\quad +\langle t\rangle ^{\gamma+2}\| \int_0^{\frac{t}{2}
}(\mathcal{G}(t-\tau)\mathcal{N}\mathbb{(}v(
\tau)\mathbb{)}-G_0(t-\tau)f(\mathcal{N}
(v(\tau))))d\tau\|_{ L^{\infty}}  \nonumber\\
&\quad +\langle t\rangle ^{\gamma+2}\| \int_{t/2}^t
\mathcal{G}(t-\tau)\mathcal{N}(v(\tau) )d\tau\|_{ L^{\infty}}
  +\langle t\rangle ^{\gamma+2}\| G_0(t)
\int_{t/2}^{\infty}f(\mathcal{N}(v(\tau)
))d\tau\|_{ L^{\infty}} \nonumber \\
&\quad +\langle t\rangle ^{\gamma+2}\|
 \int_0^{t/2}(G_0(t-\tau)-G_0(t))f(
\mathcal{N}(v(\tau)))d\tau\|_{ L^{\infty}}.  \label{General.5.6}
\end{align}
All the summands in the right-hand side of above inequality are
estimated by $C\| u_0\|_{{Z}}+C\|v\|_{{X}}^2$ via estimates
\eqref{5.1}--\eqref{5.3}.
Thus by \eqref{General.5.6} the
asymptotic formula \eqref{asv} is valid, which completes
the proof.


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\end{document}