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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 89, 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  (login: ftp)}
\thanks{\copyright 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/89\hfil Nonlinear pseudodifferential equations]
{Nonlinear pseudodifferential equations on a half-line with large
initial data}

\author[R. E. Cardiel, E. I. Kaikina\hfil EJDE-2006/89\hfilneg]
{Rosa E. Cardiel, Elena I. Kaikina}  % in alphabetical order

\address{Rosa E. Cardiel \newline
Instituto de Matem\'{a}ticas UNAM (Campus Cuernavaca),
 Av. Universidad s/n,
col. Lomas de Chamilpa, Cuernavaca, Morelos, Mexico}
\email{rosy@matcuer.unam.mx}

\address{Elena I. Kaikina \newline
Instituto de Matem\'{a}ticas\\
UNAM Campus Morelia, AP 61-3 (Xangari)\\
Morelia CP 58089, Michoac\'{a}n, Mexico}
\email{ekaikina@matmor.unam.mx}

\date{}
\thanks{Submitted February 10, 2006. Published August 9, 2006.}
\subjclass[2000]{35Q35, 35B40}
\keywords{Pseudodifferential operator; large data; asymptotic behavior}

\begin{abstract}
 We study the initial-boundary value problem for nonlinear pseudodifferential
 equations, on a half-line,
 \begin{gather*}
 u_{t}+\mathcal{\lambda}| u| ^{\sigma}u+\mathcal{L}
 u=0,\quad(x,t)\in{\mathbb{R}^{+}}\times{\mathbb{R}^{+}},\\
 u(x,0)=u_{0}(x),\quad x\in{\mathbb{R}}^{+},
 \end{gather*}
 where $\lambda>0$ and pseudodifferential operator $\mathcal{L}$ is defined by
 the inverse Laplace transform. The aim of this paper is to prove the global
 existence of solutions and to find the main term of the asymptotic
 representation in the case of the large initial data.
\end{abstract}

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

\section{Introduction}

We consider the initial-boundary value problem for a general class of the
nonlinear nonlocal equations, on a half-line,
\begin{equation}
\begin{gathered}
 u_{t}+\lambda| u|
^{\sigma}u+\mathcal{L}u=0,\quad (x,t)\in{\mathbb{R}^{+}}\times{\mathbb{R}^{+}},\\
u(x,0)=u_{0}(x),\quad x\in{\mathbb{R}}^{+},
\end{gathered}  \label{1.1}
\end{equation}
where $\lambda>0$, $\sigma>0$. The linear operator $\mathcal{L}$ is a
pseudodifferential operator defined by the inverse Laplace transform as
follows
\begin{equation}
\mathcal{L}u=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}e^{px}C_{\alpha
}p^{\alpha}\big(\hat{u}(p,t)-\frac{u(0,t)}{p}\big)\,dp, \label{1.2}
\end{equation}
where $\alpha\in(1,2)$ and
\[
\hat{u}(p)=\int_{0}^{+\infty}e^{-px}u(x)dx
\]
denotes the Laplace transform of $u$. We assume that the symbol
$K(p)=C_{\alpha}p^{\alpha}$ is dissipative, i.e.
$\Re(K(p))>0$ for $\Re(p)=0$. Here and below
$p^{\alpha}$ is the main branch of the complex analytic function in the
half-complex plane $\Re p\geq0$, so that $1^{\alpha}=1$ (we make a cut along
the negative real axis $(-\infty,0))$.

The initial-boundary value problem \eqref{1.1} is of great
interest from the physical point of view , since it describes many
physical phenomena , such as the focusing of laser beams , waves
on water (some other applications can be found \cite{ns4}) . A
great number of publications have dealt with asymptotic
representations of solutions to the Cauchy problem for nonlinear
evolution equations in the last twenty years. While not attempting
to provide a complete review of this publications, we do list some
known results
\cite{AmickBonaSchonbek}-\cite{Dix3},\cite{EscobedoKavianMatano1}
,\cite{EscobedoVazquezZuazua1},\cite{GalaktionovKurdyumovSamarskii1}
-\cite{hkn5}, \cite{Karch1}, \cite{Kavian1},
\cite{ns02}-\cite{Zuazua2}, where there were obtained optimal time
decay estimates and asymptotic formulas of solutions to different
nonlinear local and nonlocal dissipative equations. The asymptotic
theory of the initial-boundary value problems for the nonlinear
pseudodifferential equations is relatively new and traditional
questions of a general theory are far from their conclusion . A
description of the large time asymptotic behavior of solutions for
the initial-boundary value problems requires new approaches and
the reorientation of the points of view compared to the Cauchy
problem.

The initial-boundary value homogeneous problems for nonlinear
pseudodifferential equations were studied in the book \cite{hkbook}.  In the
present paper we continue the study of pseudodifferential equations on a
half-line, considering the case of a large initial data .

The aim of this paper is to prove a global existence of solutions
to the initial-boundary value problem \eqref{1.1}, and to find the
main term of the asymptotic representation of solutions. We will
obtain the a priory optimal time decay estimates of solutions in
the usual Lebesgue spaces $\mathbf{L} ^{r}$ for $1\leq
r\leq\infty$. These type of estimates enable us to consider the
critical and sub critical cases in future works.

To state our results we give some notations. The weighted Sobolev space is
\[
\mathbf{H}_{r}^{k,s}=\{f\in\mathbf{L}^{r}:\Vert
f\Vert_{\mathbf{H}_{r}^{k,s} }=\Vert\langle
i\partial_{x}\rangle ^{k}\langle x\rangle
^{s}f\Vert_{\mathbf{L}^{r}}<\infty\},
\]
where $\langle x\rangle =\sqrt{1+x^{2}}$.

We introduce the function $\Lambda_{0}(s)\in\mathbf{L}^{\infty}$,
\begin{equation}
\Lambda_{0}(s)=(-1)^{\frac{\{  \alpha\}
}{\alpha}
}\frac{C_{\alpha}}{2\pi^{2}i}\frac{A_{1}}{\pi-A_{1}}\sin(\frac{\pi\{
\alpha\}
}{\alpha})\int_{-i\infty}^{i\infty}dze^{sz}z^{\{
\alpha\}  }\int_{0}^{\infty}dq e^{-q}\frac{q^{-\frac{\{ \alpha\}
}{\alpha}}}{K(z)+q},
\end{equation}
where
\[
A_{1}=(-1)^{\{  \alpha\}  +1}(-C_{\alpha})
^{-\frac{[\alpha]}{\alpha}}\Gamma(1-\{  \alpha\}
)\Gamma(\{  \alpha\}  )\sin\pi\{  \alpha\}  .
\]
We prove the following theorem.

\begin{theorem}
\label{T1} Let $\alpha\in(1,2)$, $\sigma>\alpha$ ,$\lambda>0$ and
real valued function
$u_{0}\in\mathbf{H}_{2}^{1,0}\cap\mathbf{H}_{\infty}^{0,\mu}
\cap\mathbf{H}_{1}^{0,0}$, where
$\mu\in[0,1]$. Then there exists a unique real
valued solution of \eqref{1.1} such that
\[
u(x,t)\in C([0,\infty);\mathbf{H}_{2}^{1,0}
\cap\mathbf{H}_{\infty}^{0,\mu}\cap\mathbf{H}_{1}^{0,0}).
\]
Moreover there exists a constant $A$ such that
\begin{equation}
u(x,t)=t^{-1/\alpha} A\Lambda_{0}(xt^{-1/\alpha} )+O(t^{-\frac
{1+\mu}{\alpha}}) \label{5.5}
\end{equation}
as $t\to\infty$ uniformly with respect to $x>0$, where
\[
A=\int_{0}^{+\infty}u_{0}\,dy+\lambda\int_{0}^{+\infty}dt\int_{0}^{\infty
}| u| ^{\sigma}u\,dy.
\]
\end{theorem}

\begin{remark} \label{rmk1} \rm
We can guarantee that the coefficient $A\neq0$ in the asymptotic
representation (\ref{5.5}) if $\int_{0}^{+\infty}u_{0}\,dy\neq0$ and
$| u| ^{\sigma}u$ is small. It can occur that
$A=0$, for instance, for convective equations
$\int_{0}^{\infty}| u| ^{\sigma }u\,dy\equiv0$ if
the initial data have zero mean value $\int_{0}^{+\infty}
u_{0}\,dy=0$. In the last case formula (\ref{5.5}) gives us only
some time decay estimate for the solutions.
\end{remark}

\section{Preliminaries} \label{S2}

We introduce the function
$\kappa(\xi)=K^{-1}(-\xi)$,
such that $\operatorname{Re}\kappa(\xi)>0$ for all $\operatorname{Re}\xi
\geq0$. We denote
\[
\mathcal{G}(t_{1})\phi(t_{2})=\int_{0}^{+\infty}G(x,y,t_{1})\phi(y,t_{2})\,dy,
\]
where Green function $G(x,y,t)$ is defined by
\begin{equation}
\begin{aligned}
G(x,y,t)&=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}e^{-K(p)t+p(x-y)}
\,dp\\
&\quad  +\frac{1}{4\pi^{2}}\int_{-i\infty}^{i\infty}e^{\xi t}\kappa(\xi)\xi
^{-1}e^{-\kappa(\xi)y}\int_{-i\infty}^{i\infty}\frac{e^{px}K(p)}{p(K(p)+\xi
)}d\xi \,dp.
\end{aligned}\label{13.1}
\end{equation}
The solution $u$ of the problem \eqref{1.1} can be represented as follows (see
\cite{hkbook}, pp. 23-24)
\begin{equation}
u(x,t)=\mathcal{G}(t)u_{0}+\int_{0}^{t}d\tau\mathcal{G}(t-\tau)\mathcal{N}
(u)(\tau), \label{integral}
\end{equation}
where $\mathcal{N}(u)=\lambda| u| ^{\sigma}u$. We introduce
complete metric space
\begin{gather*}
\mathbf{X}=\{  \phi(x,t)\in\mathbf{C}([0,\infty)
;\mathbf{H}_{r}^{0,\mu})\cap\mathbf{C}((0,\infty)
;\mathbf{H}_{r}^{1,0})| \Vert\phi\Vert_{\mathbf{X}}
<+\infty,  \} ,\\
\Vert\phi\Vert_{\mathbf{X}}=\sup_{t>0}
\langle t\rangle ^{\frac{1}{\alpha}}(\langle
t\rangle ^{-\frac{1}{\alpha r}-\frac{\mu}{\alpha}}\Vert x^{\mu}\phi
\Vert_{\mathbf{L}^{r}}+t^{\frac{1}{\alpha}-\frac{1}{\alpha r}}\Vert\phi
_{x}\Vert_{\mathbf{L}^{r}}),
\end{gather*}
where $\mu\in[0,1]$, $1\leq r\leq\infty$. Let a continuous
linear functional $f(\phi):\mathbf{L}_{1}\to\mathbb{R}$ is defined as
\[
f(\phi)=\int_{0}^{+\infty}\phi(x)dx.
\]

Now we obtain some estimates for the Green operator $\mathcal{G}$
in the space $\mathbf{X}$.

\begin{lemma} \label{L4.4}
The following two estimates are valid
\begin{gather}
\Vert \mathcal{G}\phi\Vert _{\mathbf{X}}\leq C(\Vert
\phi\Vert _{\mathbf{L}^{1}}+\Vert \phi\Vert _{\mathbf{L}
^{\infty}}),\label{halp1}
\\
\sup_{t>1}t^{\frac{1}{\alpha}(1+\mu)}\Vert \mathcal{G}\phi-t^{-\frac
{1}{\alpha}}\Lambda_{0}(xt^{-1/\alpha} )f(\phi
)\Vert _{\mathbf{L}^{\infty}}\leq C\Vert \phi\Vert
_{\mathbf{L}^{1,\mu}} \label{halp2}
\end{gather}
for all $t>0$, provided that the right-hand sides are bounded.
\end{lemma}

\begin{proof}
We rewrite the Green function as
\[
G(x,y,t)=F_{1}(x-y,t)+F_{2}(x,y,t),
\]
where
\begin{gather}
F_{1}(x,t)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty
}e^{-K(p)t+px}\,dp \label{d1}
\\
F_{2}(x,y,t)=\frac{1}{4\pi^{2}}\int_{-i\infty}^{i\infty}e^{\xi
t-k(\xi)y}k(\xi)\xi^{-1}\int_{-i\infty}^{i\infty}\frac{e^{px}K(p)}
{p(K(p)+\xi)}d\xi \,dp. \label{d2}
\end{gather}
Firstly we prove the  estimate
\begin{equation}
t^{\frac{1}{\alpha}-\frac{1}{\alpha r}}\Big(\Vert F_{1}
(q,t)\Vert _{L^{r}}+t^{-\frac{\mu}{\alpha}}\Vert
q^{\mu} F_{1}(q,t)\Vert
_{L^{r}}+t^{\frac{1}{\alpha}}\Vert \partial
_{q}F_{1}(q,t)\Vert _{L^{r}}\Big)\leq C \label{aux1}
\end{equation}
for all $t>0$, where $\mu>0$ and $1\leq r\leq\infty$.
Changing variables $p^{\alpha}t=z^{\alpha}$, and
$\widetilde{q}=qt^{-1/\alpha}$ we get
\[
| F_{1}(q,t)| =\big| \frac{1}{2\pi
i}\int_{-i\infty }^{i\infty}e^{qp}e^{-K(p)t}\,dp\big|
 \leq Ct^{-1/\alpha} \big|
\int_{-i\infty}^{i\infty}e^{\widetilde{q}z}e^{-C_{\alpha}z^{\alpha}
}dz\big| .
\]
Therefore,
\[
\Vert F_{1}(q,t)\Vert _{L^{r}}\leq Ct^{-\frac{1}{\alpha}+\frac
{1}{\alpha r}}.
\]
After the  change $p^{\alpha}t=z^{\alpha}$, we have
\begin{gather*}
| q^{\mu}F_{1}(q,t)| \leq Ct^{^{-\frac{1}{\alpha}+\frac
{\mu}{\alpha}}}\big| \widetilde{q}^{\mu}\int_{-i\infty}^{i\infty
}e^{\widetilde{q}z}e^{-C_{\alpha}z^{\alpha}}dz\big|
\leq Ct^{^{-\frac {1}{\alpha}+\frac{\mu}{\alpha}}}\langle \widetilde{q}\rangle
^{\mu-1-\gamma},
\\
| \partial_{q}F_{1}(q,t)| \leq
Ct^{-\frac{2}{\alpha} }\big|
\int_{-i\infty}^{i\infty}ze^{\widetilde{q}z}e^{-C_{\alpha
}z^{\alpha}}dz\big|
\leq Ct^{-\frac{2}{\alpha}}\langle
\widetilde{q}\rangle ^{-1-\gamma},
\end{gather*}
where $0<\gamma<\alpha$. Therefore,
\begin{equation}  \label{4.15}
\begin{aligned}
\Vert q^{\mu}F_{1}(q,t)\Vert _{L^{r}}  &  \leq Ct^{^{-\frac
{1}{\alpha}+\frac{\mu}{\alpha}+\frac{1}{\alpha r}}}((\int
_{0}^{1}+\int_{1}^{\infty})d\widetilde{q}(1+\widetilde{q}
^{2})^{\frac{(\mu-1-\gamma)r}{2}})^{1/r}\\
&  \leq Ct^{^{-\frac{1}{\alpha}+\frac{\mu}{\alpha}+\frac{1}{\alpha
r}} }
\end{aligned}
\end{equation}
and
\begin{equation}
\Vert \partial_{q}F_{1}(q,t)\Vert _{L^{r}}\leq Ct^{-\frac{2}
{\alpha}+\frac{1}{\alpha r}}. \label{4.16}
\end{equation}
Estimate (\ref{aux1}) is then proved. Now we prove
\begin{equation}  \label{aux2}
\begin{aligned}
&\sup_{t>0, y>0} t^{\frac{1}{\alpha}-\frac{1}{r\alpha}+\mu
_{1}+\gamma}\langle t\rangle ^{-2\gamma}y^{-\mu_{1}\alpha}(
\Vert F_{2}(x,y,t)\Vert_{\mathbf{L}^{r}}\\
&+t^{-\frac{\mu}{\alpha}}\Vert x^{\mu}F_{2}(x,y,t)\Vert _{\mathbf{L}^{r}}
  +t^{\frac{1}{\alpha}}\Vert \partial_{x}F_{2}(x,y,t)\Vert
_{L^{r}})\leq C,
\end{aligned}
\end{equation}
where $\mu\geq0,\gamma>0$, $0\leq\mu_{1}<\frac{1}{\alpha}(
1-\frac{1} {r}-\gamma\alpha)$ and $1\leq r\leq\infty$.

We have, by definition,
$\kappa(q)=C_{1}| q| ^{\frac{1}{\alpha}}$.
Changing variables $p^{\alpha}t=z^{\alpha}$ and $\xi t=q$, we obtain the
estimate
\begin{equation}
| F_{2}(x,y,t)| \leq Ct^{-1/\alpha} \big|
\int_{-i\infty}^{i\infty}dze^{\widetilde{x}z}K(z)z^{-1}\int_{-i\infty
}^{i\infty}dq\quad q^{-1}\kappa(q)\frac{e^{q-C_{1}q^{\frac{1}{\alpha}
}\widetilde{y}}}{K(z)+q}\big| . \label{4.8.1}
\end{equation}
We move the contours of integration with respect to $z$ as follows
\begin{equation}
\mathcal{C}_{1}=\{  z=\rho e^{\pm i\beta_{1}},\rho\geq0,\beta_{1}
=\frac{\pi}{2}+\epsilon_{1}\}  \label{C1}
\end{equation}
and with respect to $q$ by
\begin{equation}
\mathcal{C}_{2}=\{  q=e^{i\phi_{2}}\quad \phi_{2}\in[-\frac
{\pi}{2},\frac{\pi}{2}]\}  \cup\{  q=\rho e^{\pm
i\phi_{2}}\rho\geq1,\quad \phi_{2}=\frac{\pi}{2}+\epsilon_{2}\}  ,
\label{4.9}
\end{equation}
where $\epsilon_{1},\epsilon_{2}>0$ are small enough. It is apparent that
\begin{equation}  \label{0}
\begin{gathered}
| e^{q}| \leq C| q|^{-\gamma} ,\quad
|K(z)+q|^{-1}\leq C|z|^{-\nu\alpha}|q|^{\nu-1},\\
| e^{z\tilde{x} }| \leq C| z\tilde{x}| ^{-\mu_{2}}, \quad
| e^{-C_{1j}q^{\frac{1}{\alpha}}\widetilde{y}}| \leq
C| q^{\frac{1}{\alpha}}\widetilde{y}| ^{-\alpha\mu_{1}}
\end{gathered}
\end{equation}
for all $q\in\mathcal{C}_{2}$ and $z\in\mathcal{C}_{1}$, where
$\nu\in[ 0,1]$, $\gamma\geq0$, and $\mu_{1}\geq0,\mu_{2}\geq0$.
Taking into account (\ref{0}) we get
\begin{equation} \label{3}
| F_{2}(x,y,t)|  \leq Ct^{-1/\alpha} y^{-\mu
_{1}\alpha}\tilde{x}^{-\mu_{2}}\int_{-i\infty}^{i\infty}|
dz| | z| ^{\alpha-1-\nu\alpha-\mu_{2}}
\int_{-i\infty}^{i\infty}| dq| \, |
q| ^{\frac{1}{\alpha}-2+\nu-\mu_{1}-\gamma}.
\end{equation}
To guarantee the convergence of this integrals we need to satisfy the
following conditions
\begin{equation}
\frac{1}{\alpha}-2+\nu-\mu_{1}>-1 \label{e1}
\end{equation}
and
\begin{equation}
\begin{gathered}
\alpha-1-\nu\alpha-\mu_{2}>-1,\\
\alpha-[\alpha]-\nu\alpha-\mu_{2}<-1.
\end{gathered} \label{e2}
\end{equation}
respectively. Under the condition
$\alpha\mu_{1}+\mu_{2}<2-[\alpha]$ there exists some $\nu\geqslant0$
such that the estimates (\ref{e2}) are valid. Therefore, we obtain
\begin{equation}
| F_{2}(p,y,t)| \leq Ct^{-\frac{1-\mu_{2}
}{\alpha}+\mu_{1}}x^{-\mu_{2}}y^{-\alpha\mu_{1}}. \label{4.11}
\end{equation}
 From (\ref{4.11}) it follows that
\begin{equation} \label{4.12}
\begin{aligned}
\Vert F_{2}(x,y,t)\Vert _{L^{r}}
&  \leq Ct^{-\frac{1}{\alpha
}+\frac{1}{r\alpha}+\mu_{1}\pm\gamma}y^{-\alpha\mu_{1}}
\big(\int_{0}^{\infty}x^{-\mu_{2}r}dx\big)^{1/r}\\
& \leq Ct^{-\frac{1}{\alpha}+\frac{1}{r\alpha}+\mu_{1}\pm\gamma}y^{-\alpha
\mu_{1}},
\end{aligned}
\end{equation}
where $0\leq\mu_{1}<\frac{1}{\alpha}(1-\frac{1}{r})$. In the
same manner we obtain
\begin{gather*}
| x^{\mu}F_{2}(x,y,t)| \leq Ct^{-\frac
{1}{\alpha}+\frac{\mu-\mu_{2}}{\alpha}+\mu_{1}}x^{\mu-\mu_{2}}y^{-\alpha
\mu_{1}}, \\
| \partial_{x}F_{2}(p,y,t)| \leq
Ct^{-\frac{2-\mu_{2}}{\alpha}+\mu_{1}}y^{-\mu_{1}\alpha}x^{-\mu_{2}}.
\end{gather*}
Thus we obtain
\begin{gather}
\Vert x^{\mu}F_{2}(x,y,t)\Vert _{L^{r}}\leq Ct^{-\frac{1+\mu
}{\alpha}+\frac{1}{r\alpha}+\mu_{1}\pm\gamma}y^{-\mu_{1}\alpha}, \label{4.13}
\\
\Vert \partial_{x}F_{2}(x,y,t)\Vert _{L^{r}}\leq Ct^{-\frac
{2}{\alpha}+\frac{1}{r\alpha}+\mu_{1}}y^{-\mu_{1}\alpha}, \label{4.14}
\end{gather}
and therefore estimate (\ref{aux2}) is proved. From estimates (\ref{aux1}) and
(\ref{aux2}) we easily get result (\ref{halp1}) of Lemma \ref{L4.4}. To obtain
(\ref{halp2}) firstly we prove the following estimate
\begin{equation}
G(x,y,t)=t^{-1/\alpha} \sin\frac{\pi\{  \alpha\}  }{\alpha
}\Lambda_{0}(xt^{-1/\alpha} )+y^{\mu}O(t^{-\frac{1+\mu
}{\alpha}}) \label{4.22}
\end{equation}
for \ $t\to\infty$, where $x,y>0$.

We write the representation  (\ref{d1}) for the function $F_{1}(x,t)$ as
\[
F_{1}(x-y,t)=F_{1}(x,t)+[F_{1}(x-y,t)-F_{1}(x,t)].
\]
After the change of variables $p^{\alpha}t=z^{\alpha}$, we easily find that
\begin{equation}
F_{1}(x,t)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}e^{xp-K(p)t}
\,dp=t^{-1/\alpha} \frac{1}{2\pi
i}\int_{-\infty}^{i\infty}e^{zs-K(z)}dz.
\label{mainf1}
\end{equation}
Using the estimate $|e^{-py}-1|\leq C|py|^{\mu}$ for $y>0$ and $p\in
(-i\infty,i\infty)$ with $\mu_{1}\in[0,1]$ we have
\begin{equation}
\begin{aligned}
|F_{1}(x-y,t)-F_{1}(x,t)|
&  \leq\big| \int_{-i\infty}^{i\infty
}e^{xp-C_{\alpha}p^{\alpha}t}(e^{-py}-1)\,dp\big| \\
&  \leq Cy^{\mu}t^{-\frac{1+\mu}{\alpha}}\frac{1}{2\pi i}\int_{-i\infty
}^{i\infty}| dz| \, |z|^{\mu}\big| e^{xt^{-\frac
{1}{\alpha}}z-C_{\alpha}z^{\alpha}}\big| \\
&  \leq Cy^{\mu}t^{-\frac{1+\mu}{\alpha}}=y^{\mu}O\big(t^{-\frac{1+\mu
}{\alpha}}\big).
\end{aligned}\label{ef1.1}
\end{equation}
Therefore, from (\ref{mainf1}), we obtain
\begin{equation}
F_{1}(x-y,t)=t^{-1/\alpha} \frac{1}{2\pi i}\int_{-\infty}^{i\infty
}e^{zs-K(z)}dz+y^{\mu}O(t^{-\frac{1+\mu}{\alpha}}). \label{asf1}
\end{equation}
Now we write the representation (\ref{d2}) of the function $F_{2}(x,y,t)$ as
\begin{equation}
F_{2}(x,y,t)=M(x,y,t)+R(x,y,t), \label{F2}
\end{equation}
where
$M(x,t)=F_{2}(x,0,t)$ % \label{M}
and $R=[F_{2}(x,y,t)-F_{2}(x,0,t)]$.
Considering that $|e^{-C\xi^{1/\alpha}y}-1|\leq C|\xi^{1/\alpha}y|^{\mu}$ and
changing variables $\xi t=q$  and $p^{\alpha}t=z^{\alpha}$, we get
\begin{align*}
&| F_{2}(x,y,t)-F_{2}(x,0,t)| \\
& \leq Ct^{-\frac{1}{\alpha
}-\frac{\mu_{1}}{\alpha}}y^{\mu_{1}}\int_{-i\infty}^{i\infty}|
dz|\, | e^{xt^{\frac{1}{\alpha}}z}|\, |z| ^{\alpha-1}
 \int_{-i\infty}^{i\infty}| dq| | q|
^{\frac{1}{\alpha}-1+\frac{\mu}{\alpha}}\frac{| e^{q}|
}{| K(z)+q| }\\
& =y^{\mu}O(t^{-\frac{1+\mu}{\alpha}}).
\end{align*}
Therefore,
\begin{equation}
F_{2}(x,y,t-\tau)=M(x,t)+y^{\mu}O(t^{-\frac{1+\mu}{\alpha}}), \label{asf2}
\end{equation}
where
\begin{equation}
M(x,t)=t^{-1/\alpha} \frac{1}{4\pi^{2}}\int_{-i\infty}^{i\infty
}dze^{sz}K(z)z^{-1}\int_{-i\infty}^{i\infty}dq e^{q}\frac{q^{-1}
\kappa(q)}{K(z)+q}, \label{formula}
\end{equation}
with $s=xt^{-1/\alpha} $.
Applying the Cauchy Theorem, we obtain
\begin{equation} \label{formula1}
\begin{aligned}
&  \int_{-i\infty}^{i\infty}dze^{sz}K(z)z^{-1}\int_{-i\infty}^{i\infty
}dq e^{q}\frac{q^{-1}\kappa(q)}{K(z)+q}\\
&  =2\pi i\int_{-i\infty}^{i\infty}dze^{sz}e^{-K(z)}
 +\int_{-i\infty}^{i\infty}dze^{sz}K(z)z^{-1}\int_{\Gamma}dq\,
e^{q}\frac{q^{-1}\kappa(q)}{K(z)+q}\\
&  =I_{1}+I_{2},
\end{aligned}
\end{equation}
where
$\Gamma=\{  z\in(-\infty e^{-i\pi},0e^{-i\pi})
\cup(0e^{i\pi},-\infty e^{i\pi})\}$.
Using
\[
\int_{-i\infty}^{i\infty}dze^{sz}\int_{\Gamma}dq e^{q}\frac{1} {K(z)+q}=0,
\]
we get
\begin{equation} \label{MFIN}
\begin{aligned}
M(x,t)  &=t^{-1/\alpha} \frac{1}{2\pi i}\int_{-i\infty}^{i\infty
}dze^{sz}e^{-C_{\alpha}z^{\alpha}}
+t^{-1/\alpha} \frac{C_{\alpha}(-1)
^{\frac{\{ \alpha\}
}{\alpha}}}{2\pi^{2}i}\frac{A_{1}}{\pi-A_{1}}\\
&\quad\times  \sin\big( \frac{\pi\{  \alpha\}  }{\alpha}\big)
\int_{-i\infty}^{i\infty }dze^{sz}z^{\{  \alpha\}
}\int_{0}^{+\infty}dq e^{-q} \frac{q^{-\frac{\{
\alpha\}  }{\alpha}}}{K(z)+q},
\end{aligned}
\end{equation}
where
\[
A_{1}=(-1)^{\{  \alpha\}  +1}(-C_{\alpha})
^{-\frac{[\alpha]}{\alpha}}\Gamma(1-\{  \alpha\}
)\Gamma(\{  \alpha\}  )\sin\pi\{  \alpha\}  .
\]

 From (\ref{asf1}), (\ref{asf2}), (\ref{MFIN}), we obtain (\ref{4.22}) and
then estimate (\ref{halp2}). This completes the proof of Lemma \ref{L4.4}.
\end{proof}

We defined the space
\[
\mathbf{W}=\{  \phi(x,t)\in C((0,\infty)
;\mathbf{L}^{1}\cap\mathbf{L}^{\infty,\mu}):\Vert\phi\Vert
_{\mathbf{W}}<\infty\}
\]
where
$\Vert\phi\Vert_{\mathbf{W}}=(\Vert\phi\Vert_{\mathbf{L}^{1}}+\Vert\phi
\Vert_{\mathbf{L}^{\infty,\mu}})$.

\begin{lemma} \label{L4.7}
The following two estimates are valid
\begin{gather*}
\Vert x^{\mu}\int_{0}^{t}\mathcal{G}(t-\tau)
\phi(\tau)d\tau\Vert _{\mathbf{L}^{r}}\leq
C\langle t\rangle ^{-\frac{1}{\alpha}+\frac{1}{\alpha
r}+\frac{\mu}{\alpha}} \Vert\langle t\rangle
^{1+\gamma}\phi\Vert_{\mathbf{W}},
\\
\Vert \partial_{x}\int_{0}^{t}\mathcal{G}(t-\tau)
\phi(\tau)d\tau\Vert _{\mathbf{L}^{r}}\leq Ct^{1-\frac
{1}{\alpha}+\frac{1}{\alpha r}}\langle t\rangle ^{-\frac{1}{\alpha
}}\Vert\langle t\rangle ^{1+\gamma}\phi\Vert_{\mathbf{W}},
\end{gather*}
for all $t>0$, $\mu\in(0,1)$, $r\geq1$ and small enough $\gamma>0$,
provided that the right-hand sides are bounded.
\end{lemma}

\begin{proof}
Let $t\in[0,1]$. By estimate (\ref{4.14}) of Lemma \ref{L4.4},
we get
\begin{equation}
\Vert \Vert\partial_{x}F_{2}(x,y,t)
\Vert_{\mathbf{L} _{x}^{r}}\Vert _{\mathbf{L}_{y}^{1}}\leq
Ct^{-\frac{2}{\alpha}+\frac {1}{\alpha
r}}\int_{0}^{\infty}\langle \widetilde{y}\rangle
^{-\mu_{1}\alpha}\,dy\leq Ct^{-\frac{1}{\alpha}+\frac{1}{\alpha r}}, \label{6.8}
\end{equation}
for all $t>0$. Therefore, using (\ref{4.15}) of Lemma \ref{L4.4} we
obtain
\begin{equation} \label{6.9}
\begin{aligned}
&\Vert x^{\mu}\int_{0}^{t}\mathcal{G}(t-\tau)
\phi(\tau)d\tau\Vert _{\mathbf{L}^{r}} \\
&\leq\int_{0}^{t} d\tau(\Vert\phi\Vert_{\mathbf{L}^{r}}\Vert
x^{\mu}F_{1}\Vert
_{\mathbf{L}^{1}}+\Vert x^{\mu}\phi\Vert_{\mathbf{L}^{r}}\Vert F_{1}
\Vert_{\mathbf{L}^{1}})
  +\int_{0}^{t}d\tau\Vert\phi\Vert_{\mathbf{L}^{\infty}}\Vert x^{\mu}
F_{2}\Vert_{\mathbf{L}_{x}^{r}\mathbf{L}_{y}^{1}}\\
& < C(t^{-\frac{1}{\alpha}+\frac{1}{r\alpha}+1}+t^{-\frac{1}{\alpha
}+\frac{1}{r\alpha}+\frac{\mu}{\alpha}+1})\Vert\phi\Vert_{\mathbf{W}
}<C\Vert\phi\Vert_{\mathbf{W}}
\end{aligned}
\end{equation}
for all $t\in[0,1]$. Using (\ref{6.8}) and (\ref{4.16}), we
have
\begin{equation} \label{6.10}
\begin{aligned}
&\Vert \partial_{x}\int_{0}^{t}\mathcal{G}(t-\tau)
\phi(\tau)d\tau\Vert _{\mathbf{L}^{r}}  \\
& \leq\int
_{0}^{t}d\tau(\Vert \phi\Vert _{\mathbf{L}^{r}}\Vert
\partial_{x}F_{1}\Vert _{\mathbf{L}^{1}}+\Vert\phi\Vert_{\mathbf{L}
^{\infty}}\Vert\partial_{x}F_{2}\Vert_{\mathbf{L}_{x}^{r}\mathbf{L}_{y}^{1}
})\\
& <Ct^{-\frac{1}{\alpha}+\frac{1}{\alpha r}+1}\Vert\phi\Vert_{\mathbf{W}}\leq
t^{-\frac{1}{\alpha}+\frac{1}{\alpha r}}\Vert\phi\Vert_{\mathbf{W}}
\end{aligned}
\end{equation}
for all $t\in[0,1]$.

We consider now the case $t>1$,
\begin{align*}
\Vert x^{\mu}\int_{0}^{t}\mathcal{G}(t-\tau)
\phi(\tau)d\tau\Vert _{\mathbf{L}^{r}}
&\leq\int_{0}^{t/2} d\tau(
\Vert\phi\Vert_{\mathbf{L}^{1}}\Vert x^{\mu}F_{1}\Vert
_{\mathbf{L}^{r}}+\Vert x^{\mu}\phi\Vert_{\mathbf{L}^{1}}\Vert F_{1}
\Vert_{\mathbf{L}^{r}})\\
&\quad  +\int_{t/2}^{t}d\tau(\Vert\phi\Vert_{\mathbf{L}^{r}}\Vert x^{\mu
}F_{1}\Vert_{\mathbf{L}^{1}}+\Vert x^{\mu}\phi\Vert_{\mathbf{L}^{r}}\Vert
F_{1}\Vert_{\mathbf{L}^{1}})\\
&\quad  +\int_{0}^{t/2}d\tau\Vert\phi\Vert_{\mathbf{L}^{1}}\Vert
x^{\mu}F_{2}
\Vert_{\mathbf{L}_{x}^{r}\mathbf{L}_{y}^{\infty}}\\
&\quad +\int_{t/2}^{t}d\tau\Vert
\phi\Vert_{\mathbf{L}^{\infty}}\Vert
x^{\mu}F_{2}\Vert_{\mathbf{L}_{x}
^{r}\mathbf{L}_{y}^{1}}.
\end{align*}
So in view of estimates (\ref{4.13}), (\ref{4.14}) , (\ref{4.15}) and
(\ref{4.16}), we attain
\begin{equation}
\Vert x^{\mu}\int_{0}^{t}\mathcal{G}(t-\tau)\phi(
\tau)d\tau\Vert _{\mathbf{L}^{r}}\leq Ct^{-\frac{1}{\alpha
}+\frac{1}{r\alpha}+\frac{\mu}{\alpha}}\Vert\langle t\rangle
^{1+\gamma}\phi\Vert_{\mathbf{W}}. \label{6.11}
\end{equation}
In the same way we estimate $\Vert\partial_{x}\mathcal{G}\phi\Vert
_{\mathbf{L}^{r}}$ for $t>1$,
\begin{equation} \label{6.12}
\begin{aligned}
\Vert \partial_{x}\int_{0}^{t}\mathcal{G}(
t-\tau)\phi(\tau)d\tau\Vert_{\mathbf{L}^{r}}
&\leq \int _{0}^{t/2}d\tau(
\Vert\phi\Vert_{\mathbf{L}^{1}}\Vert\partial_{x}
F_{1}\Vert_{\mathbf{L}^{r}}+\Vert\phi\Vert_{\mathbf{L}^{1}}\Vert\partial
_{x}F_{2}\Vert_{\mathbf{L}^{r}\mathbf{L}^{\infty}})\\
&\quad  +\int_{t/2}^{t}d\tau(\Vert\phi\Vert_{\mathbf{L}^{r}}\Vert
\partial_{x}F_{1}\Vert_{\mathbf{L}^{1}}+\Vert\phi\Vert_{\mathbf{L}^{\infty}
}\Vert\partial_{x}F_{2}\Vert_{\mathbf{L}_{x}^{r}\mathbf{L}_{y}^{1}})
\\
&\leq Ct^{-\frac{2}{\alpha}+\frac{1}{r\alpha}}\Vert\langle
t\rangle ^{1+\gamma}\phi\Vert_{\mathbf{W}}.
\end{aligned}
\end{equation}
Then, by (\ref{6.9})-(\ref{6.12}) Lemma \ref{L4.7} is proved.
\end{proof}

Denote by
\begin{gather*}
\mathbf{Z}=\{  \phi(x)\in\mathbf{H}_{2}^{1,0}\cap\mathbf{H}_{\infty
}^{0,\mu}\cap\mathbf{H}_{1}^{0,0};\Vert\phi\Vert_{\mathbf{z}}<+\infty\},
\\
\Vert\phi\Vert_{\mathbf{Z}}=(\Vert\langle
x\rangle ^{\mu}\phi\Vert_{\mathbf{L}^{\infty}}+\Vert\phi\Vert
_{\mathbf{L}^{1}}+\Vert \phi\Vert _{\mathbf{H}_{2}^{1}}),
\end{gather*}
where $\mu\in(0,1]$.
Now we prove local existence theorem. Note that the existence time $T>0$ could
be sufficiently small.

\begin{theorem} \label{General.Theorem2.1}
Let initial data $u_{0}\in\mathbf{Z}$.
Then for some time interval $T>0$ there exists a unique solution
$u\in\mathbf{C} ([0,T];\mathbf{X})$ to problem
\eqref{1.1}. Moreover the existence time $T$ can be chosen as
follows
\[
T=\Big(\Vert u_{0}\Vert _{\mathbf{Z}}\big(
1+\frac{1} {2C}\Vert u_{0}\Vert _{\mathbf{Z}}\big)
^{-\sigma-1}\Big)^{\frac{1}{\mu_{1}}},\quad \mu_{1}=1-\frac{1}{\alpha}.
\]
\end{theorem}

\begin{proof}
We apply the contraction mapping principle in a ball of a radius $\rho>0$ in a
complete metric space $\mathbf{X}_{T}$,
\[
\mathbf{X}_{T_{,}\rho}=\{  \phi\in\mathbf{C}([0,T]
;\mathbf{X}):\text{ sup}_{t\in[0,T]}\Vert u\Vert
_{\mathbf{X}}=\Vert u\Vert _{\mathbf{X}_{T}}\leq\rho\}  ,
\]
where
$\rho=\frac{1}{2C}\Vert u_{0}\Vert _{\mathbf{Z}}$.
For $v\in\mathbf{X}_{T_{,}\rho}$ we define the mapping $\mathcal{M}(
v)$ by
\begin{equation}
\mathcal{M}(v)=\mathcal{G}(t)
u_{0}-\int_{0} ^{t}\mathcal{G}(t-\tau)
\mathcal{N}(v(\tau)
)d\tau, \label{General.2.3}
\end{equation}
where $\mathcal{N}(v(\tau))=\lambda|v| ^{\sigma}v$.
We first prove that
\[
\Vert \mathcal{M}(v)\Vert _{\mathbf{X}_{T}}\leq
\rho,
\]
when $v\in\mathbf{X}_{T_{,}\rho}$. We have by Lemmas \ref{L4.4} and \ref{L4.7}
for $\mu_{1}=1-\frac{1}{\alpha}$,
\begin{equation} \label{General.2.4}
\begin{aligned}
\Vert \mathcal{M}(v)\Vert _{\mathbf{X}_{T}}  &
\leq\Vert \mathcal{G}u_{0}\Vert _{\mathbf{X}_{T}}+\Vert
\int_{0}^{t}\mathcal{G}(t-\tau)\mathcal{N}(v(
\tau))d\tau\Vert _{\mathbf{X}_{T}}\\
&  \leq C\Vert u_{0}\Vert _{\mathbf{Z}}+CT^{\mu_{1}}(
1+\Vert v\Vert _{\mathbf{X}_{T}})^{\sigma+1}\\
&  \leq\frac{\rho}{2}+CT^{\mu_{1}}(1+\rho)^{\sigma+1}\leq\rho,
\end{aligned}
\end{equation}
if $T>0$ is small enough. Therefore, the mapping $\mathcal{M}$ transforms a
ball of a radius $\rho>0$ into itself in the space $\mathbf{X}_{T}$. As in the
proof of  (\ref{General.2.4}), we have for $w,v\in\mathbf{X}_{T}$
\begin{align*}
\Vert \mathcal{M}(w)-\mathcal{M}(v)
\Vert _{\mathbf{X}_{T}}
&\leq\Vert \int_{0}^{t}\mathcal{G}(
t-\tau)(\mathcal{N}(w(\tau))
-\mathcal{N}(v(\tau)))d\tau\Vert
_{\mathbf{X}_{T}}\\
&  \leq CT^{\mu_{1}}\Vert w-v\Vert _{\mathbf{X}_{T}}(
1+\Vert w\Vert _{\mathbf{X}_{T}}+\Vert v\Vert
_{\mathbf{X}_{T}})^{\sigma}\leq\frac{1}{2}\Vert w-v\Vert
_{\mathbf{X}_{T}},
\end{align*}
since $T>0$ is small enough. Thus $\mathcal{M}$ is a contraction mapping in
$\mathbf{X}_{T_{,}\rho}$; therefore, there exists a unique solution
$u\in\mathbf{X}_{T}$ to the problem \eqref{1.1}. Theorem
\ref{General.Theorem2.1} is proved.
\end{proof}

Now we define the space $\mathbf{X}[T_{1},T_{2}]$
$=\mathbf{C}([T_{1},T_{2}];\mathbf{Z})$ with
 the norm
\[
\Vert \psi\Vert _{\mathbf{X}[T_{1},T_{2}]
} =\sup_{t\in[T_{1},T_{2}]}\Vert \psi(t)\Vert _{\mathbf{Z}}.
\]


\begin{theorem} \label{General.Theorem4.1}
Let the initial data $u_{0}\in\mathbf{Z}$ and the
following a priory estimate be valid
\begin{equation}
\| u\| _{\mathbf{X}[0,T)
}\leq C(T)\Vert u_{0}\Vert _{\mathbf{Z}},
\label{General.4.1}
\end{equation}
provided that there exists a solution $u\in\mathbf{X}[0,T)$ for
some $T>0$. Then there exists a unique global solution
$u\in\mathbf{X}[0,\infty)$ to the problem \eqref{1.1}.
\end{theorem}

\begin{proof}
 From Lemma \ref{L4.4}, we have that $\mathcal{G}(t-T_{1}):
\mathbf{Z\to X}[T_{1},T_{2}]$ for any $T_{2}>T_{1} \geq0$ and
\[
\Vert \mathcal{G}(t-T_{1})\phi\Vert
_{\mathbf{X} [T_{1},T_{2}]}\leq C\Vert
\phi\Vert _{\mathbf{Z}}.
\]
Also using Lemma \ref{L4.7} we obtain that $\int_{T_{1}}^{t}\mathcal{G}(
t-\tau)\mathcal{N}(v(\tau))d\tau
\in\mathbf{X}[T_{1},T_{2}]$ for any $v\in\mathbf{X}[
T_{1},T_{2}]$, $T_{2}>T_{1}\geq0$, and
\begin{align*}
&  \Vert \int_{T_{1}}^{t}\mathcal{G}(t-\tau)(
\mathcal{N}(w(\tau))-\mathcal{N}(v(
\tau)))d\tau\Vert _{\mathbf{X}[
T_{1},T_{2}]}\\
&  \leq C\Vert w-v\Vert _{\mathbf{X}[T_{1},T_{2}]
}(1+\Vert w\Vert _{\mathbf{X}[T_{1},T_{2}]
}+\Vert v\Vert _{\mathbf{X}[T_{1},T_{2}]})
^{\sigma},
\end{align*}
for all $v,w\in\mathbf{X}[T_{1},T_{2}]$, where
$\sigma>0$. Using a priory estimates (\ref{General.4.1}) we can prolongate
the local solution given by Theorem \ref{General.Theorem2.1} for all times
$t>0$. Indeed, by the contrary we can suppose that there exists a maximal
existence time $T>0$ such that $u\in\mathbf{X}[0,T)$. If we
choose a new initial time $T_{1}\in[0,T)$ and consider the
problem \eqref{1.1} with initial data $u(T_{1})$, then via a
priory estimate (\ref{General.4.1}) the norm
$\| u(T_{1})\| _{\mathbf{Z}}$ is bounded uniformly with
respect to $T_{1}\in[0,T)$. Then the existence time given by
the local existence Theorem \ref{General.Theorem2.1} is bounded from below
uniformly with respect to $T_{1}\in[0,T)$. Therefore if a new
initial time $T_{1}>0$ is chosen to be sufficiently close to the maximal time
$T$, then by virtue of the local existence Theorem \ref{General.Theorem2.1} we
can guarantee that there exists a unique solution $u\in\mathbf{X}[
0,T]$. Now putting $u(T)$ as a new initial data at time
$T$ we can apply the local existence Theorem \ref{General.Theorem2.1} and
prolongate the solution $u(t)$ on some bigger time interval
$[0,T+T_{2}]$. This contradicts to the fact that $T$ is a
maximal existence time. Hence there exists a unique solution $u\in
\mathbf{X}[0,\infty)$ to the problem \eqref{1.1}. Theorem
\ref{General.Theorem4.1} is proved.
\end{proof}

\section{Large initial data (proof of Theorem \ref{T1})} \label{F.S4}

To prove of Theorem \ref{T1} \ we first apply the so-called energy method to
estimate the $\mathbf{L}^{2}(\mathbb{R})$ norm: i.e. we
multiply equation \eqref{1.1} by $u$ and integrate with respect to
$x\in\mathbb{R}^{+}$, to get
\begin{equation}
\frac{d}{dt}\Vert u(t)\Vert _{\mathbf{L}^{2}}
^{2}+2\lambda\int_{0}^{+\infty}| u| ^{\sigma+1}
dx=-2\int_{0}^{+\infty}u\mathcal{L}udx. \label{F.4.1}
\end{equation}
By the Plancherel Theorem we have
\begin{align*}
2\int_{0}^{+\infty}u\mathcal{L}udx  &
=2\operatorname{Re}\frac{1}{2\pi i}
\int_{-i\infty}^{i\infty}\overline{\widehat{u}}(p)C_{\alpha}p^{\alpha
}(\widehat{u}(p)-\frac{u(0)}{p})\,dp\\
&  =\operatorname{Re}(VP\frac{1}{\pi i}\int_{-i\infty}^{i\infty
}C_{\alpha}p^{\alpha}| \widehat{u}(p)-\frac{u(0)}{p}|
^{2}\,dp \\
& \quad -\operatorname{Re}u(0)VP\frac{1}{\pi i}\int_{-i\infty}^{i\infty
}C_{\alpha}p^{\alpha-1}(\widehat{u}(p)-\frac{u(0)}{p})\,dp).
\end{align*}
By the Cauchy Theorem using the analyticity of $\widehat{u}(p)$ and
$K(p)=C_{\alpha}p^{\alpha}$ in the right-half complex plane we attain
\begin{align*}
&VP\int_{-i\infty}^{i\infty}C_{\alpha}p^{\alpha-1}(\widehat{u}
(p)-\frac{u(0)}{p})\,dp\\
&  =\frac{1}{2}\mathop{\rm res}{}_{p=0}(C_{\alpha}p^{\alpha-1}(\widehat{u}(p)-\frac{u(0)}
{p}))
  +\int_{-i\infty}^{i\infty}C_{\alpha}p^{\alpha-1}(\widehat{u}(p)-\frac
{u(0)}{p})\,dp
  =0.
\end{align*}
Thus from dissipative condition $\operatorname{Re}K(p)>0$ for
$\operatorname{Re}p=0$ we get
\[
  2\int_{0}^{+\infty}u\mathcal{L}udx
  =\frac{1}{\pi}VP\int_{-\infty}^{\infty}\operatorname{Re}K(ip)|
\widehat{u}(ip)-\frac{u(0)}{ip}| ^{2}\,dp>0.
\]
Also since $\lambda>0$ we have
\[
\lambda\int_{0}^{+\infty}| u| ^{\sigma+1}dx>0.
\]
Therefore, from equation (\ref{F.4.1}) we obtain
\begin{equation}
\frac{d}{dt}\Vert u\Vert _{\mathbf{L}^{2}}^{2}\leq0. \label{energy}
\end{equation}
Hence integrating with respect to time we see that
\begin{equation}
\sup_{t\geq0}\Vert u(t)\Vert _{\mathbf{L}^{2}}
\leq\Vert u_{0}\Vert _{\mathbf{L}^{2}}. \label{F.4.2}
\end{equation}
Now we prove that
\begin{equation}
\sup_{t\geq0}\Vert u_{x}(t)\Vert _{\mathbf{L}^{2}
}\leq\Vert u_{0x}\Vert _{\mathbf{L}^{2}}. \label{derivative}
\end{equation}
We  differentiate \eqref{1.1} with respect to the space variable, multiply
equation \eqref{1.1} by $u_{x}$ and integrate with respect to $x\in
\mathbb{R}^{+}$, to get
\begin{equation}
\frac{d}{dt}\Vert u_{x}(t)\Vert
_{\mathbf{L}^{2} }^{2}+2\sigma\lambda\int_{0}^{+\infty}|
u| ^{\sigma -1}| u|
_{x}^{2}dx=-2\int_{0}^{+\infty}u_{x}\partial
_{x}\mathcal{L}udx. \label{F.4.1a}
\end{equation}
Since
\[
2\int_{0}^{+\infty}u_{x}\partial_{x}\mathcal{L}udx=\frac{1}{\pi}
VP\int_{-\infty}^{\infty}\operatorname{Re}K(ip)|
p| ^{2}|
\widehat{u}(ip)-\frac{u(0)}{ip}| ^{2}\,dp>0
\]
and
\[
\sigma\lambda\int_{0}^{+\infty}| u| ^{\sigma-1}|
u| _{x}^{2}dx>0
\]
integrating (\ref{F.4.1a}) with respect to time we easily get
(\ref{derivative}).

Note that time decay estimates (\ref{F.4.2}) and (\ref{derivative}) are not
optimal. To get an optimal time decay estimates we need to show that the
$\mathbf{L}^{1}$ - norm of the solution does not grow with time. Using the
idea of papers \cite{BardosPenelFrischSulem}, \cite{BilerFunakiWoyczynski} we
multiply equation \eqref{1.1} by
$S\equiv\mathop{\rm sign }u\equiv\frac {u}{| u| }$ and integrate
with respect to $x$ over $\mathbb{R}^{+}$ to get
\begin{equation}
\int_{\mathbb{R}^{+}}u_{t}(x,t)\text{$S$}(x,t)
dx+\int_{\mathbb{R}^{+}}\mathcal{N}(u)(x,t)
\text{$S$}(x,t)dx=-\int_{\mathbb{R}^{+}}\text{$S$}(
x,t)\mathcal{L}udx, \label{F.4}
\end{equation}
where $\mathcal{N}(u)=\lambda| u| ^{\sigma }u$. We have
\begin{gather*}
\int_{\mathbb{R}^{+}}u_{t}(x,t)\text{$S$}(x,t)
dx=\int_{\mathbb{R}^{+}}\frac{\partial}{\partial t}| u(
x,t)| dx=\frac{d}{dt}\Vert u(t)
\Vert _{\mathbf{L}^{1}}, \\
\int_{\mathbb{R}^{+}}\mathcal{N}(u)(x,t)
\text{$S$}(x,t)dx=\lambda\int_{\mathbb{R}^{+}}|
u| ^{\sigma}dx\geq0.
\end{gather*}
Representing the operator $\mathcal{L}u$ via the Riesz potential (see
\cite{Stein}) let us show that
\begin{equation}
\int_{\mathbb{R}^{+}}\text{$S$}(x,t)\mathcal{L}udx\geq0.
\label{F.4.a}
\end{equation}
By \cite{BatemanErdelyi}, we have
\[
\frac{1}{2\pi
i}\int_{-i\infty}^{i\infty}p^{-\nu-1}e^{px}\,dp=\frac{1}
{\Gamma(\nu+1)}x^{\nu}.
\]
Thus for $\alpha\in(1,2)$, we get
\[
\mathcal{L}u=-C\partial_{x}\int_{0}^{x}(x-y)^{1-\alpha}\partial_{y}
u(y)\,dy,C>0.
\]
Denote $S(x,t)=\mathop{\rm sign}(u(x,t))$
and represent $u(x,t)=S(x,t)| u(x,t)| $. We make a regularization
\[
K_{\varepsilon}^{\prime\prime}(x)=
\begin{cases}
\partial_{x}^{2}x^{1-\alpha},& \text{for }x\geq\varepsilon\\
0, &\text{for }0\leq x<\varepsilon,
\end{cases}
\]
such that $K_{\varepsilon}'(x)\leq0$ and
$K_{\varepsilon}''(x)\geq0$ for all $x>0$. We can
easily see that
\[
\partial_{x}\int_{0}^{x}(x-y)^{1-\alpha}u_{y}(
y,t)\,dy=\lim_{\varepsilon\to0}\partial_{x}\int_{0}
^{x}K_{\varepsilon}(x-y)u_{y}(y,t)
\,dy.
\]
(To justify our calculation we note that the linear operator $\mathcal{L}$ in
equation \eqref{1.1} is strongly dissipative, therefore by smoothing effect
the solution obtain regularity $u\in\mathbf{C}^{1}(R^{+})$ (see Theorem
\ref{General.Theorem2.1} and \cite{ns4})).We have
\begin{align*}
&  \int_{\mathbb{R}^{+}}dxS(x,t)\partial_{x}\int_{0}
^{x}K_{\varepsilon}(x-y)\partial_{y}u(y,t)\,dy\\
&  =\int_{\mathbb{R}^{+}}dxS(x,t)
\partial_{x}\int_{0} ^{x}\,dyK_{\varepsilon}(x-y)
S(y,t)\partial
_{y}| u(y,t)| \\
&  =K_{\varepsilon}(0)\int_{\mathbb{R}^{+}}dxS^{2}(x,t)
\partial_{x}| u(x,t)| \\
& \quad +\int_{\mathbb{R}^{+}}\,dy\partial_{y}| u(y,t)
| \int_{y}^{+\infty}dxS(y,t)S(x,t)
\partial_{x}K_{\varepsilon}(x-y).
\end{align*}
Then via the identity $S(y,t)S(x,t)=1-\frac{1}
{2}(S(x,t)-S(y,t))^{2}$,
\begin{align*}
&  \int_{\mathbb{R}^{+}}dxS(x,t)\partial_{x}\int_{0}
^{x}K_{\varepsilon}(x-y)\partial_{y}u(y,t)\,dy\\
&  =K_{\varepsilon}(0)\int_{\mathbb{R}^{+}}dxS^{2}(x,t)
\partial_{x}| u(x,t)|
  +\int_{\mathbb{R}^{+}}dy\partial_{y}| u(y,t)
| \int_{y}^{+\infty}dx\partial_{x}K_{\varepsilon}(x-y)
\\
&\quad  -\frac{1}{2}\int_{\mathbb{R}^{+}}dy\partial_{y}| u(
y,t)| \int_{y}^{+\infty}dx(S(x,t)
-S(y,t))^{2}\partial_{x}K_{\varepsilon}(
x-y).
\end{align*}
Since
\[
\int_{\mathbb{R}^{+}}dy\partial_{y}| u(y,t)|
\int_{y}^{+\infty}dx\partial_{x}K_{\varepsilon}(x-y)
=-K_{\varepsilon}(0)\int_{\mathbb{R}^{+}}dy\partial_{y}| u(
y,t)|
\]
we obtain
\begin{equation} \label{help1}
\begin{aligned}
&  \int_{\mathbb{R}^{+}}dxS(x,t)
\partial_{x}\int_{0} ^{x}K_{\varepsilon}(x-y)
\partial_{y}u(y,t) \,dy\\
&  =-\frac{1}{2}\int_{\mathbb{R}^{+}}dy\partial_{y}| u(
y,t)| \int_{y}^{+\infty}dx(S(x,t)
-S(y,t))^{2}\partial_{x}K_{\varepsilon}(
x-y).
\end{aligned}
\end{equation}
Integrating by parts, we have
\begin{align*}
&  \int_{\mathbb{R}^{+}}dy\partial_{y}| u(y,t)
| \int_{y}^{+\infty}dx(S(x,t)-S(
y,t))^{2}\partial_{x}K_{\varepsilon}(x-y)\\
&  =-| u(0,t)| \int_{0}^{+\infty}dx(
S(x,t)-S(0,t))^{2}\partial
_{x}K_{\varepsilon}(x)\\
&\quad  +\int_{\mathbb{R}^{+}}dy| u(y,t)|
\int_{y}^{+\infty}dxK_{\varepsilon}^{\prime\prime}(x-y)(
S(x,t)-S(y,t))^{2}.
\end{align*}
Therefore, from (\ref{help1}) using
$\partial_{x}K_{\varepsilon}(x)<0$, we gain
\begin{align*}
&  \int_{\mathbb{R}^{+}}dxS(x,t)\partial_{x}\int_{0}
^{x}K_{\varepsilon}(x-y)\partial_{y}u(y,t)\,dy\\
&  \leq-\frac{1}{2}\int_{\mathbb{R}^{+}}dy| u(y,t)
| \int_{y}^{+\infty}dxK_{\varepsilon}^{\prime\prime}(
x-y)(S(x,t)-S(y,t))^{2}
\end{align*}
and therefore since $\partial_{x}^{2}K(x)>0$ for all $x>0$ we get
\[
\int_{\mathbb{R}^{+}}dxS(x,t)\partial_{x}\int_{0}
^{x}K_{\varepsilon}(x-y)\partial_{y}u(
y,t)\,dy\leq0.
\]
Hence we have
\begin{align*}
&\int_{\mathbb{R}^{+}}S(x,t)
\mathcal{L}udx\\
&\geq-\frac{C}{2} \lim_{\varepsilon\to0}\int_{\mathbb{R}^{+}}dy|
u(y,t)|
\int_{\mathbb{R}^{+}}dxK_{\varepsilon}^{\prime\prime }(
x-y)(S(x,t)-S(y,t))^{2}\geq0.
\end{align*}
Thus (\ref{F.4.a}) is true and from (\ref{F.4}) we find
$\frac{d}{dt}\Vert u(t)\Vert _{\mathbf{L}^{1}}\leq0$. %\label{F.4.3}
From this inequality,  we see that the norm
$\Vert u(t) \Vert _{\mathbf{L}^{1}}$ is bounded for all $t\geq0$.

We now prove that the norm $\Vert u_{x}(t)\Vert
_{\mathbf{L}^{2}}$ $\to0$ as $t\to\infty$. Taking $\varrho
\in(0,1)$ by the Plancherel theorem,
\begin{equation} \label{1h}
\begin{aligned}
\int_{0}^{+\infty}u_{x}\partial_{x}\mathcal{L}udx
& =\frac{1}{\pi}
VP\int_{\mathbf{-\infty}}^{\infty}\operatorname{Re}K(ip)|
p| ^{2}| \widehat{u}(ip,t)-\frac{u(0)}
{ip}| ^{2}\,dp\\
&  \geq\int_{| p| \geq\varrho}\operatorname{Re}C_{\alpha
}p^{\alpha}| \widehat{u}_{x}(p,t)|
^{2}\,dp\\
&  \geq C\varrho^{\alpha}\Vert u_{x}(t)\Vert
_{\mathbf{L}^{2}}^{2}-C\varrho^{\alpha+3}\sup_{| p|
<\varrho}| \widehat{u}(p,t)| ^{2}\\
&\quad   -C\rho^{\alpha+2}| u(0)| \sup_{| p|
<\varrho}| \widehat{u}(p,t)| -C\varrho
^{\alpha+1}| u(0)| ^{2}\\
&  \geq C\varrho^{\alpha}\Vert u(t)\Vert
_{\mathbf{L}^{2}}^{2}-C\varrho^{\alpha+3}\Vert u(t)
\Vert _{\mathbf{L}^{1}}^{2}\\
&\quad   -C\varrho^{\alpha+2}\Vert u(t)\Vert
_{\mathbf{L}^{1}}\Vert u\Vert _{\mathbf{L}^{\infty}}-C\varrho
^{\alpha+1}\Vert u\Vert _{\mathbf{L}^{\infty}}^{2}\\
&  \geq C\varrho^{\alpha}\Vert u(t)\Vert
_{\mathbf{L}^{2}}^{2}-C\varrho^{\alpha+3}\Vert u(t)
\Vert _{\mathbf{L}^{1}}^{2}\\
&\quad  -C\varrho^{\alpha+2}\Vert u(t)\Vert
_{\mathbf{L}^{1}}\Vert u\Vert
_{\mathbf{L}^{2}}^{\frac{1}{2} }\Vert u_{x}\Vert
_{\mathbf{L}^{2}}^{\frac{1}{2}}-C\varrho ^{\alpha+1}\Vert
u\Vert _{\mathbf{L}^{2}}\Vert u_{x} \Vert
_{\mathbf{L}^{2}},
\end{aligned}
\end{equation}
where we have used that
\[
\Vert u_{x}(t)\Vert _{\mathbf{L}^{\infty}}^{2}\leq
C\Vert u(t)\Vert _{\mathbf{L}^{2}}\Vert
u_{x}(t)\Vert _{\mathbf{L}^{2}}.
\]
Since the norms $\Vert u(t)\Vert _{\mathbf{L}^{1}}$,
$\Vert u(t)\Vert _{\mathbf{L}^{2}}$ and $\Vert
u_{x}(t)\Vert _{\mathbf{L}^{2}}$ are bounded, choosing
$\varrho(t)=C^{-1/\alpha} (1+t)
^{-1/\alpha} $, we obtain
\[
\frac{d}{dt}\Vert u_{x}(t)\Vert
_{\mathbf{L}^{2} }^{2}\leq-(1+t)^{-1}\Vert
u_{x}(t)\Vert _{\mathbf{L}^{2}}^{2}+C(
1+t)^{-1-\frac{1}{\alpha} }.
\]
We substitute $\Vert u_{x}(t)\Vert
_{\mathbf{L}^{2} }^{2}=\frac{1}{2}h(t)(
1+t)^{-2}$, then for $h(t)$ we have
\[
h'(t)\leq C(1+t)^{1-\frac{1}{\alpha}},
\]
hence integration with respect to time yields
$h(t)\leq C(1+t)^{2-\frac{1}{\alpha}}$.
Therefore we get the time decay estimate
\[
\Vert u_{x}(t)\Vert _{\mathbf{L}^{2}}\leq C(
1+t)^{-1/(2\alpha)}.
\]
and therefore
\[
\Vert u_{x}(t)\Vert
_{\mathbf{L}^{\infty}}\leq C\Vert u(t)
\Vert _{\mathbf{L}^{2}}^{\frac{1}{2} }\Vert u_{x}(
t)\Vert _{\mathbf{L}^{2}}^{\frac{1}{2} }\leq C(
1+t)^{-\frac{1}{4\alpha}}.
\]
We substitute this estimate in (\ref{1h}) to obtain
\[
\frac{d}{dt}\Vert u_{x}(t)\Vert
_{\mathbf{L}^{2} }^{2}\leq-(1+t)^{-1}\Vert
u_{x}(t)\Vert _{\mathbf{L}^{2}}^{2}+C(
1+t)^{-1-\frac{1}{\alpha }-\frac{1}{2\alpha}}.
\]
Again after the change $\Vert u_{x}(t)\Vert
_{\mathbf{L}^{2}}^{2}=\frac{1}{2}h(t)(1+t)^{-2}$
we get
\[
\Vert u_{x}(t)\Vert _{\mathbf{L}^{2}}\leq C(
1+t)^{-\frac{1}{2\alpha}-\frac{1}{4\alpha}}.
\]
We can repeat this consideration to get the optimal time decay estimate
\begin{equation}
\Vert u_{x}(t)\Vert _{\mathbf{L}^{2}}\leq C(
1+t)^{-\frac{3}{2\alpha}} \label{opt1}
\end{equation}
for all $t>0$.

We now prove that the norm $\Vert u(t)\Vert
_{\mathbf{L}^{2}}$ $\to0$ as $t\to\infty$. Taking $\varrho
\in(0,1)$ by the Plancherel theorem we get
\begin{align*}
&\int_{0}^{+\infty}u\mathcal{L}udx \\
&=\frac{1}{\pi}VP\int_{\mathbf{-i\infty}
}^{i\infty}\operatorname{Re}K(ip)(\widehat{u}(
ip,t)
-\frac{u(0)}{ip})\overline{\widehat{u}}\,dp\\
&  \geq\int_{| p| \geq\varrho}\operatorname{Re}C_{\alpha
}p^{\alpha}| \widehat{u}(p,t)|
^{2}\,dp-u(0)\int_{-i\infty}^{i\infty}\operatorname{Re}C_{\alpha}p^{\alpha
-1}\overline{\widehat{u}}\,dp\\
&  \geq\varrho^{\alpha}\Vert u(t)\Vert
_{\mathbf{L}^{2}}^{2}-\varrho^{\alpha+1}\sup_{| p|
<\varrho}| \widehat{u}(p,t)| ^{2}-|
u(0)| \varrho^{\alpha}\sup_{| p| <\varrho
}| \widehat{u}(p,t)| -\varrho^{\alpha
-1}| u(0)| ^{2}\\
&  \geq\varrho^{\alpha}\Vert u(t)\Vert
_{\mathbf{L}^{2}}^{2}-\varrho^{\alpha+1}\Vert u(t)
\Vert _{\mathbf{L}^{1}}^{2}-| u(0)| \varrho^{\alpha
}\Vert u(t)\Vert _{\mathbf{L}^{1}}-\varrho
^{\alpha-1}| u(0)| ^{2}.
\end{align*}
Since the norms $\Vert u(t)\Vert _{\mathbf{L}^{1}}$,
$\Vert u(t)\Vert _{\mathbf{L}^{2}}$ are bounded and
due to (\ref{opt1})
\[
| u(0)| \leq\Vert u(t)
\Vert _{\mathbf{L}^{\infty}}\leq C\Vert u(
t)\Vert _{\mathbf{L}^{2}}^{\frac{1}{2}}\Vert
u_{x}(t)\Vert
_{\mathbf{L}^{2}}^{\frac{1}{2}}\leq C(1+t)
^{-\frac{3}{4\alpha} },
\]
choosing $\varrho(t)=C^{-1/\alpha} (1+t)
^{-1/\alpha} $, we obtain
\[
\frac{d}{dt}\Vert u(t)\Vert
_{\mathbf{L}^{2}} ^{2}\leq-(1+t)^{-1}\Vert
u(t)\Vert _{\mathbf{L}^{2}}^{2}+C(
1+t)^{-1-\frac{1}{2\alpha}}.
\]
We substitute $\Vert u(t)\Vert
_{\mathbf{L}^{2}} ^{2}=h(t)(1+t)
^{-2}$, then for $h(t)$ we have
\[
h'(t)\leq C(1+t)^{1-\frac{1}{2\alpha}},
\]
hence integration with respect to time yields
\[
h(t)\leq C(1+t)^{2-\frac{1}{2\alpha}}.
\]
Therefore, we have the time decay estimates
\begin{gather*}
\Vert u(t)\Vert _{\mathbf{L}^{2}}    \leq C(
1+t)^{-\frac{1}{4\alpha}},\\
| u(0)|    \leq C(1+t)^{-\frac{3}{4\alpha}-\frac{1}{8\alpha}}.
\end{gather*}
We can repeat this consideration to get
\[
\frac{d}{dt}\Vert u(t)\Vert
_{\mathbf{L}^{2}} ^{2}\leq-(1+t)^{-1}\Vert
u(t)\Vert _{\mathbf{L}^{2}}^{2}+C(
1+t)^{-1-\frac{1}{\alpha}}.
\]
Therefore, we obtain the optimal estimate
\begin{equation}
\Vert u(t)\Vert _{\mathbf{L}^{2}}\leq C(
1+t)^{-\frac{1}{2\alpha}} \label{opt2}
\end{equation}
for all $t>0$. Also from (\ref{opt1}) and (\ref{opt2}) we get the optimal time
decay of the $\mathbf{L}^{\infty}(\mathbb{R}^{+})$-norm of
the solutions
\begin{equation}
\Vert u(t)\Vert _{\mathbf{L}^{\infty}}\leq C(
1+t)^{-1/\alpha}  \label{opt3}
\end{equation}
for all $t>0$.

Now we can estimate the norm $\mathbf{L}^{\infty,\mu}$ $(\mathbb{R}
^{+})$. By the integral formula (\ref{integral}) we have
\begin{align*}
 \Vert u(t)\Vert _{\mathbf{L}^{\infty,\mu}}
&\leq\Vert u_{0}\Vert _{\mathbf{L}^{1}}\Vert G(
t)\Vert _{\mathbf{L}^{\infty,\mu}}+\Vert u_{0}\Vert
_{\mathbf{L}^{\infty,\mu}}\Vert G(t)\Vert
_{\mathbf{L}^{1}} \\
& \quad +C\int_{0}^{t}\Vert | u(\tau)|
^{\sigma+1}\Vert _{\mathbf{L}^{1}}\Vert G(t-\tau)
\Vert _{\mathbf{L}^{\infty,\mu}}d\tau\\
&\quad +C\int_{0}^{t}\Vert | u(\tau)|
^{\sigma+1}\Vert _{\mathbf{L}^{\infty,\mu}}\Vert G(
t-\tau)\Vert _{\mathbf{L}^{1}}d\tau
\end{align*}
Hence using Lemmas \ref{L4.4} and \ref{L4.7},
\begin{align*}
&\Vert u(t)\Vert _{\mathbf{L} ^{\infty,\alpha+1}}\\
&\leq C\langle t\rangle ^{-\frac{1-\mu}{\alpha}
}+C\int_{0}^{t}\Vert u(\tau)\Vert _{\mathbf{L}
^{\infty}}^{\sigma}\langle t-\tau\rangle ^{-\frac{1-\mu}{\alpha}
}d\tau +C\int_{0}^{t}\Vert u(\tau)\Vert _{\mathbf{L}
^{\infty}}^{\sigma}\Vert u(\tau)\Vert _{\mathbf{L}
^{\infty,\mu}}d\tau\\
& \leq C\langle t\rangle
^{-\frac{1-\mu}{\alpha}}+C\int_{0} ^{t}\Vert u(
\tau)\Vert _{\mathbf{L}^{\infty,\mu} }\langle
\tau\rangle ^{-\frac{\sigma}{\alpha}}d\tau.
\end{align*}
Hence for the function $h(t)=\sup_{0\leq\tau\leq t}\Vert
u(\tau)\Vert _{\mathbf{L}^{\infty,\mu}}$, we get the
inequality
\[
h(t)\leq C\langle t\rangle
^{-\frac{1-\mu}{\alpha} }+C\int_{0}^{t}\langle
\tau\rangle ^{-\frac{\sigma}{\alpha} }h(\tau)
d\tau
\]
and since $\sigma>\alpha$ by the Gronwall's lemma it follows that
\begin{equation}
\Vert u(t)\Vert _{\mathbf{L}^{\infty,\mu}}\leq
C\langle t\rangle ^{-\frac{1-\mu}{\alpha}} \quad \forall t>0. \label{F.4.5}
\end{equation}
 From a priory estimates, due to Theorem \ref{General.Theorem4.1} \ there
exists a unique solution
\[
u(x,t)\in C([0,\infty);\mathbf{H}_{2}^{1,0}
\cap\mathbf{H}_{\infty}^{0,\mu}\cap\mathbf{H}_{1}^{0,0})
\]
to the problem \eqref{1.1}, such that
\begin{equation}
\Vert u(t)\Vert _{\mathbf{L}^{\infty}}\leq
C\langle t\rangle ^{-1/\alpha} ,\Vert u(
t)\Vert _{\mathbf{L}^{\infty,\mu}}\leq C\langle
t\rangle ^{-\frac{1-\mu}{\alpha}}\text{ and }\Vert u(
t)\Vert _{\mathbf{L}^{2}}\leq C\langle t\rangle
^{-\frac{1}{2\alpha}}. \label{main}
\end{equation}
By Lemma \ref{L4.7} we have
\[
\sup_{t>1}t^{\frac{1}{\alpha}(1+\mu)}\Vert \mathcal{G}\phi-t^{-\frac
{1}{\alpha}}\Lambda_{0}(xt^{-1/\alpha} )f(\phi
)\Vert _{\mathbf{L}^{\infty}}\leq C\Vert \phi\Vert
_{\mathbf{L}^{1,\mu}},
\]
where $\mu\in[0,1]$. Substituting this formula into
(\ref{integral}) we obtain
\begin{equation}
u(x,t)=t^{-1/\alpha} A\Lambda_{0}(\frac{x}{t^{\frac{1}{\alpha}}
})+R(x,t), \label{a}
\end{equation}
where by (\ref{main}),
\[
A=\int_{0}^{+\infty}u_{0}dy+\int_{0}^{+\infty}d\tau\int_{0}^{\infty
}\mathcal{N}(u)\,dy<\infty,
\]
and
\begin{align*}
R(x,t)  &  =O(t^{-\frac{1+\mu}{\alpha}})
\Big(\int_{0}^{+\infty}y^{\mu}
u_{0}(y)\,dy+\int_{0}^{+\infty}d\tau\int_{0}^{+\infty}y^{\mu}\mathcal{N}
(u)\,dy\Big)\\
&\quad  +\int_{0}^{+\infty}\tau^{\frac{2\mu}{\alpha}}O\big((t-\tau
)^{-\frac{1+\mu}{\alpha}}\big)d\tau\int_{0}^{+\infty}\mathcal{N}(u)\,dy\\
&  =O(t^{-\frac{1+\mu}{\alpha}}).
\end{align*}
Thus the asymptotic (\ref{5.5}) is valid. Theorem \ref{T1} is then proved.

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\end{document}