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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 109, pp. 1--25.\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 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/109\hfil Nonlinear pseudoparabolic equations]
{Initial-boundary value problems for nonlinear pseudoparabolic
equations in a critical case}

\author[E. I. Kaikina\hfil EJDE-2007/109\hfilneg]
{Elena I. Kaikina}  

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

\thanks{Submitted March 22, 2007. Published August 7, 2007.}
\subjclass[2000]{35Q35}
\keywords{Dissipative nonlinear evolution equation; Sobolev equation;
\hfill\break\indent large time asymptotic behavior}

\begin{abstract}
 We study nonlinear pseudoparabolic  equations, on the half-line
 in a critical case,
 \begin{gather*}
 \partial _{t}( u-u_{xx}) -\alpha u_{xx}=\lambda |u| u,\quad
 x\in \mathbb{R}^{+},\; t>0, \\
 u( 0,x) =u_{0}( x) , \quad x\in \mathbb{R}^{+}, \\
 u(t,0)=0,
 \end{gather*}
 where $\alpha >0$, $\lambda \in \mathbb{R}$.
 The aim of this paper is to  prove the  existence of global solutions
 to the initial-boundary value problem and to find the main term
 of the asymptotic representation of solutions.
\end{abstract}

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

\section{\label{S1}Introduction}

We study the following  nonlinear
pseudoparabolic equation  on the half line, with Dirichlet boundary condition,
\begin{equation}
\begin{gathered}
\partial _{t}( u-u_{xx}) -\alpha u_{xx}=\lambda |u| u,\quad
x\in \mathbb{R}^{+},\; t>0, \\
u( 0,x) =u_{0}( x) , \quad x\in \mathbb{R}^{+}, \\
u(t,0)=0,
\end{gathered}  \label{1.1}
\end{equation}
where $\alpha >0$ and $\lambda \in \mathbb{R}$.

The Cauchy problem for nonlinear pseudoparabolic type equations was studied
in many papers (for example, see
\cite{DemidenkoUspenskii,EgorovPyatkovPopov,FaviniYagi,Fujita,
GajewskiGregerZacharias, MitidieriPohozhaev, Showalter, Stefanelli}).
The large time asymptotics of solutions to the Cauchy problem
was obtained in papers \cite{ABS}-\cite{BL1},
\cite{kns5,Karch,Karch1,MS,PZ,Zhang}.

In this paper we study the initial boundary-value problem \eqref{1.1} in
a critical case, when the nonlinear term of equation \eqref{1.1} has the same
time decay rate as the linear terms. Recently much attention was drown to
the study of the global existence and large time asymptotic behavior of
solutions to the Cauchy problems for nonlinear equations in the critical
cases (see papers \cite{ek}-\cite{gv2,hkn4,hkn6,KaminPeletier,k}
and literature cited therein). A general theory of
nonlinear nonlocal equations on a half-line was developed in the book
\cite{bookhk1}, however the case of nonanalytic symbols $K(p)$ in the
right-hand complex plane was not studied previously.
In the present paper we fill this gap, considering as example
the pseudoparabolic type equation \eqref{1.1}
with a symbol $K(p)=\alpha p^{2}/(1-p^{2})$. We construct the Green
operator for problem \eqref{1.1} posing some necessary condition at the
singularity point $p=1$ of the symbol $K(p)$. Another difficulty which we
overcome in the present paper is in evaluating the contribution of the
boundary data into the large time asymptotic behavior of solutions. In this
paper we will prove that in the case of the initial-boundary value problem
due to boundary data the solution obtains an additional decay comparing with
the case of Cauchy problem. As a result the main term of the asymptotic
expansion does not depend on the mean value of the solution instead it
is determined by the first moment of the solution.
Thus we have to estimate the evolution
of the first moments to obtain an optimal time decay estimate of the
solution. Also we are interested in the case of large initial data. Using
the energy type a-priori estimates for the first moment of the solution we
are able to remove the smallness condition for the initial data. The
asymptotic behavior of solutions is founded by the standard way developed in
the book \cite{bookhkns}.

Below $\hat{\phi}$ is the Laplace transform of $\phi $ defined by
\[
\hat{\phi}( \xi ) =\int_{0}^{+\infty }e^{-x\xi }\phi (x) dx,
\]
Here
\[
\mathcal{L}^{-1}\hat{\phi}( \xi ) =( 2\pi i)
^{-1}\int_{\mathbf{-}i\mathbf{\infty }}^{i\infty }e^{x\xi }\hat{\phi}(
\xi ) d\xi
\]
is the inverse Laplace transform of $\phi $. By $\mathbf{C}( \mathbf{I};
\mathbf{B}) $ we denote the space of continuous functions from a time
interval $\mathbf{I}$ to the Banach space $\mathbf{B}$. The usual Lebesgue
space is denote by $\mathbf{L}^{p}(\mathbb{R}^{+})$, $1\leq p\leq \infty $,
the weighted Lebesgue space $\mathbf{L}^{p,a}(\mathbb{R}^{+})$ is defined by
\[
\mathbf{L}^{p,a}(\mathbb{R}^{+})=\left\{ \phi \in \mathbf{L}^{p}(\mathbb{R}
^{+});\| \phi \| _{\mathbf{L}^{p,a}}=\|
\langle x\rangle ^{a}\phi \| _{\mathbf{L}^{p}}<\infty
\right\} ,
\]
where $\langle x\rangle =\sqrt{1+| x| ^{2}}$,
$a\geq 0$. Weighted Sobolev spaces we define as follows
\[
\mathbf{W}_{p}^{k,a}(\mathbb{R}^{+})
=\Big\{ \phi \in \mathbf{L}^{p}(\mathbf{
R}^{+});\| \phi \| _{\mathbf{W}_{p}^{k,a}}=\sum_{j=0}^{k}
\| \partial ^{j}\phi \| _{\mathbf{L}^{p,a}}<\infty \Big\},
\]
where $k\geq 0$, $a\geq 0$, $1\leq p\leq \infty $.Also we denote by $\mathbf{
H}^{k,a}(\mathbb{R}^{+})=\mathbf{W}_{2}^{k,a}(\mathbb{R}^{+})$.
Define
\[
| \lambda | t\int_{0}^{+\infty }x( G_{0}(t,x) ) ^{2}dx=\eta ,
\]
where the heat kernel
\[
G_{0}( t,x) =( 4\pi \alpha t) ^{-1/2}\frac{x}{
\alpha t}e^{-\frac{x^{2}}{4\alpha t}}.
\]
Denote
\[
g( t) =1+| \theta | \eta \log (
1+t) ,\theta =\int_{0}^{+\infty }xu_{0}(x)dx.
\]
Now we state the results of this paper.

\begin{theorem} \label{T1}
Assume that $\lambda \theta \leq -C\varepsilon <0$. Let the
initial data $u_{0}\in \mathbf{L}^{\infty }( \mathbb{R}^{+}) \cap
\mathbf{L}^{1,1+a}( \mathbb{R}^{+}) $,
$a\in ( 0,1] $
are small $\| u_{0}\| _{\mathbf{L}^{\infty }}+\|
u_{0}\| _{\mathbf{L}^{1,1+a}}\leq \varepsilon $. Then the
initial-boundary value problem \eqref{1.1} has a unique global solution
\[
u\in \mathbf{C}( [0,\infty );\mathbf{L}^{\infty }( \mathbb{R}
^{+}) \cap \mathbf{L}^{1,1+a}( \mathbb{R}^{+}) )
\]
satisfying the time decay estimate
\begin{equation}
\| u( t) -\theta G_{0}( t) g^{-1}(
t) \| _{\mathbf{L}^{\infty }}\leq C\langle
t\rangle ^{-1}g^{-2}( t) .  \label{4.1}
\end{equation}
\end{theorem}

Using the method of papers \cite{hn23,kns5} we can remove the
smallness condition on the initial data $u_{0}( x) $.

\begin{theorem}\label{T2}
Assume  the initial data $u_{0}\in \mathbf{W}_{\infty
}^{2,0}( \mathbb{R}^{+}) \cap \mathbf{W}_{1}^{2,0}( \mathbf{R
}^{+}) \cap \mathbf{L}^{1,1+a}( \mathbb{R}^{+}) $, $0<a\leq
1$, are such that $\lambda \theta <0$. Then the initial-boundary value
problem \eqref{1.1} has a unique global solution
\[
u( t,x) \in \mathbf{C}( [0,\infty );\mathbf{L}^{\infty
}( \mathbb{R}^{+}) \cap \mathbf{L}^{1,1+a}( \mathbb{R}
^{+}) )
\]
satisfying the time decay estimate
\begin{equation}
\| u( t) -\theta G_{0}( t) g^{-1}(
t) \| _{\mathbf{L}^{\infty }}\leq C\langle
t\rangle ^{-1}g^{-2}( t) \log \log g(t).  \label{a}
\end{equation}
\end{theorem}

We start with a section where we obtain preliminary estimates for the
linearized initial boundary-value problem corresponding to \eqref{1.1}. Also
we prove a local existence theorem. In the next two sections we prove
Theorems \ref{T1} and \ref{T2} respectively.

\section{\label{S2}Preliminaries}

\subsection{\label{SS2.1}Green operator}

Consider the linear initial boundary-value problem on half-line
\begin{equation}
\begin{gathered}
\partial _{t}( u-u_{xx}) -\alpha u_{xx}=f( t,x) ,
\quad x\in {\mathbb{R}}^{+},\; t>0, \\
u( 0,x) =u_{0}( x) ,\quad x\in {\mathbb{R}}^{+} \\
u(t,0)=0.
\end{gathered}  \label{2.1}
\end{equation}
Taking Laplace transform with respect to space variable we get
\begin{equation}
\begin{gathered}
\partial _{t}\Big( \widehat{u}-p^{2}\Big( \widehat{u}-\sum
_{j=1}^{2}\frac{\partial _{x}^{j-1}u(t,0)}{p^{j}}\Big)
\Big) -\alpha p^{2}\Big(
\widehat{u}-\sum_{j=1}^{2}\frac{\partial
_{x}^{j-1}u(t,0)}{p^{j}}\Big) =\widehat{f}( t,p) ,  \\
\widehat{u}( 0,p) =\widehat{u}_{0}( p) ,\quad  \\
u(t,0)=0.
\end{gathered} \label{2.1a}
\end{equation}
We rewrite
\begin{equation}
(1-p^{2})\partial _{t}\widehat{u}-\alpha p^{2}\widehat{u}=f_{1}(t,p) ,
\label{eq}
\end{equation}
where
\[
f_{1}(t,p)=\widehat{f}( t,p) -p^{2}\sum_{j=1}^{2}\frac{
\partial _{x}^{j-1}u_{t}(0,t)}{p^{j}}-\alpha p^{2}\sum_{j=1}^{2}
\frac{\partial _{x}^{j-1}u(t,0)}{p^{j}}.
\]
Integrating \eqref{eq} with respect to  time, we obtain
\begin{equation}
\widehat{u}(t,p)=e^{-K(p)t}\widehat{u}_{0}( p)
+\int_{0}^{t}e^{-K(p)(t-\tau )}\frac{1}{1-p^{2}}f_{1}(\tau ,p)d\tau ,
\label{Lapsol}
\end{equation}
where
\[
K(p)=-\frac{\alpha p^{2}}{1-p^{2}}.
\]
Note that symbol $K(p)$ is not analytic in right-half complex plane.
Therefore we have for solution $u(t,x)$
\[
u(t,x)=\frac{1}{2\pi i}\int_{1+\varepsilon -i\infty }^{1+\varepsilon
+i\infty }e^{px}\widehat{u}(t,p)dp.
\]
To satisfy zero condition for $x\to +\infty $ we need to have
\begin{equation}
\mathop{\rm res}(\widehat{u}(t,p),1)=0.  \label{zero}
\end{equation}
We have by definition
\begin{align*}
\mathop{\rm res}(\widehat{u}(t,p),1)
&=\lim_{\rho \to 0}\int_{C_{\rho,1}}\widehat{u}(t,p)dp \\
&=\lim_{\rho \to 0}\int_{C_{\rho ,1},p\in D^{+}}\widehat{u}
(t,p)dp+\lim_{\rho \to 0}\int_{C_{\rho ,1},p\in D^{-}}\widehat{u}
(t,p)dp,
\end{align*}
where
\[
C_{\rho ,1}=\left\{ p\in \mathbb{C}: p=1+\rho e^{i\phi },\rho >0,\phi \in
[0,2\pi ) \right\} ,
\]
and by $D^{+}, D^{-}$ we denote domains where $\mathop{\rm Re}K(p)>0$
and $\mathop{\rm Re}K(p)<0$.
Since for $p\in C_{\rho ,1}$
\[
K(p)=-\frac{\alpha (1+\rho e^{i\phi })^{2}}{\rho e^{i\phi }(2+\rho e^{i\phi
})},
\]
it is easy to see that
\[
\lim_{\rho \to 0}\int_{C_{\rho ,1,}p\in D^{+}}\widehat{u}(t,p)dp=0.
\]
We rewrite formula \eqref{Lapsol} in a domain,
where $\mathop{\rm Re}K(p)<0$ in the form
\begin{align*}
\widehat{u}(t,p) &= e^{-K(p)t}( \widehat{u}_{0}( p)
+\int_{0}^{+\infty }e^{-K(p)(t-\tau )}\frac{1}{1-p^{2}}f_{1}(\tau ,p)d\tau
) \\
&\quad -\int_{t}^{+\infty }e^{-K(p)(t-\tau )}\frac{1}{1-p^{2}}f_{1}
(\tau ,p)d\tau.
\end{align*}
We have
\[
\lim_{\rho \to 0}\int_{C_{\rho ,1,}p\in D^{-}}dp\int_{t}^{+\infty
}e^{-K(p)(t-\tau )}\frac{1}{1-p^{2}}f_{1}(\tau ,p)d\tau =0.
\]
To satisfy (\ref{zero}) we have to put the following condition for all
$p\in D^{-},\mathop{\rm Re}p>0$
\begin{equation}
\widehat{u}_{0}( p) +\int_{0}^{+\infty }e^{K(p)\tau }\frac{1}{
1-p^{2}}f_{1}(\tau ,p)d\tau =0.  \label{con}
\end{equation}
Note that function $f_{1}$  includes two unknown boundary data $u(t,0)$ and
$u_{x}(0,t)$. Also we have two roots $\phi _{j}(\xi )$ of equation
$\xi =-K(p)$. By direct calculation we obtain
\[
\phi _{1}=\sqrt{\frac{\xi }{\xi +\alpha }}
\quad\text{and}\quad
\phi _{2}=-\sqrt{\frac{\xi }{\xi +\alpha }}.
\]
Since we are interested in $\mathop{\rm Re}\xi >0$ and
$\mathop{\rm Re}p>0$ making the change of variable $\xi =-K(p)$ we
rewrite condition (\ref{con}) as one
equation with two unknown boundary data $u(t,0)$ and $u_{x}(t,0)$
\begin{equation}
\widehat{u}_{0}( \phi _{1}) +\frac{1}{1-\phi _{1}^{2}}
\int_{0}^{+\infty }e^{-\xi \tau }f_{1}(\tau ,\xi )d\tau =0,  \label{pr}
\end{equation}
where
\[
f_{1}(\tau ,\xi )=\widehat{f}( \tau ,\phi _{1}) -\phi
_{1}^{2}\sum_{j=1}^{2}\frac{\partial _{x}^{j-1}u_{\tau }(\tau ,0)}{
\phi _{1}^{j}}-\alpha \phi
_{1}^{2}\sum_{j=1}^{2}\frac{\partial _{x}^{j-1}u(\tau
,0)}{\phi _{1}^{j}}.
\]
So we have to put only one boundary conditions in the problem (\ref{2.1})
and rest boundary data we will find from equation (\ref{pr}).
Putting $u(t,0)=0$ in equation (\ref{pr}) we get
\[
\mathcal{L}(\widehat{u}_{tx}(t,0)+\alpha u_{x}(t,0))=(1-\phi _{1}^{2})
\widehat{u}_{0}( \phi _{1}) +\widehat{\widehat{f}}( \xi
,\phi _{1}).
\]
So that
\begin{align*}
\widehat{u}_{tx}(t,0)+\alpha u_{x}(t,0)
&= \frac{1}{2\pi i}\int_{-i\infty
}^{i\infty }e^{\xi t}( (1-\phi _{1}^{2})\widehat{u}_{0}( \phi
_{1}) +\widehat{\widehat{f}}( \xi ,\phi _{1}) ) d\xi \\
&= \frac{1}{2\pi i}\int_{-i\infty }^{i\infty }e^{\xi t}(1-\frac{\xi }{\xi
+\alpha })(\widehat{u}_{0}( \phi _{1}) +\widehat{\widehat{f}}
( \xi ,\phi _{1}) )d\xi \\
&= \frac{1}{2\pi i}\int_{-i\infty }^{i\infty }e^{\xi t}( \frac{\alpha
\widehat{u}_{0}( \phi _{1}) }{\xi +\alpha }+\widehat{\widehat{f}}
( \xi ,\phi _{1}) ) d\xi .
\end{align*}
Substituting this representation into \eqref{Lapsol} we obtain
\begin{equation}
\widehat{u}(t,p)=I_{1}(t,p)+I_{2}(t,p),  \label{*}
\end{equation}
where
\begin{equation} \label{I1}
\begin{aligned}
I_{1}(t,p) &= e^{-K(p)t}\widehat{u}_{0}( p)
 -\frac{\alpha }{1-p^{2}}\frac{1}{2\pi i}\int_{0}^{t}d\tau e^{-K(p)(t-\tau
)}\int_{-i\infty }^{i\infty }e^{\xi \tau }\frac{\widehat{u}_{0}( \phi
_{1}) }{\xi +\alpha }d\xi   \\
&= J_{1}+J_{2}
\end{aligned}
\end{equation}
and
\begin{equation} \label{I2}
\begin{aligned}
I_{2}(t,p)
&= \frac{1}{1-p^{2}}\Big( \int_{0}^{t}e^{-K(p)(t-\tau )}f(\tau
,p)d\tau   \\
&\quad -\frac{1}{2\pi i}\int_{0}^{t}d\tau e^{-K(p)(t-\tau )}
\int_{-i\infty }^{i\infty }e^{\xi \tau }\widehat{\widehat{f}}( \xi
,\phi _{1}) d\xi \Big) .
\end{aligned}
\end{equation}
Now we consider $I_{1}$ in the representation (\ref{*}).
Changing the order of integration,
\begin{align*}
J_{2} &= \frac{\alpha }{1-p^{2}}\frac{1}{2\pi i}\int_{0}^{t}d\tau
e^{-K(p)(t-\tau )}\int_{-i\infty }^{i\infty }e^{\xi \tau }\frac{\widehat{u}
_{0}( \phi _{1}) }{\xi +\alpha }d\xi \\
&= e^{-K(p)t}\frac{\alpha }{1-p^{2}}\frac{1}{2\pi i}\int_{-i\infty
}^{i\infty }\frac{\widehat{u}_{0}( \phi _{1}) }{\xi +\alpha }
\frac{e^{(\xi +K(p))t}-1}{K(p)+\xi }d\xi .
\end{align*}
Since $\phi _{1}$ is analytic in right-half complex plane and
$\mathop{\rm Re}\phi _{1}>0$, $\mathop{\rm Re}K(p)>0$ for all
$\mathop{\rm Re}p=0$, $\mathop{\rm Re}\xi >0$ via
Cauchy Theorem we obtain
\[
\int_{-i\infty }^{i\infty }\frac{\widehat{u}_{0}( \phi _{1}) }{
\xi +\alpha }\frac{1}{K(p)+\xi }d\xi =0.
\]
Therefore,
\[
J_{2}=\frac{\alpha }{1-p^{2}}\frac{1}{2\pi i}\int_{-i\infty }^{i\infty }
\frac{\widehat{u}_{0}( \phi _{1}) }{\xi +\alpha }\frac{e^{\xi t}}{
K(p)+\xi }d\xi .
\]
Taking inverse Laplace transform with respect to space variable we obtain
\begin{align*}
\mathcal{L}_{x}^{-1}(J_{2})
&= \frac{\alpha }{(2\pi i)^{2}}\int_{-i\infty
}^{i\infty }dpe^{px}\frac{1}{1-p^{2}}\int_{-i\infty }^{i\infty }\frac{
\widehat{u}_{0}( \phi _{1}) }{\xi +\alpha }\frac{e^{\xi t}}{
K(p)+\xi }d\xi \\
&= \frac{\alpha }{(2\pi i)^{2}}\int_{-i\infty }^{i\infty }d\xi e^{\xi t}
\frac{\widehat{u}_{0}( \phi _{1}) }{\xi +\alpha }\int_{-i\infty
}^{i\infty }dpe^{px}\frac{1}{(K(p)+\xi )(1-p^{2})} \\
&= \frac{\alpha }{(2\pi i)^{2}}\int_{-i\infty }^{i\infty }d\xi e^{\xi t}
\frac{\widehat{u}_{0}( \phi _{1}) }{\xi +\alpha }\int_{-i\infty
}^{i\infty }dpe^{px}\frac{1}{-\alpha p^{2}+\xi (1-p^{2})}.
\end{align*}
Since
\begin{align*}
\int_{-i\infty }^{i\infty }dpe^{px}\frac{1}{-(\alpha +\xi )p^{2}+\xi }
&= -\frac{1}{\alpha +\xi }\int_{-i\infty }^{i\infty }dpe^{px}\frac{1}{
(p-\phi _{1})(p-\phi _{2})} \\
&= -2\pi ie^{\phi _{2}x}\frac{1}{\alpha +\xi }\frac{1}{\phi _{2}-\phi _{1}},
\end{align*}
using $\phi _{1}+\phi _{2}=0$, we obtain
\[
\mathcal{L}_{x}^{-1}(J_{2})=-\frac{\alpha }{2\pi i}\int_{-i\infty }^{i\infty
}d\xi e^{\xi t+\phi _{2}x}\frac{1}{2(\xi +\alpha )^{2}\phi _{2}}\widehat{u}
_{0}( -\phi _{2}) .
\]
We have
\[
\phi _{2}'=\frac{1}{2\phi _{2}}(\frac{\xi }{\xi +\alpha })'
=\frac{\alpha }{2\phi _{2}}\frac{1}{(\xi +\alpha )^{2}}.
\]
Making the change of the variables $p=\phi _{2}$ we obtain
\begin{align*}
\mathcal{L}_{x}^{-1}(J_{2})
&= -\frac{1}{2\pi i}\int_{-i\infty }^{i\infty
}d\xi e^{\xi t+\phi _{2}x}\phi _{2}'\widehat{u}_{0}( -\phi
_{2}) \\
&= \frac{1}{2\pi i}\int_{-i\infty }^{i\infty }dpe^{-K(p)t+px}\widehat{u}
_{0}( -p) \\
&= e^{-\alpha t}\mathcal{L}_{x}^{-1}\big\{ e^{\frac{\alpha t}{1+|
p| ^{2}}}\widehat{u}_{0}( -p) \big\} .
\end{align*}
Substituting this representation into (\ref{I1}) we get
\begin{equation}
\mathcal{L}_{x}^{-1}(I_{1})=\mathcal{G}(t)u_{0},  \label{LI1}
\end{equation}
where the Green operator $\mathcal{G}( t) $ is given by
\begin{equation}
\mathcal{G}( t) \phi =e^{-\alpha t}\mathcal{L}_{x}^{-1}
\big\{ e^{\frac{\alpha t}{1+| p| ^{2}}}( \hat{\phi}(p)-\hat{
\phi}(-p)) \big\} .  \label{Greenoper}
\end{equation}
Now we consider $I_{2}$ in the formula (\ref{I2}). In the same way we get
\begin{equation}
\mathcal{L}_{x}^{-1}(I_{2})=\int_{0}^{t}d\tau \int_{0}^{+\infty }dyf(\tau ,y)
\mathcal{L}_{x}^{-1}\big\{ e^{-K(p)(t-\tau )}\frac{1}{1-p^{2}}(
e^{-py}-e^{py}) \big\} .  \label{help}
\end{equation}
We have
\begin{align*}
&\mathcal{L}_{x}^{-1}\big\{ e^{-K(p)(t-\tau )}\frac{1}{1-p^{2}}(
e^{-py}-e^{py}) \big\} \\
&= \int_{-\infty }^{+\infty }dz\mathcal{L}_{x-z}^{-1}\left\{ e^{-K(p)(t-\tau
)}\right\} \mathcal{L}_{z}^{-1}\big\{ \frac{1}{1-p^{2}}(
e^{-py}-e^{py}) \big\} .
\end{align*}
Taking into account
\[
\mathcal{L}_{z}^{-1}\big\{ \frac{1}{1-p^{2}}( e^{-py}-e^{py})
\big\} =\frac{1}{2}( e^{-| z-y| }-e^{-|z+y| }) ,
\]
we get
\begin{align*}
&\mathcal{L}_{x}^{-1}\big\{ e^{-K(p)(t-\tau )}\frac{1}{1-p^{2}}(
e^{-py}-e^{py}) \big\} \\
&= \int_{-\infty }^{+\infty }dz\mathcal{L}_{x-z}^{-1}\left\{ e^{-K(p)(t-\tau
)}\right\} \frac{1}{2}( e^{-| z-y| }-e^{-|
z+y| }) \\
&= \int_{0}^{+\infty }( \mathcal{L}_{x-z}^{-1}\big\{ e^{-K(p)(t-\tau
)}\big\} -\mathcal{L}_{x-z}^{-1}\big\{ e^{-K(p)(t-\tau )}\big\} )
\frac{1}{2}( e^{-| z-y| }-e^{-|
z+y| }) dz.
\end{align*}
Substituting into (\ref{help}) we obtain
\begin{equation}
\mathcal{L}_{x}^{-1}(I_{2})=\int_{0}^{t}\mathcal{G}(t-\tau )\mathcal{B}
f( \tau ) d\tau ,  \label{LI2}
\end{equation}
where operator $\mathcal{G}$ was defined into (\ref{Greenoper}) and operator
$\mathcal{B}$ is defined as
\[
\mathcal{B}\phi =\frac{1}{2}\int_{0}^{+\infty }( e^{-|
x-y| }-e^{-| x+y| }) \phi ( y)
dy.
\]
Also by direct calculation we can obtain another representation of operator
$\mathcal{B}$
\[
\mathcal{B}=( 1-\partial _{x}^{2}) ^{-1}.
\]
Indeed, we put $\mathcal{B}w=u$.
Taking Laplace transform with respect to
space variable and using $u(t,0)=0$ we get
\begin{equation}
\widehat{u}(t,\xi )=\frac{1}{1-\xi ^{2}}(\widehat{w}(t,\xi )-u_{x}(0,t)).
\label{Lu}
\end{equation}
To satisfy $\lim_{x\to \infty }u(t,x)=0$ we need to put the
following condition
\[
res_{\xi =1}e^{\xi x}(\widehat{w}(t,\xi )-u_{x}(0,t))=0.
\]
Therefore we obtain
\begin{equation}
u_{x}(0,t)=\widehat{w}(t,1).  \label{bound}
\end{equation}
Substituting (\ref{bound}) into (\ref{Lu}) we find
\[
\mathcal{B}w=\int_{0}^{+\infty }B(x,y)w(y)dy,
\]
where
\[
B(x,y)=\frac{1}{2}(e^{-| x-y| }-e^{-|
x+y| }).
\]
Note that
$B(x,y)\geq 0$ for any $x\geq 0$, $y\geq 0$.
 From (\ref{*}),(\ref{LI1}) and (\ref{LI2}) we obtain integral formula for
solution of (\ref{2.1})
\begin{equation}
u(t,x)=\mathcal{G}( t) u_{0}+\int_{0}^{t}\mathcal{G}(t-\tau )
\mathcal{B}f( \tau ) d\tau .  \label{2.2}
\end{equation}
We can easily see that
\begin{equation}
\| \mathcal{B}\phi \| _{\mathbf{L}^{r}}\leq C\|
\phi \| _{\mathbf{L}^{r}}  \label{2.4}
\end{equation}
for all $1\leq r\leq \infty $ and
\begin{equation}
\| \mathcal{B}\phi \| _{\mathbf{L}^{1,b}}\leq C\|
\phi \| _{\mathbf{L}^{1,b}}\quad   \label{2.5}
\end{equation}
for any $b\geq 0$.

We first collect some preliminary estimates of the Green operator $\mathcal{G
}( t) $ in the norms $\| \phi \| _{\mathbf{L}
^{p}}$ and $\| \phi \| _{\mathbf{L}^{1,1+w}}$, where $w\in
( 0,1) $, $1\leq r\leq \infty $.

\subsection{\label{SS2.2}Preliminary lemmas}

We introduce the operator $\mathcal{G}_{0}( t) $  given by
\[
\mathcal{G}_{0}( t) \phi =\int_{0}^{+\infty }G_{1}(
t,x,y) \phi ( y) dy,
\]
where the kernel is
\[
G_{1}( t,x,y) =( 4\pi \alpha t) ^{-1/2}\Big(
e^{-\frac{( x-y) ^{2}}{4\alpha t}}-e^{-\frac{( x+y)
^{2}}{4\alpha t}}\Big) .
\]
We first prepare some preliminary estimates of the operator
$\mathcal{G}_{0}( t) $ in the Lebesgue norms
$\| \phi \| _{\mathbf{L}^{q}}$ and
$\| \phi \| _{\mathbf{L}^{1,a}}=\| \langle \cdot \rangle ^{a}\phi \| _{
\mathbf{L}^{1}}$, where $a\geq 0$, $1\leq q\leq \infty $. Denote
\[
\quad G_{0}( t,x) =\partial _{y}G_{1}( t,x,y)
| _{y=0}=( 4\pi \alpha t) ^{-1/2}\frac{x}{
\alpha t}e^{-\frac{x^{2}}{4\alpha t}}.
\]

\begin{lemma}\label{L0}
Let $\phi \in \mathbf{L}^{r}(R^{+})$. Then
\[
\| \mathcal{G}_{0}( t) \phi \| _{\mathbf{L}
^{q}}\leq C\langle t\rangle ^{\frac{1}{2}( \frac{1}{q}-\frac{
1}{r}) }\| \phi \| _{\mathbf{L}^{r}},
\]
 for all $t>0$, $1\leq q\leq \infty $, $1\leq r\leq \infty $.
Furthermore we assume that $\phi \in \mathbf{L}^{1,1+a}$,
 then the estimate
\[
\| ( \cdot ) ^{b}( \mathcal{G}_{0}( t)
\phi -\vartheta G_{0}( t) ) \| _{\mathbf{L}
^{q}}\leq Ct^{-1+\frac{1}{2q}+\frac{b-a}{2}}\| \phi \| _{
\mathbf{L}^{1,1+a}}
\]
is valid for all $t>0$, where $1\leq q\leq \infty $,
$b\in [0,1+a ] $ and
\[
\vartheta =\int_{0}^{+\infty }x\phi ( x) dx.
\]
\end{lemma}

\begin{proof}
Since
\[
| G_{1}( t,x,y) | \leq Ct^{-1/2}e^{-\frac{C}{t}| x-y| ^{2}}
\]
for all $x,y\in \mathbb{R}^{+}$, by the Young inequality we have for $p=
\frac{qr+r-q}{qr}$
\begin{align*}
\| \mathcal{G}_{0}( t) \phi \| _{\mathbf{L}^{q}}
&\leq Ct^{-1/2}\| \int_{0}^{+\infty }e^{-\frac{C}{t}
| x-y| ^{2}}\phi ( y) dy\| _{\mathbf{L
}^{q}} \\
&\leq Ct^{-1/2}\| e^{-\frac{C}{t}| x|
^{2}}\| _{\mathbf{L}^{p}}\| \phi \| _{\mathbf{L}
^{r}}\leq C\langle t\rangle ^{\frac{1}{2} ( \frac{1}{q}-
\frac{1}{r}) }\| \phi \| _{\mathbf{L}^{r}}
\end{align*}
for all $t>0$, where $1\leq q\leq \infty $, Hence the first estimate of the
lemma follows. For the second estimate we write
\[
x^{b}( {\mathcal{G}}_{0}( t) \phi -\vartheta G_{0}(
t,x) ) =\int_{0}^{+\infty }x^{b}( G_{1}( t,x,y)
-G_{0}( t,x) y) \phi ( y) dy
\]
for any $b\in [ 0,1+a] $. Applying Taylor expansion, we obtain
\[
| G_{1}( t,x,y) -G_{0}( t,x) y|
\leq Ct^{-1-\frac{a}{2}}y^{1+a}( e^{-\frac{C}{t}|
x-y| ^{2}}+e^{-\frac{C}{t}| x| ^{2}})
\]
for all $x,y\in \mathbb{R}^{+}$. Hence in the domain $y\leq \frac{x}{2}$
\begin{align*}
x^{b}| G_{1}( t,x,y) -G_{0}( t,x)y|
&\leq Ct^{-1-\frac{a}{2}}y^{a+1}x^{b}e^{-\frac{C}{t}|
x| ^{2}} \\
&\leq Ct^{-1+\frac{b-a}{2}}y^{a+1}e^{-\frac{C}{t}| x|
^{2}}.
\end{align*}
By the Lagrange finite differences Theorem we have
\[
| G_{1}( t,x,y) | \leq Ct^{-\frac{1+\nu }{2}
}y^{\nu }e^{-\frac{C}{t}| x-y| ^{2}}
\]
for all $x,y\in \mathbb{R}^{+}$, where $\nu \in [ 0,1] $. Taking
$\nu =1+a-b$, in the case $b\in [ 1,a+1] $ we get for
$y\geq \frac{x}{2}$
\begin{align*}
 x^{b}| G_{1}( t,x,y) -G_{0}( t,x) y|
&\leq x^{b}( | G_{1}( t,x,y) |+| G_{0}( t,x) y| ) \\
&\leq Ct^{-1+\frac{b-a}{2}}x^{b}y^{a+1-b}e^{-\frac{C}{t}|
x-y| ^{2}}+Ct^{-\frac{3}{2}}x^{b+1}ye^{-\frac{C}{t}|
x| ^{2}} \\
&\leq Ct^{-1+\frac{b-a}{2}}y^{a+1}( e^{-\frac{C}{t}|
x-y| ^{2}}+e^{-\frac{C}{t}| x| ^{2}}) .
\end{align*}
In the case $b\in [ 0,1] $, we write
\begin{align*}
&x^{b}| G_{1}( t,x,y) -G_{0}( t,x)y|\\
&\leq x^{b}( | G_{1}( t,x,y) |
+| G_{0}( t,x) y| ) ^{b}|
G_{1}( t,x,y) -G_{0}( t,x) y| ^{1-b} \\
&\leq Ct^{-b}| y| ^{(1+a)b}t^{-( 1+\frac{a}{2}
) ( 1-b) }| y| ^{( a+1)( 1-b) }
 \big( e^{-\frac{C}{t}| x-y| ^{2}} +e^{-\frac{C}{
t}| x| ^{2}}\big) \\
&\leq Ct^{-1+\frac{b-a}{2}}y^{1+a}( e^{-\frac{C}{t}|
x-y| ^{2}}+e^{-\frac{C}{t}| x| ^{2}}) ,
\end{align*}
for all $x,y\in \mathbb{R}^{+}$, $y\geq \frac{x}{2}$. Thus we obtain the
estimate
\[
x^{b}| G_{1}( t,x,y) -G_{0}( t,x)y|
\leq Ct^{-1+\frac{b-a}{2}}| y| ^{a+1}\Big(
e^{-\frac{C}{t}| x-y| ^{2}}+e^{-\frac{C}{t}|
x| ^{2}}\Big)
\]
for all $x,y\in \mathbb{R}^{+}$, and for any $b\in [ 0,1+a] $.
Applying the above estimate with Young inequality we find
\begin{align*}
&\| ( \cdot ) ^{b}( \mathcal{G}_{0}( t)
\phi -\vartheta G_{0}( t) ) \| _{\mathbf{L}^{p}} \\
&= \| \int_{0}^{+\infty }x^{b}( G_{1}( t,x,y)
-G_{0}( t,x) y) \phi ( y) dy\| _{
\mathbf{L}_{x}^{q}} \\
&\leq Ct^{-1+\frac{b-a}{2}}\| \int_{0}^{+\infty }( e^{-\frac{C
}{t}| x-y| ^{2}}+e^{-\frac{C}{t}| x|
^{2}}) y^{1+a}| \phi ( y) | dy\| _{\mathbf{L}_{x}^{q}} \\
&\leq Ct^{-1+\frac{1}{2q}+\frac{b-a}{2}}\| \phi \| _{
\mathbf{L}^{1,1+a}}.
\end{align*}
Thus the second estimate  follows and the lemma is proved.
\end{proof}

Denote by $\mathcal{G}( t) $ the expression
\[
\mathcal{G}( t) \phi =e^{-\alpha t}\mathcal{L}_{x}^{-1} \big\{
e^{\frac{\alpha t}{1+| p| ^{2}}}( \hat{\phi}(p)-\hat{
\phi}(-p)) \big\} .
\]

\begin{lemma}\label{L1}
Suppose that the function $\phi \in \mathbf{L}^{\infty }(
\mathbb{R}^{+}) \cap \mathbf{L}^{1,1+a}( \mathbb{R}^{+}) $,
where $a\in ( 0,1) $. Then the estimates
\begin{gather*}
\| \mathcal{G}( t) \phi \| _{\mathbf{L}^{r}}\leq
e^{-at}\| \phi \| _{\mathbf{L}^{r}}+C\langle
t\rangle ^{-\frac{1}{2}( \frac{1}{r_{1}}-\frac{1}{r})
}\| \phi \| _{\mathbf{L}^{r_{1}}},
\\
\| \mathcal{G}( t) \phi -\vartheta G_{0}( t)
\| _{\mathbf{L}^{\infty }}\leq Ct^{-1-\frac{a}{2}}\| \phi
\| _{\mathbf{L}^{1,1+a}}+e^{-\alpha t}\| \phi \| _{
\mathbf{L}^{\infty }},
\\
\| (\cdot )^{b}( \mathcal{G}( t) \phi -\vartheta
G_{0}) \| _{\mathbf{L}^{1}}\leq Ct^{-\frac{1}{2}+\frac{b-a}{2}
}\| \phi \| _{\mathbf{L}^{1,1+a}}+e^{-\alpha t}\|
(\cdot )^{b}\phi \| _{\mathbf{L}^{1}}
\end{gather*}
are valid for all $t>0$, where $1\leq r\leq \infty $.
\end{lemma}

\begin{proof}
Note that the Green operator $\mathcal{G}( t) $ can be
represented as
\begin{equation}
\mathcal{G}( t) \phi =\mathcal{G}_{0}( t) \phi
+e^{-\alpha t}\phi (x)+\mathcal{R}( t) \phi ,  \label{2.6}
\end{equation}
where the remainder is
\[
\mathcal{R}( t) \phi =\int_{0}^{+\infty }(R( t,x-y)
-R( t,x+y) )\phi ( y) dy
\]
with  kernel
\[
R( t,x) =\frac{1}{2\pi i}\int_{-i\infty }^{i\infty }e^{px}
\widehat{R}( t,p) dp,
\]
where
$\widehat{R}( t,p) =e^{\frac{\alpha tp^{2}}{1-p^{2}}}-e^{\alpha
tp^{2}}-e^{-\alpha t}$.
 From Lemma \ref{L0} the operator $\mathcal{G}_{0}( t) $ satisfies
the estimates of the lemma . Also it is easily to see that the term
$e^{-\alpha t}\phi (x)$ satisfies the estimate of the lemma.

Now we estimate the remainder $\mathcal{R}( t) $. We represent
\[
\widehat{R}( t,p) =e^{\frac{\alpha tp^{2}}{1-p^{2}}}\big(
1-e^{-\alpha t\frac{p^{4}}{1-p^{2}}}\big) -e^{-\alpha t}
\]
for all $| p| \leq 1$, and
\[
\widehat{R}( t,p) =-e^{\alpha tp^{2}}+e^{-\alpha t}
\big( e^{\frac{\alpha t}{1-p^{2}}}-1\big)
\]
for all $| p| \geq 1$. Then we see that
\[
| \partial _{p}^{j}\widehat{R}( t,p) | \leq
C\langle t\rangle ^{\frac{j}{2}-1}e^{\frac{\alpha }{2}
tp^{2}}+C\langle t\rangle ^{2}e^{-\alpha t}( 1-p^{2})
^{-3}
\]
for all $\mathop{\rm Re}p=0$, $t>0$, $0\leq j\leq 4$. Therefore,
\begin{align*}
| R( t,x) |
&\leq C\langle x\langle
t\rangle ^{-1/2}\rangle ^{-4}\langle t\rangle
^{-\frac{1}{2}-1}+C\langle x\rangle ^{-4}\langle
t\rangle ^{2}e^{-\alpha t} \\
&\leq C\langle x\langle t\rangle ^{-\frac{1}{2}
}\rangle ^{-4}\langle t\rangle ^{-\frac{1}{2}-1}
\end{align*}
for all $x\in \mathbb{R}$, $t>0$. Applying this estimate by the Young
inequality we find
\[
\| \mathcal{R}( t) \phi \| _{\mathbf{L}^{r}}\leq
C\langle t\rangle ^{-\frac{1}{2}( \frac{1}{q}-\frac{1}{r}
) -1}\| \phi \| _{\mathbf{L}^{q}}
\]
for all $1\leq q\leq r\leq \infty $ and
\[
\| \mathcal{R}( t) \phi \| _{\mathbf{L}
^{1,w}}\leq C\langle t\rangle ^{-1}( \langle
t\rangle ^{\frac{w}{2}}\| \phi \| _{\mathbf{L}
^{1}}+\| \phi \| _{\mathbf{L}^{1,w}})
\]
for all $t>0$. Now by representation (\ref{2.6}) the estimates of the lemma
follow. Lemma \ref{L1} is proved.
\end{proof}

In the next lemma we estimate the Green operator in our basic norm
\[
\| \phi \| _{\mathbf{X}}=\sup_{t>0}( \langle
t\rangle \| \phi ( t) \| _{\mathbf{L}
^{\infty }}+\langle t\rangle ^{-a/2}\| \phi
( t) \| _{\mathbf{L}^{1,1+a}}) ,
\]
where $a\in ( 0,1) $. Note that the $\mathbf{L}^{1}$-norm is
estimated by the norm in $\mathbf{X}$,
\begin{align*}
\langle t\rangle ^{\frac{1}{2}}\| \phi ( t)
\| _{\mathbf{L}^{1}} &= \int_{0}^{\langle t\rangle
}| \phi ( t,x) | dx+\int_{\langle
t\rangle }^{+\infty }| 1+x| ^{-1-\alpha }|
x| ^{1+\alpha }| \phi ( t,x) | dx \\
&\leq C\langle t\rangle \| \phi ( t)
\| _{\mathbf{L}^{\infty }}+C\langle t\rangle ^{-\frac{a}{
2}}\| \phi ( t) \| _{\mathbf{L}^{1,a}}\leq
C\| \phi \| _{\mathbf{X}}.
\end{align*}
We define
\[
g(t)=1+\kappa \log \langle t\rangle
\]
with $\kappa >0$.

\begin{lemma}\label{L2}
Let the function $f( t,x) $ have a zero first moment
$\int_{0}^{+\infty }xf( t,x) dx=0$. Then the following inequality
\[
\| g(t)\int_{0}^{t}g^{-1}(\tau )\mathcal{G}( t-\tau )
\mathcal{B}f( \tau ) d\tau \| _{\mathbf{X}}\leq
C\| \langle t\rangle f\| _{\mathbf{X}}
\]
is valid, provided that the right-hand side is finite.
\end{lemma}

\begin{proof}
Since $g^{-1}(t)<C$ in view of Lemma \ref{L1} we get
\begin{align*}
&\| \int_{0}^{t}g^{-1}(\tau ){\mathcal{G}}(t-\tau )f( \tau
) d\tau \| _{\mathbf{L}^{\infty }}+\|
\int_{0}^{t}g^{-1}(\tau ){\mathcal{G}}(t-\tau )f( \tau ) d\tau
\| _{\mathbf{L}^{1,1+a}} \\
&\leq C\| \langle t\rangle f\| _{\mathbf{X}}
\end{align*}
for all $0\leq t\leq 4$. We now consider $t>4$. From definition of function
$g(t)$ we have
\begin{gather*}
Cg^{-1}(t)\geq \langle t\rangle ^{-a/4}, \\
\sup_{\tau \in [\sqrt{t},t] }g^{-1}(\tau )\leq C( 1+\frac{
\kappa }{2}\log \langle t\rangle ) \leq Cg^{-1}(t).
\end{gather*}
Therefore by  Lemma \ref{L1} and (\ref{2.4})-(\ref{2.5}) we
obtain
\begin{align*}
&\| \int_{0}^{t}{\mathcal{G}}( t-\tau ) \mathcal{B}f(
\tau ) d\tau \| _{\mathbf{L}^{\infty }} \\
&\leq C\int_{0}^{\sqrt{t}}( t-\tau ) ^{-1-\frac{a}{2}}(
\| \mathcal{B}f( \tau ) \| _{\mathbf{L}^{\infty
}}+\| \mathcal{B}f( \tau ) \| _{\mathbf{L}
^{1,1+a}}) d\tau \\
&\quad +Cg^{-1}(t)\int_{\sqrt{t}}^{t/2}( t-\tau ) ^{-1-\frac{a}{2
}}( \| \mathcal{B}f( \tau ) \| _{\mathbf{L}
^{\infty }}+\| \mathcal{B}f( \tau ) \| _{\mathbf{
L}^{1,1+a}}) d\tau \\
&\quad +Cg^{-1}(t)\int_{\frac{t}{2}}^{t}\| \mathcal{B}f( \tau )
\| _{\mathbf{L}^{\infty }}d\tau ,
\end{align*}
hence using the definition of the norm in $\mathbf{X}$, we get
\begin{align*}
&\| \int_{0}^{t}g^{-1}(\tau ){\mathcal{G}}( t-\tau )
\mathcal{B}f( \tau ) d\tau \| _{\mathbf{L}^{\infty }} \\
&\leq C\| \langle t\rangle f\| _{\mathbf{X}
}\int_{0}^{\sqrt{t}}( t-\tau ) ^{-1-\frac{a}{2}}\langle \tau
\rangle ^{\frac{a}{2}-1}d\tau \\
&\quad +C\| \langle t\rangle f\| _{\mathbf{X}
}g^{-1}(t)\int_{\sqrt{t}}^{t/2}( t-\tau ) ^{-1-\frac{a}{2}
}\langle \tau \rangle ^{\frac{a}{2}-1}d\tau
+C\| \langle t\rangle f\| _{\mathbf{X}
}g^{-1}(t)\int_{\frac{t}{2}}^{t}\langle \tau \rangle ^{-2}d\tau \\
&\leq C( t^{-1-\frac{a}{4}}+t^{-1}g^{-1}(t)) \|
\langle t\rangle f\| _{\mathbf{X}}\leq
Ct^{-1}g^{-1}(t)\| \langle t\rangle f\| _{
\mathbf{X}}
\end{align*}
and similarly
\begin{align*}
&\| \int_{0}^{t}g^{-1}(\tau ){\mathcal{G}}(t-\tau )f( \tau
) d\tau \| _{\mathbf{L}^{1,1+a}} \\
&\leq C\int_{0}^{t}g^{-1}(\tau )\| f( \tau ) \|
_{\mathbf{L}^{1,1+a}}d\tau \\
&\leq C\| \langle t\rangle f\| _{\mathbf{X}
}\int_{0}^{\sqrt{t}}\tau ^{\frac{a}{2}-1}d\tau +C\| \langle
t\rangle f\| _{\mathbf{X}}g^{-1}(t)\int_{\sqrt{t}}^{t}\tau ^{
\frac{a}{2}-1}d\tau \\
&\leq C\varepsilon ( t^{\frac{a}{4}}+g^{-1}t^{a/2})
\| \langle t\rangle f\| _{\mathbf{X}}\leq Ct^{
\frac{a}{2}}g^{-1}\| \langle t\rangle f\| _{
\mathbf{X}}
\end{align*}
for all $t>4$. Hence the result of the lemma follows. Lemma \ref{L2} is
proved.
\end{proof}

We now prove the local existence of weak solutions to the initial
boundary-value problem \eqref{1.1}.

\begin{proposition}\label{P1}
Let $u_{0}\in \mathbf{L}^{1,1+a}( \mathbb{R}^{+}) \cap
\mathbf{L}^{\infty }( \mathbb{R}^{+}) $,
$a\geq 0$. Then for some $T>0$, there exists a unique solution
$u\in \mathbf{C}( [ 0,T] ;\mathbf{L}^{1,1+a}( \mathbb{R}^{+}) \cap \mathbf{L}
^{\infty }( \mathbb{R}^{+}) ) $ to the problem \eqref{1.1}.
\end{proposition}

\begin{proof}
We apply the contraction mapping principle. We choose a functional space
\[
\mathbf{Z}=\big\{ \phi \in \mathbf{C}( [ 0,T] ;\mathbf{L}
^{1,1+a}( \mathbb{R}^{+}) \cap \mathbf{L}^{\infty }( \mathbf{
R}^{+}) ) :\quad | | u|| _{\mathbf{Z}}<\infty \big\} ,
\]
with the  norm
\[
\| u\| _{\mathbf{Z}}=\sup_{t\in [ 0,T]}(
\| u( t) \| _{\mathbf{L}^{1,1+a}}+\|
u( t) \| _{\mathbf{L}^{\infty }}) .
\]
For $v\in \mathbf{Z}$ we define the mapping $\mathcal{M}(v)$ by
\begin{equation}
\mathcal{M}(v)=\mathcal{G}( t) u_{0}+\lambda \int_{0}^{t}d\tau
\mathcal{G}( t-\tau ) \mathcal{B}| v| ^{\sigma
}v( \tau ) .  \label{2.15}
\end{equation}
Suppose that the norm $\| v\| _{\mathbf{Z}}<\delta $. We
first prove that
\[
\| \mathcal{M}( v) \| _{\mathbf{Z}}<\delta .
\]
Applying the Lemma\ref{L1}\ we have
\begin{equation}
\begin{aligned}
\| \mathcal{M}( v) \| _{\mathbf{L}^{\infty }}
&\leq \| \mathcal{G}( t) u_{0}\| _{\mathbf{L}^{\infty }}
+| \lambda | \int_{0}^{t}\| \mathcal{G} ( t-\tau ) \mathcal{B}| v|
v( \tau) \| _{\mathbf{L}^{\infty }}d\tau   \\
&\leq C\| u_{0}\| _{\mathbf{L}^{\infty}}
 +C\int_{0}^{t}\| \mathcal{B}| v| v( \tau) \| _{\mathbf{L}^{\infty }}d\tau .
\end{aligned}\label{2.16}
\end{equation}
Similarly we obtain the estimate
\begin{equation}
\begin{aligned}
\| \mathcal{M}( v) \| _{\mathbf{L}^{1,1+a}}
&\leq \| \mathcal{G}( t) u_{0}-\vartheta_{1}G_{0}\| _{\mathbf{L}^{1,1+a}}
 +\| G_{0}\| _{\mathbf{L}^{1,1+a}}\|u_{0}\| _{\mathbf{L}^{1}}
+e^{-\alpha t}\| u_{0}\|_{\mathbf{L}^{1,1+a}}   \\
&\quad +| \lambda | \int_{0}^{t}\| \mathcal{G}(t-\tau ) \mathcal{B}| v|
 v( \tau )-\vartheta _{2}G_{0}\| _{\mathbf{L}^{1,1+a}}d\tau   \\
&\quad +| \lambda | \int_{0}^{t}\| F(t-\tau)\| _{\mathbf{L}^{1,1+a}}
 \| \mathcal{B}|v| v( \tau ) \| _{\mathbf{L}^{1}}d\tau
\\
&\quad +| \lambda | \int_{0}^{t}e^{-\alpha (t-\tau)}\| \mathcal{B}| v|
 v( \tau ) \| _{\mathbf{L}^{1,1+a}}d\tau   \\
&\leq C\| u_{0}\| _{\mathbf{L}^{1,1+a}}+C\int_{0}^{t}
\| \mathcal{B}| v| v( \tau ) \| _{\mathbf{L}^{1,1+a}}d\tau .
\end{aligned}\label{2.17}
\end{equation}
Using the estimate
\[
\| \mathcal{B}| v| v( \tau )
\| _{\mathbf{Z}}\leq \| v( \tau ) \| _{
\mathbf{L}^{\infty }}\| v( \tau ) \| _{\mathbf{Z}
}\leq \delta ^{2},
\]
 From (\ref{2.16}) and (\ref{2.17}), we obtain
\[
\| \mathcal{M}( v) \| _{\mathbf{Z}}\leq
C\| u_{0}\| _{\mathbf{Z}}+CT\delta ^{2}\leq \delta
\]
if we choose $\delta \geq 2C\| u_{0}\| _{\mathbf{Z}}$ and
$T>0$ sufficiently small. Therefore, the mapping $\mathcal{M}$ transforms a
ball of a radius $\delta >0$ into itself in the space $\mathbf{Z}$. We now
estimate the difference
\[
\| \mathcal{M}(v_{1})-\mathcal{M}(v_{2})\| _{\mathbf{Z}
}\leq \frac{1}{2}\| v_{1}-v_{2}\| _{\mathbf{Z}}.
\]
We have
\[
\| \mathcal{M}(v_{1})-\mathcal{M}(v_{2})\| _{\mathbf{Z}}
\leq C\int_{0}^{T}| | \mathcal{B}( |
v_{1}| v_{1}( \tau ) -| v_{2}|
v_{2}( \tau ) ) | | _{\mathbf{Z}
}\langle t-\tau \rangle ^{a/2}d\tau .
\]
Since
\[
| | \mathcal{B}( | v_{1}|
v_{1}( \tau ) -| v_{2}| v_{2}( \tau
) ) | | _{\mathbf{Z}}\leq C\delta |
| v_{1}-v_{2}| | _{\mathbf{Z}},
\]
it follows that
\[
\| \mathcal{M}(v_{1})-\mathcal{M}(v_{2})\| _{\mathbf{Z}
}\leq CT\delta \| v_{1}-v_{2}\| _{\mathbf{Z}}\leq \frac{1}{2
}\| v_{1}-v_{2}\| _{\mathbf{Z}}
\]
if we choose $T>0$ sufficiently small. Thus $\mathcal{M}$ is a contraction
mapping, therefore there exists a unique solution $u( t,x) \in
\mathbf{C}([ 0,T] ;\mathbf{L}^{1,1+a}( \mathbb{R}
^{+}) \cap \mathbf{L}^{\infty }( \mathbb{R}^{+}) ) $
to the problem \eqref{1.1}. Proposition \ref{P1} is proved.
\end{proof}

\section{\label{SS4.1}Proof of Theorem \ref{T1} (Small data)}

 As in reference \cite{hkn4} by making a change of the dependent variable
$u( t,x) =v( t,x) e^{-\varphi ( t) }$, for the new function $v( t,x) $
we get the  equation
\[
\partial _{t}( v-v_{xx}) -\alpha v_{xx}-\lambda e^{-\varphi
}| v| v-\varphi 'v=0.
\]
We assume that $\varphi ( t) $ is such that $\varphi (0) =1$ and
\[
\int_{0}^{+\infty }x( \lambda e^{-\varphi }| v|
v+\varphi 'v) dx=0.
\]
Since by construction
\[
\int_{0}^{+\infty }u(x)dx=0
\]
the first moment of new function $v( t,x) $ satisfies a
conservation law:
\[
\frac{d}{dt}\int_{0}^{+\infty }xv( t,x) dx=0,
\]
hence $\int_{\mathbf{0}}^{+\infty }xv( t,x) dx=\int_{\mathbf{0}
}^{+\infty }xu_{0}( x) dx$ for all $t>0$. Thus we consider the
initial-boundary value problem for the new dependent variables
$(v( t,x) ,\varphi ( t) ) $,
\begin{equation}
\begin{gathered}
\partial _{t}( v-\Delta v) -\alpha \Delta v=\lambda e^{-\varphi
}\big( | v| -\frac{1}{\theta }\int_{0}^{+\infty
}x| v| vdx\big) v, \\
\partial _{t}\varphi ( t) =-\frac{\lambda }{\theta }e^{-\varphi
}\int_{0}^{+\infty }x| v| vdx, \\
v( 0,x) =u_{0}( x) ,\quad v( t,0) =0,\quad \varphi (0)=0.
\end{gathered}  \label{4.2}
\end{equation}
We denote $h( t) =e^{\varphi ( t) }$ and write (\ref{4.2}) as
\begin{equation}
\begin{gathered}
\partial _{t}( v-v_{xx}) -\alpha v_{xx}=f( v,h) \\
v( 0,x) =u_{0}( x) ,\quad v( t,0) =0 \\
\partial _{t}h=-\frac{\lambda }{\theta }\int_{0}^{+\infty }x|
v| vdx,\quad h(0)=1,
\end{gathered}  \label{4.3}
\end{equation}
where
\[
f( v,h) =\lambda h^{-1}\Big( | v| -\frac{1}{
\theta }\int_{0}^{+\infty }x| v| vdx\Big) v.
\]
We note that the first moment of the nonlinearity is
\[
\int_{0}^{+\infty }xf( v,h) ( t,x) dx=0
\]
for all $t>0$. We now prove the existence of the solution $( v(
t,x) ,h( t) ) $ for the problem (\ref{4.3}) by the
successive approximations $( v_{m}( t,x) ,h_{m}(
t) ) $, $m=1,2,\dots$, defined as follows
\begin{equation}
\begin{gathered}
\partial _{t}( v_{m}-\partial _{x}^{2}v_{m}) -\alpha \partial
_{x}^{2}v_{m}=f( v_{m-1},h_{m-1}) , \\
\partial _{t}h_{m}=-\frac{\lambda }{\theta }\int_{0}^{+\infty }|
v_{m-1}| v_{m-1}dx,\quad  \\
v_{m}( 0,x) =u_{0}( x) ,\quad v_{m}( t,0) =0, \quad h_{m}(0)=1,
\end{gathered} \label{4.4}
\end{equation}
for all $m\geq 2$, where
$v_{1}={\mathcal{G}}( t) u_{0}$, $h_{1}=g( t)$,
$g( t) =1+| \theta | \eta \log \langle t\rangle $.

We now prove by induction the following estimates
\begin{equation}
\begin{gathered}
\| v_{m}\| _{\mathbf{X}} \leq C\varepsilon ,\\
\| v_{m}( t) -{\mathcal{G}}( t) u_{0}\| _{\mathbf{L}^{1,1}}
 \leq C\varepsilon ^{2}g^{-1}(t) ,   \\
| h_{m}( t) -g( t) | \leq C\varepsilon ( 1+\log g( t) )
\end{gathered}  \label{4.5}
\end{equation}
for all $m\geq 1$, the norm $\| \cdot \| _{\mathbf{X}}$ is
defined as above by
\[
\| \phi \| _{\mathbf{X}}=\sup_{t>0}( \langle
t\rangle \| \phi ( t) \| _{\mathbf{L}
^{\infty }}+\langle t\rangle ^{-a/2}\| \phi
( t) \| _{\mathbf{L}^{1,1+a}}) .
\]
By virtue of Lemma \ref{L1} we have
\begin{gather*}
\| {\mathcal{G}}( t) u_{0}\| _{\mathbf{L}
^{\infty }}\leq C\varepsilon \langle t\rangle ^{-1},\\
\| {\mathcal{G}}( t) u_{0}\| _{\mathbf{L}
^{1,1}}\leq C\varepsilon , \\
\| | \cdot | ^{a}( {\mathcal{G}}(
t) u_{0}-\theta G_{0}( t,x) ) \| _{\mathbf{L}
^{1,1}}\leq C\varepsilon,
\\
\| | \cdot | ^{a}G_{0}( t,x)
\| _{\mathbf{L}^{1,1}}\leq Ct^{a/2}.
\end{gather*}
Therefore, estimates (\ref{4.5}) are valid for $m=1$. We assume that
estimates (\ref{4.5}) are true with $m$ replaced by $m-1$. The integral
equations associated with (\ref{4.4}) are written as
\begin{gather*}
v_{m}( t) ={\mathcal{G}}( t) u_{0}+\int_{0}^{t}{
\mathcal{G}}(t-\tau )\mathcal{B}f( v_{m-1}( \tau )
,h_{m-1}( \tau ) ) d\tau , \\
h_{m}( t) =1-\frac{\lambda }{\theta }\int_{0}^{t}d\tau
\int_{0}^{+\infty }| v_{m-1}| v_{m-1}dx.
\end{gather*}
We have
\[
f( v,h) =\lambda h^{-1}\Big( | v| -\frac{1}{
\theta }\int_{0}^{+\infty }x| v| vdx\Big) v.
\]
\begin{equation}
\begin{aligned}
\| f( v_{m-1}( t) ,h_{m-1}( t) )\| _{\mathbf{L}^{\infty }}
&\leq Ch_{m-1}^{-1}( t) \| v_{m-1}( t)
\| _{\mathbf{L}^{\infty }}^{1+1}\Big( 1+\frac{1}{| \theta
| }\| v_{m-1}( t) \| _{\mathbf{L}
^{1,1}}\Big)   \\
&\leq C\varepsilon ^{2}\langle t\rangle ^{-2}g^{-1}(t)
\end{aligned}\label{4.6}
\end{equation}
and
\begin{equation}
\begin{aligned}
&\| f( v_{m-1}( t) ,h_{m-1}( t) )\| _{\mathbf{L}^{1,1+a}}   \\
&\leq Ch_{m-1}^{-1}( t) \| v_{m-1}( t)
\| _{\mathbf{L}^{\infty }}\| v_{m-1}( t)
\| _{\mathbf{L}^{1,1+a}}\Big( 1+\frac{1}{| \theta
| }\| v_{m-1}( t) \| _{\mathbf{L} ^{1,1}}\Big)   \\
&\leq C\varepsilon ^{2}\langle t\rangle ^{-1+\frac{a}{2}
}g^{-1}( t)
\end{aligned}\label{4.7}
\end{equation}
for all $t>0$, provided that $( v_{m-1}( t) ,h_{m-1}(t) ) $ satisfies
(\ref{4.5}). This yields the estimate
\[
\| \langle t\rangle g( t) f(
v_{m-1}( t) ,h_{m-1}( t) ) \| _{\mathbf{
X}}\leq C\varepsilon ^{2}.
\]
Since $f( v_{m-1}( \tau ) ,h_{m-1}( \tau )
) $ have the zero first moment we get via Lemma \ref{L2}
\[
\| g( t) \int_{0}^{t}g^{-1}( \tau ) {\mathcal{G
}}(t-\tau )\mathcal{B}f( v_{m-1}( \tau ) ,h_{m-1}( \tau
) ) d\tau \| _{\mathbf{X}}\leq C\varepsilon ^{2}
\]
hence it follows that
\begin{equation}
\| v_{m}\| _{\mathbf{X}}\leq C\varepsilon ,\|
v_{m}( t) -{\mathcal{G}}( t) u_{0}\| _{
\mathbf{L}^{1,1}}\leq C\varepsilon ^{2}g^{-1}( t) .  \label{4.7a}
\end{equation}

To prove the third estimate in (\ref{4.5}) we need the following lemma,
where we evaluate the large time behavior of the first moment of the
nonlinearity in equation \eqref{1.1} in the critical case. As above we take
$\theta =\int_{0}^{+\infty }xu_{0}( x) dx$.

\begin{lemma}\label{L3}
Assume that $u_{0}\in \mathbf{L}^{\infty }( \mathbb{R}
^{+}) \cap \mathbf{L}^{1,1+a}( \mathbb{R}^{+}) $, that
$\| u_{0}\| _{\mathbf{L}^{\infty }}+\|
u_{0}\| _{\mathbf{L}^{1,1+a}}=\varepsilon $ is sufficiently small,
and that $\theta \lambda \leq -C\varepsilon <0$. Let a function
$v(t,x) $ satisfy the estimates
\begin{gather*}
\langle t\rangle \| v\| _{\mathbf{L}^{\infty
}}+\| v\| _{\mathbf{L}^{1,1}}\leq C\varepsilon ,\\
\| v( t) -{\mathcal{G}}( t) u_{0}\| _{
\mathbf{L}^{1,1}}\leq C\varepsilon ^{2}g^{-1}( t)
\end{gather*}
for all $t>0$. Then the inequality
\begin{equation}
\big| 1-\frac{\lambda }{\theta }\int_{0}^{t}d\tau \int_{0}^{+\infty
}x| v| v( \tau ,x) dx-g( t)
\big| \leq C\varepsilon ( 1+\log g( t) )
\label{2.9}
\end{equation}
is valid for all $t>0$.
\end{lemma}

\begin{proof}
In view of the condition $\| v\| _{\mathbf{L}^{\infty }}$
$+\| v\| _{\mathbf{L}^{1,1}}\leq C\varepsilon $ we get
\[
| \frac{\lambda }{\theta }\int_{0}^{t}d\tau \int_{0}^{+\infty
}x| v| v( \tau ,x) dx| \leq
C\varepsilon t,
\]
hence estimate (\ref{2.9}) is true for all $0<t<1$.

We now consider the case $t\geq 1$. By the last estimate of Lemma \ref{L1}
we get
\[
\| x({\mathcal{G}}( t) u_{0}-\theta G_{0}(
t,x) )\| _{\mathbf{L}^{1}}\leq C\varepsilon t^{-a/2}.
\]
Hence we find
\begin{align*}
&\| x(| v| v-| \theta |
\theta ( G_{0}( t,x) ) ^{1+1})\| _{\mathbf{L}
^{1}} \\
&\leq C( \| x(v( t) -{\mathcal{G}}( t)
u_{0})\| _{\mathbf{L}^{1}}+\| x({\mathcal{G}}(
t) u_{0}-\theta G_{0}( t,x) )\| _{\mathbf{L}
^{1}}) \\
&\quad \times ( \| v\| _{\mathbf{L}^{\infty }}+\| {
\mathcal{G}}( t) u_{0}\| _{\mathbf{L}^{\infty
}}+| \theta | \| G_{0}( t) \|
_{\mathbf{L}^{\infty }}) \\
&\leq C\varepsilon ^{2}t^{-1}( \varepsilon g^{-1}( t) +t^{-
\frac{a}{2}})
\end{align*}
for all $t\geq 1$. Since
\[
t\int_{0}^{+\infty }x( G_{0}( t,x) ) ^{2}dx=\frac{\eta
}{| \lambda | },
\]
it follows that
\begin{align*}
\big| \int_{0}^{+\infty }x| v| v( t,x)
dx-| \theta | \theta t^{-1}\frac{\eta }{|\lambda | }\big|
&\leq C\| x(| v| v-| \theta |
\theta ( G_{0}( t,x) ) ^{2})\| _{\mathbf{L}
^{1}} \\
&\leq C\varepsilon ^{2}t^{-1}( \varepsilon g^{-1}( t) +t^{-
\frac{a}{2}})
\end{align*}
for all $t\geq 1$. Therefore,
\begin{align*}
&| \frac{| \lambda | }{\theta }
\int_{1}^{t}d\tau \int_{0}^{+\infty }x| v| v( \tau
,x) dx-| \theta | \eta \log t| \\
&\leq \int_{1}^{t}\frac{C\varepsilon ^{2}d\tau }{\tau ( 1+|
\theta | \eta \log (1+\tau )) }+C\varepsilon
\int_{1}^{t}\tau ^{-1-\frac{a}{2}}d\tau \\
&\leq C\varepsilon ( 1+\log g( t) )
\end{align*}
for all $t\geq 1$. Thus we obtain (\ref{2.9}) and complete the proof.
\end{proof}

By virtue of (\ref{4.7a}) and applying Lemma \ref{L3} we find that
\[
| h_{m}( t) -g( t) | \leq C\varepsilon ( 1+\log g( t) )
\]
for all $t>0$. Thus by induction we see that estimates (\ref{4.5}) are valid
for all $m\geq 1$. In the same way by induction we can prove that
\begin{gather*}
\| v_{m}-v_{m-1}\| _{\mathbf{X}}\leq \frac{1}{4}
\| v_{m-1}-v_{m-2}\| _{\mathbf{X}},  \\
\begin{aligned}
&\sup_{t>0}g^{-1}( t) | h_{m}( t)-h_{m-1}( t) |\\
&\leq \frac{1}{4}\| v_{m-1}-v_{m-2}\| _{\mathbf{X}}
 +\frac{1}{4}\sup_{t>0}g^{-1}( t) | h_{m-1}(t) -h_{m-2}( t) |
\end{aligned}
\end{gather*}
for all $m>2$. Therefore taking the limit $m\to \infty $, we obtain
a unique solution $\lim_{m\to \infty }v_{m}( t,x)
=v( t,x) \in \mathbf{X}$, $\lim_{m\to \infty }h_{m}(
t) =h( t) =e^{\varphi ( t) }\in \mathbf{C}(
0,\infty ) $ satisfying the equalities
\begin{equation}
\begin{gathered}
v( t) ={\mathcal{G}}( t) u_{0}+\int_{0}^{t}{\mathcal{G}
}(t-\tau )\mathcal{B}f( v( \tau ) ,h( \tau )
) d\tau , \\
h( t) =1-\frac{\lambda }{\theta }\int_{0}^{t}d\tau
\int_{0}^{+\infty }x| v| vdx,
\end{gathered}  \label{4.8}
\end{equation}
and the estimates
\begin{equation} \label{4.9}
\begin{gathered}
\| v( t) -{\mathcal{G}}( t) u_{0}\| _{
\mathbf{L}^{1,1}} \leq C\varepsilon ^{2}g^{-1}( t) ,  \\
| h_{m}( t) -g( t) | \leq C\varepsilon ( 1+\log g( t) ) .
\end{gathered}
\end{equation}
Applying (\ref{4.6}) and (\ref{4.7}) to (\ref{4.8}), we have
\begin{equation}
\| v( t) -{\mathcal{G}}( t) u_{0}\| _{
\mathbf{L}^{\infty }}\leq C\varepsilon ^{2}\langle t\rangle
^{-1}g^{-1}( t) .  \label{4.10}
\end{equation}
Then via formulas $u( t,x) =e^{-\varphi ( t) }v(
t,x) =h^{-1}( t) v( t,x) $ we find the estimates
\begin{equation}
\begin{aligned}
&\| u( t) -\theta G_{0}( t,x) e^{-\varphi
( t) }\| _{\mathbf{L}^{\infty }}   \\
&\leq \| u( t) -( {\mathcal{G}}( t)
u_{0}) e^{-\varphi ( t) }\| _{\mathbf{L}^{\infty
}}+\| ( \mathcal{G}( t) u_{0}-\theta G_{0}(
t,x) ) e^{-\varphi ( t) }\| _{\mathbf{L}
^{\infty }}   \\
&\leq  C\varepsilon ^{2}\langle t\rangle ^{-1}g^{-1}(t) ,
\end{aligned} \label{4.11}
\end{equation}
where we used the estimate
\[
\| ( \mathcal{G}( t) u_{0}-\theta G_{0}(
t,x) ) e^{-\varphi ( t) }\| _{\mathbf{L}
^{\infty }}\leq Ct^{-1-\frac{a}{2}}\| u_{0}\| _{\mathbf{L}
^{1,1+a}}
\]
and (\ref{4.10}). By (\ref{4.9}), we have
\[
\| \theta G_{0}( t) h^{-1}( t) -\theta
G_{0}( t,x) g^{-1}( t) \| _{\mathbf{L}^{\infty }}
\leq C\varepsilon t^{-1}g^{-2}( t) | h( t) -g( t) | ,
\]
hence via (\ref{4.11}) it follows that
\begin{equation}
\| u( t) -\theta G_{0}( t,x) g^{-1}(
t) \| _{\mathbf{L}^{\infty }}\leq C\varepsilon
^{2}\langle t\rangle ^{-1}g^{-2}( t) .  \label{4.12}
\end{equation}
This completes the proof of Theorem \ref{T1}.

\section{\label{SS4.2}Proof of Theorem \ref{T2} (Large data)}

Before proving Theorem \ref{T2} we need a lemma, where we compare the
solutions of the following two problems
\begin{equation}
\begin{gathered}
u_{t}-u_{xx}+| u| u=f,\quad x\in \mathbb{R}^{+},\; t>0, \\
u(0,x)=u_{0}(x), \quad u(t,0)=0,\quad x\in \mathbb{R}^{+},\; t>0
\end{gathered}   \label{4.13}
\end{equation}
and
\begin{equation}
\begin{gathered}
v_{t}-v_{xx}+\epsilon v^{2}=| f| ,\quad x\in \mathbb{R} ^{+},\; t>0, \\
v(0,x)=| u_{0}(x)| ,\quad v(t,0)=0,\quad x\in \mathbb{R}^{+},\; t>0
\end{gathered}  \label{4.14}
\end{equation}

\begin{lemma} \label{L10}
Suppose that $u_{0}\in \mathbf{L}^{\infty }( \mathbb{R}
^{+}) \cap \mathbf{C}^{0}( \mathbb{R}^{+}) $, and
$0\leq \epsilon \leq 1$. Then
$| u( t,x) | \leq v( t,x) $ for all $t\geq 0$, $x\in \mathbb{R}^{+}$.
\end{lemma}

\begin{proof}
Define $r=v-u$. Then we obtain
\begin{equation}
\begin{gathered}
r_{t}-r_{xx}+\epsilon v^{2}-| u| u=|f| -f,\quad x\in \mathbb{R}^{+},\; t>0, \\
r(0,x)=| u_{0}(x)| -u_{0}(x),\quad r(t,0)=0.
\end{gathered}  \label{4.15}
\end{equation}
We need to prove that $r\geq 0$ for all $t\geq 0$, $x\in \mathbb{R}^{+}$.
Define $\ R( t) \equiv \inf_{x\in \mathbb{R}^{+}}r(t,x) $.
On the contrary, suppose that there exists a time  $T>0$ such
that $R( T) <0$. By the continuity we can find an interval
$[T_{1},T] $ such that $R( t) \leq 0$ for all $t\in [T_{1},T] $ and
$R( T_{1}) =0$. By \cite[Theorem 2.1]{Const} there exists
a point $\zeta ( t) \in \mathbb{R}^{+}$
 such that $R( t) =r( t,\zeta ( t) ) $,
moreover $R'( t) =\frac{d}{dt}r( t,\zeta (t) ) $ almost everywhere
on $t\in [ T_{1},T] $.  We have
\[
| u| u-\epsilon v^{2}=( v-R) ^{2}-\epsilon v^{2}\geq 0
\]
for all $t\in [ T_{1},T] $. For the Laplacian $\partial _{x}^{2}$
at the point of maximum $\zeta ( t) $ we  have
\[
-\partial _{x}^{2}r(t,\zeta ( t) )\leq 0.
\]
Therefore by equation (\ref{4.15}) we get $R'( t) \geq 0$
for all $t\in [ T_{1},T] $. Integration with respect to time
yields $R( t) \geq 0$. This gives a contradiction, hence
$u(t,x) \leq | v( t,x) | $ for all $x\in \mathbb{R}^{+}$ and $t>T_{1}$.
In the same manner we prove that $v+u\geq 0$
for all $x\in \mathbb{R}^{n}$ and $t>T_{1}$. Lemma \ref{L10} is proved.
\end{proof}

\begin{lemma}\label{OPT} Let
$u_{0}\in \mathbf{W}_{\infty }^{2}( \mathbb{R}
^{+}) \cap \mathbf{W}_{1}^{2}( \mathbb{R}^{+}) $. Then we
have the following estimate for solution of initial-boundary value problem
\eqref{1.1}:
\[
\| u\| _{\mathbf{L}^{2}}\leq C\langle t\rangle ^{-3/4},\quad
\| xu\| _{\mathbf{L}^{2}}\leq C\langle t\rangle ^{-1/4}
\]
\end{lemma}

\begin{proof}
Multiplying equation \eqref{1.1} by $2u$ and integrating with respect to
$x\in \mathbb{R}^{+}$ we get
\[
\frac{d}{dt}( \| u( t) \| _{\mathbf{L}
^{2}}^{2}+\| u_{x}( t) \| _{\mathbf{L}
^{2}}^{2}) +2\alpha \| u_{x}( t) \| _{
\mathbf{L}^{2}}^{2}=2\lambda \| u( t) \| _{
\mathbf{L}^{\sigma +2}}^{\sigma +2},
\]
hence integrating we see that
\begin{align*}
&\| u( t) \| _{\mathbf{L}^{2}}^{2}+\|
u_{x}( t) \| _{\mathbf{L}^{2}}^{2}+2\alpha
\int_{0}^{t}\| u_{x}( \tau ) \| _{\mathbf{L}
^{2}}^{2}d\tau -2\lambda \int_{0}^{t}\| u( \tau )
\| _{\mathbf{L}^{\sigma +2}}^{\sigma +2}d\tau \\
&\leq \| u_{0}\| _{\mathbf{L}^{2}}^{2}+\|
u_{0x}\| _{\mathbf{L}^{2}}^{2}\\
&=\| u_{0}\| _{\mathbf{H}^{1}}^{2}
\end{align*}
for all $t\geq 0$. In particular we have
\begin{gather}
\| u\| _{\infty ,2}\equiv \sup_{t\geq 0}\| u(
t) \| _{\mathbf{L}^{2}}\leq \| u_{0}\| _{\mathbf{H}^{1}},  \label{3.19}
\\
\| u\| _{3,3}\equiv \| \| u(
t,x) \| _{\mathbf{L}_{x}^{3}}\| _{\mathbf{L}
_{t}^{3}( 0,\infty ) }\leq C\| u_{0}\| _{\mathbf{
H}^{1}}  \label{3.20}
\end{gather}
 From equation \eqref{1.1} we have
\[
u_{xx}(t,0)=0.
\]
Differentiating  \eqref{1.1}, multiplying by $2u_{x}$ and integrating
with respect to $x\in \mathbb{R}^{+}$ we get
\[
\frac{d}{dt}( \| u_{x}( t) \| _{\mathbf{L}
^{2}}^{2}+\| u_{xx}( t) \| _{\mathbf{L}
^{2}}^{2}) +2\alpha \| u_{xx}( t) \| _{
\mathbf{L}^{2}}^{2}=2\lambda \int_{0}^{+\infty }| u|
u_{x}^{2}dx,
\]
hence integrating we see that
\begin{align*}
&\| u( t) \| _{\mathbf{L}^{2}}^{2}+\|
u_{x}( t) \| _{\mathbf{L}^{2}}^{2}+2\alpha
\int_{0}^{t}\| u_{x}( \tau ) \| _{\mathbf{L}
^{2}}^{2}d\tau -2\lambda \int_{0}^{+\infty }| u|
u_{x}^{2}dx \\
&\leq \| u_{0x}\| _{\mathbf{L}^{2}}^{2}+\|
u_{0xx}\| _{\mathbf{L}^{2}}^{2}=\| u_{0}\| _{
\mathbf{H}^{2}}^{2}
\end{align*}
for all $t\geq 0$. In particular we obtain that the solution
$u(t,x)\in \mathbf{C}( [ 0,+\infty ) ,
\mathbf{C}^{1}(\mathbb{R} ^{+})) $.  Now we prove the estimates
\begin{equation}
\| u( t) \| _{\mathbf{L}^{2}}\leq C\langle t\rangle ^{-3/4}  \label{e1.2}
\end{equation}
for all $t>0$.  Denote $\Theta ( x) =1$ for all $x>0$ and
$\Theta ( x) =-1$ for all $x<0;$ $\Theta ( 0) =0$. We
multiply equation \eqref{1.1} by $x\Theta ( u( t,x) ) $
and integrate with respect to $x$ over $\mathbb{R}^{+}$ to get
\begin{equation} \label{1}
\begin{aligned}
&\partial _{t}\Big( \int_{0}^{+\infty }xu( t,x) \text{$\Theta $}
( u( t,x) ) dx-\int_{0}^{+\infty }xu_{xx}(
t,x) \text{$\Theta $}( u( t,x) ) dx\Big)\\
&-\alpha \int_{0}^{+\infty }xu_{xx}( t,x) \text{$\Theta $}(
u( t,x) ) dx   \\
&= \lambda \int_{0}^{+\infty }x| u| ^{2}dx.
\end{aligned}
\end{equation}
Since $u(t,x)\in \mathbf{C}([ 0,+\infty ) ,\mathbf{C}^{1}(\mathbb{R}^{+})) $
we get
\begin{equation}
\int_{0}^{+\infty }xu_{xx}( t,x) \text{$\Theta $}( u(
t,x) ) dx=0.  \label{3}
\end{equation}
Also we have
\begin{equation} \label{2}
\begin{gathered}
\int_{0}^{+\infty }\partial _{t}xu( t,x) \Theta (u( t,x) ) dx
= \frac{d}{dt}\| xu( t)\| _{\mathbf{L}^{1}},   \\
\lambda \int_{0}^{+\infty }x| u| ^{2}dx \leq 0.
\end{gathered}
\end{equation}
Therefore by  (\ref{1}), (\ref{2}) and (\ref{3}) we find
\begin{equation}
\frac{d}{dt}\| xu\| _{\mathbf{L}^{1}}\leq 0.  \label{3.91}
\end{equation}
Integration of inequality (\ref{3.91}) yields
\begin{equation}
\| xu\| _{\mathbf{L}^{1}}\leq C.  \label{estL1}
\end{equation}
Now we can prove the estimate (\ref{e1.2}) for all $t>0$. Indeed,
using (\ref{estL1}) in particular, we find
\begin{equation}
\sup_{\xi \in \mathbb{R}}| \widehat{u}_{\xi }( t,\xi )
| \leq C\| xu( t) \| _{\mathbf{L}
^{1}}\leq C.  \label{h2}
\end{equation}
We now multiply equation \eqref{1.1} by $2u$, then integrating with respect
to $x\in \mathbb{R}^{+}$ we get
\begin{equation}
\frac{d}{dt}\Big( \| u( t) \| _{\mathbf{L}
^{2}}^{2}+\| u_{x}( t) \| _{\mathbf{L}
^{2}}^{2}\Big) =-2\alpha \| u_{x}( t) \| _{
\mathbf{L}^{2}}^{2}+\lambda \| u( t) \| _{
\mathbf{L}^{3}}^{3}.  \label{3.11}
\end{equation}
By construction we have
\[
\widehat{u}(t,0)=\int_{0}^{+\infty }u(t,x)dx=0
\]
By the Plansherel theorem using the Fourier splitting method due
to \cite{schonbek2}, we have
\begin{align*}
\| u_{x}( t) \| _{\mathbf{L}^{2}}^{2}
&= \| \xi \widehat{u}( t) \| _{\mathbf{L}
^{2}}^{2} \\
&\geq \int_{| \xi | \geq \delta }| \widehat{u}
( t,\xi ) | ^{2}| \xi | ^{2}d\xi
\geq \delta ^{2}\int_{| \xi | \geq \delta }|
\widehat{u}( t,\xi ) | ^{2}d\xi \\
&= \delta ^{2}\int_{-\infty }^{+\infty }| \widehat{u}( t,\xi
) | ^{2}d\xi -\delta ^{2}\int_{| \xi |
<\delta }| \widehat{u}( t,\xi ) -\widehat{u}
(t,0)| ^{2}d\xi \\
&= \delta ^{2}\int_{0}^{+\infty }| \widehat{u}( t,\xi )
| ^{2}d\xi -\delta ^{2}\int_{| \xi | <\delta
}| \int_{0}^{\xi }\widehat{u}_{\xi _{1}}( t,\xi _{1})
d\xi _{1}| ^{2}d\xi \\
&\geq  \delta ^{2}\| u( t) \| _{\mathbf{L}
^{2}}^{2}-2\delta ^{5}\sup_{| \xi | \leq \delta
}| \widehat{u}_{\xi }( t,\xi ) | ^{2},
\end{align*}
where $\delta >0$. Thus from (\ref{3.11}) we have the inequality
\begin{equation}
\frac{d}{dt}\| u( t) \| _{\mathbf{H}
^{1}}^{2}\leq -\alpha \delta ^{2}\| u( t) \| _{
\mathbf{H}^{1}}^{2}+4\alpha \delta ^{5}\sup_{| \xi | \leq
\delta }| \widehat{u}_{\xi }( t,\xi ) | ^{2}.
\label{3.12}
\end{equation}
We choose $\alpha \delta ^{2}=2( 1+t) ^{-1}$ and change
$\| u( t) \| _{\mathbf{H}^{1}}^{2}=(
1+t) ^{-2}W( t) $. Then via (\ref{3.11}) we get from (\ref{3.12})
\begin{equation}
\frac{d}{dt}W( t) \leq C( 1+t) ^{-1/2}.
\label{3.13}
\end{equation}
Integration of (\ref{3.13}) with respect to time yields
\[
W( t) \leq \| u_{0}\| _{\mathbf{H}
^{1}}^{2}+C\big( ( 1+t) ^{-\frac{1}{2}+1}-1\big) .
\]
Therefore we obtain a time decay estimate in the $\mathbf{L}^{2}$-norm,
\begin{equation}
\| u( t) \| _{\mathbf{L}^{2}}\leq C(1+t) ^{-3/4}  \label{3.14}
\end{equation}
for all $t>0$. We now multiply equation \eqref{1.1} by $x^{2}u$, then
integrating with respect to $x\in \mathbb{R}^{+}$ we get
\begin{equation}
\begin{aligned}
&\frac{d}{dt}\big( \| xu( t) \| _{\mathbf{L}
^{2}}^{2}+\| xu_{x}( t) \| _{\mathbf{L}
^{2}}^{2}-2\| u( t) \| _{\mathbf{L} ^{2}}^{2}\big)  \label{3.11a} \\
&= -2\alpha \| xu_{x}( t) \| _{\mathbf{L}
^{2}}^{2}+2\alpha \| u( t) \| _{\mathbf{L}
^{2}}^{2}+\lambda \int_{0}^{+\infty }x^{2}| u| ^{3}dx.
\end{aligned}
\end{equation}
By the Plansherel theorem using the Fourier splitting method due to
\cite{schonbek2}, we have
\begin{align*}
\| xu_{x}( t) \| _{\mathbf{L}^{2}}^{2}
&= \| \partial _{\xi }\xi \widehat{u}( t) \| _{
\mathbf{L}^{2}}^{2}=\int_{-\infty }^{\infty }| \widehat{u}+\xi
\widehat{u}_{\xi }| ^{2}| d\xi | \\
& \geq \int_{-\infty }^{\infty }| \xi \widehat{u}_{\xi }|
^{2}| d\xi | -\int_{-\infty }^{\infty }|
\widehat{u}| ^{2}| d\xi | \\
&\geq \int_{| \xi | \geq \delta }| \xi \widehat{
u}_{\xi }| ^{2}d\xi -\| u( t) \| _{
\mathbf{L}^{2}}^{2} \\
&= \delta ^{2}\int_{-\infty }^{+\infty }| \widehat{u}_{\xi }(
t,\xi ) | ^{2}d\xi -\delta ^{2}\int_{| \xi
| <\delta }| \widehat{u}_{\xi }( t,\xi )
| ^{2}d\xi -\| u( t) \| _{\mathbf{L}
^{2}}^{2} \\
&\geq \delta ^{2}\| xu( t) \| _{\mathbf{L}
^{2}}^{2}-2\delta ^{3}\sup_{| \xi | \leq \delta
}| \widehat{u}_{\xi }( t,\xi ) |
^{2}-\| u( t) \| _{\mathbf{L}^{2}}^{2}
\end{align*}
where $\delta >0$. Thus from (\ref{3.11a}) we have the inequality
\begin{equation} \label{3.12a}
\begin{aligned}
&\frac{d}{dt}\big( \| xu( t) \| _{\mathbf{H}
^{1}}^{2}-2\| u( t) \| _{\mathbf{L}^{2}}^{2}\big)   \\
&\leq -\alpha \delta ^{2}\| xu( t) \| _{\mathbf{
H}^{1}}^{2}+2\alpha \delta ^{3}\sup_{| \xi | \leq \delta
}| \widehat{u}_{\xi }( t,\xi ) | ^{2}+2(\alpha
+1)\| u( t) \| _{\mathbf{L}^{2}}^{2}
\end{aligned}
\end{equation}
We choose $\alpha \delta ^{2}=2( 1+t) ^{-1}$ and change
$\| u( t) \| _{\mathbf{H}^{1}}^{2}=(
1+t) ^{-2}W( t) $. Then via (\ref{h2}),(\ref{3.14}) we get
from (\ref{3.12a})
\begin{equation}
\frac{d}{dt}W( t) \leq C( 1+t) ^{\frac{1}{2}}.
\label{3.13a}
\end{equation}
Integration of (\ref{3.13a}) with respect to time yields
\[
W( t) \leq \| xu_{0}\| _{\mathbf{H}
^{1}}^{2}+C\big( ( 1+t) ^{\frac{1}{2}+1}-1\big) .
\]
Therefore, we obtain a time decay estimate of the $\mathbf{L}^{2}$ - norm
\[
\| xu( t) \| _{\mathbf{L}^{2}}\leq C(1+t) ^{-1/4}
\]
 Lemma \ref{OPT} is proved
\end{proof}

By Lemmas \ref{L1} and  \ref{OPT}, we have
\begin{align*}
\| u( t) \| _{\mathbf{L}^{\infty }}
&\leq \| \mathcal{G}( t) u_{0}\| _{\mathbf{L}^{\infty}}
 +C\int_{\frac{t}{2}}^{t}\langle t-\tau \rangle ^{-\frac{1}{2}
}\| u( \tau ) \| _{\mathbf{L}^{2}}^{2}d\tau\\
&\quad +C\int_{0}^{t/2}\langle t-\tau \rangle ^{-1}\|
x| u| u( \tau ) \| _{\mathbf{L}^{1}}d\tau \\
&\leq C\langle t\rangle ^{-1}+C\int_{0}^{\frac{t}{2}
}\langle t-\tau \rangle ^{-1}\langle \tau \rangle
^{-1}d\tau \\
&\leq C\langle t\rangle ^{-1}\log \langle t\rangle
\end{align*}
for all $t>0$. In the same manner
\begin{align*}
&\| \partial _{x}^{2}u( t) \| _{\mathbf{L}
^{\infty }}\\
&\leq C\langle t\rangle ^{-2}+\int_{0}^{t}\|
\partial _{x}\mathcal{G}( t-\tau ) \mathcal{B}|
u| u( \tau ) \| _{\mathbf{L}^{\infty }}d\tau \\
&\leq C\langle t\rangle ^{-2}+C\int_{0}^{\frac{t}{2}
 }\langle t-\tau \rangle ^{-2}\| u( \tau )
 \| _{\mathbf{L}^{2}}^{2}d\tau
 +C\int_{\frac{t}{2}}^{t}\langle t-\tau \rangle ^{-1}\|
 u( \tau ) \| _{\mathbf{L}^{\infty }}^{2}d\tau \\
&\leq C\langle t\rangle ^{-2}+C\int_{0}^{\frac{t}{2}
 }\langle t-\tau \rangle ^{-2}\langle \tau \rangle ^{-
 \frac{3}{2}}d\tau
 +C\int_{\frac{t}{2}}^{t}\langle t-\tau \rangle
^{-1}\langle \tau \rangle ^{-2}\log ^{2}\langle \tau
\rangle d\tau \\
&\leq C\langle t\rangle ^{-2}\log ^{3}\langle t\rangle
\end{align*}
for all $t>0$. Denote $f( t,x) =u_{xxt}$. Then by Lemmas \ref{OPT}
and \ref{L1} we have the estimates
\begin{align*}
&\| f( t) \| _{\mathbf{L}^{\infty }}\\
&=\|\partial _{x}^{2}u_{t}( t) \| _{\mathbf{L}^{\infty}}\\
&\leq \| \partial _{x}^{2}\partial _{t}\mathcal{G}( t)
 u_{0}\| _{\mathbf{L}^{\infty }}+\| \partial _{x}^{2}
 \mathcal{B}| u| u( t) \| _{\mathbf{L}^{\infty }} \\
&\quad +C\int_{\frac{t}{2}}^{t}\langle t-\tau \rangle ^{-1}\|
 \partial _{x}^{2}\mathcal{B}| u| u( \tau )
 \| _{\mathbf{L}^{\infty }}d\tau
 +C\int_{0}^{t/2}\langle t-\tau \rangle ^{-3}\|
x| u| u( \tau ) \| _{\mathbf{L}^{1}}d\tau \\
&\leq C\langle t\rangle ^{-3}\log ^{4}\langle
t\rangle +C\int_{\frac{t}{2}}^{t}\langle t-\tau \rangle
^{-1}\langle \tau \rangle ^{-3}\log ^{4}\langle \tau \rangle d\tau
 +C\int_{0}^{t/2}\langle t-\tau \rangle
^{-3}\langle \tau \rangle ^{-1}d\tau \\
&\leq C\langle t\rangle ^{-3}\log ^{4}\langle t\rangle
\end{align*}
for all $t>0$. We a  sufficiently small $\varepsilon >0$ and consider the
following two auxiliary problems
\begin{equation}
\begin{gathered}
U_{t}-U_{xx}+U^{2}=\varepsilon | f| ,\quad x\in \mathbb{R}^{+},\; t>0, \\
U(0,x)=\varepsilon | u_{0}(x)|,\quad  x\in \mathbb{R} ^{+}, \\
U(x,0)=0,\quad t>0
\end{gathered}  \label{4.16}
\end{equation}
and
\begin{equation}
\begin{gathered}
V_{t}-V_{xx}+\varepsilon V^{2}=| f| ,\quad x\in \mathbb{R}^{+},\; t>0, \\
V(0,x)=| u_{0}(x)| , \quad x\in \mathbb{R}^{+}, \\
V(x,0)=0,\quad t>0.
\end{gathered} \label{4.17}
\end{equation}
Note that problem (\ref{4.17}) can be reduced to problem (\ref{4.16}) by the
change of variable $V=\varepsilon ^{-1}U$. Also note that (\ref{4.16}) has a
sufficiently small initial data and a small force
$\varepsilon |u_{xxt}| $. Applying results of paper we obtain an almost optimal
time decay estimate
\[
\| U( t) \| _{\mathbf{L}^{\infty }}\leq C\langle t\rangle ^{-1},
\]
hence by Lemma \ref{L10} we get
$\| u( t) \| _{\mathbf{L}^{\infty }}\leq C\langle t\rangle ^{-1}$.

Now we estimate the $\mathbf{L}^{1,1+a}$-norm of the solution
\begin{align*}
\| u( t) \| _{\mathbf{L}^{1,1+a}}
&\leq \| \mathcal{G}( t) u_{0}\| _{\mathbf{L}^{1,1+a}} \\
&\quad +C\int_{0}^{t}( \| \mathcal{B}| u| u(
\tau ) \| _{\mathbf{L}^{1,1+a}}+\langle t-\tau
\rangle ^{a/2}\| \mathcal{B}| u|
u( \tau ) \| _{\mathbf{L}^{1,1}}) d\tau \\
&\leq C\langle t\rangle ^{a/2}+C\int_{0}^{t}\langle
\tau \rangle ^{-1}\| u( \tau ) \| _{
\mathbf{L}^{1,a}}d\tau +C\int_{0}^{t}\langle t-\tau \rangle ^{
\frac{a}{2}}\langle \tau \rangle ^{-1}d\tau \\
&\leq C\langle t\rangle ^{a/2}+C\int_{0}^{t}\langle
\tau \rangle ^{-1}\| u( \tau ) \| _{\mathbf{L}^{1,1+a}}d\tau .
\end{align*}
Hence by Granwall's inequality we obtain
\begin{equation}
\| u( t) \| _{\mathbf{L}^{1,1+a}}\leq
C\langle t\rangle ^{a/2}  \label{4.18a}
\end{equation}
for all $t>0$. In the same manner we estimate the $\mathbf{L}^{1,1+a}$ -
norm of $f$. By Lemma \ref{L1} we have
\begin{align*}
\| f( t) \| _{\mathbf{L}^{1,1+a}}
&=\|u_{xxt}( t) \| _{\mathbf{L}^{1,1+a}} \\
&\leq \| \partial _{x}^{2}\partial _{t}\mathcal{G}( t)
u_{0}\| _{\mathbf{L}^{1,1+a}}+\| \partial _{x}^{2}\mathcal{B
}| u| u( t) \| _{\mathbf{L}^{1,1+a}}
\\
&\quad +C\int_{0}^{t}\langle t-\tau \rangle ^{-1}( \|
| u| u( \tau ) \| _{\mathbf{L}
^{1,1+a}}+\langle t-\tau \rangle ^{a/2}\|
| u| u( \tau ) \| _{\mathbf{L}
^{1,1}}) d\tau \\
&\leq C\langle t\rangle ^{-1+\frac{a}{2}}+C\int_{0}^{t}
\langle t-\tau \rangle ^{-1}\langle \tau \rangle ^{\frac{a}{
2}-1}d\tau +C\int_{0}^{t}\langle t-\tau \rangle ^{\frac{a}{2}
-1}\langle \tau \rangle ^{-1}d\tau \\
&\leq C\langle t\rangle ^{-1+\frac{a}{2}}.
\end{align*}
Thus we can apply the results of paper \cite{hkn6} to get the estimate of
the functions $U( t,x) $ and $V( t,x) $. Then by Lemma
\ref{L10} we get an optimal time decay estimate for the solution
\begin{equation}
\| u( t) \| _{\mathbf{L}^{\infty }}\leq
C\varepsilon ^{-1}\langle t\rangle ^{-1}( \log (2+t) ) ^{-1}  \label{4.18}
\end{equation}
for all $t>0$.

We make a change of the dependent variable $u( t,x) =v(
t,x) e^{-\varphi ( t) }$ as in the proof of Theorem \ref{T1}.
 Then we obtain problem (\ref{4.2}) for new functions
$( v(t,x) ,\varphi ( t) ) $. Now we prove the following
estimate
\[
\| v( t) \| _{\mathbf{L}^{1,1+a}}\leq
C\langle t\rangle ^{a/2}
\]
for all $t>0$. From (\ref{4.2}) we obtain the integral formula
\begin{equation}
v( t) = {\mathcal{G}}( t) u_{0}+\lambda
\int_{0}^{t}d\tau {\mathcal{G}}(t-\tau )\mathcal{B}\Big( | u(\tau )
| v( \tau )
 -\frac{v( \tau ) }{\theta }\int_{\mathbb{R}
^{n}}| u( \tau ) | v( \tau )dx\Big) .  \label{4.20}
\end{equation}
Using (\ref{4.18}) and Lemma \ref{L1} we have
\begin{align*}
&\big\| \int_{0}^{t}d\tau {\mathcal{G}}(t-\tau )\mathcal{B}(
| u( \tau ) | v( \tau )
 -\frac{v( \tau ) }{\theta }\int_{0}^{+\infty
}| u( \tau ) | v( \tau ) dx)
\big\| _{\mathbf{L}^{1,1+a}} \\
&\leq C\int_{0}^{t}\| | u( \tau ) |
v( \tau ) -\frac{v( \tau ) }{\theta }\int_{0}^{+\infty
}| u( \tau ) | v( \tau ) dx\| _{\mathbf{L}^{1,1+a}}d\tau \\
&\leq C\int_{0}^{t}\langle \tau \rangle ^{-1}( \log (
2+\tau ) ) ^{-1}\| v( \tau ) \| _{\mathbf{L}^{1,1+a}}d\tau
\end{align*}
for all $t>0$. Therefore, by (\ref{4.18a}) we find
\begin{align*}
\| v( t) \| _{\mathbf{L}^{1,1+a}}
&\leq \| \mathcal{G}( t) v( 0) \| _{\mathbf{L}^{1,1+a}}
 +C\int_{0}^{t}\langle \tau \rangle ^{-1}( \log (2+\tau ) ) ^{-1}
\| v( \tau ) \| _{\mathbf{L}^{1,1+a}}d\tau \\
&\leq C\langle t\rangle ^{a/2}+C\int_{0}^{t}\langle
\tau \rangle ^{-1}( \log ( 2+\tau ) )
^{-1}\| v( \tau ) \| _{\mathbf{L}^{1,1+a}}d\tau
\end{align*}
for all $t>0$. Hence by Granwall's inequality we obtain
\[
\| v( t) \| _{\mathbf{L}^{1,1+a}}\leq C\langle t\rangle ^{a/2}
\]
for all $t>0$. In the same way from (\ref{4.18}) and Lemma \ref{L1} we get
\begin{align*}
\| v( t) \| _{\mathbf{L}^{p}}
&\leq \|\mathcal{G}( t) v( 0) \| _{\mathbf{L}^{p}} \\
&\quad +\big\| \int_{0}^{t}d\tau {\mathcal{G}}(t-\tau )\mathcal{B}(
| u( \tau ) | v( \tau )
 -\frac{v( \tau ) }{\theta }\int_{0}^{+\infty
}| u( \tau ) | v( \tau ) dx) \big\| _{\mathbf{L}^{p}} \\
&\leq C\langle t\rangle ^{-\frac{1}{2}-\frac{1}{2}( 1-\frac{1}{p}) } \\
&\quad +C\int_{0}^{t/2}( t-\tau ) ^{-\frac{1}{2}-\frac{1}{2}
( 1-\frac{1}{p}) -\frac{a}{2}}\langle \tau \rangle ^{
\frac{a}{2}-1}( \log ( 2+\tau ) ) ^{-1}d\tau \\
&\quad +C\int_{\frac{t}{2}}^{t}\langle \tau \rangle ^{-1}( \log
( 2+\tau ) ) ^{-1}\| v( \tau )
\| _{\mathbf{L}^{p}}d\tau \\
&\leq C\langle t\rangle ^{-\frac{1}{2}-\frac{1}{2}( 1-\frac{
1}{p}) }+\int_{\frac{t}{2}}^{t}\langle \tau \rangle
^{-1}\| v( \tau ) \| _{\mathbf{L}^{p}}d\tau
\end{align*}
for all $t>0$. So by Granwall's inequality we have
\[
\| v( t) \| _{\mathbf{L}^{p}}\leq C\langle
t\rangle ^{-\frac{1}{2}-\frac{1}{2}( 1-\frac{1}{p}) }
\]
for all $t>0$.
Now from (\ref{4.20}) and Lemma \ref{L1} we get
\begin{equation}
\begin{aligned}
\| v( t) -{\mathcal{G}}( t) u_{0}\|_{\mathbf{L}^{1,1}}
&\leq \int_{0}^{t/2}( t-\tau ) ^{-\frac{a}{2}}
\big\| | u( \tau ) | v( \tau) -\frac{v( \tau ) }{\theta }\int_{\mathbb{R}
^{n}}| u( \tau ) | v( \tau ) dx\big\| _{\mathbf{L}^{1,1+a}}d\tau   \\
&\quad +\int_{\frac{t}{2}}^{t}\big\| | u( \tau )| v( \tau )
  -\frac{v( \tau ) }{\theta }\int_{0}^{+\infty }|
u( \tau ) | v( \tau ) dx\big\| _{\mathbf{L}^{1,1}}d\tau   \\
&\leq C\int_{0}^{t/2}( t-\tau ) ^{-\frac{a}{2}
 }\langle \tau \rangle ^{\frac{a}{2}-1}( \log ( 2+\tau) ) ^{-1}d\tau   \\
&\quad +C\int_{\frac{t}{2}}^{t}\langle \tau \rangle ^{-1}( \log
 ( 2+\tau ) ) ^{-1}d\tau \\
& \leq C\log ( 2+t) ^{-1}
\end{aligned} \label{4.21}
\end{equation}
for all $t>0$.  Therefore using Lemma\ref{L3} we find for
$h( t)=e^{\varphi ( t) }$,
\[
| h( t) -g( t) | \leq C\log g(t) ,
\]
for all $t>0$. Then from
$u( t,x) =e^{-\varphi ( t)}v( t,x) =h^{-1}( t) v( t,x) $, we have
\begin{equation}
\begin{aligned}
&\| u( t) -\theta e^{-\varphi ( t)}G_{0}( t) \| _{\mathbf{L}^{\infty }} \\
&\leq \|u( t) -e^{-\varphi ( t) }{\mathcal{G}}( t)
  u_{0}\| _{\mathbf{L}^{\infty }}
 +\| \mathcal{G}( t) u_{0}-\theta e^{-\varphi (
t) }G_{0}( t) \| _{\mathbf{L}^{\infty }}   \\
&\leq Ct^{-1}g^{-2}( t) +Ct^{-1-\frac{a}{2}}\|
u_{0}\| _{\mathbf{L}^{1,1+a}}\\
&\leq Ct^{-1}g^{-2}( t)
\end{aligned} \label{4.22}
\end{equation}
for all $t>0$. Also we have
\[
\| \theta G_{0}( t) ( h^{-1}( t)
-g^{-1}( t) ) \| _{\mathbf{L}^{\infty }}
\leq Ct^{-1}g^{-2}( t) | h( t) -g(t) | ,
\]
and so by (\ref{4.22}),
\[
\| u( t) -\theta G_{0}( t) g^{-1}(
t) \| _{\mathbf{L}^{\infty }}\leq C( 1+t)
^{-1}g^{-2}( t) \log g( t) .
\]
Theorem \ref{T2} is proved.

\subsection*{Acknowledgements}
The author would like to thank the anonymous referee for many useful
 suggestions and comments. This work is partially supported by CONACyT.


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\end{document}