\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2015 (2015), No. 281, pp. 1--24.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2015 Texas State University.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2015/281\hfil Nonlinear bvp with fractional derivative] {Nonlinear initial boundary-value problems with Riesz fractional derivative} \author[M. P. \'Arciga-Alejandre \hfil EJDE-2015/281\hfilneg] {Mart\'in P. \'Arciga-Alejandre} \address{Mart\'in P. \'Arciga-Alejandre \newline Instituto de Matem\'aticas UNAM, Campus Morelia, Apartado Postal 61-3 (Xangari), C.P. 58089, Morelia, Michoac\'an, M\'exico} \email{mparciga@gmail.com} \thanks{Submitted September 26, 2013. Published November 10, 2015.} \subjclass[2010]{35S11, 47G30, 58J40} \keywords{Riesz fractional derivative; Riemann-Hilbert problem; \hfill\break\indent dissipative nonlinear evolution equation; large time asymptotic} \begin{abstract} We consider an initial boundary-value problem for a nonlinear partial differential equation with fractional derivative of Riesz type on a half-line. We study local and global existence of solutions in time, as well as the asymptotic behavior of solutions for large time. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{definition}[theorem]{Definition} \allowdisplaybreaks \section{Introduction} \label{0s} We study the existence of local and global solutions, and the asymptotic behaviour, for the initial boundary-value problem \begin{equation}\label{222} \begin{gathered} u_t+\mathcal{N}(u)+|\partial_x|^\alpha u=0,\quad t>0,\; x>0; \\ u(x,0)=u_0(x),\quad x>0, \\ u(0,t)=h(t),\quad t>0. \end{gathered} \end{equation} where $\mathcal{N}(u)=|u|^\sigma u$, $\alpha<\sigma<\alpha+1$, $2/5<\alpha<1$, and $|\partial_x|^\alpha$ is a fractional derivative of Riesz type defined by \begin{equation}\label{2233} |\partial_x|^\alpha u=\mathcal{R}^{1-\alpha+[\alpha]}\partial_x^{[\alpha]+1}u. \end{equation} Here, $[\alpha]$ denotes the integer part of the number $\alpha>0$, $\alpha\notin\mathbb{Z}$, and $\mathcal{R}^\alpha$ is the modified Riesz Potential (see \cite{samko}), $$ \mathcal{R}^\alpha u=\frac{1}{2\Gamma(\alpha)\sin(\frac{\pi}{2}\alpha} \int_0^{+\infty}\frac{\operatorname{sgn}(x-y)}{|x-y|^{1-\alpha}}u(y)dy. $$ The Cauchy problem for nonlinear nonlocal dissipative equations has been extensively studied. In particular, the large time asymptotic behavior for the Cauchy problem for different nonlinear equations is investigated in \cite{bookhkns} and the references therein. Boundary value problems arise in many applications and play an important role in the contemporary mathematical physics. For the study of the effect of the boundary data on the qualitative properties of the solution the reader is referred to \cite{bashir1,bashir2,mohammed,mouffak,Ben,caballero,Fo} and references therein. The general theory of nonlinear nonlocal equations on a half-line was developed in \cite{hkbook}, where the pseudodifferential operator $\mathbb{K}$ ($|\partial_x|^\alpha$ in our case) defined on a half-line was introduced by virtue of the inverse Laplace transformation; it is given by $$ \mathbb{K}u=\frac{1}{2\pi i}\sum_jC_j\int_{-i\infty+\sigma}^{i\infty+\sigma}e^{px} p^{\alpha_j}\Big(\widehat{u}(p)-\sum_{l=1}^{[\alpha_j]}\partial_x^{l-1}u(0)p^{-l} \Big)dp. $$ Note that the symbol $K(p)=\sum_jC_jp^{\alpha_j}$ is analytic in the complex right half-plane. We emphasize that the operator $|\partial_x|^\alpha$ in equation \eqref{2233} has a nonanalytic nonhomogeneous symbol $K(p)=|p|^\alpha$ and the general theory in \cite{hkbook} can not be applied to problem \eqref{222} directly. As far as we know there are few results on the initial-boundary value problems with pseudodifferential equations having a nonanalytic symbol. The case of rational symbols $K(p)$ with poles in the complex right half-plane was studied in \cite{Noan,Kk}, where it was proposed a new method for constructing the Green operator based on the introduction of necessary conditions at the singular points of the symbol $K(p)$. In \cite{JFM} the initial-boundary value problem for a pseudodifferential equation with a nonanalytic homogeneous symbol $K(p)=| p|^{1/2}$ was studied, a theory of sectionally analytic functions was implemented for proving that the initial-boundary value problem is well-posed. Since the symbol $K(p)=| p|^{1/2}$ does not grow fast at infinity, no boundary data is needed. Finally, the case of $K(p)=|p|^\alpha$, $\alpha\in(1/2,1)$, was studied in \cite{a1}. In the present work we consider the nonlinear version of the problem \cite{trans}, considering $\alpha\in(2/5,1)$. In order to construct a Green operator we follow the methods used in \cite{a1,trans}. To state precisely the results of the present paper we give some notations. We denote $\langle t\rangle =1+t$, $\{ t\} =\frac{t}{\langle t\rangle }$. Here and below $p^{\alpha}$ is the main branch of the complex analytic function in the complex half-plane $\operatorname{Re}(p)\geq 0$, so that $1^{\alpha}=1$ (we make a cut along the negative real axis $(-\infty ,0)).$ Note that due to the analyticity of $p^{\alpha}$ for all $\operatorname{Re}(p)>0$ the inverse Laplace transform gives us $0$ for all $x<0$. The direct Laplace transformation $\mathcal{L}_{x\to p}$ is given by \begin{equation*} \widehat{u}(p) \equiv \mathcal{L}_{x\to p}\{u\}=\int_{0}^{+\infty }e^{-px}u( x) dx \end{equation*} and the inverse Laplace transformation $\mathcal{L}_{p\to x}^{-1}$ is defined by \begin{equation*} u( x) \equiv \mathcal{L}_{p\to x}^{-1}\{\widehat{u}\} =\frac{1}{2\pi i}\int_{\mathbf{-}i\infty }^{i\infty }e^{px}u(p) dp . \end{equation*} The norm in weighted Lebesgue space $ L^{s,\mu}(\mathbb{R}^{+}) =\{\varphi \in \mathcal{S}^{\prime };\| \varphi \| _{ \mathbf{\ L}^{s,\mu}}<\infty \}$ is given by \begin{equation*} \| \varphi \| _{ L^{s,\mu}}=( \int_{0}^{+\infty }x ^{\mu s}| \varphi ( x) | ^{s}dx) ^{\frac{1}{s}} \end{equation*} for $\mu>0$, $10}( t^{\frac{1}{\alpha}(1-\frac{1}{s})}\|\varphi\|_{ L^s}+ t^{\frac{1}{\alpha}(1-\frac{1}{s}-\mu)}\|\varphi\|_{ L^{s,\mu}}+ \{t\}^\gamma\langle t\rangle^\frac{1}{\alpha}\|\varphi\|_{ L^\infty}) $$ where $0<\gamma<1$ and $|1-\frac{1}{s}-\mu|<\alpha$. We also define the space $$ B^\alpha=\{h\in C^1(0,\infty):\|h\|_{ B^\alpha}<\infty\} $$ for the boundary data, where the norm $$ \|h\|_{ B^\alpha}=\sup_{t>0}\langle t\rangle^\frac{1}{\alpha} (|h(t)|+\langle 1+t\rangle|h'(t)|). $$ Now, to state the main results we introduce $\Lambda(s),\Upsilon(s)\in L^{\infty }(\mathbb{R}^{+})$ by the formulae \begin{equation}\label{46} \Lambda(s)=C\int_{-i\infty}^{i\infty}d\xi e^\xi \int_{-i\infty}^{i\infty}e^{ps}\frac{p^{\alpha/2}}{K(p)+\xi}L(p,\xi)dp \end{equation} and \begin{equation}\label{upsi} \begin{aligned} \Upsilon(s) &=-\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}d\xi e^\xi \int_{-i\infty}^{i\infty}e^{ps}\frac{Y^+(p,\xi)}{K(p)+\xi}I^+(p,\xi)dp \\ &\quad +\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}e^{ps-K(p)}\frac{K(p)}{p}dp, \end{aligned} \end{equation} where \begin{gather*} L(p,\xi)=(\Gamma_\xi^+-\frac{1}{K(p)+\xi})^2 +\Gamma_{\xi\xi}^++\frac{1}{(K(p)+\xi)^2}, \\ I(z,\xi)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty} \frac{1}{q-z}\frac{1}{Y^+(q,\xi)}\frac{K(q)}{q}dq. \end{gather*} Here, $Y^+(p,\xi)=(-i)^\alpha e^{\Gamma^+(p,\xi)}$, \[ \Gamma^+(p,\xi)=-\frac{1}{2\pi i}\int_{-i\infty}^{i\infty} \ln^-(q-p)(\frac{K'(q)}{K(q)+\xi}-\frac{K_1'(q)}{K_1(q)+\xi})dq, \] $K(p)=|p|^\alpha$, $K_1(p)=p^\alpha$ and $\ln^-(z)=\ln|z|+i\arg^-(z)$ for $-\frac{3}{2}\pi<\arg^-(z)\leq\frac{\pi}{2}$. \begin{theorem}\label{b5} Let the initial data $u_{0}\in \mathbb{Z}^\mu$ and the boundary value $h\in B^\alpha$ be such that the correspondent norms are sufficiently small. Also, suppose that $h(t)=bt^\beta+\mathcal{O}(t^{\beta-\delta})$, where $\beta<-\frac{1}{\alpha}-\frac{1}{2}-\gamma$ and $\delta,\gamma>0$. Then, the initial-boundary value problem \eqref{222} has unique global solution $u\in X^\mu$ and the following asymptotic is valid \begin{equation*} u(x,t)=t^{-\frac{1}{\alpha}-\frac{1}{2}}(a\Lambda(xt^{-1/\alpha}) +b\Upsilon(xt^{-1/\alpha}))+ \mathcal{O}(t^{-\frac{1}{\alpha}-\frac{1}{2}-\gamma}), \end{equation*} for $t\to \infty $ in $ L^{\infty }(\mathbb{R}^+)$, where $$ a=\int_0^{+\infty}y^{\alpha/2}u_0(y)dy-\int_{0}^{+\infty}d\tau \int_0^{+\infty}y^{\alpha/2}\mathcal{N}(u(\tau))dy. $$ \end{theorem} \section{Preliminaries} \label{S0} \begin{definition} \label{def2.1} \rm A function $\phi(q)$ is said to satisfy the H\"older condition on the curve $L$, if there exist positive constants $C$ and $\lambda$ such that \begin{equation}\label{holderc} |\phi(q_1)-\phi(q_2)|\leq C|q_1-q_2|^\lambda, \end{equation} for any $q_1, q_2 \in L$. \end{definition} The proof of the next theorem can be found in \cite{gakhov}. \begin{theorem} \label{thm2.2} Let $\phi(q)$ be a complex function, which obeys the H\"older condition \eqref{holderc} for all finite $q$ and tends to a finite limit $\phi(\infty)$ as $q\to\infty$, such that for large $q$ the following inequality holds \begin{equation*} |\phi(q)-\phi(\infty)|\leq C|q|^\mu,\quad \mu>0. \end{equation*} Then, the Cauchy integral \begin{equation*} F(z)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{\phi(q)}{q-z}dq \end{equation*} defines an analytic function in the left and right semi-planes. Here and below these functions will be denoted by $F^+(z)$ and $F^-(z)$, respectively. These functions have the limiting values $F^+(p)$ and $F^-(p)$ at all points of the imaginary axis, $\operatorname{Re}(p)=0$, on approaching the contour from the left and from the right, respectively. These limiting values are expressed by the Sokhotski-Plemelj formulae, \begin{equation}\label{sokhotski} F^\pm(p)=\frac{1}{2\pi i} \hspace{1mm}-\hspace{-4mm}\int_{-i\infty}^{i\infty} \frac{\phi(q)}{q-p}dq\pm\frac{1}{2}\phi(p). \end{equation} \end{theorem} Subtracting the formulae \eqref{sokhotski} we obtain the following equivalent formula \begin{equation} F^+(p)-F^-(p)=\phi(p), \end{equation} which will be frequently employed hereafter. Now, we consider the following linear initial-boundary value problem on half-line \begin{equation}\label{4} \begin{gathered} u_t+|\partial_x|^\alpha u=0,\quad x,t>0; \\ u(x,0)=u_0(x),\quad x>0, \\ u(0,t)=h(t),\quad t>0. \end{gathered} \end{equation} Setting $K(q)=| q|^\alpha$, $K_{1}(q)=q^\alpha$, we define \begin{gather*} \mathcal{G}(t)\phi=\int_{0}^{+\infty }G(x,y,t)\phi (y)dy,\\ \mathcal{H}(x)h=\int_0^tH(x,t-\tau)h(\tau)d\tau,\quad H(x,t)=\partial_t\partial_y^{-1}G(x,y,t)|_{y=0}, \end{gather*} where the function $G(x,y,t)$ is given by \begin{equation}\label{5} G(x,y,t)=-\frac{1}{(2\pi i)^2}\int_{-i\infty}^{i\infty}d\xi e^{\xi t}\int_{-i\infty}^{i\infty}e^{px} \frac{Y^+(p,\xi)}{K(p)+\xi}Z^-(p,\xi,y)dp, \end{equation} where $$ Z^-(p,\xi,y)=\lim_{\substack{z\to p\\ \textrm{\tiny{Re}}z>0}}\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{1}{q-z} \frac{1}{Y^+(q,\xi)}e^{-qy}dq. $$ for $x>0$, $y>0$, $t>0$. Here, $Y^+(p,\xi)=(-i)^\alpha e^{\Gamma^+(p,\xi)}$, where $$ \Gamma^+(p,\xi)=-\frac{1}{2\pi i}\int_{-i\infty}^{i\infty} \ln^-(q-p)(\frac{K'(q)}{K(q)+\xi}-\frac{K_1'(q)}{K_1(q)+\xi})dq $$ and $\ln^-(z)=\ln|z|+i\arg^-(z)$, $-\frac{3}{2}\pi<\arg^-(z)\leq\frac{\pi}{2}$, where the integrals are understood in the sense of the principal values. \begin{proposition}\label{prop2.3} % 6 If $u_0(x)\in\mathbb{Z}^\nu$, there exist a unique solution $u(x,t)$ for the initial-boundary value problem \eqref{4}, which has an integral representation \begin{equation}\label{7} u=\mathcal{G}(t)u_0+\mathcal{H}(x)h,\quad x>0,\; t>0. \end{equation} \end{proposition} \begin{proof} To obtain an integral representation for solutions of problem \eqref{4} we suppose that there exist a solution $u(x,t)$, which is continued by zero outside of $x>0$: $$ u(x,t)=0,\quad \text{for all } x<0. $$ Let $\phi(q)$ be a function of the complex variable $q$, which obeys the H\"older condition \eqref{holderc} for all $q$, such that $\operatorname{Re}(q)=0$. We define the operator $\mathbb{P}$ by $$ \mathbb{P}_{q\to z}\{\phi(q)\}=-\frac{1}{2\pi i}\int_{-i\infty}^{i\infty} \frac{1}{q-z}\phi(q)dq,\quad \operatorname{Re}(z)\neq0. $$ Applying the Laplace transform with respect to $x$ to $|\partial_x|^\alpha u$, for $\operatorname{Re}(q)>0$ we obtain \begin{equation}\label{8} \mathcal{L}_{x\to q}\{|\partial_x|^\alpha u(x,t)\}=\mathbb{P}_{p\to q}\{K(p) (\widehat{u}(p,t)-\frac{u(0,t)}{p})\}. \end{equation} Since $\widehat{u}(q,t)$ is analytic, for all $\operatorname{Re}(q)>0$, we have \begin{equation}\label{9} \widehat{u}(q,t)=\mathbb{P}_{p\to q}\{\widehat{u}(p,t)\}. \end{equation} Applying the Laplace transform with respect to $x$ to problem \eqref{4} and using \eqref{8} and \eqref{9}, for $\operatorname{Re}(p)=0$, we obtain \begin{equation}\label{10} \begin{gathered} \mathbb{P}_{p\to q}\big\{\widehat{u}_t(p,t)+K(p) \big(\widehat{u}(p,t)-\frac{h(t)}{p}\big)\big\}=0, \\ \widehat{u}(p,0)=\widehat{u}_0(p), \end{gathered} \end{equation} We rewrite \eqref{10} in the form \begin{equation}\label{11} \begin{gathered} \widehat{u}_t(p,t)+K(p) \widehat{u}(p,t)-\frac{K(p)}{p}h(t) =\Phi(p,t); \\ \widehat{u}(p,0)=\widehat{u}_0(p), \end{gathered} \end{equation} with some function $\Phi(p,t)$ such that for $\operatorname{Re}(p)=0$ \begin{equation}\label{18} \mathbb{P}_{p\to q}\{\Phi(p,t)\}=0\quad \text{and}\quad |\Phi(p,t)|\leq C|p|^{\alpha-1-\gamma}, \quad |p|>1,\;\gamma>0. \end{equation} Applying the Laplace transform with respect to time variable to \eqref{11} we find \begin{equation}\label{12} \widehat{\widehat{u}}(p,\xi)=\frac{1}{K(p)+\xi} \Big(\widehat{u}_0(p)+\frac{K(p)}{p}\widehat{h}(\xi) +\widehat{\Phi}(p,\xi)\Big), \end{equation} where $\operatorname{Re}(p)=0$ and $\operatorname{Re}(\xi)>0$. Here, the functions $\widehat{\widehat{u}}(p,\xi)$, $\widehat{h}(\xi)$, and $\widehat{\Phi}(p,\xi)$ are the Laplace transforms for $\widehat{u}(p,t)$, $h(t)$, and $\Phi(p,t)$ with respect to time, respectively. In order to obtain an integral formula for solutions to the problem \eqref{4} it is necessary to know the function $\Phi(p,t)$. We find the function $\widehat{\Phi}(p,\xi)$ using the analytic properties of the function $\widehat{\widehat{u}}$ in the right-half complex planes $\operatorname{Re}(p)>0$ and $\operatorname{Re}(\xi)>0$. The equation \eqref{9} and the Sokhotski-Plemelj formulae \eqref{sokhotski} imply, for $\operatorname{Re}(p)=0$, \begin{equation}\label{13} \widehat{\widehat{u}}(p,\xi)=-\frac{1}{\pi i} \hspace{1mm}-\hspace{-4mm}\int_{-i\infty}^{i\infty}\frac{1}{q-p} \widehat{\widehat{u}}(q,\xi)dq. \end{equation} In view of Sokhotski-Plemelj formulae \eqref{sokhotski}, via \eqref{12} the condition \eqref{13} can be written as \begin{equation}\label{14} \Theta^+(p,\xi)=-\Lambda^+(p,\xi), \end{equation} where the sectionally analytic functions $\Theta(p,\xi)$ and $\Lambda(p,\xi)$ are given by Cauchy type integrals \begin{gather}\label{15} \Theta(z,\xi)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{1}{q-z}\frac{1}{K(q)+\xi} \widehat{\Phi}(q,\xi)dq, \\ \label{16} \Lambda(z,\xi)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{1}{q-z} \frac{1}{K(q)+\xi}(\widehat{u}_0(q)+\frac{K(q)}{q}\widehat{h}(\xi))dq. \end{gather} To perform the condition \eqref{14} in the form of a nonhomogeneous Riemann-Hilbert problem we introduce the sectionally analytic function \begin{equation}\label{17} \Omega(z,\xi)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{1}{q-z}\frac{K(q)}{K(q)+\xi}\widehat{\Phi}(q,\xi)dq. \end{equation} Taking into account the assumed condition \eqref{18}, we obtain \begin{equation}\label{19} \Theta^-(p,\xi)=-\frac{1}{\xi}\Omega^-(p,\xi). \end{equation} Also observe that from \eqref{15} and \eqref{17} by Sokhotski-Plemelj formulae, \begin{equation}\label{21} K(p)(\Theta^+-\Theta^-)= \Omega^+-\Omega^-=\frac{K(p)}{K(p)+\xi}\widehat{\Phi}. \end{equation} Substituting \eqref{14} and \eqref{19} into this equation we obtain for Re $p=0$ \begin{equation}\label{20} \Omega^+(p,\xi)=W(p,\xi)\Omega^-(p,\xi)+g(p,\xi), \end{equation} where $W=\frac{K(p)+\xi}{\xi}$ and $g=-K(p)\Lambda^+$. Equation \eqref{20} is the boundary condition for a nonhomogeneous Riemann-Hilbert problem. It is required to find two functions for some fixed point $\xi$, Re $\xi>0$: $\Omega^+(z,\xi)$, analytic in the left-half complex plane Re $z<0$ and $\Omega^-(z,\xi)$, analytic in the right-half complex plane Re $z>0$, which satisfy on the contour Re $p=0$ the relation \eqref{20}. Note that bearing in mind formula \eqref{21} we can find the unknown function $\widehat{\Phi}(p,\xi)$, which involved in the formula \eqref{12}, by the relation \begin{equation}\label{22} \widehat{\Phi}(p,\xi)=\frac{K(p)+\xi}{K(p)} (\Omega^+(p,\xi)-\Omega^-(p,\xi)). \end{equation} The method for solving the Riemann problem $F^{+}(p)=\phi(p)F^{-}(p)+\varphi(p)$ is based on the following results. The proofs may be found in \cite{gakhov}. \begin{lemma} \label{lem2.4} %\label{1} An arbitrary function $\varphi(p)$ given on the contour $\operatorname{Re}p=0$, satisfying the H\"older condition, can be uniquely represented in the form $\varphi(p)=U^+(p)-U^-(p)$, where $U^\pm(p)$ are the boundary values of the analytic functions $U^\pm(z)$ and the condition $U^\pm(\infty)=0$ holds. These functions are determined by formula \begin{equation*} U(z)=\frac{1}{2\pi i}\int_{-i\infty }^{i\infty }\frac{1}{q-z}\varphi(q)dq \end{equation*} \end{lemma} \begin{lemma} \label{lem2.5} %\label{2} An arbitrary function $\phi(p)$ given on the contour $\operatorname{Re}p=0$, satisfying the H\"older condition, and having zero index, \[ \operatorname{ind}\phi(p):=\frac{1}{2\pi i}\int_{-i\infty }^{i\infty }d\ln\phi(p)=, \] is uniquely representable as the ratio of the functions $X^+(p)$ and $X^-(p)$, constituting the boundary values of functions, $X^+(z)$ and $X^-(z)$, analytic in the left and right complex semi-plane and having in these domains no zero. These functions are determined to within an arbitrary constant factor and given by formula \begin{equation*} X^{\pm }(z)=e^{\Gamma^\pm(z)},\quad \Gamma (z)=\frac{1}{2\pi i} \int_{-i\infty }^{i\infty }\frac{1}{q-z}\ln\phi(q)dq. \end{equation*} \end{lemma} In the formulations of Lemmas \ref{lem2.4} and \ref{lem2.5} the coefficient $\phi(p)$ and the free term $\varphi(p)$ of the Riemann problem are required to satisfy the H\"older condition on the contour $\operatorname{Re}p=0$. This restriction is essential. On the other hand, it is easy to observe that both functions $W(p,\xi)$ and $g(p,\xi)$ do not have limiting value as $p\to\pm i\infty$. So we can not find the solution using $\ln W(p,\xi)$. The principal task now is to get an expression equivalent to the boundary value problem \eqref{20}, such that the conditions of Lemmas are satisfied. First, we introduce the function \begin{equation}\label{23} \phi(p,\xi)=\Big(\frac{K(p)+\xi}{K_1(p)+\xi}\Big) \frac{w^-(p)}{w^+(p)},\quad w^\pm(p)=\frac{1}{(p\mp z_0)^{\alpha/2}}, \end{equation} where $K(p)=|p|^\alpha$, $K_1(p)=p^\alpha$ and $z_0>0$. We make a cut in the plane $z$: $(-\infty,-z_0]\cup[z_0,\infty)$. Owing to the manner of performing the cut the functions $w^-(z)$, $K_1(z)$ are analytic for $\operatorname{Re}z>0$ and the function $w^+(z)$ is analytic for Re $z<0$. We observe that the function $\phi(p,\xi)$ given on the contour Re $p=0$, satisfies the H\"older condition and $K_1(p)+\xi$ does not vanish for any Re $\xi>0$. Also we have $\operatorname{ind}\phi(p,\xi) =\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}d\ln\phi(p,\xi)=0$. Therefore in accordance with Lemma \ref{lem2.5}, the function $\phi(p,\xi)$ can be represented as \begin{equation}\label{25} \phi(p,\xi)=\frac{X^+(p,\xi)}{X^-(p,\xi)},\quad X^\pm(p,\xi)= \lim_{\substack{z\to p\\ \pm\textrm{\tiny{Re}}z<0}}e^{\Gamma_0(z,\xi)}, \end{equation} where $\Gamma_0(z,\xi)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{1}{q-z}\ln\phi(q,\xi)dq$. From equations \eqref{23} and \eqref{25} we obtain \begin{equation}\label{26} \frac{Y^+}{Y^-}=\frac{K(p)+\xi}{K_1(p)+\xi}, \end{equation} where $Y^\pm=e^{\Gamma^\pm}w^\pm$. Now, we show that $Y^\pm$ do not depend of $z_0$. Integrating by parts we obtain for $\Gamma_0$: \begin{equation}\label{gamma} \Gamma_0= \lim_{R\to\infty} \frac{1}{2\pi i}\ln(q-z)\ln\phi(q,\xi)\Big|_{-iR}^{iR} -\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\ln(q-z)\partial_q\ln\phi(q,\xi)dq. \end{equation} Using that $\ln\phi(\pm iR,\xi)\to0$, as $R\to\infty$, we obtain \begin{equation}\label{gamma1} \Gamma_0(z,\xi)=-\frac{1}{2\pi i} \int_{-i\infty}^{i\infty}\ln(q-z)\partial_q\ln\phi(q,\xi)dq. \end{equation} We define $\ln^\pm(z)=\ln|z|+i\arg^\pm(z)$, where $-\frac{3}{2}\pi<\arg^-(z)\leq\frac{\pi}{2}$ and $-\frac{3}{2}\pi\leq\arg^+(z)<\frac{\pi}{2}$. Then, for $\operatorname{Re}p=0$, \begin{equation}\label{gamma6} \Gamma_0^\pm(p,\xi)=-\frac{1}{2\pi i} \int_{-i\infty}^{i\infty}\ln^\mp(q-p)\partial_q\ln\phi(q,\xi)dq. \end{equation} Now, since \begin{equation*} \partial_q\ln\phi(q,\xi)=F(q,\xi)+\frac{\alpha z_0}{(q-z_0)(q+z_0)}, \end{equation*} where $F(q,\xi)=\frac{K'(q)}{K(q)+\xi}-\frac{K_1'(q)}{K_1(q)+\xi}$, and via Cauchy's Residue Theorem, \begin{equation*} -\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\ln^\pm(q-p)\frac{\alpha z_0}{(q-z_0)(q+z_0)}dq= \pm i\pi\frac{\alpha}{2}+\ln^\pm(p\pm z_0)^{\alpha/2}, \end{equation*} from \eqref{gamma6} follows that \begin{equation*} \Gamma^\pm(p,\xi)=\mp i\pi\frac{\alpha}{2}+\ln^\mp(p\mp z_0)^{\alpha/2} -\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\ln^\mp(q-p)F(q,\xi)dq. \end{equation*} Therefore, \begin{equation}\label{gamma4} Y^\pm(p,\xi)=(\mp i)^\alpha e^{\Gamma^\pm(p,\xi)}, \end{equation} where $$ \Gamma^\pm(p,\xi)=-\frac{1}{2\pi i}\int_{-i\infty}^{i\infty} \ln^\mp(q-p)F(q,\xi)dq. $$ We note that equation \eqref{26} is equivalent to $$ \frac{K(p)+\xi}{\xi}=\frac{Y^+}{Y^-} (\frac{K_1(p)+\xi}{\xi}). $$ Now, we return to the nonhomogeneous Riemann-Hilbert problem defined by the boundary condition \eqref{20}. We substitute the above equation in \eqref{20} and add $-\frac{\xi}{Y^+}\Lambda^+$ in both sides to get \begin{equation}\label{27} \frac{\Omega^+-\xi\Lambda^+}{Y^+} =\Big(\frac{K_1(p)+\xi}{\xi}\Big) \frac{\Omega^-}{Y^-}-\Big(\frac{K(p)+\xi}{Y^+}\Big)\Lambda^+. \end{equation} On the other hand, by Sokhotski-Plemelj formulae and \eqref{16}, $$ \Lambda^+-\Lambda^-=\frac{1}{K(p)+\xi} \Big(\widehat{u}_0(p)+\frac{K(p)}{p}\widehat{h}(\xi)\Big). $$ Now, we substitute $\Lambda^+$ from this equation in formula \eqref{27}, then by \eqref{26} we arrive to \begin{equation}\label{28} \frac{\Omega^+-\xi\Lambda^+}{Y^+}=\frac{K_1(p)+\xi}{\xi} \Big(\frac{\Omega^--\xi\Lambda^-}{Y^-}\Big) -\frac{1}{Y^+}\Big(\widehat{u}_0(p)+\frac{K(p)}{p}\widehat{h}(\xi)\Big). \end{equation} In subsequent consideration we shall have to use the following property of the limiting values of a Cauchy type integral, the statement of which we now quote. The proofs may be found in \cite{gakhov}. \begin{lemma} \label{G} If $L$ is a smooth closed contour and $\phi(q)$ a function that satisfies the H\"older condition on $L$, then the limiting values of the Cauchy type integral \begin{equation*} \Phi (z)=\frac{1}{2\pi i}\int_{L}\frac{1}{q-z}\phi (q)dq \end{equation*} also satisfy this condition. \end{lemma} Since $\widehat{u}_0(p)+\frac{K(p)}{p}\widehat{h}(\xi)$ satisfies on $\operatorname{Re}(p)=0$ the H\"older condition, on basis of Lemma \ref{G} the function $\frac{1}{Y^+}(\widehat{u}_0(p)+\frac{K(p)}{p}\widehat{h}(\xi))$ also satisfies this condition. Therefore, in accordance with Lemma \ref{lem2.4} it can be uniquely represented in the form of the difference of the functions $U^+(p,\xi)$ and $U^-(p,\xi)$, constituting the boundary values of the analytic function $U(z,\xi)$, given by formula \begin{equation}\label{29} U(z,\xi)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{1}{q-z}\frac{1}{Y^+(q,\xi)} \Big(\widehat{u}_0(q)+\frac{K(q)}{q}\widehat{h}(\xi)\Big)dq. \end{equation} Therefore, the equation \eqref{28} takes the form $$ \frac{\Omega^+-\xi\Lambda^+}{Y^+}+U^+=\frac{K_1(p)+\xi}{\xi} \Big(\frac{\Omega^--\xi\Lambda^-}{Y^-}\Big)+U^-. $$ The last relation indicates that the function $\frac{\Omega^+-\xi\Lambda^+}{Y^+}+U^+$, analytic in $\operatorname{Re}(z)<0$, and the function $\frac{K_1(p)+\xi}{\xi} (\frac{\Omega^--\xi\Lambda^-}{Y^-})+U^-$, analytic in $\operatorname{Re}(z)>0$, constitute the analytic continuation of each other through the contour $\operatorname{Re}(z)=0$. Consequently, they are branches of a unique analytic function in the entire plane. Moreover, this function has a zero in infinite. According to Liouville theorem this function is identically zero. Thus, we obtain the solution of the Riemann-Hilbert problem defined by the boundary condition \eqref{20}, \begin{equation}\label{31} \begin{gathered} \Omega^+(p,\xi) = -Y^+(p,\xi)U^+(p,\xi) +\xi\Lambda^+(p,\xi) \\ \Omega^-(p,\xi) = - \frac{\xi}{K_1(p)+\xi}Y^-(p,\xi)U^-(p,\xi) +\xi\Lambda^-(p,\xi). \end{gathered} \end{equation} Now, we proceed to find the unknown function $\widehat{\Phi}$ of \eqref{12} for the solution $\widehat{\widehat{u}}$ of problem \eqref{4}. First, we represent $\Omega^-$ as the limiting value of analytic functions on the left-hand side complex semi-plane. From equation \eqref{26} and Sokhotski-Plemelj formulae we obtain $\Omega^-=-\frac{\xi}{K(p)+\xi}Y^+ U^++\xi\Lambda^+$. Now, making use of \eqref{31} and the last equation, we obtain \[ \Omega^+-\Omega^-=-\frac{K(p)}{K(p)+\xi}Y^+U^+. \] Thus, by formula \eqref{22}, $\widehat{\Phi}=-Y^+U^+$. We observe that $\widehat{\Phi}$ is boundary value of a function analytic in the left-hand side complex semi-plane and therefore satisfies our basic assumption \eqref{18}. Having determined the function $\widehat{\Phi}$ and bearing in mind formula \eqref{12} we determine the required function: $\widehat{\widehat{u}}=\frac{1}{K(p)+\xi} (\widehat{u}_0(p)-Y^+U^+)$. Now we prove that, in accordance with last relation, the function $\widehat{\widehat{u}}$ constitutes the limiting value of an analytic function in $\operatorname{Re}(z)>0$. In fact, making use of Sokhotski-Plemelj formulae and using \eqref{26}, we obtain $\widehat{\widehat{u}}=-\frac{1}{K_1(p)+\xi}Y^-U^-$. Thus, the function $\widehat{\widehat{u}}$ is the limiting value of an analytic function in $\operatorname{Re}(z)>0$. We note the fundamental importance of the proven fact, the solution $\widehat{\widehat{u}}$ constitutes an analytic function in $\operatorname{Re}(z)>0$ and, as a consequence, its inverse Laplace transform vanish for all $x<0$. We now return to solution $u$ of the problem \eqref{4}. Taking inverse Laplace transform with respect to time and space variables, we obtain \eqref{7}. Proposition \ref{prop2.3} has been proved. \end{proof} In the following lemma we collect some preliminary estimates for the Green operator $\mathcal{G}(t)$ \begin{lemma}\label{39} If $\Lambda(s)$ as defined in \eqref{46}, then exists $C>0$ such that \begin{gather} \|\mathcal{G}(t)\phi-\vartheta t^{-\frac{1}{\alpha}-\frac{1}{2}} \Lambda((\cdot)t^{-1/\alpha}) \|_{ L^\infty}\leq Ct^{-\frac{1}{\alpha}-\frac{1}{2}-\gamma} \|\phi\|_{ L^{1,\frac{\alpha}{2}}},\quad\gamma>0,\label{39a} \\ \|\mathcal{G}(t)\phi\|_{ L^\infty}\leq C\{t\}^{-\gamma}\langle t\rangle^{-1/\alpha} (\|\phi\|_{ L^1}+\|\phi\|_{ L^\infty}),\quad 0<\gamma<1,\label{39b} \\ \|\mathcal{G}(t)\phi\|_{ L^{s,\mu}}\leq Ct^{-\frac{1}{\alpha} (\frac{1}{r}-\frac{1}{s}-\mu)} \|\phi\|_{ L^r}+Ct^{-\frac{1}{\alpha} (\frac{1}{r}-\frac{1}{s})} \|\phi\|_{ L^{r,\mu}},\label{39d} \end{gather} where $\vartheta=\int_0^{+\infty}y^{\alpha/2}\phi(y)dy$, $|\frac{1}{r}-\frac{1}{s}-\mu|<\alpha$ and $1\leq r\leq s\leq\infty$. \end{lemma} \begin{proof} First, we estimate the function $Y^+=(-i)^\alpha e^{\Gamma^+(p,\xi)}$, where $$ \Gamma^+(p,\xi)=-\frac{1}{2\pi i}\int_{-i\infty}^{i\infty} \ln^-(q-p)F(q,\xi)dq $$ and $F(q,\xi)=\frac{K'(q)}{K(q)+\xi}-\frac{K_1'(q)}{K_1(q)+\xi}$. Using the relation \begin{equation*} \ln^-(q-p)=\pi i+\ln^+(p)+\ln^+\big(1-\frac{q}{p}\big) +2\pi i\theta(\operatorname{Im}p)\theta(\operatorname{Im}q-\operatorname{Im}p), \end{equation*} where $\theta$ is the Heaviside step function, we obtain $$ \Gamma^+(p,\xi)=\frac{\alpha}{2}(\pi i+\ln^+(p))+\frac{1}{2\pi i} \int_{-i\infty}^{i\infty}\ln^+\big(1-\frac{q}{p}\big)F(q,\xi)dq+ \mathcal{O}(p^{-\alpha\gamma}\xi^{\gamma}), $$ where $0<\gamma<1$. Then, separating the integral, $|q|<|p|$ and $|q|<|p|$, and using $\ln(1-z)=\mathcal{O}(z^\gamma)$, for $|z|<1$ and $0<\gamma<1$, and the equation $$ \ln^+\big(1-\frac{q}{p}\big)=-\pi i+\ln^+(q)-\ln^+(p)+\ln^+(1-\frac{p}{q}) + 2\pi i\varsigma(\operatorname{Im}p,\operatorname{Im}q), $$ where $\varsigma(x,y)=\theta(x-y)(\theta(x)\theta(y)- \theta(-x)\theta(-y))+\theta(-x)$, we obtain \begin{equation}\label{gamma5} \Gamma^+(p,\xi)=\frac{\alpha}{2}(\pi i+\ln^+(p))+\mathcal{O} \big(p^{\alpha(1-\gamma)}\xi^{\gamma-1}\big),\quad 0<\gamma<1. \end{equation} Therefore, from \eqref{gamma4} and \eqref{gamma5} follows \begin{gather}\label{assy1} Y^+(p,\xi)=p^{\alpha/2}+\mathcal{O} \big(p^{\alpha(\frac{3}{2}-\gamma)}\xi^{\gamma-1}\big),\quad 0<\gamma<1, \\ \label{assy2} \frac{1}{Y^+(p,\xi)}=\mathcal{O}(p^{-\alpha/2}). \end{gather} We note that in the above formulas $p^\beta=e^{\beta\ln^+(p)}$. Now, we show that \begin{equation}\label{esty} \partial_\xi^j\Gamma^+(p,\xi)=\mathcal{O}(p^{-j\alpha\gamma}\xi^{j(\gamma-1)}), \quad 0<\gamma<1,\; j=1,2. \end{equation} In fact, from Cauchy's Theorem we obtain \[ \frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\ln^-(q-p) \frac{K_1'(q)}{(K_1(q)+\xi)^2}dq=0, \] then integrating by parts \begin{align*} \partial_\xi^j\Gamma^+(p,\xi) &= \frac{(-1)^{j+1}j}{2\pi i} \int_{-i\infty}^{i\infty}\ln^-(q-p)\frac{K'(q)}{(K(q)+\xi)^{j+1}}dq \\ &= \frac{(-1)^j}{2\pi i}\Big(\frac{\ln^-(q-p)}{(K(q)+\xi)^j} \Big|_{-i\infty}^{i\infty} -\int_{-i\infty}^{i\infty}\frac{1}{q-p}\frac{dq}{(K(q)+\xi)^j}\Big) \\ &= \mathcal{O}(p^{-j\alpha\gamma}\xi^{j(\gamma-1)}). \end{align*} As a direct consequence of \eqref{assy2} and \eqref{esty} we obtain \begin{gather}\label{esty1} \partial_\xi(\frac{1}{Y^+})=-\frac{\Gamma_\xi^+}{Y^+} =\mathcal{O}(p^{-\alpha\gamma-\frac{\alpha}{2}}\xi^{\gamma-1}),\\ \label{esty2} \partial_{\xi}^2\big(\frac{1}{Y^+}\big) =\frac{1}{Y^+}\Big((\Gamma_\xi^+)^2 -\Gamma_{\xi\xi}^+\Big)=\mathcal{O}\big(p^{-2\alpha\gamma -\frac{\alpha}{2}}\xi^{2(\gamma-1)}\big). \end{gather} Then, using $e^{-qy}-1=\mathcal{O}(q^\mu y^\mu)$, $0<\mu<1$, \eqref{esty1} and \eqref{esty2}, we obtain for $Z_0^+(p,\xi,y)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty} \frac{e^{-qy}-1}{q-p}\frac{dq}{Y^+(q,\xi)}$, \begin{equation}\label{estz1} \partial_\xi^jZ_0^+(p,\xi,y)=\mathcal{O}(y^\mu p^{\mu-\alpha(j\gamma+\frac{1}{2})} \xi^{j(\gamma-1)}),\quad0<\mu<1, \end{equation} where $j=1,2$. Moreover, using $\frac{1}{q-p}=\frac{1}{q}+\frac{p}{q(q-p)}$, we express $Z_0^+$ in the form $$ Z_0^+(p,\xi,y)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty} \frac{e^{-qy}-1}{q}\frac{dq}{Y^+(q,\xi)} +\frac{p}{2\pi i}\int_{-i\infty}^{i\infty}\frac{e^{-qy}-1}{q(q-p)} \frac{dq}{Y^+(q,\xi)}. $$ Thus, from \eqref{assy2} we arrive to \begin{equation}\label{estz2} Z_0^+(p,\xi,y)=C_0y^{\alpha/2}+\mathcal{O}(y^\mu p^{\mu-\frac{\alpha}{2}}), \quad\mu>\frac{\alpha}{2}, \end{equation} where $C_0=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty} \frac{e^{-q}-1}{q^{1+\frac{\alpha}{2}}}dq$. Now, we consider the function $\frac{Y^+}{K(p)+\xi}Z_0^+$. From \eqref{assy1}, \eqref{esty} and \eqref{estz1}, we obtain \begin{align*} &\partial_\xi^2\Big(\frac{Y^+}{K(p)+\xi}Z_0^+\Big) \\ &= \frac{Y^+}{K(p)+\xi}\Big(\Big(\Gamma_\xi^+-\frac{1}{K(p)+\xi}\Big)^2 Z_0^++2\Big(\Gamma_\xi^+-\frac{1}{K(p)+\xi}\Big)\partial_\xi Z_0^+ \\ &\quad +\Big(\Gamma_{\xi\xi}^++\frac{1}{(K(p)+\xi)^2}\Big)Z_0^+ +\partial_\xi^2Z_0^+\Big) \\ &= \frac{1}{K(p)+\xi}\big(Y^+Z_0^+L+\mathcal{O}(y^\mu p^{\mu-2\alpha\gamma} \xi^{2(\gamma-1)})\big), \end{align*} where \begin{equation*} L(p,\xi)=\Big(\Gamma_\xi^+-\frac{1}{K(p)+\xi}\Big)^2 +\Gamma_{\xi\xi}^++\frac{1}{(K(p)+\xi)^2} =\mathcal{O}\big(p^{-2\alpha\gamma} \xi^{2(\gamma-1)}\big). \end{equation*} Then, using \eqref{assy1} and \eqref{estz2} we obtain the estimate \begin{equation}\label{second} \partial_\xi^2\Big(\frac{Y^+}{K(p)+\xi}Z_0^+\Big) =\frac{C_0y^{\alpha/2}}{K(p)+\xi}\Big(p^{\alpha/2}L+ \mathcal{O}(p^{\mu-2\alpha\gamma} \xi^{2(\gamma-1)})\Big),\quad \mu>\frac{\alpha}{2}. \end{equation} On the other hand, by Sokhotski-Plemelj formulae and Cauchy's Residue Theorem, we represent the Green function $G$, \eqref{5}, as \begin{align*} G(x,y,t) &= -\frac{1}{(2\pi i)^2}\int_{-i\infty}^{i\infty}d\xi e^{\xi t} \int_{-i\infty}^{i\infty}e^{px}\frac{Y^+(p,\xi)}{K(p)+\xi}Z^+(p,\xi,y)dp \\ &\quad +\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}e^{p(x-y)-K(p)t}dp. \end{align*} Now, by Cauchy's Theorem we obtain \begin{equation*} Z=\frac{1}{Y^+}+Z_0,\quad\text{where}\quad Z_0(z,\xi,y)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty} \frac{e^{-qy}-1}{q-z}\frac{dq}{Y^+(q,\xi)}. \end{equation*} Therefore, we can represent $G$ in the form \begin{equation}\label{gest1} \begin{aligned} G &= -\frac{1}{(2\pi i)^2}\int_{-i\infty}^{i\infty}d\xi e^{\xi t} \int_{-i\infty}^{i\infty}e^{px}\frac{Y^+(p,\xi)}{K(p)+\xi}Z_0^+(p,\xi,y)dp \\ &\quad +\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}e^{px-K(p)t}(e^{-py}-1)dp=G_1+G_2. \end{aligned} \end{equation} Using $(e^{-py}-1)=\mathcal{O}(p^\mu y^\mu)$, $0<\mu<1$, we obtain for the second integral in the above equation \begin{equation}\label{gest2} G_2(x,y,t)=\mathcal{O}(t^{-\frac{1}{\alpha}(1+\mu)}y^\mu). \end{equation} To estimate $G_1$, we use the analyticity of the integrand in the left-half semiplane and integrate by parts two times with respect to $\xi$ to obtain \begin{equation*} G_1(x,y,t)=Ct^{-2}\int_{-i\infty}^{i\infty}d\xi e^{\xi t} \int_{-i\infty}^{i\infty}e^{px}\partial_\xi^2 \Big(\frac{Y^+(p,\xi)}{K(p)+\xi}Z_0^+(p,\xi,y)\Big)dp. \end{equation*} Then, by \eqref{second} and \eqref{gest2} follows \begin{equation}\label{assg} G(x,y,t)=y^{\alpha/2}t^{-\frac{1}{\alpha}(1+\frac{\alpha}{2})}\Lambda(xt^{-1/\alpha}) +\mathcal{O}(y^{\alpha/2}t^{-\frac{1}{\alpha}(1+\mu)}),\quad\mu>\frac{\alpha}{2}, \end{equation} where $2/5<\alpha<1$ and \begin{equation*} \Lambda(s)=C\int_{-i\infty}^{i\infty}d\xi e^\xi \int_{-i\infty}^{i\infty}e^{ps}\frac{p^{\alpha/2}}{K(p)+\xi}L(p,\xi)dp. \end{equation*} Multiplying by a function $\phi$ and integrating with respect to $y$ in \eqref{assg} we obtain $$ \Big|\int_0^{+\infty}(G(x,y,t)-y^{\alpha/2}t^{-\frac{1}{\alpha}(1+\frac{\alpha}{2})} \Lambda(xt^{-1/\alpha})) \phi(y)dy\Big|\leq Ct^{-\frac{1}{\alpha}(1+\mu)}\|\phi\|_{ L^{1,\frac{\alpha}{2}}}. $$ Therefore, $$ \big\|\mathcal{G}(t)\phi-t^{-\frac{1}{\alpha}(1+\frac{\alpha}{2})} \Lambda((\cdot)t^{-1/\alpha}) \int_0^{+\infty}y^{\alpha/2}\phi(y)dy\big\|_{ L^\infty} \leq Ct^{-\frac{1}{\alpha}(1+\mu)} \|\phi\|_{ L^{1,\frac{\alpha}{2}}}, $$ where $\mu>\frac{\alpha}{2}$. Thus, the first estimate in Lemma \ref{39} has been proved. Now, we are going to prove the second estimate in Lemma \ref{39}. First, for large $t$. From \eqref{gest1}, integrating by parts the first summand on the right hand side two times with respect to $\xi$ we obtain \begin{equation} \label{greenrep1} \begin{aligned} G(x,y,t) &= Ct^{-2}\int_{-i\infty}^{i\infty}d\xi e^{\xi t} \int_{-i\infty}^{i\infty}e^{px}\partial_\xi^2 \Big(\frac{Y^+(p,\xi)}{K(p)+\xi}Z^+(p,\xi,y)\Big)dp \\ &\quad +\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}e^{p(x-y)-K(p)t}dp=J_1+J_2. \end{aligned} \end{equation} We have \[ \partial_\xi^2(\frac{Y^+}{K(p)+\xi}Z^+)=\frac{1}{K(p)+\xi}\mathcal{O}( p^{-2\alpha\gamma}\xi^{2(\gamma-1)}), \] $0<\gamma<1$, then $J_1=\mathcal{O}(t^{-1/\alpha})$, provided $\frac{1}{3}<\alpha<1$, also $J_2=\mathcal{O}(t^{-1/\alpha})$. Therefore, $G=\mathcal{O}(t^{-1/\alpha})$ and \begin{equation}\label{54} \|\mathcal{G}(t)\phi\|_{ L^\infty}= \sup_{x\in\mathbb{R}^+}\big|\int_0^{+\infty}G(x,y,t)\phi(y)dy\big| \leq t^{-1/\alpha}\|\phi\|_{ L^1}. \end{equation} Now, for small $t$, using the inequality $|e^{z}|\leq|z|^{-\gamma}$, for $\operatorname{Re}(z)<0$ and $\gamma>0$, we obtain \[ \partial_\xi^2(\frac{Y^+}{K(p)+\xi}Z^+)=\frac{1}{K(p)+\xi}\mathcal{O}( y^{-\gamma}p^{-\gamma-2\alpha\gamma_1}\xi^{2(\gamma_1-1)}), \] $\gamma>0$, $0<\gamma_1<1$. Then, from \eqref{greenrep1} we obtain $J_1=\mathcal{O}(y^{-\gamma}t^{-\frac{1}{\alpha}(1-\gamma)})$, $\gamma>0$. On the other hand, we write the second summand as $$ J_2(r,t)=\frac{1}{2\pi i}\int_{\mathcal{C}_\pm}e^{pr-K(p)t}dp,\quad\pm r>0, $$ where $$ \mathcal{C}_\pm=\{p\in(\infty e^{-i(\frac{\pi}{2}\pm\varepsilon)},0) \cup(0,\infty e^{i(\frac{\pi}{2}\pm\varepsilon)})\},\quad\varepsilon>0. $$ Making the change of variable $p=zt^{-1/\alpha}$ and using the inequality $|e^{z}|\leq|z|^{-\gamma}$, for $\operatorname{Re}(z)<0$ and $\gamma>0$, we obtain $J_2=\mathcal{O}(t^{-\frac{1}{\alpha}(1-\gamma)}|x-y|^{-\gamma})$, where $x,y>0$. Therefore, $G=\mathcal{O}\big(t^{-\frac{1}{\alpha}(1-\gamma)}(|x-y|^{-\gamma}+y^{-\gamma})\big)$. Thus, \begin{align*} \|\mathcal{G}(t)\phi\|_{ L^\infty} & \leq \sup_{x\in\mathbb{R}^+}\int_0^{+\infty} |G(x,y,t)||\phi(y)|dy \\ & \leq C t^{-\frac{1}{\alpha}(1-\gamma)} \int_0^{+\infty}(|x-y|^{-\gamma}+y^{-\gamma})|\phi(y)|dy \\ & \leq Ct^{-\frac{1}{\alpha}(1-\gamma)} (\|\phi\|_{ L^1} +\|\phi\|_{ L^\infty}). \end{align*} The second estimate in Lemma \ref{39} has been proved. Let us introduce the operators \begin{gather}\label{jj1} \mathcal{J}_1(t)\phi=\theta(x)\int_0^{+\infty}J_1(x,y,t)\phi(y)dy, \\ \label{jj2} \mathcal{J}_2(t)\phi=\theta(x)\int_0^{+\infty}J_2(x-y,t)\phi(y)dy, \end{gather} Then, the operator $\mathcal{G}(t)$ can be written in the form \begin{equation}\label{a55} \mathcal{G}(t)=\mathcal{J}_1(t)+\mathcal{J}_2(t). \end{equation} Now, we are going to prove the third estimate in Lemma \ref{39}. First, we estimate the operator $\mathcal{J}_2$. Making the change of variable $p=qt^{-1/\alpha}$, we obtain \begin{equation}\label{a52} |J_2(r,t)|\leq Ct^{-1/\alpha}. \end{equation} On the other hand, making $z=t^{-1/\alpha}r$, we obtain \[ J_2=\frac{t^{-1/\alpha}}{2\pi i}\int_{-i\infty}^{i\infty}e^{qz-K(q)}dq. \] Then, integrating by parts the last equation we obtain $$ J_2(r,t)=\frac{t^{-1/\alpha}}{2\pi i}(\frac{1}{z})\int_{-i\infty}^{i\infty}e^{-K(q)}de^{qz} =\frac{t^{-1/\alpha}}{2\pi i}(\frac{\alpha}{z}) \int_{-i\infty}^{i\infty}\frac{K(q)}{q}e^{qz-K(q)}dq $$ Thus, \[ |J_2(r,t)|\leq Ct^{-1/\alpha}\frac{1}{|z|^{1+\gamma}} \int_{\mathcal{C}_\pm}|q|^{\alpha-1-\gamma} e^{-C|q|^\alpha}|dq|, \] for $\pm r>0$, where $\mathcal{C}_\pm$ are defined as above. Therefore, \begin{equation}\label{a53} |J_2(r,t)|\leq Ct^{-1/\alpha}\frac{1}{|z|^{1+\gamma}},\quad\gamma<\alpha. \end{equation} Finally, from the inequalities \eqref{a52} and \eqref{a53} we have \begin{equation}\label{a78} |J_2(r,t)|\leq C\frac{t^{-1/\alpha}} {1+(t^{-1/\alpha}|r|)^{1+\gamma}},\ \ \ \gamma<\alpha. \end{equation} Lets write some well known inequalities: $\bullet$ Young's inequality. Let $f\in L^p(\mathbb{R})$ and $g\in L^q(\mathbb{R})$, where $1\leq p,q\leq\infty$, $\frac{1}{p}+\frac{1}{q}\geq1$. Then, the convolution $h(x)\equiv\int_\mathbb{R}f(x-y)g(y)dy$ belongs to $ L^r(\mathbb{R})$, where $\frac{1}{r}=\frac{1}{p}+\frac{1}{q}-1$ and the Young's inequality \begin{equation}\label{young} \|h\|_{ L^r}\leq\|f\|_{ L^p}\|g\|_{ L^q} \end{equation} holds. $\bullet$ Minkowski's Inequality. Let $f,g\in L^p$ and $1\leq p\leq\infty$, then \begin{equation}\label{minkowski} \|f+g\|_{ L^p}\leq\|f\|_{ L^p}+\|g\|_{ L^p}. \end{equation} Now, we estimate the operator $\mathcal{J}_2$, defined in \eqref{jj1}. Using the inequality $x^\mu\leq|x-y|^\mu+y^\mu$, where $0\leq\mu\leq1$, and Minkowski's inequality \eqref{minkowski}, we obtain \begin{align*} \|\mathcal{J}_2(t)\phi\|_{ L^{s,\mu}} & \leq \Big(\int_0^{+\infty} \Big(\int_0^{+\infty}|x-y|^\mu|J_2(x-y,t)||\phi(y)|dy \Big)^sdx\Big)^{1/s} \\ &\quad + \Big(\int_0^{+\infty} \Big(\int_0^{+\infty}y^\mu|J_2(x-y,t)||\phi(y)|dy \Big)^sdx\Big)^{1/s}. \end{align*} Then, Young's inequality \eqref{young} implies \begin{equation}\label{a79} \|\mathcal{J}_2(t)\phi\|_{ L^{s,\mu}} \leq \|J_2(\cdot,t)\|_{ L^{p,\mu}} \|\phi\|_{ L^r}+\|J_2(\cdot,t)\|_{ L^p} \|\phi\|_{ L^{r,\mu}}, \end{equation} where $\frac{1}{s}=\frac{1}{p}+\frac{1}{r}-1$, $1\leq p,r\leq\infty$, $1\leq\frac{1}{p}+\frac{1}{r}\leq2$ and $0\leq\mu\leq1$. Then, by the inequality \eqref{a78} and the change of variables $x=t^{-1/\alpha}|r|$, we obtain $$ \|J_2(\cdot,t)\|_{ L^{p,\mu}}\leq Ct^{-\frac{1}{\alpha} (1-\frac{1}{p}-\mu)}\Big(\int_{-\infty}^{+\infty} (\frac{|x|^\mu}{(1+ |x|)^{1+\gamma}})^pdx\Big)^{1/p}. $$ Thus, $\|J_2(\cdot,t)\|_{ L^{p,\mu}}\leq Ct^{-\frac{1}{\alpha} (1-\frac{1}{p}-\mu)}$, provided $1+\gamma-\mu>\frac{1}{p}$. Using $\frac{1}{s}=\frac{1}{p}+\frac{1}{r}-1$, it follows \begin{equation}\label{a80} \|J_2(\cdot,t)\|_{ L^{p,\mu}}\leq Ct^{-\frac{1}{\alpha} (\frac{1}{r}-\frac{1}{s}-\mu)}, \end{equation} where $\frac{1}{r}-\frac{1}{s}-\mu+\gamma>0$. We note that $-\frac{1}{\alpha} (\frac{1}{r}-\frac{1}{s}-\mu)<1$, since $\gamma<\alpha$. Substituting \eqref{a80} in \eqref{a79}, we obtain \begin{equation}\label{a81} \|\mathcal{J}_2(t)\phi\|_{ L^{s,\mu}}\leq Ct^{-\frac{1}{\alpha} (\frac{1}{r}-\frac{1}{s}-\mu)} \|\phi\|_{ L^r}+Ct^{-\frac{1}{\alpha} (\frac{1}{r}-\frac{1}{s})} \|\phi\|_{ L^{r,\mu}}, \end{equation} where $-\frac{1}{\alpha} (\frac{1}{r}-\frac{1}{s}-\mu)<1$, $1\leq s,r\leq\infty$ and $0\leq\mu\leq1$. Now, we estimate the operator $\mathcal{J}_1$, defined in \eqref{jj2}. First, we note that using the Cauchy theorem we obtain \begin{align*} & \int_{-i\infty}^{i\infty}e^{px}\frac{Y^+(p,\xi)}{K(p)+\xi}Z^+(p,\xi,y)dp \\ &= \int_0^{+\infty} e^{-px}\Big(\frac{1}{(-ip)^\alpha+\xi}-\frac{1}{(ip)^\alpha+\xi}\Big)Y^+ (-p,\xi)Z^+(-p,\xi,y)dp. \end{align*} Then, by inequality \begin{equation}\label{hhh} \big|\frac{1}{(-ip)^\alpha+\xi}-\frac{1}{(ip)^\alpha+\xi}\big| \leq C\frac{|p|^{\alpha(1-\gamma)}}{||p|^{2\alpha}+\xi^2|^{1-\gamma}|\xi|^\gamma}, \end{equation} $0<\gamma<1$, we obtain for $J_1$ $$ |J_1|\leq C\int_{-i\infty}^{i\infty}|d\xi||e^{i\xi t}|\int_0^{+\infty}dp \frac{e^{-px}p^{\alpha(1-\gamma)+\frac{\alpha}{2}}} {|p^{2\alpha}+\xi^2|^{1-\gamma}|\xi|^\gamma} \int_{\mathcal{C}_3} \frac{e^{-C|q|y}}{|q-p|}\frac{|dq|}{|q|^{\alpha/2}}. $$ Thus, by $|\mathcal{J}_2(t)\phi|\leq\int_0^{+\infty}|J_2(x,y,t)||\phi(y)|dy$ and $$ \int_0^{+\infty}e^{-C|q|y}|\phi(y)|dy\leq \|e^{-C|q|(\cdot)}\|_{ L^l} \|\phi\|_{ L^r}=C|q|^{-\frac{1}{l}} \|\phi\|_{ L^r}, $$ where $\frac{1}{l}+\frac{1}{r}=1$, we obtain $$ |\mathcal{J}_2(t)\phi|\leq C\|\phi\|_{ L^r} \int_{-i\infty}^{i\infty}|d\xi||e^{i\xi t}|\int_0^{+\infty} \frac{e^{-px}p^{\alpha(1-\gamma)}}{|p^{2\alpha}+\xi^2| ^{1-\gamma}|\xi|^\gamma p^{\frac{1}{l}}} dp. $$ Then, using $\|e^{-p(\cdot)}\|_{ L^{s,\mu}}=Cp^{-\frac{1}{s}-\mu}$, $p>0$, and $\frac{1}{l}+\frac{1}{r}=1$, we obtain $$ \|\mathcal{J}_2(t)\phi\|_{ L^{s,\mu}}\leq C\|\phi\|_{ L^r} \int_{-i\infty}^{i\infty}|d\xi||e^{i\xi t}|\int_0^{+\infty} \frac{p^{\alpha(1-\gamma)}}{|p^{2\alpha} +\xi^2|^{1-\gamma}|\xi|^\gamma p^{1-\frac{1}{r}+\frac{1}{s}+\mu}} dp. $$ Therefore, \begin{equation}\label{a77} \|\mathcal{J}_2(t)\phi\|_{ L^{s,\mu}} \leq Ct^{-\frac{1}{\alpha} (\frac{1}{r}-\frac{1}{s}-\mu)}\|\phi\|_{ L^r}, \end{equation} where $|\frac{1}{r}-\frac{1}{s}-\mu|<\alpha$, $1\leq r\leq s\leq\infty$ and $0\leq\mu<1$. Finally, from estimates \eqref{a81} and \eqref{a77} we obtain the third estimate in Lemma \ref{39}. Then, we have proved Lemma \ref{39}. \end{proof} As a direct consequence of the above lemma, taking $r=1$, we obtain the estimate $$ \|\mathcal{G}(t)\phi\|_{ X^\mu}\leq C\|\phi\|_{\mathbb{Z}^\mu}. $$ Now, we collect some estimates for operator $\mathcal{H}$ defined by \begin{equation}\label{hhhh} \mathcal{H}(x)h=\int_0^tH(x,t-\tau)h(\tau)d\tau,\quad H(x,t)=\partial_t\partial_y^{-1}G(x,y,t)|_{y=0}. \end{equation} \begin{lemma}\label{hest} The following estimates are valid \begin{gather*} \|\mathcal{H}(\cdot)h\|_{ X^\mu} \leq C\|h\|_{ B^\alpha}, \quad\mu+\frac{1}{s}<\alpha, \\ \mathcal{H}(x)h = bt^{\beta+\gamma}\Lambda_1(xt^{-1/\alpha})+ \mathcal{O}(t^\beta), \end{gather*} for $h(t)=bt^\beta+\mathcal{O}(t^{\beta-\gamma})$, where $\beta<0$, $\gamma>0$ and \begin{align*} \Upsilon(r) &= -\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}d\xi e^\xi \int_{-i\infty}^{i\infty}e^{pr}\frac{Y^+(p,\xi)}{K(p)+\xi}I^+(p,\xi)dp \\ &\quad +\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}e^{pr-K(p)}\frac{K(p)}{p}dp. \end{align*} \end{lemma} \begin{proof} First, we note that \begin{equation}\label{hh} H(x,t)=H_1(x,t)+H_2(x,t), \end{equation} where \begin{gather*} H_1(x,t) = -\frac{1}{(2\pi i)^2}\int_{-i\infty}^{i\infty}d\xi e^{\xi t}\int_{-i\infty}^{i\infty}e^{px}\frac{Y^+(p,\xi)}{K(p)+\xi}I^+(p,\xi)dp, \\ H_2(x,t) = \frac{1}{2\pi i}\int_{-i\infty}^{i\infty}e^{px-K(p)t}\frac{K(p)}{p}dp. \end{gather*} Here, \begin{equation*} I(z,\xi)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty} \frac{1}{q-z}\frac{1}{Y^+(q,\xi)}\frac{K(q)}{q}dq. \end{equation*} For $H_2$, making the change of variable $q=pt^{-1/\alpha}$, we obtain $|H_2|\leq Ct^{-1}$. On the other hand, for $H_1$, we consider two domains: $|p|<1$ and $|p|>1$. For $|p|<1$, we use $Y^+I^+=\mathcal{O}(p^{-1+\alpha})$; and for $|p|>1$, we integrate by parts two times with respect to $\xi$ and use \[ \partial_\xi^2(\frac{Y^+}{K(p)+\xi}I^+)=\frac{p^{-1+\alpha}}{K(p)+\xi} \mathcal{O}(p^{-2\alpha\gamma}\xi^{2(\gamma-1)}) \] to obtain $|H_1|\leq Ct^{-1}$. Therefore, \begin{equation}\label{h11} |H(x,t)|\leq Ct^{-1}. \end{equation} By the Cauchy's Residue Theorem, for $\operatorname{Re}(z)<0$, we have \begin{equation}\label{h1n1} \begin{aligned} I & = \frac{1}{2\pi i}\int_{-i\infty}^{i\infty} \frac{1}{q-z}\frac{K(q)+\xi}{Y^+(q,\xi)}\frac{dq}{q} -\xi\frac{1}{2\pi i}\int_{-i\infty}^{i\infty} \frac{1}{q-z}\frac{1}{Y^+(q,\xi)}\frac{dq}{q} \\ & = -\frac{1}{2}\frac{1}{(-z)}\frac{\xi}{Y^+(0,\xi)} -\xi\Big(\frac{1}{z}\frac{1}{Y^+(z,\xi)} +\frac{1}{2}\frac{1}{(-z)}\frac{1}{Y^+(0,\xi)}\Big) \\ & = \frac{\xi}{zY^+(0,\xi)}-\frac{\xi}{zY^+(z,\xi)}. \end{aligned} \end{equation} Also, integrating $H$ with respect to $t$ and substituting \eqref{h1n1} we obtain \begin{equation}\label{h22} \partial_t^{-1}H(x,t) =-\frac{1}{(2\pi i)^2}\int_{-i\infty}^{i\infty}d\xi\frac{e^{\xi t}}{Y^+(0,\xi)} \int_{-i\infty}^{i\infty}e^{px}\frac{Y^+(p,\xi)}{K(p)+\xi}\frac{dp}{p}. \end{equation} Then, \begin{equation}\label{h12} |\partial_t^{-1}H(x,t)|\leq C \end{equation} Now, we divide the integration domain and integrate by parts \eqref{hhhh} to obtain \begin{align*} \mathcal{H}(x)h &=\int_0^{t/2}H(x,t-\tau)h(\tau)d\tau-h(\tau)\partial_\tau^{-1} H(x,t-\tau)\big|_{t/2}^t\\ &\quad +\int_{t/2}^th'(\tau)\partial_\tau^{-1}H(x,t-\tau)d\tau. \end{align*} Therefore, from \eqref{h11}, \eqref{h12} and last equation we have \begin{equation} \|\mathcal{H}(\cdot)h\|_{ L^\infty}\leq C\langle t\rangle^{-1/\alpha} \|h\|_{ B^\alpha}. \end{equation} We estimate the operator $\mathcal{H}$ in norm $ L^{s,\mu}$. First, we estimate $H_1$ from \eqref{hh}, $$ \|H_1(\cdot,t)\|_{ L^{s,\mu}}\leq C\int_{\mathcal{C}_1}|d\xi|e^{-\lambda|\xi|t} \int_{\mathcal{C}_2}\frac{1}{|K(p)+\xi|}\frac{|dp|}{|p|^{1-\alpha+\frac{1}{s}+\mu}}. $$ Making the change of variables $p=p_1t^{-1/\alpha}$, $\xi=\xi_1t^{-1}$, we obtain \begin{equation}\label{h111} \|H_1(\cdot,t)\|_{ L^{s,\mu}}\leq Ct^{\frac{1}{\alpha}(\frac{1}{s}+\mu)-1}, \quad\frac{1}{s}+\mu<\alpha. \end{equation} On the other hand for $H_2$ we have \begin{equation} \|H_2(\cdot,t)\|_{ L^{s,\mu}}\leq C\int_{-i\infty}^{i\infty}e^{-K(p)t} \frac{|dp|}{|p|^{1-\alpha+\frac{1}{s}+\mu}}. \end{equation} Thus, \begin{equation}\label{h112} \|H_2(\cdot,t)\|_{ L^{s,\mu}}\leq Ct^{\frac{1}{\alpha}(\frac{1}{s}+\mu)-1}, \quad\frac{1}{s}+\mu<\alpha. \end{equation} Therefore, from \eqref{h111} and \eqref{h112} we conclude $$ \|\mathcal{H}(\cdot)h\|_{ L^{s,\mu}}\leq C\|h\|_{ B^\alpha}\int_0^t (t-\tau)^{\frac{1}{\alpha}(\frac{1}{s}+\mu)-1}\langle t\rangle^{-1/\alpha}d\tau \leq C\langle t\rangle^{-\frac{1}{\alpha}(1-\frac{1}{s}-\mu)}\|h\|_{ B^\alpha}, $$ where $\frac{1}{s}+\mu<\alpha$. Finally, we estimate the operator $\mathcal{H}$ in norm $ L^1$. By Cauchy's Residue Theorem we have for $\operatorname{Re}(z)>0$, \begin{align*} I(z,\xi) & = \frac{1}{2\pi i}\int_{-i\infty}^{i\infty} \frac{1}{q-z}\frac{K(q)+\xi}{Y^+(q,\xi)}\frac{dq}{q}-\xi\frac{1}{2\pi i} \int_{-i\infty}^{i\infty} \frac{1}{q-z}\frac{1}{Y^+(q,\xi)}\frac{dq}{q} \\ & = -\frac{K(z)+\xi}{zY^+(z,\xi)}+\frac{\xi}{zY^+(0,\xi)}. \end{align*} Then, we obtain the formula \begin{equation}\label{hd1} H(x,t)=\frac{1}{(2\pi i)^2}\int_{-i\infty}^{i\infty}d\xi e^{\xi t} \int_{-i\infty}^{i\infty}e^{px}\Big(\frac{Y^+(p,\xi)\xi}{(K(p)+\xi)Y^+(0,\xi)}-1 \Big)\frac{dp}{p}. \end{equation} Now, we substitute the equation \begin{equation*} \frac{Y^+(p,\xi)\xi}{(K(p)+\xi)Y^+(0,\xi)}-1 =\Big(\frac{Y^+(p,\xi)-Y^+(0,\xi)}{Y^+(0,\xi)(K(p)+\xi)}\Big)\xi- \frac{K(p)}{K(p)+\xi} \end{equation*} in \eqref{hd1}, and change $\xi$ and $p$ by $\xi t^{-1}$ and $pt^{-1/\alpha}$, respectively, to obtain \begin{equation}\label{hd2} H(x,t)=t^{-1}J_1(s)+t^{-1}J_2(s),\quad s=xt^{-1/\alpha}, \end{equation} where \begin{equation}\label{hd3} J_1(s)=\frac{1}{(2\pi i)^2}\int_{-i\infty}^{i\infty}\frac{d\xi}{Y^+(0,\xi)} e^{\xi} \int_{-i\infty}^{i\infty}e^{ps}\Big(\frac{Y^+(p,\xi)-Y^+(0,\xi)}{K(p)+\xi}\Big) \xi\frac{dp}{p} \end{equation} and \begin{equation}\label{hd4} J_2(s)=-\frac{1}{(2\pi i)^2}\int_{-i\infty}^{i\infty}d\xi e^{\xi} \int_{-i\infty}^{i\infty}e^{ps}\frac{K(p)}{K(p)+\xi}\frac{dp}{p}. \end{equation} We estimate $J_2$. First, for $x<1$, we obtain \begin{equation}\label{j201} \int_0^1|J_2(xt^{-1/\alpha})|dx\leq Ct^\frac{1}{\alpha}. \end{equation} On the other hand, for $x>1$, applying Cauchy's Theorem, we write \begin{equation*} J_2(s)=-\frac{1}{(2\pi i)^2}\int_{\mathcal{C}_1}d\xi e^{\xi} \int_{-i\infty}^{i\infty}e^{ps}\frac{K(p)}{K(p)+\xi}\frac{dp}{p}, \end{equation*} where \begin{equation*} \mathcal{C}_1=\{\xi\in(-i\infty,-i)\cup\mathcal{S}\cup (i,i\infty)\}, \end{equation*} $\mathcal{S}=\{\xi=e^{i\theta}:\theta\in[-\frac{\pi}{2},-\frac{3\pi}{2}]\}$. Substituting $1=\frac{K(p)+\xi}{\xi}-\frac{K(p)}{\xi}$ in last equation, we obtain \begin{equation*} J_{2}=-\frac{1}{(2\pi i)^2}\int_{\mathcal{C}_1}\frac{d\xi}{\xi}e^{\xi} \int_{-i\infty}^{i\infty}e^{ps}K(p)\frac{dp}{p} +\frac{1}{(2\pi i)^2}\int_{\mathcal{C}_1}\frac{d\xi}{\xi} e^{\xi} \int_{-i\infty}^{i\infty}e^{ps}\frac{K(p)^2}{K(p)+\xi}\frac{dp}{p}. \end{equation*} By Cauchy's Theorem, using the analyticity with respect to $\xi$, the first summand in last equation is zero. Thus, \begin{equation}\label{hd5} J_2(s)=J_{21}(s)+J_{22}(s), \end{equation} where \begin{gather}\label{hd6} J_{21}(s)=\frac{1}{(2\pi i)^2}\int_{\mathcal{C}_1}\frac{d\xi}{\xi}e^{\xi} \int_{\mathcal{C}_2}e^{ps}\frac{K(p)^2}{K(p)+\xi}\frac{dp}{p}, \\ \label{hd7} J_{22}(s)=\frac{1}{(2\pi i)^2}\int_{\mathcal{C}_1}\frac{d\xi}{\xi}e^{\xi} \int_{-i/s^2}^{i/s^2}e^{ps}\frac{K(p)^2}{K(p)+\xi}\frac{dp}{p}. \end{gather} For $\epsilon>0$ small enough, \begin{equation*} \mathcal{C}_2=\big\{p\in\frac{1}{s^2}\big(-i+\infty e^{i(\frac{\pi}{2}+\epsilon)}, -i\big)\cup \frac{1}{s^2}\big(i,i+\infty e^{i(\frac{\pi}{2}+\epsilon)}\big)\big\}. \end{equation*} Taking the $ L^1$ norm with respect to $x$, we obtain \begin{equation}\label{hd8} \|J_{21}(\cdot)\|_{ L^1} \leq Ct^{-1/\alpha}\int_{\mathcal{C}_1}\frac{|d\xi|}{|\xi|} \int_{\mathcal{C}_2}\frac{K(p)^2}{|K(p)+\xi|}\frac{|dp|}{|p|^2} \leq Ct^{-1/\alpha}. \end{equation} Now, for $J_{22}$ we have \begin{equation}\label{hd9} |J_{22}(s)|\leq C\int_{\mathcal{C}_1}\frac{|d\xi|}{|\xi|} \int_{-i/s^2}^{i/s^2}\frac{K(p)^2}{|K(p)+\xi|}\frac{|dp|}{|p|}\leq C\frac{1}{s^2}. \end{equation} From \eqref{hd5}, \eqref{hd8} and \eqref{hd9}, we obtain for $t>0$ and $x>1$, \begin{equation}\label{j21a} \int_1^\infty|J_2(xt^{-1/\alpha})|dx\leq Ct^\frac{1}{\alpha}. \end{equation} Thus, from \eqref{j201} and \eqref{j21a} we obtain \begin{equation}\label{j2} \|J_2(\cdot)\|_{ L^1}\leq Ct^\frac{1}{\alpha}. \end{equation} We estimate $J_1$. Substituting $\frac{\xi}{p}=\frac{K(p)+\xi}{p}-\frac{K(p)}{p}$ in \eqref{hd3}, we obtain \begin{equation} J_1(s)=J_{11}(s)+J_{12}(s), \end{equation} where \begin{gather*} J_{11}(s)=\frac{1}{(2\pi i)^2}\int_{-i\infty}^{i\infty}\frac{d\xi}{Y^+(0,\xi)}e^{\xi} \int_{-i\infty}^{i\infty}e^{ps}(Y^+(p,\xi)-Y^+(0,\xi))\frac{dp}{p},\\ J_{12}(s)=-\frac{1}{(2\pi i)^2}\int_{-i\infty}^{i\infty} \frac{d\xi}{Y^+(0,\xi)} e^{\xi} \int_{-i\infty}^{i\infty}e^{ps}\Big(\frac{Y^+(p,\xi)-Y^+(0,\xi)}{K(p)+\xi}\Big) K(p)\frac{dp}{p}. \end{gather*} The integrand in $J_{11}$ is an analytic function in the left half-plane, with respect to $p$, and the residue at $p=0$ is zero, then by Cauchy's Residue Theorem we conclude that $J_{11}(s)=0$. Now, we estimate $J_{12}$. Using $$ Y^+(p,\xi)-Y^+(0,\xi)=\int_0^p\Gamma_q^+(q,\xi)Y^+(q,\xi)dq =\mathcal{O}(p^{\alpha(\frac{3}{2}-\gamma)}\xi^{\gamma-1}), $$ we obtain $\|J_{12}(\cdot)\|_{ L^1}\leq Ct^\frac{1}{\alpha}$. Therefore, \begin{equation}\label{j1} \|J_{1}(\cdot)\|_{ L^1}\leq Ct^{1/\alpha}. \end{equation} Then, from \eqref{j2} and \eqref{j1} we obtain \begin{equation}\label{j} \|H(\cdot,t)\|_{ L^1}\leq Ct^{-1+\frac{1}{\alpha}}. \end{equation} Finally, integrating with respect to time we obtain $$ \|\mathcal{H}(\cdot)h\|_{ L^1}\leq C\|h\|_{ B^\alpha}\int_0^t (t-\tau)^{\frac{1}{\alpha}-1}\langle t\rangle^{-1/\alpha}d\tau \leq C\|h\|_{ B^\alpha}. $$ Then, first estimate of Lemma \ref{hest} has been proved. Now we are going to obtain an asymptotic representation for $\mathcal{H}$. First, we split the integral, \begin{equation}\label{asyh0} \mathcal{H}(x)h=\int_0^{t-1}H(x,t-\tau)h(\tau)d\tau +\int_{t-1}^tH(x,t-\tau)h(\tau)d\tau \end{equation} For the first integral in \eqref{asyh0} we have \begin{equation}\label{asyh1} \begin{aligned} &\int_0^{t-1}H(x,t-\tau)h(\tau)d\tau \\ &= -\int_0^{t-1}H(x,t-\tau)(h(t)-h(\tau))d\tau +h(t)\int_0^{t-1}H(x,t-\tau)d\tau. \end{aligned} \end{equation} Making the change $t-\tau=s$, we obtain \begin{equation}\label{asyh2} \begin{aligned} \int_0^{t-1}H(x,t-\tau)d\tau &= \int_1^tH(x,s)ds=\int_1^t \Upsilon(xs^{-1/\alpha})\frac{ds}{s} \\ &= \int_1^t(\Upsilon(xs^{-1/\alpha})- \Upsilon(xt^{-1/\alpha}))\frac{ds}{s}+\ln(t)\Upsilon(xt^{-1/\alpha}) \\ &= \int_1^t(H(x,s)-\frac{t}{s}H(x,t))ds+\ln(t)\Upsilon(xt^{-1/\alpha}), \end{aligned} \end{equation} where $\Upsilon$ is given by \eqref{upsi}. From \eqref{asyh1} and \eqref{asyh2} we obtain \begin{equation}\label{asyh3} \begin{aligned} &\int_0^{t-1}H(x,t-\tau)h(\tau)d\tau \\ &= h(t)\ln(t)\Upsilon(xt^{-1/\alpha}) +h(t)\int_1^t\Big(H(x,s)-\frac{t}{s}H(x,t)\Big)ds \\ & \quad -\int_0^{t-1}H(x,t-\tau)(h(t)-h(\tau))d\tau. \end{aligned} \end{equation} Now, since $|H(x,t)|\leq Ct^{-1}$ and $h(t)=bt^\beta+\mathcal{O}(t^{\beta-\gamma})$, we obtain \begin{gather*} \big|\int_1^t(H(x,s)-\frac{t}{s}H(x,t) )ds\big|\leq \ln(t),\\ \int_0^{t-1}H(x,t-\tau)(h(t)-h(\tau))d\tau=bt^\beta+\mathcal{O}(t^{\beta-\delta}). \end{gather*} Then, from \eqref{asyh3} we obtain \begin{equation}\label{asyh4} \int_0^{t-1}H(x,t-\tau)h(\tau)d\tau=bt^{\beta+\gamma}\Upsilon(xt^{-1/\alpha})+ \mathcal{O}(t^\beta). \end{equation} On the other hand, for the second integral in \eqref{asyh0}, integrating by parts and using $|\partial_t^{-1}H(x,t)|\leq C$, we have \begin{equation}\label{asyh5} \int_{t-1}^tH(x,t-\tau)h(\tau)d\tau=\mathcal{O}(t^\beta). \end{equation} Thus, by \eqref{asyh0}, \eqref{asyh4} and \eqref{asyh5} we obtain \begin{equation}\label{asyh6} \mathcal{H}(x)h=bt^{\beta+\gamma}\Upsilon(xt^{-1/\alpha})+ \mathcal{O}(t^\beta). \end{equation} Therefore, Lemma \ref{hest} has been proved. \end{proof} The following theorem is a local version of Theorem \ref{b5}, its proof is similar and will be omitted. \begin{theorem}\label{b4} Let the initial data $u_0\in\mathbb{Z}^\mu$ and the boundary value $h\in B^\alpha$ such that the correspondent norms are sufficiently small. Then, for some $T>0$, there exist a unique solution $$ u\in C([0,T]; L^1)\cap C((0,T]; L^s\cap L^{s,\mu} \cap L^\infty),\quad s\geq1, $$ to the initial boundary-value problem \eqref{222}. \end{theorem} \section{Proof of Theorem \ref{b5}} By the Local Existence Theorem \ref{b4}, it follows that the global solution (if it exist) is unique. Indeed, on the contrary, we suppose that there exist two global solutions with the same initial data. And these solutions are different at some time $t>0$. By virtue of the continuity of solutions with respect to time, we can find a maximal time segment $[0,T]$, where the solutions are equal, but for $t>T$ they are different. Now, we apply the local existence theorem taking the initial time $T$ and obtain that these solutions coincide on some interval $[T,T_1]$, which give us a contradiction with the fact that $T$ is the maximal time of coincidence. So our main purpose in the proof of Theorem \ref{b5} is to show the global in time existence of solutions. First, we note that Lemma \ref{39} imply for the Green operator $\mathcal{G}:\mathbb{Z}^\mu\to X^\mu$ the inequality $\|\mathcal{G}(t)u_0\|_{ X^\mu}\leq C\|u_0\|_{\mathbb{Z}^\mu}$. Now, we are going to show the estimate \begin{equation}\label{b16} \begin{aligned} &\|\int_0^t\mathcal{G}(t-\tau)(\mathcal{N}(u(\tau))- \mathcal{N}(v(\tau)))d\tau\|_{ X^\mu} \\ &\leq C\|u-v\|_{ X^\mu} (\|u\|_{ X^\mu}+\|v\|_{ X^\mu})^\sigma, \end{aligned} \end{equation} for all $u,v\in X$, where $\mathcal{N}(u)=|u|^\sigma u$ and $\alpha<\sigma<1+\alpha$. In fact, using the inequality $$ ||u|^\sigma u-|v|^\sigma v|\leq C|u-v|(|u|^\sigma+|v|^\sigma), $$ we obtain \begin{equation}\label{b21} \begin{aligned} &\|\mathcal{N}(u)- \mathcal{N}(v)\|_{ L^{1,\nu}}\\ &\leq C\|u-v\|_{ L^\infty} \big(\|u\|_{ L^{\sigma,\frac{\nu}{\sigma}}}^\sigma+ \|v\|_{ L^{\sigma,\frac{\nu}{\sigma}}}^\sigma\big) \\ &\leq C\{\tau\}^{-\gamma}\langle\tau \rangle^{-1/\alpha}\tau^{-\frac{1}{\alpha}(\sigma-1-\nu)} \|u-v\|_{ X^\mu} (\|u\|_{ X^\mu}+\|v\|_{ X^\mu})^\sigma, \end{aligned} \end{equation} where $\mu=\nu/\sigma \geq0$, and \begin{equation}\label{b20} \begin{aligned} &\|\mathcal{N}(u)- \mathcal{N}(v)\|_{ L^\infty}\\ & \leq C\|u-v\|_{ L^\infty}(\|u\|_{ L^\infty}^\sigma+ \|v\|_{ L^\infty}^\sigma) \\ &\leq C\{\tau\}^{-\gamma(\sigma+1)} \langle\tau\rangle^{-\frac{1}{\alpha}(\sigma+1)} \|u-v\|_{ X^\mu} (\|u\|_{ X^\mu}+\|v\|_{ X^\mu})^\sigma. \end{aligned} \end{equation} Then, estimates \eqref{b21}, \eqref{b20}, and Lemma \ref{39}, imply \begin{equation}\label{b22} \begin{aligned} & \|\mathcal{G}(t-\tau)(\mathcal{N}(u(\tau))- \mathcal{N}(v(\tau)))\|_{ L^{s,\mu}} \\ &\leq C(t-\tau)^{-\frac{1}{\alpha}(1-\frac{1}{s})} \tau^{-\frac{1}{\alpha}(\sigma-1)}\{\tau\}^{-\gamma} \langle\tau\rangle^{-1/\alpha} \Big((t-\tau)^\frac{\mu}{\alpha}+\tau^\frac{\mu}{\alpha}\Big) \\ & \quad \times\|u-v\|_{ X^\mu}(\|u\|_{ X^\mu}+\|v\|_{ X^\mu})^\sigma, \end{aligned} \end{equation} where $0\leq\mu<1$, and \begin{equation}\label{b24} \begin{aligned} & \|\mathcal{G}(t-\tau)(\mathcal{N}(u)(\tau)- \mathcal{N}(v)(\tau))\|_{ L^\infty} \\ &\leq C\{t-\tau\}^{-\gamma} \langle t-\tau\rangle^{-1/\alpha} \{\tau\}^{-\gamma}\langle\tau\rangle^{-1/\alpha} \Big(\tau^{-\frac{1}{\alpha}(\sigma-1)} +\{\tau\}^{-\gamma\sigma}\langle\tau\rangle^{-\sigma/\alpha}\Big) \\ &\quad \times\|u-v\|_{ X^\mu} (\|u\|_{ X^\mu}+\|v\|_{ X^\mu})^\sigma. \end{aligned} \end{equation} Now, we integrate with respect to $\tau$ inequalities \eqref{b22} and \eqref{b24}, \begin{equation}\label{b35} \begin{aligned} & \int_0^t\|\mathcal{G}(t-\tau)(\mathcal{N}(u)(\tau)- \mathcal{N}(v)(\tau))\|_{ L^{s,\mu}} \\ &\leq C\{t\}^{1-\frac{1}{\alpha}(\sigma-\frac{1}{s}-\mu)-\gamma} \langle t\rangle^{1-\frac{1}{\alpha}(1+\sigma-\frac{1}{s}-\mu)} \|u-v\|_{ X^\mu} (\|u\|_{ X^\mu}+\|v\|_{ X^\mu})^\sigma, \end{aligned} \end{equation} provided $\sigma<1+\alpha$, $s<\frac{1}{1-\alpha}$, and \begin{equation}\label{b37} \begin{aligned} & \int_0^t\|\mathcal{G}(t-\tau)(\mathcal{N}(u)(\tau)- \mathcal{N}(v)(\tau))\|_{ L^\infty} \\ &\leq C\{t\}^{1-\frac{1}{\alpha}(\sigma-1)-2\gamma} \langle t\rangle^{1-\frac{1}{\alpha}(\sigma+1)} \|u-v\|_{ X^\mu}(\|u\|_{ X^\mu}+\|v\|_{ X^\mu})^\sigma, \end{aligned} \end{equation} whenever $\sigma<1+\alpha$. Then, for $\alpha<\sigma<1+\alpha$, the definition of the norm in the space $ X^\mu$ and the estimates \eqref{b35} and \eqref{b37}, imply \eqref{b16}. Now, we apply the Contraction Mapping Principle on a ball with ratio $\rho>0$ in the space $ X^\mu$, $ X_\rho^\mu=\{\phi\in X^\mu: \|\phi\|_{ X^\mu}\leq\rho\}$, where $\rho=3C\max\{\|u_0\|_{\mathbb{Z}^\mu},\|h\|_{ B^\alpha}\}$. First, we show that \begin{equation}\label{b38} \|\mathcal{M}(u)\|_{ X^\mu}\leq\rho, \end{equation} where $u\in X_\rho^\mu$. Indeed, from the integral formula \begin{equation}\label{223} \mathcal{M}(u)=\mathcal{G}(t)u_0-\int_0^t\mathcal{G}(t-\tau) \mathcal{N}(u(\tau))d\tau+\mathcal{H}(x)h \end{equation} and the estimate \eqref{b16} (with $v\equiv0$) we obtain \begin{align*} \|\mathcal{M}(u)\|_{ X^\mu} &\leq \|\mathcal{G}(t)u_0\|_{ X^\mu} +\|\int_0^t\mathcal{G}(t-\tau)\mathcal{N}(u)(\tau)d\tau\| _{ X^\mu}+\|\mathcal{H}(\cdot)h\|_{ X^\mu} \\ & \leq C\Big(\|u_0\|_{\mathbb{Z}^\mu}+ \|u\|_{ X^\mu}^{\sigma+1}+\|h\|_{ B^\alpha}\Big)\\ &\leq\frac{\rho}{3}+ C\rho^{\sigma+1}+\frac{\rho}{3}\leq\rho, \end{align*} provided $\rho>0$ is sufficient small. Therefore, the operator $\mathcal{M}$ transforms a ball of ratio $\rho>0$ into itself, in the space $ X^\mu$. In the same way we estimate the difference of two functions $u,v\in X_\rho^\mu$, \begin{align*} \|\mathcal{M}(u)-\mathcal{M}(v)\|_{ X^\mu} & \leq \|\int_0^t \mathcal{G}(t-\tau)(\mathcal{N}(u)(\tau)- \mathcal{N}(v)(\tau))d\tau\|_{ X^\mu} \\ & \leq C \|u-v\|_{ X^\mu} (\|u\|_{ X^\mu}+\|v\|_{ X^\mu})^\sigma \leq C(2\rho)^\sigma\|u-v\|_{ X^\mu} \\ & \leq \frac{1}{2}\|u-v\|_{ X^\mu} \end{align*} whenever $\rho>0$ is sufficient small. Thus, $\mathcal{M}$ is a contraction mapping in $ X_\rho^\mu$. Therefore, there exist a unique solution $u\in X^\mu$ to the Cauchy problem \eqref{222}. Now we prove the asymptotic formula \begin{equation} u(x,t)=t^{-\frac{1}{\alpha}-\frac{1}{2}}\big(a\Lambda(xt^{-1/\alpha})+b\Upsilon (xt^{-1/\alpha})\big)+ O(t^{-\frac{1}{\alpha}-\frac{1}{2}-\gamma}),\label{asv} \end{equation} where $\gamma>0$ and $$ a=\int_0^{+\infty}y^{\alpha/2}u_0(y)dy-\int_{0}^{+\infty}d\tau \int_0^{+\infty}y^{\alpha/2}\mathcal{N}(u(\tau))dy. $$ We denote $G_{0}(t)=t^{-\frac{1}{\alpha}-\frac{1}{2}}\Lambda(xt^{-1/\alpha})$ and $G_1(t)=bt^{\beta+\gamma}\Upsilon(xt^{-1/\alpha})$. From Lemmas \ref{39} and \ref{hest}, we have for all $t>1$, \begin{gather} \| \mathcal{G}( t) \phi -G_{0}( t) \int_0^{+\infty}y^{\alpha/2}\phi(y)dy\|_{ L^{\infty}}\leq C\langle t\rangle^{-\frac{1}{\alpha}-\frac{1}{2}-\gamma}\| \phi \|_{\mathbb{Z}^{1,\frac{\alpha}{2}}} \,, \label{General.5.11} \\ \|\mathcal{H}(\cdot)h-G_1(t) \|_{ L^{\infty}}\leq C\langle t\rangle^\beta, \label{General.5.12} \end{gather} for $h(t)=bt^\beta+\mathcal{O}(t^{\beta-\gamma})$, $\beta<0$, $\gamma>0$. Also, in view of the definition of the norm $ X^\mu$ we have \begin{align*} \big|\int_0^{+\infty}y^{\alpha/2}\mathcal{N}(u(\tau))dy \big| &\leq \|\mathcal{N}( u( \tau ) )\|_{ L^{1,\frac{\alpha}{2}}} \leq\| u( \tau )\| _{ L^\infty}^\sigma \| u( \tau )\| _{ L^{1,\frac{\alpha}{2}}} \\ &\leq C\{ \tau \} ^{-\gamma\sigma} \langle \tau \rangle ^{-\frac{\sigma}{\alpha}}\tau^\frac{1}{2} \| u\| _{ X^{\alpha/2}}^{\sigma+1}. \end{align*} By a direct calculation, for $t>1$, we have \begin{equation} \label{5.1} \begin{aligned} &\| \int_{0}^{\frac{t}{2}}d\tau( G_{0}( t-\tau ) -G_{0}( t) ) \int_0^{+\infty}y^{\alpha/2}\mathcal{N}(u(\tau))dy \| _{ L^{\infty }} \\ &\leq C\| u\| _{ X^{\alpha/2}}^{\sigma+1} \int_{0}^{\frac{t}{2}}\| G_{0}( t-\tau ) +G_{0}( t) \| _{ L^{\infty }}\{ \tau \} ^{-\gamma\sigma} \langle \tau \rangle ^{-\frac{\sigma}{\alpha}}\tau^\frac{1}{2}d\tau \\ &\leq C\langle t\rangle ^{-\frac{1}{\alpha}-\frac{1}{2}} \| u\| _{ X^{\alpha/2}}^{\sigma+1}\int_{0}^{\frac{t}{2}} \{ \tau \} ^{-\gamma\sigma}\langle \tau \rangle ^{-\frac{\sigma}{\alpha}} \tau^\frac{1}{2}d\tau \leq C\langle t\rangle ^{-\frac{1}{\alpha}-\frac{1}{2}-(\frac{\sigma}{\alpha} -\frac{3}{2})} \| u\| _{ X^{\alpha/2}}^{\sigma+1}, \end{aligned} \end{equation} where $\sigma>3\alpha/2$, provided $\gamma\sigma<\frac{3}{2}$, and in the same way \begin{equation} \| G_{0}( t) \int_{t/2}^{\infty }d \tau\int_0^{+\infty}y^{\alpha/2}\mathcal{N}(u(\tau))dy \|_{ L^{\infty }} \leq C \langle t\rangle ^{-\frac{1}{\alpha} -\frac{1}{2}-(\frac{\sigma}{\alpha}-\frac{3}{2})} \| u\| _{ X^{\alpha/2}}^{\sigma+1}, \label{5.2} \end{equation} provided $\sigma>3\alpha/2$. Also we have for all $t>1$, \begin{equation} \label{5.3} \begin{aligned} &\| \int_{0}^{\frac{t}{2}}( \mathcal{G}( t-\tau ) \mathcal{N}( u( \tau ) ) -G_{0}( t-\tau )\int_0^{+\infty}y^{\alpha/2}\mathcal{N}(u(\tau))dy ) d\tau \| _{ L^{\infty }} \\ &\quad +\| \int_{\frac{t}{2}}^{t}\mathcal{G}( t-\tau ) \mathcal{N}( u( \tau ) ) d\tau \| _{ L^{\infty }} \\ &\leq C\int_{0}^{\frac{t}{2}} ( t-\tau ) ^{-\frac{1}{\alpha}-\frac{1}{2}-\gamma} \| \mathcal{N}( u( \tau ) )\|_{ L^{1,\frac{\alpha}{2}}}d\tau \\ & \quad +C\int_{\frac{t}{2}}^{t}\{t-\tau\}^{-\gamma} \langle t-\tau\rangle^{-1/\alpha}(\| \mathcal{N} ( u( \tau ) ) \| _{ L^1}+ \| \mathcal{N}( u( \tau ) ) \| _{ L^\infty})d\tau \\ &\leq C\langle t\rangle^{-\frac{1}{\alpha}-\frac{1}{2} -(\frac{\sigma}{\alpha}-\frac{3}{2})} \| u\| _{ X^{\alpha/2}}^{\sigma+1}, \end{aligned} \end{equation} provided $\gamma\sigma<\frac{3}{2}$ and $\sigma>\frac{3}{2}\alpha$. By virtue of the integral equation \eqref{223} we obtain \begin{equation}\label{General.5.6} \begin{aligned} &\| u( t) -aG_{0}( t)-G_1(t)\| _{ L^\infty}\\ &\leq \|\mathcal{G}( t) u_{0}-G_{0}( t)\int_0^{+\infty}y^{\alpha/2}u_{0}(y)dy \| _{ L^{\infty }} \\ &\quad +\| \int_{0}^{\frac{t}{2} }( \mathcal{G}( t-\tau ) \mathcal{N}\mathbb{(} u( \tau ) \mathbb{)}-G_{0}( t-\tau )\int_0^{+\infty}y^{\alpha/2} \mathcal{N}(u(\tau))dy ) d\tau \| _{ L^{\infty }} \\ &\quad +\| \int_{\frac{t}{2}}^{t}\mathcal{G}( t-\tau ) \mathcal{N}( u( \tau ) ) d\tau \| _{ L^{\infty }}+\| G_{0}( t) \int_{\frac{t}{2}}^{\infty }d\tau\int_0^{+\infty}y^{\alpha/2}\mathcal{N}(u(\tau))dy \| _{ L^{\infty }} \\ &\quad +\| \int_{0}^{\frac{t}{2} }d\tau( G_{0}( t-\tau ) -G_{0}( t) ) \int_0^{+\infty}y^{\alpha/2}\mathcal{N}(u(\tau))dy \| _{ L^{\infty }} \\ &\quad +\|\mathcal{H}(\cdot)h-G_1(t) \|_{ L^{\infty}}. \end{aligned} \end{equation} All summands in the right-hand side of \eqref{General.5.6} are estimated, via \eqref{5.1}-\eqref{5.3}, by $$ C\langle t\rangle^{-\frac{1}{\alpha} -\frac{1}{2}-\gamma}(\| u_{0}\| _{\mathbb{Z}^{\alpha/2}} +\| u\| _{ X^{\alpha/2}}^{\sigma+1}), $$ where $0<\gamma<\frac{\sigma}{\alpha}-\frac{3}{2}$, provided $\beta<-\frac{1}{\alpha}-\frac{1}{2}-\gamma$. 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