\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2005(2005), No. 141, pp. 1--17.\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 2005 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2005/141\hfil Nonlinear decay] {Nonlinear decay and scattering of solutions to a Bretherton equation in several space dimensions} \author[A. D\'e Godefroy\hfil EJDE-2005/141\hfilneg] {Akmel D\'e Godefroy} \address{Akmel D\'e Godefroy \hfill\break Laboratoire de Mathematiques Appliqu\'ees, UFRMI, Universit\'e d'Abidjan Cocody, 22 BP 582 Abidjan 22, Cote d'Ivoire} \email{akmelde@yahoo.fr} \date{} \thanks{Submitted October 12, 2005. Published December 5, 2005.} \subjclass[2000]{35B40, 35Q10, 35Q20} \keywords{Asymptotic behavior; scattering problem; Bretherton equation} \begin{abstract} We consider a Cauchy problem for the $n$-dimensional Bretherton equation. We establish the existence of a global solution and study its long-time behavior, with small data. This is done using the oscillatory integral techniques considered in \cite{k2}. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} For the Bretherton equation, we consider the initial-value problem (I.V.P) \begin{equation} \label{e1.1} \begin{gathered} u_{tt} + u + {\triangle}u+ {\triangle}^{2}u = F(u), \quad x\in {\mathbb{R}}^{n},\; n\geq 1,\; t>0, \\ u(x,0) = f_1(x),\\ u_{t} (x,0) = f_2(x), \end{gathered} \end{equation} where $F(u)=|u|^{\alpha}u$ and $\alpha \geq 1$. Problem \eqref{e1.1} with $n=1$ was introduced by Kalantorov and Ladyzhenskaya in \cite{k1}, where they proved the blow-up of it's solutions in finite time for large data. After an investigation on the local existence of solutions to \eqref{e1.1} with $n=1$, Scialom \cite{n1} pointed out that the global existence result for ``small data" remains an open problem. Furthermore, using a new computational method called ``RATH" (Real Automated Tangent Hyperbolic function method), Zhi-bin Li \emph{et al.} \cite{z1} showed the existence of solitary-wave solutions of some partial differential equations. Yet, for the Bretherton equation, the ``RATH" method showed the non-existence of solitary-wave solution. Our scattering result here seems to confirm the computation result of Zhi-bin Li \emph{et al.} for the non-existence of solitary-wave solution to the Bretherton equation, at least for small data. Indeed it is well known that affirmative results on scattering are interpreted as the nonexistence of solitary-wave solution of arbitrary small amplitude, see \cite{b2,l1}. Our aim in this paper is to study the global existence, the uniform in $x$ decay to zero and the scattering as $t\to \infty$, for solutions of \eqref{e1.1} with sufficiently small data. More precisely, we show the following two theorems: \begin{theorem} \label{thm1.1} Let $\alpha > 5$ and $f_{1}, f_2 \in {\mathbb{H}}^{s}({\mathbb{R}}^{n}) \cap{\mathbb{L}}^{1}({\mathbb{R}}^{n})$, $n\geq 1$, with $s\geq \frac{3}{2}n$. If $ |f_1|_{1} +\|f_1\|_{3n/2} +|f_2|_{1} +\|f_2\|_{3n/2} < \delta$ with $\delta$ sufficiently small, then the solution $u$ of \eqref{e1.1} is unique in $C({\mathbb{R}}, {\mathbb{H}}^{s}({\mathbb{R}}^{n}))$ and satisfies \begin{equation} \label{e1.2} |u(x,t)|_{\infty}\leq c(1+t)^{-1/4}, \quad t \geq 0, \end{equation} where $c$ does not depend of $x$ and $t$. Moreover, there is scattering for $t\to \pm\infty$, that is, there exist $u_{+}, u_{-}$, solutions of the linear problem \eqref{e2.1}, such that $\|u(t)-u_{\pm}(t)\|_{2}$ tends to 0 as $t\to \pm\infty$. \end{theorem} \begin{theorem} \label{thm1.2} Let $\alpha > 1+\frac{4}{\theta}$ and $f_1, f_2 \in {\mathbb{H}}^{r+\frac{5}{2}n+1}({\mathbb{R}}^{n})\cap{\mathbb{L}}^{q}_{r+\frac{5}{2}n}({\mathbb{R}}^{n}),\; n\geq 1$, with $r >\frac{n}{p}$. If $\|f_1\|_{r+\frac{5}{2}n,q} +\|f_2\|_{r+\frac{5}{2}n,q} + \|f_1\|_{r+\frac{5}{2}n+1} +\|f_2\|_{r+\frac{5}{2}n+1} < \delta$ with $\delta$ small, then the solution $u$ of the I.V.P \eqref{e1.1} satisfies \begin{equation} \|u(x,t)\|_{r,p}\leq c(1+t)^{- \frac{\theta}{4}}, \quad t \geq 0, \end{equation} where $ p= 2/(1- \theta), q= 1/(1+ \theta)$, and $\theta \in ]0,1[.$ \end{theorem} \subsection*{Notation} The notation $\| \cdot \|_{r,p}$ is used to denote the norm in ${\mathbb{L}}^p_r$ such that for $u \in {\mathbb{L}}^p_r({\mathbb{R}}^{n}), \; \|u \|_{r,p} =\|u \|_{{\mathbb{L}}^p_r}=\|(1- \Delta)^{r/2}u \|_{{\mathbb{L}}^p}< \infty$. Also, $| \cdot |_p$ instead of $\| \cdot \|_{0,p}$ denotes the norm in ${\mathbb{L}}^p$, and ${\mathbb{H}}^s$ with norm $\| \cdot \|_s$ is used instead of ${\mathbb{L}}^2_s$. Throughout this paper, $c$ represents a generic constant independent of $t$ and $x$. The Fourier transform of a function $f$ is denoted by $\widehat{f}(\xi)$ or ${\mathcal{F}}(f)(\xi)$ and ${\mathcal{F}}^{-1}(f) \equiv \breve{f}$ denotes the inverse Fourier transform of $f$. For $1 \leq p,q \leq \infty$ and $f :\mathbb{R}^{n} \times \mathbb{R}\to\mathbb{R}$, $$ \| f \|_{{\mathbb{L}}^{q}(\mathbb{R}; {\mathbb{L}}^p(\mathbb{R}^{n}))} =\Big(\int^{+\infty}_{-\infty} \Big( \int_{\mathbb{R}^{n}}|f(x,t)|^pdx\Big)^{q/p}dt\Big)^{1/q}. $$ \section{Local existence result} In this section, we write the Cauchy problem associated with $\eqref{e1.1}$ in it's integral form and we prove the local existence and uniqueness of it's solution. Our method of proof is based on linear estimates and a contraction mapping argument. Thereupon, we state a locally well-posed theorem for \eqref{e1.1}. \begin{theorem} \label{thm2.1} Let $s>n/2$ be a real number, and $f_1, f_2 \in {\mathbb{H}}^s({\mathbb{R}}^{n})$, $n\geq1$. Then there exists $T_0>0$ which depends on $\|f_1\|_{s}$ and $\|f_2\|_{s}$, and a unique solution of \eqref{e1.1} in [0,T], such that $u\in {\mathcal{C}}(0,T_0;{\mathbb{H}}^{s}({\mathbb{R}}^{n}))$. \end{theorem} \begin{proof} Consider first the linear part of \eqref{e1.1}: \begin{equation} \label{e2.1} \begin{gathered} u_{tt} + u + {\triangle}u+ {\triangle}^{2}u = 0, \quad x\in {\mathbb{R}}^{n},\; n\geq 1,\; t>0, \\ u(x,0) = f_1(x), \\ u_{t} (x,0) = f_2(x), \end{gathered} \end{equation} The formal solution of \eqref{e2.1} is \begin{equation} \label{e2.2} u(x,t)=V_1(t)f_1(x) + V_2(t)f_2(x) \end{equation} where \begin{gather*} V_1(t)f_1(x)=[\frac{1}{2}(e^{it\phi(\xi)} + e^{-it\phi(\xi)})\hat{f_1}(\xi)]^{\vee}(x),\\ V_2(t)f_2(x)=[\frac{1}{2i\phi(\xi)}(e^{it\phi(\xi)} - e^{-it\phi(\xi)})\hat{f_2}(\xi)]^{\vee}(x) \end{gather*} with $\phi(\xi)=(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}$. We define \begin{gather*} S_1(t)f_1(x) = \frac{1}{{2}(2\pi)^{n}}\int_{{\mathbb{R}}^{n}}e^{ix\xi + it\phi(\xi)}\hat{f_1}(\xi) d\xi, \\ S_2(t)f_2(x) = \frac{1}{2i(2\pi)^{n}}\int_{{\mathbb{R}}^{n}}e^{ix\xi + it\phi(\xi)}\frac{\hat{f_2}(\xi)}{\phi(\xi)}d\xi. \end{gather*} Then \begin{gather*} V_1(t)f_1(x)=S_1(t)f_1(x) + S_1(-t)f_1(x), \\ V_2(t)f_2(x)=S_2(t)f_2(x) - S_2(-t)f_2(x). \end{gather*} %2.3 2.4 Note that \begin{equation} \label{e2.5} \Phi(\xi) \geq {\frac{1}{2 \sqrt{3}}}(1+ |{\xi}|^2). \end{equation} Indeed, ${\phi(\xi)}^{2}=1- |{\xi}|^2 + |{\xi}|^4=(1- \frac{1}{2}|{\xi}|^2)^{2} + \frac{3}{4}|{\xi}|^4$ so that if $|{\xi}|\leq 1$ then $$ {\phi(\xi)}^{2}\geq \frac{1}{4}+ \frac{3}{4}|{\xi}|^4 \geq \frac{1}{4}(\frac{1}{3}+\frac{2}{3}|{\xi}|^2 +\frac{1}{3}|{\xi}|^4)=\frac{1}{12}(1+|{\xi}|^2)^{2} $$ and if $|{\xi}|\geq 1$ then $$ {\phi(\xi)}^{2}\geq \frac{3}{4}|{\xi}|^4 \geq \frac{1}{12}(1+|{\xi}|^2)^{2}. $$ \begin{remark} \label{rmk2.2} \rm Since \eqref{e1.1} will not change when $t$ is switched to $-t$, the solution $u(t)$ in Theorem \ref{thm2.1} can be extended to $u\in {\mathcal{C}}([-T_0,T_0];{\mathbb{H}}^{s}({\mathbb{R}}^{n}))$. \end{remark} \begin{remark} \label{rmk2.3} \rm Note that, since the negative sign of $t$ in $S_j(-t)$ acts only on the sign of the phase function, the estimates of $S_j(t)f(x)$ below hold also for $S_j(-t)f(x)$. Hence, to estimate $V_j(t)f(x)$ one only has to estimate $S_j(t)f(x)$, $ j=1,2$. \end{remark} To prove the existence theorem for \eqref{e1.1}, we need the following inequalities. \begin{lemma} \label{lem2.4} Let $f_1, f_2 \in {\mathbb{H}}^{s}({\mathbb{R}}^{n}), s\geq 0$, and $V_1(t), V_2(t)$ defined in \eqref{e2.2}. Then \begin{gather} \|V_1(t)f_1(x)\|_{s} \leq c\|f_1\|_s \label{e2.6}\\ \|V_2(t)f_2\|_{s} \leq c\|f_2\|_{s-2} \leq c\|f_2\|_{s}. \label{e2.7} \end{gather} \end{lemma} The proof of the above lemma follows directly from the definition of $V_1(t)$ and $V_2(t)$ in \eqref{e2.2} and the use of the inequality \eqref{e2.5}. \end{proof} Thereafter, with Lemma \ref{lem2.4} in hand, one can use the contraction mapping principle to prove the local well-posedness result in Theorem \ref{thm2.1}. Then, thanks to the Duhamela principle, the solution of \eqref{e1.1} verifies the integral equation \begin{equation} \label{e2.8} u(x,t)=V_1(t)f_1(x)+ V_2(t)f_2(x) + \int^t_0V_2(t-\tau)(|u|^{\alpha }u)(\tau) d \tau. \end{equation} Let us define \begin{equation} \label{e2.9} \varphi(u)(t)=V_1(t)f_1(x)+ V_2(t)f_2(x) + \int^t_0V_2(t-\tau)(|u|^{\alpha}u)(\tau) d \tau \end{equation} and the complete metric space $$ F= \{v \in \mathcal{C}(0,T; {\mathbb{H}}^s({\mathbb{R}}^{n})), \; s>n/2, \quad \sup_{[0,T]}\|v(t)\|_{s} \leq a \}, $$ where $a$ is a positive real constant. We begin by showing that $\varphi :F\to F$ is a contraction. The use of the definition of $\varphi$ in \eqref{e2.9}, Lemma \ref{lem2.4} and the fact that ${\mathbb{H}}^s({\mathbb{R}}^{n}))$, $s>n/2$ is an Algebra, lead for all $0\leq t \leq T$, to \begin{equation} \label{e2.10} \begin{aligned} \|\varphi(u)(t)\|_{s} &\leq c(\|f_1\|_{s}+ \|f_2\|_{s}) + c\int^t_0 \|(|u|^{\alpha}u)(\tau)\|_{s} d \tau \\ &\leq c(\|f_1\|_{s}+ \|f_2\|_{s}) + c\int^t_0 \|u(\tau)\|^{\alpha+1}_{s} d \tau \\ &\leq c(\|f_1\|_{s}+ \|f_2\|_{s})+ cT(\sup_{[0,T]}\|u\|_{s})^{\alpha+1}. \end{aligned} \end{equation} Thereby, taking $\mu$ as a positive constant such that $\|f_1\|_{s}+\|f_2\|_{s}<\mu$, we get for $u \in F $, $$ \sup_{[0,T]}\|\varphi(u)(t)\|_{s} \leq c\{\mu +a^{\alpha +1}T\} $$ so that choosing $a=2c\mu$, we obtain $$ \sup_{[0,T]}\|\varphi(u)(t)\|_{s} \leq c\{\mu +2^{\alpha +1}c^{\alpha +1}{\mu}^{\alpha +1}T\} = c\mu \{1 +2^{\alpha +1}c^{\alpha +1}{\mu}^{\alpha}T\}. $$ Then, fixing $T$ such that \begin{equation} \label{e2.11} 2^{\alpha +1}c^{\alpha +1}{\mu}^{\alpha}T<1 \end{equation} we get $$ \sup_{[0,T]}\|\varphi(u)(t)\|_{s} \leq 2c\mu =a. $$ This shows that $\varphi$ maps $F$ into $F$. The next step is to prove that $\varphi$ is in fact a contraction. We consider $u$ and $v$ in $F$ with the same initial values. Thus $$ (\varphi(u)-\varphi(v))(t)= \int^t_0V_2(t-\tau)(|u|^{\alpha }u -|v|^{\alpha }v)(\tau) d \tau. $$ To estimate $\sup_{[0,T]}\|(\varphi(u)-\varphi(v))(t)\|_{s}$ we use Lemma \ref{lem2.4}, Taylor formula and the fact that ${\mathbb{H}}^s({\mathbb{R}}^{n}))$, $s>\frac{n}{2}$ is an Algebra; it follows that for all $0\leq t\leq T$, \begin{align*} \|(\varphi(u)-\varphi(v))(t)\|_{s} &\leq c\int^t_0 \|(|u|^{\alpha}+ |v|^{\alpha })(u-v)(\tau)\|_{s} d \tau \\ &\leq c\int^t_0 \|(|u|^{\alpha}+ |v|^{\alpha })(\tau)\|_{s} \|(u-v)(\tau)\|_{s} d \tau \\ &\leq c T( \sup_{[0,T]}\|u\|^{\alpha }_{s} + \sup_{[0,T]}\|v\|^{\alpha }_{s}) \sup_{[0,T]}\|u-v\|_{s}. \end{align*} which leads, with $a=2c\mu$ as above, to \begin{equation} \label{e2.12} \|(\varphi(u)-\varphi(v))(t)\|_{s} \leq 2^{\alpha +1}c^{\alpha +1}{\mu}^{\alpha}T\sup_{[0,T]}\|u-v\|_{s}. \end{equation} Hence, with the choice of $T$ as above in (2.11), we get from \eqref{e2.12} that $\varphi$ is a contraction map in $F$. Thus, the application of contraction mapping principle gives the result of local existence and uniqueness in Theorem \ref{thm2.1}. For the sequel, we need the following inequalities which are obtained by obvious computations including the inequality \eqref{e2.5}: $\forall \xi \in {\mathbb{R}}^{n}$, \begin{gather} |{\nabla}\phi(\xi)|= \frac{|{\xi}||2|{\xi}|^{2}-1|}{(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}} \label{e2.13}\\ |{D}^{2}\phi(\xi)|\leq c. \label{e2.14} \end{gather} \section{Linear Estimates} The purpose of this section is to study the linear equation associated with \eqref{e1.1} and to establish linear estimates needed for the next section. The following result is concerning the decay of solutions of the linear problem \eqref{e2.1}. \begin{lemma} \label{lem3.1} Let $V_1(t)$ and $V_2(t)$ be defined as in \eqref{e2.2}. Let $f_{1}, f_2 \in {\mathbb{H}}^{\frac{3}{2}n}({\mathbb{R}}^{n})\cap{\mathbb{L}}^{1}({\mathbb{R}}^{n}), \; n\geq 1$. Then there exists a constant $c$ independent of $f_1, f_2, t$ and $x \in {\mathbb{R}}^{n}$ such that \begin{equation} \label{e3.1} |V_j(t)f_j|_{\infty}\leq c (|f_j|_{1} +\|f_j\|_{3n/2}(1+t)^{-1/4}, \quad j=1,2, \end{equation} for all $t \geq 0$. Moreover, let $f_{1}, f_2 \in {\mathbb{H}}^{\frac{3}{2}n}({\mathbb{R}}^{n})\cap{\mathbb{L}}^{1}_{\frac{5}{2}n+k}({\mathbb{R}}^{n})$, $n\geq 1$, $ k \in {\mathbb{R}}^{+}$; then we have \begin{equation} \label{e3.2} \|V_j(t)f_j\|_{k,\infty}\leq c(\|f_j\|_{\frac{5}{2}n+k,1} +\|f_j\|_{\frac{3}{2}n +k})(1+t)^{-1/4}, \quad j=1,2. \end{equation} \end{lemma} Before proving the above lemma, we prove the following lemma. \begin{lemma} \label{lem3.2} Given $x \in {\mathbb{R}}^{n}$ and $t \in {\mathbb{R}}^{+}$, the phase function $$ \Psi(\xi)=\phi(\xi)+t^{-1}(x,\xi) $$ has a finite number of stationary points. Moreover, if ${\xi}_s$ is a stationary point of $\Psi$, then any point ${\eta}_s$ verifying $|{\eta}_s|=|{\xi}_s|$ is also a stationary point of $\Psi$. \end{lemma} \begin{proof} Since $$ \nabla{\Psi}(\xi)=\nabla{\phi}(\xi)+xt^{-1}=\frac{{\xi}(2|{\xi}|^{2}-1)}{(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}}+xt^{-1}, $$ we have \[ \nabla{\Psi}(\xi)=0\;\Leftrightarrow\; {\xi}(2|{\xi}|^{2}-1)+xt^{-1}(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}=0 \] and making the scalar product with \begin{equation}label{e3.3} {\xi}(2|{\xi}|^{2}-1)+xt^{-1}(1- |{\xi}|^2 + |{\xi}|^4)^{1/2} \end{equation} we get \begin{equation} \label{e3.4} \nabla{\Psi}(\xi)=0 \Leftrightarrow |{\xi}|^{2}(2|{\xi}|^{2}-1)^{2}+|xt^{-1}|^{2}(1- |{\xi}|^2 + |{\xi}|^4)=0 \Leftrightarrow P(|\xi|)=0 \end{equation} where $P(y)=4y^{6}-4y^{4}+y^{2}- |xt^{-1}|^{2}(1-y^{2}+y^{4}), \; y \in {\mathbb{R}}_{+}.$ The stationary points of $\Psi$ are such that their norms are the roots of $P(y)$. Then since $P(y)$ is polynomial of degree $6$ so that it has at most $6$ roots, we deduce that $\Psi$ has a finite number of stationary points in ${\mathbb{R}}^{n}$. Furthermore, since $P(0)=- |xt^{-1}|^{2}\leq 0$ and $P(y)\to + \infty$ as $y\to + \infty$, and since $P(y)$ is continuous, we deduce that there exists at least one stationary point of $\Psi$. Therefore, $\Psi$ has a finite number of stationary points. Moreover, we note that if ${\xi}_s$ is a stationary point of $\Psi$ and if ${\eta}_s$ is a point verifying $|{\eta}_s|=|{\xi}_s|$, then we have $P(|{\eta}_s|)=P(|{\xi}_s|)=0$ and consequently from \eqref{e3.4} ${\eta}_s$ is also a stationary point of $\Psi$. This completes the proof of Lemma \ref{lem3.2}. \end{proof} Next, we use Lemma \ref{lem3.2} to prove Lemma \ref{lem3.1}. Let us recall that, thanks to Remark \ref{rmk2.2}, the inequality \eqref{e3.1} of proposition \eqref{e3.1} holds for $V_1(t)$ and $V_2(t)$ whenever it holds for $S_1(t)$ and $S_2(t)$. If $0\leq t \leq 1$, we have, thanks to the Schwartz inequality, \begin{equation} \label{e3.5} \begin{aligned} \quad |S_1(t)f_1(x)| &= \frac{1}{{2}(2\pi)^{n}}|\int_{{\mathbb{R}}^{n}}e^{it\Psi(\xi)}\hat{f_1}(\xi) d\xi| \\ &\leq c\int_{{\mathbb{R}}^{n}}|\hat{f_1}(\xi)| d\xi \\ &\leq c(\int_{{\mathbb{R}}^{n}}(1+|\xi|^{2})^{-n} d\xi)^{1/2}\|f_1\|_{n}\leq c\|f_1\|_{n} \\ &\leq c(1+t)^{-1/4}\|f_1\|_{3n/2} \end{aligned} \end{equation} If $t\geq 1$, let $\Omega = \{\xi \in {\mathbb{R}}^{n}, |\xi|\leq t^{\frac{1}{4n}}\} $ and $q_t(\xi)={\chi}_{\Omega}(\xi)e^{it\phi(\xi)}$; then thanks to the Schwartz and the Young inequality, \begin{equation} \label{e3.6} \begin{aligned} &|S_1(t)f_1(x)| \\ &=\frac{1}{{2}(2\pi)^{n}}|(\int_\Omega +\int_{{\Omega}^{c}})e^{it\phi(\xi)+ix\cdot \xi}\hat{f_1}(\xi)d\xi| \\ &\leq c|\check{q_t}(x)\ast f_1(x)|_{\infty} +c(\int_{{\Omega}^{c}}(1+|\xi|^{2})^{-\frac{3}{2}n} d\xi)^{1/2}(\int_{{\Omega}^{c}}(1+|\xi|^{2})^{\frac{3}{2}n}|\hat{f_1}(\xi)|^{2} d\xi)^{1/2} \\ &\leq c|\check{q_t}(x)|_{\infty}|f_1(x)|_{1} + ct^{-1/4}\|f_1\|_{3n/2} \end{aligned} \end{equation} where the first factor in the second term of the right hand side of \eqref{e3.6} is a bound given $\forall t\geq 1$ by \begin{align*} \Big(\int_{\{|\xi|\geq t^{\frac{1}{4n}} \}}(1+|\xi|^{2})^{-\frac{3}{2}n}d\xi\Big)^{1/2} &\leq \Big(\int_{\{r\geq t^{\frac{1}{4n}} \}}r^{-3n}r^{n-1}dr\Big)^{1/2}\\ &=c(t^{-1/2})^{1/2}=ct^{-1/4}. \end{align*} It remains to estimate $\check{q_t}(x)$. We need for the sequel, the following notations: We take $\Omega= \{\xi \in {\mathbb{R}}^{n}, |\xi|\leq t^{\frac{1}{4n}}\}$ and let $\mathcal{E}_s= \{\xi \in {\mathbb{R}}^{n}, \nabla\Psi(\xi)=0\}$ be the set of stationary points of $\Psi$. Hence from Lemma \ref{lem3.2}, $\mathcal{E}_s$ has a finite number of elements. Then set $$ s(t^{-1/4})=\bigcup_{\zeta \in \mathcal{E}_s}B(\zeta , t^{-1/4})\bigcup \{\xi \in {\mathbb{R}}^{n}, |\xi|\leq t^{-1/4}\} $$ where for each $\zeta \in \mathcal{E}_s$, $B(\zeta , t^{-1/4})=\{\xi \in {\mathbb{R}}^{n}, |\xi - \zeta|\leq t^{-1/4}\}$. Let $$ \mathcal{A}=s(t^{-1/4})\bigcup \{\frac{1}{\sqrt{2}}(1-t^{-1/4})\leq|\xi|\leq \frac{1}{\sqrt{2}}(1+t^{-1/4})\}. $$ Hence \begin{equation} \label{e3.7} \check{q_t}(x) = \int_{\Omega}e^{it\phi(\xi)+ix\cdot \xi}d\xi = \Big(\int_{\Omega \cap \mathcal{A}}+\int_{\Omega \cap {\mathcal{A}}^{c}}\Big)e^{it\phi(\xi)+ix\cdot \xi} d\xi=I_{1}+I_{2}. \end{equation} Since from Lemma \ref{lem3.2}, $card(\mathcal{E}_s)<\infty$, we get \begin{equation} \label{e3.8} \begin{aligned} |I_1|&\leq \int_{\Omega \cap \mathcal{A}}d\xi \\ &\leq \sum_{\zeta \in \mathcal{E}_s}\int_{B(\zeta , t^{-1/4})}d\xi + \int_{\{|\xi|\leq t^{-1/4}\}}d\xi + \int_{\{\frac{1}{\sqrt{2}}(1-t^{-1/4})\leq|\xi|\leq \frac{1}{\sqrt{2}}(1+t^{-1/4})\}}d\xi \\ &\leq \int_{\{0 \leq r\leq t^{-1/4} \}}r^{n-1}dr + \int_{\{\frac{1}{\sqrt{2}}(1-t^{-1/4})\leq r\leq \frac{1}{\sqrt{2}}(1+t^{-1/4}) \}}r^{n-1}dr \\ &\leq ct^{-\frac{n}{4}} +\int_{\{\frac{1}{\sqrt{2}}(1-t^{-1/4})\leq r\leq \frac{1}{\sqrt{2}}(1+t^{-1/4}) \}}r^{n-1}dr \\ &\leq ct^{-\frac{n}{4}} + ct^{-1/4}\leq ct^{-1/4}. \end{aligned} \end{equation} For $I_2$, we point out that on $$ {\mathcal{A}}^{c}=\{s(t^{-1/4})\}^{c}\bigcap \{ \{|\xi| \leq \frac{1}{\sqrt{2}}(1-t^{-1/4}) \} \cup \{|\xi| \geq \frac{1}{\sqrt{2}}(1+t^{-1/4}) \} \}, $$ $\Psi$ has no stationary point so that we can integrate $I_2$ by parts as follows: \begin{equation} \begin{aligned} \label{e3.9} |I_2| &= |\int_{\Omega \cap {\mathcal{A}}^{c}}e^{it\Psi(\xi)} d\xi|\\ &= t^{-1}|\int_{\Omega \cap {\mathcal{A}}^{c}}\frac{1}{\nabla\Psi(\xi)}\nabla(e^{it\Psi(\xi)}) d\xi| \\ &\leq t^{-1}\int_{\Omega \cap {\mathcal{A}}^{c}}|\nabla(\frac{1}{\nabla\Psi(\xi)})| d\xi + t^{-1}\int_{\partial \{\Omega \cap {\mathcal{A}}^{c} \}}\frac{d\xi}{|\nabla\Psi(\xi)|} \\ &\leq ct^{-1}\int_{\Omega \cap {\mathcal{A}}^{c}} \{|\nabla(\frac{1}{\nabla\Psi(\xi)})| +|\nabla(\frac{1}{|\nabla\Psi(\xi)|})| \}d\xi \\ &\leq ct^{-1}\int_{\Omega \cap {\mathcal{A}}^{c}} \frac{|D^{2}\Psi(\xi)|}{|\nabla\Psi(\xi)|^{2}}d\xi. \end{aligned} \end{equation} For the rest of this article, we consider a point $\xi_s \in \mathcal{E}_s$; then we have \begin{align*} {\mathcal{A}}^{c}\subset &\{\xi \in {\mathbb{R}}^{n}, |\xi - \xi_s|> t^{-1/4}\}\cap \{ |\xi |> t^{-1/4}\} \cap \{\{|\xi| <\frac{1}{\sqrt{2}}(1-t^{-1/4}) \}\\ & \cup \{|\xi| > \frac{1}{\sqrt{2}}(1+t^{-1/4})\} \}. \end{align*} Hence from \eqref{e3.9}, we obtain \begin{equation} \label{e3.10} |I_2|\leq ct^{-1}\int_{\Omega \cap \{E_1 \cup E_2\}\cap \{|\xi - \xi_s|> t^{-1/4}\}\cap \{ |\xi |> t^{-1/4}\}}\frac{|D^{2}\Psi(\xi)|}{|\nabla\Psi(\xi)|^{2}}d\xi \end{equation} where $E_1=\{\xi \in {\mathbb{R}}^{n}, |\xi|<\frac{1}{\sqrt{2}}(1-t^{-1/4})\}$ and $E_2=\{\xi \in {\mathbb{R}}^{n}, |\xi| >\frac{1}{\sqrt{2}}(1+t^{-1/4})\}$. For the sequel, we need the following inequality: with $E_1$ and $E_2$ as defined in \eqref{e3.10}, we claim that on $\{E_1 \cup E_2\}\cap \{|\xi - \xi_s|> t^{-1/4}\}\cap \{ |\xi |> t^{-1/4}\}$, \begin{equation} \label{e3.11} |{\nabla}\Psi(\xi)|\geq ct^{-1/4} \frac{|{\xi}|(1+|{\xi}|)}{(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}}. \end{equation} To prove this inequality, let us give the following remark. \begin{remark} \label{rmk3.3} \rm Let $\xi_s \in \mathcal{E}_s$. Then for any $\xi \in {\mathbb{R}}^{n}$, there exists an index set $J$ empty or not, with $J \in \{1,\dots,n\}$ such that \[ \mathop{\rm sgn}({\xi}_i)= \begin{cases}- \mathop{\rm sgn}({\xi}_{si}) & \text{if } i \in J \\ \mathop{\rm sgn}({\xi}_{si}) &\text{if } i \in J^c. \end{cases} \] \end{remark} Let us prove now inequality \eqref{e3.11}. In view of Remark \ref{rmk3.3}, let $\xi_s \in \mathcal{E}_s$ and let $J$ be an index set such that \[ \mathop{\rm sgn}({\xi}_i) =\begin{cases} - \mathop{\rm sgn}({\xi}_{si}) &\text{if } i \in J \\ \mathop{\rm sgn}({\xi}_{si}) &\text{if } i \in J^c. \end{cases} \] Moreover, define a point $\eta_s$ by \[ \eta_{si}=\begin{cases} {\xi}_{si} & \text{if } i \in J \\ -{\xi}_{si} & \text{if } i \in J^c \end{cases} \] where $J$ is the same index set as above. Hence from Lemma \ref{lem3.2}, $\eta_s$ is also a stationary point and then thanks to Remark \ref{rmk3.3} and the definition of $\eta_s$, we have on $E_2=\{\xi \in {\mathbb{R}}^{n}, |\xi|>\frac{1}{\sqrt{2}}(1+t^{-1/4})\}$ and for $|\xi_s| \geq \frac{1}{\sqrt{2}}$, \begin{align*} |{\nabla}\Psi(\xi)| &= |{\nabla}\Psi(\xi)-{\nabla}\Psi(\eta_s)| = |{\nabla}\phi(\xi)-{\nabla}\phi(\eta_s)|\\ &=|{\xi}\frac{(2|{\xi}|^2-1)}{(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}}-{\eta_s}\frac{(2|{\eta_s}|^2-1)}{(1- |{\eta_s}|^2 + |{\eta_s}|^4)^{1/2}}| \\ &= (\sum_{i \in J} + \sum_{i \in J^c})|{\xi_i}\frac{(2|{\xi}|^2-1)}{(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}}-{\eta_{si}}\frac{(2|{\eta_s}|^2-1)}{(1- |{\eta_s}|^2 + |{\eta_s}|^4)^{1/2}}| \\ &= \sum_{i \in J} |{\xi_i}\frac{(2|{\xi}|^2-1)}{(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}}-{\xi_{si}}\frac{(2|{\xi_s}|^2-1)}{(1- |{\xi_s}|^2 + |{\xi_s}|^4)^{1/2}}| \\ &\quad + \sum_{i \in J^c} |{\xi_i}\frac{(2|{\xi}|^2-1)}{(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}}+{\xi_{si}}\frac{(2|{\xi_s}|^2-1)}{(1- |{\xi_s}|^2 + |{\xi_s}|^4)^{1/2}}| \\ &= \sum_{i \in J} |{\mathop{\rm sgn}(\xi_i)|\xi_i|}\frac{(2|{\xi}|^2-1)}{(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}}-{\mathop{\rm sgn}(\xi_{si})|\xi_{si}|}\frac{(2|{\xi_s}|^2-1)}{(1- |{\xi_s}|^2 + |{\xi_s}|^4)^{1/2}}| \\ &\quad + \sum_{i \in J^c} |{\mathop{\rm sgn}(\xi_i)|\xi_i|}\frac{(2|{\xi}|^2-1)}{(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}}+{\mathop{\rm sgn}(\xi_{si})|\xi_{si}|}\frac{(2|{\xi_s}|^2-1)}{(1- |{\xi_s}|^2 + |{\xi_s}|^4)^{1/2}}| \\ &= \sum_{i \in J} |{\mathop{\rm sgn}(\xi_i)|\xi_i|}\frac{(2|{\xi}|^2-1)}{(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}}+{\mathop{\rm sgn}(\xi_{i})|\xi_{si}|}\frac{(2|{\xi_s}|^2-1)}{(1- |{\xi_s}|^2 + |{\xi_s}|^4)^{1/2}}| \\ &\quad + \sum_{i \in J^c} |{\mathop{\rm sgn}(\xi_i)|\xi_i|}\frac{(2|{\xi}|^2-1)}{(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}}+{\mathop{\rm sgn}(\xi_{i})|\xi_{si}|}\frac{(2|{\xi_s}|^2-1)}{(1- |{\xi_s}|^2 + |{\xi_s}|^4)^{1/2}}| \\ &= (\sum_{i \in J}+ \sum_{i \in J^c}) (\frac{|\xi_i|(2|{\xi}|^2-1)}{(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}}+\frac{|\xi_{si}|(2|{\xi_s}|^2-1)}{(1- |{\xi_s}|^2 + |{\xi_s}|^4)^{1/2}}) \\ &\geq \frac{|\xi|(\sqrt{2}|{\xi}|-1)(\sqrt{2}|{\xi}|+1)}{(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}} \geq t^{-1/4}\frac{|\xi|(|{\xi}|+1)}{(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}}. \end{align*} Again on $E_2=\{\xi \in {\mathbb{R}}^{n}, |\xi|>\frac{1}{\sqrt{2}}(1+t^{-1/4})\}$ but now for $|\xi_s| \leq \frac{1}{\sqrt{2}}$, we write thanks to the definition of $\eta_s$, \begin{align*} |{\nabla}\Psi(\xi)| &= |{\nabla}\phi(\xi)-{\nabla}\phi(-\eta_s)|\\ &=|{\xi}\frac{(2|{\xi}|^2-1)}{(1-|{\xi}|^2 + |{\xi}|^4)^{1/2}}+{\eta_s}\frac{(2|{\eta_s}|^2-1)}{(1- |{\eta_s}|^2 + |{\eta_s}|^4)^{1/2}}| \\ &= |{\xi}\frac{(2|{\xi}|^2-1)}{(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}}-{\eta_s}\frac{(1-2|{\eta_s}|^2)}{(1- |{\eta_s}|^2 + |{\eta_s}|^4)^{1/2}}| \end{align*} so that thanks to Remark \ref{rmk3.3} and the definition of $\eta_s$, we follow the same lines as above to obtain \begin{align*} |{\nabla}\Psi(\xi)|&=|\xi|\frac{(2|{\xi}|^2-1)}{(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}}+|\xi_{s}|\frac{(1-2|{\xi_s}|^2)}{(1- |{\xi_s}|^2 + |{\xi_s}|^4)^{1/2}}\\ & \geq ct^{-1/4}\frac{|\xi|(|{\xi}|+1)}{(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}}. \end{align*} For this time, on $E_1=\{\xi \in {\mathbb{R}}^{n}, |\xi| <\frac{1}{\sqrt{2}}(1-t^{-1/4})\}$ and if $|\xi_s| \geq \frac{1}{\sqrt{2}}$, we write thanks to the definition of $\eta_s$ above, \begin{align*} |{\nabla}\Psi(\xi)| &= |{\nabla}\phi(-\eta_s)-{\nabla}\phi(\xi)|\\ &=|-{\eta_s}\frac{(2|{\eta_s}|^2-1)}{(1- |{\eta_s}|^2 + |{\eta_s}|^4)^{1/2}}-{\xi}\frac{(2|{\xi}|^2-1)}{(1- |{\xi}|^2 +|{\xi}|^4)^{1/2}}| \\ &= |-{\eta_s}\frac{(2|{\eta_s}|^2-1)}{(1- |{\eta_s}|^2 + |{\eta_s}|^4)^{1/2}}+{\xi}\frac{(1-2|{\xi}|^2)}{(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}}| \end{align*} so that thanks to Remark \ref{rmk3.3} and the definition of $\eta_s$, we follow the same lines as above to get \begin{align*} |{\nabla}\Psi(\xi)|&=|\xi_s|\frac{(2|{\xi_s}|^2-1)}{(1- |{\xi_s}|^2 + |{\xi_s}|^4)^{1/2}}+|\xi|\frac{(1-2|{\xi}|^2)}{(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}} \\ &\geq ct^{-1/4}\frac{|\xi|(|{\xi}|+1)}{(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}}. \end{align*} Finally, still on $E_1=\{\xi \in {\mathbb{R}}^{n}, |\xi| <\frac{1}{\sqrt{2}}(1-t^{-1/4})\}$ but now for $|\xi_s| \leq \frac{1}{\sqrt{2}}$, we write with the definition of $\eta_s$, \begin{align*} |{\nabla}\Psi(\xi)| &= |{\nabla}\phi(\eta_s)-{\nabla}\phi(\xi)|\\ &=|{\eta_s}\frac{(2|{\eta_s}|^2-1)}{(1- |{\eta_s}|^2 + |{\eta_s}|^4)^{1/2}}-{\xi}\frac{(2|{\xi}|^2-1)}{(1- |{\xi}|^2 +|{\xi}|^4)^{1/2}}| \\ &= |-{\eta_s}\frac{(1-2|{\eta_s}|^2)}{(1- |{\eta_s}|^2 + |{\eta_s}|^4)^{1/2}}+{\xi}\frac{(1-2|{\xi}|^2)}{(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}}| \end{align*} so that proceeding as above, we find thanks to Remark \ref{rmk3.3} and the definition of $\eta_s$, $$ |{\nabla}\Psi(\xi)| \geq ct^{-1/4}\frac{|\xi|(|{\xi}|+1)}{(1- |{\xi}|^2 + |{\xi}|^4)^{1/2}}. $$ This completes the proof of \eqref{e3.11}. For the sequel, we need the following inequality which, thanks to \eqref{e2.14} and \eqref{e3.11}, is obviously shown: That is: On $$ \Omega\cap \{E_1 \cup E_2\}\cap \{|\xi - \xi_s|> t^{-1/4}\}\cap \{ |\xi |> t^{-1/4}\} $$ where $\Omega$ is defined in \eqref{e3.5} and $E_1$ and $E_2$ are defined in \eqref{e3.10}, we have \begin{equation} \label{e3.12} \frac{|D^{2}\Psi(\xi)|}{|{\nabla}\Psi(\xi)|^{2}} \leq ct^{1/2}\frac{(1- |{\xi}|^2 + |{\xi}|^4)}{|\xi|^{2}(|{\xi}|+1)^{2}} \leq ct^{1/2}\frac{(1 + |{\xi}|)^2}{|\xi|^{2}}. \end{equation} Therefore, from \eqref{e3.10}, \eqref{e3.11}, \eqref{e3.12}, we get \begin{equation} \label{e3.13} \begin{aligned} |I_2| &\leq ct^{-1}\int_{\Omega \cap \{E_1 \cup E_2\}\cap \{|\xi - \xi_s|> t^{-1/4}\}\cap \{ |\xi |> t^{-1/4}\}}\frac{|D^{2}\Psi(\xi)|}{|\nabla\Psi(\xi)|^{2}}d\xi \\ &\leq ct^{-1}t^{1/2}\int_{\Omega \cap \{E_1 \cup E_2\}\cap \{|\xi - \xi_s|> t^{-1/4}\}\cap \{ |\xi |> t^{-1/4}\}}|\xi|^{-2}(1 + |{\xi}|)^2d\xi \\ &\leq ct^{-1/2}\int_{\Omega \cap E_1 \cap \{|\xi - \xi_s|> t^{-1/4}\}\cap \{ |\xi |> t^{-1/4}\}}|\xi|^{-2}(1 + |{\xi}|)^2d\xi \\ &\quad + ct^{-1/2}\int_{\Omega \cap E_2\cap \{|\xi - \xi_s|> t^{-1/4}\}\cap \{ |\xi |> t^{-1/4}\}}|\xi|^{-2}(1 + |{\xi}|)^2d\xi \\ &\leq ct^{-1/2} \{\int_{ \{ t^{-1/4}<|\xi | <\frac{1}{\sqrt{2}}(1-t^{-1/4})\}} |\xi|^{-2}d\xi + \int_{ \{ \frac{1}{\sqrt{2}}(1+t^{-1/4})<|\xi | < t^{\frac{1}{4n}}\}}d\xi \} \\ &\leq ct^{-1/2} \{ \int_{ \{ t^{-1/4}\frac{1}{\sqrt{2}}(1+t^{-1/4})\}$. Hence, with the estimates on $I_1$ and $I_2$ in \eqref{e3.8} and \eqref{e3.13} above, and thanks to \eqref{e3.7}, we are led to \[ %3.14 |\check{q}_t(x)|_{\infty}\leq ct^{-1/4} \quad \forall t\geq 1. \] Combining this inequality, \eqref{e3.6} and \eqref{e3.5}, we find $$ |S_1(t)f_1(x)| \leq c (1+t)^{-1/4}(|f_1|_{1}+\|f_1\|_{3n/2} \quad \forall t\geq 0, $$ which with Remark \ref{rmk2.2} leads to \eqref{e3.1} for the case $j=1$. Let us prove now the inequality \eqref{e3.1} for the case $j=2$ If $0\leq t \leq1$, we have thanks to the Schwartz inequality and the inequality \eqref{e2.5} on $\phi(\xi)$, \begin{equation} \label{e3.15} \begin{aligned} |S_2(t)f_2(x)| &= \frac{1}{{2}(2\pi)^{n}}|\int_{{\mathbb{R}}^{n}} e^{it\Psi(\xi)}\frac{\hat{f_2}(\xi)}{\phi(\xi)} d\xi| \\ &\leq c\int_{{\mathbb{R}}^{n}}\frac{|\hat{f_2}(\xi)|}{|{\phi(\xi)}|} d\xi \\ &\leq c\int_{{\mathbb{R}}^{n}}\frac{|\hat{f_2}(\xi)|}{(1 + |{\xi}|^{2})} d\xi \\ &\leq c(\int_{{\mathbb{R}}^{n}}(1+|\xi|^{2})^{-n} d\xi)^{1/2}\|f_2\|_{n-2} \\ &\leq c\|f_2\|_{n-2} \leq c(1+t)^{-1/4}\|f_2\|_{3n/2}. \end{aligned} \end{equation} If $t\geq 1$, then we have with the notation $\Omega = \{\xi \in {\mathbb{R}}^{n}, |\xi|\leq t^{\frac{1}{4n}}\}$ given above and thanks to \eqref{e2.5}, the Schwartz and the Young inequalities, \begin{equation} \label{e3.16} \begin{aligned} |S_2(t)f_2(x)| &= \frac{1}{{2}(2\pi)^{n}} \Big|\Big(\int_\Omega+\int_{{\Omega}^{c}}\Big)e^{it\phi(\xi)+ix\cdot \xi}\frac{\hat{f_2}}{\phi(\xi)}(\xi)d\xi\Big| \\ &\leq c|\check{k_t}(x)\ast f_2(x)|_{\infty} +c\Big(\int_{{\Omega}^{c}}(1+|\xi|^{2})^{-\frac{3}{2}n} d\xi\Big)^{1/2}\|f_2\|_{3n/2} \\ &\leq c|\check{k_t}(x)|_{\infty}|f_2(x)|_{1} + ct^{-1/4}\|f_2\|_{3n/2} \end{aligned} \end{equation} where the function $k_t(\xi)={\chi}_{\Omega}(\xi) e^{it\phi(\xi)}/ \phi(\xi)$. On the other hand, with the same notations of $\Omega$ and $\mathcal{A}$ given in \eqref{e3.5}, \eqref{e3.7}, \eqref{e3.9}, we write \begin{equation} \label{e3.17} \check{k_t}(x) = \frac{1}{(2\pi)^{n}}\Big(\int_{\Omega \cap \mathcal{A}}+\int_{\Omega \cap {\mathcal{A}}^{c}}\Big)e^{it\phi(\xi)+ix\cdot \xi}\frac{1}{\phi(\xi)} d\xi=J_{1}+J_{2}. \end{equation} Then, with the use of the inequality \eqref{e2.5}, we follow the same lines as the estimation of $I_1$ in \eqref{e3.8} to get \begin{equation} \label{e3.18} |J_{1}| \leq ct^{-1/4}. \end{equation} For the estimation of $J_2$, we need the following inequality which with the use of \eqref{e2.5}, \eqref{e2.13}, \eqref{e2.14}, \eqref{e3.11}, \eqref{e3.12}, is obviously proved. That is: On \begin{equation} \label{e3.19} \begin{gathered} \Omega\cap \{E_1 \cup E_2\}\cap \{|\xi - \xi_s|> t^{-1/4}\}\cap \{ |\xi |> t^{-1/4}\}, \\ \frac{|D^{2}\Psi(\xi)|}{|{\nabla}\Psi(\xi)|^{2}|\phi(\xi)|} +\frac{|{\nabla}\phi(\xi)|}{|{\nabla}\Psi(\xi)||\phi(\xi)|^{2}} \leq ct^{1/2}|\xi|^{-2}(1 + |{\xi}|^2). \end{gathered} \end{equation} Therefore, following the same lines as in the proofs of \eqref{e3.9} and \eqref{e3.13}, we find thanks to \eqref{e3.19} and integration by parts (as for $I_2$), \begin{equation} \label{e3.20} \begin{aligned} \quad |J_2| &= |\int_{\Omega \cap {\mathcal{A}}^{c}}e^{it\Psi(\xi)}\frac{1}{\phi(\xi)} d\xi|= t^{-1}|\int_{\Omega \cap {\mathcal{A}}^{c}}\frac{1}{\nabla{\Psi(\xi)}\phi(\xi)}\nabla(e^{it\Psi(\xi)}) d\xi| \\ &\leq t^{-1}\{\int_{\Omega \cap {\mathcal{A}}^{c}}|\nabla(\frac{1}{\nabla{\Psi(\xi)}\phi(\xi)})| d\xi + \int_{\partial \{\Omega \cap {\mathcal{A}}^{c} \}}\frac{d\xi}{|\nabla\Psi(\xi)||\phi(\xi)|}\} \\ &\leq ct^{-1}\int_{\Omega \cap {\mathcal{A}}^{c}} |\nabla(\frac{1}{\nabla{\Psi(\xi)}\phi(\xi)})| d\xi \\ &\leq ct^{-1}\int_{\Omega \cap {\mathcal{A}}^{c}} \{\frac{|D^{2}\Psi(\xi)|}{|\nabla\Psi(\xi)|^{2}|\phi(\xi)|} + \frac{|{\nabla}\phi(\xi)|}{|{\nabla}\Psi(\xi)||\phi(\xi)|^{2}} \} d\xi \\ &\leq ct^{-1/2}\int_{\Omega \cap E_1 \cap \{|\xi - \xi_s|> t^{-1/4}\}\cap \{ |\xi |> t^{-1/4}\}}|\xi|^{-2}(1 + |{\xi}|^2)d\xi \\ &\quad + ct^{-1/2}\int_{\Omega \cap E_2\cap \{|\xi - \xi_s|> t^{-1/4}\}\cap \{ |\xi |> t^{-1/4}\}}|\xi|^{-2}(1 + |{\xi}|^2)d\xi \\ &\leq ct^{-1/4}. \end{aligned} \end{equation} Hence \eqref{e3.17} and the estimates \eqref{e3.18} and \eqref{e3.20} on $J_1$ and $J_2$ above, give $$ |\check{k_t}(x)| \leq ct^{-1/4} \; \forall t \geq 1. $$ Then, this with \eqref{e3.16} give \[ %3.21 |S_2(t)f_2(x)| \leq c (1+t)^{-1/4}(|f_2|_{1}+\|f_2\|_{3n/2} \; \forall t\geq 1, \] Combining this inequality and \eqref{e3.15}, we get with Remark \ref{rmk2.2} the desired inequality \eqref{e3.1} for the case $j=2$. This finishes up the proof of inequality \eqref{e3.4}. In order to prove the inequality \eqref{e3.2} of Lemma \ref{lem3.1}, we set $J^k = (1-\triangle)^{k/2}$ with $k \in \mathbb{R}$, and we note that \begin{align*} J^kS_1(t)f_1(x)&= J^k{\mathcal{F}}^{-1}(\frac{1}{2}e^{it\phi(\xi)}\hat{f_1}(\xi))(x) \\ &={\mathcal{F}}^{-1}(\frac{1}{2}(1+|{\xi}|^{2})^{k/2}e^{it\phi(\xi)} \hat{f_1}(\xi))(x) \\ &= \frac{1}{{2}(2\pi)^{n}} \int_{{\mathbb{R}}^{n}}(1 +|{\xi}|^{2})^{k/2}e^{it\phi(\xi)+ix\cdot \xi}\hat{f_1}(\xi)d\xi \end{align*} and $$ J^kS_2(t)f_2(x) = \frac{1}{{2}(2\pi)^{n}} \int_{{\mathbb{R}}^{n}}(1 +|{\xi}|^{2})^{k/2}e^{it\phi(\xi)+ix\cdot \xi}\frac{\hat{f_2}}{\phi(\xi)}(\xi)d\xi. $$ Henceforth, we can prove the inequality \eqref{e3.2} of Lemma \ref{lem3.1}. We begin with the case $j=1$: For $0 \leq t \leq 1$, we follow the same lines as in \eqref{e3.5} and we get \begin{equation} \label{e3.22} |J^kS_1(t)f_1(x)| \leq c (1+t)^{-1/4}\|f_1\|_{\frac{3}{2}n+k}. \end{equation} If $t\geq 1$, let $p_t(\xi)=(1+|{\xi}|^{2})^{-\frac{5}{4}n} {\chi}_{\Omega}(\xi)e^{it\phi(\xi)}$ where $\Omega = \{\xi \in {\mathbb{R}}^{n}, |\xi|\leq t^{\frac{1}{4n}}\}$ is defined above in \eqref{e3.6}. Then thanks to the Schwartz and the Young inequalities, we have as in \eqref{e3.6}, \begin{equation} \label{e3.23} \begin{aligned} |J^{k}S_1(t)f_1(x)| &=\frac{1}{{2}(2\pi)^{n}}\Big|\Big(\int_\Omega+\int_{{\Omega}^{c}}\Big) e^{it\phi(\xi)+ix\cdot \xi}(1+|{\xi}|^{2}\Big)^{-\frac{k}{2}} \hat{f_1}(\xi)d\xi\Big| \\ &\leq c|\check{p_t}(x)\ast (1-\triangle)^{\frac{(\frac{5}{2}n+k)}{2}}f_1(x)|_{\infty} \\ &\quad +c\Big(\int_{{\Omega}^{c}}(1+|\xi|^{2})^{-\frac{3}{2}n} d\xi\Big)^{1/2}(\int_{{\Omega}^{c}}(1+|\xi|^{2})^{\frac{3}{2}n+k}|\hat{f_1}(\xi)|^{2} d\xi)^{1/2} \\ &\leq c|\check{p_t}(x)|_{\infty}\|f_1(x)\|_{\frac{5}{2}n+k,1} + ct^{-1/4}\|f_1\|_{\frac{3}{2}n+k}. \end{aligned} \end{equation} Then, with the same notation of $\mathcal{A}$ and $\Omega$ given in \eqref{e3.7} above, we write: \begin{equation} \label{e3.24} \check{p_t}(x) = \frac{1}{(2\pi)^{n}}\Big(\int_{\Omega \cap \mathcal{A}}+\int_{\Omega \cap {\mathcal{A}}^{c}}\Big) (1+|{\xi}|^{2})^{-\frac{5}{4}n}e^{it\phi(\xi)+ix\cdot \xi} d\xi=I'_{1}+I'_{2} \end{equation} and following the same lines as in \eqref{e3.8} we get \begin{equation} \label{e3.25} |I'_{1}|\leq c\int_{\Omega \cap \mathcal{A}}(1 +|{\xi}|^{2})^{-\frac{5}{4}n}d\xi\leq c\int_{\Omega \cap \mathcal{A}}d\xi \leq ct^{-1/4}. \end{equation} For the sequel, we need the following inequality which with the help of the inequalities \eqref{e2.5}, \eqref{e2.12}, \eqref{e2.13}, \eqref{e3.11}, \eqref{e3.12}, is easily proved. That is, for all $\xi \in {\mathbb{R}}^{n}$ and for any given $\gamma \geq 0$, \begin{equation} \label{e3.26} |\nabla(\frac{1}{{\nabla}\Psi(\xi)(1 + |{\xi}|^2)^{\frac{\gamma}{2}}})| +|\nabla(\frac{1}{\phi(\xi){\nabla}{\Psi(\xi)}(1 + |{\xi}|^2)^{\frac{\gamma}{2}}})| \leq ct^{1/2}|\xi|^{-2}(1 + |{\xi}|^2). \end{equation} Henceforth, thanks to the above inequality, we follow the same lines as in the proof of \eqref{e3.20}, and using integration by parts, we get \begin{equation} \label{e3.27} \begin{aligned} |I'_2| &= \frac{1}{2(2\pi)^{n}}|\int_{\Omega \cap {\mathcal{A}}^{c}}\frac{e^{it\Psi(\xi)}}{(1 + |{\xi}|^2)^{\frac{\gamma}{2}}} d\xi|\\ &= \frac{1}{2(2\pi)^{n}}t^{-1}|\int_{\Omega \cap {\mathcal{A}}^{c}}\frac{\nabla(e^{it\Psi(\xi)})}{\nabla{\Psi(\xi)}(1 + |{\xi}|^2)^{\frac{\gamma}{2}}} d\xi| \\ &\leq ct^{-1}\int_{\Omega \cap {\mathcal{A}}^{c}}|\nabla(\frac{1}{\nabla\Psi(\xi)(1 + |{\xi}|^2)^{\frac{\gamma}{2}}})| d\xi \\ &\leq ct^{-1/2}\int_{\Omega \cap E_1 \cap \{|\xi - \xi_s|> t^{-1/4}\}\cap \{ |\xi |> t^{-1/4}\}}|\xi|^{-2}(1 + |{\xi}|^2)d\xi \\ &\quad + ct^{-1/2}\int_{\Omega \cap E_2\cap \{|\xi - \xi_s|> t^{-1/4}\}\cap \{ |\xi |> t^{-1/4}\}}|\xi|^{-2}(1 + |{\xi}|^2)d\xi \\ \leq ct^{-1/4}. \end{aligned} \end{equation} Therefore, \eqref{e3.24} and the estimates in \eqref{e3.25} and \eqref{e3.27} of $I'_1$ and $I'_2$ above, give $$ |\check{p_t}(x)| \leq ct^{-1/4} \; \quad t \geq 1. $$ so that thanks to \eqref{e3.23} we get for all $t \geq 1$, \begin{equation} \label{e3.28} |J^kS_1(t)f_1(x)| \leq c (1+t)^{-1/4}(\|f_1\|_{\frac{5}{2}n+k,1}+\|f_1\|_{\frac{3}{2}n+k}), \quad k \geq 0. \end{equation} Finally, combining \eqref{e3.22} and \eqref{e3.28}, and thanks to Remark \ref{rmk2.2}, we find the case $j=1$ of the inequality \eqref{e3.2} of Lemma \ref{lem3.1}. Likewise, following the same lines as in the proof of the case $j=1$ of \eqref{e3.2}, and with the use of the inequalities \eqref{e2.5} and \eqref{e3.26}, we prove the case $j=2$ of the inequality \eqref{e3.2}. This, with Remark \ref{rmk2.2} puts an end of the proof of the inequality \eqref{e3.2} and consequently of Lemma \ref{lem3.1}. Let us give now the following lemma which will be useful for the ${\mathbb{L}}^{p}-{\mathbb{L}}^{q}$ estimates. \begin{lemma} \label{lem3.4} Let $f_1, f_2 \in {\mathbb{L}}^{1}_{\frac{5}{2}n+k}({\mathbb{R}}^{n})$, $k\geq 0$, $n\geq 1$. Then \begin{equation} \label{e3.29} \|V_j(t)f_j(x)\|_{k, \infty} \leq c (1+t)^{-1/4}\|f_j\|_{\frac{5}{2}n+k,1}, \quad j=1,2. \end{equation} \end{lemma} \begin{proof} Thanks to inequality \eqref{e3.2} of Lemma \ref{lem3.1}, it suffices to use the Sobolev embedding $W^{n,1}({\mathbb{R}}^{n}) \subset {\mathbb{L}}^{2}({\mathbb{R}}^{n})$ and we get \begin{align*} \|V_j(t)f_j(x)\|_{k, \infty} &\leq c(1+t)^{-1/4}(\|f_j\|_{\frac{5}{2}n+k,1}+ \|f_j\|_{\frac{3}{2}n+k}) \\ &= c (1+t)^{-1/4}(\|f_j\|_{\frac{5}{2}n+k,1}+ |J^{\frac{3}{2}n+k}f_j|_{2}) \\ &\leq c (1+t)^{-1/4}(\|f_j\|_{\frac{5}{2}n+k,1}+ \|J^{\frac{3}{2}n+k}f_j\|_{n,1}) \\ &= 2c(1+t)^{-1/4}\|f_j\|_{\frac{5}{2}n+k,1}, \; \; j=1,2. \end{align*} \end{proof} To end with this section, we give the following lemma. \begin{lemma} \label{lem3.5} Let $f_1, f_2 \in {\mathbb{H}}^{k+\frac{5}{2}n+1}({\mathbb{R}}^{n}) \cap{\mathbb{L}}^{q}_{k+\frac{5}{2}n}({\mathbb{R}}^{n})$, $k \geq 0$, $n\geq 1$. Then \begin{equation} \label{e3.30} \|V_j(t)f_j(x)\|_{k, p} \leq c (1+t)^{-\frac{\theta}{4}}\|f_j\|_{\frac{5}{2}n+k,q}, \quad j=1,2 \end{equation} where $p=2/(1-\theta)$, $q=2/(1+\theta)$, $\theta \in ]0,1[$. \end{lemma} \begin{proof} Thanks to \eqref{e2.6} and \eqref{e2.7}, we get for any $k \in \mathbb{R}_{+}$, \begin{equation} \label{e3.31} \|V_j(t)f_j(x)\|_{k} \leq c\|f_j(x)\|_{k} \leq c\|f_j\|_{\frac{5}{2}n+k}, \quad j=1,2; \end{equation} that is \begin{equation} \label{e3.32} |J^{k}V_j(t)f_j(x)|_{2} \leq c|J^{\frac{5}{2}n+k}f_j(x)|_{2}, \quad j=1,2. \end{equation} Moreover, from \eqref{e3.29} in Lemma \ref{lem3.4} we have \begin{equation} \label{e3.33} |J^{k}V_j(t)f_j(x)|_{\infty} \leq c(1+t)^{-1/4}|J^{\frac{5}{2}n+k}f_j(x)|_{1}, \quad j=1,2. \end{equation} We know that (see above), \[ J^{k}V_j(t)(f_j(x))=V_j(t)(J^{k}f_j(x)) =J^{-\frac{5}{2}n}V_j(t)(J^{\frac{5}{2}n+k}f_j(x)). \] Therefore, thanks to \eqref{e3.32} and \eqref{e3.33}, we apply the interpolation theorem (see \cite{b1}) for the evolution operator $J^{-\frac{5}{2}n}V_j(t)$, $j=1,2$, and we find the inequality \eqref{e3.30} of Lemma \ref{lem3.5}. This finishes up the proof of Lemma \ref{lem3.5}. \end{proof} \section{Decay and Scattering results of Solutions to the Nonlinear Equation} \begin{proof}[Proof of Theorem Theorem \ref{thm1.1}] We write \eqref{e1.1} in its integral form as given in \eqref{e2.8}: \begin{equation} \label{e4.1} u(x,t)=V_1(t)f_1(x)+ V_2(t)f_2(x) + \int^t_0V_2(t-\tau)(|u|^{\alpha }u)(\tau) d \tau. \end{equation} where $V_1(t)$ and $V_2(t)$ are defined in (2.3), (2.4). Then, taking the ${\mathbb{L}}^{\infty}$ norm of the both sides of \eqref{e4.1} we get thanks to Lemma \ref{lem3.1}, \begin{equation} \begin{aligned} \label{e4.2} |u(t)|_{\infty} &\leq c(1+t)^{-1/4}(|f_1|_{1} +\|f_1\|_{3n/2}+ |f_2|_{1} +\|f_2\|_{3n/2} \\ &\quad + c\int^t_0(1+(t-\tau))^{-1/4}(||u|^{\alpha }u|_{1} +\|{|u|}^{\alpha }u\|_{3n/2}(\tau) d\tau \\ &\leq c(1+t)^{-1/4}(|f_1|_{1} +\|f_1\|_{3n/2}+ |f_2|_{1} +\|f_2\|_{3n/2}\\ &\quad + c\int^t_0(1+(t-\tau))^{-1/4}(|u|^{\alpha -1}_{\infty}|u|^{2}_{2} +|u|^{\alpha }_{\infty}\|u\|_{3n/2}(\tau) d \tau. \end{aligned} \end{equation} Then, we define the quantity $$ Q(t)= \sup_{0\leq \tau \leq t} \{(1+\tau)^{ \frac{1}{4}} |u(\tau)|_{\infty}+\|u(\tau)\|_{3n/2} \}. $$ From \eqref{e4.2} \begin{equation} \label{e4.3} \begin{aligned} |u(t)|_{\infty} &\leq c(1+t)^{-1/4}(|f_1|_{1} +\|f_1\|_{3n/2}+ |f_2|_{1} +\|f_2\|_{3n/2} \\ &\quad + c{Q(t)}^{\alpha +1}\int^t_0(1+(t-\tau))^{-1/4}(1+\tau)^{- \frac{1}{4}(\alpha -1)} d\tau. \end{aligned} \end{equation} But since for $\alpha> 5$, \begin{align*} & \int^t_0(1+(t-\tau))^{-1/4}(1+\tau)^{- \frac{1}{4}(\alpha -1)} d\tau \\ &= (\int^{\frac{t}{2}}_{0} + \int^{t}_{\frac{t}{2}})(1+(t-\tau))^{-1/4}(1+\tau)^{- \frac{1}{4}(\alpha -1)} d \tau \leq c(1+t)^{-1/4} \end{align*} we deduce from \eqref{e4.3} that for $\alpha> 5$, \begin{equation} \label{e4.4} (1+t)^{\frac{1}{4}}|u(t)|_{\infty} \leq c\{|f_1|_{1} +\|f_1\|_{3n/2}+ |f_2|_{1} +\|f_2\|_{3n/2} + {Q(t)}^{\alpha +1}\} \end{equation} Furthermore, we get for $\alpha> 5$, thanks to the inequalities \eqref{e2.6} and \eqref{e2.7} of Lemma \ref{lem2.4} and with \eqref{e4.1}, \begin{equation} \begin{aligned} \label{e4.5} \|u(t)\|_{3n/2} &\leq c\{ \|f_1\|_{3n/2}+ \|f_2\|_{3n/2} + \int^t_0\|{|u|}^{\alpha }u\|_{3n/2}(\tau) d\tau \}\\ &\leq c\{\|f_1\|_{3n/2} +\|f_2\|_{3n/2} + \int^t_0|u|^{\alpha }_{\infty}\|u\|_{3n/2}(\tau) d \tau \} \\ &\leq c\{\|f_1\|_{3n/2} +\|f_2\|_{3n/2} + {Q(t)}^{\alpha +1}\int^t_0(1+\tau)^{- \frac{\alpha}{4}} d\tau\} \\ &\leq c\{\|f_1\|_{3n/2} +\|f_2\|_{3n/2} + {Q(t)}^{\alpha +1}\}. \end{aligned} \end{equation} Therefore, \eqref{e4.4} and \eqref{e4.5} give \begin{equation} \label{e4.6} Q(t) \leq c\{|f_1|_{1} +\|f_1\|_{3n/2}+ |f_2|_{1} +\|f_2\|_{3n/2} + {Q(t)}^{\alpha +1}\}. \end{equation} Henceforth, thanks to the inequality \eqref{e4.6}, if $ |f_1|_{1}+\|f_1\|_{3n/2}+ |f_2|_{1} +\|f_2\|_{3n/2} <\delta $ with $\delta >0$ small enough, we find that $Q(t)$ is bounded. Indeed, it is well known that inequality \eqref{e4.6} is satisfied if $Q(t) \in [0, \beta_1]\cup[\beta_2, \infty[$ with $0< \beta_1 < \beta_2 < \infty$ since $\delta$ is small. Thereby, since $Q(0) \leq 2\|f_1\|_{3n/2} <2\delta$ (because ${\mathbb{H}}^{\frac{3}{2}n}({\mathbb{R}}^{n})\subset{\mathbb{L}}^{\infty}({\mathbb{R}}^{n})$), the continuity of $Q(t)$ and the inequality \eqref{e4.6} allow us to conclude that $Q(t)$ remains bounded for all $t\geq 0$. Thus, we have obtained a bound of $Q(t)$ and consequently an a-priori estimate of the local solution which permit us to extend globally the local solution of Theorem \ref{thm2.1}. Moreover, this a-priori estimate provides the inequality \eqref{e1.2} of Theorem \ref{thm1.1}. For the proof of the scattering result in the Theorem \ref{thm1.1}, we define \begin{equation} \label{e4.7} u_{+}(x,t)=u(x,t) + \int^{+\infty}_{t}V_2(t-\tau)(|u|^{\alpha }u)(\tau) d \tau \end{equation} where $u(x,t)$ is the solution of \eqref{e1.1} given by Theorem \ref{thm1.1}. We only consider the case of $u_{+} \; (t\to +\infty)$ since the proof for the case of $u_{-} \; (t\to -\infty)$ is similar. Then, thanks to \eqref{e4.7} and with the use of the inequalities \eqref{e2.7} of Lemma \ref{lem2.4} and \eqref{e1.2} of Theorem \ref{thm1.1}, we have, \begin{align*} \|u(t)-u_{+}(t)\|_{2,2} &\leq c\int^{+\infty}_{t} |(|u|^{\alpha }u)(\tau)|_{2} d \tau \\ &\leq c\int^t_0 |u(\tau)|^{\alpha}_{\infty} |u(\tau)|_{2}d \tau \\ &\leq c\int^{+\infty}_{t}(1+\tau)^{-\frac{\alpha}{4}}d\tau \end{align*} and the integral on the right-hand side approaches to zero as $t\to + \infty$, since by hypothesis of Theorem \ref{thm1.1}, $\alpha >5$. Thereafter, set $g_{+}(x)= f_2(x) +\int^{+\infty}_{0}V_2(-\tau)(|u|^{\alpha}u)(\tau) d \tau.$ Then thanks to \eqref{e4.7} and \eqref{e4.1}, we may write $u_{+}$ as \begin{equation} u_{+}(x,t) = V_1(t)f(x)+ V_2(t)g_{+}(x). \end{equation} Therefore, we can see that $u_{+}(t)$ is a solution of the linearized equation \eqref{e2.1}. This completes the proof of Theorem \ref{thm1.1}. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1.2}] To prove Theorem \ref{thm1.2}, we need the following inequality of Gagliardo-Nirenberg type. \begin{lemma} \label{lem4.1} Let $u$ belong to ${\mathbb{L}}^{p_{2}}({\mathbb{R}}^{n})$ and its derivatives of order $m$, $D^{m}u$ belong to ${\mathbb{L}}^{r_{1}}({\mathbb{R}}^{n})$, $1\leq p_{2}$, $r_{1} \leq \infty$. For the derivatives $D^{j}u$ $0\leq j < m$, the following inequalities hold: $$ |D^{j}u|_{p_{1}} \leq c|D^{m}u|^{a}_{r_{1}}|u|^{1-a}_{p_{2}}, $$ where $$ \frac{1}{p_{1}}=\frac{j}{n}+a(\frac{1}{r_1}-\frac{m}{n})+(1-a)\frac{1}{p_2}, $$ for all $a$ in the interval $\frac{j}{m}\leq a \leq 1$. \end{lemma} The proof of the above lemma can be found in \cite{n1}. Now, we prove Theorem \ref{thm1.2}. We recall the notation $p= 2/(1- \theta)$, $q= 1/(1+ \theta)$, $\theta \in ]0,1[$. Let $r>\frac{n}{p}$ and apply the norm ${\mathbb{L}}^{p}_{r}({\mathbb{R}}^{n})$ to the two sides of \eqref{e4.1}. Then, thanks to Lemma \ref{lem3.5}, the Gagliardo-Nirenberg inequality in Lemma \ref{lem4.1} and Sobolev imbeddings theorems, \begin{equation} \label{e4.9} \begin{aligned} & \|u(x,t)\|_{r,p}\\ &\leq \|V_1(t)f(x)\|_{r,p}+\|V_2(t)g(x)\|_{r,p} + \int^t_0 \|V_2(t-\tau)(|u|^{\alpha -1}u)(\tau)\|_{r,p} d \tau \\ &\leq c(1+t)^{- \frac{\theta}{4}}(\|f_1\|_{\frac{5}{2}n+r,q} +\|f_2\|_{\frac{5}{2}n+r,q}) + c\int^t_0(1+(t-\tau))^{- \frac{\theta}{4}}\||{u}|^{\alpha }u(\tau)\|_{\frac{5}{2}n+r,q} d\tau \\ &\leq c(1+t)^{- \frac{\theta}{4}}(\|f_1\|_{\frac{5}{2}n+r,q} +\|f_2\|_{\frac{5}{2}n+r,q}) \\ &\quad + c\int^t_0(1+(t-\tau))^{- \frac{\theta}{4}}\||{u}|^{\alpha }u(\tau)\|^{a}_{\frac{5}{2}n+r+1,2}||{u}|^{\alpha }u(\tau)|^{1-a}_{1} d\tau \\ &\leq c(1+t)^{- \frac{\theta}{4}}(\|f_1\|_{\frac{5}{2}n+r,q} +\|f_2\|_{\frac{5}{2}n+r,q})\\ &\quad + c\int^t_0(1+(t-\tau))^{- \frac{\theta}{4}}(|u|^{\alpha}_{\infty} \|u\|_{\frac{5}{2}n+r+1})^{a}(|u|^{\alpha-1}_{\infty}|u|^{2}_{2} )^{1-a} d \tau \end{aligned} \end{equation} where $$ a=\frac{(\frac{5}{2}n+r)/n+(1-\theta)/2} {(\frac{5}{2}n+r+1)/n+1/2} =1-\frac{1/n+\theta/2} {(\frac{5}{2}n+r+1)/n+1/2}. $$ Set $$ K(t)= \sup_{0\leq \tau \leq t} \{(1+\tau)^{ \frac{\theta}{4}} \|u(\tau)\|_{r,p}+\|u(\tau)\|_{\frac{5}{2}n+r+1} \}. $$ Hence, since (by hypothesis above) $r>n/p$, then thanks to the Sobolev imbedding theorem ${\mathbb{L}}^{p}_{r}({\mathbb{R}}^{n})\subset{\mathbb{L}}^{\infty}({\mathbb{R}}^{n})$ and with \eqref{e4.9}, we get for $\alpha >1+ 4/\theta$, \begin{align*} %4.10 &\|u(x,t)\|_{r,p}\\ &\leq c(1+t)^{-\frac{\theta}{4}}(\|f_1\|_{\frac{5}{2}n+r,q} +\|f_2\|_{\frac{5}{2}n+r,q}) \\ &\quad + c\int^t_0(1+(t-\tau))^{- \frac{\theta}{4}}(\|u\|^{\alpha }_{r,p}\|u\|_{\frac{5}{2}n+r+1})^{a} (\|u\|^{\alpha-1}_{r,p} \|u\|^{2}_{\frac{5}{2}n+r+1} )^{1-a} d \tau. \\ &\leq c(1+t)^{- \frac{\theta}{4}}(\|f_1\|_{\frac{5}{2}n+r,q} +\|f_2\|_{\frac{5}{2}n+r,q}) \\ &\quad + c\int^t_0(1+(t-\tau))^{- \frac{\theta}{4}}({K(t)}^{\alpha +1}(1+\tau)^{- \alpha\frac{\theta}{4}})^{a} ({K(t)}^{\alpha +1}(1+\tau)^{- (\alpha-1)\frac{\theta}{4}} )^{1-a} d \tau. \\ &\leq c(1+t)^{- \frac{\theta}{4}}(\|f_1\|_{\frac{5}{2}n+r,q} +\|f_2\|_{\frac{5}{2}n+r,q})\\ &\quad + {K(t)}^{\alpha +1}\int^t_0(1+(t-\tau))^{- \frac{\theta}{4}}(1+\tau)^{- (\alpha-1)\frac{\theta}{4}} d\tau \\ &\leq c(1+t)^{- \frac{\theta}{4}}\{\|f_1\|_{\frac{5}{2}n+r,q} +\|f_2\|_{\frac{5}{2}n+r,q} + {K(t)}^{\alpha +1}\}. \end{align*} We deduce from the above inequality that for $\alpha >1+ 4/\theta$, \begin{equation} \label{e4.11} (1+t)^{ \frac{\theta}{4}}\|u(x,t)\|_{r,p} \leq c\{\|f_1\|_{\frac{5}{2}n+r,q} +\|f_2\|_{\frac{5}{2}n+r,q} + {K(t)}^{\alpha +1}\}. \end{equation} Furthermore, thanks to the inequalities \eqref{e2.6}, \eqref{e2.7} of Lemma \ref{lem2.4} and following the same lines as in the proof of the inequality \eqref{e4.5}, we find with \eqref{e4.1} and for $\alpha >1+ 4/\theta$ \begin{equation} \label{e4.12} \|u(x,t)\|_{\frac{5}{2}n+r+1} \leq c\{\|f_1\|_{\frac{5}{2}n+r+1} +\|f_2\|_{\frac{5}{2}n+r+1} + {K(t)}^{\alpha +1}\}. \end{equation} Then the combination of \eqref{e4.11} and \eqref{e4.12} leads to the inequality \begin{equation} \label{e4.13} K(t) \leq c\{\|f_1\|_{\frac{5}{2}n+r,q} +\|f_2\|_{\frac{5}{2}n+r,q} + \|f_1\|_{\frac{5}{2}n+r+1} +\|f_2\|_{\frac{5}{2}n+r+1} + {K(t)}^{\alpha +1}\}. \end{equation} Therefore, as above, we find that if \[ \|f_1\|_{\frac{5}{2}n+r,q} +\|f_2\|_{\frac{5}{2}n+r,q} + \|f_1\|_{\frac{5}{2}n+r+1} +\|f_2\|_{\frac{5}{2}n+r+1} \] is sufficiently small, then the inequality \eqref{e4.13} gives $K(t) \leq c$ for all $t \geq 0$. 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