\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{amssymb} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2018 (2018), No. 163, pp. 1--15.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2018 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2018/163\hfil Nonlinear Hartree equations] {Existence and regularity of global solutions nonlinear Hartree equations with Coulomb potentials and sublinear damping} \author[Y. Zhao, B. Feng \hfil EJDE-2018/163\hfilneg] {Yanjun Zhao, Binhua Feng} \address{Yanjun Zhao \newline College of Humanities and Sciences, Northeast Normal University, Changchun 130117, China} \email{940508086@qq.com} \address{Binhua Feng \newline Department of Mathematics, Northwest Normal University, Lanzhou 730070, China} \email{binhuaf@163.com} \dedicatory{Communicated by Jesus Ildefonso Diaz} \thanks{Submitted August 18, 2017. Published September 11, 2018.} \subjclass[2010]{35J60, 35Q55} \keywords{Nonlinear Hartree equation; Coulomb potential; sublinear damping} \begin{abstract} In this article, we consider the nonlinear Hartree equation with a sublinear damping and a time-dependent Coulomb potential in $\mathbb{R}^3$. We first prove the existence of a global solution and then obtain the $\Sigma^2$-regularity. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{definition}[theorem]{Definition} \allowdisplaybreaks \section{Introduction} Because of their important applications in physics, nonlinear Schr\"odinger equations with damping have been extensively studied; see \cite{acs,Anto,fzaa,fjmaa,mo,mjmp}. In this article, we consider the nonlinear Hartree equation with a time-dependent Coulomb potential and a sublinear damping, \begin{equation} \label{e} \begin{gathered} i\partial_tu+\Delta u=V(x)u+\frac{1}{|x-a(t)|}u +\lambda \Big( \frac{1}{|x|}\ast |u|^{2}\Big)u-ib \frac{u}{|u|^\alpha}, \\ (t,x)\in [0,\infty)\times\mathbb{R}^3,\\ u(0) = u_0 \in \Sigma, \end{gathered} \end{equation} where $u(t,x)$ is a complex-valued function in $(t,x)\in [0,\infty )\times \mathbb{R}^3$, $\lambda \in \mathbb{R}$, $b>0$, $0<\alpha \leqslant 1$, $a\in W^{1,1}((0,\infty),\mathbb{R}^3)$, $\Sigma$ denotes the energy space associated to the harmonic oscillator, i.e., \[ \Sigma:=\{u\in H^1(\mathbb{R}^3): xu\in L^2(\mathbb{R}^3)\}, \] equipped with the norm \[ \|u\|_{\Sigma}:=\|u\|_{H^1}+\|xu\|_{L^2}. \] The external potential $V$ is assumed to be harmonic, \begin{equation}\label{v} V(x)=\sum_{j=1}^3\omega_j^2x_j^2,\quad \omega_j>0. \end{equation} Equation \eqref{e} has many interesting applications in the quantum theory of large systems of non-relativistic bosonic atoms and molecules. In particular, this equation arises in the study of mean-field limit of many-body quantum systems, see, e.g., \cite{apt,miao,ss} and the references therein. An essential feature of equation \eqref{e} is that the convolution kernel $|x|^{-1}$ still retains the fine structure of micro two-body interactions of the quantum system. It is therefore of considerable interest to extend mathematical methods originally develop for nonlinear Schr\"odinger equations with local nonlinearities to the study of Hartree-type equation, see \cite{qianrwa,maoip,maoaml,shaoaml,f18cma,f18jmaa}. In this article, we are interested in the existence and $\Sigma^2$-regularity of global solutions to \eqref{e}. More precisely, we will prove that the solution $u(t)$ of \eqref{e} satisfies: $\|u(t)\|_{\Sigma}\leqslant C(\|u_0\|_{\Sigma})$ for all $t>0$. Moreover, if $u_0\in \Sigma^2$, then $u\in L^\infty((0,T);\Sigma^2)$ for any $T>0$, where \begin{equation*} \Sigma^2=\{u\in L^2: x^j\nabla^k u\in L^2,\forall \text{ multi-indices $j$ and $k$ with }|j|+|k|\leq 2\}, \end{equation*} equipped with the norm \[ \|u\|_{\Sigma^2}:=\sum_{|j|+|k|\leq 2}\|x^j\nabla^k u\|_{L^2}^2. \] To solve these problems, we mainly use the ideas from Carles et al.\ \cite{c1,c2}. Carles and Gallo \cite{c1} proved that the solution for the Schr\"odinger equation \begin{equation}\label{ae} i\partial_t u+\Delta u=-ib \frac{u}{|u|^\alpha}, \quad (t,x)\in [0,\infty)\times M, \end{equation} becomes zero in finite time, where $M$ is a compact manifold without boundary. Carles and Ozawa \cite{c2} extended this study to the equation \begin{equation}\label{aae} i\partial_t u+\Delta u=V(x)u+\lambda|u|^{2\sigma_1}u-ia|u|^{2\sigma_2}u -ib \frac{u}{|u|^\alpha}, \quad (t,x)\in [0,\infty)\times M, \end{equation} where $M$ is either a compact manifold without boundary, or the whole space in the presence of harmonic confinement $V(x)$, in space dimension one and two. However, compared with the equations \eqref{ae} and \eqref{aae} in \cite{c1,c2}, there exist some major difficulties in the analysis of the global existence and regularity of \eqref{e}. For example, due to the appearance of a time-dependent Coulomb potential $\frac{1}{|x-a(t)|}$, it is difficult to obtain the $\Sigma^2$-regularity of \eqref{e} by differentiating equation \eqref{e} two times with respect to space variable. Therefore, we use the idea due to Kato \cite{kato} (see also \cite{ca2003}), based on the general idea for Schr\"odinger equation, that two space derivatives cost the same as one time derivative. However, due to the same reason, we cannot immediately calculate the time derivative of equation \eqref{e}. To overcome this difficulty, we will use a change of variable $y=x-a(t)$ to avoid the time derivative of the time-dependent Coulomb potential. This leads to a more complicated equation \eqref{47}. For this reason, we only prove the solution $u\in L^\infty_{\rm loc}((0,\infty);\Sigma^2)$. If the initial data $u_0\neq 0$, and the corresponding solution $u\in L^\infty((0,\infty);\Sigma^2)$, then the solution of \eqref{e} becomes zero in finite time. Indeed, it follows from \eqref{gn2} that \[ \|u(t)\|_{L^2} \leqslant C\|u(t)\|_{L^{2-\alpha}}^{1-\frac{3\alpha}{8-\alpha}} \|u(t)\|_{H^2}^{\frac{3\alpha}{8-\alpha}} \leqslant C\|u\|_{L^\infty((0,\infty);\Sigma^2)}^{\frac{3\alpha}{8-\alpha}} \|u(t)\|_{L^{2-\alpha}}^{1-\frac{3\alpha}{8-\alpha}}, \] for $t\geqslant 0$. This and \eqref{2} imply that \begin{align*} \frac{d}{dt}\int_{\mathbb{R}^3}|u(t,x)|^2dx =&-2b \int_{\mathbb{R}^3} |u(t,x)|^{2-\alpha}dx \\ \leqslant &-\frac{2b}{C\|u\|_{L^\infty((0,\infty);\Sigma^2)}^{3\alpha/4}} \Big(\int_{\mathbb{R}^3}|u(t,x)|^2dx\Big)^{1-(\alpha/8)}. \end{align*} This differential inequality can be solved explicitly: \[ \|u(t)\|_{L^2}^2\leqslant \Big(\|u(0)\|_{L^2}^{\alpha/4} -\frac{b\alpha}{4C\|u\|_{L^\infty((0,\infty);\Sigma^2)}^{3\alpha/4}}t \Big)^{8/\alpha}. \] This implies that $\|u(t)\|_{L^2}$ vanishes in finite time $T$, with \[ T\leqslant \frac{4C\|u_0\|_{L^2}^{\alpha/4} \|u\|_{L^\infty((0,\infty);\Sigma^2)}^{3\alpha/4}}{b\alpha}. \] Therefore, if $u\in L^\infty((0,\infty);\Sigma^2)$, the solution of \eqref{e} becomes zero in finite time. Before stating our main results, we give the notion of weak solution of \eqref{e}. \begin{definition} \label{def1.1} \rm Assume $0< \alpha <1$ and $a\in W^{1,1}((0,\infty),\mathbb{R}^3)$. A global weak solution to \eqref{e} is a function $u \in C([0,\infty);L^2(\mathbb{R}^3))\cap L^{\infty}((0,\infty);\Sigma)$ solving \eqref{e} in $\mathcal{D}'((0,\infty)\times \mathbb{R}^3)$. \end{definition} \begin{definition} \label{def1.2} \rm Assume $\alpha =1$ and $a\in W^{1,1}((0,\infty),\mathbb{R}^3)$. A global weak solution to \eqref{e} is a function $u \in C([0,\infty);L^2(\mathbb{R}^3))\cap L^{\infty}((0,\infty);\Sigma)$ solving \[ i\partial_t u+\Delta u=V(x)u+\frac{1}{|x-a(t)|}u+\lambda \Big( \frac{1}{|x|}\ast |u|^{2}\Big)u-ibF \] in $\mathcal{D}'((0,\infty)\times \mathbb{R}^3)$, where $F$ is such that \begin{equation*} \|F\|_{L^\infty((0,\infty)\times \mathbb{R}^3)}\leqslant 1, \quad\text{and}\quad F=\frac{u}{|u|}~if~u\neq 0. \end{equation*} \end{definition} Since $a\in W^{1,1}((0,\infty),\mathbb{R}^3)\hookrightarrow L^\infty((0,\infty), \mathbb{R}^3)$, it follows that $\frac{1}{|x-a(t)|}\in L^2_{\rm loc}(\mathbb{R}^3)$. This and $u \in C([0,\infty);L^2(\mathbb{R}^3))\cap L^{\infty}((0,\infty);\Sigma)$ imply that $\frac{1}{|x-a(t)|}u\in L^1_{\rm loc}((0,\infty)\times\mathbb{R}^3)$. In addition, from Hardy's inequality, $( \frac{1}{|x|}\ast |u|^{2})\in L^{\infty}((0,\infty);L^{\infty}(\mathbb{R}^3))$. This implies that $( \frac{1}{|x|}\ast |u|^{2})u\in L^1_{\rm loc} ((0,\infty)\times\mathbb{R}^3 )$. Our main results read as follows. \begin{theorem}\label{thm1} Let $0<\alpha\leqslant 4/5$, $a\in W^{1,1}((0,\infty),\mathbb{R}^3)$ and $u_0\in \Sigma$. Then the Cauchy problem \eqref{e} has a unique, global weak solution. In addition, \begin{gather}\label{2} \frac{d}{dt}\int_{\mathbb{R}^3}|u(t,x)|^2dx =-2b \int_{\mathbb{R}^3} |u(t,x)|^{2-\alpha}dx, \\ \label{100} \|u(t)\|_{\Sigma} \leq C(\|u_0\|_{\Sigma}) \quad\forall t>0. \end{gather} \end{theorem} \begin{theorem}\label{thm2} Let $u_0 \in \Sigma^2$, $a\in W^{2,\infty}((0,\infty),\mathbb{R}^3)$, $b >0$ and $0<\alpha \leqslant\frac{1}{2}$. For every $0 0$ will stand for a constant that may be different from line to line when it does not cause any confusion. Since we exclusively deal with $\mathbb{R}^3$, we often abbreviate $L^q(\mathbb{R}^3)$, $\|\cdot\|_{L^q(\mathbb{R}^3)}$ and $H^s(\mathbb{R}^3)$ by $L^q$, $\|\cdot\|_{L^q}$ and $H^s$, respectively. We recall that a pair of exponents $(q,r)$ is Schr\"odinger-admissible if $\frac{2}{q}=3(\frac{1}{2}-\frac{1}{r})$ and $2 \leq r \leq 6$. Then, for any space-time slab $I\times \mathbb{R}^3$, we can define the Strichartz norm \begin{equation*} \|u\|_{S(I)}=\sup_{(q,r)}\|u\|_{L_t^qL_x^{r}(I)}, \end{equation*} where the supremum is taken over all admissible pairs of exponents $(q,r)$. Furthermore, $\|f\|_{S_\Sigma(I)}:=\|f\|_{S(I)}+\|\nabla f\|_{S(I)}+\|x f\|_{S(I)}$. \section{Preliminaries} In this section, we recall some useful results. Firstly, we recall the following Gagliardo-Nirenberg inequality (see \cite{ca2003}). \begin{lemma} \label{lem2.1} There exists a constant $C$ such that \begin{equation}\label{gn2} \|f\|_{L^{2}}\leqslant C\|f\|_{L^{2-\alpha}}^{1-\theta}\|f\|_{H^2}^{\theta},\quad \forall f\in L^{2-\alpha}\cap H^2, \end{equation} where $\theta=3\alpha/(8-\alpha)$ with $\alpha\in (0,1]$. \end{lemma} The following inequalities can be viewed as dual to the Gagliardo-Nirenberg inequalities. \begin{lemma} \label{lem2.2} There exist some constants $C$ such that \begin{equation}\label{gn4} \|f\|_{L^{p'}}\leqslant C\|f\|_{L^2}^{1-\delta_1(p)}\|xf\|_{L^2}^{\delta_1 (p)}, \quad \forall f\in \Sigma, \end{equation} where $\delta_1 (p)=3\big(\frac{1}{2}-\frac{1}{p}\big)$ with $p\in [2,6)$. \begin{equation}\label{gn5} \|f\|_{L^{p'}}\leqslant C\|f\|_{L^2}^{1-\delta_2(p)}\|x^2f\|_{L^2}^{\delta_2 (p)}, \quad \forall f\in \Sigma^2, \end{equation} where $\delta_2 (p)=\frac{3}{2}\big(\frac{1}{2}-\frac{1}{p}\big)$ with $p\geqslant 2$. \end{lemma} \begin{proof} The proof of \eqref{gn4} is similar to that of \cite[Lemma 6.2]{c3}. We give the proof for the reader's convenience. Let $\lambda>0$, and write \[ \int_{\mathbb{R}^3}|f(x)|^{p'}dx=\int_{|x|\leqslant \lambda}|f(x)|^{p'}dx +\int_{|x|> \lambda}|f(x)|^{p'}dx. \] Estimate the first term by H\"older's inequality \[ \int_{|x|\leqslant \lambda}|f(x)|^{p'}dx \leqslant C\lambda^{3/r'} \Big(\int_{|x|\leqslant \lambda}|f(x)|^{p'r}dx\Big)^{1/r}, \] and choose $r=2/p'$. Estimate the second term by the same H\"older's inequality, after inserting the factor $x$ as follows, \begin{align*} \int_{|x|>\lambda}|f(x)|^{p'}dx =&\int_{|x|> \lambda}|x|^{-p'}|x|^{p'}|f(x)|^{p'}dx\\ \leqslant & \Big(\int_{|x|> \lambda}|x|^{-p'r'}dx\Big)^{1/r'} \Big(\int_{|x|> \lambda}|xf(x)|^{2}dx\Big)^{1/r} \\ \leqslant & C\lambda^{\frac{3}{r'}-p'}\|xf\|_{L^2}^{2/r}. \end{align*} In summary, we have the following estimate, for any $\lambda>0$, \begin{equation}\label{0000} \|f\|_{L^{p'}}\leqslant C\lambda^{\frac{3}{r'p'}}\|f\|_{L^2} +C\lambda^{\frac{3}{r'p'}-1}\|xf\|_{L^2}. \end{equation} Notice that $\frac{3}{r'p'}=\delta (p)$, taking $\lambda=\frac{\|xf\|_{L^2}}{\|f\|_{L^2}}$, this yields \eqref{gn4}. By a similar argument, we can obtain \eqref{gn5}. \end{proof} The following lemma is vital for proving the uniqueness for \eqref{e}; it was first proved in \cite{diaz}. \begin{lemma} \label{lem2.3} Let $\sigma\geqslant -1$. For all $z_1,z_2\in \mathbb{C}$, \[ \operatorname{Re}((|z_1|^\sigma z_1-|z_2|^\sigma z_2)(\overline{z_1-z_2})) \geqslant 0. \] \end{lemma} \section{Proof of Theorem \ref{thm1}} By a similar idea as that in \cite{c1,c2}, we modify \eqref{e} by regularizing the singular potential and sublinear nonlinearity: \begin{equation} \label{be} \begin{gathered} \begin{aligned} i\partial_t u^\delta +\Delta u^\delta &=V(x)u^\delta+\frac{u^\delta}{(|x-a(t)|^2+\delta)^{1 /2}} \\ &\quad +\lambda \left(\frac{1}{|x|}\ast |u^\delta|^{2}\right)u^\delta -ib \frac{u^\delta}{(|u^\delta|^2+\delta)^{\alpha /2}}, \end{aligned}\\ u^\delta(0) = u_0. \end{gathered} \end{equation} By the energy estimates, we can infer the following global existence for \eqref{be}. \begin{lemma} \label{lem3.1} Let $\delta>0$ and $a\in W^{1,1}((0,\infty),\mathbb{R}^3)$. For every $u_0\in \Sigma$, there exists a unique global solution $u^\delta\in C([0,\infty);\Sigma)$ to \eqref{be}. In addition, for every $t>0$, we have \begin{gather}\label{31} \int_{\mathbb{R}^3}|u^\delta(t,x)|^2dx +2b \int_0^t\int_{\mathbb{R}^3}\frac{|u^\delta(s,x)|^2}{(|u^\delta(s,x)|^2 +\delta)^{\alpha /2}}dxds=\|u_0\|_{L^2}^2, \\ \label{31'} \|u^\delta(t)\|_{\Sigma} \leq C(\|u_0\|_{\Sigma}). \end{gather} \end{lemma} \begin{proof} Since the external potential $V(x)$ is quadratic, local in time Strichartz's estimates hold for the Hamiltonian $-\Delta+V(x)$. The local existence can be proved by applying a fixed-point argument in space of the type $S_\Sigma(0,T)$, which can be found for instance in \cite{ca2003} in the case of $V=0$ (see \cite{c} in the presence of a potential, see \cite{an,fna2017} for a time-dependent Coulomb potential $\frac{1}{|x-a(t)|}$). So we omit it. On the other hand, multiplying \eqref{be} by $\overline{u^\delta}$, integrating on $\mathbb{R}^3$ and taking the imaginary part, we can obtain \eqref{31}. To show \eqref{31'}, we first assume that $u(t)$ is sufficiently regular and decaying so that all of the following formal manipulations can be carried out. Once the final result is established, a standard density argument allows to conclude that it also holds for $u\in C([0, T],\Sigma)$. We deduce from \eqref{be} that \begin{equation}\label{32} \begin{aligned} &\frac{d}{dt}\|\nabla u^\delta (t)\|_{L^2}^2 \\ &=-2\operatorname{Re}\int_{\mathbb{R}^3} \Delta u^\delta (t,x)\partial_t\overline{u^\delta}(t,x)dx\\ &=-2\operatorname{Re}\int_{\mathbb{R}^3} \Big( V(x)u^\delta (t,x) +\frac{u^\delta(t,x)}{(|x-a(t)|^2+\delta)^{1 /2}}\Big) \partial_t\overline{u^\delta}(t,x)dx\\ &\quad -2\operatorname{Re}\int_{\mathbb{R}^3} \Big( \lambda \Big(\frac{1}{|x|}\ast |u^\delta(t)|^{2}\Big)u^\delta(t,x) -ibf_\delta(u^\delta)(t,x)\Big)\partial_t\overline{u^\delta}(t,x)dx\\ &=-\frac{d}{dt}\int_{\mathbb{R}^3} V(x)|u^\delta(t,x)|^2dx -\frac{d}{dt}\int_{\mathbb{R}^3}\frac{|u^\delta(t,x)|^2}{(|x-a(t)|^2 +\delta)^{1 /2}}dx \\ &\quad -\frac{\lambda}{2}\frac{d}{dt}\int_{\mathbb{R}^3} \Big(\frac{1}{|x|}\ast |u^\delta(t)|^{2}\Big)|u^\delta(t,x)|^2dx \\ &\quad -2b\operatorname{Im} \int_{\mathbb{R}^3} f_\delta(u^\delta)(t,x) \partial_t\overline{u^\delta}(t,x)dx\\ &\quad +\int_{\mathbb{R}^3} \partial_t\Big(\frac{1}{(|x-a(t)|^2+\delta)^{1 /2}}\Big) |u^\delta(t,x)|^2dx, \end{aligned} \end{equation} where \[ f_\delta(v)=\frac{v}{(|v|^2+\delta)^{\alpha /2}}. \] For the last term in \eqref{32}, we deduce from Hardy's inequality that \begin{equation} \label{33} \begin{aligned} \int_{\mathbb{R}^3} \partial_t\Big(\frac{1}{(|x-a(t)|^2+\delta)^{1 /2}}\Big) |u^\delta(t,x)|^2dx &\leqslant |\frac{da}{dt}(t)|\int_{\mathbb{R}^3} \frac{|u^\delta(t,x)|^2}{|x-a(t)|^2}dx\\ &\leqslant 4|\frac{da}{dt}(t)|\int_{\mathbb{R}^3} |\nabla u^\delta(t,x)|^2dx. \end{aligned} \end{equation} In addition, using \eqref{be} again, we have \begin{equation} \label{34} \begin{aligned} &-2b\operatorname{Im} \int_{\mathbb{R}^3} f_\delta(u^\delta)(t,x) \partial_t\overline{u^\delta}(t,x)dx\\ &=2b\operatorname{Re}\int_{\mathbb{R}^3} f_\delta(u^\delta)(t,x)\Delta \overline{u^\delta} (t,x)dx-2b\int_{\mathbb{R}^3} \frac{V(x)|u^\delta(t,x)|^2}{(|u^\delta(t,x)|^2+\delta)^{\alpha/2}}dx\\ &\quad -2b\int_{\mathbb{R}^3} \frac{1}{(|x-a(t)|^2+\delta)^{1 /2}} \frac{|u^\delta(t,x)|^2}{(|u^\delta(t,x)|^2+\delta)^{\alpha/2}}dx\\ &\quad -2b\lambda\int_{\mathbb{R}^3} \big(\frac{1}{|x|}\ast |u^\delta(t)|^{2}\big) \frac{|u^\delta(t,x)|^2}{(|u^\delta(t,x)|^2+\delta)^{\alpha/2}}dx, \end{aligned} \end{equation} where \begin{align} &2b\operatorname{Re}\int_{\mathbb{R}^3} f_\delta(u^\delta)(t,x) \Delta \overline{u^\delta}(t,x) dx \nonumber\\ &= -2b\operatorname{Re}\int_{\mathbb{R}^3} \frac{|\nabla u^\delta(t,x)|^2}{(|u^\delta(t,x)|^2 +\delta)^{\alpha/2}}dx+2b\alpha\int_{\mathbb{R}^3} \frac{|\operatorname{Re}(\overline{u^\delta}(t,x) \nabla u^\delta(t,x))|^2}{(|u^\delta(t,x)|^2+\delta)^{\alpha/2+1}}dx \label{35}\\ &=-2b\int_{\mathbb{R}^3}\frac{\delta|\nabla u^\delta(t,x)|^2 +(1-\alpha)|\operatorname{Re}(\overline{u^\delta}\nabla u^\delta)(t,x)|^2 +|\operatorname{Im}(\overline{u^\delta}\nabla u^\delta)(t,x)|^2}{(|u^\delta(t,x)|^2 +\delta)^{\alpha /2+1}}dx. \nonumber \end{align} Now, we define \begin{equation*} E^\delta(t)=\int_{\mathbb{R}^3} |\nabla u^\delta(t,x)|^2 dx +\int_{\mathbb{R}^3} V(x) |u^\delta(t,x)|^2 dx. \end{equation*} Collecting \eqref{32}-\eqref{35}, we derive \begin{align} &\frac{d}{dt}E^\delta(t) \nonumber \\ &= -\frac{d}{dt}\Big(\int_{\mathbb{R}^3} \frac{|u^\delta(t,x)|^2}{(|x-a(t)|^2+\delta)^{1 /2}}dx +\frac{\lambda}{2}\int_{\mathbb{R}^3}\Big(\frac{1}{|x|} \ast |u^\delta(t)|^{2}\Big)|u^\delta(t,x)|^2dx\Big)\nonumber \\ &\quad +\int_{\mathbb{R}^3} \partial_t\Big(\frac{1}{(|x-a(t)|^2+\delta)^{1 /2}}\Big) |u^\delta(t,x)|^2dx -2b\int_{\mathbb{R}^3} \frac{V(x)|u^\delta(t,x)|^2} {(|u^\delta(t,x)|^2+\delta)^{\alpha/2}}dx \nonumber \\ &\quad -2b\int_{\mathbb{R}^3}\frac{\delta|\nabla u^\delta(t,x)|^2 +(1-\alpha)|\operatorname{Re}(\overline{u^\delta}\nabla u^\delta)(t,x)|^2 +|\operatorname{Im}(\overline{u^\delta}\nabla u^\delta)(t,x)|^2} {(|u^\delta(t,x)|^2+\delta)^{\alpha /2+1}}dx \nonumber\\ &\quad -2b\int_{\mathbb{R}^3} \frac{1}{(|x-a(t)|^2+\delta)^{1 /2}} \frac{|u^\delta(t,x)|^2}{(|u^\delta(t,x)|^2+\delta)^{\alpha/2}}dx \label{37} \\ &\quad -2b\lambda\int_{\mathbb{R}^3} \Big(\frac{1}{|x|}\ast |u^\delta(t)|^{2}\Big) \frac{|u^\delta(t,x)|^2}{(|u^\delta(t,x)|^2+\delta)^{\alpha/2}}dx \nonumber \\ &\leqslant -\frac{d}{dt}\Big(\int_{\mathbb{R}^3}\frac{|u^\delta(t,x)|^2}{(|x-a(t)|^2 +\delta)^{1 /2}}dx+\frac{\lambda}{2} \int_{\mathbb{R}^3}\big(\frac{1}{|x|}\ast |u^\delta(t)|^{2}\big) |u^\delta(t,x)|^2dx\Big)\nonumber \\ &\quad +4|\frac{da}{dt}(t)|\int_{\mathbb{R}^3} |\nabla u^\delta(t,x)|^2dx +C\|\nabla u^\delta(t)\|_{L^2}\int_{\mathbb{R}^3} \frac{|u^\delta(t,x)|^2}{(|u^\delta(t,x)|^2+\delta)^{\alpha/2}}dx. \nonumber \end{align} Integrating this inequality on $(0,t)$ with respect to time $t$, it follows that \begin{align} &E^\delta(t) \nonumber \\ &\leqslant E^\delta(0)-\Big(\int_{\mathbb{R}^3} \frac{|u^\delta(t,x)|^2}{(|x-a(t)|^2 +\delta)^{1 /2}}dx+\frac{\lambda}{2}\int_{\mathbb{R}^3} \Big(\frac{1}{|x|}\ast |u^\delta(t)|^{2}\Big)|u^\delta(t,x)|^2dx\Big) \nonumber\\ &\quad +\Big(\int_{\mathbb{R}^3} \frac{|u_0(x)|^2}{(|x-a(0)|^2+\delta)^{1 /2}}dx +\frac{\lambda}{2}\int_{\mathbb{R}^3} \Big(\frac{1}{|x|}\ast |u_0|^{2}\Big)|u_0(x)|^2dx\Big) \nonumber\\ &\quad +4\int_0^t|\frac{da}{ds}(s)|\|\nabla u^\delta(s)\|_{L^2}^2ds +C\int_0^t\|\nabla u^\delta(s)\|_{L^2}g(s)ds \label{38}\\ &\leqslant C(\|u_0\|_{\Sigma})+\frac{|\lambda|}{2}\int_{\mathbb{R}^3} \Big(\frac{1}{|x|}\ast |u^\delta(t)|^{2}\Big)|u^\delta(t,x)|^2dx \nonumber\\ &\quad+4\int_0^t|\frac{da}{ds}(s)|\|\nabla u^\delta(s)\|_{L^2}^2ds +C\int_0^tg(s)\|\nabla u^\delta(s)\|_{L^2}^2ds+C\int_0^tg(s)\,ds, \nonumber \end{align} where \[ g(t)=\int_{\mathbb{R}^3} \frac{|u^\delta (t,x)|^2}{(|u^\delta (t,x)|^2+\delta)^{\alpha/2}}dx. \] Using Hardy's and Young's inequalities, for all $t>0$, we have \begin{equation} \label{312} \begin{aligned} \int_{\mathbb{R}^3}\Big(\frac{1}{|x|}\ast |u^\delta(t)|^{2}\Big)|u^\delta(t,x)|^2dx &\leqslant\| \frac{1}{|x|}\ast |u^\delta(t)|^{2}\|_{L^\infty} \int_{\mathbb{R}^3}|u^\delta(t,x)|^2dx\\ &\leqslant \varepsilon \|\nabla u^\delta(t)\|_{L^2}^2+C_\varepsilon\| u_0\|_{L^2}^2. \end{aligned} \end{equation} This and \eqref{38} imply \begin{equation} \label{38x} \begin{aligned} \|\nabla u^\delta(t)\|_{L^2}^2 &\leqslant C(\|u_0\|_{\Sigma})+4\int_0^t|\frac{da}{ds}(s)| \|\nabla u^\delta(s)\|_{L^2}^2ds \\ &\quad +C\int_0^tg(s)\|\nabla u^\delta(s)\|_{L^2}^2ds. \end{aligned} \end{equation} In view of \eqref{31} and $a\in W^{1,1}((0,\infty),\mathbb{R}^3)$, from Gronwall's inequality we deduce that \begin{equation*} \|\nabla u^\delta(t)\|_{L^2}^2\leqslant C(\|u_0\|_{\Sigma})\quad\forall t>0, \end{equation*} which, together with \eqref{38} imply \begin{equation} \label{313} E^\delta(t)\leqslant C(\|u_0\|_{\Sigma})\quad\forall t>0. \end{equation} This yields \begin{equation} \|u^\delta(t)\|_{\Sigma} \leqslant C(\|u_0\|_{\Sigma})\quad\forall t>0, \end{equation} which implies that the solution of \eqref{be} exists globally. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1}] Firstly, it follows from \eqref{31} and \eqref{31'} that $(u^\delta)_{0<\delta\leqslant 1}$ is uniformly bounded in $L^\infty([0,\infty);\Sigma)\cap L^{2-\alpha}([0,\infty);L^{2-\alpha})$. Therefore, there exist $u\in L^\infty([0,\infty);\Sigma)$ and a subsequence $(u^{\delta_n})$ such that \begin{equation}\label{320} u^{\delta_n}\rightharpoonup u\quad\text{in } w*~L^\infty([0,\infty);\Sigma),\text{ as }n\to\infty; \end{equation} this and Lemma \ref{lem3.1}, imply \[ \|u\|_{L^\infty([0,\infty);H^1)}\leqslant C(\|u_0\|_{\Sigma}). \] In addition, we infer from Hardy's and H\"older's inequalities that \begin{equation} \label{322} \begin{aligned} &\big\|\frac{1}{|x|}\ast |u|^{2}u-\frac{1}{|x|}\ast |v|^{2}v\big\|_{L^2}\\ &\leqslant \big\|\frac{1}{|x|}\ast |u|^{2}\big\|_{L^\infty}\|u-v\|_{L^2} +\big\|\frac{1}{|x|}\ast |u|^{2}-\frac{1}{|x|}\ast |v|^{2} \big\|_{L^\infty}\|v\|_{L^2}\\ &\leqslant C\|u\|_{L^2}\|\nabla u\|_{L^2}\|u-v\|_{L^2}\\ &\quad +C\sup_{x\in \mathbb{R}^3}\Big(\int_{\mathbb{R}^3}\frac{(|u(y)| +|v(y)|)^2}{|x-y|^2}dy\Big)^{1/2}\|v\|_{L^2}\|u-v\|_{L^2}\\ &\leqslant C\|u\|_{H^1}^2\|u-v\|_{L^2} +C(\|\nabla u\|_{L^2} +\|\nabla v\|_{L^2})\|v\|_{L^2}\|u-v\|_{L^2}\\ &\leqslant C(\|u\|_{H^1}^2+\|v\|_{H^1}^2)\|u-v\|_{L^2}. \end{aligned} \end{equation} Thus, from \eqref{320} and the compact embedding $\Sigma\hookrightarrow L^2$ it follows that \begin{equation}\label{323} \Big(\frac{1}{|x|}\ast |u^{\delta_n}(t)|^{2}\Big)u^{\delta_n}(t) \to \Big(\frac{1}{|x|}\ast |u(t)|^{2}\Big)u(t) \end{equation} in $L^2$ for a.e. $t\in[0,T]$, for every $00$. It follows from \eqref{31} that for any $t\in [0,\infty)$, \begin{align*} \frac{d}{dt}\|u^{\delta}(t)-u^{\delta}(t')\|^2_{L^2} &\leqslant \frac{d}{dt}(-2\operatorname{Re}\langle u^{\delta}(t), u^{\delta}(t')\rangle_{L^2})\\ &= 2\operatorname{Re}\langle -i\Delta u^\delta+ iV(x)u^\delta +i\frac{u^\delta}{(|x-a(t)|^2+\delta)^{1 /2}}\\ &\quad +i\lambda \big(\frac{1}{|x|}\ast |u^\delta|^{2}\big)u^\delta +b \frac{u^\delta}{(|u^\delta|^2+\delta)^{\alpha /2}},u^{\delta}\rangle_{L^2}. \end{align*} Integrating this inequality with respect to time, applying Hardy's inequality and \eqref{gn4}, we obtain \begin{align*} \|u^{\delta}(t)-u^{\delta}(t')\|^2_{L^2} \leqslant & C|t-t'|(\|\Delta u^\delta\|_{L^\infty([0,\infty);H^{-1})} \| u^\delta\|_{L^\infty([0,\infty);H^1)}\\ &+\| xu^\delta\|_{L^\infty([0,\infty);L^2)} +\| \frac{|u^\delta|^2}{|x-a(t)|}\|_{L^\infty([0,\infty);L^1)}\\ &+\| \Big(\frac{1}{|x|}\ast |u^\delta|^{2}\Big) |u^\delta|^{2}\|_{L^\infty([0,\infty);L^1)} +\| u^\delta\|^{2-\alpha}_{L^\infty([0,\infty);L^{2-\alpha})})\\ \leqslant & C|t-t'|(\|\Delta u^\delta\|_{L^\infty([0,\infty);H^{-1})} \| u^\delta\|_{L^\infty([0,\infty);H^1)}\\ &+\| xu^\delta\|_{L^\infty([0,\infty);L^2)} +\| \nabla u^\delta\|_{L^\infty([0,\infty);L^2)} \| u^\delta\|_{L^\infty([0,\infty);L^2)}\\ &+\| \nabla u^\delta\|_{L^\infty([0,\infty);L^2)}\| u^\delta\|_{L^\infty([0,\infty); L^2)}^3 \\ &+\| u^\delta\|^{\frac{4-5\alpha}{2}}_{L^\infty([0,\infty);L^2)} \| x u^\delta\|^{\frac{3\alpha}{2}}_{L^\infty([0,\infty);L^2)}), \end{align*} which, together with \eqref{31'} imply \begin{equation*} \|u^{\delta}(t)-u^{\delta}(t')\|_{L^2}\leqslant C|t-t'|^{1/2}. \end{equation*} Therefore, for any $T>0$, $(u^\delta)_{0<\delta\leqslant 1}$ is a bounded sequence in $C([0,T];\Sigma)$ and is uniformly equicontinuous from $[0,T]$ to $L^2$. In addition, since the embedding $\Sigma \hookrightarrow L^2$ is compact, the set $\{u^\delta(t)| \delta\in (0,1]\}$ is relatively compact in $L^2$. Applying Arzel\`{a}-Ascoli Theorem, it follows that $(u^{\delta_n})$ is relatively compact in $C([0,T];L^2)$. Thus, we deduce from \eqref{320} that \[ u^{\delta_n}\to u\quad\text{in } C([0,T];L^2). \] This yields that $u\in C([0,T];L^2)$ as well as $u(0)=u^\delta(0)=u_0$. Note that this holds for any $T>0$, therefore, $u\in C([0,\infty);L^2)$. Finally, up to subsequence, $u^{\delta_n}(t,x)\to u(t,x)$ in $(t,x)\in [0,\infty)\times \mathbb{R}^3$. Hence, for a.e. $(t,x)\in [0,\infty)\times \mathbb{R}^3$ such that $u(t,x)\neq 0$, it follows \[ \frac{u^{\delta_n}}{(|u^{\delta_n}|^2+\delta_n)^{\alpha/2}}(t,x) \to \frac{u}{|u|^\alpha}(t,x),\quad\text{as }n\to\infty. \] This and \eqref{315} imply that $F(t,x)=\frac{u}{|u|^\alpha}(t,x)$. This completes the proof of existence. Next we show uniqueness. Let $u$ and $v$ be two solutions to \eqref{e} with the same initial data $u_0$. We set $w=u-v$ and it satisfies \begin{equation} \label{u1} \begin{gathered} \begin{aligned} i\partial_tw+\Delta w &=V(x)w+\frac{1}{|x-a(t)|}w+\lambda \Big(\frac{1}{|x|}\ast |u|^{2}u -\frac{1}{|x|}\ast |v|^{2}v\Big) \\ &\quad -ib \Big(\frac{u}{|u|^\alpha}-\frac{v}{|v|^\alpha}\Big), \end{aligned}\\ w(0,x) = 0. \end{gathered} \end{equation} Multiplying this equation by $\overline{w}$, integrating over $\mathbb{R}^3$ and taking the imaginary part, by \eqref{322} and Lemma \ref{lem4.2} we obtain \begin{align*} \frac{1}{2}\frac{d}{dt}\int_{\mathbb{R}^3} |w(t,x)|^2dx &=\operatorname{Im}\int_{\mathbb{R}^3}\Big(\frac{1}{|x|}\ast |u|^{2}u-\frac{1}{|x|} \ast |v|^{2}v\Big)\overline{w}dx \\ &\quad -b\operatorname{Re}\int_{\mathbb{R}^3} \Big(\frac{u}{|u|^\alpha}-\frac{v}{|v|^\alpha}\Big)\overline{w}dx \\ &\leqslant \|\frac{1}{|x|}\ast |u|^{2}u-\frac{1}{|x|}\ast |v|^{2}v\|_{L^2}\|w\|_{L^2}\\ &\leqslant C\|w(t)\|_{L^2}^2. \end{align*} This and Gronwall's inequality impliy $w(t)=0$ for all $t\in [0,\infty)$. Therefore, there exists a unique global weak solution of \eqref{e}. \end{proof} \section{Proof of Theorem \ref{thm2}} Firstly, we have the following estimates. \begin{lemma} \label{lem4.1} Assume $0<\alpha\leqslant 1/2$. There exists $C>0$ such that the solution $u^\delta$ of \eqref{be} satisfies \begin{equation}\label{40} \|u^\delta(t)\|_{H^2}+\|x^2 u^\delta(t)\|_{L^2} \leqslant C\|\partial_t u^\delta(t)\|_{L^2}+C,\quad\text{for all }t>0, \end{equation} where $C$ depends only on $u_0$. \end{lemma} \begin{proof} Since $u^\delta$ is the solution of \eqref{be}, it follows that for all $t>0$ \begin{equation} \label{41} \begin{aligned} &\|u^\delta(t)\|_{H^2}+\|x^2 u^\delta(t)\|_{L^2}\\ &\leqslant C\|\Delta u^\delta(t)\|_{L^2}+C\| u^\delta(t)\|_{L^2} +C\|Vu^\delta(t)\|_{L^2}\\ &\leqslant C\|\partial_tu^\delta(t)\|_{L^2}+C\| u^\delta(t)\|_{L^2} +C\| \frac{u^\delta(t)}{|x-a(t)|}\|_{L^2}\\ &\quad +C\|\Big(\frac{1}{|x|}\ast |u^\delta(t)|^{2}\Big)u^\delta(t)\|_{L^2} +C\| |u^\delta(t)|^{1-\alpha}\|_{L^2}. \end{aligned} \end{equation} Applying Hardy's and Young's inequalities, we obtain that for any $\varepsilon>0$, there exists $C_\varepsilon>0$ such that \begin{equation}\label{42} \| \frac{u^\delta(t)}{|x-a(t)|}\|_{L^2}\leqslant 2\|\nabla u^\delta(t)\|_{L^2} \\ \leqslant \varepsilon\|u^\delta(t)\|_{H^2}+C_\varepsilon\| u^\delta(t)\|_{L^2}, \end{equation} and \begin{equation}\label{43} \|\frac{1}{|x|}\ast |u^\delta|^{2}u^\delta(t)\|_{L^2} \leqslant C\|\nabla u^\delta(t)\|_{L^2}\| u^\delta(t)\|_{L^2}^2 \leqslant \varepsilon\|u^\delta(t)\|_{H^2}+C_\varepsilon\| u^\delta(t)\|_{L^2}. \end{equation} When $\alpha\leqslant 1/2$, we deduce from \eqref{gn5} and Young's inequality that \begin{equation} \label{44} \begin{aligned} \| |u^\delta(t)|^{1-\alpha}\|_{L^2} &=\| u^\delta(t)\|_{L^{2-2\alpha}}^{1-\alpha}\\ & \leqslant C\|x^2 u^\delta(t)\|_{L^2}^{3\alpha/4} \| u^\delta(t)\|_{L^2}^{\frac{4-7\alpha}{4}}\\ & \leqslant \varepsilon\|x^2u^\delta(t)\|_{L^2}+C_\varepsilon\| u^\delta(t)\|_{L^2}. \end{aligned} \end{equation} Taking $\varepsilon =\frac{1}{6}$ in \eqref{42}-\eqref{44}, \eqref{40} follows from \eqref{41}-\eqref{44}. \end{proof} \begin{lemma} \label{lem4.2} Let $u_0 \in \Sigma^2$, $a\in W^{2,\infty}((0,\infty),\mathbb{R}^3)$, $b >0$ and $0<\alpha \leqslant\frac{1}{2}$. Then, for every $T<\infty$, there exists $C>0$ such that the solution $u^\delta$ of \eqref{be} satisfies \begin{equation*} \|\partial_tu^\delta(t)\|_{L^2}\leqslant C(\|a\|_{W^{2,\infty}((0,\infty),\mathbb{R}^3)},T,\|u_0\|_{\Sigma^2}), \quad\text{for all }t\in [0,T]. \end{equation*} \end{lemma} \begin{proof} We make the change of variables $y=x-a(t)$ and set $ u^\delta(t,x)= v^\delta(t,y)$. Then, \begin{equation}\label{45} \partial_tv^\delta(t,y)=\partial_tu^\delta(t,x) +\frac{da}{dt}(t)\cdot \nabla u^\delta(t,x), \end{equation} and $\nabla u^\delta(t,x)=\nabla v^\delta(t,y)$. Therefore, $v^\delta$ satisfies the equation \begin{equation}\label{46} \begin{gathered} \begin{aligned} i\partial_tv^\delta+\Delta v^\delta &=V(y+a(t))v^\delta+\frac{v^\delta}{(|y|^2+\delta)^{1 /2}} +\lambda \frac{1}{|x|}\ast |v^\delta|^{2}v^\delta \\ &\quad +ib \frac{v^\delta}{(|v^\delta|^2+\delta)^{\alpha /2}} +i\frac{da}{dt}(t)\cdot \nabla v^\delta, \end{aligned} \\ v^\delta(0,y) = u_0 (y+a(0)). \end{gathered} \end{equation} Now, we set $w^\delta(t,y)=\partial_tv^\delta(t,y)$ and since \[ \partial_tV(y+a(t))=\frac{da}{dt}(t)\cdot \nabla V(y+a(t)), \] it follows that $w$ satisfies \begin{equation} \label{47} \begin{gathered} \begin{aligned} i\partial_tw^\delta+\Delta w^\delta &=V(y+a(t))w^\delta+(\frac{da}{dt}(t)\cdot \nabla V(y+a(t)))v^\delta\\ &\quad +\frac{w^\delta}{(|y|^2+\delta)^{1 /2}} +\lambda \partial_t\Big(\frac{1}{|x|}\ast |v^\delta|^{2}v^\delta\Big) +ib\partial_t\Big( \frac{v^\delta}{(|v^\delta|^2 +\delta)^{\alpha /2}}\Big) \\ &\quad +i\frac{d^2a}{dt^2}(t)\cdot \nabla v^\delta +i\frac{da}{dt}(t)\cdot \nabla w^\delta, \end{aligned}\\ \begin{aligned} w^\delta(0,y) &= i\Delta v^\delta(0)-iV(y+a(0))v^\delta(0) -i\frac{v^\delta(0)}{(|y|^2+\delta)^{1 /2}} \\ &\quad -i\lambda \frac{1}{|x|}\ast |v^\delta(0)|^{2}v^\delta(0) +b \frac{v^\delta(0)}{(|v^\delta(0)|^2+\delta)^{\alpha /2}} +\frac{da}{dt}(t)\cdot \nabla v^\delta(0). \end{aligned} \end{gathered} \end{equation} Multiplying by $\overline{w^\delta}$, integrating on $\mathbb{R}^3$ and taking the imaginary part we have \begin{equation} \label{48} \begin{aligned} \frac{1}{2}\frac{d}{dt}\int_{\mathbb{R}^3} |w^\delta (t,y)|^2dy &=\operatorname{Im}\int_{\mathbb{R}^3} \Big(\frac{da}{dt}(t)\cdot \nabla V(y+a(t))\Big)v^\delta(t,y) \overline{w^\delta}(t,y)dy \\ &\quad +\lambda \operatorname{Im}\int_{\mathbb{R}^3} \partial_t \Big(\frac{1}{|x|}\ast |v^\delta|^{2}v^\delta\Big)(t,y)\overline{w^\delta}(t,y)dy \\ &\quad +b\operatorname{Re}\int_{\mathbb{R}^3} \partial_t \Big( \frac{v^\delta}{(|v^\delta|^2+\delta)^{\alpha /2}}\Big)(t,y) \overline{w^\delta}(t,y)dy\\ &\quad +\operatorname{Re}\int_{\mathbb{R}^3} \frac{d^2a}{dt^2}(t)\cdot \nabla v^\delta (t,y)\overline{w^\delta}(t,y)dy\\ &\quad + \operatorname{Re}\int_{\mathbb{R}^3} \frac{da}{dt}(t)\cdot \nabla w^\delta (t,y) \overline{w^\delta}(t,y)dy. \end{aligned} \end{equation} We deduce from the H\"older and Hardy inequalities that \begin{equation} \label{49} \begin{aligned} &\operatorname{Im}\int_{\mathbb{R}^3} \partial_t \Big(\frac{1}{|x|}\ast |v^\delta|^{2}v^\delta\Big)(t,y)\overline{w^\delta}(t,y)dy \\ &=\operatorname{Im}\int_{\mathbb{R}^3} \left(\frac{1}{|x|}\ast 2\operatorname{Re} (\overline{v^\delta}w^\delta)\right )(t,y)v^\delta(t,y) \overline{w^\delta}(t,y)dy \\ & \leqslant C\|\frac{1}{|x|}\ast \operatorname{Re}(\overline{v^\delta} (t)w^\delta(t))\|_{L^\infty} \|v^\delta(t)\|_{L^2}\|w^\delta(t)\|_{L^2}\\ & \leqslant C\|\frac{1}{|x|^2}\ast |v^\delta(t)|^2\|_{L^\infty}^{1/2} \|v^\delta(t)\|_{L^2}\|w^\delta(t)\|_{L^2}^2\\ & \leqslant C \|\nabla v^\delta(t)\|_{L^2}\|v^\delta(t)\|_{L^2} \|w^\delta(t)\|_{L^2}^2. \end{aligned} \end{equation} After some computations, we have \begin{equation} \label{410} \begin{aligned} &-b\operatorname{Re}\int_{\mathbb{R}^3} \partial_t \Big( \frac{v^\delta}{(|v^\delta|^2+\delta)^{\alpha /2}}\Big) (t,y)\overline{w^\delta}(t,y)dy\\ &=-b\operatorname{Re}\int_{\mathbb{R}^3} \Big( \frac{w^\delta}{(|v^\delta|^2+\delta)^{\alpha /2}} -\alpha v^\delta\frac{Rev^\delta\overline{w^\delta}}{(|v^\delta|^2 +\delta)^{\alpha /2+1}}\Big)(t,y)\overline{w^\delta}(t,y)dy\\ &=-b\int_{\mathbb{R}^3} \frac{(1-\alpha)|Rev^\delta\overline{w^\delta}|^2(t,y)+ |Imv^\delta\overline{w^\delta}|^2(t,y)+\delta|w^\delta|^2(t,y)}{(|v^\delta(t,y)|^2 +\delta)^{\alpha /2+1}}dy, \end{aligned} \end{equation} where we use the decomposing \[ |v^\delta|^2|w^\delta|^2=|\operatorname{Re}v^\delta\overline{w^\delta}|^2 +|\operatorname{Im}v^\delta\overline{w^\delta}|^2. \] In addition, \begin{equation}\label{411} \operatorname{Re}\int_{\mathbb{R}^3} \frac{da}{dt}(t)\cdot \nabla w^\delta(t,y) \overline{w^\delta}(t,y)dy=\frac{1}{2}\int_{\mathbb{R}^3} \frac{da}{dt}(t)\cdot \nabla |w^\delta(t,y)|^2dy=0. \end{equation} Collecting \eqref{48}-\eqref{411}, we have \begin{equation} \label{412} \begin{aligned} &\frac{1}{2}\frac{d}{dt}\int_{\mathbb{R}^3} |w^\delta (t,y)|^2dy \\ &\leqslant \operatorname{Im}\int_{\mathbb{R}^3} \Big(\frac{da}{dt}(t)\cdot \nabla V(y+a(t))\Big)v^\delta(t,y) \overline{w^\delta}(t,y)dy\\ &\quad +C\|\nabla v^\delta(t)\|_{L^2}\|v^\delta(t)\|_{L^2}\|w^\delta(t)\|_{L^2}^2\\ &\quad +\operatorname{Re}\int_{\mathbb{R}^3} \frac{d^2a}{dt^2}(t)\cdot \nabla v^\delta (t,y)\overline{w^\delta}(t,y)dy\\ &\leqslant |\frac{da}{dt}|\|xu^\delta(t)\|_{L^2}\|w^\delta(t)\|_{L^2} +C\|\nabla v^\delta(t)\|_{L^2}\|v^\delta(t)\|_{L^2}\|w^\delta(t)\|_{L^2}^2\\ &\quad +|\frac{d^2a}{dt^2}|\|\nabla v^\delta(t)\|_{L^2}\|w^\delta(t)\|_{L^2}. \end{aligned} \end{equation} Integrating in the time variable on $(0,t)$, we deduce from Lemma \ref{lem3.1} that \begin{equation} \label{413} \begin{aligned} \|w^\delta(t)\|_{L^2}^2 &\leqslant C\|u_0\|_{\Sigma^2}^2+C\int_0^t|\frac{da}{ds}(s)|\|xu^\delta(s)\|_{L^2} \|w^\delta(s)\|_{L^2}ds\\ &\quad +C\int_0^t\|\nabla v^\delta(s)\|_{L^2}\|v^\delta(s)\|_{L^2} \|w^\delta(s)\|_{L^2}^2ds \\ &\quad +\int_0^t|\frac{d^2a}{ds^2}| \|\nabla v^\delta(s)\|_{L^2}\|w^\delta(s)\|_{L^2}ds\\ &\leqslant C\|u_0\|_{\Sigma^2}^2 +C\int_0^t\Big(|\frac{da}{ds}(s)| +|\frac{d^2a}{ds^2}(s)|\Big)\|w^\delta(s)\|_{L^2}ds\\ &\quad +C\int_0^t\|w^\delta(s)\|_{L^2}^2ds. \end{aligned} \end{equation} Thus, it follows from Gronwall's inequality that \begin{equation}\label{414} \|w^\delta(t)\|_{L^2} \leqslant C(T,\|u_0\|_{\Sigma^2})\quad\forall t\in [0,T], \end{equation} for every $T<\infty$. Finally, it follows from \eqref{45} and Theorem \ref{thm1} that \begin{equation*} \|\partial_tu^\delta(t)\|_{L^2} \leqslant \|\partial_tv^\delta(t)\|_{L^2}+C\|\nabla u^\delta(t)\|_{L^2} \leqslant C(\|a\|_{W^{2,\infty}(0,\infty)},T,\|u_0\|_{\Sigma^2}). \end{equation*} \end{proof} \begin{proof}[Proof of Theorem \ref{thm2}] Combining Lemmas \ref{lem4.1} and \ref{lem4.2}, for every $00$ such that the solution $u^\delta$ of \eqref{be} satisfies for all $t\in [0,T]$ \begin{equation} \|u^\delta(t)\|_{H^2}+\|x^2 u^\delta(t)\|_{L^2} +\|\partial_tu^\delta(t)\|_{L^2}\leqslant C, \end{equation} where $C$ is independent of $\delta$. This implies that there exist $u\in L^\infty((0,T);\Sigma^2)$ and a subsequence $(u^{\delta_n})$ such that, for a.e. $t\in [0,T]$ \begin{equation} u^{\delta_n}(t)\rightharpoonup u(t),\quad\text{in $\Sigma^2$ as }n\to\infty. \end{equation} Thus, we can pass to the limit in the distributions sense in \eqref{be} as $\delta\to 0$. 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