Non-radial scattering theory for nonlinear Schrödinger equations with potential

Abstract

In this paper, we study a class of nonlinear Schrödinger equations (NLS) with potential

$$\begin{aligned} i\partial _t u +\Delta u - Vu = \pm |u|^\alpha u, \quad (t,x) \in \mathbb R\times \mathbb R^3, \end{aligned}$$

where $\frac{4}{3}<\alpha <4$ and V is a Kato-type potential including the genuine Yukawa potential as a special case. By using variational analysis and interaction Morawetz estimates, we establish a scattering criterion for the equation with non-radial initial data. As a consequence, we prove the energy scattering for the focusing problem with data below the ground state threshold. Our result extends the recent works of Hong (Commun Pure Appl Anal 15(5):1571–1601, 2016) and Hamano and Ikeda (J Evolut Equ 20:1131–1172, 2020). As a by product of the scattering criterion and the concentration-compactness lemma à la P. L. Lions, we study long time dynamics of global solutions to the focusing problem with data at the ground state threshold. Our result is robust and can be applicable to show the energy scattering for the focusing NLS with Coulomb potential.

Appendix

In this appendix, we will show Remark 1.1. Let V be as in (1.4). We first compute

$$\begin{aligned} \Vert V\Vert _{L^q}&= |c| \Vert |x|^{-\sigma }e^{-a|x|}\Vert _{L^q} \\&= |c| \left( \int _{\mathbb R^3} |x|^{-q\sigma } e^{-aq|x|} dx\right) ^{\frac{1}{q}} \\&= |c| \left( 4\pi \int _0^\infty r^{2-q\sigma } e^{-aqr} dr \right) ^{\frac{1}{q}} \\&= |c| \left[ 4\pi (aq)^{q\sigma -3} \Gamma (3-q\sigma )\right] ^{\frac{1}{q}} \end{aligned}$$

which proves (1.5).

We now compute

$$\begin{aligned} \Vert V\Vert _{\mathcal {K}} = \sup _{x\in \mathbb R^3} \int _{\mathbb R^3} \frac{|V(y)|}{|x-y|} dy. \end{aligned}$$

Consider

$$\begin{aligned} \int \frac{|V(y)|}{|x-y|} dy = |c| \int \frac{e^{-a|y|}}{|y|^\sigma |x-y|} dy. \end{aligned}$$

In the case $x=0$, we have

$$\begin{aligned} \int \frac{e^{-a|y|}}{|y|^{1+\sigma }} dy = 4\pi \int _0^\infty e^{-ar} r^{1-\sigma } dr = 4\pi a^{\sigma -2} \Gamma (2-\sigma ). \end{aligned}$$

In the case $x \ne 0$, we write

$$\begin{aligned} \int \frac{e^{-a|y|}}{|y|^\sigma |x-y|} dy&= \int _0^\infty \int _{\mathbb S^2} \frac{e^{-ar}}{r^\sigma |x-r\theta |} r^2 dr d\theta \\&=\int _0^\infty e^{-ar} r^{1-\sigma } I(x,r) dr, \end{aligned}$$

where $r = |y|$ and

$$\begin{aligned} I(x,r) = \int _{\mathbb S^2} \frac{1}{\left| \frac{x}{r} -\theta \right| } d\theta . \end{aligned}$$

Take $A \in O(3)$ such that $Ae_1 = \frac{x}{|x|}$ with $e_1=(1,0,0)$, we see that

$$\begin{aligned} I(x,r) = \int _{\mathbb S^2} \frac{1}{\left| \frac{|x|}{r} Ae_1 - \theta \right| } d\theta = \int _{\mathbb S^2} \frac{1}{\left| \frac{|x|}{r} e_1 - \theta \right| } d\theta . \end{aligned}$$

By change of variables, we arrive

$$\begin{aligned} I(x,r)&= \int _{-1}^1 \int _{\sqrt{1-s^2} \mathbb S^1} \frac{d\eta }{\sqrt{\left( \frac{|x|}{r}-s \right) ^2 +|\eta |^2}} \frac{ds}{\sqrt{1-s^2}} \\&= \int _{-1}^1 \int _{\mathbb S^1} \frac{\sqrt{1-s^2} d\zeta }{\sqrt{\left( \frac{|x|}{r}-s \right) ^2 +1-s^2}} \frac{ds}{\sqrt{1-s^2}} \\&=|\mathbb S^1| \int _{-1}^1 \frac{ds}{\sqrt{\left( \frac{|x|}{r}-s \right) ^2 +1-s^2}} \\&= 2\pi \frac{r}{|x|} \left( \frac{|x|}{r} +1 - \left| \frac{|x|}{r}-1 \right| \right) \\&= \left\{ \begin{array}{cl} 4\pi &{}\text {if } |x| \le r, \\ 4\pi \frac{r}{|x|} &{}\text {if } |x|\ge r. \end{array} \right. \end{aligned}$$

It follows that

$$\begin{aligned} \int \frac{e^{-a|y|}}{|y|^\sigma |x-y|} dy&= \frac{4\pi }{|x|} \int _0^{|x|} e^{-ar} r^{2-\sigma } dr + 4\pi \int _{|x|}^\infty e^{-ar} r^{1-\sigma } dr \\&= 4\pi a^{2-\sigma } \Gamma (2-\sigma ) + 4\pi \left( \frac{1}{|x|}\int _0^{|x|} e^{-ar} r^{2-\sigma } dr - \int _0^{|x|} e^{-ar} r^{1-\sigma } dr\right) . \end{aligned}$$

Consider

$$\begin{aligned} f(\lambda ) = \frac{1}{\lambda } \int _0^\lambda e^{-ar} r^{2-\sigma } dr - \int _0^\lambda e^{-ar} r^{1-\sigma } dr, \quad \lambda >0. \end{aligned}$$

We see that if $0<\sigma <2$, then

$$\begin{aligned} \lim _{\lambda \rightarrow 0} f(\lambda )= 0. \end{aligned}$$

Moreover,

$$\begin{aligned} f'(\lambda ) =- \frac{1}{\lambda ^2} \int _0^\lambda e^{-ar} r^{2-\sigma } dr <0, \quad \forall \lambda >0. \end{aligned}$$

This shows that f is a strictly decreasing function, hence $f(\lambda ) <0$ for all $\lambda >0$. Thus for $x\ne 0$,

$$\begin{aligned} \int \frac{e^{-a|y|}}{|y|^\sigma |x-y|} dy < 4\pi a^{2-\sigma } \Gamma (2-\sigma ). \end{aligned}$$

We conclude that

$$\begin{aligned} \Vert V\Vert _{\mathcal {K}} = 4\pi |c| a^{2-\sigma } \Gamma (2-\sigma ) \end{aligned}$$

which proves (1.6). $\square$