Scattering and Minimization Theory for Cubic Inhomogeneous Nls with Inverse Square Potential

Abstract

In this paper, we study the scattering theory for the cubic inhomogeneous Schrödinger equations with inverse square potential $iu_t+\Delta u-\frac{a}{|x|^2}u=\lambda |x|^{-b}|u|^2u$ with $a>-\frac{1}{4}$ and $0<b<1$ in dimension three. In the defocusing case (i.e. $\lambda =1$), we establish the global well-posedness and scattering for any initial data in the energy space $H^1_a(\mathbb {R}^3)$. While for the focusing case(i.e. $\lambda =-1$), we obtain the scattering for the initial data below the threshold of the ground state, by making use of the virial/Morawetz argument as in Dodson and Murphy (Proc Am Math Soc 145:4859–4867, 2017) and Campos and Cardoso (Proc Am Math Soc 150:2007–2021, 2022) that avoids the use of interaction Morawetz estimate. We also address the existence and the non-existence of normalized solutions of the above Schrödinger equation in dimension N for the focusing and defocusing cases.

Appendix

In this appendix, we will establish the interaction Morawetz estimate for the solution to (1.1) with $\lambda =1$. As an application, we give a simple proof for the scattering theory for (1.1) with $\lambda =1$ but $a>0$.

Theorem 5.9

(Interaction Morawetz estimate) Let $a>0,\;0<b<1$ and $u:\;\mathbb {R}\times \mathbb {R}^3\rightarrow \mathbb {C}$ solve (1.1) with $\lambda =1$ and $u_0\in H^1(\mathbb {R}^3)$. Then, there holds

$$\begin{aligned} \Vert u\Vert _{L_{t,x}^4(\mathbb {R}\times \mathbb {R}^3)}\le C\big (M(u_0),E(u_0)\big ). \end{aligned}$$

(5.27)

Interpolating this with $\Vert u\Vert _{L_t^\infty L_x^6}\lesssim \Vert u\Vert _{L_t^\infty \dot{H}^1}\lesssim \sqrt{E(u_0)}$ implies

$$\begin{aligned} \Vert u\Vert _{L_t^6 L_x^\frac{9}{2}(\mathbb {R}\times \mathbb {R}^3)}\le C\big (M(u_0),E(u_0)\big ). \end{aligned}$$

As a consequence, the global solution u to (1.1) with $\lambda =1$ scatters.

Remark 5.10

For the higher dimension case $N\ge 4$, one can also obtain the interaction Morawetz estimate

$$\begin{aligned} \big \Vert |\nabla |^{-\frac{N-3}{4}}u\big \Vert _{L_{t,x}^4(\mathbb {R}\times \mathbb {R}^N)}\le C\big (M(u_0),E(u_0)\big ), \end{aligned}$$

(5.28)

where $u:\mathbb {R}\times \mathbb {R}^N\rightarrow \mathbb {C}$ solves $i\partial _tu-\mathcal {L}_au=|x|^{-b}|u|^{p-1}u$ with $a>-\tfrac{(N-2)^2}{4}+\tfrac{1}{4}$ and $b>0$. Compared with dimension three case and higher dimension case, we can get the interaction Morawetz estimate for some negative a in higher dimension cases.

To prove Theorem 5.9, we first consider that the function u(t, x) solves

$$\begin{aligned} i\partial _tu+\Delta u=F(t,x),\quad (t,x)\in \mathbb {R}\times \mathbb {R}^3. \end{aligned}$$

(5.29)

Define Morawetz action

$$\begin{aligned} M_w(t):=2\textrm{Im}\int _{\mathbb {R}^3}\nabla w(x)\cdot \nabla u(x)\bar{u}(x)\;dx. \end{aligned}$$

(5.30)

A simple computation shows

Lemma 5.11

(Morawetz identity) There holds

$$\begin{aligned} \frac{d}{dt}M_w(t)=\int _{\mathbb {R}^3}(-\Delta \Delta w)|u|^2dx+4\int _{\mathbb {R}^3}w_{jk}\Re (\partial _j\bar{u}\partial _ku)dx+2\int _{\mathbb {R}^3}w_j(x)\{F,u\}_P^jdx \end{aligned}$$

where $\{f,g\}_P=\Re (f\nabla \bar{g}-g\nabla \bar{f}),$ and repeated indices are implicitly summed. Especially, if u solves (1.1) with $\lambda =1$, then

$$\begin{aligned} \nonumber \frac{d}{dt}M_w(t)=&\int _{\mathbb {R}^3} \Big [(-\Delta \Delta w)|u|^2 + 4\Re (\bar{u}_j u_k) w_{jk} +4a\frac{x\cdot \nabla w}{|x|^4}|u|^2 \Big ]\,dx\\&\quad + b\int _{\mathbb {R}^3}\frac{x\cdot \nabla w}{|x|^{2+b}}|u|^4\;dx+\int _{\mathbb {R}^3}\Delta w(x)\frac{|u|^4}{|x|^b}\;dx. \end{aligned}$$

(5.31)

Taking $w(x)=|x|$, one can obtain the standard Morawetz inequality.

Lemma 5.12

(Classical Morawetz inequality) Let $a>0,\;0<b<1$ and $u:\;\mathbb {R}\times \mathbb {R}^3\rightarrow \mathbb {C}$ solve (1.1) with $\lambda =1$ and $u_0\in H^1(\mathbb {R}^3)$. Then, there holds

$$\begin{aligned} \int _{\mathbb {R}}\int _{\mathbb {R}^3}\left( \frac{|u(t,x)|^2}{|x|^3}+\frac{|u(t,x)|^4}{|x|^{1+b}}\right) \;dx\;dt\le C\big (M(u_0),E(u_0)\big ). \end{aligned}$$

(5.32)

Define the Morawetz action center y as

$$\begin{aligned} M_w^y(t):=2\textrm{Im}\int _{\mathbb {R}^3}\nabla w(x-y)\cdot \nabla u(x)\bar{u}(x)\;dx. \end{aligned}$$

(5.33)

Now, we define the interaction Morawetz action by

$$\begin{aligned} M^{Int}(t):=&\int _{\mathbb {R}^3}|u(y)|^2 M_w^y(t)\;dy\\\nonumber =&2\textrm{Im}\iint _{\mathbb {R}^3\times \mathbb {R}^3}|u(y)|^2\nabla w(x-y)\cdot \nabla u(x)\bar{u}(x)\;dx\;dy. \end{aligned}$$

(5.34)

Note that

$$\begin{aligned} \partial _t(|u|^2)=2\textrm{Im}(-\Delta u\bar{u}+F\bar{u})=-2\partial _j\textrm{Im}(\partial _ju\bar{u})+2\textrm{Im}(F\bar{u}), \end{aligned}$$

this together with Lemma 5.11 implies

$$\begin{aligned} \frac{d}{dt}M^{Int}(t)=&\int _{\mathbb {R}^3}\partial _t(|u(y)|^2) M_w^y(t)\;dy+\int _{\mathbb {R}^3}|u(y)|^2\frac{d}{dt}M_w^y(t)\;dy\nonumber \\ =&-4\iint \textrm{Im}(\partial _ju\bar{u})(y) w_{jk}(x-y)\textrm{Im}(\partial _ku\bar{u})(x)\;dx\;dy \end{aligned}$$

(5.35)

$$\begin{aligned}&+4\iint \textrm{Im}(F\bar{u})(y)\nabla w(x-y)\cdot \textrm{Im}(\nabla u\bar{u})(x)\;dx\;dy \end{aligned}$$

(5.36)

$$\begin{aligned}&+\iint (-\Delta \Delta w)(x-y)|u(x)|^2|u(y)|^2\;dx\;dy \end{aligned}$$

(5.37)

$$\begin{aligned}&+4\iint |u(y)|^2w_{jk}(x-y)\Re (\partial _j\bar{u}\partial _ku)(x)\;dx\;dy \end{aligned}$$

(5.38)

$$\begin{aligned}&+2\iint |u(y)|^2w_j(x-y)\{F,u\}_P^j(x)\;dx\;dy. \end{aligned}$$

(5.39)

Hence, taking $w(x-y)=|x-y|$, one can obtain the interaction Morawetz identity.

Lemma 5.13

(Interaction Morawetz identity) Assume that $u:\;\mathbb {R}\times \mathbb {R}^3\rightarrow \mathbb {C}$ solves (1.1) with $\lambda =1$, then

$$\begin{aligned}&\frac{d}{dt}M^{Int}(t) =-4\iint \textrm{Im}(\partial _ju\bar{u})(y)\Big (\frac{\delta _{jk}}{|x-y|}-\frac{(x-y)_j(x-y)_k}{|x-y|^3}\Big )\textrm{Im}(\partial _ku\bar{u})(x)\;dx\;dy \end{aligned}$$

(5.40)

$$\begin{aligned}&+4\iint |u(y)|^2\Big (\frac{\delta _{jk}}{|x-y|}-\frac{(x-y)_j(x-y)_k}{|x-y|^3}\Big )\Re (\partial _j\bar{u}\partial _ku)(x)\;dx\;dy \end{aligned}$$

(5.41)

$$\begin{aligned}&+c\int _{\mathbb {R}^3} |u(x)|^4\;dx \end{aligned}$$

(5.42)

$$\begin{aligned}&+4\iint |u(y)|^2\frac{x}{|x|^4}\cdot \frac{x-y}{|x-y|}|u(x)|^2\;dx\;dy \end{aligned}$$

(5.43)

$$\begin{aligned}&+b\iint |u(y)|^2\frac{x}{|x|^{2+b}}\cdot \frac{x-y}{|x-y|}|u(x)|^4\;dx\;dy \end{aligned}$$

(5.44)

$$\begin{aligned}&+2\iint |u(y)|^2\frac{|u(x)|^4}{|x|^{b}}\;dx\;dy. \end{aligned}$$

(5.45)

By the same argument as in Killip and Visan [24], we know that

$$\begin{aligned} (5.40)+(5.41)\ge 0. \end{aligned}$$

On the other hand, by the definition of $M^{Int}(t)$, we know that

$$\begin{aligned} |M^{Int}(t)|\lesssim \Vert u(t)\Vert _{L_x^2}^3\Vert u(t)\Vert _{\dot{H}^1}\le C\big (M(u_0),E(u_0)\big ). \end{aligned}$$

Hence, we obtain

$$\begin{aligned} \int _{\mathbb {R}}\int _{\mathbb {R}^3}|u(t,x)|^4\;dx\;dt&\le 2\sup _{t\in \mathbb {R}}|M^{Int}(t)|+\Vert u_0\Vert _{L_x^2}^2\int _{\mathbb {R}}\int _{\mathbb {R}^3}\Big (\frac{|u(t,x)|^2}{|x|^3}+\frac{|u(t,x)|^4}{|x|^{1+b}}\Big )\;dx\;dt\\&\lesssim C\big (M(u_0),E(u_0)\big ), \end{aligned}$$

where we have used Lemma 5.12. Therefore, we conclude the proof of Theorem 5.9.