Quantitative Derivation and Scattering of the 3D Cubic NLS in the Energy Space

Abstract

We consider the derivation of the defocusing cubic nonlinear Schrödinger equation (NLS) on ${\mathbb {R}}^{3}$ from quantum N-body dynamics. We reformat the hierarchy approach with Klainerman-Machedon theory and prove a bi-scattering theorem for the NLS to obtain convergence rate estimates under $H^{1}$ regularity. The $H^{1}$ convergence rate estimate we obtain is almost optimal for $H^{1}$ datum, and immediately improves if we have any extra regularity on the limiting initial one-particle state.

Appendix A. Misc. Estimates

1.1 A.1 Collapsing Estimates and Strichartz Estimates

We use the original Klainerman-Machedon collapsing estimate as our iterating estimate in this paper.

Lemma A.1

([13, 19, 54]) ¹⁴There is a C independent of V, j, k, and N such that, (for $f^{(k+1)}({\mathbf {x}}_{k+1},{\mathbf {x}}_{k+1}^{\prime })$ independent of t)

$$\begin{aligned} \left\| S^{(1,k)}B_{N,j,k+1}U^{(k+1)}(t)f^{(k+1)}\right\| _{L_{t}^{2}L_{{\mathbf {x}},{\mathbf {x}}^{\prime }}^{2}}\leqslant C\left\| V\right\| _{L^{1}}\left\| S^{(1,k+1)}f^{(k+1)}\right\| _{L_{\mathbf { x},{\mathbf {x}}^{\prime }}^{2}}. \end{aligned}$$

To explore the time derivative gain by Duhamel type terms, we also need the $X_{s,b}$ version of Lemma A.1. As we are using $S^{(k)}$ to denote the space derivatives, we surpress the s notation in definition of the $X_{s,b}$ space and define the norm $X_{b}^{(k)}$ by

$$\begin{aligned} \Vert \alpha ^{(k)}\Vert _{X_{b}^{(k)}}=\left( \int \langle \tau +\left| \varvec{\xi }_{k}\right| ^{2}-\left| \varvec{\xi }_{k}^{\prime }\right| ^{2}\rangle ^{2b}\left| {\hat{\alpha }}^{(k)}(\tau ,\varvec{\xi }_{k},\varvec{\xi }_{k}^{\prime })\right| ^{2}\,d\tau \,d\varvec{\xi }_{k}\,d\varvec{\xi }_{k}^{\prime }\right) ^{1/2} \end{aligned}$$

which is essentially a $X_{0,b}$ norm. We then have the Duhamel time-derivative gain property and the $X_{s,b}$ version of Lemma A.1.

Claim A.2

([21]) Let $\frac{1}{2}<b<1$ and $\theta (t)$ be a smooth cutoff. Then

$$\begin{aligned} \left\| \theta (t)\int _{0}^{t}U^{(k)}(t-s)\beta ^{(k)}(s)\,ds\right\| _{X_{b}^{(k)}}\lesssim \Vert \beta ^{(k)}\Vert _{X_{b-1}^{(k)}} \end{aligned}$$

(A.1)

Lemma A.3

([21]) There is a C independent of j, k, and N such that (for $\alpha ^{(k+1)}(t,{\mathbf {x}}_{k+1},{\mathbf {x}}_{k+1})$ dependent on t)

$$\begin{aligned} \Vert S^{(1,k)}B_{N,j,k+1}\alpha ^{(k+1)}\Vert _{L_{t}^{2}L_{{\mathbf {x}}, {\mathbf {x}}^{\prime }}^{2}}\leqslant C\Vert S^{(1,k+1)}\alpha ^{(k+1)}\Vert _{X_{\frac{1}{2}+}^{(k+1)}} \end{aligned}$$

In the above notation, the dual Strichartz estimates we need in this paper are the following:

Lemma A.4

([21]) Let

$$\begin{aligned} \beta ^{(k)}(t,{\mathbf {x}}_{k},{\mathbf {x}}_{k}^{\prime })=N^{3\beta -1}V(N^{\beta }(x_{i}-x_{j}))\gamma ^{(k)}(t,{\mathbf {x}}_{k},{\mathbf {x}} _{k}^{\prime }) \end{aligned}$$

Then for $N\ge 1$, we have

$$\begin{aligned} \Vert \left| \nabla _{x_{i}}\right| \left| \nabla _{x_{j}}\right| \beta ^{(k)}\Vert _{X_{-\frac{1}{2}+}^{(k)}}\lesssim N^{ \frac{5}{2}\beta -1}\Vert \langle \nabla _{x_{i}}\rangle \langle \nabla _{x_{j}}\rangle \gamma ^{(k)}\Vert _{L_{t}^{2}L_{{\mathbf {x}}{\mathbf {x}} ^{\prime }}^{2}} \end{aligned}$$

(A.2)

and

$$\begin{aligned} \Vert \beta ^{(k)}\Vert _{X_{-\frac{1}{2}+}^{(k)}}\lesssim N^{\frac{1}{2} \beta -1}\Vert \langle \nabla _{x_{i}}\rangle \langle \nabla _{x_{j}}\rangle \gamma ^{(k)}\Vert _{L_{t}^{2}L_{{\mathbf {x}}{\mathbf {x}}^{\prime }}^{2}}. \end{aligned}$$

(A.3)

1.2 A.2. Convolution Estimates

Lemma A.5

Let $W_{N}(x)=N^{3\beta }V(N^{\beta }x)-b_{0}\delta (x)$, where $b_{0}=\int V(x)\,dx$. For any $0\le s\le 1$,

$$\begin{aligned} \Vert W_{N}*f\Vert _{L_{x}^{p}}\lesssim N^{-\beta s}\Vert D^{s}f\Vert _{L_{x}^{p}} \end{aligned}$$

for any $1<p<\infty$. The implicit constant depends only on $\Vert \langle x\rangle V(x)\Vert _{L^{1}}$.

Proof

The case $s=0$ is just Young’s inequality, since $\Vert V_{N}\Vert _{L^{1}}=\Vert V\Vert _{L^{1}}<\infty$, independent of N. We next establish the estimate for $s=1$. Since ${\hat{V}}(0)=b_{0}$,

$$\begin{aligned} \widehat{W_{N}}(\xi )={\hat{V}}(\xi N^{-\beta })-b_{0}=\int _{s=0}^{s=1}\frac{d }{ds}{\hat{V}}(s\xi N^{-\beta })\,ds=\int _{s=0}^{s=1}N^{-\beta }\xi \cdot \nabla {\hat{V}}(s\xi N^{-\beta })\,ds \end{aligned}$$

and thus

$$\begin{aligned} \widehat{W_{N}}(\xi ){\hat{f}}(\xi )=N^{-\beta }\int _{s=0}^{s=1}\nabla {\hat{V}} (s\xi N^{-\beta })\cdot \xi {\hat{f}}(\xi )\,ds \end{aligned}$$

Let $Y(x)=xV(x)$ so that ${\hat{Y}}(\xi )=\nabla {\hat{V}}(\xi )$. It follows that

$$\begin{aligned}&\int _{y\in {\mathbb {R}}^{3}}W_{N}(x-y)f(y)\,dy \\&\quad =N^{-\beta }\int _{s=0}^{s=1}\int _{y\in {\mathbb {R}}^{3}}s^{-3}N^{3\beta }Y(s^{-1}N^{\beta }(x-y))\,\nabla f(y)\,dy\,ds \end{aligned}$$

By Minkowski’s inequality and Young’s inequality,

$$\begin{aligned}&\left\| \int _{y\in {\mathbb {R}} ^{3}}W_{N}(x-y)f(y)\,dy\right\| _{L_{x}^{p}} \\&\quad \lesssim N^{-\beta }\int _{s=0}^{1}\left\| \int _{y\in {\mathbb {R}} ^{3}}s^{-3}N^{3\beta }Y(s^{-1}N^{\beta }(x-y))\,\nabla f(y)\,dy\right\| _{L_{x}^{p}}\,ds \\&\quad \lesssim N^{-\beta }\Vert \nabla f\Vert _{L_{x}^{p}} \end{aligned}$$

The cases $0<s<1$ follow by interpolation, as follows. Let $P_{M}$ be the Littlewood-Paley projector for frequency $0<M<\infty$. Then by the $s=0$ and $s=1$ cases,

$$\begin{aligned}&\Vert W_{N}*f\Vert _{L_{x}^{p}}=\left\| W_{N}*\sum _{M}P_{M}f\right\| _{L_{x}^{p}}\le \sum _{M}\Vert W_{N}*P_{M}f\Vert _{L_{x}^{p}} \\&\quad \lesssim \sum _{M}\min (1,N^{-\beta }M)\Vert P_{M}f\Vert _{L_{x}^{p}}\lesssim \sum _{M}\min (1,N^{-\beta }M)M^{-s}\Vert D^{s}f\Vert _{L_{x}^{p}} \end{aligned}$$

Divide the sum into the case $M\le N^{\beta }$, for which we use $\min (1,N^{-\beta }M)=N^{-\beta }M$, and the case $M\ge N^{\beta }$, for which we use $\min (1,N^{-\beta }M)=1$.

$$\begin{aligned} \lesssim \left( \sum _{M\le N^{\beta }}N^{-\beta }M^{1-s}+\sum _{M\ge N^{\beta }}M^{-s}\right) \Vert D^{s}f\Vert _{L_{x}^{p}}\lesssim N^{-\beta s}\Vert D^{s}f\Vert _{L_{x}^{p}} \end{aligned}$$

$\square$

Lemma A.6

Let $W_{N}(x)=N^{3\beta }V(N^{\beta }x)-b_{0}\delta (x)$, where $b_{0}=\int V(x)\,dx$.

$$\begin{aligned} \int (W_{N}*f_{1})f_{2}\,dx\lesssim N^{-\beta }\Vert |\nabla |^{1/2}f_{1}\Vert _{L_{x}^{2}}\Vert |\nabla |^{1/2}f_{2}\Vert _{L_{x}^{2}} \end{aligned}$$

Also, if $f_{j}$ is replaced by $P_{M_{j}}f_{j}$, then the same estimate holds but in addition we must have $M_{1}\sim M_{2}$ (or otherwise the left side is zero).

Proof

By Plancherel

$$\begin{aligned} \int (W_{N}*f_{1})f_{2}\,dx=\int _{\xi }{\hat{W}}_{N}(\xi )f_{1}(\xi )f_{2}(\xi )\,d\xi \end{aligned}$$

(A.4)

As in the proof of Lemma A.5,

$$\begin{aligned} \widehat{W_{N}}(\xi )=N^{-\beta }{\hat{Q}}_{N}(\xi )\cdot \xi \,,\qquad {\hat{Q}} _{N}(\xi )\overset{\mathrm {def}}{=}\int _{s=0}^{s=1}\nabla {\hat{V}}(s\xi N^{-\beta })\,ds \end{aligned}$$

Since

$$\begin{aligned} \Vert {\hat{Q}}_{N}\Vert _{L_{\xi }^{\infty }}\le \Vert \nabla {\hat{V}}\Vert _{L_{\xi }^{\infty }}=\Vert [xV(x)]\widehat{\;\,}\Vert _{L_{\xi }^{\infty }}\le \Vert xV(x)\Vert _{L_{x}^{1}} \end{aligned}$$

we can just complete the proof by Cauchy-Schwarz in (A.4) $\square$