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diff --git a/Jauslin_TAMU_2020.tex b/Jauslin_TAMU_2020.tex new file mode 100644 index 0000000..244fa2d --- /dev/null +++ b/Jauslin_TAMU_2020.tex @@ -0,0 +1,365 @@ +\documentclass{ian-presentation} + +\usepackage[hidelinks]{hyperref} +\usepackage{graphicx} +\usepackage{array} + +\begin{document} +\pagestyle{empty} +\hbox{}\vfil +\bf\Large +\hfil A simple equation to study interacting Bose gasses\par +\vfil +\large +\hfil Ian Jauslin +\normalsize +\vfil +\hfil\rm joint with {\bf Eric A. Carlen}, {\bf Elliott H. Lieb}\par +\vfil +arXiv:{\tt \href{https://arxiv.org/abs/1912.04987}{1912.04987}} +\hfill{\tt \href{http://ian.jauslin.org}{http://ian.jauslin.org}} +\eject + +\setcounter{page}1 +\pagestyle{plain} + +\title{Lieb's simple equation (1963)} +\vskip-10pt +\begin{itemize} + \item \href{https://doi.org/10.1103/PhysRev.130.2518}{[Lieb, 1963]}: $x\in\mathbb R^d$ + $$ + (-\Delta+v(x)+4e)u(x)=v(x)+2e\rho\ u\ast u(x) + $$ + $$ + e=\frac\rho2\int dx\ (1-u(x))v(x) + $$ + \item with + $$ + \rho>0 + ,\quad + v(x)\geqslant 0 + ,\quad + v\in L_1\cap L_{\frac d2+\epsilon}(\mathbb R^d) + $$ + \item and + $$ + u\in L_1(\mathbb R^d) + ,\quad + u\ast u(x):=\int dy\ u(x-y)u(y) + $$ +\end{itemize} +\vfill +\eject + +\title{Interacting Bose gas} +\vskip-10pt +\begin{itemize} + \item State: symmetric wave functions in a finite box of volume $V$ with periodic boundary conditions: + $$ + \psi(x_1,\cdots,x_N) + ,\quad + x_i\in \Lambda_d:=V^{\frac1d}\mathbb T^d + $$ + \item Probability distribution: $|\psi(x_1,\cdots,x_N)|^2$ + \item $N$-particle Hamiltonian: + $$ + H_N:= + -\frac12\sum_{i=1}^N\Delta_i + +\sum_{1\leqslant i<j\leqslant N}v(x_i-x_j) + $$ + with $v(x-y)\geqslant 0$ and $v\in L_1\cap L_{\frac d2+\epsilon}(\mathbb R^d)$. +\end{itemize} +\vfill +\eject + +\title{Interacting Bose gas} +\vskip-10pt +\begin{itemize} + \item Ground state: + $$ + H_N\psi_0=E_0\psi_0 + ,\quad + E_0=\min\mathrm{spec}(H_N) + $$ + \item Compute the ground state-energy per particle in the thermodynamic limit: + $$ + e_0:=\lim_{\displaystyle\mathop{\scriptstyle V,N\to\infty}_{\frac NV=\rho}}\frac{E_0}N + $$ +\end{itemize} +\vfill +\eject + +\title{Asymptotics for the Bose gas} +\vskip-10pt +\begin{itemize} + \item {\bf Theorem} \href{https://doi.org/10.1103/PhysRev.130.2518}{[Lieb, 1963]}: if $\hat v(k):=\int dx\ e^{ikx}v(x)\geqslant 0$, then + $$ + \frac{e_0}{\rho}\mathop{\longrightarrow}_{\rho\to\infty}\frac12\int dx\ v(x) + $$ + \item {\bf Theorem} \href{https://doi.org/10.1103/PhysRevLett.80.2504}{[Lieb, Yngvason, 1998]}: in 3 dimensions ($a$: scattering length) + $$ + \frac{e_0}{\rho}\mathop{\longrightarrow}_{\rho\to0}2\pi a + $$ + \href{https://doi.org/10.1103/PhysRev.106.1135}{[Lee, Huang, Yang, 1957]}, \href{https://doi.org/10.1007/s10955-009-9792-3}{[Yau, Yin, 2009]}, \href{https://arxiv.org/abs/1904.06164}{[Fournais, Solovej, 2019]}: + $$ + e_0=2\pi\rho a\left(1+\frac{128}{15\sqrt\pi}\sqrt{\rho a^3}+o(\sqrt\rho)\right) + $$ +\end{itemize} +\vfill +\eject + +\title{Simple equation for $v(x)=e^{-|x|}$ in 3 dimensions} +\hfil\includegraphics[height=6cm]{erho_effective.pdf} +\vfill +\eject + +\title{Main Theorem} +\vskip-5pt +\begin{itemize} + \item If $v(x)\geqslant 0$ and $v\in L_1\cap L_{\frac d2+\epsilon}(\mathbb R^d)$, then Lieb's simple equation + $$ + (-\Delta+4e+v)u=v+2e\rho u\ast u + ,\quad + e=\frac\rho2\int dx\ (1-u(x))v(x) + $$ + has an integrable solution (proved constructively), with $0\leqslant u\leqslant 1$. + + \item For $d=3$, + $$ + e=2\pi\rho a\left(1+\frac{128}{15\sqrt\pi}\sqrt{\rho a^3}+o(\sqrt\rho)\right) + ,\quad + \frac{e}\rho\mathop{\longrightarrow}_{\rho\to\infty}\frac12\int dx\ v(x) + . + $$ + + \item For $d=3$, if $v(x)\equiv v(|x|)$ is radially symmetric and decays exponentially, + $$ + u(|x|)\mathop\sim_{|x|\to\infty}\frac\alpha{|x|^4} + . + $$ +\end{itemize} +\vfill +\eject + +\title{Existence of a solution (sketch)} +\begin{itemize} + \item Simple equation: fixed $\rho>0$, + $$ + (-\Delta+4e+v)u=v+2e\rho u\ast u + ,\quad + e=\frac\rho2\int dx\ (1-u(x))v(x) + $$ + + \item Change the point of view: fix $e>0$, and compute $\rho$ and $u$. + + \item Iteration: $u_0=0$, + $$ + (-\Delta+4e+v)u_n=v+2e\rho_{n-1}u_{n-1}\ast u_{n-1} + ,\quad + \rho_n:=\frac{2e}{\int dx\ (1-u_n(x))v(x)} + . + $$ +\end{itemize} +\vfill +\eject + +\title{Existence of a solution (sketch)} +\begin{itemize} + $$ + u_n=(-\Delta+4e+v)^{-1}\left(v+2e\rho_{n-1}u_{n-1}\ast u_{n-1}\right) + ,\quad + \rho_n:=\frac{2e}{\int dx\ (1-u_n(x))v(x)} + . + $$ + \item $u_n(x)$ is an increasing sequence: since $v\geqslant 0$, + \begin{itemize} + \item $(-\Delta+4e+v)^{-1}$ is positivity preserving. + \item $\rho_n$ is an increasing function of $u_n$. + \end{itemize} +\end{itemize} +\vfill +\eject + +\title{Existence of a solution (sketch)} +\begin{itemize} + $$ + -\Delta u_n=(1-u_n)v-4eu_n+2e\rho_{n-1}u_{n-1}\ast u_{n-1} + ,\quad + \frac{2e}{\rho_n}=\int dx\ (1-u_n(x))v(x) + . + $$ + \item $\int dx\ u_n(x)<\frac1{\rho_n}$: integrating, + $$ + 0=\frac{2e}{\rho_n}-4e\int u_n+2e\rho_{n-1}\left(\int u_{n-1}\right)^2 + < + \frac{2e}{\rho_n}-2e\int u_n + $$ + (since $\int u_{n-1}<\frac1{\rho_{n-1}}$ and $\int u_{n-1}\leqslant\int u_n$). +\end{itemize} +\vfill +\eject + +\title{Existence of a solution (sketch)} +\begin{itemize} + $$ + -\Delta u_n=(1-u_n)v-4eu+2e\rho_{n-1}u_{n-1}\ast u_{n-1} + ,\quad + \frac{2e}{\rho_n}=\int dx\ (1-u_n(x))v(x) + . + $$ + \item $u_n(x)\leqslant 1$: since $v\geqslant 0$, + \begin{itemize} + \item for $x\in\Sigma:=\{x:\ u_n(x)>1\}$, + $$ + -\Delta u_n<-4e+2e\rho_{n-1}u_{n-1}\ast u_{n-1} + \leqslant-2e + <0 + $$ + (since $u_{n-1}\ast u_{n-1}\leqslant\|u_{n-1}\|_\infty\| u_{n-1}\|_1<\frac1{\rho_{n-1}}$). + \item Therefore $u_n$ is subharmonic on $\Sigma$, so it reaches its maximum on $\partial\Sigma$. + But, for $x\in\partial\Sigma$, $u_n(x)=1$, so $u_n(x)\leqslant 1$ in $\Sigma$, so $\Sigma=\emptyset$. + \end{itemize} +\end{itemize} +\vfill +\eject + +\title{Uniqueness} +\begin{itemize} + \item In addition, one can prove that the limiting $u$ is the unique non-negative integrable solution of the simple equation, for every fixed $e$. + + \item In addition, we prove that $e\mapsto\rho(e)$ is continuous, and $\rho(0)=0$ and $\rho(\infty)=\infty$, which allows us to compute solutions for the problem at fixed $\rho$. + + \item In order to show uniqueness of the solution of the simple equation for fixed $\rho$, one would have to show that $e\mapsto\rho(e)$ is monotone. + + \item Nevertheless, all non-negative integrable solutions are obtained by taking the limit of the sequence $u_n$ with the appropriate $e$. +\end{itemize} +\vfill +\eject + +\title{Asymptotics (sketch)} +\vskip-10pt +\begin{itemize} + $$ + -\Delta u=(1-u)v-4eu+2e\rho u\ast u + ,\quad + e=\frac\rho2\int dx\ (1-u(x))v(x) + . + $$ + \item When $\rho$ is small, $e$ is small as well, so the solution $u$ is {\it not too far from} the solution of the scattering equation + $$ + (-\Delta+v)\varphi=v + . + $$ + + \item The energy of $\varphi$ is + $$ + \frac\rho 2\int dx\ (1-\varphi(x))v(x)=2\pi\rho a + $$ + which yields the first term in the expansion. +\end{itemize} +\vfill +\eject + +\title{Asymptotics (sketch)} + $$ + -\Delta u=(1-u)v-4eu+2e\rho u\ast u + ,\quad + \frac{2e}\rho=\int dx\ (1-u(x))v(x) + . + $$ +\begin{itemize} + \item We work in Fourier space: + $$ + \rho \hat u(k)=\frac{k^2}{4e}+1-\sqrt{\left(\frac{k^2}{4e}+1\right)^2-\frac\rho{2e}\hat S(k)} + $$ + where $\hat S$ is the Fourier transform of $(1-u)v$. + + \item Small $e$ is related to small $k$. We approximate $\hat S(k)$ by $\hat S(0)=\frac{2e}\rho$, and control the error terms. +\end{itemize} +\vfill +\eject + +\title{Decay (sketch)} +$$ + u=(-\Delta+4e)^{-1}\left((1-u)v+2e\rho u\ast u\right) + ,\quad + e=\frac\rho2\int dx\ (1-u(x))v(x) +$$ +\begin{itemize} + \item $(-\Delta+4e)^{-1}$ has an exponentially decaying kernel, so $u$ cannot decay faster than $2e\rho u\ast u$. + + \item This is true for algebraically decaying functions: if $u\sim \alpha|x|^{-n}$ with $n>3$, then + $$ + u\ast u\sim \frac{2\alpha\int u}{|x|^n}. + $$ + + \item But why $|x|^{-4}$? +\end{itemize} +\vfill +\eject + +\title{Decay (sketch)} +$$ + u=(-\Delta+4e)^{-1}\left((1-u)v+2e\rho u\ast u\right) +$$ +\begin{itemize} + \item $w:=2e\rho (-\Delta+4e)^{-1}u$ satisfies + $$ + w=2e\rho (-\Delta+4e)^{-2}(1-u)v+w\ast w\geqslant w\ast w + ,\quad + \int w=\frac12 + . + $$ + \item {\bf Theorem} \href{https://arxiv.org/abs/2002.04184}{[Carlen, Jauslin, Lieb, Loss, 2020]}: for $0\leqslant \alpha<1$, + $$ + \int dx\ |x|w(x)=\infty + ,\quad + \int dx\ |x|^\alpha w(x)<\infty + . + $$ + Furthermore, $w\geqslant 0$. +\end{itemize} +\vfill +\eject + +\title{Full equation} +\hfil\includegraphics[height=5.5cm]{erho_fulleq.pdf} + +\hfil{\footnotesize Monte Carlo computation courtesy of M. Holzmann} +\vfill +\eject + +\title{Condensate fraction} +\hfil\includegraphics[height=5.5cm]{condensate.pdf} + +\hfil{\footnotesize Monte Carlo computation courtesy of M. Holzmann} +\vfill + +\title{Conclusion} +\vfill +\begin{itemize} + \item Simple equation: correct asymptotics for the ground state energy at both high and low densities. + + \item Condensate fraction seems right at low densities. + + \item Intriguing non-linear PDE. + + \item Proved existence, asymptotics, and decay rate. + + \item Full equation: does even better for the energy and condensate fraction. +\end{itemize} +\vfill +\eject + +\title{Open problems and conjectures} +\begin{itemize} + \item Monotonicity of $e\mapsto\rho(e)$, and concavity of $e\mapsto\frac1{\rho(e)}$ (would imply uniqueness). (So far, we have proofs for small and large $\rho$.) + + \item Condensate fraction: prove that $0\leqslant\eta\leqslant 1$. (Again, we have a proof for small and large $\rho$.) + + \item Other equations: interpolate between full equation and simple equation. + + \item Potentials which are not $\geqslant 0$? +\end{itemize} + +\end{document} |