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\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

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\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}
