Update to v1.0:

Fixed: Missing prefactor in the expression of f-f*f

  Fixed: 1/rho is convex, not concave.

  Fixed: Miscellaneous typos and small changes to the formatting.

Carlen_Jauslin_Lieb_2019.tex

@@ -40,7 +40,7 @@
 {\small
 In 1963 a partial differential equation with a convolution non-linearity was introduced in connection with a quantum mechanical many-body problem, namely the gas of bosonic particles.
 This equation is mathematically interesting for several reasons.
-(1) Further investigation showed that predictions based on the equation agree well over the {\it entire range} of parameters with what is expected to be true for the solution of the true many-body problem.
+(1) Although the equation was expected to be valid only for small values of the parameters, further investigation showed that predictions based on the equation agree well over the {\it entire range} of parameters with what is expected to be true for the solution of the true many-body problem.
 (2) The novel nonlinearity is easy to state but seems to have almost no literature up to now.
 (3) The earlier work did not prove existence and uniqueness of a solution, which we provide here along with properties of the solution such as decay at infinity. 
 }
@@ -117,7 +117,7 @@ Thus, any physical solutions of (\ref{simpleq})-(\ref{energy}) must necessarily
 \endtheo
 
 \bigskip
-This {\em a-priori} result  that we prove before we take up existence and uniqueness, turns on results \cite{CJLL20}   obtained in collaboration with Michael Loss on the convolution inequality $f \geqslant f\ast f$ in $L^1(\mathbb R^d)$.
+This {\em a-priori} result, which we prove before we take up existence and uniqueness, turns on results \cite{CJLL20}   obtained in collaboration with Michael Loss on the convolution inequality $f \geqslant f\ast f$ in $L^1(\mathbb R^d)$.
 While $u(x)\leqslant 1$ is a physical requirement, $u(x)\geqslant0$ is not, see section\-~\ref{sec:bosegas} for details.
 \bigskip
 
@@ -160,7 +160,7 @@ When $d=3$,
 \end{equation}
 Equation\-~(\ref{simpleq}) can be rewritten as
 \begin{equation}\label{simpleq3}
-u(x) = Y_{4e}*(\mathcal V (1- u(x)) +2e\rho Y_{4e}*u*u \ .
+u(x) = Y_{4e}*(\mathcal V (1- u(x))) +2e\rho Y_{4e}*u*u \ .
 \end{equation}
 Since $u$ and $\mathcal V$ are assumed to be integrable, and $u(x)$ is assumed to satisfy\-~(\ref{con1}), all terms in\-~(\ref{simpleq3}) are integrable, and integrating yields 
 \begin{equation}
@@ -247,7 +247,7 @@ Then there is a constructively defined continuous function $\rho(e)$ on $(0,\inf
 {\bf Remarks}:
 \begin{itemize}
  \item We do not assume here that the potential is radially symmetric.
- However, we shall see from our proof that if $\mathcal V$ is radially symmetric, then so is $u$.
+ However, the uniqueness statement implies that $u$ is radially symmetric whenever $\mathcal V$ is radially symmetric.
 
  \item The function $\rho(e)$ is the {\em density function}, which specifies the density as a function of the energy. Thus, our system together with\-~(\ref{con1}) constrains the parameters $e$ and $\rho$ to be related by a strict functional relation $\rho = \rho(e)$. In most of the early literature on the Bose gas, $\rho$ is taken as the independent parameter, as suggested by\-~(\ref{lhy}): One puts $N$ particles in a box of volume $N/\rho$, and 
 seeks to find the ground state energy per particle, $e$, as a function of $\rho$. Our theorem goes in the other direction, with $\rho$ specified as a function of $e$. We prove that $e\mapsto\rho(e)$ is continuous, and we conjecture that $\rho(e)$ is a strictly 
@@ -348,16 +348,16 @@ Let $u(x)$ be an integrable solution of the system (\ref{simpleq})-(\ref{energy}
 
 
 
-{\bf Proof} Since $Y_{4e}*(\mathcal V (1- u(x)) \geqslant 0$, it follows that 
+{\bf Proof} Since $Y_{4e}*(\mathcal V (1- u(x))) \geqslant 0$, it follows that 
 \begin{equation}
   u_- \leqslant  (f*u)_- =  (f*u_+ - f*u_-)_-  \leqslant f*u_-\ .
 \end{equation}
 Integrating, we find
-${\displaystyle \int u_- {\rm d}x \leqslant \frac12 \int u_-{\rm d}x}$,
+${\displaystyle \int u_- dx \leqslant \frac12 \int u_-dx}$,
 and this implies that $u_- = 0$. \qed
 
 
-{\bf Proof of Theorem~\ref{positivity}} Multiply (\ref{pos3}) through by $2e\rho$, and then convolve both sides with $Y_{4e}$. The result is  $f = 2e\rho Y_{4e}*(Y_{4e}*(\mathcal V (1- u)) +  f\ast f$, and since   $Y_{4e}*(Y_{4e}*(\mathcal V (1- u)) \geqslant 0$, $f$ is an integrable solution of 
+{\bf Proof of Theorem~\ref{positivity}} Multiply (\ref{pos3}) through by $2e\rho$, and then convolve both sides with $Y_{4e}$. The result is  $f = 2e\rho Y_{4e}*(Y_{4e}*(\mathcal V (1- u)) +  f\ast f$, and since   $Y_{4e}*(Y_{4e}*(\mathcal V (1- u))) \geqslant 0$, $f$ is an integrable solution of 
 \begin{equation}\label{pos5}
 f(x) \geqslant f\ast f(x)
 \end{equation}
@@ -402,7 +402,7 @@ Comparing\-~(\ref{iteration}) to\-~(\ref{iteration0}), note that the analog of $
 
 \theo{Lemma}\label{lem1} Let $\mathcal{V} \in L^1(\mathbb{R}^d)\cap L^p(\mathbb{R}^d)$, $p>\max\{\frac d2,1\}$. Both sequences $\{\rho_n\}$ and $\{u_n\}$ are well defined and increasing, and for all $n$,
 \begin{equation}\label{simple17}
-\int_{\mathbb{R}^d} u_n{\rm d}x < \frac{1}{2e}\int_{\mathbb{R}^d}\mathcal{V}(1-u_n)dx \ .
+\int_{\mathbb{R}^d} u_n dx < \frac{1}{2e}\int_{\mathbb{R}^d}\mathcal{V}(1-u_n)dx \ .
 \end{equation}
 \endtheo
 \bigskip
@@ -418,23 +418,30 @@ Since $t\mapsto t^{-1}$ is monotone decreasing on $(0,\infty)$, this shows that
 \bigskip
 
 
-\indent Now suppose that $u_{n} \geqslant u_{n-1} \geqslant 0$, $\rho_{n} \geqslant \rho_{n-1} \geqslant 0$, and $\int_{\mathbb{R}^d} u_{n}{\rm d}x < \frac{1}{2e}\int_{\mathbb{R}^d}\mathcal V(1-u_{n})$, all of which we have just verified for $n=1$. Then
+\indent Now suppose that $u_{n} \geqslant u_{n-1} \geqslant 0$, $\rho_{n} \geqslant \rho_{n-1} \geqslant 0$, and $\int_{\mathbb{R}^d} u_{n} dx < \frac{1}{2e}\int_{\mathbb{R}^d}\mathcal V(1-u_{n})$, all of which we have just verified for $n=1$. Then
 \begin{equation}
  u_{n+1} = K_e\mathcal V + 2e \rho_n K_e u_{n}*u_{n}(x) \geqslant K_e\mathcal V + 2e\rho_{n-1} K_e u_{n-1}*u_{n-1}(x) = u_n(x)\ ,
 \end{equation}
 and then 
 \begin{equation}
-\int_{\mathbb{R}^d} \mathcal V(1-u_{n+1}){\rm d}x < \int_{\mathbb{R}^d}\mathcal V(1-u_{n}){\rm d}x \ .
+\int_{\mathbb{R}^d} \mathcal V(1-u_{n+1}) dx < \int_{\mathbb{R}^d}\mathcal V(1-u_{n}) dx \ .
 \end{equation}
 Integrating both sides of $u_{n+1} = G_e \mathcal {V}(1- u_{n+1}) + 2e \rho_n G_e u_{n}*u_{n}$ yields,
 \begin{equation}\label{simple15}
-2\int_{\mathbb{R}^d} u_{n+1}{\rm d}x = \frac{1}{2e}\int_{\mathbb{R}^d}\mathcal V(1-u_{n+1}) + \rho_n \left( \int_{\mathbb{R}^d}u_{n}{\rm d}x \right)^2
+2\int_{\mathbb{R}^d} u_{n+1} dx = \frac{1}{2e}\int_{\mathbb{R}^d}\mathcal V(1-u_{n+1}) + \rho_n \left( \int_{\mathbb{R}^d}u_{n} dx \right)^2
 \end{equation}
-Then since $\int_{\mathbb{R}^d} u_{n}{\rm d}x < \frac{1}{2e}\int_{\mathbb{R}^d}\mathcal V(1-u_{n}) = \frac{1}{\rho_n}$, (\ref{simple15}) implies 
+Then since
 \begin{equation}
- 2\int_{\mathbb{R}^d} u_n{\rm d}x \leqslant \frac{1}{2e}\int_{\mathbb{R}^d}\mathcal{V}(1-u_n) + \int_{\mathbb{R}^d}u_{n-1}{\rm d}x\ .
+  int_{\mathbb{R}^d} u_{n} dx < \frac{1}{2e}\int_{\mathbb{R}^d}\mathcal V(1-u_{n}) = \frac{1}{\rho_n}
+\end{equation}
+(\ref{simple15}) implies 
+\begin{equation}
+ 2\int_{\mathbb{R}^d} u_n dx \leqslant \frac{1}{2e}\int_{\mathbb{R}^d}\mathcal{V}(1-u_n) + \int_{\mathbb{R}^d}u_{n-1} dx\ .
+\end{equation}
+Then because $\int_{\mathbb{R}^d} u_{n} dx < \int_{\mathbb{R}^d} u_{n+1} dx $, we have
+\begin{equation}
+  \int_{\mathbb{R}^d} u_{n+1} dx < \frac{1}{2e}\int_{v}\mathcal{V}(1-u_{n+1}).
 \end{equation}
-Then because $\int_{\mathbb{R}^d} u_{n}{\rm d}x < \int_{\mathbb{R}^d} u_{n+1} {\rm d}x $, we have that $\int_{\mathbb{R}^d} u_{n+1}{\rm d}x < \frac{1}{2e}\int_{v}\mathcal{V}(1-u_{n+1}) $.
 This proves\-~(\ref{simple17}) for $n+1$, and shows that 
 \begin{equation}
  0 \leqslant \frac{1}{2e}\int_{\mathbb{R}^d}\mathcal{V}(1-u_{n+1})dx \leqslant \frac{1}{2e}\int_{\mathbb{R}^d}\mathcal{V}(1-u_n)dx \ ,
@@ -454,7 +461,11 @@ and then, as before, $\rho_{n+1}\geqslant \rho_n$.
 Since $K_e$ maps $L^p(\mathbb R^d)$ into $W^{2,p}(\mathbb R^d)$, $u_1$ is continuous and vanishes at infinity. Let $A := \{x\ :\ u_1(x) > 1\}$. Then $A$ is open. If $A$ is non-empty, then $u_1$ is subharmonic on $A$, and hence takes on its maximum on the boundary of $A$. Since $u_1$ would equal $1$ on the boundary, this is impossible, and $A$ is empty. This proves the assertion for $n=1$. 
 \bigskip
 
-\indent Now make the inductive hypothesis that $0 \leqslant u_n(x) \leqslant 1$ for all $x$. Then $\|u_n\|_p^p \leqslant \|u_n\|_1 \leqslant\frac{1}{2e}\int_{\mathbb{R}^d}\mathcal{V}dx$. By Young's inequality, $\|u_n\ast u_n\|_p \leqslant \|u_n\|_p\|u_1\|_1$, and hence 
+\indent Now make the inductive hypothesis that $0 \leqslant u_n(x) \leqslant 1$ for all $x$. Then
+\begin{equation}
+  \|u_n\|_p^p \leqslant \|u_n\|_1 \leqslant\frac{1}{2e}\int_{\mathbb{R}^d}\mathcal{V}dx.
+\end{equation}
+By Young's inequality, $\|u_n\ast u_n\|_p \leqslant \|u_n\|_p\|u_1\|_1$, and hence 
 $\mathcal{V} + 2e\rho_n u_n\ast u_n \in L^p(\mathbb{R}^d)$. Therefore, 
 $u_{n+1}= K_e(\mathcal{V} + 2e\rho_n u_n\ast u_n) \in W^{2,p}(\mathbb{R}^d)$. It follows as before that $u_{n+1}$ is continuous and vanishing at infinity, and in particular, bounded, and 
 \begin{eqnarray*}
@@ -541,7 +552,7 @@ By remarks made above, this means that $u$ satisfies\-~(\ref{simpleq}) and\-~(\r
  Then since $(1-u_n(x,e))\mathcal V(x) \leqslant \mathcal{V}(x)$, the Dominated Convergence Theorem yields the continuity of $\rho_n(e)$ for each $n$.
  Writing our iteration in the equivalent form (as in\-~(\ref{simpleq3})):
  \begin{equation}\label{simpleq3B}
-u_n(x,e) = Y_{4e}*(\mathcal V (1- u_n(x,e)) +2e \frac{1}{b_{n-1}(e)} Y_{4e}\ast u_{n-1}\ast u_{n-1}(x,e) \ ,
+u_n(x,e) = Y_{4e}*(\mathcal V (1- u_n(x,e))) +2e \frac{1}{b_{n-1}(e)} Y_{4e}\ast u_{n-1}\ast u_{n-1}(x,e) \ ,
 \end{equation}
 and integrating, we obtain
  \begin{equation}\label{simpleq3C} 
@@ -571,7 +582,7 @@ Now an easy induction shows that $a_n(e)$ is continuous for each $n$. By\-~(\ref
  (\ref{con4B}). \qed
 \bigskip
  
-{\bf Remark}: Note that $\|u - u_n\|_1 = \frac{1}{\rho} - a_n$, and hence by By\-~(\ref{rate}), $\|u - u_n\|_1 \leqslant Cn^{-1/2}$. 
+{\bf Remark}: Note that $\|u - u_n\|_1 = \frac{1}{\rho} - a_n$, and hence by\-~(\ref{rate}), $\|u - u_n\|_1 \leqslant Cn^{-1/2}$. 
 In fact, numerically, we find that the rate is significantly faster than this. For example, with $\mathcal V(x)=e^{-|x|}$ and $e=10^{-4}$, $\|u - u_n\|_1$ decays at least as fast as $n^{-3.5}$.
 \bigskip
 
@@ -582,8 +593,8 @@ In fact, numerically, we find that the rate is significantly faster than this. F
 
 \bigskip
 
-We close this section by remarking that if $\mathcal{V}$ is radially symmetric, then so is each $u_1 = K_e\mathcal{V}$, and then by a simple induction, so is each $u_n$, hence also $u$, the unique solution $u$ provided by Theorem\-~\ref{theorem:existence}.
+We close this section by remarking that if $\mathcal{V}$ is radially symmetric, then so is $u_1 = K_e\mathcal{V}$, and then by a simple induction, so is $u_n$, hence also $u$, the unique solution $u$ provided by Theorem\-~\ref{theorem:existence}.
-This justifies the first remark following Theorem\-~\ref{theorem:existence}.
+This is consistent with the first remark following Theorem\-~\ref{theorem:existence}.
 
 \section{Asymptotics}\label{sec:asymptotics}
 \indent In this section, we prove Theorem\-~\ref{theorem:asymptotics}.
@@ -678,10 +689,10 @@ $R>0$ define $\mathcal{V}_R(x) = \mathcal{V}(x)$ for $|x| > R$ and otherwise $\m
 \begin{equation}\label{local}
 \|\mathcal{V}_R\|_1 < R^{-q} \quad{\rm and}\quad \|\mathcal{V}_R \|_p < R^{-q} \ .
 \end{equation}
-By the lemma below, $\lim_{x|\to\infty}|x|\varphi(x)$ exists. The {\em scattering length} $a$ is
+By the lemma below, $\lim_{|x|\to\infty}|x|\varphi(x)$ exists. The {\em scattering length} $a$ is
  defined to be (in dimension $d=3$).
  \begin{equation}\label{sl}
- a= \lim_{|x|\to \infty} |x| \phi(x).
+ a= \lim_{|x|\to \infty} |x| \varphi(x).
  \end{equation}
 For more information on the scattering length, see\-~\cite[appendix A]{LY01}.
 
@@ -707,7 +718,7 @@ ${\displaystyle \frac{1}{1+r} \leqslant \frac{|x|}{|x-y|} \leqslant \frac{1}{1-r
 It follows that for all sufficiently large $|x|$, 
 \begin{equation}
 \frac{1}{1+r}\int_{|y|< r|x|}\mathcal{V}(y)(1-\varphi(y)) dy+ o(1) \leqslant 4\pi|x|\varphi(x) \leqslant 
- \frac{1}{1-r}\int_{|y|< r|x|}\mathcal{V}(y)(1-\varphi(y) dx + o(1)\ .
+ \frac{1}{1-r}\int_{|y|< r|x|}\mathcal{V}(y)(1-\varphi(y)) dx + o(1)\ .
 \end{equation}
 Taking $|x|\to \infty$, and then $r \to 0$ proves\-~(\ref{local}).\qed
 \bigskip
@@ -919,7 +930,7 @@ This is the case if $u$ decays algebraically, but would not be so if, say, it de
 \bigskip
 
 {\bf Proof of theorem\-~\ref{theorem:decay}}:
-We begin by proving (\ref{gendecay}) in arbitrary dimension. Recall that the first part has already been proved in Theorem~\ref{positivity} without the additional assumption on the potential. For the second part, recall that by the first remark after Theorem~\ref{theorem:existence},  $u$ is also radial, and hence $\mathcal{V}(1-u)$ is non-negative and radial. It then follows from the hypotheses on $\mathcal{V}$ that $g := Y_{4e}\ast Y_{4e}*[\mathcal{V}(1-u)]$
+We begin by proving (\ref{gendecay}) in arbitrary dimension. Recall that the first part has already been proved in Theorem~\ref{positivity} without the additional assumption on the potential. For the second part, recall that by the first remark after Theorem~\ref{theorem:existence},  $u$ is also radial, and hence $\mathcal{V}(1-u)$ is non-negative and radial. It then follows from the hypotheses on $\mathcal{V}$ that $g :=2\rho eY_{4e}\ast Y_{4e}*[\mathcal{V}(1-u)]$
 satisfies
 \begin{equation}
   \int |x|^2 g(x) dx < \infty \quad{\rm and}\quad \int x g(x) d x = 0\ .
@@ -955,6 +966,7 @@ Recall that the Fourier transform of $u$ (\ref{fourieru}) satisfies\-~(\ref{hatu
   so that, taking the large $|k|$ limit in\-~(\ref{hatu}),
   \begin{equation}
     \widehat{\mathcal U}_2(|k|)=O(|k|^{-4}S^2(|k|))
+    \label{apriori_U2}
   \end{equation}
   so $\widehat{\mathcal U}_2$ is integrable.
   \bigskip
@@ -1265,7 +1277,7 @@ Recall that the Fourier transform of $u$ (\ref{fourieru}) satisfies\-~(\ref{hatu
     \leqslant 2
   \end{equation}
   provided $|x|^\tau>2C$.
-  Therefore, for large $\kappa$,
+  Therefore, for large $\kappa$, by\-~(\ref{apriori_U2}),
   \begin{equation}
     |\widehat{\mathcal U}_2(\kappa+i\eta)|=O(\kappa^{-4})
   \end{equation}
@@ -1281,7 +1293,7 @@ Recall that the Fourier transform of $u$ (\ref{fourieru}) satisfies\-~(\ref{hatu
   \begin{equation}
     \mathcal U_2(|x|)=\frac1{\pi^2\rho|x|^4}\sqrt{\frac1{2e}+\beta}+O(|x|^{-5})
   \end{equation}
-  which, using\-~(\ref{U1}), concludes the proof of the lemma.
+  which, using\-~(\ref{U1}), concludes the proof of the theorem.
   \bigskip
 \qed
 
@@ -1356,7 +1368,7 @@ for some $x$.
 
 \subsection{Numerical comparison}\label{subsec:numerics}
 \indent One of the motivations for studying the simple equation is that it provides a simple tool to approximate the ground state energy of the Bose gas.
-In\-~\cite{LL64}, it was found that in one dimension the simple equation gives a value for the energy that differs from the Bose gas ground state energy by at most 69\% (a more complete form of the equation yields even better, result, with a maximal error of 19\%).
+In\-~\cite{LL64}, it was found that in one dimension the simple equation gives a value for the energy that differs from the Bose gas ground state energy by at most 69\% (a more complete form of the equation yields an even better result, with a maximal error of 19\%).
 In one dimension, the difference is larger at high density.
 \bigskip
 
@@ -1397,7 +1409,7 @@ Using a modified iteration in which $\rho$ is fixed, we have proved that $e\rho(
 \bigskip
 
 \point{\bf Convexity}.
-Another open problem is to prove that $\rho e(\rho)$ {\it is a convex function}, or, equivalently, that $\frac1{\rho(e)}$ is concave.
+Another open problem is to prove that $\rho e(\rho)$ {\it is a convex function}, or, equivalently, that $\frac1{\rho(e)}$ is convex.
 In a physical setting, one expects $\rho e(\rho)$ to be convex.
 Indeed if $\rho e=:e_{\mathrm v}$ were {\it not convex}, there would exist $\rho_1<\rho<\rho_2$ such that $\frac{\rho_1+\rho_2}2=\rho$ and $e_{\mathrm v}(\rho_1)+e_{\mathrm v}(\rho_2)<2e_{\mathrm v}(\rho)$.
 Furthermore, $e_{\mathrm v}$ is the energy per unit volume, and, considering a volume $V$ that is split into two equal halves, we find that a configuration in which one half of the volume holds a density $\rho_1$ of particles, whereas the other holds $\rho_2$ would have energy
@@ -1443,8 +1455,8 @@ with
   \label{L}
 \end{equation}
 Note that $e$ appears only as the integral of $S$, see\-~(\ref{energy}).
-Little is known about the solutions of this equation: even proving the existence of a solution of\-~(\ref{fulleq}) is open.
+While little is known rigorously about this equation, we have been studying it numerically in collaboration with M. Holzmann \cite{CHe}, and have found it to be remarkably accurate.
-We hope to study this equation numerically in a later publication.
+These results will be detailed in a future publication.
 
 \vfill
 

Changelog

@@ -1,3 +1,12 @@
+v1.0:
+
+  * Fixed: Missing prefactor in the expression of f-f*f
+
+  * Fixed: 1/rho is convex, not concave.
+
+  * Fixed: Miscellaneous typos and small changes to the formatting.
+
+
 v0.1:
 
   * Added: Theorem on positivity of solutions.