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.
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@ -40,7 +40,7 @@
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{\small
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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.
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This equation is mathematically interesting for several reasons.
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(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.
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(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.
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(2) The novel nonlinearity is easy to state but seems to have almost no literature up to now.
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(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.
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}
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@ -117,7 +117,7 @@ Thus, any physical solutions of (\ref{simpleq})-(\ref{energy}) must necessarily
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\endtheo
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\bigskip
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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 $L^1(\mathbb R^d)$.
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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 $L^1(\mathbb R^d)$.
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While $u(x)\leqslant 1$ is a physical requirement, $u(x)\geqslant0$ is not, see section\-~\ref{sec:bosegas} for details.
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\bigskip
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@ -160,7 +160,7 @@ When $d=3$,
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\end{equation}
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Equation\-~(\ref{simpleq}) can be rewritten as
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\begin{equation}\label{simpleq3}
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u(x) = Y_{4e}*(\mathcal V (1- u(x)) +2e\rho Y_{4e}*u*u \ .
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u(x) = Y_{4e}*(\mathcal V (1- u(x))) +2e\rho Y_{4e}*u*u \ .
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\end{equation}
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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
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\begin{equation}
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@ -247,7 +247,7 @@ Then there is a constructively defined continuous function $\rho(e)$ on $(0,\inf
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{\bf Remarks}:
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\begin{itemize}
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\item We do not assume here that the potential is radially symmetric.
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However, we shall see from our proof that if $\mathcal V$ is radially symmetric, then so is $u$.
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However, the uniqueness statement implies that $u$ is radially symmetric whenever $\mathcal V$ is radially symmetric.
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\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
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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
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@ -348,16 +348,16 @@ Let $u(x)$ be an integrable solution of the system (\ref{simpleq})-(\ref{energy}
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{\bf Proof} Since $Y_{4e}*(\mathcal V (1- u(x)) \geqslant 0$, it follows that
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{\bf Proof} Since $Y_{4e}*(\mathcal V (1- u(x))) \geqslant 0$, it follows that
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\begin{equation}
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u_- \leqslant (f*u)_- = (f*u_+ - f*u_-)_- \leqslant f*u_-\ .
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\end{equation}
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Integrating, we find
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${\displaystyle \int u_- {\rm d}x \leqslant \frac12 \int u_-{\rm d}x}$,
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${\displaystyle \int u_- dx \leqslant \frac12 \int u_-dx}$,
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and this implies that $u_- = 0$. \qed
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{\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
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{\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
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\begin{equation}\label{pos5}
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f(x) \geqslant f\ast f(x)
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\end{equation}
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@ -402,7 +402,7 @@ Comparing\-~(\ref{iteration}) to\-~(\ref{iteration0}), note that the analog of $
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\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 $n$,
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\begin{equation}\label{simple17}
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\int_{\mathbb{R}^d} u_n{\rm d}x < \frac{1}{2e}\int_{\mathbb{R}^d}\mathcal{V}(1-u_n)dx \ .
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\int_{\mathbb{R}^d} u_n dx < \frac{1}{2e}\int_{\mathbb{R}^d}\mathcal{V}(1-u_n)dx \ .
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\end{equation}
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\endtheo
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\bigskip
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@ -418,23 +418,30 @@ Since $t\mapsto t^{-1}$ is monotone decreasing on $(0,\infty)$, this shows that
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\bigskip
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\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
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\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
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\begin{equation}
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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)\ ,
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\end{equation}
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and then
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\begin{equation}
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\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 \ .
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\int_{\mathbb{R}^d} \mathcal V(1-u_{n+1}) dx < \int_{\mathbb{R}^d}\mathcal V(1-u_{n}) dx \ .
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\end{equation}
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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,
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\begin{equation}\label{simple15}
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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
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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
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\end{equation}
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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
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Then since
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\begin{equation}
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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\ .
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int_{\mathbb{R}^d} u_{n} dx < \frac{1}{2e}\int_{\mathbb{R}^d}\mathcal V(1-u_{n}) = \frac{1}{\rho_n}
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\end{equation}
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(\ref{simple15}) implies
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\begin{equation}
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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\ .
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\end{equation}
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Then because $\int_{\mathbb{R}^d} u_{n} dx < \int_{\mathbb{R}^d} u_{n+1} dx $, we have
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\begin{equation}
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\int_{\mathbb{R}^d} u_{n+1} dx < \frac{1}{2e}\int_{v}\mathcal{V}(1-u_{n+1}).
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\end{equation}
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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}) $.
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This proves\-~(\ref{simple17}) for $n+1$, and shows that
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\begin{equation}
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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 \ ,
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@ -454,7 +461,11 @@ and then, as before, $\rho_{n+1}\geqslant \rho_n$.
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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 $n=1$.
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\bigskip
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\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
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\indent Now make the inductive hypothesis that $0 \leqslant u_n(x) \leqslant 1$ for all $x$. Then
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\begin{equation}
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\|u_n\|_p^p \leqslant \|u_n\|_1 \leqslant\frac{1}{2e}\int_{\mathbb{R}^d}\mathcal{V}dx.
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\end{equation}
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By Young's inequality, $\|u_n\ast u_n\|_p \leqslant \|u_n\|_p\|u_1\|_1$, and hence
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$\mathcal{V} + 2e\rho_n u_n\ast u_n \in L^p(\mathbb{R}^d)$. Therefore,
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$u_{n+1}= K_e(\mathcal{V} + 2e\rho_n u_n\ast u_n) \in W^{2,p}(\mathbb{R}^d)$. It follows as before that $u_{n+1}$ is continuous and vanishing at infinity, and in particular, bounded, and
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\begin{eqnarray*}
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@ -541,7 +552,7 @@ By remarks made above, this means that $u$ satisfies\-~(\ref{simpleq}) and\-~(\r
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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 $n$.
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Writing our iteration in the equivalent form (as in\-~(\ref{simpleq3})):
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\begin{equation}\label{simpleq3B}
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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) \ ,
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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) \ ,
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\end{equation}
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and integrating, we obtain
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\begin{equation}\label{simpleq3C}
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@ -571,7 +582,7 @@ Now an easy induction shows that $a_n(e)$ is continuous for each $n$. By\-~(\ref
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(\ref{con4B}). \qed
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\bigskip
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{\bf Remark}: Note that $\|u - u_n\|_1 = \frac{1}{\rho} - a_n$, and hence by By\-~(\ref{rate}), $\|u - u_n\|_1 \leqslant Cn^{-1/2}$.
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{\bf Remark}: Note that $\|u - u_n\|_1 = \frac{1}{\rho} - a_n$, and hence by\-~(\ref{rate}), $\|u - u_n\|_1 \leqslant Cn^{-1/2}$.
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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 $n^{-3.5}$.
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\bigskip
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@ -582,8 +593,8 @@ In fact, numerically, we find that the rate is significantly faster than this. F
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\bigskip
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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}.
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This justifies the first remark following Theorem\-~\ref{theorem:existence}.
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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}.
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This is consistent with the first remark following Theorem\-~\ref{theorem:existence}.
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\section{Asymptotics}\label{sec:asymptotics}
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\indent In this section, we prove Theorem\-~\ref{theorem:asymptotics}.
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@ -678,10 +689,10 @@ $R>0$ define $\mathcal{V}_R(x) = \mathcal{V}(x)$ for $|x| > R$ and otherwise $\m
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\begin{equation}\label{local}
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\|\mathcal{V}_R\|_1 < R^{-q} \quad{\rm and}\quad \|\mathcal{V}_R \|_p < R^{-q} \ .
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\end{equation}
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By the lemma below, $\lim_{x|\to\infty}|x|\varphi(x)$ exists. The {\em scattering length} $a$ is
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By the lemma below, $\lim_{|x|\to\infty}|x|\varphi(x)$ exists. The {\em scattering length} $a$ is
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defined to be (in dimension $d=3$).
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\begin{equation}\label{sl}
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a= \lim_{|x|\to \infty} |x| \phi(x).
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a= \lim_{|x|\to \infty} |x| \varphi(x).
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\end{equation}
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For more information on the scattering length, see\-~\cite[appendix A]{LY01}.
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@ -707,7 +718,7 @@ ${\displaystyle \frac{1}{1+r} \leqslant \frac{|x|}{|x-y|} \leqslant \frac{1}{1-r
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It follows that for all sufficiently large $|x|$,
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\begin{equation}
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\frac{1}{1+r}\int_{|y|< r|x|}\mathcal{V}(y)(1-\varphi(y)) dy+ o(1) \leqslant 4\pi|x|\varphi(x) \leqslant
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\frac{1}{1-r}\int_{|y|< r|x|}\mathcal{V}(y)(1-\varphi(y) dx + o(1)\ .
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\frac{1}{1-r}\int_{|y|< r|x|}\mathcal{V}(y)(1-\varphi(y)) dx + o(1)\ .
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\end{equation}
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Taking $|x|\to \infty$, and then $r \to 0$ proves\-~(\ref{local}).\qed
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\bigskip
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@ -919,7 +930,7 @@ This is the case if $u$ decays algebraically, but would not be so if, say, it de
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\bigskip
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{\bf Proof of theorem\-~\ref{theorem:decay}}:
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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 $g := Y_{4e}\ast Y_{4e}*[\mathcal{V}(1-u)]$
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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 $g :=2\rho eY_{4e}\ast Y_{4e}*[\mathcal{V}(1-u)]$
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satisfies
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\begin{equation}
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\int |x|^2 g(x) dx < \infty \quad{\rm and}\quad \int x g(x) d x = 0\ .
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@ -955,6 +966,7 @@ Recall that the Fourier transform of $u$ (\ref{fourieru}) satisfies\-~(\ref{hatu
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so that, taking the large $|k|$ limit in\-~(\ref{hatu}),
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\begin{equation}
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\widehat{\mathcal U}_2(|k|)=O(|k|^{-4}S^2(|k|))
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\label{apriori_U2}
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\end{equation}
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so $\widehat{\mathcal U}_2$ is integrable.
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\bigskip
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@ -1265,7 +1277,7 @@ Recall that the Fourier transform of $u$ (\ref{fourieru}) satisfies\-~(\ref{hatu
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\leqslant 2
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\end{equation}
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provided $|x|^\tau>2C$.
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Therefore, for large $\kappa$,
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Therefore, for large $\kappa$, by\-~(\ref{apriori_U2}),
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\begin{equation}
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|\widehat{\mathcal U}_2(\kappa+i\eta)|=O(\kappa^{-4})
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\end{equation}
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@ -1281,7 +1293,7 @@ Recall that the Fourier transform of $u$ (\ref{fourieru}) satisfies\-~(\ref{hatu
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\begin{equation}
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\mathcal U_2(|x|)=\frac1{\pi^2\rho|x|^4}\sqrt{\frac1{2e}+\beta}+O(|x|^{-5})
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\end{equation}
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which, using\-~(\ref{U1}), concludes the proof of the lemma.
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which, using\-~(\ref{U1}), concludes the proof of the theorem.
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\bigskip
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\qed
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@ -1356,7 +1368,7 @@ for some $x$.
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\subsection{Numerical comparison}\label{subsec:numerics}
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\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.
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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\%).
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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\%).
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In one dimension, the difference is larger at high density.
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\bigskip
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@ -1397,7 +1409,7 @@ Using a modified iteration in which $\rho$ is fixed, we have proved that $e\rho(
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\bigskip
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\point{\bf Convexity}.
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Another open problem is to prove that $\rho e(\rho)$ {\it is a convex function}, or, equivalently, that $\frac1{\rho(e)}$ is concave.
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Another open problem is to prove that $\rho e(\rho)$ {\it is a convex function}, or, equivalently, that $\frac1{\rho(e)}$ is convex.
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In a physical setting, one expects $\rho e(\rho)$ to be convex.
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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 $e_{\mathrm v}(\rho_1)+e_{\mathrm v}(\rho_2)<2e_{\mathrm v}(\rho)$.
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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
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@ -1443,8 +1455,8 @@ with
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\label{L}
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\end{equation}
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Note that $e$ appears only as the integral of $S$, see\-~(\ref{energy}).
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Little is known about the solutions of this equation: even proving the existence of a solution of\-~(\ref{fulleq}) is open.
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We hope to study this equation numerically in a later publication.
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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.
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These results will be detailed in a future publication.
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\vfill
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