From 9b3632d5a00f19aa9fdeef6e64e68af9fa0dfd27 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Marek=20Ne=C4=8Dada?= Date: Mon, 10 Sep 2018 20:01:49 +0000 Subject: [PATCH] Sum upper estimates Former-commit-id: 8644505403873e1db6c01f106e5d889da8981c62 --- notes/ewald.lyx | 124 ++++++++++++++++++++++++++++++++++++++++++++++-- 1 file changed, 119 insertions(+), 5 deletions(-) diff --git a/notes/ewald.lyx b/notes/ewald.lyx index 126cd3c..e80afa8 100644 --- a/notes/ewald.lyx +++ b/notes/ewald.lyx @@ -3144,6 +3144,120 @@ To have it complete, \end_layout +\begin_layout Subsection +Error estimates +\end_layout + +\begin_layout Standard +For the part of a 2D lattice sum that lies outside of a circle with radius + +\begin_inset Formula $R$ +\end_inset + + and +\begin_inset Formula $f(r)$ +\end_inset + + positive, radial, monotonically decreasing, we have +\end_layout + +\begin_layout Standard +\begin_inset Formula +\begin{equation} +\mathscr{A}_{\Lambda}\sum_{\begin{array}{c} +\vect R_{i}\in\Lambda\\ +\left|\vect R_{i}\right|\ge R +\end{array}}f\left(\left|\vect R_{i}\right|\right)\le2\pi\underbrace{\int_{R_{\mathrm{s}}\left(R,\Lambda\right)}^{\infty}rf(r)\,\ud r}_{\equiv B_{R_{\mathrm{s}}}\left[f\right]},\label{eq:lsum_bound} +\end{equation} + +\end_inset + +where the largest +\begin_inset Quotes eld +\end_inset + +safe radius +\begin_inset Quotes erd +\end_inset + + +\begin_inset Formula $R_{\mathrm{s}}\left(R,\Lambda\right)$ +\end_inset + + is probably something like +\begin_inset Formula $R-\left|\vect u_{\mathrm{L}}\right|$ +\end_inset + + where +\begin_inset Formula $\vect u_{\mathrm{L}}$ +\end_inset + + is the longer primitive lattice vector of +\begin_inset Formula $\Lambda$ +\end_inset + +. +\end_layout + +\begin_layout Standard +For the short-range part +\begin_inset Formula $\sigma_{n}^{m(2)}$ +\end_inset + +, the radially varying part reads +\begin_inset Formula $f_{\eta}^{\mathrm{L}}\left(R_{pq}\right)\equiv R_{pq}^{n}\int_{\eta}^{\infty}e^{-R_{pq}^{2}\xi^{2}}e^{k^{2}/4\xi^{2}}\xi^{2n}\ud\xi$ +\end_inset + + and for its integral as in +\begin_inset CommandInset ref +LatexCommand ref +reference "eq:lsum_bound" + +\end_inset + + we have +\begin_inset Formula +\begin{eqnarray*} +B_{R_{\mathrm{s}}}\left[f_{\eta}^{\mathrm{L}}\right] & = & \int_{R_{\mathrm{s}}}^{\infty}r^{n+1}\int_{\eta}^{\infty}e^{-r^{2}\xi^{2}}e^{k^{2}/4\xi^{2}}\xi^{2n}\ud\xi\,\ud r\\ + & \le & e^{k^{2}/4\eta^{2}}\int_{R_{\mathrm{s}}}^{\infty}\int_{\eta}^{\infty}r^{n+1}e^{-r^{2}\xi^{2}}\xi^{2n}\ud\xi\,\ud r\\ + & = & e^{k^{2}/4\eta^{2}}\frac{\eta^{2n+1}R_{\mathrm{s}}^{2+n}\left(E_{\frac{1}{2}-n}\left(\eta^{2}R_{\mathrm{s}}^{2}\right)-E_{-\frac{n}{2}}\left(\eta^{2}R_{\mathrm{s}}^{2}\right)\right)}{2\left(n-1\right)}\\ + & = & e^{k^{2}/4\eta^{2}}\frac{\eta^{2n+1}R_{\mathrm{s}}^{2+n}\left(\left(\eta R_{\mathrm{s}}\right)^{-2n-1}\Gamma\left(n+\frac{1}{2},\eta^{2}R_{\mathrm{s}}^{2}\right)-\left(\eta R_{\mathrm{s}}\right)^{-n-2}\Gamma\left(\frac{n}{2}+1,\eta^{2}R_{\mathrm{s}}^{2}\right)\right)}{2\left(n-1\right)}\\ + & = & \frac{e^{k^{2}/4\eta^{2}}}{2\left(n-1\right)}\left(R_{\mathrm{s}}^{1-n}\Gamma\left(n+\frac{1}{2},\eta^{2}R_{\mathrm{s}}^{2}\right)-\eta^{n-1}\Gamma\left(\frac{n}{2}+1,\eta^{2}R_{\mathrm{s}}^{2}\right)\right), +\end{eqnarray*} + +\end_inset + +where the integral is according to mathematica and the error functions were + transformed to incomplete gammas using the relation +\begin_inset Formula $\Gamma\left(s,x\right)=x^{s}E_{1-s}\left(x\right)$ +\end_inset + + from Wikipedia or equivalently +\begin_inset Formula $\Gamma\left(1-n,z\right)=z^{1-n}E_{n}\left(z\right)$ +\end_inset + + from [DLMF(8.4.13)]. + Therefore, the upper estimate for the short-range sum error is +\begin_inset Formula +\begin{eqnarray*} +\left|\sigma_{n}^{m(2)}|_{R_{pq}>R}\right| & \le & \frac{2^{n+1}}{k^{n+1}\sqrt{\pi}}\left|P_{n}^{m}\left(0\right)\right|\frac{2\pi}{\mathscr{A}_{\Lambda}}\frac{e^{k^{2}/4\eta^{2}}}{2\left(n-1\right)}\left(R_{\mathrm{s}}^{1-n}\Gamma\left(n+\frac{1}{2},\eta^{2}R_{\mathrm{s}}^{2}\right)-\eta^{n-1}\Gamma\left(\frac{n}{2}+1,\eta^{2}R_{\mathrm{s}}^{2}\right)\right)\\ + & = & \frac{2^{n+1}}{k^{n+1}}\left|P_{n}^{m}\left(0\right)\right|\frac{\sqrt{\pi}}{\mathscr{A}_{\Lambda}}\frac{e^{k^{2}/4\eta^{2}}}{n-1}\left(R_{\mathrm{s}}^{1-n}\Gamma\left(n+\frac{1}{2},\eta^{2}R_{\mathrm{s}}^{2}\right)-\eta^{n-1}\Gamma\left(\frac{n}{2}+1,\eta^{2}R_{\mathrm{s}}^{2}\right)\right). +\end{eqnarray*} + +\end_inset + +Apparently, this expression is problematic for +\begin_inset Formula $n=1$ +\end_inset + +; Mathematica gives for that case some ugly expression with +\begin_inset Formula $_{2}F_{2}$ +\end_inset + +. + Hence it might make sense to take a rougher estimate TODO +\end_layout + \begin_layout Section Major TODOs and open questions \end_layout @@ -3228,7 +3342,7 @@ where the spherical Hankel transform 2) \begin_inset Formula \[ -\bsht lg(k)\equiv\int_{0}^{\infty}\ud r\,r^{2}g(r)j_{l}\left(kr\right). +\bsht lg(k)\equiv\int_{0}^{\infty}\ud r\, r^{2}g(r)j_{l}\left(kr\right). \] \end_inset @@ -3238,7 +3352,7 @@ Using this convention, the inverse spherical Hankel transform is given by 3) \begin_inset Formula \[ -g(r)=\frac{2}{\pi}\int_{0}^{\infty}\ud k\,k^{2}\bsht lg(k)j_{l}(k), +g(r)=\frac{2}{\pi}\int_{0}^{\infty}\ud k\, k^{2}\bsht lg(k)j_{l}(k), \] \end_inset @@ -3251,7 +3365,7 @@ so it is not unitary. An unitary convention would look like this: \begin_inset Formula \begin{equation} -\usht lg(k)\equiv\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}\ud r\,r^{2}g(r)j_{l}\left(kr\right).\label{eq:unitary 3d Hankel tf definition} +\usht lg(k)\equiv\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}\ud r\, r^{2}g(r)j_{l}\left(kr\right).\label{eq:unitary 3d Hankel tf definition} \end{equation} \end_inset @@ -3305,8 +3419,8 @@ where the Hankel transform of order is defined as \begin_inset Formula \begin{eqnarray} -\pht mg\left(k\right) & = & \int_{0}^{\infty}\ud r\,g(r)J_{m}(kr)r\label{eq:unitary 2d Hankel tf definition}\\ - & = & \left(-1\right)^{m}\int_{0}^{\infty}\ud r\,g(r)J_{-m}(kr)r +\pht mg\left(k\right) & = & \int_{0}^{\infty}\ud r\, g(r)J_{m}(kr)r\label{eq:unitary 2d Hankel tf definition}\\ + & = & \left(-1\right)^{m}\int_{0}^{\infty}\ud r\, g(r)J_{-m}(kr)r \end{eqnarray} \end_inset