From cff1d2ee3c7730b672f239de5eb1aeb1a0bfd1db Mon Sep 17 00:00:00 2001
From: Ian Jauslin <jauslin@ias.edu>
Date: Fri, 12 Jan 2018 19:20:59 +0000
Subject: Simpler expression for fft term

---
 docs/nstrophy_doc/nstrophy_doc.tex | 45 ++++++++++++++++----------------------
 1 file changed, 19 insertions(+), 26 deletions(-)

(limited to 'docs')

diff --git a/docs/nstrophy_doc/nstrophy_doc.tex b/docs/nstrophy_doc/nstrophy_doc.tex
index fa5fff0..914b0b8 100644
--- a/docs/nstrophy_doc/nstrophy_doc.tex
+++ b/docs/nstrophy_doc/nstrophy_doc.tex
@@ -49,16 +49,23 @@ in terms of which, multiplying both sides of the equation by $\frac{k^\perp}{|k|
 \begin{equation}
   \partial_t\hat \varphi_k=
   -4\pi^2\nu k^2\hat \varphi_k+\frac{k^\perp}{2i\pi|k|}\cdot\hat g_k
-  -\frac{2i\pi}{|k|}\sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}}
+  +\frac{4\pi^2}{|k|}\sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}}
   (q\cdot p^\perp)(k^\perp\cdot q^\perp)\hat\varphi_p\hat\varphi_q
+  .
   \label{ins_k}
 \end{equation}
-which, since $q\cdot p^\perp=q\cdot(k^\perp-q^\perp)=q\cdot k^\perp$, is
+Furthermore
+\begin{equation}
+  (q\cdot p^\perp)(k^\perp\cdot q^\perp)
+  =
+  (q\cdot p^\perp)(q^2+p\cdot q)
+\end{equation}
+and $q\cdot p^\perp$ is antisymmetric under exchange of $q$ and $p$. Therefore,
 \begin{equation}
   \partial_t\hat \varphi_k=
   -4\pi^2\nu k^2\hat \varphi_k+\frac{k^\perp}{2i\pi|k|}\cdot\hat g_k
   +\frac{4\pi^2}{|k|}\sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}}
-  (q\cdot k^\perp)(k\cdot q)\frac{\hat\varphi_p\hat\varphi_q}{|p||q|}
+  \frac{(q\cdot p^\perp)|q|}{|p|}\hat\varphi_p\hat\varphi_q
   .
   \label{ins_k}
 \end{equation}
@@ -73,8 +80,7 @@ We truncate the Fourier modes and assume that $\hat\varphi_k=0$ if $|k_1|>K_1$ o
 \begin{equation}
   T(\hat\varphi,k):=
   \sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}}
-  \frac{\hat\varphi_p}{|p|}
-  (q\cdot k^\perp)(k\cdot q)\frac{\hat\varphi_q}{|q|}
+  \frac{(q\cdot p^\perp)|q|}{|p|}\hat\varphi_q\hat\varphi_p
 \end{equation}
 using a fast Fourier transform, defined as
 \begin{equation}
@@ -98,40 +104,27 @@ in which $\mathcal F^*$ is defined like $\mathcal F$ but with the opposite phase
   \sum_{p,q\in\mathcal K}
   \frac1{N_1N_2}
   \sum_{n\in\mathcal N}e^{-\frac{2i\pi}{N_1}n_1(p_1+q_1-k_1)-\frac{2i\pi}{N_2}n_2(p_2+q_2-k_2)}
-  \frac{\hat\varphi_p}{|p|}
-  (q\cdot k^\perp)(k\cdot q)\frac{\hat\varphi_q}{|q|}
+  (q\cdot p^\perp)\frac{|q|}{|p|}\hat\varphi_q\hat\varphi_p
 \end{equation}
 provided
 \begin{equation}
   N_i>4K_i.
 \end{equation}
 Indeed, $\sum_{n_i=0}^{N_i}e^{-\frac{2i\pi}{N_i}n_im_i}$ vanishes unless $m_i=0\%N_i$ (in which $\%N_i$ means `modulo $N_i$'), and, if $p,q\in\mathcal K$, then $|p_i+q_i|\leqslant2K_i$. Therefore,
-\begin{equation}
-  T(\hat\varphi,k)
-  =
-  \frac1{N_1N_2}
-  \mathcal F^*\left(
-    \mathcal F(|p|^{-1}\hat\varphi_p)(n)
-    \mathcal F((|q|^{-1}(q\cdot k^\perp)(k\cdot q)\hat\varphi_q)_q)(n)
-  \right)(k)
-\end{equation}
-which we expand
 \begin{equation}
   T(\hat\varphi,k)
   =
   \textstyle
-  \frac{k_x^2-k_y^2}{N_1N_2}
-  \mathcal F^*\left(
-    \mathcal F\left(\frac{\hat\varphi_p}{|p|}\right)(n)
-    \mathcal F\left(\frac{q_xq_y}{|q|}\hat\varphi_q\right)(n)
-  \right)(k)
-  -
-  \frac{k_xk_y}{N_1N_2}
+  \frac1{N_1N_2}
   \mathcal F^*\left(
-    \mathcal F\left(\frac{\hat\varphi_p}{|p|}\right)(n)
-    \mathcal F\left(\frac{q_x^2-q_y^2}{|q|}\hat\varphi_q\right)(n)
+    \mathcal F\left(\frac{p_x\hat\varphi_p}{|p|}\right)(n)
+    \mathcal F\left(q_y|q|\hat\varphi_q\right)(n)
+    -
+    \mathcal F\left(\frac{p_y\hat\varphi_p}{|p|}\right)(n)
+    \mathcal F\left(q_x|q|\hat\varphi_q\right)(n)
   \right)(k)
 \end{equation}
+\bigskip
 
 
 \vfill
-- 
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