From 75de7e03b7b73cb45ce611368a023841e530a219 Mon Sep 17 00:00:00 2001
From: Ian Jauslin <ian.jauslin@rutgers.edu>
Date: Fri, 31 Mar 2023 16:58:52 -0400
Subject: Reality in doc

---
 docs/nstrophy_doc/nstrophy_doc.tex | 53 ++++++++++++++++++++++++++++----------
 1 file changed, 40 insertions(+), 13 deletions(-)

diff --git a/docs/nstrophy_doc/nstrophy_doc.tex b/docs/nstrophy_doc/nstrophy_doc.tex
index ae571ef..54e802e 100644
--- a/docs/nstrophy_doc/nstrophy_doc.tex
+++ b/docs/nstrophy_doc/nstrophy_doc.tex
@@ -71,10 +71,16 @@ and $q\cdot p^\perp$ is antisymmetric under exchange of $q$ and $p$. Therefore,
 \begin{equation}
   \partial_t\hat u_k=
   -\frac{4\pi^2}{L^2}\nu k^2\hat u_k+\hat g_k
-  +\frac{4\pi^2}{L^2|k|}\sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}}
+  +\frac{4\pi^2}{L^2|k|}T(\hat u,k)
+  \label{ins_k}
+\end{equation}
+with
+\begin{equation}
+  T(\hat u,k):=
+  \sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}}
   \frac{(q\cdot p^\perp)|q|}{|p|}\hat u_p\hat u_q
   .
-  \label{ins_k}
+  \label{T}
 \end{equation}
 We truncate the Fourier modes and assume that $\hat u_k=0$ if $|k_1|>K_1$ or $|k_2|>K_2$. Let
 \begin{equation}
@@ -83,14 +89,30 @@ We truncate the Fourier modes and assume that $\hat u_k=0$ if $|k_1|>K_1$ or $|k
 \end{equation}
 \bigskip
 
-\point{\bf FFT}. We compute the last term in~\-(\ref{ins_k})
+\point{\bf Reality}.
+Since $U$ is real, $\hat U_{-k}=\hat U_k^*$, and so
 \begin{equation}
-  T(\hat u,k):=
-  \sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}}
-  \frac{(q\cdot p^\perp)|q|}{|p|}\hat u_q\hat u_p
-  \label{T}
+  \hat u_{-k}=\hat u_k^*
+  .
+  \label{realu}
 \end{equation}
-using a fast Fourier transform, defined as
+Similarly,
+\begin{equation}
+  \hat g_{-k}=\hat g_k^*
+  .
+  \label{realg}
+\end{equation}
+Thus,
+\begin{equation}
+  T(\hat u,-k)
+  =
+  T(\hat u,k)^*
+  .
+  \label{realT}
+\end{equation}
+\bigskip
+
+\point{\bf FFT}. We compute T using a fast Fourier transform, defined as
 \begin{equation}
   \mathcal F(f)(n):=\sum_{m\in\mathcal N}e^{-\frac{2i\pi}{N_1}m_1n_1-\frac{2i\pi}{N_2}m_2n_2}f(m_1,m_2)
 \end{equation}
@@ -279,20 +301,25 @@ To compute $\alpha$, we use the constancy of the enstrophy:
 \end{equation}
 which, in terms of $\hat u$ is
 \begin{equation}
-  \sum_{k\in\mathbb Z^2}k^2\hat u_k\partial_t\hat u_k
+  \sum_{k\in\mathbb Z^2}k^2\hat u_k^*\partial_t\hat u_k
   =0
 \end{equation}
 that is
 \begin{equation}
-  \frac{4\pi^2}{L^2}\alpha(\hat u)\sum_{k\in\mathbb Z^2}k^4\hat u_k^2
+  \frac{4\pi^2}{L^2}\alpha(\hat u)\sum_{k\in\mathbb Z^2}k^4|\hat u_k|^2
   =
-  \sum_{k\in\mathbb Z^2}k^2\hat u_k\hat g_k
-  +\frac{4\pi^2}{L^2}\sum_{k\in\mathbb Z^2}|k|\hat u_kT(\hat u,k)
+  \sum_{k\in\mathbb Z^2}k^2\hat u_k^*\hat g_k
+  +\frac{4\pi^2}{L^2}\sum_{k\in\mathbb Z^2}|k|\hat u_k^*T(\hat u,k)
 \end{equation}
 and so
 \begin{equation}
   \alpha(\hat u)
-  =\frac{\frac{L^2}{4\pi^2}\sum_k k^2\hat u_k\hat g_k+\sum_k|k|\hat u_kT(\hat u,k)}{\sum_kk^4\hat u_k^2}
+  =\frac{\frac{L^2}{4\pi^2}\sum_k k^2\hat u_k^*\hat g_k+\sum_k|k|\hat u_k^*T(\hat u,k)}{\sum_kk^4|\hat u_k|^2}
+  .
+\end{equation}
+Note that, by\-~(\ref{realu})-(\ref{realT}),
+\begin{equation}
+  \alpha(\hat u)\in\mathbb R
   .
 \end{equation}
 
-- 
cgit v1.2.3-54-g00ecf