Reality in doc

This commit is contained in:
Ian Jauslin 2023-03-31 16:58:52 -04:00
parent 2958cc0313
commit 75de7e03b7

View File

@ -71,10 +71,16 @@ and $q\cdot p^\perp$ is antisymmetric under exchange of $q$ and $p$. Therefore,
\begin{equation} \begin{equation}
\partial_t\hat u_k= \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}\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 \frac{(q\cdot p^\perp)|q|}{|p|}\hat u_p\hat u_q
. .
\label{ins_k} \label{T}
\end{equation} \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 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} \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} \end{equation}
\bigskip \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} \begin{equation}
T(\hat u,k):= \hat u_{-k}=\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{realu}
\label{T}
\end{equation} \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} \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) \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} \end{equation}
@ -279,20 +301,25 @@ To compute $\alpha$, we use the constancy of the enstrophy:
\end{equation} \end{equation}
which, in terms of $\hat u$ is which, in terms of $\hat u$ is
\begin{equation} \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 =0
\end{equation} \end{equation}
that is that is
\begin{equation} \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 \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) +\frac{4\pi^2}{L^2}\sum_{k\in\mathbb Z^2}|k|\hat u_k^*T(\hat u,k)
\end{equation} \end{equation}
and so and so
\begin{equation} \begin{equation}
\alpha(\hat u) \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} \end{equation}