%
% deepconv/arare5
% = エネルギー方程式の導出 (凝結なし, 北守さん修論の付録のトレース)
%
% 2011/12/19  追記
% 2011/11/30  追記
% 2011/11/29  修正 and 追記
% 2011/11/28  追記
% 2011/11/26  追記 ((A.9) の変形から)
% 2011/11/25  新規作成
%
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\Dtitle{deepconv/arare5}
\Dauthor{黒田美紀}        % 名前
\Ddate{2011/12/19}        % 今日の日付 (毎回変更すること)
\Dfile{energy-eq.tex}

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\section{準圧縮方程式におけるエネルギー方程式の導出}      % 節の始めからの場合はこのコマンドを使用する
\markright{A.\arabic{section}準圧縮方程式におけるエネルギー方程式の導出} %  節の題名を書き込むこと

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準圧縮方定式におけるエネルギー方程式を導出する. 
ただし, ここでは乾燥大気を想定し, 凝結に関する項を無視する. 

乾燥大気の場合, 準圧縮方程式の連続の式は
\begin{eqnarray}
  \DP{\rho}{t} + \Ddiv{\overline{\rho}\Dvect{u}} = 0
  \Deqlab{乾燥大気の連続の式}
\end{eqnarray}
である. この式より任意のスカラー量 $\phi$ に対し
\begin{eqnarray}
  \overline{\rho} \DD{\phi}{t}
  = \DP{\overline{\rho} \phi}{t}
   + \Ddiv{ (\overline{\rho} \phi \Dvect{u})}
   + \phi \DP{\rho}{t}
  \Deqlab{スカラーの保存則}
\end{eqnarray}
が成り立つ. この式から準圧縮方程式では$\phi$の時間微分を
フラックス形式で書くことができないことがわかる%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\footnote{
  $\partial \phi/\partial t + \Dvect{u} \cdot \nabla \phi = 0$ が
  成り立たないということ. }%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
.

水平方向の運動方程式 (2.2), 鉛直方向の運動方程式 (2.3) 式より, 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{align}
  \DP{\overline{\rho}K}{t} + \Ddiv{(\overline{\rho}K\Dvect{u})}
 = - c_{p} \overline{\theta} \overline{\rho} \Dvect{u} \cdot \Dgrad{\Pi}
   + \overline{\rho}\Dvect{u}\cdot\Dvect{D}
   + \overline{\rho} \frac{\theta}{\overline{\theta}}gw
   - K \DP{\rho}{t}
\end{align}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
となる. 
ここで$K = (u^2 + w^2)/2$, $\Dvect{D} = (D_u, D_w)$とおいた%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\footnote{
%  個人的めも: 導出をどこかに書いておきたいなー. }%
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. 
以降では, (A.9) の右辺第一項と右辺第三項を変形していく. 

まず, (A.9) の右辺第三項は以下のように変形される. 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{align}
 \overline{\rho} \frac{\theta}{\overline{\theta}}gw
  &= \overline{\rho} \frac{\theta}{\overline{\theta}} g \DD{z}{t}
      \nonumber \\
  &= \overline{\rho} \frac{\theta}{\overline{\theta}} \DD{(gz)}{t}
      \nonumber \\
  &= \overline{\rho}
      \left \{\frac{1}{\overline{\theta}} \DD{(\theta gz)}{t}
       - \frac{gz}{\overline{\theta}} \DD{\theta}{t}
      \right\} \nonumber \\
  &= \frac{\overline{\rho}}{\overline{\theta}} \DD{(\theta gz)}{t}
      - \frac{gz}{\overline{\theta}}
       \left\{ \DP{\overline{\rho}\theta}{t}
               + \nabla \cdot (\overline{\rho}\theta \Dvect{u})
               + \theta \DP{\rho}{t}
        \right\} . 
 \tag{A.9-1}
\end{align}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(A.9-1) の右辺第二項を変形する. 
熱力学方程式 (2.5) は, 凝結がない場合, 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{align}
 \DP{\theta^{'}}{t}
  = -\Dvect{u} \cdot \nabla \theta^{'} - w \DP{\overline{\theta}}{z}
     + \frac{\overline{\theta}}{\overline{T}}(Q_{\textmc{rad}} + Q_{\textmc{dis}}) + D_{\theta}
  \tag{2.5-1}
\end{align}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
となる. これは基本場に対しても成り立ち,  
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{align}
 \DP{\overline{\theta}}{t} = -\Dvect{u} \cdot \nabla \overline{\theta}
 \tag{2.5-2}
\end{align}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
となる. 
(2.5-1) + (2.5-2) より, 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{align}
 \DP{\theta}{t}
  = - \Dvect{u} \cdot \nabla \theta - w \DP{\overline{\theta}}{z}
    + \frac{\overline{\theta}}{\overline{T}}(Q_{\textmc{rad}} + Q_{\textmc{dis}}) + D_{\theta}. 
 \tag{2.5-3}
\end{align}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
この両辺に $\overline{\rho}$ をかけて (A.7) を用いてさらに変形すると, 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{align}
 \DP{\overline{\rho} \theta}{t}
  &= - \overline{\rho}\Dvect{u} \cdot \nabla \theta - \overline{\rho} w \DP{\overline{\theta}}{z}
    + \overline{\rho} \frac{\overline{\theta}}{\overline{T}}(Q_{\textmc{rad}} + Q_{\textmc{dis}})
    + \overline{\rho}D_{\theta} 
     \nonumber \\
  &= -\nabla \cdot (\overline{\rho} \theta \Dvect{u}) + \theta \nabla \cdot (\overline{\rho}\Dvect{u})
     - \overline{\rho} w \DP{\overline{\theta}}{z}
     + \overline{\rho} \frac{\overline{\theta}}{\overline{T}}(Q_{\textmc{rad}} + Q_{\textmc{dis}})
     + \overline{\rho}D_{\theta} 
     \nonumber　\\
  &= -\nabla \cdot (\overline{\rho} \theta \Dvect{u}) - \theta \DP{\rho}{t}
     - \overline{\rho} w \DP{\overline{\theta}}{z}
     + \overline{\rho} \frac{\overline{\theta}}{\overline{T}}(Q_{\textmc{rad}} + Q_{\textmc{dis}})
     + \overline{\rho}D_{\theta} 
\end{align}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
となる. 
また (A.9-1) 式の右辺第一項は, (A.8) より, 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{align}
 \overline{\rho} \frac{d}{dt} 
  \left(\frac{\theta gz}{\overline{\theta}}\right)
 = \frac{\partial}{\partial t}\left(\frac {\overline{\rho}\theta gz}{\overline{\theta}}\right)
  + \nabla \cdot \left(\frac{\overline{\rho}\theta gz}{\overline{\theta}}\Dvect{u}\right)
  + \frac{\theta gz}{\overline{\theta}}\DP{\rho}{t}
 \tag{A.9-2}
\end{align}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
となる. 
よって, (A.9) の右辺第三項は, 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{align}
 \overline{\rho} \frac{\theta}{\overline{\theta}}gw
  = \frac{\partial}{\partial t}\left(\frac{\overline{\rho}\theta gz}{\overline{\theta}}\right)
   + \nabla \cdot \left(\frac{\overline{\rho}\theta gz}{\overline{\theta}} \Dvect{u}\right)
   + \frac{\theta gz}{\overline{\theta}}\DP{\rho}{t}
   + \frac{\overline{\rho} gz}{\overline{\theta}} \left\{ w\DP{\overline{\theta}}{z}
    - \frac{\overline{\theta}}{\overline{T}}(Q_{\textmc{rad}} + Q_{\textmc{dis}}) - D_{\theta}\right\}
 \tag{A.9-3}
\end{align}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
と変形される. 

(A.9) の右辺第一項は以下のように変形される. 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{align}
 -c_p \overline{\theta}\overline{\rho}\Dvect{u} \cdot \nabla \Pi
  = -\nabla \cdot (c_p \overline{\theta}\overline{\rho} \Pi \Dvect{u})
   + c_p \Pi \nabla \cdot (\overline{\theta}\overline{\rho}\Dvect{u}) . 
 \tag{A.9-4}
\end{align}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
ここで, 凝結がないときの圧力方程式 (2.4) は, 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{align}
 \DP{\Pi^{'}}{t}
  = -\frac{\overline{c^{2}}}{c_p \overline{\rho}\overline{\theta}^2}
    \left[\frac{\partial}{\partial x}(\overline{\rho}\overline{\theta}\Dvect{u}) + 
         +\frac{\partial}{\partial z}(\overline{\rho}\overline{\theta}\Dvect{u})
    \right]
   + \frac{\overline{c}^{2}}{c_p \overline{\theta}^2}\DD{\theta^{'}}{t}
 \tag{2.4-1}
\end{align}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
である. これは基本場に対しても成り立ち, 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{align}
 \DP{\overline{\Pi}}{t}
  =  \frac{\overline{c}^{2}}{c_p \overline{\theta}^{2}}\DD{\overline{\theta}^{'}}{t}
 \tag{2.4-2}
\end{align}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
となる. 
(2.4-1) + (2.4-2) より, 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{align}
 \DP{\Pi}{t}
  &= -\frac{\overline{\overline{c}^2}}{c_p \overline{\rho}\overline{\theta}^2}
    \left[\frac{\partial}{\partial x}(\overline{\rho}\overline{\theta}\Dvect{u})
         +\frac{\partial}{\partial z}(\overline{\rho}\overline{\theta}\Dvect{u})
    \right]
   + \frac{\overline{c}^2}{c_p \overline{\theta}^2}\DD{\theta}{t}
  \nonumber \\
 &= -\frac{\overline{c}^2}{c_p \overline{\rho}\overline{\theta}^2}
     \nabla \cdot (\overline{\rho}\overline{\theta}\Dvect{u})
\end{align}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
となる. 
ここで, 凝結を無視しているので大気の熱膨張は無視でき, $d\theta/dt$ の項はゼロになる. 
よって (A.9-4) は, 
\setcounter{equation}{12}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{align}
 -c_p \overline{\theta}\overline{\rho}\Dvect{u} \cdot \nabla \Pi
  &= -\nabla \cdot (c_p \overline{\theta}\overline{\rho} \Pi \Dvect{u})
      - c_p \Pi \frac{c_p \overline{\rho}\overline{\theta}^2}{\overline{c}^2} \DP{\Pi}{t} \nonumber \\
  &= -\nabla \cdot (c_p \overline{\theta}\overline{\rho} \Pi \Dvect{u})
       -\frac{\partial}{\partial t} \left\{\frac{\overline{\rho}}{2}
        \left(\frac{c_p \overline{\theta}\Pi}{\overline{c}}\right)^2 \right\}
% \tag{A.13}
\end{align}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
と書き換えられる. 

したがって, (A.9) は, 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{align}
 \frac{\partial}{\partial t}&(\overline{\rho}K) + \nabla \cdot (\overline{\rho} K \Dvect{u})
  = -\nabla \cdot (c_p \overline{\theta}\overline{\theta} \Pi \Dvect{u})
     -\frac{\partial}{\partial t} \left\{\frac{\overline{\rho}}{2}
      \left(\frac{c_p \overline{\theta}\Pi}{\overline{c}}\right)^2 \right\}
     + \overline{\rho} \Dvect{u} \cdot \Dvect{D} \nonumber \\
    &+  \frac{\partial}{\partial t}\left(\frac{\overline{\rho}\theta gz}{\overline{\theta}}\right)
      + \nabla \cdot \left(\frac{\overline{\rho}\theta gz}{\overline{\theta}} \Dvect{u}\right)
      + \frac{\theta gz}{\overline{\theta}}\DP{\rho}{t}
      + \frac{\overline{\rho} gz}{\overline{\theta}} \left\{ w\DP{\overline{\theta}}{z}
      - \frac{\overline{\theta}}{\overline{T}}(Q_{\textmc{rad}} + Q_{\textmc{dis}}) - D_{\theta}\right\} 
       \nonumber \\
    &- K \DP{\rho}{t}
 \nonumber
\end{align}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
となり, これをさらに整理すると, 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{align}
% \split
 \frac{\partial}{\partial t}\left[\overline{\rho}
  \left\{K - \frac{\theta}{\overline{\theta}}gz
         + \frac{1}{2}\left(\frac{c_p \overline{\theta}\Pi}{\overline{c}}\right)^2\right\}\right]
  + \nabla \cdot \left\{\overline{\rho}\left(K - \frac{\theta}{\overline{\theta}}
                                             + c_p \overline{\theta}\Pi \right) \Dvect{u} \right\} \nonumber \\
 = \overline{\rho} \Dvect{u} \cdot \Dvect{D}
      + \frac{\overline{\rho} gz}{\overline{\theta}} \left\{ w\DP{\overline{\theta}}{z}
      - \frac{\overline{\theta}}{\overline{T}}(Q_{\textmc{rad}} + Q_{\textmc{dis}}) - D_{\theta}\right\}
   - \left(K - \frac{\theta}{\overline{\theta}}gz\right) \DP{\rho}{t}
\end{align} 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
となる. この両辺を全領域で積分し, 境界面を出入りする流れはないとすれば, 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{align}
 &  \DP{}{t} \int \overline{\rho}
     \left\{K - \frac{\theta}{\overline{\theta}} gz 
       + \frac{1}{2} \left( \frac{c_p \overline{\theta}\Pi}
             {\overline{c}} \right)^2\right\} dV
    \nonumber \\
 &= \int \overline{\rho}\Dvect{u}\cdot\Dvect{D} dV 
   + \int \frac{\overline{\rho}wgz}{\overline{\theta}} \DP{\overline{\theta}}{z} dV
   - \int \frac{\overline{\rho}gz}{\overline{T}}(Q_{rad}+Q_{dis}) dV
    \nonumber \\
 & \hspace{1em} 
   - \int \frac{\overline{\rho}gz}{\overline{\theta}} D_\theta  dV
   - \int \left(K - \frac{\theta}{\overline{\theta}} gz \right) \DP{\rho}{t} dV
\end{align}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
となる. これが準圧縮方程式のエネルギー方程式である.
左辺は全エネルギーの時間変化である. かっこの中の第1項, 第2項, 第3項は
それぞれ運動エネルギー, 浮力による位置エネルギー, 
弾性エネルギーである. 
右辺第1項, 第2項, 第3項, 第4項はそれぞれ
運動量の拡散, 基本場の鉛直温位勾配, 非断熱加熱(放射, 散逸), 
熱の拡散によるエネルギー変化率である. 
右辺最後の項は準圧縮方程式の場合スカラー量の時間微分が
フラックス形式で書けないことによることにより現れる. 
この項の存在により準圧縮方程式では外部項がない場合でも
エネルギーが保存しない. 


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% \DP{p}{z} = \rho g.
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%  \DP{u}{p} = \frac{R}{fp} \DP{T}{y} ,
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%   F_{m,n+1}^{*} &=& F_{m,n} - \frac{c \Delta t}{2\Delta x}
%   \left(F_{m+1,n} - F_{m-1,n}\right)
%   \\
%  \Deqlab{Pedlosky_1987_eq-2.4.10b}
%   F_{m,n+1} &=& F_{m,n} - \frac{c \Delta t}{2\Delta x}
%   \left(F_{m+1,n+1}^{*} - F_{m-1,n+1}^{*} \right)
% \end{eqnarray}
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%  \dfrac{u_{m+1,n} - u_{m-1,n}}{2\Delta t} &+&
%   U\dfrac{u_{m+1,n} - u_{m-1,n}}{2d} &+&
%   \dfrac{\phi_{m+1,n} - \phi_{m-1,n}}{2d} &=& 0 \\
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%%%%%%%%         参考文献 (本文に書く場合)      %%%%%%%%
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%{\bfseries 参考文献}
%\vspace{-7mm}
%\begin{description}
% \item	著者名, 2000:
%	書籍名, (章, 節).
%	出版社, 319pp.
%\end{description}
%
%{\bfseries 参考文献}
%\vspace{-7mm}
%\begin{description}
% \item	地球流体電脳倶楽部, 1998: \\
%	AGCM5マニュアル 第2部 離散化. \\
%	\texttt{http://www.gfd-dennou.org/\~{}arch/agcm5/} .
%\end{description}
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%%%%%%%%         参考文献 (脚注につける場合)    %%%%%%%%
%
%\footnote{
% 参考文献
% \begin{quote}
%  \begin{description}
%   \item 著者名, 1983:
%	 書籍名, (章, 節).
%	 出版社, 247pp.
%  \end{description}
% \end{quote}
%}
%
%
%\footnote{
% 参考文献
% \begin{quote}
%  \begin{description}
%   \item Hinkelmann, K., 1959:
%	 Der mechanismus des meteorologischen Larmes.
%	 {\itshape Tellus,} {\bfseries 3,} 285--296.
%  \end{description}
% \end{quote}
%}
%
%\footnote{
% 参考文献
% \begin{quote}
%  \begin{description}
%   \item 倉本 圭, 2000:
%	 北海道大学 理学部 地球科学科 講義
%	       『物理数学I演習, 線形代数の復習2』 \\
%	 \texttt{http://www.ep.sci.hokudai.ac.jp/\~{}pmath1ex/2000/0501/pub/No03.pdf}
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