Changeset 11331 for NEMO/trunk/doc/latex/NEMO/subfiles/annex_B.tex
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 20190723T15:02:57+02:00 (2 years ago)
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NEMO/trunk/doc/latex/NEMO/subfiles/annex_B.tex
r11151 r11331 62 62 \end{align*} 63 63 64 Equation\autoref{apdx:B2} is obtained from \autoref{apdx:B1} without any additional assumption.64 \autoref{apdx:B2} is obtained from \autoref{apdx:B1} without any additional assumption. 65 65 Indeed, for the special case $k=z$ and thus $e_3 =1$, 66 66 we introduce an arbitrary vertical coordinate $s = s (i,j,z)$ as in \autoref{apdx:A} and … … 94 94 & =\frac{1}{e_1\,e_2\,e_3 }\left[ {\left. {\;\;\;\frac{\partial }{\partial i}\left( {\frac{e_2\,e_3 }{e_1}\,A^{lT}\;\left. {\frac{\partial T}{\partial i}} \right_s } \right)} \right_s \left. \, {e_3 \frac{\partial }{\partial i}\left( {\frac{e_2 \,\sigma_1 }{e_3 }A^{lT}\;\frac{\partial T}{\partial s}} \right)} \right_s } \right. \\ 95 95 & \qquad \qquad \quad e_2 A^{lT}\;\frac{\partial \sigma_1 }{\partial s}\left. {\frac{\partial T}{\partial i}} \right_s e_1 \,\sigma_1 \frac{\partial }{\partial s}\left( {\frac{e_2 }{e_1 }A^{lT}\;\left. {\frac{\partial T}{\partial i}} \right_s } \right) \\ 96 & \qquad \qquad \quad\shoveright{ \left. { +e_1 \,\sigma_1 \frac{\partial }{\partial s}\left( {\frac{e_2 \,\sigma_1 }{e_3 }A^{lT}\;\frac{\partial T}{\partial s}} \right)+\frac{\partial }{\partial s}\left( {\frac{e_1 \,e_2 }{e_3 }A^{vT}\;\frac{\partial T}{\partial z}} \right)\;\;\;} \right] }\\96 & \qquad \qquad \quad\shoveright{ \left. { +e_1 \,\sigma_1 \frac{\partial }{\partial s}\left( {\frac{e_2 \,\sigma_1 }{e_3 }A^{lT}\;\frac{\partial T}{\partial s}} \right)+\frac{\partial }{\partial s}\left( {\frac{e_1 \,e_2 }{e_3 }A^{vT}\;\frac{\partial T}{\partial s}} \right)\;\;\;} \right] }\\ 97 97 \\ 98 98 &=\frac{1}{e_1 \,e_2 \,e_3 } \left[ {\left. {\;\;\;\frac{\partial }{\partial i} \left( {\frac{e_2 \,e_3 }{e_1 }A^{lT}\;\left. {\frac{\partial T}{\partial i}} \right_s } \right)} \right_s \left. {\frac{\partial }{\partial i}\left( {e_2 \,\sigma_1 A^{lT}\;\frac{\partial T}{\partial s}} \right)} \right_s } \right. \\ … … 105 105 \begin{array}{*{20}l} 106 106 % 107 \intertext{ using the same remark as just above, it becomes:}107 \intertext{Using the same remark as just above, it becomes:} 108 108 % 109 109 &= \frac{1}{e_1 \,e_2 \,e_3 } \left[ {\left. {\;\;\;\frac{\partial }{\partial i} \left( {\frac{e_2 \,e_3 }{e_1 }A^{lT}\;\left. {\frac{\partial T}{\partial i}} \right_s e_2 \,\sigma_1 A^{lT}\;\frac{\partial T}{\partial s}} \right)} \right_s } \right.\;\;\; \\ … … 117 117 % 118 118 \intertext{Since the horizontal scale factors do not depend on the vertical coordinate, 119 the last term of the first line and the first term of the lastline cancel, while120 the second line reduces to a single vertical derivative, so it becomes:}119 the two terms on the second line cancel, while 120 the third line reduces to a single vertical derivative, so it becomes:} 121 121 % 122 122 & =\frac{1}{e_1 \,e_2 \,e_3 }\left[ {\left. {\;\;\;\frac{\partial }{\partial i}\left( {\frac{e_2 \,e_3 }{e_1 }A^{lT}\;\left. {\frac{\partial T}{\partial i}} \right_s e_2 \,\sigma_1 \,A^{lT}\;\frac{\partial T}{\partial s}} \right)} \right_s } \right. \\ 123 123 & \qquad \qquad \quad \shoveright{ \left. {+\frac{\partial }{\partial s}\left( {e_2 \,\sigma_1 \,A^{lT}\;\left. {\frac{\partial T}{\partial i}} \right_s +A^{lT}\frac{e_1 \,e_2 }{e_3 }\;\left( {\varepsilon +\sigma_1 ^2} \right)\frac{\partial T}{\partial s}} \right)\;\;\;} \right]} \\ 124 124 % 125 \intertext{ in other words, the horizontal/vertical Laplacian operator in the ($i$,$s$) plane takes the following form:}126 \end{array} 127 } \\ 125 \intertext{In other words, the horizontal/vertical Laplacian operator in the ($i$,$s$) plane takes the following form:} 126 \end{array} 127 } \\ 128 128 % 129 129 {\frac{1}{e_1\,e_2\,e_3}} … … 169 169 \left[ {{ 170 170 \begin{array}{*{20}c} 171 {1+a_ 1 ^2} \hfill & {a_1 a_2 } \hfill & {a_1} \hfill \\172 {a_1 a_2 } \hfill & {1+a_2 ^2} \hfill & {a_2} \hfill \\173 {a_1 } \hfill & {a_2} \hfill & {\varepsilon +a_1 ^2+a_2 ^2} \hfill \\171 {1+a_2 ^2 +\varepsilon a_1 ^2} \hfill & {a_1 a_2 (1\varepsilon)} \hfill & {a_1 (1\varepsilon) } \hfill \\ 172 {a_1 a_2 (1\varepsilon) } \hfill & {1+a_1 ^2 +\varepsilon a_2 ^2} \hfill & {a_2 (1\varepsilon)} \hfill \\ 173 {a_1 (1\varepsilon)} \hfill & {a_2 (1\varepsilon)} \hfill & {\varepsilon +a_1 ^2+a_2 ^2} \hfill \\ 174 174 \end{array} 175 175 }} \right] … … 182 182 \right)\left( {\frac{\partial \rho }{\partial k}} \right)^{1} 183 183 \] 184 185 In practice, isopycnal slopes are generally less than $10^{2}$ in the ocean, 186 so $\textbf {A}_{\textbf I}$ can be simplified appreciably \citep{cox_OM87}: 184 and, as before, $\epsilon = A^{vT} / A^{lT}$. 185 186 In practice, $\epsilon$ is small and isopycnal slopes are generally less than $10^{2}$ in the ocean, 187 so $\textbf {A}_{\textbf I}$ can be simplified appreciably \citep{cox_OM87}. Keeping leading order terms\footnote{Apart from the (1,0) 188 and (0,1) elements which are set to zero. See \citet{griffies_bk04}, section 14.1.4.1 for a discussion of this point.}: 187 189 \begin{subequations} 188 190 \label{apdx:B4} … … 284 286 \] 285 287 286 To prove \autoref{apdx:B 5} by direct reexpression of \autoref{apdx:B_ldfiso} is straightforward, but laborious.288 To prove \autoref{apdx:B_ldfiso_s} by direct reexpression of \autoref{apdx:B_ldfiso} is straightforward, but laborious. 287 289 An easier way is first to note (by reversing the derivation of \autoref{sec:B_2} from \autoref{sec:B_1} ) that 288 290 the weakslope operator may be \emph{exactly} reexpressed in nonorthogonal $i,j,\rho$coordinates as … … 306 308 the (rotated,orthogonal) isoneutral axes to the nonorthogonal $i,j,\rho$coordinates. 307 309 The further transformation into $i,j,s$coordinates is exact, whatever the steepness of the $s$surfaces, 308 in the same way as the transformation of horizontal/vertical Laplacian diffusion in $z$coordinates ,310 in the same way as the transformation of horizontal/vertical Laplacian diffusion in $z$coordinates in 309 311 \autoref{sec:B_1} onto $s$coordinates is exact, however steep the $s$surfaces. 310 312 … … 316 318 \label{sec:B_3} 317 319 318 The second order momentum diffusion operator (Laplacian) in the $z$coordinateis found by320 The second order momentum diffusion operator (Laplacian) in $z$coordinates is found by 319 321 applying \autoref{eq:PE_lap_vector}, the expression for the Laplacian of a vector, 320 322 to the horizontal velocity vector: … … 361 363 \end{align*} 362 364 Using \autoref{eq:PE_div}, the definition of the horizontal divergence, 363 the third compon ant of the second vector is obviously zero and thus :365 the third component of the second vector is obviously zero and thus : 364 366 \[ 365 367 \Delta {\textbf{U}}_h = \nabla _h \left( \chi \right)  \nabla _h \times \left( \zeta \right) + \frac {1}{e_3 } \frac {\partial }{\partial k} \left( {\frac {1}{e_3 } \frac{\partial {\textbf{ U}}_h }{\partial k}} \right) … … 386 388 & = \frac{1}{e_1} \frac{\partial \left( {A^{lm}\chi } \right)}{\partial i} 387 389 \frac{1}{e_2} \frac{\partial \left( {A^{lm}\zeta } \right)}{\partial j} 388 +\frac{1}{e_3} \frac{\partial u}{\partial k}\\390 +\frac{1}{e_3} \frac{\partial}{\partial k} \left( \frac{A^{vm}}{e_3} \frac{\partial u}{\partial k} \right) \\ 389 391 D^{\textbf{U}}_v 390 392 & = \frac{1}{e_2 }\frac{\partial \left( {A^{lm}\chi } \right)}{\partial j} 391 393 +\frac{1}{e_1 }\frac{\partial \left( {A^{lm}\zeta } \right)}{\partial i} 392 +\frac{1}{e_3} \frac{\partial v}{\partial k}394 +\frac{1}{e_3} \frac{\partial}{\partial k} \left( \frac{A^{vm}}{e_3} \frac{\partial v}{\partial k} \right) 393 395 \end{align*} 394 396
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