# Diff of /trunk/doc/cookbook/example07.tex

revision 3232 by ahallam, Fri Oct 1 02:08:38 2010 UTC revision 3308 by jfenwick, Tue Oct 26 03:24:54 2010 UTC
# Line 31  calculated via the divergence of the gra Line 31  calculated via the divergence of the gra
31  example $p$. Thus we can write;  example $p$. Thus we can write;
32
33   \nabla^2 p = \nabla \cdot \nabla p =   \nabla^2 p = \nabla \cdot \nabla p =
34      \sum\hackscore{i}^n      \sum_{i}^n
35      \frac{\partial^2 p}{\partial x^2\hackscore{i}}      \frac{\partial^2 p}{\partial x^2_{i}}
36   \label{eqn:laplacian}   \label{eqn:laplacian}
37
38  For the two dimensional case in Cartesian coordinates \autoref{eqn:laplacian}  For the two dimensional case in Cartesian coordinates \autoref{eqn:laplacian}
# Line 48  intrinsic functions \textit{grad()} and Line 48  intrinsic functions \textit{grad()} and
48  For a rank 0 solution, this is of the form;  For a rank 0 solution, this is of the form;
49
50   \nabla p = \left[   \nabla p = \left[
51         \frac{\partial p}{\partial x \hackscore{0}},           \frac{\partial p}{\partial x _{0}},
52         \frac{\partial p}{\partial x \hackscore{1}}         \frac{\partial p}{\partial x _{1}}
53                    \right]                    \right]
55
56  Larger ranked solution objects will return gradient tensors. For example, a  Larger ranked solution objects will return gradient tensors. For example, a
57  pressure field which acts in the directions $p \hackscore{0}$ and $p pressure field which acts in the directions$p _{0}$and$p
58  \hackscore{1}$would return; _{1}$ would return;
59
60    \nabla p = \begin{bmatrix}    \nabla p = \begin{bmatrix}
61         \frac{\partial p \hackscore{0}}{\partial x \hackscore{0}} &         \frac{\partial p _{0}}{\partial x _{0}} &
62          \frac{\partial p \hackscore{1}}{\partial x \hackscore{0}} \\          \frac{\partial p _{1}}{\partial x _{0}} \\
63        \frac{\partial p \hackscore{0}}{\partial x \hackscore{1}} &        \frac{\partial p _{0}}{\partial x _{1}} &
64          \frac{\partial p \hackscore{1}}{\partial x \hackscore{1}}          \frac{\partial p _{1}}{\partial x _{1}}
65                    \end{bmatrix}                    \end{bmatrix}
67
68
69  \autoref{eqn:grad} corresponds to the Linear PDE general form value  \autoref{eqn:grad} corresponds to the Linear PDE general form value
70  $X$. Notice however that the general form contains the term $X$X$. Notice however that the general form contains the term$X
71  \hackscore{i,j}$\footnote{This is the first derivative in the$j^{th}$_{i,j}$\footnote{This is the first derivative in the $j^{th}$
72  direction for the $i^{th}$ component of the solution.},  direction for the $i^{th}$ component of the solution.},
73  hence for a rank 0 object there is no need to do more than calculate the  hence for a rank 0 object there is no need to do more than calculate the
74  gradient and submit it to the solver. In the case of the rank 1 or greater  gradient and submit it to the solver. In the case of the rank 1 or greater
# Line 119  be significantly less than this. Of the Line 119  be significantly less than this. Of the
119
120  The wave frequency content also plays a part in numerical stability. The  The wave frequency content also plays a part in numerical stability. The
121  nyquist-sampling theorem states that a signals bandwidth content will be  nyquist-sampling theorem states that a signals bandwidth content will be
122  accurately represented when an equispaced sampling rate $f \hackscore{n}$ is  accurately represented when an equispaced sampling rate $f _{n}$ is
123  equal to or greater than twice the maximum frequency of the signal  equal to or greater than twice the maximum frequency of the signal
124  $f\hackscore{s}$, or;  $f_{s}$, or;
125   \label{eqn:samptheorem}   \label{eqn:samptheorem}
126   f\hackscore{n} \geqslant f\hackscore{s}   f_{n} \geqslant f_{s}
127
128  For example a 50Hz signal will require a sampling rate greater than 100Hz or  For example a 50Hz signal will require a sampling rate greater than 100Hz or
129  one sample every 0.01 seconds. The wave equation relies on a spatial frequency,  one sample every 0.01 seconds. The wave equation relies on a spatial frequency,
# Line 151  second derivative; Line 151  second derivative;
151  substituting \autoref{eqn:centdiff} for $\frac{\partial ^2 p }{\partial t ^2}$  substituting \autoref{eqn:centdiff} for $\frac{\partial ^2 p }{\partial t ^2}$
152  in \autoref{eqn:acswave};  in \autoref{eqn:acswave};
153
154   \nabla ^2 p - \frac{1}{c^2h^2} \left[p\hackscore{(t+1)} - 2p\hackscore{(t)} +   \nabla ^2 p - \frac{1}{c^2h^2} \left[p_{(t+1)} - 2p_{(t)} +
155  p\hackscore{(t-1)} \right]  p_{(t-1)} \right]
156  = 0  = 0
157  \label{eqn:waveu}  \label{eqn:waveu}
158
159  Rearranging for $p_{(t+1)}$;  Rearranging for $p_{(t+1)}$;
160
161   p\hackscore{(t+1)} = c^2 h^2 \nabla ^2 p\hackscore{(t)} +2p\hackscore{(t)} -   p_{(t+1)} = c^2 h^2 \nabla ^2 p_{(t)} +2p_{(t)} -
162  p\hackscore{(t-1)}  p_{(t-1)}
163
164  this can be compared with the general form of the \modLPDE module and it  this can be compared with the general form of the \modLPDE module and it
165  becomes clear that $D=1$, $X\hackscore{i,j}=-c^2 h^2 \nabla ^2 p_{(t)}$ and  becomes clear that $D=1$, $X_{i,j}=-c^2 h^2 \nabla ^2 p_{(t)}$ and
166  $Y=2p_{(t)} - p_{(t-1)}$.  $Y=2p_{(t)} - p_{(t-1)}$.
167
168  The solution script is similar to others that we have created in previous  The solution script is similar to others that we have created in previous
# Line 275  PDE solution. \autoref{eqn:waveu} is thu Line 275  PDE solution. \autoref{eqn:waveu} is thu
275    \nabla ^2 p - \frac{1}{c^2} a = 0    \nabla ^2 p - \frac{1}{c^2} a = 0
276  \label{eqn:wavea}  \label{eqn:wavea}
277
278  and can be solved directly with $Y=0$ and $X=-c^2 \nabla ^2 p\hackscore{(t)}$.  and can be solved directly with $Y=0$ and $X=-c^2 \nabla ^2 p_{(t)}$.
279  After each iteration the displacement is re-evaluated via;  After each iteration the displacement is re-evaluated via;
280
281   p\hackscore{(t+1)}=2p\hackscore{(t)} - p\hackscore{(t-1)} + h^2a   p_{(t+1)}=2p_{(t)} - p_{(t-1)} + h^2a
282
283
284  \subsection{Lumping}  \subsection{Lumping}
# Line 306  First, we let the frequency content of t Line 306  First, we let the frequency content of t
306  the source as a cosine input, than the wavlength of the input is equal to the  the source as a cosine input, than the wavlength of the input is equal to the
307  radius of the source. Let this value be 5 meters. Now, if the maximum velocity  radius of the source. Let this value be 5 meters. Now, if the maximum velocity
308  of the model is $c=380.0ms^{-1}$ then the source  of the model is $c=380.0ms^{-1}$ then the source
309  frequency is $f\hackscore{r} = \frac{380.0}{5} = 76.0 Hz$. This is a worst case  frequency is $f_{r} = \frac{380.0}{5} = 76.0 Hz$. This is a worst case
310  scenario with a small source and the models maximum velocity.  scenario with a small source and the models maximum velocity.
311
312  Furthermore, we know from \autoref{sec:nsstab}, that the spatial sampling  Furthermore, we know from \autoref{sec:nsstab}, that the spatial sampling
# Line 315  the model mesh is a square equispaced gr Line 315  the model mesh is a square equispaced gr
315  then the sampling interval is the side length divided by the number of samples,  then the sampling interval is the side length divided by the number of samples,
316  given by $\Delta x = \frac{1000.0m}{400} = 2.5m$ and the maximum sampling  given by $\Delta x = \frac{1000.0m}{400} = 2.5m$ and the maximum sampling
317  frequency capable at this interval is  frequency capable at this interval is
318  $f\hackscore{s}=\frac{380.0ms^{-1}}{2.5m}=152Hz$ this is just equal to the  $f_{s}=\frac{380.0ms^{-1}}{2.5m}=152Hz$ this is just equal to the
319  required rate satisfying \autoref{eqn:samptheorem}.  required rate satisfying \autoref{eqn:samptheorem}.
320
321  \autoref{fig:ex07sampth} depicts three examples where the grid has been  \autoref{fig:ex07sampth} depicts three examples where the grid has been

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