| 35 |
\frac{\partial^2 p}{\partial x^2\hackscore{i}} |
\frac{\partial^2 p}{\partial x^2\hackscore{i}} |
| 36 |
\label{eqn:laplacian} |
\label{eqn:laplacian} |
| 37 |
\end{equation} |
\end{equation} |
| 38 |
For the two dimensional case in Cartesian coordinates \refEq{eqn:laplacian} |
For the two dimensional case in Cartesian coordinates \autoref{eqn:laplacian} |
| 39 |
becomes; |
becomes; |
| 40 |
\begin{equation} |
\begin{equation} |
| 41 |
\nabla^2 p = \frac{\partial^2 p}{\partial x^2} |
\nabla^2 p = \frac{\partial^2 p}{\partial x^2} |
| 66 |
\label{eqn:gradrank1} |
\label{eqn:gradrank1} |
| 67 |
\end{equation} |
\end{equation} |
| 68 |
|
|
| 69 |
\refEq{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 $j^{th}$ |
\hackscore{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 |
| 75 |
object, it is nesscary to calculate the trace also. This is the sum of the |
object, it is nesscary to calculate the trace also. This is the sum of the |
| 76 |
diagonal in \refeq{eqn:gradrank1}. |
diagonal in \autoref{eqn:gradrank1}. |
| 77 |
|
|
| 78 |
Thus when solving for equations containing the Laplacian one of two things must |
Thus when solving for equations containing the Laplacian one of two things must |
| 79 |
be completed. If the object \verb!p! is less than rank 1 the gradient is |
be completed. If the object \verb!p! is less than rank 1 the gradient is |
| 148 |
f''(x) \approx \frac{f(x+h - 2f(x) + f(x-h)}{h^2} |
f''(x) \approx \frac{f(x+h - 2f(x) + f(x-h)}{h^2} |
| 149 |
\label{eqn:centdiff} |
\label{eqn:centdiff} |
| 150 |
\end{equation} |
\end{equation} |
| 151 |
substituting \refEq{eqn:centdiff} for $\frac{\partial ^2 p }{\partial t ^2}$ |
substituting \autoref{eqn:centdiff} for $\frac{\partial ^2 p }{\partial t ^2}$ |
| 152 |
in \refEq{eqn:acswave}; |
in \autoref{eqn:acswave}; |
| 153 |
\begin{equation} |
\begin{equation} |
| 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\hackscore{(t+1)} - 2p\hackscore{(t)} + |
| 155 |
p\hackscore{(t-1)} \right] |
p\hackscore{(t-1)} \right] |
| 270 |
|
|
| 271 |
An alternative method is to solve for the acceleration $\frac{\partial ^2 |
An alternative method is to solve for the acceleration $\frac{\partial ^2 |
| 272 |
p}{\partial t^2}$ directly, and derive the displacement solution from the |
p}{\partial t^2}$ directly, and derive the displacement solution from the |
| 273 |
PDE solution. \refEq{eqn:waveu} is thus modified; |
PDE solution. \autoref{eqn:waveu} is thus modified; |
| 274 |
\begin{equation} |
\begin{equation} |
| 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} |
| 309 |
frequency is $f\hackscore{r} = \frac{380.0}{5} = 76.0 Hz$. This is a worst case |
frequency is $f\hackscore{r} = \frac{380.0}{5} = 76.0 Hz$. 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 \refSec{sec:nsstab}, that the spatial sampling |
Furthermore, we know from \autoref{sec:nsstab}, that the spatial sampling |
| 313 |
frequency must be at least twice this value to ensure stability. If we assume |
frequency must be at least twice this value to ensure stability. If we assume |
| 314 |
the model mesh is a square equispaced grid, |
the model mesh is a square equispaced grid, |
| 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 |
$f\hackscore{s}=\frac{380.0ms^{-1}}{2.5m}=152Hz$ this is just equal to the |
| 319 |
required rate satisfying \refeq{eqn:samptheorem}. |
required rate satisfying \autoref{eqn:samptheorem}. |
| 320 |
|
|
| 321 |
\reffig{fig:ex07sampth} depicts three examples where the grid has been |
\autoref{fig:ex07sampth} depicts three examples where the grid has been |
| 322 |
undersampled, sampled correctly, and over sampled. The grids used had |
undersampled, sampled correctly, and over sampled. The grids used had |
| 323 |
200, 400 and 800 nodes per side respectively. Obviously, the oversampled grid |
200, 400 and 800 nodes per side respectively. Obviously, the oversampled grid |
| 324 |
retains the best resolution of the modelled wave. |
retains the best resolution of the modelled wave. |
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