35 
\frac{\partial^2 p}{\partial x^2\hackscore{i}} 
\frac{\partial^2 p}{\partial x^2\hackscore{i}} 
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\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(xh)}{h^2} 
f''(x) \approx \frac{f(x+h  2f(x) + f(xh)}{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{(t1)} \right] 
p\hackscore{(t1)} \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. 