1 

2 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
3 
% Copyright (c) 20032012 by University of Queensland 
4 
% http://www.uq.edu.au 
5 
% 
6 
% Primary Business: Queensland, Australia 
7 
% Licensed under the Open Software License version 3.0 
8 
% http://www.opensource.org/licenses/osl3.0.php 
9 
% 
10 
% Development until 2012 by Earth Systems Science Computational Center (ESSCC) 
11 
% Development since 2012 by School of Earth Sciences 
12 
% 
13 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
14 

15 
The acoustic wave equation governs the propagation of pressure waves. Wave 
16 
types that obey this law tend to travel in liquids or gases where shear waves 
17 
or longitudinal style wave motion is not possible. An obvious example is sound 
18 
waves. 
19 

20 
The acoustic wave equation is defined as; 
21 
\begin{equation} 
22 
\nabla ^2 p  \frac{1}{c^2} \frac{\partial ^2 p}{\partial t^2} = 0 
23 
\label{eqn:acswave} 
24 
\end{equation} 
25 
where $p$ is the pressure, $t$ is the time and $c$ is the wave velocity. In this 
26 
chapter the acoustic wave equation is demonstrated. Important steps include the 
27 
translation of the Laplacian $\nabla^2$ to the \esc general form, the stiff 
28 
equation stability criterion and solving for the displacement or acceleration solution. 
29 

30 
\section{The Laplacian in \esc} 
31 
The Laplacian operator which can be written as $\Delta$ or $\nabla^2$, is 
32 
calculated via the divergence of the gradient of the object, which in this 
33 
example is the scalar $p$. Thus we can write; 
34 
\begin{equation} 
35 
\nabla^2 p = \nabla \cdot \nabla p = 
36 
\sum_{i}^n 
37 
\frac{\partial^2 p}{\partial x^2_{i}} 
38 
\label{eqn:laplacian} 
39 
\end{equation} 
40 
For the two dimensional case in Cartesian coordinates \autoref{eqn:laplacian} 
41 
becomes; 
42 
\begin{equation} 
43 
\nabla^2 p = \frac{\partial^2 p}{\partial x^2} 
44 
+ \frac{\partial^2 p}{\partial y^2} 
45 
\end{equation} 
46 

47 
In \esc the Laplacian is calculated using the divergence representation and the 
48 
intrinsic functions \textit{grad()} and \textit{trace()}. The function 
49 
\textit{grad{}} will return the spatial gradients of an object. 
50 
For a rank 0 solution, this is of the form; 
51 
\begin{equation} 
52 
\nabla p = \left[ 
53 
\frac{\partial p}{\partial x _{0}}, 
54 
\frac{\partial p}{\partial x _{1}} 
55 
\right] 
56 
\label{eqn:grad} 
57 
\end{equation} 
58 
Larger ranked solution objects will return gradient tensors. For example, a 
59 
pressure field which acts in the directions $p _{0}$ and $p 
60 
_{1}$ would return; 
61 
\begin{equation} 
62 
\nabla p = \begin{bmatrix} 
63 
\frac{\partial p _{0}}{\partial x _{0}} & 
64 
\frac{\partial p _{1}}{\partial x _{0}} \\ 
65 
\frac{\partial p _{0}}{\partial x _{1}} & 
66 
\frac{\partial p _{1}}{\partial x _{1}} 
67 
\end{bmatrix} 
68 
\label{eqn:gradrank1} 
69 
\end{equation} 
70 

71 
\autoref{eqn:grad} corresponds to the Linear PDE general form value 
72 
$X$. Notice however, that the general form contains the term $X 
73 
_{i,j}$\footnote{This is the first derivative in the $j^{th}$ 
74 
direction for the $i^{th}$ component of the solution.}, 
75 
hence for a rank 0 object there is no need to do more then calculate the 
76 
gradient and submit it to the solver. In the case of the rank 1 or greater 
77 
object, it is also necessary to calculate the trace. This is the sum of the 
78 
diagonal in \autoref{eqn:gradrank1}. 
79 

80 
Thus when solving for equations containing the Laplacian one of two things must 
81 
be completed. If the object \verb!p! is less than rank 1 the gradient is 
82 
calculated via; 
83 
\begin{python} 
84 
gradient=grad(p) 
85 
\end{python} 
86 
and if the object is greater then or equal to a rank 1 tensor, the trace of 
87 
the gradient is calculated. 
88 
\begin{python} 
89 
gradient=trace(grad(p)) 
90 
\end{python} 
91 
These values can then be submitted to the PDE solver via the general form term 
92 
$X$. The Laplacian is then computed in the solution process by taking the 
93 
divergence of $X$. 
94 

95 
Note, if you are unsure about the rank of your tensor, the \textit{getRank} 
96 
command will return the rank of the PDE object. 
97 
\begin{python} 
98 
rank = p.getRank() 
99 
\end{python} 
100 

101 

102 
\section{Numerical Solution Stability} \label{sec:nsstab} 
103 
Unfortunately, the wave equation belongs to a class of equations called 
104 
\textbf{stiff} PDEs. These types of equations can be difficult to solve 
105 
numerically as they tend to oscillate about the exact solution, which can 
106 
eventually lead to a catastrophic failure. To counter this problem, explicitly 
107 
stable schemes like the backwards Euler method, and correct parameterisation of 
108 
the problem are required. 
109 

110 
There are two variables which must be considered for 
111 
stability when numerically trying to solve the wave equation. For linear media, 
112 
the two variables are related via; 
113 
\begin{equation} \label{eqn:freqvel} 
114 
f=\frac{v}{\lambda} 
115 
\end{equation} 
116 
The velocity $v$ that a wave travels in a medium is an important variable. For 
117 
stability the analytical wave must not propagate faster then the numerical wave 
118 
is able to, and in general, needs to be much slower then the numerical wave. 
119 
For example, a line 100m long is discretised into 1m intervals or 101 nodes. If 
120 
a wave enters with a propagation velocity of 100m/s then the travel time for 
121 
the wave between each node will be 0.01 seconds. The time step, must therefore 
122 
be significantly less then this. Of the order $10E4$ would be appropriate. 
123 
This stability criterion is known as the Courant\textendash 
124 
Friedrichs\textendash Lewy condition given by 
125 
\begin{equation} 
126 
dt=f\cdot \frac{dx}{v} 
127 
\end{equation} 
128 
where $dx$ is the mesh size and $f$ is a safety factor. To obtain a time step of 
129 
$10E4$, a safety factor of $f=0.1$ was used. 
130 

131 
The wave frequency content also plays a part in numerical stability. The 
132 
Nyquistsampling theorem states that a signals bandwidth content will be 
133 
accurately represented when an equispaced sampling rate $f _{n}$ is 
134 
equal to or greater then twice the maximum frequency of the signal 
135 
$f_{s}$, or; 
136 
\begin{equation} \label{eqn:samptheorem} 
137 
f_{n} \geqslant f_{s} 
138 
\end{equation} 
139 
For example, a 50Hz signal will require a sampling rate greater then 100Hz or 
140 
one sample every 0.01 seconds. The wave equation relies on a spatial frequency, 
141 
thus the sampling theorem in this case applies to the solution mesh spacing. 
142 
This relationship confirms that the frequency content of the input signal 
143 
directly affects the time discretisation of the problem. 
144 

145 
To accurately model the wave equation with high resolutions and velocities 
146 
means that very fine spatial and time discretisation is necessary for most 
147 
problems. This requirement makes the wave equation arduous to 
148 
solve numerically due to the large number of time iterations required in each 
149 
solution. Models with very high velocities and frequencies will be the worst 
150 
affected by this problem. 
151 

152 
\section{Displacement Solution} 
153 
\sslist{example07a.py} 
154 

155 
We begin the solution to this PDE with the centred difference formula for the 
156 
second derivative; 
157 
\begin{equation} 
158 
f''(x) \approx \frac{f(x+h  2f(x) + f(xh)}{h^2} 
159 
\label{eqn:centdiff} 
160 
\end{equation} 
161 
substituting \autoref{eqn:centdiff} for $\frac{\partial ^2 p }{\partial t ^2}$ 
162 
in \autoref{eqn:acswave}; 
163 
\begin{equation} 
164 
\nabla ^2 p  \frac{1}{c^2h^2} \left[p_{(t+1)}  2p_{(t)} + 
165 
p_{(t1)} \right] 
166 
= 0 
167 
\label{eqn:waveu} 
168 
\end{equation} 
169 
Rearranging for $p_{(t+1)}$; 
170 
\begin{equation} 
171 
p_{(t+1)} = c^2 h^2 \nabla ^2 p_{(t)} +2p_{(t)}  
172 
p_{(t1)} 
173 
\end{equation} 
174 
this can be compared with the general form of the \modLPDE module and it 
175 
becomes clear that $D=1$, $X_{i,j}=c^2 h^2 \nabla ^2 p_{(t)}$ and 
176 
$Y=2p_{(t)}  p_{(t1)}$. 
177 

178 
The solution script is similar to others that we have created in previous 
179 
chapters. The general steps are; 
180 
\begin{enumerate} 
181 
\item The necessary libraries must be imported. 
182 
\item The domain needs to be defined. 
183 
\item The time iteration and control parameters need to be defined. 
184 
\item The PDE is initialised with source and boundary conditions. 
185 
\item The time loop is started and the PDE is solved at consecutive time steps. 
186 
\item All or select solutions are saved to file for visualisation later on. 
187 
\end{enumerate} 
188 

189 
Parts of the script which warrant more attention are the definition of the 
190 
source, visualising the source, the solution time loop and the VTK data export. 
191 

192 
\subsection{Pressure Sources} 
193 
As the pressure is a scalar, one need only define the pressure for two 
194 
time steps prior to the start of the solution loop. Two known solutions are 
195 
required because the wave equation contains a double partial derivative with 
196 
respect to time. This is often a good opportunity to introduce a source to the 
197 
solution. This model has the source located at it's centre. The source should 
198 
be smooth and cover a number of samples to satisfy the frequency stability 
199 
criterion. Small sources will generate high frequency signals. Here, when using 
200 
a rectangular domain, the source is defined by a cosine function. 
201 
\begin{python} 
202 
U0=0.01 # amplitude of point source 
203 
xc=[500,500] #location of point source 
204 
# define small radius around point xc 
205 
src_radius = 30 
206 
# for first two time steps 
207 
u=U0*(cos(length(xxc)*3.1415/src_radius)+1)*\ 
208 
whereNegative(length(xxc)src_radius) 
209 
u_m1=u 
210 
\end{python} 
211 

212 
\subsection{Visualising the Source} 
213 
There are two options for visualising the source. The first is to export the 
214 
initial conditions of the model to VTK, which can be interpreted as a scalar 
215 
surface in Mayavi2. The second is to take a cross section of the model which 
216 
will require the \textit{Locator} function. 
217 
First \verb!Locator! must be imported; 
218 
\begin{python} 
219 
from esys.escript.pdetools import Locator 
220 
\end{python} 
221 
The function can then be used on the domain to locate the nearest domain node 
222 
to the point or points of interest. 
223 

224 
It is now necessary to build a list of $(x,y)$ locations that specify where are 
225 
model slice will go. This is easily implemented with a loop; 
226 
\begin{python} 
227 
cut_loc=[] 
228 
src_cut=[] 
229 
for i in range(ndx/2ndx/10,ndx/2+ndx/10): 
230 
cut_loc.append(xstep*i) 
231 
src_cut.append([xstep*i,xc[1]]) 
232 
\end{python} 
233 
We then submit the output to \verb!Locator! and finally return the appropriate 
234 
values using the \verb!getValue! function. 
235 
\begin{python} 
236 
src=Locator(mydomain,src_cut) 
237 
src_cut=src.getValue(u) 
238 
\end{python} 
239 
It is then a trivial task to plot and save the output using \mpl 
240 
(\autoref{fig:cxsource}). 
241 
\begin{python} 
242 
pl.plot(cut_loc,src_cut) 
243 
pl.axis([xc[0]src_radius*3,xc[0]+src_radius*3,0.,2*U0]) 
244 
pl.savefig(os.path.join(savepath,"source_line.png")) 
245 
\end{python} 
246 
\begin{figure}[h] 
247 
\centering 
248 
\includegraphics[width=6in]{figures/sourceline.png} 
249 
\caption{Cross section of the source function.} 
250 
\label{fig:cxsource} 
251 
\end{figure} 
252 

253 

254 
\subsection{Point Monitoring} 
255 
In the more general case where the solution mesh is irregular or specific 
256 
locations need to be monitored, it is simple enough to use the \textit{Locator} 
257 
function. 
258 
\begin{python} 
259 
rec=Locator(mydomain,[250.,250.]) 
260 
\end{python} 
261 
When the solution \verb u is updated we can extract the value at that point 
262 
via; 
263 
\begin{python} 
264 
u_rec=rec.getValue(u) 
265 
\end{python} 
266 
For consecutive time steps one can record the values from \verb!u_rec! in an 
267 
array initialised as \verb!u_rec0=[]! with; 
268 
\begin{python} 
269 
u_rec0.append(rec.getValue(u)) 
270 
\end{python} 
271 

272 
It can be useful to monitor the value at a single or multiple individual points 
273 
in the model during the modelling process. This is done using 
274 
the \verb!Locator! function. 
275 

276 

277 
\section{Acceleration Solution} 
278 
\sslist{example07b.py} 
279 

280 
An alternative method to the displacement solution, is to solve for the 
281 
acceleration $\frac{\partial ^2 p}{\partial t^2}$ directly. The displacement can 
282 
then be derived from the acceleration after a solution has been calculated 
283 
The acceleration is given by a modified form of \autoref{eqn:waveu}; 
284 
\begin{equation} 
285 
\nabla ^2 p  \frac{1}{c^2} a = 0 
286 
\label{eqn:wavea} 
287 
\end{equation} 
288 
and can be solved directly with $Y=0$ and $X=c^2 \nabla ^2 p_{(t)}$. 
289 
After each iteration the displacement is reevaluated via; 
290 
\begin{equation} 
291 
p_{(t+1)}=2p_{(t)}  p_{(t1)} + h^2a 
292 
\end{equation} 
293 

294 
\subsection{Lumping} 
295 
For \esc, the acceleration solution is prefered as it allows the use of matrix 
296 
lumping. Lumping or mass lumping as it is sometimes known, is the process of 
297 
aggressively approximating the density elements of a mass matrix into the main 
298 
diagonal. The use of Lumping is motivaed by the simplicity of diagonal matrix 
299 
inversion. As a result, Lumping can significantly reduce the computational 
300 
requirements of a problem. Care should be taken however, as this 
301 
function can only be used when the $A$, $B$ and $C$ coefficients of the 
302 
general form are zero. 
303 

304 
More information about the lumping implementation used in \esc and its accuracy 
305 
can be found in the user guide. 
306 

307 
To turn lumping on in \esc one can use the command; 
308 
\begin{python} 
309 
mypde.getSolverOptions().setSolverMethod(mypde.getSolverOptions().HRZ_LUMPING) 
310 
\end{python} 
311 
It is also possible to check if lumping is set using; 
312 
\begin{python} 
313 
print mypde.isUsingLumping() 
314 
\end{python} 
315 

316 
\section{Stability Investigation} 
317 
It is now prudent to investigate the stability limitations of this problem. 
318 
First, we let the frequency content of the source be very small. If we define 
319 
the source as a cosine input, then the wavlength of the input is equal to the 
320 
radius of the source. Let this value be 5 meters. Now, if the maximum velocity 
321 
of the model is $c=380.0ms^{1}$, then the source 
322 
frequency is $f_{r} = \frac{380.0}{5} = 76.0 Hz$. This is a worst case 
323 
scenario with a small source and the models maximum velocity. 
324 

325 
Furthermore, we know from \autoref{sec:nsstab}, that the spatial sampling 
326 
frequency must be at least twice this value to ensure stability. If we assume 
327 
the model mesh is a square equispaced grid, 
328 
then the sampling interval is the side length divided by the number of samples, 
329 
given by $\Delta x = \frac{1000.0m}{400} = 2.5m$ and the maximum sampling 
330 
frequency capable at this interval is 
331 
$f_{s}=\frac{380.0ms^{1}}{2.5m}=152Hz$ this is just equal to the 
332 
required rate satisfying \autoref{eqn:samptheorem}. 
333 

334 
\autoref{fig:ex07sampth} depicts three examples where the grid has been 
335 
undersampled, sampled correctly, and over sampled. The grids used had 
336 
200, 400 and 800 nodes per side respectively. Obviously, the oversampled grid 
337 
retains the best resolution of the modelled wave. 
338 

339 
The time step required for each of these examples is simply calculated from 
340 
the propagation requirement. For a maximum velocity of $380.0ms^{1}$, 
341 
\begin{subequations} 
342 
\begin{equation} 
343 
\Delta t \leq \frac{1000.0m}{200} \frac{1}{380.0} = 0.013s 
344 
\end{equation} 
345 
\begin{equation} 
346 
\Delta t \leq \frac{1000.0m}{400} \frac{1}{380.0} = 0.0065s 
347 
\end{equation} 
348 
\begin{equation} 
349 
\Delta t \leq \frac{1000.0m}{800} \frac{1}{380.0} = 0.0032s 
350 
\end{equation} 
351 
\end{subequations} 
352 
Observe that for each doubling of the number of nodes in the mesh, we halve 
353 
the time step. To illustrate the impact this has, consider our model. If the 
354 
source is placed at the center, it is $500m$ from the nearest boundary. With a 
355 
velocity of $380.0ms^{1}$ it will take $\approx1.3s$ for the wavefront to 
356 
reach that boundary. In each case, this equates to $100$, $200$ and $400$ time 
357 
steps. This is again, only a best case scenario, for true stability these time 
358 
values may need to be halved and possibly halved again. 
359 

360 
\begin{figure}[ht] 
361 
\centering 
362 
\subfigure[Undersampled Example]{ 
363 
\includegraphics[width=0.45\textwidth,trim=0cm 6cm 5cm 6cm,clip]{figures/ex07usamp.png} 
364 
\label{fig:ex07usamp} 
365 
} 
366 
\subfigure[Just sampled Example]{ 
367 
\includegraphics[width=0.45\textwidth,trim=0cm 6cm 5cm 6cm,clip]{figures/ex07jsamp.png} 
368 
\label{fig:ex07jsamp} 
369 
} 
370 
\subfigure[Over sampled Example]{ 
371 
\includegraphics[width=0.45\textwidth,trim=0cm 6cm 5cm 6cm,clip]{figures/ex07nsamp.png} 
372 
\label{fig:ex07nsamp} 
373 
} 
374 
\caption{Sampling Theorem example for stability investigation} 
375 
\label{fig:ex07sampth} 
376 
\end{figure} 
377 
