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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 than this. Of the order $10E-4$ 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 $10E-4$, a safety factor of $f=0.1$ was used.
130
131 The wave frequency content also plays a part in numerical stability. The
132 Nyquist-sampling 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(x-h)}{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_{(t-1)} \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_{(t-1)}
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 $Y=2p_{(t)} - p_{(t-1)}$.
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(x-xc)*3.1415/src_radius)+1)*\
208 whereNegative(length(x-xc)-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/2-ndx/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 re-evaluated via;
290 \begin{equation}
291 p_{(t+1)}=2p_{(t)} - p_{(t-1)} + h^2a
292 \end{equation}
293
294 \subsection{Lumping}
295 For \esc, the acceleration solution is preferred 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 motivated 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

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