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Updates to wave equation examples. Pressure wave examples should be completed now. Working on seismic wave examples. Cookbook chapter for pressure wave examples also added.
1
2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
3 %
4 % Copyright (c) 2003-2010 by University of Queensland
5 % Earth Systems Science Computational Center (ESSCC)
6 % http://www.uq.edu.au/esscc
7 %
8 % Primary Business: Queensland, Australia
9 % Licensed under the Open Software License version 3.0
10 % http://www.opensource.org/licenses/osl-3.0.php
11 %
12 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
13
14
15
16 The acoustic wave equation governs the propagation of pressure waves. Wave
17 types that obey this law tend to travel in liquids or gases where shear waves
18 or longitudinal style wave motion is not possible. An obvious example is sound
19 waves.
20
21 The acoustic wave equation is defined as;
22 \begin{equation}
23 \nabla ^2 p - \frac{1}{c^2} \frac{\partial ^2 p}{\partial t^2} = 0
24 \label{eqn:acswave}
25 \end{equation}
26 where $p$ is the pressure, $t$ is the time and $c$ is the wave velocity.
27
28 \section{The Laplacian in \esc}
29 The Laplacian opperator which can be written as $\Delta$ or $\nabla^2$ is
30 calculated via the divergence of the gradient of the object, which is in this
31 example $p$. Thus we can write;
32 \begin{equation}
33 \nabla^2 p = \nabla \cdot \nabla p = \frac{\partial^2 p}{\partial
34 x^2\hackscore{i}}
35 \label{eqn:laplacian}
36 \end{equation}
37 For the two dimensional case in Cartesian coordinates \refEq{eqn:laplacian}
38 becomes;
39 \begin{equation}
40 \nabla^2 p = \frac{\partial^2 p}{\partial x^2}
41 + \frac{\partial^2 p}{\partial y^2}
42 \end{equation}
43
44 In \esc the Laplacian is calculated using the divergence representation and the
45 intrinsic functions \textit{grad()} and \textit{trace()}. The fucntion
46 \textit{grad{}} will return the spatial gradients of an object.
47 For a rank 0 solution, this is of the form;
48 \begin{equation}
49 \nabla p = \left[
50 \frac{\partial p}{\partial x \hackscore{0}},
51 \frac{\partial p}{\partial x \hackscore{1}}
52 \right]
53 \label{eqn:grad}
54 \end{equation}
55 Larger ranked solution objects will return gradient tensors. For example, a
56 pressure field which acts in the directions $p \hackscore{0}$ and $p
57 \hackscore{1}$ would return;
58 \begin{equation}
59 \nabla p = \begin{bmatrix}
60 \frac{\partial p \hackscore{0}}{\partial x \hackscore{0}} &
61 \frac{\partial p \hackscore{1}}{\partial x \hackscore{0}} \\
62 \frac{\partial p \hackscore{0}}{\partial x \hackscore{1}} &
63 \frac{\partial p \hackscore{1}}{\partial x \hackscore{1}}
64 \end{bmatrix}
65 \label{eqn:gradrank1}
66 \end{equation}
67
68 \refEq{eqn:grad} corresponds to the Linear PDE general form value
69 $X$. Notice however that the gernal form contains the term $X \hackscore{i,j}$,
70 hence for a rank 0 object there is no need to do more than calculate the
71 gradient and submit it to the solver. In the case of the rank 1 or greater
72 object, it is nesscary to calculate the trace also. This is the sum of the
73 diagonal in \refeq{eqn:gradrank1}.
74
75 Thus when solving for equations containing the Laplacian one of two things must
76 be completed. If the object \verb p is less than rank 1 the gradient is
77 calculated via;
78 \begin{python}
79 gradient=grad(p)
80 \end{python}
81 and if the object is greater thank or equal to a rank 1 tensor, the trace of
82 the gradient is calculated.
83 \begin{python}
84 gradient=trace(grad(p))
85 \end{python}
86 These valuse can then be submitted to the PDE solver via the general form term
87 $X$. The Laplacian is then computed in the solution process by taking the
88 divergence of $X$.
89
90 Note, if you are unsure about the rank of your tensor, the \textit{getRank}
91 command will return the rank of the PDE object.
92 \begin{python}
93 rank = p.getRank()
94 \end{python}
95
96
97 \section{Numerical Solution Stability} \label{sec:nsstab}
98 Unfortunately, the wave equation belongs to a class of equations called
99 \textbf{stiff} PDEs. These types of equations can be difficult to solve
100 numerically as they tend to oscilate about the exact solution which can
101 eventually lead to a catastrophic failure in the solution. To counter this
102 problem, explicitly stable schemes like
103 the backwards Euler method are required. There are two variables which must be
104 considered for stability when numerically trying to solve the wave equation.
105 For linear media, the two variables are related via;
106 \begin{equation} \label{eqn:freqvel}
107 f=\frac{v}{\lambda}
108 \end{equation}
109 The velocity $v$ that a wave travels in a medium is an important variable. For
110 stability the analytical wave must not propagate faster than the numerical wave
111 is able to, and in general, needs to be much slower than the numerical wave.
112 For example, a line 100m long is discretised into 1m intervals or 101 nodes. If
113 a wave enters with a propagation velocity of 100m/s then the travel time for
114 the wave between each node will be 0.01 seconds. The time step, must therefore
115 be significantly less than this. Of the order $10E-4$ would be appropriate.
116
117 The wave frequency content also plays a part in numerical stability. The
118 nyquist-sampling theorem states that a signals bandwidth content will be
119 accurately represented when an equispaced sampling rate $f \hackscore{n}$ is
120 equal to or greater than twice the maximum frequency of the signal
121 $f\hackscore{s}$, or;
122 \begin{equation} \label{eqn:samptheorem}
123 f\hackscore{n} \geqslant f\hackscore{s}
124 \end{equation}
125 For example a 50Hz signal will require a sampling rate greater than 100Hz or
126 one sample every 0.01 seconds. The wave equation relies on a spatial frequency,
127 thus the sampling theorem in this case applies to the solution mesh spacing. In
128 this way, the frequency content of the input signal directly affects the time
129 discretisation of the problem.
130
131 To accurately model the wave equation with high resolutions and velocities
132 means that very fine spatial and time discretisation is necessary for most
133 problems.
134 This requirement makes the wave equation arduous to
135 solve numerically due to the large number of time iterations required in each
136 solution. Models with very high velocities and frequencies will be the worst
137 effected by this problem.
138
139 \section{Displacement Solution}
140 \sslist{example07a.py}
141
142 We begin the solution to this PDE with the centred difference formula for the
143 second derivative;
144 \begin{equation}
145 f''(x) \approx \frac{f(x+h - 2f(x) + f(x-h)}{h^2}
146 \label{eqn:centdiff}
147 \end{equation}
148 substituting \refEq{eqn:centdiff} for $\frac{\partial ^2 p }{\partial t ^2}$
149 in \refEq{eqn:acswave};
150 \begin{equation}
151 \nabla ^2 p - \frac{1}{c^2h^2} \left[p\hackscore{(t+1)} - 2p\hackscore{(t)} +
152 p\hackscore{(t-1)} \right]
153 = 0
154 \label{eqn:waveu}
155 \end{equation}
156 Rearranging for $p_{(t+1)}$;
157 \begin{equation}
158 p\hackscore{(t+1)} = c^2 h^2 \nabla ^2 p\hackscore{(t)} +2p\hackscore{(t)} -
159 p\hackscore{(t-1)}
160 \end{equation}
161 this can be compared with the general form of the \modLPDE module and it
162 becomes clear that $D=1$, $X\hackscore{i,j}=-c^2 h^2 \nabla ^2 p_{(t)}$ and
163 $Y=2p_{(t)} - p_{(t-1)}$.
164
165 The solution script is similar to others that we have created in previous
166 chapters. The general steps are;
167 \begin{enumerate}
168 \item The necessary libraries must be imported.
169 \item The domain needs to be defined.
170 \item The time iteration and control parameters need to be defined.
171 \item The PDE is initialised with source and boundary conditions.
172 \item The time loop is started and the PDE is solved at consecutive time steps.
173 \item All or select solutions are saved to file for visualisation lated on.
174 \end{enumerate}
175
176 Parts of the script which warrant more attention are the definition of the
177 source, visualising the source, the solution time loop and the VTK data export.
178
179 \subsection{Pressure Sources}
180 As the pressure is a scalar, one need only define the pressure for two
181 time steps prior to the start of the solution loop. Two known solutions are
182 required because the wave equation contains a double partial derivative with
183 respect to time. This is often a good opportunity to introduce a source to the
184 solution. This model has the source located at it's centre. The source should
185 be smooth and cover a number of samples to satisfy the frequency stability
186 criterion. Small sources will generate high frequency signals. Here, the source
187 is defined by a cosine function.
188 \begin{python}
189 U0=0.01 # amplitude of point source
190 xc=[500,500] #location of point source
191 # define small radius around point xc
192 src_radius = 30
193 # for first two time steps
194 u=U0*(cos(length(x-xc)*3.1415/src_radius)+1)*\
195 whereNegative(length(x-xc)-src_radius)
196 u_m1=u
197 \end{python}
198 When using a rectangular domain
199
200 \subsection{Visualising the Source}
201 There are two options for visualising the source. The first is to export the
202 initial conditions of the model to VTK, which can be interpreted as a scalar
203 suface in mayavi. The second is to take a cross section of the model.
204
205 For the later, we will require the \textit{Locator} function.
206 First \verb Locator must be imported;
207 \begin{python}
208 from esys.escript.pdetools import Locator
209 \end{python}
210 The function can then be used on the domain to locate the nearest domain node
211 to the point or points of interest.
212
213 It is now necessary to build a list of $(x,y)$ locations that specify where are
214 model slice will go. This is easily implemeted with a loop;
215 \begin{python}
216 cut_loc=[]
217 src_cut=[]
218 for i in range(ndx/2-ndx/10,ndx/2+ndx/10):
219 cut_loc.append(xstep*i)
220 src_cut.append([xstep*i,xc[1]])
221 \end{python}
222 We then submit the output to \verb Locator and finally return the appropriate
223 values using the \verb getValue function.
224 \begin{python}
225 src=Locator(mydomain,src_cut)
226 src_cut=src.getValue(u)
227 \end{python}
228 It is then a trivial task to plot and save the output using \mpl.
229 \begin{python}
230 pl.plot(cut_loc,src_cut)
231 pl.axis([xc[0]-src_radius*3,xc[0]+src_radius*3,0.,2*U0])
232 pl.savefig(os.path.join(savepath,"source_line.png"))
233 \end{python}
234 \begin{figure}[h]
235 \includegraphics[width=5in]{figures/sourceline.png}
236 \caption{Cross section of the source function.}
237 \label{fig:cxsource}
238 \end{figure}
239
240
241 \subsection{Point Monitoring}
242 In the more general case where the solution mesh is irregular or specific
243 locations need to be monitored, it is simple enough to use the \textit{Locator}
244 function.
245 \begin{python}
246 rec=Locator(mydomain,[250.,250.])
247 \end{python}
248 When the solution \verb u is update we can extract the value at that point
249 via;
250 \begin{python}
251 u_rec=rec.getValue(u)
252 \end{python}
253 For consecutive time steps one can record the values from \verb u_rec in an
254 array initialised as \verb u_rec0=[] with;
255 \begin{python}
256 u_rec0.append(rec.getValue(u))
257 \end{python}
258
259 It can be useful to monitor the value at a single or multiple individual points
260 in the model during the modelling process. This is done using
261 the \verb Locator function.
262
263
264 \section{Acceleration Solution}
265 \sslist{example07b.py}
266
267 An alternative method is to solve for the acceleration $\frac{\partial ^2
268 p}{\partial t^2}$ directly, and derive the the displacement solution from the
269 PDE solution. \refEq{eqn:waveu} is thus modified;
270 \begin{equation}
271 \nabla ^2 p - \frac{1}{c^2} a = 0
272 \label{eqn:wavea}
273 \end{equation}
274 and can be solved directly with $Y=0$ and $X=-c^2 \nabla ^2 p\hackscore{(t)}$.
275 After each iteration the displacement is re-evaluated via;
276 \begin{equation}
277 p\hackscore{(t+1)}=2p\hackscore{(t)} - p\hackscore{(t-1)} + h^2a
278 \end{equation}
279
280 For \esc, the acceleration solution is prefered as it allows the use of matrix
281 lumping. Lumping or mass lumping as it is sometimes known, is the process of
282 aggressively approximating the density elements of a mass matrix into the main
283 diagonal. The use of Lumping is motivaed by the simplicity of diagonal matrix
284 inversion. As a result, Lumping can significantly reduce the computational
285 requirements of a problem.
286
287 To turn lumping on in \esc one can use the command;
288 \begin{python}
289 mypde.getSolverOptions().setSolverMethod(mypde.getSolverOptions().LUMPING)
290 \end{python}
291 It is also possible to check if lumping is set using;
292 \begin{python}
293 print mypde.isUsingLumping()
294 \end{python}
295
296 \section{Stability Investigation}
297 It is now prudent to investigate the stability limitations of this problem.
298 First, we let the frequency content of the source be very small. If we define
299 the source as a cosine input, than the wavlength of the input is equal to the
300 radius of the source. Let this value be 5 meters. Now, if the maximum velocity
301 of the model is $c=380.0ms^{-1}$ then the source
302 frequency is $f\hackscore{r} = \frac{380.0}{5} = 76.0 Hz$. This is a worst case
303 scenario with a small source and the models maximum velocity.
304
305 Furthermore, we know from \refSec{sec:nsstab}, that the spatial sampling
306 frequency must be at least twice this value to ensure stability. If we assume
307 the model mesh is a square equispaced grid,
308 then the sampling interval is the side length divided by the number of samples,
309 given by $\Delta x = \frac{1000.0m}{400} = 2.5m$ and the maximum sampling
310 frequency capable at this interval is
311 $f\hackscore{s}=\frac{380.0ms^{-1}}{2.5m}=152Hz$ this is just equal to the
312 required rate satisfying \refeq{eqn:samptheorem}.
313
314 \reffig{fig:ex07sampth} depicts three examples where the grid has been
315 undersampled, sampled correctly, and over sampled. The grids used had
316 200, 400 and 800 nodes per side respectively. Obviously, the oversampled grid
317 retains the best resolution of the modelled wave.
318
319 The time step required for each of these examples is simply calculated from
320 the propagation requirement. For a maximum velocity of $380.0ms^{-1}$,
321 \begin{subequations}
322 \begin{equation}
323 \Delta t \leq \frac{1000.0m}{200} \frac{1}{380.0} = 0.013s
324 \end{equation}
325 \begin{equation}
326 \Delta t \leq \frac{1000.0m}{400} \frac{1}{380.0} = 0.0065s
327 \end{equation}
328 \begin{equation}
329 \Delta t \leq \frac{1000.0m}{800} \frac{1}{380.0} = 0.0032s
330 \end{equation}
331 \end{subequations}
332 We can see, that for each doubling of the number of nodes in the mesh, we halve
333 the timestep. To illustrate the impact this has, consider our model. If the
334 source is placed at the center, it is $500m$ from the nearest boundary. With a
335 velocity of $380.0ms^{-1}$ it will take $\approx1.3s$ for the wavefront to
336 reach that boundary. In each case, this equates to $100$, $200$ and $400$ time
337 steps. This is again, only a best case scenario, for true stability these time
338 values may need to be halved and possibly havled again.
339
340 \begin{figure}[ht]
341 \centering
342 \subfigure[Undersampled Example]{
343 \includegraphics[width=3in]{figures/ex07usamp.png}
344 \label{fig:ex07usamp}
345 }
346 \subfigure[Just sampled Example]{
347 \includegraphics[width=3in]{figures/ex07jsamp.png}
348 \label{fig:ex07jsamp}
349 }
350 \subfigure[Over sampled Example]{
351 \includegraphics[width=3in]{figures/ex07nsamp.png}
352 \label{fig:ex07nsamp}
353 }
354 \label{fig:ex07sampth}
355 \caption{Sampling Theorem example for stability
356 investigation.}
357 \end{figure}
358
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