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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 =
34 \sum\hackscore{i}^n
35 \frac{\partial^2 p}{\partial x^2\hackscore{i}}
36 \label{eqn:laplacian}
37 \end{equation}
38 For the two dimensional case in Cartesian coordinates \refEq{eqn:laplacian}
39 becomes;
40 \begin{equation}
41 \nabla^2 p = \frac{\partial^2 p}{\partial x^2}
42 + \frac{\partial^2 p}{\partial y^2}
43 \end{equation}
44
45 In \esc the Laplacian is calculated using the divergence representation and the
46 intrinsic functions \textit{grad()} and \textit{trace()}. The fucntion
47 \textit{grad{}} will return the spatial gradients of an object.
48 For a rank 0 solution, this is of the form;
49 \begin{equation}
50 \nabla p = \left[
51 \frac{\partial p}{\partial x \hackscore{0}},
52 \frac{\partial p}{\partial x \hackscore{1}}
53 \right]
54 \label{eqn:grad}
55 \end{equation}
56 Larger ranked solution objects will return gradient tensors. For example, a
57 pressure field which acts in the directions $p \hackscore{0}$ and $p
58 \hackscore{1}$ would return;
59 \begin{equation}
60 \nabla p = \begin{bmatrix}
61 \frac{\partial p \hackscore{0}}{\partial x \hackscore{0}} &
62 \frac{\partial p \hackscore{1}}{\partial x \hackscore{0}} \\
63 \frac{\partial p \hackscore{0}}{\partial x \hackscore{1}} &
64 \frac{\partial p \hackscore{1}}{\partial x \hackscore{1}}
65 \end{bmatrix}
66 \label{eqn:gradrank1}
67 \end{equation}
68
69 \refEq{eqn:grad} corresponds to the Linear PDE general form value
70 $X$. Notice however that the general form contains the term $X
71 \hackscore{i,j}$\footnote{This is the first derivative in the $j^{th}$
72 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
74 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
76 diagonal in \refeq{eqn:gradrank1}.
77
78 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
80 calculated via;
81 \begin{python}
82 gradient=grad(p)
83 \end{python}
84 and if the object is greater thank or equal to a rank 1 tensor, the trace of
85 the gradient is calculated.
86 \begin{python}
87 gradient=trace(grad(p))
88 \end{python}
89 These valuse can then be submitted to the PDE solver via the general form term
90 $X$. The Laplacian is then computed in the solution process by taking the
91 divergence of $X$.
92
93 Note, if you are unsure about the rank of your tensor, the \textit{getRank}
94 command will return the rank of the PDE object.
95 \begin{python}
96 rank = p.getRank()
97 \end{python}
98
99
100 \section{Numerical Solution Stability} \label{sec:nsstab}
101 Unfortunately, the wave equation belongs to a class of equations called
102 \textbf{stiff} PDEs. These types of equations can be difficult to solve
103 numerically as they tend to oscilate about the exact solution which can
104 eventually lead to a catastrophic failure in the solution. To counter this
105 problem, explicitly stable schemes like
106 the backwards Euler method are required. There are two variables which must be
107 considered for stability when numerically trying to solve the wave equation.
108 For linear media, the two variables are related via;
109 \begin{equation} \label{eqn:freqvel}
110 f=\frac{v}{\lambda}
111 \end{equation}
112 The velocity $v$ that a wave travels in a medium is an important variable. For
113 stability the analytical wave must not propagate faster than the numerical wave
114 is able to, and in general, needs to be much slower than the numerical wave.
115 For example, a line 100m long is discretised into 1m intervals or 101 nodes. If
116 a wave enters with a propagation velocity of 100m/s then the travel time for
117 the wave between each node will be 0.01 seconds. The time step, must therefore
118 be significantly less than this. Of the order $10E-4$ would be appropriate.
119
120 The wave frequency content also plays a part in numerical stability. The
121 nyquist-sampling theorem states that a signals bandwidth content will be
122 accurately represented when an equispaced sampling rate $f \hackscore{n}$ is
123 equal to or greater than twice the maximum frequency of the signal
124 $f\hackscore{s}$, or;
125 \begin{equation} \label{eqn:samptheorem}
126 f\hackscore{n} \geqslant f\hackscore{s}
127 \end{equation}
128 For example a 50Hz signal will require a sampling rate greater than 100Hz or
129 one sample every 0.01 seconds. The wave equation relies on a spatial frequency,
130 thus the sampling theorem in this case applies to the solution mesh spacing. In
131 this way, the frequency content of the input signal directly affects the time
132 discretisation of the problem.
133
134 To accurately model the wave equation with high resolutions and velocities
135 means that very fine spatial and time discretisation is necessary for most
136 problems.
137 This requirement makes the wave equation arduous to
138 solve numerically due to the large number of time iterations required in each
139 solution. Models with very high velocities and frequencies will be the worst
140 affected by this problem.
141
142 \section{Displacement Solution}
143 \sslist{example07a.py}
144
145 We begin the solution to this PDE with the centred difference formula for the
146 second derivative;
147 \begin{equation}
148 f''(x) \approx \frac{f(x+h - 2f(x) + f(x-h)}{h^2}
149 \label{eqn:centdiff}
150 \end{equation}
151 substituting \refEq{eqn:centdiff} for $\frac{\partial ^2 p }{\partial t ^2}$
152 in \refEq{eqn:acswave};
153 \begin{equation}
154 \nabla ^2 p - \frac{1}{c^2h^2} \left[p\hackscore{(t+1)} - 2p\hackscore{(t)} +
155 p\hackscore{(t-1)} \right]
156 = 0
157 \label{eqn:waveu}
158 \end{equation}
159 Rearranging for $p_{(t+1)}$;
160 \begin{equation}
161 p\hackscore{(t+1)} = c^2 h^2 \nabla ^2 p\hackscore{(t)} +2p\hackscore{(t)} -
162 p\hackscore{(t-1)}
163 \end{equation}
164 this can be compared with the general form of the \modLPDE module and it
165 becomes clear that $D=1$, $X\hackscore{i,j}=-c^2 h^2 \nabla ^2 p_{(t)}$ and
166 $Y=2p_{(t)} - p_{(t-1)}$.
167
168 The solution script is similar to others that we have created in previous
169 chapters. The general steps are;
170 \begin{enumerate}
171 \item The necessary libraries must be imported.
172 \item The domain needs to be defined.
173 \item The time iteration and control parameters need to be defined.
174 \item The PDE is initialised with source and boundary conditions.
175 \item The time loop is started and the PDE is solved at consecutive time steps.
176 \item All or select solutions are saved to file for visualisation lated on.
177 \end{enumerate}
178
179 Parts of the script which warrant more attention are the definition of the
180 source, visualising the source, the solution time loop and the VTK data export.
181
182 \subsection{Pressure Sources}
183 As the pressure is a scalar, one need only define the pressure for two
184 time steps prior to the start of the solution loop. Two known solutions are
185 required because the wave equation contains a double partial derivative with
186 respect to time. This is often a good opportunity to introduce a source to the
187 solution. This model has the source located at it's centre. The source should
188 be smooth and cover a number of samples to satisfy the frequency stability
189 criterion. Small sources will generate high frequency signals. Here, the source
190 is defined by a cosine function.
191 \begin{python}
192 U0=0.01 # amplitude of point source
193 xc=[500,500] #location of point source
194 # define small radius around point xc
195 src_radius = 30
196 # for first two time steps
197 u=U0*(cos(length(x-xc)*3.1415/src_radius)+1)*\
198 whereNegative(length(x-xc)-src_radius)
199 u_m1=u
200 \end{python}
201 When using a rectangular domain
202
203 \subsection{Visualising the Source}
204 There are two options for visualising the source. The first is to export the
205 initial conditions of the model to VTK, which can be interpreted as a scalar
206 suface in mayavi. The second is to take a cross section of the model.
207
208 For the later, we will require the \textit{Locator} function.
209 First \verb Locator must be imported;
210 \begin{python}
211 from esys.escript.pdetools import Locator
212 \end{python}
213 The function can then be used on the domain to locate the nearest domain node
214 to the point or points of interest.
215
216 It is now necessary to build a list of $(x,y)$ locations that specify where are
217 model slice will go. This is easily implemeted with a loop;
218 \begin{python}
219 cut_loc=[]
220 src_cut=[]
221 for i in range(ndx/2-ndx/10,ndx/2+ndx/10):
222 cut_loc.append(xstep*i)
223 src_cut.append([xstep*i,xc[1]])
224 \end{python}
225 We then submit the output to \verb Locator and finally return the appropriate
226 values using the \verb getValue function.
227 \begin{python}
228 src=Locator(mydomain,src_cut)
229 src_cut=src.getValue(u)
230 \end{python}
231 It is then a trivial task to plot and save the output using \mpl.
232 \begin{python}
233 pl.plot(cut_loc,src_cut)
234 pl.axis([xc[0]-src_radius*3,xc[0]+src_radius*3,0.,2*U0])
235 pl.savefig(os.path.join(savepath,"source_line.png"))
236 \end{python}
237 \begin{figure}[h]
238 \centering
239 FIXME PLEASE!
240 % \includegraphics[width=6in]{figures/sourceline.png}
241 \caption{Cross section of the source function.}
242 \label{fig:cxsource}
243 \end{figure}
244
245
246 \subsection{Point Monitoring}
247 In the more general case where the solution mesh is irregular or specific
248 locations need to be monitored, it is simple enough to use the \textit{Locator}
249 function.
250 \begin{python}
251 rec=Locator(mydomain,[250.,250.])
252 \end{python}
253 When the solution \verb u is updated we can extract the value at that point
254 via;
255 \begin{python}
256 u_rec=rec.getValue(u)
257 \end{python}
258 For consecutive time steps one can record the values from \verb u_rec in an
259 array initialised as \verb u_rec0=[] with;
260 \begin{python}
261 u_rec0.append(rec.getValue(u))
262 \end{python}
263
264 It can be useful to monitor the value at a single or multiple individual points
265 in the model during the modelling process. This is done using
266 the \verb Locator function.
267
268
269 \section{Acceleration Solution}
270 \sslist{example07b.py}
271
272 An alternative method is to solve for the acceleration $\frac{\partial ^2
273 p}{\partial t^2}$ directly, and derive the displacement solution from the
274 PDE solution. \refEq{eqn:waveu} is thus modified;
275 \begin{equation}
276 \nabla ^2 p - \frac{1}{c^2} a = 0
277 \label{eqn:wavea}
278 \end{equation}
279 and can be solved directly with $Y=0$ and $X=-c^2 \nabla ^2 p\hackscore{(t)}$.
280 After each iteration the displacement is re-evaluated via;
281 \begin{equation}
282 p\hackscore{(t+1)}=2p\hackscore{(t)} - p\hackscore{(t-1)} + h^2a
283 \end{equation}
284
285 \subsection{Lumping}
286 For \esc, the acceleration solution is prefered as it allows the use of matrix
287 lumping. Lumping or mass lumping as it is sometimes known, is the process of
288 aggressively approximating the density elements of a mass matrix into the main
289 diagonal. The use of Lumping is motivaed by the simplicity of diagonal matrix
290 inversion. As a result, Lumping can significantly reduce the computational
291 requirements of a problem. Care should be taken however, as this
292 function can only be used when the $A$, $B$ and $C$ coefficients of the
293 general form are zero.
294
295 To turn lumping on in \esc one can use the command;
296 \begin{python}
297 mypde.getSolverOptions().setSolverMethod(mypde.getSolverOptions().LUMPING)
298 \end{python}
299 It is also possible to check if lumping is set using;
300 \begin{python}
301 print mypde.isUsingLumping()
302 \end{python}
303
304 \section{Stability Investigation}
305 It is now prudent to investigate the stability limitations of this problem.
306 First, we let the frequency content of the source be very small. If we define
307 the source as a cosine input, than the wavlength of the input is equal to the
308 radius of the source. Let this value be 5 meters. Now, if the maximum velocity
309 of the model is $c=380.0ms^{-1}$ then the source
310 frequency is $f\hackscore{r} = \frac{380.0}{5} = 76.0 Hz$. This is a worst case
311 scenario with a small source and the models maximum velocity.
312
313 Furthermore, we know from \refSec{sec:nsstab}, that the spatial sampling
314 frequency must be at least twice this value to ensure stability. If we assume
315 the model mesh is a square equispaced grid,
316 then the sampling interval is the side length divided by the number of samples,
317 given by $\Delta x = \frac{1000.0m}{400} = 2.5m$ and the maximum sampling
318 frequency capable at this interval is
319 $f\hackscore{s}=\frac{380.0ms^{-1}}{2.5m}=152Hz$ this is just equal to the
320 required rate satisfying \refeq{eqn:samptheorem}.
321
322 \reffig{fig:ex07sampth} depicts three examples where the grid has been
323 undersampled, sampled correctly, and over sampled. The grids used had
324 200, 400 and 800 nodes per side respectively. Obviously, the oversampled grid
325 retains the best resolution of the modelled wave.
326
327 The time step required for each of these examples is simply calculated from
328 the propagation requirement. For a maximum velocity of $380.0ms^{-1}$,
329 \begin{subequations}
330 \begin{equation}
331 \Delta t \leq \frac{1000.0m}{200} \frac{1}{380.0} = 0.013s
332 \end{equation}
333 \begin{equation}
334 \Delta t \leq \frac{1000.0m}{400} \frac{1}{380.0} = 0.0065s
335 \end{equation}
336 \begin{equation}
337 \Delta t \leq \frac{1000.0m}{800} \frac{1}{380.0} = 0.0032s
338 \end{equation}
339 \end{subequations}
340 We can see, that for each doubling of the number of nodes in the mesh, we halve
341 the timestep. To illustrate the impact this has, consider our model. If the
342 source is placed at the center, it is $500m$ from the nearest boundary. With a
343 velocity of $380.0ms^{-1}$ it will take $\approx1.3s$ for the wavefront to
344 reach that boundary. In each case, this equates to $100$, $200$ and $400$ time
345 steps. This is again, only a best case scenario, for true stability these time
346 values may need to be halved and possibly havled again.
347
348 \begin{figure}[ht]
349 \centering
350 \subfigure[Undersampled Example]{
351 \includegraphics[width=3in]{figures/ex07usamp.png}
352 \label{fig:ex07usamp}
353 }
354 \subfigure[Just sampled Example]{
355 \includegraphics[width=3in]{figures/ex07jsamp.png}
356 \label{fig:ex07jsamp}
357 }
358 \subfigure[Over sampled Example]{
359 \includegraphics[width=3in]{figures/ex07nsamp.png}
360 \label{fig:ex07nsamp}
361 }
362 \label{fig:ex07sampth}
363 \caption{Sampling Theorem example for stability
364 investigation.}
365 \end{figure}
366
367
368
369
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