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1 ahallam 3003
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6     % http://www.uq.edu.au/esscc
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9     % Licensed under the Open Software License version 3.0
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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 ahallam 3004 or longitudinal style wave motion is not possible. An obvious example is sound
19 ahallam 3003 waves.
20    
21 ahallam 3004 The acoustic wave equation is defined as;
22 ahallam 3003 \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 ahallam 3370 where $p$ is the pressure, $t$ is the time and $c$ is the wave velocity. In this
27     chapter the acoustic wave equation is demonstrated. Important steps include the
28     translation of the Laplacian $\nabla^2$ to the \esc general form, the stiff
29 ahallam 3373 equation stability criterion and solving for the displacement or acceleration solution.
30 ahallam 3003
31 ahallam 3004 \section{The Laplacian in \esc}
32 ahallam 3373 The Laplacian operator which can be written as $\Delta$ or $\nabla^2$, is
33 ahallam 3370 calculated via the divergence of the gradient of the object, which in this
34     example is the scalar $p$. Thus we can write;
35 ahallam 3004 \begin{equation}
36 ahallam 3029 \nabla^2 p = \nabla \cdot \nabla p =
37 jfenwick 3308 \sum_{i}^n
38     \frac{\partial^2 p}{\partial x^2_{i}}
39 ahallam 3004 \label{eqn:laplacian}
40     \end{equation}
41 ahallam 3232 For the two dimensional case in Cartesian coordinates \autoref{eqn:laplacian}
42 ahallam 3004 becomes;
43     \begin{equation}
44     \nabla^2 p = \frac{\partial^2 p}{\partial x^2}
45     + \frac{\partial^2 p}{\partial y^2}
46     \end{equation}
47 ahallam 3003
48 ahallam 3004 In \esc the Laplacian is calculated using the divergence representation and the
49 ahallam 3370 intrinsic functions \textit{grad()} and \textit{trace()}. The function
50 ahallam 3004 \textit{grad{}} will return the spatial gradients of an object.
51     For a rank 0 solution, this is of the form;
52     \begin{equation}
53     \nabla p = \left[
54 jfenwick 3308 \frac{\partial p}{\partial x _{0}},
55     \frac{\partial p}{\partial x _{1}}
56 ahallam 3004 \right]
57     \label{eqn:grad}
58     \end{equation}
59     Larger ranked solution objects will return gradient tensors. For example, a
60 jfenwick 3308 pressure field which acts in the directions $p _{0}$ and $p
61     _{1}$ would return;
62 ahallam 3004 \begin{equation}
63     \nabla p = \begin{bmatrix}
64 jfenwick 3308 \frac{\partial p _{0}}{\partial x _{0}} &
65     \frac{\partial p _{1}}{\partial x _{0}} \\
66     \frac{\partial p _{0}}{\partial x _{1}} &
67     \frac{\partial p _{1}}{\partial x _{1}}
68 ahallam 3004 \end{bmatrix}
69     \label{eqn:gradrank1}
70     \end{equation}
71 ahallam 3003
72 ahallam 3232 \autoref{eqn:grad} corresponds to the Linear PDE general form value
73 ahallam 3373 $X$. Notice however, that the general form contains the term $X
74 jfenwick 3308 _{i,j}$\footnote{This is the first derivative in the $j^{th}$
75 ahallam 3029 direction for the $i^{th}$ component of the solution.},
76 ahallam 3370 hence for a rank 0 object there is no need to do more then calculate the
77 ahallam 3004 gradient and submit it to the solver. In the case of the rank 1 or greater
78 ahallam 3373 object, it is also necessary to calculate the trace. This is the sum of the
79 ahallam 3232 diagonal in \autoref{eqn:gradrank1}.
80 ahallam 3004
81     Thus when solving for equations containing the Laplacian one of two things must
82 ahallam 3370 be completed. If the object \verb!p! is less then rank 1 the gradient is
83 ahallam 3004 calculated via;
84 ahallam 3025 \begin{python}
85 ahallam 3063 gradient=grad(p)
86 ahallam 3025 \end{python}
87 ahallam 3370 and if the object is greater then or equal to a rank 1 tensor, the trace of
88 ahallam 3004 the gradient is calculated.
89 ahallam 3025 \begin{python}
90 ahallam 3004 gradient=trace(grad(p))
91 ahallam 3025 \end{python}
92 ahallam 3370 These values can then be submitted to the PDE solver via the general form term
93 ahallam 3004 $X$. The Laplacian is then computed in the solution process by taking the
94     divergence of $X$.
95    
96 ahallam 3025 Note, if you are unsure about the rank of your tensor, the \textit{getRank}
97     command will return the rank of the PDE object.
98     \begin{python}
99     rank = p.getRank()
100     \end{python}
101    
102    
103     \section{Numerical Solution Stability} \label{sec:nsstab}
104 ahallam 3004 Unfortunately, the wave equation belongs to a class of equations called
105     \textbf{stiff} PDEs. These types of equations can be difficult to solve
106 ahallam 3373 numerically as they tend to oscillate about the exact solution, which can
107 ahallam 3370 eventually lead to a catastrophic failure. To counter this problem, explicitly
108 ahallam 3373 stable schemes like the backwards Euler method, and correct parameterisation of
109 ahallam 3370 the problem are required.
110    
111     There are two variables which must be considered for
112     stability when numerically trying to solve the wave equation. For linear media,
113     the two variables are related via;
114 ahallam 3004 \begin{equation} \label{eqn:freqvel}
115     f=\frac{v}{\lambda}
116     \end{equation}
117 ahallam 3025 The velocity $v$ that a wave travels in a medium is an important variable. For
118 ahallam 3370 stability the analytical wave must not propagate faster then the numerical wave
119     is able to, and in general, needs to be much slower then the numerical wave.
120 ahallam 3003 For example, a line 100m long is discretised into 1m intervals or 101 nodes. If
121     a wave enters with a propagation velocity of 100m/s then the travel time for
122     the wave between each node will be 0.01 seconds. The time step, must therefore
123 ahallam 3370 be significantly less then this. Of the order $10E-4$ would be appropriate.
124 ahallam 3003
125 ahallam 3004 The wave frequency content also plays a part in numerical stability. The
126     nyquist-sampling theorem states that a signals bandwidth content will be
127 jfenwick 3308 accurately represented when an equispaced sampling rate $f _{n}$ is
128 ahallam 3370 equal to or greater then twice the maximum frequency of the signal
129 jfenwick 3308 $f_{s}$, or;
130 ahallam 3004 \begin{equation} \label{eqn:samptheorem}
131 jfenwick 3308 f_{n} \geqslant f_{s}
132 ahallam 3004 \end{equation}
133 ahallam 3370 For example a 50Hz signal will require a sampling rate greater then 100Hz or
134 ahallam 3004 one sample every 0.01 seconds. The wave equation relies on a spatial frequency,
135 ahallam 3373 thus the sampling theorem in this case applies to the solution mesh spacing.
136     This relationship confirms that the frequency content of the input signal
137     directly affects the time discretisation of the problem.
138 ahallam 3004
139     To accurately model the wave equation with high resolutions and velocities
140     means that very fine spatial and time discretisation is necessary for most
141     problems.
142     This requirement makes the wave equation arduous to
143 ahallam 3003 solve numerically due to the large number of time iterations required in each
144 ahallam 3004 solution. Models with very high velocities and frequencies will be the worst
145 ahallam 3029 affected by this problem.
146 ahallam 3003
147     \section{Displacement Solution}
148     \sslist{example07a.py}
149    
150     We begin the solution to this PDE with the centred difference formula for the
151     second derivative;
152     \begin{equation}
153     f''(x) \approx \frac{f(x+h - 2f(x) + f(x-h)}{h^2}
154     \label{eqn:centdiff}
155     \end{equation}
156 ahallam 3232 substituting \autoref{eqn:centdiff} for $\frac{\partial ^2 p }{\partial t ^2}$
157     in \autoref{eqn:acswave};
158 ahallam 3003 \begin{equation}
159 jfenwick 3308 \nabla ^2 p - \frac{1}{c^2h^2} \left[p_{(t+1)} - 2p_{(t)} +
160     p_{(t-1)} \right]
161 ahallam 3003 = 0
162     \label{eqn:waveu}
163     \end{equation}
164     Rearranging for $p_{(t+1)}$;
165     \begin{equation}
166 jfenwick 3308 p_{(t+1)} = c^2 h^2 \nabla ^2 p_{(t)} +2p_{(t)} -
167     p_{(t-1)}
168 ahallam 3003 \end{equation}
169     this can be compared with the general form of the \modLPDE module and it
170 jfenwick 3308 becomes clear that $D=1$, $X_{i,j}=-c^2 h^2 \nabla ^2 p_{(t)}$ and
171 ahallam 3004 $Y=2p_{(t)} - p_{(t-1)}$.
172 ahallam 3003
173 ahallam 3025 The solution script is similar to others that we have created in previous
174 ahallam 3004 chapters. The general steps are;
175     \begin{enumerate}
176     \item The necessary libraries must be imported.
177     \item The domain needs to be defined.
178     \item The time iteration and control parameters need to be defined.
179     \item The PDE is initialised with source and boundary conditions.
180     \item The time loop is started and the PDE is solved at consecutive time steps.
181 ahallam 3370 \item All or select solutions are saved to file for visualisation later on.
182 ahallam 3004 \end{enumerate}
183    
184     Parts of the script which warrant more attention are the definition of the
185     source, visualising the source, the solution time loop and the VTK data export.
186    
187     \subsection{Pressure Sources}
188     As the pressure is a scalar, one need only define the pressure for two
189     time steps prior to the start of the solution loop. Two known solutions are
190     required because the wave equation contains a double partial derivative with
191     respect to time. This is often a good opportunity to introduce a source to the
192     solution. This model has the source located at it's centre. The source should
193     be smooth and cover a number of samples to satisfy the frequency stability
194 ahallam 3370 criterion. Small sources will generate high frequency signals. Here, when using
195     a rectangular domain, the source is defined by a cosine function.
196 ahallam 3025 \begin{python}
197 ahallam 3004 U0=0.01 # amplitude of point source
198     xc=[500,500] #location of point source
199     # define small radius around point xc
200     src_radius = 30
201     # for first two time steps
202 ahallam 3025 u=U0*(cos(length(x-xc)*3.1415/src_radius)+1)*\
203     whereNegative(length(x-xc)-src_radius)
204 ahallam 3004 u_m1=u
205 ahallam 3025 \end{python}
206 ahallam 3004
207 ahallam 3025 \subsection{Visualising the Source}
208     There are two options for visualising the source. The first is to export the
209     initial conditions of the model to VTK, which can be interpreted as a scalar
210 ahallam 3370 surface in Mayavi2. The second is to take a cross section of the model which
211     will require the \textit{Locator} function.
212 ahallam 3063 First \verb!Locator! must be imported;
213 ahallam 3025 \begin{python}
214     from esys.escript.pdetools import Locator
215     \end{python}
216     The function can then be used on the domain to locate the nearest domain node
217     to the point or points of interest.
218    
219     It is now necessary to build a list of $(x,y)$ locations that specify where are
220 ahallam 3370 model slice will go. This is easily implemented with a loop;
221 ahallam 3025 \begin{python}
222     cut_loc=[]
223     src_cut=[]
224     for i in range(ndx/2-ndx/10,ndx/2+ndx/10):
225     cut_loc.append(xstep*i)
226     src_cut.append([xstep*i,xc[1]])
227     \end{python}
228 ahallam 3063 We then submit the output to \verb!Locator! and finally return the appropriate
229     values using the \verb!getValue! function.
230 ahallam 3025 \begin{python}
231 ahallam 3373 src=Locator(mydomain,src_cut)
232 ahallam 3025 src_cut=src.getValue(u)
233     \end{python}
234 ahallam 3370 It is then a trivial task to plot and save the output using \mpl
235     (\autoref{fig:cxsource}).
236 ahallam 3025 \begin{python}
237 ahallam 3373 pl.plot(cut_loc,src_cut)
238 ahallam 3025 pl.axis([xc[0]-src_radius*3,xc[0]+src_radius*3,0.,2*U0])
239     pl.savefig(os.path.join(savepath,"source_line.png"))
240     \end{python}
241     \begin{figure}[h]
242 ahallam 3029 \centering
243 ahallam 3063 \includegraphics[width=6in]{figures/sourceline.png}
244 ahallam 3025 \caption{Cross section of the source function.}
245     \label{fig:cxsource}
246     \end{figure}
247    
248    
249     \subsection{Point Monitoring}
250     In the more general case where the solution mesh is irregular or specific
251     locations need to be monitored, it is simple enough to use the \textit{Locator}
252     function.
253     \begin{python}
254     rec=Locator(mydomain,[250.,250.])
255     \end{python}
256 ahallam 3029 When the solution \verb u is updated we can extract the value at that point
257 ahallam 3025 via;
258     \begin{python}
259     u_rec=rec.getValue(u)
260     \end{python}
261 ahallam 3063 For consecutive time steps one can record the values from \verb!u_rec! in an
262     array initialised as \verb!u_rec0=[]! with;
263 ahallam 3025 \begin{python}
264     u_rec0.append(rec.getValue(u))
265     \end{python}
266    
267     It can be useful to monitor the value at a single or multiple individual points
268     in the model during the modelling process. This is done using
269 ahallam 3063 the \verb!Locator! function.
270 ahallam 3025
271    
272 ahallam 3003 \section{Acceleration Solution}
273     \sslist{example07b.py}
274    
275 ahallam 3373 An alternative method to the displacement solution, is to solve for the
276     acceleration $\frac{\partial ^2 p}{\partial t^2}$ directly. The displacement can
277     then be derived from the acceleration after a solution has been calculated
278     The acceleration is given by a modified form of \autoref{eqn:waveu};
279 ahallam 3003 \begin{equation}
280     \nabla ^2 p - \frac{1}{c^2} a = 0
281     \label{eqn:wavea}
282     \end{equation}
283 jfenwick 3308 and can be solved directly with $Y=0$ and $X=-c^2 \nabla ^2 p_{(t)}$.
284 ahallam 3003 After each iteration the displacement is re-evaluated via;
285     \begin{equation}
286 jfenwick 3308 p_{(t+1)}=2p_{(t)} - p_{(t-1)} + h^2a
287 ahallam 3003 \end{equation}
288    
289 ahallam 3029 \subsection{Lumping}
290 ahallam 3004 For \esc, the acceleration solution is prefered as it allows the use of matrix
291     lumping. Lumping or mass lumping as it is sometimes known, is the process of
292     aggressively approximating the density elements of a mass matrix into the main
293     diagonal. The use of Lumping is motivaed by the simplicity of diagonal matrix
294     inversion. As a result, Lumping can significantly reduce the computational
295 ahallam 3029 requirements of a problem. Care should be taken however, as this
296     function can only be used when the $A$, $B$ and $C$ coefficients of the
297     general form are zero.
298 ahallam 3003
299 ahallam 3004 To turn lumping on in \esc one can use the command;
300 ahallam 3025 \begin{python}
301 ahallam 3004 mypde.getSolverOptions().setSolverMethod(mypde.getSolverOptions().LUMPING)
302 ahallam 3025 \end{python}
303 ahallam 3004 It is also possible to check if lumping is set using;
304 ahallam 3025 \begin{python}
305 ahallam 3004 print mypde.isUsingLumping()
306 ahallam 3025 \end{python}
307 ahallam 3004
308     \section{Stability Investigation}
309     It is now prudent to investigate the stability limitations of this problem.
310 ahallam 3025 First, we let the frequency content of the source be very small. If we define
311 ahallam 3370 the source as a cosine input, then the wavlength of the input is equal to the
312 ahallam 3025 radius of the source. Let this value be 5 meters. Now, if the maximum velocity
313 ahallam 3373 of the model is $c=380.0ms^{-1}$, then the source
314 jfenwick 3308 frequency is $f_{r} = \frac{380.0}{5} = 76.0 Hz$. This is a worst case
315 ahallam 3025 scenario with a small source and the models maximum velocity.
316    
317 ahallam 3232 Furthermore, we know from \autoref{sec:nsstab}, that the spatial sampling
318 ahallam 3025 frequency must be at least twice this value to ensure stability. If we assume
319     the model mesh is a square equispaced grid,
320     then the sampling interval is the side length divided by the number of samples,
321     given by $\Delta x = \frac{1000.0m}{400} = 2.5m$ and the maximum sampling
322     frequency capable at this interval is
323 jfenwick 3308 $f_{s}=\frac{380.0ms^{-1}}{2.5m}=152Hz$ this is just equal to the
324 ahallam 3232 required rate satisfying \autoref{eqn:samptheorem}.
325 ahallam 3004
326 ahallam 3232 \autoref{fig:ex07sampth} depicts three examples where the grid has been
327 ahallam 3025 undersampled, sampled correctly, and over sampled. The grids used had
328     200, 400 and 800 nodes per side respectively. Obviously, the oversampled grid
329     retains the best resolution of the modelled wave.
330 ahallam 3004
331 ahallam 3025 The time step required for each of these examples is simply calculated from
332     the propagation requirement. For a maximum velocity of $380.0ms^{-1}$,
333     \begin{subequations}
334     \begin{equation}
335     \Delta t \leq \frac{1000.0m}{200} \frac{1}{380.0} = 0.013s
336     \end{equation}
337     \begin{equation}
338     \Delta t \leq \frac{1000.0m}{400} \frac{1}{380.0} = 0.0065s
339     \end{equation}
340     \begin{equation}
341     \Delta t \leq \frac{1000.0m}{800} \frac{1}{380.0} = 0.0032s
342     \end{equation}
343     \end{subequations}
344 ahallam 3373 Observe that for each doubling of the number of nodes in the mesh, we halve
345     the time step. To illustrate the impact this has, consider our model. If the
346 ahallam 3025 source is placed at the center, it is $500m$ from the nearest boundary. With a
347     velocity of $380.0ms^{-1}$ it will take $\approx1.3s$ for the wavefront to
348     reach that boundary. In each case, this equates to $100$, $200$ and $400$ time
349     steps. This is again, only a best case scenario, for true stability these time
350 ahallam 3373 values may need to be halved and possibly halved again.
351 ahallam 3004
352 ahallam 3025 \begin{figure}[ht]
353     \centering
354     \subfigure[Undersampled Example]{
355 ahallam 3373 \includegraphics[width=0.45\textwidth,trim=0cm 6cm 5cm 6cm
356     ,clip]{figures/ex07usamp.png}
357 ahallam 3025 \label{fig:ex07usamp}
358     }
359     \subfigure[Just sampled Example]{
360 ahallam 3373 \includegraphics[width=0.45\textwidth,trim=0cm 6cm 5cm 6cm
361     ,clip]{figures/ex07jsamp.png}
362 ahallam 3025 \label{fig:ex07jsamp}
363     }
364     \subfigure[Over sampled Example]{
365 ahallam 3373 \includegraphics[width=0.45\textwidth,trim=0cm 6cm 5cm 6cm
366     ,clip]{figures/ex07nsamp.png}
367 ahallam 3025 \label{fig:ex07nsamp}
368     }
369     \caption{Sampling Theorem example for stability
370     investigation.}
371 ahallam 3373 \label{fig:ex07sampth}
372 ahallam 3025 \end{figure}
373 ahallam 3004
374 ahallam 3025
375    
376    
377    

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