# Contents of /release/4.0/doc/cookbook/example07.tex

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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 7 % 8 % Primary Business: Queensland, Australia 9 % Licensed under the Open Software License version 3.0 10 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} 358 359 360 361 362