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 1 ahallam 3003 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 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 3384 This stability criterion is known as the Courant–Friedrichs–Lewy 125 condition given by 126 \begin{equation} 127 dt=f\cdot \frac{dx}{v} 128 \end{equation} 129 where $dx$ is the mesh size and $f$ is a safety factor. To obtain a time step of 130 $10E-4$, a safety factor of $f=0.1$ was used. 131 ahallam 3003 132 ahallam 3004 The wave frequency content also plays a part in numerical stability. The 133 nyquist-sampling theorem states that a signals bandwidth content will be 134 jfenwick 3308 accurately represented when an equispaced sampling rate $f _{n}$ is 135 ahallam 3370 equal to or greater then twice the maximum frequency of the signal 136 jfenwick 3308 $f_{s}$, or; 137 ahallam 3004 \begin{equation} \label{eqn:samptheorem} 138 jfenwick 3308 f_{n} \geqslant f_{s} 139 ahallam 3004 \end{equation} 140 ahallam 3384 For example, a 50Hz signal will require a sampling rate greater then 100Hz or 141 ahallam 3004 one sample every 0.01 seconds. The wave equation relies on a spatial frequency, 142 ahallam 3373 thus the sampling theorem in this case applies to the solution mesh spacing. 143 This relationship confirms that the frequency content of the input signal 144 directly affects the time discretisation of the problem. 145 ahallam 3004 146 To accurately model the wave equation with high resolutions and velocities 147 means that very fine spatial and time discretisation is necessary for most 148 ahallam 3384 problems. This requirement makes the wave equation arduous to 149 ahallam 3003 solve numerically due to the large number of time iterations required in each 150 ahallam 3004 solution. Models with very high velocities and frequencies will be the worst 151 ahallam 3029 affected by this problem. 152 ahallam 3003 153 \section{Displacement Solution} 154 \sslist{example07a.py} 155 156 We begin the solution to this PDE with the centred difference formula for the 157 second derivative; 158 \begin{equation} 159 f''(x) \approx \frac{f(x+h - 2f(x) + f(x-h)}{h^2} 160 \label{eqn:centdiff} 161 \end{equation} 162 ahallam 3232 substituting \autoref{eqn:centdiff} for $\frac{\partial ^2 p }{\partial t ^2}$ 163 in \autoref{eqn:acswave}; 164 ahallam 3003 \begin{equation} 165 jfenwick 3308 \nabla ^2 p - \frac{1}{c^2h^2} \left[p_{(t+1)} - 2p_{(t)} + 166 p_{(t-1)} \right] 167 ahallam 3003 = 0 168 \label{eqn:waveu} 169 \end{equation} 170 Rearranging for $p_{(t+1)}$; 171 \begin{equation} 172 jfenwick 3308 p_{(t+1)} = c^2 h^2 \nabla ^2 p_{(t)} +2p_{(t)} - 173 p_{(t-1)} 174 ahallam 3003 \end{equation} 175 this can be compared with the general form of the \modLPDE module and it 176 jfenwick 3308 becomes clear that $D=1$, $X_{i,j}=-c^2 h^2 \nabla ^2 p_{(t)}$ and 177 ahallam 3004 $Y=2p_{(t)} - p_{(t-1)}$. 178 ahallam 3003 179 ahallam 3025 The solution script is similar to others that we have created in previous 180 ahallam 3004 chapters. The general steps are; 181 \begin{enumerate} 182 \item The necessary libraries must be imported. 183 \item The domain needs to be defined. 184 \item The time iteration and control parameters need to be defined. 185 \item The PDE is initialised with source and boundary conditions. 186 \item The time loop is started and the PDE is solved at consecutive time steps. 187 ahallam 3370 \item All or select solutions are saved to file for visualisation later on. 188 ahallam 3004 \end{enumerate} 189 190 Parts of the script which warrant more attention are the definition of the 191 source, visualising the source, the solution time loop and the VTK data export. 192 193 \subsection{Pressure Sources} 194 As the pressure is a scalar, one need only define the pressure for two 195 time steps prior to the start of the solution loop. Two known solutions are 196 required because the wave equation contains a double partial derivative with 197 respect to time. This is often a good opportunity to introduce a source to the 198 solution. This model has the source located at it's centre. The source should 199 be smooth and cover a number of samples to satisfy the frequency stability 200 ahallam 3370 criterion. Small sources will generate high frequency signals. Here, when using 201 a rectangular domain, the source is defined by a cosine function. 202 ahallam 3025 \begin{python} 203 ahallam 3004 U0=0.01 # amplitude of point source 204 xc=[500,500] #location of point source 205 # define small radius around point xc 206 src_radius = 30 207 # for first two time steps 208 ahallam 3025 u=U0*(cos(length(x-xc)*3.1415/src_radius)+1)*\ 209 whereNegative(length(x-xc)-src_radius) 210 ahallam 3004 u_m1=u 211 ahallam 3025 \end{python} 212 ahallam 3004 213 ahallam 3025 \subsection{Visualising the Source} 214 There are two options for visualising the source. The first is to export the 215 initial conditions of the model to VTK, which can be interpreted as a scalar 216 ahallam 3370 surface in Mayavi2. The second is to take a cross section of the model which 217 will require the \textit{Locator} function. 218 ahallam 3063 First \verb!Locator! must be imported; 219 ahallam 3025 \begin{python} 220 from esys.escript.pdetools import Locator 221 \end{python} 222 The function can then be used on the domain to locate the nearest domain node 223 to the point or points of interest. 224 225 It is now necessary to build a list of $(x,y)$ locations that specify where are 226 ahallam 3370 model slice will go. This is easily implemented with a loop; 227 ahallam 3025 \begin{python} 228 cut_loc=[] 229 src_cut=[] 230 for i in range(ndx/2-ndx/10,ndx/2+ndx/10): 231 cut_loc.append(xstep*i) 232 src_cut.append([xstep*i,xc]) 233 \end{python} 234 ahallam 3063 We then submit the output to \verb!Locator! and finally return the appropriate 235 values using the \verb!getValue! function. 236 ahallam 3025 \begin{python} 237 ahallam 3373 src=Locator(mydomain,src_cut) 238 ahallam 3025 src_cut=src.getValue(u) 239 \end{python} 240 ahallam 3370 It is then a trivial task to plot and save the output using \mpl 241 (\autoref{fig:cxsource}). 242 ahallam 3025 \begin{python} 243 ahallam 3373 pl.plot(cut_loc,src_cut) 244 ahallam 3025 pl.axis([xc-src_radius*3,xc+src_radius*3,0.,2*U0]) 245 pl.savefig(os.path.join(savepath,"source_line.png")) 246 \end{python} 247 \begin{figure}[h] 248 ahallam 3029 \centering 249 ahallam 3063 \includegraphics[width=6in]{figures/sourceline.png} 250 ahallam 3025 \caption{Cross section of the source function.} 251 \label{fig:cxsource} 252 \end{figure} 253 254 255 \subsection{Point Monitoring} 256 In the more general case where the solution mesh is irregular or specific 257 locations need to be monitored, it is simple enough to use the \textit{Locator} 258 function. 259 \begin{python} 260 rec=Locator(mydomain,[250.,250.]) 261 \end{python} 262 ahallam 3029 When the solution \verb u is updated we can extract the value at that point 263 ahallam 3025 via; 264 \begin{python} 265 u_rec=rec.getValue(u) 266 \end{python} 267 ahallam 3063 For consecutive time steps one can record the values from \verb!u_rec! in an 268 array initialised as \verb!u_rec0=[]! with; 269 ahallam 3025 \begin{python} 270 u_rec0.append(rec.getValue(u)) 271 \end{python} 272 273 It can be useful to monitor the value at a single or multiple individual points 274 in the model during the modelling process. This is done using 275 ahallam 3063 the \verb!Locator! function. 276 ahallam 3025 277 278 ahallam 3003 \section{Acceleration Solution} 279 \sslist{example07b.py} 280 281 ahallam 3373 An alternative method to the displacement solution, is to solve for the 282 acceleration $\frac{\partial ^2 p}{\partial t^2}$ directly. The displacement can 283 then be derived from the acceleration after a solution has been calculated 284 The acceleration is given by a modified form of \autoref{eqn:waveu}; 285 ahallam 3003 \begin{equation} 286 \nabla ^2 p - \frac{1}{c^2} a = 0 287 \label{eqn:wavea} 288 \end{equation} 289 jfenwick 3308 and can be solved directly with $Y=0$ and $X=-c^2 \nabla ^2 p_{(t)}$. 290 ahallam 3003 After each iteration the displacement is re-evaluated via; 291 \begin{equation} 292 jfenwick 3308 p_{(t+1)}=2p_{(t)} - p_{(t-1)} + h^2a 293 ahallam 3003 \end{equation} 294 295 ahallam 3029 \subsection{Lumping} 296 ahallam 3004 For \esc, the acceleration solution is prefered as it allows the use of matrix 297 lumping. Lumping or mass lumping as it is sometimes known, is the process of 298 aggressively approximating the density elements of a mass matrix into the main 299 diagonal. The use of Lumping is motivaed by the simplicity of diagonal matrix 300 inversion. As a result, Lumping can significantly reduce the computational 301 ahallam 3029 requirements of a problem. Care should be taken however, as this 302 function can only be used when the $A$, $B$ and $C$ coefficients of the 303 general form are zero. 304 ahallam 3003 305 ahallam 3384 More information about the lumping implementation used in \esc and its accuracy 306 can be found in the user guide. 307 308 ahallam 3004 To turn lumping on in \esc one can use the command; 309 ahallam 3025 \begin{python} 310 ahallam 3384 mypde.getSolverOptions().setSolverMethod(mypde.getSolverOptions().HRZ_LUMPING) 311 ahallam 3025 \end{python} 312 ahallam 3004 It is also possible to check if lumping is set using; 313 ahallam 3025 \begin{python} 314 ahallam 3004 print mypde.isUsingLumping() 315 ahallam 3025 \end{python} 316 ahallam 3004 317 \section{Stability Investigation} 318 It is now prudent to investigate the stability limitations of this problem. 319 ahallam 3025 First, we let the frequency content of the source be very small. If we define 320 ahallam 3370 the source as a cosine input, then the wavlength of the input is equal to the 321 ahallam 3025 radius of the source. Let this value be 5 meters. Now, if the maximum velocity 322 ahallam 3373 of the model is $c=380.0ms^{-1}$, then the source 323 jfenwick 3308 frequency is $f_{r} = \frac{380.0}{5} = 76.0 Hz$. This is a worst case 324 ahallam 3025 scenario with a small source and the models maximum velocity. 325 326 ahallam 3232 Furthermore, we know from \autoref{sec:nsstab}, that the spatial sampling 327 ahallam 3025 frequency must be at least twice this value to ensure stability. If we assume 328 the model mesh is a square equispaced grid, 329 then the sampling interval is the side length divided by the number of samples, 330 given by $\Delta x = \frac{1000.0m}{400} = 2.5m$ and the maximum sampling 331 frequency capable at this interval is 332 jfenwick 3308 $f_{s}=\frac{380.0ms^{-1}}{2.5m}=152Hz$ this is just equal to the 333 ahallam 3232 required rate satisfying \autoref{eqn:samptheorem}. 334 ahallam 3004 335 ahallam 3232 \autoref{fig:ex07sampth} depicts three examples where the grid has been 336 ahallam 3025 undersampled, sampled correctly, and over sampled. The grids used had 337 200, 400 and 800 nodes per side respectively. Obviously, the oversampled grid 338 retains the best resolution of the modelled wave. 339 ahallam 3004 340 ahallam 3025 The time step required for each of these examples is simply calculated from 341 the propagation requirement. For a maximum velocity of $380.0ms^{-1}$, 342 \begin{subequations} 343 \begin{equation} 344 \Delta t \leq \frac{1000.0m}{200} \frac{1}{380.0} = 0.013s 345 \end{equation} 346 \begin{equation} 347 \Delta t \leq \frac{1000.0m}{400} \frac{1}{380.0} = 0.0065s 348 \end{equation} 349 \begin{equation} 350 \Delta t \leq \frac{1000.0m}{800} \frac{1}{380.0} = 0.0032s 351 \end{equation} 352 \end{subequations} 353 ahallam 3373 Observe that for each doubling of the number of nodes in the mesh, we halve 354 the time step. To illustrate the impact this has, consider our model. If the 355 ahallam 3025 source is placed at the center, it is $500m$ from the nearest boundary. With a 356 velocity of $380.0ms^{-1}$ it will take $\approx1.3s$ for the wavefront to 357 reach that boundary. In each case, this equates to $100$, $200$ and $400$ time 358 steps. This is again, only a best case scenario, for true stability these time 359 ahallam 3373 values may need to be halved and possibly halved again. 360 ahallam 3004 361 ahallam 3025 \begin{figure}[ht] 362 \centering 363 \subfigure[Undersampled Example]{ 364 ahallam 3373 \includegraphics[width=0.45\textwidth,trim=0cm 6cm 5cm 6cm 365 ,clip]{figures/ex07usamp.png} 366 ahallam 3025 \label{fig:ex07usamp} 367 } 368 \subfigure[Just sampled Example]{ 369 ahallam 3373 \includegraphics[width=0.45\textwidth,trim=0cm 6cm 5cm 6cm 370 ,clip]{figures/ex07jsamp.png} 371 ahallam 3025 \label{fig:ex07jsamp} 372 } 373 \subfigure[Over sampled Example]{ 374 ahallam 3373 \includegraphics[width=0.45\textwidth,trim=0cm 6cm 5cm 6cm 375 ,clip]{figures/ex07nsamp.png} 376 ahallam 3025 \label{fig:ex07nsamp} 377 } 378 \caption{Sampling Theorem example for stability 379 investigation.} 380 ahallam 3373 \label{fig:ex07sampth} 381 ahallam 3025 \end{figure} 382 ahallam 3004 383 ahallam 3025 384 385 386