 # Contents of /trunk/doc/cookbook/example08.tex

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Updates to cookbook, new chapter DC Resis and IP. Using new packages hyperref, natbib

 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 \section{Seismic Wave Propagation in Two Dimensions} 15 16 \sslist{example08a.py} 17 We will now expand upon the previous chapter by introducing a vector form of 18 the wave equation. This means that the waves will have not only a scalar 19 magnitude as for the pressure wave solution, but also a direction. This type of 20 scenario is apparent in wave types that exhibit compressional and transverse 21 particle motion. An example of this would be seismic waves. 22 23 Wave propagation in the earth can be described by the elastic wave equation 24 \begin{equation} \label{eqn:wav} \index{wave equation} 25 \rho \frac{\partial^{2}u\hackscore{i}}{\partial t^2} - \frac{\partial 26 \sigma\hackscore{ij}}{\partial x\hackscore{j}} = 0 27 \end{equation} 28 where $\sigma$ is the stress given by 29 \begin{equation} \label{eqn:sigw} 30 \sigma \hackscore{ij} = \lambda u\hackscore{k,k} \delta\hackscore{ij} + \mu ( 31 u\hackscore{i,j} + u\hackscore{j,i}) 32 \end{equation} 33 and $\lambda$ and $\mu$ represent Lame's parameters. Specifically for seismic 34 waves, $\mu$ is the propagation materials shear modulus. 35 In a similar process to the previous chapter, we will use the acceleration 36 solution to solve this PDE. By substituting $a$ directly for 37 $\frac{\partial^{2}u\hackscore{i}}{\partial t^2}$ we can derive the 38 displacement solution. Using $a$ we can see that \autoref{eqn:wav} becomes 39 \begin{equation} \label{eqn:wava} 40 \rho a\hackscore{i} - \frac{\partial 41 \sigma\hackscore{ij}}{\partial x\hackscore{j}} = 0 42 \end{equation} 43 Thus the problem will be solved for acceleration and then converted to 44 displacement using the backwards difference approximation. 45 46 Consider now the stress $\sigma$. One can see that the stress consists of two 47 distinct terms: 48 \begin{subequations} 49 \begin{equation} \label{eqn:sigtrace} 50 \lambda u\hackscore{k,k} \delta\hackscore{ij} 51 \end{equation} 52 \begin{equation} \label{eqn:sigtrans} 53 \mu (u\hackscore{i,j} + u\hackscore{j,i}) 54 \end{equation} 55 \end{subequations} 56 One simply recognizes in \autoref{eqn:sigtrace} that $u\hackscore{k,k}$ is the 57 trace of the displacement solution and that $\delta\hackscore{ij}$ is the 58 kronecker delta function with dimensions equivalent to $u$. The second term 59 \autoref{eqn:sigtrans} is the sum of $u$ with its own transpose. Putting these 60 facts together we see that the spatial differential of the stress is given by the 61 gradient of $u$ and the aforementioned opperations. This value is then submitted 62 to the \esc PDE as $X$. 63 \begin{python} 64 g=grad(u); stress=lam*trace(g)*kmat+mu*(g+transpose(g)) 65 mypde.setValue(X=-stress) # set PDE values 66 \end{python} 67 The solution is then obtained via the usual method and the displacement is 68 calculated so that the memory variables can be updated for the next time 69 iteration. 70 \begin{python} 71 accel = mypde.getSolution() #get PDE solution for accelleration 72 u_p1=(2.*u-u_m1)+h*h*accel #calculate displacement 73 u_m1=u; u=u_p1 # shift values by 1 74 \end{python} 75 76 Saving the data has been handled slightly differently in this example. The VTK 77 files generated can be quite large and take a significant amount of time to save 78 to the hard disk. To avoid doing this at every iteration a test is devised which 79 saves only at specific time intervals. 80 81 To do this there are two new parameters in our script. 82 \begin{python} 83 # data recording times 84 rtime=0.0 # first time to record 85 rtime_inc=tend/20.0 # time increment to record 86 \end{python} 87 Currently the PDE solution will be saved to file $20$ times between the start of 88 the modelling and the final time step. With these parameters set, an if 89 statement is introduced to the time loop 90 \begin{python} 91 if (t >= rtime): 92 saveVTK(os.path.join(savepath,"ex08a.%05d.vtu"%n),displacement=length(u),\ 93 acceleration=length(accel),tensor=stress) 94 rtime=rtime+rtime_inc #increment data save time 95 \end{python} 96 \verb!t! is the time counter. Whenever the recording time \verb!rtime! is less 97 then \verb!t! the solution is saved and \verb!rtime! is incremented. This 98 limits the number of outputs and increases the speed of the solver. 99 100 \section{Multi-threading} 101 The wave equation solution can be quite demanding on cpu time. Enhancements can 102 be made by accessing multiple threads or cores on your computer. This does not 103 require any modification to the solution script and only comes into play when 104 escript is called from the shell. To use multiple threads \esc is called using 105 the \verb!-t! option with an interger argument for the number of threads 106 required. For example 107 \begin{verbatim} 108 $escript -t 4 example08a.py 109 \end{verbatim} 110 would call the script in this section and solve it using 4 threads. 111 112 The computation times on an increasing number of cores is outlines in 113 \autoref{tab:wpcores}. 114 115 \begin{table}[ht] 116 \begin{center} 117 \caption{Computation times for an increasing number of cores.} 118 \label{tab:wpcores} 119 \begin{tabular}{| c | c |} 120 \hline 121 Number of Cores & Time (s) \\ 122 \hline 123 1 & 691.0 \\ 124 2 & 400.0 \\ 125 3 & 305.0 \\ 126 4 & 328.0 \\ 127 5 & 323.0 \\ 128 6 & 292.0 \\ 129 7 & 282.0 \\ 130 8 & 445.0 \\ \hline 131 \end{tabular} 132 \end{center} 133 \end{table} 134 135 \section{Vector source on the boundary} 136 \sslist{example08b.py} 137 For this particular example, we will introduce the source by applying a 138 displacment to the boundary during the initial time steps. The source will again 139 be 140 a radially propagating wave but due to the vector nature of the PDE used, a 141 direction will need to be applied to the source. 142 143 The first step is to choose an amplitude and create the source as in the 144 previous chapter. 145 \begin{python} 146 U0=0.01 # amplitude of point source 147 # will introduce a spherical source at middle left of bottom face 148 xc=[ndx/2,0] 149 150 ############################################FIRST TIME STEPS AND SOURCE 151 # define small radius around point xc 152 src_length = 40; print "src_length = ",src_length 153 # set initial values for first two time steps with source terms 154 xb=FunctionOnBoundary(domain).getX() 155 y=source*(cos(length(x-xc)*3.1415/src_length)+1)*\ 156 whereNegative(length(xb-src_length)) 157 src_dir=numpy.array([0.,1.]) # defines direction of point source as down 158 y=y*src_dir 159 \end{python} 160 where \verb xc is the source point on the boundary of the model. Note that 161 because the source is specifically located on the boundary, we have used the 162 \verb!FunctionOnBoundary! call to ensure the nodes are located upon the 163 boundary only. These boundary nodes are passed to source as \verb!xb!. The 164 source direction is then defined as an$(x,y)$array and multiplied by the 165 source function. The directional array must have a magnitude of$\left| 1 166 \right| $otherwise the amplitude of the source will become modified. For this 167 example, the source is directed in the$-y$direction. 168 \begin{python} 169 src_dir=numpy.array([0.,-1.]) # defines direction of point source as down 170 y=y*src_dir 171 \end{python} 172 The function can then be applied as a boundary condition by setting it equal to 173$y$in the general form. 174 \begin{python} 175 mypde.setValue(y=y) #set the source as a function on the boundary 176 \end{python} 177 The final step is to qualify the initial conditions. Due to the fact that we are 178 no longer using the source to define our initial condition to the model, we 179 must set the model state to zero for the first two time steps. 180 \begin{python} 181 # initial value of displacement at point source is constant (U0=0.01) 182 # for first two time steps 183 u=[0.0,0.0]*wherePositive(x) 184 u_m1=u 185 \end{python} 186 187 If the source is time progressive,$y$can be updated during the 188 iteration stage. This is covered in the following section. 189 190 \begin{figure}[htp] 191 \centering 192 \subfigure[Example 08a at 0.025s ]{ 193 \includegraphics[width=3in]{figures/ex08pw50.png} 194 \label{fig:ex08pw50} 195 } 196 \subfigure[Example 08a at 0.175s ]{ 197 \includegraphics[width=3in]{figures/ex08pw350.png} 198 \label{fig:ex08pw350} 199 } \\ 200 \subfigure[Example 08a at 0.325s ]{ 201 \includegraphics[width=3in]{figures/ex08pw650.png} 202 \label{fig:ex08pw650} 203 } 204 \subfigure[Example 08a at 0.475s ]{ 205 \includegraphics[width=3in]{figures/ex08pw950.png} 206 \label{fig:ex08pw950} 207 } 208 \label{fig:ex08pw} 209 \caption{Results of Example 08 at various times.} 210 \end{figure} 211 \clearpage 212 213 \section{Time variant source} 214 215 \sslist{example08b.py} 216 Until this point, all of the wave propagation examples in this cookbook have 217 used impulsive sources which are smooth in space but not time. It is however, 218 advantageous to have a time smoothed source as it can reduce the temporal 219 frequency range and thus mitigate aliasing in the solution. 220 221 It is quite 222 simple to implement a source which is smooth in time. In addition to the 223 original source function the only extra requirement is a time function. For 224 this example the time variant source will be the derivative of a gausian curve 225 defined by the required dominant frequency (\autoref{fig:tvsource}). 226 \begin{python} 227 #Creating the time function of the source. 228 dfeq=50 #Dominant Frequency 229 a = 2.0 * (np.pi * dfeq)**2.0 230 t0 = 5.0 / (2.0 * np.pi * dfeq) 231 srclength = 5. * t0 232 ls = int(srclength/h) 233 print 'source length',ls 234 source=np.zeros(ls,'float') # source array 235 ampmax=0 236 for it in range(0,ls): 237 t = it*h 238 tt = t-t0 239 dum1 = np.exp(-a * tt * tt) 240 source[it] = -2. * a * tt * dum1 241 if (abs(source[it]) > ampmax): 242 ampmax = abs(source[it]) 243 time[t]=t*h 244 \end{python} 245 \begin{figure}[ht] 246 \centering 247 \includegraphics[width=3in]{figures/source.png} 248 \caption{Time variant source with a dominant frequency of 50Hz.} 249 \label{fig:tvsource} 250 \end{figure} 251 252 We then build the source and the first two time steps via; 253 \begin{python} 254 # set initial values for first two time steps with source terms 255 y=source 256 *(cos(length(x-xc)*3.1415/src_length)+1)*whereNegative(length(x-xc)-src_length) 257 src_dir=numpy.array([0.,-1.]) # defines direction of point source as down 258 y=y*src_dir 259 mypde.setValue(y=y) #set the source as a function on the boundary 260 # initial value of displacement at point source is constant (U0=0.01) 261 # for first two time steps 262 u=[0.0,0.0]*whereNegative(x) 263 u_m1=u 264 \end{python} 265 266 Finally, for the length of the source, we are required to update each new 267 solution in the itterative section of the solver. This is done via; 268 \begin{python} 269 # increment loop values 270 t=t+h; n=n+1 271 if (n < ls): 272 y=source[n]**(cos(length(x-xc)*3.1415/src_length)+1)*\ 273 whereNegative(length(x-xc)-src_length) 274 y=y*src_dir; mypde.setValue(y=y) #set the source as a function on the 275 boundary 276 \end{python} 277 278 \section{Absorbing Boundary Conditions} 279 To mitigate the effect of the boundary on the model, absorbing boundary 280 conditions can be introduced. These conditions effectively dampen the wave 281 energy as they approach the bounday and thus prevent that energy from being 282 reflected. This type of approach is used typically when a model only represents 283 a small portion of the entire model, which in reality may have infinite bounds. 284 It is inpractical to calculate the solution for an infinite model and thus ABCs 285 allow us the create an approximate solution with small to zero boundary effects 286 on a model with a solvable size. 287 288 To dampen the waves, the method of \citet{Cerjan1985} 289 where the solution and the stress are multiplied by a damping function defined 290 on$n\$ nodes of the domain adjacent to the boundary, given by; 291 \begin{equation} 292 \gamma =\sqrt{\frac{| -log( \gamma \hackscore{b} ) |}{n^2}} 293 \end{equation} 294 \begin{equation} 295 y=e^{-(\gamma x)^2} 296 \end{equation} 297 This is applied to the bounding 20-50 pts of the model using the location 298 specifiers of \esc; 299 \begin{python} 300 # Define where the boundary decay will be applied. 301 bn=30. 302 bleft=xstep*bn; bright=mx-(xstep*bn); bbot=my-(ystep*bn) 303 # btop=ystep*bn # don't apply to force boundary!!! 304 305 # locate these points in the domain 306 left=x-bleft; right=x-bright; bottom=x-bbot 307 308 tgamma=0.98 # decay value for exponential function 309 def calc_gamma(G,npts): 310 func=np.sqrt(abs(-1.*np.log(G)/(npts**2.))) 311 return func 312 313 gleft = calc_gamma(tgamma,bleft) 314 gright = calc_gamma(tgamma,bleft) 315 gbottom= calc_gamma(tgamma,ystep*bn) 316 317 print 'gamma', gleft,gright,gbottom 318 319 # calculate decay functions 320 def abc_bfunc(gamma,loc,x,G): 321 func=exp(-1.*(gamma*abs(loc-x))**2.) 322 return func 323 324 fleft=abc_bfunc(gleft,bleft,x,tgamma) 325 fright=abc_bfunc(gright,bright,x,tgamma) 326 fbottom=abc_bfunc(gbottom,bbot,x,tgamma) 327 # apply these functions only where relevant 328 abcleft=fleft*whereNegative(left) 329 abcright=fright*wherePositive(right) 330 abcbottom=fbottom*wherePositive(bottom) 331 # make sure the inside of the abc is value 1 332 abcleft=abcleft+whereZero(abcleft) 333 abcright=abcright+whereZero(abcright) 334 abcbottom=abcbottom+whereZero(abcbottom) 335 # multiply the conditions together to get a smooth result 336 abc=abcleft*abcright*abcbottom 337 \end{python} 338 Note that the boundary conditions are not applied to the surface, as this is 339 effectively a free surface where normal reflections would be experienced. 340 Special conditions can be introduced at this surface if they are known. The 341 resulting boundary damping function can be viewed in 342 \autoref{fig:abconds}. 343 344 \section{Second order Meshing} 345 For stiff problems like the wave equation it is often prudent to implement 346 second order meshing. This creates a more accurate mesh approximation with some 347 increased processing cost. To turn second order meshing on, the \verb!rectangle! 348 function accpets an \verb!order! keyword argument. 349 \begin{python} 350 domain=Rectangle(l0=mx,l1=my,n0=ndx, n1=ndy,order=2) # create the domain 351 \end{python} 352 Other pycad functions and objects have similar keyword arguments for higher 353 order meshing. 354 355 Note that when implementing second order meshing, a smaller timestep is required 356 then for first order meshes as the second order essentially reduces the size of 357 the mesh by half. 358 359 \begin{figure}[ht] 360 \centering 361 \includegraphics[width=5in]{figures/ex08babc.png} 362 \label{fig:abconds} 363 \caption{Absorbing boundary conditions for example08b.py} 364 \end{figure} 365 366 \begin{figure}[htp] 367 \centering 368 \subfigure[Example 08b at 0.03s ]{ 369 \includegraphics[width=3in]{figures/ex08sw060.png} 370 \label{fig:ex08pw060} 371 } 372 \subfigure[Example 08b at 0.16s ]{ 373 \includegraphics[width=3in]{figures/ex08sw320.png} 374 \label{fig:ex08pw320} 375 } \\ 376 \subfigure[Example 08b at 0.33s ]{ 377 \includegraphics[width=3in]{figures/ex08sw660.png} 378 \label{fig:ex08pw660} 379 } 380 \subfigure[Example 08b at 0.44s ]{ 381 \includegraphics[width=3in]{figures/ex08sw880.png} 382 \label{fig:ex08pw880} 383 } 384 \label{fig:ex08pw} 385 \caption{Results of Example 08b at various times.} 386 \end{figure} 387 \clearpage 388 389 \section{Pycad example} 390 \sslist{example08c.py} 391 To make the problem more interesting we will now introduce an interface to the 392 middle of the domain. Infact we will use the same domain as we did for heat flux 393 in \autoref{CHAP HEAT 2}. The domain contains a syncline with two set of 394 material properties on either side of the interface. 395 396 \begin{figure}[ht] 397 \begin{center} 398 \includegraphics[width=5in]{figures/gmsh-example08c.png} 399 \caption{Domain geometry for example08c.py showing line tangents.} 400 \label{fig:ex08cgeo} 401 \end{center} 402 \end{figure} 403 404 It is simple enough to slightly modify the scripts of the previous sections to 405 accept this domain. Multiple material parameters must now be deined and assigned 406 to specific tagged areas. Again this is done via 407 \begin{python} 408 lam=Scalar(0,Function(domain)) 409 lam.setTaggedValue("top",lam1) 410 lam.setTaggedValue("bottom",lam2) 411 mu=Scalar(0,Function(domain)) 412 mu.setTaggedValue("top",mu1) 413 mu.setTaggedValue("bottom",mu2) 414 rho=Scalar(0,Function(domain)) 415 rho.setTaggedValue("top",rho1) 416 rho.setTaggedValue("bottom",rho2) 417 \end{python} 418 Don't forget that teh source boudnary must also be tagged and added so it can be 419 referenced 420 \begin{python} 421 # Add the subdomains and flux boundaries. 422 d.addItems(PropertySet("top",tblock),PropertySet("bottom",bblock),\ 423 PropertySet("linetop",l30)) 424 \end{python} 425 It is now possible to solve the script as in the previous examples. 426 427 \begin{figure}[ht] 428 \centering 429 \includegraphics[width=4in]{figures/ex08c2601.png} 430 \caption{Modelling results of example08c.py at 0.2601 seconds. Notice the 431 refraction of the wave front about the boundary between the two layers.} 432 \label{fig:ex08cres} 433 \end{figure} 434 435

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