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1 caltinay 5293
2     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
3 jfenwick 6651 % Copyright (c) 2003-2018 by The University of Queensland
4 caltinay 5293 % http://www.uq.edu.au
5     %
6     % Primary Business: Queensland, Australia
7 jfenwick 6112 % Licensed under the Apache License, version 2.0
8     % http://www.apache.org/licenses/LICENSE-2.0
9 caltinay 5293 %
10     % Development until 2012 by Earth Systems Science Computational Center (ESSCC)
11     % Development 2012-2013 by School of Earth Sciences
12     % Development from 2014 by Centre for Geoscience Computing (GeoComp)
13     %
14     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
15    
16 gross 4537 \section{Wave Propagation}
17     \label{WAVE CHAP}
18 ksteube 1811
19 jfenwick 3295 In this next example we want to calculate the displacement field $u_{i}$ for any time $t>0$ by solving the wave equation:\index{wave equation}
20 jgs 107 \begin{eqnarray}\label{WAVE general problem}
21 jfenwick 3295 \rho u_{i,tt} - \sigma_{ij,j}=0
22 jgs 107 \end{eqnarray}
23 jfenwick 3295 in a three dimensional block of length $L$ in $x_{0}$
24     and $x_{1}$ direction and height $H$ in $x_{2}$ direction.
25 caltinay 3280 $\rho$ is the known density which may be a function of its location.
26 jfenwick 3295 $\sigma_{ij}$ is the stress field\index{stress} which in case of an
27 caltinay 3280 isotropic, linear elastic material is given by
28     \begin{eqnarray}\label{WAVE stress}
29 jfenwick 3295 \sigma_{ij} = \lambda u_{k,k} \delta_{ij} + \mu (u_{i,j} + u_{j,i})
30 jgs 107 \end{eqnarray}
31 sshaw 5284 where $\lambda$ and $\mu$ are the Lam\'e coefficients\index{Lam\'e coefficients}
32 jfenwick 3295 and $\delta_{ij}$ denotes the Kronecker symbol\index{Kronecker symbol}.
33 jgs 107 On the boundary the normal stress is given by
34 caltinay 3280 \begin{eqnarray}\label{WAVE natural}
35 jfenwick 3295 \sigma_{ij}n_{j}=0
36 jgs 107 \end{eqnarray}
37 gross 2502 for all time $t>0$.
38 lkettle 581
39 gross 2502 \begin{figure}[t!]
40 caltinay 3280 \centerline{\includegraphics[width=14cm]{waveu}}
41 jfenwick 3295 \caption{Input Displacement at Source Point ($\alpha=0.7$, $t_{0}=3$, $U_{0}=1$).}
42 gross 2502 \label{WAVE FIG 1.2}
43     \end{figure}
44    
45 caltinay 3280 \begin{figure}
46     \centerline{\includegraphics[width=14cm]{wavea}}
47 jfenwick 3295 \caption{Input Acceleration at Source Point ($\alpha=0.7$, $t_{0}=3$, $U_{0}=1$).}
48 gross 2502 \label{WAVE FIG 1.1}
49     \end{figure}
50    
51 jfenwick 3295 Here we are modelling a point source at the point $x_C$ in the
52     $x_{0}$-direction which is a negative pulse of amplitude
53     $U_{0}$ followed by the same positive pulse.
54 caltinay 3280 In mathematical terms we use
55 gross 2502 \begin{eqnarray} \label{WAVE source}
56 jfenwick 3295 u_{0}(x_C,t)= U_{0} \; \sqrt{2} \frac{(t-t_{0})}{\alpha} e^{\frac{1}{2}-\frac{(t-t_{0})^2}{\alpha^2}} \
57 jgs 107 \end{eqnarray}
58 jfenwick 3295 for all $t\ge0$ where $\alpha$ is the width of the pulse and $t_{0}$
59 caltinay 3280 is the time when the pulse changes from negative to positive.
60 jfenwick 3295 In the simulations we will choose $\alpha=0.3$ and $t_{0}=2$ (see
61 caltinay 3280 \fig{WAVE FIG 1.2}) and apply the source as a constraint in a sphere of small
62 jfenwick 3295 radius around the point $x_C$.
63 jgs 107
64 caltinay 3280 We use an explicit time integration scheme to calculate the displacement field
65     $u$ at certain time marks $t^{(n)}$, where $t^{(n)}=t^{(n-1)}+h$ with time
66     step size $h>0$.
67     In the following the upper index ${(n)}$ refers to values at time $t^{(n)}$.
68     We use the Verlet scheme\index{Verlet scheme} with constant time step size $h$
69 jgs 107 which is defined by
70     \begin{eqnarray} \label{WAVE dyn 2}
71 lkettle 575 u^{(n)}=2u^{(n-1)}-u^{(n-2)} + h^2 a^{(n)} \\
72 jgs 107 \end{eqnarray}
73 lkettle 581 for all $n=2,3,\ldots$. It is designed to solve a system of equations of the form
74 jgs 107 \begin{eqnarray} \label{WAVE dyn 2b}
75 jfenwick 3295 u_{,tt}=G(u)
76 jgs 107 \end{eqnarray}
77 gross 660 where one sets $a^{(n)}=G(u^{(n-1)})$.
78 jgs 107
79     In our case $a^{(n)}$ is given by
80     \begin{eqnarray}\label{WAVE dyn 3}
81 jfenwick 3295 \rho a^{(n)}_{i}=\sigma^{(n-1)}_{ij,j}
82 jgs 107 \end{eqnarray}
83     and boundary conditions
84 gross 593 \begin{eqnarray} \label{WAVE natural at n}
85 jfenwick 3295 \sigma^{(n-1)}_{ij}n_{j}=0
86 jgs 107 \end{eqnarray}
87     derived from \eqn{WAVE natural} where
88     \begin{eqnarray} \label{WAVE dyn 3a}
89 jfenwick 3295 \sigma_{ij}^{(n-1)} & = & \lambda u^{(n-1)}_{k,k} \delta_{ij} + \mu ( u^{(n-1)}_{i,j} + u^{(n-1)}_{j,i}).
90 jgs 107 \end{eqnarray}
91 gross 2502 We also need to apply the constraint
92     \begin{eqnarray} \label{WAVE dyn 3aa}
93 jfenwick 3295 a^{(n)}_{0}(x_C,t)= U_{0}
94     \frac{\sqrt(2.)}{\alpha^2} (4\frac{(t-t_{0})^3}{\alpha^3}-6\frac{t-t_{0}}{\alpha})e^{\frac{1}{2}-\frac{(t-t_{0})^2}{\alpha^2}}
95 gross 2502 \end{eqnarray}
96 caltinay 3280 which is derived from equation~\ref{WAVE source} by calculating the second
97     order time derivative (see \fig{WAVE FIG 1.1}).
98     Now we have converted our problem for displacement, $u^{(n)}$, into a problem
99     for acceleration, $a^{(n)}$, which depends on the solution at the previous two
100     time steps $u^{(n-1)}$ and $u^{(n-2)}$.
101 jgs 107
102 caltinay 3280 In each time step we have to solve this problem to get the acceleration $a^{(n)}$,
103     and we will use the \LinearPDE class of the \linearPDEs package to do so.
104     The general form of the PDE defined through the \LinearPDE class is discussed
105     in \Sec{SEC LinearPDE}. The form which is relevant here is
106 jgs 107 \begin{equation}\label{WAVE dyn 100}
107 jfenwick 3295 D_{ij} a^{(n)}_{j} = - X_{ij,j}\; .
108 jgs 107 \end{equation}
109 ksteube 1316 The natural boundary condition
110 jgs 107 \begin{equation}\label{WAVE dyn 101}
111 jfenwick 3295 n_{j}X_{ij} =0
112 jgs 107 \end{equation}
113 ksteube 1316 is used.
114     With $u=a^{(n)}$ we can identify the values to be assigned to $D$ and $X$:
115 caltinay 3280 \begin{equation}\label{WAVE dyn 6}
116 jgs 107 \begin{array}{ll}
117 jfenwick 3295 D_{ij}=\rho \delta_{ij}&
118     X_{ij}=-\sigma^{(n-1)}_{ij}\\
119 jgs 107 \end{array}
120     \end{equation}
121 caltinay 3280 Moreover we need to define the location $r$ where the constraint~\ref{WAVE dyn 3aa} is applied.
122     We will apply the constraint on a small sphere of radius $R$ around
123 jfenwick 3295 $x_C$ (we will use 3\% of the width of the domain):
124 caltinay 3280 \begin{equation}\label{WAVE dyn 6 1}
125 jfenwick 3295 q_{i}(x) =
126 gross 2502 \left\{
127 caltinay 3280 \begin{array}{l l}
128     1 & \quad \text{where $\|x-x_c\|\le R$}\\
129 caltinay 5296 0 & \quad \text{otherwise.}
130 gross 2502 \end{array}
131     \right.
132     \end{equation}
133 caltinay 3280 The following script defines the function \function{wavePropagation} which
134     implements the Verlet scheme to solve our wave propagation problem.
135     The argument \var{domain} which is a \Domain class object defines the domain
136     of the problem.
137     \var{h} and \var{tend} are the time step size and the end time of the simulation.
138     \var{lam}, \var{mu} and \var{rho} are material properties.
139 jgs 107 \begin{python}
140 caltinay 3280 def wavePropagation(domain,h,tend,lam,mu,rho, x_c, src_radius, U0):
141     # lists to collect displacement at point source which is returned
142     # to the caller
143     ts, u_pc0, u_pc1, u_pc2 = [], [], [], []
144    
145     x=domain.getX()
146     # ... open new PDE ...
147     mypde=LinearPDE(domain)
148 sshaw 4821 mypde.getSolverOptions().setSolverMethod(SolverOptions.HRZ_LUMPING)
149 caltinay 3280 kronecker=identity(mypde.getDim())
150     dunit=numpy.array([1., 0., 0.]) # defines direction of point source
151     mypde.setValue(D=kronecker*rho, q=whereNegative(length(x-xc)-src_radius)*dunit)
152     # ... set initial values ....
153     n=0
154     # for first two time steps
155     u=Vector(0., Solution(domain))
156     u_last=Vector(0., Solution(domain))
157     t=0
158     # define the location of the point source
159     L=Locator(domain, xc)
160     # find potential at point source
161     u_pc=L.getValue(u)
162     print("u at point charge=",u_pc)
163     ts.append(t)
164     u_pc0.append(u_pc[0])
165     u_pc1.append(u_pc[1])
166     u_pc2.append(u_pc[2])
167 gross 2580
168 caltinay 3280 while t<tend:
169     t+=h
170     # ... get current stress ....
171     g=grad(u)
172     stress=lam*trace(g)*kronecker+mu*(g+transpose(g))
173     # ... get new acceleration ....
174     amplitude=U0*(4*(t-t0)**3/alpha**3-6*(t-t0)/alpha)*sqrt(2.)/alpha**2 \
175     *exp(1./2.-(t-t0)**2/alpha**2)
176     mypde.setValue(X=-stress, r=dunit*amplitude)
177     a=mypde.getSolution()
178     # ... get new displacement ...
179     u_new=2*u-u_last+h**2*a
180     # ... shift displacements ....
181     u_last=u
182     u=u_new
183     n+=1
184     print(n,"-th time step, t=",t)
185     u_pc=L.getValue(u)
186     print("u at point charge=",u_pc)
187     # save displacements at point source to file for t > 0
188     ts.append(t)
189     u_pc0.append(u_pc[0])
190     u_pc1.append(u_pc[1])
191     u_pc2.append(u_pc[2])
192    
193     # ... save current acceleration in units of gravity and displacements
194     if n==1 or n%10==0:
195     saveVTK("./data/usoln.%i.vtu"%(n/10), \
196     acceleration = length(a)/9.81, \
197     displacement = length(u), \
198     tensor = stress, Ux = u[0])
199    
200     return ts, u_pc0, u_pc1, u_pc2
201 jgs 107 \end{python}
202 caltinay 3280 Notice that all coefficients of the PDE which are independent of time $t$ are
203     set outside the \code{while} loop.
204     This is for efficiency reasons since it allows the \LinearPDE class to reuse
205     information as much as possible when iterating over time.
206 gross 2502
207 caltinay 3280 The statement
208 gross 2502 \begin{python}
209 sshaw 4821 mypde.getSolverOptions().setSolverMethod(SolverOptions.HRZ_LUMPING)
210 gross 2502 \end{python}
211 caltinay 5296 enables the use of an aggressive approximation of the PDE operator as a
212 caltinay 3280 diagonal matrix formed from the coefficient \var{D}.
213     The approximation allows, at the cost of additional error, very fast solution
214 gross 4052 of the PDE, see also Section~\ref{LUMPING}.
215 gross 3379
216 gross 2502 There are a few new \escript functions in this example:
217 jfenwick 3295 \code{grad(u)} returns the gradient $u_{i,j}$ of $u$ (in fact \var{grad(g)[i,j]}$==u_{i,j}$).
218 caltinay 3280 There are restrictions on the argument of the \function{grad} function, for
219     instance the statement \code{grad(grad(u))} will raise an exception.
220     \code{trace(g)} returns the sum of the main diagonal elements \var{g[k,k]} of
221     \var{g} and \code{transpose(g)} returns the matrix transpose of \var{g} (i.e.
222     $\var{transpose(g)[i,j]}==\var{g[j,i]}$).
223 jgs 107
224 caltinay 3280 We initialize the values of \code{u} and \code{u_last} to be zero.
225     It is important to initialize both with the \SolutionFS as they have to be
226     seen as solutions of PDEs from previous time steps.
227     In fact, the \function{grad} does not accept arguments with a certain
228     \FunctionSpace, for more details see \Sec{SEC ESCRIPT DATA}.
229 jgs 107
230 caltinay 3280 The \class{Locator} class is designed to extract values at a given location
231 jfenwick 3295 (in this case $x_C$) from functions such as the displacement vector \code{u}.
232 caltinay 3280 Typically \class{Locator} is used in the following way:
233 gross 2502 \begin{python}
234 caltinay 3280 L=Locator(domain, xc)
235     u=...
236     u_pc=L.getValue(u)
237 gross 2502 \end{python}
238 caltinay 3280 The return value \code{u_pc} is the value of \code{u} at the location
239     \code{xc}\footnote{In fact, it is the finite element node which is closest to
240     the given position. The usage of \class{Locator} is MPI safe.}.
241     The values are collected in the lists \var{u_pc0}, \var{u_pc1} and \var{u_pc2}
242     together with the corresponding time marker in \var{ts}.
243     These values are handed back to the caller. Later we will show ways to access these data.
244 gross 2502
245 caltinay 3280 One of the big advantages of the Verlet scheme is the fact that the problem to
246     be solved in each time step is very simple and does not involve any spatial
247     derivatives (which is what allows us to use lumping in this simulation).
248     The problem becomes so simple because we use the stress from the last time
249     step rather than the stress which is actually present at the current time step.
250     Schemes using this approach are called explicit time integration schemes\index{explicit scheme}\index{time integration!explicit}.
251     The backward Euler scheme we have used in \Chap{DIFFUSION CHAP} is an example
252     of an implicit scheme\index{implicit scheme}\index{time integration!implicit}.
253     In this case one uses the actual status of each variable at a particular time
254     rather than values from previous time steps.
255     This will lead to a problem which is more expensive to solve, in particular
256     for non-linear cases.
257     Although explicit time integration schemes are cheap to finalize a single time
258     step, they need significantly smaller time steps than implicit schemes and can
259     suffer from stability problems.
260     Therefore they require a very careful selection of the time step size $h$.
261 jgs 107
262 caltinay 3280 An easy, heuristic way of choosing an appropriate time step size is the
263 sshaw 4552 Courant-Friedrichs-Lewy condition (CFL condition)\index{Courant condition}\index{explicit scheme!Courant condition}
264 caltinay 3280 which says that within a time step information should not travel further than
265     a cell used in the discretization scheme.
266     In the case of the wave equation the velocity of a (p-) wave is given as
267 jgs 107 $\sqrt{\frac{\lambda+2\mu}{\rho}}$ so one should choose $h$ from
268     \begin{eqnarray}\label{WAVE dyn 66}
269     h= \frac{1}{5} \sqrt{\frac{\rho}{\lambda+2\mu}} \Delta x
270     \end{eqnarray}
271 caltinay 3280 where $\Delta x$ is the cell diameter.
272     The factor $\frac{1}{5}$ is a safety factor considering the heuristics of the formula.
273 jgs 107
274 gross 2580 \begin{figure}[t]
275     \begin{center}
276 caltinay 3279 \includegraphics[width=2in]{Wave11}
277     \includegraphics[width=2in]{Wave22}
278     \includegraphics[width=2in]{Wave28}
279     \includegraphics[width=2in]{Wave32}
280     \includegraphics[width=2in]{Wave36}
281 gross 2580 \end{center}
282 caltinay 3280 \caption{Selected time steps ($n = 11, 22, 32, 36$) of a wave propagation over
283     a 10km $\times$ 10km $\times$ 3.125km block from a point source initially at
284     (5km, 5km, 0) with time step size $h=0.02083$. Color represents the
285     displacement. Here the view is oriented onto the bottom face.
286 gross 2580 \label{WAVE FIG 2}}
287     \end{figure}
288    
289 lkettle 581 The following script uses the \code{wavePropagation} function to solve the
290 caltinay 3280 wave equation for a point source located at the bottom face of a block.
291     The width of the block in each direction on the bottom face is 10km
292 jfenwick 3295 ($x_0$ and $x_1$ directions, i.e. \code{l0} and \code{l1}).
293     The variable \code{ne} gives the number of elements in $x_{0}$ and $x_1$ directions.
294     The depth of the block is aligned with the $x_{2}$-direction.
295     The depth (\code{l2}) of the block in the $x_{2}$-direction is chosen
296 caltinay 3280 so that there are 10 elements, and the magnitude of the depth is chosen such
297     that the elements become cubic.
298 jfenwick 3295 We chose 10 for the number of elements in the $x_{2}$-direction so
299     that the computation is faster. \code{Brick($n_0, n_1, n_2,l_0,l_1,l_2$)}
300     is an \finley function which creates a rectangular mesh with $n_0 \times n_1 \times n_2$
301     elements over the brick $[0,l_0] \times [0,l_1] \times [0,l_2]$.
302 jgs 107 \begin{python}
303 caltinay 3280 from esys.finley import Brick
304     ne = 32 # number of cells in x_0 and x_1 directions
305     width = 10000. # length in x_0 and x_1 directions
306     lam = 3.462e9
307     mu = 3.462e9
308     rho = 1154.
309     tend = 60
310     U0 = 1. # amplitude of point source
311     # spherical source at middle of bottom face
312     xc=[width/2.,width/2.,0.]
313     # define small radius around point xc
314     src_radius = 0.03*width
315     print("src_radius =",src_radius)
316     mydomain=Brick(ne, ne, 10, l0=width, l1=width, l2=10.*width/32.)
317 aellery 6686 h=(1./5.)*inf(sqrt(rho/(lam+2*mu)))*inf(domain.getSize())
318 caltinay 3280 print("time step size =",h)
319     ts, u_pc0, u_pc1, u_pc2 = \
320     wavePropagation(mydomain, h, tend, lam, mu, rho, xc, src_radius, U0)
321 jgs 107 \end{python}
322 caltinay 3280 The \function{domain.getSize()} function returns the local element size $\Delta x$.
323 gross 3379 Using \function{inf} ensures that the CFL condition~\ref{WAVE dyn 66} \index{Courant number}\index{CFL condition} holds
324 caltinay 3280 everywhere in the domain.
325 gross 2502
326 caltinay 3280 The script is available as \file{wave.py} in the \ExampleDirectory\index{scripts!\file{wave.py}}.
327     To visualize the results from the data directory:
328 gross 2580 \begin{verbatim}
329 caltinay 5296 mayavi2 -d usoln.1.vtu -m Surface
330 gross 2580 \end{verbatim}
331     You can rotate this figure by clicking on it with the mouse and moving it around.
332 caltinay 3280 Again use \emph{Configure Data} to move backward and forward in time, and
333 jfenwick 3295 also to choose the results (acceleration, displacement or $u_x$) by
334 caltinay 3280 using \emph{Select Scalar}.
335     \fig{WAVE FIG 2} shows the results for the displacement at various time steps.
336 jgs 107
337 jfenwick 2104 \begin{figure}[t!]
338 caltinay 3280 \centerline{\includegraphics[width=14cm]{WavePC}}
339     \caption{Amplitude at Point source from the Simulation}
340 gross 593 \label{WAVE FIG 1}
341     \end{figure}
342 jgs 107
343 caltinay 3280 It remains to show some possibilities to inspect the collected data
344     \var{u_pc0}, \var{u_pc1} and \var{u_pc2}.
345     One way is to write the data to a file and then use an external package such
346 caltinay 5296 as \gnuplot, LibreOffice Calc or Excel to read the data for further analysis.
347 caltinay 3280 The following code shows one possible way to write the data to the file \file{./data/U_pc.csv}:
348 gross 2580 \begin{python}
349 caltinay 3280 u_pc_data=FileWriter('./data/U_pc.csv')
350 sshaw 4777 for i in range(len(ts)):
351 caltinay 3280 u_pc_data.write("%f %f %f %f\n"%(ts[i],u_pc0[i],u_pc1[i],u_pc2[i]))
352     u_pc_data.close()
353 gross 2580 \end{python}
354 jfenwick 3295 The file \code{U_pc.csv} stores 4 columns of data: $t,u_x,u_y,u_z$
355     respectively, where $u_x,u_y,u_z$ are the
356     $x_0,x_1,x_2$ components of the displacement
357 caltinay 3280 vector $u$ at the point source.
358     These can be plotted easily using any plotting package.
359     In \gnuplot the command:
360 gross 2580 \begin{verbatim}
361 jfenwick 6678 plot 'U_pc.csv' u 1:2 title 'U_x' w l lw 2, 'U_pc.csv' \
362     u 1:3 title 'U_y' w l lw 2, \
363     'U_pc.csv' u 1:4 title 'U_z' w l lw 2
364 gross 2580 \end{verbatim}
365 caltinay 3280 will reproduce \fig{WAVE FIG 1} (As expected this is identical to the input
366     signal shown in \fig{WAVE FIG 1.2}).
367     It is pointed out that we are not using the standard \PYTHON \function{open}
368     to write to the file \code{U_pc.csv} as it is not safe when running \escript
369     under MPI, see Chapter~\ref{EXECUTION} for more details.
370 gross 2580
371 caltinay 3280 Alternatively, one can implement plotting the results at run time rather than
372     in a post-processing step.
373     This avoids the generation of an intermediate data file.
374     In {\it escript} the preferred way of creating 2D plots of time dependent data
375     is \MATPLOTLIB. The following script creates the plot and writes it into the
376 gross 2580 file \file{u_pc.png} in the PNG image format:
377     \begin{python}
378 caltinay 3280 import matplotlib.pyplot as plt
379     if getMPIRankWorld() == 0:
380     plt.title("Displacement at Point Source")
381     plt.plot(ts, u_pc0, '-', label="x_0", linewidth=1)
382     plt.plot(ts, u_pc1, '-', label="x_1", linewidth=1)
383     plt.plot(ts, u_pc2, '-', label="x_2", linewidth=1)
384     plt.xlabel('time')
385     plt.ylabel('displacement')
386     plt.legend()
387     plt.savefig('u_pc.png', format='png')
388 gross 2580 \end{python}
389 caltinay 3280 You can add \function{plt.show()} to create an interactive browser window.
390 caltinay 3331 Notice that by checking the condition \code{getMPIRankWorld()==0} the plot
391 caltinay 3280 is generated on one processor only (in this case the rank 0 processor) when
392 caltinay 3331 run under \MPI.
393 gross 2580
394 sshaw 4608 Both options for processing the point source data are included in the example
395 caltinay 3280 file \file{wave.py}. There are other options available to process these data
396     in particular through the \SCIPY package, e.g. Fourier transformations.
397     It is beyond the scope of this user's guide to document the usage of
398 caltinay 5296 \SCIPY for time series analysis but it is highly recommended to look in
399     relevant readily available documentation.
400 gross 2580

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