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Tue Sep 25 03:18:30 2007 UTC (12 years, 10 months ago) by ksteube
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Quickly edited chapters 1 and 2 of the User Guide, but it needs more work.
Ran entire document through spell checker.

1 ksteube 1316 %
2 jgs 107 % $Id$
3 gross 625 %
4 ksteube 1316 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
5 gross 625 %
6 ksteube 1316 % Copyright 2003-2007 by ACceSS MNRF
7     % Copyright 2007 by University of Queensland
8     %
9     % http://esscc.uq.edu.au
10     % Primary Business: Queensland, Australia
11     % Licensed under the Open Software License version 3.0
12     % http://www.opensource.org/licenses/osl-3.0.php
13     %
14     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
15     %
16    
17 jgs 121 \section{3D Wave Propagation}
18 jgs 107 \label{WAVE CHAP}
19    
20 ksteube 1316 In this next example we want to calculate the displacement field $u\hackscore{i}$ for any time $t>0$ by solving the wave equation
21 jgs 107 \index{wave equation}:
22     \begin{eqnarray}\label{WAVE general problem}
23     \rho u\hackscore{i,tt} - \sigma\hackscore{ij,j}=0
24     \end{eqnarray}
25     in a three dimensional block of length $L$ in $x\hackscore{0}$
26     and $x\hackscore{1}$ direction and height $H$
27     in $x\hackscore{2}$ direction. $\rho$ is the known density which may be a function of its location.
28     $\sigma\hackscore{ij}$ is the stress field \index{stress} which in case of an isotropic, linear elastic material is given by
29     \begin{eqnarray} \label{WAVE stress}
30     \sigma\hackscore{ij} & = & \lambda u\hackscore{k,k} \delta\hackscore{ij} + \mu ( u\hackscore{i,j} + u\hackscore{j,i})
31     \end{eqnarray}
32     where $\lambda$ and $\mu$ are the Lame coefficients
33     \index{Lame coefficients} and $\delta\hackscore{ij}$ denotes the Kronecker symbol\index{Kronecker symbol}.
34     On the boundary the normal stress is given by
35     \begin{eqnarray} \label{WAVE natural}
36     \sigma\hackscore{ij}n\hackscore{j}=0
37     \end{eqnarray}
38 lkettle 581 for all time $t>0$.
39    
40     At initial time $t=0$ the displacement
41     $u\hackscore{i}$ and the velocity $u\hackscore{i,t}$ are given:
42 jgs 107 \begin{eqnarray} \label{WAVE initial}
43 lkettle 581 u\hackscore{i}(0,x)= \left\{\begin{array}{cc} U\hackscore{0} & \mbox{for $x$ at point charge, $x\hackscore{C}$} \\ 0 & \mbox{elsewhere}\end{array} \right. & \mbox{ and } & u\hackscore{i,t}(0,x)=0 \;
44 jgs 107 \end{eqnarray}
45     for all $x$ in the domain.
46    
47 lkettle 581 Here we are modelling a point source at the point $x\hackscore C$, in the numerical solution we
48     set the initial displacement to be $U\hackscore 0$ in a sphere of small radius around the point
49     $x\hackscore C$.
50 jgs 107
51 ksteube 1316 We use an explicit time integration scheme to calculate the displacement field $u$ at
52 jgs 107 certain time marks $t^{(n)}$ where $t^{(n)}=t^{(n-1)}+h$ with time step size $h>0$. In the following the upper index ${(n)}$ refers to values at time $t^{(n)}$. We use the Verlet scheme \index{Verlet scheme} with constant time step size $h$
53     which is defined by
54     \begin{eqnarray} \label{WAVE dyn 2}
55 lkettle 575 u^{(n)}=2u^{(n-1)}-u^{(n-2)} + h^2 a^{(n)} \\
56 jgs 107 \end{eqnarray}
57 lkettle 581 for all $n=2,3,\ldots$. It is designed to solve a system of equations of the form
58 jgs 107 \begin{eqnarray} \label{WAVE dyn 2b}
59     u\hackscore{,tt}=G(u)
60     \end{eqnarray}
61 gross 660 where one sets $a^{(n)}=G(u^{(n-1)})$.
62 jgs 107
63     In our case $a^{(n)}$ is given by
64     \begin{eqnarray}\label{WAVE dyn 3}
65     \rho a^{(n)}\hackscore{i}=\sigma^{(n-1)}\hackscore{ij,j}
66     \end{eqnarray}
67     and boundary conditions
68 gross 593 \begin{eqnarray} \label{WAVE natural at n}
69 jgs 107 \sigma^{(n-1)}\hackscore{ij}n\hackscore{j}=0
70     \end{eqnarray}
71     derived from \eqn{WAVE natural} where
72     \begin{eqnarray} \label{WAVE dyn 3a}
73 lkettle 581 \sigma\hackscore{ij}^{(n-1)} & = & \lambda u^{(n-1)}\hackscore{k,k} \delta\hackscore{ij} + \mu ( u^{(n-1)}\hackscore{i,j} + u^{(n-1)}\hackscore{j,i}).
74 jgs 107 \end{eqnarray}
75 lkettle 581 Now we have converted our problem for displacement, $u^{(n)}$, into a problem for
76     acceleration, $a^(n)$, which now depends
77     on the solution at the previous two time steps, $u^{(n-1)}$ and $u^{(n-2)}$.
78 jgs 107
79 ksteube 1316 In each time step we have to solve this problem to get the acceleration $a^{(n)}$, and we will
80 gross 565 use the \LinearPDE class of the \linearPDEs to do so. The general form of the PDE defined through
81 ksteube 1316 the \LinearPDE class is discussed in \Sec{SEC LinearPDE}. The form which is relevant here is
82 jgs 107 \begin{equation}\label{WAVE dyn 100}
83 lkettle 581 D\hackscore{ij} a^{(n)}\hackscore{j} = - X\hackscore{ij,j}\; .
84 jgs 107 \end{equation}
85 ksteube 1316 The natural boundary condition
86 jgs 107 \begin{equation}\label{WAVE dyn 101}
87     n\hackscore{j}X\hackscore{ij} =0
88     \end{equation}
89 ksteube 1316 is used.
90 jgs 107
91 ksteube 1316 With $u=a^{(n)}$ we can identify the values to be assigned to $D$ and $X$:
92 jgs 107 \begin{equation}\label{WAVE dyn 6}
93     \begin{array}{ll}
94     D\hackscore{ij}=\rho \delta\hackscore{ij}&
95     X\hackscore{ij}=-\sigma^{(n-1)}\hackscore{ij} \\
96     \end{array}
97     \end{equation}
98 lkettle 581
99    
100     %Remember that \var{y.whereZero()} returns a function which is $1$ for those points $x$ where $y$ is zero
101     %(up to a certain tolerance, see \Sec{SEC ESCRIPT DATA}) and $0$ for those points where $y$ is not equal to zero.
102 jgs 107
103 lkettle 581 The following script defines a the function \function{wavePropagation} which
104 jgs 107 implements the Verlet scheme to solve our wave propagation problem.
105     The argument \var{domain} which is a \Domain class object
106 ksteube 1316 defines the domain of the problem. \var{h} and \var{tend} are the time step size
107     and the end time of the simulation. \var{lam}, \var{mu} and
108 lkettle 581 \var{rho} are material properties.
109 jgs 107 \begin{python}
110     from esys.linearPDEs import LinearPDE
111 lkettle 581 from numarray import identity,zeros,ones
112 jgs 107 from esys.escript import *
113 lkettle 581 from esys.escript.pdetools import Locator
114     def wavePropagation(domain,h,tend,lam,mu,rho,U0):
115 jgs 107 x=domain.getX()
116     # ... open new PDE ...
117     mypde=LinearPDE(domain)
118 lkettle 581 mypde.setSolverMethod(LinearPDE.LUMPING)
119 jgs 110 kronecker=identity(mypde.getDim())
120 lkettle 581 # spherical source at middle of bottom face
121     xc=[width/2.,width/2.,0.]
122     # define small radius around point xc
123     # Lsup(x) returns the maximum value of the argument x
124     src_radius = 0.1*Lsup(domain.getSize())
125     print "src_radius = ",src_radius
126     dunit=numarray.array([1.,0.,0.]) # defines direction of point source
127     mypde.setValue(D=kronecker*rho)
128 jgs 107 # ... set initial values ....
129 jgs 110 n=0
130 lkettle 581 # initial value of displacement at point source is constant (U0=0.01)
131     # for first two time steps
132     u=U0*whereNegative(length(x-xc)-src_radius)*dunit
133     u_last=U0*whereNegative(length(x-xc)-src_radius)*dunit
134 jgs 107 t=0
135 lkettle 581 # define the location of the point source
136     L=Locator(domain,xc)
137     # find potential at point source
138     u_pc=L.getValue(u)
139     print "u at point charge=",u_pc
140     u_pc_x = u_pc[0]
141     u_pc_y = u_pc[1]
142     u_pc_z = u_pc[2]
143     # open file to save displacement at point source
144     u_pc_data=open('./data/U_pc.out','w')
145     u_pc_data.write("%f %f %f %f\n"%(t,u_pc_x,u_pc_y,u_pc_z))
146 jgs 107 while t<tend:
147     # ... get current stress ....
148     g=grad(u)
149 jgs 110 stress=lam*trace(g)*kronecker+mu*(g+transpose(g))
150     # ... get new acceleration ....
151 lkettle 581 mypde.setValue(X=-stress)
152 jgs 107 a=mypde.getSolution()
153     # ... get new displacement ...
154     u_new=2*u-u_last+h**2*a
155     # ... shift displacements ....
156     u_last=u
157     u=u_new
158     t+=h
159 jgs 110 n+=1
160     print n,"-th time step t ",t
161 lkettle 581 L=Locator(domain,xc)
162     u_pc=L.getValue(u)
163     print "u at point charge=",u_pc
164     u_pc_x=u_pc[0]
165     u_pc_y=u_pc[1]
166     u_pc_z=u_pc[2]
167     # save displacements at point source to file for t > 0
168     u_pc_data.write("%f %f %f %f\n"%(t,u_pc_x,u_pc_y,u_pc_z))
169     # ... save current acceleration in units of gravity and displacements
170     if n==1 or n%10==0: saveVTK("./data/usoln.%i.vtu"%(n/10),acceleration=length(a)/9.81,
171     displacement = length(u), Ux = u[0] )
172     u_pc_data.close()
173 jgs 107 \end{python}
174 jgs 110 Notice that
175 ksteube 1316 all coefficients of the PDE which are independent of time $t$ are set outside the \code{while}
176     loop. This is very efficient since it allows the \LinearPDE class to reuse information as much as possible
177 jgs 110 when iterating over time.
178    
179 ksteube 1316 there are a few new \escript functions in this example:
180 jgs 107 \code{grad(u)} returns the gradient $u\hackscore{i,j}$ of $u$ (in fact \var{grad(g)[i,j]}=$=u\hackscore{i,j}$).
181 ksteube 1316 There are restrictions on the argument of the \function{grad} function, for instance
182     the statement \code{grad(grad(u))} will raise an exception.
183 jgs 107 \code{trace(g)} returns the sum of the main diagonal elements \var{g[k,k]} of \var{g}
184 ksteube 1316 and \code{transpose(g)} returns the matrix transpose of \var{g} (ie. $\var{transpose(g)[i,j]}=\var{g[j,i]}$).
185 jgs 107
186 lkettle 581 We initialize the values of \code{u} and \code{u_last} to be $U\hackscore 0$ for a small
187     sphere of radius, \code{src_radius} around the point source located at $x\hackscore C$.
188     These satisfy the initial conditions defined in \eqn{WAVE initial}.
189 jgs 107
190 lkettle 581 The output \code{U_pc.out} and \code{vtu} files are saved in a directory called \code{data}.
191     The \code{U_pc.out} stores 4 columns of data: $t,u\hackscore x,u\hackscore y,u\hackscore z$
192     respectively, where $u\hackscore x,u\hackscore y,u\hackscore z$ are the $x\hackscore 0,x\hackscore 1,x\hackscore 2$ components of
193     the displacement vector $u$ at the point source. These can be
194     plotted easily using any plotting package. In gnuplot the command:
195     \code{plot 'U_pc.out' u 1:2 title 'U_x' w l lw 2, 'U_pc.out' u 1:3 title 'U_y' w l lw 2,
196     'U_pc.out' u 1:4 title 'U_z' w l lw 2} will reproduce Figure \ref{WAVE FIG 1}.
197     %\begin{python}
198     %u=Vector(0.,ContinuousFunction(domain))
199     %\end{python}
200     %declares \var{u} as a vector-valued function of its coordinates
201     %in the domain (i.e. with two or three components in the case of
202 ksteube 1316 %a two or three dimensional domain, respectively). The first argument defines the value of the function which is equal
203 lkettle 581 %to $0$ for all vectot components. The second argument \var{ContinuousFunction(domain)}
204     %specifies the `smoothness` of the function. In our case we want the displacement field to be
205     %a continuous function over the \Domain \var{domain}. On other option would be
206 ksteube 1316 %\var{Function(domain)} in which case continuity is not guaranteed. In fact the
207 lkettle 581 %argument of the \function{grad} function has to be of the typ \var{ContinuousFunction(domain)} while
208     %returned function is of the var{Function(domain)} type. For more details see \Sec{SEC ESCRIPT DATA}.
209    
210 jgs 110 The statement \code{myPDE.setLumpingOn()}
211 ksteube 1316 switches on the use of an aggressive approximation of the PDE operator as a diagonal matrix
212     formed from the coefficient \var{D}.
213     The approximation allows, at the cost of
214     additional error, very fast
215     solution of the PDE. When using \code{setLumpingOn} the presence of \var{A}, \var{B} or \var
216     {C} will produce wrong results.
217 jgs 110
218 jgs 107
219 jgs 110 One of the big advantages of the Verlet scheme is the fact that the problem to be solved
220 ksteube 1316 in each time step is very simple and does not involve any spatial derivatives (which is what allows us to use lumping in this simulation).
221 lkettle 581 The problem becomes so simple because we use the stress from the last time step rather then the stress which is
222     actually present at the current time step. Schemes using this approach are called an explicit time integration
223 ksteube 1316 schemes \index{explicit scheme} \index{time integration!explicit}. The
224 jgs 110 backward Euler scheme we have used in \Chap{DIFFUSION CHAP} is
225 ksteube 1316 an example of an implicit scheme
226 jgs 110 \index{implicit scheme} \index{time integration!implicit}. In this case one uses the actual status of
227 lkettle 581 each variable at a particular time rather then values from previous time steps. This will lead to a problem which is
228 jgs 110 more expensive to solve, in particular for non-linear problems.
229     Although
230 lkettle 581 explicit time integration schemes are cheap to finalize a single time step, they need significantly smaller time
231     steps then implicit schemes and can suffer from stability problems. Therefore they need a
232 jgs 110 very careful selection of the time step size $h$.
233 jgs 107
234 jgs 110 An easy, heuristic way of choosing an appropriate
235 jgs 107 time step size is the Courant condition \index{Courant condition} \index{explicit scheme!Courant condition}
236     which says that within a time step a information should not travel further than a cell used in the
237     discretization scheme. In the case of the wave equation the velocity of a (p-) wave is given as
238     $\sqrt{\frac{\lambda+2\mu}{\rho}}$ so one should choose $h$ from
239     \begin{eqnarray}\label{WAVE dyn 66}
240     h= \frac{1}{5} \sqrt{\frac{\rho}{\lambda+2\mu}} \Delta x
241     \end{eqnarray}
242 lkettle 581 where $\Delta x$ is the cell diameter. The factor $\frac{1}{5}$ is a safety factor considering the heuristics of
243 jgs 107 the formula.
244    
245 lkettle 581 The following script uses the \code{wavePropagation} function to solve the
246     wave equation for a point source located at the bottom face of a block. The width of the block in
247     each direction on the bottom face is $10\mbox{km}$ ($x\hackscore 0$ and $x\hackscore 1$ directions ie. \code{l0} and \code{l1}).
248     \code{ne} gives the number of elements in $x\hackscore{0}$ and $x\hackscore 1$ directions.
249     The depth of the block is aligned with the $x\hackscore{2}$-direction.
250     The depth (\code{l2}) of the block in
251     the $x\hackscore{2}$-direction is chosen so that there are 10 elements and the magnitude of the
252     of the depth is chosen such that the elements
253     become cubic. We chose 10 for the number of elements in $x\hackscore{2}$-direction so that the
254     computation would be faster. \code{Brick($n\hackscore 0, n\hackscore 1, n\hackscore 2,l\hackscore 0,l\hackscore 1,l\hackscore 2$)} is an \finley function which creates a rectangular mesh
255     with $n\hackscore 0 \times n\hackscore 1 \times n\hackscore2$ elements over the brick $[0,l\hackscore 0] \times [0,l\hackscore 1] \times [0,l\hackscore 2]$.
256 jgs 107 \begin{python}
257 lkettle 581 from esys.finley import Brick
258     ne=32 # number of cells in x_0 and x_1 directions
259     width=10000. # length in x_0 and x_1 directions
260 jgs 110 lam=3.462e9
261     mu=3.462e9
262 jgs 107 rho=1154.
263 jgs 110 tend=60
264 lkettle 581 h=(1./5.)*sqrt(rho/(lam+2*mu))*(width/ne)
265     print "time step size = ",h
266     U0=0.01 # amplitude of point source
267 jgs 110
268 lkettle 581 mydomain=Brick(ne,ne,10,l0=width,l1=width,l2=10.*width/32.)
269     wavePropagation(mydomain,h,tend,lam,mu,rho,U0)
270 jgs 107 \end{python}
271 lkettle 581 The script is available as \file{wave.py} in the \ExampleDirectory \index{scripts!\file{wave.py}}. Before running the code make sure you have created a directory called \code{data} in the current
272     working directory.
273     To visualize the results from the data directory:
274     \begin{verbatim} mayavi -d usoln.1.vtu -m SurfaceMap &\end{verbatim} You can rotate this figure by clicking on it with the mouse and moving it around.
275     Again use \code{Configure Data} to move backwards and forward in time, and
276     also to choose the results (acceleration, displacement or $u\hackscore x$) by using \code{Select Scalar}. Figure \ref{WAVE FIG 2} shows the results for the displacement at various time steps.
277 jgs 107
278 gross 593 \begin{figure}{t!}
279 gross 599 \centerline{\includegraphics[width=3.in,angle=270]{figures/WavePC.eps}}
280 gross 593 \caption{Amplitude at Point source}
281     \label{WAVE FIG 1}
282     \end{figure}
283 jgs 107
284 lkettle 581 \begin{figure}{t}
285     \begin{center}
286 gross 599 \includegraphics[width=2in,angle=270]{figures/Wavet0.eps}
287     \includegraphics[width=2in,angle=270]{figures/Wavet1.eps}
288     \includegraphics[width=2in,angle=270]{figures/Wavet3.eps}
289     \includegraphics[width=2in,angle=270]{figures/Wavet10.eps}
290     \includegraphics[width=2in,angle=270]{figures/Wavet30.eps}
291     \includegraphics[width=2in,angle=270]{figures/Wavet288.eps}
292 lkettle 581 \end{center}
293     \caption{Selected time steps ($n = 0,1,30,100,300$ and 2880) of a wave propagation over a $10\mbox{km} \times 10\mbox{km} \times 3.125\mbox{km}$ block
294     from a point source initially at $(5\mbox{km},5\mbox{km},0)$
295     with time step size $h=0.02083$. Color represents the displacement.
296 gross 707 Here the view is oriented onto the bottom face.
297     \label{WAVE FIG 2}}
298 jgs 110 \end{figure}

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