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Fri Mar 24 08:43:21 2006 UTC (17 years ago) by gross
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that does not look to bad now although more stuff could be added.
1 jgs 107 % $Id$
2 gross 625 %
3     % Copyright © 2006 by ACcESS MNRF
4     % \url{http://www.access.edu.au
5     % Primary Business: Queensland, Australia.
6     % Licensed under the Open Software License version 3.0
7     % http://www.opensource.org/license/osl-3.0.php
8     %
9 jgs 121 \section{3D Wave Propagation}
10 jgs 107 \label{WAVE CHAP}
11    
12     We want to calculate the displacement field $u\hackscore{i}$ for any time $t>0$ by solving the wave equation
13     \index{wave equation}:
14     \begin{eqnarray}\label{WAVE general problem}
15     \rho u\hackscore{i,tt} - \sigma\hackscore{ij,j}=0
16     \end{eqnarray}
17     in a three dimensional block of length $L$ in $x\hackscore{0}$
18     and $x\hackscore{1}$ direction and height $H$
19     in $x\hackscore{2}$ direction. $\rho$ is the known density which may be a function of its location.
20     $\sigma\hackscore{ij}$ is the stress field \index{stress} which in case of an isotropic, linear elastic material is given by
21     \begin{eqnarray} \label{WAVE stress}
22     \sigma\hackscore{ij} & = & \lambda u\hackscore{k,k} \delta\hackscore{ij} + \mu ( u\hackscore{i,j} + u\hackscore{j,i})
23     \end{eqnarray}
24     where $\lambda$ and $\mu$ are the Lame coefficients
25     \index{Lame coefficients} and $\delta\hackscore{ij}$ denotes the Kronecker symbol\index{Kronecker symbol}.
26     On the boundary the normal stress is given by
27     \begin{eqnarray} \label{WAVE natural}
28     \sigma\hackscore{ij}n\hackscore{j}=0
29     \end{eqnarray}
30 lkettle 581 for all time $t>0$.
31    
32     At initial time $t=0$ the displacement
33     $u\hackscore{i}$ and the velocity $u\hackscore{i,t}$ are given:
34 jgs 107 \begin{eqnarray} \label{WAVE initial}
35 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 \;
36 jgs 107 \end{eqnarray}
37     for all $x$ in the domain.
38    
39 lkettle 581 Here we are modelling a point source at the point $x\hackscore C$, in the numerical solution we
40     set the initial displacement to be $U\hackscore 0$ in a sphere of small radius around the point
41     $x\hackscore C$.
42 jgs 107
43     We use an explicit time-integration scheme to calculate the displacement field $u$ at
44     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 $h$
45     which is defined by
46     \begin{eqnarray} \label{WAVE dyn 2}
47 lkettle 575 u^{(n)}=2u^{(n-1)}-u^{(n-2)} + h^2 a^{(n)} \\
48 jgs 107 \end{eqnarray}
49 lkettle 581 for all $n=2,3,\ldots$. It is designed to solve a system of equations of the form
50 jgs 107 \begin{eqnarray} \label{WAVE dyn 2b}
51     u\hackscore{,tt}=G(u)
52     \end{eqnarray}
53 gross 660 where one sets $a^{(n)}=G(u^{(n-1)})$.
54 jgs 107
55     In our case $a^{(n)}$ is given by
56     \begin{eqnarray}\label{WAVE dyn 3}
57     \rho a^{(n)}\hackscore{i}=\sigma^{(n-1)}\hackscore{ij,j}
58     \end{eqnarray}
59     and boundary conditions
60 gross 593 \begin{eqnarray} \label{WAVE natural at n}
61 jgs 107 \sigma^{(n-1)}\hackscore{ij}n\hackscore{j}=0
62     \end{eqnarray}
63     derived from \eqn{WAVE natural} where
64     \begin{eqnarray} \label{WAVE dyn 3a}
65 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}).
66 jgs 107 \end{eqnarray}
67 lkettle 581 Now we have converted our problem for displacement, $u^{(n)}$, into a problem for
68     acceleration, $a^(n)$, which now depends
69     on the solution at the previous two time steps, $u^{(n-1)}$ and $u^{(n-2)}$.
70 jgs 107
71     In each time step we have to solve this problem to get the acceleration $a^{(n)}$ and we will
72 gross 565 use the \LinearPDE class of the \linearPDEs to do so. The general form of the PDE defined through
73 gross 625 the \LinearPDE class is discussed in \Sec{SEC LinearPDE}. The form which is relevant here are
74 jgs 107 \begin{equation}\label{WAVE dyn 100}
75 lkettle 581 D\hackscore{ij} a^{(n)}\hackscore{j} = - X\hackscore{ij,j}\; .
76 jgs 107 \end{equation}
77 jgs 110 Implicitly, the natural boundary condition
78 jgs 107 \begin{equation}\label{WAVE dyn 101}
79     n\hackscore{j}X\hackscore{ij} =0
80     \end{equation}
81 gross 625 is assumed.
82 jgs 107
83 lkettle 581 With $u=a^{(n)}$ we can identify the values to be assigned to $D$, $X$:
84 jgs 107 \begin{equation}\label{WAVE dyn 6}
85     \begin{array}{ll}
86     D\hackscore{ij}=\rho \delta\hackscore{ij}&
87     X\hackscore{ij}=-\sigma^{(n-1)}\hackscore{ij} \\
88     \end{array}
89     \end{equation}
90 lkettle 581
91    
92     %Remember that \var{y.whereZero()} returns a function which is $1$ for those points $x$ where $y$ is zero
93     %(up to a certain tolerance, see \Sec{SEC ESCRIPT DATA}) and $0$ for those points where $y$ is not equal to zero.
94 jgs 107
95 lkettle 581 The following script defines a the function \function{wavePropagation} which
96 jgs 107 implements the Verlet scheme to solve our wave propagation problem.
97     The argument \var{domain} which is a \Domain class object
98 lkettle 581 defines the domain of the problem. \var{h} and \var{tend} are the step size
99     and the time mark after which the simulation will be terminated respectively. \var{lam}, \var{mu} and
100     \var{rho} are material properties.
101 jgs 107 \begin{python}
102     from esys.linearPDEs import LinearPDE
103 lkettle 581 from numarray import identity,zeros,ones
104 jgs 107 from esys.escript import *
105 lkettle 581 from esys.escript.pdetools import Locator
106     def wavePropagation(domain,h,tend,lam,mu,rho,U0):
107 jgs 107 x=domain.getX()
108     # ... open new PDE ...
109     mypde=LinearPDE(domain)
110 lkettle 581 mypde.setSolverMethod(LinearPDE.LUMPING)
111 jgs 110 kronecker=identity(mypde.getDim())
112 lkettle 581 # spherical source at middle of bottom face
113     xc=[width/2.,width/2.,0.]
114     # define small radius around point xc
115     # Lsup(x) returns the maximum value of the argument x
116     src_radius = 0.1*Lsup(domain.getSize())
117     print "src_radius = ",src_radius
118     dunit=numarray.array([1.,0.,0.]) # defines direction of point source
119     mypde.setValue(D=kronecker*rho)
120 jgs 107 # ... set initial values ....
121 jgs 110 n=0
122 lkettle 581 # initial value of displacement at point source is constant (U0=0.01)
123     # for first two time steps
124     u=U0*whereNegative(length(x-xc)-src_radius)*dunit
125     u_last=U0*whereNegative(length(x-xc)-src_radius)*dunit
126 jgs 107 t=0
127 lkettle 581 # define the location of the point source
128     L=Locator(domain,xc)
129     # find potential at point source
130     u_pc=L.getValue(u)
131     print "u at point charge=",u_pc
132     u_pc_x = u_pc[0]
133     u_pc_y = u_pc[1]
134     u_pc_z = u_pc[2]
135     # open file to save displacement at point source
136     u_pc_data=open('./data/U_pc.out','w')
137     u_pc_data.write("%f %f %f %f\n"%(t,u_pc_x,u_pc_y,u_pc_z))
138 jgs 107 while t<tend:
139     # ... get current stress ....
140     g=grad(u)
141 jgs 110 stress=lam*trace(g)*kronecker+mu*(g+transpose(g))
142     # ... get new acceleration ....
143 lkettle 581 mypde.setValue(X=-stress)
144 jgs 107 a=mypde.getSolution()
145     # ... get new displacement ...
146     u_new=2*u-u_last+h**2*a
147     # ... shift displacements ....
148     u_last=u
149     u=u_new
150     t+=h
151 jgs 110 n+=1
152     print n,"-th time step t ",t
153 lkettle 581 L=Locator(domain,xc)
154     u_pc=L.getValue(u)
155     print "u at point charge=",u_pc
156     u_pc_x=u_pc[0]
157     u_pc_y=u_pc[1]
158     u_pc_z=u_pc[2]
159     # save displacements at point source to file for t > 0
160     u_pc_data.write("%f %f %f %f\n"%(t,u_pc_x,u_pc_y,u_pc_z))
161     # ... save current acceleration in units of gravity and displacements
162     if n==1 or n%10==0: saveVTK("./data/usoln.%i.vtu"%(n/10),acceleration=length(a)/9.81,
163     displacement = length(u), Ux = u[0] )
164     u_pc_data.close()
165 jgs 107 \end{python}
166 jgs 110 Notice that
167     all coefficients of the PDE which are independent from time $t$ are set outside the \code{while}
168 lkettle 581 loop. This allows the \LinearPDE class to reuse information as much as possible
169 jgs 110 when iterating over time.
170    
171     We have seen most of the functions in previous examples but there are some new functions here:
172 jgs 107 \code{grad(u)} returns the gradient $u\hackscore{i,j}$ of $u$ (in fact \var{grad(g)[i,j]}=$=u\hackscore{i,j}$).
173     It is pointed out that in general there are restrictions on the argument of the \function{grad} function, for instance
174     for \finley the statement \code{grad(grad(u))} will raise an exception.
175     \code{trace(g)} returns the sum of the main diagonal elements \var{g[k,k]} of \var{g}
176     and \code{transpose(g)} returns the transposed of the matrix \var{g} (ie.
177     $\var{transpose(g)[i,j]}=\var{g[j,i]}$).
178    
179 lkettle 581 We initialize the values of \code{u} and \code{u_last} to be $U\hackscore 0$ for a small
180     sphere of radius, \code{src_radius} around the point source located at $x\hackscore C$.
181     These satisfy the initial conditions defined in \eqn{WAVE initial}.
182 jgs 107
183 lkettle 581 The output \code{U_pc.out} and \code{vtu} files are saved in a directory called \code{data}.
184     The \code{U_pc.out} stores 4 columns of data: $t,u\hackscore x,u\hackscore y,u\hackscore z$
185     respectively, where $u\hackscore x,u\hackscore y,u\hackscore z$ are the $x\hackscore 0,x\hackscore 1,x\hackscore 2$ components of
186     the displacement vector $u$ at the point source. These can be
187     plotted easily using any plotting package. In gnuplot the command:
188     \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,
189     'U_pc.out' u 1:4 title 'U_z' w l lw 2} will reproduce Figure \ref{WAVE FIG 1}.
190     %\begin{python}
191     %u=Vector(0.,ContinuousFunction(domain))
192     %\end{python}
193     %declares \var{u} as a vector-valued function of its coordinates
194     %in the domain (i.e. with two or three components in the case of
195     %a two or three dimensional domain, repectively). The first argument defines the value of the function which is equal
196     %to $0$ for all vectot components. The second argument \var{ContinuousFunction(domain)}
197     %specifies the `smoothness` of the function. In our case we want the displacement field to be
198     %a continuous function over the \Domain \var{domain}. On other option would be
199     %\var{Function(domain)} in which case continuity is not garanteed. In fact the
200     %argument of the \function{grad} function has to be of the typ \var{ContinuousFunction(domain)} while
201     %returned function is of the var{Function(domain)} type. For more details see \Sec{SEC ESCRIPT DATA}.
202    
203 jgs 110 The statement \code{myPDE.setLumpingOn()}
204 jgs 107 switches on the usage of an aggressive approximation of the PDE operator, in this case
205     formed by the coefficient \var{D} (actually the discrete
206     version of the PDE operator is approximated by a diagonal matrix). The approximation allows, at the costs of
207     an additional error, very fast
208 lkettle 581 solving of the PDE when the solution is requested. In the general case, the presence of \var{A}, \var{B} or \var
209 jgs 107 {C} \code{setLumpingOn} will produce wrong results.
210 jgs 110
211 jgs 107
212 jgs 110 One of the big advantages of the Verlet scheme is the fact that the problem to be solved
213     in each time step is very simple and does not involve any spatial derivatives (which actually allows us to use lumping).
214 lkettle 581 The problem becomes so simple because we use the stress from the last time step rather then the stress which is
215     actually present at the current time step. Schemes using this approach are called an explicit time integration
216 jgs 110 scheme \index{explicit scheme} \index{time integration!explicit}. The
217     backward Euler scheme we have used in \Chap{DIFFUSION CHAP} is
218     an example of an implicit schemes
219     \index{implicit scheme} \index{time integration!implicit}. In this case one uses the actual status of
220 lkettle 581 each variable at a particular time rather then values from previous time steps. This will lead to a problem which is
221 jgs 110 more expensive to solve, in particular for non-linear problems.
222     Although
223 lkettle 581 explicit time integration schemes are cheap to finalize a single time step, they need significantly smaller time
224     steps then implicit schemes and can suffer from stability problems. Therefore they need a
225 jgs 110 very careful selection of the time step size $h$.
226 jgs 107
227 jgs 110 An easy, heuristic way of choosing an appropriate
228 jgs 107 time step size is the Courant condition \index{Courant condition} \index{explicit scheme!Courant condition}
229     which says that within a time step a information should not travel further than a cell used in the
230     discretization scheme. In the case of the wave equation the velocity of a (p-) wave is given as
231     $\sqrt{\frac{\lambda+2\mu}{\rho}}$ so one should choose $h$ from
232     \begin{eqnarray}\label{WAVE dyn 66}
233     h= \frac{1}{5} \sqrt{\frac{\rho}{\lambda+2\mu}} \Delta x
234     \end{eqnarray}
235 lkettle 581 where $\Delta x$ is the cell diameter. The factor $\frac{1}{5}$ is a safety factor considering the heuristics of
236 jgs 107 the formula.
237    
238 lkettle 581 The following script uses the \code{wavePropagation} function to solve the
239     wave equation for a point source located at the bottom face of a block. The width of the block in
240     each direction on the bottom face is $10\mbox{km}$ ($x\hackscore 0$ and $x\hackscore 1$ directions ie. \code{l0} and \code{l1}).
241     \code{ne} gives the number of elements in $x\hackscore{0}$ and $x\hackscore 1$ directions.
242     The depth of the block is aligned with the $x\hackscore{2}$-direction.
243     The depth (\code{l2}) of the block in
244     the $x\hackscore{2}$-direction is chosen so that there are 10 elements and the magnitude of the
245     of the depth is chosen such that the elements
246     become cubic. We chose 10 for the number of elements in $x\hackscore{2}$-direction so that the
247     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
248     with $n\hackscore 0 \times n\hackscore 1 \times n\hackscore2$ elements over the brick $[0,l\hackscore 0] \times [0,l\hackscore 1] \times [0,l\hackscore 2]$.
249 jgs 107 \begin{python}
250 lkettle 581 from esys.finley import Brick
251     ne=32 # number of cells in x_0 and x_1 directions
252     width=10000. # length in x_0 and x_1 directions
253 jgs 110 lam=3.462e9
254     mu=3.462e9
255 jgs 107 rho=1154.
256 jgs 110 tend=60
257 lkettle 581 h=(1./5.)*sqrt(rho/(lam+2*mu))*(width/ne)
258     print "time step size = ",h
259     U0=0.01 # amplitude of point source
260 jgs 110
261 lkettle 581 mydomain=Brick(ne,ne,10,l0=width,l1=width,l2=10.*width/32.)
262     wavePropagation(mydomain,h,tend,lam,mu,rho,U0)
263 jgs 107 \end{python}
264 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
265     working directory.
266     To visualize the results from the data directory:
267     \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.
268     Again use \code{Configure Data} to move backwards and forward in time, and
269     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.
270 jgs 107
271 gross 593 \begin{figure}{t!}
272 gross 599 \centerline{\includegraphics[width=3.in,angle=270]{figures/WavePC.eps}}
273 gross 593 \caption{Amplitude at Point source}
274     \label{WAVE FIG 1}
275     \end{figure}
276 jgs 107
277 lkettle 581 \begin{figure}{t}
278     \begin{center}
279 gross 599 \includegraphics[width=2in,angle=270]{figures/Wavet0.eps}
280     \includegraphics[width=2in,angle=270]{figures/Wavet1.eps}
281     \includegraphics[width=2in,angle=270]{figures/Wavet3.eps}
282     \includegraphics[width=2in,angle=270]{figures/Wavet10.eps}
283     \includegraphics[width=2in,angle=270]{figures/Wavet30.eps}
284     \includegraphics[width=2in,angle=270]{figures/Wavet288.eps}
285 lkettle 581 \end{center}
286     \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
287     from a point source initially at $(5\mbox{km},5\mbox{km},0)$
288     with time step size $h=0.02083$. Color represents the displacement.
289     Here the view is oriented onto the bottom face.}
290 jgs 110 \label{WAVE FIG 2}
291     \end{figure}

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