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1 % $Id$
2 \chapter{3D Wave Propagation}
3 \label{WAVE CHAP}
4
5 We want to calculate the displacement field $u\hackscore{i}$ for any time $t>0$ by solving the wave equation
6 \index{wave equation}:
7 \begin{eqnarray}\label{WAVE general problem}
8 \rho u\hackscore{i,tt} - \sigma\hackscore{ij,j}=0
9 \end{eqnarray}
10 in a three dimensional block of length $L$ in $x\hackscore{0}$
11 and $x\hackscore{1}$ direction and height $H$
12 in $x\hackscore{2}$ direction. $\rho$ is the known density which may be a function of its location.
13 $\sigma\hackscore{ij}$ is the stress field \index{stress} which in case of an isotropic, linear elastic material is given by
14 \begin{eqnarray} \label{WAVE stress}
15 \sigma\hackscore{ij} & = & \lambda u\hackscore{k,k} \delta\hackscore{ij} + \mu ( u\hackscore{i,j} + u\hackscore{j,i})
16 \end{eqnarray}
17 where $\lambda$ and $\mu$ are the Lame coefficients
18 \index{Lame coefficients} and $\delta\hackscore{ij}$ denotes the Kronecker symbol\index{Kronecker symbol}.
19 On the boundary the normal stress is given by
20 \begin{eqnarray} \label{WAVE natural}
21 \sigma\hackscore{ij}n\hackscore{j}=0
22 \end{eqnarray}
23 for all time $t>0$. At initial time $t=0$ the displacement
24 $u\hackscore{i}$ and the velocity $u\hackscore{i,t}$ are give:
25 \begin{eqnarray} \label{WAVE initial}
26 u\hackscore{i}(0,x)=0 & \mbox{ and } & u\hackscore{i,t}(0,x)=0 \;
27 \end{eqnarray}
28 for all $x$ in the domain.
29
30 The points on the left face of the domain, ie. points $x=(x\hackscore{i})$ with $x\hackscore{0}=0$,
31 are moved into the $x\hackscore{2}$ direction by $u^{max}$. This is expressed by the constraint
32 \begin{eqnarray} \label{WAVE impact}
33 u\hackscore{2}(x,t)=s(t) \mbox{ for } x=(x\hackscore{i}) with x\hackscore{0}=0
34 \end{eqnarray}
35 where
36 \begin{eqnarray} \label{WAVE impact s}
37 s(t)=u^{max} (1-e^{-(\tau^{-1}t)^3})
38 \end{eqnarray}
39 $\tau$ measures the time needed to reach the maximum displacement $u^{max}$ (
40 actually $\tau (ln 2)^{\frac{1}{3}}) \approx 0.9 \tau$ is the time until $\frac{u^{max}}{2}$ is reached).
41 The smaller $\tau$ the faster $u^{max}$ is reached.
42
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 u^{(n)}=2u^{(n-1)}-u^{(n)} + h^2 a^{(n)} \\
48 \end{eqnarray}
49 for all $n=1,2,\ldots$. It is designed to solve a system of equations of the form
50 \begin{eqnarray} \label{WAVE dyn 2b}
51 u\hackscore{,tt}=G(u)
52 \end{eqnarray}
53 where one sets $a^{(n)}=G(u^{(n-1)})$, see \Ref{Mora94}.
54
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 \begin{eqnarray} \label{WAVE natural}
61 \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 \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 \end{eqnarray}.
67 Additionally $a^{(n)}$ has to fullfill the constraint
68 \begin{eqnarray}\label{WAVE dyn 4}
69 a^{(n)}\hackscore{2}(x)= s\hackscore{,tt}(t^{(n)})\mbox{ where } s\hackscore{,tt}(t)=\frac{u^{max}}{\tau^2}
70 (6\tau^{-1}t- 9(\tau^{-1}t)^4) e^{-(\tau^{-1}t)^3}
71 \end{eqnarray}
72 derived from \eqn{WAVE impact}.
73
74
75 In each time step we have to solve this problem to get the acceleration $a^{(n)}$ and we will
76 use the \LinearPDE class of the \linearPDEsPack to do so. The general form of the PDE defined through
77 the \LinearPDE class is discussed in \Sec{SEC LinearPDE}. The form which are relevant here are
78 \begin{equation}\label{WAVE dyn 100}
79 D\hackscore{ij} u\hackscore{j} = - X\hackscore{ij,j}\; .
80 \end{equation}
81 Implicitly, the natural boundary condition
82 \begin{equation}\label{WAVE dyn 101}
83 n\hackscore{j}X\hackscore{ij} =0
84 \end{equation}
85 is assumed. The \LinearPDE object allows defining constraints of the form
86 \begin{equation}\label{WAVE dyn 101}
87 u\hackscore{i}(x)=r\hackscore{i}(x) \mbox{ for } x \mbox{ with } q\hackscore{i}>0
88 \end{equation}
89 where $r$ and $q$ are given functions.
90
91 With $u=a^{(n)}$ we can identify the values to be assigned to $D$, $X$, $r$ and $q$:
92 \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 r\hackscore{i}=\delta\hackscore{i2} \cdot s\hackscore{,tt}(t^{(n)}) & q\hackscore{i}=\delta\hackscore{i2} \cdot x\hackscore{0}.\texttt{whereZero()}
97 \end{array}
98 \end{equation}
99 Remember that \var{y.whereZero()} returns a function which is $1$ for those points $x$ where $y$ is zero
100 (up to a certain tolerance, see \Sec{SEC ESCRIPT DATA}) and $0$ for those points where $y$ is not equal to zero.
101
102 In the following script defines a the function \function{wavePropagation} which
103 implements the Verlet scheme to solve our wave propagation problem.
104 The argument \var{domain} which is a \Domain class object
105 defines the domain of the problem. \var{h} and \var{tend} is the step size
106 and the time mark after which the simulation will be terminated. \var{lam}, \var{mu} and
107 \var{rho} are material properties. \var{s_tt} is a function which returns $s\hackscore{,tt}$ at a given time:
108 \begin{python}
109 from esys.linearPDEs import LinearPDE
110 from numarray import identity
111 from esys.escript import *
112 def wavePropagation(domain,h,tend,lam,mu,rho,s_tt):
113 x=domain.getX()
114 # ... open new PDE ...
115 mypde=LinearPDE(domain)
116 mypde.setLumpingOn()
117 kronecker=identity(mypde.getDim())
118 mypde.setValue(D=kronecker*rho, \
119 q=x[0].whereZero()*kronecker[1,:])
120 # ... set initial values ....
121 n=0
122 u=Vector(0,ContinuousFunction(domain))
123 u_last=Vector(0,ContinuousFunction(domain))
124 t=0
125 while t<tend:
126 # ... get current stress ....
127 g=grad(u)
128 stress=lam*trace(g)*kronecker+mu*(g+transpose(g))
129 # ... get new acceleration ....
130 mypde.setValue(X=-stress,r=s_tt(t+h)*kronecker[1,:])
131 a=mypde.getSolution()
132 # ... get new displacement ...
133 u_new=2*u-u_last+h**2*a
134 # ... shift displacements ....
135 u_last=u
136 u=u_new
137 t+=h
138 n+=1
139 print n,"-th time step t ",t
140 print "a=",inf(a),sup(a)
141 print "u=",inf(u),sup(u)
142 # ... save current acceleration in units of gravity
143 if n%10==0 : (length(a)/9.81).saveDX("/tmp/res/u.%i.dx"%(n/10))
144 \end{python}
145 Notice that
146 all coefficients of the PDE which are independent from time $t$ are set outside the \code{while}
147 loop. This allows the \LinearPDE class to ruse information as much as possible
148 when iterating over time.
149
150 We have seen most of the functions in previous examples but there are some new functions here:
151 \code{grad(u)} returns the gradient $u\hackscore{i,j}$ of $u$ (in fact \var{grad(g)[i,j]}=$=u\hackscore{i,j}$).
152 It is pointed out that in general there are restrictions on the argument of the \function{grad} function, for instance
153 for \finley the statement \code{grad(grad(u))} will raise an exception.
154 \code{trace(g)} returns the sum of the main diagonal elements \var{g[k,k]} of \var{g}
155 and \code{transpose(g)} returns the transposed of the matrix \var{g} (ie.
156 $\var{transpose(g)[i,j]}=\var{g[j,i]}$).
157
158 The initialization of \var{u} and \var{u_last} needs some explanation. The statement
159 \begin{python}
160 u=Vector(0.,ContinuousFunction(domain))
161 \end{python}
162 declares \var{u} as a vector-valued function of its coordinates
163 in the domain (i.e. with two or three components in the case of
164 a two or three dimensional domain, repectively). The first argument defines the value of the function which is equal
165 to $0$ for all vectot components. The second argument \var{ContinuousFunction(domain)}
166 specifies the `smoothness` of the function. In our case we want the displacement field to be
167 a continuous function over the \Domain \var{domain}. On other option would be
168 \var{Function(domain)} in which case continuity is not garanteed. In fact the
169 argument of the \function{grad} function has to be of the typ \var{ContinuousFunction(domain)} while
170 returned function is of the var{Function(domain)} type. For more details see \Sec{SEC ESCRIPT DATA}.
171
172 The statement \code{myPDE.setLumpingOn()}
173 switches on the usage of an aggressive approximation of the PDE operator, in this case
174 formed by the coefficient \var{D} (actually the discrete
175 version of the PDE operator is approximated by a diagonal matrix). The approximation allows, at the costs of
176 an additional error, very fast
177 solving of the PDE when the solution is requested. In the general case of the presents of \var{A}, \var{B} or \var
178 {C} \code{setLumpingOn} will produce wrong results.
179
180 It is pointed out that the value of the constraint \var{r} for the acceleration \var{a} at time $t=0$ is consistent
181 with the initial value of the acceleration which is identical zero. Otherwise, a discontinuity of \var{a} in time
182 direction would lead to heavy oscillations in the displacement. It is also pointed out, that at time $t=0$ the
183 constraint defined by \eqn{WAVE impact} fulfills
184 the initial conditions in \eqn{WAVE initial}.
185
186
187 One of the big advantages of the Verlet scheme is the fact that the problem to be solved
188 in each time step is very simple and does not involve any spatial derivatives (which actually allows us to use lumping).
189 The problem becomes so simple because we use the stress from the last time step rather then the stress which
190 actually is present at the current time step. Schemes using this approach are called an explicit time integration
191 scheme \index{explicit scheme} \index{time integration!explicit}. The
192 backward Euler scheme we have used in \Chap{DIFFUSION CHAP} is
193 an example of an implicit schemes
194 \index{implicit scheme} \index{time integration!implicit}. In this case one uses the actual status of
195 each variable at a particular time rather then values from previous time steps. This will lead to problem which is
196 more expensive to solve, in particular for non-linear problems.
197 Although
198 explicit time integration schemes are cheep to finalize a single time step, they need a significant smaller time
199 step then implicit schemes and can suffer from stability problems. Therefor they need a
200 very careful selection of the time step size $h$.
201
202 An easy, heuristic way of choosing an appropriate
203 time step size is the Courant condition \index{Courant condition} \index{explicit scheme!Courant condition}
204 which says that within a time step a information should not travel further than a cell used in the
205 discretization scheme. In the case of the wave equation the velocity of a (p-) wave is given as
206 $\sqrt{\frac{\lambda+2\mu}{\rho}}$ so one should choose $h$ from
207 \begin{eqnarray}\label{WAVE dyn 66}
208 h= \frac{1}{5} \sqrt{\frac{\rho}{\lambda+2\mu}} \Delta x
209 \end{eqnarray}
210 where $\Delta x$ is the cell diameter. The factor $\frac{1}{5}$ is a safety factor considering the heuristic of
211 the formula.
212
213 The following script uses the \code{wavePropagtion} function to solve the
214 wave equation over a block in crust of the earth. The depth of the block is
215 $10km$ and the width in each direction is $100km$. The depth of the block is alligned
216 with the $x\hackscore{0}$-direction. \var{ne} gives the number of elements in $x\hackscore{0}$-direction. The number
217 of elements in $x\hackscore{1}$- and $x\hackscore{2}$-direction is chosen such that the elements
218 become cubic. The impact is $2m$ acting within a time of about $10sec$:
219 \begin{python}
220 ne=10 # number of cells in x_0-direction
221 depth=10000. # length in x_0-direction
222 width=100000. # length in x_1 and x_2 direction
223 lam=3.462e9
224 mu=3.462e9
225 rho=1154.
226 tau=10.
227 umax=2.
228 tend=60
229 h=1./5.*sqrt(rho/(lam+2*mu))*(depth/ne)
230 def s_tt(t): return umax/tau**2*(6*t/tau-9*(t/tau)**4)*exp(-(t/tau)**3)
231
232 print "time step size = ",h
233 mydomain=Brick(ne,int(width/depth)*ne,int(width/depth)*ne,l0=depth,l1=width,l2=width)
234 wavePropagation(mydomain,h,tend,lam,mu,rho,s_tt)
235 \end{python}
236 The script is available as \file{wave.py} in the \ExampleDirectory \index{scripts!\file{wave.py}}.
237
238
239 % \begin{figure}
240 % \centerline{\includegraphics[width=\figwidth]{DiffusionDomain}}
241 % \caption{Temperature Diffusion Problem with Circular Heat Source}
242 % \label{WAVE FIG 1}
243 % \end{figure}
244
245 \begin{figure}
246 % \centerline{\includegraphics[width=\figwidth]{DiffusionRes1}}
247 % \centerline{\includegraphics[width=\figwidth]{DiffusionRes16}}
248 % \centerline{\includegraphics[width=\figwidth]{DiffusionRes32}}
249 %\centerline{\includegraphics[width=\figwidth]{DiffusionRes48}}
250 \caption{Selected time steps of a wave propagation over a $10km \times 100km \times 100km$
251 with time step size $h=0.0666$. Color represents the acceleration.}
252 \label{WAVE FIG 2}
253 \end{figure}
254
255

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