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

revision 3307 by ahallam, Fri Oct 1 02:08:38 2010 UTC revision 3308 by jfenwick, Tue Oct 26 03:24:54 2010 UTC
# Line 22  particle motion. An example of this woul Line 22  particle motion. An example of this woul
22
23  Wave propagation in the earth can be described by the elastic wave equation  Wave propagation in the earth can be described by the elastic wave equation
24   \label{eqn:wav} \index{wave equation}   \label{eqn:wav} \index{wave equation}
25  \rho \frac{\partial^{2}u\hackscore{i}}{\partial t^2} - \frac{\partial  \rho \frac{\partial^{2}u_{i}}{\partial t^2} - \frac{\partial
26  \sigma\hackscore{ij}}{\partial x\hackscore{j}} = 0  \sigma_{ij}}{\partial x_{j}} = 0
27
28  where $\sigma$ is the stress given by  where $\sigma$ is the stress given by
29   \label{eqn:sigw}   \label{eqn:sigw}
30   \sigma \hackscore{ij} = \lambda u\hackscore{k,k} \delta\hackscore{ij} + \mu (   \sigma _{ij} = \lambda u_{k,k} \delta_{ij} + \mu (
31  u\hackscore{i,j} + u\hackscore{j,i})  u_{i,j} + u_{j,i})
32
33  and $\lambda$ and $\mu$ represent Lame's parameters. Specifically for seismic  and $\lambda$ and $\mu$ represent Lame's parameters. Specifically for seismic
34  waves, $\mu$ is the propagation materials shear modulus.  waves, $\mu$ is the propagation materials shear modulus.
35  In a similar process to the previous chapter, we will use the acceleration  In a similar process to the previous chapter, we will use the acceleration
36  solution to solve this PDE. By substituting $a$ directly for  solution to solve this PDE. By substituting $a$ directly for
37  $\frac{\partial^{2}u\hackscore{i}}{\partial t^2}$ we can derive the  $\frac{\partial^{2}u_{i}}{\partial t^2}$ we can derive the
38  displacement solution. Using $a$ we can see that \autoref{eqn:wav} becomes  displacement solution. Using $a$ we can see that \autoref{eqn:wav} becomes
39   \label{eqn:wava}   \label{eqn:wava}
40  \rho a\hackscore{i} - \frac{\partial  \rho a_{i} - \frac{\partial
41  \sigma\hackscore{ij}}{\partial x\hackscore{j}} = 0  \sigma_{ij}}{\partial x_{j}} = 0
42
43  Thus the problem will be solved for acceleration and then converted to  Thus the problem will be solved for acceleration and then converted to
44  displacement using the backwards difference approximation.  displacement using the backwards difference approximation.
# Line 47  Consider now the stress $\sigma$. One ca Line 47  Consider now the stress $\sigma$. One ca
47  distinct terms:  distinct terms:
48  \begin{subequations}  \begin{subequations}
49   \label{eqn:sigtrace}   \label{eqn:sigtrace}
50  \lambda u\hackscore{k,k} \delta\hackscore{ij}  \lambda u_{k,k} \delta_{ij}
51
52   \label{eqn:sigtrans}   \label{eqn:sigtrans}
53  \mu (u\hackscore{i,j} + u\hackscore{j,i})  \mu (u_{i,j} + u_{j,i})
54
55  \end{subequations}  \end{subequations}
56  One simply recognizes in \autoref{eqn:sigtrace} that $u\hackscore{k,k}$ is the  One simply recognizes in \autoref{eqn:sigtrace} that $u_{k,k}$ is the
57  trace of the displacement solution and that $\delta\hackscore{ij}$ is the  trace of the displacement solution and that $\delta_{ij}$ is the
58  kronecker delta function with dimensions equivalent to $u$. The second term  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  \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  facts together we see that the spatial differential of the stress is given by the
# Line 289  To dampen the waves, the method of \cite Line 289  To dampen the waves, the method of \cite
289  where the solution and the stress are multiplied by a damping function defined  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;  on $n$ nodes of the domain adjacent to the boundary, given by;
291
292  \gamma =\sqrt{\frac{| -log( \gamma \hackscore{b} ) |}{n^2}}  \gamma =\sqrt{\frac{| -log( \gamma _{b} ) |}{n^2}}
293
294
295  y=e^{-(\gamma x)^2}  y=e^{-(\gamma x)^2}

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