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Tue Sep 8 07:11:12 2009 UTC (11 years, 4 months ago) by gross
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2D fault systems are now working. 3D still needs work.
1 gross 2654 \begin{figure} [ht]
2     \centerline{\includegraphics[width=\figwidth]{figures/Slip1}}
3     \caption{Domain $\Omega=[0,1]^2$ with a vertical fault of length $0.5$.}
4     \label{fig:slip.1}
5     \end{figure}
6    
7 gross 2647 \section{Slip on a Fault}
8     \label{Slip CHAP}
9    
10 gross 2654 In this example we illustrate how to calculate the stress distribution around a fault \index{fault} in the
11     Earth's crust caused by a slip \index{slip} through an earthquake.
12    
13     To simplify the presentation we assume a simple domain $\Omega=[0,1]^2$ with
14     a vertical fault in its center. We assume that the slip distribution $s\hackscore{i}$ on the fault is known. We want to calculate the distribution of the displacements $u\hackscore{i}$\index{displacement} and stress $\sigma_{ij}$\index{stress} in the domain. We assume an isotropic, linear elastic material model of the form
15     \begin{eqnarray} \label{Slip stress}
16 gross 2647 \sigma\hackscore{ij} & = & \lambda u\hackscore{k,k} \delta\hackscore{ij} + \mu ( u\hackscore{i,j} + u\hackscore{j,i})
17     \end{eqnarray}
18     where $\lambda$ and $\mu$ are the Lame coefficients
19     \index{Lame coefficients} and $\delta\hackscore{ij}$ denotes the Kronecker symbol\index{Kronecker symbol}.
20     On the boundary the normal stress is given by
21 gross 2654 \begin{eqnarray} \label{Slip natural fault}
22 gross 2647 \sigma\hackscore{ij}n\hackscore{j}=0
23     \end{eqnarray}
24 gross 2654 and normal displacements are set to zero:
25     \begin{eqnarray} \label{Slip constraint}
26     u\hackscore{i}n\hackscore{i} =0
27     \end{eqnarray}
28     The stress needs to fulfill the momentum equation
29     \index{momentum equation}
30     \begin{eqnarray}\label{Slip general problem}
31     - \sigma\hackscore{ij,j}=0
32     \end{eqnarray}
33     This problem is very similar to the elastic deformation problem presented in Section~\ref{ELASTIC CHAP}.
34    
35     However, we need to address an additional challenge: The displacement $u\hackscore{i}$ is in fact discontinuous across the fault
36     but we are in the lucky situation that we know the jump of the displacements across the fault. This is
37     in fact the the given slip $s\hackscore{i}$. So we can split the total distribution $u\hackscore{i}$ into a component $v\hackscore{i}$
38     which is continuous across the fault and the known slip $s\hackscore{i}$
39     \begin{eqnarray}\label{Slip Split}
40     u\hackscore{i} = v\hackscore{i} + \frac{1}{2} s^{\pm}\hackscore{i}
41     \end{eqnarray}
42     where $s^{\pm}=s$ when right of the fault and $s^{\pm}=-s$
43     when left of the fault. We assume that $s^{\pm}=0$ sufficiently away from the fault.
44    
45     We insert this into the stress definition~\ref{Slip stress}
46     \begin{eqnarray} \label{Slip stress split}
47     \sigma\hackscore{ij} & = &
48     \sigma^c\hackscore{ij} +
49     \frac{1}{2} \sigma^s\hackscore{ij}
50     \end{eqnarray}
51     with
52     \begin{eqnarray} \label{Slip stress split 1 }
53     \sigma^c\hackscore{ij} = \lambda v\hackscore{k,k} \delta\hackscore{ij} + \mu ( v\hackscore{i,j} + v\hackscore{j,i})
54     \end{eqnarray}
55     and
56     \begin{eqnarray} \label{Slip stress split 2 }
57     \sigma^s\hackscore{ij} = \lambda s^{\pm}\hackscore{k,k} \delta\hackscore{ij} + \mu ( s^{\pm}\hackscore{i,j} + s^{\pm}\hackscore{j,i})
58     \end{eqnarray}
59     In fact $\sigma^s\hackscore{ij}$ defines a stress jump across the fault. In easy way to construct this
60     function is to us a function $\chi$ which is $1$ on the right and $-1$ on the left side from the fault. One can then
61     set
62     \begin{eqnarray} \label{Slip stress split 23 }
63     \sigma^s\hackscore{ij} = \chi \cdot ( \lambda s\hackscore{k,k} \delta\hackscore{ij} + \mu ( s\hackscore{i,j} + s\hackscore{j,i}) )
64     \end{eqnarray}
65     assuming that $s$ is extended by zero away from the fault. After inserting~\ref{Slip stress split} into~\ref{Slip general problem} we get the differential equation
66     \index{momentum equation}
67     \begin{eqnarray}\label{Slip general problem 2 }
68     - \sigma^c\hackscore{ij,j}=\frac{1}{2} \sigma^s\hackscore{ij,j}
69     \end{eqnarray}
70     Together with the definition~\ref{Slip stress split 1} we have a differential equation for the
71     continuous function $v_i$. Notice that the boundary condition~\ref{Slip constraint}
72     and~\ref{Slip natural fault} transfer to $v_i$ and $\sigma^c\hackscore{ij}$ as
73     $s$ is zero away from the fault. In Section~\ref{ELASTIC CHAP} we have discussed how this problem
74     is solved using the \LinearPDE class. We refer to this section for further details.
75    
76     To define the fault we use the \class{FaultSystem} class introduced in Section~\ref{Fault System}.
77     The following statements define a fault system \var{fs} and add the fault \var{1}
78     to the system.
79     \begin{python}
80     fault_start=[0.5,0.25]
81     fault_end=[0.5,0.75]
82     fs=FaultSystem(dim=2)
83     fs.addFault(top=[fault_start, fault_end], tag=1)
84     \end{python}
85     The main purpose of the \class{FaultSystem} class is to define a parameterization
86     of the fault using a local coordinate system. One can enquire the class
87     to get the range used to parameterize a fault.
88     \begin{python}
89     p0,p1= fs.getW0Range(tag=1)
90     \end{python}
91     Typically \var{p0} is equal to zero while \var{p1} is equal to the length of the fault. The parameterization
92     is given as a mapping from a set of local coordinates onto parameter range (in our case
93     the range \var{p0} to \var{p1}). For instance to map the entire domain \var{mydomain} anto the fault
94     one can use
95     \begin{python}
96     x=mydomain.getX()
97     p, m=fs.getParametrization(x,tag=1)
98     \end{python}
99     Of course there is the problem that not all location are on the fault. For those locations which are on the
100     fault \var{m} is set to $1$ otherwise $0$ is used. So on return the values of \var{p} are defining
101     the value of the fault parameterization (typically the distance from the starting point of the fault along the fault)
102     where \var{m} is positive. On all other locations the value if \var{p} is undefined. Now \var{p} can
103     be used to define a slip distribution on the fault via
104     \begin{python}
105     s=m*(p-p0)*(p1-p)/((p1-p0)/2)**2*slip_max*[0.,1.]
106     \end{python}
107     Notice that the factor \var{m} which makes sure that \var{s} is zero away from the fault. It is important
108     that the slip is zero at the ends of the faults.
109    
110     \begin{figure} [ht]
111     \centerline{\includegraphics[width=\figwidth]{figures/Slip2}}
112     \caption{Total Displacement after slip event.}
113     \label{fig:slip.2}
114     \end{figure}
115    
116     We can now put all components together to get the script:
117     \begin{python}
118     from esys.escript import *
119     from esys.escript.linearPDEs import LinearPDE
120     from esys.escript.models import FaultSystem
121     from esys.finley import Rectangle
122     #... set some parameters ...
123     lam=1.
124     mu=1
125     slip_max=1.
126     fault_start=[0.5,0.25]
127     fault_end=[0.5,0.75]
128    
129     mydomain = Rectangle(l0=1.,l1=1.,n0=16, n1=16) # n1 need to be multiple of 4!!!
130     # .. create the fault system
131     fs=FaultSystem(dim=2)
132     fs.addFault(top=[fault_start, fault_end], tag=1)
133     # ... create a slip distribution on the fault:
134     p, m=fs.getParametrization(mydomain.getX(),tag=1)
135     p0,p1= fs.getW0Range(tag=1)
136     s=m*(p-p0)*(p1-p)/((p1-p0)/2)**2*slip_max*[0.,1.]
137     # ... calculate stress according to slip:
138     D=symmetric(grad(s))
139     chi, d=fs.getSideAndDistance(D.getFunctionSpace().getX(),tag=1)
140     sigma_s=(mu*D+lam*trace(D)*kronecker(mydomain))*chi
141     #... open symmetric PDE ...
142     mypde=LinearPDE(mydomain)
143     mypde.setSymmetryOn()
144     #... set coefficients ...
145     C=Tensor4(0.,Function(mydomain))
146     for i in range(mydomain.getDim()):
147     for j in range(mydomain.getDim()):
148     C[i,i,j,j]+=lam
149     C[j,i,j,i]+=mu
150     C[j,i,i,j]+=mu
151     # ... fix displacement in normal direction
152     x=mydomain.getX()
153     msk=whereZero(x[0])*[1.,0.] + whereZero(x[0]-1.)*[1.,0.] \
154     +whereZero(x[1])*[0.,1.] + whereZero(x[1]-1.)*[0.,1.]
155     mypde.setValue(A=C,X=-0.5*sigma_s,q=msk)
156     #... solve pde ...
157     mypde.getSolverOptions().setVerbosityOn()
158     v=mypde.getSolution()
159     # .. write the displacement to file:
160     D=symmetric(grad(v))
161     sigma=(mu*D+lam*trace(D)*kronecker(mydomain))+0.5*sigma_s
162     saveVTK("slip.vtu",disp=v+0.5*chi*s, stress= sigma)
163     \end{python}
164     The script creates the \file{slip.vtu} giving the total displacements and stress. These values are stored as cell
165     centered data. See Figure~\ref{fig:slip.2} for the result.

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