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finally there is a 3D version for the fault system class
1 \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 \section{Slip on a Fault}
8 \label{Slip CHAP}
9
10 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 \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 \begin{eqnarray} \label{Slip natural fault}
22 \sigma\hackscore{ij}n\hackscore{j}=0
23 \end{eqnarray}
24 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 fs=FaultSystem(dim=2)
81 fs.addFault(fs.addFault(V0=[0.5,0.25], strikes=90*DEG, ls=0.5, tag=1)
82 \end{python}
83 The fault added starts at point $(0.5,0.25)$ has length $0.5$ and is pointing north.
84 The main purpose of the \class{FaultSystem} class is to define a parameterization
85 of the fault using a local coordinate system. One can enquire the class
86 to get the range used to parameterize a fault.
87 \begin{python}
88 p0,p1= fs.getW0Range(tag=1)
89 \end{python}
90 Typically \var{p0} is equal to zero while \var{p1} is equal to the length of the fault. The parameterization
91 is given as a mapping from a set of local coordinates onto parameter range (in our case
92 the range \var{p0} to \var{p1}). For instance to map the entire domain \var{mydomain} anto the fault
93 one can use
94 \begin{python}
95 x=mydomain.getX()
96 p, m=fs.getParametrization(x,tag=1)
97 \end{python}
98 Of course there is the problem that not all location are on the fault. For those locations which are on the
99 fault \var{m} is set to $1$ otherwise $0$ is used. So on return the values of \var{p} are defining
100 the value of the fault parameterization (typically the distance from the starting point of the fault along the fault)
101 where \var{m} is positive. On all other locations the value if \var{p} is undefined. Now \var{p} can
102 be used to define a slip distribution on the fault via
103 \begin{python}
104 s=m*(p-p0)*(p1-p)/((p1-p0)/2)**2*slip_max*[0.,1.]
105 \end{python}
106 Notice that the factor \var{m} which makes sure that \var{s} is zero away from the fault. It is important
107 that the slip is zero at the ends of the faults.
108
109 \begin{figure} [ht]
110 \centerline{\includegraphics[width=\figwidth]{figures/Slip2}}
111 \caption{Total Displacement after slip event.}
112 \label{fig:slip.2}
113 \end{figure}
114
115 We can now put all components together to get the script:
116 \begin{python}
117 from esys.escript import *
118 from esys.escript.linearPDEs import LinearPDE
119 from esys.escript.models import FaultSystem
120 from esys.finley import Rectangle
121 from esys.escript.unitsSI import DEG
122
123 #... set some parameters ...
124 lam=1.
125 mu=1
126 slip_max=1.
127 mydomain = Rectangle(l0=1.,l1=1.,n0=16, n1=16) # n1 need to be multiple of 4!!!
128 # .. create the fault system
129 fs=FaultSystem(dim=2)
130 fs.addFault(V0=[0.5,0.25], strikes=90*DEG, ls=0.5, tag=1)
131 # ... create a slip distribution on the fault:
132 p, m=fs.getParametrization(mydomain.getX(),tag=1)
133 p0,p1= fs.getW0Range(tag=1)
134 s=m*(p-p0)*(p1-p)/((p1-p0)/2)**2*slip_max*[0.,1.]
135 # ... calculate stress according to slip:
136 D=symmetric(grad(s))
137 chi, d=fs.getSideAndDistance(D.getFunctionSpace().getX(),tag=1)
138 sigma_s=(mu*D+lam*trace(D)*kronecker(mydomain))*chi
139 #... open symmetric PDE ...
140 mypde=LinearPDE(mydomain)
141 mypde.setSymmetryOn()
142 #... set coefficients ...
143 C=Tensor4(0.,Function(mydomain))
144 for i in range(mydomain.getDim()):
145 for j in range(mydomain.getDim()):
146 C[i,i,j,j]+=lam
147 C[j,i,j,i]+=mu
148 C[j,i,i,j]+=mu
149 # ... fix displacement in normal direction
150 x=mydomain.getX()
151 msk=whereZero(x[0])*[1.,0.] + whereZero(x[0]-1.)*[1.,0.] \
152 +whereZero(x[1])*[0.,1.] + whereZero(x[1]-1.)*[0.,1.]
153 mypde.setValue(A=C,X=-0.5*sigma_s,q=msk)
154 #... solve pde ...
155 mypde.getSolverOptions().setVerbosityOn()
156 v=mypde.getSolution()
157 # .. write the displacement to file:
158 D=symmetric(grad(v))
159 sigma=(mu*D+lam*trace(D)*kronecker(mydomain))+0.5*sigma_s
160 saveVTK("slip.vtu",disp=v+0.5*chi*s, stress= sigma)
161 \end{python}
162 The script creates the \file{slip.vtu} giving the total displacements and stress. These values are stored as cell
163 centered data. See Figure~\ref{fig:slip.2} for the result.

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