[cig-commits] commit: Slip is in global coordinate system. More cleanup of equations.

[Mercurial]

changeset:   77:7191377a3793
tag:         tip
user:        Brad Aagaard <baagaard at usgs.gov>
date:        Mon Oct 03 09:57:29 2011 -0700
files:       faultRup.tex
description:
Slip is in global coordinate system. More cleanup of equations.


diff -r 3e2ec5ee23f0 -r 7191377a3793 faultRup.tex
--- a/faultRup.tex	Fri Sep 16 09:49:05 2011 -0700
+++ b/faultRup.tex	Mon Oct 03 09:57:29 2011 -0700
@@ -197,16 +197,15 @@ We solve the elasticity equation includi
   - \tensor{\nabla} \cdot \tensor{\sigma} = \vec{0} \text{ in }V, \\
   \tensor{\sigma} \cdot \vec{n} = \vec{T} \text{ on }S_T, \\
   \vec{u} = \vec{u}_0 \text{ on }S_u, \\
-  \tensor{R} \cdot \vec{d} - \left( \vec{u}_{+} - \vec{u}_{-}\right) = \vec{0}
-  \text{ on }S_f, \label{eq:faultDisp}
+  \left( \vec{u}_{+} - \vec{u}_{-}\right) - \vec{d} = \vec{0}
+  \text{ on }S_f, \label{eqn:fault:disp}
 \end{gather}
 where $\vec{u}$ is the displacement vector, $\rho$ is the mass
 density, $\vec{f}$ is the body force vector, $\tensor{\sigma}$ is the
 Cauchy stress tensor, and $t$ is time. We specify tractions $\vec{T}$
 on surface $S_T$, displacements $\vec{u_0}$ on surface $S_u$, and slip
-$\vec{d}$ on fault surface $S_f$. The rotation matrix $\tensor{R}$
-transforms vectors from the fault coordinate system to the global
-coordinate system. Because both $\vec{T}$ and $\vec{u}$ are vector
+$\vec{d}$ on fault surface $S_f$, where the tractions and fault slip
+are in global coordinates. Because both $\vec{T}$ and $\vec{u}$ are vector
 quantities, there can be some spatial overlap of the surfaces $S_T$
 and $S_u$; however, a degree of freedom at any location cannot be
 associated with both types of boundary conditions simultaneously.
@@ -274,34 +273,37 @@ surface to zero,
 surface to zero,
 \begin{equation}
   \int_{S_f} \vec{\phi} \cdot 
-  \left( \tensor{R} \cdot \vec{d} - \vec{u}_{+} + \vec{u}_{-} \right) \, dS = 0.
+  \left( \vec{u}_{+} - \vec{u}_{-} - \vec{d}\right) \, dS = 0.
 \end{equation}
 
 We express the weighting function $\vec{\phi}$, trial solution
-$\vec{u}$, and Lagrange multipliers $\vec{l}$ as linear combinations
-of basis functions,
+$\vec{u}$, Lagrange multipliers $\vec{l}$, and fault slip $\vec{d}$ as
+linear combinations of basis functions,
 \begin{gather}
 \vec{\phi} = \sum_{m} \vec{a}_m N_m, \\
 \vec{u} = \sum_{n} \vec{u}_n N_n, \\
-\vec{l} = \sum_{p} \vec{l}_p N_p.
+\vec{l} = \sum_{p} \vec{l}_p N_p, \\
+\vec{d} = \sum_{p} \vec{d}_p N_p.
 \end{gather}
 Because the weighting function is zero on $S_u$, the number of basis
 functions for the trial solution $\vec{u}$ is generally greater than
 the number of basis functions for the weighting function $\vec{\phi}$,
-i.e., $n > m$. The basis functions for the Lagrange multipliers are
-associated with the fault surface, which is a lower dimension than the
-domain, so $p \ll n$ in most cases. If we express the linear
-combination of basis functions in terms of a matrix-vector product, we
-have
+i.e., $n > m$. The basis functions for the Lagrange multipliers and
+fault slip are associated with the fault surface, which is a lower
+dimension than the domain, so $p \ll n$ in most cases. If we express
+the linear combination of basis functions in terms of a matrix-vector
+product, we have
 \begin{gather}
 \vec{\phi} = \tensor{N}_m \cdot \vec{a}_m, \\
 \vec{u} = \tensor{N}_n \cdot \vec{u}_n, \\
-\vec{l} = \tensor{N}_p \cdot \vec{l}_p.
+\vec{l} = \tensor{N}_p \cdot \vec{l}_p, \\
+\vec{d} = \tensor{N}_p \cdot \vec{d}_p.
 \end{gather}
 
-The weighting function is
-arbitrary, so the integrands must be zero for all $\vec{a}_m$, which
-leads to
+\brad{Add comment on collocated method somewhere near here.}
+
+The weighting function is arbitrary, so the integrands must be zero
+for all $\vec{a}_m$, which leads to
 \begin{gather}
   \begin{split}
 - \int_{V} \nabla \tensor{N}_m^T \cdot \tensor{\sigma} \, dV
@@ -315,10 +317,10 @@ leads to
 \end{split}
 \\
 % 
-  \int_{S_f} \tensor{N}_p^T 
-  \left( \tensor{R} \cdot \vec{d} 
-    - \tensor{N}_{n^+} \vec{u}_{n^+} + \tensor{N}_{n^-} \vec{u}_{n^-}
-     \right)
+  \int_{S_f} \tensor{N}_p^T \cdot 
+  \left( \tensor{N}_{n^+} \cdot \vec{u}_{n^+} 
+    - \tensor{N}_{n^-} \cdot \vec{u}_{n^-}
+    - \tensor{N}_p \cdot \vec{d}_p \right)
   \, dS = \vec{0}.
 \end{gather}
 We want to solve these equations for the coefficients $\vec{u}_n$ and
@@ -354,10 +356,10 @@ and the loading conditions. Considering 
   \\
 %
 \label{eqn:quasi-static:residual:fault}
-  \int_{S_f} \tensor{N}_p^T 
-  \left( \tensor{R} \cdot \vec{d}(\tpdt)
-    - \tensor{N}_{n^+} \cdot \vec{u}_{n^+}(\tpdt) + \tensor{N}_{n^-} \cdot \vec{u}_{n^-}(\tpdt)
-    \right)
+  \int_{S_f} \tensor{N}_p^T \cdot
+  \left( \tensor{N}_{n^+} \cdot \vec{u}_{n^+}(\tpdt)
+    - \tensor{N}_{n^-} \cdot \vec{u}_{n^-}(\tpdt)
+    - \tensor{N}_p \cdot \vec{d}_p(\tpdt) \right)
   \, dS = \vec{0}.
 \end{gather}
 In order to march forward in time, we simply increment time, solve the
@@ -384,9 +386,9 @@ and~(\ref{eqn:quasi-static:residual:faul
       &+ \int_{V} \tensor{N}_m^T \cdot \vec{f}(\tpdt) \, dV
     \end{aligned}
     \\
-    \int_{S_f} \tensor{N}_p^T 
-    \left( \tensor{R} \cdot \vec{d}(\tpdt)
-      - \tensor{N}_{n^+} \cdot \vec{u}_{n^+}(\tpdt) + \tensor{N}_{n^-} \cdot \vec{u}_{n^-}(\tpdt)
+    \int_{S_f} \tensor{N}_p^T \cdot
+    \left( \tensor{N}_{n^+} \cdot \vec{u}_{n^+}(\tpdt) -
+      \tensor{N}_{n^-} \cdot \vec{u}_{n^-}(\tpdt) - \vec{d}(\tpdt)
     \right)
     \, dS
   \end{array} \right).
@@ -452,10 +454,11 @@ upper portion now includes the inertial 
           \frac{\partial^2 \vec{u}_n}{\partial t^2} \, dV
         \end{aligned}
     \\
-    \int_{S_f} \tensor{N}_p^T 
-    \left( \tensor{R} \cdot \vec{d}(\tpdt)
-      - \tensor{N}_{n^+} \cdot \vec{u}_{n^+}(\tpdt) + \tensor{N}_{n^-} \cdot \vec{u}_{n^-}(\tpdt)
-    \right)
+    \int_{S_f} \tensor{N}_p^T \cdot
+    \left( \tensor{N}_{n^+} \cdot \vec{u}_{n^+}(\tpdt)
+      - \tensor{N}_{n^-} \cdot \vec{u}_{n^-}(\tpdt)
+      - \tensor{N}_p \cdot \vec{d}_p (\tpdt)
+   \right)
     \, dS
   \end{array} \right).
 \end{equation}
@@ -586,8 +589,7 @@ corresponding to a perturbation in the L
   - \tensor{L}_p^T \cdot \partial \vec{l}_p, \\
   \tensor{K}_{n^-n^-} \cdot \partial \vec{u}_{n^-} =
   \tensor{L}_p^T \cdot \partial \vec{l}_p, \\
-  \partial \vec{d} = \tensor{R}^{-1} \cdot 
-  \left( \partial \vec{u}_{n^+} - \partial \vec{u}_{n^-} \right).
+  \partial \vec{d}_p =  \partial \vec{u}_{n^+} - \partial \vec{u}_{n^-}.
 \end{gather}
 This estimate is exact if all other degrees of freedom are constrained
 (which is generally only the case for simple toy problems), quite good
@@ -639,7 +641,7 @@ In our domain decomposition approach, th
 In our domain decomposition approach, the finite-element mesh includes
 the fault as an interior surface. This forces alignment of the element
 faces along the fault. In order to impose a given fault slip, as in
-equation~(\ref{eq:faultDisp}), we must represent the displacement on
+equation~(\ref{eqn:fault:disp}), we must represent the displacement on
 both sides of the fault for any vertex on the fault. One option is to
 designate ``fault vertices'' which posses twice as many displacement
 degrees of freedom \cite{Aagaard:etal:BSSA:2001}. However, this
@@ -995,8 +997,7 @@ the Lagrange multipliers
 (equation~(\ref{eqn:spontaneous:rupture:slip:update})) are exact and
 simplify to
 \begin{equation}
-  \partial \vec{d} = - \tensor{R}^{-1} \cdot
-  \left( \tensor{K}_{n^+n^+}^{-1} + \tensor{K}_{n^-n^-}^{-1} \right)
+  \partial \vec{d}_p = - \left( \tensor{K}_{n^+n^+}^{-1} + \tensor{K}_{n^-n^-}^{-1} \right)
   \cdot \tensor{L}_p^T \cdot \partial \vec{l}_p.
 \end{equation}