[cig-commits] commit: More work on friction notes. Fixed typo in lumped solver notes (inverse of sensitivity matrix).

[Mercurial]

changeset:   36:b3db64b9d137
tag:         tip
user:        Brad Aagaard <baagaard at usgs.gov>
date:        Tue Mar 30 11:36:25 2010 -0700
files:       friction.tex lumpedsolver.tex
description:
More work on friction notes. Fixed typo in lumped solver notes (inverse of sensitivity matrix).


diff -r a6f656b56869 -r b3db64b9d137 friction.tex
--- a/friction.tex	Mon Mar 29 17:35:40 2010 -0700
+++ b/friction.tex	Tue Mar 30 11:36:25 2010 -0700
@@ -44,8 +44,8 @@ the differential displacements across th
 the differential displacements across the fault, $\vec{d}$ is the vector of
 fault slip in the fault coordinate system, and $S_f$ is the fault
 surface. For Lagrange vertex $k$, conventional vertex $i$ on the
-``positive'' side of the fault, and conventional vertex $j$ on the
-``negative'' side of the fault, the terms in the equation are
+``negative'' side of the fault, and conventional vertex $j$ on the
+``positive'' side of the fault, the terms in the equation are
 \begin{equation}
   \left( \begin{array}{ccccccccc}
     \ldots & -C_{px} & -C_{py} & -C_{pz} &  &
@@ -69,7 +69,10 @@ surface. For Lagrange vertex $k$, conven
    \vdots
     \end{array} \right).
 \end{equation}
-
+The fault coordinate system is $(p,q,r)$ and the global coordinate
+system is $(x,y,z)$. In the fault coordinate system $p$ is the
+horizontal along-strike direction, $q$ is the up-dip direction, and
+$r$ is the fault normal direction.
 
 We start with the functional for the total potential energy, while
 imposing the constraints associated with fault slip using Lagrange
@@ -221,20 +224,19 @@ and the residual,
 \fbox{Forces associated with Lagrange multiplier DOF.}
 
 The fault constitutive model places limits on the Lagrange
-multipliers. When the friction is high enough that there is no sliding
-on the fault, the fault is locked (slip is zero) and the Lagrange
-multipliers assume their usual values. The Lagrange multipliers
-correspond to the forces necessary to keep the slip zero. When the
-driving forces exceed those allowed by friction, the Lagrange
-multiplier values are set to the forces corresponding to the friction
-from the fault constitutive model. When we place limits on the
-Lagrange multipliers, we must adjust the slip accordingly to maintain
-the consistency in the system of equations. We can estimate the slip
-associated with placing limits on the Lagrange multipliers by
-computing the sensitivity of the slip to the Lagrange
-multipliers. This permits computing the increment in slip based on the
-difference between the Lagrange multiplier allowed by the fault
-constitutive model and the one needed to keep the fault locked.
+multipliers. When friction is large enough such that there is no
+sliding on the fault, the fault is locked (slip is zero) and the
+Lagrange multipliers assume their usual values. In this case the
+Lagrange multipliers correspond to the forces necessary to keep the
+slip zero. When the driving forces exceed those allowed by friction,
+we reduce the Lagrange multiplier values to those consistent with
+friction from the fault constitutive model. When we reduce the
+Lagrange multipliers, we must increment the slip accordingly to
+maintain the consistency in the system of equations. We can estimate
+the increment in slip by computing the sensitivity of the slip to the
+Lagrange multipliers. This permits computing the increment in slip
+based on the difference between the Lagrange multiplier allowed by the
+fault constitutive model and the one needed to keep the fault locked.
 
 We compute the sensitivity starting with equation ??,
 \begin{gather}
@@ -285,8 +287,8 @@ corresponding to the vertices on the ``p
 corresponding to the vertices on the ``positive'' and ``negative''
 sides of the fault. This means the structure (i.e., nonzero pattern)
 of the matrices is defined by the fault mesh, and we can use a single
-sparse matrix and extract the values from $A$ for either $A_i$ or
-$A_j$ as needed.
+sparse matrix in solving the two linear problems and extract the
+values from $A$ for either $A_i$ or $A_j$ as needed.
 
 \subsubsection{Diagonal $A$}
 
@@ -301,8 +303,8 @@ Premultiplying by $\matrix{C}$ leads to
 \end{equation}
 
 For Lagrange vertex $k$, conventional vertex $i$ on the
-``negative side of the fault, and conventional vertex $j$ on the
-``positive side of the fault, we have
+``negative'' side of the fault, and conventional vertex $j$ on the
+``positive'' side of the fault, we have
 \begin{equation}
   \left( \begin{array}{c} 
       \partial d_{kp} \\ \partial d_{kq} \\ \partial d_{kr} 
@@ -344,6 +346,25 @@ For Lagrange vertex $k$, conventional ve
       C_{rz}^2\left(\frac{1}{A_{iz}}+\frac{1}{A_{jz}}\right).
 \end{align}
 
+In the special case where A contains only intertial terms (e.g., there
+is no overlap between fault and absorbing dampers boundary
+conditions),
+\begin{gather}
+  A_{ix} = A_{iy} = A_{iz} \\
+  A_{jx} = A_{jy} = A_{jz},
+\end{gather}
+so that the sensitivity matrix, S, is diagonal with equal terms.
+\begin{gather}
+  S = \left( \begin{array}{ccc}
+      S_{pp} & 0 & 0 \\
+      0 & S_{qq} & 0 \\
+      0 & 0 & S_{rr} 
+    \end{array} \right), \\
+  S_{pp} = S_{qq} = S_{rr} = \frac{1}{A_{ix}}+\frac{1}{A_{jx}}.
+\end{gather}
+
+
+
 \fbox{Tractions associated with Lagrange multiplier DOF.}
 
 The main disadvantage of using tractions rather than forces for the
diff -r a6f656b56869 -r b3db64b9d137 lumpedsolver.tex
--- a/lumpedsolver.tex	Mon Mar 29 17:35:40 2010 -0700
+++ b/lumpedsolver.tex	Tue Mar 30 11:36:25 2010 -0700
@@ -191,7 +191,12 @@ so that the sensitivity matrix, S, is di
       0 & S_{qq} & 0 \\
       0 & 0 & S_{rr} 
     \end{array} \right), \\
-  S_{pp} = S_{qq} = S_{rr} = \frac{A_{ix} A_{jx}}{A_{ix} + A_{jx}}
+  S^{-1} = \left( \begin{array}{ccc}
+        S_{inv} & 0 & 0 \\
+        0 & S_{inv} & 0 \\
+        0 & 0 & S_{inv} 
+      \end{array} \right), \\
+    S_{inv} = \frac{A_{ix} A_{jx}}{A_{ix} + A_{jx}}
 \end{gather}
 
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