[cig-commits] r18461 - short/3D/PyLith/trunk/doc/userguide/boundaryconditions
brad at geodynamics.org
brad at geodynamics.org
Wed May 25 12:48:02 PDT 2011
Author: brad
Date: 2011-05-25 12:48:01 -0700 (Wed, 25 May 2011)
New Revision: 18461
Modified:
short/3D/PyLith/trunk/doc/userguide/boundaryconditions/boundaryconditions.lyx
Log:
Fixed typo.
Modified: short/3D/PyLith/trunk/doc/userguide/boundaryconditions/boundaryconditions.lyx
===================================================================
--- short/3D/PyLith/trunk/doc/userguide/boundaryconditions/boundaryconditions.lyx 2011-05-25 14:52:04 UTC (rev 18460)
+++ short/3D/PyLith/trunk/doc/userguide/boundaryconditions/boundaryconditions.lyx 2011-05-25 19:48:01 UTC (rev 18461)
@@ -1,21 +1,29 @@
-#LyX 1.6.4 created this file. For more info see http://www.lyx.org/
-\lyxformat 345
+#LyX 2.0 created this file. For more info see http://www.lyx.org/
+\lyxformat 413
\begin_document
\begin_header
\textclass book
\use_default_options false
+\maintain_unincluded_children false
\language english
+\language_package default
\inputencoding auto
+\fontencoding global
\font_roman default
\font_sans default
\font_typewriter default
\font_default_family default
+\use_non_tex_fonts false
\font_sc false
\font_osf false
\font_sf_scale 100
\font_tt_scale 100
\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
\paperfontsize default
\spacing single
\use_hyperref false
@@ -23,9 +31,18 @@
\use_geometry true
\use_amsmath 1
\use_esint 0
+\use_mhchem 1
+\use_mathdots 1
\cite_engine basic
\use_bibtopic false
+\use_indices false
\paperorientation portrait
+\suppress_date false
+\use_refstyle 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
\leftmargin 1in
\topmargin 1in
\rightmargin 1in
@@ -33,15 +50,16 @@
\secnumdepth 3
\tocdepth 3
\paragraph_separation indent
-\defskip medskip
+\paragraph_indentation default
\quotes_language english
\papercolumns 1
\papersides 1
\paperpagestyle default
\tracking_changes false
\output_changes false
-\author ""
-\author ""
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
\end_header
\begin_body
@@ -188,8 +206,10 @@
\begin_layout Standard
Several boundary conditions use a common formulation for the spatial and
temporal variation of the boundary condition parameters,
-\begin_inset Formula \[
-f(\vec{x})=f_{0}(\vec{x})+\dot{f}_{0}(\vec{x})(t-t_{0}(\vec{x}))+f_{1}(\vec{x})a(t-t_{1}(\vec{x})),\]
+\begin_inset Formula
+\[
+f(\vec{x})=f_{0}(\vec{x})+\dot{f}_{0}(\vec{x})(t-t_{0}(\vec{x}))+f_{1}(\vec{x})a(t-t_{1}(\vec{x})),
+\]
\end_inset
@@ -415,7 +435,7 @@
\begin_inset Tabular
<lyxtabular version="3" rows="6" columns="3">
-<features>
+<features tabularvalignment="middle">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="3in">
@@ -673,7 +693,7 @@
\begin_inset Tabular
<lyxtabular version="3" rows="4" columns="2">
-<features>
+<features tabularvalignment="middle">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="4in">
<row>
@@ -972,7 +992,7 @@
\begin_inset Tabular
<lyxtabular version="3" rows="2" columns="3">
-<features>
+<features tabularvalignment="middle">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
@@ -1153,7 +1173,7 @@
\begin_inset Tabular
<lyxtabular version="3" rows="4" columns="2">
-<features>
+<features tabularvalignment="middle">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="4in">
<row>
@@ -1423,7 +1443,7 @@
\begin_inset Tabular
<lyxtabular version="3" rows="4" columns="2">
-<features>
+<features tabularvalignment="middle">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="4in">
<row>
@@ -1602,8 +1622,10 @@
.
We can write the displacement field as
-\begin_inset Formula \begin{equation}
-\vec{u}(\vec{x},t)=\vec{u^{t}}(t-\frac{\vec{x}}{c}),\end{equation}
+\begin_inset Formula
+\begin{equation}
+\vec{u}(\vec{x},t)=\vec{u^{t}}(t-\frac{\vec{x}}{c}),
+\end{equation}
\end_inset
@@ -1644,10 +1666,12 @@
is the vertical direction tangent to the boundary, the tractions on the
boundary are
-\begin_inset Formula \begin{gather}
+\begin_inset Formula
+\begin{gather}
T_{s_{h}}=\sigma_{s_{h}n}\\
T_{s_{v}}=\sigma_{s_{v}n}\\
-T_{n}=\sigma_{nn}.\end{gather}
+T_{n}=\sigma_{nn}.
+\end{gather}
\end_inset
@@ -1658,10 +1682,12 @@
\end_inset
and we can write the tractions as
-\begin_inset Formula \begin{gather}
+\begin_inset Formula
+\begin{gather}
T_{s_{h}}=2\mu\epsilon_{s_{h}n}\\
T_{s_{v}}=2\epsilon_{s_{v}n}\\
-T_{n}=(\lambda+2\mu)\epsilon_{nn}+\lambda(\epsilon_{s_{h}s_{h}}+\epsilon_{s_{v}s_{v}}).\end{gather}
+T_{n}=(\lambda+2\mu)\epsilon_{nn}+\lambda(\epsilon_{s_{h}s_{h}}+\epsilon_{s_{v}s_{v}}).
+\end{gather}
\end_inset
@@ -1670,23 +1696,29 @@
\end_inset
and we have
-\begin_inset Formula \begin{gather}
+\begin_inset Formula
+\begin{gather}
\epsilon_{s_{h}n}=\frac{1}{2}(u_{s_{h},n}+u_{n,s_{h}})\\
\epsilon_{s_{v}n}=\frac{1}{2}(u_{s_{v},n}+u_{n,s_{v}})\\
-\epsilon_{nn}=u_{n,n}.\end{gather}
+\epsilon_{nn}=u_{n,n}.
+\end{gather}
\end_inset
For our propagating plane wave, we recognize that
-\begin_inset Formula \begin{equation}
-\frac{\partial\vec{u^{t}}(t-\frac{\vec{x}}{c})}{\partial x_{i}}=-\frac{1}{c}\frac{\partial\vec{u^{t}}(t-\frac{\vec{x}}{c})}{\partial t},\end{equation}
+\begin_inset Formula
+\begin{equation}
+\frac{\partial\vec{u^{t}}(t-\frac{\vec{x}}{c})}{\partial x_{i}}=-\frac{1}{c}\frac{\partial\vec{u^{t}}(t-\frac{\vec{x}}{c})}{\partial t},
+\end{equation}
\end_inset
so that our expressions for the tractions become
-\begin_inset Formula \begin{gather}
+\begin_inset Formula
+\begin{gather}
T_{s_{h}}=-\frac{\mu}{c}\left(\frac{\partial u_{s_{h}}^{t}(t-\frac{\vec{x}}{c})}{\partial t}+\frac{\partial u_{n}^{t}(t-\frac{\vec{x}}{c})}{\partial t}\right),\\
-T_{s_{v}}=-\frac{\mu}{c}\left(\frac{\partial u_{s_{v}}^{t}(t-\frac{\vec{x}}{c})}{\partial t}+\frac{\partial u_{n}^{t}(t-\frac{\vec{x}}{c})}{\partial t}\right).\end{gather}
+T_{s_{v}}=-\frac{\mu}{c}\left(\frac{\partial u_{s_{v}}^{t}(t-\frac{\vec{x}}{c})}{\partial t}+\frac{\partial u_{n}^{t}(t-\frac{\vec{x}}{c})}{\partial t}\right).
+\end{gather}
\end_inset
@@ -1732,16 +1764,20 @@
.
This leads to the following expressions for the tractions:
-\begin_inset Formula \begin{gather}
+\begin_inset Formula
+\begin{gather}
T_{s_{h}}=-\rho v_{s}\frac{\partial u_{s_{h}}^{t}(t-\frac{\vec{x}}{c})}{\partial t}\\
T_{s_{v}}=-\rho v_{s}\frac{\partial u_{v}^{t}(t-\frac{\vec{x}}{c})}{\partial t}\\
-T_{n}=-\rho v_{p}\frac{\partial u_{n}^{t}(t-\frac{\vec{x}}{c})}{\partial t}\end{gather}
+T_{n}=-\rho v_{p}\frac{\partial u_{n}^{t}(t-\frac{\vec{x}}{c})}{\partial t}
+\end{gather}
\end_inset
We write the weak form of the boundary condition as
-\begin_inset Formula \[
-\int_{S_{T}}T_{i}\phi_{i}\, dS=\int_{S_{T}}-\rho c_{i}\frac{\partial u_{i}}{\partial t}\phi_{i}\, dS,\]
+\begin_inset Formula
+\[
+\int_{S_{T}}T_{i}\phi_{i}\, dS=\int_{S_{T}}-\rho c_{i}\frac{\partial u_{i}}{\partial t}\phi_{i}\, dS,
+\]
\end_inset
@@ -1764,9 +1800,11 @@
is our weighting function.
We express the trial solution and weighting function as linear combinations
of basis functions,
-\begin_inset Formula \begin{gather}
+\begin_inset Formula
+\begin{gather}
u_{i}=\sum_{m}a_{i}^{m}N^{m},\\
-\phi_{i}=\sum_{n}c_{i}^{n}N^{n}.\end{gather}
+\phi_{i}=\sum_{n}c_{i}^{n}N^{n}.
+\end{gather}
\end_inset
@@ -1774,8 +1812,10 @@
\end_layout
\begin_layout Standard
-\begin_inset Formula \[
-\int_{S_{T}}T_{i}\phi_{i}\, dS=\int_{S_{T}}-\rho c_{i}\sum_{m}\dot{a}_{i}^{m}N^{m}\sum_{n}c_{i}^{n}N^{n}\, dS.\]
+\begin_inset Formula
+\[
+\int_{S_{T}}T_{i}\phi_{i}\, dS=\int_{S_{T}}-\rho c_{i}\sum_{m}\dot{a}_{i}^{m}N^{m}\sum_{n}c_{i}^{n}N^{n}\, dS.
+\]
\end_inset
@@ -1789,8 +1829,10 @@
\end_layout
\begin_layout Standard
-\begin_inset Formula \[
-r_{i}^{n}=\sum_{\text{tract cells}}\sum_{\text{quad pts}}-\rho(x_{q})c_{i}(x_{q})\sum_{m}\dot{a}_{i}^{m}N^{m}(x_{q})N^{n}(x_{q})w_{q}|J_{cell}(x_{q})|,\]
+\begin_inset Formula
+\[
+r_{i}^{n}=\sum_{\text{tract cells}}\sum_{\text{quad pts}}-\rho(x_{q})c_{i}(x_{q})\sum_{m}\dot{a}_{i}^{m}N^{m}(x_{q})N^{n}(x_{q})w_{q}|J_{cell}(x_{q})|,
+\]
\end_inset
@@ -1818,8 +1860,10 @@
\end_layout
\begin_layout Standard
-\begin_inset Formula \[
-\dot{u}_{i}(t)=\frac{1}{2\Delta t}(u_{i}(t+\Delta t)-u_{i}(t-\Delta t)).\]
+\begin_inset Formula
+\[
+\dot{u}_{i}(t)=\frac{1}{2\Delta t}(u_{i}(t+\Delta t)-u_{i}(t-\Delta t)).
+\]
\end_inset
@@ -1847,8 +1891,10 @@
\end_layout
\begin_layout Standard
-\begin_inset Formula \[
-\dot{u}_{i}(t)=\frac{1}{2\Delta t}(du_{i}(t)+u_{i}(t)-u_{i}(t-\Delta t))\]
+\begin_inset Formula
+\[
+\dot{u}_{i}(t)=\frac{1}{2\Delta t}(du_{i}(t)+u_{i}(t)-u_{i}(t-\Delta t))
+\]
\end_inset
@@ -1859,8 +1905,10 @@
\end_layout
\begin_layout Standard
-\begin_inset Formula \[
-A_{ij}^{nm}=\sum_{\text{tract cells}}\sum_{\text{quad pts}}\delta_{ij}\frac{1}{2\Delta t}\rho(x_{q})v_{i}(x_{q})N^{m}(x_{q})N^{n}(x_{q})w_{q}|J_{cells}(x_{q})|,\]
+\begin_inset Formula
+\[
+A_{ij}^{nm}=\sum_{\text{tract cells}}\sum_{\text{quad pts}}\delta_{ij}\frac{1}{2\Delta t}\rho(x_{q})v_{i}(x_{q})N^{m}(x_{q})N^{n}(x_{q})w_{q}|J_{cells}(x_{q})|,
+\]
\end_inset
@@ -2296,20 +2344,27 @@
to impose the relative motions, so they are related to the change in stress
on the fault surface associated with fault slip.
If we write the algebraic system of equations we are solving in the form
-\begin_inset Formula \begin{equation}
-\underline{A}\overrightarrow{u}=\overrightarrow{b}\,,\end{equation}
+\begin_inset Formula
+\begin{equation}
+\underline{A}\overrightarrow{u}=\overrightarrow{b}\,,
+\end{equation}
\end_inset
then including the Lagrange multiplier constraints results in
-\begin_inset Formula \begin{equation}
+\begin_inset Formula
+\begin{equation}
\left[\begin{array}{cc}
\underline{A} & \underline{C}^{T}\\
-\underline{C} & 0\end{array}\right]\left[\begin{array}{c}
+\underline{C} & 0
+\end{array}\right]\left[\begin{array}{c}
\overrightarrow{u}\\
-\overrightarrow{l}\end{array}\right]=\left[\begin{array}{c}
+\overrightarrow{l}
+\end{array}\right]=\left[\begin{array}{c}
\overrightarrow{b}\\
-\overrightarrow{d}\end{array}\right]\,,\label{eq:fault:cohesive:lagrange}\end{equation}
+\overrightarrow{d}
+\end{array}\right]\,,\label{eq:fault:cohesive:lagrange}
+\end{equation}
\end_inset
@@ -2348,9 +2403,11 @@
Because we enforce the constraints in a strong sense, the terms do not
involve integrals over the fault surface.
The additional terms in the residual are
-\begin_inset Formula \begin{gather}
+\begin_inset Formula
+\begin{gather}
r_{i}^{n}=-C_{ji}^{pn}l_{j}^{p},\\
-r_{i}^{p}=d_{i}^{p}-C_{ij}^{pn}u_{j}^{n},\end{gather}
+r_{i}^{p}=d_{i}^{p}-C_{ij}^{pn}u_{j}^{n},
+\end{gather}
\end_inset
@@ -2359,15 +2416,17 @@
\end_inset
denotes a conventional degree of freedom and
-\begin_inset Formula $k$
+\begin_inset Formula $p$
\end_inset
denotes a degree of freedom associated with a Lagrange multiplier.
The additional terms in the system Jacobian matrix are simply the direction
cosines,
-\begin_inset Formula \begin{gather}
+\begin_inset Formula
+\begin{gather}
J_{ij}^{np}=C_{ji}^{pn},\\
-J_{ij}^{pn}=C_{ij}^{pn}.\end{gather}
+J_{ij}^{pn}=C_{ij}^{pn}.
+\end{gather}
\end_inset
@@ -2390,8 +2449,10 @@
\end_inset
, we have
-\begin_inset Formula \begin{equation}
-r_{i}^{n}(t+\Delta t)=A_{ij}^{nm}(u_{j}^{m}(t)+du_{j}^{m}(t))+C_{ki}^{pn}(l_{k}^{p}(t)+dl_{k}^{p}(t)),\end{equation}
+\begin_inset Formula
+\begin{equation}
+r_{i}^{n}(t+\Delta t)=A_{ij}^{nm}(u_{j}^{m}(t)+du_{j}^{m}(t))+C_{ki}^{pn}(l_{k}^{p}(t)+dl_{k}^{p}(t)),
+\end{equation}
\end_inset
@@ -2433,9 +2494,11 @@
, and by setting the residual with all terms included to zero; thus, we
have
-\begin_inset Formula \begin{gather}
+\begin_inset Formula
+\begin{gather}
A_{ij}^{nm}(u_{j}^{m}(t)+du_{j}^{m}(t))+C_{ki}^{pn}(l_{k}^{p}(t)+dl_{k}^{p}(t))=0\text{ subject to}\\
-C_{ij}^{pn}(u_{j}^{n}(t)+du_{j}^{n}(t))=d_{i}^{p}.\end{gather}
+C_{ij}^{pn}(u_{j}^{n}(t)+du_{j}^{n}(t))=d_{i}^{p}.
+\end{gather}
\end_inset
@@ -2448,18 +2511,22 @@
\end_inset
,
-\begin_inset Formula \begin{gather}
+\begin_inset Formula
+\begin{gather}
r_{i}^{n}+A_{ij}^{nm}du_{j}^{m}+C_{ki}^{pn}dl_{k}^{p}=0\text{ subject to}\\
-C_{ij}^{pn}(u_{j}^{n}(t)+du_{j}^{n}(t))=d_{i}^{p}.\end{gather}
+C_{ij}^{pn}(u_{j}^{n}(t)+du_{j}^{n}(t))=d_{i}^{p}.
+\end{gather}
\end_inset
Explicitly writing the equations for the vertices on the negative and positive
sides of the fault yields
-\begin_inset Formula \begin{gather}
+\begin_inset Formula
+\begin{gather}
r_{i}^{n-}+A_{ij}^{nm-}du_{j}^{m-}+R_{ki}^{pn}dl_{k}^{p}=0,\\
r_{i}^{n+}+A_{ij}^{nm+}du_{j}^{m+}+R_{ki}^{pn}dl_{k}^{p}=0,\\
-R_{ij}^{pn}(u_{j}^{n+}+du_{j}^{n+}-u_{j}^{n-}-du_{j}^{n-})=d_{i}^{p}.\end{gather}
+R_{ij}^{pn}(u_{j}^{n+}+du_{j}^{n+}-u_{j}^{n-}-du_{j}^{n-})=d_{i}^{p}.
+\end{gather}
\end_inset
@@ -2472,9 +2539,11 @@
\end_inset
and combining them using the third equation leads to
-\begin_inset Formula \begin{multline}
+\begin_inset Formula
+\begin{multline}
R_{ij}^{pn}\left((A_{ij}^{nm+})^{-1}+(A_{ij}^{nm+})^{-1}\right)R_{ki}^{pn}dl_{k}^{p}=d_{i}^{p}-R_{ij}^{pn}(u_{j}^{n+}-u_{j}^{n-})\\
-+R_{ij}^{pn}\left((A_{ij}^{nm+})^{-1}r_{i}^{n+}-(A_{ij}^{nm-})^{-1}r_{i}^{n-}\right).\end{multline}
++R_{ij}^{pn}\left((A_{ij}^{nm+})^{-1}r_{i}^{n+}-(A_{ij}^{nm-})^{-1}r_{i}^{n-}\right).
+\end{multline}
\end_inset
@@ -2485,14 +2554,18 @@
is the same for all components at a vertex.
As a result the matrix on the left hand side simplifies to
-\begin_inset Formula \[
-S_{ik}^{pn}=\delta_{ik}\left(\frac{1}{A^{nm+}}+\frac{1}{A^{nm-}}\right),\]
+\begin_inset Formula
+\[
+S_{ik}^{pn}=\delta_{ik}\left(\frac{1}{A^{nm+}}+\frac{1}{A^{nm-}}\right),
+\]
\end_inset
and
-\begin_inset Formula \[
-dl_{k}^{p}=(S_{ik}^{pn})^{-1}\left(d_{i}^{p}-R_{ij}^{pn}(u_{j}^{n+}-u_{j}^{n-})+R_{ij}^{pn}\left((A_{ij}^{nm+})^{-1}r_{i}^{n+}-(A_{ij}^{nm-})^{-1}r_{i}^{n-}\right)\right).\]
+\begin_inset Formula
+\[
+dl_{k}^{p}=(S_{ik}^{pn})^{-1}\left(d_{i}^{p}-R_{ij}^{pn}(u_{j}^{n+}-u_{j}^{n-})+R_{ij}^{pn}\left((A_{ij}^{nm+})^{-1}r_{i}^{n+}-(A_{ij}^{nm-})^{-1}r_{i}^{n-}\right)\right).
+\]
\end_inset
@@ -2514,9 +2587,11 @@
\end_inset
using
-\begin_inset Formula \begin{gather*}
+\begin_inset Formula
+\begin{gather*}
\Delta du_{j}^{n-}=(A_{ij}^{nm-})^{-1}C_{ki}^{pn}dl_{k}^{p}\text{ and}\\
-\Delta du_{j}^{n+}=-(A_{ij}^{nm+})^{-1}C_{ki}^{pn}dl_{k}^{p}.\end{gather*}
+\Delta du_{j}^{n+}=-(A_{ij}^{nm+})^{-1}C_{ki}^{pn}dl_{k}^{p}.
+\end{gather*}
\end_inset
@@ -2616,7 +2691,7 @@
\begin_inset Tabular
<lyxtabular version="3" rows="8" columns="3">
-<features>
+<features tabularvalignment="middle">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="3.5in">
@@ -2979,10 +3054,13 @@
\end_layout
\begin_layout Standard
-\begin_inset Formula \begin{gather}
+\begin_inset Formula
+\begin{gather}
D(t)=\left\{ \begin{array}{cc}
0 & 0\leq t<t_{r}\\
-D_{final} & t\ge t_{r}\end{array}\right.\,,\end{gather}
+D_{final} & t\ge t_{r}
+\end{array}\right.\,,
+\end{gather}
\end_inset
@@ -3098,7 +3176,7 @@
\begin_inset Tabular
<lyxtabular version="3" rows="5" columns="3">
-<features>
+<features tabularvalignment="middle">
<column alignment="left" valignment="top" width="0">
<column alignment="left" valignment="top" width="0">
<column alignment="left" valignment="top" width="2.5in">
@@ -3296,10 +3374,13 @@
\end_layout
\begin_layout Standard
-\begin_inset Formula \begin{gather}
+\begin_inset Formula
+\begin{gather}
D(t)=\left\{ \begin{array}{cc}
0 & 0\leq t<t_{r}\\
-V(t-t_{r}) & t\ge t_{r}\end{array}\right.\,,\end{gather}
+V(t-t_{r}) & t\ge t_{r}
+\end{array}\right.\,,
+\end{gather}
\end_inset
@@ -3416,7 +3497,7 @@
\begin_inset Tabular
<lyxtabular version="3" rows="5" columns="3">
-<features>
+<features tabularvalignment="middle">
<column alignment="left" valignment="top" width="0">
<column alignment="left" valignment="top" width="0">
<column alignment="left" valignment="top" width="2.5in">
@@ -3620,11 +3701,14 @@
\end_layout
\begin_layout Standard
-\begin_inset Formula \begin{gather}
+\begin_inset Formula
+\begin{gather}
D(t)=\left\{ \begin{array}{cc}
0 & 0\leq t<t_{r}\\
-D_{final}\left(1-exp\left(-\frac{t-t_{r}}{t_{0}}\right)\left(1+\frac{t-t_{r}}{t_{0}}\right)\right) & t\ge t_{r}\end{array}\right.\,,\\
-t_{0}=0.6195t_{\mathit{rise}}\,,\end{gather}
+D_{final}\left(1-exp\left(-\frac{t-t_{r}}{t_{0}}\right)\left(1+\frac{t-t_{r}}{t_{0}}\right)\right) & t\ge t_{r}
+\end{array}\right.\,,\\
+t_{0}=0.6195t_{\mathit{rise}}\,,
+\end{gather}
\end_inset
@@ -3754,7 +3838,7 @@
\begin_inset Tabular
<lyxtabular version="3" rows="6" columns="3">
-<features>
+<features tabularvalignment="middle">
<column alignment="left" valignment="top" width="0">
<column alignment="left" valignment="top" width="0">
<column alignment="left" valignment="top" width="2.5in">
@@ -4003,15 +4087,18 @@
\end_layout
\begin_layout Standard
-\begin_inset Formula \begin{gather}
+\begin_inset Formula
+\begin{gather}
D(t)=\left\{ \begin{array}{cc}
D_{\mathit{final}}C_{n}\left(0.7t-0.7\frac{t_{1}}{\pi}\sin\frac{\pi t}{t_{1}}-1.2\frac{t_{1}}{\pi}\left(\cos\frac{\pi t}{2t_{1}}-1\right)\right) & 0\leq t<t_{1}\\
D_{\mathit{final}}C_{n}\left(1.0t-0.7\frac{t1}{\pi}\sin\frac{\pi t}{t_{1}}+0.3\frac{t2}{\pi}\sin\frac{\pi(t-t1)}{t_{2}}+\frac{1.2}{\pi}t_{1}-0.3t_{1}\right) & t_{1}\leq t<2t_{1}\\
-D_{\mathit{final}}C_{n}\left(0.7-0.7\cos\frac{\pi t}{t_{1}}+0.6\sin\frac{\pi t}{2t_{1}}\right) & 2t_{1}\leq t\leq t_{0}\end{array}\right.\,,\\
+D_{\mathit{final}}C_{n}\left(0.7-0.7\cos\frac{\pi t}{t_{1}}+0.6\sin\frac{\pi t}{2t_{1}}\right) & 2t_{1}\leq t\leq t_{0}
+\end{array}\right.\,,\\
C_{n}=\frac{\pi}{1.4\pi t_{1}+1.2t_{1}+0.3\pi t_{2}},\\
t_{0}=1.525t_{\mathit{rise}},\\
t_{1}=0.13t_{0},\\
-t_{2}=t_{0}-t_{1},\end{gather}
+t_{2}=t_{0}-t_{1},
+\end{gather}
\end_inset
@@ -4185,7 +4272,7 @@
\begin_inset Tabular
<lyxtabular version="3" rows="6" columns="3">
-<features>
+<features tabularvalignment="middle">
<column alignment="left" valignment="top" width="0">
<column alignment="left" valignment="top" width="0">
<column alignment="left" valignment="top" width="2.5in">
@@ -4441,14 +4528,19 @@
\begin_layout Standard
The algebraic systems of equations for dynamic earthquake rupture are the
same as those for kinematic rupture
-\begin_inset Formula \begin{equation}
+\begin_inset Formula
+\begin{equation}
\left[\begin{array}{cc}
\underline{A} & \underline{C}^{T}\\
-\underline{C} & 0\end{array}\right]\left[\begin{array}{c}
+\underline{C} & 0
+\end{array}\right]\left[\begin{array}{c}
\overrightarrow{u}\\
-\overrightarrow{l}\end{array}\right]=\left[\begin{array}{c}
+\overrightarrow{l}
+\end{array}\right]=\left[\begin{array}{c}
\overrightarrow{b}\\
-\overrightarrow{d}\end{array}\right].\end{equation}
+\overrightarrow{d}
+\end{array}\right].
+\end{equation}
\end_inset
@@ -4459,14 +4551,18 @@
value computed for the current slip (either zero or the slip at the previous
time step) and the value computed from the fault constitutive model.
Starting from our system of algebraic equations,
-\begin_inset Formula \begin{equation}
-A_{ij}^{nm}u_{j}^{m}+C_{ji}^{pn}l_{j}^{p}=b_{i}^{n},\end{equation}
+\begin_inset Formula
+\begin{equation}
+A_{ij}^{nm}u_{j}^{m}+C_{ji}^{pn}l_{j}^{p}=b_{i}^{n},
+\end{equation}
\end_inset
we compute the sensitivity for the given loading and boundary conditions,
-\begin_inset Formula \begin{equation}
-A_{ij}^{nm}\partial u_{j}^{m}=-C_{ji}^{pn}\partial l_{j}^{p}.\end{equation}
+\begin_inset Formula
+\begin{equation}
+A_{ij}^{nm}\partial u_{j}^{m}=-C_{ji}^{pn}\partial l_{j}^{p}.
+\end{equation}
\end_inset
@@ -4484,11 +4580,14 @@
\begin_layout Standard
In general A is a sparse matrix with off-diagonal terms of the form
-\begin_inset Formula \begin{equation}
+\begin_inset Formula
+\begin{equation}
A=\left(\begin{array}{ccc}
A_{0} & A_{1} & A_{2}\\
A_{3} & A_{n-} & 0\\
-A_{4} & 0 & A_{n+}\end{array}\right),\end{equation}
+A_{4} & 0 & A_{n+}
+\end{array}\right),
+\end{equation}
\end_inset
@@ -4496,9 +4595,11 @@
We formulate two small linear systems involving just the degrees of freedom
associated with vertices on either the positive or negative sides of the
fault,
-\begin_inset Formula \begin{gather}
+\begin_inset Formula
+\begin{gather}
A_{ij}^{nm-}\partial u_{j}^{m-}=-R_{ij}^{pn}\partial l_{j}^{p},\\
-A_{ij}^{nm+}\partial u_{j}^{m+}=R_{ij}^{pn}\partial l_{j}^{p},\end{gather}
+A_{ij}^{nm+}\partial u_{j}^{m+}=R_{ij}^{pn}\partial l_{j}^{p},
+\end{gather}
\end_inset
@@ -4525,8 +4626,10 @@
) sides of the fault.
After solving these two linear systems of equations, we compute the increment
in slip using
-\begin_inset Formula \begin{equation}
-\partial d_{i}^{p}=R_{ij}^{pn}(\partial u_{j}^{n+}-\partial u_{j}^{n-}).\end{equation}
+\begin_inset Formula
+\begin{equation}
+\partial d_{i}^{p}=R_{ij}^{pn}(\partial u_{j}^{n+}-\partial u_{j}^{n-}).
+\end{equation}
\end_inset
@@ -4584,8 +4687,10 @@
\begin_layout Standard
With a lumped Jacobian matrix, we can solve for the increment in slip directly,
-\begin_inset Formula \begin{equation}
-\partial d_{i}^{p}=-C_{ij}^{pn}(A_{jk}^{nm})^{-1}C_{lk}^{pm}\partial l_{l}^{p}.\end{equation}
+\begin_inset Formula
+\begin{equation}
+\partial d_{i}^{p}=-C_{ij}^{pn}(A_{jk}^{nm})^{-1}C_{lk}^{pm}\partial l_{l}^{p}.
+\end{equation}
\end_inset
@@ -4596,8 +4701,10 @@
for a given vertex are identical and the expression on the right-hand side
reduces to
-\begin_inset Formula \begin{equation}
-\partial d_{i}^{p}=-\left(\frac{1}{A^{n+}}+\frac{1}{A^{n-}}\right)\partial l_{i}^{p}.\end{equation}
+\begin_inset Formula
+\begin{equation}
+\partial d_{i}^{p}=-\left(\frac{1}{A^{n+}}+\frac{1}{A^{n-}}\right)\partial l_{i}^{p}.
+\end{equation}
\end_inset
@@ -4700,7 +4807,7 @@
\begin_inset Tabular
<lyxtabular version="3" rows="7" columns="3">
-<features>
+<features tabularvalignment="middle">
<column alignment="left" valignment="top" width="0">
<column alignment="left" valignment="top" width="0">
<column alignment="left" valignment="top" width="2.5in">
@@ -4989,10 +5096,13 @@
\begin_layout Standard
The static friction model produces shear tractions proportional to the fault
normal traction plus a cohesive stress,
-\begin_inset Formula \begin{equation}
+\begin_inset Formula
+\begin{equation}
T_{f}=\begin{cases}
T_{c}-\mu_{f}T_{n} & T_{n}\leq0\\
-0 & T_{n}>0\end{cases}.\end{equation}
+0 & T_{n}>0
+\end{cases}.
+\end{equation}
\end_inset
@@ -5040,7 +5150,7 @@
\begin_inset Tabular
<lyxtabular version="3" rows="3" columns="2">
-<features>
+<features tabularvalignment="middle">
<column alignment="left" valignment="top" width="0">
<column alignment="left" valignment="top" width="2.5in">
<row>
@@ -5140,11 +5250,14 @@
to the cohesive stress plus a contribution proportional to the fault normal
traction that decreases from a static value to a dynamic value as slip
progresses,
-\begin_inset Formula \begin{equation}
+\begin_inset Formula
+\begin{equation}
T_{f}=\begin{cases}
T_{c}-(\mu_{s}-(\mu_{s}-\mu_{d})\frac{d}{d_{0}})T_{n} & d\leq d_{0}\text{ and }T_{n}\leq0\\
T_{c}-\mu_{d}T_{n} & d>d_{0}\text{ and }T_{n}\leq0\\
-0 & T_{n}>0\end{cases}\end{equation}
+0 & T_{n}>0
+\end{cases}
+\end{equation}
\end_inset
@@ -5195,7 +5308,7 @@
\begin_inset Tabular
<lyxtabular version="3" rows="7" columns="3">
-<features>
+<features tabularvalignment="middle">
<column alignment="left" valignment="top" width="0">
<column alignment="left" valignment="top" width="0">
<column alignment="left" valignment="top" width="2.5in">
@@ -5464,12 +5577,15 @@
The Dietrich-Ruina rate and state friction model produces shear tractions
equal to the cohesive stress plus a contribution proportional to the fault
normal traction that depends on a state variable,
-\begin_inset Formula \begin{gather}
+\begin_inset Formula
+\begin{gather}
T_{f}=\begin{cases}
T_{c}-\mu_{f}T_{n} & T_{n}\leq0\\
-0 & T_{n}>0\end{cases}\\
+0 & T_{n}>0
+\end{cases}\\
\mu_{f}=a\sinh^{-1}\left(\frac{1}{2}\frac{V}{V_{0}}\exp\left(\frac{1}{a}\left(\mu_{0}+b\ln\left(\frac{V_{0}\theta}{L}\right)\right)\right)\right)\\
-\frac{d\theta}{dt}=1-\frac{V\theta}{L}\end{gather}
+\frac{d\theta}{dt}=1-\frac{V\theta}{L}
+\end{gather}
\end_inset
@@ -5517,8 +5633,10 @@
, we integrate the evolution equation for the state variable keeping slip
rate constant to get
-\begin_inset Formula \begin{equation}
-\theta(t+\Delta t)=\theta(t)\exp\left(\frac{-V(t)\Delta t}{L}\right)+\frac{L}{V(t)}\left(1-\exp\left(-\frac{V(t)\Delta t}{L}\right)\right).\end{equation}
+\begin_inset Formula
+\begin{equation}
+\theta(t+\Delta t)=\theta(t)\exp\left(\frac{-V(t)\Delta t}{L}\right)+\frac{L}{V(t)}\left(1-\exp\left(-\frac{V(t)\Delta t}{L}\right)\right).
+\end{equation}
\end_inset
@@ -5526,10 +5644,13 @@
1.
Using the first three terms of the Taylor series expansion of the second
exponential yields
-\begin_inset Formula \begin{equation}
+\begin_inset Formula
+\begin{equation}
\theta(t+\Delta t)=\begin{cases}
\theta(t)\exp\left(-\frac{V(t)\Delta t}{L}\right)+\Delta t-\frac{1}{2}\frac{V(t)\Delta t^{2}}{L} & \frac{V(t)\Delta t}{L}<0.00001\\
-\theta(t)\exp\left(-\frac{V(t)\Delta t}{L}\right)+\frac{L}{V(t)}\left(1-\exp\left(-\frac{V(t)\Delta t}{L}\right)\right) & \frac{V(t)\Delta t}{L}\ge0.00001\end{cases}.\end{equation}
+\theta(t)\exp\left(-\frac{V(t)\Delta t}{L}\right)+\frac{L}{V(t)}\left(1-\exp\left(-\frac{V(t)\Delta t}{L}\right)\right) & \frac{V(t)\Delta t}{L}\ge0.00001
+\end{cases}.
+\end{equation}
\end_inset
@@ -5582,7 +5703,7 @@
\begin_inset Tabular
<lyxtabular version="3" rows="8" columns="3">
-<features>
+<features tabularvalignment="middle">
<column alignment="left" valignment="top" width="0">
<column alignment="left" valignment="top" width="0">
<column alignment="left" valignment="top" width="2.5in">
@@ -5910,8 +6031,10 @@
\end_inset
, the body forces contribute to the residual,
-\begin_inset Formula \begin{equation}
-r_{i}^{n}=\int_{V}f_{i}N^{n}\: dV.\end{equation}
+\begin_inset Formula
+\begin{equation}
+r_{i}^{n}=\int_{V}f_{i}N^{n}\: dV.
+\end{equation}
\end_inset
@@ -5932,8 +6055,10 @@
\end_inset
:
-\begin_inset Formula \begin{equation}
-f_{i}=\rho ga_{i}.\end{equation}
+\begin_inset Formula
+\begin{equation}
+f_{i}=\rho ga_{i}.
+\end{equation}
\end_inset
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