[cig-commits] commit: Added commands to get line numbers to come out correct for equations.
Mercurial
hg at geodynamics.org
Thu Sep 13 14:12:46 PDT 2012
changeset: 153:6088bf21a991
tag: tip
user: Brad Aagaard <baagaard at usgs.gov>
date: Thu Sep 13 14:12:40 2012 -0700
files: faultRup.tex
description:
Added commands to get line numbers to come out correct for equations.
diff -r c86f883e07c3 -r 6088bf21a991 faultRup.tex
--- a/faultRup.tex Thu Sep 13 12:28:12 2012 -0700
+++ b/faultRup.tex Thu Sep 13 14:12:40 2012 -0700
@@ -14,15 +14,20 @@
%\setkeys{Gin}{draft=false}
%\linenumbers*[1]
+% :SUBMIT:
+% Extract figures and captions from PDF file.
+% gs -dBATCH -dNOPAUSE -q -sDEVICE=pdfwrite -dFirstPage=50 -dLastPage=64 -sOUTPUTFILE=figures.pdf faultRup.pdf
+
% ======================================================================
% TEMPORARY (must be removed for submission)
% ======================================================================
% :SUBMIT: comment out these commands (\newcommand not allowed)
-\usepackage{color}
-\newcommand\brad[1]{{\color{red}\bf [BRAD: #1]}}
-\newcommand\matt[1]{{\color{blue}\bf [MATT: #1]}}
-\newcommand\charles[1]{{\color{green}\bf [CHARLES: #1]}}
+%\usepackage{color}
+%\newcommand\brad[1]{{\color{red}\bf [BRAD: #1]}}
+%\newcommand\matt[1]{{\color{blue}\bf [MATT: #1]}}
+%\newcommand\charles[1]{{\color{green}\bf [CHARLES: #1]}}
+
% ======================================================================
% PREAMBLE
@@ -215,14 +220,14 @@ a step-by-step description of the formul
a step-by-step description of the formulation.
We solve the elasticity equation including inertial terms,
-\begin{gather}
+\begin{linenomath*}\begin{gather}
\rho \frac{\partial^2\bm{u}}{\partial t^2} - \bm{f}
- \bm{\nabla} \cdot \bm{\sigma} = \bm{0} \text{ in }V, \\
\bm{\sigma} \cdot \bm{n} = \bm{T} \text{ on }S_T, \\
\bm{u} = \bm{u}_0 \text{ on }S_u, \\
\bm{d} - (\bm{u}_{+} - \bm{u}_{-}) = \bm{0}
\text{ on }S_f, \label{eqn:fault:disp}
-\end{gather}
+\end{gather}\end{linenomath*}
where $\bm{u}$ is the displacement vector, $\rho$ is the mass
density, $\bm{f}$ is the body force vector, $\bm{\sigma}$ is the
Cauchy stress tensor, and $t$ is time. We specify tractions $\bm{T}$
@@ -237,16 +242,16 @@ fault surface for a moment), we construc
fault surface for a moment), we construct the weak form by taking the
dot product of the governing equation with a weighting function and
setting the integral over the domain equal to zero,
-\begin{equation}
+\begin{linenomath*}\begin{equation}
\int_{V} \bm{\phi} \cdot
\left( \bm{\nabla} \cdot \bm{\sigma} + \bm{f} -
\rho\frac{\partial^{2}\bm{u}}{\partial t^{2}} \right)
\, dV=0.
-\end{equation}
+\end{equation}\end{linenomath*}
The weighting function $\bm{\phi}$ is a piecewise differentiable vector
field with $\bm{\phi} = \bm{0}$ on $S_u$. After some algebra and
use of the boundary conditions, we have
-\begin{equation}
+\begin{linenomath*}\begin{equation}
\begin{split}
- \int_{V} \nabla \bm{\phi} : \bm{\sigma} \, dV
+ \int_{S_T} \bm{\phi} \cdot \bm{T} \, dS
@@ -254,7 +259,7 @@ use of the boundary conditions, we have
- \int_{V} \bm{\phi} \cdot \rho \frac{\partial^{2}\bm{u}}{\partial t^{2}} \, dV
=0.
\end{split}
-\end{equation}
+\end{equation}\end{linenomath*}
Using a domain decomposition approach, we consider the fault surface
as an interior boundary between two domains as shown in
@@ -272,7 +277,7 @@ Recognizing that the tractions on the fa
Recognizing that the tractions on the fault surface are analogous to
the boundary tractions, we add in the contributions from integrating
the Lagrange multipliers (fault tractions) over the fault surface,
-\begin{equation}
+\begin{linenomath*}\begin{equation}
\begin{split}
- \int_{V} \nabla\bm{\phi} : \bm{\sigma} \, dV
+ \int_{S_T} \bm{\phi} \cdot \bm{T} \, dS
@@ -282,27 +287,27 @@ the Lagrange multipliers (fault traction
- \int_{V} \bm{\phi} \cdot \rho \frac{\partial^{2}\bm{u}}{\partial t^{2}} \, dV
=0.
\end{split}
-\end{equation}
+\end{equation}\end{linenomath*}
Our sign convention for the fault normal and fault tractions (tension
is positive) leads to the Lagrange multiplier terms having the
opposite sign as the boundary tractions term. We also construct the
weak form for the constraint associated with slip on the fault by
taking the dot product of the constraint equation with the weighting
function and setting the integral over the fault surface to zero,
-\begin{equation}
+\begin{linenomath*}\begin{equation}
\int_{S_f} \bm{\phi} \cdot
\left(\bm{d} - \bm{u}_{+} + \bm{u}_{-} \right) \, dS = 0.
-\end{equation}
+\end{equation}\end{linenomath*}
We express the weighting function $\bm{\phi}$, trial solution
$\bm{u}$, Lagrange multipliers $\bm{l}$, and fault slip $\bm{d}$ as
linear combinations of basis functions,
-\begin{gather}
+\begin{linenomath*}\begin{gather}
\bm{\phi} = \sum_{m} \bm{a}_m N_m, \\
\bm{u} = \sum_{n} \bm{u}_n N_n, \\
\bm{l} = \sum_{p} \bm{l}_p N_p, \\
\bm{d} = \sum_{p} \bm{d}_p N_p.
-\end{gather}
+\end{gather}\end{linenomath*}
Because the weighting function is zero on $S_u$, the number of basis
functions for the trial solution $\bm{u}$ is generally greater than
the number of basis functions for the weighting function $\bm{\phi}$,
@@ -311,16 +316,16 @@ dimension than the domain, so $p \ll n$
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}
+\begin{linenomath*}\begin{gather}
\bm{\phi} = \bm{N}_m \cdot \bm{a}_m, \\
\bm{u} = \bm{N}_n \cdot \bm{u}_n, \\
\bm{l} = \bm{N}_p \cdot \bm{l}_p, \\
\bm{d} = \bm{N}_p \cdot \bm{d}_p.
-\end{gather}
+\end{gather}\end{linenomath*}
The weighting function is arbitrary, so the integrands must be zero
for all $\bm{a}_m$, which leads to
-\begin{gather}
+\begin{linenomath*}\begin{gather}
\begin{split}
- \int_{V} \nabla \bm{N}_m^T \cdot \bm{\sigma} \, dV
+ \int_{S_T} \bm{N}_m^T \cdot \bm{T} \, dS
@@ -338,7 +343,7 @@ for all $\bm{a}_m$, which leads to
- \bm{N}_{n^+} \cdot \bm{u}_{n^+}
+ \bm{N}_{n^-} \cdot \bm{u}_{n^-}
\right) \, dS = \bm{0}.
-\end{gather}
+\end{gather}\end{linenomath*}
We want to solve these equations for the coefficients $\bm{u}_n$
and $\bm{l}_p$ subject to $\bm{u} = \bm{u}_0 \text{ on
}S_u$. When we prescribe the slip, we specify $\bm{d}$ on $S_f$,
@@ -363,7 +368,7 @@ time-dependence only enters through the
time-dependence only enters through the constitutive models
and the loading conditions. Considering the deformation at time
$t+\Delta t$,
-\begin{gather}
+\begin{linenomath*}\begin{gather}
\label{eqn:quasi-static:residual:elasticity}
\begin{split}
- \int_{V} \nabla \bm{N}_m^T \cdot \bm{\sigma}(t+\Delta t) \, dV
@@ -385,7 +390,7 @@ and the loading conditions. Considering
\bm{N}_{n^-} \cdot \bm{u}_{n^-}(t+\Delta t)
\right) \, dS = \bm{0}.
\end{split}
-\end{gather}
+\end{gather}\end{linenomath*}
In order to march forward in time, we simply increment time, solve the
equations, and add the increment in the solution to the solution from
the previous time step. We solve these equations using the Portable,
@@ -411,10 +416,10 @@ increment in the stress tensor $\bm{d\si
increment in the stress tensor $\bm{d\sigma}(t)$. We approximate the
increment in the stress tensor using linear elasticity and
infinitesimal strains,
-\begin{equation}
+\begin{linenomath*}\begin{equation}
\bm{d\sigma}(t) = \frac{1}{2} \bm{C}(t) \cdot (\nabla + \nabla^T)
\bm{u}(t),
-\end{equation}
+\end{equation}\end{linenomath*}
where $\bm{C}$ is the fourth order tensor of elastic constants. For
bulk constitutive models with a linear response, $\bm{C}$ is constant
in time. For other constitutive models we form $\bm{C}(t)$ from the
@@ -422,36 +427,36 @@ in equation~(\ref{eqn:quasi-static:resid
in equation~(\ref{eqn:quasi-static:residual:elasticity}) and
expressing the displacement vector as a linear combination of basis
functions, we find this portion of the Jacobian is
-\begin{equation}\label{eqn:jacobian:implicit:stiffness}
+\begin{linenomath*}\begin{equation}\label{eqn:jacobian:implicit:stiffness}
\bm{K} = \frac{1}{4} \int_V
(\nabla^T + \nabla) \bm{N}_m^T \cdot
\bm{C} \cdot (\nabla + \nabla^T) \bm{N}_n \, dV.
-\end{equation}
+\end{equation}\end{linenomath*}
This matches the stiffness matrix in conventional solid mechanics
finite-element formulations. Following a similar procedure, we find
the portion of the Jacobian associated with the constraints,
equation~(\ref{eqn:quasi-static:residual:fault}), is
-\begin{equation}\label{eqn:jacobian:constraint}
+\begin{linenomath*}\begin{equation}\label{eqn:jacobian:constraint}
\bm{L} = \int_{S_f} \bm{N}_p^T \cdot (\bm{N}_{n^+} - \bm{N}_{n^-}) \, dS.
-\end{equation}
+\end{equation}\end{linenomath*}
Thus, the Jacobian of the entire system has the form,
-\begin{equation}\label{eqn:saddle:point}
+\begin{linenomath*}\begin{equation}\label{eqn:saddle:point}
\bm{A} =
\left( \begin{array}{cc}
\bm{K} & \bm{L}^T \\ \bm{L} & \bm{0}
\end{array} \right).
-\end{equation}
+\end{equation}\end{linenomath*}
Note that the terms in $\bm{N_{n^+}}$ and $\bm{N_{n^-}}$ are
identical, but they refer to degrees of freedom (DOF) on the positive and
negative sides of the fault, respectively. Consequently, in practice
we compute the terms for the positive side of the fault and assemble
the terms into the appropriate DOF for both sides of
the fault. Hence, we compute
-\begin{equation}\label{eqn:jacobian:constraint:code}
+\begin{linenomath*}\begin{equation}\label{eqn:jacobian:constraint:code}
\bm{L_p} = \int_{S_f} \bm{N}_p^T \cdot \bm{N}_{n^+} \, dS,
-\end{equation}
+\end{equation}\end{linenomath*}
with the Jacobian of the entire system taking the form,
-\begin{equation}\label{eqn:saddle:point:code}
+\begin{linenomath*}\begin{equation}\label{eqn:saddle:point:code}
\bm{A} =
\left( \begin{array}{cccc}
\bm{K}_{nn} & \bm{K}_{nn^+} & \bm{K}_{nn^-} & \bm{0} \\
@@ -459,7 +464,7 @@ with the Jacobian of the entire system t
\bm{K}_{n^-n} & \bm{0} & \bm{K}_{n^-n^-} & -\bm{L}_p^T \\
\bm{0} & \bm{L}_p & -\bm{L}_p & \bm{0}
\end{array} \right),
-\end{equation}
+\end{equation}\end{linenomath*}
where $n$ denotes DOF not associated with the fault,
$n^-$ denotes DOF associated with the negative side of
the fault, $n^+$ denotes DOF associated with the
@@ -479,7 +484,7 @@ simulations. Including the inertial term
simulations. Including the inertial term in
equation~(\ref{eqn:quasi-static:residual:elasticity}) for time $t$
rather than $t+\Delta t$ yields
-\begin{equation}\label{eqn:dynamic:residual:elasticity}
+\begin{linenomath*}\begin{equation}\label{eqn:dynamic:residual:elasticity}
\begin{split}
- \int_{V} \nabla \bm{N}_m^T \cdot \bm{\sigma}(t) \, dV
+ \int_{S_T} \bm{N}_m^T \cdot \bm{T}(t) \, dS \\
@@ -491,7 +496,7 @@ rather than $t+\Delta t$ yields
\frac{\partial^2 \bm{u}_n(t)}{\partial t^2} \, dV
=\bm{0}.
\end{split}
-\end{equation}
+\end{equation}\end{linenomath*}
We find the upper portion of the Jacobian of the system by considering
the action on the increment in the solution, just as we did for the
@@ -500,7 +505,7 @@ time using explicit time stepping via Ne
time using explicit time stepping via Newmark's method
\citep{Newmark:1959} with a central difference scheme wherein the
acceleration and velocity are given by
-\begin{gather}
+\begin{linenomath*}\begin{gather}
\frac{\partial^2 \bm{u}(t)}{\partial t^2} =
\frac{1}{\Delta t^2} \left(
\bm{du} - \bm{u}(t) + \bm{u}(t-\Delta t)
@@ -509,9 +514,9 @@ acceleration and velocity are given by
\frac{\partial \bm{u}(t)}{\partial t} = \frac{1}{2\Delta t} \left(
\bm{du} + \bm{u}(t) - \bm{u}(t-\Delta t)
\right).
-\end{gather}
+\end{gather}\end{linenomath*}
Expanding the inertial term yields
-\begin{equation}
+\begin{linenomath*}\begin{equation}
\begin{split}
- \int_{V} \rho \bm{N}_m^T \cdot \bm{N}_n \cdot \frac{\partial^2 \bm{u}_n(t)}{\partial
t^2} \, dV = \\
@@ -519,13 +524,13 @@ Expanding the inertial term yields
\bm{N}_n \cdot
\left( \bm{du}_n(t) - \bm{u}_n(t) + \bm{u}_n(t-\Delta t) \right) \, dV,
\end{split}
-\end{equation}
+\end{equation}\end{linenomath*}
so that the upper portion of the Jacobian is
-\begin{equation}
+\begin{linenomath*}\begin{equation}
\label{eqn:jacobian:explicit:inertia}
\bm{K} =
\frac{1}{\Delta t^2} \int_{V} \rho \bm{N}_m^T\ \cdot \bm{N}_n \, dV.
-\end{equation}
+\end{equation}\end{linenomath*}
This matches the mass matrix in conventional solid mechanics
finite-element formulations.
@@ -537,14 +542,14 @@ introduction of deformation at such shor
introduction of deformation at such short length scales we add
artificial damping via Kelvin-Voigt viscosity
\citep{Day:etal:2005,Kaneko:etal:2008} to the computation of the strain,
-\begin{gather}
+\begin{linenomath*}\begin{gather}
\bm{\varepsilon} = \frac{1}{2} \left[ \nabla \bm{u} +
(\nabla \bm{u})^T \right ], \\
\bm{\varepsilon} \approx \frac{1}{2} \left[ \nabla \bm{u}_d +
(\nabla \bm{u}_d)^T \right ], \\
\bm{u_d} = \bm{u} + \eta^* \Delta t \frac{\partial
\bm{u}}{\partial t},
-\end{gather}
+\end{gather}\end{linenomath*}
where $\eta^*$ is a nondimensional viscosity on the order of
0.1--1.0.
@@ -626,7 +631,7 @@ general form of a linear system of equat
general form of a linear system of equations ($\bm{A} \bm{u} =
\bm{b}$), our subset of equations based on
equation~(\ref{eqn:saddle:point:code}) has the form
-\begin{equation}
+\begin{linenomath*}\begin{equation}
\begin{pmatrix}
\bm{K}_{n^+n^+} & 0 & \bm{L}_p^T \\
0 & \bm{K}_{n^-n^-} & -\bm{L}_p^T \\
@@ -643,7 +648,7 @@ equation~(\ref{eqn:saddle:point:code}) h
\bm{b}_{n^-} \\
\bm{b}_p \\
\end{pmatrix},
-\end{equation}
+\end{equation}\end{linenomath*}
where $n^+$ and $n^-$ refer to the DOF associated with
the positive and negative sides of the fault,
respectively. Furthermore, we can ignore the terms $\bm{b}_{n^+}$
@@ -652,7 +657,7 @@ following system of equations to estimat
following system of equations to estimate the change in fault slip
$\partial \bm{d}$ associated with a perturbation in the Lagrange
multipliers $\partial \bm{l}_p$:
-\begin{gather}
+\begin{linenomath*}\begin{gather}
\label{eqn:spontaneous:rupture:update:lagrange}
\bm{K}_{n^+n^+} \cdot \partial \bm{u}_{n^+} =
- \bm{L}_p^T \cdot \partial \bm{l}_p, \\
@@ -660,7 +665,7 @@ multipliers $\partial \bm{l}_p$:
\bm{L}_p^T \cdot \partial \bm{l}_p, \\
\label{eqn:spontaneous:rupture:update:slip}
\partial \bm{d}_p = \partial \bm{u}_{n^+} - \partial \bm{u}_{n^-}.
-\end{gather}
+\end{gather}\end{linenomath*}
The efficiency of this iterative procedure depends on both the fault
constitutive model and how confined the deformation is to the region
@@ -678,10 +683,10 @@ increment in slip that best satisfies th
increment in slip that best satisfies the fault constitutive
model. Specifically, we search for $\alpha$ using a bilinear search in
logarithmic space to minimize
-\begin{equation}
+\begin{linenomath*}\begin{equation}
C = \| \bm{l}_p + \alpha \partial\bm{l}_p - f(\bm{d}_p +
\alpha \partial\bm{d}_p) \|_2,
-\end{equation}
+\end{equation}\end{linenomath*}
where $f(\bm{d})$ corresponds to the fault constitutive model and
$\|x\|_2$ denotes the L$^2$-norm of $x$. Performing this search in
logarithmic space rather than linear space greatly accelerates the
@@ -692,9 +697,9 @@ of which specify the shear traction on t
of which specify the shear traction on the fault $T_f$ as a function
of the cohesive stress $T_c$, coefficient of friction, $\mu_f$, and
normal traction $T_n$,
-\begin{equation}
+\begin{linenomath*}\begin{equation}
T_f = T_c - \mu_f T_n.
-\end{equation}
+\end{equation}\end{linenomath*}
$T_f$ in this equation corresponds to the magnitude of the shear
traction vector; the shear traction vector is resolved into the
direction of the slip rate. We use the sign convention that
@@ -860,18 +865,18 @@ associated with the Lagrange multipliers
associated with the Lagrange multipliers. In formulating the custom
preconditioner, we exploit the structure of the sparse Jacobian
matrix. Our system Jacobian has the form
-\begin{equation}
+\begin{linenomath*}\begin{equation}
\bm{A} = \left( \begin{array}{cc}
\bm{K} & \bm{L}^T \\
\bm{L} & \bm{0}
\end{array} \right).
-\end{equation}
+\end{equation}\end{linenomath*}
The Schur complement $\bm{S}$ of the submatrix $\bm{K}$ is given by,
-\begin{equation}
+\begin{linenomath*}\begin{equation}
\bm{S} = -\bm{L} \bm{K}^{-1} \bm{L}^T
-\end{equation}
+\end{equation}\end{linenomath*}
which leads to a simple block diagonal preconditioner for $\bm{A}$
-\begin{equation}
+\begin{linenomath*}\begin{equation}
\bm{P} = \left( \begin{array}{cc}
\bm{P}_\mathit{elasticity} & 0 \\
0 & \bm{P}_\mathit{fault}
@@ -880,7 +885,7 @@ which leads to a simple block diagonal p
\bm{K} & 0 \\
0 & -\bm{L} \bm{K}^{-1} \bm{L}^T
\end{array} \right).
-\end{equation}
+\end{equation}\end{linenomath*}
The elastic submatrix $\bm{K}$, in the absence of boundary conditions,
has three translational and three rotational null modes. These are
@@ -914,9 +919,9 @@ elements with Legendre polynomials and G
elements with Legendre polynomials and Gauss-Lobatto-Legendre
quadrature points. This leads to a diagonal matrix for the lower
portion of the conditioning matrix,
-\begin{equation}
+\begin{linenomath*}\begin{equation}
\bm{P}_\mathit{fault} = -\bm{L}_p (\bm{K}_{n+n+} + \bm{K}_{n-n-}) \bm{L}_p^{T},
-\end{equation}
+\end{equation}\end{linenomath*}
where $\bm{L}_p$ is given in
equation~(\ref{eqn:jacobian:constraint:code}) and $\bm{K}_{n+n+}$
and $\bm{K}_{n-n-}$ are the diagonal terms from
@@ -985,16 +990,16 @@ integral such that the action on rigid b
integral such that the action on rigid body motion is the same for the
diagonal approximation of the integral as it is for the original
integral,
-\begin{equation}
+\begin{linenomath*}\begin{equation}
\bm{A} \cdot \bm{u}_\mathrm{rigid} =
\bm{A}_\mathit{diagonal} \cdot \bm{u}_\mathrm{rigid}.
-\end{equation}
+\end{equation}\end{linenomath*}
Expressing the diagonal block of the Jacobian matrix as a vector and
the matrix of basis functions as a vector we have,
-\begin{equation}
+\begin{linenomath*}\begin{equation}
\bm{A} = \int_\Omega \bm{N}^T \cdot \bm{N} \, d\Omega \rightarrow
\bm{A}_\mathit{diagonal} = \int_\Omega \bm{N} \sum_i N_i \, d\Omega,
-\end{equation}
+\end{equation}\end{linenomath*}
where $N_i$ is the scalar basis function for degree of freedom $i$ and
$\Omega$ may be the domain volume (as in the case of the inertial
term) or a boundary (as in the case of the fault slip constraint
@@ -1020,20 +1025,20 @@ equations via a Schur's complement algor
equations via a Schur's complement algorithm. We compute an initial
residual assuming the increment in the solution is zero (i.e.,
$\bm{du}_n = \bm{0}$ and $\bm{dl}_p = \bm{0}$),
-\begin{equation}
+\begin{linenomath*}\begin{equation}
\bm{r}^* = \begin{pmatrix} \bm{r}_n^* \\ \bm{r}_p^* \end{pmatrix} =
\begin{pmatrix} \bm{b}_n \\ \bm{b}_p \end{pmatrix}
- \begin{pmatrix}
\bm{K} & \bm{L}^T \\ \bm{L} & 0
\end{pmatrix}
\begin{pmatrix} \bm{u}_n \\ \bm{l}_p \end{pmatrix}.
-\end{equation}
+\end{equation}\end{linenomath*}
We compute a corresponding initial solution to the system of equations
$\bm{du}_n^*$ ignoring the off-diagonal blocks in the Jacobian and
the increment in the Lagrange multipliers.
-\begin{equation}
+\begin{linenomath*}\begin{equation}
\bm{du}_n^* = \bm{K}^{-1} \cdot \bm{r}_n,
-\end{equation}
+\end{equation}\end{linenomath*}
taking advantage of the fact that we construct $\bm{K}$ so that it
is diagonal.
@@ -1041,7 +1046,7 @@ correct this initial solution so that th
correct this initial solution so that the true residual is zero.
Making use of the initial residual, the expression for the true
residual is
-\begin{equation}
+\begin{linenomath*}\begin{equation}
\label{eqn:lumped:jacobian:residual}
\bm{r} = \begin{pmatrix} \bm{r}_n \\ \bm{r}_p \end{pmatrix} =
\begin{pmatrix} \bm{r}_n^* \\ \bm{r}_p^* \end{pmatrix}
@@ -1049,49 +1054,49 @@ residual is
\bm{K} & \bm{L}^T \\ \bm{L} & 0
\end{pmatrix}
\begin{pmatrix} \bm{du}_n \\ \bm{dl}_p \end{pmatrix}.
-\end{equation}
+\end{equation}\end{linenomath*}
Solving the first row of equation~(\ref{eqn:lumped:jacobian:residual})
for the increment in the solution and accounting for the structure of
$\bm{L}$ as we write the expressions for DOF on
each side of the fault, we have
-\begin{gather}
+\begin{linenomath*}\begin{gather}
\bm{du}_{n^+} =
\bm{du}_{n^+}^* - \bm{K}_{n^+n^+}^{-1} \cdot \bm{L}_p^T \cdot \bm{dl}_p, \\
\bm{du}_{n^-} =
\bm{du}_{n^-}^* + \bm{K}_{n^-n^-}^{-1} \cdot \bm{L}_p^T \cdot \bm{dl}_p.
-\end{gather}
+\end{gather}\end{linenomath*}
Substituting into the second row of
equation~(\ref{eqn:lumped:jacobian:residual}) and isolating the term
with the increment in the Lagrange multipliers yields
-\begin{equation}
+\begin{linenomath*}\begin{equation}
\bm{L}_p \cdot
\left( \bm{K}_{n^+n^+}^{-1} + \bm{K}_{n^-n^-}^{-1} \right) \cdot
\bm{L}^T_p \cdot \bm{dl}_p =
-\bm{r}_p^* + \bm{L}_p \cdot
\left( \bm{du}_{n^+}^* - \bm{du}_{n^-}^* \right).
-\end{equation}
+\end{equation}\end{linenomath*}
Letting
-\begin{equation}
+\begin{linenomath*}\begin{equation}
\bm{S}_p = \bm{L}_p \cdot
\left( \bm{K}_{n^+n^+}^{-1} + \bm{K}_{n^-n^-}^{-1} \right) \cdot
\bm{L}^T_p,
-\end{equation}
+\end{equation}\end{linenomath*}
and recognizing that $\bm{S}_p$ is diagonal because $\bm{K}$
and $\bm{L}_p$ are diagonal allows us to solve for the increment
in the Lagrange multipliers,
-\begin{equation}
+\begin{linenomath*}\begin{equation}
\bm{dl}_p = \bm{S}_p^{-1} \cdot \left[
-\bm{r}_p^* + \bm{L}_p \cdot
\left( \bm{du}_{n^+}^* - \bm{du}_{n^-}^* \right)
\right].
-\end{equation}
+\end{equation}\end{linenomath*}
Now that we have the increment in the Lagrange multipliers, we can
correct our initial solution $\bm{du}_n^*$ so that the true residual
is zero,
-\begin{equation}
+\begin{linenomath*}\begin{equation}
\bm{du}_n =
\bm{du}_n^* - \bm{K}^{-1} \cdot \bm{L}^T \cdot \bm{dl}_p.
-\end{equation}
+\end{equation}\end{linenomath*}
Because $\bm{K}$ and $\bm{L}$ are comprised of diagonal
blocks, this expression for the updates to the solution are local to
the DOF attached to the fault and the Lagrange
@@ -1106,10 +1111,10 @@ the Lagrange multipliers
the Lagrange multipliers
(equations~(\ref{eqn:spontaneous:rupture:update:lagrange})--(\ref{eqn:spontaneous:rupture:update:slip}))
simplifies to
-\begin{equation}
+\begin{linenomath*}\begin{equation}
\partial \bm{d}_p = - \left( \bm{K}_{n^+n^+}^{-1} + \bm{K}_{n^-n^-}^{-1} \right)
\cdot \bm{L}_p^T \cdot \partial \bm{l}_p.
-\end{equation}
+\end{equation}\end{linenomath*}
Consequently, the increment in fault slip and Lagrange multipliers for
each vertex can be done independently. In dynamic simulations the time
step is small enough that the fault constitutive model is much less
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