[cig-commits] commit: Fixed typo (sign error).

[Mercurial]

changeset:   34:23bf193950cd
tag:         tip
user:        Brad Aagaard <baagaard at usgs.gov>
date:        Mon Mar 15 11:08:28 2010 -0700
files:       lumpedsolver.tex
description:
Fixed typo (sign error).


diff -r ccfecef493da -r 23bf193950cd lumpedsolver.tex
--- a/lumpedsolver.tex	Sun Jan 31 10:26:51 2010 -0800
+++ b/lumpedsolver.tex	Mon Mar 15 11:08:28 2010 -0700
@@ -40,12 +40,12 @@ equations in the form
   \left( \begin{array}{cc}
       A & C^T \\
       C & 0 \end{array} \right)
-  \left( \begin{array}{c} du \\ dl \end{array} \right) =
-  \left( \begin{array}{c} b \\ d \end{array} \right) -
+  \left( \begin{array}{c} du(t) \\ dl(t) \end{array} \right) =
+  \left( \begin{array}{c} b(t) \\ d(t+dt) \end{array} \right) -
   \left( \begin{array}{cc}
       A & C^T \\
       C & 0 \end{array} \right)
-  \left( \begin{array}{c} u \\ l \end{array} \right).
+  \left( \begin{array}{c} u(t) \\ l(t) \end{array} \right).
 \end{equation}
 
 For a lumped Jacobian matrix, we lump each cell's contribution before
@@ -61,19 +61,19 @@ Jacobian.
 
 We form the residual in the usual way,
 \begin{equation}
-  r = 
-  \left( \begin{array}{c} b \\ d \end{array} \right) -
+  r(t+dt) = 
+  \left( \begin{array}{c} b(t) \\ d(t+dt) \end{array} \right) -
   \left( \begin{array}{cc}
       A & C^T \\
       C & 0 \end{array} \right)
-  \left( \begin{array}{c} u+du \\ l+dl \end{array} \right),
+  \left( \begin{array}{c} u(t)+du(t) \\ l(t)+dl(t) \end{array} \right),
 \end{equation}
 so that we solve the system
 \begin{gather}
   \matrix{A}_\mathit{diag} \vec{du} = \vec{r}_o, \\
   \vec{r}_0 = r(du=0,dl=0).
 \end{gather}
-For convenience we will drop the distinction between $r_o$ and $r$ and simply refer to $r_0$ as $r$, because we form the residual in the first iteration at a time step assuming $du=0$ and $dl=0$. Consider Lagrange multiplier vertex $k$ associated with the slip between conventional vertex $i$ on the ``positive'' side of the fault and conventional vertex $j$ on the negative side of the fault. The system of equations is in the form \begin{gather}
+For convenience we will drop the distinction between $r_o$ and $r$ and simply refer to $r_0$ as $r$, because we form the residual in the first iteration at a time step assuming $du=0$ and $dl=0$. Consider Lagrange multiplier vertex $k$ associated with the slip between conventional vertex $i$ on the ``negative'' side of the fault and conventional vertex $j$ on the ``positive'' side of the fault. The system of equations is in the form \begin{gather}
   A_i du_i - C^T_{ki} dl_k = r_i, \\
   A_j du_j + C^T_{kj} dl_k = r_j, \\
   -C_{ki} du_i + C_{kj} du_j = d_k + C_{ki} u_i - C_{kj} u_j, \\
@@ -87,7 +87,7 @@ and substituting them into the third equ
 and substituting them into the third equation, yields
 \begin{equation}
   -C_{ki}A_i^{-1}(r_i+C^T_{ki} dl_k)
-  + C_{kj}A_j^{-1}(r_j + C_{kj}^T dl_k) = d_k + C_{ki} u_i - C_{kj} u_j.
+  + C_{kj}A_j^{-1}(r_j - C_{kj}^T dl_k) = d_k + C_{ki} u_i - C_{kj} u_j.
 \end{equation}
 Isolating the term with $dl_k$, we have
 \begin{equation}