[cig-commits] r5791 - short/3D/PyLith/trunk/doc/userguide

[willic3 at geodynamics.org]

Author: willic3
Date: 2007-01-16 08:40:20 -0800 (Tue, 16 Jan 2007)
New Revision: 5791

Added:
   short/3D/PyLith/trunk/doc/userguide/qstatic-smalldef.lyx
Log:
Started on governing equations for quasi-static small deformation
problem.  So far, all I've really done is remove acceleration terms
from Brad's equations.  Things should be pretty much OK up to about
equation 36 or 37.  Also, the equation numbers appear messed up right
now, which will need to be fixed.  I still need to get things set up
under the assumption of materially-nonlinear only deformation.  Large
strain will be dealt with separately.



Added: short/3D/PyLith/trunk/doc/userguide/qstatic-smalldef.lyx
===================================================================
--- short/3D/PyLith/trunk/doc/userguide/qstatic-smalldef.lyx	2007-01-15 21:42:04 UTC (rev 5790)
+++ short/3D/PyLith/trunk/doc/userguide/qstatic-smalldef.lyx	2007-01-16 16:40:20 UTC (rev 5791)
@@ -0,0 +1,1099 @@
+#LyX 1.4.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 245
+\begin_document
+\begin_header
+\textclass article
+\language english
+\inputencoding auto
+\fontscheme default
+\graphics default
+\paperfontsize default
+\spacing single
+\papersize default
+\use_geometry true
+\use_amsmath 1
+\cite_engine basic
+\use_bibtopic false
+\paperorientation portrait
+\leftmargin 1in
+\topmargin 1in
+\rightmargin 1in
+\bottommargin 1in
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\defskip medskip
+\quotes_language english
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tracking_changes false
+\output_changes false
+\end_header
+
+\begin_body
+
+\begin_layout Section
+Governing Equations
+\end_layout
+
+\begin_layout Standard
+\begin_inset Tabular
+<lyxtabular version="3" rows="10" columns="3">
+<features>
+<column alignment="center" valignment="top" leftline="true" width="0">
+<column alignment="center" valignment="top" leftline="true" width="0">
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+\begin_inset Text
+
+\begin_layout Standard
+Symbol
+\end_layout
+
+\end_inset
+</cell>
+<cell multicolumn="2" alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+Description
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row bottomline="true">
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+Index notation
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+Vector notation
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row topline="true">
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+\begin_inset Formula $a_{i}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+\begin_inset Formula $\overrightarrow{a}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+Vector field a
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row topline="true">
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+\begin_inset Formula $a_{ij}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+\begin_inset Formula $\underline{a}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+Second order tensor field a
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row topline="true">
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+\begin_inset Formula $u_{i}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+\begin_inset Formula $\overrightarrow{u}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+Displacement vector field
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row topline="true" bottomline="true">
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+\begin_inset Formula $f_{i}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+\begin_inset Formula $\overrightarrow{f}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+Body force vector field
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+\begin_inset Formula $T_{i}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+\begin_inset Formula $\overrightarrow{T}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+Traction vector field
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row topline="true" bottomline="true">
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+\begin_inset Formula $\sigma_{ij}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+\begin_inset Formula $\underline{\sigma}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+Stress tensor field
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row bottomline="true">
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+\begin_inset Formula $n_{i}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+\begin_inset Formula $\overrightarrow{n}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+Normal vector field
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row bottomline="true">
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+\begin_inset Formula $\rho$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+\begin_inset Formula $\rho$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Standard
+Mass density scalar field
+\end_layout
+
+\end_inset
+</cell>
+</row>
+</lyxtabular>
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Derivation of Equilibrium Equation
+\end_layout
+
+\begin_layout Subsubsection
+Index Notation
+\end_layout
+
+\begin_layout Standard
+Consider volume 
+\begin_inset Formula $V$
+\end_inset
+
+ bounded by surface 
+\begin_inset Formula $S$
+\end_inset
+
+.
+ Applying a Lagrangian description of the conservation of momentum (in the
+ absence of accelerations) gives
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula \begin{equation}
+\int_{V}f_{i}\, dV+\int_{S}T_{i}\, dS=0.\label{eqn:momentum:index}\end{equation}
+
+\end_inset
+
+The traction vector field is related to the stress tensor through
+\begin_inset Formula \begin{equation}
+T_{i}=\sigma_{ij}n_{j},\end{equation}
+
+\end_inset
+
+where 
+\begin_inset Formula $\overrightarrow{n}$
+\end_inset
+
+ is the vector normal to 
+\begin_inset Formula $S$
+\end_inset
+
+.
+ Substituting into 
+\begin_inset LatexCommand \eqref{eqn:momentum:index}
+
+\end_inset
+
+ yields
+\begin_inset Formula \begin{equation}
+\int_{V}f_{i}\, dV+\int_{S}\sigma_{ij}n_{j}\, dS=0.\end{equation}
+
+\end_inset
+
+Applying the divergence theorem,
+\begin_inset Formula \begin{equation}
+\int_{V}a_{i,j}\: dV=\int_{S}a_{j}n_{j}\: dS,\end{equation}
+
+\end_inset
+
+to the surface integral results in
+\begin_inset Formula \begin{equation}
+\int_{V}f_{i}\, dV+\int_{V}\sigma_{ij,j}\, dV=0,\end{equation}
+
+\end_inset
+
+which we can rewrite as
+\begin_inset Formula \begin{equation}
+\int_{V}\left(f_{i}+\sigma_{ij,j}\right)\, dV=0.\end{equation}
+
+\end_inset
+
+Because the volume 
+\begin_inset Formula $V$
+\end_inset
+
+ is arbitrary, the integrand must be zero at every location in the volume,
+ so that we end up with
+\begin_inset Formula \begin{gather}
+f_{i}+\sigma_{ij,j}=0\text{ in }V,\\
+\sigma_{ij}n_{j}=T_{i}\text{ on }S_{T}\text{, and}\\
+u_{i}=u_{i}^{o}\text{ on }S_{u.}\end{gather}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Vector Notation
+\end_layout
+
+\begin_layout Standard
+Consider volume 
+\begin_inset Formula $V$
+\end_inset
+
+ bounded by surface 
+\begin_inset Formula $S$
+\end_inset
+
+.
+ Applying a Lagrangian description of the conservation of momentum (in the
+ absence of accelerations) gives
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula \begin{equation}
+\int_{V}\overrightarrow{f}\, dV+\int_{S}\overrightarrow{T}\, dS=0.\label{eqn:momentum:vec}\end{equation}
+
+\end_inset
+
+The traction vector field is related to the stress tensor through
+\begin_inset Formula \begin{equation}
+\overrightarrow{T}=\underline{\sigma}\cdot\overrightarrow{n},\end{equation}
+
+\end_inset
+
+where 
+\begin_inset Formula $\overrightarrow{n}$
+\end_inset
+
+ is the vector normal to 
+\begin_inset Formula $S$
+\end_inset
+
+.
+ Substituting into 
+\begin_inset LatexCommand \eqref{eqn:momentum:vec}
+
+\end_inset
+
+ yields
+\begin_inset Formula \begin{equation}
+\int_{V}\overrightarrow{f}\, dV+\int_{S}\underline{\sigma}\cdot\overrightarrow{n}\, dS=0.\end{equation}
+
+\end_inset
+
+Applying the divergence theorem,
+\begin_inset Formula \begin{equation}
+\int_{V}\nabla\cdot\overrightarrow{a}\: dV=\int_{S}\overrightarrow{a}\cdot\overrightarrow{n}\: dS,\end{equation}
+
+\end_inset
+
+to the surface integral results in
+\begin_inset Formula \begin{equation}
+\int_{V}\overrightarrow{f}\, dV+\int_{V}\nabla\cdot\underline{\sigma}\, dV=0,\end{equation}
+
+\end_inset
+
+which we can rewrite as
+\begin_inset Formula \begin{equation}
+\int_{V}\left(\overrightarrow{f}+\nabla\cdot\overrightarrow{\sigma}\right)\, dV=0.\end{equation}
+
+\end_inset
+
+Because the volume 
+\begin_inset Formula $V$
+\end_inset
+
+ is arbitrary, the integrand must be zero at every location in the volume,
+ so that we end up with
+\begin_inset Formula \begin{gather}
+\overrightarrow{f}+\nabla\cdot\overrightarrow{\sigma}=0\text{ in }V,\\
+\underline{\sigma}\cdot\overrightarrow{n}=\overrightarrow{T}\text{ on }S_{T}\text{, and}\\
+\overrightarrow{u}=\overrightarrow{u^{o}}\text{ on }S_{u.}\end{gather}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Finite-Element Formulation of Quasi-Static Equations
+\end_layout
+
+\begin_layout Subsubsection
+Index Notation
+\end_layout
+
+\begin_layout Standard
+We start with the strong form for the quasi-static problem under the assumption
+ of infinitesimal strains,
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula \begin{gather}
+\sigma_{ij,j}+f_{i}=0\text{ in }V,\\
+\sigma_{ij}n_{j}=T_{i}\text{ on }S_{T},\\
+u_{i}=u_{i}^{o}\text{ on }S_{u},\\
+\sigma_{ij}=\sigma_{ji}\text{ (symmetric).}\end{gather}
+
+\end_inset
+
+We construct the weak form by multiplying the equation by a trial function
+ and setting the integral over the domain to zero.
+ The trial function is a piecewise differential vector field, 
+\begin_inset Formula $\phi_{i}$
+\end_inset
+
+, where 
+\begin_inset Formula $\phi_{i}=0$
+\end_inset
+
+ on 
+\begin_inset Formula $S_{u}.$
+\end_inset
+
+ Hence our weak form is
+\begin_inset Formula \begin{gather}
+\int_{V}\left(\sigma_{ij,j}+f_{i}\right)\phi_{i}\, dV=0\text{, or }\\
+\int_{V}\sigma_{ij,j}\phi_{i}\: dV+\int_{V}f_{i}\phi_{i}\: dV=0.\end{gather}
+
+\end_inset
+
+ Consider the divergence theorem applied to the dot product of the stress
+ tensor and the trial function, 
+\begin_inset Formula $\sigma_{ij}\phi_{i}$
+\end_inset
+
+,
+\begin_inset Formula \begin{equation}
+\int_{V}(\sigma_{ij}\phi_{i})_{,j}\, dV=\int_{S}(\sigma_{ij}\phi_{i})n_{i}\, dS.\end{equation}
+
+\end_inset
+
+Expanding the left hand side yields
+\begin_inset Formula \begin{gather}
+\int_{V}\sigma_{ij,j}\phi_{i}\: dV+\int_{V}\sigma_{ij}\phi_{i,j}\: dV=\int_{S}\sigma_{ij}\phi_{i}n_{i}\: dS,\text{ or}\\
+\int_{V}\sigma_{ij,j}\phi_{i}\: dV=-\int_{V}\sigma_{ij}\phi_{i,j}\, dV+\int_{S}\sigma_{ij}\phi_{i}n_{i}\, dS.\end{gather}
+
+\end_inset
+
+Substituting into the weak form gives
+\begin_inset Formula \begin{equation}
+-\int_{V}\sigma_{ij}\phi_{i,j}\, dV+\int_{S}\sigma_{ij}\phi_{i}n_{i}\, dS+\int_{V}f_{i}\phi_{i}\, dV=0.\end{equation}
+
+\end_inset
+
+Now, 
+\begin_inset Formula $\sigma_{ij}\phi_{i,j}$
+\end_inset
+
+ is a scalar, so it is symmetric,
+\begin_inset Formula \begin{equation}
+\sigma_{ij}\phi_{i,j}=\sigma_{ji}\phi_{j,i},\end{equation}
+
+\end_inset
+
+and we know that 
+\begin_inset Formula $\sigma_{ij}$
+\end_inset
+
+ is symmetric, so
+\begin_inset Formula \begin{equation}
+\sigma_{ij}\phi_{i,j}=\sigma_{ij}\phi_{j,i},\end{equation}
+
+\end_inset
+
+which means
+\begin_inset Formula \begin{equation}
+\phi_{i,j}=\phi_{j,i},\end{equation}
+
+\end_inset
+
+which we can write as
+\begin_inset Formula \begin{equation}
+\phi_{i,j}=\frac{1}{2}(\phi_{i,j}+\phi_{j,i}).\end{equation}
+
+\end_inset
+
+Substituting into the first term gives
+\begin_inset Formula \begin{equation}
+-\int_{V}\frac{1}{2}\sigma_{ij}\left(\phi_{i,j}+\phi_{j,i}\right)\, dV+\int_{S}\sigma_{ij}\phi_{i}n_{i}\, dS+\int_{V}f_{i}\phi_{i}\, dV=0.\end{equation}
+
+\end_inset
+
+Turning our attention to the second term, we separate the integration over
+ 
+\begin_inset Formula $S$
+\end_inset
+
+ into integration over 
+\begin_inset Formula $S_{T}$
+\end_inset
+
+ and 
+\begin_inset Formula $S_{u}$
+\end_inset
+
+,
+\begin_inset Formula \begin{equation}
+-\int_{V}\frac{1}{2}\sigma_{ij}(\phi_{i,j}+\phi_{j,i})\, dV+\int_{S_{T}}\sigma_{ij}\phi_{i}n_{i}\, dS+\int_{S_{u}}\sigma_{ij}\phi_{i}n_{i}\, dS+\int_{V}f_{i}\phi_{i}\, dV=0,\end{equation}
+
+\end_inset
+
+and recognize that
+\begin_inset Formula \begin{gather}
+\sigma_{ij}n_{i}=T_{i}\text{ on }S_{T}\text{ and}\\
+\phi_{i}=0\text{ on }S_{u},\end{gather}
+
+\end_inset
+
+so that the equation reduces to
+\begin_inset Formula \begin{equation}
+-\int_{V}\frac{1}{2}\sigma_{ij}(\phi_{i,j}+\phi_{j,i})\: dV+\int_{S_{T}}T_{i}\phi_{i}\, dS+\int_{V}f_{i}\phi_{i}\, dV=0.\end{equation}
+
+\end_inset
+
+This is the equation we want to solve.
+ Discretizing into finite-elements separates the integral over the domain
+ and boundaries into a sum of integrals over elements and element boundaries,
+\begin_inset Formula \begin{equation}
+\sum_{elements}(\int_{V^{e}}\frac{1}{2}\sigma_{ij}(\phi_{i,j}+\phi_{j,i})\, dV-\int_{V^{e}}f_{i}\phi_{i}\, dV-\int_{S_{t}^{e}}T_{i}\phi_{i}\, dS)=0.\end{equation}
+
+\end_inset
+
+Within an element we represent the fields as a linear combination of a set
+ of basis functions and the values of the fields at vertices of the element,
+\begin_inset Formula \begin{equation}
+a_{i}=N^{m}a_{i}^{m},\end{equation}
+
+\end_inset
+
+where 
+\begin_inset Formula $N^{m}$
+\end_inset
+
+ is the 
+\begin_inset Formula $m$
+\end_inset
+
+th basis function for an element and 
+\begin_inset Formula $a_{i}^{m}$
+\end_inset
+
+ is the field at vertex 
+\begin_inset Formula $m$
+\end_inset
+
+.
+ Rewriting the trial functions and displacement field in terms of the basis
+ functions gives
+\begin_inset Formula \begin{gather}
+\phi_{i}=N^{m},\text{ and}\\
+u_{i}=N^{m}u_{i}^{m}.\end{gather}
+
+\end_inset
+
+We force the weak form to hold for each component in the vector space.
+ For basis function 
+\begin_inset Formula $N^{p}$
+\end_inset
+
+ and component 
+\begin_inset Formula $i$
+\end_inset
+
+, we have
+\begin_inset Formula \begin{multline}
+\sum_{elements}(\int_{V^{e}}\frac{1}{2}\sigma_{ij}(N_{,j}^{p}+N_{,i}^{p})\, dV+\int_{V^{e}}\rho N_{}^{p}\sum_{q}N_{}^{q}\ddot{u}_{i}^{q}\: dV-\int_{V^{e}}N_{}^{p}f_{i}\: dV-\int_{S_{T}^{e}}N_{}^{p}T_{i}\, dS)=0.\end{multline}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Vector Notation
+\end_layout
+
+\begin_layout Standard
+We start with the wave equation (strong form),
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula \begin{gather}
+\nabla\cdot\underline{\sigma}+\overrightarrow{f}=\rho\frac{\partial^{2}\overrightarrow{u}}{\partial t^{2}}\text{ in }V,\\
+\underline{\sigma}\cdot\overrightarrow{n}=\overrightarrow{T}\text{ on }S_{T},\\
+\overrightarrow{u}=\overrightarrow{u}^{o}\text{ on }S_{u},\\
+\underline{\sigma}=\underline{\sigma}^{T}\text{ (symmetric).}\end{gather}
+
+\end_inset
+
+We construct the weak form by multiplying the wave equation by a trial function
+ and setting the integral over the domain to zero.
+ The trial function is a piecewise differential vector field, 
+\begin_inset Formula $\overrightarrow{\phi}$
+\end_inset
+
+, where 
+\begin_inset Formula $\overrightarrow{\phi}=0$
+\end_inset
+
+ on 
+\begin_inset Formula $S_{u}.$
+\end_inset
+
+ Hence our weak form is
+\begin_inset Formula \begin{gather}
+\int_{V}\left(\nabla\cdot\underline{\sigma}+\overrightarrow{f}-\rho\frac{\partial^{2}\overrightarrow{u}}{\partial t^{2}}\right)\cdot\overrightarrow{\phi}\, dV=0\text{, or }\\
+\int_{V}(\nabla\cdot\underline{\sigma})\cdot\overrightarrow{\phi}\: dV+\int_{V}\overrightarrow{f}\cdot\overrightarrow{\phi}\: dV-\int_{V}\rho\frac{\partial^{2}\overrightarrow{u}}{\partial t^{2}}\cdot\overrightarrow{\phi}\: dV=0.\end{gather}
+
+\end_inset
+
+ Consider the divergence theorem applied to the dot product of the stress
+ tensor and the trial function, 
+\begin_inset Formula $\underline{\sigma}\cdot\overrightarrow{\phi}$
+\end_inset
+
+,
+\begin_inset Formula \begin{equation}
+\int_{V}\nabla\cdot(\underline{\sigma}\cdot\overrightarrow{\phi})\, dV=\int_{S}(\underline{\sigma}\cdot\overrightarrow{\phi})\cdot\overrightarrow{n}\, dS.\end{equation}
+
+\end_inset
+
+Expanding the left hand side yields
+\begin_inset Formula \begin{equation}
+\int_{V}(\nabla\cdot\underline{\sigma})\cdot\overrightarrow{\phi}\: dV+\int_{V}\underline{\sigma}:\nabla\overrightarrow{\phi}\: dV=\int_{S}(\underline{\sigma}\cdot\overrightarrow{\phi})\cdot\overrightarrow{n}\: dS,\text{ or}\end{equation}
+
+\end_inset
+
+
+\begin_inset Formula \begin{equation}
+\int_{V}{(\nabla\cdot\underline{\sigma})\cdot\overrightarrow{\phi}}_{ij,j}\: dV=-\int_{V}\underline{\sigma}:\nabla\overrightarrow{\phi}\, dV+\int_{S}\underline{\sigma}\cdot\overrightarrow{n}\cdot\overrightarrow{\phi}\, dS.\end{equation}
+
+\end_inset
+
+Substituting into the weak form gives
+\begin_inset Formula \begin{equation}
+-\int_{V}\underline{\sigma}:\nabla\overrightarrow{\phi}\, dV+\int_{S}\underline{\sigma}\cdot\overrightarrow{n}\cdot\overrightarrow{\phi}\, dS+\int_{V}\overrightarrow{f}\overrightarrow{\phi}\, dV-\int_{V}\rho\frac{\partial^{2}\overrightarrow{u}}{\partial t^{2}}\cdot\overrightarrow{\phi}\, dV=0.\end{equation}
+
+\end_inset
+
+Now, 
+\begin_inset Formula $\underline{\sigma}:\nabla\overrightarrow{\phi}$
+\end_inset
+
+ is a scalar, so it is symmetric,
+\begin_inset Formula \begin{equation}
+{\underline{\sigma}:\nabla\overrightarrow{\phi}=(\underline{\sigma}:\nabla\overrightarrow{\phi})}^{T}=\underline{\sigma}^{T}:\overrightarrow{\phi}^{T}\nabla^{T},\end{equation}
+
+\end_inset
+
+and we know that 
+\begin_inset Formula $\underline{\sigma}$
+\end_inset
+
+ is symmetric, so
+\begin_inset Formula \begin{equation}
+\underline{\sigma}:\nabla\overrightarrow{\phi}=\underline{\sigma}:\overrightarrow{\phi}^{T}\nabla^{T},\end{equation}
+
+\end_inset
+
+which means
+\begin_inset Formula \begin{equation}
+\nabla\overrightarrow{\phi}=\overrightarrow{\phi}^{T}\nabla^{T},\end{equation}
+
+\end_inset
+
+which we can write as
+\begin_inset Formula \begin{equation}
+\nabla\overrightarrow{\phi}=\frac{1}{2}(\nabla+\nabla^{T})\overrightarrow{\phi}.\end{equation}
+
+\end_inset
+
+Substituting into the first term gives
+\begin_inset Formula \begin{equation}
+-\int_{V}\frac{1}{2}\underline{\sigma}:(\nabla+\nabla^{T})\overrightarrow{\phi}\, dV+\int_{S}\underline{\sigma}\cdot\overrightarrow{n}\cdot\overrightarrow{\phi}\, dS+\int_{V}\overrightarrow{f}\cdot\overrightarrow{\phi}\, dV-\int_{V}\rho\frac{\partial^{2}\overrightarrow{u}}{\partial t^{2}}\cdot\overrightarrow{\phi}\, dV=0.\end{equation}
+
+\end_inset
+
+Turning our attention to the second term, we separate the integration over
+ 
+\begin_inset Formula $S$
+\end_inset
+
+ into integration over 
+\begin_inset Formula $S_{T}$
+\end_inset
+
+ and 
+\begin_inset Formula $S_{u}$
+\end_inset
+
+,
+\begin_inset Formula \begin{multline}
+-\int_{V}\frac{1}{2}\underline{\sigma}:(\nabla+\nabla^{T})\overrightarrow{\phi}\, dV+\int_{S_{T}}\underline{\sigma}\cdot\overrightarrow{n}\cdot\overrightarrow{\phi}\, dS+\int_{S_{u}}\underline{\sigma}\cdot\overrightarrow{n}\cdot\overrightarrow{\phi}\, dS+\int_{V}\overrightarrow{f}\cdot\overrightarrow{\phi}\, dV\\
+-\int_{V}\rho\frac{\partial^{2}\overrightarrow{u}}{\partial t^{2}}\cdot\overrightarrow{\phi}\, dV=0,\end{multline}
+
+\end_inset
+
+and recognize that
+\begin_inset Formula \begin{gather}
+\underline{\sigma}\cdot\overrightarrow{n}=\overrightarrow{T}\text{ on }S_{T}\text{ and}\\
+\overrightarrow{\phi}=0\text{ on }S_{u},\end{gather}
+
+\end_inset
+
+so that the equation reduces to
+\begin_inset Formula \begin{equation}
+-\int_{V}\frac{1}{2}\underline{\sigma}:(\nabla+\nabla^{T})\overrightarrow{\phi}\: dV+\int_{S_{T}}\overrightarrow{T}\cdot\overrightarrow{\phi}\, dS+\int_{V}\overrightarrow{f}\cdot\overrightarrow{\phi}\, dV-\int_{V}\rho\frac{\partial^{2}\overrightarrow{u}}{\partial t^{2}}\cdot\overrightarrow{\phi}\, dV=0.\end{equation}
+
+\end_inset
+
+This is the equation we want to solve.
+ Discretizing into finite-elements separates the integral over the domain
+ and boundaries into a sum of integrals over elements and element boundaries,
+\begin_inset Formula \begin{equation}
+\sum_{elements}(\int_{V^{e}}\frac{1}{2}\underline{\sigma}:(\nabla+\nabla^{T})\overrightarrow{\phi}\, dV+\int_{V^{e}}\rho\frac{\partial^{2}\overrightarrow{u}}{\partial t^{2}}\cdot\overrightarrow{\phi}\, dV-\int_{V^{e}}\overrightarrow{f}\cdot\overrightarrow{\phi}\, dV-\int_{S_{t}^{e}}\overrightarrow{T}\cdot\overrightarrow{\phi}\, dS)=0.\end{equation}
+
+\end_inset
+
+Within an element we represent the fields as a linear combination of a set
+ of basis functions and the values of the fields at vertices of the element,
+\begin_inset Marginal
+status open
+
+\begin_layout Standard
+Is this written correctly?
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula \begin{equation}
+\overrightarrow{a}=\underline{N}\cdot\overrightarrow{a^{e}},\end{equation}
+
+\end_inset
+
+where 
+\begin_inset Formula $\underline{N}$
+\end_inset
+
+ are the basis functions for an element and 
+\begin_inset Formula $\overrightarrow{a^{e}}$
+\end_inset
+
+ is the field at an element's vertices.
+ Rewriting the trial functions and displacement field in terms of the basis
+ functions gives
+\begin_inset Marginal
+status open
+
+\begin_layout Standard
+Is the trial function expression correct?
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula \begin{equation}
+\overrightarrow{\phi}=\overrightarrow{N},\text{ and}\end{equation}
+
+\end_inset
+
+
+\begin_inset Formula \begin{equation}
+\overrightarrow{u}=\underline{N}\cdot\overrightarrow{u^{e}}.\end{equation}
+
+\end_inset
+
+Substituting into the integral equation yields
+\begin_inset Formula \begin{multline}
+\sum_{elements}(\int_{V^{e}}\frac{1}{2}\underline{\sigma}:(\nabla+\nabla^{T})\underline{N}\, dV+\int_{V^{e}}\rho\underline{N}\cdot\underline{N}\cdot\frac{\partial^{2}\overrightarrow{u^{e}}}{\partial t^{2}}\: dV-\int_{V^{e}}\underline{N}\cdot\overrightarrow{f^{e}}^{}\, dV-\int_{S_{T}}\underline{N}\cdot\overrightarrow{T^{e}}_{}\, dS)=0\end{multline}
+
+\end_inset
+
+For a linearly elastic material
+\begin_inset Formula \begin{equation}
+\underline{\sigma}=\underline{C}\cdot\underline{\varepsilon},\end{equation}
+
+\end_inset
+
+and for infinitesimal strains
+\begin_inset Formula \begin{equation}
+\underline{\varepsilon}=\frac{1}{2}(\nabla+\nabla^{T})\overrightarrow{u},\end{equation}
+
+\end_inset
+
+so in this case our integral equation becomes
+\begin_inset Formula \begin{multline}
+\sum_{elements}(\int_{V^{e}}\frac{1}{4}(\nabla+\nabla^{T})\underline{N}:C\cdot(\nabla+\nabla^{T})\underline{N}\cdot\overrightarrow{u^{e}})\, dV+\int_{V^{e}}\rho\underline{N}\cdot\underline{N}\cdot\frac{\partial^{2}\overrightarrow{u^{e}}}{\partial t^{2}}\: dV-\int_{V^{e}}\underline{N}\cdot\underline{N}\cdot\overrightarrow{f^{e}}\, dV\\
+-\int_{S_{T}}\underline{N}\cdot\underline{N}\cdot\overrightarrow{T^{e}}\, dS)=0.\end{multline}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Explicit Time Integration of Elasticity Equation
+\end_layout
+
+\begin_layout Subsubsection
+Index Notation
+\end_layout
+
+\begin_layout Standard
+Using the central difference method to approximate the acceleration (and
+ velocity),
+\begin_inset Formula \begin{gather}
+\ddot{u}_{i}=\frac{1}{\Delta t^{2}}\left(u_{i}(t+\Delta t)-2u_{i}(t)+u_{i}(t-\Delta t)\right)\\
+\dot{u}_{i}=\frac{1}{2\Delta t}\left(u_{i}(t+\Delta t)-u_{i}(t-\Delta t)\right)\end{gather}
+
+\end_inset
+
+we have
+\begin_inset Formula \begin{multline}
+\sum_{elements}(\int_{V^{e}}\frac{1}{2}\sigma_{ij}(N_{,j}^{p}+N_{,i}^{p})\, dV+\int_{V^{e}}\rho N_{}^{p}\sum_{q}N_{}^{q}(u_{i}^{q}(t+\Delta t)-2u_{i}^{q}(t)+u_{i}^{q}(t-\Delta t))\: dV\\
+-\int_{V^{e}}N^{p}f_{i}(t)\: dV-\int_{S_{T}}N^{p}T_{i}\: dS)=0,\end{multline}
+
+\end_inset
+
+which we can expand into
+\begin_inset Formula \begin{multline}
+\sum_{elements}({\frac{1}{\Delta t^{2}}\int}_{V^{e}}\rho N^{p}\sum_{q}N^{q}u_{i}^{q}(t+\Delta t)\, dV-\frac{2}{\Delta t^{2}}\int_{V^{e}}\rho N^{p}\sum_{q}N^{q}u_{i}^{q}(t)\, dV\\
++\frac{1}{\Delta t^{2}}\int_{V^{e}}\rho N^{p}\sum_{q}N^{q}u_{i}^{q}(t-\Delta t)\, dV+{\frac{1}{2}\int}_{V^{e}}\sigma_{ij}(t)(N_{,j}^{p}+N_{,i}^{p})\: dV\\
+-\int_{V^{e}}N^{p}f_{i}(t)\: dV-\int_{S_{T}}N^{p}T_{i}\: dS)=0\end{multline}
+
+\end_inset
+
+for the 
+\begin_inset Formula $i$
+\end_inset
+
+th component associated with basis function
+\begin_inset Formula $N^{p}$
+\end_inset
+
+.
+ Isolating the term containing 
+\begin_inset Formula $u_{i}^{q}(t+\Delta t)$
+\end_inset
+
+ yields
+\begin_inset Formula \begin{multline*}
+\frac{1}{\Delta t^{2}}\sum_{elements}\left(\int_{V^{e}}\rho N^{p}\sum_{q}N^{q}u_{i}^{q}(t+\Delta t)\, dV\right)=\frac{2}{\Delta t^{2}}\sum_{elements}\left(\int_{V^{e}}\rho N^{p}\sum_{q}N^{q}u_{i}^{q}(t)\, dV\right)\\
+-\frac{1}{\Delta t^{2}}\sum_{elements}\left(\int_{V^{e}}\rho N^{p}\sum_{q}N^{q}u_{i}^{q}(t-\Delta t)\, dV\right)-\frac{1}{2}\sum_{elements}\left(\int_{V^{e}}\sigma_{ij}(t)(N_{,j}^{p}+N_{,i}^{p})\: dV\right)\\
++\sum_{elements}\left(\int_{V^{e}}N^{p}f_{i}(t)\: dV\right)+\sum_{elements}\left(\int_{S_{T}}N^{p}T_{i}\: dS\right).\end{multline*}
+
+\end_inset
+
+We can rewrite the left-hand-side as a matrix-vector product where the vector
+ is the displacement field at time 
+\begin_inset Formula $t+\Delta t$
+\end_inset
+
+ and the element mass matrix is given by
+\begin_inset Formula \[
+M_{ij}^{pq}=\delta_{ij}\int_{V^{e}}\rho N^{p}N^{q}\, dV,\]
+
+\end_inset
+
+where 
+\begin_inset Formula $M_{ij}^{pq}$
+\end_inset
+
+ is a 
+\begin_inset Formula $pn$
+\end_inset
+
+ by 
+\begin_inset Formula $qn$
+\end_inset
+
+ matrix (
+\begin_inset Formula $n$
+\end_inset
+
+ is the dimension of the vector space), 
+\begin_inset Formula $p$
+\end_inset
+
+ and 
+\begin_inset Formula $q$
+\end_inset
+
+ refer to the basis functions and 
+\begin_inset Formula $i$
+\end_inset
+
+ and 
+\begin_inset Formula $j$
+\end_inset
+
+ are vector space components.
+ 
+\end_layout
+
+\begin_layout Subsubsection
+Vector Notation
+\end_layout
+
+\begin_layout Standard
+Using the central difference method to approximate the acceleration (and
+ velocity),
+\begin_inset Formula \begin{gather}
+\frac{\partial^{2}\overrightarrow{u}(t)}{\partial t^{2}}=\frac{1}{\Delta t^{2}}\left(\overrightarrow{u}(t+\Delta t)-2\overrightarrow{u}(t)+\overrightarrow{u}(t-\Delta t)\right)\\
+\frac{\partial\overrightarrow{u}(t)}{\partial t}=\frac{1}{2\Delta t}\left(\overrightarrow{u}(t+\Delta t)-\overrightarrow{u}(t-\Delta t)\right)\end{gather}
+
+\end_inset
+
+
+\end_layout
+
+\end_body
+\end_document