\end{align}
Note that the differential equation does not contain any time derivatives any
more -- all time dependence is introduced through boundary conditions and a
-possibly time-varying force function $\vec f(\vec x,t)$.
+possibly time-varying force function $\vec f(\vec x,t)$. The changes in
+configuration can therefore be considered as being stationary
+instantaneously. An alternative view of this is that $t$ is not really a time
+variable, but only a time-like parameter that governs the evolution of the
+problem.
While these equations are sufficient to describe small deformations, computing
large deformations is a little more complicated. To do so, let us first
-introduce a stress variable $\sigma$, and write the differential equations in
-terms of the stress:
+introduce a tensorial stress variable $\sigma$, and write the differential
+equations in terms of the stress:
\begin{align}
- \div \sigma &= \vec f
&&\text{in $\Omega(t)$},
Note that these equations are posed on a domain $\Omega(t)$ that
changes with time, with the boundary moving according to the
displacements $\vec u(\vec x,t)$ of the points on the boundary. To
-complete this system, we have to specify the relationship between the
-stress and the strain, as follows:
+complete this system, we have to specify the incremental relationship between
+the stress and the strain, as follows:
\begin{align}
\label{eq:stress-strain}
\dot\sigma = C \varepsilon (\dot{\vec u}),
\sigma^n = \sigma^{n-1} + C \varepsilon (\Delta \vec u^n),
\end{gather}
and $\Delta \vec u^n$ the incremental displacement for time step
-$n$. This way, if we want to solve for the displacement increment, we
+$n$. In addition, we have to specify initial data $\vec u(\cdot,0)=\vec u_0$.
+This way, if we want to solve for the displacement increment, we
have to solve the following system:
\begin{align}
- \div C \varepsilon(\Delta\vec u^n) &= \vec f + \div \sigma^{n-1}
There are differences, however:
\begin{enumerate}
- \item We have to move the mesh after each time step, in order to be
+ \item We have to move (update) the mesh after each time step, in order to be
able to solve the next time step on a new domain;
\item We need to know $\sigma^{n-1}$ to compute the next incremental
old domain, and has to be transferred to the new domain. The reason for this
can be understood as follows: locally, the incremental deformation $\Delta\vec
u$ can be decomposed into three parts, a linear translation (the constant part
-of the displacement field in the neighborhood of a point), a dilational
+of the displacement increment field in the neighborhood of a point), a
+dilational
component (that part of the gradient of the displacement field that has a
nonzero divergence), and a rotation. A linear translation of the material does
not affect the stresses that are frozen into it -- the stress values are
\nu = \frac{\lambda}{2(\lambda+\mu)},
\end{gather*}
where $\lambda,\mu$ are the Lam\'e constants of the material.
-Physical constraints indicate that $-1\le \nu\le \tfrac 12$. If $\nu$
+Physical constraints indicate that $-1\le \nu\le \tfrac 12$ (the condition
+also follows from mathematical stability considerations). If $\nu$
approaches $\tfrac 12$, then the material becomes incompressible. In that
case, pure displacement-based formulations are no longer appropriate for the
solution of such problems, and stabilization techniques have to be employed