/[escript]/trunk/doc/user/stokessolver.tex

Diff of /trunk/doc/user/stokessolver.tex

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-revision 2843 by gross,
Thu Jan 14 05:51:28 2010 UTC
+revision 2851 by gross,
Fri Jan 15 07:37:47 2010 UTC
 Line 6 
 We want to calculate the velocity \index
  \begin{equation}\label{Stokes 1}
  -\left(\eta(v\hackscore{i,j}+ v\hackscore{j,i})\right)\hackscore{,j}+p\hackscore{,i}=f\hackscore{i}-\sigma\hackscore{ij,j}
  \end{equation}
- where  $f\hackscore{i}$ defines an internal force \index{force, internal} and $\sigma\hackscore{ij}$ is an intial stress \index{stress, initial}. The viscosity $\eta$ may weakly depend on pressure and velocity. If relevant we will use the notation $\eta(v,p)$ to express this dependency.
+ where  $f\hackscore{i}$ defines an internal force \index{force, internal} and $\sigma\hackscore{ij}$ is an initial stress \index{stress, initial}. The viscosity $\eta$ may weakly depend on pressure and velocity. If relevant we will use the notation $\eta(v,p)$ to express this dependency.
  We assume an incompressible media:
  \begin{equation}\label{Stokes 2}
 Line 62 
 for the pressure and an additional corre
   \label{STOKES ITER STEP 2}
   \end{equation}
  The new velocity and pressure are then given by $v\hackscore{2}=v\hackscore{1}-dv\hackscore{2}$ and
- $p\hackscore{2}=p\hackscore{0}+dp\hackscore{2}$ which will fullfill the block system~\ref{STOKES}.
+ $p\hackscore{2}=p\hackscore{0}+dp\hackscore{2}$ which will fulfill the block system~\ref{STOKES}.
  This solution strategy is called the Uzawa scheme \index{Uzawa scheme}.
  There is a problem with this scheme: In practice we will use an iterative scheme
 Line 80 
 In the following we will use the two nor
  \|p\|\hackscore{0}^2= \int\hackscore{\Omega} p^2 \; dx.
  \label{STOKES STOP}
  \end{equation}
- for velicity $v$ and pressure $p$. The iteration is terminated if the stopping criterium
+ for velocity $v$ and pressure $p$. The iteration is terminated if the stopping criteria
   \begin{equation} \label{STOKES STOPPING CRITERIA}
  \max(\|Bv\hackscore{1}\|\hackscore{0},\|v\hackscore{2}-v\hackscore{0}\|\hackscore{1}) \le \tau \cdot \|v\hackscore{2}\|\hackscore{1}
   \end{equation}
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 norm of $B A^{-1} B^{*} dp\hackscore{2}$
  We want to optimize the tolerance choice for solving~\ref{STOKES ITER STEP 1}
  and~\ref{STOKES ITER STEP 2}. To do this we write the iteration scheme as a fixed point problem. Here
- we consider the errors produced by the iterative solvers beeing used.
+ we consider the errors produced by the iterative solvers being used.
  From Equation~\ref{STOKES ITER STEP 1} we have
  \begin{equation} \label{STOKES total V1}
  v\hackscore{1} = e\hackscore{1} + v\hackscore{0} + A^{-1} ( G - Av\hackscore{0} - B^{*} p\hackscore{0} )
 Line 123 
 the problem
  \begin{equation} \label{STOKES P OPERATOR}
  A w = B^{*} s
  \end{equation}
- with sufficient accuracy to return $q=Bw$. We assume that the choicen tolerance is
+ with sufficient accuracy to return $q=Bw$. We assume that the desired tolerance is
- sufficiently small, for instane one can take $\tau\hackscore{2}^2$
+ sufficiently small, for instance one can take $\tau\hackscore{2}^2$
  where $\tau\hackscore{2}$ is the tolerance for~\ref{STOKES ITER STEP 2}.
  In an implementation we use the fact that the residual $r$ is given as
 Line 155 
 setting the errors to zero we can write
  \begin{equation}
  v = \Phi(v,p) \mbox{ and } p = \Psi(u,p)
  \end{equation}
- with suitable functions $\Phi(v,p)$ and $ \Psi(v,p)$ representing the itartion operator without
+ with suitable functions $\Phi(v,p)$ and $ \Psi(v,p)$ representing the iteration operator without
  errors. In fact for a linear problem,  $\Phi$ and $\Psi$ are constant. With this notation we can write
  the update step in the form $p\hackscore{2}= \delta p + \Psi(v\hackscore{0},p\hackscore{0})$ and
  $v\hackscore{2}= \delta v + \Phi(v\hackscore{0},p\hackscore{0})$ where
 Line 201 
 So we choose the value for $\tau\hacksco
   M \tau\hackscore{2} \|B v\hackscore{1}\|\hackscore{0}  \le (\chi^{-})^2 \epsilon^{-}
  \end{equation}
  in order to make the perturbation for the termination of the pressure iteration a second order effect. We use a
- similar argument for the velcity:
+ similar argument for the velocity:
  \begin{equation}\label{STOKES EST 2}
  \begin{array}{rcl}
  \|v\hackscore{2}-v\|\hackscore{1} & \le & \|v\hackscore{2}-\delta v-v\|\hackscore{1} + \| \delta v\|\hackscore{1} \\
 Line 216 
 So we choose the value for $\tau\hacksco
  \begin{equation} \label{STOKES TOL1}
  K \tau\hackscore{1} \le \chi^{-}
  \end{equation}
- Assuming we have etsimates for $M$ and $K$ (if none is available, we use the value $1$.)
+ Assuming we have estimates for $M$ and $K$ (if none is available, we use the value $1$.)
  we can use~\ref{STOKES TOL1} and~\ref{STOKES TOL2} to get appropriate values for the tolerances. After
  the step has been completed we can calculate a new convergence rate $\chi =\frac{\epsilon}{\epsilon^{-}}$.
- For partical reasons we restrict $\chi$ to be less or equal a given maximum value $\chi\hackscore{max}\le 1$.
+ For partial reasons we restrict $\chi$ to be less or equal a given maximum value $\chi\hackscore{max}\le 1$.
  If we see $\chi \le \chi^{-} (1+\chi^{-})$ our choices for the tolerances was suitable. Otherwise, we need to adjust the values for $K$ and $M$. From the estimates~\ref{STOKES EST 1} and~\ref{STOKES EST 2} we establish
  \begin{equation}\label{STOKES EST 3}
  \chi \le ( 1 + \max(M \frac{\tau\hackscore{2} \|B v\hackscore{1}\|\hackscore{0}}{\chi^{-} \epsilon^{-}},K \tau\hackscore{1}  ) ) \cdot \chi^{-}
 Line 230 
 $M^{+}$ and $K^{+}$ then we get
  \chi =  ( 1 + \max(M^{+} \frac{\chi^{-}}{M},K^{+} \frac{\chi^{-}}{K}) ) \cdot \chi^{-}
  \end{equation}
  From this equation we set for
- than our choice for $K$ was not good anough. In this case we can calaculate a new value
+ than our choice for $K$ was not good enough. In this case we can calculate a new value
   \begin{equation}
  K^{+} =  \frac{\chi-\chi^{-}}{(\chi^{-})^2} K
  \end{equation}
 Line 274 
 the solution procedure is implemented as
    \begin{enumerate}
    \item update $p\hackscore{2}\leftarrow p$ and $v\hackscore{2}\leftarrow v\hackscore{1}$
     \end{enumerate}
-    \item calculate convergence messure $\epsilon$ and convergence rate $\chi$
+    \item calculate convergence measure $\epsilon$ and convergence rate $\chi$
- \item if stopping criterium~\ref{STOKES STOPPING CRITERIA} holds:
+ \item if stopping criteria~\ref{STOKES STOPPING CRITERIA} holds:
   \begin{enumerate}
   \item return $v\hackscore{2}$ and $p\hackscore{2}$
   \end{enumerate}
 Line 291 
 the solution procedure is implemented as
  \begin{classdesc}{StokesProblemCartesian}{domain}
  opens the Stokes problem\index{Stokes problem} on the \Domain domain. The domain
- needs to support LBB compliant elements for the Stokes problem, see~\cite{LBB} for detials~\index{LBB condition}.
+ needs to support LBB compliant elements for the Stokes problem, see~\cite{LBB} for details~\index{LBB condition}.
  For instance one can use second order polynomials for velocity and
- first order polynomials for the pressure on the same element. Alternativly, one can use
+ first order polynomials for the pressure on the same element. Alternatively, one can use
- macro elements\index{macro elements} using linear polynomial for both pressure and velocity bu with a subdivided
+ macro elements\index{macro elements} using linear polynomial for both pressure and velocity    with a subdivided
  element for the velocity. Typically, the macro element is more cost effective. The fact that pressure and velocity are represented in different way is expressed by
  \begin{python}
  velocity=Vector(0.0, Solution(mesh))
 Line 307 
 restoration_factor=0}}}}}}
  assigns values to the model parameters. In any call all values must be set.
  \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$,
  \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$.
- The locations and compontents where the velocity is fixed are set by
+ The locations and components where the velocity is fixed are set by
  the values of \var{fixed_u_mask}. \var{restoration_factor} defines the restoring force factor $\alpha$.
  The method will try to cast the given values to appropriate
  \Data class objects.
 Line 338 
 returns the current relative tolerance.
  \end{methoddesc}
  \begin{methoddesc}[StokesProblemCartesian]{setAbsoluteTolerance}{\optional{tolerance=0.}}
  sets the absolute tolerance for the error in the relevant norm. The tolerance must be non-negative. Typically the
- absolute talerance is set to 0.
+ absolute tolerance is set to 0.
  \end{methoddesc}
  \begin{methoddesc}[StokesProblemCartesian]{getAbsoluteTolerance}{}
- sreturns the current absolute tolerance.
+ returns the current absolute tolerance.
  \end{methoddesc}
  \begin{methoddesc}[StokesProblemCartesian]{getSolverOptionsVelocity}{}

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