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-revision 3324 by jfenwick,
Fri Oct 22 01:56:02 2010 UTC
+revision 3325 by caltinay,
Fri Oct 29 00:30:51 2010 UTC
 Line 1
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ %
+ % Copyright (c) 2003-2010 by University of Queensland
+ % Earth Systems Science Computational Center (ESSCC)
+ % http://www.uq.edu.au/esscc
+ %
+ % Primary Business: Queensland, Australia
+ % Licensed under the Open Software License version 3.0
+ % http://www.opensource.org/licenses/osl-3.0.php
+ %
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  \section{The Stokes Problem}
  \label{STOKES PROBLEM}
- In this section we discuss how to solve the Stokes problem which is defined as follows:
+ In this section we discuss how to solve the Stokes problem.
+ We want to calculate the velocity\index{velocity} field $v$ and pressure $p$
- We want to calculate the velocity \index{velocity} field $v$ and pressure $p$ of an incompressible fluid \index{incompressible fluid}. They are given as the solution of the Stokes problem\index{Stokes problem}
+ of an incompressible fluid\index{incompressible fluid}.
+ They are given as the solution of the Stokes problem\index{Stokes problem}
  \begin{equation}\label{Stokes 1}
  -\left(\eta(v_{i,j}+ v_{j,i})\right)_{,j}+p_{,i}=f_{i}-\sigma_{ij,j}
  \end{equation}
- where  $f_{i}$ defines an internal force \index{force, internal} and $\sigma_{ij}$ 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.
+ where  $f_{i}$ defines an internal force\index{force, internal} and
+ $\sigma_{ij}$ 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:
+ We assume an incompressible medium:
  \begin{equation}\label{Stokes 2}
  -v_{i,i}=0
  \end{equation}
-Line 20 
 which can be overwritten by constraints
+Line 37 
 which can be overwritten by constraints
  \begin{equation}\label{Stokes Boundary0}
  v_{i}(x)=v^D_{i}(x)
  \end{equation}
- at some locations $x$ at the boundary of the domain. $s_{i}$ defines a normal stress and
+ at some locations $x$ at the boundary of the domain.
- $\alpha\ge 0$ the spring constant for restoring normal force.
+ $s_{i}$ defines a normal stress and $\alpha\ge 0$ the spring constant for
+ restoring normal force.
  The index $i$ may depend on the location $x$ on the boundary.
  $v^D$ is a given function on the domain.
  \subsection{Solution Method \label{STOKES SOLVE}}
- If we assume that $\eta$ is independent from the velocity and pressure equations~\ref{Stokes 1} and~\ref{Stokes 2}
+ If we assume that $\eta$ is independent from the velocity and pressure,
- can be written in the block form
+ equations~\ref{Stokes 1} and~\ref{Stokes 2} can be written in the block form
  \begin{equation}
  \left[ \begin{array}{cc}
  A     & B^{*} \\
-Line 43 
 G \\
+Line 61 
 G \\
  \end{array} \right]
  \label{STOKES}
  \end{equation}
- where $A$ is coercive, self-adjoint linear operator in a suitable Hilbert space, $B$ is the $(-1) \cdot$ divergence operator and $B^{*}$ is it adjoint operator (=gradient operator).
+ where $A$ is a coercive, self-adjoint linear operator in a suitable Hilbert
- For more details on the mathematics see references \cite{AAMIRBERKYAN2008,MBENZI2005}.
+ space, $B$ is the $(-1) \cdot$ divergence operator and $B^{*}$ is the adjoint
+ operator (=gradient operator).
- If $v_{0}$ and $p_{0}$ are given initial guesses for
+ For more details on the mathematics see references \cite{AAMIRBERKYAN2008,MBENZI2005}.
- velocity and pressure we calculate a correction $dv$ for the velocity by solving the first
- equation of equation~\ref{STOKES}
+ If $v_{0}$ and $p_{0}$ are given initial guesses for velocity and pressure we
+ calculate a correction $dv$ for the velocity by solving the first equation of
+ \eqn{STOKES}
   \begin{equation}\label{STOKES ITER STEP 1}
   A dv_{1} = G - A v_{0} - B^{*} p_{0}
  \end{equation}
- We then insert the new approximation $v_{1}=v_{0}+dv_{1}$ to calculate a correction $dp_{2}$
+ We then insert the new approximation $v_{1}=v_{0}+dv_{1}$ to calculate a
- for the pressure and an additional correction $dv_{2}$ for the velocity by solving
+ correction $dp_{2}$ for the pressure and an additional correction $dv_{2}$ for
+ the velocity by solving
   \begin{equation}
   \begin{array}{rcl}
   B A^{-1} B^{*} dp_{2} & = & Bv_{1} \\
-Line 63 
 for the pressure and an additional corre
+Line 84 
 for the pressure and an additional corre
   \end{equation}
  The new velocity and pressure are then given by $v_{2}=v_{1}-dv_{2}$ and
  $p_{2}=p_{0}+dp_{2}$ which will fulfill the block system~\ref{STOKES}.
- This solution strategy is called the Uzawa scheme \index{Uzawa scheme}.
+ 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
+ There is a problem with this scheme: in practice we will use an iterative
- to solve any problem for operator $A$. So we will be unable to calculate the operator
+ scheme to solve any problem for operator $A$.
- $ B A^{-1} B^{*}$ required for $dp_{2}$ explicitly. In fact, we need to use another
+ So we will be unable to calculate the operator $ B A^{-1} B^{*}$ required for
- iterative scheme to solve the first equation in~\ref{STOKES ITER STEP 2} where in each iteration step
+ $dp_{2}$ explicitly. In fact, we need to use another iterative scheme to solve
+ the first equation in~\ref{STOKES ITER STEP 2} where in each iteration step
  an iterative solver for $A$ is applied. Another issue is the fact that the
- viscosity $\eta$ may depend on velocity or pressure and so we need to iterate over the
+ viscosity $\eta$ may depend on velocity or pressure and so we need to iterate
- three equations~\ref{STOKES ITER STEP 1} and~\ref{STOKES ITER STEP 2}.
+ over the three equations~\ref{STOKES ITER STEP 1} and~\ref{STOKES ITER STEP 2}.
  In the following we will use the two norms
  \begin{equation}
-Line 80 
 In the following we will use the two nor
+Line 102 
 In the following we will use the two nor
  \|p\|_{0}^2= \int_{\Omega} p^2 \; dx.
  \label{STOKES STOP}
  \end{equation}
- for velocity $v$ and pressure $p$. The iteration is terminated if the stopping criteria
+ for velocity $v$ and pressure $p$.
+ The iteration is terminated if the stopping criterion
   \begin{equation} \label{STOKES STOPPING CRITERIA}
  \max(\|Bv_{1}\|_{0},\|v_{2}-v_{0}\|_{1}) \le \tau \cdot \|v_{2}\|_{1}
   \end{equation}
-  for a given  given tolerance $0<\tau<1$ is meet. Notice that because of the first equation of~\ref{STOKES ITER STEP 2} we have that
+ for a given tolerance $0<\tau<1$ is met.
- $\|Bv_{1}\|_{0}$ equals the
+ Notice that because of the first equation of~\ref{STOKES ITER STEP 2} we have
- norm of $B A^{-1} B^{*} dp_{2}$ and consequently provides a norm for the pressure correction.
+ that $\|Bv_{1}\|_{0}$ equals the norm of $B A^{-1} B^{*} dp_{2}$ and
+ consequently provides a norm for the pressure correction.
  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
+ and~\ref{STOKES ITER STEP 2}. To do this we write the iteration scheme as a
- we consider the errors produced by the iterative solvers being used.
+ fixed point problem. Here we consider the errors produced by the iterative
- From Equation~\ref{STOKES ITER STEP 1} we have
+ solvers being used.
+ From \eqn{STOKES ITER STEP 1} we have
  \begin{equation} \label{STOKES total V1}
  v_{1} = e_{1} + v_{0} + A^{-1} ( G - Av_{0} - B^{*} p_{0} )
  \end{equation}
- where $ e_{1}$ is the error when solving~\ref{STOKES ITER STEP 1}.
+ where $e_{1}$ is the error when solving~\ref{STOKES ITER STEP 1}.
- We will use a sparse matrix solver so we have not full control on the norm $\|.\|_{s}$ used in the stopping criteria for this equation. In fact we will have a stopping criteria of the form
+ We will use a sparse matrix solver so we have not full control on the norm
+ $\|.\|_{s}$ used in the stopping criterion for this equation.
+ In fact we will have a stopping criterion of the form
  \begin{equation}
  \| A e_{1} \|_{s}  = \| G - A v_{1} - B^{*} p_{0} \|_{s} \le \tau_{1} \| G - A v_{0} - B^{*} p_{0} \|_{s}
  \end{equation}
- where $\tau_{1}$ is the tolerance which we need to choose. This translates into the condition
+ where $\tau_{1}$ is the tolerance which we need to choose.
+ This translates into the condition
  \begin{equation}
  \| e_{1} \|_{1} \le K \tau_{1} \| dv_{1} - e_{1} \|_{1}
  \end{equation}
- The constant $K$ represents some uncertainty combining a variety of unknown factors such as the
+ The constant $K$ represents some uncertainty combining a variety of unknown
- norm being used and the condition number of the stiffness matrix.
+ factors such as the norm being used and the condition number of the stiffness matrix.
  From the first equation of~\ref{STOKES ITER STEP 2} we have
  \begin{equation}\label{STOKES total P2}
  p_{2} =  p_{0} + (B A^{-1} B^{*})^{-1} (e_{2} + Bv_{1} )
  \end{equation}
  where $e_{2}$ represents the error when solving~\ref{STOKES ITER STEP 2}.
- We use an iterative preconditioned conjugate gradient method (PCG) \index{linear solver!PCG}\index{PCG} with iteration
+ We use an iterative preconditioned conjugate gradient method
- operator $B A^{-1} B^{*}$ using the $\|.\|_{0}$ norm. As suitable preconditioner \index{preconditioner} for the iteration
+ (PCG)\index{linear solver!PCG}\index{PCG} with iteration operator
- operator we use $\frac{1}{\eta}$ \cite{ELMAN}, ie
+ $B A^{-1} B^{*}$ using the $\|.\|_{0}$ norm.
- the evaluation of the preconditioner $P$ for a given pressure increment $q$ is the solution of
+ As suitable preconditioner\index{preconditioner} for the iteration operator we
+ use $\frac{1}{\eta}$ \cite{ELMAN}, i.e. the evaluation of the preconditioner
+ $P$ for a given pressure increment $q$ is the solution of
  \begin{equation} \label{STOKES P PREC}
  \frac{1}{\eta} (Pq) = q \; .
  \end{equation}
- Note that in each evaluation of the iteration operator $q=B A^{-1} B^{*} s$ one needs to solve
+ Note that in each evaluation of the iteration operator $q=B A^{-1} B^{*} s$
- the problem
+ one needs to solve 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 desired tolerance is
+ with sufficient accuracy to return $q=Bw$. We assume that the desired
- sufficiently small, for instance one can take $\tau_{2}^2$
+ tolerance is sufficiently small, for instance one can take $\tau_{2}^2$ where
- where $\tau_{2}$ is the tolerance for~\ref{STOKES ITER STEP 2}.
+ $\tau_{2}$ is the tolerance for~\ref{STOKES ITER STEP 2}.
  In an implementation we use the fact that the residual $r$ is given as
  \begin{equation} \label{STOKES RES }
   r= B (v_{1} -  A^{-1} B^{*} dp) =  B (v_{1} - A^{-1} B^{*} dp) = B (v_{1}-dv_{2}) = B v_{2}
  \end{equation}
- In particular we have $e_{2} = B v_{2}$
+ In particular we have $e_{2} = B v_{2}$.
- So the residual $r$ is represented by the updated velocity $v_{2}=v_{1}-dv_{2}$. In practice, if
+ So the residual $r$ is represented by the updated velocity $v_{2}=v_{1}-dv_{2}$.
- one uses the velocity to represent the residual $r$ there is no need
+ In practice, if one uses the velocity to represent the residual $r$ there is
- to recover $dv_{2}$ in~\ref{STOKES ITER STEP 2} after $dp_{2}$ has been calculated.
+ no need to recover $dv_{2}$ in~\ref{STOKES ITER STEP 2} after $dp_{2}$ has
+ been calculated.
  In PCG the iteration is terminated if
  \begin{equation} \label{STOKES P OPERATOR ERROR}
  \| P^{\frac{1}{2}}B v_{2} \|_{0} \le \tau_{2} \| P^{\frac{1}{2}}B v_{1} \|_{0}
-Line 143 
 where $\tau_{2}$ is the given tolerance.
+Line 174 
 where $\tau_{2}$ is the given tolerance.
  \begin{equation} \label{STOKES P OPERATOR ERROR 2}
  \|e_{2}\|_{0} = \| B v_{2} \|_{0} \le M \tau_{2} \| B v_{1} \|_{0}
  \end{equation}
  where $M$ is taking care of the fact that $P^{\frac{1}{2}}$ is dropped.
- As we assume that there is no significant error from solving with the operator $A$ we have
+ As we assume that there is no significant error from solving with the operator
+ $A$ we have
  \begin{equation} \label{STOKES total V2}
  v_{2} =  v_{1} - dv_{2}
  = v_{1}  - A^{-1} B^{*}dp
  \end{equation}
- Combining the equations~\ref{STOKES total V1},~\ref{STOKES total P2} and~\ref{STOKES total V2} and
+ Combining the equations~\ref{STOKES total V1},~\ref{STOKES total P2} and~\ref{STOKES total V2}
- setting the errors to zero we can write the solution process as a fix point problem
+ and setting the errors to zero we can write the solution process as a fix
+ point problem
  \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 iteration operator without
+ with suitable functions $\Phi(v,p)$ and $ \Psi(v,p)$ representing the
- errors. In fact for a linear problem,  $\Phi$ and $\Psi$ are constant. With this notation we can write
+ iteration operator without errors. In fact, for a linear problem, $\Phi$ and
- the update step in the form $p_{2}= \delta p + \Psi(v_{0},p_{0})$ and
+ $\Psi$ are constant. With this notation we can write the update step in the
- $v_{2}= \delta v + \Phi(v_{0},p_{0})$ where
+ form $p_{2}= \delta p + \Psi(v_{0},p_{0})$ and $v_{2}= \delta v + \Phi(v_{0},p_{0})$
- the total error $\delta p$ and $\delta v$ are given as
+ where the total error $\delta p$ and $\delta v$ are given as
  \begin{equation}
   \begin{array}{rcl}
  \delta p & = &  (B A^{-1} B^{*})^{-1} ( e_{2} + B e_{1} ) \\
  \delta v & = &  e_{1} -  A^{-1} B^{*}\delta p  \;.
  \end{array}\label{STOKES ERRORS}
  \end{equation}
- Notice that $B\delta v = - e_{2}=-Bv_{2}$. Our task is now to choose the tolerances
+ Notice that $B\delta v = - e_{2}=-Bv_{2}$.
- $\tau_{1}$ and $\tau_{2}$ such that the global errors $\delta p$ and $\delta v$
+ Our task is now to choose the tolerances $\tau_{1}$ and $\tau_{2}$ such that
- do not stop the convergence of the iteration process.
+ the global errors $\delta p$ and $\delta v$ do not stop the convergence of the
+ iteration process.
  To measure convergence we use
  \begin{equation}
-Line 180 
 $v_{0}$ we will use the estimate
+Line 214 
 $v_{0}$ we will use the estimate
  \begin{equation}
  \epsilon = \max(\|v_{2}-v_{0}\|_{1}, \|B v_{1}\|_{0})
  \end{equation}
  to estimate the progress of the iteration step after the step is completed.
- Note that the estimate of $\epsilon$
+ Note that the estimate of $\epsilon$ is used in the stopping
- used in the stopping criteria~\ref{STOKES STOPPING CRITERIA}. If $\chi^{-}$ is the convergence rate assuming
+ criterion~\ref{STOKES STOPPING CRITERIA}.
- exact calculations, i.e. $e_{1}=0$ and $e_{2}=0$, we are expecting
+ If $\chi^{-}$ is the convergence rate assuming exact calculations, i.e.
- to maintain $\epsilon \le \chi^{-} \cdot \epsilon^{-}$. For the
+ $e_{1}=0$ and $e_{2}=0$, we are expecting to maintain $\epsilon \le \chi^{-} \cdot \epsilon^{-}$.
- pressure increment we get:
+ For the pressure increment we get:
  \begin{equation} \label{STOKES EST 1}
  \begin{array}{rcl}
  \|B A^{-1} B^{*} (p_{2}-p)\|_{0}
-Line 196 
 pressure increment we get:
+Line 230 
 pressure increment we get:
  & \le & \chi^{-} \cdot \epsilon^{-} + M \tau_{2} \|B v_{1}\|_{0} \\
  \end{array}
  \end{equation}
  So we choose the value for $\tau_{2}$ from
  \begin{equation} \label{STOKES TOL2}
   M \tau_{2} \|B v_{1}\|_{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
+ in order to make the perturbation for the termination of the pressure
- similar argument for the velocity:
+ iteration a second order effect. We use a similar argument for the velocity:
  \begin{equation}\label{STOKES EST 2}
  \begin{array}{rcl}
  \|v_{2}-v\|_{1} & \le & \|v_{2}-\delta v-v\|_{1} + \| \delta v\|_{1} \\
-Line 216 
 So we choose the value for $\tau_{1}$ fr
+Line 250 
 So we choose the value for $\tau_{1}$ fr
  \begin{equation} \label{STOKES TOL1}
  K \tau_{1} \le \chi^{-}
  \end{equation}
- Assuming we have estimates for $M$ and $K$ (if none is available, we use the value $1$.)
+ Assuming we have estimates for $M$ and $K$\footnote{if no estimates are
- we can use~\ref{STOKES TOL1} and~\ref{STOKES TOL2} to get appropriate values for the tolerances. After
+ available, we use the value $1$} we can use~\ref{STOKES TOL1} and
- the step has been completed we can calculate a new convergence rate $\chi =\frac{\epsilon}{\epsilon^{-}}$.
+ \ref{STOKES TOL2} to get appropriate values for the tolerances.
- For partial reasons we restrict $\chi$ to be less or equal a given maximum value $\chi_{max}\le 1$.
+ After the step has been completed we can calculate a new convergence rate
- 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
+ $\chi =\frac{\epsilon}{\epsilon^{-}}$.
+ For partial reasons we restrict $\chi$ to be less or equal a given maximum
+ value $\chi_{max}\le 1$.
+ If we see $\chi \le \chi^{-} (1+\chi^{-})$ our choices for the tolerances were
+ 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_{2} \|B v_{1}\|_{0}}{\chi^{-} \epsilon^{-}},K \tau_{1}  ) ) \cdot \chi^{-}
  \end{equation}
- If we assume that this inequality would be an equation if we would have chosen the right values
+ If we assume that this inequality would be an equation if we would have chosen
- $M^{+}$ and $K^{+}$ then we get
+ the right values $M^{+}$ and $K^{+}$ then we get
  \begin{equation}\label{STOKES EST 3b}
  \chi =  ( 1 + \max(M^{+} \frac{\chi^{-}}{M},K^{+} \frac{\chi^{-}}{K}) ) \cdot \chi^{-}
  \end{equation}
- From this equation we set for
+ From this equation we see if our choice for $K$ was not good enough.
- than our choice for $K$ was not good enough. In this case we can calculate a new value
+ In this case we can calculate a new value
   \begin{equation}
  K^{+} =  \frac{\chi-\chi^{-}}{(\chi^{-})^2} K
  \end{equation}
  In practice we will use
   \begin{equation}
  K^{+}  = \max(\frac{\chi-\chi^{-}}{(\chi^{-})^2} K,\frac{1}{2}K,1)
  \end{equation}
  where the second term is used to reduce a potential overestimate of $K$.
  The same identity is used for to update $M$. The updated $M^{+}$ and $K^{+}$
  are then use in the next iteration step to control the tolerances.
- In some cases one can observe that there is a significant change
+ In some cases one can observe that there is a significant change in the
- in the velocity but the new velocity $v_{1}$ has still a
+ velocity but the new velocity $v_{1}$ has still a small divergence, i.e.
- small divergence, i.e. we have
+ we have $\|Bv_{1}\|_{0} \ll \|v_{1}-v_{0}\|_{1}$.
- $\|Bv_{1}\|_{0} \ll \|v_{1}-v_{0}\|_{1}$.
+ In this case we will get a small pressure increment and consequently only very
- In this case we will get a small pressure increment and consequently only very small changes to
+ small changes to the velocity as a result of the second update step which
- the velocity as a result of the second update step which therefore can be skipped and
+ therefore can be skipped and we can directly repeat the first update step
- we can directly repeat the first update step until the increment in velocity becomes
+ until the increment in velocity becomes significant relative to its divergence.
- significant relative to its divergence. In practice we will ignore the second half of the iteration step
+ In practice we will ignore the second half of the iteration step as long as
- as long as
   \begin{equation}\label{STOKES LARGE BV1}
  \|Bv_{1}\|_{0} \le \theta \cdot \|v_{1}-v_{0}\|
  \end{equation}
- where $0<\theta<1$ is a given factor. In this case we will also check the stopping criteria
+ where $0<\theta<1$ is a given factor. In this case we will also check the
- with $v_{1}\rightarrow v_{2}$ but we will not correct $M$ in this case.
+ stopping criterion with $v_{1}\rightarrow v_{2}$ but we will not correct $M$
+ in this case.
  Starting from an initial guess $v_{0}$ and $p_{0}$ for velocity and pressure
- the solution procedure is implemented as follows.
+ the solution procedure is implemented as follows:
  \begin{enumerate}
-  \item calculate viscosity $\eta(v_{0},p)_{0}$ and assemble operator $A$ from $\eta$.
+  \item calculate viscosity $\eta(v_{0},p)_{0}$ and assemble operator $A$ from $\eta$
-  \item calculate the tolerance $\tau_{1}$ from~\ref{STOKES TOL1}.
+  \item calculate the tolerance $\tau_{1}$ from \eqn{STOKES TOL1}
-  \item Solve~\ref{STOKES ITER STEP 1} for $dv_{1}$ with tolerance $\tau_{1}$.
+  \item solve \eqn{STOKES ITER STEP 1} for $dv_{1}$ with tolerance $\tau_{1}$
   \item update $v_{1}= v_{0}+ dv_{1}$
   \item if $Bv_{1}$ is large (see~\ref{STOKES LARGE BV1}):
   \begin{enumerate}
-  \item calculate the tolerance $\tau_{2}$ from~\ref{STOKES TOL2}.
+    \item calculate the tolerance $\tau_{2}$ from~\ref{STOKES TOL2}
-  \item Solve~\ref{STOKES ITER STEP 2} for $dp_{2}$ and $v_{2}$ with tolerance $\tau_{2}$.
+    \item solve~\ref{STOKES ITER STEP 2} for $dp_{2}$ and $v_{2}$ with tolerance $\tau_{2}$
-  \item update $p_{2}\leftarrow p_{0}+ dp_{2}$
+    \item update $p_{2}\leftarrow p_{0}+ dp_{2}$
   \end{enumerate}
   \item else:
-   \begin{enumerate}
+   \begin{itemize}
-   \item update $p_{2}\leftarrow p$ and $v_{2}\leftarrow v_{1}$
+     \item update $p_{2}\leftarrow p$ and $v_{2}\leftarrow v_{1}$
-    \end{enumerate}
+   \end{itemize}
-    \item calculate convergence measure $\epsilon$ and convergence rate $\chi$
+  \item calculate convergence measure $\epsilon$ and convergence rate $\chi$
- \item if stopping criteria~\ref{STOKES STOPPING CRITERIA} holds:
+  \item if stopping criterion~\ref{STOKES STOPPING CRITERIA} holds:
-  \begin{enumerate}
+  \begin{itemize}
-  \item return $v_{2}$ and $p_{2}$
+    \item return $v_{2}$ and $p_{2}$
-  \end{enumerate}
+  \end{itemize}
   \item else:
   \begin{enumerate}
+    \item update $M$ and $K$
-      \item update $M$ and $K$
+    \item goto step 1 with $v_{0}\leftarrow v_{2}$ and $p_{0}\leftarrow p_{2}$.
-      \item goto step 1 with $v_{0}\leftarrow v_{2}$ and $p_{0}\leftarrow p_{2}$.
+  \end{enumerate}
- \end{enumerate}
  \end{enumerate}
  \subsection{Functions}
  \begin{classdesc}{StokesProblemCartesian}{domain}
- opens the Stokes problem\index{Stokes problem} on the \Domain domain. The domain
+ opens the Stokes problem\index{Stokes problem} on the \Domain domain.
- needs to support LBB compliant elements for the Stokes problem, see~\cite{LBB} for details~\index{LBB condition}.
+ The domain 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
+ For instance one can use second order polynomials for velocity and first order
- first order polynomials for the pressure on the same element. Alternatively, one can use
+ polynomials for the pressure on the same element.
- macro elements\index{macro elements} using linear polynomial for both pressure and velocity    with a subdivided
+ Alternatively, one can use macro elements\index{macro elements} using linear
- 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
+ polynomials 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 ways is
+ expressed by
  \begin{python}
- velocity=Vector(0.0, Solution(mesh))
+   velocity=Vector(0.0, Solution(mesh))
- pressure=Scalar(0.0, ReducedSolution(mesh))
+   pressure=Scalar(0.0, ReducedSolution(mesh))
  \end{python}
  \end{classdesc}
- \begin{methoddesc}[StokesProblemCartesian]{initialize}{\optional{f=Data(), \optional{fixed_u_mask=Data(), \optional{eta=1, \optional{surface_stress=Data(), \optional{stress=Data()}, \optional{
+ \begin{methoddesc}[StokesProblemCartesian]{initialize}{\optional{f=Data(),
- restoration_factor=0}}}}}}
+     \optional{fixed_u_mask=Data(), \optional{eta=1,%
+     \optional{surface_stress=Data(), \optional{stress=Data()},%
+     \optional{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 components where the velocity is fixed are set by
+ The locations and components where the velocity is fixed are set by the values
- the values of \var{fixed_u_mask}. \var{restoration_factor} defines the restoring force factor $\alpha$.
+ of \var{fixed_u_mask}.
- The method will try to cast the given values to appropriate
+ \var{restoration_factor} defines the restoring force factor $\alpha$.
- \Data class objects.
+ The method will try to cast the given values to appropriate \Data class objects.
  \end{methoddesc}
  \begin{methoddesc}[StokesProblemCartesian]{solve}{v,p
  \optional{, max_iter=100 \optional{, verbose=False \optional{, usePCG=True }}}}
- solves the problem and return approximations for velocity and pressure.
+ solves the problem and returns approximations for velocity and pressure.
- The arguments \var{v} and \var{p} define initial guess.
+ The arguments \var{v} and \var{p} define initial guesses.
- \var{v} must have function space \var{Solution(domain)} and
+ \var{v} must have function space \var{Solution(domain)} and \var{p} must have
- \var{p} must have function space \var{ReducedSolution(domain)}.
+ function space \var{ReducedSolution(domain)}.
- The values of \var{v} marked
+ The values of \var{v} marked by \var{fixed_u_mask} remain unchanged.
- by \var{fixed_u_mask} remain unchanged.
+ If \var{usePCG} is set to \True then the preconditioned conjugate gradient
- If \var{usePCG} is set to \True
+ method (PCG)\index{preconditioned conjugate gradient method!PCG} scheme is
- reconditioned conjugate gradient method (PCG) \index{preconditioned conjugate gradient method!PCG}  scheme is used. Otherwise the problem is solved generalized minimal residual method (GMRES) \index{generalized minimal residual method!GMRES}. In most cases
+ used. Otherwise the problem is solved with the generalized minimal residual
- the PCG scheme is more efficient.
+ method (GMRES)\index{generalized minimal residual method!GMRES}.
- \var{max_iter} defines the maximum number of iteration steps.
+ In most cases the PCG scheme is more efficient.
+ \var{max_iter} defines the maximum number of iteration steps.
- If \var{verbose} is set to \True informations on the progress of of the solver are printed.
+ If \var{verbose} is set to \True information on the progress of of the solver
+ is printed.
  \end{methoddesc}
  \begin{methoddesc}[StokesProblemCartesian]{setTolerance}{\optional{tolerance=1.e-4}}
- sets the tolerance in an appropriate norm relative to the right hand side. The tolerance must be non-negative and less than 1.
+ sets the tolerance in an appropriate norm relative to the right hand side.
+ The tolerance must be non-negative and less than 1.
  \end{methoddesc}
  \begin{methoddesc}[StokesProblemCartesian]{getTolerance}{}
  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
+ sets the absolute tolerance for the error in the relevant norm.
- absolute tolerance is set to 0.
+ The tolerance must be non-negative.
+ Typically the absolute tolerance is set to 0.
  \end{methoddesc}
  \begin{methoddesc}[StokesProblemCartesian]{getAbsoluteTolerance}{}
-Line 346 
 returns the current absolute tolerance.
+Line 393 
 returns the current absolute tolerance.
  \end{methoddesc}
  \begin{methoddesc}[StokesProblemCartesian]{getSolverOptionsVelocity}{}
- returns the solver options used  solve the equations~(\ref{STOKES ITER STEP 1}) and~(\ref{STOKES P OPERATOR}) for velocity.
+ returns the solver options used to solve \eqn{STOKES ITER STEP 1} and \eqn{STOKES P OPERATOR}) for velocity.
  \end{methoddesc}
  \begin{methoddesc}[StokesProblemCartesian]{getSolverOptionsPressure}{}
- returns the solver options used solve the preconditioner equation~(\ref{STOKES P PREC}) for pressure.
+ returns the solver options used to solve the preconditioner \eqn{STOKES P PREC} for pressure.
  \end{methoddesc}
  \begin{methoddesc}[StokesProblemCartesian]{getSolverOptionsDiv}{}
- set the solver options for solving the equation to project the divergence of the velocity onto the function space of pressure.
+ sets the solver options for solving the equation to project the divergence of
+ the velocity onto the function space of pressure.
  \end{methoddesc}
+ \subsection{Example: Lid-driven Cavity}
- \subsection{Example: Lit Driven Cavity}
+ The following script \file{lid_driven_cavity.py}\index{scripts!\file{lid_driven_cavity.py}}
-  The following script \file{lit_driven_cavity.py}
+ which is available in the \ExampleDirectory illustrates the usage of the
- \index{scripts!\file{helmholtz.py}} which is available in the \ExampleDirectory
+ \class{StokesProblemCartesian} class to solve the lid-driven cavity problem:
- illustrates the usage of the \class{StokesProblemCartesian} class to solve
- the lit driven cavity problem:
  \begin{python}
- from esys.escript import *
+   from esys.escript import *
- from esys.finley import Rectangle
+   from esys.finley import Rectangle
- from esys.escript.models import StokesProblemCartesian
+   from esys.escript.models import StokesProblemCartesian
- NE=25
+   NE=25
- dom = Rectangle(NE,NE,order=2)
+   dom = Rectangle(NE,NE,order=2)
- x = dom.getX()
+   x = dom.getX()
- sc=StokesProblemCartesian(dom)
+   sc=StokesProblemCartesian(dom)
- mask= (whereZero(x[0])*[1.,0]+whereZero(x[0]-1))*[1.,0] + \
+   mask= (whereZero(x[0])*[1.,0]+whereZero(x[0]-1))*[1.,0] + \
-       (whereZero(x[1])*[0.,1.]+whereZero(x[1]-1))*[1.,1]
+         (whereZero(x[1])*[0.,1.]+whereZero(x[1]-1))*[1.,1]
- sc.initialize(eta=.1, fixed_u_mask= mask)
+   sc.initialize(eta=.1, fixed_u_mask=mask)
- v=Vector(0.,Solution(dom))
+   v=Vector(0., Solution(dom))
- v[0]+=whereZero(x[1]-1.)
+   v[0]+=whereZero(x[1]-1.)
- p=Scalar(0.,ReducedSolution(dom))
+   p=Scalar(0.,ReducedSolution(dom))
- v,p=sc.solve(v,p, verbose=True)
+   v,p=sc.solve(v, p, verbose=True)
- saveVTK("u.xml",velocity=v,pressure=p)
+   saveVTK("u.vtu", velocity=v, pressure=p)
  \end{python}

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