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-revision 2851 by gross,
Fri Jan 15 07:37:47 2010 UTC
+revision 3295 by jfenwick,
Fri Oct 22 01:56:02 2010 UTC
 Line 4 
 In this section we discuss how to solve
  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}
  \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}
+ -\left(\eta(v_{i,j}+ v_{j,i})\right)_{,j}+p_{,i}=f_{i}-\sigma_{ij,j}
  \end{equation}
- 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.
+ 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:
  \begin{equation}\label{Stokes 2}
- -v\hackscore{i,i}=0
+ -v_{i,i}=0
  \end{equation}
  Natural boundary conditions are taken in the form
  \begin{equation}\label{Stokes Boundary}
- \left(\eta(v\hackscore{i,j}+ v\hackscore{j,i})\right)n\hackscore{j}-n\hackscore{i}p=s\hackscore{i} - \alpha \cdot n\hackscore{i} n\hackscore{j} v\hackscore{j}+\sigma\hackscore{ij} n\hackscore{j}
+ \left(\eta(v_{i,j}+ v_{j,i})\right)n_{j}-n_{i}p=s_{i} - \alpha \cdot n_{i} n_{j} v_{j}+\sigma_{ij} n_{j}
  \end{equation}
  which can be overwritten by constraints of the form
  \begin{equation}\label{Stokes Boundary0}
- v\hackscore{i}(x)=v^D\hackscore{i}(x)
+ v_{i}(x)=v^D_{i}(x)
  \end{equation}
- at some locations $x$ at the boundary of the domain. $s\hackscore{i}$ defines a normal stress and
+ at some locations $x$ at the boundary of the domain. $s_{i}$ 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.
 Line 46 
 G \\
  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).
  For more details on the mathematics see references \cite{AAMIRBERKYAN2008,MBENZI2005}.
- If $v\hackscore{0}$ and $p\hackscore{0}$ are given initial guesses for
+ 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 equation~\ref{STOKES}
   \begin{equation}\label{STOKES ITER STEP 1}
-  A dv\hackscore{1} = G - A v\hackscore{0} - B^{*} p\hackscore{0}
+  A dv_{1} = G - A v_{0} - B^{*} p_{0}
  \end{equation}
- We then insert the new approximation $v\hackscore{1}=v\hackscore{0}+dv\hackscore{1}$ to calculate a correction $dp\hackscore{2}$
+ We then insert the new approximation $v_{1}=v_{0}+dv_{1}$ to calculate a correction $dp_{2}$
- for the pressure and an additional correction $dv\hackscore{2}$ for the velocity by solving
+ for the pressure and an additional correction $dv_{2}$ for the velocity by solving
   \begin{equation}
   \begin{array}{rcl}
-  B A^{-1} B^{*} dp\hackscore{2} & = & Bv\hackscore{1} \\
+  B A^{-1} B^{*} dp_{2} & = & Bv_{1} \\
-  A dv\hackscore{2} & = & B^{*} dp\hackscore{2}
+  A dv_{2} & = & B^{*} dp_{2}
  \end{array}
   \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
+ The new velocity and pressure are then given by $v_{2}=v_{1}-dv_{2}$ and
- $p\hackscore{2}=p\hackscore{0}+dp\hackscore{2}$ which will fulfill the block system~\ref{STOKES}.
+ $p_{2}=p_{0}+dp_{2}$ which will fulfill the block system~\ref{STOKES}.
  This solution strategy is called the Uzawa scheme \index{Uzawa scheme}.
  There is a problem with this scheme: In practice we will use an iterative scheme
  to solve any problem for operator $A$. So we will be unable to calculate the operator
- $ B A^{-1} B^{*}$ required for $dp\hackscore{2}$ explicitly. In fact, we need to use another
+ $ B A^{-1} B^{*}$ required for $dp_{2}$ 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
 Line 75 
 three equations~\ref{STOKES ITER STEP 1}
  In the following we will use the two norms
  \begin{equation}
- \|v\|\hackscore{1}^2 = \int\hackscore{\Omega} v\hackscore{j,k}v\hackscore{j,k} \; dx
+ \|v\|_{1}^2 = \int_{\Omega} v_{j,k}v_{j,k} \; dx
  \mbox{ and }
- \|p\|\hackscore{0}^2= \int\hackscore{\Omega} p^2 \; dx.
+ \|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
   \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}
+ \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
- $\|Bv\hackscore{1}\|\hackscore{0}$ equals the
+ $\|Bv_{1}\|_{0}$ equals the
- norm of $B A^{-1} B^{*} dp\hackscore{2}$ and consequently provides a norm for the pressure correction.
+ 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
  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} )
+ v_{1} = e_{1} + v_{0} + A^{-1} ( G - Av_{0} - B^{*} p_{0} )
  \end{equation}
- where $ e\hackscore{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 $\|.\|\hackscore{s}$ 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 criteria for this equation. In fact we will have a stopping criteria of the form
  \begin{equation}
- \| A e\hackscore{1} \|\hackscore{s}  = \| G - A v\hackscore{1} - B^{*} p\hackscore{0} \|\hackscore{s} \le \tau\hackscore{1} \| G - A v\hackscore{0} - B^{*} p\hackscore{0} \|\hackscore{s}
+ \| 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 $\tau\hackscore{1}$ 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\hackscore{1} \|\hackscore{1} \le K \tau\hackscore{1} \| dv\hackscore{1} - e\hackscore{1} \|\hackscore{1}
+ \| 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
  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\hackscore{2} =  p\hackscore{0} + (B A^{-1} B^{*})^{-1} (e\hackscore{2} + Bv\hackscore{1} )
+ p_{2} =  p_{0} + (B A^{-1} B^{*})^{-1} (e_{2} + Bv_{1} )
  \end{equation}
- where $e\hackscore{2}$ represents the error when solving~\ref{STOKES ITER STEP 2}.
+ 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
- operator $B A^{-1} B^{*}$ using the $\|.\|\hackscore{0}$ norm. As suitable preconditioner \index{preconditioner} for the iteration
+ operator $B A^{-1} B^{*}$ using the $\|.\|_{0}$ norm. As suitable preconditioner \index{preconditioner} for the iteration
  operator we use $\frac{1}{\eta}$ \cite{ELMAN}, ie
  the evaluation of the preconditioner $P$ for a given pressure increment $q$ is the solution of
  \begin{equation} \label{STOKES P PREC}
 Line 124 
 the problem
  A w = B^{*} s
  \end{equation}
  with sufficient accuracy to return $q=Bw$. We assume that the desired tolerance is
- sufficiently small, for instance one can take $\tau\hackscore{2}^2$
+ sufficiently small, for instance one can take $\tau_{2}^2$
- where $\tau\hackscore{2}$ is the tolerance for~\ref{STOKES ITER STEP 2}.
+ where $\tau_{2}$ 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\hackscore{1} -  A^{-1} B^{*} dp) =  B (v\hackscore{1} - A^{-1} B^{*} dp) = B (v\hackscore{1}-dv\hackscore{2}) = B v\hackscore{2}
+  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 $e\hackscore{2} = B v\hackscore{2}$
+ In particular we have $e_{2} = B v_{2}$
- So the residual $r$ is represented by the updated velocity $v\hackscore{2}=v\hackscore{1}-dv\hackscore{2}$. In practice, if
+ So the residual $r$ is represented by the updated velocity $v_{2}=v_{1}-dv_{2}$. In practice, if
  one uses the velocity to represent the residual $r$ there is no need
- to recover $dv\hackscore{2}$ in~\ref{STOKES ITER STEP 2} after $dp\hackscore{2}$ has been calculated.
+ 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\hackscore{2} \|\hackscore{0} \le \tau\hackscore{2} \| P^{\frac{1}{2}}B v\hackscore{1} \|\hackscore{0}
+ \| P^{\frac{1}{2}}B v_{2} \|_{0} \le \tau_{2} \| P^{\frac{1}{2}}B v_{1} \|_{0}
  \end{equation}
- where $\tau\hackscore{2}$ is the given tolerance. This translates into
+ where $\tau_{2}$ is the given tolerance. This translates into
  \begin{equation} \label{STOKES P OPERATOR ERROR 2}
- \|e\hackscore{2}\|\hackscore{0} = \| B v\hackscore{2} \|\hackscore{0} \le M \tau\hackscore{2} \| B v\hackscore{1} \|\hackscore{0}
+ \|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
  \begin{equation} \label{STOKES total V2}
- v\hackscore{2} =  v\hackscore{1} - dv\hackscore{2}
+ v_{2} =  v_{1} - dv_{2}
- = v\hackscore{1}  - A^{-1} B^{*}dp
+ = 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
  setting the errors to zero we can write the solution process as a fix point problem
 Line 157 
 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
  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
+ the update step in the form $p_{2}= \delta p + \Psi(v_{0},p_{0})$ and
- $v\hackscore{2}= \delta v + \Phi(v\hackscore{0},p\hackscore{0})$ where
+ $v_{2}= \delta v + \Phi(v_{0},p_{0})$ 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\hackscore{2} + B e\hackscore{1} ) \\
+ \delta p & = &  (B A^{-1} B^{*})^{-1} ( e_{2} + B e_{1} ) \\
- \delta v & = &  e\hackscore{1} -  A^{-1} B^{*}\delta p  \;.
+ \delta v & = &  e_{1} -  A^{-1} B^{*}\delta p  \;.
  \end{array}\label{STOKES ERRORS}
  \end{equation}
- Notice that $B\delta v = - e\hackscore{2}=-Bv\hackscore{2}$. Our task is now to choose the tolerances
+ Notice that $B\delta v = - e_{2}=-Bv_{2}$. Our task is now to choose the tolerances
- $\tau\hackscore{1}$ and $\tau\hackscore{2}$ such that the global errors $\delta p$ and $\delta v$
+ $\tau_{1}$ and $\tau_{2}$ such that the global errors $\delta p$ and $\delta v$
  do not stop the convergence of the iteration process.
  To measure convergence we use
  \begin{equation}
- \epsilon = \max(\|v\hackscore{2}-v\|\hackscore{1}, \|B A^{-1} B^{*} (p\hackscore{2}-p)\|\hackscore{0})
+ \epsilon = \max(\|v_{2}-v\|_{1}, \|B A^{-1} B^{*} (p_{2}-p)\|_{0})
  \end{equation}
- In practice using the fact that $B A^{-1} B^{*} (p\hackscore{2}-p\hackscore{0}) = B v\hackscore{1}$
+ In practice using the fact that $B A^{-1} B^{*} (p_{2}-p_{0}) = B v_{1}$
- and assuming that $v\hackscore{2}$ gives a better approximation to the true $v$ than
+ and assuming that $v_{2}$ gives a better approximation to the true $v$ than
- $v\hackscore{0}$ we will use the estimate
+ $v_{0}$ we will use the estimate
  \begin{equation}
- \epsilon = \max(\|v\hackscore{2}-v\hackscore{0}\|\hackscore{1}, \|B v\hackscore{1}\|\hackscore{0})
+ \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 $\epsilon$
  used in the stopping criteria~\ref{STOKES STOPPING CRITERIA}. If $\chi^{-}$ is the convergence rate assuming
- exact calculations, i.e. $e\hackscore{1}=0$ and $e\hackscore{2}=0$, we are expecting
+ exact calculations, i.e. $e_{1}=0$ and $e_{2}=0$, we are expecting
  to maintain $\epsilon \le \chi^{-} \cdot \epsilon^{-}$. For the
  pressure increment we get:
  \begin{equation} \label{STOKES EST 1}
  \begin{array}{rcl}
- \|B A^{-1} B^{*} (p\hackscore{2}-p)\|\hackscore{0}
+ \|B A^{-1} B^{*} (p_{2}-p)\|_{0}
-  & \le & \|B A^{-1} B^{*} (p\hackscore{2}-\delta p-p)\|\hackscore{0} +
+  & \le & \|B A^{-1} B^{*} (p_{2}-\delta p-p)\|_{0} +
- \|B A^{-1} B^{*} \delta p\|\hackscore{0} \\
+ \|B A^{-1} B^{*} \delta p\|_{0} \\
-  & = & \chi^{-} \cdot \epsilon^{-} + \|e\hackscore{2} + B e\hackscore{1}\|\hackscore{0}  \\
+  & = & \chi^{-} \cdot \epsilon^{-} + \|e_{2} + B e_{1}\|_{0}  \\
- & \approx & \chi^{-} \cdot \epsilon^{-} + \|e\hackscore{2}\|\hackscore{0} \\
+ & \approx & \chi^{-} \cdot \epsilon^{-} + \|e_{2}\|_{0} \\
- & \le & \chi^{-} \cdot \epsilon^{-} + M \tau\hackscore{2} \|B v\hackscore{1}\|\hackscore{0} \\
+ & \le & \chi^{-} \cdot \epsilon^{-} + M \tau_{2} \|B v_{1}\|_{0} \\
  \end{array}
  \end{equation}
- So we choose the value for $\tau\hackscore{2}$ from
+ So we choose the value for $\tau_{2}$ from
  \begin{equation} \label{STOKES TOL2}
-  M \tau\hackscore{2} \|B v\hackscore{1}\|\hackscore{0}  \le (\chi^{-})^2 \epsilon^{-}
+  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
  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} \\
+ \|v_{2}-v\|_{1} & \le & \|v_{2}-\delta v-v\|_{1} + \| \delta v\|_{1} \\
- & \le &  \chi^{-} \cdot \epsilon^{-}  + \| e\hackscore{1} -  A^{-1} B^{*}\delta p \|\hackscore{1} \\
+ & \le &  \chi^{-} \cdot \epsilon^{-}  + \| e_{1} -  A^{-1} B^{*}\delta p \|_{1} \\
- & \approx &  \chi^{-} \cdot \epsilon^{-}  + \| e\hackscore{1} \|\hackscore{1} \\
+ & \approx &  \chi^{-} \cdot \epsilon^{-}  + \| e_{1} \|_{1} \\
- & \le &  \chi^{-} \cdot \epsilon^{-}  +  K \tau\hackscore{1} \| dv\hackscore{1} - e\hackscore{1} \|\hackscore{1}
+ & \le &  \chi^{-} \cdot \epsilon^{-}  +  K \tau_{1} \| dv_{1} - e_{1} \|_{1}
  \\
- & \le &  ( 1  + K \tau\hackscore{1}) \chi^{-} \cdot \epsilon^{-}
+ & \le &  ( 1  + K \tau_{1}) \chi^{-} \cdot \epsilon^{-}
  \end{array}
  \end{equation}
- So we choose the value for $\tau\hackscore{1}$ from
+ So we choose the value for $\tau_{1}$ from
  \begin{equation} \label{STOKES TOL1}
- K \tau\hackscore{1} \le \chi^{-}
+ K \tau_{1} \le \chi^{-}
  \end{equation}
  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 partial 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 $\chi_{max}\le 1$.
  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^{-}
+ \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
  $M^{+}$ and $K^{+}$ then we get
 Line 243 
 The same identity is used for to update
  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 the velocity but the new velocity $v\hackscore{1}$ has still a
+ in the velocity but the new velocity $v_{1}$ has still a
  small divergence, i.e. we have
- $\|Bv\hackscore{1}\|\hackscore{0} \ll \|v\hackscore{1}-v\hackscore{0}\|\hackscore{1}$.
+ $\|Bv_{1}\|_{0} \ll \|v_{1}-v_{0}\|_{1}$.
  In this case we will get a small pressure increment and consequently only very small changes to
  the velocity as a result of the second update step which therefore can be skipped and
  we can directly repeat the first update step until the increment in velocity becomes
  significant relative to its divergence. In practice we will ignore the second half of the iteration step
  as long as
   \begin{equation}\label{STOKES LARGE BV1}
- \|Bv\hackscore{1}\|\hackscore{0} \le \theta \cdot \|v\hackscore{1}-v\hackscore{0}\|
+ \|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
- with $v\hackscore{1}\rightarrow v\hackscore{2}$ but we will not correct $M$ in this case.
+ with $v_{1}\rightarrow v_{2}$ but we will not correct $M$ in this case.
- Starting from an initial guess $v\hackscore{0}$ and $p\hackscore{0}$ for velocity and pressure
+ Starting from an initial guess $v_{0}$ and $p_{0}$ for velocity and pressure
  the solution procedure is implemented as follows.
  \begin{enumerate}
-  \item calculate viscosity $\eta(v\hackscore{0},p)\hackscore{0}$ and assemble operator $A$ from $\eta$.
+  \item calculate viscosity $\eta(v_{0},p)_{0}$ and assemble operator $A$ from $\eta$.
-  \item calculate the tolerance $\tau\hackscore{1}$ from~\ref{STOKES TOL1}.
+  \item calculate the tolerance $\tau_{1}$ from~\ref{STOKES TOL1}.
-  \item Solve~\ref{STOKES ITER STEP 1} for $dv\hackscore{1}$ with tolerance $\tau\hackscore{1}$.
+  \item Solve~\ref{STOKES ITER STEP 1} for $dv_{1}$ with tolerance $\tau_{1}$.
-  \item update $v\hackscore{1}= v\hackscore{0}+ dv\hackscore{1}$
+  \item update $v_{1}= v_{0}+ dv_{1}$
-  \item if $Bv\hackscore{1}$ is large (see~\ref{STOKES LARGE BV1}):
+  \item if $Bv_{1}$ is large (see~\ref{STOKES LARGE BV1}):
   \begin{enumerate}
-  \item calculate the tolerance $\tau\hackscore{2}$ from~\ref{STOKES TOL2}.
+  \item calculate the tolerance $\tau_{2}$ from~\ref{STOKES TOL2}.
-  \item Solve~\ref{STOKES ITER STEP 2} for $dp\hackscore{2}$ and $v\hackscore{2}$ with tolerance $\tau\hackscore{2}$.
+  \item Solve~\ref{STOKES ITER STEP 2} for $dp_{2}$ and $v_{2}$ with tolerance $\tau_{2}$.
-  \item update $p\hackscore{2}\leftarrow p\hackscore{0}+ dp\hackscore{2}$
+  \item update $p_{2}\leftarrow p_{0}+ dp_{2}$
   \end{enumerate}
   \item else:
    \begin{enumerate}
-   \item update $p\hackscore{2}\leftarrow p$ and $v\hackscore{2}\leftarrow v\hackscore{1}$
+   \item update $p_{2}\leftarrow p$ and $v_{2}\leftarrow v_{1}$
     \end{enumerate}
     \item calculate convergence measure $\epsilon$ and convergence rate $\chi$
  \item if stopping criteria~\ref{STOKES STOPPING CRITERIA} holds:
   \begin{enumerate}
-  \item return $v\hackscore{2}$ and $p\hackscore{2}$
+  \item return $v_{2}$ and $p_{2}$
   \end{enumerate}
   \item else:
   \begin{enumerate}
       \item update $M$ and $K$
-      \item goto step 1 with $v\hackscore{0}\leftarrow v\hackscore{2}$ and $p\hackscore{0}\leftarrow p\hackscore{2}$.
+      \item goto step 1 with $v_{0}\leftarrow v_{2}$ and $p_{0}\leftarrow p_{2}$.
  \end{enumerate}
  \end{enumerate}
 Line 359 
 set the solver options for solving the e
  \subsection{Example: Lit Driven Cavity}
-  The following script \file{lit\hackscore driven\hackscore cavity.py}
+  The following script \file{lit_driven_cavity.py}
  \index{scripts!\file{helmholtz.py}} which is available in the \ExampleDirectory
  illustrates the usage of the \class{StokesProblemCartesian} class to solve
  the lit driven cavity problem:

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