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a new Stokes solver added
1 gross 2719 \section{The Stokes Problem}
2     \label{STOKES PROBLEM}
3     In this section we discuss how to solve the Stokes problem which is defined as follows:
4    
5     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}
6     \begin{equation}\label{Stokes 1}
7     -\left(\eta(v\hackscore{i,j}+ v\hackscore{i,j})\right)\hackscore{,j}+p\hackscore{,i}=f\hackscore{i}-\sigma\hackscore{ij,j}
8     \end{equation}
9     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.
10    
11     We assume an incompressible media:
12     \begin{equation}\label{Stokes 2}
13     -v\hackscore{i,i}=0
14     \end{equation}
15     Natural boundary conditions are taken in the form
16     \begin{equation}\label{Stokes Boundary}
17     \left(\eta(v\hackscore{i,j}+ v\hackscore{i,j})\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}
18     \end{equation}
19     which can be overwritten by constraints of the form
20     \begin{equation}\label{Stokes Boundary0}
21     v\hackscore{i}(x)=v^D\hackscore{i}(x)
22     \end{equation}
23     at some locations $x$ at the boundary of the domain. $s\hackscore{i}$ defines a normal stress and
24     $\alpha\ge 0$ the spring constant for restoring normal force.
25     The index $i$ may depend on the location $x$ on the boundary.
26     $v^D$ is a given function on the domain.
27    
28     \subsection{Solution Method \label{STOKES SOLVE}}
29     In block form equation equations~\ref{Stokes 1} and~\ref{Stokes 2} takes the form of a saddle point problem
30     \index{saddle point problem}
31     \begin{equation}
32     \left[ \begin{array}{cc}
33     A & B^{*} \\
34     B & 0 \\
35     \end{array} \right]
36     \left[ \begin{array}{c}
37     v \\
38     p \\
39     \end{array} \right]
40     =\left[ \begin{array}{c}
41     G \\
42     0 \\
43     \end{array} \right]
44     \label{SADDLEPOINT}
45     \end{equation}
46     where $A$ is coercive (assuming $A$ is not depending on $v$ or $p$), 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}.
47     We use iterative techniques to solve this problem: Given an approximation $v$ and $p$ for
48     velocity and pressure we perform the following steps in the Uzawa scheme \index{Uzawa scheme} style:
49     \begin{enumerate}
50     \item calculate viscosity $\eta(v,p)$ and assemble operator $A$ from $\eta$.
51     \item Solve for $dv$:
52     \begin{equation}
53     A dv = G - A v - B^{*} p \label{SADDLEPOINT ITER STEP 1}
54     \end{equation}
55     \item update $v\hackscore{1}= v+ dv$
56     \item if $\max(\|Bv\hackscore{1}\|\hackscore{0},\|dv\|\hackscore{1}) \le \tau \cdot \|v\hackscore{1}\|\hackscore{1}$: return v$\hackscore{1},p$
57     \item Solve for $dp$:
58     \begin{equation}
59     \begin{array}{rcl}
60     B A^{-1} B^{*} dp & = & Bv\hackscore{1} \\
61     A dv\hackscore{2} & = & B^{*} dp
62     \end{array}
63     \label{SADDLEPOINT ITER STEP 2}
64     \end{equation}
65     \item update $p\hackscore{2}\leftarrow p+ dp$ and $v\hackscore{2}= v\hackscore{1} - dv\hackscore{2}$
66     \item goto Step 0 with $v=v\hackscore{2}$ and $p=p\hackscore{2}$.
67     \end{enumerate}
68     where $\tau$ is the given tolerance. Notice that the operator $A$ if it is depending on $v$ or $p$ is updated once only. In this algorithm
69     \begin{equation}
70     \|v\|\hackscore{1}^2 = \int\hackscore{\Omega} v\hackscore{j,k}v\hackscore{j,k} \; dx
71     \mbox{ and }
72     \|p\|\hackscore{0}^2= \int\hackscore{\Omega} p^2 \; dx.
73     \label{STOKES STOP}
74     \end{equation}
75     so the stopping criterion used is checking for convergence as well as for
76     the fact that the incompressiblity condition~\ref{Stokes 2} is fullfilled.
77    
78     To solve the update step~\ref{SADDLEPOINT ITER STEP 2} we use an iterative solver with iteration
79     operator $B A^{-1} B^{*}$, eg. using generalized minimal residual method (GMRES) \index{linear solver!GMRES}\index{GMRES}, preconditioned conjugate gradient method (PCG) \index{linear solver!PCG}\index{PCG}. As suitable preconditioner \index{preconditioner} for this operator is $\frac{1}{\eta}$, ie
80     the evaluation of the preconditioner $P$ for a given pressure increment $q$ is the solution of
81     \begin{equation} \label{P PREC}
82     \frac{1}{\eta} Pq = q \; .
83     \end{equation}
84     Note that in each evaluation of the iteration operator $q=B A^{-1} B^{*} s$ one needs to solve
85     the problem
86     \begin{equation} \label{P OPERATOR}
87     A w = B^{*} s
88     \end{equation}
89     with sufficient accuracy to return $q=Bw$. Notice that the residual $r$ is given as
90     \begin{equation} \label{STOKES RES }
91     r= B (v\hackscore{1} - B A^{-1} B^{*} dp) = B (v\hackscore{1} - A^{-1} B^{*} dp) = B (v\hackscore{1}-dv\hackscore{2}) = B v\hackscore{2}
92     \end{equation}
93     so in fact the residual $r$ is represented by the updated velocity $v\hackscore{2}$. This saves the recovery of
94     $dv\hackscore{2}$ in~\ref{SADDLEPOINT ITER STEP 2} after $dp$ has been calculated as iterative method such as PCG calculate the solution approximations along with the their residual. In PCG the iteration is terminated if
95     \begin{equation} \label{P OPERATOR}
96     \| P^{\frac{1}{2}}B v\hackscore{2} \|\hackscore{0} \le \tau\hackscore{2} \| P^{\frac{1}{2}}B v\hackscore{1} \|\hackscore{0}
97     \end{equation}
98     where $\tau\hackscore{2}$ is the given tolerane. The update step~\ref{P OPERATOR} involves the
99     solution of a sparse matrix problem. We use the tolerance $\tau\hackscore{2}^2$ in order to make sure that any
100     error from solving this problem does not interfere with the PCG iteration.
101    
102    
103    
104     We need to think about appropriate stopping criteria when solving
105     \ref{SADDLEPOINT ITER STEP 1}, \ref{SADDLEPOINT ITER STEP 2} and~\ref{P OPERATOR}. We would like to use very weak convergence criteria to reduce computational costs but they need to be tight enough to not interfere with the
106     convergence of the iteration process one level above. From Equation~\ref{SADDLEPOINT ITER STEP 1} we have
107     \begin{equation}
108     v\hackscore{1} = e\hackscore{1} + A^{-1} ( G - B^{*} p )
109     \end{equation}
110     We will use a sparse matrix solver so we have not full control on the norm $\|.\|\hackscore{s}$ used in the stopping criteria
111     \begin{equation}
112     \| G - A v\hackscore{1} - B^{*} p \|\hackscore{s} \le \tau\hackscore{1} \| G - A v - B^{*} p \|\hackscore{s}
113     \end{equation}
114     This translates into the conditoon
115     \begin{equation}
116     \| e\hackscore{1} \|\hackscore{1} \le K \tau\hackscore{1} \| dv\hackscore{1} - e\hackscore{1} \|\hackscore{1}
117     \mbox{ therefore }
118     \| e\hackscore{1} \|\hackscore{1} \le \frac{K \tau\hackscore{1}}{1-K \tau\hackscore{1}} \| dv\hackscore{1} \|\hackscore{1}
119     \end{equation}
120     The constant $K$ represents some uncertainty combining a variety of unknown factors such as the
121     solver being used and the condition number of the stiffness matrix.
122    
123     From the first equation of~\ref{SADDLEPOINT ITER STEP 2} we have
124     \begin{equation}
125     p\hackscore{2} = p + (B A^{-1} B^{*})^{-1} (e\hackscore{2} + Bv\hackscore{1} ) =
126     (B A^{-1} B^{*})^{-1} ( e\hackscore{2} + B e\hackscore{1} + B A^{-1} G)
127     \end{equation}
128     and simlar
129     \begin{equation}
130     v\hackscore{2} = v\hackscore{1} - dv\hackscore{2}
131     = v\hackscore{1} - A^{-1} B^{*}dp
132     = e\hackscore{1} + A^{-1} ( G - B^{*} p ) - A^{-1} B^{*} (p\hackscore{2}-p)
133     = e\hackscore{1} + A^{-1} ( G - B^{*} p\hackscore{2})
134     \end{equation}
135     This shows that we can write the iterative process as a fixed point iteration to solve (assume all errors are zero)
136     \begin{equation}
137     v = \Phi(v,p) \mbox{ and } p = \Psi(u,p)
138     \end{equation}
139     where
140     \begin{equation}
141     \begin{array}{rcl}
142     \Psi(v,p) & = & (B A^{-1} B^{*})^{-1} B A^{-1} G \\
143     \Phi(u,p) & = & A^{-1} ( G - B^{*} (B A^{-1} B^{*})^{-1} B A^{-1} G )
144     \end{array}
145     \end{equation}
146     Notice that if $A$ is independent from $v$ and $p$ the operators $\Phi(v,p)$ and $\Psi(u,p)$ are constant
147     and threfore the iteration will terminate - providing no termination errors in sub-iterations - after one step.
148     We also can give a formula for the error
149     \begin{equation}
150     \begin{array}{rcl}
151     \delta p & = & (B A^{-1} B^{*})^{-1} ( e\hackscore{2} + B e\hackscore{1} ) \\
152     \delta v & = & e\hackscore{1} - A^{-1} B^{*}\delta p
153     \end{array}\label{STOKES ERRORS}
154     \end{equation}
155     Notice that $B\delta v = - e\hackscore{2}=-Bv\hackscore{2}$
156     With this notation
157     \begin{equation}
158     \begin{array}{rcl}
159     v\hackscore{2} = \Phi(v,p) + \delta v \\
160     p\hackscore{2} = \Psi(v,p) + \delta p \\
161     \end{array}
162     \end{equation}
163     We use the $dv=v\hackscore{2}-v$ and $Bv\hackscore{1}=B A^{-1} B^{*} dp$ to measure the error of the
164     current approximation $v\hackscore{2}$ and $v\hackscore{2}$ towards the exact solution.
165     Assuming that the iteration does converge with a convergence rate $\chi^{-}$ we have the estimate
166     \begin{equation}
167     \max(\|dv\|\hackscore{1}, \|Bv\hackscore{1}\|\hackscore{0})
168     \le \chi^{-} \max(\|dv^{-}\|\hackscore{1}, \|Bv\hackscore{1}^{-}\|\hackscore{0})
169     + \max(\|\delta v\|\hackscore{1} + \|(B A^{-1} B^{*}) \delta p\|\hackscore{0})
170     \end{equation}
171     were the upper index $-$ referes to the increments in the last step.
172     Now we try to establish estimates for $\|\delta v\|$ and $\|Bv\hackscore{1}\|$ from
173     formulas~\ref{STOKES ERRORS} where neglect the $\delta p$ in the equation for $\delta v$ assuming that
174     $\delta p$ is controlled.
175     \begin{equation}
176     \|\delta v\|\hackscore{1} \approx \|e\hackscore{1}\|\hackscore{1} \le \frac{K \tau\hackscore{1}}{1-K \tau\hackscore{1}} \| dv\hackscore{1} \|\hackscore{1} \approx \frac{K \tau\hackscore{1}}{1-K \tau\hackscore{1}} \| dv\|\hackscore{1}
177     \end{equation}
178     and
179     \begin{equation}
180     \|(B A^{-1} B^{*}) \delta p\|\hackscore{0} \approx \| e\hackscore{2} \|\hackscore{0}
181     = \| B v\hackscore{2} \|\hackscore{0} \le M \tau\hackscore{2} \| B v\hackscore{1} \|\hackscore{0}
182     \end{equation}
183     which leads to
184     \begin{equation}
185     \max(\|dv\|\hackscore{1}, \|Bv\hackscore{1}\|\hackscore{0})
186     \le \frac{\chi^{-}}{1-max{(M \tau\hackscore{2}, \frac{K \tau\hackscore{1}}{1-K \tau\hackscore{1}})}} \max(\|dv^{-}\|\hackscore{1}, \|Bv\hackscore{1}^{-}\|\hackscore{0}) \label{STOKES ESTIMTED IMPROVEMENT}
187     \end{equation}
188     where the upper index $-$ refers to the previous iteration step.
189     If we allow require $max{(M \tau\hackscore{2}, \frac{K \tau\hackscore{1}}{1-K \tau\hackscore{1}})}\le \gamma$
190     with a given $0<\gamma<1$ we can set
191     \begin{equation} \label{STOKES SET TOL}
192     \tau\hackscore{2} = \frac{\gamma}{M} \mbox{ and } \tau\hackscore{1} = \frac{1}{K} \frac{\gamma}{1+\gamma}
193     \end{equation}
194     Problem is that we do not know $M$ and $K$ but can use the convergence rate
195     \begin{equation}
196     \chi := \frac{\max(\|dv\|\hackscore{1}, \|Bv\hackscore{1}\|\hackscore{0})}{\max(\|dv^{-}\|\hackscore{1}, \|Bv\hackscore{1}^{-}\|\hackscore{0})}
197     \end{equation}
198     to construct estimates of $M$ and $K$. We look at the two cases where our prediction and choice of
199     the tolerances was good or where we went wrong:
200     \begin{equation}
201     \chi \le \frac{\chi^{-}}{1-\gamma} \mbox{ or }
202     \chi = \frac{\chi^{-}}{1-\gamma\hackscore{0}} >\frac{\chi^{-}}{1-\gamma}.
203     \end{equation}
204     which translates to
205     \begin{equation}
206     \gamma\hackscore{0} \le \gamma \mbox{ or } \gamma\hackscore{0}=1-\frac{\chi^{-}}{ \chi}>\gamma>0
207     \end{equation}
208     In the second case use \ref{STOKES SET TOL} for $\gamma=\gamma\hackscore{0}$ to get new estimates $M^{+}$ and $K^{+}$
209     for $M$ and $K$:
210     \begin{equation} \label{TOKES CONST UPDATE}
211     M^{+} =
212     \frac{\gamma\hackscore{0}}{\gamma} M
213     \mbox{ and } K^{+} =
214     \frac{\gamma\hackscore{0}}{\gamma}
215     \frac{1+\gamma}{1+\gamma\hackscore{0}}
216     K
217     \mbox{ with } \gamma\hackscore{0}=\max(1-\frac{\chi^{-}}{ \chi},\gamma)
218     \end{equation}
219     With these updated constants we can now get better values for the tolerances via an updated value $\gamma^{+}$ for $\gamma$ and the corrected values $M^{+}$ and $K^{+}$ for $M$ and $K$. If we are in the case of convergence which we
220     meassure by
221     \begin{equation}
222     \chi < \chi\hackscore{max}
223     \end{equation}
224     where $\chi\hackscore{max}$ is given value with $0<\chi\hackscore{max}<1$. We then consider the following
225     criteria
226     \begin{itemize}
227     \item As we are in case of convergence we try to relax the tolerances by increasing $\gamma$ by a factor $\frac{1}{\omega}$ ($0<\omega<1$). So we would like to choose
228     $\gamma^{+} \ge \frac{\gamma}{\omega}$.
229     \item We need to make sure that the estimated convergence rate based on $\gamma^{+}$ will still maintain convergence
230     \begin{equation}
231     \frac{\chi}{1-\gamma^{+}} \le \chi\hackscore{max}
232     \end{equation}
233     which translates into
234     \begin{equation}
235     \gamma^{+} \le 1 - \frac{\chi}{\chi\hackscore{max}}
236     \end{equation}
237     Notice that the right hand side is positive.
238     \item We do not want to iterate more then necessary. So we would like reduce the error in the next just below the
239     desired tolerance:
240     \begin{equation}
241     \frac{\chi}{1-\gamma^{+}}\max(\|dv\|\hackscore{1}, \|Bv\hackscore{1}\|\hackscore{0}) \ge f \cdot \tau \|v\hackscore{2}\|\hackscore{1}
242     \end{equation}
243     where the left hand side provides an estimate for the error to prediced in the next step. $f$ is a
244     safty factor. This leads to
245     \begin{equation}
246     \gamma^{+} \le
247     1 -
248     \chi \frac{\max(\|dv\|\hackscore{1}, \|Bv\hackscore{1}\|\hackscore{0})} { f \cdot \tau \|v\hackscore{2}\|\hackscore{1}}
249     \end{equation}
250     \end{itemize}
251     Putting all these criteria together we get
252     \begin{equation}
253     \gamma^{+}=\min(\max(
254     \frac{\gamma}{\omega},
255     1 -
256     \chi \frac{\max(\|dv\|\hackscore{1}, \|Bv\hackscore{1}\|\hackscore{0})} { f \cdot \tau \|v\hackscore{2}\|\hackscore{1}}), 1 - \frac{\chi}{\chi\hackscore{max}})
257     \label{STOKES SET GAMMA PLUS}
258     \end{equation}
259     In case we cannot see convergence $\gamma$ is reduced by the factor $\omega$
260     \begin{equation}
261     \gamma^{+}=\max(\omega \cdot \gamma, \gamma\hackscore{min})\label{STOKES SET GAMMA PLUS2}
262     \end{equation}
263     where $\gamma\hackscore{min}$ is introduced to make sure that $\gamma^{+}$ stays away from zero.
264    
265    
266     Appling~\ref{STOKES SET TOL} for $\gamma^{+}$ and~\ref{TOKES CONST UPDATE} we get the update formula
267     \begin{equation} \label{STOKES CONST UPDATE}
268     \tau\hackscore{2}^{+} =
269     \frac{\gamma^{+}}{\gamma\hackscore{0}} \tau\hackscore{2}
270     \mbox{ and } \tau\hackscore{1}^{+} =
271     \frac{\gamma^{+}}{\gamma\hackscore{0}}
272     \frac{1+\gamma\hackscore{0}}{1+\gamma^{+}}
273     \tau\hackscore{1}
274     \end{equation}
275     to get the new tolerances $\tau\hackscore{1}^{+}$ and $\tau\hackscore{2}^{+}$ to be used in the next iteration step.
276     Notice that the coefficients $M$ and $K$ have been eliminated. The calculation of $\gamma\hackscore{0}$ requires
277     at least two iteration steps (or a previous time step).
278    
279     Typical initial values are
280     \begin{equation} \label{STOKES VALUES}
281     \tau\hackscore{1}=\tau\hackscore{1}=\frac{1}{100} ; \; \omega=\frac{1}{2} \; \gamma=\frac{1}{10} ;\gamma\hackscore{min}=10^{-6}
282     \end{equation}
283     where after the first step we set $\gamma\hackscore{0}=\gamma$ as no $\chi^{-}$ is available.
284    
285    
286     \subsection{Functions}
287    
288     \begin{classdesc}{StokesProblemCartesian}{domain}
289     opens the Stokes problem\index{Stokes problem} on the \Domain domain. The approximation
290     order needs to be two. Order two will be used to approximate the velocity and order one
291     to approximate the pressure.
292     \end{classdesc}
293    
294     \begin{methoddesc}[StokesProblemCartesian]{initialize}{\optional{f=Data(), \optional{fixed_u_mask=Data(), \optional{eta=1, \optional{surface_stress=Data(), \optional{stress=Data()}, \optional{
295     restoration_factor=0}}}}}}
296     assigns values to the model parameters. In any call all values must be set.
297     \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$,
298     \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$.
299     The locations and compontents where the velocity is fixed are set by
300     the values of \var{fixed_u_mask}. \var{restoration_factor} defines the restoring force factor $\alpha$.
301     The method will try to cast the given values to appropriate
302     \Data class objects.
303     \end{methoddesc}
304    
305     \begin{methoddesc}[StokesProblemCartesian]{solve}{v,p
306     \optional{, max_iter=100 \optional{, verbose=False \optional{, usePCG=True }}}}
307     solves the problem and return approximations for velocity and pressure.
308     The arguments \var{v} and \var{p} define initial guess. The values of \var{v} marked
309     by \var{fixed_u_mask} remain unchanged.
310     If \var{usePCG} is set to \True
311     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
312     the PCG scheme is more efficient.
313     \var{max_iter} defines the maximum number of iteration steps.
314    
315     If \var{verbose} is set to \True informations on the progress of of the solver are printed.
316     \end{methoddesc}
317    
318    
319     \begin{methoddesc}[StokesProblemCartesian]{setTolerance}{\optional{tolerance=1.e-4}}
320     sets the tolerance in an appropriate norm relative to the right hand side. The tolerance must be non-negative and less than 1.
321     \end{methoddesc}
322     \begin{methoddesc}[StokesProblemCartesian]{getTolerance}{}
323     returns the current relative tolerance.
324     \end{methoddesc}
325     \begin{methoddesc}[StokesProblemCartesian]{setAbsoluteTolerance}{\optional{tolerance=0.}}
326     sets the absolute tolerance for the error in the relevant norm. The tolerance must be non-negative. Typically the
327     absolute talerance is set to 0.
328     \end{methoddesc}
329    
330     \begin{methoddesc}[StokesProblemCartesian]{getAbsoluteTolerance}{}
331     sreturns the current absolute tolerance.
332     \end{methoddesc}
333    
334     \begin{methoddesc}[StokesProblemCartesian]{getSolverOptionsVelocity}{}
335     returns the solver options used solve the equations~(\ref{V CALC}) for velocity.
336     \end{methoddesc}
337    
338     \begin{methoddesc}[StokesProblemCartesian]{getSolverOptionsPressure}{}
339     returns the solver options used solve the equation~(\ref{P PREC}) for pressure.
340     \end{methoddesc}
341    
342     \begin{methoddesc}[StokesProblemCartesian]{getSolverOptionsDiv}{}
343     set the solver options for solving the equation to project the divergence of the velocity onto the function space of pressure.
344     \end{methoddesc}
345    
346    
347     \subsection{Example: Lit Driven Cavity}
348     The following script \file{lit\hackscore driven\hackscore cavity.py}
349     \index{scripts!\file{helmholtz.py}} which is available in the \ExampleDirectory
350     illustrates the usage of the \class{StokesProblemCartesian} class to solve
351     the lit driven cavity problem:
352     \begin{python}
353     from esys.escript import *
354     from esys.finley import Rectangle
355     from esys.escript.models import StokesProblemCartesian
356     NE=25
357     dom = Rectangle(NE,NE,order=2)
358     x = dom.getX()
359     sc=StokesProblemCartesian(dom)
360     mask= (whereZero(x[0])*[1.,0]+whereZero(x[0]-1))*[1.,0] + \
361     (whereZero(x[1])*[0.,1.]+whereZero(x[1]-1))*[1.,1]
362     sc.initialize(eta=.1, fixed_u_mask= mask)
363     v=Vector(0.,Solution(dom))
364     v[0]+=whereZero(x[1]-1.)
365     p=Scalar(0.,ReducedSolution(dom))
366     v,p=sc.solve(v,p, verbose=True)
367     saveVTK("u.xml",velocity=v,pressure=p)
368     \end{python}

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