7.1.3.3 Parallel pressure correction scheme

The saddle point problem (7.7) (linear, non-symmetric, indefinite)

$$\begin{pmatrix}{\ensuremath{\color{green}{\sf A}}} & {\ensuremath... ...een}{\sf r}}}} \ \underline{{\ensuremath{\color{green}{\sf s}}}} \end{pmatrix}$$

$$\qquad\underline{{\ensuremath{\color{green}{\sf r}}}} := \underline{{\ensuremath{\color{green}{\sf\widehat{f} \makebox[0pt... ...th{\color{green}{\sf B}}}\underline{{\ensuremath{\color{red}\mathfrak{p}}}}^{n}$$

$$\underline{{\ensuremath{\color{green}{\sf s}}}} := - {\ensuremath{\color{green}{\sf B}}}^T \underline{{\ensuremath{\color{red}\mathfrak{u}}}}^{n}$$

will be solved by a parallel pressure correction scheme.

How to solve (7.10) :

• $\widehat{C} \makebox[0pt]{} := {\ensuremath{\color{red}\mathfrak{I}}}_{N_p}$ (Uzawa) $\Longrightarrow$ $\underline{{\ensuremath{\color{red}\mathfrak{p}}}}_{\delta} = \sum\limits_{s=1... ...frak{u}}}}_{\delta} - \underline{{\ensuremath{\color{green}{\sf s}}}}_s \right)$
  If we assume constant pressure per element (discretization a) or c) ) then no vector type change is needed, i.e., no communication is necessary.
• $\widehat{A} \makebox[0pt]{} \approx {\ensuremath{\color{red}\mathfrak{A}}} = ... ...en}{\sf A}}}_s(\underline{{\ensuremath{\color{red}\mathfrak{u}}}}^n) \cdot A_s$, $\widehat{C} \makebox[0pt]{} := \left( {\ensuremath{\color{green}{\sf B}}}^T \w... ...mathfrak{A}}}} \makebox[0pt]{}^{-1}{\ensuremath{\color{green}{\sf B}}} \right)$, $\widetilde{{\ensuremath{\color{red}\mathfrak{A}}}} \makebox[0pt]{}\approx {\ensuremath{\color{red}\mathfrak{A}}}$
• If the equation (7.10) is solved via fix point iteration (7.9)

  $$\left( {\ensuremath{\color{green}{\sf B}}}^T \widetilde{{\ensurem... ...}}_{\delta} - \underline{{\ensuremath{\color{green}{\sf s}}}} \right) \enspace,$$ (7.8)

  
  
  then the calculation of the right hand side requires the of
  $$\widehat{A} \makebox[0pt]{} \cdot \underline{{\ensuremath{\color{... ... B}}} \underline{{\ensuremath{\color{red}\mathfrak{p}}}}_{\delta}^m \enspace.$$
  Again, we can use therein a parallelized iteration method from Sec. 5.
• The iteration method in the solution process of (7.11) requires in each iteration the matrix-times-vector operation
  $$\underline{{\ensuremath{\color{green}{\sf v}}}} \;:=\; {\ensurema... ...{\widehat{\underline{{\ensuremath{\color{red}\mathfrak{v}}}}} \makebox[0pt]{}}$$
  which needs the the fast parallel of
  $$\widetilde{{\ensuremath{\color{red}\mathfrak{A}}}} \makebox[0pt]{... ...r{green}{\sf B}}} \underline{{\ensuremath{\color{red}\mathfrak{w}}}} \enspace.$$
  In case of $\widetilde{{\ensuremath{\color{red}\mathfrak{A}}}} \makebox[0pt]{} := {\ensuremath{\color{red}\mathfrak{I}}}_{N_u}$ or $\widetilde{{\ensuremath{\color{red}\mathfrak{A}}}} \makebox[0pt]{} := \sum\limits_{s=1}^p\mathrm{diag}{{\ensuremath{\color{green}{\sf A}}}_s}$ this requirement is easy to fulfill and costs one vector accumulation per iteration. If one chooses a discretization with non-conform elements (a) then the diagonal property of $M$ ([Joh97],pp.27) ensures for $\widetilde{{\ensuremath{\color{red}\mathfrak{A}}}} \makebox[0pt]{} := {\ensure... ...{red}\mathfrak{M}}} = \sum\limits_{s=1}^p {\ensuremath{\color{green}{\sf M}}}_s$ the fast parallel inverting.
  We loose this property when conform elements (b)/(c) are used. Here, the choice $\widetilde{{\ensuremath{\color{red}\mathfrak{A}}}} \makebox[0pt]{}={\ensuremath{\color{red}\mathfrak{M}}}$ is parallel inefficient. Substitution of ${\ensuremath{\color{red}\mathfrak{M}}}$ by a matrix  $\widetilde{{\ensuremath{\color{red}\mathfrak{M}}}} \makebox[0pt]{}$ as described in Sec. 5.5.6 and the setting  $\widetilde{{\ensuremath{\color{red}\mathfrak{A}}}} \makebox[0pt]{}$ equal the incomplete factorization of  $\widetilde{{\ensuremath{\color{red}\mathfrak{M}}}} \makebox[0pt]{}$ (Sec. 5.5) allows an efficient parallel inverting.