From: frohne Date: Fri, 24 Feb 2012 20:34:28 +0000 (+0000) Subject: fixing the definition of G_i X-Git-Url: https://gitweb.dealii.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=4dcbf6ebb760d15d109ff6b1ab12d72596930d71;p=dealii-svn.git fixing the definition of G_i git-svn-id: https://svn.dealii.org/trunk@25166 0785d39b-7218-0410-832d-ea1e28bc413d --- diff --git a/deal.II/examples/step-41/doc/step-41-doc.tex b/deal.II/examples/step-41/doc/step-41-doc.tex index 5672f0ab28..2c1f82e565 100644 --- a/deal.II/examples/step-41/doc/step-41-doc.tex +++ b/deal.II/examples/step-41/doc/step-41-doc.tex @@ -134,7 +134,6 @@ The variational inequality above is awkward to work with. We would therefore like to reformulate it as an equivalent saddle point problem. Set $V:=H^1_0(\Omega)$. We introduce a Lagrange multiplier $\lambda$ and the convex cone $K\subset V'$, $V'$ dual space of $V$, $K:=\{\mu\in V': \langle\mu,v\rangle\geq 0,\quad 0\geq v\in V\}$ of -\marginpar{JF: Is this definition of $K$ correct?} Lagrange multipliers, where $\langle\cdot,\cdot\rangle$ denotes the duality pairing between $V'$ and $V$. This yields: @@ -169,7 +168,6 @@ inequality constraints. To get there, let us assume that we discretize both $u$ and $\lambda$ with the same finite element space, for example the usual $Q_k$ spaces. We would then get the equations -\marginpar{JF: Aren't the inequalities the wrong way around here (and below)?} \begin{eqnarray*} &A U + B\Lambda = F,&\\ &[BU-G]_i \geq 0, \quad \Lambda_i \leq 0,\quad \Lambda_i[BU-G]_i = 0 @@ -192,9 +190,7 @@ at these locations, we get a diagonal mass matrix with To define $G$ we use the same technique as for $B$ where $g_h$ is a suitable approximation of $g$ \begin{align*} - G_{ii} = \int_\Omega g_h(x) \varphi_i(\mathbf x)\ \textrm{d}x, - \qquad - G_{ij}=0 \ \text{for } i\neq j. + G_{i} = \int_\Omega g_h(x) \varphi_i(\mathbf x)\ \textrm{d}x. \end{align*} With this, the equations above can be restated as \begin{eqnarray*} @@ -209,7 +205,6 @@ Now we define for each degree of freedom $i$ the function \end{equation*} with some $c>0$. (In this program we choose $c = 100$. It is a kind of a penalty parameter which depends on the problem itself. For example there is no convergence for $c = 1$ and 7 refinements.) -\marginpar{JF: How do you choose $c$?} After some headscratching one can then convince oneself that the inequalities above can equivalently be rewritten as @@ -275,7 +270,6 @@ We can iteratively solve this by always linearizing around the previous iterate (i.e., applying a Newton method), but for this we need to linearize the function $C(\cdot,\cdot)$ that is not differentiable. That said, it is slantly differentiable, and in fact we have -\marginpar{JF: what should be in the second line? Zero or Lambda?} \begin{equation*} \dfrac{\partial}{\partial U^k_i}C([BU^k]_i,\Lambda^k_i) = \begin{cases} cB_{ii},& \text{if}\ \Lambda^k_i + c([BU^k]_i - G_i)< 0\\ @@ -370,13 +364,12 @@ system for each iteration $k$ \end{equation*} By considering the dofs in $\mathcal{A}_k$ as Dirichlet data we solve this system with a CG-method and the AMG preconditioner from Trilinos. -\marginpar{Which system do we actually solve with CG?} \section{Implementation} This tutorial is quite similar to step-4. But to solve the obstacle problem, two new methods are -implemented: assemble_mass_matrix_diagonal (TrilinosWrappers::SparseMatrix &mass_matrix) and -update_solution_and_constraints (). +implemented: assemble\_mass\_matrix\_diagonal (TrilinosWrappers::SparseMatrix \&mass\_matrix) and +update\_solution\_and\_constraints (). \end{document} \ No newline at end of file