change in Newton in a Nutshell: it makes a difference whether a matrix is multiplied from right or from left. Hint from Nicolas.

Dumux-Svn-Revison: 8950
Ported-By: Andreas Lauser <andreas.lauser@iws.uni-stuttgart.de>
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Philipp Nuske 2012-08-22 11:49:40 +00:00 committed by Andreas Lauser
parent 9c51221df8
commit 024b7857e4

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@ -24,8 +24,8 @@ One step of the \textsc{Newton} method can be formalized as follows:
\begin{align}
\label{NewtonGen}
\textbf{u}^{r+1} &= \textbf{u}^r - \left(\textbf{f}^\prime (\textbf{u}^r) \right)^{-1} \textbf{f}(\textbf{u}^r) \\
\Leftrightarrow ( \textbf{u}^{r+1}-\textbf{u}^r) {\textbf{f}^{\prime}(\textbf{u}^r)} &= -\textbf{f}(\textbf{u}^r) \\
\Leftrightarrow ( \textbf{u}^r - \textbf{u}^{r+1}) \underbrace{\textbf{f}^{\prime}(\textbf{u}^r)}_{\textnormal{Jacobian}} &= \textbf{f}(\textbf{u}^r) \label{NewtonAsUsed}
\Leftrightarrow {\textbf{f}^{\prime}(\textbf{u}^r)} ( \textbf{u}^{r+1}-\textbf{u}^r) &= -\textbf{f}(\textbf{u}^r) \\
\Leftrightarrow \underbrace{\textbf{f}^{\prime}(\textbf{u}^r)}_{\textnormal{Jacobian}} ( \textbf{u}^r - \textbf{u}^{r+1}) &= \textbf{f}(\textbf{u}^r) \label{NewtonAsUsed}
\end{align}
\end{subequations}