From: Christophe Guyeux Date: Sat, 11 Oct 2014 09:09:17 +0000 (+0200) Subject: Amélioration de la prop X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/GMRES2stage.git/commitdiff_plain/82115e26ed868a45b036ba222a0aa867a3810a9b?ds=inline Amélioration de la prop --- diff --git a/paper.tex b/paper.tex index 8764073..8adef83 100644 --- a/paper.tex +++ b/paper.tex @@ -631,12 +631,6 @@ composed by the $s$ last solutions that have been computed during the inner iterations phase. In the remainder, the $i$-th column vector of $S$ will be denoted by $S_i$. -$\|r_n\| \leq \left( 1-\frac{\lambda_{\mathrm{min}}^2(1/2(A^T + A))}{ \lambda_{\mathrm{max}}(A^T A)} \right)^{n/2} \|r_0\|,$ -In the general case, where A is not positive definite, we have - -$\|r_n\| \le \inf_{p \in P_n} \|p(A)\| \le \kappa_2(V) \inf_{p \in P_n} \max_{\lambda \in \sigma(A)} |p(\lambda)| \|r_0\|, \,$ - - At each $s$ iterations, another kind of minimization step is applied in order to compute a new solution $x$. For that, the previous residuals of $Ax=b$ are computed by the inner iterations with $(b-AS)$. The minimization of the residuals is obtained by @@ -749,34 +743,44 @@ these operations are easy to implement in PETSc or similar environment. \section{Convergence results} \label{sec:04} -Let us recall the following result, see~\cite{Saad86} for further readings. -\begin{proposition} -\label{prop:saad} -Suppose that $A$ is a positive real matrix with symmetric part $M$. Then the residual norm provided at the $m$-th step of GMRES satisfies: -\begin{equation} -||r_m|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_0|| , -\end{equation} -where $\alpha = \lambda_{min}(M)^2$ and $\beta = \lambda_{max}(A^T A)$, which proves -the convergence of GMRES($m$) for all $m$ under that assumption regarding $A$. -\end{proposition} We can now claim that, \begin{proposition} -If $A$ is a positive real matrix and GMRES($m$) is used as solver, then the TSIRM algorithm is convergent. Furthermore, +If $A$ is either a definite positive or a positive matrix and GMRES($m$) is used as solver, then the TSIRM algorithm is convergent. Furthermore, let $r_k$ be the $k$-th residue of TSIRM, then -we still have: +we have the following boundaries: +\begin{itemize} +\item when $A$ is positive: \begin{equation} ||r_k|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0|| , \end{equation} -where $\alpha$ and $\beta$ are defined as in Proposition~\ref{prop:saad}. +where $M$ is the symmetric part of $A$, $\alpha = \lambda_{min}(M)^2$ and $\beta = \lambda_{max}(A^T A)$; +\item when $A$ is definite positive: +\begin{equation} +\|r_n\| \leq \left( 1-\frac{\lambda_{\mathrm{min}}^2(1/2(A^T + A))}{ \lambda_{\mathrm{max}}(A^T A)} \right)^{n/2} \|r_0\|, +\end{equation} +\end{itemize} +%In the general case, where A is not positive definite, we have +%$\|r_n\| \le \inf_{p \in P_n} \|p(A)\| \le \kappa_2(V) \inf_{p \in P_n} \max_{\lambda \in \sigma(A)} |p(\lambda)| \|r_0\|, \,$ + \end{proposition} \begin{proof} We will prove by a mathematical induction that, for each $k \in \mathbb{N}^\ast$, $||r_k|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{mk}{2}} ||r_0||.$ +Let us first recall that, +Additionally, when $A$ is a positive real matrix with symmetric part $M$, then the residual norm provided at the $m$-th step of GMRES satisfies: +\begin{equation} +||r_m|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_0|| , +\end{equation} +where $\alpha$ and $\beta$ are defined as in Proposition~\ref{prop:saad}, which proves +the convergence of GMRES($m$) for all $m$ under that assumption regarding $A$. +Let us recall the following result, see~\cite{Saad86} for further readings. + + The base case is obvious, as for $k=1$, the TSIRM algorithm simply consists in applying GMRES($m$) once, leading to a new residual $r_1$ which follows the inductive hypothesis due to Proposition~\ref{prop:saad}. Suppose now that the claim holds for all $m=1, 2, \hdots, k-1$, that is, $\forall m \in \{1,2,\hdots, k-1\}$, $||r_m|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||$.