From: Christophe Guyeux <guyeux@gmail.com>
Date: Sat, 11 Oct 2014 09:41:46 +0000 (+0200)
Subject: Fin de la preuve
X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/GMRES2stage.git/commitdiff_plain/f2b93de905ad07c8701e6cc195e12da13406d8f3?ds=inline

Fin de la preuve
---

diff --git a/paper.tex b/paper.tex
index d0fac64..ef0321c 100644
--- a/paper.tex
+++ b/paper.tex
@@ -790,7 +790,12 @@ Suppose now that the claim holds for all $m=1, 2, \hdots, k-1$, that is, $\foral
 We will show that the statement holds too for $r_k$. Two situations can occur:
 \begin{itemize}
 \item If $k \not\equiv 0 ~(\textrm{mod}\ m)$, then the TSIRM algorithm consists in executing GMRES once. In that case and by using the inductive hypothesis, we obtain either $||r_k|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_{k-1}||\leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||$ if $A$ is positive, or $\|r_k\| \leqslant \left( 1-\frac{\lambda_{\mathrm{min}}^2(1/2(A^T + A))}{ \lambda_{\mathrm{max}}(A^T A)} \right)^{m/2} \|r_{k-1}\|$ $\leqslant$ $\left( 1-\frac{\lambda_{\mathrm{min}}^2(1/2(A^T + A))}{ \lambda_{\mathrm{max}}(A^T A)} \right)^{km/2} \|r_{0}\|$ in the positive definite case.
-\item Else, the TSIRM algorithm consists in two stages: a first GMRES($m$) execution leads to a temporary $x_k$ whose residue satisfies $||r_k|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_{k-1}||\leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||$, and a least squares resolution.
+\item Else, the TSIRM algorithm consists in two stages: a first GMRES($m$) execution leads to a temporary $x_k$ whose residue satisfies:
+\begin{itemize}
+\item $||r_k|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_{k-1}||\leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||$ in the positive case, 
+\item $\|r_k\| \leqslant \left( 1-\frac{\lambda_{\mathrm{min}}^2(1/2(A^T + A))}{ \lambda_{\mathrm{max}}(A^T A)} \right)^{m/2} \|r_{k-1}\|$ $\leqslant$ $\left( 1-\frac{\lambda_{\mathrm{min}}^2(1/2(A^T + A))}{ \lambda_{\mathrm{max}}(A^T A)} \right)^{km/2} \|r_{0}\|$ in the positive definite one,
+\end{itemize}
+and a least squares resolution.
 Let $\operatorname{span}(S) = \left \{ {\sum_{i=1}^k \lambda_i v_i \Big| k \in \mathbb{N}, v_i \in S, \lambda _i \in \mathbb{R}} \right \}$ be the linear span of a set of real vectors $S$. So,\\
 $\min_{\alpha \in \mathbb{R}^s} ||b-R\alpha ||_2 = \min_{\alpha \in \mathbb{R}^s} ||b-AS\alpha ||_2$
 
@@ -801,14 +806,16 @@ $\begin{array}{ll}
 & \leqslant \min_{\lambda \in \mathbb{R}} ||b-\lambda Ax_{k} ||_2\\
 & \leqslant ||b-Ax_{k}||_2\\
 & = ||r_k||_2\\
-& \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||,
+& \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||, \textrm{ if $A$ is positive,}\\
+& \leqslant \left( 1-\frac{\lambda_{\mathrm{min}}^2(1/2(A^T + A))}{ \lambda_{\mathrm{max}}(A^T A)} \right)^{km/2} \|r_{0}\|, \textrm{ if $A$ is}\\
+& \textrm{positive definite,} 
 \end{array}$
 \end{itemize}
 which concludes the induction and the proof.
 \end{proof}
 
-We can remark that, at each iterate, the residue of the TSIRM algorithm is lower 
-than the one of the GMRES method.
+%We can remark that, at each iterate, the residue of the TSIRM algorithm is lower 
+%than the one of the GMRES method.
 
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