X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/GMRES2stage.git/blobdiff_plain/e8b2791c920612ee2d944379890d2a6b76e0acea..82de9e14d2d33f6ed381b34feec71fa5b0964ad7:/IJHPCN/paper.tex

diff --git a/IJHPCN/paper.tex b/IJHPCN/paper.tex
index f67fa56..80a5a12 100644
--- a/IJHPCN/paper.tex
+++ b/IJHPCN/paper.tex
@@ -13,6 +13,8 @@
 \usepackage{multirow}
 \usepackage{graphicx}
 \usepackage{url}
+\usepackage{dsfont}
+
 
 \def\newblock{\hskip .11em plus .33em minus .07em}
 
@@ -407,57 +409,74 @@ little bit longer but it performs more or less the same operations.
 \section{Convergence results}
 \label{sec:04}
 
+%%NEW
+
+
+We suppose in this section that GMRES($m$) is used as solver in the TSIRM algorithm applied on a complex matrix $A$.
+Let us denote $A^\ast$ the conjugate transpose of $A$, and let $\mathfrak{R}(A)=\dfrac{1}{2} \left( A + A^\ast\right)$, $\mathfrak{I}(A)=\dfrac{1}{2i} \left( A - A^\ast\right)$. 
+
+\subsection{$\mathfrak{R}(A)$ is positive}
+
+\begin{proposition}
+\label{positiveConvergent}
+If $\mathfrak{R}(A)$ is positive, then the TSIRM algorithm is convergent.
+\end{proposition}
+
+
+\begin{proof}
+If $\mathfrak{R}(A)$ is positive, then even if $A$ is complex, it is possible to state that 
+the GMRES algorithm is convergent, see, \emph{e.g.},~\cite{Huang89}. In particular, its residual norm
+decreases to zero.
+
+At each iterate of the TSIRM algorithm, either a GMRES iteration is realized or a least square
+resolution (to find the minimum of $||b-Ax||_2$ is achieved on the linear span of the iterated approximation vectors 
+$span\left(x_{k-s+1}, x_{k-s}+2, \hdots, x_{k} \right)$
+of the last GMRES stage,
+where
+$\operatorname{span}(S) = \left \{ {\sum_{i=1}^k \lambda_i v_i \Big| k \in \mathbb{N}, v_i \in S, \lambda _i \in \mathbb{R}} \right \}$.
+
+Obviously, the minimum of $||b-Ax||_2$ on the set $span\left(x_{k-s+1}, x_{k-s}+2, \hdots, x_{k} \right)$ 
+is lower than or equal to $||b-Ax_k||_2$, which is the last obtained GMRES-residual norm. So we can
+conclude that the intermediate stage of least square resolution inserted into the GMRES algorithm
+does not break the decreasing to zero of the GMRES-residual norm. 
+
+In other words, the TSIRM algorithm is convergent.
+\end{proof}
+
 
-We can now claim that,
+Regarding the convergence speed, we can claim that,
 \begin{proposition}
 \label{prop:saad}
-If $A$ is either a definite positive or a positive matrix and GMRES($m$) is used as a solver, then the TSIRM algorithm is convergent. 
+If $A$ is a positive matrix, then the convergence of the 
+TSIRM algorithm is linear. 
 
-Furthermore, let $r_k$ be the
-$k$-th residue of TSIRM, then
+Furthermore, let $r_k$ be the $k$-th residue of TSIRM, then
 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 $M$ is the symmetric part of $A$, $\alpha = \lambda_{min}(M)^2$ and $\beta = \lambda_{max}(A^T A)$;
-\item when $A$ is positive definite:
-\begin{equation}
-\|r_k\| \leq \left( 1-\frac{\lambda_{\mathrm{min}}^2(1/2(A^T + A))}{ \lambda_{\mathrm{max}}(A^T A)} \right)^{km/2} \|r_0\|.
-\end{equation}
-\end{itemize}
-%In the general case, where A is not positive definite, we have
-%$\|r_n\| \le \inf_{p \in P_n} \|p(A)\| \le \kappa_2(V) \inf_{p \in P_n} \max_{\lambda \in \sigma(A)} |p(\lambda)| \|r_0\|, .$
+where $M$ is the symmetric part of $A$, $\alpha = \lambda_{min}(M)^2$ and $\beta = \lambda_{max}(A^T A)$.
 \end{proposition}
 
 \begin{proof}
-Let us first recall that the residue is under control when considering the GMRES algorithm on a positive definite matrix, and it is bounded as follows:
-\begin{equation*}
-\|r_k\| \leq \left( 1-\frac{\lambda_{\mathrm{min}}^2(1/2(A^T + A))}{ \lambda_{\mathrm{max}}(A^T A)} \right)^{k/2} \|r_0\| .
-\end{equation*}
-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:
+Let us first recall that, 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 such assumptions regarding $A$.
+where $\alpha$ and $\beta$ are defined as in Proposition~\ref{prop:saad}.
 These well-known results can be found, \emph{e.g.}, in~\cite{Saad86}.
 
 We will now prove by a mathematical induction that, for each $k \in \mathbb{N}^\ast$, 
-$||r_k|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{mk}{2}} ||r_0||$ when $A$ is positive, and $\|r_k\| \leq \left( 1-\frac{\lambda_{\mathrm{min}}^2(1/2(A^T + A))}{ \lambda_{\mathrm{max}}(A^T A)} \right)^{km/2} \|r_0\|$ when $A$ is positive definite.
+$||r_k|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{mk}{2}} ||r_0||$ when $A$ is positive.
 
 The base case is obvious, as for $k=1$, the TSIRM algorithm simply consists in applying GMRES($m$) once, leading to a new residual $r_1$ that follows the inductive hypothesis due to the results recalled above.
 
-Suppose now that the claim holds for all $m=1, 2, \hdots, k-1$, that is, $\forall m \in \{1,2,\hdots, k-1\}$, $||r_m|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||$ in the positive case, and $\|r_k\| \leq \left( 1-\frac{\lambda_{\mathrm{min}}^2(1/2(A^T + A))}{ \lambda_{\mathrm{max}}(A^T A)} \right)^{km/2} \|r_0\|$ in the definite positive one.
+Suppose now that the claim holds for all $m=1, 2, \hdots, k-1$, that is, $\forall m \in \{1,2,\hdots, k-1\}$, $||r_m|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||$.
 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 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 $||r_k|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_{k-1}||\leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||$.
 \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}
+$$||r_k|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_{k-1}||\leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||$$
 and a least squares resolution.
 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,\\
 $\min_{\alpha \in \mathbb{R}^s} ||b-R\alpha ||_2 = \min_{\alpha \in \mathbb{R}^s} ||b-AS\alpha ||_2$
@@ -469,20 +488,120 @@ $\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||, \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,} 
+& \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||, \\
 \end{array}$
 \end{itemize}
 which concludes the induction and the proof.
 \end{proof}
 
+
+
+\subsection{$\mathfrak{R}(A)$ is positive definite}
+
+\begin{proposition}
+\label{prop2}
+Convergence of the TSIRM algorithm is at least linear when $\mathfrak{R}(A)$ is 
+positive definite. Furthermore, the rate of convergence is lower 
+than $$\min\left( \left(1- \dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{ \lambda_{min}^{\mathfrak{R}(A)} \lambda_{max}^{\mathfrak{R}(A)} + {\lambda_{max}^{\mathfrak{I}(A)}}^2}\right)^{\frac{m}{2}}; 
+\left(1-\dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{||A||^2}\right)^{\frac{m}{2}}\right) ,$$
+where ${\lambda_{min}^{X}}$ (resp. ${\lambda_{max}^{X}}$) is the lowest (resp. largest) eigenvalue of matrix $X$.
+\end{proposition}
+
+
+\begin{proof}
+If $\mathfrak{R}(A)$ is positive definite, then it is positive, and so the TSIRM algorithm
+is convergent due to Proposition~\ref{positiveConvergent}.
+
+Furthermore, as stated in the proof of Proposition~\ref{positiveConvergent}, the GMRES residue is under control 
+when $\mathfrak{R}(A)$ is positive. More precisely, it has been proven in the literature that the residual norm 
+provided at the $m$-th step of GMRES satisfies:
+\begin{enumerate}
+\item $||r_m|| \leqslant \left(1- \dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{ \lambda_{min}^{\mathfrak{R}(A)} \lambda_{max}^{\mathfrak{R}(A)} + {\lambda_{max}^{\mathfrak{I}(A)}}^2}\right)^{\frac{mk}{2}} ||r_0||$, see, \emph{e.g.},~\cite{citeulike:2951999}, 
+\item $||r_m|| \leqslant \left(1-\dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{||A||^2}\right)^{\frac{mk}{2}} ||r_0||$, see~\cite{ANU:137201},
+\end{enumerate}
+which proves the convergence of GMRES($m$) for all $m$ under such assumptions regarding $A$.
+
+We will now prove by a mathematical induction, and following the same canvas than in the proof of Prop.~\ref{positiveConvergent}, that: for each $k \in \mathbb{N}^\ast$, the TSIRM-residual norm satisfies
+\begin{equation}
+\label{induc}
+\begin{array}{ll}
+||r_k|| \leqslant & \min\left( \left(1- \dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{ \lambda_{min}^{\mathfrak{R}(A)} \lambda_{max}^{\mathfrak{R}(A)} + {\lambda_{max}^{\mathfrak{I}(A)}}^2}\right)^{\frac{m}{2}}; \right. \\
+& \left. \left(1-\dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{||A||^2}\right)^{\frac{m}{2}}\right) ||r_0||
+\end{array}
+\end{equation}
+when $A$ is positive definite.
+
+
+The base case is obvious, as for $k=1$, the TSIRM algorithm simply consists in applying GMRES($m$) once, leading to a new residual $r_1$ that follows the inductive hypothesis due to the results recalled in the items listed above.
+
+Suppose now that the claim holds for all $u=1, 2, \hdots, k-1$, that is, $\forall u \in \{1,2,\hdots, k-1\}$, $||r_u|| \leqslant  \left(1-\dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{||A||^2}\right)^{\frac{mu}{2}} ||r_0||$.
+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 
+$||r_k|| \leqslant \left(1- \dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{ \lambda_{min}^{\mathfrak{R}(A)} \lambda_{max}^{\mathfrak{R}(A)} + {\lambda_{max}^{\mathfrak{I}(A)}}^2}\right)^{\frac{m}{2}} \leqslant \left(1- \dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{ \lambda_{min}^{\mathfrak{R}(A)} \lambda_{max}^{\mathfrak{R}(A)} + {\lambda_{max}^{\mathfrak{I}(A)}}^2}\right)^{\frac{mk}{2}} ||r_0||$, due to~\cite{citeulike:2951999}. Furthermore, we have too that: $||r_k|| \leqslant \left(1-\dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{||A||^2}\right)^{\frac{m}{2}} ||r_{k-1}|| \leqslant \left(1-\dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{||A||^2}\right)^{\frac{mk}{2}} ||r_0||$, as proven in~\cite{ANU:137201} and by using the inductive hypothesis. So we can conclude that 
+$$\begin{array}{ll}||r_k|| \leqslant & \min\left( \left(1- \dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{ \lambda_{min}^{\mathfrak{R}(A)} \lambda_{max}^{\mathfrak{R}(A)} + {\lambda_{max}^{\mathfrak{I}(A)}}^2}\right)^{\frac{mk}{2}}; \right. \\
+& \left. \left(1-\dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{||A||^2}\right)^{\frac{mk}{2}}\right) \times ||r_0|| 
+\end{array}.$$
+
+\item Else, the TSIRM algorithm consists in two stages: a first GMRES($m$) execution leads to a temporary $x_k$ whose residue satisfies, following the previous item:
+$$\begin{array}{ll}
+||r_k|| & \leqslant  \min\left( \left(1- \dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{ \lambda_{min}^{\mathfrak{R}(A)} \lambda_{max}^{\mathfrak{R}(A)} + {\lambda_{max}^{\mathfrak{I}(A)}}^2}\right)^{\frac{m}{2}}; \right. \\
+& \left. \left(1-\dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{||A||^2}\right)^{\frac{m}{2}}\right) \times ||r_{k-1}||\\
+ & \leqslant  \min\left( \left(1- \dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{ \lambda_{min}^{\mathfrak{R}(A)} \lambda_{max}^{\mathfrak{R}(A)} + {\lambda_{max}^{\mathfrak{I}(A)}}^2}\right)^{\frac{mk}{2}}; \right. \\
+& \left. \left(1-\dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{||A||^2}\right)^{\frac{mk}{2}}\right) \times ||r_0|| 
+\end{array}$$
+and the least squares resolution of $\min_{\alpha \in \mathbb{R}^s} ||b-R\alpha ||_2$.
+
+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 $S$, as defined previously. So,\\
+$\min_{\alpha \in \mathbb{R}^s} ||b-R\alpha ||_2  = \min_{\alpha \in \mathbb{R}^s} ||b-AS\alpha ||_2$
+
+$\begin{array}{ll}
+& = \min_{x \in span\left(S_{k-s+1}, S_{k-s+2}, \hdots, S_{k} \right)} ||b-AS\alpha ||_2\\
+& = \min_{x \in span\left(x_{k-s+1}, x_{k-s}+2, \hdots, x_{k} \right)} ||b-AS\alpha ||_2\\
+& \leqslant \min_{x \in span\left( x_{k} \right)} ||b-Ax ||_2\\
+& \leqslant \min_{\lambda \in \mathbb{R}} ||b-\lambda Ax_{k} ||_2\\
+& \leqslant ||b-Ax_{k}||_2\\
+& = ||r_k||_2\\
+& \leqslant \min\left( \left(1- \dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{ \lambda_{min}^{\mathfrak{R}(A)} \lambda_{max}^{\mathfrak{R}(A)} + {\lambda_{max}^{\mathfrak{I}(A)}}^2}\right)^{\frac{mk}{2}}; \right. \\
+& \left. \left(1-\dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{||A||^2}\right)^{\frac{mk}{2}}\right) \times ||r_0|| 
+\end{array} .$
+\end{itemize}
+due to the inductive hypothesis. 
+So the statement of Equation~\eqref{induc} holds too for the $k$-th iterate, which concludes the induction and the proof.
+\end{proof}
+
+\subsection{A last linear convergence}
+
+
+\begin{proposition}
+Let us define the field of values of $A$ by 
+$$\mathfrak{F}(A) = \left\{ \dfrac{x^\ast A x}{x^\ast x}, x \in \mathds{C}^n\setminus \{0\} \right\} .$$
+
+Then if $\mathfrak{F}(A)$ is included into a closed ball of radius $r$ and center $c$,
+which does not contain the origin, then the convergence of the TSIRM algorithm is at least linear. 
+
+More precisely, the rate of convergence is lower 
+than $2 \dfrac{r}{|c|}$.
+\end{proposition}
+
+\begin{proof}
+This inequality comes from the fact that, in the conditions of the proposition, the GMRES residue 
+satisfies the inequality: $|r_k| \leqslant 2 \dfrac{r}{|c|}^k |r_0|$. An induction inspired by
+the proofs of Propositions~\ref{prop:saad} and~\ref{prop2} can transfer this inequality to the
+TSIRM residue.
+\end{proof}
+
+
+
 Remark that a similar proposition can be formulated at each time
 the given solver satisfies an inequality of the form $||r_n|| \leqslant \mu^n ||r_0||$,
 with $|\mu|<1$. Furthermore, it is \emph{a priori} possible in some particular cases 
 regarding $A$, 
 that the proposed TSIRM converges while the GMRES($m$) does not.
 
+%%ENDNEW
+
+
 %%%*********************************************************
 %%%*********************************************************
 \section{Experiments using PETSc}
@@ -1045,7 +1164,7 @@ Curie and Juqueen respectively based in France and Germany.
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 
 \bibliography{biblio}
-\bibliographystyle{unsrt}
-\bibliographystyle{alpha}
+\bibliographystyle{plain}
+%\bibliographystyle{alpha}
 
 \end{document}
