X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/GMRES2stage.git/blobdiff_plain/296102e5b791e8d44dcc357426e5590f466a54e9..f2b93de905ad07c8701e6cc195e12da13406d8f3:/paper.tex diff --git a/paper.tex b/paper.tex index 8764073..ef0321c 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 @@ -749,41 +743,59 @@ 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, -let $r_k$ be the +\label{prop:saad} +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 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 +%$\|r_n\| \le \inf_{p \in P_n} \|p(A)\| \le \kappa_2(V) \inf_{p \in P_n} \max_{\lambda \in \sigma(A)} |p(\lambda)| \|r_0\|, .$ \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 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: +\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 $A$. +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. -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}. +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, $\forall m \in \{1,2,\hdots, k-1\}$, $||r_m|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||$. +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. We will show that the statement holds too for $r_k$. Two situations can occur: \begin{itemize} -\item If $k \mod m \neq 0$, then the TSIRM algorithm consists in executing GMRES once. In that case, we obtain $||r_k|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_{k-1}||\leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||$ by the inductive hypothesis. -\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 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: +\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,\\ $\min_{\alpha \in \mathbb{R}^s} ||b-R\alpha ||_2 = \min_{\alpha \in \mathbb{R}^s} ||b-AS\alpha ||_2$ @@ -794,14 +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. %%%********************************************************* %%%*********************************************************