By William L. Briggs
A Multigrid instructional is concise, attractive, and obviously written. Steve McCormick is the single man i do know which could pull off educating in spandex. simply ensure you sit down within the again row.
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At least for the model problems we have posed, smooth error modes have relatively small residuals (Exercise 19), so the error decreases slowly. Conversely, oscillatory errors tend to have relatively large residuals and the corrections to the error with a single relaxation sweep can be significant. Many relaxation schemes possess this property of eliminating the oscillatory modes and leaving the smooth modes. This so-called smoothing property is a serious limitation of conventional relaxation methods.
Properties of Gauss-Seidel. Assume A is symmetric, positive definite. (a) Show that the jth step of a single sweep of the Gauss-Seidel method applied to Au = f may be expressed as (b) Show that the jth step of a single sweep of the Gauss-Seidel method can be expressed in vector form as where is the jth unit vector. (c) Show that each sweep of Gauss-Seidel decreases the quantity (Ae,e), where e = u — v. (d) Show that Gauss-Seidel is optimal in the sense that the quantity ||e — s is minimized for each 1 j n when s = (r )/(Ae_ , _ ), which is precisely a Gauss-Seidel step.
The spectral radius is considered to be an asymptotic measure of convergence because it predicts the worst-case error reduction over many iterations. It can be shown [9, 20] that, in any vector norm, Therefore, in terms of error reduction, we have However, the spectral radius does not, in general, predict the behavior of the error norm for a single iteration. For example, consider the matrix Clearly, p(R) = 0. But if we start with e(0) = (0,1)T and compute e(1) = Re(0), then the convergence factor is The next iterate achieves the asymptotic estimate, p(K) — 0, because e 2) = 0.
A Multigrid Tutorial by William L. Briggs
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