Example 932 5 section 142 consider the solution of

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Unformatted text preview: norm was less than 10;7. The value of ! = 2=(1 + h) was used with the SOR method. Results are shown in Table 9.3.1. The two methods are comparable. Apparent convergence is at a linear rate in h, i.e., doubling h approximately 9.3. Conjugate Gradient Methods 39 doubles the number of SOR and conjugate gradient iterations to convergence. Since an SOR iteration is less costly than a conjugate gradient step, we may infer that the SOR procedure is the faster. 1=h SOR CG 10 31 27 20 64 54 40 122 107 Table 9.3.1: Number of SOR and conjugate gradient iterations to convergence for Example 9.3.2. The following theorem con rms the ndings of the previous example. Theorem 9.3.3. Let A be a symmetric and positive de nite matrix, then iterates of the conjugate gradient method satisfy kx( ) ; xk A where k kA and 2 p ;1 2 p 2 + 1 kx(0) ; xkA 2 (9.3.14) were de ned in (9.3.4a, 9.3.4c). Proof. cf. 4], Section 6.11. Examining (9.3.14) and (9.3.4c), we see that convergence is fastest when the eigenvalues of A are clustered together, i.e., when 2 (A) 1. Example 9.3.3. The factor p ;1 R = p 2+1 2 determines the convergence rate of the conjugate gradient method. Since the condition number 2 depends on the eigenvalues of A, it would seem that we have to examine the eigenvalue problem Aq = q: Using (9.2.8a), let us write this relation as (D ; L ; U)q = q 40 Solution Techniques for Elliptic Problems where D, L, and U were de ned by (9.2.8b-9.2.8c). Multiplying by D;1 and using (9.2.10b), we have MJ q = (I ; D;1 )q where MJ is the Jacobi iteration matrix. For the Laplacian operator, D = I and we have = 1 ; , where is an eigenvalue of MJ . Still con ning our attention to the Laplacian operator, we may use (9.2.17d) to evaluate and, hence, obtain n = 1 ; = 4 x sin2 mJ + 4 y sin2 2K 2 m = 1 2 ::: J ;1 n = 1 2 : : : K ; 1: For simplicity, let us focus on a square grid (J = K ) where = sin2 mJ + sin2 nJ 2 2 m n = 1 2 : : : J ; 1: The smallest eigenvalue occurs with m = n = 1 and the largest occurs with m = n = J ; 1 thus, 2 2 (J ; 1) : min = 2 sin max...
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This document was uploaded on 03/16/2014 for the course CSCI 6840 at Rensselaer Polytechnic Institute.

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