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# a6 6 6 4 l n r r 1 n 3 7 7 7 7 7 5 2 6 y6 6 4 851a

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Unformatted text preview: block tridiagonal matrix as shown in Figure 8.5.2 and used the block tridiagonal algorithm 5] to solve the problem. The block tridiagonal algorithm is a matrix extension of the scalar tridiagonal algorithm that we considered in Section 6.3. Unfortunately, this procedure may not be stable without pivoting. While the algorithm can be implemented with pivoting, it induces some matrix ll-in, which is not desirable. We will discuss a stable procedure due to Varah 8] that involves alternate row and column elimination. For the 2 2 matrix shown in Figure 8.5.1, we take a suitable combination of the rst two columns to create a zero in the (1 2) position of the matrix. The rst two columns may be interchanged so that the largest element is the pivot. We then proceed to the second row and eliminate the (3 2) element by row elimination. Interchanges of the second and third rows ensure that the largest pivot element is chosen. The process continues in this manner and the i i N i N 21 i XX XX X X XX X X XXXX A= XXXX XXXX XXXX XX Figure 8.5.1: Structure of a 2 2 block bidiagonal matrix. XX XX X X XX X X A= XXXX XXXX XXXX XXXX XX Figure 8.5.2: Embedding a 2 2 block bidiagonal matrix in a block tridiagonal matrix. resulting system is shown in Figure 8.5.3. There is no ll-in and the algorithm is stable with the partial row and column pivoting discussed. Reduction of a general system is shown in Figure 8.5.4. Recall, in Gaussian elimination row interchanges correspond to a reordering of the right hand side b while column interchanges correspond to a reordering of the unknowns y. Row interchanges correspond to a factoring PA = LU (8.5.2a) whereas column interchanges correspond to a factoring AQ = LU: (8.5.2b) The matrices L and U are lower and upper triangular matrices, respectively, and P and Q are permutation matrices containing the row and column pivot information, respectively. 22 2 6 6 6 6 6 A=6 6 6 6 6 4 0 0 0 0 0 0 0 3 7 7 7 7 7 7 7 7 7 7 5 Figure 8.5.3: Alternate row and column elimination of a 2 2 block bidiagonal matrix. 2 (0) 3 L 0 6 X U(0) X X 7 6 7 6 X 0 L(1) 0 7 6 7 (1) X 6 7 XU X 7 B=6 (2) 0 6 7 X0L 6 7 6 X U(2) X X 7 6 7 (3) 4 5 X 0L 0 X U(3) Figure 8.5.4: Alternate row and column elimination of block bidiagonal matrix with N =3 The entire factorization has the form PAQ = LBU (8.5.3) where the structure of B is shown in Figure 8.5.4. If P( ) , Q( ) , etc., are the matrices introduced at the i th stage of the process, then i P = P( ) P( N N; 1) : : : P(0) Q = Q(0) Q(1) : : : Q( N ~~ ~ L = L(0) L(1) : : : L( N ~~ U = U( ) U( N i (8.5.4a) ) (8.5.4b) ) (8.5.4c) 1) : : : U(0) : ~ (8.5.4d) N; ~ ~ The matrices L( ) and U( ) have the same structure as L( ) and U( ) , i = 0 1 : : : N , but are expanded to the dimension m(N + 1) of A. i i i 23 i Once the factorization is complete, the system (8.5.1) is solved by successively solving Lt = Pb (8.5.5a) Bz = t (8.5.5b) Us = z (8.5.5c) y = Qs: (8.5.5d) The systems (8.5.5a) and (8.5.5c) are solved by forward and backward substitution, respectively. The system (8.5.5d) is just a reordering of s. Upon examination of Figure 8.5.4, we see that we may determi...
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