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 llin, 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 llin 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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 Spring '14
 JosephE.Flaherty
 Numerical Analysis, Boundary value problem, Piecewise linear function, Infantry regiments of Sweden, BVP

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