Solution to a Linear System in a Symmetric Positive Definite Tridiagonal Matrix (Simple Driver)

Calling Sequence

CALL DPTSV	(N, NRHS, DDIAG, DSUB, DB, LDB, INFO)
CALL SPTSV	(N, NRHS, SDIAG, SSUB, SB, LDB, INFO)
CALL ZPTSV	(N, NRHS, DDIAG, ZSUB, ZB, LDB, INFO)
CALL CPTSV	(N, NRHS, SDIAG, CSUB, CB, LDB, INFO)
void dptsv	(int n, int nrhs, double *ddiag, double *dsub, double *db, int ldb, int *info)
void sptsv	(int n, int nrhs, float *sdiag, float *ssub, float *sb, int ldb, int *info)
void zptsv	(int n, int nrhs, double *ddiag, doublecomplex *dsub, doublecomplex *zb, int ldb, int *info)
void cptsv	(int n, int nrhs, float *sdiag, complex *ssub, complex *cb, int ldb, int *info)

Arguments

N	Order of the matrix A. N 0.
NRHS	Number of right-hand sides, equal to the number of columns of the matrix B. NRHS 0.
xDIAG	On entry, the N diagonal elements of the matrix A. On exit, the N diagonal elements of the diagonal matrix D from the LDL factorization of A, as computed by xPTTRF.
xSUB	On entry, the N-1 subdiagonal elements of the matrix A. On exit, the N-1 subdiagonal elements of the unit bidiagonal factor L from the LDL factorization of A, as computed by xPTTRF.
xB	On entry, the N×NRHS right-hand side matrix B. On exit, the N×NRHS solution matrix X.
LDB	Leading dimension of the array B as specified in a dimension or type statement. LDB max(1, N).
INFO	On exit:
	INFO = 0	Subroutine completed normally.
	INFO < 0	The ith argument, where i = |INFO|, had an illegal value.
	INFO > 0	The leading minor of order i of A, where i = INFO, is not positive definite. The factorization has not been completed, unless i = N, and the solution could not be computed.

Sample Program

PROGRAM TEST

IMPLICIT NONE

C

INTEGER LDB, N, NRHS

PARAMETER (N = 4)

PARAMETER (NRHS = 1)

PARAMETER (LDB = N)

C

DOUBLE PRECISION B(LDB,NRHS), DIAG(N), DLOWER(N-1)

INTEGER ICOL, INFO, IROW

C

EXTERNAL DPTSV

C

C Initialize the arrays DIAG and DLOWER to store in symmetric

C tridiagonal form the 4x4 symmetric positive definite

C coefficient matrix A shown below. Initialize the array B

C to store the right hand side vector b shown below.

C

C 2 -1 0 0 6

C A = -1 2 -1 0 b = 12

C 0 -1 2 -1 12

C 0 0 -1 2 6

C

DATA DIAG / 2.0D0, 2.0D0, 2.0D0, 2.0D0 /

DATA DLOWER / -1.0D0, -1.0D0, -1.0D0 /

DATA B / 6.0D0, 1.2D1, 1.2D1, 6.0D0 /

C

C Print the initial values of the arrays.

C

PRINT 1000

DO 100, IROW = 1, N

PRINT 1010, (0.0D0, ICOL = 1, IROW - 2),

$ (DLOWER(ICOL + 1), ICOL = ABS(IROW - 2), IROW - 2),

$ DIAG(IROW),

$ (DLOWER(IROW), ICOL = 1, MIN(1, N - IROW)),

$ (0.0D0, ICOL = IROW + 2, N)

100 CONTINUE

PRINT 1020

PRINT 1030, B

C

C Solve the system and print the results.

C

CALL DPTSV (N, NRHS, DIAG, DLOWER, B, LDB, INFO)

IF (INFO .NE. 0) THEN

PRINT 1040, INFO

STOP 1

END IF

PRINT 1050

PRINT 1030, B

C

1000 FORMAT (1X, 'A:')

1010 FORMAT (4(3X, F6.3))

1020 FORMAT (/1X, 'b:')

1030 FORMAT (1X, F6.2)

1040 FORMAT (1X, 'Error solving Ax=b, INFO = ', I5)

1050 FORMAT (/1X, 'x:')

C

END

Sample Output

A:
2.000 -1.000 0.000 0.000
-1.000 2.000 -1.000 0.000
0.000 -1.000 2.000 -1.000
0.000 0.000 -1.000 2.000
b:
6.00
12.00
12.00
6.00
x:
18.00
30.00
30.00
18.00