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|
SUBROUTINE SB03OT( DISCR, LTRANS, N, S, LDS, R, LDR, SCALE, DWORK,
$ INFO )
C
C SLICOT RELEASE 5.0.
C
C Copyright (c) 2002-2009 NICONET e.V.
C
C This program is free software: you can redistribute it and/or
C modify it under the terms of the GNU General Public License as
C published by the Free Software Foundation, either version 2 of
C the License, or (at your option) any later version.
C
C This program is distributed in the hope that it will be useful,
C but WITHOUT ANY WARRANTY; without even the implied warranty of
C MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
C GNU General Public License for more details.
C
C You should have received a copy of the GNU General Public License
C along with this program. If not, see
C <http://www.gnu.org/licenses/>.
C
C PURPOSE
C
C To solve for X = op(U)'*op(U) either the stable non-negative
C definite continuous-time Lyapunov equation
C 2
C op(S)'*X + X*op(S) = -scale *op(R)'*op(R) (1)
C
C or the convergent non-negative definite discrete-time Lyapunov
C equation
C 2
C op(S)'*X*op(S) - X = -scale *op(R)'*op(R) (2)
C
C where op(K) = K or K' (i.e., the transpose of the matrix K), S is
C an N-by-N block upper triangular matrix with one-by-one or
C two-by-two blocks on the diagonal, R is an N-by-N upper triangular
C matrix, and scale is an output scale factor, set less than or
C equal to 1 to avoid overflow in X.
C
C In the case of equation (1) the matrix S must be stable (that
C is, all the eigenvalues of S must have negative real parts),
C and for equation (2) the matrix S must be convergent (that is,
C all the eigenvalues of S must lie inside the unit circle).
C
C ARGUMENTS
C
C Mode Parameters
C
C DISCR LOGICAL
C Specifies the type of Lyapunov equation to be solved as
C follows:
C = .TRUE. : Equation (2), discrete-time case;
C = .FALSE.: Equation (1), continuous-time case.
C
C LTRANS LOGICAL
C Specifies the form of op(K) to be used, as follows:
C = .FALSE.: op(K) = K (No transpose);
C = .TRUE. : op(K) = K**T (Transpose).
C
C Input/Output Parameters
C
C N (input) INTEGER
C The order of the matrices S and R. N >= 0.
C
C S (input) DOUBLE PRECISION array of dimension (LDS,N)
C The leading N-by-N upper Hessenberg part of this array
C must contain the block upper triangular matrix.
C The elements below the upper Hessenberg part of the array
C S are not referenced. The 2-by-2 blocks must only
C correspond to complex conjugate pairs of eigenvalues (not
C to real eigenvalues).
C
C LDS INTEGER
C The leading dimension of array S. LDS >= MAX(1,N).
C
C R (input/output) DOUBLE PRECISION array of dimension (LDR,N)
C On entry, the leading N-by-N upper triangular part of this
C array must contain the upper triangular matrix R.
C On exit, the leading N-by-N upper triangular part of this
C array contains the upper triangular matrix U.
C The strict lower triangle of R is not referenced.
C
C LDR INTEGER
C The leading dimension of array R. LDR >= MAX(1,N).
C
C SCALE (output) DOUBLE PRECISION
C The scale factor, scale, set less than or equal to 1 to
C prevent the solution overflowing.
C
C Workspace
C
C DWORK DOUBLE PRECISION array, dimension (4*N)
C
C Error Indicator
C
C INFO INTEGER
C = 0: successful exit;
C < 0: if INFO = -i, the i-th argument had an illegal
C value;
C = 1: if the Lyapunov equation is (nearly) singular
C (warning indicator);
C if DISCR = .FALSE., this means that while the
C matrix S has computed eigenvalues with negative real
C parts, it is only just stable in the sense that
C small perturbations in S can make one or more of the
C eigenvalues have a non-negative real part;
C if DISCR = .TRUE., this means that while the
C matrix S has computed eigenvalues inside the unit
C circle, it is nevertheless only just convergent, in
C the sense that small perturbations in S can make one
C or more of the eigenvalues lie outside the unit
C circle;
C perturbed values were used to solve the equation
C (but the matrix S is unchanged);
C = 2: if the matrix S is not stable (that is, one or more
C of the eigenvalues of S has a non-negative real
C part), if DISCR = .FALSE., or not convergent (that
C is, one or more of the eigenvalues of S lies outside
C the unit circle), if DISCR = .TRUE.;
C = 3: if the matrix S has two or more consecutive non-zero
C elements on the first sub-diagonal, so that there is
C a block larger than 2-by-2 on the diagonal;
C = 4: if the matrix S has a 2-by-2 diagonal block with
C real eigenvalues instead of a complex conjugate
C pair.
C
C METHOD
C
C The method used by the routine is based on a variant of the
C Bartels and Stewart backward substitution method [1], that finds
C the Cholesky factor op(U) directly without first finding X and
C without the need to form the normal matrix op(R)'*op(R) [2].
C
C The continuous-time Lyapunov equation in the canonical form
C 2
C op(S)'*op(U)'*op(U) + op(U)'*op(U)*op(S) = -scale *op(R)'*op(R),
C
C or the discrete-time Lyapunov equation in the canonical form
C 2
C op(S)'*op(U)'*op(U)*op(S) - op(U)'*op(U) = -scale *op(R)'*op(R),
C
C where U and R are upper triangular, is solved for U.
C
C REFERENCES
C
C [1] Bartels, R.H. and Stewart, G.W.
C Solution of the matrix equation A'X + XB = C.
C Comm. A.C.M., 15, pp. 820-826, 1972.
C
C [2] Hammarling, S.J.
C Numerical solution of the stable, non-negative definite
C Lyapunov equation.
C IMA J. Num. Anal., 2, pp. 303-325, 1982.
C
C NUMERICAL ASPECTS
C 3
C The algorithm requires 0(N ) operations and is backward stable.
C
C FURTHER COMMENTS
C
C The Lyapunov equation may be very ill-conditioned. In particular
C if S is only just stable (or convergent) then the Lyapunov
C equation will be ill-conditioned. "Large" elements in U relative
C to those of S and R, or a "small" value for scale, is a symptom
C of ill-conditioning. A condition estimate can be computed using
C SLICOT Library routine SB03MD.
C
C CONTRIBUTOR
C
C Release 3.0: V. Sima, Katholieke Univ. Leuven, Belgium, May 1997.
C Supersedes Release 2.0 routine SB03CZ by Sven Hammarling,
C NAG Ltd, United Kingdom, Oct. 1986.
C Partly based on SB03CZ and PLYAP1 by A. Varga, University of
C Bochum, May 1992.
C
C REVISIONS
C
C Dec. 1997, April 1998, May 1999, Feb. 2004.
C
C KEYWORDS
C
C Lyapunov equation, orthogonal transformation, real Schur form.
C
C ******************************************************************
C
C .. Parameters ..
DOUBLE PRECISION ZERO, ONE, TWO
PARAMETER ( ZERO = 0.0D0, ONE = 1.0D0, TWO = 2.0D0 )
C .. Scalar Arguments ..
LOGICAL DISCR, LTRANS
INTEGER INFO, LDR, LDS, N
DOUBLE PRECISION SCALE
C .. Array Arguments ..
DOUBLE PRECISION DWORK(*), R(LDR,*), S(LDS,*)
C .. Local Scalars ..
LOGICAL CONT, TBYT
INTEGER INFOM, ISGN, J, J1, J2, J3, K, K1, K2, K3,
$ KOUNT, KSIZE
DOUBLE PRECISION ABSSKK, ALPHA, BIGNUM, D1, D2, DR, EPS, SCALOC,
$ SMIN, SMLNUM, SUM, T1, T2, T3, T4, TAU1, TAU2,
$ TEMP, V1, V2, V3, V4
C .. Local Arrays ..
DOUBLE PRECISION A(2,2), B(2,2), U(2,2)
C .. External Functions ..
DOUBLE PRECISION DLAMCH, DLANHS
EXTERNAL DLAMCH, DLANHS
C .. External Subroutines ..
EXTERNAL DAXPY, DCOPY, DLABAD, DLARFG, DSCAL, DSWAP,
$ DTRMM, DTRMV, MB04ND, MB04OD, SB03OR, SB03OY,
$ XERBLA
C .. Intrinsic Functions ..
INTRINSIC ABS, MAX, SIGN, SQRT
C .. Executable Statements ..
C
INFO = 0
C
C Test the input scalar arguments.
C
IF( N.LT.0 ) THEN
INFO = -3
ELSE IF( LDS.LT.MAX( 1, N ) ) THEN
INFO = -5
ELSE IF( LDR.LT.MAX( 1, N ) ) THEN
INFO = -7
END IF
C
IF ( INFO.NE.0 ) THEN
C
C Error return.
C
CALL XERBLA( 'SB03OT', -INFO )
RETURN
END IF
C
SCALE = ONE
C
C Quick return if possible.
C
IF (N.EQ.0)
$ RETURN
C
C Set constants to control overflow.
C
EPS = DLAMCH( 'P' )
SMLNUM = DLAMCH( 'S' )
BIGNUM = ONE / SMLNUM
CALL DLABAD( SMLNUM, BIGNUM )
SMLNUM = SMLNUM*DBLE( N*N ) / EPS
BIGNUM = ONE / SMLNUM
C
SMIN = MAX( SMLNUM, EPS*DLANHS( 'Max', N, S, LDS, DWORK ) )
INFOM = 0
C
C Start the solution. Most of the comments refer to notation and
C equations in sections 5 and 10 of the second reference above.
C
C Determine whether or not the current block is two-by-two.
C K gives the position of the start of the current block and
C TBYT is true if the block is two-by-two.
C
CONT = .NOT.DISCR
ISGN = 1
IF ( .NOT.LTRANS ) THEN
C
C Case op(M) = M.
C
KOUNT = 1
C
10 CONTINUE
C WHILE( KOUNT.LE.N )LOOP
IF ( KOUNT.LE.N ) THEN
K = KOUNT
IF ( KOUNT.GE.N ) THEN
TBYT = .FALSE.
KOUNT = KOUNT + 1
ELSE IF ( S(K+1,K).EQ.ZERO ) THEN
TBYT = .FALSE.
KOUNT = KOUNT + 1
ELSE
TBYT = .TRUE.
IF ( (K+1).LT.N ) THEN
IF ( S(K+2,K+1).NE.ZERO ) THEN
INFO = 3
RETURN
END IF
END IF
KOUNT = KOUNT + 2
END IF
IF ( TBYT ) THEN
C
C Solve the two-by-two Lyapunov equation (6.1) or (10.19),
C using the routine SB03OY.
C
B(1,1) = S(K,K)
B(2,1) = S(K+1,K)
B(1,2) = S(K,K+1)
B(2,2) = S(K+1,K+1)
U(1,1) = R(K,K)
U(1,2) = R(K,K+1)
U(2,2) = R(K+1,K+1)
C
CALL SB03OY( DISCR, LTRANS, ISGN, B, 2, U, 2, A, 2,
$ SCALOC, INFO )
IF ( INFO.GT.1 )
$ RETURN
INFOM = MAX( INFO, INFOM )
IF( SCALOC.NE.ONE ) THEN
C
DO 20 J = 1, N
CALL DSCAL( J, SCALOC, R(1,J), 1 )
20 CONTINUE
C
SCALE = SCALE*SCALOC
END IF
R(K,K) = U(1,1)
R(K,K+1) = U(1,2)
R(K+1,K+1) = U(2,2)
C
C If we are not at the end of S then set up and solve
C equation (6.2) or (10.20).
C
C Note that SB03OY returns ( u11*s11*inv( u11 ) ) in B
C and returns scaled alpha in A. ksize is the order of
C the remainder of S. k1, k2 and k3 point to the start
C of vectors in DWORK.
C
IF ( KOUNT.LE.N ) THEN
KSIZE = N - K - 1
K1 = KSIZE + 1
K2 = KSIZE + K1
K3 = KSIZE + K2
C
C Form the right-hand side of (6.2) or (10.20), the
C first column in DWORK( 1 ) ,..., DWORK( n - k - 1 )
C the second in DWORK( n - k ) ,...,
C DWORK( 2*( n - k - 1 ) ).
C
CALL DCOPY( KSIZE, R(K,K+2), LDR, DWORK, 1 )
CALL DCOPY( KSIZE, R(K+1,K+2), LDR, DWORK(K1), 1 )
CALL DTRMM( 'Right', 'Upper', 'No transpose',
$ 'Non-unit', KSIZE, 2, -ONE, A, 2, DWORK,
$ KSIZE )
IF ( CONT ) THEN
CALL DAXPY( KSIZE, -R(K,K), S(K,K+2), LDS, DWORK,
$ 1 )
CALL DAXPY( KSIZE, -R(K,K+1), S(K+1,K+2), LDS,
$ DWORK, 1)
CALL DAXPY( KSIZE, -R(K+1,K+1), S(K+1,K+2), LDS,
$ DWORK(K1), 1 )
ELSE
CALL DAXPY( KSIZE, -R(K,K)*B(1,1), S(K,K+2), LDS,
$ DWORK, 1 )
CALL DAXPY( KSIZE, -( R(K,K+1)*B(1,1) + R(K+1,K+1)
$ *B(2,1) ), S(K+1,K+2), LDS, DWORK, 1 )
CALL DAXPY( KSIZE, -R(K,K)*B(1,2), S(K,K+2), LDS,
$ DWORK(K1), 1 )
CALL DAXPY( KSIZE, -( R(K,K+1)*B(1,2) + R(K+1,K+1)
$ *B(2,2) ), S(K+1,K+2), LDS, DWORK(K1),
$ 1 )
END IF
C
C SB03OR solves the Sylvester equations. The solution
C is overwritten on DWORK.
C
CALL SB03OR( DISCR, LTRANS, KSIZE, 2, S(K+2,K+2), LDS,
$ B, 2, DWORK, KSIZE, SCALOC, INFO )
INFOM = MAX( INFO, INFOM )
IF( SCALOC.NE.ONE ) THEN
C
DO 30 J = 1, N
CALL DSCAL( J, SCALOC, R(1,J), 1 )
30 CONTINUE
C
SCALE = SCALE*SCALOC
END IF
C
C Copy the solution into the next 2*( n - k - 1 )
C elements of DWORK.
C
CALL DCOPY( 2*KSIZE, DWORK, 1, DWORK(K2), 1 )
C
C Now form the matrix Rhat of equation (6.4) or
C (10.22). Note that (10.22) is incorrect, so here we
C implement a corrected version of (10.22).
C
IF ( CONT ) THEN
C
C Swap the two rows of R with DWORK.
C
CALL DSWAP( KSIZE, DWORK, 1, R(K,K+2), LDR )
CALL DSWAP( KSIZE, DWORK(K1), 1, R(K+1,K+2), LDR )
C
C 1st column:
C
CALL DAXPY( KSIZE, -A(1,1), DWORK(K2), 1, DWORK,
$ 1 )
CALL DAXPY( KSIZE, -A(1,2), DWORK(K3), 1, DWORK,
$ 1 )
C
C 2nd column:
C
CALL DAXPY( KSIZE, -A(2,2), DWORK(K3), 1,
$ DWORK(K1), 1 )
ELSE
C
C Form v = S1'*u + s*u11', overwriting v on DWORK.
C
C Compute S1'*u, first multiplying by the
C triangular part of S1.
C
CALL DTRMM( 'Left', 'Upper', 'Transpose',
$ 'Non-unit', KSIZE, 2, ONE, S(K+2,K+2),
$ LDS, DWORK, KSIZE )
C
C Then multiply by the subdiagonal of S1 and add in
C to the above result.
C
J1 = K1
J2 = K + 2
C
DO 40 J = 1, KSIZE-1
IF ( S(J2+1,J2).NE.ZERO ) THEN
DWORK(J) = S(J2+1,J2)*DWORK(K2+J) + DWORK(J)
DWORK(J1) = S(J2+1,J2)*DWORK(K3+J) +
$ DWORK(J1)
END IF
J1 = J1 + 1
J2 = J2 + 1
40 CONTINUE
C
C Add in s*u11'.
C
CALL DAXPY( KSIZE, R(K,K), S(K,K+2), LDS, DWORK,
$ 1 )
CALL DAXPY( KSIZE, R(K,K+1), S(K+1,K+2), LDS,
$ DWORK, 1 )
CALL DAXPY( KSIZE, R(K+1,K+1), S(K+1,K+2), LDS,
$ DWORK(K1), 1 )
C
C Next recover r from R, swapping r with u.
C
CALL DSWAP( KSIZE, DWORK(K2), 1, R(K,K+2), LDR )
CALL DSWAP( KSIZE, DWORK(K3), 1, R(K+1,K+2), LDR )
C
C Now we perform the QR factorization.
C
C ( a ) = Q*( t ),
C ( b )
C
C and form
C
C ( p' ) = Q'*( r' ).
C ( y' ) ( v' )
C
C y is then the correct vector to use in (10.22).
C Note that a is upper triangular and that t and
C p are not required.
C
CALL DLARFG( 3, A(1,1), B(1,1), 1, TAU1 )
V1 = B(1,1)
T1 = TAU1*V1
V2 = B(2,1)
T2 = TAU1*V2
SUM = A(1,2) + V1*B(1,2) + V2*B(2,2)
B(1,2) = B(1,2) - SUM*T1
B(2,2) = B(2,2) - SUM*T2
CALL DLARFG( 3, A(2,2), B(1,2), 1, TAU2 )
V3 = B(1,2)
T3 = TAU2*V3
V4 = B(2,2)
T4 = TAU2*V4
J1 = K1
J2 = K2
J3 = K3
C
DO 50 J = 1, KSIZE
SUM = DWORK(J2) + V1*DWORK(J) + V2*DWORK(J1)
D1 = DWORK(J) - SUM*T1
D2 = DWORK(J1) - SUM*T2
SUM = DWORK(J3) + V3*D1 + V4*D2
DWORK(J) = D1 - SUM*T3
DWORK(J1) = D2 - SUM*T4
J1 = J1 + 1
J2 = J2 + 1
J3 = J3 + 1
50 CONTINUE
C
END IF
C
C Now update R1 to give Rhat.
C
CALL DCOPY( KSIZE, DWORK, 1, DWORK(K2), 1 )
CALL DCOPY( KSIZE, DWORK(K1), 1, DWORK(K3), 1 )
CALL DCOPY( KSIZE, DWORK(K3), 1, DWORK(2), 2 )
CALL DCOPY( KSIZE, DWORK(K2), 1, DWORK(1), 2 )
CALL MB04OD( 'Full', KSIZE, 0, 2, R(K+2,K+2), LDR,
$ DWORK, 2, DWORK, 1, DWORK, 1, DWORK(K2),
$ DWORK(K3) )
END IF
ELSE
C
C 1-by-1 block.
C
C Make sure S is stable or convergent and find u11 in
C equation (5.13) or (10.15).
C
IF ( DISCR ) THEN
ABSSKK = ABS( S(K,K) )
IF ( ( ABSSKK - ONE ).GE.ZERO ) THEN
INFO = 2
RETURN
END IF
TEMP = SQRT( ( ONE - ABSSKK )*( ONE + ABSSKK ) )
ELSE
IF ( S(K,K).GE.ZERO ) THEN
INFO = 2
RETURN
END IF
TEMP = SQRT( ABS( TWO*S(K,K) ) )
END IF
C
SCALOC = ONE
IF( TEMP.LT.SMIN ) THEN
TEMP = SMIN
INFOM = 1
END IF
DR = ABS( R(K,K) )
IF( TEMP.LT.ONE .AND. DR.GT.ONE ) THEN
IF( DR.GT.BIGNUM*TEMP )
$ SCALOC = ONE / DR
END IF
ALPHA = SIGN( TEMP, R(K,K) )
R(K,K) = R(K,K)/ALPHA
IF( SCALOC.NE.ONE ) THEN
C
DO 60 J = 1, N
CALL DSCAL( J, SCALOC, R(1,J), 1 )
60 CONTINUE
C
SCALE = SCALE*SCALOC
END IF
C
C If we are not at the end of S then set up and solve
C equation (5.14) or (10.16). ksize is the order of the
C remainder of S. k1 and k2 point to the start of vectors
C in DWORK.
C
IF ( KOUNT.LE.N ) THEN
KSIZE = N - K
K1 = KSIZE + 1
K2 = KSIZE + K1
C
C Form the right-hand side in DWORK( 1 ),...,
C DWORK( n - k ).
C
CALL DCOPY( KSIZE, R(K,K+1), LDR, DWORK, 1 )
CALL DSCAL( KSIZE, -ALPHA, DWORK, 1 )
IF ( CONT ) THEN
CALL DAXPY( KSIZE, -R(K,K), S(K,K+1), LDS, DWORK,
$ 1 )
ELSE
CALL DAXPY( KSIZE, -S(K,K)*R(K,K), S(K,K+1), LDS,
$ DWORK, 1 )
END IF
C
C SB03OR solves the Sylvester equations. The solution is
C overwritten on DWORK.
C
CALL SB03OR( DISCR, LTRANS, KSIZE, 1, S(K+1,K+1), LDS,
$ S(K,K), 1, DWORK, KSIZE, SCALOC, INFO )
INFOM = MAX( INFO, INFOM )
IF( SCALOC.NE.ONE ) THEN
C
DO 70 J = 1, N
CALL DSCAL( J, SCALOC, R(1,J), 1 )
70 CONTINUE
C
SCALE = SCALE*SCALOC
END IF
C
C Copy the solution into the next ( n - k ) elements
C of DWORK, copy the solution back into R and copy
C the row of R back into DWORK.
C
CALL DCOPY( KSIZE, DWORK, 1, DWORK(K1), 1 )
CALL DSWAP( KSIZE, DWORK, 1, R(K,K+1), LDR )
C
C Now form the matrix Rhat of equation (5.15) or
C (10.17), first computing y in DWORK, and then
C updating R1.
C
IF ( CONT ) THEN
CALL DAXPY( KSIZE, -ALPHA, DWORK(K1), 1, DWORK, 1 )
ELSE
C
C First form lambda( 1 )*r and then add in
C alpha*u11*s.
C
CALL DSCAL( KSIZE, -S(K,K), DWORK, 1 )
CALL DAXPY( KSIZE, ALPHA*R(K,K), S(K,K+1), LDS,
$ DWORK, 1 )
C
C Now form alpha*S1'*u, first multiplying by the
C sub-diagonal of S1 and then the triangular part
C of S1, and add the result in DWORK.
C
J1 = K + 1
C
DO 80 J = 1, KSIZE-1
IF ( S(J1+1,J1).NE.ZERO ) DWORK(J)
$ = ALPHA*S(J1+1,J1)*DWORK(K1+J) + DWORK(J)
J1 = J1 + 1
80 CONTINUE
C
CALL DTRMV( 'Upper', 'Transpose', 'Non-unit',
$ KSIZE, S(K+1,K+1), LDS, DWORK(K1), 1 )
CALL DAXPY( KSIZE, ALPHA, DWORK(K1), 1, DWORK, 1 )
END IF
CALL MB04OD( 'Full', KSIZE, 0, 1, R(K+1,K+1), LDR,
$ DWORK, 1, DWORK, 1, DWORK, 1, DWORK(K2),
$ DWORK(K1) )
END IF
END IF
GO TO 10
END IF
C END WHILE 10
C
ELSE
C
C Case op(M) = M'.
C
KOUNT = N
C
90 CONTINUE
C WHILE( KOUNT.GE.1 )LOOP
IF ( KOUNT.GE.1 ) THEN
K = KOUNT
IF ( KOUNT.EQ.1 ) THEN
TBYT = .FALSE.
KOUNT = KOUNT - 1
ELSE IF ( S(K,K-1).EQ.ZERO ) THEN
TBYT = .FALSE.
KOUNT = KOUNT - 1
ELSE
TBYT = .TRUE.
K = K - 1
IF ( K.GT.1 ) THEN
IF ( S(K,K-1).NE.ZERO ) THEN
INFO = 3
RETURN
END IF
END IF
KOUNT = KOUNT - 2
END IF
IF ( TBYT ) THEN
C
C Solve the two-by-two Lyapunov equation corresponding to
C (6.1) or (10.19), using the routine SB03OY.
C
B(1,1) = S(K,K)
B(2,1) = S(K+1,K)
B(1,2) = S(K,K+1)
B(2,2) = S(K+1,K+1)
U(1,1) = R(K,K)
U(1,2) = R(K,K+1)
U(2,2) = R(K+1,K+1)
C
CALL SB03OY( DISCR, LTRANS, ISGN, B, 2, U, 2, A, 2,
$ SCALOC, INFO )
IF ( INFO.GT.1 )
$ RETURN
INFOM = MAX( INFO, INFOM )
IF( SCALOC.NE.ONE ) THEN
C
DO 100 J = 1, N
CALL DSCAL( J, SCALOC, R(1,J), 1 )
100 CONTINUE
C
SCALE = SCALE*SCALOC
END IF
R(K,K) = U(1,1)
R(K,K+1) = U(1,2)
R(K+1,K+1) = U(2,2)
C
C If we are not at the front of S then set up and solve
C equation corresponding to (6.2) or (10.20).
C
C Note that SB03OY returns ( inv( u11 )*s11*u11 ) in B
C and returns scaled alpha, alpha = inv( u11 )*r11, in A.
C ksize is the order of the remainder leading part of S.
C k1, k2 and k3 point to the start of vectors in DWORK.
C
IF ( KOUNT.GE.1 ) THEN
KSIZE = K - 1
K1 = KSIZE + 1
K2 = KSIZE + K1
K3 = KSIZE + K2
C
C Form the right-hand side of equations corresponding to
C (6.2) or (10.20), the first column in DWORK( 1 ) ,...,
C DWORK( k - 1 ) the second in DWORK( k ) ,...,
C DWORK( 2*( k - 1 ) ).
C
CALL DCOPY( KSIZE, R(1,K), 1, DWORK, 1 )
CALL DCOPY( KSIZE, R(1,K+1), 1, DWORK(K1), 1 )
CALL DTRMM( 'Right', 'Upper', 'Transpose', 'Non-unit',
$ KSIZE, 2, -ONE, A, 2, DWORK, KSIZE )
IF ( CONT ) THEN
CALL DAXPY( KSIZE, -R(K,K), S(1,K), 1, DWORK, 1 )
CALL DAXPY( KSIZE, -R(K,K+1), S(1,K), 1, DWORK(K1),
$ 1 )
CALL DAXPY( KSIZE, -R(K+1,K+1), S(1,K+1), 1,
$ DWORK(K1), 1 )
ELSE
CALL DAXPY( KSIZE, -( R(K,K)*B(1,1) + R(K,K+1)
$ *B(1,2) ), S(1,K), 1, DWORK, 1 )
CALL DAXPY( KSIZE, -R(K+1,K+1)*B(1,2), S(1,K+1), 1,
$ DWORK, 1 )
CALL DAXPY( KSIZE, -( R(K,K)*B(2,1) + R(K,K+1)
$ *B(2,2) ), S(1,K), 1, DWORK(K1), 1 )
CALL DAXPY( KSIZE, -R(K+1,K+1)*B(2,2), S(1,K+1), 1,
$ DWORK(K1), 1 )
END IF
C
C SB03OR solves the Sylvester equations. The solution
C is overwritten on DWORK.
C
CALL SB03OR( DISCR, LTRANS, KSIZE, 2, S, LDS, B, 2,
$ DWORK, KSIZE, SCALOC, INFO )
INFOM = MAX( INFO, INFOM )
IF( SCALOC.NE.ONE ) THEN
C
DO 110 J = 1, N
CALL DSCAL( J, SCALOC, R(1,J), 1 )
110 CONTINUE
C
SCALE = SCALE*SCALOC
END IF
C
C Copy the solution into the next 2*( k - 1 ) elements
C of DWORK.
C
CALL DCOPY( 2*KSIZE, DWORK, 1, DWORK(K2), 1 )
C
C Now form the matrix Rhat of equation corresponding
C to (6.4) or (10.22) (corrected version).
C
IF ( CONT ) THEN
C
C Swap the two columns of R with DWORK.
C
CALL DSWAP( KSIZE, DWORK, 1, R(1,K), 1 )
CALL DSWAP( KSIZE, DWORK(K1), 1, R(1,K+1), 1 )
C
C 1st column:
C
CALL DAXPY( KSIZE, -A(1,1), DWORK(K2), 1, DWORK,
$ 1 )
C
C 2nd column:
C
CALL DAXPY( KSIZE, -A(1,2), DWORK(K2), 1,
$ DWORK(K1), 1 )
CALL DAXPY( KSIZE, -A(2,2), DWORK(K3), 1,
$ DWORK(K1), 1 )
ELSE
C
C Form v = S1*u + s*u11, overwriting v on DWORK.
C
C Compute S1*u, first multiplying by the triangular
C part of S1.
C
CALL DTRMM( 'Left', 'Upper', 'No transpose',
$ 'Non-unit', KSIZE, 2, ONE, S, LDS,
$ DWORK, KSIZE )
C
C Then multiply by the subdiagonal of S1 and add in
C to the above result.
C
J1 = K1
C
DO 120 J = 2, KSIZE
J1 = J1 + 1
IF ( S(J,J-1).NE.ZERO ) THEN
DWORK(J) = S(J,J-1)*DWORK(K2+J-2) + DWORK(J)
DWORK(J1) = S(J,J-1)*DWORK(K3+J-2) +
$ DWORK(J1)
END IF
120 CONTINUE
C
C Add in s*u11.
C
CALL DAXPY( KSIZE, R(K,K), S(1,K), 1, DWORK, 1 )
CALL DAXPY( KSIZE, R(K,K+1), S(1,K), 1, DWORK(K1),
$ 1 )
CALL DAXPY( KSIZE, R(K+1,K+1), S(1,K+1), 1,
$ DWORK(K1), 1 )
C
C Next recover r from R, swapping r with u.
C
CALL DSWAP( KSIZE, DWORK(K2), 1, R(1,K), 1 )
CALL DSWAP( KSIZE, DWORK(K3), 1, R(1,K+1), 1 )
C
C Now we perform the QL factorization.
C
C ( a' ) = Q*( t ),
C ( b' )
C
C and form
C
C ( p' ) = Q'*( r' ).
C ( y' ) ( v' )
C
C y is then the correct vector to use in the
C relation corresponding to (10.22).
C Note that a is upper triangular and that t and
C p are not required.
C
CALL DLARFG( 3, A(2,2), B(2,1), 2, TAU1 )
V1 = B(2,1)
T1 = TAU1*V1
V2 = B(2,2)
T2 = TAU1*V2
SUM = A(1,2) + V1*B(1,1) + V2*B(1,2)
B(1,1) = B(1,1) - SUM*T1
B(1,2) = B(1,2) - SUM*T2
CALL DLARFG( 3, A(1,1), B(1,1), 2, TAU2 )
V3 = B(1,1)
T3 = TAU2*V3
V4 = B(1,2)
T4 = TAU2*V4
J1 = K1
J2 = K2
J3 = K3
C
DO 130 J = 1, KSIZE
SUM = DWORK(J3) + V1*DWORK(J) + V2*DWORK(J1)
D1 = DWORK(J) - SUM*T1
D2 = DWORK(J1) - SUM*T2
SUM = DWORK(J2) + V3*D1 + V4*D2
DWORK(J) = D1 - SUM*T3
DWORK(J1) = D2 - SUM*T4
J1 = J1 + 1
J2 = J2 + 1
J3 = J3 + 1
130 CONTINUE
C
END IF
C
C Now update R1 to give Rhat.
C
CALL MB04ND( 'Full', KSIZE, 0, 2, R, LDR, DWORK,
$ KSIZE, DWORK, 1, DWORK, 1, DWORK(K2),
$ DWORK(K3) )
END IF
ELSE
C
C 1-by-1 block.
C
C Make sure S is stable or convergent and find u11 in
C equation corresponding to (5.13) or (10.15).
C
IF ( DISCR ) THEN
ABSSKK = ABS( S(K,K) )
IF ( ( ABSSKK - ONE ).GE.ZERO ) THEN
INFO = 2
RETURN
END IF
TEMP = SQRT( ( ONE - ABSSKK )*( ONE + ABSSKK ) )
ELSE
IF ( S(K,K).GE.ZERO ) THEN
INFO = 2
RETURN
END IF
TEMP = SQRT( ABS( TWO*S(K,K) ) )
END IF
C
SCALOC = ONE
IF( TEMP.LT.SMIN ) THEN
TEMP = SMIN
INFOM = 1
END IF
DR = ABS( R(K,K) )
IF( TEMP.LT.ONE .AND. DR.GT.ONE ) THEN
IF( DR.GT.BIGNUM*TEMP )
$ SCALOC = ONE / DR
END IF
ALPHA = SIGN( TEMP, R(K,K) )
R(K,K) = R(K,K)/ALPHA
IF( SCALOC.NE.ONE ) THEN
C
DO 140 J = 1, N
CALL DSCAL( J, SCALOC, R(1,J), 1 )
140 CONTINUE
C
SCALE = SCALE*SCALOC
END IF
C
C If we are not at the front of S then set up and solve
C equation corresponding to (5.14) or (10.16). ksize is
C the order of the remainder leading part of S. k1 and k2
C point to the start of vectors in DWORK.
C
IF ( KOUNT.GE.1 ) THEN
KSIZE = K - 1
K1 = KSIZE + 1
K2 = KSIZE + K1
C
C Form the right-hand side in DWORK( 1 ),...,
C DWORK( k - 1 ).
C
CALL DCOPY( KSIZE, R(1,K), 1, DWORK, 1 )
CALL DSCAL( KSIZE, -ALPHA, DWORK, 1 )
IF ( CONT ) THEN
CALL DAXPY( KSIZE, -R(K,K), S(1,K), 1, DWORK, 1 )
ELSE
CALL DAXPY( KSIZE, -S(K,K)*R(K,K), S(1,K), 1,
$ DWORK, 1 )
END IF
C
C SB03OR solves the Sylvester equations. The solution is
C overwritten on DWORK.
C
CALL SB03OR( DISCR, LTRANS, KSIZE, 1, S, LDS, S(K,K),
$ 1, DWORK, KSIZE, SCALOC, INFO )
INFOM = MAX( INFO, INFOM )
IF( SCALOC.NE.ONE ) THEN
C
DO 150 J = 1, N
CALL DSCAL( J, SCALOC, R(1,J), 1 )
150 CONTINUE
C
SCALE = SCALE*SCALOC
END IF
C
C Copy the solution into the next ( k - 1 ) elements
C of DWORK, copy the solution back into R and copy
C the column of R back into DWORK.
C
CALL DCOPY( KSIZE, DWORK, 1, DWORK(K1), 1 )
CALL DSWAP( KSIZE, DWORK, 1, R(1,K), 1 )
C
C Now form the matrix Rhat of equation corresponding
C to (5.15) or (10.17), first computing y in DWORK,
C and then updating R1.
C
IF ( CONT ) THEN
CALL DAXPY( KSIZE, -ALPHA, DWORK(K1), 1, DWORK, 1 )
ELSE
C
C First form lambda( 1 )*r and then add in
C alpha*u11*s.
C
CALL DSCAL( KSIZE, -S(K,K), DWORK, 1 )
CALL DAXPY( KSIZE, ALPHA*R(K,K), S(1,K), 1, DWORK,
$ 1 )
C
C Now form alpha*S1*u, first multiplying by the
C sub-diagonal of S1 and then the triangular part
C of S1, and add the result in DWORK.
C
DO 160 J = 2, KSIZE
IF ( S(J,J-1).NE.ZERO ) DWORK(J)
$ = ALPHA*S(J,J-1)*DWORK(K1+J-2) + DWORK(J)
160 CONTINUE
C
CALL DTRMV( 'Upper', 'No transpose', 'Non-unit',
$ KSIZE, S, LDS, DWORK(K1), 1 )
CALL DAXPY( KSIZE, ALPHA, DWORK(K1), 1, DWORK, 1 )
END IF
CALL MB04ND( 'Full', KSIZE, 0, 1, R, LDR, DWORK,
$ KSIZE, DWORK, 1, DWORK, 1, DWORK(K2),
$ DWORK(K1) )
END IF
END IF
GO TO 90
END IF
C END WHILE 90
C
END IF
INFO = INFOM
RETURN
C *** Last line of SB03OT ***
END
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