fcode 发表于 2020-7-2 10:57
矩阵不存在“求解”问题。它本身只是数据表达的一种方式，不是一个“待解的问题”。
你需要描述你的具体问 ...
!---------------------------------------------------------------------
! Routine to solve a diagonally dominant scalar tridiagonal system
! by compact Crout LU method.
!
! A(I)*x(I-1) + B(I)*x(I) + C(I)*x(I+1) = D(I)
! for 1 <= I <= N, with A(1) = C(N) = 0
!
! A : the subdiagonal coefficients, size N,
! A(1) is not used.
! B : the diagonal coefficients, size N.
! C : the above-diagonal coefficients, size N,
! C(N) is not used.
! D : the right-hand-side coefficients, size N,
! will contain the solution x on exit
! Ier : Error flag,
! = 0, normal return
! = -1, zero-sized of A, no problem to solve
! = -2, array B, C, or, D not conformable with A
!---------------------------------------------------------------------
SUBROUTINE Axb_Tri_Scalar( A, B, C, D, Ier )
IMPLICIT NONE
REAL, DIMENSION(:), INTENT(IN ) :: A
REAL, DIMENSION(:), INTENT(INOUT) :: B, C, D
INTEGER, INTENT( OUT) :: Ier
INTEGER :: N, I
!-------------------------------------
N = SIZE( A )
IF( N == 0 ) THEN
Ier = -1
RETURN
END IF
IF( ( N /= SIZE(B) ) .OR. &
( N /= SIZE(C) ) .OR. &
( N /= SIZE(D) ) ) THEN
Ier = -2
RETURN
END IF
!-----------------
C(N) = 0.0
C(1) = C(1) / B(1)
D(1) = D(1) / B(1)
DO I = 2, N
B(I) = B(I) - A(I)*C(I-1)
D(I) = ( D(I) - A(I)*D(I-1) ) / B(I)
C(I) = C(I) / B(I)
END DO
! Backward Substitution
DO I = (N-1), 1, -1
D(I) = D(I) - C(I)*D(I+1)
END DO
Ier = 0
END SUBROUTINE Axb_Tri_Scalar
!---------------------------------------------------------------------
! Routine to solve a diagonally dominant scalar pentadiagonal system
! by compact Crout LU method.
!
! A(I)*x(I-2)+B(I)*x(I-1)+C(I)*x(I)+D(I)*x(I+1)+E(I)*x(I+2)=F(I)
! for 1 <= I <= N, with A(1)=A(2)=B(1)=D(N)=E(N-1)=E(N)=0
!
! A : the 1st subdiagonal coefficients, size N,
! A(1) and A(2) are not used.
! B : the 2nd subdiagonal coefficients, size N,
! B(1) is not used.
! C : the diagonal coefficients, size N.
! D : the 1st above-diagonal coefficients, size N,
! D(N) is not used.
! E : the 2nd above-diagonal coefficients, size N,
! E(N) and E(N-1) are not used.
! F : the right-hand-side coefficients, size N,
! will contain the solution x on exit
! Ier : Error flag,
! = 0, normal return
! = -1, zero-sized of A, no problem to solve
! = -2, array B,C,D,E, or, F not conformable with A
!---------------------------------------------------------------------
SUBROUTINE Axb_Penta_Scalar( A, B, C, D, E, F, Ier )
IMPLICIT NONE
REAL, DIMENSION(:), INTENT(IN ) :: A
REAL, DIMENSION(:), INTENT(INOUT) :: B, C, D, E, F
INTEGER, INTENT( OUT) :: Ier
INTEGER :: N, N1, N2, N3, I, I1, I2
!-------------------------------------
N = SIZE(A)
IF( N == 0 ) THEN
Ier = -1
RETURN
END IF
IF( ( N /= SIZE(B) ) .OR. &
( N /= SIZE(C) ) .OR. &
( N /= SIZE(D) ) .OR. &
( N /= SIZE(E) ) .OR. &
( N /= SIZE(F) ) ) THEN
Ier = -2
RETURN
END IF
!-----------------
D(1) = D(1) / C(1)
E(1) = E(1) / C(1)
F(1) = F(1) / C(1)
C(2) = C(2) - B(2)*D(1)
D(2) = ( D(2) - B(2)*E(1) ) / C(2)
E(2) = E(2) / C(2)
F(2) = ( F(2) - B(2)*F(1) ) / C(2)
DO I = 3, (N-2)
I1 = I-1
I2 = I-2
B(I) = B(I) - A(I)*D(I2)
C(I) = C(I) - B(I)*D(I1) - A(I)*E(I2)
D(I) = ( D(I) - B(I)*E(I1) ) / C(I)
E(I) = E(I) / C(I)
F(I) = ( F(I) - B(I)*F(I1) - A(I)*F(I2) ) / C(I)
END DO
N1 = N-1
N2 = N-2
N3 = N-3
B(N1) = B(N1) - A(N1)*D(N3)
C(N1) = C(N1) - B(N1)*D(N2) - A(N1)*E(N3)
D(N1) = ( D(N1) - B(N1)*E(N2) ) / C(N1)
F(N1) = ( F(N1) - B(N1)*F(N2) - A(N1)*F(N3) ) / C(N1)
N1 = N-1
N2 = N-2
B(N) = B(N) - A(N)*D(N2)
C(N) = C(N) - B(N)*D(N1) - A(N)*E(N2)
F(N) = ( F(N) - B(N)*F(N1) - A(N)*F(N2) ) /C(N)
!----------------------
! Backward Substitution
!----------------------
I = N-1
F(I) = F(I) - D(I)*F(I+1)
DO I = (N-2), 1, -1
F(I) = F(I) - D(I)*F(I+1) - E(I)*F(I+2)
END DO
Ier = 0
END SUBROUTINE Axb_Penta_Scalar
1.88 MB, 下载次数: 28
chiangtp 发表于 2020-7-5 00:19
[mw_shl_code=fortran,false]!---------------------------------------------------------------------
! ...
Pluto 发表于 2020-7-2 15:02
如图所示。我问的其实是数值传热学中的一个问题，书中给出了求解三对角阵算法的过程，现在我要求解五对角 ...
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