在Fortran应用程序中接收NaN值

Question

我正在开发一个Fortran应用程序,用于数值求解边界值问题的二阶ODE类型:-y''+ q(x)*y = r(x).在这个应用程序中,我使用高斯消除算法来求解线性方程组并将解决方案写入文件中.但对于解决方案向量,我收到了NaN.为什么会这样？这是一些代码.

      subroutine gaussian_solve(s, c, error)

    double precision, dimension(:,:), intent(in out) :: s
    double precision, dimension(:),   intent(in out) :: c
         integer        :: error

    if(error == 0) then
        call back_substitution(s, c)
    end if

     end subroutine gaussian_solve
!=========================================================================================

!================= Subroutine gaussian_ellimination ===============================
      subroutine gaussion_ellimination(s, c, error)

    double precision, dimension(:,:), intent(in out) :: s
    double precision, dimension(:),   intent(in out) :: c
    integer,            intent(out)  :: error

    real, dimension(size(s, 1)) :: temp_array
    integer, dimension(1)       :: ksave
    integer                     :: i, j, k, n
    real                        :: temp, m

    n = size(s, 1)

    if(n == 0) then
        error = -1 
        return
    end if

    if(n /= size(s, 2)) then
        error = -2 
        return
    end if

    if(n /= size(s, 2)) then 
        error = -3 
        return
    end if

    error = 0
    do i = 1, n-1
        ksave = maxloc(abs(s(i:n, i)))
        k = ksave(1) + i - 1
        if(s(k, i) == 0) then
            error = -4
            return
        end if

        if(k /= i) then
            temp_array = s(i, :)
            s(i, :) = s(k, :)
            s(k, :) = temp_array
            temp = c(i)
            c(i) = c(k)
            c(k) = temp
        end if

        do j = i + 1, n
            m = s(j, i)/s(i, i)
            s(j, :) = s(j, :) - m*s(i, :)
            c(j) = c(j) - m*c(i)
        end do
    end do

     end subroutine gaussion_ellimination
     !==========================================================================================

   !================= Subroutine back_substitution ========================================
     subroutine back_substitution(s, c)

    double precision, dimension(:,:), intent(in) :: s
    double precision, dimension(:),   intent(in out) :: c

    real    :: w
    integer :: i, j, n

    n = size(c)

    do i = n, 1, -1
        w = c(i)
        do j = i + 1, n
            w = w - s(i, j)*c(j)
        end do
        c(i) = w/s(i, i)
    end do

      end subroutine back_substitution

其中s(i,j)是系统的系数矩阵,c(i)是解向量.

Answer 1

你永远不应该为高斯消除或类似的矩阵运算编写自己的例程.像LAPACK这样无处不在的软件包将拥有比您自己编写代码的更快,更准确的版本; 在LAPACK中,你会使用_getrf和_getrs的组合作为一般的matricies,但如果你有带状或sytrictric matricies,那么也有特殊的例程.您会发现很容易为您的系统找到并安装优化的线性代数包.(寻找像atlas或flame或gotoblas这样的包名).

在上面的代码,应该大概是call gaussion_ellimination(s, c, error)你开始接近gaussian_solve常规,你temp_array(和temp和m和w)也应该是双精度,以避免您的双精度矩阵失去精度,检查精确平等浮点零是一个危险的策略,我会检查你的输入矩阵 - 如果有任何线性退化,你将获得所有NaN(特别是如果它是最后一行与任何较早的那些线性退化的行向量).

如果这不能解决问题,你可以使用信令NaN来找出问题首先出现的地方 - 强迫gfortran首先停止程序NaN - 但你最好只使用现有的包来做这样的东西,它有多年来一直研究线性方程组数值解的人写的.