三对角线性方程组(tridiagonal systems of equations)的求解

三对角线性方程组(tridiagonal systems of equations)

三对角线性方程组,对于熟悉数值分析的同学来说，并不陌生，它经常出现在微分方程的数值求解和三次样条函数的插值问题中。三对角线性方程组可描述为以下方程组：

[a_{i}x_{i-1}+b_{i}x_{i}+c_{i}x_{i+1}=d_{i} ]

其中(1leq i leq n, a_{1}=0, c_{n}=0.) 以上方程组写成矩阵形式为(Ax=d)，即：

[{egin{bmatrix} {b_{1}}&{c_{1}}&{}&{}&{0}\ {a_{2}}&{b_{2}}&{c_{2}}&{}&{}\ {}&{a_{3}}&{b_{3}}&ddots &{}\ {}&{}&ddots &ddots &{c_{n-1}}\ {0}&&&{a_{n}}&{b_{n}}\ end{bmatrix}} {egin{bmatrix}{x_{1}}\{x_{2}}\{x_{3}}\vdots \{x_{n}}\end{bmatrix}}={egin{bmatrix}{d_{1}}\{d_{2}}\{d_{3}}\vdots \{d_{n}}\end{bmatrix}} ]

三对角线性方程组的求解采用追赶法或者Thomas算法，它是Gauss消去法在三对角线性方程组这种特殊情形下的应用，因此，主要思想还是Gauss消去法，只是会更加简单些。我们将在下面的算法详述中给出该算法的具体求解过程。
当然，该算法并不总是稳定的，但当系数矩阵(A)为严格对角占优矩阵（Strictly D iagonally Dominant, SDD）或对称正定矩阵（Symmetric Positive Definite, SPD）时，该算法稳定。对于不熟悉SDD或者SPD的读者，也不必担心，我们还会在我们的博客中介绍这类矩阵。现在，我们只要记住，该算法对于部分系数矩阵(A)是可以求解的。

算法详述

追赶法或者Thomas算法的具体步骤如下：

创建新系数(c_{i}^{*})和(d_{i}^{*})来代替原先的(a_{i},b_{i},c_{i}),公式如下：

[c^{*}_i = left{ egin{array}{lr} frac{c_1}{b_1} & ; i = 1\ frac{c_i}{b_i - a_i c^{*}_{i-1}} & ; i = 2,3,...,n-1 end{array} ight.\ d^{*}_i = left{ egin{array}{lr} frac{d_1}{b_1} & ; i = 1\ frac{d_i- a_i d^{*}_{i-1}}{b_i - a_i c^{*}_{i-1}} & ; i = 2,3,...,n-1 end{array} ight. ]

改写原先的方程组(Ax=d)如下：

[egin{bmatrix} 1 & c^{*}_1 & 0 & 0 & ... & 0 \ 0 & 1 & c^{*}_2 & 0 & ... & 0 \ 0 & 0 & 1 & c^{*}_3 & 0 & 0 \ . & . & & & & . \ . & . & & & & . \ . & . & & & & c^{*}_{n-1} \ 0 & 0 & 0 & 0 & 0 & 1 \ end{bmatrix} egin{bmatrix} x_1 \ x_2 \ x_3 \ .\ .\ .\ x_k \ end{bmatrix} = egin{bmatrix} d^{*}_1 \ d^{*}_2 \ d^{*}_3 \ .\ .\ .\ d^{*}_n \ end{bmatrix} ]

计算解向量(x)，如下：

[x_n = d^{*}_n, qquad x_i = d^{*}_i - c^{*}_i x_{i+1}, qquad i = n-1, n-2, ... ,2,1 ]

以上算法得到的解向量(x)即为原方程(Ax=d)的解。
下面，我们来证明该算法的正确性，只需要证明该算法保持原方程组的形式不变。
首先，当(i=1)时，

[1*x_{1}+c_{1}^{*}x_{2}=d_{1}^{*} Leftrightarrow 1*x_{1}+frac{c_{1}}{b_{1}}x_{2}=frac{d_{1}}{b_{1}}Leftrightarrow b_{1}*x_{1}+c_{1}x_{2}=d_{1} ]

当(i>1)时，

[1*x_{i}+c_{i}^{*}x_{i+1}=d_{i}^{*} Leftrightarrow 1*x_{i}+frac{c_{i}}{b_{i} - a_{i} c^{*}_{i-1}}x_{i+1}=frac{d_{i}- a_{i} d^{*}_{i-1}}{b_{i} - a_{i} c^{*}_{i-1}} Leftrightarrow (b_{i} - a_{i} c^{*}_{i-1})x_{i}+c_{i}x_{i+1}=d_{i}- a_{i} d^{*}_{i-1}]

结合(a_{i}x_{i-1}+b_{i}x_{i}+c_{i}x_{i+1}=d_{i})，只需要证明(x_{i-1}+c_{i-1}^{*}x_{i}=d_{i-1}^{*}),而这已经在该算法的第（3）步的中的计算(x_{i-1})中给出。证明完毕。

Python实现

我们将要求解的线性方程组如下：

[{egin{bmatrix} 4&1&{0}&{0}&{0}\ {1}&{4}&{1}&{0}&{0}\ {0}&{1}&{4}&{1}&{0}\ {0}&{0}&{1}&{4}&{1}\ {0}&{0}&{0}&{1}&{4}\ end{bmatrix}} {egin{bmatrix}{x_{1}}\{x_{2}}\{x_{3}}\{x_{4}} \{x_{5}}\end{bmatrix}}={egin{bmatrix}{1\0.5\ -1\3\2}\end{bmatrix}} ]

接下来，我们将用Python来实现该算法，函数为TDMA，输入参数为列表a,b,c,d, 输出为解向量x，代码如下：

# use Thomas Method to solve tridiagonal linear equation
# algorithm reference: https://en.wikipedia.org/wiki/Tridiagonal_matrix_algorithm

import numpy as np

# parameter: a,b,c,d are list-like of same length
# tridiagonal linear equation: Ax=d
# b: main diagonal of matrix A
# a: main diagonal below of matrix A
# c: main diagonal upper of matrix A
# d: Ax=d
# return: x(type=list), the solution of Ax=d
def TDMA(a,b,c,d):

    try:
        n = len(d)    # order of tridiagonal square matrix

        # use a,b,c to create matrix A, which is not necessary in the algorithm
        A = np.array([[0]*n]*n, dtype='float64')

        for i in range(n):
            A[i,i] = b[i]
            if i > 0:
                A[i, i-1] = a[i]
            if i < n-1:
                A[i, i+1] = c[i]

        # new list of modified coefficients
        c_1 = [0]*n
        d_1 = [0]*n

        for i in range(n):
            if not i:
                c_1[i] = c[i]/b[i]
                d_1[i] = d[i] / b[i]
            else:
                c_1[i] = c[i]/(b[i]-c_1[i-1]*a[i])
                d_1[i] = (d[i]-d_1[i-1]*a[i])/(b[i]-c_1[i-1] * a[i])

        # x: solution of Ax=d
        x = [0]*n

        for i in range(n-1, -1, -1):
            if i == n-1:
                x[i] = d_1[i]
            else:
                x[i] = d_1[i]-c_1[i]*x[i+1]

        x = [round(_, 4) for _ in x]

        return x

    except Exception as e:
        return e

def main():

    a = [0, 1, 1, 1, 1]
    b = [4, 4, 4, 4, 4]
    c = [1, 1, 1, 1, 0]
    d = [1, 0.5, -1, 3, 2]

    '''
    a = [0, 2, 1, 3]
    b = [1, 1, 2, 1]
    c = [2, 3, 0.5, 0]
    d = [2, -1, 1, 3]
    '''

    x = TDMA(a, b, c, d)
    print('The solution is %s'%x)

main()

运行该程序，输出结果为：

The solution is [0.2, 0.2, -0.5, 0.8, 0.3]

本算法的Github地址为： https://github.com/percent4/Numeric_Analysis/blob/master/TDMA.py .
最后再次声明，追赶法或者Thomas算法并不是对所有的三对角矩阵都是有效的，只是部分三对角矩阵可行。