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}

$a_{i}x_{i-1}+b_{i}x_{i}+c_{i}x_{i+1}=d_{i}$
其中

1 \leq i \leq n, a_{1} = 0, c_{n} = 0.

$1\leq i \leq n, a_{1}=0, c_{n}=0.$ 以上方程组写成矩阵形式为

A x = d

$Ax=d$ ，即：

[\begin{matrix} b_{1} & c_{1} & 0 \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & ⋱ \\ ⋱ & ⋱ & c_{n - 1} \\ 0 & a_{n} & b_{n} \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ ⋮ \\ x_{n} \end{matrix}] = [\begin{matrix} d_{1} \\ d_{2} \\ d_{3} \\ ⋮ \\ d_{n} \end{matrix}]

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

$A$ 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

$A$ 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：
$c_{i}^{*} = {\begin{array}{lr} \frac{c_{1}}{b_{1}} & ; i = 1 \\ \frac{c_{i}}{b_{i} - a_{i} c_{i - 1}^{*}} & ; i = 2, 3, . . ., n - 1 \end{array} d_{i}^{*} = {\begin{array}{lr} \frac{d_{1}}{b_{1}} & ; i = 1 \\ \frac{d_{i} - a_{i} d_{i - 1}^{*}}{b_{i} - a_{i} c_{i - 1}^{*}} & ; i = 2, 3, . . ., n - 1 \end{array}$ $c^{*}_i = \left\{ \begin{array}{lr} \frac{c_1}{b_1} & ; i = 1\\ \frac{c_i}{b_i - a_i c^{*}_{i-1}} & ; i = 2,3,...,n-1 \end{array} \right.\\ d^{*}_i = \left\{ \begin{array}{lr} \frac{d_1}{b_1} & ; i = 1\\ \frac{d_i- a_i d^{*}_{i-1}}{b_i - a_i c^{*}_{i-1}} & ; i = 2,3,...,n-1 \end{array} \right.$
改写原先的方程组 $Ax=d$ 如下：
$[\begin{matrix} 1 & c_{1}^{*} & 0 & 0 & . . . & 0 \\ 0 & 1 & c_{2}^{*} & 0 & . . . & 0 \\ 0 & 0 & 1 & c_{3}^{*} & 0 & 0 \\ . & . & . \\ . & . & . \\ . & . & c_{n - 1}^{*} \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ . \\ . \\ . \\ x_{k} \end{matrix}] = [\begin{matrix} d_{1}^{*} \\ d_{2}^{*} \\ d_{3}^{*} \\ . \\ . \\ . \\ d_{n}^{*} \end{matrix}]$ $\begin{bmatrix} 1 & c^{*}_1 & 0 & 0 & ... & 0 \\ 0 & 1 & c^{*}_2 & 0 & ... & 0 \\ 0 & 0 & 1 & c^{*}_3 & 0 & 0 \\ . & . & & & & . \\ . & . & & & & . \\ . & . & & & & c^{*}_{n-1} \\ 0 & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ .\\ .\\ .\\ x_k \\ \end{bmatrix} = \begin{bmatrix} d^{*}_1 \\ d^{*}_2 \\ d^{*}_3 \\ .\\ .\\ .\\ d^{*}_n \\ \end{bmatrix}$
计算解向量 $x$ ，如下：
$x_{n} = d_{n}^{*}, x_{i} = d_{i}^{*} - c_{i}^{*} x_{i + 1}, i = n - 1, n - 2, . . ., 2, 1$ $x_n = d^{*}_n, \qquad x_i = d^{*}_i - c^{*}_i x_{i+1}, \qquad i = n-1, n-2, ... ,2,1$

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}

$1*x_{1}+c_{1}^{*}x_{2}=d_{1}^{*} \Leftrightarrow 1*x_{1}+\frac{c_{1}}{b_{1}}x_{2}=\frac{d_{1}}{b_{1}}\Leftrightarrow b_{1}*x_{1}+c_{1}x_{2}=d_{1}$
当

i > 1

$i>1$ 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}^{*}

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

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

$a_{i}x_{i-1}+b_{i}x_{i}+c_{i}x_{i+1}=d_{i}$ ，只需要证明

x_{i - 1} + c_{i - 1}^{*} x_{i} = d_{i - 1}^{*}

$x_{i-1}+c_{i-1}^{*}x_{i}=d_{i-1}^{*}$ ,而这已经在该算法的第（3）步的中的计算

x_{i - 1}

$x_{i-1}$ 中给出。证明完毕。

Python实现

我们将要求解的线性方程组如下：

[\begin{matrix} 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{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{matrix}] = [\begin{matrix} 1 0.5 - 1 3 2 \end{matrix}]

${\begin{bmatrix} 4&1&{0}&{0}&{0}\\ {1}&{4}&{1}&{0}&{0}\\ {0}&{1}&{4}&{1}&{0}\\ {0}&{0}&{1}&{4}&{1}\\ {0}&{0}&{0}&{1}&{4}\\ \end{bmatrix}} {\begin{bmatrix}{x_{1}}\\{x_{2}}\\{x_{3}}\\{x_{4}} \\{x_{5}}\\\end{bmatrix}}={\begin{bmatrix}{1\\0.5\\ -1\\3\\2}\\\end{bmatrix}}$
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。