【数值分析】追赶法解三对角方程组，matlab实现

追赶法

设三对角矩阵 $A$ ，如果存在Doolittle分解，则有
A = [ a 1 c 1 b 2 a 2 c 2 ⋱ ⋱ ⋱ b n − 1 a n − 1 c n − 1 b n a n ] A= \begin{bmatrix} a_1 & c_1 & & & \\ b_2 & a_2 & c_2 & & \\ & \ddots & \ddots & \ddots & \\ & & b_{n-1} & a_{n-1} & c_{n-1} \\ & & & b_n & a_n \end{bmatrix} A= ​a1​b2​​c1​a2​⋱​c2​⋱bn−1​​⋱an−1​bn​​cn−1​an​​ ​

L U = A = [ 1 p 2 1 p 3 1 ⋱ ⋱ p n 1 ] [ q 1 c 1 q 2 c 2 ⋱ ⋱ q n − 1 c n − 1 q n ] LU=A= \begin{bmatrix} 1 & & & \\ p_2 & 1 & & \\ & p3 & 1& \\ & & \ddots & \ddots \\ &&&p_n&1 \end{bmatrix} \begin{bmatrix} q_1 & c_1 & & & \\ & q_2 & c_2 & & \\ & & \ddots & \ddots & \\ & & & q_{n-1} & c_{n-1} \\ & & & & q_n \end{bmatrix} LU=A= ​1p2​​1p3​1⋱​⋱pn​​1​ ​ ​q1​​c1​q2​​c2​⋱​⋱qn−1​​cn−1​qn​​ ​

步骤：
$$\begin{array}{rl}{q}_1=& a_1\\ & \\ p_i=& \frac{b_i}{q_{i-1}},i=2,3,\cdots n,我自己除以我上面\\ & \\ q_i=& a_i-p_i{c}_{i-1},我自己减去我左边乘以我上面\end{array}$$
$$\{\begin{array}{l}Ly=b\\ Ux=y\end{array}$$
$$\begin{array}{rl}{y}_1=& b_1\\ & \\ y_i=& b_i-p_i\cdot y_{i-1},i=2,3,\cdots ,n\end{array}$$
$$\begin{array}{rl}{x}_n=& \frac{y_n}{q_n}\\ & \\ x_i=& (y_i-c_i\cdot x_{i+1})/q_i,i=n-1,\cdots ,2,1\end{array}$$

[!example]-
A = [ 2 2 − 1 1 2 − 1 1 2 − 1 1 2 − 1 1 ]    ,    b = [ 6 7 9 11 1 ] A= \begin{bmatrix} 2 & 2 & & & \\ -1 & 1 & 2 & & \\ & -1 & 1 & 2 & \\ & & -1 & 1 & 2 \\ & & & -1 & 1 \end{bmatrix} \,\,,\,\, b=\begin{bmatrix} 6\\7\\9\\11\\1 \end{bmatrix} A= ​2−1​21−1​21−1​21−1​21​ ​,b= ​679111​ ​
解：
[ 2 2 − 1 2 1 − ( − 1 2 ) ⋅ 2 = 2 2 − 1 2 2 2 − 1 2 2 2 − 1 2 2 ] \begin{bmatrix} 2 & 2 & & & \\ - \frac{1}{2} & 1-(- \frac{1}{2})\cdot 2=2 & 2 & & \\ & - \frac{1}{2} & 2 & 2 & \\ & & - \frac{1}{2} & 2 & 2 \\ & & & - \frac{1}{2} & 2 \end{bmatrix} ​2−21​​21−(−21​)⋅2=2−21​​22−21​​22−21​​22​ ​
写成LU的形式
L = [ 1 − 1 / 2 1 − 1 / 2 1 − 1 / 2 1 − 1 / 2 1 ]    ,    U = [ 2 2 2 2 2 2 2 2 2 ] L= \begin{bmatrix} 1 & & & & \\ -1/2 & 1& & & \\ & -1/2 & 1 & & \\ & & -1/2 & 1 & \\ & & & -1/2 & 1 \end{bmatrix} \,\,,\,\, U= \begin{bmatrix} 2 & 2 & & & \\ & 2 & 2 & & \\ & & 2 & 2 & \\ & & & 2 & 2 \\ & & & & 2 \end{bmatrix} L= ​1−1/2​1−1/2​1−1/2​1−1/2​1​ ​,U= ​2​22​22​22​22​ ​
$Ly=b$
[ 1 − 1 / 2 1 − 1 / 2 1 − 1 / 2 1 − 1 / 2 1 ] [ y 1 y 2 y 3 y 4 y 5 ] = [ 6 7 9 11 1 ] → { y 1 = 6 − y 1 / 2 + y 2 = 7 → y 2 = 10 − y 2 / 2 + y 3 = 9 → y 2 = 14 − y 3 / 2 + y 4 = 11 → y 3 = 18 − y 4 / 2 + y 5 = 1 → y 5 = 10 \begin{bmatrix} 1 & & & & \\ -1/2 & 1& & & \\ & -1/2 & 1 & & \\ & & -1/2 & 1 & \\ & & & -1/2 & 1 \end{bmatrix} \begin{bmatrix} y_1\\y_2\\y_3\\y_4\\y_5 \end{bmatrix}= \begin{bmatrix} 6\\7\\9\\11\\1 \end{bmatrix}\to \begin{cases} y_1=6\\ -y_1/2+y_2=7\to y_2=10 \\ -y_2/2+y_3=9\to y_2=14 \\ -y_3/2+y_4=11\to y_3=18\\ -y_4/2+y_5=1\to y_5=10 \end{cases} ​1−1/2​1−1/2​1−1/2​1−1/2​1​ ​ ​y1​y2​y3​y4​y5​​ ​= ​679111​ ​→⎩ ⎨ ⎧​y1​=6−y1​/2+y2​=7→y2​=10−y2​/2+y3​=9→y2​=14−y3​/2+y4​=11→y3​=18−y4​/2+y5​=1→y5​=10​
$Ux=y$
[ 2 2 2 2 2 2 2 2 2 ] [ x 1 x 2 x 3 x 4 x 5 ] = [ 6 10 14 18 10 ] → [ 1 2 3 4 5 ] \begin{bmatrix} 2 & 2 & & & \\ & 2 & 2 & & \\ & & 2 & 2 & \\ & & & 2 & 2 \\ & & & & 2 \end{bmatrix}\begin{bmatrix} x_1\\x_2\\x_3\\x_4\\x_5 \end{bmatrix}= \begin{bmatrix} 6\\10\\14\\18\\10 \end{bmatrix} \to \begin{bmatrix} 1\\2\\3\\4\\5 \end{bmatrix} ​2​22​22​22​22​ ​ ​x1​x2​x3​x4​x5​​ ​= ​610141810​ ​→ ​12345​ ​

matlab编程实现

%% 追赶法例子
a = [2 1 1 1 1];
b = [0 -1 -1 -1 -1];
c = [2 2 2 2 0];
B = [6 7 9 11 1];
x = tridiagSolver(a,b,c,B,5)
A = diag(a)+diag(repmat([-1],1,4), -1)+ diag(repmat([2],1,4),1)
A*x'

%% 追赶法解三对角线性方程组 傻逼版
% 输入主对角元素向量，下次对角元素向量，上次对角元素向量，方程右边的向量，阶数
function x = tridiagSolver(a,b,c,B,n)
    q(1) = a(1)
    for i = 2:n
        p(i) = b(i)/q(i-1)
        q(i) = a(i)-p(i)*c(i-1)
    end
    y(1) = B(1)
    for i = 2:n
        y(i) = B(i)-p(i)*y(i-1)
    end
    x(n) = y(n)/q(n)
    for i = n-1:-1:1
        x(i)=(y(i)-c(i)*x(i+1))/q(i)
    end
end