数值分析 周维祺
y(tn+1)=ytn+∫tntn+1f(t,y(t))dt=ytn+h∫01f(tn+hx,y(tn+hx))dxy(t_{n+1})=y_{t_n}+\int_{t_n}^{t_{n+1}}f(t,y(t))dt=y_{t_n}+h\int_{0}^{1}f(t_n+hx,y(t_n+hx))dx y(tn+1)=ytn+∫tntn+1f(t,y(t))dt=ytn+h∫01f(tn+hx,y(tn+hx))dx
∫01f(tn+hx,y(tn+hx))dx≈∑j=0vbjf(tn+cjh,y(tn+cjh))⏟数值积分\int_{0}^{1}f(t_n+hx,y(t_n+hx))dx\approx \underbrace{\sum_{j=0}^vb_jf(t_n+c_jh, y(t_n+c_jh))}_{数值积分} ∫01f(tn+hx,y(tn+hx))dx≈数值积分j=0∑vbjf(tn+cjh,y(tn+cjh))
令κj\kappa_jκj为对y(tn+cjh)y(t_n+c_jh)y(tn+cjh)的估计,并令c1=0c_1=0c1=0,从而κ1=yn\kappa_1=y_nκ1=yn
κk=yn+h∑i=1vak,if(tn+cih,κi)\kappa_k=y_n+h\sum\limits_{i=1}^va_{k,i}f(t_n+c_ih,\kappa_i)κk=yn+hi=1∑vak,if(tn+cih,κi)
bjb_jbj: 权值;cjc_jcj: 节点;[ak,i]k,i[a_{k,i}]_{k,i}[ak,i]k,i:Runge-Kutta系数矩阵; v:步数v: 步数v:步数
c⃗=(c1⋮cv)\vec c=\begin{pmatrix}c_1 \\ \vdots \\ c_v \end{pmatrix}c=⎝⎛c1⋮cv⎠⎞,b⃗T=(b1…bv)\vec b^T=\begin{pmatrix}b_1 & \ldots & b_v \end{pmatrix}bT=(b1…bv), A=(a1,1…a1,v⋮⋱⋮av,1…av,v)A=\begin{pmatrix} a_{1,1} & \ldots & a_{1,v} \\ \vdots & \ddots & \vdots \\ a_{v,1} & \ldots & a_{v,v} \\ \end{pmatrix}A=⎝⎛a1,1⋮av,1…⋱…a1,v⋮av,v⎠⎞
Butcher表可以写成 c⃗Ab⃗T\begin{array}{c|c} \vec c & A \\ \hline & \vec b^T \\ \end{array}cAbT
显式:AAA是严格下三角阵(主对角线也为000)
隐式:AAA不是严格下三角阵
显式法一般还要求ci=∑j=1i−1ai,jc_i=\sum\limits_{j=1}^{i-1}a_{i,j}ci=j=1∑i−1ai,j
欧拉法(1阶): 001\begin{array}{c|c} 0 & 0 \\ \hline & 1\\ \end{array}001
中点法(2阶): 0001/21/2001\begin{array}{c|cc} 0 & 0 & 0 \\ 1/2 & 1/2 & 0 \\ \hline & 0 & 1\\ \end{array}01/201/20001
0002/32/301/43/4,\begin{array}{c|cc} 0 & 0 & 0 \\ 2/3 & 2/3 & 0 \\ \hline & 1/4 & 3/4\\ \end{array},\quad\quad02/302/31/4003/4,0001101/21/2\begin{array}{c|cc} 0 & 0 & 0 \\ 1 & 1 & 0 \\ \hline & 1/2 & 1/2\\ \end{array}01011/2001/2
若v=2,b1+b2=1,b2c2=1/2,a21=c2,c1=0v=2, b_1+b_2=1, b_2c_2=1/2, a_{21}=c_2, c_1=0v=2,b1+b2=1,b2c2=1/2,a21=c2,c1=0,则显式法具有至少2阶局部截断误差
经典法: 00001/21/2001−1201/62/31/6\begin{array}{c|ccc} 0 & 0 & 0 & 0 \\ 1/2 & 1/2 & 0 & 0 \\ 1 & -1 & 2 & 0 \\ \hline & 1/6 & 2/3 & 1/6 \end{array}01/2101/2−11/60022/30001/6
Nystro¨mNystr{\"o}mNystro¨m法:00002/32/3002/302/301/43/83/8\begin{array}{c|ccc} 0 & 0 & 0 & 0 \\ 2/3 & 2/3 & 0 & 0 \\ 2/3 & 0 & 2/3 & 0 \\ \hline & 1/4 & 3/8 & 3/8 \end{array}02/32/302/301/4002/33/80003/8
000001/21/20001/201/200100101/61/31/31/6\begin{array}{c|cccc} 0 & 0 & 0 & 0 & 0 \\ 1/2 & 1/2 & 0 & 0 & 0 \\ 1/2 & 0 & 1/2 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 \\ \hline & 1/6 & 1/3 & 1/3 & 1/6 \end{array} 01/21/2101/2001/6001/201/300011/300001/6
p≥8p\ge8p≥8时,至少需要p+3p+3p+3步才能达到p阶,存在171717及181818步的101010阶方法
隐式法
系数的设计:Collocation法