﻿功能:一阶常微分方程组初值问题求解(4阶龙格库塔法)

格式:
[y1,y2,...,yn,error]=ODESolveRK4(f,fy,t,y0,t0,Error0,n)
[y1,y2,...,yn,error]=ODESolveRK4(f,fy,t,y0,t0,Error0)
[y1,y2,...,yn,error]=ODESolveRK4(f,fy,t,y0,t0)
[y1,y2,...,yn,error]=ODESolveRK4(f,fy,t,y0)
f:用符号变量存储的每个微分表达式,每个表达式以逗号分隔。注意默认的自变量名称为"t"
fy:符号变量存储的对应f当中每个表达式的变量名称
t:为矩阵变量或者数值,表示要求的此时刻的值
y0:为矩阵变量或者数值,表示边界条件值,这个y0对应fy当中的个数
t0:数值,对应t0时的y0,此值默认为0
Error0:变步长控制的相对误差,默认1E-12
n:步长等分最大深度,默认为20

y1,y2,...,yn:对应fy里的变量,表示返回值
error:返回的残差平方和

参考:王超能.数值分析简明教程(第二版)[M].高等教育出版社,北京,2004:105

例子:

/*
已知:
y1'=3*y1+2*y2
y2'=4*y1+y2
y1(0)=0
y2(0)=1
求:t=0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1时刻的y1,y2的值
*/

f="3*y1+2*y2,4*y1+y2";
y="y1,y2";
y0=[0 1];
t0=0;
t=0.1:0.1:1;
[y1,y2,er]=ODESolveRK4(f,y,t,y0,t0)//回车后得到如下的结果
y1 =
[ 0.24796125104620    0.63318368477703    1.24695694435934    2.23957864615453    3.85865442155111    6.51224175923853    10.8729555405121    18.0496069873177    29.8701872069284    49.3484265352063 ]
y2 =
[ 1.15279866910077    1.45191443785711    1.98777516504196    2.90989869219360    4.46518508126433    7.06105339533264    11.3695408443037    18.4989359514352    30.2767568666690    49.7163059763778 ]
er =
[ 4.4295379630E-13    2.0951703532E-14    1.3179152240E-14    6.1929765516E-13    6.3764915639E-14    4.3066064996E-15    5.4322330929E-14    5.5753541199E-13    1.9682325756E-14    1.5292740451E-13 ]
