Convert higher order to first order system

25 ビュー (過去 30 日間)
kingsley
kingsley 2017 年 5 月 17 日
回答済み: Torsten 2017 年 5 月 17 日
I'm trying to solve the higher order ode by using RK4 method. Here is the code I have so far.
function [y] = rk4_high_ode(a,t0,n,h,f)
%f = @(t,y) 2*y(2)-y(3)+2*y(4);
% Do I need to define y(2) derivative first?
F=@(t,y)[y(2:end);f]; % Convert the higher order to the 1st order system
t(1)=t0;
for i=1:n
% update time
t(i+1)=t(i)+h;
k1=F(t(i) ,y(i) );
k2=F(t(i)+0.5*h,y(i)+0.5*h*k1);
k3=F(t(i)+0.5*h,y(i)+0.5*h*k2);
k2=F(t(i)+h ,y(i)+h*k1 );
y(i+1)=y(i)+h/6*(k1+2*k2+2*k3+k4);
end
end
And this is the test program:
clear
% test y=sin(t)
% y^(4) = 2*y'-y"+2*y^(3)
t0 = 0.1; n = 100; h = 1e-2;
a = [sin(t0) cos(t0) -sin(t0) -cos(t0)]';
f = @(t,y) 2*y(2)-y(3)+2*y(4);
y = rk4_high_ode(a,t0,n,h,f);
ye = sin(t0+n*h);
error2 = abs(y-ye)
There is an error " Undefined function or variable 'y'". Does that mean I need to define y(2) first? or something else.

回答 (1 件)

Torsten
Torsten 2017 年 5 月 17 日
You will have to define
f=@(t,y) [y(2) y(3) y(4) 2*y(2)-y(3)+2*y(4)];
Additionally note that "rk4_high_ode" must be modified because at the moment, it is only capable of solving a single ODE.
Best wishes
Torsten.

カテゴリ

Help Center および File ExchangeProgramming についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by