Can anyone help ?

2 次查看（过去 30 天）

显示更早的评论

diadalina 2022-4-21

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1701730-can-anyone-help

评论： Jeffrey Clark 2022-4-22

在 MATLAB Online 中打开

i want to calculate the solution of the following cauchy problem :

y'(t)=1+t-y(t) t in[0,1]

but i have this error can anyone help me:

FZERO cannot continue because user supplied function_handle ==>

@(x)dot(a,y(i:-1:i-p))+h*dot(b(1:end-1),f(t(i:-1:i-p),y(i:-1:i-p)))-x+h*b(end)*f(t(i+1),x) failed

with the error below.

Matrix dimensions must agree.

Error in adamsMp2o3 (line 14)

y(i+1)=fzero(@(x)dot(a,y(i:-1:i-p))+h*dot(b(1:end-1),f(t(i:-1:i-p),y(i:-1:i-p)))-x+h*b(end)*f(t(i+1),x),y(i));

Error in ppadamasmoultonordre3 (line 26)

[t,y]=adamsMp2o3(a,b,f,y0,t,n,af,h)

f=@(t,y)(1+t-y);
% les données 
Tf=1;
t0=0;
h=0.1;
n=(Tf-t0)/h;
t=linspace(t0,Tf,n+1);
af=3;
y0=0;
a=[1,0];
b=[8/12 -1/12 5/12]; 
%b=[2/3 -1/12 5/12] ;
[t,y]=adamsMp2o3(a,b,f,y0,t,n,af,h)
fprintf('%.af\n',y)
function [t,y]=adamsMp2o3(a,b,f,y0,t,n,af,h)
p=length(a)-1;
y(1)=y0;
t=t';
k1=f(t(1),y(1));
k2=f(t(1)+h/2,y(1)+(h/2)*k1);
k3=f(t(1)+h/2,y(1)+(h/2)*k2);
k4=f(t(1)+h,y(1)+h*k3);
y(2)=y(1)+(h/6)*(k1+2*k2+2*k3+k4);
for i=p+1:n
    y(i+1)=fzero(@(x)dot(a,y(i:-1:i-p))+h*dot(b(1:end-1),f(t(i:-1:i-p),y(i:-1:i-p)))-x+h*b(end)*f(t(i+1),x),y(i));
end

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Jeffrey Clark 2022-4-22

When looking for such problems I usually rerun useing "Pause on Errors" enabled which will allow you to examine the variables and try parts of the equation to see what is failing. You can then traceback in your code to see how the data got created incorrectly: