two-step Adams Moulton method
34 views (last 30 days)
Show older comments
Hello,
I want to use two-step Adams Moulton method to solve ODE. The code is given below.
Running this I have problems with dimensions.
Any help is greatly aprecciated.
Thank you
% initialize
f = inline('y/x-y^2/x^2','x','y');
xrange=[1,2];
h=0.1;
x=1:h:2;
n = (xrange(2)-xrange(1))/h;
y(1) = 1;
% generate starting estimates using Runge-Kutta
for i = 1
k1 = f(x(i), y(i));
k2 = f(x(i) + h/2, y(i) + h/2*k1);
k3 = f(x(i) + h/2, y(i) + h/2*k2);
k4 = f(x(i) + h, y(i) + h*k3);
y(i+1) = y(i) + h/6*(k1 + 2*k2 + 2*k3 + k4);
x(i+1) = x(i) + h;
end
% iterate
for i = 3:n+1
% Adams-Moulton -- *correct*
y(i) = y(i-1) + h/12*(5*f(x(i),y(i)) + 8*f(x(i-1),y(i-1))- f(x(i-2),y(i-2)));
end
0 Comments
Answers (2)
Torsten
on 5 Feb 2019
In the Adams-Moulton formula, y(i) appears on both sides of the equation. This means that the Adams-Moulton method is implicit. You will have to solve the equation
y(i) - (y(i-1) + h/12*(5*f(x(i),y(i)) + 8*f(x(i-1),y(i-1))- f(x(i-2),y(i-2)))) = 0
in the unknown y(i) in order to get the correct value (e.g. using MATLAB's "fzero").
Best wishes
Torsten.
0 Comments
Muhammad Sinan
on 22 Mar 2021
The only things need to correct is the discretization in iteration, here is the code
% iterate
for i = 3:length(n)-1
% Adams-Moulton -- *correct*
y(i) = y(i-1) + h/12*(5*f(x(i),y(i)) + 8*f(x(i-1),y(i-1))- f(x(i-2),y(i-2)));
end
Check it, if any thing goes wrong comment here.
Thank you!
0 Comments
See Also
Categories
Find more on Numerical Integration and Differential Equations in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!