two-step Adams Moulton method
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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
Respuestas (2)
Torsten
el 5 de Feb. de 2019
0 votos
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.
Muhammad Sinan
el 22 de Mzo. de 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!
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