Trying to make an Adams-Bashforth method with Richardson error estimate

%This program solves the initial value problem
% y' = f(x,y), x0 <= x<= b, y(x0)=y0
%Initializing vaiables
%f'(x,y)=
f = @(x,y) cos(y).^2; %derivative in question
g = @(x) atan(x); %this is the actual solution
x0 = 0; %initial value of x
x_end = 10; %end of approximation
h = 0.1; %size of decimal place (0.1,0.001,etc
y0=0; %initial value of y
n = fix((x_end-x0)/h)+1;
x = linspace(x0,x_end,n);
y = zeros(n,1);
y(1) = y0;
f1 = f(x(1),y(1));
y(2) = y(1)+h*f1;
%need to add error
for i = 3:n
f2 = f(x(i-1),y(i-1));
y(i) = y(i-1)+h*(3*f2-f1)/2;
f1 = f2;
fprintf('%5.4f %11.8f\n', x(i), y(i));
plot(x(i),y(i),'b.'); grid on;
fplot(g,[x0,x_end]);
xlabel('x values'); ylabel('y values');
hold on;
end
I'm not sure how I would add the Richardson error to this code. I see the formula in my textbook, but don't understand how I would make it work. . Like I don't really know what that means. I understand the AB method for solving DefEqs, but not ther errors

1 Kommentar

Torsten
Torsten am 10 Dez. 2022
Bearbeitet: Torsten am 10 Dez. 2022
I'm confident that after reading this article
you will know how Richardson extrapolation works.

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Numerical Integration and Differential Equations finden Sie in Hilfe-Center und File Exchange

Gefragt:

ce
am 10 Dez. 2022

Bearbeitet:

am 12 Jan. 2023

Community Treasure Hunt

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

Start Hunting!

Translated by