Trying to make an Adams-Bashforth method with Richardson error estimate
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%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
. Like I don't really know what that means. I understand the AB method for solving DefEqs, but not ther errors1 Commento
I'm confident that after reading this article
you will know how Richardson extrapolation works.
Risposte (1)
Mayank Sengar
il 12 Gen 2023
Modificato: Mayank Sengar
il 12 Gen 2023
0 voti
for understand Richardson expoitation, here is the pseudocode for it:
There is also a generalized Richardson extrapolation routine in file exchange: https://www.mathworks.com/matlabcentral/fileexchange/24388-richardson-extrapolation
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