Help with badly working loop
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Hi, hello everyone, I've been working on a method that implements the Backwartd Euler method with the Newton method, amd I have the following code:
function [t,w] = backeuler(f, dfdy, a, b, alpha, N, maxiter, tol)
h = (b-a)/N;
t = a:h:b;
w = t*0;
w(1) = alpha;
converged = false;
for i = 1:N
w0=w(i);
wj=w0;
for j=1:maxiter
wj=wj-(wj - w0 - h*f(t(i+1),wj)) / (1 - h*dfdy(t(i+1),wj));
error=(wj - w0 - h*f(t(i+1),wj)) / (1 - h*dfdy(t(i+1),wj));
fprintf('%d %g\n', j, abs(error));
if abs(error)<=tol
converged = true;
break;
end
end
if converged; break; end
end
fprintf('\n');
if ~converged
error('No Newton convergence.');
end
Now, when I terminate:
f = @(t,y) y^2 * (1-y);
dfdy = @(t,y) 2*y - 3*y^2;
a = 0; b = 2000; alpha = 0.9; maxiter = 20; tol = 1e-12;
[t3,w3] = backeuler(f, dfdy, a, b, alpha, 1, maxiter, tol);
[t4,w4] = backeuler(f, dfdy, a, b, alpha, 5, maxiter, tol);
[t5,w5] = backeuler(f, dfdy, a, b, alpha, 10, maxiter, tol);
I've been getting:
%1st solution
1 0.0270214
2 0.00149354
3 4.46627e-06
4 3.98803e-11
5 7.88716e-18
%2nd solution
1 0.0268375
2 0.00147092
3 4.32538e-06
4 3.7348e-11
5 4.40305e-17
%3rd soltuion
1 0.0266097
2 0.00144316
3 4.15576e-06
4 3.44115e-11
5 1.74336e-17
Which are the three solutions for the givens. But I should be getting more solutions, for example, I only have one solution for [t4,w4] = backeuler(f, dfdy, a, b, alpha, 5, maxiter, tol), but I should have four.
%1st solution for [t3,w3] = backeuler(f, dfdy, a, b, alpha, 1, maxiter, tol);
1 0.128469
2 0.0270214
3 0.00149354
4 4.46627e-006
5 3.98803e-011
6 7.88716e-018
%2nd(1) solution for [t4,w4] = backeuler(f, dfdy, a, b, alpha, 5, maxiter, tol);
1 0.128063
2 0.0268375
3 0.00147092
4 4.32538e-006
5 3.7348e-011
6 4.40305e-017
%2nd(2) soution for [t4,w4] = backeuler(f, dfdy, a, b, alpha, 5, maxiter, tol);
1 0.000249002
2 1.23664e-007
3 3.05022e-014
%2nd(3) solution for [t4,w4] = backeuler(f, dfdy, a, b, alpha, 5, maxiter, tol);
1 6.20646e-007
2 7.68531e-013
%2nd(4) solution for [t4,w4] = backeuler(f, dfdy, a, b, alpha, 5, maxiter, tol);
1 1.54774e-009
2 1.16283e-017
%2nd soltuoin(5)
1 3.8597e-012
2 9.69023e-018
Any help would be appreciated.
14 Kommentare
Voss
am 4 Dez. 2021
I think you need to use the value wj that has been converged on as the initial wj for the next i-iteration. So, right before or after setting converged = true, add this line: w(i+1) = wj;
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