Open to feedback of working code (please review)

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Lavorizia Vaughn
Lavorizia Vaughn 2021 年 12 月 2 日
コメント済み: Lavorizia Vaughn 2021 年 12 月 2 日
Hey whats up folks, I was you could give my code a look .I have implements both the backward Euler method and Newtons method with f=f(t,y), dfdy=f'(t,y), maxiter=maximum number of iterations, and N=the number of steps. My code is below. I am open to feedback and possible changes. Any help would surely be appreciated.
my code:
function [t,w] = backeuler_four(f, dfdy, a, b, alpha, N, maxiter, tol)
h = (b-a)/N;
t = a:h:b;
w = t*0;
w(1) = alpha;
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, break;end
end
end
fprintf('\n');
if abs(error) > tol, error('No Newton convergence.'); end
w(i+1)=wj;

採用された回答

Walter Roberson
Walter Roberson 2021 年 12 月 2 日
if abs(error)<=tol, break;end
That only breaks out of one level of for loop.
Suppose you get convergence at i = 2, j = 7. Then you leave the for j loop. But you are still inside the for i loop, and you overwrite error and so on. So your final test is really testing whether you got convergence when i == N.
  3 件のコメント
Lavorizia Vaughn
Lavorizia Vaughn 2021 年 12 月 2 日
Thank you.

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