Talk:Relaxation (iterative method)

Over-focused

This article over-focuses on the a specific application of the method, needs to be expanded to the entire general concept. —Preceding unsigned comment added by 129.97.58.55 (talk • contribs) 02:24, October 21, 2007 (UTC)

Can you give an example of the relaxation method applied to something else than the numerical solution of an elliptic p.d.e.?  --Lambiam 11:39, 21 October 2007 (UTC)

Can you give an example of a linear system of equations that does not arise from numerical partial differential equations? The article focus on a specific application in about 60% of the text body, while it does not explain at all what relaxation actually means. — Preceding unsigned comment added by 212.201.70.9 (talk) 21:42, 7 July 2012 (UTC)

Error term on second-order central difference scheme

Shouldn't the error term on the second-order central difference scheme be O(h^2) instead of O(h^4) ? —Preceding unsigned comment added by Runebarnkob (talk • contribs) 10:33, 22 November 2007 (UTC)

I don't think so. In one dimension

$h^{2}{\frac {d^{2}}{{dx}^{2}}}\varphi (x)=\varphi (x{-}h)-2\varphi (x)+\varphi (x{+}h)\,+\,{\mathcal {O}}(h^{4})\,.$

If you do this in two orthogonal directions and add to get $h^{2}{\nabla }^{2}\varphi (x,y)\,,$ as used in the equation for φ(x, y), you still have an error term of O(h⁴).  --Lambiam 14:21, 22 November 2007 (UTC)

I agree that the error term should still be the same as you expand the dimension. But I thought that the one dimensional second order three-point central difference scheme was

$h^{2}{\frac {d^{2}}{{dx}^{2}}}\varphi (x)=\varphi (x{-}h)-2\varphi (x)+\varphi (x{+}h)\,+\,{\mathcal {O}}(h^{2})\,.$

Maybe I have a lack in my understanding of the BigO-notation. --Runebarnkob (talk) 04:20, 23 November 2007 (UTC)

That formula is equivalent with

${\frac {d^{2}}{{dx}^{2}}}\varphi (x)={\frac {\varphi (x{-}h)-2\varphi (x)+\varphi (x{+}h)}{h^{2}}}\,+\,{\mathcal {O}}(1)\,,$

which is clearly too weak; it does not tell us that

$lim_{h\to 0}{\frac {\varphi (x{-}h)-2\varphi (x)+\varphi (x{+}h)}{h^{2}}}={\frac {d^{2}}{{dx}^{2}}}\varphi (x)\,.$

Use the Taylor expansion

$\varphi (x{+}h)=\varphi (x)+h\varphi '(x)+{\mathcal {O}}(h^{2})\,$

and the same with h replaced by −h, substituting it for φ(x±h) in the second-order difference formula, and you're done.  --Lambiam 08:51, 23 November 2007 (UTC)

I agree with you now. I definitely had a problem with my understanding of the O(). Thanks for your patience, Lambiam. --Runebarnkob (talk) 09:55, 23 November 2007 (UTC)

Shouldn't be in the varphi(x,y), twice the h^2{\nabla}^2\varphi(x,y)\right)? —Preceding unsigned comment added by 89.120.154.196 (talk) 18:30, 30 January 2011 (UTC)