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)