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 Incoming links: Five-point Laplacian

Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 67 / 6 / Solution 2026-10-03
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Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 68 / 7 / Solution 2026-10-03
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Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 72 / 5 / a / Solution 2026-10-03
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Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 3 / 39A / Solution Created 2026-09-24 Updated 2026-09-24
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Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 2 / 38A / iii / Solution Created 2026-09-24 Updated 2026-10-05
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The five-point Laplacian applied to f has expansion
Γ5​f=h2Δf+12h4​(fxxxx​+fyyyy​)+O(h6).
(1)
The corrected right side h2f+(h2/12)Γ5​f cancels the h4Δf/12 term in Γ9​u. Consequently
h−2(Γ9​u−h2f−12h2​Γ5​f)=O(h4).​
(2)
More precisely, the unscaled residual's h6 coefficient is −(uxxxxxx​+uyyyyyy​)/240+(uxxxxyy​+uxxyyyy​)/144. This fourth-order correction of the nine-point Poisson stencil does not require f to be harmonic.
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Total articles: 5
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