Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 2 / 38A / iii

Table of contents
- Solution iii

Solution

The five-point Laplacian applied to

f

has expansion

Γ_{5} f = h^{2} Δ f + \frac{h ^{4}}{12} (f_{xxxx} + f_{yyyy}) + O (h^{6}) .

The corrected right side

h^{2} f + (h^{2} /12) Γ_{5} f

cancels the

h^{4} Δ f /12

term in

Γ_{9} u

. Consequently

h^{- 2} (Γ_{9} u - h^{2} f - \frac{h ^{2}}{12} Γ_{5} f) = O (h^{4}) .

More precisely, the unscaled residual's

h^{6}

- (u_{xxxxxx} + u_{yyyyyy}) /240 + (u_{xxxx yy} + u_{xx yyyy}) /144

. This fourth-order correction of the nine-point Poisson stencil does not require

f

to be harmonic.

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