 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-69/3/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Expanding the symmetric spatial stencil by the Taylor theorem gives

L_{h} u = \frac{α + 2 β + 2 γ}{h ^{2}} u + (β + 4 γ) u_{xx} + \frac{h ^{2}}{12} (β + 16 γ) u_{xxxx} + \frac{h ^{4}}{360} (β + 64 γ) u_{xxxxxx} + O (h^{6}) .

h^{2}

term require

α + 2 β + 2 γ = 0

β + 4 γ = 1

, and

β + 16 γ = 0

. Solving this linear system gives

α = - \frac{5}{2}, β = \frac{4}{3}, γ = - \frac{1}{12} .

L_{h} u = u_{xx} - \frac{h ^{4}}{90} u_{xxxxxx} + O (h^{6}) .

The nonzero next coefficient proves that the highest spatial order of a numerical method is four for this three-parameter stencil.

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