 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ii/paper-1/40a/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Let

D = - 10 I / (3 h^{2})

be the diagonal part of

A

u^{(r + 1)} = (1 - ω) u^{(r)} + ω D^{- 1} [b - (A - D) u^{(r)}] .

Equivalently it is

u^{(r + 1)} = u^{(r)} + ω D^{- 1} (b - A u^{(r)})

. In grid form, all neighbours on the right are taken from the old iterate:

u_{ij}^{(r + 1)} = (1 - ω) u_{ij}^{(r)} + ω \frac{1}{5} axial \sum u^{(r)} + \frac{1}{20} diagonal \sum u^{(r)} - \frac{3 h ^{2}}{10} f_{ij} .

The boundary values stay zero at every step. The iteration matrix is

T_{ω} = I - ω D^{- 1} A

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