 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-1/41c/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The first two modified stages give

u^{n + 1} = (I + μ A_{x}) (I - μ A_{y})^{- 1} u^{n} .

The correction stage is

(I - μ A_{x}) u^{n + 1} = u^{n + 1} - μ A_{x} u^{n} .

Using

A_{x} A_{y} = A_{y} A_{x}

u^{n + 1} - μ A_{x} u^{n} = [(I + μ A_{x}) (I - μ A_{y})^{- 1} - μ A_{x}] u^{n} = (I + μ^{2} A_{x} A_{y}) (I - μ A_{y})^{- 1} u^{n} .

Therefore

u^{n + 1} = D u^{n}, D = (I - μ A_{x})^{- 1} (I + μ^{2} A_{x} A_{y}) (I - μ A_{y})^{- 1} .

On the common eigenvector

v^{(p, q)}

D

d_{pq} = \frac{1 + μ ^{2} λ _{p} λ _{q}}{( 1 - μ λ _{p} ) ( 1 - μ λ _{q} )} .

a = - μ λ_{p} > 0

and

b = - μ λ_{q} > 0

. Then

d_{pq} = \frac{1 + ab}{( 1 + a ) ( 1 + b )} .

Both numerator and denominator are positive, and

(1 + a) (1 + b) - (1 + ab) = a + b > 0.

Hence

0 < d_{pq} < 1

for every mode and every

μ > 0

∥ D ∥_{2} < 1

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