Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 341 / 3 / b

Table of contents
- Solution b

Solution

Here

Δ x = 1/ M

and periodic indexing is taken modulo

2 M

. The second-order central difference matrix is the real symmetric circulant matrix

A = M^{2} - 2 10 ⋮ 01 1 - 2 1 ⋱ 0 01 - 2 ⋱ ⋱ \dots \dots ⋱ ⋱ ⋱ 10 0 ⋱ 1 - 2 1 10 ⋮ 01 - 2 .

Since

V

is real diagonal,

H = A - V

is Hermitian. Therefore

i H

is skew-Hermitian, and

\frac{d}{d t} ∥ u ∥_{2}^{2} = u^{*} (i H) u + u^{*} (- i H) u = 0.

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