Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 1 / 41A / b

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii
- iii b
  - Solution iii

i

Solution

The matrix exponential is the absolutely convergent series

e^{X} = \sum_{j = 0}^{\infty} \frac{X ^{j}}{j !} .

Multiplying the two expansions gives

e^{k B} e^{k C} = I + k (B + C) + k^{2} (\frac{1}{2} B^{2} + BC + \frac{1}{2} C^{2}) + O (k^{3}) .

On the other hand,

e^{k (B + C)} = I + k (B + C) + \frac{k ^{2}}{2} (B^{2} + BC + CB + C^{2}) + O (k^{3}) .

Subtracting yields

e^{k B} e^{k C} = e^{k (B + C)} + \frac{k ^{2}}{2} (BC - CB) + O (k^{3}) .

This is the leading Lie-Trotter splitting commutator error.

ii

Solution

Expanding the numerical propagation matrix,

(I - k B)^{- 1} (I - k C)^{- 1} = I + k (B + C) + k^{2} (B^{2} + BC + C^{2}) + O (k^{3}) .

Its difference from

e^{k A}

is

\frac{k ^{2}}{2} (B^{2} + C^{2} + BC - CB) + O (k^{3}) .

Therefore the one-step local truncation error is

O (k^{2}),

and the method is first-order accurate globally.

iii

Solution

If

BC = CB

, the splitting commutator vanishes, but the backward-Euler resolvents still contribute

\frac{k ^{2}}{2} (B^{2} + C^{2}) + O (k^{3}) .

Thus commutativity alone does not change the order: the local error remains $O (k^{2})$ and the global method remains first order, barring an additional special cancellation.

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