Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 1 / 6D (Numerical Analysis)

Table of contents
- Solution 6D

Solution

Substituting the exact solution and expanding about

t_{n}

gives the local truncation error

y (t_{n + 1}) - y (t_{n}) - \frac{h}{2} (y^{'} (t_{n}) + y^{'} (t_{n + 1})) = - \frac{h ^{3}}{12} y^{'''} (t_{n}) + O (h^{4}) .

Thus the method is consistent of order two; as a stable one-step method it has global convergence order $k = 2$ .

The order cannot be higher. For

y^{'} = t^{2}

,

y (0) = 0

, the exact solution is

y = t^{3} /3

, and direct substitution makes the local truncation error exactly

- \frac{1}{6} h^{3} .

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