 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/ii/paper-1/38a/a/solution

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

Write

h = Δ x

k = Δ t = μ h

, and

D_{0} u = [u (x + h) - u (x - h)] / (2 h)

. The scheme is the trapezoidal time rule with a centered spatial derivative:

u^{n + 1} - u^{n} = \frac{k}{2} (D_{0} u^{n + 1} + D_{0} u^{n}) .

Insert a smooth exact solution of

u_{t} = u_{x}

. Taylor expansion of the unnormalized one-step residual gives

u (t + k) - u (t) \frac{k}{2} (D_{0} u (t + k) + D_{0} u (t)) = k u_{t} + \frac{1}{2} k^{2} u_{tt} + \frac{1}{6} k^{3} u_{ttt} + O (h^{4}), = k u_{x} + \frac{1}{2} k^{2} u_{x t} + \frac{1}{4} k^{3} u_{x tt} + \frac{1}{6} k h^{2} u_{xxx} + O (h^{4}) .

All repeated time derivatives equal the corresponding spatial derivatives. The first two orders cancel, leaving

R = - (\frac{k ^{3}}{12} + \frac{k h ^{2}}{6}) u_{xxx} + O (h^{4}) = - \frac{μ ( μ ^{2} + 2 )}{12} h^{3} u_{xxx} + O (h^{4}) .

Thus the one-step local error is

O (h^{3})

at fixed Courant number. If “local truncation error” is instead defined after division of the residual by

Δ t

, its order is

O (h^{2})

; these are two different normalizations of the same second-order method.

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