 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/ii/paper-3/31e/b/solution

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

When

b = 0

, put

D = e^{- 2 p x + 8 p^{3} t - 2 x_{0}} > 0

. Since

c = 2 i p e^{- 2 x_{0}}

i p

gives

N (i p) = 1 - D N (i p)

, hence

N (i p) = \frac{1}{1 + D}, N (k) = 1 - \frac{2 i p D}{( 1 + D ) ( k + i p )} .

Differentiate the coefficient of

1/ k

with respect to

x

, using

D_{x} = - 2 p D

, and apply the reconstruction formula. The result is

q (x, t) = \frac{8 p ^{2} D}{( 1 + D ) ^{2}} = 2 p^{2} sech^{2} (p x - 4 p^{3} t + x_{0}) .

The positive sign follows from the source's

- 2 i

reconstruction convention. The soliton has amplitude

2 p^{2}

4 p^{2}

, and centre

x = 4 p^{2} t - x_{0} / p

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