 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3/32d/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Put

c = β_{0} e^{8 χ^{3} t}, F (s) = c e^{- χs},

K (x, y) = - B (x) e^{- χy} .

- B (x) + c e^{- χ x} - B (x) c \int_{x}^{\infty} e^{- 2 χz} d z = 0,

B (x) = \frac{c e ^{- χ x}}{1 + c e ^{- 2 χ x} / ( 2 χ )}

and hence

K (x, x) = - \frac{c e ^{- 2 χ x}}{1 + c e ^{- 2 χ x} / ( 2 χ )} .

Write

q = \frac{c e ^{- 2 χ x}}{2 χ} = \frac{β _{0}}{2 χ} e^{8 χ^{3} t - 2 χ x} .

The reconstruction formula for this inverse scattering transform convention yields

u (x, t) = - 2 \frac{\partial}{\partial x} K (x, x) = - \frac{8 χ ^{2} q}{( 1 + q ) ^{2}} .

q = e^{- 2 s}

, then

4 q / (1 + q)^{2} = sech^{2} s

. Taking

s = χ (x - 4 χ^{2} t - ϕ)

therefore produces the required one-soliton of the Korteweg-De Vries equation, with

ϕ = \frac{1}{2 χ} lo g (\frac{β _{0}}{2 χ}) .

Here the usual bound-state norming constant has

β_{0} > 0

and

χ > 0

, so

ϕ

is real.

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