Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 61 / 2 / e

Solution

For a single spectral mass, the algebraic equation gives

Φ = \frac{E}{1 + c E / ( 2 κ )}, E = e^{- κ x + κ^{3} t} .

Differentiate the reconstruction, writing

r = c E / (2 κ)

q = \frac{κ c E}{[ 1 + c E / ( 2 κ ) ] ^{2}} = \frac{2 κ ^{2} r}{( 1 + r ) ^{2}} .

With

x_{0} = κ^{- 1} lo g [c / (2 κ)]

, the concise one-soliton answer is

q (x, t) = \frac{κ ^{2}}{2} sech^{2} [\frac{κ}{2} (x - κ^{2} t - x_{0})] .

Its speed is

κ^{2}

and its maximum is

κ^{2} /2

. Equivalently, setting

η = κ /2

gives

q = 2 η^{2} sech^{2} [η (x - 4 η^{2} t - x_{0})]

. To check the sign and speed independently, this profile obeys

q_{xx} = κ^{2} q - 3 q^{2}

and

q_{t} = - κ^{2} q_{x}

; differentiation gives the stated KdV equation exactly.

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