 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ii/paper-3/32e/solution

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

Integrating the vector-field equations gives the one-parameter transformations

(x, t, v) ⟼ (e^{α s} x, e^{β s} t, e^{γ s} v) .

Under this scaling,

v_{t}, v^{2} v_{x}, v_{xxx}

γ - β, 3 γ - α, γ - 3 α

. Equality of all three weights gives

β = 3 α, γ = - α .

Thus the nontrivial Lie point symmetry generator can be normalized to

x \partial_{x} + 3 t \partial_{t} - v \partial_{v}

u = v^{2} + v_{x}

, direct differentiation gives the operator identity

u_{t} - 6 u u_{x} + u_{xxx} = (\partial_{x} + 2 v) (v_{t} - 6 v^{2} v_{x} + v_{xxx}) .

For example

u_{x} = 2 v v_{x} + v_{xx}

and

u_{xxx} = 6 v_{x} v_{xx} + 2 v v_{xxx} + v_{xxxx}

; inserting them makes the

v_{x} v_{xx}

terms cancel and leaves the right side. Hence $u$ solves $u_{t} - 6 u u_{x} + u_{xxx} = 0$ whenever

v

solves the mKdV equation. Both

v^{2}

and

v_{x}

- 2 α

(x, t, u) \mapsto (e^{s} x, e^{3 s} t, e^{- 2 s} u), x \partial_{x} + 3 t \partial_{t} - 2 u \partial_{u} .

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