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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-3/32e/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 32E / b / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

For

V = 4 t^{2} \partial_{t} + 4 t x \partial_{x} - (x^{2} + 2 t) \partial_{u},

the flow equations are

\frac{d T}{d ϵ} = 4 T^{2}, \frac{d X}{d ϵ} = 4 TX, \frac{d U}{d ϵ} = - (X^{2} + 2 T),

with

(T, X, U) = (t, x, u)

at

ϵ = 0

. Solving the first two gives, with

d = 1 - 4 ϵ t

,

T = \frac{t}{d}, X = \frac{x}{d} .

Integrating the last equation then gives the one-parameter group

T = \frac{t}{1 - 4 ϵ t}, X = \frac{x}{1 - 4 ϵ t}, U = u - \frac{ϵ x ^{2}}{1 - 4 ϵ t} + \frac{1}{2} lo g (1 - 4 ϵ t),

defined locally where

1 - 4 ϵ t > 0

.

To verify the symmetry infinitesimally, use

ξ^{t} = 4 t^{2}, ξ^{x} = 4 t x, η = - (x^{2} + 2 t) .

The needed coefficients of the second prolongation of a Lie point symmetry are

η^{t} = - 2 - 8 t u_{t} - 4 x u_{x}, η^{x} = - 2 x - 4 t u_{x},

and

η^{xx} = - 2 - 8 t u_{xx} .

For the potential Burgers equation written as

F = u_{t} - u_{xx} - u_{x}^{2} = 0,

we obtain

pr^{(2)} V (F) = η^{t} - η^{xx} - 2 u_{x} η^{x} = - 8 t (u_{t} - u_{xx} - u_{x}^{2}) = - 8 tF .

Thus the prolonged generator is tangent to the solution manifold $F = 0$ , and the finite transformations above form the projective Lie symmetry of the potential Burgers equation.

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