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Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 33A / b / iii / Solution

With

x = v + w

and

y = v - w

, the equations at

r = 1

become

\overset{v}{˙} = \frac{1}{2} {a (v + w)^{3} - z (v - w)},

\overset{w}{˙} = - 2 w + \frac{1}{2} {a (v + w)^{3} + z (v - w)}, \overset{z}{˙} = - z + v^{2} - w^{2} .

The symmetry

(v, w, z) \mapsto (- v, - w, z)

suggests

w_{c} (v) = c_{3} v^{3} + O (v^{5}), z_{c} (v) = d_{2} v^{2} + d_{4} v^{4} + O (v^{6}) .

Substitute these into the invariance equations

w_{c}^{'} (v) \overset{v}{˙} = \overset{w}{˙}, z_{c}^{'} (v) \overset{v}{˙} = \overset{z}{˙} .

Comparison at orders

v^{2}, v^{3}, v^{4}

gives

d_{2} = 1, c_{3} = \frac{a + 1}{4}, d_{4} = 1 - a .

Thus the center manifold of the 2024 Cambridge cubic system is

w_{c} (v) = \frac{a + 1}{4} v^{3} + O (v^{5}), z_{c} (v) = v^{2} + (1 - a) v^{4} + O (v^{6}) .

Solved by gpt-5.6-sol high.

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