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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-77/2/ii/solution

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Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 77 / 2 / ii / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

Set the time derivatives to zero. The scalar equations give

b = \frac{a}{1 + a ^{2}}, c = \frac{a ^{2}}{1 + a ^{2}}, d = \frac{a τ}{τ ^{2} + a ^{2}}, e = \frac{a ^{2}}{τ ^{2} + a ^{2}} .

For the nonzero convection branch, division of the velocity equation by

a

gives

r (x) = (1 + x) + \frac{r _{s} τ ( 1 + x )}{τ ^{2} + x}, x = a^{2} \geq 0.

Its zero-amplitude limit is

r_{e} = 1 + r_{s} / τ

. Differentiation yields

r^{'} (x) = 1 - \frac{r _{s} τ ( 1 - τ ^{2} )}{( τ ^{2} + x ) ^{2}} .

For

0 < τ < 1

, the sharp condition for a subcritical bifurcation at the steady onset is

r_{s} > τ^{3} / (1 - τ^{2})

. The stronger bound printed in the question is sufficient, since

(1 - τ)^{2} < 1 - τ^{2}

. Under it the branch initially runs toward smaller

r

, and the unique saddle-node bifurcation has

x_{*} = a_{*}^{2} = - τ^{2} + r_{s} τ (1 - τ^{2}) .

It is a genuine positive-amplitude minimum:

r^{''} (x) = 2 r_{s} τ (1 - τ^{2}) / (τ^{2} + x)^{3} > 0

. Put

h = r_{s} τ

and

j = 1 - τ^{2}

. At the minimum,

1 + x_{*} = j (j + h)

and

τ^{2} + x_{*} = hj

, so

r_{min} = (h + j)^{2} = (r_{s} τ + 1 - τ^{2})^{2} .

These results describe the steady fold of a thermosolutal Lorenz model. They do not by themselves assert stability against every oscillatory disturbance.

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