Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 56 / 1 / c

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- Solution c

Solution

A returning orbit must have zero net change of the energy. To first order, integrate the perturbation along either unperturbed homoclinic loop:

0 = \int_{- \infty}^{\infty} (β - u_{h}^{2}) v_{h}^{2} d τ = β I_{0} - I_{2} .

For the positive loop,

v_{h} = \pm u_{h} a_{0}^{2} - u_{h}^{2} /2

; the outward and inward halves contribute equally. Therefore

I_{0} = 2 \int_{0}^{2 a_{0}} u a_{0}^{2} - u^{2} /2 d u = \frac{4}{3} a_{0}^{3},

and

I_{2} = 2 \int_{0}^{2 a_{0}} u^{3} a_{0}^{2} - u^{2} /2 d u = \frac{16}{15} a_{0}^{5} .

The integrals follow by substituting

w = a_{0}^{2} - u^{2} /2

; for the second one use

u^{2} = 2 (a_{0}^{2} - w)

. Hence

β = \frac{4}{5} a_{0}^{2} = - \frac{4}{5} α, κ_{hom} = - \frac{4}{5} λ + o (∣ λ ∣), λ < 0.

The equality in rescaled parameters is the leading Melnikov balance, not an assertion that the entire finite-parameter bifurcation curve is exactly straight. Reflection symmetry makes both saddle equilibrium loops occur together. The global curve lies below the nonzero-equilibrium Hopf line

κ = - λ

, as shown in the parameter sketch.

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