Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 56 / 2 / g / Solution

The change of time is $d\tau /dt=u$, which is positive in the half-plane under consideration. Dividing the two approximate equations by $u$ givesSet $w=v-\mu$. The inhomogeneous system becomes $(u,w{)}^{\prime }=(u-w,\lambda u)$, with eigenvalues $(1\pm \sqrt{1-4\lambda })/2$. For $\lambda >1/4$, write $\omega =\sqrt{4\lambda -1}/2$. Integrating with the prescribed initial data givesandThese expressions also verify both initial values and their initial derivatives. For fixed $\mu >0$ and small $h,v_0$ on the outgoing branch, the sine coefficient in $u$ is positive and dominates the initial cosine term. The first return to small $u$ is near the first positive zero, $\omega \tau \simeq \pi$. HenceThe factor is a half-turn of the growing spiral, not its full rotation period. After this excursion the approximate trajectory approaches the small-$u$, large-$v$ region. In the full system the term $-\lambda v$, omitted during the large excursion, allows the motion to descend along that region and return.

As $\lambda \downarrow 1/4$, the rescaled excursion time diverges and the factor $e^{\tau /2}$ produces arbitrarily large excursions. At the critical value the matrix is a nontrivial Jordan block, and the limiting formula isFor the small outgoing initial data this is positive and unbounded: the unstable manifold no longer turns back. Below $1/4$ the two eigenvalues are real and positive, so there is again no oscillatory return. This quadratic escape desingularization explains loss of the periodic orbit at infinity, together with the unstable-manifold escape, as $\lambda$ decreases through $1/4$. It is not a finite homoclinic bifurcation to the origin. The divergent $\tau$ travel time must not be identified with the physical period without integrating $dt=d\tau /u$; the time change becomes singular near $u=0$.