Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 1 / 14E (Dynamical Systems)

Linearizing about the orbit gives $\dot{\eta }=A(t)\eta$, where $A(t)=Df(X(t))$. The fundamental matrix therefore satisfiesThe Floquet multipliers are the eigenvalues of the monodromy matrix $\Phi (T)$. Differentiating the orbit equation shows that its tangent $\dot{X}(t)$ solves the variational equation. Since $\dot{X}(T)=\dot{X}(0)\ne 0$, one multiplier is one.

For an invertible fundamental matrix, differentiating the determinant gives $(\det \Phi {)}^{\prime }=\mathrm{tr}({\Phi }^{-1}\dot{\Phi })\det \Phi =\mathrm{tr}A\det \Phi$. Integration from the identity yields $\det \Phi (T)=\exp ({\int }_0^T\mathrm{tr}A(t)dt)$. The determinant is the product of the two multipliers, so the other is

Write $y=\dot{x}$ and $E=(x^2+y^2)/2$. Along a trajectory,Use the unperturbed orbit $x=R\sin t,y=R\cos t$ to compute the leading energy change over one period:Zero averaged energy change gives the nonzero amplitudeThe supplied sign condition makes this real and nonzero. The averaged amplitude equation is $\dot{R}=ϵ[\gamma R/2-(\alpha +3\beta )R^3/8]+O({ϵ}^2)$, whose derivative at that root is $-ϵ\gamma +O({ϵ}^2)$. Its nonzero derivative also makes the root persist as an isolated periodic orbit for sufficiently small positive $ϵ$.

Directly, the planar divergence is $-ϵ(\alpha x^2+3\beta y^2-\gamma )$. On the leading orbit its integral over $2\pi$ is $-ϵ[\pi R^2(\alpha +3\beta )-2\pi \gamma ]=-2\pi ϵ\gamma$. ThusThe cycle is attracting for $\gamma >0$ and repelling for $\gamma <0$. Both sign possibilities are allowed by $(\alpha +3\beta )\gamma >0$.