Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 2 / 28B (Dynamical Systems)

The linearized perturbation obeys $\dot{\eta }=D\mathbf{f}(\mathbf{X}(t))\eta$, so its fundamental matrix satisfiesThe Floquet multipliers are the eigenvalues of the monodromy matrix $\Phi (T)$. Differentiating the orbit equation shows that $\dot{\mathbf{X}}(t)=\mathbf{f}(\mathbf{X}(t))$ is a solution of the variational equation. For a nonconstant periodic orbit this tangent solution is nonzero and periodic, giving one multiplier equal to $1$.

The Liouville formula for a fundamental matrix gives $(\det \Phi {)}^{\prime }=(\mathrm{tr}D\mathbf{f})\det \Phi$. Since $\det \Phi (0)=1$, the product of the two multipliers is $\exp ({\int }_0^T\nabla \cdot \mathbf{f}(\mathbf{X}(t))dt)$. The other multiplier is therefore exactly this exponential.

Use the nearly conserved oscillator energy $E=({\dot{x}}^2+x^2)/2$. The differential equation givesFor a leading-order orbit $x=A\cos \theta$, $\dot{x}=-A\sin \theta$, the cycle averages are $⟨{\dot{x}}^4⟩=3A^4/8$, $⟨x^3\dot{x}⟩=0$, $⟨x^4{\dot{x}}^2⟩=A^6/16$, and $⟨{\dot{x}}^2⟩=A^2/2$. The energy balance method therefore givesThe two nonzero simple roots give the leading limit cycles $x\simeq \cos (t+{\theta }_0)$ and $x\simeq \sqrt{2}\cos (t+{\theta }_0)$. Below amplitude $1$ the energy increases, between $1$ and $\sqrt{2}$ it decreases, and above $\sqrt{2}$ it increases. Hence the amplitude-$1$ cycle is stable and the amplitude-$\sqrt{2}$ cycle unstable for positive small $\epsilon$.

The linearized averaged radial exponents are $-\epsilon /2$ and $\epsilon$, respectively, consistent with nontrivial Floquet multipliers $e^{-\pi \epsilon +O({\epsilon }^2)}$ and $e^{2\pi \epsilon +O({\epsilon }^2)}$. The nonzero derivatives at these averaged zeros also justify persistence of nearby cycles for sufficiently small $\epsilon$.