Past exam of the topics in applied mathematics course of the University of Oxford / 2011 / B / 5

A periodic orbit of least period $k$ is a list of distinct points $x_0,\dots ,x_{k-1}$ satisfying $f(x_i)=x_{i+1}$ for $i<k-1$ and $f(x_{k-1})=x_0$. Equivalently $f^k(x_0)=x_0$, with no smaller positive iterate fixing $x_0$. For $k=1$ this is a fixed point.

Follow a perturbation through one complete circuit. The chain rule givesAfter $m$ circuits the linear perturbation is $M^m\delta x_0$. With distinct eigenvalues, the eigenvectors span over the complex numbers, and each component is multiplied by ${\lambda }_j^m$. Thus the periodic orbit is linearly asymptotically stable exactly whenThese are the orbit's discrete Floquet multipliers. Any modulus greater than one causes instability; a unit modulus prevents decay in the linear system, and nonlinear stability then needs further analysis. The finitely many intermediate derivative factors do not affect convergence to zero between complete circuits.

A period-$k$ orbit of a Poincaré map corresponds to a closed trajectory, hence a periodic orbit of the underlying flow, meeting the chosen section at the $k$ distinct successive return points. Its time period is the sum of the $k$ return times, not necessarily $k$ times a single common time. Stability of the return-map orbit describes transverse stability of that closed trajectory; the flow itself also has the neutral time-translation direction.

There is always the fixed point $(0,0)$. For $P>0$, the first fixed-point equation gives $e^{-P}=1/r$, and the second gives $P=N(1-1/r)$. Thus for $r>1$ there is exactly one positive fixed point,For $r=1$, every point $(N,0)$, $N\ge 0$, is fixed. There are no other nonnegative fixed points.

At the origin the Jacobian matrix has eigenvalues $r,0$. It is asymptotically stable for $0<r<1$ and a saddle for $r>1$. For $r<1$, the bounds $N_{t+1}\le r{N}_t$ and $P_{t+1}\le N_t$ also prove global extinction in the nonnegative quadrant.

At the positive fixed point,Here $\log r<r-1$ and $\log r>1-1/r$ for $r>1$. The discriminant is negative because the positive trace is less than two and determinant exceeds one. The conjugate multipliers have modulus $\sqrt{N_{\ast }}>1$, so the positive fixed point is always unstable. This is the instability of the Nicholson-Bailey equilibrium.

For completeness, on the $r=1$ fixed line,For $0\le N<1$, linear perturbations are bounded but have a neutral component along the fixed line; there is no asymptotic stability of an individual point. For $N>1$ there is transverse instability. At $N=1$, the repeated unit eigenvalue has a nontrivial Jordan block, so powers have an unbounded off-diagonal entry and the linearized system is unstable. This degenerate case is not covered by a distinct-eigenvalue assumption.