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

The Poincaré-Bendixson theorem states that the omega-limit set of a forward orbit confined to a compact region of the plane, containing no equilibrium points, is a periodic orbit.

Put $x=R\cos \theta$, $y=R\sin \theta$. Direct calculation givesHere $1/2\le H\le 17/16$ and $|\cos \theta {\sin }^3\theta |\le 3\sqrt{3}/16$. Thus on the inner circle $R=\sqrt{3}/2$ the radial velocity is positive, and on the outer circle $R=3/2$ it is negative. In their closed annulus,so there is no equilibrium. The theorem gives a periodic orbit in this trapping annulus.

It is asymptotically stable: the vector-field divergence isthroughout the annulus. Its nontrivial Floquet multiplier is $\exp {\int }_0^T\mathrm{div}Fdt<1$. This proves an attracting periodic orbit exists. There can be only one there: two such nested orbits would enclose an annular region with zero outward flux, contradicting the strictly negative divergence via the divergence theorem.

A positive equilibrium must have $v=u=s$. LetIt is the unique positive root of $(1-s)(s+d)=as$. At this point the Jacobian matrix isConsequentlyIt is linearly asymptotically stable when $b>L$, and unstable when $0<b<L$. In either region it is a node or focus according as $(L-b{)}^2-4bsD/(s+d)$ is nonnegative or negative. At $b=L>0$ the linear eigenvalues are purely imaginary. If $L\le 0$, it is stable for every $b>0$.

The requested boundary follows from $1-d-2s=a-D$ and $s/(s+d)=(1-s)/a$:

To prove a limit cycle exists in the unstable case, a compact trapping set must stay away from the singular boundary $u=0$. The following construction supplies it. Choose $M>1$ large enough to contain the positive equilibrium and the orbit being considered. On $u=M$, $\dot{u}<0$, and on $v=M$ with $u\le M$, $\dot{v}\le 0$. Put $P=u{e}^{-v}$. Using $u+d\ge d$ and completing the square in $v$ givesChoose $\epsilon >0$ so small that $\epsilon e^M$ times the square bracket is less than $1/2$, and then choose $0<\eta <\epsilon$. The setis compact in the positive quadrant and forward invariant. On the curved boundary $P=\epsilon$, $u\le \epsilon e^M$ and $\dot{P}>0$. On $v=\eta$, $u\ge \epsilon >\eta$ gives $\dot{v}>0$. Its other boundaries were checked above. The constants can be chosen so the equilibrium is interior.

When $b<L$, the equilibrium is a hyperbolic repeller. Remove a sufficiently small outward-flow neighborhood of it from $\mathcal{K}$. This leaves a compact trapping region containing no equilibria. The Poincaré-Bendixson theorem gives a periodic omega-limit orbit of a nonequilibrium trajectory, hence a limit cycle. The analytic vector field excludes an accumulating family of periodic return-map zeros unless the whole local return map is the identity, which is incompatible with the repelling equilibrium and the nonperiodic approaching trajectory. This is the positive trapping construction for the Leslie-Gower model.

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.

For uniform nearest-neighbor bonds of strength $K_0$ on a ring with $N\ge 3$, retain the normalization of the given model:with indices interpreted modulo $N$. On an open chain the interior equations are the same, whileFor nonuniform bonds, replace each $K_0$ by the corresponding $K_{ij}$. A convention without the factor $1/N$ is obtained by rescaling the bond strengths; it is not silently substituted here.

For all-to-all coupling define the Kuramoto order parameterIt is the centroid of unit phase vectors: $r$ measures phase coherence and $\psi$ the mean phase. Multiplying by $e^{-i{\theta }_i}$ and taking imaginary parts givesThus only the two components of the Kuramoto order parameter mediate interaction.

For the symmetric stationary continuum branch, choose the rotating frame with $\psi =0$. Oscillators with $|\omega |\le Kr$ lock at $\theta =\arcsin (\omega /(Kr))$, choosing $\cos \theta >0$ for stability. Those with $|\omega |>Kr$ drift, with stationary phase density proportional to $|\omega -Kr\sin \theta {|}^{-1}$. Their real contribution to the order parameter vanishes, since the integral of $\cos \theta /(\omega -Kr\sin \theta )$ over a period is the change of a periodic logarithm. Their imaginary contributions cancel between $\omega$ and $-\omega$ because $g$ is symmetric.

The locked oscillators therefore give the real self-consistency relationSet $\omega =Kr\sin \theta$ and divide by $r>0$ to obtainThis is the Kuramoto self-consistency equation for partial synchronization. The incoherent state $r=0$ must be considered before this division. If $g$ is continuous at zero with $0<g(0)<\infty$, the branch meets incoherence at $K_c=2/[\pi g(0)]$.

Put $x=z$, $y=\dot{z}$. The system isIts equilibria are $\boxed{(0,0),(1,0),(-1,0)}$. At $(\pm 1,0)$, the eigenvalues are $(-A\pm \sqrt{A^2+8})/2$, of opposite signs, so these are saddles for every $A$.

At the origin, the eigenvalues are $(-A\pm \sqrt{A^2-4})/2$. It is a stable focus for $0<A<2$, a stable node for $A>2$, an unstable focus for $-2<A<0$, and an unstable node for $A<-2$. At $A=\pm 2$ it is the corresponding repeated-eigenvalue improper node. At $A=0$, linearization gives a center, but nonlinear damping matters: withsmall closed energy sublevels are invariant and $\dot{E}=-y^4$ forces convergence to the origin. Thus it is a weak attracting focus, not a nonlinear center.

Local bifurcation. Write $a=-A$. Averaging the radial equation near the origin, or eliminating nonresonant cubic terms, givesThe $x^3$ restoring term has zero average radial contribution, while $⟨y^4⟩=3R^4/8$. Thus $A=0$ is a supercritical Hopf bifurcation as $a$ increases: a stable cycle appears for small $a>0$, with leading amplitude $R\simeq \sqrt{4a/3}$.

Global changes. Two further bifurcations must be retained. Define $a_h$ as the parameter where the upper unstable branch from $(-1,0)$ reaches $(1,0)$. Central symmetry supplies the lower connection in the opposite direction, forming a heteroclinic cycle. Define $a_f>a_h$ as the subsequent collision of an inner attracting and an outer repelling periodic orbit. The complete phase portraits for $|A|<1$ are:

Here escape means leaving the central well and becoming unbounded; it is not an additional finite equilibrium. The softening Rayleigh-Duffing phase portrait in the figure includes the two critical cases.

The return-map calculation underlying this classification is useful because the two global critical parameters are very close. On the section $x=0,y>0$, let $H_a(p)$ be minus the velocity at the next crossing of $x=0$, so a symmetric cycle is a root of $H_a(p)-p$. Any periodic orbit must have its extrema inside $-1<x<1$, since at a positive maximum $y=0$ requires $x^3-x<0$. Its index is one, so it surrounds the origin and neither saddle. Every such cycle is centrally symmetric: its reflected image cannot be a distinct nonintersecting cycle of equal enclosed area. Thus this half-return test detects all central cycles. A fixed point with $H_a^{\prime }<1$ is attracting and one with $H_a^{\prime }>1$ repelling; the full-return multiplier is $(H_a^{\prime }{)}^2$.

For the saddle connection, let $p_L(a)$ be the upper velocity at $x=0$ from the left saddle's unstable branch and $p_R(a)$ the upper velocity at $x=0$ on the right saddle's stable branch. While $y>0$, their shooting equation isComparison shows $p_L$ increases with $a$ and $p_R$ decreases. At $a=0$, energy dissipation gives $p_L<1/\sqrt{2}<p_R$. At $a=1$, the lower comparison curve $y=(x+1)/2$ on $[-1,0]$ gives $p_L>1/2$, while the upper comparison curve $y=(1-x^2)/2$ on $[0,1]$ gives $p_R<1/2$. The latter follows fromand the stable-branch expansion $y=(1-x)-3(1-x{)}^2/4+\cdots$ near $x=1$. Hence there is a unique $a_h\in (0,1)$ with $p_L=p_R$.

At that connection the saddle eigenvalues are ${\lambda }_u=(a_h+\sqrt{a_h^2+8})/2$ and ${\lambda }_s=(a_h-\sqrt{a_h^2+8})/2$, so $|{\lambda }_s|/{\lambda }_u<1$. The loop is therefore repelling. In particular, the attracting Hopf cycle cannot simply disappear in that connection. Near the upper endpoint $p_R$, the half-return map hasFor $a$ just above $a_h$, this produces a second, repelling fixed point near the endpoint, in addition to the attracting inner one. Continuing the two roots of $H_a(p)-p$ makes their collision criterion explicit:This is a saddle-node bifurcation of periodic orbits, not a disappearance of the equilibrium points.

For a reproducible numerical check of the global sketches, event-based integration of this half-return map and its variational equation givesThese values are computed by shooting the stated equations; their precision is unnecessary for the requested sketches. They explain why a coarse parameter sweep can miss the two-cycle interval entirely. The three bifurcation types are therefore Hopf at $A=0$, a two-saddle heteroclinic connection at $A=-a_h$, and a fold of cycles at $A=-a_f$. There is no pitchfork of equilibria because their locations remain fixed for every $A$.