Past exam of the topics in applied mathematics course of the University of Oxford / 2010 / B / 4

Let $F(x,y)=(y,x-\Omega y-x^3-x^{2}y)$. Every term changes sign when both variables change sign, so $F(-x,-y)=-F(x,y)$. Consequently the derivative of $(-x(t),-y(t))$ equals $F(-x(t),-y(t))$, proving the stated rotational symmetry.

This is equivariance under a half-turn, not time reversibility. Reversibility of an autonomous differential equation requires a reverser $R$ satisfying $F(Rz)=-DR(z)F(z)$, so that $Rz(-t)$ is a solution. The half-turn $R=-I$ would require $F(-z)=F(z)$, the opposite of what holds. The usual mechanical candidate $R(x,y)=(x,-y)$ also fails when damping is present: its second equation would require $(\Omega +x^2)y=-(\Omega +x^2)y$ identically. The symmetry relates two orbits with the same direction of time, and does not reverse their arrows.

The planar divergence is $\nabla \cdot F=-(\Omega +x^2)<0$ for $\Omega >0$. Since the plane is simply connected, the Bendixson-Dulac theorem excludes any nonconstant periodic orbit: the flux across a closed trajectory would be zero because the vector field is tangent, whereas Green's theorem makes it the strictly negative integral of the divergence over the enclosed region.

Alternatively define $E=y^2/2-x^2/2+x^4/4$. Direct differentiation givesA nonconstant periodic solution would require the integral of this strictly dissipative expression over a period to vanish. That forces $y\equiv 0$ and hence an equilibrium, a contradiction. This energy argument identifies the damping mechanism behind the geometric criterion.

The equilibria satisfy $y=0$ and $x-x^3=0$, giving $(0,0)$ and $(\pm 1,0)$. At the origin,Their product is $-1$, so the origin is a saddle equilibrium for every real $\Omega$.

At either nonzero equilibrium,The determinant is positive and the trace is $-(\Omega +1)$. Thus both equilibria are attracting for $\Omega >-1$, repelling for $\Omega <-1$, and have purely imaginary eigenvalues $\pm i\sqrt{2}$ at $\Omega =-1$. More precisely they are stable foci for $-1<\Omega <-1+2\sqrt{2}$ and stable nodes for $\Omega >-1+2\sqrt{2}$; they are unstable foci for $-1-2\sqrt{2}<\Omega <-1$ and unstable nodes for $\Omega <-1-2\sqrt{2}$. At the two discriminant-zero endpoints they are repeated-eigenvalue degenerate nodes, with a nondiagonalizable Jacobian. At $\Omega =-1$, linearization alone does not determine nonlinear stability or prove a nonlinear centre.

At $\Omega =-2$, the origin has stable and unstable eigenvalues $1-\sqrt{2}$ and $1+\sqrt{2}$; the two nonzero equilibria are repelling foci with eigenvalues $(1\pm i\sqrt{7})/2$. A compatible portrait has one attracting closed orbit surrounding all three equilibria. Its total enclosed index is $+1+1-1=+1$, as a periodic orbit requires. Its half-turn image is also attracting, so uniqueness of the attracting cycle makes that cycle half-turn symmetric. Orbits spiral away from either focus equilibrium and approach the surrounding cycle; the stable separatrices approach the saddle instead. The figure integrates representative trajectories and shows the local saddle directions.

These trajectories cannot escape to infinity. A Liénard boundedness function supplies one explicit boundedness argument using the Liénard coordinate $z=y+F(x)$ with $F(x)=x^3/3-2x$ and $g(x)=x^3-x$. Then $\dot{x}=z-F(x)$ and $\dot{z}=-g(x)$. Put $V=x^4/4-x^2/2$ and $W=V+z^2/2-xz$. This function is coercive after addition of a constant. Its derivative obeysusing $Fz\le (z^2+F^2)/2$. The right side is negative outside a compact region, yielding bounded forward motion.

In the displayed portrait, generic non-equilibrium initial states tend to the cycle; points on the saddle's stable manifold tend to the origin; and initial data exactly at a repelling focus equilibrium remain there. The Poincaré-Bendixson theorem excludes a chaotic compact limit set for this smooth planar flow. Strictly, the assumption gives uniqueness only of an attracting periodic orbit: it does not by itself exclude exceptional nonattracting periodic or separatrix limit sets. Those would need additional global information if one wanted a classification stronger than the compatible portrait requested.

First assume ${\omega }_0\ne 0$ and regard $\theta$ modulo $2\pi$. The section $\theta =0$ has return time $T=2\pi /|{\omega }_0|$. Solving the resonantly forced oscillator from this section givesAt $t=T$, the sine vanishes and the cosine is one, while differentiating the secular term adds $\gamma T/2$ to $y$. Hence the Poincaré map isFor $\gamma >0$, every iterate has $y_n=y_0+n\gamma T/2\to \infty$. Thus every oscillator trajectory is unbounded, regardless of its initial amplitude. For an arbitrary section phase ${\theta }_0$, the return map translates by $(\gamma T\sin {\theta }_0/(2{\omega }_0),\gamma T\cos {\theta }_0/2)$, which is still nonzero. At $\gamma =0$ the map is the identity, and all $(x,y)$ motions are bounded harmonic oscillations, including the zero-amplitude equilibrium.

The zero-frequency case allowed by the printed parameter range is different. If ${\omega }_0=0$, then $\theta ={\theta }_0$ is constant and $x=x_0+y_{0}t+\frac{1}{2}\gamma \cos {\theta }_0{t}^2$. There is no phase-crossing Poincaré section. For $\gamma >0$, choosing $\cos {\theta }_0=0$ and $y_0=0$ gives bounded equilibria, contradicting an unqualified “all trajectories” statement. For $\gamma =0$ at zero frequency, free-particle motion is unbounded unless $y_0=0$. The resonance conclusion therefore requires the natural nonzero-frequency assumption.