Past exam of the topics in applied mathematics course of the University of Oxford / 2009 / B / 4

An equilibrium is asymptotically stable when it is Lyapunov stable and attracts every initial state in some neighbourhood. Lyapunov stability means that sufficiently small initial displacement stays small for all future times; attraction additionally requires convergence to the equilibrium. A Lyapunov function is a differentiable positive-definite scalar function near the equilibrium whose derivative along solutions is nonpositive. Strict negativity, or an appropriate invariant-set argument for a non-strict derivative, gives asymptotic stability.

For $\nu =-1$, equilibria satisfy $y=0$ and $x+x^3=0$, so only $(0,0)$ is real. Its Jacobian matrix is $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, with eigenvalues $\pm i$ for every $\kappa$. It is a linear centre equilibrium, so linearization alone cannot decide nonlinear attraction.

Use the positive proper energyFor $\kappa <0$, energy sublevel sets are compact and invariant. The zero-derivative set is the union of the axes, but its only invariant subset is the origin: on $x=0,y\ne 0$ one immediately leaves that axis, and on $y=0,x\ne 0$ one has $\dot{y}=-x-x^3\ne 0$. LaSalle's invariance principle therefore proves global asymptotic stability. At $\kappa =0$ the conserved energy levels are closed periodic orbits, giving a nonlinear centre equilibrium. For $\kappa >0$, time reversal together with $y\mapsto -y$ converts the system to the attracting negative-$\kappa$ case, so the origin is a nonlinear repeller and is unstable. Nonzero forward trajectories cannot remain bounded; this does not assert existence for all time if finite-time escape occurs.

For $\nu =1$, the equilibria are $(0,0)$ and $(\pm 1,0)$. At the origin the eigenvalues are $1,-1$, so it is a saddle for every $\kappa$.

At either nonzero point the Jacobian matrix is $\left(\begin{array}{cc}0& 1\\ -2& \kappa \end{array}\right)$, with eigenvaluesBoth points are attracting for $\kappa <0$ and repelling for $\kappa >0$. For $-2\sqrt{2}<\kappa <0$ they are stable foci; for $\kappa <-2\sqrt{2}$ stable nodes. For $0<\kappa <2\sqrt{2}$ they are unstable foci; for $\kappa >2\sqrt{2}$ unstable nodes. At the two repeated-root boundaries they are degenerate nodes. At $\kappa =0$, linearization has purely imaginary eigenvalues; the conserved energy in the next part shows that they are actual nonlinear centres. Thus the zero-real-part case is resolved by an additional first integral, rather than by the linear test alone.

For $\nu =1,\kappa =0$, the first integral isIts minimum $-1/4$ occurs at the two centres. Levels between $-1/4$ and $0$ give a periodic orbit around each centre equilibrium; positive levels give larger periodic orbits surrounding both wells. The zero level is the figure-eight separatrix through the saddle, with two homoclinic loops.

On $E=0$, $y^2=x^2(1-x^2/2)$. The printed parametrization $x=\pm \sqrt{2}\mathrm{sech}z$, $y=\pm \sqrt{2}\mathrm{sech}z\tanh z$, taking matching signs, satisfies this identity because $1-{\mathrm{sech}}^{2}z={\tanh }^{2}z$. It also has $dx/dz=-y$ and $dy/dz=-(x-x^3)$, so its parameter is physical reversed time, $z=-t+\text{constant}$, rather than $z=t$. It is nonetheless exactly a parametrization of the requested trajectory. In physical forward time one can use $x(t)=\sqrt{2}\mathrm{sech}t$, $y(t)=-\sqrt{2}\mathrm{sech}t\tanh t$ on the right loop.

Approach to the origin requires $z\to \pm \infty$, hence infinite physical time. Equivalently $|y|\sim x$ near the right-loop endpoint, so the elapsed time contains the divergent integral ${\int }_0^{\delta }dx/x$. The point at $x=\sqrt{2}$ is reached in finite time from any interior point; it is the asymptotic saddle endpoints, not that turning point, that cause the infinite total traversal time.

The cubic terms vanish in the linearization, leaving $J_0=\left(\begin{array}{cc}1& -2\\ 2& 1\end{array}\right)$. Its eigenvalues are $1\pm 2i$, so the origin is a repelling focus equilibrium. The nonzero imaginary part gives rotation, while the positive real part makes small amplitudes grow. In particular it cannot be the forward limit of nearby nonzero trajectories, which motivates looking for an attracting periodic orbit away from the origin.

Use $x=r\cos \theta$, $y=r\sin \theta$ with $r>0$. Dotting the vector field with the radial direction and taking its angular cross-product givesThe angular factor ranges from one to three. Hence every radial component is outward for $0<r<1/\sqrt{3}$, and every radial component is inward for $r>1$. The limiting radii areThere is a minor strictness point in the wording. At $r_1$ the outward inequality is non-strict, with tangencies where ${\sin }^2\theta =1$; at $r_2$ the inward inequality is non-strict, with tangencies where $\sin \theta =0$. Thus these are the supremum and infimum for strict crossing at all angles. The weak inequalities at the endpoints are exactly what is needed for the closed trapping annulus in the next part.

The closed annulus $1/\sqrt{3}\le r\le 1$ is positively invariant by the radial signs, and $\dot{\theta }=2$ excludes equilibria there. Poincaré-Bendixson therefore yields a periodic orbit. Its stability can be established explicitly rather than inferred merely from trapping.

For the exact anisotropic radial limit cycle, set $w=r^{-2}$. Then $\dot{w}=-2w+4-2\cos 2\theta$, or $dw/d\theta +w=2-\cos 2\theta$. Its unique periodic solution isIt is positive and stays within the trapping annulus. Any other solution at the same phase has $w-w_p=C{e}^{-2t}$, proving radial attraction and uniqueness of the positive periodic orbit. The full planar period is $\pi$, since a full angular revolution takes that time; the half-turn symmetry of the radius does not halve the period of the position vector. The nontrivial return periodic-orbit multiplier is $e^{-2\pi }<1$. Thus the periodic orbit is asymptotically stable, and all nonzero forward trajectories approach it up to phase.