Solution. The periodic orbits are the circles \(r = 1\) and \(r = 2\). With \(g(r) = r(1 - r)(2 - r)\), \(g > 0\) on \((0, 1)\), \(g < 0\) on \((1, 2)\) and \(g > 0\) on \((2, \infty )\). So orbits approach \(r = 1\) from both sides (stable limit cycle) and move away from \(r = 2\) (unstable limit cycle). The origin is an unstable focus. This is the pattern of the subcritical Hopf bifurcation of the next section.

Solution. \(Bf = e^{-2x}\bigl (y,\ -x - y + x^2 + y^2\bigr )\), so \begin {align*} \operatorname {div}(Bf) &= \partial _x(e^{-2x}y) + \partial _y\bigl (e^{-2x}(-x - y + x^2 + y^2)\bigr )\\ &= -2e^{-2x}y + e^{-2x}(-1 + 2y) = -e^{-2x} < 0 . \end {align*}

By Theorem 12.3.1 with \(D\) the whole plane, there is no periodic orbit.

Solution. \(Bf = \bigl ((a - by)/y,\ (-c + dx)/x\bigr )\), whose divergence is \(\partial _x\bigl ((a - by)/y\bigr ) + \partial _y\bigl ((-c + dx)/x\bigr ) = 0\) identically. The criterion requires the divergence not to vanish identically, so it does not apply; indeed the identically zero divergence says that the flow preserves the area form \(dx\,dy/(xy)\), which is the Hamiltonian structure behind the family of closed orbits.

Solution. \(E' = x'(x'' + x) = \mu \bigl (x'^2 - \frac 13x'^4\bigr )\). With \(x' = -r\sin t\), \(\langle x'^2\rangle = \frac {r^2}{2}\) and \(\langle x'^4\rangle = \frac {3r^4}{8}\), so \(\langle E'\rangle = \mu \bigl (\frac {r^2}{2} - \frac {r^4}{8}\bigr )\), zero at \(r = 2\), positive below and negative above. The amplitude in \(x\) is \(2\), and the cycle is stable. (Differentiating the Rayleigh equation shows that \(v = x'\) satisfies van der Pol’s equation in a scaled form, which explains the coincidence.)

Solution. \(f = (x - y - xr^2, x + y - yr^2)\) has \(\operatorname {div}f = 2 - 4r^2\) (since \(\partial _x(xr^2) = r^2 + 2x^2\) and \(\partial _y(yr^2) = r^2 + 2y^2\)), which is \(-2\) on \(r = 1\); with \(T = 2\pi \), \(\rho = e^{-4\pi } = 3.5 \times 10^{-6}\). From \(r' = r(1 - r^2)\), the linearisation at \(r = 1\) is \((r - 1)' = -2(r - 1)\), so \(r - 1\) is multiplied by \(e^{-2 \cdot 2\pi } = e^{-4\pi }\) in one revolution, the same.

Solution. By index theory a periodic orbit must enclose equilibria whose indices sum to \(+1\). A saddle has index \(-1\), and a periodic orbit enclosing no equilibrium or only the saddle would have index \(0\) or \(-1\), not \(+1\). So there is no periodic orbit.