Past exam of the mathematics course of the University of Cambridge / 2005 / ia / Paper 2 / 7B (Differential Equations)

For $r>0$, the polar coordinates satisfyA positive constant radius must therefore be one. The nonconstant periodic solutions arewith least period $2\pi$ and counterclockwise orientation. The origin is also a stationary solution, hence periodic in the degenerate sense of a constant function; it is a point rather than a nontrivial closed orbit. There are no other constant-radius solutions.

For $r_0>0$, put $v=r^2$. Then $\dot{v}=2v(1-v)$, a logistic differential equation. Solving this separable differential equation givesThe denominator is positive for every forward time. When $0<r_0<1$, the radius increases monotonically to one; when $r_0>1$, it decreases monotonically to one; and when $r_0=1$ it stays there. Thus every nonzero orbit approaches the unit circle while continuing to rotate. More precisely, its distance from the phase-matched circular solution tends to zero; its Cartesian coordinates do not converge to a single point. For $r_0=0$, the solution stays at the origin.

The origin is the only equilibrium point. Its Jacobian matrix iswhose eigenvalues are $1\pm i$. It is an unstable focus. The unit circle is an attracting limit cycle, rather than a curve of equilibria: on it the vector field remains nonzero. Linearizing the radial equation at $r=1$ gives $\dot{\delta r}=-2\delta r+O(\delta r^2)$, confirming radial attraction. All other radii are strictly monotone, so there are no additional nontrivial periodic orbits.