Question 5.4: Analyze the phase plane trajectories for the system dx/dt= −...
Analyze the phase plane trajectories for the system
\frac{dx}{dt} = −y + x(p − x^2 − y^2) (5.5.7)
\frac{dy}{dt} = x + y(p − x^2 − y^2) (5.5.8)
One can show that for all p, (0, 0) is the only equilibrium point. The Jacobian at this steady state is
J_{ss}=\begin{bmatrix} p & -1 \\ 1 & p \end{bmatrix} (5.5.9)
and its eigenvalues are λ_1 = p + ı and λ_2 = p −ı . So for p < 0 we have a stable focus and for p > 0 we have an unstable focus. Using the polar coordinates that we introduced in Example 5.2, we get the following description of the system in terms of the radial distance and the phase angle
\frac{dr}{dt} =r(p − r^2) (5.4.10)
\frac{d\theta }{dt} =1 (5.4.11)
Similar to what we saw in Example 5.3, the polar coordinates uncover the presence of an additional equilibrium solution. There is the steady state r = 0 which corresponds to y_{ss} = (0, 0), while an additional steady state emerges at r=\sqrt{p} for p > 0. This latter solution is a limit cycle.
To analyze the stability of the limit cycle, we need to analyze how the radial position changes near the limit cycle. Starting inside the limit cycle, i.e. for 0 < r <\sqrt{p}, \dot{r} is positive and the radial position increases. Starting outside the limit c, i.e. for r > \sqrt{p}, we have that \dot{r} is negative and the radial distance decreases. This implies that this limit cycle attracts the trajectories and is thus stable. As before, p = 0 corresponds to a bifurcation point, whereby a stable focus becomes unstable, and a stable limit cycle develops around it. At that bifurcation point the Jacobian has a pair of eigenvalues on the imaginary axis. As p increases, the amplitude of the oscillation of the limit cycle also increases as \sqrt{p}. Figure 5.13 shows the phase plane for Example 5.4.
