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I'm working my way through Dynamics and Bifurcations [Jack K. Hale, Huseyin Kocak] and I can't understand this probably trivial implication in Example 11.1 (page 334). We have a following system of ODEs (perturbed harmonic oscillator) \begin{aligned} \dot{x}_1 & = x_2 + a x_1 (x_1^2 + x_2^2), \\ \dot{x}_2 & = -x_1 + a x_2 (x_1^2 + x_2^2), \end{aligned} $a\in\mathbb{R}$, which in polar coordinates \begin{equation} x_1 = r \cos \theta, \quad x_2 = -r \sin \theta, \end{equation} reads \begin{aligned} \dot{r} & = a r^3, \\ \dot{\theta} & = 1. \end{aligned} The book says: "Since $\dot{\theta}>0$, the orbits spiral monotonically in $\theta$ around the origin. Therefore, the stability type of the origin of [the original planar system] is the same as that of the equilibrium point of the radial equation $\dot{r} = a r$."

I fail to see why it matters whether $\theta$ is a monotonic function of $t$ or not. Can't we draw the same conclusion that the radial equation determines the stability of the origin when this condition does not hold? (Let's say $\dot{\theta} = \sin\theta$?)

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  • $\begingroup$ Well, I fail to see too. Maybe it was confusing wording. The equation for $\theta$ has no effect on stability here. $\endgroup$ – Evgeny Jun 11 '17 at 16:44

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