Can I say this is a limit cycle?

Question

I have the folowing system of diferential equations in hands,

$$ \frac{dx}{d\lambda} = \frac{d\theta (\lambda)}{d\lambda} \left[ - y + \frac{dr}{d\lambda}\frac{d\lambda}{d\theta}\frac{x}{r}\right] \,\, ,\\ \frac{dy}{d\lambda} = \frac{d\theta (\lambda)}{d\lambda} \left[ x + \frac{dr}{d\lambda}\frac{d\lambda}{d\theta}\frac{y}{r}\right] \,\, , $$

where,

$$ \frac{d\theta}{d\lambda} = \frac{L}{r^{2}} \qquad L \in \mathbb{R}^{+} \,\, ,\\ \frac{dr}{d\lambda} = \pm \sqrt{E^{2} - \left(1-\frac{2M}{r}\right)\frac{L^{2}}{r^{2}}} \qquad E \in \mathbb{R}, \ M > 0 \,\, .\\ r^{2} = x^{2} + y^{2} \,\, . $$

Which give me the following plot when I take $\dot{r} > 0$:

And give me the following system when I take $\dot{r} < 0$:

The black line is the region with $r = 3$. The solutions of this system tends asymptotically to the circle $r = 3$ from bellow when $\dot{r} < 0$ and move away from the circle from above, and vice versa when $\dot{r} > 0$.

The parameters I used in the above were $M=1$, $E^2=\frac{25}{27}$, and $L=5$. Can I say that the circle $r = 3$ is a limit cycle?

Answer 1

Yes, you can. If you insert your parameters and $r=3$ into your differential equations, you will obtain:

• $\frac{\mathrm{d}r}{\mathrm{d}λ} = 0$. Hence, a trajectory starting on your cycle will stay there.
• $\frac{\mathrm{d}θ}{\mathrm{d}λ} ≠ 0$. Hence, a trajectory on your cycle will not converge to a fixed point (on the cycle), but forever moves around the cycle.

Therefore, you have a limit cycle.