I have the following diferential equation $$ \left( \frac{du}{d\phi}\right)^2= 2\left(u + \frac{1}{6} \right)\left( u - \frac{1}{3}\right)^{2} \,\, , \qquad (1) $$ for the region wich has the solutions: $$ u(\phi) = -\frac{1}{6}+\frac{1}{2}\tanh^{2}\left(\frac{\phi+\phi_{0}}{2}\right) \,\, , \qquad (2) $$ for $u < 1/3$ and making the susbstituion in the differential equation (1) $$ u = \frac{1}{3} + \frac{1}{2}\tan^{2}\left(\frac{1}{2}\epsilon\right) \,\, , \qquad (3) $$ we get the following new equation $$ \left( \frac{d\epsilon}{d\phi}\right)^{2} = \sin^{2}\left(\frac{1}{2}\epsilon\right) \,\, . \qquad (4) $$ Taking the following appropriate solution $$ \phi = 2 \log \left(\tan \frac{1}{4} \epsilon\right) \,\, , \qquad (5) $$ a substituing back this solution in eq. (3) we get, $$ u(\phi) = \frac{1}{3}+\frac{2e^{\phi}}{(e^{\phi}-1)^{2}} \,\, , \qquad (6) $$ for $u > 1/3$.
Plotting these curves I can see that both of them tend asymptotically to $u = 1/3$. I have an intuition that $u = 1/3$ might be a limit cycle, to show this I'm trying to apply the Poincaré-Bendixson theorem, but I'm having no insight about how to show this... Is there any way show this?
As the constant function $u=\frac13$ satisfies this implicit ODE, this gives a closed orbit. As the sign of the derivative in the explicit form $\dfrac{du}{dϕ}=\pm (u-\frac13)\sqrt{u+\frac16}$ can take both variants, you have as many solutions moving away from that orbit as you have solutions moving toward that orbit.