I have this equation

$\dot{x}=-y+\lambda x(36-9x^2-y^2)\\\dot{y}=9x+\lambda y(36-9x^2-y^2)\\\dot{z}=-6z-\lambda^2x^2y^2z^3$

and I want to find the stable and unstable manifolds associated to the singular point and to the periodic orbit depending on the values ​​of $\lambda\in\mathbb{R}$

so now I have an idea, I begin analyzing the stability of Lyapunov in the singular point and and in the periodic orbit, so I have that the eigenvalues of the matrix of monodromy are $\lambda_1=e^{-144\lambda t}$ , $\lambda_3=e^{-12\pi t}<1$ and $\lambda_{2}=1$ so by Andronov-Vitt theorem we have that the periodic orbit is Lyapunov stable.

The for the singular point I used "If there is at least one eigenvalue such that $\text{Re}(\lambda_1)>0$ then the singular point is unstable", so I considered the linear part of the equation, then I have that one of the eigenvalues is $\lambda=6>0$ then the singular point is unstable.

Now I want to find the stable and unstable manifolds using that information, but I´m really stuck in this part because this give me not to much information about the stable manifolds and I can´t find a useful theorem to help me. Thanks for any comment!

  • $\begingroup$ Actually, many proofs don't tell you how to find invariant manifolds. There are some proofs that "construct" manifolds as a limit of approximations (see any proof that uses Banach fixed point theorem). But in your case all what you've got is phase portrait and qualitative behaviour that you can guess-then-prove from it. Dynamics of whole phase space is much clearer if we'd go to "generalized" cylindric coordinates $x = a \cos \phi$, $y = b \sin \phi$, $z = h$. Also, how did you find monodromy matrix? I clearly see a limit cycle at ... $\endgroup$ – Evgeny Dec 7 '13 at 9:22
  • $\begingroup$ ... $9x^2 + y^2 = 36, \; z = 0$,but monodromy seems to require exact solution in some neighbourhood of closed trajectory? Or was it done throgh linearization? $\endgroup$ – Evgeny Dec 7 '13 at 9:24
  • $\begingroup$ in this case you can got the monodromy matrix first you have to found the canonical fundamental matrix wich is this evaluated on the period odf the solution i.e $2\pi$ $\phi (t)=\left [ \varphi _{1}(t)\lambda , \varphi _{2}(t), \varphi _{3}(t) \right ]=\begin{bmatrix} e^{-72}cos(3t)\lambda & -\frac{1}{3}sen(t) &0 \\ 3e^{-72}cos(3t)& cos(t) & 0\\ 0&0 & e^{-6t} \end{bmatrix}$ so evaluating i got $\phi (t)=\left [ \varphi _{1}(t), \varphi _{2}(t), \varphi _{3}(t) \right ]=\begin{bmatrix} e^{-144\lambda t} & 0 &0 \\ 0& 1 & 0\\ 0&0 & e^{12\pi} \end{bmatrix}$ $\endgroup$ – k73586 Dec 8 '13 at 5:13
  • $\begingroup$ $\varphi _{1}(t$) i found it by parameter variation, $\varphi _{2}(t)$ is the derived of the periodic orbit and $\varphi _{3}(t)$ got it by checking what happens with the arrow three and the column three $\endgroup$ – k73586 Dec 8 '13 at 5:17
  • $\begingroup$ Ah, ok. My advice remains the same: change of variables $x = \frac{1}{3} r \cos \phi$, $y = r \sin \phi$, $z = h$ easily shows invariant manifolds of limit cycle. $\endgroup$ – Evgeny Dec 8 '13 at 8:57

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