For a system of two differential equations, the eigenvalues determine the stability of the system. When the eigenvalues are positive, the system is unstable. In case of complex eigenvalues, the system is unstable when the real parts of the eigenvalues are positive. In the latter case, how do we mathematically show that the system enters a stable limit cycle when the fixed point is unstable?

Considering the following non-linear system, Bier et al.'s model of yeast glycolysis

\begin{align} \frac{dA}{dt} &= 2k_1GA - \frac{k_pA}{A+K_m}\\\\ \frac{dG}{dt} &= V_{in}-k_1GA\end{align}

where $G$ refers to glucose and $A$ refers to ATP. The values of the parameters are $V_{in}=0.36$, $k_1=0.02$, and $k_p=6$. When $K_m=13$, the following behavior is observedPhase plane ref.

From the Jacobian of the matrix, the eigenvalues at the fixed point are $0.0040 + 0.1132i$ and $0.0040 - 0.1132i$. Here, the real parts of both eigenvalues are positive and we observe that the phase portrait shows a limit cycle. I would like to understand how complex eigenvalues with positive real parts are mathematically related to limit cycles.

Any help would be much appreciated

  • 1
    $\begingroup$ One thing that is related to your question is an Andronov-Hopf bifurcation. If you change parameters and a stable focus equilibrium changes its stability, then in typical case this leads to an emergence of a stable limit cycle. $\endgroup$
    – Evgeny
    Commented May 8, 2018 at 14:13

2 Answers 2


The pattern for a limit cycle in a 2D dynamical system is provided by the Poincaré-Bendixon theorem. What you need is a region that contains only sources as stationary points (or no stationary points at all) while the vector field points inward on the boundary of the region.

Just a source is not sufficient as the solutions originating at the source can move towards infinity.


This affirmation is not true

To have a limit cycle, a dynamic system should be at least nonlinear

The linear system

$$ \dot x = a x + b y\\ \dot y = c x + d y $$

has as eigenvalues

$$ \mu = \frac{1}{2}\left(a+d\pm\sqrt{a^2-2 a d+4 b c+d^2}\right) $$

so we can choose $a,b,c,d$ to have an unstable complex singular point but this will not generate any stable limit cycles.

  • $\begingroup$ Could you please why limit cycles cannot be observed in linear systems ? Wouldn't the oscillations come from the imaginary part of the complex values? $\endgroup$
    – Natasha
    Commented May 7, 2018 at 3:08
  • $\begingroup$ You can have oscillations but not a limit cycle. $\endgroup$
    – jdods
    Commented May 7, 2018 at 5:26
  • 1
    $\begingroup$ @jdods Why not a limit cycle? $\endgroup$
    – user539887
    Commented May 7, 2018 at 5:49
  • $\begingroup$ One of the most famous stable limit cycles is called as Van der Pol's described by $\ddot x-\mu(1-x^2)\dot x+x = 0$. The existance of a limit cycle needs the existence of the so called attractors. Linear systems have as attractors, isolated points while Vander Pol's for instance have as attractor, deppending on $\mu$ a closed orbit path. $\endgroup$
    – Cesareo
    Commented May 7, 2018 at 7:26
  • 2
    $\begingroup$ @Natasha A limit cycle is an isolated periodic orbit. Linear systems either have no periodic orbits or a continuum of periodic orbits. That's why linear systems can't have a limit cycle. $\endgroup$
    – Evgeny
    Commented May 7, 2018 at 14:55

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