I'm asked to prove in two distinctly different ways, that the following system has no periodic solution in the phase plane:
\begin{alignat*}{2} \frac{dx}{dt} &= -x+4y \\ \frac{dy}{dt} &= -x-y^{3} \\ \end{alignat*}

For the first proof, I simply apply the Bendixson criterion.
Let \begin{equation*} f = \left[ \begin{matrix} f_{1} \\ f_{2} \end{matrix} \right]. \end{equation*} We have that \begin{alignat*}{2} f_{1} &= \dot{x} = -x+4y \\ f_{2} &= \dot{y} = -x-y^{3} \end{alignat*} $\therefore$ \begin{alignat*}{2} \frac{\partial f_{1}}{\partial x} &= -1 \\ \frac{\partial f_{2}}{\partial y} &= -3y^{2} \end{alignat*} $\therefore$ \begin{alignat*}{2} \frac{\partial f_{1}}{\partial x} + \frac{\partial f_{2}}{\partial y} &= -1 -2y^{2} \\ &< 0 &&\forall x,y \in \mathbb{R}. \end{alignat*}

By the Bendixson criterion, there are no periodic orbits (other than singular points) in a simply connected domain when $f$ is continuously differentiable and $\text{div}(f)$ does not change sign nor does it vanish identically. Thus, the system has no periodic orbits.

I find myself at an impasse for a distinct alternative proof. First, I tried a change to polar coordinates. The result I got was \begin{alignat*}{2} \left[ \begin{matrix} \dot{r}\\ \dot{\theta} \end{matrix} \right] &= \left[ \begin{matrix} 3r\cos\theta\sin\theta+r\cos^{2}\theta-r^{3}\sin^{4}\theta \\ \cos\theta\sin\theta - 4\sin^{2}\theta -\cos^{2}\theta-r^{2}\sin^{3}\theta\cos\theta\end{matrix} \right]. \end{alignat*} From here I did not see a clear way of solving for $r$ and $\theta$ or analyzing $\dot{r}$ to rule out periodicity.

Another approach that I considered was to use the theory associated with the Poincare-Bendixson theorem. For instance, if I could show that every solution crosses a transversal an infinite number of times then I could rule out a periodic solution. However, I was not sure how to construct such an argument from the given system.

I also wonder if an eigenvalue analysis of a linearization could be used here as well? I would be happy with any distinctly different approach or a hint towards one that I have tried so far. Thanks in advance.


1 Answer 1


Hint: For a simple second proof consider Lyapunov function $$ V(x,y)=x^2+4y^2. $$

I also wonder if an eigenvalue analysis of a linearization could be used here as well?

Actually it could (not always though) because there are results that state what type of an equilibrium can be inside any closed trajectory, but I do not remember them out the top of me head. But my hint is much simpler.

  • $\begingroup$ Thank you, Artem! By using the Lie derivative with your Lyapunov function, it succinctly follows that there can be no periodic solutions by the Krasovskii-LaSalle principle. Initially, this method seemed very similar to the first. So, I turned to the question of 'How are these two solutions distinct?'. Reviewing the proof of the Krasovskii-LaSalle principle, I see that we are using Lyapunov stability and Lyapunov functions, while in the method and proof of the Bendixson criterion we are using the divergence of the ODE. This makes the two methods appear to be 'distinctly' different to me. $\endgroup$
    – jpb
    May 12, 2014 at 2:10
  • $\begingroup$ hi @jpb, sorry for answering an old post, but I was curious about your use of the LaSalle principle. I can't follow your reasoning which leads to the nonexistence of periodic functions (using LaSalle of course). Thanks in advance! $\endgroup$
    – Riccardo
    Jun 9, 2014 at 10:38
  • 1
    $\begingroup$ Hi @Riccardo: Given a fixed point $x_{0}=(0,0)$ and a Lyapunov function $L$ which is not constant on any orbit lying entirely in the punctured plane, then $x_{0}$ is asymptotically stable and every orbit converges to $x_{0}$. My understanding is that this is Krasovskii-LaSalle applied to the entire plane. We also have a strict Lyanpunov function which implies nonexistence of periodic solutions since a periodic solution $\phi$ would have $L(\phi(t_{0})) = L(\phi(t_{1}))$ for some $t_{1} > t_{0}$. $\endgroup$
    – jpb
    Jun 9, 2014 at 18:06
  • $\begingroup$ @jpb clear! thanks! $\endgroup$
    – Riccardo
    Jun 9, 2014 at 20:49

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