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Exercise :

Show that the system of differential equations : $$x' = h(y)$$ $$y'=ay+g(x)$$ where $a>0$ is a constant and $h,g : \mathbb R \to \mathbb R$ smooth functions, does not have any periodic solutions.

Hint : Apply Bendixson's criteria.

Discussion/Question :

First of all, the exercise refers to Bendixson's Criteria but on our notes and previous lessons, we have explored two different theorems/criteria involving Bendixson :

Poincare-Bendixson Theorem :

Given a differentiable real dynamical system defined on an open subset of the plane, then every non-empty compact ω-limit set of an orbit, which contains only finitely many fixed points, is either :

  • a fixed point,

  • a periodic orbit, or

  • a connected set composed of a finite number of fixed points together with homoclinic and heteroclinic orbits connecting these.

Moreover, there is at most one orbit connecting different fixed points in the same direction. However, there could be countably many homoclinic orbits connecting one fixed point.


Bendixson-Dulac Theorem : Ιf there exists a $C^{1}$ function ${\displaystyle \varphi (x,y)}$ (called the Dulac function) such that the expression : $$\frac{\partial φf}{\partial x} + \frac{\partial φg}{\partial y}$$ has the same sign $( \neq 0)$ almost everywhere in a simply connected region of the plane, then the plane autonomous system : $$\frac{dx}{dt} = f(x,y)$$ $$\frac{dy}{dt} = g(x,y)$$ has no nonconstant periodic solutions lying entirely within the region."Almost everywhere" means everywhere except possibly in a set of measure 0, such as a point or line.


I wanted to clarify which of the $2$ criteria/theorems is implied to be used by the phrased used as a hint given by the exercise, as it is not clear enough.

In my opinion, I think it refers to the Bendixson - Dulac one, since it's the one regarding the non-existence (point of the exercise as well). Would the first one though lead to something that won't hold with the hypothesis that a periodic orbit exists ?

Now, applying the Bendixson-Dulac Theorem :

First of all, we're given that $h,g : \mathbb R \to \mathbb R$ are smooth functions, which means that they indeed $h,g \in C^1(\mathbb R)$. Then, for $φ(x,y) = 1 \in C^1(\mathbb R^2)$, we have :

$$\frac{\partial φf}{\partial x} + \frac{\partial φg}{\partial y} = \frac{\partial h(y) }{\partial x} + \frac{\partial (ay + g(x))}{\partial y} = a> 0$$

since $h(y)$ is a function of $y$ and $g(x)$ a function of $x$. Then, since the expression required sustains the same sign everywhere, from the Bendixson-Dulac Theorem there exists none periodic solutions to the given system of differential equations.

So, is my approach correct ? Would be letting $φ(x,y) = 1$ be correct ? It seems trivial in such case, still not sure about what exactly the hint wants us to use.

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    $\begingroup$ Seems correct to me. It's just a nice system for which it is easy to prove that divergence has constant sign everywhere, even without using complicated $\phi(x, y)$, and thus Dulac-Bendixson theorem is applicable. Bonus round: what happens when $a = 0$ ? ;) $\endgroup$ – Evgeny Jan 15 '18 at 17:55
  • $\begingroup$ By the way, Bendixson's theorem is about divergence, without using additional function $\phi(x, y)$. That's Dulac's addition, if I remember it right. $\endgroup$ – Evgeny Jan 15 '18 at 17:58
  • $\begingroup$ @Evgeny Probably then, the Bendixson-Dulac theorem applied with $φ(x,y) = 1$ is simply referred to as Bendixson. Yep, it seems really straight forward, found it just too simple as every other problem I've handled had complicated $φ$. Hmm, chaos ??? $\endgroup$ – Rebellos Jan 15 '18 at 17:58
  • $\begingroup$ @Evgeny Yeah, exactly, seems like that, that's why I mentioned it in the previous comment ! $\endgroup$ – Rebellos Jan 15 '18 at 17:59
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    $\begingroup$ No, chaos is impossible in 2D autonomous systems on plane and sphere :) if $a = 0$ the system has a first integral and certain critical points of first integrals correspond to center equilibriums surrounded by closed trajectories. Such system can also have plenty of homoclinic and heteroclinic trajectories. You can find few explanations here. $\endgroup$ – Evgeny Jan 15 '18 at 18:51

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