6
$\begingroup$

I was given the following ODE as part of an exercise:

$$\frac{dx_1}{dt} = x_2$$ $$\frac{dx_2}{dt} = -x_2 + \frac{x_1^2}{2} - \frac{1}{2}$$

The question is to show that a heteroclinic orbit exists for this system.

My attempt:

By considering the linearization around the fixed points (1,0) and (-1,0), I deduced that (1,0) is a saddle point and (-1,0) is a sink. Then I immediately concluded (by Poincare-Bendixson Theorem) that a heteroclinic orbit exists. My teacher told me that I needed to do something more before invoking the Poincare-Bendixson Theorem.

Any clues and ideas on what he meant? Or any other alternative ways to approach this problem?

$\endgroup$
4
  • $\begingroup$ Did you show the existence of a positively invariant region before applying Poincare-Bendixson? @Air Christmas $\endgroup$
    – Alex
    Oct 7 '17 at 16:33
  • $\begingroup$ Hi Alex, thanks for your reply. I think that what you meant would be the basin of attraction around (-1,0). Then I would encounter another problem. By considering a small region around (-1,0) (which is a positively invariant region), I would then need to show that a trajectory emerging from (1,0) would hit the boundary of the positively invariant region. The idea seems good, but how can I express this mathematically? $\endgroup$ Oct 8 '17 at 4:36
  • $\begingroup$ Just to know what to expect: what methods of studying systems do you know? What is expected to be used by you here? Do you have an idea? Because I have an idea how to solve it using sort of Lyapunov function. $\endgroup$
    – Evgeny
    Oct 8 '17 at 8:19
  • $\begingroup$ Hi Evgeny, funny that you mentioned Lyapunov function. I did consider using Lyapunov theory, but I just could not come up with a suitable Lyapunov function for this. Please share your answer! $\endgroup$ Oct 8 '17 at 9:41
8
$\begingroup$

Okay, let's start from the form of your equations. When I see system $\dot{x} = y, \; \dot{y} = \dots$ the first thing which I think about is that maybe system is equivalent to $\ddot{x} = F(x)$. I read the second equation and I see that it's not the case: $\dot{x} = y, \; \dot{y} = -y + \frac{x^2-1}{2}$. But how exactly it's not the case? Well, there is only $-y$ term that spoils everything. If there is no such term it is conservative system, it has first integral and there is a lot can be said about dynamics. For example, if we get rid of $-y$ term, the system beecomes $\dot{x} = y, \; \dot{y} = \frac{x^2-1}{2}$ and $\Phi(x, y) = \frac{y^2}{2} - \frac{x^3}{6} + \frac{x}{2}$ is a first integral. Let's check how badly this $-y$ term spoils the picture. For this we have to calculate derivative with respect to your original system: $$ \frac{d}{dt}(\Phi(x, y)) = \Phi_{x}(x, y) \cdot \dot{x} + \Phi_y(x, y) \cdot \dot{y} = - \frac{x^2-1}{2} \cdot y + y \cdot \left( -y + \frac{x^2-1}{2} \right ) = -y^2$$ It's a kind of Lyapunov function — you have to narrow the domain and pick some neighbourhood to apply Lyapunov theorem, but still it's useful here. For example, we get the following: when trajectory doesn't intersect $Ox$ axis the first integral decreases. From geometric point of view that means that along any level set vector field points in the same direction (or tangent).

Referring to my earlier answer we can try proving the existence of heteroclinic trajectory using the case of trapping region. Pick a level set that contains saddle equilibrium. Vector field points inward along it except at two points (they are both on $Ox$ axis). The part of this level set is a boundary of our trapping region: vector field points inward, no trajectory can escape from this region. If you want to say that there is a problem with these two points, there is no problem really: one of them is a saddle of $C^1$-system and no trajectory can escape domain through it. Why system can't escape from another point on $Ox$ axis? Well, you can check $\ddot{x}$ for a trajectory that goes through this point and find does it really go in or out; I'll stay with Bony-Brezis theorem. Now we have key ingredient for applying Poincaré-Bendixson theorem — a forward invariant domain. But Poincaré-Bendixson theorem states that in compact domain of planar system only attractors are equilibria, limit cycles and heteroclinic contours or homoclinic loops. We know all equilibria for sure, but what do we know about limit cycles, heteroclinic contours or homoclinic loops? Here is the last ingredient: Bendixson-Dulac theorem. The divergence of this system is constant and equals $-1$ — that prevents existence of limit cycles, heteroclinic contours and homoclinic loops.

So, to sum up:

  • We found forward-invariant compact region to which Poincaré-Bendixson theorem can be applied
  • In this region we know all equilibria and their stability properties
  • Also, we know that there is no limit cycles or homoclinic loops or heteroclinic contours due to Dulac-Bendixson
  • If we pick unstable separatrix of saddle, it must stay in invariant region and it must go to attractor. The only attractor is a sink, so unstable separatrix goes to it.

ADDED LATER This illustration should clarify what trapping region I was using in my answer.

enter image description here

$\endgroup$
7
  • $\begingroup$ Your answer is actually very comprehensive. Unfortunately, I lack the required knowledge to fully grasp the meaning behind what you had explained. But I was able to have a rough idea of how your answer works. There is an additional detail which I would like to mention and that is the domain derived from the 'almost Lyapunov function'. It took me a while to realise that you were referring to a region below the x-axis. $\endgroup$ Oct 8 '17 at 18:44
  • 1
    $\begingroup$ Feel free to ask questions :) I was referring to the compact domain bounded by loop of curve $\frac{y^2}{2} - \frac{x^3}{6} + \frac{x}{2} = \frac{1}{3}$. It looks like this: desmos.com/calculator/foeizb3ula . The saddle point at $(1, 0)$ is on the boundary of the domain. $\endgroup$
    – Evgeny
    Oct 8 '17 at 20:18
  • $\begingroup$ You have actually cleared up a misunderstanding that I had. Indeed, given the graph of the curve, your argument makes good sense. (The graph of the 'almost Lyapunov function' has helped to make your explanations clearer.) $\endgroup$ Oct 9 '17 at 5:17
  • $\begingroup$ Oh that's great :) I was hasty in some of the details (like mentionining what level set should be taken as a boundary of trapping region), but I still can expand them if you want me to. $\endgroup$
    – Evgeny
    Oct 9 '17 at 6:27
  • $\begingroup$ I revisited this question again for the purpose of learning. I would like to iron out the details. May I ask what is the neighbourhood that you chose to apply Lyapunov's theorem? $\endgroup$ Jan 16 '18 at 20:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.