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Suppose I have the following dynamical system $x''+x-x^2-ax'[1-3(x')^2-x^2(3-2x)]=0$. Now I want to see that for each $a>0$ we have an homoclinic orbit. Here's my attempt :

If we consider the function $V(x,y)=\frac{y^2}{2}+\frac{x^2}{6}(3-2x)$ we have that $\frac{d}{dt}V(x(t))=ay^2(1-6V(x,y))$ . The system has two equiblibrium point one being $(0,0)$ which is unstable , can be seen using this function $V$ in a small enough neighborhood of $(0,0)$ and the other one is $(1,0)$. Now I want to show that we have an homoclinic orbit for $a>0$, my idea was for example to consider the set $D:=\{(x,y)\in \mathbb{R}^2 : V(x,y)\leq \frac{1}{6}\}$ and this will be a compact set containing a finite number of equilibrium points that is $\gamma^+$-invariant , and so by the Poincare-Bendixson we have that for $x\in D$ then $\omega(x)$ is either an equilibrium point or the heteroclinic and homoclinic orbits, or a limit cycle.Assuming I could see that there are no limit cycles,I think I need to show that there exists an $x\in D$ such that $\omega(w)$ contains a point $y$ such that $f'(y)\neq 0$, where $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ is the function associated to the first order ordinary differential equation , but then it still could be that I have an heteroclinic orbit so I am not sure what to do know. I thought about considering $D$ to be $\{(x,y)\in \mathbb{R}^2 : V(x,y)< \frac{1}{6}\}$, so that I could take out the point $(1,0)$, but I don't think that this is $\gamma^+-$invariant

Another alternative would be to consider $D-\{(0,0)\}$ and then the only equilibrium point is $(1,0)$ and then I have to show that there exists an $x$ such that $\omega(x)$ contains a regular point , although I am not entirely sure how to do this. My idea was to check that $(1,0)$ is an unstable point and has unstable subspace $span\{(1,1)\}$ and so if we consider a solution starting in the unstable subspace we will never get that $f'(y)=0$ because the solution would have to be converging to the only equilibrium point in that domain , but we choose it such that doesn't happen .

I think we won't have any limit cycles in that region by looking at the divergence and seeing that it's never zero.

Any help is appreciated. Thanks in advance.

Someone
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  • Just a passing comment, but I think your statement is false. When $a = 0$, a homoclinic loop lies on a set $\frac{y^{2}}{2}+\frac{x^{2}}{6}(3-2x)=\frac{1}{6}$. Only one of the separatrices stays inside this domain when $a \neq 0$: separatrices must be tangent to respective eigenvectors of saddle equilibrium, and one of the eigenvectors "goes out of the domain" $\frac{y^{2}}{2}+\frac{x^{2}}{6}(3-2x) \leqslant \frac{1}{6}$. The picture is quite similar to the one in this answer. However, in your case the divergence does not have constant sign. – Evgeny Jan 07 '21 at 14:13
  • Ignore the last phrase about the divergence, it doesn't really matter here :) – Evgeny Jan 07 '21 at 14:22
  • Ok thanks for the answer. Just one thing out of curiosity assume the statement is true and I wanted to prove that such an homoclinic orbit existed, how would one go about proving it ? I mean one could use the Poincare-Bendixson theorem to get the 3 hypothesis but after this I don't know what could be done. If we tried to use the Dulac Criterion to show that it can't have a periodic orbit then I think this would also show we can't have an homoclinic orbit , so I guess I would have to see that there are no points other than the equiblirium where $f=(f_1,f_2)$ is zero. @Evgeny – Someone Jan 07 '21 at 15:32
  • But even after seeing that there is no periodic orbit , how could we show that there there is more than just one equilibirum point , is by doing that thing of showing that there must exist a point $y\in \omega(x)$ such that $f(y)\neq 0$, and then we get the existence of homoclinic or heteroclinic orbits , and if we only have on equilibrium point inside the region then we must have an homoclinic one. – Someone Jan 07 '21 at 15:33
  • I was just trying to understand if this would make sense, most of the times I just use the Poincare Bendixson to show the existence of a limit cycle and this is easier since we just need to check the region doesn't containg equilibrium points. – Someone Jan 07 '21 at 15:34
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    If you want to prove that there is a trajectory that connects two equilibria on a plane, most approaches were summarized here. I think I've presented all approaches that I've encountered in papers or books, but something might have been overlooked. Maybe the easiest case is when you can show that you have a trapping region with exactly one saddle equilibrium at its boundary. If there is no other equilibria inside, then limit cycles and heteroclinic cycles are excluded: both cases require extra equilibria. ... – Evgeny Jan 07 '21 at 16:08
  • ... Thus, if you can show that such trapping region contains a stable and an unstable separatrix of this saddle, you are done by Poincare-Bendixson: all options except a homoclinic loop are excluded. Such strategy relies on two key ingredients: knowing equilibria states (in some cases that could be done even analytically) and finding a trapping region. The last one is usually the hardest part. – Evgeny Jan 07 '21 at 16:18

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