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Consider the following 2D autonomous system of ODEs: $$ \left\{ \begin{array}{ll} \dot{x} = x^2 + 2y - x \\ \dot{y} = 3xy/2 - 3x^2 - y + 2x \end{array} \right. $$

How can we prove the existence and uniqueness of a homoclinic orbit (i.e. a solution $X$ of the system for which $\lim_{t \to \infty} X(t) = \lim_{t \to -\infty} X(t) = x_0,$ where $x_0$ is an equilibrium point of the system) for the system?

It is not hard to determine the equilibrium points: $(0,0), (-3, -6)$ and $(2/3, 1/9)$. The latter two are asymptotically stable equilibrium points, so there cannot be a homoclinic orbit for those points.

However, $(0,0)$ is a saddle equilibrium, so perhaps there is a homoclinic orbit here. We can also see that the function $H(x,y) = - y^2 + x^2(1-x)$ is constant on the solutions of the system. The set $H(x,y) = 0$ seems to be comprised of three solutions of the system, one of which is indeed a homoclinic orbit, so this proves existence.

However, how do we prove that this is unique? I don't really know how to approach this part.

S.T.
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  • Let me rephrase some part of my answer from here. Because your system has a smooth first integral (a function that is constant along the trajectories), the homoclinic orbit has to be in the same level set of first integral as its saddle. The proof strategy is simple: study the level sets of first integral that contain saddle-equilibria. All homoclinic orbits must lie on any of these sets, so if you showed that there is only one such orbit, then you proved what you wanted. – Evgeny Nov 14 '20 at 14:07
  • We indeed have that $(0,0)$ is the only saddle equilibrium, that $H(0,0) = 0$ and that this is the only level set of $H$ that contains $(0,0)$. However, why must all homoclinic orbits lie on the level sets of $H$? In this answer you have provided you state that this is easy to prove, but I am not sure how to prove it. – S.T. Nov 14 '20 at 15:47
  • It holds not only for homoclinic trajectories, but for all trajectories. By definition, since the first integral is constant along the trajectory $\gamma(t)$, then $H(\gamma(t)) \equiv H(\gamma(0))$. Let us consider the level set $\lbrace H(\mathbf{x}) = H(\gamma(0)) \rbrace$: just by the previous observation the whole trajectory $\gamma(t) \subseteq \lbrace H(\mathbf{x}) = H(\gamma(0))$. – Evgeny Nov 14 '20 at 16:54
  • @Evgeny Alright, I now understand this, thank you. In this case, a homoclinic orbit be a subset of the level set of the first integral which contains also a saddle equilibrium by continuity of the first integral, right? Because if $\gamma(t) \to x_0$ as $t \to \pm \infty$, then $H(\gamma(t)) \to H(x_0)$ as $t \to \pm \infty$. – S.T. Nov 14 '20 at 17:04
  • Ah, that. Pick a sequence of time moments $t_i \rightarrow +\infty$ and let $\gamma(t)$ be a homoclinic trajectory. We've already shown that $H(t_i) \equiv H_0$. As $t_i \rightarrow +\infty$, $\gamma(t_i) \rightarrow p$, where $p$ is a saddle-point. By continuity $H(\gamma(t_i)) \rightarrow H(p)$. But since $H(\gamma(t_i))$ is a constant sequence, we have that $H(p) = H_0$. – Evgeny Nov 14 '20 at 17:08
  • Thank you! Now I understand everything! Can you please write these as an answer, so I can accept it (and so the question does not remain open)? – S.T. Nov 14 '20 at 17:14

3 Answers3

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Let me explain the idea from my previous answer with few more additional details.

As it was mentioned in the question, the system has a smooth first integral $H(\mathbf{x})$ — a (locally non-constant) function that is constant along the system's trajectories. The key observation is that if such system has a trajectory $\gamma(t)$ homoclinic to a saddle $p$, then $H(p) \equiv H(\gamma(t))$. To prove that we can pick a sequence of moments $t_i \rightarrow +\infty$. By continuity, since $\gamma(t_i) \rightarrow p$ and $H(\mathbf{x})$ is smooth, then $H(\gamma(t)) \rightarrow H(p)$. But since $H(\mathbf{x})$ is constant at any point of $\gamma(t)$, we have that $H(\gamma(t)) \equiv H(p)$.

This gives us the following method to find all homoclinic trajectories for a 2D system of differential equations. First, find all saddle equilibria. Second, find the level sets of $H(\mathbf{x})$ that contain these saddles. What we have proven before means that a homoclinic trajectory must lie in the same level set as the saddle, to which it is homoclinic. Just take a look at level sets after that and you can find homoclinic (and even heteroclinic) trajectories this way.

Evgeny
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Actually, proving uniqueness of the homoclinic orbit is much easier than expected, but it requires some heavy machinery.

The Hartman-Grobman theorem says that, near a hyperbolic equilibrium point, any system of ODEs is topologically conjugate to its linearization.

In the linear case, for a saddle equilibrium point, there are $2$ trajectories going out of the equilibrium and two trajectories coming in. So, this is the behaviour for a general system as well.

Using you function $H$, you already determined your trajectories, so the above discussion tells you that there is only one homoclinic orbit.

C_M
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H(x,y) in the solution is not a first integral. It is easy to check that the trajectories along H are not constant.

ceg69y
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