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I am not sure how to solve a dynamical system with some constraint equation. For simplicity, let us consider the following system

$x'=-xy\\ y'=\frac{x}{2}\\ x+y^2=1$

The system is 1 dimensional. If I decide to get the critical points from the two first equations and then check consistency with the constraint equation, I will miss some points. In fact, from the 2 first equations, the only critical point is $x=0$, using the constraint I will get $y=\pm 1$ so we have 2 points $(0,\pm 1)$. But if I consider that the system is really 1 dimensional. I can reduce it. I decide to remove the variable $y$, we have then $x'=\mp x\sqrt{1-x}$, which gives an additional point for $x=1$ and therefore $y=0$. On the contrary if I remove $x$, I will get $y'=(1-y^2)/2$ which doesn't give the additional critical point.

So my question is, because the system is naturally 1 dimensional, should I consider all possible sub-systems (by using constraint equations) and conclude that the critical points are the union of all critical points of the sub-systems ?

ziusudra
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1 Answers1

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From the second equation, we get a fixed point for $x = 0$, period, that is, it is the "only" valid value for $x$.

However, from the first equation, we see that when $x = 0$, $y$ can be any value, so we have the critical points as $(x, y) = (0, y)$.

Now, when we look at the constraint, we want to verify if it is telling us anything differently about $x$ and $y$ and the critical points.

We know that we can only have a fixed $x = 0$, which makes the constraint equation:

$$x + y^2 = 1 \rightarrow 0 + y^2 = 1 \rightarrow y = \pm~1$$

So, instead of being wide open with the critical points $(x,y) = (0,y)$, the constraint limited the y-values to $(x,y) = (0, \pm 1)$.

What I would do with the constraint equations is:

  • $(1)$ Find the critical as you normally do.
  • $(2)$ Determine how the constraint limits those critical points from Step 1.
Amzoti
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  • Thanks but it doesn't work, I think that one point is missing. Simple proof, let us change the variable $z=y^2$, then we have

    $$x'=-x \sqrt{z}\ z'=x\sqrt{z}\ x+z=1$$

    We see clearly that we have an other point when $z=0$

     – ziusudra Dec 05 '13 at 14:27
  • Actually, no. I take that back. Recall, we 'must' satisfy the original system. If $x = 1$, that 'is not' a critical point of the second system. The constraint can only serve serve to 'constrain' the actual critical points, not add additional ones the do not satisfy the original system. – Amzoti Dec 05 '13 at 14:36
  • yes but what about the equivalent system with (x,z) ? I have the feeling that I miss a point because I consider the system 2D when it is 1D. – ziusudra Dec 05 '13 at 14:38
  • No, it is not a critical point. We must satisfy the critical points of the original system and cannot introduce new ones from the constraint, the constraint is limiting critical points and cannot introduce new ones. – Amzoti Dec 05 '13 at 14:39
  • yes but what about the equivalent system with (x,z) ? And I know from physical point of view that we should have this point – ziusudra Dec 05 '13 at 15:06
  • Draw the 2-D phase portrait, it will define your critical points. If you overlay the 2-D constraint, where do the systems critical points and the constraint equation intersect? Those are the points of interest. In other words, the constraint is limiting the entire $\mathbb{R}$ to the area inside the constraint, and thus also limiting those critical points,not introducing new ones. If I had a 3-D system, the same holds true. – Amzoti Dec 05 '13 at 15:24
  • But it doesn't say why I do not have the same phase space when I go to new variables (x,z) – ziusudra Dec 05 '13 at 15:43
  • I am not sure that there are any more ways to explain. You have a system, it has a set of fixed points. You wrap a constraint region around that, it can either keep the critical points the same or reduce them, not increase them. – Amzoti Dec 05 '13 at 15:46
  • @amWhy: I needed some ale after that one and then UV or anything. The question asked was answered and then in the ned a discussion about translating starts. I have have had many pirate moments and I think this get far field from the goals of the site. Don't mind helping, but it gets out-of-control quickly. Hope you are having a great day! – Amzoti Dec 06 '13 at 02:18
  • @Amzoti It was way simplier to tell that this change of variables is not valid because it's not a diffeomorphism at least. – Evgeny Dec 06 '13 at 08:33