The intuitive definition of an invariant set for an ode such as $\dot r = r(1-r)$.

Question

The definition of an invariant set $M$ is that $\forall x \in M, \forall t, \phi(x,t) \in M$.

For the ode $\dot r = r(1-r)$, is $A = \{(r, \theta) | 0 \leq r\leq 1\}$ an invariant set?

The cases when $r = 0$ and $r = 1$ are obvious. What if $0 < r < 1$? According to the definition of the invariant set, if $A$ is as above, then points for which $0 < r < 1$ will move forward to the circle or backward to the center and belongs to $A$. So the entire $A$ seems to be an invariant set. But intuitively, it seems to me that only the circle and the center count.

The set $A$ has $r \le 0$. Do you mean $r \ge 0$? – Matthew Cassell Oct 12 '21 at 14:45 — Matthew Cassell, Oct 12 '21 at 14:45
@mattos Thanks, it should be [0, 1] – Charlie Chang Oct 12 '21 at 14:52 — Charlie Chang, Oct 12 '21 at 14:52

score 1 · Answer 1 · answered Oct 12 '21 at 14:50

1

$\{(r,\theta) \mid r < 0\}$, $\{(r,\theta) \mid r = 0\}$, $\{(r,\theta) \mid 0 < r < 1\}$, $\{(r,\theta) \mid r = 1\}$, $\{(r,\theta) \mid r > 1\}$ are all invariant sets. So is the union of any subset of these.

answered Oct 12 '21 at 14:50

Robert Israel

448,999

I see the points in ${(r,\theta) \mid 0 < r < 1}$ perhaps do not to move to the circle and the center; it gets but infinitely close. Not sure how to prove that. – Charlie Chang Oct 12 '21 at 14:55
The uniqueness part of the Existence and Uniqueness Theorem. – Robert Israel Oct 13 '21 at 15:30

The intuitive definition of an invariant set for an ode such as $\dot r = r(1-r)$.

1 Answers1