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In one of my recent answers, I claim the following:

Since $\dot r > 0$ when $0 < r < 1,$ and $\dot r < 0$ when $r>1$, we may conclude that any solution starting in $(x_0,y_0)$ with $x_0^2+y_0^2 > 0$ will be attracted to the unit circle, so that $$\lim_{t \to \infty} r(t) = 1.$$

To briefly put this into context, we are working in polar coordinates, $r:\mathbb{R}\to [0,\infty)$ is the radius function defined by $r(t)^2 = x(t)^2+y(t)^2$ and can be assumed to be smooth for the purposes of this question.

Of course, the result is intuitively clear, but I was unable to formulate a rigorous proof of this claim. The thing that confuses me here is that we know the behaviour of the derivative depending on the value of $r$, rather than on the parameter $t$, so I am not sure how to use the information about the derivative in connection to the limit.

Could someone give me a hint on where/how to start?


EDIT: Just to be fully clear, I would like to prove the following:

Suppose that $r:\mathbb{R}\to(0,\infty)$ is smooth and satisfies \begin{align} (1) \quad \dot r > 0 \quad &\text{for} \quad r < 1\\ (2) \quad \dot r < 0 \quad &\text{for} \quad r > 1. \end{align} Then $ \lim_{t\to \infty} r(t) = 1. $


EDIT 2: As pointed out in one of the answers, the claim is not true as stated.

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  • $\begingroup$ The differential equation you are considering is $\dot{r}=r(1-r^2)$? $\endgroup$ – Fakemistake Aug 24 '18 at 8:25
  • $\begingroup$ @Fakemistake Yes. Is that relevant though? $\endgroup$ – MisterRiemann Aug 24 '18 at 8:27
  • $\begingroup$ In general: If $\dot{r}=f(r)$ is given, then the roots of $f$ are (constant) solutions of the differential equation. Any other solution converges to one of the constant solutions. $\endgroup$ – Fakemistake Aug 24 '18 at 8:31
  • $\begingroup$ @Fakemistake Yes, I am aware of that. As I said, this is intuitively clear as well. But my question concerns proving this claim e.g. using the $\epsilon$-$\delta$ definition of the limit or via other properties of limits. $\endgroup$ – MisterRiemann Aug 24 '18 at 8:33
  • $\begingroup$ Well, one way is to consider the solution of the given differential equation (wolfram alpha provides one) and then take the limit, but it guess this is not what you want. $\endgroup$ – Fakemistake Aug 24 '18 at 8:38
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This is wrong as it stands. Consider $r(t)=\frac1{10}\arctan t$. Then $r(t)<1$ for all $t$ and $\dot r(t)>0$ for all $t$ but $r(t)\not \to 1$ as $t\to\infty$.

You will need $r$ to satisfy some other conditions, for example a suitable ODE.

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  • $\begingroup$ Nice example! What if we added the condition $\dot r = 0$ at $r = 1$? $\endgroup$ – MisterRiemann Aug 24 '18 at 11:19
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    $\begingroup$ That is true in my example: whenever $r(t)=1$ (i.e. never), $\dot r=0$. $\endgroup$ – Kusma Aug 24 '18 at 11:20
  • $\begingroup$ Oh, indeed! That came as a surprise to me, thanks! $\endgroup$ – MisterRiemann Aug 24 '18 at 11:22
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One way to approach this problem might be the following: If $r(t) = \alpha \in (0, 1)$, say, you may compute the derivative at this point. It will be an $\epsilon$ away from $0$. You want to prove that for a sufficiently large interval, it remains away from $0$, so that the solution increases (by the differential equation, $r$ will stay below $1$ all the time, otherwise the derivative becomes positive).

So write down the equation $$ r(1-r^2) \le \epsilon/2. $$ What condition must $r$ satisfy so that this holds?

EDIT: For the general case, see my comment below.

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  • $\begingroup$ Should the result not hold regardless of the differential equation behind all of this? That is, if we only know that $\dot r > 0$ for $0 < r < 1$ and $\dot r < 0$ for $r > 1$, can we not still say that $\lim_{t\to \infty} r(t) = 1$? $\endgroup$ – MisterRiemann Aug 24 '18 at 8:41
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    $\begingroup$ Yes, you may use that r(t) increases monotonically until it has reached 1, so the limit exists by the convergence of monotonely increasing bounded things. If it's not one, it's strictly smaller. But then the derivative will always be strictly larger than something, and you converge to infinity. $\endgroup$ – AlgebraicsAnonymous Aug 24 '18 at 8:50
  • $\begingroup$ That seems like a good idea. Maybe you could slightly formalize your contradiction argument and edit your answer so that I can mark it as an answer? $\endgroup$ – MisterRiemann Aug 24 '18 at 8:58

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