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This question might well have an obvious affirmative answer (or an obvious counterexample!), which at present I cannot see. Suppose I have a first-order ODE $$ u'(t)=f(u) $$ whose general solution depends on one arbitrary constant $A$. Suppose that I take the large-$t$ limit of this general solution, and I find that $$ \lim_{t\to\infty}u_{gen}(t)=u^\star\ , $$ where $u^\star$ is a number (which does not depend on $A$ anymore). Furthermore, $u^\star$ is a fixed-point of my ODE, namely $f(u^\star)=0$. Can I immediately conclude that $u^\star$ is a stable fixed point? Or are there (sufficiently pathological, I guess) cases where the large-$t$ limit is an unstable fixed-point irrespective of the initial conditions? Thanks for your help, folks.

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    $\begingroup$ I think that If all initial conditions end up in $u^\star$, then, by definition, $u^\star$ is an asymptotically stable fixed point. $\endgroup$ Feb 28, 2016 at 17:44
  • $\begingroup$ you can check $f'(u^\star)$. If $f'(u^\star) < 0$, then $u^\star$ is asymptotically stable. If $f'(u^\star) > 0$, then $u^\star$ is unstable. $\endgroup$ Feb 28, 2016 at 17:51
  • $\begingroup$ What if $f(u^\star)=f'(u^\star)=0$? $\endgroup$ Feb 28, 2016 at 18:57
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    $\begingroup$ @Giuseppe Negro Actually, there examples of differential equations for which all solutions tend to an unstable equilibrium point. $\endgroup$
    – John B
    Feb 28, 2016 at 21:43
  • $\begingroup$ @Jonas could you elaborate more on this, please? $\endgroup$ Feb 28, 2016 at 21:45

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Consider the equation $$ \begin{cases} x'=x^2-y^2,\\ y'=2xy. \end{cases} $$ In order to draw the phase portrait just rewrite it in the form $z'=z^2$ on the complex plane.

This is not yet what you want (depends on what you want...) since the negative part of the horizontal axis tends to infinity. But you can see the differential equation on the $2$-sphere, in which case it has exactly one equilibrium point and all orbits tend to the equilibrium point (both when the time goes to $+\infty$ and to $-\infty$).

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  • $\begingroup$ Thanks for your answer. However, can you show explicitly that the general solution $(x(t),y(t))$ of your ODE system for $t\to\infty$ converges to the unstable fixed point [$(0,0)$, right?] irrespective of the initial conditions? I do not see it... $\endgroup$ Feb 28, 2016 at 22:16
  • $\begingroup$ Check that the solutions are $x(t)=-(a+t)/[(a + t)^2 + b^2]$, $y(t)=b/[(a + t)^2 + b^2]$, with $a,b\in\mathbb R$. $\endgroup$
    – John B
    Feb 28, 2016 at 22:20
  • $\begingroup$ Ok, many thanks for your help (+1). Any thoughts about the purely 1D case? $\endgroup$ Feb 28, 2016 at 22:23
  • $\begingroup$ Ah, if that's what you want, in the $1$-dimensional case there is no counterexample on the line but again there is a counterexample on the circle. $\endgroup$
    – John B
    Feb 28, 2016 at 22:25
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A somewhat more elementary example, where $r,\theta$ are polar coordinates: $$ \dot r=r(1-r)\\ \dot \theta=\sin^2\theta/2. $$ Point $(1,0)$ attracts all the orbits except the one starting at the origin. But is unstable.

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  • $\begingroup$ All except one, that's why I didn't give this example. $\endgroup$
    – John B
    Feb 28, 2016 at 22:08
  • $\begingroup$ @Jonas, yes, of course. $\endgroup$
    – Artem
    Feb 28, 2016 at 22:09
  • $\begingroup$ @Artem Many thanks for your example (+1). Any thoughts about the purely 1D case? $\endgroup$ Feb 28, 2016 at 22:24
  • $\begingroup$ @PierpaoloVivo In the autonomous case all the solutions are monotone, so there will be no such examples. It is quite easy to come up with an example of $x'=f(t,x)$ such that zero will be attractive at $t\to\infty$ but not asymptotically stable. For asymptotical stability solution should stay near this point for all $t>t_0$ $\endgroup$
    – Artem
    Feb 28, 2016 at 22:29

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