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Definition of an $\omega$-limit point:

Let $\phi_t(p)$ be the orbit of the solution of the ODE $\dot{x}=f(x)$ which passes through the point $p$. We know that a point $x$ in $\mathbb{ R}^n$ is called an $\omega$-limit point of the orbit through the point $p$ if there is a sequence of numbers $t_1 \le t_2 \le t_3 \le · · ·$ such that $\lim_{i \to \infty } t_i = \infty $ and $\lim_{i\to \infty} \phi_{t_i}(p) = x$. From this definition, first and foremost, we understand that an $\omega$-limit point is "the limit point of a subsequence" constructed on the orbit. Obviously, this limit point may not be on the same orbit. I appreciate it if someone can answer the following questions:

Q1- Is that possible that an $\omega$-limit point is not on any orbit (i.e., it is not in the domain of definition of the solution (flow))? I guess that if the solution is forward complete (i.e., defined for all time $t\in(0,+\infty)$), then all the omega limit points are always in the domain of definition of the solution.

Q2- How should we define the $\omega$-limit point for the case when $t$ cannot go to $\infty$, i.e., the maximal interval of existence of the solution (i.e., domain of $t$) is $(-\infty, a)$ where $a<\infty$?

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  • $\begingroup$ Regarding your second question: if the maximal interval of existence is $(-\infty,a)$, where $a<\infty$, then the solution must leave any compact set. In order to have its $\omega$-limit set nonempty one would have to take some compactification of the domain (incidentally, that idea is very fruitful in the investigation of polynomial vector fields on the Riemann sphere, dating back to Poincaré). $\endgroup$ – user539887 Jul 7 '18 at 9:11
  • $\begingroup$ @user539887 Thank you for your comment. Could you please clarify why it should leave any compact set? I think when the maximal interval of existence is $(-\infty,a)$, as $t \to a$ we have either $|\phi(t) |$ approaches $\infty$ or $\phi(t)$ approaches the boundary of $D$ ($D$ is the domain of the vector field $f$). Thus, in the latter case, the solution may remain bounded as $t\to a$. $\endgroup$ – Arthur Jul 7 '18 at 15:35
  • $\begingroup$ I forgot to write: any compact set contained in the domain. $\endgroup$ – user539887 Jul 8 '18 at 8:03
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Q1. It is quite easy to construct examples where $f$ is undefined at an $\omega$-limit point of some orbit (which is not on the orbit itself). A trivial way to do so is just to remove that point from the domain of $f$. There are also less trivial examples where $f$ does not have a limit as you approach the point in question, so it is impossible to redefine $f$ at the $\omega$-limit point to make it continuous there.

Q2. By definition, an orbit that is not defined for $t \to +\infty$ can't have an $\omega$-limit point.

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  • $\begingroup$ Thank you, Prof. Israel, for your response. Regrading Q2, my question is that why don't we generalize the definition of the $\omega$-limit point for $t \to a$ where $(-\infty,a)$ is the maximal interval of existence of solution. I am saying this because there are many ODEs whose solution is not defined as $t$ goes to $+\infty$. Is there any generalized concept to cover this issue in the literature. Thank you! $\endgroup$ – Arthur Jul 6 '18 at 21:44
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Q1: Consider $ϕ_t(p)=\frac1t$ (for instance as the solution of $\dot x=-x^2$, $x(1)=1$), then $0$ is an ω-limit without being a function value of the solution. A graphically nicer variant are stable limit cycles in 2-dimensional systems. The limit cycle is the ω-limit set for all $p$ close to it, without those solutions ever reaching the cycle itself.

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  • $\begingroup$ Thank you, @Lutzl. Your example is interesting. $0$ is not on the orbit $\phi_t(p)=1/t$. However, it is on another orbit of this ODE which is $\phi_t=0$. If $x(0)=0$, the solution is $\phi_t=0$. $\endgroup$ – Arthur Jul 6 '18 at 22:01
  • $\begingroup$ My question is that: can we construct an example where the omega limit point is not on any orbit. I think as @Robert Israel said, we should construct an example where the function $f$ is not defined at the limit point. $\endgroup$ – Arthur Jul 6 '18 at 22:11

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