In the ODE where $y'=f(y(t))$ and $y(0)=yo$.

The omega limit set $w(yo)$ is positively invariant and also negatively invariant.

I want to prove first that its positively invariant and then prove its negatively invariant.

But how do I show that using a flow function ($phi(y,t)$) given that I know only the definition and the identity of flow function and very little understanding of the concept of flow function.

Thankyou for the help!

  • 3
    $\begingroup$ What is "the omega limit set"? What does "postively invariant" and "negatively invariant" mean to you here? What is the "flow function" you speak about? $\endgroup$ – Henning Makholm Dec 16 '12 at 14:17
  • $\begingroup$ i know omega limit set is the set of limit points given the IVP for y'.+ively invariant means as t goes to infinity the function remains inside the equilibrium solution and never gets out of it same goes for -ively invariant as t goes to -infinity. i just dont get the flow function my prof. explained it briefly though! $\endgroup$ – d13 Dec 16 '12 at 14:21
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    $\begingroup$ @dl3: The set of limit points of what? What does "IVP" mean? The equilibrium solution of what? It seems that you know more about this topic than anyone reading your question here, since your prof explained it (however briefly) to you, whereas you have not even attempted to explain to the rest of us what you're speaking about. $\endgroup$ – Henning Makholm Dec 16 '12 at 14:26
  • $\begingroup$ maybe i need to refine my question here. wait! $\endgroup$ – d13 Dec 16 '12 at 14:28
  • $\begingroup$ See this link for a proof: books.google.dz/… This is a google book. $\endgroup$ – China Dec 16 '12 at 15:28

Consider the system $$ \dot x=f(x),\quad x\in U\subset\mathbf{R}^k,\,f\colon U\to\mathbf R^k, $$ with the solution $$ x=x(t;x_0). $$ By definition point $\bar x$ belongs to omega limit set is there exists a sequence $\{t_n\}$ such that $x(t_n;x_0)\to \bar x$ when $n\to\infty,\,t_n\to\infty$. This means that for any fixed $t$ one has (using the properties of the flow) $$ x(t_n+t;x_0)=x(t;x(t_n;x_0))\to x(t;\bar x), $$ which means that if $\bar x$ belongs to omega limit set then the whole orbit containing this point belongs to this set, and since the orbits are invariant this implies that omega limit set is both positive and negative invariant.

  • $\begingroup$ thanx a lot for explaining. i really appreciate ur help! $\endgroup$ – d13 Dec 16 '12 at 19:34

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