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I'm using J.Meiss -Differential dynamical systems, and have some trouble to understand a proof about Lyapunov exponents. We have a dynamical system $$ \dot{x} = f(x), $$ with the corresponding flow $ \varphi_t $. Now suppose $ \varphi_t(x) $ is a bounded orbit, and that the flow $ \varphi$ is not forward asymptotic to an equilibrium. Want to show that then it has a zero Lyapunov exponent, i.e $ \mu(x,v) = \limsup_{t \rightarrow \infty} \frac{1}{t} \ln |D_x\varphi_t(x)v| = 0 $.

So this is what they do: Consider the vector $v(t) = f( \varphi_t(x)) $. Now $$ \frac{d}{dt}v(t) = \frac{d}{dt}f( \varphi_t(x)) = Df( \varphi_t(x))\frac{d}{dt} \varphi_t(x) = Df( \varphi_t(x)) f( \varphi_t(x)) = Df( \varphi_t(x))v(t), $$ hence $$ \dot{v} = Df( \varphi_t(x)) v.$$ Thus $v$ is a solution to the linearization of the system about this orbit with initial condition $f(x)$.

Now they do some argument that I do not uderstand:

First they say that since $ \varphi $ is bounded, $v$ is also bounded. Why is that? Further they say that since $ \varphi_t$ is not asymptotic to an equilibrium $ \limsup |v(t)| > 0$. Why is that? And Therefore $ \mu(x,v) = 0$.

Thank you for your help.

Martin Sleziak
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Henrik Finsberg
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  • Are there any assumptions on $f$? For example, it's Lipschitz or continuous? – jkn May 04 '13 at 11:02
  • It's not stated explicit in the theorem, but since they are differentiating $f$ in the proof it's probably an assumption that $f$ differentiable. – Henrik Finsberg May 04 '13 at 11:51
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    Yeah of course. So, suppose that $\mathcal{I}\subseteq\mathbb{R}$ and that $g:\mathcal{I}\rightarrow [0,\infty)$ is continuous. Let $c\in[a,b]$, then $$\limsup_{t\rightarrow c} g(t)=0\Leftrightarrow \lim_{t\rightarrow c}g(t)=0.$$ So, by contradiction, $\limsup_{t\rightarrow\infty}|v(t)|>0$ (otherwise $f(\varphi_t(x))\rightarrow 0$ as $t\rightarrow\infty$, that is the trajectory converges to an equilibrium). – jkn May 04 '13 at 12:48
  • I think boundedness of $v$ follows from continuity of $f$ and that $$\varphi_t(x)=x+\int_0^tv(s)ds.$$ Thus, if $v$ is unbounded so would be $\varphi$. However, I can't seem to think of a proof for the previous sentence. – jkn May 04 '13 at 12:49
  • Thank you! That was very clarifying. – Henrik Finsberg May 04 '13 at 13:18
  • Glad I could help. – jkn May 04 '13 at 13:35

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