Why is the sum of Lyapunov exponents negative in dissipative systems?

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SUMMARY

The sum of Lyapunov exponents in dissipative systems is definitively negative, indicating the overall behavior of the system. This conclusion is supported by the relationship between the Lyapunov spectrum and the maximum Lyapunov exponent, which must be positive for chaotic systems. The discussion emphasizes that while individual Lyapunov exponents represent behavior in their respective dimensions, their sum does not equate to the exponent of the net trajectory through phase space. Key references include Pesin's formula and the Kolmogorov-Sinai entropy, which further clarify these relationships.

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waofy
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Hi, the discussions in this forum have always been a great help to me as it seems there's always someone who's answered my question. However, this time I'm still puzzled.

People often talk about the sum of the Lyapunov exponents of a dynamical system (i.e. adding the exponents from each dimension in phase space) giving an indication of the overall behaviour of the system. For example, it is said that for a chaotic system the sum of the Lyapunov exponents is positive and for a dissipative system it is negative.

However, this doesn't make sense as the sum of the exponents is not the same as the exponent for the net trajectory through phase space.

e.g. in the case of a 3-dimensional phase space (x,y,z) with the system trajectory \begin{equation}
{\bf r}(t)=[x(t),y(t),z(t)]
\end{equation}
the separation between two trajectories is given by:
\begin{equation}
|\delta{\bf r}(t)|=e^{\lambda(t-t_{o})}|\delta{\bf r}(t_{o})|
\end{equation}
where lambda is the Lyapunov exponent of the trajectory. Now considering each dimension separately,
\begin{equation}
|\delta{\bf r}(t)|=|[\delta{x}(t),\delta{y}(t),\delta{z}(t)]|=|[e^{\lambda_{x}(t-t_{o})}\delta x(t_{o}),e^{\lambda_{y}(t-t_{o})}\delta y(t_{o}),e^{\lambda_{z}(t-t_{o})}\delta z(t_{o})]|=\sqrt{e^{2\lambda_{x}(t-t_{o})}\delta x(t_{o})^{2}+e^{2\lambda_{y}(t-t_{o})}\delta y(t_{o})^{2}+e^{2\lambda_{z}(t-t_{o})}\delta z(t_{o})^{2}}
\end{equation}(1)

where the lambdas are the exponent for each dimension. However, if the exponent of the trajectory was equal to the sum of exponents from each dimension,
\begin{equation}
|\delta{\bf r}(t)|=e^{(\lambda_{x}+\lambda_{y}+\lambda_{z})(t-t_{o})}|\delta{\bf r}(t_{o})|=e^{(\lambda_{x}+\lambda_{y}+\lambda_{z})(t-t_{o})}|[\delta x(t_{o}),\delta y(t_{o}),\delta z(t_{o})]|=e^{(\lambda_{x}+\lambda_{y}+\lambda_{z})(t-t_{o})}\sqrt{\delta x(t_{o})^{2}+\delta y(t_{o})^{2}+\delta z(t_{o})^{2}}
\end{equation}(2)

Clearly (1)=/=(2).

Am I missing something here? The only thing I can think of is that I haven't understood the meaning of the Lyapunov spectrum correctly. I've always assumed each exponent represents the the behaviour of trajectories moving in a single dimension but I could be wrong?
 
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atyy said:
It's the maximum Lyapunov exponent that has to be positive, not the sum.

In addition, the system must have positive topological entropy. http://chaosbook.org/ (section 1.3.1).

The sum of the positive Lyapunov exponents is related to Kolomogorov-Sinai entropy by Pesin's formula http://www.scholarpedia.org/article/Pesin_entropy_formula.


Thanks, that makes more sense.

I think it is still true that in dissipative systems the sum of the Lyapunov exponents is negative though (http://books.google.co.uk/books?id=Uqc7zC7NULMC&pg=PA212&lpg=PA212#v=onepage&q&f=false" - first line). It makes sense that the overall exponent for the trajectory would be negative (which would also be true for a stable limit cycle) but I can't quite see why the sum should be. I'm just trying to work out what significance the sum has.
 
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