The positive Lyapunov exponent confirms chaos is a bounded, deterministic system. I suppose the boundedness of the system has to comply with the presence of one zero Lyapunov exponent. Then again why zero? Why not negative? It will be very helpful if I get a deeper insight here.
It’s not, the statement is wrong. In Hamiltonian systems the sum of the Lyapunov exponents vanishes, since the volume of phzse space is conserved.
A vanishing exponent just means the existence of a conservation law.
To see why this is the case, let x and y be the two trajectory segments, whose separation (x−y) you consider for defining or calculating the Lyapunov exponents. At every point of the attractor (or invariant manifold), we can represent this separation in a basis of Lyapunov vectors, each of which corresponds to one Lyapunov exponent. In this representation, each component of the separation grows or shrinks independently according to the respective Lyapunov exponent (on average). For example, in chaos with one positive Lyapunov exponent, the separation will quickly point in the corresponding direction because this Lyapunov exponent dominates the other ones.
Now, suppose that the trajectory segment y
is such that y(t)=x(t+ε) for some time t, i.e., it is a temporally slightly advanced version of x. The separation of these segments may grow and shrink with time, depending on the speed of the phase-space flow, but on average it should stay constant due to the following: Since the dynamics is bounded, the trajectory x will need to get close to x(t) again, i.e., there needs to be some τ such that x(t+τ)≈x(t). Due to the phase-space flow being continuous, we also have y(t+τ)=x(t+τ+ε)≈x(t+ε)=y(t) and thus: |x(t+τ)−y(t+τ)|≈|x(t)−y(t)|
Therefore, separations in the direction of time neither shrink nor grow (on average) and in this direction we get a zero Lyapunov exponent: If we consider only such separations to compute a Lyapunov exponent, we obtain:
λ=limτ→∞lim|x(t)−y(t)|→01τln(|x(t+τ)−y(t+τ)||x(t)−y(t)|)=limτ→∞lim|x(t)−y(t)|→01τln(|x(t)−y(t)||x(t)−y(t)|)=0
(We now have =
instead of ≈ due to the limits averaging everything and allowing us to consider arbitrarily close x(t) and x(t+τ).)
Therefore, all such dynamical systems must have at least one zero Lyapunov exponent.
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