Download presentation
Presentation is loading. Please wait.
Published byLilian Barnett Modified over 9 years ago
1
Analysis of the Rossler system Chiara Mocenni
2
Considering only the first two equations and assuming small z, we have: The Rossler equations
3
With eigenvalues: It is equivalent to the oscillator: For a=0 the oscilator is harmonic (centers) For a>0 the oscillator is undamped (unstable spirals) Consider 0<a<2 The reduced system
4
This fact is non sufficient by itself to produce the chaotic behavior because in the linear case this instability is not compensated by any other folding mechanism. The system is globally unstable In the 3D phase space this fact induces a stretching mechanism A nonlinear stabilizing term is necessary for maintaining the trajectories confined in a region of the 3D phase space Geometry of the phase space
5
The folding is produced by the term associated to parameter c in the third equation Consider only the equation for z and assume b>0 The equation for z has the steady state The third equation (1/2)
6
For x<c the coefficient of z in the equation is negative and the (positive) steady state z * is stable For x>c the coefficient of z in the equation is positive and the system diverges. For positive b the instability is developing along increasing z. The third equation (2/2)
7
Assume fixed b and c and vary parameter a b=2 c=4 Assume fixed b and c and vary parameter a b=2 c=4 The system has two steady states, that exist for a ≠ 0 a < c 2 /4b The system has two steady states, that exist for a ≠ 0 a < c 2 /4b The complete system
8
Assume 0 < a < c 2 /4b Let S 1 be the steady state (x 1,y 1,z 1 ) Let P 2 be the steady state (x 2,y 2,z 2 ) Let A be the Jacobian matrix Assume 0 < a < c 2 /4b Let S 1 be the steady state (x 1,y 1,z 1 ) Let P 2 be the steady state (x 2,y 2,z 2 ) Let A be the Jacobian matrix Stability of the steady states
9
For a = c 2 /4b S 1 and P 2 coincide and for a > c 2 /4b disappear The steady states
10
Steady state S 1 is a saddle because it has 3 real eigenvalues with opposite signs Linear stability of steady states (1/2) Saddle-node bifurcation
11
a < 0.125: S 2 is a stable spiral a > 0.125: S 2 is an unstable spiral a = 0.125: Hopf bifurcation a < 0.125: S 2 is a stable spiral a > 0.125: S 2 is an unstable spiral a = 0.125: Hopf bifurcation Linear stability of steady states (2/2) a = 2: S 1 and S 2 collide and disappear with a Saddle- Node bifurcation
12
The limit cycle S 2 is an unstable spiral and a stable limit cycle (with period 6.2) is formed through the Hopf bifurcation
13
The limit cycle is then destabilized through a PERIOD DOUBLING bifurcation of cycles Bifurcations of cycles
14
Explaining the mechanisms This fact is due to the mechanism shown by equation 3, that is activated for x = c (in our example c = 4) The limit cycle grows for increasing a, and when x = 4 the variable z is destabilized, producing a divergence of the trajectories along z At this point a stabilizing mechanism on x is induced by equation 1 and leading again z to enter the stable region The result of this combination of mechanisms induces that a double period limit cycle is formed This fact is due to the mechanism shown by equation 3, that is activated for x = c (in our example c = 4) The limit cycle grows for increasing a, and when x = 4 the variable z is destabilized, producing a divergence of the trajectories along z At this point a stabilizing mechanism on x is induced by equation 1 and leading again z to enter the stable region The result of this combination of mechanisms induces that a double period limit cycle is formed
15
Stretching and Folding (1/2)
16
Stretching and Folding (2/2)
17
The complete picture a Stable spiral Stable limit cycle of period T Stable limit cycle of period 2T Stable limit cycle of period 4T Stable limit cycle of period 8T CHAOS Unstable system
18
Period doubling and chaotic attractor
19
The Rossler attractor: a movie
Similar presentations
© 2024 SlidePlayer.com. Inc.
All rights reserved.