Presentation is loading. Please wait.

Presentation is loading. Please wait.

Amplitude expansion eigenvectors: (Jacobi).U=  U,  (near a bifurcation)  (Jacobi).V=– V, =O(1) Deviation from stationary point.

Similar presentations


Presentation on theme: "Amplitude expansion eigenvectors: (Jacobi).U=  U,  (near a bifurcation)  (Jacobi).V=– V, =O(1) Deviation from stationary point."— Presentation transcript:

1 Amplitude expansion eigenvectors: (Jacobi).U=  U,  (near a bifurcation)  (Jacobi).V=– V, =O(1) Deviation from stationary point = u(t) U + v (t)V d u /dt =  u +   u 2 +   uv +   v 2 +   u 3 + … slow v =  u 2 +  uv + … fast = quasistationary u =O(  ), v =  u 2 / = O(   ) d u /dt =  u +   u 2 + (    / +    u 3 + … Saddle-node: d u /dt =  u +   u 2 u  u –  /(2    d u /dt =   /(2    +   u 2 Cusp: d u /dt =   u +    u 2 + (    / +    u 3

2 Double zero eigenvalue Det(Jacobi) = Trace(Jacobi) = 0(find the condition parameterized by s.s) One eigenvector only! U = (1 0) Complete the coordinate frame by another vector: (Jacobi).V=U Deviation from stationary point = u(t) U + v (t)V Transformed Jacobi matrix – Jordan normal form d u /dt = v d v /dt = 0 Transformed linear system: u =O(  ), d/dt = O(   ), v = O(   )

3 Double zero eigenvalue – nonlinear expansion Deviation from stationary point = u(t) U + v (t)V Transformed system: Unfolding of double zero d u /dt = p d p /dt = f(u) + p g(u) f(u), g(u) – polynomials d u /dt = v +   u 2 +   uv +   u 3 + … d v /dt =  2 +  u +  u 2 +  uv + … deviation from double-0 Denote p = v +   u 2 +   uv +   u 3 + conservative subsystem dissipative correction

4 Weakly dissipative system – local analysis Dynamical system  Stationary solution Jacobi matrix Stability conditions d y /dt = p d p /dt = f(y) +  p g ( y ) f(y s )=0, p=0

5 Weakly dissipative system – global analysis Trajectories at  = 0 Conserved energy: energy change : At small  compute the rate of energy change by averaging over the period T : integration limits are values of y at turning points where p vanishes

6 d y /dt = p d p /dt =   + y 2 +  p (   + y ) sub-Hopf SL SN Unfolding of bifurcation at double zero eigenvalue no static solutions saddle + s-node saddle + u-node Det = – y s Tr=  (   + y s )  

7 d y /dt = p d p /dt =  1 y – y 3 + p (  2 – y 2 ) subcritical Hopf saddle –loop supercritical Hopf subcritical Hopf pitchfork Unfolding of bifurcation at double zero eigenvalue (symmetric to the change of signs)


Download ppt "Amplitude expansion eigenvectors: (Jacobi).U=  U,  (near a bifurcation)  (Jacobi).V=– V, =O(1) Deviation from stationary point."

Similar presentations


Ads by Google