Laurette TUCKERMAN Codimension-two points

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1 Laurette TUCKERMAN Codimension-two points

2 The 1:2 mode interaction System with O(2) symmetry with competing wavenumbersm = 1 andm = 2 Solutions approximated as: u(θ,t) = z 1 (t)e iθ +z 2 (t)e 2iθ + z 1 (t)e iθ + z 2 (t)e 2iθ with z 1 (t) = x 1 (t)+iy 1 (t) = r 1 (t)e iφ 1(t),z 2 (t) = x 2 (t)+iy 2 (t) = r 2 (t)e iφ 2(t) O(2) generated by rotation byθ 0 and reflection about θ = 0: S θ0 (z 1,z 2 ) = (e iθ 0 z 1,e 2iθ 0 z 2 ) κ(z 1,z 2 ) = ( z 1, z 2 )

3 Define ( z1 z 2 ) F ( z1 z 2 z 2 1 Show that F is equivariant with respect to O(2): ( z1 ) z 2 ) F ( z1 z 2 z 2 1 ) κ ( z1 z 2 ) z 1 2 ) ( z1 z 2 ) κ ( z1 z 2 ) Sθ0 ) F ( z1 z 2 z 1 2 ) ( z1 z 2 ( z1 z 2 ) Sθ0 ) F ( e iθ 0 z 1 e 2iθ 0 z 2 ( z1 z 2 z 2 1 ) F ( e iθ 0 z 1 z 2 e 2iθ 0 z2 1 ( e iθ 0 z 1 e 2iθ 0 z 2 e 2iθ 0 z2 1 ) Essentially 1+1 = 2 and 2 1 = 1 Dynamical system for evolution ofz 1,z 2 is: ( z 1 = z 1 z 2 +z 1 µ1 α 1 z 1 2 β 1 z 2 2) ( z 2 = z1 2 +z 2 µ2 β 2 z 1 2 α 2 z 2 2)

4 Steady states (Phase is arbitrary: z x) ( 0 = x 1 x2 + ( )) µ 1 α 1 x 2 1 β 1 x 2 2 ( ) 0 = x 2 1 +x 2 µ2 β 2 x 2 1 α 2 x 2 2 Trivial state: x 1 = x 2 = 0 Mode-two ( pure mode ) state: x 1 = 0,x 2 0: x 2 2 = µ 2 /α 2 If x 1 0 then x 2 0! Instead, have mixed-mode state : 0 = x 2 + ( ) µ 1 α 1 x 2 1 β 1 x 2 2 ( ) 0 = x 2 1 +x 2 µ2 β 2 x 2 1 α 2 x 2 2 (intersection of two conic sections)

5 Stability Jacobian in Cartesian coordinates (even ify = 0, Jacobian must includey) x 2 + µ 1 α 1 (r x2 1 ) β 1r2 2 y 2 α 1 2x 1 y 1 x 1 β 1 2x 1 x 2 y 1 β 1 2x 1 y 2 y 2 + µ 1 α 1 2x 1 y 1 x 2 + µ 1 α 1 (r y2 1 ) β 1r2 2 y 1 β 1 2y 1 x 2 x 1 β 1 2x 2 y 1 ±2x 1 β 2 2x 2 x 1 2y 1 + β 2 2x 2 y 1 µ 2 β 2 r1 2 α 2(r x2 2 ) α 22x 2 y y ±2y 1 + µ 2 β 2 2x 1 y 2 ±2y 1 y 2 α 2 2x 2 y 2 µ 2 β 2 r1 2 α 2(r y2 2 ) Trivial state: J = µ µ µ µ 2 Two 2D eigenspaces. Circle pitchforks at µ 1 = 0 and µ 2 = 0

6 Mode-two state: x 1 = y 1 = y 2 = 0,x 2 = ± µ 2 /α 2 J = = x 2 +µ 1 β 1 x µ 1 x 2 +µ 1 β 1 x µ 2 3α 2 x µ 2 α 2 x 2 2 ± µ2 α 2 +µ 1 β 1µ 2 α µ2 α 2 +µ 1 β 1µ 2 α µ Eigenvalues 2µ 2 and 0 are usual results of circle pitchfork. Other two eigenvalues concern instability to(x 1,y 1 ). They are different because mode-two state has a phase (no CP from mode-two). Mixed-mode branch bifurcates from trivial state atµ 1 = 0 and from mode-two branch at µ 1 β 1 µ 2 α 2 ± µ2 α 2 = 0

7 Evolution equations: Polar representation z 1 (t) = r 1 (t)e iφ 1(t),z 2 (t) = r 2 (t)e iφ 2(t) (ṙ 1 +ir 1 φ 1 )e iφ 1 = r 1 e iφ 1 r 2 e iφ 2 +r 1 e iφ 1 (ṙ 2 +ir 2 φ 2 )e iφ 2 = r 1 e iφ 1 r 1 e iφ 1 +r 2 e iφ 2 ( µ1 α 1 r 2 1 β 1 r 2 2) ( µ2 β 2 r 2 1 α 2 r 2 2) Dividing equations by e iφ 1 and by eiφ 2 : ( ) ṙ 1 +ir 1 φ 1 = r 1 r 2 e i(φ 2 2φ 1 ) +r 1 µ1 α 1 r1 2 β 1 r2 2 ( ) ṙ 2 +ir 2 φ 2 = r1e 2 i(2φ 1 φ 2 ) +r 2 µ2 β 2 r1 2 α 2 r2 2 Separating real and imaginary parts and dividing imaginary parts byr j 0: ( ) ṙ 1 = r 1 r 2 cos(φ 2 2φ 1 )+r 1 µ1 α 1 r1 2 β 1 r2 2 φ 1 = r 2 sin(φ 2 2φ 1 ) ( ) ṙ 2 = r1cos(2φ 2 1 φ 2 )+r 2 µ2 β 2 r1 2 α 2 r2 2 φ 2 = (r1/r 2 2 )sin(2φ 1 φ 2 )

8 Substitute Φ 2φ 1 φ 2 : ( ) ṙ 1 = r 1 r2 cosφ+µ 1 α 1 r1 2 β 1 r2 2 ( ) ṙ 2 = r1cosφ+r 2 2 µ2 β 2 r1 2 α 2 r2 2 φ 1 = r } 2sinΦ (r1/r 2 = 2 )sinφ Φ = (2r 2 r1/r 2 2 )sinφ φ 2 Suppose ṙ 1 = ṙ 2 = Φ = 0, but r 1,r = r 2 cosφ+µ 1 α 1 r1 2 β 1 r2 2 ( ) 0 = r1cosφ+r 2 2 µ2 β 2 r1 2 α 2 r2 2 0 = (2r2 2 r1)sinφ 2 Mixed modes Φ = 0,π = sinφ = 0 = φ 1 = φ 2 = 0 = steady states: 0 = ±r 2 +µ 1 α 1 r1 2 β 1 r2 2 ( ) 0 = r1 2 +r 2 µ2 β 2 r1 2 α 2 r2 2

9 Traveling Waves 0 = Φ = (2r 2 r 2 1/r 2 )sinφ sinφ 0 = 0 = 2r 2 2 r 2 1 = r 2 1 = 2r = Φ 2 φ 1 φ 2 Definition: u(θ,t) = u(θ ct,0) u(θ,t) = r 1 (t)e i(φ 1(t)+θ) +r 2 (t)e i(φ 2(t)+2θ) + complex conjugate u(θ ct,0) = r 1 (0)e i(φ 1(0)+θ ct) +r 2 (0)e i(φ 2(0)+2(θ ct)) + complex conjugate = { r1 (t) = r 1 (0) and φ 1 (t) = φ 1 (0) ct r 2 (t) = r 2 (0) and φ 2 (t) = φ 2 (0) 2ct = 2φ 1 (t) φ 2 (t) = 2φ 1 (0) φ 2 (0) = Φ(t) = Φ(0)

10 0 = 2r2 2 r1 2 = r1 2 = 2r2 2 0 = ṙ 1 = r 2 cosφ+µ 1 α 1 2r2 2 β 1 r2 2 ( ) 0 = ṙ 2 = 2r2cosΦ+r 2 2 µ2 β 2 2r2 2 α 2 r2 2 Add 2 blue equation to (1/r 2 ) green equation 0 = 2µ 1 +µ 2 (4α 1 +2β 1 +2β 2 +α 2 )r2 2 r2 2 2µ 1 +µ 2 = 4α 1 +2β 1 +2β 2 +α 2 Can also obtain: µ 1 (2α 2 +β 2 ) µ 2 (2α 1 +β 1 ) cosφ = [(2µ 1 +µ 2 )(4α 1 +2β 1 +2α 2 +β 2 )] 1/2 Traveling waves bifurcate from mixed mode branch when cosφ = 1 (µ 1 (2α 2 +β 2 ) µ 2 (2α 1 +β 1 )) 2 = (2µ 1 +µ 2 )(4α 1 +2β 1 +2α 2 +β 2 )

11 Time-dependent states Traveling waves via Hopf bif from mixed-mode branch Modulated waves via secondary Hopf bif from traveling waves Heteroclinic orbit connects two opposite-phase mode-two saddles with eigenvalues λ < 0 < λ + Can prove orbit is stable if λ > λ +, i.e. if contraction more important than expansion ( ) µ 2 µ2 µ 2 µ2 µ 1 β 1 > µ 1 β 1 + α 2 α 2 α 2 α 2 µ 2 β 1 > µ 1 α 2

12

13 Takens-Bogdanov normal form Meeting of Hopf and steady bifurcations Steady states: ẋ = y ẏ = µ 1 x+µ 2 y x 3 x 2 y 0 = y Jacobian: J = J(0,0) = 0 = µ 1 x x 3 = x = ± µ 1 ( ) 0 1 µ 1 3x 2 2xy µ 2 x 2 ( ) 0 1 = λ = µ (µ2 ) 2 2 µ 1 µ 2 2 ± µ1 2

14 J is Jordan block at codimension-two point µ 1 = µ 2 = 0 Hopf bifurcation at µ 2 = 0 forµ 1 > 0 Pitchfork bifurcation at µ 1 = 0 Real eigenvalues coalesce to form complex conjugate pair At collision, imaginary part is zero At a nearby Hopf bifurcation, frequency is near zero = period is near infinity

15

16 Heteroclinic cycles in the French washing machine Caroline Nore Laurette Tuckerman Olivier Daube Shihe Xin LIMSI CNRS, France

17 The French Washing Machine (von Karman flow) Symmetry Group: Rotations in and Combined reflection in z and Rot/Ref don't commute Douady, Brachet, Couder, Fauve et al Le Gal et al, Rabaud et al, Daviaud et al. Gelfgat et al, Lopez & Marques et al

18 Numerical Methods Time integration code for Navier Stokes eqns by Daube Spatial: finite differences in (r,z), Fourier in Temporal: 2 nd order backward difference formula Adaptations: Steady state solving via Newton for axisymmetric flows Linear stability about axisymmetric and 3D flows

19 Linear Stability of Basic Axisymmetric Flow z=1/3 z=0 z= 1/3 m=1 mixed mode Re=355 m=2 pure mode Re=410

20 Bifurcation Diagram for 1:2 mode interaction Normal Form quadratic terms Armbruster, Guckenheimer & Holmes; Proctor & Jones (1988)

21 Mixed Mode (from m=1 eigenvector)

22 Pure Mode (from m=2 eigenvector)

23 Travelling Waves (Re=415) TW = Mixed Mode + Eigenvector Reflection Symmetric Antisymmetric

24 Two types of heteroclinic cycles 4 plateaus 2 plateaus

25 Heteroclinic Cycle (Re=430) a b c d e f g h

26 Linear stability analysis about nonaxisymmetric flows Eigenvalues about pure mode Eigenvalues about mixed mode

27 Conclusion counter rotating von Karman flow with diameter=height is almost perfect realisation of 1:2 mode interaction steady states (mixed and pure modes) travelling waves robust heteroclinic cycles of two kinds possible Kelvin Helmholtz instability mechanism

28 Convection + counter rotation (Rayleigh Bénard + von Kármán) Tuckerman with Bordja, Cruz Navarro, Martin Witkowski, Barkley Pr=1 =1 axisymmetric 1 Symmetry: R z z mid z T T mid T 1 mid u z u z u u (Nore et al. 2003)

29 pitchfork Hopf

30 u z u z Re=110 near Hopf bif sinusoidal cycle u t u Re=63 near saddle node near heteroclinic cycle t

31

32 SH: subcritical secondary Hopf PF 2 : pitchfork G: gluing H: Hopf SNP: saddle node of periodic orbits PF 1 : pitchfork Takens Bogdanov normal form:

33 Eigenvalues: from real to complex

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