Symmetry Reflection Symmetry Symmetric 2D Antisymmetric 2D vector - PowerPoint PPT Presentation

Cours : Dynamique Non-Lin´ eaire Laurette TUCKERMAN laurette@pmmh.espci.fr Symmetry

Reflection Symmetry Symmetric 2D Antisymmetric 2D vector field vector field � u � � − u � κ ( x, y ) ≡ ( − x, y ) v v Group table for { I, κ } : I κ I I κ κ κ I

Other reflection operators ( κf )( x ) ≡ f ( − x ) κa ≡ − a a = g ( a ) ˙ Evolution equation: System has reflection symmetry ⇐ ⇒ g is equivariant ⇐ ⇒ gκ = κg ( gκ )( a ) = ( κg )( a ) ⇐ ⇒ g ( − a ) = − g ( a ) ⇐ ⇒ g odd a = g 1 a + g 3 a 3 = ( g 1 + g 3 a 2 ) a = ⇒ ˙ = ⇒ pitchfork bif to cubic order

� a � � − a � � 0 � � 0 � � a � � a � κ ≡ = ⇒ κ = κ = − and s s s s 0 0 � a � � g � d = G ( a, s ) ≡ ( a, s ) s h dt � g � � g � κ ( a, s ) = κ ( a, s ) h h � − g � � g � ( a, s ) = ( − a, s ) h h � g � � � g 10 a + g 11 as + g 30 a 3 + g 12 as 2 ( a, s ) = h 00 + h 01 s + h 20 a 2 + h 02 s 2 + h 21 a 2 s + h 03 s 3 h � � ( g 10 + g 11 s + g 12 s 2 + g 30 a 2 ) a = h 00 + h 01 s + h 02 s 2 + h 03 s 3 + ( h 20 + h 21 s ) a 2 � ˜ � � a � � 0 � g ( a 2 , s ) a + ˜ g ( a 2 , s ) h ( a 2 , s ) = = ˜ ˜ h ( a 2 , s ) 0 1 h = ˜ ˜ invariants: ˜ g = ˜ gκ, hκ equivariant: ( g, h ) κ = κ ( g, h )

Solutions to κ -equivariant systems are not all symmetric. (That is why bifurcation theory is interesting.) If ( a, s ) is a solution to a κ -equivariant system then κ ( a, s ) ≡ ( − a, s ) is also a solution. Asymmetric solutions ( a, s ) , a � = 0 come in pairs ( ± a, s ) . If a κ -equivariant system has a unique solution, then that solution is symmetric. e.g. linear system, or Navier-Stokes equations at low Re

Linear system G which commutes with κ : Gu = λu κGu = κλu Gκu = λκu ( λ, κu ) is also an eigenpair of G . κu could be a multiple of u : κu = cu κ 2 u = κcu u = c 2 u � 1 = ⇒ u is symmetric c = − 1 = ⇒ u is antisymmetric Or else if u and κu are linearly independent, then: u + κu is symmetric eigenvector u − κu is antisymmetric eigenvector

Verify equivariance of Navier-Stokes equations � ( κf ) ′ ( x ) = − f ′ ( − x ) ( κf )( x ) ≡ f ( − x ) = ⇒ ( κf ) ′′ ( x ) = f ′′ ( − x ) Demonstration: Let ˜ f ( x ) ≡ ( κf )( x ) ≡ f ( − x ) f ( x + ∆ x ) − ˜ ˜ f ( x ) lim ˜ f ′ ( x ) ≡ ∆ x → 0 ∆ x ∆ x → 0 f ( − x − ∆ x ) − f ( − x ) lim = ∆ x ∆ x → 0 f ( − x + ∆ x ) − f ( − x ) lim = − f ′ ( − x ) = − ∆ x

� � u �� � − ( u∂ x + v∂ y ) u − ∂ x p + ν ( ∂ 2 � x + ∂ 2 y ) u ( NS ) ( x, y ) ≡ ( x, y ) v − ( u∂ x + v∂ y ) v − ∂ y p + ν ( ∂ 2 x + ∂ 2 y ) v � � u �� � − u � κ ( x, y ) ≡ ( − x, y ) v v � � u �� � � u �� ? ( NS ) κ ( x, y ) = κ ( NS ) ( x, y ) v v � − [ − ( u∂ x + v∂ y ) u − ∂ x p + ν ( ∂ 2 � x + ∂ 2 y ) u ] ( − x, y ) − ( u∂ x + v∂ y ) v − ∂ y p + ν ( ∂ 2 x + ∂ 2 y ) v Boundary conditions & external forces determine if problem is κ -equivariant:

κ ⇓ κ ⇓ κ ⇓ Not equivalent flows

Rotations and reflections of the plane Z 2 Z 3 Z 4 . . . SO (2) D 2 D 3 D 4 . . . O (2)

Natural representation of O (2) on ( x, y ) uses z = x + iy S θ z ≡ e iθ z κz ≡ ¯ z κS θ z = κ ( e iθ z ) = e − iθ ¯ z z = e iθ ¯ S θ κz = S θ ¯ z κS θ z = S − θ κz ( S θ 0 w )( ρ, θ ) ≡ w ( ρ, θ + θ 0 ) ( κw )( ρ, θ ) ≡ w ( ρ, − θ )

Circle Pitchfork � f mn z m ¯ z n f ( z, ¯ z ) = m,n z ) = ¯ z m z n z m z n κf ( z, ¯ z ) = f ( z, ¯ f mn ¯ f ( κ ( z, ¯ z )) = f mn ¯ f real z ) = e iθ f ( z, ¯ S θ f ( z, ¯ z ) f ( S θ ( z, ¯ z )) n = f mn ( e iθ z ) m ( e iθ z ) = e iθ f mn z m ¯ z n = f mn e imθ z m e − inθ ¯ z n m − n = 1 or f mn = 0 z 2 + · · · z ) = f 10 z + f 21 z 2 ¯ z + f 32 z 3 ¯ f ( z, ¯ = ( f 10 + f 21 | z | 2 + f 32 | z | 4 + · · · ) z ˜ f ( | z | 2 ) z = | z | 2 invariant z equivariant

Circle Pitchfork z = ( µ − α | z | 2 ) z ˙

Circle Pitchfork z = ( µ − | z | 2 ) z ˙ Cartesian Form: y = µ ( x + iy ) − ( x 2 + y 2 )( x + iy ) x + i ˙ ˙ x = µx − ( x 2 + y 2 ) x ˙ y = µy − ( x 2 + y 2 ) y ˙ Polar Form: θ ) e iθ = ( µ − r 2 ) re iθ r + ir ˙ ( ˙ r = µr − r 3 ˙ ˙ θ = 0 Subcritical Form: z = ( µ + | z | 2 ) z ˙

Stability of origin (use Cartesian coordinates): �� � µ − (3 x 2 + y 2 ) � µ 0 � � − 2 xy � J ( x = 0 , y = 0) = = µ − ( x 2 + 3 y 2 ) � − 2 xy 0 µ (0 , 0) double eigenvalue µ . Stability of states on circle (use polar coordinates): �� � µ − 3 r 2 0 � − 2 µ 0 � � J ( r = √ µ, θ ) = � = � √ µ,θ 0 0 0 0 eigenvalue − 2 µ along r and marginal eigenvalue 0 along θ . dw dt ( θ ) = F ( w )( θ ) F indep of θ 0 = F ( W ) W a steady solution on circle 0 = d F ( W ) = δ F ∂W Jacobian or Frechet derivative marginal eigenvector dθ δW ∂θ

Drift Pitchfork r = ( µ − r 2 ) r ˙ ˙ θ = ζ ζ = ( r 2 − 1 − ζ 2 ) ζ ˙ µ < 0 0 ≤ µ ≤ 1 1 < µ r = √ µ r = √ µ r = 0 θ = θ 0 θ = θ 0 + ζt ζ = ±√ µ − 1 ζ = 0 ζ = 0

Drift Pitchfork Speed at onset is slow: ζ = √ µ − µ DP = √ µ − 1 Motion along the circle: group orbit Symmetry-breaking variable: ζ At µ = µ DP = 1 , Jacobian contains a Jordan block � � � 0 1 � ∂ ˙ ∂ ˙ θ θ ∂θ ∂ζ = ∂ ˙ ∂ ˙ 0 0 ζ ζ ∂θ ∂θ Function W ( θ ) is even about some θ 0 . Marginal and bifurcating eigenvectors are odd about θ 0 Drifting W is asymmetric in θ .

Drift pitchfork in reaction-diffusion equations speed From Kness, Tuckerman & Barkley, Phys. Rev. A 46, 5054 (1992).

O (2) and SO (2) Axisymmetric Taylor vortices Spiral Taylor vortices Tagg, Nonlinear Science Today 4, Antonijoan et al., Phys. Fluids 10, 1 (1994) . 829 (1995 .

No requirement of κf = fκ = ⇒ f complex z = ( µ + iω − ( α + iβ ) | z | 2 ) z ˙ θ ) e iθ = (( µ − αr 2 ) + i ( ω + βr 2 )) r e iθ r + ri ˙ ( ˙ r = ( µ − αr 2 ) r ˙ ˙ θ = ω + βr 2 Breaking of SO (2) = ⇒ motion along θ direction

Hopf bifurcation and O (2) symmetry Need four-dimensional eigenspace: u ( θ, t ) = ( z + ( t ) + z − ( t )) e iθ + (¯ z − ( t )) e − iθ z + ( t ) + ¯ where z ± are complex amplitudes (i.e. amplitude and phase) of left-going and right-going traveling waves. At linear order: � z + � � � d iωz + = z − − iωz − dt u ( θ, t ) = z + (0) e i ( θ + ωt ) + z − (0) e i ( θ − ωt ) +¯ z + (0) e − i ( θ + ωt ) +¯ z − (0) e − i ( θ − ωt ) z ± (0) arbitrary initial amplitudes. Addition of nonlinear terms compatible with O (2) symmetry greatly restricts possible equilibria.

Appropriate representation of O (2) on ( z + , z − ) : S θ 0 ( z + , z − ) = ( e iθ 0 z + , e iθ 0 z − ) κ ( z + , z − ) = (¯ z − , ¯ z + ) Simplest cubic order equivariant evolution equations: � � � � z + � � µ + iω + a | z − | 2 + b ( | z + | 2 + | z − | 2 ) d z + � � = a | z + | 2 + ¯ b ( | z + | 2 + | z − | 2 ) z − µ − iω + ¯ z − dt Substitute z ± = r ± e iφ ± : r ± = ( µ + a r r 2 ∓ + b r ( r 2 + + r 2 ˙ − )) r ± ˙ φ ± = ± ( ω + a i r 2 ∓ + b i ( r 2 + + r 2 − )) Equations do not involve phases φ ± (all phases equivalent).

Solutions with ˙ r ± = 0 : origin : r + = 0 , r − = 0 � − µ left traveling waves : r + = , r − = 0 , b r φ + = ω − µb i ˙ b r � − µ right traveling waves : r + = 0 , r − = , b r � � ω − µb i ˙ φ − = − b r � − 2 µ standing waves : r + = r − = , a r + 2 b r � � ω − 2 µ a i + 2 b i ˙ φ ± = ± a r + 2 b r

9800 0T 0.14T 9600 0.25T 0.39T 9400 0.5T h 9200 9000 8800 8600 π 2 π 0 θ Standing waves in Rayleigh-B´ enard convection in a cylinder with Γ = R/H = 1 . 47 and P r = 1 at at Ra = 26 000 . From Boro´ nska & Tuckerman, J. Fluid Mech. 559, 279 (2006).

9800 0T 0.25T 9600 0.5T 0.75T 9400 h 9200 9000 8800 8600 π 2 π 0 θ Travelling wave in Rayleigh-B´ enard convection in a cylinder with Γ = R/H = 1 . 47 and P r = 1 at Ra = 26 000 . From Boro´ nska & Tucker- man, J. Fluid Mech. 559, 279 (2006).

1.5 1.5 1.0 1.0 0.5 0.5 0.0 0.0 1.5 1.5 −0.5 −0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 1.0 1.0 0.5 0.5 0.0 0.0 2.5 2.0 5 −0.5 1.5 −0.5 4 1.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3 0.5 2 0.0 1 −0.5 0 −1 −1.0 −2 −1.5 −3 −2.0 −4 −2.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 −5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 Standing waves Travelling waves Thermosolutal convection with S = − 0 . 1 , L = 0 . 1 , P r = 10 , r ≡ Ra/Ra c = 1 . 3 .