Pattern formation in Nikolaevskiy s equation

Transcription

1 Stephen Cox School of Mathematical Sciences, University of Nottingham Differential Equations and Applications Seminar 2007 with Paul Matthews, Nottingham

2 Outline What is Nikolaevskiy s equation?

3 Outline What is Nikolaevskiy s equation? How do solutions to Nikolaevskiy s equation behave?

4 Outline What is Nikolaevskiy s equation? How do solutions to Nikolaevskiy s equation behave? Unfolding the sudden onset of spatiotemporal chaos.

5 The equation that Nikolaevskiy wrote down Nikolaevskiy (1989) derived a family of one-dimensional models for seismic waves in a viscoelastic medium: V t + V V x = n p=2 A p p V x p, where V (x, t) is the displacement velocity.

6 The equation that Nikolaevskiy wrote down Nikolaevskiy (1989) derived a family of one-dimensional models for seismic waves in a viscoelastic medium: V t + V V x = νv + n p=2 where V (x, t) is the displacement velocity. A p p V x p, In a later model (Nikolaevskiy 2007) he added a bulk dissipation term.

7 The equation that Nikolaevskiy wrote down Nikolaevskiy (1989) derived a family of one-dimensional models for seismic waves in a viscoelastic medium: V t + V V x = νv + n p=2 where V (x, t) is the displacement velocity. A p p V x p, In a later model (Nikolaevskiy 2007) he added a bulk dissipation term. The international standard Nikolaevskiy equation has no dissipation, n = 6 and no odd derivatives on the right-hand side.

8 Nikolaevskiy s equation for transverse instability of fronts Weakly nonlinear perturbations to a planar front (e.g., in a reaction diffusion system) described by the Kuramoto Sivashinsky equation φ t ( ) φ 2 = 2 φ x x 2 4 φ x 4.

9 Nikolaevskiy s equation for transverse instability of fronts Weakly nonlinear perturbations to a planar front (e.g., in a reaction diffusion system) described by the Kuramoto Sivashinsky equation φ t ( ) φ 2 = 2 φ x x 2 4 φ x 4. If the instability sets in at non-zero wavenumber then instead φ t + 1 ( ) φ 2 = 2 φ 2 x x 2 + φ α 4 x φ x 6.

10 Nikolaevskiy s equation for transverse instability of fronts Weakly nonlinear perturbations to a planar front (e.g., in a reaction diffusion system) described by the Kuramoto Sivashinsky equation φ t ( ) φ 2 = 2 φ x x 2 4 φ x 4. If the instability sets in at non-zero wavenumber then instead φ t + 1 ( ) φ 2 = 2 φ 2 x x 2 + φ α 4 x φ x 6. Put u = φ x and rescale: Nikolaevskiy s equation.

11 Nikolaevskiy s equation Nikolaevskiy s equation u t + u u x = 2 x 2 ( ru subject to u(x + L, t) = u(x, t) for all x, t. ) 2 (1 + u) 2 x 2

12 Nikolaevskiy s equation Nikolaevskiy s equation u t + u u x = 2 x 2 ( ru subject to u(x + L, t) = u(x, t) for all x, t. ) 2 (1 + u) 2 x 2 It has Galilean symmetry x x + Vt, u u + V and a reflection symmetry x x, u u.

13 Nikolaevskiy s equation Nikolaevskiy s equation u t + u u x = 2 x 2 ( ru subject to u(x + L, t) = u(x, t) for all x, t. ) 2 (1 + u) 2 x 2 It has Galilean symmetry x x + Vt, u u + V and a reflection symmetry x x, u u. It describes: seismic waves; transverse instabilities to fronts at finite wavenumber; a phase equation for reaction diffusion systems.

14 Nikolaevskiy s equation Nikolaevskiy s equation u t + u u x = 2 x 2 ( ru subject to u(x + L, t) = u(x, t) for all x, t. ) 2 (1 + u) 2 x 2 It has Galilean symmetry x x + Vt, u u + V and a reflection symmetry x x, u u. It describes: seismic waves; transverse instabilities to fronts at finite wavenumber; a phase equation for reaction diffusion systems. Note: u is constant: specify u L 0 u dx = 0 wlog.

15 Linear stability of uniform state in Nikolaevskiy s equation Uniform solution to Nikolaevskiy s equation is u = ū 0. Stability: u = ū + δe λt+ikx (δ 1) has dispersion relation λ = k 2 (r (1 k 2 ) 2 ).

16 Linear stability of uniform state in Nikolaevskiy s equation Uniform solution to Nikolaevskiy s equation is u = ū 0. Stability: u = ū + δe λt+ikx (δ 1) has dispersion relation λ = k 2 (r (1 k 2 ) 2 ). For r < r c = 0, all modes decay; for r slightly above r c, modes with k 1 grow. Large-scale modes (k 0) are only weakly damped, λ = O(k 2 ).

17 Linear stability of uniform state in Nikolaevskiy s equation Uniform solution to Nikolaevskiy s equation is u = ū 0. Stability: u = ū + δe λt+ikx (δ 1) has dispersion relation λ = k 2 (r (1 k 2 ) 2 ). For r < r c = 0, all modes decay; for r slightly above r c, modes with k 1 grow. Large-scale modes (k 0) are only weakly damped, λ = O(k 2 ). Uniform state is unstable to rolls inside the blue marginal curve, i.e., for k m(r) < k < k + m(r).

18 Typical numerical solution to Nikolaevskiy s equation Numerical solution in for 0 x 100 for 0 t 300 with r = 0.1, from a small-amplitude random initial state.

19 Typical numerical solution to Nikolaevskiy s equation Numerical solution in for 0 x 100 for 0 t 300 with r = 0.1, from a small-amplitude random initial state. Initial exponential growth of roll pattern with wavenumber k 1; then at t 80, sudden breakup of the rolls. Spatiotemporal chaos at onset.

20 Roll solutions to Nikolaevskiy s equation Perturbation expansion in ɛ 1 gives roll solutions in the form u 6ɛ(1 4q 2 )e ikx + c.c., where r = ɛ 2 and k = 1 + ɛq.

21 Roll solutions to Nikolaevskiy s equation Perturbation expansion in ɛ 1 gives roll solutions in the form u 6ɛ(1 4q 2 )e ikx + c.c., where r = ɛ 2 and k = 1 + ɛq. Determine stability of rolls by writing a perturbation expansion u ɛa(x, T )e ix + c.c., then deriving equation for envelope A(X, T ); X = ɛx,t = ɛ 2 t.

22 Roll solutions to Nikolaevskiy s equation Perturbation expansion in ɛ 1 gives roll solutions in the form u 6ɛ(1 4q 2 )e ikx + c.c., where r = ɛ 2 and k = 1 + ɛq. Determine stability of rolls by writing a perturbation expansion u ɛa(x, T )e ix + c.c., then deriving equation for envelope A(X, T ); X = ɛx,t = ɛ 2 t. But modes with small k (i.e., large-scale modes) also evolve slowly.

23 Roll solutions to Nikolaevskiy s equation Perturbation expansion in ɛ 1 gives roll solutions in the form u 6ɛ(1 4q 2 )e ikx + c.c., where r = ɛ 2 and k = 1 + ɛq. Determine stability of rolls by writing a perturbation expansion u ɛa(x, T )e ix + c.c., then deriving equation for envelope A(X, T ); X = ɛx,t = ɛ 2 t. But modes with small k (i.e., large-scale modes) also evolve slowly. If we write u ɛa(x, T )e ix + c.c. + ɛf (X, T ) + ɛ 2 u 2 + then we obtain at O(ɛ 3 ) the solvability conditions A T = A+4A XX A 2 A/36 ifa/ɛ (fa) X, f T = f XX ( A 2 ) X ff X But these are asymptotically inconsistent!

24 Stability of roll solutions A and f amplitude equations There is no way to vary the scalings associated with A and f to keep all terms consistently in A T = A+4A XX A 2 A/36 ifa/ɛ (fa) X, f T = f XX ( A 2 ) X ff X.

25 Stability of roll solutions A and f amplitude equations There is no way to vary the scalings associated with A and f to keep all terms consistently in A T = A+4A XX A 2 A/36 ifa/ɛ (fa) X, f T = f XX ( A 2 ) X ff X. Eventually we find that u ɛ 3/2 A(X, T )e ix + c.c. + ɛ 2 f (X, T ) + leads to the asymptotically consistent amplitude equations A T = A + 4A XX ifa, f T = f XX ( A 2 ) X. Note: no stabilising cubic term in the A equation!

26 Stability of roll solutions A and f amplitude equations There is no way to vary the scalings associated with A and f to keep all terms consistently in A T = A+4A XX A 2 A/36 ifa/ɛ (fa) X, f T = f XX ( A 2 ) X ff X. Eventually we find that u ɛ 3/2 A(X, T )e ix + c.c. + ɛ 2 f (X, T ) + leads to the asymptotically consistent amplitude equations A T = A + 4A XX ifa, f T = f XX ( A 2 ) X. Note: no stabilising cubic term in the A equation! RMS(u) = O(ɛ 3/2 ) is consistent with numerical simulations of the full Nikolaevskiy equation.

27 Stability of roll solutions We would usually use amplitude equations such as A T = A+4A XX A 2 A/36 ifa/ɛ (fa) X, f T = f XX ( A 2 ) X ff X. to determine the stability of roll solutions.

28 Stability of roll solutions We would usually use amplitude equations such as A T = A+4A XX A 2 A/36 ifa/ɛ (fa) X, f T = f XX ( A 2 ) X ff X. to determine the stability of roll solutions. The roll solutions are simply A = A 0, f = f 0, where A 0 = 6(1 4q 2 ) 1/2 e iqx, f 0 = 0.

29 Stability of roll solutions We would usually use amplitude equations such as A T = A+4A XX A 2 A/36 ifa/ɛ (fa) X, f T = f XX ( A 2 ) X ff X. to determine the stability of roll solutions. The roll solutions are simply A = A 0, f = f 0, where A 0 = 6(1 4q 2 ) 1/2 e iqx, f 0 = 0. Stability is determined by linearising: write A = A 0 + A 1 (X, T ), f = f 0 + f 1 (X, T ) and compute the growth rate of A 1, f 1.

30 Stability of roll solutions We would usually use amplitude equations such as A T = A+4A XX A 2 A/36 ifa/ɛ (fa) X, f T = f XX ( A 2 ) X ff X. to determine the stability of roll solutions. The roll solutions are simply A = A 0, f = f 0, where A 0 = 6(1 4q 2 ) 1/2 e iqx, f 0 = 0. Stability is determined by linearising: write A = A 0 + A 1 (X, T ), f = f 0 + f 1 (X, T ) and compute the growth rate of A 1, f 1. But... the cubic term A A 2 is missing from A T = A + 4A XX ifa, f T = f XX ( A 2 ) X. so they don t know the roll amplitude, and can t be used to determine roll stability.

31 Stability of roll solutions In the full Nikolaevskiy equation, we can compute the stability of the roll solutions by setting (with wavenumber k = 1 + ɛq) u ɛa 0 e ikx + c.c. + [ ] + δ (ɛ 1/4 b(ξ, τ) + ic(ξ, τ))e ikx + ɛ 1/2 f (ξ, τ) + where a 0 = 6(1 4q 2 ) 1/2, ξ = ɛ 3/4 x, τ = ɛ 3/2 t.

32 Stability of roll solutions In the full Nikolaevskiy equation, we can compute the stability of the roll solutions by setting (with wavenumber k = 1 + ɛq) u ɛa 0 e ikx + c.c. + [ ] + δ (ɛ 1/4 b(ξ, τ) + ic(ξ, τ))e ikx + ɛ 1/2 f (ξ, τ) + where a 0 = 6(1 4q 2 ) 1/2, ξ = ɛ 3/4 x, τ = ɛ 3/2 t. Then b τ = 4b ξξ 8qc ξ, c τ = 4c ξξ a 0 f, f τ = f ξξ 2a 0 b ξ All modes with q > 0 (i.e, k > k c = 1) are unstable to monotonically growing disturbances; all modes with q < 0 are unstable to oscillatory disturbances.

33 Stability of roll solutions In the full Nikolaevskiy equation, we can compute the stability of the roll solutions by setting (with wavenumber k = 1 + ɛq) u ɛa 0 e ikx + c.c. + [ ] + δ (ɛ 1/4 b(ξ, τ) + ic(ξ, τ))e ikx + ɛ 1/2 f (ξ, τ) + where a 0 = 6(1 4q 2 ) 1/2, ξ = ɛ 3/4 x, τ = ɛ 3/2 t. Then b τ = 4b ξξ 8qc ξ, c τ = 4c ξξ a 0 f, f τ = f ξξ 2a 0 b ξ All modes with q > 0 (i.e, k > k c = 1) are unstable to monotonically growing disturbances; all modes with q < 0 are unstable to oscillatory disturbances. All rolls are unstable at onset! This instability is responsible for spatiotemporal chaos at onset.

34 Unfolding the sudden onset of instability There is spatiotemporal chaos at onset in the Nikolaevskiy equation ( u t + u u ) 2 x = 2 x 2 ru (1 + u) 2 x 2. But many other pattern forming systems have a more gradual transition, involving a sequence of bifurcations.

35 Unfolding the sudden onset of instability There is spatiotemporal chaos at onset in the Nikolaevskiy equation ( u t + u u ) 2 x = 2 x 2 ru (1 + u) 2 x 2. But many other pattern forming systems have a more gradual transition, involving a sequence of bifurcations. Introduce damping. We can vary its strength to probe the onset of complex dynamics: the damped Nikolaevskiy equation ( u t + u u ) 2 ) 2 x = 2 x 2 ru (1 + u) 2 x 2 ν (1 + 2 x 2 u. Note that the damping affects modes away from k = 1.

36 Unfolding the sudden onset of instability There is spatiotemporal chaos at onset in the Nikolaevskiy equation ( u t + u u ) 2 x = 2 x 2 ru (1 + u) 2 x 2. But many other pattern forming systems have a more gradual transition, involving a sequence of bifurcations. Introduce damping. We can vary its strength to probe the onset of complex dynamics: the damped Nikolaevskiy equation ( u t + u u ) 2 ) 2 x = 2 x 2 ru (1 + u) 2 x 2 ν (1 + 2 x 2 u. Note that the damping affects modes away from k = 1. Damping corresponds to slight breaking of the Galilean symmetry.

37 Unfolding the sudden onset of instability The damped Nikolaevskiy equation ( u t + u u ) 2 ) 2 x = 2 x 2 ru (1 + u) 2 x 2 ν (1 + 2 x 2 u has linear spectrum λ = k 2 (r (1 k 2 ) 2 ) ν(1 k 2 ) 2.

38 Unfolding the sudden onset of instability The damped Nikolaevskiy equation ( u t + u u ) 2 ) 2 x = 2 x 2 ru (1 + u) 2 x 2 ν (1 + 2 x 2 u has linear spectrum λ = k 2 (r (1 k 2 ) 2 ) ν(1 k 2 ) 2. We set r = ɛ 2 and ν = ɛ s µ then vary the exponent s to unfold the onset of Nikolaevskiy chaos.

39 Damping ν = ɛ 0 µ If the damping is strong, ν = O(ɛ 0 ), then the usual expansion u ɛa(x, T )e ix + c.c. + ɛ 2 u 2 + gives the Ginzburg Landau equation A T = A + 4(1 + µ) 2 A X A A 2.

40 Damping ν = ɛ 0 µ If the damping is strong, ν = O(ɛ 0 ), then the usual expansion u ɛa(x, T )e ix + c.c. + ɛ 2 u 2 + gives the Ginzburg Landau equation A T = A + 4(1 + µ) 2 A X A A 2. From this it follows that rolls exist for 1 1 ɛ 4(1 + µ) < k < 1 + ɛ 1 4(1 + µ) and are stable (to the Eckhaus instability) for 1 1 ɛ 12(1 + µ) < k < 1 + ɛ 1 12(1 + µ)

41 Damping ν = ɛ 0 µ

42 Damping ν = ɛ 1 µ If damping is less strong, ν = O(ɛ 1 ), u ɛa(x, T )e ix + c.c. + ɛ 2 [ 1 36 ia2 e 2ix + c.c. + f (X, T ) ] + with the balance at O(ɛ 3 ) that f = ( A 2 ) X /µ.

43 Damping ν = ɛ 1 µ If damping is less strong, ν = O(ɛ 1 ), u ɛa(x, T )e ix + c.c. + ɛ 2 [ 1 36 ia2 e 2ix + c.c. + f (X, T ) ] + with the balance at O(ɛ 3 ) that f = ( A 2 ) X /µ. Ginzburg Landau equation becomes Mancebo, Vega JFM 2006 A T =A+4 2 A X A A 2 ifa = A A X A A 2 + i( A 2 ) X A/µ.

44 Damping ν = ɛ 1 µ If damping is less strong, ν = O(ɛ 1 ), u ɛa(x, T )e ix + c.c. + ɛ 2 [ 1 36 ia2 e 2ix + c.c. + f (X, T ) ] + with the balance at O(ɛ 3 ) that f = ( A 2 ) X /µ. Ginzburg Landau equation becomes Mancebo, Vega JFM 2006 A T =A+4 2 A X A A 2 ifa = A A X A A 2 + i( A 2 ) X A/µ. Rolls with k = 1 + ɛq: upper part of diagram consistent with ν = O(ɛ 0 ) picture; lower part of diagram requires further consideration.

45 ν = ɛ 1 µ: extended Ginzburg Landau equation for small µ In the limit µ 0 the extended Ginzburg Landau equation behaves similarly to the (undamped) Nikolaevskiy equation.

46 Damping ν = ɛ 3/2 µ If damping is even less strong, ν = O(ɛ 3/2 ), there are interactions between three types of perturbations: to the amplitude of the rolls, to the phase of the rolls and a large-scale mode.

47 Damping ν = ɛ 3/2 µ If damping is even less strong, ν = O(ɛ 3/2 ), there are interactions between three types of perturbations: to the amplitude of the rolls, to the phase of the rolls and a large-scale mode. After much trial and error, we find appropriate scalings to be u = ɛ(a 0 +ɛ 1/2 a(ξ, τ))e ix e iɛqx e iɛ1/4 φ(ξ,τ) +c.c.+ +ɛ 7/4 f (ξ, τ)+, where a 0 = 6(1 4q 2 ) 1/2, ξ = ɛ 3/4 x, τ = ɛ 3/2 t.

48 Damping ν = ɛ 3/2 µ If damping is even less strong, ν = O(ɛ 3/2 ), there are interactions between three types of perturbations: to the amplitude of the rolls, to the phase of the rolls and a large-scale mode. After much trial and error, we find appropriate scalings to be u = ɛ(a 0 +ɛ 1/2 a(ξ, τ))e ix e iɛqx e iɛ1/4 φ(ξ,τ) +c.c.+ +ɛ 7/4 f (ξ, τ)+, where a 0 = 6(1 4q 2 ) 1/2, ξ = ɛ 3/4 x, τ = ɛ 3/2 t. After substitution in the damped Nikolaevskiy equation we find φ τ = 4φ ξξ f, f τ = f ξξ µf 2a 0 a ξ, a τ = 4a ξξ 4a 0 φ 2 ξ 8qa 0φ ξ or ( τ 4 ξξ ) 2 ( τ ξξ + µ) φ = 16a 2 0 (q + φ ξ) φ ξ.

49 Damping ν = ɛ 3/2 µ Stability of rolls governed by the linearised phase equation ( τ 4 ξξ ) 2 ( τ ξξ + µ) φ = 16a 2 0qφ ξ.

50 Damping ν = ɛ 3/2 µ Stability of rolls governed by the linearised phase equation ( τ 4 ξξ ) 2 ( τ ξξ + µ) φ = 16a 2 0qφ ξ. Linear stability results: for µ > µ c 8 3, rolls with 1 1 2ɛ < k < 1 are stable; for 0 < µ < µ c, some of these rolls are unstable. This analysis suggests that two regions of stable rolls persist down to zero damping, but....

51 Damping ν = ɛ 3/2 µ... the stability of the rolls is governed by ( τ 4 ξξ ) 2 ( τ ξξ + µ) φ = 16a 2 0qφ ξ, But roll amplitude a 0 = 6(1 4q 2 ) 1/2 ; rolls exist for 1 2 < q < 1 2.

52 Damping ν = ɛ 3/2 µ... the stability of the rolls is governed by ( τ 4 ξξ ) 2 ( τ ξξ + µ) φ = 16a 2 0qφ ξ, But roll amplitude a 0 = 6(1 4q 2 ) 1/2 ; rolls exist for 1 2 < q < 1 2. So the coupling coefficient 16a 2 0 q is small in both cusps of apparent stability. Another scaling is called for when a0 2 q is small.

53 Damping ν = ɛ 2 µ near central cusp k = 1 + ɛ 2 q Tribelsky, Velarde Phys Rev E 1996 To resolve stability near the central cusp, we write u ɛ(6+ɛ 2 a(x, T ))e ix e iɛ2 qx e iɛφ(x,t ) +c.c.+ +ɛ 3 f (X, T )+. The key point here is that k 1 = O(ɛ 2 ) rather than O(ɛ).

54 Damping ν = ɛ 2 µ near central cusp k = 1 + ɛ 2 q Tribelsky, Velarde Phys Rev E 1996 To resolve stability near the central cusp, we write u ɛ(6+ɛ 2 a(x, T ))e ix e iɛ2 qx e iɛφ(x,t ) +c.c.+ +ɛ 3 f (X, T )+. The key point here is that k 1 = O(ɛ 2 ) rather than O(ɛ). After much algebra we find or φ T = 4φ XX f, f T = f XX µf 12a X, a T = 4a XX 2a + 6 ( q + 12 XX ) φx 24φ 2 X 6f X. ( T 4 XX ) ( T 4 XX + 2) ( T XX + µ) φ = 72 ( q + T + 8 X ) φxx 576φ 2 X φ XX.

55 Damping ν = ɛ 2 µ near central cusp k = 1 + ɛ 2 q With no damping (µ = 0), all rolls near the central cusp are unstable to modes like e ilx

56 Damping ν = ɛ 2 µ near central cusp k = 1 + ɛ 2 q With no damping (µ = 0), all rolls near the central cusp are unstable to modes like e ilx For increased damping (µ is marked on the graph), the rolls eventually stabilise (at µ 8.445)

57 Damping ν = ɛ 2 µ near central cusp k = 1 + ɛ 2 q With no damping (µ = 0), all rolls near the central cusp are unstable to modes like e ilx For increased damping (µ is marked on the graph), the rolls eventually stabilise (at µ 8.445) Correction: the region of stable rolls does not extend down to zero damping; below ν = 8.445ɛ 2 rolls near k = 1 are unstable.

58 Damping ν = ɛ 2 µ near left-hand marginal curve Recall: rolls exist for wavenumbers k m < k < k + m; r = ɛ 2.

59 Damping ν = ɛ 2 µ near left-hand marginal curve Recall: rolls exist for wavenumbers k m < k < k + m; r = ɛ 2. Near k = k ± m, the roll amplitude is smaller than O(ɛ).

60 Damping ν = ɛ 2 µ near left-hand marginal curve Recall: rolls exist for wavenumbers k m < k < k + m; r = ɛ 2. Near k = k ± m, the roll amplitude is smaller than O(ɛ). To resolve stability, we let u ɛ 3/2 (a 0 + a(x, T ))e ikx iφ(x,t ) e where + c.c. + + ɛ 2 f (X, T ) +, k = k ± m ɛ 2 κ, a 0 = 12 κ.

61 Damping ν = ɛ 2 µ near left-hand marginal curve Small perturbations to the rolls are governed by a 0 φ T = 4a 0 φ XX a 0 f ± 4a X, f T = f XX µf 2a 0 a X, a T = 4a XX 4a 0 φ X. For disturbances proportional to e σt +ilx, dispersion relation is ( σ + l 2 + µ ) [ ( σ + 4l 2 ) ] 2 16l 2 = ±8a0l 2 2.

62 Damping ν = ɛ 2 µ near left-hand marginal curve Small perturbations to the rolls are governed by a 0 φ T = 4a 0 φ XX a 0 f ± 4a X, f T = f XX µf 2a 0 a X, a T = 4a XX 4a 0 φ X. For disturbances proportional to e σt +ilx, dispersion relation is ( σ + l 2 + µ ) [ ( σ + 4l 2 ) ] 2 16l 2 = ±8a0l 2 2. As a 0 0, two roots are σ = 4l(±1 l), hence rolls are unstable sufficiently close to the marginal curve.

63 Damping ν = ɛ 2 µ near left-hand marginal curve Near the marginal curve, in limit l 0, σ 2 8l 2 (2 ± a 2 0/µ).

64 Damping ν = ɛ 2 µ near left-hand marginal curve Near the marginal curve, in limit l 0, σ 2 8l 2 (2 ± a 2 0/µ). Thus near k = k m + rolls are unstable since (2 + a0 2 /µ) > 0.

65 Damping ν = ɛ 2 µ near left-hand marginal curve Near the marginal curve, in limit l 0, σ 2 8l 2 (2 ± a 2 0/µ). Thus near k = k m + rolls are unstable since (2 + a0 2 /µ) > 0. Near k = k m, instability if a 2 0 < 2µ.

66 Damping ν = ɛ 2 µ near left-hand marginal curve Near the marginal curve, in limit l 0, σ 2 8l 2 (2 ± a 2 0/µ). Thus near k = k m + rolls are unstable since (2 + a0 2 /µ) > 0. Near k = k m, instability if a 2 0 < 2µ. When a 2 0 > 2µ, σ = il 8(a 2 0 /µ 2) + 4l 2 (a 2 0 /µ2 1) +

67 Damping ν = ɛ 2 µ near left-hand marginal curve Near the marginal curve, in limit l 0, σ 2 8l 2 (2 ± a 2 0/µ). Thus near k = k m + rolls are unstable since (2 + a0 2 /µ) > 0. Near k = k m, instability if a 2 0 < 2µ. When a 2 0 > 2µ, σ = il 8(a 2 0 /µ 2) + 4l 2 (a 2 0 /µ2 1) + Thus when 2µ < a 2 0 < µ2, rolls are stable, near k = k m.

68 Damping ν = ɛ 2 µ near left-hand marginal curve Near the marginal curve, in limit l 0, σ 2 8l 2 (2 ± a 2 0/µ). Thus near k = k m + rolls are unstable since (2 + a0 2 /µ) > 0. Near k = k m, instability if a 2 0 < 2µ. When a 2 0 > 2µ, σ = il 8(a 2 0 /µ 2) + 4l 2 (a 2 0 /µ2 1) + Thus when 2µ < a 2 0 < µ2, rolls are stable, near k = k m. Rolls enjoy a region of stability only for µ > µ c 2.

69 Damping ν = ɛ 2 µ near left-hand marginal curve Near the marginal curve, in limit l 0, σ 2 8l 2 (2 ± a 2 0/µ). Thus near k = k m + rolls are unstable since (2 + a0 2 /µ) > 0. Near k = k m, instability if a 2 0 < 2µ. When a 2 0 > 2µ, σ = il 8(a 2 0 /µ 2) + 4l 2 (a 2 0 /µ2 1) + Thus when 2µ < a 2 0 < µ2, rolls are stable, near k = k m. Rolls enjoy a region of stability only for µ > µ c 2. Correction: the region of stable rolls does not extend down to zero damping; below ν = 2ɛ 2 all rolls near k = k m are unstable.

70 Stability of rolls: numerical results Existence and stability of rolls in damped Nikolaevskiy equation: ɛ = 0.1 Existence and stability of rolls in damped Nikolaevskiy equation: ɛ = 0.05

71 Stability of rolls: numerical results Existence and stability of roll solutions to the damped Nikolaevskiy equation: ν = Thin wedge of stable rolls near left-hand marginal stability boundary.

72 Numerical simulations of the damped Nikolaevskiy equation Simulations of the damped Nikolaevskiy equation at r = 0.01: ν = 0.04, 0.03, 0.02, 0.

73 Numerical simulations of the damped Nikolaevskiy equation Simulations of the damped Nikolaevskiy equation at r = 0.01: ν = 0.04, 0.03, 0.02, 0.

74 Conclusions Nikolaevskiy s equation models transverse instabilities of fronts, seismic waves,....

75 Conclusions Nikolaevskiy s equation models transverse instabilities of fronts, seismic waves,.... It exhibits spatiotemporal chaos at onset.

76 Conclusions Nikolaevskiy s equation models transverse instabilities of fronts, seismic waves,.... It exhibits spatiotemporal chaos at onset. Weakly nonlinear perturbation expansions confirm that all roll solutions are unstable at onset

77 Conclusions Nikolaevskiy s equation models transverse instabilities of fronts, seismic waves,.... It exhibits spatiotemporal chaos at onset. Weakly nonlinear perturbation expansions confirm that all roll solutions are unstable at onset By adding damping, we can unfold the onset of instability.

78 Conclusions Nikolaevskiy s equation models transverse instabilities of fronts, seismic waves,.... It exhibits spatiotemporal chaos at onset. Weakly nonlinear perturbation expansions confirm that all roll solutions are unstable at onset By adding damping, we can unfold the onset of instability. A thin wedge of stable rolls controls the dynamics for moderate damping.

79 Conclusions Nikolaevskiy s equation models transverse instabilities of fronts, seismic waves,.... It exhibits spatiotemporal chaos at onset. Weakly nonlinear perturbation expansions confirm that all roll solutions are unstable at onset By adding damping, we can unfold the onset of instability. A thin wedge of stable rolls controls the dynamics for moderate damping. Increased damping increases the pattern amplitude!

80 Conclusions Nikolaevskiy s equation models transverse instabilities of fronts, seismic waves,.... It exhibits spatiotemporal chaos at onset. Weakly nonlinear perturbation expansions confirm that all roll solutions are unstable at onset By adding damping, we can unfold the onset of instability. A thin wedge of stable rolls controls the dynamics for moderate damping. Increased damping increases the pattern amplitude! What about Nikolaevskiy s real equation?