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# Nonlinear Dynamics PHYS 7224

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This 0 page Class Notes was uploaded by Kylie Bartoletti DVM on Monday November 2, 2015. The Class Notes belongs to PHYS 7224 at Georgia Institute of Technology - Main Campus taught by Staff in Fall. Since its upload, it has received 11 views. For similar materials see /class/234290/phys-7224-georgia-institute-of-technology-main-campus in Physics 2 at Georgia Institute of Technology - Main Campus.

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Date Created: 11/02/15

Example supercritical pitchfork bifurcation u k Normalform 3 L120 rlt0 uzru u gt u0iJ7 rgt0 k1 k Example subcritical pitchfork bifurcation u Normal form 3 u 0 ix r r lt 0 u ru u gt n u 0 r gt 0 X r Note There are other types of subcritical and supercritical bifurcations such as a Hopf bifurcation which occurs when two complex conjugate eigenvalues simultaneously become unstable The Hopf bifurcation leads to oscillatory nonlinear states 1 The normal forms characteristic of each bifurcation type represent a Taylor expansion of the evolution equation L2 f u r in terms of the distance to the bifurcation point in the space of parameters here r and variables here u The leading order term a constant always vanishes as the growth rate has to vanish at the bifurcation point u r 0 0 Higher order terms might vanish as well due to certain symmetries of the evolution equations e g pitchfork bifurcation has a u gt u symmetry Bifurcations Into Stripe States Confining the pattern forming system to a finite region with periodic boundary conditions reduces a problem with a continuum of unstable eigenvalues to one with discrete eigenvalues that can be treated within bifurcation theory However this is a degenerate bifurcation at which a finite number of eigenvalues go unstable together due to the rotational symmetries of the system Consider the example of a stationary transition to stripes We set up this problem in a geometry that is periodic over some long length l in the x direction which will be the direction normal to the stripes but periodic over a short distance in the ydirection so that the bifurcating solution is uniform in this direction From the parity symmetry x gt x it can be deduced that there will be two real eigenvalues that pass through zero together corresponding to wave vectors 6 and 61 giving the spatial dependences 6 From these we can form real solutions proportional to cosqx and sinqx or generally cosqx where is an arbitrary phase that represents the translational symmetry of the system the stripes may be laid down at any position along the x direction Alternatively we can write the solution in compleX notation u oc Aeiqx co Here A arequotj is a complex amplitude with the magnitude a IA giving the size of the disturbance and the phase giving the position of the stripes Thus the difference from the simple stationary bifurcation discussed previously e g a pitchfork bifurcation is that the new solution is represented by a complex rather than a real variable It is easy to see the implications of this change on the normal form Remember that changing the phase of the compleX amplitude gives a translation of the stripes through symmetryrelated states This means that the normal form must be invariant under a phase change of A ie the form of the equation must not change if we replace A gt A6 with 5 some constant Using the same idea of a Taylor eXpansion of the dynamics as for the simple bifurcations and incorporating this symmetry constraint we deduce that the normal form is atAeriA2A 1 This is analogous to the pitchfork bifurcation For the negative sign of the nonlinear term we have a supercritical bifurcation with new solutions A requot developing for small positive r with arbitrary These are easily seen to be stable within the dynamics described by E 1 For the positive sign in Eq 1 the bifurcation is subcritical and unstable solutions A rei eXist for small negative r To seek stable finite amplitude solutions we would have to investigate the right hand side of Eq 1 by extending the Taylor eXpansion to higher order 61A 2 rAA2A gA4A Again this is only a reliable predictor of the nonlinear solutions if g happens to be small 1 An important difference from the simple bifurcation is that the bifurcation to stripes is always of the pitchfork type as a consequence of translational symmetry The analogue to the transcritical bifurcation does not eXist in this case independent of whether the system shows a u gt u symmetry As we will see later the transition to a hexagonal state may be transcritical Nonlinear saturation in the SwiftHohenberg equation As we have learned previously the SwiftHohenberg equation describes a typeIS instability to a cellular structure in onedimension atu m l62u u3 As r is increased from negative values the uniform state u 0 develops an instability at r 0 to a mode with critical wave number qc 1 For small positive r we can find a perturbation growing from very small amplitude ux t oc 6 cos x We wish to understand the nonlinear saturation of this state for small r A first guess might be to look for a solution which is just a saturation of the growing linear mode ux a1 cos x with al to be determined To see if this ansatz is a solution we substitute it into the equation for stationary solutions atu 0 3 l 0 ra1 a13 cos x a13 cos 3x 4 4 where we have rewritten cos3 x generated by the L13 term as a sum of cos x and cos 3x terms Now since the Fourier modes cos nx are linearly independent the coefficient on the right hand side of each mode must be zero For instance the coefficient of cos x determines a1 ral ia1320 gt a1Oiq43r12 However equating the coefficient of cos 3x to zero selects the wrong solution 1 Zal30 gt all 0 This inconsistency shows that the original guess was incomplete We rapidly realize however that the problem is removed by adding a small amplitude of cos 3x to the original ansatz so that now ux a1 cos x a3 cos 3x With this new ansatz we obtain the following equality 3 3 3 l 3 3 0 m1 Za13 Za12a3 Eala32cos x m3 64a3 Za13 Ea12a3 Za cos 3x 3 3 3 l a1a32 a12a3 cos 5x ala32 cos 7x a33 cos 9x 4 4 4 4 where again setting the coefficients of different harmonics cos nx to zero determines a1 and a3 As we are looking for a solution close to the bifurcation point where the effect of nonlinear terms is small a3 will be small compared to al Therefore setting the coefficient of cos x to zero gives 3 3 3 3 ra1 a13 a12a3 Eala32 N ml Za13 0 4 4 ie the same equation for all as before Similarly the coefficient of cos 3x gives 3 l 3 3 l l m 64a a3 a2a a3z 64a a3 0 gt a a3ocr32 3 3 4 l l 3 3 3 l 3 l 2 4 4 256

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