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## these conditions are

these conditions are satisfied for a system that undergoes a Hopf bifurcation for values of the bifurcation parameter just below the Hopf bifurcation, i.e. in the regime in which the basic state is still linearly stable, but only weakly so (⇒ weak damping) Compare first with previous result for the linear case α = 0 Steady state (fixed point of (7,8)) 1 4 β2R 2 ∞ + Ω 2 R 2 ∞ = 1 4 f2 ⇒ R∞ = 1 2 tan φ∞ = − βR∞ ΩR∞ f � β 2 + 4Ω 2 � φ∞ = arctan − β � 2Ω The solution of (5) approaches the steady state in an oscillatory manner in the amplitude R. Recover that aspect from (7,8) by considering small perturbations around (R∞, φ∞) Insert and rewrite � r ′ ϕ ′ � � 1 − = R = R∞ + r(T) φ = φ∞ + ϕ(T) r ′ + 1 2 β (R∞ + r) = 1 2 f (sin φ∞ + cos φ∞ ϕ) R∞ϕ ′ − Ω (R∞ + r) = 1 2 f (cos φ∞ − sin φ∞ ϕ) 2β 1 2 Ω 1 − R∞ 2R∞ f cos φ∞ f sin φ∞ Using an ansatz � r ϕ 2 Ω R∞ � = � � r ϕ � r0 ϕ0 σ − 1 2 β � � 1 − β −ΩR∞ = 2 � e σT yields � ��� σ − 1 � β −ΩR∞ � � � = 0 Ω R∞ σ 2 − βσ + 1 4 β2 + Ω 2 = 0 σ = −β ± iΩ − 1 2 β � � r ϕ Previously, we got the full, exact solution y(t) for the linear problem (cf. (5)). To do that it would be easier to use multiple-scales using y(t, T) = (A0(T) + ǫA1(T) + . . .) cos (t + ΩT) + (B0(T) + ǫB1(T) + . . .)sin (t + ΩT) rather than the amplitude-phase description using (R(T), φ(T)) (the amplitude-phase equation has nonlinear functions of φ even without nonlinearities in the original equation). At O(ǫ) one gets then (−2A ′ 0 − βA0 − 2ΩB) sin(t + ΩT) + (2B ′ 0 + βB0 − 2ΩA0 − f)cos(t + ΩT) = 0 24 �

i.e. Now the nonlinear problem: Fixed points (critical points): A ′ 0 B ′ 0 + 1 2 βA0 + ΩB0 = 0 + 1 2 βB0 − ΩA0 = 1 2 f βR∞ = f sin φ∞ −2ΩR∞ + 3 4 αR3 ∞ = f cosφ∞ Start with the easier case without damping: β = 0 sin φ∞ = 0 ⇒ φ∞ = 0 or φ∞ = π and −2ΩR∞ + 3 4 αR3 ∞ = ±f For φ = 0 the oscillation is in phase with the forcing. For φ = π it is out of phase. To get an overview of the dependence on the detuning Ω it is easier to solve for Ω than for R (1,2) ∞ In-phase solution φ∞ = 0: Out-of-phase solution φ∞ = π: Notes: Ω = 3 8 αR2 ∞ − f 2R∞ Ω = 3 8 αR2 f ∞ + 2R∞ • In the absence of forcing both curves become identical and one recovers the amplitudedependence of the frequency of the Duffing oscillator ω = ω0 + 3 8 αR2 ∞ α > 0 corresponds to a hard spring: frequency increases with amplitude α < 0 corresponds to a soft spring: frequency decreases with amplitude 25

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