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Lecture Note Sketches Hermann Riecke - ESAM Home Page

Lecture Note Sketches Hermann Riecke - ESAM Home Page

Amplitude R 5 4 3 2 1 in

Amplitude R 5 4 3 2 1 in phase out of phase no forcing 0 -5 0 Frequency Omega 5 Mon Jan 11 22:34:53 2010 Characterize Ω(R∞): Amplitude R 5 4 3 2 1 out of phase in phase no forcing 0 -5 0 Frequency Omega 5 Mon Jan 11 22:37:23 2010 Figure 6: Dependence of the amplitude on the forcing frequency. dΩ dR∞ = 3 4 αR∞ ± f 2R 2 ∞ ” + ” ⇔ φ∞ = 0 For α > 0 the in-phase solution has a monotonic dependence Ω(R∞) and the out-of-phase solution has a minimum at R (min) � �1 2f 3 ∞ = 3α and vice versa for α < 0 (f ≥ 0). Thus for α > 0 there are none or two out-of-phase solutions for a given value of Ω and for α < 0 there are none or two in-phase solutions. Stability of the different solutions: � r ′ ϕ ′ � � = Ω R∞ − 1 2 β 1 2 − 9 8 αR∞ − 1 2R∞ R = R∞ + r(T) φ = φ∞ + ϕ(T) f cosφ∞ f sin φ∞ � � r ϕ � no damping � ���� = with ” + ” corresponding to φ∞ = 0 and ” − ” to φ∞ = π. The eigenvalues are σ 2 = ± 1 2 f = ∓ 1 2 f � Ω − R∞ 9 8 αR∞ � � 3 4 αR∞ ± 1 � f 2 R 2 ∞ = ± 1 2 f � 3 8 αR∞ ∓ 1 2 dΩ = ∓ f R 2 ∞ Ω R∞ f R 2 ∞ 0 ± 1 2 f − 9 8 αR∞ 0 − 9 8 αR∞ � � � r ϕ Thus both eigenvalues are either real or purely imaginary. One of the real eigenvalues is always positive. • α > 0: 26 dR∞ �

• α < 0: Notes: – in-phase solution stable: σ ∈ iR (note dΩ dR∞ – out-of-phase solution is ∗ stable for ∗ unstable for R∞ > R (min) ∞ – in-phase solution is ∗ stable for R∞ < R (min) ∞ ∗ unstable for R∞ > R (min) ∞ – out-of-phase solution is stable R∞ < � �1 3 2f 3α > 0 on this branch) = R (min) ∞ • without damping ‘stability’ means actually purely oscillatory response to small perturbations, no asymptotic approach to the fixed point. • expect that with damping the purely imaginary eigenvalues acquire a negative real part and the fixed point becomes asymptotically stable. Now with damping: β > 0 Fixed points: Can solve again for Ω Now R∞ is bounded: βR∞ = f sin φ∞ −2ΩR∞ + 3 4 αR3 ∞ = f cosφ∞ β 2 R 2 ∞ + Ω1,2 = 3 8 αR2 ∞ ± 1 2R∞ • no phase-locked solution for R∞ > f β . • For R∞ = f β � −2ΩR∞ + 3 4 αR3 ∞ � 2 = f 2 � f2 − β2R2 ∞ = 3 8 αR2 ∞ ± 1 � f 2 2 R2 − β ∞ 2 both branches merge: turning point of Ω(R∞) 27 (9)

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