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## • the result agrees in

• the result agrees in its form with that obtained using symmetry and scaling arguments (cf. (16)) • in addition to the term R 3 also the term RK 2 in the equation for the amplitude is missing: the forcing is not modifying the linear damping of the amplitude Analyze the fixed points (critical points) of (17,18) One fixed point is at R (1) ∞ = 0. It corresponds to the solution which does not excite the resonance near ω0. For R∞ �= 0 one has Thus y (1) = K cos (3 (ψ + φ)) + O(ǫ) β = − 3 4 R∞K sin 3φ∞ 2Ω − 3 4 R2 3 ∞ − 2 K2 = 3 4 R∞K cos 3φ∞ β 2 + 9 16 R4 ∞ + R2 � ∞ 2 3 4 R 4 ∞ + R2 ∞ � 2Ω − 3 4 R2 3 ∞ − 2 K2 �2 = 9 � 3 2 K2 � − 2Ω − 9 16 K2 � + � 3K 2 − 16 3 Ω � � 8 + Ω − 2K2 3 16 R2 ∞ K2 � 2Ω − 3 � 2 2 K2 � 2 + 16 9 β2 = 0 We need a real and positive solution of this bi-quadratic equation. Discriminant ∆ = � 3K 2 − 16 3 Ω �2 � �2 8 − 4 Ω − 2K2 − 3 64 9 β2 = −7K 4 + 32 3 ΩK2 − 64 9 β2 Solving for Ω we find that to have real solutions we need For Ω ≥ Ω0(K) we have − � 3K 2 − 16 3 Ω � Ω ≥ 21 32 K2 + 2 β 3 2 ≡ Ω0(K) K2 ≥ 16 21 3 � � 21 − 3 6 40 32 K2 − 3K 2 + 32 9 K 2 + 32 9 β2 = K2 β2 > 0 K2 + β 2 = 0

Thus, if ∆ > 0 both solutions R2 1,2 are positive and we conclude Ω 2 Solutions No Solutions Ω > Ω0(K) 2 positive solutions Ω < Ω0(K) no positive solution K 2 Figure 8: Phase diagram: existence of solutions as a function of Ω and K. The line marks ∆ = 0. For fixed Ω non-trivial solutions R > 0 exist only in a range Kmin ≤ K ≤ Kmax � � � K 2 min,max = 16 21 Ω ± Ω 2 − 7 4 β2 At Kmin and Kmax two solutions come into existence or disappear in a saddle-node bifurca- tion. The amplitudes at these points are Note: R 2 min,max = 1 2 � 16 3 Ω − 3K2 min,max � > 0 • These solutions come into existence at finite amplitude and do not bifurcate off the trivial solution R (1) = 0. 41

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