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Lecture Note Sketches Hermann Riecke - ESAM Home Page

Lecture Note Sketches Hermann Riecke - ESAM Home Page

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use Note: cos 3 ωt = 3 1 cosωt + cos 3ωt 4 4 cos 2 ωt cosω0t = 1 4 cos ((2ω − ω0)t) + 1 4 cos ((2ω + ω0) t) + 1 cos ω0t 2 cosωt cos 2 ω0t = 1 4 cos ((ω − 2ω0)t) + 1 4 cos ((ω + 2ω0) t) + 1 cos ωt 2 cos 3 ω0t = 3 4 cosω0t + 1 cos 3ω0t 4 • if one uses complex exponentials one does not have to deal with trig identities. Each term in f1 drives a specific part of the particular solution for y1 i.e. cos ((2ω − ω0)t) → cos ((2ω + ω0)t) → cos ((2ω0 − ω)t) → cos ((2ω0 + ω)t) → cos 3ωt → cos ˜ωt → With this forcing one can drive resonances at 1 −˜ω 2 + ω 2 0 1 4ω2 = − 4ωω0 1 1 4 ω (ω − ω0) 1 4ω2 = + 4ωω0 1 1 4 ω (ω + ω) 1 3ω2 0 − 4ω0ω 1 = + ω2 (ω − ω0) (ω − 3ω0) 1 3ω2 0 + 4ω0ω 1 = + ω2 (ω + ω0)(ω + 3ω0) 1 (ω0 − 3ω)(ω0 + 3ω) ω = 0 ω = ω0 ω = 3ω0 ω = 1 3 ω0 As our symmetry analysis showed, depending on the ratio ω/ω0 the perturbation analysis leads to different amplitude equations. 1.4.1 3:1 Forcing Consider ω0 = 1 ω = 3 (1 + Ω) 38

with O(1) forcing (cf. in Sec.1.2.1 where the forcing was O(ǫ). As in the case ω ≈ ω0 write 1 with initial conditions Again the derivative ∂ 2 t and one obtains O(1): with yields O(ǫ): y(t, T) = y0(ψ, T) + ǫy1(ψ, T) + . . . ψ = t + ΩT − φ(T) is expanded as y(0, 0) = δ dy dt = 0 ∂ ∂t = (1 + ǫ (Ω − ∂Tφ)) ∂ψ + ǫ∂T ∂2 ∂t2 = ∂2 ψ + ǫ � 2 (Ω − ∂Tφ) ∂ 2 ψ + 2∂2 � 2 ψT + O(ǫ ) ∂ 2 ψ y0 + y0 = f cos (3 (ψ + φ)) y0(0, 0) = δ ∂ty0(0, 0) = 0 y0(ψ, T) = R(T) cosψ + K cos (3 (ψ + φ)) with K = − f 8 R(0) cosψ(0) = δ R(0) sinψ(0) = 0 ⇒ R(0) = δ ψ(0) = 0 ∂ 2 ψy1 + y1 = −β∂ψy0 − � 2 (Ω − φ ′ ) ∂ 2 ψ + 2∂2 � ψT y0 − y 3 0 ≡ f1 f1 = β [R sin ψ + 3K sin (3 (ψ + φ))] + 2 (Ω − φ ′ ) [R cos ψ + 9K cos (3 (ψ + φ))] +2 [R ′ sin ψ + 9φ ′ K cos (3 (ψ + φ))] −R 3 cos 3 ψ − 3R 2 K cos 2 ψ cos (3 (ψ + φ)) − 3RK 2 cosψ cos 2 (3 (ψ + φ)) − K 3 cos 3 (3 (ψ + φ)) = β [R sin ψ + 3K sin (3 (ψ + φ))] + 2 (Ω − φ ′ ) R cosψ + 18ΩK cos (3 (ψ + φ)) + 2R ′ sin ψ − 3 4 R3 cosψ − 3 4 R2K cos (2ψ − 3 (ψ + φ)) − � �� � 3 2 RK2 cosψ + higher harmonics cos ψ cos 3φ−sinψ sin 3φ The solvability condition results then in Notes: 2R ′ + βR + 3 4 R2 K sin 3φ = 0 (17) 2 (Ω − φ ′ ) R − 3 4 R3 − 3 4 R2 K cos 3φ − 3 2 RK2 = 0 (18) 1 It might be easier to use complex exponentials as in the symmetry analysis. 39

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