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## +-sign: −-sign: � 1

+-sign: −-sign: � 1 2 + δ1c1 � 1 ∓ 2 − δ1c2 = 0 � � 1 1 − δ1 c2 = ± 2 2 c1 = The general solution is given by Notes: • for δ1 = ± 1 2 • for δ2 1 > 1 4 + δ1 u+(t) = e iǫα1t �� 1 2 − δ1 � 1 cost + 2 + δ1 � sin t = e −1 √1 2 4 −δ2 1 ǫt �� 1 2 − δ1 � 1 cost + 2 + δ1 � sin t u−(t) = e iǫα1t �� 1 2 − δ1 � 1 cos t − 2 + δ1 � sin t = e 1 √ 1 2 4 −δ2 1 ǫt �� 1 2 − δ1 � 1 cost − 2 + δ1 � sin t u(t) = A+u+(t) + A−u−(t) the solution reduces to the periodic solution obtained in Sec.1.1.1. the solutions remain bounded: they are quasiperiodic with frequencies ω = 1 � and Ω = • for δ 2 1 < 1 4 δ 2 1 − 1 4 one solution blows up exponentially: the state u(t) = 0 is unstable. • the Mathieu equation is linear ⇒ the exponential growth does not saturate • recall: δ1 characterizes the leading-order detuning between the forcing and the subharmonic resonance – weak detuning (δ2 1 < 1 4 mode – strong detuning (δ 2 1 ) ⇒ resonant driving leads to exponential growth of one 1 > ) ⇒ driving is out of resonance and is not able to pump in 4 energy to generate growth – at the border between these two regimes the solution is periodic 18

Forcing Strength ε 1 0.5 0 -0.5 n=1 unstable stable stable unstable -1 0 1 2 Frequency δ Figure 3: Stability and instability regions of Mathieu equation at the 2:1 resonance (subharmonic response). 1.2 Nonlinear Oscillators: Forced Duffing Oscillator I Nonlinearity can saturate the growth of the oscillations and can affect the resonance frequency As an example of a forced, weakly nonlinear oscillator consider the Duffing equation ¨y + ǫβ ˙y ���� + ω damping 2 0 ���� y + ǫαy natural frequency 3 ���� nonlinearity = ǫf cosωt � �� � forcing Here we already assumed that the damping, nonlinearity, and forcing are all weak: ǫ ≪ 1 Note: • here the forcing is assumed to be non-parametric; in the swing picture it would correspond to a person pushing ⇒ with forcing the solution y = 0 does not exist any more i) linear, undamped case: β = 0,α = 0 in this case one can easily consider the general case of O(1)-forcing For ω �= ω0 using initial conditions ¨y + ω 2 0 y = ˆ f cosωt y(0) = δ, ˙y(0) = 0 y(t) = c1 cosω0t + c2 sin ω0t + y(0) = c1 + ˆf ω 2 0 − ω 2 ω 2 0 ˆf cosωt − ω2 ! ���� = δ ˙y(0) = c2ω0 19 (4)

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