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Lecture Note Sketches Hermann Riecke - ESAM Home Page

Lecture Note Sketches Hermann Riecke - ESAM Home Page

1 0.5 0 -0.5 -1 0 1 2 3

1 0.5 0 -0.5 -1 0 1 2 3 4 5 6 Notes: Forcing Subharmonic response (odd) Subharmonic response (even) • in both cases the solutions have a period that is twice as long as that of the forcing: subharmonic response. • δ − δ0 = ǫδ1 + ǫ 2 δ2 + . . . characterizes the detuning between the forcing frequency and the natural frequency of the unforced oscillator • for δ = + 1 the solution contains only sin nt to the order considered, i.e. it is odd, and the solution is even. 2 for δ = −1 2 • for the forcing the transformation ǫ → −ǫ is equivalent to t → t + π . Shifting the time 2 interchanges the even and the odd solution: sin t → cost and cost → − sin t⇒ Thus: by π 2 δeven(ǫ) = δodd(−ǫ) • for any n (or δ0) the Mathieu equation is even under time reversal, i.e. the equation does not change form under the replacement t → −t ≡ ¯t since only even derivatives arise and the coefficients are even in t. I.e. if u(t) is a solution, so is u(−t) d2 dt2u(−t) + (δ + ǫ cos (2t))u(−t) = d2 d(−t) 2u(−t) + (δ + ǫ cos (−2t))u(−t) = d2 d(¯t) 2u(¯t) + (δ + ǫ cos (2¯t)) u(¯t) = 0 • if we were to find a solution u(t) that is of mixed parity (not even nor odd) then – ue ≡ u(t) + u(−t) is even – uo ≡ u(t) − u(−t) is odd and since the Mathieu equation is linear ue and uo are also solutions 10

• we can assume from the start that u(t) is either even or odd and include only sinfunctions or cos-functions Note: • the Mathieu equation is linear, therefore the amplitude of the solution is undetermined (in the absence of initial conditions) – we do not need to keep the higher-order homogenous solution since they only change the overall amplitude of the solution. – fix the amplitude of the leading-order solution by a suitable normalization condition �� 2π u(t) costdt = π for δ = +1 2 u(t) sin t dt = π for δ = −1 2 iii) case n = 2, δ = 4 Note: �0 2π 0 • the oscillation period is the same as that of the forcing: harmonic response Use the normalization conditions O(1): u0(t) = �� 2π 0 u0(t) sin 2t dt = π � 2π 0 u0(t) cos 2t dt = π � 2π 0 uk≥1(t) sin 2t dt = 0 odd � 2π 0 uk≥1(t) cos2t dt = 0 even � sin 2t odd solution, out of phase with respect to the forcing cos 2t even solution, in phase with respect to the forcing with amplitudes fixed by the normalization Odd solution: O(ǫ): thus Note: ü1 + 4u1 = −δ1 sin 2t − cos 2t sin 2t = −δ1 sin 2t − 1 sin 4t 2 δ1 = 0 u1 = 1 sin 4t 24 • because of the normalization we do not keep the homogeneous solution 11

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