CH04_NLOSC

# CH04_NLOSC - Chapter 4 Nonlinear Oscillators 4.1 Weakly...

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Unformatted text preview: Chapter 4 Nonlinear Oscillators 4.1 Weakly Perturbed Linear Oscillators Consider a nonlinear oscillator described by the equation of motion ¨ x + Ω 2 x = ǫh ( x ) . (4.1) Here, ǫ is a dimensionless parameter, assumed to be small, and h ( x ) is a nonlinear function of x . In general, we might consider equations of the form ¨ x + Ω 2 x = ǫh ( x, ˙ x ) , (4.2) such as the van der Pol oscillator, ¨ x + μ ( x 2 − 1) ˙ x + Ω 2 x = 0 . (4.3) First, we will focus on nondissipative systems, i.e. where we may write m ¨ x = − ∂ x V , with V ( x ) some potential. As an example, consider the simple pendulum, which obeys ¨ θ + Ω 2 sin θ = 0 , (4.4) where Ω 2 = g/ℓ , with ℓ the length of the pendulum. We may rewrite his equation as ¨ θ + Ω 2 θ = Ω 2 ( θ − sin θ ) = 1 6 Ω 2 θ 3 − 1 120 Ω 2 θ 5 + ... (4.5) The RHS above is a nonlinear function of θ . We can define this to be h ( θ ), and take ǫ = 1. 4.1.1 Na¨ ıve Perturbation theory and its failure Let’s assume though that ǫ is small, and write a formal power series expansion of the solution x ( t ) to equation 4.1 as x = x + ǫx 1 + ǫ 2 x 2 + ... . (4.6) 1 2 CHAPTER 4. NONLINEAR OSCILLATORS We now plug this into 4.1. We need to use Taylor’s theorem, h ( x + η ) = h ( x ) + h ′ ( x ) η + 1 2 h ′′ ( x ) η 2 + ... (4.7) with η = ǫx 1 + ǫ 2 x 2 + ... . (4.8) Working out the resulting expansion in powers of ǫ is tedious. One finds h ( x ) = h ( x ) + ǫh ′ ( x ) x 1 + ǫ 2 braceleftBig h ′ ( x ) x 2 + 1 2 h ′′ ( x ) x 2 1 bracerightBig + ... . (4.9) Equating terms of the same order in ǫ , we obtain a hierarchical set of equations, ¨ x + Ω 2 x = 0 (4.10) ¨ x 1 + Ω 2 x 1 = h ( x ) (4.11) ¨ x 2 + Ω 2 x 2 = h ′ ( x ) x 1 (4.12) ¨ x 3 + Ω 2 x 3 = h ′ ( x ) x 2 + 1 2 h ′′ ( x ) x 2 1 (4.13) et cetera , where prime denotes differentiation with respect to argument. The first of these is easily solved: x ( t ) = A cos( Ω t + ϕ ), where A and ϕ are constants. This solution then is plugged in at the next order, to obtain an inhomogeneous equation for x 1 ( t ). Solve for x 1 ( t ) and insert into the following equation for x 2 ( t ), etc. It looks straightforward enough. The problem is that resonant forcing terms generally appear in the RHS of each equation of the hierarchy past the first. Define θ ≡ Ω t + ϕ . Then x ( θ ) is an even periodic function of θ with period 2 π , hence so is h ( x ). We may then expand h ( x ( θ ) ) in a Fourier series: h ( A cos θ ) = ∞ summationdisplay n =0 h n ( A ) cos( nθ ) . (4.14) The n = 1 term leads to resonant forcing. Thus, the solution for x 1 ( t ) is x 1 ( t ) = 1 Ω 2 ∞ summationdisplay n =0 ( n negationslash =1) h n ( A ) 1 − n 2 cos( nΩ t + nϕ ) + h 1 ( A ) 2 Ω t sin( Ω t + ϕ ) , (4.15) which increases linearly with time. As an example, consider a cubic nonlinearity with h ( x ) = r x 3 , where r is a constant. Then using cos 3 θ = 3 4 cos θ + 1 4 cos(3 θ ) , (4.16)(4....
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CH04_NLOSC - Chapter 4 Nonlinear Oscillators 4.1 Weakly...

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