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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 + 2 x = h ( x ) . (4.1) Here, 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 + 2 x = h ( x, x ) , (4.2) such as the van der Pol oscillator, x + ( x 2 1) x + 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 + 2 sin = 0 , (4.4) where 2 = g/ , with the length of the pendulum. We may rewrite his equation as + 2 = 2 ( sin ) = 1 6 2 3 1 120 2 5 + ... (4.5) The RHS above is a nonlinear function of . We can define this to be h ( ), and take = 1. 4.1.1 Na ve Perturbation theory and its failure Lets assume though that is small, and write a formal power series expansion of the solution x ( t ) to equation 4.1 as x = x + x 1 + 2 x 2 + ... . (4.6) 1 2 CHAPTER 4. NONLINEAR OSCILLATORS We now plug this into 4.1. We need to use Taylors theorem, h ( x + ) = h ( x ) + h ( x ) + 1 2 h ( x ) 2 + ... (4.7) with = x 1 + 2 x 2 + ... . (4.8) Working out the resulting expansion in powers of is tedious. One finds h ( x ) = h ( x ) + h ( x ) x 1 + 2 braceleftBig h ( x ) x 2 + 1 2 h ( x ) x 2 1 bracerightBig + ... . (4.9) Equating terms of the same order in , we obtain a hierarchical set of equations, x + 2 x = 0 (4.10) x 1 + 2 x 1 = h ( x ) (4.11) x 2 + 2 x 2 = h ( x ) x 1 (4.12) x 3 + 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( t + ), where A and 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 t + . Then x ( ) is an even periodic function of with period 2 , hence so is h ( x ). We may then expand h ( x ( ) ) in a Fourier series: h ( A cos ) = summationdisplay n =0 h n ( A ) cos( n ) . (4.14) The n = 1 term leads to resonant forcing. Thus, the solution for x 1 ( t ) is x 1 ( t ) = 1 2 summationdisplay n =0 ( n negationslash =1) h n ( A ) 1 n 2 cos( n t + n ) + h 1 ( A ) 2 t sin( t + ) , (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 = 3 4 cos + 1 4 cos(3 ) , (4.16)(4....
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- Fall '11