CH04_NLOSC

CH04_NLOSC - Chapter 4 Nonlinear Oscillators 4.1 Weakly...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

Page1 / 42

CH04_NLOSC - Chapter 4 Nonlinear Oscillators 4.1 Weakly...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online