# Homework Help: Difference between the Rayleigh and Van Der Pol Oscillators?

1. Nov 29, 2011

### tjackson3

1. The problem statement, all variables and given/known data

I have two separate problems, but I get the same answer for each. I feel like this must be wrong.

Question 1: Find the leading term of a uniformly valid (for t > 0) asymptotic expansion of the solution of the IVP $$\ddot{x} + \epsilon\dot{x}(x^2-1)+x = 0 \mbox{ } x(0) = 1, \dot{x)(0) = 0$$

for the van der Pol oscillator, which describes the evolution of ICs to a limit cycle oscillation. Calculate the amplitude of the limit cycle oscillation.

Question 2: Find the leading term of a uniformly valid (for t > 0) asymptotic expansion of the solution of the IVP

$$\ddot{x} + \epsilon\dot{x}(1-\frac{1}{3}\dot{x}^3)+x = 0 \mbox{ } x(0) = 0, \dot{x)(0) = 2a$$

for the Rayleigh oscillator, which describes the evolution of ICs to a limit cycle oscillation. Calculate the amplitude of the limit cycle oscillation.

2. Relevant equations

None really

3. The attempt at a solution

There's a ton of algebra involved here. Essentially you look for 2$\pi$-periodic solutions by transforming the time t to $t' = \omega t$, using a two-time method ($\tau = \epsilon t$), and expanding x and omega in terms of epsilon ($x ~ x_0 + \epsilon x_1 + ...,\ \omega ~ 1 + \epsilon \omega_1 + ...$ where you know that the first term in the omega expansion is 1 due to the fact that you seek $2\pi$-periodic solutions of the equation). If you go through all of the algebra, you find that

$$x_0 = A(\tau)\cos t' + B(\tau)\sin t'$$

Plugging that into the expansion for the O(ε) term gives that A and B must satisfy a set of two differential equations in order to avoid secular terms. However, I get that these two differential equations are identical in both the Rayleigh and van der Pol cases, which seems wrong, since they're two different equations. The only difference between the two solutions comes from the different initial conditions, but if they had the same initial conditions, then they'd have the same solution which seems wrong. Can anyone point me in the right direction?