Difference between the Rayleigh and Van Der Pol Oscillators?

In summary, for the given two problems involving the van der Pol and Rayleigh oscillators, the leading term of the asymptotic expansion is x_0 = A(0) \cos t' + B(0)\sin t', where A(0) = 1 and B(0) = 0, and the amplitude of the limit cycle oscillation is A(0) = 1. There is a ton of algebra involved in the process, but ultimately the two problems have the same solution due to the identical differential equations that need to be satisfied for A and B.
  • #1
tjackson3
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Homework Statement



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 [tex]\ddot{x} + \epsilon\dot{x}(x^2-1)+x = 0 \mbox{ } x(0) = 1, \dot{x)(0) = 0[/tex]

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

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

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

Homework Equations



None really

The Attempt at a Solution



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

[tex]x_0 = A(\tau)\cos t' + B(\tau)\sin t'[/tex]

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?
 
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  • #2
The answer I get for both is that the leading term of the asymptotic expansion is x_0 = A(0) \cos t' + B(0)\sin t'where A(0) = 1, B(0) = 0and the amplitude of the limit cycle oscillation is A(0) = 1.
 

FAQ: Difference between the Rayleigh and Van Der Pol Oscillators?

1. What is the difference between the Rayleigh and Van Der Pol Oscillators?

The main difference between the Rayleigh and Van Der Pol Oscillators is in the type of damping they exhibit. The Rayleigh oscillator has linear damping, while the Van Der Pol oscillator has nonlinear damping. This results in different behaviors and patterns of oscillation.

2. How do the equations of motion differ between the Rayleigh and Van Der Pol Oscillators?

The Rayleigh oscillator follows the equation of motion x'' + ω^2x = -bx', where x is the displacement, ω is the natural frequency, and b is the damping coefficient. The Van Der Pol oscillator follows the equation of motion x'' + ω^2x + ε(x^2 - 1)x' = 0, where ε is a parameter related to the strength of the nonlinear damping.

3. Which oscillator is more commonly used in real-life systems?

The Van Der Pol oscillator is more commonly used in real-life systems, as it more accurately models many physical systems that exhibit nonlinear damping. This includes electrical circuits, mechanical systems, and biological systems.

4. How do the stability characteristics differ between the Rayleigh and Van Der Pol Oscillators?

The Rayleigh oscillator is a damped oscillator, meaning that its oscillations will eventually die out over time. The Van Der Pol oscillator is a self-sustaining oscillator, meaning that its oscillations will continue indefinitely without any external input. However, the oscillations of the Van Der Pol oscillator may also become chaotic, making it less stable in some cases.

5. Can the equations of motion for the Rayleigh and Van Der Pol Oscillators be combined?

Yes, the equations of motion for the Rayleigh and Van Der Pol Oscillators can be combined to create a more complex model that incorporates both linear and nonlinear damping. This can result in more realistic and accurate simulations of certain systems that exhibit both types of damping. However, this combined model may be more difficult to analyze and solve mathematically.

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