Van der Pol Oscillator

## Question:

Dear Support;
My name is *****. I am doing some research on the
characteristics and the dynamic behavior of the Van der Pol
Oscillator. I am not sure if I have the right concept here. Can
you help out?

The general equation that I have is different than the ones I
have seen on some documents. Could you please comment on
that.

This is the equation that I have:

m(d^{2}y/dt^{2})=f(y)(dy/dt)-ky

An example of a specific function f(y) is:
f(y)=c(1-y^{2}) This is the oscillator non-linear
damping coefficient. It varies with the movement of the system.
Where:

m=System Inertia

c=Damping Coefficient parameters

k=system compliance

Also, if you have some more data, text, or graphics document
please forward them to me.

I thank you in advance.

******

## Answer:

The Van der Pol (or Rayleigh - Van der Pol) oscillator is one of
the second order oscillators studied as an illustration of
nonlinear dynamics. One common form of the equation introduced by
Rayleigh in 1896 is:

y'' + a(y^{2}-1)*y' + w_{0}^{2}*y = A*sin(w*t)

where y' is the first derivative of y with respect to time,
y'' the second derivative of y with respect to time, a is
a constant damping amplitude,
w_{0} is the natural angular frequency of the
oscillator, A is the amplitude of a forcing function and w is the forcing function angular frequency.
Your equation using the same notation would be:

m*y''=f(y)*y'-k*y

Rearranging we get

m*y''-f(y)*y'+k*y=0

Now if we insert your f(y) example we see

m*y''-c*(1-y^{2})*y'+k*y=0

Next, divide through by m (not equal to zero) and reverse the
order of the terms in () to get rid of the minus sign:

y''+c/m*(y^{2}-1)*y'+k/m*y=0

This would look like the original Rayleigh version if:

a=c/m, w_{0}^{2}=k/m
and A=0;

So you equation seems to fit the Rayleigh form. Van der Pol
used electronic oscillators to investigate the Rayleigh equation.
Yours is the mechanical analog of the Van der Pol oscillator.

One of the characteristics of this sort of second order
oscillator with nonlinear damping is that it may break into self
excited oscillations for a>>1. The frequency of these
oscillations will be at w_{0}.
If we apply a small amplitude forcing function with frequency
w, not too different from w_{0}, the oscillator will lock onto
the driving frequency.

This information is brought to you by M. Casco Associates, a
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If I can be of more help, please let me know.

JDJ