Why is A(1-y²) Negative When y² < 1 in Van der Pol Equation?

[shivajikobardan]

Homework Statement

positive and negative damping in van der pol equation-simulation and modeling

Relevant Equations

Van der pol equation

The van der pol non-linear equation is given as-:

y''=A(1-y²)y'-By

y=amplitude

The analysis given by the book is this-:

When y²<1
i.e when y is small
A(1-y²) is negative.
A(1-y²) is called damping term.

I don't understand how is it negative? It obviously becomes positive in this case.

Sth similar is done for when y²>1.
The final conclusion is-:
Thus, small amplitude oscillations will build up and large amplitude oscillations will be damped out.
i.e when y is small=>there should be no damping as shown in what I'm confused with. (If I did according to me, then I'd be getting positive value i.e more damping).

 

[malawi_glenn]

The analysis given by the book

which book?

Homework Statement:: positive and negative damping in van der pol equation-simulation and modeling
Relevant Equations:: Van der pol equation

When y²<1
i.e when y is small
A(1-y²) is negative.

depends on sign of A of course

 

[shivajikobardan]

which book?

i just googled van der pol equation in google books. almost all of them say the same(any resource says the same).

depends on sign of A of course

 

[malawi_glenn]

Consider $\ddot y -\mu (1-y^2 ) \dot y + y = 0$ where $y = y(t)$ and $\mu \geq 0$
Compare with the damped oscillator equation
$\ddot y +b \dot y + y = 0$ where $b$ is the amount of "damping".
We have that $b$ plays the role of $-\mu (1-y^2 )$
Now, if $b>0$ then the solution $y$ will decay, and if $b<0$ the solution will "blow up"
We can make this analysis now, if $|y| > 1$ then $|y|^2 > 1$ and your "damping" "b" will be positive and hence solution $y$ will decay.
If $|y|< 1$ then $|y|^2 < 1$ and your "damping" "b" will be negative and hence solution $y$ will "blow up".

This is the idea of the Van Der Pol oscillator, oscilations with small amplitude will get their amplitudes increased, and vice versa. Energy is leaked out from the system when it grows to big, and energy is fed in when it grows too small.