# Van der Pol Equation: Undamped driven oscillator

differentialnumerical methodsordinary differential equations

From the van der pol equation
$$\ddot x – a(1-x^2)\dot x + x = b\cos(\omega t)$$
writing in two first order equations has the system:

$$\dot x = y$$,

$$\dot y = a(1-x^2)y-x+b\cos(\omega t).$$

In the undamped undriven case, (a=0 and b=0). I must show the general solution of equation :

$$\dot y = a(1-x^2)y-x+b\cos(\omega t)$$

is

$$x = \alpha \cos(t+\beta)$$
and $$y = -\alpha \sin(t+\beta).$$

If a=0 and b=0, then $$\dot y = -x$$ which means $$\ddot x + x = 0.$$

Solving this differential equation gives roots at $$i$$ and $$-i$$, so general solution would be $$x(t) = A\cos(-t)+ B\sin(-t).$$

I'm not sure how from here you would get the disired solution of
$$x = \alpha \cos(t+\beta)$$
and $$y = -\alpha \sin(t+\beta),$$ since only conditions given were a=0 and b=0 but no initial conditions of $$x(t)$$ were given.

You can give the solution as $$x(t)=A\cos(t)+B\sin(t)$$ or as $$x(t)=α\cos(t+β)=(α\cos(β))\cos(t)+(-α\sin(β))\sin(t).$$ Now it should be obvious that the correspondence between the parameters is $$(A,B)=(α\cos(β),-α\sin(β))$$ or $$A-iB=αe^{iβ},$$ that is, between the Cartesian and the polar representation of a point in the (complex) plane.
Note that for small positive $$a$$ and small $$x$$ the friction coefficient is negative, that is, the friction term is boosting the solution away from the zero equilibrium. For large $$x$$ the friction is positive, the solution is a cycle that spends only a short part of its period in the boosting region $$|x|<1$$. A small forcing term will thus give perturbations of the limit cycle, while a large forcing term will give rise to "chaotic" solutions. The quality of the "chaos" may also depend on the frequency $$ω$$.