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.

## Best Answer

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 $ω$.