The Van der Pol equation

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The Van der Pol equation is the second-order non-linear differential equation …

x” + εx'(x2-1)+x=0, where x=x(t) and ε>0.

If y2(t)=tan2(t)/(atan2(t)+btan(t)+c), where a and c are >0, 0<b2<4ac, and b<0,

and if we let z(t) be a ‘test’ function defined by z(t) =|y(t)| for 0<=t<=π

and z(t) = -|y(t)| for π<=t<=2π etc. then the 2π-periodic function z(t) for certain values of a, b, and c ‘seems’ to have the right kind of shape for the periodic solution to the Van der Pol equation – ignoring any scaling in the ‘t’ direction.

For example try a=0.28, b=-0.02 and c=0.004. This gives the following shape …

vdp1

Note that the ‘first’ turning point in the first quadrant has t=tan -1(-2c/b) while the z coordinate is z(t)=2√c/(√(4ac-b2).

If the amplitude is 2 then b2 = c(4a-1). This requires that a>1/4.

Note also that if b=0 and a=c>0 then z(t) is just a sine curve.

The expression for y(t) suggests that the Van der Pol equation might be written as

x² = 1 -(x + x”)/(εx’) for the purposes of obtaining a sequence of approximations to x(t).

If in the RHS of x² = 1 – (x + x”)/(εx’) we let  x = λsin(μt) then the LHS does not give an expression like that for y2(t) as above.

However, if we re-write the Van der Pol equation as x² = (εxx’ – xx”)/(1 + εxx’)

and let x=λsin(μt) in the RHS, then the resulting expression,

(εxx’ – xx”)/(1 + εxx’) = (λ²μ²tan²(μt) + ελ²μtan(μt))/(tan²(μt) + ελ²μtan(μt) +1),

is very like that for y2(t) as above.

If a = 1, b = 0.1 and c = 1 we get …

vdp5

If a = 0.25, b = 0.2 and c = 0.05 we get …

         vpd2

If a = 100, b = 0.2 and c = 0.05 we get …

vdp3

These shapes are very similar to those obtain by numerical solutions of the Van der Pol equation.

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