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

**x” + εx'(x ^{2}-1)+x=0**, where x=x(t), tεR, and ε>0 is a constant.

If **y ^{2}(t)=tan^{2}(t)/(atan^{2}(t)+btan(t)+c)**, where a and c are >0, 0<b

^{2}<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 …

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-b^{2}).

If the amplitude is 2 then b^{2 }= 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 y^{2}(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 y^{2}(t) as above.

If a = 1, b = 0.1 and c = 1 in **y ^{2}(t)=tan^{2}(t)/(atan^{2}(t)+btan(t)+c)** we get …

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

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

These shapes are very similar to those obtained by numerical solutions of the Van der Pol equation. The numerically solved solutions tend to have amplitude around 2 which these curves do not have. **A useful variation in the form for y²(t) is y²(t) = (4b – a²)tan²(t)/(1 + atan(t) + btan²(t)) where a² < 4b.** The resultant curves from this form do have amplitude 2 while exhibiting the same characteristic shapes. When a = -10 and b = 50 we get …

When a = -5 and b = 30 we get …

The Van der Pol equation in the form x² = (εxx’ – xx”)/(1 + εxx’) can be transformed a little by introducing a non-zero constant k (say).

So, it is easily checked that x² = (kεxx’ – kxx” – (k-1)x²)/(1 + kεxx’) is essentially equivalent to the Van der Pol equation and if we let x=λsin(μt) in the RHS of this equation then x² still takes the form of the ratio of 2 quadratics of the tan() function.

If x=λsin(μt) then x² = λ²tan²(μt))/(1+tan²(μt))).

**Transformation** of **x” + εx'(x ^{2}-1)+x=0**.

First multiply through by x to get ** xx” + εx’x(x ^{2}-1)+x²=0.**

If we let y = x² then y’ = 2xx’ and so (y’)² = 4y(x’)². Also, y” = 2(xx” + (x’)²). Hence, xx’ = y’/2 and xx” = y”/2 – (x’)² = y”/2 -(y’)²/(4y).

Substituting into **xx” + εx’x(x ^{2}-1)+x²=0** we get

**2yy” – (y’)² +2yy’ε(y – 1) +4y² = 0 ………. ***

If we let **z = 1/y** in * above, we get **2zz” – 3(z’)² – 2εz'(z – 1) – 4z² = 0**, which has a very similar form to that of *.

**Another approach.**

Provided that x’ ≠ 0 then x² = 1 – x/(εx’) – x”/(εx’).

If we let u (say) = x/x’, (x’ ≠ 0) then it is straightforward to show that x² = 1 + (u’ – 1 – u²)/(εu).

The term u’ – 1 – u² is again suggestive of the tan() function.

Also, if x ≠ 0, let v = x’/x, so that if in addition x’ ≠ 0 we have that v = 1/u.

It follows then that x² = 1 – (v’ + 1 + v²)/(εv).