Perturbation, straightforward expansion

Question

Consider the equation: $$\ddot{u} + \frac{\omega_0^2u}{1+u^2} = 0$$I want to determine the straightforward expansion for small but finite $u$. what form should the expansion take? Normally the nonlinearity parameter, $\epsilon$, is the perturbation parameter and the straightforward expansion takes the form : $$ u = u_0 + \epsilon u_1 + O(\epsilon^2) $$ But here, $u$ itself is the perturbation parameter. What form should the expansion take?

Answer 1

Introduce $\epsilon$ as follows:

$$\ddot{u} + \omega_0^2 \frac{u}{1+\epsilon u^2} = 0$$

where, hopefully,

$$u = u_0+\epsilon u_1+\epsilon^2 u_2 + \cdots$$

Clearly,

$$u_0(t) = A \cos{\omega_0 t} + B \sin{\omega_0 t} $$

Now expand to $O(\epsilon)$:

$$\ddot{u_0} + \omega_0^2 u_0 + \epsilon \left [\ddot{u_1} + \omega_0^2 \left (u_1- u_0^3 \right )\right ] +O(\epsilon^2) = 0$$

The term in brackets is zero to this order of approximation, and we have

$$\ddot{u_1} + \omega_0^2 u_1 = \omega_0^2 u_0^3 $$

This is an inhomogeneous equation for $u_1$ because we know $u_0$ from the previous order solution. As you go to higher orders, you will see such recursion.

To first order, the solution you seek is $u(t) = u_0(t) + u_1(t)$, as the original equation specifies that $\epsilon=1$.