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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?

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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 $\epsilon=1$.

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  • $\begingroup$ The trick to introduce $\epsilon$ was very helpful and solved my problem, thanks.But I did the same and got to this equation, which is different from you: $$\ddot{u_1} + \omega_0^2u_1 = -\ddot{u_0}u_0^2$$ $\endgroup$ Dec 30 '14 at 13:54
  • $\begingroup$ @MammadEskan: $$\frac1{1+\epsilon u^2} = 1- \epsilon u^2 + O(\epsilon^2)$$ $\endgroup$
    – Ron Gordon
    Dec 30 '14 at 13:57
  • $\begingroup$ Oh yes, the binomial expansion. I inserted $u =u_0 + \epsilon u_1$ into the equation, multiplied both sides by $(1 + \epsilon u^2)$, arranged the terms and neglected those containing $\epsilon^2$ or higher. $\endgroup$ Dec 30 '14 at 14:05
  • $\begingroup$ @MammadEskan: your RHS is exactly the same as mine, BTW. $\endgroup$
    – Ron Gordon
    Dec 30 '14 at 16:47
  • $\begingroup$ oh God you are right! Oops. But it's always exhilarating to see that two different methods yield a same answer. $\endgroup$ Dec 30 '14 at 17:05

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