Derivation of master equation

Question

In this article* I want to get the Equation(9) with comparing the equation (2). Please elaborate the left side of equation (9).

*Small amplitude quasi-breathers and oscillons by G. Fodor, et al. arXiv:0802.3525.

Answer 1

Take the left side of eqn. (2):

$-\phi_{t,t,}(x,t)+\Delta\phi(x,t)$

Since this should be a non linear wave equation, $\phi_{t,t}$ means the second derivation in time of $\phi$. Now we make a variable change. First, set $\zeta^i/\varepsilon= x^i$ (eqn. (6)). We put that into $\Delta=\sum\frac{\partial^2}{\partial x_i^2}$, which yields $\Delta=\varepsilon^2\sum\frac{\partial^2}{\partial \zeta_i^2}\equiv\tilde{\Delta}$.

Now make the second variable change: $\tau/\omega(\varepsilon)=t$ (eqn.(7)). Compute the second derivation in time as needed: $\frac{\partial^2}{\partial t^2}=[\omega(\epsilon)]^2\frac{\partial^2}{\partial \tau^2}$.

Our function will change from $\phi(x,t) \to \phi(\zeta,\tau)$. Plug the transformed operators into the left side as well and we get:

$-\frac{\partial^2}{\partial t^2}\phi(x,t)+\Delta\phi(x,t)=-[\omega(\epsilon)]^2\frac{\partial^2}{\partial \tau^2}\phi(\zeta,\tau)+\epsilon^2\tilde{\Delta}\phi(\zeta,\tau)$

This is now the left side of eqn. (9). They only renamed the operators again and left away the dependencies:

$\frac{\partial^2}{\partial \tau^2}\phi(\zeta,\tau)=\ddot{\phi} \qquad \tilde{\Delta}=\Delta$

I hope this makes it clear for you.

EDIT: Additional question

Ok. How to gain eqns. (10-12). Write down the $\varepsilon$-expansions for $\omega$ and $\phi$ up to third order in $\varepsilon$ (eqns. (5) and (8)):

$\omega^2=1+\sum\varepsilon^k \omega_k=1+\varepsilon\omega_1+\varepsilon^2\omega_2+\varepsilon^3\omega_3+\dots$

$\phi=\sum\varepsilon^k \phi_k=\varepsilon\phi_1+\varepsilon^2\phi_2+\varepsilon^3\phi_3+\dots$

Take the derivatives on $\phi$ which you need according to eqn. (9) and put everything together:

$-[1+\varepsilon\omega_1+\varepsilon^2\omega_2+\varepsilon^3\omega_3][\varepsilon\ddot\phi_1+\varepsilon^2\ddot\phi_2+\varepsilon^3\ddot\phi_3] = \varepsilon\phi_1+\varepsilon^2\phi_2+\varepsilon^3\phi_3 + \sum g_k\phi^k$

I will show you what is the principle and derive eqn. (10). You'll have to do the rest by your own. We now search for the first order solution. That means we will only recognize that part of the equation where we find the first order of $\varepsilon$ and we will throw everything else away. Means, for first order only write down what has a $\varepsilon^1$, for the second order you will have to write down everything with a $\varepsilon^2$. On the right side you will get only $\varepsilon \phi_1$ since everything else is higher order in $\epsilon$ except the sum, which is zero order. On the left side you will maintain only $-\varepsilon\ddot\phi_1$ since everything else is of higher order. So we ended up with

$-\varepsilon\ddot\phi_1=\varepsilon \phi_1 \Leftrightarrow \ddot\phi_1+\phi_1=0$

which is eqn. (10).