# Derivation of master equation

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.

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Related question by OP: physics.stackexchange.com/q/52590/2451 –  Qmechanic Jan 31 '13 at 8:06
It would be good if OP (or somebody else?) could try to make the question formulation (v3) self-contained, so one doesn't have to open the link to understand the question. –  Qmechanic Jan 31 '13 at 8:23
You can open the article from the given link to see details. –  Unlimited Dreamer Jan 31 '13 at 8:29
A reader should be able to tell whether the question is interesting to him, or not, before he has to click any links. In general, the pleasure and usefulness of browsing Phys.SE greatly diminish if one has to open documents to decide which content to read. –  Qmechanic Feb 1 '13 at 21:37

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.

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).

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Thnx you have done great. –  Unlimited Dreamer Jan 31 '13 at 8:34
Was my edit that what you were looking for? –  DaP Jan 31 '13 at 13:10
As far as I know, this site is not intended to connect it's members. –  DaP Jan 31 '13 at 17:03
Your last question, I answered in the new post. For the one before: Be careful! The right side contains a $\sum g_k \phi^k$. That $\phi$ under the sum has to be expanded as well! That is the point where the $g_k$-terms arise. –  DaP Feb 1 '13 at 8:52
Look at the definitions: The lowest order in $\phi$ is $\varepsilon^1$, in front of $\Delta$ there is $\varepsilon^2$, so the lowest order of $\varepsilon^2\Delta\phi$ will be $\varepsilon^3$. –  DaP Feb 1 '13 at 15:05