First order expansion of Euler-Lagrange equations

Question

I know that in field theory Euler Lagrange equations are $p_i-d_\mu p^\mu_i=0$. (Classical notations, $p_i=\frac{\partial L}{\partial y^i}, p_i^\mu=\frac{\partial L}{\partial y^i_\mu}$). Being a differential equation, I can expand it to first (or higher) order, but how can we do this?

I was wondering if it is right to consider the expansion of a solution as $y^i=y_0^i+\epsilon y^i_1+...$ and then substitute this expression in the E.L. equations to get the first order expansion?

I tried to do the computation but I wasn't able to get any result, any suggestion?

Answer 1

I don't recommend to use the notation $d_\mu$ since the derivative that appears in the E.L equations of motion is partial derivative $\partial_\mu$. Then if the Lagrangian density depends on the field $y$ and its derivatives $\partial_\mu y$ the E.L equations of motion read

$\begin{equation} \partial_\mu\dfrac{\partial \mathcal L}{\partial (\partial_\mu y)}-\dfrac{\partial \mathcal{L}}{\partial y} \end{equation}=0 $

Then if you want to find solutions to the E.L equations you will need to consider a specific expression for the Lagrangian density. In your case, you consider that the field can be expanded on a superposition of fields

$y=c_0y_0+c_1 y_1+c_2y_2+\ldots=\displaystyle\sum_{i=1}^{\infty}c_i y_i$

Then note that the Lagrangian density becomes a function of that many fields and its derivatives $\mathcal{L}(\partial_\mu y, y)\equiv \mathcal{L}(\partial_\mu y_0,\partial_\mu y_1,\ldots, y_0,y_1,\ldots)$. So that when you derive (through the variational principle) the Euler-Lagrange equations of motion you get a system of (possibly coupled) differential equations for the $y_0,y_1,\ldots$ fields.

$\begin{equation} \displaystyle\sum_{i=1}^{\infty}\bigg[\partial_\mu\dfrac{\partial \mathcal L}{\partial (\partial_\mu y_i)}-\dfrac{\partial \mathcal{L}}{\partial y_i}\bigg] \end{equation}=0. $