Averaged Lagrangian and the equations of motion

1. Apr 15, 2012

nickthequick

Hi,

Qualitatively: I am trying to decipher a method I've found in the literature, namely Whitham's method. It is a technique used to averaged out "fast variations" in the Lagrangian to then deduce governing equations for the system. I am trying to quantitatively deduce how accurate Whitham's method is, and whether or not it ignores relevant information at higher orders of the small parameter.

Quantitatively: Consider a Lagrangian $L=L(\phi,\eta;x,t)$ where $(\phi,\eta)$ represent the dependent variables of the system and (x,t) are the independent variables. We are going to take a WKB expansion of the dependent variables, so that

$\phi = \sum_n a_ne^{i\theta}$ and

$\eta = \sum_n b_n e^{i\theta}$.

We assume $a_n,b_n = f_n(a,a_t,a_x,...)$ so that our new dependent variables are $(a,\theta)$ and possible derivatives on these variables.

It is assumed that the coefficients are $O(\epsilon)$, for small parameter $\epsilon$ and vary slowly in space and time (for instance $a_n = a_n(\epsilon x, \epsilon t)$) while the phase is $\theta = kx-\omega t + \epsilon \sigma(\epsilon x, \epsilon t)$ for $(k,\omega) \in \mathbb{R}$, ie it has a 'fast scale' .

We now define

$\mathcal{L}=\frac{1}{2\pi} \int_0^{2\pi} L \ d\theta$.

It is conjectured that variations of this "averaged Lagrangian" $\mathcal{L}$ will then give us our governing equations.

My question is this: What, quantitatively, is the difference between the equations deduced via the condition $\delta L =0$ and $\delta \mathcal{L} = 0$ ?

Any suggestion are appreciated. Also, if this is vague or unclear, let me know and I will provide more information/examples.

Thanks,

Nick