DispersiveWiki:Sandbox: Difference between revisions

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this is the sandbox.
this is the sandbox.


== Gradient Expansion ==
== Testing MathJax ==


 
$$
Given a nonlinear equation as
\int_a^b f'(s) ds = f(b) - f(a).
 
$$
<math> \Box \phi+\lambda\phi^3=0 </math>
It works!
 
one can always build a gradient expansion by assuming <math>\lambda\rightarrow\infty</math> as
 
== Duality in perturbation theory ==
 
In this section we will show how a duality principle holds in perturbation theory
showing how to derive a strong coupling expansion with the leading order ruled by
an adiabatic dynamics in order to study the evolution of a physical system.
We consider the following perturbation problem
 
<math>
    \partial_t u = L(u) + \lambda V(u)
</math>
 
being $\lambda$ an arbitrary ordering parameter: As is well known
an expansion parameter is obtained by the computation of the series itself. The standard
approach assume the limit $\lambda\rightarrow 0$ and putting
 
<math>
    u = u_0 + \lambda u_1 +\ldots
</math>
 
one gets the equations for the series
 
<math>
    \partial_t u_0 = L(u_0) \\
</math>
 
<math>
    \partial_t u_1 = L'(u_0)u_1 + V(u_0) \\
</math>
 
<math>
    \vdots
</math>
 
where a derivative with respect to the ordering parameter is indicated by a prime. We recognize here a conventional
small perturbation theory as it should be. But the ordering parameter is just a conventional matter and so one
may ask what does it mean to consider $L(u)$ as a perturbation instead with respect to the same parameter.
Indeed one formally could write the set of equations
 
<math>
    \partial_t v_0 = V(v_0) \\
 
    \partial_t v_1 = V'(v_0)v_1 + L(v_0) \\
 
    \vdots
</math>
 
where we have interchanged $L(u)$ and $V(u)$ and renamed the solution as $v$. The question to be answered is
what is the expansion parameter now and what derivative the prime means. To answer this question we rescale the
time variable as $\tau = \lambda t$ into eq.(\ref{eq:eq1}) obtaining the equation
 
<math>
    \lambda\partial_{\tau} u = L(u) + \lambda V(u)
</math>
 
and let us introduce the small parameter $\epsilon=\frac{1}{\lambda}$. It easy to see that applying again the
small perturbation theory to the parameter $\epsilon\rightarrow 0$ we get the set of equations (\ref{eq:set})
but now the time is scaled as $t/\epsilon$, that is, at the leading order the development parameter of the
series will enter into the scale of the time evolution producing a proper slowing down ruled by the equation
 
<math>
    \epsilon\partial_t v_0 = V(v_0)
</math>
 
that we can recognize as an equation for adiabatic evolution that in the proper limit $\epsilon\rightarrow 0$
will give the static solution $V(u_0)=0$. We never assume this latter solution but rather we will study
the evolution of eq.(\ref{eq:lead}). Finally, the proof is complete as we have obtained a dual series
 
<math>
    u = v_0 + \frac{1}{\lambda} v_1 +\ldots
</math>
 
by simply interchanging the terms for doing perturbation theory. This is a strong coupling expansion
holding in the limit $\lambda\rightarrow\infty$ dual to the small perturbation theory $\lambda\rightarrow 0$
we started with and having an adiabatic equation at the leading order.
 
It is interesting to note that, for a partial differential equation,
we can be forced into a homogeneous equation because, generally, if we require
also a scaling into space variables we gain no knowledge at all on the evolution of a
physical system. On the other side, requiring a scaling on the space variables and not on
the time variable will wash away any evolution of the system. So, on most physical systems
a strong perturbation means also a homogeneous solution but this is not a general rule. As
an example one should consider fluid dynamics where two regimes dual each other can be found
depending if it is the Eulerian or the Navier-Stokes term to prevail. In general relativity
things stay in a way to get a homogeneous equation at the leading order. The reason for this
is that products of derivatives or second order derivatives in space coordinates are
the only elements forming the Einstein tensor beside time dependence.

Latest revision as of 08:37, 4 February 2011

Welcome to the sandbox! Please feel free to edit this page as you please by clicking on the "edit" tab at the top of this page. Terry 14:58, 30 July 2006 (EDT)

Some basic editing examples

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this is the sandbox.

Testing MathJax

$$ \int_a^b f'(s) ds = f(b) - f(a). $$ It works!