Csdrhrt

I have the following:

$ dot{x} = frac{dx}{dt}= Aleft( xright) + sqrt{Bleft( xright)}etaleft( tright) $

where $ Aleft( xright)=a_0 - a_1x $ and $ Bleft( xright)=b_0-b_1x+b_2x^2 $. All $ a_k,b_k geq 0 $. $ eta $ is related to a gaussian with null mean and unit variance.

Defining $ Gleft( tauright)=langle xleft( tright)xleft( t+tauright)rangle $ and supposing $ a_0 = 0 $ we have to prove that:

$ Gleft( tauright) = Gleft( 0right) e^{a_1tau} $.

I tried this:

1) Considering $ tau $ small enough to allow the use of approximation $ xleft( t+tauright)=xleft( tright)+frac{1}{2}tau dot{x}left( tright) $, I do:

begin{align*}
Gleft( tauright) &= langle xleft( tright)xleft( t+tauright) rangle
\ &= int xleft( tright)left[ xleft( tright) + frac{1}{2}tau dot{x}left( tright) right]rholeft( xright)dx
\ &= int xleft( tright)^2rholeft( xright)dx + frac{1}{2}tauint xleft( tright)dot{x}left( tright)rholeft( xright)dx
\ &= int xleft( tright)^2rholeft( xright)dx + frac{1}{4}tauint frac{dx^2}{dt}rholeft( xright)dx
\ &= langle xleft( tright)^2rangle + frac{tau}{4}langlefrac{dx^2}{dt}rangle .
end{align*}

Since the system is in thermodynamic equilibrium, $frac{drho}{dt} = 0 $ and then:

$ Gleft( tauright) = langle x^2rangle + frac{tau}{4}frac{d}{dt}langle x^2 rangle $

I don't see how this result can help me to get the proof. In this way the $ a_0=0 $ hypothesis was not required, which makes me think I'm in a way won't help me. The only thing I can see from here is something like:

$$ Gleft( tauright) = langle x^2rangleleft( 1 + frac{tau}{4}frac{d}{dt}right) Rightarrow Gleft(tau^primeright)=langle x^2rangle e^{frac{tau^prime}{4}} = Gleft( 0right) e^{frac{tau^prime}{4}} neq Gleft( 0right) e^{a_1tau} ,$$

where $ tau $ is small and $ tau^prime $ arbitrary.

2) Doing the same approximation of "1)" I decided to use the $ dot{x} $ equation:

begin{align*}
Gleft( tauright) &= langle xleft( tright)^2rangle + frac{1}{2}tauint xleft( tright)dot{x}left( tright)rholeft( xright)dx
\ &= langle x^2rangle + frac{tau}{2}int xleft( tright)left[ Aleft( xright) + sqrt{Bleft( xright)}etaleft( tright)right]rholeft( xright)dx
\ &= langle x^2rangleleft( 1 - taufrac{a_1}{2}right) + frac{tau}{2}etaleft( tright)int xleft( tright)sqrt{Bleft( xright)}rholeft( xright)dx .
end{align*}

I stopped here because the coefficients of $ B $ don't appear at the expression what I want to get. If I neglect the last integral making $ etarightarrow 0 $ I have something like:

$ Gleft( tauright) = langle x^2 rangleleft( 1 - taufrac{a_1}{2}right)^1 approx langle x^2 rangleleft( 1 - taufrac{a_1}{2}right)^{1+tau} = langle x^2 rangleleft( 1 - frac{1}{n}frac{a_1}{2}right)^{1+frac{1}{n}} .$

The last step was based on the arquimedian property of real set, $ n $ is a natural number. I almost can see the $ nrightarrow infty $ making $ Gleft( tauright) = langle x^2 rangle e^{-frac{a_1}{2}} = Gleft( 0right) e^{-frac{a_1}{2}} neq Gleft( 0right) e^{a_1tau} $ that's what I want.

This problem comes from Statistical Mechanics discipline of Mastering program on physics. As I assume $ tau $ very small to make these approximations, I think the $ Gleft(tauright) $ is something like infinitesimal generator of something in the system.

I appreciate some guidance to solve this.
I appreciate most some guidance with mathematical rigor, telling why some step can (or cannot) be taken.

First of all, there a couple of errors in your computations. For example, the average you are taking are over time so you should use $rho(t)dt$, not $rho(x)dx$!. Also the Taylor approximation should be $$x(t+tau)sim x(t)+tau dot{x}$$

Moreover, approximating $G(tau)$ for small $tau$ would just give you an hint of what would happen at small $tau$, you would not be able to recover the full $G(tau)$. If you had not made the mistakes you did, indeed, following your computations but slightly corrected and using $left< * right>$ for the average of $*$ (i.e. $left< * right> = int_t * rho(t)dt$ ):

$$G(tau)sim left< x(t)(x(t) +taudot{x} ) right>=left<x^2(t)+tau x(t)dot{x} right>=left< x^2(t) right> +tauleft<x(t)dot{x} right> $$

now, using the expression you have for $dot{x}$ and the fact that $left< x^2(t) right>=G(0)$:

$$G(tau)sim G(0) + tau left < xleft(-a_1 x+sqrt{B(x)}eta(t)right) right>$$
i.e.

$$G(tau)sim G(0)-tau a_1left<x^2right>+tauleft< xsqrt{B(x)}eta(t)right>$$

now we make the assumption that $eta(t)$ is not correlated with the $x$-terms [notice that this is the only step in which I actually have to assume. I think it is right or that any similar assumption applies, but maybe think about it), i.e. that we can write:

$$G(tau)sim G(0)-a_1tau left<x^2right>+tau left<xsqrt{B(x)}right> left< eta(t)right>$$
and now because $left< eta(t)right>=0$ we get, again because $left <x^2(t)right>=G(0)$:
$$G(tau)sim G(0)(1-a_1tau)$$
which is the small $tau$ expansion of the solution you need:
$$G(tau)=G(0)e^{-a_1tau}sim G(0)(1-a_1tau)$$

(I get a minus sign which you don't have, which I think is also right as otherwise the correlation would increase over time, which is weird... who of us made the mistake..?)

Anyways this procedure could have given you a hint, and a small-$tau$ proof of the result, but not the final solution.

What instead if try to compute

$${d G(tau)over dtau} = left< x(t){dx(t+tau)over d tau}right>$$
(where I only take the derivative of the second one because the first on has no $tau$ dependence)?. So as ${dx(t+tau)over d tau}={dx(tau)over d tau}|_{t+tau}=dot{x}|_{t+tau}$:

$${d G(tau)over dtau} = left< x(t)left(-a_1x(t+tau)+sqrt{B(x)}eta(t+tau)right)right>$$
for the exact same reasons as before $$left<eta(t+tau)right>=0$$ and we are left with
$${d G(tau)over dtau} = -a_1left< x(t)x(t+tau)right>=-a_1G(tau)$$
so that our solution is, solving the easy $dot{y}=-Ayrightarrow y(t)=y(0)e^{-At}$ differential equation

$$G(tau)=G(0)e^{-a_1tau}$$
(again with a minus sign which I trust - but I am open to discussion!)

Hope this helps not only solving it, but also showing some of your mistakes and wrong (but still not trivial!) reasoning.