I have the following:

$ \dot{x} = \frac{dx}{dt}= A\left( x\right) + \sqrt{B\left( x\right)}\eta\left( t\right) $

where $ A\left( x\right)=a_0 - a_1x $ and $ B\left( x\right)=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 $ G\left( \tau\right)=\langle x\left( t\right)x\left( t+\tau\right)\rangle $ and supposing $ a_0 = 0 $ we have to prove that:

$ G\left( \tau\right) = G\left( 0\right)\ e^{a_1\tau} $.


I tried this:

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

\begin{align*} G\left( \tau\right) &= \langle x\left( t\right)x\left( t+\tau\right) \rangle \\ &= \int x\left( t\right)\left[ x\left( t\right) + \frac{1}{2}\tau \dot{x}\left( t\right) \right]\rho\left( x\right)dx \\ &= \int x\left( t\right)^2\rho\left( x\right)dx + \frac{1}{2}\tau\int x\left( t\right)\dot{x}\left( t\right)\rho\left( x\right)dx \\ &= \int x\left( t\right)^2\rho\left( x\right)dx + \frac{1}{4}\tau\int \frac{dx^2}{dt}\rho\left( x\right)dx \\ &= \langle x\left( t\right)^2\rangle + \frac{\tau}{4}\langle\frac{dx^2}{dt}\rangle\ . \end{align*}

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

$ G\left( \tau\right) = \langle x^2\rangle + \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:

$$ G\left( \tau\right) = \langle x^2\rangle\left( 1 + \frac{\tau}{4}\frac{d}{dt}\right) \Rightarrow G\left(\tau^\prime\right)=\langle x^2\rangle e^{\frac{\tau^\prime}{4}} = G\left( 0\right) e^{\frac{\tau^\prime}{4}} \neq G\left( 0\right)\ e^{a_1\tau}\ ,$$

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*} G\left( \tau\right) &= \langle x\left( t\right)^2\rangle + \frac{1}{2}\tau\int x\left( t\right)\dot{x}\left( t\right)\rho\left( x\right)dx \\ &= \langle x^2\rangle + \frac{\tau}{2}\int x\left( t\right)\left[ A\left( x\right) + \sqrt{B\left( x\right)}\eta\left( t\right)\right]\rho\left( x\right)dx \\ &= \langle x^2\rangle\left( 1 - \tau\frac{a_1}{2}\right) + \frac{\tau}{2}\eta\left( t\right)\int x\left( t\right)\sqrt{B\left( x\right)}\rho\left( x\right)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 $ \eta\rightarrow 0 $ I have something like:

$ G\left( \tau\right) = \langle x^2 \rangle\left( 1 - \tau\frac{a_1}{2}\right)^1 \approx \langle x^2 \rangle\left( 1 - \tau\frac{a_1}{2}\right)^{1+\tau} = \langle x^2 \rangle\left( 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 $ n\rightarrow \infty $ making $ G\left( \tau\right) = \langle x^2 \rangle e^{-\frac{a_1}{2}} = G\left( 0\right) e^{-\frac{a_1}{2}} \neq G\left( 0\right)\ e^{a_1\tau} $ 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 $ G\left(\tau\right) $ 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.

up vote 0 down vote accepted

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. $\left< * \right> = \int_t * \rho(t)dt$ ):

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

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

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

$$G(\tau)\sim G(0)-\tau a_1\left<x^2\right>+\tau\left< x\sqrt{B(x)}\eta(t)\right>$$

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:

$$G(\tau)\sim G(0)-a_1\tau \left<x^2\right>+\tau \left<x\sqrt{B(x)}\right> \left< \eta(t)\right>$$ and now because $\left< \eta(t)\right>=0$ we get, again because $\left <x^2(t)\right>=G(0)$: $$G(\tau)\sim G(0)(1-a_1\tau)$$ which is the small $\tau$ expansion of the solution you need: $$G(\tau)=G(0)e^{-a_1\tau}\sim G(0)(1-a_1\tau)$$

(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

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

$${d G(\tau)\over d\tau} = \left< x(t)\left(-a_1x(t+\tau)+\sqrt{B(x)}\eta(t+\tau)\right)\right>$$ for the exact same reasons as before $$\left<\eta(t+\tau)\right>=0$$ and we are left with $${d G(\tau)\over d\tau} = -a_1\left< x(t)x(t+\tau)\right>=-a_1G(\tau)$$ so that our solution is, solving the easy $\dot{y}=-Ay\rightarrow y(t)=y(0)e^{-At}$ differential equation

$$G(\tau)=G(0)e^{-a_1\tau}$$ (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.

  • In fact your proof (with $-a_1$ signal) is right. Probably some typo from who build the exercise. This help me to clarify some points also, as you point out my mistakes. Thanks a lot! – Enrique René Dec 8 at 22:30

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.